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Ex: Square root of 25 is either 5 or -5 Principal square root means you only want the positive value. Ex: The principal square root of 25 is 5", + "video_name": "sBvRJUwXJPo", + "transcript": "We're asked to divide and simplify. And we have one radical expression over another radical expression. The key to simplify this is to realize if I have the principal root of x over the principal root of y, this is the same thing as the principal root of x over y. And it really just comes out of the exponent properties. If I have two things that I take to some power-- and taking the principal root is the same thing as taking it to the 1/2 power-- if I'm raising each of them to some power and then dividing, that's the same thing as dividing first and then raising them to that power. So let's apply that over here. This expression over here is going to be the same thing as the principal root-- it's hard to write a radical sign that big-- the principal root of 60x squared y over 48x. And then we can first look at the coefficients of each of these expressions and try to simplify that. Both the numerator and the denominator is divisible by 12. 60 divided by 12 is 5. 48 divided by 12 is 4. Both the numerator and the denominator are divisible by x. x squared divided by x is just x. x divided by x is 1. Anything we divide the numerator by, we have to divide the denominator by. And that's all we have left. So if we wanted to simplify this, this is equal to the-- make a radical sign-- and then we have 5/4. And actually, we can write it in a slightly different way, but I'll write it this way-- 5/4. And we have nothing left in the denominator other than that 4. And in the numerator, we have an x and we have a y. And now we could leave it just like that, but we might want to take more things out of the radical sign. And so one possibility that you can do is you could say that this is really the same thing as-- this is equal to 1/4 times 5xy, all of that under the radical sign. And this is the same thing as the square root of or the principal root of 1/4 times the principal root of 5xy. And the square root of 1/4, if you think about it, that's just 1/2 times 1/2. Or another way you could think about it is that this right here is the same thing as-- so you could just say, hey, this is 1/2. 1/2 times 1/2 is 1/4. Or if you don't realize it's 1/2, you say, hey, this is the same thing as the square root of 1 over the square root of 4, and the square root of 1 is 1 and the principal root of 4 is 2, so you get 1/2 once again. And so if you simplify this right here to 1/2, then the whole thing can simplify to 1/2 times the principal root-- I'll just write it all in orange-- times the principal root of 5xy. And there's nothing else that you can really take out of the radical sign here. Nothing else here is a perfect square." + }, + { + "Q": "How does 2s1 suddenly change to 2p3? Where did the \"p\" come from?", + "A": "The orbitals each have a given amount of electrons, s can have 2, and then it must move up to p. The p orbital can have up to 6 electrons, and when you run out you must move on to the d orbital, which can have 10 electrons. Lastly, you have the f orbital and that orbital can have up to 14 electrons.", + "video_name": "FmQoSenbtnU", + "transcript": "In the last few videos we learned that the configuration of electrons in an atom aren't in a simple, classical, Newtonian orbit configuration. And that's the Bohr model of the electron. And I'll keep reviewing it, just because I think it's an important point. If that's the nucleus, remember, it's just a tiny, tiny, tiny dot if you think about the entire volume of the actual atom. And instead of the electron being in orbits around it, which would be how a planet orbits the sun. Instead of being in orbits around it, it's described by orbitals, which are these probability density functions. So an orbital-- let's say that's the nucleus it would describe, if you took any point in space around the nucleus, the probability of finding the electron. So actually, in any volume of space around the nucleus, it would tell you the probability of finding the electron within that volume. And so if you were to just take a bunch of snapshots of electrons -- let's say in the 1s orbital. And that's what the 1s orbital looks like. You can barely see it there, but it's a sphere around the nucleus, and that's the lowest energy state that an electron can be in. If you were to just take a number of snapshots of electrons. Let's say you were to take a number of snapshots of helium, which has two electrons. Both of them are in the 1s orbital. It would look like this. If you took one snapshot, maybe it'll be there, the next snapshot, maybe the electron is there. Then the electron is there. Then the electron is there. Then it's there. And if you kept doing the snapshots, you would have a bunch of them really close. And then it gets a little bit sparser as you get out, as you get further and further out away from the electron. But as you see, you're much more likely to find the electron close to the center of the atom than further out. Although you might have had an observation with the electron sitting all the way out there, or sitting over here. So it really could have been anywhere, but if you take multiple observations, you'll see what that probability function is describing. It's saying look, there's a much lower probability of finding the electron out in this little cube of volume space than it is in this little cube of volume space. And when you see these diagrams that draw this orbital like this. Let's say they draw it like a shell, like a sphere. And I'll try to make it look three-dimensional. So let's say this is the outside of it, and the nucleus is sitting some place on the inside. They're just saying -- they just draw a cut-off -- where can I find the electron 90% of the time? So they're saying, OK, I can find the electron 90% of the time within this circle, if I were to do the cross-section. But every now and then the electron can show up outside of that, right? Because it's all probabilistic. So this can still happen. You can still find the electron if this is the orbital we're talking about out here. Right? And then we, in the last video, we said, OK, the electrons fill up the orbitals from lowest energy state to high energy state. You could imagine it. If I'm playing Tetris-- well I don't know if Tetris is the thing-- but if I'm stacking cubes, I lay out cubes from low energy, if this is the floor, I put the first cube at the lowest energy state. And let's say I could put the second cube But I only have this much space to work with. So I have to put the third cube at the next highest energy state. In this case our energy would be described as potential energy, right? This is just a classical, Newtonian physics example. But that's the same idea with electrons. Once I have two electrons in this 1s orbital -- so let's say the electron configuration of helium is 1s2 -- the third electron I can't put there anymore, because there's only room for two electrons. The way I think about it is these two electrons are now going to repel the third one I want to add. So then I have to go to the 2s orbital. And now if I were to plot the 2s orbital on top of this one, it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. Then you have the blue area, then the red, and the blue. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. But if you look at these, there's three ways that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon. So the electron configuration for carbon, the first two electrons go into, so, 1s1, 1s2. So then it fills-- sorry, you can't see everything. So it fills the 1s2, so carbon's configuration. It fills 1s1 then 1s2. And this is just the configuration for helium. And then it goes to the second shell, which is the second period, right? That's why it's called the periodic table. We'll talk about periods and groups in the future. And then you go here. So this is filling the 2s. We're in the second period right here. That's the second period. One, two. Have to go off, so you can see everything. So it fills these two. So 2s2. And then it starts filling up the p orbitals. So then it starts filling 1p and then 2p. And we're still on the second shell, so 2s2, 2p2. So the question is what would this look like if we just wanted to visualize this orbital right here, the p orbitals? So we have two electrons. So one electron is going to be in a-- Let's say if this is, I'll try to draw some axes. That's too thin. So if I draw a three-dimensional volume kind of axes. If I were to make a bunch of observations of, say, one of the electrons in the p orbitals, let's say in the pz dimension, sometimes it might be here, sometimes it might be there, sometimes it might be there. And then if you keep taking a bunch of observations, you're going to have something that looks like this bell shape, this barbell shape right there. And then for the other electron that's maybe in the x direction, you make a bunch of observations. Let me do it in a different, in a noticeably different, color. It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. The p orbitals. Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. And then you start filling the second energy shell. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons in the lowest energy state, will be 1s2. So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then we have 3s2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus, there's more space in between the lower energy orbitals to fill in more of these bizarro-shaped orbitals. But these are kind of the balance -- I will talk about standing waves in the future -- but these are kind of a balance between trying to get close to the nucleus and the proton and those positive charges, because the electron charges are attracted to them, while at the same time avoiding the other electron charges, or at least their mass distribution functions. Anyway, see you in the next video." + }, + { + "Q": "I'm sorry, but I don't really see what bad things deflation causes. Could someone please explain? Oh, and was it deflation that caused the Great Depression? I was under the impression that it was inflation.", + "A": "Check out the tutorial on deflation.", + "video_name": "sZRkERfzzn4", + "transcript": "I finished the last video touching on the stimulus plan and whether it's going to be big enough and what its intent was. But I was a little handwavy about things like savings and GDP and I thought it would actually be good to get a little bit more particular. This isn't just me making up things. So a good place to start is just to think about disposable income. And people talk about it without ever giving you a good definition. It's important to understand what disposable income is. So this box right here is GDP, so it's all the goods and services, essentially all the income, that we produce. The disposable income is essentially the amount that ends up in the hands of people. So the U.S. GDP, I don't know what the exact number is, it's on the order of I think $15 trillion. And so some of that money goes to taxes. I don't know. Let's say $3 trillion ends up in taxes. And these are round numbers, but it gives you a sense of Then maybe about another $1 trillion is saved by So let's say that I'm Microsoft and I make a billion Microsoft makes a lot more than that, but I make a billion dollars. And I just keep it in a bank account owned by Microsoft. I don't give it to my employees. I don't dividend it out to the shareholders. I just keep that. So corporate savings, let's say, is another trillion dollars. These are roundabout numbers. And then everything left over is, essentially, disposable income. I'm making some simplifications there, but everything left over is disposable income. And the idea is, what is in the hands of households after paying their taxes: disposable income. That check that you get after your taxes and everything else that is taken out of your paycheck, that money, that is disposable income to you. So the interesting thing is where this disposable income has been going over the last many, many years. So this right here, this chart, is a plot of the percentage of disposable income saved since 1960. I got this off the Bloomberg terminal again. And just so you get a good reference point, let me draw the line for 10%. Because I kind of view that as a reference point. That's 10% personal savings. Actually, you could go back even further than this, and you can see from the 1960s all the way until the early `80s, the go-go `80s, people were saving about 10% of their disposable income. Of what they got on that paycheck after paying taxes, they were keeping about 10% of it, putting it in the bank. And, of course, that would later be used for investment and things like that. But then starting in the early `80s, right around, let's see, about 1984, you see that people started saving less and less money. All the way to the extreme circumstance, to 2007, where, on average, people didn't save money. Where maybe I saved $5, but someone else borrowed $5. So on average, we didn't save money. Actually we went slightly negative. And all of this, and this is kind of my claim-- and it can be borne out if you look at other charts in terms of the amount of credit we took on-- is because we took on credit. I guess there's two things you could talk about. We took on credit and we started having what I call perceived savings, right? If I buy a share of IBM a lot of people say, oh, I saved I saved my money, I'm invested in the stock market. But when I bought a share of IBM, unless that dollar I paid for the share goes to IBM to build new factories-- And that's very seldom time. Most of the time, that dollar I gave to buy that IBM share, just goes to a guy who sold an IBM share. So there's no net investment. It's only investment, it's only savings, when that dollar actually goes to invest in some way, build a new factory or build a new product or something. So I think you had more and more people thinking that they were saving when they weren't. Maybe thinking that they were saving, as their home equity in their house grew, or as their stock market portfolio went up. But there wasn't actual aggregate savings going on. In fact, if anything, they were borrowing against those things, especially home equity. And the average savings rate went down and down and down. So now we're here in 2007, and maybe you could say 2008, where this is the capacity. We should probably talk about world capacity, because so much of what we consume really does come from overseas, but this is the capacity serving the U.S. And let's say, going into 2007, this was demand. Supply and demand was, let's say, pretty evenly matched. If you go to that top up here, you see that we were at 80% utilization, which isn't crazy. So if anything, you could say that demand was maybe right But that's a good level. You want to be at around 80% utilization. That's the rate at which you don't have hyperinflation, but you're utilizing things quite well. Now all of a sudden, the credit crisis hit and all this credit disappears. And I'd make the argument that the only thing that enabled us to not save this money, is more and more access to cheap credit. Every time we went through a recession, from the mid `80s, onwards, the government solution was to make financing easier. The Fed would lower interest rates. We would pump more and more money into Fannie and Freddie Mac. We would lower standards on what it We would create incentives for securitization. We would look the other way when Bear Stearns is creating collateralized debt obligations, or AIG is writing credit default swaps. And all of that enabled financing. Until we get to this point right here, where everything starts blowing up. So you essentially have forced savings. When people can't borrow anymore, the savings rate has to go to 10% because most of this is just from people taking on more credit. There were a group of people who were probably already always saving at 10% of their disposable income. But there's another group of people who were more than offsetting that by taking credit. Now that the credit crisis hits, you're going to see the savings rate-- and you already see it with this blip right there-- the savings rate is going to go back up to 10% of disposable income. Now if the savings rate goes back to 10% of disposable income, that amount of money, that 10% of disposable income, this gap, is 10% of disposable income that cannot be used for consumption. And let's think about how large of an amount that is. Right now, the U.S. disposable income, if I were to draw disposable income-- this number right here, I just looked it up-- it's around $10.7 trillion, let's say $11 trillion for roundabout. So this 10% of disposable income that we're talking about, this gap between the normal environment, the environment that was enabled from super-easy credit, that's 10% of disposable income. 10% of $11 trillion is $1.1 trillion per year of demand that will go away. That's per year. So all of a sudden, you're going to have a gap where this was the demand before and now the demand's going to drop to here. And all of this $1.1 trillion of demand is going to disappear, because credit is now not available. And then you do get a situation where you might hit a low capacity utilization. Unfortunately, the capacity utilization numbers don't go back to the Great Depression, and we could probably get a sense of, at what point does a deflationary spiral really get triggered. But because that $1.1 trillion in demand disappears, you're going to see this orange line just drop lower and lower in terms of capacity utilization. And that's going to drive prices down. And what Obama and the Fed and everyone else is trying to do, is to try to make up this gap. Now, everyone else can't borrow money. Companies can't borrow money. Homeowners can't borrow money. But the government can borrow money, because people are willing to finance it. At the bare minimum, the Fed's willing to finance the government. And so the government wants to come in and take up the slack with this demand and spend the money themselves with the stimulus. Now we just talked about, what's the magnitude of this demand shock? It's $1.1 trillion per year. While the stimulus plan is on the order of a trillion dollars. And that trillion dollars isn't per year, although I have a vague feeling that we will see more of them coming down the pipeline. This stimulus plan that just passed is expected to be spent over the next few years. So in terms of demand created over the next few years, it's going to be several hundred billion per year. So it's not going to be anywhere near large enough to make up for this demand shock. So we're still going to have low capacity utilization. So the people who argue that these stimulus plans are going to create hyperinflation, I disagree with that, at least in the near term, because in the near term it's nowhere near large enough to really soak up all of the extra capacity we have in the system. And if anything, it's going to soak up different capacity. So the stimulus plan might create inflationary-like conditions in certain markets where the stimulus plan is really focused. But in other areas, where you used to have demand, but the stimulus plan doesn't touch, like $50 spatulas from Williams Sonoma or granite countertops, that area is going to continue to see deflation. And I would say net-net, since this thing this is actually so small, even though we're talking about trillion dollars relative to the amount of demand destroyed per year, we'll probably still have deflation. And, for anyone who's paying close attention to it, and I am because I care about these things, the best thing to keep a lookout for, to know when to start running to gold maybe or being super-worried about inflation, is if you see this utilization number creeping back up into the 80% range. At least over the last 40 years, that's been the best leading indicator to say when are we And for all those goldbugs out there, who insist that all of the hyperinflation or the potential hyperinflation is caused by our not being on the gold standard, I just want to point out that you can very easily have-- we went off of Bretton Woods and completely went to a fiat currency in '73, that right around here. But our worst inflation bouts were actually while we were on the gold standard. And that's because we had very high capacity utilization. This is the war period. What happens during a war? You run your factories all-out. You run your farms all-out to feed the troops. The factories, instead of building cars, build planes. Everyone was working. Wages go up. Everything goes up and you have inflation. A lot of people say, oh, the war was the solution to the And it is true in that it got us out of that negative deflationary spiral that I talked about in the last video. And it did it by creating an unbelievable amount of inflation. And then after the war, you could argue what allowed-- I don't have GDP here, but the U.S. GDP really did do well in the postwar period-- it wasn't the war per se, although the war kind of did take us out of the deflationary spiral. So after the war, you had all of this capacity, after World War II. So you had all this capacity that was being used up by the war, and then, all of the sudden, the war ends. And you're like, well, we don't have to build planes anymore, we don't have to feed the troops anymore, and now we have all these unemployed troops who come back home. What are we going to do with them? You'd say maybe, capacity utilization would come back down there. But what a lot of people don't talk about is, the rest of the industrial world's capacity was blown to smithereens. At that point in time, the U.S. was a smaller part of GDP and the other major players were Germany, Western Europe and Japan. In the U.S., we didn't have any factories bombed. The entire war took place in these areas. And whenever people go on bombing raids, the ideal thing that they want to bomb is factories. So what you had in the postwar period, is you had all of these countries that had their capacity blown to smithereens. The U.S. was essentially the only person left with any capacity, and so all the demand from the rest of the world picked up the slack in the U.S. capacity. And that's why, even though there was a demand shock in the U.S. after that, you also had a supply shock from the war where you had all of this capacity that was blown to smithereens. In this situation that we are in now, we have a major demand shock. Financing just disappears and the savings rate's going to skyrocket because people can't borrow anymore. But there's no counteracting supply shock. Supply of factories making widgets and all the rest is So the Obama administration is trying to create a stimulus to sop some of this up, but it's not going to be enough. And my only fear is, with all the printing money and all that goes, once we do get back to the 80% capacity utilization, once we do go back here, how quickly they can unwind all of this printing press money and all of the stimulus plan. Because that's going to be the key. Because, if we do get to 80% capacity utilization or we start pushing up there, and we continue to run the printing press-- because that's what, essentially, government's incentive is to do, because they always feel better when we're flush with money-- then, and only then, you might see a hyperinflation scenario. But I don't see that at least for the next few years." + }, + { + "Q": "At 4:29 What is the exact difference between obtuse and acute triangles", + "A": "in the obtuse triangle, it has one obtuse angle (bigger than a right angle) and in an acute angle, all angles are smaller that right angle. Ex: acute:all angles 60 ( 60+60+60=180) Ex: obtuse: one angle 120 another 35 and another 25 Hope this helped! Please vote this", + "video_name": "D5lZ3thuEeA", + "transcript": "What I want to do in this video is talk about the two main ways that triangles are categorized. The first way is based on whether or not the triangle has equal sides, or at least a few equal sides. Then the other way is based on the measure of the angles of the triangle. So the first categorization right here, and all of these are based on whether or not the triangle has equal sides, is scalene. And a scalene triangle is a triangle where none of the sides are equal. So for example, if I have a triangle like this, where this side has length 3, this side has length 4, and this side has length 5, then this is going to be a scalene triangle. None of the sides have an equal length. Now an isosceles triangle is a triangle where at least two of the sides have equal lengths. So for example, this would be an isosceles triangle. Maybe this has length 3, this has length 3, and this has length 2. Notice, this side and this side are equal. So it meets the constraint of at least two of the three sides are have the same length. Now an equilateral triangle, you might imagine, and you'd be right, is a triangle where all three sides have the same length. So for example, this would be an equilateral triangle. And let's say that this has side 2, 2, and 2. Or if I have a triangle like this where it's 3, 3, and 3. Any triangle where all three sides have the same length is going to be equilateral. Now you might say, well Sal, didn't you just say that an isosceles triangle is a triangle has at least two sides being equal. Wouldn't an equilateral triangle be a special case of an isosceles triangle? And I would say yes, you're absolutely right. An equilateral triangle has all three sides equal, so it meets the constraints for an isosceles. So by that definition, all equilateral triangles are also isosceles triangles. But not all isosceles triangles are equilateral. So for example, this one right over here, this isosceles triangle, clearly not equilateral. All three sides are not the same. Only two are. But both of these equilateral triangles meet the constraint that at least two of the sides are equal. Now down here, we're going to classify based on angles. An acute triangle is a triangle where all of the angles are less than 90 degrees. So for example, a triangle like this-- maybe this is 60, let me draw a little bit bigger so I can draw the angle measures. That's a little bit less. I want to make it a little bit more obvious. So let's say a triangle like this. If this angle is 60 degrees, maybe this one right over here is 59 degrees. And then this angle right over here is 61 degrees. Notice they all add up to 180 degrees. This would be an acute triangle. Notice all of the angles are less than 90 degrees. A right triangle is a triangle that has one angle that is exactly 90 degrees. So for example, this right over here would be a right triangle. Maybe this angle or this angle is one that's 90 degrees. And the normal way that this is specified, people wouldn't just do the traditional angle measure and write 90 degrees here. They would draw the angle like this. They would put a little, the edge of a box-looking thing. And that tells you that this angle right over here is 90 degrees. And because this triangle has a 90 degree angle, and it could only have one 90 degree angle, this is a right triangle. So that is equal to 90 degrees. Now you could imagine an obtuse triangle, based on the idea that an obtuse angle is larger than 90 degrees, an obtuse triangle is a triangle that has one angle that is larger than 90 degrees. So let's say that you have a triangle that looks like this. Maybe this is 120 degrees. And then let's see, let me make sure that this would make sense. Maybe this is 25 degrees. Or maybe that is 35 degrees. And this is 25 degrees. Notice, they still add up to 180, or at least they should. 25 plus 35 is 60, plus 120, is 180 degrees. But the important point here is that we have an angle that is a larger, that is greater, than 90 degrees. Now, you might be asking yourself, hey Sal, can a triangle be multiple of these things. Can it be a right scalene triangle? Absolutely, you could have a right scalene triangle. In this situation right over here, actually a 3, 4, 5 triangle, a triangle that has lengths of 3, 4, and 5 actually is a right triangle. And this right over here would be a 90 degree angle. You could have an equilateral acute triangle. In fact, all equilateral triangles, because all of the angles are exactly 60 degrees, all equilateral triangles are actually acute. So there's multiple combinations that you could have between these situations and these situations right over here." + }, + { + "Q": "What is the meaning of AP Physics? Does AP stand for Applied Physics\nIf so the class is called Applied Physics - Physics (:<))", + "A": "Nope. AP is for A dvanced P lacement physics, not applied physics.", + "video_name": "IMC3Uqv5yYc", + "transcript": "- I have taught AP Physics classes for the last seven years, AP Physics one, AP Physics B, back in the day, and AP Physics C now. I try to make my lessons personable, relate to the students, offer them real life examples where things happen. But also focus on, instead of a bunch of little memorization facts, what's the big idea that helps you figure out all these different situations for like, allows you to be more of a robust problem solver. So we're focusing on the big core ideas of our subject, being physics, and I think you'll find that across all the different AP content subjects. We are finely tuned to what the AP exam is gonna be like, directly, and we are building in videos for instructions specific to that and also we have practice exercises that I'm really excited about, that you can assign it to students and there's hints and there's feedback why a certain problem is wrong, and why a certain problem is right, and then it shows you the solution path of how you would get to the solution. So, when you assign a student a problem, they don't have to wait a whole 24 hours until they get back to class, they can get it wrong, fail early, try another one, fail often, and do a couple times on that, and I actually think my might actually do more homework because they're getting instant feedback on the process rather than just trying a couple and getting stuck and like, ah I'm frustrated, I'm out of here. They can really get in the seat and stay in there because they're getting instant feedback and help as their working through problems. I think that's gonna be really powerful." + }, + { + "Q": "i have learned that the denser the medium, the faster the waves travel in it...", + "A": "True for sound waves and mechanical waves.", + "video_name": "yF4cvbAYjwI", + "transcript": "- [Voiceover] To change the speed of sound you have to change the properties of the medium that sound wave is traveling through. There's two main factors about a medium that will determine the speed of the sound wave through that medium. One is the stiffness of the medium or how rigid it is. The stiffer the medium the faster the sound waves will travel through it. This is because in a stiff material, each molecule is more interconnected to the other molecules around it. So any disturbance gets transmitted faster down the line. The other factor that determines the speed of a sound wave is the density of the medium. The more dense the medium, the slower the sound wave will travel through it. This makes sense because if a material is more massive it has more inertia and therefore it's more sluggish to changes in movement or oscillations. These two factors are taken into account with this formula. V is the speed of sound. Capital B is called the bulk modulus of the material. The bulk modulus is the official way physicists measure how stiff a material is. The bulk modulus has units of pascals because it's measuring how much pressure is required to compress the material by a certain amount. Stiff, rigid materials like metal would have a large bulk modulus. More compressible materials like marshmallows would have a smaller bulk modulus. Row is the density of the material since density is the mass per volume, the density gives you an idea of how massive a certain portion of the material would be. For example, let's consider a metal like iron. Iron is definitely more rigid and stiff than air so it has a much larger bulk modulus than air. This would tend to make sound waves travel faster through iron than it does through air. But iron also has a much higher density than air, which would tend to make sound waves travel slower through it. So which is it? Does sound travel faster though iron or slower? Well it turns out that the higher stiffness of iron more than compensates for the increased density and the speed of sound through iron is about 14 times faster than through air. This means that if you were to place one ear on a railroad track and someone far away struck the same railroad track with a hammer, you should hear the noise 14 times faster in the ear placed on the track compared to the ear just listening through the air. In fact, the larger bulk modulus of more rigid materials usually compensates for any larger densities. Because of this fact, the speed of sound is almost always faster through solids than it is through liquids and faster through liquids than it is through gases because solids are more rigid than liquids and liquids are more rigid than gases. Density is important in some aspects too though. For instance, if you heat up the air that a sound wave is travelling through, the density of the air decreases. This explains why sound travels faster through hotter air compared to colder air. The speed of sound at 20 degrees Celsius is about 343 meters per second, but the speed of sound at zero degrees Celsius is only about 331 meters per second. Remember, the only way to change the speed of sound is to change the properties of the medium it's travelling in and the speed of sound is typically faster through solids than it is through liquids and faster through liquids than it is through gases." + }, + { + "Q": "so if there are two supermassive black holes in the universe, and their gravitational force is the same, what would happen to the mass in between those two black holes?", + "A": "It would probably be more likely that the two super massive black holes joined each other and suck in everything in between but the black hole the matter got sucked into would depend on the distance to the black holes and the size of the black holes.", + "video_name": "DxkkAHnqlpY", + "transcript": "In the videos on massive stars and on black holes, we learned that if the remnant of a star, of a massive star, is massive enough, the gravitational contraction, the gravitational force, will be stronger than even the electron degeneracy pressure, even stronger than the neutron degeneracy pressure, even stronger than the quark degeneracy pressure. And everything would collapse into a point. And we called these points black holes. And we learned there's an event horizon around these black holes. And if anything gets closer or goes within the boundary of that event horizon, there's no way that it can never escape from the black hole. All it can do is get closer and closer to the black hole. And that includes light. And that's why it's called a black hole. So even though all of the mass is at the central point, this entire area, or the entire surface of the event horizon, this entire surface of the event horizon-- I'll do it in purple because it's supposed to be black-- this entire thing will appear black. It will emit no light. Now these type of black holes that we described, we call those stellar black holes. And that's because they're formed from collapsing massive stars. And the largest stellar black holes that we have observed are on the order of 33 solar masses, give or take. So very massive to begin with, let's just be clear. And this is what the remnant of the star has to be. So a lot more of the original star's mass might have been pushed off in supernovae. That's plural of supernova. Now there's another class of black holes here and these are somewhat mysterious. And they're called supermassive black holes. And to some degree, the word \"super\" isn't big enough, supermassive black holes, because they're not just a little bit more massive than stellar black holes. They're are a lot more massive. They're on the order of hundreds of thousands to billions of solar masses, hundred thousands to billions times the mass of our Sun, solar masses. And what's interesting about these, other than the fact that there are super huge, is that there doesn't seem to be black holes in between or at least we haven't observed black holes in between. The largest stellar black hole is 33 solar masses. And then there are these supermassive black holes that we think exist. And we think they mainly exist in the centers of galaxies. And we think most, if not all, centers of galaxies actually have one of these supermassive black holes. But it's kind of an interesting question, if all black holes were formed from collapsing stars, wouldn't we see things in between? So one theory of how these really massive black holes form is that you have a regular stellar black hole in an area that has a lot of matter that it can accrete around it. So I'll draw the-- this is the event horizon around it. The actual black hole is going to be in the center of it, or rather the mass of the black hole will be in the center of it. And then over time, you have just more and more mass just falling into this black hole. Just more and more stuff just keeps falling into this black hole. And then it just keeps growing. And so this could be a plausible reason, or at least the mass in the center keeps growing and so the event horizon will also keep growing in radius. Now this is a plausible explanation based on our current understanding. But the reason why this one doesn't gel that well is if this was the explanation for supermassive black holes, you expect to see more black holes in between, maybe black holes with 100 solar masses, or a 1,000 solar masses, or 10,000 solar masses. But we're not seeing those right now. We just see the stellar black holes, and we see the supermassive black holes. So another possible explanation-- my inclinations lean towards this one because it kind of explains the gap-- is that these supermassive black holes actually formed shortly after the Big Bang, that these are primordial black holes. These started near the beginning of our universe, primordial black holes. Now remember, what do you need to have a black hole? You need to have an amazingly dense amount of matter or a dense amount of mass. If you have a lot of mass in a very small volume, then their gravitational pull will pull them closer, and closer, and closer together. And they'll be able to overcome all of the electron degeneracy pressures, and the neutron degeneracy pressures, and the quark degeneracy pressures, to really collapse into what we think is a single point. I want to be clear here, too. We don't know it's a single point. We've never gone into the center of a black hole. Just the mathematics of the black holes, or at least as we understand it right now, have everything colliding into a single point where the math starts to break down. So we're really not sure what happens at that very small center point. But needless to say, it will be an unbelievably, maybe infinite, maybe almost infinitely, dense point in space, or dense amount of matter. And the reason why I kind of favor this primordial black hole and why this would make sense is right after the formation of the universe, all of the matter in the universe was in a much denser space because the universe was smaller. So let's say that this is right after the Big Bang, some period of time after the Big Bang. Now what we've talked about before when we talked about cosmic background is that at that point, the universe was relatively uniform. It was super, super dense but it was relatively uniform. So a universe like this, there's no reason why anything would collapse into black holes. Because if you look at a point here, sure, there's a ton of mass very close to it. But it's very close to it in every direction. So the gravitational force would be the same in every direction if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly nonuniform. So let's say it becomes slightly nonuniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly nonuniform, but extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you could imagine in that primordial universe, that very shortly after the Big Bang when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough, so that you could kind of have this snowballing effect, so that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the Sun. And, this is maybe even the more interesting part, those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going pass it fast enough, it'll just start going in orbit around the black hole. And so you could imagine that this is how the early galaxies or even our galaxy formed. And so you might be wondering, well, what about the black hole at the center of the Milky Way? And we think there is one. We think there is one because we've observed stars orbiting very quickly around something at the center of the universe-- sorry, at the center of our Milky Way. I want to be very clear, not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the Sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are and somehow they grow into supermassive black holes and that everything in between we just can't observe. Or that they really are a different class of black holes. They're actually formed different ways. Maybe they formed near the beginning of the actual universe. When the density of things was a little uniform, things condensed into each other. And what we're going to talk about in the next video is how these supermassive black holes can help generate unbelievable sources of radiation, even though the black holes themselves aren't emitting them. And those are going to be quasars." + }, + { + "Q": "If you were to have an element in the 5th period and in the d-block, what is the reason for subtracting 1 when writing out the electon conginguration.", + "A": "Because, the period actually starts a row higher than the chart pictured.", + "video_name": "YURReI6OJsg", + "transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. Then we fill out 2p6. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. We're going to go and backfill the third shell. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that they have a higher probability of being further from the nucleus than these right here. Now, another way to figure out the electron configuration for nickel-- and this is covered in some chemistry classes, although I like the way we just did it because you look at the periodic table and you gain a familiarity with it, which is important, because then you'll start having an intuition for how different elements react with each other -- is to just say, oK, nickel has 28 electrons, if it's neutral. It has 28 electrons, because that's the same number of protons, which is the atomic number. Remember, 28 just tells you how many protons there are. This is the number of protons. We're assuming it's neutral. So it has the same number of electrons. That's not always going to be the case. But when you do these electron configurations, that tends to be the case. So if we say nickel has 28, has an atomic number of 28, so it's electron configuration we can do it this way, too. We can write the energy shells. So one, two, three, four. And then on the top we write s, p, d. Well we're not going to get to f. But you could write f and g and h and keep going. What's going to happen is you're going to fill this one first, then you're going to fill this one, then that one, then this one, then this one. Let me actually draw it. So what you do is, these are the shells that exist, period. These are the shells that exist, in green. What I'm drawing now isn't the order that you fill them. This is just, they exist. So there is a 3d subshell. There's not a 3f subshell. There is a 4f subshell. Let me draw a line here, just so it becomes a little bit neater. And the way you fill them is you make these diagonals. So first you fill this s shell like that, then you fill this one like that. Then you do this diagonal down like that. Then you do this diagonal down like that. And then this diagonal down like that. And you just have to know that there's only two can fit in s, six in p, in this case, 10 in d. And we can worry about f in the future, but if you look at the f-block on a periodic table, you know how many there are in f. So you fill it like that. So first you just say, OK. For nickel, 28 electrons. So first I fill this one out. So that's 1s2. 1s2. Then I go, there's no 1p, so then I go to 2s2. Let me do this in a different color. So then I go right here, 2s2. That's that right there. Then I go up to this diagonal, and I come back down. And then there's 2p6. And you have to keep track of how many electrons you're dealing with, in this case. So we're up to 10 now. So we used that one up. Then the arrow tells us to go down here, so now we do the third energy shell. So 3s2. And then where do we go next? 3s2. Then we follow the arrow. We start there, there's nothing there, there's something here. So we go to 3p6. And then the next thing we fill out is 4s2. So then we go to 4s2. And then what's the very next thing we fill out? We have to go back to the top. We come here and then we fill out 3d. And then how many electrons do we have left to fill out? So we're going to be in 3d. So 3d. And how many have we used so far? 2 plus 2 is 4. 4 plus 6 is 10. 10 plus 2 is 12. 18. 20. We've used 20, so we have 8 more electrons to configure. And the 3d subshell can fit the 8 we need, so we have 3d8. And there you go, you've got the exact same answer that we had when we used the first method. Now I like the first method because you're looking at the periodic table the whole time, so you kind of understand an intuition of where all the elements are. And you also don't have to keep remembering, OK, how many have I used up as I filled the shells? Right? Here you have to say, i used two, then I used two more. And you have to draw this kind of elaborate diagram. Here you can just use the periodic table. And the important thing is you can work backwards. Here there's no way of just eyeballing this and saying, OK, our most energetic electrons are going to be and our highest energy shell is going to be 4s2. There's no way you could get that out of this without going through this fairly involved process. But when do you use this method, you can immediately say, OK, if I'm worried about element Zr, right here. If I'm worried about element Zr. I could go through the whole exercise of filling out the entire electron configuration. But usually the highest shell, or the highest energy electrons, are the ones that matter the most. So you immediately say, OK, I'm filling in 2 d there, but remember, d, you go one period below. So this is 4d2. Right? Because the period is five. So you say, 4d2. 4d2. And then, before that, you filled out the 5s2 electrons. The 5s2 electrons. And then you could keep going backwards. And you filled out the 4p6. 4p6. And then, before you filled out the 4p6. then you had 10 in the d here. But what is that? It's in the fourth period, but d you subtract one from it, so this is 3d10. So 3d10. And then you had 4s2. This is getting messy. Let me just write that. So you have 4d2. That's those two there. Then you have 5s2. 5s2. Then we had 4p6. That's over here. Then we had 3d10. Remember, 4 minus 1, so 3d10. And then you had 4s2. And you just keep going backwards like that. But what's nice about going backwards is you immediately know, OK, what electrons are in my highest energy shell? Well I have this five as the highest energy shell I'm at. And these two that I filled right there, those are actually the electrons in the highest energy shell. They're not the highest energy electrons. These are. But these are kind of the ones that have the highest probability of being furthest away from the nucleus. So these are the ones that are going to react. And these are the ones that matter for most chemistry purposes. And just a little touchpoint here, and this isn't covered a lot, but we like to think that electrons are filling these buckets, and they stay in these buckets. But once you fill up an atom with electrons, they're not just staying in this nice, well-behaved way. They're all jumping between orbitals, and doing all sorts of crazy, unpredictable things. But this method is what allows us to at least get a sense of what's happening in the electron. For most purposes, they do tend to react or behave in ways that these orbitals kind of stay to themselves. But anyway, the main point of here is really just to teach you how to do electron configurations, because that's really useful for later on knowing how things will interact. And what's especially useful is to know what electrons are in the outermost shell, or what are the valence electrons." + }, + { + "Q": "Should their be a multiplication symbol in between 4 and 5??", + "A": "Use of parentheses are one way of showing multiplication, so you do not need a second symbol, that would be extraneous (not needed).", + "video_name": "GiSpzFKI5_w", + "transcript": "We're asked to simplify 8 plus 5 times 4 minus, and then in parentheses, 6 plus 10 divided by 2 plus 44. Whenever you see some type of crazy expression like this where you have parentheses and addition and subtraction and division, you always want to keep the order of operations in mind. Let me write them down over here. So when you're doing order of operations, or really when you're evaluating any expression, you should have this in the front of your brain that the top priority goes to parentheses. And those are these little brackets over here, or however Those are the parentheses right there. That gets top priority. Then after that, you want to worry about exponents. There are no exponents in this expression, but I'll just write it down just for future reference: exponents. One way I like to think about it is parentheses always takes top priority, but then after that, we go in descending order, or I guess we should say in-- well, yeah, in descending order of how fast that computation is. When I say fast, how fast it grows. When I take something to an exponent, when I'm taking something to a power, it grows really fast. Then it grows a little bit slower or shrinks a little bit slower if I multiply or divide, so that comes next: multiply or divide. Multiplication and division comes next, and then last of all comes addition and subtraction. So these are kind of the slowest operations. This is a little bit faster. This is the fastest operation. And then the parentheses, just no matter what, always take priority. So let's apply it over here. Let me rewrite this whole expression. So it's 8 plus 5 times 4 minus, in parentheses, 6 plus 10 divided by 2 plus 44. So we're going to want to do the parentheses first. We have parentheses there and there. Now this parentheses is pretty straightforward. Well, inside the parentheses is already evaluated, so we could really just view this as 5 times 4. So let's just evaluate that right from the get go. So this is going to result in 8 plus-- and really, when you're evaluating the parentheses, if your evaluate this parentheses, you literally just get 5, and you evaluate that parentheses, you literally just get 4, and then they're next to each other, so you multiply them. So 5 times 4 is 20 minus-- let me stay consistent with the colors. Now let me write the next parenthesis right there, and then inside of it, we'd evaluate this first. Let me close the parenthesis right there. And then we have plus 44. So what is this thing right here evaluate to, this thing inside the parentheses? Well, you might be tempted to say, well, let me just go left to right. 6 plus 10 is 16 and then divide by 2 and you would get 8. But remember: order of operations. Division takes priority over addition, so you actually want to do the division first, and we could actually write it here like this. You could imagine putting some more parentheses. Let me do it in that same purple. You could imagine putting some more parentheses right here to really emphasize the fact that you're going to do the division first. So 10 divided by 2 is 5, so this will result in 6, plus 10 divided by 2, is 5. 6 plus 5. Well, we still have to evaluate this parentheses, so this results-- what's 6 plus 5? Well, that's 11. So we're left with the 20-- let me write it all down again. We're left with 8 plus 20 minus 6 plus 5, which is 11, plus 44. And now that we have everything at this level of operations, we can just go left to right. So 8 plus 20 is 28, so you can view this as 28 minus 11 plus 44. 28 minus 11-- 28 minus 10 would be 18, so this is going to be 17. It's going to be 17 plus 44. And then 17 plus 44-- I'll scroll down a little bit. 7 plus 44 would be 51, so this is going to be 61. So this is going to be equal to 61. And we're done!" + }, + { + "Q": "Aren't there 2 forces her, at least? 1. The car pushes against the wall of the loop, creating a centripetal force that pushes inward, and 2. To this we add the pull exercised by gravity? If the loop was horizontal, and the wheels on the side, the car would also rotate against the loop, without the need for gravity, so gravity is an amplifying force here, and the push inwards needs to be compounded?", + "A": "The push inward needs to be biggest at the bottom, where it has to also counteract gravity to get sufficient centripetal force, and it can be lowest at the top, where gravity is providing a lot of the centripetal force.", + "video_name": "4SQDybFjhRE", + "transcript": "What I want to do now is figure out, what's the minimum speed that the car has to be at the top of this loop de loop in order to stay on the track? In order to stay in a circular motion. In order to not fall down like this. And I think we can all appreciate that is the most difficult part of the loop de loop, at least in the bottom half right over here. The track itself is actually what's providing the centripetal force to keep it going in a circle. But when you get to the top, you now have gravity that is pulling down on the car, almost completely. And the car will have to maintain some minimum speed in order to stay in this circular path. So let's figure out what that minimum speed is. And to help figure that out, we have to figure out what the radius of this loop de loop actually is. And it actually does not look like a perfect circle, based on this little screen shot that I got here. It looks a little bit elliptical. But it looks like the radius of curvature right over here is actually smaller than the radius of the curvature of the entire loop de loop. That if you made this into a circle, it would actually be maybe even a slightly smaller circle. But let's just assume, for the sake of our arguments right over here, that this thing is a perfect circle. And it was a perfect circle, let's think about what that minimum velocity would have to be up here at the top of the loop de loop. So we know that the magnitude of your centripetal acceleration is going to be equal to your speed squared divided by the radius of the circle that you are going around. Now at this point right over here, at the top, which is going to be the hardest point, the magnitude of our acceleration, this is going to be 9.81 meters per second squared. And the radius, we can estimate-- I copied and pasted the car, and it looks like I can get it to stack on itself four times to get the radius of this circle right over here. And I looked it up on the web, and a car about this size is going to be about 1.5 meters high from the bottom of the tires to the top of the car. And so it looks like-- just eyeballing it based on these copying and pasting of the cars, that the radius of this loop de loop right over here is 6 meters. So this right over here is 6 meters. So you multiply both sides by 6 meters. Or actually, we could keep it just in the variables. So let me just rewrite it-- just to manipulate it so we can solve for v. We have v squared over r is equal to a. And then you multiply both sides by r. You get v squared is equal to a times r. And then you take the principal square root of both sides. You get v is equal to the principal square root of a times r. And then if we plug in these numbers, this velocity that we have to have in order to stay in the circle is going to be the square root of 9.81 meters per second squared, times 6 meters. And you can verify that these units work out. Meters times meters is meter squared, per second squared. You take the square root of that, you're going to get meters per second. But let's get our calculator out to actually calculate this. So we are going to get the principal square root of 9.81 times 6 meters. It gives us-- now here's our drum roll-- 7.67. I'll just round to three significant digits, 7.67 meters per second squared. And significant digits is a whole conversation, because this is just a very, very rough approximation. I'm not able to measure this that accurately at all. But I get roughly 7 point-- I'll just round, 7.7 meters per second. So this is approximately 7.7 meters per second. And just to give a sense of how that translates into units that we're used to when we're driving cars, we can convert 7.7 meters per second. If we want to say how many meters we go to an hour, well, there's 3,600 seconds in an hour. And then if you want to convert that into kilometers-- this will be in meters-- you divide by 1,000. One kilometer is equal to 1,000 meters. And you see here, the units cancel out. You have meters, meters, seconds, seconds. You're left with kilometers per hour. So let's actually calculate this. And so we get our previous answer. We want to multiply it times 3,600 to figure out how many meters in an hour. And then you divide by 1,000 to convert it to kilometers per hour. So you divide by 1,000. And we get 27.6 kilometers per hour. So this is equal to 27.6 kilometers per hour, which is surprisingly slow. I would have thought it would have to be much, much, much faster. But it turns out, it does not have to be much, much faster. Only 27.6 kilometers per hour. Now the important thing to keep in mind is this is just fast enough, at this point, to maintain the circular motion. But if this were a perfect circle right over here, and you were going at exactly 27.6 kilometers per hour, you would not have much traction with the road. And if you don't have much traction with the road, the car might slip and might not be able to actually maintain its speed. So you definitely want your speed to be a good bit larger than this in order to keep a nice margin of safety-- in order to especially have traction with the actual loop de loop, and to be able to maintain your speed. Now what I want to do in the next video is actually time the car to figure out how long does it take it to do this loop de loop. And we're going to assume that it's a circle. And we're going to figure it out. And we're going to figure out how fast it's actual average velocity was over the course of this loop de loop." + }, + { + "Q": "Why is it called quadratic if it is raised to the second power. Shouldn't it be called bidratic or something?", + "A": "It actually comes from the Latin word quadratum , which means square . The reason lies in geometry: x^2 is the surface of a square with sides of length x.", + "video_name": "eF6zYNzlZKQ", + "transcript": "In this video I want to do a bunch of examples of factoring a second degree polynomial, which is often called a quadratic. Sometimes a quadratic polynomial, or just a quadratic itself, or quadratic expression, but all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, in all of the examples we'll do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x, plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a, and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, which is bx, plus a times x, plus a times b-- plus ab. Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. So can we somehow pattern match this to that? Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9. So we could factor this as being x plus 1, times x plus 9. And if you multiply these two out, using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x, plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form, so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples. I think the more examples we do the more sense this'll make. Let's say we had x squared plus 10x, plus-- well, I already did 10x, let's do a different number-- x squared plus 15x, plus 50. And we want to factor this. Well, same drill. We have an x squared term. We have a first degree term. This right here should be the sum of two numbers. And then this term, the constant term right here, should be the product of two numbers. So we need to think of two numbers that, when I multiply them I get 50, and when I add them, I get 15. And this is going to be a bit of an art that you're going to develop, but the more practice you do, you're going to see that it'll start to come naturally. So what could a and b be? Let's think about the factors of 50. It could be 1 times 50. 2 times 25. Let's see, 4 doesn't go into 50. It could be 5 times 10. I think that's all of them. Let's try out these numbers, and see if any of these add up to 15. So 1 plus 50 does not add up to 15. 2 plus 25 does not add up to 15. But 5 plus 10 does add up to 15. So this could be 5 plus 10, and this could be 5 times 10. So if we were to factor this, this would be equal to x plus 5, times x plus 10. And multiply it out. I encourage you to multiply this out, and see that this is indeed x squared plus 15x, plus 10. In fact, let's do it. x times x, x squared. x times 10, plus 10x. 5 times x, plus 5x. 5 times 10, plus 50. Notice, the 5 times 10 gave us the 50. The 5x plus the 10x is giving us the 15x in between. So it's x squared plus 15x, plus 50. Let's up the stakes a little bit, introduce some negative signs in here. Let's say I had x squared minus 11x, plus 24. Now, it's the exact same principle. I need to think of two numbers, that when I add them, need to be equal to negative 11. a plus b need to be equal to negative 11. And a times b need to be equal to 24. Now, there's something for you to think about. When I multiply both of these numbers, I'm getting a positive number. I'm getting a 24. That means that both of these need to be positive, or both of these need to be negative. That's the only way I'm going to get a positive number here. Now, if when I add them, I get a negative number, if these were positive, there's no way I can add two positive numbers and get a negative number, so the fact that their sum is negative, and the fact that their product is positive, tells me that both a and b are negative. a and b have to be negative. Remember, one can't be negative and the other one can't be positive, because the product would be negative. And they both can't be positive, because when you add them it would get you a positive number. So let's just think about what a and b can be. So two negative numbers. So let's think about the factors of 24. And we'll kind of have to think of the negative factors. But let me see, it could be 1 times 24, 2 times 11, 3 times 8, or 4 times 6. Now, which of these when I multiply these-- well, obviously when I multiply 1 times 24, I get 24. When I get 2 times 11-- sorry, this is 2 times 12. I get 24. So we know that all these, the products are 24. But which two of these, which two factors, when I add them, should I get 11? And then we could say, let's take the negative of both of those. So when you look at these, 3 and 8 jump out. 3 times 8 is equal to 24. 3 plus 8 is equal to 11. But that doesn't quite work out, right? Because we have a negative 11 here. But what if we did negative 3 and negative 8? Negative 3 times negative 8 is equal to positive 24. Negative 3 plus negative 8 is equal to negative 11. So negative 3 and negative 8 work. So if we factor this, x squared minus 11x, plus 24 is going to be equal to x minus 3, times x minus 8. Let's do another one like that. Actually, let's mix it up a little bit. Let's say I had x squared plus 5x, minus 14. So here we have a different situation. The product of my two numbers is negative, right? a times b is equal to negative 14. My product is negative. That tells me that one of them is positive, and one of them is negative. And when I add the two, a plus b, it'd be equal to 5. So let's think about the factors of 14. And what combinations of them, when I add them, if one is positive and one is negative, or I'm really kind of taking the difference of the two, do I get 5? So if I take 1 and 14-- I'm just going to try out things-- 1 and 14, negative 1 plus 14 is negative 13. Negative 1 plus 14 is 13. So let me write all of the combinations that I could do. And eventually your brain will just zone in on it. So you've got negative 1 plus 14 is equal to 13. And 1 plus negative 14 is equal to negative 13. So those don't work. That doesn't equal 5. Now what about 2 and 7? If I do negative 2-- let me do this in a different color-- if I do negative 2 plus 7, that is equal to 5. That worked! I mean, we could have tried 2 plus negative 7, but that'd be equal to negative 5, so that wouldn't have worked. But negative 2 plus 7 works. And negative 2 times 7 is negative 14. So there we have it. We know it's x minus 2, times x plus 7. That's pretty neat. Negative 2 times 7 is negative 14. Negative 2 plus 7 is positive 5. Let's do several more of these, just to really get well honed this skill. So let's say we have x squared minus x, minus 56. So the product of the two numbers have to be minus 56, have to be negative 56. And their difference, because one is going to be positive, and one is going to be negative, right? Their difference has to be negative 1. And the numbers that immediately jump out in my brain-- and I don't know if they jump out in your brain, we just learned this in the times tables-- 56 is 8 times 7. I mean, there's other numbers. It's also 28 times 2. It's all sorts of things. But 8 times 7 really jumped out into my brain, because they're very close to each other. And we need numbers that are very close to each other. And one of these has to be positive, and one of these has to be negative. Now, the fact that when their sum is negative, tells me that the larger of these two should probably be negative. So if we take negative 8 times 7, that's equal to negative 56. And then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8, times x plus 7. This is often one of the hardest concepts people learn in algebra, because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors when one is positive, one is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more. Let's say we had negative x squared-- everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x, plus 24. How do we do this? Well, the easiest way I can think of doing it is factor out a negative 1, and then it becomes just like the problems we've been doing before. So this is the same thing as negative 1 times positive x squared, plus 5x, minus 24. Right? I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1. And you get that right there. Now, same game as before. I need two numbers, that when I take their product I get negative 24. So one will be positive, one will be negative. When I take their sum, it's going to be 5. So let's think about 24 is 1 and 24. Let's see, if this is negative 1 and 24, it'd be positive 23, if it was the other way around, it'd be negative 23. Doesn't work. What about 2 and 12? Well, if this is negative-- remember, one of these has to be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works! So if we pick negative 3 and 8, negative 3 and 8 work. Because negative 3 plus 8 is 5. Negative 3 times 8 is negative 24. So this is going to be equal to-- can't forget that negative 1 out front, and then we factor the inside. Negative 1 times x minus 3, times x plus 8. And if you really wanted to, you could multiply the negative 1 times this, you would get 3 minus x if you did. Or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared plus 18x, minus 72. So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared, minus 18x, plus 72. Now we just have to think of two numbers, that when I multiply them I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18. So they're the same sign, and their sum is a negative number, they both must be negative. And we could go through all of the factors of 72. But the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9, doesn't work. That turns into 17. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17. Close, but no cigar. We have 6 and 12. That actually seems pretty good. If we have negative 6 plus negative 12, that is equal to negative 18. Notice, it's a bit of an art. You have to try the different factors here. So this will become negative 1-- don't want to forget that-- times x minus 6, times x minus 12." + }, + { + "Q": "What if, in the second question, DE is known but CD isn't? Can you still solve this?", + "A": "You can, but it would get a little more complicated because if CD = x. then CE = x + DE. so just for fun if you put 4 for DE and all else is the same, you would get 5/8 = x/x+4, cross multiply to get 5(x+4) = 8x, distribute 5x + 20 = 8x, subtract 5x, 20= 3x, divide by 3, x = 20/3. More likely if you were given DE, the numbers would be more compatible to not end up with a fraction.", + "video_name": "R-6CAr_zEEk", + "transcript": "In this first problem over here, we're asked to find out the length of this segment, segment CE. And we have these two parallel lines. AB is parallel to DE. And then, we have these two essentially transversals that form these two triangles. So let's see what we can do here. So the first thing that might jump out at you is that this angle and this angle are vertical angles. So they are going to be congruent. The other thing that might jump out at you is that angle CDE is an alternate interior angle with CBA. So we have this transversal right over here. And these are alternate interior angles, and they are going to be congruent. Or you could say that, if you continue this transversal, you would have a corresponding angle with CDE right up here and that this one's just vertical. Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles and two of the corresponding angles are the same. And that by itself is enough to establish similarity. We actually could show that this angle and this angle are also congruent by alternate interior angles, but we don't have to. So we already know that they are similar. And actually, we could just say it. Just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar, even before doing that. So we already know that triangle-- I'll color-code it so that we have the same corresponding vertices. And that's really important-- to know what angles and what sides correspond to what side so that you don't mess up your, I guess, your ratios or so that you do know what's corresponding to what. So we know triangle ABC is similar to triangle-- so this vertex A corresponds to vertex E over here. It's similar to vertex E. And then, vertex B right over here corresponds to vertex D. EDC. Now, what does that do for us? Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it in just the way that we've written down the similarity. If this is true, then BC is the corresponding side to DC. So we know that the length of BC over DC right over here is going to be equal to the length of-- well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to-- what's the corresponding side to CE? The corresponding side over here is CA. It's going to be equal to CA over CE. This is last and the first. Last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. And now, we can just solve for CE. Well, there's multiple ways that you could think about this. You could cross-multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE. 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2/5, or 2.4. So this is going to be 2 and 2/5. And we're done. We were able to use similarity to figure out this side just knowing that the ratio between the corresponding sides are going to be the same. Now, let's do this problem right over here. Let's do this one. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also, in both triangles-- so I'm looking at triangle CBD and triangle CAE-- they both share this angle up here. Once again, we could have stopped at two angles, but we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other. And once again, this is an important thing to do, is to make sure that you write it in the right order when you write your similarity. We now know that triangle CBD is similar-- not congruent-- it is similar to triangle CAE, which means that the ratio of corresponding sides are going to be constant. So we know, for example, that the ratio between CB to CA-- so let's write this down. We know that the ratio of CB over CA is going to be equal to the ratio of CD over CE. And we know what CB is. CB over here is 5. We know what CA is. And we have to be careful here. It's not 3. CA, this entire side is going to be 5 plus 3. So this is going to be 8. And we know what CD is. CD is going to be 4. And so once again, we can cross-multiply. We have 5CE. 5 times CE is equal to 8 times 4. 8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to think about that, 6 and 2/5. Now, we're not done because they didn't ask for what CE is. They're asking for just this part right over here. They're asking for DE. So we know that this entire length-- CE right over here-- this is 6 and 2/5. And so DE right over here-- what we actually have to figure out-- it's going to be this entire length, 6 and 2/5, minus 4, minus CD right over here. So it's going to be 2 and 2/5. 6 and 2/5 minus 4 and 2/5 is 2 and 2/5. DE is 2 and 2/5." + }, + { + "Q": "At 1:36 what does he mean by \"third variable S\"", + "A": "I think he meant to say the fourth variable (three sides of the triangle being the first three), but the ordinal number used to refer to it is irrelevant. s = the semi-perimeter of the triangle. That would be (a + b + c) / 2.", + "video_name": "-YI6UC4qVEY", + "transcript": "I think it's pretty common knowledge how to find the area of the triangle if we know the length of its base and its height. So, for example, if that's my triangle, and this length right here-- this base-- is of length b and the height right here is of length h, it's pretty common knowledge that the area of this triangle is going to be equal to 1/2 times the base times the height. So, for example, if the base was equal to 5 and the height was equal to 6, then our area would be 1/2 times 5 times 6, which is 1/2 times 30-- which is equal to 15. Now what is less well-known is how to figure out the area of a triangle when you're only given the sides of the triangle. When you aren't given the height. So, for example, how do you figure out a triangle where I just give you the lengths of the sides. Let's say that's side a, side b, and side c. a, b, and c are the lengths of these sides. How do you figure that out? And to do that we're going to apply something called Heron's Formula. And I'm not going to prove it in this video. I'm going to prove it in a future video. And really to prove it you already probably have the tools necessary. It's really just the Pythagorean theorem and a lot of hairy algebra. But I'm just going to show you the formula now and how to apply it, and then you'll hopefully appreciate that it's pretty simple and pretty easy to remember. And it can be a nice trick to impress people with. So Heron's Formula says first figure out this third variable S, which is essentially the perimeter of this triangle divided by 2. a plus b plus c, divided by 2. Then once you figure out S, the area of your triangle-- of this triangle right there-- is going to be equal to the square root of S-- this variable S right here that you just calculated-- times S minus a, times S minus b, times S minus c. That's Heron's Formula right there. This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18. And then the area by Heron's Formula is going to be equal to the square root of S-- 18-- times S minus a-- S minus 9. 18 minus 9, times 18 minus 11, times 18 minus 16. And then this is equal to the square root of 18 times 9 times 7 times 2. Which is equal to-- let's see, 2 times 18 is 36. So I'll just rearrange it a bit. This is equal to the square root of 36 times 9 times 7, which is equal to the square root of 36 times the square root of 9 times the square root of 7. The square root of 36 is just 6. This is just 3. And we don't deal with the negative square roots, because you can't have negative side lengths. And so this is going to be equal to 18 times the square root of 7. So just like that, you saw it, it only took a couple of minutes to apply Heron's Formula, or even less than that, to figure out that the area of this triangle right here is equal to 18 square root of seven. Anyway, hopefully you found that pretty neat." + }, + { + "Q": "Cant mushy or sticky go in material?\nThey are both physical traits.....", + "A": "I m a bit confused about these categories myself, but I think that the material category refers more to what something is made of, such as the wooden spoon or the glass vase. Mushy and sticky are more relative terms, and so they would fit in the opinion category. But I totally understand your confusion, though. I had never even heard of DOSASCOMP before watching this video. Also, WOW! 5 million energy points? That s impressive.", + "video_name": "OfxiZdsqGeA", + "transcript": "- [David] Hey grammarians, hey Paige. - [Paige] Hi David. - [David] Hey, so Paige, I went to the grocery store yesterday and I got this apple and I put it in the fridge. And this morning, when I opened the fridge, the apple was all gross and sticky and mushy and I really wanna write a letter to the grocery store and say, \"Hey, you sold me a gross apple.\" - [Paige] Yeah. - [David] But I'm stuck. - [Paige] Okay? - [David] I feel like I ought to put a comma in here. So, here's the sentence I've got so far. \"You sold me a mushy, sticky apple!\" - [Paige] Gross. - [David] Gross. - [David] Unacceptable. - [Paige] Totally. - [David] \"So, I would like my $1.38 back.\" (Paige laughs) But in another draft of the letter, I wrote, \"You sold me a mushy green apple!\" - [Paige] Okay, so mushy, sticky or mushy green. - [David] Yeah, and I feel like, and I don't know how to put this into words yet, but we're gonna put it into words 'cause that's our job, (Paige laughs) whether or not there should be a comma between mushy and green or mushy and sticky, 'cause these feel different to me. - [Paige] Right, right. Yeah, so there's actually a couple tests that you can do on these sentences to see if you need a comma between the adjectives or not. - [David] So, Paige, what distinguishes a pair like mushy and sticky from mushy and green? - [Paige] There's this sort of hierarchy or order that you put adjectives in when you have multiple of them in a sentence, and it is called DOSASCOMP. - [David] DOSASCOMP? All together now, D - [Both] Determiner! - [David] O. - [Both] Opinion! - [David] S. - [Both] Size! - [David] A. - [Both] Age! - [David] S again. - [Both] Shape! - [David] C. - [Both] Color! - [David] O. - [Both] Origin! - [David] M. - [Both] Material! - [David] And P. - [Both] Purpose! - [David] Oh, okay. So, mushy and sticky are both opinion adjectives. - [Paige] Yeah. - [David] So, these are kinda in the same class, whereas mushy and green, that's an opinion and a color. - [Paige] Yeah, they're in two different classes. - [David] Right, so DOSASCOMP helps determine adjective order, right? - [David] So, it's the order in which you would use, if in which you were stacking these adjectives you would use them in this DOSASCOMP order. - [Paige] Right, so like in the second sentence, mushy is an opinion, like you said, and green is a color. And in DOSASCOMP, opinion comes before color, so that's the order that you write them in. So, like in the first sentence, when you have mushy and sticky, which are both opinion adjectives, these are called coordinate adjectives. They're in the same category of DOSASCOMP. And if you wanted to, you could reverse the order. You could say, \"Sticky, mushy apple.\" - [David] Well, let's try that. - [Paige] Yeah. - [David] Looks good to me. Sticky, mushy, mushy, sticky. - [Paige] Same thing. - [David] Okay, so if we're trying to determine whether or not we have coordinate adjectives, I guess the first thing we would do is consult DOSASCOMP, right? So that's step one. And what is a dosa, Paige? - [Paige] Oh, it's like a pancake, right? - [David] Yeah, it's like a South Indian pancake. - [Paige] That's pretty cool. - [David] And to scomp is a word we made up that means, \"to eat.\" - [Paige] Sure, eat some pancakes. - [David] Scomp on 'em. So, that's step one. Step two, try the reversal method. - [Paige] Right, and that's like changing it from mushy, sticky to sticky, mushy. - [David] And step three is stick an \"and\" in there. - [Paige] Mm-hmm. If you can reverse the order of the adjectives and you can put \"and\" in between the two of them, then they're coordinate adjectives. - [David] And if they're coordinate adjectives, you need to separate them with a comma. - [Paige] Exactly. And that's why, in the case of the second sentence, with mushy green apple, you don't put a comma between them. They're in different categories in DOSASCOMP, and so they must stay in that order and there's no need for a comma. - [David] Thanks, Paige. I'm gonna get back to drafting my angry letter. (Paige laughs) You can learn anything. - [David] David out. - [Paige] Paige out." + }, + { + "Q": "How can you tell if an angle is going clockwise or counterclockwise if there is no angle arc there to show?", + "A": "You can t unless there is some indication of the direction it is going. Later on when you deal with more complicated mathematics angles usually go counterclockwise.", + "video_name": "wJ37GJyViU8", + "transcript": "This is the video for the measuring angles module because, clearly, at the time that I'm doing this video, there is no video for the measuring angles module. And this is a pretty neat module. This was made by Omar Rizwan, one of our amazing high school interns that we had this past summer. This is the summer of 2011. And what it really is, is it makes you measure angles. And he made this really cool protractor tool here so that you actually use this protractor to measure the angles there. And so the trick here is you would actually measure it the way you would measure any angle using an actual physical protractor. You'd want to put the center of the protractor right at the vertex of where the two lines intersect. You can view it as the vertex of the angle. And then you'd want to rotate it so that, preferably, this edge, this edge at 0 degrees, is at one of these sides. So let's do it so that this edge right over here is right along this line. So let me rotate it. So then-- I've got to rotate it a little bit further, maybe No, that's too far. So that looks about right. And then if you look at it this way, you can see that the angle-- and I don't have my Pen tool here. I'm just using my regular web browser-- if you look at the angle here, you see that the other line goes to 130 degrees. So this angle that we need to measure here is 130 degrees, assuming you can read sideways. So that is 130 degrees. Let me check my answer. Very good, I got it right. It would have been embarrassing if I didn't. Let's do the next question. I'll do a couple of examples like this. So once again, let us put the center of the protractor right at the vertex right over there. And let's get this 0 degrees side to be on one of these sides so that this angle will be within the protractor. So let me rotate it this way. And this really is pretty cool what Omar did with this module. So let's see. Let's do it one more time. That's too far. And so that looks about right. And then you can see that the angle right over here, if we look at where the other line points to, it is 40 degrees. Check answer-- very good. Let's do another one. This is fun. So let's get our protractor right over there. And you don't always have to do it in that same order. You could rotate it first so that the 0 degrees is-- and what you want to do is you want to rotate the 0 degrees to one of the sides so that the angle is still within the protractor. So let's rotate it around. So if you did it like that-- so you don't always have to do it in that same order. Although I think it's easier to rotate it when you have the center of the protractor at the vertex of the angle. So we have to rotate it a little bit more. So 0 degrees is this line. And then as we go further and further up, I guess, since this is on its side, it looks like this other line gets us to 150 degrees. And hopefully you're noticing that the higher the degrees, the more open this angle is. And so this one right over here is 150 degrees. And so let's do that-- 150. Let's do one more. Now let me show you what not to do. So what not to do is-- so you could put the center right over there. And you might say, OK, let me make the 0 go right over on this side, right over here. So if you did that, notice your angle would not be within the protractor. So you won't be able to measure it. And what you're attempting to do is measure this outer angle over here, which is an angle, but that's not the angle that this question is asking us to measure. This little arc over here is telling us that that's the angle that we need to measure. So that arc has to be within the protractor. So let's rotate this protractor a little bit more. I overdid it. And so this looks like this is 0 degrees, and then this right over here is 60 degrees. 60 degrees-- we got that one right, too. So hopefully that helps you with this module. It's kind of fun." + }, + { + "Q": "Can we simply break the numerator (X^3 -1) as X^2 (X-1) ??\nThat way also we can get a X-1 to cancel out with the denominator right?", + "A": "Consider your suggestion, and do the math. You want to rewrite (x\u00c2\u00b3-1) as x\u00c2\u00b2(x-1). That means you are asserting that (x\u00c2\u00b3-1) = x\u00c2\u00b2(x-1). When you distribute the x\u00c2\u00b2 in x\u00c2\u00b2(x-1) you get x\u00c2\u00b3-x\u00c2\u00b2 . . . Oops x\u00c2\u00b3-1 \u00e2\u0089\u00a0 x\u00c2\u00b3-x\u00c2\u00b2. It was an interesting observation, and an opportunity to test your intuition, which you should have tried. Getting into the habit of investigating your ideas and trying them out is an important aspect to getting really good at math! Keep Studying!", + "video_name": "rU222pVq520", + "transcript": "Let's try to find the limit as x approaches 1 of x to the third minus 1 over x squared minus 1. And at first when you just try to substitute x equals 1, you get 0/0 1 minus 1 over 1 minus 1. So that doesn't help us. So let's see if we can try to simplify this in some way. So you might immediately recognize-- so let's rewrite this expression right over here so it's x to the third minus 1 over x squared minus 1. This on the bottom immediately jumps out as a difference of squares. So we know on the bottom that this could be factored as x minus 1 times x plus 1. And so if somehow this thing on the top also has an x minus 1 as a factor, then that x minus 1 will cancel with this, and then we're not going to have an issue of dividing by 0. The reason why I care about the x minus 1 term is that this is what's making our denominator equal 0. When you say x equals 1, you have 1 minus 1 times 1 plus 1. So 0 times 2, it's this 0 that's making our denominator 0. So if we can have an x minus 1 up here, then we can cancel these out for any x not equal to 1. And then we might have a much simpler thing to find the limit of. So let's think about whether x to the third minus 1 is the product of x minus 1 and something else. And to do that we can do a little bit of algebraic long division. Some of you guys might already recognize a pattern here, but we'll try to do-- well, let's divide x minus 1 into it to see whether it divides evenly into x to the third minus 1. So x minus 1-- we just look at the highest degree term-- x goes into x to the third x squared times. Goes x squared times. Actually, let me do it this way so that way we can keep track of the place. So this would be x-- this would be the second degree place, first degree place, and this would be the constant. So x to the third minus 1. x goes into x to the third x squared times. x squared times x is x to the third. x squared times negative 1 is minus x squared. And now we're going to want to subtract this. So we are then left with x squared. x goes into x squared x times plus x. x times x is x squared. x times minus 1 is minus x. And once again we're going to subtract this. We'll swap the signs, negative and positive. And so these cancel out, and we're left with x. And then we bring down a minus 1. x minus 1 goes into x minus 1 exactly one time. 1 times x minus 1 is x minus 1. And then you subtract, and then you have no remainder. So this numerator right over here can be factored as x minus 1 times x squared plus x plus 1. And so we can say that this is the same exact thing. We can have these cancel out if we assume x does not equal 1. So that is equal to x squared plus x plus 1 over x plus 1, for x does not equal 1. And that's completely fine, because we're not evaluating x equals 1. We're evaluating as x approaches 1. So this is going to be the same thing as the limit as x approaches 1 of x squared plus x plus 1 over x plus 1. And now this is much easier to find. You could literally just say, well, what happens as we get right to x equals 1? Then you have 1 squared, which is 1 plus 1 plus 1, which is 3, over 1 plus 1, which is 2. So we get that equaling 3/2." + }, + { + "Q": "I have a question about demand curves in general. If the quantity supplied cannot be increased, is the price more likely to go up?", + "A": "I don t know a lot about microeconomics, but i do know that if the demand for a product is large, and the quantity of the product cannot be increased as you said, the product will become more expensive. For example, if a new Samsung phone was released and it became crazy popular, but there were only 20,000 of them manufactured, the price would skyrocket because of the limited quantity. Not to mention there would be huge fights at stores.", + "video_name": "do1HDIdfQkU", + "transcript": "So we've been going through all of the other things that we were assuming are held constant in order to be moving along one demand curve. And now let's list a few other. And before I do any more of them, let's talk about the ones we already talked about. So one, we said that one of the things we held constant-- let me write this down. So held constant. One of the things that we held constant to move along one demand curve for the demand itself to not shift, for the curve to not shift, is price of related goods. The other thing we assumed that's being held constant is price expectations for our good. And now we'll list a couple of them that are fairly intuitive, but you'll see in the next few videos that there are often special cases even to this. So the other thing that we've been holding constant to stay on one demand curve is income. And this one is fairly intuitive. What happens if everyone's income were to increase? And in real terms, it were to actually increase. Well then, all of a sudden, they have more disposable income, maybe to spend on something like e-books. And so for any given price point, the demand would increase. And so it would increase the demand. And once again, when we talk about increasing demand, we're talking about shifting the entire curve. We're not talking about a particular quantity of demand. So income goes up, then it increases demand. Demand goes up. And remember, when we're talking about when demand goes up, we're talking about the whole curve shifting to the right. At any given price point, we are going to have a larger quantity demanded. So the whole curve, this whole demand schedule would change. And likewise if income went down, demand would go down. And we're going to see in a future video-- it's actually quite interesting-- that's not always the case. This is only true for normal goods. And in a future video we'll see goods called inferior goods where this is not necessarily the case. Or by definition for an inferior good, it would not be the case. Now the other ones that are somewhat intuitive are population-- once again, if population goes up, obviously, at any given price point, more people will want it. So it would shift the demand curve to the right, or it would increase demand. If population were to go down, it would decrease demand, which means shifting the whole curve to the left. And then the last one we'll talk about-- and remember, we're holding all of these things constant in order for demand not to change. The last thing is just preferences. We're assuming that people's tastes and preferences don't change while we move along a specific demand curve. If preferences actually change, then it will change the curve. So for example, if all of a sudden, the author of the book is on some very popular talk show that tells everyone that this is the best book that was ever written, then preferences would go up, and that would increase the total demand. At any given price point, more people will be willing to buy the book. If, on the other hand, on that same talk show, it turns out that they do an expose on the author having this sordid past, and the author plagiarized the whole book, then the demand will go down. The entire curve, regardless of the price point-- at any given price point, the quantity demanded will actually go down." + }, + { + "Q": "So the video 'Price of Related Products and Demand' comes before 'Change in Expected Future Prices and Demand'?", + "A": "yes change in expected future prices and demand will effect price of related products and demand", + "video_name": "-oClpRv7msg", + "transcript": "We've talked a little bit about the law of demand which tells us all else equal, if we raise the price of a product, then the quantity demanded for that product will go down. Common sense. If we lower the price, than the quantity demanded will go up, and we'll see a few special cases for this. But what I want to do in this video is focus on these other things that we've been holding equal, the things that allow us to make this statement, that allow us to move along this curve, and think about if we were to change one of those things, that we were otherwise considering equal, how does that change the actual curve? How does that actually change the whole quantity demanded price relationship? And so the first of these that I will focus on, the first is the price of competing products. So if you assume that the price of-- actually I shouldn't say competing products, I'll say the price of related products, because we'll see that they're not competing. The price of related products is one of the things that we're assuming is constant when we, it's beheld equal when we show this relationship. We're assuming that these other things aren't changing. Now, what would happen if these things changed? Well, imagine we have, say, other ebooks-- books is price-- price goes up. The price of other ebooks go up. So what will that do to our price quantity demanded relationship? If other ebooks prices go up, now all of a sudden, my ebook, regardless of what price point we're at, at any of the price points, my ebook is going to look more desirable. At $2, it's more likely that people will want it, because the other stuff's more expensive. At $4 more people will want it, at $6 more people will want it, $8 more people will want it, at $10 more people will want it. So if this were to happen, that would actually shift the entire demand curve to the right. So it would start to look something like this. That is scenario one. And these other ebooks, we can call them substitutes for my product. So this right over here, these other ebooks, these are substitutes. People might say, oh, you know, that other book looks kind of comparable, if one is more expensive or one is cheaper, maybe I'll read one or the other. So in order to make this statement, in order to stay along this curve, we have to assume that this thing is constant. If this thing changes, this is going to move the curve. If other ebooks prices go up, it'll probably shift our curve to the right. If other ebooks prices go down, that will shift our entire curve to the left. So this is actually changing our demand. It's changing our whole relationship. So it's shifting demand to the right. So let me write that. So this is going to shift demand. So the entire relationship, demand, to the right. I really want to make sure that you have this point clear. When we hold everything else equal, we're moving along a given demand curve. We're essentially saying the demand, the price quantity demanded relationship, is held constant, and we can pick a price and we'll get a certain quantity demanded. We're moving along the curve. If we change one of those things, we might actually shift the curve. We'll actually change this demand schedule, which will change this curve. Now, there other related products, they don't just have to be substitutes. So, for example, let's think about scenario two. Or maybe the price of a Kindle goes up. Let me write this this way. Kindle's price goes up. Now, the Kindle is not a substitute. People don't either buy an ebook or they won't either buy my ebook or a Kindle. Kindle is a compliment. You actually need a Kindle or an iPad or something like it in order to consume my ebook. So this right over here is a complement. So if a complement's price becomes more expensive, and this is one of the things people might use to buy my book, then it would actually, for any given price, lower the quantity demanded. So in this situation, if my book is $2, since fewer people are going to have Kindles, or since maybe they used some of their money already to buy the Kindle, they're going to have less to buy my book or just fewer people will have the Kindle, for any given price is going to lower the quantity demanded. And so it'll essentially will shift, it'll change the entire demand curve will shift the demand curve to the left. So this right over here is scenario two. And you could imagine the other way, if the Kindle's price went down, then that would shift my demand curve to the right. If the price of substitutes went down, then that would shift my entire curve to the left. So you can think about all the scenarios, and actually I encourage you to. Think about drawing yourself, think about for products, that could be an ebook or could be some other type of product, and think about what would happen. Well, one, think about what the related products are, the substitutes and potentially complements, and then think about what happen as those prices change. And always keep in mind the difference between demand, which is this entire relationship, the entire curve that we can move along if we hold everything else equal and only change price, and quantity demanded, which is a particular quantity for our particular price holding everything else equal." + }, + { + "Q": "why is the soviet union so dangerous in the winter", + "A": "It is because the climate of Soviet Union is so cold that the army can t live.", + "video_name": "X3bqQI7-sCg", + "transcript": "- [Voiceover] As we've already seen in the last few videos, with the war officially starting in September of 1939, the Axis powers get momentum through the end of 1939, all the way into 1940. That was the last video that we covered and that takes us to 1941 and what we're gonna see in 1941 which is the focus of this video is that the Axis powers only seem to gain more momentum. Because of all of that momentum they perhaps gets a little bit overconfident and stretch themselves or begin to stretch themselves too thin. So let's think about what happens in 1941. So, if we talk about early 1941 or the Spring of 1941, in March, Bulgaria decides to join the Axis powers. You can imagine there's a lot of pressuring applied to them and they kind of see where the momentum is. Let's be on that side. Bulgaria joins the Axis and then in North Africa you might remember that in 1940, the Allies, in particular, the British, were able to defeat the Italians and push them back into Libya but now in March of 1941, the Italians get reinforcements, Italian reinforcements and also German reinforcements under the command of Rommel the Desert Fox, famous desert commander and they are able to push the British back to the Egyptian border and they also take siege of the town of Tobruk. Now, you might have noticed something that I just drew. The supply lines in the North Africa campaign are very, very, very long and that's part of the reason why there's one side. One side has supply lines and as they start to make progress and as the Allies make progress and push into Libya, their supply lines got really long and so the other side has an easier timely supply. Then as the Axis pushes the Allies back into Egypt, then their supply lines get really long and the other side...it makes it easier for them to resupply and so North Africa is kind of defined by this constant back and forth. But, by early 1941, it looks like the Axis is on the offensive, able to push the British back into Egypt lay siege to the town of Tobruk. So, let me write this down as North Africa. So, I'll just say North Africa over here or I'll could say Rommel in North Africa pushing the British back. And then we can start talking about what happens in the Balkans and this is still in Spring as we go into April of 1941 and just as a little bit of background here, and frankly I should have covered it a couple of videos ago. As far back as 1939, actually before World War II officially started, in Spring of 1939, Italy actually occupies Albania so this actually should have already been red. This is in 1939 that this happens and then at the end of 1940, Italy uses Albania as a base of operation to try to invade Greece but they are pushed back. Actually one of the reasons why the British we able to be pushed back in North Africa is after they were successful against the Italians, most of the bulk of the British forces we sent to Greece to help defend Greece at the end of 1940. So, in 1939, Albania gets taken over by Italy and at the end of 1940...October 1940, Greece is invaded by Italy but they are then pushed back but to help the Greeks, the Allies send many of the forces that were in North Africa after they were successful against the Italians in Libya. Now, as we go into April of 1941... that was all background, remember Albania before the war started in April 1939, October 1940 was Italy's kind of first push into Greece and it was unsuccessful. Then the Greeks get support from the Allies in North Africa and now as we go into 1941, the Germans start supporting and really take charge in Balkans and in Greece and so with the help of the Germans the Axis is able to take over Yugoslavia and Greece and start aerial bombardment of Crete. So, once again, we're not even halfway through the year in 1941 and we see a huge swath of Europe is under the control of the Axis powers. And now we go into the summer of 1941. This is actually a pivotal move, what's about to happen. Now you can imagine that the Axis powers, in particular, Hitler, are feeling pretty confident. We are only about that far into the war. So we're not even two years into the war yet and it looks like the Axis is going to win. Now you might remember that they have a pact with the Soviet Union. Hey, we're gonna split a lot of Eastern Europe into our spheres of influence so to speak, but now Hitler's like, well, I think I'm ready to attack and when you attack the Soviet Union really matters. You do not want to attack the Soviet Union in the winter...or Russia in the winter. Russia's obviously at the heart of the Soviet Union. That something that Napoleon learned. Many military commanders have learnt. You do not want to be fighting in Russia over the winter, so summer of 1941, Hitler figures, hey, this is the Axis chance. And so, in June, he decides to attack the Soviet Union. So, this is a very, very, very bold move because now they're fighting the British. Remember, the British are kind of not a joke to be battling out here in Western Europe and now they're going to be taking on the Soviet Union in the east, a major, major world power. But at first, like always, it seems like it's going well for the Germans. By September, they're able to push up all the way to Leningrad. So, this is September of 1941 and lay lay siege and begin laying siege to that town. This is kind of a long bloody siege that happens there. So, we're right now, right about there. And most historians would tell you that this was one of the mistakes of Adolph Hitler because now he is stretched very, very, very thin. He has to fight two world powers, Soviet Union and Great Britain and the United States hasn't entered into the war yet and that's what we're about to get into because if we go into Asia it was still in 1941 what happens in July. So, little bit after Hitler decides to start invading the Soviet Union going back on the pact, the non- aggression pact. In July, you could imagine the US, they were never pleased with what's been happening, what the Empire of Japan has been doing in the Pacific, what they've been doing in China, in Manchuria or even in terms of the war in China, the second Sino Japanese War. They weren't happy of the Japanese taking over French Indochina. There's a big world power here, the Empire of Japan. There's a big world power here, the United States, that has a lot of possessions in the Pacific and so, the United States in July of 1941...So remember, this is still all 1941, this is the same year...decides to freeze the assets of Japan and probably the most important part of that was an oil embargo of Japan. This is a big, big deal. Japan is fighting a major conflict with the Chinese. It's kind of flexing it's imperial muscles but it does not have many natural resources in and of itself and in fact, that's one of the reasons why it's trying to colonize other places to get more control of natural resources. And now if it's fighting a war it doesn't have it's own oil resources and now there's an oil embargo of Japan and the United States at the time was major oil producer and even today, it's major oil producer. This was a big deal to the Japanese because some estimates say they only had about two years of reserves and they were fighting a war where they might have to touch their reserves even more. So, you could imagine the Japanese, they want to have their imperial ambitions. They probably want, especially now with this oil embargo, they probably want to take over more natural resources and they probably want to knock out the US or at least keep the US on its heels so the US can't stop Japan from doing what it wants to do. So, all it wants in December 1941, that's over the course of December 7th and 8th, and it gets a little confusing because a lot of this happens across the International Date Line. But over the course of December 7th and 8th, Japan goes on the offensive in a major way in the Pacific. Over the course of several hours, at most, a day, Japan is able to attack Malaya, which is a British possession. It's able to attack Pearl Harbor, where the US Pacific fleet is in hope to knock out the US Pacific fleet so the US will have trouble stopping Japan from doing whatever Japan wants to do. In the US, we focus a lot on Pearl Harbor but this was just one of the attacks in this whole kind of several hours of attacks where Japan went on the offensive. So, we have Malaya, we have Pearl Harbor, we have Singapore, we have Guam, we have (which was the US military base), Wake Island. was a US possession ever since the Spanish American war. You have Hong Kong, which is a British possession and then shortly after that as you get further into December, so this is kind of when you have Japan offensive. Then as you go on into later Decemeber, the kind of real prize for Japan was what we would now call Indonesia but the Dutch East Indies. On this map it says Netherlands East Indies. You have to remember the Netherlands had been overrun. They're the low countries they were already overrun by German forces so the Japanese say hey, look there are a lot of resources here, natural resources, especially oil. Let's go for this and so by the end of 1941, they're also going for the Dutch East Indies and for Burma so you could imagine it's a very aggressive, very, very bold move on Japan but they kind of had imperial ambitions. They were afraid of they access to natural resources so they went for it but obviously one of the major consequence of this is the United States was not happy about this and they were already sympathetic to the Allies. They didn't like what was going on in Europe either. They didn't like what was going on in China and so that causes the United States to enter into World War II on the side of the Allies and then the Axis powers to declare war on the United States, which was a big deal." + }, + { + "Q": "At 7:48, it says there's a decrease in Sn2 mechanism. I understand that, but if a strong nucleophile is put in a polar protic solvent, it becomes a weak nucleophile. Then, shouldn't Sn1 be favoured over Sn2 as a minor product ?", + "A": "A polar protic solvent doesn t make a strong nucleophile into a weak nucleophile.It makes it into a weaker nucleophile. Ethoxide ion is a strong nucleophile in an aprotic solvent. It is weaker in a protic solvent, but it is still a pretty strong nucleophile even there.", + "video_name": "vFSZ5PU0dIY", + "transcript": "- [Instructor] Let's look at elimination versus substitution for a secondary substrate. And these are harder than for a primary or tertiary substrate because all four of these are possible to start with. So, if we look at the structure of our substrate and we say it's secondary, we next need to look at the reagent. So, we have NaCl which we know is Na plus and Cl minus and the chloride anion functions only as a nucleophile. So, we would expect a substitution reaction, nucleophilic substitution. So, E1 and E2 are out. Between SN1 and SN2 with the secondary substrate, we're not sure until we look at the solvent and DMSO is a polar aprotic solvent, which we saw in an earlier video, favors an SN2 mechanism. So, SN1 is out and we're gonna think about our chloride anion functioning as a nucleophile. So, let me draw it in over here. So, this is with a negative one formal charge. And an SN2 mechanism are nucleophile attacks the same time we get loss of a leaving group and our nucleophile is going to attack this carbon in red. So, we're gonna form a bond between the chlorine and this carbon in red and when the nucleophile attacks, we also get loss of our leaving group. So, these electrons come off onto the oxygen and we know that tosylate is a good leaving group. So, when we draw our product, let's draw this in here, and the carbon in red is this one, we know an SN2 mechanism means inversion of configuration. The nucleophile has to attack from the side opposite of the leaving group. So, we had a wedge here for our leaving groups, so that means we're gonna have a dash for our chlorines. We're gonna put the chlorine right here and that's the product of our SN2 reaction. For our next problem, we have a secondary alkyl halide. So, just looking at our reactions, we can't really rule any out here. So, all four are possible, until we look at our reagent. Now, we saw in an earlier video, that DBN is a strong base, it does not act like a nucleophile. So SN1 and SN2 are out. And a strong base means an E2 reaction. So, E1 is out. Now that we know we're doing an E2 mechanism, let's analyze the structure of our alkyl halide. The carbon that's directly bonded to our halogen is our alpha carbon and the carbons directly bonded to the alpha carbon are the beta carbons. So, I'll just do the beta carbon on the right since they are the same essentially. And we know that our base is gonna take a proton from that beta carbon. So, let me just draw in a hydrogen here. And DBN is a neutral base, so I'll just draw a generic base here. Our base is going to take this proton at the same time these electrons move in to form a double bond and these electrons come off to form our bromide anion. So, our final product is an alkyne and our electrons in magenta in here moved in to form our double bond. we have another secondary alkyl halide, so right now all four of these are possible until we look at our reagent which is sodium hydroxide, Na plus, OH minus, and we know that the hydroxide ion can function as a strong nucleophile or a strong base. So a strong nucleophile makes us think an SN2 reaction and not an SN1. The strong base makes us think about an E2 reaction and not an E1 reaction. Since we have heat, heat favors an elimination reaction over a substitution, so E2 should be the major reaction here. So, when we analyze our alkyl halide, the carbon bonded to the halogen is our alpha carbon and the carbons directly bonded to that would be our beta carbons. So, we have two beta carbons here and let me number this ring. I'm gonna say the alpha carbon is carbon one, I'm gonna go round clockwise, so that's one, two, three, carbon four, carbon five and then carbon six. And next we're going to translate this to our chair confirmation over here. So, carbon one would be this carbon and then carbon two would be this one. This'll be carbon three, four, five and six. The bromine is coming out at us in space at carbon one which means it's going up. So, if I look at carbon one, we would have the bromine going up, which would be up axial. At carbon two, I have a methyl group going away from me in space, so that's going down, so at carbon two we must have a methyl group going down which makes it down axial. So, we care about carbon two. Let me highlight these again. So, we care about carbon two which is a beta carbon. We also care about carbon six which is another beta carbon. So, let's put in the hydrogens on those beta carbons. At carbon two, we would have a hydrogen that's up equatorial and at carbon six we would have a hydrogen that's down axial and one that is up equatorial. So, when we think about our E2 mechanism, we know our strong base is going to take a proton and that proton must be antiperiplanar to our halogen. So, our halogen, let me highlight our halogen here which is bromine, that is in the axial position, so we need to take a proton that is antiperiplanar to that bromine, so that carbon two, and let's look at carbon two first. At carbon two I do not have a hydrogen that's antiperiplanar to my halogen but I do have one at carbon six. It's the one that is down axial. So, our base is gonna take that proton, so let's draw in the hydroxide ion which is a strong base and the hydroxide ion is going to take this proton and then these electrons are gonna move in to form a double bond at the same time we get these electrons coming off onto the bromine to form the bromide ion. So, let's draw the product for this reaction. We would have our ring and a double bond forms between carbon one and carbon six. So, that means a double bond forms in here. And then at carbon two, we still have a methyl group going away from us in space. So, let me draw that in like that. So, the electrons in red, hard to see, but if you think about these electrons in red back here, are gonna move in to form our double bond between what I've labeled as carbon one and carbon six. Let me label those again here. So, carbon one and carbon six. Again not IUPAC nomenclature just so we can think about our product compared to our starting material. So, this would be the major product of our reaction which is an E2 reaction. It would also be possible to get some products from an SN2 mechanism, but since heat is here, an elimination reaction is favored over a substitution. Next we have a secondary alcohol with phosphoric acid and heat. And we saw a lot of these types of problems in the videos on elimination reactions. So, it's not gonna be SN1 or SN2 and we don't have a strong base, so don't think E2, think E1. And our first step would be to protonate our alcohol to form a better leaving group. So phosphoric acid is a source of protons and we're going to protonate this oxygen for our first step. So, let's draw in our ring and we protonate our oxygen, so now our oxygen has two bonds to hydrogen, one lone pair of electrons and a plus one formal charge on the oxygen. So, this lone pair of electrons on the oxygen picked up a proton from phosphoric acid to form this bond. And now we have a better leaving group than the hydroxide ion. These electrons come off onto the oxygen and we remove a bond from this carbon in red which would give us a secondary carbocation. So, let's draw in our secondary carbocation and the carbon in red is this one and that carbon would have a plus one formal charge. So, let me draw in a plus one formal charge here. And now we have water which can function as a weak base in our E1 reaction and take a proton from a carbon next to our carbon with a positive charge. So, let's say this carbon right here. It has two hydrogens on it. I'll just draw one hydrogen in and water functions as a base, takes this proton and these electrons move in to form a double bond. So, let's draw our final product here. We would have a ring, we would have a double bond between these two carbons, so our electrons in, let's use magenta, electrons in magenta moved in to form our double bond. So, our product is cyclohexane. So, a secondary alcohol undergoes an E1 reaction if you use something like sulfuric acid or phosphoric acid and you heat it up. For this reaction we have this secondary alkyl halide reacting with an aqueous solution of formic acid. Formic acid is a weak nucleophile and water is a polar protic solvent. A weak nucleophile and a polar protic solvent should make us think about an SN1 type mechanism because water as a polar protic solvent can stabilize the formation of a carbocation. So, let's draw the carbocation that would result. These electrons would come off onto our bromine and we're taking a bond away from this carbon in red. So, the carbon in red gets a plus one formal charge and let's draw our carbocation. So, we have our benzine ring here. I'll put in my pi electrons and the carbon in red is this one, so that carbon gets a plus one formal charge. This is a secondary carbocation but it's also a benzylic carbocation. So, the positive charge is actually de-localized because of the pi electrons on the ring. So, this is more stable than most secondary carbocations. Next, if we're thinking an SN1 type mechanism, this would be our electrophile, our carbocation is our electrophile and our nucleophile would be formic acid. And we saw in an earlier video how the carbonyl oxygen is actually more nucleophilic than this oxygen. So, a lone pair of electrons on the carbonyl oxygen would attack our carbon in red. And we would end up with, let's go ahead and draw in the result of our nucleophilic attack, and I won't go through all the steps of the mechanisms since I cover this in great detail in an earlier video. So, this is from our SN1, SN2 final summary video. So, let me draw in what we would form in here. So, this would be carbon double bonded to an oxygen and this would be a hydrogen. So, this is our product and this carbon is a chiral center and because this is an SN1 type mechanism and we have planer geometry in our carbocation, our nucleophile can attack from either side and we're gonna end up with a mix of enantiomers. So again, for more details on this mechanism, I skipped a few steps here, please watch the SN1, SN2 final summary video. Next, let's think about what else could possibly happen. So, SN2 is out, we formed a carbocation, E1 is possible because we have a carbocation here and we also have a weak base present. So, our weak base could be something like water, and I'll just draw a generic base in here, and let's draw in a proton on this carbon. So, our base could take this proton here and these electrons would move in to form a double bond. So, another possible product, we would have our benzine ring, so I'll draw that in, and then we would have a double bond. So, another possibility is an E1 mechanism." + }, + { + "Q": "Why do they have so many \"spheres\" like the asthenosphere or the atmosphere", + "A": "sphere is not only the round thing, but this term is used frequently to mean something like realm", + "video_name": "f2BWsPVN7c4", + "transcript": "What I want to do in this video is talk a little bit about plate tectonics. And you've probably heard the word before, and are probably, or you might be somewhat familiar with what it discusses. And it's really just the idea that the surface of the Earth is made up of a bunch of these rigid plates. So it's broken up into a bunch of rigid plates, and these rigid plates move relative to each other. They move relative to each other and take everything that's on them for a ride. And the things that are on them include the continents. So it literally is talking about the movement of these plates. And over here I have a picture I got off of Wikipedia of the actual plates. And over here you have the Pacific Plate. Let me do that in a darker color. You have the Pacific Plate. You have a Nazca Plate. You have a South American Plate. I could keep going on. You have an Antarctic Plate. It's actually, obviously whenever you do a projection onto two dimensions of a surface of a sphere, the stuff at the bottom and the top look much bigger than they actually are. Antarctica isn't this big relative to say North America or South America. It's just that we've had to stretch it out to fill up the rectangle. But that's the Antarctic Plate, North American Plate. And you can see that they're actually moving relative to each other. And that's what these arrows are depicting. You see right over here the Nazca Plate and the Pacific Plate are moving away from each other. New land is forming here. We'll talk more about that in other videos. You see right over here in the middle of the Atlantic Ocean the African Plate and the South American Plate meet each other, and they're moving away from each other, which means that new land, more plate material I guess you could say, is somehow being created right here-- we'll talk about that in future videos-- and pushing these two plates apart. Now, before we go into the evidence for plate tectonics or even some of the more details about how plates are created and some theories as to why the plates might move, what I want to do is get a little bit of the terminology of plate tectonics out of the way. Because sometimes people call them crustal plates, and that's not exactly right. And to show you the difference, what I want to do is show you two different ways of classifying the different layers of the Earth and then think about how they might relate to each other. So what you traditionally see, and actually I've made a video that goes into a lot more detail of this, is a breakdown of the chemical layers of the Earth. And when I talk about chemical layers, I'm talking about what are the constituents of the different layers? So when you talk of it in this term, the top most layer, which is the thinnest layer, is the crust. Then below that is the mantle. Actually, let me show you the whole Earth, although I'm not going to draw it to scale. So if I were to draw the crust, the crust is the thinnest outer layer of the Earth. You can imagine the blue line itself is the crust. Then below that, you have the mantle. So everything between the blue and the orange line, this over here is the mantle. So let me label the crust. The crust you can literally view as the actual blue pixels over here. And then inside of the mantle, you have the core. And when you do this very high level division, these are chemical divisions. This is saying that the crust is made up of different types of elements. Its makeup is different than the stuff that's in the mantle, which is made up of different things than what's inside the core. It's not describing the mechanical properties of it. And when I talk about mechanical properties I'm talking about whether something is solid and rigid. Or maybe it's so hot and melted it's kind of a magma, or kind of a plastic solid. So this would be the most brittle stuff. If it gets warmed up, if rock starts to melt a little bit, then you have something like a magma, or you can view it as like a deformable or a plastic solid. When we talk about plastic, I'm not talking about the stuff that the case of your cellphone is made of. I'm talking about it's deformable. This rock is deformable because it's so hot and it's somewhat melted. It kind of behaves like a fluid. It actually does behave like a fluid, but it's much more viscous. It's much thicker and slower moving than what we would normally associate with a fluid like water. So this a viscous fluid. And then the most fluid would, of course, be the liquid state. This is what we mean when we talk about the mechanical properties. And when you look at this division over here, the crust is solid. The mantle actually has some parts of it that are solid. So the uppermost part of the mantle is solid. Then below that, the rest of the mantle is kind of in this magma, this deformable, somewhat fluid state, and depending on what depth you go into the mantle there are kind of different levels of fluidity. And then the core, the outer level layer of the core, the outer core is liquid, because the temperature is so high. The inner core is made up of the same things, and the temperature is even higher, but since the pressure is so high it's actually solid. So that's why the mantle, crust, and core differentiations don't tell you whether it's solid, whether it's magma, or whether it's really a liquid. It just really tells you what the makeup is. Now, to think about the makeup, and this is important for plate tectonics, because when we talk about these plates we're not talking about just the crust. We're talking about the outer, rigid layer. Let me just zoom in a little bit. Let's say we zoomed in right over there. So now we have the crust zoomed in. This right here is the crust. And then everything below here we're actually talking about the upper mantle. We haven't gotten too deep in the mantle right here. So that's why we call it the upper mantle. Now, right below the crust, the mantle is cool enough that it is also in real solid form. So this right here is solid mantle. And when we talk about the plates were actually talking about the outer solid layer. So that includes both the crust and the solid part of the mantle. And we call that the lithosphere. When people talk about plate tectonics, they shouldn't say crustal plates. They should call these lithospheric plates. And then below the lithosphere you have the least viscous part of the mantle, because the temperature is high enough for the rock to melt, but the pressure isn't so large as what will happen when you go into the lower part of the mantle that the fluid can actually kind of move past each other, although still pretty viscous. This still a magma. So this is still kind of in its magma state. And this fluid part of the mantle, we can't quite call it a liquid yet, but over large periods of time it does have fluid properties. This, that essentially the lithosphere is kind of riding on top of, we call this the asthenosphere. So when we talk about the lithosphere and asthenosphere we're really talking about mechanical layers. The outer layer, the solid layers, the lithosphere sphere. The more fluid layer right below that is the asthenosphere. When we talk about the crust, mantle, and core, we are talking about chemical properties, what are the things actually made up of." + }, + { + "Q": "Why is it not ln|ln|x||?", + "A": "The natural log on the inside is just u , which came from the original expression. You can see that one didn t have an absolute value on it.", + "video_name": "OLO64d4Y1qI", + "transcript": "We are faced with a fairly daunting-looking indefinite integral of pi over x natural log of x dx. Now, what can we do to address this? Is u-substitution a possibility here? Well for u-substitution, we want to look for an expression and its derivative. Well, what happens if we set u equal to the natural log of x? Now what would du be equal to in that scenario? du is going to be the derivative of the natural log of x with respect to x, which is just 1/x dx. This is an equivalent statement to saying that du dx is equal to 1/x. So do we see a 1/x dx anywhere in this original expression? Well, it's kind of hiding. It's not so obvious, but this x in the denominator is essentially a 1/x. And then that's being multiplied by a dx. Let me rewrite this original expression to make a little bit more sense. So the first thing I'm going to do is I'm going to take the pi. I should do that in a different color since I've already used-- let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral. And so this becomes the integral of-- and let me write the 1 over natural log of x first. 1 over the natural log of x times 1/x dx. Now it becomes a little bit clearer. These are completely equivalent statements. But this makes it clear that, yes, u-substitution will work over here. If we set our u equal to natural log of x, then our du is 1/x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of 1/u. Natural log of x is u-- we set that equal to natural log of x-- times du. Now this becomes pretty straightforward. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log of the absolute value of u so that we can handle even negative values of u. The natural log of the absolute value of u plus c, just in case we had a constant factor out here. And we're almost done. We just have to unsubstitute for the u. u is equal to natural log of x. So we end up with this kind of neat-looking expression. The anti of this entire indefinite integral we have simplified. We have evaluated it, and it is now equal to pi times the natural log of the absolute value of u. But u is just the natural log of x. And then we have this plus c right over here. And we could have assumed that, from the get go, this original expression was only defined for positive values of x because you had to take the natural log here, and it wasn't an absolute value. So we can leave this as just a natural log of x, but this also works for the situations now because we're doing the absolute value of that where the natural log of x might have been a negative number. For example, if it was a natural log of 0.5 or, who knows, whatever it might be. But then we are all done. We have simplified what seemed like a kind of daunting expression." + }, + { + "Q": "Sal uses the world orthogonal, could someone define it for me?", + "A": "Orthogonal is a generalisation of the geometric concept of perpendicular. When dealing with vectors it means that the vectors are all at 90 degrees from each other.", + "video_name": "9kW6zFK5E5c", + "transcript": "I want to bring everything we've learned about linear independence and dependence, and the span of a set of vectors together in one particularly hairy problem, because if you understand what this problem is all about, I think you understand what we're doing, which is key to your understanding of linear algebra, these two concepts. So the first question I'm going to ask about the set of vectors s, and they're all three-dimensional vectors, they have three components, Is the span of s equal to R3? It seems like it might be. If each of these add new information, it seems like maybe I could describe any vector in R3 by these three vectors, by some combination of these three vectors. And the second question I'm going to ask is are they linearly independent? And maybe I'll be able to answer them at the same time. So let's answer the first one. Do they span R3? To span R3, that means some linear combination of these three vectors should be able to construct any vector in R3. So let me give you a linear combination of these vectors. I could have c1 times the first vector, 1, minus 1, 2 plus some other arbitrary constant c2, some scalar, times the second vector, 2, 1, 2 plus some third scaling vector times the third vector minus 1, 0, 2. I should be able to, using some arbitrary constants, take a combination of these vectors that sum up to any vector in R3. And I'm going to represent any vector in R3 by the vector a, b, and c, where a, b, and c are any real numbers. So if you give me any a, b, and c, and I can give you a formula for telling you what your c3's, your c2's and your c1's are, then than essentially means that it spans R3, because if you give me a vector, I can always tell you how to construct that vector with these three. So Let's see if I can do that. Just from our definition of scalar multiplication of a vector, we know that c1 times this vector, I could rewrite it if I want. I normally skip this step, but I really want to make it clear. So c1 times, I could just rewrite as 1 times c-- it's each of the terms times c1. Similarly, c2 times this is the same thing as each of the terms times c2. And c3 times this is the same thing as each of the terms times c3. I want to show you that everything we do it just formally comes from our definition of multiplication of a vector times a scalar, which is what we just did, or vector addition, which is what we're about to do. So vector addition tells us that this term plus this term plus this term needs to equal that term. So let me write that down. We get c1 plus 2c2 minus c3 will be equal to a. Likewise, we can do the same thing with the next row. Minus c1 plus c2 plus 0c3 must be equal to b. So we get minus c1 plus c2 plus 0c3-- so we don't even have to write that-- is going to be equal to b. And then finally, let's just do that last row. 2c1 plus 3c2 plus 2c3 is going to be equal to c. Now, let's see if we can solve for our different constants. I'm going to do it by elimination. I think you might be familiar with this process. I think I've done it in some of the earlier linear algebra videos before I started doing a formal presentation of it. And I'm going to review it again in a few videos from now, but I think you understand how to solve it this way. What I'm going to do is I'm going to first eliminate these two terms and then I'm going to eliminate this term, and then I can solve for my various constants. If I want to eliminate this term right here, what I could do is I could add this equation to that equation. Or even better, I can replace this equation with the sum of these two equations. Let me do that. I'm just going to add these two equations to each other and replace this one with that sum. So minus c1 plus c1, that just gives you 0. I can ignore it. Then c2 plus 2c2, that's 3c2. And then 0 plus minus c3 is equal to minus c3. Minus c3 is equal to-- and I'm replacing this with the sum of these two, so b plus a. It equals b plus a. Let me write down that first equation on the top. So the first equation, I'm not doing anything to it. So I get c1 plus 2c2 minus c3 is equal to a. Now, in this last equation, I want to eliminate this term. Let's take this equation and subtract from it 2 times this top equation. You can also view it as let's add this to minus 2 times this top equation. Since we're almost done using this when we actually even wrote it, let's just multiply this times minus 2. So this becomes a minus 2c1 minus 4c2 plus 2c3 is equal to minus 2a. If you just multiply each of these terms-- I want to be very careful. I don't want to make a careless mistake. Minus 2 times c1 minus 4 plus 2 and then minus 2. And now we can add these two together. And what do we get? 2c1 minus 2c1, that's a 0. I don't have to write it. 3c2 minus 4c2, that's a minus c2. And then you have your 2c3 plus another 2c3, so that is equal to plus 4c3 is equal to c minus 2a. All I did is I replaced this with this minus 2 times that, and I got this. Now I'm going to keep my top equation constant again. I'm not going to do anything to it, so I'm just going to move it to the right. So I get c1 plus 2c2 minus c3 is equal to a. I'm also going to keep my second equation the same, so I get 3c2 minus c3 is equal to b plus a. Let me scroll over a good bit. And then this last equation I want to eliminate. My goal is to eliminate this term right here. What I want to do is I want to multiply this bottom equation times 3 and add it to this middle equation to eliminate this term right here. So if I multiply this bottom equation times 3-- let me just do-- well, actually, I don't want to make things messier, so this becomes a minus 3 plus a 3, so those cancel out. This becomes a 12 minus a 1. So this becomes 12c3 minus c3, which is 11c3. And then this becomes a-- oh, sorry, I was already done. When I do 3 times this plus that, those canceled out. And then when I multiplied 3 times this, I get 12c3 minus a c3, so that's 11c3. And I multiplied this times 3 plus this, so I get 3c minus 6a-- I'm just multiplying this times 3-- plus this, plus b plus a. So what can I rewrite this by? Actually, I want to make something very clear. This c is different than these c1's, c2's and c3's that I had up here. I think you realize that. But I just realized that I used the letters c twice, and I just didn't want any confusion here. So this c that doesn't have any subscript is a different constant then all of these things over here. Let's see if we can simplify this. We have an a and a minus 6a, so let's just add them. So let's get rid of that a and this becomes minus 5a. If we divide both sides of this equation by 11, what do we get? We get c3 is equal to 1/11 times 3c minus 5a. So you give me any a or c and I'll already tell you what c3 is. What is c2? c2 is equal to-- let me simplify this equation right here. Let me do it right there. So if I just add c3 to both sides of the equation, I get 3c2 is equal to b plus a plus c3. And if I divide both sides of this by 3, I get c2 is equal to 1/3 times b plus a plus c3. I'll just leave it like that for now. Then what is c1 equal to? I could just rewrite this top equation as if I subtract 2c2 and add c3 to both sides, I get c1 is equal to a minus 2c2 plus c3. What have I just shown you? You can give me any vector in R3 that you want to find. So you can give me any real number for a, any real number for b, any real number for c. And if you give me those numbers, I'm claiming now that I can always tell you some combination of these three vectors that will add up to those. And I've actually already solved for what I have to multiply each of those vectors by to add up to this third vector. So you give me your a's, b's and c's, I just have to substitute into the a's and the c's right here. Oh, sorry. I forgot this b over here. There's also a b. It was suspicious that I didn't have to deal with a b. So there was a b right there. So this is 3c minus 5a plus b. Let me write that. There's a b right there in a parentheses. But I think you get the general idea. You give me your a's, b's and c's, any real numbers can apply. There's no division over here, so I don't have to worry about dividing by zero. So this is just a linear combination of any real numbers, so I can clearly get another real number. So you give me your a's, b's and c's, I'm going to give you a c3. Now, you gave me a's, b's and c's. I got a c3. This is just going to be another real number. I'm just going to take that with your former a's and b's and I'm going to be able to give you a c2. We were already able to solve for a c2 and a c3, and then I just use your a as well, and then I'm going to give you a c1. Hopefully, you're seeing that no matter what a, b, and c you give me, I can give you a c1, c2, or c3. There's no reason that any a's, b's or c's should break down these formulas. We're not doing any division, so it's not like a zero would break it down. I can say definitively that the set of vectors, of these three vectors, does indeed span R3. Let me ask you another question. I already asked it. Are these vectors linearly independent? We said in order for them to be linearly independent, the only solution to c1 times my first vector, 1, minus 1, 2, plus c2 times my second vector, 2, 1, 3, plus c3 times my third vector, minus 1, 0, 2. If something is linearly independent that means that the only solution to this equation-- so I want to find some set of combinations of these vectors that add up to the zero vector, and I did that in the previous video. If they are linearly dependent, there must be some non-zero solution. One of these constants, at least one of these constants, would be non-zero for this solution. You can always make them zero, no matter what, but if they are linearly dependent, then one of these could be non-zero. If they're linearly independent then all of these have to be-- the only solution to this equation would be c1, c2, c3. All have to be equal to 0. c1, c2, c3 all have to be equal to 0. Linear independence implies this, this implies linear independence. Now, this is the exact same thing we did here, but in this case, I'm just picking my a's, b's and c's to be zero. This is a, this is b and this is c, right? I can pick any vector in R3 for my a's, b's and c's. I'm now picking the zero vector. So let's see what our c1's, c2's and c3's are. So my a equals b is equal to c is equal to 0. I'm setting it equal to the zero vector. What linear combination of these three vectors equal the zero vector? Well, if a, b, and c are all equal to 0, that term is 0, that is 0, that is 0. You have 1/11 times 0 minus 0 plus 0. That's just 0. So c3 is equal to 0. Now, if c3 is equal to 0, we already know that a is equal to 0 and b is equal to 0. C2 is 1/3 times 0, so it equals 0. Well, it's c3, which is 0. c2 is 0, so 2 times 0 is 0. So c1 is just going to be equal to a. I just said a is equal to 0. So the only solution to this equation right here, the only linear combination of these three vectors that result in the zero vector are when you weight all of them by zero. So I just showed you that c1, c2 and c3 all have to be zero. And because they're all zero, we know that this is a linearly independent set of vectors. Or that none of these vectors can be represented as a combination of the other two. I have exactly three vectors that span R3 and they're linearly independent. And linearly independent, in my brain that means, look, I don't have any redundant vectors, anything that could have just been built with the other vectors, and I have exactly three vectors, and it's spanning R3. So in general, and I haven't proven this to you, but I could, is that if you have exactly three vectors and they do span R3, they have to be linearly independent. If they weren't linearly independent, then one of these would be redundant. Let's say that that guy was a redundant one. I always pick the third one, but let's say this guy would be redundant, which means that the span of this would be equal to the span of these two, right? Because if this guy is redundant, he could just be part of the span of these two guys. And the span of two of vectors could never span R3. Or the other way you could go, if you have three linear independent-- three tuples, and they're all independent, then you can also say that that spans R3. I haven't proven that to you, but hopefully, you get the sense that each of these is contributing new directionality, right? One is going like that. They're not completely orthogonal to each other, but they're giving just enough directionality that you can add a new dimension to what's going on. Hopefully, that helped you a bit, and I'll see you in the next video." + }, + { + "Q": "is it possible to borrow the stock and then sell it without buying it later when it goes down. You will get the entire price of the stock without buying it in the beginning.", + "A": "No, that would be stealing, not borrowing.", + "video_name": "jAOtWm_WZiE", + "transcript": "Let's say you don't like company ABCD very much and you're convinced that the stock is going to go down. So in that situation, you can actually short the stock, which in a very high level is a bet that the stock is going to go down. And the way that you do that mechanically is that you borrow the stock from someone else who owns it, and then you immediately sell that stock that you don't even own. You sell the stock that you borrowed from someone else, and you'll sell it at the current price. So, for example, in this situation, you would sell it at the current trading price of $50. You would then hope that the stock price goes down. Let's think about the situation where the stock price goes down. So if you shorted it right over here, you borrowed the stock and you sold it for $50. And then if the stock were to actually go down-- let's say it goes all the way down to $20, and you think that's about how far it's going to go down, then you can buy back the stock for $20 in this situation. And then give it back to the owner. You had borrowed the stock. Now you can hand back the stock to the owner. So you could give it back and you've essentially unwound it. And what it allowed you to do is it allowed you to do the buying and selling in reverse order. Normally before you sell something you have to buy something. But here you were able to sell it and buy it later for a lower price. So the situation where the end price is $20 you had sold it for $50. So you got $50. And then you had to use $20 of it to buy it back. So in that situation, shorting the stock, you would have made $30. Let me write this column here. This is the short option. You sold at $50. You borrowed and sold at $50. Then when it went down to $20 you bought it back for $20. So you had $50 of proceeds. You had to use $20 of it to buy it back. So you had a $30 profit. But that's only in the good scenario. What happens if your bet is wrong? What happens if the stock price goes up to $80? And over here you get so scared. You're like, oh my god, I have to buy the stock back by $80. What if it keeps going up? I could lose an unlimited amount of money. So over here you get scared and you unwind your situation. You say, OK, I'll go and buy the stock for $80, so I can give it back. So in this situation where the stock goes up you actually could lose a lot of money. You had sold it for $50. So you only have $50 that you have from the transaction. But now you have to buy the stock for $80. So if you sold for $50 and you buy for $80 you've now lost $30. You're $30 in the hole. So now you are at negative $30. And really shorting is the riskiest of all of the things you can do, because a stock price and go unbelievably high. What happens if the stock price goes to $800 or goes to $8,000? All of a sudden, you've sold something for $50 and you have the obligation at some point in the future, because you have to give the stock back, of paying $500, or $800, or $8,000. You don't know how much you'll have to lose. So it's really the riskiest thing you can do. But it is one way to bet that a stock price will go down, or profit from a stock price going down." + }, + { + "Q": "What is the difference between Equilateral and Isosceles triangles?", + "A": "An equilateral triangle has 3 congruent sides. An isosceles triangle has 2 congruent sides. A scalene triangle has no congruent sides.", + "video_name": "7FTNWE7RTfQ", + "transcript": "Let's do some example problems using our newly acquired knowledge of isosceles and equilateral triangles. So over here, I have kind of a triangle within a triangle. And we need to figure out this orange angle right over here and this blue angle right over here. And we know that side AB or segment AB is equal to segment BC, which is equal to segment CD. Or we could also call that DC. So first of all, we see that triangle ABC is isosceles. And because it's isosceles, the two base angles are going to be congruent. This is one leg. This is the other leg right over there. So the two base angles are going to be congruent. So we know that this angle right over here is also 31 degrees. Well, if we know two of the angles in a triangle, we can always figure out the third angle. They have to add up to 180 degrees. So we could say 31 degrees plus 31 degrees plus the measure of angle ABC is equal to 180 degrees. You can subtract 62. This right here is 62 degrees. You subtract 62 from both sides. You get the measure of angle ABC is equal to-- let's see. 180 minus 60 would be 120. You subtract another 2. You get 118 degrees. So this angle right over here is 118 degrees. Let me just write it like this. This is 118 degrees. Well, this angle right over here is supplementary to that 118 degrees. So that angle plus 118 is going to be equal to 180. We already know that that's 62 degrees. 62 plus 118 is 180. So this right over here is 62 degrees. Now, this angle is one of the base angles for triangle BCD. I didn't draw it that way, but this side and this side are congruent. BC has the same length as CD. Those are the two legs of an isosceles triangle. You can kind of imagine it was turned upside down. This is the vertex. This is one base angle. This is the other base angle. Well, the base angles are going to be congruent. So this is going to be 62 degrees, as well. And then finally, if you want to figure out this blue angle, the blue angle plus these two 62-degree angles are going to have to add up to 180 degrees. So you get 62 plus 62 plus the blue angle, which is the measure of angle BCD, is going to have to be equal to 180 degrees. These two characters-- let's see. 62 plus 62 is 124. You subtract 124 from both sides. You get the measure of angle BCD is equal to-- let's see. If you subtract 120, you get 60, and then you have to subtract another 4. So you get 56 degrees. So this is equal to 56 degrees. And we're done. Now, we could do either of these. Let's do this one right over here. So what is the measure of angle ABE? So they haven't even drawn segment BE here. So let me draw that for us. And so we have to figure out the measure of angle ABE. So we have a bunch of congruent segments here. And in particular, we see that triangle ABD, all of its sides are equal. So it's an equilateral triangle, which means all of the angles are equal. And if all of the angles are equal in a triangle, they all have to be 60 degrees. So all of these characters are going to be 60 degrees. Well, that's part of angle ABE, but we have to figure out this other part right over here. And to do that, we can see that we're actually dealing with an isosceles triangle kind of tipped over to the left. This is the vertex angle. This is one base angle. This is the other base angle. And the vertex angle right here is 90 degrees. And once again, we know it's isosceles because this side, segment BD, is equal to segment DE. And once again, these two angles plus this angle right over here are going to have to add up to 180 degrees. So you call that an x. You call that an x. You've got x plus x plus 90 is going to be 180 degrees. So you get 2x plus-- let me just write it out. Don't want to skip steps here. We have x plus x plus 90 is going to be equal to 180 degrees. x plus x is the same thing as 2x, plus 90 is equal to 180. And then we can subtract 90 from both sides. You get 2x is equal to 90. Or divide both sides by 2. You get x is equal to 45 degrees. And then we're done because angle ABE is going to be equal to the 60 degrees plus the 45 degrees. So it's going to be this whole angle, which is what we care about. Angle ABE is going to be 60 plus 45, which is 105 degrees. And now we have this last problem over here. This one looks a little bit simpler. I have an isosceles triangle. This leg is equal to that leg. This is the vertex angle. I have to figure out B. And the trick here is like, wait, how do I figure out one side of a triangle if I only know one other side? Don't I need to know two other sides? And we'll do it the exact same way we just did that second part of that problem. If this is an isosceles triangle, which we know it is, then this angle is going to be equal to that angle there. And so if we call this x, then this is x as well. And we get x plus x plus 36 degrees is equal to 180. The two x's, when you add them up, you get 2x. And then-- I won't skip steps here. 2x plus 36 is equal to 180. Subtract 36 from both sides, we get 2x-- that 2 looks a little bit funny. We get 2x is equal to-- 180 minus 30 is 150. And then you want to subtract another 6 from 150, gets us to 144. Did I do that right? 180 minus 30 is 150, yep, 144. Divide both sides by 2. You get x is equal to 72 degrees. So this is equal to 72 degrees. And we are done." + }, + { + "Q": "When we say \"surface of the sun\" is there actually a surface? Let's say, hypothetically, if a spaceship could survive the heat of the sun would it just travel straight through the sun without obstacle? It is just gas, after all.", + "A": "If you re thinking of a solid surface, like that or Earth, then no. The sun s surface is made up of gas. So it s a gaseous surface.", + "video_name": "EdYyuUUY-nc", + "transcript": "In the last video, we started with a star in its main sequence, like the sun. And inside the core of that star, you have hydrogen fusion going on. So that is hydrogen fusion, and then outside of the core, you just had hydrogen. You just hydrogen plasma. And when we say plasma, it's the electrons and protons of the individual atoms have been disassociated because the temperatures and pressures are so high. So they're really just kind of like this soup of electrons and protons, as opposed to proper atoms that we associate with at lower temperatures. So this is a main sequence star right over here. And we saw in the last video that this hydrogen is fusing into helium. So we start having more and more helium here. And as we have more and more helium, the core becomes more and more dense, because helium is a more massive atom. It is able to pack more mass in a smaller volume. So this gets more and more dense. So core becomes more dense. And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster. Because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing, so it starts to fuse hotter. So let me write this, so the fusion, so hydrogen fuses faster. And actually, we even see this in our sun. Our sun today is brighter and hotter. It's fusing faster than it was when it was born 4.5 or 4.6 billion years ago. But eventually you're going to get to the point so that the core, you only have helium. So there's going to be some point where the entire core is all helium. And it's going to be way denser than this core over here. All of that mass over there has now been turned into helium. A lot of it has been turned into energy. But most of it is now in helium, and it's going to be at a much, much smaller volume. And the whole time, the temperature is increasing, the fusion is getting faster and faster. And now there's this dense volume of helium that's not fusing. You do have, and we saw this in this video, a shell around it of hydrogen that is fusing. So this right here is hydrogen fusion going on. And then this over here is just hydrogen plasma. Now the unintuitive thing, or at least this was unintuitive to me at first, is what's going on the core is that the core is getting more and more dense. It's fusing at a faster rate. And so it's getting hotter and hotter. So the core is hotter, fusing faster, getting more and more dense. I kind of imagine it's starting to collapse. Every time it collapses, it's getting hotter and more dense. But at the same time that's happening, the star itself is getting bigger. And this is actually not drawn to scale. Red giants are much, much larger than main sequence stars. But the whole time that this is getting more dense, the rest of the star is, you could kind of view it as getting less dense. And that's because this is generating so much energy that it's able to more than offset, or better offset the gravitational pull into it. So even though this is hotter, it's able to disperse the rest of the material in the sun over a larger volume. And so that volume is so big that the surface, and we saw this in the last video, the surface of the red giant is actually cooler-- let me write that a little neater-- is actually cooler than the surface of a main sequence star. This right here is hotter. And just to put things in perspective, when the sun becomes a red giant, and it will become a red giant, its diameter will be 100 times the diameter that it is today. Or another way to be put it, it will have the same diameter as the Earth's orbit around the current sun. Or another way to view it is, where we are right now will be on the surface or near the surface or maybe even inside of that future sun. Or another way to put it, when the sun becomes a red giant, the Earth's going to be not even a speck out here. And it will be liquefied and vaporized at that point in time. So this is super, super huge. And we've even thought about it. Just for light to reach the current sun to our point in orbit, it takes eight minutes. So that's how big one of these stars are. To get from one side of the star to another side of the star, it'll take 16 minutes for light to travel, if it was traveling that diameter, and even slightly longer if it was to travel it in a circumference. So these are huge, huge, huge stars. And we'll talk about other stars in the future. They're even bigger than this when they become supergiants. But anyway, we have the hydrogen in the center-- sorry. We have the helium in the center. Let me write this down. We have a helium core in the center. We're fusing faster and faster and faster. We're now a red giant. The core is getting hotter and hotter and hotter until it gets to the temperature for ignition of helium. So until it gets to 100 million Kelvin-- remember the ignition temperature for hydrogen was 10 million Kelvin. So now we're at 100 million Kelvin, factor of 10. And now, all of a sudden in the core, you actually start to have helium fusion. And we touched on this in the last video, but the helium is fusing into heavier elements. And some of those heavier elements, and predominately, it will be carbon and oxygen. And you may suspect this is how heavier and heavier elements form in the universe. They form, literally, due to fusion in the core of stars. Especially when we're talking about elements up to iron. But anyway, the core is now experiencing helium fusion. It has a shell around it of helium that is not quite there, does not quite have the pressures and temperatures to fuse yet. So just regular helium. But then outside of that, we do have the pressures and temperatures for hydrogen to continue to fuse. So out here, you do have hydrogen fusion. And then outside over here, you just have the regular hydrogen plasma. So what just happened here? When you have helium fusion all of a sudden-- now this is, once again, providing some type of energetic outward support for the core. So it's going to counteract the ever-increasing contraction of the core as it gets more and more dense, because now we have energy going outward, energy pushing things outward. But at the same time that that is happening, more and more hydrogen in this layer is turning into helium, is fusing into helium. So it's making this inert part of the helium core even larger and larger and denser, even larger and larger, and putting even more pressure on this inside part. And so what's actually going to happen within a few moments, I guess, especially from a cosmological point of view, this helium fusion is going to be burning super-- I shouldn't use-- igniting or fusing at a super-hot level. But it's contained due to all of this pressure. But at some point, the pressure won't be able to contain it, and the core is going to explode. But it's not going to be one of these catastrophic explosions where the star is going to be destroyed. It's just going to release a lot of energy all of a sudden into the star. And that's called a helium flash. But once that happens, all of a sudden, then now the star is going to be more stable. And I'll use that in quotes without writing it down because red giants, in general, are already getting to be less stable than a main sequence star. But once that happens, you now will have a slightly larger volume. So it's not being contained in as small of a tight volume. That helium flash kind of took care of that. So now you have helium fusing into carbon and oxygen. And there's all sorts of other combinations of things. Obviously, there's many elements in between helium and carbon and oxygen. But these are the ones that dominate. And then outside of that, you have helium forming. You have helium that is not fusing. And then outside of that, you have your fusing hydrogen. Over here, you have hydrogen fusing into helium. And then out here in the rest of the radius of our super-huge red giant, you just have your hydrogen plasma out here. Now what's going to happen as this star ages? Well, if we fast forward this a bunch-- and remember, as a star gets denser and denser in the core, and the reactions happen faster and faster, and this core is expelling more and more energy outward, the star keeps growing. And the surface gets cooler and cooler. So if we fast forward a bunch, and this is what's going to happen to something the mass of our sun, if it's more massive, then at some point, the core of carbon and oxygen that's forming can start to fuse into even heavier elements. But in the case of the sun, it will never get to that 600 million Kelvin to actually fuse the carbon and the oxygen. And so eventually you will have a core of carbon and oxygen, or mainly carbon and oxygen surrounded by fusing helium surrounded by non-fusing helium surrounded by fusing hydrogen, which is surrounded by non-fusing hydrogen, or just the hydrogen plasma of the sun. But eventually all of this fuel will run out. All of the hydrogen will run out in the stars. All of this hydrogen, all of this fusing hydrogen will run out. All of this fusion helium will run out. This is the fusing hydrogen. This is the inert helium, which will run out. It'll be used in kind of this core, being fused into the carbon and oxygen, until you get to a point where you literally just have a really hot core of carbon and oxygen. And it's super-dense. This whole time, it will be getting more and more dense as heavier and heavier elements show up in the course. So it gets denser and denser and denser. But the super dense thing will not, in the case of the sun-- and if it was a more massive star, it would get there-- but in the case of the sun, it will not get hot enough for the carbon and the oxygen to form. So it really will just be this super-dense ball of carbon and oxygen and all of the other material in the sun. Remember, it was superenergetic. It was releasing tons and tons of energy. The more that we progressed down this, the more energy was releasing outward, and the larger the radius of the star became, and the cooler the outside of the star became, until the outside just becomes this kind of cloud, this huge cloud of gas around what once was the star. And in the center-- so I could just draw it as this huge-- this is now way far away from the star, much even bigger than the radius or the diameter of a red giant. And all we'll have left is a mass, a superdense mass of, I would call it, inert carbon or oxygen. This is in the case of the sun. And at first, when it's hot, and it will be releasing radiation because it's so hot. We'll call this a white dwarf. This right here is called a white dwarf. And it'll cool down over many, many, many, many, many, many, many, years, until it becomes, when it's completely cooled down, lost all of its energy-- it'll just be this superdense ball of carbon and oxygen, at which point, we would call it a black dwarf. And these are obviously very hard to observe because they're not emitting light. And they don't have quite the mass of something like a black hole that isn't even emitting light, but you can see how it's affecting things around it. So that's what's going to happen to the sun. In the next few videos, we're going to talk about what would happen to things less massive than the sun and what would happen to things more massive can imagine the more massive. There would be so much pressure on these things, because you have so much mass around it, that these would begin to fuse into heavier and heavier elements until we get to iron." + }, + { + "Q": "At 6:16 why did he subtract (36-9)!?", + "A": "36! means 36*35*34*33*32*31*30*29*28*27*26*25*24*23*22....*1 But since we want it to stop at 28, we have to cancel the rest out by dividing with 27!(27*26*25*24*23*22*21*20*19*18*17....*1)", + "video_name": "SbpoyXTpC84", + "transcript": "A card game using 36 unique cards, four suits, diamonds, hearts, clubs and spades-- this should be spades, not spaces-- with cards numbered from 1 to 9 in each suit. A hand is chosen. A hand is a collection of 9 cards, which can be sorted however the player chooses. Fair enough. How many 9 card hands are possible? So let's think about it. There are 36 unique cards-- and I won't worry about, you know, there's nine numbers in each suit, and there are four suits, 4 times 9 is 36. But let's just think of the cards as being 1 through 36, and we're going to pick nine of them. So at first we'll say, well look, I have nine slots in my hand, right? 1, 2, 3, 4, 5, 6, 7, 8, 9. I'm going to pick nine cards for my hand. And so for the very first card, how many possible cards can I pick from? Well, there's 36 unique cards, so for that first slot, there's 36. But then that's now part of my hand. Now for the second slot, how many will there be left to pick from? Well, I've already picked one, so there will only be 35 to pick from. And then for the third slot, 34, and then Then 33 to pick from, 32, 31, 30, 29, and 28. So you might want to say that there are 36 times 35, times 34, times 33, times 32, times 31, times 30, times 29, times 28 possible hands. Now, this would be true if order mattered. This would be true if I have card 15 here. Maybe I have a-- let me put it here-- maybe I have a 9 of spades here, and then I have a bunch of cards. And maybe I have-- and that's one hand. And then I have another. So then I have cards one, two, three, four, five, six, seven, eight. I have eight other cards. Or maybe another hand is I have the eight cards, 1, 2, 3, 4, 5, 6, 7, 8, and then I have the 9 of spades. If we were thinking of these as two different hands, because we have the exact same cards, but they're in different order, then what I just calculated would make a lot of sense, because we did it based on order. But they're telling us that the cards can be sorted however the player chooses, so order doesn't matter. So we're overcounting. We're counting all of the different ways that the same number of cards can be arranged. So in order to not overcount, we have to divide this by the ways in which nine cards can be rearranged. So we have to divide this by the way nine cards can be rearranged. So how many ways can nine cards be rearranged? If I have nine cards and I'm going to pick one of nine to be in the first slot, well, that means I have 9 ways to put something in the first slot. Then in the second slot, I have 8 ways of putting a card in the second slot, because I took one to put it in the first, so I have 8 left. Then 7, then 6, then 5, then 4, then 3, then 2, then 1. That last slot, there's only going to be 1 card left to put in it. So this number right here, where you take 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1, or 9-- you start with 9 and then you multiply it by every number less than 9. Every, I guess we could say, natural number less than 9. This is called 9 factorial, and you express it as an exclamation mark. So if we want to think about all of the different ways that we can have all of the different combinations for hands, this is the number of hands if we cared about the order, but then we want to divide by the number of ways we can order things so that we don't overcount. And this will be an answer and this will be the correct answer. Now this is a super, super duper large number. Let's figure out how large of a number this is. We have 36-- let me scroll to the left a little bit-- 36 times 35, times 34, times 33, times 32, times 31, times 30, times 29, times 28, divided by 9. Well, I can do it this way. I can put a parentheses-- divided by parentheses, 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1. Now, hopefully the calculator can handle this. And it gave us this number, 94,143,280. Let me put this on the side, so I can read it. So this number right here gives us 94,143,280. So that's the answer for this problem. That there are 94,143,280 possible 9 card hands in this situation. Now, we kind of just worked through it. We reasoned our way through it. There is a formula for this that does essentially the exact same thing. And the way that people denote this formula is to say, look, we have 36 things and we are going to choose 9 of them. And we don't care about order, so sometimes it'll be written as n choose k. Let me write it this way. So what did we do here? We have 36 things. We chose 9. So this numerator over here, this was 36 factorial. But 36 factorial would go all the way down to 27, 26, 25. It would just keep going. But we stopped only nine away from 36. So this is 36 factorial, so this part right here, that part right there, is not just 36 factorial. It's 36 factorial divided by 36, minus 9 factorial. What is 36 minus 9? It's 27. So 27 factorial-- so let's think about this-- 36 factorial, it'd be 36 times 35, you keep going all the way, times 28 times 27, going all the way down to 1. That is 36 factorial. Now what is 36 minus 9 factorial, that's 27 factorial. So if you divide by 27 factorial, 27 factorial is 27 times 26, all the way down to 1. Well, this and this are the exact same thing. This is 27 times 26, so that and that would cancel out. So if you do 36 divided by 36, minus 9 factorial, you just get the first, the largest nine terms of 36 factorial, which is exactly what we have over there. And then we divided it by 9 factorial. And this right here is called 36 choose 9. And sometimes you'll see this formula written like this, n choose k. And they'll write the formula as equal to n factorial over n minus k factorial, and also in the denominator, k factorial. And this is a general formula that if you have n things, and you want to find out all of the possible ways you can pick k things from those n things, and you don't care about the order. All you care is about which k things you picked, you don't care about the order in which you picked those k things. So that's what we did here." + }, + { + "Q": "hi Khan Academy,\nI was just thinking if there a subset for the X which can make this statement true\nA \u00e2\u0088\u00aa x = B when B\u00e2\u008a\u0084 A\nThank you!", + "A": "What if x=B where A is a proper subset of B?", + "video_name": "1wsF9GpGd00", + "transcript": "Let's define ourselves some sets. So let's say the set A is composed of the numbers 1. 3. 5, 7, and 18. Let's say that the set B-- let me do this in a different color-- let's say that the set B is composed of 1, 7, and 18. And let's say that the set C is composed of 18, 7, 1, and 19. Now what I want to start thinking about in this video is the notion of a subset. So the first question is, is B a subset of A? And there you might say, well, what does subset mean? Well, you're a subset if every member of your set is also a member of the other set. So we actually can write that B is a subset-- and this is a notation right over here, this is a subset-- B is a subset of A. B is a subset. So let me write that down. B is subset of A. Every element in B is a member of A. Now we can go even further. We can say that B is a strict subset of A, because B is a subset of A, but it does not equal A, which means that there are things in A that are not in B. So we could even go further and we could say that B is a strict or sometimes said a proper subset of A. And the way you do that is, you could almost imagine that this is kind of a less than or equal sign, and then you kind of cross out this equal part of the less than or equal sign. So this means a strict subset, which means everything that is in B is a member A, but everything that's in A is not a member of B. So let me write this. This is B. B is a strict or proper subset. So, for example, we can write that A is a subset of A. In fact, every set is a subset of itself, because every one of its members is a member of A. We cannot write that A is a strict subset of A. This right over here is false. So let's give ourselves a little bit more practice. Can we write that B is a subset of C? Well, let's see. C contains a 1, it contains a 7, it contains an 18. So every member of B is indeed a member C. So this right over here is true. Now, can we write that C is a subset? Can we write that C is a subset of A? Can we write C is a subset of A? Let's see. Every element of C needs to be in A. So A has an 18, it has a 7, it has a 1. But it does not have a 19. So once again, this right over here is false. Now we could have also added-- we could write B is a subset of C. Or we could even write that B is a strict subset of C. Now, we could also reverse the way we write this. And then we're really just talking about supersets. So we could reverse this notation, and we could say that A is a superset of B, and this is just another way of saying that B is a subset of A. But the way you could think about this is, A contains every element that is in B. And it might contain more. It might contain exactly every element. So you can kind of view this as you kind of have the equals symbol there. If you were to view this as greater than or equal. They're note quite exactly the same thing. But we know already that we could also write that A is a strict superset of B, which means that A contains everything B has and then some. A is not equivalent to B. So hopefully this familiarizes you with the notions of subsets and supersets and strict subsets." + }, + { + "Q": "At 2:30, when he says that the absolute value could also work in ensuring that the value is positive, can someone explain to me why the absolute value isn't used in the equation?", + "A": "In the video, Sal does some algebraic manipulation to achieve the formula of the parabola. It is much easier to derive the parabola if he were to square the expression and take the square root than to take the absolute value of the expression. Both methods yield the same value of the expression; however, the latter method (squaring then taking the square root) allows for more easier manipulation. Hope this helps!", + "video_name": "okXVhDMuGFg", + "transcript": "- [Voiceover] What I have attempted to draw here in yellow is a parabola, and as we've already seen in previous videos, a parabola can be defined as the set of all points that are equidistant to a point and a line, and the point is called the focus of the parabola, and the line is called the directrix of the parabola. What I want to do in this video, it's gonna get a little bit of hairy algebra, but given that definition, I want to see, and given that definition, and given a focus at the point x equals a, y equals b, and a line, a directrix, at y equals k, to figure out what is the equation of that parabola actually going to be, and it's going to be based on a's, b's, and k's, so let's do that. So let's take a arbitrary point on the parabola. Let's say we take this point right over here, and its x-coordinate is x, and its y-coordinate is y, and by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix, so what does that mean? That means that the distance to the directrix, which I'm drawing here in blue, has to be the same as the distance to the focus, which I am drawing in magenta, and when we take the distance to the directrix, we literally just drop a perpendicular, that is, that's going to be the shortest distance to that line, but the distance to the focus, well we see that's at a bit of an angle, and we might have to use the distance formula, which is really just the Pythagorean Theorem. So let's do that. This distance has to be the same as that distance. So, what's this blue distance? Well, that's just gonna be our change in y. It's going to be this y, minus k. It's just this distance. So it's going to be y minus k. Now we have to be careful. The way I've just drawn it, yes, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances, but you can definitely have a parabola where the y-coordinate of the focus is lower than the y-coordinate of the directrix, in which case this would be negative. So what we really want is the absolute value of this, or, we could square it, and then we could take the square root, the principle root, which would be equivalent to taking the absolute value of y minus k. So that's this distance right over here, and by the definition of a parabola, in order for (x,y) to be sitting on the parabola, that distance needs to be the same as the distance from (x,y) to (a,b), to the focus. So what's that going to be? Well, we just apply the distance formula, or really, just the Pythagorean Theorem. It's gonna be our change in x, so, x minus a, squared, plus the change in y, y minus b, squared, and the square root of that whole thing, the square root of all of that business. Now, this right over here is an equation of a parabola. It doesn't look like it, it looks really hairy, but it IS the equation of a parabola, and to show you that, we just have to simplify this, and if you get inspired, I encourage you to try to simplify this on your own, it's just gonna be a little bit of hairy algebra, but it really is not too bad. You're gonna get an equation for a parabola that you might recognize, and it's gonna be in terms of a general focus, (a,b), and a gerneral directrix, y equals k, so let's do that. So the simplest thing to start here, is let's just square both sides, so we get rid of the radicals. So if you square both sides, on the left-hand side, you're gonna get y minus k, squared is equal to x minus a, squared, plus y minus b, squared. Fair enough? Now what I want to do is, I just want to end up with just a y on the left-hand side, and just x's, ab's, and k's on the right-hand side, so the first thing I might want to do, is let's expand each of these expressions that involve with y, so this blue one on the left-hand side, that is going to be y squared minus 2yk, plus k squared, and that is going to be equal to, I'm gonna keep this first one the same, so it's gonna be x minus a, squared, and now let me expand, I'm gonna find a color, expand this in green, so plus y squared, minus 2yb, plus b squared. All I did, is I multiplied y minus b, times y minus b. Now let's see if we can simplify things. So, I have a y squared on the left, I have a y squared on the right, well, if I subtract y squared from both sides, so I can do that. Well, that simplified things a little bit, and now I can, let's see what I can do. Well let's get the k squared on this side, so let's subtract k squared from both sides, so, subtract k squared from both sides, so that's gonna get rid of it on the left-hand side, and now let's add 2yb to both sides, so we have all the y's on the left-hand side, so, plus 2yb, that's gonna give us a 2yb on the left-hand side, plus 2yb. So what is this going to be equal to? And I'm starting to run into my graph, so let me give myself a little bit more real estate over here. So on the left-hand side, what am I going to have? This is the same thing as 2yb minus 2yk, which is the same thing, actually let me just write that down. That's going to be 2y-- Do it in green, actually, well, yeah, why not green? That's going to be-- Actually, let me start a new color. (chuckles) That's going to be 2yb minus 2yk. You can factor out a 2y, and it's gonna be 2y times, b minus k. So let's do that. So we could write this as 2 times, b minus k, y if you factor out a 2 and a y, so that's the left-hand side, so that's that piece right over there. These things cancel out. Now, on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise. On the right-hand side, you have x minus a, squared, and then, let's see, these characters cancel out, and you're left with b squared minus k squared, so these two are gonna be b squared minus k squared, plus b squared minus k squared. Now, I said all I want is a y on the left-hand side, so let's divide everything by two times, b minus k. So, let's divide everything, two times, b minus k, so, two times, b minus k. And I'm actually gonna divide this whole thing by two times, b minus k. Now, obviously on the left-hand side, this all cancels out, you're left with just a y, and then it's going to be y equals, y is equal to one over, two times, b minus k, and notice, b minus k is the difference between the y-coordinate of the focus, and the y-coordinate, I guess you could say, of the line, y equals k, so it's one over, two times that, times x minus a, squared. So if you knew what b minus k was, this would just simplify to some number, some number that's being multiplied times x minus a, squared, so hopefully this is starting to look like the parabolas that you remember from your childhood, (chuckles) if you do remember parabolas from your childhood. Alright, so then let's see if we could simplify this thing on the right, and you might recognize, b squared minus k squared, that's a difference of squares, that's the same thing as b plus k, times b minus k, so the b minus k's cancel out, and we are just left with, and we deserve a little bit of a drum roll, we are just left with 1/2 times, b plus k. So, there you go. Given a focus at a point (a,b), and a directrix at y equals k, we now know what the formula of the parabola is actually going to be. So, for example, if I had a focus at the point, I don't know, let's say the point (1,2), and I had a directrix at y is equal to, I don't know, let's make it y is equal to -1, what would the equation of this parabola be? Well, it would be y is equal to one over, two times, b minus k, so two minus -1, that's the same thing as two plus one, so that's just three, two minus -1 is three, times x minus one, squared, plus 1/2 times, b plus k. Two plus -1 is one, so one, and so what is this going to be? You're gonna get y is equal to 1/6, x minus one, squared, plus 1/2. There you go. That is the parabola with a focus at (1,2) and a directrix at y equals -1. Fascinating." + }, + { + "Q": "difference between arcsin and inverse sin", + "A": "inverse sin x, arcsin x and sin\u00e2\u0081\u00bb\u00c2\u00b9 x all mean exactly the same thing.", + "video_name": "JGU74wbZMLg", + "transcript": "If I were to walk up to you on the street and say you, please tell me what-- so I didn't want to write that thick --please tell me what sine of pi over 4 is. And, obviously, we're assuming we're dealing in radians. You either have that memorized or you would draw the unit circle right there. That's not the best looking unit circle, but you get the idea. You'd go to pi over 4 radians, which is the same thing as 45 degrees. You would draw that unit radius out. And the sine is defined as a y-coordinate on the unit circle. So you would just want to know this value right here. And you would immediately say OK. This is a 45 degrees. Let me draw the triangle a little bit larger. The triangle looks like this. This is 45. That's 45. This is 90. And you can solve a 45 45 90 triangle. The hypotenuse is 1. This is x. They're going to be the same values. This is an isosceles triangle, right? Their base angles are the same. So you say, look. x squared plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1/2. x is equal to the square root of 1/2, which is one over the square root of 2. I can put that in rational form by multiplying that by the square root of 2 over 2. And I get x is equal to the square root of 2 over 2. So the height here is square root of 2 over 2. And if you wanted to know this distance too, it would also be the same thing. But we just cared about the height. Because the sine value, the sine of this, is just this height right here. The y-coordinate. And we got that as the square root of 2 over 2. This is all review. We learned this in the unit circle video. But what if someone else-- Let's say on another day, I come up to you and I say you, please tell me what the arcsine of the square root of 2 over 2 is. What is the arcsine? You're like I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize, when they have this word arc in front of it-- This is also sometimes referred to as the inverse sine. This could have just as easily been written as: what is the inverse sine of the square root of 2 over 2? All this is asking is what angle would I have to take the sine of in order to get the value square root of 2 over 2. This is also asking what angle would I have to take the sine of in order to get square root of 2 over 2. I could rewrite either of these statements as saying square-- Let me do it. I could rewrite either of these statements as saying sine of what is equal to the square root of 2 over 2. And this, I think, is a much easier question for you to answer. Sine of what is square root of 2 over 2? Well I just figured out that the sine of pi over 4 is square root of 2 over 2. So, in this case, I know that the sine of pi over 4 is equal to square root of 2 over 2. So my question mark is equal to pi over 4. Or, I could have rewritten this as, the arcsine-- sorry --arcsine of the square root of 2 over 2 is equal to pi over 4. Now you might say so, just as review, I'm giving you a value and I'm saying give me an angle that gives me, when I take the sine of that angle that gives me that value. But you're like hey Sal. Let me go over here. You're like, look pi over 2 worked. 45 degrees worked. But I could just keep adding 360 degrees or I could keep just adding 2 pi. And all of those would work because those would all get me to that same point of the unit circle, right? And you'd be correct. And so all of those values, you would think, would be valid answers for this, right? Because if you take the sine of any of those angles-- You could just keep adding 360 degrees. If you take the sine of any of them, you would get square root of 2 over 2. And that's a problem. You can't have a function where if I take the function-- I can't have a function, f of x, where it maps to multiple values, right? Where it maps to pi over 4, or it maps to pi over 4 plus 2 pi or pi over 4 plus 4 pi. So in order for this to be a valid function-- In order for the inverse sine function to be valid, I have to restrict its range. And the way that-- We'll just restrict its range to the most natural place. So let's restrict its range. Actually, just as a side note, what's its domain restricted to? So if I'm taking the arcsine of something. So if I'm taking the arcsine of x, and I'm saying that that is equal to theta, what's the domain restricted to? What are the valid values of x? x could be equal to what? Well if I take the sine of any angle, I can only get values between 1 and negative 1, right? So x is going to be greater than or equal to negative 1 and then less than or equal to 1. That's the domain. Now, in order to make this a valid function, I have to restrict the range. The possible values. I have to restrict the range. Now for arcsine, the convention is to restrict it to the first and fourth quadrants. To restrict the possible angles to this area right here along the unit circle. So theta is restricted to being less than or equal to pi over 2 and then greater than or equal to minus pi over 2. So given that, we now understand what arcsine is. Let's do another problem. Clear out some space here. Let me do another arcsine. So let's say I were to ask you what the arcsine of minus the square root of 3 over 2 is. Now you might have that memorized. And say, I immediately know that sine of x, or sine of theta is square root of 3 over 2. And you'd be done. But I don't have that memorized. So let me just draw my unit circle. And when I'm dealing with arcsine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I? If the sine of something is minus square root of 3 over 2, that means the y-coordinate on the unit circle is minus square root of 3 over 2. So it means we're right about there. So this is minus the square root of 3 over 2. This is where we are. Now what angle gives me that? Let's think about it a little bit. My y-coordinate is minus square root of 3 over 2. This is the angle. It's going to be a negative angle because we're going below the x-axis in the clockwise direction. And to figure out-- Let me just draw a little triangle here. Let me pick a better color than that. That's a triangle. Let me do it in this blue color. So let me zoom up that triangle. Like that. This is theta. That's theta. And what's this length right here? Well that's the same as the y-height, I guess Which is square root of 3 over 2. It's a minus because we're going down. But let's just figure out this angle. And we know it's a negative angle. So when you see a square root of 3 over 2, hopefully you recognize this is a 30 60 90 triangle. The square root of 3 over 2. This side is 1/2. And then, of course, this side is 1. Because this is a unit circle. So its radius is 1. So in a 30 60 90 triangle, the side opposite to the square root of 3 over 2 is 60 degrees. This side over here is 30 degrees. So we know that our theta is-- This is 60 degrees. That's its magnitude. But it's going downwards. So it's minus 60 degrees. So theta is equal to minus 60 degrees. But if we're dealing in radians, that's not good enough. So we can multiply that times 100-- sorry --pi radians for every 180 degrees. Degrees cancel out. And we're left with theta is equal to minus pi over 3 radians. And so we can say-- We can now make the statements that the arcsine of minus square root of 3 over 2 is equal to minus pi over 3 radians. Or we could say the inverse sign of minus square root of 3 over 2 is equal to minus pi over 3 radians. And to confirm this, let's just-- Let me get a little calculator out. I put this in radian mode already. You can just check that. Per second mode. I'm in radian mode. So I know I'm going to get, hopefully, the right answer. And I want to figure out the inverse sign. So the inverse sine-- the second and the sine button --of the minus square root of 3 over 2. It equals minus 1.04. So it's telling me that this is equal to minus 1.04 radians. So pi over 3 must be equal to 1.04. Let's see if I can confirm that. So if I were to write minus pi divided by 3, what do I get? I get the exact same value. So my calculator gave me the exact same value, but it might have not been that helpful because my calculator doesn't tell me that this is minus pi over 3." + }, + { + "Q": "How do you know weather to multiply, divide, add, or subtract?", + "A": "The rule is that you do the opposite function. Examples: 1) y+5 = 7. The 5 is being added to the y . To move it, we do the opposite operation which is subtraction. 2) 5x = 20. The 5 is being multiplied with the 20. To eliminate the 5, we do the opposite of multiplication, which is division.", + "video_name": "kbqO0YTUyAY", + "transcript": "So once again, we have three equal, or we say three identical objects. They all have the same mass, but we don't know what the mass is of each of them. But what we do know is that if you total up their mass, it's the same exact mass as these nine objects And each of these nine objects have a mass of 1 kilograms. So in total, you have 9 kilograms on this side. And over here, you have three objects. They all have the same mass. And we don't know what it is. We're just calling that mass x. And what I want to do here is try to tackle this a little bit more symbolically. In the last video, we said, hey, why don't we just multiply 1/3 of this and multiply 1/3 of this? And then, essentially, we're going to keep things balanced, because we're taking 1/3 of the same mass. This total is the same as this total. That's why the scale is balanced. Now, let's think about how we can represent this symbolically. So the first thing I want you to think about is, can we set up an equation that expresses that we have these three things of mass x, and that in total, their mass is equal to the total mass Can we express that as an equation? And I'll give you a few seconds to do it. Well, let's think about it. Over here, we have three things with mass x. So their total mass, we could write as-- we could write their total mass as x plus x plus x. And over here, we have nine things with mass of 1 kilogram. I guess we could write 1 plus 1 plus 1. That's 3. Plus 1 plus 1 plus 1 plus 1. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9. And actually, this is a mathematical representation. If we set it up as an equation, it's an algebraic representation. It's not the simplest possible way we can do it, but it is a reasonable way to do it. If we want, we can say, well, if I have an x plus another x plus another x, I have three x's. So I could rewrite this as 3x. And 3x will be equal to? Well, if I sum up all of these 1's right over here-- 1 We're doing that. We have 9 of them, so we get 3x is equal to 9. And let me make sure I did that. 1, 2, 3, 4, 5, 6, 7, 8, 9. So that's how we would set it up. And so the next question is, what would we do? What can we do mathematically? Actually, to either one of these equations, but we'll focus on this one right now. What can we do mathematically in order to essentially solve for the x? In order to figure out what that mystery mass actually is? And I'll give you another second or two to think about it. Well, when we did it the last time with just the scales we said, OK, we've got three of these x's here. We want to have just one x here. So we can say, whatever this x is, if the scale stays balanced, it's going to be the same as whatever we have there. There might be a temptation to subtract two of the x's maybe from this side, but that won't help us. And we can even see it mathematically over here. If we subtract two x's from both sides, on the left-hand side you're going to have 3x minus 2x. And on the right-hand side, you're going to have 9 minus 2x. And you're just going to be left with 3 of something minus 2 something is just 1 of something. So you will just have an x there if you get rid of two of them. But on the right-hand side, you're going to get 9 minus 2 So the x's still didn't help you out. You still have a mystery mass on the right-hand side. So that doesn't help. So instead, what we say is-- and we did this the last time. We said, well, what if we took 1/3 of these things? If we take 1/3 of these things and take 1/3 of these things, we should still get the same mass on both sides because the original things had the same mass. And the equivalent of doing that mathematically is to say, why don't we multiply both sides by 1/3? Or another way to say it is we could divide both sides by 3. Multiplying by 1/3 is the same thing as dividing by 3. So we're going to multiply both sides by 1/3. When you multiply both sides by 1/3-- visually over here, if you had three x's, you multiply it by 1/3, you're only going to have one x left. If you have nine of these one-kilogram boxes, you multiply it by 1/3, you're only going to have three left. And over here, you can even visually-- if you divide by 3, which is the same thing as multiplying by 1/3, you divide by 3. So you divide by 3. You have an x is equal to a 1 plus 1 plus 1. An x is equal to 3. Or you see here, an x is equal to 3. Over here you do the math. 1/3 times 3 is 1. You're left with 1x. So you're left with x is equal to 9 times 1/3. Or you could even view it as 9 divided by 3, which is equal to 3." + }, + { + "Q": "If 6.0 g of NO2 came in contact with a cloud containing 2.0 g of H2O, what mass in grams of acid would be produced?", + "A": "You need to take the molar masses of the two substances, find the number of moles of both and then look at which of the substances is the limiting reagent, depending on the stoichiometry of the reaction. Once you have the limiting reagent, the molar ratio will give you the number of moles of reactants and products. Moles of product is easily converted into mass (grams) by using the molar mass formula. Your reaction (I assume) is 8NO2 + 3H2O = 6HNO3 + N2O", + "video_name": "SjQG3rKSZUQ", + "transcript": "We know what a chemical equation is and we've learned how to balance it. Now, we're ready to learn about stoichiometry. And this is an ultra fancy word that often makes people think it's difficult. But it's really just the study or the calculation of the relationships between the different molecules in a reaction. This is the actual definition that Wikipedia gives, stoichiometry is the calculation of quantitative, or measurable, relationships of the reactants and the products. And you're going to see in chemistry, sometimes people use the word reagents. For most of our purposes you can use the word reagents and reactants interchangeably. They're both the reactants in a reaction. The reagents are sometimes for special types of reactions where you want to throw a reagent in and see if something happens. And see if your belief about that substance is true or things like that. But for our purposes a reagent and reactant is the same thing. So it's a relationship between the reactants and the products in a balanced chemical equation. So if we're given an unbalanced one, we know how to get to the balanced point. A balanced chemical equation. So let's do some stoichiometry. Just so we get practice balancing equations, I'm always going to start with unbalanced equations. Let's say we have iron three oxide. Two iron atoms with three oxygen atoms. Plus aluminum, Al. And it yields Al2 O3 plus iron. So remember when we're doing stoichiometry, first of all we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation. But I think it's good practice to actually balance the equations ourselves. So let's try to balance this one. We have two iron atoms here in this iron three oxide. How many iron atoms do we have on the right hand side? We only have one. So let's multiply this by 2 right here. All right, oxygen, we have three on this side. We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams. So my question to you is how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So we have two irons and three oxygens. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. Is that right? That's 48 plus 112, right, 160. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? Well 85 grams of iron three oxide is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. I just took 0.53 times 2. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units. So one mole of aluminium is going to be 27 grams. Or 6.02 times 10 to 23 aluminium atoms is going to be 27 grams. So if we need 1.06 moles, how many is that going to be? So 1.06 moles of aluminium is equal to 1.06 times 27 grams. And what is that? What is that? Equals 28.62. So we need 28.62 grams of aluminium, I won't write the whole thing there, in order to essentially use up our 85 grams of the iron three oxide. And if we had more than 28.62 grams of aluminium, then they'll be left over after this reaction happens. Assuming we keep mixing it nicely and the whole reaction happens all the way. And we'll talk more about that in the future. And in that situation where we have more than 28.63 grams of aluminium, then this molecule will be the limiting reagent. Because we had more than enough of this, so this is what's going to limit the amount of this process from happening. If we have less than 28.63 grams of, I'll start saying aluminum, then the aluminum will be the limiting reagent, because then we wouldn't be able to use all the 85 grams of our iron molecule, or our iron three oxide molecule. Anyway, I don't want to confuse you in the end with that limiting reagents. In the next video, we'll do a whole problem devoted to limiting reagents." + }, + { + "Q": "is there such thing as a 400 degree angle?", + "A": "yes, but the reference angle is 40 degrees. Hope this helps", + "video_name": "92aLiyeQj0w", + "transcript": "Now that we know what an angle is, let's think about how we can measure them. And we already hinted at one way to think about the measure of angle in the last video where we said, look, this angle XYZ seems more open than angle BAC. So maybe the measure of angle XYZ should be larger than the angle of the BAC, and that is exactly the way we think about the measures of angles. But what I want to do in this video is come up with an exact way to measure an angle. So what I've drawn over here is a little bit of a half-circle, and it looks very similar to a tool that you can buy at your local school supplies store to measure angles. So this is actually a little bit of a drawing of a protractor. And what we do in something like a protractor-- you could even construct one with a piece of paper-- is we've taken a half-circle right here, and we've divided it into a 180 sections, and each of these marks marks 10 of those sections. And what you do for any given angle is you put one of the sides of the angle. So each of the rays of an angle are considered one of its sides. So you put the vertex of the angle at the center of this half-circle-- or if you're dealing with an actual protractor, at the center of that protractor-- and then you put one side along the 0 mark. So I'm going redraw this angle right over here at the center of this protractor. So if we said this is Y, then the Z goes right over here. And then the other ray, ray YX in this circumstance, will go roughly in that direction. And so it is pointing on the protractor to the-- let's see. This looks like this is the 70th of section. This is the 80th section. So maybe this is, I would guess, the 77th section. So this is pointing to 77 right over here. Assuming that I drew it the right way right over here, we could say the measure of angle XYZ-- sometimes they'll just say angle XYZ is equal to, but this is a little bit more formal-- the measure of angle XYZ is equal to 77. Each of these little sections, we call them \"degrees.\" So it's equal to 77-- sometimes it's written like that, the same way you would write \"degrees\" for the temperature outside. So you could write \"77 degrees\" like that or you could actually write out the word right over there. So each of these sections are degrees, so we're measuring in degrees. And I want to be clear, degrees aren't the only way to measure angles. Really, anything that measures the openness. So when you go into trigonometry, you'll learn that you can measure angles, not only in degrees, but also using something called \"radians.\" But I'll leave that to another day. So let's measure this other angle, angle BAC. So once again, I'll put A at the center, and then AC I'll put along the 0 degree edge of this half-circle or of this protractor. And then I'll point AB in the-- well, assuming that I'm drawing it exactly the way that it's Normally, instead of moving the angle, you could actually move the protractor to the angle. So it looks something like that, and you could see that it's pointing to right about the 30 degree mark. So we could say that the measure of angle BAC is equal to 30 degrees. And so you can look just straight up from evaluating these numbers that 77 degrees is clearly larger than 30 degrees, and so it is a larger angle, which makes sense because it is a more open angle. And in general, there's a couple of interesting angles to think about. If you have a 0 degree angle, you actually have something that's just a closed angled. It really is just a ray at that point. As you get larger and larger or as you get more and more open, you eventually get to a point where one of the rays is completely straight up and down while the other one is left to right. So you could imagine an angle that looks like this where one ray goes straight up down like that and the other ray goes straight right and left. Or you could imagine something like an angle that looks like this where, at least, the way you're looking at it, one doesn't look straight up down or one does it look straight right left. But if you rotate it, it would look just like this thing right over here where one is going straight up and down and one is going straight right and left. And you can see from our measure right over here that that gives us a 90 degree angle. It's a very interesting angle. It shows up many, many times in geometry and trigonometry, and there's a special word for a 90 degree angle. It is called a \"right angle.\" So this right over here, assuming if you rotate it around, would look just like this. We would call this a \"right angle.\" And there is a notation to show that it's a right angle. You draw a little part of a box right over there, and that tells us that this is, if you were to rotate it, exactly up and down while this is going exactly right and left, if you were to rotate it properly, or vice versa. And then, as you go even wider, you get wider and wider and wider and wider until you get all the way to an angle that looks like this. So you could imagine an angle where the two rays in that angle form a line. So let's say this is point X. This is point Y, and this is point Z. You could call this angle ZXY, but it's really so open that it's formed an actual line here. Z, X, and Y are collinear. This is a 180 degree angle where we see the measure of angle ZXY is 180 degrees. And you can actually go beyond that. So if you were to go all the way around the circle so that you would get back to 360 degrees and then you could keep going around and around and around, and you'll start to see a lot more of that when you enter a trigonometry class. Now, there's two last things that I want to introduce in this video. There are special words, and I'll talk about more types of angles in the next video. But if an angle is less than 90 degrees, so, for example, both of these angles that we started our discussion with are less than 90 degrees, we call them \"acute angles.\" So this is acute. So that is an acute angle, and that is an acute angle right over here. They are less than 90 degrees. What does a non-acute angle look like? And there's a word for it other than non-acute. Well, it would be more than 90 degrees. So, for example-- let me do this in a color I haven't used-- an angle that looks like this, and let me draw it a little bit better than that. An angle it looks like this. So that's one side of the angle or one of the rays and then I'll put the other one on the baseline right Clearly, this is larger than 90 degrees. If I were to approximate, let's see, that's 100, 110, 120, almost 130. So let's call that maybe a 128-degree angle. We call this an \"obtuse angle.\" The way I remember it as acute, it's kind of \"a cute\" angle. It's nice and small. I believe acute in either Latin or Greek or maybe both means something like \"pin\" or \"sharp.\" So that's one way to think about it. An acute angle seems much sharper. Obtuse, I kind of imagine something that's kind of lumbering and large. Or you could think it's not acute. It's not nice and small and pointy. So that's one way to think about it, but this is just general terminology for different types of angles. Less than 90 degrees, you have an acute angle. At 90 degrees, you have a right angle. Larger than 90 degrees, you have an obtuse angle. And then, if you get all the way to 180 degrees, your angle actually forms a line." + }, + { + "Q": "how is direct and inverse variation the same or how is different ?", + "A": "They are both functions that make two parameters increase or decrease. The first makes both decrease or increase, while the second makes one parameter increase while the other decreases.", + "video_name": "92U67CUy9Gc", + "transcript": "I want to talk a little bit about direct and inverse variations. So I'll do direct variation on the left over here. And I'll do inverse variation, or two variables that vary inversely, on the right-hand side over here. So a very simple definition for two variables that vary directly would be something like this. y varies directly with x if y is equal to some constant with x. So we could rewrite this in kind of English as y varies directly with x. And if this constant seems strange to you, just remember this could be literally any constant number. So let me give you a bunch of particular examples of y varying directly with x. You could have y is equal to x. Because in this situation, the constant is 1. We didn't even write it. We could write y is equal to 1x, then k is 1. We could write y is equal to 2x. We could write y is equal to 1/2 x. We could write y is equal to negative 2x. We are still varying directly. We could have y is equal to negative 1/2 x. We could have y is equal to pi times x. We could have y is equal to negative pi times x. I don't want to beat a dead horse now. I think you get the point. Any constant times x-- we are varying directly. And to understand this maybe a little bit more tangibly, let's think about what happens. And let's pick one of these scenarios. Well, I'll take a positive version and a negative version, just because it might not be completely intuitive. So let's take the version of y is equal to 2x, and let's explore why we say they vary directly with each other. So let's pick a couple of values for x and see what the resulting y value would have to be. So if x is equal to 1, then y is 2 times 1, or is 2. If x is equal to 2, then y is 2 times 2, which is going to be equal to 4. So when we doubled x, when we went from 1 to 2-- so we doubled x-- the same thing happened to y. We doubled y. So that's what it means when something varies directly. If we scale x up by a certain amount, we're going to scale up y by the same amount. If we scale down x by some amount, we would scale down y by the same amount. And just to show you it works with all of these, let's try the situation with y is equal to negative 2x. I'll do it in magenta. y is equal to negative-- well, let me do a new example that I haven't even written here. Let's try y is equal to negative 3x. So once again, let me do my x and my y. When x is equal to 1, y is equal to negative 3 times 1, which is negative 3. When x is equal to 2, so negative 3 times 2 is negative 6. So notice, we multiplied. So if we scaled-- let me do that in that same green color. If we scale up x by 2-- it's a different green color, but it serves the purpose-- we're also scaling up y by 2. To go from 1 to 2, you multiply it by 2. To go from negative 3 to negative 6, you're also multiplying by 2. So we grew by the same scaling factor. And if you wanted to go the other way-- let's try, I don't know, let's go to x is 1/3. If x is 1/3, then y is going to be-- negative 3 times 1/3 is negative 1. So notice, to go from 1 to 1/3, we divide by 3. To go from negative 3 to negative 1, we also divide by 3. We also scale down by a factor of 3. So whatever direction you scale x in, you're going to have the same scaling direction as y. That's what it means to vary directly. Now, it's not always so clear. Sometimes it will be obfuscated. So let's take this example right over here. y is equal to negative 3x. And I'm saving this real estate for inverse variation in a second. You could write it like this, or you could algebraically manipulate it. You could maybe divide both sides of this equation by x, and then you would get y/x is equal to negative 3. Or maybe you divide both sides by x, and then you divide both sides by y. So from this, so if you divide both sides by y now, you could get 1/x is equal to negative 3 times 1/y. These three statements, these three equations, are all saying the same thing. So sometimes the direct variation isn't quite in your face. But if you do this, what I did right here with any of these, you will get the exact same result. Or you could just try to manipulate it back to this form over here. And there's other ways we could do it. We could divide both sides of this equation by negative 3. And then you would get negative 1/3 y is equal to x. And now, this is kind of an interesting case here because here, this is x varies directly with y. Or we could say x is equal to some k times y. And in general, that's true. If y varies directly with x, then we can also say that x varies directly with y. It's not going to be the same constant. It's going to be essentially the inverse of that constant, but they're still directly varying. Now with that said, so much said, about direct variation, let's explore inverse variation a little bit. Inverse variation-- the general form, if we use the same variables. And it always doesn't have to be y and x. It could be an a and a b. It could be a m and an n. If I said m varies directly with n, we would say m is equal to some constant times n. Now let's do inverse variation. So if I did it with y's and x's, this would be y is equal to some constant times 1/x. So instead of being some constant times x, it's some constant times 1/x. So let me draw you a bunch of examples. It could be y is equal to 1/x. It could be y is equal to 2 times 1/x, which is clearly the same thing as 2/x. It could be y is equal to 1/3 times 1/x, which is the same thing as 1 over 3x. it could be y is equal to negative 2 over x. And let's explore this, the inverse variation, the same way that we explored the direct variation. So let's pick-- I don't know/ let's pick y is equal to 2/x. And let me do that same table over here. So I have my table. I have my x values and my y values. If x is 1, then y is 2. If x is 2, then 2 divided by 2 is 1. So if you multiply x by 2, if you scale it up by a factor of 2, what happens to y? y gets scaled down by a factor of 2. You're dividing by 2 now. Notice the difference. Here, however we scaled x, we scaled up y by the same amount. Now, if we scale up x by a factor, when we have inverse variation, we're scaling down y by that same. So that's where the inverse is coming from. And we could go the other way. If we made x is equal to 1/2. So if we were to scale down x, we're going to see that it's going to scale up y. Because 2 divided by 1/2 is 4. So here we are scaling up y. So they're going to do the opposite things. They vary inversely. And you could try it with the negative version of it, as well. So here we're multiplying by 2. And once again, it's not always neatly written for you like this. It can be rearranged in a bunch of different ways. But it will still be inverse variation as long as they're algebraically equivalent. So you can multiply both sides of this equation right here by x. And you would get xy is equal to 2. This is also inverse variation. You would get this exact same table over here. You could divide both sides of this equation by y. And you could get x is equal to 2/y, which is also the same thing as 2 times 1/y. So notice, y varies inversely with x. And you could just manipulate this algebraically to show that x varies inversely with y. So y varies inversely with x. This is the same thing as saying-- and we just showed it over here with a particular example-- that x varies inversely with y. And there's other things. We could take this and divide both sides by 2. And you would get y/2 is equal to 1/x. There's all sorts of crazy things. And so in general, if you see an expression that relates to variables, and they say, do they vary inversely or directly or maybe neither? You could either try to do a table like this. If you scale up x by a certain amount and y gets scaled up by the same amount, then it's direct variation. If you scale up x by some-- and you might want to try a couple different times-- and you scale down y, you do the opposite with y, then it's probably inverse variation. A surefire way of knowing what you're dealing with is to actually algebraically manipulate the equation so it gets back to either this form, which would tell you that it's inverse variation, or this form, which would tell you that it is direct variation." + }, + { + "Q": "why does liquid change to steam?", + "A": "because when molecules absorb energy (thermal energy ) its kinetic energy increase and it collide faster then spaces between molecules increase and its density decrease and it turns into steam", + "video_name": "pKvo0XWZtjo", + "transcript": "I think we're all reasonably familiar with the three states of matter in our everyday world. At very high temperatures you get a fourth. But the three ones that we normally deal with are, things could be a solid, a liquid, or it could be a gas. And we have this general notion, and I think water is the example that always comes to at least my mind. Is that solid happens when things are colder, relatively colder. And then as you warm up, you go into a liquid state. And as your warm up even more you go into a gaseous state. So you go from colder to hotter. And in the case of water, when you're a solid, you're ice. When you're a liquid, some people would call ice water, but let's call it liquid water. I think we know what that is. And then when it's in the gas state, you're essentially vapor or steam. So let's think a little bit about what, at least in the case of water, and the analogy will extend to other types of molecules. But what is it about water that makes it solid, and when it's colder, what allows it to be liquid. And I'll be frank, liquids are kind of fascinating because you can never nail them down, I guess is the best way to view them. Or a gas. So let's just draw a water molecule. So you have oxygen there. You have some bonds to hydrogen. And then you have two extra pairs of valence electrons in the oxygen. And a couple of videos ago, we said oxygen is a lot more electronegative than the hydrogen. It likes to hog the electrons. So even though this shows that they're sharing electrons here and here. At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules. Let me draw some more molecules. When we talk about the whole state of the whole matter, we actually think about how the molecules are interacting with Not just how the atoms are interacting with each other within a molecule. I just drew one oxygen, let me copy and paste that. But I could do multiple oxygens. And let's say that that hydrogen is going to want to be near this oxygen. Because this has partial negative charge, this has a partial positive charge. And then I could do another one right there. And then maybe we'll have, and just to make the point clear, you have two hydrogens here, maybe an oxygen wants to hang out there. So maybe you have an oxygen that wants to be here because it's got its partial negative here. And it's connected to two hydrogens right there that have their partial positives. But you can kind of see a lattice structure. Let me draw these bonds, these polar bonds that start forming between the particles. These bonds, they're called polar bonds because the molecules themselves are polar. And you can see it forms this lattice structure. And if each of these molecules don't have a lot of kinetic energy. Or we could say the average kinetic energy of this matter is fairly low. And what do we know is average kinetic energy? Well, that's temperature. Then this lattice structure will be solid. These molecules will not move relative to each other. I could draw a gazillion more, but I think you get the point that we're forming this kind of fixed structure. And while we're in the solid state, as we add kinetic energy, as we add heat, what it does to molecules is, it just makes them vibrate around a little bit. If I was a cartoonist, they way you'd draw a vibration is to put quotation marks there. That's not very scientific. But they would vibrate around, they would buzz around a little bit. I'm drawing arrows to show that they are vibrating. It doesn't have to be just left-right it could be up-down. But as you add more and more heat in a solid, these molecules are going to keep their structure. So they're not going to move around relative to each other. But they will convert that heat, and heat is just a form of energy, into kinetic energy which is expressed as the vibration of these molecules. Now, if you make these molecules start to vibrate enough, and if you put enough kinetic energy into these molecules, what do you think is going to happen? Well this guy is vibrating pretty hard, and he's vibrating harder and harder as you add more and more heat. This guy is doing the same thing. At some point, these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations. And once that happens, the molecules-- let me draw a couple more. Once that happens, the molecules are going to start moving past each other. So now all of a sudden, the molecule will start shifting. But they're still attracted. Maybe this side is moving here, that's moving there. You have other molecules moving around that way. But they're still attracted to each other. Even though we've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules. Our vibration, or our kinetic energy for each molecule, still isn't strong enough to completely separate them. They're starting to slide past each other. And this is essentially what happens when you're in a liquid state. You have a lot of atoms that want be touching each other but they're sliding. They have enough kinetic energy to slide past each other and break that solid lattice structure here. And then if you add even more kinetic energy, even more heat, at this point it's a solution now. They're not even going to be able to stay together. They're not going to be able to stay near each other. If you add enough kinetic energy they're going to start looking like this. They're going to completely separate and then kind of bounce around independently. Especially independently if they're an ideal gas. But in general, in gases, they're no longer touching They might bump into each other. But they have so much kinetic energy on their own that they're all doing their own thing and they're not touching. I think that makes intuitive sense if you just think about what a gas is. For example, it's hard to see a gas. Why is it hard to see a gas? Because the molecules are much further apart. So they're not acting on the light in the way that a liquid or a solid would. And if we keep making that extended further, a solid-- well, I probably shouldn't use the example with ice. Because ice or water is one of the few situations where the solid is less dense than the liquid. That's why ice floats. And that's why icebergs don't just all fall to the bottom of the ocean. And ponds don't completely freeze solid. But you can imagine that, because a liquid is in most cases other than water, less dense. That's another reason why you can see through it a little Or it's not diffracting-- well I won't go into that too much, than maybe even a solid. But the gas is the most obvious. And it is true with water. The liquid form is definitely more dense than the gas form. In the gas form, the molecules are going to jump around, not touch each other. And because of that, more light can get through the substance. Now the question is, how do we measure the amount of heat that it takes to do this to water? And to explain that, I'll actually draw a phase change diagram. Which is a fancy way of describing something fairly straightforward. Let me say that this is the amount of heat I'm adding. And this is the temperature. We'll talk about the states of matter in a second. So heat is often denoted by q. Sometimes people will talk about change in heat. They'll use H, lowercase and uppercase H. They'll put a delta in front of the H. Delta just means change in. And sometimes you'll hear the word enthalpy. Let me write that. Because I used to say what is enthalpy? It sounds like empathy, but it's quite a different concept. At least, as far as my neural connections could make it. But enthalpy is closely related to heat. It's heat content. For our purposes, when you hear someone say change in enthalpy, you should really just be thinking of change in heat. I think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary. The best way to think about it is heat content. Change in enthalpy is really just change in heat. And just remember, all of these things, whether we're talking about heat, kinetic energy, potential energy, enthalpy. You'll hear them in different contexts, and you're like, I thought I should be using heat and they're talking about enthalpy. These are all forms of energy. And these are all measured in joules. And they might be measured in other ways, but the traditional way is in joules. And energy is the ability to do work. And what's the unit for work? Well, it's joules. Force times distance. But anyway, that's a side-note. But it's good to know this word enthalpy. Especially in a chemistry context, because it's used all the time and it can be very confusing and non-intuitive. Because you're like, I don't know what enthalpy is in my everyday life. Just think of it as heat contact, because that's really But anyway, on this axis, I have heat. So this is when I have very little heat and I'm increasing my heat. And this is temperature. Now let's say at low temperatures I'm here and as I add heat my temperature will go up. Temperature is average kinetic energy. Let's say I'm in the solid state here. And I'll do the solid state in purple. No I already was using purple. I'll use magenta. So as I add heat, my temperature will go up. Heat is a form of energy. And when I add it to these molecules, as I did in this example, what did it do? It made them vibrate more. Or it made them have higher kinetic energy, or higher average kinetic engery, and that's what temperature is a measure of; average kinetic energy. So as I add heat in the solid phase, my average kinetic energy will go up. And let me write this down. This is in the solid phase, or the solid state of matter. Now something very interesting happens. Let's say this is water. So what happens at zero degrees? Which is also 273.15 Kelvin. Let's say that's that line. What happens to a solid? Well, it turns into a liquid. Ice melts. Not all solids, we're talking in particular about water, about H2O. So this is ice in our example. All solids aren't ice. Although, you could think of a rock as solid magma. Because that's what it is. I could take that analogy a bunch of different ways. But the interesting thing that happens at zero degrees. Depending on what direction you're going, either the freezing point of water or the melting point of ice, something interesting happens. As I add more heat, the temperature does not to go up. As I add more heat, the temperature does not go up for a little period. Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or which water will boil. But something happens. And they're really getting kinetically active. But just like when you went from solid to liquid, there's a certain amount of energy that you have to contribute to the system. And actually, it's a good amount at this point. Where the water is turning into vapor, but it's not getting any hotter. So we have to keep adding heat, but notice that the temperature didn't go up. We'll talk about it in a second what was happening then. And then finally, after that point, we're completely vaporized, or we're completely steam. Then we can start getting hot, the steam can then get hotter as we add more and more heat to the system. So the interesting question, I think it's intuitive, that as you add heat here, our temperature is going to go up. But the interesting thing is, what was going on here? We were adding heat. So over here we were turning our heat into kinetic energy. Temperature is average kinetic energy. But over here, what was our heat doing? Well, our heat was was not adding kinetic energy to the system. The temperature was not increasing. But the ice was going from ice to water. So what was happening at that state, is that the kinetic energy, the heat, was being used to essentially break these bonds. And essentially bring the molecules into a higher energy state. So you're saying, Sal, what does that mean, higher energy state? Well, if there wasn't all of this heat and all this kinetic energy, these molecules want to be very close to each other. For example, I want to be close to the surface of the earth. When you put me in a plane you have put me in a higher energy state. I have a lot more potential energy. I have the potential to fall towards the earth. Likewise, when you move these molecules apart, and you go from a solid to a liquid, they want to fall towards each other. But because they have so much kinetic energy, they never quite are able to do it. But their energy goes up. Their potential energy is higher because they want to fall towards each other. By falling towards each other, in theory, they could do some work. So what's happening here is, when we're contributing heat-- and this amount of heat we're contributing, it's called the heat of fusion. Because it's the same amount of heat regardless how much direction we go in. When we go from solid to liquid, you view it as the heat of melting. It's the head that you need to put in to melt the ice into liquid. When you're going in this direction, it's the heat you have to take out of the zero degree water to turn it into ice. So you're taking that potential energy and you're bringing the molecules closer and closer to each other. So the way to think about it is, right here this heat is being converted to kinetic energy. Then, when we're at this phase change from solid to liquid, that heat is being used to add potential energy into the system. To pull the molecules apart, to give them more potential energy. If you pull me apart from the earth, you're giving me potential energy. Because gravity wants to pull me back to the earth. And I could do work when I'm falling back to the earth. A waterfall does work. It can move a turbine. You could have a bunch of falling Sals move a turbine as well. And then, once you are fully a liquid, then you just become a warmer and warmer liquid. Now the heat is, once again, being used for kinetic energy. You're making the water molecules move past each other faster, and faster, and faster. To some point where they want to completely disassociate from each other. They want to not even slide past each other, just completely jump away from each other. And that's right here. This is the heat of vaporization. And the same idea is happening. Before we were sliding next to each other, now we're pulling apart altogether. So they could definitely fall closer together. And then once we've added this much heat, now we're just heating up the steam. We're just heating up the gaseous water. And it's just getting hotter and hotter and hotter. But the interesting thing there, and I mean at least the interesting thing to me when I first learned this, whenever I think of zero degrees water I'll say, oh it must be ice. But that's not necessarily the case. If you start with water and you make it colder and colder and colder to zero degrees, you're essentially taking heat out of the water. You can have zero degree water and it hasn't turned into ice yet. And likewise, you could have 100 degree water that hasn't turned into steam yeat. You have to add more energy. You can also have 100 degree steam. You can also have zero degree water. Anyway, hopefully that gives you a little bit of intuition of what the different states of matter are. And in the next problem, we'll talk about how much heat exactly it does take to move along this line. And maybe we can solve some problems on how much ice we might need to make our drink cool." + }, + { + "Q": "in moon 1day=how many hours", + "A": "That is an interesting fact. Approximately, 29.5 days on earth is equal to 1 moon day. So 1 moon day = 708 hours", + "video_name": "TvPsESzCYis", + "transcript": "Do you feel that? Do you feel the weight of the atmosphere pushing down on you? Actually, you, like most people, have probably not thought about how heavy the atmosphere actually is and about the pressure it pushes on you every day. In fact, you have about 14.7 pounds of force pushing on you per every square inch of you. But where does this pressure come from? For the answer to that question, science had to wait till the middle of the 17th century, when Pascal had an hypothesis. You see he thought that the atmospheric pressure that we experience every day is actually due to the massive column of air that towers above us and the weight that it exerts by gravity. To test this, he devised a very simple experiment. He took an instrument called a barometer that measures atmospheric pressure and made a measurement at several locations in the lowest part of town, at the top of a church tower. And he even climbed a local mountain to measure the pressure at the top with the idea that if the height of the column of air stayed the same pressure, the pressure should stay the same in each of those locations. But if the height of the column decreases at each location, then the pressure should decrease as well. I'm going to show you how to build a simple pressure measuring device called a manometer. First, I'll need two relatively stiff tubes. I used two plastic pipe heads I had lying around the lab. I got one end off to make room for the plug. You'll also need a plug to plug one end and a piece of relatively stiff tube. Now to make the manometer, you need to connect the tubing to each of the pipe heads or plastic tubes. Once they're connected, you'll have a U-shaped device. You can then fill it with water, add some food coloring if you like, and then plug one end to fix the pressure. You have a manometer, and you're ready to make pressure measurements. At the beginning, the pressure exerted on each column of liquid is equal so the liquid levels in tube A and B are equal as well. If the pressure on tube A increases, the water in tube A will fall, and the water in tube B will rise. If the pressure over tube A decreases, then the water in tube B will fall, and the water in tube A will rise. For our first measurement, we went to Constitution Park Here we show that the pressure is greater than it was in the lab but using our manometer to show that column A is 12 millimeters below column B. But what's the pressure going to be up on Bunker Hill Monument? Well, let's find out. After climbing 275 stairs, we were able to take a measurement at the top of Bunker Hill Monument. Here we are able to show that the column A is just six millimeters below column B. For our final measurement, we drove about 20 miles south to the great Blue Hill. After climbing for about a mile and a half, we reached the top with our stunning views of Boston and the ocean. Here we get our lowest measurement yet. Column A is actually three millimeters above column B. Now for some resulting calculations-- getting from changes in water levels to pressure calculations. For starters, the air in Tube B is constrained by the plug. Because of this, it's an ideal gas and obeys the ideal gas law, which states the pressure times the volume equals nRT. Now nRT remains constant through all measurements. Because of this, we can then produce a nice relationship, and that is, P1 times V1 equals P2 times V2. Or in other words, the pressure times the volume at one state must equal the pressure and the volume at another state. But how do we get from differences in the water levels that we measured in the field to changes in the volume of Tube B, which is what we need for our actual calculations. Well, from my device, I've calculated that for each 1 millimeter change in the water level that that corresponds to a change in volume of 0.133 ml, with half of that volume change occurring in column A and the other half occurring in column B. So to calculate the change in volume of Tube B only, we can take my conversion factor of 0.133 ml per millimeter and multiply it by half of the height change. Therefore, the final volume in Tube B that we saw in the field is V2 equals the initial volume, V1 of 22 ml minus the change in volume. So using this formula, I went ahead and calculated the final volume to be at each of our locations. At Constitution Park, the final volume of Tube B was 20.4 ml, at Bunker Hill Monument, 21.2 ml, and at Great Blue Hill, 22.4 ml. Now we can take this relationship and rearrange it to actually solve for the pressure in each of these locations. Using the initial conditions of the air in Tube B, of P1 equals 101.6 kilopascals, which is the pressure the day that I built my manometer and the initial volume of Tube B, that is the air volume of 22 miles. Using this equation, we get an actual atmospheric pressure at Constitution Park of 109.6 kilopascals, at Bunker Hill Monument 105.5 pascals, and at the Great Blue Hill of 99.8% kilopascals. But how do these atmospheric pressures relate to the elevations that we measured them. Well just as pascal showed, we see that the pressure decreases with increased elevation at a very linear relationship over the elevations that we measured. Finally, I'd like to leave you with some tips when you make your own measurement. First, be patient. it can take time for the manometer to reach equilibrium as you change pressures. Second, take multiple measurements. This will greatly increase your accuracy. And third, pay attention to weather as things like changes in temperature. Or incoming weather systems can have a dramatic effect on the pressures that you measure. Some final questions to think about based on what you've learned today. First, what would happen to the water in the manometer if it went to 5,000 meters, or what about the moon? And second, what would happen if you made a measurement on an airplane at 10,000 meters? Think about these things as you think about pressure." + }, + { + "Q": "At 1:48 what he's doing looks a little like Fermat's Little Theory. Am I correct?", + "A": "Kind of, but it s not exactly Fermat s Little Theorem. It certainly looks like it, though. It s more about redundancy and how consecutive a s are more common than consecutive b s or c s in this language.", + "video_name": "WyAtOqfCiBw", + "transcript": "Voiceover: Shannon had just finished developing his theories related to cryptography and therefore was well aware that human communication was a mix of randomness and statistical dependencies. Letters in our messages were obviously dependent on previous letters to some extent. In 1949, he published a groundbreaking paper, \"A Mathematical Theory of Communication\". In it, he uses Markov models as the basis for how we can think about communication. He starts with a toy example. Imagine you encounter a bunch of text written in an alphabet of A, B, and C. Perhaps you know nothing about this language, though you notice As seem to clump together, while Bs and Cs do not. He then shows that you could design a machine to generate similar-looking text, using a Markov chain. He starts off with a zeroth-order approximation, which means we just independently select each symbol A, B, or C at random, and form a sequence However, notice that this sequence doesn't look like the original. He shows then you could do a bit better with a first-order approximation, where the letters are chosen independently, but according to the probability of each letter in the original sequence. This is slightly better as As are now more likely, but it still doesn't capture much structure. The next step is key. A second-order approximation takes into account each pair of letters which can occur. In this case, we need three states. The first state represents all pairs which begin with A, the second all pairs that begin with B, and the third state all pairs that begin with C. Notice now that the A cup has many AA pairs, which makes sense, since the conditional probability of an A after an A is higher in our original message. We can generate a sequence using this second-order model easily as follows. We start anywhere and pick a tile, and we write down our output the first letter, and move to the cup defined by the second letter. Then we pick a new tile, and repeat this process indefinitely Notice that this sequence is starting to look very similar to the original message, because this model is capturing the conditional dependencies between letters. If we wanted to do even better, we could move to a third-order approximation, which takes into account groups of three letters, or \"trigrams\". In this case, we would need nine states. But next, Shannon applies this exact same logic to actual English text, using statistics that were known for letters, pairs, and trigrams, etc. He shows the same progression from zeroth-order random letters to first-order, second-order and third-order sequences. He then goes on and tries the same thing using words instead of letters, and he writes \"the resemblance to ordinary English text \"increases quite noticeably at each depth.\" Indeed, these machines were producing meaningless text, though they contained approximately the same statistical structure you'd see in actual English. Shannon then proceeds to define a quantitative measure of information, as he realizes that the amount of information in some message must be tied up in the design of the machine which could be used to generate similar-looking sequences. Which brings us to his concept of entropy." + }, + { + "Q": "velocity = displacement/ time\nvelocity inversely proportional to time\nthis implies that velocity-time graph should be a rectangular hyperbola but here it is a straight line curve..?why?", + "A": "Just because you have a function of the form F(t) = y/t doesn t always indicate a hyperbolic curve. If in the function above y = 2t it would simplify to F(t) = 2.", + "video_name": "MAS6mBRZZXA", + "transcript": "The goal of this video is to explore some of the concept of formula you might see in introductional physics class but more importantly to see they are really just common sense ideas So let's just start with a simple example Let's say that and for the sake of this video keep things that magnitudes and velocities that's the direction of velocity etc. let's just assume that if I have a positive number that it means for example postive velocity that it means I'm going to the right let's say I have a negative number we won't see in this video let's assume we are going to the left In that way I can just write a number down only operating in one dimension you know that by specifying the magnitude and the direction if I say velocity is 5m/s that means 5m/s to the right if I say negative 5m/s that means 5m/s to the left let's say just for simplicitiy, say that we start with initial velocity we start with an initial velocity of 5m/s once again I specify the magnitude and the direction because of this convention here, we know it is to the right let's say we have a constant acceleration we have a constant acceleration 2m/s^2 or 2m per second square and once again since this is positive it is to the right and let's say that we do this for a duration so my change in time, let's say we do this for a duration of 4 I will just use s, second and s different places so s for this video is seconds So I want to do is to think about how far do we travel? and there is two things how fast are we going? after 4 seconds and how far have we travel over the course of those 4 seconds? so let's draw ourselves a little diagram here So this is my velocity axis, and this over here is time axis we have to draw a straighter line than that So that is my time axis, time this is velocity This is my velocity right over there and I'm starting off with 5m/s, so this is 5m/s right over here So vi is equal to 5m/s And every second goes by it goes 2m/s faster that's 2m/s*s every second that goes by So after 1 second when it goes 2m/s faster it will be at 7 another way to think about it is the slope of this velocity line is my constant accleration, my constant slope here so it might look something like that So what has happend after 4s? So 1 2 3 4 this is my delta t So my final velocity is going to be right over there I'm writing it here because this get into the way of veloctiy so this is v this is my final velocity what would it be? Well I'm starting at 5m/s So we are doing this both using the variable and concretes Some starting with some initial velocity I'm starting with some initial velocity Subscript i said i for initial and then each second that goes by I'm getting this much faster so if I gonna see how much faster have I gone I multiply the number of second, I will just multiply the number second it goes by times my acceleration, times my acceleration and once again, this right here, subscript c saying that is a constant acceleration, so that will tell my how fast I have gone If I started at this point and multiply the duration time with slope I will get this high, I will get to my final velocity just to make it clear with the numbers, this number can really be anything I'm just taking this to make it concrete in your mind you have 5m/s plus 4s plus, I wanna do it in yellow plus 4s times our acceleration with 2m per second square and what is this going to be equal to? you have a second that is cancelling out one of the second down here You have 4 times, so you have 5m/s plus 4 times 2 is 8 this second gone, we just have 8m/s or this is the same thing as 13m/s which is going to be our final velocity and I wanna take a pause here, you can pause and think about it yourself this whole should be intuitive, we are starting by going with 5m/s every second goes by we are gonna going 2m/s faster so after 1s it would be 7 m/s, after 2s we will be 9m/s after 3s we will be 11m/s, and after 4s we will be at 13 m/s so you multiply how much time pass times acceleration this is how much faster we are gonna be going, we are already going 5m/s 5 plus how much faster? 13 m/s so this right up here is 13m/s So I will take a little pause here hopefully intuitive and the whole play of that is to show you this formula you will see in many physics book is not something that randomly pop out of there it just make complete common sense Now the next thing I wanna talk about is what is the total distance that we would have travel? and we know from the last video that distance is just the area under this curve right over here, so it's just the area under this curve you see this is kind of a strange shape here how do I caculate this area? and we can use a little symbol of geometry to break it down into two different areas, it's very easy to calculate their areas two simple shapes, you can break it down to two, blue part is the rectangle right over here, easy to figure out the area of a rectangle and we can break it down to this purple part, this triangle right here easy to figure out the area of a triangle and that will be the total distance we travel even this will hopefully make some intuition because this blue area is how far we would have travel if we are not accelerated, we just want 5m/s for 4s so you goes 5m/s 1s 2s 3s 4s so you are going from 0 to 4 you change in time is 4s so if you go 5m/s for 4s you are going to go 20 m this right here is 20m that is the area of this right here 5 times 4 this purple or magentic area tells you how furthur than this are you going because you are accelerating because kept going faster and faster and faster it's pretty easy to calculate this area the base here is still 5(4) because that's 5(4) second that's gone by what's the height here? The height here is my final velocity minus my initial velocity minus my initial velocity or it's the change in velocity due to the accleration 13 minus 5 is 8 or this 8 right over here it is 8m/s so this height right over here is 8m/s the base over here is 4s that's the time that past what's this area of the triangle? the area of this triangle is one half times the base which is 4s 4s times the height which is 8m/s times 8m/s second cancel out one half time 4 is 2 times 8 is equal to 16m So the total distance we travel is 20 plus 16 is 36m that is the total I could say the total displacement and once again is to the right, since it's positive so that is our displacement What I wanna do is to do the exact the same calculation keep it in variable form, that will give another formula many people often memorize You might understand this is completely intuitive formula and that just come out of the logical flow of reasoning that we went through this video what is the area once again if we just think about the variables? well the area of this rectangle right here is our initial velocity times our change in time, times our change in time So that is the blue rectangle right over here, and plus what do we have to do? we have the change in time once again we have the change in time times this height which is our final velocity which is our final velocity minus our initial velocity these are all vectors, they are just positive if going to the right we just multiply the base with the height that will just be the area of the entire rectangle I will take it by half because triangle is just half of that rectangle so times one half, so times one half so this is the area, this is the purple area right over here this is the area of this, this is the area of that and let's simplify this a little bit let's factor out the delta t, so you factor out the delta t you get delta t times a bunch of stuff v sub i your initial velocity we factor this out plus this stuff, plus this thing right over here and we can distribute the one half we factor the one half, we factor the delta t out, taking it out and let's multiply one half by each of these things so it's gonna be plus one half times vf, times our final velocity that's not the right color, I will use the right color so you would understand what I am doing, so this is the one half so plus one half times our final velocity final velocity minus one half, minus one half times our initial velocity I'm gonna do that in blue, sorry I have trouble in changing color today minus one half times our initial velocity, times our initial velocity and what is this simplify do? we have something plus one half times something else minus one half of the original something so what is vi minus one half vi? so anything minus its half is just a positive half left so these two terms, this term and this term will simplify to one half vi one half initial velocity plus one half times the final velocity plus one half times the final velocity and all of that is being multiplied with the change in time the time that has gone by and this tells us the distance, the distance that we travel another way to think about it, let's factor out this one half you get distance that is equal to change in time times factoring out the one half vi plus vf, vi no that's not the right color vi plus vf so this is interesting, the distance we travel is equal to one half of the initial velocity plus the final velocity so this is really if you just took this quantity right over here it's just the arithmetic, I have trouble saying that word it's the arithmetic mean of these two numbers, so I'm gonna define, this is something new, I'm gonna call this the average velocity we have to be very careful with this this right here is the average velocity but the only reason why I can just take the starting velocity and ending velocity and adding them together and divide them by two since you took an average of two thing it's some place over here and I take that as average velocity it's because my acceleration is constant which is usually an assumption in introductory physics class but it's not always the assumption but if you do have a constant acceleration like this you can assume that the average velocity is gonna be the average of the initial velocity and the final velocity if this is a curve and the acceleration is changing you could not do that but what is useful about this is if you wanna figure out the distance that was travelled, you just need to know the initial velocity and the final velocity, average their two and multiply the times it goes by so in this situation our final velocity is 13m/s our initial velocity is 5m/s so you have 13 plus 5 is equal to 18 you divided that by 2, you average velocity is 9m/s if you take the average of 13 and 5 and 9m/s times 4s gives you 36m so hopefully it doesn't confuse you I just wannt show you some of these things you will see in your physics class but you shouldn't memorize they can all be deduced" + }, + { + "Q": "can you multiply 3 digit number times 2 digit number", + "A": "Yes, you can, the method is just the same as that of a 3 digit multiplication by another 3 digit multiplication.", + "video_name": "t8m0NalQtEk", + "transcript": "Let's start with a warm-up problem to avoid getting any mental cramps as we learn new things. So this is a problem that hopefully, if you understood what we did in the last video, you can kind of understand what we're about to do right now. And I'm going to escalate it even more. In the last video I think we finished what a four-digit number times a one-digit number. Let's up the stakes to a five-digit number. Let's do 64,329 times-- let me think of a nice number. Times 4. I'm going to show you right now that we're going to do the exact same process that we did in the last video. We just have to do it a little bit longer than we did before. So we just start off saying, OK, what's 4 times 9? 4 times 9 is equal to 36. 18 times 2. Yep, 36. So we write the 6 down here, carry the 3 up there. Just put the 3 up there, then you got 4 times 2. And they're going to have to add the 3. So let me just write that there. Plus 3 is equal to-- you do the multiplication first. You can even think of it as order of operations, but you just should know that you do the multiplication first. So 4 times 2 is 8. Plus 3 is equal to 11. Put this 1 down here and put the one 10 and 11 up there. Then you got 4 times 3. You got that one up there, so you're going to have to add that plus 1 is equal to-- that's going to equal 12. Plus 1, which is equal to 13. So it's 13. Then you have 4 times 4. You have this little one hanging out here from the previous multiplications, so you're going to to have to add that. And that's equal to 16 plus 1. It's equal to 17. Stick the 7 down here, put the 1 up there. We're almost done. And then we have 4 times 6. Plus 1. What is that? 4 times 6 is 24. Plus 1 is 25. Put the 5 down here. There's no where to put the 2. There's no more multiplications to do, so we just put the 2 down there. So 64,329 is 257,316. And in case you're wondering, these commas don't mean much. They just help me read the number. So I put it after every 3 digits, so I know that for example, that everything after this-- so s is 7,000. If I had another comma here, this is millions. So it just helps me read the problem a bit. So if you got that you're now ready to escalate to a slightly more complicated situation. Although the first way that we're going to do it it's actually not going to look any more complicated. It's just going to involve one more step. So everything we've done so far are a bunch of digits times a one-digit number. Now let's do a bunch of digits times a two-digit number. So let's say we want to multiply 36 times-- instead of putting a one-digit number here I'm going to put a two-digit number. So times 23. So you start off doing this problem exactly the way you would have done it if there was just a 3 down here. You can kind of ignore the 2 for a little bit. So 3 times 6 is equal to 18. So you just put the 8 here, put the 10 there, or the 1 there because it's 10 plus 8. 3 times 3 is 9. Plus 1, so 3 times 3 plus 1 is equal to-- that's 9 plus 1 is equal to 10. So you put the 10 there. There's nothing left. You put the 0 there. There's nothing left to put the 1, so you put the 10 there. So you essentially have solved the problem that 36-- let me do this is another color. That 36 times 3 is equal to 108. That's what we've solved so far, but we have this 20 sitting out here. We have this 20. We have to figure out what 20 times 360 is. Or sorry, what 20 times 36 is. This 2 is really a 20. And to make it all work out like that, what we do is we throw a 0 down here. We throw a 0 right there. In a second I'm going to explain why exactly we did that. So let's just do the same process as we did before with the 3. Now we do it with a 2, but we start filling up here and move to the left. So 2 times 6. That's 12. So 2 times 6 is 12. We put the 1 up here and we have to be very careful because we had this 1 from our previous problem, which doesn't apply anymore. So we could erase it or that 1 we could get rid of. If you have an eraser get rid of it, or you can just keep track in your head that the 1 you're about to write is a different 1. So what were we doing? We wrote 2 times 6 is 12. Put the 2 here. Put the 1 up here. And I got rid of the previous 1 because that would've just messed me up. Now I have 2 times 3. 2 times 3 is equal to 6. But then I have this plus 1 up here, so I have to add plus 1. So I get 7. So that is equal to 7. 2 times 3 plus 1 is equal to 7. So this 720 we just solved, that's literally-- let me write that down. That is 36 times 20. 36 times 20 is equal to 720. And hopefully that should explain why we had to throw this 0 here. If we didn't throw that 0 here we would have just a 2-- we would just have a 72 here, instead of 720. And 72 is 36 times 2. But this isn't a 2. This is a 2 in the 10's place. This is a 20. So we have to multiply 36 times 20, and that's why we got 720 there. So 36 times 23. Let's write it this way. Let me get some space up here. So we could write 30-- well, actually, let me just finish the problem and then I'll explain to you why it worked. So now to finish it up we just add 108 to 720. So 8 plus 0 is 8. 0 plus 2 is 2. 1 plus 7 is 8. So 36 times 23 is 828. Now you're saying Sal, why did that work? why were we able to figure out separately 36 times 3 is equal to 108, and then 36 times 20 is equal to 720, and then add them up like that? Because we could have rewritten the problem like this. We could have rewritten the problem as 36-- the original problem was this. We could have rewritten this as 36 times 20 plus 3. And this, and I don't know if you've learned the distributive property yet, but this is just the distributive property. This is just the same thing as 36 times 20 plus 36 times 3. If that confuses you, then you don't have to worry about it. But if it doesn't, then this is good. It's actually teaching you something. 36 times 20 we saw was 720. We learned that 36 times 3 was 108. And when you added them together we got what? 828? We got 828. And you could expand it even more like we did in the previous video. You could write this out as 30 plus 6 times 20 plus 3. Actually, let me just do it that way because I think that could help you out a little bit. If it confuses you, ignore it. If it doesn't, that's good. So we could do it 3 times 6. 3 times 6 is 18. 18 is just 10 plus 8. So it's 8, then we put a 10 up here. And ignore all this up here. 3 times 30. 3 times 30 is 90. 90 plus 10 is 100. So 100 is zero 10's plus 100. I don't know if this confuses you or not. If it does, ignore it. If it doesn't, well I don't want to complicate the issue. And now we can multiply 20. We can ignore this thing that we had before. 20 times 6 is 120. So that's 20 plus 100. So I'll put that 100 up here. 20 times 30 is 2 times 3 and you have two 0's there. And I think I'm maybe jumping the gun a little bit, assuming a little bit too much of what you may or may not know. But 20 times 30 is going to be 600. and you add another hundred there, that's 700. And then you add them all up. You get 800. 100 plus 700. Plus 20 plus 8, which is equal to 828. My point here is to show you why that system we did worked. Why we added a 0 here to begin with. But if it confuses you, don't worry about that right now. Learn how to do it and then maybe re-watch this video. Let's just do a bunch of more examples because I think the examples are what really, hopefully, explain the situation. So let's do 77. 77 times 77. 7 times 7 is 49. Put the 1 up here. 7 times 7, well, that's 49. Plus 4 is 53. There's no where to put the 5, so we put it down here. 7 times 7 is 49. Plus 4 is 53. Stick a 0 here. Now we're going to do this 7. So stick a 0 here. Let's get rid of this right there because that'll just mess us up. 7 times 7 is 49. Stick a 9 there. Put a 4 there. 7 times 7 is 49. Plus 4, which is 53. So notice, when we multiplied 7 times 77 we got 539. When we multiplied 70 times 77 we got 5,390. And it makes sense. They just differ by a 0. By a factor of 10. And now we can just add them up, and what do we get? 9 plus 0 is 9. 3 plus 9 is 12. Carry the 1. 1 plus 5 is 6. 6 plus 3 is 9. So it's 5,929." + }, + { + "Q": "In 5:02 he says AB is congruent to segment AC but he wrote AC before AB", + "A": "In Geometry the order only matters when you say AC and AB individually. a is congruent to a, while C is congruent to B. saying CA is congruent to AB would be incorrect. Tell me if it is still unclear, I m not the best explainer", + "video_name": "7UISwx2Mr4c", + "transcript": "So we're starting off with triangle ABC here. And we see from the drawing that we already know that the length of AB is equal to the length of AC, or line segment AB is congruent to line segment AC. And since this is a triangle and two sides of this triangle are congruent, or they have the same length, we can say that this is an isosceles triangle. Isosceles triangle, one of the hardest words for me to spell. I think I got it right. And that just means that two of the sides are equal to each other. Now what I want to do in this video is show what I want to prove. So what I want to prove here is that these two-- and they're sometimes referred to as base angles, these angles that are between one of the sides, and the side that isn't necessarily equal to it, and the other side that is equal and the side that's not equal to it. I want to show that they're congruent. So I want to prove that angle ABC, I want to prove that that is congruent to angle ACB. And so for an isosceles triangle, those two angles are often called base angles. And this might be called the vertex angle over here. And these are often called the sides or the legs of the isosceles triangle. And these are-- obviously they're sides. These are the legs of the isosceles triangle and this one down here, that isn't necessarily the same as the other two, you would call the base. So let's see if we can prove that. So there's not a lot of information here, just that these two sides are equal. But we have, in our toolkit, a lot that we know about triangle congruency. So maybe we can construct two triangles here that are congruent. And then we can use that information to figure out whether this angle is congruent to that angle there. And the first step, if we're going to use triangle congruency, is to actually construct two triangles. So one way to construct two triangles is let's set up another point right over here. Let's set up another point D. And let's just say that D is the midpoint of B and C. So it's the midpoint. So the distance from B to D is going to be the same thing as the distance-- let me do a double slash here to show you it's not the same as that distance. So the distance from B to D is going to be the same thing as the distance from D to C. And obviously, between any two points, you have a midpoint. And so let me draw segment AD. And what's useful about that is that we have now constructed two triangles. And what's even cooler is that triangle ABD and triangle ACD, they have this side is congruent, this side is congruent, and they actually share this side right over here. So we know that triangle ABD we know that it is congruent to triangle ACD. And we know it because of SSS, side-side-side. You have two triangles that have three sides that are congruent, or they have the same length. Then the two triangles are congruent. And what's useful about that is if these two triangles are congruent, then their corresponding angles are congruent. And so we've actually now proved our result. Because the corresponding angle to ABC in this triangle is angle ACD in this triangle right over here. So that we then know that angle ABC is congruent to angle ACB. So that's a pretty neat result. If you have an isosceles triangle, a triangle where two of the sides are congruent, then their base angles, these base angles, are also going to be congruent. Now let's think about it the other way. Can we make the other statement? If the base angles are congruent, do we know that these two legs are going to be congruent? So let's try to construct a triangle and see if we can prove it the other way. So I'll do another triangle right over here. Let me draw another one just like that. That's not that pretty of a triangle, so let me draw it a little nicer. I'm going to draw it like this. Let me do that in a different color. So I'll call that A. I will call this B. I will call that C right over there. And now we're going to start off with the idea that this angle, angle ABC, is congruent to angle ACB. So they have the same exact measure. And what we want to do in this case-- we want to prove-- so let me draw a little line here to show that we're doing a different idea. Here we're saying if these two sides are the same, then the base angles are going to be the same. We've proved that. Now let's go the other way. If the base angles are the same, do we know that the two sides are the same? So we want to prove that segment AC is congruent to AB. Or you could say that the length of segment AC, which we would denote that way, is equal to the length of segment AB. These are essentially equivalent statements. Once again in our toolkit, we have our congruency theorems. But in order to apply them, you really do need to have two triangles. So let's construct two triangles here. And this time, instead of defining another point as the midpoint, I'm going to define D this time as the point that if I were to go straight up, the point that is essentially-- if you view BC as straight horizontal, the point that goes straight down from A. And the reason why I say that is there's some point-- you could call it an altitude-- that intersects BC at a right angle. And there will definitely be some point like that. And so if it's a right angle on that side, if that's 90 degrees, then we know that this is 90 degrees as well. Now, what's interesting about this? And let me write this down. So I've constructed AD such that AD is perpendicular to BC. And you can always construct an altitude. Essentially, you just have to make BC lie flat on the ground. And then you just have to drop something from A, and that will give you point D. You can always do that with a triangle like this. So what does this give us? So over here, we have an angle, an angle, and then a side in common. And over here, you have an angle that corresponds to that angle, an angle that corresponds to this angle, and the same side in common. And so we know that these triangles are congruent by AAS, angle-angle-side, which we've shown is a valid congruent postulate. So we can say now that triangle ABD is congruent to triangle ACD. And we know that by angle-angle-side. This angle and this angle and this side. This angle and this angle and this side. And once we know these two triangles are congruent, we know that every corresponding angle or side of the two triangles are also going to be congruent. So then we know that AB is a corresponding side to AC. So these two sides must be congruent. And so you get AB is going to be congruent to AC, and that's because these are congruent triangles. And we've proven what we wanted to show. If the base angles are equal, then the two legs are going to be equal. If the two legs are equal, then the base angles are equal. It's a very, very, very useful tool in geometry. And in case you're curious, for this specific isosceles triangle, over here we set up D so it was the midpoint. Over here we set up D so it was directly below A. We didn't say whether it was the midpoint. But here, we can actually show that it is the midpoint just as a little bit of a bonus result, because we know that since these two triangles are congruent, BD is going to be congruent to DC because they are the corresponding sides. So it actually turns out that point D for an isosceles triangle, not only is it the midpoint but it is the place where, it is the point at which AD-- or we could say that AD is a perpendicular bisector of BC. So not only is AD perpendicular to BC, but it bisects it. That D is the midpoint of that entire base." + }, + { + "Q": "How would I do a cube root on a scientific calculator?", + "A": "Typically there is a root symbol with a y in the valley of the symbol.", + "video_name": "87_qIofPwhg", + "transcript": "- [Voiceover] We already know a little bit about square roots. For example, if I were to tell you that seven squared is equal to 49, that's equivalent to saying that seven is equal to the square root of 49. The square root essentially unwinds taking the square of something. In fact, we could write it like this. We could write the square root of 49, so this is whatever number times itself is equal to 49. If I multiply that number times itself, if I square it, well I'm going to get 49. And that's going to be true for any number, not just 49. If I write the square root of X and if I were to square it, that's going to be equal to X and that's going to be true for any X for which we can evaluate the square root, evaluate the principle root. Now typically and as you advance in math you're going to see that this will change, but typically you say, okay if I'm going to take the square root of something, X has to be non-negative. X has to be non-negative. This is going to change once we start thinking about imaginary and complex numbers, but typically for the principle square root, we assume that whatever's under the radical, whatever's under here, is going to be non-negative because it's hard to square a number at least the numbers that we know about, it's hard to square them and get a negative number. So for this thing to be defined, for it to make sense, it's typical to say that, okay we need to put a non-negative number in here. But anyway, the focus of this video is not on the square root, it's really just to review things so we can start thinking about the cube root. And as you can imagine, where does the whole notion of taking a square of something or a square root come from? Well it comes from the notion of finding the area of a square. If I have a square like this and if this side is seven, well if it's a square, all the sides are going to be seven. And if I wanted to find the area of this, it would be seven times seven or seven squared. That would be the area of this. Or if I were to say, well what is if I have a square, if I have, and that doesn't look like a perfect square, but you get the idea, all the sides are the same length. If I have a square with area X. If the area here is X, what are the lengths of the sides going to be? Well it's going to be square root of X. All of the sides are going to be the square root of X, so it's going to be the square root of X by the square root of X and this side is going to be the square root of X as well and that's going to be the square root of X as well. So that's where the term square root comes from, where the square comes from. Now what do you think cube root? Well same idea. If I have a cube. If I have a cube. Let me do my best attempt at drawing a cube really fast. If I have a cube and a cube, all of it's dimensions have the same length so this is a two, by two, by two cube, what's the volume over here. Well the volume is going to be two, times two, times two, which is two to the third power or two cubed. This is two cubed. That's why they use the word cubed because this would be the volume of a cube where each of its sides have length two and this of course is going to be equal to eight. But what if we went the other way around? What if we started with the cube? What if we started with this volume? What if we started with a cube's volume and let's say the volume here is eight cubic units, so volume is equal to eight and we wanted to find the lengths of the sides. So we wanted to figure out what X is cause that's X, that's X, and that's X. It's a cube so all the dimensions have the same length. Well there's two ways that we could express this. We could say that X times X times X or X to the third power is equal to eight or we could use the cube root symbol, which is a radical with a little three in the right place. Or we could write that X is equal to, it's going to look very similar to the square root. This would be the square root of eight, but to make it clear, they were talking about the cube root of eight, we would write a little three over there. In theory for square root, you could put a little two over here, but that'd be redundant. If there's no number here, people just assume that it's the square root. But if you're figuring out the cube root or sometimes you say the third root, well then you have to say, well you have to put this little three right over here in this little notch in the radical symbol right over here. And so this is saying X is going to be some number that if I cube it, I get eight. So with that out of the way, let's do some examples. Let's say that I have... Let's say that I want to calculate the cube root of 27. What's that going to be? Well if say that this is going to be equal to X, this is equivalent to saying that X to the third or that 27 is equal to X to the third power. So what is X going to be? Well X times X times X is equal to 27, well the number I can think of is three, so we would say that X, let me scroll down a little bit, X is equal to three. Now let me ask you a question. Can we write something like... Can we pick a new color? The cube root of, let me write negative 64. I already talked about that if we're talking the square root, it's fairly typical that hey you put a negative number in there at least until we learn about imaginary numbers, we don't know what to do with it. But can we do something with this? Well if I cube something, can I get a negative number? Sure. So if I say this is equal to X, this is the same thing as saying that negative 64 is equal to X to the third power. Well what could X be? Well what happens if you take negative four times negative four times negative four? Negative four times four is positive 16, but then times negative four is negative 64 is equal to negative 64. So what could X be here? Well X could be equal to negative four. X could be equal to negative four. So based on the math that we know so far you actually can take the cube root of a negative number. And just so you know, you don't have to stop there. You could take a fourth root and in this case you'd have a four here, a fifth root, a sixth root, a seventh root of numbers and we'll talk about that later in your mathematical career. But most of what you're going to see is actually going to be square root and every now and then you're going to see a cube root. Now you might be saying, well hey look, you know, you just knew that three to the third power is 27, you took the cube root, you get X, is there any simple way to do this? And like you know if i give you an arbitrary number. If I were to just say, I don't know, if I were to say cube root of 125. And the simple answer is, well the easiest way to actually figure this out is actually just to do a factorization and particular prime factorization of this thing right over here and then you would figure it out. So you would say, okay well 125 is five times 25, which is five times five. Alright, so this is the same thing as the cube root of five to the third power, which of course, is going to be equal to five. If you have a much larger number here, yes, there's no very simple way to compute what a cube root or a fourth root or a fifth root might be and even square root can get quite difficult. There's no very simple way to just calculate it the way that you might multiply things or divide it." + }, + { + "Q": "What happens when the steam gets hotter and hotter ?", + "A": "Water molecules move, vibrate, and rotate in several ways. As temperatures increase, their kinetic, vibrational, and rotational energies increase. At extremely high temperatures, these energies of motion are greater than the strength of the covalent O-H bonds that hold the molecules together. The molecules fall apart. We end up with a plasma, a gas that consists of H\u00e2\u0081\u00ba and O\u00c2\u00b2\u00e2\u0081\u00bb ions.", + "video_name": "pKvo0XWZtjo", + "transcript": "I think we're all reasonably familiar with the three states of matter in our everyday world. At very high temperatures you get a fourth. But the three ones that we normally deal with are, things could be a solid, a liquid, or it could be a gas. And we have this general notion, and I think water is the example that always comes to at least my mind. Is that solid happens when things are colder, relatively colder. And then as you warm up, you go into a liquid state. And as your warm up even more you go into a gaseous state. So you go from colder to hotter. And in the case of water, when you're a solid, you're ice. When you're a liquid, some people would call ice water, but let's call it liquid water. I think we know what that is. And then when it's in the gas state, you're essentially vapor or steam. So let's think a little bit about what, at least in the case of water, and the analogy will extend to other types of molecules. But what is it about water that makes it solid, and when it's colder, what allows it to be liquid. And I'll be frank, liquids are kind of fascinating because you can never nail them down, I guess is the best way to view them. Or a gas. So let's just draw a water molecule. So you have oxygen there. You have some bonds to hydrogen. And then you have two extra pairs of valence electrons in the oxygen. And a couple of videos ago, we said oxygen is a lot more electronegative than the hydrogen. It likes to hog the electrons. So even though this shows that they're sharing electrons here and here. At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules. Let me draw some more molecules. When we talk about the whole state of the whole matter, we actually think about how the molecules are interacting with Not just how the atoms are interacting with each other within a molecule. I just drew one oxygen, let me copy and paste that. But I could do multiple oxygens. And let's say that that hydrogen is going to want to be near this oxygen. Because this has partial negative charge, this has a partial positive charge. And then I could do another one right there. And then maybe we'll have, and just to make the point clear, you have two hydrogens here, maybe an oxygen wants to hang out there. So maybe you have an oxygen that wants to be here because it's got its partial negative here. And it's connected to two hydrogens right there that have their partial positives. But you can kind of see a lattice structure. Let me draw these bonds, these polar bonds that start forming between the particles. These bonds, they're called polar bonds because the molecules themselves are polar. And you can see it forms this lattice structure. And if each of these molecules don't have a lot of kinetic energy. Or we could say the average kinetic energy of this matter is fairly low. And what do we know is average kinetic energy? Well, that's temperature. Then this lattice structure will be solid. These molecules will not move relative to each other. I could draw a gazillion more, but I think you get the point that we're forming this kind of fixed structure. And while we're in the solid state, as we add kinetic energy, as we add heat, what it does to molecules is, it just makes them vibrate around a little bit. If I was a cartoonist, they way you'd draw a vibration is to put quotation marks there. That's not very scientific. But they would vibrate around, they would buzz around a little bit. I'm drawing arrows to show that they are vibrating. It doesn't have to be just left-right it could be up-down. But as you add more and more heat in a solid, these molecules are going to keep their structure. So they're not going to move around relative to each other. But they will convert that heat, and heat is just a form of energy, into kinetic energy which is expressed as the vibration of these molecules. Now, if you make these molecules start to vibrate enough, and if you put enough kinetic energy into these molecules, what do you think is going to happen? Well this guy is vibrating pretty hard, and he's vibrating harder and harder as you add more and more heat. This guy is doing the same thing. At some point, these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations. And once that happens, the molecules-- let me draw a couple more. Once that happens, the molecules are going to start moving past each other. So now all of a sudden, the molecule will start shifting. But they're still attracted. Maybe this side is moving here, that's moving there. You have other molecules moving around that way. But they're still attracted to each other. Even though we've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules. Our vibration, or our kinetic energy for each molecule, still isn't strong enough to completely separate them. They're starting to slide past each other. And this is essentially what happens when you're in a liquid state. You have a lot of atoms that want be touching each other but they're sliding. They have enough kinetic energy to slide past each other and break that solid lattice structure here. And then if you add even more kinetic energy, even more heat, at this point it's a solution now. They're not even going to be able to stay together. They're not going to be able to stay near each other. If you add enough kinetic energy they're going to start looking like this. They're going to completely separate and then kind of bounce around independently. Especially independently if they're an ideal gas. But in general, in gases, they're no longer touching They might bump into each other. But they have so much kinetic energy on their own that they're all doing their own thing and they're not touching. I think that makes intuitive sense if you just think about what a gas is. For example, it's hard to see a gas. Why is it hard to see a gas? Because the molecules are much further apart. So they're not acting on the light in the way that a liquid or a solid would. And if we keep making that extended further, a solid-- well, I probably shouldn't use the example with ice. Because ice or water is one of the few situations where the solid is less dense than the liquid. That's why ice floats. And that's why icebergs don't just all fall to the bottom of the ocean. And ponds don't completely freeze solid. But you can imagine that, because a liquid is in most cases other than water, less dense. That's another reason why you can see through it a little Or it's not diffracting-- well I won't go into that too much, than maybe even a solid. But the gas is the most obvious. And it is true with water. The liquid form is definitely more dense than the gas form. In the gas form, the molecules are going to jump around, not touch each other. And because of that, more light can get through the substance. Now the question is, how do we measure the amount of heat that it takes to do this to water? And to explain that, I'll actually draw a phase change diagram. Which is a fancy way of describing something fairly straightforward. Let me say that this is the amount of heat I'm adding. And this is the temperature. We'll talk about the states of matter in a second. So heat is often denoted by q. Sometimes people will talk about change in heat. They'll use H, lowercase and uppercase H. They'll put a delta in front of the H. Delta just means change in. And sometimes you'll hear the word enthalpy. Let me write that. Because I used to say what is enthalpy? It sounds like empathy, but it's quite a different concept. At least, as far as my neural connections could make it. But enthalpy is closely related to heat. It's heat content. For our purposes, when you hear someone say change in enthalpy, you should really just be thinking of change in heat. I think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary. The best way to think about it is heat content. Change in enthalpy is really just change in heat. And just remember, all of these things, whether we're talking about heat, kinetic energy, potential energy, enthalpy. You'll hear them in different contexts, and you're like, I thought I should be using heat and they're talking about enthalpy. These are all forms of energy. And these are all measured in joules. And they might be measured in other ways, but the traditional way is in joules. And energy is the ability to do work. And what's the unit for work? Well, it's joules. Force times distance. But anyway, that's a side-note. But it's good to know this word enthalpy. Especially in a chemistry context, because it's used all the time and it can be very confusing and non-intuitive. Because you're like, I don't know what enthalpy is in my everyday life. Just think of it as heat contact, because that's really But anyway, on this axis, I have heat. So this is when I have very little heat and I'm increasing my heat. And this is temperature. Now let's say at low temperatures I'm here and as I add heat my temperature will go up. Temperature is average kinetic energy. Let's say I'm in the solid state here. And I'll do the solid state in purple. No I already was using purple. I'll use magenta. So as I add heat, my temperature will go up. Heat is a form of energy. And when I add it to these molecules, as I did in this example, what did it do? It made them vibrate more. Or it made them have higher kinetic energy, or higher average kinetic engery, and that's what temperature is a measure of; average kinetic energy. So as I add heat in the solid phase, my average kinetic energy will go up. And let me write this down. This is in the solid phase, or the solid state of matter. Now something very interesting happens. Let's say this is water. So what happens at zero degrees? Which is also 273.15 Kelvin. Let's say that's that line. What happens to a solid? Well, it turns into a liquid. Ice melts. Not all solids, we're talking in particular about water, about H2O. So this is ice in our example. All solids aren't ice. Although, you could think of a rock as solid magma. Because that's what it is. I could take that analogy a bunch of different ways. But the interesting thing that happens at zero degrees. Depending on what direction you're going, either the freezing point of water or the melting point of ice, something interesting happens. As I add more heat, the temperature does not to go up. As I add more heat, the temperature does not go up for a little period. Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or which water will boil. But something happens. And they're really getting kinetically active. But just like when you went from solid to liquid, there's a certain amount of energy that you have to contribute to the system. And actually, it's a good amount at this point. Where the water is turning into vapor, but it's not getting any hotter. So we have to keep adding heat, but notice that the temperature didn't go up. We'll talk about it in a second what was happening then. And then finally, after that point, we're completely vaporized, or we're completely steam. Then we can start getting hot, the steam can then get hotter as we add more and more heat to the system. So the interesting question, I think it's intuitive, that as you add heat here, our temperature is going to go up. But the interesting thing is, what was going on here? We were adding heat. So over here we were turning our heat into kinetic energy. Temperature is average kinetic energy. But over here, what was our heat doing? Well, our heat was was not adding kinetic energy to the system. The temperature was not increasing. But the ice was going from ice to water. So what was happening at that state, is that the kinetic energy, the heat, was being used to essentially break these bonds. And essentially bring the molecules into a higher energy state. So you're saying, Sal, what does that mean, higher energy state? Well, if there wasn't all of this heat and all this kinetic energy, these molecules want to be very close to each other. For example, I want to be close to the surface of the earth. When you put me in a plane you have put me in a higher energy state. I have a lot more potential energy. I have the potential to fall towards the earth. Likewise, when you move these molecules apart, and you go from a solid to a liquid, they want to fall towards each other. But because they have so much kinetic energy, they never quite are able to do it. But their energy goes up. Their potential energy is higher because they want to fall towards each other. By falling towards each other, in theory, they could do some work. So what's happening here is, when we're contributing heat-- and this amount of heat we're contributing, it's called the heat of fusion. Because it's the same amount of heat regardless how much direction we go in. When we go from solid to liquid, you view it as the heat of melting. It's the head that you need to put in to melt the ice into liquid. When you're going in this direction, it's the heat you have to take out of the zero degree water to turn it into ice. So you're taking that potential energy and you're bringing the molecules closer and closer to each other. So the way to think about it is, right here this heat is being converted to kinetic energy. Then, when we're at this phase change from solid to liquid, that heat is being used to add potential energy into the system. To pull the molecules apart, to give them more potential energy. If you pull me apart from the earth, you're giving me potential energy. Because gravity wants to pull me back to the earth. And I could do work when I'm falling back to the earth. A waterfall does work. It can move a turbine. You could have a bunch of falling Sals move a turbine as well. And then, once you are fully a liquid, then you just become a warmer and warmer liquid. Now the heat is, once again, being used for kinetic energy. You're making the water molecules move past each other faster, and faster, and faster. To some point where they want to completely disassociate from each other. They want to not even slide past each other, just completely jump away from each other. And that's right here. This is the heat of vaporization. And the same idea is happening. Before we were sliding next to each other, now we're pulling apart altogether. So they could definitely fall closer together. And then once we've added this much heat, now we're just heating up the steam. We're just heating up the gaseous water. And it's just getting hotter and hotter and hotter. But the interesting thing there, and I mean at least the interesting thing to me when I first learned this, whenever I think of zero degrees water I'll say, oh it must be ice. But that's not necessarily the case. If you start with water and you make it colder and colder and colder to zero degrees, you're essentially taking heat out of the water. You can have zero degree water and it hasn't turned into ice yet. And likewise, you could have 100 degree water that hasn't turned into steam yeat. You have to add more energy. You can also have 100 degree steam. You can also have zero degree water. Anyway, hopefully that gives you a little bit of intuition of what the different states of matter are. And in the next problem, we'll talk about how much heat exactly it does take to move along this line. And maybe we can solve some problems on how much ice we might need to make our drink cool." + }, + { + "Q": "What are capillaries and where are they found?", + "A": "Capillaries are the smallest blood vessels that bring nutriments and oxygen to the tissues and absorbs carbon dioxide and waste products. They are located in many parts of the body such as lungs, muscles, endocrine glands, liver, central nervous system, etc. Hope this helped! :)", + "video_name": "QhiVnFvshZg", + "transcript": "Where I left off in the last video, we talked about how the hemoglobin in red blood cells is what sops up all of the oxygen so that it increases the diffusion gradient-- or it increases the incentive, we could say, for the oxygen to go across the membrane. We know that the oxygen molecules don't know that there's less oxygen here, but if you watch the video on diffusion you know how that process happens. If there's less concentration here than there, the oxygen will diffuse across the membrane and there's less inside the plasma because the hemoglobin is sucking it all up like a sponge. Now, one interesting question is, why does the hemoglobin even have to reside within the red blood cells? Why aren't hemoglobin proteins just freely floating in the blood plasma? That seems more efficient. You don't have to have things crossing through, in and out of, these red blood cell membranes. You wouldn't have to make red blood cells. What's the use of having these containers of hemoglobin? It's actually a very interesting idea. If you had all of the hemoglobin sitting in your blood plasma, it would actually hurt the flow of the blood. The blood would become more viscous or more thick. I don't want to say like syrup, but it would become thicker than blood is right now-- and by packaging the hemoglobin inside these containers, inside the red blood cells, what it allows the blood to do is flow a lot better. Imagine if you wanted to put syrup in water. If you just put syrup straight into water, what's going to happen? The water's going to become a little syrupy, a little bit more viscous and not flow as well. So what's the solution if you wanted to transport syrup in water? Well, you could put the syrup inside little containers or inside little beads and then let the beads flow in the water and then the water wouldn't be all gooey-- and that's exactly what's happening inside of our blood. Instead of having the hemoglobin sit in the plasma and make it gooey, it sits inside these beads that we call red blood cells that allows the flow to still be non-viscous. So I've been all zoomed in here on the alveolus and these capillaries, these pulmonary capillaries-- let's zoom out a little bit-- or zoom out a lot-- just to understand, how is the blood flowing? And get a better understanding of pulmonary arteries and veins relative to the other arteries and veins that are in the body. So here-- I copied this from Wikipedia, this diagram of the human circulatory system-- and here in the back you can see the lungs. Let me do it in a nice dark color. So we have our lungs here. You can see the heart is sitting right in the middle. And what we learned in the last few videos is that we have our little alveoli and our lungs. Remember, we get to them from our bronchioles, which are branching off of the bronchi, which branch off of the trachea, which connects to our larynx, which connects to our pharynx, which connects to our mouth and nose. But anyway, we have our little alveoli right there and then we have the capillaries. So when we go away from the heart-- and we're going to delve a little bit into the heart in this video as well-- so when blood travels away from the heart, it's de-oxygenated. It's this blue color. So this right here is blood. This right here is blood traveling away from the heart. It's going behind these two tubes right there. So this is the blood going away from the heart. So this blue that I've been highlighting just now, these are the pulmonary arteries and then they keep splitting into arterials and all of that and eventually we're in capillaries-- super, super small tubes. They run right past the alveoli and then they become oxygenated and now we're going back to the heart. So we're talking about pulmonary veins. So we go back to the heart. So these capillaries-- in the capillaries we get oxygen. Now we're going to go back to the heart. Hope you can see what I'm doing. And we're going to enter the heart on this side. You actually can't even see where we're entering the heart. We're going to enter the heart right over here-- and I'm going to go into more detail on that. Now we have oxygenated blood. And then that gets pumped out to the rest of the body. Now this is the interesting thing. When we're talking about pulmonary arteries and veins-- remember, the pulmonary artery was blue. As we go away from the heart, we have de-oxygenated blood, but it's still an artery. Then as we go towards the heart from the lungs, we have a vein, but it's oxygenated. So that's this little loop here that we start and I'm going to keep going over the circulation pattern because the heart can get a little confusing, especially because of its three-dimensional nature. But what we have is, the heart pumps de-oxygenated blood from the right ventricle. You're saying, hey, why is it the right ventricle? That looks like the left side of the drawing, but it's this dude's right-hand side, right? This is this guy's right hand. And this is this dude's left hand. He's looking at us, right? We don't care about our right or left. We care about this guy's right and left. And he's looking at us. He's got some eyeballs and he's looking at us. So this is his right ventricle. Actually, let me just start off with the whole cycle. So we have de-oxygenated blood coming from the rest of the body, right? The name for this big pipe is called the inferior vena cava-- inferior because it's coming up below. Actually, you have blood coming up from the arms and the head up here. They're both meeting right here, in the right atrium. Let me label that. I'm going to do a big diagram of the heart in a second. And why are they de-oxygenated? Because this is blood returning from our legs if we're running, or returning from our brain, that had to use respiration-- or maybe we're working out and it's returning from our biceps, but it's de-oxygenated blood. It shows up right here in the right atrium. It's on our left, but this guy's right-hand side. From the right atrium, it gets pumped into the right ventricle. It actually passively flows into the right ventricle. The ventricles do all the pumping, then the ventricle contracts and pumps this blood right here-- and you don't see it, but it's going behind this part right here. It goes from here through this pipe. So you don't see it. I'm going to do a detailed diagram in a second-- into the pulmonary artery. We're going away from the heart. This was a vein, right? This is a vein going to the heart. This is a vein, inferior vena cava vein. This is superior vena cava. They're de-oxygenated. Then I'm pumping this de-oxygenated blood away from the heart to the lungs. Now this de-oxygenated blood, this is in an artery, right? This is in the pulmonary artery. It gets oxygenated and now it's a pulmonary vein. And once it's oxygenated, it shows up here in the left-- let me do a better color than that-- it shows up right here in the left atrium. Atrium, you can imagine-- it's kind of a room with a skylight or that's open to the outside and in both of these cases, things are entering from above-- not sunlight, but blood is entering from above. On the right atrium, the blood is entering from above. And in the left atrium, the blood is entering-- and remember, the left atrium is on the right-hand side from our point of view-- on the left atrium, the blood is entering from above from the lungs, from the pulmonary veins. Veins go to the heart. Then it goes into-- and I'll go into more detail-- into the left ventricle and then the left ventricle pumps that oxygenated blood to the rest of the body via the non-pulmonary arteries. So everything pumps out. Let me make it a nice dark, non-blue color. So it pumps it out through there. You don't see it right here, the way it's drawn. It's a little bit of a strange drawing. It's hard to visualize, but I'll show it in more detail and then it goes to the rest of the body. Let me show you that detail right now. So we said, we have de-oxygenated blood. Let's label it right here. This is the superior vena cava. This is a vein from the upper part of our body from our arms and heads. This is the inferior vena vaca. This is veins from our abdomen and from our legs and the rest of our body. So it it first enters the right atrium. Remember, we call the right atrium because this is someone's heart facing us, even though this is on the left-hand side. It enters through here. It's de-oxygenated blood. It's coming from veins. the body used the oxygen. Then it shows up in the right ventricle, right? These are valves in our heart. And it passively, once the right ventricle pumps and then releases, it has a vacuum and it pulls more blood from the It pumps again and then it pushes it through here. Now this blood right here-- remember, this one still is de-oxygenated blood. De-oxygenated blood goes to the lungs to become oxygenated. So this right here is the pulmonary-- I'm using the word pulmonary because it's going to or from the lungs. It's dealing with the lungs. And it's going away from the heart. It's the pulmonary artery and it is de-oxygenated. Then it goes to the heart, rubs up against some alveoli and then gets oxygenated and then it comes right back. Now this right here, we're going to the heart. So that's a vein. It's in the loop with the lungs so it's a pulmonary vein and it rubbed up against the alveoli and got the oxygen diffused into it so it is oxygenated. And then it flows into your left atrium. Now, the left atrium, once again, from our point of view, is on the right-hand side, but from the dude looking at it, it's his left-hand side. So it goes into the left atrium. Now in the left ventricle, after it's done pumping, it expands and that oxygenated blood flows into the left ventricle. Then the left ventricle-- the ventricles are what do all the pumping-- it squeezes and then it pumps the blood into the aorta. This is an artery. Why is it an artery? Because we're going away from the heart. Is it a pulmonary artery? No, we're not dealing with the lungs anymore. We dealt with the lungs when we went from the right ventricle, went to the lungs in a loop, back to the left atrium. Now we're in the left ventricle. We pump into the aorta. Now this is to go to the rest of the body. This is an artery, a non-pulmonary artery-- and it is oxygenated. So when we're dealing with non-pulmonary arteries, we're oxygenated, but a pulmonary artery has no oxygen. It's going away from the heart to get the oxygen. Pulmonary vein comes from the lungs to the heart with oxygen, but the rest of the veins go to the heart without oxygen because they want to go into that loop on the pulmonary loop right there. So I'll leave you there. Hopefully that gives-- actually, let's go back to that first diagram. I think you have a sense of how the heart is dealing, but let's go look at the rest of the body and just get a sense of things. You can look this up on Wikipedia if you like. All of these different branching points have different names to them, but you can see right here you have kind of a branching off, a little bit below the heart. This is actually the celiac trunk. Celiac, if I remember correctly, kind of refers to an abdomen. So this blood that-- your hepatic artery. Hepatic deals with the liver. Your hepatic artery branches off of this to get blood flow to the liver. It also gives blood flow to your stomach so it's very important in digestion and all that. And then let's say this is the hepatic trunk. Your liver is sitting like that. Hepatic trunk-- it delivers oxygen to the liver. The liver is doing respiration. It takes up the oxygen and then it gives up carbon dioxide. So it becomes de-oxygenated and then it flows back in and to the inferior vena cava, into the vein. I want to make it clear-- it's a loop. It's a big loop. The blood doesn't just flow out someplace and then come back someplace else. This is just one big loop. And if you want to know at any given point in time, depending on your size, there's about five liters of blood. And I looked it up-- it takes the average red blood cell to go from one point in the circulatory system and go through the whole system and come back, 20 seconds. That's an average because you can imagine there might be some red blood cells that get stuck someplace and take a little bit more time and some go through the completely perfect route. Actually, the 20 seconds might be closer to the perfect route. I've never timed it myself. But it's an interesting thing to look at and to think about what's connected to what. You have these these arteries up here that they first branch off the arteries up here from the aorta into the head and the neck and the arm arteries and then later they go down and they flow blood to the rest of the body. So anyway, this is a pretty interesting idea. In the next video, what I want to do is talk about, how does the hemoglobin know when to dump the oxygen? Or even better, where to dump the oxygen-- because maybe I'm running so I need a lot of oxygen in the capillaries around my thigh muscles. I don't need them necessarily in my hands. How does the body optimize where the oxygen is actually It's actually fascinating." + }, + { + "Q": "Is there an actuall unit that can measure our weight and length accurately the way we are,like we never know how big the universe is,it may can be just some dust out there in a bigger world,we cant never know what exactly how big we are right?", + "A": "Excellent question! We can measure things very, very precisely according to the definitions we make up. But if I know that something is 1 meter, how can I say if that is big or small? The only way is by comparing it to other things, right? You have just discovered the essence of the theory of relativity. Good thinking!", + "video_name": "5FEjrStgcF8", + "transcript": "The purpose of this video is to just begin to appreciate how vast and enormous the universe is. And frankly, our brains really can't grasp it. What we'll see in this video is that we can't even grasp things that are actually super small compared to the size of the universe. And we actually don't even know what the entire size of the universe is. But with that said, let's actually just try to appreciate how small we are. So this is me right over here. I am 5 foot 9 inches, depending on whether I'm wearing shoes-- maybe 5 foot 10 with shoes. But for the sake of this video, let's just roughly approximate around 6 feet, or around roughly-- I'm not to go into the details of the math-- around 2 meters. Now, if I were to lie down 10 times in a row, you'd get about the length of an 18-wheeler. That's about 60 feet long. So this is times 10. Now, if you were to put an 18-wheeler-- if you were to make it tall, as opposed to long-- somehow stand it up-- and you were to do that 10 times in a row, you'll get to the height of roughly a 60-story skyscraper. So once again, if you took me and you piled me up 100 times, you'll get about a 60-story skyscraper. Now, if you took that skyscraper and if you were to lie it down 10 times in a row, you'd get something of the length of the Golden Gate Bridge. And once again, I'm not giving you the exact numbers. It's not always going to be exactly 10. But we're now getting to about something that's a little on the order of a mile long. So the Golden Gate Bridge is actually longer than a mile. But if you go within the twin spans, it's roughly about It's actually a little longer than that. But that gives you a sense of a mile. Now, if you multiply that by 10, you get to the size of a large city. And this right here is a satellite photograph of San Francisco. This is the actual Golden Gate Bridge here. And when I copy and pasted this picture, I tried to make it roughly 10 miles by 10 miles just so you appreciate the scale. And what's interesting here-- and this picture's interesting. Because this is the first time we can relate to cities. But when you look at a city on this scale, it's starting to get larger than what we're used to processing on a daily basis. A bridge-- we've been on a bridge. We know what a bridge looks like. We know that a bridge is huge. But it doesn't feel like something that we can't comprehend. Already, a city is something that we can't comprehend all at once. We can drive across a city. We can look at satellite imagery. But if I were to show a human on this, it would be unbelievably, unbelievably small. You wouldn't actually be able to see it. It would be less than a pixel on this image. A house is less than a pixel on this image. But let's keep multiplying by 10. If you multiply by 10 again, you get to something roughly the size of the San Francisco Bay Area. This whole square over here is roughly that square right over there. Let's multiply by 10 again. So this square is about 100 miles by 100 miles. So this one would be about 1,000 miles by 1,000 miles. And now you're including a big part of the Western United States. You have California here. You Nevada here. You have Arizona and New Mexico-- so a big chunk of a big continent we're already including. And frankly, this is beyond the scale that we're used to operating. We've seen maps, so maybe we're a little used to it. But if you ever had to walk across this type of distance, it would take you a while. To some degree, the fact that planes goes so fast-- almost unimaginably fast for us-- that it's made it feel like things like continents aren't as big. Because you can fly across them in five or six hours. But these are already huge, huge, huge distances. But once again, you take this square that's about 1,000 miles by 1,000 miles, and you multiply that by 10. And you get pretty close-- a little bit over-- the diameter of the Earth-- a little bit over the diameter of the Earth. But once again, we're on the Earth. We kind of relate to the Earth. If you look carefully at the horizon, you might see a little bit of a curvature, especially if you were to get into the plane. So even though this is, frankly, larger than my brain can really grasp, we can kind of relate to the Earth. Now you multiply the diameter of Earth times 10. And you get to the diameter of Jupiter. And so if you were to sit Earth right next to Jupiter-- obviously, they're nowhere near that close. That would destroy both of the planets. Actually, it would definitely destroy Earth. It would probably just be merged into Jupiter. So if you put Earth next to Jupiter, it would look something like that right over there. So I would say that Jupiter is definitely-- on this diagram that I'm drawing here-- is definitely the first thing that I have I can't comprehend. The Earth, itself, is so vastly huge. Jupiter is-- it's 10 times bigger in diameter. It's much larger in terms of mass, and volume, and all the rest. But just in terms of diameter, it is 10 times bigger. But let's keep going. 10 times Jupiter gets us to the sun. This is times 10. So if this is the Sun-- and if I were to draw Jupiter, it would look something like-- I'll do Jupiter in pink-- Jupiter would be around that big. And then the Earth would be around that big if you were to put them all next to each other. So the Sun, once again, is huge. Even though we see it almost every day, it is unimaginably huge. Even the Earth is unimaginably huge. And the Sun is 100 times more unimaginably bigger. Now we're going to start getting really, really, really wacky. You multiply the diameter of the Sun, which is already 100 times the diameter of the Earth-- you multiply that times 100. And that is the distance from the Earth to the Sun. So I've drawn the Sun here as a little pixel. And I didn't even draw the Earth as a pixel. Because a pixel would be way too large. It would have to be a hundredth of a pixel in order to draw the Earth properly. So this is a unbelievable distance between the Earth and the Sun. It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star-- so remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times-- and once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball-- if the average star was about the size of a basketball-- in our part of the galaxy in a volume the size of the Earth-- so if you had a big volume the size of the Earth, if the stars were the sizes of basketballs, in our part of the galaxy, you would only have a handful of basketballs per that volume. So unbelievably sparse. Even though, when you look at the galaxy-- and this is just an artist's depiction of it-- it looks like something that has the spray of stars, and it looks reasonably dense, there is actually a huge amount of space that the great, great, great, great, great majority of the volume in the galaxy is just empty, empty space. There's no stars, no planets, no nothing. I mean, this is a huge jump that I'm talking about. And then if you really want to realize how large a galaxy, itself, can be, you take this distance between the Sun, or between our solar system and the nearest star-- so that's 200,000 times the distance between the Earth and the Sun-- and you multiply that distance by 25,000. So if the Sun is right here, our nearest star will be in that same pixel. They'll actually be within-- you'd actually get a ton of stars within that one pixel, even though they're so far apart. And then this whole thing is 100,000 light years. It's 25,000 times the distance than the distance between the Sun and the nearest star. So we're talking about unimaginable, unfathomable distances, just for a galaxy. And now we're going to get our-- frankly, my brain is already well beyond anything that it can really process. At this point, it almost just becomes abstract thinking. It just becomes playing with numbers and mathematics. But to get a sense of the universe, itself, the observable universe-- and we have to be clear. Because we can only observe light that started leaving from its source 13.7 billion years ago. Because that's how old the universe is. The observable universe is about 93 billion light years across. And the reason why it's larger than 13.7 billion is that the points in space that emitted light 13.7 billion years ago, those have been going away from us. So now they're on the order of 40 billion light years away. But this isn't about cosmology. This is just about scale and appreciating how huge the universe is. Just in the part of the universe that we can theoretically observe, you have to get-- and that we can observe, just because we're getting electromagnetic radiation from those parts of the universe-- you would have to multiply this number. So let me make this clear. 100,000 light years-- that's the diameter of the Milky Way. You would have to multiply not by 1,000. 1,000 would get you to 100 million light years. This is 100,000 times 1,000 is 100 million. You have to multiply by 1,000 again to get to 100 billion light years. And the universe, for all we know, might be much, much, much, much, larger. It might even be infinite. Who knows? But to get from just the diameter of the Milky Way to the observable universe, you have to multiply by a million. And already, this is an unfathomable distance. So in the whole scheme of things, not only are we pretty small, and not only are the things we build pretty small, and not only is our planet ultra small, and not only is our Sun ultra small, and our solar system ultra small, but our galaxy is really nothing compared to the vastness of the universe." + }, + { + "Q": "So I was working on a christmas themed hangman game, and wanted to make a candycane. Is there a way to make a striped pattern without making a ton of rectangles right next to eachother? I don't really like the idea of long that would take.", + "A": "You can try using quads and for loops.", + "video_name": "4VqHGULLA4o", + "transcript": "Voiceover: Congratulations! You now understand the JavaScript language. Variables, loops, strings, functions, objects, arrays, even object-oriented design. But what good is a language if you can't make something cool with it? There are lots of ways you can use your new knowledge, but one of the most popular ways to use JavaScript with Processing.js is to make games and visualizations, which you probably know, if you've ever looked at our hot programs list. A game is something interactive, where you get some reward. There is usually a win state, a lose state, a score. A visualization is also highly interactive, but without the game mechanics. Let's look at some of the common components we'll need. We'll need UI controls for anything we're doing. Like buttons, and sliders, and menus. And some of these will be simple buttons, other times we'll need multiple buttons, other times we'll need sliders, and drop-downs, and it all builds upon the same basic principles, though. Besides interaction with a mouse, we'll also want keyboard control, like being able to use the arrows to move our characters up and down, or to change the angle of our visualization. We often also want the notion of scenes. A scene is like your start screen, and your options screen, and your main screen, and your end screen. And they're usually very different, and at any given point we want to be showing one of them or the other, so you have to really organize your code in order to know the difference between the scenes, and have a good way of switching between them. And now let's talk about a few things specific to games. The environment of a game. Is it a side-scroller, which means it's kinda a character moving forward through a space? Is it a bird's eye view, like going through a maze? Is it a 3D environment? It's crazy, but you can do it. Are there multiple levels, and each of them have different environments? What are the characters in the game? They'll probably have different behaviors, and emotions, and states, like a happy state and a dead state. And they'll might be user controlled or sometimes they'll be programatically controlled. And your program gives it some sort of logic to follow. There could be one of them, there could be lots of them. They could get spawned during the game as it's played. Now once we've got characters in an environment, we usually also add in some items, and then we have a lot of things colliding with each other. And we usually want to know when things collide, because things are typically trying to attract each other, or avoid each other, like when you're picking up gems, or avoiding nasty turtles. So we need to be able to detect collision between objects, and sometimes it's very simple collision, and other times it's more complex if the objects are all different shapes and sizes. Finally, if it's a game, it's usually got a score. So how do you measure how well the user is doing? When do you tell them if they've lost or won? How spectacular can you make the win screen or the lose screen? So, as you can see, there are a lot of aspects to think about when making games and visualizations. We'll walk through some of them here, but we don't know what's in your head, and most likely, you'll have to just combine the knowledge you learned here to make whatever really cool thing is in your head right now." + }, + { + "Q": "What happens when one of the things is negative? example: 3x - 45\nWouldn't the answer be negative/ That wouldn't make sense.", + "A": "3x-45=0 3x = 45 (-ive becomes +ive) x = 45/3 x= 15", + "video_name": "Ld7Vxb5XV6A", + "transcript": "So I've got two parallel lines. So that's the first line right over there and then the second line right over here. Let me denote that these are parallel. These are parallel lines. Actually, I can do that a little bit neater. And let me draw a transversal, so a line that intersects both of these parallel lines, so something like that. And now let's say that we are told that this angle right over here is 9x plus 88. And this is in degrees. And we're also told that this angle right over here is 6x plus 182, once again, in degrees. So my goal here and my question for you is, can we figure out what these angles actually are, given that these are parallel lines and this is a transversal line? And I encourage you to pause this video to try this on your own. Well, the key here to realize is that these right over here are related by the fact that they're formed from a transversal intersecting parallel lines. And we know, for example, that this angle corresponds to this angle right over here. They're going to be congruent angles. And so this is 6x plus 182. This is also going to be 6x plus 182. And then that helps us realize that this blue angle and this orange angle are actually going to be supplementary. They're going to add up to 180 degrees, because put together, when you make them adjacent, their outer rays form a line right over here. So we know that 6x plus 182 plus 9x plus 88 is going to be equal to 180 degrees. And now we just have to simplify this thing. So 6x plus 9x is going to give us 15x. And then we have 182 plus 88. Let's see, 182 plus 8, would get us to 190. And then we add another 80. It gets us to 270-- plus 270-- is equal to 180. If we subtract 270 from both sides, we get 15x is equal to negative 90. And now we can divide both sides by 15. And we get x is equal to-- what is this? Let's see, 6 times 15 is 60 plus 30 is 90. So x is going to be equal to negative 6. So far, we've made a lot of progress. We figured out what x is equal to. x is equal to negative 6, but we still haven't figured out what these angles are equal to. So this angle right over here, 9x plus 88, this is going to be equal to 9 times negative 6 plus 88. 9 times negative 6 is negative 54. Let me write this down before I make a mistake. Negative 54 plus 88 is going to be-- let's see, to go from 88 minus 54 will give us 34 degrees. So this is equal to 34, and it's in degrees. So this orange angle right here is 34 degrees. The blue angle is going to be 180 minus that. But we can verify that by actually evaluating 6x plus 182. So this is going to be equal to 6 times negative 6 is negative 36 plus 182. So this is going to be equal to-- let's see, if I subtract the 6 first, I get to 176. So this gets us to 146 degrees. And you can verify-- 146 plus 34 is equal to 180 degrees. Now, we could also figure out the other angles from this as well. We know that if this is 34 degrees, then this must be 34 degrees as well. Those are opposite angles. This angle also corresponds to this angle so it must also be 34 degrees, which is opposite to this angle, which is going to be 34 degrees. Similarly, if this one right over here is 146 degrees, we already know that this one is going to be 146. This one's going to be 146 since it's opposite. And that's going to be 146 degrees as well." + }, + { + "Q": "why is the magnetic field always perpendicular to velocity? are they related?", + "A": "Yes, they are related. Go to youtube and type veritasium how do magnets work Watch both videos.", + "video_name": "NnlAI4ZiUrQ", + "transcript": "We know a little bit about magnets now. Let's see if we can study it further and learn a little bit about magnetic field and actually the effects that they have on moving charges. And that's actually really how we define magnetic field. So first of all, with any field it's good to have a way to visualize it. With the electrostatic fields we drew field lines. So let's try to do the same thing with magnetic fields. Let's say this is my bar magnet. This is the north pole and this is the south pole. Now the convention, when we're drawing magnetic field lines, is to always start at the north pole and go towards the south pole. And you can almost view it as the path that a magnetic north monopole would take. So if it starts here-- if a magnetic north monopole, even though as far as we know they don't exist in nature, although they theoretically could, but let's just say for the sake of argument that we do have a magnetic north monopole. If it started out here, it would want to run away from this north pole and would try to get to the south pole. So it would do something, its path would look something like this. If it started here, maybe its path would look something like this. Or if it started here, maybe its path would look something like this. I think you get the point. Another way to visualize it is instead of thinking about a magnetic north monopole and the path it would take, you could think of, well, what if I had a little compass here? Let me draw it in a different color. Let's say I put the compass here. That's not where I want to do it. Let's say I do it here. The compass pointer will actually be tangent to the field line. So the pointer could look something like this at this point. It would look something like this. And this would be the north pole of the pointer and this would be the south pole of the pointer. Or you could-- that's how north and south were defined. People had compasses, they said, oh, this is the north seeking pole, and it points in that direction. But it's actually seeking the south pole of the larger magnet. And that's where we got into that big confusing discussion of that the magnetic geographic north pole that we're used to is actually the south pole of the magnet that we call Earth. And you could view the last video on Introduction to Magnetism to get confused about that. But I think you see what I'm saying. North always seeks south the same way that positive seeks negative, and vice versa. And north runs away from north. And really the main conceptual difference-- although they are kind of very different properties-- although we will see later they actually end up being the same thing, that we have something called an electromagnetic force, once we start learning about Maxwell's equations and relativity and all that. But we don't have to worry about that right now. But in classical electricity and magnetism, they're kind of a different force. And the main difference-- although you know, these field lines, you can kind of view them as being similar-- is that magnetic forces always come in dipoles, soon. while you could have electrostatic forces that are monopoles. You could have just a positive or a negative charge. So that's fine, you say, Sal, that's nice. You drew these field lines. And you've probably seen it before if you've ever dropped metal filings on top of a magnet. They kind of arrange themselves But you might say, well, that's kind of useful. But how do we determine the magnitude of a magnetic field at any point? And this is where it gets interesting. The magnitude of a magnetic field is really determined, or it's really defined, in terms of the effect that it has on a moving charge. So this is interesting. I've kind of been telling you that we have this different force called magnetism that is different than the electrostatic force. But we're defining magnetism in terms of the effect that it has on a moving charge. And that's a bit of a clue. And we'll learn later, or hopefully you'll learn later as you advance in physics, that magnetic force or a magnetic field is nothing but an electrostatic field moving at a very high speed. At a relativistic speed. Or you could almost view it as they are the same thing, just from different frames of reference. I don't want to confuse you right now. But anyway, back to what I'll call the basic physics. So if I had to find a magnetic field as B-- so B is a vector and it's a magnetic field-- we know that the force on a moving charge could be an electron, a proton, or some other type of moving charged particle. And actually, this is the basis of how they-- you know, when you have supercolliders-- how they get the particles to go in circles, and how they studied them by based on how they get deflected by the magnetic field. But anyway, the force on a charge is equal to the magnitude of the charge-- of course, this could be positive or negative-- times, and this is where it gets interesting, the velocity of the charge cross the magnetic field. So you take the velocity of the charge, you could either multiply it by the scalar first, or you could take the cross product then multiply it by the scalar. Doesn't matter because it's just a number, this isn't a vector. But you essentially take the cross product of the velocity and the magnetic field, multiply that times the charge, and then you get the force vector on that particle. Now there's something that should immediately-- if you hopefully got a little bit of intuition about what the cross product was-- there's something interesting going on here. The cross product cares about the vectors that are perpendicular to each other. So for example, if the velocity is exactly perpendicular to the magnetic field, then we'll actually get a number. If they're parallel, then the magnetic field has no impact on the charge. That's one interesting thing. And then the other interesting thing is when you take the cross product of two vectors, the result is perpendicular to both of these vectors. So that's interesting. A magnetic field, in order to have an effect on a charge, has to be perpendicular to its you velocity. And then the force on it is going to be perpendicular to both the velocity of the charge and the magnetic field. I know I'm confusing you at this point, so let's play around with it and do some problems. But before that, let's figure out what the units of the magnetic field are. So we know that the cross product is the same thing as-- so let's say, what's the magnitude of the force? The magnitude of the force is equal to? Well, the magnitude of the charge-- this is just a scalar quantity, so it's still just the charge-- times the magnitude of the velocity times the magnitude of the field times the sine of the angle between them. This is the definition of a cross product and then we could put-- if we wanted the actual force vector, we can just multiply this times the vector we get using the We'll do that in a second. Anyway we're just focused on units. Sine of theta has no units so we can ignore it for this discussion. We're just trying to figure out the units of the magnetic field. So force is newtons-- so we could say newtons equals-- charge is coulombs, velocity is meters per second, and then this is times the-- I don't know what we'll call this-- the B units. We'll call it unit sub B. So let's see. If we divide both sides by coulombs and meters per second, we get newtons per coulomb. And then if we divide by meters per second, that's the same thing as multiplying by seconds per meter. Equals the magnetic field units. So the magnetic field in SI terms, is defined as newton seconds per coulomb meter. And that might seem a little disjointed, and they've come up with a brilliant name. And it's named after a deserving fellow, and that's Nikolai Tesla. And so the one newton second per coulomb meter is equal to one tesla. And I'm actually running out of time in this video, because I want to do a whole problem here. But I just want you to sit and think about it for a second. Even though in life we're used to dealing with magnets as we have these magnets-- and they're fundamentally maybe different than what at least we imagine electricity to be-- but the magnitude or actually the units of magnetism is actually defined in terms of the effect that it would have on a moving charge. And that's why the unit-- one tesla, or a tesla-- is defined as a newton second per coulomb. So the electrostatic charge per coulomb meter. Well, I'll leave you now in this video. Maybe you can sit and ponder that. But it'll make a little bit more sense when we do some actual problems with some actual numbers in the next video. See" + }, + { + "Q": "Why did Hitler hate the Jews, weren't they good people", + "A": "He believed that the Jewish leaders during WW1 had lost the war by surrendering. He was a soldier on the German side in WW1, so that devastated him greatly.", + "video_name": "EtZnPoYbRyA", + "transcript": "Narrator: Where we left off in the last video, in 1924, Hitler was in jail, his famous coup d'\u00e9tat in 1923, his famous Beer Hall Putsched in Munich had failed. He's now in jail, he's writing Mein Kampf. When he gets out of jail, so this is when he's in jail, the Nazi party is banned and a lot of the economic turmoil that made the possibility of overthrowing the government more likely, that we saw in the early 20's, that hyperinflation in Weimar Germany, this was now under control by the time Hitler comes out of jail. They had issued new currency, it was far more stable. To a certain degree the Nazi's and Hitler were starting from scratch, although even at this point Hitler continues to be an ever growing influence. He's a famous speaker, there are more and more people who are knowing about him and who are following him. Over the next few years his book does get published and it sells, actually, tens of thousands of copies over the next several years, but for the most part he's still a relatively small actor in German politics. But then we fast forward as we get to the late 20s, the Nazi's are gaining some influence, but then in 1929, (writing) in 1929, you have a global change for the economy of the world and that's the beginning of The Great Depression. In particular, what's often the first sign that The Great Depression was at hand is you have the U.S. stock market crashes in October of 1929, famous Black Tuesday. That was the mark of the beginning of a, not just American Depression, but a global depression. So you have the whole world going into a depression. Anytime you have economic turmoil it tends give more energy to the more extreme parties, whether it is the parties like the Nazi's, who one could consider maybe to be on the extreme right, or often considered to be on the extreme right, or maybe you could say very nationalistic, or even the extreme left parties who are obviously against capitalist systems and whatever else. So, by the election of 1930, now we're talking about Parliamentary elections and the Parliament in Germany is the Reichstag. (writing) The Reichstag, and I know I'm mispronouncing it. In the Reichstag elections, the Nazi party, for the first time is able to have a significant showing. It gets 18, it gets roughly 18% of the vote and a proportional representation in the Parliament. Now all of a sudden, this kind of marks the beginning of the Nazi's being significant, significant players in German politics. Then we get to 1932 and the economy is not improving, it is only getting worse. (writing) 1932. Adolf Hitler actually makes a run for President. The current President at that point is Paul von Hindenburg, famous for the Hindenburg line, later for the Hindenburg, the Zeppelin, the famous exploding Zeppelin disaster. He was, with Ludendorff, one of the two leaders of the German military effort during World War I. He's President of the Weimar Republic since 1925 and in 1932 he is able to get re-election, but Hitler has a fairly good showing. Hitler is able to get 35% of the vote. (writing) Hitler gets 35% of the presidential election votes, (writing) of the vote. The Weimar Republic had this strange system. It wasn't quite a Presidential system like the U.S. and it wasn't quite a pure Parliamentary system like the current-day Germany. The President was independently elected and had some powers, and then the Parliament was also independently elected and then they would try to build coalitions to have a ruling government. Needless to say, 1932 Hitler is now a major actor, the Nazi's also have a many, many, many seats in Parliament. Now, you have several Parliamentary elections as well in 1932 and as we just talked about two in particular. In order for a government to form in Parliament, in order to find the Cabinet and the Chancellor, who essentially is the Prime Minister, you have an election and the different parties get different amounts of votes. If no party has a majority, the parties have to form a coalition that can make a majority. There's a lot of horsetrading going on with parties negotiating, hey why don't we form a coalition with each other, if we do that maybe someone from my party can be Minister of the Interior, someone of your party could be the Chancellor and maybe we can get a coalition together to rule over the government. But you have two Parliamentary elections and no majority coalition forms. (writing) So, two, two elections. So this is Parliamentary. So this is in the Presidential election, Hindenburg is still President, but Hitler has a good showing and then you have two Parliamentary elections. (writing) Parliament elections, or Reichstag elections where you have no majority, no coalition. (writing) no majority, majority coalition. The Nazi's continue to be a major actor here, they continue to have more and more of a showing inside the Reichstag. Then by 1933 it's a bit of crisis. So as we get in to early 1933 we have a little bit of a crisis. We have no government, we have no Chancellor, we have no Cabinet to essentially be the executive, the government of the country because there's been no major coalitions. The Weimar Constitution allowed a strange thing, it allowed the President to appoint a government, appoint a Cabinet, a Chancellor that might not even be representative of what's going on in Parliament. So, Paul von Hindenburg is convinced that ... hey look, he was no fan, he was no fan of Adolf Hitler but he's convinced that look, Adolf Hitler was your opponent if you make Adolf Hitler the head of an interim Government, the head of an interim Cabinet then that might be a way to create some national unity and then maybe we could have some Parliamentary elections that there can be a majority coalition and you could have, I guess you could say, a more legitimate government take hold. So, Paul von Hindenburg is convinced and so he does, even though the Nazi's are still a minority party, even though they weren't part of any type of a majority coalition, Paul von Hindenburg who is not a fan of Adolf Hitler appoints him as Chancellor. This is in January. So in January, Hitler, (writing) Hitler is appointed Chancellor, Chancellor, which is essentially the Prime Minister of the Reichstag of Germany. Then we get to February and events get really, really, really interesting. In February of 1933 you have a fire in the Reichstag building in Berlin. This is the Reichstag building right over here and it is on fire. They find this gentleman here on the scene, Marinus van der Lubbe, he is a Dutch communist. It is essentially the blame is placed as this was some type of a, the beginning of some type of a communist revolution. This is used as a pretext. Hitler then advises Paul von Hindenburg to essentially use some of his emergency powers as President, which is another strange thing that the Weimar Constitution allowed for, it allowed the President under emergency conditions to start to suspend civil rights. This was an emergency situation and so Paul von Hindenburg does that. He essentially issues ... once you have the Reichstag fire (writing) Reichstag fire, and then Hindenburg is convinced by the Nazi's to pass the Reichstag Fire Decree. (writing) Fire decree, which essentially suspends, it gives the government emergency powers and it suspends civil liberties, which everything up to this point now is actually legal, this was actually allowed for in the Weimar Constitution. (writing) Suspends, suspends civil, civil liberties. And since there's no coalition, the whole point that Hitler's Cabinet was going to be an interim one, you have another Parliamentary election coming in March with the hope of maybe a majority coalition forms, but that March election, especially with civil liberties suspended you could imagine that the Nazi's ... and they have their paramilitary troopers started intimidating other parties, making sure that they had a better showing at the polls, they started intimidating other candidates. The March election start to swing hugely in the Nazi's favor, so in the March election they're able to get 44% of the vote, which is still not enough, by themselves, to form a government. It's still not a majority, but they're able now ... they're now the largest part in the Reichstag, in the Parliament. They're able to now form a majority coalition, and I guess you could say more legitimately ... although this was a election of intimidation, they were able to now form a government, they're able to now form a government based on a majority coalition and Hitler remains Chancellor. But then, this new Parliament passes the Enabling Act in March. (writing) Enabling Act, Enabling Act, which is essentially an amendment to the Weimar Constitution which gives the Cabinet, especially the Chancellor, effectively the Chancellor who's the head of the Cabinet, legislative powers, unlimited legislative powers for the next four years. So, it gives legislative powers and remember we already have suspended civil rights. So, the Reichstag is essentially giving over the legislative powers, (writing) legislative powers, to the Chancellor who happens to be, who happens to be Hitler. There was some check on this by the President, but then we have Hindenburg dying the next year. After this, after the suspension of civil rights and then the Enabling Act shortly afterwards, Hitler is essentially in full control, Hitler and the Nazi's are essentially in full control of the German government. At this point, Hitler is the dictator, (writing) the dictator of, he is the dictator of Germany. They start to act fast, they start to intimidate other parties, they use violence, they start to imprison people and by July of 1933 ... so they're acting very, very fast, by July of 1933 Nazi's are the only legal party. (writing) only legal Pot party and they essentially have full control. Now, this is how Hitler came to power and the question that's probably circling in your mind is, \"Who did this fire?\" This fire was the catalyst, although Hitler was already Chancellor and maybe he would have found some way to get to power regardless, but this fire, even though there was evidence that it looked like maybe Marinus van der Lubbe did it, it was blamed on the communist, it was the pretext that was used to give the government even more power, especially the Nazi's even more power. This is an open question, one of those great open questions, one of those great open questions of history. Some people feel that maybe it was just a communist plot, maybe it was Marinus van der Lubbe acting on his own and maybe it just happened to fall into the hands of Hitler and they were able to use it, while other historians think that this was actually a plot by the Nazi's to create this emergency state and Marinus van der Lubbe was kind of a puppet in this whole plot. So, open question of history, but needless to say as we go from 1919 to 1933, Hitler goes from a fairly unknown individual to full dictator of Germany." + }, + { + "Q": "How to use vector addition in real life or application development? In general, I understand how it is done but why should it is important?", + "A": "Understanding vectors is essential to the understanding of more advanced topics which also have no application! ; )", + "video_name": "r4bH66vYjss", + "transcript": "In the last video I was a little formal in defining what Rn is, and what a vector is, and what vector addition or scalar multiplication is. In this video I want to kind of go back to basics and just give you a lot of examples. And give you a more tangible sense for what vectors are and how we operate with them. So let me define a couple of vectors here. And I'm going to do, most of my vectors I'm going to do in this video are going to be in R2. And that's because they're easy to draw. Remember R2 is the set of all 2-tuples. Ordered 2-tuples where each of the numbers, so you know you could have x1, my 1 looks like a comma, x1 and x2, where each of these are real numbers. So you each of them, x1 is a member of the reals, and x2 is a member of the reals. And just to give you a sense of what that means, if this right here is my coordinate axes, and I wanted a plot all my x1's, x2's. You know you could view this as the first coordinate. We always imagine that as our x-axis. And then our second coordinate we plotted on the vertical axis. That traditionally is our y-axis, but we'll just call that the second number axis, whatever. You could visually represent all of R2 by literally every single point on this plane if we were to continue off to infinity in every direction. That's what R2 is. R1 would just be points just along one of these number lines. That would be R1. So you could immediately see that R2 is kind of a bigger space. But anyway, I said that I wouldn't be too abstract, that I would show you examples. So let's get some vectors going in R2. So let me define my vector a. I'll make it nice and bold. My vector a is equal to, I'll make some numbers up, negative 1, 2. And my vector b, make it nice and bold, let me make that, I don't know, 3, 1. Those are my two vectors. Let's just add them up and see what we get. Just based on my definition of vector addition. I'll just stay in one color for now so I don't have to keep switching back and forth. So a, nice deep a, plus bolded b is equal to, I just add up each of those terms. Negative 1 plus 3. And then 2 plus 1. That was my definition of vector addition. So that is going to be equal to 2 and 3. Fair enough that just came out of my definition of vector addition. But how can we represent this vector? So we already know that if we have coordinates, you know, if I have the coordinate, and this is just a convention. It's just the way that we do it. The way we visualize things. If I wanted to plot the point 1, 1, I go to my coordinate axes. The first point I go along the horizontal, what we traditionally call our x-axis. And I go 1 in that direction. And then convention is, the second point I go 1 in the vertical direction. So the point 1, 1. Oh, sorry, let me be very clear. This is 2 and 2, so one is right here, and one is right there. So the point 1, 1 would be right there. That's just the standard convention. Now our convention for representing vectors are, you might be tempted to say, oh, maybe I just represent this vector at the point minus 1, 2. And on some level you can do that. I'll show you in a second. But the convention for vectors is that you can start at any point. Let's say we're dealing with two dimensional vectors. You can start at any point in R2. So let's say that you're starting at the point x1, and x2. This could be any point in R2. To represent the vector, what we do is we draw a line from that point to the point x1. And let me call this, let's say that we wanted to draw a. So x1 minus 1. So this is, I'm representing a. So this is, I want to represent the vector a. x1 minus 1, and then x1 plus 2. Now if that seems confusing to you, when I draw it, it'll be very obvious. So let's say I just want to start at the point, let's just say for quirky reasons, I just pick a random point here. I just pick a point. That one right there. That's my starting point. So minus 4, 4. Now if I want to represent my vector a, what I just said is that I add the first term in vector a to my first So x1 plus minus 1 or x1 minus 1. So my new one is going to be, so this is my x1 minus 4. So now it's going to be, let's see, I'm starting at the point minus 4 comma 4. If I want to represent a, what I do is, I draw an arrow to minus 4 plus this first term, minus 1. And then 4 plus the second term. 4 plus 2. And so this is what? This is minus 5 comma 6. So I go to minus 5 comma 6. So I go to that point right there and I just draw a line. So my vector will look like this. I draw a line from there to there. And I draw an arrow at the end point. So that's one representation of the vector minus 1, 2. Actually let me do it a little bit better. Because minus 5 is actually more, a little closer to right here. Minus 5 comma 6 Is right there, so I draw my vector like that. But remember this point minus 4 comma 4 was an arbitrary place to draw my vector. I could have started at this point here. I could have started at the point 4 comma 6 and done the same thing. I could have gone minus 1 in the horizontal direction, that's my movement in the horizontal direction. And then plus 2 in the vertical direction. So I could have drawn, so minus 1 in the horizontal and plus 2 in the vertical gets me right there. So I could have just as easily drawn my vector like that. These are both interpretations of the same vector a. I should draw them in the color of vector a. So vector a was this light blue color right there. So this is vector a. This is vector a. Sometimes there'll be a little arrow notation over the vector. I could draw an infinite number of vector a's. I could draw vector a here. I could draw it like that. Vector a, it goes back 1 and up 2. So vector a could be right there. Similarly vector b. What does vector b do? I could pick some arbitrary point for vector b. It goes to the right 3, so it goes to the right 1, 2, 3 and then it goes up 1. So vector b, one representation of vector b, looks like this. Another represention. I can start it right here. I could go to the right 3, 1, 2, 3, and then up 1. This would be another representation of my vector b. There's an infinite number of representations of them. But the convention is to often put them in what's called the standard position. And that's to start them off at 0, 0. So your initial point, let me write this down. Standard position is just to start the vectors at 0, 0 and then draw them. So vector a in standard position, I'd start at 0, 0 like that and I would go back 1 and then up 2. So this is vector a in standard position right there. And then vector b in standard position. Let me write that. That's a. And then vector b in standard position is 3, go to the 3 right and then up 1. These are the vectors in standard position, but any of these other things we drew are just as valid. Now let's see if we can get an interpretation of what happened when we added a plus b. Well if I draw that vector in standard position, I just calculated, it's 2, 3. So I go to the right 2 and I go up 3. So if I just draw it in standard position it looks like this. This vector right there. And at first when you look at it, this vector right here is the vector a plus b in standard position. When you draw it like that, it's not clear what the relationship is when we added a and b. But to see the relationship what you do is, you put a and b head to tails. What that means is, you put the tail end of b to the front end of a. Because remember, all of these are valid representations of b. All of the representations of the vector b. They all have, they're all parallel to each other, but they can start from anywhere. So another equally valid representation of vector b is to start at this point right here, kind of the end point of vector a in standard position, and then draw vector b So you go 3 to the right. So you go 1, 2, 3. And then you go up 1. So vector b could also be drawn just like that. And then you should see something interesting had happened. And remember, this vector b representation is not in standard position, but it's just an equally valid way to represent my vector. Now what do you see? When I add a, which is right here, to b what do I get if I connect the starting point of a with the end point of b? I get the addition. I have added the two vectors. And I could have done that anywhere. I could have started with a here. And then I could have done the end point. I could have started b here and gone 3 to the right, 1, 2, 3 and then up 1. And I could have drawn b right there like that. And then if I were to add a plus b, I go to the starting point of a, and then the end point of b. And that should also be the visual representation of a plus b. Just to make sure it confirms with this number, what I did here was I went 2 to the right, 1, 2 and then I went 3 up. 1, 2, 3 and I got a plus b. Now let's think about what happens when we scale our vectors. When we multiply it times some scalar factor. So let me pick new vectors. Those have gotten monotonous. Let me define vector v. v for vector. Let's say that it is equal to 1, 2. So if I just wanted to draw vector v in standard position, I would just go 1 to the horizontal and then 2 to the vertical. That's the vector in standard position. If I wanted to do it in a non standard position, I could do it right here. 1 to the right up 2, just like that. Equally valid way of drawing vector v. Equally valid way of doing it. Now what happens if I multiply vector v. What if I have, I don't know, what if I have 2 times v? 2 times my vector v is now going to be equal to 2 times each of these terms. So it's going to be 2 times 1 which is 2, and then 2 times 2 which is 4. Now what does 2 times vector v look like? Well let me just start from an arbitrary position. Let me just start right over here. So I'm going to go 2 to the right, 1, 2. And I go up 4. 1, 2, 3, 4. So this is what 2 times vector v looks like. This is 2 times my vector v. And if you look at it, it's pointing in the exact same direction but now it's twice as long. And that makes sense because we scaled it by a factor of 2. When you multiply it by a scalar, or you're not changing its direction. Its direction is the exact same thing as it was before. You're just scaling it by that amount. And I could draw this anywhere. I could have drawn it right here. I could have drawn 2v right on top of v. Then you would have seen it, I don't want to cover it. You would have seen that it goes, it's exactly, in this case when I draw it in standard position, it's colinear. It's along the same line, it's just twice as far. it's just twice as long but they have the exact same direction. Now what happens if I were to multiply minus 4 times our vector v? Well then that will be equal to minus 4 times 1, which is minus 4. And then minus 4 times 2, which is minus 8. So this is on my new vector. Minus 4, minus 8. This is minus 4 times our vector v. So let's just start at some arbitrary point. Let's just do it in standard position. So you go to the right 4. Or you go to the left 4. So so you go to the left 4, 1, 2, 3, 4. And then down 8. Looks like that. So this new vector is going to look like this. Let me try and draw a relatively straight line. There you go. So this is minus 4 times our vector v. I'll draw a little arrow on it to make sure you know it's a vector. Now what happened? Well we're kind of in the same direction. Actually we're in the exact opposite direction. But we're still along the same line, right? But we're just in the exact opposite direction. And it's this negative right there that flipped us around. If we just multiplied negative 1 times this, we would have just flipped around to right there, right? But we multiplied it by negative 4. So we scaled it by 4, so you make it 4 times as long, and then it's negative, so then it flips around. It flips backwards. So now that we have that notion, we can kind of start understanding the idea of subtracting vectors. Let me make up 2 new vectors right now. Let's say my vector x, nice and bold x, is equal to, and I'm doing everything in R2, but in the last part of this video I'll make a few examples in R3 or R4. Let's say my vector x is equal to 2, 4. And let's say I have a vector y. y, make it nice and bold. And then that is equal to negative 1, minus 2. And I want to think about the notion of what x minus y is equal to. Well we can say that this is the same thing as x plus minus 1 times our vector y. So x plus minus 1 times our vector y. Now we can use our definitions. We know how to multiply by a scalar. So we'll say that this is equal to, let me switch colors. I don't like this color. This is equal to our x vector is 2, 4. And then what's minus 1 times y? So minus 1 times y is minus 1 times minus 1 is 1. And then minus 1 times minus 2 is 2. So x minus y is going to be these two vectors added to I'm just adding the minus of y. This is minus vector y. So this x minus y is going to be equal to 3 and 3 and 6. So let's see what that looks like when we visually represent them. Our vector x was 2, 4. So 2, 4 in standard position it looks like this. That's my vector x. And then vector y in standard position, let me do it in a different color, I'll do y in green. Vector y is minus 1, minus 2. It looks just like this. And actually I ended up inadvertently doing collinear vectors, but, hey, this is interesting too. So this is vector y. So then what's their difference? This is 3, 6. So it's the vector 3, 6. So it's this vector. Let me draw it someplace else. If I start here I go 1, 2, 3. And then I go up 6. So then up 6. It's a vector that looks like this. That's the difference between the two vectors. So at first you say, this is x minus y. Hey, how is this the difference of these two? Well if you overlay this. If you just shift this over this, you could actually just start here and go straight up. And you'll see that it's really the difference between the end points. You're kind of connecting the end points. I actually didn't want to draw collinear vectors. Let me do another example. Although that one's kind of interesting. You often don't see that one in a book. Let me to define vector x in this case to be 2, 3. And let me define vector y to be minus 4, minus 2. So what would be x in standard position? It would be 2, 3. It'd look like that. That is our vector x if we start at the origin. So this is x. And then what does vector y look like? I'll do y in orange. Minus 4, minus 2. So vector y looks like this. Now what is x minus y? Well you know, we could view this, 2 plus minus 1 times this. We could just say 2 minus minus 4. I think you get the idea now. But we just did it the first way the last time because I wanted to go from my basic definitions of scalar multiplication. So x minus y is just going to be equal to 2 plus minus 1 times minus 4, or 2 minus minus 4. That's the same thing as 2 plus 4, so it's 6. And then it's 3 minus minus 2, so it's 5. So the difference between the two is the vector 6, 5. So you could draw it out here again. So you could go, add 6 to 4, go up there, then to 5, you'd go like that. So the vector would look something like this. It shouldn't curve like that, so that's x minus y. But if we drew them between, like in the last example, I showed that you could draw it between their two heads. So if you do it here, what does it look like? Well if you start at this point right there and you go 6 to the right and then up 5, you end up right there. So the difference between the two vectors, let me make sure I get it, the difference between the two vectors looks like that. It looks just like that. Which kind of should make sense intuitively. x minus y. That's the difference between the two vectors. You can view the difference as, how do you get from one vector to another vector, right? Like if, you know, let's go back to our kind of second grade world of just scalars. If I say what 7 minus 5 is, and you say it's equal to 2, well that just tells you that 5 plus 2 is equal to 7. Or the difference between 5 and 7 is 2. And here you're saying, look the difference between x and y is this vector right there. It's equal to that vector right there. Or you could say look, if I take 5 and add 2 I get 7. Or you could say, look, if I take vector y, and I add vector x minus y, then I get vector x. Now let's do something else that's interesting. Let's do what y minus x is equal to. y minus x. What is that equal to? Do it in another color right here. Well we'll take minus 4, minus 2 which is minus 6. And then you have minus 2, minus 3. It's minus 5. So y minus x is going to be, let's see, if we start here we're going to go down 6. 1, 2, 3, 4, 5, 6. And then back 5. So back 2, 4, 5. So y minus x looks like this. It's really the exact same vector. Remember, it doesn't matter where we start. It's just pointing in the opposite direction. So if we shifted it here. I could draw it right on top of this. It would be the exact as x minus y, but just in the opposite direction. Which is just a general good thing to know. So you can kind of do them as the negatives of each other. And actually let me make that point very clear. You know we drew y. Actually let me draw x, x we could draw as 2, 3. So you go to the right 2 and then up 3. I've done this before. This is x in non standard position. That's x as well. What is negative x? Negative x is minus 2 minus 3. So if I were to start here, I'd go to minus 2, then I'd go minus 3. So minus x would look just like this. Minus x. It looks just like x. It's parallel. It has the same magnitude. It's just pointing in the exact opposite direction. And this is just a good thing to kind of really get seared into your brain is to have an intuition for these things. Now just to kind of finish up this kind of idea of adding and subtracting vectors. Everything I did so far was in R2. But I want to show you that we can generalize them. And we can even generalize them to vector spaces that aren't normally intuitive for us to actually visualize. So let me define a couple of vectors. Let me define vector a to be equal to 0, minus 1, 2, and 3. Let me define vector b to be equal to 4, minus 2, 0, 5. We can do the same addition and subtraction operations with them. It's just it'll be hard to visualize. We can keep them in just vector form. So that it's still useful to think in four dimensions. So if I were to say 4 times a. This is the vector a minus 2 times b. What is this going to be equal to? This is a vector. What is this going to be equal to? Well we could rewrite this as 4 times this whole column vector, 0, minus 1, 2, and 3. Minus 2 times b. Minus 2 times 4, minus 2, 0, 5. And what is this going to be equal to? This term right here, 4 times this, you're going to get, the pen tablet seems to not work well there, so I'm going to do 4 times this, you're going to get 4 times 0, 0, minus 4, 8. 4 times 3 is 12. And then minus, I'll do it in yellow, minus 2 times 4 is 8. 2 times minus 2 is minus 4. 2 times 0 is 0. 2 times 5 is 10. This isn't a good part of my board, so let me just. It doesn't write well right over there. I haven't figured out the problem, but if I were just right it over here, what do we get? With 0 minus 8? Minus 8. Minus 4, minus 4. Minus negative 4. So that's minus 4 plus 4, so that's 0. 8 minus 0 is 8. 12 minus, what was this? I can't even read it, what it says. Oh, this is a 10. Now you can see it again. Something is very bizarre. 2 times 5 is 10. So it's 12 minus 10, so it's 2. So when we take this vector and multiply it by 4, and subtract 2 times this vector, we just get this vector. And even though you can't represent this in kind of an easy kind of graph-able format, this is a useful concept. And we're going to see this later when we apply some of these vectors to multi-dimensional spaces." + }, + { + "Q": "What's the difference between a prophage and a provirus?\n\nIs it that a prophage is the integration of a chromosome in a bacteria, whereas, a provirus is the integration of a chromosome in an animal cell.", + "A": "A prophage is the genetic material of a bacteriophage, that becomes incorporated into the genome of a bacterium, and is able to produce more phages if stimulated for the specific activity. A provirus is similar to a prophage, in that it is the genetic material of a virus that becomes incorporated into the genome of a host cell, and has the ability to replicate with that cell.", + "video_name": "0h5Jd7sgQWY", + "transcript": "Considering that I have a cold right now, I can't imagine a more appropriate topic to make a video on than a virus. And I didn't want to make it that thick. A virus, or viruses. And in my opinion, viruses are, on some level, the most fascinating thing in all of biology. Because they really blur the boundary between what is an inanimate object and what is life? I mean if we look at ourselves, or life as one of those things that you know it when you see it. If you see something that, it's born, it grows, it's constantly changing. Maybe it moves around. Maybe it doesn't. But it's metabolizing things around itself. It reproduces and then it dies. You say, hey, that's probably life. And in this, we throw most things that we see-- or we throw in, us. We throw in bacteria. We throw in plants. I mean, I could-- I'm kind of butchering the taxonomy system here, but we tend to know life when we see it. But all viruses are, they're just a bunch of genetic information inside of a protein. Inside of a protein capsule. So let me draw. And the genetic information can come in any form. So it can be an RNA, it could be DNA, it could be single-stranded RNA, double-stranded RNA. Sometimes for single stranded they'll write these two little S's in front of it. Let's say they are talking about double stranded DNA, they'll put a ds in front of it. But the general idea-- and viruses can come in all of these forms-- is that they have some genetic information, some chain of nucleic acids. Either as single or double stranded RNA or single or double stranded DNA. And it's just contained inside some type of protein structure, which is called the capsid. And kind of the classic drawing is kind of an icosahedron type looking thing. Let me see if I can do justice to it. It looks something like this. And not all viruses have to look exactly like this. There's thousands of types of viruses. And we're really just scratching the surface and understanding even what viruses are out there and all of the different ways that they can essentially replicate themselves. We'll talk more about that in the future. And I would suspect that pretty much any possible way of replication probably does somehow exist in the virus world. But they really are just these proteins, these protein capsids, are just made up of a bunch of little proteins put together. And inside they have some genetic material, which might be DNA or it might be RNA. So let me draw their genetic material. The protein is not necessarily transparent, but if it was, you would see some genetic material inside of there. So the question is, is this thing life? It seems pretty inanimate. It doesn't grow. It doesn't change. It doesn't metabolize things. This thing, left to its own devices, is just It's just going to sit there the way a book on a table just sits there. It won't change anything. But what happens is, the debate arises. I mean you might say, hey Sal, when you define it that way, just looks like a bunch of molecules put together. That isn't life. But it starts to seem like life all of a sudden when it comes in contact with the things that we normally consider life. So what viruses do, the classic example is, a virus will attach itself to a cell. So let me draw this thing a little bit smaller. So let's say that this is my virus. I'll draw it as a little hexagon. And what it does is, it'll attach itself to a cell. And it could be any type of cell. It could be a bacteria cell, it could be a plant cell, it could be a human cell. Let me draw the cell here. Cells are usually far larger than the virus. In the case of cells that have soft membranes, the virus figures out some way to enter it. Sometimes it can essentially fuse-- I don't want to complicate the issue-- but sometimes viruses have their own little membranes. And we'll talk about in a second where it gets their membranes. So a virus might have its own membrane like that. That's around its capsid. And then these membranes will fuse. And then the virus will be able to enter into the cell. Now, that's one method. And another method, and they're seldom all the same way. But let's say another method would be, the virus convinces-- just based on some protein receptors on it, or protein receptors on the cells-- and obviously this has to be kind of a Trojan horse type of thing. The cell doesn't want viruses. So the virus has to somehow convince the cell that it's a non-foreign particle. We could do hundreds of videos on how viruses work and it's a continuing field of research. But sometimes you might have a virus that just gets consumed by the cell. Maybe the cell just thinks it's something that it needs to consume. So the cell wraps around it like this. And these sides will eventually merge. And then the cell and the virus will go into it. This is called endocytosis. I'll just talk about that. It just brings it into its cytoplasm. It doesn't happen just to viruses. But this is one mechanism that can enter. And then in cases where the cell in question-- for example in the situation with bacteria-- if the cell has a very hard shell-- let me do it in a good color. So let's say that this is a bacteria right here. And it has a hard shell. The viruses don't even enter the cell. They just hang out outside of the cell like this. Not drawing to scale. And they actually inject their genetic material. So there's obviously a huge-- there's a wide variety of ways of how the viruses get into cells. But that's beside the point. The interesting thing is that they do get into the cell. And once they do get into the cell, they release their genetic material into the cell. So their genetic material will float around. If their genetic material is already in the form of RNA-- and I could imagine almost every possibility of different ways for viruses to work probably do exist in nature. We just haven't found them. But the ones that we've already found really do kind of do it in every possible way. So if they have RNA, this RNA can immediately start being used to essentially-- let's say this is the nucleus of the cell. That's the nucleus of the cell and it normally has the DNA in it like that. Maybe I'll do the DNA in a different color. But DNA gets transcribed into RNA, normally. So normally, the cell, this a normal working cell, the RNA exits the nucleus, it goes to the ribosomes, and then you have the RNA in conjunction with the tRNA and it produces these proteins. The RNA codes for different proteins. And I talk about that in a different video. So these proteins get formed and eventually, they can form the different structures in a cell. But what a virus does is it hijacks this process here. Hijacks this mechanism. This RNA will essentially go and do what the cell's own RNA would have done. And it starts coding for its own proteins. Obviously it's not going to code for the same things there. And actually some of the first proteins it codes for often start killing the DNA and the RNA that might otherwise compete with it. So it codes its own proteins. And then those proteins start making more viral shells. So those proteins just start constructing more and more viral shells. At the same time, this RNA is replicating. It's using the cell's own mechanisms. Left to its own devices it would just sit there. But once it enters into a cell it can use all of the nice machinery that a cell has around to replicate itself. And it's kind of amazing, just the biochemistry of it. That these RNA molecules then find themselves back in these capsids. And then once there's enough of these and the cell has essentially all of its resources have been depleted, the viruses, these individual new viruses that have replicated themselves using all of the cell's mechanisms, will find some way to exit the cell. The most-- I don't want to say, typical, because we haven't even discovered all the different types of viruses there are-- but one that's, I guess, talked about the most, is when there's enough of these, they'll release proteins or they'll construct proteins. Because they don't make their own. That essentially cause the cell to either kill itself or its membrane to dissolve. So the membrane dissolves. And essentially the cell lyses. Let me write that down. The cell lyses. And lyses just means that the cell's membrane just And then all of these guys can emerge for themselves. Now I talked about before that have some of these guys, that they have their own membrane. So how did they get there, these kind of bilipid membranes? Well some of them, what they do is, once they replicate inside of a cell, they exit maybe not even killing-- they don't have to lyse. Everything I talk about, these are specific ways that a virus might work. But viruses really kind of explore-- well different types of viruses do almost every different combination you could imagine of replicating and coding for proteins and escaping from cells. Some of them just bud. And when they bud, they essentially, you can kind of imagine that they push against the cell wall, or the membrane. I shouldn't say cell wall. The cell's outer membrane. And then when they push against it, they take some of the membrane with them. Until eventually the cell will-- when this goes up enough, this'll pop together and it'll take some of the membrane with it. And you could imagine why that would be useful thing to have with you. Because now that you have this membrane, you kind of look like this cell. So when you want to go infect another cell like this, you're not going to necessarily look like a foreign particle. So it's a very useful way to look like something that you're not. And if you don't think that this is creepy-crawly enough, that you're hijacking the DNA of an organism, viruses can actually change the DNA an organism. And actually one of the most common examples is HIV virus. Let me write that down. HIV, which is a type of retrovirus, which is fascinating. Because what they do is, so they have RNA in them. And when they enter into a cell, let's say that they got into the cell. So it's inside of the cell like this. They actually bring along with them a protein. And every time you say, where do they get this protein? All of this stuff came from a different cell. They use some other cell's amino acids and ribosomes and nucleic acids and everything to build themselves. So any proteins that they have in them came from another cell. But they bring with them, this protein reverse transcriptase. And the reverse transcriptase takes their RNA and codes it into DNA. So its RNA to DNA. Which when it was first discovered was, kind of, people always thought that you always went from DNA to RNA, but this kind of broke that paradigm. But it codes from RNA to DNA. And if that's not bad enough, it'll incorporate that DNA into the DNA of the host cell. So that DNA will incorporate itself into the DNA of the host cell. Let's say the yellow is the DNA of the host cell. And this is its nucleus. So it actually messes with the genetic makeup of what it's infecting. And when I made the videos on bacteria I said, hey for every one human cell we have twenty bacteria cells. And they live with us and they're useful and they're part of us and they're 10% of our dry mass and all of that. But bacteria are kind of along for the ride. They don't change who we are. But these retroviruses, they're actually changing our I mean, my genes, I take very personally. They define who I am. But these guys will actually go in and change my genetic makeup. And then once they're part of the DNA, then just the natural DNA to RNA to protein process will code their actual proteins. Or their-- what they need to-- so sometimes they'll lay dormant and do nothing. And sometimes-- let's say sometimes in some type of environmental trigger, they'll start coding for themselves again. And they'll start producing more. But they're producing it directly from the organism's cell's DNA. They become part of the organism. I mean I can't imagine a more intimate way to become part of an organism than to become part of its DNA. I can't imagine any other way to actually define an organism. And if this by itself is not eerie enough, and just so you know, this notion right here, when a virus becomes part of an organism's DNA, this is called a provirus. But if this isn't eerie enough, they estimate-- so if this infects a cell in my nose or in my arm, as this cell experiences mitosis, all of its offspring-- but its offspring are genetically identical-- are going to have this viral DNA. And that might be fine, but at least my children won't get it. You know, at least it won't become part of my species. But it doesn't have to just infect somatic cells, it could infect a germ cell. So it could go into a germ cell. And the germ cells, we've learned already, these are the ones that produce gametes. For men, that's sperm and for women it's eggs. But you could imagine, once you've infected a germ cell, once you become part of a germ cell's DNA, then I'm passing on that viral DNA to my son or my daughter. And they are going to pass it on to their children. And just that idea by itself is, at least to my mind. vaguely creepy. And people estimate that 5-8%-- and this kind of really blurs, it makes you think about what we as humans really are-- but the estimate is 5-8% of the human genome-- so when I talked about bacteria I just talked about things that were along for the ride. But the current estimate, and I looked up this a lot. I found 8% someplace, 5% someplace. I mean people are doing it based on just looking at the DNA and how similar it is to DNA in other organisms. But the estimate is 5-8% of the human genome is from viruses, is from ancient retroviruses that incorporated themselves into the human germ line. So into the human DNA. So these are called endogenous retroviruses. Which is mind blowing to me, because it's not just saying these things are along for the ride or that they might help us or hurt us. It's saying that we are-- 5-8% of our DNA actually comes from viruses. And this is another thing that speaks to just genetic variation. Because viruses do something-- I mean this is called horizontal transfer of DNA. And you could imagine, as a virus goes from one species to the next, as it goes from Species A to B, if it mutates to be able to infiltrate these cells, it might take some-- it'll take the DNA that it already has, that makes it, it with it. But sometimes, when it starts coding for some of these other guys, so let's say that this is a provirus right here. Where the blue part is the original virus. The yellow is the organism's historic DNA. Sometimes when it codes, it takes up little sections of the other organism's DNA. So maybe most of it was the viral DNA, but it might have, when it transcribed and translated itself, it might have taken a little bit-- or at least when it translated or replicated itself-- it might take a little bit of the organism's previous DNA. So it's actually cutting parts of DNA from one organism and bringing it to another organism. Taking it from one member of a species to another member of But it can definitely go cross-species. So you have this idea all of a sudden that DNA can jump between species. It really kind of-- I don't know, for me it makes me appreciate how interconnected-- as a species, we kind of imagine that we're by ourselves and can only reproduce with each other and have genetic variation within But viruses introduce this notion of horizontal transfer via transduction. Horizontal transduction is just the idea of, look when I replicate this virus, I might take a little bit of the organism that I'm freeloading off of, I might take a little bit of their DNA with me. And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. So when a virus lyses it like this, this is called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia. One is from the CDC. These little green dots you see right here all over the surface, this big thing you see here, this is a white blood cell. Part of the human immune system. This is a white blood cell. And what you see emerging from the surface, essentially budding from the surface of this white blood cell-- and this gives you a sense of scale too-- these are HIV-1 viruses. And so you're familiar with the terminology, the HIV is a virus that infects white blood cells. AIDS is the syndrome you get once your immune system is weakened to the point. And then many people suffer infections that people with a strong immune system normally won't suffer from. But this is creepy. These things went inside this huge cell, they used the cell's own mechanism to reproduce its own DNA or its own RNA and these protein capsids. And then they bud from the cell and take a little bit of the membrane with it. And they can even leave some of their DNA behind in this cell's own DNA. So they really change what the cell is all about. This is another creepy picture. These are bacteriaphages. And these show you what I said before. This is a bacteria right here. This is its cell wall. And it's hard. So it's hard to just emerge into it. Or you can't just merge, fuse membranes with it. So they hang out on the outside of this bacteria. And they are essentially injecting their genetic material into the bacteria itself. And you could imagine, just looking at the size of these things. I mean, this is a cell. And it looks like a whole planet or something. Or this is a bacteria and these things are so much smaller. Roughly 1/100 of a bacteria. And these are much less than 1/100 of this cell we're talking about. And they're extremely hard to filter for. To kind of keep out. Because they are such, such small particles. If you think that these are exotic things that exist for things like HIV or Ebola , which they do cause, or SARS, you're right. But they're also common things. I mean, I said at the beginning of this video that I have a cold. And I have a cold because some viruses have infected the tissue in my nasal passage. And they're causing me to have a runny nose and whatnot. And viruses also cause the chicken pox. They cause the herpes simplex virus. Causes cold sores. So they're with us all around. I can almost guarantee you have some virus with you as you speak. They're all around you. But it's a very philosophically puzzling question. Because I started with, at the beginning, are these life? And at first when I just showed it to you, look they are just this protein with some nucleic acid molecule in it. And it's not doing anything. And that doesn't look like life to me. It's not moving around. It doesn't have a metabolism. It's not eating. It's not reproducing. But then all of a sudden, when you think about what it's doing to cells and how it uses cells to kind of reproduce. It kind of like-- in business terms it's asset light. It doesn't need all of the machinery because it can use other people's machinery to replicate itself. You almost kind of want to view it as a smarter form of life. Because it doesn't go through all of the trouble of what every other form of life has. It makes you question what life is, or even what we are. Are we these things that contain DNA or are we just transport mechanisms for the DNA? And these are kind of the more important things. And these viral infections are just battles between different forms of DNA and RNA and whatnot. Anyway I don't want to get too philosophical on you. But hopefully this gives you a good idea of what viruses are and why they really are, in my mind, the most fascinating pseudo organism in all of biology." + }, + { + "Q": "Wait... at 0:43 Sal says that 0 to the first power is 0*1, but I thought it was 0*0, can anyone explain why? Thanks!", + "A": "Ah, 0^1 = 0 and 0^0 is undefined. Does that help?", + "video_name": "PwDnpb_ZJvc", + "transcript": "So let's think a little bit about powers of 0. So what do you think 0 to the first power is going to be? And I encourage you to pause this video. Well, let's just think about it. One definition of exponentiation is that you start with a 1, and then, you multiply this number times a 1 one time. So this is literally going to be 1 times-- let me do it in the right color-- it's 1 times 0. You're multiplying the 1 by 0 one time. 1 times 0, well, that's just going to be equal to 0. Now, what do you think 0 squared or 0 to the second power is going to be equal to? Well, once again, one way of thinking about this is that you start with a 1, and we're going to multiply it by 0 two times. So times 0 times 0. Well, what's that going to be? Well, you multiply anything times 0, once again, you are going to get 0. And I think you see a pattern here. If I take 0 to any non-zero number-- so to the power of any non-zero, so this is some non-zero number, then this is going to be equal to 0. Now, this raises a very interesting question. What happens at 0 to the 0-th power? So here, 0 to the millionth power is going to be 0. 0 to the trillionth power is going to be 0. Even negative or fractional exponents, which we haven't talked about yet, as long as they're non-zero, this is just going to be equal to 0, kind of makes sense. But now, let's think about what 0 to the 0-th power is, because this is actually a fairly deep question. And I'll give you a hint. Well, actually, why don't you pause the video and think a little bit about what 0 to the 0-th power should be. Well, there's two trains of thought here. You could say, look, 0 to some non-zero number is 0. So why don't we just extend this to all numbers and say 0 to any number should be 0. And so maybe you should say that 0 to the 0-th power is 0. But then, there was another train of logic that we've already learned, that any non-zero number, if you take any non-zero number, and you raise it to the 0-th power. We've already established that you start with a 1, and you multiply it times that non-zero number 0 times. So this is always going to be equal to 1 for non-zero numbers. So maybe say, hey, maybe we should extend this to all numbers, including 0. So maybe 0 to the 0-th power should be 1. So we could make the argument that 0 to the 0-th power should be equal to 1. So you see a conundrum here, and there's actually really good cases, and you can get actually fairly sophisticated with your mathematics. And there's really good cases for both of these, for 0 to 0-th being 0, and 0 to the 0-th power being 1. And so when mathematicians get into this situation, where they say, well, there's good cases for either. There's not a completely natural one. Either of these definitions would lead to difficulties in mathematics. And so what mathematicians have decided to do is, for the most part-- and you'll find people who will dispute this; people will say, no, I like one more than the other-- but for the most part, this is left undefined. 0 to the 0-th is not defined by at least just kind of more conventional mathematics. In some use cases, it might be defined to be one of these two things. So 0 to any non-zero number, you're going to get 0. Any non-zero number to the 0-th power, you're going to get 1. But 0 to the 0, that's a little bit of a question mark." + }, + { + "Q": "When I learn how to use negative numbers, would it be strange for me to not regroup, but to allow the ones place to go negative?", + "A": "Yes and no. If you were doing 86 - 19, you would not want to write the answer as 7-3 or 7negative3. However, because 80 - 10 = 70 and 6 - 9 = -3, you can think of the answer as 3 less than 70, which would be 67. Have a blessed, wonderful day!", + "video_name": "9T3AAn-Cw3g", + "transcript": "Let's try to subtract 659 from 971. And as soon as you start trying to do it, you face a problem. You go to the ones place, and you say, how am I going to subtract a 9 from a 1? And the answer lies in regrouping, taking value from one of the other places here and giving it to the ones place. And to understand that a little bit better, let me rewrite these two numbers. Let me expand it out. So this 9 is in the hundreds place, so it represents 900. The 7 is in the tens place, so it represents 7 tens. And then, this 1 is in the ones place, so it just represents 1. And then down here, this 6 represents 600. This 5 represents 5 tens, or 50. And then, this 9-- well, it still just represents 9 ones, or 9. And we're subtracting this. We're subtracting 600 plus 50 plus 9. Or another way of thinking about it, we're subtracting 600, we're subtracting 50, we are subtracting 9. So let's work it out over here. So this is the exact same problem, just written a little bit differently. And we still have the same issue. How do we subtract a larger number from a smaller number? And the solution lies in trying to take value from one of the other places. And the easiest place to go is-- look, we've got 70 here. Why don't we take 10 from here, and we'll be left with 60, and give that 10 to the ones place. So if you add 10 to 1, what do we have? Well, then we're going to have 11. Notice, I have not changed the value of the number. 971 is the same thing as 900 plus 60 plus 11. It's still 971. And now we can actually subtract. 11 minus 9 is 2. 60 minus 50 is 10. And 900 minus 600 is 300. So this subtraction should result in 300 plus 10 plus 2, which is 312. Now, let's do the exact same thing here, but we're going to do it without expanding it out. So same issue-- how do we subtract a 9 from a 1? Well, let's take a 10 from the tens place. We're going to regroup. So we're going to get rid of one of these tens, so we're only going to have 6 tens left in the tens place. And we're going to give that 10 to the ones place. So 10 plus 1 is 11. Now we are ready to subtract. 11 minus 9 is 2. 6 minus 5 is 1. 9 minus 6 is 3. We get-- let me do that same color-- 312." + }, + { + "Q": "so any number can follow this law?", + "A": "Yes that is correct", + "video_name": "5RzDVNob0-0", + "transcript": "Use the associative law of multiplication to write-- and here they have 12 times 3 in parentheses, and then they want us to multiply that times 10-- in a different way. Simplify both expressions to show they have identical results. So the way that they wrote it is-- let me just rewrite it. So they have 12 times 3 in parentheses, and then they multiply that times 10. Now whenever something is in parentheses, that means do that first. So this literally says let's do the 12 times 3 first. Now, what is 12 times 3? It's 36. So this evaluates to 36, and then we still have that times 10 over there. Whenever we multiply something times a power of ten, we just add the number of zeroes that we have at the back of it, so this is going to be 360. This is going to be equal to 360. Now, the associative law of multiplication, once again, it sounds like a very fancy thing. All that means is it doesn't matter how we associate the multiplication or it doesn't matter how we put the parentheses, we're going to get the same answer, so let me write it down again. If we were to do 12 times 3 times 10, if we just wrote it like this without parentheses, if we just went left to right, that would essentially be exactly what we just did here But the associative law of multiplication says, you know what? We can multiply the 3 times 10 first and then multiply the 12, and we're going to get the exact same answer as if we multiplied the 12 times the 3 and then the 10. So let's just verify it for ourselves. So 3 times 10 is 30, and we still want to multiply the 12 times that. Now, what's 12 times 30? And we've seen this several times before. You can view it as a 12 times 3, which is 36, but we still have this 0 here. So that is also equal to 360. So it didn't matter how we associated the multiplication. You can do the 12 times 3 first or you can do the 3 times 10 first. Either way, they both evaluated to 360." + }, + { + "Q": "What would happen if the shape of the car changed when it hit the truck and the system didn't move?", + "A": "If the car were to change shape, there would be energy lost in the collision. This would mean that the final velocity after the collision would be a bit less.", + "video_name": "XFhntPxow0U", + "transcript": "Welcome back. I will now introduce you to the concept of momentum. And the letter for momentum is, in physics, or at least in mechanics, it's the letter P. P for momentum. And I assume that's because the letter M has already been used for mass, which is I guess an even more fundamental idea. So P for momentum. So what is momentum? Well, you probably have a general idea of it. If you see a big guy running really fast, they'll say, he has a lot of momentum. And if there's a big guy running really fast and a small guy running really fast, most people would say, well, the big guy has more momentum. Maybe they don't have a quantitative sense of why they're saying that, but they just feel that And if we look at the definition of momentum, it'll The definition of momentum is equal to mass times velocity. So something with, say, a medium mass and a huge velocity is going to have a big momentum. Or something with maybe a medium mass, but-- the other I forgot what I just said. So medium mass and big velocity, huge momentum, or Huge mass, medium velocity would have maybe the same momentum, but it would still have a big momentum. Or another way of doing momentum is how little you would like to be in the way of that object as it passes by. How unpleasant would it be to be hit by that object? That's a good way of thinking about momentum. So momentum is mass times velocity. So how does it relate to everything we've been learning so far? So we know that force is equal to mass times acceleration. And what's acceleration? Well acceleration is just change in velocity. So we also know that force is equal to mass times change in velocity per unit of time, right? Per change in time. T for time. So force is also equal to-- well, mass times change in velocity. Mass, let's assume that mass doesn't change. So that could also be viewed as the change in mass times velocity in the unit amount of time. And this is a little tricky here, I said, you know, the mass times the change in velocity, that's the same thing as the change in the mass times the velocity, assuming the mass doesn't change. And here we have mass times velocity, which is momentum. So force can also be viewed as change in momentum per unit of time. And I'll introduce you to another concept called impulse. And impulse kind of means that you think it means. An impulse is defined as force times time. And I just want to introduce this to you just in case you see it on the exam or whatever, show you it's not a difficult concept. So force times change in time, or time, if you assume time starts at time 0. But force times change in time is equal to impulse. I actually don't know-- I should look up what letters they use for impulse. But another way of viewing impulse is force times change in time. Well that's the same thing as change in momentum over change in time times change in time. Right? Because this is just the same thing as force. And that's just change in momentum, so that's impulse as well. And the unit of impulse is the joule. And we'll go more into the joule when we do work in all of that. And if this confuses you, don't worry about it too much. The main thing about momentum is that you realize it's mass times velocity. And since force is change in momentum per unit of time, if you don't have any external forces on a system or, on say, on a set of objects, their combined, or their net momentum won't change. And that comes from Newton's Laws. The only way you can get a combined change in momentum is if you have some type of net force acting on the system. So with that in mind, let's do some momentum problems. Whoops. Invert colors. OK. So let's say we have a car. Say it's a car. Let me do some more interesting colors. A car with a magenta bottom. And it is, let's see, what does this problem say? It's 1,000 kilograms. So a little over a ton. And it's moving at 9 meters per second east. So its velocity is equal to 9 meters per second east, or to the right in this example. And it strikes a stationary 2, 000 kilogram truck. So here's my truck. Here's my truck and this is a 2,000 kilogram truck. And it's stationary, so the velocity is 0. And when the car hits the truck, let's just say that it somehow gets stuck in the truck and they just both keep moving together. So they get stuck together. The question is, what is the resulting speed of the combination truck and car after the collision? Well, all we have to do is think about what is the combined momentum before the collision? The momentum of the car is going to be the mass times the car-- mass of the car. Well the total momentum is going to the mass of the car times the velocity of the car plus the mass of the truck times the velocity of the truck. And this is before they hit each other. So what's the mass of the car? That's 1,000. What's the velocity of the car? It's 9 meters per second. So as you can imagine, a unit of momentum would be kilogram meters per second. So it's 1,000 times 9 kilogram meters per second, but I won't write that right now just to keep things simple, or so I save space. And then the mass of the truck is 2,000. And what's its velocity? Well, it's 0. It's stationary initially. So the initial momentum of the system-- this is 2,000 times 0-- is 9,000 plus 0, which equals 9,000 kilogram meters per second. That's the momentum before the car hits the back of the truck. Now what happens after the car hits the back of the truck? So let's go to that situation. So we have the truck. I'll draw it a little less neatly. And then you have the car and it's probably a little bit-- well, I won't go into whether it's banged up and whether it released heat and all of that. Let's assume that there was nothing-- if this is a simple problem that we can do. So if we assume that, there would be no change in momentum. Because we're saying that there's no net forces acting on the system. And when I say system, I mean the combination of the car and the truck. So what we're saying is, is this combination, this new vehicle called a car truck, its momentum will have to be the same as the car and the truck's momentum when they were separate. So what do we know about this car truck object? Well we know its new mass. The car truck object, it will be the combined mass of the two. So it's 1,000 kilograms plus 2,000 kilograms. So it's 3,000 kilograms. And now we can use that information to figure out its velocity. How? Well, its momentum-- this 3,000 kilogram object's momentum-- has to be the same as the momentum of the two objects before the collision. So it still has to be 9,000 kilogram meters per second. So once again, mass times velocity. So mass is 3,000 times the new velocity. So we could call that, I don't know, new velocity, v sub n. That will equal 9,000. Because momentum is conserved. That's what you always have to remember. Momentum doesn't change unless there's a net force acting on the system. Because we saw a force is change in momentum per time. So if you have no force in it, you have no change in momentum. So let's just solve. Divide both sides of this by 3,000 and you get the new velocity is 3 meters per second. And that kind of makes sense. You have a relatively light car moving at 9 meters per second and a stationary truck. Then it smacks the truck and they move together. The combined object-- and it's going to be to the east. And we'll do more later, but we assume that a positive velocity is east. If somehow we ended up with a negative, it would have been west. But it makes sense because we have a light object and a stationery, heavy object. And when the light object hits the stationery, heavy object, the combined objects still keeps moving to the right, but it moves at a relatively slower speed. So hopefully that gives you a little bit of intuition for momentum, and that was not too confusing of a problem. And in the next couple of videos, I'll do more momentum problems and then I'll introduce you to momentum problems in two dimensions. I will see you soon." + }, + { + "Q": "I always wonder how does a company like SpaceX make money? It certainly does not sell a product or provide people with services...Could someone explain how it works? :D", + "A": "In the interview, Mr. Musk mentioned that the company was started using a portion of his resources accumulated from the success of PayPal. He went on to say that SpaceX now stays in business providing services in space. They launch everything from communication and Global Positioning System (GPS) satellites to supplies for the International Space Station (ISS) into orbit around Earth. SpaceX makes money through the efficient transportation services they offer. I hope this helps!", + "video_name": "vDwzmJpI4io", + "transcript": "0:00 0:01 0:04 0:05 0:06 0:08 0:10 36:32 0:12 0:14 0:15 0:16 0:20 0:23 0:26 0:27 0:29 0:31 0:34 0:39 0:40 0:42 0:45 26:02 0:49 0:53 0:56 0:57 0:59 1:01 1:04 1:10 1:12 1:14 1:17 1:22 1:27 1:29 1:35 1:36 1:38 1:39 1:39 1:40 1:42 1:44 1:47 1:50 1:52 1:57 2:00 2:03 2:07 2:11 2:13 2:15 2:20 2:21 2:25 2:27 2:30 2:32 2:34 2:36 2:39 2:43 2:46 2:49 2:52 2:54 2:55 3:00 3:02 3:05 3:08 48:11 3:11 3:13 3:16 3:17 3:22 3:23 34:58 3:25 3:27 3:32 3:34 3:35 3:37 3:39 3:44 3:45 3:47 3:49 3:53 3:56 3:57 3:59 4:04 4:06 4:07 4:10 4:13 4:16 34:06 4:17 4:19 4:22 4:25 4:28 4:30 4:32 4:34 4:38 4:40 4:42 4:44 4:48 4:49 4:51 4:55 5:00 5:02 5:04 5:07 5:09 5:11 5:12 5:15 It's been 700,000, Right. SAL KHAN: Super volcano for those of you who don't know. It would envelop, but well-- I know exactly what you're talking about. SAL KHAN: We read the same books. I can tell. ELON MUSK: Absolutely. I mean something bad is bound to happen if you give it enough time. And civilization has been around for such a very short period of time that these time scales seem like very long, but on an evolutionary time scale, they're very short. A million years on an evolutionary time scale is really not very much. And Earth's been around for four and a half billion years, so that's a very tiny, tiny amount of time, But for us that would be-- can you can imagine if human civilization continued at anything remotely like the current pace of technology ad advancement for a million years? Where would we be? I think we're either extinct or on a lot of planets. We should-- SAL KHAN: But given that-- I mean, one, that's kind of as epic as one can think about things, literally. How did you make that concrete? How does that turn into SpaceX, Tesla and Paypal? 6:32 6:35 6:36 6:38 6:40 6:45 6:47 6:49 6:52 6:55 6:58 6:59 7:00 7:02 7:03 7:04 7:05 7:07 7:09 7:12 7:14 7:16 7:18 7:21 7:23 7:25 7:29 7:31 7:37 7:38 7:38 7:40 7:41 25:43 7:44 7:45 7:45 7:48 7:50 7:52 7:55 7:57 7:59 8:02 8:06 And they actually were pretty good. They had like the energy density of a lead-acid battery, which for a capacitor, that's a big deal. But they used ruthenium tantalum oxide. And I think at the time, there was maybe like one or two tons of ruthenium mined per year in the world. So it's not a scalable solution. But I thought there could be some solid-state solution, like just using chip-making equipment. That was going to be the basic idea. But it was one of those things where I wasn't sure if success was one of possible outcomes. 8:43 8:46 8:48 8:50 8:50 8:51 8:52 8:54 8:56 35:06 8:59 9:02 9:06 9:08 9:09 9:11 9:13 9:16 9:22 9:25 9:28 9:32 9:34 9:36 9:39 9:42 9:43 9:46 9:50 9:56 9:58 And there goes seven years of my life. So that was one path. And I was prepared to do that. But then the internet came along. And it was like, oh, OK, the Internet, I'm pretty sure success is one of the outcomes, and it seemed like I could either do a PhD and watch the Internet happen, or I could participate and help build in some fashion. Like, I was just concerned with the idea of watching it happen. So I decided to put things on hold and start an Internet company. And we worked on internet publishing software, maps and directions, yellow pages, those kind of things. And we had as investors and customers the media companies. So like the New York Times Company, Knight Ridder. SAL KHAN: And this is just at the early stages. I mean this was like-- So it's really early stages, so it's really out the gate. Absolutely. And so then we-- the reason we worked with the media companies was because we needed to have money. There was no advertising money in '95. In fact, the idea of advertising on the internet seemed like a ridiculous idea to people. Obviously, not so ridiculous anymore. But, at the time, it seemed like a very unlikely proposition. And a lot of the media companies weren't even sure that they should be online. Like, what's the point of that? SAL KHAN: And did you all think that PayPal was just going to be a simple, little internet way to-- or did you think it was going to turn into the major kind of transaction processing engine that it is right now? ELON MUSK: I didn't expect PayPal's growth rate to be what it was. And that actually created major problems. So we started Paypal on University Avenue. After the first month or so of the website being active, we 100,000 customers. SAL KHAN: Really? Wow, I didn't realize it was-- ELON MUSK: Yeah, it was nutty. SAL KHAN: And how did it start? How did people just even know to use it? I mean, obviously, both buyer and seller have to be involved. ELON MUSK: Yeah. Well, we started off first by offering people $20 if they opened an account. And $20 if they referred anyone. And then we dropped it to $10. And we dropped it to $5. As the network got bigger and bigger, the value of the network itself exceeded any sort of carrot SAL KHAN: So much money did you all spend with that kind of $5, $10, $20 incentive to get that critical mass going? ELON MUSK: It was a fair amount. I think it was probably $60 or $70 million. SAL KHAN: Oh, wow, OK. So it was substantial. OK. So we're not talking peanuts here. ELON MUSK: It depends on your relative scale. 12:34 12:35 12:37 12:37 12:39 12:41 12:44 12:45 12:47 12:49 12:50 12:52 12:55 12:56 12:57 12:57 13:00 39:06 13:05 13:06 It's just like bacteria in a Petri dish. So what you want to do is try to have one customer generate like two customers. Or something like that. Maybe three customers, ideally. And then you want that to happen really fast. And you could probably model it just like bacteria growth in a Petri dish. And then it'll just expand very quickly until it hits the side of the Petri dish and then it slows down. SAL KHAN: And then after Paypal, then I mean-- to some degree, especially us in Silicon Valley, we kind of understand the Internet. We know people. PayPal's obviously of the scale that is noteworthy, but then SpaceX just seems really, you know-- well, one, how did you decide that I'm definitely going to do that? And then like what's the first thing that you do? How do you even go out-- I don't even know how to start trying to make a rocket company. ELON MUSK: Well, neither did I really. And in fact, the first three launches failed. So it's not as though it was like spot on. It's like, did not hit the bull's eye. But you're launching rockets. I don't even how do you get there? One, how did you decide? And then what did you do on day one? Did you write a plan? Did you start-- I don't even know. ELON MUSK: Actually, the origin of SpaceX is that I was trying to figure out why we'd not sent any people to Mars. Because the obvious next step after Apollo was to send people to Mars. But what in fact happened was that we sent a few people to the moon and then we didn't send anyone after that to the moon or Mars or anything. But if you'd asked people in 1969, what would 2013 look like, they would have said, there will be a base on the moon. We would have least sent some people to Mars. And maybe there'd even be a base on Mars. There'd be like orbiting space hotels. And there'd be all this awesome stuff in space. And that's what people expected. And if you said, well, actually, the United States in 2013 will not be able to send anyone to orbit. But I'll tell you what will exist is that there'll be this device in your pocket that's like the size of-- smaller than a deck of cards that has access to all the world's information, and you can talk to any one on planet Earth. And even if you're like in some remote village somewhere so long as there's something called the Internet-- they wouldn't know what that means, of course-- then you would you be able to communicate with anyone instantly and have access to all of humanity's knowledge. They would have said, like bullshit. There's no way that that's going to be true. Right. ELON MUSK: And yet we all have that. And space is not happening. So I was trying to figure out like what was the deal here. And this was 2001. And it was just a friend of mine asked me, what am I going to do after Paypal. And I said, well, you know, I've always been interested in space, but I don't think there's anything that an individual could do in space, because it's the province of government, and usually a large government. But, I am curious as to when we're going to send some one to Mars. So I went to the NASA website to try to figure out where is the place that tells you that. And I couldn't find that. So I was like, either I'm bad at looking at the website, or they have a terrible website, because surely there must be a date. SAL KHAN: That should be a big date. ELON MUSK: Yeah. This should be on the front page. And then I discovered actually that NASA had no plans to send people to Mars, or even really back to the moon. So this was really was disappointing. I thought well, maybe this is a question of national will. Like do we to get people excited about space again? And try to get NASA a bigger budget, and then we would send people to Mars. And so I started researching the area, becoming more familiar with space, reading lots of books. And I came up with this idea to do so-called Mars oasis, which was to send a small greenhouse with seeds in dehydrated gel that upon landing, you hydrate the gel. You have green plants on a red background. The public responses to precedents and superlatives. So it would be the first life on Mars. The furthest that life's ever traveled. And you'd have this money shot of green plants on a red background. So that seemed like it would get people pretty excited. 17:34 17:36 17:39 17:42 17:45 17:47 17:51 17:53 17:54 17:57 17:58 18:04 18:08 18:09 18:13 18:15 18:19 18:20 18:22 18:24 18:27 18:29 18:32 18:37 18:39 18:41 18:42 18:44 And there were just lots of people that thought it was a really crazy idea. And there was some people that had tried to start rocket companies, not succeeded. And they tried to talk me out of it. But the thing is that-- their premise for talking me out of it was, well, we think you're going to lose the money that you invest. I was like, well, that was my expectation anyway, so I don't really mind if I lose-- you I mean, I mind, but I mean it's not like I was trying to figure out the rank-ordered best way to invest money and on that basis chose space. It's not like that's-- I thought, wow-- SAL KHAN: You weren't looking at like money-market bonds, AAA bonds, rocket company. You weren't like-- ELON MUSK: I could do real estate. I could invest in shoe making. Anything. And, whoa, space is the highest ROI. That is not what-- it wasn't the premise. I just thought that it was important that humanity expand beyond Earth, and we weren't doing that, so maybe there was something I could do to spur that on. And then I was able to compress the costs of the spacecraft and everything down to a relatively manageable number. And I got stuck on the rocket. The US rockets were way too expensive. I ended up going to Russia-- I flew to Russia three times to negotiate a purchase of an ICBM. I tried to buy two of the biggest ICBMs in the Russian fleet in 2001 and 2002. And I actually negotiated a price. SAL KHAN: I'll just let that statement stand. I'm not even going to-- Well, actually, I have to-- like who did you call? ELON MUSK: You open the yellow pages. Go to ICBMs. SAL KHAN: How does this-- I don't want to get too much in to it but I'm curious about this one particular thing. You decide at some point you need to buy an ICBM? ELON MUSK: Yeah. Well, actually at first I tried to buy just a normal launch program that they use to launch satellites, but those are too expensive. I see. ELON MUSK: The Boeing Delta II would have cost $65 million each, so two would have been $130 million. And then I was like, woah, OK, that breaks my budget right there. 20:54 20:56 20:57 20:59 21:01 21:02 21:04 21:05 21:07 21:11 21:13 21:18 21:20 21:24 21:29 21:32 21:34 21:37 21:38 21:39 21:45 21:50 21:52 21:53 21:58 22:01 22:03 22:05 22:06 22:08 22:10 22:12 22:16 22:17 22:20 22:22 22:24 22:25 22:28 22:33 22:35 22:38 22:39 22:40 Anyways, so I thought, OK, it's not really going to maybe matter that much if I do this mission, because what really matters is having a way. So I was wrong-- I thought there wasn't enough will, but there actually was plenty of will, if people thought there was a way. So then I said, OK, well, I need to work on the way. How hard is it really to make a rocket? Historically, all rockets have been expensive, so therefore, in the future, all rockets will be expensive. But actually that's not true. If you say, what is a rocket made of. And say, OK, it's made of aluminum, titanium, copper, carbon fiber, if you want to go that direction. And you can break down and say, what is the raw material cost of all these components. And if you have them stacked on the floor and could wave a magic wand so that the cost of rearranging the atoms was zero, then what would the cost of the rocket And I was like, wow, OK, it's really small. It's like 2% of what a rocket costs. So clearly it would be in how the atoms are arranged. 23:50 23:51 23:56 23:59 24:04 24:06 24:08 24:10 24:11 24:13 24:14 24:16 24:18 24:21 24:23 24:27 24:28 24:31 24:33 24:38 24:43 24:45 24:50 24:54 25:00 25:02 25:04 25:06 25:07 But there's an even better step beyond that which is to make rockets reusable. Right now that is around what our comparison price is-- excluding the refurbished ICBMs. So, if you say building a rocket from new, how does the SpaceX rocket compare to a rocket from Boeing or Lockheed? It's about a quarter of the price. 25:34 25:37 25:39 25:41 25:42 For you. SAL KHAN: Only today. Memorial day sale. 25:52 25:55 25:59 26:01 26:04 26:06 26:08 26:10 26:12 26:13 26:15 26:17 26:18 26:18 26:21 26:22 26:24 26:27 We've been working on it for a long time. I should say, SpaceX has been around for 11 years, and thus far we have not recovered any rockets. We recovered the spacecraft from orbit. So that was great. But none of our attempts to recover the rocket stages have been successful. The rocket stages have always blown up essentially on reentry. Now, we think we've figured out why that was the case. And it's a tricky thing, because Earth's gravity is really quite strong. And with an advanced rocket, you can do maybe 2% to 3% of your lift-off mass to orbit, typically. And then reusability subtracts 2% to 3% So then you've got like nothing to orbit or negative. And that's obviously not helpful. And so the trick is to try to shift that from say 2%, 3% in an expendable configuration to make the rocket mass efficiency, engine efficiency, and so forth, so much better that it moves to maybe around 3.5% to 4% in expendable configuration. And then try to get clever about the reusability elements and try to drop that to around the 1.5% to 2% level. So you have a net payload to orbit of about 2%. SAL KHAN: But you're doing it at one, two orders of magnitude cheaper. Absolutely, because our Falcon 9 rocket cost about $60 million. But the propellant cost-- which is mostly oxygen-- it's two-thirds oxygen, one-third fuel-- is only about $200,000. SAL KHAN: Wow. ELON MUSK: And it's much like a 747. It costs about as much to refuel our rocket as it does to refuel a 747 within-- well, pretty close, essentially. SAL KHAN: So assuming you all are successful, and you all have proven yourself to be successful on these audacious things in the past, I mean, what happens? I mean that seems like it's-- what happens in the next 5, 10 years in the space industry, if you all are successful there? I mean do we get to Mars? Do we have kind of market forces, commercialization of space starting to happen? ELON MUSK: Yeah. Let's see. Well, the first step is that we need to earn enough money to keep going as a company. So we have to make sure that we're launching satellites. 28:55 29:00 29:06 29:09 29:09 29:11 29:13 29:15 29:16 29:20 29:26 29:29 29:34 29:39 29:42 29:45 29:47 29:49 29:51 29:52 29:53 29:56 29:59 30:01 It's not for sure. SAL KHAN: I could talk about this for-- people know, I'm-- ELON MUSK: Aspirational it'd be a round trip. SAL KHAN: This is mind blowing. And then on Tesla. I mean Tesla's obviously, from my vantage, it's a huge success. What do you think in that industry-- well, one, I'll ask kind of the same question. What did you think-- this is something that GM and Toyota and these massive multi-billion dollar organizations have been trying. What gave you the confidence to pursue it? And now that it seems to be a huge success, where do you think this industry's going to be the next 5, 10 years? ELON MUSK: Yes. So with Tesla, the goal is try to accelerate the advent of sustainable transport. I think it would happen anyway, just out of necessity. But because we have an un-priced externality in the cost of gasoline. We weren't pricing in the environmental effects of CO2 in the oceans and atmosphere. That's causing the normal market forces to not function properly. And so the goal of Tesla is to try to act as a catalyst to accelerate those sort of normal forces. The normal sort of market reaction that would occur. We're trying to have a catalytic effect on that. And try to make it happen, I don't know, maybe 10 years sooner than it would otherwise occur. That's the goal of Tesla. So that's the reason we're making electric cars and not any other kind of car. And we also supply power trains to Toyota and to Mercedes and maybe to other car companies in the future to accelerate their production of electric vehicles. So that's the goal there. And so far, it's working out pretty well. SAL KHAN: I mean, I just saw a news report earlier today that you all sold more Model S's than-- you all are leading that segment of the industry. The Mercedes S class, the BMW 7 Series, or the Lexus LS400, or whatever it is. ELON MUSK: Yeah, actually, that seems to be the case. I didn't realize they sold so few cars in that segment. 32:16 32:17 32:19 32:22 32:26 32:29 32:31 32:34 32:35 32:37 32:38 32:42 32:45 32:47 32:49 32:53 32:59 33:01 33:05 33:08 33:13 33:15 33:17 33:18 33:19 33:22 33:23 33:27 33:28 33:31 33:33 33:38 33:42 33:44 33:47 33:49 33:54 34:02 34:05 34:10 34:12 34:12 34:13 34:15 The nature of new technology adoption is it tends to follow an S-curve. So what usually happens is people under-predict it in in the beginning, because people tend to extrapolate in a straight line. And then they'll over-predict it at the midpoint, because there's late adopters. And then it'll actually take longer than people think at the mid-point, but much shorter than people think at the beginning. 34:44 34:47 34:51 34:55 34:57 35:00 35:01 35:02 35:03 35:03 35:05 35:08 35:09 35:11 35:12 35:14 35:16 35:17 ELON MUSK: Yeah. It seems like you're doing an amazing job of-- really super leveraged. I mean, obviously, a small team, and you're having a dramatic effect-- SAL KHAN: Yeah, half these people don't even work here. There just like-- so it's like it's even-- ELON MUSK: Right. So it's, I think very impressive thing you're doing to spread knowledge and understanding throughout the world. SAL KHAN: The universe soon, if you hold up your end of the bargain. ELON MUSK: It's actually kind of funny. If you think, what is education? Like you're basically downloading data and algorithms into your brain. And it's actually amazingly bad in conventional education. Because like it shouldn't be like this huge chore. So you're making it way, way better. 36:10 36:12 36:16 36:19 36:23 36:25 36:27 36:28 36:32 36:35 36:37 36:38 ELON MUSK: So to the degree that you can make somehow learning like a game, then it's better. And I think, unfortunately, a lot of education is very vaudevillian. You've got someone standing up there kind of lecturing at people. And they've done the same lecture 20 years in a row, and they're not very excited about it. And that lack of enthusiasm is conveyed to the students. They're not very excited about it. They don't know why they're there. Like why are we letting this stuff. We don't even know why. In fact, I think a lot of things that people learn that probably there's no point in learning them. Because they never use them in the future. SAL KHAN: Because who's going to launch a rocket into space? I mean, that's just like-- exactly, that never happens. ELON MUSK: Well, you have to say-- people don't stand back and say, well, why are we teaching people these things. And we should tell them, probably, why we're teaching these things. Because a lot of kids are probably just in school, probably puzzled as to why they're there. 37:40 37:43 37:44 37:46 37:48 So I think that's pretty important. And just make it entertaining. But I think just in general conventional education should be massively overhauled. And I'm sure you pretty much agree with that. I mean the analogy I sometimes use is, have you seen like Batman, the Chris Nolan movie, the recent one. And it's pretty freaking awesome. And you've got incredible special effects, great script, multiple takes, amazing actors, and great sound, and it's very engaging. But if you were to instead say, OK-- even if you had the same script, so at least it's same script. And you said, OK, now that script, instead of having movies, we're going to have that script performed by the local town troop. OK, and so in every small town in America, if movies didn't exist, they'd have to recreate The Dark Night. With like home-sewn costumes and like jumping across the stage. And not really getting their lines quite right. And not really looking like the people in the movie. And no special effects. And I mean that would suck. 39:02 39:03 39:06 39:07 SAL KHAN: So with that-- and I apologize to all of you guys for hogging up all of the time, because, obviously, I could talk for hours about this stuff. But we do have time, probably 5 or 10 minutes for a handful of questions. If none of you all have any, I have about nine more. But, yes. SPEAKER 1: So I noticed-- I picked up two kind of themes from what you were discussing. One was somewhat audacious goals. And the other was I don't think I heard you use the word profit in anything that you spoke about. You seem to be-- each thing is pointed at like re-invigorating an industry or bringing back space missions. How much of your success do you attribute to having really audacious goals or versus just not being focused on the short term, money coming in, or I don't know, investors? ELON MUSK: Unfortunately, one does have to be focused on the short time and money coming in when creating a company, because otherwise the company will die. So I think that a lot of times people think like creating company is going to be fun. I would say it's really not that fun. I mean there are periods of fun. And there are periods where it's just awful. And, particularly, if you're the CEO of the company, you actually have a distillation of all the worst problems in the company. There's no point in spending your time on things that are going right. So you're only spending your time on things that are going wrong. And there are things that are going wrong that other people can't take care of. So you have like the worst-- you have a filter for the crappest problems in the company. The most pernicious and painful problem. 40:51 40:57 40:59 41:01 41:03 41:06 41:10 41:11 41:13 41:17 41:19 41:25 41:28 41:32 41:37 41:41 41:44 41:46 41:48 41:53 41:55 41:59 42:00 42:03 42:05 42:06 ELON MUSK: Well, it's just a very small percentage of mental energy is on the big picture. Like you know where you're generally heading for and the actual path is going to be some sort of zigzaggy thing in that direction. You're trying not to deviate too far from the path that you want to be on, but you're going to have to that to some degree. 42:33 42:35 42:39 42:40 42:42 42:45 42:47 42:50 42:53 42:56 42:58 43:02 43:05 43:09 43:12 43:16 43:17 SAL KHAN: I think we have time for one more question. Joel. JOEL: Yeah, I have an important one. SAL KHAN: OK, very good. Yes, please. SPEAKER 3: No. JOEL: OK, so few months ago, you teased Hyperloop, and we haven't heard anything since. So, first of all, a few of us engineers were talking about it, and I think we have a few ideas, if you need help. But, if you feel comfortable, maybe you could tell us a little bit more. ELON MUSK: I was reading about the California high-speed rail, and it was quite depressing. Because California taxpayers are going to be on the hook to build the most expensive high-speed rail per mile in the world-- and the slowest. 44:03 44:05 And, it's like, damn, we're in California, we make super high-tech stuff. Why are we going to be spending-- now the estimates are around $100 billion-- for something that will take two hours to go from LA to San Francisco? I'm like, OK, well, I can get on a plane and do that it 45 minutes. It doesn't make much sense. And isn't there some better way to do it than that. So if you just say, OK, well what would you ideally want in a transportation system? You'd say, OK, well you'd want something that relative to existing modes of transportation is faster-- let's say twice as fast-- costs half as much per ticket, can't crash, is immune to weather, and is-- you can make the whole thing like self-powering with like solar panels or something like that. That would be pretty-- SAL KHAN: That would be great, yes. ELON MUSK: --a good outcome. And so what would do that? And what's the fastest way short of inventing teleportation that you could do something like that? And some of the elements of that solution are fairly obvious, and some of them are not so obvious. And then the details-- the devil's in the details of actually making something like that work. But I came to the conclusion that there is something like that that could work. And would be practical. SAL KHAN: Is this around the evacuated tubes? The vacuum tubes? Like the old bank-- ELON MUSK: It's something like that. SAL KHAN: But you haven't been more public with what this is? ELON MUSK: No. Although I did say that once Tesla was profitable that I would talk more about it. But, we haven't done our earnings call yet. So I think I'll probably do it after the earnings call. And the thing is I'm kind of strung out on things that I'm already doing. So adding another thing-- it's like doesn't-- it's a lot SAL KHAN: Learning the guitar You could pick up all sorts of things. ELON MUSK: Right. I tried learning the violin. That's, by the way, a hard thing to learn. SAL KHAN: Yeah. Launching rockets, electric cars, revolutionizing transportation. Yeah, it's easy. ELON MUSK: I cannot play the violin at all. Very horrible. If you think about the future, you want a future that's better than the past, and so if we had something like the Hyperloop, I think that would be like cool. You'd look forward to the day that was working. And if something like that, even if it was only in one place-- from LA to San Francisco, or New York to DC or something like that-- then it would be cool enough that it would be like a tourist attraction. It would be like a ride or something. So even if some of the initial assumptions didn't work out, the economics didn't work out quite as one expected, it would be cool enough that like, I want to journey to that place just to ride on that thing. That would be pretty cool. And so that's I think how-- if you come with a new technology, it should feel like that. You should really-- if you told it to an objective person, would they look forward to the day that that thing became available. And it would be pretty exciting to do something like that. Or an aircraft. Like I thought it was really disappointing when the Concorde was taking out of commission, and there was no supersonic transport available. And of course the 787 has had some issues. 47:37 47:40 47:44 47:46 47:49 47:51 47:54 47:57 47:58 48:00 48:04 48:05 48:08 48:09 48:14 48:17 48:20 48:22 48:24 48:25 48:28 48:29 48:31 48:33 48:34 48:34 48:35" + }, + { + "Q": "why is the variable always X?", + "A": "It s not always x. But x is the most common one to use in simple problems.", + "video_name": "Ye13MIPv6n0", + "transcript": "So we have our scale again. And we've got some masses on the left hand side and some masses on the right hand side. And we see that our scale is balanced. We have the same total mass on the left hand side that we have on the right hand side. Instead of labeling the mystery masses as question mark, I've labeled them all x. And since they all have an x on it, we know that each of these have the same mass. But what I'm curious about is, what is that mass? What is the mass of each of these mystery masses, I guess we could say? And so I'll let think about that for a second. How would you figure out what this x value actually is? How many kilograms is the mass of each of these things? What could you do to either one or both sides of this scale? I'll give you a few seconds to think about that. So you might be tempted to say, well if I could end up with just one mystery mass on the left hand side, and if I keep my scale balanced, then that thing's going to be equal to whatever I have on the right hand side. And that part would actually be a true statement. But then to get only one of these mystery masses on the left hand side, you might say, well why don't I just remove two of them? You might just say, well why don't I just remove-- let me do it a good color for removing-- why don't I just remove that one and that one? And then I'll just be left with that right over there. But if you just removed these two, then the left hand side is going to become lighter or it's going to have a lower mass than the right hand side. So it's going to move up and the right hand side is going to move down. And then you might say, OK, I understand. Whatever I have to do to the left hand side, I have to do to the right hand side in order to keep my scale balanced. So you might say, well why don't I remove two of these mystery masses from the right hand side? But that's a problem too because you don't know what this mystery mass is. You could try to remove two from this, but how many of these blocks represent a mystery mass? We actually don't know. But you might then say, well let's see, I've got three of these things here. If I essentially multiply what I have here by 1/3 or if I only leave a 1/3 of the stuff here, and if I only leave a 1/3 of the stuff here, then the scale should be balanced. If this has the total mass as this, then 1/3 of this total mass is going to be the same thing as 1/3 of that total mass. So let's just keep only 1/3 of this here. So that's the equivalent to multiplying by 1/3. So if we're only going to keep 1/3 there, we're going to be left with only one of the masses. And if we only keep 1/3 here, let's see, we have one, two, three, four, five, six, seven, eight, nine masses. If we multiply this by 1/3, or if we only keep 1/3 of it there, 1/3 times 9 is 3. So we're going to remove these . And so we have 1/3 of what we originally had on the right hand side and 1/3 of what we originally had on the left hand side. And they will be balanced because we took 1/3 of the same total masses. And so what you're left with is just one of these mystery masses, this x thing right over here, whatever x might be. And you have three kilograms on the right hand side. And so you can make the conclusion, and the whole time you kept this thing balanced, that x is equal to 3." + }, + { + "Q": "Question:\nwhat happens if the show people pulled a trick and didn't paint anyone blue? How would they figure that out?", + "A": "Because they said that at least one person s forehead was painted blue", + "video_name": "-xYkTJFbuM0", + "transcript": "So we had the hundred logicians. All of their foreheads were painted blue. And before they entered the room, they were told that at least one of you hundred logicians has your forehead painted blue. And then every time that they turned on the lights, so that they could see each other, they said OK, once you've determined that you have a blue forehead, when the lights get turned off again, we want you to leave the room. And then once that's kind of settled down, they'll turn the lights on again. And people will look at each other again. And then they'll turn them off again. And maybe people will leave the room. And so forth and so on. And they're also all told that everyone in the room is a perfect logician. They have infallible logic. So the question was, what happens? And actually maybe an even more interesting question is why does it happen? So I'll answer the first, what happens? And if just take the answer, and you don't know why, it almost seems mystical. That essentially the light gets turned on and off 100 times, and then after the hundredth time that the light gets turned on, and the lights get turned off again, all of them leave. They all leave. So I mean, it's kind of weird, right? Let's say I'm one of them. Or you're one of them. I go into this room. The lights get turned on. And I see 99 people with blue foreheads. And I can't see my own forehead. They see my forehead, of course. But to any other person, I'm one of the 99, right? But I see 99 blue foreheads. So essentially what happens if we were to watch the show is, the lights get turned on. You see 99 blue foreheads. Then the lights get turned off again. And then the lights get turned on again. And everyone's still sitting there. And I still see 99 blue foreheads. And that happens 100 times. And after the hundredth time the light gets turned on, everyone leaves the room. And at first glance, that seems crazy, because nothing changes. Nothing changes between every time we turn on the light. But the way you need to think about this-- and this is what makes it interesting-- is what happens instead of 100, let's say there was one person in the room. So before the show starts-- they never told me that there were going to be 100 people in the room. They just said, at least one of you, at least one of the people in the room, has your forehead painted blue. And as soon as you know that your forehead is painted blue, you leave the room. And that everyone's a perfect logician. So imagine the situation where instead of 100 there's only one perfect logician. Let's say it's me. So that's the room. I walk in. And I sit down. And maybe I should do it with blue. And then they turn the lights on, and say, look around the room. And I look around the room, and I see nobody else, right? And remember, even in the case of one, we've painted everyone's forehead blue. So in this case, this one dude, or me or whoever you want to call him. His forehead is painted blue. So he looks around and he sees no one in the room. But he remembers the statement, and maybe it's even written down on a card for him in case he forgets. That at least one of you has your forehead painted blue. So if he looks around the room and he says, well I'm the only dude in the room. And they told me that at least one of the dudes in the room is going to have their foreheads blue. Well, I'm the only dude in the room. So I must have a blue forehead. So as soon as they turn the lights off, he's going to leave. Fair enough. That's almost trivially simple. And you might say, so how does this apply to 100? Well what happens when there are two people. And once again, both of them have their foreheads painted blue. So let me draw another. I don't want to keep drawing the blue forehead room. Let's say there's two people now. So let's put ourselves in the head of this guy. Right behind the blue forehead. That's where we're sitting. So when he enters the room. He says, I either have a blue forehead. I either have a blue forehead, or I don't have a blue forehead. No blue. Right? This is what this guy's thinking. Let me draw him. And he has a blue forehead. But he doesn't know it. He can't see it. That's the whole point about painting the forehead blue, as opposed to another part of the body. So he says, I either have a blue forehead or I don't have a blue forehead. He walks in. Let's say this is this guy. He walks in. The first time the lights get turned on, he sees this other dude there who has a blue forehead. And he says OK, now let me think about it. How will this guy respond depending on each of these states? So let's say that I don't have a blue forehead. Let's go into this reality. If I don't have a blue forehead, what is this guy going to see? When the lights get turned on, he's going to see that I don't have a blue forehead. And we were both told that at least one of us has a blue forehead. So this guy, because he's a perfect logician, will deduce that he has to have a blue forehead. Remember, this is in a situation, if I assume that I don't have a blue forehead. We're in this world. I'm a perfect logician, so if I can assume, if I'm simulating the reality where I don't have a blue forehead, then this guy will see, I don't have a blue forehead. And then he'll say, I must have a blue forehead. And so when the lights get turned off, this guy will leave. He'll exit the room. And vice versa. The other guy will make the same logic. But since both of them have blue foreheads, what happens? The lights get turned off, then the lights get turned back on. When the light gets turned back on, this guy's still sitting here. And I just determined that if I didn't have a blue forehead-- and this is me-- this guy would have left. Because he could've said, oh I must be the only guy with a blue forehead. So he would have left. But since he didn't leave, I now know that I have a blue forehead. So then the second time that the lights get turned on, I can deduce that I have a blue forehead. And then when the lights get turned off again, I'll leave. And this guy, he's a perfect logician, so he makes the exact same conclusion. Because he also simulated in his head, OK if I didn't have a blue forehead, then this guy will leave as soon as the lights get turned off the first time. If this guy doesn't have a blue forehead, then this guy will say, well I see no-one else in the room with a blue forehead, and since I know that there's at least one with a blue forehead, I'll have to leave. But since he didn't leave, this guy will also know that he must have a blue forehead as well, so they'll actually leave together. Maybe they'll bump into each other on the way out. Fair enough. Now what happens if you extend it to three people? So we already said, if you have one person in the room, he'll come to the conclusion the first time that the lights And then he'll leave right when they're turned off. If you have two people, it takes them essentially two times for the light to get turned on to reach that conclusion. Now if you have three people, and I think you see where this is going. One, two and three. Now remember, no one knows if their foreheads are painted blue, but the producers of the show actually did paint everyone's forehead blue. So so once again, let's get into this guy's head. So this guy says, he's either blue or he's not blue. So in the reality when he's not blue, what's going to happen? Well, this guy-- and this gets a little bit confusing, but if you think about it from the previous example, it makes a lot of sense. A person who has a not-blue forehead actually shouldn't affect the outcome of what the blue people do. Because let's say that this guy says, well what's going to happen if I'm not blue? Well, this guy's going to look at that guy, and say, oh he has a blue forehead. He doesn't have a blue forehead. So if I don't have a blue forehead, this guy's going to see two people without blue foreheads. And he's going to leave the room the first time that the lights are turned on. He'll come to the conclusion. Now the second time that the light's turned on, this guy will say, gee, this guy didn't leave the room. That guy doesn't have a blue forehead. And this guy didn't leave the room because he must have seen someone with a blue forehead. Therefore I must have a blue forehead. And so, if this guy doesn't have a blue forehead, both of these guys would leave the room the second time that the light gets turned on. Now what happens if they don't leave the room the second time that the lights are turned on? Well if I was not blue, they would have left. So if they haven't left by the third showing of the light, then I know that I'm blue. So when you have three people, they're all perfect logicians, they all have their foreheads painted blue. The light will be shown three times. Or the light will be turned on three times. And then when the light gets turned off, they'll all leave together. And so this logic applies for any. You could have 1,000 people. You can keep extending it. The fourth person will have the exact same logic. If he's not blue, then these guys are going to leave after three turnings on of the light. But if they don't leave after three turnings on of the light, then he must be blue. And so they're all going to leave together. All four of them on the fourth showing of everyone's foreheads. Anyway, and you can keep extending this all the way to 100. And 100 is arbitrary. You could do this with a million people, and they would just keep looking at each other a million times. And then on the millionth showing, they would all reach the conclusion that they all have blue foreheads, and they would leave the room. And if you think about it, it's fairly straightforward logic. But it leads to kind of a very almost eerie result. Hopefully that satisfies you. See you in the next video." + }, + { + "Q": "solve the equation by extracting square root (x+6)^2 = 5", + "A": "(x + 6)\u00c2\u00b2 = 5 We take the square root of both sides: x + 6 = \u00c2\u00b1\u00e2\u0088\u009a5 Subtract 6 from both sides: x = (\u00c2\u00b1\u00e2\u0088\u009a5) - 6 So: x = -6 - (\u00e2\u0088\u009a5) AND x = (\u00e2\u0088\u009a5) + 6", + "video_name": "55G8037gsKY", + "transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? Then the left-hand side of the equation becomes 4x plus 1 squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. So we get 4x plus 1 is equal to-- we can factor out the 4, or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously. So if I subtract 1 from both sides of this equation, what do I have? On the left-hand side, I'm just left with 4x. On the right-hand side, I have-- you can't really mathematically, I mean, you could do them if you had a calculator, but I'll just leave it as negative 1 plus or minus the square root, or 2 times the square root of 2. That's what 4x is equal to. If we did it here, as two separate equations, same idea. Subtract 1 from both sides of this equation, you get 4x is equal to negative 1 plus 2, times the square root of 2. This equation, subtract 1 from both sides. 4x is equal to negative 1 minus 2, times the square root of 2. This statement right here is completely equivalent to these two statements. Now, last step, we just have to divide both sides by 4, so you divide both sides by 4, and you get x is equal to negative 1 plus or minus 2, times the square root of 2, over 4. Now, this statement is completely equivalent to dividing each of these by 4, and you get x is equal to negative 1 plus 2, times the square root 2, over 4. This is one solution. And then the other solution is x is equal to negative 1 minus 2 roots of 2, all of that over 4. That statement and these two statements are equivalent. And if you want, I encourage you to-- let's substitute one of these back in, just so you feel confident that something as bizarro as one of these expressions can be a solution to a nice, vanilla-looking equation like this. So let's substitute it back in. 4 times x, or 4 times negative 1, plus 2 root 2, over 4, plus 1 squared, minus 8 is equal to 0. Now, these 4's cancel out, so you're left with negative 1 plus 2 roots 2, plus 1, squared, minus 8 is equal to 0. This negative 1 and this positive 1 cancel out, so you're left with 2 roots of 2 squared, minus 8 is equal to 0. And then what are you going to have here? So when you square this, you get 4 times 2, minus 8 is equal to 0, which is true. 8 minus 8 is equal to 0. And if you try this one out, you're going to get the exact same answer. Let's do another one like this. And remember, these are special forms where we have squares of binomials in our expression. And we're going to see that the entire quadratic formula is actually derived from a notion like this, because you can actually turn any, you can turn any, quadratic equation into a perfect square equalling something else. We'll see that two videos from now. But let's get a little warmed up just seeing this type of thing. So let's say you have x squared minus 10x, plus 25 is equal to 9. Now, once again your temptation-- and it's not a bad temptation-- would be to subtract 9 from both sides, so you get a 0 on the right-hand side, but before you do that, just inspect this really fast. And say, hey, is this just maybe a perfect square of a binomial? And you see-- well, what two numbers when I multiply them I get positive 25, and when I add them I get negative 10? And hopefully negative 5 jumps out at you. So this expression right here is x minus 5, times x minus 5. So this left-hand side can be written as x minus 5 squared, and the right-hand side is still 9. And I want to really emphasize. I don't want this to ruin all of the training you've gotten on factoring so far. We can only do this when this is a perfect square. If you got, like, x minus 3, times x plus 4, and that would be equal to 9, that would be a dead end. You wouldn't be able to really do anything constructive with that. Only because this is a perfect square, can we now say x minus 5 squared is equal to 9, and now we can take the square root of both sides. So we could say that x minus 5 is equal to plus or minus 3. Add 5 to both sides of this equation, you get x is equal to 5 plus or minus 3, or x is equal to-- what's 5 plus 3? Well, x could be 8 or x could be equal to 5 minus 3, or x is equal to 2. Now, we could have done this equation, this quadratic equation, the traditional way, the way you were tempted to do it. What happens if you subtract 9 from both sides of this equation? You'll get x squared minus 10x. And what's 25 minus 9? 25 minus 9 is 16, and that would be equal to 0. And here, this would be just a traditional factoring problem, the type that we've seen in the last few videos. What two numbers, when you take their product, you get positive 16, and when you sum them you get negative 10? And maybe negative 8 and negative 2 jump into your brain. So we get x minus 8, times x minus 2 is equal to 0. And so x could be equal to 8 or x could be equal to 2. That's the fun thing about algebra: you can do things in two completely different ways, but as long as you do them in algebraically-valid ways, you're not going to get different answers. And on some level, if you recognize this, this is easier because you didn't have to do that little game in your head, in terms of, oh, what two numbers, when you multiply them you get 16, and when you add them you get negative 10? Here, you just said, OK, this is x minus 5-- oh, I guess you did have to do it. You had to say, oh, 5 times 5 is 25, and negative 10 is negative 5 plus negative 5. So I take that back, you still have to do that little game in your head. So let's do another one. Let's do one more of these, just to really get ourselves nice and warmed up here. So, let's say we have x squared plus 18x, plus 81 is equal to 1. So once again, we can do it in two ways. We could subtract 1 from both sides, or we could recognize that this is x plus 9, times x plus 9. This right here, 9 times 9 is 81, 9 plus 9 is 18. So we can write our equation as x plus 9 squared is equal to 1. Take the square root of both sides, you get x plus 9 is equal to plus or minus the square root of 1, which is just 1. So x is equal to-- subtract 9 from both sides-- negative 9 plus or minus 1. And that means that x could be equal to-- negative 9 plus 1 is negative 8, or x could be equal to-- negative 9 minus 1, which is negative 10. And once again, you could have done this the traditional way. You could have subtracted 1 from both sides and you would have gotten x squared plus 18x, plus 80 is equal to 0. And you'd say, hey, gee, 8 times 10 is 80, 8 plus 10 is 18, so you get x plus 8, times x plus 10 is equal to 0. And then you'd get x could be equal to negative 8, or x could be equal to negative 10. That was good warm up. Now, I think we're ready to tackle completing the square." + }, + { + "Q": "What is a model?", + "A": "a diagram or a 2d picture", + "video_name": "Rd4a1X3B61w", + "transcript": "Here some picture of what most people associate when they think of chemistry. They think of scientists working on a bench with the different vials of different chemicals. They might think of a mad scientist. Some of them boiling and changing colors. They might associate chemistry with chemical equations. Thinking about how different things will react together to form other things. They might think about models of the different molecules that can be depicted different ways. They might associate it with the periodic table of elements. And all of these things are a big part of chemistry. But I want you to do in this video is appreciate what at its essence chemistry is all about. And chemistry is one of the sciences that really just helps us understand and make models and make predictions about our reality. And even something like the periodic table of elements, which you'll see at the front of any chemistry classroom, you take it for granted. But this is the product of, frankly, thousands of years of human beings trying to get to an understanding of all of the different complexity in the world. If you look at the world around us, and it doesn't even have to be our planet, it could be the universe around us, you see all these different substances that seem to be different in certain ways. You see things like fire and rock and water. Even in the planets, you see meteorological patterns. In life, you see all of this complexity and all of these different things and it looks like there's just like a infinite spectrum of differentness out of there. Of different substances. Even in things like our human brain. The complexity and the electrochemical interactions. And you could imagine as a species, this is kind of overwhelming. How do you make the sense of all of this? And it was not an easy path, but over thousands of years, we did start to make sense of it. And why it's very lucky for all of us to be born when we are now or to be around when we are now. To be able to learn chemistry where we are now is that we get the answer. And it's a partial answer, which is also exciting, cause we don't want the full answer. But it's a partial answer that takes us a long way. We realize that the periodic table of elements, that all of this complexity that we're seeing before, that at the end of the day, things are made of basic building blocks. Kind of you could imagine the legos that really make up everything. And there aren't an infinite number of legos. There's actually a finite number of them. We're discovering more all of the time, well not all of the time, now new elements are not discovered that frequently, but there's a few of these elements that are disproportionately showing up in a lot of what we see here. These things that seem so different. Well a lot of this is different compositions of elements like carbon and oxygen and hydrogen. And even the elements themselves are made of things like protons and electrons and neutrons that are just rearranged in different ways to give us these elements that have all of these different properties. So when you think about chemistry, yes, it might visually look something like this. These are obviously much older pictures. But at its essence, it's how do we create models and understand the models that describe a lot of the complexity in the universe around us? And just to put chemistry in, I guess you could say, in context with some of the other sciences, many people would say at the purest level, you would have mathematics. That math, you're studying ideas, which could even be independent, you're seeing logical ideas that could be even independent of anything that you've ever observed or experienced. And a lot of folks that say if we ever communicate with another intelligent species that could be completely different than us, math might be that common language. Because even if we perceive the world differently and think differently in certain ways, math might be that common language. But on top of math, we start to say, well how is our reality actually structured? At the most basic level, what are the constituents of matter and what are the mathematical properties that describe how they react together? And then, or interact with each other? Then you go one level above that, you get to the topic of this video, which is chemistry. Which is very closely related to physics. When we talk about these chemical equations and we create these molecular structures, the interactions between these atoms, these are quantum mechanical interactions which we do not fully understand at the deepest level yet. But with chemistry, we can start to make use of the math and they physics to start to think about how all of these different building blocks can interact to explain all sorts of different phenomena. This chemical equation you see right here, this is combustion. This is hydrogen combusting with oxygen to produce a lot of energy. To produce energy. When we imagine combustion, we think of fire. But what even is fire at its most fundamental level? How do we get, why do we perceive this thing here? And chemistry is super important because on top of that, we build biology. We build biology. And as you'll see as you study all of these things, there's points where these things start to bleed together. But the biology in, say, a human being, or really in any species, it's based on molecular interactions. Interactions between molecules, between atoms, which, at the end of the day, is all about chemistry. As I speak, the only reason why I'm able to speak is because of really, hard to imagine the number of chemical interactions happening in me right now to create this soundness. To create this thing that thinks it exists that wants to make a video about how awesome and amazing chemistry is. And then from biology, you can build out on all of everything else. So sciences like psychology and economics, which of course, these things also leverage math and other things. But this gives you kind of a sense of how we build up and how we explain the reality around us. And not one of these is more important than the other. These are all studying incredibly fascinating things that as humans beings first became thoughtful about their environment, said, \"Gee, why are we here? \"What is this place? \"Why do we exist? \"How do we exist?\" And chemistry builds models for us to understand interactions at a scale and a speed that we can't directly observe, but nonetheless, we can to start to make predictions. So that's what's really cool about this. When you study chemistry, you should not view this as some type of a chore that the school system is forcing you through. There are people who would've done anything 100 years ago to get the answers that are in your chemistry book today or that you can learn from your chemistry teacher or that you can learn from a Khan Academy video. There are people in the world in the past and today who'd do anything to be able to understand deeply what this is. That they consider it a privilege to be able to learn at this level. And then to think about where this could go because none of these fields are complete. We have very partial knowledge of all of these fields. Arguably, there's an infinite more that we could learn relative to what we know. But what's exciting is that we have such a strong start. We're starting to make sense of it. To really describe everything in our reality." + }, + { + "Q": "does anybody know how many numbers are in pie?", + "A": "Actually, pie is irrational as well as transcendental, which means that it has a infinite number of digits. However many people are racing to see who can find the most digits of pi.", + "video_name": "ZyOhRgnFmIY", + "transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? well, you might remember that the radius is half of the diameter. Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. You might well say, that what number it is close to. I want a decimal representation of this. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi. In which case, we will say 64 times, and than we have to look for the pi on the calculator, it's up here this yellow, so I'll use the 2nd function to get the pi there. Now, we are using the calculator's internal representation of pi which is going to be more precise than what I had in the last one. And you can 201.06 (to the nearest hundred) So, more precise is 201.06 square millimeters. So, this is closer to the actual answer, because the calculator's representation is more precise than this very rough approximation of what pi is." + }, + { + "Q": "Where would fascism fit in on the graph?", + "A": "The Middle of the graph", + "video_name": "MmRgMAZyYN0", + "transcript": "Thought I would do a video on communism just because I've been talking about it a bunch in the history videos, and I haven't given you a good definition of what it means, or a good understanding of what it means. And to understand communism-- let me just draw a spectrum here. So I'm going to start with capitalism. And this is really just going to be an overview. People can do a whole PhD thesis on this type of thing. Capitalism, and then I'll get a little bit more-- and then we could progress to socialism. And then we can go to communism. And the modern versions of communism are really kind of the brainchild of Karl Marx and Vladimir Lenin. Karl Marx was a German philosopher in the 1800s, who, in his Communist Manifesto and other writings, kind of created the philosophical underpinnings for communism. And Vladimir Lenin, who led the Bolshevik Revolution in the-- and created, essentially, the Soviet Union-- he's the first person to make some of Karl Marx's ideas more concrete. And really every nation or every country which we view as communist has really followed the pattern of Vladimir Lenin. And we'll talk about that in a second. But first, let's talk about the philosophical differences between these things, and how you would move. And Karl Marx himself viewed communism as kind of a progression from capitalism through socialism to communism. So what he saw in capitalism-- and at least this part of what he saw was right-- is that you have private property, private ownership of land. That's the main aspect of capitalism. And this is the world that most of us live in today. The problem that he saw with capitalism is he thought, well, look, when you have private property, the people who start accumulating some capital-- and when we talk about capital, we could be talking about land, we could be talking about factories, we could be talking about any type of natural resources-- so the people who start getting a little bit of them-- so let me draw a little diagram here. So let's say someone has a little bit of capital. And that capital could be a factory, or it could be land. So let me write it. Capital. And let's just say it's land. So let's say someone starts to own a little bit of land. And he owns more than everyone else. So then you just have a bunch of other people who don't own land. But they need, essentially-- and since this guy owns all the land, they've got to work on this guy's land. They have to work on this guy's land. And from Karl Marx's point of view, he said, look, you have all of these laborers who don't have as much capital. This guy has this capital. And so he can make these laborers work for a very small wage. And so any excess profits that come out from this arrangement, the owner of the capital will be able to get it. Because these laborers won't be able to get their wages to go up. Because there's so much competition for them to work on this guy's farm or to work on this guy's land. He really didn't think too much about, well, maybe the competition could go the other way. Maybe you could have a reality eventually where you have a bunch of people with reasonable amounts of capital, and you have a bunch of laborers. And the bunch of people would compete for the laborers, and maybe the laborers could make their wages go up, and they could eventually accumulate their own capital. They could eventually start their own small businesses. So he really didn't think about this reality too much over here. He just saw this reality. And to his defense-- and I don't want to get in the habit of defending Karl Marx too much-- to his defense, this is what was happening in the late 1800s, especially-- we have the Industrial Revolution. Even in the United States, you did have kind of-- Mark Twain called it the Gilded Age. You have these industrialists who did accumulate huge amounts of capital. They really did have a lot of the leverage relative to the laborers. And so what Karl Marx says, well, look, if the guy with all the capital has all the leverage, and this whole arrangement makes some profits, he's going to be able to keep the profits. Because he can keep all of these dudes' wages low. And so what's going to happen is that the guy with the capital is just going to end up with more capital. And he's going to have even more leverage. And he'll be able to keep these people on kind of a basic wage, so that they can never acquire capital for themselves. So in Karl Marx's point of view, the natural progression would be for these people to start organizing. So these people maybe start organizing into unions. So they could collectively tell the person who owns the land or the factory, no, we're not going to work, or we're going to go on strike unless you increase our wages, or unless you give us better working conditions. So when you start talking about this unionization stuff, you're starting to move in the direction of socialism. The other element of moving in the direction of socialism is that Karl Marx didn't like this kind of high concentration-- or this is socialists in general, I should say-- didn't like this high concentration of wealth. That you have this reality of not only do you have these people who could accumulate all of this wealth-- and maybe, to some degree, they were able to accumulate it because they were innovative, or they were good managers of land, or whatever, although the Marxists don't give a lot of credit to the owners of capital. They don't really give a lot of credit to saying maybe they did have some skill in managing some type of an operation. But the other problem is is that it gets handed over. It gets handed over to their offspring. So private property, you have this situation where it just goes from maybe father to son, or from parent to a child. And so it's not even based on any type of meritocracy. It's really just based on this inherited wealth. And this is a problem that definitely happened in Europe. When you go back to the French Revolution, you have generation after generation of nobility, regardless of how incompetent each generation would be, they just had so much wealth that they were essentially in control of everything. And you had a bunch of people with no wealth having to work for them. And when you have that type of wealth disparity, it does lead to revolutions. So another principle of moving in the socialist direction is kind of a redistribution of wealth. So let me write it over here. So redistribution. So in socialism, you can still have private property. But the government takes a bigger role. So you have-- let me write this. Larger government. And one of the roles of the government is to redistribute wealth. And the government also starts having control of the major factors of production. So maybe the utilities, maybe some of the large factories that do major things, all of a sudden starts to become in the hands of the government, or in the words of communists, in the hands of the people. And the redistribution is going on, so in theory, you don't have huge amounts of wealth in the hands of a few people. And then you keep-- if you take these ideas to their natural conclusion, you get to the theoretical communist state. And the theoretical communist state is a classless, and maybe even a little bit-- a classless society, and in Karl Marx's point of view-- and this is a little harder to imagine-- a stateless society. So in capitalism, you definitely had classes. You had the class that owns the capital, and then you had the labor class, and you have all of these divisions, and they're different from each other. He didn't really imagine a world that maybe a laborer could get out of this, they could get their own capital, then maybe they could start their own business. So he just saw this tension would eventually to socialism, and eventually a classless society where you have a central-- Well, he didn't even go too much into the details but you have kind of equal, everyone in society has ownership over everything, and society somehow figures out where things should be allocated, and all of the rest. And it's all stateless. And that's even harder to think about in a concrete fashion. So that's Karl Marx's view of things. But it never really became concrete until Vladimir Lenin shows up. And so the current version of communism that we-- The current thing that most of us view as communism is sometimes viewed as a Marxist-Leninist state. These are sometimes used interchangeably. Marxism is kind of the pure, utopian, we're eventually going to get to a world where everyone is equal, everyone is doing exactly what they want, there's an abundance of everything. I guess to some degree, it's kind of describing what happens in Star Trek, where everyone can go to a replicator and get what they want. And if you want to paint part of the day, you can paint part of the day, and you're not just a painter, you can also do whatever you want. So it's this very utopian thing. Let me write that down. So pure Marxism is kind of a utopian society. And just in case you don't know what utopian means, it's kind of a perfect society, where you don't have classes, everyone is equal, everyone is leading these kind of rich, diverse, fulfilling lives. And it's also, utopian is also kind of viewed as unrealistic. It's kind of, if you view it in the more negative light, is like, hey, I don't know how we'll ever be able to get there. Who knows? I don't want to be negative about it. Maybe we will one day get to a utopian society. But Leninist is kind of the more practical element of communism. Because obviously, after the Bolshevik Revolution, 1917, in the Russian Empire, the Soviet Union gets created, they have to actually run a government. They have to actually run a state based on these ideas of communism. And in a Leninist philosophy-- and this is where it starts to become in tension with the ideas of democracy-- in a Leninist philosophy, you need this kind of a party system. So you need this-- and he calls this the Vanguard Party. So the vanguard is kind of the thing that's leading, the one that's leading the march. So this Vanguard Party that kind of creates this constant state of revolution, and its whole job is to guide society, is to kind of almost be the parent of society, and take it from capitalism through socialism to this ideal state of communism. And it's one of those things where the ideal state of communism was never-- it's kind of hard to know when you get there. And so what happens in a Leninist state is it's this Vanguard Party, which is usually called the Communist Party, is in a constant state of revolution, kind of saying, hey, we're shepherding the people to some future state without a real clear definition of what that future state is. And so when you talk about Marxist-Leninist, besides talking about what's happening in the economic sphere, it's also kind of talking about this party system, this party system where you really just have one dominant party that it will hopefully act in the interest of the people. So one dominant communist party that acts in the interest of the people. And obviously, the negative here is that how do you know that they actually are acting in the interest of people? How do you know that they actually are competent? What means are there to do anything if they are misallocating things, if it is corrupt, if you only have a one-party system? And just to make it clear, the largest existing communist state is the People's Republic of China. And although it is controlled by the Communist Party, in economic terms it's really not that communist anymore. And so it can be confusing. And so what I want to do is draw a little bit of a spectrum. On the vertical axis, over here, I want to put democratic. And up here, I'll put authoritarian or totalitarian. Let me put-- well, I'll put authoritarian. I'll do another video on the difference. And they're similar. And totalitarian is more an extreme form of authoritarian, where the government controls everything. And you have a few people controlling everything and it's very non-democratic. But authoritarian is kind of along those directions. And then on this spectrum, we have the capitalism, socialism, and communism. So the United States, I would put-- I would put the United States someplace over here. I would put the United States over here. It has some small elements of socialism. You do have labor unions. They don't control everything. You also have people working outside of labor unions. It does have some elements of redistribution. There are inheritance taxes. There are-- I mean it's not an extreme form of redistribution. You can still inherit private property. You still have safety nets for people, you have Medicare, Medicaid, you have welfare. So there's some elements of socialism. But it also has a very strong capitalist history, private property, deep market, so I'd stick the United States over there. I would put the USSR-- not current Russia, but the Soviet Union when it existed-- I would put the Soviet Union right about there. So this was the-- I would put the USSR right over there. I would put the current state of Russia, actually someplace over here. Because they actually have fewer safety nets, and they kind of have a more-- their economy can kind of go crazier, and they actually have a bigger disparity in wealth than a place like the United States. So this is current Russia. And probably the most interesting one here is the People's Republic of China, the current People's Republic of China, which is at least on the surface, a communist state. But in some ways, it's more capitalist than the United States, in that they don't have strong wealth redistribution. They don't have kind of strong safety nets for people. So you could put some elements of China-- and over here, closer to the left. And they are more-- less democratic than either the US or even current Russia, although some people would call current Russia-- well, I won't go too much into it. But current China, you could throw it here a little bit. So it could be even a little bit more capitalist than the United States. Definitely they don't even have good labor laws, all the rest. But in other ways, you do have state ownership of a lot, and you do have state control of a lot. So in some ways, they're kind of spanning this whole range. So this right over here is China. And even though it is called a communist state, in some ways, it's more capitalist than countries that are very proud of their capitalism. But in a lot of other ways, especially with the government ownership and the government control of things, and this one dominant party, so it's kind of Leninist with less of the Marxist going on. So in that way, it is more in the communist direction. So hopefully that clarifies what can sometimes be a confusing topic." + }, + { + "Q": "This video series jumps straight in to exercise examples. Is there a video somewhere on introducing the concepts of dependent and independent variables? If not, explain it generically...?", + "A": "The difference is that the independent can easily be found without doing much work, but the dependent variable can only be found by using the independent, so the independent can work alone, but the dependent has to be found by using the other which is why its called the dependent variable, because its depending on the other.", + "video_name": "0eWm-LY23W0", + "transcript": "On your math quiz, you earn 5 points for each question that you answer correctly. In the table below, x represents the number of questions that you answer correctly, and y represents the total number of points that you score on your quiz. Fair enough? The relationship between these two variables can be expressed by the following equation-- y is equal to 5x. Graph the equation below. So you could look at a couple of your points. You could say, well, look, if I got 0 questions right I'm going to have 0 points. So you could literally graph that point-- if I got 0 questions right, I get 0 points. And then if I get one question right, and the table tells us that, or we could logically think about it, every question I would get right I'm going to get 5 more points. So if I get one question right, I'm going to get 5 more points, and we saw that in our table as well. We saw that right over here-- one question, 5 points. We could also plot it two questions, 10 points, But you only have to do two points to define a line. So it looks like we are actually done here." + }, + { + "Q": "I don't understand the graph that he made at 2:35. What does it mean that the ions like to keep the cell at certain potentials? Does that mean that their gated pathways open up at these potentials? Are the potentials just for the individual ions or the sum of the positive and negative charge differences between the intracellular ad extra-cellular environment?", + "A": "The way I understand it is that those voltages are Resting Potentials for each type of Ion. In other words, each type will want to come into or out of the cell until the cell reaches a specific voltage. If the voltage happens to be at the Resting Potential voltage of that ion, then that type of ion will stop moving across the membrane (rest). Watching the earlier videos in the series should be very helpful, if you haven t yet watched them", + "video_name": "rIVCuC-Etc0", + "transcript": "Let's figure out how a heart squeezes, exactly. And to do that, we have to actually get down to the cellular level. We have to think about the heart muscle cells. So we call them cardiac myocytes. These are the cells within the heart muscle. And these are the cells that actually do that squeezing. So if you actually were to go with a microscope and look down at one of these cells, it might look a little bit like this, with proteins inside of it. And when it's relaxed, these proteins are all kind of spread apart. And when it's squeezing down, because each cell has to squeeze for the overall heart to squeeze, these proteins look completely different. They're totally overlapped. And that overlapping is really what we call squeezing. So this is a squeezed version of the cell. And the first one was a relaxed version. And the trigger that kind of gets it from squeezing-- and of course actually, I should probably draw that too. The fact that, of course, at some point it has to go back to relaxed to do it again, to beat again. But the trigger for squeezing is calcium. So it's easy to get confused when you're thinking about all this kind of squeezing relaxing all this kind of stuff. But if you just keep your eye on calcium and think about the fact that calcium is the trigger, then you'll never get confused. You'll always be able to kind of find your way in terms of where the heart is in its cycle. So I'm going to draw for you the heart cycle, and specifically the cycle of an individual cell. This is what one cell is going to kind of go through over time. And the heart cycle, or the cycle for a cell, a heart cell, is going to be measured in millivolts. We're going to use millivolts to think about this. And you could use, I guess, a lot of different things. But this is probably one of the simplest things to kind of summarize what's happening with all of the different ions that are moving back and forth across that cell. Now the major ions, the ones that are going to mostly influence our heart cell, are going to be calcium, sodium, and potassium. So I'll put those three on here. And I'm putting them really just as benchmarks just so you can kind of keep track of where things would like to be. So calcium would like to be at 123 millivolts. Sodium at 67. And what that means is that if these were the only ions moving through, then sodium would like to keep things positive. And potassium, on the other hand, would like to make the membrane potential negative. So this scale is actually the scale for the membrane potential. And if we move up the scale, if we go from negative to something positive, this process would be called depolarization. That just means going from some negative number up towards something positive. And if you were to do the reverse, if you were going to from something positive to something negative, you'd call that repolarization. So these are just a couple of terms I wanted to make sure that we're familiar with, because we're going to be able to then get at some of the interesting things that happen. I'm going to make some space here inside of this cell. So let's start with a little picture of the cell. So let's say that this is our cell here. And I'm going to draw in little gap junctions, which are little connections between cells. So maybe a couple there. Maybe one there and maybe one over here. And let me label that. So these are the gap junctions. And also let's draw in some channels. So we have, let's say, a potassium channel right here. We know the potassium likes to leave cells. So this is going to be the way that potassium's going to flow. And it's going to leave behind a negative membrane potential, And let's say potassium is the main ion for this cell, which it is. Then our membrane potential is going to be really, really negative. In fact, if it was the only ion, it would be negative 92. But it's not. It's actually just the dominant ion. So it's over here and our membrane potential is around negative 90. And it continues around negative 90. So let's say nothing changes over a bit of time. So we stay at negative 90. So this is what things look like with the dominant ion that our cell is permeable to being potassium. Now a neighboring cell, let's say now, has a little bit of a depolarization. So it goes positive and through the gap junctions leak a little bit of sodium and some calcium. So this stuff starts leaking through the gap junctions, right? Now what will happened to our membrane potential? Well it was negative 90, but now that we've got some positive ions sitting inside of our cell, our cell becomes a little bit more positive, right? So it goes up to, let's say, here. And it happens pretty quickly. So now it's at negative 70 up from negative 90. So at this point, you actually get-- I'm going to erase gap functions-- but now that you're at negative 70, you actually get new channels opening up. And I haven't drawn them yet, and I'm going to erase sodium and calcium just to make some space. But you get new channels opening up. And these are going to be the sodium channels. So let me draw those in. Sodium channels. And there's so many of them. Lots and lots of these fast sodium channels open up. And I say fast because the sodium can flow through very quickly. So the sodium starts gushing in. And you know that's going to happen because there's a lot more sodium on the outside of a cell than the inside of a cell. And so sodium gushes in, and it's going to drive the membrane potential very quickly up to a very positive range. Now it would go all the way, let's say close to 67, maybe not exactly 67, because you still have those potassium ions leaving. But close to it, if not for the fact that these voltage-gated channels actually close down. So these sodium channels are voltage-gated. And they will actually close down just as quickly as they opened up. To show that, I'm actually going to do a little cut paste. I'm going to just draw this cell here. And I'm going to move it down here. So we've got our cell just as before. And now these voltage-gated channels, they close down. So let me get rid of all these arrow heads. But we're already now in positive range. So at this point, you could say our channels have caused a depolarization. And let me just quickly show these shut downs so that you don't get confused. There's no more sodium flowing through. You still have some potassium leaking out, but that's kind of as it's always been. And in addition to those potassium channels, that little channel I've drawn here, you have new potassium channels that open up down here. And these are actually voltage-gated potassium So you had them before. They existed. But they were actually not open. So let me just draw little x's. And the only reason they flipped open is because the depolarization happened. You had a negative go to a positive. So now that our cell is in positive territory, actually let me write in positive 20 or so, our potassium voltage-gated channels open up. So these voltage-gated channels open up. And you can guess what's going to happen. Like which direction do you think that the membrane potential will go? Well, if the sodium channels aren't gushing the sodium inwards and potassium is leaking outwards, now you're going to have a downwards repolarization. So now potassium is causing the membrane potential to go back down. And let's say it gets to about positive 5. And if it continued, again, it would go all the way back down to negative 90. But an interesting new development occurs. At this point, I'm going to actually cut paste again. And I'll show you what happens next, which is that calcium-- this is the thing I said keep your eye on the whole time, right?-- calcium finally kind of starts leaking in. So let me get rid of this. And this is the key idea, right? I don't want to forget that this is potassium. So you still have potassium in the same over here. But now calcium leaks in. And let's draw that over here. So you have these calcium voltage-gated channel that allow calcium to come it. So you've got calcium coming in, potassium leaving. Now think about what will happen in this situation. So calcium is going to want to rise the membrane potential this way. Potassium leaving is going to want it to continue going down this way. And because both are happening simultaneously, you basically get something like this. You get kind of a flatline. So because both events are happening, both potassium leaving the cell and calcium entering the cell, you get this kind of flatline. And the membrane potential stays kind of around the same. And so it can just write something similar, something like positive 5. Just so we're clear, these are also voltage-gated calcium channels. So to round this out, then what happens after that? So you have so far, so good. We have all these channels coming in to our cells and allowing different ions passage. And now we get to something like this. And I'm going to try to clean this up a little bit. And what happens is that the calcium channels actually close just as suddenly as they opened. So now you don't have any more calcium coming in. And if calcium was the only thing that was keeping this membrane potential going flat-- you know, I said that the potassium makes it want to go down, but the calcium was making it flat-- well, what will happen now? Well, if again you have just those potassium channels open, well then you're going to have the membrane potential go back down. It's going to go back down to negative 90 or so. So this is kind of the last stage, where those potassium channels are going back down. And those voltage-gated potassium channels also close at this point. So finally, they close down as well. And so now that they're closed, you're going to finally get back to just your initial state, which was having a little bit of potassium kind of leaking out of this cell. And those voltage-gated channel have shut down now. So now that you're at negative 90, you stay down there. And this process is ready to begin again. The last thing I want to say is the stages, how they're named. So this is state four, this kind of baseline negative state that the relaxed muscle cell is. And then this action potential, when it finally fires and it hits that negative 70, this is actually considered a threshold. This is our threshold. When it gets to that point, we call that stage 0. And then on the other side of stage 0 you have stage 1, 2, and 3. So stage 1 is that point when just the potassium channels first open up, the voltage-gated ones. And then stage 2 is when they're balanced with the calcium And stage 3 is again when you have just potassium channels, voltage-gated ones that are open. And then you get back to stage 4 again. So this would be stage 4. And because stage 0 is happening so rapidly, because this is so fast, we actually call this a fast action potential. So compare that to how the action potential goes in the pacemaker cells, where it's much slower. This fast action potential is a result of those really, really amazingly quick voltage-gated sodium channels." + }, + { + "Q": "how do you say it actully?", + "A": "For example, 5.5 is 5 and 5 tenths.", + "video_name": "qSPwUDmpnJ4", + "transcript": "- [Voiceover] Let's say that I had the number zero point one seven. How could I say this number? I said it one way, I said zero point one seven, but what are other ways that I could say it, especially if I wanted to express it in terms of tenths or hundredths or other places? And like always, try to pause the video and try think about it on your own. Alright, so there's actually a couple of ways that we could say this number. One is just to say zero point one seven. Other ways are to say look, I have a one in the tenths place, so that's going to be one tenth, one tenth and one tenth and I have a seven in the hundredths place, so this is a seven right over here in the hundredths place, so I can say one tenth and seven hundredths. Hun- Hundredths. And there you go. That's one way to say this number. Now another to think about it is just say the whole thing in terms of hundredths. So a tenth is how many hundredths? Well a tenth is the same thing as 10 hundredths, so you could say, you could say instead of a tenth, you could say this is 10 hundredths, and the way I'm writing it right now, very few people would actually do it this way. 10 hundredths and and seven hundredths. And seven hundredths. Well not I could just add these hundredths, if I have 10 hundredths and I have another seven hundredths, that's going to be 17 hundredths. So I could just write this down as 17 hundredths. Hundredths. And to make that intuition of how we could just call this 17 hundredths instead of just calling it one tenth and seven hundredths, let's actually count by hundredths. So that is one hundredth, that is two hundredth, and actually, let me just go straight to nine hundredths. So I skipped a bunch right over here. And what would be the next, how would I say 10 hundredths? Well 10 hundredths, let me write it this way, 10 hundredths is the same thing as one tenth. So if we go from nine hundredths, the next, if I'm counting by hundredths, the next one's going to be 10 hundredths. Now once again, 10 hundredths is the same thing as one tenth, just the same way that 10 ones is the same thing as one 10. I hope that doesn't confuse you, but we could keep counting. 10 hundredths, 11 hundredths, 12 hundredths, 13 hundredths, 14 hundredths, 15 hundredths, 16 hundredths, and then finally 17 hundredths. So hopefully that gives you a little intuition for why we can call this number, instead of just calling it zero point one seven, or one tenth and seven hundredths, we could call this 17 hundredths. So with that out of the way, let's do a couple of examples going the other way. Let's say we're given a name of a number or the words, and we wanna write it down as a decimal. So this is four tens and three hundredths. Alright, so four tens. So the four tens right over here. So actually let me just put some places over here, so this would be, if this is our tens place, and then this is our ones place, and then you're gonna have your decimal, and then you're gonna have your tenths place, tenths place, and then you're going to have, and I'll do this in a different color, your hundredths place. Hun- Hundredths place. So we're going to have four tens. Not tenths, four tens. So four tens, zero ones, zero ones, we got our decimal, they don't have any tenths over here so zero tenths, and then we have three hundredths. Three hundredths, so three hundredths. So four tens, which is the same thing as 40, and three hundredths right over here, so 40 point zero three. Let's do another one of these, this is kinda fun. So we have 24 hundredths. 24 hundredths. So by the logic that we saw in the first one, in the first one, we could just write this as, remember, this would be nine hundredths, and if we want one more hundredth, this would be 10 hundredths. So if we want to do 24 hundredths, it would just be zero point two four. And if you're saying well wait a minute, this looks like two tenths and four hundredths, you'd be right! But remember, so actually let me, I could re-write this as, I could re-write this as two tenths, two tenths and four hundredths. Let me say and four hundredths. Hundredths. But remember, a tenth is equal to 10 hundredths, so two tenths is the same thing as 20 hundredths. Hundredths. I have trouble saying- So this is the same thing as 20 hundredths and four hundredths. And four, let me write it neatly. Four hundredths. And of course 20 hundredths and four hundredths is the same thing as 24 hundredths. So hopefully that makes sense." + }, + { + "Q": "Is half a circle Pi?", + "A": "Half of the circumference of a circle isn t pi, unless the radius is 1. (2pi*r = 1) The angle formed by going halfway around the unit circle, however, is pi radians, or 180 degrees.", + "video_name": "1jDDfkKKgmc", + "transcript": "What I want to do in this video is revisit a little bit of what we know about pi, and really how we measure angles in radians. And then think about whether pi is necessarily the best number to be paying attention to. So let's think a little bit about what I just said. So pi, we know, is defined-- and I'll write defined as a triple equal sign, I guess you could call it that way-- pi is defined as the ratio of the circumference of a circle to its diameter, which is the same thing as the ratio of the circumference of the circle to two times the radius. And from that, we get all these interesting formulas that you get in geometry class that, hey, if you have the radius and you want to calculate the circumference, multiply both sides of this definition, or this equation, by two times the radius. And you get two times the radius times pi is equal to the circumference, or more familiarly, it would be circumference is equal to 2 pi r. This is one of those fundamental things that you learn early on in your career and you use it to find circumferences, usually, or figure out radiuses if you know circumference. And from that comes how we measure our angles in radians once we get to trigonometry class. And just as a review here, let me draw myself a circle. Let me draw myself a better circle. So there is my-- it'll do the job. And here is the positive x-axis, and let me make some angle here. I'll make the angle kind of obvious just so that it-- so let me make this angle. And the way that we measure angles, when we talk about radians, we're really talking about the angle subtended by something of a certain arc length. And we measure the arc length in-- well, the way I like to think about it, is the angle is in radians and the arc length itself is in radiuses, which isn't really a word. But that's how I think about it. How many radiuses is this arc length that subtends the angle in radians? So let me show you what I'm talking about. So this arc length right here, if the radius is r, what is the length of this arc length? Well, we know from basic geometry the entire circumference over here is going to be 2 pi r, right? This entire circumference, that's really by definition, this entire circumference is going to be 2 pi r. So what is just this arc length here? And I'm assuming this is a fourth of the circle. So it's going to be 2 pi r over 4. So this arc length over here is going to be 2 pi r over 4, which is the same thing as pi over 2 r. Or you could say this is the same thing as pi over 2 radiuses. One of those-- you know, not a real word, but that's how I like to think about it. Or you could say it subtends an angle of pi over 2 radians. So over here, theta is pi over 2 radians. And so really, when you're measuring angles in radians, it's really you're saying, OK, that angle subtended by an arc that has a length of how many radiusi, or I don't even know what the plural of radius is. Actually, I think it's radii but it's fun to try to say radiuses. Radii, actually let me do that just so no one says, Sal, you're teaching people the wrong plural form of radius. Radii. So this arc length is pi over 2 radii and it subtends an angle of pi over 2 radians. We could do another one just for the sake of making the point clear. If you went all the way around the circle and you got back to the positive x-axis here, what is the arc length? Well now, all of a sudden the arc length is the entire circumference of the circle. It would be 2 pi r, which is the same thing as 2 pi radii. And we would say that the angle subtended by this arc length, the angle that we care about going all the way around the circle, is 2 pi radians. And so, out of this comes all of the things that we know about how to graph trigonometric functions or at least how we measure the graph on the x-axis. And I'll also touch on Euler's formula, which is the most beautiful formula, I think, in all of mathematics. And let's visit those right now, just to remind ourselves of how pi fits in to all of that. So if I think about our trigonometric functions. Remember, if this was-- so on trigonometric functions, we assume we have a unit circle here. So in the trig functions, this is the unit circle definition of the trig function. So this is a nice review of all of that. You assume you have a unit circle, a circle of radius 1. And then the trig functions are defined as, for any angle you have here, for any angle, theta, cosine of theta is how far you have to move in-- or the x-coordinate of the point along the arc that subtends this angle. So that's cosine of theta. And then sine of theta is the y value of that point. Let me make that clear. Cosine of theta is the x value, sine of theta is the y value. And so if you were to graph one of these functions, and I'll just do sine of theta for convenience. But you could try it with cosine of theta. So let's graph sine of theta. Let's do one revolution of sine of theta. And we tend to label it-- so let's do sine. When the angle is 0, sine of theta is 0. Let me draw the x- and y-axis just so you remember. This is the y-axis and this is the x-axis. So when the angle is 0, we're right here on the unit circle. The y value there is 0. So sine of theta is going to be right like that. Let me draw it like this. So this is our theta, and this is-- I'm going to graph sine of theta along the y-axis. So we'll say y is equal to sine of theta in this graph that I'm drawing right over here. And then we could do-- well, I'll just do the simple points here. And then if we make the angle go-- if we did it in degrees, 90 degrees, or if we do it in radians, pi over 2 radians. What is sine of theta? Well, now it is 1. This is the unit circle. It has a radius 1. So when theta is equal to pi over 2, then sine of theta is equal to 1. So if this is 1 right here, sine of theta is equal to 1. If theta-- and then if we go 180 degrees or halfway around the circle-- theta is now equal to pi. Let me do this in a color, orange. I have already used orange. Theta is now equal to pi. When theta is equal to pi, the y value of this point right here is once again 0. So we go back to 0. Remember, we're talking about sine of theta. And then we can go all the way down here, where you can see there's 270 degrees. Or you could view this as 3 pi over 2 radians. So this is in radians, this axis. So 3 pi over 2 radians, sine of theta is the y-coordinate on the unit circle right over here. So it's going to be negative 1. So this is negative 1. And then finally, when you go all the way around the circle, you've gone 2 pi radians, and you're back where you began. And the sine of theta, or the y-coordinate, is now 0 once again. And if you connect the dots, or if you'd plotted more points, you would see a sine curve over just the part that we've graphed right over here. So that's another application. You say hey, Sal, where is this going? Well I'm showing you-- I'm reminding you of all of these things because we're going to revisit it with a different number other than pi. And so I want to do one last visit with pi. You say, look, pi is powerful because-- or one of the reasons why pi seems to have some type of mystical power, and we've shown this in the calculus playlist-- is Euler's formula, that e to the i theta is equal to cosine of theta plus i sine of theta. This, by itself, is just a crazy-- it's just one of those mind boggling formulas. But it sometimes looks even more mind boggling when you put pi in for theta because then, from Euler's formula, you would get e to the i pi is equal to-- well, what's cosine of pi? Cosine of pi is negative 1. And then sine of pi is 0. So 0 times i. So you get this formula, which is pretty profound, and then you say, OK, if I want to put all of the fundamental numbers together in one formula, I can add 1 to both sides of this. And you get e to the i pi plus 1 is equal to 0. Sometimes this is called Euler's Identity, the most beautiful formula or equation in all of mathematics. And it is pretty profound. You have all of the fundamental numbers in one equation. e, i, pi, 1, 0, although for my aesthetic taste it would have been even more powerful if this was a 1 right over here. Because then this would have said, look, e to the i pi, this bizarre thing, would have equaled unity. That would have been super duper profound to me. It seems a little bit of a hack to add 1 to both sides and say, oh, look, now I have 0 here. But this is pretty darn good. But with that, I'm going to make-- well, I'm not going to argue for it. I'm going to show an argument for another number, a number different than pi. And I want to make it clear that these ideas are not my own. It comes from-- well, it's inspired by-- many people are on this movement now, the Tau Movement, but these are kind of the people that gave me the thinking on this. And the first is Robert Palais on \"Pi is Wrong.\" And he doesn't argue that pi is calculated wrong. He still agrees that it is the ratio of the circumference to the diameter of a circle that is 3.14159. But what he's saying is that we're paying attention to the wrong number. And also, you have Michael Hartl, \"The Tau Manifesto.\" All of this is available online. And what they argue for is a number called tau, or what they call tau. And they define tau, and it's a very simple change from pi. They define tau not as the ratio of the circumference the diameter, or the ratio of the circumference to 2 times the radius. They say, hey, wouldn't it be natural to define some number, the ratio of the circumference to the radius? And as you see here, this pi is just one half times this over here, right? Circumference over 2 r, this the same thing as one half times circumference over r. So pi is just half of tau. Or another way to think about it is that tau is just 2 times pi. Or, and I'm sure you probably don't have this memorized, because you're like, wait, I spent all my life memorizing pi, but it's 6.283185 and keeps going on and on and on, never repeating just like pi. It's 2 times pi. And so you're saying, hey Sal, pi has been around for millennia, really. Why mess with such a fundamental number, especially when you just spent all this time showing how profound it is? And the argument that they'd make, and it seems like a pretty good argument, is that actually things seem a little bit more elegant when you pay attention to this number instead of half of this number, when you pay attention to tau. And to see that, let's revisit everything that we did here. Now, all of a sudden, if you pay attention to 2 pi, as opposed to pi. Or we should call it-- if you pay attention to tau instead of tau over 2. What is this angle that we did in magenta? Well first of all, let's think about this formula right What is the circumference in terms of the radius? Well, now we could say the circumference is equal to tau times the radius because tau is the same thing as 2 pi. So it makes that formula a little bit neater, although it does make the pi r squared a little bit messier. So you could argue both sides of that. But it makes the measure of radians much more intuitive because you could say that this is pi over 2 radians, or you could say that this is pi over 2 radians is the same thing as to tau over 4 radians. And where did I get that from? Remember, if you go all the way around the circle, that is the circumference. The arc length would be the circumference. It would be tau radii, or it would be to tau radians would be the angle subtended by that arc length. It would be tau radians. All the way around is tau radians. So that by itself is intuitive. One revolution is one tau radians. If you go only one fourth of that, it's going to be tau over four radians. So the reason why tau is more intuitive here is because it immediately-- you don't have to do this weird conversion where you saying, oh, divide by 2, multiply by 2, all that. You're just like, look, however many radians in terms of tau, that's really how many revolutions you've gone around the circle. And so, if you've gone one fourth around, that's tau over four radians. If you've gone halfway around, that'd be tau over two radians. If you go 3/4 around that'd be three tau over four radians. If you go all the way around that would be tau radians. If someone tells you that they have an angle of 10 tau radians, you'd go around exactly 10 times. It would be much more intuitive. You wouldn't have to do this little mental math, converting, saying, do I multiply or divide by 2 when I convert to radians in terms of pi? No, when you do it in terms of tau radians, it's just natural. One revolution is one tau radians. So that makes-- and it makes a sine function over here. Instead of writing pi over 2-- well, when you look at a graph like this you're like, where was this on the unit circle? Was this one fourth around the circle? Was this one half? And this is actually one fourth around the circle, right over here? But now it becomes obvious if you write it in tau. Pi over 2 is the same thing as to tau over 4. Pi is the same thing as to tau over 2. 3 pi over 2 is 3 pi-- oh, sorry, 3 tau over 4, 3/4 tau. And then one revolution is tau. And then immediately, now when you look at it this way, you know exactly where you are in the unit circle. You're one fourth around the unit circle, you're halfway around the unit circle, you're 3/4 of the way around the unit circle. And then you're all the way around the unit circle. And so the last thing that I think the strong pi defenders would say is well, look Sal, you just pointed out one of the most beautiful identities or formulas in mathematics. How does tau hold up to this? Well, let's just try it out and see what happens. So if we take e to the i tau, that will give us cosine of tau plus i sine of tau. And once again, let's just think about what this is. Tau radians means we've gone all the way around the unit circle. So cosine of tau-- remember, we're back at the beginning of the unit circle right over here-- so cosine of tau is going to be equal to 1. And then sine of tau is equal to 0. So e to the i tau is equal to 1. And I'll leave it up to you to decide which one seems to be more aesthetically profound." + }, + { + "Q": "Hi friends :-) I have a question for you-\n1> at 6:27 \"Sal\" proved us that alternate interior angles are equal , right? so, here is my question ?\nwe know that :- b=c,f=g\nnow b and c are vertical angles and we also know that b = f (corresponding angle) so why can't we say that c = f ,right? and why not alternate exterior angles?", + "A": "Ok, so you said b=c,f=g. Which says b=c and f=g. If you look back at Sals blackboard on the video there is no comma, it simply says b=c=f=g which means they are ALL equal, so yes , c does equal f and c equals g, and b equals g etc. etc.. As for alternate exterior angles, the same rule applies. a=d=e=h. They are all equal.", + "video_name": "H-E5rlpCVu4", + "transcript": "Let's say we have two lines over here. Let's call this line right over here line AB. So A and B both sit on this line. And let's say we have this other line over here. We'll call this line CD. So it goes through point C and it goes through point D. And it just keeps on going forever. And let's say that these lines both sit on the same plane. And in this case, the plane is our screen, or this little piece of paper that we're looking at right over here. And they never intersect. So they're on the same plane, but they never intersect each other. If those two things are true, and when they're not the same line, they never intersect and they can be on the same plane, then we say that these lines are parallel. They're moving in the same general direction, in fact, the exact same general direction. If we were looking at it from an algebraic point of view, we would say that they have the same slope, but they have different y-intercepts. They involve different points. If we drew our coordinate axes here, they would intersect that at a different point, but they would have the same exact slope. And what I want to do is think about how angles relate to parallel lines. So right over here, we have these two parallel lines. We can say that line AB is parallel to line CD. Sometimes you'll see it specified on geometric drawings like this. They'll put a little arrow here to show that these two lines are parallel. And if you've already used the single arrow, they might put a double arrow to show that this line is parallel to that line right over there. Now with that out of the way, what I want to do is draw a line that intersects both of these parallel lines. So here's a line that intersects both of them. Let me draw a little bit neater than that. So let me draw that line right over there. Well, actually, I'll do some points over here. Well, I'll just call that line l. And this line that intersects both of these parallel lines, we call that a transversal. This is a transversal line. It is transversing both of these parallel lines. This is a transversal. And what I want to think about is the angles that are formed, and how they relate to each other. The angles that are formed at the intersection between this transversal line and the two parallel lines. So we could, first of all, start off with this angle right over here. And we could call that angle-- well, if we made some labels here, that would be D, this point, and then something else. But I'll just call it this angle right over here. We know that that's going to be equal to its vertical angles. So this angle is vertical with that one. So it's going to be equal to that angle right over there. We also know that this angle, right over here, is going to be equal to its vertical angle, or the angle that is opposite the intersection. So it's going to be equal to that. And sometimes you'll see it specified like this, where you'll see a double angle mark like that. Or sometimes you'll see someone write this to show that these two are equal and these two are equal right over here. Now the other thing we know is we could do the exact same exercise up here, that these two are going to be equal to each other and these two are going to be equal to each other. They're all vertical angles. What's interesting here is thinking about the relationship between that angle right over there, and this angle right up over here. And if you just look at it, it is actually obvious what that relationship is-- that they are going to be the same exact angle, that if you put a protractor here and measured it, you would get the exact same measure up here. And if I drew parallel lines-- maybe I'll draw it straight left and right, it might be a little bit more obvious. So if I assume that these two lines are parallel, and I have a transversal here, what I'm saying is that this angle is going to be the exact same measure as that angle there. And to visualize that, just imagine tilting this line. And as you take different-- so it looks like it's the case over there. If you take the line like this and you look at it over here, it's clear that this is equal to this. And there's actually no proof for this. This is one of those things that a mathematician would say is intuitively obvious, that if you look at it, as you tilt this line, you would say that these angles are the same. Or think about putting a protractor here to actually measure these angles. If you put a protractor here, you'd have one side of the angle at the zero degree, and the other side would specify that point. And if you put the protractor over here, the exact same thing would happen. One side would be on this parallel line, and the other side would point at the exact same point. So given that, we know that not only is this side equivalent to this side, it is also equivalent to this side over here. And that tells us that that's also equivalent to that side over there. So all of these things in green are equivalent. And by the same exact argument, this angle is going to have the same measure as this angle. And that's going to be the same as this angle, because they are opposite, or they're vertical angles. Now the important thing to realize is just what we've deduced here. The vertical angles are equal and the corresponding angles at the same points of intersection are also equal. And so that's a new word that I'm introducing right over here. This angle and this angle are corresponding. They represent kind of the top right corner, in this example, of where we intersected. Here they represent still, I guess, the top or the top right corner of the intersection. This would be the top left corner. They're always going to be equal, corresponding angles. And once again, really, it's, I guess, for lack of a better word, it is a bit obvious. Now on top of that, there are other words that people will see. We've essentially just proven that not only is this angle equivalent to this angle, but it's also equivalent to this angle right over here. And these two angles-- let me label them so that we can make some headway here. So I'm going to use lowercase letters for the angles themselves. So let's call this lowercase a, lowercase b, lowercase c. So lowercase c for the angle, lowercase d, and then let me call this e, f, g, h. So we know from vertical angles that b is equal to c. But we also know that b is equal to f because they are corresponding angles. And that f is equal to g. So vertical angles are equivalent, corresponding angles are equivalent, and so we also know, obviously, that b is equal to g. And so we say that alternate interior angles are equivalent. So you see that they're kind of on the interior of the intersection. They're between the two lines, but they're on all opposite sides of the transversal. Now you don't have to know that fancy word, alternate interior angles, you really just have to deduce what we just saw over here. Know that vertical angles are going to be equal and corresponding angles are going to be equal. And you see it with the other ones, too. We know that a is going to be equal to d, which is going to be equal to h, which is going to be equal to e." + }, + { + "Q": "At 4:14 where did you get the plus sign from? Wouldn't you put a subtraction sign there? (because the previous color coded terms had a subtraction sign at the beginning?)", + "A": "At that point in the video, Sal is combining: -2x^2 + 3x^2. He does this by adding/subtracting the coefficients: -2 + 3 = +1, not a -1. So, once combined, the 2 terms creates +x^2. Hope this helps.", + "video_name": "FNnmseBlvaY", + "transcript": "Now we have a very, very, very hairy expression. And once again, I'm going to see if you can simplify this. And I'll give you little time to do it. So this one is even crazier than the last few we've looked at. We've got y's and xy's, and x squared and x's, well more just xy's and y squared and on and on and on. And there will be a temptation, because you see a y here and a y here to say, oh, maybe I can add this negative 3y plus this 4xy somehow since I see a y and a y. But the important thing to realize here is that a y is different than an xy. Think about it they were numbers. If y was 3 and an x was a 2, then a y would be a 3 while an xy would have been a 6. And a y is very different than a y squared. Once again, if the why it took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same-- I guess you cannot add these two or subtract these two terms. A y is different than a y squared, is different than an xy. Now with that said, let's see if there is anything that we can simplify. So first, let's think about the y terms. So you have a negative 3y there. Do we have any more y term? We have this 2y right over there. So I'll just write it out-- I'll just reorder it. So we have negative 3y plus 2y. Now, let's think about-- and I'm just going in an arbitrary order, but since our next term is an xy term-- let's think about all of the xy terms. So we have plus 4xy right over here. So let me just write it down-- I'm just reordering the whole expression-- plus 4xy. And then I have minus 4xy right over here. Then let's go to the x squared terms. I have negative 2 times x squared, or minus 2x squared. So let's look at this. So I have minus 2x squared. Do I have any other x squared? Yes, I do. I have this 3x squared right over there. So plus 3x squared. And then let's see, I have an x term right over here, and that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term-- I'll circle that in orange-- so plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have two of something and I subtract 3 of that, what am I left with? Well, I'm left with negative 1 of that something. So I could write negative 1y, or I could just write negative y. And another way you could think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. What's the coefficient here? It's 2. Where obviously both are dealing-- they're both y terms, not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1, or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms. If I have 4 of this, 4 xy's and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say add the coefficients, 4 plus negative 4, gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. And so I'm left with no xy's. And then I have right over here-- I could have written 0xy, but that seems unnecessary-- then right over here I have my x squared terms. Negative 2 plus 3 is 1. Or another way of saying it, if I have 3x squared and I were to take away 2 of those x squared, so I'm left with the 1x squared. So this right over here simplifies to 1x squared. Or I could literally just write x squared. 1x squared is the same thing as x squared. So plus x squared, and then these there's nothing really left to simplify. So plus 2x plus y squared. And obviously you might have gotten an answer in some other order, but the order in which I write these terms don't matter. It just matters that you were able to simplify it to these four terms." + }, + { + "Q": "At, 8:40 - 8:42, there is repulsion between the lone pairs & the bond pairs but what about the lone pairs? Don't they repel themselves too?", + "A": "Of course they do. And they repel one another the most.", + "video_name": "BM-My1AheLw", + "transcript": "Voiceover: The concept of steric number is very useful, because it tells us the number of hybridized orbitals that we have. So to find the steric number, you add up the number of stigma bonds, or single-bonds, and to that, you add the number of lone pairs of electrons. So, let's go ahead and do it for methane. So, if I wanted to find the steric number, the steric number is equal to the number of sigma bonds, so I look around my carbon here, and I see one, two, three, and four sigma, or single-bonds. So I have four sigma bonds; I have zero lone pairs of electrons around that carbon, so four plus zero gives me a steric number of four. In the last video, we saw that SP three hybridized situation, we get four hybrid orbitals, and that's how many we need, the steric number tells us we need four hybridized orbitals, so we took one S orbital, and three P orbitals, and that gave us four, SP three hybrid orbitals, so this carbon must be SP three hybridized. So let's go ahead, and draw that in here. So this carbon is SP three hybridized, and in the last video, we also drew everything out, so we drew in those four, SP three hybrid orbitals, for that carbon, and we had one valence electron in each of those four, SP three hybrid orbitals, and then hydrogen had one valence electron, in an un-hybridized S orbital, so we drew in our hydrogens, and the one valence electron, like that. This head-on overlap; this is, of course, a sigma bond, so we talked about this in the last video. And so now that we have this picture of the methane molecule, we can think about these electron pairs, so these electron pairs are going to repel each other: like charges repel. And so, the idea of the VSEPR theory, tell us these electron pairs are going to repel, and try to get as far away from each other as they possibly can, in space. And this means that the arrangement of those electron pairs, ends up being tetrahedral. So let's go ahead and write that. So we have a tetrahedral arrangement of electron pairs around our carbon, like that. When we think about the molecular geometry, so that's like electron group geometry, you wanna think about the geometry of the entire molecule. I could think about drawing in those electrons, those bonding electrons, like that. So we have a wedge coming out at us in space, a dash going away from us in space, and then, these lines mean, \"in the plane of the page.\" And so, we can go ahead a draw in our hydrogens, and this is just one way to represent the methane molecule, which attempts to show the geometry of the entire molecule. So the arrangement of the atoms turns out to also be tetrahedral, so let's go ahead and write that. So, tetrahedral. And, let's see if we can see that four-sided figure, so a tetrahedron is a four-sided figure, so we can think about this being one face, and then let's go ahead and draw a second face. And if I draw a line back here, that gives us four faces to our tetrahedron. So our electron group geometry is tetrahedral, the molecular geometry of methane is tetrahedral, and then we also have a bond angle, let me go ahead and draw that in, so a bond angle, this hydrogen-carbon-hydrogen bond angle in here, is approximately 109 point five degrees. All right, let's go ahead and do the same type of analysis for a different molecule, here. So let's do it for ammonia, next. So we have NH three, if I want to find the steric number, the steric number is equal to the number of sigma bonds, so that's one, two, three; so three sigma bonds. Plus number of lone pairs of electrons, so I have one lone pair of electrons here, so three plus one gives me a steric number of four. So I need four hybridized orbitals, and once again, when I need four hybridized orbitals, I know that this nitrogen must be SP three hybridized, because SP three hybridization gives us four hybrid orbitals, and so let's go ahead and draw those four hybrid orbitals. So we would have nitrogen, and let's go ahead and draw in all four of those. So, one, two, three, and four; those are the four hybrid orbitals. When you're drawing the dot structure for nitrogen, you would have one electron, another electron, another electron, and then you'd have two in this one, like that. And then you'd go ahead, and put in your hydrogens, so, once again, each hydrogen has one electron, in a hybridized S orbital, so we go ahead and draw in those hydrogens, so our overlap of orbitals, so here's a sigma bond, here's a sigma bond, and here's a sigma bond; so three sigma bonds in ammonia, and then we have this lone pair up here. So the arrangement of these electron pairs, is just what we talked about before: So we have this tetrahedral arrangement of electron pairs, or electron groups, so the VSEPR theory tells us that's how they're going to repel. However, that's not the shape of the molecule, so if I go ahead and draw in another picture over here, to talk about the molecular geometry, and go ahead and draw in the bonding electrons, like that, and then I'll put in my non-bonding electrons, up here: this lone pair right here, housed in an SP three hybridized orbital. So, the arrangement of the atoms turns out not to be tetrahedral, and that has to do with this lone pair of electrons up here, at the top. So, this lone pair of electrons is going to repel these bonding electrons more strongly than in our previous example, and because it's going to repel those electrons a little bit more strongly, you're not gonna get a bond angle of 109 point five; it's going to decrease the bond angle. So let me go ahead, and use the same color we used before, so this bond angle is not 109 point five; it goes down a bit, because of the extra repulsion, so it turns out to be approximately 107 degrees. And in terms of the shape of the molecule, we don't say \"tetrahedral\"; we say \"trigonal-pyramidal.\" So let me go ahead, and write that here, so the geometry of the ammonia molecule is trigonal-pyramidal, and let's analyze that a little bit. So, \"trigonal\" refers to the fact that nitrogen is bonded to three atoms here, so nitrogen is bonded to three hydrogens, so that takes care of the trigonal part. The \"pyramidal\" part comes in, because when you're doing molecular geometry, you ignore lone pairs of electrons. So if you ignore that lone pair of electrons, and just look at this nitrogen, at the top of this pyramid right here, so that's where the \"pyramidal\" term comes in. So bonded to three other atoms, like this, this, and this; for our pyramid. So trigonal-pyramidal is the geometry of the ammonia molecule, but the nitrogen is SP three hybridized. All right, let's do one more example; let's do water. So, first we calculate the steric number, so the steric number is equal to the number of sigma bonds, so that's one, two; so, two sigma bonds. Plus numbers of lone pairs of electrons, so here's a lone pair, here's a lone pair; so we have two plus two, which is equal to four, so we need four hybridized orbitals. As we've seen in the previous two examples, when you need four hybridized orbitals, that's an SP three hybridization situation; you have four SP three hybridized orbitals. So this oxygen is SP three hybridized, so I'll go ahead and write that in here, so oxygen is SP three hybridized. So we can draw that out, showing oxygen with its four SP three hybrid orbitals; so there's four of them. So I'm gonna go ahead and draw in all four. In terms of electrons, this orbital gets one, this orbital gets one, and these orbitals are going to get two, like that; so that takes care of oxygen's six valence electrons. When you're drawing in your hyrdogens, so let's go ahead and put in the hydrogen here, so, once again, each hydrogen with one electron, in a un-hybridized S orbital, like that. So in terms of overlap of bonds, here's one sigma bond, and here's another sigma bond; so that's our two sigma bonds for water. Once again, the arrangement of these electron pairs is tetrahedral, so VSEPR theory says the electrons repel, and so the electron group geometry, you could say, is tetrahedral, but that's not the geometry of the entire molecule, 'cause I was just thinking about electron groups, and these hybrid orbitals. The geometry of the molecule is different, so we'll go ahead and draw that over here. So we have our water molecule, and draw in our bonding electrons, and now let's put in our non-bonding electrons, like that, so we have a different situation than with ammonia. With ammonia, we had one lone pair of electrons repelling these bonding electrons up here; for water, we have two lone pairs of electrons repelling these bonding electrons, and so that's going to change the bond angle; it's going to short it even more than in the previous example. So the bond angle decreases, so this bond angle in here decreases to approximately 105 degrees, rounded up a little bit. So, thinking about the molecular geometry, or the shape of the water molecule, so we actually call this \"bent,\" or \"angular,\" so this is, \"bent geometry,\" because you ignore the lone pairs of electrons, and that would just give you this oxygen here, and then this angle; so you could also call this, \"angular.\" So we have this bent molecular geometry, like that, or angular, and once again, for molecular geometry, ignore your lone pairs of electrons. So these are examples of three molecules, and the central atom in all three of these molecules is SP three hybridized, and so, this is one way to figure out your overall molecular geometry, and to think about bond angles, and to think about how those hybrid orbitals affect the structure of these molecules." + }, + { + "Q": "When you divide 13 by 93 on a calculator you get .1397. Did I do something wrong? Sal's answer doing it without a calculator is .138R.", + "A": "I think it is because it is a repeating decimal with about 15 repeating numbers. I did not watch the video but it might mean that it is a repeating decimal and it has rounded to 3 decimal places.", + "video_name": "Llt-KkHugRQ", + "transcript": "Welcome to the presentation on ordering numbers. Let's get started with some problems that I think, as you go through the examples hopefully, you'll understand how to do these problems. So let's see. The first set of numbers that we have to order is 35.7%, 108.1% 0.5, 13/93, and 1 and 7/68. So let's do this problem. The important thing to remember whenever you're doing this type of ordering of numbers is to realize that these are all just different ways to represent-- these are all a percent or a decimal or a fraction or a mixed-- are all just different ways of representing numbers. It's very hard to compare when you just look at it like this, so what I like to do is I like to convert them all to decimals. But there could be someone who likes to convert them all to percentages or convert them all to fractions and then compare. But I always find decimals to be the easiest way to compare. So let's start with this 35.7%. Let's turn this into a decimal. Well, the easiest thing to remember is if you have a percent you just get rid of the percent sign and put it over 100. So 35.7% is the same thing as 35.7/100. Like 5%, that's the same thing as 5/100 or 50% is just the same thing as 50/100. So 35.7/100, well, that just equals 0.357. If this got you a little confused another way to think about percentage points is if I write 35.7%, all you have to do is get rid of the percent sign and move the decimal to the left two spaces and it becomes 0.357. Let me give you a couple of more examples down here. Let's say I had 5%. That is the same thing as 5/100. Or if you do the decimal technique, 5%, you could just move the decimal and you get rid of the percent. And you move the decimal over 1 and 2, and you put a 0 here. It's 0.05. And that's the same thing as 0.05. You also know that 0.05 and 5/100 are the same thing. So let's get back to the problem. I hope that distraction didn't distract you too much. Let me scratch out all this. So 35.7% is equal to 0.357. Similarly, 108.1%. Let's to the technique where we just get rid of the percent and move the decimal space over 1, 2 spaces to the left. So then that equals 1.081. See we already know that this is smaller than this. Well the next one is easy, it's already in decimal form. 0.5 is just going to be equal to 0.5. Now 13/93. To convert a fraction into a decimal we just take the denominator and divide it into the numerator. So let's do that. 93 goes into 13? Well, we know it goes into 13 zero times. So let's add a decimal point here. So how many times does 93 go into 130? Well, it goes into it one time. 1 times 93 is 93. Becomes a 10. That becomes a 2. Then we're going to borrow, so get 37. Bring down a 0. So 93 goes into 370? 4 times 93 would be 372, so it actually goes into it only three times. 3 times 3 is 9. 3 times 9 is 27. So this equals? Let's see, this equals-- if we say that this 0 becomes a 10. This become a 16. This becomes a 2. 81. And then we say, how many times does 93 go into 810? It goes roughly 8 times. And we could actually keep going, but for the sake of comparing these numbers, we've already gotten to a pretty good level of accuracy. So let's just stop this problem here because the decimal numbers could keep going on, but for the sake of comparison I think we've already got a good sense of what this decimal looks like. It's 0.138 and then it'll just keep going. So let's write that down. And then finally, we have this mixed number here. And let me erase some of my work because I don't Actually, let me keep it the way it is right now. The easiest way to convert a mixed number into a decimal is to just say, OK, this is 1 and then some fraction that's less than 1. Or we could convert it to a fraction, an improper fraction like-- oh, actually there are no improper fractions here. Actually, let's do it that way. Let's convert to an improper fraction and then convert that into a decimal. Actually, I think I'm going to need more space, so let me clean up this a little bit. We have a little more space to work with now. So 1 and 7/68. So to go from a mixed number to an improper fraction, what you do is you take the 68 times 1 and add it to the numerator here. And why does this make sense? Because this is the same thing as 1 plus 7/68. 1 and 7/68 is the same thing as 1 plus 7/68. And that's the same thing as you know from the fractions module, as 68/68 plus 7/68. And that's the same thing as 68 plus 7-- 75/68. So 1 and 7/68 is equal to 75/68. And now we convert this to a decimal using the technique we did for 13/93. So we say-- let me get some space. We say 68 goes into 75-- suspicion I'm going to run out of space. 68 goes into 75 one time. 1 times 68 is 68. 75 minus 68 is 7. Bring down the 0. Actually, you don't have to write the decimal there. Ignore that decimal. 68 goes into 70 one time. 1 times 68 is 68. 70 minus 68 is 2, bring down another 0. 68 goes into 20 zero times. And the problem's going to keep going on, but I think we've already once again, gotten to enough accuracy that we can compare. So 1 and 7/68 we've now figured out is equal to 1.10-- and if we kept dividing we'll keep getting more decimals of accuracy, but I think we're now ready to compare. So all of these numbers I just rewrote them as decimals. So 35.7% is 0.357. 108.1%-- ignore this for now because we just used It's 108.1% is equal to 1.081. 0.5 is 0.5. 13/93 is 0.138. And 1 and 7/68 is 1.10 and it'll keep going on. So what's the smallest? So the smallest is 0.-- actually, no. The smallest is right here. So I'm going to rank them from smallest to largest. So the smallest is 0.138. Then the next largest is going to be 0.357. Then the next largest is going to be 0.5. Then you're going to have 1.08. And then you're going to have 1 and 7/68. Well, actually, I'm going to do more examples of this, but for this video I think this is the only one I have time for. But hopefully this gives you a sense of doing these problems. I always find it easier to go into the decimal mode to compare. And actually, the hints on the module will be the same for you. But I think you're ready at least now to try the problems. If you're not, if you want to see other examples, you might just want to either re-watch this video and/or I might record some more videos with more examples right now. Anyway, have fun." + }, + { + "Q": "What is the definition of pizzicato?", + "A": "Pizzicato is when string instruments pluck the strings instead of using the bow. Pizzicato is the Latin word for pluck.", + "video_name": "mVFbGjnysP0", + "transcript": "- [Voiceover] Here is an example of 6/4 time from American composer Joseph Schwantner's The Poet's Hour. Six beats in a measure with quarter note getting one beat. (The Poet's Hour) The next meter that we will discuss is 6/8, six beats in a measure with the eighth note getting one beat. This one is a little more complicated because at a slow tempo, or a slow speed, we can think of each measure with six beats in a bar, but if the tempo was fast, we'd divide the bar into two beats with each beat worth three eighth notes, or a dotted quarter. for a slow version of 6/8, let's listen to this beautiful passage for the oboes and English horn from Ravel's Ballet Daphnis et Chloe. (Daphnis et Chloe) For a fast version, let's listen to part of the Firebird's Variation from Stravinsky's Ballet, The Firebird. (Firebird's Variation) 9/8 continues with the same pattern. In a slow speed, or tempo, each eighth note receives one beat with nine beats in a bar. In a faster tempo, the pulse is three with a dotted quarter note receiving one pulse or one beat. 12/8 continues in the same pattern. In a slow tempo, 12 beats in a measure with an eighth note receiving one beat. If we look at the opening of Stravinsky's Firebird, we see 12/8 at the slow tempo with a feeling of 12 beats in a measure. (Firebird) If we look at the same introduction a few bars later, the pulse moves to four beats with a dotted quarter note getting one beat. (Firebird)" + }, + { + "Q": "why did Hitler do all this stuff and are there still survivors from the holocaust.", + "A": "Hitler didn t like the way that Jewish shops were running so much of Germany, that s why he wanted to wipe them out. He just didn t like them, like, at all. Think about it, though, what if you were a terrorist (hypothetically, of course!) and you wanted to wipe out a whole entire race. How easy would that be? Not very easy. We have survivors from the Holocaust because it s next to virtually impossible to wipe out a whole race, even if they are in capitivity.", + "video_name": "QCkn5bu8GgM", + "transcript": "Adolf Hitler got his start in the military during World War I. He was a dispatch runner on the Western Front. He actually gets fairly decorated. And by most accounts, this is where he finds meaning. He finds meaning in being part of the military. He finds meaning in frankly, the war itself. But then in 1918, we, of course, have the end of the war. Well, first you have the abdication of the Kaiser. You have the Republican government, people who want to form a republic, take control. And then they sign an armistice with the Allies in November. And this is not well received by Hitler. And frankly, it's not well received by many in the military. From their point of view, they somewhat delusionally believed that Germany would have won World War I if they weren't stabbed in the back by these November Criminals, by the folks who had taken over after the Kaiser. So you have this whole stab in the back theory by those who had taken over and signed the armistice. And this wasn't just believed by folks like Hitler. This was believed even by very senior people in the military. This right over here is General Ludendorff, one of two people-- the other gentleman, Hindenburg-- who were in charge of the entire German military. He believed in the stab in the back theory. He thought that they would have won if they didn't sign the armistice, if these November Criminals, these people who had taken control of the government, did not sign this with the Allies. And then you go to 1919. From the point of view of people of like Hitler, things only got worse. You have the Treaty of Versailles that applied all the war guilt to the Germans. You have these huge reparations that would even be paid in resources. You have the former German Empire, a significant amount of its territory is given over to the Allies, or to form new states. Then, you also have the formal establishment of the Weimar Republic. It's called the Weimar Republic because the new German constitution is drafted in the town of Weimar. And it sets it up as a parliamentary democracy. That's why it's called a republic. But it's a little bit of a bizarre parliamentary democracy. It actually gave a good bit of power, a directly elected president that had a reasonable amount of power, especially in emergencies. And that would become relevant later on when Hitler actually comes to power over a decade later. But then in 1919, Hitler was still looking-- he was very upset about the war ending. He stays part of the military. And part of the military he's assigned to start infiltrating or spying on the German Workers' Party. And the acronym in German is the DAP. But the English translation would be the German Workers' Party. But he actually gets quite impressed by the German Workers' Party, which is really ultra-nationalist. And when we talk about ultra-nationalist it's all about German race superiority. It's in line with this whole idea that they would have won the war if they weren't stabbed in the back. And it's also anti-communist. It's anti-capitalist. And it's anti ethnic minorities, in particular anti-Jewish. And all of these ideas Hitler found very impressive. And just to be clear, a lot of times when people talk about ultra-nationalist groups they often will call them as ultra-right wing. And this bears some clarification because the right wing is also often viewed as very capitalist. But ultra-nationalists really put the nation, and the race that they view as indicative of the nation, above all other concerns. So yes, they were anti-communist. They were anti-distribution of wealth. Communists believe in no classes, as little private property as possible. The German Workers' party didn't believe in that. They were anti that. But they were also anti unfettered capitalism, especially capitalism that might get in the way of the nation's interests. But he becomes very impressed with them. And he actually joins as the 55th member. So you can imagine, this, at this point, is a very, very small party. But then we fast forward to 1920. By 1920, the party leadership has taken note of Hitler. They actually notice him when he's arguing with people and that other people are listening. He's actually a really great orator. And so they allow him to give more and more talks. He has more and more authority. And in order for the party to have more of an appeal, especially to nationalists in general, they change their name. To German Workers' Party they add the Nationalist Socialist German Workers' Party, DAP, German Workers' Party, or the NSDAP. And if you pronounce nationalist in German, it sounds something like-- and I'm going to butcher it right now-- Nazionalist. And so, if you were to shorten it, they called themselves the Nazis. And Hitler actually designed the logo for the Nazis, which included this symbol right over here, the swastika. And the swastika is worth talking about because it was really this bizarre corruption of a very ancient symbol. Hitler and the Nazis created this entire mythology around the Germans being the descendants of the Aryans, or being the purest example of the Aryans. And the Aryans are the superior race that's responsible for all of civilization's advancement. It was a delusion because frankly, there was an ancient Aryan race. But the most indicative descendants of them are frankly, the Persians or the Indians. And actually, the swastika symbol here, you might actually even see it at a Hindu temple. It does not mean all of what we associate with Nazism now. It actually is an ancient Hindu symbol of auspiciousness, of good luck. But the Nazis usurped it. But for them, this was a very important idea to create this mythology around race superiority and to even have a symbol like this as opposed to say something like a cross that's a religious symbol. Anyone could believe in Christianity and say, hey, I'm a Christian. But the swastika, at least in Hitler's mind and the Nazi's mind, was a racial symbol. So it represented their superior race. And obviously, if their race was superior, a lot of what they consider the ills of Germany were caused by being infiltrated with what they considered less pure races, like Jews, and also infiltrated with less pure ideas, like the ideas of communism. But Hitler gets more and more recognition with the party. The party membership continues to grow. And by 1921, you have some disagreements in the party. Some people threatened to splinter off. And when Hitler says, hey look, if this is going to happen, I'm going to leave the party, they realize that he has so much value to the party that the party would just dissolve if Hitler leaves. And so they make him the chairman. Hitler takes control. Hitler is the chairman of the Nazi party. And by this point, he's becoming more and more well known on the speaking circuit. And we now have several thousand members of the Nazi party. Although, it's still a fairly small group. But then, things start to get a lot worse in Weimar, Germany. You start having hyperinflation. The government keeps printing more and more currency. The economy is weak. It's trying to pay reparations. And so what you have here-- and this is actually one of the most famous cases of hyperinflation in world history-- you see the value of their currency, it devalues from 1919 to 1923 not by a factor of 1,000 or a million or a billion, but nearly a trillion. So the currency becomes, frankly, worthless. The hyperinflation is happening this entire time. And you see, it accelerates through 1922 and then 1923. But then in 1922, you have Mussolini comes to power in Italy. And he comes to power through his March on Rome. And Mussolini is a fascist. That's where the word comes from. He's a member of the Fascist Party. And the Fascists' ideas were very similar to the Nazis. It was all about extreme nationalism, all about racial superiority, a very strong government. And this Hitler finds quite inspiring. The Weimar Republic is having economic difficulty. You have many other groups, including the communists, attempt their own coup d'etats. They fail. But things are getting less and less stable. And then you fast forward to 1923. The inflation is getting super bad, about as bad as inflation can get. The currency is worthless. The economy is going into a tailspin. And on top of that, because Germany can't pay the reparations to France anymore, France occupies the Ruhr. So the Ruhr region is occupied by France. And you might remember from the terms of the Treaty of Versailles, the Saar region was already being occupied. And all that coal was being shipped out to France. The Ruhr region was another significant region of coal and steel production. And now the French are fully occupying this. They're forcing a lot of the civilians out of the region. They're forcing a lot of the workers to work in the mines and the factories. And then they're shipping all of that supply out to France. So this further debilitates the economy, but it's a huge humiliation. The Treaty of Versailles, in the minds of Germans, especially in the minds of nationalists, was bad enough. But now you have this huge humiliation by the French. And this isn't just amongst the nationalists. The general German population is getting very, very, very, very upset about this. And so this gives a lot of fuel to extreme nationalist groups, like the Nazis. So this fuels the Nazis. And based on the estimates I've seen, entering into the year they're starting to have in excess of 10,000 members, starting to be several tens of thousands. And as we get into the later part of the year, we're approaching, I've seen estimates of 40,000 to 55,000 members of the Nazi party. And that's just formal members. And then on top of that, you might have non-members who are growing increasingly sympathetic. And so what we'll see in the next video, at the end of 1923, Hitler sees this as his chance. He's inspired by Mussolini, the economy is in a tailspin, the Germans have been further humiliated by the French, and the Nazis, in particular, are starting to get quite popular." + }, + { + "Q": "what exactly is a \"special product\"?", + "A": "A special product is the product of the binomials of the form (a+b)(a-b). The product is a^2 - b^2. This is useful to remember because it shows up a lot in rational equations, but you can always prove it to yourself by using FOIL (first-outer-inner-last) or the distributive property twice: (a+b)(a-b) = (a^2) + (-ab) + (ab) + (-b^2) = a^2 - b^2", + "video_name": "4fQeHtSdw80", + "transcript": "- Find the area of a square with side (6x-5y). Let me draw our square and all of the sides of a square are going to have the same length, and they're telling us that the length for each of the sides which is the same for all of them is (6x-5y). So the height would be 6x-5y, and so would the width, 6x-5y, and if we wanted to find the area of the square we just have to multiply the width times the height. So the area for this square is just going to be the width, which is (6x-5y) times the height, times the height which is also (6x-5y), so we just have to multiply these two binomials. To do this, you could either do FOIL if you like memorizing things or you could just remember this is just applying the distributive property twice. So what we could do is distribute this entire magenta, (6x-5y), distribute it over each of these terms, in the yellow (6x-5y). If we do that, we will get this 6x times this entire (6x-5y), so (6x-5y), and then we have -5y, - 5y times once again, the entire magenta, (6x-5y). And what does this give us? So we have, we have 6x times 6x, so when I distributed just this, I'm now doing the distributive property for the second time, 6x times 6x is 36x squared, and then when I take 6x times -5y, I get 6 times -5 is -30, and then I have an x times y, -30xy. And then I want to take, I'm trying to introduce many colors here, so I have this -5y times this 6x right over here so -5 times 6 is -30. - 30 and I have a y and an x or an x and a y, and then finally I have my last distribution to do, let me do that maybe in white, I have -5y times another -5y, so the negative times a negative is a positive so it is positive, 5 times 5 is 25, y times y is y squared. And then we are almost done. Right over here, we could say, we can just add these two terms in the middle right over here, - 30xy-30xy is going to be -60xy. So you get 36x squared -60xy +25y squared. Now, there is a faster way to do this if you recognize. If you recognize that if I'm squaring a binomial, which is essentially what we're doing here, this is the exact same thing as 6x-5y squared. So you might recognize a pattern. If I have (a+b) squared, this is the same thing as (a+b) times (a+b) and if you were to multiply it out this exact same way we just did it here, the pattern here is it's a times a which is a squared, plus a times b, +ab, plus b times a, which is also +ab, we just switched the order, plus b squared, +b squared so this is equal to a squared +2ab +b squared. This is kind of the fast way to look, if you're squaring any binomial, it will be a+b squared, it will be a squared +2ab + b squared. And if you knew this ahead of time, then you could have just applied that to this squaring of the binomial right up here so let's do it that way as well. So if we 6x, (6x-5y) squared, we could just say well, this is going to be a squared. It's going to be a squared in which in this case is 6x squared +2ab, so that's +2 times a which is (6x) times b which is (-5y), - 5y +b squared, which is +(-5y), everything is squared. And then this will simplify too, 6x squared is 36x squared plus, actually it's going to be a negative here because it's going to be 2 times 6 is 12 times -5 is -60, we have x and a y, x and a y, and the -5y squared is +25y squared. So hopefully you saw multiple ways to do this, if you saw this pattern immediately, and if you knew this pattern immediately, you could just cut to the chase and go straight here, you wouldn't have to do distributive property twice, although, this will never be wrong." + }, + { + "Q": "If we see light from stars that are 500 million years old. Then how many years old would be the light from the sun?", + "A": "It depends on what you mean by how old the light is. It takes about 8 minutes for light to get to earth from the surface of the sun but the energy for a photon created inside the sun can take anywhere from a few thousand to 100 s of thousands of years to make it to the surface.", + "video_name": "rcLnMe1ELPA", + "transcript": "In the last video, I hinted that things were about to get wacky. And they are. So if we start where we left off in the last video, we started right over here, looking at the distance to the nearest star. And just as a reminder, in this drawing right here, this depiction right here, this circle right here, this solar system circle, it's not the size of the Sun. It's not the size of the orbits of the Earth, or Pluto, or the Kuiper Belt. This is close to the size of the Oort Cloud. And the actual orbit of Earth is about-- well, the diameter of the orbit of Earth is about 1/50,000 of this. So you wouldn't even see it on this. It would not even make up a pixel on this screen right here, much less the actual size of the Sun or something much smaller. And just remember that orbit of the Earth, that was at a huge distance. It takes eight minutes for light to get from the Sun to the Earth, this super long distance. If you shot a bullet at the Sun from Earth, it would take you that 17 years to actually get to the Sun. So one thing, this huge distance wouldn't even show up on this picture. Now, what we saw in the last video is that if you travel at unimaginably fast speeds, if you travel at 60,000 kilometers per hour-- and I picked that speed because that's how fast Voyager 1 actually is traveling. That's I think the fastest object we have out there in space right here. And it's actually kind of leaving the solar system as we speak. But even if you were able to get that fast, it would still take 80,000 years, 75,000 or 80,000 years, to travel the 4.2 light years to the Alpha Centauri cluster of stars. To the nearest star, it would take 80,000 years. And that scale of time is already an amount of time that I have trouble comprehending. As you can imagine, all of modern civilization has occurred, well, definitely in the last 10,000 years. But most of recorded history is in the last 4,000 or 5,000 years. So this is 80,000 years to travel to the nearest star. So it's a huge distance. Another way to think about it is if the Sun were the size of a basketball and you put that basketball in London, if you wanted to do it in scale, the next closest star, which is actually a smaller basketball, right over here, Proxima Centauri, that smaller basketball you would have to put in Kiev, Ukraine in order to have a similar scale. So these are basketballs sitting in these cities. And you would have to travel about 1,200 miles to place the next basketball. And these basketballs are representing these super huge things that we saw in the first video. The Sun, if you actually made the Earth relative to these basketballs, these would be little grains of sand. So there are any little small planets over here, they would have to be grains of sand in Kiev, Ukraine versus the grain of sand in London. So this is a massive, massive distance, already, at least in my mind, unimaginable. And when it gets really wacky is when you start really looking at this. Even this is a super, super small distance relative to the galactic scale. So this whole depiction of kind of our neighborhood of stars, this thing over here is about, give or take-- and we're doing rough estimates right here-- it's about 30 light years. I'll just do LY for short. So that's about 30 light years, And once again, you can take pictures of our galaxy from our point of view. But you actually can't take a picture of the whole galaxy from above it. So these are going to be artists' depictions. But if this is 30 light years, this drawing right here of kind of our local neighborhood of the galaxy, this right here is roughly-- and these are all approximations. Let me do this in a darker color. This is about 1,000 light years. And this is the 1,000 light years of our Sun's neighborhood, if you can even call it a neighborhood anymore. Even this isn't really a neighborhood if it takes you 80,000 years to get to your nearest neighbor. But this whole drawing over here-- now, it would take forever to get anywhere over here-- it would be 1/30 of this. So it would be about that big, this whole drawing. And what's really going to blow your mind is this would be roughly a little bit more than a pixel on this drawing right here, that spans a 1,000 light years. But then when you start to really put it into perspective-- so now, let's zoom out a little bit-- so this drawing right here, this 1,000 light years is now this 1,000 light years over here. So this is the local vicinity of the Sun. And once again, the word \"local\" is used in a very liberal way at this point. So this right here is 1,000 light years. If you're sitting here and you're looking at an object that's sitting-- let me do this in a darker color-- if we're sitting here on Earth and we're looking at an object out here that's 500 light years away, we're looking at it as it was 500 years ago because the light that is reaching our eyeballs right now, or our telescopes right now, left this guy over here 500 years ago. In fact, he's not going to even be there anymore. He probably has moved around a little bit. So just even on this scale, we're talking about these unimaginably huge distances. And then when we zoom out, this is kind of our local part of the galaxy right over here. This piece right here, this is called the Orion Spur. And people are still trying to work out exactly the details of the actual shape the Milky Way Galaxy, the galaxy that we're in. But we're pretty sure-- actually, we're very sure-- we have these spiral arms and we have these spurs off of them. But it's actually very hard to come up with the actual shape, especially because you can't see a lot of the galaxy, because it's kind of on the other side, on the other side of the center. But really just to get a sense of something that at least-- I mean it blows my mind if you really think about what it's saying-- these unbelievable distances show up as a little dot here. This whole drawing shows up as a dot here. Now when we zoom out, over here that dot would no longer even show up. It wouldn't even register a pixel on this drawing right over here. And then this whole drawing, this whole thing right over here, this whole picture is this grid right over here. It is this right over here. So hopefully, that gives you a sense of how small even our local neighborhood is relative to the galaxy as a whole. And the galaxy as a whole, just to give you a sense, has 200 to 400 billion stars. Or maybe I should say solar systems just to give you a sense that when we saw the solar system, it's not just the Sun. There's all this neat, dynamic stuff. And there are planets, and asteroids, and solar winds. And so there's 200 to 400 billion stars and for the most part, 200 and 400 billion solar systems. So it's an unimaginably, I guess, complex or huge place. And just to make it clear, even when we zoom in to this picture right here, and I think it was obvious based on telling you about this, that these little white pockets right here, This isn't two stars. These are thousands of stars here. So when you go over here, each little blotch of white that you see, that's not a star, that's not a thousand stars. We're starting to talk in the millions of stars when you look at certain blotches here and there. I mean, maybe it might be one star that's closer to you or might be a million stars that are far apart and that are just relatively close together. And everything has to be used in kind of loose terms here. And we'll talk more about other galaxies. But even this isn't the upper bound of galaxies. People believe the Andromeda Galaxy has a trillion stars in it, a trillion solar systems. We're talking about these huge, huge, immense distances. And so just to give you a sense of where we fit in the picture, this is a rough location of our Sun. And remember, that little dot I drew just now is including millions of stars, millions of solar systems, already unimaginable the distances. But if you really want to get at the sense relative to the whole galaxy, this is an artist's depiction. Once again, we could never obviously get this perspective on the galaxy. It would take us forever to travel this far so that you could see the galaxy from above. But this is our best guess looking at things from our vantage point. And we actually can't even see this whole area over here because it's on the other side of the center of the galaxy, which is super, super dense and super bright. And so it's very hard to see things on the other side. We think-- or actually there's a super massive black hole at the center of the galaxy. And we think that they're at the center of all or most galaxies. But you know the whole point of this video, actually this whole series of videos, this is just kind of-- I don't know-- to put you in awe a little bit of just how huge Because when you really think about the scale-- I don't know-- no words can really describe it. But just to give you a sense, we're about 25,000 light years from the center of the galaxy. So even when we look at things in the center of the galaxy, that's as they were 25,000 years ago. It took 25,000 years for that light to get to us. I mean when that light left the center of the galaxy, I won't even guess to think what humanity was like at that point in time. So it's these huge distances in the whole galaxy over here. And once again, like solar systems, it's hard to say the edge of the galaxy, because there's always going to be a few more stars and other things orbiting around the galaxy as you go further and further out, but it gets less dense with stars. But the main density, the main disk, is about 100,000 light years. 100,000 light years is the diameter roughly of the main part of the galaxy. And it's about 1,000 light years thick. So you can kind of imagine it as this disk, this thing that's fairly flat. But it's 1,000 light years thick. It's 1,000 light years thick. You would have to do this distance 250 times just to go from the top part of the galaxy to the bottom part, much less going across the galaxy. So it might seem relatively flat. But it's still immensely, immensely thick. And just as another way to visualize it, if this thing right over here that includes the Oort Cloud, roughly a light year in diameter, is a grain of sand, a millimeter in diameter grain of sand, then the universe as a whole is going to be the diameter of a football field. And that might tell you, OK, those are two tractable things. I can imagine a grain of sand, a millimeter wide grain of sand in a football field. But remember, that grain of sand is still 50,000 or 60,000 times the diameter of Earth's orbit. And Earth's orbit, it would still take a bullet or something traveling as fast as a jet plane 15 hours to just go half of that-- or sorry, not-- 15 years or 17 years, I forgot the exact number. But it was 15, 16, 17 years to even cover half of that distance. So 30 years just to cover the diameter of Earth's orbit. That's 1/60,000 of our little grain of sand in the football field. And just to kind of really, I don't know, have an appreciation for how mind-blowing this really is, this is actually a picture of the Milky Way Galaxy, our galaxy, from our vantage point. As you can see, we're in the galaxy and this is looking towards the center. And even this picture, you start to appreciate the complexity of what 100 billion stars are. But what I really want to point out is even in this picture, when you're looking at these things, some of these things that look like stars, those aren't stars. those are thousands of stars or millions of stars. Maybe it could be one star closer up. But when we're starting to approach the center of the galaxy, these are thousands and thousands and millions of stars or solar systems that we're actually looking at. So really, it starts to boggle the mind to imagine what might actually be going on over there." + }, + { + "Q": "I don't understand the claims at 10:37 :( . There is an equation written :\ndelta P * V = P * delta V\nand Sal says if the pressure is constant, these will cancel out.\nPressure being constant means: delta P = 0. But P still has a value.\nSo filling it in gives:\n0 * V = P * delta V\n0 = P * delta V (because P does have a value, even though it's constant)\nBut P has a value, so the right term IS NOT 0 (unless delta V is 0, or P = 0).\nSo i dont know how both terms cancel out if only the pressure is constant T_T, help?", + "A": "It s delta(PV) = P(delta V) not (delta P)*V = P(delta V)", + "video_name": "fucyI7Ouj2c", + "transcript": "Let me draw a good old PV diagram. That's my pressure axis, this is my volume axis. Just like that, I have pressure and volume. I showed several videos ago that if we start at some state here in the PV diagram, right there, and that I change the pressure and the volume to get to another state, and I do it in a quasistatic way, so essentially I'm always close to equilibrium, so my state variables are always defined, I could have some path that takes me to some other state right there. And this is my path. I'm going from this state to that state. And we showed that if I just did this, the work done by the system is the area under this curve. And then if I were to move back to the previous state, and then if I were, you know, by some path, just some random path that I happened to be drawing, the work done to the system would be the area under this light blue curve. So the net work done by the system ended up being the area inside of this path. So this is-- let me do it in different color. The net work done would be the area inside of this path when I go in this clockwise type of direction. So this is the net work done by the system. And now, we also know that if we're at some point on this PV diagram, that our state is the same as it was before. So if we go all the way here, and then go all the way back, all of our state variables will not have changed. Our pressure is the same as it was before. Our volume was the same as it was before because we went all the way back to that same point on the PV diagram. And our internal energy is also the same point as it was before, so our change in internal energy over this path, you're going to have a different internal energy here than you had here, but when you go around the circle and you get back, your change in internal energy is equal to zero. And we know that our change in internal energy is defined as, and this is from the first law of thermodynamics, the heat added to the system minus the work done by the system. Now, if we go on a closed loop on our PV diagram, then what's our change in internal energy? It's zero. So we get zero change in internal energy, because we're at the same state is equal to the heat applied to the system minus the work done or-- and I've done this little exercise multiple times. I think is probably the fourth or fifth time I'm doing it. We get that the heat added to the system, if we just add w to both sides, is equal to the work done by the system. So this area inside of this path, I already said, it's the work done by the system, and if you don't remember even where that came from, it was, remember, pressure times volume times change in volume is a little incremental change in work, and that's why it relates to the area. But we've done that multiple times. I won't go there just yet. So if you have any area here, some heat was added to the system, some net heat, right? Some heat was added here, and some heat was probably taken out here, but you have some net heat that's added to the system. And I use that argument to say why heat isn't a good state variable. Because-- and I had a whole video on this-- that if I define some state variable, let's just say, heat content. Let's say I want to define some state variable heat content. And I would say that the change in heat content would, of course, be equal to the change in heat. That's what I'm defining. If I'm adding heat to the system, my heat content should go up. But the problem with that heat content state variable was that, let's say over here, I say that the heat content is equal to 5. Now, I just showed you that if we go on some path here and we come back, and there's some area in this little path that I took, that some heat was added. So let's say that this area right here, so this is q is equal to the work done by the system, let's say it's equal to 2. So every time, if I start at heat content is equal to 5, that's just an arbitrary number, and I were to do this entire path, when I go back, the heat content would have to be 7. And then when I go back and do it again, my heat content would have to be 9. And it would have to increment by 2 every time I do this It would have to increment by the amount of area that this path goes around. So heat content can't be a state variable, because it's dependent on how you got there. A state variable-- and remember this. In order to be a state variable, if you're at this point, you have to have the same value. If your internal energy was 10 here, when you do the path and you come back, your internal energy will be 10 again. That's why internal energy is a valid state variable. It's dependent only on your state. If your entropy was 50 here, when you soon. go back and you do all sorts of crazy things, and you come back to this point, your entropy is once again 50. If your pressure here is 5 atmospheres, when you come back here, your pressure will be 5 atmospheres. Your state variable cannot change based on what path you took. If you're at a certain state, that's all that matters to the state variable. Now this heat content didn't work, and that's why we actually led into some videos where I divided by t and we got entropy, which was an interesting variation. But that's still not satisfying. What if we really wanted to develop something that could in some way be a state variable, but at the same time measure heat? So obviously, we're going to have to make some compromises, because if we just do a very arbitrary kind of heat content variable, then every time you go around this, it's going to change. That's not a valid state variable. So let's see if we can make up one. So let's just make up a definition. Let's call my new thing that I'm going to try to maybe approximate heat, let's call it h, and just as a little bit of a preview, we're going to call it enthalpy. And let's just define it. I'm just playing around. Let's just define it as the internal energy plus my pressure times my volume. So then what would my change in enthalpy be? So my change in enthalpy will be, of course, the change of these things. But I could just say, that's my change in my internal energy plus my change in pressure times volume. Now, this is interesting. And I want to make a point here. This, by definition, is a valid state variable. Why is it? Because it's the addition of other state variables, right? At any point in my PV diagram, and it's also true if I did diagrams that were entropy in temperature or anything that dealt with state variables, at any point on my diagram, u is going to be the same, no matter how I got there. p is by definition going to be the same. That's why it's at that point. v is definitely going to be the same point. So if I just add them up, this is a valid state variable because it's just the sum of a bunch of other valid state variables. So let's see if we can somehow relate this thing that we've already established as a valid state variable. From the get-go, from our definition, this works because it's just the sum of completely valid state variables. So let's see if we can relate this somehow to heat. So we know what delta u is. If we're dealing with all of the internal energy or the change in internal energy, and I'm not going to deal with all the other chemical potentials and all of that, it's equal to the heat applied to the system minus the work done by the system, right? Let me put everything else there. The change in enthalpy is equal to the heat applied minus the work done-- that's just the change in internal energy-- plus delta PV. This is just from the definition of my enthalpy. Now this is starting to look interesting. What's the work done by a system? So I could write change in h, or enthalpy, is equal to the heat applied to the system minus-- what's the work done If I have some system here, it's got some piston on it, you know, if we're doing it in a quasistatic, I have those classic pebbles that I've talked about in multiple videos. When I apply heat or let's say I remove some of these pebbles, so I'm at a different equilibrium, but what's actually happening? When is the work being done? You have some pressure being applied up here, and this piston is going to be moving up, and your volume is going And we showed multiple videos ago that the work done by the system can be, and you can kind of view this as the volume expansion work, it's equal to pressure times change in volume. And let's add the other part. So this was our change in internal energy. And I had several videos where I show this. And now let me add the other part of the equation. So our enthalpy, our change in enthalpy, can be defined by this. Something interesting is going on. I said that I wanted to define something, because I wanted to somehow measure heat content. My change in enthalpy will be equal to the heat added to the system, if these last two terms cancel out. If I can somehow get these last two terms to cancel out, then my change in enthalpy will be equal to this, if somehow these are equal to each other. So under what conditions are these equal to each other? Or another way, under what conditions is delta pressure times volume equal to pressure times delta volume? When does this happen? When can I make this statement? Because if I can make this statement, then these two terms are equivalent right here, and then you my change in enthalpy will be equal to the heat added. Well, the only way I can make this statement is if pressure is constant. Now why is that? Let's just think about it mathematically. If this is a constant, then if I just change-- you know, if this is just 5, 5 times a change in something is the same thing as the change in 5 times that thing, so it just mathematically works out. Or if you view it another way, if this is a constant, you can just factor it out, right? Well, if I said, the change in 5x, that would be equal to 5 times x final minus 5 times x initial. And you could say, well, that's just equal to 5 times x final minus x initial. Well, that's just equal to 5 times the change in x. It's kind of almost too obvious for me to explain. I think sometimes when you overexplain things, it might become more confusing. So this applies-- and the 5 I'm just doing as the analogy So if pressure is constant, then this equation is true. So if pressure is constant-- so this is a key assumption-- then if heat is being applied in a constant pressure system-- so we could write it this way. I'll write it multiple times, because this is key. If pressure is constant, then our definition, our little thing we made up, this enthalpy thing, which we defined as internal energy plus pressure and volume, then in a constant pressure system, our change in enthalpy we just showed is equal to the heat added to the system because all of these two things become equivalent under constant pressure, so I should write that. This is only true when heat is added in a constant pressure system. So how does this gel with what we did up here on our PV diagram? What's happening in a constant pressure system? Let me draw our PV diagram. That's P, that's V. So what's happening in constant pressure? We're at some pressure right there. So if we're under constant pressure, that means we can only move along this line. So we could go from here to there and back to there, or we could go from there to there, back to there. So we could go there, all the way there, and then go back. But what do we see about this? Is there any area in this curve? I mean, there is no curve to speak of, because we're staying in a constant pressure. We've kind of squeezed out this diagram. We've made the forward path and the return path the same exact path. So because of this, you don't have that state problem because no net heat is being added to the system when you go from this point all the way to this point and then back to this point. So because of that, you can kind of see visually that enthalpy in a constant pressure, when you're not moving up and down in pressure, is the same thing as heat added. So you might say, hey Sal, this was a bit of a compromise, constant pressure, you know, that's a big assumption to make. Why is this useful at all? Well, it's useful, because most chemical reactions, especially ones that occur in an open beaker, or that might occur at sea level, and that should be a big clue, they occur at constant pressure. You know, if I'm sitting at the beach, and I have my chemistry set, and I have some beaker of something, and I'm throwing other stuff into it, and I'm looking for a reaction or something, it's a constant pressure system. This is going to be atmospheric pressure. I'm sitting at sea level. So this is actually a very useful concept for everyday It might not be so useful for engines, because engines always have pressure changing, but it's very useful for actual chemistry, for actually dealing with what's going to happen to a reaction at a constant pressure. So what we're going to see is that this enthalpy, you can kind of view it as the heat content when pressure is constant. In fact, it is the heat content when pressure is constant. So somehow-- well, not somehow, I showed you how-- we were able to make this definition, which by definition was a state variable, because it was the sum of other state variables, and if we just make that one assumption of constant pressure, it all of a sudden reduces to the heat content of that system. So we'll talk more in the future of measuring enthalpy, but you just have to say, if pressure is constant, enthalpy is the same thing as-- and it's really only useful when we're dealing with a constant pressure. But if we have a pressure constant, enthalpy can be imagined as heat content. And it's very useful for understanding whether chemical reactions need heat to occur or whether they release heat, so on and so forth. See" + }, + { + "Q": "Around 5:05 he describes the occupation of Czechslovakia. Didn't they occupy Bohemia+Moravia while creating a puppet state in Slovakia?", + "A": "To start with, yes - but this was quickly enveloped by the greater German mass.", + "video_name": "-kKCjwNvNkQ", + "transcript": "World War II was the largest conflict in all of human history. The largest and bloodiest conflict And so you can imagine it is quite complex My goal in this video is to start giving us a survey, an overview of the war. And I won't even be able to cover it all in this video. It is really just a think about how did things get started. Or what happened in the lead up? And to start I am actually going to focus on Asia and the Pacific. Which probably doesn't get enough attention when we look at it from a western point of view But if we go back even to the early 1900s. Japan is becoming more and more militaristic. More and more nationalistic. In the early 1900s it had already occupied... It had already occupied Korea as of 1910. and in 1931 it invades Manchuria. It invades Manchuria. So this right over here, this is in 1931. And it installs a puppet state, the puppet state of Manchukuo. And when we call something a puppet state, it means that there is a government there. And they kind of pretend to be in charge. But they're really controlled like a puppet by someone else. And in this case it is the Empire of Japan. And we do remember what is happening in China in the 1930s. China is embroiled in a civil war. So there is a civil war going on in China. And that civil war is between the Nationalists, the Kuomintang and the Communists versus the Communists The Communists led by Mao Zedong. The Kuomintang led by general Chiang Kai-shek. And so they're in the midst of the civil war. So you can imagine Imperial Japan is taking advantage of this to take more and more control over parts of China And that continues through the 30s until we get to 1937. And in 1937 the Japanese use some pretext with, you know, kind of a false flag, kind of... well, I won't go into the depths of what started it kind of this Marco-Polo Bridge Incident But it uses that as justifications to kind of have an all-out war with China so 1937...you have all-out war and this is often referred to as the Second Sino-Japanese War ...Sino-Japanese War Many historians actually would even consider this the beginning of World War II. While, some of them say, ok this is the beginning of the Asian Theater of World War II of the all-out war between Japan and China, but it isn't until Germany invades Poland in 1939 that you truly have the formal beginning, so to speak, of World War II. Regardless of whether you consider this the formal beginning or not, the Second Sino-Japanese War, and it's called the second because there was another Sino-Japanese War in the late 1800s that was called the First Sino-Japanese War, this is incredibly, incredibly brutal and incredibly bloody a lot of civilians affected we could do a whole series of videos just on that But at this point it does become all-out war and this causes the civil war to take a back seat to fighting off the aggressor of Japan in 1937. So that lays a foundation for what's happening in The Pacific, in the run-up to World War II. And now let's also remind ourselves what's happening... what's happening in Europe. As we go through the 1930s Hitler's Germany, the Nazi Party, is getting more and more militaristic. So this is Nazi Germany... Nazi Germany right over here. They're allied with Benito Mussolini's Italy. They're both extremely nationalistic; they both do not like the Communists, at all You might remember, that in 1938... 1938, you have the Anschluss, which I'm sure I'm mispronouncing, and you also have the takeover of the Sudetenland in Czechoslovakia. So the Anschluss was the unification with Austria and then you have the Germans taking over the of Sudetenland in Czechoslovakia and this is kind of the famous, you know, the rest of the, what will be called the Allied Powers kind of say, \"Okay, yeah, okay maybe Hitler's just going to just do that... well we don't want to start another war. We still all remember World War I; it was really horrible. And so they kind of appease Hitler and he's able to, kind of, satisfy his aggression. so in 1938 you have Austria, Austria and the Sudetenland ...and the Sudetenland... are taken over, are taken over by Germany and then as you go into 1939, as you go into 1939 in March they're able to take over all of Czechoslovakia they're able to take over all of Czechoslovakia and once again the Allies are kind of, they're feeling very uncomfortable, they kind of, have seen something like this before they would like to push back, but they still are, kind of, are not feeling good about starting another World War so they're hoping that maybe Germany stops there. So let me write this down... So all of Czechoslovakia... ...Czechoslovakia... is taken over by the Germans. This is in March of 1939. And then in August you have the Germans, and this is really in preparation for, what you could guess is about to happen, for the all-out war that's about to happen the Germans don't want to fight the Soviets right out the gate, as we will see, and as you might know, they do eventually take on the Soviet Union, but in 1939 they get into a pact with the Soviet Union. And so this is, they sign the Molotov\u2013Ribbentrop Pact with the Soviet Union, this is in August, which is essentially mutual non-aggression \"Hey, you know, you do what you need to do, we know what we need to do.\" and they secretly started saying \"Okay were gonna, all the countries out here, we're going to create these spheres of influence where Germany can take, uh, control of part of it and the Soviet Union, and Stalin is in charge of the Soviet Union at this point, can take over other parts of it. And then that leads us to the formal start where in September, let me write this in a different color... so September of 1939, on September 1st, Germany invades Poland Germany invades Poland on September 1st, which is generally considered the beginning of World War II. and then you have the Great Britain and France declares war on Germany so let me write this World War II... starts everyone is declaring war on each other, Germany invades Poland, Great Britain and France declare war on Germany, and you have to remember at this point Stalin isn't so concerned about Hitler he's just signed the Molotov-Ribbentrop Pact and so in mid-September, Stalin himself invades Poland as well so they both can kind of carve out... ...their spheres of influence... so you can definitely sense that things are not looking good for the world at this point you already have Asia in the Second Sino-Japanese War, incredibly bloody war, and now you have kind of, a lot of very similar actors that you had in World War I and then they're starting to get into a fairly extensive engagement." + }, + { + "Q": "Is there fusion going on in the core?", + "A": "Nuclear fusion happens in the core of a star once it has reached a certain temperature (around 15 million degrees). The maximum temperature of the inner core is around 6,000 degrees, so definitely not!", + "video_name": "f2BWsPVN7c4", + "transcript": "What I want to do in this video is talk a little bit about plate tectonics. And you've probably heard the word before, and are probably, or you might be somewhat familiar with what it discusses. And it's really just the idea that the surface of the Earth is made up of a bunch of these rigid plates. So it's broken up into a bunch of rigid plates, and these rigid plates move relative to each other. They move relative to each other and take everything that's on them for a ride. And the things that are on them include the continents. So it literally is talking about the movement of these plates. And over here I have a picture I got off of Wikipedia of the actual plates. And over here you have the Pacific Plate. Let me do that in a darker color. You have the Pacific Plate. You have a Nazca Plate. You have a South American Plate. I could keep going on. You have an Antarctic Plate. It's actually, obviously whenever you do a projection onto two dimensions of a surface of a sphere, the stuff at the bottom and the top look much bigger than they actually are. Antarctica isn't this big relative to say North America or South America. It's just that we've had to stretch it out to fill up the rectangle. But that's the Antarctic Plate, North American Plate. And you can see that they're actually moving relative to each other. And that's what these arrows are depicting. You see right over here the Nazca Plate and the Pacific Plate are moving away from each other. New land is forming here. We'll talk more about that in other videos. You see right over here in the middle of the Atlantic Ocean the African Plate and the South American Plate meet each other, and they're moving away from each other, which means that new land, more plate material I guess you could say, is somehow being created right here-- we'll talk about that in future videos-- and pushing these two plates apart. Now, before we go into the evidence for plate tectonics or even some of the more details about how plates are created and some theories as to why the plates might move, what I want to do is get a little bit of the terminology of plate tectonics out of the way. Because sometimes people call them crustal plates, and that's not exactly right. And to show you the difference, what I want to do is show you two different ways of classifying the different layers of the Earth and then think about how they might relate to each other. So what you traditionally see, and actually I've made a video that goes into a lot more detail of this, is a breakdown of the chemical layers of the Earth. And when I talk about chemical layers, I'm talking about what are the constituents of the different layers? So when you talk of it in this term, the top most layer, which is the thinnest layer, is the crust. Then below that is the mantle. Actually, let me show you the whole Earth, although I'm not going to draw it to scale. So if I were to draw the crust, the crust is the thinnest outer layer of the Earth. You can imagine the blue line itself is the crust. Then below that, you have the mantle. So everything between the blue and the orange line, this over here is the mantle. So let me label the crust. The crust you can literally view as the actual blue pixels over here. And then inside of the mantle, you have the core. And when you do this very high level division, these are chemical divisions. This is saying that the crust is made up of different types of elements. Its makeup is different than the stuff that's in the mantle, which is made up of different things than what's inside the core. It's not describing the mechanical properties of it. And when I talk about mechanical properties I'm talking about whether something is solid and rigid. Or maybe it's so hot and melted it's kind of a magma, or kind of a plastic solid. So this would be the most brittle stuff. If it gets warmed up, if rock starts to melt a little bit, then you have something like a magma, or you can view it as like a deformable or a plastic solid. When we talk about plastic, I'm not talking about the stuff that the case of your cellphone is made of. I'm talking about it's deformable. This rock is deformable because it's so hot and it's somewhat melted. It kind of behaves like a fluid. It actually does behave like a fluid, but it's much more viscous. It's much thicker and slower moving than what we would normally associate with a fluid like water. So this a viscous fluid. And then the most fluid would, of course, be the liquid state. This is what we mean when we talk about the mechanical properties. And when you look at this division over here, the crust is solid. The mantle actually has some parts of it that are solid. So the uppermost part of the mantle is solid. Then below that, the rest of the mantle is kind of in this magma, this deformable, somewhat fluid state, and depending on what depth you go into the mantle there are kind of different levels of fluidity. And then the core, the outer level layer of the core, the outer core is liquid, because the temperature is so high. The inner core is made up of the same things, and the temperature is even higher, but since the pressure is so high it's actually solid. So that's why the mantle, crust, and core differentiations don't tell you whether it's solid, whether it's magma, or whether it's really a liquid. It just really tells you what the makeup is. Now, to think about the makeup, and this is important for plate tectonics, because when we talk about these plates we're not talking about just the crust. We're talking about the outer, rigid layer. Let me just zoom in a little bit. Let's say we zoomed in right over there. So now we have the crust zoomed in. This right here is the crust. And then everything below here we're actually talking about the upper mantle. We haven't gotten too deep in the mantle right here. So that's why we call it the upper mantle. Now, right below the crust, the mantle is cool enough that it is also in real solid form. So this right here is solid mantle. And when we talk about the plates were actually talking about the outer solid layer. So that includes both the crust and the solid part of the mantle. And we call that the lithosphere. When people talk about plate tectonics, they shouldn't say crustal plates. They should call these lithospheric plates. And then below the lithosphere you have the least viscous part of the mantle, because the temperature is high enough for the rock to melt, but the pressure isn't so large as what will happen when you go into the lower part of the mantle that the fluid can actually kind of move past each other, although still pretty viscous. This still a magma. So this is still kind of in its magma state. And this fluid part of the mantle, we can't quite call it a liquid yet, but over large periods of time it does have fluid properties. This, that essentially the lithosphere is kind of riding on top of, we call this the asthenosphere. So when we talk about the lithosphere and asthenosphere we're really talking about mechanical layers. The outer layer, the solid layers, the lithosphere sphere. The more fluid layer right below that is the asthenosphere. When we talk about the crust, mantle, and core, we are talking about chemical properties, what are the things actually made up of." + }, + { + "Q": "Why we have 2 different enthalpy (combustion&reaction enthalpy) but not just reaction enthalpy?", + "A": "They are all just enthalpy changes. It is common to refer to an enthalpy change according to the process involved. Thus, we talk about enthalpy of combustion, enthalpy of neutralization, enthalpy of vaporization, etc. But they are all just enthalpy changes.", + "video_name": "8bCL8TQZFKo", + "transcript": "This problem is from chapter five of the Kotz, Treichel, Townsend Chemistry and Chemical Reactivity textbook. So they tell us, suppose you want to know the enthalpy change-- so the change in total energy-- for the formation of methane, CH4, from solid carbon as a graphite-- that's right there-- and hydrogen gas. So we want to figure out the enthalpy change of this reaction. How do we get methane-- how much energy is absorbed or released when methane is formed from the reaction of-- solid carbon as graphite and hydrogen gas? So they tell us the enthalpy change for this reaction cannot to be measured in the laboratory because the reaction is very slow. So normally, if you could measure it you would have this reaction happening and you'd kind of see how much heat, or what's the temperature change, of the surrounding solution. Maybe this is happening so slow that it's very hard to measure that temperature change, or you can't do it in any meaningful way. We can, however, measure enthalpy changes for the combustion of carbon, hydrogen, and methane. So they're giving us the enthalpy changes for these combustion reactions-- combustion of carbon, combustion of hydrogen, combustion of methane. And they say, use this information to calculate the change in enthalpy for the formation of methane from its elements. So any time you see this kind of situation where they're giving you the enthalpies for a bunch of reactions and they say, hey, we don't know the enthalpy for some other reaction, and that other reaction seems to be made up of similar things, your brain should immediately say, hey, maybe this is a Hess's Law problem. Hess's Law. And all Hess's Law says is that if a reaction is the sum of two or more other reactions, then the change in enthalpy of this reaction is going to be the sum of the change in enthalpies of those reactions. Now, when we look at this, and this tends to be the confusing part, how can you construct this reaction out of these reactions over here? And what I like to do is just start with the end product. So I like to start with the end product, which is methane in a gaseous form. And when we look at all these equations over here we have the combustion of methane. So this actually involves methane, so let's start with this. But this one involves methane and as a reactant, not a product. But what we can do is just flip this arrow and write it as methane as a product. So if we just write this reaction, we flip it. So now we have carbon dioxide gas-- let me write it down here-- carbon dioxide gas plus-- I'll do this in another color-- plus two waters-- if we're thinking of these as moles, or two molecules of water, you could even say-- two molecules of water in its liquid state. That can, I guess you can say, this would not happen spontaneously because it would require energy. But if we just put this in the reverse direction, if you go in this direction you're going to get two waters-- or two oxygens, I should say-- I'll do that in this pink color. So two oxygens-- and that's in its gaseous state-- plus a gaseous methane. CH4. CH4 in a gaseous state. And all I did is I wrote this third equation, but I wrote it in reverse order. I'm going from the reactants to the products. When you go from the products to the reactants it will release 890.3 kilojoules per moles of the reaction going on. But if you go the other way it will need 890 kilojoules. So the delta H here-- I'll do this in the neutral color-- so the delta H of this reaction right here is going to be the reverse of this. So it's positive 890.3 kilojoules per mole of the reaction. All I did is I reversed the order of this reaction right there. The good thing about this is I now have something that at least ends up with what we eventually want to end up with. This is where we want to get. This is where we want to get eventually. Now, if we want to get there eventually, we need to at some point have some carbon dioxide, and we have to have at some point some water to deal with. So how can we get carbon dioxide, and how can we get water? Well, these two reactions right here-- this combustion reaction gives us carbon dioxide, this combustion reaction gives us water. So we can just rewrite those. Let me just rewrite them over here, and I will-- let me use So if I start with graphite-- carbon in graphite form-- carbon in its graphite form plus-- I already have a color for oxygen-- plus oxygen in its gaseous state, it will produce carbon dioxide in its gaseous form. It will produce carbon-- that's a different shade of green-- it will produce carbon dioxide in its gaseous form. And this reaction, so when you take the enthalpy of the carbon dioxide and from that you subtract the enthalpy of these reactants you get a negative number. Which means this had a lower enthalpy, which means energy Because there's now less energy in the system right here. So this is essentially how much is released. But our change in enthalpy here, our change in enthalpy of this reaction right here, that's reaction one. I'll just rewrite it. Minus 393.5 kilojoules per mole of the reaction occurring. So the reaction occurs a mole times. This would be the amount of energy that's essentially released. This is our change in enthalpy. So if this happens, we'll get our carbon dioxide. Now we also have-- and so we would release this much energy and we'd have this product to deal with-- but we also now need our water. And this reaction right here gives us our water, the combustion of hydrogen. So we have-- and I haven't done hydrogen yet, so let me do hydrogen in a new color. That's not a new color, so let me do blue. So right here you have hydrogen gas-- I'm just rewriting that reaction-- hydrogen gas plus 1/2 O2-- pink is my color for oxygen-- 1/2 O2 gas will yield, will it give us some water. Will give us H2O, will give us some liquid water. Now, before I just write this number down, let's think about whether we have everything we need. To make this reaction occur, because this gets us to our final product, this gets us to the gaseous methane, we need a mole. Or we can even say a molecule of carbon dioxide, and this reaction gives us exactly one molecule of carbon dioxide. So that's a check. And we need two molecules of water. Now, this reaction only gives us one molecule of water. So let's multiply both sides of the equation to get two molecules of water. So this is a 2, we multiply this by 2, so this essentially You multiply 1/2 by 2, you just get a 1 there. And then you put a 2 over here. So I just multiplied this second equation by 2. So I just multiplied-- this is becomes a 1, this becomes a 2. And if you're doing twice as much of it, because we multiplied by 2, the delta H now, the change enthalpy of the reaction, is now going to be twice this. Let's get the calculator out. It's now going to be negative 285.8 times 2. Because we just multiplied the whole reaction times 2. So negative 571.6. So it's negative 571.6 kilojoules per mole of the reaction. Now, let's see if the combination, if the sum of these reactions, actually is this reaction up here. And to do that-- actually, let me just copy and paste this top one here because that's kind of the order that we're going to go in. You don't have to, but it just makes it hopefully a little bit easier to understand. So let me just copy and paste this. Actually, I could cut and paste it. Cut and then let me paste it down here. That first one. And let's see now what's going to happen. To see whether the some of these reactions really does end up being this top reaction right here, let's see if we can cancel out reactants and products. Let's see what would happen. So this produces carbon dioxide, but then this mole, or this molecule of carbon dioxide, is then used up in this last reaction. So this produces it, this uses it. Let me do it in the same color so it's in the screen. This reaction produces it, this reaction uses it. Now, this reaction right here produces the two molecules of water. And now this reaction down here-- I want to do that same color-- these two molecules of water. Now, this reaction down here uses those two molecules of water. Now, this reaction right here, it requires one molecule of molecular oxygen. This one requires another molecule of molecular oxygen. So these two combined are two molecules of molecular oxygen. So those are the reactants. And in the end, those end up as the products of this last reaction. So those, actually, they go into the system and then they leave out the system, or out of the sum of reactions unchanged. So they cancel out with each other. So we could say that and that we cancel out. And so what are we left with? What are we left with in the reaction? Well, we have some solid carbon as graphite plus two moles, or two molecules of molecular hydrogen yielding-- all we have left on the product side is some methane. So it is true that the sum of these reactions is exactly what we want. All we have left on the product side is the graphite, the solid graphite, plus the molecular hydrogen, plus the gaseous hydrogen-- do it in that color-- plus two hydrogen gas. And all we have left on the product side is the methane. All we have left is the methane in the gaseous form. So it is true that the sum of these reactions-- remember, we have to flip this reaction around and change its sign, and we have to multiply this reaction by 2 so that the sum of these becomes this reaction that we really care about. So this is the sum of these reactions. Its change in enthalpy of this reaction is going to be the sum of these right here. That is Hess's Law. So this is the fun part. So we just add up these values right here. So we have negative 393.-- no, that's not what I wanted to do. Let me just clear it. So I have negative 393.5, so that step is exothermic. And then we have minus 571.6. That is also exothermic. Those were both combustion reactions, which are, as we know, very exothermic. And we have the endothermic step, the reverse of that last combustion reaction. So plus 890.3 gives us negative 74.8. It gives us negative 74.8 kilojoules for every mole of the reaction occurring. Or if the reaction occurs, a mole time. So there you go. We figured out the change in enthalpy. And it is reasonably exothermic. Nowhere near as exothermic as these combustion reactions right here, but it is going to release energy. And we're done." + }, + { + "Q": "the problem asks for the absolute maximum and the |-8 - 2pi| is clearly the largest so should the value be x = -4 and not x = 5/2", + "A": "No. Absolute maximum means the highest point on the graph of g(x), not the point where the absolute value of g(x) is the highest.", + "video_name": "OvMBNVi5bLY", + "transcript": "Part b. \"Determine the x-coordinate of the point at which g has an absolute maximum on the interval negative 4 is less than or equal to x, is less than or equal to 3. And justify your answer.\" So let's just think about it in general terms. If we just think about a general function over an interval, where it could have an absolute maximum. So let me draw some axes over here. And I'm speaking in general terms first, and then we can go back to our function g, which is derived from this function f right over here. So let's say that these are my coordinate axes, and let's say we care about some interval here. So let's say this is the interval that I care about. A function could look something like this. And in this case, its absolute maximum is going to occur at the beginning of the interval. Or a function could look something like this. And then the absolute maximum could occur at the end point of the interval. Or the other possibility is that the function looks something like this. At which point, the maximum would be at this critical point. And I say critical point, as opposed to just a point where the slope is zero, because it's possible to the functions not differentiable there. You could imagine a function that looks like this, and maybe wouldn't be differentiable there. But that point there still would be the absolute maximum. So what we really just have to do is evaluate g at the different endpoints of this interval, to see how high it gets, or how large of a value we get for the g at the end points. And then we have to see if g has any critical points in between. And then evaluate it there to see if that's a candidate for the global maximum. So let's just evaluate g of the different points. So let's start off, let's evaluate g at negative 4-- at kind of the lowest end, or the starting point of our interval. So g of negative 4 is equal to 2 times negative 4 plus the integral from 0 to negative 4 f of t dt. The first part is very easy, 2 times negative 4 is negative 8. Let me do it over here so I have some real estate. So this is equal to negative 8. And instead of leaving this as 0 to negative 4 f of t dt, let's change the bounds of integration here. Especially so that we can get the lower number as the lower bound. And that way, it becomes a little bit more natural to think of it in terms of areas. So this expression right here can be rewritten as the negative of the integral between negative 4 and 0 f of t dt. And now this expression right over here is the area under f of t, or in this case, f of x-- or the area under f between negative 4 and 0. So it's this area right over here. And we have to be careful, because this part over here is below the x-axis. So this we would consider negative area when we think of it in integration terms. And this would be positive area. So the total area here is going to be this positive area minus this area right over here. So let's think about what this is. So this area-- This section over here we did this in part a, actually, This section-- this is a quarter circle, so it's 1/4-- so these are both quarter circles. So we could multiply 1/4 times the area of this entire circle, if we were to draw the entire thing all the way around. It has a radius of 3. So the area of the entire circle would be pi times 3 squared or it would be 9 pi. And of course we're going to divide it by 4-- multiply it by 1/4 to just get this quarter circle right over there. And then this area right over here, the area of the entire circle, we have a radius of 1. So it's going to be pi times 1 squared. So it's going to be pi, and then we're going to divide it by four, because it's only one fourth of that circle. And we're going to subtract that. So we have negative pi and we were multiplying it times 1/4 out here, because it's just a quarter circle in either case. And we're subtracting it, because the area is below the x-axis. And so this simplifies to-- this is equal to 1/4 times 8 pi, which is the same thing as 2 pi. Did I do that right? 1/4 times 8 pi. So this all simplifies to 2 pi. So g of negative 4 is equal to negative 8 minus 2 pi. So clearly it is a negative number here. More negative than negative 8. So let's try the other bounds. So let's see what g of positive 3 is. I'll do it over here so I have some more space. So g of positive 3, when x is equal to 3, that-- we go back to our definition-- that is 2 times 3 plus the integral from 0 to 3 f of t dt. And this is going to be equal to 2 times 3 is 6. And the integral from 0 to 3 f of t dt, that's this entire area. So we have positive area over here. And then we have an equal negative area right over here because it's below the x-axis. So the integral from 0 to 3 is just going to be 0, you're going to have this positive area, and then this negative area right over here is going to completely cancel it out. Because it's symmetric right over here. So this thing is going to evaluate to 0. So g of 3 is 6. So we already know that our starting point, g of negative 4-- that when x is equal to negative 4-- that is not where g hits a global maximum. Because that's a negative number. And we already found the end point, where g hits a positive value. So negative 4 is definitely not a candidate. x is equal to 3 is still in the running for the x-coordinate where g has a global absolute maximum. Now what we have to do is figure out any critical points that g has in between. So points in there where it's either undifferentiable or its derivative is equal to 0. So let's look at this derivative. So g prime of x-- we just take the derivative of this business up here. Derivative of 2x is 2. Derivative of this definition going from 0 to x of f of t dt-- we did that in part a, this is just the fundamental theorem of calculus-- this is just going to be plus f of x right over there. So it actually turns out that g is differentiable over the entire interval. You give any x value over this interval, we have a value for f of x. f of x isn't differentiable everywhere, but definitely f of x is defined everywhere, over the interval. So you'll get a number here, and obviously two is just two, and you add two to it, and you get the derivative of g at that point in the interval. So g is actually differentiable throughout the interval. So the only critical points would be where this derivative is equal to 0. So let's set this thing equal to 0. So we want to solve the equation-- I'll just rewrite it actually-- So we want to solve the equation g prime of x is equal to 0, or 2 plus f of x is equal to 0. You can subtract 2 from both sides, and you get f of x is equal to negative 2. So any x that satisfies f of x is equal to negative 2 is a point where the derivative of g is equal to 0. And let's see if f of x is equal to negative 2 at any point. So let me draw a line over here at negative 2. We have to look at it visually, because there's only given us this visual definition of f of x. Doesn't equal negative 2, doesn't equal to negative 2, only equals negative 2 right over there. And it looks like we're at about 2 and 1/2, but let's get exact. Let's actually figure out the slope of the line, and figure out what x value actually gives f of x equal to negative 2. And we could figure out the slope of this line fairly visually-- or figure out the equation of this line fairly visually, we can figure out its slope. If we run 3-- if our change in x is 3, then our change in y, our rise, is negative 6. Change in y is equal to negative 6. Slope is rise over run, or change in y over change in x. So negative 6 divided by 3 is negative 2. It has a slope of negative 2. And actually, I could have done that easier. Where if we go forward one, we go down by 2. So we could have seen that the slope is negative 2. So this part of f of x, we have y is equal to negative 2x plus-- and then the y-intercept is pretty straightforward. This is at 3-- 1, 2, 3. Negative 2x plus 3. So part of f of x where clearly it equals negative 2 at some point of that-- this part of f of x is defined by this line. Obviously this part of f of x is not defined by that line. But to figure out the exact value, we just have to figure out when this line is equal to negative 2. So we have negative 2x plus 3 is equal to negative 2. And remember, this isn't-- this is what f of x is equal to, over the interval that we care about. If we were talking about f of x over there, we wouldn't be able to put a negative 2x plus 3, we would have to have some form of equation for these circles. But right over here, this is what f of x is, and now we can solve this pretty straightforwardly. So we can subtract 3 from both sides, and we get negative 2x is equal to negative 5. Divide both sides by negative 2, you get x is equal to negative 5 over negative 2, which is equal to 5/2. Which is exactly what we thought it was when we looked at it visually. It looked like we were at about 2 and 1/2, which is the same thing as 5/2. Now we don't know what this is. We don't know if this is an inflection point. Is this a maximum? Is this a minimum? So really we just want to evaluate g at this point to see if it gets higher than when we evaluate g at 3. So let's evaluate g at 5/2. So g at 5/2 is going to be equal to 2 times 15/2 plus the integral from 0 to 5/2 of f of t dt. So this first part right over here, the 2's cancel out. So this is going to be equal to 5. And then plus the integral from 0 to 5/2. Now, you might be able to do it visually, but we know what the value is of f of t over this interval, we already figured out the equation for it over this interval. So it's the integral of negative 2x plus 3 dt. And then let's just evaluate this. Let me get some real estate. So this is-- let me draw a line here., so we don't get confused. So this is going to be equal to 5 plus and then I take the antiderivative. The antiderivative of negative 2x is negative x squared. So we have negative x squared. And the antiderivative of 3 is just going to be 3x. So plus 3x. And we're going to evaluate it from 0 to 5/2. So this is going to be equal to 5 plus-- and I'll do all this stuff right over here. I'll do this stuff in green. So when we evaluate it at 5/2, this is going to be negative 5/2 squared. So it's going to be negative 25 over 4 plus 3 times 5/2, which is 15 over 2. And then from that, we're going to subtract this evaluated at 0. But negative 0 squared plus 3 times 0 is just 0. So this is what it simplifies to. And so what do we have right over here? So let's get our ourselves a common denominator. Looks like a common denominator right over here could be 4 So this is equal to-- 5 is the same thing as 20 over 4 minus 25 over 4 and then plus 30 over 4. So 20 plus 30 is 50 minus 25 is 25. So this is equal to 25 over 4. And 25 over 4 is the same thing as 6 and 1/4. So when we evaluate our function at this critical point, at this thing where the slope, or the derivative, is equal to 0, we got 6 and 1/4, which is higher than six, which is what g was at this end point. And it's definitely higher than what g was a negative 4. So the x-coordinate of the point at which g has an absolute maximum, on the interval negative 4 to three is x is equal to 5/2." + }, + { + "Q": "What happens when you approach the freezing temperature of something? If you cooled it equally, would it all freeze at the same time?", + "A": "Theoretically yes but most elements or compounds won t cool evenly but instead they form crystal lattices well cooling which eventually connect to form the solid that you would see.", + "video_name": "pKvo0XWZtjo", + "transcript": "I think we're all reasonably familiar with the three states of matter in our everyday world. At very high temperatures you get a fourth. But the three ones that we normally deal with are, things could be a solid, a liquid, or it could be a gas. And we have this general notion, and I think water is the example that always comes to at least my mind. Is that solid happens when things are colder, relatively colder. And then as you warm up, you go into a liquid state. And as your warm up even more you go into a gaseous state. So you go from colder to hotter. And in the case of water, when you're a solid, you're ice. When you're a liquid, some people would call ice water, but let's call it liquid water. I think we know what that is. And then when it's in the gas state, you're essentially vapor or steam. So let's think a little bit about what, at least in the case of water, and the analogy will extend to other types of molecules. But what is it about water that makes it solid, and when it's colder, what allows it to be liquid. And I'll be frank, liquids are kind of fascinating because you can never nail them down, I guess is the best way to view them. Or a gas. So let's just draw a water molecule. So you have oxygen there. You have some bonds to hydrogen. And then you have two extra pairs of valence electrons in the oxygen. And a couple of videos ago, we said oxygen is a lot more electronegative than the hydrogen. It likes to hog the electrons. So even though this shows that they're sharing electrons here and here. At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules. Let me draw some more molecules. When we talk about the whole state of the whole matter, we actually think about how the molecules are interacting with Not just how the atoms are interacting with each other within a molecule. I just drew one oxygen, let me copy and paste that. But I could do multiple oxygens. And let's say that that hydrogen is going to want to be near this oxygen. Because this has partial negative charge, this has a partial positive charge. And then I could do another one right there. And then maybe we'll have, and just to make the point clear, you have two hydrogens here, maybe an oxygen wants to hang out there. So maybe you have an oxygen that wants to be here because it's got its partial negative here. And it's connected to two hydrogens right there that have their partial positives. But you can kind of see a lattice structure. Let me draw these bonds, these polar bonds that start forming between the particles. These bonds, they're called polar bonds because the molecules themselves are polar. And you can see it forms this lattice structure. And if each of these molecules don't have a lot of kinetic energy. Or we could say the average kinetic energy of this matter is fairly low. And what do we know is average kinetic energy? Well, that's temperature. Then this lattice structure will be solid. These molecules will not move relative to each other. I could draw a gazillion more, but I think you get the point that we're forming this kind of fixed structure. And while we're in the solid state, as we add kinetic energy, as we add heat, what it does to molecules is, it just makes them vibrate around a little bit. If I was a cartoonist, they way you'd draw a vibration is to put quotation marks there. That's not very scientific. But they would vibrate around, they would buzz around a little bit. I'm drawing arrows to show that they are vibrating. It doesn't have to be just left-right it could be up-down. But as you add more and more heat in a solid, these molecules are going to keep their structure. So they're not going to move around relative to each other. But they will convert that heat, and heat is just a form of energy, into kinetic energy which is expressed as the vibration of these molecules. Now, if you make these molecules start to vibrate enough, and if you put enough kinetic energy into these molecules, what do you think is going to happen? Well this guy is vibrating pretty hard, and he's vibrating harder and harder as you add more and more heat. This guy is doing the same thing. At some point, these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations. And once that happens, the molecules-- let me draw a couple more. Once that happens, the molecules are going to start moving past each other. So now all of a sudden, the molecule will start shifting. But they're still attracted. Maybe this side is moving here, that's moving there. You have other molecules moving around that way. But they're still attracted to each other. Even though we've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules. Our vibration, or our kinetic energy for each molecule, still isn't strong enough to completely separate them. They're starting to slide past each other. And this is essentially what happens when you're in a liquid state. You have a lot of atoms that want be touching each other but they're sliding. They have enough kinetic energy to slide past each other and break that solid lattice structure here. And then if you add even more kinetic energy, even more heat, at this point it's a solution now. They're not even going to be able to stay together. They're not going to be able to stay near each other. If you add enough kinetic energy they're going to start looking like this. They're going to completely separate and then kind of bounce around independently. Especially independently if they're an ideal gas. But in general, in gases, they're no longer touching They might bump into each other. But they have so much kinetic energy on their own that they're all doing their own thing and they're not touching. I think that makes intuitive sense if you just think about what a gas is. For example, it's hard to see a gas. Why is it hard to see a gas? Because the molecules are much further apart. So they're not acting on the light in the way that a liquid or a solid would. And if we keep making that extended further, a solid-- well, I probably shouldn't use the example with ice. Because ice or water is one of the few situations where the solid is less dense than the liquid. That's why ice floats. And that's why icebergs don't just all fall to the bottom of the ocean. And ponds don't completely freeze solid. But you can imagine that, because a liquid is in most cases other than water, less dense. That's another reason why you can see through it a little Or it's not diffracting-- well I won't go into that too much, than maybe even a solid. But the gas is the most obvious. And it is true with water. The liquid form is definitely more dense than the gas form. In the gas form, the molecules are going to jump around, not touch each other. And because of that, more light can get through the substance. Now the question is, how do we measure the amount of heat that it takes to do this to water? And to explain that, I'll actually draw a phase change diagram. Which is a fancy way of describing something fairly straightforward. Let me say that this is the amount of heat I'm adding. And this is the temperature. We'll talk about the states of matter in a second. So heat is often denoted by q. Sometimes people will talk about change in heat. They'll use H, lowercase and uppercase H. They'll put a delta in front of the H. Delta just means change in. And sometimes you'll hear the word enthalpy. Let me write that. Because I used to say what is enthalpy? It sounds like empathy, but it's quite a different concept. At least, as far as my neural connections could make it. But enthalpy is closely related to heat. It's heat content. For our purposes, when you hear someone say change in enthalpy, you should really just be thinking of change in heat. I think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary. The best way to think about it is heat content. Change in enthalpy is really just change in heat. And just remember, all of these things, whether we're talking about heat, kinetic energy, potential energy, enthalpy. You'll hear them in different contexts, and you're like, I thought I should be using heat and they're talking about enthalpy. These are all forms of energy. And these are all measured in joules. And they might be measured in other ways, but the traditional way is in joules. And energy is the ability to do work. And what's the unit for work? Well, it's joules. Force times distance. But anyway, that's a side-note. But it's good to know this word enthalpy. Especially in a chemistry context, because it's used all the time and it can be very confusing and non-intuitive. Because you're like, I don't know what enthalpy is in my everyday life. Just think of it as heat contact, because that's really But anyway, on this axis, I have heat. So this is when I have very little heat and I'm increasing my heat. And this is temperature. Now let's say at low temperatures I'm here and as I add heat my temperature will go up. Temperature is average kinetic energy. Let's say I'm in the solid state here. And I'll do the solid state in purple. No I already was using purple. I'll use magenta. So as I add heat, my temperature will go up. Heat is a form of energy. And when I add it to these molecules, as I did in this example, what did it do? It made them vibrate more. Or it made them have higher kinetic energy, or higher average kinetic engery, and that's what temperature is a measure of; average kinetic energy. So as I add heat in the solid phase, my average kinetic energy will go up. And let me write this down. This is in the solid phase, or the solid state of matter. Now something very interesting happens. Let's say this is water. So what happens at zero degrees? Which is also 273.15 Kelvin. Let's say that's that line. What happens to a solid? Well, it turns into a liquid. Ice melts. Not all solids, we're talking in particular about water, about H2O. So this is ice in our example. All solids aren't ice. Although, you could think of a rock as solid magma. Because that's what it is. I could take that analogy a bunch of different ways. But the interesting thing that happens at zero degrees. Depending on what direction you're going, either the freezing point of water or the melting point of ice, something interesting happens. As I add more heat, the temperature does not to go up. As I add more heat, the temperature does not go up for a little period. Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or which water will boil. But something happens. And they're really getting kinetically active. But just like when you went from solid to liquid, there's a certain amount of energy that you have to contribute to the system. And actually, it's a good amount at this point. Where the water is turning into vapor, but it's not getting any hotter. So we have to keep adding heat, but notice that the temperature didn't go up. We'll talk about it in a second what was happening then. And then finally, after that point, we're completely vaporized, or we're completely steam. Then we can start getting hot, the steam can then get hotter as we add more and more heat to the system. So the interesting question, I think it's intuitive, that as you add heat here, our temperature is going to go up. But the interesting thing is, what was going on here? We were adding heat. So over here we were turning our heat into kinetic energy. Temperature is average kinetic energy. But over here, what was our heat doing? Well, our heat was was not adding kinetic energy to the system. The temperature was not increasing. But the ice was going from ice to water. So what was happening at that state, is that the kinetic energy, the heat, was being used to essentially break these bonds. And essentially bring the molecules into a higher energy state. So you're saying, Sal, what does that mean, higher energy state? Well, if there wasn't all of this heat and all this kinetic energy, these molecules want to be very close to each other. For example, I want to be close to the surface of the earth. When you put me in a plane you have put me in a higher energy state. I have a lot more potential energy. I have the potential to fall towards the earth. Likewise, when you move these molecules apart, and you go from a solid to a liquid, they want to fall towards each other. But because they have so much kinetic energy, they never quite are able to do it. But their energy goes up. Their potential energy is higher because they want to fall towards each other. By falling towards each other, in theory, they could do some work. So what's happening here is, when we're contributing heat-- and this amount of heat we're contributing, it's called the heat of fusion. Because it's the same amount of heat regardless how much direction we go in. When we go from solid to liquid, you view it as the heat of melting. It's the head that you need to put in to melt the ice into liquid. When you're going in this direction, it's the heat you have to take out of the zero degree water to turn it into ice. So you're taking that potential energy and you're bringing the molecules closer and closer to each other. So the way to think about it is, right here this heat is being converted to kinetic energy. Then, when we're at this phase change from solid to liquid, that heat is being used to add potential energy into the system. To pull the molecules apart, to give them more potential energy. If you pull me apart from the earth, you're giving me potential energy. Because gravity wants to pull me back to the earth. And I could do work when I'm falling back to the earth. A waterfall does work. It can move a turbine. You could have a bunch of falling Sals move a turbine as well. And then, once you are fully a liquid, then you just become a warmer and warmer liquid. Now the heat is, once again, being used for kinetic energy. You're making the water molecules move past each other faster, and faster, and faster. To some point where they want to completely disassociate from each other. They want to not even slide past each other, just completely jump away from each other. And that's right here. This is the heat of vaporization. And the same idea is happening. Before we were sliding next to each other, now we're pulling apart altogether. So they could definitely fall closer together. And then once we've added this much heat, now we're just heating up the steam. We're just heating up the gaseous water. And it's just getting hotter and hotter and hotter. But the interesting thing there, and I mean at least the interesting thing to me when I first learned this, whenever I think of zero degrees water I'll say, oh it must be ice. But that's not necessarily the case. If you start with water and you make it colder and colder and colder to zero degrees, you're essentially taking heat out of the water. You can have zero degree water and it hasn't turned into ice yet. And likewise, you could have 100 degree water that hasn't turned into steam yeat. You have to add more energy. You can also have 100 degree steam. You can also have zero degree water. Anyway, hopefully that gives you a little bit of intuition of what the different states of matter are. And in the next problem, we'll talk about how much heat exactly it does take to move along this line. And maybe we can solve some problems on how much ice we might need to make our drink cool." + }, + { + "Q": "How can you find the y-int by only them giving you the x-int?", + "A": "To find the y-intercept, all you have to do is set the x value to 0. To find the x intercept, you just have to set the y value to 0. This makes sense because a y intercept is always (0,something) and on a graph, it doesn t go horizontally in either direction, it just stays on the axis. Hope this helped :)", + "video_name": "uk7gS3cZVp4", + "transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line must intersect the y-axis at y is equal to negative 2, so it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y would change by 1. So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done." + }, + { + "Q": "What are negative numbers?", + "A": "negative numbers are numbers below zero", + "video_name": "tJrSILRXOUc", + "transcript": "Voiceover:Which numbers are greater than six? Select all that apply. We see six here on the number line, so the numbers that are greater than six are going to be the ones that are to the right of six on the number line. We see that we're increasing beyond six as we go to the right. Six, seven, eight, nine, ten. Seven is greater than six, eight is greater than six, nine is greater than six, and 10 is greater than six, and we could keep going. 11 is greater than six, and 12, and on and on and on. Which of these are greater than six? Well we see 10 is to the right, is on the right hand side of six, eight is also to the right of six but four is to the left of six. Four is less than six. These are the two numbers that are greater than, the two choices that are greater than six. Which numbers are less than six? Well that's all of these numbers right over here. The numbers to the left of six. Nine is definitely not less than six but four is. Notice four is to the left of six and three is even more to the left of six, so four and three are definitely less than six. When every day life when you're thinking of well if you have four things, or you have three things you have less than someone who has six things. In every day life if you have 10 bananas, you have a greater number of bananas than someone who has six bananas." + }, + { + "Q": "So do you just multiply the denominator of the first fraction and the numerator of the second fraction", + "A": "Unlike adding you multiply the denominator and the numerator and same with division.", + "video_name": "yb7lVnY_VCY", + "transcript": "Tommy is studying for final exams this weekend. He will spend 1/5 of the weekend studying. What fraction of the weekend will he spend studying for each of his 4 subjects if he spends the same amount of time studying for each subject? So the total amount of time he's going to spend studying this weekend is 1/5 of the weekend. And he has to divide that into 4 equal sections. And he's going to spend that much time on each subject. So he's going to divide this by 4. Now, we've already seen that dividing by a number is the same thing as multiplying by its reciprocal. You might say, hey, well, what's the reciprocal of 4? You just have to remind yourself that 4 is the same thing as 4/1. So 1/5 divided by 4/1 is the same thing as 1/5 times 1/4. And you could also view this as 1/4 of 1/5 or 1/5 of 1/4, either way. But here we multiply our numerators to get 1. And then we multiply our denominators, 4 times 5 is 20. So you get 1/20 of the weekend will be spent studying for each subject. Now, let's also try to think about this visually. Let's imagine that this is his entire weekend. And I've divided it into 5 equal sections. And so we already know that the total amount of his weekend spent studying is 1/5. So that's the total amount studying for the weekend is 1/5. Now, he has to divide this into 4 equals section. So let's do that. He's got four subjects, and he's going to spend the same amount of time on each of the 4 subjects. So he's going to divide this into 4 equal sections. So how much time does he spend on one subject? Well, in each subject, that would be this little area that I'm doing in yellow right over here. And what is that? Well, that's 1 over-- how many equal sections are there of that size in the weekend? Well, I've just drawn out the grid. You had 5 rows, and now you have 4 columns. So 5 rows times 4 columns, you have 20 equal sections. So once again, looking at it visually, he's spending 1/20 of his weekend on each of the 4 subjects. And then if you do this for 4 subjects, that means that in this whole weekend, 1/5 will be spent studying. But the question that they're asking, he's spending 1/20 of the weekend on each subject." + }, + { + "Q": "Hi :)\n\nAt 7:24, it is mentioned that the molecules have the same kinetic energy. How so? If the volume of the box was decreased then the molecules kinetic energy would increase, wouldn't it?\n\nOr does it mean that all the molecules inside this smaller box all have the same average kinetic energy? But this kinetic energy is still higher than the kinetic energy of the molecules on the bigger box, isn't it?\n\nThank you!", + "A": "Me thinks the average K.E of molecules remains the same but the extra energy they gained by virtue of reduced volume is transferred as increase in heat of collision which is very minute.", + "video_name": "tQcB9BLUoVI", + "transcript": "After all the work we've been doing with fluids, you probably have a pretty good sense of what pressure is. Now let's think a little bit about what it really means, especially when we think about it in terms of a gas in a volume. Remember, what was the difference between a gas and a liquid? They're both fluids, they both take the shape of their containers, but a gas is compressible, while a liquid is incompressible. Let's start focusing on gases. Let's say I have a container, and I have a bunch of gas in it. What is a gas made of? It's just made up of a whole bunch of the molecules of the gas itself, and I'll draw each of the molecules with a little dot-- it's just going to have a bunch of molecules in it. There's many, many, many more than what I've drawn, but that's indicative, and they'll all be going in random directions-- this one might be going really fast in that direction, and that one might be going a little bit slower in that direction. They all have their own little velocity vectors, and they're always constantly bumping into each other, and bumping into the sides of the container, and ricocheting here and there and changing velocity. In general, especially at this level of physics, we assume that this is an ideal gas, that all of the bumps that occur, there's no loss of energy. Or essentially that they're all elastic bumps between the different molecules. There's no loss of momentum. Let's keep that in mind, and everything you're going to see in high school and on the AP test is going to deal with ideal gases. Let's think about what pressure means in this context. A lot of what we think about pressure is something pushing on an area. If we think about pressure here-- let's pick an arbitrary area. Let's take this side. Let's take this surface of its container. Where's the pressure going to be generated onto this surface? It's going to be generated by just the millions and billions and trillions of little bumps every time-- let me draw a side view. If this is the side view of the container, that same side, every second there's always these little molecules of gas moving around. If we pick an arbitrary period of time, they're always ricocheting off of the side. We're looking at time over a super-small fraction of time. And over that period of time, this one might end up here, this one maybe bumped into it right after it ricocheted and came here, this one changes momentum and goes like that. This one might have already been going in that direction, and that one might ricochet. But what's happening is, at any given moment, since there's so many molecules, there's always going to be some molecules that are bumping into the side of the wall. When they bump, they have a change in momentum. All force is change in momentum over time. What I'm saying is that in any interval of time, over any period or any change in time, there's just going to be a bunch of particles that are changing their momentum on the side of this wall. That is going to generate force, and so if we think about how many on average-- because it's hard to keep track of each particle individually, and when we did kinematics and stuff, we'd keep track of the individual object at play. But when we're dealing with gases and things on a macro level, you can't keep track of any individual one, unless you have some kind of unbelievable supercomputer. We can say, on average, this many particles are changing momentum on this wall in this amount of time. And so the force exerted on this wall or this surface is going to be x. If we know what that force is, and we you know the area of the wall, we can figure out pressure, because pressure is equal to force divided by area. What does this help us with? I wanted to give you that intuition first, and now I'm just going to give you the one formula that you really just need to know in thermodynamics. And then as we go into the next few videos, I'll prove to you why it works, and hopefully give you more of an intuition. Now you understand, hopefully, what pressure means in the context of a gas in a container. With that out of the way, let me give you a formula. I hope by the end of this video you have the intuition for why this formula works. In general, if I have an ideal gas in a container, the pressure exerted on the gas-- on the side of the container, or actually even at any point within the gas, because it will all become homogeneous at some point-- and we'll talk about entropy in future videos-- but the pressure in the container and on its surface, times the volume of the container, is equal to some constant. We'll see in future videos that that constant is actually proportional to the average kinetic energy of the molecules bouncing around. That should make sense to you. If the molecules were moving around a lot faster, then you would have more kinetic energy, and then they would be changing momentum on the sides of the surface a lot more, so you would have more pressure. Let's see if we can get a little bit more intuition onto why pressure times volume is a constant. Let's say I have a container now, and it's got a bunch of molecules of gas in it. Just like I showed you in that last bit right before I erased, these are bouncing off of the sides at a certain rate. Each of the molecules might have a different kinetic energy-- it's always changing, because they're always transferring momentum to each other. But on average, they all have a given kinetic energy, they keep bumping at a certain rate into the wall, and that determines the pressure. What happens if I were able to squeeze the box, and if I were able to decrease the volume of the box? I just take that same box with the same number of molecules in it, but I squeeze. I make the volume of the box smaller-- what's going to happen? I have the same number of molecules in there, with the same kinetic energy, and on average, they're moving with the same velocities. So now what's going to happen? They're going to be hitting the sides more often-- at the same time here that this particle went bam, bam, now it could go bam, bam, bam. They're going to be hitting the sides more often, so you're going to have more changes in momentum, and so you're actually going to have each particle exert more force on each surface. Because it's going to be hitting them more often in a given amount of time. The surfaces themselves are smaller. You have more force on a surface, and on a smaller surface, you're going to have higher pressure. Hopefully, that gives you an intuition that if I had some amount of pressure in this situation-- if I squeeze the volume, the pressure increases. Another intuition-- if I have a balloon, what blows up a balloon? It's the internal air pressure of the helium, or your own exhales that you put into the balloon. The more and more you try to squeeze a balloon-- if you squeeze it from all directions, it gets harder and harder to do it, and that's because the pressure within the balloon increases as you decrease the volume. If volume goes down, pressure goes up, and that makes sense. That follows that when they multiply each other, you have to have a constant. Let's take the same example again, and what happens if you make the volume bigger? Let's say I have-- it's huge like that, and I should have done it more proportionally, but I think you get the idea. You have the same number of particles, and if I had a particle here, in some period of time it could have gone bam, bam, bam-- it could have hit the walls twice. Now, in this situation, with larger walls, it might just go bam, and in that same amount of time, it will maybe get here and won't even hit the other wall. The particles, on average, are going to be colliding with the wall less often, and the walls are going to have a larger area, as well. So in this case, when our volume goes up, the average pressure or the pressure in the container goes down. Hopefully, that gives you a little intuition, and so you'll never forget that pressure times volume is constant. And then we can use that to do some pretty common problems, which I'll do in the next video. See you soon." + }, + { + "Q": "What about when working with decimals in the equation? Do you get rid of the decimals first? Example problem - -2+.3x-.1x+x=6x HELP...we still cannot get this one right...thanks.", + "A": "you can get rid of the decimals by multiplying the decimals by 10", + "video_name": "PL9UYj2awDc", + "transcript": "We have the equation 3/4x plus 2 is equal to 3/8x minus 4. Now, we could just, right from the get go, solve this the way we solved everything else, group the x terms, maybe on the left-hand side, group the constant terms on the right-hand side. But adding and subtracting fractions are messy. So what I'm going to do, right from the start of this video, is to multiply both sides of this equation by some number so I can get rid of the fractions. And the best number to do it by-- what number is the smallest number that if I multiply both of these fractions by it, they won't be fractions anymore, they'll be whole numbers? That smallest number is going to be 8. I'm going to multiply 8 times both sides of this equation. You say, hey, Sal, how did you get 8? And I got 8 because I said, well, what's the least common multiple of 4 and 8? Well, the smallest number that is divisible by 4 and 8 is 8. So when you multiply by 8, it's going to get rid of the fractions. And so let's see what happens. So 8 times 3/4, that's the same thing as 8 times 3 over 4. Let me do it on the side over here. That's the same thing as 8 times 3 over 4, which is equal to 8 divided by 4 is just 2. So it's 2 times 3, which is 6. So the left-hand side becomes 8 times 3/4x is 6x. And then 8 times 2 is 16. You have to remember, when you multiply both sides, or a side, of an equation by a number, you multiply every term by that number. So you have to distribute the 8. So the left-hand side is 6x plus 16 is going to be equal to-- 8 times 3/8, that's pretty easy, the 8's cancel out and you're just left with 3x. And then 8 times negative 4 is negative 32. And now we've cleaned up the equation a good bit. Now the next thing, let's try to get all the x terms on the left-hand side, and all the constant terms on the right. So let's get rid of this 3x from the right. Let's subtract 3x from both sides to do it. That's the best way I can think of of getting rid of the 3x from the right. The left-hand side of this equation, 6x minus 3x is 3x. 6 minus 3 is 3. And then you have a plus 16 is equal to-- 3x minus 3x, that's the whole point of subtracting 3x, is so they cancel out. So those guys cancel out, and we're just left with a negative 32. Now, let's get rid of the 16 from the left-hand side. So to get rid of it, we're going to subtract 16 from both sides of this equation. Subtract 16 from both sides. The left-hand side of the equation just becomes-- you have this 3x here; these 16's cancel out, you don't have to write anything-- is equal to negative 32 minus 16 is negative 48. So we have 3x is equal to negative 48. To isolate the x, we can just divide both sides of this equation by 3. So let's divide both sides of that equation by 3. The left-hand side of the equation, 3x divided by 3 is just an x. That was the whole point behind dividing both sides by 3. And the right-hand side, negative 48 divided by 3 is negative 16. And we are done. x equals negative 16 is our solution. So let's make sure that this actually works by substituting to the original equation up here. And the original equation didn't have those 8's out front. So let's substitute in the original equation. We get 3/4-- 3 over 4-- times negative 16 plus 2 needs to be equal to 3/8 times negative 16 minus 4. So 3/4 of 16 is 12. And you can think of it this way. What's 16 divided by 4? It is 4. And then multiply that by 3, it's 12, just multiplying fractions. So this is going to be a negative 12. So we get negative 12 plus 2 on the left-hand side, negative 12 plus 2 is negative 10. So the left-hand side is a negative 10. Let's see what the right-hand side is. You have 3/8 times negative 16. If you divide negative 16 by 8, you get negative 2, times 3 is a negative 6. So it's a negative 6 minus 4. Negative 6 minus 4 is negative 10. So when x is equal to negative 16, it does satisfy the Both sides of the equation become negative 10. And we are done." + }, + { + "Q": "why didn't he just use sin (3pi/12 and 4pi/12) to solve this instead of sin(pi/4 and pi/3)?", + "A": "Those are actually the same, because 3pi/12=pi/4 and 4pi/12=pi/3 due to the cancellation of common factors(3 in the first, 4 in the second equation).", + "video_name": "2RbKfRfzD-M", + "transcript": "Voiceover:What I want to attempt to do in this video is figure out what the sine of seven pi over 12 is without using a calculator. And so let's just visualize seven pi over 12 in the unit circle. One side of the angle is going along the positive x-axis if we go straight up, that's pi over two, which is the same thing as six pi over 12, so then we essentially just have another pi over 12 to get right over there. This is the angle that we're talking about that is seven pi over 12 radians, by the unit circle definition of sine, it's the y-coordinate of where this ray intersects the unit circle. This is the unit circle, has radius one where it intersects the unit. The y-coordinate is the sine. Another way to think about it, it's the length of this line right over here. I encourage you to pause the video right now and try to think about it on your own. See if you can use your powers of trigonometry to figure out what sine of seven pi over 12 is or essentially the length of this magenta line. I'm assuming you've given a go at it, and if you're like me, your first temptation might have been just to focus on this triangle right over here that I drew for you. The triangle looks like this. It looks like this, where that's what you're trying to figure out, this length right over here, sine of seven pi over 12. We know the length of the hypotenuse is one. It's a radius of the unit circle. It's a right triangle right over there. We also know this angle right over here, which is this angle right over here, this gets us six pi over 12, and then we have another pi over 12, so we know that that is pi over 12, not pi over 16. We know that this angle right over here is pi over 12. Given this information, we can figure out this, or we can at least relate this side to this other side using a trig function relative to this angle. This is the adjacent side. The cosine of pi over 12 is going to be this magenta side over one, or you could just say it's equal to this magenta side. You could say that this is cosine of pi over 12. We just figured out that sine of seven pi over 12 is the same thing as cosine of pi over 12, but that still doesn't help me. I don't know offhand what the cosine of pi over 12 radians is without using a calculator. Instead of thinking about it this way, let's see if we can compose this angle or if we can decompose it into some angles for which we do know the sine and cosine. What angles are those? Those are the angles in special right triangles. For example, we are very familiar with 30-60-90 triangles. 30-60-90 triangles look something like this. This is my best attempt at hand drawing it. Instead of writing 30-degree side, since we're thinking in radians, I'll write that as pi over six radians. The 60-degree side, I'm going to write that as pi over three radians, and of course, this is the right angle. If the hypotenuse here is one, then the side opposite the 30-degree side, or the pi over six radian side, is going to be half the hypotenuse, which, in this case is 1/2. Then the other side that's opposite the 60-degree side or the pi over three radian side, is going to be square root of three times the shorter side. It's going to be square root of three over two. We've used these types of triangles in the past to figure out the sine or cosine of 30 or 60, or in this case, pi over six or pi over three. We know about pi over six and pi over three. We also know about 45-45-90 triangles. We know that they're isosceles right triangles. They look like this, my best attempt at drawing it. That one actually doesn't look that isosceles, so let me make it a little bit more ... I don't know. That looks closer to being an isosceles right triangle. We know if the length of the hypotenuse is one, and this comes straight out of Pythagorean theorem, then the length of each of the other two sides are going to be square root of two over two times the hypotenuse, which, in this case, is the square root of two over two. Instead of describing these as 45-degree angles, we know that's the same thing as pi over four, pi over four radians. If you give me pi over six, pi over three, pi over four, I can use these triangles either using the classic definition, SOHCAHTOA definitions, or I could stick them on the unit circle here to use the unit circle definition of trig functions to figure out what the sine, cosine, or tangent of these angles are. Can I decompose seven pi over 12 into some combination of pi over sixes, pi over threes, or pi over fours? Think about that. Let me rewrite pi over six, pi over three, and pi over four with a denominator over 12. Let me write that. Pi over six is equal to two pi over 12, pi over three is equal to four pi over 12, and pi over four is equal to three pi over 12. Let's see. Two plus four is not seven, two plus three is not seven, but four plus three is seven. So I could use this and this. Four pi over 12 plus three pi over 12 is seven pi over 12. I could rewrite this. This is the same thing as sine of three pi over 12 plus 4 pi over 12, which, of course, is the same thing, sine of pi over 4, I'll do this in another color, sine of pi over 4 plus ... let me do this ... plus pi over three, Now we can use our angle addition formula for sine in order to write this as the sum of products of cosines and sines of these angles. Let's actually do that. This right over here is going to be equal to, this is going to be equal to the sine, the sine of pi over four times the cosine of pi over three plus the other way around, cosine of pi over four times the sine of pi over three, sine of pi over three. Now we just have to figure out these things, and I've already set up the triangles to do it. What is sine of pi over four? Sine of pi over four, well, let's think about ... This is pi over four right over here. Sine is opposite over hypotenuse. That's just going to be square root of two over two. This is square root of two over two, square root of two over two. What is cosine of pi over three? This is a pi over three radian angle right over here. Cosine is adjacent over a hypotenuse. It's adjacent over a hypotenuse, so this is going to be 1/2. What is cosine of pi over four? Go back to pi over four. It's adjacent over a hypotenuse. It's square root of two over two. It is also square root of two over two, square root of two over two. What's sine of pi over three? Sine is opposite over a hypotenuse, so square root of three over two over one. Square root of three over two divided by one, which is square root of three over two. Now we just have to simplify all of this business. This is going to be equal to the sum of this, or the product, I should say, is just square root of two over four, and then plus the product of these. Let's see. We could write that as square root of six over four, square root of six over four, or we could just rewrite this whole thing as, and we deserve a little bit of a drum roll at this point, this is equivalent to, let me just scroll over to the right a little bit. This is equivalent to square root of two plus square root of six, all of that over four. That's what sine of seven pi over 12 is, or cosine of pi over 12, what that is equal to." + }, + { + "Q": "What is the difference b/w oscillation and vibration?", + "A": "Oscillation refers to some sort of back and forth movement. Vibration is not really a word we would use in this context. You would not say a pendulum vibrates, right? but vibration refers vaguely to some sort of jiggling movement.", + "video_name": "ZcZQsj6YAgU", + "transcript": "- [Instructor] Alright, we should talk about oscillators. And what an oscillator is is an object or variable that can move back and forth or increase and decrease, go up and down, left and right, over and over and over. So for instance, a mass on a spring here is an oscillator if we pull this mass back, it's gonna oscillate back and forth, and that's what we mean by an oscillator. Or another common example is a pendulum, and a pendulum is just a mass connected to a string, and you pull the mass back and then it swings back and forth. So you've got something going back and forth, that's an oscillator. These are the two most common types. Masses on springs, pendulum, but there's many other examples and all those examples share one common feature of why they're an oscillator. So you could ask why do these things oscillate in the first place, and it's because they all share this common fact, that they all have a restoring force. And a restoring force, like the name suggests, tries to restore this system, but restore it to what? Restore the system to the equilibrium position. So every oscillator has an equilibrium position, and that would be the point at which there's no net force on the object that's oscillating. So for instance, for this mass, if this mass on the spring was sitting at the equilibrium position, the net force on that mass would be 0 because that's what we mean by the equilibrium position. In other words, if you just sat the mass there it would just stay there because there's no net force on it. However, if I pull this mass to the right, the spring's like uh uh, now I'm gonna try and restore this mass back to the equilibrium position, the spring would pull to the left. If I push this mass to the left, the spring's like uh uh, we're movin' this thing back to the equilibrium position, we're trying to push it back there. So if I push left, the spring pushes right. And if I pull the mass right, the spring pulls left. It tries to restore always, it tries to restore mass back to the equilibrium position. Sam for the pendulum. If I pull the pendulum to the right, gravity is the restoring force trying to bring it back to the left. But if I pull the mass to the left, gravity tries to pull it back to the right, always trying to restore this mass back to the equilibrium position. That's what we mean by a restoring force. Now there's lots of oscillators, but only some of those oscillators are really special, and we give those a special name. We call them Simple Harmonic Oscillators. And you might be thinking, that's a pretty dumb name because that doesn't sound very simple. But they're something called the Simple Harmonic Oscillator. So what makes Simple Harmonic Oscillator's so special is that even though all oscillators have a restoring force, Simple Harmonic Oscillators have a restoring force that's proportional to the amount of displacement. So what that means is if I pull this mass to the right there will be a restoring force, but if it's proportional to the displacement, if I pulled this mass back twice as much, I'd get twice the restoring force. And if I pulled it back three times as much, I'd get three times the restoring force. Same down here. If I pulled this pendulum back with two times the angle, I'd get two times the restoring force. If that's the case, then you've got what we call a Simple Harmonic Oscillator. And you still might not be impressed, you might be like who cares if the restoring force is proportional to the displacement. Why should I care about that? You should care about that because these satisfy some very special rules that I'll show you throughout this video and it turns out that even though this doesn't sound very simple, they are much simpler than the alternative of Non-Simple Harmonic Oscillators. So these are what we typically study in introductory physics classes, and it turns out a mass on a spring is a Simple Harmonic Oscillator, and a pendulum also for small oscillations, here you have to make a caveat, you have to say only for small angles, but for those small angles, the pendulum is a Simple Harmonic Oscillator as well. Now in this video, we're just going to look at the mass on the spring to make it simple. We could look at the pendulum later. So I'm going to get rid of the pendulum so we can focus on this mass on a spring. Now you might not be convinced, you might be like how do we know this mass on a spring is really a Simple Harmonic Oscillator? Well we can prove it because the force that's providing the restoring force in this case is the spring. So the spring is the restoring force in this case, and we know the formula for the force from a spring, that's given by Hooke's Law. And Hooke's Law says that the spring force, the force provided by the spring, is going to be negative. The spring constant times x, the spring displacement, so x is going to be positive if the spring has been displaced to the right because the spring's going to get longer. So this would be a positive x amount. And if you compress the spring, the length of the spring gets smaller, that's going to count as a negative x value. But think about it, if I compress the spring to the left, my x is going to be negative, and that negative combines with this negative to be a positive so I'd get a positive force. That means the spring is there's a force to the right. And that makes sense. Restoring, it means it opposes what you do. If you push the mass to the left, the spring is going to push to the right. And if we did it the other way, if we pulled the mass to the right, now that would be a positive x value. If I have a positive x value in here and combine that with a negative, I'd get a negative spring force. And that means the spring would be pulling to the left, it's restoring this mass back to the equilibrium position. And that's exactly what an oscillator does. And look at it up here, this spring force, this restoring force, is proportional to the displacement. So x is the displacement, this is a force that's proportional to the displacement. And that's the definition. That was what we meant by Simple Harmonic Oscillator. So that's why masses on springs are going to be Simple Harmonic Oscillators, because the restoring force is proportional to the displacement. Now to be completely honest, it has to be negatively proportional to the displacement. If you just had f equals kx with no negative, then if you displaced it to the right, the force would be to the right which would displace it more to the right, which would create a larger force to the right, this would be a runaway solution, this thing would blow up, that wouldn't be good. So it's really forces that have a negative proportionality to the displacement. That way it's going to restore back to the equilibrium position and if this is proportional, you get a Simple Harmonic Oscillator. And so we should talk about this, what the heck do we mean by simple? Like what is simple about this? It turns out that what's simple is that these types of oscillators are going to be described by sin and cosin functions. So Simple Harmonic Oscillators will be described by sin and cosin and that should make sense because think about sin and cosin, what do those look like? Sin and cosin look like this. So here's what sin looks like, it's a function that oscillates back and forth. And cosin looks like this, it starts up here, so it's also a function that oscillates back and forth. And so these are simple, turns out those are very simple functions that oscillate back and forth. And because of that, we like those. In physics, we love things that are described by sin and cosin, it turns out they're pretty easy to deal with mathematically. Maybe you don't feel that way, but they're much easier than the alternatives of other things that could oscillate. So that's what Simple Harmonic Oscillators mean. But let's try to get some intuition, what is really going on for this mass on a spring? So let's imagine we pull the mass back, right? So the mass, if the mass just continues to sit at the equilibrium position, it's a pretty boring problem because the net force right there would be 0 and it would just continue to sit there. So let's say we pull the mass back, we pull it back by a certain amount. Say we pull it back this far, and then we let go. So since we let go of the mass, we've released it at rest. So it started at rest. And that means the speed initially over here is 0. So it starts off with 0 speed, but the spring has been stretched. And so the spring is going to restore, right, the spring is always trying to restore the mass back to the equilibrium position. So the spring pulling the mass to the left, speeding it up, speeds the mass up until it gets to the equilibrium position, and then the spring realizes, oh crud, I messed up. I wanted to get the mass here but I pulled it so much this mass has a huge speed to the left now. And masses don't just stop on their own, They need some force to do that. So this mass has inertia, and according to Newton's First Law, it's going to try and keep moving. So even though the spring got the mass back to the equilibrium position, that was its goal, it got it back there with this huge speed and the mass continues straight through the equilibrium position and the spring starts getting compressed and the spring's like oh no, I've gotta start pushing this thing to the right. I want to get the mass back to the equilibrium position. So now the spring's pushing to the right, slowing the mass down until it stops it, but the spring is compressed, so it's going to keep pushing to the right. Now it's pushing in the direction the mass is moving. Now it's got it going back to the equilibrium position again, which is good, but again, same mistake, the spring gets this mass back to the equilibrium position with a huge speed to the right, and now the spring's like oh great, I did it again, I got this mass back where I wanted it, but this mass had a huge speed and it's got inertia, and so this mass is going to keep moving to the right, past the equilibrium position. And this is why the oscillation happens. It's a constant fight between inertia of the mass wanting to keep moving because it's got mass and it's got velocity, and the restoring force that is desperately trying to get this mass back to the equilibrium position and they can never quite figure it out because they keep overshooting each other and this oscillation happens over and over and over. So just knowing the story, let's you say some really important things about the oscillation. One of them is that at these end points, at these points of maximum compression or extension, the speed is 0. So this mass is moving the slowest, i.e. it's not moving at all at these maximum points of compression or extension because that's where the spring has stopped the mass and started bringing it back in the other direction. Whereas in the middle, at the equilibrium position, you get the most speed. So this is where the mass is moving fastest, when the spring has got it back to the equilibrium position and the spring at that point realizes oh crap, this mass is going really fast, and the mass is coming at it or going away from it too fast for the spring to stop it immediately. So if the equilibrium point this mass has the most speed during the oscillation. So we could also ask where will the magnitude of the restoring force be biggest and where will it be least during this oscillation? And we've got a formula for that. Look at, the spring force is the restoring force. So we could just ask where will the spring force be biggest? That's going to be where this x is biggest or smallest. So if we wanted to know where the magnitude of this f is largest, we could just ask where will the magnitude of the x be largest? If we don't care about which way the force is, we just want to know where we'll get a really big force, we just try to figure out where will I get the biggest x? X is displacement. So the x value at the equilibrium position is 0. So there's no displacement of the spring right here, that's what it means to be the equilibrium position, this is the natural length of the spring. That's the length that the spring wants to be. If the spring has that shape right there, it doesn't push or pull. But if you've displaced it this way, or the other way, this would be positive displacement, and this would be negative displacement, now the spring's going to exert a force. So where will the force be greatest? It's where the spring has been compressed or stretched the most. So at these points here, at the points of maximum extension or compression, you're going to have the greatest amount of force. So greatest magnitude of force, because the spring is really stretched, it's going to pull with a great amount of force back toward the equilibrium position. And we can say which way it points, right? This spring's going to be pulling to the left, so there's going to be a great spring force to the left. Technically that'd be a negative force, so I mean, if you're taking sins into account, you could say that that's the least force because it's really negative. But if you're just worried about magnitude, that would be a great magnitude of force. And then also over here, at the maximum compression, this spring is really pushing the mass to the right, you get a great amount of force this way because your x, even though it's very negative at that point, it's going to give you a large amount of force. And so here you would also have a great amount of magnitude of force which can be confusing because look it. At these end points, you have the least speed, but the greatest force. Sometimes that freaks people out. They're like, how can you have a great force and your speed be so small? Well that's the point where the spring has stopped the mass and started pulling it in the other direction. So even though the speed is 0, the force is greatest. So, be careful, force does not have to be proportional to the speed. The force has to be proportional to the acceleration, right? Because we know net force, we could say that the net force is equal to ma. So wherever you have the largest amount of force, you'll have the largest amount of acceleration. So we could also say at these endpoints, you'll have not only the greatest magnitude of the force, but the greatest magnitude of acceleration as well. Because where you're pulling or pushing on something with the greatest amount of force, you're going to get the greatest amount of acceleration according to Newton's Second Law. So at these endpoints, the force is greatest, the acceleration is also greatest. The magnitude, the acceleration is also greatest even though the speed is 0 at those points. So those are the points where you get the greatest force and greatest acceleration. Where will you get the least amount of magnitude of force and magnitude of acceleration? Well look at up here. The least force will happen where you get the least possible displacement. And the least possible displacement's right here in the middle, this equilibrium position is when x equals 0. That's when the spring is not pushing or pulling. When it's at this point here. So when the mass is passing through the equilibrium position, there is 0 force. Right, that's the point where the mass got back there and the spring was like I'm glad I got it back to the equilibrium position and then the spring quickly realized, oh no, this mass, I got it back there, but the mass was moving really fast, so it shot straight through that point. But right at that moment, the spring had this glorious moment where it thought it had done it and it stopped exerting any force because at that point, the x is 0. And if x is 0, we know from up here, the force is 0. So this would be the least possible force. And I guess I should say it's actually 0 force, it's not just the least, there is 0 force exerted at this point. And if there's 0 force, by the same argument, we could say that there's 0 acceleration at that point. Hopefully that gives you some intuition about why oscillators do what they do and where you might find the largest speed or force at any given point. So recapping, objects with a restoring force that's negatively proportional to the displacement will be a Simple Harmonic Oscillator and for all Simple Harmonic Oscillators, at the equilibrium position you'll get the greatest speed but 0 restoring force and 0 acceleration. Whereas at the points of maximum displacement, you'll get the maximum magnitude of restoring force and acceleration but the least possible speed." + }, + { + "Q": "Is drawing a star with more than 10 and odd number of points possible", + "A": "Maybe. Try it.", + "video_name": "CfJzrmS9UfY", + "transcript": "Let's say you're me, and you're in math class, and you're supposed to be learning about factoring. Trouble is, your teacher is too busy trying to convince you that factoring is a useful skill for the average person to know, with real-world applications ranging from passing your state exams all the way to getting a higher SAT score. And unfortunately, does not have the time to show you why factoring is actually interesting. It's perfectly reasonable for you to get bored in this situation. So like any reasonable person, you start doodling. Maybe it's because your teacher's soporific voice reminds you of a lullaby, but you're drawing stars. And because you're me, you quickly get bored of the usual five-pointed star and get to wondering, why five? So you start exploring. It seems obvious that a five-pointed star is the simplest one, the one that takes the least number of strokes to draw. Sure, you can make a start with four points, but that's not really a star the way you're defining stars. Then there's a six-pointed star, which is also pretty familiar, but totally different from the five-pointed star because it takes two separate lines to make. And then you're thinking about how, much like you can put two triangles together to make a six-pointed star, you can put two squares together to make an eight-pointed star. And any even-numbered star with p points can be made out of two p/2-gons. It is at this point that you realize that if you wanted to avoid thinking about factoring, maybe drawing stars was not the greatest idea. But wait, four would be an even number of points, but that would mean you could make it out of two 2-gons. Maybe you were taught polygons with only two sides But for the purposes of drawing stars, it works out rather well. Sure, the four-pointed star doesn't look too star-like. But then you realize you can make the six-pointed star out of three of these things, and you've got an asterisk, which is definitely a legitimate star. In fact, for any star where the number of points is divisible by 2, you can draw it asterisk style. But that's not quite what you're looking for. What you want is a doodle game, and here it is. Draw p points in a circle, evenly spaced. Pick a number Q. Starting at one point, go around the circle and connect to the point two places over. Repeat. If you get to the starting place before you've covered all the points, jump to a lonely point, and keep going. That's how you draw stars. And it's a successful game, in that previously you were considering running screaming from the room. Or the window was open, so that's an option, too. But now, you're not only entertained but beginning to become curious about the nature of this game. The interesting thing is that the more points you have, the more different ways there is to draw the star. I happen to like seven-pointed stars because there's two really good ways to draw them, but they're still simple. I would like to note here that I've never actually left a math class by the window, not that I can say the same for other subjects. Eight is interesting, too, because not only are there a couple nice ways to draw it, but one's a composite of two polygons, while another can be drawn without picking up the pencil. Then there's nine, which, in addition to a couple of other nice versions, you can make out of three triangles. And because you're me, and you're a nerd, and you like to amuse yourself, you decide to call this kind of star a square star because that's kind of a funny name. So you start drawing other square stars. Four 4-gons, two 2-gons, even the completely degenerate case of one 1-gon. Unfortunately, five pentagons is already difficult to discern. And beyond that, it's very hard to see and appreciate the structure of square stars. So you get bored and move on to 10 dots in a circle, which is interesting because this is the first number where you can make a star as a composite of smaller stars-- that is, two boring old five-pointed stars. Unless you count asterisk stars, in which case 8 was two 4s's or four 2's or two 2's and a 4. But 10 is interesting because you can make it as a composite in more than one way because it's divisible by 5, which itself can be made in two ways. Then there's 11, which can't be made out of separate parts at all because 11 is prime. Though here you start to wonder how to predict how many times around the circle we'll go before getting back to start. But instead of exploring the exciting world of modular arithmetic, you move on to 12, which is a really cool number because it has a whole bunch of factors. And then something starts to bother you. Is a 25-pointed star composite made of five five-pointed stars a square star? You had been thinking only of pentagons because the lower numbers didn't have this question. How could you have missed that? Maybe your teacher said something interesting about prime numbers, and you accidentally lost focus for a moment. I don't know. It gets even worse. 6 squared would be a 36-pointed star made of six hexagons. But if you allow use of six-pointed stars, then it's the same as a composite of 12 triangles. And that doesn't seem in keeping with the spirit of square stars. You'll have to define square stars more strictly. But you do like the idea that there's three ways to make the seventh square star. Anyway, the whole theory of what kind of stars can be made with what numbers is quite interesting. And I encourage you to explore this during your math class." + }, + { + "Q": "why does carbon only bond with hydrogen??", + "A": "It doesn t! It also bonds with oxygen, nitrogen, sulfur, phosphorus and the halogens, along with some other elements. Organic chemistry is a vast subject and this video is concerned solely with alkanes, which are hydrocarbons. You will encounter different types of bonds as you progress but this course starts with the simple stuff and builds up.", + "video_name": "pMoA65Dj-zk", + "transcript": "The one thing that probably causes some of the most pain in chemistry, and in organic chemistry, in particular, is just the notation and the nomenclature or the naming that we use. And what I want to do here in this video and really the next few videos is to just make sure we have a firm grounding in the notation and in the nomenclature or how we name things, and then everything else will hopefully not be too difficult. So just to start off, and this is really a little bit of review of regular chemistry, if I just have a chain of carbons, and organic chemistry is dealing with chains of carbons. Let me just draw a one-carbon chain, so it's really kind of ridiculous to call it a chain, but if we have one carbon over here and it has four valence electrons, it wants to get to eight. That's the magic number we learned in just regular chemistry. For all molecules, that's the stable valence structure, I guess you could say it. A good partner to bond with is hydrogen. So it has four valence electrons and then hydrogen has one valence electron, so they can each share an electron with each other and then they both look pretty happy. I said eight's the magic number for everybody except for hydrogen and helium. Both of them are happy because they're only trying to fill their 1s orbital, so the magic number for those two guys is two. So all of the hydrogens now feel like they have two electrons. The carbon feels like it has eight. Now, there's several ways to write this. You could write it just like this and you can see the electrons explicitly, or you can draw little lines here. So I could also write this exact molecule, which is methane, and we'll talk a little bit more about why it's called methane later in this video. I can write this exact structure like this: a carbon bonded to four hydrogens. And the way that I've written these bonds right here you could imagine that each of these bonds consists of two electrons, one from the carbon and one from the hydrogen. Now let's explore slightly larger chains. So let's say I have a two-carbon chain. Well, let me do a three-carbon chain so it really looks like a chain. So if I were to draw everything explicitly it might So I have a carbon. It has one, two, three, four electrons. Maybe I have another carbon here that has-- let me do the carbons in slightly different shades of yellow. I have another carbon here that has one, two, three, four electrons. And then let me do the other carbon in that first yellow. And then I have another carbon so we're going to have a three-carbon chain. It has one, two, three, four valence electrons. Now, these other guys are unpaired, and if you don't specify it, it's normally going to be hydrogen, so let me draw some hydrogens over here. So you're going to have a hydrogen there, a hydrogen over there, a hydrogen over here, a hydrogen over here, a hydrogen over there, a hydrogen over here, almost done, a hydrogen there, and then a hydrogen there. Now notice, in this molecular structure that I've drawn, I have three carbons. They were each able to form four bonds. This guy has bonds with three hydrogens and another carbon. This guy has a bond with two hydrogens and two carbons. This guy has a bond with three hydrogens and then this carbon And so this is a completely valid molecular structure, but it was kind of a pain to draw all of these valence electrons here. So what we typically would want to do is, at least in this structure, and we're going to see later in this video there's even simpler ways to write it, so if we want at least do it with these lines, we can draw it like this. So you have a carbon, carbon, carbon, and then they are bonded to the hydrogens. So you'll almost never see it written like this because this is just kind of crazy. Hyrdrogen, hydrogen-- at least crazy to write. It takes forever. And it might be messy, like it might not be clear where these electrons belong. I didn't write it as clearly as I could. So they have two electrons there. They share with these two guys. Hopefully, that was reasonably clear. But if we were to draw it with the lines, it looks just like that. So it's a little bit neater, faster to draw, same exact idea here and here. And in general, and we'll go in more detail on it, this three-carbon chain, where everything is a single bond, is propane. Let me write these words down because it's helpful to get. This is methane. And you're going to see the rhyme-- you're going to see the reason to this naming soon enough. This is methane; this is propane. And there's an even simpler way to write propane. You could write it like this. Instead of explicitly drawing these bonds, you could say that this part right here, you could write that that part right there, that is CH3, so you have a CH3, connected to a-- this is a CH2, that is CH2 which is then connected to another CH3. And the important thing is, no matter what the notation, as long as you can figure out the exact molecular structure, as long as you can-- so there's this last CH3. Whether you have this, this, or this, you know what the molecular structure is. You could draw any one of these given any of the others. Now, there's an even simpler way to write this. You could write it just like this. Let me do it in a different color. You literally could write it so we have three carbons. So one, two, three. Now, this seems ridiculously simple and you're like, how can this thing right here give you the same information as all of these more complicated ways to draw it? Well, in chemistry, and in organic chemistry in particular, any of these-- let me call it a line diagram or a line angle diagram. It's the simplest way and it's actually probably the most useful way to show chains of carbons or to show organic molecules. Once they start to get really, really complicated, because then it's a pain to draw all of the H's, but when you see something like this, you assume that the end points of any lines have a carbon on it. So if you see something like that, you assume that there's a carbon at that end point, a carbon at that end point, and a carbon at that end point. And then you know that carbon makes four bonds. There are no charges here. All the carbons are going to make four bonds, and each of the carbons here, this carbon has two bonds, so the other two bonds are implicitly going to be with hydrogens. If they don't draw them, you assume that they're going to be with hydrogens. This guy has one bond, so the other three must be with hydrogen. This guy has one bond, so the other three must be hydrogens. So just drawing that little line angle thing right there, I actually did convey the exact same information as this depiction, this depiction, or this depiction. So you're going to see a lot of this. This really simplifies things. And sometimes you see things that are in between. You might see someone draw it like this, where they'll write CH3, and then they'll draw it like that. So that's kind of combining this way of writing the molecule where you write the CH3's for the end points, but then you implicitly have the CH2 on the inside. You assume that this end point right here is a C and it's bonded to two hydrogens. So these are all completely valid ways of drawing the molecular structures of these carbon chains or of these organic compounds." + }, + { + "Q": "When would you use vectors/vector notation/vector addition & subtraction like this in \"real life\"? My son asked and I had to admit that I have no idea.", + "A": "Pilots use vector addition. To figure out which direction and how fast they are actually traveling they add two vectors. One for the air, and one for the plane. The wind is usually not that strong compared to a jet, but to go straight , you may end up having to point the plane a little to the right or left.", + "video_name": "9ylUcCOTH8Y", + "transcript": "We've already seen that you can visually represent a vector as an arrow, where the length of the arrow is the magnitude of the vector and the direction of the arrow is the direction of the vector. And if we want to represent this mathematically, we could just think about, well, starting from the tail of the vector, how far away is the head of the vector in the horizontal direction? And how far away is it in the vertical direction? So for example, in the horizontal direction, you would have to go this distance. And then in the vertical direction, you would have to go this distance. Let me do that in a different color. You would have to go this distance right over here. And so let's just say that this distance is 2 and that this distance is 3. We could represent this vector-- and let's call this vector v. We could represent vector v as an ordered list or a 2-tuple of-- so we could say we move 2 in the horizontal direction and 3 in the vertical direction. So you could represent it like that. You could represent vector v like this, where it is 2 comma 3, like that. And what I now want to introduce you to-- and we could come up with other ways of representing this 2-tuple-- is another notation. And this really comes out of the idea of what it means to add and scale vectors. And to do that, we're going to define what we call unit vectors. And if we're in two dimensions, we define a unit vector for each of the dimensions we're operating in. If we're in three dimensions, we would define a unit vector for each of the three dimensions that we're operating in. And so let's do that. So let's define a unit vector i. And the way that we denote that is the unit vector is, instead of putting an arrow on top, we put this hat on top of it. So the unit vector i, if we wanted to write it in this notation right over here, we would say it only goes 1 unit in the horizontal direction, and it doesn't go at all in the vertical direction. So it would look something like this. That is the unit vector i. And then we can define another unit vector. And let's call that unit vector-- or it's typically called j, which would go only in the vertical direction and not in the horizontal direction. And not in the horizontal direction, and it goes 1 unit in the vertical direction. So this went 1 unit in the horizontal. And now j is going to go 1 unit in the vertical. So j-- just like that. Now any vector, any two dimensional vector, we can now represent as a sum of scaled up versions of i and j. And you say, well, how do we do that? Well, you could imagine vector v right here is the sum of a vector that moves purely in the horizontal direction that has a length 2, and a vector that moves purely in the vertical direction that has length 3. So we could say that vector v-- let me do it in that same blue color-- is equal to-- so if we want a vector that has length 2 and it moves purely in the horizontal direction, well, we could just scale up the unit vector i. We could just multiply 2 times i. So let's do that-- is equal to 2 times our unit vector i. So 2i is going to be this whole thing right over here or this whole vector. Let me do it in this yellow color. This vector right over here, you could view as 2i. And then to that, we're going to add 3 times j-- so plus 3 times j. Let me write it like this. Let me get that color. Once again, 3 times j is going to be this vector right over here. And if you add this yellow vector right over here to the magenta vector, you're going to get-- notice, we're putting the tail of the magenta vector at the head of the yellow vector. And if you start at the tail of the yellow vector and you go all the way to the head of the magenta vector, you have now constructed vector v. So vector v, you could represent it as a column vector like this, 2 3. You could represent it as 2 comma 3, or you could represent it as 2 times i with this little hat over it, plus 3 times j, with this little hat over it. i is the unit vector in the horizontal direction, in the positive horizontal direction. If you want to go the other way, you would multiply it by a negative. And j is the unit vector in the vertical direction. As we'll see in future videos, once you go to three dimensions, you'll introduce a k. But it's very natural to translate between these two Notice, 2, 3-- 2, 3. And so with that, let's actually do some vector operations using this notation. So let's say that I define another vector. Let's say it is vector b. I'll just come up with some numbers here. Vector b is equal to negative 1 times i-- times the unit vector i-- plus 4 times the unit vector in the horizontal direction. So given these two vector definitions, what would the would be the vector v plus b be equal to? And I encourage you to pause the video and think about it. Well once again, we just literally have to add corresponding components. We could say, OK, well let's think about what we're doing in the horizontal direction. We're going 2 in the horizontal direction here, and now we're going negative 1. So our horizontal component is going to be 2 plus negative 1-- 2 plus negative 1 in the horizontal direction. And we're going to multiply that times the unit vector i. And this, once again, just goes back to adding the corresponding components of the vector. And then we're going to have plus 4, or plus 3 plus 4-- And let me write it that way-- times the unit vector j in the vertical direction. And so that's going to give us-- I'll do this all in this one color-- 2 plus negative 1 is 1i. And we could literally write that just as i. Actually, let's do that. Let's just write that as i. But we got that from 2 plus negative 1 is 1. 1 times the vector is just going to be that vector, plus 3 plus 4 is 7-- 7j. And you see, this is exactly how we saw vector addition in the past, is that we could also represent vector b like this. We could represent it like this-- negative 1, 4. And so if you were to add v to b, you add the corresponding terms. So if we were to add corresponding terms, looking at them as column vectors, that is going to be equal to 2 plus negative 1, which is 1. 3 plus 4 is 7. So this is the exact same representation as this. This is using unit vector notation, and this is representing it as a column vector." + }, + { + "Q": "how do you solve\n3y-x=-9\n2y+5x=11", + "A": "I did it like this: 3y-x=-9 2y+5x=11 (3y-x=-9) x 5 15y-5x=-45 2y+5x=11 Add the two together. 17y=34 Divide by 17 y=2 Then just go back, substitute for y and solve for x Hope that helped", + "video_name": "uzyd_mIJaoc", + "transcript": "Use substitution to solve for x and y. And we have a system of equations here. The first equation is 2y is equal to x plus 7. And the second equation here is x is equal to y minus 4. So what we want to do, when they say substitution, what we want to do is substitute one of the variables with an expression so that we have an equation and only one variable. And then we can solve for it. Let me show you what I'm talking about. So let me rewrite this first equation. 2y is equal to x plus 7. And we have the second equation over here, that x is equal to y minus 4. So if we're looking for an x and a y that satisfies both constraints, well we could say, well look, at the x and y have to satisfy both constraints, both of these constraints have to be true. So x must be equal to y minus 4. So anywhere in this top equation where we see an x, anywhere we see an x, we say well look, that x by the second constraint has to be equal to y minus 4. So everywhere we see an x, we can substitute it with a y minus 4. So let's do that. So if we substitute y minus 4 for x in this top equation, the top equation becomes 2y is equal to instead of an x, the second constraint tells us that x needs to be equal to y minus 4. So instead of an x, we'll write a y minus 4, and then we have a plus 7. All I did here is I substituted y minus 4 for x. The second constraint tells us that we need to do it. y minus 4 needs to be equal to x or x needs to be equal to y minus 4. The value here is now we have an equation, one equation with one variable. We can just solve for y. So we get 2y is equal to y, and then we have minus 4 plus 7. So y plus 3. We can subtract y from both sides of this equation. The left hand side, 2y minus y is just y. y is equal to-- these cancel out. y is equal to 3. And then we could go back and substitute into either of these equations to solve for x. This is easier right over here, so let's substitute right over here. x needs to be equal to y minus 4. So we could say that x is equal to 3 minus 4 which is equal to negative 1. So the solution to this system is x is equal to negative 1 and y is equal to 3. And you can verify that it works in this top equation 2 times 3 is 6 which is indeed equal to negative 1 plus 7. Now I want to show you that over here we substituted-- we had an expression that, or we had an equation, that explicitly solved for x. So we were able to substitute the x's. What I want to show you is we could have done it the other way around. We could have solved for y and then substituted for the y's. So let's do that. And we could have substituted from one constraint into the other constraint or vice versa. Either way, we would have gotten the same exact answer. So instead of saying x is equal to y minus 4, in that second equation, if we add 4 to both sides of this equation, we get x plus 4 is equal to y. This and this is the exact same constraint. I just added 4 to both sides of this to get this constraint over here. And now since we've solved this equation explicitly for y, we can use the first constraint, the first equation. And everywhere where we see a y, we can substitute it with x plus 4. So it's 2 times-- instead of 2 times y, we can write 2 times x plus 4. 2 times x plus 4 is equal to x plus 7. We can distribute this 2. So we get 2x plus 8 is equal to x plus 7. We can subtract x from both sides of this equation. And then we can subtract 8 from both sides of this equation, subtract 8. The left hand side, that cancels out. We're just left with an x. On the right hand side, that cancels out, and we are left with a negative 1. And then we can substitute back over here we have y is equal to x plus 4, or so y is equal to negative 1 plus 4 which is equal to 3. So once again, we got the same answer even though this time we substituted for y instead of substituting for x. Hopefully you found that interesting." + }, + { + "Q": "Ask a question...can you cancel out the y instead of the x?", + "A": "Yes you can. That would mean that you solve for x first and then substitute the value to get y. Same method.", + "video_name": "u5dPUHjagSI", + "transcript": "We never know when we might have to do a little bit more party planning. So it doesn't hurt to have some practice. And that's what this exercise is doing for us, is generating problems so that we can try solving systems of equations with elimination. And so in this first problem, it says solve for x and y using elimination. And then this is what they have-- 6x minus 6y is equal to negative 24. Negative 5x minus 5y is equal to negative 60. So let me get my scratch pad out to solve this. Let me rewrite it. So they gave us 6x minus 6y is equal to negative 24. And negative 5x minus 5y is equal to negative 60. So what we have to think about, and we saw this in several of the other videos, is when we want to eliminate a variable, we want to manipulate these two equations. And if we were to add the corresponding sides, that variable might disappear. So if we just added a 6x to a negative 5x, that's not going to cancel it out. If this was a negative 6x, that would work out. Or if this was a positive 5x, that would work out. But this isn't exactly right. So if I want to eliminate the x, I have to manipulate these equations so that these two characters might cancel out. And one thing that pops into my brain is it looks like all of this stuff up here is divisible by 6, and all of this stuff down here is divisible by 5. And if we were to divide all this stuff up here by 6, we'd be left with an x over here. And if we were to divide all this bottom stuff by 5, we'd be left with a negative x right over here. And then they just might cancel out. So let's try that out. Let's take this first equation. And we're going to multiply both sides by 1/6. Or another way you could think about it is we're dividing both sides by 6. And as long as we do the same thing to both sides, the equation holds. The equality holds. So if you multiply everything by 1/6, 6x times 1/6 is just going to be x. 6y times 1/6 is just y. So it's negative y. Negative 24 times 1/6 is negative 4. Or you could just view it as negative 24 divided by 6 is negative 4. So this equation, the blue one, we've simplified as x minus y is equal to negative 4. Let's do something similar with the second one. Here we could say we're going to multiply everything times 1/5. Or you could say that we're dividing everything by 5. If we do that, negative 5x divided by 5 is just negative x. Negative 5y divided by 5 is negative y. And then negative 60 divided by 5 is negative 12. And now, this looks pretty interesting. If we add the two left-hand sides-- and remember, we can keep the equality, because we're essentially adding the same thing to both sides. You can imagine we're starting with the blue equation. And on the left-hand side, we're adding negative x minus y. And on the right-hand side, we're adding negative 12. But the second equation tells us that those two things So we're doing the same principle that we saw when we first started looking to algebra, that you can maintain your equality as long as you add the same thing. On the left-hand side, we're going to add this. And on the right-hand side, we're going to add this. But this second equation tells us that those two things are equal. So we can maintain our equality. So let's do that. What do we get on the left-hand side? Well, you have a positive x and a negative x. They cancel out. That was the whole point behind manipulating them in this way. And then you have negative y minus y, which is negative 2y. And then on the right-hand side, you have negative 4 minus 12, which is negative 16. And these are going to be equal to each other. Once again, we're adding the same thing to both sides. To solve for y, we can divide both sides by negative 2. And we are left with y is equal to positive 8. But we are not done yet. We want to go and substitute back into one of the equations. And we can substitute back into this one and to this one, or this one and this one. The solutions need to satisfy all of these essentially. This blue one is another way of expressing this blue equation. This green equation is another way of expressing this green equation. So I'll go for whichever one seems to be the simplest. And this one seems to be pretty simple right over here. So let's take x minus y-- we just solved that y would be positive 8-- is equal to negative 4. And now to solve for x, we just have to add 8 to both sides. And we are left with, on the left-hand side, negative 8 plus 8 cancels out. You're just left with an x. And negative 4 plus 8 is equal to positive 4. So you get x is equal to 4, y is equal to 8. And you can verify that it would work with either one of these equations. 6 times 4 is 24 minus 6 times 8-- so it's 24 minus 48-- is, indeed, negative 24. Negative 5 times 4 is negative 20, minus negative 40, if y is equal to 8, does, indeed, get you negative 60. So it works out for both of these. And we can try it out by inputting our answers. So x is 4, y is 8. So let's do that. So let me type this in. x is going to be equal to 4. y is going to be equal to 8. And let's check our answer. It is correct. Very good." + }, + { + "Q": "Was it already a clear plan for Germans, by the year 1939, to spread over the new countries as a 'superior race'?", + "A": "Not by the Germans- as in their people, nor even the majority of those in the military. But, yes it WAS the plan of the top echelon of Nazi Party Leadership.", + "video_name": "VTdV9JaHiIA", + "transcript": "As we get into the second half of the 1930s, we see an increasingly aggressive Nazi Germany. In 1935, they publicly announce their intent to rearm their military. The reason why this is significant is not that they were all of a sudden building their military. They, in fact, were doing this as soon as they had taken power, in 1933. But now, they felt confident enough to publicly state their intention -- which is another way of publicly stating that they [could n't] care less about the Treaty of Versailles, which had said that Germany was limited to a 100,000-soldier military. Then, we get into 1936. 1936, you might remember -- another term of the Treaty of Versailles was that Germany was not allowed to occupy the Rhineland -- this area in yellow right over here. And then that was actually reaffirmed in 1925 by the Treaties of Locarno, where Germany, itself, agreed to not occupy the Rhineland. But by 1936, Hitler decides to ignore all of those, and occupies the Rhineland. But once again the allies -- The French are not so happy about this. The UK, in particular, once again, [was] not super happy about this. But they decided this is not reason to potentially start another war over. So they really don't push back on Germany. Then, we get into 1938, and German aggression really goes into full gear. In March of 1938, you have a coup d'\u00e9tat, orchestrated by the Nazis in Austria, that really overthrows the Austrian government and allows the Germans to unify the two countries. So, you have the Germans come into Austria -- really a bloodless takeover. And there was already popular support for the Nazis in Austria. There was a Nazi party in Austria. There had been popular sentiment for many years, amongst many Austrians, to possibly be unified with the Germans. Austria [was and] is, fundamentally, a German-speaking nation. And so in March, this actually happens. This 'Anschluss' -- or unification. And, if you remember, that was also another forbidden term of the Treaty of Versailles. So now, the Germans are pretty much ignoring the Treaty of Versailles and the Treaty of St. Germain, which was the equivalent of the Treaty of Versailles -- but with the Austrians. So, you have the unification of Germany and Austria. Then, as we get into late 1938 -- in September in particular -- Hitler and the Nazis are interested in bringing the German-speaking populations of Czechoslovakia under German control. And this region, right over here in magenta, this is where you have large populations of German speakers. These regions are collectively referred to as the 'Sudetenland.' And really, just continuing the policy of not wanting to rock the boat with Germany, you have France, Great Britain, and Italy agreeing -- And Italy was an ally of the Germans. But France and Great Britain, in particular, are not interested in rocking the boat with the Germans. And so, in September of 1938, they sign the Munich Agreement, which did not actually -- where they actually did not consult the Czechoslovakian government -- where they allowed Germany to take over this region right over here -- the Sudetenland. And that, frankly, with the Germans taking over this significant part of the population of Czechoslovakia, a significant part of the industrial capacity of Czechoslovakia, this eventually leads to early 1939, where all of what we would now consider the Czech Republic -- this area right over here, all [of] this -- becomes a protectorate of Germany. So, they call it the 'Protectorate of Bohemia and Moravia.' So, Bohemia and Moravia go to Germany. And, so this is 1939. So, [by] 1939, [we have seen this pattern repeat itself during the previous] four years -- Nazi Germany ignoring the Treaty of Versailles, by rearming, by occupying the Rhineland, by unifying with Austria. Now, they're expanding their territory. They are actively allowed to take over the German-speaking areas of Czechoslovakia, under the Munich agreement. And eventually, they're able to take over Bohemia and Moravia -- all of what we would currently call the 'Czech Republic.' And this general pattern of German aggression, [allowed] by the other powers in Europe essentially allowing it to happen -- and, in particular, Great Britain allowing it to happen -- has been referred to as a 'policy of appeasement.' Obviously, the word 'appeasement' means there is someone who is angry about something, and you just don't want to make them any angrier -- you just let them do whatever they want -- this is, essentially, what was happening over here. And in hindsight, it might be easy to say, \"Hey, look!. They were allowing Germany to take over more and more -- to become more aggressive, which made [Germany] more and more confident. And this would eventually lead to World War II.\" But at the time, you do have to remember [that[ everyone still had a very strong memory of what had happened in World War I. And no one was interested in starting another [pan-European] war. And so, even [though] in hindsight, it's easy to say that the British -- in particular Neville Chamberlain, who was the Prime Minister from 1937 on -- were weak and allowed German[y] -- Hitler -- to gain confidence, which eventually led to the Nazi invasion of Poland in the fall of 1939. But it's easy to say that in hindsight. But what we see, as we get into 1939, is an aggressive Germany, a Germany that's not being checked by the other powers of Europe. And this is what eventually leads to September of 1939, where, actually, the Germans and the Soviets agree to partition Poland into their own spheres of influence, which allows Germany to invade Poland in early September [of] 1939 -- which is, you could kind of say, 'the straw that broke the camel's back,' and is the beginning of -- So, Poland invasion. The invasion of Poland. -- which is the beginning of World War II." + }, + { + "Q": "what is a fibonacci", + "A": "Fibonacci was a person for whom the series is named. The series is F2 = F1 + F0 as in 0 0 + 1 = 1 1 + 1 = 2 2 + 1 = 3 3 + 2 = 5 5 + 3 = 8", + "video_name": "ahXIMUkSXX0", + "transcript": "Voiceover:Say [unintelligible], you're in math class and your teacher's talking about ... Well, who knows what your teacher's talking about. Probably a good time to start doodling. And you're feeling spirally today, so yeah. Oh, and because of overcrowding in your school, your math class is taking place in greenhouse number three. Plants. You've decided there are three basic types of spirals. There's the kind where, as you spiral out, you keep the same distance. Or you could start big but make it tighter and tighter as you go around, in which case the spiral ends. Or you could start tight but make the spiral bigger as you go out. The first kind is good if you really want to fill up a page with lines. Or if you want to draw curled up snakes. You can start with a wonky shape to spiral around but you've noticed that, as you spiral out, it gets rounder and rounder. Probably something to do with how the ratio between two different numbers approaches one as you repeatedly add the same number to both. But you can bring the wonk back by exaggerating the bumps and it gets all optical illusiony. Anyway, you're not sure what the second kind of spiral is good for, but I guess it's a good way to draw snuggled up slug cats, which are a species you've invented just to keep this kind of spiral from feeling useless. This third spiral, however, is good for all sorts of things. You could draw a snail or a nautilus shell. And elephant with a curled up trunk, the horns of a sheep, a fern frond, a cochlea in an inner ear diagram, an ear itself. Those other spirals can't help but be jealous of this clearly superior kind of spiral. But I draw more slug cats. Here's one way to draw a really perfect spiral. Start with one square and draw another next to it that is the same height. Make the next square fit next to both together, that is each side is length two. The next square has length three. The entire outside shape will always be a rectangle. Keep spiraling around, adding bigger and bigger squares. This one has side length one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13. And now 21. Once you do that you can add a curve going through each square, arcing from one corner to the opposite corner. Resist the urge to zip quickly across the diagonal, if you want a nice smooth spiral. Have you ever looked at the spirally pattern on a pine cone and thought, \"Hey, sure are \"spirals on this pine cone?\" I don't know why there's pine cones in your greenhouse. Maybe the greenhouse is in a forest. Anyway, there's spirals and there's not There's one, two, three, four, five, six, seven, eight going this way. Or you could look at the spirals going the other way and there's one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13. Look familiar? Eight and 13 are both numbers in the Fibonacci series. That's the one where you start by adding one and one to get two, then one and two to get three, two and three to get five. Three plus five is eight, five plus eight is 13, and so on. Some people think that instead of starting with one plus one you should start with zero and one. Zero plus one is one, one plus one is two, one plus two and three, and it continues on the same way as starting with one and one. Or, I guess you could start with one plus zero and that would work too. Or why not go back one more to negative one and so on? Anyway, if you're into the Fibonacci series, you probably have a bunch memorized. I mean, you've got to know one, one, two, three, five. Finish off the single digits with eight and, ooh with 13, how spooky. And once you're memorizing double digits, you might as well know 21, 34, 55, 89 so that whenever someone turns a Fibonacci number you can say, \"Happy Fib Birthday.\" And then, isn't it interesting that 144, 233, 377? But 610 breaks that pattern, so you'd better know that one too. And oh my goodness, 987 is a neat number and, well, you see how these things get out of hand. Anyway, 'tis the season for decorative scented pine cones and if you're putting glitter glue spirals on your pine cones during math class, you might notice that the number of spirals are five and eight or three and five or three and five again. Five and eight. This one was eight and thirteen and one Fibonacci pine cone is one thing, but all of them? What is up with that? This pine cone has this wumpy weird part. Maybe that messes it up. Let's count the top. Five and eight. Now let's check out the bottom. Eight and 13. If you wanted to draw a mathematically realistic pine cone, you might start by drawing five spirals one way and eight going the other. I'm going to mark out starting and ending points for my spirals first as a guide and then draw the arms. Eight one way and five the other. Now I can fill in the little pine coney things. So there's Fibonacci numbers in pine cones but are there Fibonacci numbers in other things that start with pine? Let's count the spirals on this thing. One, two, three, four, five, six, seven, eight. And one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13. The leaves are hard to keep track of, but they're in spirals too. Of Fibonacci numbers. What if we looked at these really tight spirals going almost straight up? One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21. A Fibonacci number. Can we find a third spiral on this pine cone? Sure, go down like this. And one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13 (muttering) 19, 20, 21. But that's only a couple examples. How about this thing I found on the side of the road? I don't know what it is. It probably starts with pine, though. Five and eight. Let's see how far the conspiracy goes. What else has spirals in it? This artichoke has five and eight. So does this artichoke looking flower thing. And this cactus fruit does too. Here's an orange cauliflower with five and eight and a green one with five and eight. I mean, five and eight. Oh, it's actually five and eight. Maybe plants just like these numbers though. Doesn't mean it has anything to do with Fibonacci, does it? So let's go for some higher numbers. We're going to need some flowers. I think this is a flower. It's got 13 and 21. These daisies are hard to count, but they have 21 and 34. Now let's bring in the big guns. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34. And one, two, three, four, five, six, seven, eight, nine, 10, 11, (muttering) 17, 24, (muttering) 42, 53, 54, 55. I promise, this is a random flower and I didn't pick it out specially to trick you into thinking there's Fibonacci numbers in things, but you should really count for yourself next time you see something spirally. There's even Fibonacci numbers in how the leaves are arranged on this stalk, or this one, or the Brussels sprouts on this stalk are a beautiful delicious three and five. Fibonacci is even in the arrangement of the petals on this rose, and sunflowers have shown Fibonacci numbers as high as 144. It seems pretty cosmic and wondrous, but the cool thing about the Fibonacci series and spiral is not that it's this big complicated mystical magical super math thing beyond the comprehension of our puny human minds that shows up mysteriously everywhere. We'll find that these numbers aren't weird at all. In fact, it would be weird if they weren't there. The cool thing about it is that these incredibly intricate patterns can result from utterly simple beginnings." + }, + { + "Q": "what is 0 times 0", + "A": "The general rule is 0 times anything is 0. So 0 * 0 = 0", + "video_name": "lR_kUUPL8YY", + "transcript": "We have the number 7,346,521.032. And what I want to think about is if I look at the same digit in two different places, in particular, I'm going to look at the digit 3 here and the digit 3 here, how much more value does this left 3 represent than this right 3? In order to think about that, we have to think about place value. So let's write down all the place values. So this right over here, this is the ones place. Now, we could move to the right. And as we move to the right in place values, each place represents 1/10 of the place before it. Or you could divide by 10 as we're moving to the right. So this is the ones place. This is divide by 10. This is the 1/10 place, or the tenths place. Divide by 10 again, this is the hundredths place. Divide by 10 again, this is the thousandths place. And that \"s\" I'm just saying to be plural-- hundredths, thousandths. Now, if we go to the left, now each place represents a factor of 10 more. So if this is ones, multiply by 10, this is the tens place. This is the hundreds place. This is the thousands place. This is the ten thousands place. I'm going to have to write a little bit smaller. This is the hundred thousands place. And then the 7 is in the millions place. So what does this number, what does this 3 represent? Well, it's in the hundred thousands place. It literally represents 3 hundred thousands, or you could say 300,000, 3 followed by five zeroes. Now, what does this 3 represent? It's in the hundredths place. It literally represents 3 hundredths. It represents 3 times 1/100, which is the same thing as 3, which is equal to 3 over-- let me do the 3 in that purple color. Which is the same thing as 3/100, which is the same thing as 0.03. These are all equivalent statements. Now let's try to answer our original question. How much larger is this 3 than that 3 there? Well, one way to think about it is how much would you have to multiply this 3 by to get to this 3 over here? Well, one way to think about is to look directly at place value. So we got to multiply by 10. Every time we multiply by 10, that's equivalent to thinking about shifting it to one place to the left. So we would have to multiply by 10 one, two, three, four, five, six, seven times. So multiplying by 10 seven times. Let me write this down. So this multiplied by 10 seven times should be equal to this. Let me rewrite this. 300,000 should be equal to 3/100-- let me write it the same way. 3/100 multiplied by 10 seven times, so times 10 times 10 times 10 times 10 times 10-- let's see, that's five times-- times 10 times 10. Now, multiplying by 10 seven times is the same thing as multiplying by 1 followed by seven zeroes. Every time you multiply by 10, you're going to get another zero here. So this is the same thing as 3/100 times 1 followed by one, two, three, four, five, six, seven zeroes. So this is literally 3/100 times 10 million. So let's see if this actually is the case. Does this actually equal 300,000? Well, if you divide 10 million by 100, dividing 10 million by 100, or I guess you'd say in the numerator, you have 10 million and the denominator you have 100, if you were to just multiply it like this, if you view this as 3 over 100 times 10 million over 1. Well, you divide the numerator by 100, you're going to get rid of two of these zeroes. Divide the denominator by 100, you're going to get rid of this 100 here. And so you're going to be left with 3 times-- now we got to be careful with the commas here, because since I removed two zeroes, the commas are going to be different. It's going to be 3 times-- we put our commas in the right place, so just like that. So this simplified to 3 times 100,000, which is indeed 300,000. So it did work out. Shifting the 3 one, two, three, four, five, six, seven decimal places makes that 3 worth 1 followed by seven zeroes more, or it essentially makes that 3 worth 10 million more. So this 3 represents 10 million times the value of this 3. Let me write down the numbers. So this 3 is 10 million times the value of that 3." + }, + { + "Q": "So do weight and gravitational force mean the same ?explain?", + "A": "Weight is the force of gravity on an object.", + "video_name": "IuBoeDihLUc", + "transcript": "In this video, I want to clarify two ideas that we talk about on a regular basis, but are really muddled up in our popular language. And these are the ideas of mass and weight. And first, I'll tell you what they are. And then, we'll talk about how they are muddled up. So mass is literally-- there's a couple of ways to view mass. One way to view mass is-- and this is not a technical definition, but it will give you a sense of it-- is how much the stuff there is. So this is similar to saying matter. So if I have more molecules of a given mass, I will have a total of more mass. Or if I have more atoms, I will have more mass. So how much stuff there is of something. And I want to be careful with this definition right here, because there are other things that aren't what we would traditionally associate with matter, once we start going into more fancy physics, that still will exhibit mass. So another way to define mass is, how does something react to a specific force? And we already learned from Newton's second law that if you have a given force and you have more mass, you'll accelerate less. If you have less mass, you'll accelerate more. So how something responds to a given force. Something with lower mass will accelerate more for a given force. Something with higher mass will accelerate less. Now weight is the force of gravity on a mass, or on an object. So this is the force of gravity on an object. And just to think about the difference here, let's think about, I guess, myself sitting on Earth. So if I'm on Earth, my mass is 70 kilograms. My mass-- let me do this in a new color-- so my mass is 70 kilograms. There's 70 kilograms of stuff that constitute Sal. But my weight is not 70 kilograms. I mean, you'll often hear people say, I weighs 70 kilograms. And that's all right in just conversational usage, but that is not technically correct. Because weight is the force that Earth is pulling down-- or I should say-- the force of gravity on my mass. And so my weight-- let me think about the weight for a second-- the weight is going to be equal to the gravitational field at Earth. Hopefully you've watched the video on gravity. Or if you haven't, feel free to watch it. But the gravitational attraction between two objects, so the force of gravity between two objects is going to big G, the universal gravitational constant, times the mass of the first object-- let me actually-- times the mass of the second object, divided by the distance that separates the two objects squared. And if you're on the Earth, and if you take all of this stuff right over here combined-- so if you say that this right here is the mass of Earth. If you say this, right here, is the distance from the center to the surface of Earth, because that's where I'm sitting right now. So distance from center to surface of Earth. Then all of this stuff over here simplifies to what's sometimes called as lower case g. And lower case g is-- and I'm just rounding it here-- 9.8 meters per second squared. So the force of gravity for something near the surface of the Earth is going to be this quantity right over here times the mass. So my weight on the surface of the Earth is this 9.8 meters per second squared times my mass, times 70 kilograms. And so this is going to be-- I won't do, well, I could get my calculator out. Why don't I just get my calculator out and do the math? I was going to round it to 10 and say it's about 700, but let's just actually calculate it. So we have 9.8 times 70 kilograms. So we have 686. So this gives me 686. And then the units are kilogram meters per second squared. And these units, kilogram meters per second squared, are the same thing as a newton. So my weight-- and you'll never hear people say this-- but my weight on the surface of the Earth is 686 newtons. And notice, I just said that is my weight on the surface of the Earth. Because as you could imagine, weight is the force due to gravity on an object, on a mass. So if I go someplace else, if I go to the moon, for example, my weight will change. But my mass will not. So let's write this. This is the weight on Earth. If I were to take my weight on the moon-- and I haven't looked this up before the video. And you can verify this for me if you like. But I've been told that the gravitational force on the moon, or the gravitational attraction at the surface of the moon, is about 1/6 that of the surface of the Earth. So my weight on the moon will be roughly 1/6 of my weight on the Earth. Times 686 newtons. So that gives me a little bit over-- what is that-- 114 maybe? I'll just get the calculator out. My brain operates a little bit slower while I'm recording videos. Yeah. 114. So that gets us 114, approximately 114 newtons. So this is the thing I really want to emphasize then. Weight is a force due to gravity on an object. Your weight changes from planet to planet you go on. Your weight would actually even change if you went to a very high altitude because you're getting slightly further-- it would be immeasurably small-- but you're getting slightly further from the center of the Earth. Your weight would change an imperceptible amount. In fact, because the Earth is not a perfect sphere-- it's often referred to as an oblique spheroid-- your weight is actually slightly different on different parts If you went to the poles verses the equator, you would have a slightly different weight. Your mass does not change. It doesn't matter where you go, assuming that you don't have some type of nuclear reaction going on inside of you. Your mass does not change. So your mass does not change depending on where you are. Now, you might be saying, hey, look, I don't deal with kilograms, and newtons, and all of this. I operate in America. And in America, we talk about pounds. Is pounds appropriate? And yes, pounds is a unit of weight. So if I say that I weigh 160 pounds, this is indeed weight. I'm saying that the force of gravity on me is 160 pounds. But then you might say, well, what is mass then if you're talking about the English system or sometimes called the imperial system? And here, I will introduce you to a concept that very few people know. It's kind of a good trivial concept. The unit of mass-- so let's just be clear here-- the unit of mass in the imperial system, mass is called the slug. So if you wanted to figure out how many slugs you are, so your weight-- the force of gravity on you is 160 pounds. This is going to be equal to-- if you were to calculate all of this stuff-- the force of gravity on the surface of Earth. But if you were to do it in imperial units, instead of getting 9.8 meters per second squared, you would get 32 feet per second squared, which is also the acceleration near the surface of the Earth due to gravity in feet and seconds, as opposed to meters and seconds. And then, this is times your mass in slugs. So to figure it out, you divide both sides by 32 feet per second squared. So let's do that. Let's divide both sides by 32 feet per second squared. It cancels out. And then let me get my calculator out. So I have 160 pounds divided by 32 feet per second squared. And I get exactly 5. I should have been able to do that in my head. So I get 5. And the units here-- in the numerator, I have pounds. And then I'm dividing by feet per second squared. That's the same thing as multiplying by second squared per feet. And these units, 5 pounds second squareds over feet, this is the same thing as a slug. So if I weigh 160 pounds, my mass is going to be equal to 5 slugs. If my mass is 70 kilograms, my weight is 686 newtons. So hopefully that clarifies things a little bit." + }, + { + "Q": "Solve x5 y5=x-y", + "A": "If you re saying x = 5 and y = 5, the answer to x - y (or even y - x) is 0. Hope this helps! :D", + "video_name": "p5e5mf_G3FI", + "transcript": "We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then that is equal to x over 4. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. I could write just a plus 0, but I think that's a little unnecessary. And so we have negative 18 is equal to x over 4. And our whole goal here is to isolate the x, to solve for the x. And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4, times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4, you're just going to be left with an x. And on the other side, 4 times negative 18. Let's see, that's 40. Well, let's just write it out. So 18 times 4. If we were to multiply 18 times 4, 4 times 8 is 32. 4 times 1 is 4, plus 1 is 72. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to-- instead of writing x, I'm going to write negative 72-- is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true. This right-hand side, when x is equal to negative 72, does indeed equal negative 16." + }, + { + "Q": "I think it would be helpful to define what you mean by d.v. and i.v. before launching into the question.", + "A": "Thats easy D.v means Dependent variable and I.v means independent variable", + "video_name": "i9j_VUMq5yg", + "transcript": "On your math quiz, you earn 5 points for each question that you answer correctly. In the table above, q represents the number of questions that you answer correctly on your math quiz, and p represents the total number of points that you score on your quiz. The relationship between these two variables can be expressed by the following equation-- p is equal to 5q, where p is the points you get and q is the number of questions you answered correctly. And you could see that in the table. If q is 0, if you got no questions right, you get 0 points. If you got no questions right, well, 5 times 0 is going to be 0. If you get one question right, well, 1 times 5 is 5. You get 5 points per question. Two questions right, well, 2 times 5 is 10. 3 times 5 is 15. So this all makes sense. So then they ask us, which of the following statements Check all that apply. So let's think about this. They say the dependent variable is the number of points you score. So when you think about what's happening here, is your number of points you score is being driven by how many questions you get right. It's not like somehow the teacher says you got 15 points and now you have to get exactly three questions right. It's the other way around. The number of questions you get right is the independent variable, and that's driving the number of points you score. So the number of points you score is the dependent variable. And typically, the convention is to have the dependent variable be equal to some expression involving the independent variable. And you see that right over here. p is dependent on what happens to q. Depending on the number of questions, you multiply it by 5, and you get p. So the dependent variable is the number of points you score. The dependent variable is the number of questions you answer correctly. No, we've already talked about that. That's the independent variable. The independent variable is the number of points you score. That's the dependent variable. The independent variable is a number of questions you answer correctly. That's what's driving the dependent variable. And we can check our answer." + }, + { + "Q": "does sound travel fast in solid, liquid or gases and why", + "A": "think about the distance beween molecules and the stiffness of the bonds. Then ask again if you are not sure", + "video_name": "yF4cvbAYjwI", + "transcript": "- [Voiceover] To change the speed of sound you have to change the properties of the medium that sound wave is traveling through. There's two main factors about a medium that will determine the speed of the sound wave through that medium. One is the stiffness of the medium or how rigid it is. The stiffer the medium the faster the sound waves will travel through it. This is because in a stiff material, each molecule is more interconnected to the other molecules around it. So any disturbance gets transmitted faster down the line. The other factor that determines the speed of a sound wave is the density of the medium. The more dense the medium, the slower the sound wave will travel through it. This makes sense because if a material is more massive it has more inertia and therefore it's more sluggish to changes in movement or oscillations. These two factors are taken into account with this formula. V is the speed of sound. Capital B is called the bulk modulus of the material. The bulk modulus is the official way physicists measure how stiff a material is. The bulk modulus has units of pascals because it's measuring how much pressure is required to compress the material by a certain amount. Stiff, rigid materials like metal would have a large bulk modulus. More compressible materials like marshmallows would have a smaller bulk modulus. Row is the density of the material since density is the mass per volume, the density gives you an idea of how massive a certain portion of the material would be. For example, let's consider a metal like iron. Iron is definitely more rigid and stiff than air so it has a much larger bulk modulus than air. This would tend to make sound waves travel faster through iron than it does through air. But iron also has a much higher density than air, which would tend to make sound waves travel slower through it. So which is it? Does sound travel faster though iron or slower? Well it turns out that the higher stiffness of iron more than compensates for the increased density and the speed of sound through iron is about 14 times faster than through air. This means that if you were to place one ear on a railroad track and someone far away struck the same railroad track with a hammer, you should hear the noise 14 times faster in the ear placed on the track compared to the ear just listening through the air. In fact, the larger bulk modulus of more rigid materials usually compensates for any larger densities. Because of this fact, the speed of sound is almost always faster through solids than it is through liquids and faster through liquids than it is through gases because solids are more rigid than liquids and liquids are more rigid than gases. Density is important in some aspects too though. For instance, if you heat up the air that a sound wave is travelling through, the density of the air decreases. This explains why sound travels faster through hotter air compared to colder air. The speed of sound at 20 degrees Celsius is about 343 meters per second, but the speed of sound at zero degrees Celsius is only about 331 meters per second. Remember, the only way to change the speed of sound is to change the properties of the medium it's travelling in and the speed of sound is typically faster through solids than it is through liquids and faster through liquids than it is through gases." + }, + { + "Q": "aren't sound waves supposed to be longitudinal or am I missing something", + "A": "yep but when theyre displayed on an oscilloscope, they still come up the same way trasverse waves would- the amplitudes represent rarefaction and compression", + "video_name": "oTjTXS40pqs", + "transcript": "- [Instructor] So imagine you've got a wave source. This could be a little oscillator that's creating a wave on a string, or a little paddle that goes up and down that creates waves on water, or a speaker that creates sound waves. This could be any wave source whatsoever creates this wave, a nice simple harmonic wave. Now let's say you've got a second wave source. If we take this wave source, the second one, and we put it basically right on top of the first one, we're gonna get wave interference because wave interference happens when two waves overlap. And if we want to know what the total wave's gonna look like we add up the contributions from each wave. So if I put a little backdrop in here and I add the contributions, if the equilibrium point is right here, so that's where the wave would be zero, the total wave can be found by adding up the contributions from each wave. So if we add up the contributions from wave one and wave two wave one here has a value of one unit, wave two has a value of one unit. One unit plus one unit is two units. And then zero units and zero units is still zero. Negative one and negative one is negative two, and you keep doing this and you realize wait, you're just gonna get a really big cosine looking wave. I'm just gonna drop down to here. We say that these waves are constructively interfering. We call this constructive interference because the two waves combined to construct a wave that was twice as big as the original wave. So when two waves combine and form a wave bigger than they were before, we call it constructive interference. And because these two waves combined perfectly, sometimes you'll hear this as perfectly constructive or totally constructive interference. You could imagine cases where they don't line up exactly correct, but you still might get a bigger wave. In that case, it's still constructive. It might not be totally constructive. So that was constructive interference. And these waves were constructive? Think about it because this wave source two looked exactly like wave source one did, and we just overlapped them and we got double the wave, which is kinda like alright, duh. That's not that impressive. But check this out. Let's say you had another wave source. A different wave source two. This one is what we call Pi shifted 'cause look at it. Instead of starting at a maximum, this one starts at a minimum compared to what wave source one is at. So it's 1/2 of a cycle ahead of or behind of wave source one. 1/2 of a cycle is Pi because a whole cycle is two Pi. That's why people often call this Pi shifted, or 180 degrees shifted. Either way, it's out of phase from wave source one by 1/2 of a cycle. So what happens if we overlap these two? Now I'm gonna take these two. Let's get rid of that there, let's just overlap these two and see what happens. I'm gonna overlap these two waves. We'll perform the same analysis. I don't even really need the backdrop now because look at. I've got one and negative one. One and negative one, zero. Zero and zero, zero. Negative one and one, zero. Zero and zero, zero and no matter where I'm at, 1/2, a negative 1/2, zero. These two waves are gonna add up to zero. They add up to nothing, so we call this destructive interference because these two waves essentially destroyed each other. This seems crazy. Two waves add up to nothing? How can that be the case? Are there any applications of this? Well yeah. So imagine you're sitting on an airplane and you're listening to the annoying roar of the airplane engine in your ear. It's very loud and it might be annoying. You put on your noise canceling headphones, and what those noise canceling headphones do? They sit on your ear, they listen to the wave coming in. This is what they listen to. This sound wave coming in, and they cancel off that sound by sending in their own sound, but those headphones Pi shift the sound that's going into your ear. So they match that roar of the engine's frequency, but they send in a sound that's Pi shifted so that they cancel and your ear doesn't hear anything. Now it's often now completely silent. They're not perfect, but they work surprisingly well. They're essentially fighting fire with fire. They're fighting sound with more sound, and they rely on this idea of destructive interference. They're not perfectly, totally destructive, but the waves I've drawn here are totally destructive. If they were to perfectly cancel, we'd call that total destructive interference, or perfectly destructive interference. And it happens because this wave we sent in was Pi shifted compared to what the first wave was. So let me show you something interesting if I get rid of all this. Let me clean up this mess. If I've got wave source one, let me get wave source two back. So this was the wave that was identical to wave source one. We overlap 'em, we get constructive interference because the peaks are lining up perfectly with the peaks, and these valleys or troughs are matching up perfectly with the other valleys or troughs. But as I move this wave source too forward, look at what happens. They start getting out of phase. When they're perfectly lined up we say they're in phase. They're starting to get out of phase, and look at when I move it forward enough what was a constructive situation, becomes destructive. Now all the peaks are lining up with the valleys, they would cancel each other out. And if I move it forward a little more, it lines up perfectly again and you get constructive, move it more I'm gonna get destructive. Keep doing this, I go from constructive to destructive over and over. So in other words, one way to get constructive interference is to take two wave sources that start in phase, and just put them right next to each other. And a way to get destructive is to take two wave sources that are Pi shifted out of phase, and put them right next to each other, and that'll give you destructive 'cause all the peaks match the valleys. But another way to get constructive or destructive is to start with two waves that are in phase, and make sure one wave gets moved forward compared to the other, but how far forward should we move these in order to get constructive and destructive? Well let's just test it out. When they're right next to each other we get constructive. If I move this second wave source that was initially in phase all the way to here, I get constructive again. How far did I move it? I moved it this far. The front of that speaker moved this far. So how far was that? Let me get rid of this. That was one wavelength. So look at this picture. From peak to peak is exactly one wavelength. We're assuming these waves have the same wavelength. So notice that essentially what we did, we made it so that the wave from wave source two doesn't have to travel as far to whatever's detecting the sound. Maybe there's an ear here, or some sort of scientific detector detecting the sound. Wave source two is now only traveling this far to get to the detector, whereas wave source one is traveling this far. In other words, we made it so that wave source one has to travel one wavelength further than wave source two does, and that makes it so that they're in phase and you get constructive interference again. But that's not the only option, we can keep moving wave source two forward. We move it all the way to here, we moved it another wavelength forward. We again get constructive interference, and at this point, wave source one is having to make its wave travel two wavelengths further than wave source two does. And you could probably see the pattern. No matter how many wavelengths we move it forward, as long as it's an integer number of wavelengths we again get constructive interference. So something that turns out to be useful is a formula that tells us alright, how much path length difference should there be? So if I'm gonna call this X two, the distance that the wave from wave source two has to travel to get to whatever's detecting that wave. And the distance X one, that wave source one has to travel to get to that detector. So we could write down a formula that relates the difference in path length, I'll call that delta X, which is gonna be the distance that wave one has to travel minus the distance that wave two has to travel. And given what we saw up here, if this path length difference is ever equal to an integer number of wavelengths, so if it was zero that was when they were right next to each other, you got constructive. When this difference is equal to one wavelength, we also got constructive. When it was two wavelengths, we got constructive. It turns out any integer wavelength gives us constructive. So how would we get destructive interference then? Well let's continue with this wave source that originally started in phase, right? So these two wave sources are starting in phase. How far do I have to move it to get destructive? Well let's just see. I have to move it 'til it's right about here. So how far did the front of that speaker move? It moved about this far, which if I get rid of that speaker you could see is about 1/2 of a wavelength. From peak to valley, is 1/2 of a wavelength, I can keep moving it forward. Let's just see, that's constructive. My next destructive happens here which was an extra this far. How far was that? Let's just see. That's one wavelength, so notice at this point, wave source one is having to go one and 1/2 wavelengths further than wave source two does. So let's just keep going. Move wave source two, that's constructive. We get another destructive here which is an extra this far forward, and that's equal to one more wavelength. So if we get rid of this you could see valley to valley is a whole nother wavelength. So in this case, wave source two has to travel two and 1/2 wavelengths farther than wave source two. Any time wave source one has to travel 1/2 integer more wavelengths than wave source two, you get destructive interference. In other words, if this path length difference here is equal to lambda over two, three lambda over two, which is one and 1/2 wavelengths. Five lambda over two, which is two and 1/2 wavelengths, and so on, that leads to destructive interference. So this is how the path length differences between two wave sources can determine whether you're gonna get constructive or destructive interference. But notice we started with two wave sources that were in phase. These started in phase. So this whole analysis down here assumes that the two sources started in phase with each other, i.e. neither of them are Pi shifted. What would this analysis give you if we started with one that was Pi shifted? So let's get rid of this wave two. Let's put this wave two back in here. Remember this one? This one was Pi shifted relative to relative to wave source one. So if we put this one in here, and we'll get rid of this, now when these two wave sources are right next to each other you're getting destructive interference. So this time for a path length difference of zero, right? These are both traveling the same distance to get to the detector. So X one and X two are gonna be equal. You subtract them, you'd get zero. This time the zero's giving us destructive instead of constructive. So let's see what happens if we move this forward, let's see how far we've gotta move this forward to again get destructive. We'd have to move it over to here. How far did we move it? Let's just check. We moved the front of this speaker that far, which is one whole wavelength. So if we get rid of this, we had to move the front of the speaker one whole wavelength, and look at again it's destructive. So again, zero gave us destructive this time, and the lambda's giving us destructive, and you realize oh wait, all of these integer wavelengths. If I move it another integer wavelength forward, I'm again gonna get destructive interference because all these peaks are lining up with valleys. So interestingly, if two sourcese started Pi out of phase, so I'm gonna change this. Started Pi out of phase, then path length differences of zero, lambda, and two lambda aren't gonna give us constructive, they're gonna give us destructive. And so you could probably guess now, what are these path length differences of 1/2 integer wavelengths gonna give us? Well let's just find out. Let's start here, and we'll get rid of these. Let's just check. We'll move this forward 1/2 of a wavelength and what do I get? Yup, I get constructive. So if I move this Pi shifted source 1/2 a wavelength forward instead of giving me destructive, it's giving me constructive now. And if I move it so it goes another wavelength forward over to here, notice this time wave source one has to move one and 1/2 wavelengths further than wave source two. That's 3/2 wavelengths. But instead of giving us destructive, look. These are lining up perfectly. It's giving us constructive, and you realize oh, all these 1/2 integer wavelength path length differences, instead of giving me destructive are giving me constructive now because one of these wave sources was Pi shifted compared to the other. So I can take this here, and I could say that when the two sources start Pi out of phase, instead of leading to destructive this is gonna lead to constructive interference. And these two ideas are the foundation of almost all interference patterns you find in the universe, which is kind of cool. If there's an interference pattern you see out there, it's probably due to this. And if there's an equation you end up using, it's probably fundamentally based on this idea if it's got wave interference in it. So I should say one more thing, that sources don't actually have to start out of phase. Sometimes they travel around. Things happen, it's a crazy universe. Maybe one of the waves get shifted along its travel. Regardless, if any of them get a Pi shift either at the beginning or later on, you would use this second condition over here to figure out whether you get constructive or destructive. If neither of them get a phase shift, or interestingly, if both of them get a phase shift, you could use this one 'cause you could imagine flipping both of them over, and it's the same as not flipping any of them over. So recapping, constructive interference happens when two waves are lined up perfectly. Destructive interference happens when the peaks match the valleys and they cancel perfectly. And you could use the path length difference for two wave sources to determine whether those waves are gonna interfere constructively or destructively. The path length difference is the difference between how far one wave has to travel to get to a detector compared to how far another wave has to travel to get to that same detector, assuming those two sources started in phase and neither of them got a Pi shift along their travels. Path length differences of integer wavelengths are gonna give you constructive interference, and path length differences of 1/2 integer wavelengths are gonna give you destructive interference. Whereas if the two sources started Pi out of phase, or one of the got a Pi phase shift along its travel, integer wavelengths for the path length difference are gonna give you destructive interference. And 1/2 integer wavelengths for the path length difference are gonna give you constructive interference." + }, + { + "Q": "What is a constant?", + "A": "A constant is a specific number, no variable. For example, these are constants: 5; -8; 11.2; -2/5", + "video_name": "bAerID24QJ0", + "transcript": "Welcome to level one linear equations. So let's start doing some problems. So let's say I had the equation 5-- a big fat 5, 5x equals 20. So at first this might look a little unfamiliar for you, but if I were to rephrase this, I think you'll realize this is a pretty easy problem. This is the same thing as saying 5 times question mark equals 20. And the reason we do the notation a little bit-- we write the 5 next to the x, because when you write a number right next to a variable, you assume that you're multiplying them. So this is just saying 5 times x, so instead of a question mark, we're writing an x. So 5 times x is equal to 20. Now, most of you all could do that in your head. You could say, well, what number times 5 is equal to 20? Well, it equals 4. But I'll show you a way to do it systematically just in case that 5 was a more complicated number. So let me make my pen a little thinner, OK. So rewriting it, if I had 5x equals 20, we could do two things and they're essentially the same thing. We could say we just divide both sides of this equation by 5, in which case, the left hand side, those two 5's will cancel out, we'll get x. And the right hand side, 20 divided by 5 is 4, and we would have solved it. Another way to do it, and this is actually the exact same way, we're just phrasing it a little different. If you said 5x equals 20, instead of dividing by 5, we could multiply by 1/5. And if you look at that, you can realize that multiplying by 1/5 is the same thing as dividing by 5, if you know the difference between dividing and multiplying fractions. And then that gets the same thing, 1/5 times 5 is 1, so you're just left with an x equals 4. I tend to focus a little bit more on this because when we start having fractions instead of a 5, it's easier just to think about multiplying by the reciprocal. Actually, let's do one of those right now. So let's say I had negative 3/4 times x equals 10/13. Now, this is a harder problem. I can't do this one in my head. We're saying negative 3/4 times some number x is equal to 10/13. If someone came up to you on the street and asked you that, I think you'd be like me, and you'd be pretty stumped. But let's work it out algebraically. Well, we do the same thing. We multiply both sides by the coefficient on x. So the coefficient, all that is, all that fancy word means, is the number that's being multiplied by x. So what's the reciprocal of minus 3/4. Well, it's minus 4/3 times, and dot is another way to use times, and you're probably wondering why in algebra, there are all these other conventions for doing times as opposed to just the traditional multiplication sign. And the main reason is, I think, just a regular multiplication sign gets confused with the variable x, so they thought of either using a dot if you're multiplying two constants, or just writing it next to a variable to imply you're multiplying a variable. So if we multiply the left hand side by negative 4/3, we also have to do the same thing to the right hand side, minus 4/3. The left hand side, the minus 4/3 and the 3/4, they cancel out. You could work it out on your own to see that they do. They equal 1, so we're just left with x is equal to 10 times minus 4 is minus 40, 13 times 3, well, that's equal to 39. So we get x is equal to minus 40/39. And I like to leave my fractions improper because it's easier to deal with them. But you could also view that-- that's minus-- if you wanted to write it as a mixed number, that's minus 1 and 1/39. I tend to keep it like this. Let's check to make sure that's right. The cool thing about algebra is you can always get your answer and put it back into the original equation to make sure you are right. So the original equation was minus 3/4 times x, and here we'll substitute the x back into the equation. Wherever we saw x, we'll now put our answer. So it's minus 40/39, and our original equation said that equals 10/13. Well, and once again, when I just write the 3/4 right next to the parentheses like that, that's just another way of writing times. So minus 3 times minus 40, it is minus 100-- Actually, we could do something a little bit simpler. This 4 becomes a 1 and this becomes a 10. If you remember when you're multiplying fractions, you can simplify it like that. So it actually becomes minus-- actually, plus 30, because we have a minus times a minus and 3 times 10, over, the 4 is now 1, so all we have left is 39. And 30/39, if we divide the top and the bottom by 3, we get 10 over 13, which is the same thing as what the equation said we would get, so we know that we've got the right answer. Let's do one more problem. Minus 5/6x is equal to 7/8. And if you want to try this problem yourself, now's a good time to pause, and I'm going to start doing the problem right now. So same thing. What's the reciprocal of minus 5/6? Well, it's minus 6/5. We multiply that. If you do that on the left hand side, we have to do it on the right hand side as well. Minus 6/5. The left hand side, the minus 6/5 and the minus 5/6 cancel out. We're just left with x. And the right hand side, we have, well, we can divide both the 6 and the 8 by 2, so this 6 becomes negative 3. This becomes 4. 7 times negative 3 is minus 21/20. And assuming I haven't made any careless mistakes, that should be right. Actually, let's just check that real quick. Minus 5/6 times minus 21/20. Well, that equals 5, make that into 1. Turn this into a 4. Make this into a 2. Make this into a 7. Negative times negative is positive. So you have 7. 2 times 4 is 8. And that's what we said we would get. So we got it right. I think you're ready at this point to try some level one equations. Have fun." + }, + { + "Q": "is there any other way to find cube roots", + "A": "Some other ways to find cube roots are long division, succesive subtraction, etc. but these are the easiest", + "video_name": "DKh16Th8x6o", + "transcript": "We are asked to find the cube root of negative 512. Or another way to think about it is if I have some number, and it is equal to the cube root of negative 512, this just means that if I take that number and I raise it to the third power, then I get negative 512. And if it doesn't jump out at you immediately what this is the cube of, or what we have to raise to the third power to get negative 512, the best thing to do is to just do a prime factorization of it. And before we do a prime factorization of it to see which of these factors show up at least three times, let's at least think about the negative part a little bit. So negative 512, that's the same thing-- so let me rewrite the expression-- this is the same thing as the cube root of negative 1 times 512, which is the same thing as the cube root of negative 1 times the cube root of 512. And this one's pretty straightforward to answer. What number, when I raise it to the third power, do I get negative 1? Well, I get negative 1. This right here is negative 1. Negative 1 to the third power is equal to negative 1 times negative 1 times negative 1, which is equal to negative 1. So the cube root of negative 1 is negative 1. So it becomes negative 1 times this business right here, times the cube root of 512. And let's think what this might be. So let's do the prime factorization. So 512 is 2 times 256. 256 is 2 times 128. 128 is 2 times 64. We already see a 2 three times. 64 is 2 times 32. 32 is 2 times 16. We're getting a lot of twos here. 16 is 2 times 8. 8 is 2 times 4. And 4 is 2 times 2. So we got a lot of twos. So essentially, if you multiply 2 one, two, three, four, five, six, seven, eight, nine times, you're going to get 512, or 2 to the ninth power is 512. And that by itself should give you a clue of what the cube root is. But another way to think about it is, can we find-- there's definitely three twos here. But can we find three groups of twos, or we could also find-- let me look at it this way. We can find three groups of two twos over here. So that's 2 times 2 is 4. 2 times 2 is 4. So definitely 4 multiplied by itself three times is divisible into this. But even better, it looks like we can get three groups of three twos. So one group, two groups, and three groups. So each of these groups, 2 times 2 times 2, that's 8. That is 8. This is 2 times 2 times 2. That's 8. And this is also 2 times 2 times 2. So that's 8. So we could write 512 as being equal to 8 times 8 times 8. And so we can rewrite this expression right over here as the cube root of 8 times 8 times 8. So this is equal to negative 1, or I could just put a negative sign here, negative 1 times the cube root of 8 times 8 times 8. So we're asking our question. What number can we multiply by itself three times, or to the third power, to get 512, which is the same thing as 8 times 8 times 8? Well, clearly this is 8. So the answer, this part right over here, is just going to simplify to 8. And so our answer to this, the cube root of negative 512, is negative 8. And we are done. And you could verify this. Multiply negative 8 times itself three times. Negative 8 times negative 8 times negative 8. Negative 8 times negative 8 is positive 64. You multiply that times negative 8, you get negative 512." + }, + { + "Q": "So i have been confused. Why is pluto a dwarf planet?", + "A": "It is very small and it has an irregular orbit for a planet.", + "video_name": "kJSOqlcFpJw", + "transcript": "In the last video, we had a large cloud of hydrogen atoms eventually condensing into a high pressure, high mass, I guess you could say, ball of hydrogen atoms. And when the pressure and the temperature got high enough-- and so this is what we saw the last video-- when the pressure and temperature got high enough, we were able to get the hydrogen protons, the hydrogen nucleuses close enough to each other, or hydrogen nuclei close enough to each other, for the strong force to take over and fusion to happen and release energy. And then that real energy begins to offset the actual gravitational force. So the whole star-- what's now a star-- does not collapse on itself. And once we're there, we're now in the main sequence of a star. What I want to do in this video is to take off from that starting point and think about what happens in the star next. So in the main sequence, we have the core of the star. So this is the core-- star's core. And you have hydrogen fusing into helium. And it's releasing just a ton of energy. And that energy is what keeps the core from imploding. It's kind of the outward force to offset the gravitational force that wants to implode everything, that wants to crush everything. And so you have the core of a star, a star like the sun, and that energy then heats up all of the other gas on the outside of the core to create that really bright object that we see as a star, or in our case, in our sun's case, the sun. Now, as the hydrogen is fusing into helium, you could imagine that more and more helium is forming in the core. So I'll do the helium as green. So more, more, and more helium forms in the core. It'll especially form-- the closer you get to the center, the higher the pressures will be, and the faster that this fusion, this ignition, will happen. In fact, the bigger the mass of the star, the more the pressure, the faster the fusion occurs. And so you have this helium building up inside of the core as this hydrogen in the core gets fused. Now what's going to happen there? Helium is a more dense atom. It's packing more mass in a smaller space. So as more and more of this hydrogen here turns into helium, what you're going to have is the core itself is going to shrink. So let me draw a smaller core here. So the core itself is going to shrink. And now it has a lot more helium in it. And let's just take it to the extreme point where it's all helium, where it's depleted. But it's much denser. That same amount of mass that was in this sphere is now in a denser sphere, in a helium sphere. So it's going to have just as much attraction to it, gravitational attraction. But things can get even closer to it. And we know that the closer you are to a mass, the stronger the pull of gravity. So then instead of having just the hydrogen fusion occurring at the core, you're now going to have hydrogen fusion in a shell around the core. So now you're going to have hydrogen fusing in a shell around the core. Let me just be clear. This isn't just happens all of a sudden. It is a gradual process. As we have more and more helium in the core, the core gets denser and denser and denser. And so the pressures become even larger and larger near the core because you're able to get closer to a more massive core since it is now more dense. And as that pressure near the core increases even more and more, the fusion reaction happens faster and faster and faster until you get to this point. So here, let me be clear. You have a helium core. All of the hydrogen in the core has been used up. And then you have the hydrogen right outside of the core is now under enormous pressure. It's actually under more pressure than it was when it was just a pure hydrogen core. Because it's-- there's so much mass on the outside here, trying to, I guess you could say, exerting downwards, or gravitational force trying to get to that even denser helium core because everything is able to get closer in. And so now you have fusion occurring even faster. And it's occurring over a larger radius. So this faster fusion over a larger radius, the force is now going to expel-- the energy that's released from this fusion is now going to expel these outer layers of the star even further. So the whole time, this gradual process as the hydrogen turns into helium, or fuses into helium in the core, the hydrogen right outside of the core, right outside that area, starts to burn faster and faster. I shouldn't say burn. It starts to fuse faster and faster and over a larger and larger radius. The unintuitive thing is the fusion is happening faster over a larger radius. And the reason that is is because you have even a denser core that is causing even more gravitational pressure. And as that's happening, the star's getting brighter. And it's also-- the fusion reactions, since they're happening in a more intense way and over a larger radius, are able to expel the material of the star even larger. So the radius of the star itself is getting bigger and bigger and bigger. So if this star looked like this-- maybe let me draw it in white-- That's not white. Now what's happening to my color changer? There you go. OK, this star looked like this right over here. Now, this star over here, since a faster fusion reaction is happening over a larger radius, is going to be far larger. And I'm not even drawing it to scale. In the case of our sun, when it gets to this point, it's going to be 100 times the diameter. And at this point, it is a red giant. And the reason why it's redder than this one over here is that even though the fusion is happening more furiously, that energy is being dissipated over a larger surface area. So the actual surface temperature of the red giant, at this point, is actually going to be cooler. So it's going to emit a light at a larger wavelength, a redder wavelength than this thing over here. This thing, the core, was not burning as furiously as this thing over here. But that energy was being dissipated over a smaller volume. So this has a higher surface temperature. This over here, the core is burning more-- sorry, the core is no longer burning. The core is now helium that's not burning. It's getting denser and denser as the helium packs in on itself. But the hydrogen fusion over here is occurring more intensely. It's occurring in a hotter way. But the surface here is less hot because it's just a larger surface area. So it doesn't make-- the increased heat is more than mitigated by how large the star has become. Now, this is going to keep happening. And this core is keep-- the pressures keep intensifying because more and more helium is getting And this core keeps collapsing. And the temperature here keeps going up. So we said that the first ignition, the first fusion, occurs at around 10 million Kelvin. This thing will keep heating up until it gets to 100 million Kelvin. And now I'm talking about a star that's about as massive as the sun. Some stars will never even be massive enough to condense the core so that its temperature reaches But let's just talk about the case in which it does. So eventually, you'll get to a point-- so we're still sitting in the red giant phase, so we're this huge star over here. We have this helium core. And that helium core keeps getting condensed and condensed and condensed. And then we have a shell of hydrogen that keeps fusing into helium around it. So this is our hydrogen shell. Hydrogen fusion is occurring in this yellow shell over here that's expelling, that's allowed-- that's causing the radius of the star to get bigger and bigger, to expand. But when the temperature get sufficiently hot-- and now I think you're going to get a sense of how heavier and heavier elements form in the universe, and all of the heavy elements that you see around us, including the ones that are in you, were formed it this way from, initially, hydrogen-- when it gets hot enough at 100 million Kelvin, in this core, because of such enormous pressures, then the helium itself will start to fuse. So then we're going to have a core in here where the helium itself will start to fuse. And now we're talking about a situation. You have helium, and you had hydrogen. And all sorts of combinations will form. But in general, the helium is mainly going to fuse into carbon and oxygen. And it'll form into other things. And it becomes much more complicated. But I don't want to go into all of the details. But let me just show you a periodic table. I didn't have this in the last one. I had somehow lost it. But we see hydrogen here has one proton. It actually has no neutrons. It was getting fused in the main sequence into helium, two protons, two neutrons. You need four of these to get one of those. Because this actually has an atomic mass of 4 if we're talking about helium-4. And then the helium, once we get to 100 million Kelvin, can start being fused. If you get roughly three of them-- and there's all of these other things that are coming and leaving the reactions-- you can get to a carbon. You get four of them, four of them at least as the starting raw material. You get to an oxygen. So we're starting to fuse heavier and heavier elements. So what happens here is this helium is fusing into carbon and oxygen. So you start building a carbon and oxygen core. So I'm going to leave you there. I realize I'm already past my self-imposed limit of 10 minutes. But what I want you to think about is what is likely to happen. What is likely to happen here if this star will never have the mass to begin to fuse this carbon and oxygen? If it does have the mass, if it is a super massive star, it eventually will be able to raise even this carbon and oxygen core to 600 million Kelvin and begin to fuse that into even heavier elements. But let's think about what's going to happen for something like the sun, where it'll never have the mass, it'll never have the pressure, to start to fuse carbon and oxygen. And that'll be the topic of the next video." + }, + { + "Q": "is it possible to find the median except the mode?", + "A": "With only two points, the mean and median will both work.", + "video_name": "Ez_-RwV9WVo", + "transcript": "Let's say I have the point 3 comma negative 4. So that would be 1, 2, 3, and then down 4. 1, 2, 3, 4. So that's 3 comma negative 4. And I also had the point 6 comma 1. So 1, 2, 3, 4, 5, 6 comma 1. So just like that. 6 comma 1. In the last video, we figured out that we could just use the Pythagorean theorem if we wanted to figure out the distance between these two points. We just drew a triangle there and realized that this was the In this video, we're going to try to figure out what is the coordinate of the point that is exactly halfway between this point and that point? So this right here is kind of the distance, the line that connects them. Now what is the coordinate of the point that is exactly halfway in between the two? What is this coordinate right here? It's something comma something. And to do that-- let me draw it really big here. Because I think you're going to find out that it's actually pretty straightforward. At first it seems like a really tough problem. Gee, let me use the distance formula with some variables. But you're going to see, it's actually one of the simplest things you'll learn in algebra and geometry. So let's say that this is my triangle right there. This right here is the point 6 comma 1. This down here is the point 3 comma negative 4. And we're looking for the point that is smack dab in between those two points. What are its coordinates? It seems very hard at first. But it's easy when you think about it in terms of just the x and the y coordinates. What's this guy's x-coordinate going to be? This line out here represents x is equal to 6. This over here-- let me do it in a little darker color-- this over here represents x is equal to 6. This over here represents x is equal to 3. What will this guy's x-coordinate be? Well, his x-coordinate is going to be smack dab in between the two x-coordinates. This is x is equal to 3, this is x is equal to 6. He's going to be right in between. This distance is going to be equal to that distance. His x-coordinate is going to be right in between the 3 and the 6. So what do we call the number that's right in between the 3 and the 6? Well we could call that the midpoint, or we could call it the mean, or the average, or however you want to talk about it. We just want to know, what's the average of 3 and 6? So to figure out this point, the point halfway between 3 and 6, you literally just figure out, 3 plus 6 over 2. Which is equal to 4.5. So this x-coordinate is going to be 4.5. Let me draw that on this graph. 1, 2, 3, 4.5. And you see, it's smack dab in between. That's its x-coordinate. Now, by the exact same logic, this guy's y-coordinate is going to be smack dab between y is equal to negative 4 and y is equal to 1. So it's going to be right in between those. So this is the x right there. The y-coordinate is going to be right in between y is equal to negative 4 and y is equal to 1. So you just take the average. 1 plus negative 4 over 2. That's equal to negative 3 over 2 or you could say negative 1.5. So you go down 1.5. It is literally right there. So just like that. You literally take the average of the x's, take the average of the y's, or maybe I should say the mean to be a little bit more specific. A mean of only two points. And you will get the midpoint of those two points. The point that's equidistant from both of them. It's the midpoint of the line that connects them. So the coordinates are 4.5 comma negative 1.5. Let's do a couple more of these. These, actually, you're going to find are very, very straightforward. But just to visualize it, let me graph it. Let's say I have the point 4, negative 5. So 1, 2, 3, 4. And then go down 5. 1, 2, 3, 4, 5. So that's 4, negative 5. And I have the point 8 comma 2. So 1, 2, 3, 4, 5, 6, 7, 8 comma 2. 8 comma 2. So what is the coordinate of the midpoint The point that is smack dab in between them? Well, we just average the x's, average the y's. So the midpoint is going to be-- the x values are 8 and 4. It's going to be 8 plus 4 over 2. And the y value is going to be-- well, we have a 2 and a negative 5. So you get 2 plus negative 5 over 2. And what is this equal to? This is 12 over 2, which is 6 comma 2 minus 5 is negative 3. Negative 3 over 2 is negative 1.5. So that right there is the midpoint. You literally just average the x's and average the y's, or So let's graph it, just to make sure it looks like midpoint. 6, negative 5. 1, 2, 3, 4, 5, 6. Negative 1.5. Negative 1, negative 1.5. Yep, looks pretty good. It looks like it's equidistant from this point and that point up there. Now that's all you have to remember. Average the x, or take the mean of the x, or find the x that's right in between the two. Average the y's. You've got the midpoint. What I'm going to show you now is what's in many textbooks. They'll write, oh, if I have the point x1 y1, and then I have the point-- actually, I'll just stick it in yellow. It's kind of painful to switch colors all the time-- and then I have the point x2 y2, many books will give you something called the midpoint formula. Which once again, I think is kind of silly to memorize. Just remember, you just average. Find the x in between, find the y in between. So midpoint formula. What they'll really say is the midpoint-- so maybe we'll say the midpoint x-- or maybe I'll call it this way. I'm just making up notation. The x midpoint and the y midpoint is going to be equal to-- and they'll give you this formula. x1 plus x2 over 2, and then y1 plus y2 over 2. And it looks like something you have to memorize. But all you have to say is, look. That's just the average, or the mean, of these two numbers. I'm adding the two together, dividing by two, adding these two together, dividing by two. And then I get the midpoint. That's all the midpoint formula is." + }, + { + "Q": "what is the lower quartile and the upper quartile?", + "A": "The lower and upper quartiles are sections of a box and whisker plot. To find the lower quartile, find the middle of the numbers to the left of the median without including the median in this computation. The area in between that middle number and the median is the lower quartile. You can also find the upper quartile in a similar fashion.", + "video_name": "b2C9I8HuCe4", + "transcript": "An ecologist surveys the age of about 100 trees in a local forest. He uses a box-and-whisker plot to map his data shown below. What is the range of tree ages that he surveyed? What is the median age of a tree in the forest? So first of all, let's make sure we understand what this box-and-whisker plot is even about. This is really a way of seeing the spread of all of the different data points, which are the age of the trees, and to also give other information like, what is the median? And where do most of the ages of the trees sit? So this whisker part, so you could see this black part is a whisker, this is the box, and then this is another whisker right over here. The whiskers tell us essentially the spread of all of the data. So it says the lowest to data point in this sample is an eight-year-old tree. I'm assuming that this axis down here is in the years. And it says at the highest-- the oldest tree right over here is 50 years. So if we want the range-- and when we think of range in a statistics point of view we're thinking of the highest data point minus the lowest data point. So it's going to be 50 minus 8. So we have a range of 42. So that's what the whiskers tell us. It tells us that everything falls between 8 and 50 years, including 8 years and 50 years. Now what the box does, the box starts at-- well, let me explain it to you this way. This line right over here, this is the median. And so half of the ages are going to be less than this median. We see right over here the median is 21. So this box-and-whiskers plot tells us that half of the ages of the trees are less than 21 and half are older than 21. And then these endpoints right over here, these are the medians for each of those sections. So this is the median for all the trees that are less than the real median or less than the main median. So this is in the middle of all of the ages of trees that are less than 21. This is the middle age for all the trees that are greater than 21 or older than 21. And so we're actually splitting all of the data into four groups. This we would call the first quartile. So I'll call it Q1 for our first quartile. Maybe I'll do 1Q. This is the first quartile. Roughly a fourth of the tree, because the way you calculate it, sometimes a tree ends up in one point or another, about a fourth of the trees A fourth of the trees are between 14 and 21. A fourth are between 21 and it looks like 33. And then a fourth are in this quartile. So we call this the first quartile, the second quartile, the third quartile, and the fourth quartile. So to answer the question, we already did the range. There's a 42-year spread between the oldest and the youngest And then the median age of a tree in the forest is at 21. So even though you might have trees that are as old as 50, the median of the forest is actually closer to the lower end of our entire spectrum of all of the ages. So if you view median as your central tendency measurement, it's only at 21 years. And you can even see it. It's closer to the left of the box and closer to the end of the left whisker than the end of the right whisker." + }, + { + "Q": "How do u turn 8.84 into a fraction", + "A": "Put 84 over 100, which is 84/100. Then, add 8.", + "video_name": "EGr3KC55sfU", + "transcript": "Let's see if we can write 0.0727 as a fraction. Now, let's just think about what places these are in. This is in the tenths place. This is in the hundredths place. This 2 is in the thousandths place. And this 7 right here, this last 7, is in the ten-thousandths place. So there's a couple of ways we can do this. The way I like to think of this, this term over here is in the ten-thousandths place. We can view this whole thing right over here as 727 ten-thousandths because this is the smallest place right over here. So let's just rewrite it. This is equal to 727 over 10,000. And we've already written it as a fraction. And I think that's about as simplified as we can get. This number up here is not divisible by 2. It's not divisible by 5. In fact, it's not divisible by 3, which means it wouldn't be divisible by 6 or 9. It doesn't even seem to be divisible by 7. It might be a prime number. But I think we are done." + }, + { + "Q": "At 6:02, why does the hydrocarbon chain have no polarity?", + "A": "Carbon and Hydrogen have very similar electronegativity, so the elements don t pull each other to a specific direction. Thus the hydrocarbon chain is not polar.", + "video_name": "Pk4d9lY48GI", + "transcript": "- DNA gets a lot of attention as the store of our genetic information, and it deserves that. If we didn't have DNA, there would be no way of keeping the information that makes us us, and other organisms what those organisms are. And DNA has some neat properties, it can replicate itself, and we go into a lot of depth on that in other videos. So DNA producing more DNA, we call that, we call that replication, but just being able to replicate yourself on its own isn't enough to actually produce an organism. And to produce an organism, you somehow have to take that information in the DNA, and then produce things like a structural molecules, enzymes, transport molecules, signaling molecules, that actually do the work of the organism. And that process, the first step, and this is all a review that we've seen in other videos. The first step is to go from DNA to RNA, and in particular, messenger RNA. \"Messenger RNA,\" and this process right over here, this is called transcription. \"Transcription,\" we go into a lot of detail on this in other videos. And then you wanna go from that messenger RNA, it goes to the ribosomes and then tRNA goes and grabs amino acids, and they form actual proteins. So you go from messenger RNA, and then in conjunction, so this is all, this is in conjunction with tRNA and amino acids, so let me say \"+tRNA,\" and \"amino acids.\" And I'll write \"amino acids\" in, I'll write it in a brighter color, since that's going to be the focus of this video. So tRNA and amino acids, you're able to construct proteins. You are able to construct proteins, which are made up of chains of amino acids, and it's the proteins that do a lot of the work of the organism. Proteins, which are nothing but chains of amino acids, or they're made up of, sometimes multiple chains of amino acids. So you can image, I'm just going to, that's an amino acid. That's another amino acid. This is an amino acid. This is an amino acid, you could keep going. So these chains of amino acids, based on how these different, based on the properties of these different amino acids, and how the protein takes shape and how it might interact with its surrounding, these proteins can serve all sorts of different functions. Anything from part of your immune system, antibodies, they can serve as enzymes, they can serve as signaling hormones, like insulin. They're involved in muscle contraction. Actin and myosin, we actually have a fascinating video on that. Transport of oxygen. Hemoglobin. So proteins, the way at least my brain of it, is they do a lot of the work. DNA says, well, what contains the information, but a lot of the work of organism is actually done, is actually done by the proteins. And as I just said, the building blocks of the proteins are the amino acids. So let's focus on that a little bit. So up here are some examples of amino acids. And there are 20 common amino acids, there are a few more depending on what organism you look at, and theoretically there could be many more. But in most biological systems, there are 20 common amino acids that the DNA is coding for, and these are two of them. So let's just first look at what is common. So, we see that both these, and actually all three of this, this is just a general form, you have an amino group. You have an amino group, and this where, this is why we call it an \"amino,\" an amino acid. So you have an amino group. Amino group right over here. Now you might say, \"well, it's called an amino acid,\" \"so where is the acid?\" And that comes from this carboxyl group right over here. So that's why we call it an acid. This carboxyl group is acidic. It likes to donate this proton. And then in between, we have a carbon, and we call that the alpha carbon. We call that the alpha carbon. Alpha carbon, and that alpha carbon is bonded, it has a covalent bond to the amino group, covalent bond to the carboxyl group, and a covalent bond to a hydrogen. Now, from there, that's where you get the variation in the different amino acids, and actually, there's even some exceptions for how the nitrogen is, but for the most part, the variation between the amino acids is what this fourth covalent bond from the alpha carbon does. So you see in serine, you have this, what you could call it an alcohol. You could have an alcohol side chain. In valine right over here, you have a fairly pure hydrocarbon, hydrocarbon side chain. And so in general, we refer to these side chains as an R group, and it's these R groups that play a big role in defining the shape of the proteins, and how they interact with their environment and the types of things they can do. And you can even see, just from these examples how these different sides chains might behave differently. This one has an alcohol side chain, and we know that oxygen is electronegative, it likes to hog electrons, it's amazing how much of chemistry or even biology you can deduce from just pure electronegativity. So, oxygen likes to hog electrons, so you're gonna have a partially-negative charge there. Hydrogen has a low electronegativity relative to oxygen, so it's gonna have its electrons hogged, so you're gonna have a partially positive charge, just like that, and so this has a polarity to it, and so it's going to be hydrophilic, it's going to, at least this part of the molecule is going to be able to be attracted and interact with water. And that's in comparison to what we have over here, this hydrocarbon side chain, this has no polarity over here, so this is going to be hydrophobic. So this is going to be hydrophobic. And so when we start talking about the structures of proteins, and how the structures of proteins are influenced by its side chains, you could image that parts of proteins that have hydrophobic side chains, those are gonna wanna get onto the inside of the proteins if we're in an aqueous solution, while the ones that are more hydrophilic will wanna go onto the outside, and you might have some side chains that are all big and bulky, and so they might make it hard to tightly pack, and then you might have other side chains that are nice and small that make it very easy to pack, so these things really do help define the shape, and we're gonna talk about that a lot more when we talk about the structure. But how do these things actually connect? And we're gonna go into much more detail in another video, but if you have... If you have serine right over here, and then you have valine right over here, they connect through what we call peptide bonds, and a peptide is the term for two or more amino acids connected together, so this would be a dipeptide, and the bond isn't this big, I just, actually let me just, let me draw it a little bit smaller. So... That's serine. This is valine. They can form a peptide bond, and this would be the smallest peptide, this would be a dipeptide right over here. \"Peptide,\" \"peptide bond,\" or sometimes called a peptide linkage. And as this chain forms, that polypeptide, as you add more and more things to it, as you add more and more amino acids, this is going to be, this can be a protein or can be part of a protein that does all of these things. Now one last thing I wanna talk about, this is the way, the way these amino acids have been drawn is a way you'll often see them in a textbook, but at physiological pH's, the pH's inside of your body, which is in that, you know, that low sevens range, so it's a pH of roughly 7.2 to 7.4. What you have is this, the carboxyl group right over here, is likely to be deprotonated, it's likely to have given away its hydrogen, you're gonna find that more likely than when you have... It's gonna be higher concentrations having been deprotonated than being protonated. So, at physiological conditions, it's more likely that this oxygen has taken both of those electrons, and now has a negative charge, so it's given, it's just given away the hydrogen proton but took that hydrogen's electron. So it might be like this, and then the amino group, the amino group at physiological pH's, it's likely to actually grab a proton. So nitrogen has an extra loan pair, so it might use that loan pair to grab a proton, in fact it's physiological pH's, you'll find a higher concentration of it having grabbed a proton than not grabbing a proton. So, the nitrogen will have grabbed a proton, use its loan pairs to grab a proton, and so it is going to have... So it is going to have a... It is going to have a positive charge. And so sometimes you will see amino acids described this way, and this is actually more accurate for what you're likely to find at physiological conditions, and these molecules have an interesting name, a molecule that is neutral even though parts of it have charge, like this, this is called a zwitterion. That's a fun, fun word. Zwitterion. And \"zwitter\" in German means \"hybrid,\" and \"ion\" obviously means that it's going to have charge, and so this has hybrid charge, even though it has charges at these ends, the charges net out to be neutral." + }, + { + "Q": "How do you do the standard form of a polynomial?", + "A": "Standard form for a polynomial has the terms listed from highest degree to lowest. For example: 2 + 6x^2 - 7x +9x^3 written in standard form becomes: 9x^3 + 6x^2 - 7x + 2", + "video_name": "vN0aL-_vIKM", + "transcript": "simply 3x squared minus 8x plus 7 plus 2x to the third minus x squared plus eight x minus 3 so when we simplify this we're essentially going to add up like terms and just as a reminder we can only add or subtract like terms or simplify like terms and just a reminder and what I mean by that if I had an x squared to an x squared these are like terms they're both x terms raised to the same power the same degree so if I have one x squared and another x squared well then I have 2x squared - this is 2x squared If I have an x to the third - let's say I have 3x to thirds plus another 4x to the thirds Well that means I have 7x to the thirds - 7x to the thirds I can't take an x squared and add it to an x to the third I cannot simplify this in any way, so this you cannot simplify cannot simply - these are not like terms just because they both have Xs The Xs are not to - they are not to the same degree With that in mind let's look at the Xs to the same degree Let's start with the highest degree So the highest degree or the highest exponent on an x here - is actually this x to the third here but it looks like the only one it's the only place where we're raising x to the third power so that can't be merged or added or subtracted to anything else so let's just write that down. so we have 2x to the third and let's look at the x squared terms. We have 3x squared over there and we have a minus or we can view as a negative x squared over there so if we want to simplify we can add these two terms - we can add- so let me just write it down we can add 3x squared to negative x squared so I'm just rearranging it really right now I'm putting the like terms next to each other so it'll be easy to simplify now let's just worry about the x to the first terms or just the x terms you have a negative 8x term right over here so let me write it over here negative 8x and then you have a positive 8x term right over here, so let me write that down so positive 8x and then finally let's look at the constant terms you can view those as times x to the zeroth power and the constant terms are - you have a positive 7 over here - so plus 7 and then you have a negative 3 over here you have a negative 3 so all I'll I've done is I've really just used the communative property of addition to just change the order - or addition and subtraction - to change the order in which I'm doing this I've just rearranged the things so the like terms are next to each other but now we can simplify so we have 2x to the third - nothing to simplify that with but then if we subtract - if we have - let me do that in the same blue color if we have 3x squared and from that we're taking away an x squared well, we're only gonna have 2x squared left so that's gonna be plus 2x squared and then over here if we have negative 8 Xs and then we add 8 Xs to it or you can actually swap these around you could view it as you are subtracting 8 x from positive 8x we'll those are just going to cancel out so it's just going to be zero - I could just write plus zero here, but that'd just be redundant it wouldn't change the value and then finally I have a plus 7 minus 3 well that is just clearly 7 minus 3 is 4 so I have plus 4 And we're done! We've simplified it! 2x to the third plus 2 x squared plus 4" + }, + { + "Q": "is are blood actually blue but with the mix of oxygen it turns red", + "A": "No, blood is never blue. When it is deoxygenated, it is dark red. When it is oxygenated, it is a brighter red.", + "video_name": "7b6LRebCgb4", + "transcript": "- [Instructor] Let's talk a little bit about arteries and veins and the roles they play in the circulatory system. So I want you to pause this video and first think to yourself, Do you have a sense of what arteries and veins are? Well one idea behind arteries and veins are that well, in most of these drawings, arteries are drawn in red, and I even made the artery word here in red. And veins are drawn in blue. And so maybe that represents how much oxygen they have. And so one possible explanation is that arteries carry oxygenated blood, oxygenated, oxygenated blood, while veins carry deoxygenated blood. So blood that has less oxygen now. Now this is actually incorrect. It is, many times, the case that arteries are the ones carrying oxygenated blood and veins are carrying the deoxygenated blood. But as we will see, this is not always the case. And since we're already talking about oxygenated blood and deoxygenated blood and the colors red and blue, it's worth addressing another misconception. Many times it is said that deoxygenated blood looks blue, and the reason why people believe that is if you look at your wrist and you're able to see some of the vessels in there, you will see some blue vessels. And those, or at least they look blue when you're looking from the outside of your skin. And those, indeed, are veins. And so that's where the misconception has come from, that veins, which, in your arm, are carrying deoxygenated blood. That that deoxygenated blood is blue. It turns out that it is not blue. It is just a deeper red. And the reason why the veins look blue is because of the optics of light going through your skin and then seeing the outside of the veins and then reflecting back. That is not the color of the actual blood. So so far I have not given you a clear definition of what arteries versus veins are. A better definition, so let me cross these two out, are that arteries carry blood away from the heart. Away from the heart. And veins carry blood towards the heart. Towards the heart. And I can get a zoomed in image of the heart right here and that will make it a little bit clearer. And you can also see, or we're about to see, why this first definition, or this first distinction between arteries and veins does not always hold. So let's just imagine some blood that is being pumped away from the heart. So right when it gets pumped away from the heart, it'll be right over here. It gets pumped through the aorta, and you can see the aorta branches, so some blood can go up towards your head, and if it didn't, you would pass out and die. And then a lot of the blood goes down towards the rest of your body. And that, indeed, is the most oxygenated blood. And so it'll flow through your body. And these arteries will keep branching and branching into smaller vessels, all the way until they form these very small branches. And it's that place, especially, where they will lose a lot of their oxygen to the fluid and the cells around them. And then the blood is less oxygenated. And then even though deoxygenated blood is not blue, it often gets depicted as blue in a lot of diagrams. So I will do the same. And these vessels start building into your veins. And these really small vessels that really bridge between arteries and veins, where a lot of the gas and nutrient exchange occurs, these are called capillaries. And so after going through the capillaries, the blood will then come back to the heart and now it's coming towards the heart through the veins. It comes into the right atrium, then the right ventricle. Then that gets pumped towards the lungs. And this is the exception to the first incorrect definition of arteries and veins that we looked at. This right over here, is an artery. Even though it's carrying less oxygenated or deoxygenated blood, it's an artery because it's carrying blood away from the heart. But in this case, it's not carrying it to the rest of the body, it is carrying it to the lungs. That is why it is called the pulmonary artery, even though it's carrying less oxygenated blood. So that it goes to the lungs and then, in the lungs, there's more gas exchange that occurs. The blood gets oxygenated and then it comes back to the heart. And so it comes back to the heart in these vessels right over here, and that even though these are carrying highly oxygenated blood, these are considered veins because they're carrying blood towards the heart. So these are pulmonary veins. And then the cycle starts again. The pulmonary veins bring the oxygenated blood into the left atrium and the left ventricle, and then that pumps it to the rest of the body to the aorta, for your systemic circulation. You have your pulmonary circulation, which circulates the blood to, through and from the lungs. And you have your systemic circulation, which takes the blood to and from the rest of the body. So now that we have this main distinction between arteries and veins, what are some other interesting things that we know about it? Well one thing to keep in mind is that since arteries are being pumped directly by the heart towards the rest of the body, they have high pressure. I'll write that in caps. High pressure. And so if you were to have an accident of some type, which you do not want to have, and you were to accidentally cut an artery, because of that high pressure, it would actually spurt blood, a lot more than if you were to cut a vein. And most of the times where you get a cut, you're really just cutting capillaries. Like if you were to prick your finger, it's usually a series of capillaries that get cut, and that's why the blood would come out very very slowly. Now if arteries are high pressure, veins are low pressure. Low, low pressure. And one way to think about it is the arteries, the blood is being pumped directly by the heart. But then once it goes through the capillaries and comes back through the veins, it's kind of sluggishly making its way back to the heart. It's not being directly pumped. And that's why in veins, because you don't have that high pressure to bring everything back to the heart, you have these valves that make sure that for the most part, the blood is going in one direction. I'm going to draw the blood in red in the veins, just so we don't keep going with that misconception, that blood in the veins is blue somehow. Now related to the fact that the blood in the arteries is under higher pressure, in order to transport a fixed volume of blood in a certain amount of time, you need less volume. And so that's why arteries are low volume. And on the other hand, veins are high volume. And to appreciate the difference, the blood volume in arteries are only approximately 15% of the entire blood volume in your body, while the blood volume in veins are closer to 65%, approximately 65%. And if you're wondering where the rest of the blood is, about five percent is in capillaries, five percent is in your heart, and about 10% is in your lungs. So I will leave you there. The big take away: arteries are the vessels that take blood away from the heart. Veins are the vessels that take blood towards the heart." + }, + { + "Q": "Where is the setting of the portrait of the Mona Lisa.", + "A": "Leonardo began the painting in Florence, Italy but the background landscape in the painting is presumed to be imaginary.", + "video_name": "3kQ_p2EZX4Q", + "transcript": "[MUSIC PLAYING] BETH HARRIS: We thought we would start by looking at what is perhaps the most famous painting in the world, and whether we can actually even really still see. SAL KHAN: Right. Because I have seen this before. And I've even visited it at the Louvre-- I know I'm pronouncing it wrong. Yes, you're right. This is probably the most famous painting world. BETH HARRIS: And I just read that most people spend about 15 seconds in the Louvre looking at the painting, which is a funny statistic. SAL KHAN: Well, it's stressful, because there's people behind you. And on top of that, it's actually surprisingly small when you see it in real life. I mean, now that I'm able to take my time, and not worry about the tourists behind me, and I'm looking at it for real, I'm already-- things are jumping out at me that I actually had never noticed before. BETH HARRIS: Like what? SAL KHAN: Well, it looks like the scenery is some kind of like Vulcan territory or something. [LAUGHING] There's this-- it's like mountainous, and well, I guess, there's a little bridge in there. There's a road. I guess I never paid much attention to that before. Yeah, actually, I'd never even noticed this chair she was on before either. You can see hand resting on it. Actually-- and I never noticed that there's a ledge, right behind her, where's there jars. I could probably keep going. BETH HARRIS: I like your analogy to Vulcan territory, as a Star Trek fan myself. That landscape is otherworldly and very mysterious. But it's Interesting, isn't it, how the bottom part of the landscape at her neck and below looks like an inhabited landscape with a winding road and a bridge, but the landscape that's at her neck and head is more mysterious and looks very much like another planet? SAL KHAN: That's right. And actually, when you point that out and how that painting is divided based on where those landscapes and the ledge divide the painting, I don't have my ruler out, but I would guess that it's pretty close to the golden mean. BETH HARRIS: I think you're probably right. Those things that look like jars are actually the bottom of columns cut off on either side of the painting. SAL KHAN: So Leonardo da Vinci actually painted the columns, and it was cropped? BETH HARRIS: That's right. And so the space that she's in would have made a lot more sense as a balcony. SAL KHAN: Well, you know, all of this-- actually, just take a step back. I mean, we started with this presumption that it's-- and it's true-- that it's probably the most famous painting in the world, but I guess, I've never quite gotten why. I mean, is this just a case of marketing? BETH HARRIS: I think it happened in 1911, when the painting was stolen from the Louvre and disappeared for a couple of years and became notorious. At that point in the 19th century, the \"Mona Lisa\" was not the most popular painting at the Louvre. Paintings by other artists, like Titian and Raphael, were much more popular and even valued more highly for insurance purposes. So it really probably is only in the 20th century that she became as important as she is now. SAL KHAN: If you go back 150 years ago, \"Mona Lisa\" was not something that was just ingrained in our culture. BETH HARRIS: She was important. People were interested in her, and people were writing about her and they said some interesting things. But she wasn't as famous as she is now. And also, don't forget that the technology to reproduce her existed only, really, in the 20th century in terms of mass color reproductions. And so her currency has certainly increased, I think, in the last 100 years or so. SAL KHAN: I see. If you go back 150 years, there was probably no such thing as super famous paintings. BETH HARRIS: I think that might be true, actually. There were paintings that were famous, or important, but not celebrities in the way that the \"Mona Lisa\" is. SAL KHAN: Right. Not something that every person on the street would recognize. BETH HARRIS: Yeah. And of course, now I think most people would say that what's so interesting about her is her look and her smile, which have been interpreted in many different ways. SAL KHAN: Yeah, I know. And I know that's kind of, I guess, one of the claims to fame of the painting. And you see that. I mean, people like to look at it-- is she smirking, is she happy, is she sad. All of these things. Is she looking at you. All of these things that people try to-- but, I guess, trying to look at it without all of the social programming that I've had around this painting, it strikes me is an interesting painting. And it seems very technically well done. And there's something very bright, and just kind of an aura around her face. I don't know if I wasn't programmed to really know this painting and if I were to see this in the museum amongst many, many others, that I would-- it really jump out at me. BETH HARRIS: Portraits really took off during the Renaissance beginning in the 1400s in Italy. And Leonardo painted this in Florence. And that's because of humanism. One way that we define humanism is taking an interest in human beings, and the things of this world, and human achievement, and individuality. All of those values becoming more important in the 15th century. And so we begin to see a lot more portraits. Also with the beginnings of a wealthy merchants class in Florence in the 15th century, people could afford portraits and begin to want them. At first, portraits were painted with the figure in profile. But later, especially in northern Europe, artists like Durer or Memling started to put their figures in believable spaces. SAL KHAN: Right. BETH HARRIS: And so, Leonardo's really the first artist in Italy to do those things. To make an oil painting, which is a relatively new medium SAL KHAN: What did people use before oil? BETH HARRIS: They used fresco and tempera painting. Tempera for panel painting. So this is oil on wood, whereas before, artists would paint tempera on wood. Tempera tends to look more flat than oil paint, where you can really get a sense of modeling and light and dark. So Leonardo's making this three-dimensional figure, and he's using another technique called sfumato, which means a kind of smokey haziness. So he obscures the hard outlines around the forms, which tend to flatten them. One of the things that's fun to talk about with the Mona Lisa, too, is all the things that people have said about her over the years. You might not be aware of the fact that Sigmund Freud actually had a particular interpretation of the Mona Lisa. SAL KHAN: Yes, I'm sure he did. [LAUGHTER] I'm somewhat skeptical of him. I would like to interpret his interpretations someday. But yes. BETH HARRIS: Freud said that the \"Mona Lisa's\" smile combined the two ways that we tend to look at women in our culture. In one way, she's very mothering and nurturing. And in the other way, she seems very seductive. SAL KHAN: I think that says more about Freud than about Leonardo. BETH HARRIS: You could be right. [LAUGHTER] And later artists, another artist that you already know, Duchamp-- SAL KHAN: Duchamp, my favorite. BETH HARRIS: Your favorite. He took a reproduction of the \"Mona Lisa\" and drew a mustache on her. SAL KHAN: I could imagine him doing that. [LAUGHTER] BETH HARRIS: I think the moustache is interesting, because there is something not entirely feminine about her. Something a little bit masculine. SAL KHAN: Do you think it's that? Or I mean, I guess there is a certain-- I mean, it's kind of old now, especially because Duchamp did it, I'm guessing, 80, 90 years ago. But there is something hilarious about drawing a mustache on a feminine form. We all remember doing it as school kids-- just getting a kick out of it. And I could see it's especially funny for this painting. BETH HARRIS: Taking something that's so high art and making it silly, you know? SAL KHAN: Exactly. BETH HARRIS: Recently, the Prado in Madrid, found what turns out to be, after some scientific testing, a copy of the \"Mona Lisa,\" which in and of itself is not that unusual, but it turns out that their copy was made by another artist sitting right next to Leonardo copying what he did stroke for stroke. And they can tell this by analyzing the under drawing. SAL KHAN: Yeah, she looks much younger. BETH HARRIS: She has eyebrows. SAL KHAN: Oh, that's right. That's where the creepiness comes from, because the \"Mona Lisa\" we see looks jaundiced-- it's yellow. And so, the painting is a little bit different. The face is a little bit different, but we can assume that the colors might have not been that different. BETH HARRIS: Exactly. And it's a really interesting thing to think about. What she would look like if she was cleaned. And if she would still mean what she means to us. SAL KHAN: Oh, I don't think she would, because when I look at this cleaned painting, it loses a lot of the mystery. BETH HARRIS: Yeah, I agree. And you can then understand the Louvre's decision not to clean her. SAL KHAN: I mean, the cleaned one, she looks better. She looks younger. She loses a lot of the motherly aspects that Freud seems to want to ascribe to her. Yeah, because the colors are brighter, they're more vibrant, it's not as muted as the one that we've learned to like. BETH HARRIS: Yeah. Although, her reputation has grown over the years, who's to say that we won't care so much about her again. SAL KHAN: There might be a post-celebrity world at some point. [MUSIC PLAYING]" + }, + { + "Q": "At one point Jay says \"a positively charged oxygen\" at another he says \"an oxygen with a plus one formal charge\". Are both the terms equivalent?", + "A": "Yes, they mean the same thing.", + "video_name": "dJhxphep_gY", + "transcript": "So in a hydration reaction, water is added across a double bond. And the OH adds in a Markovnikov way. So according to Markovnikov's rule. So let's go ahead and write that down here. So you have to think about Markovnikov when you're doing this reaction. And this is an acid catalyzed reaction. So technically, this reactions is at equilibrium, and we will cover that at the very end of the video here. So let's look at the mechanism for the acid catalyzed addition of water across a double bond. So here I have my alkene, and I have water present with sulfuric acid. So sulfuric acid, being a strong acid, will donate a proton in solution. And let's say the water molecule picks up that proton. So H2O would go to H3O+. 3 So I'm just jumping ahead to the H3O+ ion called the hydronium ion. So here's my H3O+ ion. So put my one pair of electrons on there, and it's positively charged. So what's going to happen is the pi electrons, the electrons in this pi bond here are going to function as a base. They're going to abstract a proton. They're going to accept a proton. They're going to take, let's say, this proton right here, which would cause these two electrons in this bond to kick off onto your oxygen. So acid base equilibrium. So I'll go ahead and make this an equilibrium arrow here. And what are we going to get if that happens? Well, let's say that the proton added to the carbon on the right-- so the proton added to the carbon on the right here. So I'm saying that the blue electrons on the left are going to be these electrons right here, like that. And that would mean that I took a bond away from the carbon on the left. This carbon over here on the left, this carbon right here, used to have four bonds to it. Now it has only three bonds to it. So it ends up with a plus 1 formal charge. So we have a carbocation in our mechanism. So what's left? In this acid base reaction, we took a proton away from H3O+, which leaves us H2O. So here we have H2O over here, so I'll go ahead and put lone pairs of electrons in on our water molecule. And we know that water can act as a nucleophile here. So this lone pair of electrons is going to be attracted to something that's positively charged. So nucleophilic attack on our carbocation. And this is technically at equilibrium as well, depending on the concentrations of your reactants. So let's go ahead and show that water molecule adding on to the carbon on the left. So the carbon on the right already had a hydrogen or proton added onto it, and the carbon on the left is going to have an oxygen now bonded to that carbon. Two hydrogens bonded to that oxygen, and there was a lone pair of electrons on that oxygen that did not participate in any kind of bonding. This gives this oxygen right here a plus 1 formal charge. So our oxygen is now positively charged, like that. And we're almost to our product. So we're almost there. We need one more acid base reaction to get rid of that proton on our oxygen. So water can function as a base this time. So water comes along, and this time it's going to act as a Bronsted Lowry base and accept a proton. So let's get those electrons in there. So this lone pair of electrons, let's say it takes that proton, leaving these electrons behind on my oxygen. Once again, I'll draw my equilibrium arrows here, acid base reaction. And I'm going to end up with an OH on the carbon on the left, and the carbon on the right there is a hydrogen, like that. So I added water. I ended up adding water across my double bond. And to be complete, this would regenerate my hydronium ion. I'd get H2O plus H+ would give me H3O+. So hydronium is regenerated. And so there you go. So remember, a carbocation is present, so you have to think about Markovnikov addition. And since a carbocation is present, you have to think about possible rearrangements. So Markovnikov and rearrangements. Let's take a look at an example where you have a rearrangement here. So let's look at a reaction. So let's look at this as our starting alkene. And let's go ahead and think about the mechanisms. So we know H3O+ is going to be present. So H3O+ right here. So we're adding our alkene to a solution of water and sulfuric acid. And our first step in the mechanism, the pi electrons are going to function as a base and take a proton from our hydronium ion, leaving these electrons in here letting them kick back off onto the oxygen. So let's see what we would have from that acid base reaction. And I realize I didn't draw an equilibrium arrow here. I'm more concerned with getting the right product. So this is our carbon skeleton. And which side do we add the hydrogen? Which side of the double bond-- do we add the proton to the left side, or do we add the proton to the right side? Well, we want to form the most stable carbocation we possibly can. And if we add the proton to the left side of our double bond, we end up with a secondary carbocation. This carbocation right here is secondary, because this carbon that has the positive charge is bonded to two other carbons. So this is a secondary carbocation. If we had added the proton on to the other side of the double bond, would have a primary carbocation. So secondary carbocation is more stable. But can we form something that's even more stable than a secondary carbocation? Of course we can. We can form a tertiary carbocation if we think about the possibility of a hydride shift. So right here there is a hydrogen attached to that carbon. And if the proton and these two electrons are going to move over here and form a new bond with our positively charged carbon, so we get a hydride shift at this point. So let's draw what would result from that hydride shift. We moved a hydrogen over here. That took a bond away from this carbon. So that is the carbon that's going to end up with the positive charge now. We added a bond to what used to be our secondary carbocation carbon. And so that formal charge goes away. The formal charge moves to this carbon right here, which is now a tertiary carbocation. If you look at the carbons connected to that carbon, this is a tertiary carbocation. So we know tertiary carbocations are more stable than secondary. So now we're at the step of the mechanism where a water molecule is going to come along. So we have a water molecule, which is going to function as a nucleophile and attack our positively charged carbon, like that. So let's go ahead and draw what the result of that nucleophilic attack would look like. So we have our carbon skeleton, and we have an oxygen atom now bonded to that carbon. So two hydrogens here, and once again, one lone pair of electrons now participate in that reaction, giving this oxygen a plus 1 formal charge. And then finally, instead of showing the last step, a water molecule comes along, takes one of the protons off of our positively charged oxygen and gives us our major product with the OH adding on to this carbon right here. So this is a major product. This is our major product. And we would get some of the alcohol that forms from the secondary carbocation. So a minor product, that's what we would get if this oxygen had attacked our secondary carbocation. And you will get some of that. But if your test asks for the major product, you should show the product of this rearrangement. Now, we're lucky in this instance because our product here, this carbon right here, ends up not being a chirality center. Because I have two methyl groups attached to that carbon, so I don't have to worry about my stereochemistry here. Let's do a reaction where we do have to worry about stereochemistry. OK. So let's look at this reaction right here. So we take this as our starting alkene. So I'll put the double bond right there. And let's make that a little bit more clear here. So the double bond is between these two carbons right here. And once again, we're going to add water and sulfuric acid. So H2SO4. And when we think about the mechanism, we know that we're going to add a proton to one side of the double bond and the other side of the double bond is going to end up being our carbocation. So the first thing to think about is OK, which one of these two sides is going to get the proton. We want to form the most stable carbocation we can. So the proton's going to add on to the carbon on the left. So if I can go ahead and show the intermediate here. So for an intermediate-- don't need that arrow. We'll just go ahead and show the proton adding on to the carbon on the left. So we get an H here. And then the carbon on the right of the double bond now ends up being our carbocation. So this is now positively charged, like that. And remember, when we have a carbocation, this carbon is bonded to three other atoms. You have to think about what that looks like. So remember, a carbocation-- when something is bonded to three other carbons, you get this situation where everything is in the same plane. Sp2 hybridized carbon exhibits trigonal planar geometry. Also with your unhybridized p orbital, like that. So this is your sp2 hybridized carbocation situation here. So when your water molecule comes along and acts as a nucleophile, your water molecule could end up attacking from the top here. Or it could end up attacking from below here. So that's where the stereochemistry comes in. So let's go ahead and take our carbocation and let's see if we can draw the products that would result from our nucleophilic attack of water. And then we'll just go ahead and think about the proton going away in our heads for the mechanism. So when you have enough practice, you can do steps of the mechanism in your head. So let's see. What would we get for our two possible products? Well, the OH could've added this way, which would push that methyl group there away from us. So the methyl group would be going away from us. Or the OH could've added from the opposite side. The OH could've been the one back here, and that would've pushed the methyl group out like this. OK. And we know that these are chirality centers. We know that this is our chirality center on these guys. So we get enantiomers here. 50% racemic mixture for our products. And we see that the OH adds on in a Markovnikov fashion. It adds on to the side that's the most stable carbocation here. So one more thing about this reaction. So let's just do one that doesn't have any stereochemistry to worry about. And we'll try to make a different point about this reaction here. So this is our reaction. So we're going to add water to this. And we'll put sulfuric acid up here. And we'll make our arrow a little bit different to illustrate the point here. So we could go-- let's go ahead and draw an equilibrium arrow here. So let's say this whole reaction is at equilibrium. So let me get this equilibrium arrow in here. And our product. So if we don't have to worry about stereochemistry, we think OK, really all I have to do is think about which side of the double bond do I put my OH. And again, It's Markovnikov addition. The more substituted carbon is the one that's going to get your OH. So the more substituted carbon would be the one on the left here. So if you were a product, you would say OK, I know all I have to do is really just go ahead and put my OH in there on the more substituted carbon and I'm done. I don't have to worry about stereochemistry for this reaction. I don't have to worry about rearrangement, since it's the tertiary carbocation. So that takes care of it. Now, this reaction is technically at equilibrium. And you could think about water as being one of your reactants. So if water is one of your reactants and you think about general chemistry, Le Chatelier's principle, how do you shift in equilibrium? If you want to make more of your product, if you wanted to make more of this, your product or your alcohol, one way to do it would be to increase the concentration of water. So if you increase the concentration of water, the equilibrium will shift to decrease the stress that was put on the system. So you're going to get a shift to the right, and you're going to form more and more of your product here. But remember, if you have an alcohol for a product, and if you react this alcohol with sulfuric acid, that's an E1 elimination reaction that we saw in earlier videos. So acid catalyzed dehydration, the addition of concentrated sulfuric acid to your alcohol can actually form your alkene. So that's a reaction that we saw earlier, an E1 elimination acid catalyzed dehydration. Which your major product would be your most substituted alkene here. So you could go back the other way. You could go back to the left. Let's say you decreased the concentration of water to shift the equilibrium to the left, and you'd actually form your alkene here. So the way to control your equilibrium is if you want to go to the right, you just dilute your sulfuric acid. You add more water to it, which would increase this concentration. If you want to shift the equilibrium to the left, you decrease your concentration of water, which means using concentrated sulfuric acid. So I could just write concentrated sulfuric acid here. And that would shift your equilibrium to the left and make more of your alkene. So it all depends on what you're trying to make. And so you have to remember all that general chemistry, shifting equilibrium stuff." + }, + { + "Q": "Is it possible that a an issurance company can be insured but another?", + "A": "Yes, this is called reinsurance, and it is common.", + "video_name": "neAFEvNsiqw", + "transcript": "So let's see if we can get a big picture of everything that's happening in this credit default swap market. I'll speak in generalities. Let's say we have Corporation A, Corporation B, Corporation C. And let's say we have a bunch of people who write the credit default swaps, and I'll call them insurers. Because that's essentially what a credit default swap is, it's insurance on debt. If someone doesn't pay the debt, then the insurance company will pay it for you. In exchange, you're essentially giving some of the interest on the debt. So let's say we have Insurer 1, let's say we have Insurer 2. And some of these were insurance companies, some of these were banks. Some of these may have even been hedge funds. So these are the people who write the credit default swaps, and then there are the people who would actually buy the credit default swaps. In the previous example, I had Pension Fund 1, that was my pension fund. Then you could have another pension fund, Pension Fund 2. Let's re-draw some of the connections between the organizations. Let's say Pension Fund 1 were to lend $1 billion to A. A will pay Pension Fund 1 10%. But Pension Fund 1 wants to make sure that they'll definitely get the money, because they can't lend money to people with anything less than stellar credit ratings. So they get some insurance from Insurer 1. So what they do is out of this 10%, they pay them some of the basis points. So let's say they pay them 100 basis points. And in exchange, they get-- I'll call it Insurance On A. This is this new notation that I'm creating. They get Insurance On A. Fair enough. And the reason why this I1, this first insurer was able to do that is because Moody's has given them a very high credit rating. And so when they insure something, you're essentially the total package, right? The loan to this guy, plus the insurance, kind of is like you're lending the money to this guy, but you're just getting more insurance-- I mean you're getting more interest, right? So this bond becomes a Double A bond. Because the odds that you are not going to get your money are not the odds that this guy defaults, but it's now the odds that this guy defaults. And Moody's or the standard is poor, as I've already said. Hey, these guys are good for their money, they're Double A or whatever. So now your risk is really a Double A risk and not a Double B risk, or whatever. But anyway, this happens. This is Corporation B, and maybe Pension Fund 2 wants to lend to Corporation B. Maybe they lend them $2 billion. They get, I don't know, they get 12%, maybe Corporation B is a little bit more dangerous. But once again, they go to this first insurer. And maybe they get some of it-- well let's just say they get Insurance On B. And B is a little bit riskier, so they have to pay 200 basis points. 200 basis points goes from Pension Fund 2 to B. Now this, already, this is a little bit dangerous, right? Because you can think about what's happening. One, as long as this insurer does not get a downgrade from their credit ratings from S&P or Moody's or whoever, they can just keep it issuing this insurance. There's no limit for how much insurance they can issue. There's no law that says, you know what, if you insure a billion dollars of debt, you have to put a billion dollars aside. So that if that debt defaults, you definitely have that billion dollars there. Or if you insure 2 billion here, you don't have to put that 2 billion aside. What you have is a bunch of people who statistically say, oh, you know, what's the probability that all of this debt defaults? So I just have to keep enough capital so that probabilistically, whatever debt defaults, I can pay it. But you don't keep enough capital to pay all of the defaulting debt. So you already see an interesting risk forming. What if all of these corporations, for whatever reason, do start defaulting simultaneously? Then all of a sudden this insurance company has to pay more out in insurance then it might even have. So you have to wonder whether it even deserves this Double A rating, because it actually is taking on a lot of risk. But in the short term, while these companies are-- everyone is doing well and the economy's doing well, it's a great business for these guys. These guys are just collecting premiums essentially on the insurance, without having to pay out anything. Now let's add another twist on it. These pension funds, P1 and P2, it was reasonable for them to get insurance, because they were giving out these loans and then they got the insurance. So they were essentially hedging the default risk by buying these credit default swaps, which was essentially just an insurance policy from this Insurer 1. But you can have another party. This is no less legitimate, really. But you could call them-- I don't know-- let's call it Hedge Fund 1. And they've done a lot of work, and frankly, they often are much more sophisticated than the pension fund-- in fact, they almost always are. And they say, you know what? Company B looks really, really, really, really shady. I think 200 basis points for the chance that Company B defaults is frankly cheap. Because I think there's a huge probability that Company B defaults. So what I'm going to do, I'm not going to lend Company B money, because if anything, I think that they're maybe about to go out of business. But what I can do is I can buy a credit default swap on Company B's debt. Which is, essentially, I'm getting insurance that they fail without actually lending the money. So let's say I do that from Insurer 2. So I can go and I'll pay Insurer 2 200 basis points a year, or 2% on the notional value of the insurance I'm getting. So let's say it's 200 basis points, and let's say that's Insurance On-- I'm making a big bet-- so they're going to give me Insurance On B for-- I don't know-- $10 billion. And something interesting is going on here already. B might not have even borrowed $10 billion, right? So all of a sudden you have this hedge fund that is getting insurance on more debt than B has even borrowed money on, right? And it's essentially, you just kind of have this side bet between these two parties. This party says, you know what? I think it's a good deal. I get 200 basis points on the 10 billion every year, as long as B doesn't default. And this guy says, I think B's going to default. So I think that's a good deal on that insurance. And just so you understand the math, so the notional value is $10 billion. So what's 2% of 10 billion? 2% on a billion is 20 million, so it's $200 million. 200 if I did my math correct. So they'll pay $200 million a year to this insurer. So the 200 basis points on 10 billion is equal to 200 million. These numbers maybe are a little bit on the big side, but who knows? Actually, this could be a huge hedge fund. This could be a $10 billion hedge fund. Or even worse, maybe it's a billion dollar hedge fund, or maybe it's a $20 million hedge fund, but they've taken a $180 million loan to essentially buy this insurance because they think that B's collapse is imminent. So they're willing to take that bet right now. You know, it might be a good bet. If B collapses tomorrow, what's going to happen? They only dished out maybe 200 million for maybe that first year, although you normally pay it on a quarterly basis. So they'll pay 50 million every three months. Let's say they pay the first payment, 50 million, right? And then over the next three months, B just goes bankrupt and people realize that debt was worth nothing. Then these guys get $10 billion. Right? But something else is interesting here. They probably did insurance to a lot of other people too, maybe on B's debt. Or maybe they also insured A's debt. So maybe they gave some insurance on A's debt, as well. So what happens? Let's say B all of a sudden defaults. So a couple of things happen. I1 is going to owe P2 $2 billion, right? I2, the second insurer, is going to owe this hedge fund $10 billion. Now let's just assume I2's good for the money. They have $10 billion they pay to this hedge fund. This hedge fund is great, they get great bonuses for the year and they go buy yachts, et cetera. But this insurer right here, they pay the money they were good for but something interesting might happen. All of a sudden Moody's finally wakes up, these ratings agencies, and says, oh, my God. Well, there's a couple of things that might make them say, oh, my God. First of all, they might say, oh, look. You have to pay out $10 billion. And I doubt that was the only person you have to pay, maybe they have to pay out a lot of money. Now I2, Insurance Company 2, you are undercapitalized. I am now going to downgrade your rating. So, you were Double A, but since you had to give out all of this capital, Moody's is now going to downgrade you to, I don't know, B+. I'm just making these ratings up. But that's the sound of how these ratings happen, right. A is better, B is worse. The more A's you have, the better it is. But all of a sudden, when this guy is B+, and this guy insured, let's say, some other corporation's debt for this pension fund, now all of a sudden this insurance that this pension fund had is no longer Double A Insurance. It's now B+ Insurance, and maybe this pension fund, by its charter, can't hold something that has a B+ credit rating. So they're going to have to unwind the transaction, or maybe they'll have to unload the debt that was insured. So one, just by Company B defaulting, maybe this guy was holding some of Company A's debt, and it was insured by Insurance Company 1. Now they're going to have to unload that debt. So just one default creates this chain reaction, right? This one default happens, this guy has to pay this guy money, then this guy gets undercapitalized since they have to pay out money. Then Moody's says, oh, my God, you're undercapitalized. We're going to reduce your ratings. Maybe this guy was insuring some of A's debt, but now since he was insuring some of A's debt, all of a sudden that insurance is worth less because it has a lower rating. And now A's debt, less people want to hold it, because there are less people to insure it. I know that's very confusing, but this is really the point that Warren Buffett was saying when he said that the credit defaults swap market, or in general, the derivative market, are financial weapons of mass destruction. Because you have so many people who didn't have to set aside a capital, right? This guy could insure $10 billion worth of debt without having to set aside $10 billion. And you have so many people making all of these side bets, but they're all making two core assumptions. One, that these rating agencies's ratings are valid. And two, that the other person is good for the money. But if all of a sudden you have one failure someplace in the system, you could have this cascade where one, there's just a lot of downgrades. And then a lot of the people end up not being good for the money." + }, + { + "Q": "In my lecture notes, it states that the ATP synthase produces 32 ATP and not 34.\nThis gives a total of 32 (from ATP synthase) + 4 (from glycolysis and the Krebs cycle) = 36 ATP\nand the extra 2 ATP comes from something else, (I think it says fermentation)", + "A": "Different text books and notes vary in the information, but most(if not all) have Cellular Respiration making 36-38 ATP. I hope you found this helpful!", + "video_name": "mfgCcFXUZRk", + "transcript": "After being done with glycolysis and the Krebs Cycle, we're left with 10 NADHs and 2 FADH2s. And I told you that these are going to be used in the electron transport chain. And they're all sitting in the matrix of our mitochondria. And I said they're going to be used in the electron transport chain in order to actually generate ATP. So that's what I'm going to focus on in this video. The electron transport chain. And just so you know, a lot of this stuff is known. But some of the details are actually current areas of research. People have models and they're trying to substantiate the models. But things are happening at such a small scale here that people can just look at the evidence, some of which is indirect, and say, this is probably what's happening. Most of this is very well established, but some of the exact mechanisms-- for example, how exactly some of the proteins work-- aren't completely understood. So I think it's very important for you to understand that this is at the cutting edge, that you're already there. So the basic idea here is that the NADHs-- and that's where FADH2 is kind of the same idea. Although its electrons are just at slightly lower energy state. So they won't produce quite as many ATPs. Each NADH is going to be-- as you'll see-- indirectly responsible for the production of three ATPs. And each FADH2, in a very efficient cell, in both of these cases, will be indirectly responsible for the production of two ATPs. And the reason why this guy produces fewer ATPs is because the electrons that he has going into the electron transport chain are at a slightly lower energy level than the ones from NADH. So in general, I just said indirectly. How does this whole business work? Well in general, NADH, when it gets oxidized-- remember, oxidation is the losing of electrons or the losing of hydrogens that happen to have electrons. We can write its half reaction like this. Its oxidation reaction like this. You'll have some NAD plus, which you can then go and use back in the Krebs Cycle and in glycolysis. You have some NAD plus, you'll have a proton, a positive hydrogen ion is just a proton. And then you'll have two electrons. This is the oxidation of NADH. It's losing these two electrons. Oxidation is losing electrons. OIL RIG. Oxidation is losing electrons. Or you can imagine it's losing hydrogens, from which it can hog electrons. Either one of those is the case. Now this is really the first step of the electron transport chain. These electrons are transported out of the NADH. Now, the last step of the electron transport chain is you have two electrons-- and you could view it as the same two electrons if you like-- two electrons plus two hydrogen protons. And obviously if you just add these two together, you're just going to have two hydrogen atoms, which is just a proton and an electron. Plus one oxygen atom, so I could say one half of molecular oxygen. That's the same thing as saying one oxygen atom. And you're going to produce-- if I have one oxygen and two complete hydrogens, I'm left with water. And you could view this, we're adding electron or we're gaining electrons to oxygen. OIL RIG. Reduction is gaining electrons. So this is the reduction of oxygen to water. This is the oxidation of NADH to NAD plus. Now, these electrons that are popping out of-- these electrons right here-- that are popping out of this NADH. And when they're in NADH they're at a very high energy state. And what happens over the course of the electron transport chain is that these electrons get transported to a series of, I guess you could call them transition molecules. But these transition molecules, as the electrons go from one to the other, they go into slightly lower energy states. And I won't even go into the details of these molecules. One is coenzyme Q, and cytochrome C. And then they eventually end up right here and they are used to reduce your oxygen into water. Now every time an electron goes from a higher energy state to a lower energy state-- and that's what it's doing over the course of this electron transport chain-- it's releasing energy. So energy is released when you go from a higher state to a lower state. When these electrons were in NADH, they were at a higher state than they are when they bond to coenzyme Q. So they release energy. Then they go to cytochrome C and release energy. Now that energy is used to pump protons across the cristae across the inner membrane of the mitochondria. And I know this is all very complicated sounding. And this is the cutting edge. So it maybe should sound a little complicated. Let me draw a mitochondria. So let me draw a small mitochondria just so you know where we're operating. That's its outer membrane. And then its inner membrane, or its cristae, would look like that. And let me zoom in on the membrane. So let's say if I were to zoom in right there. So if I were to zoom that out, that box would look like this. You have your crista here. And I'm going to draw it thick. So I'm zooming in. This green line right here, I'm going to draw it really thick. I'm going to color it in with the green, just like that. And then you have your outer membrane. This outer membrane, I can do it up here. And I'll just color it in. You don't even have to see the outside of the outer membrane. Right here, this space right here, this is the outer compartment. And then we learned in the last video, this space right here is the matrix. This is where our Krebs cycle occurred. And where a lot of our NADH, or really all of our NADH, is sitting. So what happens is, every time NADH gets oxidized to NAD plus, and the electrons keep transferring from one molecule to another, it's occurring in these big protein complexes. And I'm not going to go into the details on this. So each of these protein complexes span-- so let's say that's a protein complex where this first oxidation reaction is occurring and releasing energy. And then let's say there's another protein complex here, where the second oxidation reaction is occurring and releasing energy. And these proteins are able to use that energy to essentially pump-- this might all seem very complicated-- to essentially pump hydrogens into the outer membrane. It actually pumps hydrogen protons. Hydrogen protons into the outer membrane. And every one of these reactions pump out a certain number of hydrogen protons. So by the end of the electron transport chain, or if we just followed one set of electrons, by the time that they've gone from their high energy state in NADH to their lower energy state in water, by the time they've done that, they've supplied the energy to these protein complexes that span our cristae to pump hydrogen from the matrix into the outer membrane. So really the only byproduct of the oxidation of NADH into, eventually, water, or the oxidation of NADH and the reduction of oxygen into water, isn't ATPs yet. It's just this gradient where we have a lot higher hydrogen proton concentration in the outer compartment than we do in the matrix. Or you could say that the outer compartment becomes a lot more acidic. Remember acidity is just hydrogen proton concentration, the concentration of hydrogen protons. So the byproduct of all of this energy is used to really just pump these protons into the outer membrane. So you have two things. The outer membrane becomes more acidic than the matrix inside. Maybe we could call that basic. And obviously these are all positively charged particles. So there's actually an electric gradient, an electric potential between the outer membrane and the inner membrane. This becomes slightly negative, that becomes slightly positive. These guys wouldn't naturally do this on their own. If this is already acidic and it's already positive, left to its own devices, these more protons wouldn't be entering. And the energy to do that is supplied by electrons going from high energy state in NADH to going to a lower energy state, eventually, on the oxygen in the water. That's what's happening. But essentially all that's happening is protons being pumped from the matrix into the outer compartment. Now once that gradient forms, these guys want to get back in. These guys want to get back into the matrix. And that is where the ATPs are formed. So there's a protein that also spans this. Let me draw. Remember this is all this inner membrane right here. Let me just draw it a little bit bigger right here. So that's our inner membrane, our cristae right there. There's a special protein called-- and I'll show you actually a better diagram of what looks like in a second-- called ATP synthase. And what happens is, remember because of the electron transport chain, we have all of these hydrogen ions up here, all of these protons really. All they are is a proton. That really want to get back into the matrix down here. This crista is impermeable to them so they have to find a special way to get through. They were able to go the reverse direction through the special protein complexes. Now they're going to go back into the matrix through ATP synthase. So they're going to go back, but something interesting happens. And this is really an area of current research where people think they know how it works but they're not sure. Because you can't just take these proteins apart and watch them operate like you can a regular mechanical engine. These are ultra-small and they have to be in a living system. And they have to have the right conditions. And you can't even-- it's hard to see hydrogen protons. These are ultra-small things that are, pretty much, you can't see them. But what happens, the current model is, as these enter, as these go through my ATP synthase there's actually an axle. So you can kind of view this as a housing. And then there's an axle. And this is all just a big protein. And there's an axle and then there's another part of the synthase down here. So you can imagine, this is kind of mind-blowing. That something this fancy is occurring on the membranes of pretty much all living systems' cells. It's not just eukaryotes. Even prokaryotes do this. They don't do it in their mitochondria; they do it in their cellular membrane. But it's a pretty neat thing. And what happens is, as these go through, you can kind of imagine as water flowing through a turbine. It mechanically causes this structure in the middle to spin. To actually spin. This is the current thinking. And this thing is all uneven. It's not, like, this nice tube. It'll look all crazy like that. And what happens is that an ADP molecule-- let's say that this is the A part of the ADP. And then you have two phosphate groups. It'll attach to one part of this protein. And maybe a phosphate will just randomly attach to another part of this protein. Just like that. So right now it's just ADP and a phosphate. But as this inner axle turns-- because it's not a symmetrical tube, it has different things sticking out that have different amounts of atomic charge and it's going to play with this outer housing right here. And so as this turns, the outer housing, because of just the proteins bumping against each other and electrical charge and whatever else, it's going to squeeze the ADP and the phosphate together to form, actually form, ATP. And actually the current thinking is that it does it on three different sites simultaneously. So as this spins around, ADP and phosphate groups kind of show up on the inside of this housing. You could imagine it like that. And I don't even know if it's on the inside. But they show up on the housing. And as this thing spins around, it stretches and pulls on this outer part and pushes these two things together. So it's using the energy from this proton gradient to drive this axle. And because it's all strange, it does all these distortions on this outer part and actually pushes the two ATPs together. So when you start off with your 10 NADHs, it'll provided just enough energy and just enough protons to put into the outer membrane that when they go back through our ATP synthase-- you could almost view it as an ATP synthase motor-- just based on people's observations they see that this will produce, on a per-NADH level, roughly three ATPs. On a per-FADH2 level, roughly two ATPs. And I've said multiple times in the videos, this is kind of an ideal. That a lot of times, maybe you'll have some protons leak, so their energy can't be captured properly. Or maybe some of these electrons might somehow jump the gun or jump some steps, so some of the energy gets lost. So you don't always have a completely efficient system. And just so you believe that this is actually occurring on our membrane, there's actual visual depictions of these proteins. This is the actual protein structure of ATP synthase right here. That is actually ATP synthase. And as you can see, there's this piece right here that holds this part and that part. You can kind of imagine it relatively stationary. The hydrogen comes through here. The axle gets spun. And as the axle gets spun, ADP and phosphate groups that are lodged inside this F1 part of the protein, get pushed together. You have to put energy into the reaction in order to make them stick together. But they get pushed together by the protein itself as this axle turns around. And this axle turns around from the energy of the hydrogen going. I don't even know what the mechanics would look like. But you could imagine-- in my head I imagine, the simplest thing is a windmill. Or not a windmill, as maybe some type of water turbine or maybe the simplest thing is, if you have something like that. I don't know if that's what the protein If you have any kind of thing passing by, it's going to spin it. It's going to spin it like that. And you could be more creative if you want to change the angle of the spin and whatnot. And that's all, people are really still trying to understand this at a deeper and deeper level. But for your purposes, especially in an introductory biology level, you just have to realize that two things are happening in the electron transport chain. Electrons are moving from the NADHs and the FADH2s to eventually show up and reduce the oxygen. And as they do that, they're releasing energy as they go from one molecule to another. They're going to lower energy states. That energy is used to pump hydrogen protons into the outer compartment of the mitochondria. And then that gradient, those hydrogen protons want to get back into the matrix of the mitochondria. So as they go back in, that drives this ATP synthase engine, which actually produces the ATP. So just like we said in the past, when you have 10 of these, on average-- let me say this way-- on average each NADH is going to produce 3 ATPs. Not directly. It produces enough of a gradient of hydrogen protons to produce 2 ATPs in the ATP synthase. And each FADH2, on average, produces enough of a hydrogen gradient to produce 2 ATPs. So if we come in with 10 NADH, they're going to produce-- in this ideal world-- 30 ATP. And then our 2 FADH2s are going to produce 4 ATP. And then if you remember from glycolysis, we had 2 net ATPs directly produced. And from the Krebs cycle we had 2 ATPs directly produced. So then you have 4 from glycolysis and Krebs, and that gets us, once again, to our magic 38 ATPs from one molecule of glucose. And now, I think you have a pretty good grasp of cellular respiration." + }, + { + "Q": "I have a question how do i find the percent of a fraction ? please help", + "A": "or if it is simple like 40/100 than it will be .40 = 40%", + "video_name": "X2jVap1YgwI", + "transcript": "Let's do some more percentage problems. Let's say that I start this year in my stock portfolio with $95.00. And I say that my portfolio grows by, let's say, 15%. How much do I have now? I think you might be able to figure this out on your own, but of course we'll do some example problems, just in case it's a little confusing. So I'm starting with $95.00, and I'll get rid of the dollar sign. We know we're working with dollars. 95 dollars, right? And I'm going to earn, or I'm going to grow just because I was an excellent stock investor, that 95 dollars is going to grow by 15%. So to that 95 dollars, I'm going to add another 15% of 95. So we know we write 15% as a decimal, as 0.15, so 95 plus 0.15 of 95, so this is times 95-- that dot is just a times sign. It's not a decimal, it's a times, it's a little higher than a decimal-- So 95 plus 0.15 times 95 is what we have now, right? Because we started with 95 dollars, and then we made another 15% times what we started with. Hopefully that make sense. Another way to say it, the 95 dollars has grown by 15%. So let's just work this out. This is the same thing as 95 plus-- what's 0.15 times 95? Let's see. So let me do this, hopefully I'll have enough space here. 95 times 0.15-- I don't want to run out of space. Actually, let me do it up here, I think I'm about to run out of space-- 95 times 0.15. 5 times 5 is 25, 9 times 5 is 45 plus 2 is 47, 1 times 95 is 95, bring down the 5, 12, carry the 1, 15. And how many decimals do we have? 1, 2. 15.25. Actually, is that right? I think I made a mistake here. See 5 times 5 is 25. 5 times 9 is 45, plus 2 is 47. And we bring the 0 here, it's 95, 1 times 5, 1 times 9, then we add 5 plus 0 is 5, 7 plus 5 is 12-- oh. I made a mistake. It's 14.25, not 15.25. So I'll ask you an interesting question? How did I know that 15.25 was a mistake? Well, I did a reality check. I said, well, I know in my head that 15% of 100 is 15, so if 15% of 100 is 15, how can 15% of 95 be more than 15? I think that might have made sense. The bottom line is 95 is less than 100. So 15% of 95 had to be less than 15, so I knew my answer of 15.25 was wrong. And so it turns out that I actually made an addition error, and the answer is 14.25. So the answer is going to be 95 plus 15% of 95, which is the same thing as 95 plus 14.25, well, that equals what? 109.25. Notice how easy I made this for you to read, especially this 2 here. 109.25. So if I start off with $95.00 and my portfolio grows-- or the amount of money I have-- grows by 15%, I'll end up with $109.25. Let's do another problem. Let's say I start off with some amount of money, and after a year, let's says my portfolio grows 25%, and after growing 25%, I now have $100. How much did I originally have? Notice I'm not saying that the $100 is growing by 25%. I'm saying that I start with some amount of money, it grows by 25%, and I end up with $100 after it grew by 25%. To solve this one, we might have to break out a little bit of algebra. So let x equal what I start with. So just like the last problem, I start with x and it grows by 25%, so x plus 25% of x is equal to 100, and we know this 25% of x we can just rewrite as x plus 0.25 of x is equal to 100, and now actually we have a level-- actually this might be level 3 system, level 3 linear equation-- but the bottom line, we can just add the coefficients on the x. x is the same thing as 1x, right? So 1x plus 0.25x, well that's just the same thing as 1 plus 0.25, plus x-- we're just doing the distributive property in reverse-- equals 100. And what's 1 plus 0.25? That's easy, it's 1.25. So we say 1.25x is equal to 100. Not too hard. And after you do a lot of these problems, you're going to intuitively say, oh, if some number grows by 25%, and it becomes 100, that means that 1.25 times that number is equal to 100. And if this doesn't make sense, sit and think about it a little bit, maybe rewatch the video, and hopefully it'll, over time, start to make a lot of sense to you. This type of math is very very useful. I actually work at a hedge fund, and I'm doing this type of math in my head day and night. So 1.25 times x is equal to 100, so x would equal 100 divided by 1.25. I just realized you probably don't know what a hedge fund is. I invest in stocks for a living. Anyway, back to the math. So x is equal to 100 divided by 1.25. So let me make some space here, just because I used up too much space. Let me get rid of my little let x statement. Actually I think we know what x is and we know how we got to there. If you forgot how we got there, you can I guess rewatch the video. Let's see. Let me make the pen thin again, and go back to the orange color, OK. X equals 100 divided by 1.25, so we say 1.25 goes into 100.00-- I'm going to add a couple of 0's, I don't know how many I'm going to need, probably added too many-- if I move this decimal over two to the right, I need to move this one over two to the right. And I say how many times does 100 go into 100-- how many times does 125 go into 100? None. How many times does it go into 1000? It goes into it eight times. I happen to know that in my head, but you could do trial and error and think about it. 8 times-- if you want to think about it, 8 times 100 is 800, and then 8 times 25 is 200, so it becomes 1000. You could work out if you like, but I think I'm running out of time, so I'm going to do this fast. 8 times 125 is 1000. Remember this thing isn't here. 1000, so 1000 minus 1000 is 0, so you can bring down the 0. 125 goes into 0 zero times, and we just keep getting 0's. This is just a decimal division problem. So it turns out that if your portfolio grew by 25% and you ended up with $100.00 you started with $80.00. And that makes sense, because 25% is roughly 1/4, right? So if I started with $80.00 and I grow by 1/4, that means I grew by $20, because 25% of 80 is 20. So if I start with 80 and I grow by 20, that gets me to 100. Makes sense. So remember, all you have to say is, well, some number times 1.25-- because I'm growing it by 25%-- is equal to 100. Don't worry, if you're still confused, I'm going to add at least one more presentation on a couple of more examples like this." + }, + { + "Q": "at 11:41, why is the average velocity in the horizontal direction is 5 square roots of 3 metres per second? I know Sal said it is because it doesn't change, but why does it not change?", + "A": "Gravity only affects the velocity in the vertical direction, and since we are assuming that there is no air resistance, there is nothing to change the horizontal velocity.", + "video_name": "ZZ39o1rAZWY", + "transcript": "- [Voiceover] So I've got a rocket here. And this rocket is going to launch a projectile, maybe it's a rock of some kind, with the velocity of ten meters per second. And the direction of that velocity is going to be be 30 degrees, 30 degrees upwards from the horizontal. Or the angle between the direction of the launch and horizontal is 30 degrees. And what we want to figure out in this video is how far does the rock travel? We want to figure out how, how far does it travel? Does it travel? And to simplify this problem, what we're gonna do is we're gonna break down this velocity vector into its vertical and horizontal components. We're going to use a vertical component, so let me just draw it visually. So this velocity vector can be broken down into its vertical and its horizontal components. And its horizontal components. So we're gonna get some vertical component, some amount of velocity in the upwards direction, and we can figure, we can use that to figure out how long will this rock stay in the air. Because it doesn't matter what its horizontal component is. Its vertical component is gonna determine how quickly it decelerates due to gravity and then re-accelerated, and essentially how long it's going to be the air. And once we figure out how long it's in the air, we can multiply it by, we can multiply it by the horizontal component of the velocity, and that will tell us how far it travels. And, once again, the assumption that were making this videos is that air resistance is negligible. Obviously, if there was significant air resistance, this horizontal velocity would not stay constant while it's traveling through the air. But we're going to assume that it does, that this does not change, that it is negligible. We can assume that were doing this experiment on the moon if we wanted to have a, if we wanted to view it in purer terms. But let's solve the problem. So the first that we want to do is we wanna break down this velocity vector. We want to break down this velocity vector that has a magnitude of ten meters per second. And has an angle of 30 degrees with the horizontal. We want to break it down it with x- and y-components, or its horizontal and vertical components. so that's its horizontal, let me draw a little bit better, that's its horizontal component, and that its vertical component looks like this. This is its vertical component. So let's do the vertical component first. So how do we figure out the vertical component given that we know the hypotenuse of this right triangle and we know this angle right over here. And the angle, and the side, this vertical component, or the length of that vertical component, or the magnitude of it, is opposite the angle. So we want to figure out the opposite. We have to hypotenuse, so once again we write down so-cah, so-ca-toh-ah. Sin is opposite over hypotenuse. So we know that the sin, the sin of 30 degrees, the sin of 30 degrees, is going to be equal to the magnitude of our vertical component. So this is the magnitude of velocity, I'll say the velocity in the y direction. That's the vertical direction, y is the upwards direction. Is equal to the magnitude of our velocity of the velocity in the y direction. Divided by the magnitude of the hypotenuse, or the magnitude of our original vector. Divided by ten meters per second. Ten meters per second. And then, to solve for this quantity right over here, we multiply both sides by 10. And you get 10, sin of 30. 10, sin of 30 degrees. 10 sin of 30 degrees is going to be equal to the magnitude of our, the magnitude of our vertical component. And so what is the sin of 30 degrees? And this, you might have memorized this from your basic trigonometry class. You can get the calculator out if you want, but sin of 30 degrees is pretty straightforward. It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation where we're starting in the ground and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, We're gonna be stationary at some point. And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information to figure out how long it's in the air? Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air, we just divide both sides by negative 9.8 meters per second squared. So we get, lets just do that, I wanna do that in the same color. So I do it in, that's not, well, that close enough. So we get negative 9.8 meters per second squared. Negative 9.8 meters per second squared. That cancels out, and I get my change in time. And I'll just get the calculator. I have a negative divided by a negative so that's a positive, which is good, because we want to go in positive time. We assume that the elapsed time is a positive one. And so what we get? If I get my calculator out, I get my calculator out. I have, this is the same thing as positive 10 divided by 9.8. 10, divided by 9.8. Gives me 1.02. I'll just round to two digits right over there. So that gives me 1.02 seconds So our change in time, so this right over here is 1.02. So our change in time, delta t, I'm using lowercase now but I can make this all lower case. Is equal to 1.02 1.02 seconds. Now how do we use this information to figure out how far this thing travels? Well if we assume that it retains its horizontal component of its velocity the whole time, we just assume we can this multiply that times our change in time and we'll get the total displacement in the horizontal direction. So to do that, we need to figure out this horizontal component, So this is the component of our velocity in the x direction, or the horizontal direction. Once again, we break out a little bit of trigonometry. This side is adjacent to the angle, so the adjacent over hypotenuse is the cosine of the angle. Cosine of an angle is adjacent over hypotenuse. So we get cosine. Cosine of 30 degrees, I just want to make sure I color-code it right, cosine of 30 degrees is equal to the adjacent side. Is equal to the adjacent side, which is the magnitude of our horizontal component, is equal to the adjacent side over the hypotenuse. Over 10 meters per second. multiply both sides by 10 meters per second, you get the magnitude of our adjacent side, color transitioning is difficult, the magnitude of our adjacent side is equal to 10 meters per second. Is equal to 10 meters per second. Times the cosine, times the cosine of 30 degrees. And you might not remember the cosine of 30 degrees, you can use a calculator for this. Or you can just, if you do remember it, you know that it's the square root of three over two. Square root of three over two. So to figure out the actual component, I'll stop to get a calculator out if I want, well I don't have to use it, do it just yet, because I have 10 times the square root of three over two. Which is going to be 10 divided by two is five. So it's going to be five times the square root of three meters per second. So if I wanna figure out the entire horizontal displacement, so let's think about it this way, the horizontal displacement, we're trying to figure out, the horizontal displacement, a S for displacement, is going to be equal to the average velocity in the x direction, or the horizontal direction. And that's just going to be this five square root of three meters per second because it doesn't change. So it's gonna be five, I don't want to do that same color, is going to be the five square roots of 3 meters per second times the change in time, times how long it is in the air. And we figure that out! Its 1.02 seconds. Times 1.02 seconds. The seconds cancel out with seconds, and we'll get that answers in meters, and now we get our calculator out to figure it out. so we have five time the square root of three, times 1.02. It gives us 8.83 meters, So this is going to be equal to, this is going to be equal to, this is going to be oh, sorry. this is going to be equal to 8.8, is that the number I got? 8.83, 8.83 meters. And we're done. And the next video, I'm gonna try to, I'll show you another way of solving for this delta t. To show you, really, that there's multiple ways to solve this. It's a little bit more complicated but it's also a little bit more powerful if we don't start and end at the same elevation." + }, + { + "Q": "So is size a noun", + "A": "Watch this: What is your size? ( Size in that form is a noun) and this: Can I get your size? ( That way it is a verb!) So @volpeana000 there you go!", + "video_name": "ETzngG8N3AU", + "transcript": "- Hello grammarians! Let's talk about singular and plural nouns. Nouns, as we discussed previously, are a type of word. They are a part of speech. A noun is any word that is a person, a place, a thing, or an idea. In English, we can figure out just by looking at a noun whether or not there is one of something, whether it's a singular, or whether or not there is more than one of something. There's an easy way to tell the difference between singular and plural. If you write the words down, singular contains the word single. Single, means there's only one of it. Plural is maybe a little bit less obvious, but it comes to us from Latin. It comes to us from this word plus, which means more, which you might recognize plus, as we call it in English, from mathematics, from arithmetic. We usually says it looks like this little plus symbol. So, whenever you think, whenever you see plural, just think more; just think plus. There is more than one. Singular is one thing. Plural, more than one thing; there is more. Let's go through it. Let's do some examples. I'll show you how you make the plural in English, how you indicate using your language that there is more than one thing. So let's just throw out a couple of words. Dog, cat, dinosaur, and whale. All you need to do to make it plural is very simply just take an s and you add it onto the end like so. Dogs, cats, dinosaurs, whales. If you want to make something plural, think about plus, more. All you have to do is add an s like that: add an s. This is what we call the regular plural. This is the regular plural. What that means is it obeys this one rule. All you have to do to say that there's more than one dog is throw on an s, and we're lucky because most English nouns behave that way. Most nouns are regular. However, here's the bad news. There are some irregular plurals. They are not regular, thus irregular, not. Now we have words like leaf, child, and fungus, which is like a mushroom, mouse, and sheep. How would you, you know you can't just add an s to these? That's unfortunately not how these nouns work in English. You can't say leafs, childs, and funguses, and mouses, and sheeps. This is how you do it. Each one of these words corresponds to a class of words that has its own unique pluralization standards. So, leaf becomes leaves. Child becomes children. Fungus becomes fungi. Mouse becomes mice. And sheep stays sheep, believe it or not. These are the irregular plurals, and we'll be covering each of these in turn in later videos, but for now I just want you to focus on the regular plural, which again we can sum up in this way. All you have to do is add an s. Here's a good example, right? We have one elephant here. Down here we have two elephants. The only difference between this word and this word is that this one has an s on the end of it. So if we wanted to say that this elephant here was not, in fact, one elephant, and was two elephants, all we have to do is add an s, changing it from singular to plural. Remember, plural comes from plus. Add an s. So one elephant becomes two elephants. (humming) World's fastest elephant drawing, go! (humming) It's kind of an elephant monkey, but you get the vague idea. If you're ever in need of more than one thing, for the regular plural, just add an s. You can learn anything. David, out!" + }, + { + "Q": "Is reflection actually refraction on a denser media?", + "A": "no", + "video_name": "jxptCXHLxKQ", + "transcript": "Before doing more examples with Snell's Law which essentially amounts to math problems what I do is give you an intuitive understanding for why this straw looks bent in this picture right over here To do that, let me just do a simplified version of that picture This is the side profile of the cup, or glass right over here The best I can draw it And then let me draw the actual straw. I'll first draw the straw where it actually is coming in off the side of the cup and the straw is actually not bending goes to the bottom of the cup just like that and then it goes up like that and then it goes slightly above. Then it actually does bent up here It's irrelevant to what we want to talk about What I want to do in this video is talk about when we look over here why does it look like the straw got bent? It all comes out of the refraction of the light As the light from the straw down here changes as it go from one medium to another Now we know from refraction indices or just in general the light moves slower in water than it does in air slower in water; faster in air Let's think about what's going to happen Let me draw 2 rays that are coming from this point on the straw right over here I draw one ray right over here. I'm gonna take an arbitrary direction. Like that Now when it goes from the slower medium to the faster medium, what's going to happen to it? Until this light go here so the left side of the ray is going to end up in the air before the right side and I'm using the car example to think about which way this light's going to bend So if you visualize it as a car--sometimes people visualize it as a marching band The left side of the marching band is gonna get out before the right side and start moving faster So this is going to turn to the right Let me do another ray Let the ray come from the same point Right along the straw, so another ray just like that It will also turn to the right Now if someone's eye is right over here-- Draw their nose and all the rest If they're looking down where does it look like this 2 light rays? Let's say his eye's big enough to capture both of these rays Where does it look like this 2 light rays are coming from? So if you trace both of these rays back if you just assume that there's a line here--that's what our eyes and brains do-- if you assume whatever direction this ray is currently going it's the direction it came from and same thing for this magenta ray It would look to this observer that this point on the straw is actually right over there And if you kept doing that for bunch of points on the straw it would look like this point on the straw is actually right over here It would look like this point on the straw is actually right over here So to this observer, the straw would look like this. It would look like bent Even though the light from here is going up and it moves out to because it gets bent, when you convert it back, it would converge to this just like we saw with that first point The light from this point when it goes out and gets bent. If you extrapolate backwards from their new directions, you get to that point So to this observer this point on the straw will look to be right over here even though the light was emitted down here And that's why the straw actually looks bent So this is all really just because of refraction from going from a slower medium to a faster one So hopefully you find that interesting. In next video, I'll do some examples of Snell's law just to get ourselves comfortable with the mathematics" + }, + { + "Q": "he says change in y but the way i learned is rise over run so which one is it", + "A": "Those mean the same thing. Rise corresponds to change in y , and run corresponds to change in x . See, if we denote change in y by \u00e2\u0088\u0086y and change in x by \u00e2\u0088\u0086x, rise over run refers to \u00e2\u0088\u0086y / \u00e2\u0088\u0086x, which is the slope.", + "video_name": "5fkh01mClLU", + "transcript": "In this video I'm going to do a bunch of examples of finding the equations of lines in slope-intercept form. Just as a bit of a review, that means equations of lines in the form of y is equal to mx plus b where m is the slope and b is the y-intercept. So let's just do a bunch of these problems. So here they tell us that a line has a slope of negative 5, so m is equal to negative 5. And it has a y-intercept of 6. So b is equal to 6. So this is pretty straightforward. The equation of this line is y is equal to negative 5x plus 6. That wasn't too bad. Let's do this next one over here. The line has a slope of negative 1 and contains the point 4/5 comma 0. So they're telling us the slope, slope of negative 1. So we know that m is equal to negative 1, but we're not 100% sure about where the y-intercept is just yet. So we know that this equation is going to be of the form y is equal to the slope negative 1x plus b, where b is the y-intercept. Now, we can use this coordinate information, the fact that it contains this point, we can use that information to solve for b. The fact that the line contains this point means that the value x is equal to 4/5, y is equal to 0 must satisfy this equation. So let's substitute those in. y is equal to 0 when x is equal to 4/5. So 0 is equal to negative 1 times 4/5 plus b. I'll scroll down a little bit. So let's see, we get a 0 is equal to negative 4/5 plus b. We can add 4/5 to both sides of this equation. So we get add a 4/5 there. We could add a 4/5 to that side as well. The whole reason I did that is so that cancels out with that. You get b is equal to 4/5. So we now have the equation of the line. y is equal to negative 1 times x, which we write as negative x, plus b, which is 4/5, just like that. Now we have this one. The line contains the point 2 comma 6 and 5 comma 0. So they haven't given us the slope or the y-intercept explicitly. But we could figure out both of them from these So the first thing we can do is figure out the slope. So we know that the slope m is equal to change in y over change in x, which is equal to-- What is the change in y? Let's start with this one right here. So we do 6 minus 0. Let me do it this way. So that's a 6-- I want to make it color-coded-- minus 0. So 6 minus 0, that's our change in y. Our change in x is 2 minus 2 minus 5. The reason why I color-coded it is I wanted to show you when I used this y term first, I used the 6 up here, that I have to use this x term first as well. So I wanted to show you, this is the coordinate 2 comma 6. This is the coordinate 5 comma 0. I couldn't have swapped the 2 and the 5 then. Then I would have gotten the negative of the answer. But what do we get here? This is equal to 6 minus 0 is 6. 2 minus 5 is negative 3. So this becomes negative 6 over 3, which is the same thing as negative 2. So that's our slope. So, so far we know that the line must be, y is equal to the slope-- I'll do that in orange-- negative 2 times x plus our y-intercept. Now we can do exactly what we did in the last problem. We can use one of these points to solve for b. We can use either one. Both of these are on the line, so both of these must satisfy this equation. I'll use the 5 comma 0 because it's always nice when you have a 0 there. The math is a little bit easier. So let's put the 5 comma 0 there. So y is equal to 0 when x is equal to 5. So y is equal to 0 when you have negative 2 times 5, when x is equal to 5 plus b. So you get 0 is equal to -10 plus b. If you add 10 to both sides of this equation, let's add 10 to both sides, these two cancel out. You get b is equal to 10 plus 0 or 10. So you get b is equal to 10. Now we know the equation for the line. The equation is y-- let me do it in a new color-- y is equal to negative 2x plus b plus 10. We are done. Let's do another one of these. All right, the line contains the points 3 comma 5 and negative 3 comma 0. Just like the last problem, we start by figuring out the slope, which we will call m. It's the same thing as the rise over the run, which is the same thing as the change in y over the change in x. If you were doing this for your homework, you wouldn't I just want to make sure that you understand that these are all the same things. Then what is our change in y over our change in x? This is equal to, let's start with the side first. It's just to show you I could pick either of these points. So let's say it's 0 minus 5 just like that. So I'm using this coordinate first. I'm kind of viewing it as the endpoint. Remember when I first learned this, I would always be tempted to do the x in the numerator. No, you use the y's in the numerator. So that's the second of the coordinates. That is going to be over negative 3 minus 3. This is the coordinate negative 3, 0. This is the coordinate 3, 5. We're subtracting that. So what are we going to get? This is going to be equal to-- I'll do it in a neutral color-- this is going to be equal to the numerator is negative 5 over negative 3 minus 3 is negative 6. So the negatives cancel out. You get 5/6. So we know that the equation is going to be of the form y is equal to 5/6 x plus b. Now we can substitute one of these coordinates in for b. So let's do. I always like to use the one that has the 0 in it. So y is a zero when x is negative 3 plus b. So all I did is I substituted negative 3 for x, 0 for y. I know I can do that because this is on the line. This must satisfy the equation of the line. Let's solve for b. So we get zero is equal to, well if we divide negative 3 by 3, that becomes a 1. If you divide 6 by 3, that becomes a 2. So it becomes negative 5/2 plus b. We could add 5/2 to both sides of the equation, plus 5/2, plus 5/2. I like to change my notation just so you get familiar with both. So the equation becomes 5/2 is equal to-- that's a 0-- is equal to b. b is 5/2. So the equation of our line is y is equal to 5/6 x plus b, which we just figured out is 5/2, plus 5/2. We are done. Let's do another one. We have a graph here. Let's figure out the equation of this graph. This is actually, on some level, a little bit easier. What's the slope? Slope is change in y over change it x. So let's see what happens. When we move in x, when our change in x is 1, so that is our change in x. So change in x is 1. I'm just deciding to change my x by 1, increment by 1. What is the change in y? It looks like y changes exactly by 4. It looks like my delta y, my change in y, is equal to 4 when my delta x is equal to 1. So change in y over change in x, change in y is 4 when change in x is 1. So the slope is equal to 4. Now what's its y-intercept? Well here we can just look at the graph. It looks like it intersects y-axis at y is equal to negative 6, or at the point 0, negative 6. So we know that b is equal to negative 6. So we know the equation of the line. The equation of the line is y is equal to the slope times x plus the y-intercept. I should write that. So minus 6, that is plus negative 6 So that is the equation of our line. Let's do one more of these. So they tell us that f of 1.5 is negative 3, f of negative 1 is 2. What is that? Well, all this is just a fancy way of telling you that the point when x is 1.5, when you put 1.5 into the function, the function evaluates as negative 3. So this tells us that the coordinate 1.5, negative 3 is on the line. Then this tells us that the point when x is negative 1, f of x is equal to 2. This is just a fancy way of saying that both of these two points are on the line, nothing unusual. I think the point of this problem is to get you familiar with function notation, for you to not get intimidated if you see something like this. If you evaluate the function at 1.5, you get negative 3. So that's the coordinate if you imagine that y is equal to f of x. It would be equal to negative 3 when x is 1.5. Anyway, I've said it multiple times. Let's figure out the slope of this line. The slope which is change in y over change in x is equal to, let's start with 2 minus this guy, negative 3-- these are the y-values-- over, all of that over, negative 1 minus this guy. Let me write it this way, negative 1 minus that guy, minus 1.5. I do the colors because I want to show you that the negative 1 and the 2 are both coming from this, that's why I use both of them first. If I used these guys first, I would have to use both the x and the y first. If I use the 2 first, I have to use the negative 1 first. That's why I'm color-coding it. So this is going to be equal to 2 minus negative 3. That's the same thing as 2 plus 3. So that is 5. Negative 1 minus 1.5 is negative 2.5. 5 divided by 2.5 is equal to 2. So the slope of this line is negative 2. Actually I'll take a little aside to show you it doesn't matter what order I do this in. If I use this coordinate first, then I have to use that coordinate first. Let's do it the other way. If I did it as negative 3 minus 2 over 1.5 minus negative 1, this should be minus the 2 over 1.5 minus the negative 1. This should give me the same answer. This is equal to what? Negative 3 minus 2 is negative 5 over 1.5 minus negative 1. That's 1.5 plus 1. That's over 2.5. So once again, this is equal the negative 2. So I just wanted to show you, it doesn't matter which one you pick as the starting or the endpoint, as long as If this is the starting y, this is the starting x. If this is the finishing y, this has to be the finishing x. But anyway, we know that the slope is negative 2. So we know the equation is y is equal to negative 2x plus some y-intercept. Let's use one of these coordinates. I'll use this one since it doesn't have a decimal in it. So we know that y is equal to 2. So y is equal to 2 when x is equal to negative 1. Of course you have your plus b. So 2 is equal to negative 2 times negative 1 is 2 plus b. If you subtract 2 from both sides of this equation, minus 2, minus 2, you're subtracting it from both sides of this equation, you're going to get 0 on the left-hand side is equal to b. So b is 0. So the equation of our line is just y is equal to negative 2x. Actually if you wanted to write it in function notation, it would be that f of x is equal to negative 2x. I kind of just assumed that y is equal to f of x. But this is really the equation. They never mentioned y's here. So you could just write f of x is equal to 2x right here. Each of these coordinates are the coordinates of x and f of x. So you could even view the definition of slope as change in f of x over change in x. These are all equivalent ways of viewing the same thing." + }, + { + "Q": "At 0:19, when Sal said 8/2 is not a perfect square, what's the difference between a perfect square and a non-perfect square?", + "A": "A perfect square is when a whole number is the square root of the number. like \u00e2\u0088\u009a49=7 But \u00e2\u0088\u009a74 is a non-perfect square.", + "video_name": "d9pO2z2qvXU", + "transcript": "Which of the following real numbers are irrational? Well, irrational just means it's not rational. It means that you cannot express it as the ratio of two integers. So let's see what we have here. So we have the square root of 8 over 2. If you take the square root of a number that is not a perfect square, it is going to be irrational. And then if you just take that irrational number and you multiply it, and you divide it by any other numbers, you're still going to get an irrational number. So square root of 8 is irrational. You divide that by 2, it is still irrational. So this is not rational. Or in other words, I'm saying it is irrational. Now, you have pi, 3.14159-- it just keeps going on and on and on forever without ever repeating. So this is irrational, probably the most famous of all of the irrational numbers. 5.0-- well, I can represent 5.0 as 5/1. So 5.0 is rational. It is not irrational. 0.325-- well, this is the same thing as 325/1000. So I can clearly represent it as a ratio of integers. So this is rational. Just as I could represent 5.0 as 5/1, both of these are rational. They are not irrational. Here I have 7.777777, and it just keeps going on and on and on forever. And the way we denote that, you could just say these dots that say that the 7's keep going. Or you could say 7.7. And this line shows that the 7 part, the second 7, just keeps repeating on forever. Now, if you have a repeating decimal-- in other videos, we'll actually convert them into fractions-- but a repeating decimal can be represented as a ratio of two integers. Just as 1/3 is equal to 0.333 on and on and on. Or I could say it like this. I could say 3 repeating. We can also do the same thing for that. I won't do it here, but this is rational. So it's not irrational. 8 and 1/2? Well, that's the same thing. 8 and 1/2 is the same thing as 17/2. So it's clearly rational. So the only two irrational numbers are the first two right over here." + }, + { + "Q": "Why can't the answer also be 14 because when 14 is plugged into the inequality it becomes 3(14) - 6 > 8 which is in fact true.", + "A": "It can be, the inequality states that l is greater than or equal to 14/3. 14 is greater than 14/3, so it is included in the solutions.", + "video_name": "EkBUTZe_SiM", + "transcript": "- [Voiceover] Three L minus six is greater than or equal to eight. Which of the following best describes the solutions to the inequality shown above? So it's three times L minus six is greater than or equal to eight. Well all of these choices, these are in terms of L. They've said L on one side is greater than or equal to, actually all of these choices are greater than or equal to something else. So let's see what we can do to get just an L on the left-hand side. So the first thing we might want to do is let's get rid of this subtracting the six, and the best way we can do that is we can add six. Let's add six to both sides. This six and this six are going to add to zero, and then we are going to be left with, we are going to be left with three L on the left hand side is greater than or equal to eight plus six is 14. Now to just get an L on the left-hand side we can divide both sides by three, and if you divide both sides by three, you're not going to change the sign, you're not going to change the inequality. If you're dividing by a negative number, you would swap the inequality. Greater than or equal to would turn into less than or equal to, but we're dividing by a positive number, so this is going to be L is greater than or equal to 14 over three which is that choice right there." + }, + { + "Q": "At the end shouldn't the numbering be 1,2,4 instead of 1,4,6 and why?", + "A": "I m sure Sal did that sort of naming because this numbering would be favoring the double bonds.", + "video_name": "KWv5PaoHwPA", + "transcript": "Everything we've named so far has been an alkane. We've seen all single bonds. Let's see if we can expand our repertoire a little bit and do some alkenes. So let's look at this first carbon chain right here. And actually, here I drew out all of the hydrogens just to remind you that everything we were doing before with just the lines, it really was representing something like this. When you start having the double bonds, and we'll explain it in more detail later on, it actually starts to matter a little bit more to draw the constituents, because there's actually different ways that you can arrange it. Because these double bonds, you can imagine, they're more rigid, you can't rotate around them as much. But don't think about that too much right now. Let's just try to name these things. So like we always do, let's try to find the longest chain of carbons. And there's only one chain of carbons here. There's one, two, three, four, five, six, seven carbons in that chain. So we're going to be dealing with hept, that is seven carbons. But it's not going to be a heptane. Heptane would mean that we have all single bonds. Here we have a double bond, so this is going to be an alkene. So this tells us right here that we're dealing with an alkene, not an alkane. If you have a double bond, it's an alkene. Triple bond, alkyne. We'll talk about that in future videos. This is hept, and we'll put an ene here, but we haven't specified where the double bond is and we haven't numbered our carbons. When you see an alkene like this, you start numbering closest to the double bond, just like as if it was a alkyl group, as if it was a side chain of carbons. So this side is closest to the double bond, so let's start numbering there. One, two, three, four, five, six, seven. The double bond is between two and three, and to specify its location, you start at the lowest of these numbers. So this double bond is at two. This is actually hept-2-ene. So this tells us that we have a seven carbon chain that has a double bond starting-- the ene tells us a double bond. Let me write that down. So this double bond right there, that's what the ene tells us. Double bond between two carbons, it's an alkene. The double bond starts-- if you start at this point-- the double bond starts at number two carbon, and then it will go to the number three carbon. Now you might be asking, well, what if I had more than one double bond here? So let me draw a quick example of that. Let's say I have something like, one, two, three, four, five, six, seven. So this is the same molecule again. One, two, three, four, five, six, seven. The way we drew it up here, it would look something like this. What if I had another double bond sitting right here? How would we specify this? Well, once again we have seven carbons. One, two, three, four, five, six, seven. So we're still going to have a hept here. It's still going to be an alkene, so we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three, so it would be hept-2. And we also have another double bond starting from four and going to five, so hept-2,4-ene. That's what this molecule right there is. Sometimes, this is the-- I guess-- proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene, they'll write that as 2-heptene, probably because it's easier to say 2-heptene. And from this, you would be able to draw this thing over here, so it's giving you the same amount of information. Similarly over here, they might say 2,4-heptene. But this is the specific, this is the correct way to write it. It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon. So we're dealing with a hept again. We have a double bond starting from the second carbon to the third carbon. So this thing right here, this double bond from the second carbon to the third carbon. So it's hept-2,3-ene-- sorry, not 2,3, 2-ene. You don't write both endpoints. If there was a three, then there would have been another double bond there. It's hept-2-ene. And then we have this methyl group here, which is also sitting on the second carbon. So this methyl group right there on the second carbon. So we would say 2-methyl-hept-2-ene. It's a hept-2-ene, that's all of this part over here, double bonds starting on the two if we're numbering from the right. And then the methyl group is also attached to that second carbon. Let's do one more of these. So we have a cycle here, and once again the root is going to be the largest chain or the largest ring here. Our main ring is the largest one, and we have one, two, three, four, five, six, carbon. So we are dealing with hex as our root for kind of the core of our structure. It's in a cycle, so it's going to be cyclohex. So let me write that. So it's going to be cyclohex. But it has a double bond in it. So it's cyclohex ene, cyclohexene. Let me do this in a different color. So we have this double bond here, and that's why we know it is an ene. Now you're probably saying, Hey Sal, how come we didn't have to number where the ene is? So if you only have one double bond in a ring, it's assumed that one end point of the double bond is your 1-carbon. When you write just cyclohexene, you know-- so cyclohexene would look just like this. Just like that. You don't have to specify where it is. It's just, one of these are going to be the double bond. Now when you have other constituents on it, by definition or I guess the proper naming mechanism, is one of the endpoints of the double bond will be the 1-carbon, and if any of those endpoints have something else on it, that will definitely be the 1-carbon. So these both are kind of the candidates for the 1-carbon, but this point right here also has this methyl group. We will start numbering there, one, and then you want to number in the direction of the other side of the double bond. One, two, three, four, five, six. So we have three methyl groups, one on one. So these are the-- let me circle the methyl groups. That's a methyl group right there. That's a methyl group right there. That's just one carbon. So we have three methyl groups, so this is going to be-- it's at the one, the four, and the six. So it is 1, 4, 6. We have three methyl groups, so it's trimethyl cyclohexene. 1, 4, 6-trimethylcyclohexene. That's what that is, hopefully you found that useful." + }, + { + "Q": "what is the area of a heptagon", + "A": "That depends on the heptagon in question. You need to be more specific.", + "video_name": "qG3HnRccrQU", + "transcript": "We already know that the sum of the interior angles of a triangle add up to 180 degrees. So if the measure of this angle is a, the measure of this angle over here is b, and the measure of this angle is c, we know that a plus b plus c is equal to 180 degrees. But what happens when we have polygons with more than three sides? So let's try the case where we have a four-sided polygon-- a quadrilateral. And I am going to make it irregular just to show that whatever we do here it probably applies to any quadrilateral with four sides. Not just things that have right angles, and parallel lines, Actually, that looks a little bit too close to being parallel. So let me draw it like this. So the way you can think about it with a four sided quadrilateral, is well we already know about this-- the measures of the interior angles of a triangle add up to 180. So maybe we can divide this into two triangles. So from this point right over here, if we draw a line like this, we've divided it into two triangles. And so if the measure this angle is a, measure of this is b, measure of that is c, we know that a plus b plus c is equal to 180 degrees. And then if we call this over here x, this over here y, and that z, those are the measures of those angles. We know that x plus y plus z is equal to 180 degrees. And so if we want the measure of the sum of all of the interior angles, all of the interior angles are going to be b plus z-- that's two of the interior angles of this polygon-- plus this angle, which is just going to be a plus x. a plus x is that whole angle. The whole angle for the quadrilateral. Plus this whole angle, which is going to be c plus y. And we already know a plus b plus c is 180 degrees. And we know that z plus x plus y is equal to 180 degrees. So plus 180 degrees, which is equal to 360 degrees. So I think you see the general idea here. We just have to figure out how many triangles we can divide something into, and then we just multiply by 180 degrees since each of those triangles will have 180 degrees. Let's do one more particular example. And then we'll try to do a general version where we're just trying to figure out how many triangles can we fit into that thing. So let me draw an irregular pentagon. So one, two, three, four, five. So it looks like a little bit of a sideways house there. Once again, we can draw our triangles inside of this pentagon. So that would be one triangle there. That would be another triangle. So I'm able to draw three non-overlapping triangles that perfectly cover this pentagon. This is one triangle, the other triangle, and the other one. And we know each of those will have 180 degrees if we take the sum of their angles. And we also know that the sum of all of those interior angles are equal to the sum of the interior angles of the polygon as a whole. And to see that, clearly, this interior angle is one of the angles of the polygon. This is as well. But when you take the sum of this one and this one, then you're going to get that whole interior angle of the polygon. And when you take the sum of that one and that one, you get that entire one. And then when you take the sum of that one plus that one plus that one, you get that entire interior angle. So if you take the sum of all of the interior angles of all of these triangles, you're actually just finding the sum of all of the interior angles of the polygon. So in this case, you have one, two, three triangles. So three times 180 degrees is equal to what? 300 plus 240 is equal to 540 degrees. Now let's generalize it. And to generalize it, let's realize that just to get our first two triangles, we have to use up four sides. We have to use up all the four sides in this quadrilateral. We had to use up four of the five sides-- right here-- in this pentagon. One, two, and then three, four. So four sides give you two triangles. And it seems like, maybe, every incremental side you have after that, you can get another triangle out of it. Let's experiment with a hexagon. And I'm just going to try to see how many triangles I get out of it. So one, two, three, four, five, six sides. I get one triangle out of these two sides. One, two sides of the actual hexagon. I can get another triangle out of these two sides of the actual hexagon. And it looks like I can get another triangle out of each of the remaining sides. So one out of that one. And then one out of that one, right over there. So in general, it seems like-- let's say. So let's say that I have s sides. s-sided polygon. And I'll just assume-- we already saw the case for four sides, five sides, or six sides. So we can assume that s is greater than 4 sides. Let's say I have an s-sided polygon, and I want to figure out how many non-overlapping triangles will perfectly cover that polygon. How many can I fit inside of it? And then I just have to multiply the number of triangles times 180 degrees to figure out what are the sum of the interior angles of that polygon. So let's figure out the number of triangles as a function of the number of sides. So once again, four of the sides are going to be used to make two triangles. So those two sides right over there. And then we have two sides right over there. I can draw one triangle over-- and I'm not even going to talk about what happens on the rest of the sides of the polygon. You could imagine putting a big black piece of construction paper. There might be other sides here. I'm not going to even worry about them right now. So out of these two sides I can draw one triangle, just like that. Out of these two sides, I can draw another triangle right over there. So four sides used for two triangles. And then, no matter how many sides I have left over-- so I've already used four of the sides, but after that, if I have all sorts of craziness here. I could have all sorts of craziness here. Let me draw it a little bit neater than that. So I could have all sorts of craziness right over here. It looks like every other incremental side I can get another triangle out of it. So that's one triangle out of there, one triangle out of that side, one triangle out of that side, one triangle out of that side, and then one triangle out of this side. So for example, this figure that I've drawn is a very irregular-- one, two, three, four, five, six, seven, eight, nine, 10. Is that right? One, two, three, four, five, six, seven, eight, nine, 10. It is a decagon. And in this decagon, four of the sides were used for two triangles. So I got two triangles out of four of the sides. And out of the other six sides I was able to get a triangle each. These are six. This is one, two, three, four, five. Actually, let me make sure I'm counting the number of sides right. So I have one, two, three, four, five, six, seven, eight, nine, 10. So let me make sure. Did I count-- am I just not seeing something? Oh, I see. I actually didn't-- I have to draw another line right over These are two different sides, and so I have to draw another line right over here. I can get another triangle out of that right over there. And so there you have it. I have these two triangles out of four sides. And out of the other six remaining sides I get a triangle each. So plus six triangles. I got a total of eight triangles. And so we can generally think about it. The first four, sides we're going to get two triangles. So let me write this down. So our number of triangles is going to be equal to 2. And then, I've already used four sides. So the remaining sides I get a triangle each. So the remaining sides are going to be s minus 4. So the number of triangles are going to be 2 plus s minus 4. 2 plus s minus 4 is just s minus 2. So if I have an s-sided polygon, I can get s minus 2 triangles that perfectly cover that polygon and that don't overlap with each other, which tells us that an s-sided polygon, if it has s minus 2 triangles, that the interior angles in it are going to be s minus 2 times 180 degrees. Which is a pretty cool result. So if someone told you that they had a 102-sided polygon-- so s is equal to 102 sides. You can say, OK, the number of interior angles are going to be 102 minus 2. So it's going to be 100 times 180 degrees, which is equal to 180 with two more zeroes behind it. So it'd be 18,000 degrees for the interior angles of a 102-sided polygon." + }, + { + "Q": "What is recursive", + "A": "Repeating, going on in a pattern from the first case (number) on.", + "video_name": "ayRpoJgph0E", + "transcript": "The following table contains the first five terms of the given Pattern A. Generate Pattern B according to this rule. For every term of Pattern A-- so they give us the terms of Pattern A here-- multiply the term by 3 and add 1 to get the corresponding term of Pattern B. Then graph the pairs of corresponding terms. So for every term in Pattern A, we want to multiply by 3 and 1. So if we multiply 0 by 3, we get 0. And you add 1, you get 1. If you multiply 1 by 3, you get 3. And then you add 1, you get 4. 2 times 3 is 6, plus 1 is 7. 3 times 3 is 9, plus 1 is 10. Remember, we're just multiplying by 3 and adding 1. 4 times 3 is 12, plus 1 is 13. So those are the corresponding terms for Pattern B. And then they ask us to graph them. So let's try to graph these points. So when Pattern A is 0, Pattern B is 1. When Pattern A is 0-- so this is Pattern A equaling 0. That's our horizontal axis, the value of Pattern A-- Pattern B is the value of our vertical axis. Pattern B is 1. When Pattern A is 1, Pattern B is 4. So when Pattern A is 1, Pattern B is 4. Pattern B is on the vertical axis. When Pattern A is 2, Pattern B is 7. When Pattern A is 3, Pattern B is 10, so 3 in the horizontal direction. That's our Pattern A value. And our Pattern B value is 10. And then, finally, when Pattern A is 4, Pattern B is 13. Now, let's just look at these patterns. We see Pattern A is increasing by 1 each time, while Pattern B is increasing by it's-- well, Pattern A starts at 0 and increases by 1, while Pattern B starts at 1 and increases by 3, which makes complete sense. It makes sense that it starts at 1, because all of these, you multiply by 3 and add 1. So you start at 1. And then, the fact that we're multiplying by 3, that's what's leading to the distance between these points being 3. So let's check our answer to make sure we got this right, and we did." + }, + { + "Q": "I think \"9:50\" does not need a proof as they're just i j k l unit vectors.", + "A": "9:54 A proof may be simple, but still needed. That is the case here.", + "video_name": "JUgrBkPteTg", + "transcript": "We've seen in several videos that the column space of a matrix is pretty straightforward to find. In this situation the column space of A is just equal to all of the linear combinations of the column vectors of A. Another way of saying all of the linear combinations is just the span of each of these column vectors. So if we call this one right here a1. This is a2, a3, a4. This is a5. Then the column space of A is just equal to the span of a1, a2, a3, a4, and a5. Fair enough. But a more interesting question is whether these guys form a basis for the column space. Or even more interesting, what is the basis for the column space of A? And in this video I'm going to show you a method for determining the basis, and along the way we'll get an intuition for maybe why it works. And if I have time, actually I probably won't have time in In the next video I'll prove to you why it works. So we want to figure out the basis for the column space of A. Remember the basis just means that vectors span, C, A. Clearly these vectors span our column space. I mean the span of these vectors is the column space. But in order to be a basis, the vectors also have to be linearly, let me just write, linearly independent. And we don't know whether these guys or what subset of these guys are linearly independent. So what you do-- and I'm just really going to describe the process here, as opposed to the proof-- is you put this guy in reduced row echelon form. So let's do that. So let me see if we can do that. Let's keep our first row the same. 1, 0. Let me do it actually in the right side right here. So let's keep the first row the same. 1, 0, minus 1, 0, 4. And then let's replace our second row with the second row minus 2 times the first row. So then our second row. 2 minus 2 times 1 is 0. 1 minus 2 times 0 is 1. 0 minus 2 times negative 1, so that's 0 plus 2. 0 minus 2 times 0 is just 0. And then 9 minus 2 times 4 is 1. Fair enough. Now we want to zero out this guy. Well it seems like a pretty straightforward way. Just replace this row with this row plus the first row. So minus 1 plus 1 is 0. 2 plus 0 is 2. 5 minus 1 is 4. 1 plus 0 is 1. Minus 5 plus 4 is minus 1. And then finally we got this guy right here, and in order to zero him out, let's replace him with him minus the first row. So 1 minus 1 is 0. Minus 1 minus 0 is minus 1. Minus 3 minus negative 1, that's minus 3 plus 1, so that's minus 2. Minus 2 minus 0 is minus 2. And then 9 minus 4 is 5. So we did one round. We got our first pivot column going. Now let's do another round of row operations. Well we want to zero all of these guys out. Luckily this is already 0. So we don't have to change our first row or our second row. So we get 1, 0, minus 1, 0, 4. Our second row becomes 0, 1, 2, 0, 1. And now let us see if we can eliminate this guy right here. And let's do it by replacing our blue row, our third row, with the third row minus 2 times the second row. So 0 minus 2 times 0 is 0. 2 minus 2 times 1 is 0. 4 minus 2 times 2 is 0. 1 minus 2 times 0 is 1. Minus 1 minus 2 times 1 is minus 3. All right. Now this last guy we want to eliminate him, and we want turn this into a 0. Let's replace this fourth row with the fourth row plus the second row. So 0 plus 0 is 0. Minus 1 plus minus 1 is 0. Minus 2 plus minus 2 is 0. Minus 2 plus 0 is minus 2. And then 5 plus 1 is 6. We're getting close. So let's look at our pivot entries. We have this is a pivot entry. That's a pivot entry. And this is not a pivot entry, because it's following obviously another. This guy is a pivot entry right here, or will be. Zero this minus 2 out, and I think we'll be done. So let me write my first row just the way it is, because everything above it is 0, so we don't have So my first row I can just write as 1, 0, minus 1, 0, 4. I can write my second row, 0, 1, 2, 0, 1. I can write my third row as 0, 0, 0, 1 minus 3. And now let's replace my fourth row. Let's replace it with it plus 2 times the second row. So 0 plus 2 times 0, 0 plus 2 times 0, 0 plus 2 times 0, minus 2 plus 2 times 1 is just 0. 6 plus 2 times minus 3, that's 6 minus 6, that's just 0. And there we've actually put our matrix in reduced row echelon form. So let me put brackets around it. It's not so bad if you just kind of go and just do the manipulations. And sometimes you kind of get a headache thinking about doing something like this, but this wasn't too bad. So this is let me just say the reduced row echelon form of A. Let me just call that matrix R. So this is matrix R right there. Now what do we see about matrix R? Well it has 3 pivot entries, or 3 pivot columns. Let me square them out, or circle them out. Column 1 is a pivot column, column 2 is a pivot column, and column 3 is a pivot column. And we've done this in previous videos. There's two things that you can see. These three columns are clearly linearly independent. How do we know that? And that's just with respect to each other. If we just took a set of, let's call this r1, r2, and this would be r3, this would be r4 right here. It's clear that the set r1, r2, and r4 is linearly independent. And you say why is that? Well look, our one's got a 1 here, while the other two have a 0 in that entry, right? And this is by definition of pivot entries. Pivot entries have 0's, or pivot columns have 0's everywhere except for where they have a 1. For any pivot column, it will be the only pivot column that has 0's there. Or it'll be the only pivot column that has a 1 there. So there's no way that you can add up combinations of these guys to get a 1. You can say 100 times 0, minus 3, times 0. You're just going to get a bunch of 0's. So no combination of these two guys is going to be equal to that guy. By the same reasoning, no combination of that and that is going to equal this. This is by definition of a pivot entry. When you put it in reduced row echelon form, it's very clear that any pivot column will be the only one to have 1 in that place. So it's very clear that these guys are linearly independent. Now it turns out, and I haven't proven it to you, that the corresponding columns in A-- this is r1, but it's A before we put it in reduced row echelon form-- that these guys right here, so a1, a2, and a4 are also linearly independent. So a1-- let me circle it-- a2, and a4. So if I write it like this, a1, a2, and a4. Let me write it in set notation. These guys are also linearly independant, which I haven't proven. But I think you can kind of get a sense that these row operations really don't change the sense of the matrix. And I'll do a better explanation of this, but I really just wanted you to understand how to develop a basis for the column space. So they're linearly independent. So the next question is do they span our column space? And in order for them to span, obviously all of these 5 vectors, if you have all of them, that's going to span your column space by definition. But if we can show, and I'm not going to show it in this video, but it turns out that you can always represent the non-pivot columns as linear combinations of the pivot columns. And we've kind of touched on that in previous videos where we find a solution for the null space and all that. So these guys can definitely be represented as linear combinations of these guys. I haven't shown you that, but if you take that on faith, then you don't need that column and that column to span. If you did then, or I guess a better way to think it, you don't need them to span, although they are part of the span. Because if you needed this guy, you can just construct him with linear combinations of these guys. So if you wanted to figure out a basis for the column space of A, you literally just take A into reduced row echelon form. You look at the pivot entries in the reduced row echelon form of A, and that's those three. And then you look at the corresponding columns to those pivot columns in your original A. And those form the basis. Because any linear combination of them, or linear combinations of them can be used to construct the non-pivot columns, and they're linearly independant. So I haven't shown you that. But for this case, if you want to know the basis, it's just a1, a2, and a4. And now we can answer another question. So a1, a2, and a4 form a basis for the column space of A, because you can construct the other two guys with linear combinations of our basis vectors, and they're also linearly independent. Now the next question is what is the dimension of the basis? Or what is the dimension-- not the dimension of the basis-- what is the dimension of the column space of A? Well the dimension is just the number of vectors in any basis for the column space. And all bases have the same number of vectors for any given subspace. So we have 1, 2, 3 vectors. So the dimension of our column space is equal to 3. And the dimension of a column space actually has a specific term for it, and that's called the rank. So the rank of A, which is the exact same thing as the dimension of the column space, it is equal to 3. And another way to think about it is, the rank of A is the number of linearly independent column vectors that you have that can span your entire column space. Or the number of linearly independent column vectors that can be used to construct all of the other column vectors. But hopefully this didn't confuse you too much, because the idea is very simple. Take A, put it into reduced row echelon form, see which columns are pivot columns. The corresponding columns are going to be a basis for your column space. If you want to know the rank for your matrix, you can just count them. Or if you don't want to count those, you could literally just count the number of pivot columns you have in your reduced row echelon form. So that's how you do it. In the next video I'll explain why this worked." + }, + { + "Q": "How to represent 13/-5 on a number line ?", + "A": "That would be -2.6 so you would put it at 2.6 to the left of zero. Hope this helped!", + "video_name": "uC09taczvOo", + "transcript": "- [Voiceover] Plot the following numbers on the number line. The first number we have here is five, and so five is five to the right of zero, five is right over there. That's our five. Then we get 1/3. 1/3. So 1/3 is between zero and one. We can actually split this into thirds. So that would be 1/3, 2/3, and then 3/3, which is one, so 1/3 is going to sit right over there. It's 1/3 of the way from zero to one, that's 1/3. Let me write that. That's 1/3 right over there. Then we have negative 1.2. I'll do that in this blue color. Negative 1.2. So, negative one is right over here. This is more negative than negative one. It's negative 1.2. It's negative one, and then another .2, so it's going to be right over here. This is negative 1.2. Zero is pretty straight forward. Zero is right over there. It's even labeled for us at zero. Five was labeled for us too at five. Then we have negative two and 1/4. So let's go to negative two. Negative two is here, and it's going to be more negative than negative two. It's negative two and then another another negative 1/4. So it's negative two, and then we go 1/4 of the way to negative three. So negative 2 and 1/4 is going to be right over here. So negative two and 1/4. And then finally we have 4.1. 4.1. So four is right over here. .1 is another tenth greater than four, another tenth on the way to five. So four and 1/10 is going to be right over here. 4.1. 4.1. And we are done." + }, + { + "Q": "is there any formula to get sum of terms in geometric sequences?", + "A": "Yes, Sum = a(1-r^n)/(1-r)", + "video_name": "Iq7a2vEsT-o", + "transcript": "So I have the function g of x is equal to 9 times 8 to the x minus 1 power. And it's defined for x being a positive-- or if x is a positive-- integer. If x is a positive integer. So we could say the domain of this function, or all the valid inputs here are positive integers. So 1, 2, 3, 4, 5, on and on and on. So this is an explicitly defined function. What I now want to do is to write a recursive definition of this exact same function. That given an x, it'll give the exact same outputs. So let's first just try to understand the inputs and outputs here. So let's make a little table. Let's make a table here. And let's think about what happens when we put in various x's into this function definition. So the domain is positive integers. So let's try a couple of them. 1, 2, 3, 4. And then see what the corresponding g of x is. g of x. So when x is equal to 1, g of x is 9 times 8 to the 1 minus 1 power, 9 times 8 to the 0 power, or 9 times 1. So g of x is going to be just 9. When x is 2, what's going to happen? It'll be 9 times 8 to the 2 minus 1. So that's the same thing as 9 times 8 to the 1st power. And that's just going to be 9 times 8. So that is 72. Actually let me just write it that way. Let me write it as just 9 times 8. 9 times 8. Then when x is equal to 3, what's going on here? Well this is going to be 3 minus 1 is 2. So it's going to be 8 squared. So it's going to be 9 times 8 squared. So we could write that as 9 times 8 times 8. I think you see a little bit of a pattern forming. When x is 4, this is going to be 8 to the 4 minus 1 power, or 9 to the 3rd power. So that's 9 times 8 times 8 times 8. So this gives a good clue about how we would define this recursively. Notice, if our first term, when x equals 1 is 9, every term after that is 8 times the preceding term. Is 8 times the preceding term. 8 times the preceding term. 8 times the preceding term. So let's define that as a recursive function. So first define our base case. So we could say g of x-- and I'll do this is a new color because I'm overusing the red. I like the blue. g of x. Well we can define our base case. It's going to be equal to 9 if x is equal to 1. g of x equals 9 if x equals 1. So that took care of that right over there. And then if it equals anything else it equals the previous g of x. So if we're looking at-- let's go all the way down to x minus 1, and then an x. So if this entry right over here is g of x minus 1, however many times you multiply the 8s and we have a 9 in front, so this is g of x minus 1. We know that g of x-- we know that this one right over here is going to be the previous entry, g of x minus 1. The previous entry times 8. So we could write that right here. Times 8. So for any other x other than 1, g of x is equal to the previous entry-- so it's g of-- I'll do that in a blue color-- g of x minus 1 times 8. If x is greater than 1, or x is integer greater than 1. Now let's verify that this actually works. So let's draw another table here. So once again, we're going to have x and we're going to have g of x. But this time we're going to use this recursive definition And the reason why it's recursive is it's referring to itself. In its own definition, it's saying hey, g of x, well if x doesn't equal 1 it's going to be g of x minus 1. It's using the function itself. But we'll see that it actually does work out. So let's see... When x is equal to 1, so g of 1-- well if x equals 1, it's equal to 9. It's equal to 9. So that was pretty straightforward. What happens when x equals 2? Well when x equals 2, this case doesn't apply anymore. We go down to this case. So when x is equal to 2 it's going to be equivalent to g of 2 minus 1. Let me write this down. It's going to be equivalent to g of 2 minus 1 times 8, which is the same thing as g of 1 times 8. And what's g of 1? Well g of 1 is right over here. g of 1 is 9. So this is going to be equal to 9 times 8. Exactly what we got over here. And of course this was equivalent to g of 2. So let me write this. This is g of 2. Let me scroll over a little bit so I don't get all scrunched up. So now let's go to 3. Let's go to 3. And right now I'll write g of 3 first. So g of 3 is equal to-- we're going to this case-- it's equal to g of 3 minus 1 times 8. So that's equal to g of 2 times 8. Well what's g of 2? Well g of 2, we already figured out is 9 times 8. So it's equal to 9 times 8-- that's g of 2-- times 8 again. And so you see we get the exact same results. So this is the recursive definition of this function." + }, + { + "Q": "Is atomic mass essentially atomic weight then?", + "A": "Atomic mass is the mass of a specific isotope Relative atomic mass ( atomic weight ) is the average mass of an atom of an element taking into account the percentage of each isotope there is on earth", + "video_name": "NG-rrorZcM8", + "transcript": "Let's have a little bit of a primer on weight and mass, especially if we start talking about atomic weight and atomic mass. If we're sitting in a physics class, weight and mass mean something very, very ... well, they mean different things. It might be a discovery, or a new learning, for some of you, because in everyday life, when we say something's mass, we think, \"Well, the more mass it has, the more weight it has.\" Or, if we think something has more or less weight, we think, \"Okay, that relates to its mass.\" But in physics class, we see that these actually represent two different ideas, albeit related ideas. Mass is a notion of how much of something there is, or you could say, how hard is it to accelerate or decelerate it. Or you could view it as a measure of an object's inertia. We typically, it, kind of a human scale, might measure mass in terms of grams or kilograms. What's confusing is, if you go to Europe, and you ask someone their weight, they'll often give you their weight in terms of kilogram, even though that is a unit of mass. Now weight, on the other hand, is not ... it's different than mass. Weight is a force, it's how much the Earth, or whatever planet you happen to be on, is pulling on you. This right over here is a force. And, in the metric system, you measure weight, not with grams or kilograms, but with Newtons. Newtons. Really, when you ask someone their weight in Europe, they should give it to you in Newtons. If you ask them their mass, what they're telling you is actually their mass. They should say, \"My mass is 60 kilograms,\" or, \"70 kilograms,\" or whatever they might be. It's a very important difference in physics. If I go from Earth to the Moon, my mass does not change, but my weight does change because the force with which the Moon is pulling on me, or that we're pulling on each other, is less than it would be on Earth. In fact, even on the surface of the Earth, if you were to even go to the top of a building, you're just so ... Yeah, it would be very hard to measure it, but you're just slightly further from the center of the Earth, so there's a different gravitational force. Your weight will change ever so slightly, but your mass does not change. You go to deep space, and there's very little gravitational influence, you have pretty much, or close to, zero weight. But you're in deep space, and if there's no planets nearby, but your mass is still going to be whatever your mass happens to be. That's a primer on mass and weight. Now, with that out of the way, I might confuse you because, as we go into a chemistry context, it starts getting a little bit more muddled again. Let me go to chemistry, chemistry. And in any science, if people just talk generally about mass or weight, this is what they're talking about. They're talking about a measure of inertia for mass, and they're talking about a force when they're talking about weight. But in chemistry, we start thinking about things on an atomic scale. You'll hear ... You'll hear the term \"Atomic,\" \"Atomic mass.\" Atomic mass is, literally, a measure of mass. It is measured in atomic mass units. Atomic mass units, which is, and we'll talk in the future videos, a very, very, very, very, small fraction. One atomic mass unit is a very, very, very, very, very, very small fraction of a gram. It is actually defined using the most common isotope of carbon. It's defined using carbon-12. The current definition is carbon-12. Carbon-12 has a mass, has a mass, has a mass of exactly, exactly, exactly 12 atomic mass units. So they can then, you or chemists, use that as the benchmark to figure out what the atomic mass, or the mass of any other atom. And you might say, \"Oh, why didn't they just do a hydrogen, \"and just say that's one atomic mass unit, and all that,\" and actually, they had started there. They had been there at an earlier stage, but for a whole set of reasons, carbon-12 is kind of being the benchmark, as having 12 atomic mass units, is what people went with. Now, what is atomic weight, then? Atomic weight. Let me write this in a different color. I'll do it in blue. Atomic weight. So if you draw the same analogy that we did up here, you might say, \"Okay, this must be a ... \"This must be a force. \"It should maybe, you know, \"an atomic weight unit would be a small fraction of, \"very small fraction of a unit.\" But it turns out in chemistry, when we talk about atomic weight, we're still measuring in atomic mass units. This is still a mass. Atomic mass units. But it's not the mass of just one atom or just one molecule. It's a weighted average across many, many ... of how typically, what you would see, or the makeup of what you would see on Earth. What do I mean by that? Well, on Earth, there are two ... The primary isotope of carbon is carbon-12. Carbon-12, which is defined as having a mass of exactly 12 atomic mass units. But there's also some carbon-14. Carbon-14. What do these numbers mean, just as a reminder? Well, carbon-12 has six protons, and the six protons are what make it carbon. Carbon-14 is also going to have six protons. But carbon-12, carbon-12 also has six neutrons. Six neutrons. While carbon-14 has eight neutrons. I know what you're already thinking. You're, like, \"Well, wait. \"Why don't we say that a proton or a neutron \"weighs one atomic mass unit? \"Because it looks like this is 12, \"and I'm guessing that this, \"that this, the mass of this is going to be \"pretty close to 14.\" If you're thinking that way, that's not an unreasonable way to think. In fact, when I'm kind of just working through chemistry, that is how I think about it. But they don't weigh exactly one atomic mass unit by this definition. Remember, the electron is ever so small, it has very small mass, but it is contributing, or the electrons are contributing, something to the mass. So, a proton or a neutron have very, very, very close ... They are close to one atomic mass unit. Let me write this down. One proton, one proton, or one neutron, one neutron, very close to one atomic mass unit, but not exactly. But anyway, going back to what atomic weight is, right over here, the most common isotope of carbon ... Remember, when we're saying \"isotopes,\" we're saying the same element, we have the same number of protons, but we have different number of neutrons. The most common isotope on Earth is carbon-12, but there's also some carbon-14. If you were to take a weighted average, as found on the Earth, of all the carbon-12 and all of the carbon-14, the weighted average of the atomic masses is the atomic weight. And the atomic weight of carbon ... And you'll see this on a periodic table. In fact, I have one right over here. Notice, the six protons, this is what defines it to be carbon. But then they write 12.011, which is the weighted average of the masses of all of the carbons. Now, it's very close to 12, as opposed to being closer to 14, because most of the carbon on Earth is carbon-12. We could write this down. This is the atomic weight. This is the atomic weight of carbon on Earth. This is 12.011. Typically, if people are telling you, \"Hey, I'm talking about the isotope. \"I am talking about the isotope carbon-12,\" you say, \"Okay, if you're talking about a particular atom \"of carbon-12,\" you would say, \"Okay, that's going to be 12 atomic mass units.\" But then if you said, \"Hey, you know, \"I'm randomly ... \"If I have a big bag of carbon, you could say, \"on average, the weighted average of those carbon atoms, \"they're going to have an atomic mass, \"or a weighted average atomic mass,\" which is atomic weight, \"of 12.011 atomic mass units.\" AMU. Hopefully, this clarified it more than it confused. All right, see you in the next video." + }, + { + "Q": "for the first right hand rule> i know your fingers first point in the direction and velocity and curl towards the magnetic field. But which way do you know the magnetic field is flowing? do you curl your fingers towards the south of the magnetic field.", + "A": "like electric field, magnetic field also has a particular direction.by convention its taken to be originating frm north and end on south. but in the above problems since the the source generating the magnetic field is not specified, its would be better to consider the direction only using the vectors drawn", + "video_name": "b1QFKLZC11U", + "transcript": "In the last video we figured out that if we had a proton coming into the right at a velocity of 6 times 10 to the seventh meters per second. So the magnitude of the velocity is 1/5 the speed of light. And if it were to cross this magnetic field, we used this formula to figure out that the magnitude of the force on this proton would be 4.8 times 10 to the negative 12 newtons. And then the direction, we used our right hand rule because it's a cross product. And we figured out that it would be perpendicular-- well, it has to be perpendicular to both, because we're taking the cross product-- and right when it enters, the net force will be downwards. But then think about what happens. If you have a downward force right there, then the particle will be deflected downward a little bit, so its velocity vector will then look something like that. But it's still in the magnetic field, right? And not only is it still in the magnetic field, but since the particle is still moving within the plane of your video screen, it's still completely perpendicular to the magnetic field. And so the magnitude of the force on the moving particle won't change, just the direction will. Because if we do the right hand rule here, but if we just move our hand down a little bit, if we tilt it down, then our thumb's going to be pointing in this direction. And that just keeps happening. It gets deflected that way a little bit. So the magnitude of the velocity doesn't ever change. It always stays perpendicular to the magnetic field because it's always staying in this plane. But the orientation does change within the plane. And because of that, because the orientation of the velocity changes, the orientation of the force changes. So when the velocity is here, the force is perpendicular. So it acts as kind of a centripetal force, and so the particle will start moving in a circle. So let's see if we can break out our toolkit from what we've learned before in classical mechanics, and figure out what the radius of that circle is. And that might seem more daunting than it really is. Well, what do we know about centripetal forces and radiuses of circles, et cetera? So, what is the formula for centripetal force? And we proved it many, many videos ago, early in the physics playlist. Well, centripetal acceleration is the magnitude of the velocity vector squared over the radius of the circle. And since this is acceleration, if we want to know the centripetal force, it's just the mass times acceleration. So it's the mass of the particle, or the object in question, times the magnitude of its velocity squared divided by the radius of the circle. In this case, this is the radius of the circle. And that's what we're going to try to solve for. And what do we know about the centripetal force? What is causing the centripetal force? Well, it's the magnetic field and we've figured that out. This is going to be equal to this, which we figured out is going to be equal to-- at least the magnitudes-- the magnitude of this is equal to the magnitude of this. And that magnitude is 4.8 times 10 to the minus 12 newtons. And so the radius is going to be-- let's see, if we flip both sides of this equation, we get radius over mass velocity squared is equal to 1 over 4.8 times 10 to the minus 12. I could just figure out what that number is, but I won't Then we can multiply both sides times this mv squared. And we get that the radius of the circle is going to be equal to the mass of the proton times the magnitude of its velocity squared divided by the force from the magnetic field. The centripetal force. 4.8 times 10 to the minus 12 newtons. And the radius should be in meters, since everything is kind of in the standard SI units. And let's see if we can figure this out. Get our calculator. And this is where that constant function is useful again, because what is the mass of a proton? Well, that's something that I personally don't have memorized. But if we go into the built-in constants on the TI-85-- let's see more. Mass of a proton. This is mass of an electron. This is mass of a proton. So mass of a proton-- that's what we care about-- times the magnitude of the velocity squared. What was the velocity? It was 6 times 10 to the seventh meters per second. So times 6 times 10 to the seventh meters per second squared. And all of that divided by the magnitude of the centripetal force. Which is the force that's being generated by the That's 4.8 times 10 to the negative 12. Divided by 4.8 E minus 12. Hopefully we don't get something funky. There we go. That's actually a pretty neat number. 1.25 meters. That's actually a number that we can imagine. So if you have a proton going in this direction at 1/5 the speed of light through a-- what'd I say it was? It was a 0.5 tesla magnetic field, where the vectors are pointing out of the video. We have just shown that this proton will go in a circle of radius 1.25 meters. Which is neat because it's a number that I can actually visualize. And so this whole business of magnetic fields making charged particles go into circles, this is one of the few times that I can actually say has a direct application into things that you've seen. Namely, your TV. Or at least the old-school TVs. The non-plasma or LCD TVs, your cathode ray TVs, take advantage of this. Where you essentially have a beam of not protons but electrons. And a magnet-- if you take apart a TV, which I don't think you should do, because you're more likely to hurt yourself because there's a vacuum in there that can implode, and all that-- but essentially, you have a magnet that deflects this electron beam and does it really fast so it scans your entire screen of different intensities, and that's what forms the image. I won't go into that detail. Maybe one day I'll do a whole video on how TVs work. So that's one application of a magnetic field causing a beam of charged particles to curve. And then the other application, and this is actually one where it's actually useful to make the particle go in a circle, is these cyclotrons that you read about, where they take these protons and they make them go in circles really, really fast, and then they smash them together. Well, have you ever wondered, how do they even make a proton go in a circle? It's not like you could hold it and guide it around in a circle. They pass it through an appropriate strength magnetic field, and it curves the path of the proton so that it can keep going through the same field over and over again. And then they can actually use those electric fields. I don't claim to have any expertise in this, but then they can keep speeding it up using the same devices, because it keeps passing through the same part of the collider. And then once it collides, you've probably seen those pictures. You know, that you spend billions of dollars on supercolliders, and you end up with these pictures. And somehow these physicists are able to take these pictures and say, oh, this is some new particle because of the way it moved. Well, what they're actually talking about is these are moving at relativistic speeds. And since they're at relativistic speeds as they move at different velocities, their mass is changing, and all that. But the basic idea is what we just learned. They move in circles. They move in circles because they're going through a magnetic field. But their radiuses are different because their charges and their velocities are going to be different. And actually some will move to the left and some will move to the right. And that might be because they're positive or negative, and then the radius will be dependent on their masses. Anyway, I don't want to confuse you. But I just wanted to show you that we actually are touching on some physics that a physicist would actually care about. Now with that said, what would have happened if this wasn't a proton but if this was an electron moving at this velocity at 6 times 10 to the seventh meters per second through a 0.5 tesla magnetic field popping out of this video. What would have happened? Well, this formula would have still been safe. The magnitude of the force is the charge-- but it wouldn't have been the charge of a proton, it would have been the charge of an electron, times 6 times 10 to the seventh meters per second times 0.5 teslas. So what's the difference between the charge of a proton and the charge of an electron? Well, the charge of an electron is negative. So if this was an electron, then the net force would actually end up being a negative number. So what does that mean? Well, when we used the right hand rule with the proton example, we said that the-- at least when the proton is moving in this direction-- that the net force would be downwards. But now, all of a sudden, if we reverse the charge, if we say we have a negative charge-- the same magnitude but it's negative, because it's an electron-- what happens? The force is now in this direction, using the right hand rule, but it is a negative. So really it's going to be a positive force of the same magnitude in this direction. So if we have a proton, it'll go in a circle in this direction. It'll go like this. But if we have an electron, it'll go in a circle of the other direction. Now let me ask a question. Is that circle going to be a tighter circle or a wider circle? Well, the mass of an electron is a lot smaller than the mass of a proton. And we had the radius is equal to the mass times the velocity squared divided by the centripetal force. So this mass is smaller and the radius is going to be smaller. So the electron's path would actually move up and it would be a smaller radius. Actually proportional to the difference in their radiuses is the difference in their masses, actually. But that would be the path of the electron. Anyway, I thought you'd be interested in that, as well. I have run out of time. I will see you in the next video." + }, + { + "Q": "What is the electron configuration of Cd2+ and how?", + "A": "The electron configuration for Cd is: [Kr] 4d^10 5s^2 In Cd^2+ the highest energy electrons are lost which are the two 5s electrons, leaving: [Kr] 4d^10", + "video_name": "YURReI6OJsg", + "transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. Then we fill out 2p6. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. We're going to go and backfill the third shell. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that they have a higher probability of being further from the nucleus than these right here. Now, another way to figure out the electron configuration for nickel-- and this is covered in some chemistry classes, although I like the way we just did it because you look at the periodic table and you gain a familiarity with it, which is important, because then you'll start having an intuition for how different elements react with each other -- is to just say, oK, nickel has 28 electrons, if it's neutral. It has 28 electrons, because that's the same number of protons, which is the atomic number. Remember, 28 just tells you how many protons there are. This is the number of protons. We're assuming it's neutral. So it has the same number of electrons. That's not always going to be the case. But when you do these electron configurations, that tends to be the case. So if we say nickel has 28, has an atomic number of 28, so it's electron configuration we can do it this way, too. We can write the energy shells. So one, two, three, four. And then on the top we write s, p, d. Well we're not going to get to f. But you could write f and g and h and keep going. What's going to happen is you're going to fill this one first, then you're going to fill this one, then that one, then this one, then this one. Let me actually draw it. So what you do is, these are the shells that exist, period. These are the shells that exist, in green. What I'm drawing now isn't the order that you fill them. This is just, they exist. So there is a 3d subshell. There's not a 3f subshell. There is a 4f subshell. Let me draw a line here, just so it becomes a little bit neater. And the way you fill them is you make these diagonals. So first you fill this s shell like that, then you fill this one like that. Then you do this diagonal down like that. Then you do this diagonal down like that. And then this diagonal down like that. And you just have to know that there's only two can fit in s, six in p, in this case, 10 in d. And we can worry about f in the future, but if you look at the f-block on a periodic table, you know how many there are in f. So you fill it like that. So first you just say, OK. For nickel, 28 electrons. So first I fill this one out. So that's 1s2. 1s2. Then I go, there's no 1p, so then I go to 2s2. Let me do this in a different color. So then I go right here, 2s2. That's that right there. Then I go up to this diagonal, and I come back down. And then there's 2p6. And you have to keep track of how many electrons you're dealing with, in this case. So we're up to 10 now. So we used that one up. Then the arrow tells us to go down here, so now we do the third energy shell. So 3s2. And then where do we go next? 3s2. Then we follow the arrow. We start there, there's nothing there, there's something here. So we go to 3p6. And then the next thing we fill out is 4s2. So then we go to 4s2. And then what's the very next thing we fill out? We have to go back to the top. We come here and then we fill out 3d. And then how many electrons do we have left to fill out? So we're going to be in 3d. So 3d. And how many have we used so far? 2 plus 2 is 4. 4 plus 6 is 10. 10 plus 2 is 12. 18. 20. We've used 20, so we have 8 more electrons to configure. And the 3d subshell can fit the 8 we need, so we have 3d8. And there you go, you've got the exact same answer that we had when we used the first method. Now I like the first method because you're looking at the periodic table the whole time, so you kind of understand an intuition of where all the elements are. And you also don't have to keep remembering, OK, how many have I used up as I filled the shells? Right? Here you have to say, i used two, then I used two more. And you have to draw this kind of elaborate diagram. Here you can just use the periodic table. And the important thing is you can work backwards. Here there's no way of just eyeballing this and saying, OK, our most energetic electrons are going to be and our highest energy shell is going to be 4s2. There's no way you could get that out of this without going through this fairly involved process. But when do you use this method, you can immediately say, OK, if I'm worried about element Zr, right here. If I'm worried about element Zr. I could go through the whole exercise of filling out the entire electron configuration. But usually the highest shell, or the highest energy electrons, are the ones that matter the most. So you immediately say, OK, I'm filling in 2 d there, but remember, d, you go one period below. So this is 4d2. Right? Because the period is five. So you say, 4d2. 4d2. And then, before that, you filled out the 5s2 electrons. The 5s2 electrons. And then you could keep going backwards. And you filled out the 4p6. 4p6. And then, before you filled out the 4p6. then you had 10 in the d here. But what is that? It's in the fourth period, but d you subtract one from it, so this is 3d10. So 3d10. And then you had 4s2. This is getting messy. Let me just write that. So you have 4d2. That's those two there. Then you have 5s2. 5s2. Then we had 4p6. That's over here. Then we had 3d10. Remember, 4 minus 1, so 3d10. And then you had 4s2. And you just keep going backwards like that. But what's nice about going backwards is you immediately know, OK, what electrons are in my highest energy shell? Well I have this five as the highest energy shell I'm at. And these two that I filled right there, those are actually the electrons in the highest energy shell. They're not the highest energy electrons. These are. But these are kind of the ones that have the highest probability of being furthest away from the nucleus. So these are the ones that are going to react. And these are the ones that matter for most chemistry purposes. And just a little touchpoint here, and this isn't covered a lot, but we like to think that electrons are filling these buckets, and they stay in these buckets. But once you fill up an atom with electrons, they're not just staying in this nice, well-behaved way. They're all jumping between orbitals, and doing all sorts of crazy, unpredictable things. But this method is what allows us to at least get a sense of what's happening in the electron. For most purposes, they do tend to react or behave in ways that these orbitals kind of stay to themselves. But anyway, the main point of here is really just to teach you how to do electron configurations, because that's really useful for later on knowing how things will interact. And what's especially useful is to know what electrons are in the outermost shell, or what are the valence electrons." + }, + { + "Q": "why in sigma complex carbon contains +charge\n+charge appear on it at that time when it loses its one electron.Does it lose its electron?\nI am confused over here can you please explain it for me ?", + "A": "The two electrons in the \u00cf\u0080 bond are used to attack the electrophile. The carbon that is bonded to the electrophile doesn t lose any electrons, but the carbon ortho to it does lose an electron, and that gives it a positive charge.", + "video_name": "eQzbpL0uWVA", + "transcript": "Let's look at the general reaction for electrophilic aromatic substitution. So we start with the benzene ring, and we react benzene with a molecule that contains an electrophile in there. And what happens in electrophilic aromatic substitution. We're going to substitute the electrophile for a proton on our benzene ring. And so over here, we can see the electrophile is now in place of that proton. So that's where that that's where the electrophilic part comes in this. And that's where the substitution parts comes in. You're substituting an electrophile for a proton. The aromatic comes in because you are going to reform an aromatic ring in your mechanism. Electrophilic aromatic substitution requires a catalyst. And the point of a catalyst is to generate your electrophile. So down here, you can see that the catalyst is going to react to produce the positively charged electrophile. So remember, electrophile means loving electron. So if something is positively charged, it's going to love electrons. We also formed this catalyst complex over here, which is going to factor into our mechanism. So now that we formed our electrophile, let's look in more detail as to what happens in electrophilic, aromatic substitution. So we start with our benzene ring. And I'm showing one of the hydrogens on the benzene ring. It could be any of the six, since they are all equivalent. And now we formed our electrophile from our catalyst. So the pi electrons in the benzene ring can be attracted to the positively charged electrophile. Because negative charges are attracted to positive charges. And so pi electrons in your benzene ring are going to function as a nucleophile, and those electrons are going to attack the electrophile. So this is a nucleophile, electrophile attack, where those pi electrons are going to bond to that electrophile there. So those pi electrons are going to form a covalent bond with your electrophile. So let's go ahead and show that. So these pi electrons didn't do anything. The hydrogen stays there. Now, I could show the electrophile adding to either of the two carbons on the side of the double bonds. So it could be that carbon. Or it could be this carbon. Since I've drawn this hydrogen up here at the top, I'm going to go ahead and say that the electrophile adds to the top carbon there. So there's my electrophile there. Let me go ahead and highlight the electrons that are forming that covalent bond. So these pi electrons here are the ones that are functioning as a nucleophile. And those pi electrons are going to form this bond right here. Now in forming that bond, we're taking a bond away from this bottom carbon here. And so that bottom carbon is going to be left with a positive one formal charge. Therefore, we can draw a resonance structure for this cation. So let's go ahead and show a possible resonance structure here. So these pi electrons could move over to here. And let's go ahead and draw what would result if that happened. So now, we have these pi electrons up here. We have our hydrogen. We have our electrophile. And the electrons moved over to this position. Let me go ahead and highlight those in magenta. So I'm saying that these pi electrons right here moved over to here. And when those electrons moved over to there, we're taking a bond away from this carbon this time. So that is the carbon that's going to get a plus 1 formal charge like that. So we can draw another resonance structure. So let's go ahead and do that. So we could take these pi electrons and move them into here. So let's go ahead and show what that would look like. So if those pi electrons moved into there, we would now have, again, our hydrogen, our electrophile, these pi electrons, and then these pi electrons right here. So once again, let me go ahead and highlight those. This time I'll use blue. These pi electrons are going to move over to here. And once again, we're taking a bond away from a carbon. This time, it's this top carbon up here. So that's the carbon that's going to get the plus one formal charge like that. So these are all resonance structures. And remember, the actual cation would be a hybrid of these resonance structures. And we call we call that hybrid a sigma complex. So you have a positive one formal charge de-localized over three carbons in your sigma complex. So the next step in the mechanism-- I'm just going to redraw the first resonance structure that we did here. So I'm going to go and redraw that down here. So let's go ahead and show the first resonance structure. So in our first resonance structure, we had our hydrogen here, our electrophile already bonded to our ring. And we had a positive one formal charge on this carbon right here. Well remember, the catalyst had formed a complex. And I represent it like this. So something bonded to your catalyst like that. So let's just go up here and refresh our memory. So right up here, when we generated our electrophile, we also generated this catalyst complex up here. So y bonded to a catalyst, so I have y bonded to a catalyst down here. And you could think about this as functioning as a base. Or it's going to accept a proton. So I could show these electrons in here taking this proton. And if it takes that proton, that leaves these electrons behind. And those electrons are going to move in here to reform your benzene ring and take away that positive one formal charge. So let's go ahead and show that. So we now have our benzene ring back. And our electrophile is now bonded to our ring. And the proton has left. So the electrophile has completely substituted for that proton. Let's follow those electrons again. So the electrons in magenta in here, so those are the ones that are going to move in here to reform your aromatic ring. So deproteination of the sigma complex restores the aromatic ring. And so we have a stable product here. So the other product you could think about this y here is now going to be bonded to that proton. So you could have the y here bonded to that proton. And you could highlight those electrons. You could say that these electrons right here are now these electrons. And, taking those electrons away from the catalyst would of course regenerate your catalyst. And so it's free to then catalyze another reaction. And so this is the general mechanism for electrophilic aromatic substitution, which the reactions that we're going to see are pretty much going to follow this general mechanism." + }, + { + "Q": "Don't you learn about parabolas in algebra 1? Or does it vary depending on the curriculum?", + "A": "I m pretty sure that it varries upon the curriculum and how fast your class moves.", + "video_name": "0A7RR0oy2ho", + "transcript": "Let's see if we can learn a thing or two about conic sections. So first of all, what are they and why are they called conic sections? Actually, you probably recognize a few of them already, and I'll write them out. They're the circle, the ellipse, the parabola, and the hyperbola. Hyperbola. And you know what these are already. When I first learned conic sections, I was like, oh, I know what a circle is. I know what a parabola is. And I even know a little bit about ellipses and hyperbolas. Why on earth are they called conic sections? So to put things simply because they're the intersection of a plane and a cone. And I draw you that in a second. But just before I do that it probably makes sense to just draw them by themselves. And I'll switch colors. Circle, we all know what that is. Actually let me see if I can pick a thicker line for my circles. so a circle looks something like that. It's all the points that are equidistant from some center, and that distance that they all are that's the radius. So if this is r, and this is the center, the circle is all the points that are exactly r away from this center. We learned that early in our education what a circle is; it makes the world go round, literally. Ellipse in layman's terms is kind of a squished circle. It could look something like this. Let me do an ellipse in another color. So an ellipse could be like that. Could be like that. It's harder to draw using the tool I'm drawing, but it could also be tilted and rotated around. But this is a general sense. And actually, circles are a special case of an ellipse. It's an ellipse where it's not stretched in one dimension It's kind of perfectly symmetric in every way. Parabola. You've learned that if you've taken algebra two and you probably have if you care about conic sections. But a parabola-- let me draw a line here to separate things. A parabola looks something like this, kind of a U shape and you know, the classic parabola. I won't go into the equations right now. Well, I will because you're probably familiar with it. y is equal to x squared. And then, you could shift it around and then you can even have a parabola that goes like this. That would be x is equal to y squared. You could rotate these things around, but I think you know the general shape of a parabola. We'll talk more about how do you graph it or how do you know what the interesting points on a parabola actually are. And then the last one, you might have seen this before, is a hyperbola. It almost looks like two parabolas, but not quite, because the curves look a little less U-ish and a little more open. But I'll explain what I mean by that. So a hyperbola usually looks something like this. So if these are the axes, then if I were to draw-- let me draw some asymptotes. I want to go right through the-- that's pretty good. These are asymptotes. Those aren't the actual hyperbola. But a hyperbola would look something like this. They get to be right here and they get really close to the asymptote. They get closer and closer to those blue lines like that and it happened on this side too. The graphs show up here and then they pop over and they show up there. This magenta could be one hyperbola; I haven't done true justice to it. Or another hyperbola could be on, you could kind of call it a vertical hyperbola. That's not the exact word, but it would look something like that where it's below the asymptote here. It's above the asymptote there. So this blue one would be one hyperbola and then the magenta one would be a different hyperbola. So those are the different graphs. So the one thing that I'm sure you're asking is why are they called conic sections? Why are they not called bolas or variations of circles or whatever? And in fact, wasn't even the relationship. It's pretty clear that circles and ellipses are somehow related. That an ellipse is just a squished circle. And maybe it even seems that parabolas and hyperbolas are somewhat related. This is a P once again. They both have bola in their name and they both kind of look like open U's. Although a hyperbola has two of these going and kind of opening in different directions, but they look related. But what is the connection behind all these? And that's frankly where the word conic comes from. So let me see if I can draw a three-dimensional cone. So this is a cone. That's the top. I could've used an ellipse for the top. Looks like that. Actually, it has no top. It would actually keep going on forever in that direction. I'm just kind of slicing it so you see that it's a cone. This could be the bottom part of it. So let's take different intersections of a plane with this cone and see if we can at least generate the different shapes that we talked about just now. So if we have a plane that goes directly-- I guess if you call this the axis of this three-dimensional cone, so this is the axis. So if we have a plane that's exactly perpendicular to that axis-- let's see if I can draw it in three dimensions. The plane would look something like this. So it would have a line. This is the front line that's closer to you and then they would have another line back here. That's close enough. And of course, you know these are infinite planes, so it goes off in every direction. If this plane is directly perpendicular to the axis of these and this is where the plane goes behind it. The intersection of this plane and this cone is going to look like this. We're looking at it from an angle, but if you were looking straight down, if you were listening here and you look at this plane-- if you were looking at it right above. If I were to just flip this over like this, so we're looking straight down on this plane, that intersection would be a circle. Now, if we take the plane and we tilt it down a little bit, so if instead of that we have a situation like this. Let me see if I can do it justice. We have a situation where it's-- whoops. Let me undo that. Undo. Where it's like this and has another side like this, and I connect them. So that's the plane. Now the intersection of this plane, which is now not orthogonal or it's not perpendicular to the axis of this three-dimensional cone. If you take the intersection of that plane and that cone-- and in future videos, and you don't do this in your But eventually we'll kind of do the three-dimensional intersection and prove that this is definitely the case. You definitely do get the equations, which I'll show you in the not too far future. This intersection would look something like this. I think you can visualize it right now. It would look something like this. And if you were to look straight down on this plane, if you were to look right above the plane, this would look something-- this figure I just drew in purple-- would look something like this. Well, I didn't draw it that well. It'd be an ellipse. You know what an ellipse looks like. And if I tilted it the other way, the ellipse would squeeze the other way. But that just gives you a general sense of why both of these are conic sections. Now something very interesting. If we keep tilting this plane, so if we tilt the plane so it's-- so let's say we're pivoting around that point. So now my plane-- let me see if I can do this. It's a good exercise in three-dimensional drawing. Let's say it looks something like this. I want to go through that point. So this is my three-dimensional plane. I'm drawing it in such a way that it only intersects this bottom cone and the surface of the plane is parallel to the side of this top cone. In this case the intersection of the plane and the cone is going to intersect right at that point. You can almost view that I'm pivoting around this point, at the intersection of this point and the plane and the cone. Well this now, the intersection, would look something like this. It would look like that. And it would keep going down. So if I were to draw it, it would look like this. If I was right above the plane, if I were to just draw the plane. And there you get your parabola. If you keep kind of tilting-- if you start with a circle, tilt a little bit, you get an ellipse. You get kind of a more and more skewed ellipse. And at some point, the ellipse keeps getting more and more skewed like that. It kind of pops right when you become exactly parallel to the side of this top cone. And I'm doing it all very inexact right now, but I think I want to give you the intuition. It pops and it turns into a parabola. So you can kind of view a parabola-- there is this relationship. Parabola is what happens when one side of an ellipse pops open and you get this parabola. And then, if you keep tilting this plane, and I'll do it another color-- so it intersects both sides of the cone. Let me see if I can draw that. So if this is my new plane-- whoops. That's good enough. So if my plane looks like this-- I know it's very hard to read now-- and you wanted the intersection of this plane, this green plane and the cone-- I should probably redraw it all, but hopefully you're not getting overwhelmingly confused-- the intersection would look like this. It would intersect the bottom cone there and it would intersect the top cone over there. And then you would have something like this. This would be intersection of the plane and the bottom cone. And then up here would be the intersection of the plane and the top one. Remember, this plane goes off in every direction infinitely. So that's just a general sense of what the conic sections are and why frankly they're called conic sections. And let me know if this got confusing because maybe I'll do another video while I redraw it a little bit cleaner. Maybe I can find some kind of neat 3D application that can do it better than I can do it. This is kind of just the reason why they all are conic sections, and why they really are related to each other. And will do that a little more in depth mathematically in a few videos. But in the next video, now that you know what they are and why they're all called conic sections, I'll actually talk about the formulas about these and how do you recognize the formulas. And given a formula, how do you actually plot the graphs of these conic sections? See you in the next video." + }, + { + "Q": "At 1:19- 1:24 the video says that the ft would cancel out. I don't see how they cancel out. In class I'm doing this and my teacher says use a chart to do this, but how do i get things to cancel out. i really want to pass the 9th eoct.", + "A": "It is known as simplifying the equation . Suppose you are given 9/3*27/6. You could simplify it and you will get 3/1*9/2, that is 27/2. If you do it the long way you will the same answer. Sal did the same thing.", + "video_name": "F0LLR7bs7Qo", + "transcript": "A squirrel is running across the road at 12 feet per second. It needs to run 9 feet to get across the road. How long will it take the squirrel to run 9 feet? Round to the nearest hundredth of a second. Fair enough. A car is 50 feet away from the squirrel-- OK, this is a high-stakes word problem-- driving toward it at a speed of 100 feet per second. How long will it take the car to drive 50 feet? Round to the nearest hundredth of a second. Will the squirrel make it 9 feet across the road before the car gets there? So this definitely is high stakes, at least for the squirrel. So let's answer the first question. Let's figure out how long will it take the squirrel to run 9 feet. So let's think about it. So the squirrel's got to go 9 feet, and we want to figure out how many seconds it's going to take. So would we divide or multiply this by 12? Well, to think about that, you could think about the units where we want to get an answer in terms of seconds. We want to figure out time, so it'd be great if we could multiply this times seconds per foot. Then the feet will cancel out, and I'll be left with seconds. Now, right over here, we're told that the squirrel can run at 12 feet per second, but we want seconds per foot. So the squirrel, every second, so they go 12 feet per second, then we could also say 1 second per every 12 feet. So let's write it that way. So it's essentially the reciprocal of this because the units are the reciprocal of this. So, it's 1 second for every 12 feet. Notice, all I did is I took this information right over here, 12 feet per second, and I wrote it as second per foot-- 12 feet for every 1 second, 1 second for every 12 feet. What's useful about this is this will now give me the time it takes for the squirrel in seconds. So the feet cancel out with the feet, and I am left with 9 times 1/12, which is 9/12 seconds. And 9/12 seconds is the same thing as 3/4 seconds, which is the same thing as 0.75 seconds for the squirrel to cross the street. Now let's think about the car. So now let's think about the car. And it's the exact same logic. They tell us that the car is 50 feet away. So the squirrel is trying to cross the road like that, and the car is 50 feet away coming in like that, and we want to figure out if the squirrel will survive. So the car is 50 feet away. So it's 50 feet away. We want to figure out the time it'll take to travel that 50 feet. Once again, we would want it in seconds. So we would want seconds per feet. So we would want to multiply by seconds per foot. They give us the speed in feet per second, 100 feet per second. And so we just have to realize that this is 100 feet for every 1 second, or 1/100 seconds per feet. This is once again just this information, but we took the reciprocal of it, because we don't want feet per second, we want seconds per feet. And if we do that, that cancels with that, and we're left with 50/100 seconds. So this is 50/100 is 0.50 seconds. And so now let's answer the question, this life and death situation for the squirrel. Will the squirrel make it 9 feet across the road before the car gets there? Well, it's going to take the squirrel 0.75 seconds to cross, and it's going to take the car only half a second. So the car is going to get to where the squirrel is crossing before the squirrel has a chance to get all the way across the road. So unfortunately for the squirrel, the answer is no." + }, + { + "Q": "How do I solve x+6 divides by 5 is greater than or equal to 10", + "A": "(x + 6) / 5 \u00e2\u0089\u00a5 10 1. Multiply both sides by 5 --> (x + 6) \u00e2\u0089\u00a5 10 x 5 2. Subtract 6 from both sides --> x \u00e2\u0089\u00a5 50 - 6 3. Therefore, x \u00e2\u0089\u00a5 44", + "video_name": "Yh4TXMVq9eg", + "transcript": "- [Voiceover] We have two inequalities here, the first one says that x plus two is less than or equal to two x. This one over here in I guess this light-purple-mauve color, is three x plus four is greater than five x. Over here we have four numbers and what I want to do in this video is test whether any of these four numbers satisfy either of these inequalities. I encourage you to pause this video and try these numbers out, does zero satisfy this inequality? Does it satisfy this one? Does one satisfy this one? Does it satisfy that one? I encourage you to try these four numbers out on these two inequalities. Assuming you have tried that, let's work through this together. Let's say, if we try out zero on this inequality right over here, let's substitute x with zero. So, we'll have zero plus two needs to be less than or equal to two times zero. Is that true? Well, on the left hand side, this is two needs to be less than or equal to zero. Is that true, is two less than or equal to zero? No, two is larger than zero. So this is not going to be true, this does not satisfy the left hand side inequality, let's see if it satisfies this inequality over here. In order to satisfy it, three times zero plus four needs to be greater than five times zero. Well three times zero is just zero, five times zero is zero. So four needs to be greater than zero, which is true. So it does satisfy this inequality right over here so zero does satisfy this inequality. Let's try out one. To satisfy this one, one plus two needs to be less than or equal to two. One plus two is three, is three less than or equal than two? No, three is larger than two. This does not satisfy the left hand inequality. What about the right hand inequality right over here? Three times one plus four needs to be greater than five times one. So three times one is three, plus four. So seven needs to be greater than five, well that's true. Both zero and one satisfy three x plus four is greater than five x, neither of them satisfy x plus two is less than or equal to two x. Now let's go to the two. I know it's getting a little bit unaligned, but I'll just do it all in the same color so you can tell. Let's try out two here, two plus two needs to be less than or equal to two times two. Four needs to be less than or equal to four. Well four is equal to four and it just has to be less than or equal, so this satisfies. This satisfies this inequality. What about this purple inequality? Let's see, three times two plus four needs to be greater than five times two. Three times two is six plus four is ten, needs to be greater than 10. 10 is equal to 10, it's not greater than 10. It does not satisfy this inequality. If this was a greater than or equal to it would have satisfied but it's not. 10 is not greater than 10. It would satisfy greater than or equal to because 10 is equal to 10. So two satisfies the left hand one but not the right hand one. Let's try out five. Five plus two needs to be less than or equal to two times five, once again everywhere we see an x, we replace it with a five. Seven needs to be less than or equal to 10. Which is absolutely true, seven is less than 10. So it satisfies less than or equal to. Five satisfies this inequality and what you're probably noticing now is that an inequality can have many numbers that satisfy. In fact they sometimes will have nothing that satisfies it and sometimes they might have an infinite number of numbers that satisfy it and you see that right over here. We're just testing out a few numbers. For this left one, zero and one didn't work, two and five did work. This right one, zero and one worked, two didn't work. Let's see what five does. In order for five to satisfy it, three times x. Now we're gonna try x being five. Three times five plus four needs to be greater than five times five. Three times five is fifteen, fifteen plus four is nineteen. Nineteen is to be greater than 25, it is not. So five does not satisfy this inequality right over here. Anyway, hopefully you found that fun." + }, + { + "Q": "Is an isoprene ever a functional unit? Would I be correct to describe geraniol as 2 isoprene units and an alcohol?", + "A": "Yes, structurally, the skeleton of geraniol consists of two isoprene units. But an isoprene unit is not a functional group. A functional group would be a double bond or an alcohol group. The functional groups in geraniol are the two C=C double bonds and the alcohol group.", + "video_name": "esJ5MbAHswc", + "transcript": "- [Voiceover] Let's practice identifying functional groups in different compounds. So this molecule on the left is found in perfumes, and let's look for some of the functional groups that we've talked about in the previous videos. Well here is a carbon-carbon double bond, and we know that a carbon-carbon double bond is an alkene. So here is an alkene functional group. Here's another alkene, right, here's another carbon-carbon double bond. What is this functional group? We have an OH and then we have the rest of the molecule, so we have ROH. ROH is an alcohol, so there's also an alcohol present in this compound. Next let's look at aspirin. So what functional groups can we find in aspirin? Well, here is an aromatic ring. So this is an arene, so there is an arene functional group present in aspirin. What about this one up here? We have an OH, and the oxygen is directly bonded to a carbonyl, so let's go ahead and write that out. We have an OH where the oxygen is directly bonded to a carbon double bonded to an oxygen, and then we have the rest of the molecule, so hopefully you recognize this as being a carboxylic acid. So let me go ahead and write that out here. So this is a carboxylic acid. All right, our next functional group. We have an oxygen, and that oxygen is directly bonded to a carbonyl. So here's a carbon double bonded to an oxygen, so let's write this out. We have an oxygen directly bonded to a carbonyl, and then for this oxygen, we have the rest of the molecule so that's all of this stuff over here, and then, on the other side of the carbonyl we have another R group. So I'll go ahead and write that in, so that is an ester. RO, C double bond O, R, is an ester. So there's an ester functional group present in the aspirin molecule. Let's look at some of the common mistakes that students make. One of them is, students will say a carboxylic acid is an alcohol. So let me write out here a carboxylic acid, so we can talk about that. So sometimes the students will look at that and say, oh, well I see an OH, and then I see the rest of the molecule, so isn't that an alcohol? But since this oxygen is right next to this carbonyl, this is a carboxylic acid. So this is an example of a carboxylic acid. If we moved the OH further away, from the carbonyl, let's go ahead and draw one out like that. So here is our carbonyl, and now the OH is moved further away, now we do have an alcohol, now we have an OH and then the rest of the molecule. So this would be, we can go ahead and use a different color here. So now we are talking about an alcohol, so this is an alcohol. And what would this one be? We have a carbonyl and then we have an R group on one side, an R group on the other side. That is a ketone, let me draw this out. So when you have a carbonyl and an R group on one side, an R group on the other side, they could be the same R group, they could be a different R group. Sometimes you'll see R prime drawn for that. So this is a ketone. So now we have a ketone and an alcohol, so two functional groups present in the same compound. So hopefully you can see the difference between this compound and this compound. This one is a carboxylic acid, and this one is a ketone and an alcohol. Another common mistake that I've seen a lot is on this functional group right here, on aspirin, students will look at this oxygen here, and say, okay, I have an oxygen, and then I have an R group on one side, and I have an R group on the other side. So an R group on one side of the oxygen, an R group on the other side of the oxygen, isn't that an ether? Well, this is, ROR would represent an ether, however, we have this carbonyl here. So this carbonyl right next to this oxygen is what makes this an ester. How could we turn that into an ether? Let me go ahead and redraw this molecule here. So I'll first put in our ring, so I drew the double bonds a little bit differently from how I drew it up here but it doesn't really matter, and then I'll put in our carboxylic acid up here, and now, when I draw in this oxygen, I'm gonna take out the carbonyl. So now the carbonyl is gone, and now we do have an ether. So this actually is an ether now, we have an oxygen, we have an R group on one side, and we have the rest of the molecule over here on the other side, so now this is an ether. So hopefully you see the difference there. Look for the carbonyl right next to the oxygen, that makes it an ester. All right, so more common mistakes that students make is they mix up these two functional groups, so let's look at the functional groups in these two molecules here. And we start with benzaldehyde, and the name is a dead giveaway as to the functional group, we're talking about an aldehyde here. So first, we have our aromatic ring, our arene, and then we have an aldehyde. We have a carbonyl and we have a hydrogen that's directly bonded to the carbonyl carbon. So we have an R group, and then we have a carbonyl, and then we have a hydrogen directly bonded to our carbonyl carbon, that is an aldehyde. If we took off that hydrogen, and we put a CH3 instead, that would be the compound on the right so now we have a CH3 directly bonded to this carbonyl carbon. So now we have an R group on one side, a carbonyl, and then another R group, so we have R, C double bond O, R, and that is a ketone. And you can tell by the ending of our name here that we have a ketone present in this compound. So again, this difference is subtle, but it's important, and a lot of students mess this up. An aldehyde has a hydrogen directly bonded to this carbonyl carbon, but if there's no hydrogen, we're talking about a ketone here, so R, C double bond O, R, is a ketone. Finally, let's look at one giant compound with lots of different functional groups, and let's see if we can identify all the functional groups present in this molecule. This molecule, it is called atenolol. This is a beta blocker. So this is a heart medication. Let's look for some functional groups we've seen before. Here is that aromatic ring, so we know that an arene is present in atenolol, so let me go ahead and write this in here. Next, we have an oxygen, and there's an R group on one side of the oxygen, and an R group on the other side of the oxygen, so ROR, we know that's an ether. So there's an ether present in this compound. Next, we have an OH, and then the rest of the molecule. So ROH would be an alcohol. So there's an alcohol present. All right, next we have a nitrogen with a lone pair of electrons. There's an R group on one side, there's an R group on the other side. So this is an amine. So we have an amine, and finally, over here on the left, so this is one that is messed up a lot. We do have a nitrogen with a lone pair of electrons on it, so it's tempting to say we have an amine here. But this nitrogen is right next to a carbonyl, so it's not an amine. It's an amide, or amid. So this is an amide, so a lot of people pronounce this \"amid\", all right, so it's not an amine. So let's talk more about the difference between an amide and an amine. So let me go ahead and draw out another compound here, so we can see we have our NH2, and then we have our carbonyl. So for this one, we have our nitrogen, directly bonded to the carbonyl carbon. And that's what makes this an amide. We can move these electrons into here, and push these electrons off onto the oxygen. So resonance is possible with this compound. So this is an amide, or an \"amid\" If we move the nitrogen further away from the carbonyl, let's go ahead and do that over here. So we have our carbonyl, and now our nitrogen is further away. Now we don't have anymore resonance right? You can't draw a resonance structure showing the delocalization of the lone pair of electrons on the nitrogen. So now, now we do have an amine, so this over here, this would be an amine. Let me change colors, let me do blue. This is an amine. And then, what would this functional group be? We have a carbonyl and then we have an R group on one side, R group on the other side, that is a ketone. So this is a ketone and an amine. And then over here, we have an amide, or an \"amid\", so make sure to know the difference between these. I've see a lot of very smart students mess up the difference between these two functional groups." + }, + { + "Q": "when I typed in 81 for 9 x 8 it said it was incorrect how is it incorrect?", + "A": "9 x 9 actually equals 81. 9 x 8 is 72.", + "video_name": "NehkLV77ITk", + "transcript": "You're just walking down the street and someone comes up to you and says \"Quick! Quick!-- 4792. Is this divisible by 3? This is an emergency! Tell me as quickly as possible! And luckily you have a little tool in your toolkit where you know how to test for divisibility by 3 Well, you say I can just add up the digits If the sum of that is a multiple of 3 then this whole thing is a multiple of 3 So you say 4 plus 7 plus 9 plus 2 That's 11. Plus 9, it's 20. Plus 2 is 22 That's not divisible by 3 If you're unsure, you can even add the digits of that 2 plus 2 is 4. Clearly not divisible by 3 So this thing right over here is not divisible by 3 And so luckily that emergency was saved But then you walk down the street a little bit more and someone comes up to you--- \"Quick! Quick! Quick! 386,802-- Is that divisible by 3?\" Well, you employ the same tactic You say, what's 3 plus 8 plus 6 plus 8 plus 0 plus 2? 3 plus 8 is 11. Plus 6 is 17. Plus 8 is 25. Plus 2 is 27 Well, 27 is divisible by 3 And if you're unsure, you could add these digits right over here 2 plus 7 is equal to 9. Clearly divisible by 3 So this is divisible by 3 as well So now you feel pretty good You've helped two perfect strangers with their emergencies You figured out if these numbers were divisible by 3 very very very very quickly But you have a nagging feeling Because you're not quite sure why that worked You've just kind of always known it And so, let's think about why it worked To think about it, I'll just pick a random number But we could do this really for any number But I don't want to go too puffy on it just so you can see it's pretty common sense here And the number we'll use is 498 I can literally use any number in this situation And to think about why this whole little tool this little system works we just have to rewrite 498 We can rewrite the 4- since it's in the hundred's place we can write that as 4 times 100 Or 4 times 100, that's the same thing as 4 times 1 plus 99 That's all this 4 is 400, which is the same thing as 4 times 100 which is the same thing as 4 times 1 plus 99 And the little trick here is I want to write- instead of writing 100, I want to write this as the sum of 1 plus something that is divisible by 3 And 99 is divisible by 3 If I add more digits here- 999, 9999-- they're all divisible by 3 And this is why you can do the same reasoning for divisibility by 9 Because they are divisible by 9 as well Anyway, that's what the 4 in the hundred's place represents This 9 in the ten's place- well that represents 90 or 9 times 10, or 9 times 1 plus 9 And then finally this 8. That's in the one's place 8 times 1, or we just write plus 8 Now we can distribute this 4 This is 4 times 1 plus 4 times 99. So it's 4 plus 4 times 99 Actually let me write it like this. I'm going to write-- Actually let me write it first like 4 plus 4 times 99 Do the same thing over here This is the same thing as plus 9-- do that magenta color- plus 9 plus 9 times 9 And then finally I have this 8 right over here And I can rearrange everything These terms right over here, the 4 times 99, and the 9 times 9 I can write over here 4 times 99- I'll write what's like a different notation plus the 9 times the 9, that's those two terms and then we have the plus 4 plus 9 plus 8 Well, can we now tell whether this is divisible by 3? These terms, these first two terms are definitely divisible by 3 This's divisible by 3 because 99 is divisible by 3 regardless of what we have already you don't even have to look at this This is divisible by 3, so if you're multiplying it it's still going to be divisible by 3 This is divisible by 3, so if you're multiplying this whole thing it's still going to be divisible by 3 If you add two things that are divisible by 3 the whole thing is going to be divisible by 3 So all of this is divisible by 3 And if you have another digit here, you'd done the same exact thing Instead of having 1 plus 99, you'd had 1 plus 999, 1 plus 9999, etc So the only thing you have to really worry about is this part right over here you have to ask yourself in order for this whole thing to be divisible by 3 this part is- well that part is, then this part in order for the whole thing has to be divisible by 3 that also has to be divisible by 3 But what is this right over here? These are just our original digits 498. 4 and 9 and 8 We just have to make sure that when we take the sum it's divisible by 3" + }, + { + "Q": "I still do not see how the parabola equation helps when trying to solve for a parabola with a given set of foci and a directrix, the previous activity just equated the directrix and foci using the distance formulas. So what are you supposed to use the parabola equation for?", + "A": "He s showing how to convert between the two forms, focus/directrix and vertex , and how these two different ways of thinking about parabolas give the same set of functions.", + "video_name": "w56Vuf9tHfA", + "transcript": "- This right here is an equation for a parabola and the role of this video is to find an alternate or to explore an alternate method for finding the focus and directrix of this parabola from the equation. So the first thing I like to do is solve explicitly for y. I don't know, my brain just processes things better that way. So, let's get this 23 over four to the right hand side. So let's add 23 over four to both sides and then we'll get y is equal to negative one-third times x minus one squared plus 23 over four. Now let's remind ourselves what we've learned about foci and directrixes, I think is how to say it. So, the focus. If the focus of a parabola is at the point a, b and the directrix, the directrix, directrix is the line y equals k. We've shown in other videos with a little bit of hairy algebra that the equation of the parabola in a form like this is going to be y is equal to one over two times b minus k. This b minus k is then the difference between this y coordinate and this y value, I guess you could say. Times x minus one squared plus b plus k. I'm sorry, not x minus one. I'm getting confused with this. x minus a squared. x minus a squred plus b plus k over two. The focus is a,b and the directrix is y equals k and this is gonna be the equation of the parabola. Well, we've already seen the technique where, look, we can see the different parts. We can see that, okay, this x minus one squared. Actually, let me do this in a different color. This x minus one squared corresponds to the x minus a squared and so one corresponds to a, so just like that, we know that a is going to be equal to one and actually let me just write that down. a is equal to one in this example right over here. And then you could see that the negative one-third over here corresponds to the one over two b minus k and you would see that the 23 over four corresponds to the b plus k over two. Now the first technique that we explored, we said, \"Okay, let's set negative one-third \"to this thing right over here. \"Solve for b minus k.\" We're not solving for b or k, we're solving for the expression b minus k. So you got b minus k equals something. And then you could use 23 over four and this to solve for b plus k. So you get b plus k equals something and then you have two equations, two unknowns, you can solve for b and k. What I wanna do in this video is explore a different method that really uses our knowledge of the vertex of a parabola to be able to figure out where the focus and the directrix is going to be. So let's think about the vertex of this parabola right over here. Remember, the vertex, if the parabola is upward opening like this, the vertex is this minimum point. If it is downward opening, it's going to be this maximum point. And so when you look over here, you see that you have a negative one-third in front of the x minus one squared. So this quantity over here is either going to be zero or negative. It's not going to add to 23 over four, it's either gonna add nothing or take away from it. So this thing's going to hit a maximum point, when this thing is zero, when this thing is zero, and that's just gonna go down from there and when this thing is zero, y is going to be equal to 23 over four. So our vertex is going to be that maximum point. Well, when does this equal zero? Well, when x equals one. When x equals one, you get one minus one squared. So zero squared times negative one-third, this is zero. So when x is equal to one, we're at our maximum y value of 23 over four which five and three-fourths. Actually, let me write that as a . Actually, I'll leave just that's our vertex. and it is a downward opening parabola. So actually, let me start to draw this. So we'd get some axis here. So we have to go all the way up to five and three-fourths. So. Let's make this our y, this is our y axis. This is the x axis. That's the x axis. We're gonna see, we're gonna go to one. Let's call that one. Let's call that two. And then I wanna get, let's see, if I go to five and three-fourths, let's go up to, let's see one, two, three, four five, six, seven. We can label 'em. One, two, three, four five, six and seven and so our vertex is right over here. One comma 23 over four, so that's five and three-fourths. So it's gonna be right around right around there and as we said, since we have a negative value in front of this x minus one squared term, I guess we could call it, this is going to be a downward opening parabola. This is going to be a maximum point. So our actual parabola is going to look is going to look something it's gonna look something like this. It's gonna look something like this and we could, obviously, I'm hand drawing it, so it's not going to be exactly perfect, but hopefully you get the general idea of what the parabola is going look like and actually, let me just do part of it, 'cause I actually don't know that much information about the parabola just yet. I'm just gonna draw it like that. So we don't know just yet where the directrix and focus is, but we do know a few things. The focus is going to sit on the same, I guess you could say, the same x value as the vertex. So if we draw, this is x equals one, if x equals one, we know from our experience with focuses, foci, (laughs) I guess, that they're going to sit on the same axis as the vertex. So the focus might be right over here and then the directrix is going to be equidistant on the other side, equidistant on the other side. So the directrix might be something like this. Might be right over here. And once again, I haven't figured it out yet, but what we know is that because this point, the vertex, sits on the parabola, by definition has to be equidistant from the focus and the directrix. So. This distance has to be the same as this distance right over here and what's another way of thinking about this entire distance? Remember, this coordinate right over here is a, b and this is the line y is equal to k. This is y equals k. So what's this distance in yellow? What's this difference in y going to be? Well, you could call that, in this case, the directrix is above the focus, so you could say that this would be k minus b or you could say it's the absolute value of b minus k. This would actually always work. It'll always give you kind of the positive distance. So if we knew what the absolute value of b minus k is, if we knew this distance, then just split it in half with the directrix is gonna be that distance, half the distance above and then the focus is gonna be half the distance below. So let's see if we can figure this out. And we can figure this out because we see in this, I guess you could say, this equation, you can see where b minus k is involved. One over two times b minus k needs to be equal to negative one-third. So let's solve for b minus k. So we get we get one over two times b minus k is going to be equal to negative one-third. Once again, this corresponds to that. It's going to be equal to negative one-third. We could take the reciprocal of both sides and we get two times b minus k is equal to, is equal to three, is equal to three. Now we can divide both sides we can divide both sides by two and so we're gonna get we're gonna get b b minus k is equal to is equal to, what is that, three-halves, three-halves. b minus k is equal to, oh, let me make sure that has to be a negative three, so this has to be negative three-halves. And so if you took the absolute value of b minus k you're gonna get positive three-halves, or if you took k minus b, you're going to get positive three-halves. So just like that, using this part, just actually matching the negative one-third to this part of this equation, we're able to solve for the absolute value of b minus k which is going to be the distance between the y axis in the y direction between the focus and the directrix. So this distance right over here is three-halves. So what is half that distance? And the reason why I care about half that distance is because then I can calculate where the focus is, because it's going to be half that distance below the vertex and I could say, whatever that distance is is going to be that distance also above the directrix. So half that distance, so one half times three-halves is equal to three-fourths. So just like that, we're able to figure out the directrix is going to be three-fourths above this. So I could say the directrix, so let me see, I'm running out of space, the directrix is gonna be y is equal to the y coordinate of the focus. Sorry, the y coordinate of the vertex. I might be careful with my language. It's gonna be equal to the y coordinate of the vertex plus three-fourths, plus three- fourths. So plus three-fourths, which is equal to 26 over four, which is equal to, what is that, that's equal to six and a half. So this right over here, actually I got pretty close when I drew it is actually going to be the directrix. Y is equal to six and a half and the focus, well, we know the x coordinate of the focus, a is going to be equal to one and b is going to be three-fourths less than the y coordinate of the directrix. So 23 over four minus three-fourths. Gonna be 23 over four 23 over four minus three-fourths which is 20 over four, which is just equal to which is just equal to five. And we are done. That's the focus, one comma five. Directrix is y is equal to six and a half." + }, + { + "Q": "what do i do if i dont have an x value for example 4x cubed plus 2x squared minus 8", + "A": "Write 0x as a placeholder. --> 4x^3 + 2x^2 + 0x - 8. Do the long division and treat the zero as any other number (e.g. 0x - 5x = -5x). Does that make sense?", + "video_name": "MwG6QD352yc", + "transcript": "- [Voiceover] So let's introduce ourselves to the Polynomial Remainder Theorem. And as we'll see a little, you'll feel a little magical at first. But in future videos, we will prove it and we will see, well, like many things in Mathematics. When you actually think it through, maybe it's not so much magic. So what is the Polynomial Remainder Theorem? Well it tells us that if we start with some polynomial, f of x. So this right over here is a polynomial. Polynomial. And we divide it by x minus a. Then the remainder from that essentially polynomial long division is going to be f of a. It is going to be f of a. I know this might seem a little bit abstract right now. I'm talking about f of x's and x minus a's. Let's make it a little bit more concrete. So let's say that f of x is equal to, I'm just gonna make up a, let's say a second degree polynomial. This would be true for any polynomial though. So three x squared minus four x plus seven. And let's say that a is, I don't know, a is one. So we're gonna divide that by, we're going to divide by x minus one. So a, in this case, is equal to one. So let's just do the polynomial long division. I encourage you to pause the video. If you're unfamiliar with polynomial long division, I encourage you to watch that before watching this video because I will assume you know how to do a polynomial long division. So divide three x squared minus four x plus seven. Divide it by x minus one. See what you get as the remainder and see if that remainder really is f of one. So assuming you had a go at it. So let's work through it together. So let's divide x minus one into three x squared minus four x plus seven. All right, little bit of polynomial long division is never a bad way to start your morning. It's morning for me. I don't know what it is for you. All right, so I look at the x term here, the highest degree term. And then I'll start with the highest degree term here. So how many times does x going to three x squared? What was three x times? Three x times x is three x squared. So I'll write three x over here. I'll write it in the, I could say the first degree place. Three x times x is three x squared. Three x times negative one is negative three x. And now we want to subtract this thing. It's just the way that you do traditional long division. And so, what do we get? Well, three x squared minus three x squared. That's just going to be a zero. So this just add up to zero. And this negative four x, this is going to be plus three x, right? And negative of a negative. Negative four x plus three x is going to be negative x. I'm gonna do this in a new color. So it's going to be negative x. And then we can bring down seven. Complete analogy to how you first learned long division in maybe, I don't know, third or fourth grade. So all I did is I multiplied three x times this. You get three x squared minus three x and then I subtract to that from three x squared minus four x to get this right over here or you could say I subtract it from this whole polynomial and then I got negative x plus seven. So now, how many times does x minus one go to negative x plus seven? Well x goes into negative x, negative one times x is negative x. Negative one times negative one is positive one. But then we're gonna wanna subtract this thing. We're gonna wanna subtract this thing and this is going to give us our remainder. So negative x minus negative x. Just the same thing as negative x plus x. These are just going to add up to zero. And then you have seven. This is going to be seven plus one. Remember you have this negative out so if you distribute the negative, this is going to be a negative one. Seven minus one is six. So your remainder here is six. One way to think about it, you could say that, well (mumbles). I'll save that for a future video. This right over here is the remainder. And you know when you got to the remainder, this is just all review of polynomial long division, is when you get something that has a lower degree. This is, I guess you could call this a zero degree polynomial. This has a lower degree than what you are actually dividing into or than the x minus one than your divisor. So this a lower degree so this is the remainder. You can't take this into this anymore times. Now, by the Polynomial Remainder Theorem, if it's true and I just picked a random example here. This is by no means a proof but just kinda a way to make it tangible of Polynomial (laughs) Remainder Theorem is telling us. If the Polynomial Remainder Theorem is true, it's telling us that f of a, in this case, one, f of one should be equal to six. It should be equal to this remainder. Now let's verify that. This is going to be equal to three times one squared, which is going to be three, minus four times one, so that's just going to be minus four, plus seven. Three minus four is negative one plus seven is indeed, we deserve a minor drumroll, is indeed equal to six. So this is just kinda, at least for this particular case, looks like okay, it seems like the Polynomial Remainder Theorem worked. But the utility of it is if someone said, \"Hey, what's the remainder if I were to divide \"three x squared minus four x plus seven \"by x minus one if all I care about is the remainder?\" They don't care about the actual quotient. All they care about is the remainder, you could, \"Hey, look, I can just take that, in this case, a is one. \"I can throw that in. \"I can evaluate f of one and I'm gonna get six. \"I don't have to do all of this business. \"All I had, would have to do is this \"to figure out the remainder of three x squared.\" Well you take three x squared minus four plus seven and divide by x minus one." + }, + { + "Q": "can you write that congruency sign (at 1:39) on a keyboard?", + "A": "In Windows, like in Microsoft Word or Microsoft Excel, you can your font to Symbol. Then, that symbol is SHIFT+2 (or the @ key on your keyboard.). As a side note, with the Symbol font you get most of the Greek letters..alpha, beta, delta, pi, etc.", + "video_name": "CJrVOf_3dN0", + "transcript": "Let's talk a little bit about congruence, congruence And one to think about congruence, it's really kind of equivalence for shapes So, when in algebra when something is equal to another thing it means that their quantities are the same But when we're all of the sudden talking about shapes and we say that those shapes are the same, the shapes are the same size and shape then we say that they're congruent And just to see a simple example here: I have this triangle, right over there and let's say I have this triangle right over here And if you are able to shift, you are able to shift this triangle and flip this triangle, you can make it look exactly like this triangle As long as you're not changing the lengths of any of the sides or the angles here But you can flip it, you can shift it, you can rotate it So you can shift, let me write this, you can shift it, you can flip it and you can rotate If you can do those three procedures to make these the exact same triangle, then they are congruent And if you say that a triangle is congruent, let me label this So, let's call this triangle ABC Now let's call this D, let me call it XYZ XY and Z So, if we were to say, if we make the claim that both of these triangles are congruent So, if we say triangle ABC is congruent And the way you specify it, it almost look like an equal sign But it's equal sign with a curly thing on top Let me write it a little bit either So, we would write it like this If we know that triangle ABC is congruent to triangle XYZ That means their corresponding sides have the same length And their corresponding angles have the same measure So, if we make this assumption or someone tells us that this is true then we know, for example, that AB is going to equal to XY The length of segment AB is gonna be equal to the segment of XY And we could do this like this, and I'm assuming this are the corresponding sides And you can see that actually we've defined these triangles A corresponds to X, B corresponds to Y and C corresponds to Z right over there So, side AB is gonna have the same length as XY Then you can sometimes if you don't have the colors you can denote it just like that These two length are- or this two lines segments have the same length And you can actually say this, you don't always see this written this way You could also make the statement that line segment AB is congruent to line segment XY But congruence of line segments really just means that their lengths are equivalent So, these two things mean the same thing If one line segment is congruent to another line segment that just means the measure of one line segment is equal to the measure of the other line segment And so we can go thru all the corresponding sides If these two characters are congruent, we also know that BC, we also know that the length BC is gonna be the length of YZ Assuming those are the corresponding sides And we can put these double hash marks right over here to show that these lengths are the same And when we go the third side, we also know that these are going to be has same length or the line segments are going to be congruent So, we also know that the length of AC is going to be equal to the length of XZ Not only do we know that all of the sides, the corresponding sides are gonna have the same length If someone tells that a triangle is congruent We also know that all the corresponding angles are going to have the same measure So, for example: we also know that this angle's measure is going to be the same as the corresponding angle's measure, and the corresponding angle is right over It's between these orange side and blue side Or orange side and purple side, I should say And between the orange side and this purple side And so it also tells us that the measure of angle is BAC is equal to the measure of angle of YXZ Let me write that angle symbol, a little less like that, measure of angle of YXZ YXZ We can also write that as angle BAC is congruent angle YXZ And once again, like line segment, if one line segment is congruent to another line segment It just means that their lengths are equal And if one angle is congruent to another angle it just means that their measures are equal So, we know that those two corresponding angles have the same measure, they're congruent We also know that these two corresponding angles I'll use a double arch to specify that this has the same measure as that So, we also know the measure of angle ABC is equal to the measure of angle XYZ And then finally we know that this angle, if we know that these two characters are congruent, then this angle is gonna have the same measure as this angle as a corresponding angle So, we know that the measure of angle ACB is gonna be equal to the measure of angle XZY Now what we're gonna concern ourselves a lot with is how do we prove congruence? 'Cause it's cool, 'cause if you can prove congruence of 2 triangles then all of the sudden you can make all of these assumptions And what we're gonna find out, and this is going to be, we're gonna assume it for the sake of introductory geometry course This is an axiom or a postulate or just something you assume So, an axiom, very fancy word Postulate, also a very fancy word It really just means things we are gonna assume are true An axiom is sometimes, there's a little bit of distinction sometimes where someone would say \"an axiom is something that is self-evident\" or it seems like a universal truth that is definitely true and we just take it for granted You can't prove an axiom A postulate kinda has that same role but sometimes let's just assume this is true and see if we assume that it's true what can we derive from it, what we can prove if we assume its true But for the sake of introductory geometry class and really most in mathematics today, these two words are use interchangeably An axiom or a postulate, just very fancy words that things we take as a given Things that we'll just assume, we won't prove them, we will start with this assumptions and then we're just gonna build up from there And one of the core ones that we'll see in geometry is the axiom or the postulate That if all of the sides are congruent, if the length of all the sides of the triangle are congruent, then we are dealing with congruent triangles So, sometimes called side, side, side postulate or axiom We're not gonna prove it here, we're just gonna take it as a given So this literally stands for side, side, side And what it tells is, if we have two triangles and So I say that's another triangle right over there And we know that corresponding sides are equal So, we know that this side right over here is equal into, like, that side right over there Then we know and we're just gonna take this as an assumption and we can build off of this We know that they are congruent, the triangle, that these two triangles are congruent to each other I didn't put any labels there so it's kinda hard for me to refer to them But these two are congruent triangles And what's powerful there is we know that the corresponding sides are equal Then we know they're congruent and we can make all the other assumptions Which means that the corresponding angles are also equal So, that we know, is gonna be congruent to that or have the same measure That's gonna have the same measure as that and then that is gonna have the same measure as that right over there And to see why that is a reasonable axiom or a reasonable assumption or a reasonable postulate to start off with Let's take one, let's start with one triangle So, let's say I have this triangle right over here So, it has this side and then it has this side and then it has this side right over here And what I'm gonna do is see if I have another triangle that has the exact same line, side lengths is there anyway for me to construct a triangle with the same side lengths that is different, that can't be translated to this triangle thru flipping, shifting or rotating So, we assume this other triangle is gonna have the same size, the same length as that one over there So, I'll try to draw it like that Roughly the same length We know that it's going to have a size that's that length So, it's gonna have a side that is that length Let me put it on this side just to make it look a little bit more interesting So, we know that it's gonna have a side like that So, I'm gonna draw roughly the same length but I'm gonna try to do it in a different angle Now we know that's it's gonna have that looks like that So, let me, I'll put it right over here It's about that length right over there And so clearly this isn't a triangle, in order to make it a triangle, I'll have to connect this point to that point right over there And really there's only two ways to do it I can rotate it around that little hinge right over there If I connect them over here then I'm going to get a triangle that looks likes this Which is really a just a flip, am I visualizing it right? Yeah, just a flip version You can rotate it a little back this way, and you'd have a magenta on this side and a yellow one on this side And you can flip it, you could flip it vertically and it'll look exactly like this Our other option to make these two points connect is to rotate them out this way And the yellow side is gonna be here And then the magenta side is gonna be here and that's not magenta The magenta side is gonna be just like that And if we do that, then we actually just have to rotate it We just have to rotate it around to get that exact triangle So, this isn't a proof, and actually we're gonna start assuming that his is an axiom But hopefully you'll see that it's a pretty reasonable starting point that all of the sides, all of the corresponding sides of two different triangles are equal Then we are going to- we know that they are congruent We are just gonna assume that it's an axiom for that we're gonna build off, that they are congruent And we also know that he corresponding angles are going to be equivalent" + }, + { + "Q": "if i = radical(-1), then i^2 = radical(-1) * radical(-1), which means i^2 should equal radical(-1 * -1) which simplies to radical(1). Making i^2 = 1", + "A": "No, because the property\u00e2\u0088\u009a(ab) = (\u00e2\u0088\u009aa)(\u00e2\u0088\u009ab) does not hold for imaginary numbers. Thus, (\u00e2\u0088\u009a-1)(\u00e2\u0088\u009a-1) \u00e2\u0089\u00a0 \u00e2\u0088\u009a(\u00e2\u0088\u00921 \u00c3\u0097 \u00e2\u0088\u00921)", + "video_name": "s03qez-6JMA", + "transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. If we have the principal square root of the product of two things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. You cannot do this, right over here. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i. But the principal square root of negative 1 is i. And then we're going to multiply that times the square root of 4 times 13. And this is going to be equal to i times the square root of 4. i times the square root of 4, or the principal square root of 4 times the principal square root of 13. The principal square root of 4 is 2. So this all simplifies, and we can switch the order, over here. This is equal to 2 times the square root of 13. 2 times the principal square root of 13, I should say, times i. And I just switched around the order. It makes it a little bit easier to read if I put the i after the numbers over here. But I'm just multiplying i times 2 times the square root of 13. That's the same thing as multiplying 2 times the principal square root of 13 times i. And I think this is about as simplified as we can get here." + }, + { + "Q": "Where can I find the \"rigorous proof\" of these properties?", + "A": "A rigorous proof can usually be found in any old calculus text, in the section on limits. A fun exercise might be to write down the epsilon-delta definition of limits then try to figure out exactly how one would prove these statements!", + "video_name": "lSwsAFgWqR8", + "transcript": "What I want to do in this video is give you a bunch of properties of limits. And we're not going to prove it rigorously here. In order to have the rigorous proof of these properties, we need a rigorous definition of what a limit is. And we're not doing that in this tutorial, we'll do that in the tutorial on the epsilon delta definition of limits. But most of these should be fairly intuitive. And they are very helpful for simplifying limit problems in the future. So let's say we know that the limit of some function f of x, as x approaches c, is equal to capital L. And let's say that we also know that the limit of some other function, let's say g of x, as x approaches c, is equal to capital M. Now given that, what would be the limit of f of x plus g of x as x approaches c? Well-- and you could look at this visually, if you look at the graphs of two arbitrary functions, you would essentially just add those two functions-- it'll be pretty clear that this is going to be equal to-- and once again, I'm not doing a rigorous proof, I'm just really giving you the properties here-- this is going to be the limit of f of x as x approaches c, plus the limit of g of x as x approaches c. Which is equal to, well this right over here is-- let me do that in that same color-- this right here is just equal to L. It's going to be equal to L plus M. This right over here is equal to M. Not too difficult. This is often called the sum rule, or the sum property, of limits. And we could come up with a very similar one with differences. The limit as x approaches c of f of x minus g of x, is just going to be L minus M. It's just the limit of f of x as x approaches c, minus the limit of g of x as x approaches c. So it's just going to be L minus M. And we also often call it the difference rule, or the difference property, of limits. And these once again, are very, very, hopefully, reasonably intuitive. Now what happens if you take the product of the functions? The limit of f of x times g of x as x approaches c. Well lucky for us, this is going to be equal to the limit of f of x as x approaches c, times the limit of g of x, as x approaches c. Lucky for us, this is kind of a fairly intuitive property of limits. So in this case, this is just going to be equal to, this is L times M. This is just going to be L times M. Same thing, if instead of having a function here, we had a constant. So if we just had the limit-- let me do it in that same color-- the limit of k times f of x, as x approaches c, where k is just some constant. This is going to be the same thing as k times the limit of f of x as x approaches c. And that is just equal to L. So this whole thing simplifies to k times L. And we can do the same thing with difference. This is often called the constant multiple property. We can do the same thing with differences. So if we have the limit as x approaches c of f of x divided by g of x. This is the exact same thing as the limit of f of x as x approaches c, divided by the limit of g of x as x approaches c. Which is going to be equal to-- I think you get it now-- this is going to be equal to L over M. And finally-- this is sometimes called the quotient property-- finally we'll look at the exponent property. So if I have the limit of-- let me write it this way-- of f of x to some power. And actually, let me even write it as a fractional power, to the r over s power, where both r and s are integers, then the limit of f of x to the r over s power as x approaches c, is going to be the exact same thing as the limit of f of x as x approaches c raised to the r over s power. Once again, when r and s are both integers, and s is not equal to 0. Otherwise this exponent would not make much sense. And this is the same thing as L to the r over s power. So this is equal to L to the r over s power. So using these, we can actually find the limit of many, many, many things. And what's neat about it is the property of limits kind of are the things that you would naturally want to do. And if you graph some of these functions, it actually turns out to be quite intuitive." + }, + { + "Q": "let the objects be A and B and let us assume that we are going to throw the objects from the tower of some height then which object would reach the ground earlier or both the objects reach the ground at same time? in other words the speed would be same or different depending on mass?", + "A": "Acceleration due to gravity will always be the same, so with no air resistance the two objects would hit the ground at the same time. If there was air resistance, the lighter object would be slowed more.", + "video_name": "VYgSXBjEA8I", + "transcript": "So I'm curious about how much acceleration does a pilot, or the pilot and the plane, experience when they need to take off from an aircraft carrier? So I looked up a few statistics on the Internet, this right here is a picture of an F/A-18 Hornet right over here. It has a take-off speed of 260 kilometers per hour. If we want that to be a velocity, 260 km/hour in this direction, if it's taking off from this Nimitz class carrier right over here. And I also looked it up, and I found the runway length, or I should say the catapult length, because these planes don't take off just with their own power. They have their own thrusters going, but they also are catapulted off, so they can be really accelerated quickly off of the flight deck of this carrier. And the runway length of a Nimitz class carrier is about 80 meters. So this is where they take off from. This right over here is where they take off from. And then they come in and they land over here. But I'm curious about the take-off. So to do this, let's figure out, well let's just figure out the acceleration, and from that we can also figure out how long it takes them to be catapulted off the flight deck. So, let me get the numbers in one place, so the take-off velocity, I could say, is 260 km/hour, so let me write this down. So that has to be your final velocity when you're getting off, of the plane, if you want to be flying. So your initial velocity is going to be 0, and once again I'm going to use the convention that the direction of the vector is implicit. Positive means going in the direction of take-off, negative would mean going the other way. My initial velocity is 0, I'll denote it as a vector right here. My final velocity over here has to be 260 km/hour. And I want to convert everything to meters and seconds, just so that I can get my, at least for meters, so that I can use my runway length in meters. So let's just do it in meters per second, I have a feeling it'll be a little bit easier to understand when we talk about acceleration in those units as well. So if we want to convert this into seconds, we have, we'll put hours in the numerator, 1 hour, so it cancels out with this hour, is equal to 3600 seconds. I'll just write 3600 s. And then if we want to convert it to meters, we have 1000 meters is equal to 1 km, and this 1 km will cancel out with those kms right over there. And whenever you're doing any type of this dimensional analysis, you really should see whether it makes sense. If I'm going 260 km in an hour, I should go much fewer km in a second because a second is so much shorter amount of time, and that's why we're dividing by 3600. If I can go a certain number of km in an hour a second, I should be able to go a lot, many many more meters in that same amount of time, and that's why we're multiplying by 1000. When you multiply these out, the hours cancel out, you have km canceling out, and you have 260 times 1000 divided by 3600 meters per second. So let me get my trusty TI-85 out, and actually calculate that. So I have 260 times 1000 divided by 3600 gets me, I'll just round it to 72, because that's about how many significant digits I can assume here. 72 meters per second. So all I did here is I converted the take-off velocity, so this is 72 m/s, this has to be the final velocity after accelerating. So let's think about what that acceleration could be, given that we know the length of the runway, and we're going to assume constant acceleration here, just to simplify things a little bit. But what does that constant acceleration have to be? So let's think a little bit about it. The total displacement, I'll do that in purple, the total displacement is going to be equal to our average velocity while we're accelerating, times the difference in time, or the amount of time it takes us to accelerate. Now, what is the average velocity here? It's going to be our final velocity, plus our initial velocity, over 2. It's just the average of the initial and final. And we can only do that because we are dealing with a constant acceleration. And what is our change in time over here? What is our change in time? Well our change in time is how long does it take us to get to that velocity? Or another way to think about it is: it is our change in velocity divided by our acceleration. If we're trying to get to 10 m/s, or we're trying to get 10 m/s faster, and we're accelerating at 2 m/s squared, it'll take us 5 seconds. Or if you want to see that explicitly written in a formula, we know that acceleration is equal to change in velocity over change in time. You multiply both sides by change in time, and you divide both sides by acceleration, so let's do that, multiply both sides by change in time and divide by acceleration. Multiply by change in time and divide by acceleration. And you get, that cancels out, and then you have that cancels out, and you have change in time is equal to change in velocity divided by acceleration. Change in velocity divided by acceleration. So what's the change in velocity? Change in velocity, so this is going to be change in velocity divided by acceleration. Change in velocity is the same thing as your final velocity minus your initial velocity, all of that divided by acceleration. So this delta t part we can re-write as our final velocity minus our initial velocity, over acceleration. And just doing this simple little derivation here actually gives us a pretty cool result! If we just work through this math, and I'll try to write a little bigger, I see my writing is getting smaller, our displacement can be expressed as the product of these two things. And what's cool about this, well let me just write it this way: so this is our final velocity plus our initial velocity, times our final velocity minus our initial velocity, all of that over 2 times our acceleration. Our assumed constant acceleration. And you probably remember from algebra class this takes the form: a plus b times a minus b. And so this equal to -- and you can multiply it out and you can review in our algebra playlist how to multiply out two binomials like this, but this numerator right over here, I'll write it in blue, is going to be equal to our final velocity squared minus our initial velocity squared. This is a difference of squares, you can factor it out into the sum of the two terms times the difference of the two terms, so that when you multiply these two out you just get that over there, over 2 times the acceleration. Now what's really cool here is we were able to derive a formula that just deals with the displacement, our final velocity, our initial velocity, and the acceleration. And we know all of those things except for the acceleration. We know that our displacement is 80 meters. We know that this is 80 meters. We know that our final velocity, just before we square it, we know that our final velocity is 72 meters per second. And we know that our initial velocity is 0 meters per second. And so we can use all of this information to solve for our acceleration. And you might see this formula, displacement, sometimes called distance, if you're just using the scalar version, and really we are thinking only in the scalar, we're thinking about the magnitudes of all of these things for the sake of this video. We're only dealing in one dimension. But sometimes you'll see it written like this, sometimes you'll multiply both sides times the 2 a, and you'll get something like this, where you have 2 times, really the magnitude of the acceleration, times the magnitude of the displacement, which is the same thing as the distance, is equal to the final velocity, the magnitude of the final velocity, squared, minus the initial velocity squared. Or sometimes, in some books, it'll be written as 2 a d is equal to v f squared minus v i squared. And it seems like a super mysterious thing, but it's not that mysterious. We just very simply derived it from displacement, or if you want to say distance, if you're just thinking about the scalar quantity, is equal to average velocity times the change in time. So, so far we've just derived ourselves a kind of a neat formula that is often not derived in physics class, but let's use it to actually figure out the acceleration that a pilot experiences when they're taking off of a Nimitz class carrier. So we have 2 times the acceleration times the distance, that's 80 meters, times 80 meters, is going to be equal to our final velocity squared. What's our final velocity? 72 meters per second. So 72 meters per second, squared, minus our initial velocity. So our initial velocity in this situation is just 0. So it's just going to be minus 0 squared, which is just going to be 0, so we don't even have to write it down. And so to solve for acceleration, to solve for acceleration, you just divide, so this is the same thing as 160 meters, well, let's just divide both sides by 2 times 80, so we get acceleration is equal to 72 m/s squared over 2 times 80 meters. And what we're gonna get is, I'll just write this in one color, it's going to be 72 divided by 160, times, we have in the numerator, meters squared over seconds squared, we're squaring the units, and then we're going to be dividing by meters. So times, I'll do this in blue, times one over meters. Right? Because we have a meters in the denominator. And so what we're going to get is this meters squared divided by meters, that's going to cancel out, we're going to get meters per second squared. Which is cool because that's what acceleration should be in. And so let's just get the calculator out, to calculate this exact acceleration. So we have to take, oh sorry, this is 72 squared, let me write that down. So this is, this is going to be 72 squared, don't want to forget about this part right over here. 72 squared divided by 160. So we have, and we can just use the original number right over here that we calculated, so let's just square that, and then divide that by 160, divided by 160. And if we go to 2 significant digits, we get 33, we get our acceleration is, our acceleration is equal to 33 meters per second squared. And just to give you an idea of how much acceleration that is, is if you are in free fall over Earth, the force of gravity will be accelerating you, so g is going to be equal to 9.8 meters per second squared. So this is accelerating you 3 times more than what Earth is making you accelerate if you were to jump off of a cliff or something. So another way to think about this is that the force, and we haven't done a lot on force yet, we'll talk about this in more depth, is that this pilot would be experiencing more than 3 times the force of gravity, more than 3 g's. 3 g's would be about 30 meters per second squared, this is more than that. So an analogy for how the pilot would feel is when he's, you know, if this is the chair right here, his pilot's chair, that he's in, so this is the chair, and he's sitting on the chair, let me do my best to draw him sitting on the chair, so this is him sitting on the chair, flying the plane, and this is the pilot, the force he would feel, or while this thing is accelerating him forward at 33 meters per second squared, it would feel very much to him like if he was lying down on the surface of the planet, but he was 3 times heavier, or more than 3 times heavier. Or if he was lying down, or if you were lying down, like this, let's say this is you, this is your feet, and this is your face, this is your hands, let me draw your hands right here, and if you had essentially two more people stacked above you, roughly, I'm just giving you the general sense of it, that's how it would feel, a little bit more than two people, that squeezing sensation. So his entire body is going to feel 3 times heavier than it would if he was just laying down on the beach or something like that. So it's very very very interesting, I guess, idea, at least to me. Now the other question that we can ask ourselves is how long will it take to get catapulted off of this carrier? And if he's accelerating at 33 meters per second squared, how long would it take him to get from 0 to 72 meters per second? So after 1 second, he'll be going 33 meters per second, after 2 seconds, he'll be going 66 meters per second, so it's going to take, and so it's a little bit more than 2 seconds. So it's going to take him a little bit more than 2 seconds. And we can calculate it exactly if you take 72 meters per second, and you divide it by 33, it'll take him 2.18 seconds, roughly, to be catapulted off of that carrier." + }, + { + "Q": "What is the purpose of meiosis II? There are already cells with a diploid number of chromosomes so why not just keep them and not undergo meiosis II?", + "A": "Suppose we just say that meiosis 2 is not necessary than what will the offspring be normal for example we just have 46 chromosome than our child is going to be 92 which quite obviously may have abnormal effects on child.In case of Downs syndrome one chromosome can change the mental and physical state.We know that every cell got 46 chromosome which in each one there are genes for function many genes are involved in one function but addition of one chromosome can have large effect on cell", + "video_name": "IQJ4DBkCnco", + "transcript": "- [Voiceover] Before we go in-depth on meiosis, I want to do a very high level overview comparing mitosis to meiosis. So, in mitosis, this is all a review, if you've watched the mitosis video, in mitosis, we start with a cell, that has a diploid number of chromosomes. I'll just write 2n to show it has a diploid number. For human beings, this would be 46 chromosomes. 46 for humans, you get 23 chromosomes from your mother, 23 chromosomes from your father or you can say you have 23 homologous pairs, which leads to 46 chromosomes. Now after the process of mitosis happens and you have your cytokinesis and all the rest, you end up with two cells that each have the same genetic information as the original. So you now have two cells that each have the diploid number of chromosomes. So, 2n and 2n. And now each of these cells are just like this cell was, it can go through interphase again. It grows and it can replicate its DNA and centrosomes and grow some more then each of these can go through mitosis again. And this is actually how most of the cells in your body grow. This is how you turn from a single cell organism into you, or for the most part, into you. So that is mitosis. It's a cycle. After each of these things go through mitosis, they can then go through the entire cell cycle again. Let me write this a little bit neater. Mitosis, that s was a little bit hard to read. Now what happens in meiosis? What happens in meiosis? I'll do that over here. In meiosis, something slightly different happens and it happens in two phases. You will start with a cell that has a diploid number of chromosomes. So you will start with a cell that has a diploid number of chromosomes. And in it's interphase, it also replicates its DNA. And then it goes through something called Meiosis One. And in Meiosis One, what you end up with is two cells that now have haploid number of chromosomes. So you end up with two cells, You now have two cells that each have a haploid number of chromosomes. So you have n and you have n. So if we're talking about human beings, you have 46 chromosomes here, and now you have 23 chromosomes in this nucleus. And now you have 23 in this nucleus. But you're still not done. Then each of these will go through a phase, which I'll talk about in a second, which is very similar to mitosis, which will duplicate this entire cell into two. So actually, let me do it like this. So now, this one, you're going to have four cells that each have the haploid number that each have the haploid number of chromosomes. And they don't all necessarily have the same genetic informatioin anymore. Because as we go through this first phase, right over here of meiosis, and this first phase here you go from diploid to haploid, right over here, this is called Meiosis One. Meiosis One, you're essentially splitting the homologous pairs and so this one might get some of the ones that you originally got from your father, and some that you originally got from your mother, some that you originally got from your father, some that you originally got from your mother, they split randomly, but each homogolous pair gets split up. And then in this phase, Meiosis Two, so this phase right over here is called Meiosis Two, it's very similar to mitosis, except your now dealing with cells that start off with the haploid number. It's important to realize that meiosis is not a cycle. These cells that you have over here, these are gametes. This are sex cells. These are gametes. This can now be used in fertilization. If we're talking about, if you're male, this is happening in your testes, and these are going to be sperm cells If you are female, this is happening in your ovaries and these are going to be egg cells. If you a tree, this could be pollen or it could be an ovul. But these are used for fertilization. These will fuse together in sexual reproduction to get to a fertilized egg, which then can undergo mitosis to create an entirely new organism. So not a cycle here, although these will find sex cells from another organism and fuse with them and those can turn into another organism. And I guess the whole circle of life starts again. But it's not the case with mitosis where this can keep going and going, going. This cell is just like this cell, while these sex cells are differeent than this one right over here. Now, where does this happen in the body? We've talked about this in previous videos. These are your somatic cells right over here. These are the ones that make up the bulk of your body, somatic cells. And where is this happening? Well, this is happening in germ cells, As we mentioned, if you're male it's in your tesis and if you're female it's in your ovaries. And germ cells actually can undergo mitosis to produce other germ cells that have a diploid number of chromosomes, or they can undergo meiosis in order to produce sperm or egg cells in order to produce gametes." + }, + { + "Q": "At 0:16, How would you turn that into a linear equation?", + "A": "Good Question. first take the change of your y and x points. which is y( 7) and x(4). now we need to find the slope. to find the slope lest divide our y difference by our x difference: 7/4 now we have our slope! now so far we have Y=7/4x+b at 2:30 Sal was confirming about the dotted line. as we know we are now trying to find the y-Intercept. looking at the graph we can see that the y-Intercept is -7 so now we get!( Drum roll!) Y=7/4X+(-7) or: Y=7/4X-7 Hope This Helps! =)", + "video_name": "wl2iQAuQl7Y", + "transcript": "f is a linear function whose table of values is shown below. So they give us different values of x and what the function is for each of those x's. Which graphs show functions which are increasing at the same rate as f? So what is the rate at which f is increasing? When x increases by 4, we have our function increasing by 7. So we could just look for which of these lines are increasing at a rate of 7/4, 7 in the vertical direction every time we move 4 in the horizontal direction. And an easy way to eyeball that would actually be just to plot two points for f, and then see what that rate looks like visually. So if we see here when x is 0, f is negative 1. When x is 0, f is negative 1. So when x is 0, f is negative 1. And when x is 4, f is 6, so 1, 2, 3, 4, 5, 6, so just like that. And two points specify a line. We know that it is a linear function. You can even verify it here. When we increase by 4 again, we increase our function by 7 again. We know that these two points are on f and so we get a sense of the rate of change of f. Now, when you draw it like that, it immediately becomes pretty clear which of these has the same rate of change of f. A is increasing faster than f. C is increasing slower. A is increasing much faster than f. C is increasing slower than f. B is decreasing, so that's not even close. But D seems to have the exact same inclination, the exact same slope, as f. So D is what we would go with. And we could even verify it, even if we didn't draw it in this way. Our change in f for a given change in x is equal to-- when x changed plus 4, our function changed plus 7. It is equal to 7/4. And we can verify that on D, if we increase in the x-direction by 4, so we go from 4 to 8, then in the vertical direction we should increase by 7, so 1, 2, 3, 4, 5, 6, 7. And it, indeed, does increase at the exact same rate." + }, + { + "Q": "Can expressions also be polynomials?", + "A": "Yes, a polynomial is just an expression with certain requirements.", + "video_name": "BXHNzUaIRR0", + "transcript": "We're asked to evaluate the expression a squared plus 10b minus 8 when a is equal to 7 and b is equal to 4. So to evaluate the expression, we really just have to substitute a with 7 and substitute b with negative 4 because they're saying evaluate it when a is equal to 7 and b is equal to negative 4. So let's do that. So a everywhere we see an a in the expression, we should put a 7 there. So instead of a squared, we should write 7 squared plus-- I'll do it in that same color-- plus 10 times b. But instead of a b there, we are now going to substitute it with b is equal to negative 4. So 10 times negative 4 instead of the b right over there. And then we have the minus 8. And now we just have to evaluate this thing. 7 squared is 49. And then 10 times negative 4. Remember, order of operations, multiplication comes before addition. So we have to multiply this. 10 times negative 4 is negative 40. So it's negative 40. And then we have minus 8 back over here. And so we get 49 plus negative 40, which is really the same thing as 49 minus 40 is going to be 9. And then we're going to subtract 8 from that. And so we get 1. 49 minus 40 is 9 minus 8 is 1. And we are done. We've evaluated the expression when a is equal to 7 and b is equal to negative 4." + }, + { + "Q": "how many wars did the romans have", + "A": "Lets just say a lot they had battles from 8th century BC all the way to the 4th century and that s a long time.", + "video_name": "kiMNT18c4Ko", + "transcript": "DR. STEVEN ZUCKER: We're standing in the marvelous new museum that was just done by Richard Meier to hold the Ara Pacis, one of the most important monuments from Augustan Rome. DR. BETH HARRIS: Ara Pacis means altar of peace. Augustus was the first emperor of Rome. DR. STEVEN ZUCKER: And the person who established the Pax Romana, that is, the Roman peace. The event that prompted the building of this altar to peace under Augustus was Augustus' triumphal return from military campaigns in what is now Spain and France. DR. BETH HARRIS: And when he returned, the Senate vowed to create an altar commemorating the peace that he established in the empire. DR. STEVEN ZUCKER: And apparently, on July 4 in the year 13, the sacred precinct was marked out on which the altar itself would be built. It's really kind of wonderful because today, it's July 4, 2012. DR. BETH HARRIS: Now we're talking about the Ara Pacis, but of course, this has been reconstructed from many, many fragments that were discovered, some in the 17th century, mostly in the 20th century. DR. STEVEN ZUCKER: Actually, it's a small miracle that we've been able to reconstruct this at all. It had been lost to memory. DR. BETH HARRIS: The remains of it lay under someone's palace. When it was recognized what these fragments were, it became really important to excavate them and to reconstruct the altar. DR. STEVEN ZUCKER: That was finally done under Mussolini, the fascist leader in the years leading up to the Second World War, and during the Second World War. And that was important to Mussolini, because Mussolini identified himself with Augustus, the first emperor of Rome. Mussolini was very much trying to reestablish a kind of Italian empire. We should talk a little bit about what an altar is. We talk about the altar, really what we're looking at are the walls of the precinct around what is in the middle, the altar where sacrifices would have occurred. is interesting and important when we think about Augustus. Augustus is establishing a centralized power. Rome had been, since its earliest founding years when it was under the rule of kings, it had been controlled by the Senate. It had been a republic. DR. BETH HARRIS: That's right. And the Senate was basically a group of the leading elder citizens of Rome. So Rome was a republic, and it really was a republic until Julius Caesar, who was a dictator and Augustus' uncle. And then Caesar is assassinated, there's civil war, and then peace is established by Augustus. DR. STEVEN ZUCKER: Right. Augustus, whose real name was Octavian, is given the term Augustus as a kind of honorific as a way of representing his power. And it's interesting the kind of politics that Augustus involved himself with. He gave great power back to the Senate, but by doing so, he established real and central authority for himself. DR. BETH HARRIS: He made himself princeps, or first among But of course he controlled everything. DR. STEVEN ZUCKER: He also held the title of pontificus maximus, that is, the head priest of the state religion, and so he held tremendous power. DR. BETH HARRIS: Now don't forget, too, that his uncle Julius Caesar had been made a god, and so he also represented himself as the son of a god. DR. STEVEN ZUCKER: And so the idea of establishing this altar has a political as well as spiritual significance. DR. BETH HARRIS: He's looking back to the golden age of Greece of the fifth century BC, but he's also looking back to the Roman republic. He is reestablishing some of the ancient rituals of traditional Roman religion. He is embracing traditional Roman values. DR. STEVEN ZUCKER: But even as he's doing that, he's remaking Rome radically. He's changing Rome from a city of brick to a city of marble, and the Ara Pacis is a spectacular example of that. DR. BETH HARRIS: And when we look closely at the Ara Pacis, what we're going to see is that this speaks to the sense of a golden age that Augustus brought about in the Roman Empire. DR. STEVEN ZUCKER: One of the most remarkable elements of the Ara Pacis is all of the highly decorative relief carving in the lower frieze. DR. BETH HARRIS: And that goes all the way around. It apparently shows more than 50 different species of plants. They're very natural in that we can identify the species, but they're also highly abstracted, and they form these beautiful symmetrical and linear patterns. DR. STEVEN ZUCKER: There is a real order that's given to the complexity of nature here. Let me just describe quickly what I'm seeing. This massive, elegant acanthus leaf, which is a native plant, which were made famous in Corinthian capitals. And then almost like a candelabra growing up from it, we see these tendrils of all kinds of plants that spiral. DR. BETH HARRIS: And there are also animal forms within these leaves and plants. We find frogs and lizards and birds. DR. STEVEN ZUCKER: And the carving is quite deep, so that there is this sharp contrast between the brilliance of the external marble and then the shadows that are cast as it seems to lift off the surface. DR. BETH HARRIS: And art historians interpret all of this as a symbol of fertility, of the abundance of the golden age that Augustus brought about. DR. STEVEN ZUCKER: You also see that same pattern repeated in the pilasters that frame these panels. And then we also have meander that moves horizontally around the entire exterior. And it's above that meander that we see the narrative phrases. DR. BETH HARRIS: These panels relate again to this golden age that Augustus establishes. These refer back to Aeneas, Rome's founder and Augustus' ancestor. We see other allegorical figures representing Rome and peace. DR. STEVEN ZUCKER: We have to be a little bit careful when we try to characterize what precisely is being represented. There are lots of conflicting interpretations. DR. BETH HARRIS: And these allegorical or mythological scenes appear on the front and back of the altar. And then on the sides of the altar we see a procession. DR. STEVEN ZUCKER: We've walked around the outer wall, and we're now looking at a panel that's actually in quite good condition. But that doesn't mean we really know what's going on. DR. BETH HARRIS: No, there's a lot of argument about what the figure in the center represents. Some art historians think this figure represents Venus, some think it represents a figure of peace, some the figure of Tellus, or Mother Earth. In any case, she is clearly a figure that suggests fertility and abundance. DR. STEVEN ZUCKER: She's beautifully rendered. Look at the way the drapery clings to her torso so closely as to really review the flesh underneath, like the goddesses on the Parthenon on the Acropolis in Greece. DR. BETH HARRIS: And on her lap sit two children, one of whom offers her some fruit. There's fruit on her lap. On either side of her sit two mythological figures who art historians think represent the winds of the earth and the sea. DR. STEVEN ZUCKER: Well, look at the way the drapes that they're holding whip up, creating these beautiful almost halos around their bodies. DR. BETH HARRIS: And at her feet we see an ox and a sheep. So there's a sense of harmony, of peace and fertility. DR. STEVEN ZUCKER: And that must have been such a rare thing in the ancient world. DR. BETH HARRIS: Well, Augustus reigns after decades of civil war after the assassination of Julius Caesar. So I think there is a powerful sense that this was the golden age. DR. STEVEN ZUCKER: So, let's walk to the sides now, and take a look at the procession. The frieze moves from the back wall of the precinct up towards the very front on both sides, and the figures are also facing towards the main staircase. DR. BETH HARRIS: Art historians are not really clear what event is being depicted here. DR. STEVEN ZUCKER: Art historians aren't clear about any of this, are we? DR. BETH HARRIS: No. There are a couple of possibilities that have been raised. One is that what we are seeing is the procession that would have taken place at the time that the altar was inaugurated. The figures that we see here are priests, and we can identify those figures because of the veils on their heads, and there also seem to be members of Augustus' family, although their identities are not quite firmly established. DR. STEVEN ZUCKER: We think we know which figure is Augustus, although the marble itself is not in especially good condition, and we've lost the front of his body. And we also think we can identify one of his most important ministers. DR. BETH HARRIS: And that would be Agrippa. If we think about this as looking back to the frieze on the Parthenon from the golden age of Greece, those figures are all ideally beautiful. They don't represent anyone specific so much as the Athenian people generally. DR. STEVEN ZUCKER: But these are portraits. DR. BETH HARRIS: That's right. And we can't always identify them for certain, but they really are specific individuals on a specific date taking part in a specific event. DR. STEVEN ZUCKER: It's interesting to think about it, because of course throughout the republic, portraiture in stone was something that the Romans were extremely good at. And so it doesn't surprise me that they would not look to the idealized so much as look to the specific. DR. BETH HARRIS: We also notice those differences in the depths of the carving. Some figures are represented in high relief. Other figures that are supposed to be in the background are represented in low relief. So there's a real illusion of space and of a crowd, here at the procession. DR. STEVEN ZUCKER: Another way that the specificity of the Romans is expressed is through the inclusion of children. This is a sacred event, and a formal event. And yet there are children doing what children do. That is to say, they're not always paying attention. DR. BETH HARRIS: There are a couple of interpretations that have been offered about the presence of children here. Augustus was actually worried about the birth rate and passed laws that encouraged marriage and the birth of children. It originally was painted. We would have seen pinks and blues and greens, and it's very difficult to imagine that when we look at the marble today. DR. STEVEN ZUCKER: Well, it's true, especially in Meier's building, which is so stark and modern. It's almost a little garish to imagine how brightly painted this would have been. They were pretty bright. DR. BETH HARRIS: They were. So one of the things that Augustus said of himself was that he found Rome a city of brick, and he left it a city of marble. Augustus created an imperial city. And here we are 2,000 years later in the room that Augustus created." + }, + { + "Q": "so you just mutiply the fractions right because he was just checking it.", + "A": "He was trying to show what your are doing by multiplying the fraction.", + "video_name": "hr_mTd-oJ-M", + "transcript": "Let's think a little bit about what it means to multiply fractions. Say I want to multiply 1/2 times 1/4. Well, one way to think about this is we could view this as 1/2 of a 1/4. And what do I mean there? Well let me take a whole, let me take a whole here, and let me divide it into fourths. So let me divide it into fourths, so I'll divided into 4 equal sections. And so 1/4 would be 1 of these 4 equal sections. But we want to take 1/2 of that. So how do we take half of that? Well, we could divide this into 2 equal sections, and then just take 1 of them. So divide it into 2 equal sections, and then take 1 of them. So we're taking this pink area, this whole pink area is 1/4, and now we're going to take 1/2 of it. We're now going to take 1/2 of it. So that's this yellow square right over here. But what fraction of the whole does this yellow represent? Well, it now represents 1 out of 1, 2, 3, 4, 5, 6, 7, 8 equal sections. So this right over here, this represents 1/8 of the whole. And so we see conceptually that 1/2 times 1/4, it completely makes sense, that 1/2 of 1/4 should be 1/8. And it hopefully makes sense that you get this 8 by multiplying the 2 times the 4. You started with 4 equal sections, but then you divided each of those 4 equal sections into 2 equal sections. So then you have 8 total equal sections that you split your whole into. Let's do another example, but now let's multiply two fractions that don't have 1's in the numerator. So let's multiply, let's multiply 2/3 times 4/5. And I encourage you now to pause the video and do something very similar to what I just did. Try to represent 4/5 of a whole and then try to represent 2/3 of that 4/5 and see what fraction of the whole you actually have. So pause now. So let's think about this. Let's represent 4/5. So if I have a whole like this, let me try to divide it into 5 equal sections. 5 equal sections, so let's say that is 1 equal section, that is 2 equal sections, that is 3, 4, and 5-- I can do a better This is always the hard part. I'm trying my best to make them look, at least, like equal sections-- 2, 3, 4, and 5. I think you get the point here. I'm trying to make them equal sections. And we want 4/5. So we want 4 of these 5 equal sections. So this would be 1 of the 5 equal sections, 2 of them, 3 of them, and then 4 of them. So that right over there is 4/5. Now we can view this as 2/3 of the 4/5. So how can we think about that? Well, we could take this section and divide it into thirds. So let's do that. Divide it into thirds. So we're going divide it into 3 equal sections. So that's 1/3, and then 2/3. So we took each of the 5 equal sections, and we divided them into 3 equal sections. Now what's going to be 2/3 of the 4/5? Well, that's going to be this part right over here. So let me make this clear. This is 1/3 of the 4/5. And then this would be 2/3 of the 4/5. So this right over here, would be 2/3 of the 4/5, or 2/3 times 4/5. But what fraction of the whole does that represent? Well, how many total, how many total equal sections do we now have? Well, we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. So we have 15 equal sections. I'm using a new color. We have 15 equal sections, and that make sense. We started with 5 equal sections, but then we divided each of those into 3 equal sections. So now we have 5 times 3 total equal sections. And then how many of those are now colored in? Well, we see it's 2 times 4. 1, 2, 3, 4, 5, 6, 7, 8. How many of them are in the 2/3 of the 4/5, I should say. And there's 8 of them, 8 of the 15 equals sections. And so there you have it. It should hopefully now make visual sense, or it makes conceptual sense, that 2/3 times 4/5-- you can obviously compute it by just multiplying the numerators, 2 times 4 is 8. And then multiplying the denominators, 3 times 5 is 15-- but hopefully this now makes conceptual sense as 2/3 of 4/5." + }, + { + "Q": "What if he stated \"There were sheep on the hill\". Would the statement still be correct?", + "A": "That is a correct statement. Maybe you re getting confused because the singular and plural form of sheep is the same. It can either mean one sheep or more than one sheep depending on the context. In your sentence, we can tell more than one sheep are on the hill.", + "video_name": "UnJmPywSSvg", + "transcript": "- [Voiceover] Hello grammarians. I wanted to talk today about a different kind of irregular plural. So we've been talking about regular plurals where you take a word and you add an S. So for example, the word dog becomes dogs. You add an S. And this is the regular plural here. But I've been talking about the irregular plural, the plural, the multiple form of a verb that is not regular, irregular. But today I figured we'd talk about something called the base plural. Which I will illustrate for you using our friend the sheep. Now, sheep is a very strange word in that it doesn't matter whether not there's more than one of them. The form of the word always looks the same whether it's one sheep or two sheep. It's an irregular plural you don't add an S to. This is called a base plural 'cause the base sheep, the thing that you would normally add this particle S to doesn't change whether it's singular sheep or plural sheep. So that's you know. There was one sheep... on the hill. There's a sentence. What if we put another little baby sheep on that hill? A little lamb. Well now the sentence looks like this. Two sheep... on the hill. Now the only difference between these two sentences is that there's one sheep and two sheep and therefore that means that the verb changes to a plural conjugation. So there was one sheep. There were two sheep on the hill. But everything else stays exactly the same. One sheep, two sheep. This is very strange, it's a base plural. So in standard English, the form is two sheep... and not two sheeps. Now, there are more words that do behave this way. So let's go investigate. So there are a small number of words that also behave this way, the way sheep does, these weird sheep plurals, these base plurals. One of them is fish. So you could say the fish are plentiful this season, but you could also say, you know, the fish... is delicious. You could say the bison migrate west or you could also say the bison migrates west indicating a single bison, you see. Bison can be singular or plural. Fish can be singular or plural. As is so frequently the case, there is a special exception regarding the word fishes, which you may have heard before, and fishes is a word that we would use when we're talking about individual species of fish. And fish is the word that we would use to refer to individual fish. So let's say your uncle Marty is a prodigious fisherman and he catches, he goes fly fishing one weekend, he comes back, he has 30 fish. Marty caught... 30 fish. But let's say on the other hand your aunt Marta is a prodigious marine biologist and she studies 30 different types of fish. You would say Marta studies... 30 fishes. And that doesn't mean that she studies 30 individual fish. That means she studies 30 types of fish. That's the difference. Fishes is referring to species. Fish refers to individuals. That's how you'd use them in the plural. So to review, there's this entire class of words called base plurals where the word itself, the base, doesn't take an S for the plural, it's just the same. The singular is the same as the plural. So that gives us words like sheep, fish, and bison. There aren't a ton of English words that behave this way where the plural is the same as the singular. I just wanted to make you aware of some of the most common ones. There are also more examples in the exercises. So I just wanted you to be aware of them. You can learn anything. David out." + }, + { + "Q": "If both C's are \"equally\" substituted but one has an ethyl and the other a methyl, which will the H2O attack?", + "A": "You can consider the Br\u00e2\u0081\u00ba in the cyclic bromonium ion to be an enhanced leaving group. So the reaction will be similar to an SN1 reaction, and the carbon that forms the more stable carbocation will be the one that is preferentially attacked. An ethyl group provides a little more electron density and stabilization than a methyl group (for example, propionic acid is weaker than acetic acid). You will get attack on both carbons, but the major product will have the OH on the carbon with the ethyl group.", + "video_name": "FaOOx6IZxV8", + "transcript": "Here's the general reaction to make halohydrins from alkenes. So if I start with my alkene on the left, and I add a halogen to it and some water, you can see that an OH and a halogen are added anti to each other. So anti, or on opposite sides of where the double bond used to be. The mechanism for this reaction starts off the exact same way the halogenation reaction did. And so we have our halogen approaching our alkene. And we saw in that video that the halogen is usually, of course, nonpolar, because those two atoms have the exact same electronegativity. So these blue electrons in here are pulled with equal force to either halogen, so it's a nonpolar molecule. However, if the pi electrons in my alkene here get too close to the electrons in blue, we saw how that could induce a temporary dipole on the halogen molecule. So those electrons in blue are repelled by the electrons in magenta and push closer to the top halogen, which gives the top halogen a partial negative charge and leaves the bottom halogen with a partial positive charge. The bottom halogen is now an electrophile, so it wants electrons. It's going to get electrons from those pi electrons here, which are going to move out and nucleophilic attack that partially positively charged halogen atom. And then this lone pair of electrons is going to form a bond with this carbon at the same time these blue electrons move out onto the halogen. So when we draw the result of all those electrons moving around, we're going to form a bond between the carbon on the right and the halogen, and we use the magenta electrons to show that. So there's now a bond there. And so we used red electrons before to show these electrons in here forming a bond with the carbon on the left. That halogen had two lone pairs of electrons still on it, like that, which gives that halogen a plus 1 formal charge. We called this our cyclic halonium ion in an earlier video. And if I think about that cyclic halonium ion, I think about the halogen being very electronegative. It's going to attract, I'll say the electrons in magenta again just to be consistent, closer towards it. So it's going to take away a little bit of electron density from this carbon right down here. So I'm going to say this carbon is partially positive. It's going to have some partial carbocationic character. So in the next step of the mechanism, water's going to come along. And water's going to function as a nucleophile. So one of the lone pairs of electrons on water is going to nucleophilic attack our electrophile, which is this carbon right here. And so when that lone pair of electrons on oxygen attacks this carbon, that's going to kick the electrons in magenta off onto your halogen. And so let's go ahead and draw the product. We're going to have, on the left carbon, this halogen now used to have two lone pairs of electrons. It picked up the ones in magenta, so now it looks like that. On the right, we still have the carbon on the right bonded to other things, except now it's bonded to what used to be our water molecule. So the oxygen is now bonded to the carbon. And there's still one lone pair of electrons on that oxygen, giving it a plus 1 formal charge. So let's go ahead and highlight these electrons here in blue. Those electrons in blue are the ones that formed this bond between the carbon and the oxygen. So we're almost done. The last step of the mechanism would just be an acid-base reaction. So another water molecule comes along, and one of the lone pairs of electrons on the water molecule is going to function as a base and take this proton, leaving these two electrons behind on the oxygen. And we are finally done. We have formed our halohydrin, right? So I have my halogen on one side. And then I now have my OH on the opposite side, like that. Let's go ahead and do an example so we can examine the stereochemistry a little bit more here. So if I start with an alkene, and to this alkene we are going to add bromine and water. And we're going to think about doing this two different ways here. So we'll start with the way on the left. So Br2 and H2O. And then we'll come back and we'll go ahead and do this on the right. So BR2 and H2O. So on the left side, I'm going to think about the formation of that bromonium ion here. So I'm going to once again look at this molecule a little bit from above, so looking down. And I'm going to say the bromonium ion is going to form this way. So the bromine's going to form on top here. And so there's going to be two lone pairs of electrons It's going to have a plus 1 formal charge. And if I look at this carbon right here, that's this carbon. So I'm going to say that my methyl group is now going down in space. So with the addition of my bromonium ion, that would be my intermediate. And so now, when I think about water coming along and acting as a nucleophile, so here is H2O, and I think about which carbon will the oxygen attack? So I have two options, right? This oxygen could attack the carbon on the left, or it could attack the carbon on the right. It's been proven that the option is going to attack the most substituted carbon. So if I look at the carbon on the left, and if I think about what sort of carbocation would that be, the carbon on the left is bonded to two other carbons. So this would be similar to a secondary carbocation, or a partial carbocation in character. So you could think about it as being like a partial secondary carbocation on the left here, if this was a partial positive. Or on the right, if I think about this carbon right here, the one in red. And if I think about that being a carbocation, that would be bonded to one, two, three other carbons. So it's like a tertiary carbocation. And we know that tertiary carbocations are more stable than secondary. So even though this isn't a full carbocation, this carbon in red exhibits some partial carbocation character, and that is where our water is going to attack. So the nucleophile is going to attack the electrophile. And it's more stable for it to attack this one on the right, since it has partial carbocation character similar to a tertiary carbocation. And if it attacks that carbon on the right, these electrons here we kick off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. OK, so what would we have here? Let's go ahead and draw our ring. And the bromine is going to swing over to the carbon on the left. It's now going to have three lone pairs of electrons around it, like that. And the methyl group that was down relative to the plane is going to be pushed up when that water nucleophilic attacks. And now the methyl group is up, and this oxygen is now going to be bonded to this carbon. And so we still have our two hydrogens attached to it, like that. And there's a lone pair of electrons on this oxygen, giving it a plus 1 formal charge. So once again, let's go ahead and highlight those electrons. I'll draw them in blue here. These electrons right here, those are the ones that formed this bond. So let's go back and let's think about the formation of that cyclic bromonium ion in a different way here. So on the left I showed the bromine adding from the top. If I think about the alkene portion of my starting material, well, there's a chance that the bromonium ion could form from below that plane as well. So let's go ahead and draw the result of that over here on the right. So I'm showing another bromonium ion that is possible. And this time the bromine is going to add from below the plane, like that. It's going to get two lone pairs of electrons and have a plus 1 formal charge just like usual. And I will say that this is the carbon that has the methyl on it, OK? So let's go ahead and draw that in as well. So let's see, I'll put it like that. So that's my CH3. And this time, when I think about where will water attack-- so let's go ahead and think about water as my nucleophile-- it's the same idea if I compare the carbons on either side. So I compare this carbon with this carbon. It's the carbon on the right that's going to be the most stable partial carbocation, right? So I'll draw a partial carbocation here. It would be the most stable one. So that's where my nucleophile is going to attack. So I can think about this lone pair of electrons on oxygen that are going to attack right here, which would kick these electrons off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. So let's see what that would look like. So I have my ring like that. And now I'm going to have my oxygen bonded here. It still has two hydrogens attached to it. It has a lone pair of electrons on it, which give it a plus 1 formal charge. And when the oxygen attacked, that is going to push down this methyl group. So this methyl group is going to be pushed down relative to the plane of the ring. So now we have a methyl group down at this carbon. And the bromine is going to swing over to the carbon on the left. And so that's the position of my bromine now. And so, in the last step-- now I have these two guys right here-- they're both going to lose a proton in the next step, right? So it's an acid-base reaction. And we could show water coming along. So for the molecule on the left, water comes along. And water's going to take this proton. These electrons are going to kick off on to that oxygen. And we can draw that product. So if we were to draw that product, we would look down this way. And we would treat this as being the top carbon here. So there's a methyl group going up at that carbon. So I can say that this is going to, after it loses a proton, so that carbon in blue is going to have a methyl group up relative to the plane of the ring. And it's going to have an OH group down. So this OH is going to be down relative to the plane. So I can go ahead and put OH going away from me in space. And then this bromine over here. This bromine is going to be coming out at me. So I can go ahead and show a wedge for that bromine. So that's one of our possible products. Over here on the right, if this is what happens on the right, I do the same thing. I put my eye right here and I stare down, with this carbon being the top carbon. So I can go ahead and draw my cyclohexane ring. And I can see that this time my OH will be coming out at me after it loses a proton. So if I wanted to, I could go ahead and draw water in here and show the last step of the mechanism. Lone pair of electrons taking this proton, leaving these electrons behind, giving me an OH coming out at me in space. And then this methyl group would therefore be going away from me in space. So I can go ahead and show that methyl group as a dash. And then finally, this bromine over here would be going away from me. So that would be a dash on my ring, like that. When I finally get to my two products, I can analyze them. And I can see that they are enantiomers to each other. So they're different molecules. And we can look at the absolute configurations really fast. I can see that I have a carbon coming out at me on a wedge and an oxygen going away from me. And it's been reversed over here on the right. This time the oxygen is coming out at me on a wedge, and the carbon is going away from me on a dash. When I look at the bromine, it's coming out at me on the left, and the bromine's going away from me on the right. So I can see that I have different absolute configurations at both chirality centers. And so these two will be enantiomers to each other." + }, + { + "Q": "What is the relationship between acetyl CoA and the intermediates in the Krebs cycle? What are the effects of depleted intermediates on the oxydation of acetyl CoA?", + "A": "two answer the first question, Acetyl CoA donates an acetyl group to oxaloacetate to make citrate, an intermediate in the Krebs cycle. Since it is a cycle, I presume that depleted intermediates would eventually lead to a lack of oxaloacetate to oxidise acetyl CoA, leading to a buildup of the unoxidised molecule. Hope that helps! Ryan, 2nd biochemistry", + "video_name": "juM2ROSLWfw", + "transcript": "So we already know that if we start off with a glucose molecule, which is a 6-carbon molecule, that this essentially gets split in half by glycolysis and we end up 2 pyruvic acids or two pyruvate molecules. So glycolysis literally splits this in half. It lyses the glucose. We end up with two pyruvates or pyruvic acids. ruby And these are 3-carbon molecules. There's obviously a lot of other stuff going on in the carbons. You saw it in the past. And you could look up their chemical structures on the internet or on Wikipedia and But this is kind of the important thing. Is that it was lysed, it was cut in half. And this is what happened in glycolysis. And this happened in the absence of oxygen. Or not necessarily. It can happen in the presence or in the absence of oxygen. It doesn't need oxygen. And we got a net payoff of two ATPs. And I always say the net there, because remember, it used two ATPs in that investment stage, and then it generated four. So on a net basis, it generated four, used two, it gave us two ATPs. And it also produced two NADHs. That's what we got out of glycolysis. And just so you can visualize this a little bit better, let me draw a cell right here. Maybe I'll draw it down here. Let's say I have a cell. That's its outer membrane. Maybe its nucleus, we're dealing with a eukaryotic cell. That doesn't have to be the case. It has its DNA and its chromatin form all spread around like that. And then you have mitochondria. And there's a reason why people call it the power We'll look at that in a second. So there's a mitochondria. It has an outer membrane and an inner membrane just like that. I'll do more detail on the structure of a mitochondria, maybe later in this video or maybe I'll do a whole video on them. That's another mitochondria right there. And then all of this fluid, this space out here that's between the organelles-- and the organelles, you kind of view them as parts of the cell that do specific things. Kind of like organs do specific things within our own bodies. So this-- so between all of the organelles you have this fluidic space. This is just fluid of the cell. And that's called the cytoplasm. And that's where glycolysis occurs. So glycolysis occurs in the cytoplasm. Now we all know-- in the overview video-- we know what the next step is. The Krebs cycle, or the citric acid cycle. And that actually takes place in the inner membrane, or I should say the inner space of these mitochondria. Let me draw it a little bit bigger. Let me draw a mitochondria here. So this is a mitochondria. It has an outer membrane. It has an inner membrane. If I have just one inner membrane we call it a crista. If we have many, we call them cristae. This little convoluted inner membrane, let me give it a label. So they are cristae, plural. And then it has two compartments. Because it's divided by these two membranes. This compartment right here is called the outer compartment. This whole thing right there, that's the outer compartment. And then this inner compartment in here, is called the matrix. Now you have these pyruvates, they're not quite just ready for the Krebs cycle, but I guess-- well that's a good intro into how do you make them ready for the Krebs cycle? They actually get oxidized. And I'll just focus on one of these pyruvates. We just have to remember that the pyruvate, that this happens twice for every molecule of glucose. So we have this kind of preparation step for the Krebs Cycle. We call that pyruvate oxidation. And essentially what it does is it cleaves one of these carbons off of the pyruvate. And so you end up with a 2-carbon compound. You don't have just two carbons, but its backbone of carbons is just two carbons. Called acetyl-CoA. And if these names are confusing, because what is acetyl coenzyme A? These are very bizarre. You could do a web search on them But I'm just going to use the words right now, because it will keep things simple and we'llget the big picture. So it generates acetyl-CoA, which is this 2-carbon compound. And it also reduces some NAD plus to NADH. And this process right here is often given credit-- or the Krebs cycle or the citric acid cycle gets credit for this step. But it's really a preparation step for the Krebs cycle. Now once you have this 2-carbon chain, acetyl-Co-A right here. you are ready to jump into the Krebs cycle. This long talked-about Krebs cycle. And you'll see in a second why it's called a cycle. Acetyl-CoA, and all of this is catalyzed by enzymes. And enzymes are just proteins that bring together the constituent things that need to react in the right way so that they do react. So catalyzed by enzymes. This acetyl-CoA merges with some oxaloacetic acid. A very fancy word. But this is a 4-carbon molecule. These two guys are kind of reacted together, or merged together, depending on how you want to view it. I'll draw it like that. It's all catalyzed by enzymes. And this is important. Some texts will say, is this an enzyme catalyzed reaction? Everything in the Krebs cycle is an enzyme catalyzed reaction. And they form citrate, or citric acid. Which is the same stuff in your lemonade or your orange juice. And this is a 6-carbon molecule. You have a 2-carbon and a 4-carbon. You get a 6-carbon molecule. And then the citric acid is then oxidized over a bunch of steps. And this is a huge simplification here. But it's just oxidized over a bunch of steps. Again, the carbons are cleaved off. Both 2-carbons are cleaved off of it to get back to oxaloacetic acid. And you might be saying, when these carbons are cleaved off, like when this carbon is cleaved off, what happens to it? It becomes CO2. It gets put onto some oxygen and leaves the system. So this is where the oxygen or the carbons, or the carbon dioxide actually gets formed. And similarly, when these carbons get cleaved off, it forms CO2. And actually, for every molecule of glucose you have six carbons. When you do this whole process once, you are generating three molecules of carbon dioxide. But you're going to do it twice. You're going to have six carbon dioxides produced. Which accounts for all of the carbons. You get rid of three carbons for every turn of this. Well, two for every turn. But really, for the steps after glycolysis you get rid of three carbons. But you're going to do it for each of the pyruvates. You're going to get rid of all six carbons, which will have to exhale eventually. But this cycle, it doesn't just generate carbons. The whole idea is to generate NADHs and FADH2s and ATPs. So we'll write that here. And this is a huge simplification. I'll show you the detailed picture in a second. We'll reduce some NAD plus into NADH. We'll do it again. And of course, these are in separate steps. There's intermediate compounds. I'll show you those in a second. Another NAD plus molecule will be reduced to NADH. It will produce some ATP. Some ADP will turn into ATP. Maybe we have some-- and not maybe, this is what happens-- some FAD gets-- let me write it this way-- some FAD gets oxidized into FADH2. And the whole reason why we even pay attention to these, you might think, hey cellular respiration is all about ATP. Why do we even pay attention to these NADHs and these FADH2s that get produced as part of the process? The reason why we care is that these are the inputs into the electron transport chain. These get oxidized, or they lose their hydrogens in the electron transport chain, and that's where the bulk of the ATP is actually produced. And then maybe we'll have another NAD get reduced, or gain in hydrogen. Reduction is gaining an electron. Or gaining a hydrogen whose electron you can hog. NADH. And then we end up back at oxaloacetic acid. And we can perform the whole citric acid cycle over again. So now that we've written it all out, let's account for what we have. So depending on-- let me draw some dividing lines so we know what's what. So this right here, everything to the left of that line right there is glycolysis. We learned that already. And then most-- especially introductory-- textbooks will give the Krebs cycle credit for this pyruvate oxidation, but that's really a preparatory stage. The Krebs cycle is really formally this part where you start with acetyl-CoA, you merge it with oxaloacetic acid. And then you go and you form citric acid, which essentially gets oxidized and produces all of these things that will need to either directly produce ATP or will do it indirectly in the electron transport chain. But let's account for everything that we have. Let's account for everything that we have so far. We already accounted for the glycolysis right there. Two net ATPs, two NADHs. Now, in the citric acid cycle, or in the Krebs cycle, well first we have our pyruvate oxidation. That produced one NADH. But remember, if we want to say, what are we producing for every glucose? This is what we produced for each of the pyruvates. This NADH was from just this pyruvate. But glycolysis produced two pyruvates. So everything after this, we're going to multiply by two for every molecule of glucose. So I'll say, for the pyruvate oxidation times two means that we got two NADHs. And then when we look at this side, the formal Krebs cycle, what do we get? We have, how many NADHs? One, two, three NADHs. So three NADHs times two, because we're going to perform this cycle for each of the pyruvates produced from glycolysis. So that gives us six NADHs. We have one ATP per turn of the cycle. That's going to happen twice. Once for each pyruvic acid. So we get two ATPs. And then we have one FADH2. But it's good, we're going to do this cycle twice. This is per cycle. So times two. We have two FADHs. Now, sometimes in a lot of books these two NADHs, or per turn of the Krebs cycle, or per pyruvate this one NADH, they'll give credit to the Krebs cycle for that. So sometimes instead of having this intermediate step, they'll just write four NADHs right here. And you'll do it twice. Once for each puruvate. So they'll say eight NADHs get produced from the Krebs cycle. But the reality is, six from the Krebs cycle two from the preparatory stage. Now the interesting thing is we can account whether we get to the 38 ATPs promised by cellular respiration. We've directly already produced, for every molecule of glucose, two ATPs and then two more ATPs. So we have four ATPs. Four ATPs. How many NADHs do we have? 2, 4, and then 4 plus 6 10. We have 10 NADHs. And then we have 2 FADH2s. I think in the first video on cellular respiration I said FADH. It should be FADH2, just to be particular about things. And these, so you might say, hey, where are our 38 ATPs? We only have four ATPs right now. But these are actually the inputs in the electron transport chain. These molecules right here get oxidized in the electron transport chain. Every NADH in the electron transport chain produces three ATPs. So these 10 NADHs are going to produce 30 ATPs in the electron transport chain. And each FADH2, when it gets oxidized and gets turned back into FAD in the electron transport chain, will produce two ATPs. So two of them are going to produce four ATPs in the electron transport chain. So we now see, we get four from just what Glycolysis, the preparatory stage and the Krebs or citric acid cycle. And then eventually, these outputs from glycolysis and the citric acid cycle, when they get into the electron transport chain, are going to produce another 34. So 34 plus 4, it does get us to the promised 38 ATP that you would expect in a super-efficient cell. This is kind of your theoretical maximum. In most cells they really don't get quite there. But this is a good number to know if you're going to take the AP bio test or in most introductory biology courses. There's one other point I want to make here. Everything we've talked about so far, this is carbohydrate metabolism. Or sugar catabolism, we could call it. We're breaking down sugars to produce ATP. Glucose was our starting point. But animals, including us, we can catabolize other things. We can catabolize proteins. We can catabolize fats. If you have any fat on your body, you have energy. In theory, your body should be able to take that fat and you should be able to do things with that. You should be able to generate ATP. And the interesting thing, the reason why I bring it up here, is obviously glycolysis is of no use to these things. Although fats can be turned into glucose in the liver. But the interesting thing is that the Krebs cycle is the entry point for these other catabolic mechanisms. Proteins can be broken down into amino acids, which can be broken down into acetyl-CoA. Fats can be turned into glucose, which actually could then go the whole cellular respiration. But the big picture here is acetyl-CoA is the general catabolic intermediary that can then enter the Krebs cycle and generate ATP regardless of whether our fuel is carbohydrates, sugars, proteins or fats. Now, we have a good sense of how everything works out right now, I think. Now I'm going to show you a diagram that you might see in your biology textbook. Or I'll actually show you the actual diagram from Wikipedia. I just want to show you, this looks very daunting and very confusing. And I think that's why many of us have trouble with cellular respiration initially. Because there's just so much information. It's hard to process what's important. But I want to just highlight the important steps here. Just so you see it's the same thing that we talked about. From glycolysis you produce two pyruvates. That's the pyruvate right there. They actually show its molecular structure. This is the pyruvate oxidation step that I talked about. The preparatory step. And you see we produce a carbon dioxide. And we reduce NAD plus into NADH. Then we're ready to enter the Krebs cycle. The acetyl-CoA and the oxaloacetate or oxaloacetic acid, they are reacted together to create citric acid. They've actually drawn the molecule there. And then the citric acid is oxidized through the Krebs cycle right there. All of these steps, each of these steps are facilitated by enzymes. And it gets oxidized. But I want to highlight the interesting parts. Here we have an NAD get reduced to NADH. We have another NAD get reduced to NADH. And then over here, another NAD gets reduced to NADH. So, so far, if you include the preparatory step, we've had four NADHs formed, three directly from the Krebs cycle. That's just what I told you. Now we have, in this diagram they say GDP. GTP gets formed from GDP. The GTP is just guanosine triphosphate. It's another purine that can be a source of energy. But then that later can be used to form an ATP. So this is just the way they happen to draw it. But this is the actual ATP that I drew in the diagram on the top. And then they have this Q group. And I won't go into it. And then it gets reduced over here. It gets those two hydrogens. But that essentially ends up reducing the FADH2s. So this is where the FADH2 gets produced. So as promised, we produced, for each pyruvate that inputted-- remember, so we're going to do it twice-- for each pyruvate we produced one, two, three, four NADHs. We produced one ATP and one FADH2. That's exactly what we saw up here. I'll see you in the next video." + }, + { + "Q": "Why do you need to have one side of the equation equal 0 to solve the equation?", + "A": "The Zero Factor Principle tells you that at least one of the factors must equal zero in your equation. So if one side equals zero, you can then make both sides zero because that makes it easy and quicker to solve.", + "video_name": "STcsaKuW-24", + "transcript": "The height of a triangle is four inches less than the length of the base. The area of the triangle is 30 inches squared. Find the height and base. Use the formula area equals one half base times height for the area of a triangle. OK. So let's think about it a little bit. We have the-- let me draw a triangle here. So this is our triangle. And let's say that the length of this bottom side, that's the base, let's call that b. And then this is the height. This is the height right over here. And then the area is equal to one half base times height. Now in this first sentence they tell us at the height of a triangle is four inch is less than the length of the base. So the height is equal to the base minus 4. That's what that first sentence tells us. The area of the triangle is 30 inches squared. So if we take one half the base times the height we'll get 30 inches squared. Or we could say that 30 inches squared is equal to one half times the base, times the height. Now instead of putting an h in for height, we know that the height is the same thing as 4 less than the base. So let's put that in there. 4 less than the base. And then let's see what we get here. We get-- let me do this in yellow. We get 30 is equal to one half times-- let's distribute the b-- times b, let me make it clear. So let's do it this way. Times b over 2, times b, minus 4. I just multiplied the one half times the b. Now let's distribute the b over 2. So 30 is equal to b squared over 2, be careful. b over 2 times b is just b squared over 2. And then b over 2, times negative 4 is negative 2b. Now just to get rid of this fraction here let's multiply both sides of this equation by 2. So let's multiply that side by 2. And let's multiply that side by 2. On the left hand side you get 60. On the right hand side 2 times b squared over 2 is just b squared. Negative 2b times 2 is negative 4b. And now we have a quadratic here. And the best way to solve a quadratic-- we have a second degree term right here-- is to get all of the terms on one side of the equation, having them equal 0. So let's subtract 60 from both sides of this equation. And we get 0 equal to b squared, minus 4b, minus 60. And so what we need to do here is just factor this thing right now, or factor it. And then, no-- if I have the product of some things, and that equals 0, that means that either one or both of those things need to be equal to 0. So we need to factor b squared, minus 4b, minus 60. So what we want to do, we want to find two numbers whose sum is negative 4 and whose product is negative 60. Now, given that the product is negative, we know there are different signs. And this tells us that their absolute values are going to be four apart. That one is going to be four less than the others. So you could look at the products of the factors of 60. 1 and 60 are too far apart. Even if you made one of the negative, you would either get positive 59 as the sum or negative 59 as the sum. 2 and 30, still too far apart. 3 and 20, still too far apart. If you had made one negative you'd either get negative 17 or positive 17. Then you could have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and a 12, still seems too far apart One of them is negative, then you either have their sum being positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They are four apart. So if we make-- and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10 their sum will be negative 4, and their product is negative 60. So that works. So you could literally say that this is equal to b plus 6, times b, minus 10. b plus the a, plus b minus the b. And let me be very careful here. This b over here, I want to make it very clear, is different than the b that we're using in the equation. I just used this b here to say, look, we're looking for two numbers that add up to this second term It's a different b. I could have said x plus y is equal to negative 4, and x times y is equal to negative 60. In fact, let me do it that way just so we don't get confused. So we could write x plus y is equal to negative 4. And then we have x times y is equal to negative 60. So we have b plus 6, times b plus y. x is 6, y is negative 10. And that is equal to 0. Let's just solve this right here. And then we'll go back and show you. You could also factor this by grouping. But just from this, we know that either one of these is equal to zero. Either b plus 6 is equal to 0, or b minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get b is equal to negative 6. Or if you add 10 to both sides of this equation, you get b is equal to 10. And those are our two solutions. You could put them back in and verify that they satisfy our constraints. Now the other way that you could solve this, and we're going to get exact same answer. Is you could just break up this negative 4b into its constituents. So you could have broken this up into 0 is equal to b squared. And then you could have broken it up into plus 6b, minus 10b, minus 60. And then factor it by grouping. Group these first two terms. Group these second two terms. Just going to add them together. The first one you could factor out a b. So you have b times b, plus 6. The second one you can factor out a negative 10. So minus 10 times b, plus 6. All that's equal to 0. And now you can factor out a b plus 6. So if you factor out a b plus 6 here, you get 0 is equal to b minus 10, times b, plus 6. We're literally just factoring out this out of the expression. You're just left with a b minus 10. You get the same thing that we did in one step over here. Whatever works for you. But either way, the solutions are either b is equal to negative 6, or b is equal to 10. And we have to be careful here. Remember, this is a word problem. We can't just state, oh b could be negative 6 or b could be 10. We have to think about whether this makes sense in the context of the actual problem. We're talking about lengths of triangles, or lengths of the sides of triangles. We can't have a negative length. So because of that, the base of a triangle can't have length of negative 6. So we can cross that out. So we actually only have one solution here. Almost made a careless mistake. Forgot that we were dealing with the word problem. The only possible base is 10. And let's see, they say find the height and the base. Once again, done. So the base we're saying is 10. The height is four inches less. It's b minus 4. So the height is 6. And then you can verify. The area is 6 times 10 times one half, which is 30." + }, + { + "Q": "can verocity and speed be used intechangebly", + "A": "Speed is a scalar and tells us how fast we are going. Velocity is a vector and tells us how fast we are going (speed) AND direction. While they are similar, the best practice is not to use them interchangeably.", + "video_name": "MAS6mBRZZXA", + "transcript": "The goal of this video is to explore some of the concept of formula you might see in introductional physics class but more importantly to see they are really just common sense ideas So let's just start with a simple example Let's say that and for the sake of this video keep things that magnitudes and velocities that's the direction of velocity etc. let's just assume that if I have a positive number that it means for example postive velocity that it means I'm going to the right let's say I have a negative number we won't see in this video let's assume we are going to the left In that way I can just write a number down only operating in one dimension you know that by specifying the magnitude and the direction if I say velocity is 5m/s that means 5m/s to the right if I say negative 5m/s that means 5m/s to the left let's say just for simplicitiy, say that we start with initial velocity we start with an initial velocity of 5m/s once again I specify the magnitude and the direction because of this convention here, we know it is to the right let's say we have a constant acceleration we have a constant acceleration 2m/s^2 or 2m per second square and once again since this is positive it is to the right and let's say that we do this for a duration so my change in time, let's say we do this for a duration of 4 I will just use s, second and s different places so s for this video is seconds So I want to do is to think about how far do we travel? and there is two things how fast are we going? after 4 seconds and how far have we travel over the course of those 4 seconds? so let's draw ourselves a little diagram here So this is my velocity axis, and this over here is time axis we have to draw a straighter line than that So that is my time axis, time this is velocity This is my velocity right over there and I'm starting off with 5m/s, so this is 5m/s right over here So vi is equal to 5m/s And every second goes by it goes 2m/s faster that's 2m/s*s every second that goes by So after 1 second when it goes 2m/s faster it will be at 7 another way to think about it is the slope of this velocity line is my constant accleration, my constant slope here so it might look something like that So what has happend after 4s? So 1 2 3 4 this is my delta t So my final velocity is going to be right over there I'm writing it here because this get into the way of veloctiy so this is v this is my final velocity what would it be? Well I'm starting at 5m/s So we are doing this both using the variable and concretes Some starting with some initial velocity I'm starting with some initial velocity Subscript i said i for initial and then each second that goes by I'm getting this much faster so if I gonna see how much faster have I gone I multiply the number of second, I will just multiply the number second it goes by times my acceleration, times my acceleration and once again, this right here, subscript c saying that is a constant acceleration, so that will tell my how fast I have gone If I started at this point and multiply the duration time with slope I will get this high, I will get to my final velocity just to make it clear with the numbers, this number can really be anything I'm just taking this to make it concrete in your mind you have 5m/s plus 4s plus, I wanna do it in yellow plus 4s times our acceleration with 2m per second square and what is this going to be equal to? you have a second that is cancelling out one of the second down here You have 4 times, so you have 5m/s plus 4 times 2 is 8 this second gone, we just have 8m/s or this is the same thing as 13m/s which is going to be our final velocity and I wanna take a pause here, you can pause and think about it yourself this whole should be intuitive, we are starting by going with 5m/s every second goes by we are gonna going 2m/s faster so after 1s it would be 7 m/s, after 2s we will be 9m/s after 3s we will be 11m/s, and after 4s we will be at 13 m/s so you multiply how much time pass times acceleration this is how much faster we are gonna be going, we are already going 5m/s 5 plus how much faster? 13 m/s so this right up here is 13m/s So I will take a little pause here hopefully intuitive and the whole play of that is to show you this formula you will see in many physics book is not something that randomly pop out of there it just make complete common sense Now the next thing I wanna talk about is what is the total distance that we would have travel? and we know from the last video that distance is just the area under this curve right over here, so it's just the area under this curve you see this is kind of a strange shape here how do I caculate this area? and we can use a little symbol of geometry to break it down into two different areas, it's very easy to calculate their areas two simple shapes, you can break it down to two, blue part is the rectangle right over here, easy to figure out the area of a rectangle and we can break it down to this purple part, this triangle right here easy to figure out the area of a triangle and that will be the total distance we travel even this will hopefully make some intuition because this blue area is how far we would have travel if we are not accelerated, we just want 5m/s for 4s so you goes 5m/s 1s 2s 3s 4s so you are going from 0 to 4 you change in time is 4s so if you go 5m/s for 4s you are going to go 20 m this right here is 20m that is the area of this right here 5 times 4 this purple or magentic area tells you how furthur than this are you going because you are accelerating because kept going faster and faster and faster it's pretty easy to calculate this area the base here is still 5(4) because that's 5(4) second that's gone by what's the height here? The height here is my final velocity minus my initial velocity minus my initial velocity or it's the change in velocity due to the accleration 13 minus 5 is 8 or this 8 right over here it is 8m/s so this height right over here is 8m/s the base over here is 4s that's the time that past what's this area of the triangle? the area of this triangle is one half times the base which is 4s 4s times the height which is 8m/s times 8m/s second cancel out one half time 4 is 2 times 8 is equal to 16m So the total distance we travel is 20 plus 16 is 36m that is the total I could say the total displacement and once again is to the right, since it's positive so that is our displacement What I wanna do is to do the exact the same calculation keep it in variable form, that will give another formula many people often memorize You might understand this is completely intuitive formula and that just come out of the logical flow of reasoning that we went through this video what is the area once again if we just think about the variables? well the area of this rectangle right here is our initial velocity times our change in time, times our change in time So that is the blue rectangle right over here, and plus what do we have to do? we have the change in time once again we have the change in time times this height which is our final velocity which is our final velocity minus our initial velocity these are all vectors, they are just positive if going to the right we just multiply the base with the height that will just be the area of the entire rectangle I will take it by half because triangle is just half of that rectangle so times one half, so times one half so this is the area, this is the purple area right over here this is the area of this, this is the area of that and let's simplify this a little bit let's factor out the delta t, so you factor out the delta t you get delta t times a bunch of stuff v sub i your initial velocity we factor this out plus this stuff, plus this thing right over here and we can distribute the one half we factor the one half, we factor the delta t out, taking it out and let's multiply one half by each of these things so it's gonna be plus one half times vf, times our final velocity that's not the right color, I will use the right color so you would understand what I am doing, so this is the one half so plus one half times our final velocity final velocity minus one half, minus one half times our initial velocity I'm gonna do that in blue, sorry I have trouble in changing color today minus one half times our initial velocity, times our initial velocity and what is this simplify do? we have something plus one half times something else minus one half of the original something so what is vi minus one half vi? so anything minus its half is just a positive half left so these two terms, this term and this term will simplify to one half vi one half initial velocity plus one half times the final velocity plus one half times the final velocity and all of that is being multiplied with the change in time the time that has gone by and this tells us the distance, the distance that we travel another way to think about it, let's factor out this one half you get distance that is equal to change in time times factoring out the one half vi plus vf, vi no that's not the right color vi plus vf so this is interesting, the distance we travel is equal to one half of the initial velocity plus the final velocity so this is really if you just took this quantity right over here it's just the arithmetic, I have trouble saying that word it's the arithmetic mean of these two numbers, so I'm gonna define, this is something new, I'm gonna call this the average velocity we have to be very careful with this this right here is the average velocity but the only reason why I can just take the starting velocity and ending velocity and adding them together and divide them by two since you took an average of two thing it's some place over here and I take that as average velocity it's because my acceleration is constant which is usually an assumption in introductory physics class but it's not always the assumption but if you do have a constant acceleration like this you can assume that the average velocity is gonna be the average of the initial velocity and the final velocity if this is a curve and the acceleration is changing you could not do that but what is useful about this is if you wanna figure out the distance that was travelled, you just need to know the initial velocity and the final velocity, average their two and multiply the times it goes by so in this situation our final velocity is 13m/s our initial velocity is 5m/s so you have 13 plus 5 is equal to 18 you divided that by 2, you average velocity is 9m/s if you take the average of 13 and 5 and 9m/s times 4s gives you 36m so hopefully it doesn't confuse you I just wannt show you some of these things you will see in your physics class but you shouldn't memorize they can all be deduced" + }, + { + "Q": "How to write 230078 in expanded form", + "A": "In the number 230078, the 2 is in the hundred thousands place, the 3 is in the ten thousands place, the 0 s are in the thousands and hundreds places, the 7 is in the tens place, and the 8 is in the ones place. So 230078 represents 2 hundred thousands, 3 ten thousands, 7 tens, and 8 ones. So the expanded form of 230078 is 200000 + 30000 + 70 + 8, or (2 x 100000) + (3 x 10000) + (7 x 10) + (8 x 1). Have a blessed, wonderful day!", + "video_name": "iK0y39rjBgQ", + "transcript": "Write 14,897 in expanded form. Let me just rewrite the number, and I'll color code it, and that way, we can keep track of our digits. So we have 14,000. I don't have to write it-- well, let me write it that big. 14,000, 800, and 97-- I already used the blue; maybe I should use yellow-- in expanded form. So let's think about what place each of these digits are in. This right here, the 7, is in the ones place. The 9 is in the tens place. This literally represents 9 tens, and we're going to see this in a second. This literally represents 7 ones. The 8 is in the hundreds place. The 4 is in the thousands place. It literally represents 4,000. And then the 1 is in the ten-thousands place. And you see, every time you move to the left, you move one place to the left, you're multiplying by 10. Ones place, tens place, hundreds place, thousands place, ten-thousands place. Now let's think about what that really means. If this 1 is in the ten-thousands place, that means that it literally represents-- I want to do this in a way that my arrows don't get mixed up. Actually, let me start at the other end. Let me start with what the 7 represents. The 7 literally represents 7 ones. Or another way to think about it, you could say it represents 7 times 1. All of these are equivalent. They represent 7 ones. Now let's think about the 9. That's why I'm doing it from the right, so that the arrows don't have to cross each other. So what does the 9 represent? It represents 9 tens. You could literally imagine you have 9 actual tens. You could have a 10, plus a 10, plus a 10. Do that nine times. That's literally what it represents: 9 actual tens. 9 tens, or you could say it's the same thing as 9 times 10, or 90, either way you want to think about it. So let me write all the different ways to think about it. It represents all of these things: 9 tens, or 9 times 10, or 90. So then we have our 8. Our 8 represents-- we see it's in the hundreds place. It represents 8 hundreds. Or you could view that as being equivalent to 8 times 100-- a hundred, not a thousand-- 8 times 100, or 800. That 8 literally represents 8 hundreds, 800. And then the 4. I think you get the idea here. This represents the thousands place. It represents 4 thousands, which is the same thing as 4 times 1,000, which is the same thing as 4,000. 4,000 is the same thing as 4 thousands. Add it up. And then finally, we have this 1, which is sitting in the ten-thousands place, so it literally represents 1 ten-thousand. You can imagine if these were chips, kind of poker chips, that would represent one of the blue poker chips and each blue poker chip represents 10,000. I don't know if that helps you or not. And 1 ten-thousand is the same thing as 1 times 10,000 which is the same thing as 10,000. So when they ask us to write it in expanded form, we could write 14,897 literally as the sum of these numbers, of its components, or we could write it as the sum of these numbers. Actually, let me write this. This top 7 times 1 is just equal to 7. So 14,897 is the same thing as 10,000 plus 4,000 plus 800 plus 90 plus 7. So you could consider this expanded form, or you could use this version of it, or you could say this the same thing as 1 times 10,000, depending on what people consider to be expanded form-- plus 4 times 1,000 plus 8 times 100 plus 9 times 10 plus 7 times 1. I'll scroll to the right a little bit. So either of these could be considered expanded form." + }, + { + "Q": "What is the fraction thirteen thirds reduced to? Why doesn't the person reduce the fractions?", + "A": "The fraction thirteen thirds is the simplest form we can simplify it to. The reason we don t convert it into a decimal is because we d have the monstrosity 4.33333333333333....... Also, in math, it s preferred not to have decimals in a fraction.", + "video_name": "XMJ72mtMn4Y", + "transcript": "A line goes through the points (-1, 6) and (5, 4). What is the equation of the line? Let's just try to visualize this. So that is my x axis. And you don't have to draw it to do this problem but it always help to visualize That is my y axis. And the first point is (-1,6) So (-1, 6). So negative 1 coma, 1, 2, 3, 4 ,5 6. So it's this point, rigth over there, it's (-1, 6). And the other point is (5, -4). So 1, 2, 3, 4, 5. And we go down 4, So 1, 2, 3, 4 So it's right over there. So the line connects them will looks something like this. Line will draw a rough approximation. I can draw a straighter than that. I will draw a dotted line maybe Easier do dotted line. So the line will looks something like that. So let's find its equation. So good place to start is we can find its slope. Remember, we want, we can find the equation y is equal to mx plus b. This is the slope-intercept form where m is the slope and b is the y-intercept. We can first try to solve for m. We can find the slope of this line. So m, or the slope is the change in y over the change in x. Or, we can view it as the y value of our end point minus the y value of our starting point over the x-value of our end point minus the x-value of our starting point. Let me make that clear. So this is equal to change in y over change in x wich is the same thing as rise over run wich is the same thing as the y-value of your ending point minus the y-value of your starting point. This is the same exact thing as change in y and that over the x value of your ending point minus the x-value of your starting point This is the exact same thing as change in x. And you just have to pick one of these as the starting point and one as the ending point. So let's just make this over here our starting point and make that our ending point. So what is our change in y? So our change in y, to go we started at y is equal to six, we started at y is equal to 6. And we go down all the way to y is equal to negative 4 So this is rigth here, that is our change in y You can look at the graph and say, oh, if I start at 6 and I go to negative 4 I went down 10. or if you just want to use this formula here it will give you the same thing We finished at negative 4, we finished at negative 4 and from that we want to subtract, we want to subtract 6. This right here is y2, our ending y and this is our beginning y This is y1. So y2, negative 4 minus y1, 6. or negative 4 minus 6. That is equal to negative 10. And all it does is tell us the change in y you go from this point to that point We have to go down, our rise is negative we have to go down 10. That's where the negative 10 comes from. Now we just have to find our change in x. So we can look at this graph over here. We started at x is equal to negative 1 and we go all the way to x is equal to 5. So we started at x is equal to negative 1, and we go all the way to x is equal to 5. So it takes us one to go to zero and then five more. So are change in x is 6. You can look at that visually there or you can use this formula same exact idea, our ending x-value, our ending x-value is 5 and our starting x-value is negative 1. 5 minus negative 1. 5 minus negative 1 is the same thing as 5 plus 1. So it is 6. So our slope here is negative 10 over 6. wich is the exact same thing as negative 5 thirds. as negative 5 over 3 I devided the numerator and the denominator by 2. So we now know our equation will be y is equal to negative 5 thirds, that's our slope, x plus b. So we still need to solve for y-intercept to get our equation. And to do that, we can use the information that we know in fact we have several points of information We can use the fact that the line goes through the point (-1,6) you could use the other point as well. We know that when is equal to negative 1, So y is eqaul to 6. So y is equal to six when x is equal to negative 1 So negative 5 thirds times x, when x is equal to negative 1 y is equal to 6. So we literally just substitute this x and y value back into this and know we can solve for b. So let's see, this negative 1 times negative 5 thirds. So we have 6 is equal to positive five thirds plus b. And now we can subtract 5 thirds from both sides of this equation. so we have subtracted the left hand side. From the left handside and subtracted from the rigth handside And then we get, what's 6 minus 5 thirds. So that's going to be, let me do it over here We take a common denominator. So 6 is the same thing as Let's do it over here. So 6 minus 5 over 3 is the same thing as 6 is the same thing as 18 over 3 minus 5 over 3 6 is 18 over 3. And this is just 13 over 3. And this is just 13 over 3. And then of course, these cancel out. So we get b is equal to 13 thirds. So we are done. We know We know the slope and we know the y-intercept. The equation of our line is y is equal to negative 5 thirds x plus our y-intercept which is 13 which is 13 over 3. And we can write these as mixed numbers. if it's easier to visualize. 13 over 3 is four and 1 thirds. So this y-intercept right over here. this y-intercept right over here. That's 0 coma 13 over 3 or 0 coma 4 and 1 thirds. And even with my very roughly drawn diagram it those looks like this. And the slope negative 5 thirds that's the same thing as negative 1 and 2 thirds. You can see here the slope is downward because the slope is negative. It's a little bit steeper than a slope of 1. It's not quite a negative 2. It's negative 1 and 2 thirds. if you write this as a negative, as a mixed number. So, hopefully, you found that entertaining." + }, + { + "Q": "I can not answer this problem ? a^-b = 1/(a^b) (and why a^0 =1)", + "A": "Fill in some values for your variables. Say a = 4, b = 2. 4^-2 = 1/(4^2) = 1/(4 * 4) = 1/16. Anything to the power of zero is one. That s just the rules of math. 51256^0 is 1.", + "video_name": "Tqpcku0hrPU", + "transcript": "I have been asked for some intuition as to why, let's say, a to the minus b is equal to 1 over a to the b. And before I give you the intuition, I want you to just realize that this really is a definition. The inventor of mathematics wasn't one person. It was, you know, a convention that arose. But they defined this, and they defined this for the reasons that I'm going to show you. Well, what I'm going to show you is one of the reasons, and then we'll see that this is a good definition, because once you learned exponent rules, all of the other exponent rules stay consistent for negative exponents and when you raise something to the zero power. So let's take the positive exponents. Those are pretty intuitive, I think. So the positive exponents, so you have a to the 1, a squared, a cubed, a to the fourth. What's a to the 1? a to the 1, we said, is a, and then to get to a squared, what did we do? We multiplied by a, right? a squared is just a times a. And then to get to a cubed, what did we do? We multiplied by a again. And then to get to a to the fourth, what did we do? We multiplied by a again. Or the other way, you could imagine, is when you decrease the exponent, what are we doing? We are multiplying by 1/a, or dividing by a. And similarly, you decrease again, you're dividing by a. And to go from a squared to a to the first, you're dividing by a. So let's use this progression to figure out what a to the 0 is. So this is the first hard one. So a to the 0. So you're the inventor, the founding mother of mathematics, and you need to define what a to the 0 is. And, you know, maybe it's 17, maybe it's pi. I don't know. It's up to you to decide what a to the 0 is. But wouldn't it be nice if a to the 0 retained this pattern? That every time you decrease the exponent, you're dividing by a, right? So if you're going from a to the first to a to the zero, wouldn't it be nice if we just divided by a? So let's do that. So if we go from a to the first, which is just a, and divide by a, right, so we're just going to go-- we're just going to divide it by a, what is a divided by a? Well, it's just 1. So that's where the definition-- or that's one of the intuitions behind why something to the 0-th power is equal to 1. Because when you take that number and you divide it by itself one more time, you just get 1. So that's pretty reasonable, but now let's go into So what should a to the negative 1 equal? Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent we're dividing by a. So let's divide by a again, so 1/a. So we're going to take a to the 0 and divide it by a. a to the 0 is one, so what's 1 divided by a? It's 1/a. Now, let's do it one more time, and then I think you're going to get the pattern. Well, I think you probably already got the pattern. What's a to the minus 2? Well, we want-- you know, it'd be silly now to Every time we decrease the exponent, we're dividing by a, so to go from a to the minus 1 to a to the minus 2, let's just divide by a again. And what do we get? If you take 1/2 and divide by a, you get 1 over a squared. And you could just keep doing this pattern all the way to the left, and you would get a to the minus b is equal to 1 over a to the b. Hopefully, that gave you a little intuition as to why-- well, first of all, you know, the big mystery is, you know, something to the 0-th power, why does that equal 1? First, keep in mind that that's just a definition. Someone decided it should be equal to 1, but they had a good reason. And their good reason was they wanted to keep this pattern going. And that's the same reason why they defined negative exponents in this way. And what's extra cool about it is not only does it retain this pattern of when you decrease exponents, you're dividing by a, or when you're increasing exponents, you're multiplying by a, but as you'll see in the exponent rules videos, all of the exponent rules hold. All of the exponent rules are consistent with this definition of something to the 0-th power and this definition of something to the negative power. Hopefully, that didn't confuse you and gave you a little bit of intuition and demystified something that, frankly, is quite mystifying the first time you learn it." + }, + { + "Q": "do you now of there is a world war 2? if so, when?", + "A": "Yes World War II happened. It was between 1939 and 1945", + "video_name": "eIfQ4GfSz3U", + "transcript": "Narrator: As we'll see in this video and in others, the roots of a lot of the current disagreements in the Middle East and a lot of the conflict in the Middle East can actually be traced back to World War I. I realize this is an incredibly touchy subject that there are people who have very strong feelings on either side of it and my goal here is to really give my best attempt at what really happened. I encourage you to doubt any of this and look it up yourself and come, frankly, to your own conclusions. Let's rewind back to October of 1915, or 1915 in particular. The British were already at war with the Ottoman's. Just as a reminder of some of what happened in 1915, the Gallipoli campaign, by the end of 1915 it was pretty clear that this was a disaster for the allies. The Ottoman's were able to fend off the allies, they were in retreat. The British were able to fend off the Ottoman's when they tried to attack the Suez canal in 1915. This is the background, you can imagine the British are eager to get any other allies they can in their battle against the Ottoman's. In particular, they are eager to get the help of the Arab's who have been under the rule of the Ottoman's for hundreds of years. That's the backdrop where you have this correspondence between the high commissioner in Egypt, the British high commissioner, Sir Henry McMahon and the Sharif of Mecca, Hussein bin \u02bfAli, who had his own aspirations to essentially be the king of an independent Arab state. They kept going back and forth from mid 1915 to early 1916 talking about what the state could be. Obviously the British want his support, wants him to lead a revolt against the Ottoman's. He's already articulated the boundaries for a state that he would like to see. So, that gives us a context for this correspondence in October of 1915. This is from Sir Henry McMahon to Hussein. \"... it is with great pleasure that I communicate to you \"on their behalf,\" the British government's behalf, \"the following statement, which I am confident \"you will receive with satisfaction. \"The two districts of Mersina and Alexandretta \"and portions of Syria lying to the west \"of the districts of Damascus, Homs, Hama, \"and Aleppo cannot be said to be purely Arab, \"and should be excluded from the limits demanded.\" This is referring to the limits that Hussein bin Ali had demanded in previous correspondence. \"With the above modifications,\" so just that region right over there, this right over here is Mersina, Alexandretta, this is Hama, Homs, Damascus, so really what he's referring to is this region, the west, the west of those cities right over here. He's saying look, you can't really consider this to be purely Arab, I'm going to exclude this out of the boundaries of this potential independent Arab state. \"With the above modification, and without prejudice \"to our existing treaties with Arab chiefs \"we accept those limits,\" we accept those limits. \"As for those regions lying within those frontiers \"wherein Great Britain is free to act \"without determinant to the interest of her ally, \"France,\" so as long as I'm not getting in trouble with France, \"I'm empowered in the name of \"the Government of Great Britain to give the following \"assurances and make the following reply \"to your letter; Subject to the above modifications,\" so taking this part out, \"Great Britain is prepared \"to recognize and support the independence \"of the Arabs in all the regions within the limits \"demanded by the Sharif of Mecca.\" So, essentially it included all of this region and actually much beyond what I'm showing here, kind of present day Syria, Jordan, Iraq, parts of present day Saudi Arabia. All of that is essentially, the British are saying, yeah we're going to allow you to have that, an independent state there. \"Great Britain will guarantee the Holy Places \"against all external aggression \"and will recognize their inviolability. \"... I am convinced that this declaration will assure \"you beyond all possible doubt,\" beyond all possible doubt, \"of the sympathy \"of Great Britain towards the aspiration of her friends \"the Arabs, and will result in a firm \"and lasting alliance, the immediate results \"of which will be the expulsion of the Turks \"from the Arab countries and the freeing \"of the Arab peoples from the Turkish yoke, \"which for so many years has pressed heavily \"upon them.\" This actually does help to convince the Arab's to rise up against the Turks, against the Ottoman Empire, they play a significant role in the Palestine Campaign, they rise up in June of 1916. Now, the video that I did on the Palestine Campaign, I got several comments of people being cynical about Britain's intentions and it does look like the British were, indeed, cynical. T.E. Lawrence famous for Lawrence of Arabia was often depicted as this mystical fellow, this guy who had this kinship with the Arab's. His actual correspondence with the British government actually do show that he did have a kind of ... he was doing, I guess, in the words of George W. Bush, a little bit of strategery, he had a more cynical view of this relationship with the Arab's. This is some correspondence that he wrote in early 1916, so right about the same time that all of this was going on. This says he's referring to a possible Arab revolt, or Hussein's activity. \"Hussein's activity seems beneficial to us, \"because it matches with our immediate aims, \"the break-up of the Islamic 'bloc' \"and the defeat and disruption of the \"Ottoman Empire.\" Assuming he didn't really talk about this, this being one of the ... the British didn't talk about that when they were talking to Hussein. \"If we can arrange that this political change \"shall be a violent one, we will have abolished \"the threat of Islam, by dividing it against itself, \"in its very heart.\" \"There will then be a Khalifa,\" kind of a seat of Islam, \"in Turkey \"and a Khalifa in Arabia, in theological warfare.\" This is T.E. Lawrence, I got this from The Golden Warrior: The Life and Legend of Lawrence of Arabia. Even this, somewhat portrayed as a heroic figure, was doing things in very strategic, strategic terms. To make things worse for the Arab's, while the British were trying to convince them to revolt, they were also in secret negotiations with the French on how they would divide the Middle East if they were able to beat the Ottoman's. At this point in the war the British were already making some progress in Mesopotamia, but they really hadn't really started on the Palestine Campaign right here. So, this was all conjecture. The British representatives was Sykes, the French representative was Picot, this was done with the consent of the Russian's. You didn't have a revolution in Russia as of now, so in early 1916, in May this agreement was concluded, this secret agreement. You have the Sykes-Picot Agreement, it's secret. Let me write that, it is a secret agreement between Britain and France and essentially they are carving up the entire Middle East between them. This blue area right here, this would be occupied by the French, part of eastern Turkey or modern day eastern Turkey would be given to the Russian's. The British would be able to occupy, would occupy southern Mesopotamia essentially insuring protection of the oil that is coming out of Persia. Oil is becoming more and more of a relevant factor in kind of global power. Then you have these two protectorates right over here, which in theory could be independent or an independent Arab state, or two independent Arab states under the protection. Let me put that in quotes, because \"protectorate\" is always not as nice as it sounds, under the protection of the French or the British which means, \"Hey you're an independent state, but we will \"protect you in case anyone wants to invade.\" The reality of protectorate is that it usually involves the people doing the protecting have all the real power and all the real influence. The Sykes-Picot Agreement also give this little carve out to Britain so they would have access to the Mediterranean. Palestine, or the Roman Kingdom of Judea, this is carved out as a separate international property something that would be administered by multiple states and I guess the argument would be, this is where the Holy Land's are, multiple religions have some of their holiest sites within here and so they carved it out like this. Once again, this is all in secret, they obviously don't want the Arab's to find out because they're about to convince the Arab's to join in a revolt against the Ottoman's. Now, to make things ... once again, this was all secret up to this point in 1916 when it was all agreed on. Then you forward to 1917 where we have the famous Balfour Declaration. This right over here is the Balfour Declaration and it was essentially a letter from the Foreign Secretary of the U.K., Balfour, to Lord Rothschild who was a leading [Briticizen] , a leading member of the Jewish community. In it he writes, \"Dear Lord Rothschild, \"I have much pleasure in conveying to you, \"on behalf of His Majesty's Government, \"the following declaration of sympathy \"with Jewish Zionist aspirations which has been \"submitted to, and approved by, the Cabinet. \"His Majesty's Government view with favor \"the establishment in Palestine of a national home,\" of a national home, \"for the Jewish people, \"and will use their best endeavors to facilitate \"the achievement of this objective. \"It being clearly understood that nothing shall be done \"which may prejudice the civil and religious rights \"of existing non-Jewish communities in Palestine, \"or the rights and political status enjoyed by Jews \"in any other country. \"I should be grateful if you would bring \"this declaration to the knowledge of the \"Zionist Federation.\" Signed Artur Balfour. In here, he's not explicitly saying ... and they're being very careful here, he's not saying we're supporting a state for the Jewish people, but he's saying he is supporting the return of national home for the Jewish people, but at the same time, he's saying that it being clearly understood that nothing shall be done which may prejudice the civil and religious rights of the existing non-Jewish communities in Palestine. Needless to say, you can imagine that this is making the Arab's fairly uncomfortable. On one side it seems, based on some of the McMahon-Hussein correspondences that were ... especially in 1915, that they were being promised an independent Arab state which included much of this territory, but at the same time, in the Balfour Declaration the British were promising to, kind of the Jewish diaspora, that they could have a homeland there and it might one day, who knows, it might one day turn in to some type of a state. To make the Arab's even more uncomfortable, this was in November 2, 1917. By the end of November, you have to remember that 1917 you first had a revolution, in Russia the Czar was overthrown in February and in March of 1917, and October the Bolshevik's take over. They want to get out of the war, they don't like all these secret deals, not clear that they would even get what they were entitled to these secret deals, so they actually release all the entire text of the Sykes-Picot Agreement. They released this, so in the same month you have the Arab's and the Ottoman's and the Ottoman's were very happy to see this because it would undermine the Arab's belief in maybe supporting the allies, but in one month you have the Arab's finding out about the Balfour Declaration, which was a pulbic declaration and then later that month because of the Russian release of it, the formally secret Sykes-Picot Agreement, so it makes them very, or at least a little bit more suspicious. So you can imagine the British Empire trying to have it both ways, to kind of have support from the Jewish Diaspora while at the same time have support from the Arab's in their revolt against the Ottoman's would lead to very significant conflicts over the decades to come. Regardless of which side of the issue you fall on, a lot of the seed is happening right around now, right around World War I. This has been admitted by the British government. This is right here, this was the then Secretary, or Foreign Secretary Jack Straw, U.K Foreign Secretary in 2002. This is a statement he made to the News Statesman Magazine in 2002. \"A lot of the problems we are having to deal with now, \"I have to deal with now,\" he's the Foreign Secretary, \"are a consequence \"of our colonial past ...\" Consequence of our colonial past. \"The Balfour Declaration and the contradictory assurances,\" \"and the contradictory assurances \"which were being given to Palestinian's \"in private at the same time as they were \"being given to the Israelis ... \"again, an interesting history for us, \"but not an honorable one.\" This is really just the beginning as we'll see in future videos as we go to the Interwar period, the British kind of go back and forth on this issue over, over, and over again, but needless to say, it's lead to a very messy situation in the modern Middle East." + }, + { + "Q": "what does linear mean", + "A": "it means straight like a line, hence LINEar. a wobbly string isn t linear, but a straight string is. A problem that is linear basically means if you graph it, it will be a straight line.", + "video_name": "AZroE4fJqtQ", + "transcript": "Deirdre is working with a function that contains the following points. These are the x values, these are y values. They ask us, is this function linear or non-linear? So linear functions, the way to tell them is for any given change in x, is the change in y always going to be the same value. For example, for any one-step change in x, is the change in y always going to be 3? Is it always going to be 5? If it's always going to be the same value, you're dealing with a linear function. If for each change in x--so over here x is always changing by 1, so since x is always changing by 1, the change in y's have to always be the same. If they're not, then we're dealing with a non-linear function. We can actually show that plotting out. If the changes in x-- we're going by different values, if this went from 1 to 2 and then 2 to 4-- what you'd want to do, then, is divide the change in y by the change in x, and that should always be a constant. In fact, let me write that down. If something is linear, then the change in y over the change in x always constant. Now, in this example, the change in x's are always 1, right? We go from 1 to 2, 2 to 3, 3 to 4, 4 to 5. So in this example, the change in x is always going to be 1. So in order for this function to be linear, our change in y needs to be constant because we're just going to take that and divide it by 1. So let's see if our change in y is constant. When we go from 11 to 14, we go up by 3. When we go from 14 to 19, we go up by 5, so I already see that it is not constant. We didn't go up by 3 this time, we went up by 5. And here, we go up by 7. And here, we're going up by 9. So we're actually going up by increasing amounts, so we're definitely dealing with a non-linear function. And we can see that if we graph it out. So let me draw-- I'll do a rough graph here. So let me make that my vertical axis, my y-axis. And we go all the way up to 35. So I'll just do 10, 20, 30. Actually, I can it do a little bit more granularly than that. I could do 5, 10, 15, 20, 25, 30, and then 35. And then our values go 1 through 5. I'll do it on this axis right here. They're not obviously the exact same scale, so I'll do 1, 2, 3, 4, and 5. So let's plot these points. So the first point is 1, 11, when x is 1, y is 11. This is our x-axis. When x is 1, y is 11, that's right about there. When x is 2, y is 14, that's right about there. When x is 3, y is 19, right about there. When x is 4, y is 26, right about there. And then finally, when x is 5, y is 35, right up there. So you can immediately see that this is not tracing out a line. If this was a linear function, then all the points would be on a line that looks something like that. That's why it's called a linear function. In this case, it's not, it's non-linear. The rate of increase as x changes is going up." + }, + { + "Q": "Why are we allowed to take the natural log of e^x when finding the derivative. Isn't that changing the function?", + "A": "Try looking at it this way, maybe this will answer your question. d/dx [ln(e^x)] = d/dx [ln(e^x)] I don t need to prove that, right? It s obvious, everything is equal to itself. And then he differentiates the right hand side and gets d/dx [x*lne] = dx/dx * 1 = 1 and then differentiates the other side of the equation and gets d/dx[e^x]* 1/(e^x) But because we said that d/dx [ln(e^x)] = d/dx [ln(e^x)] we can set our two answers equal to each other. Did I help? Make sure to watch the HD version of the video.", + "video_name": "sSE6_fK3mu0", + "transcript": "Let's prove with the derivative of e to the x's, and I think that this is one of the most amazing things, depending on how you view it about either calculus or math or the universe. Well we're essentially going to prove-- I've already told you before that the derivative of e to the x is equal to e to the x, which is amazing. The slope at any point of that line is equal to the x value-- is equal to the function at that point, not the x value. The slope at any point is equal to e. That is mind boggling. And that also means that the second derivative at any point is equal to the function of that value or the third derivative, or the infinite derivative, and that never ceases to amaze me. But anyway back to work. So how are we going to prove this? Well we already proved-- I actually just did it right before starting this video-- that the derivative-- and some people actually call this the definition of e. They go the other way around. They say there is some number for which this is true, and we call that number e. So it could almost be viewed as a little bit circular, but be we said that e is equal to the limit as n approaches infinity of 1 over 1 plus n to the end. And then using this we actually proved that derivative of ln of x is equal to 1/x. The derivative of log base e of x is equal to 1/x. So now that we prove this out, let's use this to prove this. Let me keep switching colors to keep it interesting. Let's take the derivative of ln of e to the x. This is almost trivial. This is equal to the logarithm of a to the b is equal to b times the logarithm of a, so this is equal to the derivative of x ln of e. And this is just saying e to what power is equal to e. Well, to the first power, right? So this just equals the derivative of x, which we have shown as equal to 1. I think we have shown it, hopefully we've shown it. If we haven't, that's actually a very easy one to prove. OK fair enough. We did that. But let's do this another way. Let's use the chain rule. So what doe the chain rule say? If we have f of g of x, where we have one function embedded in another one, the chain rule say we take the derivative of the inside function, so d/dx of e to the x. And then we take the derivative of the outside function or the derivative of the outside function with respect to the inner function. You can almost view it that way. So the derivative ln of e to the x with respect to e to the x. I know that's a little confusing. You could have written a d e to the x down over here, but I think you know the chain rule by now. That is equal to 1 over e to the x. And that just comes from this. But instead of an x, we have e to the x. So this is just a chain rule. Well what else do we know? We know that this is equal to this, and we also know that this is equal to this. So this must be equal to this. So this must be equal to 1. Well let's just multiply both sides of this equation by e to the x. We get on the left hand side, we're just left with this expression. The derivative of e to the x times- we're multiplying both sides by e to the x, times e to the x over e to the x. I just chose to put the e to the x on this term, is equal to e to the x. This is 1. Scratch it out. We're done. That might not have been completely satisfying for you, but it works. The derivative of e to the x is equal to e to the x. I think the school or the nation should take a national holiday or something, and people should just ponder this, because it really is fascinating. But then actually this will lead us to I would say even more dramatic results in the not too far off future. Anyway, I'll see in the next video." + }, + { + "Q": "What is an example of a quadratic equation with two imaginary solutions?", + "A": "x^2+2x+2=0 (x+1)^2= -1 x+1 = i or -i x = i-1 or x=-i-1", + "video_name": "dnjK4DPqh0k", + "transcript": "We're asked to solve 2x squared plus 5 is equal to 6x. And so we have a quadratic equation here. But just to put it into a form that we're more familiar with, let's try to put it into standard form. And standard form, of course, is the form ax squared plus bx plus c is equal to 0. And to do that, we essentially have to take the 6x and get rid of it from the right hand side. So we just have a 0 on the right hand side. And to do that, let's just subtract 6x from both sides of this equation. And so our left hand side becomes 2x squared minus 6x plus 5 is equal to-- and then on our right hand side, these two characters cancel out, and we just are left with 0. And there's many ways to solve this. We could try to factor it. And if I was trying to factor it, I would divide both sides by 2. If I divide both sides by 2, I would get integer coefficients on the x squared in the x term, but I would get 5/2 for the constant. So it's not one of these easy things to factor. We could complete the square, or we could apply the quadratic formula, which is really just a formula derived from completing the square. So let's do that in this scenario. And the quadratic formula tells us that if we have something in standard form like this, that the roots of it are going to be negative b plus or minus-- so that gives us two roots right over there-- plus or minus square root of b squared minus 4ac over 2a. So let's apply that to this situation. Negative b-- this right here is b. So negative b is negative negative 6. So that's going to be positive 6, plus or minus the square root of b squared. Negative 6 squared is 36, minus 4 times a-- which is 2-- times 2 times c, which is 5. Times 5. All of that over 2 times a. a is 2. So 2 times 2 is 4. So this is going to be equal to 6 plus or minus the square root of 36-- so let me just figure this out. 36 minus-- so this is 4 times 2 times 5. This is 40 over here. So 36 minus 40. And you already might be wondering what's going to happen here. All of that over 4. Or this is equal to 6 plus or minus the square root of negative 4. 36 minus 40 is negative 4 over 4. And you might say, hey, wait Sal. Negative 4, if I take a square root, I'm going to get an imaginary number. And you would be right. The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. So we're essentially going to get two complex numbers when we take the positive and negative version of this root. So let's do that. So the square root of negative 4, that is the same thing as 2i. And we know that's the same thing as 2i, or if you want to think of it this way. Square root of negative 4 is the same thing as the square root of negative 1 times the square root of 4, I could even do it one step-- that's the same thing as negative 1 times 4 under the radical, which is the same thing as the square root of negative 1 times the square root of 4. And the principal square root of negative 1 is i times the principal square root of 4 is 2. So this is 2i, or i times 2. So this right over here is going to be 2i. So we are left with x is equal to 6 plus or minus 2i over 4. And if we were to simplify it, we could divide the numerator and the denominator by 2. And so that would be the same thing as 3 plus or minus i over 2. Or if you want to write them as two distinct complex numbers, you could write this as 3 plus i over 2, or 3/2 plus 1/2i. That's if I take the positive version of the i there. Or we could view this as 3/2 minus 1/2i. This and these two guys right here are equivalent. Those are the two roots. Now what I want to do is a verify that these work. Verify these two roots. So this one I can rewrite as 3 plus i over 2. These are equivalent. All I did-- you can see that this is just dividing both of these by 2. Or if you were to essentially factor out the 1/2, you could go either way on this expression. And this one over here is going to be 3 minus i over 2. Or you could go directly from this. This is 3 plus or minus i over 2. So 3 plus i over 2. Or 3 minus i over 2. This and this or this and this, or this. These are all equal representations of both of the roots. But let's see if they work. So I'm first going to try this character right over here. It's going to get a little bit hairy, because we're going to have to square it and all the rest. But let's see if we can do it. So what we want to do is we want to take 2 times this quantity squared. So 2 times 3 plus i over 2 squared plus 5. And we want to verify that that's the same thing as 6 times this quantity, as 6 times 3 plus i over 2. So what is 3 plus i squared? So this is 2 times-- let me just square this. So 3 plus i, that's going to be 3 squared, which is 9, plus 2 times the product of three and i. So 3 times i is 3i, times 2 is 6i. So plus 6i. And if that doesn't make sense to you, I encourage you to kind of multiply it out either with the distributive property or FOIL it out, and you'll get the middle term. You'll get 3i twice. When you add them, you get 6i. I And then plus i squared, and i squared is negative 1. Minus 1. All of that over 4, plus 5, is equal to-- well, if you divide the numerator and the denominator by 2, you get a 3 here and you get a 1 here. And 3 distributed on 3 plus i is equal to 9 plus 3i. And what we have over here, we can simplify it just to save some screen real estate. 9 minus 1 is 8. So if I get rid of this, this is just 8 plus 6i. We can divide the numerator and the denominator right here by 2. So the numerator would become 4 plus 3i, if we divided it by 2, and the denominator here is just going to be 2. This 2 and this 2 are going to cancel out. So on the left hand side, we're left with 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works. Now let's try 3 minus i. So once again, just looking at the original equation, 2x squared plus 5 is equal to 6x. Let me write it down over here. Let me rewrite the original equation. We have 2x squared plus 5 is equal to 6x. And now we're going to try this root, verify that it works. So we have 2 times 3 minus i over 2 squared plus 5 needs to be equal to 6 times this business. 6 times 3 minus i over 2. Once again, a little hairy. But as long as we do everything, we put our head down and focus on it, we should be able to get the right result. So 3 minus i squared. 3 minus i times 3 minus i, which is-- and you could get practice taking squares of two termed expressions, or complex numbers in this case actually-- it's going to be 9, that's 3 squared, and then 3 times negative i is negative 3i. And then you're going to have two of those. So negative 6i. So negative i squared is also negative 1. That's negative 1 times negative 1 times i times i. So that's also negative 1. Negative i squared is also equal to negative 1. Negative i is also another square root. Not the principal square root, but one of the square roots of negative 1. So now we're going to have a plus 1, because-- oh, sorry, we're going to have a minus 1. Because this is negative i squared, which is negative 1. And all of that over 4. All of that over-- that's 2 squared is 4. Times 2 over here, plus 5, needs to be equal to-- well, before I even multiply it out, we could divide the numerator and the denominator by 2. So 6 divided by 2 is 3. 2 divided by 2 is 1. So 3 times 3 is 9. 3 times negative i is negative 3i. And if we simplify it a little bit more, 9 minus 1 is going to be-- I'll do this in blue. 9 minus 1 is going to be 8. We have 8 minus 6i. And then if we divide 8 minus 6i by 2 and 4 by 2, in the numerator, we're going to get 4 minus 3i. And in the denominator over here, we're going to get a 2. We divided the numerator and the denominator by 2. Then we have a 2 out here. And we have a 2 in the denominator. Those two characters will cancel out. And so this expression right over here cancels or simplifies to 4 minus 3i. Then we have a plus 5 needs to be equal to 9 minus 3i. I We have a negative 3i on the left, a negative 3i on the right. We have a 4 plus 5. We could evaluate it. This left hand side is 9 minus 3i, which is the exact same complex number as we have on the right hand side, 9 minus 3i. So it also checks out. It is also a root. So we verified that both of these complex roots, satisfy this quadratic equation." + }, + { + "Q": "What is a rational number??", + "A": "Rational numbers is which u convert decimal n fraction into like if a fraction is given to u 1_3 so u will getr it 1:3 understood", + "video_name": "VZOHWaw5dqM", + "transcript": "- We're told to look at the rational numbers below, order them from least to greatest. They really didn't have to tell us this first sentence, I would have known to look at the rational numbers to order them from least to greatest. Well anyway, they tell us 1/2, negative five, three point three, zero, 21 over 12, negative five point five, and two and 1/8ths. So the easiest way to visualize this might just be to make a number line that's long enough that it actually can contain all of these numbers, and then we can think about how we can compare them. So let me just draw a huge number line over here. So, take up almost all the entire screen. I'll stick with, we have some negative numbers here, we go as low as negative five point five, and we have some positive numbers here, looks like we go as high as three point three. This thing is still a little less than two, so we go about as high as three point three, so I can put, I can safely I think put zero right here in the middle, I can go a little bit to the right since we have our negative numbers go more negative. So zero, and let's make this negative one, negative two, negative three, negative four, negative five, and well that should be enough. Negative five, and then in the positive direction we have, one, two, three, in the positive direction. And let's see if we can plot these. So, to start off, to start off let's look at 1/2. Where does 1/2 sit, so it sits, let me actually make the scale a little bit better. So this is one, two and three, and four. Alright, so let's start with 1/2. 1/2 is directly in-between zero and one, it is half of a whole. This right here would be one whole. This would be one whole, let me label that. This over here is one. So 1/2 is directly between zero and one. So 1/2 is gonna sit right over here. So that is, let me write that a little bit bigger, you probably have trouble reading that, Alright, one over two, which is also zero point five. So this is also zero point five, anyway that's where it sits. Then we have negative five. Negative five, well this is negative one, negative two, negative three, negative four, negative five. Negative five sits right over there. And then we have three point three. Positive three point three, I'll do that in blue. Positive three point three. So this is one, two, three, and then we want to do another point three. So point three is about a third of the way, a little less than a third of the way, it would be three point three three three forever, if it was a third of the way. So a third of the way, that looks like about right over here. This is three, this right over here would be three point three, let me label. What I'm gonna do is I'm gonna label the numbers on the number line up here so it's one, two, three, four, this is zero, negative one, negative two, negative three, negative four, negative five, and so on and so forth. And then we get to zero, which is one of the numbers that we've already written down. Zero is obviously right over there on the number line, so I'll just write this zero in orange to make it clear, it's this zero. Then we have 21 over 12, which is an improper fraction, and to think about where we should place that on the number line, to think about where to place it on the number line, let me do this in this blue color. To think about where to place this on the number line let's change it into a mixed number, makes it a little bit easier to visualize, at least for my brain. So 12 goes into 21, well it goes into it one time. One times 12 is 12. If you subtract you get a remainder of, well we could actually regroup here, or borrow, if you don't want to do this in your head, you would get nine, but let's do this. So if we borrow one from the two, the two becomes a one, this becomes 11, or we're really regrouping a 10. Anyway, 11 minus two is nine, one minus one is zero. So we have a remainder of nine. So this thing, written as a mixed number, 21 over 12 written as a mixed number is one and 9/12ths. You get one 12/12ths in there and then you get 9/12ths left over. So one and 9/12ths we can also write that, actually we could've simplified this right from the get go, cause both 21 and 12 are divisible by three, but now we can just divide nine, we can simplify 9/12ths, divide both the numerator and the denominator by three, we then get one and three over four, one and 3/4ths. And just to make it clear, I could have simplified this right from the get go, 21 divided by three, is equal to seven, and 12 divided by three, is equal to four. So this is the same thing as 7/4ths, and if you were to divide four into seven, four goes into seven one time, subtract, one times four is four, subtract to get a remainder of three, one and 3/4ths. So going back to where do we plot this? Well it is, it's one, and then we have 3/4ths, we're going to go three fourths of the way. This is half way, this is one fourths, two fourths, three fourths, would be right over there. So this is our 21 over 12, which is the same thing as 7/4ths, which is the same thing as one and 3/4ths. And then we have negative five point five. Negative five point five, I'll do that in magenta again, running out of colors. Negative five point five, well this is negative five, so negative five point five is going to be between negative five and negative six. So let me add negative six to our number line, right here just to make it clear. So let me go a little bit further, let's say that this is negative six. Negative six, and our number line will keep going to smaller values. Let me scroll to the left a little bit. Negative six, so if we go to negative five point five, it's smack dab in-between negative five and negative six. So this is negative five point five, right over there. And then finally we have two and 1/8ths. I'll do that in orange again, or I'll do it in blue. Two and 1/8ths, so it's two and then 1/8th. And so if we want to find the exact place we could divide this into eighths, this would be 4/8ths, this would be 2/8ths, and that would be 6/8ths, and then 1/8th would sit right over here. So that right over there is two and 1/8th. So we've actually plotted, as best as we could, the exact locations. You didn't have to plot the exact locations if you were just trying to order them, but it doesn't hurt to see exactly where they sit when we order them. So now we've essentially ordered them cause we stuck them all on this number line. The order is negative five point five is the smallest, then negative five, then a zero, and then positive 1/2, then 21 over 12, then two and 1/8th, and then three point three. And we're done." + }, + { + "Q": "Are du and dx able to just be treated like normal variables?\n@5:30 he treats the dx as another variable, to be multiplied with the 7", + "A": "Yes, when using Leibnitz notation, you can think of them as quantifiable numbers. What they really mean is this - an infinitely small change in the given variable. For example, du is an infinitely small change in u. dx is an infinitely small change in x. Then, if we think of it this way, du/dx as giving the slope makes sense! The change in u over the change in x gives slope! Long story short, you can treat them as actual numbers - infinitely small changes in a variable.", + "video_name": "oqCfqIcbE10", + "transcript": "Let's take the indefinite integral of the square root of 7x plus 9 dx. So my first question to you is, is this going to be a good case for u-substitution? Well, when you look here, maybe the natural thing to set to be equal to u is 7x plus 9. But do I see its derivative anywhere over here? If we set u to be equal to 7x plus 9, what is the derivative of u with respect to x going to be? Derivative of u with respect to x is just going to be equal to 7. Derivative of 7x is 7. Derivative of 9 is 0. So do we see a 7 lying around anywhere over here? Well, we don't. But what could we do in order to have a 7 lying around, but not change the value of the integral? Well, the neat thing-- and we've seen this multiple times-- is when you're evaluating integrals, scalars can go in and outside of the integral very easily. Just to remind ourselves, if I have the integral of let's say some scalar a times f of x dx, this is the same thing as a times the integral of f of x dx. The integral of the scalar times a function is equal to the scalar times the integral of the functions. So let me put this aside right over here. So with that in mind, can we multiply and divide by something that will have a 7 showing up? Well, we can multiply and divide by 7. So imagine doing this. Let's rewrite our original integral. So let me draw a little arrow here just to go around that aside. We could rewrite our original integral as being 9 to the integral of times 1/7 times 7 times the square root of 7x plus 9 dx. And if we want to, we could take the 1/7 outside We don't have to, but we can rewrite this as 1/7 times the integral of 7, times the square root of 7x plus 9 dx. So now if we set u equal to 7x plus 9, do we have its derivative laying around? Well, sure. The 7 is right over here. We know that du-- if we want to write it in differential form-- du is equal to 7 times dx. So du is equal to 7 times dx. That part right over there is equal to du. And if we want to care about u, well, that's just going to be the 7x plus 9. That is are u. So let's rewrite this indefinite integral in terms of u. It's going to be equal to 1/7 times the integral of-- and I'll just take the 7 and put it in the back. So we could just write the square root of u du, 7 times dx is du. And we can rewrite this if we want as u to the 1/2 power. It makes it a little bit easier for us to kind of do the reverse power rule here. So we can rewrite this as equal to 1/7 times the integral of u to the 1/2 power du. And let me just make it clear. This u I could have written in white if I want it the same color. And this du is the same du right over here. So what is the antiderivative of u to the 1/2 power? Well, we increment u's power by 1. So this is going to be equal to-- let me not forget this 1/7 out front. So it's going to be 1/7 times-- if we increment the power here, it's going to be u to the 3/2, 1/2 plus 1 is 1 and 1/2 or 3/2. So it's going to be u to the 3/2. And then we're going to multiply this new thing times the reciprocal of 3/2, which is 2/3. And I encourage you to verify the derivative of 2/3 u to the 3/2 is indeed u to the 1/2. And so we have that. And since we're multiplying 1/7 times this entire indefinite integral, we could also throw in a plus c right over here. There might have been a constant. And if we want, we can distribute the 1/7. So it would get 1/7 times 2/3 is 2/21 u to the 3/2. And 1/7 times some constant, well, that's just going to be some constant. And so I could write a constant like that. I could call that c1 and then I could call this c2, but it's really just some arbitrary constant. Oh, actually, no we aren't done. We still just have our entire thing in terms of u. So now let's unsubstitute it. So this is going to be equal to 2/21 times u to the 3/2. And we already know what u is equal to. u is equal to 7x plus 9. Let me put a new color here just to ease the monotony. So it's going to be 2/21 times 7x plus 9 to the 3/2 power plus c. And we are done. We were able to take a kind of hairy looking integral and realize that even though it wasn't completely obvious at first, that u-substitution is applicable." + }, + { + "Q": "What about the boiling point of ethers? Are they generally low or are they high as compared to the others?", + "A": "The boiling point of ethers is generally low, the most common ether, diethyl ether (C2H5-O-C2H5), having a bp of 35\u00c2\u00b0C.", + "video_name": "pILGRZ0nT4o", + "transcript": "- [Voiceover] A liquid boils when its molecules have enough energy to break free of the attractions that exist between those molecules. And those attractions between the molecules are called the intermolecular forces. Let's compare two molecules, pentane on the left and hexane on the right. These are both hydrocarbons, which means they contain only hydrogen and carbon. Pentane has five carbons, one, two, three, four, five, so five carbons for pentane. And pentane has a boiling point of 36 degrees Celsius. Hexane has six carbons, one, two, three, four, five, and six. So six carbons, and a higher boiling point, of 69 degrees C. Let's draw in another molecule of pentane right here. So there's five carbons. Let's think about the intermolecular forces that exist between those two molecules of pentane. Pentane is a non-polar molecule. And we know the only intermolecular force that exists between two non-polar molecules, that would of course be the London dispersion forces, so London dispersion forces exist between these two molecules of pentane. London dispersion forces are the weakest of our intermolecular forces. They are attractions between molecules that only exist for a short period of time. So I could represent the London dispersion forces like this. So I'm showing the brief, the transient attractive forces between these two molecules of pentane. If I draw in another molecule of hexane, so over here, I'll draw in another one, hexane is a larger hydrocarbon, with more surface area. And more surface area means we have more opportunity for London dispersion forces. So I can show even more attraction between these two molecules of hexane. So the two molecules of hexane attract each other more than the two molecules of pentane. That increased attraction means it takes more energy for those molecules to pull apart from each other. More energy means an increased boiling point. So hexane has a higher boiling point than pentane. So as you increase the number of carbons in your carbon chain, you get an increase in the boiling point of your compound. So this is an example comparing two molecules that have straight chains. Let's compare, let's compare a straight chain to a branched hydrocarbon. So on the left down here, once again we have pentane, all right, with a boiling point of 36 degrees C. Let's write down its molecular formula. We already know there are five carbons. And if we count up our hydrogens, one, two, three, four, five, six, seven, eight, nine, 10, 11 and 12. So there are 12 hydrogens, so H12. C5 H12 is the molecular formula for pentane. What about neopentane on the right? Well, there's one, two, three, four, five carbons, so five carbons, and one, two, three, four, five, six, seven, eight, nine, 10, 11 and 12 hydrogens. So C5 H12. So these two compounds have the same molecular formula. All right? So the same molecular formula, C5 H12. The difference is, neopentane has some branching, right? So neopentane has branching, whereas pentane doesn't. It's a straight chain. All right. Let's think about the boiling points. Pentane's boiling point is 36 degrees C. Neopentane's drops down to 10 degrees C. Now, let's try to figure out why. If I draw in another molecule of pentane, all right, we just talk about the fact that London dispersion forces exist between these two molecules of pentane. So let me draw in those transient attractive forces between those two molecules. Neopentane is also a hydrocarbon. It's non-polar. So if I draw in another molecule of neopentane, all right, and I think about the attractive forces between these two molecules of neopentane, it must once again be London dispersion forces. Because of this branching, the shape of neopentane in three dimensions resembles a sphere. So it's just an approximation, but if you could imagine this molecule of neopentane on the left as being a sphere, so spherical, and just try to imagine this molecule of neopentane on the right as being roughly spherical. And if you think about the surface area, all right, for an attraction between these two molecules, it's a much smaller surface area than for the two molecules of pentane, right? We can kind of stack these two molecules of pentane on top of each other and get increased surface area and increased attractive forces. But these two neopentane molecules, because of their shape, because of this branching, right, we don't get as much surface area. And that means that there's decreased attractive forces between molecules of neopentane. And because there's decreased attractive forces, right, that lowers the boiling point. So the boiling point is down to 10 degrees C. All right. I always think of room temperature as being pretty close to 25 degrees C. So most of the time, you see it listed as being between 20 and 25. But if room temperature is pretty close to 25 degrees C, think about the state of matter of neopentane. Right? We are already higher than the boiling point of neopentane. So at room temperature and room pressure, neopentane is a gas, right? The molecules have enough energy already to break free of each other. And so neopentane is a gas at room temperature and pressure. Whereas, if you look at pentane, pentane has a boiling point of 36 degrees C, which is higher than room temperature. So we haven't reached the boiling point of pentane, which means at room temperature and pressure, pentane is still a liquid. So pentane is a liquid. And let's think about the trend for branching here. So we have the same number of carbons, right? Same number of carbons, same number of hydrogens, but we have different boiling points. Neopentane has more branching and a decreased boiling point. So we can say for our trend here, as you increase the branching, right? So not talk about number of carbons here. We're just talking about branching. As you increase the branching, you decrease the boiling points because you decrease the surface area for the attractive forces. Let's compare three more molecules here, to finish this off. Let's look at these three molecules. Let's see if we can explain these different boiling points. So once again, we've talked about hexane already, with a boiling point of 69 degrees C. If we draw in another molecule of hexane, our only intermolecular force, our only internal molecular force is, of course, the London dispersion forces. So I'll just write \"London\" here. So London dispersion forces, which exist between these two non-polar hexane molecules. Next, let's look at 3-hexanone, right? Hexane has six carbons, and so does 3-hexanone. One, two, three, four, five and six. So don't worry about the names of these molecules at this point if you're just getting started with organic chemistry. Just try to think about what intermolecular forces are present in this video. So 3-hexanone also has six carbons. And let me draw another molecule of 3-hexanone. So there's our other molecule. Let's think about electronegativity, and we'll compare this oxygen to this carbon right here. Oxygen is more electronegative than carbon, so oxygen withdraws some electron density and oxygen becomes partially negative. This carbon here, this carbon would therefore become partially positive. And so this is a dipole, right? So we have a dipole for this molecule, and we have the same dipole for this molecule of 3-hexanone down here. Partially negative oxygen, partially positive carbon. And since opposites attract, the partially negative oxygen is attracted to the partially positive carbon on the other molecule of 3-hexanone. And so, what intermolecular force is that? We have dipoles interacting with dipoles. So this would be a dipole-dipole interaction. So let me write that down here. So we're talk about a dipole-dipole interaction. Obviously, London dispersion forces would also be present, right? So if we think about this area over here, you could think about London dispersion forces. But dipole-dipole is a stronger intermolecular force compared to London dispersion forces. And therefore, the two molecules here of 3-hexanone are attracted to each other more than the two molecules of hexane. And so therefore, it would take more energy for these molecules to pull apart from each other. And that's why you see the higher temperature for the boiling point. 3-hexanone has a much higher boiling point than hexane. And that's because dipole-dipole interactions, right, are a stronger intermolecular force compared to London dispersion forces. And finally, we have 3-hexanol over here on the right, which also has six carbons. One, two, three, four, five, six. So we're still dealing with six carbons. If I draw in another molecule of 3-hexanol, let me do that up here. So we sketch in the six carbons, and then have our oxygen here, and then the hydrogen, like that. We know that there's opportunity for hydrogen bonding. Oxygen is more electronegative than hydrogen, so the oxygen is partially negative and the hydrogen is partially positive. The same setup over here on this other molecule of 3-hexanol. So partially negative oxygen, partially positive hydrogen. And so hydrogen bonding is possible. Let me draw that in. So we have a hydrogen bond right here. So there's opportunities for hydrogen bonding between two molecules of 3-hexanol. So let me use, let me use deep blue for that. So now we're talking about hydrogen bonding. And we know that hydrogen bonding, we know the hydrogen bonding is really just a stronger dipole-dipole interaction. So hydrogen bonding is our strongest intermolecular force. And so we have an increased attractive force holding these two molecules of 3-hexanol together. And so therefore, it takes even more energy for these molecules to pull apart from each other. And that's reflected in the higher boiling point for 3-hexanol, right? 3-hexanol has a higher boiling point than 3-hexanone and also more than hexane. So when you're trying to figure out boiling points, think about the intermolecular forces that are present between two molecules. And that will allow you to figure out which compound has the higher boiling point." + }, + { + "Q": "So, it doesn't matter if I add or subtract either 2 equation ?", + "A": "That is right the main aim is to cancel out one of the variables and you add and subtract to get there.", + "video_name": "vA-55wZtLeE", + "transcript": "Let's explore a few more methods for solving systems of equations. Let's say I have the equation, 3x plus 4y is equal to 2.5. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. If you think of it graphically, this would be the intersection of the lines that represent the solution sets to both of these equations. So how can we proceed? We saw in substitution, we like to eliminate one of the variables. We did it through substitution last time. But is there anything we can add or subtract-- let's focus on this yellow, on this top equation right here-- is there anything that we can add or subtract to both sides of this equation? Remember, any time you deal with an equation you have to add or subtract the same thing to both sides. But is there anything that we could add or subtract to both sides of this equation that might eliminate one of the variables? And then we would have one equation in one variable, and we can solve for it. And it's probably not obvious, even though it's sitting right in front of your face. Well, what if we just added this equation to that equation? What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you're probably saying, Sal, hold on, how can you just add two equations like that? And remember, when you're doing any equation, if I have any equation of the form-- well, really, any equation-- Ax plus By is equal to C, if I want to do something to this equation, I just have to add the same thing to both sides of the equation. So I could, for example, I could add D to both sides of the equation. Because D is equal to D, so I won't be changing the equation. You would get Ax plus By, plus D is equal to C plus D. And we've seen that multiple, multiple times. Anything you do to one side of the equation, you have to do to the other side. But you're saying, hey, Sal, wait, on the left-hand side, you're adding 5x minus 4y to the equation. On the right-hand side, you're adding 25.5 to the equation. Aren't you adding two different things to both sides of the equation? And my answer would be no. We know that 5x minus 4y is 25.5. This quantity and this quantity are the same. They're both 25.5. This second equation is telling me that explicitly. So I can add this to the left-hand side. I'm essentially adding 25.5 to it. And I could add 25.5 to the right-hand side. So let's do that. If we were to add the left-hand side, 3x plus 5x is 8x. And then what is 4y minus 4y? And this was the whole point. When I looked at these two equations, I said, oh, I have a 4y, I have a negative 4y. If you just add these two together, they are going to cancel out. They're going to be plus 0y. Or that whole term is just going to go away. And that's going to be equal to 2.5 plus 25.5 is 28. So you divide both sides. So you get 8x is equal to 28. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. That's equal to 7 over 2. That's our x value. Now we want to solve for our y value. And we could substitute this back into either of these two equations. Let's use the top one. You could do it with the bottom one as well. So we know that 3 times x, 3 times 7 over 2-- I'm just substituting the x value we figured out into this top equation-- 3 times 7 over 2, plus 4y is equal to 2.5. Let me just write that as 5/2. We're going to stay in the fraction world. So this is going to be 21 over 2 plus 4y is equal to 5/2. Subtract 21 over 2 from both sides. So minus 21 over 2, minus 21 over 2. The left-hand side-- you're just left with a 4y, because these two guys cancel out-- is equal to-- this is 5 minus 21 over 2. That's negative 16 over 2. So that's negative 16 over 2, which is the same thing-- well, I'll write it out as negative 16 over 2. Or we could write that-- let's continue up here-- 4y-- I'm just continuing this train of thought up here-- 4y is equal to negative 8. Divide both sides by 4, and you get y is equal to negative 2. So the solution to this equation is x is equal to 7/2, y is equal to negative 2. This would be the coordinate of their intersection. And you could try it out on both of these equations right here. So let's verify that it also satisfies this bottom equation. 5 times 7/2 is 35 over 2 minus 4 times negative 2, so minus negative 8. That's equivalent to-- let's see, this is 17.5 plus 8. And that indeed does equal 25.5. So this satisfies both equations. Now let's see if we can use our newly found skills to tackle a word problem, our newly found skills in elimination. So here it says, Nadia and Peter visit the candy store. Nadia buys 3 candy bars and 4 Fruit Roll-Ups for $2.84. Peter also buys 3 candy bars, but can only afford 1 additional Fruit Roll-Up. His purchase costs $1.79. What is the cost of each candy bar and each Fruit Roll-Up? So let's define some variables. Let's just use x and y. Let's let x equal cost of candy bar-- I was going to do a c and a f for Fruit Roll-Up, but I'll just stick with x and y-- cost of candy bar. And let y equal the cost of a Fruit Roll-Up. All right. So what does this first statement tell us? Nadia buys 3 candy bars, so the cost of 3 candy bars is going to be 3x. And 4 Fruit Roll-Ups. Plus 4 times y, the cost of a Fruit Roll-Up. This is how much Nadia spends. 3 candy bars, 4 Fruit Roll-Ups. And it's going to cost $2.84. That's what this first statement tells us. It translates into that equation. The second statement. Peter also buys 3 candy bars, but could only afford 1 additional Fruit Roll-Up. So plus 1 additional Fruit Roll-Up. His purchase cost is equal to $1.79. What is the cost of each candy bar and each Fruit Roll-Up? And we're going to solve this using elimination. You could solve this using any of the techniques we've seen so far-- substitution, elimination, even graphing, although it's kind of hard to eyeball things with the graphing. So how can we do this? Remember, with elimination, you're going to add-- let's focus on this top equation right here. Is there something we could add to both sides of this equation that'll help us eliminate one of the variables? Or let me put it this way, is there something we could add or subtract to both sides of this equation that will help us eliminate one of the variables? Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? And remember, by doing that, I would be subtracting the same thing from both sides of the equation. This is $1.79. Because it says this is equal to $1.79. So if we did that we would be subtracting the same thing from both sides of the equation. So let's subtract 3x plus y from the left-hand side of the equation. And let me just do this over on the right. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So let's subtract it. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. I'm just taking the second equation. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. And what do we get? When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. 3x minus 3x is 0x. I won't even write it down. You get 4x minus-- sorry, 4y minus y. That is 3y. And that is going to be equal to $2.84 minus $1.79. What is that? That's $1.05. So 3y is equal to $1.05. Divide both sides by 3. y is equal to-- what's $1.05 divided by 3? So 3 goes into $1.05. It goes into 1 zero times. 0 times 3 is 0. 1 minus 0 is 1. Bring down a 0. 3 goes into 10 three times. 3 times 3 is 9. Subtract. 10 minus 9 is 1. Bring down the 5. 3 goes into 15 five times. 5 times 3 is 15. Subtract. We have no remainder. So y is equal to $0.35. So the cost of a Fruit Roll-Up is $0.35. Now we can substitute back into either of these equations to figure out the cost of a candy bar. So let's use this bottom equation right here. Which was originally, if you remember before I multiplied it by negative 1, it was 3x plus y is equal to $1.79. So that means that 3x plus the cost of a Fruit Roll-Up, 0.35 is equal to $1.79. If we subtract 0.35 from both sides, what do we get? The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. That's $1.44. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. 1 times 3 is 0. Bring down the 1. Subtract. Bring down the 4. 3 goes into 14 four times. 4 times 3 is 12. I'm making this messy. 14 minus 12 is 2. Bring down the 4. 3 goes into 24 eight times. 8 times 3 is 24. No remainder. So x is equal to 0.48. So there you have it. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35." + }, + { + "Q": "im a fourth grader it dosent make any sense i need help", + "A": "Don t worry about this now. College will come after 12th grade in high school, which is 8 years in the future. But if you want to look into college, please fully understand this topic before you do so you don t waste any of your time. Even if you don t, you will most probably have to start thinking about college in the 8th or 9th grade.", + "video_name": "cGg1j1ZCCOs", + "transcript": "- Hi, I'm Sal Khan, founder of the Khan Academy. I just wanted to welcome you to this resource on navigating the college of, I guess, application, admissions, and paying-for process. You know, for me this is a super-important thing. College was a big part of my life. Obviously, we all know you learn a lot at college, but it also opens up your mind. You have, it's important to be in a community that really challenges you, between your peers, your professors, I met my wife in college, many of the same people that I worked with in college, I remember I was in a entrepreneurship competition my junior year, and many of the people who were on my team then, I'm still working with now, 20 years later, on Khan Academy. And I think it would be fair to say if I hadn't had that experience, Khan Academy might not exist. And because college can be such a mind-expanding, an eye-opening experience, it's really important that everyone realizes that it's actually more accessible than most people think. When I was a kid, my mom raised us as a single mother and I just assumed that we would never be able to afford a selective, private four-year university. But my sister, who was three years older, when she applied to college, she got into actually several really good schools, and her first choice, when she applied, you know, when I looked at the actual dollar amount for the tuition, there's no way. That was actually, the tuition at that university was more than my mom made in a year. But they gave my sister a financial aid package, some was contributed by my mother, but my sister had to take on a loan, do some work study, but they made it so that she could go. And that opened the possibilities for me. When it was my time, I applied where I thought that I could thrive the most, and I applied for financial aid, and like my sister, they made it possible for me to have that very rich and important experience in my own life. So my biggest hope is that, as you go through these resources, you look at it from your own lens, and it opens up the same type of possibilities that I was lucky enough to have in my own life." + }, + { + "Q": "How to weite 770,070 in expanded form", + "A": "if you cannot solve this problem, you need to re-learn the skill. You should be able to find this topic right in Khan Academy, watch the videos, work on the practice problems until you master the skills. That will help you for the long run. That is what I have been doing for myself too. Sincerely.", + "video_name": "iK0y39rjBgQ", + "transcript": "Write 14,897 in expanded form. Let me just rewrite the number, and I'll color code it, and that way, we can keep track of our digits. So we have 14,000. I don't have to write it-- well, let me write it that big. 14,000, 800, and 97-- I already used the blue; maybe I should use yellow-- in expanded form. So let's think about what place each of these digits are in. This right here, the 7, is in the ones place. The 9 is in the tens place. This literally represents 9 tens, and we're going to see this in a second. This literally represents 7 ones. The 8 is in the hundreds place. The 4 is in the thousands place. It literally represents 4,000. And then the 1 is in the ten-thousands place. And you see, every time you move to the left, you move one place to the left, you're multiplying by 10. Ones place, tens place, hundreds place, thousands place, ten-thousands place. Now let's think about what that really means. If this 1 is in the ten-thousands place, that means that it literally represents-- I want to do this in a way that my arrows don't get mixed up. Actually, let me start at the other end. Let me start with what the 7 represents. The 7 literally represents 7 ones. Or another way to think about it, you could say it represents 7 times 1. All of these are equivalent. They represent 7 ones. Now let's think about the 9. That's why I'm doing it from the right, so that the arrows don't have to cross each other. So what does the 9 represent? It represents 9 tens. You could literally imagine you have 9 actual tens. You could have a 10, plus a 10, plus a 10. Do that nine times. That's literally what it represents: 9 actual tens. 9 tens, or you could say it's the same thing as 9 times 10, or 90, either way you want to think about it. So let me write all the different ways to think about it. It represents all of these things: 9 tens, or 9 times 10, or 90. So then we have our 8. Our 8 represents-- we see it's in the hundreds place. It represents 8 hundreds. Or you could view that as being equivalent to 8 times 100-- a hundred, not a thousand-- 8 times 100, or 800. That 8 literally represents 8 hundreds, 800. And then the 4. I think you get the idea here. This represents the thousands place. It represents 4 thousands, which is the same thing as 4 times 1,000, which is the same thing as 4,000. 4,000 is the same thing as 4 thousands. Add it up. And then finally, we have this 1, which is sitting in the ten-thousands place, so it literally represents 1 ten-thousand. You can imagine if these were chips, kind of poker chips, that would represent one of the blue poker chips and each blue poker chip represents 10,000. I don't know if that helps you or not. And 1 ten-thousand is the same thing as 1 times 10,000 which is the same thing as 10,000. So when they ask us to write it in expanded form, we could write 14,897 literally as the sum of these numbers, of its components, or we could write it as the sum of these numbers. Actually, let me write this. This top 7 times 1 is just equal to 7. So 14,897 is the same thing as 10,000 plus 4,000 plus 800 plus 90 plus 7. So you could consider this expanded form, or you could use this version of it, or you could say this the same thing as 1 times 10,000, depending on what people consider to be expanded form-- plus 4 times 1,000 plus 8 times 100 plus 9 times 10 plus 7 times 1. I'll scroll to the right a little bit. So either of these could be considered expanded form." + }, + { + "Q": "how do input cot ,csc and sec on my calculator (fx82) ?", + "A": "Cot= 1/tan CSC = 1/Sin SEC = 1/Cos", + "video_name": "Q7htxHDN8LE", + "transcript": "Determine the six trigonometric ratios for angle A in the right triangle below. So this right here is angle A, its in vertex A and help me remember the definitions of the trig ratios, these are human constructed definitions that have ended up being very, very useful for analyzing a whole series of things in the world and to help me remember them, I use the word, SOH-CAH-TOA let me write that down, so SOH-CAH-TOA sometimes you can think of it as one word, but its really the three parts that define atleast three of the trig functions for you, then we can get the other three by looking at the first three, so SOH tells us that sin of an angle, in this case its sin of A, so sin of A is equal to the opposite opposite, thats the o over the hypotenuse so opposite over the hypotenuse, well in this context what is the opposite sin to angle A? so we go across the triangle it opens up onto side BC, it has length 12, so that is the opposite side. So this is going to be equal to 12, and whats the hypotenuse? well, the hypotenuse is the longest side of the triangle; it's opposite to the 90 degree angle, and so we go opposite the 90 degree angle the longest side is side AB, it has length 13, so this side right over here is the hypotenuse, and so the sin of A is 12/13ths. now lets go to CAH CAH defines, cosine for us. it tells us that cosine of an angle, in this case, cosine of A is equal to the adjacent side, the adjacent side to the angle over the hypotenuse, over the hypotenuse. so whats the adjacent side to angle A? well if we look at angle A, there is 2 sides next to it. One of them is the hypotenuse the other one has length 5, the adjacent one is side CA so its 5, and what is the hypotenuse, well we've already figured that out, the hypotenuse right over here its opposite the 90 degree angle, it the longest side of the right triangle, it has length 13, so the cosine of A is 5/13ths and let me label this, this right over here is the adjacent side and this is all specific to angle A, the hypotenuse would be the same regardless of what angle you pick, but the opposite and adjacent is dependent on the angle we choose in the right triangle, now lets go to TOA. TOA defines tangent for us, it tells us that the tangent the tangent of an angle is equal to the opposite, equal to the opposite side over the adjacent side, so given this definition what is the tangent of A? well the opposite we already figured out has length 12, and the adjacent side we already figured out has length 5 so the tangent of A, which is opposite over adjacent is 12/5ths now we'll go to the other three trig ratios which you can think of as the reciprocals of these right over here, but I'll define them so first you have cosecant, and cosecant; its always a little bit unintuitive, why cosecant is the reciprocal of sine of A even though it starts with a co like cosine but cosecant is the reciprocal of the sin of A, so sin of A is opposite over hypotenuse. Cosecant of A is hypotenuse over opposite and so whats the hypotenuse over the opposite, well hypotenuse is 13 and the opposite side is 12 and notice 13/12ths is the reciprocal of 12/13ths now, secant of A is the reciprocal so instead of it being adjacent over hypotenuse, which we got out of the CAH part of SOH CAH TOA, its hypotenuse over adjacent so what is the secant of A? well the hypotenuse, we figured out multiple times is 13 and what is the adjacent side? it is 5, so 13 13/5ths, which is once again the reciprocal of cosine of A, 5/13ths finally lets get the cotangent, and the cotangent is the reciprocal of the tangent of A, instead of being opposite over adjacent it is adjacent over opposite, so what is the cotangent of A? well we figured out the adjacent side multiple times for angle A, its 5 and the opposite side to angle A is 12, so 5/12ths which is once again the reciprocal of the tangent of A which is 12/5ths" + }, + { + "Q": "what is the point of the scientific method", + "A": "It is a method of reasoning that tries to ensure that its conclusions are logical and accurate descriptions of nature.", + "video_name": "N6IAzlugWw0", + "transcript": "Let's explore the scientific method. Which at first might seem a bit intimidating, but when we walk through it, you'll see that it's actually almost a common-sense way of looking at the world and making progress in our understanding of the world and feeling good about that progress of our understanding of the world. So, let's just use a tangible example here, and we'll walk through what we could consider the steps of the scientific method, and you'll see different steps articulated in different ways, but they all boil down to the same thing. You observe something about reality, and you say, well, let me try to come up with a reason for why that observation happens, and then you try to test that explanation. It's very important that you come up with explanations that you can test, and then you can see if they're true, and then based on whether they're true, you keep iterating. If it's not true, you come up with another explanation. If it is true, but it doesn't explain everything, well once again, you try to explain more of it. So, as a tangible example, let's say that you live in, in I don't know, northern Canada or something, and let's say that you live near the beach, but there's also a pond near your house, and you notice that the pond, it tends to freeze over sooner in the Winter than the ocean does. It does that faster and even does it at higher temperatures than when the ocean seems to freeze over. So, you could view that as your observation. So, the first step is you're making an observation. Observation. In our particular case is that the pond freezes over at higher temperatures than the ocean does, and it freezes over sooner in the Winter. Well, the next question that you might wanna, or the next step you could view as a scientific method. It doesn't have to be this regimented, but this is a structured way of thinking about it. Well, ask yourself a question. Ask a question. Why does, so in this particular question, or in this particular scenario, why does the pond tend to freeze over faster and at higher temperatures than the ocean does? Well, you then try to answer that question, and this is a key part of the scientific method, is what you do in this third step, is that you try to create an explanation, but what's key is that it is a testable explanation. So, you try to create a testable explanation. Testable explanation, and this is kind of the core, one of the core pillars of the scientific method, and this testable explanation is called your hypotypothesis. Your hypothesis. And so, in this particular case, a testable explanation could be that, well the ocean is made up of salt water, and this pond is fresh water, so your testable explanation could be salt water, salt water has lower freezing point. Has lower freezing, freezing point. Lower freezing point, so it takes colder temperatures to freeze it than fresh water. Than fresh water. So, this, right over here, this would be a good hypothesis. It doesn't matter whether the hypothesis is actually true or not. We haven't actually run the experiment, but it's a good one, because we can construct an experiment that tests this very well. Now, what would be an example of a bad hypothesis or of something that you couldn't even necessarily consider as part of the scientific method? Well, you could say that there is a fairy that blesses that blesses, let's say that performs magic, performs magic on the pond to freeze it faster. Freeze it faster. And, the reason why this isn't so good is that this is not so testable, because it's depending on this fairy, and you don't know how to convince the fairy to try to do it again. You haven't seen the fairy. You haven't observed the fairy. It's not based on any observation, and so this one right over here, this would not be a good hypothesis for the scientific method, so we would wanna rule that one out. So, let's go back to our testable explanation, our hypothesis. Salt water has a lower freezing point than fresh water. Well, the next step would be to make a prediction based on that, and this is the part where we're really designing an experiment. So, you could just view all of this as designing. Let me do this in a different color. Where we wanna design an experiment. Design an experiment. And in that experiments lets say, and let's see, the next two steps I will put as part of this experimental. Whoops. I messed up. Let me, I did my undo step. So, the next part that I will do is the experiment. Experiment. And there you go. So, the first thing is, we'll say I take, you know, there's all sorts of things that are going on outside. The ocean has waves. You know, maybe there are boats going by that might potentially break up the ice. So, I just wanna isolate that one variable that I care about, whether something is salt water or not, and I want a control for everything else. So, I want a control for whether there's waves or not or whether there's wind or any other possible explanation for why the pond freezes over faster. So, what I do, in a very controlled environment I take two cups. I take two cups. That's one cup and two cups, and I put water in those cups. I put water in those cups. Now, let's say I start with distilled water, but then this one stays, the first one right over here stays distilled, and distilled means that through evaporation I've taken out all of the impurities of that water, and in the second one I take that distilled water, and I throw a bunch of salt in it. So, this one is fresh, very fresh, and in fact, far fresher than you would find in a pond. It's distilled water. And then this is over here, this is salt water. So, you wouldn't see the salt, but just for our visuals, you depict it. Then we would make a prediction, and we could even view this as step 4, our prediction. We predict that the fresh water will freeze at a higher temperature than the salt water. So, our prediction, let's say the fresh freezes at zero degrees Celsius, but salt doesn't. Salt water doesn't. Salt water doesn't. So, what you then do is that you test your prediction. So, then you test it. And how would you test it? Well, you could have a very accurate freezer that is exactly at zero degrees Celsius, and you put both of these cups into it, and you wanna make sure that they're identical and everything where you control for everything else. You control for the surface area. You control for the material of the glass. You control for how much water there is. But, then you test it. Then you see what happened from your test. Leave it in overnight, and if you see that the fresh water has frozen over, so it's frozen over, but the salt water hasn't, well then that seems to validate your testable explanation. That salt water has a lower freezing point than fresh water, and if it didn't freeze, well it's like, okay, well maybe that, or if there isn't a difference, maybe either both of them didn't freeze or both of them did freeze, then you might say, well, okay, that wasn't a good explanation. I have to find another explanation for why the ocean seems to freeze at a lower temperature. Or, you might say, well that's part of the explanation, but that by itself doesn't explain it, or you might now wanna ask even further questions about, well, when does salt water freeze, and what else is it dependent on? Do the waves have an impact? Does the wind have an impact? So, then you can go into the process of iterating and refining. So, you then refine, refine, refine and iterate on the process. When I'm talking about iterate, you're doing it over again, but then, based on the things that you've learned. So, you might come up with a more refined testable explanation, or you might come up with more experiments that could get you a better understanding of the difference between fresh and salt water, or you might try to come up with experiments for why exactly, what is it about the salt that makes this water harder to freeze? So, that's essentially the essence of the scientific method, and I wanna emphasize this isn't some, you know, bizarre thing. This is logical reasoning. Make a testable explanation for something that you're observing in the world, and then you test it, and you see if your explanation seems to hold up based on the data from your test. And then whether or not it holds up, you then keep going, and you keep refining. And you keep learning more about the world, and the reason why this is better than just saying, oh well, look, okay, I see the pond has frozen over and the ocean hasn't, it must be the salt water, and you know, I just feel good about that, is that you can't feel good about that. There's a million different reasons, and you shouldn't just go on your gut, 'cause at some point, your gut might be right 90% of the time, but that 10% that it's wrong, you're going to be passing on knowledge or assumptions about the world that aren't true, and then other people are going to build on that, and then all of our knowledge is going to be built on kind of a shaky foundation, and so the scientific method ensures that our foundation is strong. And I'll leave you with the gentleman who's often considered to be the father, or one of the fathers of the scientific method. He lived in Cairo, and in what is now Egypt, nearly 1,000 or roughly 1,000 years ago. And he was a famous astronomer and phycisist and mathematician. And his quote is a pretty powerful one, 'cause I think it even stands today: \"The duty of the man who investigates the writings of scientists, if learning the truth is his goal, ...\" Let me start over, just so I can get the dramatic effect right. \"The duty of the man who investigates the writings of scientists, if learning the truth is his goal, is to make himself an enemy of all that he reads, and ... attack it from every side. He should also suspect himself as he performs his critical examination of it, so that he may avoid falling into either prejudice or leniency.\" Hasan Ibn al-Haytham, and his Latinized name is Alhazen. So, he's saying be skeptical, and not just skeptical of what other people write and read, but even of yourself. And another aspect of the scientific method which is super important is, if someone says they made a hypothesis and they tested and they got a result, in order for that to be a good test and in order for that to be a good hypothesis, that experiment has to be reproducible. Someone can't say, oh it's only, you know, a certain time that only happens once every 100 years and not, that that's why it happened that day. It has to be reproducible, and reproducible is key, because then another skeptical scientist like yourself can say, let me see if I can Let me not just believe it, because that person looks like they're smart, and they said that it is true." + }, + { + "Q": "what does co planar mean?", + "A": "4 or more points are coplanar if they lie on the same plane.", + "video_name": "J2Qz-7ZWDAE", + "transcript": "We've already been exposed to points and lines. Now let's think about planes. And you can view planes as really a flat surface that exists in three dimensions, that goes off in every direction. So for example, if I have a flat surface like this, and it's not curved, and it just keeps going on and on and on in every direction. Now the question is, how do you specify a plane? Well, you might say, well, let's see. Let's think about it a little bit. Could I specify a plane with a one point, right over here? Let's call that point, A. Would that, alone, be able to specify a plane? Well, there's an infinite number of planes that could go through that point. I could have a plane that goes like this, where that point, A, sits on that plane. I could have a plane like that. Or, I could have a plane like this. I could have a plane like this where point A sits on it, as well. So I could have a plane like that. And I could just keep rotating around A. So one point by itself does not seem to be sufficient to define a plane. Well, what about two points? Let's say I had a point, B, right over here. Well, notice the way I drew this, point A and B, they would define a line. For example, they would define this line right over here. So they would define, they could define, this line right over here. But both of these points and in fact, this entire line, exists on both of these planes that I just drew. And I could keep rotating these planes. I could have a plane that looks like this. I could have a plane that looks like this, that both of these points actually sit on. I'm essentially just rotating around this line that is defined by both of these points. So two points does not seem to be sufficient. Let's try three. So there's no way that I could put-- Well, let's be careful here. So I could put a third point right over here, point C. And C sits on that line, and C sits on all of these planes. So it doesn't seem like just a random third point is sufficient to define, to pick out any one of these planes. But what if we make the constraint that the three points are not all on the same line. Obviously, two points will always define a line. But what if the three points are not collinear. So instead of picking C as a point, what if we pick-- Is there any way to pick a point, D, that is not on this line, that is on more than one of these planes? We'll, no. If I say, well, let's see, the point D-- Let's say point D is right over here. So it sits on this plane right over here, one of the first ones that I drew. So point D sits on that plane. Between point D, A, and B, there's only one plane that all three of those points sit on. So a plane is defined by three non-colinear points. So D, A, and B, you see, do not sit on the same line. A and B can sit on the same line. D and A can sit on the same line. D and B can sit on the same line. But A, B, and D does not sit on-- They are non-colinear. So for example, right over here in this diagram, we have a plane. This plane is labeled, S. But another way that we can specify plane S is we could say, plane-- And we just have to find three non-collinear points on that plane. So we could call this plane AJB. We could call it plane JBW. We could call it plane-- and I could keep going-- plane WJA. But I could not specify this plane, uniquely, by saying plane ABW. And the reason why I can't do this is because ABW are all on the same line. And this line sits on an infinite number of planes. I could keep rotating around the line, just as we did over here. It does not specify only one plane." + }, + { + "Q": "what about negative c , can the area be negative? integration of -f(x) is equal to the negative of integration of f(x) . So definite integrals are area or something else ?please help.", + "A": "Good question, this is a tricky topic. A negative c value will produce a negative answer. The results of integrals are allowed to be negative. However, you also mention area, which is not allowed to be negative. I teach my students that the area is the absolute value of the integral and that speed is the absolute value of velocity.", + "video_name": "JyArK4jw3XU", + "transcript": "- [Voiceover] We've already seen and you're probably getting tired of me pointing it out repeatedly, that this yellow area right over here, this area under the curve y is equal to f of x and above the positive x-axis or I guess I can say just above the x-axis between x equals a and x equals b, that we can denote this area right over here as the definite integral of from a to b of f of x dx. Now what I want to explore in this video and it'll come up with kind of an answer that you probably could have guessed on your own, but at least get an intuition for it, is that I want to start thinking about the area under the curve that's a scaled version of f of x. Let's say it's y is equal to c times f of x. Y is equal to some number times f of x, so it's scaling f of x. And so I want this to be kind of some arbitrary number, but just to help me visualize, you know I have to draw something so I'm just gonna kind of in my head let's just pretend the c is a three for visualization purposes. So it's going to be three times, so instead of one, instead of this far right over here it's going to be about this far. For right over here, instead of this far right over here it's going to be that and another right over there. And then instead of it's going to be about there. And then instead of it being like that it's going to be one, two and then three, right around there. So I'm starting to get a sense of what this curve is going to look like, a scaled version of f of x. And at least what I'm drawing is pretty close to three times f of x, but just to give you an idea is going to look something like, and let's see over here if this distance, do a second one, a third one, is gonna be up here. It's gonna look something like this. It's gonna look something like that. So this is a scaled version and the scale I did right here I assumed a positive c greater than zero, but this is just for visualization purposes. Now what do we think the area under this curve is going to be between a and b? So what do we think this area right over here is going to be? Now we already know how we can denote it. That area right over there is equal to the definite integral from a to b of the function we're integrating is c f of x dx. I guess to make the question a little bit clearer, how does this relate to this? How does this green area relate to this yellow area? Well one way to think about it is we just scaled the vertical dimension up by c, so one way that you could reason it is if I'm finding the area of something, if I have the area of a rectangle and I have the vertical dimension is let's say I don't want to use those same letters over and over again. Well let's say the vertical dimension is alpha and the horizontal dimension is beta. We know that the area is going to be alpha times beta. Now if I scale up the vertical dimension by c, so instead of alpha this is c times alpha and this is, the width is beta, if I scale up the vertical dimension by c so this is now c times alpha, what's the area going to be? Well it's going to be c alpha times beta, or another way to think of it, when I scale one of the dimensions by c I take my old area and I scale up my old area up by c. And that's what we're doing, we're scaling up the vertical dimension by c. When you multiply c times f of x, f of x is giving us the vertical height. Now obviously that changes as our x changes, but when you think back to the Reimann sums the f of x was what gave us the height of our rectangles. We're now scaling up the height or scaling I should say because we might be scaling down depending on the c. We're scaling it, we're scaling one dimension by c. If you scale one dimension by c you're gonna scale the area by c. So this right over here, the integral, let me just rewrite it. The integral from a to b of c f of x dx, that's just going to be the scaled, we're just going to take the area of f of x, so let me do that in the same color. We're going to take the area under the curve f of x from a to b f of x dx and we're just going to scale it up by this c. So you might say, \"Okay maybe I could have felt \"that was, you know, if I have a c inside the integral \"now I can take the c out of the integral\", and once again this is not a rigorous proof based on the definition of the definite integral, but it hopefully gives you a little bit of intuition why you can do this. If you scale up the function, you're essentially scaling up the vertical dimension, so the area under this is going to just be a scaled up version of the area under the original function f of x. And once again really, really, really useful property of definite integrals that's going to help us solve a bunch of definite integrals. And kind of clarify what we're even doing with them." + }, + { + "Q": "how did sal reduce the original number/ 8:20 to 2:5", + "A": "common factor of 8 and 20 is 4 and 8/4 = 2 and 20/4 = 5 so 2:5. does this help you?", + "video_name": "UK-_qEDtvYo", + "transcript": "Voiceover:Let's think about another scenario involving ratios. In this case, let's think about the ratio of the number of apples. Number of apples to ... Instead of taking the ratio of the number of apples to the number of oranges, let's take the number of apples to the number of fruit. The number of fruit that we have over here. And I encourage you to pause the video and think about that on your own. Well, how many total apples do we have? We have 2, 4, 6, 8 apples. So we're going to have 8 apples. And then how much total fruit do we have? Well we have 8 apples and we have 3, 6, 9, 12 oranges. So our total fruit is 8 plus 12. We have 20 pieces of fruit. So this ratio is going to be 8 to ... 8 to 20. Or, if we want to write this in a more reduced form, we can divide both of these by 4. 4 is their greatest common divisor. And so this is the same thing as a ratio. 8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as 2 over 5. As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to. 2 to 5. Or we could say it's 2/5, the fraction 2/5, which would sometimes be read as 2 to 5. This is also, when it's written this way, you could also read that as a ratio, depending on the context. In a sentence like this I would read this as 2/5 of the fruit are apples." + }, + { + "Q": "At 3:38,sal says atooms are joined by covalent bonds ......so my first question is what are covalent bonds and how can we find radius if atom is not joined to anyone??", + "A": "Covalent bonds are bonds between two atoms sharing electrons.", + "video_name": "q--2WP8wXtk", + "transcript": "Voiceover: Let's think a little bit about the notion of atomic size or atomic radius in this video. At first thought, you might think well this might be a fairly straight-forward thing. If I'm trying to calculate the radius of some type of circular object I'm just thinking about well what's the distance between the center of that circular object and the edge of it. So the length of this line right over here. That would be the radius. And so a lot of people when they conceptualize an atom they imagine a positive nucleus with the protons in the center right over here then they imagine the electrons on these fixed orbits around that nucleus so they might imagine some electrons in this orbit right over here, just kind of orbiting around and then there might be a few more on this orbit out here orbiting around, orbiting around out here. And you might say, \"well okay, that's easy to figure out the atomic radius. I just figure out the distance between the nucleus and the outermost electron and we could call that the radius.\" That would work except for the fact that this is not the right way to conceptualize how electrons or how they move or how they are distributed around a nucleus. Electrons are not in orbits the way that planets are in orbit around the sun and we've talked about this in previous videos. They are in orbitals which are really just probability distributions of where the electrons can be, but they're not that well defined. So, you might have an orbital, and I'm just showing you in 2 dimensions. It would actually be in 3 dimensions, where maybe there's a high probability that the electrons where I'm drawing it in kind of this more shaded in green. But there's some probability that the electrons are in this area right over here and some probability that the electrons are in this area over here, and let's say even a lower probability that the electrons are over this, like this over here. And so you might say, well at a moment the electron's there. The outermost electron we'd say is there. You might say well that's the radius. But in the next moment, there's some probability it might be likely that it ends up here. But there's some probability that it's going to be over there. Then the radius could be there. So electrons, these orbitals, these diffuse probability distributions, they don't have a hard edge, so how can you say what the size of an atom actually is? There's several techniques for thinking about this. One technique for thinking about this is saying, okay, if you have 2 of the same atom, that are- 2 atoms of the same element that are not connected to each other, that are not bonded to each other, that are not part of the same molecule, and you were able to determine somehow the closest that you could get them to each other without them bonding. So, you would kind of see, what's the closest that they can, they can kind of get to each other? So let's say that's one of them and then this is the other one right over here. And if you could figure out that distance, that closest, that minimum distance, without some type of, you know, really, I guess, strong influence happening here, but just the minimum distance that you might see between these 2 and then you could take half of that. So that's one notion. That's actually called the Van der Waals radius. Another way is well what about if you have 2 atoms, 2 atoms of the same element that are bonded to each other? They're bonded to each other through a covalent bond. So a covalent bond, we've already- we've seen this in the past. The most famous of covalent bonds is well, a covalent bond you essentially have 2 atoms. So that's the nucleus of one. That's the nucleus of the other. And they're sharing electrons. So their electron clouds actually, their electron clouds actually overlap with each other, actually overlap with each other so the covalent bond, there the electrons in that bond could spend some of their time on this atom and some of their time on this atom right over here. And so when you have a covalent bond like this, you can then find the distance between the 2 nuclei and take half of that and call that call that the atomic radius. So these are all different ways of thinking about it. Now, with that out of the way, let's think about what the trends for atomic size or atomic radii would be in the periodic table. So the first thing to think about is what do you think will be the trend for atomic radii as we move through a period. So let's say we're in the fourth period and we were to go from potassium to krypton. What do you think is going to be the trend here? And if you want to think about the extremes, how do you think potassium is going to compare to krypton in terms of atomic radius. I encourage you to pause this video and think about that on your own. Well, when you're in the fourth period, the outermost electrons are going to be in your fourth shell. Here, you're filling out 4S1, 4S2. Then you start back filling into the 3D subshell and then you start filling again in 4P1 and so forth. You start filling out the P subshell. So as you go from potassium to krypton, you're filling out that outermost fourth shell. Now what's going on there? Well, when you're at potassium, you have 19- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. You have 19 protons and you have 19 electrons. Well I'll just draw those. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, but you only have 1 electron in that outermost, in that fourth shell, so let's just say that's that electron at a moment, just for visually. It doesn't necessarily have to be there but just to visualize that. So that 1 electron right over there, you have 19, yeah, you have 19 protons. So, you have some, I guess you could say Coulom force that is attracting it, that is keeping it there. But if you go to krypton, all of a sudden you have much more positive charge in the nucleus. So you have 1, 2, 3, 4, 5, 6, 7, 8- I don't have to do them all. You have 36. You have a positive charge of 36. Let me write that, you have plus 36. Here you have plus 19. And you have 36 electrons, you have 36 electrons- I don't know, I've lost track of it, but in your outermost shell, in your fourth, you're going to have the 2S and then you're going to have the 6P. So you have 8 in your outermost shell. So that'd be 1, 2, 3, 4, 5, 6, 7, 8. So one way to think about it, if you have more positive charge in the center, and you have more negative charge on that outer shell, so that's going to bring that outer shell inward. It's going to have more I guess you could imagine one way, more Coulomb attraction right over there. And because of that, that outer most shell is going to drawn in. Krypton is going to be smaller, is going to have a smaller atomic radius than potassium. So the trend, as you go to the right is that you are getting, and the general trend I would say, is that you are getting smaller as you go to the right in a period. That's the reason why the smallest atom of all, the element with the smallest atom is not hydrogen, it's helium. Helium is actually smaller than hydrogen, depending on how you, depending on what technique you use to measure it. That's because, if we take the simplest case, hydrogen, you have 1 proton in the nucleus and then you have 1 electron in that 1S shell, and in helium you have 2, 2 protons in the nucleus and I'm not drawing the neutrons and obviously there's different isotopes, different numbers of neutrons, but you have 2 electrons now in your outer most shell. So you have more, I guess you could say, you could have more Coulomb attraction here. This is plus 2 and then these 2 combined are negative 2. They're going to be drawn inward. So, that's the trend as we go to the right, as we go from the left to the right of the periodic table, we're getting smaller. Now what do you think is going to happen as we go down the period table? As we go down the periodic table in a given group? Well, as we go down a group, each new element down the group, we're adding, we're in a new period. We're adding a new shell. So you're adding more and more and more shells. Here you have just the first shell, now the second shell and each shell is getting further and further and further away. So as you go down the periodic table, you are getting, you are getting larger. You're having a larger atomic radius depending on how you are measuring it. So what's the general trend? Well if you get larger as you go down, that means you're getting smaller as you go up. You get smaller, smaller as you go up. So, what are going to be, what's going to be the smallest ones? Well, we've already said helium is the smallest. So what are going to be some of the largest? What are going to be some of the largest atoms? Well that's going to be the atoms down here at the bottom left. So, these are going to be large, these are going to be small. So, large over here, small over here and the general trend, as you go from the bottom left to the top right you are getting, you are getting smaller." + }, + { + "Q": "Do you think 4th grade and 3rd grade are easy?I do", + "A": "im 7th grade and 5th and the rest are easy", + "video_name": "NWJinKmWzx8", + "transcript": "- [Voiceover] What I hope to do in this video is get a little more practice and intuition when we're multiplying multi-digit numbers. So let's say that we wanted to calculate what 7,000 times six is. 7,000 times six. Now, for some of you, it might just jump out at you. That, hey look, if I have seven of anything, and here I have seven thousands, and I multiply that by six, I'm now going to have seven times six of that thing, or 42 of that thing, and in this case we have 42 thousands. So you might just be able to cut to the chase and say, hey look, six times seven thousands is going to be 42 thousands. And that's great if you can just cut to the chase like that, and another way to think about it is like, look, six times seven is 42, and then since we're talking about thousands, we're not just talking about seven, we're talking about seven thousands. I have three zeros here, so I'm going to have 42 thousands, three zeros there. But I want to make sure that we really understand what is going on here. This will also help us with a little bit of practice of our multiplication properties. So 7,000 is the same thing as 1,000 times seven, or seven times 1,000. It's seven thousands. Or you could view it as a thousand sevens, either way. So this is the same thing as 1,000 times seven times six. And so you could view it as, you could do 1,000 times seven first, which would be 7,000, and then times six. Or you could do the seven times six first, and this is, this right over here is the associative property of multiplication. It sounds very fancy, but it just says that, hey look, we can multiply the seven times six first, before we multiply by the thousand. So we could rewrite this as 1,000 times, and if we're going to do the seven times six first we can put the parentheses around that, times seven, times six. Seven times six. Notice, it's 1,000 times seven times six. I could do 1,000 times seven first to get 7,000, or I could do the seven times six first to get, and you know where there is going, so if you multiply the seven times six first, you're going to get 42, and you're gong to have 1,000 times 42. 1,000 times 42. So you can view this as a thousand 42's or maybe a little bit more intuitively you could view this as 42 thousands. So, once again, we get to 42,000. And so the whole reason, some of y'all might have just been able to do this immediately in your head and that's all good, but it's good to understand what's actually going on here. And the reason why I also broke it up that way, this way, is that the exercises on Kahn Adacemy make you do this to make sure that you really are understanding how to break up these numbers and how you could re-associate when you multiply. Let's do another one. Let's say that we wanted to figure out, let's say that we wanted to figure out, let me give ourselves some space, let's say that we wanted to figure out what 56 times eight is. And there is a bunch of different ways that you could do it. You could say that, look, 56, this is the same thing as 50, five tens, that's 50 plus six ones, so 50 plus six, and all of that times eight. And then you could distribute the eight and you could say, look, this is going to be 50 times eight, So it's going to be 50 times eight, plus six times eight. Plus six times eight. And 50 times eight? Well, five times eight is 40, but we're not just saying five, we're saying five tens, so five tens times eight is going to be 40 tens, or it's going to be 400. Another way to think about it, five times eight is 40, but we're not talking about five, we're talking about five tens, so it's going to be 40 tens. So 50 times eight is 400, and then six times eight is, of course, equal to 48. So this is going to be equal to 448. This is actually how I do things in my head. When I do it in my head I obviously am not writing things down like this but I think, okay, 56 times eight, I could break that up into 50 and six, and eight times 50, well that's 400, or 40 tens you could say. Eight times five tens is going to be 40 tens, or 400. And then eight times six is going to be 48, so it's going to be 400 plus 48. Once you get some practice you're going to be able to do things like this in your head. And if it helps, we can also visualize this looking at an area of a rectangle. So imagine this rectangle right over here, and let's say that this dimension right over here is eight, it is eight units tall. So that's the eight. And this entire dimension, this entire length here is 56. So the area of this rectangle is going to be 56 times eight, which is what we set to figure out, and to do that, well we could break it up into 50 and six, so this first section right over here, this has length, we could say this has length 50, that has length 50, and then this second section, this has length six. This has length six. And the reason why we broke it up this way is cause we can, maybe in our heads, or without too much work, figure out what eight times 50 is and then separately figure out what eight times six is. So separately figure out the areas of these two pieces of the big rectangle and then add them together. So what's eight times five tens? Well it's going to be 40 tens, or 400. This is going to be 400 square units, is going to be the area of this yellow part. And then what's eight times six? Well we know that's going to be 48 square units. So the entire rectangle is going to be the eight times the 50, or the 50 times the eight, the 400, this area, the yellow area, plus the magenta area, plus the eight times the six, the 48, which is 448. 448 is going to be the area of the whole thing. Eight times 56." + }, + { + "Q": "At 0:21, what does Sal mean by \" arbrutary\"?", + "A": "Any old An arbitrary angle measure is a random angle or just any old angle", + "video_name": "0gzSreH8nUI", + "transcript": "Thought I would do some more example problems involving triangles. And so this first one, it says the measure of the largest angle in a triangle is 4 times the measure of the second largest angle. The smallest angle is 10 degrees. What are the measures of all the angles? Well, we know one of them. We know it's 10 degrees. Let's draw an arbitrary triangle right over here. So let's say that is our triangle. We know that the smallest angle is going to be 10 degrees. And I'll just say, let's just assume that this right over here is the measure of the smallest angle. It's 10 degrees. Now let's call the second largest angle-- let's call that x. So the second largest angle, let's call that x. So this is going to be x. And then the first sentence, they say the measure of the largest angle in a triangle is 4 times the measure of the second largest angle. So the second largest angle is x. 4 times that measure is going to be 4x. So the largest angle is going to be 4x. And so the one thing we know about the measures of the angles inside of a triangle is that they add up to 180 degrees. So we know that 4x plus x plus 10 degrees is going to be equal to 180 degrees. It's going to be equal to 180. And 4x plus x, that just gives us 5x. And then we have 5x plus 10 is equal to 180 degrees. Subtract 10 from both sides. You get 5x is equal to 170. And so x is equal to 170/5. And let's see, it'll go into it-- what is that, 34 times? Let me verify this. So 5 goes into-- yeah, it should be 34 times because it's going to go into it twice as many times as 10 would go into it. 10 would go into 170 17 times. 5 would go into 170 34 times. So we could verify it. Go into 170. 5 goes into 17 three times. 3 times 5 is 15. Subtract, you get 2. Bring down the 0. 5 goes into 20 four times, and then you're not going to have a remainder. 4 times 5 is 20. No remainder. So it's 34 times. So x is equal to 34. So the second largest angle has a measure of 34 degrees. This angle up here is going to be 4 times that. So 4 times 34-- let's see, that's going to be 120 degrees plus 16 degrees. This is going to be 136 degrees. Is that right? 4 times 4 is 16, 4 times 3 is 120, 16 plus 120 is 136 degrees. So we're done. The three measures, or the sizes of the three angles, are 10 degrees, 34 degrees, and 136 degrees. Let's do another one. So let's see. We have a little bit of a drawing here. And what I want to do is-- and we could think about different things. We could say, let's solve for x. I'm assuming that 4x is the measure of this angle. 2x is the measure of that angle right over there. We can solve for x. And then if we know x, we can figure out what the actual measures of these angles are, assuming that we can figure out x. And the other thing that they tell us is that this line over here is parallel to this line over here. And it was very craftily drawn. Because it's parallel, but one stops here, and then one starts up there. So the first thing I want to do-- if they're telling us that these two lines are parallel, there's probably going to be something involving transversals or something. It might be something involving-- the other option is something involving triangles. And at first, you might say, wait, is this angle and that angle vertical angles? But you have to be very careful. This is not the same line. This line is parallel to that line. This line, it's bending right over there, so we can't make any type of assumption like that. So the interesting thing-- and I'm not sure if this will lead in the right direction-- is to just make it clear that these two are part of parallel lines. So I could continue this line down like this. And then I can continue this line up like that. And then that starts to look a little bit more like we're used to when we're dealing with parallel lines. And then this line segment, BC-- or we could even say line BC, if we were to continue it on. If we were to continue it on and on, even pass D, then this is clearly a transversal of those two parallel lines. This is clearly a transversal. And so if this angle right over here is 4x, it has a corresponding angle. Half of the-- or maybe most of the work on all of these is to try to see the parallel lines and see the transversal and see the things that might be useful for you. So that right there is the transversal. These are the parallel lines. That's one parallel line. That is the other parallel line. You can almost try to zone out all of the other stuff in the diagram. And so if this angle right over here is 4x, it has a corresponding angle where the transversal intersects the other parallel line. This right here is its corresponding angle. So let me draw it in that same yellow. This right over here is a corresponding angle. So this will also be 4x. And we see that this angle-- this angle and this angle, this angle that has measure 4x and this angle that measures 2x-- we see that they're supplementary. They're adjacent to each other. Their outer sides form a straight angle. So they're supplementary, which means that their measures add up to 180 degrees. They kind of form-- they go all the way around like that if you add the two adjacent angles together. So we know that 4x plus 2x needs to be equal to 180 degrees, or we get 6x is equal to 180 degrees. Divide both sides by 6. You get x is equal to 30, or x is equal to-- well, I shouldn't say-- well, x could be 30. And then this angle right over here is 2 times x. So it's going to be 60 degrees. So this angle right over here is going to be 60 degrees. And this angle right over here is 4 times x. So it is 120 degrees, and we're done." + }, + { + "Q": "Is hydrogen an balanced element??", + "A": "What is the difference between the diatomic elemental molecule H2 and an atom of He?", + "video_name": "FmQoSenbtnU", + "transcript": "In the last few videos we learned that the configuration of electrons in an atom aren't in a simple, classical, Newtonian orbit configuration. And that's the Bohr model of the electron. And I'll keep reviewing it, just because I think it's an important point. If that's the nucleus, remember, it's just a tiny, tiny, tiny dot if you think about the entire volume of the actual atom. And instead of the electron being in orbits around it, which would be how a planet orbits the sun. Instead of being in orbits around it, it's described by orbitals, which are these probability density functions. So an orbital-- let's say that's the nucleus it would describe, if you took any point in space around the nucleus, the probability of finding the electron. So actually, in any volume of space around the nucleus, it would tell you the probability of finding the electron within that volume. And so if you were to just take a bunch of snapshots of electrons -- let's say in the 1s orbital. And that's what the 1s orbital looks like. You can barely see it there, but it's a sphere around the nucleus, and that's the lowest energy state that an electron can be in. If you were to just take a number of snapshots of electrons. Let's say you were to take a number of snapshots of helium, which has two electrons. Both of them are in the 1s orbital. It would look like this. If you took one snapshot, maybe it'll be there, the next snapshot, maybe the electron is there. Then the electron is there. Then the electron is there. Then it's there. And if you kept doing the snapshots, you would have a bunch of them really close. And then it gets a little bit sparser as you get out, as you get further and further out away from the electron. But as you see, you're much more likely to find the electron close to the center of the atom than further out. Although you might have had an observation with the electron sitting all the way out there, or sitting over here. So it really could have been anywhere, but if you take multiple observations, you'll see what that probability function is describing. It's saying look, there's a much lower probability of finding the electron out in this little cube of volume space than it is in this little cube of volume space. And when you see these diagrams that draw this orbital like this. Let's say they draw it like a shell, like a sphere. And I'll try to make it look three-dimensional. So let's say this is the outside of it, and the nucleus is sitting some place on the inside. They're just saying -- they just draw a cut-off -- where can I find the electron 90% of the time? So they're saying, OK, I can find the electron 90% of the time within this circle, if I were to do the cross-section. But every now and then the electron can show up outside of that, right? Because it's all probabilistic. So this can still happen. You can still find the electron if this is the orbital we're talking about out here. Right? And then we, in the last video, we said, OK, the electrons fill up the orbitals from lowest energy state to high energy state. You could imagine it. If I'm playing Tetris-- well I don't know if Tetris is the thing-- but if I'm stacking cubes, I lay out cubes from low energy, if this is the floor, I put the first cube at the lowest energy state. And let's say I could put the second cube But I only have this much space to work with. So I have to put the third cube at the next highest energy state. In this case our energy would be described as potential energy, right? This is just a classical, Newtonian physics example. But that's the same idea with electrons. Once I have two electrons in this 1s orbital -- so let's say the electron configuration of helium is 1s2 -- the third electron I can't put there anymore, because there's only room for two electrons. The way I think about it is these two electrons are now going to repel the third one I want to add. So then I have to go to the 2s orbital. And now if I were to plot the 2s orbital on top of this one, it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. Then you have the blue area, then the red, and the blue. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. But if you look at these, there's three ways that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon. So the electron configuration for carbon, the first two electrons go into, so, 1s1, 1s2. So then it fills-- sorry, you can't see everything. So it fills the 1s2, so carbon's configuration. It fills 1s1 then 1s2. And this is just the configuration for helium. And then it goes to the second shell, which is the second period, right? That's why it's called the periodic table. We'll talk about periods and groups in the future. And then you go here. So this is filling the 2s. We're in the second period right here. That's the second period. One, two. Have to go off, so you can see everything. So it fills these two. So 2s2. And then it starts filling up the p orbitals. So then it starts filling 1p and then 2p. And we're still on the second shell, so 2s2, 2p2. So the question is what would this look like if we just wanted to visualize this orbital right here, the p orbitals? So we have two electrons. So one electron is going to be in a-- Let's say if this is, I'll try to draw some axes. That's too thin. So if I draw a three-dimensional volume kind of axes. If I were to make a bunch of observations of, say, one of the electrons in the p orbitals, let's say in the pz dimension, sometimes it might be here, sometimes it might be there, sometimes it might be there. And then if you keep taking a bunch of observations, you're going to have something that looks like this bell shape, this barbell shape right there. And then for the other electron that's maybe in the x direction, you make a bunch of observations. Let me do it in a different, in a noticeably different, color. It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. The p orbitals. Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. And then you start filling the second energy shell. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons in the lowest energy state, will be 1s2. So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then we have 3s2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus, there's more space in between the lower energy orbitals to fill in more of these bizarro-shaped orbitals. But these are kind of the balance -- I will talk about standing waves in the future -- but these are kind of a balance between trying to get close to the nucleus and the proton and those positive charges, because the electron charges are attracted to them, while at the same time avoiding the other electron charges, or at least their mass distribution functions. Anyway, see you in the next video." + }, + { + "Q": "Can 2 sperms fertilize one egg?", + "A": "The correct answer is if that happens hydatiform mole will occur. either partial or complete.", + "video_name": "VYSFNwTUkG0", + "transcript": "- We're gonna talk about the ovarian cycle. The ovaries are two structures in a female's reproductive system that produce her eggs. Each month her eggs go through a maturation process called the ovarian cycle, and that cycle creates a secondary oocyte than can be then fertilized by a sperm to result in a pregnancy. The ovarian cycle is also responsible for what we commonly know as the menstrual cycle. Basically, the primary oocytes that are destined to be ovulated will develop in the ovaries, complete meiosis one just before ovulation, and then they'll be ejected out of the ovary as a secondary oocyte to be picked up by the fimbriae and swept into the uterine tube to hope for fertilization. So let's start from the beginning. Inside the ovaries, eggs develop in structures called follicles, these purple circles here. And they start off as primordial follicles. And so what a follicle is- I'll just blow that up for you- It's one primary oocyte, so an egg cell, surrounded by a layer of cells called granulosa cells. And the granulosa cells develop and become more numerous as the follicle matures. Now the granulosa cells also secrete a few hormones. Estrogen, a little progesterone and some inhibin, and we'll talk about the functions of those a little bit later on. So let's put a timeline on this. Now the ovarian cycle lasts 28 days. This is day zero here at the primordial follicle, where we're going counter-clockwise. All the way over here, this is day 13. Here, where the secondary oocyte gets ejected, or ovulated, that's day 14. And then the rest of the time spent getting back to the primordial follicle stage are days 15 through 28. So now you have an idea of about how long this all takes. So you remember when we said that the granulosa cells produce hormones? Well, as the follicles develop over the first 13 days, and you can see the changes between the one here and the one here. It's got a lot more purple cells around here. Those are granulosa cells. So the number of granulosa cells goes up, and since they produce hormones, what do you think happens to the hormone levels in the blood? They go up. So that's sort of just a general point. So keep that in mind, but first we'll jump back to these. We know these are primordial follicles here. The next stage of development are these guys here, and these are called primary follicles. And in the primary follicles, the layers of granulosa cells and the oocyte, the egg, start to be separated by this other layer that starts to form between them. That's called the zona pellucida, and I'll draw it here in light blue. And even though the egg I've drawn in blue, there's still a layer of zona pellucida, even though the egg is originally drawn in blue because I wanted to draw the egg in blue. There's still a layer of zone pellucida around it. Now even though the zona pellucida is there separating the granulosa cells from the actual egg, the granulosa cells can still nourish the egg through gap junctions that go through the zona pellucida and into the egg. Gap junctions are just little passageways from one cell to another cell where they can exchange nutrients or other signals. And actually, through those gap junctions, the granulosa cells send through little chemicals that keep those primary oocytes stuck at that meiosis one stage, 'cause you remember at this point all of these primary oocytes are stuck in meiotic arrest. They're not dividing and reducing their chromosome copy number. So as we develop from our primordial to our primary to our next follicle here, called our pre-antral follicle, and you'll see why it's called that in a minute, the granulosa cells are actually starting to divide and become a lot greater in number. You can see that there's a pretty big difference in granulosa cell number from our primary follicles to our pre-antral follicle here. And remember the granulosa cells are shaded in in purple here. So while the granulosa cells are proliferating, this wall on the outside of the follicle called the theca starts to form. Theca cells have receptors for luteinizing hormone from the anterior pituitary, and when luteinizing hormone, or LH, binds these theca cells, they produce a hormone called androstenedione. And when the thecas get androstenedione, they give it to the granulosa cells, who then convert it to estrogen and release it into the blood. So the blood estrogen levels start to go up at this point. And so that's what these red and blue bits running down the middle of the ovary are, blood vessels, arteries and veins. And if they look a little bit weird to you, or unusual, that's just because they're cut in cross-section as well. Now you might be wondering what an antral refers to, like what you see in the pre-antral follicle and this early antral follicle here. It actually refers to the antrum, which will be formed in the next step. This space here is called an antrum. And the antrum is just basically fluid that's being produced by the granulosa cells. And it's that antrum and the fluid in the antrum that pushes against the edges of the follicle and causes it to expand. Now just so you're aware, during this ovarian cycle, multiple follicles are actually forming. It's not just this one pre-antral follicle, and then this one early antral follicle, and this one mature follicle. You're getting a lot of these happening at one time. But only one of the biggest ones is the one that eventually gets ovulated, because you only ovulate one egg every 28 days. And that one that gets ovulated is called the dominant follicle. So let's just say that what we're seeing here is an example of the dominant follicle's development. Because the rest of the ones that were developing along this pathway sort of degenerate and die off in a process called atresia. So I'll write that at the bottom here. And atresia just means to degenerate. So another note. In the ones that undergo atresia, both the follicle and the eggs they contain die off. And that means that a woman loses anywhere between 15 to 25 eggs per menstrual cycle to atresia, while only one gets ovulated. So you can kind of imagine how you go from two to four million eggs when you were born to having zero after about 35-ish years of ovulation. It's not just that one egg you lose by ovulation. You lose quite a few. So anyway, back to the development of the dominant follicle. It enlarges mostly due to the expanding antrum, as I mentioned earlier. And granulosa cells actually start to form this bit of a mound here that protrudes into the middle of the antrum. This mound of granulosa cells is called the cumulus oophorus. As part of the development of the dominant follicle, the cumulus oophorus and the egg sort of separate together from the wall of the follicle and float around in the middle of the antrum, like a little island. And the follicle increases in size. So the actual follicle is increasing in size as it gets filled with more and more fluid from the granulosa cells. And the granulosa cells are just producing fluid as a by-product of their metabolism and creation of hormones. Eventually this dominant follicle, which at this point is called the mature follicle, it starts to balloon out the side of the ovary, kind of like this. Just starts to push out against the edge of the ovary. And then because the edge of the ovary and the wall of the mature follicle are in such close proximity, enzymes within the follicle break down that common wall between them, and the egg pops out onto the surface of the ovary, because now this wall is broken down. And by the way, an enzyme is a protein that carries out a specific task. The task here is to break down the wall between the mature follicle and the ovary, and that happens on day 14. So it takes day zero to 13 of build up to get to this event. When this happens, some women feel a little bit of pelvic pain. And actually sometimes, by chance, two or more follicles reach maturity, and they all pop out. And that's how you get twins or triplets or quadruplets or octuplets, when they all pop out and get fertilized by different sperm each. Because they're all subsequently swept up into the uterine tubes where sperm can fertilize them. So now you have the egg out here, but what about the old follicle it was in? The follicle actually collapses a little and transforms into a structure called the corpus luteum. And in this transformation the granulosa cells get a lot bigger and start to produce more estrogen, progesterone and that other hormone, inhibin, that we mentioned before. Just briefly, inhibin lowers the amount of FSH, follicle stimulating hormone, that comes from the anterior pituitary. And it does that because follicle stimulating hormone actually propagates this whole process of follicle maturation, as you can imagine from the name. So if you didn't know this before, these are the exact follicles that follicle stimulating hormone refers to. At least in a female. Anyway, if the egg doesn't get fertilized, then the corpus luteum reaches a maximum size in about 10 days. So that's about day 25, which it's probably sitting at in this diagram. And then it degenerates by apoptosis. That's a process that cells use to sort of self-destruct and die off. And here I'm abbreviating corpus luteum as CL, just so you know what I mean. But if the egg is fertilized, i.e., it travels into the uterine tubes and gets fertilized by a sperm, then the corpus luteum persists, I mean it keeps living, because we want it to keep producing estrogen and progesterone. That's because estrogen and progesterone prepare the inner lining of the uterus, that's called the endometrium, for implantation, which would be really handy since we have a fertilized egg now that needs to develop. And that's where it does it, by implanting in the endometrium of the uterus. So just a final note. Ovulation doesn't happen forever. At about age 50 to 51, females undergo something called menopause. First menstrual cycles become less and less regular. In other words, they don't happen every 28 days like they do when you're under the age of 50. And then ultimately, they stop happening entirely. And that cessation of ovulation is called menopause. The main cause of menopause is sometimes referred to as ovarian failure. Basically the ovaries lose the ability to respond to signalling hormones from the brain called gonadotropins. And we know these as LH and FSH. And this happens because most, or all of the follicles and eggs have already gone through that process that we talked about called atresia. In other words, they've degenerated." + }, + { + "Q": "The solution to IXI=5 ARE THE same distance from 0. What can you say abou the solutions to Ix-2I=5?", + "A": "They are the same distance from whatever makes the expression inside the | | zero. |x| - the expression is 0 when x=0, so the solutions are the same distance from 0. |x - 2| - the expression is 0 when x - 2 = 0. That happens when x = 2, so the solutions are the same distance from 2.", + "video_name": "D1cKk48kz-E", + "transcript": "Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate the c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. And we are doing that. We are multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. And you can verify that it works in the original inequality. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. We would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so it's good that we did not include negative 4. So this is our solution. And this is that solution graphed. And I wanted to do that in that other green color. That's what it looks like." + }, + { + "Q": "I was taught that a double bond is a pi bond (and only a pi bond) not a sigma bond + a pi bond, which is correct? Plus I'm not sure I understand even after reading some of the comments how to get the hybridisation from the normal electronic configuration", + "A": "Whoever taught you was not correct, a double bond does have both a sigma bond and a pi bond. You can t find the hybridisation from the electron configuration. You can tell from steric number, which is the number of bonded atoms plus the number of lone pairs. 4 = sp3 3 = sp2 2 = sp", + "video_name": "ROzkyTgscGg", + "transcript": "Voiceover: In an earlier video, we saw that when carbon is bonded to four atoms, we have an SP3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees. If you look at one of the carbons in ethenes, let's say this carbon right here, we don't see the same geometry. The geometry of the atoms around this carbon happens to be planar. Actually, this entire molecule is planar. You could think about all this in a plane here. And the bond angles are close to 120 degrees. Approximately, 120 degree bond angles and this carbon that I've underlined here is bonded to only three atoms. A hydrogen, a hydrogen and a carbon and so we must need a different hybridization for each of the carbon's presence in the ethylene molecule. We're gonna start with our electron configurations over here, the excited stage. We have carbons four, valence electron represented. One, two, three and four. In the video on SP3 hybridization, we took all four of these orbitals and combined them to make four SP3 hybrid orbitals. In this case, we only have a carbon bonded to three atoms. We only need three of our orbitals. We're going to promote the S orbital. We're gonna promote the S orbital up and this time, we only need two of the P orbitals. We're gonna take one of the P's and then another one of the P's here. That is gonna leave one of the our P orbitals unhybridized. Each one of these orbitals has one electron and it's like that. This is no longer an S orbital. This is an SP2 hybrid orbital. This is no longer a P orbital. This is an SP2 hybrid orbital and same with this one, an SP2 hybrid orbital. We call this SP2 hybridization. Let me go and write this up here. and use a different color here. This is SP2 hybridization because we're using one S Orbital and two P orbitals to form our new hybrid orbitals. This carbon right here is SP2 hybridized and same with this carbon. Notice that we left a P orbital untouched. We have a P orbital unhybridized like that. In terms of the shape of our new hybrid orbital, let's go ahead and get some more space down here. We're taking one S orbital. We know S orbitals are shaped like spheres. We're taking two P orbitals. We know that a P orbital is shaped like a dumbbell. We're gonna take these orbitals and hybridized them to form three SP2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that. Once again, when we draw the pictures, we're going to ignore the smaller back lobe. This gives us our SP2 hybrid orbitals. In terms of what percentage character, we have three orbitals that we're taking here and one of them is an S orbital. One out of three, gives us 33% S character in our new hybrid SP2 orbital and then we have two P orbitals. Two out of three gives us 67% P character. 33% S character and 67% P character. There's more S character in an SP2 hybrid orbital than an SP3 hybrid orbital and since the electron density in an S orbital is closer to the nucleus. We think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that's gonna have an effect on the length of the bonds that we're gonna be forming. Let's go ahead and draw the picture of the ethylene molecule now. We know that each of the carbons in ethylenes. Just going back up here to emphasize the point. Each of these carbons here is SP2 hybridized. That means each of those carbons is going to have three SP2 hybrid orbitals around it and once unhybridized P orbital. Let's go ahead and draw that. We have a carbon right here and this is an SP2 hybridized orbitals. We're gonna draw in. There's one SP2 hybrid orbital. Here's another SP2 hybrid orbital and here's another one. Then we go back up to here and we can see that each one of those orbitals. Let me go ahead and mark this. Each one of those SP2 hybrid orbitals has one electron in it. Each one of these orbitals has one electron. I go back down here and I put in the one electron in each one of my orbitals like that. I know that each of those carbons is going to have an unhybridized P orbital here. An unhybridized P orbital with one electron too. Let me go ahead and draw that in. I'll go ahead and use a different color. We have our unhybridized P orbital like that and there's one electron in our unhybridized P orbital. Each of the carbons was SP2 hybridized. Let me go ahead and draw the dot structure right here again so we can take a look at it. The dot structure for ethylene. Let's do the other carbon now. The carbon on the right is also SP2 hybridized. We can go ahead and draw in an SP2 hybrid orbital and there's one electron in that orbital and then there's another one with one electron and then here's another one with one electron. This carbon being SP2 hybridized also has an unhybridized P orbital with one electron. Go ahead and draw in that P orbital with its one electron. We also have some hydrogens. We have some hydrogens to think about here. Each carbon is bonded to two hydrogens. Let me go ahead and put in the hydrogens. The hydrogen has a valance electron in an unhybridized S orbital. I'm going ahead and putting in the S orbital and the one valance electron from hydrogen like this. When we take a look at what we've drawn here, we can see some head on overlap of orbitals, which we know from our earlier video is called a sigma bond. Here's the head on overlap of orbitals. That's a sigma bond. here's another head on overlap of orbitals. The carbon carbon bond, here's also a head on overlap of orbitals and then we have these two over here. We have a total of five sigma bonds in our molecules. Let me go ahead and write that over here. There are five sigma bonds. If I would try to find those on my dot structure this would be a sigma bond. This would be a sigma bond. One of these two is a sigma bond and then these over here. A total of five sigma bonds and then we have a new type of bonding. These unhybridized P orbitals can overlap side by side. Up here and down here. We get side by side overlap of our P orbitals and this creates a pi bond. A pi bond, let me go ahead and write that here. A pi bond is side by side overlap. There is overlap above and below this sigma bond here and that's going to prevent free rotation. When we're looking at the example of ethane, we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here, this pi bond is going to prevent rotations so we don't get different confirmations of the ethylene molecules. No free rotation due to the pi bonds. When you're looking at the dot structure, one of these bonds is the pi bonds, I'm just gonna say it's this one right here. If you have a double bond, one of those bonds, the sigma bond and one of those bonds is a pi bond. We have a total of one pi bond in the ethylene molecule. If you're thinking about the distance between the two carbons, let me go ahead and use a different color for that. The distance between this carbon and this carbon. It turns out to be approximately 1.34 angstroms, which is shorter than the distance between the two carbons in the ethane molecule. Remember for ethane, the distance was approximately 1.54 angstroms. A double bond is shorter than a single bond. One way to think about that is the increased S character. This increased S character means electron density is closer to the nucleus and that's going to make this lobe a little bit shorter than before and that's going to decrease the distance between these two carbon atoms here. 1.34 angstroms. Let's look at the dot structure again and see how we can analyze this using the concept of steric number. Let me go ahead and redraw the dot structure. We have our carbon carbon double bond here and our hydrogens like that. If you're approaching this situation using steric number remember to find the hybridization. We can use this concept. Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. If my goal was to find the steric number for this carbon. I count up my number of sigma bonds. That's one, two and then I know when I double bond one of those is sigma and one of those is pi. One of those is a sigma bond. A total of three sigma bonds. I have zero lone pairs of electrons around that carbon. Three plus zero, gives me a steric number of three. I need three hybrid orbitals and we've just seen in this video that three SP2 hybrid orbitals form if we're dealing with SP2 hybridization. If we get a steric number of three, you're gonna think about SP2 hybridization. One S orbital and two P orbitals hybridizing. That carbon is SP2 hybridized and of course, this one is too. Both of them are SP2 hybridized. Let's do another example. Let's do boron trifluoride. BF3. If you wanna draw the dot structure of BF3, you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine. I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here. What is the hybridization stage of this boron? Let's use the concept of steric number. Once again, let's use steric number. Find the hybridization of this boron. Steric number is equal to number of sigma bonds. That's one, two, three. Three sigma bonds plus lone pairs of electrons. That's zero. Steric number of three tells us this boron is SP2 hybridized. This boron is gonna have three SP2 hybrid orbitals and one P orbital. One unhybridized P orbital. Let's go ahead and draw that. We have a boron here bonded to three flourines and also it's going to have an unhybrized P orbital. Now, remember when you are dealing with Boron, it has one last valance electron and carbon. Carbon have four valance electrons. Boron has only three. When you're thinking about the SP2 hybrid orbitals that you create. SP2 hybrid orbital, SP2, SP2 and then one unhybridized P orbital right here. Boron only has three valance electrons. Let's go ahead and put in those valance electrons. One, two and three. It doesn't have any electrons in its unhybridized P orbital. Over here when we look at the picture, this has an empty orbital and so boron can accept a pair of electrons. We're thinking about its chemical behavior, one of the things that BF3 can do, the Boron can accept an electron pair and function as a lewis acid. That's one way in thinking about how hybridizational allows you to think about the structure and how something might react. This boron turns out to be SP2 hybridized. This boron here is SP2 hybridized and so we can also talk about the geometry of the molecule. It's planar. Around this boron, it's planar and so therefore, your bond angles are 120 degrees. If you have boron right here and you're thinking about a circle. A circle is 360 degrees. If you divide a 360 by 3, you get 120 degrees for all of these bond angles. In the next video, we'll look at SP hybridization." + }, + { + "Q": "I believe the title of this section is a little misleading. I sort of was led to think I was just going to rotate the shape on a plane and find the same shape somewhere else on the plane (rotations do not affect area, etc.) but this video is talking about what shape you would create if you have a constant cross-section of the original shape.", + "A": "Reported it as a mistake in the video, Title should be somewhere along the lines of revolutions not rotations", + "video_name": "vdpyWeiHXmU", + "transcript": "- What I want to do in this video is get some practice visualizing what happens if we were to try to rotate two dimensional shapes in three dimensions. Well what do I mean by that? Let's say I started with a right triangle. So let's say my right triangle looks like this. So let's say it looks like that. Right over there. And so this is a right angle. And let's say that this width right over here is three units and let's say that this length is five units and now I'm gonna do something interesting. I'm gonna take this two dimensional right triangle and I'm gonna try to rotate it in three dimensions around this line, around the line that I'm doing as a dotted magenta line. So I'm gonna rotate it around this line right over there. So if I were to rotate it around this line, what type of a shape am I going to get? And I encourage you -- It's going to be a three dimensional shape. I encourage you to think about it, maybe take out a piece of paper, draw it, or just try to imagine it in your head. Well to think about it in three dimensions, what I'm going to do is try to look at this thing in three dimensions. So let me draw this same line but I'm gonna draw it at an angle so we can visualize the whole thing in three dimensions. So imagine if this was sitting on the ground. So that's our magenta line, and then I can draw my triangle. So my triangle would look something like this. So it would look like this. So once again this is five units, this is three units, this a right triangle. I'm gonna rotate it around the line, so what's it gonna look like? Well this and this right over here is gonna rotate around and it's gonna form a circle with a radius of three, right? So it's gonna form, so it intersects, if that was on the ground it's gonna be three again. And let me draw it down so it's gonna keep going down. Whoops. We don't want to press the wrong button. So it's gonna look something like this. That's what the base is gonna look like. But then this end right over here is just gonna stay at a point because this is right on that magenta line. So it's gonna stay at a point. And so if you were to look at the intersect so it would look something like this. So it would look like this and then you'd have another thing that goes like this and so if you were to take a section like this it would have a little smaller circle here based on what this distance is. So what is the shape, what is the shape that I am drawing? Well what you see, what it is, it's a cone. It's a cone and if I shade it in you might see the cone a little bit better. So let me shade it in so you see the cone. So what you end up getting is a cone where it's base, so I'm shading it in so that hopefully helps a little bit, so what you end up getting is a cone where the base has a radius of three units. So let me draw this. This right over here is the radius of the base and it is three units. I could also draw it like this. So the cone is gonna look like this. And this is the tip of the cone and it's gonna look just like this. And once again let me shade it a little bit so that you can appreciate that this is a three dimensional shape. So draw the cone so you can shade it and we can even construct the original so that, well or we can construct the original shape so you see how it constructs so it makes this, the line, that magenta line, is gonna do this type of thing. It's gonna go through the center of the base, it's gonna go through the center of the base just like that. And our original shape, our original right triangle, if you just took a cross section of it that included that line you would have your original shape. Let me do this in orange. So the original shape is right over there. So what do you get? You get a cone where the radius of the base is three units. Interesting." + }, + { + "Q": "is Maurice Ravel French?", + "A": "Yes. Joseph Maurice Ravel (7 March 1875 \u00e2\u0080\u0093 28 December 1937) was a French composer, pianist and conductor. Hope this helps! Take care", + "video_name": "qfqbomqKZqY", + "transcript": "(Maurice Ravel's \"Daphnis et Chloe\") Gerard: Maurice Ravel, one of the great composers of the 20th century. He was commissioned by Diaghilev to do a ballet, \"Daphnis et Chloe.\" Premiered two years after The Firebird. Very much the same cast of characters, the same choreographers, the same dancers, the same huge orchestra except Ravel added a chorus besides the huge orchestra. A kind of similar type story, in this case it was the story about the love between Daphnis and Chloe. Chloe was abducted and then through the intervention of Pan, who was the God of playing the flute, and his love for the Syrinx. Chloe was found, and Daphnis and Chloe lived happily ever after. More or less something like that. From this great ballet that Ravel wrote, he created two suites. Ravel used basically the first part of the ballet for Suite One and the second part of the ballet for Suite Two. Suite No. Two is the one that's done the most often. The piece begins in the most remarkable way. The woodwinds, two flutes and then two clarinets, play these incredibly fast notes. One of the most important orchestral excerpts for these instruments, and hard and fast and soft. The harp playing glissandos underneath that and the double basses just underpinning the whole thing with a melodic gesture, not a melody certainly. It is an incredible moment. It creates such an atmosphere. (Suite Two, Maurice Ravel's \"Daphnis et Chloe\") It is just about to be daybreak. You can just feel how the sun is about to come up. Ravel uses his melody. (piano playing) This melody can go on forever. (piano playing) It's a kind of melody that you can repeat over and over again, it just keeps going. If you think of that little gesture, this little melody, and if you orchestrate it and put it in different instruments and do it in interesting ways, it can be a glorious moment. (orchestral music) The little bird comes in the piccolo, and the solo violins. (Maurice Ravel's \"Daphnis et Chloe\") As the daybreak and the sun is coming up and birds are there, shepherds start to be seen coming through. Obviously, they're getting up and getting ready to take care of their sheep and doing their job in a sense. You can hear the little solo from the piccolo and the little solo from the E-flat clarinet. We have a group of herdsmen come in. For the herdsmen, Ravel uses a slightly different melody. You can hear the similarities. (piano playing) (orchestra playing) We have all of these filigree, the glissandos of the harp, these melodies that are being developed but nothing, no great gestures yet. Some new material comes in, played by the viola and the clarinet. You can feel that what's happening now, of course, is that Daphnis is trying to find Chloe. If you didn't know that, it wouldn't matter. It's what was needed at the moment. You can tell musically, after all these beautiful gestures, you can't just do beautiful gestures forever, and he gets a little agitated section. In the story, Daphnis is looking for Chloe and she appears surrounded by the shepherds. (Maurice Ravel's \"Daphnis et Chloe\") In the next juncture of this ballet, we get to the part where, of course, Chloe has been abducted. Now, Pan is helping Daphnis to find her. In fact, they reverse roles. Daphnis becomes Pan, and Chloe becomes the Syrinx that Pan loved. It leads to the central section of the piece which is this incredible flute solo. The greatest flute solo probably ever written. Very simple accompaniment, pizzicato strings, second and fourth horns, and harp, and this beautiful flute solo. (flute playing) The flute solo becomes more and more agitated, and eventually Chloe falls into Daphnis' arms. It's this mini-concerto for flute section. You see the first flute playing a cascading scale, then the second flute picks it up. The alto flute is the final one. Starts with the piccolo and works its way down in the section. In fact, what Ravel does is he continues that and it continues to be a little flute section concerto right in the middle of this piece. A phenomenal use of the instruments. (Maurice Ravel's \"Daphnis et Chloe\") Finally it comes to an end and at this moment, the nymphs are falling in love and they pledge their love for each other and they dedicate some sheep to their joy together. It is represented by this somewhat gorgeous chorale. (orchestral chorale) This leads us to the first inkling of the fast section. Everything to now has been relatively slow. A build-up for the flutes and then it comes back again and then we have this little chorale for the sheep and the shepherds. The women of the company, in this case the dance company, enter to do a special general dance. It starts out where the music that is going to permeate the rest of the piece is sounded but only a few bars because immediately it comes back down and we hear that same beautiful chorale, the solo for the alto flute, and then the general dance begins. This is the dance that's basically in five. There are a few moments that are in three but basically in five. It's five beats per bar, accent on two, and that in itself is unusual. (Maurice Ravel's \"Daphnis et Chloe\") At one point, after this tremendous build-up, like any great composer, he could have ended it just there. Instead, he brings everything back down. Everything back down to the essence, which is the rhythm and in a very soft way, the snare drum and the double basses play this bom-ba-dom-ba-dom-ba-dom-ba-dom, ba-dom-ba-dom-ba-dom-ba-dom. It starts over again. It's so interesting that composers do this because you know it could end. He brings it back and then when it does end, it's even more exciting. The end of \"Daphnis et Chloe,\" Second Suite, is among the most exciting pieces of music one could ever hear. Using seven percussion, four trumpets, four flutes, oboes, English horn, clarinet, it's a huge orchestra. Full of what we know of Ravel, one of the greatest orchestrators of all time. (Second Suite of Maurice Ravel's \"Daphnis et Chloe\")" + }, + { + "Q": "find the derivatives of f(x)=arc sin 1/x", + "A": "f(x) = arcsin(1/x) f (x) = d/dx (1/x)/\u00e2\u0088\u009a[1 - (1/x)\u00c2\u00b2] f (x) = -1/x\u00c2\u00b2/\u00e2\u0088\u009a[1 - (1/x)\u00c2\u00b2] f (x) = -1/x\u00c2\u00b2\u00e2\u0088\u009a[1 - (1/x)\u00c2\u00b2]`", + "video_name": "FJ7AMaR9miI", + "transcript": "In terms of k, where k does not equal 0, what is the y-intercept of the line tangent to the curve f of x is equal to 1/x at the point of the curve where x is equal to k? So let's just think about what they're asking. So if I were to draw myself-- let me draw some quick axes right over here. So that's my y-axis. This is my x-axis right over here. And the graph f of x is equal to 1/x would look something like this. So it looks something like this. So it kind of spikes up there, and then it comes down, and then it goes like this. And I'm just doing a rough approximation of it. So it would look something like that. And then on the negative side, it looks something like this. So this is my hand-drawn version of roughly what this graph looks like. So this right over here is f of x is equal to 1/x. Now, we are concerning ourselves with the point x equals k. So let's say-- I mean, it could be anything that's non-zero, but let's just say that this is k right over here. So that is the point k 1/k. We can visualize the line tangent to the curve there. So it might look something like this. And we need to figure out its y-intercept. Where does it intercept the y-axis? So we need to figure out this point right over here. Well, the best way to do it, if we can figure out the slope of the tangent line here, the slope of the tangent line is just the derivative of the line at that point. If we could figure out the slope of the tangent line, we already know that line contains the point. Let me do this in a different color. We know it contains the point k comma 1/k. So if we know its slope, we know what point it contains, we can figure out what its y-intercept is. So the first step is just, well, what's the slope of the tangent line? Well, to figure out the slope of the tangent line, let's take the derivative. So if we write f of x, instead writing it as 1/x, I'll write it as x to the negative 1 power. That makes it a little bit more obvious that we're about to use the power rule here. So the derivative of f at any point x is going to be equal to-- well, it's going to be the exponent here is negative 1. So negative 1 times x to the-- now we decrement the exponent to the negative 2 power. Or I could say it's negative x to the negative 2. Now, what we care about is the slope when x equals k. So f prime of k is going to be equal to negative k to the negative 2 power. Or another way of thinking about it, this is equal to negative 1 over k squared. So this right over here is the slope of the tangent line at that point. Now, let's just think about what the equation of the tangent line is. And we could think about it in slope-intercept form. So we know the equation of a line in slope-intercept form is y is equal to mx plus b, where m is the slope and b is the y-intercept. So if we can get it in this form, then we know our answer. We know what the y-intercept is going to be. It's going to be b. So let's think about it a little bit. This equation, so we could say y is equal to our m, our slope of the tangent line, when x is equal to k, we just figure out to be this business. It equals this thing right over here. So let me write that in blue. Negative 1 over k squared times x plus b. So how do we solve for b? Well, we know what y is when x is equal to k. And so we can use that to solve for b. We know that y is equal to 1/k when x is equal to k. So this is going to be equal to negative 1-- that's not the same color. Negative 1 over k squared times k plus b. Now, what does this simplify to? See, k over k squared is the same thing as 1/k, so this is going to be negative 1/k. So this part, all of this simplifies to negative 1/k. So how do we solve for b? Well, we could just add 1/k to both sides and we are left with-- if you add 1/k here-- actually, let me just do that. Plus 1/k, left-hand side, you're left with 2/k, and on the right-hand side, you're just left with b, is equal to b. So we're done. The y-intercept of the line tangent to the curve when x equals k is going to be 2/k. If we wanted the equation of the line, well, we've done all the work. Let's write it out. It'll be satisfying. It's going to be y is equal to negative 1/k squared x plus our y-intercept, plus 2/k. And we're done." + }, + { + "Q": "This came up on the practice section can anyone tell me what I am doing wrong?\n'Which of the following ordered pairs represents a solution to the equation below?'\n(\u00e2\u0088\u00922,\u00e2\u0088\u00922)(\u00e2\u0088\u00921,\u00e2\u0088\u00922)(0,\u00e2\u0088\u00921)(1,3)(2,7)\ny=2x+1\n\nEvery time I put the first number in the y place and the second number in the x place they come up different.\nI understood the question as meaning something like is y the same as two ? plus 1\nIs that wrong?", + "A": "What the question is asking you is, when you put the first number of the pair in for x, and the second number in for y, does the equation make a true statement? I ll do one for you: -2 = 2(-2) + 1? -2 =-4 +1? -2 = -3? NO. So the pair (-2,-2) does not represent a solution to the equation.", + "video_name": "HUn4XwV7o9I", + "transcript": "Is 3 comma negative 4 a solution to the equation 5x plus 2y is equal to 7? So they're saying, does x equal 3, y equal negative 4, satisfy this equation, or this relationship right here? So one way to do it is just to substitute x is equal to 3 and y equals negative 4 into this and see if 5 times x plus 2 times y does indeed equal 7. So we have 5 times 3 plus 2 times negative 4. This is equal to 15. 15 plus negative 8, which does indeed equal 7. So it does satisfy the equation. So it is on the line. It is a solution. x equals 3, y equals negative 4 is a solution to this equation. So we've essentially answered our question. It is. Now, another way to do it, and I'm not going to go into the details here, is you could actually graph the line. So maybe the line might look something like this, I'm not going to do it in detail. And you see, if you have a very good drawing of it, you see whether the point lies on the line. If the point, when you graph the point, does lie on the line, it would be a solution. If the point somehow ends up not being on the line then you'd know it isn't a solution. But to do this, you would have to have a very good drawing and so you could very precisely determine whether it's on the line. If you do the substitution method, if you just substitute the values into the equation and see if it comes out mathematically, this will always be exact. So this is all we really had to do in this example. So it definitely is a solution to the equation." + }, + { + "Q": "I have two Questions:\n\nDoes the production of lactic acid in skeletal muscle contribute to lowering the pH and thus increase O2 delivery?\n\nAlso, because there is low CO2 in the lungs and therefore less HCO3- and H2CO3 because of Le'Chatelier's principle, does that mean that the lungs have a more basic environment than the rest of the body?\n\nThanks for any info on this!", + "A": "I think that you raise a good point for #2. In the lungs HCO3- and H2CO3 are converted back into CO2 and H2O, which removes protons from the blood, which should raise the pH of the blood. So I guess yes, the blood in the lungs will be more alkali than the blood in the muscles. Great question!", + "video_name": "dHi9ctwDUnc", + "transcript": "So we've talked a little bit about the lungs and the tissue, and how there's an interesting relationship between the two where they're trying to send little molecules back and forth. The lungs are trying to send, of course, oxygen out to the tissues. And the tissues are trying to figure out a way to efficiently send back carbon dioxide. So these are the core things that are going on between the two. And remember, in terms of getting oxygen across, there are two major ways, we said. The first one, the easy one is just dissolved oxygen, dissolved oxygen in the blood itself. But that's not the major way. The major way is when oxygen actually binds hemoglobin. In fact, we call that HbO2. And the name of that molecule is oxyhemoglobin. So this is how the majority of the oxygen is going to get delivered to the tissues. And on the other side, coming back from the tissue to the lungs, you've got dissolved carbon dioxide. A little bit of carbon dioxide actually, literally comes just right in the plasma. But that's not the majority of how carbon dioxide gets back. The more effective ways of getting carbon dioxide back, remember, we have this protonated hemoglobin. And actually remember, when I say there's a proton on the hemoglobin, there's got to be some bicarb floating around in the plasma. And the reason that works is because when they get back to the lungs, the proton, that bicarb, actually meet up again. And they form CO2 and water. And this happens because there's an enzyme called carbonic anhydrase inside of the red blood cells. So this is where the carbon dioxide actually gets back. And of course, there's a third way. Remember, there's also some hemoglobin that actually binds directly to carbon dioxide. And in the process, it forms a little proton as well. And that proton can go do this business. It can bind to a hemoglobin as well. So there's a little interplay there. But the important ones I want you to really kind of focus in on are the fact that hemoglobin can bind to oxygen. And also on this side, that hemoglobin actually can bind to protons. Now, the fun part about all this is that there's a little competition, a little game going on here. Because you've got, on the one side, you've got hemoglobin binding oxygen. And let me draw it twice. And let's say this top one interacts with a proton. Well, that protons going to want to snatch away the hemoglobin. And so there's a little competition for hemoglobin. And here, the oxygen gets left out in the cold. And the carbon dioxide does the same thing, we said. Now, we have little hemoglobin bound to carbon dioxide. And it makes a proton in the process. But again, it leave oxygen out in the cold. So depending on whether you have a lot of oxygen around, if that's the kind of key thing going on, or whether you have a lot of these kinds of products the proton or the carbon dioxide. Depending on which one you have more of floating around in the tissue in the cell, will determine which way that reaction goes. So keeping this concept in mind, then I could actually step back and say, well, I think that oxygen is affected by carbon dioxide and protons. I could say, well, these two, carbon dioxide and protons, are actually affecting, let's say, are affecting the, let's say, the affinity or the willingness of hemoglobin to bind, of hemoglobin for oxygen. That's one kind of statement you could make by looking at that kind of competition. And another person come along and they say, well, I think oxygen actually is affecting, depending on which one, which perspective you take. You could say, oxygen is affecting maybe the affinity of hemoglobin for the carbon dioxide and proton of hemoglobin for CO2 and protons. So you could say it from either perspective. And what I want to point out is that actually, in a sense, both of these are true. And a lot of times we think, well, maybe it's just saying the same thing twice. But actually, these are two separate effects. And they have two separate names. So the first one, talking about carbon dioxide and protons, their effect is called the Bohr effect. So you might see that word or this description. This is the Bohr effect. And the other one, looking at it from the other prospective, looking at it from oxygen's perspective, this would be the Haldane effect. That's just the name of it, Haldane effect. So what is the Bohr effect and the Haldane effect? Other than simply saying that the things compete for hemoglobin. Well, let me actually bring up a little bit of the canvas. And let's see if I can't diagram this out. Because sometimes I think a little diagram would really go a long way in explaining these things. So let's see if I can do that. Let's use a little graph and see if we can illustrate the Bohr effect on this graph. So this is the partial pressure of oxygen, how much is dissolved in the plasma. And this is oxygen content, which is to say, how much total oxygen is there in the blood. And this, of course, takes into account mostly the amount of oxygen that's bound to hemoglobin. So as I slowly increase the partial pressure of oxygen, see how initially, not too much is going to be binding to the hemoglobin. But eventually as a few of the molecules bind, you get cooperativity. And so then, slowly the slope starts to rise. And it becomes more steep. And this is all because of cooperativity. Oxygen likes to bind where other oxygens have already bound. , And then it's going to level off. And the leveling off is because hemoglobin is starting to get saturated. So there aren't too many extra spots available. So you need lots and lots of oxygen dissolved in the plasma to be able to seek out and find those extra remaining spots on hemoglobin. So let's say we choose two spots. One spot, let's say, is a high amount of oxygen dissolved in the blood. And this, let's say, is a low amount of oxygen dissolved in the blood. I'm just kind of choosing them arbitrarily. And don't worry about the units. And if you were to think of where in the body would be a high location, that could be something like the lungs where you have a lot of oxygen dissolved in blood. And low would be, let's say, the thigh muscle where there's a lot of CO2 but not so much oxygen dissolved in the blood. So this could be two parts of our body. And you can see that. Now, if I want to figure out, looking at this curve how much oxygen is being delivered to the thigh, then that's actually pretty easy. I could just say, well, how much oxygen was there in the lungs, or in the blood vessels that are leaving the lungs. And there's this much oxygen in the blood vessels leaving the lungs. And there's this much oxygen in the blood vessels leaving the thigh. So the difference, whenever oxygen is between these two points, that's the amount of oxygen that got delivered. So if you want to figure out how much oxygen got delivered to any tissue you can simply subtract these two values. So that's the oxygen delivery. But looking at this, you can see an interesting point which is that if you wanted to increase the oxygen delivery. Let's say, you wanted for some reason to increase it, become more efficient, then really, the only way to do that is to have the thigh become more hypoxic. As you move to the left on here, that's really becoming hypoxic, or having less oxygen. So if you become more hypoxic, then, yes, you'll have maybe a lower point here, maybe a point like this. And that would mean a larger oxygen delivery. But that's not ideal. You don't want your thighs to become hypoxic. That could start aching and hurting. So is there another way to have a large oxygen delivery without having any hypoxic tissue, or tissue that has a low amount of oxygen in it. And this is where the Bohr effect comes into play. So remember, the Bohr effect said that, CO2 and protons affect the hemoglobin's affinity for oxygen. So let's think of a situation. I'll do it in green. And in this situation, where you have a lot of carbon dioxide and protons, the Bohr effect tells us that it's going to be harder for oxygen to bind hemoglobin. So if I was to sketch out another curve, initially, it's going to be even less impressive, with less oxygen bound to hemoglobin. And eventually, once the concentration of oxygen rises enough, it will start going up, up, up. And it does bind hemoglobin eventually. So it's not like it'll never bind hemoglobin in the presence of carbon dioxide and protons. But it takes longer. And so the entire curve looks shifted over. These conditions of high CO2 and high protons, that's not really relevant to the lungs. The lungs are thinking, well, for us, who cares. We don't really have these conditions. But for the thigh, it is relevant because the thigh has a lot of CO2. And the thigh has a lot of protons. Again, remember, high protons means low pH. So you can think of it either way. So in the thigh, you're going to get, then, a different point. It's going to be on the green curve not the blue curve. So we can draw it at the same O2 level, actually being down here. So what is the O2 content in the blood that's leaving the thigh? Well, then to do it properly, I would say, well, it would actually be over here. This is the actual amount. And so O2 deliver is actually much more impressive. Look at that. So O2 delivery is increased because of the Bohr effect. And if you want to know exactly how much it's increased, I could even show you. I could say, well, this amount from here down to here. Literally the vertical distance between the green and the blue lines. So this is the extra oxygen delivered because of the Bohr effect. So this is how the Bohr effect is so important at actually helping us deliver oxygen to our tissues. So let's do the same thing, now, but for the Haldane effect. And to do this, we actually have to switch things around. So our units and our axes are going to be different. So we're going to have the amount of carbon dioxide there. And here, we'll do carbon dioxide content in the blood. So let's think through this carefully. Let's first start out with increasing the amount of carbon dioxide slowly but surely. And see how the content goes up. And here, as you increase the amount of carbon dioxide, the content is kind of goes up as a straight line. And the reason it doesn't take that S shape that we had with the oxygen is that there's no cooperativity in binding the hemoglobin. It just goes up straight. So that's easy enough. Now, let's take two points like we did before. Let's take a point, let's say up here. This will be a high amount of CO2 in the blood. And this will be a low amount of CO2 in the blood. So you'd have a low amount, let's say right here, in what part of the tissue? Well, low CO2, that sounds like the lungs because there's not too much CO2 there. But high CO2, it probably is the thighs because the thighs like little CO2 factories. So the thigh has a high amount and the lungs have a low amount. So if I want to look at the amount of CO2 delivered, we'd do it the same way. We say, OK, well, the thighs had a high amount. And this is the amount of CO2 in the blood, remember. And this is the amount of CO2 in the blood when it gets to the lungs. So the amount of CO2 that was delivered from the thigh to the lungs is the difference. And so this is how much CO2 delivery we're actually getting. So just like we had O2 delivery, we have this much CO2 delivery. Now, read over the Haldane effect. And let's see if we can actually sketch out another line. In the presence of high oxygen, what's going to happen? Well, if there's a lot of oxygen around, then it's going to change the affinity of hemoglobin for carbon dioxide and protons. So it's going to allow less binding of protons and carbon dioxide directly to the hemoglobin. And that means that you're going to have less CO2 content for any given amount of dissolved CO2 in the blood. So the line still is a straight line, but it's actually, you notice, it's kind of slope downwards. So where is this relevant? Where do you have a lot of oxygen? Well, it's not really relevant for the thighs because the thighs don't have a lot of oxygen. But it is relevant for the lungs. It is very relevant there. So now you can actually say, well, let's see what happens. Now that you have high O2, how much CO2 delivery are you getting? And you can already see it. It's going to be more because now you've got this much. You've got going all the way over here. So this is the new amount of CO2 delivery. And it's gone up. And in fact, you can even show exactly how much it's gone up by, by simply taking this difference. So this difference right here between the two, this is the Haldane effect. This is the visual way that you can actually see that Haldane effect. So the Bohr effect and the Haldane effect, these are two important strategies our body has for increasing the amount of O2 delivery and CO2 delivery going back and forth between the lungs and the tissues." + }, + { + "Q": "16.88 rounded to the nearest hundredth is", + "A": "The answer is 16.90", + "video_name": "19yOv4P2ccw", + "transcript": "We're asked to round 152, 137, 245, and 354 to the nearest 100, which is another way of saying round each of these numbers to the nearest multiple of 100. So let's think about them one by one. So let's draw a number line here. And here I'm counting up by hundreds. What I've marked here, 100, 200, 300, 400, these are all multiples of 100. I could keep going up. I could go to 500, 600, so on and so forth. Now, let's start with 152. Well, where does 152 sit? So halfway in between is 150. 152 is going to be right to the right of that. So that's 152 right over here. So what are our two options? We might round up. The multiple of 100 above 152 is 200. The multiple of 100 below 152 is 100. So which direction do we go in? Do we round up to 200, or do we round down to 100? Well, if we're rounding to the nearest 100, we want to look at one place to the right of that. We want to look at the tens place to decide which multiple of 100 it is closer to. And the rules are very similar to when we're rounding to the nearest tens place or really any place. We look one place to the right of it. So in this case, we look at the tens place. And we say, if this is 5 or larger, we round up. And this is 5 or larger, so we're going to round up to 200. So this we're going to round. 152, we're going to round to 200, which also makes sense. 152 is a little bit closer to 200. It's 48 away from to 200 than it to 100. It's 52 away from 100. So it makes sense to go up to the nearest multiple of 100. Now, let's think about 137. And I encourage you to now pause the video and try to round each of these other three numbers to the nearest 100. Well, 137 is going to sit someplace right over here. 137 is going to be right over there. So two options-- we can round down to 100. That's the multiple of 100 below 137. Or we could round up to 200. Well, 137, just looking at it, is clearly closer to 100. Or we could apply our rule. If we're rounding to the nearest 100, we want to look at one place to the right of that. We want to look at the tens place. If this is 5 or larger, we round up. If it's less than 5, we round down. So in this case, we would round down to 100. Let's do the same thing with 245. If you haven't paused it and tried it yourself, once again, I want to emphasize. That'll make it really valuable for you to try it on your own. So let's plot where 245 is. So 245 is right around-- this is 250, so 245 might be right around here. Now, let's apply that rule. If we're trying to round to the nearest 100, we would look to one place to the right. We'll look at the tens place. We could ignore the ones place. We look at the tens place here. If it is greater than or equal to 5, we round up. If it is less than 5, we round down. So here, we're clearly going to round down. And when we round down, we're going to round down to the multiple of 100 that's directly below 245. Well, we're going to round down to 200. We had two options. If we rounded up, we would have gone to 300. If we rounded down, we would go to 200. We're clearly closer to 200. And we can verify that with the rule. The tens place, we're in the 40's here. The tens place is a 4. We're going to round down. Now let's think about 354. If we were to plot that, this is 350. 354 might be right over here. So if we rounded down, we would go to 300. If we rounded up, we would go to 400. Now let's apply our rule, and let's also think about it on the number line. This is all about finding the multiple of 100 that it's closest to. If you're trying to round to the nearest 100, you want to look at the tens place, a place to the right of the place you're rounding to. If the tens place is a 5 or larger, you're going to round up. This is a 5 or larger so we're going to round up to 400. And that also makes sense. The rule is really valuable. If you're right at 350, right in between, then you would need the rule to say, hey, let's look at this 5 But 354 is also closer to 400 than it is to 300. It's 54 away from 300. It's 46 away from 400. So it makes sense that we would round up to 400." + }, + { + "Q": "So the numbers were 13 and 15... but they could just as well have been 29 and 31, and the shared secret would be the same... so how could you ever find the original numbers?", + "A": "They don t need to find the original numbers. The whole point is to generate the shared secret (which they can to generate a key for a symmetric key cipher).", + "video_name": "M-0qt6tdHzk", + "transcript": "Now this is our solution. First Alice and Bob agree publicly on a prime modulus and a generator, in this case 17 and 3. Then Alice selects a private random number, say 15, and calculates three to the power 15 mod 17 and sends this result publicly to Bob. Then Bob selects his private random number, say 13, and calculates 3 to the power 13 mod 17 and sends this result publicly to Alice. And now the heart of the trick; Alice takes Bob's public result and raises it to the power of her private number to obtain the shared secret, which in this case is 10. Bob takes Alice's public result and raises it to the power of his private number resulting in the same shared secret. Notice they did the same calculation, though it may not look like it at first. Consider Alice, the 12 she received from Bob was calculated as 3 to the power 13 mod 17. So her calculation was the same as 3 to the power 13 to the power 15 mod 17. Now consider Bob, the 6 he received from Alice was calculated as 3 to the power 15 mod 17. So his calculation was the same as 3 to the power 15 to the power 13. Notice they did the same calculation with the exponents in a different order. When you flip the exponent the result doesn't change. So they both calculated 3 raised to the power of their private numbers. Without one of these private numbers, 15 or 13, Eve will not be able to find the solution. And this is how it's done; While Eve is stuck grinding away at the discrete logarithm problem, and with large enough numbers, we can say it's practically impossible for her to break the encryption in a reasonable amount of time. This solves the Key Exchange problem. It can be used in conjunction with a pseudorandom generator to encrypt messages between people who have never met." + }, + { + "Q": "What is the oxidation state of hydrogen in acids? Is it +1 or -1 ?", + "A": "The oxidation state of hydrogen in acids is +1. In fact, in the majority of its compounds hydrogen is in the +1 state (other than, of course, hydrogen gas). The -1 state of hydrogen in pretty much limited to a class of chemicals called hydrides. For example, NaH, or NaBH\u00e2\u0082\u0084. Since hydrides react with water, I don t know of any that you would find in aqueous solution.", + "video_name": "CCsNJFsYSGs", + "transcript": "Let's see if we can come up with some general rules of thumb or some general trends for oxidation states by looking at the periodic table. So first, let's just focus on the alkali metals. And I'll box them off. We'll think about hydrogen in a second. Well, I'm going to box-- I'm going to separate hydrogen because it's kind of a special case. But if we look at the alkali metals, the Group 1 elements right over here, we've already talked about the fact they're not too electronegative. They have that one valence electron. They wouldn't mind giving away that electron. And so for them, that oxidation state might not even be a hypothetical charge. These are very good candidates for actually forming ionic bonds. And so it's very typical that when these are in a molecule, when these form bonds, that these are the things that are being oxidized. They give away an electron. So they get to-- a typical oxidation state for them would be positive 1. If we go one group over right over here to the alkaline earth metals, two valence electrons, still not too electronegative. So they're likely to fully give or partially give away two electrons. So if you're forced to assign an ionic-- if you were to say, well, none of this partial business, just give it all away or take it, you would say, well, these would typically have an oxidation state of positive 2. In a hypothetical ionic bonding situation, they would be more likely to give the two electrons because they are not too electronegative, and it would take them a lot to complete their valence shell to get all the way to 8. Now, let's go to the other side of the periodic table to Group 7, the halogens. The halogens right over here, they're quite electronegative, sitting on the right-hand side of the periodic table. They're one electron away from being satisfied from a valence electron point of view. So these are typically reduced. They typically have an oxidation state of negative 1. And I keep saying typically, because these are not going to always be the case. There are other things that could happen. But this is a typical rule of thumb that they're likely to want to gain an electron. If we move over one group to the left, Group 6-- and that's where the famous oxygen sits-- we already said that oxidizing something is doing to something what oxygen would have done, that oxidation is taking electrons away from it. So these groups are typically oxidized. And oxygen is a very good oxidizing agent. Or another way of thinking about it is oxygen normally takes away electrons. These like to take away electrons, typically two electrons. And so their oxidation state is typically negative 2-- once again, just a rule of thumb-- or that their charge is reduced by two electrons. So these are typically reduced. These are typically oxidized. Now, we could keep going. If we were to go right over here to the Group 5 elements, typical oxidation state is negative 3. And so you see a general trend here. And that general trend-- and once again, it's not even a hard and fast rule of thumb, even for the extremes, but as you get closer and closer to the middle of the periodic table, you have more variation in what these typical oxidation states could be. Now, I mentioned that I put hydrogen aside. Because if you really think about it, hydrogen, yes, hydrogen only has one electron. And so you could say, well, maybe it wants to give away that electron to get to zero electrons. That could be a reasonable configuration for hydrogen. But you can also view hydrogen kind of like a halogen. So you could kind of view it kind of like an alkali metal. But in theory, it could have been put here on the periodic table as well. You could have put hydrogen here, because hydrogen, in order to complete its first shell, it just needs one electron. So in theory, hydrogen could have been put there. So hydrogen actually could typically could have a positive or a negative 1 oxidation state. And just to see an example of that, let's think about a situation where hydrogen is the oxidizing agent. And an example of that would be lithium hydride right Now, in lithium hydride, you have a situation where hydrogen is more electronegative. A lithium is not too electronegative. It would happily give away an electron. And so in this situation, hydrogen is the one that's oxidizing the lithium. Lithium is reducing the hydrogen. Hydrogen is the one that is hogging the electron. So the oxidation state on the lithium here is a positive 1. And the oxidation state on the hydrogen here is a negative. So just, once again, I really want to make sure we get the notation. Lithium has been oxidized by the hydrogen. Hydrogen has been reduced by the lithium. Now, let's give an example where hydrogen plays the other role. Let's imagine hydroxide. So the hydroxide anion-- so you have a hydrogen and an oxygen. And so essentially, you could think of a water molecule that loses a hydrogen proton but keeps that hydrogen's electron. And this has a negative charge. This has a negative 1 charge. But what's going on right over here? And actually, let me just draw that, because it's fun to think about it. So this is a situation where oxygen typically has-- 1, 2, 3, 4, 5, 6 electrons. And when it's water, you have 2 hydrogens like that. And then you share. And then you have covalent bond right over there sharing that pair, covalent bond sharing that right over there. To get to hydroxide, the oxygen essentially nabs both of these electrons to become-- so you get-- that pair, that pair. Now you have-- let me do this in a new color. Now, you have this pair as well. And then you have that other covalent bond to the other hydrogen. And now this hydrogen is now just a hydrogen proton. This one now has a negative charge. So this is hydroxide. And so the whole thing has a negative charge. And oxygen, as we have already talked about, is more electronegative than the hydrogen. So it's hogging the electrons. So when you look at it right over here, you would say, well, look, hydrogen, if we had to, if we were forced to-- remember, oxidation states is just an intellectual tool which we'll find useful. If you had to pretend this wasn't a covalent bond, but an ionic bond, you'd say, OK, then maybe this hydrogen would fully lose an electron, so it would get an oxidation state of plus 1. It would be oxidized by the oxygen. And that the oxygen actually has fully gained one electron. And you could say, well, if we're forced to, we could say-- if we're forced to think about this is an ionic bond, we'll say it fully gains two electrons. So we'll have an oxidation state of negative 2. And once again, the notation, when you do the superscript notation for oxidation states and ionic charge, you write the sign after the number. And this is just the convention. And now, with these two examples, the whole point of it is to show that hydrogen could have a negative 1 or a positive 1 oxidation state. But there's also something interesting going on here. Notice, the oxidation states of the molecules here, they add up to the whole-- or the oxidation state of each of the atoms in a molecule, they add up to the entire charge of the molecule. So if you add a positive 1 plus negative 1, you get 0. And that makes sense because the entire molecule lithium hydride is neutral. It has no charge. Similarly, hydrogen, plus 1 oxidation state; oxygen, negative 2 oxidation number or oxidation state-- you add those two together, you have a negative 1 total charge for the hydroxide anion, which is exactly the charge that we have right over there." + }, + { + "Q": "hey i am studying for a algebra test why am i having to study in trig and geo?", + "A": "Usually you learn algebra to apply it in different math topics such as trigonometry and geometry. Connecting topics can help you understand them more.", + "video_name": "Ei54NnQ0FKs", + "transcript": "Voiceover:Let's say you're studying some type of a little hill or rock formation right over here. And you're able to figure out the dimensions. You know that from this point to this point along the base, straight along level ground, is 60 meters. You know the steeper side, steeper I guess surface or edge of this cliff or whatever you wanna call it, is 20 meters. And then the longer side here, I guess the less steep side, is 50 meters long. So you're able to measure that. But now what you wanna do is use your knowledge of trigonometry, given this information, to figure out how steep is this side. What is the actual inclination relative to level ground? Or another way of thinking about it, what is this angle theta right over there? And I encourage you to pause the video and think about it on your own. Well it might be ringing a bell. Well you know three sides of a triangle and then we want to figure out an angle. And so the thing that jumps out in my head, well maybe the law of cosines could be useful. Let me just write out the law of cosines, before we try to apply it to this triangle right over here. So the law of cosines tells us that C-squared is equal to A-squared, plus B-squared, minus two A B, times the cosine of theta. And just to remind ourselves what the A, B's, and C's are, C is the side that's opposite the angle theta. So if I were to draw an arbitrary triangle right over here. And if this is our angle theta, then this determines that C is that side, and then A and B could be either of these two sides. So A could be that one and B could be that one. Or the other way around. As you can see, A and B essentially have the same role in this formula right over here. This could be B or this could be A. So what we wanna do is somehow relate this angle... We wanna figure out what theta is in our little hill example right over here. So if this is going to be theta, what is C going to be? Well C is going to be this 20 meter side. And then we could set either one of these to be A or B. We could say that this A is 50 meters and B is 60 meters. And now we could just apply the law of cosines. So the law of cosines tells us that 20-squared is equal to A-squared, so that's 50 squared, plus B-squared, plus 60 squared, minus two times A B. So minus two times 50, times 60, times 60, times the cosine of theta. This works out well for us because they've given us everything. There's really only one unknown. There's theta here. So let's see if we can solve for theta. So 20 squared, that is 400. 50 squared is 2,500. 60 squared is 3,600. And then 50 times... Let's see, two times 50 is 100, times 60, this is all equal to 6,000. So let's see, if we simplify this a little bit we're going to get 400 is equal to 2,500 plus 3,600. Let's see, that'd be 6,100. That's equal to 6,000... Let me do this in a new color. So when I add these two, I get 6,100. Did I do that right? So it's 2,000 plus 3,000, plus 5,000. 500 plus 600 is 1,100. So I get 6,100 minus 6,000, times the cosine of theta. And let's see, now we can subtract 6,100 from both sides. So I'm just gonna subtract 6,100 from both sides so that I get closer to isolating the theta. So let's do that. So this is going to be negative 5,700. Is that right? 5,700 plus... Yes, that is right. Right, because if this was the other way around, if this was 6,100 minus 400 it would be positive 5,700. Alright. And then these two of course cancel out. And this is going to be equal to negative 6,000 times the cosine of theta. Now we can divide both sides by negative 6,000. And we get... I'm just gonna swap the sides. We get cosine of theta is equal to... Let's see we could divide the numerator and the denominator by essentially negative 100. So these are both going to become positive. So cosine of theta is equal to 57 over 60. And actually that can be simplified even more. Three goes into 57, is that 19 times? Yep, so this is actually... This could be simplified. This is equal to 19 over 20. We actually didn't have to do that simplification step because we're about to use our calculators, but that makes the math a little more tractable. Right, 3 goes into 57, yeah, 19 times. And so now we can take the inverse cosine of both sides. So we could get theta is equal to the inverse cosine, or the arc cosine, of 19 over 20. So let's get our calculator out and see if we get something that makes sense. So we wanna do the inverse cosine of 19 over 20. And we deserve a drum roll. We get 18.19 degrees. And I already verified that my calculator is in degree mode. So it gets 18.19 degrees. So if we wanted to round, this is approximately equal to 18.2 degrees, if we wanna round to the nearest tenth. So that essentially gives us a sense of how steep this slope actually is." + }, + { + "Q": "Does the \"Angular Momentum\" in the term \"Angular Moment Quantum Number\" share any similar concept at all to the \"Angular Momentum\" in Physics?", + "A": "Yes angular momentum in quantum theories is called angular momentum because it acts just like angular momentum in the macroscopic world.", + "video_name": "KrXE_SzRoqw", + "transcript": "- [Voiceover] In the Bohr model of the hydrogen atom, the one electron of hydrogen is in orbit around the nucleus at a certain distance, r. So in the Bohr model, the electron is in orbit. In the quantum mechanics version of the hydrogen atom, we don't know exactly where the electron is, but we can say with high probability that the electron is in an orbital. An orbital is the region of space where the electron is most likely to be found. For hydrogen, imagine a sphere, a three-dimensional volume, a sphere, around the nucleus. Somewhere in that region of space, somewhere in that sphere, we're most likely to find the one electron of hydrogen. So we have these two competing visions. The Bohr model is classical mechanics. The electron orbits the nucleus like the planets around the sun, but quantum mechanics says we don't know exactly where that electron is. The Bohr model turns out to be incorrect, and quantum mechanics has proven to be the best way to explain electrons in orbitals. We can describe those electrons in orbitals using the four quantum numbers. Let's look at the first quantum number here. This is called the principal quantum number. The principal quantum number is symbolized by n. n is a positive integer, so n could be equal to one, two, three, and so on. It indicates the main energy level occupied by the electron. This tells us the main energy level. You might hear this referred to as a shell sometimes, so we could say what kind of shell the electron is in. As n increases, the average distance of the electron from the nucleus increases, and therefore so does the energy. For example, if this was our nucleus right here, and let's talk about n is equal to one. For n is equal to one, let's say the average distance from the nucleus is right about here. Let's compare that with n is equal to two. n is equal to two means a higher energy level, so on average, the electron is further away from the nucleus, and has a higher energy associated with it. That's the idea of the principal quantum number. You're thinking about energy levels or shells, and you're also thinking about average distance from the nucleus. All right, our second quantum number is called the angular momentum quantum number. The angular momentum quantum number is symbolized by l. l indicates the shape of the orbital. This will tell us the shape of the orbital. Values for l are dependent on n, so the values for l go from zero all the way up to n minus one, so it could be zero, one, two, or however values there are up to n minus one. For example, let's talk about the first main energy level, or the first shell. n is equal to one. There's only one possible value you could get for the angular momentum quantum number, l. n minus one is equal to zero, so that's the only possible value, the only allowed value of l. When l is equal to zero, we call this an s orbital. This is referring to an s orbital. The shape of an s orbital is a sphere. We've already talked about that with the hydrogen atom. Just imagine this as being a sphere, so a three-dimensional volume here. The angular momentum quantum number, l, since l is equal to zero, that corresponds to an s orbital, so we know that we're talking about an s orbital here which is shaped like a sphere. So the electron is most likely to be found somewhere in that sphere. Let's do the next shell. n is equal to two. If n is equal to two, what are the allowed values for l? l goes zero, one, and so on all the way up to n minus one. l is equal to zero. Then n minus one would be equal to one. So we have two possible values for l. l could be equal to zero, and l could be equal to one. Notice that the number of allowed values for l is equal to n. So for example, if n is equal to one, we have one allowed value. If n is equal to two, we have two allowed values. We've already talked about what l is equal to zero, what that means. l is equal to zero means an s orbital, shaped like a sphere. Now, in the second main energy level, or the second shell, we have another value for l. l is equal to one. When l is equal to one, we're talking about a p orbital. l is equal to one means a p orbital. The shape of a p orbital is a little bit strange, so I'll attempt to sketch it in here. You might hear several different terms for this. Imagine this is a volume. This is a three-dimensional region in here. You could call these dumbbell shaped or bow-tie, whatever makes the most sense to you. This is the orbital, this is the region of space where the electron is most likely to be found if it's found in a p orbital here. Sometimes you'll hear these called sub-shells. If n is equal to two, if we call this a shell, then we would call these sub-shells. These are sub-shells here. Again, we're talking about orbitals. l is equal to zero is an s orbital. l is equal to one is a p orbital. Let's look at the next quantum number. Let's get some more space down here. This is the magnetic quantum number, symbolized my m sub l here. m sub l indicates the orientation of an orbital around the nucleus. This tells us the orientation of that orbital. The values for ml depend on l. ml is equal to any integral value that goes from negative l to positive l. That sounds a little bit confusing. Let's go ahead and do the example of l is equal to zero. l is equal to zero up here. Let's go ahead and write that down here. If l is equal to zero, what are the allowed values for ml? There's only one, right? The only possible value we could have here is zero. When l is equal to zero ... Let me use a different color here. If l is equal to zero, we know we're talking about an s orbital. When l is equal to zero, we're talking about an s orbital, which is shaped like a sphere. If you think about that, we have only one allowed value for the magnetic quantum number. That tells us the orientation, so there's only one orientation for that orbital around the nucleus. And that makes sense, because a sphere has only one possible orientation. If you think about this as being an xyz axis, (clears throat) excuse me, and if this is a sphere, there's only one way to orient that sphere in space. So that's the idea of the magnetic quantum number. Let's do the same thing for l is equal to one. Let's look at that now. If we're considering l is equal to one ... Let me use a different color here. l is equal to one. Let's write that down here. If l is equal to one, what are the allowed values for the magnetic quantum number? ml is equal to -- This goes from negative l to positive l, so any integral value from negative l to positive l. Negative l would be negative one, so let's go ahead and write this in here. We have negative one, zero, and positive one. So we have three possible values. When l is equal to one, we have three possible values for the magnetic quantum number, one, two, and three. The magnetic quantum number tells us the orientations, the possible orientations of the orbital or orbitals around the nucleus here. So we have three values for the magnetic quantum number. That means we get three different orientations. We already said that when l is equal to one, we're talking about a p orbital. A p orbital is shaped like a dumbbell here, so we have three possible orientations for a dumbbell shape. If we went ahead and mark these axes here, let's just say this is x axis, y axis, and the z axis here. We could put a dumbbell on the x axis like that. Again, imagine this as being a volume. This would be a p orbital. We call this a px orbital. It's a p orbital and it's on the x axis here. We have two more orientations. We could put, again, if this is x, this is y, and this is z, we could put a dumbbell here on the y axis. There's our second possible orientation. Finally, if this is x, this is y, and this is z, of course we could put a dumbbell on the z axis, like that. This would be a pz orbital. We could write a pz orbital here, and then this one right here would be a py orbital. We have three orbitals, we have three p orbitals here, one for each axis. Let's go to the last quantum number. The last quantum number is the spin quantum number. The spin quantum number is m sub s here. When it says spin, I'm going to put this in quotations. This seems to imply that an electron is spinning on an axis. That's not really what's happening, but let me just go ahead and draw that in here. I could have an electron ... Let me draw two different versions here. I could have an electron spin like a top, if you will, this way, or I could have an electron spin around that axis going this way. Again, this is not actually what's happening in reality. The electrons don't really spin on an axis like a top, but it does help me to think about the fact that we have two possible values for this spin quantum number. You could spin one way, so we could say the spin quantum number is equal to a positive one-half. Usually you hear that called spin up, so spin up, and we'll symbolize this with an arrow going up in later videos here. Then the other possible value for the spin quantum number, so the spin quantum number is equal to a negative one-half. You usually hear that referred to as spin down, and you could put an arrow going down. Again, electrons aren't really spinning in a physical sense like this, but, again, if you think about two possible ways for an electron to spin, then you get these two different, these two possible spin quantum numbers, so positive one-half or negative one-half. Those are the four quantum numbers, and we're going to use those to, again, think about electrons in orbitals." + }, + { + "Q": "How do I work this problem x^2-7x+10", + "A": "x\u00c2\u00b2-7x+10 = x\u00c2\u00b2-5x-2x+10 = x\u00c2\u00b2+(-5-2)x + (-5*-2) = (x-5)(x-2) (Since, x+(a+b)x+ab = (x+a)(x+b))", + "video_name": "2ZzuZvz33X0", + "transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be positive, one has to be negative. And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common factor of negative 7, so let's factor that out. So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers is equal to zero. If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5. That is one solution to the equation, or you can add 7 to both sides of that equation, and you get s is equal to 7. So if s is equal to negative 5, or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is minus 35. That does equal zero. If you have 7, 49 minus 14 minus 35 does equal zero. So we've solved for s. Now, I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what is that equal to? x times x is x squared, x times b is bx. a times x is plus ax. a times b is ab. So you get x squared plus-- these two can be added-- plus a plus bx plus ab. And that's the pattern that we have right here. We have 1 as a leading coefficient here, we have 1 as a leading coefficient here. So once we have our two numbers that add up to negative 2, that's our a plus b, and we have our product that gets to negative 35, then we can straight just factor it into the product of those two things. So it will be-- or the product of the binomials, where those will be the a's and the b's. So we figured it out. It's 5 and negative 7. 5 plus negative 7 is negative 2. 5 times negative 7 is negative 35. So we could have just straight factored at this point. 2, well, actually this was the case of s. So we could have factored it straight to the case of s plus 5 times s minus 7. We could have done that straight away and would've gotten to that right there. And, of course, that whole thing was equal to zero. So that would've been a little bit of a shortcut, but factoring by grouping is a completely appropriate way to do it as well." + }, + { + "Q": "What is in-between the pericardial walls", + "A": "Not very much, just the regular fluid that is all around your body. The interesting thing is that there are no cells between the walls, it is just that small amount of fluid", + "video_name": "bm65xCS5ivo", + "transcript": "So you're probably feeling pretty comfortable with the diagram of the heart, but let me just go ahead and label a few things just to make sure we're all on the same page. So blood flows from the right atrium to the right ventricle and then goes to the lungs and then the left atrium to the left ventricle. So that's usually the flow of blood. And one of the things that keeps the blood flowing in the right direction, we know, is the valves. And two of the valves I'm actually going to give you new names, something slightly different from what we have been referring to them by. These are the atrioventricular valves, and you can take a guess as to which ones I'm referring to. Atrioventricular valves are the two valves between the atria and the ventricles. So one will be the tricuspid valve, and the other is the mitral valve. And just to orient us, this is the tricuspid, the T. And this is our mitral, or M. And the atrioventricular valves, these two valves, if you look at them, they're both facing downwards. And one of the things that you might be wondering is, well, how is it that they aren't just flopping back and forth? And these valves, in particular, have a very interesting strategy. And that is that they actually are tethered to the walls. So they're held down here like that, and they have on the other end of those tethers a little muscle there. Now, this makes perfect sense if you think about it, because the ventricles are very strong. We know the ventricles are really, really strong. And so if the ventricles are squeezing, there's a good chance that that blood is going to shoot up in any direction it can go. It's going to go back perhaps through the mitral valve if it can go there, or it'll go through that tricuspid valve if it can go there. But the reason that it won't is that these papillary muscles are basically kind of sending out little lifelines, these chordae tendineae lifelines, to keep the valve from flipping backwards. So these chordae tendineae, these cords, are important for that reason. They're keeping the valve from flipping backwards. So these are all chordae tendineae, and these are all the papillary muscles. And these are particularly important, then, we can tell, for when you're trying to make sure that the ventricles don't screw up the valves. And now let's say that by accident our ventricle is just too strong, too powerful. Let's say that it broke one of these cords. Let's say it broke this one right here. And that's because our ventricle was just forcing too much blood back, and it just snapped the cord. What would happen? Well, this would basically kind of start flipping back and forth. It would flip this way and this way. And then on the next heartbeat, blood would start going the wrong direction, because this valve is not able to keep that nice tight seal. And so blood would basically kind of go this way when it wasn't supposed to. And all of a sudden, our flow of blood is now going in the wrong direction. So the chordae tendineae and the papillary muscles do a really, really important job in preventing that from happening. So let's move our attention to another area. Let's focus on this right here, which is the interventricular septum. And you can think of septum as basically a wall, interventricular septum. In this interventricular septum, the one thing I want to point out, which is maybe fairly obvious when you look at it-- you might think, well, why did you even have to say it? That's pretty obvious. This area is really thin, and this area is really thick by comparison. So the two areas are not equal in size. This is much thicker. And the reason I wanted to bring that up is because this first area in blue is called the membranous part, literally like a membrane. And the bottom, the red part, is the muscular part. This is the strong muscular part. So you have two different areas in that interventricular septum, the wall between the ventricles. And one of the interesting things about the membranous part, in particular, is that a lot of babies are born with holes in that membranous part. So when I say a lot, I don't mean the majority of babies, by any means. But one of the most common defects, if there is going to be a defect, would be that you would actually have a communication between these two so that blood could actually, again, flow from a place that it's not supposed to go, the left ventricle, into a place it shouldn't be going, the right ventricle. So blood can actually flow through those holes, and that is a problem. That is called a VSD. And actually, you might hear that term at some point. So I just wanted to point out where that happens. And while I'm writing VSD, you can take a stab at guessing what it might stand for. Ventricular, and S is septal. Again, septal just means wall. And D is defect. So a VSD is most common in that membranous part, more so than that muscular part. Now, let's move on again to one final thing I want to point out, which is I want to zoom in on the walls. So here in a gray box I'm going to kind of highlight what's going on this wall and how many layers there are in this wall. Let me draw out a little rectangle to correspond to that little rectangle I drew on the heart itself. So there are three layers to the heart muscle. And actually, I'm going to go through all three layers. And we'll start from the inside and work our way out. So on the inside, you have what's called the endocardium. And I'm actually going to draw the endocardium all the way around here. It goes all the way around the valves, so now you already learned that the valves now have endocardium. It goes around the ventricle and, as I showed you in the beginning, also around the atrium. And it goes all the way up and covers both the right and left side. The endocardium is very, very similar in many ways to the inner lining of the blood vessels, actually. So it's a really thin layer. It's not a very thick layer. It's the layer that all the red blood cells are kind of bumping up against. So when the red blood cells are entering the chambers of the heart, the part that they're going to see is going to be the endocardium. So this is what it looks like, and this is that green layer all the way around that I've drawn now. So if I was to draw it kind of in a blown-up version, it might look like this. And it's a few cell layers thick. And like I said, on the inside you have some red blood cells bumping along. So maybe this is one red blood cell, and this is maybe another one. And they would bump into that endocardium. Now, if you go a little bit deeper to the endocardium, what do you get to next? Well, next you get to myocardium. And that would be, let's say, the biggest chunk of our wall. And that would look something like this. And that myocardium you can kind of appreciate without even having me point it out, because it's the most common part of this entire thing. So this is our myocardium, and let me go back and actually label the endocardium as well. And on the other side-- and actually, just notice that the words are all pretty similar. Myo means muscle. And actually, while I'm on myocardium, let me just point out one more thing. The myocardium is where all of the contractile muscle is going to be, so that's where a lot of the work is being done. And that's also where a lot of the energy is being used up. So when the heart needs oxygen, it's usually the myocardium, because that's the part that's doing all of the work. OK. Now, on the other side of myocardium, what do we have on the outside? Well, we have a layer called the pericardium, and let me try to draw that in for you. Pericardium is something like this, kind of a thin layer. And the interesting thing about pericardium is that there's actually two layers to it. So there's actually something like this where you have two layers, an inner layer and an outer layer. And between the two layers you have literally a gap. There's a gap right there. And in that gap, you might have a little bit of fluid. But it's not actually cells. I guess that's the biggest point. It's not actually cells. It's more just a little bit of fluid that hangs out there. So this whole thing is called the pericardium. Now, you may be wondering how in the world do you get a layer that has a gap within it. So let me actually try to show you what happens in a fetus. So let's say you have a little fetus heart, a tiny little heart like this, and it gets a little bit bigger like this. And then it finally gets into an adult heart, something like that. So this would be the adult heart, right? Well, at the same time that the heart is growing, you actually also have a sac, almost like a little balloon. And this balloon actually begins to envelope the heart, so this growing heart kind of grows right into the balloon. And so this balloon kind of starts going around it like that, and you get something like this. And then eventually, as the heart gets really big, you get something like this. You basically have this kind of inner layer of the balloon that's pancaked out that doesn't even look like a balloon anymore. It's very flat, and then it kind of folds back on itself like that. And it comes all the way around. And now you can see why even though it's continuous-- it's not like it breaks. It is continuous here-- you can see how if you actually just were to look at one chunk of it, like we're looking at right here, you can see how it would actually look like a pancake. And so on our heart actually, it literally would be something like this, like a very thin kind of pancake. And I'm not doing a very, very good job making it look thin, but you can imagine what it is that it could look like if I was to zoom in on it. Basically, something like that, where you have two layers that are basically just kind of turned in on themselves. And both layers put together are called your pericardium. Now, there are actually separate names for the two layers. So for example, the layer that's kind of hugging up against the heart, this layer that I'm drawing right now, this layer is called the visceral pericardium. So you call that the visceral pericardium. And the name visceral, this right here, would be visceral. And the reason it's called visceral is because viscera refers to organs, so that's called the visceral pericardium. And then this outer layer, the one I'm drawing now, is called the parietal pericardium. And that's the layer that actually is on the outside, so let me label that as well. So that's this guy. That would be the parietal pericardium. So now you can actually see the layers of the heart-- the endocardium, myocardium, and pericardium. And actually, just to throw you a curve ball, because I'm pretty sure you can handle it, this visceral pericardium, another name for it, just because you might see it sometime, is the epicardium. Sometimes you might see the name epicardium. And don't get thrown off. It's really just the visceral pericardium. It's just the outermost layer of that heart before you get to the parietal layer." + }, + { + "Q": "100% of 100 squares is 1 whole right ?", + "A": "I m pretty sure this is what you re asking: 100% of 100 squares is 100 squares, just as 20% of 100 squares is 20 squares. 100% is the total amount so out of 100 squares it s 100. So yes, I think you re correct.", + "video_name": "Lvr2YsxG10o", + "transcript": "We're asked to shade 20% of the square below. Before doing that, let's just even think about what percent means. Let me just rewrite it. 20% is equal to-- I'm just writing it out as a word-- 20 percent, which literally means 20 per cent. And if you're familiar with the word century, you might already know that cent comes from the Latin for the word hundred. This literally means you can take cent, and that literally means 100. So this is the same thing as 20 per 100. If you want to shade 20%, that means, if you break up the square into 100 pieces, we want to shade 20 of them. 20 per 100. So how many squares have they drawn here? So if we go horizontally right here, we have one, two, three, four, five, six, seven, eight, nine, ten squares. If we go vertically, we have one, two, three, four, five, six, seven, eight, nine, ten. So this is a 10 by 10 square. So it has 100 squares here. Another way to say it is that this larger square-- I guess that's the square that they're talking about. This larger square is a broken up into 100 smaller squares, so it's already broken up into the 100. So if we want to shade 20% of that, we need to shade 20 of every 100 squares that it is broken into. So with this, we'll just literally shade in 20 squares. So let me just do one. So if I just do one square, just like that, I have just shaded 1 per 100 of the squares. 100 out of 100 would be the whole. I've shaded one of them. That one square by itself would be 1% of the entire square. If I were to shade another one, if I were to shade that and that, then those two combined, that's 2% of the It's literally 2 per 100, where 100 would be the entire square. If we wanted to do 20, we do one, two, three, four-- if we shade this entire row, that will be 10%, right? One, two, three, four, five, six, seven, eight, nine, ten. And we want to do 20, so that'll be one more row. So I can shade in this whole other row right here. And then I would have shaded in 20 of the 100 squares. Or another way of thinking about it, if you take this larger square, divide it into 100 equal pieces, I've shaded in 20 per 100, or 20%, of the entire larger square. Hopefully, that makes sense." + }, + { + "Q": "would you use the same formula for an isoscles triangle", + "A": "No. Area is still 1/2 s * h but it has a base length a, and two sides length b. So a + 2b = x/3.", + "video_name": "IFU7Go6Qg6E", + "transcript": "Let's say that I have a 100 meter long wire. So that is my wire right over there. And it is 100 meters. And I'm going to make a cut someplace on this wire. And so let's say I make the cut right over there. With the left section of wire-- I'm going to obviously cut it in two-- with the left section, I'm going to construct an equilateral triangle. And with the right section, I'm going to construct a square. And my question for you and for me is, where do we make this cut in order to minimize the combined areas of this triangle and this square? Well, let's figure out. Let's define a variable that we're trying to minimize, or that we're trying to optimize with respect to. So let's say that the variable x is the number of meters that we decide to cut from the left. So if we did that, then this length for the triangle would be x meters, and the length for the square would be, well, if we use x up for the left hand side, we're going to have 100 minus x for the right hand side. And so what would the dimensions of the triangle and the square Well, the triangle sides are going to be x over 3, x over 3, and x over 3 as an equilateral triangle. And the square is going to be 100 minus x over 4 by 100 minus x over 4. Now it's easy to figure out an expression for the area of the square in terms of x. But let's think about what the area of an equilateral triangle might be as a function of the length of its sides. So let me do a little bit of an aside right over here. So let's say we have an equilateral triangle. Just like that. And its sides are length s, s, and s. Now we know that the area of a triangle is 1/2 times the base times the height. So in this case, the height we could consider to be altitude, if we were to drop an altitude just like this. This length right over here, this is the height. And this would be perpendicular, just like that. So our area is going to be equal to one half times our base is s. 1/2 times s times whatever our height is, times our height. Now how can we express h as a function of s? Well, to do that we just have to remind ourselves that what we've drawn over here is a right triangle. It's the left half of this equilateral triangle. And we know what this bottom side of this right triangle is. This altitude splits this side exactly into two. So this right over here has length s over 2. So to figure out what h is, we could just use the Pythagorean theorem. We would have h squared plus s over 2 squared plus s over 2 squared is going to be equal to the hypotenuse squared, is going to be equal to s squared. So you would get h squared plus s squared over 4 is equal to s squared. Subtract s squared over 4 from both sides, and you get h squared is equal s squared minus s squared over 4. Now, to do this I could call it s squared. I could call this 4s squared over 4, just to be able to have a common denominator. And 4s squared minus s squared over 4 is going to be equal to 3s squared over 4. So we get h squared is equal to 3s squared over 4. Now we can take the principal root of both sides, and we get h is equal to the square root of 3 times s over 2. So now we can just substitute back right over here and we get our area. Our area is equal to 1/2 s times h. Well h is this business. So it's s times this. So it's 1/2 times s times the square root of 3s over 2, which is going to be equal to s times s is s squared. So this is going to be square root of 3 s squared over 2 times 2 over 4. So this is the area of an equilateral triangle as the function of the length of its sides. So what's the area of this business going to be? So the area of our little equilateral triangle-- let me write a combined area. And let me do that in a neutral color. So let me do that in white. So the combined area, I'll write it A sub c is going to be equal to the area of my triangle A sub t plus the area of my square. Well, the area of my triangle, we know what it's going to be. It's going to be square root of 3 times the length of a side squared divided by 4. So it's going to be square root of 3. Let me do that in that same yellow color. It's going to be-- make sure I switch colors. It's going to be square root of 3 over 4 times a side squared, times x over 3 squared. All I did is the length of a side is x over 3. We already know what the area is. It's the square root of 3 over 4 times the length of a side squared. And then the area of this square right over here, the area of the square is just going to be 100 minus x over 4 squared. So our area, our combined area-- and maybe I can write it like this-- our combined area as a function of where we make the cut is all of this business right over here. And this is what we need to minimize. So we need to minimize that right over there. And I will do that in the next video." + }, + { + "Q": "If a white dwarf isn't undergoing fusion, why does it radiate anything at all? Shouldn't it just skip straight to a black dwarf?", + "A": "Leftover heat.", + "video_name": "EdYyuUUY-nc", + "transcript": "In the last video, we started with a star in its main sequence, like the sun. And inside the core of that star, you have hydrogen fusion going on. So that is hydrogen fusion, and then outside of the core, you just had hydrogen. You just hydrogen plasma. And when we say plasma, it's the electrons and protons of the individual atoms have been disassociated because the temperatures and pressures are so high. So they're really just kind of like this soup of electrons and protons, as opposed to proper atoms that we associate with at lower temperatures. So this is a main sequence star right over here. And we saw in the last video that this hydrogen is fusing into helium. So we start having more and more helium here. And as we have more and more helium, the core becomes more and more dense, because helium is a more massive atom. It is able to pack more mass in a smaller volume. So this gets more and more dense. So core becomes more dense. And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster. Because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing, so it starts to fuse hotter. So let me write this, so the fusion, so hydrogen fuses faster. And actually, we even see this in our sun. Our sun today is brighter and hotter. It's fusing faster than it was when it was born 4.5 or 4.6 billion years ago. But eventually you're going to get to the point so that the core, you only have helium. So there's going to be some point where the entire core is all helium. And it's going to be way denser than this core over here. All of that mass over there has now been turned into helium. A lot of it has been turned into energy. But most of it is now in helium, and it's going to be at a much, much smaller volume. And the whole time, the temperature is increasing, the fusion is getting faster and faster. And now there's this dense volume of helium that's not fusing. You do have, and we saw this in this video, a shell around it of hydrogen that is fusing. So this right here is hydrogen fusion going on. And then this over here is just hydrogen plasma. Now the unintuitive thing, or at least this was unintuitive to me at first, is what's going on the core is that the core is getting more and more dense. It's fusing at a faster rate. And so it's getting hotter and hotter. So the core is hotter, fusing faster, getting more and more dense. I kind of imagine it's starting to collapse. Every time it collapses, it's getting hotter and more dense. But at the same time that's happening, the star itself is getting bigger. And this is actually not drawn to scale. Red giants are much, much larger than main sequence stars. But the whole time that this is getting more dense, the rest of the star is, you could kind of view it as getting less dense. And that's because this is generating so much energy that it's able to more than offset, or better offset the gravitational pull into it. So even though this is hotter, it's able to disperse the rest of the material in the sun over a larger volume. And so that volume is so big that the surface, and we saw this in the last video, the surface of the red giant is actually cooler-- let me write that a little neater-- is actually cooler than the surface of a main sequence star. This right here is hotter. And just to put things in perspective, when the sun becomes a red giant, and it will become a red giant, its diameter will be 100 times the diameter that it is today. Or another way to be put it, it will have the same diameter as the Earth's orbit around the current sun. Or another way to view it is, where we are right now will be on the surface or near the surface or maybe even inside of that future sun. Or another way to put it, when the sun becomes a red giant, the Earth's going to be not even a speck out here. And it will be liquefied and vaporized at that point in time. So this is super, super huge. And we've even thought about it. Just for light to reach the current sun to our point in orbit, it takes eight minutes. So that's how big one of these stars are. To get from one side of the star to another side of the star, it'll take 16 minutes for light to travel, if it was traveling that diameter, and even slightly longer if it was to travel it in a circumference. So these are huge, huge, huge stars. And we'll talk about other stars in the future. They're even bigger than this when they become supergiants. But anyway, we have the hydrogen in the center-- sorry. We have the helium in the center. Let me write this down. We have a helium core in the center. We're fusing faster and faster and faster. We're now a red giant. The core is getting hotter and hotter and hotter until it gets to the temperature for ignition of helium. So until it gets to 100 million Kelvin-- remember the ignition temperature for hydrogen was 10 million Kelvin. So now we're at 100 million Kelvin, factor of 10. And now, all of a sudden in the core, you actually start to have helium fusion. And we touched on this in the last video, but the helium is fusing into heavier elements. And some of those heavier elements, and predominately, it will be carbon and oxygen. And you may suspect this is how heavier and heavier elements form in the universe. They form, literally, due to fusion in the core of stars. Especially when we're talking about elements up to iron. But anyway, the core is now experiencing helium fusion. It has a shell around it of helium that is not quite there, does not quite have the pressures and temperatures to fuse yet. So just regular helium. But then outside of that, we do have the pressures and temperatures for hydrogen to continue to fuse. So out here, you do have hydrogen fusion. And then outside over here, you just have the regular hydrogen plasma. So what just happened here? When you have helium fusion all of a sudden-- now this is, once again, providing some type of energetic outward support for the core. So it's going to counteract the ever-increasing contraction of the core as it gets more and more dense, because now we have energy going outward, energy pushing things outward. But at the same time that that is happening, more and more hydrogen in this layer is turning into helium, is fusing into helium. So it's making this inert part of the helium core even larger and larger and denser, even larger and larger, and putting even more pressure on this inside part. And so what's actually going to happen within a few moments, I guess, especially from a cosmological point of view, this helium fusion is going to be burning super-- I shouldn't use-- igniting or fusing at a super-hot level. But it's contained due to all of this pressure. But at some point, the pressure won't be able to contain it, and the core is going to explode. But it's not going to be one of these catastrophic explosions where the star is going to be destroyed. It's just going to release a lot of energy all of a sudden into the star. And that's called a helium flash. But once that happens, all of a sudden, then now the star is going to be more stable. And I'll use that in quotes without writing it down because red giants, in general, are already getting to be less stable than a main sequence star. But once that happens, you now will have a slightly larger volume. So it's not being contained in as small of a tight volume. That helium flash kind of took care of that. So now you have helium fusing into carbon and oxygen. And there's all sorts of other combinations of things. Obviously, there's many elements in between helium and carbon and oxygen. But these are the ones that dominate. And then outside of that, you have helium forming. You have helium that is not fusing. And then outside of that, you have your fusing hydrogen. Over here, you have hydrogen fusing into helium. And then out here in the rest of the radius of our super-huge red giant, you just have your hydrogen plasma out here. Now what's going to happen as this star ages? Well, if we fast forward this a bunch-- and remember, as a star gets denser and denser in the core, and the reactions happen faster and faster, and this core is expelling more and more energy outward, the star keeps growing. And the surface gets cooler and cooler. So if we fast forward a bunch, and this is what's going to happen to something the mass of our sun, if it's more massive, then at some point, the core of carbon and oxygen that's forming can start to fuse into even heavier elements. But in the case of the sun, it will never get to that 600 million Kelvin to actually fuse the carbon and the oxygen. And so eventually you will have a core of carbon and oxygen, or mainly carbon and oxygen surrounded by fusing helium surrounded by non-fusing helium surrounded by fusing hydrogen, which is surrounded by non-fusing hydrogen, or just the hydrogen plasma of the sun. But eventually all of this fuel will run out. All of the hydrogen will run out in the stars. All of this hydrogen, all of this fusing hydrogen will run out. All of this fusion helium will run out. This is the fusing hydrogen. This is the inert helium, which will run out. It'll be used in kind of this core, being fused into the carbon and oxygen, until you get to a point where you literally just have a really hot core of carbon and oxygen. And it's super-dense. This whole time, it will be getting more and more dense as heavier and heavier elements show up in the course. So it gets denser and denser and denser. But the super dense thing will not, in the case of the sun-- and if it was a more massive star, it would get there-- but in the case of the sun, it will not get hot enough for the carbon and the oxygen to form. So it really will just be this super-dense ball of carbon and oxygen and all of the other material in the sun. Remember, it was superenergetic. It was releasing tons and tons of energy. The more that we progressed down this, the more energy was releasing outward, and the larger the radius of the star became, and the cooler the outside of the star became, until the outside just becomes this kind of cloud, this huge cloud of gas around what once was the star. And in the center-- so I could just draw it as this huge-- this is now way far away from the star, much even bigger than the radius or the diameter of a red giant. And all we'll have left is a mass, a superdense mass of, I would call it, inert carbon or oxygen. This is in the case of the sun. And at first, when it's hot, and it will be releasing radiation because it's so hot. We'll call this a white dwarf. This right here is called a white dwarf. And it'll cool down over many, many, many, many, many, many, many, years, until it becomes, when it's completely cooled down, lost all of its energy-- it'll just be this superdense ball of carbon and oxygen, at which point, we would call it a black dwarf. And these are obviously very hard to observe because they're not emitting light. And they don't have quite the mass of something like a black hole that isn't even emitting light, but you can see how it's affecting things around it. So that's what's going to happen to the sun. In the next few videos, we're going to talk about what would happen to things less massive than the sun and what would happen to things more massive can imagine the more massive. There would be so much pressure on these things, because you have so much mass around it, that these would begin to fuse into heavier and heavier elements until we get to iron." + }, + { + "Q": "what dose the little 2 mean", + "A": "The little 2 is an exponent. The exponent is another way to write a number, like 5^2(5 with a little two on the top), it means 5 * 5, when you see a tiny number next to a big number(has to be on the top), it means the big numbers times itself that many times(the small number). So 250^600 would mean 250 multiplied by 250 600 times.", + "video_name": "aoXUWSwiDzE", + "transcript": "Let's do some practice problems dealing with variable expressions. So these first problems say write the following in a more condensed form by leaving out the multiplication symbol or leaving out a multiplication symbol. So here we have 2 times 11x, so if we have 11 x's and then we're going to have 2 times those 11 x's, we're going to have 22 x's. So another way you could view this, 2 times 11x, you could view this as being equal to 2 times 11, and all of that times x, and that's going to be equal to 22 x's. You had 11 x's, you're going to have 2 times as many x's, so you're going to have 22 x's. Let's see, you have 1.35 times y. Now here we're just going to do a straight simplifying how we write it. So 1.35 times y-- I'll do it in a different color-- 1.35 times y-- that's a little dot there. In algebra we can just get rid of that dot symbol. If we have a variable following a number, we know that means 1.35 times that variable. So that, we could rewrite as just being equal to 1.35y. We've condensed it by getting rid of the multiplication sign. Let's see, here we have 3 times 1/4. Well, this is just straight up multiplying a fraction. So in problem 3-- this was problem 1, this is problem 2, problem 3-- 3 times 1/4, that's the same thing as 3 over 1 times 1/4. Multiply the numerators, you get 3. Multiply the denominators, 1 times 4, you get 4. So number 3, I got 3/4. And then finally, you have 1/4 times z. We could do the exact same thing we did up here in problem number 2. This was the same thing as 1.35y. That's the same thing as 1.35 times y. So down here we could rewrite this as either being equal to 1/4z, or we could view this as being equal to 1 over 4 times z over 1, which is the same thing as z times 1, over 4 times 1, or the same thing as z over 4. So all of these are equivalent. Now, what do they want us to do down here? Evaluate the following expressions for a is equal to 3, b is equal to 2, c is equal to 5, and d is equal to minus 4-- or, actually, I should say negative 4 is the correct terminology. Negative 4. So we just substitute. Every time we see an a, we're going to put a minus 3 there, or a negative 3 there. Every time we see a b, we'll put a positive 2 there. Every time we see a c, we'll put a 5 there. And every time we see a d, we'll put a minus 4 there. And I'll do a couple of these. I won't do all of them, just for the sake of time. So let's say problem number 5. They gave us 2 times a plus 3 times b. Well, this is the same thing as 2 times-- instead of an a, we know that a is going to be equal to negative 3. So 2 times minus 3, plus 3 times b-- what's b? They're telling us that b is equal to 2-- so 3 times 2. And what is this equal to? 2 times minus 3-- let me do it in a different color-- 2 times negative 3 is negative 6, plus 3 times 2. 3 times 2 is 6. That's positive 6. So that is equal to 0. And notice the order of operations. We did the multiplications, we did the two multiplications before we added the two numbers. Multiplication and division takes precedence over addition and subtraction. Let's do problem 6. I'll do that right here. So you have 4 times c. 4 times-- now what's c equal to? They tell us c is equal to 5. So 4 times 5, that's our c, plus d. d is minus or negative 4. So we have 4 times 5 is 20, plus negative 4-- that's the same thing as minus 4, so that is equal to 16. Problem 6. Now, let's do one of the harder ones down here. This problem 10 looks a little bit more daunting. Problem 10 right there. So we have a minus 4b in the numerator, if you can read it, it's kind of small. a is minus 3. So we have minus 3-- or negative 3-- minus 4 times b. b is 2. So 4 times 2. Remember, this right here is a, that right there is b. They're telling me up here. And then all of that over-- all of that is over 3c plus 2d. So 3 times-- what was c? c is 5 plus 2 times d. What is d? d is negative 4. So let's figure this out. So we have to do order of operations. Multiplication comes first before addition and subtraction. So this is going to be equal to minus 3 minus 4 times 2, minus 8, all of that over-- 3 times 5 is 15, plus 2 times negative 4 is negative 8, or 15 plus negative 8 is 15 minus 8. And now, the numerator becomes negative 3 minus 8, which is negative 11. And the denominator is 15 minus 8, which is 7. So problem 10, we simplified it to negative 11 over 7. Right there. Let's do a couple of these over here. OK, we see some exponents. I'll pick one of the harder ones. Let's do this one over here, problem 18. So 2x squared minus 3x squared, plus 5x minus 4. OK, well, this wasn't that hard. All of them are dealing with x. But what we could do here-- let me write this down. 2x squared minus 3x squared, plus 5x minus 4. And they tell us that x is equal to negative 1. One thing we could do is simplify this before we even substitute for negative 1. So what's 2 of something minus 3 of something? This is 2x squareds minus 3x squareds. So 2 of something minus 3 of something, that's going to be minus 1 of that something. So that right there-- or negative 1 of that something-- that would be negative 1x squared plus 5x minus 4. And they tell us x is equal to negative 1. So this is negative 1 times x squared, negative 1 squared, plus 5 times x, which is negative 1, minus 4. So what is this? Negative 1 squared is just 1. That's just 1. So this whole expression simplifies to negative 1 plus 5 times negative 1-- we do the multiplication first, of course. So that's minus 5, or negative 5 minus 4. So negative 1 minus 5 is negative 6, minus 4 is equal to negative 10. And I'll do these last two just to get a sample of all of the types of problems in this variable expression section. The weekly cost, c, of manufacturing x remote controls-- so the cost is c, x is the remote controls-- is given by this formula. The cost is equal to 2,000 plus 3 times the number of remote controls, where cost is given in dollars. Question a, what is the cost of producing 1,000 remote controls? Well, the number of remote controls is x. So for part a-- I could write it over here-- the cost is going to be equal to-- just use this formula-- 2,000 plus 3 times the number of remote controls. x is the number of remote controls. So 3 times 1,000. So it's going to be equal to 2,000 plus 3 times 1,000 is 3,000, which is equal to $5,000. So that's part a. $5,000. Now part b. What is the cost of producing 2,000 remote controls? Well, the cost-- just use the same formula-- is equal to 2,000 plus 3 times the number of remote controls. So 3 times 2,000. So that's equal to 2,000 plus 3 times 2,000 is 6,000. So that's equal to $8,000. Now we're at problem 22. The volume of a box without a lid is given by the formula, volume is equal to 4x times 10 minus x squared, where x is a length in inches, and v is the volume in cubic inches. What is the volume when x is equal to 2? So part a, x is equal to 2. We just substitute 2 wherever we see an x here. So the volume is going to be equal to 4 times x, which is 2, times 10 minus 2 squared. And so this is going to be equal to 4 times 2 is 8, times 10 minus 2 is 8 squared. So that's 8 times-- so this is equal to 8 times-- 8 squared is 64. You could say this is going to be 8 to the third power. And 64 times 8-- 4 times 8 is 32. 6 times 8 is 48, plus 3 is 51. So it's 512. Now, what is the volume when x is equal to 3? I'll do it here in pink. b, when is equal to 3, then the volume is equal to 4 times 3-- x is equal to 3 now-- times 10 minus 3 squared. 4 times 3 is 12, times 10 minus 3 is 7 squared. So it's equal to 12 times 49. And just to get the exact answer, let's multiply that. So 49 times 12-- 2 times 9 is 18, 2 times 4 is 8, plus 1 is 9. Put a 0. 1 times 9 is 9. 1 times 4 is 4. And then we add the two. We get 8, 9 plus 9 is 18, and then 1 plus 4 is 588." + }, + { + "Q": "But what if you choose something like S and choose your number shift as 7.Then what do you do?", + "A": "It wraps around the alphabet. So if you shift A by 1 to get B, you would shift Z by 1 to get A.", + "video_name": "sMOZf4GN3oc", + "transcript": "SPEAKER 1: The first well known cipher, a substitution cipher, was used by Julius Caesar around 58 BC. It is now referred to as the Caesar Cipher. Caesar shifted each letter in his military commands in order to make them appear meaningless should the enemy intercept it. Imagine Alice and Bob decided to communicate using the Caesar Cipher First, they would need to agree in advance on a shift to use-- say, three. So to encrypt her message, Alice would need to apply a shift of three to each letter in her original message. So A becomes D, B becomes E, C becomes F, and so on. This unreadable, or encrypted message, is then sent to Bob openly. Then Bob simply subtracts the shift of three from each letter in order to read the original message. Incredibly, this basic cipher was used by military leaders for hundreds of years after Caesar. JULIUS CAESAR: I have fought and won. But I haven't conquered over man's spirit, which is indomitable. SPEAKER 1: However, a lock is only as strong as its weakest point. A lock breaker may look for mechanical flaws. Or failing that, extract information in order to narrow down the correct combination. The process of lock breaking and code breaking are very similar. The weakness of the Caesar Cipher was published 800 years later by an Arab mathematician named Al-Kindi. He broke the Caesar Cipher by using a clue based on an important property of the language a message is written in. If you scan text from any book and count the frequency of each letter, you will find a fairly consistent pattern. For example, these are the letter frequencies of English. This can be thought of as a fingerprint of English. We leave this fingerprint when we communicate without realizing it. This clue is one of the most valuable tools for a codebreaker. To break this cipher, they count up the frequencies of each letter in the encrypted text and check how far the fingerprint has shifted. For example, if H is the most popular letter in the encrypted message instead of E, then the shift was likely three. So they reverse the shift in order to reveal the original message. This is called frequency analysis, and it was a blow to the security of the Caesar cipher." + }, + { + "Q": "I think this is a really dumb question, but I don't get how the sun goes through the atmosphere.", + "A": "(facepalm) The Sun is 93 million miles away. It s not in our atmosphere at all.", + "video_name": "05qDIjKevJo", + "transcript": "In the last video, we talk about how seasons on Earth are not caused by how close Earth is to the sun in its orbit. And we also hint at the fact that it's actually caused by the tilt of the earth. And so in this video, I want to show you how the tilt of the earth causes the seasons to happen. So let's draw-- so I'm going to try to draw as many diagrams as possible here. Because at least for my brain they help me visualize what's actually going on. So we could imagine a top view first. So let's have a top view. That is the sun right over there. And let me draw the earth's orbit. So Earth's orbit maybe looks something like that. Let me draw it almost, it is almost circular. So I'll draw it as something that's pretty close to a circle right over here. And I'm going to draw Earth at different points in its orbit. And I'm going to try to depict the tilt of its rotational axis. And obviously, this is not drawn anywhere near close to scale. Earth is much further away from the sun, and much, much smaller than the sun as well. So I'll draw the earth at that point. And at this point, the earth will be tilted away from the sun. So Earth's tilt does not change if you think about the direction, or at least over the course of a year, if we think about relatively small periods of time. It does not change relative to the direction that it's pointing at in the universe. And we'll talk about that in a second. So let's say right over here we are pointed away from the sun. So we're up and out of this page. So if I wanted to put some perspective on an arrow it would be up and-- actually it would be more like up and out of this page. So that's the direction. If you were to come straight out of the North Pole. And if you were to go straight out of the South Pole you'd go below that circle right over there. And if I wanted to draw the same position, but if we're looking sideways along the plane, the orbital plane, or the plane of Earth's orbit. So if we're looking at it from that direction. So let me do it this way. If we're looking at directly sideways, this is the sun right over here, and this is Earth at that position. This is Earth right over there. If I were to draw an arrow pointing straight out of the North Pole it would look something like this. So this arrow and this arrow they are both popping straight out of the North Pole. And so when we talk about the tilt of the earth, we're talking about the tilt of its orbital axis, kind of this pole that could go straight between the South Pole and the North Pole. The angle between that and a pole that would actually be at a 90-degree angle, or perpendicular, to the plane of its orbit. And so compared to if it was just straight up and down, relative to the plane of the orbit. So this right here is the angle of Earth's tilt. Let me draw that a little bit bigger just so it becomes a little bit clearer. So if this is the plane of the orbit, we're looking sideways along the plane of the orbit. And this is Earth right over here. My best attempt to draw a circle. That is Earth. Earth does not rotate. Its axis of rotation is not perpendicular to the plane of the orbit. So this is how Earth would orbit. This is how Earth would rotate if it was. Earth rotates, Earth's rotational axis is at an angle to that vertical relative to the plane of its orbit, I guess you could say it. It rotates at an angle like this. So this would be the North Pole. That is the South Pole. And so it rotates like this. And that angle relative to being vertical with respect to the orbital plane, this angle right here for Earth right now is 23.4 degrees. And if we're talking about relatively short periods of time, like our lifespans, that is constant. But it is actually changing over long periods of time. That is changing between-- and these are rough numbers-- it is changing between 22.1 degrees and 24.5 degrees, if my sources are correct. But that gives a rough estimate of what it's changing between. But I want to make it clear, this is not happening overnight. The period to go from roughly a 22-degree angle to a 24 and 1/2-degree angle and back to a 22-degree angle is 41,000 years. And this long-term change in the tilt, this might play into some of the long-term climactic change. Maybe it might contribute, on some level, to some of the ice ages that have formed over Earth's past. But for the sake of thinking about our annual seasons you don't have to worry too much, or you don't have to worry at all really about this variation. You really just have to know that it is tilted. And right now it is tilted at an angle of 23.4 degrees. Now you might say OK, I understand what the tilt is. But how does that change the seasons in either the Northern or that Southern Hemisphere? And to do that, I'm going to imagine the earth when the Northern Hemisphere is most tilted away from the sun, and when it is most tilted towards the sun. So remember this tilt, the direction this arrow points into relative to the rest of the universe, if we assume that this tilt is at 23.4%, it's not changing throughout the year. But depending on where it is in the orbit it's either going to be tilting away from the sun, as it is in this example right over here. Or it will be tilting towards the sun. I'll do the towards the sun in this magenta color, or it would be tilting towards the sun. So six months later when the earth is over here, it's going to, relative to the rest of the universe, it will be tilted in that same direction, up out of this page and to the right. Just like it was over here. But now that it's on the other side of the sun that makes it tilt a little bit more towards the sun. If I were to draw it right over here, it is now tilted towards the sun. And what I want to think about is how much sunlight will different parts of the planet receive. And I'll focus on the Northern Hemisphere. But you can make a similar argument for the Southern Hemisphere. I want to think about how much sunlight they receive when it's tilted away or tilted towards the sun. And so let's think about those two situations. So first of all, let's think about this situation here where we are tilted away from the sun. So let me zoom in a little bit. So this is the situation, where we're tilted away from the sun. So if this is the vertical, so let me draw it. I could actually just use this diagram. But let me make it. So we're tilted away from the sun like this. I'm going to do this in a different color. So if we have an arrow coming straight out of the North Pole it would look like this. And we are rotating around like that. So we're out of the page on the left-hand side, and then into the page on the right-hand side. And so we're rotating towards the east, constantly. So this arrow is in the direction of the east. So when we're at this point in Earth's orbit, and actually let me copy and paste this. And I'm going to use the same exact diagram for the different seasons. So let me copy. And then let me paste this exact diagram. I'll do it over here for two different points. So when we are here in Earth's orbit where is the sunlight coming from? Well, it's going to be coming from the left, at least the way I've drawn the diagram right over here. So the sunlight is coming from the left in this situation. And so if you think about it, what part of the earth is being lit by sunlight? Or what part of the earth is in daylight, the way I've drawn it right over here? Well, the part that is facing the sun. So all of this right over here is going to be in daylight. As we rotate whatever part of the surface of the earth enters into this yellow part right over here will be in daylight. But let's think about what's happening So let me draw the equator, which separates our Northern and Southern Hemispheres. So this is the equator. And then let me go into the Northern Hemisphere. And I want to show you why when the North Pole is pointed away from the sun why this is our winter. So when we're pointed away from the sun-- Well, if we go to the Arctic Circle-- so let me go right over here. Let me go to some point in the Arctic Circle. As it goes, as the earth rotates every 24 hours, this point on the globe will just rotate around just like that. It will just keep rotating around just like that. And so my question is, that point in the Arctic Circle, as it rotates will it ever see sunlight? Well, no, it will never see sunlight. Because the North Pole is tilted away from the sun. So what I'm drawing, what I'm shading here in purple, that part of the earth, when it's completely tilted away, will never see sunlight. Or at least it won't see sunlight while it's tilted away, while it's in this position, or in this position in the orbit. I won't say never, because once it becomes summer they will be able to see it. So no sunlight, no day, I guess you could say, no daylight. If you go to slightly more southern latitudes, so let's say you go over here. So maybe that's the latitude of something like, I don't know, New York or San Francisco or something like that. Let's think about what it would see as the earth rotates every 24 hours. So this would be daylight, daylight, daylight, daylight, then nighttime, nighttime, nighttime, nighttime, nighttime. This is now going behind the globe nighttime, nighttime, nighttime, nighttime, nighttime, daylight, daylight, daylight, daylight. So if you just compare this. So let me do the daylight in orange. And then nighttime I will do in this bluish purplish color. So night time over here. So if you go to really northern latitudes, like the Arctic Circle, they don't get any daylight when we are tilted away from the earth. And if we go to slightly still northern latitudes, but not as north as the Arctic Circle, it does get daylight. But it gets a lot less daylight. It spends a lot less time in the daylight than in the night time. So notice if you say that this circumference represents the positions over 24 hours, it spends much less time in the daylight than it does in the nighttime. So because, while the Northern Hemisphere is tilted away from the earth, the latitudes in the northern hemisphere are getting less daylight. They are also getting less energy from the sun. And so that's what leads to winter, or just being generally colder. And to see what happens in the summer let's just go the other side. So now we're going to the other side of our orbit This is going to be six months later. And notice the actual direction, relative to the rest of the universe, has not changed. We're still pointed in that same direction. We still have a 23.4 degree tilt relative to, I guess, being straight up and down. But now once we're over here the light from the sun is going to be coming from the right. Just like that. And now, if on this diagram at least, this is the side of the earth that is going to be getting the sunlight. And let me draw the equator again, or my best attempt to draw the equator. I'll draw the equator in that same color actually, in that green color. So this separates the Northern and the Southern Hemisphere. And now let's think about the Arctic Circle. So let's say I'm sitting here in the Arctic Circle. As the day goes on, as 24 hours go around, I'll keep rotating around here. But notice the whole time I am inside of the sun. I'm getting no nighttime. There is no night in the Arctic Circle while we are tilted towards the sun. And if we still do that fairly northern latitude, but not as far as the Arctic Circle, maybe in San Francisco or New York, or something like that . If we go to that latitude, notice how much time we spend in the sun. So maybe we just enter. So this is right at sunrise. And then as the day goes on we're in sunlight, sunlight, sunlight, sunlight, sunlight, sunlight, sunlight, sunlight. Then we hit sunset. Then we hit nighttime, nighttime, then we hit nighttime, and then we get sunrise again. And so when you look at the amount of time that something in the Northern Hemisphere spends in the daylight versus sunlight, you'll see it spends a lot more time in the daylight when the Northern Hemisphere is tilted towards the sun. So this is more day, less night. So it is getting more energy from the sun. So when it is tilted towards the sun it is getting more energy from the sun. So things will generally be warmer. And so you are now talking about summer in the Northern Hemisphere. And the arguments for the Southern Hemisphere are identical. You could even play it right over here. When the Northern Hemisphere is tilted away from the sun, then the Southern Hemisphere is tilted towards the sun. And so for example, the South Pole will have all daylight and no nighttime. And southern latitudes will have more daylight than nighttime. And so the south will have summer. So this is summer in the south, in the Southern Hemisphere. And it's winter in the north. And then down here the Southern Hemisphere is pointed away from the sun. So this is winter in the Southern Hemisphere. And you might be saying, hey Sal, what about you haven't talked a lot about spring and fall. Well let's think about it. Well, if we're talking about the Northern Hemisphere, this over here, we decided, was winter in the Northern Hemisphere. And we're going to rotate around the sun. And at some point, we're going to get over here. And then because of this tilt we aren't pointed away or towards the sun. We're kind of pointed I guess sideways relative to the direction of the sun. But this doesn't favor one hemisphere over the other. So when we're over here in-- and this will actually be the spring now- when we're in the spring, both hemispheres are getting the equal amount of daylight and sunlight, or for a given latitude above or below the equator, they're getting the same amount. And the same thing is true over here when we get to-- so this is spring. This is the summer in the Northern Hemisphere. Now this will be the fall in the Northern Hemisphere. And once again, we're tilted in this direction. And so the Northern Hemisphere isn't tilted away or towards the sun. And so both hemispheres are going to get the same amount of radiation from the sun. So you really see the extremes in the winters and the summers. Now one thing I do want to make clear, and I started off with just the length of day and nighttime. Because frankly, that's maybe a little bit, or at least in my brain, a little bit easier to visualize. But that by itself does not account for all of the difference between summer and winter. Another cause, and actually this is probably the biggest cause, is if you think about the total amount of sun. So let's talk about the Northern Hemisphere winter. And let's say there's a certain amount of sunlight that is reaching the earth. So this is the total amount of sunlight that's reaching the earth at any point in time. You see that much more of that is hitting the Southern Hemisphere than the Northern Hemisphere here. All of these, if you imagine it, all of these rays right over here are hitting the Southern Hemisphere. So a majority of the rays are hitting the Southern Hemisphere. And much fewer are hitting the Northern Hemisphere. So actually a smaller amount of the radiation period, at even a given period in time, not even talking about the amount of time you are facing the sun. But at any given moment in time more energy is hitting the Southern Hemisphere than the Northern. And the opposite is true when the tilt is then towards the sun. And now a disproportionate amount of the sun's energy is hitting the Northern Hemisphere. So if you draw a bunch of, if you just think that this is all of the energy from the sun, most of it, all of these rays up here, are hitting the Northern Hemisphere. And only these down here are hitting the Southern And on top of that, what makes it even more extreme is that the actual angle, and of course, this is to some degree is due to the fact that where the angle of the sun relative to the horizon, or where you are on Earth. But even more than that if you are on, let's say that this is the land, and we're talking about the winter in the Northern Hemisphere. So let's say you're talking about, let's say we're up over here at this northern latitude. And we're just looking at the sun here. And over here, you could see even when we are closest to the sun the sun is not directly overhead. When we're closest to the sun the sun still is pretty low on the horizon. So it may be right over here when we're closest to the sun in the winter the sun might be right over here. But if you look at that same latitude in the summer when it is closest to the sun, the sun is more close to being directly overhead. It still won't be directly overhead. Because we are still at a relatively northern latitude. But the sun is going to be much higher in the sky. And these are all related to each other. It's kind of connected with this idea that more energy is hitting one hemisphere or the other. But also, when you have a, I guess you could say, a steeper angle from the rays of the sun with the earth, it's actually going to be dissipated less by the atmosphere. And let me just make it clear how this is. So in the summer-- so let's say that that's the land. And let's say that-- let me draw the atmosphere in white-- so all of this area right over here, this is the atmosphere. And obviously there's not a hard boundary for the atmosphere. But let's just say this is the densest part of the atmosphere. In the summer, when the sun is higher in the sky, the rays from the sun are dissipated by less atmosphere. So they have to get through this much atmosphere. And they're bounced off. And they heat some of that atmosphere. And they're absorbed before they get to the ground. In the winter when the sun is lower in the sky, so maybe the sun is out here. Let me draw it a little bit. So when the sun is lower in the sky relative to this point, you see that the rays of sunlight have to travel through a lot more atmosphere. So they get dissipated much more before they get to this point on the planet. So all in all it is the tilt that is causing the changes in the season. But it's causing it for multiple reasons. One is when you're tilted, we'll say when you're tilted towards the sun, you're getting more absolute hours of daylight. Not only are you getting more absolute hours of daylight, but at any given moment, most or more of the sun's total rays that are hitting the earth are hitting the Northern Hemisphere as opposed to the Southern Hemisphere. And the stuff that's hitting the places that have summer, it has to go through less atmosphere. So it gets dissipated less." + }, + { + "Q": "Give the geometric interpretation of equation 3y+15=0 as in\n1) One Variable &\n2) Two Varibles", + "A": "The equation 3y + 15 = 0 simplifies to y = -5. This is linear equation in one variable, so the geometric interpretation of this is a straight line. In this case it is parallel to the x-axis and crosses the y-axis at the point (0, -5).", + "video_name": "2VeqrZ_PMiY", + "transcript": "Use substitution to solve for x and y. And they give us a system of equations here. y is equal to negative 5x plus 8 and 10x plus 2y is equal to negative 2. So they've set it up for us pretty well. They already have y explicitly solved for up here. So they tell us, this first constraint tells us that y must be equal to negative 5x plus 8. So when we go to the second constraint here, every time we see a y, we say, well, the first constraint tells us that y must be equal to negative 5x plus 8. So everywhere we see a y, we can substitute it with negative 5x plus 8. Because that's what the first constraint tells us. y is equal to that. I don't want to be repetitive, but I really want you to internalize that's all it's saying. y is that. So every time we see a y in the second constraint, we can substitute it with that. So let's do it. So the second equation over here is 10x plus 2. And instead of writing a y there, and I've said it multiple times already, we can write a negative 5x plus 8. The first constraint tells us that's what y is. So negative 5x plus 8 is equal to negative 2. Now, we have one equation with one unknown. We can just solve for x. We have 10x plus. So we can multiply it. We can distribute this 2 onto both of these terms. So we have 2 times negative 5x is negative 10x. And then 2 times 8 is 16. So plus 16 is equal to negative 2. Now we have 10x minus 10x. Those guys cancel out. 10x minus 10x is equal to 0. So these guys cancel out. And we're just left with 16 equals negative 2, which is crazy. We know that 16 does not equal negative 2. This is an inconsistent result. And that's because these two lines actually don't intersect. And we could see that by actually graphing these lines. Whenever you get something like some number equalling some other number that they're clearly not equal to, that means it's an inconsistent result, It's an inconsistent system, and that these lines actually don't intersect. So let me just graph these just to make it clear. This first equation is already in slope y-intercept form. So it looks something like this. That's our x-axis. This is our y-axis. And it's negative 5x plus 8, so 1, 2, 3, 4, 5, 6, 7, 8. And then it has a very steep downward slope. Every time you move forward 1, you have to go down 5. So it looks something like that. That's this first equation right over there. The second equation, let me rewrite it in slope y-intercept form. So it's 10x plus 2y is equal to negative 2. Let's subtract 10x from both sides. You get 2y is equal to negative 10x minus 2. Let's divide both sides by 2. You get y is equal to negative 5x, negative 5x minus 1. So it's y-intercept is negative 1. It's right over there. And it has the same slope as this first line. So it looks like this. It's parallel. It's just shifted down a bit. So it just looks like that. So they're parallel lines. They have the same slope, different y-intercepts. We get an inconsistent result. They don't intersect. And the telltale sign of that, when you're doing it algebraically, is you get something wacky like this. This is why it's called inconsistent. It's not consistent for 16 to be equal to negative 2. These don't intersect. There's no solution to both of these constraints, no x and y that satisfies both of them." + }, + { + "Q": "Isn't FOILing basically just doing the distributive property?", + "A": "Yes... It is an expansion of the distributive property. If you want to multiply: (x+3)(x+5), the distributive property allows us to distribute the whole binomial (x+3) across the 2nd binomial. We get: x(x+3) +5(x+3). We then apply the distributive property to distribute both the x and the 5, and we get: x^2 + 3x + 5x + 15. This is the same sequence of terms as if you apply FOIL. FOIL let s us skip the 1st step of creating x(x+3) +5(x+3). Hope this helps.", + "video_name": "ZMLFfTX615w", + "transcript": "Multiply (3x+2) by (5x-7). So we are multiplying two binomials. I am actually going to show you two really equivalent ways of doing this. One that you might hear in a classroom and it is kind of a more mechanical memorizing way of doing it which might be faster but you really don't know what you are doing and then there is the one where you are essentially just applying something what you already know and kind of a logical way. So I will first do the memorizing way that you might be exposed to and they'll use something called FOIL. So let me write this down here. So you can immediately see that whenever someone gives you a new mnemonic to memorize, that you are doing something pretty mechanical. So FOIL literally stands for First Outside, let me write it this way.....F O I L where the F in FOIL stands for First, the O in FOIL stand for Outside, the I stands for Inside and then the L stands for Last. The reason why I don't like these things is that when you are 35 years old, you are not going to remember what FOIL stood for and then you are not going to remember how to multiply this binomial. But lets just apply FOIL. So First says just multiply the first terms in each of these binomials. So just multiply the 3x times the 5x. So (3x. 5x). The Outside part tells us to multiply the outside terms. So in this case, you have 3x on the outside and you have -7 on the outside. So that is +3x(-7). The inside, well the inside terms here are 2 and 5x. So, (+2.5x) and then finally you have the last terms. You have the 2 and the -7. So the last terms are 2 times -7. 2(-7). So what you are essentially doing is just making sure that you are multipying each term by every other term here. What we are essentially doing is multiplying, doing the distributive property twice. We are multiplying the 3x times (5x-7). So 3x times (5x-7) is (3x . 5x) plus (3x - 7). And we are multiplying the 2 times (5x-7) to give us these terms. But anyway, lets just multiply these out just to get our answer. 3x times 5x is same thing as (3 times 5) ( x times x) which is the same thing as 15x square. You can just do this x to the first time to x to the first. You multiply the x to get x squared. 3 times 5 is 15. This term right here 3 times -7 is -21 and then you have your x right over here. And then you have this term which is 2 times 5 which is 10 times x. So +10x. And then finally you have this term here in blue. 2 times -7 is -14. And we aren't done yet, we can simplify this a little bit. We have two like terms here. We have this...let me find a new color. We have 2 terms with a x to the first power or just an x term right over here. So we have -21 of something and you add 10 or in another way, you have 10 of something and you subtract 21 of them, you are going to have -11 of that something. We put the other terms here, you have 15... 15x squared and then you have your -14 and we are done. Now I said I would show you another way to do it. I want to show you why the distributive property can get us here without having to memorize FOIL. So the distributive property tells us that if we 're... look if we are multipying something times an expression, you just have to multiply times every term in the expression. So we can distribute, we can distribute the 5x onto the 3..., or actually we could...well, let me view it this way... we could distribute the 5x-7, this whole thing onto the 3x+2. Let me just change the order since we are used to distributing something from the left. So this is the same thing as (5x-7)(3x+2). I just swapped the two expressions. And we can distribute this whole thing times each of these terms. Now what happens if I take (5x-7) times 3x? Well, thats just going to be 3x times (5x-7). So I have just distributed the 5x-7 times 3x and to that I am going to add 2 times 5x-7. I have just distributed the 5x-7 onto the 2. Now, you can do the distributive property again. We can distribute the 3x onto the 5x. We can distribute the 3x onto the 5x. And we can distribute the 3x onto the -7. We can distribute the 2 onto the 5x, over here and we can distribute the 2 on that -7. Now if we do it like this what do we get ? 3x times 5x, that's this right over here. If we do 3x times -7, that's this term right over here. If you do 2 times 5x, that's this term right over here. If you do 2 times -7, that is this term right over here. So we got the exact same result that we got with FOIL. Now, FOIL can be faster if you just wanted to do it and kind of skip to this step. I think its important that you know that this is how it actually works. Just in case you do forget this when you are 35 or 45 years old and you are faced with multiplying binomial, you just have to remember the distributive property." + }, + { + "Q": "what was the Bernoulli Principle", + "A": "ok that was the Bernoulli Principle! very helpful now i know what this guy is talking about!", + "video_name": "aLJzEl5st8s", + "transcript": "So this is an airplane here. OK, so you probably already knew that. If you've flown in one, or maybe just seen them fly. But even if you've seen them or been in one, do you know how they work? Is it magic? [CHANTS GIBBERISH] Are there invisible fairies that hold the plane aloft? All right, men. We've got a busy morning and lots of flights to carry. Or is it science? Well, you can guess that the answer is indeed science. That's ridiculous. What? So to discuss how an airplane flies, we first have to talk about the forces on an airplane which push it around in all sorts of different directions. Now we're going to focus on airplanes today because they're awesome, but most of these forces apply to any other vehicle. The first force acts on all these vehicles-- really, it acts on everything. It's the weight force, which points down towards the center of Earth. Weights is equal to the mass of the airplane-- m right here-- times the acceleration due to gravity. Here on Earth, g is equal to 9.81 meters per second squared. Now that's only for Earth. The acceleration due to gravity really depends on the mass of the planet that your are on. The larger the planet, the higher the gravity. So 9.81 meters per second squared here on Earth. The moon, however-- it's smaller than Earth. So the acceleration due to gravity is only 1/6 that on Earth-- 1.6 meters per second squared. This is why astronauts can bounce high on the moon, but not on Earth. This isn't nearly as much fun. Obviously, there has to be another force opposing the weight and pushing the airplane up. This force is called lift. Lift operates perpendicular to the airplane's wings, which are right here in this side view. Now, if these are our only two forces, our aircraft will be able to go up and down, but it won't go anywhere. So we have to have a force that pushes the airplane forward, and this is called thrust. All vehicles have thrust, otherwise they wouldn't go anywhere like our airplane. Why didn't you buy a car with thrust? I'm sorry. We can at least roll down the hill. On an aircraft, this thrust is produced by engines. There are two main types of engines. We have propellers, like this little guy right here. And jet engines, like our first model. Whatever the type of engines, they all work by the same principle. Let's draw a little side view of an engine here. The engines accelerate air out the back this direction. And by Newton's third law, there's an equal and opposite reaction, and that's the thrust force pushing the aircraft forward. This is really the same thing that happens when you blow up a balloon and you let it go. The air come out the back and the balloon moves forward. We have a force that opposes the thrust. It's called drag. It points opposite the direction of flight. The major type of drag is pressure drag, which is the force caused by the air smacking into the airplane. So we try to minimize this type of drag by making the airplane as aerodynamic as possible. That means that it has smooth lines in the air flows nice and cleanly over the front here. You can feel the pressure drag when you stick your hand out the window of a moving car. Uh, honey, honey. Your hand-- your hand, please. When your hand this horizontal, it's aerodynamic and you really don't feel a lot of drag. But if you slowly turn your hand vertical, you really feel the drag increasing. So these are our four forces on the airplane, but perhaps you're thinking-- So this really cool and everything, but how do we increase and decrease the airplanes lift to move up and down? That's a great question. Let's look at the equation for the magnitude of lift per unit wing area. We'll call that L. L equals 1/2 times rho times cl times v squared. That simple. OK, OK, I'll tell you what each of these things mean. So rho-- it's not a P. It's the Greek letter, rho. Rho is the density of the air, which is a measure of the number of air molecules in a certain volume. Density of the air varies with altitude and temperature, so you go higher up. There, the air is thinner, and so the density is lower. If we want to simplify things, we generally use the standard density, which is 1.2754 kilograms per meters cubed. v here is the speed of the aircraft, or how fast it's traveling. And cl is something called the coefficient of lift. It's a number that gives us some information about the shape of the aircraft's wings-- these things right here. The coefficient of lift changes with the angle of attack. Angle of what? Aircraft can pitch up and down, and even if they're pitched up, they're still traveling in a horizontal direction like that. Now the angle formed here by the horizontal direction of travel and the direction of the aircraft's nose is called the angle of attack, and we denote that with the Greek letter alpha. So we can make a little plot here of that. We're going to put coefficient of lift up on the y-axis, and the angle of attack down on the x-axis. So as the airplane starts to pitch up-- if I can get a little hand here-- thank you. As the aircraft starts to pitch up, the coefficient of lift increases. This is a good thing because we have more lift. As we continue to increase, we eventually reach a point where we keep pitching up but the lift starts decreasing. This is something called stall, and it's not a good thing. So we generally avoid try to pitching up this much. There's a similar equation for the drag per unit wing area, D. D equals 1/2 rho. Not cl-- that wouldn't make any sense. cd, as you can guess, is the coefficient of drag times the velocity squared. The coefficient of drag is-- it's another number that tells us something about the wings, and it also varies with the angle of attack. So as the angle of attack increases-- oh, thank you-- the coefficient of drag increases as well. Thank you very much. This is because as the aircraft is pitching up, there is more wing area perpendicular to the flow. Now, this reminds me of something that we talked about earlier. Exactly. This is very similar to whenever you hold your hand out the window of a car. And so, that's pretty much everything you need know about how an aircraft flies. So the next time you're on an airplane or you just see one, you can really know exactly what it is that's keeping it up in the air. Nope. No, it's not them either. Ah, there you are. Now you got it." + }, + { + "Q": "What about something like p over 5 is equal to 230", + "A": "p/5=230 p/5*5=230*5 p=1,150", + "video_name": "a3acutLstF8", + "transcript": "Let's get some practice solving some equations, and we're gonna set up some equations that are a little bit hairier than normal, they're gonna have some decimals and fractions in them. So let's say I had the equation 1.2 times c is equal to 0.6. So what do I have to multiply times 1.2 to get 0.6? And it might not jump out immediately in your brain but lucky for us we can think about this a little bit methodically. So one thing I like to do is say okay, I have the c on the left hand side, and I'm just multiplying it by 1.2, it would be great if this just said c. If this just said c instead of 1.2c. So what can I do there? Well I could just divide by 1.2 but as we've seen multiple times, you can't just do that to the left hand side, that would change, you no longer could say that this is equal to that if you only operate on one side. So you have to divide by 1.2 on both sides. So on your left hand side, 1.2c divided by 1.2, well that's just going to be c. You're just going to be left with c, and you're going to have c is equal to 0.6 over 1.2 Now what is that equal to? There's a bunch of ways you could approach it. The way I like to do it is, well let's just, let's just get rid of the decimals. Let's just multiply the numerator and denominator by a large enough number so that the decimals go away. So what happens if we multiply the numerator and the denominator by... Let's see if we multiply them by 10, you're gonna have a 6 in the numerator and 12 in the denominator, actually let's do that. Let's multiply the numerator and denominator by 10. So once again, this is the same thing as multiplying by 10 over 10, it's not changing the value of the fraction. So 0.6 times 10 is 6, and 1.2 times 10 is 12. So it's equal to six twelfths, and if we want we can write that in a little bit of a simpler way. We could rewrite that as, divide the numerator and denominator by 6, you get 1 over 2, so this is equal to one half. And if you look back at the original equation, 1.2 times one half, you could view this as twelve tenths. Twelve tenths times one half is going to be equal to six tenths, so we can feel pretty good that c is equal to one half. Let's do another one. Let's say that we have 1 over 4 is equal to y over 12. So how do we solve for y here? So we have a y on the right hand side, and it's being divided by 12. Well the best way I can think of of getting rid of this 12 and just having a y on the right hand side is multiplying both sides by 12. We do that in yellow. So if I multiply the right hand side by 12, I have to multiply the left hand side by 12. And once again, why did I pick 12? Well I wanted to multiply by some number, that when I multiply it by y over 12 I'm just left with y. And so y times 12 divided by 12, well that's just going to be 1. And then on the left hand side you're going to have 12 times one fourth, which is twelve fourths. So you get 12 over 4, is equal to y. Or you could say y is equal to 12 over 4, y is equal to, let me do that just so you can see what I'm doing, just flopping the sides, doesn't change what's being said, y is equal to 12 over 4. Now what is twelve fourths? Well, you can view this as 12 divided by 4, which is 3, or you could view this as twelve fourths which would be literally, 3 wholes. So you could say this would be equal to 3. Y is equal to 3, and you can check that. One fourth is equal to 3 over 12, so it all works out. That's the neat thing about equations, you can always check to see if you got the right answer. Let's do another one, can't stop. 4.5 is equal to 0.5n So like always, I have my n already on the right hand side. But it's being multiplied by 0.5, it would be great if it just said n. So what can I do? Well I can divide both sides, I can divide both sides by 0.5, once again, if I do it to the right hand side I have to do it to the left hand side. And why am I dividing by 0.5? So I'm just left with an n on the right hand side. So this is going to be, so on the left hand side, I have 4.5 over 0.5, let me just, I don't want to skip too many steps. 4.5 over 0.5, is equal to n, because you have 0.5 divided by 0.5, you're just left with an n over here. So what does that equal to? Well 4.5 divided by 0.5, there's a couple ways to view this. You could view this as forty-five tenths divided by five tenths, which would tell you okay, this is going to be 9. Or if that seems a little bit confusing or a little bit daunting, you can do what we did over here. You could multiply the numerator and the denominator by the same number, so that we get rid of the decimals. And in this case, if you multiply by 10 you can move the decimal one to the right. So once again, it has to be multiplying the numerator and the denominator by the same thing. We're multiplying by 10 over 10, which is equivalent to 1, which tells us that we're not changing the value of this fraction. So let's see, this is going to be 45 over 5, is equal to n. And some of you might say wait wait wait, hold on a second, you just told us whatever we do to one side of the equation, we have to do to the other side of the equation and here you are, you're just multiplying the left hand side of this equation by 10 over 10. Now remember, what is 10 over 10? 10 over 10 is just 1. Yes, if I wanted to, I could multiply the left hand side by 10 over 10, and I could multiply the right hand side by 10 over 10, but that's not going to change the value of the right hand side. I'm not actually changing the values of the two sides. I'm just trying to rewrite the left hand side by multiplying it by 1 in kind of a creative way. But notice, n times 10 over 10, well that's still going to just be n. So I'm not violating this principle of whatever I do to the left hand side I do to the right hand side. You can always multiply one side by 1 and you can do that as many times as you want. Like the same way you can add 0 or subtract 0 from one side, without necessarily having to show you're doing it to the other side, because it doesn't change the value. But anyways, you have n is equal to 45 over 5, well what's 45 over 5? Well that's going to be 9. So we have 9 is equal to, why did I switch to green? We have 9 is equal to n, or we could say n is equal to 9. And you could check that: 4.5 is equal to 0.5 times 9, yup half of 9 is 4.5 Let's do one more, because once again I can't stop. Alright, let me get some space here, so we can keep the different problems apart that we had. So let's do, let's have a different variable now. Let's say we have g over 4 is equal to 3.2. Well I wanna get rid of this dividing by 4, so the easiest way I can think of doing that is multiplying both sides by 4. So I'm multiplying both sides by 4, and the whole reason is 4 divided by 4 gives me 1, so I'm gonna have g is equal to, what's 3.2 times 4? Let's see 3 times 4 is 12, and two tenths times 4 is eight tenths, so it's gonna be 12 and eight tenths. G is going to be 12.8, and you can verify this is right. 12.8 divided by 4 is 3.2." + }, + { + "Q": "Does the heart pump the same blood back to the body? If it does, why does it have to go get oxygen from the lungs when it already went through the process? Also, if it is the same blood, why does it turn blue when blood already went through the pulmonary circulation?", + "A": "As the blood travels through the body, oxygen is transferred to the tissues. Blood has to pass through the lungs to get more oxygen. You can think of the hemoglobin in the blood like buckets that hold oxygen - the buckets keep going around and around. When the buckets travel through the body they drop off oxygen and when they go to the lungs they pick more up.", + "video_name": "7b6LRebCgb4", + "transcript": "- [Instructor] Let's talk a little bit about arteries and veins and the roles they play in the circulatory system. So I want you to pause this video and first think to yourself, Do you have a sense of what arteries and veins are? Well one idea behind arteries and veins are that well, in most of these drawings, arteries are drawn in red, and I even made the artery word here in red. And veins are drawn in blue. And so maybe that represents how much oxygen they have. And so one possible explanation is that arteries carry oxygenated blood, oxygenated, oxygenated blood, while veins carry deoxygenated blood. So blood that has less oxygen now. Now this is actually incorrect. It is, many times, the case that arteries are the ones carrying oxygenated blood and veins are carrying the deoxygenated blood. But as we will see, this is not always the case. And since we're already talking about oxygenated blood and deoxygenated blood and the colors red and blue, it's worth addressing another misconception. Many times it is said that deoxygenated blood looks blue, and the reason why people believe that is if you look at your wrist and you're able to see some of the vessels in there, you will see some blue vessels. And those, or at least they look blue when you're looking from the outside of your skin. And those, indeed, are veins. And so that's where the misconception has come from, that veins, which, in your arm, are carrying deoxygenated blood. That that deoxygenated blood is blue. It turns out that it is not blue. It is just a deeper red. And the reason why the veins look blue is because of the optics of light going through your skin and then seeing the outside of the veins and then reflecting back. That is not the color of the actual blood. So so far I have not given you a clear definition of what arteries versus veins are. A better definition, so let me cross these two out, are that arteries carry blood away from the heart. Away from the heart. And veins carry blood towards the heart. Towards the heart. And I can get a zoomed in image of the heart right here and that will make it a little bit clearer. And you can also see, or we're about to see, why this first definition, or this first distinction between arteries and veins does not always hold. So let's just imagine some blood that is being pumped away from the heart. So right when it gets pumped away from the heart, it'll be right over here. It gets pumped through the aorta, and you can see the aorta branches, so some blood can go up towards your head, and if it didn't, you would pass out and die. And then a lot of the blood goes down towards the rest of your body. And that, indeed, is the most oxygenated blood. And so it'll flow through your body. And these arteries will keep branching and branching into smaller vessels, all the way until they form these very small branches. And it's that place, especially, where they will lose a lot of their oxygen to the fluid and the cells around them. And then the blood is less oxygenated. And then even though deoxygenated blood is not blue, it often gets depicted as blue in a lot of diagrams. So I will do the same. And these vessels start building into your veins. And these really small vessels that really bridge between arteries and veins, where a lot of the gas and nutrient exchange occurs, these are called capillaries. And so after going through the capillaries, the blood will then come back to the heart and now it's coming towards the heart through the veins. It comes into the right atrium, then the right ventricle. Then that gets pumped towards the lungs. And this is the exception to the first incorrect definition of arteries and veins that we looked at. This right over here, is an artery. Even though it's carrying less oxygenated or deoxygenated blood, it's an artery because it's carrying blood away from the heart. But in this case, it's not carrying it to the rest of the body, it is carrying it to the lungs. That is why it is called the pulmonary artery, even though it's carrying less oxygenated blood. So that it goes to the lungs and then, in the lungs, there's more gas exchange that occurs. The blood gets oxygenated and then it comes back to the heart. And so it comes back to the heart in these vessels right over here, and that even though these are carrying highly oxygenated blood, these are considered veins because they're carrying blood towards the heart. So these are pulmonary veins. And then the cycle starts again. The pulmonary veins bring the oxygenated blood into the left atrium and the left ventricle, and then that pumps it to the rest of the body to the aorta, for your systemic circulation. You have your pulmonary circulation, which circulates the blood to, through and from the lungs. And you have your systemic circulation, which takes the blood to and from the rest of the body. So now that we have this main distinction between arteries and veins, what are some other interesting things that we know about it? Well one thing to keep in mind is that since arteries are being pumped directly by the heart towards the rest of the body, they have high pressure. I'll write that in caps. High pressure. And so if you were to have an accident of some type, which you do not want to have, and you were to accidentally cut an artery, because of that high pressure, it would actually spurt blood, a lot more than if you were to cut a vein. And most of the times where you get a cut, you're really just cutting capillaries. Like if you were to prick your finger, it's usually a series of capillaries that get cut, and that's why the blood would come out very very slowly. Now if arteries are high pressure, veins are low pressure. Low, low pressure. And one way to think about it is the arteries, the blood is being pumped directly by the heart. But then once it goes through the capillaries and comes back through the veins, it's kind of sluggishly making its way back to the heart. It's not being directly pumped. And that's why in veins, because you don't have that high pressure to bring everything back to the heart, you have these valves that make sure that for the most part, the blood is going in one direction. I'm going to draw the blood in red in the veins, just so we don't keep going with that misconception, that blood in the veins is blue somehow. Now related to the fact that the blood in the arteries is under higher pressure, in order to transport a fixed volume of blood in a certain amount of time, you need less volume. And so that's why arteries are low volume. And on the other hand, veins are high volume. And to appreciate the difference, the blood volume in arteries are only approximately 15% of the entire blood volume in your body, while the blood volume in veins are closer to 65%, approximately 65%. And if you're wondering where the rest of the blood is, about five percent is in capillaries, five percent is in your heart, and about 10% is in your lungs. So I will leave you there. The big take away: arteries are the vessels that take blood away from the heart. Veins are the vessels that take blood towards the heart." + }, + { + "Q": "How will this help me later in life?", + "A": "Whichever job you are in, you are sure to find your self using math to solve stuff. So almost everything in math will help you later in life.", + "video_name": "Tm98lnrlbMA", + "transcript": "I'm here with Jesse Ro, whose a math teacher at Summit San Jose and a Khan Academy teaching fellow and you had some interesting ideas or questions. Yeah, one question that students ask a lot when they start Algebra is why do we need letters, why can't we just use numbers for everything? Why letters? So why do we have all these Xs and Ys and Zs and ABCs when we start dealing with Algebra? Yeah, exactly. That's interesting, well why don't we let people think about that for a second. So Sal, how would you answer this question? Why do we need letters in Algebra? So why letters. So there are a couple of ways I'd think about it. One is if you have an unknown. So if I were to write X plus three is equal to ten the reason why we're doing this is that we don't know what X is It's literally an unknown. And so we're going to solve for it in some way. But it did not have to be the letter X. We could have literally written blank plus three is equal to ten. Or we could have written Question Mark plus three is equal to ten. So it didn't have to be letters, but we needed some type of symbol. It literally could've been Smiley Face plus three is equal to ten. But until you know it, you need some type of a symbol to represent whatever that number is. Now we can go and solve this equation and then know what that symbol represents. But if we knew it ahead of time, it wouldn't be an unknown. It wouldn't be something that we didn't know. So that's one reason why I would use letters and where just numbers by itself wouldn't be helpful. The other is when you're describing relationships between numbers. So I could do something like - I could say - that whenever you give me a three, I'm going to give you a four. And I could say, if you give me a five, I'm going to give you a six. And i could keep going on and on forever. If you give me a 7.1, I'm going to give you an 8.1. And I could keep listing this on and on forever. Maybe you could give me any number, and I could tell you what I'm going to give you. But I would obviously run out of space and time if I were to list all of them. And we could do that much more elegantly if we used letters to describe the relationship. Maybe what you give me we call X, and what I give you we call Y. And so I say, look, whatever you give me, I'm going to add one to it. And that's what I'm going to give back to you. And so now, this very simple equation here can describe an infinite number of relationships between X or an infinite number of corresponding Ys and Xs. So now someone knows whatever X you give me you give me three, I add one to it, and I'm going to give you four. You give me 7.1, I'm going to add one to it and give you 8.1. So there is no more elegant way that you could've done it than by using symbols. With that said, I didn't have to use Xs and Ys. This is just a convention that kind of comes to use from history. I could've defined what you give me as Star and what I give you as Smiley Face and this also would've been a valid way to express this. So the letters are really just symbols. Nothing more." + }, + { + "Q": "And wouldn't it be safer for you to both deny? then you would both keep the same sentences. You are taking the risk of taking on another year if you both confess.", + "A": "But they cannot talk together... remember that.", + "video_name": "UkXI-zPcDIM", + "transcript": "On the same day, police have made two at first unrelated arrests. They arrest a gentleman named Al. And they caught him red handed selling drugs. So it's an open and shut case. And the same day, they catch a gentleman named Bill. And he is also caught red handed, stealing drugs. And they bring them separately to the police station. And they tell them, look, this is an open and shut case. You're going to get convicted for drug dealing and you're going to get two years. And they tell this to each of them individually. They were selling the same type of drugs, just happened to be that. But they were doing it completely independently. Two years for drugs is what's going to happen assuming nothing else. But then the district attorney has a chance to chat with each of these gentleman separately and while he's chatting with them he reinforces the idea this is an open and shut case for the drug dealing. They're each going to get two years, if nothing else happens. But then he starts to realize that these two characters look like-- he starts to have a suspicion, for whatever reason, that these were the two characters that actually committed a much more serious offense. That they had committed a major armed robbery a few weeks ago. And all the district attorney has to go on is his hunch, his suspicion. He has no hard evidence. So what he wants to do is try to get a deal with each of these guys so that they have an incentive to essentially snitch on each other. So what he tells each of them is, look, you're going to get two years for drug dealing. That's kind of guaranteed. But he says, look, if you confess and the other doesn't then you will get 1 year. And the other guy will get 10 years. So he's telling Al, look, we caught Bill, too, just If you confess that it was you and Bill who performed that armed robbery your term is actually going to go down from two years to one year. But Bill is obviously going to have to spend a lot more time in jail. Especially because he is not cooperating with us. He is not confessing. But then, the other statement is also true. If you deny and the other confesses now it switches around. You will get 10 years, because you're not cooperating. And the other, your co-conspirator, will get a reduced sentence-- will get the one year. So this is like telling Al, look, if you deny that you were the armed robber and Bill snitches you out, then you're going to get 10 years in prison. And Bill's only going to get one year in prison. And if both of you essentially confess, you will both get three years. So this scenario is called the prisoner's dilemma. Because we'll see in a second there is a globally optimal scenario for them where they both deny and they both get two years. But we'll see, based on their incentives, assuming they don't have any unusual loyalty to each other-- and these are hardened criminals here. They're not brothers or related to each other in any way. They don't have any kind of loyalty pact. We'll see that they will rationally pick, or they might rationally pick, a non-optimal scenario. And to understand that I'm going to draw something called a payoff matrix. So let me do it right here for Bill. So Bill has two options. He can confess to the armed robbery or he can deny that he had anything-- that he knows anything about the armed robbery. And Al has the same two options. Al can confess and Al can deny. And since it's called a payoff matrix, let me draw some grids here. Let me draw some grids and let's think about all of the different scenarios and what the payoffs would be. If Al confesses and Bill confesses then we're in scenario four. They both get three years in jail. So they both will get three for Al and three for Bill. Now, if Al confesses and Bill denies, then we are in scenario two from Al's point of view. Al is only going to get one year. But Bill is going to get 10 years. Now, if the opposite thing happens, if Bill confesses and Al denies, then it goes the other way around. Al's going to get 10 years for not cooperating. And Bill's going to have a reduced sentence of one year for cooperating. And then if they both deny, they're in scenario one, where they're both just going to get their time for the drug dealing. So Al will get two years, and Bill will get two years. Now, I alluded to this earlier in the video. What is the globally optimal scenario for them? Well, it's this scenario, where they both deny having anything to do with the armed robbery. Then they both get two years. But what we'll see is actually somewhat rational, assuming that they don't have any strong loyalties to each other, or strong level of trust with the other party, to not go there. And it's actually rational for both of them to confess. And the confession is actually a Nash equilibrium. And we'll talk more about this, but a Nash equilibrium is where each party has picked a choice given the choices of the other party. So when we think of, or each party has to pick the optimal choice, given whatever choice the other party picks. And so from Al's point of view, he says, well, look, I don't know whether Bill is confessing or denying. So let's say he confesses. What's better for me to do? If he confesses, and I confess, then I get three years. If he confesses and I deny I get 10 years. So if he confesses, it's better for me to confess as well. So this is a preferable scenario to this one down here. Now, I don't know that Bill confessed. He might deny. If I assume Bill denied, is it better for me to confess and get one year or deny and get two years? Well, once again, it's better for me to confess. And so regardless of whether Bill confesses or denies, so this once again, the optimal choice for Al to pick, taking into account Bill's choices, is to confess. If Bill confesses, Al is better off confessing. And if Bill denies, Al is better off confessing. Now, we look at it from Bill's point of view. And it's completely symmetric. If Bill says, well, I don't know if Al is confessing or denying. If Al confesses, I can confess and get three years or I can deny and get 10 years. Well, three years in prison is better than 10. So I will go-- I would go for the three years if I know Al is confessing. But I don't know that Al is definitely confessing. He might deny. If Al is denying, I could confess and get one year or I could deny and get two years. Well, once again, I would want to confess and get the one year. So Bill, taking into account each of the scenarios that Al might take, it's always better for him to confess. And so this is interesting. They are rationally deducing that they should get to this scenario, this Nash equilibrium state, as opposed to this globally optimal state. They're both getting three years by both confessing as opposed to both of them getting two years by both denying. The problem with this one is this is an unstable state. If one of them assumes that the other one has-- if one of them assumes that they're somehow in that state temporarily, they say, well, I can always improve my scenario by changing what I want to do. If Al thought that Bill was definitely denying, Al could improve his circumstance by moving out of that state and confessing and only getting one here. Likewise, if Bill thought that maybe Al is likely to deny, he realizes that he can optimize by moving in this direction. Instead of denying, getting, two and two, he could move in that direction right over there. So this is an unstable optimal scenario. But this Nash equilibrium, this state right over here, is actually very, very, very stable. If they assume, it's better for each of them to confess regardless of what the other ones does. And assuming all of the other actors have chosen their strategy, there's no incentive for Bill. So if assuming everyone else has changed their strategy, you can only move in that direction. If you're Bill, you can go from the Nash equilibrium of confessing to denying, but you're worse off. So you won't want to do that. Or you could move in this direction, which would be Al changing his decision. But once again, that gives a worse outcome for Al. You're going from three years to 10 years. So this is the equilibrium state, the stable state, that both people will pick something that is not optimal globally." + }, + { + "Q": "I am confused about the way you write electron configuration. ex. 1s^2 2s^2 and so on. I'm pretty sure the \"1s\" means first shell and \"2s\" means second shell. (correct me if I'm wrong) but what is the exponent? (^2) what does that represent?", + "A": "The number in front tells you the shell The letter tells you what type of orbital it is The superscript number tells you how many electrons are in that orbital 1s^2 means 2 electrons are in the 1s orbital 2s^2 means 2 electrons are in the 2s orbital Note the s doesn t stand for shell", + "video_name": "FmQoSenbtnU", + "transcript": "In the last few videos we learned that the configuration of electrons in an atom aren't in a simple, classical, Newtonian orbit configuration. And that's the Bohr model of the electron. And I'll keep reviewing it, just because I think it's an important point. If that's the nucleus, remember, it's just a tiny, tiny, tiny dot if you think about the entire volume of the actual atom. And instead of the electron being in orbits around it, which would be how a planet orbits the sun. Instead of being in orbits around it, it's described by orbitals, which are these probability density functions. So an orbital-- let's say that's the nucleus it would describe, if you took any point in space around the nucleus, the probability of finding the electron. So actually, in any volume of space around the nucleus, it would tell you the probability of finding the electron within that volume. And so if you were to just take a bunch of snapshots of electrons -- let's say in the 1s orbital. And that's what the 1s orbital looks like. You can barely see it there, but it's a sphere around the nucleus, and that's the lowest energy state that an electron can be in. If you were to just take a number of snapshots of electrons. Let's say you were to take a number of snapshots of helium, which has two electrons. Both of them are in the 1s orbital. It would look like this. If you took one snapshot, maybe it'll be there, the next snapshot, maybe the electron is there. Then the electron is there. Then the electron is there. Then it's there. And if you kept doing the snapshots, you would have a bunch of them really close. And then it gets a little bit sparser as you get out, as you get further and further out away from the electron. But as you see, you're much more likely to find the electron close to the center of the atom than further out. Although you might have had an observation with the electron sitting all the way out there, or sitting over here. So it really could have been anywhere, but if you take multiple observations, you'll see what that probability function is describing. It's saying look, there's a much lower probability of finding the electron out in this little cube of volume space than it is in this little cube of volume space. And when you see these diagrams that draw this orbital like this. Let's say they draw it like a shell, like a sphere. And I'll try to make it look three-dimensional. So let's say this is the outside of it, and the nucleus is sitting some place on the inside. They're just saying -- they just draw a cut-off -- where can I find the electron 90% of the time? So they're saying, OK, I can find the electron 90% of the time within this circle, if I were to do the cross-section. But every now and then the electron can show up outside of that, right? Because it's all probabilistic. So this can still happen. You can still find the electron if this is the orbital we're talking about out here. Right? And then we, in the last video, we said, OK, the electrons fill up the orbitals from lowest energy state to high energy state. You could imagine it. If I'm playing Tetris-- well I don't know if Tetris is the thing-- but if I'm stacking cubes, I lay out cubes from low energy, if this is the floor, I put the first cube at the lowest energy state. And let's say I could put the second cube But I only have this much space to work with. So I have to put the third cube at the next highest energy state. In this case our energy would be described as potential energy, right? This is just a classical, Newtonian physics example. But that's the same idea with electrons. Once I have two electrons in this 1s orbital -- so let's say the electron configuration of helium is 1s2 -- the third electron I can't put there anymore, because there's only room for two electrons. The way I think about it is these two electrons are now going to repel the third one I want to add. So then I have to go to the 2s orbital. And now if I were to plot the 2s orbital on top of this one, it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. Then you have the blue area, then the red, and the blue. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. But if you look at these, there's three ways that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon. So the electron configuration for carbon, the first two electrons go into, so, 1s1, 1s2. So then it fills-- sorry, you can't see everything. So it fills the 1s2, so carbon's configuration. It fills 1s1 then 1s2. And this is just the configuration for helium. And then it goes to the second shell, which is the second period, right? That's why it's called the periodic table. We'll talk about periods and groups in the future. And then you go here. So this is filling the 2s. We're in the second period right here. That's the second period. One, two. Have to go off, so you can see everything. So it fills these two. So 2s2. And then it starts filling up the p orbitals. So then it starts filling 1p and then 2p. And we're still on the second shell, so 2s2, 2p2. So the question is what would this look like if we just wanted to visualize this orbital right here, the p orbitals? So we have two electrons. So one electron is going to be in a-- Let's say if this is, I'll try to draw some axes. That's too thin. So if I draw a three-dimensional volume kind of axes. If I were to make a bunch of observations of, say, one of the electrons in the p orbitals, let's say in the pz dimension, sometimes it might be here, sometimes it might be there, sometimes it might be there. And then if you keep taking a bunch of observations, you're going to have something that looks like this bell shape, this barbell shape right there. And then for the other electron that's maybe in the x direction, you make a bunch of observations. Let me do it in a different, in a noticeably different, color. It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. The p orbitals. Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. And then you start filling the second energy shell. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons in the lowest energy state, will be 1s2. So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then we have 3s2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus, there's more space in between the lower energy orbitals to fill in more of these bizarro-shaped orbitals. But these are kind of the balance -- I will talk about standing waves in the future -- but these are kind of a balance between trying to get close to the nucleus and the proton and those positive charges, because the electron charges are attracted to them, while at the same time avoiding the other electron charges, or at least their mass distribution functions. Anyway, see you in the next video." + }, + { + "Q": "I am having trouble converting 2x=12+2y into slope intercept form", + "A": "All we have to do is isolate the variable y : 2x = 12 + 2y Remember that slope-intercept form is: y = mx + b Let s flip our equation to make it easier to work with: 12 + 2y = 2x Subtract 12 from both sides: 2y = 2x - 12 Now divide both sides by 2: y = (2x - 12)/2 y = x - 6 Comment if you have questions.", + "video_name": "V6Xynlqc_tc", + "transcript": "We're asked to convert these linear equations into slope-intercept form and then graph them on a single coordinate plane. We have our coordinate plane over here. And just as a bit of a review, slope-intercept form is a form y is equal to mx plus b, where m is the slope and b is the intercept. That's why it's called slope-intercept form. So we just have to algebraically manipulate these equations into this form. So let's start with line A, so start with a line A. So line A, it's in standard form right now, it's 4x plus 2y is equal to negative 8. The first thing I'd like to do is get rid of this 4x from the left-hand side, and the best way to do that is to subtract 4x from both sides of this equation. So let me subtract 4x from both sides. The left hand side of the equation, these two 4x's cancel out, and I'm just left with 2y is equal to. And on the right-hand side I have negative 4x minus is 8, or negative 8 minus 4, however you want to do it. Now we're almost at slope-intercept form. We just have to get rid of this 2, and the best way to do that that I can think of is divide both sides of this equation by 2. So let's divide both sides by 2. So we divide the left-hand side by 2 and then divide the right-hand side by 2. You have to divide every term by 2. And then we are left with y is equal to negative 4 divided by 2 is negative 2x. Negative 8 divided by 2 is negative 4, negative 2x minus 4. So this is line A, let me graph it right now. So line A, its y-intercept is negative 4. So the point 0, negative 4 on this graph. If x is equal to 0, y is going to be equal to negative 4, you can just substitute that in the graph. So 0, 1, 2, 3, 4. That's the point 0, negative 4. That's the y-intercept for line A. And then the slope is negative 2x. So that means that if I change x by positive 1 that y goes down by negative 2. So let's do that. So if I go over one in the positive direction, I have to go down 2, that's what a negative slope's going to do, negative 2 slope. If I go over 2, I'm going to have to go down 4. If I go back negative 1, so if I go in the x direction negative 1, that means in the y direction I go positive two, because two divided by negative one is still negative two, so I go over here. If I go back 2, I'm going to go up 4. Let me just do that. Back 2 and then up 4. So this line is going to look like this. Do my best to draw it, that's a decent job. That is line A right there. All right, let's do line B. So line B, they say 4x is equal to negative 8, and you might be saying hey, how do I get that into slope-intercept form, I don't see a y. And the answer is you won't be able to because you this can't be put into slope-intercept form, but we can simplify it. So let's divide both sides of this equation by 4. So you divide both sides of this equation by 4. And you get x is equal to negative 2. So this just means, I don't care what your y is, x is just always going to be equal to negative 2. So x is equal to negative 2 is right there, negative 1, negative 2, and x is just always going to be equal to negative 2 in both directions. And this is the x-axis, that's the y-axis, I forgot to label them. Now let's do this last character, 2y is equal to negative eight. So line C, we have 2y is equal to negative 8. We can divide both sides of this equation by 2, and we get y is equal to negative 4. So you might say hey, Sal, that doesn't look like this form, slope-intercept form, but it is. It's just that the slope is 0. We can rewrite this as y is equal to 0x minus 4, where the y-intercept is negative 4 and the slope is 0. So if you move an arbitrary amount in the x direction, the y is not going to change, it's just going to stay at negative 4. Let me do a little bit neater. y is just going to stay at negative 4. Or you can just interpret it as y is equal to negative 4 no matter what x is. So then we are done." + }, + { + "Q": "The H2CO3 is carbonic acid, right?", + "A": "Correct, H2CO3 is carbonic acid.", + "video_name": "OCD4Dr3kmmA", + "transcript": "Let me do a little experiment. Let's say I have oxygen here, and we know that oxygen is about 21% of the atmosphere. And I decide to take a cup, let's say a cup like this-- simple cup of water. And I leave it out on the counter. And it's about room temperature, about 25 degrees Celsius here. And I want to know, how much oxygen is really going to enter that cup at that surface layer? So let's say I want to measure the concentration of oxygen in that surface layer of water. Well, you know, I say 21% so, of course, there's some molecules of oxygen here. And it's only 21%, it's not like it's the majority. So I've got to draw some other molecules. This could be nitrogen or some other molecule, let's say. But I'm focused on the blue dots, because the blue dots are the oxygen dots. And so over time I let this kind of sit out. And maybe I come back and check, and a little bit of oxygen has entered my surface layer of water. In fact, if I measured it, I could say, well, the concentration, C, at that level is 0.27 millimoles per liter. And this number is literally just something that I would have to measure, right? I would actually measure the concentration there, and that's the measure of oxygen. So I've learned about Henry's law, and I can think well, you know, I know the partial pressure now. And I can rearrange the formula so that it looks something like this. I can say, well Henry's law basically is like that. So if I know the pressure and I know the concentration, I should be able to figure out the constant for myself. I can figure it out and kind of give it in units that I like. So I'm going to write the units of the KH down here. I can say, well, 769 liters times atmospheres over moles. And that's something that I've just calculated. I've just taken two numbers and I've divided them by each other. So this is my calculation for oxygen. And so far, so good, right? But now I decide to challenge myself and say, let's do this again. But instead of with oxygen, I'm going to create an environment that's 21% carbon dioxide, which is way more carbon dioxide than we actually have. But imagine I could actually do that. I actually find a way to crank up the carbon dioxide, and I do the exact same thing. I take a cup of water and I keep it out at room temperature, 25 degrees Celsius. And I say, OK, let's see how much carbon dioxide goes into my cup. I've got my carbon dioxide out here. And over time more and more molecules kind of settle in here. And, of, course the atmosphere is not going to run out of carbon dioxide molecules. They're just going to keep replacing them. But they keep settling into this top layer, this surface layer, of my water. So it's actually looking already really different than what was happening on the other side. We only had a little bit of oxygen but now I've got tons of carbon dioxide. And I don't want to make it uneven. I mentioned before, we have nitrogen-- so let me still draw a bunch of nitrogen-- that will outnumber the carbon dioxide dramatically. Because we have here about, let's say, 79% nitrogen and we only had 21% carbon dioxide. So it'll look something like that. But there's lots and lots of carbon dioxide there. In fact, if I was to calculate the concentration on this side, the concentration would be pretty high. It would be 7.24 millimoles per liter. Again, these numbers, I'm assuming that I'm doing the experiment. This is the number I would find if I actually did the experiment. So it's a much bigger number than I had over here. On the oxygen side, the number was actually pretty small, not very impressive. And yet on the carbon oxide side, much, much higher. Now that's kind of funny. It might strike you as kind of a funny thing. Because look, these partial pressures are basically the same. I mean, not even basically, they're exactly the same. There's no difference in the partial pressure. And yet the concentrations are different. So if you keep the P the same, the only way to make for different concentrations is if you have a different constant. So let me actually move on and figure out what the constant is. So what do you think the constant on this side would be, higher or lower? Let's see if we can figure it out together. The K sub H on this side is going to be lower. It's going to be lower. It's 29 liters times atmosphere divided by moles. So it's a much lower number. And I don't want you to get so distracted by this bit. This is kind of irrelevant to what we're talking about. It's just the units, and we can change the units to whatever we want. But it's this part-- it's the fact that the number itself on the carbon dioxide side is lower. Now let's think back to this idea of Henry's law. Henry's law told us that the partial pressure, this number, tells you about what's going to be going into the water, and that the K sub H tells you about what's going out of the water. And so if what's going in on both sides is equivalent, then really the difference is going to be what's the leaving. And on this side, on the first side of our experiment, we had lots of oxygens leaving this water. They didn't like being in water. They were leaving readily. And so you didn't see that, but they were actually constantly leaving. And on the carbon dioxide side, you had maybe a little bit of leaving, but not very much. The carbon dioxide was actually very comfortable with the water. In fact, to see that as a chemical formula, you might recall this. Remember, there's this formula where CO2 binds with water and it forms H2CO3. Well, think about that. If it's a binding to the water then it's not going to want to leave. It's pretty comfortable being in the water. And so the moment that carbon dioxide goes into water, it does something like this. It binds to the water. It turns into bicarbonate and protons. And so it's a very comfortable being in water, and that's why it's not leaving. In fact, I can take this one step further and even compare I could say, well, 769 divided by 29 equals about 26. So that's another way of saying that carbon dioxide is 26 times more soluble than oxygen. I'll put that in parentheses-- than oxygen. And I should make sure I make it very clear. This is at 25 degrees Celsius, and this is in water. Now, you might say, well, that's fine for 25 degrees Celsius. But what about body temperature? What's happening in our actual body? What's happening in our lungs? So in our lungs, we have 37 degrees Celsius. And instead of water-- actually, I shouldn't be writing water-- instead of water, it's blood, which is slightly different than water. The consistency is different. And so these K sub H values are actually temperature dependent. And they're going to change as you increase the temperature. So at this new temperature, it turns out that carbon dioxide is about 22 times more soluble than oxygen. So it's still pretty impressive. Sometimes you might even see 24 times, depending on what numbers you read. But this is an impressive difference. And actually, what I wanted to get to is the fact that it goes back to the idea of what's going in and what's coming out. And the net difference is why you end up with a huge difference in concentrations between carbon dioxide and oxygen." + }, + { + "Q": "What is Hemophelia?", + "A": "Hemophilia is an inherited blood disorder in which the blood s clotting ability is impaired. Therefore there is no mechanism in the body to stop the bleeding if a hemophiliac get a cut or even bruise (bleeding under the skin). The primary gene that codes for hemophilia is x-linked, or carried on the x-chromosome so it affects males more often than females as they don t have another x-chromosome to mask the expression of the gene.", + "video_name": "-ROhfKyxgCo", + "transcript": "By this point in the biology playlist, you're probably wondering a very natural question, how is gender determined in an organism? And it's not an obvious answer,because throughout the animal kingdom, it's actually determined in different ways. In some creatures,especially some types of reptiles,it's environmental Not all reptiles,but certain cases of it. It could be maybe the temperature in which the embryo develops will dictate whether it turns into a male or female or other environmental factors. And in other types of animals,especially mammals, of which we are one example,it's a genetic basis. And so your next question is,hey,Sal,so-- let me write this down,in mammals it's genetic-- so,OK,maybe they're different alleles,a male or a female allele. But then you're like,hey,but there's so many different characteristics that differentiate a man from a woman. Maybe it would have to be a whole set of genes that have to work together. And to some degree,your second answer would be more correct. It's even more than just a set of genes. It's actually whole chromosomes determine it. So let me draw a nucleus.That's going to be my nucleus. And this is going to be the nucleus for a man. So 22 of the pairs of chromosomes are just regular non-sex-determining chromosomes. So I could just do,that's one of the homologous,2,4,6,8,10,12,14. I can just keep going.And eventually you have 22 pairs. So these 22 pairs right there,they're called autosomal. And those are just our standard pairs of chromosomes that code for different things. Each of these right here is a homologous pair,homologous, which we learned before you get one from each of your parents. They don't necessarily code for the same thing, for the same versions of the genes,but they code for the same genes. If eye color is on this gene,it's also on that gene, on the other gene of the homologous pair. Although you might have different versions of eye color on either one and that determines what you display. But these are just kind of the standard genes that have nothing to do with our gender. And then you have these two other special chromosomes. I'll do this one. It'll be a long brown one,and then I'll do a short blue one. And the first thing you'll notice is that they don't look homologous. How could they code for the same thing when the blue one is short and the brown one's long? And that's true.They aren't homologous. And these we'll call our sex-determining chromosomes. And the long one right here, it's been the convention to call that the x chromosome. Let me scroll down a little bit. And the blue one right there,we refer to that as the y chromosome. And to figure out whether something is a male or a female, it's a pretty simple system. If you've got a y chromosome,you are a male. So let me write that down. So this nucleus that I drew just here-- obviously you could have the whole broader cell all around here-- this is the nucleus for a man. So if you have an x chromosome-- and we'll talk about in a second why you can only get that from your mom-- an x chromosome from your mom and a y chromosome from your dad, you will be a male. If you get an x chromosome from your mom and an x chromosome from your dad,you're going to be a female. And so we could actually even draw a Punnett square. This is almost a trivially easy Punnett square, but it kind of shows what all of the different possibilities are. So let's say this is your mom's genotype for her sex-determining chromosome. She's got two x's. That's what makes her your mom and not your dad. And then your dad has an x and a y-- I should do it in capital--and has a Y chromosome. And we can do a Punnett square. What are all the different combinations of offspring? Well,your mom could give this X chromosome, in conjunction with this X chromosome from your dad. This would produce a female. Your mom could give this other X chromosome with that X chromosome. That would be a female as well. Well,your mom's always going to be donating an X chromosome. And then your dad is going to donate either the X or the Y. So in this case,it'll be the Y chromosome. So these would be female,and those would be male. And it works out nicely that half are female and half are male. But a very interesting and somewhat ironic fact might pop out at you when you see this. what determines whether someone is or Who determines whether their offspring are male or female? Is it the mom or the dad? Well,the mom always donates an X chromosome, so in no way does- what the haploid genetic makeup of the mom's eggs of the gamete from the female, in no way does that determine the gender of the offspring. It's all determined by whether--let me just draw a bunch of-- dad's got a lot of sperm,and they're all racing towards the egg. And some of them have an X chromosome in them and some of them have a Y chromosome in them. And obviously they have others. And obviously if this guy up here wins the race. Or maybe I should say this girl.If she wins the race, then the fertilized egg will develop into a female. If this sperm wins the race, then the fertilized egg will develop into a male. And the reason why I said it's ironic is throughout history, and probably the most famous example of this is Henry the VIII. I mean it's not just the case with kings. It's probably true,because most of our civilization is male dominated, that you've had these men who are obsessed with producing a male heir to kind of take over the family name. And,in the case of Henry the VIII,take over a country. And they become very disappointed and they tend to blame their wives when the wives keep producing females,but it's all their fault. Henry the VIII,I mean the most famous case was with Ann Boleyn. I'm not an expert here,but the general notion is that he became upset with her that she wasn't producing a male heir. And then he found a reason to get her essentially decapitated, even though it was all his fault. He was maybe producing a lot more sperm that looked like that than was looking like this. He eventually does produce a male heir so he was-- and if we assume that it was his child-- then obviously he was producing some of these,but for the most part, it was all Henry the VIII's fault. So that's why I say there's a little bit of irony here. Is that the people doing the blame are the people to blame for the lack of a male heir? Now one question that might immediately pop up in your head is, Sal,is everything on these chromosomes related to just our sex-determining traits or are there other stuff on them? So let me draw some chromosomes. So let's say that's an X chromosome and this is a Y chromosome. Now the X chromosome,it does code for a lot more things, although it is kind of famously gene poor. It codes for on the order of 1,500 genes. And the Y chromosome,it's the most gene poor of all the chromosomes. It only codes for on the order of 78 genes. I just looked this up,but who knows if it's exactly 78. But what it tells you is it does very little other than determining what the gender is. And the way it determines that, it does have one gene on it called the SRY gene. You don't have to know that. SRY,that plays a role in the development of testes or the male sexual organ.So if you have this around, this gene right here can start coding for things that will eventually lead to the development of the testicles. And if you don't have that around,that won't happen, so you'll end up with a female. And I'm making gross oversimplifications here. But everything I've dealt with so far,OK, this clearly plays a role in determining sex. But you do have other traits on these genes. And the famous cases all deal with specific disorders. So,for example,color blindness. The genes,or the mutations I should say. So the mutations that cause color blindness. Red-green color blindness,which I did in green, which is maybe a little bit inappropriate. Color blindness and also hemophilia. This is an inability of your blood to clot. Actually, there's several types of hemophilia. But hemophilia is an inability for your blood to clot properly. And both of these are mutations on the X chromosome. And they're recessive mutations.So what does that mean? It means both of your X chromosomes have to have-- let's take the case for hemophilia-- both of your X chromosomes have to have the hemophilia mutation in order for you to show the phenotype of having hemophilia. So,for example,if there's a woman, and let's say this is her genotype. She has one regular X chromosome and then she has one X chromosome that has the-- I'll put a little superscript there for hemophilia-- she has the hemophilia mutation.She's just going to be a carrier. Her phenotype right here is going to be no hemophilia. She'll have no problem clotting her blood. The only way that a woman could be a hemophiliac is if she gets two versions of this, because this is a recessive mutation. Now this individual will have hemophilia. Now men,they only have one X chromosome. So for a man to exhibit hemophilia to have this phenotype, he just needs it only on the one X chromosome he has. And then the other one's a Y chromosome. So this man will have hemophilia. So a natural question should be arising is, hey, you know this guy-- let's just say that this is a relatively infrequent mutation that arises on an X chromosome-- the question is who's more likely to have hemophilia? A male or a female? All else equal,who's more likely to have it? Well if this is a relatively infrequent allele,a female, in order to display it,has to get two versions of it. So let's say that the frequency of it-- and I looked it up before this video-- roughly they say between 1 in 5,000 to 10,000 men exhibit hemophilia. So let's say that the allele frequency of this is 1 in 7,000, the frequency of Xh,the hemophilia version of the X chromosome. And that's why 1 in 7,000 men display it, because it's completely determined whether-- there's a 1 in 7,000 chance that this X chromosome they get is the hemophilia version. Who cares what the Y chromosome they get is, cause that essentially doesn't code at all for the blood clotting factors and all of the things that drive hemophilia. Now,for a woman to get hemophilia,what has to happen? She has to have two X chromosomes with the mutation. Well the probability of each of them having the mutation is 1 in 7,000 So the probability of her having hemophilia is 1 in 7,000 times 1 in 7,000,or that's 1 in what,49 million. So as you can imagine,the incidence of hemophilia in women is much lower than the incidence of hemophilia in men. And in general for any sex-linked trait, if it's recessive, if it's a recessive sex-linked trait,which means men, if they have it,they're going to show it, because they don't have another X chromosome to dominate it. Or for women to show it, she has to have both versions of it. The incidence in men is going to be, so let's say that m is the incidence in men.I'm spelling badly. Then the incidence in women will be what? You could view this as the allele frequency of that mutation on the X chromosome. So women have to get two versions of it. So the woman's frequency is m squared. And you might say,hey,that looks like a bigger number. I'm squaring it. But you have to remember that these numbers, the frequency is less than 1, so in the case of hemophilia,that was 1 in 7,000. So if you square 1 in 7,000,you get 1 in 49 million. Anyway,hopefully you found that interesting and now you know how we all become men and women. And even better you know whom to blame when some of these, I guess, male-focused parents are having trouble getting their son." + }, + { + "Q": "I don't mean to be inappropriate, but was Ernst Rohm a homosexual?", + "A": "Yes, he was gay", + "video_name": "ZrbbKMnPDUk", + "transcript": "We have now seen the Nazis come to power in 1933 - and by mid 1933 they are the only allowed party in Germany. And Hitler is fundamentally the dictator of Germany. But they aren't happy with just that consolidated power, they want to ensure that they stay in power. And so as we get into 1934 they continue to consolidate their power and now more directly start eliminating their opponents. And when I say eliminating, they are actually directly killing these individuals. And there is an entire Nazi power apparatus that is at play, but the three figures here are the ones that are most notable other than Hitler. This is Joseph Goebbels who is the head of Nazi Propaganda, Hermann G\u00f6ring, who is the head of the Nazi's secret police, the Gestapo, which later goes under the control of the SS, under Heinrich Himmler. He is a major player in the rearmament of Germany for war footing. He eventually helds the Luftwaffe which is the German airforce during the World War II. We have Heinrich Himmler who is the head of Schutzstaffel, more famously known as the SS, which is the paramilitary group of the Nazi party, used to intimidate and eliminate (as we will see) their opponents and they are also responsible for the execution of the actual the planning and the execution - of the actual holocaust. And so as we get into 1934 Hitler is eager to eliminate more of his opponents - and we are not just talking about opponents outside of the Nazi party, we are also talking about rivals within the Nazi party. And the most notable of these was Ernst R\u00f6hm who as you see at this picture was clearly a Nazi. He was head of the Storm Battalion or the SA, which was the paramilitary group that the SS splintered out of and it was far more independent than Hitler would have liked and it was not popular amongst the German people because of its violence - and Hitler and his lieutenants viewed R\u00f6hm as a potential rival to Hitler's authority within the Nazi party. And so in 1934, June 30th in particular - or we could say June 30th to July 2nd - you have what's called the Night of the Long Knives and it really should be called the Nights of the Long Knives but the Night of the Long Knives, where under the pretext of a supposed a coup d'\u00e9tat or a putsch on the part of Ernst R\u00f6hm The SS and the Gestapo start rounding up R\u00f6hm, his allies and any perceived enemies of Hitler. Start rounding them up and eliminating them and shooting them... And Ernst R\u00f6hm is arrested and when he refuses to kill himself in his jail cell he is shot at point-blank range. Gregor Strasser, who is a former rival of Hitler within the Nazi party, he is eliminated. Kurt von Schleicher who is a previous chancellor of Germany, perceived as a rival to Hitler - he and his wife are gunned down. Gustav Ritter von Kahr who is in no way anymore a rival to Hitler but he was one of the major actors for putting down the Beer Hall Putsch - the fail the Beer Hall Putch of 1923. He is hacked - he is hacked to get dead. And these are just four of the more notable folks. On those few nights over 85 major officials in Germany were eliminated. And as we will see this is just the beginning of the consolidation of Nazi power and now they are doing it through extra legal means." + }, + { + "Q": "09:00 You're placing the oxygen on the same side as Br but you're saying that it's inversion. How come?\nAlso, I really don't understand that triangle type bond, what's that?", + "A": "See videos on stereochemistry. This video is way too advanced if you don t understand chirality or wedge and dash representations of bonds.", + "video_name": "3LiyCxCTrqo", + "transcript": "- [Instructor] Let's look at the mechanism for an SN2 reaction. On the left we have an alkyl halide and we know that this bromine is a little bit more electronegative than this carbon so the bromine withdraws some electron density away from that carbon which makes this carbon a little bit positive, so we say partially positive. That's the electrophilic center so this on the left is our electrophile. On the right, we know that this hydroxide ion which we could get from something like sodium hydroxide, is a negative one formal charge on the oxygen which makes it a good nucleophile. Let me write down here. This is our nucleophile on the right and on the left is our electrophile which I'm also gonna refer to as a substrate in this video. This alkyl halide is our substrate. We know from an earlier video that the nucleophile will attack the electrophile because opposite charges attract. This negative charge is attracted to this partially positive charge. Lone pair of electrons on the oxygen will attack this partially positive carbon. At the same time, the two electrons in this bond come off onto the bromine. Let me draw the bromine over here. The bromine had three lone pairs of electrons on it and it's gonna pick up another lone pair of electrons. Let me show those electrons in magenta. This bond breaks and these two electrons come off onto the bromine which gives the bromine a negative one formal charge. This is the bromide anion. And we're also forming a bond between the oxygen and this carbon and this bond comes from this lone pair of electrons which I've just marked in blue here. Those two electrons in blue form this bond and we get our product which is an alcohol. The SN2 mechanism is a concerted mechanism because the nucleophile attacks the electrophile, at the same time we get loss of a leaving groups. There's only one step in this mechanism. Let's say we did a series of experiments to determine the rate law for this reaction. Remember from general chemistry, rate laws are determined experimentally. Capital R is the rate of the reaction and that's equal to the rate constant k times the concentration of our alkyl halide. And it's determined experimentally this is to the first power times the concentration of the hydroxide ion also to the first power. So what does this mean? This means if we increased the concentration of our alkyl halide. If we increase the concentration of our alkyl halide by a factor of two, what happens to the rates of the reaction? Well, the rate of the reaction is proportional to the concentration of the alkyl halide to the first power. Two to the first is equal to two which means the overall rate of the reaction would increase by a factor of two. Doubling the concentration of your alkyl halide while keeping this concentration, the hydroxide ion concentration the same should double the rate of the reaction. And also, if we kept the concentration of alkyl halide the same and we double the concentration of hydroxide, that would also increase the rate by a factor of two. And this experimentally determined rate law makes sense with our mechanism. If we increase the concentration of the nucleophile or we increase the concentration of the electrophile, we increase the frequency of collisions between the two which increases the overall rate of the reaction. The fact that our rate law is proportional to the concentration of both the substrate and the nucleophile fits with our idea of a one step mechanism. Finally, let's take a look at where this SN2 comes from. We keep on saying an SN2 mechanism, an SN2 reaction. The S stands for substitution. Let me write in here substitution because our nucleophile is substituting for our leaving group. We can see in our final products here, the nucleophile has substituted for the leaving group. The N stands for nucleophilic because of course it is our nucleophile that is doing the substituting. And finally the two here refers to the fact that this is bimolecular which means that the rate depends on the concentration of two things. The substrate and the nucleophile. That's different from an SN1 mechanism where the rate is dependent only on the concentration of one thing. The rate of the reaction also depends on the structure of the alkyl halide, on the structure of the substrate. On the left we have a methyl halide followed by a primary alkyl halide. The carbon bonded to our bromine is directly attached to one alkyl group followed by a secondary alkyl halide, the carbon bonded to the bromine is bonded to two alkyl groups, followed by a tertiary alkyl halide. This carbon is bonded to three alkyl groups. Turns out that the methyl halides and the primary alkyl halide react the fastest in an SN2 mechanism. Secondary alkyl halides react very slowly and tertiary alkyl halides react so, so slowly that we say they are unreactive toward an SN2 mechanism. And this makes sense when we think about the mechanism because remember, the nucleophile has to attack the electrophile. The nucleophile needs to get close enough to the electrophilic carbon to actually form a bond and steric hindrance would prevent that from happening. Something like a tertiary alkyl halide has this big bulky methyl groups which prevent the nucleophile for attacking. Let's look at a video so we can see this a little bit more clearly. Here's our methyl halide with our carbon directly bonded to a halogen which I'm seeing as yellow. And here's our nucleophile which could be the hydroxide ion. The nucleophile approaches the electrophile for the side opposite of the leaving group and you can see with the methyl halide there's no steric hindrance. When we move to a primary alkyl halide, the carbon bonded to the halogen has only one alkyl group bonded to it, it's still easy for the nucleophile to approach. When we move to a secondary alkyl halide, so for a secondary you can see that the carbon bonded to the halogen has two methyl groups attached to it now. It gets a little harder for the nucleophile to approach in the proper orientation. These bulky methyl groups make it more difficult for the nucleophile to get close enough to that electrophilic carbon. When we go to a tertiary alkyl halide, so three alkyl groups. There's one, there's two and there's three. There's a lot more steric hindrance and it's even more difficult for our nucleophile to approach. As we saw on the video, for an SN2 reaction we need decreased steric hindrance. So, if we look at this alkyl halide, the carbon that is directly bonded to our halogen is attached to only one alkyl group. This is a primary alkyl halide and that makes this a good SN2 reaction. The decreased steric hindrance allows the nucleophile to attack the electrophile." + }, + { + "Q": "Help with this is greatly appreciated. I got lost when he divided by -3 by 3 and 5/6 by 3. Also, how come its just 6/3 giving us 5/2 and not 5/6 divided by 3?", + "A": "Can you give me a time-stamp or the exact equation? I can t seem to understand your writing.", + "video_name": "5fkh01mClLU", + "transcript": "In this video I'm going to do a bunch of examples of finding the equations of lines in slope-intercept form. Just as a bit of a review, that means equations of lines in the form of y is equal to mx plus b where m is the slope and b is the y-intercept. So let's just do a bunch of these problems. So here they tell us that a line has a slope of negative 5, so m is equal to negative 5. And it has a y-intercept of 6. So b is equal to 6. So this is pretty straightforward. The equation of this line is y is equal to negative 5x plus 6. That wasn't too bad. Let's do this next one over here. The line has a slope of negative 1 and contains the point 4/5 comma 0. So they're telling us the slope, slope of negative 1. So we know that m is equal to negative 1, but we're not 100% sure about where the y-intercept is just yet. So we know that this equation is going to be of the form y is equal to the slope negative 1x plus b, where b is the y-intercept. Now, we can use this coordinate information, the fact that it contains this point, we can use that information to solve for b. The fact that the line contains this point means that the value x is equal to 4/5, y is equal to 0 must satisfy this equation. So let's substitute those in. y is equal to 0 when x is equal to 4/5. So 0 is equal to negative 1 times 4/5 plus b. I'll scroll down a little bit. So let's see, we get a 0 is equal to negative 4/5 plus b. We can add 4/5 to both sides of this equation. So we get add a 4/5 there. We could add a 4/5 to that side as well. The whole reason I did that is so that cancels out with that. You get b is equal to 4/5. So we now have the equation of the line. y is equal to negative 1 times x, which we write as negative x, plus b, which is 4/5, just like that. Now we have this one. The line contains the point 2 comma 6 and 5 comma 0. So they haven't given us the slope or the y-intercept explicitly. But we could figure out both of them from these So the first thing we can do is figure out the slope. So we know that the slope m is equal to change in y over change in x, which is equal to-- What is the change in y? Let's start with this one right here. So we do 6 minus 0. Let me do it this way. So that's a 6-- I want to make it color-coded-- minus 0. So 6 minus 0, that's our change in y. Our change in x is 2 minus 2 minus 5. The reason why I color-coded it is I wanted to show you when I used this y term first, I used the 6 up here, that I have to use this x term first as well. So I wanted to show you, this is the coordinate 2 comma 6. This is the coordinate 5 comma 0. I couldn't have swapped the 2 and the 5 then. Then I would have gotten the negative of the answer. But what do we get here? This is equal to 6 minus 0 is 6. 2 minus 5 is negative 3. So this becomes negative 6 over 3, which is the same thing as negative 2. So that's our slope. So, so far we know that the line must be, y is equal to the slope-- I'll do that in orange-- negative 2 times x plus our y-intercept. Now we can do exactly what we did in the last problem. We can use one of these points to solve for b. We can use either one. Both of these are on the line, so both of these must satisfy this equation. I'll use the 5 comma 0 because it's always nice when you have a 0 there. The math is a little bit easier. So let's put the 5 comma 0 there. So y is equal to 0 when x is equal to 5. So y is equal to 0 when you have negative 2 times 5, when x is equal to 5 plus b. So you get 0 is equal to -10 plus b. If you add 10 to both sides of this equation, let's add 10 to both sides, these two cancel out. You get b is equal to 10 plus 0 or 10. So you get b is equal to 10. Now we know the equation for the line. The equation is y-- let me do it in a new color-- y is equal to negative 2x plus b plus 10. We are done. Let's do another one of these. All right, the line contains the points 3 comma 5 and negative 3 comma 0. Just like the last problem, we start by figuring out the slope, which we will call m. It's the same thing as the rise over the run, which is the same thing as the change in y over the change in x. If you were doing this for your homework, you wouldn't I just want to make sure that you understand that these are all the same things. Then what is our change in y over our change in x? This is equal to, let's start with the side first. It's just to show you I could pick either of these points. So let's say it's 0 minus 5 just like that. So I'm using this coordinate first. I'm kind of viewing it as the endpoint. Remember when I first learned this, I would always be tempted to do the x in the numerator. No, you use the y's in the numerator. So that's the second of the coordinates. That is going to be over negative 3 minus 3. This is the coordinate negative 3, 0. This is the coordinate 3, 5. We're subtracting that. So what are we going to get? This is going to be equal to-- I'll do it in a neutral color-- this is going to be equal to the numerator is negative 5 over negative 3 minus 3 is negative 6. So the negatives cancel out. You get 5/6. So we know that the equation is going to be of the form y is equal to 5/6 x plus b. Now we can substitute one of these coordinates in for b. So let's do. I always like to use the one that has the 0 in it. So y is a zero when x is negative 3 plus b. So all I did is I substituted negative 3 for x, 0 for y. I know I can do that because this is on the line. This must satisfy the equation of the line. Let's solve for b. So we get zero is equal to, well if we divide negative 3 by 3, that becomes a 1. If you divide 6 by 3, that becomes a 2. So it becomes negative 5/2 plus b. We could add 5/2 to both sides of the equation, plus 5/2, plus 5/2. I like to change my notation just so you get familiar with both. So the equation becomes 5/2 is equal to-- that's a 0-- is equal to b. b is 5/2. So the equation of our line is y is equal to 5/6 x plus b, which we just figured out is 5/2, plus 5/2. We are done. Let's do another one. We have a graph here. Let's figure out the equation of this graph. This is actually, on some level, a little bit easier. What's the slope? Slope is change in y over change it x. So let's see what happens. When we move in x, when our change in x is 1, so that is our change in x. So change in x is 1. I'm just deciding to change my x by 1, increment by 1. What is the change in y? It looks like y changes exactly by 4. It looks like my delta y, my change in y, is equal to 4 when my delta x is equal to 1. So change in y over change in x, change in y is 4 when change in x is 1. So the slope is equal to 4. Now what's its y-intercept? Well here we can just look at the graph. It looks like it intersects y-axis at y is equal to negative 6, or at the point 0, negative 6. So we know that b is equal to negative 6. So we know the equation of the line. The equation of the line is y is equal to the slope times x plus the y-intercept. I should write that. So minus 6, that is plus negative 6 So that is the equation of our line. Let's do one more of these. So they tell us that f of 1.5 is negative 3, f of negative 1 is 2. What is that? Well, all this is just a fancy way of telling you that the point when x is 1.5, when you put 1.5 into the function, the function evaluates as negative 3. So this tells us that the coordinate 1.5, negative 3 is on the line. Then this tells us that the point when x is negative 1, f of x is equal to 2. This is just a fancy way of saying that both of these two points are on the line, nothing unusual. I think the point of this problem is to get you familiar with function notation, for you to not get intimidated if you see something like this. If you evaluate the function at 1.5, you get negative 3. So that's the coordinate if you imagine that y is equal to f of x. It would be equal to negative 3 when x is 1.5. Anyway, I've said it multiple times. Let's figure out the slope of this line. The slope which is change in y over change in x is equal to, let's start with 2 minus this guy, negative 3-- these are the y-values-- over, all of that over, negative 1 minus this guy. Let me write it this way, negative 1 minus that guy, minus 1.5. I do the colors because I want to show you that the negative 1 and the 2 are both coming from this, that's why I use both of them first. If I used these guys first, I would have to use both the x and the y first. If I use the 2 first, I have to use the negative 1 first. That's why I'm color-coding it. So this is going to be equal to 2 minus negative 3. That's the same thing as 2 plus 3. So that is 5. Negative 1 minus 1.5 is negative 2.5. 5 divided by 2.5 is equal to 2. So the slope of this line is negative 2. Actually I'll take a little aside to show you it doesn't matter what order I do this in. If I use this coordinate first, then I have to use that coordinate first. Let's do it the other way. If I did it as negative 3 minus 2 over 1.5 minus negative 1, this should be minus the 2 over 1.5 minus the negative 1. This should give me the same answer. This is equal to what? Negative 3 minus 2 is negative 5 over 1.5 minus negative 1. That's 1.5 plus 1. That's over 2.5. So once again, this is equal the negative 2. So I just wanted to show you, it doesn't matter which one you pick as the starting or the endpoint, as long as If this is the starting y, this is the starting x. If this is the finishing y, this has to be the finishing x. But anyway, we know that the slope is negative 2. So we know the equation is y is equal to negative 2x plus some y-intercept. Let's use one of these coordinates. I'll use this one since it doesn't have a decimal in it. So we know that y is equal to 2. So y is equal to 2 when x is equal to negative 1. Of course you have your plus b. So 2 is equal to negative 2 times negative 1 is 2 plus b. If you subtract 2 from both sides of this equation, minus 2, minus 2, you're subtracting it from both sides of this equation, you're going to get 0 on the left-hand side is equal to b. So b is 0. So the equation of our line is just y is equal to negative 2x. Actually if you wanted to write it in function notation, it would be that f of x is equal to negative 2x. I kind of just assumed that y is equal to f of x. But this is really the equation. They never mentioned y's here. So you could just write f of x is equal to 2x right here. Each of these coordinates are the coordinates of x and f of x. So you could even view the definition of slope as change in f of x over change in x. These are all equivalent ways of viewing the same thing." + }, + { + "Q": "what are the communist countries that still exist today", + "A": "The main that comes to mind is of course China. But China actually implement some forms of liberalization and since the 70s permits private property, capital and business to function, so it is goign in the direction of being some kind of hybrid version. Then we have Cuba, which is communist since 1959. North Korea is also a communist country with maybe the tightest regime in the world. And also we have Laos and Vietnam, which are communist since the mid 70s after there bloody conflicts there.", + "video_name": "MmRgMAZyYN0", + "transcript": "Thought I would do a video on communism just because I've been talking about it a bunch in the history videos, and I haven't given you a good definition of what it means, or a good understanding of what it means. And to understand communism-- let me just draw a spectrum here. So I'm going to start with capitalism. And this is really just going to be an overview. People can do a whole PhD thesis on this type of thing. Capitalism, and then I'll get a little bit more-- and then we could progress to socialism. And then we can go to communism. And the modern versions of communism are really kind of the brainchild of Karl Marx and Vladimir Lenin. Karl Marx was a German philosopher in the 1800s, who, in his Communist Manifesto and other writings, kind of created the philosophical underpinnings for communism. And Vladimir Lenin, who led the Bolshevik Revolution in the-- and created, essentially, the Soviet Union-- he's the first person to make some of Karl Marx's ideas more concrete. And really every nation or every country which we view as communist has really followed the pattern of Vladimir Lenin. And we'll talk about that in a second. But first, let's talk about the philosophical differences between these things, and how you would move. And Karl Marx himself viewed communism as kind of a progression from capitalism through socialism to communism. So what he saw in capitalism-- and at least this part of what he saw was right-- is that you have private property, private ownership of land. That's the main aspect of capitalism. And this is the world that most of us live in today. The problem that he saw with capitalism is he thought, well, look, when you have private property, the people who start accumulating some capital-- and when we talk about capital, we could be talking about land, we could be talking about factories, we could be talking about any type of natural resources-- so the people who start getting a little bit of them-- so let me draw a little diagram here. So let's say someone has a little bit of capital. And that capital could be a factory, or it could be land. So let me write it. Capital. And let's just say it's land. So let's say someone starts to own a little bit of land. And he owns more than everyone else. So then you just have a bunch of other people who don't own land. But they need, essentially-- and since this guy owns all the land, they've got to work on this guy's land. They have to work on this guy's land. And from Karl Marx's point of view, he said, look, you have all of these laborers who don't have as much capital. This guy has this capital. And so he can make these laborers work for a very small wage. And so any excess profits that come out from this arrangement, the owner of the capital will be able to get it. Because these laborers won't be able to get their wages to go up. Because there's so much competition for them to work on this guy's farm or to work on this guy's land. He really didn't think too much about, well, maybe the competition could go the other way. Maybe you could have a reality eventually where you have a bunch of people with reasonable amounts of capital, and you have a bunch of laborers. And the bunch of people would compete for the laborers, and maybe the laborers could make their wages go up, and they could eventually accumulate their own capital. They could eventually start their own small businesses. So he really didn't think about this reality too much over here. He just saw this reality. And to his defense-- and I don't want to get in the habit of defending Karl Marx too much-- to his defense, this is what was happening in the late 1800s, especially-- we have the Industrial Revolution. Even in the United States, you did have kind of-- Mark Twain called it the Gilded Age. You have these industrialists who did accumulate huge amounts of capital. They really did have a lot of the leverage relative to the laborers. And so what Karl Marx says, well, look, if the guy with all the capital has all the leverage, and this whole arrangement makes some profits, he's going to be able to keep the profits. Because he can keep all of these dudes' wages low. And so what's going to happen is that the guy with the capital is just going to end up with more capital. And he's going to have even more leverage. And he'll be able to keep these people on kind of a basic wage, so that they can never acquire capital for themselves. So in Karl Marx's point of view, the natural progression would be for these people to start organizing. So these people maybe start organizing into unions. So they could collectively tell the person who owns the land or the factory, no, we're not going to work, or we're going to go on strike unless you increase our wages, or unless you give us better working conditions. So when you start talking about this unionization stuff, you're starting to move in the direction of socialism. The other element of moving in the direction of socialism is that Karl Marx didn't like this kind of high concentration-- or this is socialists in general, I should say-- didn't like this high concentration of wealth. That you have this reality of not only do you have these people who could accumulate all of this wealth-- and maybe, to some degree, they were able to accumulate it because they were innovative, or they were good managers of land, or whatever, although the Marxists don't give a lot of credit to the owners of capital. They don't really give a lot of credit to saying maybe they did have some skill in managing some type of an operation. But the other problem is is that it gets handed over. It gets handed over to their offspring. So private property, you have this situation where it just goes from maybe father to son, or from parent to a child. And so it's not even based on any type of meritocracy. It's really just based on this inherited wealth. And this is a problem that definitely happened in Europe. When you go back to the French Revolution, you have generation after generation of nobility, regardless of how incompetent each generation would be, they just had so much wealth that they were essentially in control of everything. And you had a bunch of people with no wealth having to work for them. And when you have that type of wealth disparity, it does lead to revolutions. So another principle of moving in the socialist direction is kind of a redistribution of wealth. So let me write it over here. So redistribution. So in socialism, you can still have private property. But the government takes a bigger role. So you have-- let me write this. Larger government. And one of the roles of the government is to redistribute wealth. And the government also starts having control of the major factors of production. So maybe the utilities, maybe some of the large factories that do major things, all of a sudden starts to become in the hands of the government, or in the words of communists, in the hands of the people. And the redistribution is going on, so in theory, you don't have huge amounts of wealth in the hands of a few people. And then you keep-- if you take these ideas to their natural conclusion, you get to the theoretical communist state. And the theoretical communist state is a classless, and maybe even a little bit-- a classless society, and in Karl Marx's point of view-- and this is a little harder to imagine-- a stateless society. So in capitalism, you definitely had classes. You had the class that owns the capital, and then you had the labor class, and you have all of these divisions, and they're different from each other. He didn't really imagine a world that maybe a laborer could get out of this, they could get their own capital, then maybe they could start their own business. So he just saw this tension would eventually to socialism, and eventually a classless society where you have a central-- Well, he didn't even go too much into the details but you have kind of equal, everyone in society has ownership over everything, and society somehow figures out where things should be allocated, and all of the rest. And it's all stateless. And that's even harder to think about in a concrete fashion. So that's Karl Marx's view of things. But it never really became concrete until Vladimir Lenin shows up. And so the current version of communism that we-- The current thing that most of us view as communism is sometimes viewed as a Marxist-Leninist state. These are sometimes used interchangeably. Marxism is kind of the pure, utopian, we're eventually going to get to a world where everyone is equal, everyone is doing exactly what they want, there's an abundance of everything. I guess to some degree, it's kind of describing what happens in Star Trek, where everyone can go to a replicator and get what they want. And if you want to paint part of the day, you can paint part of the day, and you're not just a painter, you can also do whatever you want. So it's this very utopian thing. Let me write that down. So pure Marxism is kind of a utopian society. And just in case you don't know what utopian means, it's kind of a perfect society, where you don't have classes, everyone is equal, everyone is leading these kind of rich, diverse, fulfilling lives. And it's also, utopian is also kind of viewed as unrealistic. It's kind of, if you view it in the more negative light, is like, hey, I don't know how we'll ever be able to get there. Who knows? I don't want to be negative about it. Maybe we will one day get to a utopian society. But Leninist is kind of the more practical element of communism. Because obviously, after the Bolshevik Revolution, 1917, in the Russian Empire, the Soviet Union gets created, they have to actually run a government. They have to actually run a state based on these ideas of communism. And in a Leninist philosophy-- and this is where it starts to become in tension with the ideas of democracy-- in a Leninist philosophy, you need this kind of a party system. So you need this-- and he calls this the Vanguard Party. So the vanguard is kind of the thing that's leading, the one that's leading the march. So this Vanguard Party that kind of creates this constant state of revolution, and its whole job is to guide society, is to kind of almost be the parent of society, and take it from capitalism through socialism to this ideal state of communism. And it's one of those things where the ideal state of communism was never-- it's kind of hard to know when you get there. And so what happens in a Leninist state is it's this Vanguard Party, which is usually called the Communist Party, is in a constant state of revolution, kind of saying, hey, we're shepherding the people to some future state without a real clear definition of what that future state is. And so when you talk about Marxist-Leninist, besides talking about what's happening in the economic sphere, it's also kind of talking about this party system, this party system where you really just have one dominant party that it will hopefully act in the interest of the people. So one dominant communist party that acts in the interest of the people. And obviously, the negative here is that how do you know that they actually are acting in the interest of people? How do you know that they actually are competent? What means are there to do anything if they are misallocating things, if it is corrupt, if you only have a one-party system? And just to make it clear, the largest existing communist state is the People's Republic of China. And although it is controlled by the Communist Party, in economic terms it's really not that communist anymore. And so it can be confusing. And so what I want to do is draw a little bit of a spectrum. On the vertical axis, over here, I want to put democratic. And up here, I'll put authoritarian or totalitarian. Let me put-- well, I'll put authoritarian. I'll do another video on the difference. And they're similar. And totalitarian is more an extreme form of authoritarian, where the government controls everything. And you have a few people controlling everything and it's very non-democratic. But authoritarian is kind of along those directions. And then on this spectrum, we have the capitalism, socialism, and communism. So the United States, I would put-- I would put the United States someplace over here. I would put the United States over here. It has some small elements of socialism. You do have labor unions. They don't control everything. You also have people working outside of labor unions. It does have some elements of redistribution. There are inheritance taxes. There are-- I mean it's not an extreme form of redistribution. You can still inherit private property. You still have safety nets for people, you have Medicare, Medicaid, you have welfare. So there's some elements of socialism. But it also has a very strong capitalist history, private property, deep market, so I'd stick the United States over there. I would put the USSR-- not current Russia, but the Soviet Union when it existed-- I would put the Soviet Union right about there. So this was the-- I would put the USSR right over there. I would put the current state of Russia, actually someplace over here. Because they actually have fewer safety nets, and they kind of have a more-- their economy can kind of go crazier, and they actually have a bigger disparity in wealth than a place like the United States. So this is current Russia. And probably the most interesting one here is the People's Republic of China, the current People's Republic of China, which is at least on the surface, a communist state. But in some ways, it's more capitalist than the United States, in that they don't have strong wealth redistribution. They don't have kind of strong safety nets for people. So you could put some elements of China-- and over here, closer to the left. And they are more-- less democratic than either the US or even current Russia, although some people would call current Russia-- well, I won't go too much into it. But current China, you could throw it here a little bit. So it could be even a little bit more capitalist than the United States. Definitely they don't even have good labor laws, all the rest. But in other ways, you do have state ownership of a lot, and you do have state control of a lot. So in some ways, they're kind of spanning this whole range. So this right over here is China. And even though it is called a communist state, in some ways, it's more capitalist than countries that are very proud of their capitalism. But in a lot of other ways, especially with the government ownership and the government control of things, and this one dominant party, so it's kind of Leninist with less of the Marxist going on. So in that way, it is more in the communist direction. So hopefully that clarifies what can sometimes be a confusing topic." + }, + { + "Q": "Do we borrow money from the Federal Reserve?", + "A": "Well, where do you think government treasury bonds come from?", + "video_name": "-05OfTp6ZEE", + "transcript": "Before we talk about the debt ceiling, it's important to realize the difference between the deficit and the debt. Because these words are thrown around and it's clear that they're related, but sometimes people might confuse one for the other. The deficit is how much you overspend in a given year, while the debt is the total amount, the cumulative amount, of debt you you've gotten over many, many years. So let's take a look, I guess a very simplified example, let's say you have some type of a country. And that country spends, in a given year, $10. But it's only bringing in $6 in tax revenue. So it's bringing in taxes. It's only bringing in $6. So this country in this year, where it spends $10, even though it only has $6 to spend, it has a $4 deficit. Def is the short for deficit. And well, let me just write it out. You might think it's defense or something. It has a $4 deficit. And you might say, well, how does it spend more money that it brings in? How can it actually continue to spend this much? Where will it get the $4 from? And the answer is, it will borrow that $4. Our little country will borrow it. And so the debt, maybe going into this year, the country already had some debt. Maybe it already had $100 of debt. And so in this situation, it would have to borrow another $4 of debt. And so exiting this year, it would have $104 of debt. If the country runs the same $4 deficit the year after this, then the debt will increase to $108. If it runs another $4 deficit, than the debt will increase to $112. Now that we have that out of the way, let's think about what the debt ceiling is. So you could imagine, the United States actually does. It's continuing to run a deficit. It's continuing to spend more than it brings in. And actually, for the United States, these ratios are appropriate. For every dollar that the United States spends right now, 40% is borrowed. Or another way to think about it, it only has 60% of every dollar that it needs to spend right now. So it has to go out into the debt markets and borrow 40% to keep spending at its current rate. And so if it's continuing to borrow, you could imagine that the debt keeps on increasing. So let me draw a little graph here. So that axis is time. This axis right over here is the total cumulative amount of debt that we have. We continue to have to borrow 40% of every dollar And so our debt is continuing to increase. And Congress has the power, or Congress has the authority, to essentially limit how much debt we have. So right now we have a current debt limit of $14.3 trillion. And even though Congress has this authority, the way that it's worked in the past, is this kind of just a rubber stamp. Congress has just always allowed the debt ceiling to go up and up and up to fund our borrowing costs. And if you think about it, that kind of makes sense because right now Congress is the one that decides where to spend the money. What are the obligations. And so the debt ceiling is like, OK, we've already agreed what you have to spend your money on. Congress is the one that figures out what we spend our money on, and what our taxes are. And so they say, look, we've already determined how much you have to borrow. It would seem kind of ridiculous for us after we've determined how much you borrow to say that you cannot borrow it. You cannot you cannot actually do what we've told you to do. And so historically, Congress has just kind of gone with the flow. They said, OK, yeah we've told you we need to borrow more money to execute-- the executive branch has to run the government-- for you to actually run the government based on the budget we told you. So they just keep upping it. And the last time the debt ceiling was raised was actually very recently, February 12, 2010. It was raised from $12.3 trillion, point actually $12.4 trillion to the $14.3 trillion. And this happens pretty regularly. It's happened 10 times since 2001, 74 times since 1962. So it's just a regular operating thing. And right now the Obama administration says, look, we've actually come up against our debt ceiling. We want to raise it, and ideally for the Obama administration, they want to raise it by about $2.4 trillion. So they want to raise it to $16.7 trillion, which will kind of put it off the table for a little bit. Put it past the elections so that we don't have to debate this anymore. The Republicans on the other the side, want to essentially use this, and this is a little bit unusual, to use this as leverage to essentially reduce the deficit. And not only to reduce the deficit, but it's in particular to reduce the deficit through spending cuts. And so that's why it's become this big game of chicken and why we're going up against this limit. Now, one thing that you may or may not realize is that we've actually already hit the debt limit, the current debt limit. And we hit that debt limit on May 16, 2011. I'm making this video at the end of July in 2011. And the only reason why the country's continuing to operate, and the only reason why the country has been able to continue to pay interest on its obligations, and pay issue social security checks, and support Medicare, and buy fuel for aircraft carriers, and all the rest, is that Geithner, who's the Treasury Secretary, has been able to find cash in other places, cash normally set aside for employee pensions and all the rest. And has essentially done a little bit of a bookkeeping, taking money from one place to feed another. But what he said, what he's publicly said, is that he won't be able to do that anymore as of August 2, So this right here is the date that everyone is paying attention to, August 2, 2011. According to Geithner, at that point, he won't be able to find random pockets of cash here and there and shuffle it around. And what he calls extraordinary measures. And at that point, the United States will not be able to fulfill all of its obligations. And so if you think about all of the obligations of the United States, this is a huge oversimplification here. So this bar represent all of the obligations. Some of those obligations are things like interest on the debt that it already owes. It already owes a huge amount of debt, $14.3 trillion. And things like social security, Medicare, defense, and then all of the other stuff that the country has to support, all of their other obligations. So if as of August 2, 2011, we cannot issue any more debt, and Geithner doesn't have any extra cash laying around with these extraordinary measures, then, if those are the only options on the table, The only option is to somehow reduce some of these things by 40%. Because 40% of every dollar we used to spend on all of these obligations, 40% are borrowed. And so something over here is going to give. We're not going to fulfill our obligations to one or more of these things, all of these things that we are legally obligated to fulfill. That Congress has said, these are the things that the United States should be spending its money on. And so at that point, it is perceived that we would have to default. And a default actually would be on any of its obligations. But in particular, we could be, especially if we have to cut everything by 40%. And we don't want to see retirees not be able to get evicted from their houses, or aircraft carriers not have fuel, or whatever else. We might defer, or try to restructure, or do something weird with our debt. In which case, we would be defaulting. And I want to be clear, a default, it's usually referred to not fully paying the interest on debt that you owe. But a default would be any of its obligations. The United States has this AAA rating. If the United States says it's going to give you a Social Security check, you trust that. If the United States says that it's going to pay for that Medicare payment, you trust that. If it says it's going to give you an interest payment, you trust that. All of a sudden, if United States does not fulfill any of those obligations, then all of the obligations becomes suspect. And the reason why this is a big deal, as you can imagine, if you borrow money, you've always been good at paying back that money, you're going to pay lower interest than other people would have to pay. But all of a sudden, for whatever reason, one day you default. You either delay your payment, or you say you don't have the cash to pay your payments, then people's like, wow you're a much riskier person to lend money to. So now I'm going to increase the interest rates on you. And so the perception is if the United States were to default on its debt, or any of its obligations, that interest rates would go up. And the reason why this would really not great is because it would make the debt and the deficit even worse. Then this chunk is going to have to grow. Our obligations are on debt. As new debt gets issued, we're gonna have to pay more and more interest. So it's going to just make matters worse. It's going to make the deficit worse. And on top of that, it's not just that the government's debt, the interest on the government's debt will go up, but interest on all debt in the United States will probably go up. Because government debt is perceived to be the safest, it's the benchmark. A lot of other debt contractors are actually tied to government debt. So you'll have interest rates throughout the economy go up, which is exactly what you do not want to happen when you are either in a recession, or when you are recovering from a recession." + }, + { + "Q": "I was confused by the answer because I did the equation as (fx - gx) and got a different answer. The equation must be (gx -fx) to answer the problem.\n(fx - gx) does not equal (g - f)(x).", + "A": "I think this is mainly because of the commutative property. In addition you can move things around, but you can t do this when it comes to subtraction. That s why you got completely different answers in both of your expressions. It s interesting that you took the time to figure that out :) Good job.", + "video_name": "KvMyZY9upuA", + "transcript": "- [Instructor] We're told that f of x is equal to two x times the square root of five minus four. And we're also told that g of x is equal to x squared plus two x times the square root of five minus one. And they want us to find g minus f of x. So pause this video, and see if you can work through that on your own. So the key here is to just realize what this notation means. G minus f of x is the same thing as g of x minus f of x. And so, again, if this was helpful to you, once again I encourage you to pause the video. All right, now let's work through this again. So this is going, or, I guess the first time, but now that we know that this is equal to g of x minus f of x. So what is g of x? Well, that's the same thing as x squared plus two x times the square root of five minus one. And what is f of x? Well, it's going to be two x times the square root of five minus four. And we are subtracting f of x from g of x. So let's subtract, this is f of x, from g of x. And so now it's just going to be a little bit of algebraic simplification. So this is going to be equal to, this is equal to x squared plus two x times the square root of five minus one. And now we just have to distribute this negative sign. So negative one times two x times the square root of five is, we're gonna have minus two x times the square root of five. And then the negative of negative four is positive four. Now let's see if we can simplify this some. So this is going to be equal to, we only have one x squared term, so that's that one there. So we have x squared. Now let's see, we have two x times the square root of five. And then we have another, oh, and then we subtract two x times the square root of five. So these two cancel out with each other. So those cancel out. And then we have minus one plus four. So if we have negative one and then we add four to it, we're going to have positive three. So if we just fact, if we take this and this into consideration, four minus one is going to be equal to three, and we're done. That's what g minus f of x is equal to, x squared plus three." + }, + { + "Q": "Where can I get a printed pattern for my hexaflexagon?", + "A": "There are lots of templates online, just print one out.", + "video_name": "AmN0YyaTD60", + "transcript": "Hexaflexagons-- they're cool, hip, and hexa-fun to play with, right? Wrong. Hexaflexagons are not toys. With the increasing number of hexaflexagons finding their way into homes and schools, it's important to be aware of proper flexagation regulations when engaging in flexagon construction and use. Taking proper precautions can help avoid a flexa-catastrophe. Do not wear loose clothing when engaging in flexagation. If you have long hair, tie it back, so it doesn't get caught in a flexagation device. Ties are also a common source of incidents. Stay alert. Never flexagate while under the influence. When using a hexaflexagon, sudden unexpected sides may appear, and drugs like alcohol can slow reaction time. If you aren't sure what kind of flexagon you're dealing with, it's safer to temporarily disable the flexagon. Flexagons can be disarmed by using scissors to cut them apart. You can cut across the original seam where the paper strip was taped together, which may appear on the edge or through the face of the flexagon. In an emergency, however, flexagons can be cut apart right through a triangle, or on three edges if you want to retain symmetry, or into nine separate triangles if you really want to be safe. You can even cut them in half down the length of the paper strip like this, into two separate-- Once you cut your flexagon apart, you can figure out what kind it is. If it has nine triangles, that's 18 triangle sides. So at six triangles per hexagon side, that's three sides of trihexaflexagon. Note that some flexagons might be made from a double strip of triangles that have been folded in half, so that marker doesn't bleed through. Don't let yourself be fooled by the extra triangles. Avoid danger during hexaflexagon construction. If you're not working from a printed pattern, you might start your flexagon by picking a point on the edge of a strip of paper, folding that 180 degree angle into thirds to create 360 degree angles, and then using the equilateral triangle that results as a guide to fold the rest of the strip of paper, zigzagging back and forth. Without proper attention and focus, this could easily lead to becoming unreasonably amused with the springy spring of happy triangles that results. Always keep your hexaflexagon in good working order. Pre-creasing all the triangles both ways before configuring them into hexaflexagonal formation will help your flexagon operate properly and avoid accidents. Keep a close watch on the chirality of your hexaflexagon. That is, whether it is right or left handed. Notice how in this hexaflexagon, water flows clockwise down under the flaps, even if you flip it over or flex it. Well, in this hexaflexagon, it flows counter-clockwise. They're mirror images. The chirality is decided when you fold and tape your triangles into a twisty loop, and once taped, it is impossible to change from one to the other without cutting it apart, at least in three-dimensional euclidean space. A change in chirality could be a sign that your flexagon has been flipped through four-dimensional space and is possibly a highly dangerous multi-dimensional portal. With experience, a hexaflexagon master can construct a hexaflexagon in mere seconds. Some forgo tape and scissors entirely by folding a double strip that's too long and tucking the extra in. This is an advanced technique that should not be attempted without prior training. Beware topological changes. This family seems safe from this philosoraptor, because they live on separate planets with a cold, empty vacuum of space between them. But after a single flex, the unfortunate victims are now doomed, protected only by the inconsequential barrier of their domicile. Your stars might explode, your frowns may become smiles, your most pointy of triangles might become the roundest of circles. Perfectly healthy snakes may turn into snake loops, or worse, become decapitated. Either state is fatal for the snake, as having no head can lead to starvation. This can be avoided by simply marking where connections will be across neighboring triangles first. Afterwards, the lines can be filled in however you like. Be aware that with the trihexaflexagon, there are two variations to each face. So you can simply draw one side where triangles connect, and flip and draw the other. But in the hexa-hexaflexagon, the main three faces each appear four different ways. If you use hexaflexagons, keep an eye out for signs of dependency. Overuse can lead to addiction and possibly an overdose. Some users of hexaflexagons report confusion, mind-blown syndrome, hexaflexaperplexia, hexaflexadyslexia, hexaflexaperfectionism, and hexaflexa-Mexican-food-cravings. If you find yourself experiencing any of these symptoms, stop flexagon use immediately, and see the head of your math department. With proper precautions, flexagating can be a great part of your life. Follow these simple safety guidelines, and you should be ready for a fun and safe hexaflexagon experience." + }, + { + "Q": "Wow I thought violas were smaller and higher pitched than the violins! Are their more violins or violas in an orchestra? And how many if you know?", + "A": "I think you just got the two mixed up! Violas are bigger and lower, as you noticed, and there are more violins in an orchestra. There are two violin sections - the violin 1 and violin 2. There would probably be about 20 violins and then 8 or so violas. Hope this helps!", + "video_name": "ivbT-wvzK58", + "transcript": "(\"Symphony No. 4 in F minor\" by Pyotr Ilyich Tchaikovsky) - This large instrument is a viola. It's actually larger than most violas. Most violas are about this, the body part comes to about here. This one is here and it doesn't seem like a lot but it's huge. When you need to reach something, not many people can play this instrument. Violins are pretty much all the same size. Violas do come in many different shapes and sizes. If you look around in an orchestra, some of the mark cut down like this and they look awfully pretty. And some of the mark kinda big down here and small up here which is kinda nice because it's easier to get up here. You don't have to worry about this shoulder here. It's easier. (\"Symphony No. 3 in E-flat major\" by Robert Schumann) Violins are tuned like this. (tune playing) Okay, violas sound like this. (tune playing) They're much lower. It's like if you have a singing group and you have sopranos and you have altos, the violins and then the violas are the alto voice, and the cello's the tenor and base is the bass. (\"Symphony No. 9 in E minor\" by Antonin Dvorak) A bunch of years ago, I was playing a double viola concerto that was written for me and my stand partner of (mumbles). She had just acquired a million-dollar instrument, Gaspar Cassado, from Italy. I was playing something that got me my job. I mean, I played and I got it myself but the instrument that I was playing when I got in Philharmonic, I also got a major job in another orchestra at some point. On that instrument, the sound just didn't compare to this Gaspar. So the orchestra, administrators I guess, and the powers that be said, \"Go find yourself an instrument.\" So the orchestra owns this instrument. I went out and I asked around and the people put the word out and I tried many instruments. There was one that I had found and loved it but there was... Once we went to find out the background about it, there was little suspect. Anyway, a lot of politics and instruments that I didn't know about, but found this one. It's so big that it is... I said, \"Don't show me anything 17 inches or over.\" And this is actually 17 1/2 plus but the guy who was joining to make (mumbles) said, \"You have to hear this instrument.\" So I picked it up and went, \"Oh no (laughs), I really love it.\" (\"Symphony No. 9 in E minor\" by Antonin Dvorak) I always knew I was gonna be a musician. I started because I used to go to the young people's concert when I was two 1/2. I saw people playing and the story goes. I don't remember but my mom said I used to roll up the programs and put one here and put one here and pretend and then when I came time to play, my mom said, \"Would you like to play?\" I asked her, I remember like it was yesterday. We were standing in the garage and she was putting garbage in the garbage can and she had one, the can in her hand, the top in her hand like this and I said, \"I want to play violin.\" And she said, \"Well, don't you want to start with piano?\" And I said, \"No, I want to do this.\" So she found me a teacher shortly thereafter and got a feel and I started playing and then, I guess I got kinda good. I auditioned for Juilliard Pre-College at eight but I didn't get in but they sent me to a teacher, Fannie Chase, who is no longer with us, but she taught me every technical thing that I know on the violin or viola. Then I reauditioned five years later and I got in when I was 13 to Juilliard. But by the time I was done I guess I was a little bit bored. I used to like to play the second violin parts which are always the harmony parts, not the melody parts. I even, with my own children, I would always make them sing the melody of something so that I could sing the harmony. Like a commercial on TV, I would always make them sing the high things, so I could sing the middle thing. But it was the same thing for me in string quartets, I always wanted to play the middle voice. So when I was ready, I was not happy with it. I would guess I was a little bored. Somebody suggested to my mom, \"\"In Pre-College Juilliards, we need some more violas. \"And she's tall and she got long arms \"and maybe she could play the viola.\" So I said, \"All right, I'll try it.\" And I tried it and I loved it. My teacher was very inspiring. It was only nine months with that one particular teacher, Eugene Becker. (\"Symphony No. 4 in F minor\" by Pyotr Ilyich Tchaikovsky) That's what I always did. It's what I always knew I would do. Not only did I always know I was gonna be a musician but I always knew I was gonna be in the Philharmonic. Don't ask me. It's just the way it always was. It wasn't, \"Oh, I hope I get this audition\", or, \"I hope I can get in to Juilliards.\" \"I hope\", it's just, \"Okay, now I'm gonna take \"this audition and now I'm gonna do this.\" It just kinda... I mean, I'm lucky I guess. (\"Of Paradise and Light\" by Augusta Read Thomas)" + }, + { + "Q": "I don't understand how Bob decrypts the message.", + "A": "He is given a key beforehand.", + "video_name": "FlIG3TvQCBQ", + "transcript": "For over 400 years, the problem remained. How could Alice design a cipher that hides her fingerprint, thus stopping the leak of information? The answer is randomness. Imagine Alice rolled a 26 sided die to generate a long list of random shifts, and shared this with Bob instead of a code word. Now, to encrypt her message, Alice uses the list of random shifts instead. It is important that this list of shifts be as long as the message, as to avoid any repetition. Then she sends it to Bob, who decrypts the message using the same list of random shifts she had given him. Now Eve will have a problem, because the resulting encrypted message will have two powerful properties. One, the shifts never fall into a repetitive pattern. And two, the encrypted message will have a uniform frequency distribution. Because there is no frequency differential and therefore no leak, it is now impossible for Eve to break the encryption. This is the strongest possible method of encryption, and it emerged towards the end of the 19th century. It is now known as the one-time pad. In order to visualize the strength of the one-time pad, we must understand the combinatorial explosion which takes place. For example, the Caesar Cipher shifted every letter by the same shift, which was some number between 1 and 26. So if Alice was to encrypt her name, it would result in one of 26 possible encryptions. A small number of possibilities, easy to check them all, known as brute force search. Compare this to the one-time pad, where each letter would be shifted by a different number between 1 and 26. Now think about the number of possible encryptions. It's going to be 26 multiplied by itself five times, which is almost 12 million. Sometimes it's hard to visualize, so imagine she wrote her name on a single page, and on top of it stacked every possible encryption. How high do you think this would be? With almost 12 million possible five-letter sequences, this stack of paper would be enormous, over one kilometer high. When Alice encrypts her name using the one-time pad, it is the same as picking one of these pages at random. From the perspective of Eve, the code breaker, every five letter encrypted word she has is equally likely to be any word in this stack. So this is perfect secrecy in action." + }, + { + "Q": "I don't really get that. Dose that mean when your simplifeying a number like 3/8 whatever number comes like suppose you have to multiply the 3 by a 2 will you have to multiply 8 by a 2 as well?", + "A": "To create an equivalent fraction, you must multiply both top and bottom by the same value. So, yes... if you multiply the 3 in 3/8 by 2, you would also multiply the 8 by 2. 3/8 * 2/2 = 6/16", + "video_name": "dCQbfaQZtaY", + "transcript": "You are filling up trays to make ice cubes. You notice that each tray holds the exact same amount of water but has a different number of ice cubes that it makes. The blue tray makes 8 equally sized ice cubes. The pink tray makes 16 equally sized ice cubes. So let me draw the blue tray here. The blue tray, I'll draw it like this. That is not blue. Let me draw it in blue. So the blue tray makes 8. So this is the blue tray right over here. And let me draw it and divide it into 8 sections to represent the 8 equally sized ice cubes. And I'll try to do it. That's about in half, and then I'm going to put each of those in half, and then each of those in half. So there we go. This is pretty close to 8 equally sized cubes, except for this one there. Let me actually make it a little bit cleaner. So there's the half, half, half, half, half. All right, so this is looking a little bit better, so 8 equally sized ice cubes. This is the blue tray. Now, the pink tray takes the same amount of water, so I'm going to make it the exact same length. So the pink tray is the exact same length, but it has 16 equally sized ice cubes. So what I'm going to do is I'm going to make these same sections here, but then I'm going to cut these sections in two. So that's 8, and to 16, I got to divide these into two. So almost done. This is the hard part. That last one wasn't that neat. Almost done. All right. There we go. Set up the problem. Now, you put 3 ice cubes from the blue tray into your drink. So, let's do that. So, 1, 2, 3 ice cubes from the blue tray into your drink. How many ice cubes from the pink tray would you need to equal the same exact amount of ice? So there's a bunch of ways to do it. We can think about it with numbers, or we can think about it visually. Let's first think about it with numbers. So how much of this tray have I pulled out? Well, I have 8 equally sized cubes and I took 3 of them out, so this literally represents 3/8. So the question is 3 over 8, if I take 3 out of 8 equally sized cubes, that's the same quantity as taking what? I want to do that in white. That's the same quantity as taking how many cubes out of 16, out of 16 equally sized cubes? Well, let's look at it over here visually. So if we want the exact same amount of ice, so we're gonna have 1, 2, 3, 4, 5, 6. We have 6/16, so this is equal to 6/16. Now, does that make actual sense? Well, sure. To go from 3/8 to 6/16, you multiply the numerator by 2, and you multiply the denominator by 2. Now, does that actually make sense? Well, sure it does. Because for the pink ice tray, you have 2 ice cubes for every 1 that you have in the blue ice So the blue ice tray, you have 8 equally sized cubes. Well, for each of those, you're going to have 2 in the pink ice tray, so you multiply by 2 to have 16 equally sized cubes. And out of the blue tray, if you take 3, well, that equivalent amount for each of those cubes, you would get 2 from the pink ice tray. So you're multiplying by 2 right over there. So the answer to the question, how many ice cubes from the pink tray would you need to equal the same amount of ice? Well, that is you would need 6 cubes." + }, + { + "Q": "why didn't the Germans go through Switzerland or Italy then??\nthey would have a better chance of winning the war .", + "A": "Tanks can t go through the Alps. Not to mention it is SUPER COLD there. If your mindset is not Hannibal or Napoleon, it would be harder than death doing it", + "video_name": "huOnuYAyv6w", + "transcript": "In the last video, we left off with the assassination of Archduke Franz Ferdinand of Austria, the heir to the Austro-Hungarian empire. And so you could imagine, the Austro-Hungarian empire did not take that well. They already did not enjoy the kingdom of Serbia trying to, essentially, provoke this nationalistic movement. And they viewed them as this small, little, weak country right below them. And so they use this to, essentially, issue an ultimatum. Essentially say, look, immediately bring all the people who did this to justice, all the people who might have conspired with the Gavrilo Princips to allow this assassination attempt to occur, and do it or else, and accept full responsibility. And actually, the kingdom of Serbia was in no mood to get into a war with Austria-Hungary, and so they tried their best to comply, but their best wasn't enough. So then July 28, a month later, you have the Austrians declaring war on the Serbians. Now up until this point, the Austro-Hungarian empire is thinking, OK, Serbia is a small, little kingdom here. It has some ties and obviously, it's a Slavic nation so it has some linguistic ties with the Russian empire. The Russian empire also had some political ties to it, but Russia is not going to get into a war with us over this. We're justified in attacking them. They've just killed the heir to our throne. So we're going to go in there and Russia is probably not interested in actually creating a larger skirmish here. That was a severe miscalculation on the part of the Austrians. The Russians were not happy about this. They felt close ties to the Serbians and they felt a need to protect the Serbians, or you could argue, that maybe they wanted to mobilize just to scare the Austrians. Whatever it might be, whether it was Russia wanted to get into a war, whether they were really looking to protect the Serbians, or whether they were looking to mobilize to scare the Austrians from actually attacking, the Russians began to mobilize. So the Russians began to gather their troops. So the Russians mobilize. And now, this is where all of the alliances start to come into effect. If you remember about the alliances we talked about several videos ago, if you go to 1879, you have the Dual Alliance Treaty between Germany and Austria-Hungary to protect each other if Russia attacks and actually, if Russia attacks or mobilizes. So now, Germany is like, hey, maybe I am obligated to protect Austria-Hungary from Russia. You also have to remember in 1892, the Franco-Russian Military Convention. Military assistance both ways in the event of attack. So Germany is thinking, look, we signed this treaty and we, to some degree, are maybe eager for war because we've been militarizing so much. And I can't just fight Russia. I also know that France and Russia have this alliance right over here. So Germany in quite surprising quickness decides to declare war on both. So on August 1, Germany declares war on Russia. And then on August 3, Germany declares war on France because they know or at least they feel that they can't declare war on only one of them. And they wanted to do it very quickly because they didn't want Russia a chance to mobilize too much. And the fact that they were able to do this so quickly-- we're talking three days after Austria declares war on Serbia and then another two days declare war on France-- kind of shows that Germany was already in a war footing. It's not a joke to all of a sudden invade or declare war on countries. So Germany was, essentially, preparing for this. And this right over here gave them the excuse to, essentially, declare war. So they declare war on both of these characters, Germany against Russia. Germany is declaring war on France. Now, the easiest way for them to move into France-- so they're literally going on the offense here-- is for them to go through Belgium. But the Germans were aware. They're aware that there's this 75-year-old treaty, the Treaty of London in 1839, Article 7 said that Britain was to protect the neutrality of Belgium. And Germany was in no interest to get into a war with the British. The British had a powerful military, especially a very powerful navy. The Germans said, hey, let's just take on the Russians and the French for now. And so they actually reached out to the British and said, hey, this little treaty that you got here from 1839, this 75-year-old treaty, you don't really take this seriously, right? I mean if we have to go through Belgium, you're not really going to hold true to this treaty? And the British said, no, we actually take that very seriously. Obviously, the British didn't want the Germans to be aggressive here. The British didn't want the Germans to be able to invade France. And so on August 4 when Germany, essentially, rolls through Belgium, Germany invades Belgium to get to France, this gave the legal justification for the British to declare war on Germany. And so in a matter of only-- I mean not even, we're talking a few months here from the assassination of Franz Ferdinand of Austria, you essentially have all of the major powers of Europe. And then as we'll see because they had these empires, in not too long most of the world is at war with each other." + }, + { + "Q": "What is relative permeability? Does any one knows have brief expalanation", + "A": "It s just a constant needed to make the equations work, like G or epsilon naught . It relates the proportionality of the current to the magnetic field and distance.", + "video_name": "Ri557hvwhcM", + "transcript": "So not only can a magnetic field exert some force on a moving charge, we're now going to learn that a moving charge or a current can actually create a magnetic field. So there is some type of symmetry here. And as we'll learn later when we learn our calculus and we do this in a slightly more rigorous way, we'll see that magnetic fields and electric fields are actually two sides of the same coin, of electromagnetic fields. But anyway, we won't worry about that now. And I think it's enough to ponder right now that a current can actually induce a magnetic field. And actually, just a moving electron creates a magnetic field. And it does it in a surface of a sphere-- I won't go into all Because the math gets a little bit fancy there. But what you might encounter in your standard high school physics class-- that's not getting deeply into vector calculus-- is that if you just have a wire-- let me draw a wire. That's my wire. And it's carrying some current I, it turns out that this wire will generate a magnetic field. And the shape of that magnetic field is going to be co-centric circles around this wire. Let me see if I can draw that. So here I'll draw it just like how I do when I try to do rotations of solids in the calculus video. So the magnetic field would go behind and in front and it goes like that. Or another way you can think about it is if-- let's go down here-- is on the left side of this wire. If you say that the wire's in the plane of this video, the magnetic field is popping out of your screen. And on this side, on the right side, the magnetic field is popping into the screen. It's going into the screen. And you could imagine that, right? You could imagine if, on this drawing up here, you could say this is where it intersects the screen. All of this is kind of behind the screen. And all of this is in front of the screen. And this is where it's popping out. And this is where it's popping into the screen. Hopefully that makes a little bit of sense. And how did I know that it's rotating this way? Well, it actually does come out of the cross product when you do it with a regular charge and all of that. But we're not going to go into that right now. And so there's a different right hand rule that you can use. And it's literally you hold this wire, or you imagine holding this wire, with your right hand with your thumb going in the direction of the current. And if you hold this wire with your thumb going in the direction of the current, your fingers are going to go in the direction of the magnetic field. So let me see if I can draw that. I will draw it in blue. So if this is my thumb, my thumb is going along the top of the wire. And then my hand is curling around the wire. Those are the veins on my hand. This is my nail. So as you can see, if I was holding that same wire-- let me draw the wire. So if I was holding that same wire, we see that my thumb is going in the direction of the current. So this is a slightly new thing to memorize. And what is the magnetic field doing? Well, it's going in the direction of my fingers. My fingers are popping out on this side of the wire. They're coming out on this side of the wire. And they're going in, or at least my hand is going in, on that side. It's going into the screen. Hopefully that makes sense. Now, how can we quantify? Well, before we quantify, let's get a little bit more of the intuition of what's happening. It turns out that the closer you get to the wire, the stronger the magnetic field, and the further you get out, the weaker the magnetic field. And that kind of makes sense if you imagine the magnetic field spreading out. I don't want to go into too sophisticated analogies. But if you imagine the magnetic field spreading out, and as it goes further and further out it kind of gets distributed over a wider and wider circumference. And actually the formula I'm going to give you kind of has a taste for that. So the formula for the magnetic field-- and it really is defined with a cross product and things like that, but for our purposes we won't worry about that. You'll just have to know that this is the shape if the current is going in that direction. And, of course, if the current was going downwards then the magnetic field would just reverse directions. But it would still be in co-centric circles around the wire. But anyway, what is the magnitude of that field? The magnitude of that magnetic field is equal to mu-- which is a Greek letter, which I will explain in a second-- times the current divided by 2 pi r. So this has a little bit of a feel for what I was saying before. That the further you go out-- where r is how far you are from the wire-- the further you go out, if r gets bigger, the magnitude of the magnetic field is going to get weaker. And this 2 pi r, that looks a lot like the circumference of a circle. So that gives you a taste for it. I know I haven't proved anything rigorously. But it at least gives you a sense of, look there's a little formula for circumference of a circle here. And that kind of makes sense, right? Because the magnetic field at that point is kind of a circle. The magnitude is equal at an equal radius around that wire. Now what is this mu, this thing that looks like a u? Well, that's the permeability of the material that the wire's in. So the magnetic field is actually going to have a different strength depending on whether this wire is going through rubber, whether it's going through a vacuum, or air, or metal, or water. And for the purposes of your high school physics class, we assume that it's going through air normally. And the value for air is pretty close to the value for a vacuum. And it's called the permeability of a vacuum. And I forget what that value is. I could look it up. But it actually turns out that your calculator has that value. So let's do a problem, just to put some numbers to the formula. So let's say I had this current and it is-- I don't know, the current is equal to-- I'm going 2 amperes. And let's say that I just pick a point right here that is-- let's say that that's 3 meters away from the wire in question. So my question to you is what is the magnitude in the direction of the magnetic field right there? Well, the magnitude is easy. We just substitute in this equation. So the magnitude of the magnetic field at this point is equal to-- and we assume that the wire's going through air or a vacuum-- the permeability of free space-- that's just a constant, though it looks fancy-- times the current times 2 amperes divided by 2 pi r. What's r? It's 3 meters. So 2 pi times 3. So it equals the permeability of free space. The 2 and the 2 cancel out over 3 pi. So how do we calculate that? Well, we get out our trusty TI-85 calculator. And I think you'll be maybe pleasantly surprised or shocked to realize that-- I deleted everything just so you can see how I get there-- that it actually has the permeability of free space stored in it. So what you do is you go to second and you press constant, which is the 4 button. It's in the built-in constants. Let's see, it's not one of those. You press more. It's not one of those, press more. Oh look at that. Mu not. The permeability of free space. That's what I need. And I have to divide it by 3 pi. Divide it by 3-- and then where is pi? There it is. It's over the power sign. Divided by 3 pi. It equals 1.3 times 10 to the negative seventh. It's going to be teslas. The magnetic field is going to be equal to 1.3 times 10 to the minus seventh teslas. So it's a fairly weak magnetic field. And that's why you don't have metal objects being thrown around by the wires behind your television set. But anyway, hopefully that gives you a little bit-- and just so you know how it all fits together. We're saying that these moving charges, not only can they be affected by a magnetic field, not only can a current be affected by a magnetic field or just a moving charge, it actually creates them. And that kind of creates a little bit of symmetry in your head, hopefully. Because that was also true of electric field. A charge, a stationary charge, is obviously pulled or pushed by a static electric field. And it also creates its own static electric field. So it's always in the back of your mind. Because if you keep studying physics, you're going to actually prove to yourself that electric and magnetic fields are two sides of the same coin. And it just looks like a magnetic field when you're in a different frame of reference, When something is whizzing past you. While if you were whizzing along with it, then that thing would look static. And then it might look a little bit more like an electric field. But anyway, I'll leave you there now. And in the next video I will show you what happens when we have two wires carrying current parallel to each other. And you might guess that they might actually attract or repel each other. Anyway, I'll see you in the next video." + }, + { + "Q": "When in history were we first able to calculate the size of the sun?", + "A": "The distance to the sun was accurately determined in the late 1700 s. Once the distance is known, the diameter is easy to calculate using basic geometry.", + "video_name": "GZx3U0dbASg", + "transcript": "My goal in this video and the next video is to start giving a sense of the scale of the earth and the solar system. And as we see, as we start getting into to the galaxy and the universe, it just becomes almost impossible to imagine. But we'll at least give our best shot. So I think most of us watching this video know that this right here is earth. And just to get a sense of scale here, I think probably the largest distance that we can somehow relate to is about 100 miles. You can get into a car for an hour, hour and a half, and go about 100 miles. And on the earth that would be about this far. It would be a speck that would look something like that. That is 100 miles. And also to get us a bit of scale, let's think about a speed that at least we can kind of comprehend. And that would be, maybe, the speed of a bullet. Maybe we can't comprehend it, but I'll say this is the fastest thing that we could maybe comprehend. It goes about-- and there are different types of bullets depending on the type of gun and all of that-- about 280 meters per second, which is about 1,000 kilometers per hour. And this is also roughly the speed of a jet. So just to give a sense of scale here, the earth's circumference-- so if you were to go around the planet-- is about 40,000 kilometers. So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the earth. And I think none of this information is too surprising. You might have taken a 12- or 15-hour flight that gets you-- not all the way around the earth-- but gets you pretty far. San Francisco to Australia, or something like that. So right now these aren't scales that are too crazy. Although, even for me, the earth itself is a pretty mind-blowingly large object. Now, with that out of the way let's think about the sun. Because the sun starts to approach something far huger. So this obviously here is the sun. And I think most people appreciate that the sun is much larger than the earth, and that it's pretty far away from the earth. But I don't think most people, including myself, fully appreciate how large the sun is or how far it is away from the earth. So just to give you a sense, the sun is 109 times the circumference of the earth. So if we do that same thought exercise there-- if we said, OK, if I'm traveling at the speed of a bullet or the speed of a jetliner, it would take me 40 hours to go around the earth. Well, how long would it take to go around the sun? So if you were to get on a jet plane and try to go around the sun, or if you were to somehow ride a bullet and try to go around the sun-- do a complete circumnavigation of the sun-- it's going to take you 109 times as long as it would have taken you to do the earth. So it would be 100 times-- I could do 109, but just for approximate-- it's roughly 100 times the circumference of the earth. So 109 times 40 is equal to 4,000 hours. And just to get a sense of what 4,000 is-- actually, since I have the calculator out, let's do the exact calculation. It's 109 times the circumference of the earth times 40 hours. That's what it would take to do the circumference of the Earth. So it's 4,360 hours to circumnavigate the sun, going at the speed of a bullet or a jetliner. And so that is-- 24 hours in the day-- that is 181 days. It would take you roughly half a year to go around the sun at the speed of a jetliner. Let me write this down. Half a year. The sun is huge. Now, that by itself may or may not be surprising--and actually let me give you a sense of scale here, because I have this other diagram of a sun. And we'll talk more about the rest of the solar system in the next video. But over here, at this scale, the sun, at least on my screen-- if I were to complete it, it would probably be about 20 inches in diameter. The earth is just this little thing over here, smaller than a raindrop. If I were to draw it on this scale, where the sun is even smaller, the earth would be about that big. Now, what isn't obvious, because we've all done our science projects in third and fourth grade--or we always see these diagrams of the solar system that look something like this-- is that these planets are way further away. Even though these are depicted to scale, they're way further away from the sun than this makes it look. So the earth is 150 million kilometers from the sun. So if this is the sun right here, at this scale you wouldn't even be able to see the earth. It wouldn't even be a pixel. But it would be 150 million kilometers from the earth. And this distance right here is called an astronomical unit-- and we'll be using that term in the next few videos just because it's an easier way to think about distance-- sometimes abbreviated AU, astronomical unit. And just to give a sense of how far this is, light, which is something that we think is almost infinitely fast and that is something that looks instantaneous, that takes eight minutes to travel from the sun to the earth. If the sun were to disappear, it would take eight minutes for us to know that it disappeared on earth. Or another way, just to put it in the sense of this jet airplane-- let's get the calculator back out. So we're talking about 150 million kilometers. So if we're going at 1,000 kilometers an hour, it would take us 150,000 hours at the speed of a bullet or at the speed of a jet plane to get to the sun. And just to put that in perspective, if we want it in days, there's 24 hours per day. So this would be 6,250 days. Or, if we divided by 365, roughly 17 years. If you were to shoot a bullet straight at the sun it would take 17 years to get there, if it could maintain its velocity somehow. So this would take a bullet or a jet plane 17 years to get to the sun. Or another way to visualize it-- this sun right over here, on my screen it has about a five- or six-inch diameter. If I were to actually do it at scale, this little dot right here, which is the earth, this speck-- I would have to put this back about 50 feet away from the sun. 50 or 60 feet away from the sun. If you were to look at the solar system-- and obviously there's other things in the solar system, and we'll talk more about them in the next video-- you wouldn't even notice this speck. This is a little dust thing flying around this sun. And as we go further and further out of this solar system, you're going to see even this distance starts to become ridiculously small. Or another way to think about it-- if the sun was about this size, then the earth at this scale would be about 200 feet away from it. So you could imagine, if you had a football field-- let me draw a football field. These are the end zones-- one end zone, another end zone. And if you were to stick something-- maybe the size of a medicine ball, a little bit bigger than a basketball, at one end zone-- this little speck would be about 60 yards away, roughly 60 meters away. So it's this little speck. You wouldn't even notice it on the scale of a football field, something this size. Anyway, I'm going to leave you there. Hopefully that just starts to blow your mind when you think about the scale of the sun, the earth, and how far the earth is away from the sun. And then we're going to see even those distances, even those scales, are super small when you start thinking about the rest of the solar system. And especially when we start going beyond the solar system." + }, + { + "Q": "Hi! I was wondering: what happens if there is a negative number inside the square root when you -4ac from b^2? I have had problems like that before on math homework. I've tried many ways to get answers but non of them work\u00e2\u0080\u00a6 here's and example!\n3x^2 \u00e2\u0080\u0093 5x + 4 = 0\nThanks!", + "A": "If there is a negative number inside the radical, then there is NO real solutions. This would only be solvable through complex numbers (imaginary numbers).", + "video_name": "iulx0z1lz8M", + "transcript": "Use the quadratic formula to solve the equation, 0 is equal to negative 7q squared plus 2q plus 9. Now, the quadratic formula, it applies to any quadratic equation of the form-- we could put the 0 on the left hand side. 0 is equal to ax squared plus bx plus c. And we generally deal with x's, in this problem we're dealing with q's. But the quadratic formula says, look, if you have a quadratic equation of this form, that the solutions of this equation are going to be x is going to be equal to negative b plus or minus the square root of b squared minus 4ac-- all of that over 2a. And this is actually two solutions here, because there's one solution where you take the positive square root and there's another solution where you take the negative root. So it gives you both roots of this. So if we look at the quadratic equation that we need to solve here, we can just pattern match. We're dealing with q's, not x's, but this is the same general idea. It could be x's if you like. And if we look at it, negative 7 corresponds to a. That is our a. It's the coefficient on the second degree term. 2 corresponds to b. It is the coefficient on the first degree term. And then 9 corresponds to c. It's the constant. So, let's just apply the quadratic formula. The quadratic formula will tell us that the solutions-- the q's that satisfy this equation-- q will be equal to negative b. b is 2. Plus or minus the square root of b squared, of 2 squared, minus 4 times a times negative 7 times c, which is 9. And all of that over 2a. All of that over 2 times a, which is once again negative 7. And then we just have to evaluate this. So this is going to be equal to negative 2 plus or minus the square root of-- let's see, 2 squared is 4-- and then if we just take this part right here, if we just take the negative 4 times negative 7 times 9, this negative and that negative is going to cancel out. So it's just going to become a positive number. And 4 times 7 times 9. 4 times 9 is 36. 36 times 7. Let's do it up here. 36 times 7. 7 times 6 is 42. 7 times 3, or 3 times 7 is 21. Plus 4 is 25. 252. So this becomes 4 plus 252. Remember, you have a negative 7 and you have a minus out front. Those cancel out, that's why we have a positive 252 for that part right there. And then our denominator, 2 times negative 7 is negative 14. Now what does this equal? Well, we have this is equal to negative 2 plus or minus the square root of-- what's 4 plus 252? It's just 256. All of that over negative 14. And what's 256? What's the square root of 256? It's 16. You can try it out for yourself. This is 16 times 16. So the square root of 256 is 16. So we can rewrite this whole thing as being equal to negative 2 plus 16 over negative 14. Or negative 2 minus-- right? This is plus 16 over negative 14. Or minus 16 over negative 14. If you think of it as plus or minus, that plus is that plus right there. And if you have that minus, that minus is that minus right there. Now we just have to evaluate these two numbers. Negative 2 plus 16 is 14 divided by negative 14 is negative 1. So q could be equal to negative 1. Or negative 2 minus 16 is negative 18 divided by negative 14 is equal to 18 over 14. The negatives cancel out, which is equal to 9 over 7. So q could be equal to negative 1, or it could be equal to 9 over 7. And you could try these out, substitute these q's back into this original equation, and verify for yourself that they satisfy it. We could even do it with the first one. So if you take q is equal to negative 1. Negative 7 times negative 1 squared-- negative 1 squared is just 1-- so this would be negative 7 times 1, right? That's negative 1 squared. Negative 1 times 2 is minus 2 plus 9. So it's negative 7 minus 2, which is negative 9, plus 9, does indeed equal 0. So this checks out. And I'll leave it up to you to verify that 9 over 7 also works out." + }, + { + "Q": "how is (2,-1) a solution for -3x-4y=-2 and y=2x-5", + "A": "If a point is a solution to the system of equations, then it will make both equations be true. Use substitution to test the ordered pair in each equation. The ordered pair tells you X = 2 and Y = -1 1st equation: -3(2) - 4(-1)=-2 -6 + 4=-2 -2=-2 (the ordered pair makes this equation true) 2nd equation: -1 = 2(2) - 5 -1 = 4 - 5 -1 = -1 (the ordered pair makes this equation true as well) Since the ordered pair satisfies both equations, it is a solution to the system.", + "video_name": "GWZKz4F9hWM", + "transcript": "So that it's less likely that we get shown up by talking birds in the future, we've set a little bit of exercise for solving systems of equations with substitution. And so this is the first exercise or the first problem that they give us. -3x-4y=-2 and y=2x-5 So let me get out my little scratch pad and let me rewrite the problem. So this is -3x-4y=-2 and then they tell us y=2x-5. So, what's neat about this is that they've already solved the second equation. They've already made it explicitly solved for y which makes it very easy to substitute for. We can take this constraint, the constraint on y in terms of x and substitute it for y in this first blue equation and then solve for x. So let's try it out. So this first blue equation would then become -3x-4 but instead of putting a y there the second constraint tells us that y needs to be equal to 2x-5. So it's 4(2x-5) and all of that is going to be equal to -2. So now we get just one equation with one unknown. and now we just have to solve for x. So, let's see if we can do that. So, it's -3x and then this part right over here we have a -4, be careful, we have a -4 we want to distribute. We are going to multiply -4*2x which is -8x and -4*-5 is positive 20 and thats going to equal -2. And now we can combine all the x terms so -3x-8x, that's going to be -11x and then we have -11x+20=-2. Now to solve for x, we'll subtract 20 from both sides to get rid of the 20 on the left hand side. On the left hand side, we're just left with the -11x and then on the right hand side we are left with -22. Now we can divide both sides by -11. And we are left with x is equal to 22 divided by 11 is 2, and the negatives cancel out. x = 2. So we are not quite done yet. We've done, I guess you can say the hard part, we have solved for x but now we have to solve for y. We could take this x value to either one of these equations and solve for y. But this second one has already explicitly solved for y so let's use that one, so it says y = 2 times and instead of x, we now know that the x value where these two intersect, you could view it that way is going to be equal to 2, so 2 * 2 - 5 let's figure out the corresponding y value. So you get y=2(2)-5 and y = 4 - 5 so y = -1. And you can verify that it'll work in this top equation If y = -1 and x=2, this top equation becomes -3(2) which is -6-4(-1) which would be plus 4. And -6+4 is indeed -2. So it satisfies both of these equations and now we can type it in to verify that we got it right, although, we know that we did, so x=2 and y=-1. So, let's type it in... x=2 and y=-1. Excellent, now we're much less likely to be embarassed by talking birds." + }, + { + "Q": "I still don't understand why at the end, the answer is negative.Why did the triangle double towards the negative axis? Also, how is the x co-ordinate the cosine of the angle? Thanks.", + "A": "As Sal mentions at 3:16, he s using the unit circle definition of the trigonometric functions. I could try to explain that in this answer, but I think you re better off watching the series of videos on the unit circle definition under the Basic Trigonometry section of Trigonometry and Precalculus. Hopefully those ll help you make more sense of this.", + "video_name": "D_smr0GBPvA", + "transcript": "We have triangle ABC here, which looks like a right triangle. And we know it's a right triangle because 3 squared plus 4 squared is equal to 5 squared. And they want us to figure out what cosine of 2 times angle So that's this angle-- ABC. Well, we can't immediately evaluate that, but we do know what the cosine of angle ABC is. We know that the cosine of angle ABC-- well, cosine is just adjacent over hypotenuse. It's going to be equal to 3/5. And similarly, we know what the sine of angle ABC is. That's opposite over hypotenuse. That is 4/5. So if we could break this down into just cosines of ABC and sines of ABC, then we'll be able to evaluate it. And lucky for us, we have a trig identity at our disposal that does exactly that. We know that the cosine of 2 times an angle is equal to cosine of that angle squared minus sine of that angle squared. And we've proved this in other videos, but this becomes very helpful for us here. Because now we know that the cosine-- Let me do this in a different color. Now, we know that the cosine of angle ABC is going to be equal to-- oh, sorry. It's the cosine of 2 times the angle ABC. That's what we care about. 2 times the angle ABC is going to be equal to the cosine of angle ABC squared minus sine of the angle ABC squared. And we know what these things are. This thing right over here is just going to be equal to 3/5 squared. Cosine of angle a ABC is 3/5. So we're going to square it. And this right over here is just 4/5 squared. So it's minus 4/5 squared. And so this simplifies to 9/25 minus 16/25, which is equal to 7/25. Sorry. It's negative. Got to be careful there. 16 is larger than 9. Negative 7/25. Now, one thing that might jump at you is, why did I get a negative value here when I doubled the angle here? Because the cosine was clearly a positive number. And there you just have to think of the unit circle-- which we already know the unit circle definition of trig functions is an extension of the Sohcahtoa definition. Y-axis. Let me draw a unit circle here. My best attempt. So that's our unit circle. So this angle right over here looks like something like this. And so you see its x-coordinate-- which is the cosine of that angle-- looks positive. But then, if you were to double this angle, it would take you out someplace like this. And then, you see-- by the unit circle definition-- the x-coordinate, we are now sitting in the second quadrant. And the x-coordinate can be negative. And that's essentially what happened in this problem." + }, + { + "Q": "What is displacement?", + "A": "It s the distance from and object, just with a direction as well.", + "video_name": "oRKxmXwLvUU", + "transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. And he did it in 1 hour in his car. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. And my best sense of that is, once you start doing calculus, you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time. So these two, you could call them formulas, or you could call them definitions, although I would think that they're pretty intuitive for you. How fast something is going, you say, how far did it go over some period of time. These are essentially saying the same thing. This is when you care about direction, so you're dealing with vector quantities. This is where you're not so conscientious about direction. And so you use distance, which is scalar, and you use rate or speed, which is scalar. Here you use displacement, and you use velocity. Now with that out of the way, let's figure out what his average velocity was. And this key word, average, is interesting. Because it's possible that his velocity was changing over that whole time period. But for the sake of simplicity, we're going to assume that it was kind of a constant velocity. What we are calculating is going to be his average velocity. But don't worry about it, you can just assume that it wasn't changing over that time period. So his velocity is, his displacement was 5 kilometers to the north-- I'll write just a big capital. Well, let me just write it out, 5 kilometers north-- over the amount of time it took him. And let me make it clear. This is change in time. This is also a change in time. Sometimes you'll just see a t written there. Sometimes you'll see someone actually put this little triangle, the character delta, in front of it, which explicitly means \"change in.\" It looks like a very fancy mathematics when you see that, but a triangle in front of something literally means \"change in.\" So this is change in time. So he goes 5 kilometers north, and it took him 1 hour. So the change in time was 1 hour. So let me write that over here. So over 1 hour. So this is equal to, if you just look at the numerical part of it, it is 5/1-- let me just write it out, 5/1-- kilometers, and you can treat the units the same way you would treat the quantities in a fraction. 5/1 kilometers per hour, and then to the north. Or you could say this is the same thing as 5 kilometers per hour north. So this is 5 kilometers per hour to the north. So that's his average velocity, 5 kilometers per hour. And you have to be careful, you have to say \"to the north\" if you want velocity. If someone just said \"5 kilometers per hour,\" they're giving you a speed, or rate, or a scalar quantity. You have to give the direction for it to be a vector quantity. You could do the same thing if someone just said, what was his average speed over that time? You could have said, well, his average speed, or his rate, would be the distance he travels. The distance, we don't care about the direction now, is 5 kilometers, and he does it in 1 hour. His change in time is 1 hour. So this is the same thing as 5 kilometers per hour. So once again, we're only giving the magnitude here. This is a scalar quantity. If you want the vector, you have to do the north as well. Now, you might be saying, hey, in the previous video, we talked about things in terms of meters per second. Here, I give you kilometers, or \"kil-om-eters,\" depending on how you want to pronounce it, kilometers per hour. What if someone wanted it in meters per second, or what if I just wanted to understand how many meters he travels in a second? And there, it just becomes a unit conversion problem. And I figure it doesn't hurt to work on that right now. So if we wanted to do this to meters per second, how would we do it? Well, the first step is to think about how many meters we are traveling in an hour. So let's take that 5 kilometers per hour, and we want to convert it to meters. So I put meters in the numerator, and I put kilometers in the denominator. And the reason why I do that is because the kilometers are going to cancel out with the kilometers. And how many meters are there per kilometer? Well, there's 1,000 meters for every 1 kilometer. And I set this up right here so that the kilometers cancel out. So these two characters cancel out. And if you multiply, you get 5,000. So you have 5 times 1,000. So let me write this-- I'll do it in the same color-- 5 times 1,000. So I just multiplied the numbers. When you multiply something, you can switch around the order. Multiplication is commutative-- I always have trouble pronouncing that-- and associative. And then in the units, in the numerator, you have meters, and in the denominator, you have hours. Meters per hour. And so this is equal to 5,000 meters per hour. And you might say, hey, Sal, I know that 5 kilometers is the same thing as 5,000 meters. I could do that in my head. And you probably could. But this canceling out dimensions, or what's often called dimensional analysis, can get useful once you start doing really, really complicated things with less intuitive units than something But you should always do an intuitive gut check right here. You know that if you do 5 kilometers in an hour, that's a ton of meters. So you should get a larger number if you're talking about meters per hour. And now when we want to go to seconds, let's do an intuitive gut check. If something is traveling a certain amount in an hour, it should travel a much smaller amount in a second, or 1/3,600 of an hour, because that's how many seconds there are in an hour. So that's your gut check. We should get a smaller number than this when we want to say meters per second. But let's actually do it with the dimensional analysis. So we want to cancel out the hours, and we want to be left with seconds in the denominator. So the best way to cancel this hours in the denominator is by having hours in the numerator. So you have hours per second. So how many hours are there per second? Or another way to think about it, 1 hour, think about the larger unit, 1 hour is how many seconds? Well, you have 60 seconds per minute times 60 minutes per hour. The minutes cancel out. 60 times 60 is 3,600 seconds per hour. So you could say this is 3,600 seconds for every 1 hour, or if you flip them, you would get 1/3,600 hour per second, or hours per second, depending on how you want to do it. So 1 hour is the same thing as 3,600 seconds. And so now this hour cancels out with that hour, and then you multiply, or appropriately divide, the numbers right here. And you get this is equal to 5,000 over 3,600 meters per-- all you have left in the denominator here is second. Meters per second. And if we divide both the numerator and the denominator-- I could do this by hand, but just because this video's already getting a little bit long, let me get my trusty calculator out. I get my trusty calculator out just for the sake of time. 5,000 divided by 3,600, which would be really the same thing as 50 divided by 36, that is 1.3-- I'll just round it over here-- 1.39. So this is equal to 1.39 meters per second. So Shantanu was traveling quite slow in his car. Well, we knew that just by looking at this. 5 kilometers per hour, that's pretty much just letting the car roll pretty slowly." + }, + { + "Q": "A small question, I think I get how an increase in volume increases entropy, but now I understand a little more this concept... can somebody explain to me how come an increase in temperature increases this combination of states?", + "A": "When the temperature of a gas is increased, the molecules have higher kinetic energies. When the molecules move around faster they can configure themselves in more ways than they could have if they had less kinetic energy. More specifically, molecules in an ideal gas follow Maxwell-Boltzmann statistics which imply that the molecules velocity distribution broadens as the temperature of the system is increased.", + "video_name": "xJf6pHqLzs0", + "transcript": "I've now supplied you with two definitions of the state variable entropy. And it's S for entropy. The thermodynamic definition said that the change in entropy is equal to the heat added to the system divided by the temperature at which the heat is added. So obviously, if the temperature is changing while we add the heat, which is normally the case, we're going to have to do a little bit of calculus. And then you can view this as the mathematical, or the statistical, or the combinatorical definition of entropy. And this essentially says that entropy is equal to some constant times the natural log of the number of states the system can take on. And this is the case when all the states are equally probable, which is a pretty good assumption. If you have just a gazillion molecules that could have a gazillion gazillion states, you can assume they're all roughly equally likely. There's a slightly more involved definition if they had different probabilities, but we won't worry about that now. So given that we've seen these two definitions, it's a good time to introduce you to the second law of thermodynamics. And that's this. And it's a pretty simple law, but it explains a whole range of phenomena. It tells us that the change in entropy for the universe when any process is undergone is always greater than or equal to 0. So that tells us that when anything ever happens in the universe, the net effect is that there's more entropy in the universe itself. And this seems very deep, and it actually is. So let's see if we can apply it to see why it explains, or why it makes sense, relative to some examples. So let's say I have two reservoirs that are in contact with each other. So I have T1. And let's call this our hot reservoir. And then I have T2. I'll call this our cold reservoir. Well, we know from experience. What happens if I put a hot cup of water, and it's sharing a wall with a cold glass of water, or cold cube of water, what happens? Well, their temperatures equalize. If these are the same substance, we'll end up roughly in between, if they're in the same phase. So essentially, we have a transfer of heat from the hotter substance to the colder substance. So we have some heat, Q, that goes from the hotter substance to the colder substance. You don't see, in everyday reality, heat going from a colder substance to a hotter substance. If I put an ice cube in, let's say, some hot tea, you don't see the ice cube getting colder and the hot tea getting hotter. You see them both getting to some equal temperature, which essentially the T is giving heat to the ice cube. Now in this situation there are reservoirs, so I'm assuming that their temperatures stay constant. Which would only be the case if they were both infinite, which we know doesn't exist in the real world. In the real world, T1's temperature as it gave heat would go down, and T2's temperature would go up. But let's just see whether the second law of thermodynamics says that this should happen. So what's happening here? What's the net change in entropy for T1? So the second law of thermodynamics says that the change in entropy for the universe is greater than 0. But in this case, that's equal to the change in entropy for T1 plus the change in entropy for-- oh, I shouldn't-- instead of T1, let me call it just 1. For system 1, that's this hot system up here, plus the change in entropy for system 2. So what's the change in entropy for system 1? It loses Q1 at a high temperature. So this equals minus the heat given to the system is Q over some hot temperature T1. And then we have the heat being added to the system T2. So plus Q over T2. This is the change in entropy for the system 2, right? This guy loses the heat, and is at temperature 1, which is a higher temperature. This guy gains the heat, and he is at a temperature 2, which is a colder temperature. Now, is this going to be greater than 0? Let's think about it a little bit. If I divide-- let me rewrite this. So I can rearrange them, so that we can write this as Q over T2 minus this one. I'm just rearranging it. Minus Q over T1. Now, which number is bigger? T2 to T1? Well, T1 is bigger, right? This is bigger. Now, if I have a bigger number, bigger than this-- when we use the word bigger, you have to compare it to something. Now, T1 is bigger than this. We have the same number in the numerator in both cases, right? So if I take, let's say, 1 over some, let's say, 1/2 minus 1/3, we're going to be bigger than 0. This is a larger number than this number, because this has a bigger denominator. You're dividing by a larger number. That's a good way to think about it. You're dividing this Q by some number here to get something, and then you're subtracting this Q divided by a larger number. So this fraction is going to be a smaller absolute number. So this is going to be greater than 0. So that tells us the second law of thermodynamics, it verifies this observation we see in the real world, that heat will flow from the hot body to the cold body. Now, you might say, hey, Sal. I have a case that will show you that you are wrong. You could say, look. If I put an air conditioner in a room-- Let's say this is the room, and this is outside. You'll say, look what the air conditioner does. The room is already cold, and outside is already hot. But what the air conditioner does, is it makes the cold even colder, and it makes the hot even hotter. It takes some Q and it goes in that direction. Right? It takes heat from the cold room, and puts it out into the hot air. And you're saying, this defies the second law of thermodynamics. You have just disproved it. You deserve a Nobel Prize. And I would say to you, you're forgetting one small fact. This air conditioner inside here, it has some type of a compressor, some type of an engine, that's actively doing this. It's putting in work to make this happen. And this engine right here-- I'll do it in magenta-- it's also expelling some more heat. So let's call that Q of the engine. So if you wanted to figure out the total entropy created for the universe, it would be the entropy of the cold room plus the change in entropy for outside-- I'll call it outside, maybe I'll call this, for the room. So you might say, OK. This change in entropy for the room, it's giving away heat-- let's see the room is roughly at a constant temperature for that one millisecond we're looking at it. It's giving away some Q at some temperature T1. And then-- so that's a minus. And then this the outside is gaining some heat at some temperature T2. And so you'll immediately say, hey. This number right here is a smaller number than this one. Because the denominator is higher. So if you just look at this, this would be negative entropy, and you'd say hey, this defies the second law of thermodynamics. But what you have to throw in here is another notion. You have to throw in here the notion that the outside is also getting this heat from the engine over the outside temperature. And this term, I can guarantee you-- I'm not giving you numbers right now-- will make this whole expression positive. This term will turn the total net entropy to the universe to be positive. Now let's think a little bit how about what entropy is and what entropy isn't in terms of words. So when you take an intro chemistry class, the teacher often says, entropy equals disorder. Which is not incorrect. It is disorder, but you have to be very careful what we mean by disorder. Because the very next example that's often given is that they'll say, look. A clean room-- let's say your bedroom is clean, and then it And they'll say, look. The universe became more disordered. The dirty room has more disorder than the clean room. And this is not a case of entropy increase. So this is not a good example. Why is that? Because clean and dirty are just states of the room. Remember, entropy is a macro state variable. It's something you use to describe a system where you're not in the mood to sit there and tell me what exactly every particle is doing. And this is a macro variable that actually tells me how much time would it take for me to tell you what every particle is doing. It actually tells you how many states there are, or how much information I would have to give you to tell you the exact state. Now, when you have a clean room and a dirty room, these are two different states of the same room. If the room has the same temperature, and it has the same number of molecules in it and everything, then they have the same entropy. So clean to dirty, it's not more entropy. Now, for example, I could have a dirty, cold room. And let's say I were to go into that room and, you know, I work really hard to clean it up. And by doing so, I add a lot of heat to the system, and my sweat molecules drop all over the place, and so there's just more stuff in that room, and it's all warmed up to me-- so to a hot, clean room with sweat in it-- so it's got more stuff in here that can be configured in more ways, and because it's hot, every molecule in the room can take on more states, right? Because the average kinetic energy is up, so they can kind of explore the spaces of how many kinetic energies it can have. There's more potential energies that each molecule can take on. This is actually an increase in entropy. From a dirty, cold room to a hot, clean room. And this actually goes well with what we know. I mean, when I go into room and I start cleaning it, I am in putting heat into the room. And the universe is becoming more-- I guess we could say it's the entropy is increasing. So where does the term disorder apply? Well, let's take a situation where I take a ball. I take a ball, and it falls to the ground. And then it hits the ground. And there should have been a question that you've been asking all the time, since the first law of thermodynamics. So the ball hits the ground, right? It got thrown up, it had some potential energy at the top, then that all gets turned into kinetic energy and it hits the ground, and then it stops. And so your obvious question is, what happened to all that energy, right? Law of conservation of energy. Where did all of it go? It had all that kinetic energy right before it hit the ground, then it stopped. It seems like it disappeared. But it didn't disappear. So when the ball was falling, it had a bunch of-- you know, everything had a little bit of heat. But let's say the ground was reasonably ordered. The ground molecules were vibrating with some kinetic energy and potential energies. And then our ball molecules were also vibrating a little bit. But most of their motion was downwards, right? Most of the ball molecules' motion was downwards. Now, when it hits the ground, what happens-- let me show you the interface of the ball. So the ball molecules at the front of the ball are going to look like that. And there's a bunch of them. It's a solid. It will maybe be some type of lattice. And then it hits the ground. And when it hits the ground-- so the ground is another solid like that-- All right, we're looking at the microstate. What's going to happen? These guys are going to rub up against these guys, and they're going to transfer their-- what was downward kinetic energy, and a very ordered downward kinetic energy-- they're going to transfer it to these ground particles. And they're going to bump into the ground particles. And so when this guy bumps into that guy, he might start moving in that direction. This guy will start oscillating in that direction, and go back and forth like that. That guy might bounce off of this guy, and go in that direction, and bump into that guy, and go into that direction. And then, because that guy bumped here, this guy bumps here, and because this guy bumps here, this guy bumps over there. And so what you have is, what was relatively ordered motion, especially from the ball's point of view, when it starts rubbing up against these molecules of the ground, it starts making the kinetic energy, or their movement, go in all sorts of random directions. Right? This guy's going to make this guy go like that, and that guy go like that. And so when the movement is no longer ordered, if I have a lot of molecules-- let me do it in a different color-- if I have a lot of molecules, and they're all moving in the exact same direction, then my micro state looks like my macro state. The whole thing moves in that direction. Now, if I have a bunch of molecules, and they're all moving in random directions, my ball as a whole will be stationary. I could have the exact same amount of kinetic energy at the molecular level, but they're all going to be bouncing into each other. And in this case, we described the kinetic energy as internal energy, or we describe it as temperature, where temperature is the average kinetic energy. So in this case, when we talk about, the world is becoming more disordered, you think about the order of maybe the velocities or the energies of the molecules. Before they were reasonably ordered, the molecules-- they might have been vibrating a little bit, but they were mainly going down in the ball. But when they bump into the ground, all of a sudden they start vibrating in random directions a little bit more. And they make the ground vibrate in more random directions. So it makes-- at the microstate-- everything became just that much more disordered. Now there's an interesting question here. There is some probability you might think-- Look, this ball came down and hit the ground. Why doesn't the ball just-- isn't there some probability that if I have a ground, that these molecules just rearrange themselves in just the right way to just hit these ball molecules in just the right way? There's some probability, just from the random movement, that at get some second, all the ground molecules just hit the ball molecules just right to send the ball back up. And the answer is yes. There's actually some infinitesimally small chance that that happens. That you could have a ball that's sitting on the ground-- and this is interesting-- could have a ball that's sitting on the ground, and while you're looking, you'll probably have to wait a few gazillion years for it to happen, if it happens at all-- it could just randomly pop up. And there's some random, very small chance that these molecules just randomly vibrate in just the right way to be ordered for a second, and then the ball will pop up. But the probability of this happening, relative to everything else, is essentially 0. So when people talk about order and disorder, the disorder is increasing, because now these molecules are going in more random directions, and they can take on more potential states. And we saw that here. And you know, on some level, entropy seems something kind of magical, but on some level, it seems relatively common sense. In that video-- I think was the last video-- I had a case where I had a bunch of molecules, and then I had this extra space here, and then I removed the wall. And we saw that these molecules will-- we know, there's always some modules that are bouncing off this wall before, because we probably had some pressure associated with it. And then as soon as we remove that wall, the molecule that would have bounced there just keeps going. There's nothing to stop it from there. In that direction, there's a lot of stuff. It could bump into other molecules, and it could bumping into these walls. But in this direction, the odds of it bumping into everything is, especially for these leading molecules, is So it's going to expand to fill the container. So that's kind of common sense. But the neat thing is that the second law of thermodynamics, as we saw in that video, also says that this will happen. That the molecules will all expand to fill the container. And that the odds of this happening are very low. That they all come back and go into a ordered state. Now there is some chance, just from the random movements once they fill, that they all just happen to come back here. But it's a very, very small probability. And even more-- and I want to make this very clear-- S is a macro state. We never talk about the entropy for an individual molecule. If we know what an individual molecule is doing, we shouldn't be worrying about entropy. We should be worrying about the system as a whole. So even if we're looking at the system, if we're not looking directly at the molecules, we won't even know that this actually happened. All we can do is look at the statistical properties of the molecules. How many molecules they are, what their temperature is, all their macro dynamics, the pressure, and say, you know what? A box that has these molecules has more state than a smaller box, than the box when we had the wall there. Even if, by chance, all of the molecules happened to be collecting over there, we wouldn't know that that happened, because we're not looking at the micro states. And that's a really important thing to consider. When someone says that a dirty room has a higher entropy than a clean room, they're looking at the micro states. And entropy essentially is a macro state variable. You could just say that a room has a certain amount of entropy. So entropy is associated with the room, and it's only useful when you really don't know exactly what's going on in the room. You just have a general sense of how much stuff there is in the room, what's the temperature of the room, what's the pressure in the room. Just the general macro properties. And then entropy will essentially tell us how many possible micro states that macro system can actually have. Or how much information-- and there's a notion of information entropy-- how much information would I have to give you to tell you what the exact micro state is of a system at that point in time. Hopefully you found this discussion a little bit useful, and it clears up some misconceptions about entropy, and gives you a little bit more intuition about what it actually is. See you in the next video." + }, + { + "Q": "what is 700.75 compared to 700.756", + "A": "700.75 is less because 700.756 has 6 thousandths more", + "video_name": "JJawhaMqaXg", + "transcript": "We're asked to order the following numbers from least to greatest, and I encourage you to pause this video and try to think of it on your own. Order these numbers from least to greatest. Well, let's work through it together now. So, none of these numbers have any places, any value to the left of the decimal point. They have no ones here. 0 ones, 0 ones, 0 ones, 0 ones, 0 ones. So let me then go to the next decimal place to the right. So I'm starting with the largest decimal places, and then I'm going to successively smaller decimal places. So I'll go to the tenths place. So this one right over here has 0 tenths. This has 0 tenths. This has 0 tenths. This one has 0 tenths. This one has 7 tenths, so this one actually has tenths. This has seven of them, so I'm going to leave this one here as the greatest. We're ordering from least to greatest. Now let's move to the hundredths place. So this one, this one, this one, this one all have 0 tenths. Let's look at the hundredths place. This has 7 hundredths. This has 7 hundredths. This one has 7 hundredths. This has no hundredths as well. This has 0 hundredths, so this one has neither tenths nor hundredths. So this one is going to be the smallest. This one has no tenths, no hundredths. This one actually has tenths. All of these-- these three in the middle-- have no tenths, but they have some hundredths, and they all have the same number of hundredths. 7 hundredths, 7 hundredths, 7 hundredths. So now let's look at the thousandths place. This one has 9 thousandths. This one has 0 thousandths. And this one has 0 thousandths as well. So out of these three, this one is the largest, because this one actually had some thousandths out of these three. Now let's go look at-- we have to pick between these two. Both of these have no tenths. Both of them have exactly 7 hundredths. Both of them have no thousandths, but this one has 9 ten thousandths while this one has no ten thousandths. So this one is less than that one. And now I'm done. I think I have ordered the numbers from least to greatest. And the key here is go to the place value that's most significant-- I guess you could say-- that has the most value. So that was the ones place. Compare them on that. Then go to successively smaller place values. Keep going to the right, and keep comparing them, and then you'll be able to order them." + }, + { + "Q": "so if you are at x and you take away 1/3x you have 2/3x?", + "A": "yes 1x - 1/3 x = 3/3 x - 1/3 x = 2/3 x", + "video_name": "wo7DSaPP8hQ", + "transcript": "We're told that for the past few months, Old Maple Farms has grown about 1,000 more apples than their chief rival in the region, River Orchards. Due to cold weather this year, the harvests at both farms were down by about a third. However, both farms made up for some of the shortfall by purchasing equal quantities of apples from farms in neighboring states. What can you say about the number of apples available at each farm? Does one farm have more than the other, or do they have the same amount? How do I know? So let's define some variables here. Let's let M be equal to number of apples at Maple Farms. And then who's the other guy? River Orchards. So let's let R be equal to the number of apples at River Orchards. So this first sentence, they say-- let me do this in a different color-- they say for the past few years, Old Maple Farms has grown about 1,000 more apples than their chief rival in the region, River Orchards. So we could say, hey, Maple is approximately Old River, or M is approximately River plus 1,000. Or since we don't know the exact amount-- it says it's about 1,000 more, so we don't know it's exactly 1,000 more-- we can just say that in a normal year, Old Maple Farms, which we denote by M, has a larger amount of apples than River Orchard. So in a normal year, M is greater than R, right? It has about 1,000 more apples than Old Maple Farms. Now, they say due to cold weather this year-- so let's talk about this year now-- the harvests at both farms were down about a third. So this isn't a normal year. Let's talk about what's going to happen this year. In this year, each of these characters are going to be down by 1/3. Now if I go down by 1/3, that's the same thing as being 2/3 of what I was before. Let me do an example. If I'm at x, and I take away 1/3x, I'm left with 2/3x. So going down by 1/3 is the same thing as multiplying the quantity by 2/3. So if we multiply each of these quantities by 2/3, we can still hold this inequality, because we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number. If we were multiplying by a negative number, we would have to swap the inequality. So we can multiply both sides of this by 2/3. So 2/3 of M is still going to be greater than 2/3 of R. And you could even draw that in a number line if you like. Let's do this in a number line. This all might be a little intuitive for you, and if it is, I apologize, but if it's not, it never hurts. So that's 0 on our number line. So in a normal year, M is has 1,000 more than R. So in a normal year, M might be over here and maybe R is over here. I don't know, let's say R is over there. Now, if we take 2/3 of M, that's going to stick us some place around, oh, I don't know, 2/3 is right about there. So this is M-- let me write this-- this is 2/3 M. And what's 2/3 of R going to be? Well, if you take 2/3 of this, you get to right about there, that is 2/3R. So you can see, 2/3R is still less than 2/3M, or 2/3M is greater than 2/3R. Now, they say both farms made up for some of the shortfall by purchasing equal quantities of apples from farms in neighboring states. So let's let a be equal to the quantity of apples both purchased. So they're telling us that they both purchased the same amount. So we could add a to both sides of this equation and it will not change the inequality. As long as you add or subtract the same value to both sides, it will not change the inequality. So if you add a to both sides, you have a plus 2/3M is a greater than 2/3R plus a. This is the amount that Old Maple Farms has after purchasing the apples, and this is the amount that River Orchards has. So after everything is said and done, Old Maple Farms still has more apples, and you can see that here. Maple Farms, a normal year, this year they only had 2/3 of the production, but then they purchased a apples. So let's say a is about, let's say that a is that many apples, so they got back to their normal amount. So let's say they got back to their normal amount. So that's how many apples they purchased, so he got back to M. Now, if R, if River Orchards also purchased a apples, that same distance, a, if you go along here gets you to right about over there. So once again, this is-- let me do it a little bit different, because I don't like it overlapping, so let me So let's say this guy, M-- I keep forgetting their names-- Old Maple Farms purchases a apples, gets them that far. So that's a apples. But River Orchards also purchases a apples, so let's add that same amount. I'm just going to copy and paste it so it's the exact same amount. So River Orchards also purchases a, so it also purchases that same amount. So when all is said and done, River Orchards is going to have this many apples in the year that they had less production but they went and purchased it. So this, right here, is-- this value right here is 2/3R plus a. That's what River Orchards has. And then Old Maple Farms has this value right here, which is 2/3M plus a. Everything said and done, Old Maple Farms still has more apples." + }, + { + "Q": "how do you find the sin cos or tan of an angle without a triangle or calculator?", + "A": "(1) If the angle is 30, 45, or 60 degrees, then the ratios can be figured out exactly. (The special right triangles .) (2) You can use a Taylor series or the CORDIC algorithm to compute a specific value. Neither is something you want to attempt manually. (3) Back in the days... we used printed tables to look them up. Presumably those tables were computed using one of the techniques in (2).", + "video_name": "l5VbdqRjTXc", + "transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly solving for a, we could just multiply both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. I say approximately because I rounded it down. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse. So this side WY is the hypotenuse. And so what trigonometric ratios-- or we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that XY is exactly 5 and we don't have to deal with this approximation, let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well we see from SohCahToa cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees is equal to the length of the adjacent side, which is 5 over the length of the hypotenuse, which has a length of b. And then we can try to solve for b. You multiply both sides times b, you're left with b times cosine of 65 degrees is equal to 5. And then to solve for b, you could divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing-- we have to figure it out what our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees. And we're left with b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is 5 divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that 5 squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle W right over here. So I'll give you a few seconds to think about what the measure of angle W is. Well here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is-- well we can simplify the left-hand side right over here. 65 plus 90 is 155. So angle W plus 155 degrees is equal to 180 degrees. And then we get angle W-- if we subtract 155 from both sides-- angle W is equal to 25 degrees. And we are done solving the right triangle shown below." + }, + { + "Q": "When a boy jumps from bus there is a danger for him to fall\nA)in the direction of motion\nB)towards the moving bus\nC)opposite in the direction of motion\nD)away from the bus", + "A": "At the moment that he jumps from the bus, he would have velocity pushing him forward in the direction of the motion of the bus at the same magnitude of the velocity of the bus, and (assuming he jumps straight forward), velocity straight outwards perpendicular to the bus. This means that he would fall in both the direction of the motion and away from the bus. Hope this helps, and please correct if im wrong, as i often am :)", + "video_name": "jmSWImPs6fQ", + "transcript": "- [Instructor] Let's talk about how to handle a horizontally launched projectile problem. These, technically speaking, if you already know how to do projectile problems, there is nothing new, except that there's one aspect of these problems that people get stumped by all of the time. So I'm gonna show you what that is in a minute so that you don't fall into the same trap. What we mean by a horizontally launched projectile is any object that gets launched in a completely horizontal velocity to start with. So if something is launched off of a cliff, let's say, in this straight horizontal direction with no vertical component to start with, then it's a horizontally launched projectile. What could that be? I mean a boring example, it's just a ball rolling off of a table. If you just roll the ball off of the table, then the velocity the ball has to start off with, if the table's flat and horizontal, the velocity of the ball initially would just be horizontal. So if the initial velocity of the object for a projectile is completely horizontal, then that object is a horizontally launched projectile. A more exciting example. People do crazy stuff. Let's say this person is gonna cliff dive or base jump, and they're gonna be like \"whoa, let's do this.\" We're gonna do this, they're pumped up. They're gonna run but they don't jump off the cliff, they just run straight off of the cliff 'cause they're kind of nervous. Let's say they run off of this cliff with five meters per second of initial velocity, straight off the cliff. And let's say they're completely crazy, let's say this cliff is 30 meters tall. So that's like over 90 feet. That is kind of crazy. So 30 meters tall, they launch, they fly through the air, there's water down here, so they initially went this way, and they start to fall down, and they do something like pschhh, and then they splash in the water, hopefully they don't hit any boats or fish down here. That fish already looks like he got hit. He or she. Alright, fish over here, person splashed into the water. We want to know, here's the question you might get asked: how far did this person go horizontally before striking the water? This is a classic problem, gets asked all the time. And if you were a cliff diver, I mean don't try this at home, but if you were a professional cliff diver you might want to know for this cliff high and this speed how fast do I have to run in order to avoid maybe the rocky shore right here that you might want to avoid. Maybe there's this nasty craggy cliff bottom here that you can't fall on. So how fast would I have to run in order to make it past that? Alright, so conceptually what's happening here, the same thing that happens for any projectile problem, the horizontal direction is happening independently of the vertical direction. And what I mean by that is that the horizontal velocity evolves independent to the vertical velocity. Let me get the velocity this color. So say the vertical velocity, or the vertical direction is pink, horizontal direction is green. This vertical velocity is gonna be changing but this horizontal velocity is just gonna remain the same. These do not influence each other. In other words, this horizontal velocity started at five, the person's always gonna have five meters per second of horizontal velocity. So this horizontal velocity is always gonna be five meters per second. The whole trip, assuming this person really is a freely flying projectile, assuming that there is no jet pack to propel them forward and no air resistance. This person's always gonna have five meters per second of horizontal velocity up onto the point right when they splash in the water, and then at that point there's forces from the water that influence this acceleration in various ways that we're not gonna consider. How about vertically? Vertically this person starts with no initial velocity. So this person just ran horizontally straight off the cliff and then they start to gain velocity. So they're gonna gain vertical velocity downward and maybe more vertical velocity because gravity keeps pulling, and then even more, this might go off the screen but it's gonna be really big. So a lot of vertical velocity, this should keep getting bigger and bigger and bigger because gravity's influencing this vertical direction but not the horizontal direction. So how do we solve this with math? Let's write down what we know. What we know is that horizontally this person started off with an initial velocity. V initial in the x, I could have written i for initial, but I wrote zero for v naught in the x, it still means initial velocity is five meters per second. And we don't know anything else in the x direction. You might think 30 meters is the displacement in the x direction, but that's a vertical distance. This is not telling us anything about this horizontal distance. This horizontal distance or displacement is what we want to know. This horizontal displacement in the x direction, that's what we want to solve for, so we're gonna declare our ignorance, write that here. We don't know how to find it but we want to know that we do want to find so I'm gonna write it there. How about in the y direction, what do we know? We know that the, alright, now we're gonna use this 30. You might want to say that delta y is positive 30 but you would be wrong, and the reason is, this person fell downward 30 meters. Think about it. They started at the top of the cliff, ended at the bottom of the cliff. It means this person is going to end up below where they started, 30 meters below where they started. So this has to be negative 30 meters for the displacement, assuming you're treating downward as negative which is typically the convention shows that downward is negative and leftward is negative. So if you choose downward as negative, this has to be a negative displacement. What else do we know vertically? Well, for a freely flying object we know that the acceleration vertically is always gonna be negative 9.8 meters per second squared, assuming downward is negative. Now, here's the point where people get stumped, and here's the part where people make a mistake. They want to say that the initial velocity in the y direction is five meters per second. I mean people are just dying to stick these five meters per second into here because that's the velocity that you were given. But this was a horizontal velocity. That's why this is called horizontally launched projectile motion, not vertically launched projectile motion. So think about it. The initial velocity in the vertical direction here was zero, there was no initial vertical velocity. This person was not launched vertically up or vertically down, this person was just launched straight horizontally, and so the initial velocity in the vertical direction is just zero. People don't like that. They're like \"hold on a minute.\" They're like, this person is gonna start gaining, alright, this person is gonna start gaining velocity right when they leave the cliff, this starts getting bigger and bigger and bigger in the downward direction. But that's after you leave the cliff. We're talking about right as you leave the cliff. That moment you left the cliff there was only horizontal velocity, which means you started with no initial vertical velocity. So this is the part people get confused by because this is not given to you explicitly in the problem. The problem won't say, \"Find the distance for a cliff diver \"assuming the initial velocity in the y direction was zero.\" Now, they're just gonna say, \"A cliff diver ran horizontally off of a cliff. \"Find this stuff.\" And you're just gonna have to know that okay, if I run off of a cliff horizontally or something gets shot horizontally, that means there is no vertical velocity to start with, I'm gonna have to plug this initial velocity in the y direction as zero. So that's the trick. Don't fall for it now you know how to deal with it. So we want to solve for displacement in the x direction, but how many variables we know in the y direction? I mean we know all of this. But we can't use this to solve directly for the displacement in the x direction. We need to use this to solve for the time because the time is gonna be the same for the x direction and the y direction. So I find the time I can plug back in over to there, because think about it, the time it takes for this trip is gonna be the time it takes for this trip. It doesn't matter whether I call it the x direction or y direction, time is the same for both directions. In other words, the time it takes for this displacement of negative 30 is gonna be the time it takes for this displacement of whatever this is that we're gonna find. So let's solve for the time. Now, how will we do that? Think about it. We know the displacement, we know the acceleration, we know the initial velocity, and we know the time. But we don't know the final velocity and we're not asked to find the final velocity, we don't want to know it. So let's use a formula that doesn't involve the final velocity and that would look like this. So if we use delta y equals v initial in the y direction times time plus one half acceleration in the y direction times time squared. Alright, now we can plug in values. My displacement in the y direction is negative 30. My initial velocity in the y direction is zero. This is where it would happen, this is where the mistake would happen, people just really want to plug that five in over here. But don't do it, it's a trap. So, zero times t is just zero so that whole term is zero. Plus one half, the acceleration is negative 9.8 meters per second squared. And then times t squared, alright, now I can solve for t. I'm gonna solve for t, and then I'd have to take the square root of both sides because it's t squared, and what would I get? I'd have to multiply both sides by two. So I get negative 30 meters times two, and then I have to divide both sides by negative 9.8 meters per second squared, equals, notice if you would have forgotten this negative up here for negative 30, you come down here, this would be a positive up top. You'd have a negative on the bottom. You'd have to plug this in, you'd have to try to take the square root of a negative number. Your calculator would have been all like, \"I don't know what that means,\" and you're gonna be like, \"Er, am I stuck?\" So you'd start coming back here probably and be like, \"Let's just make stuff positive and see if that works.\" It would work because look at these negatives canceled but it's best to just know what you're talking about in the first place. So be careful: plug in your negatives and things will work out alright. So if you solve this you get that the time it took is 2.47 seconds. It's actually a long time. It might seem like you're falling for a long time sometimes when you're like jumping off of a table, jumping off of a trampoline, but it's usually like a fraction of a second. This is actually a long time, two and a half seconds of free fall's a long time. So we could take this, that's how long it took to displace by 30 meters vertically, but that's gonna be how long it took to displace this horizontal direction. We can use the same formula. We can say that well, if delta x equals v initial in the x direction, I'm just using the same formula but in the x direction, plus one half ax t squared. So the same formula as this just in the x direction. Delta x is just dx, we already gave that a name, so let's just call this dx. So I'm gonna scooch this equation over here. Dx is delta x, that equals the initial velocity in the x direction, that's five. Alright, this is really five. In the x direction the initial velocity really was five meters per second. How about the initial time? Oh sorry, the time, there is no initial time. The time here was 2.47 seconds. This was the time interval. The time between when the person jumped, or ran off the cliff, and when the person splashed in the water was 2.4, let me erase this, 2.47 seconds. So 2.47 seconds, and this comes over here. How about this ax? This ax is zero. Remember there's nothing compelling this person to start accelerating in x direction. If they've got no jet pack, there is no air resistance, there is no reason this person is gonna accelerate horizontally, they maintain the same velocity the whole way. So what do we get? If we solve this for dx, we'd get that dx is about 12.4, I believe. Let's see, I calculated this. 12.4-ish meters. Okay, so if these rocks down here extend more than 12 meters, you definitely don't want to do this. I mean if it's even close you probably wouldn't want do this. In fact, just for safety don't try this at home, leave this to professional cliff divers. I'm just saying if you were one and you wanted to calculate how far you'd make it, this is how you would do it. So, long story short, the way you do this problem and the mistakes you would want to avoid are: make sure you're plugging your negative displacement because you fell downward, but the big one is make sure you know that the initial vertical velocity is zero because there is only horizontal velocity to start with. That's not gonna be given explicitly, you're just gonna have to provide that on your own and your own knowledge of physics." + }, + { + "Q": "can someone please help me solve y-3x=2\ny=-2x+7", + "A": "So I you put y=-2x+7 in the first eq. you get -2x+7-3x=2 put the x s together and the # s together you get: -2x-3x = 2-7 -5x=-5 then divide by -5 you get x=1 Check your answer by plugging in the 1 for x and y= 5", + "video_name": "2VeqrZ_PMiY", + "transcript": "Use substitution to solve for x and y. And they give us a system of equations here. y is equal to negative 5x plus 8 and 10x plus 2y is equal to negative 2. So they've set it up for us pretty well. They already have y explicitly solved for up here. So they tell us, this first constraint tells us that y must be equal to negative 5x plus 8. So when we go to the second constraint here, every time we see a y, we say, well, the first constraint tells us that y must be equal to negative 5x plus 8. So everywhere we see a y, we can substitute it with negative 5x plus 8. Because that's what the first constraint tells us. y is equal to that. I don't want to be repetitive, but I really want you to internalize that's all it's saying. y is that. So every time we see a y in the second constraint, we can substitute it with that. So let's do it. So the second equation over here is 10x plus 2. And instead of writing a y there, and I've said it multiple times already, we can write a negative 5x plus 8. The first constraint tells us that's what y is. So negative 5x plus 8 is equal to negative 2. Now, we have one equation with one unknown. We can just solve for x. We have 10x plus. So we can multiply it. We can distribute this 2 onto both of these terms. So we have 2 times negative 5x is negative 10x. And then 2 times 8 is 16. So plus 16 is equal to negative 2. Now we have 10x minus 10x. Those guys cancel out. 10x minus 10x is equal to 0. So these guys cancel out. And we're just left with 16 equals negative 2, which is crazy. We know that 16 does not equal negative 2. This is an inconsistent result. And that's because these two lines actually don't intersect. And we could see that by actually graphing these lines. Whenever you get something like some number equalling some other number that they're clearly not equal to, that means it's an inconsistent result, It's an inconsistent system, and that these lines actually don't intersect. So let me just graph these just to make it clear. This first equation is already in slope y-intercept form. So it looks something like this. That's our x-axis. This is our y-axis. And it's negative 5x plus 8, so 1, 2, 3, 4, 5, 6, 7, 8. And then it has a very steep downward slope. Every time you move forward 1, you have to go down 5. So it looks something like that. That's this first equation right over there. The second equation, let me rewrite it in slope y-intercept form. So it's 10x plus 2y is equal to negative 2. Let's subtract 10x from both sides. You get 2y is equal to negative 10x minus 2. Let's divide both sides by 2. You get y is equal to negative 5x, negative 5x minus 1. So it's y-intercept is negative 1. It's right over there. And it has the same slope as this first line. So it looks like this. It's parallel. It's just shifted down a bit. So it just looks like that. So they're parallel lines. They have the same slope, different y-intercepts. We get an inconsistent result. They don't intersect. And the telltale sign of that, when you're doing it algebraically, is you get something wacky like this. This is why it's called inconsistent. It's not consistent for 16 to be equal to negative 2. These don't intersect. There's no solution to both of these constraints, no x and y that satisfies both of them." + }, + { + "Q": "I have literally been looking for over a week on how to do this through the quadratic formula and completing the square but I just don't get it.. I can't logicize this no matter how hard i try...", + "A": "If you just substitute the values into the quadratic formula then you will get the solutions for the equation(what x will equal to make the equation equal zero) is you do it on a calculator, then the answer may be an irrational number because the calculator tries to put the roots into decimal notation so it may be easier to do it by hand", + "video_name": "TV5kDqiJ1Os", + "transcript": "We're asked to complete the square to solve 4x squared plus 40x minus 300 is equal to 0. So let me just rewrite it. So 4x squared plus 40x minus 300 is equal to 0. So just as a first step here, I don't like having this 4 out front as a coefficient on the x squared term. I'd prefer if that was a 1. So let's just divide both sides of this equation by 4. So let's just divide everything by 4. So this divided by 4, this divided by 4, that divided by 4, and the 0 divided by 4. Just dividing both sides by 4. So this will simplify to x squared plus 10x. And I can obviously do that, because as long as whatever I do to the left hand side, I also do the right hand side, that will make the equality continue to be valid. So that's why I can do that. So 40 divided by 4 is 10x. And then 300 divided by 4 is what? That is 75. Let me verify that. 4 goes into 30 seven times. 7 times 4 is 28. You subtract, you get a remainder of 2. Bring down the 0. 4 goes into 20 five times. 5 times 4 is 20. Subtract zero. So it goes 75 times. This is minus 75 is equal to 0. And right when you look at this, just the way it's written, you might try to factor this in some way. But it's pretty clear this is not a complete square, or this is not a perfect square trinomial. Because if you look at this term right here, this 10, half of this 10 is 5. And 5 squared is not 75. So this is not a perfect square. So what we want to do is somehow turn whatever we have on the left hand side into a perfect square. And I'm going to start out by kind of getting this 75 out You'll sometimes see it where people leave the 75 on the left hand side. I'm going to put on the right hand side just so it kind of clears things up a little bit. So let's add 75 to both sides to get rid of the 75 from the left hand side of the equation. And so we get x squared plus 10x, and then negative 75 plus 75. Those guys cancel out. And I'm going to leave some space here, because we're going to add something here to complete the square that is equal to 75. So all I did is add 75 to both sides of this equation. Now, in this step, this is really the meat of completing the square. I want to add something to both sides of this equation. I can't add to only one side of the equation. So I want to add something to both sides of this equation so that this left hand side becomes a perfect square. And the way we can do that, and saw this in the last video where we constructed a perfect square trinomial, is that this last term-- or I should say, what we see on the left hand side, not the last term, this expression on the left hand side, it will be a perfect square if we have a constant term that is the square of half of the coefficient on the first degree So the coefficient here is 10. Half of 10 is 5. 5 squared is 25. So I'm going to add 25 to the left hand side. And of course, in order to maintain the equality, anything I do the left hand side, I also have to do to the right hand side. And now we see that this is a perfect square. We say, hey, what two numbers if I add them I get 10 and when I multiply them I get 25? Well, that's 5 and 5. So when we factor this, what we see on the left hand side simplifies to, this is x plus 5 squared. x plus 5 times x plus 5. And you can look at the videos on factoring if you find that confusing. Or you could look at the last video on constructing perfect square trinomials. I encourage you to square this and see that you get exactly this. And this will be equal to 75 plus 25, which is equal to 100. And so now we're saying that something squared is equal to 100. So really, this is something right over here-- if I say something squared is equal to 100, that means that that something is one of the square roots of 100. And we know that 100 has two square roots. It has positive 10 and it has negative 10. So we could say that x plus 5, the something that we were squaring, that must be one of the square roots of 100. So that must be equal to the plus or minus square root of 100, or plus or minus 10. Or we could separate it out. We could say that x plus 5 is equal to 10, or x plus 5 is equal to negative 10. On this side right here, I can just subtract 5 from both sides of this equation and I would get-- I'll just write it out. Subtracting 5 from both sides, I get x is equal to 5. And over here, I could subtract 5 from both sides again-- I subtracted 5 in both cases-- subtract 5 again and I can get x is equal to negative 15. So those are my two solutions that I got to solve this equation. We can verify that they actually work, and I'll do that in blue. So let's try with 5. I'll just do one of them. I'll leave the other one for you. I'll leave the other one for you to verify that it works. So 4 times x squared. So 4 times 25 plus 40 times 5 minus 300 needs to be equal to 0. 4 times 25 is 100. 40 times 5 is 200. We're going to subtract that 300. 100 plus 200 minus 300, that definitely equals 0. So x equals 5 worked. And I think you'll find that x equals negative 15 will also work when you substitute it into this right over here." + }, + { + "Q": "Who is actually learning things from this video? I know I am.", + "A": "It depends on your skill level, but yes i am sure lots of people are learning from this. Khan Academy has very good videos :)", + "video_name": "Ye13MIPv6n0", + "transcript": "So we have our scale again. And we've got some masses on the left hand side and some masses on the right hand side. And we see that our scale is balanced. We have the same total mass on the left hand side that we have on the right hand side. Instead of labeling the mystery masses as question mark, I've labeled them all x. And since they all have an x on it, we know that each of these have the same mass. But what I'm curious about is, what is that mass? What is the mass of each of these mystery masses, I guess we could say? And so I'll let think about that for a second. How would you figure out what this x value actually is? How many kilograms is the mass of each of these things? What could you do to either one or both sides of this scale? I'll give you a few seconds to think about that. So you might be tempted to say, well if I could end up with just one mystery mass on the left hand side, and if I keep my scale balanced, then that thing's going to be equal to whatever I have on the right hand side. And that part would actually be a true statement. But then to get only one of these mystery masses on the left hand side, you might say, well why don't I just remove two of them? You might just say, well why don't I just remove-- let me do it a good color for removing-- why don't I just remove that one and that one? And then I'll just be left with that right over there. But if you just removed these two, then the left hand side is going to become lighter or it's going to have a lower mass than the right hand side. So it's going to move up and the right hand side is going to move down. And then you might say, OK, I understand. Whatever I have to do to the left hand side, I have to do to the right hand side in order to keep my scale balanced. So you might say, well why don't I remove two of these mystery masses from the right hand side? But that's a problem too because you don't know what this mystery mass is. You could try to remove two from this, but how many of these blocks represent a mystery mass? We actually don't know. But you might then say, well let's see, I've got three of these things here. If I essentially multiply what I have here by 1/3 or if I only leave a 1/3 of the stuff here, and if I only leave a 1/3 of the stuff here, then the scale should be balanced. If this has the total mass as this, then 1/3 of this total mass is going to be the same thing as 1/3 of that total mass. So let's just keep only 1/3 of this here. So that's the equivalent to multiplying by 1/3. So if we're only going to keep 1/3 there, we're going to be left with only one of the masses. And if we only keep 1/3 here, let's see, we have one, two, three, four, five, six, seven, eight, nine masses. If we multiply this by 1/3, or if we only keep 1/3 of it there, 1/3 times 9 is 3. So we're going to remove these . And so we have 1/3 of what we originally had on the right hand side and 1/3 of what we originally had on the left hand side. And they will be balanced because we took 1/3 of the same total masses. And so what you're left with is just one of these mystery masses, this x thing right over here, whatever x might be. And you have three kilograms on the right hand side. And so you can make the conclusion, and the whole time you kept this thing balanced, that x is equal to 3." + }, + { + "Q": "where are some easy practice questions", + "A": "Go to the left tab and you ll find a practice skill.", + "video_name": "ory05j2jgBM", + "transcript": "- [Voiceover] What I wanna do in this video is compare the fractions 3/4 and 4/5, and I wanna do this visually. So what I'm gonna do is I'm gonna have two copies of the same whole, so let me just draw that, but I'm gonna divide the first one, so this is one whole right over here, this rectangle, when we draw the whole thing. So this is a whole, and right below that, we have the same whole. We have a rectangle of exactly the same size. Now you might notice that I've divided them into a different number of equal sections. In the top one, I've divided it into four equal sections because I am concerned with fourths so I've divided this top whole into fourths and I've divided this bottom whole, or this bottom bar or this bottom rectangle, into fifths, or five equal sections. So let's think about what 3/4 represent. So that's gonna be one of the fourths, right over here, two of the fourths, and then three of the fourths. And what is 4/5 going to be? Well, 4/5 is going to be one fifth, two fifths, three fifths, and four fifths. So when you look at them visually, remember, we're taking fractions of the same whole. This is 3/4 of that rectangle, this is 4/5 of a same-sized rectangle. It wouldn't make any sense if you're doing it for different shapes or different sized rectangles. We just divided them into different sections and you see that if you have four of the fifths, that that is going to be more than three of the fourths, and so 4/5 is greater than 3/4 or you could say 3/4 is less than 4/5, or any way you wanna think about it. The symbol you wanna use always opens to the larger number. 4/5 is larger than 3/4, so the large end of our symbol is facing the 4/5, so we would say 3/4 is less than 4/5." + }, + { + "Q": "what omega \"w\" '\"signifies\" in simple harmonic motion", + "A": "it represents angular velocity", + "video_name": "oqBHBO8cqLI", + "transcript": "And if you were covering your eyes because you didn't want to see calculus, I think you can open your eyes again. There shouldn't be any significant displays of calculus in this video. But just to review what we went over, we just said, OK if we have a spring-- and I drew it vertically this time-- but pretend like there's no gravity, or maybe pretend like we're viewing-- we're looking at the top of a table, because we don't want to look at the effect of We just want to look at a spring by itself. So this could be in deep space, or something else. But we're not thinking about gravity. But I drew it vertically just so that we can get more intuition for this curve. Well, we started off saying is if I have a spring and 0-- x equals 0 is kind of the natural resting point of the spring, if I just let this mass-- if I didn't pull on the spring at all. But I have a mass attached to the spring, and if I were to stretch the spring to point A, we said, well what happens? Well, it starts with very little velocity, but there's a restorative force, that's going to be pulling it back towards this position. So that force will accelerate the mass, accelerate the mass, accelerate the mass, until it gets right here. And then it'll have a lot of velocity here, but then it'll start decelerating. And then it'll decelerate, decelerate, decelerate. Its velocity will stop, and it'll come back up. And if we drew this as a function of time, this is what happens. It starts moving very slowly, accelerates. At this point, at x equals 0, it has its maximum speed. So the rate of change of velocity-- or the rate of change of position is fastest. And we can see the slope is very fast right here. And then, we start slowing down again, slowing down, until we get back to the spot of A. And then we keep going up and down, up and down, like that. And we showed that actually, the equation for the mass's position as a function of time is x of t-- and we used a little bit of differential equations to prove it. But this equation-- not that I recommend that you memorize anything-- but this is a pretty useful equation to memorize. Because you can use it to pretty much figure out anything-- about the position, or of the mass at any given time, or the frequency of this oscillatory motion, or anything else. Even the velocity, if you know a little bit of calculus, you can figure out the velocity at anytime, of the object. And that's pretty neat. So what can we do now? Well, let's try to figure out the period of this oscillating system. And just so you know-- I know I put the label harmonic motion on all of these-- this is simple harmonic motion. Simple harmonic motion is something that can be described by a trigonometric function like this. And it just oscillates back and forth, back and forth. And so, what we're doing is harmonic motion. And now, let's figure out what this period is. Remember we said that after T seconds, it gets back to its original position, and then after another T seconds, it gets back to its original position. Let's figure out with this T is. And that's essentially its period, right? What's the period of a function? It's how long it takes to get back to your starting point. Or how long it takes for the whole cycle to happen once. So what is this T? So let me ask you a question. What are all the points-- that if this is a cosine function, right? What are all of the points at which cosine is equal to 1? Or this function would be equal to A, right? Because whenever cosine is equal to 1, this whole function is equal to A. And it's these points. Well cosine is equal to 1 when-- so, theta-- let's say, when is cosine of theta equal to 1? So, at what angles is this true? Well it's true at theta is equal to 0, right? Cosine of 0 is 1. Cosine of 2 pi is also 1, right? We could just keep going around that unit circle. You should watch the unit circle video if this makes no sense to you. Or the graphing trig functions. It's also true at 4 pi. Really, any multiple of 2 pi, this is true. Right? Cosine of that angle is equal to 1. So the same thing is true. This function, x of t, is equal to A at what points? x of t is equal to A whenever this expression-- within the cosines-- whenever this expression is equal to 0, 2 pi, 4 pi, et cetera. And this first time that it cycles, right, from 0 to 2 pi-- from 0 to T, that'll be at 2 pi, right? So this whole expression will equal A, when k-- and that's these points, right? That's when this function is equal to A. It'll happen again over here someplace. When this little internal expression is equal to 2 pi, or really any multiple of 2 pi. So we could say, so x of t is equal to A when the square root of k over m times t, is equal to 2 pi. Or another way of thinking about it, is let's multiply both sides of this equation times the inverse of the square root of k over m. And you get, t is equal to 2 pi times the square root-- and it's going to be the inverse of this, right? Of m over k. And there we have the period of this function. This is going to be equal to 2 pi times the square root of m over k. So if someone tells you, well I have a spring that I'm going to pull from some-- I'm going to stretch it, or compress it a little bit, then I let go-- what is the period? How long does it take for the spring to go back to its It'll keep doing that, as we have no friction, or no gravity, or any air resistance, or Air resistance really is just a form of friction. You could immediately-- if you memorize this formula, although you should know where it comes from-- you could immediately say, well I know how long the period is. It's 2 pi times m over k. That's how long it's going to take the spring to get back-- to complete one cycle. And then what about the frequency? If you wanted to know cycles per second, well that's just the inverse of the period, right? So if I wanted to know the frequency, that equals 1 over the period, right? Period is given in seconds per cycle. So frequency is cycles per second, and this is seconds per cycle. So frequency is just going to be 1 over this. Which is 1 over 2 pi times the square root of k over m. That's the frequency. But I have always had trouble memorizing this, and this. k over m, and m over k, and all of that. All you have to really memorize is this. And even that, you might even have an intuition as to why it's true. You can even go to the differential equations if you want to reprove it to yourself. Because if you have this, you really can answer any question about the position of the mass, at any time. The velocity of the mass, at any time, just by taking the Or the period, or the frequency of the function. As long as you know how to take the period and frequency of trig functions. You can watch my videos, and watch my trig videos, to get a refresher on that. One thing that's pretty interesting about this, is notice that the frequency and the period, right? This is the period of the function, that's how long it takes do one cycle. This is how many cycles it does in one second-- both of them are independent of A. So it doesn't matter, I could stretch it only a little bit, like there, and it'll take the same amount of time to go back, and come back like that, as it would if I stretch it a lot. It would just do that. If I stretched it just a little bit, the function would look like this. Make sure I do this right. I'm not doing that right. Edit, undo. If I just do it a little bit, the amplitude is going to be less, but the function is going to essentially do the same thing. It's just going to do that. So it's going to take the same amount of time to complete the cycle, it'll just have a lower amplitude. So that's interesting to me, that how much I stretch it, it's not going to make it take longer or less time to complete one cycle. That's interesting. And so if I just told you, that I actually start having objects compressed, right? So in that case, let's say my A is minus 3. I have a spring constant of-- let's say k is, I don't know, 10. And I have a mass of 2 kilograms. Then I could immediately tell you what the equation of the position as a function of time at any point is. It's going to be x of t will equal-- I'm running out of space-- so x of t would equal-- this is just basic subsitution-- minus 3 cosine of 10 divided by 2, right? k over m, is 5. So square root of 5t. I know that's hard to read, but you get the point. I just substituted that. But the important thing to know is this-- this is, I think, the most important thing-- and then if given a trig function, you have trouble remembering how to figure out the period or frequency-- although I always just think about, when does this expression equal 1? And then you can figure out-- when does it equal 1, or when does it equal 0-- and you can figure out its period. If you don't have it, you can memorize this formula for period, and this formula for frequency, but I think that might be a waste of your brain space. Anyway, I'll see you in the next video." + }, + { + "Q": "At 2:05 Sal puts in the -9y^2x at the end of the simplified polynomial equation. Could he have put -9y^2x at the beginning? Why did he put the -9y^2x where he did?", + "A": "The terms in the polynomial can be listed in any order. For example - these are all the same: 4x^2y - 10xy + 45 - 9y^2x (this is Sal s version) 4x^2y - 10xy - 9y^2x + 45 45 - 9y^2x - 10xy + 4x^2y - 9y^2x + 45 + + 4x^2y - 10xy etc.", + "video_name": "AqMT_zB9rP8", + "transcript": "We've got 4x squared y minus 3xy plus 25 minus the entire expression 9y squared x plus 7xy minus 20. So when we're subtracting this entire expression, that's equivalent to subtracting each of these terms individually if we didn't have the parentheses. Or another way of thinking about it-- we could distribute this negative sign. Or you could view this as a negative 1 times this entire expression. And we can distribute it. So let's do that. So let me write this first expression here. I'm going to write it unchanged. So it is 4x squared y minus 3xy plus 25. And now let me distribute the negative 1, or the negative sign times all of this stuff. So negative 1 times 9y squared x is negative 9y squared x. Negative 1 times 7xy is negative 7xy. And then negative 1 times 20 is positive 20. And now we just have to add these terms. And we just want to group like terms. So let's see, is there another x squared y term anywhere? No, I don't see one. So I'll just rewrite this. So we have 4x squared y. Now, is there another xy term? Yeah, there is. So we can group negative 3xy and negative 7xy. Negative 3 of something minus another 7 of that something is going to be negative 10 of that something. So it's negative 10xy. And then we have a 25, which is just a constant term. Or an x to the 0 term. It's 25x to the 0. You could view it that way. And there's another constant term right over here. We can always add 25 to 20. That gives us 45. And then we have this term right over here, which clearly can't be merged with anything else. So minus 9y squared. Let me do that in that original color. Minus 9-- I'm having trouble shifting colors-- minus 9y squared x. And we are done." + }, + { + "Q": "How to balance Pb3O4 ---> PbO + O2? sorry i know the answer, but i don't understand why and how to get it. can someone explain step-by-step to me please? Thank you!", + "A": "Balancing Pb3O4 ---> PbO + O2 _ Pb3O4 --> _ PbO + __O2 Starting off. 1 Pb3O4 --> 3 PbO + __O2 . The Pb atoms are now balanced with 3 Pb atoms on the reactant side and 3 Pb atoms on the product side. 1 Pb3O4 --> 3 PbO + 1/2 O2 Balancing the oxygen atoms with 4 on each side. Since it s not possible to have 1/2 a O2 molecule, all the coefficients would need to be multiplied by 2. The final equation would be: 2 Pb3O4 --> 6 PbO + O2", + "video_name": "8KXWJCmshEE", + "transcript": "- [Voiceover] All right, let's see what's going on in this chemical reaction. On the reactant side, we have an iron oxide. This is ferric oxide right over here, reacting with sulfuric acid, and it's producing, this is ferric sulfate and water. We want to balance this chemical equation. I encourage you to pause this video and try to balance this. I'm assuming you've had a go at it, and you might have been able to successfully balance it. If you didn't, if you weren't able to successfully balance it, one theory of why you weren't able to is because this sulfate group made things really confusing. You have four oxygens, three oxygens here, you have seven on this side, then you have four oxygens in the sulfate group here, but you have three sulfate groups. This is 12 oxygens here, and then you have another oxygen here. This seems really, really, really, really confusing, especially with all this sulfate group business. The key here is to appreciate that the sulfate group is kind of staying together. You can kind of treat it ... Instead of just saying, \"Hey, let's try to balance \"all of the oxygens,\" you can say, \"Let's balance the oxygens \"that are outside of the sulfate groups separately, \"and let's balance the sulfate groups separately.\" To help our brains grapple with that, I'm going to rewrite this chemical equation with a substitution. I'm going to say, let's say that x is equal to a sulfate, a sulfate group. Let me rewrite all of this business. I have ... It's nice to have, I guess this is close to a rust color, which seems appropriate. Let's say I have some ferric oxide. I'm just rewriting what I have above right over here. Ferric oxide plus some sulfuric acid. But instead of writing H2, and then writing the sulfate group, I'm going to write H2 and then x. H2 and then x is going to yield ... Is going to yield ferric sulfate. Ferric sulfate has three sulfate groups. So x is a sulfate group. It's going to have three of them. Ferric sulfate plus molecular, plus molecular water. Now, even though x represents an entire group, let's treat it like an element and just ... make sure we have the same number of x's, or the same number of sulfate groups on both sides. Let's balance this chemical equation. Let's start with the iron. Over here, I have two irons. And over here, on the right-hand side, I have two irons. It doesn't seem like I have to tweak the irons at all. Now let's move on to the oxygens. I have three oxygens here, on the left-hand side. On the right-hand side, I only have one oxygen. But I can change that by saying, \"Let's have three water molecules.\" Then this is going to be three, right over here. Now let's focus on the hydrogens. Focus on the hydrogens, I have two hydrogens here, and I have six hydrogens right over here. If I have six hydrogens right over there, how do I get six hydrogens here? Well, I'll have three molecules of sulfuric acid. Each of them have two hydrogen atoms. Now I have six hydrogens. My hydrogens, my irons, and my oxygens that are not part of the sulfate group are all balanced. Now let's see if we can balance the sulfate groups. On the left-hand side, I have three sulfate groups. Let me do that in that magenta color. I have three sulfate groups. On the right-hand side, I also have three sulfate groups. I'm all balanced. And if we want to un-substitute, we just go back up here. Okay, we didn't change the coefficient on this molecule, on the ferric oxide. We did change the coefficient on the sulfuric acid. We say we have, for every molecule of ferric oxide, we have three molecules of sulfuric acid. We didn't change this. They're going to yield one for every one molecule of ferric oxide, and three molecules of sulfuric acid. It's going to yield, the product, are going to be one molecule of ferric sulfate and three molecules of water. And we are done." + }, + { + "Q": "I'm confused, when he divided 5 from 4 how did he go from getting 0.8 to 1.8?", + "A": "The original number before he divided as 1 4/5. He did the division of the 4/5 = 0.8 (temporarily ignoring the 1). Then add the 1 with the decimal to get the decimal equivalent of 1 4/5 = 1 + 0.8 = 1.8", + "video_name": "-lUEWEEpmIo", + "transcript": "Let's see if we can figure out what 30% of 6 is. So one way of thinking about 30%-- this literally means 30 per 100. So you could view this as 30/100 times 6 is the same thing as 30% of 6. Or you could view this as 30 hundredths times 6, so 0.30 times 6. Now we could solve both of these, and you'll see that we'll get the same answer. If you do this multiplication right over here, 30/100-- and you could view this times 6/1-- this is equal to 180/100. We can simplify. We can divide the numerator and the denominator by 10. And then we can divide the numerator and the denominator by 2. And we will get 9/5, which is the same thing as 1 and 4/5. And then if we wanted to write this as a decimal, 4/5 is 0.8. And if you want to verify that, you could verify that 5 goes into 4-- and there's going to be a decimal. So let's throw some decimals in there. It goes into 4 zero times. So we don't have to worry about that. It goes into 40 eight times. 8 times 5 is 40. Subtract. You have no remainder, and you just have 0's left here. So 4/5 is 0.8. You've got the 1 there. This is the same thing as 1.8, which you would have gotten if you divided 5 into 9. You would've gotten 1.8. So 30% of 6 is equal to 1.8. And we can verify it doing this way as well. So if we were to multiply 0.30 times 6-- let's do that. And I could just write that literally as 0.3 times 6. Well, 3 times 6 is 18. I have only one digit behind the decimal amongst both of these numbers that I'm multiplying. I only have the 3 to the right of the decimal. So I'm only going to have one number to the right of the decimal here. So I just count one number. It's going to be 1.8. So either way you think about it or calculate it, 30% of 6 is 1.8." + }, + { + "Q": "Is mean and arithmetic mean the same ?", + "A": "Usually yes, if left as mean the arithmetic mean is implied.", + "video_name": "h8EYEJ32oQ8", + "transcript": "We will now begin our journey into the world of statistics, which is really a way to understand or get our head around data. So statistics is all about data. And as we begin our journey into the world of statistics, we will be doing a lot of what we can call descriptive statistics. So if we have a bunch of data, and if we want to tell something about all of that data without giving them all of the data, can we somehow describe it with a smaller set of numbers? So that's what we're going to focus on. And then once we build our toolkit on the descriptive statistics, then we can start to make inferences about that data, start to make conclusions, start to make judgments. And we'll start to do a lot of inferential statistics, make inferences. So with that out of the way, let's think about how we can describe data. So let's say we have a set of numbers. We can consider this to be data. Maybe we're measuring the heights of our plants in our garden. And let's say we have six plants. And the heights are 4 inches, 3 inches, 1 inch, 6 inches, and another one's 1 inch, and another one is 7 inches. And let's say someone just said-- in another room, not looking at your plants, just said, well, you know, how tall are your plants? And they only want to hear one number. They want to somehow have one number that represents all of these different heights of plants. How would you do that? Well, you'd say, well, how can I find something that-- maybe I want a typical number. Maybe I want some number that somehow represents the middle. Maybe I want the most frequent number. Maybe I want the number that somehow represents the center of all of these numbers. And if you said any of those things, you would actually have done the same things that the people who first came up with descriptive statistics said. They said, well, how can we do it? And we'll start by thinking of the idea of average. And in every day terminology, average has a very particular meaning, as we'll see. When many people talk about average, they're talking about the arithmetic mean, which we'll see shortly. But in statistics, average means something more general. It really means give me a typical, or give me a middle number, or-- and these are or's. And really it's an attempt to find a measure of central tendency. So once again, you have a bunch of numbers. You're somehow trying to represent these with one number we'll call the average, that's somehow typical, or middle, or the center somehow of these numbers. And as we'll see, there's many types of averages. The first is the one that you're probably most familiar with. It's the one-- and people talk about hey, the average on this exam or the average height. And that's the arithmetic mean. Just let me write it in. I'll write in yellow, arithmetic mean. When arithmetic is a noun, we call it arithmetic. When it's an adjective like this, we call it arithmetic, arithmetic mean. And this is really just the sum of all the numbers divided by-- this is a human-constructed definition that we've found useful-- the sum of all these numbers divided by the number of numbers we have. So given that, what is the arithmetic mean of this data set? Well, let's just compute it. It's going to be 4 plus 3 plus 1 plus 6 plus 1 plus 7 over the number of data points we have. So we have six data points. So we're going to divide by 6. And we get 4 plus 3 is 7, plus 1 is 8, plus 6 is 14, plus 1 is 15, plus 7. 15 plus 7 is 22. Let me do that one more time. You have 7, 8, 14, 15, 22, all of that over 6. And we could write this as a mixed number. 6 goes into 22 three times with a remainder of 4. So it's 3 and 4/6, which is the same thing as 3 and 2/3. We could write this as a decimal with 3.6 repeating. So this is also 3.6 repeating. We could write it any one of those ways. But this is kind of a representative number. This is trying to get at a central tendency. Once again, these are human-constructed. No one ever-- it's not like someone just found some religious document that said, this is the way that the arithmetic mean must be defined. It's not as pure of a computation as, say, finding the circumference of the circle, which there really is-- that was kind of-- we studied the universe. And that just fell out of our study of the universe. It's a human-constructed definition that we found useful. Now there are other ways to measure the average or find a typical or middle value. The other very typical way is the median. And I will write median. I'm running out of colors. I will write median in pink. So there is the median. And the median is literally looking for the middle number. So if you were to order all the numbers in your set and find the middle one, then that is your median. So given that, what's the median of this set of numbers going to be? Let's try to figure it out. Let's try to order it. So we have 1. Then we have another 1. Then we have a 3. Then we have a 4, a 6, and a 7. So all I did is I reordered this. And so what's the middle number? Well, you look here. Since we have an even number of numbers, we have six numbers, there's not one middle number. You actually have two middle numbers here. You have two middle numbers right over here. You have the 3 and the 4. And in this case, when you have two middle numbers, you actually go halfway between these two numbers. You're essentially taking the arithmetic mean of these two numbers to find the median. So the median is going to be halfway in-between 3 and 4, which is going to be 3.5. So the median in this case is 3.5. So if you have an even number of numbers, the median or the middle two, the-- essentially the arithmetic mean of the middle two, or halfway between the middle two. If you have an odd number of numbers, it's a little bit easier to compute. And just so that we see that, let me give you another data set. Let's say our data set-- and I'll order it for us-- let's say our data set was 0, 7, 50, I don't know, 10,000, and 1 million. Let's say that is our data set. Kind of a crazy data set. But in this situation, what is our median? Well, here we have five numbers. We have an odd number of numbers. So it's easier to pick out a middle. The middle is the number that is greater than two of the numbers and is less than two of the numbers. It's exactly in the middle. So in this case, our median is 50. Now, the third measure of central tendency, and this is the one that's probably used least often in life, is the mode. And people often forget about it. It sounds like something very complex. But what we'll see is it's actually a very straightforward idea. And in some ways, it is the most basic idea. So the mode is actually the most common number in a data set, if there is a most common number. If all of the numbers are represented equally, if there's no one single most common number, then you have no mode. But given that definition of the mode, what is the single most common number in our original data set, in this data set right over here? Well, we only have one 4. We only have one 3. But we have two 1's. We have one 6 and one 7. So the number that shows up the most number of times here is our 1. So the mode, the most typical number, the most common number here is a 1. So, you see, these are all different ways of trying to get at a typical, or middle, or central tendency. But they do it in very, very different ways. And as we study more and more statistics, we'll see that they're good for different things. This is used very frequently. The median is really good if you have some kind of crazy number out here that could have otherwise skewed the arithmetic mean. The mode could also be useful in situations like that, especially if you do have one number that's showing up a lot more frequently. Anyway, I'll leave you there. And we'll-- the next few videos, we will explore statistics even deeper." + }, + { + "Q": "Why did you multiply by 20 to get numerators?", + "A": "It s the LCM (Lowest Common Multiple)", + "video_name": "N8dIOmk_lHs", + "transcript": "What I want to do in this video is order these fractions from least to greatest. And the easiest way and the way that I think we can be sure we'll get the right answer here is to find a common denominator, because if we don't find a common denominator, these fractions are really 4/9 versus 3/4 versus 4/5, 11/12, 13/15. You can try to estimate them, but you'll be able to directly compare them if they all had the same denominator. So the trick here, or at least the first trick here, is to try to find that common denominator. And there's many ways to do it. You could just pick one of these numbers and keep taking its multiples and find that multiple that is divisible by all the rest. Another way to do it is look at the prime factorization of each of these numbers. And then the least common multiple of them will have to have at least all of those prime numbers in it. It has to be composed of all of these numbers. So let's do it that second way. And then let's verify that it definitely is divisible. So 9 is the same thing as 3 times 3. So our least common multiple is going to have at least one 3 times 3 in it. And then 4 is the same thing as 2 times 2. So we're going to also have to have a 2 times 2 in our prime factorization of our least common multiple. 5 is a prime number. So we're going to need to have a 5 in there. And then 12-- I'm going to do that in yellow. 12 is the same thing as 2 times 6, which is the same thing as 2 times 3. And so in our least common multiple, we have to have two 2's. But we already have two 2's right over here from our 4. And we already have one 3 right over here. Another way to think about it is something that is divisible by both 9 and 4 is going to be divisible by 12, because you're going to have the two 2's. And you're going to have that one 3 right over there. And then, finally, we need to be divisible by 15's prime factors. So let's look at 15's prime factors. 15 is the same thing as 3 times 5. So once again, this number right over here already has a 3 in it. And it already has a 5 in it. So we're cool for 15, for 12, and, obviously, for the rest of them. So this is our least common multiple. And we can just take this product. And so this is going to be equal to 3 times 3 is 9. 9 times 2 is 18. 18 times 2 is 36. 36 times 5, you could do that in your head if you're like. But I'll do it on the side just in case. 36 times 5, just so that we don't mess up. 6 times 5 is 30. 3 times 5 is 15 plus 3 is 180. So our least common multiple is 180. So we want to rewrite all of these fractions with 180 in the denominator. So this first fraction, 4/9, is what over 180? To go from 9 to 180, we have to multiply the denominator by 20. So let me do it this way. So if we do 4/9, to get the denominator of 9 to be 180, you have to multiply it by 20. And since we don't want to change the value of the fraction, we should also multiply the 4 by 20. So we're just really multiplying by 20/20. And so 4/9 is going to be the same thing as 80/180. Now, let's do the same thing for 3/4. Well, what do we have to multiply the denominator by to get us to 180? So it looks like 45. You could divide 4 into 180 to figure that out. But if you take 4 times 45, 4 times 40 is 160. 4 times 5 is 20. You get 180. So if you multiply the denominator by 45, you also have to multiply the numerator by 45. 3 times 45 is 120 plus 15. So it's 135. And the denominator here is 180. Now, let's do 4/5. To get our denominator to be 180, what do you have to multiply 5 by? Let's see. If you multiply 5 by 30, you'll get to 150. But then you have another 30. Actually, we know it right over here. You have to multiply it by 36. Well, then you have to multiply the numerator by 36 as well. And so our denominator is going to be 180. Our numerator, 4 times 30 is 120. 4 times 6 is 24. So it's 144/180. And then we have only two more to do. So we have our 11/12. So to get the denominator to be 180, we have to multiply 12 by-- so 12 times 10 is 120. Then you have 60 left. So you have to multiply it by 15, 15 In the denominator, and 15 in the numerator. And so the denominator gives us 180. And 11 times 15. So 10 times 15 is 150. And then you have one more 15. So it's going to be 165. And then, finally, we have 13/15. To get our denominator to be 180, have to multiply it by 12. We already figured out that 12 times 15 is 180. So you have to multiply it by 12. That will give us 180 in the denominator. And so you have to also multiply the numerator by 12, so that we don't change the value of the fraction. We know 12 times 12 is 144. You could put one more 12 in there. You get 156. Did I do that right? 12 plus 144 is going to be 156. So we've rewritten each of these fractions with that new common denominator of 180. And now, it's very easy to compare them. You really just have to look at the numerators. So the smallest of the numerators is this 80 right over here. So 4/9 is the smallest. 4/9 is the least of these numbers. So let me just write it over here. So this is our ordering. We have 4/9 comes first, which is the same thing as 80/180. Let me write it both ways-- 80/180. Then the next the smallest number looks like it's this 135 right over here. I want to do it in that same color. The next one is going to be that 135/180, which is the same thing as 3/4. And then the next one is going to be-- let's see, we have the 144/180. So this is going to be the 144/180, which is the same thing as 4/5. And then we have two more. The next is this 156/180. So then we have our 156/180, which is the same thing as 13/15. And then we have one left over, the 165/180, which is the same thing-- I want to do that in yellow. We have our 165/180, which is the same thing as 11/12. And we're done. We have finished our ordering. So if you're doing the Khan Academy module on this, this is what you would input into that little box there." + }, + { + "Q": "This video confused me. What does this video have to do with polynomials?", + "A": "It s basically one polynomial divided by another.", + "video_name": "S-XKGBesRzk", + "transcript": "Let's see if we can learn a thing or two about partial fraction expansion, or sometimes it's called partial fraction decomposition. The whole idea is to take rational functions-- and a rational function is just a function or expression where it's one expression divided by another-- and to essentially expand them or decompose them into simpler parts. And the first thing you've got to do, before you can even start the actual partial fraction expansion process, is to make sure that the numerator has a lower degree than the denominator. In the situation, the problem, that I've drawn right here, I've written right here, that's not the case. The numerator has the same degree as the denominator. So the first step we want to do to simplify this and to get it to the point where the numerator has a lower degree than the denominator is to do a little bit of algebraic division. And I've done a video on this, but it never hurts to get a review here, so to do that, we divide the denominator into the numerator to figure out the remainder, so we divide x squared minus 3x minus 40 into x squared minus 2x minus 37. So how many times? You look at the highest degree term, so x squared goes into x squared one time, one times this whole thing is x squared minus 3x minus 40, and now you want to subtract this from that to get the remainder. And see, if I'm subtracting, so I'm going to subtract, and then minus minus is a plus, a plus, and then you can add them. These cancel out. Minus 2x plus 3x, that's x. Minus 37 plus 40, that's plus 3. So this expression up here can be rewritten as-- let me scroll down a little bit-- as 1 plus x plus 3 over x squared minus 3x minus 40. This might seem like some type of magic thing I just did, but this is no different than what you did in the fourth or fifth grade, where you learned how to convert improper regular fractions into mixed numbers. Let me just do a little side example here. If I had 13 over 2, and I want to turn it into a mixed number, what you do-- you can probably do this in your head now-- but what you did is, you divide the denominator into the numerator, just like we did over here. 2 goes into 13. We see 2 goes into 13 six times, 6 times 2 is 12, you subtract that from that, you get a remainder of 1. 2 doesn't go into 1, so that's just the remainder. So if you wanted to rewrite this, it would be the number of times the denominator goes into the numerator, that's 6, plus the remainder over the denominator. Plus 6-- plus 1 over 2. And when you did it in elementary school, you would just write 6 1/2, but 6 1/2 is the same thing as 6 plus 1/2. That's exactly the same thing we did here. The denominator went to the numerator one time, and then there was a remainder of x plus 3 left over, so it's 1 plus x plus 3 over this expression. Now we see that that numerator in this rational expression does have a lower degree than the denominator. The highest degree here is 1, the highest degree here is 2, so we're ready to commence our partial fraction decomposition. And all that is, is taking this expression up here and turning it into two simpler expressions where the denominators are the factors of this lower term. So given that, let's factor this lower term. So let's see. What two numbers add up to minus 3, and when you multiply them, you get minus 40? So let's see. They have to be different signs, because when you multiply them you get a negative, so it has to be minus 8 and plus 5. So we can rewrite this up here as-- I'll switch colors-- 1 plus x plus 3 over x plus 5 times x minus 8. 5 times 8 is minus 40-- 5 times negative 8 is minus 40, plus 5 minus 8 is minus 3, so we're all set. Now I'll just focus on this part right now. We can just remember that that 1 is sitting out there out front. This is the expression we want to decompose or expand. And we're going to expand it into two simpler expressions where each of these are the denominator-- and I will make the claim, and if the numbers work out then the claim is true-- I'll make the claim that I can expand this, or decompose this, into two fractions where the first fraction is just some number a over the first factor, over x plus 5, plus some number b over the second factor, over x minus 8. That's my claim, and if I can solve for a and b in a way that it actually does add up to this, then I'm done and I will have fully decomposed this fraction. I guess is the way-- I don't know if that's the correct terminology. So let's try to do that. So if I were to add these two terms, what do I get? When you add anything, you find the common denominator, and the common denominator, the easiest common denominator, is to multiply the two denominators, so let me write this here. So a over x plus 5 plus b over x minus 8 is equal to-- well, let's get the common denominator-- it's equal to x plus 5 times x minus 8. And then the a term, we would-- a over x plus 5 is the same thing as a times x minus 8 over this whole thing. I mean, if I just wrote this right here, you would just cancel these two terms out and you would get a over x plus 5. And then you could add that to the common denominator, x plus 5 times x minus 8, and it would be b times x plus 5. Important to realize, that, look. This term is the exact same thing as this term if you just cancel the x minus 8 out, and this term is the exact same thing as this term if you just cancel the x plus 5 out. But now that we have an actual common denominator, we can add them together, so we get-- let me just write the left side here over-- a over x plus 5-- I'm sorry. I want to write this over here. I want to write x plus 3 over plus 5 times x minus 8 is equal to is equal to the sum of these two things on top. a times x minus 8 plus b times x plus 5, all of that over their common denominator, x plus 5 times x minus eight. So the denominators are the same, so we know that this, when you add this together, you have to get this. So if we want to solve for a and b, let's just set that equality. We can ignore the denominators. So we can say that x plus 3 is equal to a times x minus 8 plus b times x plus 5. Now, there's two ways to solve for a and b from this point going forward. One is the way that I was actually taught in the seventh or eighth grade, which tends to take a little longer, then there's a fast way to do it and it never hurts to do the fast way first. If you want to solve for a, let's pick an x that'll make this term disappear. So what x would make this term disappear? Well, if I say x is minus 5, then this becomes 0, and then the b disappears. So if we say x is minus 5-- I'm just picking an arbitrary x to be able to solve for this-- then this would become minus 5 plus 3-- let me just write it out, minus 5 plus 3-- is equal to a times minus 5 minus 8-- let me just write it out, minus 5 minus 8-- plus b times minus 5 plus 5. And I picked the minus 5 to make this expression 0. So then you get-- pick a brighter color-- minus 5 plus 3 is minus 2, is equal to-- what is this?-- minus 13a plus-- this is 0, right? That's 0. Minus 5 plus 5 is 0, 0 times b is 0, and then you divide both sides by minus 13, you get-- negatives cancel out-- you get 2 over 13 is equal to a, and now we can do the same thing up here and get rid of the a terms by making x is equal to 8. If x is equal to 8, you get x plus 3 is equal to 11, is equal to a times 0 plus b times-- what's 5-- 8 plus 5 is-- plus b times 13. Their b looks a bit like a 13. And then you get 11 is equal to 13b, divide both sides by 13, you get b is equal to 11 over 13. So we were able to solve for our a's and our b's. And so we can go back to our original equation and we could say, wow. This just has to be equal to 2 over 13, and this just has to be equal to 11 over 13. So our original, our very original thing we wrote up here, can be decomposed into 1, that's this 1 over here, plus this, which is 2 over 13-- I'll just write it like this for now-- 2 over 13, over x plus 5. You could bring the 13 down here if you want to write it so you don't have a fraction over a fraction. Plus 11 over 13 times-- over x minus 8. And once again, you could bring the 13 down so you don't have a fraction over a fraction. But we have just successfully decomposed this pretty-- I don't want to say that we necessarily simplified it, because you could say, oh, we only have one expression here, now I have three-- but I've reduced the degree of both the numerators and the denominators. And you might say, well, Sal, why would I ever have to do this? And you're right. In algebra you probably won't. But this is actually a really useful technique later on when you get to calculus, and actually, differential equations, because a lot of times it's much easier-- and I'll throw out a word here that you don't understand-- to take the integral or the antiderivative of something like this, then something like this. And later, when you do inverse Laplace transforms and differential equations, it's much easier take an inverse Laplace transform of something like this than something like that. So anyway, hopefully I've given you another tool kit in your-- or another tool in your tool kit, and I'll probably do a couple more videos because we haven't exhausted all of the examples that we could we could show for partial fraction decomposition." + }, + { + "Q": "e^(x+1)=25 please solve (for Nana)", + "A": "e^(x + 1) = 25 x + 1 = ln(25) x = ln(25) - 1 x \u00e2\u0089\u0088 2.2189", + "video_name": "sBhEi4L91Sg", + "transcript": "I would guess that you're reasonably familiar with linear scales. These are the scales that you would typically see in most of your math classes. And so just to make sure we know what we're talking about, and maybe thinking about in a slightly different way, let me draw a linear number line. Let me start with 0. And what we're going to do is, we're going to say, look, if I move this distance right over here, and if I move that distance to the right, that's equivalent to adding 10. So if you start at 0 and you add 10, that would obviously get you to 10. If you move that distance to the right again, you're going to add 10 again, that would get you to 20. And obviously we could keep doing it, and get to 30, 40, 50, so on and so forth. And also, just looking at what we did here, if we go the other direction. If we start here, and move that same distance to the left, we're clearly subtracting 10. 10 minus 10 is equal to 0. So if we move that distance to the left again, we would get to negative 10. And if we did it again, we would get to negative 20. So the general idea is, however many times we move that distance, we are essentially adding-- or however many times you move that distance to the right-- we are essentially adding that multiple of 10. If we do it twice, we're adding 2 times 10. And that not only works for whole numbers, it would work for fractions as well. Where would 5 be? Well to get to 5, we only have to multiply 10-- or I guess one way to think about it is 5 is half of 10. And so if we want to only go half of 10, we only have to go half this distance. So if we go half this distance, that will get us to 1/2 times 10. In this case, that would be 5. If we go to the left, that would get us to negative 5. And there's nothing-- let me draw that a little bit more centered, negative 5-- and there's nothing really new here. We're just kind of thinking about it in a slightly novel way that's going to be useful when we start thinking about logarithmic. But this is just the number line that you've always known. If we want to put 1 here, we would move 1/10 of the distance, because 1 is 1/10 of 10. So this would be 1, 2, 3, 4, I could just put, I could label frankly, any number right over here. Now this was a situation where we add 10 or subtract 10. But it's completely legitimate to have an alternate way of thinking of what you do when you move this distance. And let's think about that. So let's say I have another line over here. And you might guess this is going to be the logarithmic number line. Let me give ourselves some space. And let's start this logarithmic number line at 1. And I'll let you think about, after this video, why I didn't start it at 0. And if you start at 1, and instead of moving that, so I'm still going to define that same distance. But instead of saying that that same distance is adding 10 when I move to the right, I'm going to say when I move the right that distance on this new number line that I have created, that is the same thing as multiplying by 10. And so if I move that distance, I start at 1, I multiply by 10. That gets me to 10. And then if I multiply by 10 again, if I move by that distance again, I'm multiplying by 10 again. And so that would get me to 100. I think you can already see the difference that's happening. And what about moving to the left that distance? Well we already have kind of said what happens. Because if we start here, we start at 100 and move to the left of that distance, what happens? Well I divided by 10. 100 divided by 10 gets me 10. 10 divided by 10 get me 1. And so if I move that distance to the left again, I'll divide by 10 again. That would get me to 1/10. And if I move that distance to the left again, that would get me to 1/100. And so the general idea is, is however many times I move that distance to the right, I'm multiplying my starting point by 10 that many times. And so for example, when I move that distance twice, so this whole distance right over here, I went that distance twice. So this is times 10 times 10, which is the same thing as times 10 to the second power. And so really I'm raising 10 to what I'm multiplying it times 10 to whatever power, however many times I'm jumping to the right. Same thing if I go to the left. If I go to the left that distance twice-- let me do that in a new color-- this will be the same thing as dividing by 10 twice. Dividing by 10, dividing by 10, which is the same thing as multiplying by-- one way to think of it-- 1/10 squared. Or dividing by 10 squared is another way of thinking about it. And so that might make a little, that might be hopefully a little bit intuitive. And you can already see why this is valuable. We can already on this number line plot a much broader spectrum of things than we can on this number line. We can go all the way up to 100, and then we even get some nice granularity if we go down to 1/10 and 1/100. Here we don't get the granularity at small scales, and we also don't get to go to really large numbers. And if we go a little distance more, we get to 1,000, and then we get to 10,000, so on and so forth. So we can really cover a much broader spectrum on this line right over here. But what's also neat about this is that when you move a fixed distance, so when you move fixed distance on this linear number line, you're adding or subtracting that amount. So if you move that fixed distance you're adding 2 to the right. If you go to the left, you're subtracting 2. When you do the same thing on a logarithmic number line, this is true of any logarithmic number line, you will be scaling by a fixed factor. And one way to think about what that fixed factor is is this idea of exponents. So if you wanted to say, where would 2 sit on this number line? Then you would just think to yourself, well, if I ask myself where does 100 sit on that number line-- actually, that might be a better place to start. If I said, if I didn't already plot it and said where does 100 sit on that number line? I would say, how many times would we have to multiply 10 by itself to get 100? And that's how many times I need to move this distance. And so essentially I'll be asking 10 to the what power is equal to 100? And then I would get that question mark is equal to 2. And then I would move that many spaces to plot my 100. Or another way of stating this exact same thing is log base 10 of 100 is equal to question mark. And this question mark is clearly equal to 2. And that says, I need to plot the 100 2 of this distance to the right. And to figure out where do I plot the 2, I would do the exact same thing. I would say 10 to what power is equal to 2? Or log base 10 of 2 is equal to what? And we can get the trusty calculator out, and we can just say log-- and on most calculators if there's a log without the base specified, they're assuming base 10-- so log of 2 is equal to roughly 0.3. 0.301. So this is equal to 0.301. So what this tells us is we need to move this fraction of this distance to get to 2. If we move this whole distance, it's like multiplying times 10 to the first power. But since we only want to get 10 to the 0.301 power, we only want to do 0.301 of this distance. So it's going to be roughly a third of this. It's going to be roughly-- actually, a little less than a third. 0.3, not 0.33. So 2 is going to sit-- let me do it a little bit more to the right-- so 2 is going to sit right over here. Now what's really cool about it is this distance in general on this logarithmic number line means multiplying by 2. And so if you go that same distance again, you're going to get to 4. If you multiply that same distance again, you're going to multiply by 4. And you go that same distance again, you're going to get to 8. And so if you said where would I plot 5? Where would I plot 5 on this number line? Well, there's a couple ways to do it. You could literally figure out what the base 10 logarithm of 5 is, and figure out where it goes on the number line. Or you could say, look, if I start at 10 and if I move this distance to the left, I'm going to be dividing by 2. So if I move this distance to the left I will be dividing by 2. I know it's getting a little bit messy here. I'll maybe do another video where we learn how to draw a clean version of this. So if I start at 10 and I go that same distance I'm dividing by 2. And so this right here would be that right over there would be 5. Now the next question, you said well where do I plot 3? Well we could do the exact same thing that we did with 2. We ask ourselves, what power do we have to raise 10 to to get to 3? And to get that, we once again get our calculator out. log base 10 of 3 is equal to 0.477. So it's almost halfway. So it's almost going to be half of this distance. So half of that distance is going to look something like right over there. So 3 is going to go right over here. And you could do the logarithm-- let's see, we're missing 6, 7, and 8. Oh, we have 8. We're missing 9. So to get 9, we just have to multiply by 3 again. So this is 3, and if we go that same distance, we multiply by 3 again, 9 is going to be squeezed in right over here. 9 is going to be squeezed in right over there. And if we want to get to 6, we just have to multiply by 2. And we already know the distance to multiply by 2, it's this thing right over here. So you multiply that by 2, you do that same distance, and you're going to get to 6. And if you wanted to figure out where 7 is, once again you could take the log base-- let me do it right over here-- so you'll take the log of 7 is going to be 0.8, roughly 0.85. So 7 is just going to be squeezed in roughly right over there. And so a couple of neat things you already appreciated. One, we can fit more on this logarithmic scale. And, as I did with the video with Vi Hart, where she talked about how we perceive many things with logarithmic scales. So it actually is a good way to even understand some of human perception. But the other really cool thing is when we move a fixed distance on this logarithmic scale, we're multiplying by a fixed constant. Now the one kind of strange thing about this, and you might have already noticed here, is that we don't see the numbers lined up the way we normally see them. There's a big jump from 1 to 2, then a smaller jump from 3 to 4, then a smaller jump from that from 3 to 4, then even smaller from 4 to 5, then even smaller 5 to 6 it gets. And then 7, 8, 9, you know 7's going to be right in there. They get squeeze, squeeze, squeezed in, tighter and tighter and tighter, and then you get 10. And then you get another big jump. Because once again, if you want to get to 20, you just have to multiply by 2. So this distance again gets us to 20. If you go this distance over here that will get you to 30, because you're multiplying by 3. So this right over here is a times 3 distance. So if you do that again, if you do that distance, then that gets you to 30. You're multiplying by 3. And then you can plot the whole same thing over here again. But hopefully this gives you a little bit more intuition of why logarithmic number lines look the way they do. Or why logarithmic scale looks the way it does. And also, it gives you a little bit of appreciation for why it might be useful." + }, + { + "Q": "I'm stuck on a problem.\nHow would you simplify the following: (x^3)^(2/3)\n\nMy first thought would be to multiply the exponents: 3/1 * 2/3 which would leave me with an exponent of 2. Can anyone confirm this answer for me?", + "A": "If my brain does not fail me I think that s correct. The answer is x^2.", + "video_name": "S34NM0Po0eA", + "transcript": "We've already seen how to think about something like 64 to the 1/3 power. We saw that this is the exact same thing as taking the cube root of 64. And because we know that 4 times 4 times 4, or 4 to the third power, is equal to 64, if we're looking for the cube root of 64, we're looking for a number that that number times that number times that same number is going to be equal to 64. Well, we know that number is 4, so this thing right over here is going to be 4. Now we're going to think of slightly more complex fractional exponents. The one we see here has a 1 in the numerator. Now we're going to see something different. So what I want to do is think about what 64 to the 2/3 power is. And here I'm going to use a property of exponents that we'll study more later on. But this property of exponents is the idea that-- let's say with a simple number-- if I raise something to the third power and then I were to raise that to, say, the fourth power, this is going to be the same thing as raising it to the 2 to the 3 times 4 power, or 2 to the 12th power, which you could also write as raising it to the fourth power and then the third power. All this is saying is, if I raise something to a power and then raise that whole thing to a power, it's the same thing as multiplying the two exponents. This is the same thing as 2 to the 12th. So we could use that property here to say, well, 2/3 is the same thing as 1/3 times 2. So we could go in the other direction. We could say, hey look, well this is going to be the same thing as 64 to the 1/3 power and then that thing squared. Notice, I'm raising something to a power and then raising that to a power. If I were to multiply these two things, I would get 64 to the 2/3 power. Now, why did I do this? Well, we already know what 64 to the 1/3 power is. We just calculated it. That's equal to 4. So we could say that-- and I'll write it in that same yellow color-- this is equal to 4 squared, which is equal to 16. So 64 to the 2/3 power is equal to 16. The way I think of it, let me find the cube root of 64, which is 4. And then let me square it. And that is going to get me to 16. Now I'll give you in even hairier problem. And I encourage you to try this one on your own before I work through it. So we're going to work with 8/27. And we're going to raise this thing to the-- and I'll try to color code it-- negative 2 over 3 power, to the negative 2/3 power. I encourage you to pause and try this on your own. Well the first thing I do whenever I see a negative exponent is to say, well, how can I get rid of that negative exponent? And I just remind myself, well, the negative exponent really just says, take the reciprocal of this to the positive exponent. I'm using a different color. I'm going to use that light mauve color. So this is going to be equal to 27/8. I just took the reciprocal of this right over here. It's equal to 27/8 to the positive 2/3 power. So notice, all I did, I got rid of the exponent and took the reciprocal of the base right over here. 8/27 is the base. Negative 2/3 is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over a denominator to some power-- and this is another very powerful exponent property-- this is going to be the exact same thing as raising 27 to the 2/3 power-- to the 2 over 3 power-- over 8 to the 2/3 power. This is another very powerful exponent property. Notice, if I have something divided by something and I'm raising the whole thing to a power, I can essentially raise the numerator to that power over the denominator raised to that power. Now, let's think about what this is. Well just like we saw before, this is going to be the same thing. This is going to be the same thing as 27 to the 1/3 power and then that squared because 1/3 times 2 is 2/3. So I'm going to raise 27 to the 1/3 power and then square whatever that is. All this color coding is making this have to switch a lot of colors. This is going to be over 8 to the 1/3 power. And then that's going to be raised to the second power. Same thing we were doing in the denominator-- we raise 8 to the 1/3 and then square that. So what's this going to be? Well, 27 to the 1/3 power is the cube root of 27. It's some number-- that number times that same number times that same number is going to be equal to 27. Well, it might jump out at you already that 3 to the third is equal to 27 or that 27 to the 1/3 is equal to 3. So the numerator, we're going to end up with 3 squared. And then in the denominator, we are going to end up with-- well, what's 8 to the 1/3 power? Well, 2 times 2 times 2 is 8. So this is 8 to the 1/3 third is 2. Let me do that same orange color. 8 to the 1/3 is 2, and then we're going to square that. So this is going to simplify to 3 squared over 2 squared, which is just going to be equal to 9/4. So if you just break it down step by step, it actually is not too daunting." + }, + { + "Q": "What does it mean when a slope is undefined?", + "A": "If a slope is undefined that means the line is perfectly vertical.", + "video_name": "CFSHq099Mx0", + "transcript": "Find the slope of the line that goes through the ordered pairs 7, negative 1 and negative 3, negative 1. Let me just do a quick graph of these just so we can visualize what they look like. So let me draw a quick graph over here. So our first point is 7, negative 1. So 1, 2, 3, 4, 5, 6, 7. This is the x-axis. 7, negative 1. So it's 7, negative 1 is right over there. 7, negative 1. This, of course, is the y-axis. And then the next point is negative 3, negative 1. So we go back 3 in the horizontal direction. Negative 3 for the y-coordinate is still negative 1. So the line that connects these two points will look like this. It will look like that. Now, they're asking us to find the slope of the line that goes through the ordered pairs. Find the slope of this line. And just to give a little bit of intuition here, slope is a measure of a line's inclination. And the way that it's defined-- slope is defined as rise over run, or change in y over change in x, or sometimes you'll see it defined as the variable m. And then they'll define change in y as just being the second y-coordinate minus the first y-coordinate and then the change in x as the second x-coordinate minus the first x-coordinate. These are all different variations in slope, but hopefully you'll appreciate that these are measuring inclination. If I rise a ton when I run a little bit, if I move a little bit in the x direction, and I rise a bunch, then I have a very steep line. I have a very steep upward-sloping line. If I don't change at all when I run a bit, then I have a very low slope. And that's actually what's happening here. I'm going from-- you could either view this as the starting point or view this as the starting point. But let's view this as the starting point. So this negative 3, 1. If I go from negative 3, negative 1 to 7, negative 1, I'm running a good bit. I'm going from negative 3. My x value is negative 3 here, and it goes all the way to 7. So my change in x here is 10. To go from negative 3 to 7, I changed my x value by 10. But what's my change in y? Well, my y value here is negative 1, and my y value over here is still negative 1. So my change in y is a 0. My change in y is going to be 0. My y value does not change no matter how much I change my x value. So the slope here is going to be-- when we run 10, what was our rise? How much did we change in y? Well, we didn't rise at all. We didn't go up or down. So the slope here is 0. Or another way to think about is this line has no inclination. It's a completely flat-- it's a completely horizontal line. So this should make sense. This is a 0. The slope here is 0. And just to make sure that this gels with all of these other formulas that you might know-- but I want to make it very clear. These are all just telling you rise over run or change in y over change in x, a way to measure inclination. But let's just apply them just so, hopefully, it all makes sense to you. So we could also say slope is change in y over change in x. If we take this to be our start and if we take this to be our end point, then we would call this over here x1. And then this is over here. This is y1. And then we would call this x2 and we would call this y2, if this is our start point and that is our end point. And so the slope here, the change in y, y2 minus y1. So it's negative 1 minus negative 1, all of that over x2, negative 3, minus x1, minus 7. So the numerator, negative 1 minus negative 1, that's the same thing as negative 1 plus 1. And our denominator is negative 3 minus 7, which is negative 10. So once again, negative 1 plus 1 is 0 over negative 10. And this is still going to be 0. And the only reason why we got a negative 10 here and a positive 10 there is because we swapped the starting and the ending point. In this example right over here, we took this as the start point and made this coordinate over here as the end point. Over here, we swapped them around. 7, negative 1 was our start point, and negative 3, negative 1 is our end point. So if we start over here, our change in x is going to be negative 10. But our change in y is still going to be 0. So regardless of how you do it, the slope of this line is 0. It's a horizontal line." + }, + { + "Q": "what would we be if non of that hapend", + "A": "nothing we wouldn t be here", + "video_name": "VbNXh0GaLYo", + "transcript": "What I'm going to attempt to do in the next two videos is really just give an overview of everything that's happened to Earth since it came into existence. We're going start really at the formation of Earth or the formation of our Solar system or the formation of the Sun, and our best sense of what actually happened is that there was a supernova in our vicinity of the galaxy, and this right here is a picture of a supernova remnant, actually, the remnant for Kepler's supernova. The supernova in this picture actually happened four hundred years ago in 1604, so right at the center a star essentially exploded and for a few weeks was the brightest object in the night sky, and it was observed by Kepler and other people in 1604, and this is what it looks like now. What we see is kinda the shockwave that's been traveling out for the past 400 years, so now it must be many light years across. It wasn't, obviously, matter wasn't traveling at the speed of light, but it must've been traveling pretty, pretty fast, at least relativistic speeds, a reasonable fraction of the speed of light. This has traveled a good bit out now, but what you can imagine is when you have the shockwave traveling out from a supernova, let's say you had a cloud of molecules, a cloud of gas, that before the shockwave came by just wasn't dense enough for gravity to take over, and for it to accrete, essentially, into a solar system. When the shockwave passes by it compresses all of this gas and all of this material and all of these molecules, so it now does have that critical density to form, to accrete into a star and a solar system. We think that's what's happened, and the reason why we feel pretty strongly that it must've been caused by a supernova is that the only way that the really heavy elements can form, or the only way we know that they can form is in kind of the heat of a supernova, and our uranium, the uranium that seems to be in our solar system on Earth, seems to have formed roughly at the time of the formation of Earth, at about four and a half billion years ago, and we'll talk in a little bit more depth in future videos on exactly how people figure that out, but since the uranium seems about the same age as our solar system, it must've been formed at around the same time, and it must've been formed by a supernova, and it must be coming from a supernova, so a supernova shockwave must've passed through our part of the universe, and that's a good reason for gas to get compressed and begin to accrete. So you fast-forward a few million years. That gas would've accreted into something like this. It would've reached the critical temperature, critical density and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium, and this right here is our early sun. Around the sun you have all of the gases and particles and molecules that had enough angular velocity to not fall into the sun, to go into orbit around the sun. They were actually supported by a little bit of pressure, too, because you can kinda view this as kind of a big cloud of gas, so they're always bumping into each other, but for the most part it was their angular velocity, and over the next tens of millions of years they'll slowly bump into each other and clump into each other. Even small particles have gravity, and they're gonna slowly become rocks and asteroids and, eventually, what we would call \"planetesimals,\" which are, kinda view them as seeds of planets or early planets, and then those would have a reasonable amount of gravity and other things would be attracted to them and slowly clump up to them. This wasn't like a simple process, you know, you could imagine you might have one planetesimal form, and then there's another planetesimal formed, and instead of having a nice, gentle those two guys accreting into each other, they might have huge relative velocities and ram into each other, and then just, you know, shatter, so this wasn't just a nice, gentle process of constant accretion. It would actually have been a very violent process, actually happened early in Earth's history, and we actually think this is why the Moon formed, so at some point you fast-forward a little bit from this, Earth would have formed, I should say, the mass that eventually becomes our modern Earth would have been forming. Let me draw it over here. So, let's say that that is our modern Earth, and what we think happened is that another proto-planet or another, it was actually a planet because it was roughly the size of Mars, ran into our, what it is eventually going to become our Earth. This is actually a picture of it. This is an artist's depiction of that collision, where this planet right here is the size of Mars, and it ran into what would eventually become Earth. This we call Theia. This is Theia, and what we believe happened, and if you look up, if you go onto the Internet, you'll see some simulations that talk about this, is that we think it was a glancing blow. It wasn't a direct hit that would've just kinda shattered each of them and turned into one big molten ball. We think it was a glancing blow, something like this. This was essentially Earth. Obviously, Earth got changed dramatically once Theia ran into it, but Theia is right over here, and we think it was a glancing blow. It came and it hit Earth at kind of an angle, and then it obviously the combined energies from that interaction would've made both of them molten, and frankly they probably already were molten because you had a bunch of smaller collisions and accretion events and little things hitting the surface, so probably both of them during this entire period, but this would've had a glancing blow on Earth and essentially splashed a bunch of molten material out into orbit. It would've just come in, had a glancing blow on Earth, and then splashed a bunch of molten material, some of it would've been captured by Earth, so this is the before and the after, you can imagine, Earth is kind of this molten, super hot ball, and some of it just gets splashed into orbit from the collision. Let me just see if I can draw Theia here, so Theia has collided, and it is also molten now because huge energies, and it splashes some of it into orbit. If we fast-forward a little bit, this stuff that got splashed into orbit, it's going in that direction, that becomes our Moon, and then the rest of this material eventually kind of condenses back into a spherical shape and is what we now call our Earth. So that's how we actually think right now that the Moon actually formed. Even after this happened, the Earth still had a lot more, I guess, violence to experience. Just to get a sense of where we are in the history of Earth, we're going to refer to this time clock a lot over the next few videos, this time clock starts right here at the formation of our solar system, 4.6 billion years ago, probably coinciding with some type of supernova, and as we go clockwise on this diagram, we're moving forward in time, and we're gonna go all the way forward to the present period, and just so you understand some of the terminology, \"Ga\" means \"billions of years ago\" 'G' for \"Giga-\" \"Ma\" means \"millions of years ago\" 'M' for \"Mega-\" So where we are right now, the Moon has formed, and we're in what we call the Hadean period or actually I shouldn't say \"period.\" It's the Hadean eon of Earth. \"Period\" is actually another time period, so let me make this very clear. It's the Hadean, we are in the Hadean eon, and an eon is kind of the largest period of time that we talk about, especially relative to Earth, and it's roughly 500 million to a billion years is an eon, and what makes the Hadean eon distinctive, well, from a geological point of view what makes it distinctive is really we don't have any rocks from the Hadean period. We don't have any kind of macroscopic-scale rocks from the Hadean period, and that's because at that time, we believe, the Earth was just this molten ball of kind of magma and lava, and it was molten because it was a product of all of these accretion events and all of these collisions and all this kinetic energy turning into heat. If you were to look at the surface of the Earth, if you were to be on the surface of the Earth during the Hadean eon, which you probably wouldn't want to be because you might get hit by a falling meteorite or probably burned by some magma, whatever, it would look like this, and you wouldn't be able to breathe anyway; this is what the surface of the Earth would look like. It would look like a big magma pool, and that's why we don't have any rocks from there because the rocks were just constantly being recycled, being dissolved and churned inside of this giant molten ball, and frankly the Earth still is a giant molten ball, it's just we live on the super-thin, cooled crust of that molten ball. If you go right below that crust, and we'll talk a little bit more about that in future videos, you will get magma, and if you go dig deeper, you'll have liquid iron. I mean, it still is a molten ball. And this whole period is just a violent, not only was Earth itself a volcanic, molten ball, it began to harden as you get into the late Hadean eon, but we also had stuff falling from the sky and constantly colliding with Earth, and really just continuing to add to the heat of this molten ball. Anyway, I'll leave you there, and, as you can imagine, at this point there was no, as far as we can tell, there was no life on Earth. Some people believe that maybe some life could've formed in the late Hadean eon, but for the most part this was just completely inhospitable for any life forming. I'll leave you there, and where we take up the next video, we'll talk a little bit about the Archean eon." + }, + { + "Q": "Sal, aren't we meant to name a molecule on the basis of having the smallest number of branches on the main chain? (I'm confused because my chemistry book says that :s) So for the first example, isn't it meant to be 4-(1-methylpropyl) tridecane instead of 3-methyl-4-propyltridecane? because the first name would mean that there's only one branch instead of two.", + "A": "As you noted, there is more than one possible choice of main chains. When this happens, you do not name the molecule based on the smallest number of branches on the main chain. Instead, you choose the main chain whose substituents are least substituted. Both methyl and propyl are simpler than methylpropyl, so the name in the video is correct. In effect, it boils down to choosing the main chain based on the largest number of substituents.", + "video_name": "Se-ekDNhCDk", + "transcript": "We've got a few more molecular structures to name, so let's look at this first one right here. The first thing you always want to do is identify the longest chain. If we start over here, we have one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen carbons, looks pretty long. Now what if we start over here? This looks like it could also be a long chain: one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen. So we have two different chains, depending on whether we want to go up here or whether we go over here, that have a length of thirteen. So you're probably asking, which chain do you choose? You should always, if we have two chains of equal lengths and they're the longest chains, you pick the one that will have more branches or more alkyl groups on it. So this group right here, if we pick this from here to here as our chain, we only have one group on it, that group up there. If we pick this chain, starting over here and then going over here, we have two groups on it. We would have this group over here and then we would have this group over here. This is the better chain to use because it has more groups on it. It has more groups, but the groups are smaller and simpler. So let's start counting. And the direction we want to count, we always want to start on the side of the chain where we're going to encounter something first, and everything is closer this end of the chain so we'll start counting here. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, so we have thirteen carbons on our main chain. Let me draw our main chain. So our main chain is this thing in orange I'm drawing right here. That is our main chain. Thirteen, that's three and ten. The prefix is tridec- and it's tridecane because we have all single bonds here, so it's tridecane. Tridecane is the main chain. Then we have two groups over here. This one in green, this only has one the carbon branching off of the main chain, so it's prefix will be meth- and it'll be a methyl group, so that is methyl. That is a methyl group right there. Then this one down here, we have one, two, three carbons. the prefix is prop-, so this is a propyl group. And the methyl is sitting on the three carbon of our main chain and the propyl group is sitting on the four carbon: one, two, three, four. Now when we figure out what order to list them in when we actually write out the name, M, we just do it in alphabetical order. M comes before P, so we right 3-methyl before 4-propyl. The entire compound here is or the entire molecule is 3-methyl-4-propyltridecane And this is actually all going to be one word. You use dashes to separate when you have numbers, but if you have a word followed by a word, it just becomes propyltridecane, so 3-methyl-4-propyltridecane. And we're done. Let's do this one down here. Now this one seems a little bit more complex. The first thing to see is what is the largest chain or the largest ring that we have in our structure? The two candidates, we have this chain over here. This has one, two, three, four, five, six, seven carbons. Let's see how many carbons our ring has. Our ring has one, two, three, four, five, six, seven, eight, nine carbons. So the ring is the largest core structure in this molecule, so that will be our core structure. We have nine carbons-- let me highlight it-- so the ring is this right here. It has nine carbons. The prefix for nine is non-. It's all single bonds, so it's nonane, and it is in a cycle. It's a ring, so it's cyclononane. This part right here, that part right there, is cyclononane. We have several things that branch off of the cyclononane, so let's look at them one at a time, and then we'll think about how we're going to number them on the ring. We looked already at this at this chain that has seven carbons: one, two, three, four, five, six, seven. So that is a heptyl group. This over here, let's see what we're dealing with. We have one, two, three carbons, so that is just a standard propyl group. Then here, we have one, two, three, four carbons, so we could say this is a butyl group. But this isn't just any butyl group. If we use the common naming, the carbon we immediately touch on or that we immediately get to when we go off of our main ring, that branches off into three other carbons, so this is tert-, tert- for three. The tert- is usually written in italics. It's hard to differentiate that when you see it. I'll write it in cursive. This is tert. This is a tertbutyl group. Now, the next question is how do we specify where these different groups sit on this main ring? If you just had one group, you wouldn't have to specify it, but when you have more than one, what you actually do is you figure out which one would be alphabetically first, and that would be number one. Now, we have an H in heptyl, a P in propyl, and tert-butyl, you might say, well, do I use the T or do I use the B? And this is just the convention, you use the B. If you have sec-butyl or tert-butyl, you ignore the sec- or the tert-. If this was an isobutyl or an isopropyl, you actually would use the I, so it's a little bit-- I guess the best way to think about it is there's a dash here so you can kind of ignore it, but if this was isobutyl it would all be one word, so you would consider the I. So in this situation, we would consider the B, and B comes before P or an H, so that is where we will start numbering one. Then to figure out which direction to keep numbering in, we just go in the direction where we're likely to encounter the first or where we will encounter the first side chain, so we'll go in this direction because we get right to the propyl group. One, two, three, four, five, six, seven, eight, nine. So this compound, we're going to start with the alphabetically first side chain, so it's 1-tert-butyl. I'll write this in cursive. 1-tert-butyl. Then the next one alphabetically is the heptyl group. That's H for heptyl, so then it is 5-heptyl. And then we have the propyl, and then it is 2-propyl, and then finally cyclononane, and we're done!" + }, + { + "Q": "Hi,\n\nI have a question about commas and the word 'AND'. If my memory serves me correctly, the rule was that the comma would replace the word 'and' in a list etc. E.g. If the sentence is -: I have an apple and a banana and a mango. We would replace the \"and's\" with commas and it would now read, 'I have an apple, a banana and a mango'. How is it that in the example above, you guys are placing a comma before the word 'and'? Isn't that like saying \"and and\"?", + "A": "The comma you re questioning is called the Oxford comma. For more information, check out the bonus video titled The Oxford Comma.", + "video_name": "DBMQOK64VQY", + "transcript": "- [Voiceover] Hey Paige! - [Voiceover] What's up David? Okay, so I'm about to go to the grocery store, and it looks like it says, \"I need to get squid pickles and chocolate at the grocery store.\" - [Voiceover] Yeah. - [Voiceover] Did you want squid pickles? - [Voiceover] No, I wanted squid and pickles. - [Voiceover] I must have written it down wrong, okay. So I think what we need to do in order to fix this list and avoid confusion like this in the future is using commas to punctuate a list. 'Cause right now this just looks like squid pickles, which, I mean probably delicious, pickled squid. - [Voiceover] Yeah. - [Voiceover] Sure. - [Voiceover] But not what we were looking to get today. - [Voiceover] Right. - [Voiceover] If we don't want to get pickled squid today, then we have to put commas in between the elements of the list. - [Voiceover] Right. - [Voiceover] Right, because this is what commas do. The separate elements of every everything. So let's put in those dividers. I need to get squid, comma, pickles, comma, and chocolate at the grocery store. - [Voiceover] Exactly. - [Voiceover] Okay, so we can punctuate a list by separating out nouns, and I see from the second sentence here, Paige could you give me a read for that? - [Voiceover] I'm going to go for a run, read a chapter of my book and go see the new Colonel Justice movie. - [Voiceover] Oh I hear that's good. So right now it says that, but it could also just be \"I'm going to go for a run read a chapter of my book, and go see...\" you know there's like no-- - [Voiceover] It's a little confusing. - [Voiceover] It's a little confusing, right? There could be such a thing as a run read. - [Voiceover] There probably is. - [Voiceover] You know, like where you go for a jog while holding a book. - [Voiceover] Sounds difficult. - [Voiceover] And so we can also use commas in lists to separate not just nouns like in this first one, but also verb phrases. So I'm going for a run, comma, read a chapter of my book, comma, and go see the new Colonel Justice movie. - [Voiceover] Perfect. - [Voiceover] Cool. So that's how you punctuate a list with commas. - [Voiceover] Yeah, you got it. - [Voiceover] You can learn anything. David out. - [Voiceover] Paige out." + }, + { + "Q": "Would this apply to an equation like f(g(x))? Would that be the same as f(x) * g(x)? or would it be adding? I was told I would add but that just doesn't make sense. I would love a clarification.", + "A": "No, when finding f(g(x)), you first find what g(x) equals, then using the answer to g(x) as x, you find f(x). For example, f(x)=2x g(x)=4x-1 Find f(g(1)) You would first find g(1), which would be 4(1)-1=3. Then, you would use the 3 to find f(x), which would be 2(3)=6. So, f(g(1))=6. Hope that clarifies things :)", + "video_name": "JKvmAexeMgY", + "transcript": "f of x is equal to 7x minus 5. g of x is equal to x to the third power plus 4x. And then they ask us to find f times g of x So the first thing to realize is that this notation f times g of x is just referring to a function that is a product of f of x and g of x. So by definition, this notation just means f of x times g of x. And then we just have to substitute f of x with this definition, g of x with this definition, and then multiply out these algebraic expressions. f of x is right over there. And g of x, is right over there. So let's do it. So this is going to be equal to-- switch back to the orange color. It's going to be equal to f of x, which is 7x minus 5 times g of x, and g of x is x to the third power plus 4x. And you could-- we're multiplying two expressions that each have two terms. You could use FOIL if you like. I don't like using FOIL because you might forget what it's even about. Foil is really just using the distributive property twice. So for example, you take this expression. Whatever you have out here, if you had a 9 out here, or an a, or an x, or anything. Now you have 7x minus 5. If you're multiplying it times this expression, you would multiply this times each term over here. So when you multiply 7x minus 5 times x to the third, you get-- I'll write it this way. You get x to the third times-- actually, let me write it the other way. You get 7x minus 5 times x to the third. And then you have plus 7x minus 5 times 4x. And now we can do the distributive property again. We're not normally used to seeing the things we distribute on the right hand side. It's the same exact idea. We could put the x to the third here as well. And when we distribute, you multiply x to the third times 7x and times negative 5. x to the third times 7x is 7x to the fourth power. X to the third times negative 5 is minus 5x to the third. And then you do it over here. You distribute the 4x over the 7x. 4x times 7x is plus 28x squared. 4x times negative 5 is minus 20x. And let's see if we can simplify this. We only have one fourth degree term, one third degree term, one second degree term, and one first degree term. Actually, we can't simplify this anymore. And we're done. This is the product of those two function definitions. This is f times g of x. It is a new function created by multiplying the other two functions." + }, + { + "Q": "What if you use the current year as the base year, and calculate real GDP for the previous years using the current year's prices? Would that give you any usable information?", + "A": "When you compare the result of this calculation with that obtained from a normal kind of calculation for the actual year involved (that is as a percentage), it will show you how much the prices of goods taken all together have changed.", + "video_name": "lBDT2w5Wl84", + "transcript": "Let's say we're studying a very small and oversimplified country that only sells apples, and we measure the GDP in year one. And we measure that GDP as $1,000. And all of that is due to apples. And we also know that the price of apples in year one was $0.50 a pound. So I'll write it as $0.50 per pound. And let's say that now year one has gone by and even year two has gone by, and we're able to measure the GDP in year two. So the GDP in year two is $1,200. And the price of apples in year two, let's just say it is $0.55 a pound. So my question to you is, GDP, the whole point of measuring GDP, is measuring the productivity of a country. I mean we are measuring in terms of dollars, but we care more about just the dollar amount. We really care is, was this country more productive? And if it was more productive, how much more productive was it? And if we just look at these GDP numbers right over here, this $1,000 versus this $1,200, it gives you the sense that-- well, at least if you just look at the numbers-- $1,200 is 20% larger than $1,000. So if you just look at those numbers right over there, it looks like the GDP grew by 20%. But is that an accurate representation of the productivity of this country? Did it actually produce 20% more goods? And a big clue is looking at this price here. Because some of this GDP actually might have increased just due to price. But that doesn't actually make the country more productive. The quantity, the extra quantity of apples that the country produces, is actually what adds to the total productivity. One way to think about it-- Let me draw a little diagram over here. On this axis, I'll do quantity. On this axis, I will do price. And P1, so if I want to figure out the GDP in year one, I would have the price of apples in year one-- that's the only good or service, just to simplify things-- times the quantity of apples in year one. And then this right over here, the area of this green rectangle, would GDP in year one. And then GDP in year two would be the price in year two. So we're going to go from $0.50 to $0.55. The price in year two times the quantity in year two-- we'll assume some growth as occurred-- times the quantity in year two. And so GDP in year two would be the area of this entire rectangle. And if we want to find the difference between GDP in year two and GDP in year one, it would be the difference in area. So it would be what I am shading in, in blue right over here. And based on the numbers that we went over right over here, the area that I'm shading in, in blue-- so the difference between GDP in year two and GDP in year one, the area I'm shading in blue-- would be this 200, the 200 increment. So this area right over here would be that 200. Now when you look at it over here, you see that that 200, some of it is due to an increase in quantity. But a lot of it is also due to an increase in price. So if we really wanted to figure out how much more productive the country got, and we still want to measure GDP in dollars, maybe we can take a measure of GDP that measures year two's GDP, but it does it in year one's prices. So if we could somehow multiply-- if we could multiply year two's quantity by year one's prices, then we would get this rectangle right over here. And then the difference between that and year one, would give us the incremental GDP in year one prices due to quantity. And that's what we care about. We care about total productivity. When we're thinking about GDP one, we say how much more productive did the country get? So let's try to do it with these numbers right over here. So we can figure out quantity two, we could figure out the quantity in year two just by dividing the GDP by the price. Just by dividing this area of the entire blue rectangle and dividing it by the price, that will give us the quantity. So if we divide 1,200 divided by $0.55-- let me get my calculator out. So if I do 1,200 divided by $0.55, this is my quantity of apples and in pounds in year two. And I'll just round it, 2,182. So this is 2,182. So the quantity in year two is 2,182 pounds. So this is equal to that. And then I could multiply this times the price. So this is this quantity. It's 2,182 pounds. And then I could multiply it times the price in year one at year one's price. So I'm going to multiply it times-- P1 is equal to $0.50 a pound, $0.50 per pound. And this will give me-- so let me just get my calculator out. I should be able to do that one in my head. But let's see 0.5. And I get 1,090. Obviously, I'll round it to 1,091. So this is equal to 1,091. And this is an interesting number. So this is-- you could view this as year two's GDP. In year-- or adjusted for-- I'll write it, adjusted for prices, or adjusted for price increases. Or you could say in year one prices. And what's useful about this is, this says, look, if prices had remained constant, this is what our GDP would have gotten to. If prices did not increase, our GDP would have gotten to this 1,091. 1,091 is this area that I drew in pink here. And so now, you could say if prices were held constant, the growth in GDP would have been $91 not $200. So this area right over here that I'm-- actually, let me do it in a color. Let me do it in orange, maybe. This area right over here, the actual growth, if prices were held constant, would have been $91. We would have gone from $1,000 of GDP to $1,091. So this right over here, that area, is $91 of-- and we could even call it real growth. It really measures the productivity. Now this gives us an interesting, I guess, set of ideas. One idea is to just measure your GDP in the current year's dollars. So this was GDP measured in year two's dollars. It was year two GDP measured in year two dollars, year two prices. So we could call that year two's nominal GDP. Nominal, in name. So it's GDP in name, in that year's prices. But this right over here, where we measured year two's GDP, in some base year's prices-- so it allows a real comparison of how much did our productivity actually increase. Our productivity actually increased by 9%. We produced 9% more apples. This, we call real GDP. Because it gives you a measure of real productivity. It tries to take out price increases. What we'll see in the future, or we might not do it in an introductory course, but in practice, it's kind of hard to really measure what the absolute-- this was a simple economy, where we only had one product. But if you have many, many, many products-- actually gazillions of products in a real economy and the prices are adjusting and the quantities are adjusting, it's not so easy to figure out how to adjust for price. But the folks running the national income accounts do try to do this. So they get a sense of how much was the actual real growth." + }, + { + "Q": "4:05 12 is a number that is in the problem!\n\nAll numbers divisible by both 12 and 20 are also divisible by 12?", + "A": "Yup, and all numbers divisible by both 12 and 20 are also divisible by 20.", + "video_name": "zWcfVC-oCNw", + "transcript": "- [Voiceover] In this video I wanna do a bunch of example problems that show up on standardized exams and definitely will help you with our divisibility module because it's asking you questions like this. And this is just one of the examples. All numbers divisible by both 12 and 20 are also divisible by: And the trick here is to realize that if a number is divisible by both 12 and 20, it has to be divisible by each of these guy's prime factors. So let's take the prime factorization, the prime factorization of 12, let's see, 12 is 2 times 6. 6 isn't prime yet so 6 is 2 times 3. So that is prime. So any number divisible by 12 needs to be divisible by 2 times 2 times 3. So its prime factorization needs to have a 2 times a 2 times a 3 in it, any number that's divisible by 12. Now any number that's divisible by 20 needs to be divisible by, let's take it's prime factorization. 2 times 10 10 is 2 times 5. So any number divisible by 20 needs to also be divisible by 2 times 2 times 5. Or another way of thinking about it, it needs to have two 2's and a 5 in its prime factorization. If you're divisible by both, you have to have two 2's, a 3, and a 5, two 2's and a 3 for 12, and then two 2's and a 5 for 20, and you can verify this for yourself that this is divisible by both. Obviously if you divide it by 20, let me do it this way. Dividing it by 20 is the same thing as dividing by 2 times 2 times 5, so you're going to have the 2's are going to cancel out, the 5's are going to cancel out. You're just going to have a 3 left over. So it's clearly divisible by 20, and if you were to divide it by 12, you'd divide it by 2 times 2 times 3. This is the same thing as 12. And so these guys would cancel out and you would just have a 5 left over so it's clearly divisible by both, and this number right here is 60. It's 4 times 3, which is 12, times 5, it's 60. This right here is actually the least common multiple of 12 and 20. Now this isn't the only number that's divisible by both 12 and 20. You could multiply this number right here by a whole bunch of other factors. I could call them a, b, and c, but this is kind of the smallest number that's divisible by 12 and 20. Any larger number will also be divisible by the same things as this smaller number. Now with that said, let's answer the questions. All numbers divisible by both 12 and 20 are also divisible by: Well we don't know what these numbers are so we can't really address it. They might just be 1's or they might not exist because the number might be 60. It might be 120. Who knows what this number is? So the only numbers that we know can be divided into this number, well we know 2 can be, we know that 2 is a legitimate answer. 2 is obviously divisible into 2 times 2 times 3 times 5. We know that 2 times 2 is divisible into it, cuz we have the 2 times 2 over there. We know that 3 is divisible into it. We know that 2 times 3 is divisible into it. So that's 6. Let me write these. This is 4. This is 6. We know that 2 times 2 times 3 is divisible into it. I could go through every combination of these numbers right here. We know that 3 times 5 is divisible into it. We know that 2 times 3 times 5 is divisible into it. So in general you can look at these prime factors and any combination of these prime factors is divisible into any number that's divisible by both 12 and 20, so if this was a multiple choice question, and the choices were 7 and 9 and 12 and 8. You would say, well let's see, 7 is not one of these prime factors over here. 9 is 3 times 3 so I need to have two 3's here. I only have one 3 here so 9 doesn't work. 7 doesn't work, 9 doesn't work. 12 is 4 times 3 or another way to think about it, 12 is 2 times 2 times 3. Well there is a 2 times 2 times 3 in the prime factorization, of this least common multiple of these two numbers, so this is a 12 so 12 would work. 8 is 2 times 2 times 2. You would need three 2's in the prime factorization. We don't have three 2's, so this doesn't work. Let's try another example just so that we make sure that we understand this fairly well. So let's say we wanna know, we ask the same question. All numbers divisible by and let me think of two interesting numbers, all numbers divisible by 12 and let's say 9, and I don't know, let's make it more interesting, 9 and 24 are also divisible by are also divisible by And once again we just do the prime factorization. We essentially think about the least common multiple of 9 and 24. You take the prime factorization of 9, it's 3 times 3 and we're done. Prime factorization of 24 is 2 times 12. 12 is 2 times 6, 6 is 2 times 3. So anything that's divisible by 9 has to have a 9 in it's factorization or if you did its prime factorization would have to be a 3 times 3, anything divisible by 24 has got to have three 2's in it, so it's gotta have a 2 times a 2 times a 2, and it's got to have at lease one 3 here. And we already have at least one 3 from the 9, so we have that. So this number right here is divisible by both 9 and 24. And this number right here is actually 72. This is 8 times 9, which is 72. So if the choices for this question, let's assume that it was multiple choice. Let's say the choices here were 16 27 5 11 and 9. So 16, if you were to do its prime factorization, is 2 times 2 times 2 times 2. It's 2 to the 4th power. So you would need four 2's here. We don't have four 2's over here. I mean there could be some other numbers here but we don't know what they are. These are the only numbers that we can assume are in the prime factorization of something divisible by both 9 and 24. So we can rule out 16. We don't have four 2's here. 27 is equal to 3 times 3 times 3, so you need three 3's in the prime factorization. We don't have three 3's. So once again, cancel that out. 5's a prime number. There are no 5's here. Rule that out. 11, once again, prime number. No 11's here. Rule that out. 9 is equal to 3 times 3. And actually I just realized that this is a silly answer because obviously all numbers divisible by 9 and 24 are also divisible by 9. So obviously 9 is going to work but I shouldn't have made that a choice cuz that's in the problem, but 9 would work, and what also would work if we had a, if 8 was one of the choices, because 8 is equal to 2 times 2 times 2, and we have a 2 times 2 times 2 here. 4 would also work. That's 2 times 2. That's 2 times 2. 6 would work since that's 2 times 3. 18 would work cuz that's 2 times 3 times 3. So anything that's made up of a combination of these prime factors will be divisible into something divisible by both 9 and 24. Hopefully that doesn't confuse you too much." + }, + { + "Q": "I know that tan(theta) = opp/adj, so, if I want to calculate the opposite side (the length traveled by the globe) at the given time, and at the given time+1 min, shouldn't I be able to:\ntan( pi/4+.2 ) * 500 - tan( pi/4 ) * 500 = 254.25 meters/min\n\n\nThat all divided by 1 minute of course. However, that clearly doesn't give the expected result, which is 200 meters/min... where am I wrong?", + "A": "To calculate [tan(\u00cf\u0080/4 + 0.2) - tan(\u00cf\u0080/4)] * 500 will give you the result whose unit is meters, not meters/min. We need to find the rate of change.", + "video_name": "_kbd6troMgA", + "transcript": "You're watching some type of hot air balloon show, and you're curious about how quickly one hot air balloon in particular is rising. And you have some information at your disposal. You know the spot on the ground that is directly below the hot air balloon. Let's say it took off from that point, it's just been going straight up ever since. And you know, you've measured it out, that you're 500 meters away from there. So you know that you are 500 meters away from that. And you're also able to measure the angle between the horizontal and the hot air balloon. You could do that with, I don't know, I'm not exactly a surveyor, but I guess a viewfinder or something like that. So you're able to-- and I'm not sure if that's the right tool, but there are tools that you can measure the angles between the horizontal and something that's not on the horizontal. So you know that this angle right over here is pi over 4 radians, or 45 degrees. We're going to keep it pi over 4, because when you take derivatives of trig functions, you assume that you're dealing with radians. So right over here, this is pi over 4 radians. And you also are able to measure the rate at which this angle is changing. So this is changing at 0.2 radians per minute. Now my question to you, or the question that you're trying to figure out as you watch this hot air balloon, is how fast is it rising right now? How fast is it rising just as the angle between the horizontal and kind of the line between you and the hot air balloon is pi over 4 radians, and that angle is changing at 0.2 radians per minute? So let's think about what we know and what we're trying to figure out. So we know a couple of things. We know that theta is equal to pi over 4 if we call theta the angle right over here. So this is theta. We also know the rate at which data is changing. We know d theta. Let me do this in yellow. We know d theta dt is equal to 0.2 radians per minute. Now what are we trying to figure out? Well, we're trying to figure out the rate at which the height of the balloon is changing. So if you call this distance right over here h, what we want to figure out is dh dt. That's what we don't know. So what we'd want to come up with is a relationship between dh dt, d theta dt, and maybe theta, if we need it. Or another way to think about it, if we can come up with the relationship between h and theta, then we could take the derivative with respect to t, and we'll probably get a relationship between all of this stuff. So what's the relationship between theta and h? Well, it's a little bit of trigonometry. We know we're trying to figure out h. We already know what this length is right over here. We know opposite over adjacent. That's the definition of tangent. So let's write that down. So we know that the tangent of theta is equal to the opposite side-- the opposite side is equal to h-- over the adjacent side, which we know is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between d theta dt, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of both sides of this with respect to t implicitly. So let's do that. And actually, let me move over this h over 500 a little bit. So let me move it over a little bit so I have space to show the derivative operator. So let's write it like that. And now let's take the derivative with respect to t. So d dt. I'm going to take the derivative with respect to t on the left. We're going to take the derivative with respect to t on the right. So what's the derivative with respect to t of tangent of theta? Well, we're just going to apply the chain rule here. It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to t, times d theta dt. Once again, this is just the derivative of the tangent, the tangent of something with respect to that something times the derivative of the something with respect to t. Derivative of tangent theta with respect to theta times the derivative theta with respect to t gives us the derivative of tangent of theta with respect to t, which is what we want when we use this type of a derivative operator. We're taking the derivative with respect to t, not just the derivative with respect to theta. So this is the left hand side. And then the right hand side becomes, well, it's just going to be 1 over 500 dh dt. So 1 over 500 dh dt. We're literally saying it's just 1 over 500 times the derivative of h with respect to t. But now we have our relationship. We have the relationship that we actually care about. We have a relationship between the rate at which the height is changing with respect to time and the rate at which the angle is changing with respect to time and our angle at any moment. So we can just take these values up here, throw it in here, and then solve for the unknown. So let's do that. Let's do that right over here. So we get secant squared of theta. So we get secant squared. Right now our theta is pi over 4. Secant squared of pi over 4. Let me write those colors in to show you that I'm putting these values in. Secant squared of pi over 4. Secant times d theta dt. Well, that is just 0.2. So times 0.2. And then this is going to be equal to 1 over 500. And we want to make sure. Since this is in radians per minute, we're going to get meters per. And this is meters right over here. We're going to get meters per minute right over here. We just want to make sure we know what our units are doing. I haven't written the units here to save some space. But we get 1 over 500 times dh dt. So if we want to solve for dh dt, we can multiply both sides by 500, and you get the rate at which our height is changing is equal to 500 times, let's see, secant squared of pi over 4. That is 1 over cosine squared of pi over 4. Let me write this over here. Cosine of pi over 4 is square root of 2 over 2. Cosine squared of pi over 4 is going to be equal to 2 over 4, which is equal to 1/2. And so secant squared of pi over 4 is just 1 over that, is equal to 2. So this is going to be equal to-- let me rewrite this instead of-- so the secant squared of pi over 4-- let me erase this right over here. Secant squared of pi over 4, all of this business right over here simplifies to 2. So times 2 times 0.2 times 0.2. So what is this going to be? This is going to be 500 times 0.4. So this is equal to 500 times-- let me just write a dot instead-- times 0.4, which is equal to-- let me make sure I get this right. This would be with two 0's and one behind the decimal. Yep, there you go. It would be 200. So the rate at which our height is changing with respect to time right at that moment is 200 meters per minute. dh dt is equal to 200 meters per minute." + }, + { + "Q": "If the plates are still moving, will there be a super continent in the future?", + "A": "yes.", + "video_name": "axB6uhEx628", + "transcript": "We know that new plate material is being formed, and these lithosphere plates on the surface of the Earth are moving around. And that might raise the question in your brain-- what happens if we kind of reverse things? We know the direction they're moving in. What does that tell us about where they came from? So let's just do the thought experiment. Right now, South America and Africa are moving away from each other, because of new plate material being created at the mid-Atlantic rift. Let's rewind it. Let's bring them back together. We know that India is jamming into the Eurasian Plate right now, causing the Himalayas to get higher and higher. What if we rewind that? Let's bring India back down towards Antarctica. Same thing with Australia. We have new plate material being formed between Australia and Antarctica that's making the continents move apart. Let's bring them back together. Let's rewind the clock. Even North America-- it's not as obvious from this diagram, but if you actually look at the GPS data, it becomes pretty obvious that North America, right now is moving in a counterclockwise rotation. So let's rewind it into a-- let's go back, moving it in a clockwise direction. Let's, instead of Eurasia going further away from North America, let's bring it back together. And so what you could imagine is a reality where India, Australia are jammed down into South America-- sorry, into Antarctica. South America and Africa are jammed together. North America is jammed in there. And essentially, Eurasia is also jammed in there. So it looks like they all would clump together if you go back a few hundred million years. And based on, literally-- based on just that thought experiment, you could imagine at one point, all of the continents on the world were merged into one supercontinent. And that supercontinent is called Pangaea-- pan for entire, or whole, and gaea, coming from Gaia, for the world. And it turns out that all of the evidence we've seen actually does make us believe that there was a supercontinent called-- well, we call it Pangaea, now. Obviously, there probably weren't things on the planet calling it anything back then. Or, there were things back then, but not things that would actually go and try to label continents that we know of. But all of the evidence tells us that Pangaea existed about 200 to 300 million years ago, roughly maybe 250 million, give or take, years ago. And I want to be clear. This was not the first supercontinent. To a large degree, it's kind of the most recent supercontinent. And it's easiest for us to construct because it was the most recent one. But we believe that there were other supercontinents before this. That if you rewind even more that you would have to break up Pangaea and it would reform. But we're now going back in time. Or that there were several supercontinents in the past that broke up, reformed, broke up, reformed. And the last time we had a supercontinent was Pangaea, about 250 million years ago. And now it's broken up into our current day geography. Now, I won't go into all of the detail why we believe that there was a Pangaea about 250 million years ago-- or, this diagram tells us, about 225 million years ago, give or take. But I'll go into some of the interesting evidence. On a very high level, you have a lot of rock commonalities between things that would have had to combine during Pangaea. And probably the most interesting thing is the fossil evidence. There are a whole bunch of fossils. And here are examples of it, from species that were around between 200 and 300 million years ago. And their fossils are found in a very specific place. This animal right here, cynognathus-- I hope I'm pronouncing that right-- cynognathus. This animal's fossils are only found in this area of South America on a nice clean band here, and in this part of Africa. So not only does South America look like it fits very nicely into Africa. But the fossil evidence also makes it look like there was a nice clean band where this animal lived and where we find the fossils. So it really makes it seem like these were connected, at least when this animal lived, maybe on the order of 250 million years ago. This species right over here, its fossils are found in this area-- let me do it in a color that has more contrast-- in this area right over here. This plant, its fossils-- now, this starts to connect to a lot of dots between a lot of cont-- its fossils are found in this entire area, across South America, Africa, Antarctica, India, and Australia. And so not only does it look like the continents fit together in a puzzle piece, not only do we get it to a configuration like this if we essentially just rewind to the movement that we're seeing now-- but the fossil evidence also kind of confirms that they fit together in this way, This animal right here, we find fossils on this nice stripe that goes from Africa through India, all the way to Antarctica. Now, this only gives us evidence of the Southern Hemisphere of Pangaea. But there is other evidence. We find kind of continuing mountain chains between North America and Europe. We find rock evidence, where just the way we see the fossils line up nicely. We see common rock that lines up nicely between South America and Africa and other continents that were at once connected. So all the evidence, as far as we can tell now, does make us think that there at one time was a Pangaea. And, for all we know, all the continents are going to keep moving. And maybe in a few hundred million years, we'll have another supercontinent. Who knows?" + }, + { + "Q": "why we only took 5 from 20 what about 2,2?", + "A": "we already have the 2 and 2 from prime factorization of 12. We don t want to repeat them.", + "video_name": "zWcfVC-oCNw", + "transcript": "- [Voiceover] In this video I wanna do a bunch of example problems that show up on standardized exams and definitely will help you with our divisibility module because it's asking you questions like this. And this is just one of the examples. All numbers divisible by both 12 and 20 are also divisible by: And the trick here is to realize that if a number is divisible by both 12 and 20, it has to be divisible by each of these guy's prime factors. So let's take the prime factorization, the prime factorization of 12, let's see, 12 is 2 times 6. 6 isn't prime yet so 6 is 2 times 3. So that is prime. So any number divisible by 12 needs to be divisible by 2 times 2 times 3. So its prime factorization needs to have a 2 times a 2 times a 3 in it, any number that's divisible by 12. Now any number that's divisible by 20 needs to be divisible by, let's take it's prime factorization. 2 times 10 10 is 2 times 5. So any number divisible by 20 needs to also be divisible by 2 times 2 times 5. Or another way of thinking about it, it needs to have two 2's and a 5 in its prime factorization. If you're divisible by both, you have to have two 2's, a 3, and a 5, two 2's and a 3 for 12, and then two 2's and a 5 for 20, and you can verify this for yourself that this is divisible by both. Obviously if you divide it by 20, let me do it this way. Dividing it by 20 is the same thing as dividing by 2 times 2 times 5, so you're going to have the 2's are going to cancel out, the 5's are going to cancel out. You're just going to have a 3 left over. So it's clearly divisible by 20, and if you were to divide it by 12, you'd divide it by 2 times 2 times 3. This is the same thing as 12. And so these guys would cancel out and you would just have a 5 left over so it's clearly divisible by both, and this number right here is 60. It's 4 times 3, which is 12, times 5, it's 60. This right here is actually the least common multiple of 12 and 20. Now this isn't the only number that's divisible by both 12 and 20. You could multiply this number right here by a whole bunch of other factors. I could call them a, b, and c, but this is kind of the smallest number that's divisible by 12 and 20. Any larger number will also be divisible by the same things as this smaller number. Now with that said, let's answer the questions. All numbers divisible by both 12 and 20 are also divisible by: Well we don't know what these numbers are so we can't really address it. They might just be 1's or they might not exist because the number might be 60. It might be 120. Who knows what this number is? So the only numbers that we know can be divided into this number, well we know 2 can be, we know that 2 is a legitimate answer. 2 is obviously divisible into 2 times 2 times 3 times 5. We know that 2 times 2 is divisible into it, cuz we have the 2 times 2 over there. We know that 3 is divisible into it. We know that 2 times 3 is divisible into it. So that's 6. Let me write these. This is 4. This is 6. We know that 2 times 2 times 3 is divisible into it. I could go through every combination of these numbers right here. We know that 3 times 5 is divisible into it. We know that 2 times 3 times 5 is divisible into it. So in general you can look at these prime factors and any combination of these prime factors is divisible into any number that's divisible by both 12 and 20, so if this was a multiple choice question, and the choices were 7 and 9 and 12 and 8. You would say, well let's see, 7 is not one of these prime factors over here. 9 is 3 times 3 so I need to have two 3's here. I only have one 3 here so 9 doesn't work. 7 doesn't work, 9 doesn't work. 12 is 4 times 3 or another way to think about it, 12 is 2 times 2 times 3. Well there is a 2 times 2 times 3 in the prime factorization, of this least common multiple of these two numbers, so this is a 12 so 12 would work. 8 is 2 times 2 times 2. You would need three 2's in the prime factorization. We don't have three 2's, so this doesn't work. Let's try another example just so that we make sure that we understand this fairly well. So let's say we wanna know, we ask the same question. All numbers divisible by and let me think of two interesting numbers, all numbers divisible by 12 and let's say 9, and I don't know, let's make it more interesting, 9 and 24 are also divisible by are also divisible by And once again we just do the prime factorization. We essentially think about the least common multiple of 9 and 24. You take the prime factorization of 9, it's 3 times 3 and we're done. Prime factorization of 24 is 2 times 12. 12 is 2 times 6, 6 is 2 times 3. So anything that's divisible by 9 has to have a 9 in it's factorization or if you did its prime factorization would have to be a 3 times 3, anything divisible by 24 has got to have three 2's in it, so it's gotta have a 2 times a 2 times a 2, and it's got to have at lease one 3 here. And we already have at least one 3 from the 9, so we have that. So this number right here is divisible by both 9 and 24. And this number right here is actually 72. This is 8 times 9, which is 72. So if the choices for this question, let's assume that it was multiple choice. Let's say the choices here were 16 27 5 11 and 9. So 16, if you were to do its prime factorization, is 2 times 2 times 2 times 2. It's 2 to the 4th power. So you would need four 2's here. We don't have four 2's over here. I mean there could be some other numbers here but we don't know what they are. These are the only numbers that we can assume are in the prime factorization of something divisible by both 9 and 24. So we can rule out 16. We don't have four 2's here. 27 is equal to 3 times 3 times 3, so you need three 3's in the prime factorization. We don't have three 3's. So once again, cancel that out. 5's a prime number. There are no 5's here. Rule that out. 11, once again, prime number. No 11's here. Rule that out. 9 is equal to 3 times 3. And actually I just realized that this is a silly answer because obviously all numbers divisible by 9 and 24 are also divisible by 9. So obviously 9 is going to work but I shouldn't have made that a choice cuz that's in the problem, but 9 would work, and what also would work if we had a, if 8 was one of the choices, because 8 is equal to 2 times 2 times 2, and we have a 2 times 2 times 2 here. 4 would also work. That's 2 times 2. That's 2 times 2. 6 would work since that's 2 times 3. 18 would work cuz that's 2 times 3 times 3. So anything that's made up of a combination of these prime factors will be divisible into something divisible by both 9 and 24. Hopefully that doesn't confuse you too much." + }, + { + "Q": "At about 7:12 sal says \"for every one molecule of this, we need one molecule of that.\" I thought Fe2O3 would have an ionic bond and therefore technically wouldn't be a molecule. Am I mistaken?", + "A": "You are not mistaken, but we are often a little sloppy in terminology. We technically should be saying formula unit when speaking of Fe\u00e2\u0082\u0082O\u00e2\u0082\u0083 or any other ionic compound instead of molecule .", + "video_name": "SjQG3rKSZUQ", + "transcript": "We know what a chemical equation is and we've learned how to balance it. Now, we're ready to learn about stoichiometry. And this is an ultra fancy word that often makes people think it's difficult. But it's really just the study or the calculation of the relationships between the different molecules in a reaction. This is the actual definition that Wikipedia gives, stoichiometry is the calculation of quantitative, or measurable, relationships of the reactants and the products. And you're going to see in chemistry, sometimes people use the word reagents. For most of our purposes you can use the word reagents and reactants interchangeably. They're both the reactants in a reaction. The reagents are sometimes for special types of reactions where you want to throw a reagent in and see if something happens. And see if your belief about that substance is true or things like that. But for our purposes a reagent and reactant is the same thing. So it's a relationship between the reactants and the products in a balanced chemical equation. So if we're given an unbalanced one, we know how to get to the balanced point. A balanced chemical equation. So let's do some stoichiometry. Just so we get practice balancing equations, I'm always going to start with unbalanced equations. Let's say we have iron three oxide. Two iron atoms with three oxygen atoms. Plus aluminum, Al. And it yields Al2 O3 plus iron. So remember when we're doing stoichiometry, first of all we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation. But I think it's good practice to actually balance the equations ourselves. So let's try to balance this one. We have two iron atoms here in this iron three oxide. How many iron atoms do we have on the right hand side? We only have one. So let's multiply this by 2 right here. All right, oxygen, we have three on this side. We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams. So my question to you is how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So we have two irons and three oxygens. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. Is that right? That's 48 plus 112, right, 160. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? Well 85 grams of iron three oxide is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. I just took 0.53 times 2. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units. So one mole of aluminium is going to be 27 grams. Or 6.02 times 10 to 23 aluminium atoms is going to be 27 grams. So if we need 1.06 moles, how many is that going to be? So 1.06 moles of aluminium is equal to 1.06 times 27 grams. And what is that? What is that? Equals 28.62. So we need 28.62 grams of aluminium, I won't write the whole thing there, in order to essentially use up our 85 grams of the iron three oxide. And if we had more than 28.62 grams of aluminium, then they'll be left over after this reaction happens. Assuming we keep mixing it nicely and the whole reaction happens all the way. And we'll talk more about that in the future. And in that situation where we have more than 28.63 grams of aluminium, then this molecule will be the limiting reagent. Because we had more than enough of this, so this is what's going to limit the amount of this process from happening. If we have less than 28.63 grams of, I'll start saying aluminum, then the aluminum will be the limiting reagent, because then we wouldn't be able to use all the 85 grams of our iron molecule, or our iron three oxide molecule. Anyway, I don't want to confuse you in the end with that limiting reagents. In the next video, we'll do a whole problem devoted to limiting reagents." + }, + { + "Q": "What about 'investing' in a ponzi scheme and withdrawing in the early days before it busts?", + "A": "Lots of people think they are going to be the ones to get out in time! Surprise! A typical characteristic of Ponzi schemes is that when they go, they go fast. If you suspect something is a Ponzi scheme, how will you know whether it is early days or not? You won t. Don t do it.", + "video_name": "5UVpLPtgdF4", + "transcript": "You've probably heard the term Ponzi scheme before, let me write it down, and what we're going to do in this video is explain what it really is. You might have a sense it's some type of scam, that you're taking one person's money and giving it to another, but we're going to do a tangible example of how it actually works. And right here, I have two pictures of the probably the two most famous perpetrators of Ponzi schemes, this is Charles Ponzi right here. It was obviously named after him and then, more recently, this is Bernie Madoff, who pulled off probably the longest-lasting and largest Ponzi scheme of all time. Who knows, maybe there's a longer-lasting and larger one out there that we have still haven't figured out yet, but this is the largest one to date. And Ponzi wasn't the first person to come up with the Ponzi scheme, but they decided to name it after him because he was the first person to really make it famous. This mugshot was taken in the early 1900s when he was finally caught for perpetrating his scheme. So how does it work? So let's start with some investors here. We could get rid of pictures of these two gentlemen. So let's say that I've got a scheme, and what I'm going to do is I'm going to set up my investors, so these are my investors. And then, let's say, we have several time periods. So we can see why the investors think, at least initially, that my scheme is legitimate. So, let's say, that this is the time period, so we have period one, maybe this is years, year one, year two, year three, year four, year five. What I'm going to do is I'm going to write each investor-- I'm going to write down how much money they think they have with me, the person operating the Ponzi scheme and then I'll show you exactly how much money I have and how I can even get away with having the first investors think that I'm legitimate. So let's say I have investor A. So this is going to be their investor's perceived-- let me do this in a different color-- investor's perceived holdings or perceived value. And over here, I'm going to write total actual value. So you can imagine, this is the actual amount of cash that I have. So we have the investor A in blue, and let's say in year one, he gives me $10,000. And I say I I've got a surefire way of doubling his money in the second year. So in the second year, I actually do nothing with the money and you know, I might actually be spending it on my own yachts and you know, fancy suits and whatnot. But let's just say I'm just keeping it in a bank account so you know, he gives it to me, $10,000 and I do nothing with that money. I don't even get interest on it. It's not even in a bank account, I stuff it into my mattress. So the reality is, after a year, it's still only $10,000, but I promised him that I had some type of a genius scheme that could double his money in a year, so I send him a statement that says that his $10,000 is now $20,000. And I feel good about it because I know that he's going to be so excited that his money doubled that he's going to want to keep his money with me, because he'll hope that it can double again. And not only is he going to do that, but he's going to go to the country club and show off to all of his other friends how he was able to do way better than they did with their investments. So he's going to essentially convince other people to join in. So let's say he convinces investor B to join in. Investor B looks at the statement says, hey, this guy running this-- well he doesn't know it's a scheme, let's say he says , Sal seems to know what he's doing, he doubled investor A's money in a year. I'm going to give him a bunch of money, let's say I'm going to give him $15,000 in year two. So how much total actual value do I have in year two now? I have the $10,000 from investor A plus this $15,000, so I have a total of $25,000. This is the actual amount that I have in my bank account, assuming that I'm not spending it on my yacht or my fancy suits. But the total perceived value, let me write that down another line, if everyone actually wanted the amount of money that they thought they had back, I would have to pay out $35,000. But we know that's not going to happen, because people think I'm such a good investor, they want their money to ride as long as possible. So you already see this discrepancy. There's only $25,000 in this little pile of money that I'm collecting, but people think that there should be $35,000. This is their perceived value, because the this guy thought his money doubled, although it didn't, it just sat there. Now, let's say that the next year I sent them statements that say, look, I made super-awesome investments again, the money doubled again. So this guy's money, his perceived value, he gets a statement that says you now have $40,000. This guy down here, investor B, gets a statement that says you now have $30,000. And then they go back to the country club, and they get investor C onboard. They're like, look, both of us have tried out this guy, he's doubled our money two years in a row for both of us, you probably want in on this as well. And investor C is like yeah, well you know, my two buddies, they look like legitimate guys, let me put my money there. Let's say it gets bigger every time because that's usually how these things go, you know, you normally don't don't have just three investors, you'll have hundreds of investors. And the more fake positive returns you get, the more people that want to put their money in. So investor C, let's say he comes in and he puts in $20,000. So what's the reality? Let's focus on it. So there's a perception on year 3, this guy just put his $20,000. A think he's got $40,000, B thinks he has $30,000, so the perceived value here's $40,000 plus $30,000, $70,000 plus $20,000 is $90,000. That's the perceived value. But the actual value is just going to be this $25,000 we had in period two, assuming we didn't spend the money or do anything with it plus the $20,000 that this guy just deposited. So the reality is $45,000. Now let's say that as we go from period three to period four, or let's say right when this guy gets his statement for $40,000, and it's the exact same time period that this guy had put in his $20,000. Let's say person A, he says, you know what, I felt like my ride has gone long enough, I don't want to test fate, let me take my money out and you might say, oh you the scheme will be ruined, but it's going to work because enough money is coming in from new investors to pay this guy off. We now have $45,000 actual value, even though the people think there's $90,000. So if this guy withdraws all of his money, so if he withdraws all of his money so it goes to zero, I have the cash to pay them. I have $40,000 even though people think there's $90,000. So I subtract out $40,000 right here, and there's only $5,000 dollars left in the bank account and since this guy withdrew his money, the perceived value-- this $40,000 is no longer there, I gave the guy the cash-- the perceived value now is $50,000. So I essentially owe people $50,000, investors B and C think that they have $50,000 invested with me, but the reality is that I only have $5,000 in my bank account. And probably a more realistic reality is I was probably spending a lot of this money on my own little luxuries the whole time. But let's continue another way, once again this guy, not only did he double his money for two years, they're all the same country club. This guy doubled his money again, this guy just invested his money. And now this guy says, look, the scheme is legitimate. This legitimized the scheme This is legit. Because, look, I doubled my money for two years and I was able to withdraw the money. So he was able to withdraw his money, so when he goes back to the country club, he gives this guy and that guy more conviction that this is all on the up and up. And then investor D will probably jump in too and say, wow, now that he's doubled money two years in a row, I see it's legitimate. Investor A was able to withdraw his money, I'll give even more money, I'll give $100,000. Maybe this is a ton of people who are now going to put in a $100,000. And then the next year, I double the money again. And obviously, I won't make it exactly double, I'll make it, you know 40% one year and 30% percent the next year so doesn't look too suspicious. I want to make it look like real returns, but for the sake of our math, let's say I double it again. So now and we're in year four. And this guy withdrew all of his money, but investor B now thinks he has $60,000. Investor C thinks he has $40,000. And investor D thinks he has $200,000. Oh, and I forgot to put investor D's deposit here. So when he put a $100,000, I only had $5,000 in my bank account, but then if I add $100,000 I'll now have $105,000 in my bank account after this guy comes in at, you know, at the end of period three, we can imagine. And, with the perceived amount, I owe is $150,000. So the green is what happens after D comes in, so these are no longer valid. But you can see that as more money comes in, I have more and more money to pay out, even though I'm not doing anything. Even though all of these returns are fake. So now I actually have $105,000 even though people think, well that's at the end of period three, at the end of period four, what do people think? People think that I have $300,000 of holdings. Let me write that down. But the reality is I still only have $105,000 of holdings. This is the total actual value. But notice, if this guy or this guy, some of the early investors, wanted to pull out some of their money, although they probably don't want to, because where else can you double your money every year and this guy already showed that I'm good for paying back the money. But even if this guy or this guy wanted to pull out their money, I would be able to give it to them because I have at least enough for those withdrawals. Now, everything would be ruined if everyone gets freaked out or scared and if I have mass withdrawals or if more people withdraw money than there is in the bank plus the amount of money that comes in. So in order for a Ponzi scheme to keep going, and Bernie Madoff was able to do this for very long time, you have to have good, believable, legitimate returns. Although they're not legitimate, they just need to look legitimate. So that you have more money coming in that out. And the whole point of the doing this Ponzi scheme, if you're a stylish criminal, isn't just to keep the cash there, you know, the $10,000 from one period to the next. The whole point of it is to take a lot of that for yourself, for you to live off of and put into some Swiss bank account to be able to escape the country at some point. Anyway, hopefully you found that enjoyable." + }, + { + "Q": "At 2:20 sir, you said that Concrete nouns are those which we can see, count, or measure. We can measure the velocity of air, do we? Is it Concrete or Abstract Noun sir? Thanks!", + "A": "Good question, Jibran! Let s break it down: you can measure the velocity of wind. But is velocity abstract or concrete? I d say it s abstract, because it is about the idea of directional speed. Is air abstract or concrete? Well, I d say it s concrete, because it s nitrogen and oxygen that we can feel and inhale and touch.", + "video_name": "3AF_rN-yN-Y", + "transcript": "- [Voiceover] Hello grammarians. So today I'd like to talk to you about the idea of concrete and abstract nouns, and before we do that, I'd like to get into some word origins or etymology. So let's take each of these words in turn, because I think by digging into what these words mean, literally what they mean and where they come from, we'll get a better understanding of this concept. So both of these words come to us from Latin. Concrete comes to us from the Latin concretus, which means to grow together. So this part of it means grown. And this part means together. It refers to something that, you know, has grown together and become thick and kind of hard to get through and physical. The connotation here is that this is a physical thing. Something that is concrete is physical. Abstract, on the other hand, means to draw something away. So something that is abstract is drawn away from the real, from the concrete, from the physical. So this is not physical. And we make this distinction in English when we're talking about nouns. Is it something that is concrete, is it something you can look at or pick up or smell or sense or something that is abstract, something that isn't physical, but can still be talked about. So for example, the word sadness... Is a noun, right? This is definitely a noun. It's got this noun-making ending, this noun-forming suffix, ness. You know, we take the adjective sad and we toss this ness part onto it, we've got a noun. But can you see sadness? Is it something you can pick up? Sure, you can tell by being, you know observant and empathetic that your friend is sad, but it's not something you can pick up. You can't be like a measurable degree of sad. You couldn't take someone's sadness, put it under a microscope and say \"Oh, Roberta, you are 32 degrees microsad.\" You know, it's not something physical. Concrete things, on the other hand, are things that we can see or count or measure. Just parts of the physical world. So anything you look at, like a dog is concrete, a ball is concrete, a cliff is concrete. Happiness... Is abstract. The idea of freedom... Is abstract. Though the presence of freedom in your life may manifest in physical objects, like \"Oh, my parents let me have the freedom to eat ice cream.\" Ice cream is, you know, a concrete noun. But freedom, the thing that allows you, you know, the permission that you get from your parents to have ice cream. That's not a physical object. So that's basically the difference. So a concrete noun is a physical object and an abstract noun is not. This is why I really wanted to hit the idea that a noun can be a person, place, thing or idea, because nouns can be ideas, and those ideas tend to be abstract. Sadness, happiness, freedom, permission, liberty, injustice. All of these are abstract ideas. That's the difference. You can learn anything. David out." + }, + { + "Q": "at 3:46 , Can you really just square the 1/3 and then plug it into the radical? I have never seen that before.", + "A": "Yes as you re not changing the equation in any way. You re basically squaring a positive multiple then when you place it inside the radical you re square rooting it again so it s exactly the same multiple. (1/3)^2 = 1/9 sqrt(1/9) = 1/3", + "video_name": "WAoaBTWKLoI", + "transcript": "In the last video, in order to evaluate this indefinite integral, we first made the substitution that x is equal to 3 sine theta. And then this got us to an integral of this form. Then we were able to break up these sines and cosines and use a little bit of our trig identities. To get it into the form where we could do u substitution, we did another substitution where we said that u is equal to cosine of theta. And then finally, we were able to get it into a form using that second round of substitution. And this time, it was u substitution. We were able to get it into a form that we could actually take the antiderivative. And we got the final answer here in terms of u. But now we've got to go and undo everything. We have to undo the substitutions. So the last substitution we had done-- we're now going to go in reverse order-- was that u was equal to cosine theta. So you might just want to substitute u with cosine theta here. But then we're going to have everything in terms of cosine theta, which still doesn't get us to x. So the ideal is if I can somehow express u in terms of x. So let's think how we can do it. We know that u is equal to cosine theta. We know the relationship between x and theta is right over here. x is equal to 3 sine theta. So let's write that over here. So we know that x-- let me write it over here-- we know that x is equal to 3 sine theta. So if we could somehow write cosine-- let me rewrite this a little bit differently. Or we could also say that x over 3 is equal to sine theta. I just divided both sides by 3. So if we could somehow re-express this in terms of sine theta, then we can replace all the sine thetas with x over 3's, and we are done. So how can we do that? And I'll actually show you two techniques for doing it. So the first one is to make the realization, OK, u is equal to cosine of theta. If I want to write this in terms of sine of theta, I can just say that this is equal to-- straight up, this is the most fundamental trigonometric identity. Cosine theta is the square root of 1 minus sine squared theta. And we see sine of theta is equal to x over 3. So this is the square root of 1 minus x over 3 squared. So this is u in terms of x. So everywhere we see a u up here we can replace it with this expression. And we are essentially done. We would have written this in terms of x. Now, there's another technique you might sometimes see in a calculus class where someone says, OK, we know that u is equal to cosine theta. We know this relationship. How can we express u in terms of x? And we'll say, let's draw a right triangle. They'll draw a right triangle like this. They'll draw a right triangle, and they'll say, OK, look, sine of theta is x over 3. So if we say that this is theta right over here, sine of theta is the same thing as opposite over hypotenuse. Opposite over hypotenuse is equal to x over 3. So let's say that this is x and then this right over here is 3. Then the sine of theta will be x over 3. So we look at that first substitution right over here. But in order to figure out what u is in terms of x, we need to figure out what cosine of theta is. Well, cosine is adjacent over hypotenuse. So we have to figure out what this adjacent side is. Well, we can just use the Pythagorean theorem for that. Pythagorean theorem would tell us that this is going to be the square root of the hypotenuse squared, which is 9, minus the other side squared, minus x squared. So from this, we fully solved the right triangle in terms of x. We can realize that cosine of theta is going to be equal to the adjacent side, square root of 9 minus x squared, over the hypotenuse, over 3, which is the same thing as 1/3 times the square root of 9 minus x squared, which is the same thing if we square 1/3 and put it into the radical. So we're essentially going to take the square-- 1/3 is the same thing as the square root of 1/9. So can rewrite this as the square root of 1/9 times 9 minus x squared. Essentially, we just brought the 1/3 third into the radical. Now it's 1/9. And so now this is going to be the same thing as the square root of 1 minus x squared over 9, which is exactly this thing right over here. x squared over 9 is the same thing as x over 3 squared. So either way, you get the same result. I find using the trig identity right over here to express cosine of theta in terms of sine theta and then just do the substitution to be a little bit more straightforward. But now we can just substitute into the original thing. So either of these-- I can write it as either way-- this thing right over here, this is the same thing as 1 minus x squared over 9 to the 1/2 power. That's what u is equal to. And everywhere we see u, we just substitute it with this thing. So our final answer in terms of x is going to be equal to 243 times u to the fifth, this to the fifth power over 5. This to the fifth power is 1 minus x squared over 9. It was to the 1/2, but if we raise that to the fifth power, it's now going to be to the 5/2 power over 5 minus this to the third power, 1 minus x squared over 9 to the 3/2 raising this to the third power-- that's this right over here-- over 3, and then all of that plus c. And we're done. It's messy, but using first trig substitution then u substitution, or trig substitution then rearranging using a couple of our techniques for manipulating these powers of trig functions, we're able to get into a form where we could use u substitution, and then we were able to unwind all the substitutions and actually evaluate the indefinite integral." + }, + { + "Q": "I couldn't seem to find any videos explaining triangle altitudes and how to use them. I have a math problem that I can't seem to figure out:\n\n~An equilateral triangle has an altitude length of 10 feet. Determine the length of a side of a triangle.\n\nWould the sides be 10 feet as well?", + "A": "They would not, because the height of the equilateral triangle is not the length of a side. Before giving you the answer, let me give you a few hints, and you try it out. 1. Draw an equilateral triangle and mark the altitude as a dashed line on the triangle. 2. Let each side be of length l Can you solve for the side length? Hint: Think Pythagoras", + "video_name": "aGwT2-RERXY", + "transcript": "What I want to do in this video is to show that if we start with any arbitrary triangle-- and this will be the arbitrary triangle that we're starting with-- that we can always make this the medial triangle of a larger triangle. And when we say the medial triangle, we mean that each of the vertices of this triangle will be the midpoint of the sides of a larger triangle. And I wanted to show that you can always construct that. If you start with this triangle, you can always have this be the medial triangle of a larger triangle. So to do that, let's draw a line that goes through this point right over here, but that's parallel to this line down here. So this line and this line up here are going to be parallel. So just like that. And immediately we can start to say some interesting things about the angles. So if we have a transversal right over here, we could view this side as a transversal of these two parallel lines, or of this line in the segment. We know that alternate interior angles are congruent. So that angle is going to be congruent to that angle. And we also know that this angle in blue, is going to be congruent to that angle right over there. Now, let's do that for the other two sides. So let's create a line that is parallel to this side of the triangle, but that goes through this point right over here. So let me draw it as well as possible. And so these two characters are going to be parallel, and you could always construct a line that's parallel to another line that goes to a point that's not on that line. And so once again, we can use alternate interior angles. We know that if this angle right over here-- let's say we have this orange angle-- it's alternate interior angle is this angle right over there. We also have corresponding angles. This blue angle corresponds to this angle right over here. So it will correspond to that angle right over there. And now let's draw another line that is parallel to this line right over here, but it goes through this vertex. It goes through the vertex that's opposite that line. And so let me just draw it. And you can always construct these parallel lines just like that. And let's see what happens. So once again, these two lines are parallel. So you could view this green line as a transversal. If this green line is a transversal, this corresponding angle is this angle right over here. If we view this green line as a transversal of both of these pink lines, then this angle corresponds to this angle right over here. If we view this yellow line as a transversal of both of these pink lines-- actually, let's look at it this way. View the pink line as a transversal of these two yellow lines, then we know that this angle corresponds to this angle right over here. And if you view this yellow line as a transversal of these two pink lines, then this angle corresponds to this angle right over here. And then the last thing we need to think about is if we think about the two green parallel lines and you view this yellow line as a transversal, then this corresponding angle in orange is right over here. This corresponds to that angle, because this yellow line is a transversal on both of these green lines. So what I've just shown starting with this inner triangle right over here is that if I construct these parallel lines in this way, that I now have four triangles if I include the original one, and they're all going to be similar to each other. And we know that they're all similar because they all have the exact same angles. You just need two angles to prove similarity. But all four of these triangles have the exact three angles. Now, the other thing we can show is that they're congruent. So all of these four are similar. And we also know they're congruent. For example, this side right over here in yellow is the side in this triangle, between the orange and the green side, is the side between the orange and the green side on this triangle right over here. So these two-- we have an angle, a side, and an angle. Angle-side-angle congruency. So these two are going to be congruent to each other. Then over here, on this inner triangle, our original triangle, the side that's between the orange and the blue side is going to be congruent to the side between the orange and the blue side on that triangle. Once again, we have angle-side-angle congruency. So this is congruent to this, which is congruent to that. All of these are going to be congruent. And by the same exact argument, this middle triangle is going to be congruent to this bottom triangle. You have an angle, blue angle, purple side, green angle. Blue angle, purple side, green angle. They're congruent to each other. So you have all of these triangles are congruent to each other. So their corresponding sides are equal. So if you look at this triangle over here, we know that the side between the blue angle and the green angle is going to be equal to this angle right over here. Sorry, equal to this length. So it's going to be equal to this length. Between the blue and the green we have this length, between the blue and the green we have that length, between the blue and the green we have that length right over there. So you immediately see that this point-- and let me label it now, maybe I should've labeled it before. If we call that point A, we see that A is the midpoint of-- let's call this point B, and call this point C right over here. So A is the midpoint of BC. So that's fair enough. So I was able to construct it in that way. Now let's look at the other sides. So this green side on all the triangles is the side between the blue and the orange angle. So between the blue and the orange angle, you have the green side, between the blue and the orange angle you have the green side. So once again, this length is equal to this length. And so if we call this point over here D, and maybe this point over here E, you see that D is the midpoint of BE. And then finally, the yellow side is between the green and the orange. So between the green and the orange, we have a yellow side. Between the green and the orange you have a yellow side. All of these triangles are congruent. So once again, let me call this F. We see that F is the midpoint of EC. So we've done what we wanted to do. We've shown that if you start with any arbitrary triangle, triangle ADF, we can construct a triangle BCE so that ADF is triangle BCE's medial triangle. And all that means is that the vertices of ADF sit on the midpoints of BCE. So you might say Sal, that by itself is interesting, but what's the whole point of this? The whole point of this is actually, I wanted to use this fact that if you give me any triangle, I can make it the medial triangle of the larger one to prove that the altitudes of this triangle are concurrent. And to see that, let me first draw the altitudes. So an altitude from vertex A looks like this. It starts at the vertex, goes to the opposite side, and is perpendicular to the opposite side. If I draw an altitude from vertex D, it would look like this. And if I draw an altitude from vertex F, it will look like this. And what I did, this whole set up of this video is to show, to prove that these will always be concurrent. And you might say, wait how do we know that they are concurrent? Well all you have to do is think about how they interact with the larger triangle. What are these altitudes to the larger triangle? Well, this yellow altitude to the larger triangle. Remember, these two yellow lines, line AD and line CE are parallel. So if this is a 90-degree angle, so its alternate interior angle is also going to be 90 degrees. So this right over here is perpendicular to CE, and it bisects CE, because we know that ADE is the medial triangle. This is the midpoint. So this right over here is perpendicular bisector. This is a perpendicular bisector for the larger triangle, for triangle BCE. So this altitude for the smaller one is a perpendicular bisector for the larger one. We can do that for all of them. If this angle right over here is 90 degrees, then this angle right over there is going to be 90 degrees, because this line is parallel to this, this is a transversal, alternate interior angles are the same. So this line right over here, this altitude of the smaller triangle, it bisects right at the midpoint of the larger one, on this side, and it's also a perpendicular bisector. So it's a perpendicular bisector of the larger triangle. And then finally, the same thing is true of this altitude right over here. It bisects this side of the larger triangle at a 90-degree angle. We know that because these two magenta lines the way we constructed the larger triangle, they're going to be parallel. So once again, this is a perpendicular bisector. So this whole reason, if you just give me any triangle, I can take its altitudes and I know that its altitude are going to intersect in one point. They're going to be concurrent. Because for any triangle, I can make it the medial triangle of a larger one, and then it's altitudes will be the perpendicular bisector for the larger And we already know that the perpendicular bisectors for any triangle are concurrent. They do intersect in exactly one point." + }, + { + "Q": "Did she ever have arms or a head or did she lose them ?", + "A": "Marble is very fragile material. Sculptors often carved torsos separately from heads (sometimes) and from arms, particularly arms held out and away from the body. The elements can become detached. They also can break off.", + "video_name": "TPM1LuW3Y5w", + "transcript": "STEVEN ZUCKER: We're in the Louvre at the top of one of the grand staircases. And we're looking at the \"Nike of Samothrace,\" that dates to the second century CE, or after Christ. BETH HARRIS: So we're in the Hellenistic period. And the sculpture is nine feet high, so it's really large. STEVEN ZUCKER: It's called the \"Nike of Samothrace\" because it was found on the island in the north of the Aegean which is called Samothrace. It was found in a sanctuary in the harbor that actually faces in such a way the predominant wind that blows off the coast actually seems to be enlivening her drapery. BETH HARRIS: So she never stood on the prow of a real boat. STEVEN ZUCKER: No, she stood on the prow of a stone ship that was within a temple environment. BETH HARRIS: So she's the goddess of victory. She's a messenger goddess who spreads the news of victory. STEVEN ZUCKER: In fact, there are some reconstructions of what the sculpture would've originally looked like that show her as literally a herald with a horn. This is an image that will have an enormous impact on Western art. But you had mentioned the Hellenistic before. And so gone is all of that very reserved, high classical style. And in its place is a kind of voluptuousness. is a kind of windswept energy that is full of motion and full of emotion. BETH HARRIS: I feel as though she moves in several directions at the same time. She's grounded by her legs but strides forward. Her torso lifts up. Her abdomen twists. Her wings move back. One can almost feel the wind around her, whipping her, pulling back that drapery that flows out behind her, swirling around her abdomen, where it really reminds us of, actually, the sculptures of hundreds of years earlier on the Parthenon frieze. STEVEN ZUCKER: Yes, exactly. But instead of the quiet, relaxed attitude of the gods on Mount Olympus, you have instead this sense of energy and a goddess that's responding, in this case, to actually natural forces. BETH HARRIS: The environment. STEVEN ZUCKER: Absolutely, just as we would stand there, very likely having the wind whip around us. BETH HARRIS: And that drapery that clings to her body and creates so many creases and folds that play against the light, and the different texture of her wings-- the marble is really made to do so many different things in terms of texture. STEVEN ZUCKER: So here is a culture that has studied the body, celebrated the body, and then is willing then to use the body for tremendous expressive force." + }, + { + "Q": "how dose it make oxygen is only trees do?", + "A": "Bacteria also make oxygen.", + "video_name": "pBZ-RiT5nEE", + "transcript": "In the video on electronegativity, we learned how to determine whether a covalent bond is polar or nonpolar. In this video, we're going to see how we figure out whether molecules are polar or nonpolar and also how to apply that polarity to what we call intermolecular forces. Intermolecular forces are the forces that are between molecules. And so that's different from an intramolecular force, which is the force within a molecule. So a force within a molecule would be something like the covalent bond. And an intermolecular force would be the force that are between molecules. And so let's look at the first intermolecular force. It's called a dipole-dipole interaction. And let's analyze why it has that name. If I look at one of these molecules of acetone here and I focus in on the carbon that's double bonded to the oxygen, I know that oxygen is more electronegative than carbon. And so we have four electrons in this double bond between the carbon and the oxygen. So I'll try to highlight them right here. And since oxygen is more electronegative, oxygen is going to pull those electrons closer to it, therefore giving oxygen a partial negative charge. Those electrons in yellow are moving away from this carbon. So the carbon's losing a little bit of electron density, and this carbon is becoming partially positive like that. And so for this molecule, we're going to get a separation of charge, a positive and a negative charge. So we have a polarized double bond situation here. We also have a polarized molecule. And so there's two different poles, a negative and a positive pole here. And so we say that this is a polar molecule. So acetone is a relatively polar molecule. The same thing happens to this acetone molecule down here. So we get a partial negative, and we get a partial positive. So this is a polar molecule as well. It has two poles. So we call this a dipole. So each molecule has a dipole moment. And because each molecule is polar and has a separation of positive and negative charge, in organic chemistry we know that opposite charges attract, So this negatively charged oxygen is going to be attracted to this positively charged carbon. And so there's going to be an electrostatic attraction between those two molecules. And that's what's going to hold these two molecules together. And you would therefore need energy if you were to try to pull them apart. And so the boiling point of acetone turns out to be approximately 56 degrees Celsius. And since room temperature is between 20 and 25, at room temperature we have not reached the boiling point of acetone. And therefore, acetone is still a liquid. So at room temperature and pressure, acetone is a liquid. And it has to do with the intermolecular force of dipole-dipole interactions holding those molecules together. And the intermolecular force, in turn, depends on the electronegativity. Let's look at another intermolecular force, and this one's called hydrogen bonding. So here we have two water molecules. And once again, if I think about these electrons here, which are between the oxygen and the hydrogen, I know oxygen's more electronegative than hydrogen. So oxygen's going to pull those electrons closer to it, giving the oxygen a partial negative charge like that. The hydrogen is losing a little bit of electron density, therefore becoming partially positive. The same situation exists in the water molecule down here. So we have a partial negative, and we have a partial positive. And so like the last example, we can see there's going to be some sort of electrostatic attraction between those opposite charges, between the negatively partially charged oxygen, and the partially positive hydrogen like that. And so this is a polar molecule. Of course, water is a polar molecule. And so you would think that this would be an example of dipole-dipole interaction. And it is, except in this case it's an even stronger version of dipole-dipole interaction that we call hydrogen bonding. So at one time it was thought that it was possible for hydrogen to form an extra bond. And that's where the term originally comes from. But of course, it's not an actual intramolecular force. We're talking about an intermolecular force. But it is the strongest intermolecular force. The way to recognize when hydrogen bonding is present as opposed to just dipole-dipole is to see what the hydrogen is bonded to. And so in this case, we have a very electronegative atom, hydrogen, bonded-- oxygen, I should say-- bonded to hydrogen. And then that hydrogen is interacting with another electronegative atom like that. So we have a partial negative, and we have a partial positive, and then we have another partial negative over here. And this is the situation that you need to have when you have hydrogen bonding. Here's your hydrogen showing intermolecular force here. And what some students forget is that this hydrogen actually has to be bonded to another electronegative atom in order for there to be a big enough difference in electronegativity for there to be a little bit extra attraction. And so the three electronegative elements that you should remember for hydrogen bonding are fluorine, oxygen, and nitrogen. And so the mnemonics that students use is FON. So if you remember FON as the electronegative atoms that can participate in hydrogen bonding, you should be able to remember this intermolecular force. The boiling point of water is, of course, about 100 degrees Celsius, so higher than what we saw for acetone. And this just is due to the fact that hydrogen bonding is a stronger version of dipole-dipole interaction, and therefore, it takes more energy or more heat to pull these water molecules apart in order to turn them into a gas. And so, of course, water is a liquid at room temperature. Let's look at another intermolecular force. And this one is called London dispersion forces. So these are the weakest intermolecular forces, and they have to do with the electrons that are always moving around in orbitals. And even though the methane molecule here, if we look at it, we have a carbon surrounded by four hydrogens for methane. And it's hard to tell in how I've drawn the structure here, but if you go back and you look at the video for the tetrahedral bond angle proof, you can see that in three dimensions, these hydrogens are coming off of the carbon, and they're equivalent in all directions. And there's a very small difference in electronegativity between the carbon and the hydrogen. And that small difference is canceled out in three dimensions. So the methane molecule becomes nonpolar as a result of that. So this one's nonpolar, and, of course, this one's nonpolar. And so there's no dipole-dipole interaction. There's no hydrogen bonding. The only intermolecular force that's holding two methane molecules together would be London dispersion forces. And so once again, you could think about the electrons that are in these bonds moving in those orbitals. And let's say for the molecule on the left, if for a brief transient moment in time you get a little bit of negative charge on this side of the molecule, so it might turn out to be those electrons have a net negative charge on this side. And then for this molecule, the electrons could be moving the opposite direction, giving this a partial positive. And so there could be a very, very small bit of attraction between these two methane molecules. It's very weak, which is why London dispersion forces are the weakest intermolecular forces. But it is there. And that's the only thing that's holding together these methane molecules. And since it's weak, we would expect the boiling point for methane to be extremely low. And, of course, it is. So the boiling point for methane is somewhere around negative 164 degrees Celsius. And so since room temperature is somewhere around 20 to 25, obviously methane has already boiled, if you will, and turned into a gas. So methane is obviously a gas at room temperature and pressure. Now, if you increase the number of carbons, you're going to increase the number of attractive forces that are possible. And if you do that, you can actually increase the boiling point of other hydrocarbons dramatically. And so even though London dispersion forces are the weakest, if you have larger molecules and you sum up all those extra forces, it can actually turn out to be rather significant when you're working with larger molecules. And so this is just a quick summary of some of the intermolecular forces to show you the application of electronegativity and how important it is." + }, + { + "Q": "what is a propaganda", + "A": "Information, especially of a biased or misleading nature, used to promote a political cause or point of view.", + "video_name": "dHXzusNSF60", + "transcript": "Despite the fact that Wilson had just won reelection in 1916 based on a platform of keeping the United States out of war, by April of 1917, the administration had decided that Germany had gone too far. And in particular, had gone too far with the unrestricted submarine warfare. So this right over here is a picture of President Wilson on April 2, 1917, giving a war message to Congress as to why the US needs to declare war on Germany. And April 4, Congress passes the resolution to declare war. And then the President approves it on April 6. So by early April, the United States was at war with Germany. Which is a good time to start thinking about, why did all of this happen. Now, the things that are typically cited, and these are the things that are inflamed public opinion in the US and that many of which were cited by President Woodrow Wilson. And in this tutorial that this is part of on khanacademy.org, I put the entire text of his speech, which I highly recommend reading to see all of the things the President Wilson cited in his speech. But just as a summary of that, the things that tend to get cited most often are the unrestricted submarine warfare on the part of Germany. And particular cases or the most cited example of that is the sinking of the Lusitania. The Germans had stopped doing that for a little under two years. But then, as we enter into 1917, they began doing it again. And it also made the Americans quite angry to realize that the Germans were trying to incite the Mexicans against them. So you have the Zimmerman telegram. Zimmerman telegram is also a reason that the Wilson administration, and why people in general, were fairly angry about things. Now, on top of that, there were atrocities committed by the Germans in their march through Belgium as they were trying to execute on the Schlieffen Plan. So Belgian atrocities. And these were earlier in the war in 1914, which immediately made many Americans not like what's going on. Belgian atrocities. And to put on top of that, the British were able to leverage the Belgian atrocities to fairly, to execute a fairly effective propaganda campaign in America. Now on top of that-- and this is something that Wilson speaks very strongly about in his speech-- is the notion of fighting for democracy. And what you have here, in the First World War, the Central Powers. You're talking about the German Empire, you're talking about the Austro-Hungarians. These are monarchies. These are emperors who are controlling it. And even though the UK, the United Kingdom, was nominally a kingdom, it was really a democracy. At least for those who could vote. We're not talking about the entire British Empire. So UK is functionally a democracy, democratic. And so was the Third French Republic. And so was France. So there's this argument that the US is fighting for the representation of people. Now, there is a more cynical argument that some people have made. And I think it's reasonable to give that to due time. And one of the cynical arguments, or more cynical arguments, is that the US had close financial and trade ties to Britain, not to mention cultural ties. Financial ties to the British. On top of that, you had very successful British propaganda. One, talking about the atrocities in Belgium, which did actually happen. But the British were able to exploit this as a propaganda machine. Successful propaganda. But they also spread rumors that after the sinking of the Lusitania that the Germans had their school children celebrating. And these were all made up propaganda. And then, more cynical view of why the US entered the war-- and this is true of probably most wars-- is that there was a lot of lobbying on the part of war profiteers. In fact, in \"Little Orphan Annie,\" Daddy Warbucks, the name, the reason why his last name is Warbucks is because he made his fortune as a war profiteer during World War I. And war profiteers, these are people who might be selling arms to the Allies. Or who might sell arms to the US government if the US were to get into a war that might somehow supply the troops. And it includes, potentially, folks on Wall Street. There were significant lending to the Allies, and mainly the Allies, not the Central Power. And so the view is if the Allies win, those loans are going to be made good. And I had the entire text of the speech from Senator George Norris who was one of five senators, or sorry, one of six senators to vote against the resolution There were 50 representatives who also voted against it. This is a little excerpt but also in this tutorial, I have the full text of his speech. And I highly, highly, highly recommend reading that along with Wilson's text of his speech to Congress in his war message. But I'll just read this part because it does, I think, point out that the US, from the beginning, did have biases that were more pro-British. And so this is part of his speech. \"The reason given by the President in asking Congress to declare war against Germany is that the German government has declared certain war zones, within which by the use of submarines, she sinks, without notice, American ships and destroys American lives. The first war zone was declared by Great Britain. She gave us and the world notice of it on the 4th day of November 1914. The zone became effective November 5, 1914. This zone, so declared by Great Britain, covered the whole of the North Sea. The first German war zone was declared on the 4th day of February, 1915, just three months after the British war zone was declared. Germany gave 15 days notice of the establishment of her zone, which became effective on the 18th day of February, 1915. The German war zone cover the English Channel and the high seawaters around the British Isles. It is unnecessary to cite authority to show that both of these orders declaring military zones were illegal and contrary to international law. It is sufficient to say that our government has officially declared both of them to be illegal and has officially protested against both of them. The only difference is that, in the case of Germany we have persisted in our protest, while in the case of England, we have submitted.\" And I encourage you, once again, to read the text of both Wilson's speech and Senator Norris' speech and come up to your, with your own decisions. And it might be a little bit of both." + }, + { + "Q": "on minute 2:50, Sal says US government buys treasury securities, or other kind of securities...\nthis means treasury bonds, that they issued themselves right?\nwhat other kind of securities is Sal referring to?\nAlso, who are those people government buys securities from?", + "A": "Under normal operations, the Fed (not the US Government) buys treasury securities. In the last few years, in the wake of the credit crisis, the Fed has taken to purchasing other types of securities as well (MBS, for example). They buy these on the open market, through broker/dealers.", + "video_name": "wDuCOxDxMzY", + "transcript": "Let's say we have two banks, bank A and bank B, and you might already know that banks, all banks, lend out the great majority of the money that they get in as deposits, but they keep some of the money as reserves. One, just in case their depositor comes and hey, can I have some of my money back and two because the central bank, the Federal Reserve, says you have to keep a certain amount of your deposits in reserve. There is a reserve requirement, but you can imagine over the course of doing transactions, thousands of transactions a day, millions maybe, maybe bank B more of its depositors come by and say hey, give me some of my deposits back. Obviously he's lent out a lot of that money and so he starts running low on reserves. Maybe bank A, the depositors haven't asked for the money or for whatever reason Bank A is sitting on a lot of cash. In this situation, what you're going to have happen is bank A is going to lend some reserves, is going to lend some cash to bank B. This is lending some cash and they'll charge an interest rate for lending that cash. Maybe it will be 5% interest and that won't be 5% per day and usually these loans are on a per day basis and then the next day it'll be renegotiated on a per day basis, but it's not 5% per day. It'll be 5% per year, so it will be a much smaller fraction, but usually as I mentioned this lending takes place on a per day basis. We'll say hey, this is your cash for just tonight. If you need it for the next night, we'll talk again and maybe it will be another 5% or maybe the interest rate can change again. Let's say the Federal Reserve is sitting over here and for whatever reason wants to stimulate the economy. This is the Fed and they want to stimulate the economy so they start printing some money. I should do money in green. The Federal Reserve here, they're starting to print some money and they want to do two things. They want to inject this money into the banking system which essentially, hopefully, will find its way into the economy and they also want to lower the interest rate, especially the short term interest rate. This overnight borrowing. Remember this is the annual interest rate, but this is an overnight loan. Overnight loan. When I talk about the short term interest rate, I'm talking about the interest rate on loans that are made over very short periods of time. What the Federal Reserve will do is what's called open market operations. They will go to the market and maybe directly to these banks or some other banks and they will buy treasuries. They will give this money to the market and in exchange, they will usually buy treasury securities. Sometimes something slightly different, but usually very safe securities and maybe it's temporarily buy. I'll take about repurchase agreements in the future. What happens is that this cash goes in the hands of the people who just sold the treasury securities and they have to deposit it in banks. They might deposit it in this bank over here. They might deposit it in this bank over here or other banks, but the net-net effect is that there's more cash now in the banking system. If there's more cash in the banking system, this guy right over here needs less. This guy needs less cash, so it lowers demand. It lowers the demand for cash and then this guy has more to give, so it raises supply. Raises the supply because some people maybe just took some of this cash and deposited with them. If it raises the supply of cash and this guy needs it less, then the rate to borrow this cash is going to go down. Maybe instead of 5%, it's goes down to 4%. What that would do is it would lower the short term of the yield curve, the short end of the yield curve. Let me draw a yield curve right over here. This is maturity on this axis, maturity, and this is yield. Let's say the yield curve before looked like this. Let's say it looked like this where this right over here is 5% and this overnight. Overnight. This might be yield on, I don't know, one year debt. This might be yield on, I don't know, maybe it's five year debt. Whatever, I could keep going, but by doing this open market operation, the Fed was able to do both of its goals. It was able to inject cash, printed cash, into the economy and it's also able to lower the interest rate. It took it from being 5% to down to 4%. Now because of this open market operation, the Fed, the yield curve might start to look something like that." + }, + { + "Q": "why is velocity zero when no external forces are acting on body?", + "A": "Velocity may or may not be zero. Acceleration is zero because F = ma (Newton s second law)", + "video_name": "VrflZifKIuw", + "transcript": "I will now do a presentation on the center of mass. And the center mass, hopefully, is something that will be a little bit intuitive to you, and it actually has some very neat applications. So in very simple terms, the center of mass is a point. Let me draw an object. Let's say that this is my object. Let's say it's a ruler. This ruler, it exists, so it has some mass. And my question to you is what is the center mass? And you say, Sal, well, in order to know figure out the center mass, you have to tell me what the center of mass is. And what I tell you is the center mass is a point, and it actually doesn't have to even be a point in the object. I'll do an example soon where it won't be. But it's a point. And at that point, for dealing with this object as a whole or the mass of the object as a whole, we can pretend that the entire mass exists at that point. And what do I mean by saying that? Well, let's say that the center of mass is here. And I'll tell you why I picked this point. Because that is pretty close to where the center of mass will be. If the center of mass is there, and let's say the mass of this entire ruler is, I don't know, 10 kilograms. This ruler, if a force is applied at the center of mass, let's say 10 Newtons, so the mass of the whole ruler is 10 kilograms. If a force is applied at the center of mass, this ruler will accelerate the same exact way as would a point mass. Let's say that we just had a little dot, but that little dot had the same mass, 10 kilograms, and we were to push on that dot with 10 Newtons. In either case, in the case of the ruler, we would accelerate upwards at what? Force divided by mass, so we would accelerate upwards at 1 meter per second squared. And in this case of this point mass, we would accelerate that point. When I say point mass, I'm just saying something really, really small, but it has a mass of 10 kilograms, so it's much smaller, but it has the same mass as this ruler. This would also accelerate upwards with a magnitude of 1 meters per second squared. So why is this useful to us? Well, sometimes we have some really crazy objects and we want to figure out exactly what it does. If we know its center of mass first, we can know how that object will behave without having to worry about the shape of that object. And I'll give you a really easy way of realizing where the center of mass is. If the object has a uniform distribution-- when I say that, it means, for simple purposes, if it's made out of the same thing and that thing that it's made out of, its density, doesn't really change throughout the object, the center of mass will be the object's geometric center. So in this case, this ruler's almost a one-dimensional object. We just went halfway. The distance from here to here and the distance from here to here are the same. This is the center of mass. If we had a two-dimensional object, let's say we had this triangle and we want to figure out its center of mass, it'll be the center in two dimensions. So it'll be something like that. Now, if I had another situation, let's say I have this square. I don't know if that's big enough for you to see. I need to draw it a little thicker. Let's say I have this square, but let's say that half of this square is made from lead. And let's say the other half of the square is made from something lighter than lead. It's made of styrofoam. That is lighter than lead. So in this situation, the center of mass isn't going to be the geographic center. I don't know how much denser lead is than styrofoam, but the center of mass is going to be someplace closer to the right because this object does not have a uniform density. It'll actually depend on how much denser the lead is than the styrofoam, which I don't know. But hopefully, that gives you a little intuition of what the center of mass is. And now I'll tell you something a little more interesting. Every problem we have done so far, we actually made the simplifying assumption that the force acts on the center of mass. So if I have an object, let's say the object that looks like a horse. Let's say that object. If this is the object's center of mass, I don't know where the horse's center of mass normally is, but let's say a horse's center of mass is here. If I apply a force directly on that center of mass, then the object will move in the direction of that force with the appropriate acceleration. We could divide the force by the mass of the entire horse acceleration in that direction. But now I will throw in a twist. And actually, every problem we did, all of these Newton's Law's problems, we assumed that the force acted at the center of mass. But something more interesting happens if the force acts away from the center of mass. Let me actually take that ruler example. I don't know why I even drew the horse. If I have this ruler again and this is the center of mass, as we said, any force that we act on the center of mass, the whole object will move in the direction of the force. It'll be shifted by the force, essentially. Now, this is what's interesting. If that's the center of mass and if I were to apply a force someplace else away from the center of mass, let' say I apply a force here, I want you to think about for a second what will probably happen to the object. Well, it turns out that the object will rotate. And so think about if we're on the space shuttle or we're in deep space or something, and if I have a ruler, and if I just push at one end of the ruler, what's going to happen? Am I just going to push the whole ruler or is the whole ruler is going to rotate? And hopefully, your intuition is correct. The whole ruler will rotate around the center of mass. And in general, if you were to throw a monkey wrench at someone, and I don't recommend that you do, but if you did, and while the monkey wrench is spinning in the air, it's spinning around its center of mass. Same for a knife. If you're a knife catcher, that's something you should think about, that the object, when it's free, when it's not fixed to any point, it rotates around its center of mass, and that's very interesting. So you can actually throw random objects, and that point at which it rotates around, that's the object's center of mass. That's an experiment that you should do in an open field around no one else. Now, with all of this, and I'll actually in the next video tell you what this is. When you have a force that causes rotational motion as opposed to a shifting motion, that's torque, but we'll do that in the next video. But now I'll show you just a cool example of how the center of mass is relevant in everyday applications, like high jumping. So in general, let's say that this is a bar. This is a side view of a bar, and this is the thing holding the bar. And a guy wants to jump over the bar. His center of mass is-- most people's center of mass is around their gut. I think evolutionarily that's why our gut is there, because it's close to our center of mass. So there's two ways to jump. You could just jump straight over the bar, like a hurdle jump, in which case your center of mass would have to cross over the bar. And we could figure out this mass, and we can figure out how much energy and how much force is required to propel a mass that high because we know projectile motion and we know all of Newton's laws. But what you see a lot in the Olympics is people doing a very strange type of jump, where, when they're going over the bar, they look something like this. Their backs are arched over the bar. Not a good picture. But what happens when someone arches their back over the bar like this? I hope you get the point. This is the bar right here. Well, it's interesting. If you took the average of this person's density and figured out his geometric center and all of that, the center of mass in this situation, if someone jumps like that, actually travels below the bar. Because the person arches their back so much, if you took the average of the total mass of where the person is, their center of mass actually goes below the bar. And because of that, you can clear a bar without having your center of mass go as high as the bar and so you need less force to do it. Or another way to say it, with the same force, you could clear a higher bar. , Hopefully, I didn't confuse you, but that's exactly why these high jumpers arch their back, so that their center of mass is actually below the bar and they don't have to exert as much force. Anyway, hopefully you found that to be a vaguely useful introduction to the center of mass, and I'll see you in the next video on torque." + }, + { + "Q": "This might be a stupid question but, I'm confused on how sadness is a noun? when I look at sadness in a sentence to me its more of a verb word. Can you help me understand why its a noun and maybe help me understand how I can find nouns in a sentence that aren't as normal as person, place, or thing.", + "A": "Sadness isn t an action or verb. You can t say He sadness today because he failed an exam or The dog sadness because he can t have a treat. Sadness is an emotion, a thing. Not something you do . There are plenty of things you can t touch, but they are still considered nouns. These are called abstract nouns. There was sadness in her voice . The verb here is was and the noun here is sadness . Just like the sentence There was water everywhere . The verb is was and the noun is water .", + "video_name": "3AF_rN-yN-Y", + "transcript": "- [Voiceover] Hello grammarians. So today I'd like to talk to you about the idea of concrete and abstract nouns, and before we do that, I'd like to get into some word origins or etymology. So let's take each of these words in turn, because I think by digging into what these words mean, literally what they mean and where they come from, we'll get a better understanding of this concept. So both of these words come to us from Latin. Concrete comes to us from the Latin concretus, which means to grow together. So this part of it means grown. And this part means together. It refers to something that, you know, has grown together and become thick and kind of hard to get through and physical. The connotation here is that this is a physical thing. Something that is concrete is physical. Abstract, on the other hand, means to draw something away. So something that is abstract is drawn away from the real, from the concrete, from the physical. So this is not physical. And we make this distinction in English when we're talking about nouns. Is it something that is concrete, is it something you can look at or pick up or smell or sense or something that is abstract, something that isn't physical, but can still be talked about. So for example, the word sadness... Is a noun, right? This is definitely a noun. It's got this noun-making ending, this noun-forming suffix, ness. You know, we take the adjective sad and we toss this ness part onto it, we've got a noun. But can you see sadness? Is it something you can pick up? Sure, you can tell by being, you know observant and empathetic that your friend is sad, but it's not something you can pick up. You can't be like a measurable degree of sad. You couldn't take someone's sadness, put it under a microscope and say \"Oh, Roberta, you are 32 degrees microsad.\" You know, it's not something physical. Concrete things, on the other hand, are things that we can see or count or measure. Just parts of the physical world. So anything you look at, like a dog is concrete, a ball is concrete, a cliff is concrete. Happiness... Is abstract. The idea of freedom... Is abstract. Though the presence of freedom in your life may manifest in physical objects, like \"Oh, my parents let me have the freedom to eat ice cream.\" Ice cream is, you know, a concrete noun. But freedom, the thing that allows you, you know, the permission that you get from your parents to have ice cream. That's not a physical object. So that's basically the difference. So a concrete noun is a physical object and an abstract noun is not. This is why I really wanted to hit the idea that a noun can be a person, place, thing or idea, because nouns can be ideas, and those ideas tend to be abstract. Sadness, happiness, freedom, permission, liberty, injustice. All of these are abstract ideas. That's the difference. You can learn anything. David out." + }, + { + "Q": "does a euro pound count as a pound in England", + "A": "No, at the current exchange rate, a british pound is worth about 1.3 euros.", + "video_name": "s6NOa1KTCxQ", + "transcript": "- [Voiceover] Zhang Tao has two one dollar bills. So two one dollar bills. So let's just write that as one plus one, because each one dollar bill represents one dollar. So, one dollar plus one dollar. That's the two one dollar bills. One five dollar bill. One five dollar bill. So that represents five dollars. And three 10 dollar bills. Three 10 dollar bills. So that's plus ten plus ten plus ten. How much money does he have in all? Well, let's see. One plus, let's see if we can calculate it. One plus one is two. Two plus five is seven. So, these are going to be seven dollars. And then 10 plus 10 plus 10, that's three 10's. So that's going to be 30 dollars. Now what's seven ones plus three 10's? Or what's seven plus 30? Well, that's going to be 37 dollars. So Zhang Tao has 37 dollars in all. Let's do another one of these. Diya has six one dollar bills. And actually, let's draw it out. Diya has six one dollar bills. So that's one, two, three, four, five and six. And these are one dollar bills. So, one dollar bills. This is my rough drawing of one dollar bills. Have to draw a lot of one dollar bills here. So, she has six one dollar bills. She has three five dollar bills. Can we do that in a different color? So, three five dollar bills. So one, so that's a five dollar bill. Two, that's a five dollar bill. And three five dollar bills. That little circle in .... That's supposed to be the picture of someone. Let me make it clear these are pictures of folks. Of famous historical figures. So, these are pictures of famous historical figures here. And then finally she has one 10 dollar bill. One 10 dollar bill. So one 10 dollar bill. So it's worth 10 dollars. A picture of a famous historical figure right over there. So, how much money does she have in all? Well, the six one dollar bills that's going to be six dollars. The three five dollar bills that's going to be worth five, 10, 15 dollars. So plus 15 dollars. And then the one 10 dollar bill. That's going to be worth 10 dollars. 10 dollars. So, what's six plus 15 plus 10? Well, let's see, six plus fifteen is going to be 21. 21 plus 10 is going to be equal to 31 dollars. Is equal to 31 dollars. And you could have also added it this way, You could have said 15 plus 10 plus 10 plus six plus six. Add them together and we would have gotten five plus zero plus six is 11. That's one one and one 10. And then you have three 10's here. So, 31 dollars." + }, + { + "Q": "why is the electric field of a point charge is not uniform ?\nPlease explain.\nThank you.", + "A": "Because the electric field generated by a point charge extends radially from the point iteself. This means that the field decreases with the square of theradius of distance from the point.", + "video_name": "0YOGrTNgGhE", + "transcript": "Let's imagine that instead of having two charges, we just have one charge by itself, sitting in a vacuum, sitting in space. So that's this charge here, and let's say its charge is Q. That's some number, whatever it is. That's it's charge. And I want to know, if I were to place another charge close to this Q, within its sphere of influence, what's going to happen to that other charge? What's going to be the net impact on it? And we know if this has some charge, if we put another charge here, if this is 1 coulomb and we put another charge here that's 1 coulomb, that they're both positive, they're going to repel each other, so there will be some force that pushes the next charge away. If it's a negative charge and I put it here, it'll be even a stronger force that pulls it in because it'll be closer. So in general, there's this notion of what we can call an electric field around this charge. And what's an electric field? We can debate whether it really exists, but what it allows us to do is imagine that somehow this charge is affecting the space around it in some way that whenever I put-- it's creating a field that whenever I put another charge in that field, I can predict how the field will affect that charge. So let's put it in a little more quantitative term so I stop confusing you. So Coulomb's Law told us that the force between two charges is going to be equal to Coulomb's constant times-- and in this case, the first charge is big Q. And let's say that the second notional charge that I eventually put in this field is small q, and then you divide by the distance between them. Sometimes it's called r because you can kind of view the distance as the radial distance between the two charges. So sometimes it says r squared, but it's the distance between them. So what we want to do if we want to calculate the field, we want to figure out how much force is there placed per charge at any point around this Q, so, say, at a given distance out here. At this distance, we want to know, for a given Q, what is the force going to be? So what we can do is we could take this equation up here and divide both sides by this small 1, and say, OK, the force-- and I will arbitrarily switch colors. The force per charge at this point-- let's call that d1-- is equal to Coulomb's constant times the charge of the particle that's creating the field divided by-- well, in this case, it's d1-- d1 squared, right? Or we could say, in general-- and this is the definition of the electric field, right? Well, this is the electric field at the point d1, and if we wanted a more general definition of the electric field, we'll just make this a general variable, so instead of having a particular distance, we'll define the field for all distances away from the point Q. So the electric field could be defined as Coulomb's constant times the charge creating the field divided by the distance squared, the distance we are away from the charge. So essentially, we've defined-- if you give me a force and a point around this charge anywhere, I can now tell you the exact force. For example, if I told you that I have a minus 1 coulomb charge and the distance is equal to-- oh, I don't know. The distance is equal to let's say-- let's make it easy. Let's say 2 meters. So first of all, we can say, in general, what is the electric field 2 meters away from? So what is the electric field out here? This is 2, right? And it's going to be 2 meters away. It's radial so it's actually along this whole circle. What is the electric field there? Well, the electric field at that point is going to be equal to what? And it's also a vector quantity, right? Because we're dividing a vector quantity by a scalar quantity charge. So the electric field at that point is going to be k times whatever charge it is divided by 2 meters, so divided by 2 meters squared, so that's 4, right, distance squared. And so if I know the electric field at any given point and then I say, well, what happens if I put a negative 1 coulomb charge there, all I have to do is say, well, the force is going to be equal to the charge that I place there times the electric field at that point, right? So in this case, we said the electric field at this point is equal to-- and the units for electric field are newtons per coulomb, and that makes sense, right? Because it's force divided by charge, so newtons per coulomb. So if we know that the electric charge-- well, let me put some real numbers here. Let's say that this is-- I don't know. It's going to be a really large number, but let's say this-- let me pick a smaller number. Let's say this is 1 times 10 to the minus 6 coulombs, right? If that's 1 times 10 to the minus 6 coulombs, what is the electric field at that point? Let me switch colors again. What's the electric field at that point? Well, the electric field at that point is going to be equal to Coulomb's constant, which is 9 times 10 to the ninth-- times the charge generating the field-- times 1 times 10 to the minus 6 coulombs. And then we are 2 meters away, so 2 squared. So that equals 9 times 10 to the third divided by 4. So I don't know, what is that? 2.5 times 10 to the third or 2,500 newtons per coulomb. So we know that this is generating a field that when we're 2 meters away, at a radius of 2 meters, so roughly that circle around it, this is generating a field that if I were to put-- let's say I were to place a 1 coulomb charge here, the force exerted on that 1 coulomb charge is going to be equal to 1 coulomb times the electric fields, times 2,500 newtons per coulomb. So the coulombs cancel out, and you'll have 2,500 newtons, which is a lot, and that's because 1 coulomb is a very, very large charge. And then a question you should ask yourself: If this is 1 times 10 to the negative 6 coulombs and this is 1 coulomb, in which direction will the force be? Well, they're both positive, so the force is going to be outwards, right? So let's take this notion and see if we can somehow draw an electric field around a particle, just to get an intuition of what happens when we later put a charge anywhere near the particle. So there's a couple of ways to visualize an electric field. One way to visualize it is if I have a-- let's say I have a point charge here Q. What would be the path of a positive charge if I placed it someplace on this Q? Well, if I put a positive charge here and this Q is positive, that positive charge is just going to accelerate outward, right? It's just going to go straight out, but it's going to accelerate at an ever-slowing rate, right? Because here, when you're really close, the outward force is very strong, and then as you get further and further away, the electrostatic force from this charge becomes weaker and weaker, or you could say the field becomes weaker and weaker. But that's the path of a-- it'll just be radially outward-- of a positive test charge. And then if I put it here, well, it would be radially outward that way. It wouldn't curve the way I drew it. It would be a straight line. I should actually use the line tool. If I did it here, it would be like that, but then I can't draw the arrows. If I was here, it would out like that. I think you get the picture. At any point, a positive test charge would just go straight out away from our charge Q. And to some degree, one measure of-- and these are called electric field lines. And one measure of how strong the field is, is if you actually took a unit area and you saw how dense So here, they're relatively sparse, while if I did that same area up here-- I know it's not that obvious. I'm getting more field lines in. But actually, that's not a good way to view it because I'm covering so much area. Let me undo both of them. You can imagine if I had a lot more lines, if I did this area, for example, in that area, I'm capturing two of these field lines. Well, if I did that exact same area out here, I'm only capturing one of the field lines, although you could have a bunch more in between here. And that makes sense, right? Because as you get closer and closer to the source of the electric field, the charge gets stronger. Another way that you could have done this, and this would have actually more clearly shown the magnitude of the field at any point, is you could have-- you could say, OK, if that's my charge Q, you could say, well, really close, the field is strong. So at this point, the vector, the newtons per coulomb, is that strong, that strong, that strong, that strong. We're just taking sample points. You can't possibly draw them at every single point. So at that point, that's the vector. That's the electric field vector. But then if we go a little bit further out, the vector is going to be-- it falls off. This one should be shorter, then this one should be even shorter, right? You could pick any point and you could actually calculate the electric field vector, and the further you go out, the shorter and shorter the electric field vectors get. And so, in general, there's all sorts of things you can draw the electric fields for. Let's say that this is a positive charge and that this is a negative charge. Let me switch colors so I don't have to erase things. If I have to draw the path of a positive test charge, it would go out radially from this charge, right? But then as it goes out, it'll start being attracted to this one the closer it gets to the negative, and then it'll curve in to the negative charge and these arrows go like this. And if I went from here, the positive one will be repelled really strong, really strong, it'll accelerate fast and it's rate of acceleration will slow down, but then as it gets closer to the negative one, it'll speed up again, and then that would be its path. Similarly, if there was a positive test charge here, its path would be like that, right? If it was here, its path would be like that. If it was here, it's path would be like that. If it was there, maybe its path is like that, and at some point, its path might never get to that-- this out here might just go straight out that way. That one would just go straight out, and here, the field lines would just come in, right? A positive test charge would just be naturally attracted to that negative charge. So that's, in general, what electric field lines show, and we could use our little area method and see that over here, if we picked a given area, the electric field is much weaker than if we picked that same area right here. We're getting more field lines in than we do right there. So that hopefully gives you a little sense for what an electric field is. It's really just a way of visualizing what the impact would be on a test charge if you bring it close to another charge. And hopefully, you know a little bit about Coulomb's constant. And let's just do a very simple-- I'm getting this out of the AP Physics book, but they say-- let's do a little simple problem: Calculate the static electric force between a 6 times 10 to the negative sixth coulomb charge. So 6 times-- oh, no, that's not on an electric field. Oh, here it says: What is the force acting on an electron placed in an external electric field where the electric field is-- they're saying it is 100 newtons per coulomb at that point, wherever the electron is. So the force on that, the force in general, is just going to be the charge times the electric field, and they say it's an electron, so what's the Well, we know it's negative, and then in the first video, we learned that its charge is 1.6 times 10 to the negative nineteenth coulombs times 100 newtons per coulomb. The coulombs cancel out. And this is 10 squared, right? This is 10 to the positive 2, so it'll be 10 to the minus 19 times 10 to the positive 2. The force will be minus 1.6 times 10 to the minus 17 newtons. So the problems are pretty simple. I think the more important thing with electric fields is to really understand intuitively what's going on, and kind of how it's stronger near the point charges, and how it gets weaker as it goes away, and what the field lines depict, and how they can be used to at least approximate the strength of the field. I will see you in the next video." + }, + { + "Q": "if we compress H2O will it turn in to ice ?", + "A": "If you mean normal ice, then the answer is no. But there some other kinds of ice that you can form by exposing water to EXTREME pressures. If you crush the water with over 10000 times normal atmospheric pressure (and there are some conditions that have to be just right for this to work), then it is possible to convert the water to one of the other kinds of ice such as Ice VI or Ice VII. These other kinds of ice are structurally different than normal ice.", + "video_name": "pKvo0XWZtjo", + "transcript": "I think we're all reasonably familiar with the three states of matter in our everyday world. At very high temperatures you get a fourth. But the three ones that we normally deal with are, things could be a solid, a liquid, or it could be a gas. And we have this general notion, and I think water is the example that always comes to at least my mind. Is that solid happens when things are colder, relatively colder. And then as you warm up, you go into a liquid state. And as your warm up even more you go into a gaseous state. So you go from colder to hotter. And in the case of water, when you're a solid, you're ice. When you're a liquid, some people would call ice water, but let's call it liquid water. I think we know what that is. And then when it's in the gas state, you're essentially vapor or steam. So let's think a little bit about what, at least in the case of water, and the analogy will extend to other types of molecules. But what is it about water that makes it solid, and when it's colder, what allows it to be liquid. And I'll be frank, liquids are kind of fascinating because you can never nail them down, I guess is the best way to view them. Or a gas. So let's just draw a water molecule. So you have oxygen there. You have some bonds to hydrogen. And then you have two extra pairs of valence electrons in the oxygen. And a couple of videos ago, we said oxygen is a lot more electronegative than the hydrogen. It likes to hog the electrons. So even though this shows that they're sharing electrons here and here. At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules. Let me draw some more molecules. When we talk about the whole state of the whole matter, we actually think about how the molecules are interacting with Not just how the atoms are interacting with each other within a molecule. I just drew one oxygen, let me copy and paste that. But I could do multiple oxygens. And let's say that that hydrogen is going to want to be near this oxygen. Because this has partial negative charge, this has a partial positive charge. And then I could do another one right there. And then maybe we'll have, and just to make the point clear, you have two hydrogens here, maybe an oxygen wants to hang out there. So maybe you have an oxygen that wants to be here because it's got its partial negative here. And it's connected to two hydrogens right there that have their partial positives. But you can kind of see a lattice structure. Let me draw these bonds, these polar bonds that start forming between the particles. These bonds, they're called polar bonds because the molecules themselves are polar. And you can see it forms this lattice structure. And if each of these molecules don't have a lot of kinetic energy. Or we could say the average kinetic energy of this matter is fairly low. And what do we know is average kinetic energy? Well, that's temperature. Then this lattice structure will be solid. These molecules will not move relative to each other. I could draw a gazillion more, but I think you get the point that we're forming this kind of fixed structure. And while we're in the solid state, as we add kinetic energy, as we add heat, what it does to molecules is, it just makes them vibrate around a little bit. If I was a cartoonist, they way you'd draw a vibration is to put quotation marks there. That's not very scientific. But they would vibrate around, they would buzz around a little bit. I'm drawing arrows to show that they are vibrating. It doesn't have to be just left-right it could be up-down. But as you add more and more heat in a solid, these molecules are going to keep their structure. So they're not going to move around relative to each other. But they will convert that heat, and heat is just a form of energy, into kinetic energy which is expressed as the vibration of these molecules. Now, if you make these molecules start to vibrate enough, and if you put enough kinetic energy into these molecules, what do you think is going to happen? Well this guy is vibrating pretty hard, and he's vibrating harder and harder as you add more and more heat. This guy is doing the same thing. At some point, these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations. And once that happens, the molecules-- let me draw a couple more. Once that happens, the molecules are going to start moving past each other. So now all of a sudden, the molecule will start shifting. But they're still attracted. Maybe this side is moving here, that's moving there. You have other molecules moving around that way. But they're still attracted to each other. Even though we've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules. Our vibration, or our kinetic energy for each molecule, still isn't strong enough to completely separate them. They're starting to slide past each other. And this is essentially what happens when you're in a liquid state. You have a lot of atoms that want be touching each other but they're sliding. They have enough kinetic energy to slide past each other and break that solid lattice structure here. And then if you add even more kinetic energy, even more heat, at this point it's a solution now. They're not even going to be able to stay together. They're not going to be able to stay near each other. If you add enough kinetic energy they're going to start looking like this. They're going to completely separate and then kind of bounce around independently. Especially independently if they're an ideal gas. But in general, in gases, they're no longer touching They might bump into each other. But they have so much kinetic energy on their own that they're all doing their own thing and they're not touching. I think that makes intuitive sense if you just think about what a gas is. For example, it's hard to see a gas. Why is it hard to see a gas? Because the molecules are much further apart. So they're not acting on the light in the way that a liquid or a solid would. And if we keep making that extended further, a solid-- well, I probably shouldn't use the example with ice. Because ice or water is one of the few situations where the solid is less dense than the liquid. That's why ice floats. And that's why icebergs don't just all fall to the bottom of the ocean. And ponds don't completely freeze solid. But you can imagine that, because a liquid is in most cases other than water, less dense. That's another reason why you can see through it a little Or it's not diffracting-- well I won't go into that too much, than maybe even a solid. But the gas is the most obvious. And it is true with water. The liquid form is definitely more dense than the gas form. In the gas form, the molecules are going to jump around, not touch each other. And because of that, more light can get through the substance. Now the question is, how do we measure the amount of heat that it takes to do this to water? And to explain that, I'll actually draw a phase change diagram. Which is a fancy way of describing something fairly straightforward. Let me say that this is the amount of heat I'm adding. And this is the temperature. We'll talk about the states of matter in a second. So heat is often denoted by q. Sometimes people will talk about change in heat. They'll use H, lowercase and uppercase H. They'll put a delta in front of the H. Delta just means change in. And sometimes you'll hear the word enthalpy. Let me write that. Because I used to say what is enthalpy? It sounds like empathy, but it's quite a different concept. At least, as far as my neural connections could make it. But enthalpy is closely related to heat. It's heat content. For our purposes, when you hear someone say change in enthalpy, you should really just be thinking of change in heat. I think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary. The best way to think about it is heat content. Change in enthalpy is really just change in heat. And just remember, all of these things, whether we're talking about heat, kinetic energy, potential energy, enthalpy. You'll hear them in different contexts, and you're like, I thought I should be using heat and they're talking about enthalpy. These are all forms of energy. And these are all measured in joules. And they might be measured in other ways, but the traditional way is in joules. And energy is the ability to do work. And what's the unit for work? Well, it's joules. Force times distance. But anyway, that's a side-note. But it's good to know this word enthalpy. Especially in a chemistry context, because it's used all the time and it can be very confusing and non-intuitive. Because you're like, I don't know what enthalpy is in my everyday life. Just think of it as heat contact, because that's really But anyway, on this axis, I have heat. So this is when I have very little heat and I'm increasing my heat. And this is temperature. Now let's say at low temperatures I'm here and as I add heat my temperature will go up. Temperature is average kinetic energy. Let's say I'm in the solid state here. And I'll do the solid state in purple. No I already was using purple. I'll use magenta. So as I add heat, my temperature will go up. Heat is a form of energy. And when I add it to these molecules, as I did in this example, what did it do? It made them vibrate more. Or it made them have higher kinetic energy, or higher average kinetic engery, and that's what temperature is a measure of; average kinetic energy. So as I add heat in the solid phase, my average kinetic energy will go up. And let me write this down. This is in the solid phase, or the solid state of matter. Now something very interesting happens. Let's say this is water. So what happens at zero degrees? Which is also 273.15 Kelvin. Let's say that's that line. What happens to a solid? Well, it turns into a liquid. Ice melts. Not all solids, we're talking in particular about water, about H2O. So this is ice in our example. All solids aren't ice. Although, you could think of a rock as solid magma. Because that's what it is. I could take that analogy a bunch of different ways. But the interesting thing that happens at zero degrees. Depending on what direction you're going, either the freezing point of water or the melting point of ice, something interesting happens. As I add more heat, the temperature does not to go up. As I add more heat, the temperature does not go up for a little period. Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or which water will boil. But something happens. And they're really getting kinetically active. But just like when you went from solid to liquid, there's a certain amount of energy that you have to contribute to the system. And actually, it's a good amount at this point. Where the water is turning into vapor, but it's not getting any hotter. So we have to keep adding heat, but notice that the temperature didn't go up. We'll talk about it in a second what was happening then. And then finally, after that point, we're completely vaporized, or we're completely steam. Then we can start getting hot, the steam can then get hotter as we add more and more heat to the system. So the interesting question, I think it's intuitive, that as you add heat here, our temperature is going to go up. But the interesting thing is, what was going on here? We were adding heat. So over here we were turning our heat into kinetic energy. Temperature is average kinetic energy. But over here, what was our heat doing? Well, our heat was was not adding kinetic energy to the system. The temperature was not increasing. But the ice was going from ice to water. So what was happening at that state, is that the kinetic energy, the heat, was being used to essentially break these bonds. And essentially bring the molecules into a higher energy state. So you're saying, Sal, what does that mean, higher energy state? Well, if there wasn't all of this heat and all this kinetic energy, these molecules want to be very close to each other. For example, I want to be close to the surface of the earth. When you put me in a plane you have put me in a higher energy state. I have a lot more potential energy. I have the potential to fall towards the earth. Likewise, when you move these molecules apart, and you go from a solid to a liquid, they want to fall towards each other. But because they have so much kinetic energy, they never quite are able to do it. But their energy goes up. Their potential energy is higher because they want to fall towards each other. By falling towards each other, in theory, they could do some work. So what's happening here is, when we're contributing heat-- and this amount of heat we're contributing, it's called the heat of fusion. Because it's the same amount of heat regardless how much direction we go in. When we go from solid to liquid, you view it as the heat of melting. It's the head that you need to put in to melt the ice into liquid. When you're going in this direction, it's the heat you have to take out of the zero degree water to turn it into ice. So you're taking that potential energy and you're bringing the molecules closer and closer to each other. So the way to think about it is, right here this heat is being converted to kinetic energy. Then, when we're at this phase change from solid to liquid, that heat is being used to add potential energy into the system. To pull the molecules apart, to give them more potential energy. If you pull me apart from the earth, you're giving me potential energy. Because gravity wants to pull me back to the earth. And I could do work when I'm falling back to the earth. A waterfall does work. It can move a turbine. You could have a bunch of falling Sals move a turbine as well. And then, once you are fully a liquid, then you just become a warmer and warmer liquid. Now the heat is, once again, being used for kinetic energy. You're making the water molecules move past each other faster, and faster, and faster. To some point where they want to completely disassociate from each other. They want to not even slide past each other, just completely jump away from each other. And that's right here. This is the heat of vaporization. And the same idea is happening. Before we were sliding next to each other, now we're pulling apart altogether. So they could definitely fall closer together. And then once we've added this much heat, now we're just heating up the steam. We're just heating up the gaseous water. And it's just getting hotter and hotter and hotter. But the interesting thing there, and I mean at least the interesting thing to me when I first learned this, whenever I think of zero degrees water I'll say, oh it must be ice. But that's not necessarily the case. If you start with water and you make it colder and colder and colder to zero degrees, you're essentially taking heat out of the water. You can have zero degree water and it hasn't turned into ice yet. And likewise, you could have 100 degree water that hasn't turned into steam yeat. You have to add more energy. You can also have 100 degree steam. You can also have zero degree water. Anyway, hopefully that gives you a little bit of intuition of what the different states of matter are. And in the next problem, we'll talk about how much heat exactly it does take to move along this line. And maybe we can solve some problems on how much ice we might need to make our drink cool." + }, + { + "Q": "how to know the quadratic formula does not have answer?", + "A": "I think you mean how to know if the quadratic equation has no real solutions, which just means it does not touch the x-axis. You have to look at the discriminant which is the b^2-4ac part. If it is negative then there is no real solutions and the parabola does not touch the x-axis, since there is no real number that gives you the square root of a negative.", + "video_name": "BGz3pkoGPag", + "transcript": "- [Instructor] In this video, we are going to talk about one of the most common types of curves you will see in mathematics, and that is the parabola. The word parabola sounds quite fancy, but we'll see it's describing something that is fairly straightforward. Now in terms of why it is called the parabola, I've seen multiple explanations for it. It comes from Greek para, that root word, similar to parable. You could view of something beside, alongside, something in parallel. Bola, same root as when we're talking about ballistics, throwing something. So you could interpret it as beside, alongside, something that is being thrown. Now how does that relate to curves like this? Well my brain immediately imagines this is the trajectory, this is the path that is a pretty good approximation for the path of things that are actually thrown. When you study physics, you will see the path, you'll approximate, the paths of objects being thrown, as parabolas, so maybe that's where it comes from, but there are other potential explanations for why it is actually called the parabola. It has been lost to history. But what exactly is a parabola? In future videos, we're gonna describe it a little bit more algebraically. In this one, we just wanna get a sense for what parabolas look like and introduce ourselves to some terminology around a parabola. These three curves, they are all hand-drawn versions of a parabola, and so you immediately notice some interesting things about them. Some of them are opened upwards like this yellow one and this pink one, and some of them are open downwards. You will hear people say things like open, opened down, open downwards or open down or open upwards, so it's good to know what they are talking about, and it's, hopefully, fairly self-explanatory. Open upwards, the parabola is open towards the top of our graph paper. Here it's open towards the bottom of our graph paper. This looks like a right-side up U. This looks like an upside down U right over here. This pink one would be open upwards. Now another term that you'll see associated with the parabola, and once again, in the future, we'll learn how to calculate these things and find them precisely, is the vertex. The vertex you should view as the maximum or minimum point on a parabola. So if a parabola opens upwards like these two on the right, the vertex is the minimum point. The vertex is the minimum point right over there, and so if someone said what is the vertex of this yellow parabola? Well it looks like the x, looks like the x coordinate is three and a half, so it is three and a half. It looks like the y coordinate, it looks like it is about negative three and a half. Once again, once we start representing these things with equations, we'll have techniques for calculating them more precisely, but the vertex of this other upward-opening parabola, it is the minimum point. It is the low point. There is no maximum point on an upward-opening parabola. It just keeps increasing as x gets larger in the positive or the negative direction. Now if your parabola opens downward, then your vertex is going to be your maximum point. Now related to the idea of a vertex is the idea of an axis of symmetry. In general when we're talking about, well not just three, two dimensions but even three dimensions, but especially in two dimensions, you can imagine a line over which you can flip the graph, and so it meets, it folds onto itself. The axis of symmetry for this yellow graph right over here, for this yellow parabola, it would be this line. I'm gonna have to draw it a little bit better. It would be that line right over there. You could fold the parabola over that line, and it would meet itself. And that line, I didn't draw it as neat as I should, that should go directly through the vertex, so to describe that line you'd say that line is x is equal to 3.5. Similarly the axis of symmetry for this pink parabola, it should go through the line x equals negative one, so let me do that. That's the axis of symmetry. It goes through the vertex, and if you were to fold the parabola over it, it would meet itself. The axis of symmetry for this green one? It should, once again, go through the vertex. It looks like it is x is equal to negative six. This is, let me write that down, that is the axis of symmetry. Now another concept that isn't unique to parabolas, but we'll talk a lot about it in the context of parabolas, are intercepts, so when people say y-intercept, and you saw this when you first graphed lines, they're saying where is the graph, where does the curve intercept or intersect the y-axis? So the y-intercept of this yellow line would be right there. It looks like it's the point zero comma three, zero comma three. The y-intercept for the pink one is right over there. At least on this graph paper, we don't see the y-intercept, but it eventually will intersect the y-axis. It just will be way off of this screen. You might also be familiar with the term x-intercept, and that's especially interesting with parabolas as we'll see in the future. X-intercept is where do you intercept or intersect the x-axis? Here this yellow one you see it does it two places, and this is where it gets interesting. Lines will only intersect the x-axis once at most, but here we see that a parabola can intersect the x-axis twice, because it curves back around to intersect it again, and so for here the x-intercepts are going to be the point one comma zero and six comma zero. You might already notice something interesting. The x-intercepts are symmetric around the axis of symmetry, so they should be equal distant from that axis of symmetry, and you can see they indeed are. They are both exactly two and a half away from that axis of symmetry, and so if you know where the intercepts are, you just take, you could say, the midpoint of the x coordinates, and then you're going to have the axis of symmetry, the x coordinate of the axis of symmetry and the x coordinate of the actual vertex. Similarly the x-intercept here looks like it's negative, the points are negative seven comma zero and negative five comma zero, and the x coordinate of the vertex, or the line of symmetry, is right in between those two points. It's worth noting not every parabola is going to intersect the x-axis. Notice this pink upward-opening parabola, it's low point is above the x-axis, so it's never going to intersect the actual x-axis, so this is actually not going to have any x-intercepts. I'll leave you there. Those are actually the core ideas or the core visual themes around parabolas, and we're going to discuss them in a lot more detail when we represent them with equations. As you'll see, these equations are going to involve second-degree terms. So the most simple parabola is going to be y is equal to x squared, but then you can complicate it a little bit. You could have things like y is equal to two x squared minus five x plus seven. These types that we'll talk about in more general terms, these types of equations sometimes called quadratics, they are represented, generally, by parabolas." + }, + { + "Q": "Even though I do everything right in the practice, I still get it wrong! What can I do to fix that?", + "A": "Study and watch more videos on the topic and pause them and try to do the questions before Sal does.", + "video_name": "xXIG8ouHcsc", + "transcript": "Let's divide 7,182 by 42. And what's different here is we're now dividing by a two-digit number, not a one-digit number, but the same idea holds. So we say, hey, how many times does 42 go into 7? Well, it doesn't really go into 7 at all, so let's add one more place value. How many times does 42 go into 71? Well, it goes into 71 one time. Just a reminder, whoever's doing the process where you say, hey, 42 goes into 71 one time. But what we're really saying, 42 goes into 7,100 100 times because we're putting this one in the hundreds place. But let's put that on the side for a little bit and focus on the process. So 1 times 42 is 42, and now we subtract. Now, you might be able to do 71 minus 42 in your head, knowing, hey, 72 minus 42 would be 30. So 71 minus 42 would be 29, but we could also do it by regrouping. To regroup, you want to subtract a 2 from a 1. You can't really do that in any traditional way. So let's take a 10 from the 70, so that it becomes a 60, and give that 10 to the ones place, and then that becomes an 11. And so 11 minus 2 is 9, and 6 minus 4 is 2. So you get 29. And we can bring down the next place value. Bring down an 8. And now, this is where the art happens when we're dividing by a multi-digit number right over here. We have to estimate how many times does 42 go into 298. And sometimes it might involve a little bit of trial and error. So you really just kind of have to eyeball it. If you make a mistake, try again. The way you know you make a mistake is, if say it goes into it 9 times, and you do 9 times 42 and you get a number larger than 298, then you overestimated. If you say it goes into it three times, you do 3 times 42, you get some number here. When you subtract, you get something larger than 42, then you also made a mistake, and you have to adjust upwards. Well, let's see if we can eyeball it. So this is roughly 40. This is roughly 300. 40 goes into 300 the same times as 4 goes into the 30, so it's going to be roughly 7. Let's see if that's right. 7 times 2 is 14. 7 times 1 is 28, plus 1 is 29. So I got pretty close. My remainder here-- notice 294 is less than 298. So I'm cool there. And my remainder is less than 42, so I'm cool as well. So now let's add another place value. Let's bring this 2 down. And here we're just asking ourselves, how many times does 42 go into 42? Well, 42 goes into 42 exactly one time. 1 times 42 is 42, and we have no remainder. So this one luckily divided exactly. 42 goes into 7,182 exactly 171 times." + }, + { + "Q": "why does a magnet always points towards south and north pole??", + "A": "The earth has a magnetic field, as though it has a giant bar magnet in it, with one pole near Earth s North Pole and the other pole near what we call the South Pole. The north end of a magnet attracts the south end of another magnet, and south attracts north. So a magnet that is free to spin in earth s magnetic field points its north pole toward the earth s North Pole. (this tells you that the magnetic pole that is near the North Pole is really a south magnetic pole)", + "video_name": "8Y4JSp5U82I", + "transcript": "We've learned a little bit about gravity. We've learned a little bit about electrostatic. So, time to learn about a new fundamental force of the universe. And this one is probably second most familiar to us, And that's magnetism. Where does the word come from? Well, I think several civilizations-- I'm no historian-- found these lodestones, these objects that would attract other objects like it, other magnets. Or would even attract metallic objects like iron. Ferrous objects. And they're called lodestones. That's, I guess, the Western term for it. And the reason why they're called magnets is because they're named after lodestones that were found near the Greek province of Magnesia. And I actually think the people who lived there were called Magnetes. But anyway, you could Wikipedia that and learn more about it than I know. But anyway let's focus on what magnetism is. And I think most of us have at least a working knowledge of what it is; we've all played with magnets and we've dealt with compasses. But I'll tell you this right now, what it really is, is pretty deep. And I think it's fairly-- I don't think anyone has-- we can mathematically understand it and manipulate it and see how it relates to electricity. We actually will show you the electrostatic force and the magnetic force are actually the same thing, just viewed from different frames of reference. I know that all of that sounds very complicated and all of that. But in our classical Newtonian world we treat them as two different forces. But what I'm saying is although we're kind of used to a magnet just like we're used to gravity, just like gravity is also fairly mysterious when you really think about what it is, so is magnetism. So with that said, let's at least try to get some working knowledge of how we can deal with magnetism. So we're all familiar with a magnet. I didn't want it to be yellow. I could make the boundary yellow. No, I didn't want it to be like that either. So if this is a magnet, we know that a magnet always has two poles. It has a north pole and a south pole. And these were just labeled by convention. Because when people first discovered these lodestones, or they took a lodestone and they magnetized a needle with that lodestone, and then that needle they put on a cork in a bucket of water, and that needle would point to the Earth's north pole. They said, oh, well the side of the needle that is pointing to the Earth's north, let's call that the north pole. And the point of the needle that's pointing to the south pole-- sorry, the point of the needle that's pointing to the Earth's geographic south, we'll call that the south pole. Or another way to put it, if we have a magnet, the direction of the magnet or the side of the magnet that orients itself-- if it's allowed to orient freely without friction-- towards our geographic north, we call that the north pole. And the other side is the south pole. And this is actually a little bit-- obviously we call the top of the Earth the north pole. You know, this is the north pole. And we call this the south pole. And there's another notion of magnetic north. And that's where-- I guess, you could kind of say-- that is where a compass, the north point of a compass, will point to. And actually, magnetic north moves around because we have all of this moving fluid inside of the earth. And a bunch of other interactions. It's a very complex interaction. But magnetic north is actually roughly in northern Canada. So magnetic north might be here. So that might be magnetic north. And magnetic south, I don't know exactly where that is. But it can kind of move around a little bit. It's not in the same place. So it's a little bit off the axis of the geographic north pole and the south pole. And this is another slightly confusing thing. Magnetic north is the geographic location, where the north pole of a magnet will point to. But that would actually be the south pole, if you viewed the Earth as a magnet. So if the Earth was a big magnet, you would actually view that as a south pole of the magnet. And the geographic south pole is the north pole of the magnet. You could read more about that on Wikipedia, I know it's a little bit confusing. But in general, when most people refer to magnetic north, or the north pole, they're talking about the geographic north area. And the south pole is the geographic south area. But the reason why I make this distinction is because we know when we deal with magnets, just like electricity, or electrostatics-- but I'll show a key difference very shortly-- is that opposite poles attract. So if this side of my magnet is attracted to Earth's north pole then Earth's north pole-- or Earth's magnetic north-- actually must be the south pole of that magnet. And vice versa. The south pole of my magnet here is going to be attracted to Earth's magnetic south. Which is actually the north pole of the magnet we call Earth. Anyway, I'll take Earth out of the equation because it gets a little bit confusing. And we'll just stick to bars because that tends to be a little bit more consistent. Let me erase this. There you go. I'll erase my Magnesia. I wonder if the element magnesium was first discovered in Magnesia, as well. Probably. And I actually looked up Milk of Magnesia, which is a laxative. And it was not discovered in Magnesia, but it has magnesium in it. So I guess its roots could be in Magnesia if magnesium was discovered in Magnesia. Anyway, enough about Magnesia. Back to the magnets. So if this is a magnet, and let me draw another magnet. Actually, let me erase all of this. All right. So let me draw two more magnets. We know from experimentation when we were all kids, this is the north pole, this is the south pole. That the north pole is going to be attracted to the south pole of another magnet. And that if I were to flip this magnet around, it would actually repel north-- two north facing magnets would repel each other. And so we have this notion, just like we had in electrostatics, that a magnet generates a field. It generates these vectors around it, that if you put something in that field that can be affected by it, it'll be some net force acting on it. So actually, before I go into magnetic field, I actually want to make one huge distinction between magnetism and electrostatics. Magnetism always comes in the form of a dipole. It means that we have two poles. A north and a south. In electrostatics, you do have two charges. You have a positive charge and a negative charge. So you do have two charges. But they could be by themselves. You could just have a proton. You don't have to have an electron there right next to it. You could just have a proton and it would create a positive electrostatic field. And our field lines are what a positive point charge would do. And it would be repelled. So you don't always have to have a negative charge there. Similarly you could just have an electron. And you don't have to have a proton there. So you could have monopoles. These are called monopoles, when you just have one charge when you're talking about electrostatics. But with magnetism you always have a dipole. If I were to take this magnet, this one right here, and if I were to cut it in half, somehow miraculously each of those halves of that magnet will turn into two more magnets. Where this will be the south, this'll be the north, this'll be the south, this will be the north. And actually, theoretically, I've read-- my own abilities don't go this far-- there could be such a thing as a magnetic monopole, although it has not been observed yet in nature. So everything we've seen in nature has been a dipole. So you could just keep cutting this up, all the way down to if it's just one electron left. And it actually turns out that even one electron is still a magnetic dipole. It still is generating, it still has a north pole and a south pole. And actually it turns out, all magnets, the magnetic field is actually generated by the electrons within it. By the spin of electrons and that-- you know, when we talk about electron spin we imagine some little ball of charge spinning. But electrons are-- you know, it's hard to-- they do have mass. But it starts to get fuzzy whether they are energy or mass. And then how does a ball of energy spin? Et cetera, et cetera. So it gets very almost metaphysical. So I don't want to go too far into it. And frankly, I don't think you really can get an intuition. It is almost-- it is a realm that we don't normally operate in. But even these large magnets you deal with, the magnetic field is generated by the electron spins inside of it and by the actual magnetic fields generated by the electron motion around the protons. Well, I hope I'm not overwhelming you. And you might say, well, how come sometimes a metal bar can be magnetized and sometimes it won't be? Well, when all of the electrons are doing random different things in a metal bar, then it's not magnetized. Because the magnetic spins, or the magnetism created by the electrons are all canceling each other out, because it's random. But if you align the spins of the electrons, and if you align their rotations, then you will have a magnetically charged bar. But anyway, I'm past the ten-minute mark, but hopefully that gives you a little bit of a working knowledge of what a magnet is. And in the next video, I will show what the effect is. Well, one, I'll explain how we think about a magnetic field. And then what the effect of a magnetic field is on an electron. Or not an electron, on a moving charge. See you in the next video." + }, + { + "Q": "Why is the tan of pi + 0.46 also a positive slope? If you start at the origin, isn't it then going downward since the arrow is going from right to left in a downward direction?", + "A": "It might help not to think of the terminal ray of pi+.46 as going from the origin outward. On the unit circle there are 3 variables, x, y, and the angle t. All the variation is along the circumference of the unit circle, which determines x, y, and t. The radius doesn t change. The slope of any line is delta y over delta x, which you can verify is y/x = tan(t) (In the 3rd quadrant, as x increases, so does y).", + "video_name": "C3HFAyigqoY", + "transcript": "Voiecover:One angle whose tangent is half is 0.46 radians. So we're saying that the tangent right over here is... So the tangent... So we're gonna write this down. So we're saying that the tangent of 0.46 radians is equal to half. And another way of thinking about the tangent of an angle is that's the slope of that angle's terminal ray. So it's the slope of this ray right over here. Yeah that makes sense that that slope is about half. Now what other angles have a tangent of 1 half? So let's look at these choices. So this is our original angle, 0.46 radians, plus pi over 2. If you think in degrees, pi is 180. pi over 2 is 90 degrees. So this one... Actually let me do in a color you're more likely to see. This one is gonna look like this. Where this is an angle of pi over 2. And just eyeballing it, you immediately see that the slope of this ray is very different than the slope of this ray right over here. In fact they look like they are. They are perpendicular because they have an angle of pi over 2 between them. But they're definitely not going to have the same tangent. They don't have the same slope. Let's think about pi minus 0.46. So that's essentially pi is going along the positive x axis. You go all the way around. Or half way around to your pi radians. But then we're gonna subtract 0.46. So it's gonna look something like this. It's gonna look something like that where this is 0.46 that we have subtracted. Another way to think about it, if we take our original terminal ray and we flip it over the y axis, we get to this terminal ray right over here. And you could immediately see that the slope of the terminal ray is not the same as the slope of this one, of our first one, of our original, in fact they look like the negatives of each other. So we can rule that one out as well. 0.46 plus pi or pi plus 0.46. So that's going to take us... If you add pi to this you're essentially going half way around the unit circle and you're getting to a point that is... Or you're forming a ray that is collinear with the original ray. So that's that angle right over here. So pi plus 0.46 is this entire angle right over there. And when you just look at this ray, you see its collinear is going to have the exact same slope as the terminal ray for the 0.46 radion. So just that tells you that the tangent is going to be the same. So I could check that there. And in previous videos when we explore the symmetries of the tangent function, we in fact saw that. That if you took an angle and you add pi to it, you're going to have the same tangent. And if you wanna dig a little bit deeper, I encourage you to look at that video on the symmetries of unit circle symmetries for the tangent function. So let's look at these other choices. 2 pi minus 0.46. So 2 pi... If this is 0 degrees, 2 pi gets you back to the positive x axis and then you're going to subtract 0.46. So that's going to be this angle right over here. And that looks like it has the negative slope of this original ray right up here. So these aren't going to have the same tangent. Now this one, you're taking 0.46 and you're adding 2 pi. So you're taking 0.46 and then you're adding 2 pi which essentially is just going around the unit circle once and you get to the exact same point. So you add 2 pi to any angle measure, you're going to not only have the same tangent value, you're gonna have the same sine value, cosine value because you're essentially going back around to the exact same angle when you add 2 pi. So this is definitely also going to be true." + }, + { + "Q": "If they used stem cells to cure diseases, wouldn't that stop one of natures ways to lower human population? I don't want to take sides on this one, but I think it would be unfriendly to the environment as a whole,", + "A": "There is no Nature s way that we can know of. That s a religious question. It is true though that overpopulation is a serious problem, and we haven t even cured that many diseases yet. The earth s biosphere is currently undergoing a mass extinction period almost certainly because of the rise of the human species.", + "video_name": "-yCIMk1x0Pk", + "transcript": "Where we left off after the meiosis videos is that we had two gametes. We had a sperm and an egg. Let me draw the sperm. So you had the sperm and then you had an egg. Maybe I'll do the egg in a different color. That's the egg, and we all know how this story goes. The sperm fertilizes the egg. And a whole cascade of events start occurring. The walls of the egg then become impervious to other sperm so that only one sperm can get in, but that's not the focus of this video. The focus of this video is how this fertilized egg develops once it has become a zygote. So after it's fertilized, you remember from the meiosis videos that each of these were haploid, or that they had-- oh, I added an extra i there-- that they had half the contingency of the DNA. As soon as the sperm fertilizes this egg, now, all of a sudden, you have a diploid zygote. Let me do that. So now let me pick a nice color. So now you're going to have a diploid zygote that's going to have a 2N complement of the DNA material or kind of the full complement of what a normal cell in our human body would have. So this is diploid, and it's a zygote, which is just a fancy way of saying the fertilized egg. And it's now ready to essentially turn into an organism. So immediately after fertilization, this zygote starts experiencing cleavage. It's experiencing mitosis, that's the mechanism, but it doesn't increase a lot in size. So this one right here will then turn into-- it'll just split up via mitosis into two like that. And, of course, these are each 2N, and then those are going to split into four like that. And each of these have the same exact genetic complement as that first zygote, and it keeps splitting. And this mass of cells, we can start calling it, this right here, this is referred to as the morula. And actually, it comes from the word for mulberry because it looks like a mulberry. So actually, let me just kind of simplify things a little bit because we don't have to start here. So we start with a zygote. This is a fertilized egg. It just starts duplicating via mitosis, and you end up with a ball of cells. It's often going to be a power of two, because these cells, at least in the initial stages are all duplicating all at once, and then you have this morula. Now, once the morula gets to about 16 cells or so-- and we're talking about four or five days. This isn't an exact process-- they started differentiating a little bit, where the outer cells-- and this kind of turns into a sphere. Let me make it a little bit more sphere like. So it starts differentiating between-- let me make some outer cells. This would be a cross-section of it. It's really going to look more like a sphere. That's the outer cells and then you have your inner cells on the inside. These outer cells are called the trophoblasts. Let me do it in a different color. Let me scroll over. I don't want to go there. And then the inner cells, and this is kind of the crux of what this video is all about-- let me scroll down a little bit. The inner cells-- pick a suitable color. The inner cells right there are called the embryoblast. And then what's going to happen is some fluid's going to start filling in some of this gap between the embryoblast and the trophoblast, so you're going to start having some fluid that comes in there, and so the morula will eventually look like this, where the trophoblast, or the outer membrane, is kind of this huge sphere of cells. And this is all happening as they keep replicating. Mitosis is the mechanism, so now my trophoblast is going to look like that, and then my embryoblast is going to look like this. Sometimes the embryoblast-- so this is the embryoblast. Sometimes it's also called the inner cell mass, so let me write that. And this is what's going to turn into the organism. And so, just so you know a couple of the labels that are involved here, if we're dealing with a mammalian organism, and we are mammals, we call this thing that the morula turned into is a zygote, then a morula, then the cells of the morula started to differentiate into the trophoblast, or kind of the outside cells, and then the embryoblast. And then you have this space that forms here, and this is just fluid, and it's called the blastocoel. A very non-intuitive spelling of the coel part of blastocoel. But once this is formed, this is called a blastocyst. That's the entire thing right here. Let me scroll down a little bit. This whole thing is called the blastocyst, and this is the case in humans. Now, it can be a very confusing topic, because a lot of times in a lot of books on biology, you'll say, hey, you go from the morula to the blastula or the blastosphere stage. Let me write those words down. So sometimes you'll say morula, and you go to blastula. Sometimes it's called the blastosphere. And I want to make it very clear that these are essentially the same stages in development. These are just for-- you know, in a lot of books, they'll start talking about frogs or tadpoles or things like that, and this applies to them. While we're talking about mammals, especially the ones that are closely related to us, the stage is the blastocyst stage, and the real differentiator is when people talk about just blastula and blastospheres. There isn't necessarily this differentiation between these outermost cells and these embryonic, or this embryoblast, or this inner cell mass here. But since the focus of this video is humans, and really that's where I wanted to start from, because that's what we are and that's what's interesting, we're going to focus on the blastocyst. Now, everything I've talked about in this video, it was really to get to this point, because what we have here, these little green cells that I drew right here in the blastocysts, this inner cell mass, this is what will turn into the organism. And you say, OK, Sal, if that's the organism, what's all of these purple cells out here? This trophoblast out there? That is going to turn into the placenta, and I'll do a future video where in a human, it'll turn into a placenta. So let me write that down. It'll turn into the placenta. And I'll do a whole future video about I guess how babies are born, and I actually learned a ton about that this past year because a baby was born in our house. But the placenta is really kind of what the embryo develops inside of, and it's the interface, especially in humans and in mammals, between the developing fetus and its mother, so it kind of is the exchange mechanism that separates their two systems, but allows the necessary functions to go on between them. But that's not the focus of this video. The focus of this video is the fact that these cells, which at this point are-- they've differentiated themselves away from the placenta cells, but they still haven't decided what they're going to become. Maybe this cell and its descendants eventually start becoming part of the nervous system, while these cells right here might become muscle tissue, while these cells right here might become the liver. These cells right here are called embryonic stem cells, and probably the first time in this video you're hearing a term that you might recognize. So if I were to just take one of these cells, and actually, just to introduce you to another term, you know, we have this zygote. As soon as it starts dividing, each of these cells are called a blastomere. And you're probably wondering, Sal, why does this word blast keep appearing in this kind of embryology video, these development videos? And that comes from the Greek for spore: blastos. So the organism is beginning to spore out or grow. I won't go into the word origins of it, but that's where it comes from and that's why everything has So these are blastomeres. So when I talk what embryonic stem cells, I'm talking about the individual blastomeres inside of this embryoblast or inside of this inner cell mass. These words are actually unusually fun to say. So each of these is an embryonic stem cell. Let me write this down in a vibrant color. So each of these right here are embryonic stem cells, and I wanted to get to this. And the reason why these are interesting, and I think you already know, is that there's a huge debate around these. One, these have the potential to turn into anything, that they have this plasticity. That's another word that you might hear. Let me write that down, too: plasticity. And the word essentially comes from, you know, like a plastic can turn into anything else. When we say that something has plasticity, we're talking about its potential to turn into a lot of different things. So the theory is, and there's already some trials that seem to substantiate this, especially in some lower organisms, that, look, if you have some damage at some point in your body-- let me draw a nerve cell. Let me say I have a-- I won't go into the actual mechanics of a nerve cell, but let's say that we have some damage at some point on a nerve cell right there, and because of that, someone is paralyzed or there's some nerve dysfunction. We're dealing with multiple sclerosis or who knows what. The idea is, look, we have these cell here that could turn into anything, and we're just really understanding how it knows what to turn into. It really has to look at its environment and say, hey, what are the guys around me doing, and maybe that's what helps dictate what it does. But the idea is you take these things that could turn to anything and you put them where the damage is, you layer them where the damage is, and then they can turn into the cell that they need to turn into. So in this case, they would turn into nerve cells. They would turn to nerve cells and repair the damage and maybe cure the paralysis for that individual. So it's a huge, exciting area of research, and you could even, in theory, grow new organs. If someone needs a kidney transplant or a heart transplant, maybe in the future, we could take a colony of these embryonic stem cells. Maybe we can put them in some type of other creature, or who knows what, and we can turn it into a replacement heart or a So there's a huge amount of excitement about what these can do. I mean, they could cure a lot of formerly uncurable diseases or provide hope for a lot of patients who might otherwise die. But obviously, there's a debate here. And the debate all revolves around the issue of if you were to go in here and try to extract one of these cells, you're going to kill this embryo. You're going to kill this developing embryo, and that developing embryo had the potential to become a human being. It's a potential that obviously has to be in the right environment, and it has to have a willing mother and all of the rest, but it does have the potential. And so for those, especially, I think, in the pro-life camp, who say, hey, anything that has a potential to be a human being, that is life and it should not be killed. So people on that side of the camp, they're against the destroying of this embryo. I'm not making this video to take either side to that argument, but it's a potential to turn to a human being. It's a potential, right? So obviously, there's a huge amount of debate, but now, now you know in this video what people are talking about when they say embryonic stem cells. And obviously, the next question is, hey, well, why don't they just call them stem cells as opposed to embryonic stem cells? And that's because in all of our bodies, you do have what are called somatic stem cells. Let me write that down. Somatic or adults stem cells. And we all have them. They're in our bone marrow to help produce red blood cells, other parts of our body, but the problem with somatic stem cells is they're not as plastic, which means that they can't form any type of cell in the human body. There's an area of research where people are actually maybe trying to make them more plastic, and if they are able to take these somatic stem cells and make them more plastic, it might maybe kill the need to have these embryonic stem cells, although maybe if they do this too good, maybe these will have the potential to turn into human beings as well, so that could become a debatable issue. But right now, this isn't an area of debate because, left to their own devices, a somatic stem cell or an adult stem cell won't turn into a human being, while an embryonic one, if it is implanted in a willing mother, then, of course, it will turn into a human being. And I want to make one side note here, because I don't want to take any sides on the debate of-- well, I mean, facts are facts. This does have the potential to turn into a human being, but it also has the potential to save millions of lives. Both of those statements are facts, and then you can decide on your own which side of that argument you'd like to or what side of that balance you would like to kind of put your own opinion. But there's one thing I want to talk about that in the public debate is never brought up. So you have this notion of when you-- to get an embryonic stem cell line, and when I say a stem cell line, I mean you take a couple of stem cells, or let's say you take one stem cell, and then you put it in a Petri dish, and then you allow it to just duplicate. So this one turns into two, those two turn to four. Then someone could take one of these and then put it in their own Petri dish. These are a stem cell line. They all came from one unique embryonic stem cell or what initially was a blastomere. So that's what they call a stem cell line. So the debate obviously is when you start an embryonic stem cell line, you are destroying an embryo. But I want to make the point here that embryos are being destroyed in other processes, and namely, in-vitro fertilization. And maybe this'll be my next video: fertilization. And this is just the notion that they take a set of eggs out of a mother. It's usually a couple that's having trouble having a child, and they take a bunch of eggs out of the mother. So let's say they take maybe 10 to 30 eggs out of the mother. They actually perform a surgery, take them out of the ovaries of the mother, and then they fertilize them with semen, either it might come from the father or a sperm donor, so then all of these becomes zygotes once they're fertilized with semen. So these all become zygotes, and then they allow them to develop, and they usually allow them to develop to the blastocyst stage. So eventually all of these turn into blastocysts. They have a blastocoel in the center, which is this area of fluid. They have, of course, the embryo, the inner cell mass in them, and what they do is they look at the ones that they deem are healthier or maybe the ones that are at least just not unhealthy, and they'll take a couple of these and they'll implant these into the mother, so all of this is occurring in a Petri dish. So maybe these four look good, so they're going to take these four, and they're going to implant these into a mother, and if all goes well, maybe one of these will turn into-- will give the couple a child. So this one will develop and maybe the other ones won't. But if you've seen John & Kate Plus 8, you know that many times they implant a lot of them in there, just to increase the probability that you get at least one child. But every now and then, they implant seven or eight, and then you end up with eight kids. And that's why in-vitro fertilization often results in kind of these multiple births, or reality television shows on cable. But what do they do with all of these other perfectly-- well, I won't say perfectly viable, but these are embryos. They may or may not be perfectly viable, but you have these embryos that have the potential, just like this one right here. These all have the potential to turn into a human being. But most fertility clinics, roughly half of them, they either throw these away, they destroy them, they allow them to die. A lot of these are frozen, but just the process of freezing them kills them and then bonding them kills them again, so most of these, the process of in-vitro fertilization, for every one child that has the potential to develop into a full-fledged human being, you're actually destroying tens of very viable embryos. So at least my take on it is if you're against-- and I generally don't want to take a side on this, but if you are against research that involves embryonic stem cells because of the destruction of embryos, on that same, I guess, philosophical ground, you should also be against in-vitro fertilization because both of these involve the destruction of zygotes. I think-- well, I won't talk more about this, because I really don't want to take sides, but I want to show that there is kind of an equivalence here that's completely lost in this debate on whether embryonic stem cells should be used because they have a destruction of embryos, because you're destroying just as many embryos in this-- well, I won't say just as many, but you are destroying embryos. There's hundreds of thousands of embryos that get destroyed and get frozen and obviously destroyed in that process as well through this in-vitro fertilization process. So anyway, now hopefully you have the tools to kind of engage in the debate around stem cells, and you see that it all comes from what we learned about meiosis. They produce these gametes. The male gamete fertilizes a female gamete. The zygote happens or gets created and starts splitting up the morula, and then it keeps splitting and it differentiates into the blastocyst, and then this is where the stem cells are. So you already know enough science to engage in kind of a very heated debate." + }, + { + "Q": "This question is not about the video but one I have had for awhile, What is the best way to write this: 3/4 or .75, I prefer decimals I do not know why but i do, Khan Academy normally asks for fractions, I have the right decimal for the fraction but it is considered wrong. What is best fraction or decimals or is it all opinion?", + "A": "Using fraction will help you in higher level math, and you will need to be able to use them well. They are just preparing you now for what s ahead.", + "video_name": "iI_2Piwn_og", + "transcript": "I now want to solve some inequalities that also have absolute values in them. And if there's any topic in algebra that probably confuses people the most, it's this. But if we kind of keep our head on straight about what absolute value really means, I think you will find that it's not that bad. So let's start with a nice, fairly simple warm-up problem. Let's start with the absolute value of x is less than 12. So remember what I told you about the meaning of absolute value. It means how far away you are from 0. So one way to say this is, what are all of the x's that are less than 12 away from 0? Let's draw a number line. So if we have 0 here, and we want all the numbers that are less than 12 away from 0, well, you could go all the way to positive 12, and you could go all the way to negative 12. Anything that's in between these two numbers is going to have an absolute value of less than 12. It's going to be less than 12 away from 0. So this, you could say, this could be all of the numbers where x is greater than negative 12. Those are definitely going to have an absolute value less than 12, as long as they're also-- and, x has to be less than 12. So if an x meets both of these constraints, its absolute value is definitely going to be less than 12. You know, you take the absolute value of negative 6, that's only 6 away from 0. The absolute value of negative 11, only 11 away from 0. So something that meets both of these constraints will satisfy the equation. And actually, we've solved it, because this is only a one-step equation there. But I think it lays a good foundation for the next few problems. And I could actually write it like this. In interval notation, it would be everything between negative 12 and positive 12, and not including those numbers. Or we could write it like this, x is less than 12, and is greater than negative 12. That's the solution set right there. Now let's do one that's a little bit more complicated, that allows us to think a little bit harder. So let's say we have the absolute value of 7x is greater than or equal to 21. So let's not even think about what's inside of the absolute value sign right now. In order for the absolute value of anything to be greater than or equal to 21, what does it mean? It means that whatever's inside of this absolute value sign, whatever that is inside of our absolute value sign, it must be 21 or more away from 0. Let's draw our number line. And you really should visualize a number line when you do this, and you'll never get confused then. You shouldn't be memorizing any rules. So let's draw 0 here. Let's do positive 21, and let's do a negative 21 here. So we want all of the numbers, so whatever this thing is, that are greater than or equal to 21. They're more than 21 away from 0. Their absolute value is more than 21. Well, all of these negative numbers that are less than negative 21, when you take their absolute value, when you get rid of the negative sign, or when you find their distance from 0, they're all going to be greater than 21. If you take the absolute value of negative 30, it's going to be greater than 21. Likewise, up here, anything greater than positive 21 will also have an absolute value greater than 21. So what we could say is 7x needs to be equal to one of these numbers, or 7x needs to be equal to one of these numbers out here. So we could write 7x needs to be one of these numbers. Well, what are these numbers? These are all of the numbers that are less than or equal to negative 21, or 7x-- let me do a different color here-- or 7x has to be one of these numbers. And that means that 7x has to be greater than or equal to positive 21. I really want you to kind of internalize what's going on here. If our absolute value is greater than or equal to 21, that means that what's inside the absolute value has to be either just straight up greater than the positive 21, or less than negative 21. Because if it's less than negative 21, when you take its absolute value, it's going to be more than 21 away from 0. Hopefully that make sense. We'll do several of these practice problems, so it really gets ingrained in your brain. But once you have this set up, and this just becomes a compound inequality, divide both sides of this equation by 7, you get x is less than or equal to negative 3. Or you divide both sides of this by 7, you get x is greater than or equal to 3. So I want to be very clear. This, what I drew here, was not the solution set. This is what 7x had to be equal to. I just wanted you to visualize what it means to have the absolute value be greater than 21, to be more than 21 away from 0. This is the solution set. x has to be greater than or equal to 3, or less than or equal to negative 3. So the actual solution set to this equation-- let me draw a number line-- let's say that's 0, that's 3, that is negative 3. x has to be either greater than or equal to 3. That's the equal sign. Or less than or equal to negative 3. Let's do a couple more of these. Because they are, I think, confusing, but if you really start to get the gist of what absolute value is saying, they become, I think, intuitive. So let's say that we have the absolute value-- let me get a good one. Let's say the absolute value of 5x plus 3 is less than 7. So that's telling us that whatever's inside of our absolute value sign has to be less than 7 away from 0. So the ways that we can be less than 7 away from 0-- let me draw a number line-- so the ways that you can be less than 7 away from 0, you could be less than 7, and greater than negative 7. You have to be in this range. So in order to satisfy this thing in this absolute value sign, it has to be-- so the thing in the absolute value sign, which is 5x plus 3-- it has to be greater than negative 7 and it has to be less than 7, in order for its absolute value to be less than 7. If this thing, this 5x plus 3, evaluates anywhere over here, its absolute value, its distance from 0, will be less than 7. And then we can just solve these. You subtract 3 from both sides. 5x is greater than negative 10. Divide both sides by 5. x is greater than negative 2. Now over here, subtract 3 from both sides. 5x is less than 4. Divide both sides by 5, you get x is less than 4/5. And then we can draw the solution set. We have to be greater than negative 2, not greater than or equal to, and less than 4/5. So this might look like a coordinate, but this is also interval notation, if we're saying all of the x's between negative 2 and 4/5. Or you could write it all of the x's that are greater than negative 2 and less than 4/5. These are the x's that satisfy this equation. And I really want you to internalize this visualization here. Now, you might already be seeing a bit of a rule here. And I don't want you to just memorize it, but I'll give it to you just in case you want it. If you have something like f of x, the absolute value of f of x is less than, let's say, some number a. So this was the situation. We have some f of x less than a. That means that the absolute value of f of x, or f of x has to be less than a away from 0. So that means that f of x has to be less than positive a or greater than negative a. That translates to that, which translates to f of x greater than negative a and f of x less than a. But it comes from the same logic. This has to evaluate to something that is less than a away from 0. Now, if we go to the other side, if you have something of the form f of x is greater than a. That means that this thing has to evaluate to something that is further than a away from 0. So that means that f of x is either just straight up greater than positive a, or f of x is less than negative a. If it's less than negative a, maybe it's negative a minus another 1, or negative 5 plus negative a. Then, when you take its absolute value, it'll become a plus 5. So its absolute value is going to be greater than a. So I just want to-- you could memorize this if you want, but I really want you to think about this is just saying, OK, this has to evaluate, be less than a away from 0, this has to be more than a away from 0. Let's do one more, because I know this can be a little bit confusing. And I encourage you to watch this video over and over and over again, if it helps. Let's say we have the absolute value of 2x-- let me do another one over here. Let's do a harder one. Let's say the absolute value of 2x over 7 plus 9 is greater than 5/7. So this thing has to evaluate to something that's more than 5/7 away from 0. So this thing, 2x over 7 plus 9, it could just be straight up greater than 5/7. Or it could be less than negative 5/7, because if it's less than negative 5/7, its absolute value is going to be greater than 5/7. Or 2x over 7 plus 9 will be less than negative 5/7. We're doing this case right here. And then we just solve both of these equations. See if we subtract-- let's just multiply everything by 7, just to get these denominators out of the way. So if you multiply both sides by 7, you get 2x plus 9 times 7 is 63, is greater than 5. Let's do it over here, too. You'll get 2x plus 63 is less than negative 5. Let's subtract 63 from both sides of this equation, and you get 2x-- let's see. 5 minus 63 is 58, 2x is greater than 58. If you subtract 63 from both sides of this equation, you get 2x is less than negative 68. Oh, I just realized I made a mistake here. You subtract 63 from both sides of this, 5 minus 63 is negative 58. I don't want to make a careless mistake there. And then divide both sides by 2. You get, in this case, x is greater than-- you don't have to swap the inequality, because we're dividing by a positive number-- negative 58 over 2 is negative 29, or, here, if you divide both sides by 2, or, x is less than negative 34. 68 divided by 2 is 34. And so, on the number line, the solution set to that equation will look like this. That's my number line. I have negative 29. I have negative 34. So the solution is, I can either be greater than 29, not greater than or equal to, so greater than 29, that is that right there, or I could be less than negative 34. So any of those are going to satisfy this absolute value inequality." + }, + { + "Q": "Doesn't he mean cross product at 2:00?", + "A": "Yes, he meant cross product.", + "video_name": "b7JTVLc_aMk", + "transcript": "What I want to do with this video is cover something called the triple product expansion-- or Lagrange's formula, sometimes. And it's really just a simplification of the cross product of three vectors, so if I take the cross product of a, and then b cross c. And what we're going to do is, we can express this really as sum and differences of dot products. Well, not just dot products-- dot products scaling different vectors. You're going to see what I mean. But it simplifies this expression a good bit, because cross products are hard to take. They're computationally intensive and, at least in my mind, they're confusing. Now this isn't something you have to know if you're going to be dealing with vectors, but it's useful to know. My motivation for actually doing this video is I saw some problems for the Indian Institute of Technology entrance exam that seems to expect that you know Lagrange's formula, or the triple product expansion. So let's see how we can simplify this. So to do that, let's start taking the cross product of b and c. And in all of these situations, I'm just going to assume-- let's say I have vector a. That's going to be a, the x component of vector a times the unit of vector i plus the y component of vector a times the unit vector j plus the z component of vector a times unit vector k. And I could do the same things for b and c. So if I say b sub y, I'm talking about what's scaling the j component in the b vector. So let's first take this cross product over here. And if you've seen me take cross products, you know that I like to do these little determinants. Let me just take it over here. So b cross c is going to be equal to the determinant. And I put an i, j, k up here. This is actually the definition of the cross product, so no proof necessary to show you why this is true. This is just one way to remember the dot product, if you remember how to take determinants of three-by-threes. And we'll put b's x term, b's y coefficient, and b's z component. And then you do the same thing for the c, cx, cy, cz. And then this is going to be equal to-- so first you have the i component. So it's going to be the i component times b. So you ignore this column and this row. So bycz minus bzcy. So I'm just ignoring all of this. And I'm looking at this two-by-two over here, minus bzcy. And then we want to subtract the j component. Remember, we alternate signs when we take our determinant. Subtract that. And then we take out that column and that row, so it's going to be bxcz-- this is a little monotonous, but hopefully, it'll have an interesting result-- bxcz minus bzcx. And then finally, plus the k component. OK, we're going to have bx times cy minus bycx. We just did the dot product, and now we want to take the-- oh, sorry, we just did the cross product. I don't want to get you confused. We just took the cross product of b and c. And now we need take the cross product of that with a, or the cross product of a with this thing right over here. Instead of rewriting the vector, let me just set up another matrix here. So let me write my i j k up here. And then let me write a's components. So we have a sub x, a sub y, a sub z. And then let's clean this up a little bit. We're just looking at-- no, I want to do that in black. Let's do this in black, so that we can kind of erase that. Now this is a minus j times that. So what I'm going to do is I'm going to get rid of the minus and the j, but I am going to rewrite this with the signs swapped. So if you swap the signs, it's actually bzcx minus bxcz. So let me delete everything else. So I just took the negative and I multiplied it by this. I hope I'm not making any careless mistakes here, so let me just check and make my brush size little bit bigger, so I can erase that a little more efficiently. And then we also want to get rid of that right over there. Now let me get my brush size back down to normal size. All right. So now let's just take this cross product. So once again, set it up as a determinant. And what I'm only going to focus on-- because it'll take the video, or it'll take me forever if I were to do the i, j, and k components-- I'm just going to focus on the i component, just on the x component of this cross product. And then we can see that we'll get the same result for the j and the k. And then we can see what, hopefully, this simplifies down to. So if we just focus on the i component here, this is going to be i times-- and we just look at this two-by-two matrix right over here. We ignore i's column, i's row. And we have ay times all of this. So let me just multiply it out. So it's ay times bxcy, minus ay times by, times bycx. And then we're going to want to subtract. We're going to have minus az times this. So let's just do that. So it's minus, or negative, azbzcx. And then we have a negative az times this, so it's plus azbxcz. And now what I'm going to do-- this is a little bit of a trick for this proof right here, just so that we get the results that I want. I'm just going to add and subtract the exact same thing. So I'm going to add an axbxcx. And then I'm going to subtract an axbxcx, minus axbxcx. So clearly, I have not changed this expression. I've just added and subtracted the same thing. And let's see what we can simplify. Remember, this is just the x component of our triple product. Just the x component. But to do this, let me factor out. I'm going to factor out a bx. So let me do this, let me get the bx. So if I were to factor it out-- I'm going to factor it out of this term that has a bx. I'm going to factor it out of this term. And then I'm going to factor it out of this term. So if I take the bx out, I'm going to have an aycy. Actually, let me write it a little bit differently. Let me factor it out of this one first. So then it's going to have an axcx. a sub x, c sub x. So I used this one up. And then I'll do this one now. Plus, if I factor the bx out, I get ay cy. I've used that one now. And now I have this one. I'm going to factor the bx out. So I'm left with a plus az, cz. So that's all of those. So I've factored that out. And now, from these right over here, let me factor out a negative cx. And so, if I do that-- let me go to this term right over here-- I'm going to have an axbx when I factor it out. So an axbx, cross that out. And then, over here, I'm going to have an ayby. Remember, I'm factoring out a negative cx, so I'm going to have a plus ay, sub by. And then, finally, I'm going to have a plus az, az bz. And what is this? Well, this right here, in green, this is the exact same thing as the dot products of a and c. This is the dot product of the vectors a and c. It's the dot product of this vector and that vector. So that's the dot of a and c times the x component of b minus-- I'll do this in the same-- minus-- once again, this is the dot product of a and b now, minus a dot b times the x component of c. And we can't forget, all of this was multiplied by the unit vector i. We're looking at the x component, or the i component of that whole triple product. So that's going to be all of this. All of this is times the unit vector i. Now, if we do this exact same thing-- and I'm not going to do it, because it's computationally intensive. But I think it won't be a huge leap of faith for you. This is for the x component. If I were to do the exact same thing for the y component, for the j component-- so it'll be plus-- if I do the same thing for the j component, we can really just pattern match. We have bx, cx, that's for the x component. We'll have by and c y for the j component. And then this is not component-specific, so it will be a dot c over here, and minus a dot b over here. You can verify any of these for yourself, if you don't believe me. But it's the exact same process we just did. And then, finally, for the z component, or the k component-- let me put parentheses over here-- same idea. You're going to have bz, cz. And then you're going to have a dot b over there. And then you're going to have a dot c over here. Now what does this become? How can we simplify this? Well, this right over here, we can expand this out. We can factor out an a dot c from all of these terms over here. And remember, this is going to be multiplied times i. Actually, let me not skip too many steps, just because I want you to believe what I'm doing. So if we expand the i here-- instead of rewriting it, let me just do it like this. It's a little bit messier, but let me just-- so I could write this i there and that i there. I'm kind of just distributing that x unit vector, or the i unit vector. And let me do the same thing for j. So I could put the j there. And I could put the j right over there. And then I could do the same thing for the k, put the k there, and then put the k there. And now what are these? Well, this part right over here is exactly the same thing as a dot c times-- and I'll write it out here-- bx times i plus by times j, plus bz times k. And then, from that, we're going to subtract all of this, a dot b. We're going to subtract a dot b times the exact same thing. And you're going to notice, this right here is the same thing as vector b. That is vector b. When you do it over here, you're going to get vector c. So I'll just write it over here. You're just going to get vector c. So just like that, we have a simplification for our triple product. I know it took us a long time to get here, but this is a simplification. It might not look like one, but computationally it is. It's easier to do. If I have-- I'll try to color-code it-- a cross b cross-- let me do it in all different colors-- c, we just saw that this is going to be equivalent to-- and one way to think about it is, it's going to be, you take the first vector times the dot product of-- the first vector in this second dot product, the one that we have our parentheses around, the one we would have to do first-- you take your first vector there. So it's vector b. And you multiply that times the dot product of the other two vectors, so a dot c. And from that, you subtract the second vector multiplied by the dot product of the other two vectors, of a dot b. And we're done. This is our triple product expansion. Now, once again, this isn't something that you really have to know. You could always, obviously, multiply it. You could actually do it by hand. You don't have to know this. But if you have really hairy vectors, or if this was some type of math competition, and sometimes it simplifies real fast when you reduce it to dot products, this is a useful thing to know, Lagrange's formula, or the triple product expansion." + }, + { + "Q": "What are new operators useful for?", + "A": "They are just definitions. Using these arbitrary operators helps us to be more flexible and comfortable in using algebra, and gives us a better understanding of numerical reasoning and computer science.", + "video_name": "ND-Bbp_q46s", + "transcript": "We're all used to the traditional operators like addition and subtraction and multiplication and division. And we've seen there's multiple ways to represent this. But what we're going to do in this video is a little fun. We're actually going to define our own operators. And what's neat about this is it kind of shows how broad mathematics can be. And on a more practical sense, it's actually something that you might see on some standardized tests. And the reason why they do that is so that you can appreciate that these aren't the only operators out there-- plus exponentiation and all those-- that in mathematics, you can define a whole new set of operators. So let's just do that. So let me just define x diamond y. And I'm going to define that as 5x minus y. So you could view this as defining it a function. But we're defining it using an operator. So if I have x diamond y, by definition, we have defined this operator. That means that's going to be equal to 5x minus y. So given that definition, what would 7 diamond 11 be? Well, you just go to the definition. 7 diamond 11. Instead of an x we have a 7. So it's going to be 5 times 7. So let me do it 5 times 7 minus, and instead of a y, we have an 11. So one way to think about it is, in our definition, every place you saw an x, you could replace with a 7, every place you saw a y, you replace with an 11. So you have minus 11 over here. Let me make the number-- So this is the 7. This 7 is this 7. And this 11 is this 11 right over here. And then we just evaluate that. So 5 times 7 is 35. So this is equal to 35 minus 11, which is equal to 24. So 7 diamond 11 is equal to 24. We can define other things. We can define something crazy like, let me define a-- well, I mentioned a star, let me use a star-- a star-- let me write it this way-- a star b. Let's say that that is the same thing as-- I don't know-- a over a plus b. And so same idea. What would 5 star 6 be? Well, you go back to the definition. By definition, every place where you see the a, you would now replace with a 5. Every time you saw the b, you would now replace with a 6. So this is going to be equal to 5 over 5 plus 6. a plus b. a is 5, b is 6 over 5 plus 6. So this would be 5/11. And then you can compound them. And we haven't defined any order of operations for these particular operations that we've just defined. So we're going to be careful to use parentheses when we put some of these together, but you can do something like, something interesting, like negative 1 diamond 0 star 5. And once again, we just focus on parentheses, because that's the only thing that's telling us what to start on first. Because we haven't figured out, we haven't defined whether diamond takes precedence over star, or star takes precedence over diamond the way that we have that saying that, hey, you do multiplication before you do addition. We haven't defined it for those operations, but that's what the parentheses helps us do. So we want to evaluate these parentheses first. 0 star 5 that is 0-- because you could view this 0 as the a and the 5 as the b--so it's going to be 0 over 0 plus 5, which is just going to be 0. So this over here, 0 or 5 just goes to 0. So this whole expression simplifies to negative 1 diamond-- this diamond right over here-- diamond 0. And now we go to the definition of the diamond operator. Well, that's five times the first number in our operator, or the first term that we're giving the operator. I guess you could think of it that way. So 5 times that. So it'll be 5 times negative 1, x is negative 1 minus y. Well, y here is the 0. Minus 0. So 5 times negative 1 is negative 5. And you will see-- and the idea here is just to make you feel comfortable defining new operators like this. And not being daunted if all of a sudden you see a diamond, and they're defining the diamond for you. And you're like, wait, I never saw a diamond. They're actually defining it for you, so you shouldn't say, I never saw a diamond. You should say, well, they've defined a diamond for me. This is how I use that operator. And sometimes you'll see even wackier things. You'll see things like this. Let me draw. So they'll define. I don't know if you would even consider this an operator. But you'll see something like this, that by definition, if someone writes a symbol like this, and they put-- a, b, c-- let me write it this way-- a, b, c, d. They'll say this is the same thing as ad minus b, all of that over c. And once again, this is just a definition. They have this weird symbolic way of representing these variables in all this. But they're defining how do you evaluate this crazy expression. And so, if someone were to give you, were to say, evaluate this diamond. Let me evaluate the diamond. So evaluate the diamond where, in my little sections of the diamond, I have a negative 1, a 5, a 3, and a 2. We would just use the definition of how to evaluate this diamond. And we'd say, OK, every time we see an a, that is going to be negative 1. So we have a negative 1 times d. Well, d is whatever is in the bottom right section of this diameter or this kite. So d is going to be 2. Let me write it this way. This is b. This is c. And this is d. So it's going to be negative 1 times 2 minus b-- well b is 5-- minus 5, all of that over c, which is 3. So this is going to be equal to negative 2 minus 5. So that is negative 7 over 3. And you could go crazy like this. And it might be a fun thing, actually, if you have some spare time. Define your own operators and see how creative you can get with those operators." + }, + { + "Q": "Why do we need parametric equations to draw lines in 3 dimensions? I'm not really familiar with the 3-dimensional coordinate plane, so this is new to me.", + "A": "Because we have 3 directions. With parametric equations you can describe all the coordinates of any point on the line. You just have to split the equation into x,y and z to get the coordinates of a point.", + "video_name": "hWhs2cIj7Cw", + "transcript": "Everything we've been doing in linear algebra so far, you might be thinking, it's kind of a more painful way of doing things that you already knew how to do. You've already dealt with vectors. I'm guessing that some of you all have already dealt with vectors in your calculus or your pre-calculus or your physics classes. But in this video I hope to show you something that you're going to do in linear algebra that you've never done before, and that it would have been very hard to do had you not been exposed to these videos. Well I'm going to start with, once again, a different way of doing something you already know how to do. So let me just define some vector here, instead of making them bold, I'll just draw it with the arrow on top. I'm going to define my vector to be-- I can do with the arrow on top or I can just make it super bold. I'm just going to define my vectors, it's going to be a vector in R2. Let's just say that my vector is the vector 2, 1. If I were to draw it in standard position, it looks like this. You go two to the right, up one, like that. That's my, right there, that is my vector v. Now, if I were to ask you, what are all of the possible vectors I can create? So let me define a set. Let me define a set, s, and it's equal to-- all of the vectors I can create, if I were to multiply v times some constant, so I multiply some constant, some scalar, times my vector v, and just to maybe be a little bit formal, I'll say such that c is a member of the real numbers. Now what would be a graphical representation of this set? Well, if we draw them all in standard position, c could be any real number. So if I were to multiply, c could be 2. If c is 2, let me do it this way. If I do 2 times our vector, I'm going to get the vector 4, 2. Let me draw that in standard position, 4, 2. It's right there. It's this vector right there. It's collinear with this first vector. It's along the same line, but just goes out further 2. Now I could've done another. I could have done 1.5 times our vector v. Let me do that in a different color. And maybe that would be, that would be what? That'd be 1.5 times 2, which is 3, 1.5. Where would that vector get me? I'd go one 1.5 and then I'd go 3, and then 1.5, I'd get right there. And I can multiply by anything. I can multiply 1.4999 times vector v, and get right over here. I could do minus 0.0001 times vector v. Let me write that down. I could do 0.001 times our vector v. And where were that put me? It would put me little super small vector right there. If I did minus 0.01, it would make a super small vector right there pointing in that direction. If I were to do minus 10, I would get a vector going in this direction that goes way like that. But you can imagine that if I were to plot all of the vectors in standard position, all of them that could be represented by any c in real numbers, I'll essentially get-- I'll end up drawing a bunch of vectors where their arrows are all lined up along this line right there, and all lined up in even in negative direction-- let me make sure I draw it properly-- along that line, like that. I think you get the idea. So it's a set of collinear vectors. So let me write that down. And if we view these vectors as position vectors, that this vector represents a point in space in R2-- this R2 is just our Cartesian coordinate plane right here in every direction-- if we view this vector as a position vector-- let me write that down-- if we view it as kind of a coordinate in R2, then this set, if we visually represent it as a bunch of position vectors, it'll be represented by this whole line over here. And I want to make that point clear because it's essentially a line, of slope 2. Right? Sorry, slope 1/2. Your rise is 1. Your rise is 1 for going over 2. But I don't want to go back to our Algebra 1 But I want to make this point that this line of slope 2 that goes through the origin, this is if we draw all of the vectors in the set as in their standard form, or if we draw them all as position vectors. If I didn't make that clarification, or that qualification, I could have drawn these vectors anywhere. Because this 4, 2 vector, I could have drawn over here. And then, to say that it's collinear probably wouldn't have made as much visual sense to you. But I think this collinearity of it makes more sense to you if you say, oh, let's draw them all in standard form. All of them start at the origin, and then their tails are at the origin, and their heads go essentially to the coordinate they represent. That's what I mean by their position vectors. They don't necessarily have to be position vectors, but for the visualization in this video, let's stick to that. Now I was only able to represent something that goes through the origin with this slope. So you can almost view that this vector kind of represented its slope. You almost want to view it as a slope vector, if you wanted to tie it in to what you learned in Algebra 1. What if we wanted to represent other lines that had that slope? What if we wanted to represent the the same line, or I guess a parallel line-- that goes through that point over there, the point 2 comma 4? Or if we're thinking in position vectors, we could say that point is represented by the vector, and we will call that x. It's represented by the vector x. And the vector x is equal to 2, 4. That point right there. What if I want to represent the line that's parallel to this that goes through that point 2, 4? So I want to represent this line right here. I'll draw it as parallel to this as I can. I think you get the idea, and it just keeps going like that in every direction. These two lines are parallel. How can I represent the set of all of these vectors, drawn in standard form, or all of the vectors, that if I were to draw them in standard form, would show this line? Well, you can think about it this way. If every one of the vectors that represented this line, if I start with any vector that was on this line, and I add my x vector to it, I'll show up at a corresponding point on this line that I want to be at. Right? Let's say I do negative 2 times my original, so minus 2 times my vector v, that equaled what? Minus 4, minus 2, so that's that vector there. But if I were to add x to it, if I were to add my x vector. So if I were to do minus 2 times my vector v, but I were to add x to it, so plus x. I'm adding this vector 2 comma 4 to it, so from here I'd go right 2 and up 4, so I'd go here. Or visually you could just say, heads to tails, so I would go right there. So I would end up at a corresponding point over there. So when I define my set, s, as the set of all points where I just multiply v times the scalar, I got this thing that went through the origin. But now let me define another set. Let me define a set l, maybe l for line, that's equal to the set of all of vectors where the vector x, I could do it bold or I'll just draw an arrow on it, plus some scalar-- I could use c, but let me use t, because I'm going to call this a parametrization of the line-- so plus some scalar, t times my vector v such that t could be any member of the real numbers. So what is this going to be? This is going to be this blue line. If I were to draw all of these vectors in standard position, I'm going to get my blue line. For example, if I do minus 2, this is minus 2, times my vector v, I get here. Then if I add x, I go there. So this vector right here that has its endpoint right there-- its endpoint sits on that line. I can do that with anything. If I take this vector, this is some scalar times my vector v, and I add x to it, I end up with this vector, whose endpoint, if I view it as a position vector, it's endpoint dictates some coordinate in the xy plane. So it will [UNINTELLIGIBLE] that point. So I can get to any of these vectors. This is a set of vectors right here, and all of these vectors are going to point-- they're essentially going to point to something-- when I draw them in standard form, if I draw them in standard form-- they're going to point to a point on that blue line. Now you might say, hey Sal, this was a really obtuse way of defining a line. I mean we do it in Algebra 1, where we just say, hey you know, y is equal to mx plus b. And we figure out the slope by figuring out the difference of two points, and then we do a little substitution. And this is stuff you learned in seventh or eighth grade. This was really straightforward. Why am I defining this obtuse set here and making you think in terms of sets and vectors and adding vectors? And the reason is, is because this is very general. This worked well in R2. So in R2, this was great. I mean, we just have to worry about x's and y's. But what about the situation, I mean notice, in your algebra class, your teacher never really told you much, at least in the ones I took, about how do you present lines in three dimensions? Maybe some classes go there, but they definitely didn't tell you how do you represent lines in four dimensions, or a hundred dimensions. And that's what this is going to do for us. Right here, I defined x and v as vectors in R2. They're two-dimensional vectors, but we can extend it to an arbitrary number of dimensions. So just to kind of hit the point home, let's do one more example in R2, where, it's kind of the classic algebra problem where you need to find the equation for the line. But here, we're going to call it the set definition for the line. Let's say we have two vectors. Let's say we have the vector a, which I'll define as-- let me just says it's 2, 1. So if I were draw it in standard form, it's 2, 1. That's my vector a, right there. And let's say I have vector b, let me define vector b. I'm going to define it as, I don't know, let me define it as 0, 3. So my vector b, 0-- I don't move to the right at all and I go up. So my vector b will look like that. Now I'm going to say that these are position vectors, that we draw them in standard form. When you draw them in standard form, their endpoints represent some position. So you can almost view these as coordinate points in R2. This is R2. All of these coordinate axes I draw are going be R2. Now what if I asked you, give me a parametrization of the line that goes through these two points. So essentially, I want the equation-- if you're thinking in Algebra 1 terms-- I want the equation for the line that goes through these two points. So the classic way, you would have figured out the slope and all of that, and then you would have substituted back in. But instead, what we can do is, we can say, hey look, this line that goes through both of those points-- you could almost say that both of those vectors lie on-- I guess that's a better-- Both of these vectors lie on this line. Now, what vector can be represented by that line? Or even better, what vector, if I take any arbitrary scalar-- can represent any other vector on that line? Now let me do it this way. What if I were to take-- so this is vector b here-- what if I were to take b minus a? We learned in, I think it was the previous video, that b minus a, you'll get this vector right here. You'll get the difference in the two vectors. This is the vector b minus the vector a. And you just think about it. What do I have to add to a to get to b? I have to add b minus a. So if I can get the vector b minus a-- right, we know how We just subtract the vectors, and then multiply it by any scalar, then we're going to get any point along that line. We have to be careful. So what happens if we take t, so some scalar, times our vector, times the vectors b minus a? What will we get then? So b minus a looks like that. But if we were to draw it in standard form-- remember, in standard form b minus a would look something like this. It would start at 0, it would be parallel to this, and then from 0 we would draw its endpoint. So if we just multiplied some scalar times b minus a, we would actually just get points or vectors that lie on this line. Vectors that lie on that line right there. Now, that's not what we set out to do. We wanted to figure out an equation, or parametrization, if you will, of this line, or this set. Let's call this set l. So we want to know what that set is equal to. So in order to get there, we have to start with this, which is this line here, and we have to shift it. And we could shift it either by shifting it straight up, we could add vector b to it. So we could take this line right here, and add vector b to it. And so any point on here would have its corresponding point there. So when you add vector b, it essentially shifts it up. That would work. So we could, say, we could add vector b to it. And now all of these points for any arbitrary-- t is a member of the real numbers, will lie on this green line. Or the other option we could have done is we could have added vector a. Vector a would have taken any arbitrary point here and shifted it that way. You would be adding vector a to it. But either way, you're going to get to the green line that we cared about, so you could have also defined it as the set of vector a plus this line, essentially, t times vector b minus a, where t is a member of the reals. So my definition of my line could be either of these things. The definition of my line could be this set, or it could be this set. And all of this seems all very abstract, but when you actually deal with the numbers, it actually becomes very simple. It becomes arguably simpler than what we did in Algebra 1. So this l, for these particular case of a and b, let's figure it out. My line is equal to-- let me just use the first example. It's vector b, so it's the vector 0, 3 plus t, times the vector b minus a. Well what's b minus a? 0 minus 2 is minus 2, 3, minus 1 is 2, for t is a member of the reals. Now, if this still seems kind of like a convoluted set definition for you, I could write it in terms that you might recognize better. If we want to plot points, if we call this the y-axis, and we call this the x-axis, and if we call this the x-coordinate, or maybe more properly that the x-coordinate and call this the y-coordinate, then we can set This actually is the x-slope. This is the x-coordinate, that's the y-coordinate. Or actually, even better, whatever-- actually, let me be very careful there. This is always going to end up becoming some vector, l1, l2. This is a set of vectors, and any member of this set is going to look something like this. So this could be li. So, this is the x-coordinate, and this is the y-coordinate. And just to get this in a form that you recognize, so we're saying that l is the set of this vector x plus t times this vector b minus a here. If we wanted to write it in kind of a parametric form, we can say, since this is what determines our x-coordinate, we would say that x is equal to 0 plus t times minus 2, or minus 2 times t. And then we can say that y, since this is what determines our y-coordinate, y is equal to 3 plus t times 2 plus 2t. So we could have rewritten that first equation as just x is equal to minus 2t, and y is equal to 2t plus 3. So if you watch the videos on parametric equations, this is just a traditional parametric definition of this line right there. Now, you might have still viewed this as, Sal, this was a waste of time, this was convoluted. You have to define these sets and all that. But now I'm going to show you something that you probably-- well, unless you have done this before, but I guess that's true of anything. But you probably haven't seen in your traditional algebra class. Let's say I have two points, and now I'm going to deal in three dimensions. So let's say I have one vector. I'll just call it point 1, because these are position vectors. We'll just call it position 1. This is in three dimensions. Just make up some numbers, negative 1, 2, 7. Let's say I have Point 2. Once again, this is in three dimensions, so you have to specify three coordinates. This could be the x, the y, and the z coordinate. Point 2, I don't know. Let's make it 0, 3, and 4. Now, what if I wanted to find the equation of the line that passes through these two points in R3? So this is in R3. Well, I just said that the equation of this line-- so I'll just call that, or the set of this line, let me just call this l. It's going to be equal to-- we could just pick one of these guys, it could be P1, the vector P1, these are all vectors, be careful here. The vector P1 plus some random parameter, t, this t could be time, like you learn when you first learn parametric equations, times the difference of the two vectors, times P1, and it doesn't matter what order you take it. So that's a nice thing too. P1 minus P2. It could be P2 minus P1-- because this can take on any positive or negative value-- where t is a member of the real numbers. So let's apply it to these numbers. Let's apply it right here. What is P1 minus P2? P1 minus P2 is equal to-- let me get some space here. P1 minus P2 is equal, minus 1 minus 0 is minus 1. 2 minus 3 is minus 1. 7 minus 4 is 3. So that thing is that vector. And so, our line can be described as a set of vectors, that if you were to plot it in standard position, it would be this set of position vectors. It would be P1, it would be-- let me do that in green-- it would be minus 1, 2, 7. I could've put P2 there, just as easily-- plus t times minus 1, minus 1, 3, where, or such that, t is a member of the real numbers. Now, this also might not be satisfying for you. You're like, gee, how do I plot this in three dimensions? Where's my x, y's, and z's? And if you want to care about x, y's, and z's, let's say that this is the z-axis. This is the x-axis, and let's say the y-axis. It kind of goes into our board like this, so the y-axis comes out like that. So what you can do, and actually I probably won't graph, so the determinate for the x-coordinate, just our convention, is going to be this term right here. So we can write that x-- let me write that down. So that term is going to determine our x-coordinate. So we can write that x is equal to minus 1-- be careful with the colors-- minus 1, plus minus 1 times t. That's our x-coordinate. Now, our y-coordinate is going to be determined by this part of our vector addition because these are the y-coordinates. So we can say the y-coordinate is equal to-- I'll just write it like this-- 2 plus minus 1 times t. And then finally, our z-coordinate is determined by that there, the t shows up because t times 3-- or I could just put this t into all of this. So that the z-coordinate is equal to 7 plus t times 3, or I could say plus 3t. And just like that, we have three parametric equations. And when we did it in R2, I did a parametric equation, but we learned in Algebra 1, you can just have a regular y in terms x. You don't have to have a parametric equation. But when you're dealing in R3, the only way to define a line is to have a parametric equation. If you have just an equation with x's, y's, and z's, if I just have x plus y plus z is equal to some number, this is not a line. And we'll talk more about this in R3. This is a plane. The only way to define a line or a curve in three dimensions, if I wanted to describe the path of a fly in three dimensions, it has to be a parametric equation. Or if I shoot a bullet in three dimensions and it goes in a straight line, it has to be a parametric equation. So these-- I guess you could call it-- these are the equations of a line in three dimensions. So hopefully you found that interesting. And I think this will be the first video where you have an appreciation that linear algebra can solve problems or address issues that you never saw before. And there's no reason why we have to just stop at three, three coordinates, right here. We could have done this with fifty dimensions. We could have defined a line in fifty dimensions-- or the set of vectors that define a line, that two points sit on, in fifty dimensions-- which is very hard to visualize, but we can actually deal with it mathematically." + }, + { + "Q": "At 4:00, Sal says \"should.\" Can some cancer cells be so mutated that they do not produce MHC I at all?", + "A": "Yes. Sometimes cancer cells are so mutated or too close to your actual cells that if your cyto cells attack them, they will also attack your healthy cells. That s why to be safe they don t attack them. They also don t recognize them so they don t kill them. That s why all people need help to fight off cancers since your body can t kill most cancers.", + "video_name": "YdBXHm3edL8", + "transcript": "Voiceover: Everything we've discussed so far involved recognizing and tagging or engulfing shady things that were found outside of cells. We've seen things like a B cell. A B cell has its membrane bound antibodies. Maybe one of these might recognize something shady out in the outside of the cell and of course this part over here as we know they all have a variable portion right over here. This is specific to this shady thing then this will be engulfed and then parts of it will be attached to an MHC two complex. Let me do that in a different color. An MHC two complex and then that will go to the surface, that will go to the membrane of the cell to present itself. That's an MHC two complex. It has little bit of the little piece of this shady thing out here we call this little shady piece, this is an antigen presenting cell here. We've seen as especially if this is a B cell then a helper T cell that it also has a variable portion that corresponds to this specific antigen. This would be helper T cell right over here. This is a helper, T helper cell. When it recognizes then it will start dividing into memory helper T cells and effector helper T cells. The effector helper T cells essentially ring the alarm bells and start kind of accelerating the B cell replication or I guess you could say that the B cell activation. The theory is that this is kind of a double handshake process. Once again this is what's occurring outside of cells. When we found stuff outside of cells, we engulf them and then we presented them on MHC two complexes. Now you're probably thinking well, I mean that's the outside of cells but there's MHC two, there's this helper T cells but we've also referred to cytotoxic T cells. What do those do? We've also, if there's MHC two, there's probably an MHC one complex. What does the MHC one complex do? We can recognize shady things that are happening outside of cells but don't shady things sometimes happen inside cells and how does our immune system respond to that? Actually as you can imagine all of those things will be answered in the rest of this video. Let's think about what happens when shady things start to happen inside the cell. For example it might not even be due to a virus or due to some type of bacteria could be the cell itself is gone awry. Let's say that this right over here is a cancer cell. It's had some mutations. It's starting to multiply like crazy. This is a cancer cell and a cancer cell because it had mutations are going to produce, it's going to produce some weird proteins. These cancer cells are going to produce some weird proteins. Every cell with a nucleus in your body and that's pretty much every cell except for red blood cells has MHC one complexes. The whole point of the MHC one complex is to bind two shady things that are produced inside of the cell, and then present them to the membrane. Even a malfunctioning cancer cell should be doing this. This MHC one complex is bound to this strange proteins that are produced by the mutations inside of the nucleus and then it can present them. You could imagine what the appropriate immune response should be. These cancer cells should be killed and actually let me label this properly. That was MHC two, you're presenting an antigen that was found, those initially found outside of the cells engulfing and taken out. MHC one, it's binding to shady things inside the cell and then presenting it out. This thing should be killed. Now as you can imagine, what's going to kill it? Well that's where the cytotoxic T cell comes into the picture. The cytotoxic T cell is going to have, that's a receptor right there. It will have a variable portion that's specific to this type of an antigen and so it will bind there. Once it does that, it says, oh boy, there's all this shady stuff here, this shady proteins that are being produce. This cell and all the other ones like it need to be killed. The cytotoxic T cell will begin to replicate once again like other types of immune cells is going to replicate into the memory cells just in case this type of thing shows up 10 years in the future and also the effector themselves. This is memory and also effector cells, effector cytotoxic T cells. As we always know the effector version is the thing that actual does something, it starts to actually affect things. What it is going to affect, it is going to start binding two things that are presenting the same antigen as part of their MR on top of their MHC one complex. This character right over here, so it's presenting that same antigen on this MHC one complex. Remember, the variable portions need to match up. Let's say that this is an effector cytotoxic T cell and actually let me draw in a little bit different. Let me draw it like this. This is effector cytotoxic T cell. Its receptor, its variable portion is the one that's compatible with this antigen that's being presented right over here. Let me just label this again. This is the MHC one complex. This is an effector cytotoxic T cell. We'll put the C there for cytotoxic and what it does is essentially kind of latches on to the cell that needs to die and it does it, not only have this receptor interfacing with the MHC one complex but actually has a whole series of proteins and I'm not drawing this to scale really. This would be much smaller or relative to the scale cell. It essentially latches on between the two and I'm not going to go in detail but essentially forms what you can call an immuno synapse which is kind of where the two things are interacting with each other. When it identifies this, it says, okay, I need to kill this thing or essentially I need to make this thing kill itself. It starts releasing all of these molecules. It can release molecules like preference so to release this preference which will essentially cause gaps or holes to form in the membrane of the cell that needs to die and it could release other things like granzymes that can go in and essentially cause this thing to kill itself. The whole point of this video is to appreciate I guess what we haven't talked about yet. We had already talked about what happens when you identify shady things outside of the cells and then how you can kind of bring them in and then present them and then use that to further activate the immune response. Now we're talking about identifying shady things inside the cells. Those get presented by MHC one complexes and then the cytotoxic T cells recognize them and then force the cell to kill themselves. This wouldn't just be cancer cells, this could also apply to a cell that has already been affected by a virus. For example a cell like this all ready, so that's its nucleus. It's already been infected by some virus so the virus has hijacked the cell's replication machinery in order to replicate itself. The proper immune response is hey look, I'm a virus making machine. I should kill myself. It will wreck some of the antigens that are being produced inside by the viruses. They're going to bind to MHC one complexes. Pieces of the virus are going to bind to MHC one complexes and then they're going to be presented on the surface. Let me do it this way, presented on the surface. This exact process can happen again." + }, + { + "Q": "using the squeeze theorem how can I evaluate: lim x-> infinity e^-8x cos x\n\ncos x will always be bound between -1 and 1 so do I set it up so do I set it up where\n\n-e^-8x (x - y)^3 + 3xy(x-y) = x^3 - y^3 => x^3 - y^3 = (x - y) [(x - y)^2 + 3xy] => x^3 - y^3 = (x - y)(x^2 + y^2 - 2xy + 3xy) => x^3 - y^3 = (x - y)(x^2 + xy + y^2 )", + "video_name": "rU222pVq520", + "transcript": "Let's try to find the limit as x approaches 1 of x to the third minus 1 over x squared minus 1. And at first when you just try to substitute x equals 1, you get 0/0 1 minus 1 over 1 minus 1. So that doesn't help us. So let's see if we can try to simplify this in some way. So you might immediately recognize-- so let's rewrite this expression right over here so it's x to the third minus 1 over x squared minus 1. This on the bottom immediately jumps out as a difference of squares. So we know on the bottom that this could be factored as x minus 1 times x plus 1. And so if somehow this thing on the top also has an x minus 1 as a factor, then that x minus 1 will cancel with this, and then we're not going to have an issue of dividing by 0. The reason why I care about the x minus 1 term is that this is what's making our denominator equal 0. When you say x equals 1, you have 1 minus 1 times 1 plus 1. So 0 times 2, it's this 0 that's making our denominator 0. So if we can have an x minus 1 up here, then we can cancel these out for any x not equal to 1. And then we might have a much simpler thing to find the limit of. So let's think about whether x to the third minus 1 is the product of x minus 1 and something else. And to do that we can do a little bit of algebraic long division. Some of you guys might already recognize a pattern here, but we'll try to do-- well, let's divide x minus 1 into it to see whether it divides evenly into x to the third minus 1. So x minus 1-- we just look at the highest degree term-- x goes into x to the third x squared times. Goes x squared times. Actually, let me do it this way so that way we can keep track of the place. So this would be x-- this would be the second degree place, first degree place, and this would be the constant. So x to the third minus 1. x goes into x to the third x squared times. x squared times x is x to the third. x squared times negative 1 is minus x squared. And now we're going to want to subtract this. So we are then left with x squared. x goes into x squared x times plus x. x times x is x squared. x times minus 1 is minus x. And once again we're going to subtract this. We'll swap the signs, negative and positive. And so these cancel out, and we're left with x. And then we bring down a minus 1. x minus 1 goes into x minus 1 exactly one time. 1 times x minus 1 is x minus 1. And then you subtract, and then you have no remainder. So this numerator right over here can be factored as x minus 1 times x squared plus x plus 1. And so we can say that this is the same exact thing. We can have these cancel out if we assume x does not equal 1. So that is equal to x squared plus x plus 1 over x plus 1, for x does not equal 1. And that's completely fine, because we're not evaluating x equals 1. We're evaluating as x approaches 1. So this is going to be the same thing as the limit as x approaches 1 of x squared plus x plus 1 over x plus 1. And now this is much easier to find. You could literally just say, well, what happens as we get right to x equals 1? Then you have 1 squared, which is 1 plus 1 plus 1, which is 3, over 1 plus 1, which is 2. So we get that equaling 3/2." + }, + { + "Q": "i have a question from my exam.\n\"In what way will the temperature of water at the bottom of a waterfall be different from the temperature at the top? Give a reason for your answer.\"\n\nIs water at the bottom hotter?\nI think that as water falls down- Potential.E. is converted to Kinetic E. and some of PE is also converted to heat.\nAm I right?", + "A": "That s correct, but you could also argue that evaporation takes place during the fall and the splashing at the bottom. The evaporation would have a cooling effect. Can t tell how it balances out.", + "video_name": "kw_4Loo1HR4", + "transcript": "Welcome back. At the end of the last video, I left you with a bit of a question. We had a situation where we had a 1 kilogram object. This is the 1 kilogram object, which I've drawn neater in this video. That is 1 kilogram. And we're on earth, and I need to mention that because gravity is different from planet to planet. But as I mentioned, I'm holding it. Let's say I'm holding it 10 meters above the ground. So this distance or this height is 10 meters. And we're assuming the acceleration of gravity, which we also write as just g, let's assume it's just 10 meters per second squared just for the simplicity of the math instead of the 9.8. So what we learned in the last video is that the potential energy in this situation, the potential energy, which equals m times g times h is equal to the mass is 1 kilogram times the acceleration of gravity, which is 10 meters per second squared. I'm not going to write the units down just to save space, although you should do this when you do it on your test. And then the height is 10 meters. And the units, if you work them all out, it's in newton meters or joules and so it's equal to 100 joules. That's the potential energy when I'm holding it up there. And I asked you, well when I let go, what happens? Well the block obviously will start falling. And not only falling, it will start accelerating to the ground at 10 meters per second squared roughly. And right before it hits the ground-- let me draw that in brown for ground-- right before the object hits the ground or actually right when it hits the ground, what will be the potential energy of the object? Well it has no height, right? Potential energy is mgh. The mass and the acceleration of gravity stay the same, but the height is 0. So they're all multiplied by each other. So down here, the potential energy is going to be equal to 0. And I told you in the last video that we have the law of conservation of energy. That energy is conserved. It cannot be created or destroyed. It can just be converted from one form to another. But I'm just showing you, this object had 100 joules of energy or, in this case, gravitational potential energy. And down here, it has no energy. Or at least it has no gravitational potential energy, and that's the key. That gravitational potential energy was converted into something else. And that something else it was converted into is kinetic energy. And in this case, since it has no potential energy, all of that previous potential energy, all of this 100 joules that it has up here is now going to be converted into kinetic energy. And we can use that information to figure out its velocity right before it hits the ground. So how do we do that? Well what's the formula for kinetic energy? And we solved it two videos ago, and hopefully it shouldn't be too much of a mystery to you. It's something good to memorize, but it's also good to know how we got it and go back two videos if you forgot. So first we know that all the potential energy was converted into kinetic energy. We had 100 joules of potential energy, so we're still going to have 100 joules, but now all of it's going to be kinetic energy. And kinetic energy is 1/2 mv squared. So we know that 1/2 mv squared, or the kinetic energy, is now going to equal 100 joules. What's the mass? The mass is 1. And we can solve for v now. 1/2 v squared equals 100 joules, and v squared is equal to 200. And then we get v is equal to square root of 200, which is something over 14. We can get the exact number. Let's see, 200 square root, 14.1 roughly. The velocity is going to be 14.1 meters per second squared downwards. Right before the object touches the ground. Right before it touches the ground. And you might say, well Sal that's nice and everything. We learned a little bit about energy. I could have solved that or hopefully you could have solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it. And then this is the bottom. This is the ground down here. This is the ground. So what's going to happen this time? I'm still 10 meters in the air, so let me draw that. That's still 10 meters. I should switch colors just so not everything is ice. So that's still 10 meters, but instead of the object going straight down now, it's going to go down here and then start It's going to go sliding along this hill. And then at this point it's going to be going really fast in the horizontal direction. And right now we don't know how fast. And just using our kinematics formula, this would have been a really tough formula. This would have been difficult. I mean you could have attempted it and it actually would have taken calculus because the angle of the slope changes continuously. We don't even know the formula for the angle of the slope. You would have had to break it out into vectors. You would have to do all sorts of complicated things. This would have been a nearly impossible problem. But using energy, we can actually figure out what the velocity of this object is at this point. And we use the same idea. Here we have 100 joules of potential energy. We just figured that out. Down here, what's the height above the ground? Well the height is 0. So all the potential energy has disappeared. And just like in the previous situation, all of the potential energy is now converted into kinetic energy. And so what is that kinetic energy going to equal? It's going to be equal to the initial potential energy. So here the kinetic energy is equal to 100 joules. And that equals 1/2 mv squared, just like we just solved. And if you solve for v, the mass is 1 kilogram. So the velocity in the horizontal direction will be, if you solve for it, 14.1 meters per second. Instead of going straight down, now it's going to be going in the horizontal to the right. And the reason why I said it was ice is because I wanted this to be frictionless and I didn't want any energy lost to heat or anything like that. And you might say OK Sal, that's kind of interesting. And you kind of got the same number for the velocity than if I just dropped the object straight down. And that's interesting. But what else can this do for me? And this is where it's really cool. Not only can I figure out the velocity when all of the potential energy has disappeared, but I can figure out the velocity of any point-- and this is fascinating-- along this slide. So let's say when the box is sliding down here, so let's say the box is at this point. It changes colors too as it falls. So this is the 1 kilogram box, right? It falls and it slides down here. And let's say at this point it's height above the ground is 5 meters. So what's its potential energy here? So let's just write something. All of the energy is conserved, right? So the initial potential energy plus the initial kinetic energy is equal to the final potential energy plus the final kinetic energy. I'm just saying energy is conserved here. Up here, what's the initial total energy in the system? Well the potential energy is 100 and the kinetic energy is 0 because it's stationary. I haven't dropped it. I haven't let go of it yet. It's just stationary. So the initial energy is going to be equal to 100 joules. That's cause this is 0 and this is 100. So the initial energy is 100 joules. At this point right here, what's the potential energy? Well we're 5 meters up, so mass times Mass is 1, times gravity, 10 meters per second squared. Times height, times 5. So it's 50 joules. That's our potential energy at this point. And then we must have some kinetic energy with the velocity going roughly in that direction. Plus our kinetic energy at this point. And we know that no energy was destroyed. It's just converted. So we know the total energy still has to be 100 joules. So essentially what happened, and if we solve for this-- it's very easy, subtract 50 from both sides-- we know that the kinetic energy is now also going to be equal to 50 joules. Halfway down, essentially half of the potential energy got converted to kinetic energy. And we can use this information that the kinetic energy is 50 joules to figure out the velocity at this point. 1/2 mv squared is equal to 50. The mass is 1. Multiply both sides by 2. You get v squared is equal to 100. The velocity is 10 meters per second along this crazy, icy slide. And that is something that I would have challenged you to solve using traditional kinematics formulas, especially considering that we don't know really much about the surface of this slide. And even if we did, that would have been a million times harder than just using the law of conservation of energy and realizing that at this point, half the potential energy is now kinetic energy and it's going along the direction of the slide. I will see you in the next video." + }, + { + "Q": "What kind of change occurs when one scratches a steel surface. Is it chemical or physical? I understand that rust/tarnishing a steel surface is a chemical change, I would imagine this holds true for a scratch, correct?", + "A": "No, if you scratch something, it doesn t change chemically. The part that you scratch off is still steel. Anything something breaks physically, it is a physical change.", + "video_name": "pKvo0XWZtjo", + "transcript": "I think we're all reasonably familiar with the three states of matter in our everyday world. At very high temperatures you get a fourth. But the three ones that we normally deal with are, things could be a solid, a liquid, or it could be a gas. And we have this general notion, and I think water is the example that always comes to at least my mind. Is that solid happens when things are colder, relatively colder. And then as you warm up, you go into a liquid state. And as your warm up even more you go into a gaseous state. So you go from colder to hotter. And in the case of water, when you're a solid, you're ice. When you're a liquid, some people would call ice water, but let's call it liquid water. I think we know what that is. And then when it's in the gas state, you're essentially vapor or steam. So let's think a little bit about what, at least in the case of water, and the analogy will extend to other types of molecules. But what is it about water that makes it solid, and when it's colder, what allows it to be liquid. And I'll be frank, liquids are kind of fascinating because you can never nail them down, I guess is the best way to view them. Or a gas. So let's just draw a water molecule. So you have oxygen there. You have some bonds to hydrogen. And then you have two extra pairs of valence electrons in the oxygen. And a couple of videos ago, we said oxygen is a lot more electronegative than the hydrogen. It likes to hog the electrons. So even though this shows that they're sharing electrons here and here. At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules. Let me draw some more molecules. When we talk about the whole state of the whole matter, we actually think about how the molecules are interacting with Not just how the atoms are interacting with each other within a molecule. I just drew one oxygen, let me copy and paste that. But I could do multiple oxygens. And let's say that that hydrogen is going to want to be near this oxygen. Because this has partial negative charge, this has a partial positive charge. And then I could do another one right there. And then maybe we'll have, and just to make the point clear, you have two hydrogens here, maybe an oxygen wants to hang out there. So maybe you have an oxygen that wants to be here because it's got its partial negative here. And it's connected to two hydrogens right there that have their partial positives. But you can kind of see a lattice structure. Let me draw these bonds, these polar bonds that start forming between the particles. These bonds, they're called polar bonds because the molecules themselves are polar. And you can see it forms this lattice structure. And if each of these molecules don't have a lot of kinetic energy. Or we could say the average kinetic energy of this matter is fairly low. And what do we know is average kinetic energy? Well, that's temperature. Then this lattice structure will be solid. These molecules will not move relative to each other. I could draw a gazillion more, but I think you get the point that we're forming this kind of fixed structure. And while we're in the solid state, as we add kinetic energy, as we add heat, what it does to molecules is, it just makes them vibrate around a little bit. If I was a cartoonist, they way you'd draw a vibration is to put quotation marks there. That's not very scientific. But they would vibrate around, they would buzz around a little bit. I'm drawing arrows to show that they are vibrating. It doesn't have to be just left-right it could be up-down. But as you add more and more heat in a solid, these molecules are going to keep their structure. So they're not going to move around relative to each other. But they will convert that heat, and heat is just a form of energy, into kinetic energy which is expressed as the vibration of these molecules. Now, if you make these molecules start to vibrate enough, and if you put enough kinetic energy into these molecules, what do you think is going to happen? Well this guy is vibrating pretty hard, and he's vibrating harder and harder as you add more and more heat. This guy is doing the same thing. At some point, these polar bonds that they have to each other are going to start not being strong enough to contain the vibrations. And once that happens, the molecules-- let me draw a couple more. Once that happens, the molecules are going to start moving past each other. So now all of a sudden, the molecule will start shifting. But they're still attracted. Maybe this side is moving here, that's moving there. You have other molecules moving around that way. But they're still attracted to each other. Even though we've gotten the kinetic energy to the point that the vibrations can kind of break the bonds between the polar sides of the molecules. Our vibration, or our kinetic energy for each molecule, still isn't strong enough to completely separate them. They're starting to slide past each other. And this is essentially what happens when you're in a liquid state. You have a lot of atoms that want be touching each other but they're sliding. They have enough kinetic energy to slide past each other and break that solid lattice structure here. And then if you add even more kinetic energy, even more heat, at this point it's a solution now. They're not even going to be able to stay together. They're not going to be able to stay near each other. If you add enough kinetic energy they're going to start looking like this. They're going to completely separate and then kind of bounce around independently. Especially independently if they're an ideal gas. But in general, in gases, they're no longer touching They might bump into each other. But they have so much kinetic energy on their own that they're all doing their own thing and they're not touching. I think that makes intuitive sense if you just think about what a gas is. For example, it's hard to see a gas. Why is it hard to see a gas? Because the molecules are much further apart. So they're not acting on the light in the way that a liquid or a solid would. And if we keep making that extended further, a solid-- well, I probably shouldn't use the example with ice. Because ice or water is one of the few situations where the solid is less dense than the liquid. That's why ice floats. And that's why icebergs don't just all fall to the bottom of the ocean. And ponds don't completely freeze solid. But you can imagine that, because a liquid is in most cases other than water, less dense. That's another reason why you can see through it a little Or it's not diffracting-- well I won't go into that too much, than maybe even a solid. But the gas is the most obvious. And it is true with water. The liquid form is definitely more dense than the gas form. In the gas form, the molecules are going to jump around, not touch each other. And because of that, more light can get through the substance. Now the question is, how do we measure the amount of heat that it takes to do this to water? And to explain that, I'll actually draw a phase change diagram. Which is a fancy way of describing something fairly straightforward. Let me say that this is the amount of heat I'm adding. And this is the temperature. We'll talk about the states of matter in a second. So heat is often denoted by q. Sometimes people will talk about change in heat. They'll use H, lowercase and uppercase H. They'll put a delta in front of the H. Delta just means change in. And sometimes you'll hear the word enthalpy. Let me write that. Because I used to say what is enthalpy? It sounds like empathy, but it's quite a different concept. At least, as far as my neural connections could make it. But enthalpy is closely related to heat. It's heat content. For our purposes, when you hear someone say change in enthalpy, you should really just be thinking of change in heat. I think this word was really just introduced to confuse chemistry students and introduce a non-intuitive word into their vocabulary. The best way to think about it is heat content. Change in enthalpy is really just change in heat. And just remember, all of these things, whether we're talking about heat, kinetic energy, potential energy, enthalpy. You'll hear them in different contexts, and you're like, I thought I should be using heat and they're talking about enthalpy. These are all forms of energy. And these are all measured in joules. And they might be measured in other ways, but the traditional way is in joules. And energy is the ability to do work. And what's the unit for work? Well, it's joules. Force times distance. But anyway, that's a side-note. But it's good to know this word enthalpy. Especially in a chemistry context, because it's used all the time and it can be very confusing and non-intuitive. Because you're like, I don't know what enthalpy is in my everyday life. Just think of it as heat contact, because that's really But anyway, on this axis, I have heat. So this is when I have very little heat and I'm increasing my heat. And this is temperature. Now let's say at low temperatures I'm here and as I add heat my temperature will go up. Temperature is average kinetic energy. Let's say I'm in the solid state here. And I'll do the solid state in purple. No I already was using purple. I'll use magenta. So as I add heat, my temperature will go up. Heat is a form of energy. And when I add it to these molecules, as I did in this example, what did it do? It made them vibrate more. Or it made them have higher kinetic energy, or higher average kinetic engery, and that's what temperature is a measure of; average kinetic energy. So as I add heat in the solid phase, my average kinetic energy will go up. And let me write this down. This is in the solid phase, or the solid state of matter. Now something very interesting happens. Let's say this is water. So what happens at zero degrees? Which is also 273.15 Kelvin. Let's say that's that line. What happens to a solid? Well, it turns into a liquid. Ice melts. Not all solids, we're talking in particular about water, about H2O. So this is ice in our example. All solids aren't ice. Although, you could think of a rock as solid magma. Because that's what it is. I could take that analogy a bunch of different ways. But the interesting thing that happens at zero degrees. Depending on what direction you're going, either the freezing point of water or the melting point of ice, something interesting happens. As I add more heat, the temperature does not to go up. As I add more heat, the temperature does not go up for a little period. Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or which water will boil. But something happens. And they're really getting kinetically active. But just like when you went from solid to liquid, there's a certain amount of energy that you have to contribute to the system. And actually, it's a good amount at this point. Where the water is turning into vapor, but it's not getting any hotter. So we have to keep adding heat, but notice that the temperature didn't go up. We'll talk about it in a second what was happening then. And then finally, after that point, we're completely vaporized, or we're completely steam. Then we can start getting hot, the steam can then get hotter as we add more and more heat to the system. So the interesting question, I think it's intuitive, that as you add heat here, our temperature is going to go up. But the interesting thing is, what was going on here? We were adding heat. So over here we were turning our heat into kinetic energy. Temperature is average kinetic energy. But over here, what was our heat doing? Well, our heat was was not adding kinetic energy to the system. The temperature was not increasing. But the ice was going from ice to water. So what was happening at that state, is that the kinetic energy, the heat, was being used to essentially break these bonds. And essentially bring the molecules into a higher energy state. So you're saying, Sal, what does that mean, higher energy state? Well, if there wasn't all of this heat and all this kinetic energy, these molecules want to be very close to each other. For example, I want to be close to the surface of the earth. When you put me in a plane you have put me in a higher energy state. I have a lot more potential energy. I have the potential to fall towards the earth. Likewise, when you move these molecules apart, and you go from a solid to a liquid, they want to fall towards each other. But because they have so much kinetic energy, they never quite are able to do it. But their energy goes up. Their potential energy is higher because they want to fall towards each other. By falling towards each other, in theory, they could do some work. So what's happening here is, when we're contributing heat-- and this amount of heat we're contributing, it's called the heat of fusion. Because it's the same amount of heat regardless how much direction we go in. When we go from solid to liquid, you view it as the heat of melting. It's the head that you need to put in to melt the ice into liquid. When you're going in this direction, it's the heat you have to take out of the zero degree water to turn it into ice. So you're taking that potential energy and you're bringing the molecules closer and closer to each other. So the way to think about it is, right here this heat is being converted to kinetic energy. Then, when we're at this phase change from solid to liquid, that heat is being used to add potential energy into the system. To pull the molecules apart, to give them more potential energy. If you pull me apart from the earth, you're giving me potential energy. Because gravity wants to pull me back to the earth. And I could do work when I'm falling back to the earth. A waterfall does work. It can move a turbine. You could have a bunch of falling Sals move a turbine as well. And then, once you are fully a liquid, then you just become a warmer and warmer liquid. Now the heat is, once again, being used for kinetic energy. You're making the water molecules move past each other faster, and faster, and faster. To some point where they want to completely disassociate from each other. They want to not even slide past each other, just completely jump away from each other. And that's right here. This is the heat of vaporization. And the same idea is happening. Before we were sliding next to each other, now we're pulling apart altogether. So they could definitely fall closer together. And then once we've added this much heat, now we're just heating up the steam. We're just heating up the gaseous water. And it's just getting hotter and hotter and hotter. But the interesting thing there, and I mean at least the interesting thing to me when I first learned this, whenever I think of zero degrees water I'll say, oh it must be ice. But that's not necessarily the case. If you start with water and you make it colder and colder and colder to zero degrees, you're essentially taking heat out of the water. You can have zero degree water and it hasn't turned into ice yet. And likewise, you could have 100 degree water that hasn't turned into steam yeat. You have to add more energy. You can also have 100 degree steam. You can also have zero degree water. Anyway, hopefully that gives you a little bit of intuition of what the different states of matter are. And in the next problem, we'll talk about how much heat exactly it does take to move along this line. And maybe we can solve some problems on how much ice we might need to make our drink cool." + }, + { + "Q": "The simplest way I can see to solve this problem is to take the first valve: 15 subtract it by the second valve 15-9=6. Now you have the common difference, then you do this:\n15-6(100-1) then you compute that should give you get the answer.", + "A": "PEMDAS rules always apply. You can t do the subtraction 1st. 1) Do the parentheses 1st: 100-1 = 99 2) Muliply: 99 * (-6) = -594 3) Then subtract: 15 - 594 = - 579", + "video_name": "JtsyP0tnVRY", + "transcript": "We are asked, what is the value of the 100th term in this sequence? And the first term is 15, then 9, then 3, then negative 3. So let's write it like this, in a table. So if we have the term, just so we have things straight, and then we have the value. and then we have the value of the term. I'll do a nice little table here. So our first term we saw is 15. Our second term is 9. Our third term is 3. I'm just really copying this down, but I'm making sure we associate it with the right term. And then our fourth term is negative 3. And they want us to figure out what the 100th term of this sequence is going to be. So let's see what's happening here, if we can discern some type of pattern. So when we went from the first term to the second term, what happened? 15 to 9. Looks like we went down by 6. It's always good to think about just how much the numbers changed by. That's always the simplest type of pattern. So we went down by 6, we subtracted 6. Then to go from 9 to 3, well, we subtracted 6 again. And then to go from 3 to negative 3, well, we subtracted 6 again. So it looks like, every term, you subtract 6. So the second term is going to be 6 less than the first term. The third term is going to be 12 from the first term, or negative 6 subtracted twice. So in the third term, you subtract negative 6 twice. In the fourth term, you subtract negative 6 three times. So whatever term you're looking at, you subtract negative 6 one less than that many times. Let me write this down just so-- Notice when your first term, you have 15, and you don't subtract negative 6 at all. Or you could say you subtract negative 6 0 times. So you can say this is 15 minus negative 6 times-- or let me write it better this way --minus 0 times negative 6. That's what that first term is right there. What's the second term? This is 15. We just subtracted negative 6 once, or you could say, minus 1 times 6. Or you could say plus 1 times negative 6. Either way, we're subtracting the 6 once. Now what's happening here? This is 15 minus 2 times negative 6-- or, sorry --minus 2 times 6. We're subtracting a 6 twice. What's the fourth term? This is 15 minus-- We're subtracting the 6 three times from the 15, so minus 3 times 6. So, if you see the pattern here, when we have our fourth term, we have the term minus 1 right there. The fourth term, we have a 3. The third term, we have a 2. The second term, we have a 1. So if we had the nth term, if we just had the nth term here, what's this going to be? It's going to be 15 minus-- You see it's going to be n minus 1 right here. When n is 4, n minus 1 is 3. When n is 3, n minus 1 is 2. When n is 2, n minus 1 is 1. When n is 1, n minus 1 is 0. So we're going to have this term right here is n minus 1. So minus n minus 1 times 6. So if you want to figure out the 100th term of this sequence, I didn't even have to write it in this general term, you can just look at this pattern. It's going to be-- and I'll do it in pink --the 100th term in our sequence-- I'll continue our table down --is going to be what? It's going to be 15 minus 100 minus 1, which is 99, times 6. I just follow the pattern. 1, you had a 0 here. 2, you had a 1 here. 3, you had a 2 here. 100, you're going to have a 99 here. So let's just calculate what this is. What's 99 times 6? So 99 times 6-- Actually you can do this in your head. You could say that's going to be 6 less than 100 times 6, which is 600, and 6 less is 594. But if you didn't want to do it that way, you just do it the old-fashioned way. 6 times 9 is 54. Carry the 5. 9 times 6, or 6 times 9 is 54. 54 plus 5 is 594. So this right here is 594. And then to figure out what 15-- So we want to figure out what 15 minus 594 is. And this can sometimes be confusing, but the way I always process this in my head is, I say that this is the exact same thing as the negative of 594 minus 15. And if you don't believe me, distribute out this negative sign. Negative 1 times 594 is negative 594. Negative 1 times negative 15 is positive 15. So these two statements are equivalent. This is much easier for my brain to understand. So what's 594 minus 15? We can do this in our heads. 594 minus 14 would be 580, and then 580 minus 1 more would be 579. So that right there is 579, and then we have this negative sign sitting out there. So the 100th term in our sequence will be negative 579." + }, + { + "Q": "Can a Van Der Graaf Generator light a light bulb up if you put it on it's metallic top because isn't it in a way generating a steady current of negative charge which is what a light bulb requires.", + "A": "Hello Alex, Yes it could (in theory) but it would take a special light bulb. The bulb would need to operate at thousands of volts with a very small current. This is nearly impossible to construct with a traditional filament. There is another option, a fluorescent tube operates with high voltage and minimal current. Google de graaf fluorescent and you will find many examples. Regards, APD", + "video_name": "ZRLXDiiUv8Q", + "transcript": "- [Voiceover] All right, now we're gonna talk about the idea of an electric current. The story about currents starts with the idea of charge. We've learned that we have two kinds of charges, positive and negative charge. We'll just make up two little charges like that. And we know if they're the opposite sign, that there'll be a force of attraction between them. And if they have two like signs, here's two charges that are both positive, and these charges are gonna repel each other. So this is the basic electrostatics idea, and the same thing for two minus charges. They also repel. So like charges repel, and unlike charges attract. That's one idea. We have the idea of charge. And now we need a place to get some charge. One of the places we like to get charge from is copper, copper wires. A copper atom looks like this. Copper atom has a nucleus with some protons in it, and it also has electrons flying around the outside, electrons in orbits around the outside. So we'll draw the electrons like this. There'll be orbits around this nucleus. Pretty good circles. And there'll be electrons in these. Little minus signs. There's electrons stacked up in this. And even farther out, there's electrons. So there's kind of a interesting looking copper atom. Copper, the symbol for copper is Cu, and its atomic number is 29. That means there's 29 protons inside here, and there's 29 electrons outside. It turns out, just as a coincidence for copper, that the last orbital out here has just one electron in it, that guy right there. And that's the one that is the easiest to pull away from copper and have it go participate in conduction, in electric current. If I have a chunk of copper, every copper atom will have the opportunity to contribute one, this one lonely electron out here. If we look at another element, like for instance silver, silver has this same kind of electron configuration, where there's just one out here. And that's why silver and copper are such good, good conductors. Now we're gonna build, let's build a copper wire. Here's sort of a copper wire. It's just made of solid copper. It's all full of copper atoms. And I'm gonna put a voltage across this. There's our little battery. This is the minus sign, this is the plus side. And we'll hook up a battery to this. What's going on in here? Inside this copper is a whole bunch of electrons that are associated with atoms. It's a neutral piece of metal. There's the same number of protons as there is electrons. But these electrons are a little bit loose. So if I put a plus over there, that's this situation right here, where a plus is attracting a minus. So an electron is gonna sort of wander over this way and go like that. And that's gonna leave a net positive charge in this region. So these electrons are all gonna start moving in this direction. And down at the end, here, an electron is gonna come out of this battery, travel in here, and it's gonna go in there and make up the difference. So if I had a net positive charge here from the electrons leaving and going to the left, this battery would fill those in. And I'm gonna get a net movement of charge, of negative charge, around in this direction, like this. The question is, how do I measure that? How do I measure or give a number to that amount of stuff that's going on? So we wanna quantify that, we wanna assign a number to the amount of current happening here. What we do is, in our heads, we put a boundary across here. So just make that up in your head. And it cuts all the way through the copper. And what we know, we're gonna stand right here. We're gonna keep our eye right on this boundary down in here. As we watch, what we're gonna do is, we're gonna count the number of electrons that move by here, and we're gonna have a stop watch and we're gonna time that. So we're gonna get, basically, this is charge, it's negative charge, and it's moving to the side. What we're gonna do is, at one little spot right here, we're just gonna count the number that go by in one second. So we're gonna get charge per second. It's gonna be a negative charge moving by. That's what we call current. It's the same as water flowing by in a river. That's the same idea. Now I'm gonna set up a different situation that also produces a current. And this time, we're gonna do it with water, water and salt. Let's build a tube of salt, of salt water, like this. We're gonna pretend this is some tube that's all full of water. I'm also gonna put a battery here. Let's put another battery. And we'll stick the wire into there. We'll stick the wire into there. This is the plus side of the battery, and this is the minus side of the battery. Water is H20, and this does not conduct. There's no free electrons available here. But what I'm gonna do is, I'm gonna put some table salt in it. This is ordinary salt that you put on your food. It's made of sodium, that's the symbol for sodium, and chloride, Cl is chloride. Sodium chloride is table salt. If we sprinkle some table salt into water, what happens is, these dissolve and we get a net plus charge here and a net minus charge on the chlorine. So out here is floating around Na's with plus signs and Cl's nearby, really close nearly, with minuses. Let's keep it even. Now, when I dip my battery wires into this water, what's gonna happen is, this plus charge, this plus charge over here from the battery is gonna attract the minus Cl's. So the Cl's gonna move that way a little bit. And over here, the same thing is happening. There's a minus sign here. There's a minus from the battery. That's going to attract this, and it's also gonna repel Cl minuses. So what we get is a net motion of positive charge, plus q going this way, and we get minus q going this way. How do we measure that current? How do we measure that current? Well, we do it the same way as we did up there with copper. We put a boundary through here in our heads. We stand here and we watch the charges moving by. What we're gonna get is some sodiums, Na's, moving this way, and chlorines moving this way. Just like we showed here. Na moving this way. So there's gonna be plus charges moving through the boundary and minus charges moving through the boundary in the opposite direction. If I take the total sum of that. For example, if I see one Na go this way and one Cl go this way, that's equivalent to two charges moving through the boundary. Hope that makes sense. It's equivalent to two charges, one going this way and one going this way. Because they have opposite signs, they add together and make two charges. In this case, current is equal, again, to charge per second." + }, + { + "Q": "Are there still more printed copies of Mein Kampf in the world?", + "A": "Yes india has lots of copies", + "video_name": "5qWI2pEv0wg", + "transcript": "Narrator: By the end of 1923, Hitler sees it as his chance to seize power in Germany. He's getting this popularity, the Nazis are getting this follower-ship because the Weimar Republic is falling apart. You have the hyperinflation, the German people feel insulted by this French occupation of the Ruhr region. It isn't just regular people who are starting to support the Nazis, it's very notable people as well. This right over here is General Ludendorff, we've already talked about him, he's one of the believers in the 'stab-in-the-back' theory that Germany would have won World War I if it wasn't stabbed in the back by the November Criminals who had taken control of the government during the revolution in October and November. He becomes a supporter of Hitler as well. In 1922 you have Mussolini come to power, this inspires Hitler. So, as we get in to November, Hitler sees this as his chance and the way that he wants to take control, is he wants to abduct or kidnap the leaders of the Bavarian region, and there's three of them, in particular at this time. Then from there, try to take control of the nation as a whole. So, in November of 1923 you have a gathering of the three gentlemen who are essentially in charge of Bavaria, a gathering of them and several thousand officials in Bavaria at a local beer hall in Munich. (writing) at a beer hall. Hitler sees this as the opportunity to take control. This is where he launches his Beer Hall Putsch, and I know I'm mispronouncing it, but Putsch literally means coup d'\u00e9tat, to try to overthrow the government. So, Hitler and his Nazi's the go to that Beer Hall meeting of the government officials, they surround it with their paramilitary group, their storm troopers, Hitler enters into the hall, gets on stage, shoots into the air twice and says, look this is the revolution, it is beginning. He forces the three leaders of Bavaria at gunpoint to pledge allegiance to the Nazi party and to this Putsch and to Hitler, in particular. Then things start to go a little bit ... get a little bit ... start to dissolve. As Hitler tries to address some issues that are going on outside, the members who they were going to kidnap are allowed to leave, you have chaos in the area amongst the Nazi's and, frankly, amongst the government throughout that evening, into that morning, at which point Hitler and his followers, and Ludendorff is one of them, decide to march (writing) decide to march into central Munich. All of this is happening ... all of this is happening in Munich, which is in Bavaria. They decide to march, and it's during that march that they have a confrontation with the official government troops. It's unclear who fired the first shot, but you do have an exchange of fire and during that exchange of fire, I've seen estimates of about 14-16 Nazi's are shot. A few days later ... and a few policemen, or a few soldiers are shot as well, and then a few days later Hitler is arrested. (writing) Hitler, Hitler is arrested. He's tried in early 1924 and then he is sentenced to jail, so all of his ambitions were lead to nothing. In jail, he still continued to develop his philosophy. He actually continued to develop his following. He spent roughly the second two-thirds of 1924, in 1924, he spent it in jail. (writing) 1924 was spent primarily in jail, but while he was in jail he had dictated his autobiography and his, frankly, his belief system in Mein Kampf, which literally means 'My struggle.' It's actually banned in many countries, it's not banned in the U.S. It does make for interesting reading because you get a sense for, on one level, how bizarre Hitler's brain was and how disturbed Hitler's brain was, but on the other side, you can appreciate that he was a very, he was a strong communicator. Even before any of this people would talk about how transfixing his eyes were, how much attention people paid to him when he would give a speech. You can even see this in his writing, and you can do a web search on it and you can get the entire text of Mein Kampf. It's disturbing and fascinating at the same time, but this is a little passage. In this passage, it gives you an idea of Hitler's view of why Germany was having these failures and what he, in his bizarrely disturbed mind, thought what the solution was. \"If we pass all the causes of the German collapse \"in review, the ultimate and most decisive \"remains the failure to recognize the racial problem, \"and especially the Jewish menace.\" He's blaming all of Hitler's difficulty on a racial problem and in particular on Jews. \"The defeats on the battlefield in August 1918 \"would have been child's play to bear. \"They stood in no proportion to the victories \"of our people. \"It was not they that cause our downfall, \"no, it was brought about by that power which \"prepared these defeats by systematically, \"over many decades, robbing our people of \"the political and moral instincts and forces \"which alone make nations capable, \"and hence worthy of existence.\" If you read a lot of the other text, what he's talking about is this decades of, essentially, watering down their society, watering down their society with other people. If they didn't water it down, they say the defeats in the battlefield would've been child's play to bear. \"In heedlessly ignoring the question \"of the preservation of the racial foundations of our \"nation, the old Reich disregarded the sole right \"which gives life in this world.\" He views this racial, in his mind, racial impurity as the reason why Germany was facing all of this difficulty. As we'll see over the next few videos, this leads to one of the ugliest and bloodiest periods of human history." + }, + { + "Q": "Is this about video showing about the value of american coins or teaching word problems of money?", + "A": "It is definitely teaching you about the value of American coins. Remember: 1 dollar = 100 cents 1 quarter = 25 cents 1 dime = 10 cents 1 nickel = 5 cents 1 penny = 1 cent As for your second question, there are many annoyances with currency; for example, the value of currency fluctuates, and there is not one universal currency. That sure would make things a whole lot simpler!", + "video_name": "pJ8KwRztfF0", + "transcript": "- Let's get some practice counting money! So I have six coins right over here, and these are all United States coins, we're counting money in the United States for these examples, and what is this first coin? Well, this is called a quarter, or a quarter-dollar, so it represents 25 cents, and we could write it out as 25 cents, but I'll just keep that one like that. Now this one's another quarter, so it is also going to be 25 cents. Now this one looks different, but it's just the other side of these coins. This is what the other side looks like. So this is also going to be 25 cents. These are three quarters right over here. So how much money do these three quarters represent? Well, it's going to be 25 plus 25, which is 50, plus 25, which is going to be 75, so these three quarters are going to be 75 cents. And remember, 100 cents make a dollar, so this is still less than one dollar. But we're not done yet. We have this nickel, this is a nickel right over here, that represents five cents, and then we have another nickel here, it looks different, but it's just the other side, this is the head side, this is the tail side. So this is also another five cents, and so these two nickels, if you add them together, they are going to represent 10 cents, and then finally, you have a penny, and a penny, and it even says it right over here, is one cent, in fact, they all say it here, this is five cents, this is one cent. So this right over here's gonna be one cent. So what is 75 plus 10 plus 1? Well 75 plus 10 is 85, plus 1 is 86, so this is equal to 86 cents. And if it felt a little bit too fast to count up 25, 50, 75 in your head, you could also add them up. 25, 25, 25, plus 5, another 5, plus 1. You could add them up this way, and then what would you get? 5 plus 5 is 10, plus 5 is 15, plus 5 is 20, plus 5 is 25, plus 1 is 26, so that's two tens and one six, put the two tens up here. 2 plus 2 is 4, 4 plus 2 is 6, 6 plus 2 is 8. So you could get 86 cents, either way. Let's do one that has even more coins in it. So here we go, we have, so what is it, what's going on here? This right over here is a quarter, that's going to be 25 cents. 25 cents, and then, we have one that we didn't see in the previous example, we have two dimes. A dime represents 10 cents, so we have two dimes, where we can have those each represent 10 cents, then we have two nickels, we've already seen those each represent five cents, so 5 and 5, and then we have four pennies, one, two, three, four. Now we could put each penny separately, like that, or we could say, look, four pennies, each of them represent one cent, so that's going to be four cents. So let me do it that way. So, this one, two, three, four, that's going to be four cents. And then we could just add everything up, so 5 plus 0 plus 0 plus 5 is 10, plus 5 is 15, plus 4 is 19. 19 is one ten and nine ones, so I could put the one ten in the tens place, 1 plus 2 is 3, plus 1 is 4, plus 1 is 5. So it's five tens, five tens and nine ones, so 59 cents. This right over here is 59 cents. And notice, we have more coins, but it represents less value than the previous example. That's because we had a lot of coins that didn't represent a lot of values, like, we had these four pennies here, while the previous example, we had three quarters! Each of these quarters is equivalent to 25 pennies, so we're able to represent more money with fewer coins in the first example." + }, + { + "Q": "At 18:45 How does khan figure out that when derivative of -x_0-1/2x_0 is equal to 0 is actually minimum or maximum of the function?", + "A": "derivative = 0 ==> (implies) a maximum or minimum", + "video_name": "viaPc8zDcRI", + "transcript": "I just got sent this problem, and it's a pretty meaty problem. A lot harder than what you'd normally find in most textbooks. So I thought it would help us all to work it out. And it's one of those problems that when you first read it, your eyes kind of glaze over, but when you understand what they're talking about, it's reasonably interesting. So they say, the curve in the figure above is the parabola y is equal to x squared. So this curve right there is y is equal to x squared. Let us define a normal line as a line whose first quadrant intersection with the parabola is perpendicular to the parabola. So this is the first quadrant, right here. And they're saying that a normal line is something, when the first quadrant intersection with the parabola is normal to the parabola. So if I were to draw a tangent line right there, this line is normal to that tangent line. That's all that's saying. So this is a normal line, right there. Normal line. Fair enough. 5 normal lines are shown in the figure. 1, 2, 3, 4, 5. Good enough. And these all look perpendicular, or normal to the parabola in the first quadrant intersection, so that makes sense. For a while, the x-coordinate of the second quadrant intersection of the normal line of the parabola gets smaller, as the x-coordinate of the first quadrant intersection gets smaller. So let's see what happens as the x-quadrant of the first intersection gets smaller. So this is where I left off in that dense text. So if I start at this point right here, my x-coordinate right there would look something like this. Let me go down. my x-coordinate is right around there. And then as I move to a smaller x-coordinate to, say, this one right here, what happened to the normal line? Or even more important, what happened to the intersection of the normal line in the second quadrant? This is the second quadrant, right here. So when I had a larger x-value here, my normal line intersected here, in the second quadrant. Then when I brought my x-value in, when I lowered my x-value, my x-value here, because this is the next point, right here, my x-value at the intersection here, went-- actually, their wording is bad. They're saying that the second quadrant intersection gets smaller. But actually, it's not really getting smaller. It's getting less negative. I guess smaller could be just absolute value or magnitude, but it's just getting less negative. It's moving there, but it's actually becoming a larger number, right? It's becoming less negative, but a larger number. But if we think in absolute value, I guess it's getting smaller, right? As we went from that point to that point, as we moved the x in for the intersection of the first quadrant, the second quadrant intersection also moved in a bit, from that line to that line. Fair enough. But eventually, a normal line second quadrant intersection gets as small as it can get. So if we keep lowering our x-value in the first quadrant, so we keep on pulling in the first quadrant, as we get to this point. And then this point intersects the second quadrant, right there. And then, if you go even smaller x-values in the first quadrant then your normal line starts intersecting in the second quadrant, further and further negative numbers. So you can kind of view this as the highest value, or the smallest absolute value, at which the normal line can intersect in the second quadrant Let me make that clear. Up here, you were intersecting when you had a large x in the first quadrant, you had a large negative x in the second quadrant intersection. And then as you lowered your x-value, here, you had a smaller negative value. Up until you got to this point, right here, you got this, which you can view as the smallest negative value could get, and then when you pulled in your x even more, these normal lines started to push out again, out in the second quadrant. That's, I think, what they're talking about. The extreme normal line is shown as a thick line in the figure. This is the extreme normal line, right there. So this is the extreme one, that deep, bold one. Extreme normal line. After this point, when you pull in your x-values even more, the intersection in your second quadrant starts to push out some. And you can think of the extreme case, if you draw the normal line down here, your intersection with the second quadrant is going to be way out here someplace, although it seems like it's kind of asymptoting a little bit. Let's read the rest of the problem. Once the normal line passes the extreme normal line, the x-coordinates of their second quadrant intersections what the parabola start to increase. And they're really, when they say they start to increase, they're actually just becoming more negative. That wording is bad. I should change this to more, more negative. Or they're becoming larger negative numbers. Because once you get below this, then all of a sudden the x-intersections start to push out more in the second quadrant. Fair enough. The figures show 2 pairs of normal lines. Fair enough. The 2 normal lines of a pair have the same second quadrant intersection with the parabola, but 1 is above the extreme normal line, in the first quadrant, the other is below it. Right, fair enough. For example, this guy right here, this is when we had a large x-value. He intersects with the second quadrant there. Then if you lower and lower the x-value, if you lower it enough, you pass the extreme normal line, and then you get to this point, and then this point, he intersects, or actually, you go to this point. So if you pull in your x-value enough, you once again intersect at that same point in the second quadrant. So hopefully I'm making some sense to you, as I try to make some sense of this problem. Now what do they want to know? And I think I only have time for the first part of this. Maybe I'll do the second part in the another video. Find the equation of the extreme normal line. Well, that seems very daunting at first, but I think our toolkit of derivatives, and what we know about equations of a line, should be able to get us there. So what's the slope of the tangent line at any point on this curve? Well, we just take the derivative of y equals x squared, and y prime is just equal to 2x. This is the slope of the tangent at any point x. So if I want to know the slope of the tangent at x0, at some particular x, I would just say, well, let me just say, slope, it would be 2 x0. Or let me just say, f of x0 is equal to 2 x0. This is the slope at any particular x0 of the tangent line. Now, the normal line slope is perpendicular to this. So the perpendicular line, and I won't review it here, but the perpendicular line has a negative inverse slope. So the slope of normal line at x0 will be the negative inverse of this, because this is the slope of the tangent line x0. So it'll be equal to minus 1 over 2 x0. Fair enough. Now, what is the equation of the normal line at x0 let's say that this is my x0 in question. What is the equation of the normal line there? Well, we can just use the point-slope form of our equation. So this point right here will be on the normal line. And that's the point x0 squared. Because this the graph of y equals x0, x squared. So this normal line will also have this point. So we could say that the equation of the normal line, let me write it down, would be equal to, this is just a point-slope definition of a line. You say, y minus the y-point, which is just x0 squared, that's that right there, is equal to the slope of the normal line minus 1 over 2 x0 times x minus the x-point that we're at. Minus x, minus x0. This is the equation of the normal line. So let's see. And what we care about is when x0 is greater than 0, right? We care about the normal line when we're in the first quadrant, we're in all of these values right there. So that's my equation of the normal line. And let's solve it explicitly in terms of x. So y is a function of x. Well, if I add x0 squared to both sides, I get y is equal to, actually, let me multiply this guy out. I get minus 1/2 x0 times x, and then I have plus, plus, because I have a minus times a minus, plus 1/2. The x0 and the over the x0, they cancel out. And then I have to add this x0 to both sides. So all I did so far, this is just this part right there. That's this right there. And then I have to add this to both sides of the equation, so then I have plus x0 squared. So this is the equation of the normal line, in mx plus b form. This is its slope, this is the m, and then this is its y-intercept right here. That's kind of the b. Now, what do we care about? We care about where this thing intersects. We care about where it intersects the parabola. And the parabola, that's pretty straightforward, that's just y is equal to x squared. So to figure out where they intersect, we just have to set the 2 y's to be equal to each other. So they intersect, the x-values where they intersect, x squared, this y would have to be equal to that y. Or we could just substitute this in for that y. So you get x squared is equal to minus 1 over 2 x0 times x, plus 1/2 plus x0 squared. Fair enough. And let's put this in a quadratic equation, or try to solve this, so we can apply the quadratic equation. So let's put all of this stuff on the lefthand side. So you get x squared plus 1 over 2 x0 times x minus all of this, 1/2 plus x0 squared is equal to 0. All I did is, I took all of this stuff and I put it on the lefthand side of the equation. Now, this is just a standard quadratic equation, so we can figure out now where the x-values that satisfy this quadratic equation will tell us where our normal line and our parabola intersect. So let's just apply the quadratic equation here. So the potential x-values, where they intersect, x is equal to minus b, I'm just applying the quadratic equation. So minus b is minus 1 over 2 x0, plus or minus the square root of b squared. So that's that squared. So it's one over four x0 squared minus 4ac. So minus 4 times 1 times this minus thing. So I'm going to have a minus times a minus is a plus, so it's just 4 times this, because there was one there. So plus 4 times this, right here. 4 times this is just 2 plus 4 x0 squared. All I did is, this is 4ac right here. Well, minus 4ac. The minus and the minus canceled out, so There's a 1. So 4 times c is just 2 plus 4x squared. I just multiply this by two, and of course all of this should be over 2 times a. a is just 2 there. So let's see if I can simplify this. We're just figuring out where the normal line and the parabola intersect. Now, what do we get here. This looks like a little hairy beast here. Let me see if I can simplify this a little bit. So let us factor out-- let me rewrite this. I can just divide everything by 1/2, so this is minus 1 over 4 x0, I just divided this by 2, plus or minus 1/2, that's just this 1/2 right there, times the square root, let me see what I can simplify out of here. So if I factor out a 4 over x0 squared, then what does my expression become? This term right here will become an x to the fourth, x0 to the fourth, plus, now, what does this term become? This term becomes a 1/2 x0 squared. And just to verify this, multiply 4 times 1/2, you get 2, and then the x0 squares cancel out. So write this term times that, will equal 2, and then you have plus-- now we factored a four out of this and the x0 squared, so plus 1/16. Let me scroll over a little bit. And you can verify that this works out. If you were to multiply this out, you should get this business right here. I see the home stretch here, because this should actually factor out quite neatly. So what does this equal? So the intersection of our normal line and our parabola is equal to this. Minus 1 over 4 x0 plus or minus 1/2 times the square root of this business. And the square root, this thing right here is 4 over x0 squared. This is actually, lucky for us, a perfect square. And I won't go into details, because then the video will get too long, but I think you can recognize that this is x0 squared, plus 1/4. If you don't believe me, square this thing right here. You'll get this expression right there. And luckily enough, this is a perfect square, so we can actually take the square root of it. And so we get, the point at which they intersect, our normal line and our parabola, and this is quite a hairy problem. The points where they intersect is minus 1 over 4 x0, plus or minus 1/2 times the square root of this. The square root of this is the square root of this, which is just 2 over x0 times the square root of this, which is x0 squared plus 1/4. And if I were to rewrite all of this, I'd get minus 1 over 4 x0 plus, let's see, this 1/2 and this 2 cancel out, right? So these cancel out. So plus or minus, now I just have a one over x0 times x0 squared. So I have 1 over x0-- oh sorry, let me, we have to be very careful there-- x0 squared divided by x0 is just x0, let me do that in a yellow color so you know what I'm dealing with. This term multiplied by this term is just x0, and then you have a plus 1/4 x0. And this is all a parentheses here. So these are the two points at which the normal curve and our parabola intersect. Let me just be very clear. Those 2 points are, for if this is my x0 that we're dealing with, right there. It's this point and this point. And we have a plus or minus here, so this is going to be the plus version, and this is going to be the minus version. In fact, the plus version should simplify into x0. Let's see if that's the case. Let's see if the plus version actually simplifies to x0. So these are our two points. If I take the plus version, that should be our first quadrant intersection. So x is equal to minus 1/4 x0 plus x0 plus 1/4 x0. And, good enough, it does actually cancel out. That cancels out. So x0 is one of the points of intersection, which makes complete sense. Because that's how we even defined the problem. But, so this is the first quadrant intersection. So that's the first quadrant intersection. The second quadrant intersection will be where we take the minus sign right there. So x, I'll just call it in the second quadrant intersection, it'd be equal to minus 1/4 x0 minus this stuff over here, minus the stuff there. So minus x0 minus 1 over 4 mine x0. Now what do we have? So let's see. We have a minus 1 over 4 x0, minus 1 over 4 x0. So this is equal to minus x0, minus x0, minus 1 over 2 x0. So if I take minus 1/4 minus 1/4, I get minus 1/2. And so my second quadrant intersection, all this work I did got me this result. My second quadrant intersection, I hope I don't run out of space. My second quadrant intersection, of the normal line and the parabola, is minus x0 minus 1 over 2 x0. Now this by itself is a pretty neat result we just got, but we're unfortunately not done with the problem. Because the problem wants us to find that point, the maximum point of intersection. They call this the extreme normal line. The extreme normal line is when our second quadrant intersection essentially achieves a maximum point. I know they call it the smallest point, but it's the smallest negative value, so it's really a maximum point. So how do we figure out that maximum point? Well, we have our second quadrant intersection as a function of our first quadrant x. I could rewrite this as, my second quadrant intersection as a function of x0 is equal to minus x minus 1 over 2 x0. So this is going to reach a minimum or a maximum point when its derivative is equal to 0. This is a very unconventional notation, and that's probably the hardest thing about this problem. But let's take this derivative with respect to x0. So my second quadrant intersection, the derivative of that with respect to x0, is equal to, this is pretty straightforward. It's equal to minus 1, and then I have a minus 1/2 times, this is the same thing as x to the minus 1. So it's minus 1 times x0 to the minus 2, right? I could have rewritten this as minus 1/2 times x0 to the minus 1. So you just put its exponent out front and decrement it by 1. And so this is the derivative with respect to my first quadrant intersection. So let me simplify this. So x, my second quadrant intersection, the derivative of it with respect to my first quadrant intersection, is equal to minus 1, the minus 1/2 and the minus 1 become a positive when you multiply them, and so plus 1/2 over x0 squared. Now, this'll reach a maximum or minimum when it equals 0. So let's set that equal to 0, and then solve this problem right there. Well, we add one to both sides. We get 1 over 2 x0 squared is equal to 1, or you could just say that that means that 2 x0 squared must be equal to 1, if we just invert both sides of this equation. Or we could say that x0 squared is equal to 1/2, or if we take the square roots of both sides of that equation, we get x0 is equal to 1 over the square root of 2. So we're really, really, really close now. We've just figured out the x0 value that gives us our extreme normal line. This value right here. Let me do it in a nice deeper color. This value right here, that gives us the extreme normal line, that over there is x0 is equal to 1 over the square root of 2. Now, they want us to figure out the equation of the extreme normal line. Well, the equation of the extreme normal line we already figured out right here. It's this. The equation of the normal line is that thing, right there. So if we want the equation of the normal line at this extreme point, right here, the one that creates the extreme normal line, I just substitute 1 over the square root of 2 in for x0. So what do I get? I get, and this is the home stretch, and this is quite a beast of a problem. y minus x0 squared. x0 squared is 1/2, right? 1 over the square root of 2 squared is 1/2. Is equal to minus 1 over 2 x0. So let's be careful here. So minus 1/2 times 1 over x0. One over x0 is the square root of 2, right? All of that times x minus x0. So that's 1 over the square root of 2. x0 is one So let's simplify this a little bit. So the equation of our normal line, assuming I haven't made any careless mistakes, is equal to, so y minus 1/2 is equal to, let's see. If we multiply this minus square root of 2 over 2x, and then if I multiply these square root of 2 over this, it becomes one. And then I have a minus and a minus, so that I have a plus 1/2. I think that's right. Yeah, plus 1/2, this times this times that is equal to plus 1/2. And then, we're at the home stretch. So we just add 1/2 to both sides of this equation, and we get our extreme normal line equation, which is y is equal to minus square root of 2 over 2x. If you add 1/2 to both sides of this equation, you get plus 1. And there you go. That's the equation of that line there, assuming I haven't made any careless mistakes. But even if I have, I think you get the idea of hopefully how to do this problem, which is quite a beastly one." + }, + { + "Q": "so from 00:01 to 23:17 he talking about the common cold and flu (influenza)? :|", + "A": "He s talking about Viruses in general and not about a specific one. At the beginning he says that because he has a cold that he s going to talk about Viruses. In between those times he covers how a most viruses interact with living cells and also how a retrovirus might.", + "video_name": "0h5Jd7sgQWY", + "transcript": "Considering that I have a cold right now, I can't imagine a more appropriate topic to make a video on than a virus. And I didn't want to make it that thick. A virus, or viruses. And in my opinion, viruses are, on some level, the most fascinating thing in all of biology. Because they really blur the boundary between what is an inanimate object and what is life? I mean if we look at ourselves, or life as one of those things that you know it when you see it. If you see something that, it's born, it grows, it's constantly changing. Maybe it moves around. Maybe it doesn't. But it's metabolizing things around itself. It reproduces and then it dies. You say, hey, that's probably life. And in this, we throw most things that we see-- or we throw in, us. We throw in bacteria. We throw in plants. I mean, I could-- I'm kind of butchering the taxonomy system here, but we tend to know life when we see it. But all viruses are, they're just a bunch of genetic information inside of a protein. Inside of a protein capsule. So let me draw. And the genetic information can come in any form. So it can be an RNA, it could be DNA, it could be single-stranded RNA, double-stranded RNA. Sometimes for single stranded they'll write these two little S's in front of it. Let's say they are talking about double stranded DNA, they'll put a ds in front of it. But the general idea-- and viruses can come in all of these forms-- is that they have some genetic information, some chain of nucleic acids. Either as single or double stranded RNA or single or double stranded DNA. And it's just contained inside some type of protein structure, which is called the capsid. And kind of the classic drawing is kind of an icosahedron type looking thing. Let me see if I can do justice to it. It looks something like this. And not all viruses have to look exactly like this. There's thousands of types of viruses. And we're really just scratching the surface and understanding even what viruses are out there and all of the different ways that they can essentially replicate themselves. We'll talk more about that in the future. And I would suspect that pretty much any possible way of replication probably does somehow exist in the virus world. But they really are just these proteins, these protein capsids, are just made up of a bunch of little proteins put together. And inside they have some genetic material, which might be DNA or it might be RNA. So let me draw their genetic material. The protein is not necessarily transparent, but if it was, you would see some genetic material inside of there. So the question is, is this thing life? It seems pretty inanimate. It doesn't grow. It doesn't change. It doesn't metabolize things. This thing, left to its own devices, is just It's just going to sit there the way a book on a table just sits there. It won't change anything. But what happens is, the debate arises. I mean you might say, hey Sal, when you define it that way, just looks like a bunch of molecules put together. That isn't life. But it starts to seem like life all of a sudden when it comes in contact with the things that we normally consider life. So what viruses do, the classic example is, a virus will attach itself to a cell. So let me draw this thing a little bit smaller. So let's say that this is my virus. I'll draw it as a little hexagon. And what it does is, it'll attach itself to a cell. And it could be any type of cell. It could be a bacteria cell, it could be a plant cell, it could be a human cell. Let me draw the cell here. Cells are usually far larger than the virus. In the case of cells that have soft membranes, the virus figures out some way to enter it. Sometimes it can essentially fuse-- I don't want to complicate the issue-- but sometimes viruses have their own little membranes. And we'll talk about in a second where it gets their membranes. So a virus might have its own membrane like that. That's around its capsid. And then these membranes will fuse. And then the virus will be able to enter into the cell. Now, that's one method. And another method, and they're seldom all the same way. But let's say another method would be, the virus convinces-- just based on some protein receptors on it, or protein receptors on the cells-- and obviously this has to be kind of a Trojan horse type of thing. The cell doesn't want viruses. So the virus has to somehow convince the cell that it's a non-foreign particle. We could do hundreds of videos on how viruses work and it's a continuing field of research. But sometimes you might have a virus that just gets consumed by the cell. Maybe the cell just thinks it's something that it needs to consume. So the cell wraps around it like this. And these sides will eventually merge. And then the cell and the virus will go into it. This is called endocytosis. I'll just talk about that. It just brings it into its cytoplasm. It doesn't happen just to viruses. But this is one mechanism that can enter. And then in cases where the cell in question-- for example in the situation with bacteria-- if the cell has a very hard shell-- let me do it in a good color. So let's say that this is a bacteria right here. And it has a hard shell. The viruses don't even enter the cell. They just hang out outside of the cell like this. Not drawing to scale. And they actually inject their genetic material. So there's obviously a huge-- there's a wide variety of ways of how the viruses get into cells. But that's beside the point. The interesting thing is that they do get into the cell. And once they do get into the cell, they release their genetic material into the cell. So their genetic material will float around. If their genetic material is already in the form of RNA-- and I could imagine almost every possibility of different ways for viruses to work probably do exist in nature. We just haven't found them. But the ones that we've already found really do kind of do it in every possible way. So if they have RNA, this RNA can immediately start being used to essentially-- let's say this is the nucleus of the cell. That's the nucleus of the cell and it normally has the DNA in it like that. Maybe I'll do the DNA in a different color. But DNA gets transcribed into RNA, normally. So normally, the cell, this a normal working cell, the RNA exits the nucleus, it goes to the ribosomes, and then you have the RNA in conjunction with the tRNA and it produces these proteins. The RNA codes for different proteins. And I talk about that in a different video. So these proteins get formed and eventually, they can form the different structures in a cell. But what a virus does is it hijacks this process here. Hijacks this mechanism. This RNA will essentially go and do what the cell's own RNA would have done. And it starts coding for its own proteins. Obviously it's not going to code for the same things there. And actually some of the first proteins it codes for often start killing the DNA and the RNA that might otherwise compete with it. So it codes its own proteins. And then those proteins start making more viral shells. So those proteins just start constructing more and more viral shells. At the same time, this RNA is replicating. It's using the cell's own mechanisms. Left to its own devices it would just sit there. But once it enters into a cell it can use all of the nice machinery that a cell has around to replicate itself. And it's kind of amazing, just the biochemistry of it. That these RNA molecules then find themselves back in these capsids. And then once there's enough of these and the cell has essentially all of its resources have been depleted, the viruses, these individual new viruses that have replicated themselves using all of the cell's mechanisms, will find some way to exit the cell. The most-- I don't want to say, typical, because we haven't even discovered all the different types of viruses there are-- but one that's, I guess, talked about the most, is when there's enough of these, they'll release proteins or they'll construct proteins. Because they don't make their own. That essentially cause the cell to either kill itself or its membrane to dissolve. So the membrane dissolves. And essentially the cell lyses. Let me write that down. The cell lyses. And lyses just means that the cell's membrane just And then all of these guys can emerge for themselves. Now I talked about before that have some of these guys, that they have their own membrane. So how did they get there, these kind of bilipid membranes? Well some of them, what they do is, once they replicate inside of a cell, they exit maybe not even killing-- they don't have to lyse. Everything I talk about, these are specific ways that a virus might work. But viruses really kind of explore-- well different types of viruses do almost every different combination you could imagine of replicating and coding for proteins and escaping from cells. Some of them just bud. And when they bud, they essentially, you can kind of imagine that they push against the cell wall, or the membrane. I shouldn't say cell wall. The cell's outer membrane. And then when they push against it, they take some of the membrane with them. Until eventually the cell will-- when this goes up enough, this'll pop together and it'll take some of the membrane with it. And you could imagine why that would be useful thing to have with you. Because now that you have this membrane, you kind of look like this cell. So when you want to go infect another cell like this, you're not going to necessarily look like a foreign particle. So it's a very useful way to look like something that you're not. And if you don't think that this is creepy-crawly enough, that you're hijacking the DNA of an organism, viruses can actually change the DNA an organism. And actually one of the most common examples is HIV virus. Let me write that down. HIV, which is a type of retrovirus, which is fascinating. Because what they do is, so they have RNA in them. And when they enter into a cell, let's say that they got into the cell. So it's inside of the cell like this. They actually bring along with them a protein. And every time you say, where do they get this protein? All of this stuff came from a different cell. They use some other cell's amino acids and ribosomes and nucleic acids and everything to build themselves. So any proteins that they have in them came from another cell. But they bring with them, this protein reverse transcriptase. And the reverse transcriptase takes their RNA and codes it into DNA. So its RNA to DNA. Which when it was first discovered was, kind of, people always thought that you always went from DNA to RNA, but this kind of broke that paradigm. But it codes from RNA to DNA. And if that's not bad enough, it'll incorporate that DNA into the DNA of the host cell. So that DNA will incorporate itself into the DNA of the host cell. Let's say the yellow is the DNA of the host cell. And this is its nucleus. So it actually messes with the genetic makeup of what it's infecting. And when I made the videos on bacteria I said, hey for every one human cell we have twenty bacteria cells. And they live with us and they're useful and they're part of us and they're 10% of our dry mass and all of that. But bacteria are kind of along for the ride. They don't change who we are. But these retroviruses, they're actually changing our I mean, my genes, I take very personally. They define who I am. But these guys will actually go in and change my genetic makeup. And then once they're part of the DNA, then just the natural DNA to RNA to protein process will code their actual proteins. Or their-- what they need to-- so sometimes they'll lay dormant and do nothing. And sometimes-- let's say sometimes in some type of environmental trigger, they'll start coding for themselves again. And they'll start producing more. But they're producing it directly from the organism's cell's DNA. They become part of the organism. I mean I can't imagine a more intimate way to become part of an organism than to become part of its DNA. I can't imagine any other way to actually define an organism. And if this by itself is not eerie enough, and just so you know, this notion right here, when a virus becomes part of an organism's DNA, this is called a provirus. But if this isn't eerie enough, they estimate-- so if this infects a cell in my nose or in my arm, as this cell experiences mitosis, all of its offspring-- but its offspring are genetically identical-- are going to have this viral DNA. And that might be fine, but at least my children won't get it. You know, at least it won't become part of my species. But it doesn't have to just infect somatic cells, it could infect a germ cell. So it could go into a germ cell. And the germ cells, we've learned already, these are the ones that produce gametes. For men, that's sperm and for women it's eggs. But you could imagine, once you've infected a germ cell, once you become part of a germ cell's DNA, then I'm passing on that viral DNA to my son or my daughter. And they are going to pass it on to their children. And just that idea by itself is, at least to my mind. vaguely creepy. And people estimate that 5-8%-- and this kind of really blurs, it makes you think about what we as humans really are-- but the estimate is 5-8% of the human genome-- so when I talked about bacteria I just talked about things that were along for the ride. But the current estimate, and I looked up this a lot. I found 8% someplace, 5% someplace. I mean people are doing it based on just looking at the DNA and how similar it is to DNA in other organisms. But the estimate is 5-8% of the human genome is from viruses, is from ancient retroviruses that incorporated themselves into the human germ line. So into the human DNA. So these are called endogenous retroviruses. Which is mind blowing to me, because it's not just saying these things are along for the ride or that they might help us or hurt us. It's saying that we are-- 5-8% of our DNA actually comes from viruses. And this is another thing that speaks to just genetic variation. Because viruses do something-- I mean this is called horizontal transfer of DNA. And you could imagine, as a virus goes from one species to the next, as it goes from Species A to B, if it mutates to be able to infiltrate these cells, it might take some-- it'll take the DNA that it already has, that makes it, it with it. But sometimes, when it starts coding for some of these other guys, so let's say that this is a provirus right here. Where the blue part is the original virus. The yellow is the organism's historic DNA. Sometimes when it codes, it takes up little sections of the other organism's DNA. So maybe most of it was the viral DNA, but it might have, when it transcribed and translated itself, it might have taken a little bit-- or at least when it translated or replicated itself-- it might take a little bit of the organism's previous DNA. So it's actually cutting parts of DNA from one organism and bringing it to another organism. Taking it from one member of a species to another member of But it can definitely go cross-species. So you have this idea all of a sudden that DNA can jump between species. It really kind of-- I don't know, for me it makes me appreciate how interconnected-- as a species, we kind of imagine that we're by ourselves and can only reproduce with each other and have genetic variation within But viruses introduce this notion of horizontal transfer via transduction. Horizontal transduction is just the idea of, look when I replicate this virus, I might take a little bit of the organism that I'm freeloading off of, I might take a little bit of their DNA with me. And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. So when a virus lyses it like this, this is called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia. One is from the CDC. These little green dots you see right here all over the surface, this big thing you see here, this is a white blood cell. Part of the human immune system. This is a white blood cell. And what you see emerging from the surface, essentially budding from the surface of this white blood cell-- and this gives you a sense of scale too-- these are HIV-1 viruses. And so you're familiar with the terminology, the HIV is a virus that infects white blood cells. AIDS is the syndrome you get once your immune system is weakened to the point. And then many people suffer infections that people with a strong immune system normally won't suffer from. But this is creepy. These things went inside this huge cell, they used the cell's own mechanism to reproduce its own DNA or its own RNA and these protein capsids. And then they bud from the cell and take a little bit of the membrane with it. And they can even leave some of their DNA behind in this cell's own DNA. So they really change what the cell is all about. This is another creepy picture. These are bacteriaphages. And these show you what I said before. This is a bacteria right here. This is its cell wall. And it's hard. So it's hard to just emerge into it. Or you can't just merge, fuse membranes with it. So they hang out on the outside of this bacteria. And they are essentially injecting their genetic material into the bacteria itself. And you could imagine, just looking at the size of these things. I mean, this is a cell. And it looks like a whole planet or something. Or this is a bacteria and these things are so much smaller. Roughly 1/100 of a bacteria. And these are much less than 1/100 of this cell we're talking about. And they're extremely hard to filter for. To kind of keep out. Because they are such, such small particles. If you think that these are exotic things that exist for things like HIV or Ebola , which they do cause, or SARS, you're right. But they're also common things. I mean, I said at the beginning of this video that I have a cold. And I have a cold because some viruses have infected the tissue in my nasal passage. And they're causing me to have a runny nose and whatnot. And viruses also cause the chicken pox. They cause the herpes simplex virus. Causes cold sores. So they're with us all around. I can almost guarantee you have some virus with you as you speak. They're all around you. But it's a very philosophically puzzling question. Because I started with, at the beginning, are these life? And at first when I just showed it to you, look they are just this protein with some nucleic acid molecule in it. And it's not doing anything. And that doesn't look like life to me. It's not moving around. It doesn't have a metabolism. It's not eating. It's not reproducing. But then all of a sudden, when you think about what it's doing to cells and how it uses cells to kind of reproduce. It kind of like-- in business terms it's asset light. It doesn't need all of the machinery because it can use other people's machinery to replicate itself. You almost kind of want to view it as a smarter form of life. Because it doesn't go through all of the trouble of what every other form of life has. It makes you question what life is, or even what we are. Are we these things that contain DNA or are we just transport mechanisms for the DNA? And these are kind of the more important things. And these viral infections are just battles between different forms of DNA and RNA and whatnot. Anyway I don't want to get too philosophical on you. But hopefully this gives you a good idea of what viruses are and why they really are, in my mind, the most fascinating pseudo organism in all of biology." + }, + { + "Q": "I still don't get how to divide notes how dose that work?", + "A": "Dividing notes is basically just dividing a note into smaller subsections of notes to get shorter duration notes. (different rhythms) (Depends on the time signature and such.) Just keep practicing, watching and researching more! It ll help! ;)", + "video_name": "esbqpgAD-EM", + "transcript": "- [Voiceover] At times, especially in popular dance forms the meter will remain constant. All ballroom dancing fits into this category. A march would also fit into this category, remaining constant, usually in two-four. It is also the case with most music from the 18th and 19th centuries. In the latter part of the 1800's and into the 1900's, composers started to feel free to change meters during a movement or work, sometimes quite often. The actual meters remain as we have discussed. If we look at the last movement of the Sam Jones \"Cello Concerto\", we can see some simple changes of meter, from two-four to three-four, back to two-four, then three-four and four-four. (\"Cello Concerto\" by Sam Jones) Another simple example is in Phillip Glass' \"Harmonium Mountain\". At this excerpt, he mixes the meters two-four, three-four, and four-four. (\"Harmonium Mountain\" by Phillip Glass) If we look at David Stock's work called \"Blast\" written in 2010, we find a more complicated section of meter changers, using five-eight, seven-eight, three-four, and four-four. (\"Blast\" by David Stock) If we look at a four-four measure, we have learned that the measure can easily be divided by using various note lengths. Half notes, quarter notes, sixteenth notes, thirty-second notes and so forth. But what if a composer would like to divide one of those quarter notes into three equal parts to create more rhythmic interest? We accomplish this by adding a three above or below a group of three eighth notes. The three signifies that three notes are performed during the time of one quarter note. Let's again look at Phillip Glass' \"Harmonium Mountain\". In this passage we see the violins playing the groups of three called triplets, and the violas and cellos are playing quarter notes. Then the violas join the violins playing triplets, the cellos play the eighth notes and the double basses play the quarter notes. (\"Harmonium Mountain\" by Phillip Glass) This method of changing duple notes to triple notes can work in any duple meter. The composer can also divide the triple beat in different ways. For example, instead of three eighth notes in a beat, we could see an eighth note and a quarter note, or a quarter note and an eighth note. We still need the number three above or below the notes. If we look at Ravel's \"Daphnis et Chloe\" the meter is five-four, but Ravel adds the three for each part to create triplets. You will see that he doesn't continue to write the three during the continuation of the excerpt, assuming that the performer understands the pattern. (\"Daphnis et Chloe\" by Ravel) The triplet is the most common variation within a meter, but there could also be, for example, five notes within a quarter, again with a five above or below, or six, or quite frankly any number that is not common to the meter. In a triple meter like six-eight, one could do the same. Six-eight can be one dotted half note or two dotted quarter notes, or six eighths, or twelve sixteenths. We could also have a rhythm of quarter, eighth, or eighth, quarter. Or any combination that adds up to six eighth notes. If a composer wanted four notes during a dotted quarter note, the number four would go above or below the group of notes. As you can see, the notation of rhythm can become very complicated. We will discuss this in later lessons." + }, + { + "Q": "what does circumscribed mean?", + "A": "2. GEOMETRY draw (a figure) around another, touching it at points but not cutting it", + "video_name": "KjQ1KN5GgoE", + "transcript": "Line AC is tangent to circle O at point C. So this is line AC, tangent to circle O at point C. What is the length of segment AC? What is this distance right over here, between point A and point C? And I encourage you now to pause this video and try this out on your own. So I'm assuming you've given a go at it. So the key thing to realize here, since AC is tangent to the circle at point C, that means it's going to be perpendicular to the radius between the center of the circle and point C. So this right over here is a right angle. And the reason why that is useful is now we know that triangle AOC is a right triangle. So if we know two of its sides, we could use the Pythagorean theorem to figure out the third. Now, we clearly know OC. Now OA, we don't know the entire side. They only give us that AB is equal to 2. But the thing that might jump out in your mind is OB is a radius. It's going to be the same length as any radius. So this is going to be 3 as well. It's the distance between the center of the circle and a point on the circle, just like the distance between O and C. So this is going to be 3 as well. And so now we are able to figure out that the hypotenuse of this triangle has length 5. And so we need to figure out what the length of segment AC is. So let's just call that, I don't know. I'll call that x. And so we know that x squared plus 3 squared-- I'm just applying the Pythagorean theorem here-- is going to be equal to the length of the hypotenuse squared, is going to be equal to 5 squared. And I know this is the hypotenuse. It's the side opposite the 90-degree angle. It's the longest side of the right triangle. So x squared plus 9 is equal to 25. Subtract 9 from both sides, and you get x squared is equal to 16. And so it should jump out at you that x is going to be equal to 4. So x is equal to 4. x is the same thing as the length of segment AC, so the length of segment AC is 4." + }, + { + "Q": "at 5:07 i got confused.", + "A": "There, 15 goes into 75 ----> 5 times r 1. To review it, when he was writing 5*5 = 25 in the division table, after writing 5, he takes 2 to add with (1*5). But, instead of writing 2 there, he faultily wrote it as 7. Then, after finding the error, he rectified it.", + "video_name": "gHTH6PKfpMc", + "transcript": "Welcome to the presentation on level 4 division. So what makes level 4 division harder than level 3 division is instead of having a one-digit number being divided into a multi-digit number, we're now going to have a two-addition number divided into a multi-digit number. So let's get started with some practice problems. So let's start with what I would say is a relatively straightforward example. The level 4 problems you'll see are actually a little harder than this. But let's say I had 25 goes into 6,250. So the best way to think about this is you say, OK, I have 25. Does 25 go into 6? Well, no. Clearly 6 is smaller than 25, so 25 does not go into 6. So then ask yourself, well, then if 25 doesn't go into 6, does 25 go into 62? Well, sure. 62 is larger than 25, so 25 will go into 62? Well, let's think about it. 25 times 1 is 25. 25 times 2 is 50. So it goes into 62 at least two times. And 25 times 3 is 75. So that's too much. So 25 goes into 62 two times. And there's really no mechanical way to go about figuring this out. You have to kind of think about, OK, how many times do I think 25 will go into 62? And sometimes you get it wrong. Sometimes you'll put a number here. Say if I didn't know, I would've put a 3 up here and then I would've said 3 times 25 and I would've gotten a 75 here. And then that would have been too large of a number, so I would have gone back and changed it to a 2. Likewise, if I had done a 1 and I had done 1 tmes 25, when I subtracted it out, the difference I would've gotten would be larger than 25. And then I would know that, OK, 1 is too small. I have to increase it to 2. I hope I didn't confuse you too much. I just want you to know that you shouldn't get nervous if you're like, boy, every time I go through the step it's kind of like- I kind of have to guess what the numbers is as opposed to kind of a method. And that's true; everyone has to do that. So anyway, so 25 goes into 62 two times. Now let's multiply 2 times 25. Well, 2 times 5 is 10. And then 2 times 2 plus 1 is 5. And we know that 25 times 2 is 50 anyway. Then we subtract. 2 minus 0 is 2. 6 minus 5 is 1. And now we bring down the 5. So the rest of the mechanics are pretty much just like a level 3 division problem. Now we ask ourselves, how many times does 25 go into 125? Well, the way I think about it is 25-- it goes into 100 about four times, so it will go into 125 one more time. It goes into it five times. If you weren't sure you could try 4 and then you would see that you would have too much left over. Or if you tried 6 you would see that you would actually get 6 times 25 is a number larger than 125. So you can't use 6. So if we say 25 goes into 125 five times then we just multiply 5 times 5 is 25. 5 times 2 is 10 plus 2, 125. So it goes in exact. So 125 minus 125 is clearly 0. Then we bring down this 0. And 25 goes into 0 zero times. 0 times 25 is 0. Remainder is 0. So we see that 25 goes into 6,250 exactly 250 times. Let's do another problem. Let's say I had-- let me pick an interesting number. Let's say I had 15 and I want to know how many times it goes into 2,265. Well, we just do the same thing we did before. We say OK, does 15 go into 2? No. So does 15 go into 22? Sure. 15 goes into 22 one time. Notice we wrote the 1 above the 22. If it go had gone into 2 we would've written the 1 here. But 15 goes into 22 one time. 1 times 15 is 15. 22 minus 15-- we could do the whole carrying thing-- 1, 12. 12 minus 5 is 7. 1 minus 1 is 0. 22 minus 15 is 7. Bring down the 6. OK, now how many times does 15 go into 76? Once again, there isn't a real easy mechanical way to do it. You can kind of eyeball it and estimate. Well, 15 times 2 is 30. 15 times 4 is 60. 15 times 5 is 75. That's pretty close, so let's say 15 goes into 76 five times. So 5 times 5 once again, I already figured it out in my head, but I'll just do it again. 5 times 1 is 5. Plus 7. Oh, sorry. 5 times 5 is 25. 5 times 1 is 5. Plus 2 is 7. Now we just subtract. 76 minus 75 is clearly 1. Bring down that 5. Well, 15 goes into 15 exactly one time. 1 times 15 is 15. Subtract it and we get a remainder of 0. So 15 goes into 2,265 exactly 151 times. So just think about what we're doing here and why it's a little bit harder than when you have a one-digit number here. Is that you have to kind of think about, well, how many times does this two-digit number go into this larger number? And since you don't know two-digit multiplication tables-- very few people do-- you have to do a little bit of guesswork. Sometimes you can look at this first digit and look at the first digit here and make an estimate. But sometimes it's trial and error. You'll try and when you multiply it out you might get it wrong on the first try. Let's do another problem. And actually, I'm going to pick numbers at random, so it might not have an easy remainder. But I think you'll get the point. I won't teach you decimals now, so I'll just leave the remainder if there is one. Let's say I had 67 going into 5,978. So I just picked these numbers randomly out of my head, so I'll show you that I also sometimes have to do a little bit of guesswork to figure out how many times one of these two-digit numbers go into a larger number. So 67 goes into 5 zero times. 67 goes into 59 zero times. 67 goes into 597-- so let's see. 67 is almost 70 and 597 is almost 600. So if it was 70 goes into-- 70 times 9 to 630. Because 7 times 9 is 63. So I'm going to just eyball approximate. I'm going to say that it goes into it eight times. I might be wrong. And you can always check, but well, we're going to actually check in this step essentially. 8 times 7-- well that's 56. And then 8 times 6 is 48. Plus 2 is 53. 7 minus 6 is 1. 9 minus 9 is 6. 5 minus 5 is 0. 61. So good. I got it right because if I got a number here that was larger than-- 67 or larger, than that means that this number up here wasn't large enough. But here, I got a number that's positive because 536 is less than 597. And it's less than 67, so I did that step right. So now we bring down this 8. Now this one might be a little bit trickier this time. Once again, we have almost 70 and here we have almost 630. So maybe it will go into it 9 times. Well, let's give it a try and see if it does. 9 times 7 is 63. 9 times 6 is 54. Plus 6 is 60. Good. So it did actually go into it nine times because 603 is less than 618. 8 minus 3 is 5. 1 minus 0 is 1. And 6 minus 6 is 0. We have a remainder of 15, which is smaller than 67. So I'm not going to teach you decimals right now, so we can just leave that remainder. So what we could say is that 67 goes into 5,978 89 times. And when it goes into it 89 times, you're left with a remainder of 15. hopefully you're ready now to try some level 4 division problems. Have fun." + }, + { + "Q": "I have a question: if the quantity supplied of an item in the market decreases, according to the law of supply, the price would decrease too. But wouldn't the fact that it is less abundant, mean that the price actually goes up, as there is less of it.", + "A": "Well, it depends on the demand for the product. If it is a private good, often highly demanded, a reduction in supply will result in higher demand, which increases the inelasticity of the good, meaning the firm can confidently raise their prices. So yea, it comes down to the product and it s demand in home and overseas markets.", + "video_name": "3xCzhdVtdMI", + "transcript": "We've talked a lot about demand. So now let's talk about supply, and we'll use grapes as this example. We'll pretend to be grape farmers of some sort. So I will start by introducing you-- and maybe I'll do it in purple in honor of the grapes-- to the law of supply, which like the law of demand, makes a lot of intuitive sense. If we hold all else equal-- in the next few videos, we'll talk about what happens when we change some of those things that we're going to hold equal right now-- but if you hold all else equal and the only thing that you're doing is you're changing price, then the law of supply says that if the price goes up-- I'll just say p for price-- if the price goes up, then the supply-- now, let me be careful-- the quantity supplied goes up. And then you can imagine, if the price goes down, the quantity supplied goes down. And you might already notice that I was careful to say quantity supplied. And it's just like we saw with demand. When we talk about demand going up or down, we're talking about the entire price-quantity relationship shifting. When we talk about a particular quantity demanded, we say quantity demanded. We don't just say demand. This is the exact same thing for supply. When we're talking about a particular quantity, we'll be careful to say quantity. If we talk about supply increasing, we're talking about the entire relationship shifting either up or down. So let's just make sure that this makes intuitive sense for us. And I think it probably does. Let's think about ourselves as grape farmers. And I'll make a little supply schedule right over here. So Grape Supply Schedule, which is really just a table showing the relationship between, all else equal, the price and the quantity supplied. So let's label some scenarios over here, just like we did with the demand schedule. Scenarios. And then let's put our Price over here. This will be in price per pound, the per pound price of grapes. And then this is the quantity produced over the time period. And whenever we do any of these supply or demand schedules, we're talking over a particular time period. It could be per day, it could be per month, it could be per year. But that's the only way to make some sense of, OK, what is the quantity per day going to be produced if that's the price? So if we didn't say per day, we don't know what we're really talking about. Quantity Supplied. And so let's just say Scenario A, if the price per pound of grapes is $0.50-- if it's $0.50 per pound-- actually, let me just do round numbers, but you get the idea. If the price per pound is $1, let's just say for us, we consider that to be a relatively low price. And so we'll only kind of do the easiest land, our most fertile land, where it's easy to produce grapes. And maybe the fertile-- and sheep land. So no one else wants to use that land for other things. It's only good for growing grapes. And so we will provide-- so this is price per pound. And in that situation, we can produce 1,000 pounds in this year. And I've never been a grape farmer, so I actually don't know if that's a reasonable amount or not, but I'll just go with it, 1,000 pounds. Now, let's take Scenario B. Let's say the price goes up to $2. Well now, not only would we produce what we were producing before, but we might now want to maybe buy some more land, land that might have had other uses, land that's maybe not as productive for grapes. But we would, because now we can get more for grapes. And so maybe now we are willing to produce 2,000 pounds. And we can keep going. The same dynamics keep happening. So let's say the price-- if the price were $3 per pound, now we do want to produce more. Maybe we're even willing to work a little harder or plant things closer to each other, or maybe I'll get even more land involved than I would have otherwise used for other crops. And so then I'm going to produce 2,500 pounds. And I'll do one more scenario. Let's say Scenario D, the price goes to $4 a pound. Same dynamic, I will stop planting other crops, use them now for grapes, because grape prices are so high. And so I will produce 2,750 pounds. And so we can draw a supply curve just like we have drawn demand curves. And it's the same exact convention, which I'm not a fan of, putting price on the vertical axis. Because as you see, we tend to talk about price as the independent variable. We don't always talk about it that way. And in most of math and science, you put the independent variable on the horizontal axis. But the convention in economics is to put it on the vertical axis. So price on the vertical axis. So then this is really Price per pound. And then in the horizontal axis, Quantity Produced, or-- let me just write it. Quantity Produced, I'll say in the next year. We're assuming all of this is for the next year, so next year. And it's in thousands of pounds, so I'll put it in thousands of pounds. And so let's see, we go all the way from 1,000 to close to 3,000. So let's say this is 1,000, that's 1 for 1,000, that's 2,000, and that is 3,000. And then the price goes all the way up to 4. So it's 1, 2, 3, and then 4. So we can just plot these points. These are specific points on the supply curve. So at $1, we would supply 1,000 pounds, at $1, 1,000 pounds. That's Scenario A. At $2, we would supply 2,000 pounds, $2, we'd supply 2,000 pounds. That's scenario B. At $3, we'd supply 2,500 pounds, $3-- oh, Now, when we look up-- See, now notice, I get my axes confused. This is Price. This isn't, when we talk about it this way, that we're viewing the thing that's changing. Although, you don't always have to do it that way. So at one $1, 1,000 pounds. $1, 1,000 pounds. $2, 2,000 pounds. $2, 2,000 pounds. $3-- this isn't $3, this is $3. $3, 2,500 pounds. So right about there. That's about 2,500. But I want to do it in that blue color, so we don't get confused. So $3, 2,500 pounds. That's about right. So this is Scenario C. And then Scenario D, at $4-- actually, let me be a little bit clearer with that, because we're getting close. So this is 2,500 pounds, gets us right over here. This is Scenario C. And then Scenario D at $4, 2,750. So 2,750 is like right over there. So that is $4. That is Scenario D. And if we connect them, they should all be on our supply curve. So they will all be-- it will look something like that. And there's some minimum price we would need to supply some grapes at all. We wouldn't give them away for free. So maybe that's something-- that minimum price is over here, that just even gets started producing grapes. So this right over here is what our supply curve would look like. Now remember, the only thing we're varying here is the price. So if the price were to change, all else equal, we would move along this curve here. Now, in the next few videos, I'll talk about all those other things we've been holding equal and what they would do at any given price point to this curve or, in general, what they would do to the curve." + }, + { + "Q": "If Hitler hated communists so much, why didn't he alliance with France, Great Britain, and the United States to to attack the USSR?", + "A": "First off, France, Great Britain and the United States wanted peace in Europe, they would NEVER support his future Operation Barbarossa which would launch Europe into war. Second, Hitler wanted to conquer all of Europe, he didn t want to be allies with them. He was fueled more by his desire to expand than his hatred of Communism. Hope this helped :) Happy holidays!!", + "video_name": "X3bqQI7-sCg", + "transcript": "- [Voiceover] As we've already seen in the last few videos, with the war officially starting in September of 1939, the Axis powers get momentum through the end of 1939, all the way into 1940. That was the last video that we covered and that takes us to 1941 and what we're gonna see in 1941 which is the focus of this video is that the Axis powers only seem to gain more momentum. Because of all of that momentum they perhaps gets a little bit overconfident and stretch themselves or begin to stretch themselves too thin. So let's think about what happens in 1941. So, if we talk about early 1941 or the Spring of 1941, in March, Bulgaria decides to join the Axis powers. You can imagine there's a lot of pressuring applied to them and they kind of see where the momentum is. Let's be on that side. Bulgaria joins the Axis and then in North Africa you might remember that in 1940, the Allies, in particular, the British, were able to defeat the Italians and push them back into Libya but now in March of 1941, the Italians get reinforcements, Italian reinforcements and also German reinforcements under the command of Rommel the Desert Fox, famous desert commander and they are able to push the British back to the Egyptian border and they also take siege of the town of Tobruk. Now, you might have noticed something that I just drew. The supply lines in the North Africa campaign are very, very, very long and that's part of the reason why there's one side. One side has supply lines and as they start to make progress and as the Allies make progress and push into Libya, their supply lines got really long and so the other side has an easier timely supply. Then as the Axis pushes the Allies back into Egypt, then their supply lines get really long and the other side...it makes it easier for them to resupply and so North Africa is kind of defined by this constant back and forth. But, by early 1941, it looks like the Axis is on the offensive, able to push the British back into Egypt lay siege to the town of Tobruk. So, let me write this down as North Africa. So, I'll just say North Africa over here or I'll could say Rommel in North Africa pushing the British back. And then we can start talking about what happens in the Balkans and this is still in Spring as we go into April of 1941 and just as a little bit of background here, and frankly I should have covered it a couple of videos ago. As far back as 1939, actually before World War II officially started, in Spring of 1939, Italy actually occupies Albania so this actually should have already been red. This is in 1939 that this happens and then at the end of 1940, Italy uses Albania as a base of operation to try to invade Greece but they are pushed back. Actually one of the reasons why the British we able to be pushed back in North Africa is after they were successful against the Italians, most of the bulk of the British forces we sent to Greece to help defend Greece at the end of 1940. So, in 1939, Albania gets taken over by Italy and at the end of 1940...October 1940, Greece is invaded by Italy but they are then pushed back but to help the Greeks, the Allies send many of the forces that were in North Africa after they were successful against the Italians in Libya. Now, as we go into April of 1941... that was all background, remember Albania before the war started in April 1939, October 1940 was Italy's kind of first push into Greece and it was unsuccessful. Then the Greeks get support from the Allies in North Africa and now as we go into 1941, the Germans start supporting and really take charge in Balkans and in Greece and so with the help of the Germans the Axis is able to take over Yugoslavia and Greece and start aerial bombardment of Crete. So, once again, we're not even halfway through the year in 1941 and we see a huge swath of Europe is under the control of the Axis powers. And now we go into the summer of 1941. This is actually a pivotal move, what's about to happen. Now you can imagine that the Axis powers, in particular, Hitler, are feeling pretty confident. We are only about that far into the war. So we're not even two years into the war yet and it looks like the Axis is going to win. Now you might remember that they have a pact with the Soviet Union. Hey, we're gonna split a lot of Eastern Europe into our spheres of influence so to speak, but now Hitler's like, well, I think I'm ready to attack and when you attack the Soviet Union really matters. You do not want to attack the Soviet Union in the winter...or Russia in the winter. Russia's obviously at the heart of the Soviet Union. That something that Napoleon learned. Many military commanders have learnt. You do not want to be fighting in Russia over the winter, so summer of 1941, Hitler figures, hey, this is the Axis chance. And so, in June, he decides to attack the Soviet Union. So, this is a very, very, very bold move because now they're fighting the British. Remember, the British are kind of not a joke to be battling out here in Western Europe and now they're going to be taking on the Soviet Union in the east, a major, major world power. But at first, like always, it seems like it's going well for the Germans. By September, they're able to push up all the way to Leningrad. So, this is September of 1941 and lay lay siege and begin laying siege to that town. This is kind of a long bloody siege that happens there. So, we're right now, right about there. And most historians would tell you that this was one of the mistakes of Adolph Hitler because now he is stretched very, very, very thin. He has to fight two world powers, Soviet Union and Great Britain and the United States hasn't entered into the war yet and that's what we're about to get into because if we go into Asia it was still in 1941 what happens in July. So, little bit after Hitler decides to start invading the Soviet Union going back on the pact, the non- aggression pact. In July, you could imagine the US, they were never pleased with what's been happening, what the Empire of Japan has been doing in the Pacific, what they've been doing in China, in Manchuria or even in terms of the war in China, the second Sino Japanese War. They weren't happy of the Japanese taking over French Indochina. There's a big world power here, the Empire of Japan. There's a big world power here, the United States, that has a lot of possessions in the Pacific and so, the United States in July of 1941...So remember, this is still all 1941, this is the same year...decides to freeze the assets of Japan and probably the most important part of that was an oil embargo of Japan. This is a big, big deal. Japan is fighting a major conflict with the Chinese. It's kind of flexing it's imperial muscles but it does not have many natural resources in and of itself and in fact, that's one of the reasons why it's trying to colonize other places to get more control of natural resources. And now if it's fighting a war it doesn't have it's own oil resources and now there's an oil embargo of Japan and the United States at the time was major oil producer and even today, it's major oil producer. This was a big deal to the Japanese because some estimates say they only had about two years of reserves and they were fighting a war where they might have to touch their reserves even more. So, you could imagine the Japanese, they want to have their imperial ambitions. They probably want, especially now with this oil embargo, they probably want to take over more natural resources and they probably want to knock out the US or at least keep the US on its heels so the US can't stop Japan from doing what it wants to do. So, all it wants in December 1941, that's over the course of December 7th and 8th, and it gets a little confusing because a lot of this happens across the International Date Line. But over the course of December 7th and 8th, Japan goes on the offensive in a major way in the Pacific. Over the course of several hours, at most, a day, Japan is able to attack Malaya, which is a British possession. It's able to attack Pearl Harbor, where the US Pacific fleet is in hope to knock out the US Pacific fleet so the US will have trouble stopping Japan from doing whatever Japan wants to do. In the US, we focus a lot on Pearl Harbor but this was just one of the attacks in this whole kind of several hours of attacks where Japan went on the offensive. So, we have Malaya, we have Pearl Harbor, we have Singapore, we have Guam, we have (which was the US military base), Wake Island. was a US possession ever since the Spanish American war. You have Hong Kong, which is a British possession and then shortly after that as you get further into December, so this is kind of when you have Japan offensive. Then as you go on into later Decemeber, the kind of real prize for Japan was what we would now call Indonesia but the Dutch East Indies. On this map it says Netherlands East Indies. You have to remember the Netherlands had been overrun. They're the low countries they were already overrun by German forces so the Japanese say hey, look there are a lot of resources here, natural resources, especially oil. Let's go for this and so by the end of 1941, they're also going for the Dutch East Indies and for Burma so you could imagine it's a very aggressive, very, very bold move on Japan but they kind of had imperial ambitions. They were afraid of they access to natural resources so they went for it but obviously one of the major consequence of this is the United States was not happy about this and they were already sympathetic to the Allies. They didn't like what was going on in Europe either. They didn't like what was going on in China and so that causes the United States to enter into World War II on the side of the Allies and then the Axis powers to declare war on the United States, which was a big deal." + }, + { + "Q": "Did Sal make this website. We'll anyway, Salman is awesome!", + "A": "Yes he did", + "video_name": "jxA8MffVmPs", + "transcript": "Find the place value of 3 in 4,356. Now, whenever I think about place value, and the more you do practice problems on this it'll become a little bit of second nature, but whenever I see a problem like this, I like to expand out what 4,356 really is, so let me rewrite So if I were to write it-- and I'll write it in different colors. So 4,356 is equal to-- and just think about how I just said it. It is equal to 4,000 plus 300 plus 50 plus 6. And you could come up with that just based on how we said it: four thousand, three hundred, and fifty-six. Now another way to think about this is this is just like saying this is 4 thousands plus-- or you could even think of \"and\"-- so plus 3 hundreds plus 50, you could think of it as 5 tens plus 6. And instead of 6, we could say plus 6 ones. And so if we go back to the original number 4,356, this is the same thing as 4-- I'll write it down. Let me see how well I can-- I'll write it up like this. This is the same thing is 4 thousands, 3 hundreds, 5 tens and then 6 ones. So when they ask what is the place value of 3 into 4,356, we're concerned with this 3 right here, and it's place value. It's in the hundreds place. If there was a 4 here, that would mean we're dealing with 4 hundreds. If there's a 5, 5 hundreds. It's the third from the right. This is the ones place. That's 6 ones, 5 tens, 3 hundreds. So the answer here is it is in the hundreds place." + }, + { + "Q": "Hi\nWhen I have substituted the value of vi in the eq s= ut +1/2at^2, I have got the answer correct with 2.5 sec, but when I am doing it with t= 5sec I am getting zero as answer and with your equation taken t=5s I get72.5 m kindly clarify", + "A": "The equation he uses only works for the first half of the trajectory. To work for the second half, you have to find the average velocity for the ball going down, now. The average velocity will be the same value, but now in the negative direction (Vavg = -12.25 m/s). Using his equation, you have to use t = 2.5 s ( cause it s the time for the second half), and the displacement will be s = -30.625 m. Since the ball was at the height of 30.625 m, the final position would be 0 m.", + "video_name": "IYS4Bd9F3LA", + "transcript": "Let's say you and I are playing a game or I'm trying to figure out how high a ball is being thrown in the air How fast would we throwing that ball in the air? And what we do is one of us has a ball and the other one has a stop watch over here So this is my best attempt to make it more like a cat than a stop watch but I think you get the idea And what we do is one of us throw the ball the other one times how long the ball is in the air And what we do is gonna use that time in the air to figure out how fast the ball was thrown straight up and how long it was in the air or how high it got And there is going to be one assumption I make here frankly that's an assumption we are gonna make in all of these projectile motion type problem is that air resistance is negligible And for something like it, this is a baseball or something like that That's a pretty good approximation So when can I get the exact answer, I encourage you experiment it on your own or even to see what air resistance does to your calculations We gonna assume for this projectile motion in future one at least in the basic Physics playlist We gonna assume air resistance is negligible And what that does for us is we can assume that the time up That the time for the ball to go up to its peak height is the same thing as the time that takes it to go down If you look at this previous video, we've plot it displacement verse time You see after 2 seconds the ball went from being on the ground or I guess the thrower's hand all the way to its peak height And then the next 2 seconds it took the same amount of time to go back down to the ground which makes sense whatever the initial velocity is, it take half the time to go to zero and it takes the same amount of time to now be accelerated into downward direction back to that same magnitude of velocity but now in the downward direction So let's play around with some numbers here Just so you get a little bit more of concrete sense So let's say I throw a ball in the air And you measure using the stop watch and the ball is in the air for 5s So how do we figure out how fast I threw the ball? Well the first thing we could do is we could say look at the total time in the air was 5 seconds that mean the time, let me write it, that means the change in time to go up during the first half, I guess the ball time in the air is going to be 2.5 seconds and which tells us that over this 2.5 seconds we went from our initial velocity, whatever it was We went from our initial velocity to our final velocity which is a velocity of 0 m/s in the 2.5 seconds And this is a graph for that example, This is the graph for the previous one, The previous example we knew the initial velocity but in whatever the time is you are going from you initial velocity to be stationery at the top, right with the ball being stationery and then start getting increasing velocity in the downward direction So it takes 2.5 seconds to go from some initial velocity to 0 seconds So we do know what the acceleration of the gravity is We know that the acceleration We know the acceleration of gravity here, we are assuming it's constant or slightly not constant but we are gonna assume it's constant We are just dealing close to the surface of the earth is negative 9.8 m/s*s, so let's think about it This change in velocity, are change in velocity Their change in velocity is the final velocity minus the initial velocity which is the same thing as zero minus the initial velocity which is the negative of the initial velocity That's another way to think about change in velocity We just shown the definition of acceleration change in velocity is equal to acceleration, is equal to acceleration negative 9.8 m/s*s times time or times change in time, our change in time, we are just talking about the first half of the ball's time in the air So the change in time is 2.5 s, times 2.5 s So what is our change in velocity which is also the same thing as negative of our initial velocity Get the calculator out, let me get my calculator, bring it on to the screen, so it is negative 9.8 m/s times 2.5 s Times 2.5 s, it gives us negative 24.5, so this gives us I will write it in new color This gives us negative 24.5 m/s, this seconds cancels out With one of these seconds in the denominator we only have one of the denominator out m/s, and this is the same thing as the negative, as the negative initial velocity Negative initial velocity that's the same thing as change in velocity So you multiply both side by a negative, you will get our initial velocity So that simply we are able to figure out what our velocity is So literally you take the time, the total time in the air divide by two And multiply that by acceleration of gravity and if you take I guess you can take the absolute value of that or take the positive version of that And that gives you your initial velocity So your initial velocity here is literally 24.5 m/s Since it's a positive quantity it is upwards in this example So that's my initial velocity, so we already figure out part of this game The initial velocity that threw upward That's also going to be, we gonna have the same magnitude of velocity The balls about to hit the ground although is gonna be in the other direction So what is the distance or let me make it clear what is the displacement of the ball from its lowest point right when it leaves your hand all the way to the peak, all the way to the peak? We just have to remember, all of these come from very straight forward ideas Change in velocity is equal to acceleration times change in time And the other simple idea is that displacement is equal to average velocity, average velocity times change in time Now what is our average velocity? Our average velocity is your initial velocity plus your final velocity Divided by 2, or we assume that acceleration is constant So literally just the arithmetic mean of your initial and final velocity So what is that? That's gonna be 24.5 m/s plus our final velocity In this situation we are just going over to the first 2.5 s So our final velocity is once again 0 m/s We are just talking about when we get to his point over here So our final velocity is just 0 m/s And we just gonna divide that by 2 This will give us the average velocity And we wanna multiply that by 2.5 s, times 2.5 s So we get this part right over here 24.5 divided by 2 When you go with the 0, it is still 24.5 It gives us 12.25 times 2.5, and remember this right over here is in seconds let me write the units down So this is 12.25 m/s times 2.5 seconds And just to remind ourselves We are calculating the displacement over the first 2.5 seconds So this gives us, I get the calculator out once again We have 12.25 times 2.5 seconds gives us 30.625 So this gives us, this gives us, the displacement is 30.625 m The seconds cancel out This is actually a ton, you know, roughly give or take about 90 feet throw into the air, this looks like a nine stories building And I frankly do not have the arm for that But if someone is able to throw the ball for 5 seconds in the air They have thrown 30 meters in the air Hopefully you will find that entertaining In my next video I'll generalize this maybe we can get a little bit of formula so maybe you can generalize it So regardless of the measurement of time you can get the displacement in the air Or even better, try to derive it yourself And you will see how, at least how I tackle it in the next video" + }, + { + "Q": "Is it possible in a real life situation for an atom like chlorine to give 7 valence electrons to Sodium to create a new element ClNa", + "A": "No, that s not possible! But, even if it were, it would not be a new element because what you describe is two atoms and the term element refers to a single atom with a single nucleus.", + "video_name": "Rr7LhdSKMxY", + "transcript": "Voiceover: What I want to talk about in this video are the notions of Electronegativity, electro, negati, negativity, and a closely, and a closely related idea of Electron Affinity, electron affinity. And they're so closely related that in general, if something has a high electronegativity, they have a high electron affinity, but what does this mean? Well, electron affinity is how much does that atom attract electrons, how much does it like electrons? Does it want, does it maybe want more electrons? Electronegativity is a little bit more specific. It's when that atom is part of a covalent bond, when it is sharing electrons with another atom, how likely is it or how badly does it want to hog the electrons in that covalent bond? Now what do I mean by hogging electrons? So let me make, let me write this down. So how badly wants to hog, and this is an informal definition clearly, hog electrons, keep the electrons, to spend more of their time closer to them then to the other party in the covalent bond. And this is how, how much they like electrons, or how much affinity they have towards electrons. So how much they want electrons. And you can see that these are very, these are very related notions. This is within the context of a covalent bond, how much electron affinity is there? Well this, you can think of it as a slightly broader notion, but these two trends go absolutely in line with each other. And to think about, to just think about electronegativity makes it a little bit more tangible. Let's think about one of the most famous sets of covalent bonds, and that's what you see in a water molecule. Water, as you probably know, is H two O, you have an oxygen atom, and you have two hydrogens. Each of the hydrogen's have one valence electron, and the oxygen has, we see here, at it's outermost shell, it has one, two, three, four, five, six valence electrons. One, two, three, four, five, six valence electrons. And so you can imagine, hydrogen would be happy if it was able to somehow pretend like it had another electron then it would have an electron configuration a stable, first shell that only requires two electrons, the rest of them require eight, hydrogen would feel, hey I'm stable like helium if it could get another electron. And oxygen would feel, hey I'm stable like neon if I could get two more electrons. And so what happens is they share each other's electrons. This, this electron can be shared in conjunction with this electron for this hydrogen. So that hydrogen can kind of feel like it's using both and it gets more stable, it stabilizes the outer shell, or it stabilizes the hydrogen. And likewise, that electron could be, can be shared with the hydrogen, and the hydrogen can kind of feel more like helium. And then this oxygen can feel like it's a quid pro quo, it's getting something in exchange for something else. It's getting the electron, an electron, it's sharing an electron from each of these hydrogens, and so it can feel like it's, that it stabilizes it, similar to a, similar to a neon. But when you have these covalent bonds, only in the case where they are equally electronegative would you have a case where maybe they're sharing, and even there what happens in the rest of the molecule might matter, but when you have something like this, where you have oxygen and hydrogen, they don't have the same electronegativity. Oxygen likes to hog electrons more than hydrogen does. And so these electrons are not gonna spend an even amount of time. Here I did it kind of just drawing these, you know, these valence electrons as these dots. But as we know, the electrons are in this kind of blur around, around the, around the actual nuclei, around the atoms that make up the atoms. And so, in this type of a covalent bond, the electrons, the two electrons that this bond represents, are going to spend more time around the oxygen then they are going to spend around the hydrogen. And these, these two electrons are gonna spend more time around the oxygen, then are going to spend around the hydrogen. And we know that because oxygen is more electronegative, and we'll talk about the trends in a second. This is a really important idea in chemistry, and especially later on as you study organic chemistry. Because, because we know that oxygen is more electronegative, and the electrons spend more time around oxygen then around hydrogen, it creates a partial negative charge on this side, and partial positive charges on this side right over here, which is why water has many of the properties that it does, and we go into much more in depth in that in other videos. And also when you study organic chemistry, a lot of the likely reactions that are going to happen can be predicted, or a lot of the likely molecules that form can be predicted based on elecronegativity. And especially when you start going into oxidation numbers and things like that, electronegativity will tell you a lot. So now that we know what electronegativity is, let's think a little bit about what is, as we go through, as we start, as we go through, as we go through a period, as say as we start in group one, and we go to group, and as we go all the way all the way to, let's say the halogens, all the way up to the yellow column right over here, what do you think is going to be the trend for electronegativity? And once again, one way to think about it is to think about the extremes. Think about sodium, and think about chlorine, and I encourage you to pause the video and think about that. Assuming you've had a go at it, and it's in some ways the same idea, or it's a similar idea as ionization energy. Something like sodium has only one electron in it's outer most shell. It'd be hard for it to complete that shell, and so to get to a stable state it's much easier for it to give away that one electron that it has, so it can get to a stable configuration like neon. So this one really wants to give away an electron. And we saw in the video on ionization energy, that's why this has a low ionization energy, it doesn't take much energy, in a gaseous state, to remove an electron from sodium. But chlorine is the opposite. It's only one away from completing it's shell. The last thing it wants to do is give away electron, it wants an electron really, really, really, really badly so it can get to a configuration of argon, so it can complete it's third shell. So the logic here is that sodium wouldn't mind giving away an electron, while chlorine really would love an electron. So chlorine is more likely to hog electrons, while sodium is very unlikely to hog electrons. So this trend right here, when you go from the left to the right, your electronegativity, let me write this, your getting more electronegative. More electro, electronegative, as you, as you go to the right. Now what do you think the trend is going to be as you go down, as you go down in a group? What do you think the trend is going to be as you go down? Well I'll give you a hint. Think about, think about atomic radii, and given that, what do you think the trend is? Are we gonna get more or less electronegative as we move down? So once again I'm assuming you've given a go at it, so as we know, from the video on atomic radii, our atom is getting larger, and larger, and larger, as we add more and more and more shells. And so cesium has one electron in it's outer most shell, in the sixth shell, while, say, lithium has one electron. Everything here, all the group one elements, have one electron in it's outer most shell, but that fifty fifth electron, that one electron in the outer most shell in cesium, is a lot further away then the outer most electron in lithium or in hydrogen. And so because of that, it's, well one, there's more interference between that electron and the nucleus from all the other electrons in between them, and also it's just further away, so it's easier to kind of grab it off. So cesium is very likely to give up, it's very likely to give up electrons. It's much more likely to give up electrons than hydrogen. So, as you go down a given group, you're becoming less, less electronegative, electronegative. So what, what are, based on this, what are going to be the most electronegative of all the atoms? Well they're going to be the ones that are in the top and the right of the periodic table, they're going to be these right over here. These are going to be the most electronegative, Sometimes we don't think as much about the noble gases because they aren't, they aren't really that reactive, they don't even form covalent bond, because they're just happy. While these characters up here, they sometimes will form covalent bonds, and when they do, they really like to hog those electrons. Now what are the least electronegative, sometimes called very electropositive? Well these things down here in the bottom left. These, over here, they have only, you know in the case of cesium, they have one electron to give away that would take them to a stable state like, like xenon, or in the case of these group two elements they might have to give away two, but it's much easier to give away two then to gain a whole bunch of them. And they're big, they're big atoms. So those outer most electrons are getting less attracted to the positive nucleus. So the trend in the periodic table as you go from the bottom left, to the top right, you're getting more, more electro, electronegative." + }, + { + "Q": "So what do you do if you're given a hypotenuse that, when squared, has no combination of natural numbers squared to equal itself? How do you find the length of the two missing legs?", + "A": "In a 45-45-90, you have an isosceles triangle, which means the two legs are equal. Taking the Pythagorean formula a^2 + b^2 = c^2, we will now be able to simplify it to a^2 + a^2 = c^2 or 2\u00e2\u0080\u00a2a^2 = c^2. Given the hypotenuse, isolate equation for a: a^2 = c^2/2 a = sqrt(c^2/2)", + "video_name": "McINBOFCGH8", + "transcript": "In the last video, we showed that the ratios of the sides of a 30-60-90 triangle are-- if we assume the longest side is x, if the hypotenuse is x. Then the shortest side is x/2 and the side in between, the side that's opposite the 60 degree side, is square root of 3x/2. Or another way to think about it is if the shortest side is 1-- Now I'll do the shortest side, then the medium size, then the longest side. So if the side opposite the 30 degree side is 1, then the side opposite the 60 degree side is square root of 3 times that. So it's going to be square root of 3. And then the hypotenuse is going to be twice that. In the last video, we started with x and we said that the 30 degree side is x/2. But if the 30 degree side is 1, then this is going to be twice that. So it's going to be 2. This right here is the side opposite the 30 degree side, opposite the 60 degree side, and then the hypotenuse opposite the 90 degree side. And so, in general, if you see any triangle that has those ratios, you say hey, that's a 30-60-90 triangle. Or if you see a triangle that you know is a 30-60-90 triangle, you could say, hey, I know how to figure out one of the sides based on this ratio right over here. Just an example, if you see a triangle that looks like this, where the sides are 2, 2 square root of 3, and 4. Once again, the ratio of 2 to 2 square root of 3 is 1 to square root of 3. The ratio of 2 to 4 is the same thing as 1 to 2. This right here must be a 30-60-90 triangle. What I want to introduce you to in this video is another important type of triangle that shows up a lot in geometry and a lot in trigonometry. And this is a 45-45-90 triangle. Or another way to think about is if I have a right triangle that is also isosceles. You obviously can't have a right triangle that is equilateral, because an equilateral triangle has all of their angles have to be 60 degrees. But you can have a right angle, you can have a right triangle, that is isosceles. And isosceles-- let me write this-- this is a right isosceles triangle. And if it's isosceles, that means two of the sides are equal. So these are the two sides that are equal. And then if the two sides are equal, we have proved to ourselves that the base angles are equal. And if we called the measure of these base angles x, then we know that x plus x plus 90 have to be equal to 180. Or if we subtract 90 from both sides, you get x plus x is equal to 90 or 2x is equal to 90. Or if you divide both sides by 2, you get x is equal to 45 degrees. So a right isosceles triangle can also be called-- and this is the more typical name for it-- it can also be called a 45-45-90 triangle. And what I want to do this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle. And this one's actually more straightforward. Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x. And then we can use the Pythagorean Theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the square of length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. We can take the principal root of both sides of that. I wanted to just change it to yellow. Last, take the principal root of both sides of that. The left-hand side you get, principal root of 2 is just square root of 2, and then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2 is equal to c. So if you have a right isosceles triangle, whatever the two legs are, they're going to have the same length. That's why it's isosceles. The hypotenuse is going to be square root of 2 times that. So c is equal to x times the square root of 2. So for example, if you have a triangle that looks like this. Let me draw it a slightly different way. It's good to have to orient ourselves in different ways every time. So if we see a triangle that's 90 degrees, 45 and 45 like this, and you really just have to know two of these angles to know what the other one is going to be, and if I tell you that this side right over here is 3-- I actually don't even have to tell you that this other side's going to be 3. This is an isosceles triangle, so those two legs are going to be the same. And you won't even have to apply the Pythagorean theorem if you know this-- and this is a good one to know-- that the hypotenuse here, the side opposite the 90 degree side, is just going to be square root of 2 times the length of either of the legs. So it's going to be 3 times the square root of 2. So the ratio of the size of the hypotenuse in a 45-45-90 triangle or a right isosceles triangle, the ratio of the sides are one of the legs can be 1. Then the other leg is going to have the same measure, the same length, and then the hypotenuse is going to be square root of 2 times either of those. 1 to 1, 2 square root of 2. So this is 45-45-90. That's the ratios. And just as a review, if you have a 30-60-90, the ratios were 1 to square root of 3 to 2. And now we'll apply this in a bunch of problems." + }, + { + "Q": "you can use remainders", + "A": "Yeah, what s the question?", + "video_name": "xXIG8ouHcsc", + "transcript": "Let's divide 7,182 by 42. And what's different here is we're now dividing by a two-digit number, not a one-digit number, but the same idea holds. So we say, hey, how many times does 42 go into 7? Well, it doesn't really go into 7 at all, so let's add one more place value. How many times does 42 go into 71? Well, it goes into 71 one time. Just a reminder, whoever's doing the process where you say, hey, 42 goes into 71 one time. But what we're really saying, 42 goes into 7,100 100 times because we're putting this one in the hundreds place. But let's put that on the side for a little bit and focus on the process. So 1 times 42 is 42, and now we subtract. Now, you might be able to do 71 minus 42 in your head, knowing, hey, 72 minus 42 would be 30. So 71 minus 42 would be 29, but we could also do it by regrouping. To regroup, you want to subtract a 2 from a 1. You can't really do that in any traditional way. So let's take a 10 from the 70, so that it becomes a 60, and give that 10 to the ones place, and then that becomes an 11. And so 11 minus 2 is 9, and 6 minus 4 is 2. So you get 29. And we can bring down the next place value. Bring down an 8. And now, this is where the art happens when we're dividing by a multi-digit number right over here. We have to estimate how many times does 42 go into 298. And sometimes it might involve a little bit of trial and error. So you really just kind of have to eyeball it. If you make a mistake, try again. The way you know you make a mistake is, if say it goes into it 9 times, and you do 9 times 42 and you get a number larger than 298, then you overestimated. If you say it goes into it three times, you do 3 times 42, you get some number here. When you subtract, you get something larger than 42, then you also made a mistake, and you have to adjust upwards. Well, let's see if we can eyeball it. So this is roughly 40. This is roughly 300. 40 goes into 300 the same times as 4 goes into the 30, so it's going to be roughly 7. Let's see if that's right. 7 times 2 is 14. 7 times 1 is 28, plus 1 is 29. So I got pretty close. My remainder here-- notice 294 is less than 298. So I'm cool there. And my remainder is less than 42, so I'm cool as well. So now let's add another place value. Let's bring this 2 down. And here we're just asking ourselves, how many times does 42 go into 42? Well, 42 goes into 42 exactly one time. 1 times 42 is 42, and we have no remainder. So this one luckily divided exactly. 42 goes into 7,182 exactly 171 times." + }, + { + "Q": "What is vector xstar? What is its significance, and magnitude? I mean aside from what is explained in the video... I noticed that if you draw it, it is just there...", + "A": "xstar is the vector we need to multiply A by to find our solution, representing slope and y-intercept. Doing so multiplies the slope (x*0) by the x value of our points, and adds it (add in dot product is equivalent to add in mx + b in this case) to the y intercept (X* 1) times 1 to equal our y points.", + "video_name": "QkepM8Vv3kw", + "transcript": "So I've got four Cartesian coordinates here. This first one is minus 1, 0. I tried to draw them ahead of time. So minus 1, 0 is this point right there. Doing this in these new colors. The next point is a 0, 1, which is that point right there. Then the next point is 1, 2, which is that point right up there. And then the last point is 2, 1, which is that point there. Now my goal in this video is to find some line, y equals mx plus v, that goes through these points. Now the first thing I'd say is, hey Sal, there is not going to be any line that goes through these points, and you can see that immediately. You could find a line that maybe goes through these points, but it's not going to go through this point over here. If you try to make a line to goes through these two points, it's not going to go through those points there. So you're not going to be able to find a solution that goes through those points. Let's set up the equation that we know we can't find the solution to and maybe we can use our least squares approximation to find a line that almost goes through all these points. Or it's at least the best approximation for a line that goes through those points. So this first one, I can express my line, y Let me just express it as f of x is equal to mx plus b, or y is equal to f of x. We can write it that way. So our first point right there -- let me do it in that color, that orange -- that tells us that f of minus 1, which is equal to m times -- let me just write this way -- minus 1 times m, it's minus m plus b, that that is going to be equal to 0. That's what that first equation tells us. The second equation tells us that f of 0, which is equal to 0 times m, which is just 0 plus b is equal to 1. f of 0 is 1. This is f of x. The next one -- let me do it in this yellow color -- tells us that f of 1, which is equal to 1 times m, or just m, plus b, is going to be equal to 2. And then this last one down here tells us that f of 2, which is of course 2 times m plus b, that that is going to be equal to 1. These are the constraints. If we assume that our line can go through all of these points, then all of these things must be true. Now you could immediately, if you wish, try to solve this equation, but you'll find that you won't find a solution. We want to find some m's and b's that satisfy all of these equations. Or another way of writing this -- We want to write it as a matrix vector or a matrix equation . We could write it like this. Minus 1, 1, 0, 1, 1, 1, 2, 1, times the vector mv has got to be equal to the vector 0, 1, 2, 1. These two systems, this system and this system right here, are equivalent statements, right? Minus 1 times m plus 1 times b has got to be equal to that 0. 0 times m plus 1 times b has got to be equal to that 1 That's equivalent to that statement right here. And this isn't going to have a solution. The solution would have to go through all of those points. So let's at least try to find a least squares solution. So if we call this a, if we call that x, and let's call this b, there is no solution to ax is equal to b. Now maybe we can find a least -- Well, we can definitely find a least squares solution. So let's find our least squares solution such that a transpose a times our least squares solution is equal to a transpose times b. Our least squares solution is the one that satisfies this equation. We proved it two videos ago. So let's figure out what a transpose a is and what a transpose b is, and then we can solve. So a transpose will look like this. b minus 1, 1, 0, 1, 1, 1, and then 2, 1. This first column becomes this first row; this second column becomes this second row. So we're going to take the product of a transpose and then a-- a is that thing right there --minus 1, 0, 1, 2, and we just get a bunch of 1's. So what does this equal to? We have a 2 by 4 times a 4 by 2. So we're going to have a 2 by 2 matrix. So this is going to be -- Let's do it this way. Well, we're going to have minus 1 times minus 1, which is 1, plus 0 times 0, which is 0 -- so we're at 1 right now --plus 1 times 1. So that's 1 plus the other 1 up there, so that's 2, plus 2 times 2. 2 times 2 is 4, so we get 6. That's that row, dotted with that column, was equal to 6. Now let's take this row dotted with this column. So it's going to be negative 1 times 1, plus 0 times 1, so all of these guys times 1 plus each other. So minus 1 plus 0 plus 1 -- that's all 0's --plus 2. So it's going to get a 2. I just dotted that guy with that guy. Now I need to take the dot of this guy with this column. So it's just going to be 1 times minus 1 plus 1 times 0 plus 1 times 1 plus 1 times 2. Well, these are all 1 times everything, so it's minus 1 plus 0 plus 1, which is 0 plus 2. It's going to be 2. And then finally -- Well. I mean, I think you see some symmetry here. We're going to have to take the dot of this guy and this guy over here. That's 1 times 1, which is 1, plus 1 times 1, which is 2, plus 1 times 1. So we're going to have 1 plus itself four times. So we're going to get that it's equal to 4. So this is a transpose a. And let's figure out what a transpose b looks like. Scroll down a little bit. So a transpose is this matrix again-- let me switch colors --minus 1, 0, 1, 2. We get all of our 1's just like that. And then the matrix b is 0, 1, 2, 1. We have a 2 by 4 times a 4 by 1, so we're just going to get a 2 by 1 matrix. So this is going to be equal to a 2 by 1 matrix. We have here, let's see, minus 1 times 0 is 0, plus 0 times 1 is still 0. Plus 1 times 2, which is 2, plus 2 times 1, which is 4, right? This is 2 plus 2, so it's going to be 4 right there. And then we have 1 times 0, plus 1 times 2, plus-- So 1 times all of these guys added up. So 0 plus 1 is 1, 1 plus 2 is 3, 3 plus 1 is 4. So this right here is a transpose b. So just like that, we know that the least squares solution will be the solution to this system. 6, 2, 2, 4, times our least squares solution, is going to be equal to 4, 4. Or we could write it this way. We could write it 6, 2, 2, 4, times our least squares solution, which I'll write-- Remember, the first entry was m . I'll write it as m star. That's our least square m, and this is our least square b, is equal to 4, 4. And I can do this as an augmented matrix or I could just write this as a system of two unknowns, which is actually probably easier. So let's do it that way. So this, if I were to write it as a system of equations, is 6 times m star plus 2 times b star, is equal to 4. And then I get 2 times m star plus 4 times b star is equal to this 4. So let me solve for my m stars and my b stars. So let's multiply this second equation, actually let's multiply that top equation by 2. This is just straight Algebra 1. So times 2, what do we get? We get 12m star plus 4b star is equal to 8. We just multiplied that top guy by 2. Now let's multiply this magenta 1 by negative 1. So this becomes a minus, this becomes a minus, that becomes a minus, and now we can add these two equations. So we get minus 2 plus 12m star, that's 10m star. And then the minus 4b and the 4b cancel out, is equal to 4, or m star is equal to 4 over 10, which is equal to 2/5. Now we can just go and back-substitute into this. We can say 6 times m star-- This is just straight Algebra 1. So 6 times our m star, so 6 times 2 over 5, plus 2 times our b star is equal to 4. Enough white, let me use yellow. So we get 12 over 5 plus 2b star is equal to 4, or we could say 2b star-- let me scroll down a little bit --2b star is equal to 4. Which is the same thing as 20 over 5, minus 12 over 5, which is equal to-- I'm just subtracting the 12 over 5 from both sides --which is equal to 8 over 5. And you divide both sides of the equation by 2, you get b star is equal to 4/5. And just like that, we got our m star and our b star. Our least squares solution is equal to 2/5 and 4/5. So m is equal to 2/5 and b is equal to 4/5. And remember, the whole point of this was to find an equation of the line. y is equal to mx plus b. Now we can't find a line that went through all of those points up there, but this is going to be our least squares solution. This is the one that minimizes the distance between a times our vector and b. No vector, when you multiply times that matrix a-- that's not a, that's transpose a --no other solution is going to give us a closer solution to b than when we put our newly-found x star into this equation. This is going to give us our best solution. It's going to minimize the distance to b. So let's write it out. y is equal to mx plus b. So y is equal to 2/5 x plus 2/5. Let's graph that out. y is equal to 2/5 x plus 2/5. So its y-intercept is 2/5, which is about there . This is at 1. 2/5 is right about there. And then its slope is 2/5. Let's think of it this way: for every 2 and 1/2 you go to the right, you're going to go up 1. So if you go 1, 2 and 1/2, we're going to go up 1. We're going to go up 1 like that. So our line-- and obviously this isn't precise --but our line is going to look something like this. I want to do my best shot at drawing it because this is the fun part. It's going to look something like that. And that right there is my least squares estimate for a line that goes through all of those points. And you're not going to find a line that minimizes the error in a better way, at least when you measure the error as the distance between this vector and the vector a times our least squares estimate. Anyway, thought you would find that neat." + }, + { + "Q": "Magnets need energy to apply force in the form of attracting and repelling.\nFrom where does they get this energy?\nDoes it violates the law of conservation of energy?", + "A": "Your first statement is false, so the first question is moot and the answer to the second one is no . Consider an analogy: If you drop a book, do you say the earth needs energy to apply force to the book to attract it ? Does it violate the law of conservation of energy when the book gets attracted to the earth? No, right? Why not? Now see if you can apply the same thinking to your magnet question.", + "video_name": "8Y4JSp5U82I", + "transcript": "We've learned a little bit about gravity. We've learned a little bit about electrostatic. So, time to learn about a new fundamental force of the universe. And this one is probably second most familiar to us, And that's magnetism. Where does the word come from? Well, I think several civilizations-- I'm no historian-- found these lodestones, these objects that would attract other objects like it, other magnets. Or would even attract metallic objects like iron. Ferrous objects. And they're called lodestones. That's, I guess, the Western term for it. And the reason why they're called magnets is because they're named after lodestones that were found near the Greek province of Magnesia. And I actually think the people who lived there were called Magnetes. But anyway, you could Wikipedia that and learn more about it than I know. But anyway let's focus on what magnetism is. And I think most of us have at least a working knowledge of what it is; we've all played with magnets and we've dealt with compasses. But I'll tell you this right now, what it really is, is pretty deep. And I think it's fairly-- I don't think anyone has-- we can mathematically understand it and manipulate it and see how it relates to electricity. We actually will show you the electrostatic force and the magnetic force are actually the same thing, just viewed from different frames of reference. I know that all of that sounds very complicated and all of that. But in our classical Newtonian world we treat them as two different forces. But what I'm saying is although we're kind of used to a magnet just like we're used to gravity, just like gravity is also fairly mysterious when you really think about what it is, so is magnetism. So with that said, let's at least try to get some working knowledge of how we can deal with magnetism. So we're all familiar with a magnet. I didn't want it to be yellow. I could make the boundary yellow. No, I didn't want it to be like that either. So if this is a magnet, we know that a magnet always has two poles. It has a north pole and a south pole. And these were just labeled by convention. Because when people first discovered these lodestones, or they took a lodestone and they magnetized a needle with that lodestone, and then that needle they put on a cork in a bucket of water, and that needle would point to the Earth's north pole. They said, oh, well the side of the needle that is pointing to the Earth's north, let's call that the north pole. And the point of the needle that's pointing to the south pole-- sorry, the point of the needle that's pointing to the Earth's geographic south, we'll call that the south pole. Or another way to put it, if we have a magnet, the direction of the magnet or the side of the magnet that orients itself-- if it's allowed to orient freely without friction-- towards our geographic north, we call that the north pole. And the other side is the south pole. And this is actually a little bit-- obviously we call the top of the Earth the north pole. You know, this is the north pole. And we call this the south pole. And there's another notion of magnetic north. And that's where-- I guess, you could kind of say-- that is where a compass, the north point of a compass, will point to. And actually, magnetic north moves around because we have all of this moving fluid inside of the earth. And a bunch of other interactions. It's a very complex interaction. But magnetic north is actually roughly in northern Canada. So magnetic north might be here. So that might be magnetic north. And magnetic south, I don't know exactly where that is. But it can kind of move around a little bit. It's not in the same place. So it's a little bit off the axis of the geographic north pole and the south pole. And this is another slightly confusing thing. Magnetic north is the geographic location, where the north pole of a magnet will point to. But that would actually be the south pole, if you viewed the Earth as a magnet. So if the Earth was a big magnet, you would actually view that as a south pole of the magnet. And the geographic south pole is the north pole of the magnet. You could read more about that on Wikipedia, I know it's a little bit confusing. But in general, when most people refer to magnetic north, or the north pole, they're talking about the geographic north area. And the south pole is the geographic south area. But the reason why I make this distinction is because we know when we deal with magnets, just like electricity, or electrostatics-- but I'll show a key difference very shortly-- is that opposite poles attract. So if this side of my magnet is attracted to Earth's north pole then Earth's north pole-- or Earth's magnetic north-- actually must be the south pole of that magnet. And vice versa. The south pole of my magnet here is going to be attracted to Earth's magnetic south. Which is actually the north pole of the magnet we call Earth. Anyway, I'll take Earth out of the equation because it gets a little bit confusing. And we'll just stick to bars because that tends to be a little bit more consistent. Let me erase this. There you go. I'll erase my Magnesia. I wonder if the element magnesium was first discovered in Magnesia, as well. Probably. And I actually looked up Milk of Magnesia, which is a laxative. And it was not discovered in Magnesia, but it has magnesium in it. So I guess its roots could be in Magnesia if magnesium was discovered in Magnesia. Anyway, enough about Magnesia. Back to the magnets. So if this is a magnet, and let me draw another magnet. Actually, let me erase all of this. All right. So let me draw two more magnets. We know from experimentation when we were all kids, this is the north pole, this is the south pole. That the north pole is going to be attracted to the south pole of another magnet. And that if I were to flip this magnet around, it would actually repel north-- two north facing magnets would repel each other. And so we have this notion, just like we had in electrostatics, that a magnet generates a field. It generates these vectors around it, that if you put something in that field that can be affected by it, it'll be some net force acting on it. So actually, before I go into magnetic field, I actually want to make one huge distinction between magnetism and electrostatics. Magnetism always comes in the form of a dipole. It means that we have two poles. A north and a south. In electrostatics, you do have two charges. You have a positive charge and a negative charge. So you do have two charges. But they could be by themselves. You could just have a proton. You don't have to have an electron there right next to it. You could just have a proton and it would create a positive electrostatic field. And our field lines are what a positive point charge would do. And it would be repelled. So you don't always have to have a negative charge there. Similarly you could just have an electron. And you don't have to have a proton there. So you could have monopoles. These are called monopoles, when you just have one charge when you're talking about electrostatics. But with magnetism you always have a dipole. If I were to take this magnet, this one right here, and if I were to cut it in half, somehow miraculously each of those halves of that magnet will turn into two more magnets. Where this will be the south, this'll be the north, this'll be the south, this will be the north. And actually, theoretically, I've read-- my own abilities don't go this far-- there could be such a thing as a magnetic monopole, although it has not been observed yet in nature. So everything we've seen in nature has been a dipole. So you could just keep cutting this up, all the way down to if it's just one electron left. And it actually turns out that even one electron is still a magnetic dipole. It still is generating, it still has a north pole and a south pole. And actually it turns out, all magnets, the magnetic field is actually generated by the electrons within it. By the spin of electrons and that-- you know, when we talk about electron spin we imagine some little ball of charge spinning. But electrons are-- you know, it's hard to-- they do have mass. But it starts to get fuzzy whether they are energy or mass. And then how does a ball of energy spin? Et cetera, et cetera. So it gets very almost metaphysical. So I don't want to go too far into it. And frankly, I don't think you really can get an intuition. It is almost-- it is a realm that we don't normally operate in. But even these large magnets you deal with, the magnetic field is generated by the electron spins inside of it and by the actual magnetic fields generated by the electron motion around the protons. Well, I hope I'm not overwhelming you. And you might say, well, how come sometimes a metal bar can be magnetized and sometimes it won't be? Well, when all of the electrons are doing random different things in a metal bar, then it's not magnetized. Because the magnetic spins, or the magnetism created by the electrons are all canceling each other out, because it's random. But if you align the spins of the electrons, and if you align their rotations, then you will have a magnetically charged bar. But anyway, I'm past the ten-minute mark, but hopefully that gives you a little bit of a working knowledge of what a magnet is. And in the next video, I will show what the effect is. Well, one, I'll explain how we think about a magnetic field. And then what the effect of a magnetic field is on an electron. Or not an electron, on a moving charge. See you in the next video." + }, + { + "Q": "4:48 Hey, does anyone know why Sal puts the = sign like a smiley face? =D", + "A": "It s not an equal sign, but rather an arrow. He just draws it in a way that it doesn t look quite connected. This is actually a common way to note progression of steps in mathematics.", + "video_name": "X2jVap1YgwI", + "transcript": "Let's do some more percentage problems. Let's say that I start this year in my stock portfolio with $95.00. And I say that my portfolio grows by, let's say, 15%. How much do I have now? I think you might be able to figure this out on your own, but of course we'll do some example problems, just in case it's a little confusing. So I'm starting with $95.00, and I'll get rid of the dollar sign. We know we're working with dollars. 95 dollars, right? And I'm going to earn, or I'm going to grow just because I was an excellent stock investor, that 95 dollars is going to grow by 15%. So to that 95 dollars, I'm going to add another 15% of 95. So we know we write 15% as a decimal, as 0.15, so 95 plus 0.15 of 95, so this is times 95-- that dot is just a times sign. It's not a decimal, it's a times, it's a little higher than a decimal-- So 95 plus 0.15 times 95 is what we have now, right? Because we started with 95 dollars, and then we made another 15% times what we started with. Hopefully that make sense. Another way to say it, the 95 dollars has grown by 15%. So let's just work this out. This is the same thing as 95 plus-- what's 0.15 times 95? Let's see. So let me do this, hopefully I'll have enough space here. 95 times 0.15-- I don't want to run out of space. Actually, let me do it up here, I think I'm about to run out of space-- 95 times 0.15. 5 times 5 is 25, 9 times 5 is 45 plus 2 is 47, 1 times 95 is 95, bring down the 5, 12, carry the 1, 15. And how many decimals do we have? 1, 2. 15.25. Actually, is that right? I think I made a mistake here. See 5 times 5 is 25. 5 times 9 is 45, plus 2 is 47. And we bring the 0 here, it's 95, 1 times 5, 1 times 9, then we add 5 plus 0 is 5, 7 plus 5 is 12-- oh. I made a mistake. It's 14.25, not 15.25. So I'll ask you an interesting question? How did I know that 15.25 was a mistake? Well, I did a reality check. I said, well, I know in my head that 15% of 100 is 15, so if 15% of 100 is 15, how can 15% of 95 be more than 15? I think that might have made sense. The bottom line is 95 is less than 100. So 15% of 95 had to be less than 15, so I knew my answer of 15.25 was wrong. And so it turns out that I actually made an addition error, and the answer is 14.25. So the answer is going to be 95 plus 15% of 95, which is the same thing as 95 plus 14.25, well, that equals what? 109.25. Notice how easy I made this for you to read, especially this 2 here. 109.25. So if I start off with $95.00 and my portfolio grows-- or the amount of money I have-- grows by 15%, I'll end up with $109.25. Let's do another problem. Let's say I start off with some amount of money, and after a year, let's says my portfolio grows 25%, and after growing 25%, I now have $100. How much did I originally have? Notice I'm not saying that the $100 is growing by 25%. I'm saying that I start with some amount of money, it grows by 25%, and I end up with $100 after it grew by 25%. To solve this one, we might have to break out a little bit of algebra. So let x equal what I start with. So just like the last problem, I start with x and it grows by 25%, so x plus 25% of x is equal to 100, and we know this 25% of x we can just rewrite as x plus 0.25 of x is equal to 100, and now actually we have a level-- actually this might be level 3 system, level 3 linear equation-- but the bottom line, we can just add the coefficients on the x. x is the same thing as 1x, right? So 1x plus 0.25x, well that's just the same thing as 1 plus 0.25, plus x-- we're just doing the distributive property in reverse-- equals 100. And what's 1 plus 0.25? That's easy, it's 1.25. So we say 1.25x is equal to 100. Not too hard. And after you do a lot of these problems, you're going to intuitively say, oh, if some number grows by 25%, and it becomes 100, that means that 1.25 times that number is equal to 100. And if this doesn't make sense, sit and think about it a little bit, maybe rewatch the video, and hopefully it'll, over time, start to make a lot of sense to you. This type of math is very very useful. I actually work at a hedge fund, and I'm doing this type of math in my head day and night. So 1.25 times x is equal to 100, so x would equal 100 divided by 1.25. I just realized you probably don't know what a hedge fund is. I invest in stocks for a living. Anyway, back to the math. So x is equal to 100 divided by 1.25. So let me make some space here, just because I used up too much space. Let me get rid of my little let x statement. Actually I think we know what x is and we know how we got to there. If you forgot how we got there, you can I guess rewatch the video. Let's see. Let me make the pen thin again, and go back to the orange color, OK. X equals 100 divided by 1.25, so we say 1.25 goes into 100.00-- I'm going to add a couple of 0's, I don't know how many I'm going to need, probably added too many-- if I move this decimal over two to the right, I need to move this one over two to the right. And I say how many times does 100 go into 100-- how many times does 125 go into 100? None. How many times does it go into 1000? It goes into it eight times. I happen to know that in my head, but you could do trial and error and think about it. 8 times-- if you want to think about it, 8 times 100 is 800, and then 8 times 25 is 200, so it becomes 1000. You could work out if you like, but I think I'm running out of time, so I'm going to do this fast. 8 times 125 is 1000. Remember this thing isn't here. 1000, so 1000 minus 1000 is 0, so you can bring down the 0. 125 goes into 0 zero times, and we just keep getting 0's. This is just a decimal division problem. So it turns out that if your portfolio grew by 25% and you ended up with $100.00 you started with $80.00. And that makes sense, because 25% is roughly 1/4, right? So if I started with $80.00 and I grow by 1/4, that means I grew by $20, because 25% of 80 is 20. So if I start with 80 and I grow by 20, that gets me to 100. Makes sense. So remember, all you have to say is, well, some number times 1.25-- because I'm growing it by 25%-- is equal to 100. Don't worry, if you're still confused, I'm going to add at least one more presentation on a couple of more examples like this." + }, + { + "Q": "Why couldnt they just settle it peacefully instead of aggrivating each other to the point of war?", + "A": "After decades of trying to compromise and to settle their differences, the South saw no alternative than to leave the Union. The remaining states in the US did not recognize the secession of the southern states and the formation of the Confederacy. Once the attack was made on Fort Sumter, there was no turning back.", + "video_name": "L87VpmRLAPg", + "transcript": "- [Voiceover] All right Kim, so we left off in I guess, early-mid 1861, you have Lincoln gets inagurated, Fort Sumter which is kind of the first real conflict of the war, if not the first major battle. Lincoln forms his volunteer army, and then the rest of the southern states secede, four more states secede. - [Kim] Right. - [Voiceover] And then what was the first major conflict? - [Kim] So the first major conflict comes after a number of months. There are a couple of little skirmishes here and there, but nothing super large until about 60,000 troops meet outside of Manassas, Virginia, at a place called Bull Run. An interesting fact, I think, is that Union armies and Confederate armies actually named battles different things, if you've ever been confused about this. The Union armies tended to name battles after bodies of water, whereas the Confederate armies tended to name them by nearby towns. So if you've ever heard the Battle of Manassas and the Battle of Bull Run, they're the same thing, it's just the Union officers are talking about this creek, Bull Run, whereas the Confederates are talking about the town nearby. - [Voiceover] I see, and the 60,000 troops between the two of them. - [Kim] Right. So they meet, and this is very close to Washington, D.C., so much so that people go out and they bring picnics to watch this battle. - [Voiceover] They think it's going to be entertaining. - [Kim] Yeah, they think it's going to be like a football game. And it is not like a football game. It is a gigantic battle, 800 people die that day, which doesn't sound like a lot to us, but it was the most deadly battle ever in American history up until that point. So it's a Confederate victory, which is very surprising to the Union, because they think that they have such superior forces that this is really going to be a very short war. And this is a quick rebellion, in 90 days we're going to be able to, you know, suppress this rebellion and that'll be it. But Bull Run is really the first sign that this is going to be a major war. It's not going to be quick and it is going to be very deadly. - [Voiceover] This was July of... - [Both] 1861. - [Voiceover] Okay, so now it's clear to both sides, especially, I guess you could say the North, that this is not going to be a short war. So they need to prepare. How are they approaching this? - [Kim] Well, so both sides have some advantages and disadvantages. For the South, they have some of the same advantages that the United States would have had during the war for independence. They have home court advantage, we could say, which is that they know the territory very well and also there's a real incentive for people to protect their homes, right. You're gonna care more about a war that's happening on your property than a war that's gonna happen very far away. The other advantage that they have is just really, really terrific military leadership. So they have Robert E. Lee, who is widely considered the greatest general of his era. He's truly a military genius. He, in fact, was offered a commission in the Union army but when Virginia seceded, he went with Virginia. He preferred his home state. So he is a terrific general. The Union is gonna really struggle to come up with the kind of military leadership that the South has. - [Voiceover] Who is in charge of the Union or the You said, the United States Army. - [Kim] The United States Army. The first general that Lincoln puts in charge is George B. McClellan. This is problematic for a lot of reasons. One is that George McClellan is a Democrat, so he doesn't agree politically with Lincoln. I think he would have preferred peace, in fact in 1864 he runs against Lincoln for President on a platform of letting the South go, basically. And so Lincoln is struggling to match the South when it comes to military leadership, but he does have other advantages. For one thing, there are four times as many free people in the North as there are in the South. - [Voiceover] And you made the point, free people. - [Kim] Right. - [Voiceover] Because the South, as you mentioned, it has a majority of the population was not free. - [Kim] I wouldn't say a majority of the population, in many states, - [Voiceover] In Deep South. - [Kim] In the Deep South states, right. But so there are only about 9,000,000 people living in the South, and of those 9,000,000 people 3,500,000 to 4,000,000 of them are enslaved. So they're not going to be fighting to continue the institution of slavery. By contrast, the North has 22,000,000 people and it also has a terrific industrial base. One of the major cultural differences between the North and South that leads to the Civil War is that the South is primarily agrarian, and the North becomes very industrial. But industry is really helpful in a war. They've got miles and miles of railroad tracks which means that they can move supplies very quickly, and they also have hundreds and hundreds of factories that make it easy for them to make munitions. - [Voiceover] This is the middle of the Industrial Revolution - [Kim] Right. - [Voiceover] and an industrial base matters a lot. And so what's, given the North's advantages and the South's advantages, what's their strategies, how do they try to play to their strengths? - [Kim] Right, so the South, they are basically trying to outlast the North. They know that they have this territory, and if the North wants them to come back into the Union, they're going to have to conquer this territory. And even though it's hard to kind of tell, the territory of the South is actually larger than Western Europe. - [Voiceover] Wow. - [Kim] So in a way, the North has a bigger job to conquer the South than the Allies did in World War II, to conquer Europe. So they know that the North is gonna have to fight a war to conquer them, whereas the South just needs to win the war of waiting. - [Voiceover] Of attrition. - [Kim] Yeah, they're hoping that the North will get tired of fighting. - [Voiceover] Fighting in another person's land, you're not defending your own land. - [Kim] Right, and they know that there are plenty of whites in the North who don't care about slavery. It's not in their.. - [Voiceover] They're indifferent, what do they care. - [Kim] Yeah, what do they care, in fact some people are afraid that if the slaves are freed in the South, they're all gonna come up North and they're going to compete for labor with poor white people. So there are plenty of whites in the North who have no interest in the slaves in the South being free, even if that's not an early war aim of the North. So the South is hoping that maybe they can win a couple of really big battles that show this isn't gonna be a big war. - [Voiceover] Or it'd be so painful for the North to try to conquer the South, so to speak. - [Kim] And they're also trying to show that they're serious, to an international audience, particularly England, because the South is producing 3/4 of the world's supply of cotton at this point, and England is an industrial nation which is built in many cases around textile manufacturing. So they're hoping that if they show that they are serious about their own nationhood that they're going to win this war against the North that England will intercede on their behalf to protect their supply of cotton. - [Voiceover] So this would be an appeal to England on purely economic grounds. - [Kim] Right. Because England, they didn't have slavery. - [Kim] No. - [Voiceover] But purely economically, at least, appeal to them. - [Kim] So on the other hand, the North's strategy is what they call \"The Anaconda Plan\". And the idea of the Anaconda Plan is that they are going to squeeze the South, economically. What they want to do, - [Voiceover] Like an anaconda. - [Kim] Like an anaconda, right. So they want to blockade the Atlantic ocean because they don't want the South to be able to sell their cotton to get money, and they also don't want the South to be able to buy the kinds of things that they're going to need to make a war happen. They also want to control the Mississippi River cuz that's the real main artery of commerce in the West. Anyone who is gonna be shipping their grain or their cotton is gonna be shipping it down the Mississippi to the port of New Orleans. So the Union hopes that if they can basically surround the South, and make sure nothing gets in or out, then eventually they're just gonna starve to death. - [Voiceover] This also goes through the industrial bays, it can also produce more ships and etc. - [Kim] Right, and it takes them a while to do that, in fact at the start of the war, the Union only has 90 ships. I've heard it compared to \"Five leaky boats\". Right, we're not a naval power at this point and so it's gonna take them a while to build up the kind of naval power they need to do that, cuz this is 3500 miles of coastline that they're gonna need to patrol. - [Voiceover] I'm just looking at this map, not getting too much into details, it looks like a lot of the battles are concentrated right in this Virginia/Maryland area, and then there's more, it's a little bit more sparse but you have a few that are in the Deep South and along this Mississippi corridor. - [Kim] There are two major theaters of the war. We'd say the Eastern Theater, and this is that 100-mile corridor between Washington and Richmond, where a huge amount of the fighting takes place. It's important to remember that the capital of the Confederacy and the capital of the United States are only 100 miles apart. - [Voiceover] This capital is, you can't see it on this map but it's someplace in the middle of Virginia, and then D.C., literally, as you mentioned you said it was 100 miles apart? - [Kim] Yep. - [Voiceover] Fascinating." + }, + { + "Q": "So , if A goes 5km and then goes back, is tht mean the displacement is 0?", + "A": "yes because the forward movement is positive and the backward movement is negative and the same in length so adding them up gives you zero", + "video_name": "oRKxmXwLvUU", + "transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. And he did it in 1 hour in his car. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. And my best sense of that is, once you start doing calculus, you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time. So these two, you could call them formulas, or you could call them definitions, although I would think that they're pretty intuitive for you. How fast something is going, you say, how far did it go over some period of time. These are essentially saying the same thing. This is when you care about direction, so you're dealing with vector quantities. This is where you're not so conscientious about direction. And so you use distance, which is scalar, and you use rate or speed, which is scalar. Here you use displacement, and you use velocity. Now with that out of the way, let's figure out what his average velocity was. And this key word, average, is interesting. Because it's possible that his velocity was changing over that whole time period. But for the sake of simplicity, we're going to assume that it was kind of a constant velocity. What we are calculating is going to be his average velocity. But don't worry about it, you can just assume that it wasn't changing over that time period. So his velocity is, his displacement was 5 kilometers to the north-- I'll write just a big capital. Well, let me just write it out, 5 kilometers north-- over the amount of time it took him. And let me make it clear. This is change in time. This is also a change in time. Sometimes you'll just see a t written there. Sometimes you'll see someone actually put this little triangle, the character delta, in front of it, which explicitly means \"change in.\" It looks like a very fancy mathematics when you see that, but a triangle in front of something literally means \"change in.\" So this is change in time. So he goes 5 kilometers north, and it took him 1 hour. So the change in time was 1 hour. So let me write that over here. So over 1 hour. So this is equal to, if you just look at the numerical part of it, it is 5/1-- let me just write it out, 5/1-- kilometers, and you can treat the units the same way you would treat the quantities in a fraction. 5/1 kilometers per hour, and then to the north. Or you could say this is the same thing as 5 kilometers per hour north. So this is 5 kilometers per hour to the north. So that's his average velocity, 5 kilometers per hour. And you have to be careful, you have to say \"to the north\" if you want velocity. If someone just said \"5 kilometers per hour,\" they're giving you a speed, or rate, or a scalar quantity. You have to give the direction for it to be a vector quantity. You could do the same thing if someone just said, what was his average speed over that time? You could have said, well, his average speed, or his rate, would be the distance he travels. The distance, we don't care about the direction now, is 5 kilometers, and he does it in 1 hour. His change in time is 1 hour. So this is the same thing as 5 kilometers per hour. So once again, we're only giving the magnitude here. This is a scalar quantity. If you want the vector, you have to do the north as well. Now, you might be saying, hey, in the previous video, we talked about things in terms of meters per second. Here, I give you kilometers, or \"kil-om-eters,\" depending on how you want to pronounce it, kilometers per hour. What if someone wanted it in meters per second, or what if I just wanted to understand how many meters he travels in a second? And there, it just becomes a unit conversion problem. And I figure it doesn't hurt to work on that right now. So if we wanted to do this to meters per second, how would we do it? Well, the first step is to think about how many meters we are traveling in an hour. So let's take that 5 kilometers per hour, and we want to convert it to meters. So I put meters in the numerator, and I put kilometers in the denominator. And the reason why I do that is because the kilometers are going to cancel out with the kilometers. And how many meters are there per kilometer? Well, there's 1,000 meters for every 1 kilometer. And I set this up right here so that the kilometers cancel out. So these two characters cancel out. And if you multiply, you get 5,000. So you have 5 times 1,000. So let me write this-- I'll do it in the same color-- 5 times 1,000. So I just multiplied the numbers. When you multiply something, you can switch around the order. Multiplication is commutative-- I always have trouble pronouncing that-- and associative. And then in the units, in the numerator, you have meters, and in the denominator, you have hours. Meters per hour. And so this is equal to 5,000 meters per hour. And you might say, hey, Sal, I know that 5 kilometers is the same thing as 5,000 meters. I could do that in my head. And you probably could. But this canceling out dimensions, or what's often called dimensional analysis, can get useful once you start doing really, really complicated things with less intuitive units than something But you should always do an intuitive gut check right here. You know that if you do 5 kilometers in an hour, that's a ton of meters. So you should get a larger number if you're talking about meters per hour. And now when we want to go to seconds, let's do an intuitive gut check. If something is traveling a certain amount in an hour, it should travel a much smaller amount in a second, or 1/3,600 of an hour, because that's how many seconds there are in an hour. So that's your gut check. We should get a smaller number than this when we want to say meters per second. But let's actually do it with the dimensional analysis. So we want to cancel out the hours, and we want to be left with seconds in the denominator. So the best way to cancel this hours in the denominator is by having hours in the numerator. So you have hours per second. So how many hours are there per second? Or another way to think about it, 1 hour, think about the larger unit, 1 hour is how many seconds? Well, you have 60 seconds per minute times 60 minutes per hour. The minutes cancel out. 60 times 60 is 3,600 seconds per hour. So you could say this is 3,600 seconds for every 1 hour, or if you flip them, you would get 1/3,600 hour per second, or hours per second, depending on how you want to do it. So 1 hour is the same thing as 3,600 seconds. And so now this hour cancels out with that hour, and then you multiply, or appropriately divide, the numbers right here. And you get this is equal to 5,000 over 3,600 meters per-- all you have left in the denominator here is second. Meters per second. And if we divide both the numerator and the denominator-- I could do this by hand, but just because this video's already getting a little bit long, let me get my trusty calculator out. I get my trusty calculator out just for the sake of time. 5,000 divided by 3,600, which would be really the same thing as 50 divided by 36, that is 1.3-- I'll just round it over here-- 1.39. So this is equal to 1.39 meters per second. So Shantanu was traveling quite slow in his car. Well, we knew that just by looking at this. 5 kilometers per hour, that's pretty much just letting the car roll pretty slowly." + }, + { + "Q": "think what can go into that # so break it up so we have 2x256 then 2x 128 then 2x64 then 2x32 then 2x16 then 2x8 then 2x4 then 2x2. he could just divide by 4 or 8 or 16 or 32 but he just adds more work but all those things make up 512", + "A": "2 comments in response to your observation. 1. It is easier to divide by 2 in your head. 2. The goal is to get to prime factors so any method that ends up with 2^9 will be useful. You could factor like this. 512 = 16*32 = 4*4 * 8*4 = 2^2*2^2 * 2^3*2^2 = 2^9 It s less steps but the steps took me longer.", + "video_name": "DKh16Th8x6o", + "transcript": "We are asked to find the cube root of negative 512. Or another way to think about it is if I have some number, and it is equal to the cube root of negative 512, this just means that if I take that number and I raise it to the third power, then I get negative 512. And if it doesn't jump out at you immediately what this is the cube of, or what we have to raise to the third power to get negative 512, the best thing to do is to just do a prime factorization of it. And before we do a prime factorization of it to see which of these factors show up at least three times, let's at least think about the negative part a little bit. So negative 512, that's the same thing-- so let me rewrite the expression-- this is the same thing as the cube root of negative 1 times 512, which is the same thing as the cube root of negative 1 times the cube root of 512. And this one's pretty straightforward to answer. What number, when I raise it to the third power, do I get negative 1? Well, I get negative 1. This right here is negative 1. Negative 1 to the third power is equal to negative 1 times negative 1 times negative 1, which is equal to negative 1. So the cube root of negative 1 is negative 1. So it becomes negative 1 times this business right here, times the cube root of 512. And let's think what this might be. So let's do the prime factorization. So 512 is 2 times 256. 256 is 2 times 128. 128 is 2 times 64. We already see a 2 three times. 64 is 2 times 32. 32 is 2 times 16. We're getting a lot of twos here. 16 is 2 times 8. 8 is 2 times 4. And 4 is 2 times 2. So we got a lot of twos. So essentially, if you multiply 2 one, two, three, four, five, six, seven, eight, nine times, you're going to get 512, or 2 to the ninth power is 512. And that by itself should give you a clue of what the cube root is. But another way to think about it is, can we find-- there's definitely three twos here. But can we find three groups of twos, or we could also find-- let me look at it this way. We can find three groups of two twos over here. So that's 2 times 2 is 4. 2 times 2 is 4. So definitely 4 multiplied by itself three times is divisible into this. But even better, it looks like we can get three groups of three twos. So one group, two groups, and three groups. So each of these groups, 2 times 2 times 2, that's 8. That is 8. This is 2 times 2 times 2. That's 8. And this is also 2 times 2 times 2. So that's 8. So we could write 512 as being equal to 8 times 8 times 8. And so we can rewrite this expression right over here as the cube root of 8 times 8 times 8. So this is equal to negative 1, or I could just put a negative sign here, negative 1 times the cube root of 8 times 8 times 8. So we're asking our question. What number can we multiply by itself three times, or to the third power, to get 512, which is the same thing as 8 times 8 times 8? Well, clearly this is 8. So the answer, this part right over here, is just going to simplify to 8. And so our answer to this, the cube root of negative 512, is negative 8. And we are done. And you could verify this. Multiply negative 8 times itself three times. Negative 8 times negative 8 times negative 8. Negative 8 times negative 8 is positive 64. You multiply that times negative 8, you get negative 512." + }, + { + "Q": "At 2:25 Sal says that anatomically modern humans have been on earth for about 200,000 years. Then what kinds of characteristics of human anatomy count as the characteristics of modern humans?", + "A": "Mainly it means a smaller jaw, bigger brain, walking upright almost all of the time, Ability to use hands more like modern day hands instead of really weird feet.", + "video_name": "13E90TAtZ30", + "transcript": "LeBron: If the history of the earth was a basketball game, at what point in the game will the humans show up? Voiceover: Let's first think about how long a basketball game is in the NBA. We have 4 quarters that each lasts 12 minutes, so we're talking about 48 minutes. 48 minutes of regulation play. I'm not considering half time and the time outs and the commercial breaks and potential overtime. I am just taking about regulation play. We could think about how the numbers might change if you think about the total duration of the game, including time outs and commercial breaks and everything that might get you closer to 2 or 2 1/2 hours. But 48 minutes we can actually convert to seconds because we know that there are 60 seconds per minute, times 60 seconds, 60 seconds per minute and this is pretty straight forward multiplication. We can just say 48 times 60 gives us, so we got this 0 here and then 6 times 48, 6 times 8 is 48 and then 6 times 4 is 24 plus 4 is 28. So there's 2,880 seconds during regulation play. Now let's think about the actual age of earth. We estimate that the earth, and actually the entire solar system, which was all formed roughly at the same time, is 4.54 billion years old. So let me draw that here. so I am going to draw it as the same length. So 4.54 billion years old and just to give a sense of how large of a number that is, a billion is a 1,000 million. So we could also write this as 4,540 million years. or we could write it as 4,540,000 thousand years, or millennia, or we could just write out the number as 4,540,000,000 years which seems kind of old but let's think about how long anatomically, modern humans have been roaming the surface of the earth and here we estimate that anatomically modern humans have been on the surface of the earth for about 200,000 years, which seems like a reasonable amount of time but we'll see it's a very small fraction when you compare it to 4.54 billion years. So let's say that that's that little there and I am actually overdoing it when I'm drawing the diagram. So that right over there is the amount of time humans have been 200,000 years and I am actually drawing this way too big. But what we want to figure out is what is that equivalent length in seconds on a basketball game or another way to think about it, 200,000 years is to 4.54 billion years as the number of seconds. Let's call this thing right over here x as x is to the total number of seconds in a game. So 2,880 seconds, 2,880 seconds just like that. And one way that we can do this, to solve for x and this is kind of a more basic algebra, but just as a reminder if we want to solve for x here, the easiest way is to multiply by 2,880 and that will cancel with this right over here. but we can't just do it to the right hand side, we also have to do it to the left hand side and so if you multiply both sides by 2,880, so multiply both sides by 2,880, you get that x, the number of the equivalent number of seconds in a basketball game. If the history of the earth was a basketball game, when the humans would show up is equal to the fraction of earth's history that humans have been around. That's this part right over here, times the number of seconds in a game. So let's think about what we get there. We are going to have the fraction of earth's history, so 200,000 divided by 4.45 billion. There's a couple of ways I could write it. I could write it 4.54e9, which literally means 4.54 times 10 to the 9th or 4.54 times 1 followed with 9 zeros or 4.54 times a billion, which is exactly 4.54 billion. So I could write it like that or I could just write it out. I could write 4, 5 so 4 billion, we'll be careful, 4, 540,000,000 and now I am doing the thousands 2, 3 and now I am just doing the 1, 2, 3. So this is 4 billion and then I should have 9 places after that 1,2,3,4,5,6,7,8,9. So this expression right here is a fraction of earth's history that anatomically modern humans have been around and then I am going to multiply that, I am going to multiply that, times I am going to multiply that times the number of seconds of regulation play, 2,880 and now drum roll, we get .12, we could round up .13 seconds. So we get x is equal to, or I could say maybe approximately equal to 0.13 seconds and so, just to imagine if the history of earth were regulation play of an NBA game and let's say this game is kind of, at the end of this game there is a buzzer beater shot that wins the game, the humans don't show up in the game until the ball has left that final shot taker's hands and it has just about to enter the basket that's when the first humans, 200,000 years ago will show up. A little over a tenth of a second before the end of the game and since we already have our brains in this mindset, I will throw out another interesting question. Okay, humans are you know, just a flash that they've actually shown up. Just as the ball is about to go into the basket, we have a little over a tenth of a second left in the game. Think about when the dinosaurs went extinct. I will give you a hint here. The dinosaurs went extinct. So we believe an asteroid hit the earth. So, this kind of a meteorite. This is a current theory, meteorite hit the earth, mass extinction event 65.5 million years ago. So if we think of that 65.5 million years ago relative to the history of the earth and if the history of the earth were just regulation play of a basketball game, when did that happen? Well, once again let's get our calculator out. What fraction of the earth's history ago was this? 65.5 million, I could write this as 65.5 times 10 to the 6th, which is that and then I'll just divide that by 4.54 billion years. So times 10 to the 9th and so this is the fraction of earth's history that has happened since the extinction of the dinosaurs. So a little more, about a percent and a half of earth's history has happened since the extinction of the dinosaurs 65.5 million years ago and if we are talking about a basketball game, let's just multiply that times the number of seconds in a basketball game, 2,880 and we get about 41 seconds. So with less than a minute left in the 4th quarter is when the meteorite hits the earth" + }, + { + "Q": "32 plus 32 equals 64", + "A": "Yes because you solve addition problems on top of each other from right to left: 2+2=4, there s your last number. 3+3=6, there s your first. They collide to make 64 so that is true, 32+32=64", + "video_name": "nFsQA2Zvy1o", + "transcript": "Most of us are familiar with the equal sign from our earliest days of arithmetic. You might see something like 1 plus 1 is equal to 2. Now, a lot of people might think when they see something like this that somehow equal means give me the answer. 1 plus 1 is the problem. Equal means give me the answer and 1 plus 1 is 2. That's not what equal actually means. Equal is actually just trying to compare two quantities. When I write 1 plus 1 equals 2, that literally means that what I have on the left hand side of the equal sign is the exact same quantity as what I have on the right hand side of the equal sign. I could have just as easily have written 2 is equal to 1 plus 1. These two things are equal. I could have written 2 is equal to 2. This is a completely true statement. These two things are equal. I could have written 1 plus 1 is equal to 1 plus 1. I could have written 1 plus 1 minus 1 is equal to 3 minus 2. These are both equal quantities. What I have here on the left hand side, this is 1 plus 1 minus 1 is 1 and this right over here is 1. These are both equal quantities. Now I will introduce you to other ways of comparing numbers. The equal sign is when I have the exact same quantity on both sides. Now we'll think about what we can do when we have different quantities on both sides. So let's say I have the number 3 and I have the number 1 and I want to compare them. So clearly 3 and 1 are not equal. In fact, I could make that statement with a not equal sign. So I could say 3 does not equal 1. But let's say I want to figure out which one is a larger and which one is smaller. So if I want to have some symbol where I can compare them, where I can tell, where I can state which of these is larger. And the symbol for doing that is the greater than symbol. This literally would be read as 3 is greater than 1. 3 is a larger quantity. And if you have trouble remembering what this means-- greater than-- the larger quantity is on the opening. I guess if you could view this as some type of an arrow, or some type of symbol, but this is the bigger side. Here, you have this little teeny, tiny point and here you have the big side, so the larger quantity is on the big side. This would literally be read as 3 is greater than-- so let me write that down-- greater than, 3 is greater than 1. And once again, it just doesn't have to be numbers like this. I could write an expression. I could write 1 plus 1 plus 1 is greater than, let's say, well, just one 1 right over there. This is making a comparison. But what if we had things the other way around. What if I wanted to make a comparison between 5 and, let's say, 19. So now the greater than symbol wouldn't apply. It's not true that 5 is greater than 19. I could say that 5 is not equal to 19. So I could still make this statement. But what if I wanted to make a statement about which one is larger and which one is smaller? Well, as in plain English, I would want to say 5 is less than 19. So I would want to say-- let me write that down-- I want to write 5 is less than 19. That's what I want to say. And so we just have to think of a mathematical notation for writing \"is less than.\" Well, if this is greater than, it makes complete sense that let's just swap it around. Let's make, once again, the point point towards the smaller quantity and the big side of the symbol point to the larger quantity. So here 5 is a smaller quantity so I'll make the point point there. And 19 is a larger quantity, so I'll make it open like this. And so this would be read as 5 is less than 19. 5 is a smaller quantity than 19. I could also write this as 1 plus 1 is less than 1 plus 1 plus 1. It's just saying that this statement, this quantity, 1 plus 1 is less than 1 plus 1 plus 1." + }, + { + "Q": "Does anybody have an idea of the approximate cost of the parts necessary for this project, assuming one doesn't just have any of them lying around but needs to buy everything new?", + "A": "Depends where you live, robot part are about twice as expensive in Australia then they are in America. You re probably looking at $40. (he might tell you in another video).", + "video_name": "VnfpSf6YxuU", + "transcript": "This is all our parts all laid out for the Bit-zee bot, the things that you'll need to make one. Now you can make yours out of a broad variety of things, and we highly recommend that you do that. The only thing that you really have to have is the Arduino. Everything else you can switch out for other things. You can use different types of batteries. You can use different motors, et cetera. I'm going to go through what I've got here and where the products came from, or where the parts came from, I should say. And then we're going to start to put a Bit-zee together on this board so you can see how it's all wired up. But if you don't happen to have two hair dryers that you can take apart, you can either go and buy two electric motors and get some wheels for them. Or there's a variety of things you can do to solve that problem. So again, these are two motors from our hair dryer. You can see the hair dryer blower fan there. And underneath that is a sheet of Lexan. It's a stiff plastic that's really resilient. And that sheet of Lexan, it's easy to machine. You can drill holes in it and do stuff like that. So it's going to be used for mounting some of our devices. And you can get that at a hardware store for a few dollars. And this is a universal remote, and it can be gotten at Target for around $8. And we're going to use that to control our Bit-zee bot. And then we've got some electrical tape and different 22-gauge wire. And then we have some solder here. We'll use that to make our solder connections. Just like if you saw the video for the motor controller, it was used to solder that together. And this is a motor controller, which will allow us to control the speed and direction of our motors. And this is our Arduino. It's our microprocessor that we can plug into-- I should say a microcontroller that we can plug into our computer and download code to it to get the motor controller and other things to function the way we want them to. So this is a breadboard, and it's used for prototyping. And we're going to show you how to wire it up and how to connect different electronic components to that. And this is our digital recording module, and it's for basically recording sounds and playing them back. And we're going to use the Arduino to trigger that so that when the little bot drives around it can make some sounds. Of course, these are just double-A batteries, and they're going to go in this battery holder. The double-As are 1.5 volts. But when we connect them in series together, they're going to be 12 volts. So that'll be great for powering our motors, because they want to run on a higher voltage than 1.5. And so we have our different transistors here that we're going to use to do some switching in our circuits. And we've got some three-color LED and some screws and nuts and then a bunch of resistors. And these are 330 ohm, 10K ohm, 220 ohm. We'll go into the details on that kind of stuff later. And then I have some of the board of our alarm clock radio, so we're going to use some components off of that board. And we've got our coffee maker here. Or I should say our coffee carafe; it's just the holder for the coffee. And we're going to use some of the components on this for the Bit-zee. And then we have some-- over here you can see some zip ties. And so we also have an-- we're going to need a infrared sensor for the Bit-zee. And that infrared sensor will be used for sensing from the remote control. And so it looks like this, and we'll have one of those as well." + }, + { + "Q": "How do you convert a fractional percent into a percent", + "A": "convert the fraction to decimal and multiply with 100", + "video_name": "FaDtge_vkbg", + "transcript": "Let's give ourselves a little bit of practice with percentages. So let's ask ourselves, what percent of-- I don't know, let's say what percent of 16 is 4? And I encourage you to pause this video and to try it out yourself. So when you're saying what percent of 16 is 4, percent is another way of saying, what fraction of 16 is 4? And we just need to write it as a percent, as per 100. So if you said what fraction of 16 is 4, you would say, well, look, this is the same thing as 4/16, which is the same thing as 1/4. But this is saying what fraction 4 is of 16. You'd say, well, 4 is 1/4 of 16. But that still doesn't answer our question. What percent? So in order to write this as a percent, we literally have to write it as something over 100. Percent literally means \"per cent.\" The word \"cent\" you know from cents and century. It relates to the number 100. So it's per 100. So you could say, well, this is going to be equal to question mark over 100, the part of 100. And there's a bunch of ways that you could think about this. You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25. In the numerator to go from-- I need to also multiply by 25 in order to have an equivalent fraction. So I'm also going to multiply by 25. So 1/4 is the same thing as 25/100. And another way of saying 25/100 is this is 25 per 100, or 25%. So this is equal to 25%. Now, there's a couple of other ways you could have thought about it. You could have said well, 4/16, this is literally 4 divided by 16. Well, let me just do the division and convert to a decimal, which is very easy to convert to a percentage. So let's try to actually do this division right over here. So we're going to literally divide 4 by 16. Now, 16 goes into 4 zero times. 0 times 16 is 0. You subtract, and you get a 4. And we're not satisfied just having this remainder. We want to keep adding zeroes to get a decimal answer right over here. So let's put a decimal right over here. We're going into the tenths place. And let's throw some zeroes right over here. The decimal makes sure we keep track of the fact that we are now in the tenths, and in the hundredths, and in the thousandths place if we have to go that far. But let's bring another 0 down. 16 goes into 40 two times. 2 times 16 is 32. If you subtract, you get 8. And you could bring down another 0. And we have 16 goes into 80. Let's see, 16 goes into 80 five times. 5 times 16 is 80. You subtract, you have no remainder, and you're done. 4/16 is the same thing as 0.25. Now, 0.25 is the same thing as twenty-five hundredths. Or, this is the same thing as 25/100, which is the same thing as 25%." + }, + { + "Q": "You said that the diastole occurs between the dub and lub but isnt the dub sound or the 2nd heart sound supposed to be diastole?", + "A": "The dub and lub sounds are actually the sounds created by the valves closing; the first heart sound (S1) are the atrioventricular (AV) valves closing, and the second heart sounds (S2) are the semilunar valves closing. So diastole (the heart filling with blood) occurs when the AC valves are open and the semilunar are closed, and vice versa for systole.", + "video_name": "-4kGMI-qQ3I", + "transcript": "If you take a good long listen to your heart, you'll actually notice that it makes sounds. And those sounds are usually described as lub dub, lub dub, lub dub. And if you actually try to figure out what that would spell out like, usually it's something L- U- B, D- U- B. And it just repeats over and over and over. And to sort of figure out where those sounds come from, what I did is I took that diagram of the heart that we've been using and actually exaggerated the valves, made them really, really clear to see in this picture. And we'll use those valves to kind of talk through where those sounds are coming from. So let's start by labeling our heart. So we've got at the top, blood is coming into the right atrium and going to the right ventricle. It goes off to the lungs, comes back into the left atrium and then the left ventricle. So these are the chambers of our heart. Now, keep your eye on the valves. And we'll actually talk about them as the blood moves through. So let's start with blood going from the right atrium this way into the right ventricle. Now, at the same moment that blood is actually going from the right atrium to the right ventricle, blood is actually also going from the left atrium to the left ventricle. Now, you might think, well, how's that possible? How can blood be in two places at one time? But now remember that blood is constantly moving through the heart. So in a previous cycle, you actually had some blood that was coming back from the lungs, and that's what's dumping into the left ventricle. And in a new cycle, you have a bit of blood that's going from the right atrium to the right ventricle. So you have simultaneously two chambers that are full of blood-- the right and left ventricle. Now, to get the blood into those ventricles, the valves had to open. And specifically, let's label all the valves now. So here you have our tricuspid valve, and I'm going to label that as just a T. And then up here, you have the pulmonary valve, and this'll be just a P. And on the other side, you've got the mitral valve, which separates the left atrium from the left ventricle. And you've got the aortic valve. So these are the four valves of the heart. And as the blood is now in the ventricles, you can see that the tricuspid and the mitral valve are open. So far, so good. Now, I've actually drawn the pulmonary valve as being open. But is that really the case? And the answer is no, because what happens is that as blood is moving down from the right atrium to the right ventricle, let's say that-- and I'm going to draw it in black. Black arrows represent the bad or the wrong direction of flow. So let's say some blood is actually trying to go that way, which is not the way it should be going. What happens is that these two valves, they, based on their shape, are actually not-- they're going to jam up. They're going to basically just jam up like this, and they're not going to let the blood pass through. So this is what happens as that valve closes down. And the same thing happens on this side. Let's imagine you have some backwards flow of blood by accident, meaning that it's going in the wrong direction. Well, then these valves are going to close down. So the white arrows represent the correct flow of blood, and the black arrows represent the incorrect flow of blood. So these valves shut down like that. So now you can see how the valves, the aortic and pulmonary valve, are actually closed when the mitral and tricuspid valve are open. So what happens after this? So now our ventricles are full of blood, right? They're full of blood. And let's say they squeeze down, and they jettison all the blood into those arteries. Well, now you're going to have-- this is actually going to close down. Let's say this arrow flips around. These arrows become white, because the direction of flow is going to be in the direction we want it. It's going to go this way and this way And to allow that, of course, I need to show you that these open up. And they allow the blood to go the way that we want it to go, so now blood is going to flow through those two valves. But similar to before, you could have some backflow here. You could have backflow here. And you can have backflow here. So you can imagine now, let's say you have a little bit of backflow that wants to go this way, which is the wrong direction. Right? Well, then these valves are going to close up. They're going to say, no, you can't go that way. They're going to close right up, and they're going to not allow blood to go that way. So this is going to happen on both sides, both ventricles. And the valves shut. And so basically the backflow of blood is not allowed, because the valves keep shutting. And when the valves snap shut-- so for example, right now the tricuspid valve and the mitral valve snapped shut. Well, that makes a noise. So when T and M snap shut, that makes a noise that we call lub. That's that first noise, that first heart sound. In fact, sometimes people don't even call it lub dub. They say, well, it's the first heart sound. And to make that even shorter, sometimes people call that S1. So if you hear S1, you know they're talking about that same exact thing. And this dub is called the second heart sound. And, no surprise, just as before, if that's S1, this is S2. So you'll hear S1 when the tricuspid and mitral valve snap shut. So far, so good. But you also know that if that's what's making noise, you can kind of guess-- and it's a very smart guess-- that at the same time, the pulmonic valve and the aortic valve just opened. So if the other valves snap shut, these just opened. You can kind of assume that, although the noise you're hearing is actually from here. So what's happening with dub? Well, the opposite. And what I mean by that is-- let me now show you what happens a moment later. Well, after the ventricles are done squeezing, then we get to a point where you might have a little bit of flow that way and that way, just as I drew before. And these valves snap shut as well. So now these snap shut. And as these snap shut-- because they don't want to allow backflow, right? They're going to snap shut like that. They make noise. And so when you have dub, you actually have noise coming from the pulmonic and aortic valve snapping shut. And that must mean that then the other two valves just opened up-- the tricuspid and mitral just opened. You can assume that, right? And I didn't draw that in the picture. Let me update my picture now to show that. So now these two have opened up, and blood is coming into the ventricles again. So it's actually a nice little rhythm that you get going. And every time these valves go open and shut, you hear noise. So you can kind of figure out what's happening based on-- and these actually-- let me erase that. And now you have white arrows going this way. And we've returned to where we started from. So you basically have a full cycle, and between these two-- so let's say from lub to dub, because there's a little bit of space there. If you were to follow it over time, over time, this is what it might look like if this is a little timeline. You might hear lub here, or the first heart sound. I'll just call it S1. And you might hear S2 here, the second heart sound. And then you'll hear S1 again over here and S2. And what's happening between the two-- so between these two, this time lag here-- is that blood is actually squeezing out, because the pulmonary and aortic valves just opened. It is squeezing out and going out to the whole body. So this is when blood is going to the body, and sometimes we call that systole. And between dub and the next lub-- so in this area right here-- well, at that point, blood is kind of refilling from the atriums into the ventricles, and we call that diastole. So now you can actually listen to your heart. And you can actually figure out, well, if you're listening to the sound between lub and dub or the space in time between lub and dub, that's when you're having systole. And if you're listening to or waiting for the sound to start up again-- so you just heard dub, and you're waiting for lub again-- then that space in time is diastole." + }, + { + "Q": "can someone explain how 5 /5\u00e2\u0088\u009a61/61 = \u00e2\u0088\u009a61? I don't understand how the bottom 61 cancels out. Thanks : )", + "A": "What is 61 divided by square root of 61?", + "video_name": "l5VbdqRjTXc", + "transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly solving for a, we could just multiply both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. I say approximately because I rounded it down. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse. So this side WY is the hypotenuse. And so what trigonometric ratios-- or we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that XY is exactly 5 and we don't have to deal with this approximation, let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well we see from SohCahToa cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees is equal to the length of the adjacent side, which is 5 over the length of the hypotenuse, which has a length of b. And then we can try to solve for b. You multiply both sides times b, you're left with b times cosine of 65 degrees is equal to 5. And then to solve for b, you could divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing-- we have to figure it out what our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees. And we're left with b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is 5 divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that 5 squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle W right over here. So I'll give you a few seconds to think about what the measure of angle W is. Well here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is-- well we can simplify the left-hand side right over here. 65 plus 90 is 155. So angle W plus 155 degrees is equal to 180 degrees. And then we get angle W-- if we subtract 155 from both sides-- angle W is equal to 25 degrees. And we are done solving the right triangle shown below." + }, + { + "Q": "what would happen if point a is underneath the x-axis?", + "A": "You would follow the same procedure.", + "video_name": "0rlNHYHhrWs", + "transcript": "Find the point B on segment AC, such that the ratio of AB to BC is 3 to 1. And I encourage you to pause this video and try this on your own. So let's think about what they're asking. So if that's point C-- I'm just going to redraw this line segment just to conceptualize what they're asking for. And that's point A. They're asking us to find some point B that the distance between C and B, so that's this distance right over here. So if this distance is x, then the distance between B and A is going to be 3 times that. So this will be 3x. That the ratio of AB to BC is 3 to 1. So that would be the ratio-- let me write this down. It would be AB-- that looks like an HB-- it would be AB to BC is going to be equal to 3x to x, which is the same thing as 3 to 1, if we wanted to write it a slightly different way. So how can we think about it? You might be tempted to say, oh, well, you could use the distance formula to find the distance, which by itself isn't completely uncomplicated. And then this will be 1/4 of the way. Because if you think about it, this entire distance is going to be 4x. Let me draw that a little bit neater. This entire distance, if you have an x plus a 3x, is going to be 4x. So you'd say, well, this is 1 out of the 4 x's along the way. This is going to be 1/4 of the distance between the two points. Let me write that down. This is 1/4 of the way between C and B, going from C to A. B is going to be 1/4 of the way. So maybe you try to find the distance. And you say, well, what are all the points that are 1/4 of the way? But it has to be 1/4 of that distance away. But then it has to be on that line. But that makes it complicated, because this line is at an incline. It's not just horizontal. It's not just vertical. What we can do, however, is break this problem down into the vertical change between A and C, and the horizontal change between A and C. So for example, the horizontal change between A and C, A is at 9 right over here, and C is at negative 7. So this distance right over here is 9 minus negative 7, which is equal to 9 plus 7, which is equal to 16. And you see that here. 9 plus 7, this total distance is 16. That's the horizontal distance change going from A to C, or going from C to A. And the vertical change, and you could even just count that, that's going to be 4. C is at 1. A is at 5. Going from 1 to 5, you've changed vertically 4. So what we can say, going from C to B in each direction, in the vertical direction and the horizontal direction, we're going to go 1/4 of the way. So if we go 1/4 in the vertical direction, we're going to end up at y is equal to 2. So I'm just going, starting at C, 1/4 of the way. 1/4 of 4 is 1. So I've just moved up 1. So our y is going to be equal to 2. And if we go 1/4 in the horizontal direction, 1/4 of 16 So we go 1, 2, 3, 4. So we end up right over here. Our x is negative 3. So we end up at that point right over there. We end up at this point. This is the point negative 3 comma 2. And if you were really careful with your drawing, you could have actually just drawn-- well, actually you don't have to be that careful, since this is graph paper. You actually could have just said, hey, we're going to go 1/4 this way. Where does that intersect the line? Hey, it intersects the line right over there. Or you could have said, we're going to go 1/4 this way. Where does that intersect the line? And that would have let you figure it out either way. So this point right over here is B. It is 1/4 of the way between C and A. Or another way of thinking about the distance between C and B, which we haven't even figured out. We could do that using the distance formula or the Pythagorean theorem, which it really is. This distance, the distance CB, is 1/3 the distance BA. The ratio of AB to BC is 3 to 1." + }, + { + "Q": "what if the numerator is larger than the denominator how would you solve it?", + "A": "Same thing just it would be 1.", + "video_name": "NM8qTo361ic", + "transcript": "Let's see if we can express 16/21 as a decimal. Or we could call this 16 twenty-firsts. This is also 16 divided by 21. So we can literally just divide 21 into 16. And because 21 is larger than 16, we're going to get something less than 1. So let's just literally divide 21 into 16. And we're going to have something less than 1. So let's add some decimal places here. We're going to round to the nearest thousandths in case our digits keep going on, and on, and on. And let's start dividing. 21 goes into 1 zero times. 21 goes into 16 zero times. 21 goes into 160-- well, 20 would go into 160 eight times. So let's try 7. Let's see if 7 is the right thing. So 7 times 1 is 7. 7 times 2 is 14. And then when we subtract it, we should get a remainder less than 21. If we pick the largest number here where, if I multiply it by 21, I get close to 160 without going over. And so if we subtract, we do get 13. So that worked. 13 is less than 21. And you could just subtract it. I did it in my head right there. But you could regroup. You could say this is a 10. And then this would be a 5. 10 minus 7 is 3. 5 minus 4 is 1. 1 minus 1 is 0. Now let's bring down a 0. 21 goes into 130. Would 6 work? It looks like 6 would work. 6 times 21 is 126. So that looks like it works. So let's put a 6 there. 6 times 1 is 6. 6 times 2 is 120. There's a little bit of an art to this. All right, now let's subtract. And once again, we can regroup. This would be a 10. We've taken 10 from essentially this 30. So now this becomes a 2. 10 minus 6 is 4. 2 minus 2 is 0. 1 minus 1 is 0. Now let's bring down another 0. 21 goes into 40, well, almost two times, but not quite, so only one time. 1 times 21 is 21. And now let's subtract. This is a 10. This becomes a 3. 10 minus 1 is 9. 3 minus 2 is 1. And we're going have to get this digit. Because we want to round to the nearest thousandth. So if this is 5 or over, we're going round up. If this is less than 5, we're going to round down. So let's bring another 0 down here. And 21 goes into 190. Let's see, I think 9 will work. 9 times 1 is 9. 9 times 2 is 18. When you subtract, 190 minus 189 is 1. And we could keep going on, and on, and on. But we already have enough digits to round to the nearest thousandth. This digit right over here is greater than or equal to 5. So we will round up in the thousandths place. So if we round to the nearest thousandths, we can say that this is 0.76. And then this is where we're going around up-- 762." + }, + { + "Q": "is there a formula to con vert the remainder of a problem into a fraction in an easy way?", + "A": "take the remainder and put it over the divisor in a fraction then simplify into lowest terms. if there is a remainder or 2 and a divisor of 10 the fraction is 2/10 with both numbers reduced to 1/5", + "video_name": "NcADzGz3bSI", + "transcript": "It never hurts to get a lot of practice, so in this video I'm just going to do a bunch more of essentially, what we call long division problems. And so if you have 4 goes into 2,292. And I don't know exactly why they call it long division, and we saw this in the last video a little bit. I didn't call it long division then, but I think the reason why is it takes you a long time or it takes a long piece of your paper. As you go along, you kind of have this thing, this long tail that develops on the problem. So all of those are, at least, reasons in my head why it's called long division. But we saw in the last video there's a way to tackle any division problem while just knowing your multiplication tables up to maybe 10 times 10 or 12 times 12. But just as a bit of review, this is the same thing as 2,292 divided by 4. And it's actually the same thing, and you probably haven't seen this notation before, as 2,292 divided by 4. This, this, and this are all equivalent statements on some level. And you could say, hey Sal, that looks like a fraction in case you have seen fractions already. And that is exactly what it is. It is a fraction. But anyway, I'll just focus on this format and in future videos we'll think about other ways to represent division. So let's do this problem. So 4 goes into 2 how many times? It goes into 2 no times, so let's move on to-- let me just switch colors. So let's move on to the 22. 4 goes into 22 how many times? 4 times 5 is equal to 20. 4 times 6 is equal to 24. So 6 is too much. So 4 goes into 22 five times. 5 times 4 is 20. There's going to be a little bit of a leftover. And then we subtract 22 minus 20. Well that's just 2. And then you bring down this 9. And you saw in the last video exactly what this means. When you wrote this 5 up here-- notice we wrote So this is really a 500. But in this video I'm just going to focus more on the process, and you can think more about what it actually means in terms of where I'm writing the numbers. But I think the process is going to be crystal clear hopefully, by the end of this video. So we brought down the 9. 4 goes into 29 how many times? It goes into at least six times. What's 4 times 7? 4 times 7 is 28. So it goes into it at least seven times. What's 4 times 8? 4 times 8 is 32, so it can't go into it eight times so it's going to go into it seven. 4 goes into 29 nine seven times. 7 times 4 is 28. 29 minus 28 to get our remainder for this step in the problem is 1. And now we're going to bring down this 2. We're going to bring it down and you get a 12. 4 goes into 12? That's easy. 4 times 3 is 12. 4 goes into 12 three times. 3 times 4 is 12. 12 minus 12 is 0. We have no remainder. So 4 goes into 2,292 exactly 573 times. So this 2,292 divided by 4 we can say is equal to 573. Or we could say that this thing right here is equal to 573. Let's do a couple of more. Let's do a few more problems. So I'll do that red color. Let's say we had 7 going into 6,475. Maybe it's called long division because you write it nice and long up here and you have this line. I don't know. There's multiple reasons why it could be called long division. So you say 7 goes into 6 zero times. So we need to keep moving forward. So then we go to 64. 7 goes into 64 how many times? Let's see. 7 times 7 is? Well, that's way too small. Let me think about it a little bit. Well 7 times 9 is 63. That's pretty close. And then 6 times 10 is going to be too big. 7 times 10 is 70. So that's too big. So 7 goes into 64 nine times. 9 times 7 is 63. 64 minus 63 to get our remainder of this stage 1. Bring down the 7. 7 goes into 17 how many times? Well, 7 times 2 is 14. And then 7 times 3 is 21. So 3 is too big. So 7 goes into 17 two times. 2 times 7 is 14. 17 minus 14 is 3. And now we bring down the 5. And 7 goes into 35? That's in our 7 multiplication tables, five times. 5 times 7 is 35. And there you go. So the remainder is zero. So all the examples I did so far had no remainders. Let's do one that maybe might have a remainder. And to ensure it has a remainder I'll just make up the problem. It's much easier to make problems that have remainders than the ones that don't have remainders. So let's say I want to divide 3 into-- I'm going to divide it into, let's say 1,735,092. This will be a nice, beastly problem. So if we can do this we can handle everything. So it's 1,735,092. That's what we're dividing 3 into. And actually, I'm not sure if this will have a remainder. In the future video I'll show you how to figure out whether something is divisible by 3. Actually, we can do it right now. We can just add up all these digits. 1 plus 7 is 8. 8 plus 3 is 11. 11 5 five is 16. 16 plus 9 is 25. 25 plus 2 is 27. So actually, this number is divisible by 3. So if you add up all of the digits, you get 27. And then you can add up those digits-- 2 plus 7 is 9. So that is divisible by 9. That's a trick that only works for 3. So this number actually is divisible by 3. So let me change it a little bit, so it's not divisible by 3. Let me make this into a 1. Now this number will not be divisible by 3. I definitely want a number where I'll end up with a remainder. Just so you see what it looks like. So let's do this one. 3 goes into 1 zero times. You could write a 0 here and multiply that out, but that just makes it a little bit messy in my head. So we just move one to the right. 3 goes into 17 how many times? Well, 3 times 5 is equal to 15. And 3 times 6 is equal to 18 and that's too big. So 3 goes into 17 right here five times. 5 times 3 is 15. And we subtract. 17 minus 15 is 2. And now we bring down this 3. 3 goes into 23 how many times? Well, 3 times 7 is equal to 21. And 3 times 8 is too big. That's equal to 24. So 3 goes into 23 seven times. 7 times 3 is 21. Then we subtract. 23 minus 21 is 2. Now we bring down the next number. We bring down the 5. I think you can appreciate why it's called long division now. We bring down this 5. 3 goes into 25 how many times? Well, 3 times 8 gets you pretty close and 3 times 9 is too big. So it goes into it eight times. 8 times 3 is 24. I'm going to run out of space. You subtract, you get 1. 25 minus 24 is 1. Now we can bring down this 0. And you get 3 goes into 10 how many times? It goes into it three times. 3 times 3 is 9. That's about as close to 10 as we can get. 3 times 3 is 9. 10 minus 9, I'm going to have to scroll up and down here a little bit. 10 minus 9 is 1, and then we can bring down the next number. I'm running out of colors. I can bring down that 9. 3 goes into 19 how many times? Well, 6 is about as close as we can get. That gets us to 18. 3 goes into 19 six times. 6 times 3-- let me scroll down. 6 times 3 is 18. 19 minus 18-- we subtract it up here too. 19 minus 18 is 1 and then we're almost done. I can revert back to the pink. We bring down this 1 right there. 3 goes into 11 how many times? Well, that's three times because 3 times 4 is too big. 3 times 4 is 12, so that's too big. So it goes into it three times. So 3 goes into 11 three times. 3 times 3 is 9. And then we subtract and we get a 2. And there's nothing left to bring down. When we look up here there's nothing left to bring down, so we're done. So we're left with the remainder of 2 after doing this entire problem. So the answer, 3 goes into 1,735,091-- it goes into it 578,363 remainder 2. And that remainder 2 was what we got all the way down there. So hopefully you now appreciate and you can tackle pretty much any division problem. And you also, through this exercise, can appreciate why it's called long division." + }, + { + "Q": "Did you notice that the numbers that have an odd number of factors all are perfect square numbers! I wonder if it has to do with the extra one factor...He explains that later in the video", + "A": "You mean the loss of one factor, since you can t count the square root of the number twice.", + "video_name": "WNhxkpmVQYw", + "transcript": "Let's say we have 100 light bulbs. Well, let me actually just draw them. So I have one light bulb there. I have another light bulb. And I have 100 of them. 100 light bulbs. And what I'm going to do is-- Well actually, before I even start turning these light bulbs on and off, let me let you know that they are all off. So they start off. Now, the next thing I'm going to do is I'm going to number the hundred light bulbs. I'm going to number them one through 100. So the first light bulb is light bulb one. The second light bulb is light bulb two. All the way to light bulb 100. And what I'm going to do is, first I'm going to go and I'm going to switch essentially every light bulb. So if they all start off, I'm going to turn them all on. So let me do it. So on my first pass-- let's call this pass one-- so in pass one, I'm going to turn all of these on. On, on, on. They're all going to be turned on. On. And then in pass two, what I'm going to do is I'm going to switch only every other light bulb. So, for example I'll say, OK, I won't switch light bulb one. I'll only switch light bulb two. So light bulb one will stay on. Light bulb two will be off. Light bulb three will be on. Light bulb four will be off. And so essentially, every light bulb-- if you look at their numbers-- that is a multiple of two, will be switched. So 100 will be switched, so that'll also be off. Then I'm going to come-- and ignore this right here-- and I'm going to switch every third light bulb. So what's going to happen? This one's going to-- let me switch colors arbitrarily-- this one's going to stay on. This one's going to stay off. And I'm going to switch the third light bulb. So this one was on. Now this one will be off. The fourth light bulb will stay off, because I'm not touching it. The fifth light bulb would have been on and it'll stay on. Now the sixth light bulb, in this case we switched it off, and now it'll be on again. But I think you get the point. Every third light bulb, or if we look at the numbers of the light bulb, every numbered light bulb would that is a multiple of three is going to be switched. And if it was a multiple of three and two, it would have been switched on the first time and then off the second time. But I think you're getting the point. But what I'm going to do is, I'm going to do 100 passes. So the first pass, I switch every light bulb. They all started off, so they're all going to be turned on. The second pass, I switch every other light bulb or every second light bulb. The third pass, I do every third one, or that's a multiple of three. And my question to you is, after 100 passes, how many light bulbs are still on? Or how many are on, period? And that is the brain teaser? How do you figure out, of the hundred, which ones are going to be on? You should be able to do this in your head. You don't have to make an Excel spreadsheet and actually do all the on and off switches. So the first question is, how many of these are going to be on after I do 100 passes? And just to make it clear, what's the 100th path going to be? Well, I'm only going to switch every 100th light bulb. So whatever this light bulb was already doing, I'm just going to switch it. If it was off, it'll come on. If it was on, it'll become off. So the first question is, how many of these are going to be on after 100 passes? And then the bonus question is, which of these are going to be on? And so that's the question. Pause it if you don't want the answer. And then try to solve it. I think it shouldn't take you too much time. But now I'm going to give you the answer. Or maybe I'll start with a couple of hints. So, when do we know that a light bulb is being switched? So if I'm on the second pass, I will-- I don't want to say turn on. I will switch every light bulb that's divisible by two. And then if I'm on the third pass, I'll switch every light bulb that's divisible by three. So on every pass, what am I doing? If I'm on pass n-- this is a hint if you want it-- what do I know? I know all the light bulbs that are numbered where n is a factor of that light bulb, that will get switched. So we know that switched if n is a factor of the light bulb number. And that's just a fancy way of saying, look, if I'm on pass 17, I'm going to switch all the multiples of 17. Or I could say, I know that I'm going to switch light bulb 51, because 17 is a factor of 51. So that tells you that we're always going to be switching one of these light bulbs on or off when one of its factors is our pass. So for example, if we're looking at light bulb eight. This is light bulb eight. So when will it be switched? So on pass one, we're definitely going So pass one, it's going to be switched on. Pass two, it'll be switched off. I know that because two is divisible into eight. Pass three, nothing's going to happen. On pass three, nothing's going to happen because this isn't a multiple of three. Pass four, what'll happen? It will be switched. It'll be switched back on. And then pass eight is the next time we'll touch this light bulb, and it'll be switched back. So every time one of its factors go by, we're going to switch this thing. And as you can see, in order for it to be on at the end, you have to have an odd number of factors. So that's an interesting thing. So in order for a light bulb to be on, it has to have odd number of factors. Now that's an interesting question. What numbers have an odd number of factors. And this is something that I think they should teach you in grade school, and they never do. But it's a really interesting kind of number theory. It's a simple one, but it's interesting to think about. So what numbers are true? Let's do all the factors for some of the starting numbers. So all the factors of one. Well one, the only factor is just one. So one works? One has an odd number of factors. So that means that one will remain on. Because you're only going to turn it on in the first pass. Makes sense. Two. What are all the factors of two. Well you have one and two. So two has an even number. You're going to switch it on the first time, then off the second time. Then you're never going to touch it again. So this is going to stay off. Your factors are one and three. Four. Your factors are one, two and four. Interesting. Here we have three factors. We have an odd number of factors. So four is going to stay on. We're going to turn it on in our first pass, we're going to turn it off on our second pass. And we're going to turn it on again in our fourth pass. Let's keep going. So five. The factors of five are one and five. The factors are one, two, three and six. It's an even number, so they're going to be off when we're done with it. Seven. it's one and seven. We just did that. It's one, two, four and eight. Still going to be off. Nine. Let's see. The factors are one, three and nine. Interesting. Once again we have an odd number of factors. So the light bulb number nine is also going to be on when everything's done. Let's keep going. I don't know, I actually did this at our mental boot camp with some of the kids. And they immediately said, the distance between one and four is three. The distance between four and nine is five. And maybe the distance between nine and the next number is going to be seven. It increases by odd numbers. What's nine plus seven? 16. What are the factors of 16? They're one, two, four, eight and 16. Interesting. From nine to 16, you incremented by seven. From four to nine, you incremented by five. So it seems like we have a pattern. But can you see something even more interesting about the numbers one, four, nine and 16. And you could try all the numbers between nine and 16, and you'll see that they have an even number of factors. But what's interesting about all of these numbers? Why do they have an odd number of factors? In all of these other cases, every factor is paired with another number. One times two is two. One time six is six. Two times three is six. There's always a pair. Except for these numbers. There's no pair. Why isn't there? One times four is four. But two times two is also four. So we only write the two once. Three times three is nine. Four times four is 16. So all the lights that are going to be on are actually perfect squares. That's why they have an odd number of factors. So what's our question? So this problem of what light bulbs are going to be on, boils down to how many perfect squares are there between one and 100? And you could just list them out. And you could say, oh well the perfect squares are one, four, nine, 16. And you could try to think of them all. Or you could say, well how many numbers can I square and get a number less than or equal to 100? Well 100 is equal to 10 squared. So you can only square the numbers between one and 10 to get a perfect square. So there's only 10 perfect squares between one and 100. Hopefully you didn't lose that, but if that confuses you, just list them all out. Given that 100 is the largest number, it's the whole largest perfect square there. And that's 10 squared. The only other perfect squares in our range we're talking about are one squared, two squared, three squared, all And we could do that. Four squared is 16. Five squared is 25. 36, 49, 64, 81, and 100. So the light bulbs with these numbers on it are the ones that will stay lit when they're all done. Anyway, hope you enjoyed that." + }, + { + "Q": "At 5:17 why do you have to times everything with -2?", + "A": "Because by doing so, we can get 200m in one equation and -200m in the other, which allows us to solve using elimination. We can add the two equations and be left with only one variable, w.", + "video_name": "VuJEidLhY1E", + "transcript": "Everyone in the kingdom is very impressed with your ability to help with the party planning, everyone except for this gentleman right over here. This is Arbegla, and he is the king's top adviser and also chief party planner, And he seems somewhat threatened by your ability to solve these otherwise unsolvable problems or, at least, from his point of view because he keeps over-ordering or under-ordering things like cupcakes, and so he asks... he says, \"King, that cupcake problem was easy.\" \"Ask them about the potato chip issue...\" \"because we can never get the potato chips right.\" And so the king says,\"Arbegla, that's a good idea. We need to get the potato chips right.\" So he comes to you and says \"How do we figure out, on average, how many potato chips we need to order?\" And to do that, we have to figure out, how much, on average does each man and each woman eat? You say, \"Well, what about the children?\" He says, the king says,\"In our kingdom, we forbid the potato chips for children.\" You say, \"Oh well, that's...that's always good. So tell me what happened at the previous parties.\" And so, the king says ,\" You might remember, at the last party, in fact the last two parties we had 500 adults, at the last party, 200 of them were men and 300 of them were women, and in total they ate 1200 bags of potato chips.\" You say, \"Well, what about the party before that?\" He says,\" That we had a bigger sque towards women. We only had a 100 men, and 400 women. And THAT time, we actually had fewer bags consumed; 1100 bags of potato chips.\" So you say, \"King and Arbegla, this seems like a fairly straightfoward thing, let me define some variables to represent our unknowns. So you go ahead, and you say, \"Lets let m=the number of bags eaten by each man. You could think of it on average, or maybe everyone, all the men in that kingdom are completely identical, or its its the average bags eaten by each man.\" \"And lets let w= the number of bags eaten by each woman. And so with that, these definitions of our variables, lets think about how we can represent this first piece of information. \" \" In green. Well, let's think about the total number of bags that the men ate. You had 200 men, [Let me scroll over a little bit] and they each ate m bags per man. \" \"So the man at this first party collectively ate 200 times m bags. If m is 10 bags per man, then this would be 2000. If m was 5 bags per man, then this would be 5000. We don't know what m is, but 200 times m is the total eaten by the man.\" \"How do we figure out, on average, how many potato chips we need to order?\"" + }, + { + "Q": "Is there a differents if you say something like 7.2 does it matter if you put the point or do you have to only put the R because when i was taught division with remainders we put the point.", + "A": "If you put a point your answer could be confusing: it looks like the decimal number 7.2. For example: 5 / 2 = 2, with remainder 1 or: 5 / 2 = 2.5 Writing 2.1 in this case would be very confusing. Note that there is no universally accepted way to write remainders - nor for decimal numbers (e.g. many European countries use a decimal comma instead of a point). Explicitly using the word remainder makes it immediately clear.", + "video_name": "8Ft5iHhauJ0", + "transcript": "Let's now see if we can divide into larger numbers. And just as a starting point, in order to divide into larger numbers, you at least need to know your multiplication tables from the 1-multiplication tables all the way to, at least, the 10-multiplication. So all the way up to 10 times 10, which you know is 100. And then, starting at 1 times 1 and going up to 2 times 3, all the way up to 10 times 10. And, at least when I was in school, we learned through 12 times 12. But 10 times 10 will probably do the trick. And that's really just the starting point. Because to do multiplication problems like this, for example, or division problems like this. Let's say I'm taking 25 and I want to divide it by 5. So I could draw 25 objects and then divide them into groups of 5 or divide them into 5 groups and see how many elements are in each group. But the quick way to do is just to think about, well 5 times what is 25, right? 5 times question mark is equal to 25. And if you know your multiplication tables, especially your 5-multiplication tables, you know that 5 times 5 is equal to 25. So something like this, you'll immediately just be able to say, because of your knowledge of multiplication, that 5 goes into 25 five times. And you'd write the 5 right there. Not over the 2, because you still want to be careful of the place notation. You want to write the 5 in the ones place. It goes into it 5 ones times, or exactly five times. And the same thing. If I said 7 goes into 49. That's how many times? Well you say, that's like saying 7 times what-- you could even, instead of a question mark, you could put a blank there --7 times what is equal to 49? And if you know your multiplication tables, you know that 7 times 7 is equal to 49. All the examples I've done so far is a number multiplied by itself. Let me do another example. Let me do 9 goes into 54 how many times? Once again, you need to know your multiplication tables to do this. 9 times what is equal to 54? And sometimes, even if you don't have it memorized, you could say 9 times 5 is 45. And 9 times 6 would be 9 more than that, so that would be 54. So 9 goes into 54 six times. So just as a starting point, you need to have your multiplication tables from 1 times 1 all the way up the 10 times 10 memorized. In order to be able to do at least some of these more basic problems relatively quickly. Now, with that out of the way, let's try to do some problems that's might not fit completely cleanly into your multiplication tables. So let's say I want to divide-- I am looking to divide 3 into 43. And, once again, this is larger than 3 times 10 or 3 times 12. Well, let me do another problem. Let me do 3 into 23. And, if you know your 3-times tables, you realize that there's 3 times nothing is exactly 23. 3 times 1 is 3. 3 times 2 is 6. Let me just write them all out. 3 times 3 is 9, 12, 15, 18, 21, 24, right? There's no 23 in the multiples of 3. So how do you do this division problem? Well what you do is you think of what is the largest multiple of 3 that does go into 23? And that's 21. And 3 goes into 21 how many times? Well you know that 3 times 7 is equal to 21. So you say, well 3 will go into 23 seven times. But it doesn't go into it cleanly because 7 times 3 is 21. So there's a remainder left over. So if you take 23 minus 21, you have a remainder of 2. So you could write that 23 divided by 3 is equal to 7 remainder-- maybe I'll just, well, write the whole word out --remainder 2. So it doesn't have to go in completely cleanly. And, in the future, we'll learn about decimals and fractions. But for now, you just say, well it goes in cleanly 7 times, but that only gets us to 21. But then there's 2 left over. So you can even work with the division problems where it's not exactly a multiple of the number that you're dividing into the larger number. But let's do some practice with even larger numbers. And I think you'll see a pattern here. So let's do 4 going into-- I'm going to pick a pretty large number here --344. And, immediately when you see that you might say, hey Sal, I know up to 4 times 10 or 4 times 12. 4 times 12 is 48. This is a much larger number. This is way out of bounds of what I know in my 4-multiplication tables. And what I'm going to show you right now is a way of doing this just knowing your 4-multiplication tables. So what you do is you say 4 goes into this 3 how many times? And you're actually saying 4 goes into this 3 how many hundred times? So this is-- Because this is 300, right? This is 344. But 4 goes into 3 no hundred times, or 4 goes into-- I guess the best way to think of it --4 goes 3 0 times. So you can just move on. 4 goes into 34. So now we're going to focus on the 34. So 4 goes into 34 how many times? And here we can use our 4-multiplication tables. 4-- Let's see, 4 times 8 is equal to 32. 4 times 9 is equal to 36. So 4 goes into 34-- 30-- 9 is too many times, right? 36 is larger than 34. So 4 goes into 34 eight times. There's going to be a little bit left over. 4 goes in the 34 eights times. So let's figure out what's left over. And really we're saying 4 goes into 340 how many ten times? We're actually saying 4 goes into 340 eighty times. Because notice we wrote this 8 in the tens place. But just for our ability to do this problem quickly, you just say 4 goes into 34 eight times, but make sure you write the 8 in the tens place right there. 8 times 4. 8 times 4 is 32. And then we figure out the remainder. 34 minus 32. Well, 4 minus 2 is 2. And then these 3's cancel out. So you're just left with a 2. But notice we're in the tens column, right? This whole column right here, that's the tens column. So really what we said is 4 goes into 340 eighty times. 80 times 4 is 320, right? Because I wrote the 3 in the hundreds column. And then there is-- and I don't want to make this look like a-- I don't want to make this look like a-- Let me clean this up a little bit. I didn't want to make that line there look like a-- when I was dividing the columns --to look like a 1. But then there's a remainder of 2, but I wrote the 2 in the tens place. So it's actually a remainder of 20. But let me bring down this 4. Because I didn't want to just divide into 340. I divided into 344. So you bring down the 4. Let me switch colors. And then-- So another way to think about it. We just said that 4 goes into 344 eighty times, right? We wrote the 8 in the tens place. And then 8 times 4 is 320. The remainder is now 24. So how many times does 4 go into 24? Well we know that. 4 times 6 is equal to 24. So 4 goes into 24 six times. And we put that in the ones place. 6 times 4 is 24. And then we subtract. 24 minus 24. That's-- We subtract at that stage, either case. And we get 0. So there's no remainder. So 4 goes into 344 exactly eighty-six times. So if your took 344 objects and divided them into groups of 4, you would get 86 groups. Or if you divided them into groups of 86, you would get 4 groups. Let's do a couple more problems. I think you're getting the hang of it. Let me do 7-- I'll do a simple one. 7 goes into 91. So once again, well, this is beyond 7 times 12, which is 84, which you know from our multiplication tables. So we use the same system we did in the last problem. 7 goes into 9 how many times? 7 goes into 9 one time. 1 times 7 is 7. And you have 9 minus 7 is 2. And then you bring down the 1. And remember, this might seem like magic, but what we really said was 7 goes into 90 ten times-- 10 because we wrote the 1 in the tens place --10 times 7 is 70, right? You can almost put a 0 there if you like. And 90 with the remainder-- And 91 minus 70 is 21. So 7 goes into 91 ten times remainder 21. And then you say 7 goes into 21-- Well you know that. 7 times 3 is 21. So 7 goes into 21 three times. 3 times 7 is 21. You subtract these from each other. Remainder 0. So 91 divided by 7 is equal to 13. And I won't take that little break to explain the places and all of that. I think you understand that. I want, at least, you to get the process down really really well in this video. So let's do 7-- I keep using the number 7. Let me do a different number. Let me do 8 goes into 608 how many times? So I go 8 goes into 6 how many times? It goes into it 0 time. So let me keep moving. 8 goes into 60 how many times? Let me write down the 8. Let me draw a line here so we don't get confused. Let me scroll down a little bit. I need some space above the number. So 8 goes into 60 how many times? We know that 8 times 7 is equal to 56. And that 8 times 8 is equal to 64. So 8 goes into-- 64 is too big. So it's not this one. So 8 goes into 60 seven times. And there's going to be a little bit left over. So 8 goes into 60 seven times. Since we're doing the whole 60, we put the 7 above the ones place in the 60, which is the tens place in the whole thing. 7 times 8, we know, is 56. 60 minus 56. That's 4. Or if we wanted, we can borrow. That be a 10. That would be a 5. 10 minus 6 is 4. Then you bring down this 8. 8 goes into 48 how many times? Well what's 8 times 6? Well, 8 times 6 is exactly 48. So 8 times-- 8 goes into 48 six times. 6 times 8 is 48. And you subtract. We subtracted up here as well. 48 minus 48 is 0. So, once again, we get a remainder of 0. So hopefully, that gives you the hang of how to do these larger division problems. And all we really need to know to be able to do these, to tackle these, is our multiplication tables up to maybe 10 times 10 or 12 times 12." + }, + { + "Q": "what is hemoglobin?", + "A": "Hemoglobin is the iron-containing oxygen-transport metalloprotein in the red blood cells. And is responsible for aiding in the transportation of gases to and from the body. :)", + "video_name": "QhiVnFvshZg", + "transcript": "Where I left off in the last video, we talked about how the hemoglobin in red blood cells is what sops up all of the oxygen so that it increases the diffusion gradient-- or it increases the incentive, we could say, for the oxygen to go across the membrane. We know that the oxygen molecules don't know that there's less oxygen here, but if you watch the video on diffusion you know how that process happens. If there's less concentration here than there, the oxygen will diffuse across the membrane and there's less inside the plasma because the hemoglobin is sucking it all up like a sponge. Now, one interesting question is, why does the hemoglobin even have to reside within the red blood cells? Why aren't hemoglobin proteins just freely floating in the blood plasma? That seems more efficient. You don't have to have things crossing through, in and out of, these red blood cell membranes. You wouldn't have to make red blood cells. What's the use of having these containers of hemoglobin? It's actually a very interesting idea. If you had all of the hemoglobin sitting in your blood plasma, it would actually hurt the flow of the blood. The blood would become more viscous or more thick. I don't want to say like syrup, but it would become thicker than blood is right now-- and by packaging the hemoglobin inside these containers, inside the red blood cells, what it allows the blood to do is flow a lot better. Imagine if you wanted to put syrup in water. If you just put syrup straight into water, what's going to happen? The water's going to become a little syrupy, a little bit more viscous and not flow as well. So what's the solution if you wanted to transport syrup in water? Well, you could put the syrup inside little containers or inside little beads and then let the beads flow in the water and then the water wouldn't be all gooey-- and that's exactly what's happening inside of our blood. Instead of having the hemoglobin sit in the plasma and make it gooey, it sits inside these beads that we call red blood cells that allows the flow to still be non-viscous. So I've been all zoomed in here on the alveolus and these capillaries, these pulmonary capillaries-- let's zoom out a little bit-- or zoom out a lot-- just to understand, how is the blood flowing? And get a better understanding of pulmonary arteries and veins relative to the other arteries and veins that are in the body. So here-- I copied this from Wikipedia, this diagram of the human circulatory system-- and here in the back you can see the lungs. Let me do it in a nice dark color. So we have our lungs here. You can see the heart is sitting right in the middle. And what we learned in the last few videos is that we have our little alveoli and our lungs. Remember, we get to them from our bronchioles, which are branching off of the bronchi, which branch off of the trachea, which connects to our larynx, which connects to our pharynx, which connects to our mouth and nose. But anyway, we have our little alveoli right there and then we have the capillaries. So when we go away from the heart-- and we're going to delve a little bit into the heart in this video as well-- so when blood travels away from the heart, it's de-oxygenated. It's this blue color. So this right here is blood. This right here is blood traveling away from the heart. It's going behind these two tubes right there. So this is the blood going away from the heart. So this blue that I've been highlighting just now, these are the pulmonary arteries and then they keep splitting into arterials and all of that and eventually we're in capillaries-- super, super small tubes. They run right past the alveoli and then they become oxygenated and now we're going back to the heart. So we're talking about pulmonary veins. So we go back to the heart. So these capillaries-- in the capillaries we get oxygen. Now we're going to go back to the heart. Hope you can see what I'm doing. And we're going to enter the heart on this side. You actually can't even see where we're entering the heart. We're going to enter the heart right over here-- and I'm going to go into more detail on that. Now we have oxygenated blood. And then that gets pumped out to the rest of the body. Now this is the interesting thing. When we're talking about pulmonary arteries and veins-- remember, the pulmonary artery was blue. As we go away from the heart, we have de-oxygenated blood, but it's still an artery. Then as we go towards the heart from the lungs, we have a vein, but it's oxygenated. So that's this little loop here that we start and I'm going to keep going over the circulation pattern because the heart can get a little confusing, especially because of its three-dimensional nature. But what we have is, the heart pumps de-oxygenated blood from the right ventricle. You're saying, hey, why is it the right ventricle? That looks like the left side of the drawing, but it's this dude's right-hand side, right? This is this guy's right hand. And this is this dude's left hand. He's looking at us, right? We don't care about our right or left. We care about this guy's right and left. And he's looking at us. He's got some eyeballs and he's looking at us. So this is his right ventricle. Actually, let me just start off with the whole cycle. So we have de-oxygenated blood coming from the rest of the body, right? The name for this big pipe is called the inferior vena cava-- inferior because it's coming up below. Actually, you have blood coming up from the arms and the head up here. They're both meeting right here, in the right atrium. Let me label that. I'm going to do a big diagram of the heart in a second. And why are they de-oxygenated? Because this is blood returning from our legs if we're running, or returning from our brain, that had to use respiration-- or maybe we're working out and it's returning from our biceps, but it's de-oxygenated blood. It shows up right here in the right atrium. It's on our left, but this guy's right-hand side. From the right atrium, it gets pumped into the right ventricle. It actually passively flows into the right ventricle. The ventricles do all the pumping, then the ventricle contracts and pumps this blood right here-- and you don't see it, but it's going behind this part right here. It goes from here through this pipe. So you don't see it. I'm going to do a detailed diagram in a second-- into the pulmonary artery. We're going away from the heart. This was a vein, right? This is a vein going to the heart. This is a vein, inferior vena cava vein. This is superior vena cava. They're de-oxygenated. Then I'm pumping this de-oxygenated blood away from the heart to the lungs. Now this de-oxygenated blood, this is in an artery, right? This is in the pulmonary artery. It gets oxygenated and now it's a pulmonary vein. And once it's oxygenated, it shows up here in the left-- let me do a better color than that-- it shows up right here in the left atrium. Atrium, you can imagine-- it's kind of a room with a skylight or that's open to the outside and in both of these cases, things are entering from above-- not sunlight, but blood is entering from above. On the right atrium, the blood is entering from above. And in the left atrium, the blood is entering-- and remember, the left atrium is on the right-hand side from our point of view-- on the left atrium, the blood is entering from above from the lungs, from the pulmonary veins. Veins go to the heart. Then it goes into-- and I'll go into more detail-- into the left ventricle and then the left ventricle pumps that oxygenated blood to the rest of the body via the non-pulmonary arteries. So everything pumps out. Let me make it a nice dark, non-blue color. So it pumps it out through there. You don't see it right here, the way it's drawn. It's a little bit of a strange drawing. It's hard to visualize, but I'll show it in more detail and then it goes to the rest of the body. Let me show you that detail right now. So we said, we have de-oxygenated blood. Let's label it right here. This is the superior vena cava. This is a vein from the upper part of our body from our arms and heads. This is the inferior vena vaca. This is veins from our abdomen and from our legs and the rest of our body. So it it first enters the right atrium. Remember, we call the right atrium because this is someone's heart facing us, even though this is on the left-hand side. It enters through here. It's de-oxygenated blood. It's coming from veins. the body used the oxygen. Then it shows up in the right ventricle, right? These are valves in our heart. And it passively, once the right ventricle pumps and then releases, it has a vacuum and it pulls more blood from the It pumps again and then it pushes it through here. Now this blood right here-- remember, this one still is de-oxygenated blood. De-oxygenated blood goes to the lungs to become oxygenated. So this right here is the pulmonary-- I'm using the word pulmonary because it's going to or from the lungs. It's dealing with the lungs. And it's going away from the heart. It's the pulmonary artery and it is de-oxygenated. Then it goes to the heart, rubs up against some alveoli and then gets oxygenated and then it comes right back. Now this right here, we're going to the heart. So that's a vein. It's in the loop with the lungs so it's a pulmonary vein and it rubbed up against the alveoli and got the oxygen diffused into it so it is oxygenated. And then it flows into your left atrium. Now, the left atrium, once again, from our point of view, is on the right-hand side, but from the dude looking at it, it's his left-hand side. So it goes into the left atrium. Now in the left ventricle, after it's done pumping, it expands and that oxygenated blood flows into the left ventricle. Then the left ventricle-- the ventricles are what do all the pumping-- it squeezes and then it pumps the blood into the aorta. This is an artery. Why is it an artery? Because we're going away from the heart. Is it a pulmonary artery? No, we're not dealing with the lungs anymore. We dealt with the lungs when we went from the right ventricle, went to the lungs in a loop, back to the left atrium. Now we're in the left ventricle. We pump into the aorta. Now this is to go to the rest of the body. This is an artery, a non-pulmonary artery-- and it is oxygenated. So when we're dealing with non-pulmonary arteries, we're oxygenated, but a pulmonary artery has no oxygen. It's going away from the heart to get the oxygen. Pulmonary vein comes from the lungs to the heart with oxygen, but the rest of the veins go to the heart without oxygen because they want to go into that loop on the pulmonary loop right there. So I'll leave you there. Hopefully that gives-- actually, let's go back to that first diagram. I think you have a sense of how the heart is dealing, but let's go look at the rest of the body and just get a sense of things. You can look this up on Wikipedia if you like. All of these different branching points have different names to them, but you can see right here you have kind of a branching off, a little bit below the heart. This is actually the celiac trunk. Celiac, if I remember correctly, kind of refers to an abdomen. So this blood that-- your hepatic artery. Hepatic deals with the liver. Your hepatic artery branches off of this to get blood flow to the liver. It also gives blood flow to your stomach so it's very important in digestion and all that. And then let's say this is the hepatic trunk. Your liver is sitting like that. Hepatic trunk-- it delivers oxygen to the liver. The liver is doing respiration. It takes up the oxygen and then it gives up carbon dioxide. So it becomes de-oxygenated and then it flows back in and to the inferior vena cava, into the vein. I want to make it clear-- it's a loop. It's a big loop. The blood doesn't just flow out someplace and then come back someplace else. This is just one big loop. And if you want to know at any given point in time, depending on your size, there's about five liters of blood. And I looked it up-- it takes the average red blood cell to go from one point in the circulatory system and go through the whole system and come back, 20 seconds. That's an average because you can imagine there might be some red blood cells that get stuck someplace and take a little bit more time and some go through the completely perfect route. Actually, the 20 seconds might be closer to the perfect route. I've never timed it myself. But it's an interesting thing to look at and to think about what's connected to what. You have these these arteries up here that they first branch off the arteries up here from the aorta into the head and the neck and the arm arteries and then later they go down and they flow blood to the rest of the body. So anyway, this is a pretty interesting idea. In the next video, what I want to do is talk about, how does the hemoglobin know when to dump the oxygen? Or even better, where to dump the oxygen-- because maybe I'm running so I need a lot of oxygen in the capillaries around my thigh muscles. I don't need them necessarily in my hands. How does the body optimize where the oxygen is actually It's actually fascinating." + }, + { + "Q": "is tension always occurring when on a string or can it occur otherwise?", + "A": "any material that can withstand being stretched", + "video_name": "_UrfHFEBIpU", + "transcript": "I will now introduce you to the concept of tension. So tension is really just the force that exists either within or applied by a string or wire. It's usually lifting something or pulling on something. So let's say I had a weight. Let's say I have a weight here. And let's say it's 100 Newtons. And it's suspended from this wire, which is right here. Let's say it's attached to the ceiling right there. Well we already know that the force-- if we're on this planet that this weight is being pull down by gravity. So we already know that there's a downward force on this weight, which is a force of gravity. And that equals 100 Newtons. But we also know that this weight isn't accelerating, It also has no velocity. But the important thing is it's not accelerating. But given that, we know that the net force on it must be 0 by Newton's laws. So what is the counteracting force? You didn't have to know about tension to say well, the string's pulling on it. The string is what's keeping the weight from falling. So the force that the string or this wire applies on this weight you can view as the force of tension. Another way to think about it is that's also the force that's within the wire. And that is going to exactly offset the force of gravity on And that's what keeps this point right here stationery and keeps it from accelerating. That's pretty straightforward. Tension, it's just the force of a string. And just so you can conceptualize it, on a guitar, the more you pull on some of those higher-- what was it? The really thin strings that sound higher pitched. The more you pull on it, the higher the tension. It actually creates a higher pitched note. So you've dealt with tension a lot. I think actually when they sell wires or strings they'll probably tell you the tension that that wire or string can support, which is important if you're going to build a bridge or a swing or something. So tension is something that should be hopefully, a little bit intuitive to you. So let's, with that fairly simple example done, let's create a slightly more complicated example. So let's take the same weight. Instead of making the ceiling here, let's add two more strings. Let's add this green string. Green string there. And it's attached to the ceiling up here. That's the ceiling now. And let's see. This is the wall. And let's say there's another string right here attached to the wall. So my question to you is, what is the tension in these two strings So let's call this T1 and T2. Well like the first problem, this point right here, this red point, is stationary. It's not accelerating in either the left/right directions and it's not accelerating in the up/down directions. So we know that the net forces in both the x and y dimensions must be 0. My second question to you is, what is going to be the offset? Because we know already that at this point right here, there's going to be a downward force, which is the force of gravity again. The weight of this whole thing. We can assume that the wires have no weight for simplicity. So we know that there's going to be a downward force here, this is the force of gravity, right? The whole weight of this entire object of weight plus wire is pulling down. So what is going to be the upward force here? Well let's look at each of the wires. This second wire, T2, or we could call it w2, I guess. The second wire is just pulling to the left. It has no y components. It's not lifting up at all. So it's just pulling to the left. So all of the upward lifting, all of that's going to occur from this first wire, from T1. So we know that the y component of T1, so let's call-- so if we say that this vector here. Let me do it in a different color. Because I know when I draw these diagrams it starts to get confusing. Let me actually use the line tool. So I have this. Let me make a thicker line. So we have this vector here, which is T1. And we would need to figure out what that is. And then we have the other vector, which is its y component, and I'll draw that like here. This is its y component. We could call this T1 sub y. And then of course, it has an x component too, and I'll do that in-- let's see. I'll do that in red. Once again, this is just breaking up a force into its component vectors like we've-- a vector force into its x and y components like we've been doing in the last several problems. And these are just trigonometry problems, right? We could actually now, visually see that this is T sub 1 x and this is T sub 1 sub y. Oh, and I forgot to give you an important property of this problem that you needed to know before solving it. Is that the angle that the first wire forms with the ceiling, this is 30 degrees. So if that is 30 degrees, we also know that this is a parallel line to this. So if this is 30 degrees, this is also going to be 30 degrees. So this angle right here is also going to be 30 degrees. And that's from our-- you know, we know about parallel lines and alternate interior angles. We could have done it the other way. We could have said that if this angle is 30 degrees, this angle is 60 degrees. This is a right angle, so this is also 30. But that's just review of geometry that you already know. But anyway, we know that this angle is 30 degrees, so what's its y component? Well the y component, let's see. What involves the hypotenuse and the opposite side? Let me write soh cah toa at the top because this is really just trigonometry. soh cah toa in blood red. So what involves the opposite and the hypotenuse? So opposite over hypotenuse. So that we know the sine-- let me switch to the sine of 30 degrees is equal to T1 sub y over the tension in the string going in this direction. So if we solve for T1 sub y we get T1 sine of 30 degrees is equal to T1 sub y. And what did we just say before we kind of dived into the math? We said all of the lifting on this point is being done by the y component of T1. Because T2 is not doing any lifting up or down, it's only pulling to the left. So the entire component that's keeping this object up, keeping it from falling is the y component of this tension vector. So that has to equal the force of gravity pulling down. This has to equal the force of gravity. That has to equal this or this point. So that's 100 Newtons. And I really want to hit this point home because it might be a little confusing to you. We just said, this point is stationery. It's not moving up or down. It's not accelerating up or down. And so we know that there's a downward force of 100 Newtons, so there must be an upward force that's being provided by these two wires. This wire is providing no upward force. So all of the upward force must be the y component or the upward component of this force vector on the first wire. So given that, we can now solve for the tension in this first wire because we have T1-- what's sine of 30? Sine of 30 degrees, in case you haven't memorized it, sine of 30 degrees is 1/2. So T1 times 1/2 is equal to 100 Newtons. Divide both sides by 1/2 and you get T1 is equal to 200 Newtons. So now we've got to figure out what the tension in this second wire is. And we also, there's another clue here. This point isn't moving left or right, it's stationary. So we know that whatever the tension in this wire must be, it must be being offset by a tension or some other force in the opposite direction. And that force in the opposite direction is the x component of the first wire's tension. So it's this. So T2 is equal to the x component of the first wire's tension. And what's the x component? Well, it's going to be the tension in the first wire, 200 Newtons times the cosine of 30 degrees. It's adjacent over hypotenuse. And that's square root of 3 over 2. So it's 200 times the square root of 3 over 2, which equals 100 square root of 3. So the tension in this wire is 100 square root of 3, which completely offsets to the left and the x component of this wire is 100 square root of 3 Newtons to the right. Hopefully I didn't confuse you. See you in the next video." + }, + { + "Q": "In the equation:\n2KNO3-->2KNO2+O2\nwould the reduction agent be just oxygen or the entire conpound 2KNO3?", + "A": "It would be the compound as the compound itself is getting oxidised, not just the oxygen. Hope this helps!", + "video_name": "TOdHMORp4is", + "transcript": "Let's see how to identify the oxidizing and reducing agents in a redox reaction. So here, we're forming sodium chloride from sodium metal and chlorine gas. And so before you assign oxidizing and reducing agents, you need to assign oxidation states. And so let's start with sodium. And so the sodium atoms are atoms in their elemental form and therefore have an oxidation state equal to 0. For chlorine, each chlorine atom is also an atom in its elemental form, and therefore, each chlorine atom has an oxidation state equal to 0. We go over here to the right, and the sodium cation. A plus 1 charge on sodium, and for monatomic ions, the oxidation state is equal to the charge on the ion. And since the charge on the ion is plus 1, that's also the oxidation state. So plus 1. We're going to circle the oxidation state to distinguish it from everything else we have on the board here. And for chloride anion, a negative 1 charge. Therefore, the oxidation state is equal to negative 1. And so let's think about what happened in this redox reaction. Sodium went from an oxidation state of 0 to an oxidation state of plus 1. That's an increase in the oxidation state. 0 to plus 1 is an increase in oxidation state, so therefore, sodium, by definition, is being oxidized. So sodium is being oxidized in this reaction. We look at chlorine. Chlorine is going from an oxidation state of 0 to an oxidation state of negative 1. That's a decrease in the oxidation state, and therefore, chlorine is being reduced. So each chlorine atom is being reduced here. Now, before we assign oxidizing and reducing agents, let's just go ahead and talk about this one more time, except showing all of the valence electrons. So let's also assign some oxidation states using this way because there are two ways to assign oxidation states. So let's assign an oxidation state to sodium over here. So if you have your electrons represented as dots, you can assign an oxidation state by thinking about how many valence electrons the atom normally has and subtracting from that how many electrons you have in your picture here. So for sodium, being in group one, one valence electron normally, and that's exactly what we have in our picture. Each sodium has a valence electron right here. So 1 minus 1 gives us an oxidation state equal to 0, which is what we saw up here, as well. So sodium has an oxidation state equal to 0. Notice that I have two sodium atoms drawn here, and that's just what the two reflects in the balanced equation up here. Let's assign an oxidation state to each chlorine atom in the chlorine molecule. And so we have a bond between the two chlorine atoms, and we know that bond consists of two electrons. Now, when you're assigning oxidation states and dot structures, you want to give those electrons to the more electronegative elements. In this case, it's the same element, so there's no difference. And so we give one electron to one atom and the other electron to the other atom, like that. And so assigning an oxidation state, you would say chlorine normally has seven valence electrons, and in our picture here, this chlorine atom has seven electrons around it. So 7 minus 7 gives us an oxidation state equal to 0. And of course, that's what we saw up here as well, when we were just using the memorized rules. And so it's the same for this chlorine atom over here, an oxidation state equal to 0. So sometimes it just helps to see the electrons. We'll go over here for our products. We had two sodium chlorides, so here are two sodium chlorides. And let's see what happened with our electrons. So the electron in magenta, this electron over here in magenta on this sodium, added onto one of these chlorines here. And then this electron on this sodium added onto the other chlorine, like that, and so sodium lost its valence electron. Each sodium atom lost its valence electron, forming a cation. And when we calculate the oxidation state, we do the same thing. Sodium normally has one valence electron, but it lost that valence electron. So 1 minus 0 is equal to plus 1 for the oxidation state, which is also what we saw up here. And then when we do it for chlorine, chlorine normally has seven valence electrons, but it gained the one in magenta. So now it has eight around it. So 7 minus 8 gives us an oxidation state equal to negative 1. And so maybe now it makes more sense as to why these oxidation states are equal to the charge on the polyatomic monatomic ion here. And so now that we've figured out what exactly is happening to the electrons in magenta, let's write some half reactions and then finally talk about what's the oxidizing agent and what's the reducing agent. So let's break down the reaction a little bit more in a different way. So you can see we have two sodium atoms over here. So we're going to write two sodiums. And when we think about what's happening, those two sodium atoms are turning into two sodium ions over here on the right. And so we have two sodium ions on the right. Now, those sodium atoms turned into the ions by losing electrons, so each sodium atom lost one electron. So we have a total of two electrons that are lost. I'm going to put it in magenta here. So those 2 electrons are lost, and this is the oxidation half reaction. You know it's the oxidation half reaction, because you're losing electrons here. So remember, LEO the lion. So Loss of Electrons is Oxidation. So this is the oxidation half reaction. We're going to write the reduction half reaction. The chlorine molecule gained those two electrons in magenta. So those two electrons in magenta we're going to put over here this time. The chlorine molecule gained them, and that turned the chlorine atoms into chloride anions. And so we have two chloride anions over here. And so those are, of course, over here on the right, our two chloride anions. And so here we have those two electrons being added to the reactant side. That's a gain of electrons, so this is our reduction half reaction, because LEO the lion goes GER. Gain of Electrons is Reduction. And so if we add those two half reactions together, we should get back the original redox reaction, because those two electrons are going to cancel out. It's actually the same electrons. These two electrons in magenta that are lost by sodium are the same electrons that are gained by chlorine, and so when we add all of our reactants that are left, we get 2 sodiums and Cl2, so we get 2 sodiums plus chlorine gas. And then for our products, we would make 2 NaCl, so we get 2 NaCl for our products, which is, of course, our original balanced redox reaction. So finally, we're able to identify our oxidizing and reducing agents. I think it was necessary to go through all of that, because thinking about those electrons and the definitions are really the key to not being confused by these terms here. And so sodium is undergoing oxidation, and by sodium undergoing oxidation, it's supplying the two electrons for the reduction of chlorine. Therefore, you could say that sodium is the agent for the reduction of chlorine, or the reducing agent. So let's go ahead and write that here. So sodium, even though it is being oxidized, is the reducing agent. It is allowing chlorine to be reduced by supplying these two electrons. And chlorine, by undergoing reduction, is taking the electrons from the 2 sodium atoms. That allows sodium to be oxidized, so chlorine is the agent for the oxidation of sodium, or the oxidizing agent. Let me go ahead and write that in red here. Chlorine is the oxidizing agent. And so this is what students find confusing sometimes, because sodium is itself being oxidized, but it is actually the reducing agent. And chlorine itself is being reduced, but it is actually the oxidizing agent. But when you think about it by thinking about what happened with those electrons, those are the exact same electrons. The electrons that are lost by sodium are the same electrons gained by chlorine, and that allows sodium to be the reducing agent for chlorine, and that is allowing chlorine at the same time to oxidize sodium. And so assign your oxidation states, and then think about these definitions, and then you can assign oxidizing and reducing agents." + }, + { + "Q": "if we have a switch parallel to a resistor in a circuit, if the switch is closed will current pass through the resistor parallel to it? if yes, why? if no, why ?", + "A": "Hello Ram, If you have an ideal switch with zero resistance than all current will flow through the switch. If you have a real switch the current will flow through ALL parallel branches. Most of the current will flow through the switch. Here we assume that the switch has a low resistance relative to the resistor. Regards, APD", + "video_name": "3NcIK0s3IwU", + "transcript": "Let's see if we can apply what we've learned to a particularly hairy problem that I have constructed. So let me see how I can construct this. So let's say in parallel, I have this resistor up here. And I try to make it so the numbers work out reasonably neat. That is 4 ohms. Then I have another resistor right here. That is 8 ohms. Then I have another resistor right here. That is 16 ohms. And then, I have another resistor here, that's ohms. Actually, I'm now making it up on the fly. I think the numbers might work out OK. 16 ohms. And let's say that now here in series, I have a resistor that is 1 ohm, and then in parallel to this whole thing-- now you can see how hairy it's getting-- I have a resistor that is 3 ohms. And let's say I have a resistor here. Let's just make it simple: 1 ohm. And just to make the numbers reasonably easy-- I am doing this on the fly now-- that's the positive terminal, negative terminal. Let's say that the voltage difference is 20 volts. So what I want us to do is, figure out what is the current flowing through the wire at that point? Obviously, that's going to be different than the current at that point, that point, that point, that point, all of these different points, but it's going to be the same as the current flowing at this point. So what is I? So the easiest way to do this is try to figure out the equivalent resistance. Because once we know the equivalent resistance of this big hairball, then we can just use Ohm's law and be done. So first of all, let's just start at, I could argue, the simplest part. Let's see if we could figure out the equivalent resistance of these four resistors in parallel. Well, we know that that resistance is going to be equal to 1/4 plus 1/8 plus 1/16 plus 1/16. So that resistance-- and now it's just adding fractions-- over 16. 1/4 is 4/16 plus 2/16 plus 1 plus 1, so 1/R is equal to 4 plus 2 is equal to 8/16-- the numbers are working out-- is equal to 1/2, so that equivalent resistance is 2. So that, quickly, we just said, well, all of these resistors combined is equal to 2 ohms. So let me erase that and simplify our drawing. Simplify it. So that whole thing could now be simplified as 2 ohms. I lost some wire here. I want to make sure that circuit can still flow. So that easily, I turned that big, hairy mess into something that is a lot less hairy. Well, what is the equivalent resistance of this resistor and this resistor? Well, they're in series, and series resistors, they just add up together, right? So the combined resistance of this 2-ohm resistor and this 1-ohm resistor is just a 3-ohm resistor. So let's erase and simplify. So then we get that combined resistor, right? We had the 2-ohm that we had simplified and then we had a 1-ohm. So we had a 2-ohm and a 1-ohm in series, so those simplify to 3 ohms. Well, now this is getting really simple. So what do these two resistors simplify to? Well, 1 over their combined resistance is equal to 1/3 plus 1/3. 2/3. 1/R is equal to 2/3, so R is equal to 3/2, or we could say 1.5, right? So let's erase that and simplify our drawing. So this whole mess, the 3-ohm resistor in parallel with the other 3-ohm resistor is equal to one resistor with a 1.5 resistance. And actually, this is actually a good point to give you a little intuition, right? Because even though these are 3-ohm resistors, we have two of them, so you're kind of increasing the pipe that the electrons can go in by a factor of two, right? So it's actually decreasing the resistance. It's giving more avenues for the electrons to go through. Actually, they're going to be going in that direction. And that's why the combined resistance of both of these in parallel is actually half of either one of these I encourage you to think about that some more to give you some intuition of what's actually going on with the electrons, although I'll do a whole video on resistivity. OK so we said those two resistors combined-- I want to delete all of that. Those two resistors combined equal to a 1.5-ohm resistor. That's 1.5 ohms. And now all we're left with is two resistors in parallel, so the whole circuit becomes this, which is the very basic one. This is a resistor: 1.5 ohms, 1 ohm in series. Did I say parallel just now? No, they're in series. 1.5 plus 1, that's 2.5 ohms. The voltage is 20 volts across them. So what is the current? Ohm's law. V is equal to IR. Voltage is 20 is equal to current times our equivalent resistance times 2.5 ohms. Or another way to write 2.5 five is 5/2, right? So 20 is equal to I times 5/2. Or I is equal to 2/5 times 20, and what is that? 2/5 is equal to I is equal to 8. 8 amperes. That was not so bad, I don't think. Although when you saw it initially, it probably looked extremely intimidating. Anyway, if you understood that, you can actually solve fairly complicated circuit problems. I will see you in future videos." + }, + { + "Q": "Is Papal States Pronounced Pay-pal or Pa-pal", + "A": "Probably the closest is Pay-Pull with the emphasis on the first syllable.", + "video_name": "ALJGz4r_VF0", + "transcript": "In the video on the Fourth Coalition, I forgot to add to one super important consequence of the Treaties of Tilsit. And especially the Treaty of Tilsit with Prussia. I already talked about that it was all about carving up Prussia, and humiliating Prussia. And really removing it from the status of one the preeminent powers. And all I talked about was the loss of the territories of Prussia west of the Elbe. And that's about that area right there. But just as important as that, the Polish holdings of Prussia. So all of this area right over here, this also was removed from Prussia and became a French satellite state. It became the Duchy of Warsaw. So I just really want to emphasize. The Treaty of Tilsit, I only emphasized kind of what happened on the western side of Prussia, but the eastern side of Prussia also got carved up. And Prussia essentially lost half of its size. So it's very dramatic humiliation for Prussia at the end of the Treaty of Tilsit. Or the Treaties of Tilsit. Now with that out of the way, we talked about in the last video, that at the end of the Fourth Coalition, Napoleon was kind of near the peak of his power. He'd kind of done everything right. He had this kind of steady upward momentum, or France had a steady upward momentum in its power. But what we're going to see in this video, at least the beginnings of the downfall of Napoleon. And it's not going to be obvious when you look at the territory. Because from a territorial point of view, you're going to see in this video that he's actually gaining territory. But he is going to start doing some of the actions that end up undermining him. So we talked about in the last video, we talked about this whole notion of the Continental System, where Napoleon was obsessed with people on the continent of Europe boycotting England, not trading with England. And he figured this is the only way that he could really undermine England's dominance on the ocean. Or eventually maybe even undermine England generally. So as we said, in the Treaties of Tilsit he got Russia to participate in the Continental System. So he wanted everyone to buy into it. And one party that, at this point, we're talking about-- we're in 1807 now-- one party that wasn't all that keen in participating in the Continental System was Portugal. That's Portugal right there. So Napoleon goes and chats-- well they didn't chat directly-- but he gets the agreement of the King of Spain. This is Charles IV, and he's going to look like a bit of a fool and this video. And Napoleon says, hey Charles, let's go in there, let's go into Portugal, that little upstart country that doesn't want to participate in the Continental System. You and me, we'll invade together. We'll bring them into kind of our realm of influence. And we can both kind of pillage the lands and get the wealth of Portugal. Charles IV, he's all up for this. So a combined French and Spanish force invade Portugal. So in 1807, this is the end of 1807, it's actually in October. In October, you have a combined French and Spanish invasion of Portugal. And they are able to take Portugal, but we're going to see that it's reasonably temporary. Now I just mentioned that this guy is going to look like the fool of this video. And the reason is, because with the excuse of reinforcements, obviously to get to Portugal, you have to go through Spain. So with the excuse of sending in reinforcements, Napoleon in 1808-- and now we're talking about early 1808, in particular in March. So with the excuse of sending in reinforcements to support the Portugal campaign, and Spain is like your my ally, sure, send those hundreds of thousands of troops right through our territory. We're not going to worry about it. And with that excuse, Napoleon was able to send 100,000 troops and occupy Madrid. So this is one of those lessons of never get too greedy. This guy got greedy, wanted to help Napoleon. Or I guess the other lesson is be careful who your friends are. This guy wanted to invade Portugal, but the side effect of it is that Madrid gets occupied. And that actually he gets dethrowned. And so you have this situation here, the French are now in control of Spain. In May of 1808-- and this is really going to be the first little spark that is kind of the downfall of Napoleon. In May 2, 1808, a popular uprising starts in Madrid. Dos de Mayo. So a popular uprising in Madrid. And at the same time, a little bit after that-- So you can imagine, this is a hugely tumultuous time. You have this occupation of Portugal with the excuse of reinforcements in March. The French troops occupy Madrid. Then in May-- so a couple of months later-- a popular uprising starts in Madrid. This leads to popular uprisings throughout Spain. But at the very same time as this-- this is a little bit after the uprising in May-- Napoleon says, oh, this is just a little uprising, I'm still in control of Spain. He appoints his other brother-- remember, there's this whole business he's putting his brothers in charge of different parts of the Empire. He puts his brother Joseph, he appoints his brother Joseph-- or you could kind of say-- he inserts his brother Joseph as the King of Spain. So this is all in kind of early, mid-1808. Spain is in all of this turmoil. A new king has been appointed, who is Napoleon's brother. The old king is no longer in charge. You have this ongoing battle in Portugal. They don't have a firm hold on Portugal just yet. And in the rest of 1808, the uprising that occurs throughout Spain is actually pretty successful in enforcing the French troops to retreat. And a major, I guess, aspect of this uprising is it's one of the first real national uprisings in history. It's people saying we are Spanish, we do not like being controlled by the French. We do not like how they have treated our royalty. We as a nation are going to rise up. And the other interesting aspect of this whole uprising that starts in Madrid with Dos de Mayo, but then it starts continuing throughout the whole nation, is the idea of guerrilla warfare. Not gorilla warfare. And this comes from the Spanish for little war. Not from the large ape. And what it implies, you probably heard the word on the news before, is kind of a non-conventional style of fighting, where small little groups kind of engage their enemy in very nontraditional styles. So it becomes a very painful-- at least for it Napoleon's forces-- it became very difficult fighting these non-conventional battles all over Spain. So they were able to force the French to retreat. Napoleon says, gee, you know what? If you want a job well done, you've got to do it yourself. So Napoleon comes in at the end of the year, and then he retakes Madrid. So December of 1808, Napoleon back in Madrid. Now, you might say all is fine and well. Now Napoleon is back here. He has firm control of Spain. But not everything is good. Because as you could imagine, there's all these other characters here that keep forming coalitions for and against Napoleon. Even when they say that they're allied, you know that in the back of their minds they can't wait until they can declare the next war on Napoleon. So in 1809-- let me write this down-- Austria declares war. And since Great Britain was in-- at this point in time-- perpetual war with France, this becomes the Fifth Coalition. But this one is fairly short-lived. Napoleon says gee, I got these guys on my eastern front. Austria is re-declaring war on me. So he leaves Spain to go lead that fight. And he leaves 300,000 of his best troops in Spain to hold Spain. And frankly, this is the most important side effect of the Fifth Coalition, is that it makes Napoleon go to fight Austria, to lead that effort, as opposed to worrying about Spain. And essentially by doing that-- and I don't know if it's necessarily the fact that Napoleon wasn't there. But it could be because Napoleon wasn't there-- is that Spain just becomes a major thorn in Napoleon's side. This guerrilla warfare just continues on and on and on. And it just goes back and forth. And the French will win a battle and they'll win another battle. But they still don't have control. And these guerrillas will kind of peck at them and continue the uprising. And this really just drains the French army. And really just gets at them little bit by little bit, really over the remainder of Napoleon's reign. So all the way until 1814. We haven't gone over that yet. But this occurs all the way to 1814. So I said at the beginning the video, this is one of the starting points of Napoleon's downfall. That's just because he was just stuck in Spain from 1808 on, just continuing to have to send troops and supplies and reinforcements and wealth to support what they called the Peninsular Campaign. And it just drains him. It drains his resources, it drains his energy. And it really hurts his ability to fight wars with all of the other people who he needs to fight wars with. This is one of the major downfalls. The other one, which we'll probably talk about in the next video or video after that, is his invasion of Russia. Which he does in 1812. One could debate which one drains France's resources more. But the invasion of Russia really decimates Napoleon's forces. And really makes him susceptible to really conquest by England and all of the other allies. And we're going to see that in a couple of videos. So you have this been Peninsular Campaign continuing to drain Napoleon. It all started because he wanted to enforce the Continental System on Portugal. And he got a little bit greedy. And he also wanted to conquer Spain. And just to highlight why it's called the Peninsular Campaign. This right here, a little bit of geography, this is called the Iberian Peninsula right there that I'm circling. So you could call it the Iberian Peninsular Campaign, because it's everything that's going on in this Peninsula in Spain and Portugal. Now if we back up a little bit back to 1808, where we had this uprising in Spain, and they were able to push the French back. At the same time, you also had a popular uprising in Portugal roughly in the fall, late summer or fall of 1808. The British got excited. They saw it as their chance to push Napoleon out of Portugal. So you have this gentleman right here, Sir Arthur Wellesley. He's a future Duke of Wellington. And he's eventually going to be responsible for pushing Napoleon out of Spain entirely. Or at least out of Madrid. Him and along with the British and along with the Portuguese are able to push the French out in August of 1808. So let me put this in my not so neatly drawn timeline here. So in December, Napoleon is back, so right before that in August, out of Portugal. And this is another motivation for Napoleon to say, gee, you know what? Things aren't going well on the Iberian Peninsula, I have to take charge of things myself. Now, at the very same time as all of this is happening, and this is really just kind of out of interest. Well it's more than out of interest, because actually it has huge global repercussions. You might say, OK, well you have this Iberian Peninsula. Spain is going back and forth between the French and the guerillas. And Portugal has this whole situation where their king was dethrowned, but then the British help and take it back. But you could imagine, these nations are in just a super state of flux. Now you could also imagine, the King of Spain wasn't just the King of Spain, he was King of the Spanish Empire. And the Spanish Empire, the main land mass of the Spanish Empire was in the Americas. So this right here. That was the Spanish Empire at the time. This was a 400 year old Spanish Empire. Starting with Columbus sailing the ocean blue in 1492. You had this huge Spanish Empire. And one of the really important side effects of Napoleon invading Spain and having this long protracted engagement in Spain, is it catalyzed the ability of these colonies at the time to start looking for their And we're going to do whole videos on that in the future. But this really is one of the things that allowed them to get independence. Obviously, if the empire is in flux, these guys can say hey, gee, why do we have to listen to that nation anymore? We don't even know who's in charge there. At the same time, same thing in Portugal. Brazilian independence didn't come until a little bit later after this period, but Napoleon's invasion is what really sparked the beginning of a lot of turmoil in Portugal. And that eventually is one of the causes that leads to the eventual independence of Brazil. That doesn't happen for another 10 or 15 years. But you could imagine, this is where a lot of that action can be traced back. Now, another interesting point that occurred around this time. And actually, I didn't tell you what happened on the Fifth Coalition. I said Austria declared war. Obviously Britain was already at war. So it was the Fifth Coalition. Napoleon had to leave, that maybe made Spain a little bit harder to hold for France. And that's why it kind of bled France slowly. But Napoleon was able to take care of Austria. And then he was able to take a little bit more land from them. Actually Galicia, this area of Austria was given to the Duchy of Warsaw, which was a French satellite state. And then Austria once again, had to say oh Napoleon, we're your friend, we're going to do whatever you ask us to. So you can imagine at this time landwise, the empire of Napoleon seemed pretty dramatic. You could include Spain here. Although he had to spend a lot of resources to keep Spain. And then now we had Austria, at least it was in the fold. You know Prussia was not really happy about it. But this whole area here, the western half of Poland was under French control. Germany-- the Confederacy of the Rhine-- which is now Germany. And then a good bit of Italy, the Kingdom of Italy was also a French satellite state. But Napoleon, of course, he wanted everyone to participate in the Continental System. That's the only way to really strangle England. And the Papal States were not participating in the Continental System. So he sent some people over to kind of try to convince them to. And when they didn't, they occupied the Papal States. So French troops occupied the Papal States. And then once again, this was still back in 1808. This is actually early 1808, it's just on a different front. So in February, up here, in 1808. Actually, that's before they even occupied Madrid. So in 1808, February, French troops occupy Papal States. They essentially give them over to the Kingdom of Italy, which at that time was a French satellite state. So it's almost like annexing it to France. And then once the Fifth Coalition was done with, Napoleon felt so good about himself, that he formally annexed the Papal States. Now we're in 1809. In 1809, he formally-- The Papal States are actually annexed into the French Empire. Now you can imagine that the Pope wasn't that happy about this. This is the Pope at the time, this is Pius VII. He wasn't so happy about it. So he excommunicates Napoleon. And I'll do a whole video on excommunication. But it's really about as bad as something you could do to someone within the powers of the Catholic Church. And by implication, you're no longer part of the church, and you'll probably go to hell now, at least if the Pope has anything to do with it. So Napoleon wasn't happy about this. He sent some people, some officers, once again to talk to the Pope about it. To say hey, gee, why do you want to excommunicate Napoleon? Why don't you just play nice? Why don't you just agree to whatever Napoleon says? The Pope doesn't agree, and so he gets abducted. This is why it's interesting. Napoleon, he's not afraid to take some serious action. So he gets abducted in 1809 by French officers. And it's not clear, it's not obvious that Napoleon told But once he was abducted, and they actually started shuttling him all around France depending on who needed to talk to him. Or if they were afraid that the British might try to free him from one port, they would send them some place else. But it wasn't clear that Napoleon ordered this. But he never ordered his release. So in some ways, you got to say that it was sanctioned by Napoleon. So all of this mess starts, you know, Napoleon is messing with the Pope. He has this ongoing bleeding going on in Spain. And that ends actually in 1812 where Sir Arthur Wellesley finally retakes Madrid. But during this whole period, you can imagine it's really draining into the resources of the French Empire." + }, + { + "Q": "What does R^2 represent? Can we just use c or d? what's the significance of using R^2 as obtaining the members a and b?", + "A": "\u00e2\u0084\u009d\u00c2\u00b2 represents the set of all 2 dimensional vectors composed of real numbers. When Sal writes a\u00e2\u0083\u0097, b\u00e2\u0083\u0097 \u00e2\u0088\u0088 \u00e2\u0084\u009d\u00c2\u00b2 all he s saying is that the vectors a\u00e2\u0083\u0097 and b\u00e2\u0083\u0097 are 2 dimensional vectors composed of real numbers. I don t know what you mean by Can we just use c or d . If you mean as labels for the vectors, then yes, you can name the vectors with whatever symbol you want, Sal used a\u00e2\u0083\u0097 and b\u00e2\u0083\u0097 just to use the first letters of the alphabet, but there is no special significance in that.", + "video_name": "8QihetGj3pg", + "transcript": "So I have two 2-dimensional vectors right over here, vector a and vector b. And what I want to think about is how can we define or what would be a reasonable way to define the sum of vector a plus vector b? Well, one thing that might jump at your mind is, look, well, each of these are two dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just a sum of the first two components of these two vectors. So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two 2-dimensional vectors. You add them together, you get another two 2-dimensional vectors. If you think about it in terms of real coordinates bases, both of these are members of R2-- I'll write this down here just so we get used to the notation. So vector a and vector b are both members of R2, which is just another way of saying that these are both two tuples. They are both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it, but how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector a, we could visualize, this tells us how far this vector moves in each of these directions-- horizontal direction and vertical direction. So if we put the, I guess you could say the tail of the vector at the origin-- remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction. 1, 2, 3, 4, 5, 6. And then negative 2 in the vertical. So negative 2. So vector a could look like this. Vector a looks like that. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector. And the direction is the direction that it is pointed in. And also just to emphasize, I could have drawn vector a like that or I could have put it over here. These are all equivalent vectors. These are all equal to vector a. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector b. Vector b in the horizontal direction goes negative 4-- 1, 2, 3, 4, and in the vertical direction goes 4-- 1, 2, 3, 4. So its tail if we start at the origin, if its tail is at the origin, its head would be at negative 4, 4. So let me draw that just like that. So that right over here is vector b. And once again, vector b we could draw it like that or we could draw it-- let me copy and let me paste it-- so this would also be another way to draw vector b. Once again, what I really care about is its magnitude and its direction. All of these green vectors have the same magnitude. They all have the same length and they all have the same direction. So how does the way that I drew vector a and b gel with what its sum is? So let me draw its sum like this. Let me draw its sum in this blue color. So the sum based on this definition we just used, the vector addition would be 2, 2. So 2, 2. So it would look something like this. So how does this make sense that the sum, that this purple vector plus this green vector is somehow going to be equal to this blue vector? I encourage you to pause the video and think about if that even makes sense. Well, one way to think about it is this first purple vector, it shifts us this much. It takes us from this point to that point. And so if we were to add it, let's start at this point and put the green vector's tail right there and see where it ends up putting us. So the green vector, we already have a version. So once again, we start the origin. Vector a takes us there. Now, let's start over there with the green vector and see where green vector takes us. And this makes sense. Vector a plus vector b. Put the tail of vector b at the head of vector a. So if you were to start at the origin, vector a takes you there then if you add on what vector b takes you, it takes you right over there. So relative to the origin, how much did you-- I guess you could say-- shift? And once again, vectors don't only apply to things like displacement. It can apply to velocity. It can apply to actual acceleration. It can apply to a whole series of things, but when you visualize it this way, you see that it does make complete sense. This blue vector, the sum of the two, is what results where you start with vector a. At that point right over there, vector a takes you there, then you take vector b's tail, start over there and it takes you to the tip of the sum. Now, one question you might be having is well, vector a plus vector b is this, but what is vector b plus vector a? Does this still work? Well, based on the definition we had where you add the corresponding components, you're still going to get the same sum vector. So it should come out the same. So this will just be negative 4 plus 6 is 2. 4 plus negative 2 is 2. But does that make visual sense? So if we start with vector b. So let's say you start right over here. Vector b takes you right over there. And then if you were to go there and you were to start with vector a-- so let's do that. So actually, let me make this a little bit-- actually, let me start with a new vector b. So let's say that that's our vector b right over there. And then-- actually, let me give this a place where I'll have some space to work with. So let's say that's my vector b right over there. And then let me get a copy of the vector a. That's a good one. So copy and let me paste it. So I could put vector a's tail at the tip of vector b, and then it'll take me right over there. So if I start right over here, vector b takes me there. And now I'm adding to that vector a, which starting here will take me there. And so from my original starting position, I have gone this far. Now, what is this vector? Well, this is exactly the vector 2, 2. Or another way of thinking about it, this vector shifts you 2 in the horizontal direction and 2 in the vertical direction. So either way, you're going to get the same result, and that should, hopefully, make visual or conceptual sense as well." + }, + { + "Q": "How many times cn u actually borrow?", + "A": "you can borrow (total number of digits - 1) times if needed. Its not necessary to borrow every time though.", + "video_name": "OJ-wajo6oa4", + "transcript": "We've got 9,601 minus 8,023. And immediately when we try to start subtracting in our ones place, we have a problem. This 3 is larger than this 1. And we also have that problem in the tens place. This 2 is larger than this 0. So we're going to have to do some type of borrowing or regrouping. And so the way I like to think about it-- I like to go to the first place value that has something to give. Obviously, the tens place is in no position to give anything to the ones place. It needs things itself. And so we're going to go to the hundreds place. And the hundreds place has an abundance of value that it can regroup into the tens and ones place. This 6 right over here represents 600. So why don't we take 100 from that 600-- so then this will become 500-- and then give that 100 over to the tens place. Now, if we give 100 to the tens place, how would I represent that in the tens place? Well, I have zero 10's. And now I'm going to give 100. 100 is the same thing as 10 10's. It's going to be 0 plus 100. 100 in the tens place is just 10. So let me write it this way. So this right over here is now going to be rewritten as 10. Now, you might be saying, wait, wait, wait. What's going on here, Sal? You took 100 from the hundreds place. That's why it became 500. Now, why did this become 10 and not 100? Now remember, this is 10 10's. So this is still representing 100. You have not changed the value of this top number. Before, the value was 9,000 plus 600 plus 1. Now it's 9,000 plus 500 plus 100-- 10 10's is 100-- plus 1. I have not changed the value here. Now, we're still not done yet. We don't want to just subtract because we still have the problem with the ones place. The ones place still doesn't have enough value. Now, the good thing is we've given some value to the tens So why don't we take 10 from the tens place? So if you have 10 10's, and you take one 10 away, you're going to be left with nine 10's, or 90. And then we can take that 10 and give it to the ones place So let's do that. You take that 10 we just took from there, and you give it to the ones place. You now have 11 here. And now we are ready to subtract. 11 minus 3 is 8. 9 minus 2-- and this is really 90 minus 20-- is 70. But in the tens place, we represent that as a 7. 500 minus zero hundred is 500, represented as a 5 in the hundreds place. 9,000 minus 8,000 is 1,000. And we're done. And just to make things really clear, I'm going to redo this problem now but with things expanded out. So this first number is 9,000 plus 600 plus zero 10's plus 1. And this number right here, we're subtracting 8,000. We're subtracting zero 100's. We're subtracting two 10's, which is 20. Subtracting 20. And subtracting three 1's. So I have just rewritten this exact same statement. But the regrouping and the borrowing is going to become a little bit clearer now. So the same exact thing-- we said, hey, we can't subtract the 3 from the 1 or the 20 from the 0. But we have a lot of value right over here in the 600. So why don't we take 100 from that? So this becomes 500. And we give that 100 to the tens place. So this becomes 100. Notice, the value has not changed. This is 9,000 plus 500 plus 100 plus 1. That's the same thing as 9,000 plus 600 plus 1. We've just put the value in different places. And here we have explicitly written 100. But when we represent it in the tens place, 10 10's is the same thing as 100. Now, we aren't done regrouping just yet. We want to give some value to the ones place. So we can take 10 from the tens place-- and this becomes a 90-- and give that 10 to the ones place. 10 plus 1 is 11. So notice, I did the exact borrowing, the exact regrouping, that I did here. I just represented it a little bit different. This 500 was represented by a 5 in the hundreds place. This 90 was represented by a 9 in the tens place. But either way, we're ready to subtract now. 11 minus 3 is 8. 90 minus 20 is 70. Write a plus there. 500 minus 0 is 500. And then 9,000 minus 8,000 is 1,000. And we got the same result because 1,000 plus 500 plus 70 plus 8 is 1,578." + }, + { + "Q": "I'm solving stuff like -1/4x+3=-3/2x-2 and i want to know how to solve it but i can't seem to find any thing that will help me.", + "A": "Its a linear equation, i would do something like this: -1/4x + 3 = -3/2x -2 i d take -3/2x by adding 3/2x to both sides of the equation to get -1.75x + 3 = -2 the i d take away 3 to get -1.75x = -5 then divide both sides to get x =2.8571428571428571428571428571429", + "video_name": "5a6zpfl50go", + "transcript": "Let's say I have the equation y is equal to x plus 3. And I want to graph all of the sets, all of the coordinates x comma y that satisfy this equation right there. And we've done this many times before. So we draw our axis, our axes. That's my y-axis. This is my x-axis. And this is already in mx plus b form, or slope-intercept form. The y-intercept here is y is equal to 3, and the slope here is 1. So this line is going to look like this. We intersect at 0 comma 3-- 1, 2, 3. At 0 comma 3. And we have a slope of 1, so every 1 we go to the right, we go up 1. So the line will look something like that. It's a good enough approximation. So the line will look like this. And remember, when I'm drawing a line, every point on this line is a solution to this equation. Or it represents a pair of x and y that satisfy this equation. So maybe when you take x is equal to 5, you go to the line, and you're going to see, gee, when x is equal to 5 on that line, y is equal to 8 is a solution. And it's going to sit on the line. So this represents the solution set to this equation, all of the coordinates that satisfy y is equal to x plus 3. Now let's say we have another equation. Let's say we have an equation y is equal to negative x plus 3. And we want to graph all of the x and y pairs that satisfy this equation. Well, we can do the same thing. This has a y-intercept also at 3, right there. But its slope is negative 1. So it's going to look something like this. Every time you move to the right 1, you're going to move down 1. Or if you move to the right a bunch, you're going to move down that same bunch. So that's what this equation will look like. Every point on this line represents a x and y pair that will satisfy this equation. Now, what if I were to ask you, is there an x and y pair that satisfies both of these equations? Is there a point or coordinate that satisfies both equations? Well, think about it. Everything that satisfies this first equation is on this green line right here, and everything that satisfies this purple equation is on the purple line right there. So what satisfies both? Well, if there's a point that's on both lines, or essentially, a point of intersection of the lines. So in this situation, this point is on both lines. And that's actually the y-intercept. So the point 0, 3 is on both of these lines. So that coordinate pair, or that x, y pair, must satisfy both equations. When x is 0 here, 0 plus 3 is equal to 3. When x is 0 here, 0 plus 3 is equal to 3. It satisfies both of these equations. So what we just did, in a graphical way, is solve a system of equations. Let me write that down. And all that means is we have several equations. Each of them constrain our x's and y's. So in this case, the first one is y is equal to x plus 3, and then the second one is y is equal to negative x plus 3. This constrained it to a line in the xy plane, this constrained our solution set to another line in the xy plane. And if we want to know the x's and y's that satisfy both of these, it's going to be the intersection of those lines. So one way to solve these systems of equations is to graph both lines, both equations, and then look at their intersection. And that will be the solution to both of these equations. In the next few videos, we're going to see other ways to solve it, that are maybe more mathematical and less graphical. But I really want you to understand the graphical nature of solving systems of equations. Let's do another one. Let's say we have y is equal to 3x minus 6. That's one of our equations. And let's say the other equation is y is equal to negative x plus 6. And just like the last video, let's graph both of these. I'll try to do it as precisely as I can. There you go. Let me draw some. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And then 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I should have just copied and pasted some graph paper here, but I think this'll do the job. So let's graph this purple equation here. Y-intercept is negative 6, so we have-- let me do another 1, 2, 3, 4, 5, 6. So that's y is equal to negative 6. And then the slope is 3. So every time you move 1, you go up 3. You moved to the right 1, your run is 1, your rise is 1, 2, 3. That's 3, right? 1, 2, 3. So the equation, the line will look like this. And it looks like I intersect at the point 2 comma 0, which is right. 3 times 2 is 6, minus 6 is 0. So our line will look something like that right there. That's that line there. What about this line? Our y-intercept is plus 6. 1, 2, 3, 4, 5, 6. And our slope is negative 1. So every time we go 1 to the right, we go down 1. And so this will intersect at-- well, when y is equal to 0, x is equal to 6. 1, 2, 3, 4, 5, 6. So right over there. So this line will look like that. The graph, I want to get it as exact as possible. And so we're going to ask ourselves the same question. What is an x, y pair that satisfies both of these equations? Well, you look at it here, it's going to be this point. This point lies on both lines. And let's see if we can figure out what that point is. Just eyeballing the graph here, it looks like we're at 1, 2, 3 comma 1, 2, 3. It looks like this is the same point right there, that this is the point 3 comma 3. I'm doing it just on inspecting my hand-drawn graphs, so maybe it's not the exact-- let's check this answer. Let's see if x is equal to 3, y equals 3 definitely satisfies both these equations. So if we check it into the first equation, you get 3 is equal to 3 times 3, minus 6. This is 9 minus 6, which is indeed 3. So 3 comma 3 satisfies the top equation. And let's see if it satisfies the bottom equation. You get 3 is equal to negative 3 plus 6, and negative 3 plus 6 is indeed 3. So even with our hand-drawn graph, we were able to inspect it and see that, yes, we were able to come up with the point 3 comma 3, and that does satisfy both of these equations. So we were able to solve this system of equations. When we say system of equations, we just mean many equations that have many unknowns. They don't have to be, but they tend to have more than And you use each equation as a constraint on your variables, and you try to find the intersection of the equations to find a solution to all of them. In the next few videos, we'll see more algebraic ways of solving these than drawing their two graphs and trying to find their intersection points." + }, + { + "Q": "when is it appropriate to use the prepositions at and in", + "A": "Use at to indicate where you are, like heidi mentioned - but never where are you at? the at in that usage is redundant. Use in to indicate position such as I am in the house or The hamster is in its cage or Your books are in the desk .", + "video_name": "O-6q-siuMik", + "transcript": "- [Voiceover] Hi, everyone. My name is David, and I'm here to introduce you to Grammar on Khan Academy. I'm so glad you could join me. So, let's start by asking the question, \"What is grammar?\" What is this thing, why is it worthwhile to study it, why would you wanna put up with listening to me? Well, first of all, grammar is a set of conventions and rules that govern language. So what's the difference between a convention and a rule? Well, a rule is kind of the bare minimum of what it takes to make your language understandable by other people, right? So in order to make a car work, for example, in order to make it move forward as intended, the wheels have to go on the bottom instead of the roof. That's a rule. The idea that all cars should be painted teal, for example, is a convention. Now is that true that all cars should be teal? No, not necessarily, but that leads me to my second point, that grammar is context dependent. The kind of grammar that you use throughout your day changes. It depends on who you're talking to, what you're trying to say, and how you're trying to say it. And so we use multiple kinds of grammar throughout our days and throughout our lives. Another thing you need to know is that you already know so much grammar. Just from living and existing in the world and talking to other people. You know how to put a sentence together. If you can understand me, then you know so much about grammar. You know more than I can teach you. What these videos are for is to give names to the things you already know. To give you a greater command of them. And I want to say, too, that these videos are only about a very specific kind of grammar. It's called Standard American English. But I want you to know that there are many Englishes. And you know what? They're all great. They are all wonderful and vibrant and important and special. And what I do not want for you to take away from these videos is that I'm trying to teach you what is right and what is wrong. If the kind of English you speak doesn't sound like the kind of English I speak, that is okay, you know? You are great. What I want to do is give you the tools to harness language. To harness English and use it any way you want. I mean, I'm saying I don't care what color your car is. It could be pink, it could be green, it could be purple, it could be paisley, you know. I'm just trying to make sure your wheels are on straight. You are a grammarian. You have made a study of grammar throughout your entire speaking and reading life. And I firmly believe that you can learn anything. Welcome to Grammar on Khan Academy. David out." + }, + { + "Q": "if on angel is 90 degrees and the other is 80.54 I really then the other angel should be 9.46 degrees and according to the picture its not", + "A": "Do not let pictures fool you. Unless they are indicated as drawn to scale, then you must assume that they are sketches for calculation purposes, The upper angle is obviously very small, but it does not have to be exactly 9.46 since we have no indication that the drawing is to scale. That is why architects get paid the big bucks, they almost always have to draw and model to scale.", + "video_name": "aHzd-u35LuA", + "transcript": "A tiny but horrible alien is standing at the top of the Eiffel Tower-- so this is where the tiny but horrible alien is-- which is 324 meters tall-- and they label that, the height of the Eiffel Tower-- and threatening to destroy the city of Paris. A Men In Black-- or a Men In Black agent. I was about to say maybe it should be a man in black. A Men In Black agent is standing at ground level, 54 meters across the Eiffel square. So 54 meters from, I guess you could say the center of the base of the Eiffel Tower, aiming his laser gun at the alien. So this is him aiming the laser gun. At what angle should the agent shoot his laser gun? Round your answer, if necessary, to two decimal places. So if we construct a right triangle here, and we can. So the height of this right triangle is 324 meters. This width right over here is 54 meters. It is a right triangle. What they're really asking us is what is this angle right over here. And they've given us two pieces of information. They gave us the side that is opposite the angle. And they've given us the side that is adjacent to the angle. So what trig function deals with opposite and adjacent? And to remind ourselves, we can write, like I always like to do, soh, cah, toa. And these are really by definition. So you just have to know this, and soh cah toa helps us. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. We can write that the tangent of theta is equal to the length of the opposite side-- 324 meters-- over the length of the adjacent side-- over 54 meters. Now you might say, well, OK, that's fine. What angle, when I take its tangent, gives me 324/54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the Inverse Tan Function. So we could rewrite this as we're going to take the inverse tangent-- and sometimes it's written as tangent with this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324/54. And just to be clear, what is this inverse tangent? This just literally says, this will return what is the angle that, when I take the tangent of it, gives me 324/54. This says, what is the angle that, when I take the tangent of it, gives me tangent of theta? So this right over here, this just simplifies to theta. Theta is the angle that when you get the tangent of it gets you tangent of theta. And so we get theta is equal to inverse tangent of 324/54. Once again, this inverse tangent thing you might find confusing. But all this is saying is, over here, we're saying tangent of some angle is 324/54. This is just saying my angle is whatever angle I need so that when I take the tangent of it, I get 324/54. It's how we will solve for theta. So let's get our calculator out. And let's say that we want our answer in degrees. Well, I'm just going to assume that they want our answers in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the 2nd mode right over here. And actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees. Now let me exit out of here. And let me just type in the inverse tangent-- so it's in this yellow color right here-- inverse tangent of 324 divided by 54 is going to be-- and they told us to round to two decimal places-- 80.54 degrees. So theta is equal to 80.54 degrees. That's the angle at which you should shoot the gun to help defeat this horrible alien." + }, + { + "Q": "what is the meaning of sub x and sub y ...\n\n\nplease tell me ..", + "A": "When you are dealing with a vector like velocity you often break it up into the amounts along each axis. So if you have the velocity vector V the amount along the X axis is usually V sub x and along the Y axes is V sub y. It is just a way of indicating that you are dealing with a portion of the vector.", + "video_name": "FaF3v-ezbSk", + "transcript": "Good afternoon. We've done a lot of work with vectors. In a lot of the problems, when we launch something into--- In the projectile motion problems, or when you were doing the incline plane problems. I always gave you a vector, like I would draw a vector like this. I would say something has a velocity of 10 meters per second. It's at a 30 degree angle. And then I would break it up into the x and y components. So if I called this vector v, I would use a notation, v sub x, and the v sub x would have been this vector right here. v sub x would've been this vector down here. The x component of the vector. And then v sub y would have been the y component of the vector, and it would have been this vector. So this was v sub x, this was v sub y. And hopefully by now, it's second nature of how we would figure these things out. v sub x would be 10 times cosine of this angle. 10 cosine of 30 degrees, which I think is square root of 3/2, but we're not worried about that right now. And v sub y would be 10 times the sine of that angle. This hopefully should be second nature to you. If it's not, you can just go through SOH-CAH-TOA and say, well, the sine of 30 degrees is the opposite of the And you would get back to this. But we've reviewed all of that, and you should review the initial vector videos. But what I want you to do now, because this is useful for simple projectile motion problems-- But once we start dealing with more complicated vectors-- and maybe we're dealing with multi-dimensional of vectors, three-dimensional vectors, or we start doing linear algebra, where we do end dimensional factors --we need a coherent way, an analytical way, instead of having to always draw a picture of representing vectors. So what we do is, we use something I call, and I think everyone calls it, unit vector notation. So what does that mean? So we define these unit vectors. Let me draw some axes. And it's important to keep in mind, this might seem a little confusing at first, but this is no different than what we've been doing in our physics problem so far. Let me draw the axes right there. Let's say that this is 1, this is 0, this is 2. 0, 1, 2. I don't know if must been writing an Arabic or something, going backwards. This is 0, 1, 2, that's not 20. And then let's say this is 1, this is 2, in the y direction. I'm going to define what I call the unit vectors in two dimensions. So I'm going to first define a vector. I'll call this vector i. And this is the vector. It just goes straight in the x direction, has no y component, and it has the magnitude of 1. And so this is i. We denote the unit vector by putting this little cap on top of it. There's multiple notations. Sometimes in the book, you'll see this i without the cap, and it's just boldface. There's some other notations. But if you see i, and not in the imaginary number sense, you should realize that that's the unit vector. It has magnitude 1 and it's completely in the x direction. And I'm going to define another vector, and that one is called j. And that is the same thing but in the y direction. That is the vector j. You put a little cap over it. So why did I do this? Well, if I'm dealing with two dimensions. And as later we'll see in three dimensions, so there will actually be a third dimension and we'll call that k, but don't worry about that right now. But if we're dealing in two dimensions, we can define any vector in terms of some sum of these two vectors. So how does that work? Well, this vector here, let's call it v. This vector, v, is the sum of its x component plus its y component. When you add vectors, you can put them head to tail like this. And that's the sum. So hopefully knowing what we already know, we knew that the vector, v, is equal to its x component plus its y component. When you add vectors, you essentially just put them head to tails. And then the resulting sum is where you end up. It would be if you added this vector, and then you put this tail to this head. So you end up there. So that's the vector. So can we define v sub x as some multiple of i, of this unit vector? Well, sure. v sub x completely goes in the x direction. But it doesn't have a magnitude of 1. It has a magnitude of 10 cosine 30 degrees. So its magnitude is ten. Let me draw the unit vector up here. This is the unit vector i. It's going to look something like this and this. So v sub x is in the exact same direction, and it's just a scaled version of this unit vector. And what multiple is it of that unit vector? Well, the unit vector has a magnitude of 1. This has a magnitude of 10 cosine of 30 degrees. I think that's like, 5 square roots of 3, or something like that. So we can write v sub x-- I keep switching colors to keep things interesting. We can write v sub x is equal to 10 cosine of 30 degrees times-- that's the degrees --times the unit vector i-- let me stay in that color, so you don't confused --times the unit vector i. Does that make sense? Well, the unit vector i goes in the exact same direction. But the x component of this vector is just a lot longer. It's 10 cosine 30 degrees long. And that's equal to-- cosine of 30 degrees is square root of 3/2 --so that's 5 square roots of 3 i. Similary, we can write the y component of this vector as some multiple of j. So we could say v sub y, the y component-- Well, what is sine of 30 degrees? Sine of 30 degrees is 1/2. 1/2 times 10, so this is 5. So the y component goes completely in the y direction. So it's just going to be a multiple of this vector j, of the unit vector j. And what multiple is it? Well, it has length 5, while the unit vector has just length 1. So it's just 5 times the unit vector j. So how can we write vector v? Well, we know the vector v is the sum of its x component and its y component. And we also know, so this is a whole vector v. What's its x component? Its x component can be written as a multiple of the x unit vector. That's that right there. So you can write it as 5 square roots of 3 i plus its y component. So what's its y component? Well, its y component is just a multiple of the y unit vector, which is called j, with the little funny hat on top. And that's just this. It's 5 times j. So what we've done now, by defining these unit vectors-- And I can switch this color just so you remember this is i. This unit vector is this. Using unit vectors in two dimensions, and we can eventually do them in multiple dimensions, we can analytically express any two dimensional vector. Instead of having to always draw it like we did before, and having to break out its components and always do it visually. We can stay in analytical mode and non graphical mode. And what makes this very useful is that if I can write a vector in this format, I can add them and subtract them without having to resort to visual means. And what do I mean by that? So if I had to find some vector a, is equal to, I don't know, 2i plus 3j. And I have some other vector b. This little arrow just means it's a vector. Sometimes you'll see it as a whole arrow. As, I don't know, 10i plus 2j. If I were to say what's the sum of these two vectors a plus b? Before we had this unit vector notation, we would have to draw them, and put them heads to tails. And you had to do it visually, and it would take But once you have it broken up into the x and y components, you can just separately add the x and y components. So vector a plus vector b, that's just 2 plus 10 times i plus 3 plus 2 times j. And that's equal to 12i plus 5j. And something you might want to do, maybe I'll do it in the future video, is actually draw out these two vectors and add them visually. And you'll see that you get this exact answer. And as we go into further videos, or future videos, you'll see how this is super useful once we start doing more complicated physics problems, or once we start doing physics with calculus. Anyway, I'm about to run out of time on the ten minutes. So I'll see you in the next video." + }, + { + "Q": "Why didn't any of the native American tribes help Spain?", + "A": "One, Spain wasn t really involved in this particular war. Two, at first they did but would really keep helping the people that is conquering and killing you?", + "video_name": "UAhbwYBAoe0", + "transcript": "- [Instructor] When we're talking about major wars in colonial North America, we tend to think about the American Revolution, not its earlier iteration, the Seven Years' War, and I think that's a shame because the Seven Years' War was incredibly influential not only on the American Revolution, but on the complexion of the world. Thanks to the Seven Years' War, Canada became a British country, not a French country. The Acadians moved down to Louisiana and became known as the Cajuns, and most importantly, England became the world's preeminent empire. So if you've been following along this far, you may have noticed two things. One, that the people who named this war seem to be very bad at math because 1754 to 1763 is nine years, not seven, and that this war seems to have two names, both the Seven Years' War and the French and Indian War, which is a name you perhaps have heard before. Well, lemme tackle those two oddities in reverse order. So not only does the Seven Years' War have two names, it has a whole number of names. It's called the Seven Years' War, the French and Indian War, the War of the Conquest, the Pomeranian War, the Third Silesian War, the Third Carnatic War. This is a war with a whole bunch of names, and the reason that it has a whole bunch of names is that it was fought in a whole bunch of places. The Seven Years' War was really the first global war, and we're talking 150 years before World War One. Aspects of the Seven Years' War, as you can kinda see from this map, were fought in Europe, in South America, the coast of Africa, in India, the Philippines, and of course, in North America. The many different names come from the many different fronts of this war, and I would say that French and Indian War is actually the name for the North American front of this war, or theater of this war. So there are two reasons why I think Seven Years' War is a better name than French and Indian War. One is that Seven Years' War gets at the idea that it was not just happening in North America. It was happening all over the world, so it shows that it was a global war, but I also think Seven Years' War is a better name than French and Indian War because I think French and Indian War is kind of confusing because you would think that it means that the principal parties in this war were the English versus the French and the Indians, when in fact it was the English and their Indian allies versus the French and their Indian allies. Native Americans fought on both sides of this conflict, so rather than the English and Indian versus French and Indian War, let's go with the shorter Seven Years' War, which brings us back to our awkward date range. So the reason that it's called the Seven Years' War is because the English didn't actually declare war on the French until 1756. So even though fighting started a little bit earlier in North America, the true range of dates, at least in legal terms, is from 1756 to 1763, or seven years. It's a complicated name for a complicated war, but really what it came down to was England and France duking it out over who was going to be the supreme imperial power in the world, and they were concerned about who was going to have the most territory in the world, therefore, their concern over who was going to control North America and their competing claims here, and also access to trade. So who was going to be able to trade with North Americans? Who was going to be able to trade with the lucrative Indian subcontinent, and who would be the leading power in Europe? So let's dial in a little closer on the North American theater of this war, which will have the most effect on the future United States. Alright, so here is map of territorial claims by European powers in North America before the Seven Years' War. Now you can see that there are some places where they overlap, which is really gonna be the heart of the problem in this conflict. So England, shown here in red, I'm gonna outline it a bit, was, as you know from your early American history, here along the eastern seaboard of what's today the United States, and also up into Canada. France claimed this interior region of Canada and today of the territorial United States, and Spain was in the mix here. Remember Spain has still been a fairly influential colonial power in Florida and in contemporary Mexico, and also down here in Cuba and South America. Alright, so we've got three major European powers in the mix here in North America, England, France, and Spain, but what this map doesn't show is the American Indian powers, who are also in this area. So most of this region really west of the Appalachian mountains, is Indian country, and the majority of inhabitants were Native Americans, and they really held the majority of power in this region as well. So major Native American groups that are in play in this conflict are Iroquois Confederacy, and also Cherokees, Hurons, Algonquians, Abenakis, and Mi'kmaqs, and that's just a small sampling. So you can see that there are a number of important Native American tribes who are specifically in this area of Canada, which is disputed, and also moving in the greater Appalachian region. So what does each of these groups want? Well, England definitely wants territory. They want to make sure that they're English settlers along the eastern seaboard, whom we'll soon be calling Americans, have room to expand. The French wanna make sure that they still have access to trade with Native Americans because their main concern is fur, which is a very valuable commodity in Europe, and Spain wants to make sure that they have access to their sugar islands and also their precious metals in the Caribbean and in South America. Now it's worth noting, 'cause I think this is really interesting to students of American history, that all of this territory, all of North America, was way less valuable than all of this territory because we're not talking about just value in land. We're talking about value in commodities, and what the Caribbean had was sugar, and sugar is the most valuable crop in this time period. So a tiny island down here in the Bahamas is probably worth more to a European power than the entire interior of North America, and what do these Native American groups want? Well, some of them want help with revenge on each other. Many other smaller Native American groups have been displaced by the Iroquois, who are here in upstate New York, kind of Quebec region. So the Iroquois is actually expanding and really defending their claim as the largest Native American empire, but the other thing that they want is to make sure that their territory is no longer encroached upon by English settlers in particular. Now one mistake I see early students of U.S. history making is thinking that all Native Americans kind of shared a cultural and political bond, right? That they saw themselves as one larger people who had to unite against the encroachment of Europeans, and that was definitely not the case. Native Americans had been living in this territory for thousands of years, and they had enemies and beef with other groups that went back way longer than the arrival of Europeans in North America. So when nations like England and France arrived with their weapons and their trade goods, the American Indians didn't look at each other and say, \"Oh wait, now we're all one race. \"We need to join together against \"the encroachment of whites.\" They saw England and France and Spain as possible avenues to getting one up on their older enemies. So when an English trader sold a gun to, say, a Huron, he was way more likely to go after, say, the Iroquois with that gun than he was to go after a French trader. So another reason why the Seven Years' War is a better name for the French and Indian War than French and Indian War is because these Native American groups did not ally all with France. In fact, the Iroquois and Cherokee ended up allied with England, and most of the other Native American groups ended up allied with France, but they were fighting each other in addition to fighting England. Alright, so the stage is set for this conflict with all of these competing groups in this unclear territory, and how this turns into a war, we'll get to in the next video." + }, + { + "Q": "Where's the manger animals? This is an encouraging subject for Leonardo...to use his anatomy studies..:)", + "A": "Wrong gospel. The Magi are in Matthew, where there is neither stable nor animals. Those participants are found in Luke, where there are no Magi, no star, and not even any camels. Leonardo was doing everyone a favor by depicting merely ONE of the stories, and not conflating or confusing the two.", + "video_name": "QxNqWZPzsGw", + "transcript": "SPEAKER 1: We're standing in front of an unfinished painting by Leonardo da Vinci. It's a big painting. And interestingly, it's almost a perfect square. SPEAKER 2: It's really unfinished. It's not just that it has parts that are unfinished, but it's really just the underpainting. SPEAKER 1: This is The Adoration of the Magi, a moment in the Christian story when Christ has been born and three kings from the east, guided by the star of Bethlehem, come to Mary and offer Christ three gifts, frankincense, myrrh, and gold. What's revealed to us here is Leonardo's working method-- not only his brilliant drawing, but the way in which he constructs figures. Remember that Leonardo is first and foremost not a painter. He's really a scientist. He's really an engineer. He's somebody who looks and understands nature. SPEAKER 2: We have a sense of Leonardo's deep understanding of human anatomy. Even when he's painting clothed figures, he's really understanding the skeletal structure. He's understanding the musculature of the body. SPEAKER 1: If you look at the group of figures to the right, about mid-level, you see one figure that almost looks like it's a skull. And it's as if Leonardo is literally constructing the bones before he'll put flesh on them, before he'll put clothes on them, before he'll add color. That's a group of figures that's often referred to as the philosophers. But maybe we should discuss the central group first. SPEAKER 2: So we have Mary and the Christ child front and center, forming a pyramid shape together with the Magi in front. And that's a shape that we see very often in paintings of the High Renaissance that provide a stable form. And you see that right here in the foreground with Mary and Christ. SPEAKER 1: That's especially important in this painting, which is so chaotic, where there's so much going on. On the upper right, for instance, there's actually a battle. You have two horses rearing up. On the upper left, you have the fragments of what look like some sort of classical architecture. You can see these wonderful steps in perfect linear perspective. Leonardo actually did some brilliant drawings in preparation for this painting. But let's look a little more closely at what you just said, and see if we can define those lines a little more exactly. If you start with the Virgin Mary and you look at her face, she's glancing across the top of her son's head, down his arm. He picks up, actually, her glance and brings our eye down until it's met by one of the Magi who's offering a gift. We can actually run that line down past his toes to the corner of the painting. Or we can actually pick up from Mary again and go the other way. If we go down the bridge of her nose, across her shoulder, picked up by the kneeling figure in the foreground at the left. What's interesting, as you said, this is not simply a triangle. But this is a pyramid that actually comes forward as it moves down. SPEAKER 2: --and exists in space. I'm struck by the way that Leonardo is paying attention to all of these human reactions to what's going on. And we've got lots of faces half hidden in the darkness and a lot of gestures. And it really reminds me, in a way, of Leonardo's Last Supper, where you have Christ in the middle forming a kind of pyramid shape with his outstretched arms. And all the chaos, and the reactions of the apostles around him, but this real sense of stability in the center. SPEAKER 1: That's such a characteristic of the High Renaissance-- this notion of balance, of a kind of perfection, of a sense of the eternal. But then, of course, how do we as humans, react? There's another element here, which is important and very characteristic of Leonardo. And even though this is just the underpainting, we can make it out. And that's this technique of sfumato, which in Italian is smoke. And it tends to create a kind of visual glue that creates a kind of harmony between forms within the paint, brings things together, and keeps paintings from having that sense of the isolated, so much a characteristic of the early Renaissance. SPEAKER 2: Right, so instead of figures being defined by lines, the figures are enveloped in atmospheric [? haziness ?] or softness, that kind of smokiness. And they almost seem to emerge out of the darkness into light and fade back into the darkness again. And so Leonardo's unifying the figures in yet another way. Not only in the pyramid composition and through their glances and gestures, but also into that smoky atmosphere. SPEAKER 1: We see this beautiful chiaroscuro, this beautiful smoke, this beautiful line, this beautiful composition, this complex sense of emotion. I'd love to know what this painting would have looked like had it been finished. SPEAKER 2: Me too." + }, + { + "Q": "So you need to have the same y change the same way each time? Example:\nx=1|2|3|\ny=10|20|30\nthat would be a direct variation right?", + "A": "Yes as x increases y should also increase at a constant interval", + "video_name": "rSadG6EtJmY", + "transcript": "We're told that the total cost of filling up your car with gas varies directly with the number of gallons of gasoline you are purchasing. So this first statement tells us that if x is equal to the number of gallons purchased, and y is equal to the cost of filling up the car, this first statement tells us that y varies directly with the number of gallons, with x. So that means that y is equal to some constant, we'll just call that k, times x. This is what it means to vary directly. If x goes up, y will go up. We don't know what the rate is. k tells us the rate. If x goes down, y will be down. Now, they give us more information, and this will help us figure out what k is. If a gallon of gas costs $2.25, how many gallons could you purchase for $18? So if x is equal to 1-- this statement up here, a gallon of gas-- that tells us if we get 1 gallon, if x is equal to 1, then y is $2.25, right? y is what it costs. They tell us 1 gallon costs $2.25, so you could write it right here, $2.25 is equal to k times x, times 1. Well, I didn't even have to write the times 1 there. It's essentially telling us exactly what the rate is, what k is. We don't even have to write that 1 there. k is equal to 2.25. That's what this told us right there. So the equation, how y varies with x, is y is equal to 2.25x, where x is the number of gallons we purchase. y is the cost of that purchase, so it's $2.25 a gallon. And then they ask us, how many gallons could you purchase for $18? So $18 is going to be our total cost. It is y cost of filling the car. So 18 is going to be equal to 2.25x. Now if we want to solve for x, we can divide both sides by 2.25, so let's do that. You divide 18 by 2.25, divide 2.25x by 2.25, and what do we get? Let me scroll down a little bit. The right-hand side, the 2.25's cancel out, you get x. And then what is 18 divided by 2.25? So let me write this down. So first of all, I just like to think of it as a fraction. 2.25 is the same thing-- let me write over here-- 2.25 is equal to 2 and 1/4, which is the same thing as 9 over 4. So 18 divided by 2.25 is equal to 18 divided by 9 over 4, which is equal to 18 times 4 over 9, or 18 over 1 times 4 over 9. And let's see, 18 divided by 9 is 2, 9 divided by 9 is 1. That simplifies pretty nicely into 8. So 18 divided by 2.25 is 8, so we can buy 8 gallons for $18." + }, + { + "Q": "What if it says negative 2/3? (Like it's not -2/3 or 2/-3, the negative sign is in the middle)", + "A": "-2/3 or 2/-3, they both are negative 2/3.", + "video_name": "pi3WWQ0q6Lc", + "transcript": "Let's do a few examples multiplying fractions. So let's multiply negative 7 times 3/49. So you might say, I don't see a fraction here. This looks like an integer. But you just to remind yourself that the negative 7 can be rewritten as negative 7/1 times 3/49. Now we can multiply the numerators. So the numerator is going to be negative 7 times 3. And the denominator is going to be 1 times 49. 1 times 49. And this is going to be equal to-- 7 times 3 is 21. And one of their signs is negative, so a negative times a positive is going to be a negative. So this is going to be negative 21. You could view this as negative 7 plus negative 7 plus negative 7. And that's going to be over 49. And this is the correct value, but we can simplify it more because 21 and 49 both share 7 as a factor. That's their greatest common factor. So let's divide both the numerator and the denominator by 7. Divide the numerator and the denominator by 7. And so this gets us negative 3 in the numerator. And in the denominator, we have 7. So we could view it as negative 3 over 7. Or, you could even do it as negative 3/7. Let's do another one. Let's take 5/9 times-- I'll switch colors more in this one. That one's a little monotonous going all red there. 5/9 times 3/15. So this is going to be equal to-- we multiply the numerators. So it's going to be 5 times 3. 5 times 3 in the numerator. And the denominator is going to be 9 times 15. 9 times 15. We could multiply them out, but just leaving it like this you see that there is already common factors in the numerator and the denominator. Both the numerator and the denominator, they're both divisible by 5 and they're both divisible by 3, which essentially tells us that they're divisible by 15. So we can divide the numerator and denominator by 15. So divide the numerator by 15, which is just like dividing by 5 and then dividing by 3. So we'll just divide by 15. Divide by 15. And this is going to be equal to-- well, 5 times 3 is 15. Divided by 15 you get 1 in the numerator. And in the denominator, 9 times 15 divided by 15. Well, that's just going to be 9. So it's equal to 1/9. Let's do another one. What would negative 5/9 times negative 3/15 be? Well, we've already figured out what positive 5/9 times positive 3/15 would be. So now we just have to care about the sign. If we were just multiplying the two positives, it would be 1/9. But now we have to think about the fact that we're multiplying by a negative times a negative. Now, we remember when you multiply a negative times a negative, it's a positive. The only way that you get a negative is if one of those two numbers that you're taking the product of is negative, not two. If both are positive, it's positive. If both are negative, it's positive. Let's do one more example. Let's take 5-- I'm using the number 5 a lot. So let's do 3/2, just to show that this would work with improper fractions. 3/2 times negative 7/10. I'm arbitrarily picking colors. And so our numerator is going to be 3 times negative 7. 3 times negative 7. And our denominator is going to be 2 times 10. 2 times 10. So this is going to be the numerator. Positive times a negative is a negative. 3 times negative 7 is negative 21. Negative 21. And the denominator, 2 times 10. Well, that is just 20. So this is negative 21/20. And you really can't simplify this any further." + }, + { + "Q": "how does Sal explain it so thoroughly?", + "A": "Because he has had much practice with mathmathics", + "video_name": "XkRD9lv_y44", + "transcript": "- [Voiceover] Let's give ourselves some practice substituting positive and negative values for variables. So we're told to evaluate X, we're told to evaluate X minus negative Y, where X is equal to negative two and Y is equal to five. So everywhere we see an X, we can replace with a negative two. Everywhere we see a Y, we replace with a five. So this is a Y right over here, and then of course, and then of course this is an X. Let's do that. So instead of that X, let's write negative two. So we have negative two minus and then we have in the parentheses a negative Y. Y is equal to, Y is equal to five. Now what is this going to evaluate to? Well, this is the same thing as negative two, now subtracting a negative five, so all of this business here, subtracting a negative five, that's the same thing as adding a five. So it's going to be negative two plus five, which is equal to three. This is going to be equal to three. And there's several ways to think about this. my brain thinks okay, I'm starting at negative two. If I add two I get to zero and then I would have to add another, and then I have to add another three so that gets me to three. You can even view this as, you can even view this, my brain kind of says negative two, well let's see, I have to add, i have to add two to get back to zero and then I have to add another three if I want to add a total of five. So this is going to be zero plus three gets me to three. Another way to think about it, negative two plus five is the same thing as five minus two. Five minus two, which is of course equal to three. Or you could of course draw it on a number line. And you would say if you start at negative two and you take five steps to the right, you get to positive three. Let's do another one of these. This one's a little bit more, a little bit more complex. So let's see. We're told to evaluate three minus negative six plus negative H plus negative four, where H is equal to negative seven. Well there's two ways you could do it. I could just take the negative seven and replace it, the H with that negative seven, or I could actually try to simplify this expression first and then do the substitution for H. Let's actually do that. My brain feels like doing that. So this expression, I have the three, but instead of subtracting a negative six, instead of subtracting a negative six, that's gonna be the same thing as just adding six. So three plus six. And adding a negative H, adding a negative H, adding a negative H, that's the same thing as just subtracting H. So three plus six minus H, and then adding a negative four, adding a negative four, that's the same thing as subtracting four. That's the same thing as subtracting four. And now of course we can do this in any, you know, addition and subtraction we have we can change the order in which we do it. So let's do that just to kind of simplify all of this. So, actually first of all, I can figure out what three plus six is. Three plus six is equal to nine. And then I have this minus four here. So I could say nine minus four, actually I want to be careful not to skip any steps. So three plus six, three plus six, in a color that you can see, so three plus six is nine. So that's nine minus H minus four, minus four. Now I could change the order in which I do this addition or subtraction, so this is going to be the equivalent of nine minus four. Nine minus four minus H, minus H. And I just did that so I can simplify and figure out what nine minus four is. Nine minus four, of course, let me do this in blue, navy blue. Nine minus four is five. So this whole thing simplified to five minus H before I even did the substitution. And now I can substitute H with negative seven. So this is going to be equal to, this is going to be equal to, when I do the substitution, I'll write it up here, it's going to be five minus, I'll do the minus in that magenta color, minus and now where I see an H, I'm gonna replace it with negative seven. Five minus negative seven. You want to be very careful there, you might be tempted to say, oh I have a negative here, negative here, let me just replace H with a seven. Remember, H is negative seven, so you're subtracting H. You're gonna subtract negative seven. So this is five minus negative seven, which is the same thing, which is the same thing as five plus seven. Five plus seven, which we all know is equal to 12. And we're done. Let's do a few more of these. You can't really get enough practice here, this is some important foundational skills for the rest of your mathematical lives. (laughing) Alright. So consider, I don't make you too stressed about it. Consider the following number line. Alright, so we've got a number line here and let's see, they didn't mark off all the numbers here. This is negative four, E is at this point, this is then we go to two. So it looks like we're counting by twos here, that this is negative two, this is zero, yep. Negative four, if you increase by two, negative four, negative two, zero, two, four, this would be a six. This would be a negative six. They intentionally left those numbers off, so we had to figure out that hey, look, between negative four and two, to go from negative four to two, you have to increase by six and we only have one, two, three hashmarks. So each of those hashmarks must be increasing by two. Well anyway, now we know, now we know what all the points in the number line are. Evaluate E minus F. Well we know, we know that E is equal to negative two. And we know, we know that F is equal to four. So this is going to be the same thing, E is equal to negative two minus, minus positive four. Minus positive four. Let me do it in that same blue color. Minus four. Well negative two minus four, we have the number line in front of us, this is just going to be negative six. If you start at negative two, you subtract, if you subtract two you get to negative four, you subtract another two, you get to negative six. So we are done. Let's do one more of these. Let's evaluate T plus negative U. Where once again T and U are on this number line and it looks like each of these hashmarks we're incrementing by three as we go up. Zero, three, six. And if we go down, zero, negative three, negative six. So it's clear that U is equal to three. U is equal to three and it is also clear that T is equal to negative six. T is equal to negative six. So T plus negative U is going to be, so it's going to be negative six plus, negative six plus negative U, U is three. U is three. You want to be very careful, you might say oh well U is positive so I'm just gonna put a positive number here, but remember, it's negative U. Wherever you see the U, replace it with the three. So it's negative three. So this is gonna be six plus negative three, or sorry, negative six plus negative three, which we can rewrite as negative six minus three. All of this business can be rewritten as negative six minus three. Negative six minus three. If you're at negative six and you go three to the left, you go three more negative, you're going to end up at negative nine. And we're all done." + }, + { + "Q": "How do you find a slope when you only have an x coordinate?", + "A": "You can t find a slope with only an X coordinate. You need an X and Y to find a slope", + "video_name": "xR9r38mZjK4", + "transcript": "what i want to do with this video is think about the relationship between variables and then think about what the graph of that relationship should look like. So let's say these two axis, the horizontal axis over here I plot the price of a product and lets say this vertical axis over here i plot the demand for the product, and i'm only plotting the first quadrant here because i'm assuming that the price can only be positive and i'm assuming that the demand can only be positive, that people aren't going to pay someone to take the product away from them. So let's think about what would happen for the price and demand for most normal products. So, if the price is low, you would expect that a lot of poeple are willing to buy that thing they're like \"Oh it's a good price, i would like to buy it.\" So, if the price is low, then the demand would be high so maybe it would be somplace over here, all the way that you would have really high demand if the price was zero. so if the price was low the demand would be high. Now what happens is the price -- so right here the price is low, demand is high, if the price were to go up a little bit then maybe the demand goes down a little bit, right? price went up a little bit demand went down a little bit. if the price went up a little bit more then maybe demand goes down a little bit more. as the price went up a bunch then demand would go down a bunch and so the line that represents how the demand relates to price might look something like this, and i'm just going to assume it's it a line. It might not be a line, it might be a curve. It might look something like that. Or it might look something like that. But in general is someone were to ask you, if you saw this magenta curve that as price increases what happens to demand. You just say \"Well look price increases, as price increases what happens to demand?\" Well demand is decreasing. Now let's think about a different scenario. Let's talk about the demand for real estate. For actual property, and lets say that on this axis that we plot the population. The population in the area, and this right over here this is demand for land. So when the population is very low, you can imagine, if the population is zero there is no one there that would want to buy land. So if the population is very low the demand is going to be very low. And as population increases, demand should increase. If the population increases, more people are going to want to buy land. And if the population goes up a bunch then a lot of people are going to want to buy land. So you'll see a line that looks something like this. And once again I drew a line, it doesn't have to be a line it could be a curve of some kind. It could be a curve that looks something like that, or a curve that looks something like that. We don't know but the general idea is that if someone showed you a graph that looked like this. And as population increases what happens to demand. We'll you'd say \"Look, this is population increasing, what happens to demand?\" Demand is going up. Where as price increased the demand went down. Here as population is increasing demand went up. And you can just make that more general with variables. We're talking about specific cases here. But if I were to plot something like this, if you were to see a graph that looked like that and this is the variable x and this is the variable y. And someone were to ask you what happens to y as x increases. Well you take any x that's the y that we have for that x. And as you increase x, as you move in the positive horizontal directions. As you increase x what is happening to y? Y is going up. So for this example, as x increases y is increasing. If we had a graph that looked like this. Let's call that the a axis and this is the b axis and maybe our graph looked like this. What happens as A increases? If you pick an A right over here. We're at that A and that B. As A increases what's happening to B? Well as A increases our B is lower. As A increases here B is decreasing. So, just wanting to give you a general idea, when X and Y increased together the line goes from the bottom left to the top right, we would call this an upward sloping line. We would call this a positive slope. Everytime X is increased Y also increases is upwards sloping. When our independant variable increases and our dependent variable decreases. When the independent variable is increasing, then you say it has a downward slope, when you go from the top left to the bottom right." + }, + { + "Q": "0:47 Why is it weak?\nIs it weak because both oxygens have the same electronegitivity?", + "A": "Not because of the electronegativity but because of how alcohols are more stable than cyclic peroxides and can easily form from a reaction with water.", + "video_name": "KfTosrMs5W0", + "transcript": "If you start with an alkene and add to that alkene a percarboxylic acid, you will get epoxide. So this is an epoxide right here, which is where you have oxygen in a three-membered ring with those two carbons there. You can open up this ring using either acid or base catalyzed, and we're going to talk about an acid catalyzed reaction in this video. And what ends up happening is you get two OH groups that add on anti, so anti to each other across from your double bond. So the net result is you end up oxidizing your alkene. So you could assign some oxidation numbers on an actual problem and find out that this is an oxidation reaction. All right. Let's look at the mechanism to form our epoxide. So we start with our percarboxylic acid here, which looks a lot like a carboxylic acid except it has an extra oxygen. And the bond between these two oxygen atoms is weak, so this bond is going to break in the mechanism. The other important thing to note about the structure of our percarboxylic acid is the particular confirmation that it's in. So this hydrogen ends up being very close to this oxygen because there's a source of attraction between those atoms. There's some intramolecular hydrogen bonding that keeps it in this conformation. When the percarboxylic acid approaches the alkene, when it gets close enough in this confirmation, the mechanism will begin. This is a concerted eight electron mechanism, which means that eight electrons are going to move at the same time. So the electrons in this bond between oxygen and hydrogen are going to move down here to form a bond with this carbon. The electrons in this pi bond here are going to move out and grab this oxygen. That's going to break this weak oxygen-oxygen bond, and those electrons move into here. And then finally, the electrons in this pi bond are going to move to here to form an actual bond between that oxygen and that hydrogen. So let's see if we can draw the results of this concerted eight electron mechanism. So, of course, at the bottom here we're going to form our epoxide. So we draw in our carbons, and then we can put in our oxygen here. And then we show the bond between those like that. And then up at the top here, here's my carbonyl carbon. So now there's only one bond between that carbon and this oxygen. There is a new bond that formed between that oxygen and that hydrogen, and there is an R group over here. And then there used to be only one bond to this oxygen, but another lone pair of electrons moved in to form a carbonyl here. So this is our other product, which you can see is a carboxylic acid. Let's color code these electrons so we can follow them a little bit better. So let's make these electrons in here, those electrons are going to form the bond on the left side between the carbon and the oxygen like that. Let's follow these electrons next. So now let's look at these electrons in here, the electrons in this pi bond. Those are the ones that are going to form this side of our epoxide ring like that. And let's make our oxygen-oxygen bond blue here. So the electrons in this bond, those are the ones that moved in here to form our carbonyl like that. And then let's go ahead and make these green right here, the electrons in this bond right here. These are the ones that moved out here to form the bond between our oxygen and our hydrogen. So our end result is to form a carboxylic acid and our epoxide. Let's look at a reaction, an actual reaction for the formation of epoxide, and then we'll talk about how to form a diol from that. So if we start with cyclohexene-- let's go ahead and draw cyclohexene in here. Let's do another one. That one wasn't very good. So we draw our cyclohexene ring like that. And to cyclohexene, we're going to add peroxyacetic acid. So what does peroxyacetic acid look like? Well, it's based on acetic acid. But it has one extra oxygen in there, so it looks like that. So that's our peroxyacetic acid. So we add cyclohexene to peroxyacetic acid, we're going to form an epoxide. So we're going to form a three-membered ring, including oxygen. I'm going to say the oxygen adds to the top face of our ring. It doesn't really matter for this example, but we'll go ahead and put in our epoxide using wedges here. And that must mean going away from us, those are hydrogens in space. So that's the epoxide that would form using the mechanism that we put above there. Let's go ahead and open this up epoxide using acid. So just to refresh everyone's memory, go back up here. Now we're going to look at this second part where we add H3O plus to form our diol. So let's take a look at that now. So we're going to add H3O plus to this epoxide. And I'm going to redraw our epoxide to give us a better view point here. So I'm going to put my oxygen right here, and then that's bonded to our two carbons like this. And then we see if we can draw the rest of the ring. And so in the back here, here is the rest of my cyclohexane ring like that. And we'll go ahead and put in our lone pairs of electrons. So this is the same exact drawing above here. Now I have my H3O plus in here like this, so my hydronium ion is present with a lone pair of electrons, giving us a plus one formal charge like that. So the oxygen on our epoxide is going to act as a base. It's going to take a proton. So this lone pair of electrons is going to take this proton right here, which would kick these electrons in here off onto my oxygen. So let's draw the result of that acid-base reaction. So I'm going to make a protonated epoxide. So let's go ahead and draw our oxygen here, and it's connected to those carbons down here. So I'll go ahead and draw the rest of my ring in the back here like that. And then one lone pair of electrons didn't do anything, so it's still there. One lone pair of electrons is the one that formed the bond on that proton, so this is my structure now. And this would give this oxygen a plus one formal charge, so it's positively charged now. So this is the same structure that we saw in the earlier videos, like with our cyclic halonium ion. And just like the cyclic halonium ion in those earlier videos-- check out the halohydrin video-- you're going to get a partial carbocation character with these carbons down here. So the resonance hybrid is going to give these carbons some partial positive character. So when water comes along as a nucleophile, the lone pair of electrons on water are going to be attracted to those carbons. So opposite charges attract. These two blue carbons are partially positive. The negative electrons are attracted to the partially positive carbon, and you're going to get nucleophilic attacks. So let's say this lone pair of electrons attacks right here. Well, that would kick the electrons in this bond off onto your oxygen. So let's go ahead and draw the result of an attack on the carbon on the left. So let's get some more room here. So what would happen in that instance? Well, let's go ahead and draw our cyclohexane ring back here. So here is our cyclohexane ring. The oxygen attacked the carbon on the left. So there is the oxygen that did the nucleophilic attack, so it has two hydrogens on it. It has one lone pair of electrons now, and it formed a plus one formal charge. Our epoxide opened. The electrons kicked off onto the top oxygen and that means that the top oxygen moves over here like that. So that would be your structure. Well, this lone pair electrons could have attacked this carbon, right? This carbon could have been the partially positive one in the resonance hybrid, which would kick these electrons off onto this oxygen. So let's go ahead and draw the result of that nucleophilic attack. So I'll go ahead and put in my cyclohexane ring like that. This time our oxygen is going to bond with a carbon on the left, two hydrogens attach to it, a lone pair of electrons, a positive one formal charge. And then this time the oxygen on top is going to kick off onto this oxygen over here in the left like that. So you're going to get an OH over there. So in the next step of the mechanism-- we're almost done, we've almost formed our diol. We're going to have water come along. And this time, instead of water acting as a nucleophile, water is going to act as a base. It's going to take a proton. So let's look at the product on the left, here. So this lone pair of electrons would take one of these protons, kick these electrons off onto your oxygen like that. So let's go ahead and draw the result of that acid-base reaction. So let's draw our cyclohexane ring like that, and now we have an OH down here. So this is now an OH, and this was an OH. So we've achieved our product. We've added 2 OHs anti to each other. Same thing can happen over here. You could grab these. You could grab this proton, kick these electrons off onto your oxygen like that. And so on the right, after we draw our cyclohexane ring, we're going to have an OH right here. And then we're going to have an OH over here like that. So we have two products. And if you look at them, they are mirror images of each other. If I were to put a mirror right here, you would see that they would be reflected in a mirror. And they are nonsuperimposable. So nonsuperimposable mirror images, enantiomers. So you're going to get two products for this reaction. So just to summarize this reaction, let's do it one more time. Let's start with the cyclohexene. And in the first step of our mechanism, the first of our reaction here, we added peroxyacetic acid. So we added CH3CO3H. And in the second step of our mechanism, in the second step of our reactions here, we added H3O plus. And so that opened up the epoxide that formed to form our diol, and we get two products. So this product over here on the left, let's go ahead and redraw this product over here on the left in a way that's a little bit more familiar. So once again, I put my eye right here. I stare down at this top carbon. If this is the top of my head, this OH is coming out at me. So I would draw that product. I would draw my cyclohexane ring, and at that top carbon I would show the OH coming out at me. And then, of course, at this carbon down here, the OH would be going away from me. So I go ahead and draw my OH as a dash here, down here. And then I do the same thing with this one right here. So if I stare down at that molecule, once again, if I stare down here and look this way, this time at this top carbon-- if my head's right here, top of my head, this OH would be down. So I go ahead and put a dash right here and put my OH. And then, over here, at this carbon it would be going up. So it might be easier to see that these are enantiomers when you look at them drawn like this. A different absolute configuration at both carbons. So this is how to form an epoxide and one way to make a diol. In the next video we will see another way to make a diol, although it will add in a different way to give you a slightly different product." + }, + { + "Q": "who's idea was the declaration of indepence to be made?", + "A": "It was Benjamin Franklin s idea to create a document that declared that the colonies wanted independence and freedom from Great Britain. Thomas Jefferson created the document with the advice of Benjamin Franklin and John Adams.", + "video_name": "er1L9BB6UoE", + "transcript": "Male: I'm here with Walter Isaacson and what are we about to talk about? Walter: We're talking about the Declaration of Independence, which happens, as it says up top, on July 4, 1776, but what we have to remember is that for more than a year, since April 1775 there had been a lot of fighting going on. There was a revolution happening, but up until this point the fighting was mainly against what they considered to be the acts of Parliament and Parliament's ministers and taxing the colonies. Finally, with this document they decide they're going to become ... The American colonies are going to become free and independent, a separate country, which means rebelling against the King himself, George the III. Male: And this is George the III in all of his regality. (laughing) Walter: Right. You know up until they met in Congress with the Continental Congress gathering themselves together they pretended at least to respect George the III, and they were blaming everything on the British Parliament, but it was a pretty difficult thing to decide you were going to overthrow the King himself. Male: And just to be clear, this Continental Congress it's easy now for us in hindsight. It seems like a very official thing, but this was really a rebel congress. It wasn't sanctioned in any way by kind of the formal government, by the government of Great Britain. Walter: Right, and as you see it says the 13 United States of America. This is the first time they really start using the phrase United States of America. They weren't really a country yet. They were 13 different colonies, and not all of them wanted to come to this Continental Congress. Getting them all together was quite difficult. They do so partly to help George Washington's troops get funded because they've started the skirmishes up in Lexington and Concord and Massachusetts, but by 1776 George Washington has a real army and they have to fund him, and eventually they'd figure out well this congress ought to decide are we really having a revolution? Are we trying to break away from the King himself? And the answer here is yes. Male: Wow. And so this is the beginning of the Declaration of Independence and these three fellows on the right-hand side look very familiar. Walter: They're on the committee that the Congress appoints to draft the Declaration. Actually there were five people on the committee, but these of course are the three most important. Thomas Jefferson only 33 years old, by far the youngest person on the committee and he's chosen to write the first draft. Then Benjamin Franklin, who is a mentor of Jefferson's, a printer from Philadelphia. Franklin had just been spending the past two decades almost going back and forth to England to try to prevent a revolution. Then John Adams, the very passionate sage from the State of Massachusetts who was the one who was most in favor of revolution. In fact, when Franklin comes back to Philadelphia in early 1776 after having tried to hold Britain and America together most of these people didn't know whether he would be on the side of revolution or not. In fact, his own son, William Franklin, is, at this time of this Declaration, the Royal Governor of New Jersey and is staying loyal to King George. Male: And just to, once again emphasize the context, the Royal Governor of New Jersey. This wasn't like the Governor of New Jersey He wasn't elected by the people. He was appointed. Walter: No, he was appointed by the King. He was the Royal Governor, and you know Franklin was proud of his son, but they have this incredible split starting in 1776 where William Franklin remains loyal to the crown and loyal to King George the III who had made him Governor of New Jersey. Male: Now, one thing that you had mentioned a few seconds ago that I think is surprising is when you mentioned that Jefferson was 33 years old when he wrote the Declaration of Independence. How normal was that? You know back then it seemed like people did I guess mature faster, but he was still perceived as a fairly young person. Walter: Yes. They all loved to be thinking of themselves as young rebels too. In fact, Jefferson I think is the second youngest person in the Continental Congress, but there was a third person who lied about his age to pretend to be younger and actually wasn't. Jefferson was a good wordsmith. He was from Virginia and it was very important of this person from ... Franklin from Pennsylvania and Adams from Massachusetts to make sure we got Virginia in because Virginia, there was a chance it would remain loyal and ... Male: It was a large wealthy ... Walter: A large wealthy land-owning colony and so getting the Virginians in, and there were very strong rebels from Virginia. The Lees of Virginia as well as Jefferson were in favor of declaring independence so they decide they want to make sure that Jefferson gets to write the first draft. Male: Interesting. And so what we see here, this is the final text. This is the official Declaration of Independence. Male: In the future videos we can talk about previous drafts. Walter: Right. They went through five drafts to get to this draft and this is the one that they do after unaninously all 13 colonies, now called the 13 United States in this document, declare this to be the cause of the colonies. And what you can see in the first paragraph is they have to explain why are we writing this document. They say well if you're going to have a revolution, if you're going to dissolve the political bands which have connected you with another state, then the laws of nature and of nature's God entitle them a decent respect. This is the equal station they get because of nature and nature's God. It's a pretty interesting phrase. Male: Let's read the whole thing. Male: \"When in the course of human events it becomes necessary for one people to dissolve the political bands,\" this is what you were talking about, \"which have connected them with another and to assume among the powers of the earth the separate and equal station which the laws of nature and of nature's God entitle them.\" Walter: What they're saying right there is that the laws of nature and the fact that nature's God created us all equally means that one set of people don't have to be subserviant to or occupied by or colonies of another set of people. They want to be free and independent, and it's interesting that they use the phrase, \"Laws of nature and nature's God,\" because this is sort of the beginning of the enlightenment where we're supposed to understand that nature gives us our rights and reasons. John Locke, the great British philosopher, believed in the laws of nature, and these were deists. They kind of were religious a bit, but they didn't subscribe to any particular religious dogma, and so they just talk about nature's God allowing us all to be free and equal. Male: Right. \"A decent respect to the opinions of mankind requires that they should declare the causes which impel them to a separation.\" So that's really just saying ... Walter: What they're saying is we care what the rest of you think. And by the way, it's directed at one particular people, the French, because we're not going to win this revolution in the United States unless the French help support us. The French, by the way, were already at war with the British. There's been a long set of wars throughout the 18th century where France and Britian were fighting each other. So, this document is particularly aimed at saying to the French, you've been fighting Britain for a long time so we have a decent respect for all the opinions of mankind, but yours in particular and we're going to tell you why we're fighting this and of course in France at the time this notion of liberty, equality, fraternity, that's bubbling up as well. So, the document is to try to persuade other nations please support us. We're explaining why we're having this revolution. Male: And that's I think important to remember for someone in 21st century America. It's obviously a major world power now, the major world power. But back then this was like a little colony. It's kind of a sideshow. Walter: Right, right. And it's important for us to remember now too that whenever we do something, whenever we get involved in the world we should have a decent respect to the opinions of mankind. Walter: That's how we started as a nation saying when we do something we're going to be open. We're going to be honest. We're going to explain to you why." + }, + { + "Q": "I tried to solve this using the following equation:\n60^2 = 50^2 + 20^2 - 2(50)(20)(cos\u00ce\u00b8)\nCan anybody explain why this didn't work? Are the values for a, b, and c in the video the only ones that work in the equation?", + "A": "your equation solves for the angle between the purple and yellow segments, not the angle in the video", + "video_name": "Ei54NnQ0FKs", + "transcript": "Voiceover:Let's say you're studying some type of a little hill or rock formation right over here. And you're able to figure out the dimensions. You know that from this point to this point along the base, straight along level ground, is 60 meters. You know the steeper side, steeper I guess surface or edge of this cliff or whatever you wanna call it, is 20 meters. And then the longer side here, I guess the less steep side, is 50 meters long. So you're able to measure that. But now what you wanna do is use your knowledge of trigonometry, given this information, to figure out how steep is this side. What is the actual inclination relative to level ground? Or another way of thinking about it, what is this angle theta right over there? And I encourage you to pause the video and think about it on your own. Well it might be ringing a bell. Well you know three sides of a triangle and then we want to figure out an angle. And so the thing that jumps out in my head, well maybe the law of cosines could be useful. Let me just write out the law of cosines, before we try to apply it to this triangle right over here. So the law of cosines tells us that C-squared is equal to A-squared, plus B-squared, minus two A B, times the cosine of theta. And just to remind ourselves what the A, B's, and C's are, C is the side that's opposite the angle theta. So if I were to draw an arbitrary triangle right over here. And if this is our angle theta, then this determines that C is that side, and then A and B could be either of these two sides. So A could be that one and B could be that one. Or the other way around. As you can see, A and B essentially have the same role in this formula right over here. This could be B or this could be A. So what we wanna do is somehow relate this angle... We wanna figure out what theta is in our little hill example right over here. So if this is going to be theta, what is C going to be? Well C is going to be this 20 meter side. And then we could set either one of these to be A or B. We could say that this A is 50 meters and B is 60 meters. And now we could just apply the law of cosines. So the law of cosines tells us that 20-squared is equal to A-squared, so that's 50 squared, plus B-squared, plus 60 squared, minus two times A B. So minus two times 50, times 60, times 60, times the cosine of theta. This works out well for us because they've given us everything. There's really only one unknown. There's theta here. So let's see if we can solve for theta. So 20 squared, that is 400. 50 squared is 2,500. 60 squared is 3,600. And then 50 times... Let's see, two times 50 is 100, times 60, this is all equal to 6,000. So let's see, if we simplify this a little bit we're going to get 400 is equal to 2,500 plus 3,600. Let's see, that'd be 6,100. That's equal to 6,000... Let me do this in a new color. So when I add these two, I get 6,100. Did I do that right? So it's 2,000 plus 3,000, plus 5,000. 500 plus 600 is 1,100. So I get 6,100 minus 6,000, times the cosine of theta. And let's see, now we can subtract 6,100 from both sides. So I'm just gonna subtract 6,100 from both sides so that I get closer to isolating the theta. So let's do that. So this is going to be negative 5,700. Is that right? 5,700 plus... Yes, that is right. Right, because if this was the other way around, if this was 6,100 minus 400 it would be positive 5,700. Alright. And then these two of course cancel out. And this is going to be equal to negative 6,000 times the cosine of theta. Now we can divide both sides by negative 6,000. And we get... I'm just gonna swap the sides. We get cosine of theta is equal to... Let's see we could divide the numerator and the denominator by essentially negative 100. So these are both going to become positive. So cosine of theta is equal to 57 over 60. And actually that can be simplified even more. Three goes into 57, is that 19 times? Yep, so this is actually... This could be simplified. This is equal to 19 over 20. We actually didn't have to do that simplification step because we're about to use our calculators, but that makes the math a little more tractable. Right, 3 goes into 57, yeah, 19 times. And so now we can take the inverse cosine of both sides. So we could get theta is equal to the inverse cosine, or the arc cosine, of 19 over 20. So let's get our calculator out and see if we get something that makes sense. So we wanna do the inverse cosine of 19 over 20. And we deserve a drum roll. We get 18.19 degrees. And I already verified that my calculator is in degree mode. So it gets 18.19 degrees. So if we wanted to round, this is approximately equal to 18.2 degrees, if we wanna round to the nearest tenth. So that essentially gives us a sense of how steep this slope actually is." + }, + { + "Q": "During WWII Was china still in the civil war with it self or has it stopped when WWII happened?", + "A": "The Chinese civil war began in 1927 and was both before and after World War II. It wasn t considered over until 1950. Aggression between Japan and China was an ongoing series of incidents that became a total war in 1937 and ended with the surrender of Japan to the Allies in 1945.", + "video_name": "-kKCjwNvNkQ", + "transcript": "World War II was the largest conflict in all of human history. The largest and bloodiest conflict And so you can imagine it is quite complex My goal in this video is to start giving us a survey, an overview of the war. And I won't even be able to cover it all in this video. It is really just a think about how did things get started. Or what happened in the lead up? And to start I am actually going to focus on Asia and the Pacific. Which probably doesn't get enough attention when we look at it from a western point of view But if we go back even to the early 1900s. Japan is becoming more and more militaristic. More and more nationalistic. In the early 1900s it had already occupied... It had already occupied Korea as of 1910. and in 1931 it invades Manchuria. It invades Manchuria. So this right over here, this is in 1931. And it installs a puppet state, the puppet state of Manchukuo. And when we call something a puppet state, it means that there is a government there. And they kind of pretend to be in charge. But they're really controlled like a puppet by someone else. And in this case it is the Empire of Japan. And we do remember what is happening in China in the 1930s. China is embroiled in a civil war. So there is a civil war going on in China. And that civil war is between the Nationalists, the Kuomintang and the Communists versus the Communists The Communists led by Mao Zedong. The Kuomintang led by general Chiang Kai-shek. And so they're in the midst of the civil war. So you can imagine Imperial Japan is taking advantage of this to take more and more control over parts of China And that continues through the 30s until we get to 1937. And in 1937 the Japanese use some pretext with, you know, kind of a false flag, kind of... well, I won't go into the depths of what started it kind of this Marco-Polo Bridge Incident But it uses that as justifications to kind of have an all-out war with China so 1937...you have all-out war and this is often referred to as the Second Sino-Japanese War ...Sino-Japanese War Many historians actually would even consider this the beginning of World War II. While, some of them say, ok this is the beginning of the Asian Theater of World War II of the all-out war between Japan and China, but it isn't until Germany invades Poland in 1939 that you truly have the formal beginning, so to speak, of World War II. Regardless of whether you consider this the formal beginning or not, the Second Sino-Japanese War, and it's called the second because there was another Sino-Japanese War in the late 1800s that was called the First Sino-Japanese War, this is incredibly, incredibly brutal and incredibly bloody a lot of civilians affected we could do a whole series of videos just on that But at this point it does become all-out war and this causes the civil war to take a back seat to fighting off the aggressor of Japan in 1937. So that lays a foundation for what's happening in The Pacific, in the run-up to World War II. And now let's also remind ourselves what's happening... what's happening in Europe. As we go through the 1930s Hitler's Germany, the Nazi Party, is getting more and more militaristic. So this is Nazi Germany... Nazi Germany right over here. They're allied with Benito Mussolini's Italy. They're both extremely nationalistic; they both do not like the Communists, at all You might remember, that in 1938... 1938, you have the Anschluss, which I'm sure I'm mispronouncing, and you also have the takeover of the Sudetenland in Czechoslovakia. So the Anschluss was the unification with Austria and then you have the Germans taking over the of Sudetenland in Czechoslovakia and this is kind of the famous, you know, the rest of the, what will be called the Allied Powers kind of say, \"Okay, yeah, okay maybe Hitler's just going to just do that... well we don't want to start another war. We still all remember World War I; it was really horrible. And so they kind of appease Hitler and he's able to, kind of, satisfy his aggression. so in 1938 you have Austria, Austria and the Sudetenland ...and the Sudetenland... are taken over, are taken over by Germany and then as you go into 1939, as you go into 1939 in March they're able to take over all of Czechoslovakia they're able to take over all of Czechoslovakia and once again the Allies are kind of, they're feeling very uncomfortable, they kind of, have seen something like this before they would like to push back, but they still are, kind of, are not feeling good about starting another World War so they're hoping that maybe Germany stops there. So let me write this down... So all of Czechoslovakia... ...Czechoslovakia... is taken over by the Germans. This is in March of 1939. And then in August you have the Germans, and this is really in preparation for, what you could guess is about to happen, for the all-out war that's about to happen the Germans don't want to fight the Soviets right out the gate, as we will see, and as you might know, they do eventually take on the Soviet Union, but in 1939 they get into a pact with the Soviet Union. And so this is, they sign the Molotov\u2013Ribbentrop Pact with the Soviet Union, this is in August, which is essentially mutual non-aggression \"Hey, you know, you do what you need to do, we know what we need to do.\" and they secretly started saying \"Okay were gonna, all the countries out here, we're going to create these spheres of influence where Germany can take, uh, control of part of it and the Soviet Union, and Stalin is in charge of the Soviet Union at this point, can take over other parts of it. And then that leads us to the formal start where in September, let me write this in a different color... so September of 1939, on September 1st, Germany invades Poland Germany invades Poland on September 1st, which is generally considered the beginning of World War II. and then you have the Great Britain and France declares war on Germany so let me write this World War II... starts everyone is declaring war on each other, Germany invades Poland, Great Britain and France declare war on Germany, and you have to remember at this point Stalin isn't so concerned about Hitler he's just signed the Molotov-Ribbentrop Pact and so in mid-September, Stalin himself invades Poland as well so they both can kind of carve out... ...their spheres of influence... so you can definitely sense that things are not looking good for the world at this point you already have Asia in the Second Sino-Japanese War, incredibly bloody war, and now you have kind of, a lot of very similar actors that you had in World War I and then they're starting to get into a fairly extensive engagement." + }, + { + "Q": "what are radicals and free radicals?", + "A": "Radicals and free radicals are the same thing. In organic chemistry, they generally refer to atoms or molecules with an unpaired electron.", + "video_name": "nQ7QSV4JRSs", + "transcript": "We've already seen alcohols in many of these videos, but I thought it was about time that I actually made a video on alcohols. Now, alcohols is the general term for any molecule that fits the pattern some type of functional group or chain of carbons OH. And they use the letter R. And I've used it before. R stands for radical. And I don't want you to confuse this R with free radical. It means completely different things. R in this form really just means a functional group or a chain of carbons here. It doesn't mean a free radical. This just means it could be just something attached to this OH right there. Now another point of clarification, do not think that anything that fits this pattern is drinkable. Do not associate it with the traditional alcohol that you may or may not have been exposed to. Traditional drinking alcohol is actually ethanol. Alcohol is actually-- let me write out the molecular formula. CH3, CH2, and then OH. This is what is inside of wine and beer and hard liquor, or whatever you might want. You do not want to drink and maybe you might not actually want to drink this either, but you definitely do not want to drink something like methanol. It might kill you. So you do not want to do something like this. You do not want to ingest that. Might kill or blind you. This might do it in a more indirect way. So I want to get that out of the way and just so that we get kind of a little bit more comfortable with alcohols, and we've seen them involved in other reactions. We've seen hydroxides act as nucleophiles and Sn2 substitution reactions create alcohols. But I want to do is just learn to get comfortable and really make sure we know how to name these things. So let's just name these molecules that I drew right before I pressed record right over here. So over here, like everything else, we always want to define the longest carbon chain. We have 1, 2, 3, 4, 5 carbons. So it's going to be pent. And there's no double bonds. So I'll just write pentane right then. And we're not going to just write a pentane because actually, the fact that makes it an alcohol, that takes precedence over the fact that it is an alkane. So it actually, the suffix of the word will involve the alcohol part. So it is pentanol. That tells us that's an alcohol. And to know where the OH is grouped, we'll start numbering closest to the OH. So 1, 2, 3, 4, 5. Sometimes it'll be called 2-pentanol. And this is pretty clear because we only have one group here, only one OH. So we know that that is what the 2 applies to. But a lot of times, if people want to be a little bit more particular, they might write pentan-2-ol. And this way is more useful, especially if you have multiple functional groups. So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try the name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. We want to call it an octyne. But because we have an alcohol there, we want to call this an octyne-- let me make it very clear. So oct tells us that we have 8 carbons. Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that So it is oct-5-yn. That tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-al. And now we have these two bromo groups here on the 7 carbon. So it's 7,7-dibromo oct-5-yn-4-al. And this would all be one word. Let me make sure that you realize that I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex. And they're all single bonds, so it's a hexane. It's a cyclohexane. But then of course, the hydroxide or the hydroxy group I should call it, takes dominance. It's a hexanol. So this is a cyclohexanol. And once again, that comes from the OH right there. And you don't have to number it. Because no matter what carbon it's on, it's on the same one. If you had more than one of these OH groups, then we would have to worry about numbering them. Let's just do this one right over here. So once again, what is our carbon chain? We have 1, 2, 3 carbons. And we have the hydroxy group attached to the 1 and the 3 carbon. Prop is our prefix. It is an alkane. So we would call this-- and there's a couple of ways to do this. We could call this 1 comma 3 propanediol. Actually, I don't have to put a dash their. Propanediol. And over here, we would add the E because we have the D right there. So it's propanediol. If it wasn't diol, it would be propanal. You wouldn't have the E, D and the I there. So this would specify we're at the 1 and the 3 carbons. We have the hydroxy group. Or this could also be written as propane- 1, 3- diol. And once again, the di is telling us that we have two of the hydroxy groups attached to this thing. But either of these things are ways that you would see this molecule named." + }, + { + "Q": "What about Canada?", + "A": "This video focuses on US Native American history because this is US history. This does sometimes cut off important knowledge about First Nations in Canada (regional term for aboriginal peoples), since many tribes moved across what we now think of as borders for those two countries. Because the two countries and colonizing forces had different approaches to treaties, aggression and settlement, other than acknowledging a connection, the history relating to the two countries diverges in many areas.", + "video_name": "o2XjXFvruIM", + "transcript": "- It is believed that the first humans settled North and South America, or began to settle it, about 15 to 16,000 years ago and the mainstream theory is that they came across from northeast Asia, across the Bering Strait, during the last glaciation period, when sea levels were lower and there was a land bridge, the famous Bering Land Bridge connecting the two continents and we have archeological evidence of humans in southern Chile as early as 14,500 years ago and as well in Florida as early as 14,500 years ago. So humans had migrated into, and settled in the Americas many, many, many thousands of years ago. And like other places in the world, they followed similar development patterns. The first evidence we have of the development of agriculture in the Americas is about seven to eight or nine thousand years ago so once again, it coincides with when agriculture, we believe, started to emerge in other parts of the world. And the more archeological evidence we find, we'll probably find dates that go even further back than that, in fact, I've seen some that go eight, nine thousand years ago. Now one misconception, significant misconception, about the Americas is that when the Europeans colonized, remember Columbus comes sailing in 1492, looking for the East Indies and then he bumps into this, he actually doesn't bump into the whole continent, he bumps into an island that's close to the continent, but with that you start having the beginning of the European colonization of the Americas, roughly the last 500 years, and one misconception that folks often have is, well it was maybe sparsely populated, mainly by hunter-gatherer nomadic people and nothing could be further from the truth. The modern estimates of the population of the Americas at the time of the European colonization, roughly around 1500 is 50-100 million people and to put that in perspective, so that's right around there, that's about 10 to 20% of the world population at that time. The world population at that time was about 500 million people and given that the Americas is about one third of the land, if you don't count Antarctica, it's not that different of a population density than the other continents, and we have significant cities that were in place in the pre-Columbian era, in the era before Columbus and the European colonization. For example, you might have heard of the Aztecs, this really, the core, the Mexica people, the Mexica tribe, in many ways the foundations, of the Mexican people pre-European colonization. You might also be familiar with the Mayan civilization, one of the longest lasting civilizations in, actually, in history, they're famous for one of the earliest cultures where we have the hieroglyphics, where we have writing. You're probably familiar with the Inca Empire and yes, that is me on a recent trip and at the time of the Inca Empire it is believed that it was possibly the largest empire on the earth at that time, incredibly complex structures and social structures, they had. Now what's often less talked about are things like the Mississippian culture, which was in North America right over here. The Mississippi River is named for them. This if their famous city of Cahokia near St. Louis and in there, in that peak, it would have 40,000 people in it. Around the world at that time, at the time of the Mississippian culture, there weren't many cities in the world that had 40,000 people, so it wasn't these, just hunter-gatherers and people who were nomadic, there were sophisticated civilizations, with sophisticated cultures and dense population centers and it had also been in place for a long time, similar, in timeframe, to some of the great ancient civilizations that we see in Mesopotamia, the Indus Valley, and in China. For example, the oldest civilization we know of, in Mesoamerica, is the Olmec civilization, right over here, here's a few of their artifacts they have. If we go into the Andes, near modern day Peru, we have the Chavin culture, right over there. As you can see, a lot of these cultures, at least the ones that I'm putting here, and this is just a sample, I'm sampling some around North America, some around Mesoamerica, and some in the Andes, and then you can even go further back and you can go to the Caral civilization, and what's really interesting about the Caral civilization is some archeologists call this the first civilization and it's unclear whether they had, whether they farmed grains and cereals that we often associate with civilizations, they use their surplus crops to have a more specialized labor force, but they were a maritime culture, even today the coast of Peru is a significant source of all of the, or a good chunk of the seafood in the world but a significant culture developed there, these are the remnants of their pyramids, and they developed, we believe, in the 4th millennium BCE, so this is around the same time as when Egypt first got unified around Menes, by Menes, or you have the first Sumerians in Mesopotamia and as far back as them you have these Quipus knots, which many archeologists view as a form of writing, it was a form of record-keeping and it was even used later on by the Incas. So the big take away here, is to challenge that misconception that the Americas somehow were not as, has populations and civilizations like everything else, it was only when the Europeans came in that all of that started to happen, no. Well before the Europeans came in, North and South America had been settled, agriculture developed at a similar timescale, significant, complex civilizations, writing developed on a similar timescale, but once you have the European colonization, some people say it was intentional, it was probably a combination of intentional and just diseases that were unfamiliar to the people here, within 150 years, that 50 to 100 million population, so now we're talking about, roughly by 1650, so you move a little bit forward in time, the population had gone to roughly six million people, some people refer it to a genocide, some people would say it's a combination of an intentional extermination of people plus just inadvertent disease, whatever it is, this was the significant decline of a complex and diverse set of populations. This is just a small sample of the major civilizations that were there, you had thousands of tribes across North and South America that had different cultures, different languages, different traditions and different religions." + }, + { + "Q": "Why can't 9.564 round up to 9.600 instead of 9.6? Doesn't 564 round up to 600?\nI'm confused!", + "A": "It can round up even if it has those 0s, but the 0s don t add anything since they are place holders.", + "video_name": "_MIn3zFkEcc", + "transcript": "0.710 Round 9.564, or nine and five hundred sixty-four Thousandths, to the nearest tenth. So lie me write it a bit larger, 9.564 And we need to round to the nearest tenth. So what's the tenth place? The tenths place is right here This represents 5 tenths. This is the ones place, this is the tenths place, this is The hundredths place, and this is the Thousandths place right here So we need to round to the nearest tenth. So if we round up, this will be 9.6 If we round down, this will be 9.5 And just like regular rounding, when we're not Dealing with decimals, you move to one spot, or you look At one place to the right or one place lower, I guess, and You say is that 5 larger If it is, you round if it isn't,you round down 6 is definitely 5 or larger, so we want to round up. So this 9.564 becomes 9.6, or we can call this Nine and sixth tenths. And then we're done!" + }, + { + "Q": "Many times in this video, Sal says the word \"annex,\" or \"annexed.\" What does that mean?", + "A": "annex - to colonize or add on If a country annexes another that is next to it, it adds that country to itself.", + "video_name": "heKuwogLwnk", + "transcript": "We're now ready to talk about one of the most famous events in all of world history that really was the trigger for World War I, or the Great War, as it was called back then. So just as a little bit of backdrop, in 1908, the Austro-Hungarian Empire formally annexes Bosnia and Herzegovina. It had already been occupying it since the late 1800s, since the Ottomans were being pushed out. But then in 1908, it formally annexes it. And just as a little bit more backdrop, as the Ottomans were being pushed out of the Balkans, it helped rekindle or bring about more hope of unifying the Yugoslavic people, the southern Slavic people. When people talk about Yugoslav, they're literally talking about the southern Slavs. So that literally means southern. So you had these nationalistic hopes. But now in 1908, it was already being occupied. A significant state, that would be part of a potential future Yugoslav, was now formally annexed by the Austro-Hungarians. Now, you also had an independent kingdom of Serbia right here. And you can imagine that this was the home base of the nationalistic movement. If only they could add the other southern Slavic states to this, it could one day turn into a greater Yugoslavia. So in that context, we get to 1914. So let me draw a little line here. So we're getting to 1914. June 28, which is one of the most famous dates in all of history. And you have the Archduke Franz Ferdinand and his wife, Sophie. They're visiting Sarajevo which is now in annexed Bosnia. And when they are there, there is a ploy. There is a scheme to assassinate them, from a group-- they're called the Young Bosnians. They have ties to the Black Hand, which is this nationalistic group. That has ties, many, many people say-- all these things are all very shady and behind the back, behind the scenes. But it has ties to elements in the kingdom of Serbia. They attempt to assassinate Archduke Franz Ferdinand. And it's actually a fascinating story because the initial assassination attempt is completely, completely botched. There's even one case of a guy, one of the guys who tried to be an assassin when it gets botched, he tries to bite on a cyanide capsule and then jump into a river. The cyanide capsule had gone bad. The river was only 10 inches deep. And so they were able to get their hands on him. And one of the conspirators, Gavrilo Princip-- at this point, once the whole thing was botched, he gives up on the whole assassination attempt. And he's having, literally, a sandwich at a cafe in Sarajevo, thinking about how botched their whole attempt was. And while that was happening, a mistake on the part of those planning Archduke Franz Ferdinand's route as he was traveling within Sarajevo has them driving right near Gavrilo Princip. So he sees, all of a sudden, that they've taken the wrong route, that they're driving right by him again. Remember, his people already knew that there was an assassination attempt on him earlier in the day. So they should have been more careful. Now, Gavrilo Princip gets up, puts his sandwich down, and starts walking over to where he sees Archduke Franz Ferdinand and Sophie's car going. Now, the drivers, once they realized that they had made a mistake, they had taken a less safe route. They tried to back up, which makes things even worse because then the car starts stalling. And Gavrilo Princip literally walks up to the car and is able to shoot Archduke Franz Ferdinand and Sophie. And just to give you a sense of how important this is, Archduke Franz Ferdinand of Austria is the heir. He's the nephew of Franz Josef, who was the ruler of Austria-Hungary. And so he is the heir to the empire. And so he gets assassinated by Gavrilo Princip. So Franz Ferdinand assassinated by Gavrilo Princip. And we have right over here a picture right after Gavrilo Princip-- I believe this is Gavrilo Princip right over here, right after he was arrested. And just to get a little sense of how this was tied to this whole Yugoslavian nationalistic movement. This is what he said once he was arrested. \"I am a Yugoslav nationalist, aiming for the unification of all Yugoslavs, and I do not care what form of state, but it must be free of Austria.\" So this act, this assassination motivated by a nationalistic movement, motivated by a desire to maybe merge Bosnia and Herzegovina with Serbia and maybe eventually Croatia, with Bosnia and Herzegovina and Serbia. This assassination, as we'll see in the next video, is the trigger for all of World War I. And the reason why it triggers it is because, well, there's many things you can cite. You could argue that many of the empires in Europe were already militarizing, already had a desire for conflict. But then you also had all of these alliances that essentially allowed the dominoes to fall in all of Europe. And because they had these empires, essentially much of the world to be at war with each other." + }, + { + "Q": "Why do we need proofs in the first place, and why do they have to be in a certain format?", + "A": "Without proofs there is no way to know if these theories are valid or applicable. Proofs are like a mathematical way to show that things work 100% or don t work at all. Proofs are a pain in the butt, but they are an essential way of knowing what is real. Proofs have to be in a certain format, because proofs are a formal way to display mathematical knowledge in a clear and concise format.", + "video_name": "rcBaqkGp7CA", + "transcript": "In case you haven't noticed, I've gotten somewhat obsessed with doing as many proofs of the Pythagorean theorem as I can do. So let's do one more. And like how all of these proofs start, let's construct ourselves a right triangle. So I'm going to construct it so that its hypotenuse sits on the bottom. So that's the hypotenuse of my right triangle. Try to draw it as big as possible, so that we have space to work with. So that's going to be my hypotenuse. And then let's say that this is the longer side that's not the hypotenuse. We can have two sides that are equal. But I'll just draw it so that it looks a little bit longer. Let's call that side length a. And then let's draw this side right over here. It has to be a right triangle. So maybe it goes right over there. That's side of length b. Let me extend the length a a little bit. So it definitely looks like a right triangle. And this is our 90-degree angle. So the first thing that I'm going to do is take this triangle and then rotate it counterclockwise by 90 degrees. So if I rotate it counterclockwise by 90 degrees, I'm literally just going to rotate it like that and draw another completely congruent version of this one. So I'm going to rotate it by 90 degrees. And if I did that, the hypotenuse would then sit straight up. So I'm going to do my best attempt to draw it almost to scale as much as I can eyeball it. This side of length a will now look something like this. It'll actually be parallel to this over here. So let me see how well I can draw it. So this is the side of length a. And if we cared, this would be 90 degrees. The rotation between the corresponding sides are just going to be 90 degrees in every case. That's going to be 90 degrees. That's going to be 90 degrees. Now, let me draw side b. So it's going to look something like that or the side that's length b. And this and the right angle is now here. So all I did is I rotated this by 90 degrees counterclockwise. Now, what I want to do is construct a parallelogram. I'm going to construct a parallelogram by essentially-- and let me label. So this is height c right over here. Let me do that white color. This is height c. Now, what I want to do is go from this point and go up c as well. Now, so this is height c as well. And what is this length? What is the length over here from this point to this point going to be? Well, a little clue is this is a parallelogram. This line right over here is going to be parallel to this line. It's maintained the same distance. And since it's traveling the same distance in the x direction or in the horizontal direction and the vertical direction, this is going to be the same length. So this is going to be of length a. Now, the next question I have for you is, what is the area of this parallelogram that I have just constructed? Well, to think about that, let's redraw this part of the diagram so that the parallelogram is sitting on the ground. So this is length a. This is length c. This is length c. And if you look at this part right over here, it gives you a clue. I'll use this green color. The height of the parallelogram is given right over here. This side is perpendicular to the base. So the height of the parallelogram is a as well. So what's the area? Well, the area of a parallelogram is just the base times the height. So the area of this parallelogram right over here is going to be a squared. Now, let's do the same thing. But let's rotate our original right triangle. Let's rotate it the other way. So let's rotate it 90 degrees clockwise. And this time, instead of pivoting on this point, we're going to pivot on that point right over there. So what are we going to get? So the side of length c if we rotate it like that, it's going to end up right over here. I'll try to draw it as close to scale as possible. So that side has length c. Now, the side of length of b is going to pop out and look something like this. It's going to be parallel to that. This is going to be a right angle. So let me draw it like that. That looks pretty good. And then the side of length a is going to be out here. So that's a. And then this right over here is b. And I wanted to do that b in blue. So let me do the b in blue. And then this right angle once we've rotated is just sitting right over here. Now, let's do the same exercise. Let's construct a parallelogram right over here. So this is height c. This is height c as well. So by the same logic we used over here, if this length is b, this length is b as well. These are parallel lines. We're going the same distance in the horizontal direction. We're rising the same in the vertical direction. We know that because they're parallel. So this is length b down here. This is length b up there. Now, what is the area of this parallelogram right over there? What is the area of that parallelogram going to be? Well, once again to help us visualize it, we can draw it sitting flat. So this is that side. Then you have another side right over here. They both have length b. And you have the sides of length c. So that's c. That's c. What is its height? Well, you see it right over here. Its height is length b as well. It gives us right there. We know that this is 90 degrees. We did a 90-degree rotation. So this is how we constructed the thing. So given that, the area of a parallelogram is just the base times the height. The area of this parallelogram is b squared. So now, things are starting to get interesting. And what I'm going to do is I'm going to copy and paste this part right over here, because this is, in my mind, the most interesting part of our diagram. Let me see how well I can select it. So let me select this part right over here. So let me copy. And then I'm going to scroll down. And then let me paste it. So this diagram that we've constructed right over here, it's pretty clear what the area of it is, the combined diagram. And let me delete a few parts of it. I want to do that in black so that it cleans it up. So let me clean this thing up, so we really get the part that we want to focus on. So cleaning that up and cleaning this up, cleaning this up right over there. And actually, let me delete this right down here as well, although we know that this length was c. And actually, I'll draw it right over here. This was from our original construction. We know that this length is c. We know this height is c. We know this down here is c. But my question for you is, what is the area of this combined shape? Well, it's just a squared plus b squared. Let me write that down. The area is just a squared plus b squared, the area of those two parallelograms. Now, how can we maybe rearrange pieces of this shape so that we can express it in terms of c? Well, it might have jumped out at you when I drew this line right over here. I want to do that in white. We know that this part right over here is of length c. This comes from our original construction. I lost my diagram. This is of length c. That's of length c. And then this right over here is of length c. And so what we could do is take this top right triangle, which is completely congruent to our original right triangle, and shift it down. So remember, the entire area, including this top right triangle, is a squared plus b squared. But we're excluding this part down here, which was our original triangle. But what happens if we take that? So let me actually cut. And then let me paste it. And all I'm doing is I'm moving that triangle down here. So now, it looks like this. So I've just rearranged the area that was a squared b squared. So this entire area of this entire square is still a squared plus b squared. a squared is this entire area right over here. It was before a parallelogram. I just shifted that top part of the parallelogram down. b squared is this entire area right over here. Well, what's this going to be in terms of c? Well, we know that this entire thing is a c by c square. So the area in terms of c is just c squared. So a squared plus b squared is equal to c squared. And we have, once again, proven the Pythagorean theorem." + }, + { + "Q": "Unfortunately, he did the right hand rule incorrectly. The fact that the dot with the circle means magnetic field facing \"out\" (which means our fingers point outward) and the velocity, according to what he has down, is to the right (which means we'd have our thumb facing to the right if we were looking at our own hand). That leaves us with a palm facing upward, which means our force is upward. Just remember the following: thumb is direction of 'velocity' or 'charge', fingers are mag. field...", + "A": "You sir, are incorrect. He used it correctly. Right hand rule for force on an electron. Left hand rule for force on a conventional current.", + "video_name": "LTuGQy4rmmo", + "transcript": "In the last video we learned-- or at least I showed you, I don't know if you've learned it yet, but we'll learn it in But we learned that the force on a moving charge from a magnetic field, and it's a vector quantity, is equal to the charge-- on the moving charge-- times the cross product of the velocity of the charge and the magnetic field. And we use this to show you that the units of a magnetic field-- this is not a beta, it's a B-- but the units of a magnetic field are the tesla-- which is abbreviated with a capital T-- and that is equal to newton seconds per coulomb meters. So let's see if we can apply that to an actual problem. So let's say that I have a magnetic field, and let's say it's popping out of the screen. I'm making this up on the fly, so I hope the numbers turn out. It's inspired by a problem that I read in Barron's AP calculus book. So if I want to draw a bunch of vectors or a vector field that's popping out of the screen, I could just do the top of the arrowheads. I'll draw them in magenta. So let's say I have a vector field. So you can imagine a bunch of arrows popping out of the screen. I'll just draw a couple of them just so you get the sense that it's a field. It pervades the space. These are a bunch of arrows popping out. And the field is popping out. And the magnitude of the field, let's say it is, I don't know, let's say it is 0.5 teslas. Let's say I have some proton that comes speeding along. And it's speeding along at a velocity-- so the velocity of the proton is equal to 6 times 10 to the seventh meters per second. And that is actually about 1/5 of the velocity or 1/5 of the speed of light. So we're pretty much in the relativistic realm, but we won't go too much into relativity because then the mass of the proton increases, et cetera, et cetera. We just assume that the mass hasn't increased significantly at this point. So we have this proton going at a 1/5 of the speed of light and it's crossing through this magnetic field. So the first question is what is the magnitude and direction of the force on this proton from this magnetic field? Well, let's figure out the magnitude first. So how can we figure out the magnitude? Well, first of all, what is the charge on a proton? Well, we don't know it right now, but my calculator has And if you have a TI graphing calculator, your calculator would also have it stored in it. So let's just write that down as a variable right now. So the magnitude of the force on the particle is going to be equal to the charge of a proton-- I'll call it Q sub p-- times the magnitude of the velocity, 6 times 10 to the seventh meters per second. We're using all the right units. If this was centimeters we'd probably want to convert it to meters. 6 times 10 to the seventh meters per second. And then times the magnitude of the magnetic field, which is 0.5 soon. teslas-- I didn't have to write the units there, but I'll do it there-- times sine of the angle between them. I'll write that down right now. But let me ask you a question. If the magnetic field is pointing straight out of the screen-- and you're going to have to do a little bit of three-dimensional visualization now-- and this particle is moving in the plane of the field, what is the angle between them? If you visualize it in three dimensions, they're actually orthogonal to each other. They're at right angles to each other. Because these vectors are popping out of the screen. They are perpendicular to the plane that defines the screen, while this proton is moving within this plane. So the angle between them, if you can visualize it in three dimensions, is 90 degrees. Or they're perfectly perpendicular. And when things are perfectly perpendicular, what is the sine of 90 degrees? Or the sine of pi over 2? Either way, if you want to deal in radians. Well, it's just equal to 1. The whole-- hopefully-- intuition you got about the cross product is we only want to multiply the components of the two vectors that are perpendicular to each other. And that's why we have the sine of theta. But if the entire vectors are perpendicular to each other, then we just multiply the magnitude of the vector. Or if you even forget to do that, you say, oh well, they're perpendicular. They're at 90 degree angles. Sine of 90 degrees? Well, that's just 1. So this is just 1. So the magnitude of the force is actually pretty easy to calculate, if we know the charge on a proton. And let's see if we can figure out the charge on a proton. Let me get the trusty TI-85 out. Let me clear there, just so you can appreciate the TI-85 store. If you press second and constant-- that's second and then the number 4. They have a little constant above it. You get their constant functions. Or their values. And you say the built-in-- I care about the built-in functions, so let me press F1. And they have a bunch of-- you know, this is Avogadro's number, they have a bunch of interesting-- this is the charge of an electron. Which is actually the same thing as the charge of a proton. So let's use that. Electrons-- just remember-- electrons and protons have offsetting charges. One's positive and one's negative. It's just that a proton is more massive. And of course, it's positive. Let's just confirm that that's the charge of an electron. But that's also the charge of a proton. And actually, this positive value is the exact charge of a proton. They should have maybe put a negative number here, but all we care about is the value. So let's use that again. The charge of an electron-- and it is positive, so that's the same thing as the charge for a proton-- times 6 times 10 to the seventh-- 6 E 7, you just press that EE button on your calculator-- times 0.5 teslas. Make sure all your units are in teslas, meters, and coulombs, and then your result will be in newtons. And you get 4.8 times 10 to the negative 12 newtons. Let me write that down. So the magnitude of this force is equal to 4.8 times 10 to the minus 12 newtons. So that's the magnitude. Now what is the direction? What is the direction of this force? Well, this you is where we break out-- we put our pens down if we're right handed, and we use our right hand rule to figure out the direction. So what do we have to do? So when you take something crossed something, the first thing in the cross product is your index finger on your right hand. And then the second thing is your middle finger pointed at a right angle with your index finger. Let's see if I can do this. So I want my index finger on my right hand to point to the right. But I want my middle finger to point upwards. Let me see if I can pull that off. So my right hand is going to look something like this. And my hand is brown. So my right hand is going to look something like this. My index finger is pointing in the direction of the velocity vector, while my middle finger is pointing the direction of the magnetic field. So my index finger is going to point straight up, so all you see is the tip of it. And then my other fingers are just going to go like that. And then my thumb is going to do what? My thumb is going-- this is the heel of my thumb-- and so my thumb is going to be at a right angle to both of them. So my thumb points down like this. This is often the hardest part. Just making sure you get your hand visualization right with the cross product. So just as a review, this is the direction of v. This is the direction of the magnetic field. It's popping out. And so if I arrange my right hand like that, my thumb points down. So this is the direction of the force. So as this particle moves to the right with some velocity, there's actually going to be a downward force. Downward on this plane. So the force is going to move in this direction. So what's going to happen? Well, what happens-- if you remember a little bit about your circular motion and your centripetal acceleration and all that-- what happens when you have a force perpendicular to velocity? If you have a force here and the velocity is like that, if the particles-- it'll be deflected a little bit to the right. And then since the force is always going to be perpendicular to the velocity vector, the force is going to charge like that. So the particle is actually going to go in a circle. As long as it's in the magnetic field, the force applied to the particle by the magnetic field is going to be perpendicular to the velocity of the particle. So the velocity of the particle-- so it's going to actually be like a centripetal force on the particle. So the particle is going to go into a circle. And in the next video we'll actually figure out the radius of that circle. And just one thing I want to let you think about. It's kind of weird or spooky to me that the force on a moving particle-- it doesn't matter about the particle's mass. It just matters the particle's velocity and charge. So it's kind of a strange phenomenon that the faster you move through a magnetic field-- or at least if you're charged, if you're a charged particle-- the faster you move through a magnetic field, the more force that magnetic field is going to apply to you. It seems a little bit, you know, how does that magnetic field know how fast you're moving? But anyway, I'll leave you with that. In the next video we'll explore this magnetic phenomenon a little bit deeper. See" + }, + { + "Q": "what is an intial value ?", + "A": "In the context of exponential functions, it means the function evaluated at 0.", + "video_name": "G2WybA4Hf7Y", + "transcript": "- [Voiceover] So let's think about a function. I'll just give an example. Let's say, h of n is equal to one-fourth times two to the n. So, first of all, you might notice something interesting here. We have the variable, the input into our function. It's in the exponent. And a function like this is called an exponential function. So this is an exponential. Ex-po-nen-tial. Exponential function, and that's because the variable, the input into our function, is sitting in its definition of what is the output of that function going to be. The input is in the exponent. I could write another exponential function. I could write, f of, let's say the input is a variable, t, is equal to is equal to five times times three to the t. Once again, this is an exponential function. Now there's a couple of interesting things to think about in exponential function. In fact, we'll explore many of them, but I'll get a little used to the terminology, so one thing that you might see is a notion of an initial value. In-i-tial Intitial value. And this is essentially the value of the function when the input is zero. So, for in these cases, the initial value for the function, h, is going to be, h of zero. And when we evaluate that, that's going to be one-fourth times two to the zero. Well, two to the zero power, is just one. So it's equal to one-fourth. So the initial value, at least in this case, it seems to just be that number that sits out here. We have the initial value times some number to this exponent. And we'll come up with the name for this number. Well let's see if this was true over here for, f of t. So, if we look at its intial value, f of zero is going to be five times three to the zero power and, the same thing again. Three to the zero is just one. Five times one is just five. So the initial value is once again, that. So if you have exponential functions of this form, it makes sense. Your initial value, well if you put a zero in for the exponent, then the number raised to the exponent is just going to be one, and you're just going to be left with that thing that you're multiplying by that. Hopefully that makes sense, but since you're looking at it, hopefully it does make a little bit. Now, you might be saying, well what do we call this number? What do we call that number there? Or that number there? And that's called the common ratio. The common common ratio. And in my brain, we say well why is it called a common ratio? Well, if you thought about integer inputs into this, especially sequential integer inputs into it, you would see a pattern. For example, h of, let me do this in that green color, h of zero is equal to, we already established one-fourth. Now, what is h of one going to be equal to? It's going to be one-fourth times two to the first power. So it's going to be one-fourth two. What is h of two going to be equal to? Well, it's going to be one-fourth times two squared, so it's going to be times two times two. Or, we could just view this as this is going to be two times h of one. And actually I should have done this when I wrote this one out, but this we can write as two times h of zero. So notice, if we were to take the ratio between h of two and h of one, it would be two. If we were to take the ratio between h of one it would be two. That is the common ratio between successive whole number inputs into our function. So, h of I could say plus one over h of n is going to be equal to is going to be equal to actually I can work it out mathematically. One-fourth times two to the n plus one over one-fourth two to the n. Two to the n plus one, divided by two to the n is just going to be equal to two. That is your common ratio. So for the function h. For the function f, our common ratio is three. If we were to go the other way around, if someone said, hey, I have some function whose initial value, so let's say, I have some function, I'll do this in a new color, I have some function, g, and we know that its initial initial value is five. And someone were to say its common ratio its common ratio is six, what would this exponential function look like? And they're telling you this is an exponential function. Well, g of let's say x is the input, is going to be equal to our initial value, which is five. That's not a negative sign there, Our initial value is five. I'll write equals to make that clear. And then times our common ratio to the x power. So once again, initial value, right over there, that's the five. And then our common ratio is the six, right over there. So hopefully that gets you a little bit familiar with some of the parts of an exponential function, why they are called what they are called." + }, + { + "Q": "Shouldn't NAD+ turn into NAD^2+ when combined with H+", + "A": "The whole point of NAD+ is that it is a high energy electron carrier. When it picks up H+, it also picks up two high energy electrons. The net result is a neutral NADH: NAD+ + 2e- ---> NADH", + "video_name": "GR2GA7chA_c", + "transcript": "In the last video we learned a little bit about photosynthesis. And we know in very general terms, it's the process where we start off with photons and water and carbon dioxide, and we use that energy in the photons to fix the carbon. And now, this idea of carbon fixation is essentially taking carbon in the gaseous form, in this case carbon dioxide, and fixing it into a solid structure. And that solid structure we fix it into is a carbohydrate. The first end-product of photosynthesis was this 3-carbon chain, this glyceraldehyde 3-phosphate. But then you can use that to build up glucose or any other carbohydrate. So, with that said, let's try to dig a little bit deeper and understand what's actually going on in these stages of photosynthesis. Remember, we said there's two stages. The light-dependent reactions and then you have the light independent reactions. I don't like using the word dark reaction because it actually occurs while the sun is outside. It's actually occurring simultaneously with the light reactions. It just doesn't need the photons from the sun. But let's focus first on the light-dependent reactions. The part that actually uses photons from the sun. Or actually, I guess, even photons from the heat lamp that you might have in your greenhouse. And uses those photons in conjunction with water to produce ATP and reduce NADP plus to NADPH. Remember, reduction is gaining electrons or hydrogen atoms. And it's the same thing, because when you gain a hydrogen atom, including its electron, since hydrogen is not too electronegative, you get to hog its electron. So this is both gaining a hydrogen and gaining electron. But let's study it a little bit more. So before we dig a little deeper, I think it's good to know a little bit about the anatomy of a plant. So let me draw some plant cells. So plant cells actually have cell walls, so I can draw them a little bit rigid. So let's say that these are plant cells right here. Each of these squares, each of these quadrilaterals is a plant cell. And then in these plant cells you have these organelles called chloroplasts. Remember organelles are like organs of a cell. They are subunits, membrane-bound subunits of cells. And of course, these cells have nucleuses and DNA and all of the other things you normally associate with cells. But I'm not going to draw them here. I'm just going to draw the chloroplasts. And your average plant cell-- and there are other types of living organisms that perform photosynthesis, but we'll focus on plants. Because that's what we tend to associate it with. Each plant cell will contain 10 to 50 chloroplasts. I make them green on purpose because the chloroplasts contain chlorophyll. Which to our eyes, appear green. But remember, they're green because they reflect green light and they absorb red and blue and other wavelengths of light. Because it's reflecting. But it's absorbing all the other wavelengths. But anyway, we'll talk more about that in detail. But you'll have 10 to 50 of these chloroplasts right here. And then let's zoom in on one chloroplast. So if we zoom in on one chloroplast. So let me be very clear. This thing right here is a plant cell. That is a plant cell. And then each of these green things right here is an organelle called the chloroplast. And let's zoom in on the chloroplast itself. If we zoom in on one chloroplast, it has a membrane like that. And then the fluid inside of the chloroplast, inside of its membrane, so this fluid right here. All of this fluid. That's called the stroma. The stroma of the chloroplast. And then within the chloroplast itself, you have these little stacks of these folded membranes, These little folded stacks. Let me see if I can do justice here. So maybe that's one, two, doing these stacks. Each of these membrane-bound-- you can almost view them as pancakes-- let me draw a couple more. Maybe we have some over here, just so you-- maybe you have some over here, maybe some over here. So each of these flattish looking pancakes right here, these are called thylakoids. So this right here is a thylakoid. That is a thylakoid. The thylakoid has a membrane. And this membrane is especially important. We're going to zoom in on that in a second. So it has a membrane, I'll color that in a little bit. The inside of the thylakoid, so the space, the fluid inside of the thylakoid, right there that area. This light green color right there. That's called the thylakoid space or the thylakoid lumen. And just to get all of our terminology out of the way, a stack of several thylakoids just like that, that is called a grana. That's a stack of thylakoids. That is a grana. And this is an organelle. And evolutionary biologists, they believe that organelles were once independent organisms that then, essentially, teamed up with other organisms and started living inside of their cells. So there's actually, they have their own DNA. So mitochondria is another example of an organelle that people believe that one time mitochondria, or the ancestors of mitochondria, were independent organisms. That then teamed up with other cells and said, hey, if I produce your energy maybe you'll give me some food or whatnot. And so they started evolving together. And they turned into one organism. Which makes you wonder what we might evolve-- well anyway, that's a separate thing. So there's actually ribosomes out here. That's good to think about. Just realize that at one point in the evolutionary past, this organelle's ancestor might have been an independent organism. But anyway, enough about that speculation. Let's zoom in again on one of these thylakoid membranes. So I'm going to zoom in. Let me make a box. Let me zoom in right there. So that's going to be my zoom-in box. So let me make it really big. Just like this. So this is my zoom-in box. So that little box is the same thing as this whole box. So we're zoomed in on the thylakoid membrane. So this is the thylakoid membrane right there. That's actually a phospho-bilipd layer. It has your hydrophilic, hydrophobic tails. I mean, I could draw it like that if you like. The important thing from the photosynthesis point of view is that it's this membrane. And on the outside of the membrane, right here on the outside, you have the fluid that fills up the entire chloroplast. So here you have the stroma. And then this space right here, this is the inside of your thylakoid. So this is the lumen. So if I were to color it pink, right there. This is your lumen. Your thylakoid space. And in this membrane, and this might look a little bit familiar if you think about mitochondria and the electron transport chain. What I'm going to describe in this video actually is an electron transport chain. Many people might not consider it the electron transport chain, but it's the same idea. Same general idea. So on this membrane you have these proteins and these complexes of proteins and molecules that span this membrane. So let me draw a couple of them. So maybe I'll call this one, photosystem II. And I'm calling it that because that's what it is. Photosystem II. You have maybe another complex. And these are hugely complicated. I'll do a sneak peek of what photosystem II actually looks like. This is actually what photosystem II looks like. So, as you can see, it truly is a complex. These cylindrical things, these are proteins. These green things are chlorophyll molecules. I mean, there's all sorts of things going here. And they're all jumbled together. I think a complex probably is the best word. It's a bunch of proteins, a bunch of molecules just jumbled together to perform a very particular function. We're going to describe that in a few seconds. So that's what photosystem II looks like. Then you also have photosystem I. And then you have other molecules, other complexes. You have the cytochrome B6F complex and I'll draw this in a different color right here. I don't want to get too much into the weeds. Because the most important thing is just to understand. So you have other protein complexes, protein molecular complexes here that also span the membrane. But the general idea-- I'll tell you the general idea and then we'll go into the specifics-- of what happens during the light reaction, or the light dependent reaction, is you have some photons. Photons from the sun. They've traveled 93 million miles. so you have some photons that go here and they excite electrons in a chlorophyll molecule, in a chlorophyll A molecule. And actually in photosystem II-- well, I won't go into the details just yet-- but they excite a chlorophyll molecule so those electrons enter into a high energy state. Maybe I shouldn't draw it like that. They enter into a high energy state. And then as they go from molecule to molecule they keep going down in energy state. But as they go down in energy state, you have hydrogen atoms, or actually I should say hydrogen protons without the electrons. So you have all of these hydrogen protons. Hydrogen protons get pumped into the lumen. They get pumped into the lumen and so you might remember this from the electron transport chain. In the electron transport chain, as electrons went from a high potential, a high energy state, to a low energy state, that energy was used to pump hydrogens through a membrane. And in that case it was in the mitochondria, here the membrane is the thylakoid membrane. But either case, you're creating this gradient where-- because of the energy from, essentially the photons-- the electrons enter a high energy state, they keep going into a lower energy state. And then they actually go to photosystem I and they get hit by another photon. Well, that's a simplification, but that's how you can think of it. Enter another high energy state, then they go to a lower, lower and lower energy state. But the whole time, that energy from the electrons going from a high energy state to a low energy state is used to pump hydrogen protons into the lumen. So you have this huge concentration of hydrogen protons. And just like what we saw in the electron transport chain, that concentration is then-- of hydrogen protons-- is then used to drive ATP synthase. So the exact same-- let me see if I can draw that ATP synthase here. You might remember ATP synthase looks something like this. Where literally, so here you have a huge concentration of hydrogen protons. So they'll want to go back into the stroma from the lumen. And they go through the ATP synthase. Let me do it in a new color. So these hydrogen protons are going to make their way back. Go back down the gradient. And as they go down the gradient, they literally-- it's like an engine. And I go into detail on this when I talk about respiration. And that turns, literally mechanically turns, this top part-- the way I drew it-- of the ATP synthase. And it puts ADP and phosphate groups together. It puts ADP plus phosphate groups together to produce ATP. So that's the general, very high overview. And I'm going to go into more detail in a second. But this process that I just described is called photophosphorylation. Let me do it in a nice color. Why is it called that? Well, because we're using photons. That's the photo part. We're using light. We're using photons to excite electrons in chlorophyll. As those electrons get passed from one molecule, from one electron acceptor to another, they enter into lower and lower energy states. As they go into lower energy states, that's used to drive, literally, pumps that allow hydrogen protons to go from the stroma to the lumen. Then the hydrogen protons want to go back. They want to-- I guess you could call it-- chemiosmosis. They want to go back into the stroma and then that drives ATP synthase. Right here, this is ATP synthase. ATP synthase to essentially jam together ADPs and phosphate groups to produce ATP. Now, when I originally talked about the light reactions and dark reactions I said, well the light reactions have two byproducts. It has ATP and it also has-- actually it has three. It has ATP, and it also has NADPH. NADP is reduced. It gains these electrons and these hydrogens. So where does that show up? Well, if we're talking about non-cyclic oxidative photophosphorylation, or non-cyclic light reactions, the final electron acceptor. So after that electron keeps entering lower and lower energy states, the final electron acceptor is NADP plus. So once it accepts the electrons and a hydrogen proton with it, it becomes NADPH. Now, I also said that part of this process, water-- and this is actually a very interesting thing-- water gets oxidized to molecular oxygen. So where does that happen? So when I said, up here in photosystem I, that we have a chlorophyll molecule that has an electron excited, and it goes into a higher energy state. And then that electron essentially gets passed from one guy to the next, that begs the question, what can we use to replace that electron? And it turns out that we use, we literally use, the electrons in water. So over here you literally have H2O. And H2O donates the hydrogens and the electrons with it. So you can kind of imagine it donates two hydrogen protons and two electrons to replace the electron that got excited by the photons. Because that electron got passed all the way over to photosystem I and eventually ends up in NADPH. So, you're literally stripping electrons off of water. And when you strip off the electrons and the hydrogens, you're just left with molecular oxygen. Now, the reason why I want to really focus on this is that there's something profound happening here. Or at least on a chemistry level, something profound is happening. You're oxidizing water. And in the entire biological kingdom, the only place where we know something that is strong enough of an oxidizing agent to oxidize water, to literally take away electrons from water. Which means you're really taking electrons away from oxygen. So you're oxidizing oxygen. The only place that we know that an oxidation agent is strong enough to do this is in photosystem II. So it's a very profound idea, that normally electrons are very happy in water. They're very happy circulating around oxygens. Oxygen is a very electronegative atom. That's why we even call it oxidizing, because oxygen is very good at oxidizing things. But all of a sudden we've found something that can oxidize oxygen, that can strip electrons off of oxygen and then give those electrons to the chlorophyll. The electron gets excited by photons. Then those photons enter lower and lower and lower energy states. Get excited again in photosystem I by another set of photons and then enter lower and lower and lower energy states. And then finally, end up at NADPH. And the whole time it entered lower and lower energy states, that energy was being used to pump hydrogen across this membrane from the stroma to lumen. And then that gradient is used to actually produce ATP. So in the next video I'm going to give a little bit more context about what this means in terms of energy states of electrons and what's at a higher or lower energy state. But this is essentially all that's happening. Electrons get excited. Those electrons eventually end up at NADPH. And as the electron gets excited and goes into lower and lower energy states, it pumps hydrogen across the gradient. And then that gradient is used to drive ATP synthase, to generate ATP. And then that original electron that got excited, it had to be replaced. And that replaced electron is actually stripped off of H2O. So the hydrogen protons and the electrons of H2O are stripped away and you're just left with molecular oxygen. And just to get a nice appreciation of the complexity of all of this-- I showed you this earlier in the video-- but this is literally a-- I mean this isn't a picture of photosystem II. You actually don't have cylinders like this. But these cylinders represent proteins. Right here, these green kind of scaffold-like molecules, that's chlorophyll A. And what literally happens, is you have photons hitting-- actually it doesn't always have to hit chlorophyll A. It can also hit what's called antenna molecules. So antenna molecules are other types of chlorophyll, and actually other types of molecules. And so a photon, or a set of photons, comes here and maybe it excites some electrons, it doesn't have to be in It could be in some of these other types of chlorophyll. Or in some of these other I guess you could call them, pigment molecules that will absorb these photons. And then their electrons get excited. And you can almost imagine it as a vibration. But when you're talking about things on the quantum level, vibrations really don't make sense. But it's a good analogy. They kind of vibrate their way to chlorophyll A. And this is called resonance energy. They vibrate their way, eventually, to chlorophyll A. And then in chlorophyll A, you have the electron get excited. The primary electron acceptor is actually this molecule right here. Pheophytin. Some people call it pheo. And then from there, it keeps getting passed on from one molecule to another. I'll talk a little bit more about that in the next video. But this is fascinating. Look how complicated this is. In order to essentially excite electrons and then use those electrons to start the process of pumping hydrogens across a membrane. And this is an interesting place right here. This is the water oxidation site. So I got very excited about the idea of oxidizing water. And so this is actually where it occurs in the photosystem II complex. And you actually have this very complicated mechanism. Because it's no joke to actually strip away electrons and hydrogens from an actual water molecule. I'll leave you there. And in the next video I'll talk a little bit more about these energy states. And I'll fill in a little bit of the gaps about what some of these other molecules that act as hydrogen acceptors. Or you can also view them as electron acceptors along the way." + }, + { + "Q": "how do you answer this question its quite puzzling\n3 (6 + 3 \u00c3\u00b7 3 ) +2", + "A": "Order of Operations: Divide first, so 3( 6 + 1) + 2 Then You can do distributive property or add to finish the parenthesis 3(7) + 2 21 + 2 23", + "video_name": "a-e8fzqv3CE", + "transcript": "While I'm working on some more ambitious projects, I wanted to quickly comment on a couple of mathy things that have been floating around the internet, just so you know I'm still alive. So there's this video that's been floating around about how to multiply visually like this. Pick two numbers, let's say, 12 times 3. And then you draw these lines. 12, 31. Then you start counting the intersections-- 1, 2, 3 on the left; 1, 2, 3, 4, 5, 6, 7 in the middle; 1, 2 on the right, put them together, 3, 7, 2. There's your answer. Magic, right? But one of the delightful things about mathematics is that there's often more than one way to solve a problem. And sometimes these methods look entirely different, but because they do the same thing, they must be connected somehow. And in this case, they're not so different at all. Let me demonstrate this visual method again. This time, let's do 97 times 86. So we draw our nine lines and seven lines time eight lines and six lines. Now, all we have to do is count the intersections-- 1,2, 3, 4, 5, 6, 7, 8, 9, 10. This is boring. How about instead of counting all the dots, we just figure out how many intersections there are. Let's see, there's seven going one way and six Hey, let's do 6 times 7, which is-- huh. Forget everything I ever said about learning a certain amount of memorization in mathematics being useful, at least at an elementary school level. Because apparently, I've been faking my way through being a mathematician without having memorized 6 times 7. And now I'm going to have to figure out 5 times 7, which is half of 10 times 7, which is 70, so that's 35, and then add the sixth 7 to get 42. Wow, I really should have known that one. OK, but the point is that this method breaks down the two-digit multiplication problem into four one-digit multiplication problems. And if you do have your multiplication table memorized, you can easily figure out the answers. And just like these three numbers became the ones, tens, and hundreds place of the answer, these do, too-- ones, tens, hundreds-- and you add them up and voila! Which is exactly the same kind of breaking down into single-digit multiplication and adding that you do during the old boring method. The whole point is just to multiply every pair of digits, make sure you've got the proper number of zeroes on the end, and add them all up. But of course, seeing that what you're actually doing is multiplying every possible pair is not something your teachers want you to realize, or else you might remember the every combination concept when you get to multiplying binomials, and it might make it too easy. In the end, all of these methods of multiplication distract from what multiplication really is, which for 12 times 31 is this. All the rest is just breaking it down into well-organized chunks, saying, well, 10 times 30 is this, 10 times 1 this, 30 times 2 is that, and 2 times 1 is that. Add them all up, and you get the total area. Don't let notation get in the way of your understanding. Speaking of notation, this infuriating bit of nonsense has been circulating around recently. And that there has been so much discussion of it is sign that we've been trained to care about notation way too much. Do you multiply here first or divide here first? The answer is that this is a badly formed sentence. It's like saying, I would like some juice or water with ice. Do you mean you'd like either juice with no ice or water with ice? Or do you mean that you'd like either juice with ice or water with ice? You can make claims about conventions and what's right and wrong, but really the burden is on the author of the sentence to put in some commas and make things clear. Mathematicians do this by adding parenthesis and avoiding this divided by sign. Math is not marks on a page. The mathematics is in what those marks represent. You can make up any rules you want about stuff as long as you're consistent with them. The end." + }, + { + "Q": "My parents have brown eyes but I have hazel eyes, how did that occur, ? My grandfather on my dads side has blue eyes my grandmother, my mom's mother has brown.", + "A": "probably becasue the next generation that you create will have hazel eyes just like you", + "video_name": "eEUvRrhmcxM", + "transcript": "Well, before we even knew what DNA was, much less how it was structured or it was replicated or even before we could look in and see meiosis happening in cells, we had the general sense that offspring were the products of some traits that their parents had. That if I had a guy with blue eyes-- let me say this is the blue-eyed guy right here --and then if he were to marry a brown-eyed girl-- Let's say this is the brown-eyed girl. Maybe make it a little bit more like a girl. If he were to marry the brown-eyed girl there, that most of the time, or maybe in all cases where we're dealing with the brown-eyed girl, maybe their kids are brown-eyed. Let me do this so they have a little brown-eyed baby here. And this is just something-- I mean, there's obviously thousands of generations of human beings, and we've observed this. We've observed that kids look like their parents, that they inherit some traits, and that some traits seem to dominate other traits. One example of that tends to be a darker pigmentation in maybe the hair or the eyes. Even if the other parent has light pigmentation, the darker one seems to dominate, or sometimes, it actually ends up being a mix, and we've seen that all around us. Now, this study of what gets passed on and how it gets passed on, it's much older than the study of DNA, which was really kind of discovered or became a big deal in the middle of the 20th century. This was studied a long time. And kind of the father of classical genetics and heredity is Gregor Mendel. He was actually a monk, and he would mess around with plants and cross them and see which traits got passed and which traits didn't get passed and tried to get an understanding of how traits are passed from one generation to another. So when we do this, when we study this classical genetics, I'm going to make a bunch of simplifying assumptions because we know that most of these don't hold for most of our genes, but it'll give us a little bit of sense of how to predict what might happen in future generations. So the first simplifying assumption I'll make is that some traits have kind of this all or nothing property. And we know that a lot of traits don't. Let's say that there are in the world-- and this is a gross oversimplification --let's say for eye color, let's say that there are two alleles. Now remember what an allele was. An allele is a specific version of a gene. So let's say that you could have blue eye color or you could have brown eye color. That we live in a universe where someone could only have one of these two versions of the eye color gene. We know that eye color is far more complex than that, so this is just a simplification. And let me just make up another one. Let me say that, I don't know, maybe for tooth size, that's a trait you won't see in any traditional biology textbook, and let's say that there's one trait for big teeth and there's another allele for small teeth. And I want to make very clear this distinction between a gene and an allele. I talked about Gregor Mendel, and he was doing this in the 1850s well before we knew what DNA was or what even chromosomes were and how DNA was passed on, et cetera, but let's go into the microbiology of it to understand the So I have a chromosome. Let's say on some chromosome-- let me pick some chromosome here. Let's say this is some chromosome. Let's say I got that from my dad. And on this chromosome, there's some location here-- we could call that the locus on this chromosome where the eye color gene is --that's the location of the eye color gene. Now, I have two chromosomes, one from my father and one from my mother, so let's say that this is the chromosome from my mother. We know that when they're normally in the cell, they aren't nice and neatly organized like this in the chromosome, but this is just to kind of show you the idea. Let's say these are homologous chromosomes so they code for the same genes. So on this gene from my mother on that same location or locus, there's also the eye color gene. Now, I might have the same version of the gene and I'm saying that there's only two versions of this gene in the world. Now, if I have the same version of the gene-- I'm going to make a little shorthand notation. I'm going to write big B-- Actually, let me do it the other way. I'm going to write little b for blue and I'm going to write big B for brown. There's a situation where this could be a little b and this could be a big B. And then I could write that my genotype-- I have the allele, I have one big B from my mom and I have one small b from my dad. Each of these instances, or ways that this gene is expressed, is an allele. So these are two different alleles-- let me write that --or versions of the same gene. And when I have two different versions like this, one version from my mom, one version from my dad, I'm called a heterozygote, or sometimes it's called a heterozygous genotype. And the genotype is the exact version of the alleles I have. Let's say I had the lowercase b. I had the blue-eyed gene from both parents. So let's say that I was lowercase b, lowercase b, then I would have two identical alleles. Both of my parents gave me the same version of the gene. And this case, this genotype is homozygous, or this is a homozygous genotype, or I'm a homozygote for this trait. Now, you might say, Sal, this is fine. These are the traits that you have. I have a brown from maybe my mom and a blue from my dad. In this case, I have a blue from both my mom and dad. How do we know whether my eyes are going to be brown or blue? And the reality is it's very complex. It's a whole mixture of things. But Mendel, he studied things that showed what we'll call dominance. And this is the idea that one of these traits dominates the other. So a lot of people originally thought that eye color, especially blue eyes, was always dominated by the other traits. We'll assume that here, but that's a gross oversimplification. So let's say that brown eyes are dominant and blue are recessive. I wanted to do that in blue. Blue eyes are recessive. If this is the case, and this is a-- As I've said repeatedly, this is a gross oversimplification. But if that is the case, then if I were to inherit this genotype, because brown eyes are dominant-- remember, I said the big B here represents brown eye and the lowercase b is recessive --all you're going to see for the person with this genotype is brown eyes. So let me do this here. Let me write this here. So genotype, and then I'll write phenotype. Genotype is the actual versions of the gene you have and then the phenotypes are what's expressed or what do you see. So if I get a brown-eyed gene from my dad-- And I want to do it in a big-- I want to do it in brown. Let me do it in brown so you don't get confused. So if I've have a brown-eyed gene from my dad and a blue-eyed gene from my mom, because the brown eye is recessive, the brown-eyed allele is recessive-- And I just said a brown-eyed gene, but what I should say is the brown-eyed version of the gene, which is the brown allele, or the blue-eyed version of the gene from my mom, which is the blue allele. Since the brown allele is dominant-- I wrote that up here --what's going to be expressed are brown eyes. Now, let's say I had it the other way. Let's say I got a blue-eyed allele from my dad and I get a brown-eyed allele for my mom. Same thing. The phenotype is going to be brown eyes. Now, what if I get a brown-eyed allele from both my mom and my dad? Let me see, I keep changing the shade of brown, but they're all supposed to be the same. So let's say I get two dominant brown-eyed alleles from my mom and my dad. Then what are you going to see? Well, you could guess that. I'm still going to see brown eyes. So there's only one last combination because these are the only two types of alleles we might see in our population, although for most genes, there's more than two types. For example, there's blood types. There's four types of blood. But let's say that I get two blue, one blue allele from each of my parents, one from my dad, one from my mom. Then all of a sudden, this is a recessive trait, but there's nothing to dominate it. So, all of a sudden, the phenotype will be blue eyes. And I want to repeat again, this isn't necessarily how the alleles for eye color work, but it's a nice simplification to maybe understand how heredity works. There are some traits that can be studied in this simple way. But what I wanted to do here is to show you that many different genotypes-- so these are all different genotypes --they all coded for the same phenotype. So just by looking at someone's eye color, you didn't know exactly whether they were homozygous dominant-- this would be homozygous dominant --or whether they were heterozygotes. This is heterozygous right here. These two right here are heterozygotes. These are also sometimes called hybrids, but the word hybrid is kind of overloaded. It's used a lot, but in this context, it means that you got different versions of the allele for that gene. So let's think a little bit about what's actually happening when my mom and my dad reproduced. Well, let's think of a couple of different scenarios. Let's say that they're both hybrids. My dad has the brown-eyed dominant allele and he also has the blue-eyed recessive allele. Let's say my mom has the same thing, so brown-eyed dominant, and she also has the blue-eyed recessive allele. Now let's think about if these two people, before you see what my eye color is, if you said, look, I'm giving you what these two people's genotypes are. Let me label them. Let me make this the mom. I think this is the standard convention. And let's make this right here, this is the dad. What are the different genotypes that their children could have? So let's say they reproduce. I'm going to draw a little grid here. So let me draw a grid. So we know from our study of meiosis that, look, my mom has this gene on-- Let me draw the genes again. So there's a homologous pair, right? This is one chromosome right here. That's another chromosome right there. On this chromosome in the homologous pair, there might be-- at the eye color locus --there's the brown-eyed gene. And at this one, at the eye color locus, there's a blue-eyed gene. And similarly from my dad, when you look at that same chromosome in his cells-- Let me do them like this. So this is one chromosome there and this is the other chromosome here. When you look at that locus on this chromosome or that location, it has the brown-eyed allele for that gene, and on this one, it has the blue-eyed allele on this gene. And we learn from meiosis when the chromosomes-- Well, they replicate first, and so you have these two chromatids on a chromosome. But they line up in meiosis I during the metaphase. And we don't know which way they line up. For example, my dad might give me this chromosome or might give me that chromosome. Or my mom might give me that chromosome or might give me that chromosome. So I could have any of these combinations. So, for example, if I get this chromosome from my mom and this chromosome from my dad, what is the genotype going to be for eye color? Well, it's going to be capital B and capital B. If I get this chromosome from my mom and this chromosome from my dad, what's it going to be? Well, I'm going to get the big B from my dad and then I'm going to get the lowercase b from my mom. So this is another possibility. Now, this is another possibility here where I get the brown-eyed allele from my mom and I get the blue eye allele from my dad. And then there's a possibility that I get this chromosome from my dad and this chromosome from my mom, so it's this situation. Now, what are the phenotypes going to be? Well, we've already seen that this one right here is going to be brown, that one's going to be brown, this one's going to be brown, but this one is going to be blue. I already showed you this. But if I were to tell you ahead of time that, look, I have two people. They're both hybrids, or they're both heterozygotes for eye color, and eye color has this recessive dominant situation. And they're both heterozygotes where they each have one brown allele and one blue allele, and they're going to have a child, what's the probability that the child has brown eyes? What's the probability? Well, each of these scenarios are equally likely, right? There's four equal scenarios. So let's put that in the denominator. Four equal scenarios. And how many of those scenarios end up with brown eyes? Well, it's one, two, three. So the probability is 3/4, or it's a 75% probability. Same logic, what's the probability that these parents produce an offspring with blue eyes? Well, that's only one of the four equally likely possibilities, so blue eyes is only 25%. Now, what is the probability that they produce a heterozygote? So what is the probability that they produce a heterozygous offspring? So now we're not looking at the phenotype anymore. We're looking at the genotype. So of these combinations, which are heterozygous? Well, this one is, because it has a mix. It's a hybrid. It has a mix of the two alleles. And so is this one. So what's the probability? Well, there's four different combinations. All of those are equally likely, and two of them result in a heterozygote. So it's 2/4 or 1/2 or 50%. So using this Punnett square, and, of course, we had to make a lot of assumptions about the genes and whether one's dominant or one's a recessive, we can start to make predictions about the probabilities of different outcomes. And as we'll see in future videos, you can actually even You can say, hey, given that this couple had five kids with brown eyes, what's the probability that they're both heterozygotes, or something like that. So it's a really interesting area, even though it is a bit of oversimplification. But many traits, especially some of the things that Gregor Mendel studied, can be studied in this way." + }, + { + "Q": "so why are they all equal vectors regardless of where you put them?", + "A": "Because a vector is described by its length and direction. It doesn t matter where it starts. So if we have a vector in component form given by v=<1,4>, it could start at (0,0) (in the Cartesian plane) and run to (1,4), or it could start at (3, -1) and run to (4, 3)...or any other such starting and ending positions. The vector itself is unchanged by being moved around. Does that help?", + "video_name": "8QihetGj3pg", + "transcript": "So I have two 2-dimensional vectors right over here, vector a and vector b. And what I want to think about is how can we define or what would be a reasonable way to define the sum of vector a plus vector b? Well, one thing that might jump at your mind is, look, well, each of these are two dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just a sum of the first two components of these two vectors. So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two 2-dimensional vectors. You add them together, you get another two 2-dimensional vectors. If you think about it in terms of real coordinates bases, both of these are members of R2-- I'll write this down here just so we get used to the notation. So vector a and vector b are both members of R2, which is just another way of saying that these are both two tuples. They are both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it, but how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector a, we could visualize, this tells us how far this vector moves in each of these directions-- horizontal direction and vertical direction. So if we put the, I guess you could say the tail of the vector at the origin-- remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction. 1, 2, 3, 4, 5, 6. And then negative 2 in the vertical. So negative 2. So vector a could look like this. Vector a looks like that. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector. And the direction is the direction that it is pointed in. And also just to emphasize, I could have drawn vector a like that or I could have put it over here. These are all equivalent vectors. These are all equal to vector a. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector b. Vector b in the horizontal direction goes negative 4-- 1, 2, 3, 4, and in the vertical direction goes 4-- 1, 2, 3, 4. So its tail if we start at the origin, if its tail is at the origin, its head would be at negative 4, 4. So let me draw that just like that. So that right over here is vector b. And once again, vector b we could draw it like that or we could draw it-- let me copy and let me paste it-- so this would also be another way to draw vector b. Once again, what I really care about is its magnitude and its direction. All of these green vectors have the same magnitude. They all have the same length and they all have the same direction. So how does the way that I drew vector a and b gel with what its sum is? So let me draw its sum like this. Let me draw its sum in this blue color. So the sum based on this definition we just used, the vector addition would be 2, 2. So 2, 2. So it would look something like this. So how does this make sense that the sum, that this purple vector plus this green vector is somehow going to be equal to this blue vector? I encourage you to pause the video and think about if that even makes sense. Well, one way to think about it is this first purple vector, it shifts us this much. It takes us from this point to that point. And so if we were to add it, let's start at this point and put the green vector's tail right there and see where it ends up putting us. So the green vector, we already have a version. So once again, we start the origin. Vector a takes us there. Now, let's start over there with the green vector and see where green vector takes us. And this makes sense. Vector a plus vector b. Put the tail of vector b at the head of vector a. So if you were to start at the origin, vector a takes you there then if you add on what vector b takes you, it takes you right over there. So relative to the origin, how much did you-- I guess you could say-- shift? And once again, vectors don't only apply to things like displacement. It can apply to velocity. It can apply to actual acceleration. It can apply to a whole series of things, but when you visualize it this way, you see that it does make complete sense. This blue vector, the sum of the two, is what results where you start with vector a. At that point right over there, vector a takes you there, then you take vector b's tail, start over there and it takes you to the tip of the sum. Now, one question you might be having is well, vector a plus vector b is this, but what is vector b plus vector a? Does this still work? Well, based on the definition we had where you add the corresponding components, you're still going to get the same sum vector. So it should come out the same. So this will just be negative 4 plus 6 is 2. 4 plus negative 2 is 2. But does that make visual sense? So if we start with vector b. So let's say you start right over here. Vector b takes you right over there. And then if you were to go there and you were to start with vector a-- so let's do that. So actually, let me make this a little bit-- actually, let me start with a new vector b. So let's say that that's our vector b right over there. And then-- actually, let me give this a place where I'll have some space to work with. So let's say that's my vector b right over there. And then let me get a copy of the vector a. That's a good one. So copy and let me paste it. So I could put vector a's tail at the tip of vector b, and then it'll take me right over there. So if I start right over here, vector b takes me there. And now I'm adding to that vector a, which starting here will take me there. And so from my original starting position, I have gone this far. Now, what is this vector? Well, this is exactly the vector 2, 2. Or another way of thinking about it, this vector shifts you 2 in the horizontal direction and 2 in the vertical direction. So either way, you're going to get the same result, and that should, hopefully, make visual or conceptual sense as well." + }, + { + "Q": "Round to the nearest 0.1:\n5.312", + "A": "When you round to the nearest tenth, we need to check the hundredths place to see what we should round to. For example, your example 5.312 has a hundredths digit of 1. Remembering our rounding rules, the digits 0 through 4 round down and the digits 5 through 9 roundup. Since the 1 rounds down, the tenths place remains as 0.3. The answer is 5.3.", + "video_name": "_MIn3zFkEcc", + "transcript": "0.710 Round 9.564, or nine and five hundred sixty-four Thousandths, to the nearest tenth. So lie me write it a bit larger, 9.564 And we need to round to the nearest tenth. So what's the tenth place? The tenths place is right here This represents 5 tenths. This is the ones place, this is the tenths place, this is The hundredths place, and this is the Thousandths place right here So we need to round to the nearest tenth. So if we round up, this will be 9.6 If we round down, this will be 9.5 And just like regular rounding, when we're not Dealing with decimals, you move to one spot, or you look At one place to the right or one place lower, I guess, and You say is that 5 larger If it is, you round if it isn't,you round down 6 is definitely 5 or larger, so we want to round up. So this 9.564 becomes 9.6, or we can call this Nine and sixth tenths. And then we're done!" + }, + { + "Q": "Where did the 10,000 shares amount come from?", + "A": "This is the total number of shares held by the shareholders of the company. If you know the par/face value of a share and total (par/face) value of all the shares you can derive the number of share easily by dividing the total value with the par value of a share.", + "video_name": "5lmHzAHbtzg", + "transcript": "Let's say that both Ben's Shoes and Jason's Shoes are publicly traded companies. And all that means is that both of their shares are traded on exchanges. Maybe it's the NASDAQ, or the New York Stock Exchange, or some other exchange. And the going price on those exchanges-- the last closing price for Ben's stock was $21.50 per share, and the last closing price for Jason's stock is $12.00 per What I want to explore in this video, and probably the next few is, what is that saying for what the market thinks these businesses are worth? So in both of these situations, they have 10,000 shares. And remember, the shares are a split of the owner's equity. It's not a split of the assets. It's a split of just the equity part, right over here. So if shareholders are willing to pay 21.50 per share for Ben's stock, and there are 10,000 shares in Ben's company. So you take 21.50 times-- I'll do it this way-- times 10,000 shares gives us a market cap. So 21.50 times 10,000 gives us $215,000. And what this says is, look, if each of those 10,000 slices of the equity is worth 21.50, then the entire equity portion is going to be-- the market is valuing it at $215,000. And this calculation, this multiplication of the market price per share times the number of shares, this is called the market cap. Short for market capitalization. The market cap of the company. And all it is, is what is the market valuing the equity part of Ben's company worth? Let's do the same thing for Jason's company. You have $12.00 per share times 10,000 shares. That gives us $120,000 market cap. So the market is telling us that even though on the books, Ben's equity-- based on how he valued his assets and his liabilities-- is 135,000. The market is actually valuing this at 215,000. And in the next video, we'll think about what that means for how the market is actually valuing the business. In the case of Jason's business, instead of $35,000 of equity-- just straight up from what's on the books-- the market is valuing this piece right here at $120,000. So hopefully that gives you a little sense of one, what shares are a share of. They're a share of the owner's equity, not of the assets. And also gives you a good sense of what market cap is. It's the market's value of the owner's equity. And notice, in both cases-- and it's usually the case-- it's going to be a different number than the book value, the number that's actually on the books." + }, + { + "Q": "How can we find the eigen values of a matrix A,Where a11=B,a12=1 and a21=2 ,a22=5?", + "A": "Hi Sal, in my notes I have the characteristic equation : | A-hI |= 0 which is the reverse of yours. It doesn t seem to matter in this case which way they go A-hI or hI-A do you know if this is always the case?", + "video_name": "pZ6mMVEE89g", + "transcript": "In the last video we were able to show that any lambda that satisfies this equation for some non-zero vectors, V, then the determinant of lambda times the identity matrix minus A, must be equal to 0. Or if we could rewrite this as saying lambda is an eigenvalue of A if and only if-- I'll write it as if-- the determinant of lambda times the identity matrix minus A is equal to 0. Now, let's see if we can actually use this in any kind of concrete way to figure out eigenvalues. So let's do a simple 2 by 2, let's do an R2. Let's say that A is equal to the matrix 1, 2, and 4, 3. And I want to find the eigenvalues of A. So if lambda is an eigenvalue of A, then this right here tells us that the determinant of lambda times the identity matrix, so it's going to be the identity matrix in R2. So lambda times 1, 0, 0, 1, minus A, 1, 2, 4, 3, is going to be equal to 0. Well what does this equal to? This right here is the determinant. Lambda times this is just lambda times all of these terms. So it's lambda times 1 is lambda, lambda times 0 is 0, lambda times 0 is 0, lambda times 1 is lambda. And from that we'll subtract A. So you get 1, 2, 4, 3, and this has got to equal 0. And then this matrix, or this difference of matrices, this is just to keep the determinant. This is the determinant of. This first term's going to be lambda minus 1. The second term is 0 minus 2, so it's just minus 2. The third term is 0 minus 4, so it's just minus 4. And then the fourth term is lambda minus 3, just like that. So kind of a shortcut to see what happened. The terms along the diagonal, well everything became a negative, right? We negated everything. And then the terms around the diagonal, we've got a lambda out front. That was essentially the byproduct of this expression right there. So what's the determinant of this 2 by 2 matrix? Well the determinant of this is just this times that, minus this times that. So it's lambda minus 1, times lambda minus 3, minus these two guys multiplied by each other. So minus 2 times minus 4 is plus eight, minus 8. This is the determinant of this matrix right here or this matrix right here, which simplified to that matrix. And this has got to be equal to 0. And the whole reason why that's got to be equal to 0 is because we saw earlier, this matrix has a non-trivial null space. And because it has a non-trivial null space, it can't be invertible and its determinant has to be equal to 0. So now we have an interesting polynomial We can multiply it out. We get what? We get lambda squared, right, minus 3 lambda, minus lambda, plus 3, minus 8, is equal to 0. Or lambda squared, minus 4 lambda, minus 5, is equal to 0. And just in case you want to know some terminology, this expression right here is known as the characteristic polynomial. Just a little terminology, polynomial. But if we want to find the eigenvalues for A, we just have to solve this right here. This is just a basic quadratic problem. And this is actually factorable. Let's see, two numbers and you take the product is minus 5, when you add them you get minus 4. It's minus 5 and plus 1, so you get lambda minus 5, times lambda plus 1, is equal to 0, right? Minus 5 times 1 is minus 5, and then minus 5 lambda plus 1 lambda is equal to minus 4 lambda. So the two solutions of our characteristic equation being set to 0, our characteristic polynomial, are lambda is equal to 5 or lambda is equal to minus 1. So just like that, using the information that we proved to ourselves in the last video, we're able to figure out that the two eigenvalues of A are lambda equals 5 and lambda equals negative 1. Now that only just solves part of the problem, right? We know we're looking for eigenvalues and eigenvectors, right? We know that this equation can be satisfied with the lambdas equaling 5 or minus 1. So we know the eigenvalues, but we've yet to determine the actual eigenvectors. So that's what we're going to do in the next video." + }, + { + "Q": "Why Christ child looks bald?", + "A": "You can see the hair on the side of his head.", + "video_name": "jHN0BfowL7s", + "transcript": "MAN: We're in the Uffizi. And we're looking at a Raphael. And this is the \"Madonna of the Goldfinch,\" which is a really funny title. WOMAN: It is a funny title. And John-- who we see here on the left-- is holding out a goldfinch, the bird, to the Christ child, who strokes its head. And the goldfinch is a symbol of the passion of Christ, of Christ's suffering. And so we have that idea that we often have, of the foretelling Christ's terrible future. MAN: At the same time, this is a painting of two children and a mother. And so it exists in several different planes, because they're children doing childlike things-- one showing a pet to another, one wanting to touch it, the mother looking down protectively. WOMAN: And even a kind of tenderness between the mother and son-- look at the way that Christ puts his foot on his mother's. So there's that skin-to-skin moment of human contact there that's really lovely. But to me, Christ doesn't look like a child having fun. He looks very much all-knowing. I suppose if you were looking at a painting from the 1300s, Christ would look-- instead of looking like a baby, he would look like a little man, in order to indicate his sense of wisdom. But here I think Raphael communicates that through the elegance of Christ's body. Look at the way he lifts his arm up, strokes the goldfinch, and tilts his head back. He stands in this incredibly elegant contrapposto that no child would ever stand in. I mean, it's such a pose. MAN: It's true. And it's a beautiful foreshortening of his head, of his face as he leans back. But then there's a kind of energy and child-likeness that we see in John. John seems so engaged-- look what I can show you. WOMAN: And yet it's this symbol, this really potent symbol, of Christ's suffering. MAN: What's so interesting is that, unlike the 1300s as you mentioned before, we don't have the Madonna on the throne. Here, nature itself is the throne. We have this verdant environment, this beautiful atmospheric perspective. And she sits on a rock. That is, divinity is all around us. By the time we get to the late 15th century through the early 16th century, in the High Renaissance, nature itself has taken on the expression of God. We don't need, in a sense, those kingly symbols. WOMAN: Look at how composed it is, it in a way that we don't even notice immediately. We have a pyramid composition, with Mary at the top, and Saint John and Christ on either side, and that sense of real stability and balance that's also so much a part of the High Renaissance. MAN: Even as the figures are so engaged with each other-- and there's real dialogue that's taking place with them-- there is also that sense, that High Renaissance sense, you're right, of balance, of perfection, of the eternal. WOMAN: That interlocking of gestures and glances-- Mary looking at down at John, John looking at Christ, Christ looking back at John-- all of them enclosed within the pyramid structure of Mary's body, that unified composition that brings everything together in this really lovely landscape. MAN: I'm intrigued by the book. Mary had been reading. She's kept her place. And of course, that reminds us of an earlier scene in the Annunciation, when Gabriel interrupts as she's been piously reading the Bible. But here she's been reading, and now she's interrupted by her charges. She's doing a little bit of babysitting." + }, + { + "Q": "Why is around 2:30 mins in... (1/3) dividing 3x but on the other side it is multiplying!", + "A": "He s saying that multiplying both sides of the equation by the fraction (1/3) is the same as dividing both sides of the equation by 3.", + "video_name": "kbqO0YTUyAY", + "transcript": "So once again, we have three equal, or we say three identical objects. They all have the same mass, but we don't know what the mass is of each of them. But what we do know is that if you total up their mass, it's the same exact mass as these nine objects And each of these nine objects have a mass of 1 kilograms. So in total, you have 9 kilograms on this side. And over here, you have three objects. They all have the same mass. And we don't know what it is. We're just calling that mass x. And what I want to do here is try to tackle this a little bit more symbolically. In the last video, we said, hey, why don't we just multiply 1/3 of this and multiply 1/3 of this? And then, essentially, we're going to keep things balanced, because we're taking 1/3 of the same mass. This total is the same as this total. That's why the scale is balanced. Now, let's think about how we can represent this symbolically. So the first thing I want you to think about is, can we set up an equation that expresses that we have these three things of mass x, and that in total, their mass is equal to the total mass Can we express that as an equation? And I'll give you a few seconds to do it. Well, let's think about it. Over here, we have three things with mass x. So their total mass, we could write as-- we could write their total mass as x plus x plus x. And over here, we have nine things with mass of 1 kilogram. I guess we could write 1 plus 1 plus 1. That's 3. Plus 1 plus 1 plus 1 plus 1. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9. And actually, this is a mathematical representation. If we set it up as an equation, it's an algebraic representation. It's not the simplest possible way we can do it, but it is a reasonable way to do it. If we want, we can say, well, if I have an x plus another x plus another x, I have three x's. So I could rewrite this as 3x. And 3x will be equal to? Well, if I sum up all of these 1's right over here-- 1 We're doing that. We have 9 of them, so we get 3x is equal to 9. And let me make sure I did that. 1, 2, 3, 4, 5, 6, 7, 8, 9. So that's how we would set it up. And so the next question is, what would we do? What can we do mathematically? Actually, to either one of these equations, but we'll focus on this one right now. What can we do mathematically in order to essentially solve for the x? In order to figure out what that mystery mass actually is? And I'll give you another second or two to think about it. Well, when we did it the last time with just the scales we said, OK, we've got three of these x's here. We want to have just one x here. So we can say, whatever this x is, if the scale stays balanced, it's going to be the same as whatever we have there. There might be a temptation to subtract two of the x's maybe from this side, but that won't help us. And we can even see it mathematically over here. If we subtract two x's from both sides, on the left-hand side you're going to have 3x minus 2x. And on the right-hand side, you're going to have 9 minus 2x. And you're just going to be left with 3 of something minus 2 something is just 1 of something. So you will just have an x there if you get rid of two of them. But on the right-hand side, you're going to get 9 minus 2 So the x's still didn't help you out. You still have a mystery mass on the right-hand side. So that doesn't help. So instead, what we say is-- and we did this the last time. We said, well, what if we took 1/3 of these things? If we take 1/3 of these things and take 1/3 of these things, we should still get the same mass on both sides because the original things had the same mass. And the equivalent of doing that mathematically is to say, why don't we multiply both sides by 1/3? Or another way to say it is we could divide both sides by 3. Multiplying by 1/3 is the same thing as dividing by 3. So we're going to multiply both sides by 1/3. When you multiply both sides by 1/3-- visually over here, if you had three x's, you multiply it by 1/3, you're only going to have one x left. If you have nine of these one-kilogram boxes, you multiply it by 1/3, you're only going to have three left. And over here, you can even visually-- if you divide by 3, which is the same thing as multiplying by 1/3, you divide by 3. So you divide by 3. You have an x is equal to a 1 plus 1 plus 1. An x is equal to 3. Or you see here, an x is equal to 3. Over here you do the math. 1/3 times 3 is 1. You're left with 1x. So you're left with x is equal to 9 times 1/3. Or you could even view it as 9 divided by 3, which is equal to 3." + }, + { + "Q": "What does carry mean, at 0:16?", + "A": "When you add, you carry by putting numbers more than 10 to the top so it can be easier to solve.", + "video_name": "Wm0zq-NqEFs", + "transcript": "Let's add 536 to 398. And we're going to do it two different ways so that we really understand what this carrying is all about. So first, we'll do it in the more traditional way. We start in the ones place. We say, \"Well, what's 6 + 8?\" Well, we know that 6 + 8 is equal to 14. And so when we write it down here in the sum, we could say, \u201c \"Well look. The 4 is in the ones place.\u201d So it's equal to 4 + 1 ten.\" So let's write that 1 ten in the tens place. And now we focus on the tens place. We have 1 ten + 3 tens + 9 tens. So, what's that going to get us? 1 + 3 + 9 is equal to 13. Now we have to remind ourselves that this is 13 tens. Or another way of thinking about it, this is 3 tens and 1 hundred. You might say, \"Wait, wait! How does that make sense?\" Remember, this is in the tens place. When we're adding 1 ten + 3 tens + 9 tens, we're actually adding 10 + 30 + 90, and we're getting 130. And so we're putting the 30 (the 3 in the tens place represents the 30) \u2013 So this is the 3. The 3 represents the 30. And then we're placing this 1 in the hundreds place. 10 tens is equal to 100. And now we're adding up the numbers in the hundreds place. 1 + 5 + 3 is equal to \u2013 let's see. 1 + 5 is equal to 6, + 3 is equal to 9. But we have to remind ourselves: this is 9 hundreds. This is in the hundreds place. So this is actually 1 hundred. So this is actually 1 hundred + 5 hundreds + 3 hundreds, is equal to 9 hundreds. And that's exactly what we got here. 100 + 500 + 300 is equal to 900. And we're done. This is equal to 934." + }, + { + "Q": "is mirrored the same as flipped?", + "A": "yes. Think about it when you look in a mirror. your reflection is flipped from normal state", + "video_name": "CJrVOf_3dN0", + "transcript": "Let's talk a little bit about congruence, congruence And one to think about congruence, it's really kind of equivalence for shapes So, when in algebra when something is equal to another thing it means that their quantities are the same But when we're all of the sudden talking about shapes and we say that those shapes are the same, the shapes are the same size and shape then we say that they're congruent And just to see a simple example here: I have this triangle, right over there and let's say I have this triangle right over here And if you are able to shift, you are able to shift this triangle and flip this triangle, you can make it look exactly like this triangle As long as you're not changing the lengths of any of the sides or the angles here But you can flip it, you can shift it, you can rotate it So you can shift, let me write this, you can shift it, you can flip it and you can rotate If you can do those three procedures to make these the exact same triangle, then they are congruent And if you say that a triangle is congruent, let me label this So, let's call this triangle ABC Now let's call this D, let me call it XYZ XY and Z So, if we were to say, if we make the claim that both of these triangles are congruent So, if we say triangle ABC is congruent And the way you specify it, it almost look like an equal sign But it's equal sign with a curly thing on top Let me write it a little bit either So, we would write it like this If we know that triangle ABC is congruent to triangle XYZ That means their corresponding sides have the same length And their corresponding angles have the same measure So, if we make this assumption or someone tells us that this is true then we know, for example, that AB is going to equal to XY The length of segment AB is gonna be equal to the segment of XY And we could do this like this, and I'm assuming this are the corresponding sides And you can see that actually we've defined these triangles A corresponds to X, B corresponds to Y and C corresponds to Z right over there So, side AB is gonna have the same length as XY Then you can sometimes if you don't have the colors you can denote it just like that These two length are- or this two lines segments have the same length And you can actually say this, you don't always see this written this way You could also make the statement that line segment AB is congruent to line segment XY But congruence of line segments really just means that their lengths are equivalent So, these two things mean the same thing If one line segment is congruent to another line segment that just means the measure of one line segment is equal to the measure of the other line segment And so we can go thru all the corresponding sides If these two characters are congruent, we also know that BC, we also know that the length BC is gonna be the length of YZ Assuming those are the corresponding sides And we can put these double hash marks right over here to show that these lengths are the same And when we go the third side, we also know that these are going to be has same length or the line segments are going to be congruent So, we also know that the length of AC is going to be equal to the length of XZ Not only do we know that all of the sides, the corresponding sides are gonna have the same length If someone tells that a triangle is congruent We also know that all the corresponding angles are going to have the same measure So, for example: we also know that this angle's measure is going to be the same as the corresponding angle's measure, and the corresponding angle is right over It's between these orange side and blue side Or orange side and purple side, I should say And between the orange side and this purple side And so it also tells us that the measure of angle is BAC is equal to the measure of angle of YXZ Let me write that angle symbol, a little less like that, measure of angle of YXZ YXZ We can also write that as angle BAC is congruent angle YXZ And once again, like line segment, if one line segment is congruent to another line segment It just means that their lengths are equal And if one angle is congruent to another angle it just means that their measures are equal So, we know that those two corresponding angles have the same measure, they're congruent We also know that these two corresponding angles I'll use a double arch to specify that this has the same measure as that So, we also know the measure of angle ABC is equal to the measure of angle XYZ And then finally we know that this angle, if we know that these two characters are congruent, then this angle is gonna have the same measure as this angle as a corresponding angle So, we know that the measure of angle ACB is gonna be equal to the measure of angle XZY Now what we're gonna concern ourselves a lot with is how do we prove congruence? 'Cause it's cool, 'cause if you can prove congruence of 2 triangles then all of the sudden you can make all of these assumptions And what we're gonna find out, and this is going to be, we're gonna assume it for the sake of introductory geometry course This is an axiom or a postulate or just something you assume So, an axiom, very fancy word Postulate, also a very fancy word It really just means things we are gonna assume are true An axiom is sometimes, there's a little bit of distinction sometimes where someone would say \"an axiom is something that is self-evident\" or it seems like a universal truth that is definitely true and we just take it for granted You can't prove an axiom A postulate kinda has that same role but sometimes let's just assume this is true and see if we assume that it's true what can we derive from it, what we can prove if we assume its true But for the sake of introductory geometry class and really most in mathematics today, these two words are use interchangeably An axiom or a postulate, just very fancy words that things we take as a given Things that we'll just assume, we won't prove them, we will start with this assumptions and then we're just gonna build up from there And one of the core ones that we'll see in geometry is the axiom or the postulate That if all of the sides are congruent, if the length of all the sides of the triangle are congruent, then we are dealing with congruent triangles So, sometimes called side, side, side postulate or axiom We're not gonna prove it here, we're just gonna take it as a given So this literally stands for side, side, side And what it tells is, if we have two triangles and So I say that's another triangle right over there And we know that corresponding sides are equal So, we know that this side right over here is equal into, like, that side right over there Then we know and we're just gonna take this as an assumption and we can build off of this We know that they are congruent, the triangle, that these two triangles are congruent to each other I didn't put any labels there so it's kinda hard for me to refer to them But these two are congruent triangles And what's powerful there is we know that the corresponding sides are equal Then we know they're congruent and we can make all the other assumptions Which means that the corresponding angles are also equal So, that we know, is gonna be congruent to that or have the same measure That's gonna have the same measure as that and then that is gonna have the same measure as that right over there And to see why that is a reasonable axiom or a reasonable assumption or a reasonable postulate to start off with Let's take one, let's start with one triangle So, let's say I have this triangle right over here So, it has this side and then it has this side and then it has this side right over here And what I'm gonna do is see if I have another triangle that has the exact same line, side lengths is there anyway for me to construct a triangle with the same side lengths that is different, that can't be translated to this triangle thru flipping, shifting or rotating So, we assume this other triangle is gonna have the same size, the same length as that one over there So, I'll try to draw it like that Roughly the same length We know that it's going to have a size that's that length So, it's gonna have a side that is that length Let me put it on this side just to make it look a little bit more interesting So, we know that it's gonna have a side like that So, I'm gonna draw roughly the same length but I'm gonna try to do it in a different angle Now we know that's it's gonna have that looks like that So, let me, I'll put it right over here It's about that length right over there And so clearly this isn't a triangle, in order to make it a triangle, I'll have to connect this point to that point right over there And really there's only two ways to do it I can rotate it around that little hinge right over there If I connect them over here then I'm going to get a triangle that looks likes this Which is really a just a flip, am I visualizing it right? Yeah, just a flip version You can rotate it a little back this way, and you'd have a magenta on this side and a yellow one on this side And you can flip it, you could flip it vertically and it'll look exactly like this Our other option to make these two points connect is to rotate them out this way And the yellow side is gonna be here And then the magenta side is gonna be here and that's not magenta The magenta side is gonna be just like that And if we do that, then we actually just have to rotate it We just have to rotate it around to get that exact triangle So, this isn't a proof, and actually we're gonna start assuming that his is an axiom But hopefully you'll see that it's a pretty reasonable starting point that all of the sides, all of the corresponding sides of two different triangles are equal Then we are going to- we know that they are congruent We are just gonna assume that it's an axiom for that we're gonna build off, that they are congruent And we also know that he corresponding angles are going to be equivalent" + }, + { + "Q": "Ayesha cycles 13 km east,then turns around and goes 9km west of the starting point, then she again turns back and returns to the starting point, what is the total distance she covers?", + "A": "She cycles 44 kilometers.", + "video_name": "Oo2vGhVkvDo", + "transcript": "One of the coldest temperatures ever recorded outside was negative 128 degrees Fahrenheit in Antarctica. One of the warmest temperatures ever recorded outside was 134 degrees Fahrenheit in Death Valley, California. How many degrees difference are there between the coldest and warmest recorded outside temperatures? So let's think about this a little bit. Now, what I'll do is I'll plot them on a number line. But I'm going to plot it on a vertical number line that has a resemblance to a thermometer, since we're talking about temperature. So I'm going to make my number line vertical right over here. So there's my little vertical number line. And this right over here is 0 degrees Fahrenheit, which really is of no significance. If it was Celsius, we'd be talking about the freezing point. But for Fahrenheit, that happens at 32 degrees. But let's say this is 0 degrees Fahrenheit. And let's plot these two points. So one of the coldest ever recorded temperatures was negative 128 degrees Fahrenheit. So let's say that's right over here. This is negative 128 degrees Fahrenheit. And one of the warmest temperatures ever recorded was 134 degrees. This is a positive 134. So it's about that far and a little bit further. So it's a positive 134 degrees Fahrenheit. So when they're asking us how many degrees difference are there between the coldest and the warmest, they're essentially saying, well, what is this distance between the coldest and the warmest right over here? What is this distance? And there's a couple of ways you could think about it. You could say, hey, if I started at the coldest temperature and I wanted to go all the way up to the warmest, how much would I have to add? Or you could say, well, what's the difference between the coldest and the warmest? So you could take the larger number. So it's, say, 134. And from that, you could subtract the smaller number, which is negative 128. So this essentially saying what's the difference between these two numbers? It's going to be positive, because we're subtracting the smaller one from the larger one. This is going to give you the exact same thing as this. Now, there's several ways to think about it. One is we know that if you subtract a negative number, that's the same thing as adding the positive of that number, or adding the absolute value. So this is the same thing. This is going to be equal to 134 plus positive 128 degrees. And what's the intuition behind that? Why does this happen? Well, look at this right over here. We're trying to figure out this distance. This distance is 134 minus negative 128. And if you look at that, it's going to be the absolute value of 134. It's going to be this distance right over here, which is just 134-- which is just that right over there-- plus this distance right over here. Now, what is this distance? Well, it's the absolute value of negative 128. It's just 128. So it's going to be that distance, 134, plus 128. And that's why it made sense. This way, you're thinking of what's the difference between a larger number and a smaller number. But since it's a smaller number and you're subtracting a negative, it's the same thing as adding a positive. And hopefully this gives you a little bit of that intuition. But needless to say, we can now figure out what's going to be. And this is going to be equal to-- let me figure this out separately over here. So if I were to add 134 plus 128, I get 4 plus 8 is 12, 1 plus 3 plus 2 is 6. It's 262. This right over here is equal to 262. How many degrees difference are there between the coldest and warmest recorded outside temperature? 262 degrees Fahrenheit difference." + }, + { + "Q": "Whats a fraction", + "A": "They represent pieces of a whole. Special if a pizza is cut into 8 slices, and you eat 3... then you can say you ate 3/8 of the pizza", + "video_name": "fvtv2uYjo_E", + "transcript": "- [Voiceover] Let's see if we can calculate what 5/6 plus 1/4 is, and to help us, I have a visual representation of 5/6, and a visual representation of 1/4. Notice I have this whole whole, I guess you could say, broken up into one, two, three, four, five, six sections, and we've shaded in five of them, so this is 5/6, and then down here, we have another whole, and we have one out of the four equal sections shaded in so this is 1/4, and I want to add them, and I encourage you at any point, pause the video, and see if you could figure it out on your own. Well, whenever we're adding fractions, we like to think in terms of fractions that have the same denominator, and these clearly don't have the same denominator, but in order to rewrite them, with a common denominator, we just have to think of a common multiple of six and four, and ideally, the smallest common multiple of six and four, and the way that I like to do that is I like to take the larger of the two, which is six, and then think about its multiples. So I could first think about six itself. Six is clearly divisible by six, but it's not perfectly divisible by four, so now, let's multiply by two, so then we get to 12. 12 is divisible by both six and four. So 12 is a good common denominator here. It's the least common multiple of six and four. So we can rewrite both of these fractions as something over 12. So, something over 12 plus something, plus something over 12 is equal to. Now, there's a bunch of ways to tackle it, but what I want to do is I just want to visualize it here on this drawing. So, if I go, if I were to go from, if I were to go from six equal sections to 12 equal sections, which is what I'm doing if I'm going from six in the denominator to 12 in the denominator. I'm essentially multiplying each of these sections by, or, I'm essentially multiplying the number of sections I have by two, or I'm taking each of these existing sections and I'm turning them into two sections, so let's do that. Let's do that. Let me see if I can do it pretty neatly, so, I can do it a little bit neater than that. So, it'll look like that. And, whoops. Let me do this one. I want to divide them fairly close to evenly. I'm doing it by eye so it's not going to be perfect. So, and you have that one. And then last not, last but not least, you have that one there, and then notice, I had six sections, but now I've doubled the number of sections. I've turned the six sections into 12 sections by turning each of the original six into two, so now I have one, two, three, four, five, six, seven, eight, nine, 10, 11, 12 sections. So if I have 12 sections now, how many of those 12 are now shaded in? Instead of having five of the six, I now have 10 of the 12 that are shaded in. So I now have 10/12. 5/6 is the same thing as 10/12. Another way you could have thought about that, to go from six to 12, I had to multiply by two, so then I have to do the same thing in the numerator. Five times two is 10. But hopefully you see that those two fractions are equivalent, that I didn't change how much is shaded in, I just took each of the original six and I turned it into two, or I multiplied the total number of sections by two to get 12, and then instead of having 5/6, I now have 10/12 shaded in. Now let's do the same thing with the four, with the 1/4. Right here, I've depicted 1/4, but I want to turn this into something over 12. So to turn it into something over 12, each section has to be turned into three sections. So let's do that. Let's turn each section into three sections. So, that's one, two, and three. So then I have one, two, and three. I have, I think you can see where this is going. One, two and three. I have one, two, and three. And so notice, all I did is I multiplied, before I had four equal sections. Now I turned each of those four sections into three sections, so now I have 12 equal sections. And I did that, essentially, by multiplying the number of sections I had by three. So now what fraction is shaded in? Well, now, this original that was one out of the four, we can now see is three out of the 12 equal sections. It's now three out of the 12 equal sections, and so what is this going to be? Well, if I have 10/12, and I'm adding it to 3/12, well how many twelfths do I have? I'm going to have 13/12. And you could see it visually over here as well. Up here in green, I have 10/12 shaded in. Each of these boxes are a twelfth. Let me write that down. Each of these boxes are 1/12. That's 1/12. This is 1/12. So how many twelfths do I have shaded in? I have the 10 that are shaded in in green, and then I have an 11/12, a 12/12 and then finally, the 13/12 is one way to think about it." + }, + { + "Q": "What if there are two chemicals like Mg that have no number with them? What do I do then?", + "A": "Then there is an understood 1 after it", + "video_name": "xqpYeiefZl8", + "transcript": "- [Voiceover] Let's now see if we can balance a chemical equation with slightly more complex molecules. So, here we have a chemical equation, describing a chemical reaction. This is actually a combustion reaction. You have some ethylene right over here, in the presence of oxygen, and you need to get a little bit of energy to get this going, but then you're going to have this reaction that's actually going to release energy as well, but we're not accounting for the energy, at least the way we've written it. Right over here, you have some ethylene, and this little g in parentheses, says it's in the gas form or gaseous form, so gaseous ethylene plus some dioxygen molecule, which is the most prevalent form of oxygen molecule that you would find in the atmosphere. And so, that's also in the gas form. Put them together, you end up with some carbon dioxide gas and some liquid water. This is the classic combustion reaction. But now let's think about, how do we balance this thing? Let's make sure we have the same number of each atom on both sides. And when you see something more complicated like this, where, you know, here I have an oxygen and two different molecules over here, and a lot of these molecules have multiple elements in it. It might be very daunting. Where do I start? And this is where the art of balancing chemical equations starts to come into play. The general idea is, Try to balance the... try to balance the molecules that have multiple elements in them first, and leave the... molecules that only have one element in them for last. And the idea there is, is that these are harder. They're going to have all sorts of implications, and then, at the end of the day, you can just set a number here for the number of dioxygens. If you saved, say the ethylene for last, then every time, and you're trying to balance the carbons, you try to change the number of carbons, you're going to change the number of hydrogens, which is going to change the... You're going to have to balance over and over, and you're going to go into this really really really confusing circle. So, the best thing to do, try to balance the complex molecules first, and then save the single element molecules for last. So let's do that. So, let's start with the carbons. So, over here, I have two carbons. Over here, I only have one carbon. I only have one carbon. So, it seems like the best way to balance it is, I should have two molecules of carbon dioxide, and I haven't even thought about the oxygens yet. By putting that two there, that's going to change the number of oxygens I have on the righthand side. But at least it balances my carbons. I now have two carbons on the lefthand side, and I have two carbons on the righthand side. I\u2019m no longer magically destroying a carbon atom, all right. Now, let's move on to the hydrogens, and remember, what I said is, let's wait to do the oxygens last, because we have a molecule that only contains oxygen right over here, so we'll save oxygen for last. So, let's do hydrogen next. So, hydrogen, right over here, we have four hydrogens. And on the righthand side, we have two hydrogens. So, it seems like the easiest thing to do to balance the hydrogens is to have two of these water molecules. Now I have four hydrogens here, and I have four hydrogens there. Now, let's do the oxygen. Now, let's do the oxygen. I've balanced the carbons and the hydrogens. And the reason why oxygen's going to be interesting, I can just count the amount of oxygen I now have here, after changing the amount of molecules I have. And then I can adjust this accordingly, because this is only going to affect the number of oxygens that I have on the lefthand side. Right now, on the lefthand side, I have two oxygens, and on the righthand side, let me count this, I have two O two's, really. So, this is going to be four oxygens here, and then I have, each of these water molecules has one oxygen, but I have two water molecules, so this is going to be two oxygens, two oxygens here. So, on the righthand side, I have four plus two oxygens. So, I have six oxygens on the righthand side. I need six oxygens on the lefthand side. I need this number to be six. So, how do I do that? Well, I just need three of these molecules. If I have three molecules, each of them have two oxygens, I'm going to have a total of six oxygens. And just like that, we have balanced this combustion reaction, this chemical equation." + }, + { + "Q": "Is it possible for galaxies to crash into one another ?", + "A": "Yes, but there is lots of space between the stars in a galaxy, so a crash is really a gravitational interaction in which the galaxies pass through one another. The result is somewhat unpredictable. The galaxies could merge, with some stars being thrown out , or they could distort each other and then mostly go their separate ways, each leaving some stars with the other, or they could go into a sort of dance with each other that could last a long time before settling down.", + "video_name": "JiE_kNk3ucI", + "transcript": "Where we left off in the last video, we were just kind of staring, amazed, at this Earth's view of the Milky Way galaxy, just making sure we understood how enormous and how many stars we were looking at. And even if each of these dots were a star, this is a huge amount of stars. But a lot of these dots are thousands of stars. So our mind was already blown. But what we're going to see in this video is that in some ways, this is kind of just the beginning. And to some degree, I'm going to stop doing these particles of sand and a football field analogy because at some point, the particles of sand become so vast that are our minds can't even grasp it to begin with. But let's just start with our Milky Way. And we saw in the last video, the Milky Way right here, we're sitting here about 25,000 light years from the center. It's roughly 100,000 light years in diameter. And then, let's put it in perspective of its local neighborhood. So let's look at the Local Group. And when we talk about Local Group, we're talking about the local group of galaxies. So this right here is the Milky Way's Local Group. That's us right there, sitting right over here, about 25,000 light years from the center of the Milky Way. You have some of these \"small\"-- and I use \"small\" in quotation marks-- because these are also vast entities, also unimaginable entities. But we have these satellite galaxies around, under the gravitational influence, some of them, of the Milky Way. But the nearest large galaxy to us is Andromeda right over here. And this distance right over here. And now, we're going to start talking in the millions of light years. So this distance right here is 2.5 million light years. And just as a bit of reference, if that's any reference at all, one light year is roughly the radius of the Oort Cloud. Or another way to think about it, one radius of the Oort Cloud is about 50,000 or 60,000 astronomical units. And that's the distance from the Sun to the Earth. So you could view this as 2.5 million times 60,000 or so times the distance from the Sun to the Earth. So this is an unbelievably large distance we're talking about here. And that's to get to the next big galaxy over here. But even these things are huge things with many-- I mean, just unfathomably many-- stars. But Andromeda, in particular, we said that the Milky Way has 200 to 400 billion stars. Andromeda, people believe, has on the order of 1 trillion stars. So these just start to become numbers. It's hard to grasp. But we're not going to stop here. So in this, over here, this whole diagram right here, it's about four light years across, if you go from point to point. If you go from one side to the other side, this is about-- not four light years. This is 4 million light years. Four light years is just the distance from us to the Alpha Centauri. So that was nothing. That would only take that a Voyager 1 80,000 years to get. This is 4 million light years. So 4 million times the distance to the nearest star. But even this is-- I mean I'm starting to stumble on my words because there's really no words to describe it-- even this is small on an intergalactic scale. Because when you zoom out more, you can see our Local Group. Our local group is right over here. And this right over here is the Virgo Super Cluster. And each dot here is at least one galaxy. But it might be more than one galaxy. And the diameter here is 150 million light years. So what we saw in the Local Group, in the last diagram, the distance from the Milky Way to Andromeda, which was 2 and 1/2 million light years, which would be just a little dot just like that, that would be the distance between the Milky Way and Andromeda. And now, we're looking at the Virgo Super Cluster that is 150 million light years. But we're not done yet. We can zoom out even more. We can zoom out even more, and over here. So you had your Virgo Super Cluster, 150 million light years was that last diagram, this diagram right over here. I want to keep both of them on the screen if I can. This diagram right here, 150 million light years across. That would fit right about here on this diagram. So this is all of the super clusters that are near us. And once again, \"near\" has to be used very, very, very loosely. Here, this distance is about 150 million light years. A billion light years is-- two, three, four, five-- a billion light years is about from here to there. So we're starting to talk on a fairly massive-- I guess we've always been talking on a massive scale. But now, it's an even more massive scale. But we're still not done. Because this whole diagram-- now these dots that you're seeing now, I want to make it very clear. These aren't stars. These aren't even clusters of stars. These aren't even clusters of millions or even billions of stars. Each of these dots are clusters of galaxies, each of those galaxies having hundreds of billions to trillions of stars. So we're just on an unbelievably massive scale at this point. But we're still not done. We're still not done. This is roughly about a billion light years across. But right here is actually the best estimate of the visible universe. And in future videos, we're going to talk a lot more about what the visible universe means. So if you were to zoom out enough, this entire diagram right here, about a billion light years, would fit just like that. So we're talking about a super small amount of this part right here. And this is just the visible universe. I want to make it clear. This is not the entire universe. And we say it's the visible universe because think about what's happening. When we think about the a point out here, and we're observing it, and that's let's say 13 billion light years away. Let's say that point 13 billion. We're going to talk more about this in future videos, 13 billion light years. And I feel it's almost a sacrilege to be writing on this because this complexity that we're seeing here is just mind boggling. But this 13 billion light year away object, the light is just getting to us. This light left some point 13 billion light years ago. So what we're actually doing is observing that object close to the beginning of the actual universe. And the reason why it's the visible universe is there might have been something a little bit further out. Maybe it's light hasn't reached us yet or maybe the universe itself, and we'll talk more about this, it's expanding so fast that the light will never, ever reach us. So it's actually a huge question mark on how big the actual universe is. And then some people might say, well, does it even matter? Because this by itself is a huge distance. And I want to make it clear, you might say, OK, if this light over here, if this is coming from 13 billion light years away, or if this is 13 billion light years away, then you could say, hey, so everything that we can observe or that we can even observe the past of, the radius is about 26 billion light years. But even there, we have to be careful because remember the universe is expanding. When this light was emitted-- and I'll do a whole video on this because the geometry of it is kind of hard to visualize-- when this light was emitted, where we are in the Virgo Super Cluster, inside of the Milky Way Galaxy, where we are was much closer to that point. It was on the order of-- and I want to make sure I get this right-- 36 million light years. So we were super close by, I guess, astronomical scales. We were super close, only 36 million light years, to this object, when that light was released. But that light was coming to us and the whole time the universe So we were also moving away from it, if you just think about all of the space, that everything is expanding away from each other, And only 13 billion years later did it finally catch up with us. But the whole time that that was happening, this object has also been moving. This object has also been moving away from us. And so our best estimate of where this object is now, based on how space is expanding, is on the order a 40 or 45 billion light years away. We're just observing where that light was emitted 13 billion years ago. And I want to be very clear. What we are observing, this light is coming from something very, very, very primitive. That object or that area of space where that light was emitted from has now condensed into way more, I guess, mature astronomical structures. If you take it from the other point of view, people sitting where in this point of space now, and they've now moved 46 billion light years out, when they observe our region of space, they're not going to see us. They're not going to see Earth as it is now. They're going to see the region of space where Earth is at a super primitive stage, shortly after the Big Bang. And when I use words like \"shortly,\" I use that also loosely. We're still talking about hundreds of thousands or even millions of years. So we'll talk more about that in a future video. But the whole point of this video is it's beyond mind numbing. I would say the last video, about the Milky Way, that alone was mind numbing. But now, we're going in a reality where just the Milky Way becomes something that's almost unbelievably insignificant when you think about this picture right here. And the really mind numbing thing is, if someone told me that this is the entire universe, this by itself would certainly put things in perspective. But it's unknown what's beyond it. There's some estimates that this might be only be 1 times 10 to the 23rd of the entire universe. And it might even be the reality that the entire universe is smaller than this. And that's an interesting thing to think about. But I'll leave you there because I think no matter how you think about it, it's just-- I don't know. I actually, before doing this video, I stared at some of these photos for half an hour. This is my least productive day just because it's just so awe inspiring to think about what these dots and dots of the dots really are." + }, + { + "Q": "whut about the organisysms at the bottom of the ochin that do not depend on the sun", + "A": "It s a generalization that holds true 99.9% of the time. But you re right, at geothermal vents and deep in the earth there are bacteria independent of light.", + "video_name": "-rsYk4eCKnA", + "transcript": "Let's talk about one of the most important biological processes. Frankly, if this process didn't occur, we probably wouldn't have life on Earth, and I wouldn't be making this video for you, because there'd be no place for me to actually get food. And the process is called photosynthesis. And you're probably reasonably familiar with the idea. The whole idea is plants, and actually bacteria and algae and other things, but we normally associate it with plants. Let me make it in very simple terms. So we normally associate it with plants. And it's a process that plants use, and we might have learned this when we were very young. It's the process that plants use to take carbon dioxide plus some water plus some sunlight and turn it into some sugars or some maybe carbohydrates. Carbohydrates or sugars plus oxygen. Obviously, this has two very profound pieces to it for us as a living species. One, we need carbohydrates or we need sugars in order to fuel our bodies. You saw that in the cellular respiration videos. We generate all of our ATP by performing cellular respiration on glucose, which is essentially a byproduct, or a broken down carbohydrate. It's the simplest one for us to process in cellular respiration. And the second hugely important part is getting the oxygen. Once again, we need to breathe oxygen in order for us to break down glucose, in order to respire, in order to perform cellar respiration. So these two things are key for life, especially for life that breathes oxygen. So this process, other than the fact that it's interesting, that there are organisms around us, mostly plants, that are able to harness actual sunlight. You have these fusion reactions in the sun 93 million miles away, and it's releasing these photons, and some small subset of those photons reach the surface of Earth. They make their way through clouds and whatever else. And then these plants and bacteria and algae are able to harness that somehow and turn them into sugars that we can then eat or maybe the cow eats them and we eat the cow if we're not vegetarians, and we can then use that for energy. Not that the cow is all carbohydrates, but this is essentially what is used as the fuel or the energy for all of the other important compounds that we eat. This is where we get all of our fuel. So this is fuel for animals. Or you know, if you eat a potato directly, you are directly getting your carbohydrates. But anyway, this is a very simple notion of photosynthesis, but it's not incorrect. I mean, if you had to know one thing about photosynthesis, this would be it. But let's delve a little bit deeper and try to get into the guts of it and see if we can understand a little bit better how this actually happens. I find it amazing that somehow photons of sunlight are used to create these sugar molecules or these carbohydrates. So let's delve a little bit deeper. So we can write the general equation for photosynthesis. Well, I've almost written it here. But I'll write it a little bit more scientifically specific. You start off with some carbon dioxide. You add to that some water, and you add to that-- instead of sunlight, I'm going to say photons because these are what really do excite the electrons in the chlorophyll that go down, and you'll see this process probably in this video, and we'll go in more detail in the next few videos. But that excited electron goes to a high energy state, and as it goes to a lower energy state, we're able to harness that energy to produce ATPs, and you'll see NADPHs, and those are used to produce carbohydrates. But we'll see that in a little bit. But the overview of photosynthesis, you start off with these constituents, And then you end up with a carbohydrate. And a carbohydrate could be glucose, doesn't have to be glucose. So the general way we can write a carbohydrates is CH2O. And we'll put an n over here, that we could have n multiples of these, and normally, n will be at least three. In the case of glucose, n is 6. You have 6 carbons, 12 hydrogens and 6 oxygens. So this is a general term for carbohydrates, but you could have many multiples of that. You could have these long-chained carbohydrates, so you end up with a carbohydrate and then you end up with some oxygen. So this right here isn't so different than what I wrote up here in my first overview of how we always imagined photosynthesis in our heads. In order to make this equation balance-- let's see, I have n carbons so I need n carbons there. Let's see, I have two n hydrogens here. Two hydrogens and I have n there, so I need two n hydrogens here. So I'll put an n out there. And lets see how many oxygens. I have two n oxygens, plus another n, so I have three n oxygens. So let's see, I have one n, and you put an n here, and then I have two n, and I think this equation balances out. So this is a 30,000-foot view of what's going on in photosynthesis. But when you dig a little deeper, you'll see that this doesn't happen directly, that this happens through a bunch of steps that eventually gets us to the carbohydrate. So in general, we can break down photosynthesis. I'll rewrite the word. We can break down photosynthesis-- and we'll delve deeper into future videos, but I want to get you the overview first-- into two stages. We can call one the light reactions. Or sometimes they are called the light-dependent reactions, and that actually would probably be a better Let me write it like that. Light dependent means that they need light to occur. Light-dependent reactions. And then you have something called the dark reactions, and that's actually a bad name, because it also occurs in the light. Dark reactions, I wrote in a slightly darker color. And the reason why I said it's a bad name is because it still occurs in the light. But the reason why they probably called it the dark reaction is that you don't need light, or that part of photosynthesis isn't dependent on photons to occur. So a better term for it would have been light-independent reaction. So just to be clear, the light reactions actually need sunlight. They actually need photons for them to proceed. The dark reactions do not need photons for them to happen, although they do occur when the sun is out. They don't need those photons, but they need the byproducts from the light reaction to occur, so that's why it's called the light-independent reaction. They occur while the sun is out, but they don't need the sun. This needs the sun, so let me make it very clear. So this requires sunlight. This requires photons. And let me just make a very brief overview of this. This'll maybe let us start building a scaffold from which we can dig deeper. So the light reactions need photons, and then it needs water. So water goes into the light reactions and out of the other side of the light reactions. We end up with some molecular oxygen. So that's what happens in the light reactions, and I'm going to go much deeper into what actually occurs. And what the light the actions produce is ATP, which we know is the cellular or the biological currency of energy. It produces ATP and it produces NADPH. Now, when we studied cellular respiration, we saw the molecule NADH. NADPH is very similar. You just have this P there. You just have this phosphate group there, but they really perform similar mechanisms. That this agent right here, this molecule right here, is able to give away-- now let's think about what this means-- it's able to give away this hydrogen and the electron associated with this hydrogen. So if you give away an electron to someone else or someone else gains an electron, that something else is being reduced. Let me write that down. This is a good reminder. OIL RIG. Oxidation is losing an electron. Reduction is gaining an electron. Your charge is reduced when you gain an electron. It has a negative charge. So this is a reducing agent. It gets oxidized by losing the hydrogen and the electron with it. I have a whole discussion on the biological versus chemistry view of oxidation, but it's the same idea. When I lose a hydrogen, I also lose the ability to hog that hydrogen's electron. So this right here, when it reacts with other things, it's a reducing agent. It gives away this hydrogen and the electron associated with it, and so the other thing gets reduced. So this thing is a reducing agent. And what's useful about it is when this hydrogen, and especially the electron associated with that hydrogen, goes from the NADPH to, say, another molecule and goes to a lower energy state, that energy can also be used in the dark reactions. And we saw in cellular respiration the very similar molecule, NADH, that through the Kreb Cycle, or actually more importantly, that through the electron transport chain, was able to help produce ATP as it gave away its electrons and they went to lower energy states. But I don't want to confuse you too much. So the light reactions, you take in photons, you take in water, it spits out oxygen, and it spits out ATP and NADPH that can then be used in the dark reactions. And the dark reactions, for most plants we talk about, it's called the Calvin Cycle. And I'll go into a lot more detail of what actually occurs in the Calvin Cycle, but it takes in the ATP, the NADPH, and it produces-- it doesn't directly produce glucose. It produces-- oh, you probably saw this. You could call it PGAL. You could call it G3P. These all stand for-- let me write these down-- this is phosphoglyceraldehyde. My handwriting broke down. Or you could call it glyceraldehyde 3-phosphate. Same exact molecule. You can almost imagine it as-- this is a very gross oversimplification-- as three carbons with a phosphate group attached to it. But this can then be used to produce other carbohydrates, including glucose. If you have two of these, you can use those two to produce glucose. So let's just take a quick overview again because this is super important. I'm going to make videos on the light reactions and the dark reactions. Those will be the next two videos I make. So photosynthesis, you start with photons. All of these occur when the sun is out, but only the light reactions actually need the photons. The light reactions take photons-- we're going to go into more detail about what actually occurs-- and it takes in water. Oxygen gets spit out. ATP and NADPH get spit out, which are then used by the dark reaction, or the Calvin Cycle, or the light-independent reaction, because these still occur in the light. They just don't need photons. So they're the light-independent reaction. And it uses that in conjunction-- and we'll talk about other molecules that are used in conjunction. Oh, and I forgot a very important constituent of the dark reaction. It needs carbon dioxide. That's where you get your carbons to keep producing these phosphoglyceraldehydes, or glyceraldehyde 3-phosphate. So that's super important. It takes in the carbon dioxide, the products from the light reactions, and then uses that in the Calvin Cycle to produce this very simple building block of other And if you remember from glycolysis, you might remember that this PGAL molecule, or this G3P-- same thing-- this was actually the first product when we split glucose in two when we performed the glycolysis. So now we're going the other way. We're building glucose so that we can split it later for energy. So this is an overview of photosynthesis, and in the next couple of videos, I'm actually going to delve a little bit deeper and tell you about the light reactions and the dark reactions and how they actually occur." + }, + { + "Q": "I'm baffled as to why we keep using our original sample standard deviations as estimates for the population SDs (c. 6:50) once we're assuming the null hypothesis. If (and I might be barking up the wrong tree here) the hypothesis is that there's no meaningful difference whatsoever in weight loss effect between the two diets, why should their SDs remain distinct when imagined across the whole population? If the two groups' data are basically identical when viewed globally, shouldn't their SDs be identical too?", + "A": "because it would lead to same answer. if you sample twice from the same population then the best variance estimator is ((n1-1)var(x1) + (n2-1)var(x2))/(n1+n2-2) ... i know you understand which symbol means what here .. now calculate for variance of difference of means of two iid samples from this population using the just calculated estimate of variance. It is the same thing as what sal does", + "video_name": "N984XGLjQfs", + "transcript": "In the last video, we came up with a 95% confidence interval for the mean weight loss between the low-fat group and the control group. In this video, I actually want to do a hypothesis test, really to test if this data makes us believe that the low-fat diet actually does anything at all. And to do that let's set up our null and alternative hypotheses. So our null hypothesis should be that this low-fat diet does nothing. And if the low-fat diet does nothing, that means that the population mean on our low-fat diet minus the population mean on our control should be equal to zero. And this is a completely equivalent statement to saying that the mean of the sampling distribution of our low-fat diet minus the mean of the sampling distribution of our control should be equal to zero. And that's because we've seen this multiple times. The mean of your sampling distribution is going to be the same thing as your population mean. So this is the same thing is that. That is the same thing is that. Or, another way of saying it is, if we think about the mean of the distribution of the difference of the sample means, and we focused on this in the last video, that that should be equal to zero. Because this thing right over here is the same thing as that right over there. So that is our null hypothesis. And our alternative hypothesis, I'll write over here. It's just that it actually does do something. And let's say that it actually has an improvement. So that would mean that we have more weight loss. So if we have the mean of Group One, the population mean of Group One minus the population mean of Group Two should be greater then zero. So this is going to be a one tailed distribution. Or another way we can view it, is that the mean of the difference of the distributions, x1 minus x2 is going to be greater then zero. These are equivalent statements. Because we know that this is the same thing as this, which is the same thing as this, which is what I wrote right over here. Now, to do any type of hypothesis test, we have to decide on a level of significance. What we're going to do is, we're going to assume that our null hypothesis is correct. And then with that assumption that the null hypothesis is correct, we're going to see what is the probability of getting this sample data right over here. And if that probability is below some threshold, we will reject the null hypothesis in favor of the alternative hypothesis. Now, that probability threshold, and we've seen this before, is called the significance level, sometimes called alpha. And here, we're going to decide for a significance level of 95%. Or another way to think about it, assuming that the null hypothesis is correct, we want there to be no more than a 5% chance of getting this result here. Or no more than a 5% chance of incorrectly rejecting the null hypothesis when it is actually true. Or that would be a type one error. So if there's less than a 5% probability of this happening, we're going to reject the null hypothesis. Less than a 5% probability given the null hypothesis is true, then we're going to reject the null hypothesis in favor of the alternative. So let's think about this. So we have the null hypothesis. Let me draw a distribution over here. The null hypothesis says that the mean of the differences of the sampling distributions should be equal to zero. Now, in that situation, what is going to be our critical region here? Well, we need a result, so we're going to need some critical value here. Because this isn't a normalized normal distribution. But there's some critical value here. The hardest thing is statistics is getting the wording right. There's some critical value here that the probability of getting a sample from this distribution above that value is only 5%. So we just need to figure out what this critical value is. And if our value is larger than that critical value, then we can reject the null hypothesis. Because that means the probability of getting this is less than 5%. We could reject the null hypothesis and go with the alternative hypothesis. Remember, once again, we can use Z-scores, and we can assume this is a normal distribution because our sample size is large for either of those samples. We have a sample size of 100. And to figure that out, the first step, if we just look at a normalized normal distribution like this, what is your critical Z value? We're getting a result above that Z value, only has a 5% chance. So this is actually cumulative. So this whole area right over here is going to be 95% chance. We can just look at the Z table. We're looking for 95% percent. We're looking at the one tailed case. So let's look for 95%. This is the closest thing. We want to err on the side of being a little bit maybe to the right of this. So let's say 95.05 is pretty good. So that's 1.65. So this critical Z value is equal to 1.65. Or another way to view it is, this distance right here is going to be 1.65 standard deviations. I know my writing is really small. I'm just saying the standard deviation of that distribution. So what is the standard deviation of that We actually calculated it in the last video, and I'll recalculate it here. The standard deviation of our distribution of the difference of the sample means is going to be equal to the square root of the variance of our first population. Now, the variance of our first population, we don't know it. But we could estimate it with our sample standard deviation. If you take your sample standard deviation, 4.67 and you square it, you get your sample variance. And so this is the variance. This is our best estimate of the variance of the population. And we want to divide that by the sample size. And then plus our best estimate of the variance of the population of group two, which is 4.04 squared. The sample standard deviation of group two squared. That gives us variance divided by 100. I did before in the last. Maybe it's still sitting on my calculator. Yes, it's still sitting on the calculator. It's this quantity right up here. 4.67 squared divided by 100 plus 4.04 squared divided by 100. So it's 0.617. So this right here is going to be 0.617. So this distance right here, is going to be 1.65 times 0.617. So let's figure out what that is. So let's take 0.617 times 1.65. So it's 1.02. This distance right here is 1.02. So what this tells us is, if we assume that the diet actually does nothing, there's a only a 5% chance of having a difference between the means of these two samples to have a difference of more than 1.02. There's only a 5% chance of that. Well, the mean that we actually got is 1.91. So that's sitting out here someplace. So it definitely falls in this critical region. The probability of getting this, assuming that the null hypothesis is correct, is less than 5%. So it's smaller probability than our significance level. Actually, let me be very clear. The significance level, this alpha right here, needs to be 5%. Not the 95%. I think I might have said here. But I wrote down the wrong number there. I subtracted it from one by accident. Probably in my head. But anyway, the significance level is 5%. The probability given that the null hypothesis is true, the probability of getting the result that we got, the probability of getting that difference, is less than our significance level. It is less than 5%. So based on the rules that we set out for ourselves of having a significance level of 5%, we will reject the null hypothesis in favor of the alternative that the diet actually does make you lose more weight." + }, + { + "Q": "What exactly is the factorial exclamation point thingy?", + "A": "Before I begin, I feel obligated to say that you are probably the 5 gazillionth person to ask that. sigh Very well. That just is a short way of saying something like this: 6*5*4*3*2*1 Instead, you just say: 6! If that doesn t clear it up, then one isn t a prime number. (Logic people, please do not take advantage of that statement.)", + "video_name": "SbpoyXTpC84", + "transcript": "A card game using 36 unique cards, four suits, diamonds, hearts, clubs and spades-- this should be spades, not spaces-- with cards numbered from 1 to 9 in each suit. A hand is chosen. A hand is a collection of 9 cards, which can be sorted however the player chooses. Fair enough. How many 9 card hands are possible? So let's think about it. There are 36 unique cards-- and I won't worry about, you know, there's nine numbers in each suit, and there are four suits, 4 times 9 is 36. But let's just think of the cards as being 1 through 36, and we're going to pick nine of them. So at first we'll say, well look, I have nine slots in my hand, right? 1, 2, 3, 4, 5, 6, 7, 8, 9. I'm going to pick nine cards for my hand. And so for the very first card, how many possible cards can I pick from? Well, there's 36 unique cards, so for that first slot, there's 36. But then that's now part of my hand. Now for the second slot, how many will there be left to pick from? Well, I've already picked one, so there will only be 35 to pick from. And then for the third slot, 34, and then Then 33 to pick from, 32, 31, 30, 29, and 28. So you might want to say that there are 36 times 35, times 34, times 33, times 32, times 31, times 30, times 29, times 28 possible hands. Now, this would be true if order mattered. This would be true if I have card 15 here. Maybe I have a-- let me put it here-- maybe I have a 9 of spades here, and then I have a bunch of cards. And maybe I have-- and that's one hand. And then I have another. So then I have cards one, two, three, four, five, six, seven, eight. I have eight other cards. Or maybe another hand is I have the eight cards, 1, 2, 3, 4, 5, 6, 7, 8, and then I have the 9 of spades. If we were thinking of these as two different hands, because we have the exact same cards, but they're in different order, then what I just calculated would make a lot of sense, because we did it based on order. But they're telling us that the cards can be sorted however the player chooses, so order doesn't matter. So we're overcounting. We're counting all of the different ways that the same number of cards can be arranged. So in order to not overcount, we have to divide this by the ways in which nine cards can be rearranged. So we have to divide this by the way nine cards can be rearranged. So how many ways can nine cards be rearranged? If I have nine cards and I'm going to pick one of nine to be in the first slot, well, that means I have 9 ways to put something in the first slot. Then in the second slot, I have 8 ways of putting a card in the second slot, because I took one to put it in the first, so I have 8 left. Then 7, then 6, then 5, then 4, then 3, then 2, then 1. That last slot, there's only going to be 1 card left to put in it. So this number right here, where you take 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1, or 9-- you start with 9 and then you multiply it by every number less than 9. Every, I guess we could say, natural number less than 9. This is called 9 factorial, and you express it as an exclamation mark. So if we want to think about all of the different ways that we can have all of the different combinations for hands, this is the number of hands if we cared about the order, but then we want to divide by the number of ways we can order things so that we don't overcount. And this will be an answer and this will be the correct answer. Now this is a super, super duper large number. Let's figure out how large of a number this is. We have 36-- let me scroll to the left a little bit-- 36 times 35, times 34, times 33, times 32, times 31, times 30, times 29, times 28, divided by 9. Well, I can do it this way. I can put a parentheses-- divided by parentheses, 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1. Now, hopefully the calculator can handle this. And it gave us this number, 94,143,280. Let me put this on the side, so I can read it. So this number right here gives us 94,143,280. So that's the answer for this problem. That there are 94,143,280 possible 9 card hands in this situation. Now, we kind of just worked through it. We reasoned our way through it. There is a formula for this that does essentially the exact same thing. And the way that people denote this formula is to say, look, we have 36 things and we are going to choose 9 of them. And we don't care about order, so sometimes it'll be written as n choose k. Let me write it this way. So what did we do here? We have 36 things. We chose 9. So this numerator over here, this was 36 factorial. But 36 factorial would go all the way down to 27, 26, 25. It would just keep going. But we stopped only nine away from 36. So this is 36 factorial, so this part right here, that part right there, is not just 36 factorial. It's 36 factorial divided by 36, minus 9 factorial. What is 36 minus 9? It's 27. So 27 factorial-- so let's think about this-- 36 factorial, it'd be 36 times 35, you keep going all the way, times 28 times 27, going all the way down to 1. That is 36 factorial. Now what is 36 minus 9 factorial, that's 27 factorial. So if you divide by 27 factorial, 27 factorial is 27 times 26, all the way down to 1. Well, this and this are the exact same thing. This is 27 times 26, so that and that would cancel out. So if you do 36 divided by 36, minus 9 factorial, you just get the first, the largest nine terms of 36 factorial, which is exactly what we have over there. And then we divided it by 9 factorial. And this right here is called 36 choose 9. And sometimes you'll see this formula written like this, n choose k. And they'll write the formula as equal to n factorial over n minus k factorial, and also in the denominator, k factorial. And this is a general formula that if you have n things, and you want to find out all of the possible ways you can pick k things from those n things, and you don't care about the order. All you care is about which k things you picked, you don't care about the order in which you picked those k things. So that's what we did here." + }, + { + "Q": "By that high energy - low energy metaphor, what exactly does he mean?", + "A": "The bonds contain a large amount of energy, and when the bonds are broken, the energy stored in the bonds is released.", + "video_name": "PK6HmIe2EAg", + "transcript": "Sal: ATP or adenosine triposphate is often referred to as the currency of energy, or the energy store, adenosine, the energy store in biological systems. What I want to do in this video is get a better appreciation of why that is. Adenosine triposphate. At first this seems like a fairly complicated term, adenosine triphosphate, and even when we look at its molecular structure it seems quite involved, but if we break it down into its constituent parts it becomes a little bit more understandable and we'll begin to appreciate why, how it is a store of energy in biological systems. The first part is to break down this molecule between the part that is adenosine and the part that is the triphosphates, or the three phosphoryl groups. The adenosine is this part of the molecule, let me do it in that same color. This part right over here is adenosine, and it's an adenine connected to a ribose right over there, that's the adenosine part. And then you have three phosphoryl groups, and when they break off they can turn into a phosphate. The triphosphate part you have, triphosphate, you have one phosphoryl group, two phosphoryl groups, two phosphoryl groups and three phosphoryl groups. One way that you can conceptualize this molecule which will make it a little bit easier to understand how it's a store of energy in biological systems is to represent this whole adenosine group, let's just represent that as an A. Actually let's make that an Ad. Then let's just show it bonded to the three phosphoryl groups. I'll make those with a P and a circle around it. You can do it like that, or sometimes you'll see it actually depicted, instead of just drawing these straight horizontal lines you'll see it depicted with essentially higher energy bonds. You'll see something like that to show that these bonds have a lot of energy. But I'll just do it this way for the sake of this video. These are high energy bonds. What does that mean, what does that mean that these are high energy bonds? It means that the electrons in this bond are in a high energy state, and if somehow this bond could be broken these electrons are going to go into a more comfortable state, into a lower energy state. As they go from a higher energy state into a lower, more comfortable energy state they are going to release energy. One way to think about it is if I'm in a plane and I'm about to jump out I'm at a high energy state, I have a high potential energy. I just have to do a little thing and I'm going to fall through, I'm going to fall down, and as I fall down I can release energy. There will be friction with the air, or eventually when I hit the ground that will release energy. I can compress a spring or I can move a turbine, or who knows what I can do. But then when I'm sitting on my couch I'm in a low energy, I'm comfortable. It's not obvious how I could go to a lower energy state. I guess I could fall asleep or something like that. These metaphors break down at some point. That's one way to think about what's going on here. The electrons in this bond, if you can give them just the right circumstances they can come out of that bond and go into a lower energy state and release energy. One way to think about it, you start with ATP, adenosine triphosphate. And one possibility, you put it in the presence of water and then hydrolysis will take place, and what you're going to end up with is one of these things are going to be essentially, one of these phosphoryl groups are going to be popped off and turn into a phosphate molecule. You're going to have adenosine, since you don't have three phosphoryl groups anymore, you're only going to have two phosphoryl groups, you're going to have adenosine diphosphate, often known as ADP. Let me write this down. This is ATP, this is ATP right over here. And this right over here is ADP, di for two, two phosphoryl groups, adenosine diphosphate. Then this one got plucked off, this one gets plucked off or it pops off and it's now bonded to the oxygen and one of the hydrogens from the water molecule. Then you can have another hydrogen proton. The really important part of this I have not drawn yet, the really important part of it, as the electrons in this bond right over here go into a lower energy state they are going to release energy. So plus, plus energy. Here, this side of the reaction, energy released, energy released. And this side of the interaction you see energy, energy stored. As you study biochemistry you will see time and time again energy being used in order to go from ADP and a phosphate to ADP, so that stores the energy. You'll see that in things like photosynthesis where you use light energy to essentially, eventually get to a point where this P is put back on, using energy putting this P back on to the ADP to get ATP. Then you'll see when biological systems need to use energy that they'll use the ATP and essentially hydrolysis will take place and they'll release that energy. Sometimes that energy could be used just to generate heat, and sometimes it can be used to actually forward some other reaction or change the confirmation of a protein somehow, whatever might be the case." + }, + { + "Q": "what is a covalent bond?", + "A": "Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability.", + "video_name": "T2DaaGuKOTo", + "transcript": "Hank: Hello, I'm Hank. I assume that you are here because you are interested in biology. If you are, that makes sense, because like any good 50 Cent song, Biology is just about sex and not dying, and everyone watching this should be interested in sex and not dying, being that you are, I assume, a human being. I'm gonna teach this biology course a little differently than most courses you've ever experienced. For example, I'm not going to spend the first class talking about how I'm going to teach the class. I'm just going to start teaching the class. Starting right after this next cut. First, I just wanted to say if I'm going to fast for you, the great thing about me being a video and not a person is that you can always go back and listen to what I've said again. I promise I will not mind. You are encouraged to do this often. A great professor of mine once told me that in order to understand any topic, you only really need to understand a bit of the level of complexity just below that topic. The level of complexity just below biology is chemistry, or if you're a biochemist, you would probably argue that it's biochemistry, so we need to know a little bit more about chemistry, and that is where we're gonna start. (lively intro music) I'm a collection of organic compounds called Hank Green. An organic compound is more or less any chemical that contains carbon, and carbon is awesome. Why? Lots of reasons. I'm gonna give you three. First, carbon is small. It doesn't have that many protons and neutrons. Almost always 12, rarely it has some extra neutrons making it C-13 or C-14. Because of that, carbon does not take up a lot of space and can form itself into elegant shapes. It can form rings. It can form double or even triple bonds. It can form spirals and sheets and all kinds of really awesome things that bigger molecules would never manage to do. Basically, carbon is like an olympic gymnast. It can only do the remarkable and beautiful things it can do because it's petite. Second, carbon is kind. It's not like other elements that desperately want to gain or lose or share electrons to get the exact number they want. No, carbon knows what it's like to be lonely, so it's not all, \"I can't live without your electrons.\" Needy, like chlorine or sodium is. This is why chlorine tears apart your insides if you breathe it in gaseous form, and why sodium metal, if ingested, will explode. Carbon, though, eh. It wants more electrons, but it's not going to kill for them. It's easy to work with. It makes and breaks bonds like a 13-year-old mall rat, but it doesn't ever really hold a grudge. Third, carbon loves to bond because it needs 4 extra electrons, so it will bond with whoever happens to be nearby. Usually, it will bond with 2 or 3 or 4 of them at the same time. Carbon can bond with lots of different elements. Hydrogen, oxygen, phosphorus, nitrogen, and other atoms of carbon. It can do this in infinite configurations, allowing it to be the core element of the complicated structures that make living things like ourselves. Because carbon is small, kind, and loves to bond, life is pretty much built around it. Carbon is the foundation of biology. So fundamental that scientists have a hard time even conceiving of life that is not carbon-based. Silicon, which is analogous to carbon in many ways, is often cited as a potential element for alien life to be based on, but it's bulkier, so it doesn't form the same elegant shapes as carbon. It's also not found in any gases, meaning that life would have to be formed by eating solid silicon, whereas life here on earth is only possible because carbon is constantly floating around in the air in the form of carbon dioxide. Carbon, on its own, is an atom with 6 protons, 6 electrons, and 6 neutrons. Atoms have electron shells, and they need or want to have these shells filled, in order to be happy, fulfilled atoms. The first electron shell called the S-orbital needs 2 electrons to be full. Then there's the 2nd S-orbital, which also needs 2, carbon has this filled as well. Then we have the first P-orbital, which needs 6 to be full. Carbon only has 2 left over, so it wants 4 more. Carbon forms a lot of bonds that we call \"covalent\". These are bonds where the atoms actually share electrons, so the simplest carbon compound ever, methane, is carbon sharing 4 electrons with 4 hydrogen atoms. Hydrogen only has 1 electron, so it wants its first S-orbital full. Carbon shares its 4 electrons with those 4 hydrogens, and those 4 hydrogens each share 1 electron with carbon, This can all be represented with what we call Lewis dot structures. Gilbert Lewis, also the guy behind Lewis acids and bases, was nominated for the Nobel Prize 35 times and won none. This is more nominations than anyone else in history, and roughly the same number of wins as everyone else. Lewis disliked this a great deal. He may have been the most influential chemist of his time. He coined the term photon. He revolutionized how we think about acids and bases. He produced the first the first molecule of heavy water, and he was the first person to conceptualize the covalent bond that we're talking about right now. But, he was extremely difficult to work with. He was forced to resign from many important posts, and was also passed up for the Manhattan Project, so while all of his colleagues worked to save his country, Lewis wrote a horrible novel. Lewis died alone in his laboratory while working on cyanide compounds after having had lunch with a younger, more charismatic colleague who had won the Nobel prize and worked on the Manhattan Project. Many suspect that he killed himself with the cyanide compounds that he was working on, but the medical examiner said heart attack without really looking into it. I told you all that because, well, the little Lewis structure that I'm about to show you was created by a deeply troubled genius. It's not some abstract scientific thing that has always existed. Someone, somewhere, thought it up, and it was such a marvelously useful tool, that we've been using it ever since. In biology, most compounds can be shown in Lewis structure form. One of the rules of thumb when making these diagrams is that some elements tend to react with each other in such a way that each atom ends up with 8 electrons in its outermost shell. That's called the octet rule, because these atoms want to complete their octets of electrons to be happy and satisfied. Oxygen has 6 electrons in its outer shell, and needs 2, which is why we get H2O. It can also bond with carbon, which needs 4, so 2 double bonds to 2 different oxygen atoms, you end up with CO2, that pesky global warming gas, and also the stuff that plants and, thus, all life are made of. Nitrogen has 5 electrons in its outer shell. Here's how we count them. There are four placeholders. Each wants two atoms, and like people getting on a bus, they prefer to start out not sitting next to each other. I'm not kidding about this. They really don't double up until they have to. We count it out. 1, 2, 3, 4, 5. So, for maximum happiness, nitrogen bonds with 3 hydrogens, forming ammonia, or with 2 hydrogens, sticking off another group of atoms which we call an amino group. And if that amino group is bonded to a carbon that is bonded to a carboxylic acid group, you have an amino acid. Sometimes electrons are shared equally within a covalent bond like with O2. That's called a non-polar covalent bond, but often one of the participants is more greedy. In water, for example, the oxygen molecule sucks the electrons in, and they spend more time around the oxygen than around the hydrogens. This creates a slight positive charge around the hydrogens and a slight negative charge around the oxygen. When something has a charge, we say that it's polar. It has a positive and negative pole. This is a polar covalent bond. Ionic bonds occur when instead of sharing electrons, atoms just donate or accept an electron from another atom completely and then live happily as a charged atom or ion. Atoms would, in general, prefer to be neutral but compared with having the full electron shells is not that big of a deal. The most common ionic compound in our daily lives? that would be good old table salt, NaCl, sodium chloride, but don't be fooled by its deliciousness. Sodium chloride, as I previously mentioned, is made of 2 very nasty elements. Chlorine is a halogen, or an element that only needs one proton to fill its octet, while sodium is an alkali metal, an element that only has one electron in its octet. They will happily tear apart any chemical compound they come in contact with, searching to satisfy the octet rule. No better outcome could occur than sodium meeting chlorine. They immediately transfer electrons so sodium doesn't have its extra and the chlorine fills its octet. They become Na+ and Cl-, and are so charged that they stick together, and that stickiness is what we call an ionic bond. These chemical changes are a big deal, remember? Sodium and chlorine just went from being deadly to being delicious. They're also hydrogen bonds, which aren't really bonds, so much. So, you remember water? I hope you didn't forget about water. Water is important. Since water is stuck together with a polar covalent bond, the hydrogen bit of it is a little bit positively-charged and the oxygen is a little negatively-charged. When water molecules move around, they actually stick together a little bit, hydrogen side to oxygen side. This kind of bonding happens in all sorts of molecules, particularly in proteins. It plays an extremely important role in how proteins fold up to do their jobs. bonds, even when they're written with dashes or solid lines, or no lines at all, are not the same strength. Sometimes ionic bonds are stronger than covalent bonds, though that's the exception rather than the rule, and covalent bond strength varies hugely. The way that those bonds get made and broken is intensely important to how life and our lives operate. Making and breaking bonds is the key to life itself. It's also like if you were to swallow some sodium metal, the key to death. Keep all of this in mind as you move forward in biology. Even the hottest person you have ever met is just a bunch of chemicals rambling around in a bag of water. That, among many other things, is what we're gonna talk about next time." + }, + { + "Q": "i do not understand 'jerk'. please explain", + "A": "Jerk, also known as jolt, surge, or lurch, is the rate of change of acceleration; that is, the derivative of acceleration with respect to time, and as such the second derivative of velocity, or the third derivative of position.", + "video_name": "DD58B2siDv0", + "transcript": "- [Instructor] All right, I wanna talk to you about acceleration versus time graphs because as far as motion graphs go, these are probably the hardest. One reason is because acceleration just naturally is an abstract concept for a lot of people to deal with and now it's a graph and people don't like graphs either particularly often times. Another reason is, if you wanted to know the motion of the object, let's say it was this doggie. This is my doggie Daisy. Let's say Daisy was accelerating. If you wanted to know the velocity that daisy had, you can't figure it out directly from this graph unless you have some extra information. You have to know information about the velocity Daisy had at some moment in order to figure out from this graph the velocity Daisy had at some other moment. So, what can this graph tell you about the motion of Daisy? Well, let's say this graph described Daisy's acceleration. So Daisy can be accelerating. Maybe we're playing catch. We'll give her a ball. We'll throw the ball. Hopefully she actually lets go and she brings it back. This graph is gonna represent her acceleration. So this graph, we just read it, it says that Daisy had two meters per second squared of acceleration for the first four seconds and then her acceleration dropped to zero at six seconds and then her acceleration came negative until it was negative three at nine seconds. But, from this we can't tell if she's speeding up or slowing down. what can we figure out? Well, we can figure out some stuff because acceleration is related to velocity and we can figure out how it's related to velocity by remembering that it is defined to be the change in velocity over the change in time. So this is how we make our link to velocity. So if we solve this for delta v, we get that the delta v, the change in velocity over some time interval, will be the acceleration during that time interval times interval itself, how long did that take. This is the key to relating this graph to velocity. In other words, let's consider this first four seconds. Let's go between zero and four seconds. Daisy had an acceleration of two meters per second squared. So that means, well, two was the acceleration meters per second squared, times the accel, times the time, excuse me, the time was four seconds. So there was four seconds worth of acceleration. You get positive eight. What are the units? This second cancels with that second. You get positive eight meters per second. So the change in velocity for the first four seconds was positive eight. This isn't the velocity. It's the change in velocity. How would you ever find that for this diagonal region. This is as problem. Look at this. If I wanted to find, let's say the velocity at six seconds, well the acceleration at this point is two but then the acceleration at this point is one. The acceleration at this point is zero. That acceleration we keep changing. How would I ever figure this out? What acceleration would I plug in during this portion? But we're in luck. This formula allows us to say something really important. A geometric aspect of these graphs that are gonna make our life easier and the way it makes our life easier is that, look at what this is. This is saying acceleration times delta t, but look it. The acceleration we plug in was this, two. So for the first four seconds, the acceleration was two. The time, delta t, was four. We took this two multiplied by that four and got a number, positive eight, but this is a height times the width. If you take height times width, that just represents the area of a rectangle. So all we found was the area of this rectangle. The area is giving us our delta v because area, right, of a rectangle is height times width. We know that the height is gonna represent the acceleration here and the width is gonna represent delta t. Just by the definition of acceleration we arranged, we know that a times delta t has to just be the change in velocity. So area and change in velocity are representing the exact same thing on this graph. Area is the change in velocity. That's gonna be really useful because when you come over to here the area is still gonna be the change in velocity. That's useful because I know how to easily find the area of a triangle. The area of a triangle is just 1/2 base times height. I don't easily know how to deal with an acceleration that's varying within this formula but I do know how to find the area. For instance the area here, though I have 1/2, the base is two seconds, the height is gonna be positive two meters per second squared. What are we gonna get? One of the halves, cancel. Well, the half cancels one of the twos and I'm gonna get that this is gonna be equal to two meters per second. That's gonna be the area that represents the change in velocity. So Daisy's velocity changed by two meters per second during this time. Now you might object. You might say, \"Wait a minute. \"I'll buy this over here because height times width \"is just a times delta t, \"but triangle, that has an extra factor of a half in it, \"and there's no half up here. \"How does this, I mean, how can we still make this claim?\" We can make this claim because we'll do the same thing we always do. We can imagine, all right, imagine a rectangle here. We're gonna estimate the area with a bunch of rectangles. Then this rectangle, and this rectangle in your line like that looks horrible. That doesn't look like the area of a triangle at all. It's got all these extra pieces right here, right? You don't want all of that. And okay, I agree. That didn't work so well. Let's make them even smaller, right? Smaller width. So we'll do a rectangle like that. We'll do this one. You see we're getting better. This is definitely closer. This is not as bad as the other one but it's still not exact. And I agree, that is not exact so we'll make it even smaller rectangle and an even smaller rectangle here all of these at the same width but they're even smaller than the ones before. Now we're getting really close. This area is really gonna get close to the area of the triangle. The point is if you make them infinite testable small, they'll exactly represent the area of a triangle. Each one of them can be found with this formula. The delta v for each one will be the area, or sorry, the acceleration of the height of that rectangle times the small infinite testable width and you'll get the total delta v which is so gonna be the total area. Long story short, area on a, acceleration versus time graphs represents the change in velocity. This is one you got to remember. this is the most important aspect of an acceleration graph, oftentimes the most useful aspect of it, the way you analyze it. So why do we care about change in velocity? Because it will allow us to find the velocity. We just need to know the velocity at one point then we can find the velocity at any other point. For instance, let's say I gave you the velocity Daisy had. For some reason I'm gonna stopwatch. I start my stopwatch at right at that moment. At t equals zero, Daisy had a velocity of, let's say positive one meter per second. So Daisy was traveling that fast at t equals zero. That was her velocity at t equals zero seconds. Now I can get the velocity wherever I want. If I want the velocity at four, let's figure this out. To get the velocity at four, I can say that the delta v during this time period right here, this four seconds. I know what that delta v was. That delta v was positive eight. We found that area, height times width. So positive eight is what the delta v is gotta equal. What's delta v? That's v at four seconds minus v at zero seconds. That's gotta be positive eight. I know what v at zero second was. That was one. So we can get that v at four minus one meter per second is equal to positive eight meters per second. So I get the velocity at four was positive nine meters per second. And you're like, phew, that was hard. I don't wanna do that every time. Yeah, I wouldn't wanna do that every time either so there's a quick way to do it. We can just do this. What's the velocity we had to start with? That was one. What was our change in velocity? That was positive eight. So what's our final velocity? Well, one plus eight gives us our final velocity. It's positive nine. Well it's just gonna take this change in velocity of this area which represents the change in velocity which is gonna add our initial velocity to it when we solve for this final velocity. for instance, if I didn't make sense, for instance, if we want to find the velocity at six, well, we can just say we started at t equals four seconds with a velocity of positive nine. We start here with positive nine. Our change was positive two so we're gonna end with positive 11 meters per second. You might object. You might say, \"Wait a minute, hold on now. \"If we want delta v, \"right, and that's positive two, \"shouldn't delta v be the whole thing \"from like zero to six seconds? \"Shouldn't I say v at six seconds minus v at zero \"is positive two meters per second?\" I can't do that. The reason I can't do that is because look at what I did on the left hand side, my time interval goes from zero to six but on the right hand side, I only included the area from four to six. That's the area, there's a yellow triangle right here. If I wanted to put six and zero on this left hand side, I could do that but from my total area, I wouldn't use that. I have to use the total area. In other words, the total are from zero all the way to six because that's what I define on this side. These sides have to agree with each other. So from zero to six, my total area would be, this area here was eight, right? We found that rectangle was eight. This area here was two. So my total area would be 10. I can do that if I want. I could say v at six minus v at zero was, well v at zero we said was one because I just gave you that, equals 10 meters per second. I get that the v at six would be 11 meters per second just like we got it before. So you can still do it mathematically like this but make sure your time intervals agree on those sides. Now let's do the last part here. So we can find this area. This area and the area always represents the area from the curve to the horizontal axis. So in this case it's below the horizontal axis. That means it can negative area. The reason is it's a triangle again. So 1/2 base times height. So 1/2, the base is one, two, three seconds. The height is negative three, negative now, negative three meters per second squared. I get that the total area is gonna be negative 4.5 meters per second. All right, now Daisy's gonna have a change in velocity of negative 4.5. If we want to get the velocity at nine, there's a few ways we can do it. Right, just conceptually, we can say that Daisy started at six with a velocity of 11. Her change during this period was negative 4.5. If you just add the two, you add the change to the value she started with. Well you're gonna get positive 6.5 if I add 11 and negative 4.5 meters per second or, if that sounded like mathematical witchcraft, you can say that, all right, delta v equals, what, negative 4.5 meters per second. Delta v would be, all right, you gotta be careful, this negative 4.5 represents this triangle so it's gotta be the delta v between six and nine. So v at nine minus v at six has to be negative 4.5 meters per second. V at nine minus the v at six we know, v at six was 11. So I've got minus 11 meters per second equals negative 4.5. Wow, we ran out of room. V at nine would be negative 4.5 plus 11. That's what we did up here. We got that it was just 6.5 meters per second and that agrees with what we said earlier. So finding the area can get you the change in velocity and then knowing the velocity at one unknown at a time can get you the velocity at any other moment in time. Just be careful. Make sure you're associating the right time interval on both the length and the right side. They have to agree. One more thing before you go. The slope on these graphs often represents something meaningful. That's the same in this graph. So the slope of this graph, let's try to interpret what this means. The slope on an acceleration versus time graph. Well the slope is always represented as the rise over the run and the rise is y two minus y one over x two minus x one except instead of y and x, we have a and t. So we're gonna have a two minus a one over t two minus t one. This is gonna be delta a, the change in a over the change in time. What is that? It's the rate of change of the acceleration. That is even one more layer removed from what we're used dealing with, right? Velocity, velocity is the change in position with respect to time. Acceleration is the change in velocity with respect to time. Now we're saying that the something is the change in acceleration with respect to time. What is it? It's the jerk. So this is often called the jerk. That's the name of it. It's not used all that often. It's quite honestly not the most useful motion variable you'll ever meet and you won't get asked for that often most likely on test and whatnot but it has its application sometimes that exist and it has a name that's called the jerk. So recapping, the area, the important fact here is that the area under acceleration versus time graphs gives you the change in velocity. Once you know the velocity at one point, you could find the velocity at any other point. The slope of an acceleration versus time graph gives you the jerk." + }, + { + "Q": "The following is also cool, and I wonder why he didn't mention it in the video. Since\ne^(i*pi) = -1\n, and since\ni^2 = -1\n, then\ne^(i*pi) = i^2\n, so,\ni = sqrt(e^(i*pi))\n. Hence another statement for what i really means.", + "A": "That s just Euler s formula with \u00ce\u00b8 = \u00cf\u0080/2, so Sal kind of did talk about it.", + "video_name": "mgNtPOgFje0", + "transcript": "Voiceover: In the last video, we took the Maclaurin expansion of E to the X and we saw that it looked like it was some type a combination of the polynomial approximations of cosine of X and of sine of X. But it's not quite, because there's a couple of negatives in there. If we were to add these two together that we did not have when we took the representation of E to the X. But to reconcile these, I'll do a little bit of a I don't know if you can even call it a trick. Let's see if we take this polynomial expansion of E to the X, this approximation, what happens if we say E to the X is equal to this, especially as this becomes an infinite number of terms and becomes less of an approximation and more of an equality. What happens if I take E to the IX and before that might have been kind of a weird thing to do. Let me write it down, E to the IX. Because before, I said how do you define E to the Ith power? That's a very bizarre thing to do, take something to the XI power. How do you even comprehend some type of a function like that. But now that we can have a polynomial expansion of E to the X, we can maybe make sense of it. Because we can take I to different amounts to different powers and we know what that gives, I squared is negative one, I to the third is negative I, and so on and so forth. So what happens if we take E to the IX. So once again, just like taking the X up here and replacing it with an IX, so every where we see the X in it's polynomial approximation, we would write an IX. So let's do that. So E to the IX should be approximately equal to, and it will become more and more equal, and this is more to give you an intuition I'm not doing a rigorous proof here, but it's still profound. Not to oversell it, but I don't think I can oversell what is about to be discovered or seen in this video, it would be equal to one plus instead of an X will have an IX, plus IX, plus, so what's IX squared? So let me write this down. What is IX squared over two factorial? Well I squared is going to be negative one, and then you'd have X squared over two factorials. It's going to be minus X squared over two factorial, I think you might see where this is going to go. And then what is IX? Remember everywhere we saw an X, we're going to replace it with an IX. So what is IX to the third power? Actually let me write this out, let me not skip some steps over here. So this is going to be IX squared over two factorial, actually let me, I want to do it just the way. So plus IX squared over two factorial, plus IX to the third over three factorial, plus IX to the fourth over four factorial, and we can keep going, plus IX to the fifth over five factorial, and we can just keep going so on and so forth. Let's evaluate if these IXs raised to these different powers. So this will be equal to one plus IX, IX squared, that's the same thing as I squared times X squared, I squared is negative one. So this is negative X squared over two factorial, and this is going to be the same thing as I to the third times X to the third. I to the third is the same thing as I squared times I, so it's going to be negative I. So this is going to be minus I times X to the third over three factorial. So then plus, you're going to have, what's I to the fourth power? So that's I squared squared. So that's negative one squared, that's just So I to the fourth is one and then you have X to the fourth. Plus X to the fourth over four factorial. And then you're going to have, I don't even write the plus yet, I to the fifth. So I to the fifth is going to be one times I, so it's going to be I times X to the fifth over five factorials plus I times X to the fifth over five factorial. I think you might see a pattern here. Coefficient is one, then I, then negative one, then negative I, then one, then I, then negative one, X to the sixth over six factorial, and then negative IX to the seventh over seven factorial. So we have some terms, some of them are imaginary, they're being multiplied by I. Some of them are real. Why don't we separate them out? So once again, E to the IX, is going to be equal to this thing, especially as we add an infinite number of terms. Let's separate out the real and the non-real terms. Or the real and the imaginary terms I should say. So this is real, this is real, this is real, and this right over here is real. And obviously we could keep going on with that. So the real terms here are one minus X squared over two factorial plus X to the fourth over four factorial, you might be getting excited now, minus X to the sixth over six factorial and that's all I've done here, but they would keep going so plus so on and so forth. So that's all of the real terms. And what are the imaginary terms here. And I'll just factor out the I over here. Actually, let me just factor out. So it's going to be plus I times, well this is IX, so this will be X. And then the next, so that's an imaginary term, this is an imaginary term. We're factoring out the I, so minus X to the third over three factorial, and then the next imaginary term is right over there. We factor out the I, plus X to the fifth over five factorial and then the next imaginary term is right there, we factored out the I. So it's minus X to the seventh over seven factorial. And then we would obviously keep going. So plus minus keep going, so on and so forth, preferably to infinite so that we get as good of a approximation as possible. So we have a situation where E to the IX is equal to all of this business here. But you probably remember from the last two videos the real part, this was the polynomial, this was the Maclaurin approximation of cosine of X around I should say the Taylor approximation around zero or you could also call it the Maclaurin approximation. So this and this are the same thing. So this is cosine of X, especially when you added an infinite power of terms, cosine of X. This over here is sine of X, the exact same thing. So it looks like we're able to reconcile how you can add up cosine of X and sine of X to get something that's like E of the X. This right here is sine of X. And so if we take it for granted, I'm not rigorously proving it to you, and if you were to take an infinite number of terms here that this will essentially become cosine of X and if you take an infinite number of terms here this will become sine of X. It leads to a fascinating formula. We could say that E to the IX, is the same thing as cosine of X, and you should be getting goose pimples right around now. is equal to cosine of X, plus I times sine of X, This is Euler's Formula. And this right here is Euler's Formula. And if that by itself isn't exciting and crazy enough for you, because is really should be. Because we've already done some pretty cool things. We're involving E, which we get from continuous compounding interest. We have cosine and sine of X, which are ratios of right triangles, it comes out of the unit circle. And some how we've thrown in the square root of negative one. There seems to be this cool relationship here. But it becomes extra cool, and we're going to assume we're operating in radians here. If we assume Euler's Formula, what happens when X is equal to pi? Just to throw in another wacky number in there. The ratio between the circumference and the diameter of a cirle. What happens when we throw in pi? We get E to the Ipi is equal to cosine of pi. Cosine of pi is what? Pi is half way around the unit circle, so cosine of pi is negative one and then sine of pi is zero, so this term goes away. So if you evaluated at pi you get something amazing, it's called Euler's Identity. I always have trouble pronouncing Euler's. Euler's Identity, which we could write like this, or we could add one to both sides and we could write it like this. And I'll write it in different colors for emphasis. E to the I times pi plus one is equal to , I'll do that in a neutral color, is equal to, I'm just adding one to both sides of this thing right over here, is equal to zero. And this is thought provoking. I mean here we have, this tells you that there's some connectedness to the universe that we don't fully understand or at least I don't fully understand. I is defined by engineers for simplicity so that they can find the roots of all sorts of polynomials. As you could say the square root of negative one. Pi is the ratio between the circumference of a circle and it's diameter. Once again another interesting number but seems like it comes from a different place as I. E comes from a bunch of different places. E you could either think of it comes out of continuing compounding interest, super valuable for finance. It also comes from the notion that the derivative of E to the X is also E to the X, another fascinating number. But once again seemingly unrelated to how we came up with I and seemingly unrelated with how we came up with pi. And then of course you have some of the most profound basic numbers right over here. You have one, I don't have to explain why one is a cool number. And I shouldn't have to explain why zero is a cool number. And so this right here connects all of these fundamental numbers in some mystical way that shows that there's some connectedness to the universe. So frankly, if this does not blow your mind, you really have no emotion." + }, + { + "Q": "Um, slightly confused here. It says Newtons first law is the law of motion but there are three laws of motion. Basicaly what I'm asking is which law is the first law or is Newtons first law all three laws of motions. Am I just being stupid or have I misunderstood something?\nHELP!", + "A": "Yep. There are 3 laws of motion. I m not sure is the 1st law should be called THE law of motion, since the three are importantand connected to each other.", + "video_name": "D1NubiWCpQg", + "transcript": "Now that we know a little bit about Newton's First Law, let's give ourselves a little quiz. And what I want you to do is figure out which of these statements are actually true. And our first statement is, \"If the net force on a body is zero, its velocity will not change.\" Interesting. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" Also an interesting statement. Statement number three, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" And statement four, \"An unbalanced force on an object will always change the object's direction.\" So I'll let you think about that. So let's think about these statement by statement. So our first statement right over here, \"If the net force on a body is zero, its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. An unbalanced force on a body will always impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope, they'll start going in whatever direction they were traveling in when they let go. They'll keep going on in that direction. And if we assume very, very, very small frictions from the ice skating rink, they'll actually have the same speed. So the force, the inward force, the tension from the rope pulling on the skater in this situation, would have only changed the skater's direction. So and unbalanced force doesn't necessarily have to impact the object's speed. It often does. But in that situation, it would have only impacted the skater's direction. Another situation like this-- and once again, this involves centripetal acceleration, inward forces, inward acceleration-- is a satellite in orbit, or any type of thing in orbit. So if that is some type of planet, and this is one of the planet's moons right over here, the reason why it stays in orbit is because the pull of gravity keeps making the object change its direction, but not its speed. Its speed is the exact right speed. So this was its speed right here. If the planet wasn't there, it would just keep going on in that direction forever and forever. But the planet right over here, there's an inward force of gravity. And we'll talk more about the force of gravity in the future. But this inward force of gravity is going to accelerate this object inwards while it travels. And so after some period of time, this object's velocity vector-- if you add the previous velocity with how much it's changed its new velocity vector. Now this is after its traveled a little bit-- its new velocity vector might look something like this. And it's traveling at the exact right speed so that the force of gravity is always at a right angle to its actual trajectory. It's the exact right speed so it doesn't go off into deep space and so it doesn't plummet into the earth. And we'll cover that in much more detail. But the simple answer is, unbalanced force on a body will always impact its velocity. It could be its speed, its direction, or both, but it doesn't have to be both. It could be just the speed or just the direction. So this is an incorrect statement. Now the third statement, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" This is absolutely true. And this is the example we gave. If I take an object, if I take my book and I try to slide it across the desk, the reason why it eventually comes to stop is because we have the unbalanced force of friction-- the grinding of the surface of the book with the grinding of the table. If I'm inside of a pool or even if there's absolutely no current in the pool, and if I were to try to push some type of object inside the water, it eventually comes to stop because of all of the resistance of the water itself. It's providing an unbalanced force in a direction opposite it's motion. That is what's slowing it down. So in our everyday life, the reason why we don't see these things go on and on forever is that we have these frictions, these air resistants, or the friction with actual surfaces. And then the last statement, \"An unbalanced force on an object will always change the object's direction.\" Well, this one actually is maybe the most intuitive. We always have this situation. Let's say I have a block right over here, and it's traveling with some velocity in that direction-- five meters per second. If I apply an unbalanced force in that same direction-- so that's my force right over there. If I apply it in that same direction, I'm just going to accelerate it in that same direction. So I won't necessarily change it. Even if I were to act against it, I might decelerate it, but I won't necessarily change its direction. I could change its direction by doing something like this, but I don't necessarily. I'm not always necessarily changing the object's direction. So this is not true. An unbalanced force on an object will not always change the object's direction. It can, like these circumstances, but not always. So \"always\" is what makes this very, very, very wrong." + }, + { + "Q": "Could someone please help me understand from 1:59 to 2:13 better. If we used the product rule, then (sin)(2x) will be cos(2x)+ 2sin. I am really lost.", + "A": "You may be accustomed to your instructor using more parentheses. Note that the function in question is meant to be sin(2x), NOT sin(x) (2x). This function is a composition function (double x, then apply sin). As such, you should use the chain rule, not the product rule.", + "video_name": "BiVOC3WocXs", + "transcript": "Let's say we need to evaluate the limit as x approaches 0 of 2 sine of x minus sine of 2x, all of that over x minus sine of x. Now, the first thing that I always try to do when I first see a limit problem is hey, what happens if I just try to evaluate this function at x is equal to 0? Maybe nothing crazy happens. So let's just try it out. If we try to do x equals 0, what happens? We get 2 sine of 0, which is 0. Minus sine of 2 times 0. Well, that's going to be sine of 0 again, which is 0. So our numerator is going to be equal to 0. Sine of 0, that's 0. And then we have another sine of 0 there. That's another 0, so all 0's. And our denominator, we're going to have a 0 minus sine of 0. Well that's also going to be 0. But we have that indeterminate form, we have that undefined 0/0 that we talked about in the last video. So maybe we can use L'Hopital's rule here. In order to use L'Hopital's rule then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist. So let's just apply L'Hopital's rule and let's just take the derivative of each of these and see if we can find the limit. If we can, then that's going to be the limit of this thing. So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here. And so what's the derivative of the numerator going to be? I'll do it in a new color. I'll do it in green. Well, the derivative of 2 sine of x is 2 cosine of x. And then, minus-- well, the derivative of sine of 2x is 2 cosine of 2x. So minus 2 cosine of 2x. Just use the chain rule there, derivative of the inside is just 2. That's the 2 out there. Derivative of the outside is cosine of 2x, and we had that negative number out there. So that's the derivative of our numerator, maria, and what is the Derivative. of our denominator? Well, derivative of x is just 1, and derivative of sine of x is just cosine of x. So 1 minus cosine of x. So let's try to evaluate this limit. What do we get? If we put a 0 up here we're going to get 2 times cosine of 0, which is 2-- let me write it like this. So this is 2 times cosine of 0, which is 1. So it's 2 minus 2 cosine of 2 times 0. Let me write it this way. Actually, let me just do it this way. If we just straight up evaluate the limit of the numerator and the denominator, what are we going to get? We get 2 cosine of 0, which is 2. Minus 2 times cosine of-- well, this 2 times 0 is still going to be 0. So minus 2 times cosine of 0, which is 2. All of that over 1 minus the cosine of 0, which is 1. So once again, we get 0/0. So does this mean that the limit doesn't exist? No, it still might exist, we might just want to do L'Hopital's rule again. Let me take the derivative of that and put it over the derivative of that. And then take the limit and maybe L'Hopital's rule will help us on the next [INAUDIBLE]. So let's see if it gets us anywhere. So this should be equal to the limit if L'Hopital's We're not 100% sure yet. This should be equal to the limit as x approaches 0 of the derivative of that thing over the derivative of that thing. So what's the derivative of 2 cosine of x? Well, derivative of cosine of x is negative sine of x. So it's negative 2 sine of x. And then derivative of cosine of 2x is negative 2 sine of 2x. So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times the 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times-- the negative right there's a plus. You have a positive sine, so it's the sine of 2x. That's the numerator when you take the derivative. And the denominator-- this is just an exercise in What's the derivative of the denominator? Derivative of 1 is 0. And derivative negative cosine of x is just-- well, that's just sine of x. So let's take this limit. So this is going to be equal to-- well, immediately if I take x is equal to 0 in the denominator, I know that sine of 0 is just 0. Let's see what happens in the numerator. Negative 2 times sine of 0. That's going to be 0. And then plus 4 times sine of 2 times 0. Well, that's still sine of 0, so that's still going to be 0. So once again, we got indeterminate form again. Do we give up? Do we say that L'Hopital's rule didn't work? No, because this could have been our first limit problem. And if this is our first limit problem we say, hey, maybe we could use L'Hopital's rule here because we got an indeterminate form. Both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to-- if the limit exist, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then, plus the derivative of 4 sine of 2x. Well, it's 2 times 4, which is 8. Times cosine of 2x. Derivative of sine of 2x is 2 cosine of 2x. And that first 2 gets multiplied by the 4 to get the 8. And then the derivative of the denominator, derivative of sine of x is just cosine of x. So let's evaluate this character. So it looks like we've made some headway or maybe L'Hopital's rule stop applying here because we take the limit as x approaches 0 of cosine of x. That is 1. So we're definitely not going to get that indeterminate form, that 0/0 on this iteration. Let's see what happens to the numerator. We get negative 2 times cosine of 0. Well that's just negative 2 because cosine of 0 is 1. Plus 8 times cosine of 2x. Well, if x is 0, so it's going to be cosine of 0, which is 1. So it's just going to be an 8. So negative 2 plus 8. Well this thing right here, negative 2 plus 8 is 6. 6 over 1. This whole thing is equal to 6. So L'Hopital's rule-- it applies to this last step. If this was the problem we were given and we said, hey, when we tried to apply the limit we get the limit as this numerator approaches 0 is 0. Limit as this denominator approaches 0 is 0. As the derivative of the numerator over the derivative of the denominator, that exists and it equals 6. So this limit must be equal to 6. Well if this limit is equal to 6, by the same argument, this limit is also going to be equal to 6. And by the same argument, this limit has got to also be equal to 6. And we're done." + }, + { + "Q": "why does the heart keep us living?", + "A": "It pumps our blood, and the continues flow of blood is what ensures that our cells gets what they need (like oxygen) and that their waste is removed.", + "video_name": "7XaftdE_h60", + "transcript": "So what you're looking at is one of the most amazing organs in your body. This is the human heart. And it's shown with all the vessels on it. And you can see the vessels coming into it and out of it. But the heart, at its core, is a pump. And this pump is why we call it the hardest working organ Because it starts pumping blood from the point where you're a little fetus, maybe about eight weeks old, all the way until the point where you die. And so this organ, I think, would be really cool to look at in a little bit more detail. But it's hard to do that looking just at the outside. So what I did is I actually drew what it might look like on the inside. So let me actually just show you that now. And we'll follow the path of blood through the heart using this diagram. Let me start with a little picture in the corner. So let's say we have a person here. And this is their face, and this is their neck. I'm going to draw their arms. And they have, in the middle of their chest, their heart. And so the whole goal is to make sure that blood from all parts of their body, including their legs, can make its way back to the heart, first of all, and then get pumped back out to the body. So blood is going to come up from this arm, let's say, and dump into there. And the same on this side. And it's going to come from their head. And all three sources, the two arms and the head, are going to come together into one big vein. And that's going to be dumping into the top of the heart. And then separately, you've got veins from the legs meeting up with veins from the belly, coming into another opening into the heart. So that's how the blood gets back to the heart. And any time I mention the word vein, I just want you to make sure you think of blood going towards the heart. Now if blood is going towards the heart, then after the blood is pumped by the heart, it's going to have to go out to the heart. It's going to have to go away from the heart. So that's the aorta. And the aorta actually has a little arch, like that. We call it the aortic arch. And it sends off one vessel to the arm, one vessel up this way, a vessel over this way. And then this arch kind of goes down, down, down and splits like that. So this is kind of a simplified version of it. But you can see how there are definitely some parallels between how the veins and the arteries are set up. And arteries, anytime I mention the word artery, I want you to think of blood going away from the heart. And an easy way to remember that is that they both start with the letter A. So going to our big diagram now. We can see that blood coming in this way and blood coming in this way is ending up at the same spot. It's going to end up at the-- actually, maybe I'll draw it here-- is ending up at the right atrium. That's just the name of the chamber that the blood ends up in. And it came into the right atrium from a giant vessel up top called the superior vena cava. And this is a vein, of course, because it's bringing blood towards the heart. And down here, the inferior vena cava. So these are the two directions that blood is going to be flowing. And once blood is in the right atrium, it's going to head down into the right ventricle. So this is the right ventricle, down here. This is the second chamber of the heart. And it gets there by passing through a valve. And this valve, and all valves in the heart, are basically there to keep blood moving in the right direction. So it doesn't go in the backwards direction. So this valve is called the tricuspid valve. And it's called that because it's basically got three little flaps. That's why they call it tri. And I know you can only see two in my drawing, and that's just because my drawing is not perfect. And it's hard to show a flap coming out at you, but you can imagine it. So blood goes into the right ventricle. And where does it go next? Well after that, it's going to go this way. It's going to go into this vessel, and it's going to split. But before it goes there, it has to pass through another valve. So this is a valve, right here, called the pulmonary valve. And it gives you a clue as to where things are going to go next. Because the word pulmonary means lungs. And so, if this is my lung, on this side, this is my left lung. And this is my right lung, on this side. Then these vessels-- and I'll let you try to guess what they would be called-- these vessels. This would be my-- I want to make sure I get my right and left straight. This is my left pulmonary artery. And I hesitated there just to make sure you got that because it's taking blood away from the heart. And this is my right pulmonary artery. So this is my right and left pulmonary artery. And so blood goes, now, into my lungs. These are the lungs that are kind of nestled into my thorax, where my heart is sitting. It goes into my lungs. And remember, this blood is blue. Why is it blue? Well, it's blue because it doesn't have very much oxygen. And so one thing that I need to pick up is oxygen. And so that's one thing that the lungs are going to help me pick up. And I'm going to write O2 for oxygen. And it's also blue. And that reminds us that it's full of carbon dioxide. It's full of waste because it's coming from the body. And the body's made a lot of carbon dioxide that it's trying to get rid of. So in the lungs, you get rid of your carbon dioxide and you pick up oxygen. So that's why I switch, at this point, from a blue-colored vessel to a red-colored vessel. So now blood comes back in this way and this way and dumps into this chamber. So what is that? This is our left atrium. So just like our right atrium, we have one on the left. And it goes down into-- and you can probably guess what this one is called-- it's our left ventricle. So just like before, where it went from the right atrium to the right ventricle, now we're going from the left atrium to the left ventricle. And it passes through a valve here. So this valve is called the mitral valve. And its job is, of course, to make sure that blood does not go from the left ventricle back to the left atrium by accident. It wants to make sure that there's forward flow. And then the final valve-- I have to find a nice spot to write it, maybe right here. This final valve that it passes through is called the aortic valve. And the aortic valve is going to be what divides the left ventricle from this giant vessel that we talked about earlier. And this is, of course, the aorta. This is my aorta. So now blood is going to go through the aorta to the rest of the body. So you can see how blood now flows from the body into the four chambers. First into the right atrium-- this is chamber number one. And then it goes into the right ventricle. This is chamber number two. It goes to the lungs and then back out to the left atrium. So this is chamber number three. And then the left ventricle. And this happens every moment of every day. Every time you hear your heart beating, this process is going on." + }, + { + "Q": "I simply love your videos. I learn A LOT from them. However, 91*3600/5280=62, not 62.05. Your calculator's been making a mistake. Keep it up Sal! You're benifiting me alot.", + "A": "i think what you did was something called not following the order of operations then you wouldve got 62 if you didnt follow if you did you would have gotten 62.05 wither way you get 62.05 or 62.04545454545454545454545454545454545454545454545454545454545454545454545454545454545454545454 repeating", + "video_name": "aTjNDKlz8G4", + "transcript": "Welcome to the presentation on units. Let's get started. So if I were to tell you -- let me make sure my pen is set up right -- if I were to tell you that someone is, let's say they're driving at a speed of -- let's say it's Zack. So let's say I have Zack. And they're driving at a speed of, let me say, 28 feet per minute. So what I'm going to ask you is if he's going 28 feet in every minute, how many inches will Zack travel in 1 second? So how many inches per second is he going to be going? Let's try to figure this one out. So let's say if I had 28, and I'll write ft short for feet, feet per minute, and I'll write min short for a minute. So 28 feet per minute, let's first figure out how many inches per minute that is. Well, we know that there are 12 inches per foot, right? If you didn't know that you do now. So we know that there are 12 inches per foot. So if you're going 28 feet per minute, he's going to be going 12 times that many inches per minute. So, 12 times 28 -- let me do the little work down here -- 28 times 12 is 16, 56 into 280. I probably shouldn't be doing it this messy. And this kind of stuff it would be OK to use a calculator, although it's always good to do the math yourself, it's good practice. So that's 6, 5 plus 8 is 13. 336. So that equals 336 inches per minute. And something interesting happened here is that you noticed that I had a foot in the numerator here, and I had a foot in the denominator here. So you can actually treat units just the same way that you would treat actual numbers or variables. You have the same number in the numerator and you have the same number in the denominator, and your multiplying not adding, you can cancel them out. So the feet and the feet canceled out and that's why we were left with inches per minute. I could have also written this as 336 foot per minute times inches per foot. Because the foot per minute came from here, and the inches per foot came from here. Then I'll just cancel this out and I would have gotten inches per minute. So anyway, I don't want to confuse you too much with all of that unit cancellation stuff. The bottom line is you just remember, well if I'm going 28 feet per minute, I'm going to go 12 times that many inches per minute, right, because there are 12 inches per foot. So I'm going 336 inches per minute. So now I have the question, but we're not done, because the question is how many inches am I going to be traveling in 1 second. So let me erase some of the stuff here at the bottom. So 336 inches -- let's write it like that -- inches per minute, and I want to know how many inches per second. Well what do we know? We know that 1 minute -- and notice, I write it in the numerator here because I want to cancel it out with this minute here. 1 minute is equal to how many seconds? It equals 60 seconds. And this part can be confusing, but it's always good to just take a step back and think about what I'm doing. If I'm going to be going 336 inches per minute, how many inches am I going to travel in 1 second? Am I going to travel more than 336 or am I going to travel less than 336 inches per second. Well obviously less, because a second is a much shorter period of time. So if I'm in a much shorter period of time, I'm going to be traveling a much shorter distance, if I'm going the same speed. So I should be dividing by a number, which makes sense. I'm going to be dividing by 60. I know this can be very confusing at the beginning, but that's why I always want you to think about should I be getting a larger number or should I be getting a smaller number and that will always give you a good reality check. And if you just want to look at how it turns out in terms of units, we know from the problem that we want this minutes to cancel out with something and get into seconds. So if we have minutes in the denominator in the units here, we want the minutes in the numerator here, and the seconds in the denominator here. And 1 minute is equal to 60 seconds. So here, once again, the minutes and the minutes cancel out. And we get 336 over 60 inches per second. Now if I were to actually divide this out, actually we could just divide the numerator and the denominator by 6. 6 goes into 336, what, 56 times? 56 over 10, and then we can divide that again by 2. So then that gets us 28 over 5. And 28 over 5 -- let's see, 5 goes into 28 five times, 25. 3, 5.6. So this equals 5.6. So I think we now just solved the problem. If Zack is going 28 feet in every minute, that's his speed, he's actually going 5.6 inches per second. Hopefully that kind of made sense. Let's try to see if we could do another one. If I'm going 91 feet per second, how many miles per hour is that? Well, 91 feet per second. If we want to say how many miles that is, should we be dividing or should we be multiplying? We should be dividing because it's going to be a smaller number of miles. We know that 1 mile is equal to -- and you might want to just memorize this -- 5,280 feet. It's actually a pretty useful number to know. And then that will actually cancel out the feet. Then we want to go from seconds to hours, right? So, if we go from seconds to hours, if I can travel 91 feet per second, how many will I travel in an hour, I'm going to be getting a larger number because an hour's a much larger period of time than a second. And how many seconds are there in an hour? Well, there are 3,600 seconds in an hour. 60 seconds per minute and 60 minutes per hour. So 3,600 over 1 seconds per hour. And these seconds will cancel out. Then we're just left with, we just multiply everything out. We get in the numerator, 91 times 3,600, right? 91 times 1 times 3,600. In the denominator we just have 5,280. This time around I'm actually going to use a calculator -- let me bring up the calculator just to show you that I'm using the calculator. Let's see, so if I say 91 times 3,600, that equals a huge number divided by 5,280. Let me see if I can type it. 91 times 3,600 divided by 5,280 -- 62.05. So that equals 62.05 miles per hour." + }, + { + "Q": "What is a Chaperone and how is it related to protein folding?", + "A": "A chaperone protein is a protein whose function it is to assist other proteins in folding into their correct tertiary shape.", + "video_name": "dNHtdiVjQbM", + "transcript": "Let's talk about conformational stability and how this relates to protein-folding and denaturation. So first, let's review a couple of terms just to make sure we're all on the same page. And first, let's start out with the term conformation. And the term confirmation just refers to a protein's folded, 3D structure, or in other words, the active form of a protein. And next we can review what the term denatured means when you're talking about proteins. And denatured proteins just refer to proteins that have become unfolded, or inactive. So all conformational stability is really talking about are the various forces that help to keep a protein folded in the right way. And these various forces are the four different levels of protein structure, and we can review those briefly right here. So recall that the primary structure of a protein just refers to the actual sequence of amino acids in that protein, and this is determined by a protein's peptide bonds. And then next, you have secondary structure, which just refers to the local substructures in a protein. And they are determined by backbone interactions held together by hydrogen bonds. Then you have tertiary structure, which just talks about the overall 3D structure of a single protein molecule. And this is described by distant interactions between groups within a single protein. And these interactions are stabilized by van der Waals interactions, hydrophobic packing, and disulfide bonding, in addition to the same hydrogen bonding that helps to determine secondary structure. And then quaternary structure just describes the different interactions between individual protein subunits. So you have the folded-up proteins that then come together to assemble the completed, overall protein. And the interaction of these different protein subunits are stabilized by the same kinds of bonds that help to determine tertiary structure. So all of these levels of protein structure help to stabilize the folded-up, active confirmation of a protein. So why is it so important to know about the different levels of protein structure and how they contribute to conformational stability? Well, like I said, a protein is only functional when they are in their proper conformation, in their proper 3D form. And an improperly folded or degraded, denatured, protein is inactive. So in addition to the four levels of protein structure that I just reviewed, there is also another force that helps to stabilize a protein's conformation. And that force is called the solvation shell. Now, the solvation shell is just a fancy way of describing the layer of solvent that is surrounding a protein. So say I have a protein who has all these exterior residues that are overall positively charged. And picture this protein in the watery environment of the interior of one of our cells, then the solvation shell is going to be the layer of water right next to this protein molecule. And remember that water is a polar molecule, so you have the electronegative oxygen atom with a predominantly negative charge leaving a positive charge over next to the hydrogen atoms. The same is true for each of these water molecules. So now, as you can see, the electronegative oxygen atoms are stabilizing all the positively charged amino acid residues on the exterior of this protein. So as you can see, the conformational stability of a protein depends not only on all of these interactions that contribute to primary, secondary, tertiary, and quaternary structure, but also what sort of environment that protein is in. And all of these interactions are very crucial for keeping a protein folded properly, so that it can do its job. Now, what happens when things go wrong? How does a protein become unfolded and thus inactive? Well, remember that this is called denaturation. And this can be done by changing a lot of different parameters within a protein's environment, including changing the temperature, the pH, adding chemical denaturants, or even adding enzymes. So let's start with what happens if you alter the temperature around a protein. And we can use the example of an egg when we put it into a pot of boiling water, because an egg, especially the white part, is full of protein, and this pot of boiling water is representing heat. And remember that heat is really just a form of energy. So when you heat an egg, the proteins gain energy and literally shake apart the bonds between the parts of the amino acid chains. And this causes the proteins to unfold. So increased temperature destroys the secondary, tertiary, and quaternary structure of a protein, but the primary structure is still preserved. So the take-away point is that when you change the temperature of a protein by heating it up, you destroy all of the different levels of protein structure except for the primary structure. So now, let's say you were to take an egg and then add vinegar, which is really just an acid. The acid in the vinegar will break all the ionic bonds that contribute to tertiary and quaternary structure. So the take-away point when you change the pH surrounding a protein is that you have disruption of ionic bonds. And if we think about this a little bit more deeply, it kind of makes sense, because ionic bonds are dependent upon the interaction of positive and negative charges. So when you add either an acid or a base, which in the case of an acid is just like adding a bunch of positive charges, you kind of disrupt the balance between all these interactions between the positive and negative charges within the protein. So now let's look at how chemicals denature proteins. Chemical denaturants often disrupt the hydrogen bonding within a protein. And remember that hydrogen bonds contribute to secondary, tertiary, all the way up to quaternary structure. So all these levels of protein structure will be disrupted if you add a chemical denaturant. So let's take our same example of a protein with an egg, and say, if you were 21 years or older, you got your hands on some alcohol, and you added this to the egg. Then, all the hydrogen bonds would be broken up, leaving you with just linear polypeptide chains. And then finally, let's take our hard-boiled egg from the temperature example, and let's eat it. So here's my beautiful drawing of a person, representing you, eating this hard boiled egg. Once the egg enters our digestive tract we have enzymes that break down the already denatured proteins in the egg even further. They take the linear polypeptide chain, whose primary structure is still intact. And they break the bonds between the individual amino acids, the peptide bonds, so that we can absorb these amino acids from our intestines into our bloodstream. And then we can use them as building blocks for our own protein synthesis. And that's how enzymes can alter a protein's primary structure and thus the protein's overall conformational stability. So what did we learn? Well, we learned that the conformational stability refers to all the forces that keep a protein properly folded in its active form. And this includes all the different levels of protein structure, as well as the salvation shell. And we also learned that a protein can be denatured into its inactive form by changing a variety of factors in its environment, including changing the temperature, the pH, adding chemicals, or enzymes." + }, + { + "Q": "how did you get the 6 to make 48", + "A": "A=1/2bh. The base= 12. H=8. When substituting, A=1/2(12)(8). 1/2 of 12= 6 as 12/2=6. So A=6(8). A= 48 sq units", + "video_name": "mtMNvnm71Z0", + "transcript": "- What I want to do in this video is get some practice finding surface areas of figures by opening them up into what's called nets. And one way to think about it is if you had a figure like this, and if it was made out of cardboard, and if you were to cut it, if you were to cut it right where I'm drawing this red, and also right over here and right over there, and right over there and also in the back where you can't see just now, it would open up into something like this. So if you were to open it up, it would open up into something like this. And when you open it up, it's much easier to figure out the surface area. So the surface area of this figure, when we open that up, we can just figure out the surface area of each of these regions. So let's think about it. So what's first of all the surface area, what's the surface area of this, right over here? Well in the net, that corresponds to this area, it's a triangle, it has a base of 12 and height of eight. So this area right over here is going to be one half times the base, so times 12, times the height, times eight. So this is the same thing as six times eight, which is equal to 48 whatever units, or square units. This is going to be units of area. So that's going to be 48 square units, and up here is the exact same thing. That's the exact same thing. You can't see it in this figure, but if it was transparent, if it was transparent, it would be this backside right over here, but that's also going to be 48. 48 square units. Now we can think about the areas of I guess you can consider them to be the side panels. So that's a side panel right over there. It's 14 high and 10 wide, this is the other side panel. It's also this length over here is the same as this length. It's also 14 high and 10 wide. So this side panel is this one right over here. And then you have one on the other side. And so the area of each of these 14 times 10, they are 140 square units. This one is also 140 square units. And then finally we just have to figure out the area of I guess you can say the base of the figure, so this whole region right over here, which is this area, which is that area right over there. And that's going to be 12 by 14. So this area is 12 times 14, which is equal to let's see. 12 times 12 is 144 plus another 24, so it's 168. So the total area is going to be, let's see. If you add this one and that one, you get 96. 96 square units. The two magenta, I guess you can say, side panels, 140 plus 140, that's 280. 280. And then you have this base that comes in at 168. We want it to be that same color. 168. One, 68. Add them all together, and we get the surface area for the entire figure. And it was super valuable to open it up into this net because we can make sure we got all the sides. We didn't have to kinda rotate it in our brains. Although you could do that as well. So, with six plus zero plus eight is 14. Regroup the one ten to the tens place, there's now one ten. So one plus nine is ten, plus eight is 18, plus six is 24, and then you have two plus two plus one is five. So the surface area of this figure is 544. 544 square units." + }, + { + "Q": "How long was the famine as the people seemed to be starving since a long time?", + "A": "well france had really no money to plant food because war was every where", + "video_name": "0t4MF9ZoppM", + "transcript": "We left off the last video at the end of 1789. The Bastille had been stormed in July as Parisians wanted to get the weapons from the Bastille and free a few political prisoners to, in their minds, protect themselves from any tyranny from Louis XVI. Louis XVI had reluctantly kind of gone behind the scenes and said, OK, National Assembly, I'm not going to get in your Because he's seen the writing on the walls that every time he's done something, it's only led to even more extreme counteraction. So at the end of 1789, already chaos has broken loose in a lot of France. The National Assembly, they're in process of creating a constitution, which won't fully happen until 1791. But they're starting to bring things together in order to draft that constitution. But in August of 1789, they've already done their version of the Declaration of Independence. The Declaration of the Rights of Man and the Citizen. So if everything was well, we would just wait until a few years, we'd get a constitution, and maybe we would have some type of a constitutional monarchy. But unfortunate, especially for Louis XVI, things weren't all well. As we mentioned, all of this was propagated, all of this was started to begin with because people were hungry. We have this fiscal crisis, we have a famine. And so in October of 1789-- we're still in 1789-- October of 1789-- rumors started to spread that Marie-Antoinette, the king's wife, that she was hoarding grain at Versailles. So people started imagining these big stacks of grain at Versailles, and this is in a time where people couldn't get And bread was the main staple of the diet. So there was actually a march of peasant women onto And they were armed. This is a depiction of the peasant women marching on And they went to Versailles, and they actually were able to get into the building itself. And they demanded-- because they were suspicious of what Louis XVI and Marie-Antoinette were up to at Versailles-- they demanded that they move to Paris. So the women's march. And they were able to get their demands. It resulted in Louis XVI and wife, Marie-Antoinette, moving back to Paris, where they couldn't do things like hoard grain. And they'll be surrounded by all of the maybe not-so-friendly people who could watch what they're doing. I think the main factor was that people are hungry, rumors are spreading that the king is hoarding grain. But there were also rumors that the king was being very disrespectful to some of the symbols of the new France, of the new National Assembly. So that also made people angry. And across the board everyone kind of knew, and including Louis XVI, that he wasn't really into what was going on. He wasn't into this kind of constitutional monarchy that was forming, this power that was being lost to the National Assembly. So we have this very uncomfortable situation entering into 1790, where the king and queen are essentially in house arrest in a building called the Tuileries in Paris. You have this National Assembly drafting this constitution. They're charted to draft the constitution up there. They all pledged at the Tennis Court Oath. And at the same time, throughout France, you have some counter insurgencies. This is France right here. Throughout France you have counter insurgencies, people who don't like the Revolution that's going on. And then those would be subdued. And people are all plotting one against each other. And then you have some nobility, that says, gee, you know what? I don't like the way that this is going. We've seen already a lot of violence. People are angry. I'm just going to take my money and whatever I can pack, and I'm just going to get out of the country. I'm going to emigrate away from the country. So you start having nobility starting to leave France. They're called the Emigres. I know I'm not pronouncing it correctly. But you see, you have this notion of gee, I had it good in France, I'm not going to have it good much longer, I'd better leave. And this same idea, now that we get to 1791. So 1790 was just kind of a bunch of unease. Now that we're at 1791, the same idea of trying to get away from the danger got into the heads of Louis XVI and But they couldn't leave the country. They didn't trust Great Britain. They didn't trust any of these other countries to safely give them shelter. So one of their generals, who was sympathetic to their cause, said, hey, at least come here to the frontier areas and you could hide from all of the unrest that's going on. So dressed as actual servants-- and it shows you what type of people they were-- they dressed as servants. And they actually made their servants dress as nobility to make them the targets in case they were ambushed anyway on their way trying to escape from Paris. Dressed as servants, the king and queen-- the king tried to escape to this general's estate. But when they were in Varennes, right here, they were actually spotted. And then the people essentially took them captive and brought them back to Paris. So this is called, or you could imagine this is the flight to Varennes, or the flight away from Paris, or however you want to do it. So already, Louis XVI started to see the writing on the wall. They tried to get away. But people brought them back. Now you can imagine, a lot of people already did not like the king. They didn't like the notion of even having a king. And the most revolutionary, the most radical elements, were called the Jacobins. And after the king and queen tried to escape and came back, they were like hey, gee, what's the use of even having a king? You National Assembly, why are you even trying to write some type of constitution that gives any power whatsoever to a king? We should have a republic. Which is essentially-- there's a lot of kind of nuanced definitions of what a republic is, but the most simple one is it's a state without a king, without an emperor, without a queen. So they're saying, we don't need, you know-- you National Assembly, you think you're being radical. But you're not being radical enough. We want to eliminate the idea of having a monarchy altogether. And the fact that Louis XVI actually tried to run away, we view that as him abdicating the throne. Abdication, or essentially quitting. And they actually started to organize in Paris. This right here is the Champ-de-Mars. I know I'm saying it completely wrong. This is a current picture of it. And so they started taking signatures in this kind of public park in Paris to essentially say, we don't need a king. We want to essentially create our own republic. That this National Assembly, they're not radical enough. And so people started gathering over here in the Champ-de-Mars and things got a little ugly. So the actual troops were sent in to kind of calm everyone down. And these were actually troops controlled by the National Assembly. The people who are mainly controlled by the Third Estate. But things got a little crazy. Rocks were thrown at some of the troops. Some of the troops, at first, they started firing in the air. But eventually when things got really crazy, they fired into the crowd. And about 50 people died. And this was the massacre. Or the Champ-de-Mars Massacre. I know I'm saying it wrong. This isn't a video on French pronunciation. But you could imagine, now people are even angrier. People are still starving. That problem has not gone away. The king and queen has been kind of very reluctantly-- everyone is suspicious of the fact that they're probably going to try to come back to power. They tried to run away. When the Jacobins, or in general kind of revolutionaries, but they're led by the Jacobins, when they start to suggest that, hey, we should have a republic. We shouldn't even have a king. And they gather people here, all of a sudden, the troops that are controlled by the current National Assembly actually fire on the crowd, and actually kill civilians for throwing rocks. And they might have been big rocks. But you can imagine, this is going to anger already hungry and already suppressed people even more. And to make people even more paranoid that the king and queen might eventually come back to power, you had two major powers all of a sudden trying to insert certain themselves into the French Revolution. I'm going to do a little bit of an aside here. Because this is something, at least you when I first learned European history, I found the most confusing. You have these states, you can call them. You have Austria, which I've highlighted in orange. The kind of map here is a modern map. But in orange, I've kind of shown what Austria was at that point in time. Around 1789, 1790, 1791. In this red color, I have Prussia. I want to show you that these are very different than our current notions of one, Austria. Austria today is this modern country right here. And Prussia doesn't even exist as a modern country. And then you had this notion of the Holy Roman Empire, which overlaps with these other kingdoms, or empires, or whatever you want to call them. And I want to do a little bit of an aside here. The Holy Roman Empire, as Voltaire famously said, is neither holy nor Roman nor an empire. And he was right. It was really kind of a very loose confederation of German kingdoms and states-- mainly German kingdoms and states. As you can see, it kind of coincides with modern Germany. And the two most influential powers in the Holy Roman Empire, or actually the most influential power in the Holy Roman Empire, was the Austrians. And the ruler of the Austrians had the title of Holy Roman Emperor. And that was Leopold II. But it's not like he was like the Roman Emperors of old. The Roman Emperors of old actually came out of Rome. Notice, nothing in the Holy Roman Empire at that time, it had no control of Rome. So it was not Roman, we're not talking about people who spoke Latin. We're talking about people who spoke Germanic languages. And it wasn't an empire. That it wasn't a tightly knit kind of governance structure. It was this loose confederation of states. But what was the most influential was the region that was under control of the Habsburgs of Austria, or Leopold II. And not only was he in control, or not only did he have the title of Holy Roman Empire, and essentially had control of the Austrian, I guess you could say Empire, at that point in time. He was also Marie-Antoinette's brother. Leopold II, that's her brother. So Leopold II and Frederick William II of Prussia, which is another mainly Germanic state. Let me do that in a better color. They issued the Declaration of Pillnitz. Let me write this down. So this is going to add even more insult to injury to just the general population of France. The Declaration of Pillnitz. And this was done in August. so I just want to make it very clear what happened. In June of 1791, they tried to escape, they were captured at Varennes. Then in July of 1791, you have the Champ-de-Mars Massacre. So already, people are very wary of the royals. The idea that we don't need them is spreading. And people are getting angrier. And then you have the Declaration of Pillnitz by these foreign powers, one of whom is essentially the brother of the current French royalty. And that declaration is essentially saying that they intend to bring the French monarchy back to power. They don't say that they're definitely going to do it in military terms or whatever. But it's a declaration that they do not approve of what's going on in France. And even though they themselves might have not taken it too seriously, the people of France took it really seriously. You have these huge powers on their border right here. You had the Austrians and the Prussians. So this wasn't anything that people could take very lightly. So it only increased the fear that the royals were going to do something to come back to full power and really suppress people. And it really gave even more fuel for the Jacobins to kind of argue for some type of a republic. So I'm going to leave you there in this video. As you can see, we saw in the first video, things got bad. Now they're getting really worst. Chaos is breaking out in France. People are questioning whether they even need a king or queen. Foreign powers are getting involved, saying hey, they don't like what they're seeing there, with kings and queens getting overthrown. Maybe that'll give ideas to their people. And by the way, I'm your brother, so I want to help you out too. That scares people even more. The current National Assembly, which is kind of the beginning of the Revolution, they themselves are on some level massacring people. So it's really leading to a really tense and ugly time in French history. And you're going to see that that's going to culminate with what's called the Reign of Terror. And we're going to see that in the next video." + }, + { + "Q": "I am wondering what college or university do you think I should go to? Writing/language arts", + "A": "That depends If you want to stay local. If you don t mind traveling you could go to Harvard or Berkeley, but those are top of the line.", + "video_name": "jPrEKz1rAno", + "transcript": "- So to give you a big picture of what will be presented to a college, from your point of view as a student, you'll be submitting an application. That application will have in it biographical information, your extracurricular activities, essays that you've written. Basically giving an overview of who you are and what you're doing, but there'll also be lots of other information that's submitted that an admissions office will use. From your high school, they will submit a transcript with your grades and courses you've taken. Hopefully they'll send along a profile that sort of details that school and gives some information about the school you're in. There also will be probably a guidance counselor or college counselor letter of recommendation. A lot of schools also require one or potentially two letters of recommendation from teachers, so that could possibly be in the file as well. If the school requires any kind of standardized testing, from SAT, ACT, potentially AP or IB scores, those sorts of things, they could be in your file as well. You may be required to do an interview for that school. And there are also opportunities to submit... If you have special talents, in things like the arts, theatre, music, athletics, there may be special ways to submit samples of your work to also be evaluated. So, in general, I would say most selective and highly-selective schools are going to put the most emphasis on your programming grades, number one, just to make sure they feel you can do the work at their school. How they use the rest of those components will vary greatly by school." + }, + { + "Q": "is the number line you are trying to show that you divide the number line just like you divide normally?", + "A": "There s no difference. I don t get what difference you see.", + "video_name": "Z3qRkxzmYnU", + "transcript": "PROBLEM: \"Omar rode his boat for a total of 50 miles over the past 5 days. And he rowed the same amount each day. How many miles did Omar row his boat each day?\" So let\u2019s just visualize what's going on. He is able to travel 50 miles. So let's make a \u2013 Let\u2019s say this line represents the 50 miles that he travels. So this whole distance right over here is 50 miles. And he does it \u2013 They tell us that he does it over 5 days, and that each day, he does the same amount. So this 50 miles is if you were to add together all of what he did over the 5 days. And so, if you want to know how much he did each day, you essentially want to divide this 50 miles into 5 equal sections. And the length of each of those sections is the amount he did each day. So if we just visualize it \u2013 So that's one section \u2013 second section \u2013 third section \u2013 fourth \u2013 and fifth section. And actually, I didn't do that very well. It should look a little bit more equal than that. First, second \u2013 (And that's not going to \u2013 Let's see.) First, second, third, fourth, and fifth. And you don't have to actually do this. This is just to help visualize. So essentially, what we want to figure out is what is one of these distances? And as you can see in our visualization, this is really just taking our 50 miles and dividing it into 5 equal chunks. So we're essentially just taking 50, and we're going to divide it by 5. So 50 divided by 5 is going to be equal to 10. So if he goes 50 miles over 5 days, and you divide by the 5 days, he goes 10 miles each day. And we're done." + }, + { + "Q": "How would you graph something like x0) cosx/x ?", + "A": "cosx is approaching 1 and so the graph of cosx/x looks like the graph of 1/x around zero. The line x=0 is a vertical assymptote. From the right (positive) side, the values increase without bound (go to infinity) and from the left (negative) side, they decrease without bound (go to negative infinity). Since the one sided limits are different, we say that there is no limit or that the limit does not exist sometimes abbreviated D.N.E.", + "video_name": "riXcZT2ICjA", + "transcript": "In this video, I want to familiarize you with the idea of a limit, which is a super important idea. It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really, really, really simple idea. So let me draw a function here, actually, let me define a function here, a kind of a simple function. So let's define f of x, let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal look, I have the same thing in the numerator and denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true, the difference between f of x equals 1 and this thing right over here, is that this thing can never equal-- this thing is undefined when x is equal to 1. Because if you set, let me define it. Let me write it over here, if you have f of, sorry not f of 0, if you have f of 1, what happens. In the numerator, we get 1 minus 1, which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here is my x-axis. And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here. And so once again, if someone were to ask you what is f of 1, you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. So you could say, and we'll get more and more familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, let me call it g of x. Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. So it's going to be a parabola, looks something like this, let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that. So this, on the graph of f of x is equal to x squared, this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function x squared. So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1, what's g of 2.01, what's g of 2.001, what is that approaching as we get closer and closer to it. And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1. The limit of g of x as x approaches 2 is equal to 4. And you could even do this numerically using a calculator, and let me do that, because I think that will be interesting. So let me get the calculator out, let me get my trusty TI-85 out. So here is my calculator, and you could numerically say, OK, what's it going to approach as you approach x equals 2. So let's try 1.94, for x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared. And so you get 3.61, well what if you get even closer to 2, so 1.99, and once again, let me square that. Well now I'm at 3.96. What if I do 1.999, and I square that? I'm going to have 3.996. Notice I'm going closer, and closer, and closer to our point. And if I did, if I got really close, 1.9999999999 squared, what am I going to get to. It's not actually going to be exactly 4, this calculator just rounded things up, but going to get to a number really, really, really, really, really, really, really, really, really close to 4. And we can do something from the positive direction too. And it actually has to be the same number when we approach from the below what we're trying to approach, and above what we're trying to approach. So if we try to 2.1 squared, we get 4.4. let me go a couple of steps ahead, 2.01, so this is much closer to 2 now, squared. Now we are getting much closer to 4. So the closer we get to 2, the closer it seems like we're getting to 4. So once again, that's a numeric way of saying that the limit, as x approaches 2 from either direction of g of x, even though right at 2, the function is equal to 1, because it's discontinuous. The limit as we're approaching 2, we're getting closer, and closer, and closer to 4." + }, + { + "Q": "so what is the affect on macroeconomics equilibrium after a shift in aggregate demand?", + "A": "Nothing will really happen. All you will get is inflation or deflation, depending on which way demand shifts.", + "video_name": "8W0iZk8Yxhs", + "transcript": "Narrator: We've talked a lot about aggregate demand over the last few videos, so in this video, I thought I would talk a little bit about aggregate supply. In particular, we're going to think about aggregate supply in the long-run. In economics, whether it's in micro or macro economics, when we think about long-run, we're thinking about enough time for a lot of fixed costs and a lot of fixed contracts to expire. In the short-term, you might be stuck into some labor contract, or stuck into your using some factory that you've already paid money for, so it was a fixed cost, but over the long-run you'll have a chance that factory will wear down and you'll have a chance to decide whether you want another factory or the price of the factory might change; or in the long-run, you'll have a chance contracts will expire, and you'll have a chance to renegotiate those contracts at a new price. That's what we really mean when we talk about the long-run. I'm going to plot aggregate supply on the same axis as we plotted aggregate demand, and we're going to focus on the long-run now, and then we're going to think about what actually might happen in the short-run while we are in fixed-price contracts, or we already have spent money on something, or we have already, in some ways, there are sticky things that can't adjust as quickly. But, we'll first focus on the long-run. On this axis, I'm just going to plot price, and remember, we're thinking in macro-economic terms. This is some measure of the prices of the goods and services in our economy. This axis right over here, the horizontal axis is going to be real GDP. Once again, this is just a model, you should take everything in economics with a huge grain of salt. These are over-simplifications of a highly, highly complex thing, the economy. Millions and millions of actors doing complex things, human beings, each of them and their brain have billions and billions and billions of neurons, doing all sorts of unpredictable things. But economists like to make really simplifying, super-simplifying assumptions, so that we can deal with it in a attractable way, and in a even dealing in a mathematical way. The assumtion that economists often make when we think about aggregate supply and aggregate demand is, in the long-run, real GDP actually does not depend on prices in the long-run; so, what you have is, regardless of what the price is, you're going to have the same real GDP. You can view this as a natural level of productivity for the economy. This is some level right over here. It's important to realize this is just a snap shot in time, and this is all else things equal, so we're not assuming that we're having changes in productivity overtime; this is just a snap shot if we did have any of those things that change. For example, if the population increased, then that would cause this level to shift to the right, then we would have a higher natural level of productivity. If, for whatever reason, we were able to create tools so that it was easier to find people jobs, there's always a natural rate of unemployment. There's frictions, people have to look for jobs, some people have to retrain to get their skills, but maybe we improve that in some way so that there's some website where people can find jobs easier, or easier ways to train for jobs, and the natural level of unemployment goes down, more people can produce, that would also shift this curve to the right. You could have a reality where there's technological improvements that would also, and then all of a sudden, on an average, people would become more productive; that could shift things to the right. You could have discovery of natural resources, new land that is super fertile, and everything else; that could also shift things to the right. You could have a war, and maybe your factories get bombed, or bad things happen in a war, especially if the war is on your soil, and that could actually shift things to the left. So, it's important to realize that this is just taking a snap shot in time, and a lot of these other things that we think about would just shift it in 1 direction or another. I'm going to leave you there, and this is a kind of it might not seem intuitive at first, because you're saying, \"Wait, look, if prices were to change dramatically, if all of a sudden everything in the economy got twice as expensive, that would have some impact on peoples' minds and that they would behave differently and all the rest, and that might affect how much they can produce.\" We did think a little about that when we thought about aggregate demand, but when we think about aggregate supply, we're just thinking about their capability to produce. We're saying all else equal. We're saying that peoples' mind-shifts aren't changing, their willingness to work isn't changing, nothing else is changing, technology isn't changing. Given that, price really is just a numeric thing. If you just looked at the resources and the productive capability of a country, the factors of production, the people and all the rest, regardless of what the prices are, they in theory, should be able to produce the same level of goods and services." + }, + { + "Q": "do all of the arteries branch off of the Aorta?", + "A": "all the systemic arteries branch out from there", + "video_name": "iqRTd1NY-pU", + "transcript": "I want to figure out how blood gets from my heart, which I'm going to draw here, all the way to my toe. And I'm going to draw my foot over here and show you which toe I'm talking about. Let's say this toe right here. Now, to start the journey, it's going to have to go out of the left ventricle and into the largest artery of the body. This is going to be the aorta. And the aorta is very, very wide across. And that's why I say it's a large artery. And from the aorta-- I'm actually not drawing all the branches of the aorta. But from the aorta, it's going to go down into my belly. And it's going to branch towards my left leg and my right leg. So let's say we follow just the left leg. So this artery over here on the top, it's going to get a little bit smaller. And maybe I'd call this a medium-sized artery by this point. This is actually now getting down towards my ankle. Let's say we've gone quite a distance down in my ankle. And then there are, of course, little branches. And let's just follow the branch that goes towards my foot, which is this top one. Let's say this one goes towards my foot, and this is going to be now an even smaller artery. Let's call it small artery. From there, we're actually going to get into what we call arterioles, so it's going to get even tinier. It's going to branch. Now, these are very, very tiny branches coming off my small artery. And let's follow this one right here, and this one is my arteriole. So these are all the different branches I have to go through. And finally, I'm going to get into tiny little branches. I'm going to have to draw them very, very skinny just to convince you that we're getting smaller and smaller. Let me draw three of them. Let's draw four just for fun. And this is actually going to now get towards my little toe cells. So let me draw some toes cells in here to convince you that I actually have gotten there. Let's say one, two over here, and maybe one over here. These are my toes cells. And after the toe cells have kind of taken out whatever they need-- maybe they need glucose or maybe they need some oxygen. Whatever they've taken out, they're also going to put in their waste. So they have, of course, some carbon dioxide waste that we need to drag back. This is now going to dump into what we call a venule. And this venule is going to basically then feed into many, many other venules. Maybe there's a venule down here coming in, and maybe a venule up here coming in maybe from the second toe. And it's going to basically all kind of gather together, and again, to a giant, giant set of veins. Maybe veins are dumping in here now, maybe another vein dumping in here. And these veins are all going to dump into an enormous vein that we call the inferior vena cava. I'll write that right here, inferior vena cava. And this is the large vein that brings back all the blood from the bottom half of the body. There's also another one over here called the superior vena cava, and this is bringing back blood from the arms and head. So these two veins, the superior vena cava and the inferior vena cava, are dragging the blood back to the heart. And generally speaking, these are all considered, of course, veins. Let's back up now and start with the large and medium arteries. These guys together are sometimes referred to as elastic arteries. And the reason they're called elastic arteries, one of the good reasons why they're called that is that they have a protein in the walls of the blood vessel called elastin. They have a lot of this elastin protein. And if you think about the word elastin or elastic-- obviously very similar words-- you might think of something like a rubber band or a balloon. And that's probably the easiest way to think about it. If you have a blood vessel, one of these large arteries, for example, and let's say blood is under a lot of pressure because the heart is squeezing out a lot of high pressure blood, this artery is literally going to balloon out. And if you actually looked at it from the outside, it would look like a little sausage, something like this where it's puffed out. So what's happened there between the first and second picture is that the pressure energy-- so the heart is squeezing out a lot of pressurized blood. And, of course, there's energy in that blood. That pressure energy has been converted over into elastic energy. It's actually converting energy. We don't really always think about it that way, but that's exactly what's happening. And when you convert from pressure energy to elastic energy, what you're really then doing is you're balancing out those high pressures. So you're balancing out high pressures. And this is actually very important, because the blood that's coming into our arteries is under, let's not forget, high pressure. So the arterial system we know is a high-pressure system. So this makes perfect sense that the first few arteries, those large arteries and even those medium-sized arteries, are going to be able to deal with the pressure really well. Now, let me draw a little line here just to keep it straight. The small artery and the arteriole, these two are actually sometimes called the muscular arteries. And the reason, again, if you just want to look at the wall of the artery, you'll get the answer. The wall of the artery is actually very muscular. In fact, specifically, it's smooth muscle. So not the kind of muscle you have in your heart or in your biceps, but this is smooth muscle that's in the wall of the artery. And there's lots of it. So again, if you have a little blood vessel like this, if you imagine tons and tons of smooth muscle on the outside-- so let's draw it like this, little bands of smooth muscle. If those bands decide that they want to contract down, that they want to squeeze down, you're going to get something that looks like a little straw, because those muscles are now tight. They're tightly wound, so you're going to create like a little straw. And this process is called vasoconstriction. Vaso just means blood vessel. And constriction is kind of tightening down. So vasoconstriction, tightening down of the blood vessel. And what that does is it increases resistance. Just like if you're trying to blow through a tiny, tiny little straw, there's a lot of resistance. Well, it's the same idea here. And actually, a lot of that resistance and change in the vasoconstriction is happening at the arteriole level. So that's why they're very special and I want you to remember them. From there, blood is going to go through the capillaries. I didn't actually label them the first time, but let me just write that here. Some, as they call them, capillary beds. I'll write that out. And then it's going to go and get collected in the venules and eventually into the veins. And the important thing about the veins-- I'm going to stop right here and just talk about it very briefly-- is that they have these little valves. And these valves make sure that the blood continues to flow in one direction. So one important thing here is the valves. And remember, the other important thing is that they are able to deal with large volumes. So unlike the arterial side where it was all about large pressure, down here with the vein side, we have to think about large volumes. Remember about 2/3 of your blood at any point in time is sitting in some vein or venule somewhere." + }, + { + "Q": "can parentheses be in parentheses", + "A": "Yes, you do the inner parenthesis first.", + "video_name": "GiSpzFKI5_w", + "transcript": "We're asked to simplify 8 plus 5 times 4 minus, and then in parentheses, 6 plus 10 divided by 2 plus 44. Whenever you see some type of crazy expression like this where you have parentheses and addition and subtraction and division, you always want to keep the order of operations in mind. Let me write them down over here. So when you're doing order of operations, or really when you're evaluating any expression, you should have this in the front of your brain that the top priority goes to parentheses. And those are these little brackets over here, or however Those are the parentheses right there. That gets top priority. Then after that, you want to worry about exponents. There are no exponents in this expression, but I'll just write it down just for future reference: exponents. One way I like to think about it is parentheses always takes top priority, but then after that, we go in descending order, or I guess we should say in-- well, yeah, in descending order of how fast that computation is. When I say fast, how fast it grows. When I take something to an exponent, when I'm taking something to a power, it grows really fast. Then it grows a little bit slower or shrinks a little bit slower if I multiply or divide, so that comes next: multiply or divide. Multiplication and division comes next, and then last of all comes addition and subtraction. So these are kind of the slowest operations. This is a little bit faster. This is the fastest operation. And then the parentheses, just no matter what, always take priority. So let's apply it over here. Let me rewrite this whole expression. So it's 8 plus 5 times 4 minus, in parentheses, 6 plus 10 divided by 2 plus 44. So we're going to want to do the parentheses first. We have parentheses there and there. Now this parentheses is pretty straightforward. Well, inside the parentheses is already evaluated, so we could really just view this as 5 times 4. So let's just evaluate that right from the get go. So this is going to result in 8 plus-- and really, when you're evaluating the parentheses, if your evaluate this parentheses, you literally just get 5, and you evaluate that parentheses, you literally just get 4, and then they're next to each other, so you multiply them. So 5 times 4 is 20 minus-- let me stay consistent with the colors. Now let me write the next parenthesis right there, and then inside of it, we'd evaluate this first. Let me close the parenthesis right there. And then we have plus 44. So what is this thing right here evaluate to, this thing inside the parentheses? Well, you might be tempted to say, well, let me just go left to right. 6 plus 10 is 16 and then divide by 2 and you would get 8. But remember: order of operations. Division takes priority over addition, so you actually want to do the division first, and we could actually write it here like this. You could imagine putting some more parentheses. Let me do it in that same purple. You could imagine putting some more parentheses right here to really emphasize the fact that you're going to do the division first. So 10 divided by 2 is 5, so this will result in 6, plus 10 divided by 2, is 5. 6 plus 5. Well, we still have to evaluate this parentheses, so this results-- what's 6 plus 5? Well, that's 11. So we're left with the 20-- let me write it all down again. We're left with 8 plus 20 minus 6 plus 5, which is 11, plus 44. And now that we have everything at this level of operations, we can just go left to right. So 8 plus 20 is 28, so you can view this as 28 minus 11 plus 44. 28 minus 11-- 28 minus 10 would be 18, so this is going to be 17. It's going to be 17 plus 44. And then 17 plus 44-- I'll scroll down a little bit. 7 plus 44 would be 51, so this is going to be 61. So this is going to be equal to 61. And we're done!" + }, + { + "Q": "The path of the ice-sock is a great circle which indicates a continuous change in direction. Why isn't acceleration present; since acceleration is a vector having both direction and magnitude? Thanks.", + "A": "Acceleration is present. It s centripetal acceleration. That s the point of the video.", + "video_name": "CEdXvoAv_oM", + "transcript": "This is a picture of the planet Lubricon-VI. And Lubricon-VI is a very special planet because it's made up of a yet to be discovered element called Lubrica. And Lubrica is special because if anything glides across the surface of Lubrica, it will experience absolutely no friction. So if this right over here is a sheet of Lubrica-- we're looking at it from the side. And if we have a brick on top of it, maybe gliding on top of it like that, it experiences absolutely no friction. Now, the other things we know about Lubricon-VI is it's drifting in deep space and it does not have an atmosphere. In fact, it is a complete vacuum outside of it. It's in such deep space, such a remote part of space, that there aren't even a few hydrogen atoms right over here. It is a complete, absolute vacuum. And it's also an ancient planet. The star that it used to orbit around has long since died away. So it's just this lonely planet drifting in deep space without an atmosphere. The other thing we know about Lubricon-VI is that it is a perfect sphere. It is a perfect, perfect sphere. Now, my question to you. For some bizarre reason there happens to be, on the surface of Lubricon-VI-- so this right over here is the surface of Lubricon-VI. There happens to be a sock that is frozen in a block of ice. So this is my sock and its frozen in this block of ice. And it happens to be traveling at 1 kilometer per hour in that direction. If we were to look at it from this kind of macro scale when we're looking at the planet, let's say then that is the frozen sock, and it is traveling along the equator. It is traveling along the equator of Lubricon-VI. So my question to you, given all of the assumptions we made that it has absolutely no atmosphere, it's a perfect sphere, and Lubrica has absolutely no friction regardless of what's traveling on top of it-- what will happen to this frozen sock over time? To answer that question, we need to think about all of the forces that are acting on this, I guess, frozen block of ice and sock. And first of all, let's think about these forces that are acting in the radial direction, inward or outward, of the center of the planet. Well, this planet has a mass. And so you have an inward force towards the planet's center of mass. And so you have the force of gravity acting on this block going radially inward to the center of the planet. So I'll draw it like this. So we have our force of gravity. We have our force of gravity going radially inward, just like that. But then we know that the block is not just spiraling towards the center of the earth. We have the surface here. It's not going to go through the surface of Lubrica. We can also assume that Lubrica is a very, very, very strong material. And so you also have a normal force. You also have a normal force that is keeping the block from spiraling towards the center of the earth. So this is a normal force. And one thing we'll think about now, and we'll address it directly in another tutorial, is whether this normal force is equal to the force of gravity. We'll think about that in a future video. But these are all the forces that are acting in the radial direction, either inward towards the center of the planet or outward. But if we think about in the tangential direction, along the surface of the planet, there are no net forces. And because there are no net forces in this tangential direction right over here, this block will not either accelerate nor decelerate. There is no air friction. Or I should say air resistance, which is really just friction with the particles if you had an atmosphere. It's a complete vacuum, so there's nothing there. There is no friction with the surface of the planet. So there's no friction there, which could have been a force in the tangential direction. So there's absolutely no forces in the tangential direction. So this block of ice will actually continue to travel at one kilometer per hour for all of eternity. So it'll just continue to do it given the assumptions that we've just made." + }, + { + "Q": "in the video at 4:24 Sal says that we can get infinitely closer to one. If we can get infinitely closer to one doesn't that mean that we can never approach one?", + "A": "True. Yes, we can always get closer and closer to one but the function actually never reaches one.", + "video_name": "riXcZT2ICjA", + "transcript": "In this video, I want to familiarize you with the idea of a limit, which is a super important idea. It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really, really, really simple idea. So let me draw a function here, actually, let me define a function here, a kind of a simple function. So let's define f of x, let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal look, I have the same thing in the numerator and denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true, the difference between f of x equals 1 and this thing right over here, is that this thing can never equal-- this thing is undefined when x is equal to 1. Because if you set, let me define it. Let me write it over here, if you have f of, sorry not f of 0, if you have f of 1, what happens. In the numerator, we get 1 minus 1, which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here is my x-axis. And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here. And so once again, if someone were to ask you what is f of 1, you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. So you could say, and we'll get more and more familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, let me call it g of x. Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. So it's going to be a parabola, looks something like this, let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that. So this, on the graph of f of x is equal to x squared, this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function x squared. So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1, what's g of 2.01, what's g of 2.001, what is that approaching as we get closer and closer to it. And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1. The limit of g of x as x approaches 2 is equal to 4. And you could even do this numerically using a calculator, and let me do that, because I think that will be interesting. So let me get the calculator out, let me get my trusty TI-85 out. So here is my calculator, and you could numerically say, OK, what's it going to approach as you approach x equals 2. So let's try 1.94, for x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared. And so you get 3.61, well what if you get even closer to 2, so 1.99, and once again, let me square that. Well now I'm at 3.96. What if I do 1.999, and I square that? I'm going to have 3.996. Notice I'm going closer, and closer, and closer to our point. And if I did, if I got really close, 1.9999999999 squared, what am I going to get to. It's not actually going to be exactly 4, this calculator just rounded things up, but going to get to a number really, really, really, really, really, really, really, really, really close to 4. And we can do something from the positive direction too. And it actually has to be the same number when we approach from the below what we're trying to approach, and above what we're trying to approach. So if we try to 2.1 squared, we get 4.4. let me go a couple of steps ahead, 2.01, so this is much closer to 2 now, squared. Now we are getting much closer to 4. So the closer we get to 2, the closer it seems like we're getting to 4. So once again, that's a numeric way of saying that the limit, as x approaches 2 from either direction of g of x, even though right at 2, the function is equal to 1, because it's discontinuous. The limit as we're approaching 2, we're getting closer, and closer, and closer to 4." + }, + { + "Q": "Whats a \"reciprocal\"? (4:07)", + "A": "For any fraction, its reciprocal is created by flipping the fraction. Example: 3/4: its reciprocal is 4/3 -5/2: its reciprocal is -2/5 6: Note 6 as a fraction is 6/1. Its reciprocal = 1/6 In the video, Sal is using the reciprocal of (5x^4)/4, which would be 4/(5x^4). Hope this helps.", + "video_name": "6nALFmvvgds", + "transcript": "- [Voiceover] So up here, we are multiplying two rational expressions. And here, we're dividing one rational expression by another one. Now what I encourage you do is pause this video and think about what these become when you multiply them. I don't know, maybe you simplify it a little bit, and I also want you to think about what constraints do you have to put on the x values in order for your resulting expression to be algebraically equivalent to your original expression. So let's work it out together just so you realize what I'm talking about. So this is going to be, in our numerator, we are going to get six x to the third power times two, and our denominator, we're going to have five times three x. And we can see both the numerator and the denominator are divisible by x, so let's divide the denominator by x. We get one there. Let's divide x to the third by x. We get x squared. And we can also see the both the numerator and denominator are divisible by three, so divide six by three, you get two. Divide three by three, you get one. And we are left with two x squared times two, which is going to be four x squared over five times one times one over five. And we can also write that as 4/5 x squared. Now someone just presented you on the street with the expression 4/5 x squared and say, for what x is this defined? I could put any x here, x could be zero because zero squared is zero times 4/5 is just going to be zero, so it does seems to be defined for zero, and that is true. But if someone says, how would I have to constrain this in order for it to be algebraically equivalent to this first expression? Well then, you'd have to say, well, this first expression is not defined for all x. For example, if x were equal to zero, then you would be dividing by zero right over here, which would make this undefined. So you could explicitly call it out, x can not be equal to zero. And so if you want this to be algebraically equivalent, you would have to make that same condition, x cannot be equal to zero. Another way to think about it, if you had a function defined this way, if you said, if you said f of x is equal to six x to the third over five times two over, times two over three x; and if someone said, well what is f of zero, you would say f of zero is undefined. Undefined. Why is that? Because you put x equals zero there, you're going to get two divided by zero and it's undefined. But if you said, okay, well, can I simplify this a little bit to get the exact same function? Well, we're saying you can say f of x is equal to 4/5 times x squared. But if you just left it at that, you would get f of zero is equal to zero. So now it would be defined at zero, but then this would make it a different function. These are two different functions the way they're written right over here. Instead, to make them, to make it clear that this is equivalent to that one, you would have to say x cannot be equal to zero. Now these functions are equivalent because now, if u said f of zero, you'd say all right, x cannot be equal to zero, you know? This would be the case if x is anything other than zero and it's not defined for zero, and so you would say f or zero is undefined. So now, these two functions are equivalent, or these two expressions are algebraically equivalent. So thinking about that, let's tackle this division situation here. So immediately, when you look at this, you say, woah, what are constrains here? Well, x cannot be equal to zero because if x was a zero, this second, this five x to the fourth over four would be zero and you'd be dividing, you'd be dividing by zero. So we can explicitly call out that x cannot be equal to zero. And so if x cannot be equal to zero in the original expression, if the result, whatever we get for the resulting expression, in order for it to be algebraically equivalent, we have to give this same constraint. So let's multiply this, or let's do the division. So this is going to be the same thing as two x to the fourth power over seven times the reciprocal, times... The reciprocal of this is going to be four over five x to the fourth, which is going to be equal to in the numerator, we're going to have eight x to the fourth. So we're going to have eight x to the fourth, four times two x to the fourth, over seven times five x to the fourth is 35 x to the fourth. And now, there's something. We can do a little bit of simplification here, both the numerator and the denominator are divisible by x to the fourth, so let's divide by x to the fourth and we get eight over 35. So once again, you just look at eight 30, Well, this is going to be defined for any x. X isn't even involved in the expression. But if we want this to be algebraically equivalent to this first expression, then we have to make the same constraint, x does not, cannot be equal to zero. And to see, you know, this even seems a little bit more nonsensical to say x cannot be equal to zero for an expression that does not even involve x. But one way to think about it is imagine a function that was defined as g of x is equal to, is equal to all of this business here. Well, g of zero would be undefined. But if you said g of x is equal to 8/35, well now, g of zero would be defined as 8/35, which would make it a different function. So to make them algebraically equivalent, you could say g of x is equal to 8/35, as long as x does not equal zero. And you could say it's undefined if you want. Undefined for x equals zero. Or you don't even have to include that second row, and that will literally just make it undefined. But now, this expression, this algebraic expression, is equivalent to our original one even though we had simplified it." + }, + { + "Q": "Which trig identities should I have memorized?", + "A": "You can have the basic trig identies memorised 1) sin^2 x + cos^2 x=1 2)tan^2 x + 1 = sec^2 x 3) 1+ cot^2 x = csc^2 x", + "video_name": "rElAJA9GyL4", + "transcript": "- [Voiceover] Let's see if we can take the indefinite integral of cosine of X to the third power. I encourage you to pause the video and see if you can figure this out on your own. You have given it a go and you might have gotten stuck. Some of you all might have been able to figure it out, but some of you all might have gotten stuck. You're like, \"Okay, cosine to the third power. \"Well, gee, if I only had a derivative of cosine here, \"if I had a negative sign of X or a sin of X here, \"maybe I could've used U substitution, \"but how do I take the anti-derivative \"of cosine of X to the third power?\" The key here is, is to use some basic trigonometric identities. What do I mean by that? We know that sin squared X plus cosine squared X is equal to one, or if we subtract sin squared from both sides, we know that cosine squared X is equal to one, write it this way, is equal to one minus sin squared X. What would happen if cosine to the third power, that's cosine squared times cosine. What happens if we were to take that cosine squared? Let me just rewrite it. This is the same thing as cosine of X times cosine squared of X, DX. What if we were to take this thing right over here, let me do that magenta color. What if we were to take this right over here and replace it with this. I now what you're thinking. \"Sal, what's that going to do for me? \"This feels like I'm making this integral \"even more convoluted.\" What I would tell you, I would say, \"This might seem like it's getting more complicated, \"but as you explore and you play with it, \"you'll see that this actually makes \"the integral more solvable.\" Let's try it out. If we do that, this is going to be equal to the indefinite integral, cosine of X times one minus sin squared X, DX. What is this going to be equal to? This is going to be equal to, let me do this in that green color. This is going to be equal to the indefinite integral of cosine X. I'm just going to distribute the cosine of X. Cosine of X minus, minus cosine of X, cosign of X sin squared of X, sin squared, sin squared X and then I can close the parentheses, DX. This, of course, is going to be equal to the integral of cosine of X, DX, and we know what that's going to be, minus the integral. I'll switch to one color now, of cosine of X, sin squared X, sin squared X, DX. Now, this is where it gets interesting. This part right over here is pretty straight forward. The anti-derivative of cosine of X is just sin of X. This right over here is going to be sin of X. I'll worry about the plus C at the end because both of these are going to have a plus C, so might as well just put one big plus c at the end. That's sin of X and then what do we have going on over here? Well, you might recognize, I have a function of sin of X. I'm taking sin of X and I'm squaring it and then I have sin of X's derivative right over here. This fits the, I have some derivative of a function and then I have another and then I have a, I guess you could say, a function of that function. G of F of X. That's a sin that maybe U substitution is in order, or we've seen the pattern, we've seen this show multiple times already, that you could just say, \"Okay, if I have a function of a function \"and I have that functions derivative, \"then essentially I can just take the anti-derivative \"with respect to this function.\" This would be equal to, say, capital G is the anti-derivative of lower case G. Capital G of F of X plus C. Now, if what I said didn't make sense, then we could do U substitution and go through it a little bit more step by step. Let's just do that because we want things to make sense. That's the whole point of these videos. We could say U is equal to sin of X and then DU is going to be equal to cosine of X, DX. This part and that part is going to be DU and then this is going to be U squared. This is going to be minus. We have the integral of U squared, DU. What is this going to be? This is going to be, we're going to have negative U to the third power over three. Then, we know what U is. The U is equal to sin of X. We have our sin of X here for the first part of the integral, for the first integral. We have the sin of X and then this is going to be minus. Let me just write it this way. Minus 1/3 minus 1/3. Instead of U to the third, we know U is sin of X. Sin of X to the third power. Then now, we can throw that plus C there. We're done. We've just evaluated that indefinite integral. The key to it is to just play around a little bit with trigonometric identities so that you can get the integral to a point that you can use the reverse chain rule or you can use U substitution, which is just really another way of expressing the reverse chain rule." + }, + { + "Q": "how did they know it what to name our planets?", + "A": "The names for the planets in our solar system came from the ancient Romans. The Romans named the planets after their gods.", + "video_name": "VbNXh0GaLYo", + "transcript": "What I'm going to attempt to do in the next two videos is really just give an overview of everything that's happened to Earth since it came into existence. We're going start really at the formation of Earth or the formation of our Solar system or the formation of the Sun, and our best sense of what actually happened is that there was a supernova in our vicinity of the galaxy, and this right here is a picture of a supernova remnant, actually, the remnant for Kepler's supernova. The supernova in this picture actually happened four hundred years ago in 1604, so right at the center a star essentially exploded and for a few weeks was the brightest object in the night sky, and it was observed by Kepler and other people in 1604, and this is what it looks like now. What we see is kinda the shockwave that's been traveling out for the past 400 years, so now it must be many light years across. It wasn't, obviously, matter wasn't traveling at the speed of light, but it must've been traveling pretty, pretty fast, at least relativistic speeds, a reasonable fraction of the speed of light. This has traveled a good bit out now, but what you can imagine is when you have the shockwave traveling out from a supernova, let's say you had a cloud of molecules, a cloud of gas, that before the shockwave came by just wasn't dense enough for gravity to take over, and for it to accrete, essentially, into a solar system. When the shockwave passes by it compresses all of this gas and all of this material and all of these molecules, so it now does have that critical density to form, to accrete into a star and a solar system. We think that's what's happened, and the reason why we feel pretty strongly that it must've been caused by a supernova is that the only way that the really heavy elements can form, or the only way we know that they can form is in kind of the heat of a supernova, and our uranium, the uranium that seems to be in our solar system on Earth, seems to have formed roughly at the time of the formation of Earth, at about four and a half billion years ago, and we'll talk in a little bit more depth in future videos on exactly how people figure that out, but since the uranium seems about the same age as our solar system, it must've been formed at around the same time, and it must've been formed by a supernova, and it must be coming from a supernova, so a supernova shockwave must've passed through our part of the universe, and that's a good reason for gas to get compressed and begin to accrete. So you fast-forward a few million years. That gas would've accreted into something like this. It would've reached the critical temperature, critical density and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium, and this right here is our early sun. Around the sun you have all of the gases and particles and molecules that had enough angular velocity to not fall into the sun, to go into orbit around the sun. They were actually supported by a little bit of pressure, too, because you can kinda view this as kind of a big cloud of gas, so they're always bumping into each other, but for the most part it was their angular velocity, and over the next tens of millions of years they'll slowly bump into each other and clump into each other. Even small particles have gravity, and they're gonna slowly become rocks and asteroids and, eventually, what we would call \"planetesimals,\" which are, kinda view them as seeds of planets or early planets, and then those would have a reasonable amount of gravity and other things would be attracted to them and slowly clump up to them. This wasn't like a simple process, you know, you could imagine you might have one planetesimal form, and then there's another planetesimal formed, and instead of having a nice, gentle those two guys accreting into each other, they might have huge relative velocities and ram into each other, and then just, you know, shatter, so this wasn't just a nice, gentle process of constant accretion. It would actually have been a very violent process, actually happened early in Earth's history, and we actually think this is why the Moon formed, so at some point you fast-forward a little bit from this, Earth would have formed, I should say, the mass that eventually becomes our modern Earth would have been forming. Let me draw it over here. So, let's say that that is our modern Earth, and what we think happened is that another proto-planet or another, it was actually a planet because it was roughly the size of Mars, ran into our, what it is eventually going to become our Earth. This is actually a picture of it. This is an artist's depiction of that collision, where this planet right here is the size of Mars, and it ran into what would eventually become Earth. This we call Theia. This is Theia, and what we believe happened, and if you look up, if you go onto the Internet, you'll see some simulations that talk about this, is that we think it was a glancing blow. It wasn't a direct hit that would've just kinda shattered each of them and turned into one big molten ball. We think it was a glancing blow, something like this. This was essentially Earth. Obviously, Earth got changed dramatically once Theia ran into it, but Theia is right over here, and we think it was a glancing blow. It came and it hit Earth at kind of an angle, and then it obviously the combined energies from that interaction would've made both of them molten, and frankly they probably already were molten because you had a bunch of smaller collisions and accretion events and little things hitting the surface, so probably both of them during this entire period, but this would've had a glancing blow on Earth and essentially splashed a bunch of molten material out into orbit. It would've just come in, had a glancing blow on Earth, and then splashed a bunch of molten material, some of it would've been captured by Earth, so this is the before and the after, you can imagine, Earth is kind of this molten, super hot ball, and some of it just gets splashed into orbit from the collision. Let me just see if I can draw Theia here, so Theia has collided, and it is also molten now because huge energies, and it splashes some of it into orbit. If we fast-forward a little bit, this stuff that got splashed into orbit, it's going in that direction, that becomes our Moon, and then the rest of this material eventually kind of condenses back into a spherical shape and is what we now call our Earth. So that's how we actually think right now that the Moon actually formed. Even after this happened, the Earth still had a lot more, I guess, violence to experience. Just to get a sense of where we are in the history of Earth, we're going to refer to this time clock a lot over the next few videos, this time clock starts right here at the formation of our solar system, 4.6 billion years ago, probably coinciding with some type of supernova, and as we go clockwise on this diagram, we're moving forward in time, and we're gonna go all the way forward to the present period, and just so you understand some of the terminology, \"Ga\" means \"billions of years ago\" 'G' for \"Giga-\" \"Ma\" means \"millions of years ago\" 'M' for \"Mega-\" So where we are right now, the Moon has formed, and we're in what we call the Hadean period or actually I shouldn't say \"period.\" It's the Hadean eon of Earth. \"Period\" is actually another time period, so let me make this very clear. It's the Hadean, we are in the Hadean eon, and an eon is kind of the largest period of time that we talk about, especially relative to Earth, and it's roughly 500 million to a billion years is an eon, and what makes the Hadean eon distinctive, well, from a geological point of view what makes it distinctive is really we don't have any rocks from the Hadean period. We don't have any kind of macroscopic-scale rocks from the Hadean period, and that's because at that time, we believe, the Earth was just this molten ball of kind of magma and lava, and it was molten because it was a product of all of these accretion events and all of these collisions and all this kinetic energy turning into heat. If you were to look at the surface of the Earth, if you were to be on the surface of the Earth during the Hadean eon, which you probably wouldn't want to be because you might get hit by a falling meteorite or probably burned by some magma, whatever, it would look like this, and you wouldn't be able to breathe anyway; this is what the surface of the Earth would look like. It would look like a big magma pool, and that's why we don't have any rocks from there because the rocks were just constantly being recycled, being dissolved and churned inside of this giant molten ball, and frankly the Earth still is a giant molten ball, it's just we live on the super-thin, cooled crust of that molten ball. If you go right below that crust, and we'll talk a little bit more about that in future videos, you will get magma, and if you go dig deeper, you'll have liquid iron. I mean, it still is a molten ball. And this whole period is just a violent, not only was Earth itself a volcanic, molten ball, it began to harden as you get into the late Hadean eon, but we also had stuff falling from the sky and constantly colliding with Earth, and really just continuing to add to the heat of this molten ball. Anyway, I'll leave you there, and, as you can imagine, at this point there was no, as far as we can tell, there was no life on Earth. Some people believe that maybe some life could've formed in the late Hadean eon, but for the most part this was just completely inhospitable for any life forming. I'll leave you there, and where we take up the next video, we'll talk a little bit about the Archean eon." + } +] \ No newline at end of file diff --git a/Full-data/train.json b/Full-data/train.json new file mode 100644 index 0000000000000000000000000000000000000000..728f332c13db646f0dcbe69192357eebea76ccd1 --- /dev/null +++ b/Full-data/train.json @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:369a0c367ccbddedd249738585e4ca39cd20afd4f8118348fbea7790cebc5d10 +size 1010353396 diff --git a/MathSc-Timestamp/dev.json b/MathSc-Timestamp/dev.json new file mode 100644 index 0000000000000000000000000000000000000000..2c9bf321303c7c81a473edc761f87e192b5eada5 --- /dev/null +++ b/MathSc-Timestamp/dev.json @@ -0,0 +1,25229 @@ +[ + { + "Q": "\nAt 4:30,Of what physical quantity or what measure does the Boyle's Law P.V=K gives or indicates us?Like momentum gives us a measure of the force to be applied on a body in motion to oppose its motion.", + "A": "Look at the units: Pressure = force / area = N/m^2 Volume = m^3 P*V = N/m^2 * m^3 = N*m N*m is the unit of force*displacement. What is force*displacement?", + "video_name": "x34OTtDE5q8", + "timestamps": [ + 270 + ], + "3min_transcript": "telling you that the box is squeezed, I'm not telling you whether it did any work, or anything like that-- the same constant is going to be equal to the new pressure times the new volume, which is equal to P2 times V2. You could just have the general relationship: P1 times V1 is equal to P2 times V2, assuming that no work was done, and there was no exchange of energy from outside of the system. In most of these cases, when you see this on an exam, that is the case. The old pressure was 500 pascals times 50 meters cubed. One thing to keep in mind, because this equivalence is not equal, and we're not saying it has to equal some necessary absolute number-- for example, we don't know exactly what this K is, although we could figure it out right now-- as long as you're using one unit for you just have to use the same units. We could have done this same exact problem the exact same way, if instead of meters cubed, they said liters, as long as we had liters here. You just have to make sure you're using the same units on both sides. In this case, we have 500 pascals as the pressure, and the volume is 50 meters cubed. That's going to be equal to the new pressure, P2, times the new volume, 20 meters cubed. Let's see what we can do: we can divide both sides by 10, so we can take the 10 out of there, and we could divide both sides by 2, so that becomes a 250. We we get 250 times 5 is equal to P2, and so P2 is equal to 1250 pascals, and if we kept with the units, you would have seen that. pressure actually increased by 2 1/2, so that gels with what we talked about before. Let's add another variable into this mix-- let's talk about temperature. Like pressure, volume, work, and a lot of concepts that we talk about in physics, temperature is something that you probably are at least reasonably familiar with. How do you view temperature? A high temperature means something is hot, and a low temperature means something is cold, and I think that also gives you intuition that a higher temperature object has more energy. The sun has more energy than an ice cube-- I think that's" + }, + { + "Q": "At 19:05, what is the pinkish stuff on the white blood cell (first picture)? Thanks.\n", + "A": "It is a phagocyte (a kind of white blood cells) which helps kill bacteria.", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 1145 + ], + "3min_transcript": "And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia. These little green dots you see right here all over the surface, this big thing you see here, this is a white blood cell. Part of the human immune system. This is a white blood cell. And what you see emerging from the surface, essentially budding from the surface of this white blood cell-- and this gives you a sense of scale too-- these are HIV-1 viruses. And so you're familiar with the terminology, the HIV is a virus that infects white blood cells. AIDS is the syndrome you get once your immune system is weakened to the point. And then many people suffer infections that people with a strong immune system normally won't suffer from." + }, + { + "Q": "at 11:11 sal says the RNA goes to DNA. How does that happen? Also, how does DNA go to RNA in the first place?\n", + "A": "These are things you learn more about in Biology class. But basically, DNA and RNA are the instructions of the cell. DNA can be copied inside a cell and create more RNA or DNA molecules. RNA can go to DNA in a reverse method - in Sal s video he mentions reverse transcriptase.", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 671 + ], + "3min_transcript": "Everything I talk about, these are specific ways that a virus might work. But viruses really kind of explore-- well different types of viruses do almost every different combination you could imagine of replicating and coding for proteins and escaping from cells. Some of them just bud. And when they bud, they essentially, you can kind of imagine that they push against the cell wall, or the membrane. I shouldn't say cell wall. The cell's outer membrane. And then when they push against it, they take some of the membrane with them. Until eventually the cell will-- when this goes up enough, this'll pop together and it'll take some of the membrane with it. And you could imagine why that would be useful thing to have with you. Because now that you have this membrane, you kind of look like this cell. So when you want to go infect another cell like this, you're not going to necessarily look like a foreign particle. So it's a very useful way to look like something that And if you don't think that this is creepy-crawly enough, that you're hijacking the DNA of an organism, viruses can actually change the DNA an organism. And actually one of the most common examples is HIV virus. Let me write that down. HIV, which is a type of retrovirus, which is fascinating. Because what they do is, so they have RNA in them. And when they enter into a cell, let's say that they got into the cell. So it's inside of the cell like this. They actually bring along with them a protein. And every time you say, where do they get this protein? All of this stuff came from a different cell. They use some other cell's amino acids and ribosomes and nucleic acids and everything to build themselves. So any proteins that they have in them came from another cell. But they bring with them, this protein reverse transcriptase. codes it into DNA. So its RNA to DNA. Which when it was first discovered was, kind of, people always thought that you always went from DNA to RNA, but this kind of broke that paradigm. But it codes from RNA to DNA. And if that's not bad enough, it'll incorporate that DNA into the DNA of the host cell. So that DNA will incorporate itself into the DNA of the host cell. Let's say the yellow is the DNA of the host cell. And this is its nucleus. So it actually messes with the genetic makeup of what it's infecting. And when I made the videos on bacteria I said, hey for every one human cell we have twenty bacteria cells. And they live with us and they're useful and they're part of us and they're 10% of our dry mass and all of that. But bacteria are kind of along for the ride." + }, + { + "Q": "\nIn about 12:09 how can they change our genetic makeup", + "A": "it can affect a lot of things..like for example the way are mouth is shaped our body or eye color...ect i hope this helped a little x)", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 729 + ], + "3min_transcript": "And if you don't think that this is creepy-crawly enough, that you're hijacking the DNA of an organism, viruses can actually change the DNA an organism. And actually one of the most common examples is HIV virus. Let me write that down. HIV, which is a type of retrovirus, which is fascinating. Because what they do is, so they have RNA in them. And when they enter into a cell, let's say that they got into the cell. So it's inside of the cell like this. They actually bring along with them a protein. And every time you say, where do they get this protein? All of this stuff came from a different cell. They use some other cell's amino acids and ribosomes and nucleic acids and everything to build themselves. So any proteins that they have in them came from another cell. But they bring with them, this protein reverse transcriptase. codes it into DNA. So its RNA to DNA. Which when it was first discovered was, kind of, people always thought that you always went from DNA to RNA, but this kind of broke that paradigm. But it codes from RNA to DNA. And if that's not bad enough, it'll incorporate that DNA into the DNA of the host cell. So that DNA will incorporate itself into the DNA of the host cell. Let's say the yellow is the DNA of the host cell. And this is its nucleus. So it actually messes with the genetic makeup of what it's infecting. And when I made the videos on bacteria I said, hey for every one human cell we have twenty bacteria cells. And they live with us and they're useful and they're part of us and they're 10% of our dry mass and all of that. But bacteria are kind of along for the ride. But these retroviruses, they're actually changing our I mean, my genes, I take very personally. They define who I am. But these guys will actually go in and change my genetic makeup. And then once they're part of the DNA, then just the natural DNA to RNA to protein process will code their actual proteins. Or their-- what they need to-- so sometimes they'll lay dormant and do nothing. And sometimes-- let's say sometimes in some type of environmental trigger, they'll start coding for themselves again. And they'll start producing more. But they're producing it directly from the organism's cell's DNA. They become part of the organism. I mean I can't imagine a more intimate way to become part of an organism than to become part of its DNA. I can't imagine any other way to actually define an organism. And if this by itself is not eerie enough, and just so you know, this notion right here, when a virus becomes part of" + }, + { + "Q": "\nWhat's the red thing at 19:20?", + "A": "I believe that is part of the infection of the HIV infection. It might be a rupture or infection of the white blood cell from the HIV infection.", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 1160 + ], + "3min_transcript": "And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia. These little green dots you see right here all over the surface, this big thing you see here, this is a white blood cell. Part of the human immune system. This is a white blood cell. And what you see emerging from the surface, essentially budding from the surface of this white blood cell-- and this gives you a sense of scale too-- these are HIV-1 viruses. And so you're familiar with the terminology, the HIV is a virus that infects white blood cells. AIDS is the syndrome you get once your immune system is weakened to the point. And then many people suffer infections that people with a strong immune system normally won't suffer from." + }, + { + "Q": "In 13:10 what is a provirus?\n", + "A": "The genetic material of a virus as incorporated into, and able to replicate with, the genome of a host cell.", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 790 + ], + "3min_transcript": "codes it into DNA. So its RNA to DNA. Which when it was first discovered was, kind of, people always thought that you always went from DNA to RNA, but this kind of broke that paradigm. But it codes from RNA to DNA. And if that's not bad enough, it'll incorporate that DNA into the DNA of the host cell. So that DNA will incorporate itself into the DNA of the host cell. Let's say the yellow is the DNA of the host cell. And this is its nucleus. So it actually messes with the genetic makeup of what it's infecting. And when I made the videos on bacteria I said, hey for every one human cell we have twenty bacteria cells. And they live with us and they're useful and they're part of us and they're 10% of our dry mass and all of that. But bacteria are kind of along for the ride. But these retroviruses, they're actually changing our I mean, my genes, I take very personally. They define who I am. But these guys will actually go in and change my genetic makeup. And then once they're part of the DNA, then just the natural DNA to RNA to protein process will code their actual proteins. Or their-- what they need to-- so sometimes they'll lay dormant and do nothing. And sometimes-- let's say sometimes in some type of environmental trigger, they'll start coding for themselves again. And they'll start producing more. But they're producing it directly from the organism's cell's DNA. They become part of the organism. I mean I can't imagine a more intimate way to become part of an organism than to become part of its DNA. I can't imagine any other way to actually define an organism. And if this by itself is not eerie enough, and just so you know, this notion right here, when a virus becomes part of But if this isn't eerie enough, they estimate-- so if this infects a cell in my nose or in my arm, as this cell experiences mitosis, all of its offspring-- but its offspring are genetically identical-- are going to have this viral DNA. And that might be fine, but at least my children won't get it. You know, at least it won't become part of my species. But it doesn't have to just infect somatic cells, it could infect a germ cell. So it could go into a germ cell. And the germ cells, we've learned already, these are the ones that produce gametes. For men, that's sperm and for women it's eggs. But you could imagine, once you've infected a germ cell, once you become part of a germ cell's DNA, then I'm passing on that viral DNA to my son or my daughter." + }, + { + "Q": "\nat 18:38, he says something about a nuclear membrane? Please explain", + "A": "Nuclear membrane is the membrane that envelops the DNA and nucleolus. In lysogenic cycle, the viral DNA becomes a part of the host s genome.", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 1118 + ], + "3min_transcript": "And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia. These little green dots you see right here all over the surface, this big thing you see here, this is a white blood cell. Part of the human immune system. This is a white blood cell. And what you see emerging from the surface, essentially budding from the surface of this white blood cell-- and this gives you a sense of scale too-- these are HIV-1 viruses. And so you're familiar with the terminology, the HIV is a virus that infects white blood cells. AIDS is the syndrome you get once your immune system is weakened to the point. And then many people suffer infections that people with a strong immune system normally won't suffer from." + }, + { + "Q": "That example with the plane at about 9:15 in the video- I don't exactly get it. I've been on a plane multiple times, and I always know if it's moving or not. Am I missing something here?\n", + "A": "Once you are at cruising altitude, how do you know the plane is moving and the world is not? Maybe the world is moving and you are standing still.", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 555 + ], + "3min_transcript": "The grass is going to stop it from going. And even while it's in the air, it's going to slow down. It's not going to have a constant velocity. Because you have all of these air particles that are going to bump into it and exert force to slow it down. So what was really brilliant about these guys is that they could imagine a reality where you didn't have gravity, where you did not have air slowing things down. And they could imagine that in that reality, something would just keep persisting in its motion. And the reason why Galileo, frankly, was probably good at thinking about that is that he studied the orbits of planets. And he could, or at least he's probably theorized that, hey, maybe there's no air out there. And that maybe that's why these planets can just keep going round and round in orbit. And I should say their speed, because their direction is changing, but their speed never slows down, because there's nothing in the space to actually slow down those planets. Because on some level, it's super-duper obvious. But on a whole other level, it's completely not obvious, especially this moving uniformly straightforward. And just to make the point clear, if gravity disappeared, and you had no air, and you threw a ball, that ball literally would keep going in that direction forever, unless some other unbalanced force acted to stop it. And another way to think about it-- and this is an example that you might see in everyday life-- is, if I'm in an airplane that's going at a completely constant velocity and there's absolutely no turbulence in the airplane. So if I'm sitting in the airplane right over here. And it's going at a constant velocity, completely smooth, no turbulence. There's really no way for me to tell whether that airplane is moving without looking out the window. Let's assume that there's no windows in that airplane. It's going at a constant velocity. And let's say, I can't hear anything. So I can't even hear the engines. There's no way for me to sense that the plane is moving. Because from my frame of reference, it looks completely identical to if I was in that same plane that was resting on the ground. And that's another way to think about it. That it's actually very intuitive that they're similar states, moving at a constant velocity or being at rest. And you really can't tell whether you are one or the other." + }, + { + "Q": "\nat 6:04 after the 3rd resonance structure, why isn't the proton next to NO2 shift over to the positively charged C to make the 4th resonance structure and give positive charge on the Carbon that NO2 is attached to?", + "A": "You can move only electrons to form resonance strictures. If you move the H, you are creating a tautomeric isomer.", + "video_name": "rC165FcI4Yg", + "timestamps": [ + 364 + ], + "3min_transcript": "And then over here, we would have an oxygen with three lone pairs of electrons, giving that a negative 1 formal charge. And the nitrogen, of course, is still going to have a plus 1 formal charge like that. All right, let me go ahead and highlight those electrons. So once again, these pi electrons are going to be attracted to the positive charge, nucleophile-electrophile. And those pi electrons are going to form this bond right here to our nitro group. Well, once again, as we've seen several times before, we took away a bond from this carbon. So that's where our plus 1 the formal charge is going to go like that. And so we can draw some resonance structures. So let's go ahead and show a resonance structure for this. We could move these pi electrons in over here. So let's go ahead and draw that. So we had a hydrogen up here. And you could just show a nitro group as NO2. So I'm just going to go ahead and do that to save some time. And I'm saying that those pi electrons moved over to here. So let me go ahead and highlight those. So these pi electrons in blue move over to here, took a bond away from that carbon. So now we can put a plus 1 formal charge at that carbon like that. We can draw yet another resonance structure. So I could show these electrons over here moving to here. So let me go ahead and draw that. So we have our ring. We have our nitro group already on our ring. We have some pi electrons right here. And we have some more pi electrons moving from here to here, which, of course, takes a bond away from this top carbon. So that's where our positive 1 formal charge is now. So now we have our three resonance structures. And remember, once again, that the sigma complex is a hybrid of these three. And we're now ready for our last step, So if we go back up to here, we think, what could function as a base? Well, the water molecule here could function as a base. So a lone pair of electrons on our water molecule are going to take that proton, which would cause these electrons to move in here to reform your aromatic ring. So let's go ahead and show that. So we're going to reform our benzene ring here. And we took off the proton. So deprotonation of the sigma complex yields our product with a nitro group substituted in. So let me go ahead and highlight those electrons again. So this time I'll use green. So these electrons right in here, when that sigma complex is deprotonated, those electrons are going to move in here to restore the aromatic ring, and we have created our product. We have added in our nitro group." + }, + { + "Q": "\nAt 10:06, I'm confused as to how creating an ova is different through meiosis then creating sperm. How does the cell know if it is creating an ova or a sperm, and wouldn't only one type of sex cell be made inside each gender? Why would the cell create smaller non-functional ova cells that are not useful? What exactly is the difference between the process of making an ova and a sperm? I am very confused about almost the whole last two minutes of the video. Any help appreciated!", + "A": "Meiosis is basically the same for males and females, both males and females undergo the process but males make sperm and females make eggs (ova). There is effectively just a second cell division without replicating the DNA so that the germ cells end up with one copy of each chromosome instead of two (as happens in mitosis). The egg cell divides dispropotionately so that the one big cell has enough resources to support itself once it is fertilized, plus you don t need a lot of eggs.", + "video_name": "TX7-Kdn6lJQ", + "timestamps": [ + 606 + ], + "3min_transcript": "and were then pulled in half, but not here. In meiosis, each chromosome lines up next to it's homologous pair partner that it's already swapped a few genes with. Now, the homologous pairs get pulled apart and migrate to either end of the cell and that's anaphase I. The final phase of the first round, telophase I rolls out in pretty much the same way as mitosis. The nuclear membrane reforms, the nucleoli form within them, the chromosomes fray out back into chromatin, a crease forms between the two new cells called cleavage and then the two new nuclei move apart from each other, the cells separate in a process called cytokinesis, literally again, cell movement and that is the end of round one. We now have two haploid cells, each with 23 double chromosomes that are new, unique combinations of the original chromosome pairs. In these new cells, the chromosomes are still duplicated and still connected at the centromeres. They still look like X's, but remember, the aim is to end up with four cells. Here, the process is exactly the same as mitosis, except that the aim here isn't to duplicate the double chromosomes, but instead to pull them apart into separate single strand chromosomes. Because of this, there's no DNA replication involved in prophase II. Instead, the DNA just clumps up again into chromosomes and the infrastructure for moving them, the microtubules are put back in place. In metaphase II, the chromosomes are moved into alignment into the middle of the cell and in anaphase II, the chromotids are pulled apart into separate single chromosomes. The chromosomes uncoil into chromatin, the crease form in cleavage and the final separation of cytokinesis then mark the end of telophase II. From one original cell with 46 chromosomes, we now have four new cells with 23 single chromosomes each. If these are sperm, all four of the resulting cells are the same size, but they each have slightly different genetic information and half will be for making girls and half will be for making boys, but if this is the egg making process, and the result is only one egg. To rewind a little, during telophase I, more of the inner goodness of a cell, the cytoplasm, the organelles heads into one of the cells that gets split off then to the other one. In telophase II, when it's time to split again, the same thing happens with more stuff going into one of the cells than the other. This big ol' fat remaining cell becomes the egg with more of the nutrients and cytoplasm and organelles that it will take to make a new embryo. The other three cells that were produced, the little ones, are called polar bodies and they're totally useless in people, though they are useful in plants. In plants, those polar bodies actually, also get fertilized too and they become the endosperm. That's the starchy, protein-ey stuff that we grind into wheat, or pop into popcorn and it's basically the nutrients that feed the plant embryo, the seed. And that's all there is to it. I know you were probably were really excited when I started talking about reproduction, but then I rambled on for a long time" + }, + { + "Q": "At 7:17, why are the X chromosomes shown to have crossed over if they can't cross over?\n", + "A": "Hank explains that since the chromosomes are the same, they can cross over. The XY pair can t cross over because they are different chromosomes.", + "video_name": "TX7-Kdn6lJQ", + "timestamps": [ + 437 + ], + "3min_transcript": "Each double chromosome has two chromotids. Here in meiosis prophase I includes two additional and very important steps, crossover and homologous recombination. Remember, the point here, is to end up with four sex cells that each have just one single chromosome from each of the homologous pairs, but unlike in mitosis where all the copies end up the same, here, every copy is going to be different from the rest. Each double chromosome lines up next to it's homologue, so there's your mother's version lined up right next to your father's version of the same chromosome. Now, if you look, you'll see that these two double chromosomes, each with two chromotids, add up to four chromotids. Now watch, one chromotid from each X gets tangled up with the other X. That's crossover and while they're all tangled up, they trade sections of DNA. That's the recombination. The sections that they're trading are from the same location on each chromosome, so one is giving up its genetic code for, it's getting the other chromosomes genes for that trait. Now this is important, what just happened here, creating new gene combinations on a single chromosome. It's the whole point of reproducing this way. Life might be a lot less stressful if we could just clone ourselves, but then we'd also clone all our bad gene combinations and we wouldn't be able to change and adapt to our environment. Remember, that one of the pillars of natural selection is variation and this is a major source of that variation. What's more? Since all of the four chromotids have swapped some DNA segment at random, that means that all four chromotids are now different. Later on in the process, each chromotid will end up in a separate sex cell and that's why all eggs produced by the same woman have a slightly different genetic code. The same for sperm in men and that's why my brother John and I look different even though we're both made from the same two sets of DNA, because of the luck of the genetic draw that happens in recombination, I got this mane of luscious hair and John was stuck with his trash brown puff and don't forget But then, of course, there is that one pair of chromosomes that doesn't always go through the crossover or recombination and that's the 23rd pair and those are your sex chromosomes. If you're female, you have two matched, beautiful, fully capable chromosomes. They are your X chromosomes. Since they are the same, they can do the whole crossover and recombination thing, but if you, like me, are a male, you get one of those X chromosomes and another from your dad that's kind of ugly and short and runted and doesn't have a lot of genetic information on it. During prophase, the X wants nothing to do with the little Y because they're not homologous, so they don't match up and because the XY pairs on these chromosomes will split later into single chromotids, half of the four resulting sperm will be X, leading to female offspring and half will be Y, leading to male offspring. Now, what comes next is another kind of amazing feat of alignment. This is metaphase I and in mitosis you might recall that all of the chromosomes" + }, + { + "Q": "\nat 4:39 when Sal tells us the magnetic field direction is it always like that? is the magnetic field given which hasn't been drawn in the videos ?", + "A": "the magnetic field doesn t have to be to the left, it depends on where the source of it is, he s just arbitrarily chosen it to come from the left for this example.. is that what you meant?", + "video_name": "XMkUDyl1ZRo", + "timestamps": [ + 279 + ], + "3min_transcript": "Now this is when we're looking at top on, where it's popping out of the screen, where it's above the screen. And now we've rotated 180 degrees and this side is on this side, right? Let me pick a suitable color. If this side was green. Now this side, we flip the whole thing over 180 degrees. And now something interesting happens. Remember, before we had this commutator and everything, if we just flipped it over, the current, because before when we didn't have the commutator, the current here was flowing down here, up here. And before the commutator, we had the current flowing down here and up here. And so we were switching directions. And so you would have had this thing that would never completely rotate. It would just keep flipping over, right? Which may be useful for, I don't know, if you wanted to flip things. But it's not useful as a motor. So what happens here? Now this side, all of a sudden instead of being connected to And this green side is now connected to this lead. So something interesting happens. Now the current on the left side is still flowing down, right, and the current on the right side is still flowing up. So we're back to this configuration except that this contraption has flipped over. The brown side is now on the left and the green side is now on the right. And what that allows is that those net torques are still going in that same rotational direction. Use your right hand rule. The current is flowing down here. So if your magnetic field is coming to the left, then the net force is going to be down there and it's going to be up there. And so we can continue indefinite, and we solve our other problem. That we will never keep twisting these wires here. So now using the commutator, we have essentially created an electric motor. And remember I drew that little thing, that could be like the axle. Maybe that turns the wheels or something. So if you have a constant magnetic field and you just by using this commutator which, as soon as you get to that when you go a little bit past vertical, a little bit past 90 degrees, it switches the direction of the current. So on the left hand side you always have the current coming down, and on the right hand side you always have the current going up. So that the net torque is always going to be pushing, is always going to be rotating this contraption down on the left hand side and up on the right hand side. Coming out of the page on the right hand side and then down And you could actually turn a wheel now. You could create an electric car. So that is the basics really of how electric motors are created. Well, there's another way you could have done it. You didn't have to use the commutator. One methodology you could have used is you could have had the magnetic field going until you get to this point, and then you turn off the magnetic field, right? And maybe you wait for this situation to go all the way" + }, + { + "Q": "At 7:50 when he's talking about fluid being pushed in the cochlea, does he just mean that there's a pressure wave in the fluid filling the cochlea? Or is liquid actually sloshing back and forth in it?\n", + "A": "There really is fluid in there! It s called endolymph (in the middle part [scala media] where the hair cells project) and perilymph (in the scalae vestibuli and tympani).", + "video_name": "6GB_kcdVMQo", + "timestamps": [ + 470 + ], + "3min_transcript": "and the next thing that they hit is the eardrum, or tympanic membrane. So, the next thing that they hit is the eardrum. What the eardrum does is it actually starts to vibrate. As this pressurized sound wave hits the eardrum, the eardrum starts vibrating back and forth. When it's vibrating back and forth, it actually causes these little bones - there are three little bones here, one, two and three - it causes these three little bones to vibrate. The first bone is known as the malleus. The second bone is known as the incus, and the third little bone over here is known as the stapes. Let's just recap real quick. The sound waves come in, get funneled by the pinna into the external auditory meatus, otherwise known as the auditory canal, then hit the eardrum, otherwise known as the tympanic membrane, This vibration causes three little bones, known as the malleus, incus and stapes to vibrate back and forth accordingly. The next thing that happens is the stapes is attached to this oval window over here. It's known as the elliptical window, which I'm underlining here. It's also known as the oval window. This oval window starts to vibrate back and forth as well. The next thing that happens is there's actually fluid, so this structure that the oval window is attached to is known as the cochlea. This round structure right here is known as the cochlea. Inside the cochlea is a bunch of fluid. As the oval window gets pushed inside and outside of the cochlea by the stapes, it actually pushes the fluid. It causes the fluid to be pushed this way, and causes the fluid to go all the way around the cochlea. It keeps going all the way around the cochlea, until it reaches the tip of the cochlea. what does it do? The only thing it can do is go back. Now the fluid is gonna have to go back. Let's just follow this green line over here. The fluid moves back towards where it came, but it actually doesn't go back to the oval window. It actually goes to this other window known as the circular, or round, window. Let me just fix that, so it goes to this circular or round window. It causes the round window to get pushed out. This basically keeps happening, so the fluid moves all the way to the tip of the cochlea, all the way back out, and back and forth, and back and forth, until the energy of this sound wave - eventually the fluid stops moving - all that energy is dissipated. Meanwhile, hair cells inside the cochlea are being pushed back and forth, and that transmits an electrical impulse" + }, + { + "Q": "At 9:20, when we are try to figure out the direction of the induced current are we referring to the conventional current or the actual electron flow? This becomes a point of confusion in a number of my homework problems for my AP class, do physicists generally stick to conventional current for all problems having to do with inducing a current?\n", + "A": "Current always refers to conventional current unless it says electron current.", + "video_name": "9q-T8o1HUcw", + "timestamps": [ + 560 + ], + "3min_transcript": "amperes, 2.5 amperes. So we now know the magnitude of the current that's going to be induced while we have this change in flux, remember this is going to happen while, over the course of those four seconds, as we have this rate of change of flux, this average rate of change of flux, which we'll assume is the actual rate of change of flux, we're assuming that it's changing at a constant rate and so while it is changing we were just able to figure out that it would induce a current of 2.5 amperes. Now the next question we should ask ourselves and this is where this little negative comes in, is a reminder for us to use Lenz's law is, well which direction is that current going to go in? Is it going to go in, let me pick two orientations, is it going to go in a, is it going to go in a, in a clockwise direction, is it going to go that way over the course of this change in flux or is it going to go in a counterclockwise direction, And to think about that we just have to use the right hand rule, take our right hand, point our thumb in the direction of the proposed direction of the current and so if we went with this one, our right hand, our right hand would look like this, I'm literally taking my left hand out and-- I mean my right hand out and I'm drawing it and I'm looking at it to think about what would happen, so that's my right hand so if I use the right hand if the current went in this direction then it would induce a magnetic field that went, that went like this and so if the current went in this direction the magnetic field it induces inside the surface would only reinforce the change in flux so it would only add to the flux so, and it's going in the same direction as the change in flux, which would just keep us, you know as we talked about in the Lenz's law video, that would turn into just this source of energy that comes out of nowhere and defies the law of conservation of energy so this absolutely not, is not going to be the direction is going to be in a clockwise one. So the current, the 2.5 ampere current is going to flow, is going to flow like that, and we're done! By thinking about our change in flux and how long it's taking us, we were able to figure out not only the magnitude of the current, we were able to figure out the orientation of the direction that it's actually going to flow in." + }, + { + "Q": "Hmm im not sure if the explanation at 9:20 is sound. Would there be a magnetic flux outside of the circle also, then the circuit could not \"countertact (lenz)\" in any way. If the B Field created on the inside counteracts it will strengthen in on the outside ?\n", + "A": "We are only talking about the flux THROUGH the coil. The field lines (flux) are passing only inside the coil in this case. This is just a theoretical problem. You can t really do this irl", + "video_name": "9q-T8o1HUcw", + "timestamps": [ + 560 + ], + "3min_transcript": "amperes, 2.5 amperes. So we now know the magnitude of the current that's going to be induced while we have this change in flux, remember this is going to happen while, over the course of those four seconds, as we have this rate of change of flux, this average rate of change of flux, which we'll assume is the actual rate of change of flux, we're assuming that it's changing at a constant rate and so while it is changing we were just able to figure out that it would induce a current of 2.5 amperes. Now the next question we should ask ourselves and this is where this little negative comes in, is a reminder for us to use Lenz's law is, well which direction is that current going to go in? Is it going to go in, let me pick two orientations, is it going to go in a, is it going to go in a, in a clockwise direction, is it going to go that way over the course of this change in flux or is it going to go in a counterclockwise direction, And to think about that we just have to use the right hand rule, take our right hand, point our thumb in the direction of the proposed direction of the current and so if we went with this one, our right hand, our right hand would look like this, I'm literally taking my left hand out and-- I mean my right hand out and I'm drawing it and I'm looking at it to think about what would happen, so that's my right hand so if I use the right hand if the current went in this direction then it would induce a magnetic field that went, that went like this and so if the current went in this direction the magnetic field it induces inside the surface would only reinforce the change in flux so it would only add to the flux so, and it's going in the same direction as the change in flux, which would just keep us, you know as we talked about in the Lenz's law video, that would turn into just this source of energy that comes out of nowhere and defies the law of conservation of energy so this absolutely not, is not going to be the direction is going to be in a clockwise one. So the current, the 2.5 ampere current is going to flow, is going to flow like that, and we're done! By thinking about our change in flux and how long it's taking us, we were able to figure out not only the magnitude of the current, we were able to figure out the orientation of the direction that it's actually going to flow in." + }, + { + "Q": "at 11:05 can a system ever truly have zero internal energy. Wouldn't the system itself contain energy? This sounds like a black hole situation. Is it possible for the balloon to have an absence of molecules and atoms...no internal energy at all? Or is this just hypothetical to make the equation simpler?\n", + "A": "At 11:05 this was referring to no change in energy, not no energy.", + "video_name": "aOSlXuDO4UU", + "timestamps": [ + 665 + ], + "3min_transcript": "W is work done. Now, the system did work to something else. It didn't have work done to it. So if this is work done to the system, and it did work, then this is going to be a minus 10. Minus 10 joules. And then you solve both sides of this. You add 10 to both sides and you get 10 is equal to Q, which is exactly what we got up here. But this can get confusing sometimes, because you're like, oh, is this heat that the system did? Is this heat that was added to the system or taken away? The convention tends to be that this is heat added. But then sometimes it's confusing. Is this the work done to the system or work done by? And that's why I like just doing it this way. If the system does work it loses energy. If the system has work done to it, it gains energy. So let's do another problem. other formula that you'll sometimes see. Delta U is equal to Q minus the work that the system does by the system-- Work done by the system. And in this case, once again, change in internal energy was 0. That is equal to heat added to the system, minus the work done. So minus-- I told you at the beginning of the problem that the system did 10 joules of work-- so minus the work done. Minus 10 joules. We get the same situation up here from two different formulas. And we got 10 is equal to Q. Either way, the heat added to the system is 10 joules. Let's do one more. Let's say that, I don't know, 5 joules of heat taken away And let's say that 1 joule of work done on system. So maybe we're compressing the balloon on the system. What is our change in internal energy? Let's just figure out our change. So the way I think of it is 5 joules of heat taken away from the system, that's going to reduce our internal energy by minus 5. And if 1 joule work is done onto the system, we're putting energy into it so that'll be plus 1. So minus 5, plus 1, is going to be minus 4. Or enter our change in internal energy is minus 4 joules. Now we could have done that a little bit more formally with the formula, change in internal energy is equal to heat added to the system, plus work done on the system." + }, + { + "Q": "\nAt 6:43 Sal said that change in interal energy=Q+W is the definition for internal energy can you give me a more precise definition for it?", + "A": "Internal energy is all the energy contained in an object, including both kinetic energy and potential energy.", + "video_name": "aOSlXuDO4UU", + "timestamps": [ + 403 + ], + "3min_transcript": "transfer amount? No, wire transfer was how money was deposited or taken Likewise, it didn't have a checking deposit account, so I can't really-- it just seems weird to me to say change in wire transfer is $10, or change in check deposits is $20 or minus $20. Would you say, I made a $10 wire transfer and I paid $20 in checks. So I had a net change in my bank account of $10. Likewise, I say how much work was done to me or I did, which is essentially a deposit or withdrawal of energy. Or I could say much heat was given to me or how much heat was released, which is another way of depositing or withdrawing energy from my energy bank account. So that's why I like to stick to this. And I like to stay away from this. And just like I said, you can't say how much heat is in the system. So someone will say, oh, how much heat is in this? There's no heat state variable for that system. You have internal energy. The closest thing to heat, we'll talk about it in a future video, is enthalpy. Enthalpy is essentially a way of measuring how much heat is in a system, but we won't talk about that just now. And you can't say how much work is in a system. You can't say, oh I have x amount of work in a system. The system can do work or have work done to it, but there's not a certain amount of work, because that energy in the system could be all used for work, it could all be used for heat, it could all be used for a ton of different things. So you can't say those things. And that's why I don't like treating them like state variables, or state functions. So with that said, this is our definition. Let's do a couple of simple problems, just to give you intuition. And I really want to make you comfortable. My real goal is to make you comfortable with, when to know to use plus or minus on the work. And the best way to do it is not to memorize a formula, but just to kind of think about what's happening. So let's say that I have some system here and, I don't know, And let's say that I have no change in internal energy. Internal energy is 0. And for our purposes we can kind of view it as the kinetic energy of the particles haven't changed. And let's say by expanding a bit, by my balloon expanding a bit-- I did some work. I'll do this in more detail in the next video. So the system does 10 joules of work. My question is, how much heat was added or taken away from the system? So the way I can think about it-- you don't even have to write the formula down, or you can write it-- you could say, look, the internal energy, the amount of energy in the system hasn't changed. The system did 10 joules of work. So that's energy going out of the system. It did 10 joules. So 10 joules went out in the form of work." + }, + { + "Q": "\nAt 2:14 Sal says that the egg cell is also a gamete. Isn't he wrong because a gamete is haploid while the egg cell, as a result of the fertilization, is diploid ? He confirms what I say at 4:45.", + "A": "The egg is not diploid. The egg and the sperm cell are gametes and haploid. The zygote is the result of the fertilization of the egg cell by the sperm cell. The zygote is diploid.", + "video_name": "dNp7vErqlaA", + "timestamps": [ + 134, + 285 + ], + "3min_transcript": "Voiceover: So let's talk a little bit about how we all came into being. What we see right over here, this is a picture of a sperm cell fusing with an egg cell. So that's a sperm cell and this is an egg cell or we could call this an ovum. And even this scene depicted right over here, this is the end of an epic competition because this sperm cell is one of one of two to three hundred million that is vying for this ovum. So there's two to three hundred million of these characters and they're all vying for this ovum and the one that you see that's about to fuse for it, this is the winner of this incredibly - remember two to three hundred, 200 million to 300 million sperm are trying to get here so this is a major victory and to some degree we should all feel pretty good about ourselves because we are all the by-product of that that won this race getting to our mother's ovum. So the sperm cell came from our father and the egg cell, this is all happening inside of our mothers, the egg cell is from our mother. Now, once this happens, let's talk a little bit about the terminology. So once these two fuse, or the process of them fusing, we call that fertilization. Fertilization. And it produces a cell that then differentiates into all of the cells of our body, so you can imagine that this is an important process. So let's make sure that we understand the different terminology, the different words for the different things that are acting in this process. So each of these sex cells, I guess we could say, So this right over here is a gamete and the ovum is a gamete, the egg cell is also a gamete. And as we'll see, each gamete has half the number of chromosomes as your body cells or most of the somatic cells of your body so outside of your sex cells that might be in your ovaries or your testes, depending on whether you're male or female, these have half the number so let's dig a little bit deeper into what I mean there. So let's just do a blow up of this sperm cell right over here, so a blow up of a sperm cell and I'm not going to draw it to scale, you see the sperm cell is much smaller than the egg cell but just to get a sense, so let me draw the nucleus" + }, + { + "Q": "At 4:40, is it ever possible for an ovum to carry a Y chromosome?\n", + "A": "Adding on to what Mubashra Igbal said, because the female has two X chromosomes, it is only possible to get an X from the female. The X chromosome is required for life so even if they could give a Y and if the male gave a Y, the baby would not survive.", + "video_name": "dNp7vErqlaA", + "timestamps": [ + 280 + ], + "3min_transcript": "If we're talking about a human being, and I'm assuming you are a human being, so that might be of interest to you, this will have 23 chromosomes from your father so let's do them. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and for the 23rd one, that's going to be your sex-determining chromosome so if your father contributes an x, you are going to be female, if your father contributes a y, you are going to be male. So these are the chromosomes in the male gamete or I guess I should say the gamete that your father's contributing, the sperm. So this is a gamete right over here and that's going to fuse with the egg, the ovum that your mother is contributing and once again, I'm not drawing that to scale. So this is the egg, and let me draw it's nucleus. none of this is drawn to scale. And your mother is also going to contribute 23 chromosomes. So one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and then she will contribute an x chromosome for the sex determining so your sex determining chromosomes are going to be xy, you're going to be male, if this was xx, you're going to be female so this is also a gamete here. So a gamete is the general term for either a sperm or an egg. Now once these two things are fused, what do we have? Once they're fused, then we're going to have you could say a fertilized egg but we are going to call that a zygote so let me draw that. I'm going to do this in a new color, and I'm running out of space and I want this all to fit and so let me draw the nucleus of the zygote, I'm going to make the nucleus fairly large so that we can focus on the chromosomes in it, once again none of this is drawn to scale. So you're going to have the 23 chromosomes from your father, so let me do that. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and 23, and then the 23 chromosomes from your mother. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and 23" + }, + { + "Q": "At 7:06 The drawing shows that we have 46 chromosomes, but do we los 23 of those chromosomes when sperm connects with the ovum\n", + "A": "Normal cells have 46 chromosomes or 23 pairs of chromosomes. Normal cell devision, mitosis, produces 2 cells with 46 chromosomes but during the creation of sex cells, meiosis, cells with 23 chromosomes are produced so that when the two sex cells come together they produce a cell with 46 chromosomes.", + "video_name": "dNp7vErqlaA", + "timestamps": [ + 426 + ], + "3min_transcript": "and so let me draw the nucleus of the zygote, I'm going to make the nucleus fairly large so that we can focus on the chromosomes in it, once again none of this is drawn to scale. So you're going to have the 23 chromosomes from your father, so let me do that. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and 23, and then the 23 chromosomes from your mother. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 and 23 And as you might have notice, I've drawn them in pairs so you now have a total, let me make it clear, you have 23 chromosomes here, 23 chromosomes in the sperm, you have 23 chromosomes in the egg and now you have 46 chromosomes in the fertilized egg, 46 chromosomes, and now that we have a full contingent of chromosomes and then this cell can now keep replicating, keep splitting and differentiating into all of what makes you, you, we call this right over here, we call this a zygote. So one way to think about it, the gametes are the sex cells that have half the number of chromosomes and the zygote is the cell that's now ready to differentiate into an actual organism that has double the number that has 46 chromosomes, and you see that I've made them in pairs and these pairs, we call these homologous pairs and in each of these pairs, this is a pair of homologous chromosomes. So what does that mean? Well that means that in general, these two chromosomes, you got one from your father, one from your mother, they code for the same things, they code for the same proteins but there are different variants of how they code for those proteins, those traits that you have so gross oversimplification is, let's say that there is a gene on, that one from your father that helps code for hair color well there would be a similar, there would be another variant of that gene on the chromosome from your mother" + }, + { + "Q": "\nAt 2:10, Sal talks about how if you had a mono-North pole on top of a mono-South pole, the mono-North pole would go around and get under the mono-South pole. He does this to describe magnetic fields, but what I don't get is why the mono-North pole doesn't just go strait down-strait to the mono-South pole. I don't understand why the field goes around rather than strait down. Somebody please explain.", + "A": "magnetic field lines actually pass through the object, they do not stop at the poles. some field lines are actually passing straight down.", + "video_name": "NnlAI4ZiUrQ", + "timestamps": [ + 130 + ], + "3min_transcript": "We know a little bit about magnets now. Let's see if we can study it further and learn a little bit about magnetic field and actually the effects that they have on moving charges. And that's actually really how we define magnetic field. So first of all, with any field it's good to have a way to visualize it. With the electrostatic fields we drew field lines. So let's try to do the same thing with magnetic fields. Let's say this is my bar magnet. This is the north pole and this is the south pole. Now the convention, when we're drawing magnetic field lines, is to always start at the north pole and go towards the south pole. And you can almost view it as the path that a magnetic north monopole would take. So if it starts here-- if a magnetic north monopole, even though as far as we know they don't exist in nature, although they theoretically could, but let's just say for the sake of argument that we do have a magnetic north monopole. If it started out here, it would want to run away from this north pole and would try to get to the south pole. something like this. If it started here, maybe its path would look something like this. Or if it started here, maybe its path would look something like this. I think you get the point. Another way to visualize it is instead of thinking about a magnetic north monopole and the path it would take, you could think of, well, what if I had a little compass here? Let me draw it in a different color. Let's say I put the compass here. That's not where I want to do it. Let's say I do it here. The compass pointer will actually be tangent to the field line. So the pointer could look something like this at this point. It would look something like this. And this would be the north pole of the pointer and this would be the south pole of the pointer. Or you could-- that's how north and south were defined. People had compasses, they said, oh, this is the north seeking pole, and it points in that direction. of the larger magnet. And that's where we got into that big confusing discussion of that the magnetic geographic north pole that we're used to is actually the south pole of the magnet that we call Earth. And you could view the last video on Introduction to Magnetism to get confused about that. But I think you see what I'm saying. North always seeks south the same way that positive seeks negative, and vice versa. And north runs away from north. And really the main conceptual difference-- although they are kind of very different properties-- although we will see later they actually end up being the same thing, that we have something called an electromagnetic force, once we start learning about Maxwell's equations and relativity and all that. But we don't have to worry about that right now. But in classical electricity and magnetism, they're kind of a different force. And the main difference-- although you know, these field lines, you can kind of view them as being similar-- is that magnetic forces always come in dipoles, soon. while you could have electrostatic forces that are monopoles." + }, + { + "Q": "\nIs the formula he used, the same as the formula for the Lorentz power? At 8:10 he defined the formula, and indicated the unit for B as Tesla. In my book, the formula for the Lorentz power is given as:\nF=B*I*L.\nF= Lorentz power\nB= magnetic induction\nI= electric current\nL= length of the magnetic field\n\nIt's a bit confusing which formula I should use. Can somebody explain it please?", + "A": "I think you mean Lorentz force, not power. This is a specific application of the Lorentz force.", + "video_name": "NnlAI4ZiUrQ", + "timestamps": [ + 490 + ], + "3min_transcript": "I know I'm confusing you at this point, so let's play around with it and do some problems. But before that, let's figure out what the units of the magnetic field are. So we know that the cross product is the same thing as-- so let's say, what's the magnitude of the force? The magnitude of the force is equal to? Well, the magnitude of the charge-- this is just a scalar quantity, so it's still just the charge-- times the magnitude of the velocity times the magnitude of the field times the sine of the angle between them. This is the definition of a cross product and then we could put-- if we wanted the actual force vector, we can just multiply this times the vector we get using the We'll do that in a second. Anyway we're just focused on units. Sine of theta has no units so we can ignore it for this discussion. We're just trying to figure out the units of the magnetic field. So force is newtons-- so we could say newtons equals-- this is times the-- I don't know what we'll call this-- the B units. We'll call it unit sub B. So let's see. If we divide both sides by coulombs and meters per second, we get newtons per coulomb. And then if we divide by meters per second, that's the same thing as multiplying by seconds per meter. Equals the magnetic field units. So the magnetic field in SI terms, is defined as newton seconds per coulomb meter. And that might seem a little disjointed, and they've come up with a brilliant name. And it's named after a deserving fellow, and that's Nikolai Tesla. And so the one newton second per coulomb meter is And I'm actually running out of time in this video, because I want to do a whole problem here. But I just want you to sit and think about it for a second. Even though in life we're used to dealing with magnets as we have these magnets-- and they're fundamentally maybe different than what at least we imagine electricity to be-- but the magnitude or actually the units of magnetism is actually defined in terms of the effect that it would have on a moving charge. And that's why the unit-- one tesla, or a tesla-- is defined as a newton second per coulomb. So the electrostatic charge per coulomb meter. Well, I'll leave you now in this video. Maybe you can sit and ponder that. But it'll make a little bit more sense when we do some actual problems with some actual numbers in the next video. See" + }, + { + "Q": "\nAt 7:30, the generic acid, A, was used to deprotonate the H from the nitrogen group. Could we have used H2O to deprotonate, because that way it would become H3O+ in the end?", + "A": "Yes, the generic base A: could have been a water molecule.", + "video_name": "Gl7bQNm92fE", + "timestamps": [ + 450 + ], + "3min_transcript": "So this intermediate is called a carbinolamine. So let me go ahead and write that. So this is called a carbinolamine. And for the next step we can protonate the O-H group. So we take our generic acid once again so H-A plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let's go ahead and show that. So let's get once again some more space in here. So if we protonate our carbinolamine, we would have our carbon here, our nitrogen, lone pair of electrons on our nitrogen, our Y group, our hydrogen here, our alkyl groups. And then if we protonate the O-H, we would form water as a good leaving group. So right in here, this would be a plus one formal charge So in here, lone pair of electrons on the oxygen, pick up this proton and so forming this bond right here and now you can see we have a good leaving group. So we have water as a good leaving group. So if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen. So let's go ahead and show that. So our next step here, this is where we lose water. So we're going to minus H2O and let's go ahead and show our next structure here. We have our carbon double bonded to our nitrogen this time, our nitrogen is still bonded to our Y group. And our hydrogen over here. That gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups And so let's show those electrons here on our nitrogen. So I'm going to go ahead and make those magenta. So these magenta electrons move in here to form And this is an important structure, this is called an iminium ion, so let's go ahead and draw that. So an iminium ion and then we lose water of course. So minus water at this stage. So we're almost to our final product, we would just have to deprotonate our iminium ion. And so we can do that with our imine. Could take that proton and leave these electrons behind on the nitrogen. So that's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our Y group here. A lone pair of electrons on that nitrogen and then we have our alkyl groups and so let's show those electrons here in blue moving in here, off onto our nitrogen. And then once again, if our Y is a hydrogen an alkyl group we have formed an imine. Alright, so if that's formation of an imine," + }, + { + "Q": "At 3:35, How can we go around the sphere if its a 2 dimensional object? We can only go back and forth.\n", + "A": "The surface of the Earth is pretty much two dimensional... how do you travel on it?", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 215 + ], + "3min_transcript": "And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that Let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitudinal lines on this sphere. On this sphere, all of a sudden-- and I'll shade it in a little bit, make it look nice-- this type of a sphere, you have a finite area. You could imagine the surface of a balloon, or the surface of a bubble, or the surface of the Earth. You have a finite area, but you have no edge. If you keep going forever in one direction, you're going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and-- and I don't want to say finite area anymore, because we're not talking about a three-dimensional space. Let me draw it over here. So let's think about a three-dimensional space, so a three-dimensional space. I want to talk about a finite volume and no edge. How do I do that? And when you think about it superficially, well, look, if I have a finite volume, maybe it'll be contained in some type of a cube. And then we clearly have edges in those situations. Or you could even think about a finite volume as being the inside of a sphere. And that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has a finite volume and no edge? And that I'm going to tell you right now, it's very hard for us to visualize it. But in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy, unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere." + }, + { + "Q": "\nHow is a sphere a 2D object? There's not much more to say other than it was at 3:21 in the video, so there's my question.", + "A": "A sphere is a 3D object but you can imagine its surface as a curved 2 dimensional space.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 201 + ], + "3min_transcript": "has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that Let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitudinal lines on this sphere. On this sphere, all of a sudden-- and I'll shade it in a little bit, make it look nice-- this type of a sphere, you have a finite area. You could imagine the surface of a balloon, or the surface of a bubble, or the surface of the Earth. You have a finite area, but you have no edge. If you keep going forever in one direction, you're going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and-- and I don't want to say finite area anymore, because we're not talking about a three-dimensional space. Let me draw it over here. So let's think about a three-dimensional space, so a three-dimensional space." + }, + { + "Q": "\nat 3:10 you are drawing a sphere while explaining about two dimensional space. please explain me this concept.", + "A": "The surface of the sphere has only two dimensions to it. You can think of those dimensions as latitude/longitude or right ascension/declination.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 190 + ], + "3min_transcript": "has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that Let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitudinal lines on this sphere. On this sphere, all of a sudden-- and I'll shade it in a little bit, make it look nice-- this type of a sphere, you have a finite area. You could imagine the surface of a balloon, or the surface of a bubble, or the surface of the Earth. You have a finite area, but you have no edge. If you keep going forever in one direction, you're going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and-- and I don't want to say finite area anymore, because we're not talking about a three-dimensional space. Let me draw it over here. So let's think about a three-dimensional space, so a three-dimensional space." + }, + { + "Q": "\nAround 6:22 Sal mentions a \"toroid would fit the bill\" Could someone explain this shape to me. I dd look up the definition by the way. I'm looking for something a bit more intuitive.", + "A": "A toroid is kind of like a donut shape. Its basically just a cylinder bend so thats its two ends connect.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 382 + ], + "3min_transcript": "I want to talk about a finite volume and no edge. How do I do that? And when you think about it superficially, well, look, if I have a finite volume, maybe it'll be contained in some type of a cube. And then we clearly have edges in those situations. Or you could even think about a finite volume as being the inside of a sphere. And that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has a finite volume and no edge? And that I'm going to tell you right now, it's very hard for us to visualize it. But in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy, unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere. This is a two-dimensional surface. On the surface of the sphere, you can only move into directions, two perpendicular directions. You could move like that or you could move like that. You could move left and right or you could move up and down. So it's a two-dimensional surface of a three-dimensional sphere. So if we take it by analogy, let's imagine, and it's hard to imagine, a three-dimensional surface. And you can do it mathematically. The math here is actually not that difficult. It's a three-dimensional surface of a four-dimensional sphere. And I'm going to draw it the same way. are just these two dimensions of the surface, the same thing. It's the same thing. And if you imagine that-- I'm not saying that this is actually the shape of the universe. We don't know the actual shape. But we do know that it does have a slight curvature. We don't know the actual shape, but a sphere is the simplest. There's other ones we could do. A toroid would also fit the bill of having a finite volume with no edge. And another thing, I want to make it clear, we actually don't even know whether it has just a finite volume. That's still an open question. But what I want to do is show you that it can have a finite volume and also have no edge. And most people believe-- and I want to say \"believe\" here because we can just go based on evidence and all that-- that we are talking about something with a finite volume, especially when you talk about the Big Bang theory. That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite." + }, + { + "Q": "At 6:30 sal talks about a torroid. Meaning, please ?!?!?\n", + "A": "A torroid is something that has the properties of a torus. A torus is something shaped like a donut.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 390 + ], + "3min_transcript": "I want to talk about a finite volume and no edge. How do I do that? And when you think about it superficially, well, look, if I have a finite volume, maybe it'll be contained in some type of a cube. And then we clearly have edges in those situations. Or you could even think about a finite volume as being the inside of a sphere. And that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has a finite volume and no edge? And that I'm going to tell you right now, it's very hard for us to visualize it. But in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy, unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere. This is a two-dimensional surface. On the surface of the sphere, you can only move into directions, two perpendicular directions. You could move like that or you could move like that. You could move left and right or you could move up and down. So it's a two-dimensional surface of a three-dimensional sphere. So if we take it by analogy, let's imagine, and it's hard to imagine, a three-dimensional surface. And you can do it mathematically. The math here is actually not that difficult. It's a three-dimensional surface of a four-dimensional sphere. And I'm going to draw it the same way. are just these two dimensions of the surface, the same thing. It's the same thing. And if you imagine that-- I'm not saying that this is actually the shape of the universe. We don't know the actual shape. But we do know that it does have a slight curvature. We don't know the actual shape, but a sphere is the simplest. There's other ones we could do. A toroid would also fit the bill of having a finite volume with no edge. And another thing, I want to make it clear, we actually don't even know whether it has just a finite volume. That's still an open question. But what I want to do is show you that it can have a finite volume and also have no edge. And most people believe-- and I want to say \"believe\" here because we can just go based on evidence and all that-- that we are talking about something with a finite volume, especially when you talk about the Big Bang theory. That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite." + }, + { + "Q": "\nAt 6:22 Sal mentiones a \"toroid\" (not sure of the spelling). What is a toroid, what does it look like?", + "A": "A toroid is a donut shape.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 382 + ], + "3min_transcript": "I want to talk about a finite volume and no edge. How do I do that? And when you think about it superficially, well, look, if I have a finite volume, maybe it'll be contained in some type of a cube. And then we clearly have edges in those situations. Or you could even think about a finite volume as being the inside of a sphere. And that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has a finite volume and no edge? And that I'm going to tell you right now, it's very hard for us to visualize it. But in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy, unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere. This is a two-dimensional surface. On the surface of the sphere, you can only move into directions, two perpendicular directions. You could move like that or you could move like that. You could move left and right or you could move up and down. So it's a two-dimensional surface of a three-dimensional sphere. So if we take it by analogy, let's imagine, and it's hard to imagine, a three-dimensional surface. And you can do it mathematically. The math here is actually not that difficult. It's a three-dimensional surface of a four-dimensional sphere. And I'm going to draw it the same way. are just these two dimensions of the surface, the same thing. It's the same thing. And if you imagine that-- I'm not saying that this is actually the shape of the universe. We don't know the actual shape. But we do know that it does have a slight curvature. We don't know the actual shape, but a sphere is the simplest. There's other ones we could do. A toroid would also fit the bill of having a finite volume with no edge. And another thing, I want to make it clear, we actually don't even know whether it has just a finite volume. That's still an open question. But what I want to do is show you that it can have a finite volume and also have no edge. And most people believe-- and I want to say \"believe\" here because we can just go based on evidence and all that-- that we are talking about something with a finite volume, especially when you talk about the Big Bang theory. That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite." + }, + { + "Q": "At 00:41 Sal said that evertything is packed in together? i dont understand whats that everything? And where this everything came from?\n", + "A": "We don t know where it came from. We just know that everything in the universe now was once much, much closer together before it was flung apart by the big bang.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 41 + ], + "3min_transcript": "Right now, the prevailing theory of how the universe came about is commonly called the Big Bang theory. And really is just this idea that the universe started as kind of this infinitely small point, this infinitely small singularity. And then it just had a big bang or it just expanded from that state to the universe that we know right now. And when I first imagined this-- and I think if it's also a byproduct of how it's named-- Big Bang, you kind of imagine this type of explosion, that everything was infinitely packed in together and then it exploded. And then it exploded outward. And then as all of the matter exploded outward, it started to condense. And then you have these little galaxies and super clusters of galaxies. And they started to condense. And then within them, planets condensed and stars condensed. And then we have the type of universe that we have right now. has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that" + }, + { + "Q": "\n@2:31 sal says finite area, what does that mean?", + "A": "No infinite or has a definable distance.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 151 + ], + "3min_transcript": "has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that Let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitudinal lines on this sphere. On this sphere, all of a sudden-- and I'll shade it in a little bit, make it look nice-- this type of a sphere, you have a finite area. You could imagine the surface of a balloon, or the surface of a bubble, or the surface of the Earth. You have a finite area, but you have no edge. If you keep going forever in one direction, you're going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and-- and I don't want to say finite area anymore, because we're not talking about a three-dimensional space. Let me draw it over here. So let's think about a three-dimensional space, so a three-dimensional space." + }, + { + "Q": "\nAt 5:15 when he is dividing by .3639, is he dividing by the molar mass of Hg?", + "A": "No, he is dividing by the number of moles of Hg.", + "video_name": "NM0WycKCCDU", + "timestamps": [ + 315 + ], + "3min_transcript": "so that's why I said roughly. So how many moles is this? This is going to be 73 over 200.59 of a mole. If a mole is 200.59 we have 73, this is the fraction of a mole that we have, moles of mercury. Remember, moles are just a number, Avogadro's number of something, but let's just figure out what it is. If we take 73 divided by 200.59 we get .36, I'll just say 0.364, and once again this is, so approximately 0.364. That's how many moles of mercury that we have. We can do the same thing for chlorine. This is going to be 27 over 35.453 moles of chlorine, 27 divided by 35.453 is equal to .76, I'll just say two. So 0.762 moles of chlorine. What's going to be the ratio of mercury to chlorine? Or I guess we could say since chlorine, there's more of that, chlorine to mercury. Remember, this is just a number. When I say 0.762 moles, this is just 0.762 times Avogadro's number of chlorine atoms. This is 0.364 times Avogadro's number of mercury atoms. We can literally think of this as the ratio. This is a certain number of moles, Well what's the ratio, let's see. What's the ratio of chlorine to mercury? Well you can eyeball it, it looks like it's roughly two to one, and you can verify that if you take that number and you divide it by .3639, and once again I'm just going to get the rough approximate. You can see it's pretty close to two. So when you see something like this the simplest explanation is often the best. Okay, there probably will be some measuring error right over here, but you can say that it looks like roughly, this is what I'm talking about when you're trying to find the empirical formula for mass composition it tends to be a rough science. You can say roughly the ratio of chlorine to mercury is two to one. You have two chlorines for every mercury. And because of that you can say this is likely to be, so likely, for every mercury you have two chlorines, you have two chlorines." + }, + { + "Q": "6:50 how is there more chlorine than mercury if there is .73% and 200 moles of Hg and only .27% and 36 moles of Cl?\n", + "A": "Because if you do 73/200.5(Mercury s Atomic Mass), you get .3640897756 as the amount of moles of Mercury. If you do 25/35.453, you get .7051589428 as the amount of moles of Chlorine. So, .7051589428 is greater than .3640897756. therefore, there are more moles of Chlorine than there are moles of Mercury. Is that clear?", + "video_name": "NM0WycKCCDU", + "timestamps": [ + 410 + ], + "3min_transcript": "Well what's the ratio, let's see. What's the ratio of chlorine to mercury? Well you can eyeball it, it looks like it's roughly two to one, and you can verify that if you take that number and you divide it by .3639, and once again I'm just going to get the rough approximate. You can see it's pretty close to two. So when you see something like this the simplest explanation is often the best. Okay, there probably will be some measuring error right over here, but you can say that it looks like roughly, this is what I'm talking about when you're trying to find the empirical formula for mass composition it tends to be a rough science. You can say roughly the ratio of chlorine to mercury is two to one. You have two chlorines for every mercury. And because of that you can say this is likely to be, so likely, for every mercury you have two chlorines, you have two chlorines. very likely that you have mercury two chloride. The reason why it's called mercury two chloride is because, well I won't go into too much detail right over here but chlorine is highly electronegative, it's an oxidizing agent, it likes to take other people's electrons or hog other people electrons. In this case it's hogging, since each of the chlorine likes to hog one electron, this case two chlorines are going to hog two electrons, so it's hogging two electrons from the mercury. When you lose electrons or when your electrons are being hogged you're being oxidized, so the oxidation state on mercury right over here is two. Two of its electrons are being hogged, one by each of the two chlorines. This is mercury two chloride, where the two is the oxidation state of the mercury. The ratio, we have two chlorines for every one mercury, roughly." + }, + { + "Q": "\nAt 5:42, it is said that the ratio of Chlorine to Mercury is 2:1, because 0.762/0.364 = ~2. and therefor stating that for each Mercury, we have 2 Chlorine.\nBut what about the ratio of Mercury to Chorine? Wouldn't that be 0.364/0.762= ~0.5, stating that for each Chlorine, we have 0.5 Mercury?", + "A": "You cannot have half of an atom.", + "video_name": "NM0WycKCCDU", + "timestamps": [ + 342, + 121 + ], + "3min_transcript": "27 divided by 35.453 is equal to .76, I'll just say two. So 0.762 moles of chlorine. What's going to be the ratio of mercury to chlorine? Or I guess we could say since chlorine, there's more of that, chlorine to mercury. Remember, this is just a number. When I say 0.762 moles, this is just 0.762 times Avogadro's number of chlorine atoms. This is 0.364 times Avogadro's number of mercury atoms. We can literally think of this as the ratio. This is a certain number of moles, Well what's the ratio, let's see. What's the ratio of chlorine to mercury? Well you can eyeball it, it looks like it's roughly two to one, and you can verify that if you take that number and you divide it by .3639, and once again I'm just going to get the rough approximate. You can see it's pretty close to two. So when you see something like this the simplest explanation is often the best. Okay, there probably will be some measuring error right over here, but you can say that it looks like roughly, this is what I'm talking about when you're trying to find the empirical formula for mass composition it tends to be a rough science. You can say roughly the ratio of chlorine to mercury is two to one. You have two chlorines for every mercury. And because of that you can say this is likely to be, so likely, for every mercury you have two chlorines, you have two chlorines. very likely that you have mercury two chloride. The reason why it's called mercury two chloride is because, well I won't go into too much detail right over here but chlorine is highly electronegative, it's an oxidizing agent, it likes to take other people's electrons or hog other people electrons. In this case it's hogging, since each of the chlorine likes to hog one electron, this case two chlorines are going to hog two electrons, so it's hogging two electrons from the mercury. When you lose electrons or when your electrons are being hogged you're being oxidized, so the oxidation state on mercury right over here is two. Two of its electrons are being hogged, one by each of the two chlorines. This is mercury two chloride, where the two is the oxidation state of the mercury." + }, + { + "Q": "\nAt 1:28 something was said about shining a wavelength of light is specifically sensitive to the solute. my question is why is the spectrophotemeter set at certain wavelengths, why cant it just be set at any wavelength of our choice?", + "A": "That is because chemicals only absorb very specific wavelengths of light. This is why we can use a spectrophotometer to measure the concentration of a specific chemical. If the wavelength of the light is wrong, then the light won t be absorbed.", + "video_name": "qbCZbP6_j48", + "timestamps": [ + 88 + ], + "3min_transcript": "What I want to do in this video is to talk a little bit about spectrophotometry. Spectrophotometry sounds fairly sophisticated, but it's really based on a fairly simple principle. So let's say we have two solutions that contain some type of solute. So that is solution one, and then this is solution two. And let's just assume that our beakers have the same width. Now let's say solution 1-- let me put it right here, number 1, and number 2. Now let's say that solution 1 has less of the solute in it. So that's the water line right there. So this guy has less of it. And let's say it's yellow or to our eyes it looks yellow. So this has less of it. Actually, let me do it this way. Let me shade it in like this. So it has less of it. And let's say solution number 2 has more of the solute. So it's more. So I'll just kind of represent that as more So the concentration of the solute is higher here. So let me write higher concentration. And let's say this is a lower concentration. Now let's think about what will happen if we shine some light through each of these beakers. And let's just assume that we are shining at a wavelength of light that is specifically sensitive to the solute that we have dissolved in here. I'll just leave that pretty general right now. So let's say I have some light here of some intensity. So let's just call that the incident intensity. I'll say that's I0. So it's some intensity. What's going to happen as the light exits the other side of this beaker right here? Well, some of it is going to be at absorbed. Some of this light, at certain frequencies, is going to be And so you're actually going to have less light coming out from the other side. Especially less of those specific frequencies that these molecules in here like to absorb. So your're going to have less light come out the other side. I'll call this I1. Now in this situation, if we shine the same amount of light-- so I0-- that's supposed to be an arrow there, but my arrow is kind of degrading. If we shined the same amount of light into this beaker-- so it's the same number, that and that is the same-- the same intensity of light, what's going to happen? Well more of those specific frequencies of light are going to be absorbed as the light travels through this beaker. It's just going to bump into more molecules because it's a higher concentration here." + }, + { + "Q": "at 1:18 why is #1 a lower concentration?\n", + "A": "For example s sake", + "video_name": "qbCZbP6_j48", + "timestamps": [ + 78 + ], + "3min_transcript": "What I want to do in this video is to talk a little bit about spectrophotometry. Spectrophotometry sounds fairly sophisticated, but it's really based on a fairly simple principle. So let's say we have two solutions that contain some type of solute. So that is solution one, and then this is solution two. And let's just assume that our beakers have the same width. Now let's say solution 1-- let me put it right here, number 1, and number 2. Now let's say that solution 1 has less of the solute in it. So that's the water line right there. So this guy has less of it. And let's say it's yellow or to our eyes it looks yellow. So this has less of it. Actually, let me do it this way. Let me shade it in like this. So it has less of it. And let's say solution number 2 has more of the solute. So it's more. So I'll just kind of represent that as more So the concentration of the solute is higher here. So let me write higher concentration. And let's say this is a lower concentration. Now let's think about what will happen if we shine some light through each of these beakers. And let's just assume that we are shining at a wavelength of light that is specifically sensitive to the solute that we have dissolved in here. I'll just leave that pretty general right now. So let's say I have some light here of some intensity. So let's just call that the incident intensity. I'll say that's I0. So it's some intensity. What's going to happen as the light exits the other side of this beaker right here? Well, some of it is going to be at absorbed. Some of this light, at certain frequencies, is going to be And so you're actually going to have less light coming out from the other side. Especially less of those specific frequencies that these molecules in here like to absorb. So your're going to have less light come out the other side. I'll call this I1. Now in this situation, if we shine the same amount of light-- so I0-- that's supposed to be an arrow there, but my arrow is kind of degrading. If we shined the same amount of light into this beaker-- so it's the same number, that and that is the same-- the same intensity of light, what's going to happen? Well more of those specific frequencies of light are going to be absorbed as the light travels through this beaker. It's just going to bump into more molecules because it's a higher concentration here." + }, + { + "Q": "\nAt 2:24, if the same amount of light shine through the same beaker, why does the concentration of each beaker differ and why does one have a higher concentration from another? Somebody please explain.", + "A": "The concentration is not dependent on the amount of light shining through. It would be similar to having to glasses of water that have different amounts of food coloring in them. The concentrations differ for example s sake. He is simply showing how transmittance differs in varying conditions.", + "video_name": "qbCZbP6_j48", + "timestamps": [ + 144 + ], + "3min_transcript": "What I want to do in this video is to talk a little bit about spectrophotometry. Spectrophotometry sounds fairly sophisticated, but it's really based on a fairly simple principle. So let's say we have two solutions that contain some type of solute. So that is solution one, and then this is solution two. And let's just assume that our beakers have the same width. Now let's say solution 1-- let me put it right here, number 1, and number 2. Now let's say that solution 1 has less of the solute in it. So that's the water line right there. So this guy has less of it. And let's say it's yellow or to our eyes it looks yellow. So this has less of it. Actually, let me do it this way. Let me shade it in like this. So it has less of it. And let's say solution number 2 has more of the solute. So it's more. So I'll just kind of represent that as more So the concentration of the solute is higher here. So let me write higher concentration. And let's say this is a lower concentration. Now let's think about what will happen if we shine some light through each of these beakers. And let's just assume that we are shining at a wavelength of light that is specifically sensitive to the solute that we have dissolved in here. I'll just leave that pretty general right now. So let's say I have some light here of some intensity. So let's just call that the incident intensity. I'll say that's I0. So it's some intensity. What's going to happen as the light exits the other side of this beaker right here? Well, some of it is going to be at absorbed. Some of this light, at certain frequencies, is going to be And so you're actually going to have less light coming out from the other side. Especially less of those specific frequencies that these molecules in here like to absorb. So your're going to have less light come out the other side. I'll call this I1. Now in this situation, if we shine the same amount of light-- so I0-- that's supposed to be an arrow there, but my arrow is kind of degrading. If we shined the same amount of light into this beaker-- so it's the same number, that and that is the same-- the same intensity of light, what's going to happen? Well more of those specific frequencies of light are going to be absorbed as the light travels through this beaker. It's just going to bump into more molecules because it's a higher concentration here." + }, + { + "Q": "\nAt 1:30, what do you mean by \"sigma\"?", + "A": "sigma usually stands for sum", + "video_name": "24vtg9Ehr0Q", + "timestamps": [ + 90 + ], + "3min_transcript": "- [Instructor] If you're face to face with a sophisticated Newton's Second Law problem, you're gonna need a sophisticated understanding of Newton's Second Law. That's what I'm gonna try to provide you with here, so that no matter what scenario you're faced with you can apply this law in a correct way. Most people know Newton's Second Law is F equals MA, which is fine, it's a simple way to understand it, and it's fine for simple problems, if I had an asteroid for instance, of mass m, out in outer space so there's no air resistance or friction, and there was only one force on it, a force F, and that force pointed to the right, let's say that force was 50 newtons, well I could plug the 50 newtons into the force, I could plug the mass of the asteroid, let's just say it's 10 kilograms, into the mass, and I'd find the acceleration of the asteroid, in this case, 50 over 10 would give me five meters per second squared. But, what if we had extra forces on this asteroid? What if there was another force that pointed to the left, that was 30 newtons? So let's call, let's name these now, let's call this F1, this 50 newtons, let's say F2 was the magnitude of the 30 newton force, it points to the left, yes, that's the negative direction, but let's just say these forces here are just giving the magnitude of it and then the direction is specified by the direction of the arrow. Now what would I do? Well, to handle this we need to understand that the left hand side here isn't just force, it's the net force. Or, you can call it the sum of the forces. So to denote the net force, we often write this Greek letter sigma, and sigma is a mathematical symbol that represents the sum of whatever comes after it. So this is the sum of the forces. Because F comes afterward. If I had G it would be the sum of the G's, and if I had H it would be the sum of the H's. And this is a little confusing already. People hear sum of, phonetically, and they think oh, sometimes they're like oh, so, some of? Like a few of? No no no, we mean all of, all of the forces. You add up all of the forces, that will equal So in this case, we'd take this 50 newtons, I can take 50 newtons because it goes to the right, and, I mean we can call leftward positive if we really wanted to, if there was a good reason, but unless otherwise specified, we're gonna just choose rightward as positive and upward as positive, so this 50 has to be positive. And I can't now, by sum of, I add up the forces, but I have to add them up like vectors. This force here is a vector. Forces are vectors, and so I have to add them up as vectors. This is a vector equation. I can't just take 50 plus 30 to get the answer, because vectors that point to the left we're gonna consider negative, and vectors that point to the right we'll consider positive, and so I'll take 50 newtons minus 30 newtons. That's what's gonna be equal to the mass times the acceleration, so I could plug in 10 kilograms if I wanted to, multiply by A, and in this case I'd get 20 over 10 is two, meters per second squared. So you have to add these up like vectors, and if I had more forces it'd be just as easy to deal with," + }, + { + "Q": "I don't get something at 10:00. It doesn't intuitively make sense to me that on one side the force is lesser and on the other its more, shouldn't force be conserved? I know if I look at the mathematical equations, it makes sense there, but how could the force be larger on one side, even though its being applied on the same amount of volume, and just a different surface area? Shouldn't it be smaller since its on a larger surface area?\n", + "A": "In this system, and for that matter all systems, it s the work that is conserved. work being force x distance means that the larger piston will travel a shorter distance than the smaller piston with a greater amount of force. It took me several times reviewing this video to understand the conclusion that he was coming to. he probably should have put that in the end of the video but we are all only human.", + "video_name": "lWDtFHDVqqk", + "timestamps": [ + 600 + ], + "3min_transcript": "it's a very natural unit. Let's say my pressure in is 10 pascals, and let's say that my input area is 2 square meters. If I looked the surface of the water there it would be 2 square meters, and let's say that my output area is equal to 4 meters squared. What I'm saying is that I can push on a piston here, and that the water's going to push up with some piston here. First of all, I told you what my input pressure is-- what's my input force? Input pressure is equal to input force divided by input area, so 10 pascals is equal to my input force divided by I get input force is equal to 20 newtons. My question to you is what is the output force? How much force is the system going to push upwards at this end? We know that must if my input pressure was 10 pascals, my output pressure would also be 10 pascals. So I also have 10 pascals is equal to my out force over my out cross-sectional area. So I'll have a piston here, and it goes up like that. That's 4 meters, so I do 4 times 10, and so I get 40 newtons is equal to my output force. So what just happened here? I inputted-- so my input force is equal to 20 newtons, and my output force is equal to 40 newtons, so I just doubled my force, or essentially I had a mechanical advantage of 2. it's a hydraulic machine. Anyway, I've just run out of time. I'll see you in the next video." + }, + { + "Q": "At 4:31 , he said about external pressure is distributed through out the fluid, That means if we jump into the ocean, then we give external pressure for throughout the ocean.?\n", + "A": "An EXTREMELY minuscule amount, but yes.", + "video_name": "lWDtFHDVqqk", + "timestamps": [ + 271 + ], + "3min_transcript": "is exerting on this piston. So that's the output pressure, P2. And what's area 2 times D2? The cross sectional area, times the height at which how much the water's being displaced upward, that is equal to volume 2. But what do we know about these two volumes? I went over it probably redundantly in the previous video-- those two volumes are equal, V1 is equal to V2, so we could just divide both sides by that equation. You get the pressure input is equal to the pressure output, so P1 is equal to P2. I did all of that just to show you that this isn't a new concept: this is just the conservation of energy. The only new thing I did is I divided-- we have this notion of the cross-sectional area, and we have this notion of This actually tells us-- and you can do this example in multiple situations, but I like to think of if we didn't have gravity first, because gravity tends to confuse things, but we'll introduce gravity in a video or two-- is that when you have any external pressure onto a liquid, onto an incompressible fluid, that pressure is distributed evenly throughout the fluid. That's what we essentially just proved just using the law of conservation of energy, and everything we know about work. What I just said is called Pascal's principle: if any external pressure is applied to a fluid, that pressure is distributed throughout the fluid equally. Another way to think about it-- we proved it with this little drawing here-- is, let's say that I have a tube, and at the end of the tube is a balloon. It's saying that if I increase-- say I have some piston here. This is stable, and I have water throughout this whole thing. Let me see if I can use that field function again-- oh no, there must have been a hole in my drawing. Let me just draw the water. I have water throughout this whole thing, and all Pascal's principle is telling us that if I were to apply some pressure here, that that net pressure, that extra pressure I'm applying, is going to compress this little bit. That extra compression is going to be distributed through the whole balloon. Let's say that this right here is rigid-- it's some kind of middle structure." + }, + { + "Q": "Why does Sal divide at 00:45\n", + "A": "because the force is being applied over an area A, in this case A1 which is just pressure right? When we compress something, we apply a pressure.", + "video_name": "lWDtFHDVqqk", + "timestamps": [ + 45 + ], + "3min_transcript": "Welcome back. To just review what I was doing on the last video before I ran out of time, I said that conservation of energy tells us that the work I've put into the system or the energy that I've put into the system-- because they're really the same thing-- is equal to the work that I get out of the system, or the energy that I get out of the system. That means that the input work is equal to the output work, or that the input force times the input distance is equal to the output force times the output distance-- that's just the definition of work. Let me just rewrite this equation here. If I could just rewrite this exact equation, I could say-- the input force, and let me just divide it by this area. The input here-- I'm pressing down this piston that's pressing down on this area of water. So this input force-- times the input area. Let's call the input 1, and call the output 2 for simplicity. Let me do this in a good color-- brown is good color. I have another piston here, and there's going to be some outward force F2. The general notion is that I'm pushing on this water, the water can't be compressed, so the water's going to push up on this end. The input force times the input distance is going to be equal to the output force times the output distance right-- this is just the law of conservation of energy and everything we did with work, et cetera. I'm rewriting this equation, so if I take the input force and divide by the input area-- let me switch back to green-- then I multiply by the area, and then I just multiply times D1. You see what I did here-- I just multiplied and divided by A1, which you can do. You can multiply and divide by any number, and these two cancel out. It's equal to the same thing on the other side, which is over A2 times A2 times D2. Hopefully that makes sense. What's this quantity right here, this F1 divided by A1? Force divided by area, if you haven't been familiar with it already, and if you're just watching my videos there's no reason for you to be, is defined as pressure. Pressure is force in a given area, so this is pressure-- we'll call this the pressure that I'm inputting into the system. What's area 1 times distance 1? That's the area of the tube at this point, the cross-sectional area, times this distance. That's equal to this volume that I calculated in the previous video-- we could say that's the input volume, or V1. Pressure times V1 is equal to the output pressure-- force 2" + }, + { + "Q": "\nAt 6:02, how can two different orbitals have same sub shell.\nHow can 1st shell have s-sub shell and 2nd shell also have s-sub shell?", + "A": "The are not the same subshell, they are the same TYPE of subshell. So, 1s is the s-subshell belonging to the first shell. 2s is the s-subshell belonging to the second shell.", + "video_name": "KrXE_SzRoqw", + "timestamps": [ + 362 + ], + "3min_transcript": "So the electron is most likely to be found somewhere in that sphere. Let's do the next shell. n is equal to two. If n is equal to two, what are the allowed values for l? l goes zero, one, and so on all the way up to n minus one. l is equal to zero. Then n minus one would be equal to one. So we have two possible values for l. l could be equal to zero, and l could be equal to one. Notice that the number of allowed values for l is equal to n. So for example, if n is equal to one, we have one allowed value. If n is equal to two, we have two allowed values. We've already talked about what l is equal to zero, what that means. l is equal to zero means an s orbital, shaped like a sphere. Now, in the second main energy level, or the second shell, we have another value for l. l is equal to one. l is equal to one means a p orbital. The shape of a p orbital is a little bit strange, so I'll attempt to sketch it in here. You might hear several different terms for this. Imagine this is a volume. This is a three-dimensional region in here. You could call these dumbbell shaped or bow-tie, whatever makes the most sense to you. This is the orbital, this is the region of space where the electron is most likely to be found if it's found in a p orbital here. Sometimes you'll hear these called sub-shells. If n is equal to two, if we call this a shell, then we would call these sub-shells. These are sub-shells here. Again, we're talking about orbitals. l is equal to zero is an s orbital. l is equal to one is a p orbital. Let's look at the next quantum number. This is the magnetic quantum number, symbolized my m sub l here. m sub l indicates the orientation of an orbital around the nucleus. This tells us the orientation of that orbital. The values for ml depend on l. ml is equal to any integral value that goes from negative l to positive l. That sounds a little bit confusing. Let's go ahead and do the example of l is equal to zero. l is equal to zero up here. Let's go ahead and write that down here. If l is equal to zero, what are the allowed values for ml? There's only one, right? The only possible value we could have here is zero." + }, + { + "Q": "\nat 8:50, from the three values of ml: -1 0 1, how can we know that the orientation of the 3 types p orbital should be on the three axis but not somewhere else like in between the axis? And one more thing is that the spin quantum number indicates the way electrons spin around themselves, so why can't we use -1 and 1 or any different number instead of -1/2 and 1/2?", + "A": "It s complicated! The answer comes from a combination of quantum mechanics and the theory of special relativity.", + "video_name": "KrXE_SzRoqw", + "timestamps": [ + 530 + ], + "3min_transcript": "Let me use a different color here. If l is equal to zero, we know we're talking about an s orbital. When l is equal to zero, we're talking about an s orbital, which is shaped like a sphere. If you think about that, we have only one allowed value for the magnetic quantum number. That tells us the orientation, so there's only one orientation for that orbital around the nucleus. And that makes sense, because a sphere has only one possible orientation. If you think about this as being an xyz axis, (clears throat) excuse me, and if this is a sphere, there's only one way to orient that sphere in space. So that's the idea of the magnetic quantum number. Let's do the same thing for l is equal to one. Let's look at that now. If we're considering l is equal to one ... Let me use a different color here. l is equal to one. If l is equal to one, what are the allowed values for the magnetic quantum number? ml is equal to -- This goes from negative l to positive l, so any integral value from negative l to positive l. Negative l would be negative one, so let's go ahead and write this in here. We have negative one, zero, and positive one. So we have three possible values. When l is equal to one, we have three possible values for the magnetic quantum number, one, two, and three. The magnetic quantum number tells us the orientations, the possible orientations of the orbital or orbitals around the nucleus here. So we have three values for the magnetic quantum number. That means we get three different orientations. We already said that when l is equal to one, we're talking about a p orbital. A p orbital is shaped like a dumbbell here, so we have three possible orientations If we went ahead and mark these axes here, let's just say this is x axis, y axis, and the z axis here. We could put a dumbbell on the x axis like that. Again, imagine this as being a volume. This would be a p orbital. We call this a px orbital. It's a p orbital and it's on the x axis here. We have two more orientations. We could put, again, if this is x, this is y, and this is z, we could put a dumbbell here on the y axis. There's our second possible orientation. Finally, if this is x, this is y, and this is z, of course we could put a dumbbell on the z axis, like that. This would be a pz orbital. We could write a pz orbital here, and then this one right here would be a py orbital." + }, + { + "Q": "4:22 How does it hold the hot water, wouldn't it melt? Or is it not plastic? Also on 6:03 wouldn't the water come out?\n", + "A": "Some plastics can handle tempatures well about the boiling poit of water.", + "video_name": "XQTIKNXDAao", + "timestamps": [ + 262, + 363 + ], + "3min_transcript": "Oh wow, it pops apart there. You can see there's a little bracket on the inside that the screw goes into and holds this steel band in place. That's what the handle is help on by. Then at the top, there's just a little lip that holds the top of the handle there in place. Just a little piece that folds over the glass and snaps on. So that's how that's held. This is made out of one molded part, this is made out of another, and this is made out of another. So that's how they made the handle, there's three different molded plastic parts there and the molds came together like this. You can tell that because you can feel the mold seam on the inside of the handle, there. It would make it easy to pull the mold out this way. They probably also had, it was probably a three part mold and there was a section that also came out in this direction. Then this is just another injection molded part that snaps onto this one, as we've seen. This is the part that holds the handle on. Very important part. I think they definitely paid the extra money for a stainless piece there because it's really important that that doesn't come loose and it probably gets fairly wet, so if it was made out of regular steel or another material it might rust and could potentially come apart. We wouldn't want hot coffee on us, now would we. Alright, so that's the coffee kettle. So, inside, here's our coffee maker. We know that hot water... We've got a container here and in this container, is a space where we put our coffee filter and then we put our coffee grounds and we fill this with water and then we close the top and we turn it on and we wait. What happens is that water that we pour in drains down a little hole on the inside there, you can see it right there. Let me point to it with the screw driver. It drains down that hole and it goes down into this underside, so we'll take a look at the underside and see what happens down there. Okay, so, I've modified a screw driver. This was a low-cost screw driver. It was a 99 cent one, so I modified the end of it so I could take out these safety screws. Don't do this at home unless you have a professional with you because this is not meant to be taken apart. That's why they use these special screw heads, so you won't take it apart." + }, + { + "Q": "\nat 6:09,he said \"The heat to get in\" why do we need heat to get in?", + "A": "I think that was just a slip up as he fixed it with allowing air to get in its so that as it heats up inside it doesn t pressurize the air and and blow up.", + "video_name": "XQTIKNXDAao", + "timestamps": [ + 369 + ], + "3min_transcript": "is a space where we put our coffee filter and then we put our coffee grounds and we fill this with water and then we close the top and we turn it on and we wait. What happens is that water that we pour in drains down a little hole on the inside there, you can see it right there. Let me point to it with the screw driver. It drains down that hole and it goes down into this underside, so we'll take a look at the underside and see what happens down there. Okay, so, I've modified a screw driver. This was a low-cost screw driver. It was a 99 cent one, so I modified the end of it so I could take out these safety screws. Don't do this at home unless you have a professional with you because this is not meant to be taken apart. That's why they use these special screw heads, so you won't take it apart. There we go. Again, this is an injection molded part. This is a co-molded part, it looks like. Which means that there were two different materials molded together. Let's see if I can knock that screw out. Okay, it wants to stay, that's fine. This material here is, these feet are made out of a softer material and this is a polypropylene material. So it's a plastic, a low-cost plastic. So the mold comes together and they injection mold this material and then once this material has begun to harden, they injection mold the softer material, so it's co-molded or it's a dual molded part. You can see other parts are done like this, like sometimes you'll see toothbrushes that have soft saniprene and then the hard toothbrush and they're molded in one mold. It's a dual shot mold. In any case, so that's the bottom. heater chamber to vent out, I think a little heat in there. So here is the heater. This is where all the magic happens. The water comes down this tube, and it goes around this horse shoe shape, and then it comes up here. What causes the water to raise back up and go all the way up this tube. It goes all the way up the tube, here and then it comes out this apparatus here and then it goes and drains out of these holes right into here. There's just a little piece of plastic that causes this part to line up right over the top of this, when you close it. The water just comes up that and drains right out and into your coffee grounds and then makes your coffee. So the heater does two things. It causes the water to go down this pipe, expand out, and to drip into your coffee maker. But it also heats up the plate, this plate right here." + }, + { + "Q": "At 3:05, you say Gibbs Free Energy is only when there's constant pressure and TEMPERATURE. Gibbs Free Energy formula has temperature as a constant so how does that work?\n", + "A": "It means that the Gibb s Free Energy value can change if temperature changes. That kinda makes sense, since a reaction that may not be spontaneous with low inputs of energy may become spontaneous if there is a greater input of energy into the system.", + "video_name": "J2L-X2sUigs", + "timestamps": [ + 185 + ], + "3min_transcript": "assumption if you're doing something in a beaker that's open to the air or if you're thinking about a lot of different biological systems. Now based on that logic what do you think this word means, endothermic. Well endothermic, therm same root and now your prefix is endo so this is a process that absorbs heat. Absorbs heat. Or if you're thinking of a constant pressure, you can say your enthalpy after the reaction is gonna be higher than the enthalpy before the reaction. So your delta H is going to be greater than zero. All right, fair enough. Now let's look at these two characters over here. Exergonic and endergonic so exergonic the root here is ergon and you might not be as familiar with that as you are with therm but you might have heard the word ergonomic. Say, hey that's a nice ergonomic desk. or it's a nice ergonomic chair. An ergon does indeed come from the Greek for work. And so exergonic is a reaction that releases work energy or at least that's what the word implies. Releases, let me do that in the same color. This is something that is going to release work energy. And endergonic, same logic, well that's gonna be something based on just the way the word is setup that absorbs work energy or uses work energy. Now one of our variables or properties that we can use to think about energy that can be used for work is Gibbs free energy and the formula for Gibbs free energy, if we're thinking about constant pressure and temperature, so let me write that down. So if we're talking about constant pressure for Gibbs free energy or you can even view this as a definition of Gibbs free energy. The change in Gibbs free energy, let me do this in another color. The change in Gibbs free energy is equal to our change in enthalpy minus, use in the different color. Minus our temperature times our change in entropy and if this looks completely foreign to you, I encourage you to watch the video on Gibbs free energy but the reason why this is related to energy for work is okay, look I have my, whether I'm absorbing or I'm releasing heat and I'm subtracting out entropy which is kind of the energy that is going to the disorder of the universe and what's left over is the energy that I can do for work." + }, + { + "Q": "\nFrom 7:45 on, a intermolecular Fischer esterification that forms lactone is shown. Is it possible that at the same time of forming lactone, an OH part of one molecule attack the COOH part of another molecule, and form a ester with two 5-carbon chain (and no ring) in it?", + "A": "yes, you can also get the two 5 carbon chain product and you could essentially form a polymer from it because the OH portion of one can keep attacking the COOH on the other end of the molecule. Then theres also the possibility of a ring forming at any point in that process as well.", + "video_name": "ynBuPEmcjp4", + "timestamps": [ + 465 + ], + "3min_transcript": "And so our end result is to form our ester and water. Ok, so that's a little bit of a long mechanism. Let's take a look at some reactions to form esters using the Fischer esterification reactions. So, let's start with this molecule over here on the left. So this is salicylic acid, and if we add methanol, and we use sulfuric acid as our source of protons, we're going to form an ester. And this is one of those famous labs that's always done in undergraduate organic chemistry. So if we think about the mechanism, remember that this oxygen on our alcohol, and in this case it's methyl, are going to ag. So we're going to lose this OH on our carboxylic acid, and we're going to put this oxygen and this methyl group on in place. So let's go ahead and draw the product. So we would have our benzene ring right here, to our oxygen, and we would have the oxygen from the alcohol, from methanol, and then our methyl group like that, and then we still have our OH right here. The reason why this is one of those classic undergraduate labs, is this is wintergreen. So this is an incredible smell. It's always a lot of fun to do this in an undergraduate lab, because the lab smells great when you're done, so the synthesis of wintergreen. Alright, let's look at another Fischer esterification. This one is a little bit different. This one is an intramolecular Fischer esterification. So if we look at our starting molecule on the left. This time we have our carboxylic acid and our alcohol in the exact same molecule. And we have all these single bonds in here, which we know we can have some free rotation. So if we draw the molecule in a different conformation, so let's go ahead and do that, so we have our carboxylic acid up here, so let me use red for that, so we have one, we have carbon one, two, three, four, and five. So we have five carbons, so let's go ahead and draw them in, so there's carbon one, two, three, four, five, and then we have our OH. So let me go ahead and number those carbons. So this is carbon one, carbon two, carbon three, carbon four, and carbon five. And so in a different conformation, we can think about this oxygen attacking this carbonyl in the mechanism. So we know that we're going to lose this OH, we know we're gonna lose this Hydrogen, and so we can stick those together and think about our final product. So we're going to form an ester, but it's a different ester than what we've seen before. So we have our carbonyl right here, and then we have this oxygen." + }, + { + "Q": "\non the 4:40 timestamp he talks about coronary artery disease but my question about the arteries is that since it caries deoxygenated blood to the lungs it would make it blue from the deoxygenated blood but why does it all just look red in the real life photo", + "A": "Deoxygenated blood isn t actually blue, it is dark red. Blue is used on diagrams to make them easier to read.", + "video_name": "vYnreB1duro", + "timestamps": [ + 280 + ], + "3min_transcript": "And the plaques, the material inside of them are lipids, so things like fat, cholesterol and also dead white blood cells, which is this kind of messy substance right over here. This is what we call a plaque. And the formation of these plaques that obstruct the actual blood vessel, that actually obstruct the artery. We call it.....make it clear you see that. This is kind of tube over here. Let me draw the blood So this formation of these plaques we call atherosclerosis. So you can imagine if you have these things build up, So it would be destructing the blood flow downstream right over there. In that general process we talk about the restriction of blood flow, that is ischemia that's happening. So ischemia is deprivation of blood flow and oxygen downstream from this right over there. That's what we call coronary artery disease, or heart disease. So this causes coronary artery disease, which is sometimes called heart disease. Coronary heart disease would be redundant, because coronary is already referring to the heart. This is also sometimes called heart disease. And so you can imagine if downstream the muscle tissue is when they are exerting themselves, they need more oxygen. The heart needs to pump a little harder. If downstream the cells are not getting all of the oxygen they need, you can imagine that the heart maybe not able to provide all of the functions, whoever's heart this is, that they needed to do. And when that happens that's called heart failure. So heart disease is one of the causes of heart failure. Now I want to be clear, heart failure does not mean that the heart is stopping. That the heart is stopped and the person is dead, it literally just means that the heart is failing to do what it should be doing. It's failing to provide the needs of that person. So it's not pumping hard enough or well enough to provide adequate function for that person." + }, + { + "Q": "Is it just my imagination, or are the absolute configurations of the alcohol and bromoalkane incorrect at 5:48? I thought the alcohol (with OH on a wedge and thus H on a dash) would be S and the bromoalkane (with Br on a dash and thus H on a wedge) would be R. :)\n", + "A": "The video is correct, perhaps take a second look at it? O is #1 priority, the right carbon is #2 and the left carbon is #3. Going from 1->2->3 is clockwise so it s R.", + "video_name": "KPh60w6McPI", + "timestamps": [ + 348 + ], + "3min_transcript": "to give us a much better leaving group, and here we have an excellent leaving group. So that's one of the reasons for using tosylates here. So in the second step we are going to add sodium bromide and we are going to get again a SN-two type mechanism. So a nucleophile is going to attack our electrophile. So we can identify our electrophile. It's the carbon bonded to the oxygen. So the oxygen is withdrawing some electron density from this carbon, so this the electrophilic portion of the molecule. And then once again our bromide anion is going to function as a nucleophile. So here is our bromide, and I am going to highlight this lone pair of electrons right here in blue. And so our bromide anion attacks our electrophilic carbon and forms a bond. At the same time, these electrons come off onto the oxygen. So once again a concerted SN-two type mechanism. And this time we do have to worry about stereochemistry. So we have this wedge in here. is coming out at us in space. And so the bromide anion has to attack from the opposite side. And so if it is attacking from the opposite side when you draw the final product, you would have to show this is a dash. So the bromide had to attack from the opposite side which gives us inversion of configuration. Inversion of absolute configuration here. So when you assign your absolute configuration, for your starting alcohol, this would be R. So our chiral center would be right here. So this carbon is chiral and so this carbon is chiral for our products, and for our product we would form the S in the [an-te-um] right here. So we formed S in the [an-te-um] row and we started with a R in the [an-te-um] row. So a SN-two type mechanism inversion of configuration because a nucleophile has to attack from the opposite side. So lets look at one more example here. Lets do a SN-one mechanism using a tertiary alcohol. So lets do that. reacting with concentrated hydrochloric acid. So concentrated hydrochloric acid is going to function as an acid. Our alcohol is going to function as a base. And let me just highlight the fact that this is going to be a SN-one type mechanism because we have a tertiary alcohol. This carbon bonds to the OH bond to three other carbons. And so the alcohol functions as a base, and is protonated and we would form the chloride anion here. So lets go ahead and draw that in. So we would have the chloride anion, so negative one formal charge. And lets go ahead and put those electrons in blue like we did before. So these electrons in here, lets say that those are these electrons. So the chloride anion. We protonate the oxygen, so lets go ahead and draw that. So if we protonate the oxygen, now we have our tertbutyl group over here, protonate the oxygen so a plus one formal charge" + }, + { + "Q": "At 7:43, Sal integrated the entire equation. The right hand side of equation is zero. The integration of zero must be a constant as derivative of a constant is zero. But, instead after integrating the equation was written still equal to zero.\nCan anyone explain? It will be very helpful. Thanks.\n", + "A": "I did not watch the video but probably the answer is that Sal was doing a definite integral rather than an indefinite integral. When you do a definite integral, you are integrating the same function and calculating its value at two different points, and then subtract the starting point from the ending point. When you do that subtraction, the constant cancels out, so you can just say it is zero.", + "video_name": "ixRtSV3CXPA", + "timestamps": [ + 463 + ], + "3min_transcript": "Well, on this term, the n's cancel out, the R cancels out. Over here, this nRT cancels out with this nRT. And what are we left with? We're left with 3/2-- we have this 1 over T left-- times 1 over T delta T plus 1 over V delta V is equal to-- well, zero divided by anything is just equal to 0. Now we're going to integrate over a bunch of really small delta T's and delta V's. So let me just change those to our calculus terminology. We're going to do an infinite sum over infinitesimally small changes in delta T and delta V. So I'll rewrite this as 3/2 1 over T dt plus 1 over V dv is small change in volume. This is a very, very, very, small change, an infinitesimally small change, in temperature. And now I want to do the total change in temperature. I want to integrate over the total change in temperature and the total change in volume. So let's do that. So I want to go from always temperature start to temperature finish. And this will be going from our volume start to volume finish. Fair enough. Let's do these integrals. This tends to show up a lot in thermodynamics, these antiderivatives. The antiderivative of 1 over T is natural log of T. So this is equal to 3/2 times the natural log of T. We're going to evaluate it at the final temperature and then the starting temperature, plus the natural log-- the plus the natural log of V, evaluated from our final velocity, and we're going to subtract out the starting velocity. This is just the calculus here. And this is going to be equal to 0. I mean, we could integrate both sides-- well, if every infinitesimal change is equal to the sum is equal to 0, the sum of all of the infinitesimal changes are also So this is still equal to 0. See what we can do here. So we could rewrite this green part as-- so it's 3/2 times the natural log of TF minus the natural log of TS, which is just, using our log properties, the natural log of TF over the natural log of TS. Right? When you evaluate, you get natural log of TF minus the natural log of TS." + }, + { + "Q": "\nMin 3:40 How does he know that delta U and delta T have a linear relationship with teach other? Of course they are related, but the nature of this relationship could be non-linear...", + "A": "No it can t because U is internal energy and T is average energy of the molecules. If the average energy increases then U has to increase proportionally.", + "video_name": "ixRtSV3CXPA", + "timestamps": [ + 220 + ], + "3min_transcript": "small change in volume. So that's what we have there. Now, if it's adiabatic, we know that this is 0. And if that's 0, we can add P delta V to both sides of this equation, and we will get that-- this is only true if it were adiabatic-- that delta U, our change in internal energy, plus our pressure times our change in volume, is equal to 0. And let's see if we can do this somehow, we can do something with this equation to get to that result that I'm trying to get to. So a few videos ago, I proved to you that U, the internal energy at any point in time-- let me write it here. The internal energy at any point in time is equal to 3/2 times n times R times T. Which is also equal to 3/2 times PV. change on this side? Something must have changed. Well, 3/2 can't change. n can't change. We're not going to change the number of molecules we have. The universal gas constant can't change. So the temperature must change. You have delta U could be rewritten as delta-- let me do it in a different color-- delta U could be written as 3/2 n times R times our change in temperature. And that's why I keep saying in this-- especially when we're dealing with the situation where all of the internal energy is essentially kinetic energy-- that if you don't have a change in temperature, you're not going to have a change in internal energy. Likewise, if you don't have a change in internal energy, you're not going to have a change in temperature. So let me put this aside right here. I'm going to substitute it back there. But let's see if we can do something with this P here. Well, we'll just resort to our ideal gas equation. Because we're dealing with an ideal gas, we might as well. This should be emblazoned in your mind, at this point. So if we want to solve for P, we get O is equal to nRT over V. Fair enough. So let's put both of these things aside, and substitute them into this formula. So delta U is equal to this thing. So that means that 3/2 nR delta T plus P-- P is this thing-- plus nRT over V times delta V is equal to 0. Interesting. So what can we do further here? And I'll kind of tell you where I'm going with this. So that tells me, my change in internal energy over a very small delta T-- this tells me my work done by the system" + }, + { + "Q": "\nReferring to 7:04, how do you know that all of the base will react with the acid but not vice-versa?", + "A": "I would assume that this is because we have much less base than acid. In other words, the base is our limiting reagent. There are 0.005 mol of -OH and 0.0100 mol of H3O+. All of the available H3O+ cannot react with the added -OH because there is much more (double actually) of it than there is of -OH available to react to form water. Hopefully this makes sense.", + "video_name": "JoGQYSTlOKo", + "timestamps": [ + 424 + ], + "3min_transcript": "that we have, and what's the volume. We have 10 milliliters, so in liters that would be, move our decimal place three, so that's .01. So this is equal to .01 for our liters. Solve for moles. So we would just multiply .5 by .01 and we would get our moles equal to .005. So that's how many moles of hydroxide ions we have. Now that we've found moles of both our acid and our base, we can think about the neutralization reaction that occurs. The base that we add is going to neutralize the acid that we had present. So we had hydronium ions present, and we added some base. So the acid donates a proton to the base. If OH minus picks up an H plus, then we get H2O. molecule of H2O, so we end up with two H2O over here. Or, if you prefer, instead of writing H3O plus, you could have just written H plus as your proton, and your proton reacts with hydroxide to give you water. So either way of representing the neutralization reaction is fine. So let's plug in our moles here. We know that we started with .01 moles of H3O plus, so let me write that down here. .01 moles of H3O plus, and we started with, and we added I should say, .005 moles of hydroxide. We added .005 moles of hydroxide ions. All of the hydroxide is going to react. It's going to neutralize the same amount of hydronium ions. So we're going to lose all of our hydroxide ions so we're left with zero moles of our base. If we're losing that much hydroxide, we're also going to lose that much hydronium, so that much is reacting with the hydroxide ions. We're losing the same amount of hydronium ions. So .01 minus .005 is of course equal to .005. That's how many moles of hydronium are left over. We've neutralized half of the hydronium ions present, and so we have another half left over. So one half of the acid has been neutralized, so one half of the acid is left. Let's think about the new concentration of hydronium ions in solution, so the concentration of hydronium is equal to moles over liters." + }, + { + "Q": "\nSal mentioned the f-block at 2:14 . Why don't universities ask questions about the f-block on the chemistry exams?", + "A": "Sal is awesome", + "video_name": "qkLzAXUP_K0", + "timestamps": [ + 134 + ], + "3min_transcript": "Voiceover: What I want to do in this video is think about the electron configuration for an atom whose highest energy electron is not in the S block or the P block but the D block. And to help us think about that with the periodic table, I'm going to rearrange the periodic table a little bit. I'm gonna take Helium, which it makes sense that's it's right over here in the top right because it's a noble gas, it's inert like the other noble gases, has very similar properties. But for the purpose of electron configurations, because Helium's highest energy electron is in the S sub-shell it's electron configuration is essentially 1S2. I'm gonna put it over here and then that allows us to construct, essentially, an S block. So let me cut and paste that here. So what do I mean by an S block? Well that means that any of these, there are some exceptions, but the general rule is that energy electron is going to be in an S sub-shell. So this is the S block right over there. Now, for example, Potassium's highest energy electron is going to be in the 4S sub-shell. While Hydrogen's highest energy electron, of course, is in the 1S sub-shell. But it's going to be in an S sub-shell. Now that allows us to divide the rest of the periodic table to other blocks. This right over here is the P block, same idea. The general rule is that the highest energy electron in the electron configuration of these elements is going to be in a P sub-shell. So this is the P block right over there. And, as you might have guessed, this in the middle is the D block. the F block in a future video. This is the F block. And you can actually place it here and kind of push over the D and the P blocks to make space for this. But we'll do that in a future video. So this right over here is the D block. Now why is this helpful? Well, if something is in the D block we can say that it's highest energy electron and there's going to be some special cases, but we can say, in general, the highest energy electron is going to be sitting in a D sub-shell. So, for example, if we were to focus on Iron, it's highest energy electron is going to be in a D sub-shell. Now the D sub-shell starts to become interesting. For something like the S and P sub-shells, you can say look at Oxygen, and you can say \"OK, Oxygen's in the P block it's highest \"energy electron is going to be in a P sub-shell.\"" + }, + { + "Q": "why is Andromeda then moving towards us?\nAlso was sal trying to explain relativity at 4:18\n", + "A": "Andromeda is moving toward us because it is drawn in by the mutual gravitational attraction between the mass in the Milky Way and the mass in Andromeda.", + "video_name": "1V9wVmO0Tfg", + "timestamps": [ + 258 + ], + "3min_transcript": "of this-- one way to think of this, if you think of the universe as an infinite flat sheet. You can imagine that we're just taking a sheet of, I don't know, some type of sheet of stretchy material and just stretching it out. We're just stretching it out. That's if we kind of imagine a more infinite universe that just goes off in every direction. We're just stretching that infinite sheet out. So it has no boundaries, but we're still stretching it out. Another way to visualize it-- and this what we did earlier on-- Is you can imagine that the universe is the three dimensional surface of a four dimensional sphere. Or the three dimensional surface of a hyper-sphere. So at an early stage in the universe, the sphere looked like this. And these points here-- that magenta point is right over here. The green point is right over there. Then we add the blue point up here. And then let me just draw the rest of the yellow points. They're all on the surface of this sphere. Obviously I'm only dealing with two dimensions right now and it's nearly impossible, or maybe impossible, to imagine a three dimensional surface of a four dimensional sphere. But the analogy holds. If this is a surface the balloon, or the surface of a bubble, if the bubble were to expand over a few billion years-- and once again, not drawn to scale. So now we have a bigger bubble here. This part of the surface is all going to expand. So once again, you have your magenta. You have your blue dot. You have your green dot right over here. And then let me just draw the rest in yellow. So they will have all expanded away from each other on the surface of this sphere. And just to make it clear that this is a sphere, let me draw some contour lines. So this is a contour line. Just to make it clear that we are on the surface of an actual sphere. about what is the apparent velocity with which things And remember, we're going have to say, not only how far things are moving away, but we're going to say how far they are moving away from-- if the observer is us-- depending on how far they already are. So what we're going to do-- we could say is-- let me write this down. All objects moving away from each other. And the apparent relative velocity is proportional to distance." + }, + { + "Q": "Why have the points all moved away from each other during 0:41\n", + "A": "Here s a good experiment. Draw a couple dots on a deflated balloon. When you blow it up, all the dots move away from each other, like Sal explained in 4:26.", + "video_name": "1V9wVmO0Tfg", + "timestamps": [ + 41 + ], + "3min_transcript": "Over many videos now, we've been talking about how every interstellar object is moving away from Earth. And we've also been talking about how the further something is away from Earth, the faster it's moving. What I want to do in this video is to put a little bit of numbers behind it, or even better conceptualize what we've been talking about. So one way to think about it is that, if at an early stage in the universe, I were to pick some points. So that's one point, another point, another point, another point. Let me just pick nine points so that I have a proper grid. So this at an early stage in the universe. If we fast forward a few billion years-- and I'm clearly not drawing it to scale-- all of these points have all moved away from each other. So this point is over here-- actually, let me draw another column, just to make it clear. So if we fast forward a few billion years, the universe has expanded. And so everything has moved away from everything. Let me make this point magenta. So this point, the magenta point is now here. This green point has now moved away from the magenta point. And now this blue point has now moved away from the magenta point in that direction. And we could keep going. This yellow point is maybe over here now. I think you get the general idea. And I'll just draw the other yellow points. So they've all moved away from each other. So there's no center here. Everything is just expanding away from things next to it. And what you can see here is not only did this thing expand away from this, but this thing expanded away from this even further. Because it had this expansion, plus this expansion. Or another way to think about it is, the apparent velocity with which something is expanding is going to be proportional to how far it is. Because every point in between is also expanding away. of this-- one way to think of this, if you think of the universe as an infinite flat sheet. You can imagine that we're just taking a sheet of, I don't know, some type of sheet of stretchy material and just stretching it out. We're just stretching it out. That's if we kind of imagine a more infinite universe that just goes off in every direction. We're just stretching that infinite sheet out. So it has no boundaries, but we're still stretching it out. Another way to visualize it-- and this what we did earlier on-- Is you can imagine that the universe is the three dimensional surface of a four dimensional sphere. Or the three dimensional surface of a hyper-sphere. So at an early stage in the universe, the sphere looked like this. And these points here-- that magenta point is right over here. The green point is right over there. Then we add the blue point up here. And then let me just draw the rest of the yellow points." + }, + { + "Q": "At around 4:30 Sal demonstrates the universe be a sphere with small dots in it. Will the dots ever stop moving away from each other? Will the sphere ever stop growing?\n", + "A": "We currently think the growth will not ever stop.", + "video_name": "1V9wVmO0Tfg", + "timestamps": [ + 270 + ], + "3min_transcript": "of this-- one way to think of this, if you think of the universe as an infinite flat sheet. You can imagine that we're just taking a sheet of, I don't know, some type of sheet of stretchy material and just stretching it out. We're just stretching it out. That's if we kind of imagine a more infinite universe that just goes off in every direction. We're just stretching that infinite sheet out. So it has no boundaries, but we're still stretching it out. Another way to visualize it-- and this what we did earlier on-- Is you can imagine that the universe is the three dimensional surface of a four dimensional sphere. Or the three dimensional surface of a hyper-sphere. So at an early stage in the universe, the sphere looked like this. And these points here-- that magenta point is right over here. The green point is right over there. Then we add the blue point up here. And then let me just draw the rest of the yellow points. They're all on the surface of this sphere. Obviously I'm only dealing with two dimensions right now and it's nearly impossible, or maybe impossible, to imagine a three dimensional surface of a four dimensional sphere. But the analogy holds. If this is a surface the balloon, or the surface of a bubble, if the bubble were to expand over a few billion years-- and once again, not drawn to scale. So now we have a bigger bubble here. This part of the surface is all going to expand. So once again, you have your magenta. You have your blue dot. You have your green dot right over here. And then let me just draw the rest in yellow. So they will have all expanded away from each other on the surface of this sphere. And just to make it clear that this is a sphere, let me draw some contour lines. So this is a contour line. Just to make it clear that we are on the surface of an actual sphere. about what is the apparent velocity with which things And remember, we're going have to say, not only how far things are moving away, but we're going to say how far they are moving away from-- if the observer is us-- depending on how far they already are. So what we're going to do-- we could say is-- let me write this down. All objects moving away from each other. And the apparent relative velocity is proportional to distance." + }, + { + "Q": "2:07. Why does the lone pair of the oxygen attack the carbon?\n", + "A": "Because it is more nucleophilic than the carbon. This is covered earlier in the organic chemistry lectures.", + "video_name": "bFj3HpdC4Uk", + "timestamps": [ + 127 + ], + "3min_transcript": "In this video, we're going to look at the cleavage of alkenes using a reaction called ozonolysis. So over here on the left I have my generic alkene, and to that alkene we're going to add O3 in the first step, which is ozone. And in the second step, we're going to add DMS, which is dimethyl sulfide. And be careful because there are different regions you could add in the second step, so make sure to learn the one that your professor wants you to use. If you use it DMS, you're going to get to a mixture of aldehydes and or ketones for your product, depending on what is attached to your initial alkene. So let's start by looking at a dot structure for ozone. So over here on the left is a possible structure for ozone. And we can draw a resonant structure by taking these electrons and moving them in here to form a double bond between those two oxygens. And now it pushed these electrons in here off onto the oxygen on the left. So let's go ahead and draw the other resonant structure. So now I would have a double bond between my two oxygens on the right. The oxygen on the far right has two lone pairs The oxygen in the center still has a lone pair of electrons on it, and the oxygen on the far left now has three lone pairs of electrons on it. So the oxygen on the far left now has a negative 1 formal charge, and the oxygen on the top here still has a plus 1 formal charge. And so those are our two resonant structures. Remember that the actual molecule is a hybrid of these two resonant structures. So let's go ahead and pick one of those resonant structures. I'm just going to take the one on the right. So let me just go ahead and redraw the resonant structure on the right. And so we're going to do the one that has the negative charge on the oxygen on the far left. And this top oxygen is a plus 1 formal charge. And the oxygen on the right has no charge in this resonant structure. So let's go ahead and draw in our alkene. And so here is our alkene with unknown substituents at the moment. on the oxygen-- this lone pair of electrons here-- is going to attack this carbon which would push these pi electrons off. And those pi electrons are actually going go to this oxygen right here which would push these pi electrons in here off onto this oxygen. So it's a concerted mechanism here. And so let's go ahead and draw the results of those electrons moving. So the oxygen on the left is now bonded to the carbon on the left. The carbon on the left now has a single bond to the carbon on the right. The carbon on the right is now bonded to this oxygen. And then these two oxygens are bonded to an oxygen in the center. For lone pairs of electrons, all of our oxygens are going to have two lone pairs of electrons. Like that. And so we had so many electrons moving, let's say if we can follow them. So let's color coordinate some electrons here. So I'm going to say that these electrons in blue, those" + }, + { + "Q": "\nLa and Ac belongs to d block, 3:05 as unlike Lanthanides and Actinides they possess no electron in f orbital.", + "A": "These two of the four elements whose position in the periodic table is disputed. You will see periodic tables with the disputed elements in either the d or f block. In the official periodic table maintained by IUPAC, La, Ac, Lu, and Lr all appear in the f block. But this is far from a settled matter amongst chemists.", + "video_name": "L-0FkEPPdXE", + "timestamps": [ + 185 + ], + "3min_transcript": "And remember, the D block, even though we're in the fourth period, we're going to actually fill in the three D subshell. We take the period minus one, so it's going to be three D, and to fill it up we have one, two, three, four, five, six, seven, eight, nine, 10, three D 10. Then you're back in the P block and since you're in the fourth period, it's gonna be four P, and once again to fill it up, you have six electrons. And then we go to the fifth period. Five S two and then we're going to go... The period is five but we're in the D block so we're going to go back to the four D subshell. Four D 10 and then we come back and Xenon perfectly fills out the five P subshell. Five P six. So now let's think about what the electron configuration of Neodymium is. So one thing that you're saying, \"Hey, \"what's going on with this F block right over here? \"Just kind of this island of elements.\" And it's really not an island. The reason that they put these colors here is because the F block actually belongs right here and if people had more width on pages to display these periodic tables, they would shove the D and the P blocks over to the right and stuff the F blocks right over here. And we actually did that and I actually like to look at it that way because it makes it clear it's not really an island and it actually kind of fits in the pattern of the periodic table. That is right over here. This right over here is our F block. So that is our... This right over here is our f block. the electron configuration for Neodymium. And I'll give you a hint. In the D block you go back and back fill one subshell below the one shell below the period that you're in. In the F block you go two, you go back two shells. So here, we're not filling out the six F subshell, not even the five F subshell, but the four F subshell. So given that, figure out how much more do we have to add. How much more do we have to go past Xenon in order to get its electron configuration. So I encourage you to pause the video and figure that out right now for Neodymium. So I'm assuming you've had a go at it. So Neodymium is going to have the electron configuration of Xenon so Xenon gets us right over." + }, + { + "Q": "At 4:10, Sal says that Xenon is the last noble gas with a lower atomic number than Neodymium. Why doesn't he just use Barium (56) instead.??\n\nThx in advance!!\nClarissa\n", + "A": "Barium is not a noble gas so can be changed at any time. In order to start electron configuration with an element you have to start with a element that cannot be changed, so we always start with a noble gas.", + "video_name": "L-0FkEPPdXE", + "timestamps": [ + 250 + ], + "3min_transcript": "So now let's think about what the electron configuration of Neodymium is. So one thing that you're saying, \"Hey, \"what's going on with this F block right over here? \"Just kind of this island of elements.\" And it's really not an island. The reason that they put these colors here is because the F block actually belongs right here and if people had more width on pages to display these periodic tables, they would shove the D and the P blocks over to the right and stuff the F blocks right over here. And we actually did that and I actually like to look at it that way because it makes it clear it's not really an island and it actually kind of fits in the pattern of the periodic table. That is right over here. This right over here is our F block. So that is our... This right over here is our f block. the electron configuration for Neodymium. And I'll give you a hint. In the D block you go back and back fill one subshell below the one shell below the period that you're in. In the F block you go two, you go back two shells. So here, we're not filling out the six F subshell, not even the five F subshell, but the four F subshell. So given that, figure out how much more do we have to add. How much more do we have to go past Xenon in order to get its electron configuration. So I encourage you to pause the video and figure that out right now for Neodymium. So I'm assuming you've had a go at it. So Neodymium is going to have the electron configuration of Xenon so Xenon gets us right over. has a smaller atomic number than Neodymium. So we can start with that, which we already saw is fairly involved. And then to that we just add the incremental higher energy electrons. So now we're in the sixth period so we're gonna go six S two, that's that right over there. That's six S two and now we're filling, we're gonna start going into the F subshell but it's not gonna be the sixth shell, it's gonna go two. It's gonna be our period minus two so it's gonna be four. Let me use a color you can see. So this gonna be, we are four electrons into that subshell. One, two, three, four so it's gonna be four F four. So we're in the sixth period but since we're in the F block, we're gonna go back, fill it to the fourth shell. That's the fourth shell right over there. And we are one, two, three, four," + }, + { + "Q": "0:47, How did you know it was a first order reaction?\n", + "A": "He is saying, IF we know that the reaction is first order, this is how to do the math.", + "video_name": "Bt0mz4mGddk", + "timestamps": [ + 47 + ], + "3min_transcript": "- [Voiceover] Let's say we have a first order reaction where A turns into our products, and when time is equal to zero we have our initial concentration of A, and after some time T, we have the concentration of A at that time T, and let's go ahead and write out the rate for our reaction. In an early video, we said that we could express the rate of our reaction in terms of the disappearance of A, so we said that's the change in the concentration of A over the change in time, and we put a negative in sign in here to give us a positive value for the rate. We could also write out the rate law, so for the rate law, the rate is equal to the rate constant K times the concentration of A, and since this is a first order reaction, this would be to the first power, so we talked about this in an early video. We can set these equal to each other, right? Since they're both equal to the rates, we can say that the rate of disappearance of A, so the negative change in the concentration of A times the concentration of A to the first power. This is the average rate of reaction over here on the left, and so if we wanted to write this as the instantaneous rate, we need to think about calculus, all right? So the instantaneous rate would be at negative D A, right? D A, the negative rate of change of A with respect to time, so this would be D T, and this is equal to K A to the first power, and now we have a differential equation, and when you're solving a differential equation eventually you get to a function, and our function would be the concentration as a function of time, so we will eventually get there, but your first step for solving a differential equation is to separate your variables, so you need to put all the A's on one side, and we'll put the T on the opposite side, so we need to divide through by A, We divide both sides by A, and we get D A over the concentration of A here. We're going to multiply both sides by D T to get the T on the right side, so we would get K D T over here on the right, and I'll go ahead and put the negative sign on the right as well, so we just rearranged a few things to get us ready for integration, right? After you separate your variables you need to integrate, so we're going to integrate the left, and since K is a constant I can pull it out of my integral, and I can go like that, and let's see, what would be integrating from? Well, for time, let's go back up to here. All right, time we would be integrating from time is equal to zero to time is equal to T, and for concentration we'd be integrating from the initial concentration to the concentration at some time T, so let's go ahead and plug those in, so we'd be integrating from zero to T on the right side, and the left side we'd be integrating from" + }, + { + "Q": "\nAt 3:17, Sal states that the cell takes in nutrients from the environment in order to grow. What sort of nutrients does the cell take in? And doesn't this intake of nutrients affect the concentration of cytosol which in turn may affect the organelles?", + "A": "The nutrition are what you eat. Everything you eat is broken down and divided up into carbohydrates, portein, fats, etc. which get sent off to the different parts of the body including the protein, which is sent to the cells. The proteins do affect the concentration of the cytosol, which does affect the organelles, but it will not affect it negatively. If you have any more questions or need more detail, comment below...", + "video_name": "VXLSTd_dlKg", + "timestamps": [ + 197 + ], + "3min_transcript": "But what I wanna focus on in this video is interphase. To do that, let's draw ourselves a cell. So let me copy and paste. So this right over here, actually let me, I did that just to save time. So let's say this is a cell, so green. I have it's nuclear membrane, or not nuclear membrane, I have its cell membrane. Inside of that, of course, you have all of the, all of the cytosol, and then this, in this orangeish color, I have the nuclear membrane that defines the nucleus. And then inside of that I have the DNA. And you might be used to seeing DNA all tightly bound, or chromosomes all tightly bound like that and like that or like this, this would be another chromosome right over here in magenta. But during interphase, the chromosomes aren't tightly bound like that so that they're easy to see For most of a cell's life, the chromosomes are completely unwound. They are in their chromatin form. So they are in their chromatin form. It's actually hard to see if you have just a simple microphone (laughing) a simple microscope. It's all unwound, you just have the proteins and the DNA, it's all tangled together. Now there's one other thing that I drew here. You might say, why am I drawing it when I haven't drawn most of the other organelles? But I'm drawing this thing, which is called a centrosome, 'cause it's going to be important for, it's going to be important for when we go into mitosis. Now, this drawing as well, you might say, wait, doesn't a cell, at least a human cell that has a diploid number of chromosomes, and once again, if we're not talking about sex cells, we're talking about just our somatic cells, doesn't it have to have 46 chromosomes? It looks like you only drew two. And it is true, I only drew two chromosomes that this is the cell of some organism that's much simpler, that it only has two chromosomes. So anyway, this is the new cell right over here. It is going to grow. So it is going to grow, it's going to take in nutrients from its environment, and it's going to grow as we would expect it to. So that's that right over there. And then let me give it its nucleus and its centrosome just like that. And this phase, this phase, where it is just growing from this new cell, this is, this phase right over here, is the G1 phase, the G1, actually I'm gonna do that in a different color since I'm already using that green so much. This is the G1 phase and so that might look something like this, different cells are going to do this for different periods of time, the G1 phase. But then you can imagine, well look, it's going to need to replicate some of the, or, it's gonna replicate the information inside of, or that's coded by the DNA" + }, + { + "Q": "\nI don't get what the G1 and G2 phases are. This is mentioned in 3:40.", + "A": "Interphase: G1 phase -First phase cell enters if it\u00e2\u0080\u0099s going to divide / Normal cell functions, cell growth, and protein synthesis / Organelles replicate / May take anywhere from 8-12 hrs to months Interphase: G2 Phase -Protein synthesis / Preparation for division / Lasts 2-5 hours", + "video_name": "VXLSTd_dlKg", + "timestamps": [ + 220 + ], + "3min_transcript": "For most of a cell's life, the chromosomes are completely unwound. They are in their chromatin form. So they are in their chromatin form. It's actually hard to see if you have just a simple microphone (laughing) a simple microscope. It's all unwound, you just have the proteins and the DNA, it's all tangled together. Now there's one other thing that I drew here. You might say, why am I drawing it when I haven't drawn most of the other organelles? But I'm drawing this thing, which is called a centrosome, 'cause it's going to be important for, it's going to be important for when we go into mitosis. Now, this drawing as well, you might say, wait, doesn't a cell, at least a human cell that has a diploid number of chromosomes, and once again, if we're not talking about sex cells, we're talking about just our somatic cells, doesn't it have to have 46 chromosomes? It looks like you only drew two. And it is true, I only drew two chromosomes that this is the cell of some organism that's much simpler, that it only has two chromosomes. So anyway, this is the new cell right over here. It is going to grow. So it is going to grow, it's going to take in nutrients from its environment, and it's going to grow as we would expect it to. So that's that right over there. And then let me give it its nucleus and its centrosome just like that. And this phase, this phase, where it is just growing from this new cell, this is, this phase right over here, is the G1 phase, the G1, actually I'm gonna do that in a different color since I'm already using that green so much. This is the G1 phase and so that might look something like this, different cells are going to do this for different periods of time, the G1 phase. But then you can imagine, well look, it's going to need to replicate some of the, or, it's gonna replicate the information inside of, or that's coded by the DNA So let's depict that. So let me draw, let me draw the nucleus and the centrosome again. Let me draw that again. Let me draw the cellular membrane. This nice healthy growing cell. And now, its DNA is actually going to replicate. So instead of having one copy of its DNA, it's essentially going to go to two copies. But I wanna be very very careful now. So if I draw that magenta chromosome up here, so once again it's all unwound like that. When it replicates, it's going to create a copy of its DNA, and once again, I'm not doing justice for how much DNA, how much DNA there actually is. But it was one chromosome before, it was one chromosome when it was just like this, and it's still one chromosome, even though it's copied its genetic material." + }, + { + "Q": "\nAt 4:08, are you saying that over time, carbon will become nitrogen?", + "A": "Carbon-14 will decay into Nitrogen-14 over time.", + "video_name": "gqrh8wbPXVE", + "timestamps": [ + 248 + ], + "3min_transcript": "so equal numbers of protons and neutrons turns out to be stable. So for this example, the helium-four nucleus is stable. Thinking about that, let's look at carbon-14 next. We have carbon-14, so let's get a little space right down here. So carbon-14 atomic number of six. Therefore, carbon has six protons in the nucleus. So there are six protons. Number of neutrons will be 14 minus six, so eight neutrons. So what's the neutron to proton ratio? So what's the N to Z ratio here? Well the N to Z ratio would be eight neutrons and six protons, and obviously that number is greater than one, so we have an unstable nucleus. The carbon-14 nucleus is unstable, it's radioactive, it's going to undergo spontaneous decay. It's going to try to get a better neutron to proton ratio. which represents the spontaneous decay of carbon-14. So here is our nuclear equation. And when you're writing nuclear equations, you're representing only the nuclei here, so for example, on the left side of my nuclear equation, I have carbon-14, we're talking about only the nucleus, so we're talking about six protons and eight neutrons in the nucleus. And so let's look and see what happens here. So carbon-14, the nucleus, the carbon-14 nucleus is actually going to give off an electron, and so that's pretty weird, and we'll talk about more why in the next video. It's a conversion that's governed by the weak nuclear force. But we know that an electron has a negative one charge, and so that's what we're talking about here. Here for carbon, we have six protons, let me go and write that, six protons here. let's write a negative one charge here for the electron. The carbon-14 nucleus is turning into the nucleus for nitrogen here. Let's look at what we have. Our atomic number is seven, so we have seven protons, let's go ahead and write that here. Seven protons, and 14 minus 7 gives us seven neutrons. So we look at the mass number here, so 14 minus seven gives us seven neutrons. And so that ratio, the ratio of neutrons to protons is seven over seven, which is equal to one. That implies that we have a stable nucleus here. That's the reason why carbon-14 undergoes radioactive decay. Let's look at more details about a nuclear equation, because that's really what I'm most concerned about here in this video. The number of nucleons is conserved. Let's use a different color here." + }, + { + "Q": "\nBy 2:02 can we say that Deuterium and Tritium are radioactive? What about elements like Boron and Chlorine? Are they also radioactive as they do not follow this rule?", + "A": "Deuterium isn t radioactive, tritium is. What about boron and chlorine exactly? They both have two stable isotopes so I m not sure what you are getting at. If you mean because their N/Z isn t exactly equal to 1, that s a general rule only. There are exceptions to almost every rule in chemistry.", + "video_name": "gqrh8wbPXVE", + "timestamps": [ + 122 + ], + "3min_transcript": "- [Voiceover] In the last video, we talked about the helium nucleus, which contains two protons and two neutrons. Protons and neutrons in the nucleus are called nucleons, and so I'll use that term a few times in this video. Here's a picture of the nucleus, with two protons and two neutrons, and we know it's stable, even though we know like charges repel. And so these two protons are repelling each other, and that's the electrostatic force. So let me go ahead and write that here. The electrostatic force says like charges repel. We know that this nucleus is stable, so there must be something else holding the nucleus together, which we call the strong force. So the nuclear strong force is stronger than the electrostatic force. The strong force acts only over short distances though, but it does act between all nucleons. For example, a proton-proton interaction is the same as a proton-neutron interaction, which is the same as a neutron-neutron interaction. about the strong force. That's not really the point of this video. The point is that this nucleus is stable. And let's think about why. We have equal numbers of protons and neutrons, and so that's interesting. So let's think about the atomic number, which tells us the number of protons, which we represent by Z. And the number of neutrons we could say is capital N. So if we're concerned with the ratio, the ratio of neutrons to protons, so the N to Z ratio. In this example, we have two protons and two neutrons. So two neutrons over two protons is equal to one. We have N to Z ratio of one. It turns out that nuclei that have small numbers of protons, so if we're talking about Z is less than 20, they have stable nuclei when the N to Z ratio is equal to one. So when N over Z is equal to one, so equal numbers of protons and neutrons turns out to be stable. So for this example, the helium-four nucleus is stable. Thinking about that, let's look at carbon-14 next. We have carbon-14, so let's get a little space right down here. So carbon-14 atomic number of six. Therefore, carbon has six protons in the nucleus. So there are six protons. Number of neutrons will be 14 minus six, so eight neutrons. So what's the neutron to proton ratio? So what's the N to Z ratio here? Well the N to Z ratio would be eight neutrons and six protons, and obviously that number is greater than one, so we have an unstable nucleus. The carbon-14 nucleus is unstable, it's radioactive, it's going to undergo spontaneous decay. It's going to try to get a better neutron to proton ratio." + }, + { + "Q": "From 5:35 to 6:50, it seems more reasonable to me that water molecule should attach to secondary carbocation instead of tertiary carbocation because I think since secondary carbocation is more positive than the tertiary one, secondary carbocation would be more strongly attracted to a lone pair of electrons on the water molecule. Why is it not the case in the video?\n", + "A": "Yes, both C1 and C2 have \u00ce\u00b4\u00e2\u0081\u00ba character, but C1 has more of the + charge. The cyclic bromonium ion is a resonance hybrid of two contributors: one with the + on C1, and the other with the + on C2. The second one is the minor contributor, so its C-Br bond is closer to an ordinary covalent bond. The major contributor is closer to having a tertiary carbocation at C1, so that is where the water attacks.", + "video_name": "FaOOx6IZxV8", + "timestamps": [ + 335, + 410 + ], + "3min_transcript": "can examine the stereochemistry a little bit more here. So if I start with an alkene, and to this alkene we are going to add bromine and water. And we're going to think about doing this two different ways here. So we'll start with the way on the left. So Br2 and H2O. And then we'll come back and we'll go ahead and do this on the right. So BR2 and H2O. So on the left side, I'm going to think about the formation of that bromonium ion here. So I'm going to once again look at this molecule a little bit from above, so looking down. And I'm going to say the bromonium ion is going to form this way. So the bromine's going to form on top here. And so there's going to be two lone pairs of electrons It's going to have a plus 1 formal charge. So I'm going to say that my methyl group is now going down in space. So with the addition of my bromonium ion, that would be my intermediate. And so now, when I think about water coming along and acting as a nucleophile, so here is H2O, and I think about which carbon will the oxygen attack? So I have two options, right? This oxygen could attack the carbon on the left, or it could attack the carbon on the right. It's been proven that the option is going to attack the most substituted carbon. So if I look at the carbon on the left, and if I think about what sort of carbocation would that be, the carbon on the left is bonded to two other carbons. So this would be similar to a secondary carbocation, or a partial carbocation in character. So you could think about it as being if this was a partial positive. Or on the right, if I think about this carbon right here, the one in red. And if I think about that being a carbocation, that would be bonded to one, two, three other carbons. So it's like a tertiary carbocation. And we know that tertiary carbocations are more stable than secondary. So even though this isn't a full carbocation, this carbon in red exhibits some partial carbocation character, and that is where our water is going to attack. So the nucleophile is going to attack the electrophile. And it's more stable for it to attack this one on the right, since it has partial carbocation character similar to a tertiary carbocation. And if it attacks that carbon on the right, these electrons here we kick off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. OK, so what would we have here?" + }, + { + "Q": "Please refer instants 00:34 & 02:25 seconds of video Fluids (Part 5)\nIn the example you delt with, assume there is water in the jar, upto height h in the jar without any objects. Now we know, the pressure is going to vary as per formula (density times the (g * h)).\nPlease educate me on how the pressure at each and every point of the jar is going to get effected, if a cube denser than the water is suspended (and immersed) in the water using a thread. Is the profile of pressure with respect to the\n", + "A": "the density (or weight/mass) of the cube will not matter beyond causing it to sink because it will be off set by the tension in the string suspending it. putting it in the jar will be no different than augmenting the shape of the jar", + "video_name": "vzID7ds600c", + "timestamps": [ + 34, + 145 + ], + "3min_transcript": "Let's say we have a cup of water. Let me draw the cup. This is one side of the cup, this is the bottom of the cup, and this is the other side of the cup. Let me say that it's some liquid. It doesn't have to be water, but some arbitrary liquid. It could be water. That's the surface of it. We've already learned that the pressure at any point within this liquid is dependent on how deep we go into the liquid. One point I want to make before we move on, and I touched on this a little bit before, is that the pressure at some point isn't just acting downwards, or it isn't just acting in one direction. It's acting in all directions on that point. So although how far we go down determines how much pressure there is, the pressure is actually acting in all directions, including up. The reason why that makes sense is because I'm assuming that this is a static system, or that the fluids in this down here, and it's stationary. The fact that it's stationary tells us that the pressure in every direction must be equal. Let's think about a molecule of water. A molecule of water, let's say it's roughly a sphere. If the pressure were different in one direction or if the pressure down were greater than the pressure up, then the object would start accelerating downwards, because its surface area pointing upwards is the same as the surface area pointing downwards, so the force upwards would be more. It would start accelerating downwards. Even though the pressure is a function of how far down we go, at that point, the pressure is acting in every direction. Let's remember that, and now let's keep that in mind to learn a little bit about Archimedes' principle. Let's say I submerge a cube into this liquid, and let's What I want to do is I want to figure out if there's any force or what is the net force acting on this cube due to the water? Let's think about what the pressure on this cube is at different points. At the depths along the side of the cube, we know that the pressures are equal, because we know at this depth right here, the pressure is going to be the same as at that depth, and they're going to offset each other, and so these are going to be the same. But one thing we do know, just based on the fact that pressure is a function of depth, is that at this point the pressure is going to be higher-- I don't know how much higher-- than at this point, because this point is deeper into the water. Let's call this P1. Let's call that pressure on top, PT, and let's call this" + }, + { + "Q": "\n@2:03, why did Jay say that he \"has to\" make chlorine point up (axial)? Could't he make it point equatorial?", + "A": "He randomly chose that first carbon to be #1, and the up one on that carbon happened to be axial. He could have chosen the next carbon along instead and then it would have been equatorial, but then the problem is this mechanism wouldn t work.", + "video_name": "uCW6154hPkc", + "timestamps": [ + 123 + ], + "3min_transcript": "- [Instructor] We're doing E two mechanism for a substituted cyclohexane. The key is to think about your chair conformations. For example, let's look at this substituted cyclohexane on the left here, and let's first try to solve this problem without looking at a chair conformation. So I would start by saying my chlorine is attached to my alpha carbon, and the carbons directly bonded to the alpha carbon would be the beta carbons, so there's a beta carbon on the right, which I call beta one, there's a beta carbon on the left which I will call beta two. I know that sodium methoxide will be my strong base. The ethoxide anion will take a proton from one of those beta carbons and a double bond would form between the alpha carbon and one of those beta carbons, and the chlorine would leave as the chloride anion, so a double bond forms between the alpha and the beta one carbon, so I'll draw in a double bond here, and the chlorine is now gone, but I would still have this methyl group going away from us in space, so CH3, so that's one possible product, and you might think oh I might also form a double bond between the alpha and the beta two carbon, and so my methyl group would just be on a straight line this time, and so you might think that could be one of the products, but when you actually think about chair conformations, you'll realize that you would not realize this product, so this product is not observed, so you would only get this one, and let's look at why. So first, we need to number our cyclohexane ring, and remember, when you number a cyclohexane ring for a chair conformation, you're not necessarily doing IUPAC nomenclature. You're just using whatever numbers you need to help you draw a proper chair conformation, so I'm going to start with the chlorine being number one, then I'm going to go around to the right, so two, three, carbon four, carbon five, and carbon six, so we're not naming this. We're just trying to get a chair conformation, so on my chair conformation down here, I always say this is carbon one, and so I need to have the chlorine, the chlorine up at carbon one, The chlorine has to be up axial, and so if I go around to carbon six, so this would be carbon one here, two, three, four, five, and six, I have a methyl group going away from me in space, so this would be going down, so I'm gonna draw in a Me for a methyl group right here, so it's down axial. So now let's draw in some hydrogens on our beta carbons, so let me highlight our beta carbons here. I'll use red, so this would be, what I've marked is being beta one, so I have two hydrogens on that carbon, so I'll draw those in here, and then my other beta carbon which I called beta two up here, so I only have, so this is beta two, I have only one hydrogen, and it is equatorial. So let's go to a video, so we can analyze which one of these protons will participate in our E two mechanism. Here's our chair conformation," + }, + { + "Q": "\nAt 2:09, does anyone know more about Robert Hooke's microscope structure?", + "A": "When Hooke viewed a thin cutting of cork using his self designed microscope, he discovered empty spaces contained by walls, and termed them cells. They looked like honeycombs. In latin, cell means tiny rooms. Thus, as Robert Hooke thought the compartments looked like small rooms. Thus, he called them cells.", + "video_name": "1aJBToJrlvA", + "timestamps": [ + 129 + ], + "3min_transcript": "Man: This is an animal. This is also an animal. Animal. Animal carcass. Animal. Animal carcass again. Animal. The thing that all of these other things have in common is that they're made out of the same basic building block, the animal cell. (music) Animals are made up of your run-of-the-mill eukaryotic cells. These are called eukaryotic because they have a true kernel in the Greek, a good nucleus. That contains the DNA and calls the shots for the rest of the cell, also containing a bunch of organelles. There's a bunch of different kinds of organelles and they all have very specific functions. All of this is surrounded by the cell membrane. Of course, plants are eukaryotic cells too, but theirs are set up a little bit differently. Of course, they have oranelles that allow them to make their own food, which is super nice. We don't have those. Also, their cell membrane is actually a cell wall that's made of cellulose. It's rigid which is why plants can't dance. we did a whole video on it and you can click on it hit here, if it's online yet; it might not be. A lot of the stuff in this video is going to apply to all eukaryotic cells, which includes plants and fungi and protists. Rigid cell walls, that's cool and all, but one of the reasons that animals have been so successful is that their flexible membrane, in addition to allowing them the ability to dance, gives animals the flexibility to create a bunch of different cell types and organ types and tissue types that could never be possible in a plant. Cell walls that protect plants and give them structure prevent them from evolving complicated nerve structures and muscle cells that allow animals to be such a powerful force for eating plants. Animals can move around, find shelter and food, find things to mate with, all that good stuff. In fact, the ability to move oneself around using specialized muscle tissue has been 100% trademarked by kindgom animalia. Voiceover: What about protozoans? Man: Excellent point. What about protozoans? They don't have specialized muscle tissue. They move around with cilia and flagella and that kind of thing. Robert Hooke discovered cells with his kind of crude beta version microscope. He called them cells because they looked like bare, spartan Monk's bedrooms with not much going on inside. Hooke was a smart guy and everything, but he could not have been more wrong about what was going on inside of a cell. There is a whole lot going on inside of a eukaryotic cell. It's more like a city than a Monk's cell. In fact, let's go with that. A cell is like a city. It has defined geographical limits, a ruling government, power plants, roads, waste treatment plants, a police force, industry, all the things a booming metropolis needs to run smoothly. But, this city does not have one of those hippy governments where everybody votes on stuff and talks things out at town hall meetings Nope. Think fascist Italy circa 1938. Think Kim Jong-Il's, I mean Kim Jong-Un's North Korea and you might be getting a closer idea of how eukaryotic cells do their business. Let's start out with city limits. As you approach the city of Eukaryopolis, there's a chance that you will notice something that a traditional city never has, which is either cilia or flagella." + }, + { + "Q": "at 3:28 Hank says microtubules are called (9+2) structure. Can anyone tell me that why are they called so?\n", + "A": "They are called the 9+2 structure because as he says at 3:24 there are made of 9 microtubules forming a ring around 2 microtubules.", + "video_name": "1aJBToJrlvA", + "timestamps": [ + 208 + ], + "3min_transcript": "we did a whole video on it and you can click on it hit here, if it's online yet; it might not be. A lot of the stuff in this video is going to apply to all eukaryotic cells, which includes plants and fungi and protists. Rigid cell walls, that's cool and all, but one of the reasons that animals have been so successful is that their flexible membrane, in addition to allowing them the ability to dance, gives animals the flexibility to create a bunch of different cell types and organ types and tissue types that could never be possible in a plant. Cell walls that protect plants and give them structure prevent them from evolving complicated nerve structures and muscle cells that allow animals to be such a powerful force for eating plants. Animals can move around, find shelter and food, find things to mate with, all that good stuff. In fact, the ability to move oneself around using specialized muscle tissue has been 100% trademarked by kindgom animalia. Voiceover: What about protozoans? Man: Excellent point. What about protozoans? They don't have specialized muscle tissue. They move around with cilia and flagella and that kind of thing. Robert Hooke discovered cells with his kind of crude beta version microscope. He called them cells because they looked like bare, spartan Monk's bedrooms with not much going on inside. Hooke was a smart guy and everything, but he could not have been more wrong about what was going on inside of a cell. There is a whole lot going on inside of a eukaryotic cell. It's more like a city than a Monk's cell. In fact, let's go with that. A cell is like a city. It has defined geographical limits, a ruling government, power plants, roads, waste treatment plants, a police force, industry, all the things a booming metropolis needs to run smoothly. But, this city does not have one of those hippy governments where everybody votes on stuff and talks things out at town hall meetings Nope. Think fascist Italy circa 1938. Think Kim Jong-Il's, I mean Kim Jong-Un's North Korea and you might be getting a closer idea of how eukaryotic cells do their business. Let's start out with city limits. As you approach the city of Eukaryopolis, there's a chance that you will notice something that a traditional city never has, which is either cilia or flagella. of these structures; cilia being a bunch of tiny little arms that wriggle around, flagella being one long whip-like tail. Some cells have neither. Sperm cells, for instance, have flagella and our lungs and throat cells have cilia that push mucus up and out of our lungs. Cilia and flagella are made out of long protein fibers called microtubules, and they both have the same basic structure; nine pairs of microtubules forming a ring around two central microtubles, this is often called the 9+2 structure. Anyway, that's just so you know when you're approaching the city, watch out for the cilia and flagella. If you make it past the cilia, you will encounter what is called a cell membrane, which is a kind of squishy, not rigid plant cell wall, which totally encloses the city and all of its contents. It's also in charge of monitoring what comes in and out of the cell, kind of like the fascist border police. The cell membrane has selective pemeability, meaning that it can choose what molecules come in and out of the cells, for the most part. I did an entire video on this which you can check out right here." + }, + { + "Q": "\nat 4:20 , aren't centrosomes also called centrioles", + "A": "Centrosomes are made up of a pair of centrioles", + "video_name": "1aJBToJrlvA", + "timestamps": [ + 260 + ], + "3min_transcript": "Robert Hooke discovered cells with his kind of crude beta version microscope. He called them cells because they looked like bare, spartan Monk's bedrooms with not much going on inside. Hooke was a smart guy and everything, but he could not have been more wrong about what was going on inside of a cell. There is a whole lot going on inside of a eukaryotic cell. It's more like a city than a Monk's cell. In fact, let's go with that. A cell is like a city. It has defined geographical limits, a ruling government, power plants, roads, waste treatment plants, a police force, industry, all the things a booming metropolis needs to run smoothly. But, this city does not have one of those hippy governments where everybody votes on stuff and talks things out at town hall meetings Nope. Think fascist Italy circa 1938. Think Kim Jong-Il's, I mean Kim Jong-Un's North Korea and you might be getting a closer idea of how eukaryotic cells do their business. Let's start out with city limits. As you approach the city of Eukaryopolis, there's a chance that you will notice something that a traditional city never has, which is either cilia or flagella. of these structures; cilia being a bunch of tiny little arms that wriggle around, flagella being one long whip-like tail. Some cells have neither. Sperm cells, for instance, have flagella and our lungs and throat cells have cilia that push mucus up and out of our lungs. Cilia and flagella are made out of long protein fibers called microtubules, and they both have the same basic structure; nine pairs of microtubules forming a ring around two central microtubles, this is often called the 9+2 structure. Anyway, that's just so you know when you're approaching the city, watch out for the cilia and flagella. If you make it past the cilia, you will encounter what is called a cell membrane, which is a kind of squishy, not rigid plant cell wall, which totally encloses the city and all of its contents. It's also in charge of monitoring what comes in and out of the cell, kind of like the fascist border police. The cell membrane has selective pemeability, meaning that it can choose what molecules come in and out of the cells, for the most part. I did an entire video on this which you can check out right here. is kind of wet and squishy; it's a bit of a swampland. Each eukaryotic cell is filled with a solution of water and nutrients called cytoplasm. Inside of the cytoplasm is a scaffolding called the cytoskeleton. It's basically just a bunch of protein strands that reinforce the cell. Centrosomes are a special part of this reinforcement. They assemble long microtubules out of proteins that act like steel girders that hold all the cities' building together. The cytoplasm provides the infrastructure necessary for all the organelles to do all of their awesome amazing business, with the notable exception of the nucleus, which has its own kind of cytoplasm called the nucleoplasm, which is a more luxurious, premium environment befitting the cell's beloved leader. First, let's talk about the cell's highway system. The endoplasmic reticulum, or just ER, are organelles that create a network of membranes that carry stuff around the cell. These membranes are phospolipid bilayers same as in the cell membrane. There are two types of ER. There's the rough and the smooth; fairly similar, but slightly different shapes, slightly different functions. The rough ER looks all bumpy because it has" + }, + { + "Q": "At 4:23, what are centrosomes, are they centrioles?\n", + "A": "An organelle near the nucleus of a cell that contains the centrioles (in animal cells) and from which the spindle fibers develop in cell division.", + "video_name": "1aJBToJrlvA", + "timestamps": [ + 263 + ], + "3min_transcript": "Robert Hooke discovered cells with his kind of crude beta version microscope. He called them cells because they looked like bare, spartan Monk's bedrooms with not much going on inside. Hooke was a smart guy and everything, but he could not have been more wrong about what was going on inside of a cell. There is a whole lot going on inside of a eukaryotic cell. It's more like a city than a Monk's cell. In fact, let's go with that. A cell is like a city. It has defined geographical limits, a ruling government, power plants, roads, waste treatment plants, a police force, industry, all the things a booming metropolis needs to run smoothly. But, this city does not have one of those hippy governments where everybody votes on stuff and talks things out at town hall meetings Nope. Think fascist Italy circa 1938. Think Kim Jong-Il's, I mean Kim Jong-Un's North Korea and you might be getting a closer idea of how eukaryotic cells do their business. Let's start out with city limits. As you approach the city of Eukaryopolis, there's a chance that you will notice something that a traditional city never has, which is either cilia or flagella. of these structures; cilia being a bunch of tiny little arms that wriggle around, flagella being one long whip-like tail. Some cells have neither. Sperm cells, for instance, have flagella and our lungs and throat cells have cilia that push mucus up and out of our lungs. Cilia and flagella are made out of long protein fibers called microtubules, and they both have the same basic structure; nine pairs of microtubules forming a ring around two central microtubles, this is often called the 9+2 structure. Anyway, that's just so you know when you're approaching the city, watch out for the cilia and flagella. If you make it past the cilia, you will encounter what is called a cell membrane, which is a kind of squishy, not rigid plant cell wall, which totally encloses the city and all of its contents. It's also in charge of monitoring what comes in and out of the cell, kind of like the fascist border police. The cell membrane has selective pemeability, meaning that it can choose what molecules come in and out of the cells, for the most part. I did an entire video on this which you can check out right here. is kind of wet and squishy; it's a bit of a swampland. Each eukaryotic cell is filled with a solution of water and nutrients called cytoplasm. Inside of the cytoplasm is a scaffolding called the cytoskeleton. It's basically just a bunch of protein strands that reinforce the cell. Centrosomes are a special part of this reinforcement. They assemble long microtubules out of proteins that act like steel girders that hold all the cities' building together. The cytoplasm provides the infrastructure necessary for all the organelles to do all of their awesome amazing business, with the notable exception of the nucleus, which has its own kind of cytoplasm called the nucleoplasm, which is a more luxurious, premium environment befitting the cell's beloved leader. First, let's talk about the cell's highway system. The endoplasmic reticulum, or just ER, are organelles that create a network of membranes that carry stuff around the cell. These membranes are phospolipid bilayers same as in the cell membrane. There are two types of ER. There's the rough and the smooth; fairly similar, but slightly different shapes, slightly different functions. The rough ER looks all bumpy because it has" + }, + { + "Q": "\nAt 13:14 how do you get 2p6? I thought each orbital only had 2 electrons, so how does this one have 6?", + "A": "each orbital can hold two electrons. in the p orbital shell, it has three orbitals, each holding two electrons, therefore holding a maximum of six", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 794 + ], + "3min_transcript": "The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus," + }, + { + "Q": "7:05 (carbon) I'm confused about the P orbitals.. so in carbon the 1s2 and 2s2 are filled and then he goes to the P orbitals and why is it 2p2? I thought it filled only one of the first P orbital and then did one in the second so why isn't it 1p1 and 2p1 instead of 2p2?\n", + "A": "The base number measures the energy level, or the row # of the periodic table. The p block starts on level 2, so there is no 1p. Carbon is also the 2nd element in the p block of that row, so the exponent is 2.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 425 + ], + "3min_transcript": "And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon. the first two electrons go into, so, 1s1, 1s2. So then it fills-- sorry, you can't see everything. So it fills the 1s2, so carbon's configuration. It fills 1s1 then 1s2. And this is just the configuration for helium. And then it goes to the second shell, which is the second period, right? That's why it's called the periodic table. We'll talk about periods and groups in the future. And then you go here. So this is filling the 2s. We're in the second period right here. That's the second period. One, two. Have to go off, so you can see everything. So it fills these two. So 2s2. And then it starts filling up the p orbitals. So then it starts filling 1p and then 2p. And we're still on the second shell, so 2s2, 2p2." + }, + { + "Q": "\nAt 6:10, he refers to the \"z\" direction as up and down, and the \"y\" direction as forward and backward, but isn't that reversed? I thought, that \"x\" was left-right, \"y\" up-down, and \"z\" forward-backward (representing 3 dimensions). Are these interchangeable, now? I'm only asking because it has been 18 years since I last took a math course.", + "A": "They are just labels for axes. Up and down depends on how you are oriented when you are looking at something.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 370 + ], + "3min_transcript": "it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon." + }, + { + "Q": "\nAt 1:30,would those be snapshots of different atoms or the same one?", + "A": "it s the same atom. It is just the electron, that is represented in different snapshots. When you overlay the different snapshots it becomes apparent, that the electron is more likely to be near the nucleus.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 90 + ], + "3min_transcript": "In the last few videos we learned that the configuration of electrons in an atom aren't in a simple, classical, Newtonian orbit configuration. And that's the Bohr model of the electron. And I'll keep reviewing it, just because I think it's an important point. If that's the nucleus, remember, it's just a tiny, tiny, tiny dot if you think about the entire volume of the actual atom. And instead of the electron being in orbits around it, which would be how a planet orbits the sun. Instead of being in orbits around it, it's described by orbitals, which are these probability density functions. So an orbital-- let's say that's the nucleus it would describe, if you took any point in space around the nucleus, the probability of finding the electron. So actually, in any volume of space around the nucleus, it would tell you the probability of finding the electron within that volume. And so if you were to just take a bunch of snapshots of electrons -- let's say in the 1s orbital. You can barely see it there, but it's a sphere around the nucleus, and that's the lowest energy state that an electron can be in. If you were to just take a number of snapshots of electrons. Let's say you were to take a number of snapshots of helium, which has two electrons. Both of them are in the 1s orbital. It would look like this. If you took one snapshot, maybe it'll be there, the next snapshot, maybe the electron is there. Then the electron is there. Then the electron is there. Then it's there. And if you kept doing the snapshots, you would have a bunch of them really close. And then it gets a little bit sparser as you get out, as you get further and further out away from the electron. But as you see, you're much more likely to find the electron close to the center of the atom than further out. Although you might have had an observation with the electron sitting all the way out there, or sitting over here. So it really could have been anywhere, but if you take multiple observations, It's saying look, there's a much lower probability of finding the electron out in this little cube of volume space than it is in this little cube of volume space. And when you see these diagrams that draw this orbital like this. Let's say they draw it like a shell, like a sphere. And I'll try to make it look three-dimensional. So let's say this is the outside of it, and the nucleus is sitting some place on the inside. They're just saying -- they just draw a cut-off -- where can I find the electron 90% of the time? So they're saying, OK, I can find the electron 90% of the time within this circle, if I were to do the cross-section. But every now and then the electron can show up outside of that, right? Because it's all probabilistic. So this can still happen. You can still find the electron if this is the orbital we're talking about out here. Right? And then we, in the last video, we said, OK, the electrons fill up the orbitals from lowest energy state to high energy state." + }, + { + "Q": "I don't understand why the P orbital has Pz, Px and Py? 6:26\n", + "A": "Because there s 3 p orbitals in a shell (that s why a total p subshell can hold 6 electrons). There s one p orbital aligned on each 3-dimensional axis.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 386 + ], + "3min_transcript": "it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon." + }, + { + "Q": "At 13:26 how does he get the configuration for Silicon? Can someone please summarize this whole video. I'm confused\n", + "A": "That s a good explanation, but you explained how to get to Sulfur, not Silicon. So 3p2, rather than 3p4, because Silicon has 14 electrons.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 806 + ], + "3min_transcript": "The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus," + }, + { + "Q": "At 13:37, why does one write out the whole electron configuration, rather than just the last \"shell\"? In other words, wouldn't it be enough to write 3p^2 for silicon, since all the lower energy shells are always filled up?\n", + "A": "When you get to heavier elements you ll find what you say about the lower energy orbitals always being full isn t always true.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 817 + ], + "3min_transcript": "So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus, to fill in more of these bizarro-shaped orbitals. But these are kind of the balance -- I will talk about standing waves in the future -- but these are kind of a balance between trying to get close to the nucleus and the proton and those positive charges, because the electron charges are attracted to them, while at the same time avoiding the other electron charges, or at least their mass distribution functions. Anyway, see you in the next video." + }, + { + "Q": "\nat around 12:08. How does he know 3 electrons go in the p orbital. Is it because that is what is left over after filling the s orbitals?", + "A": "Yes, when determining the electron configuration, you count up from the bottom (1s) orbital and add electrons one at a time until all are used up for that atom or ion. The electrons are first put in singly, then double up with an opposite spin. You cannot move up to the next higher level until the lower orbitals are filled. When the level or sub-shells has more than one orbital (as there are 3 p orbitals), then each is filled up with a single electron first before they are doubled up.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 728 + ], + "3min_transcript": "Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2." + }, + { + "Q": "at 5:19, do the electrons fill the orbital from lowest energy orbital to highest energy orbital?\n", + "A": "Yes, nature always wants to minimize energy.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 319 + ], + "3min_transcript": "well I don't know if Tetris is the thing-- but if I'm stacking cubes, I lay out cubes from low energy, if this is the floor, I put the first cube at the lowest energy state. And let's say I could put the second cube But I only have this much space to work with. So I have to put the third cube at the next highest energy state. In this case our energy would be described as potential energy, right? This is just a classical, Newtonian physics example. But that's the same idea with electrons. Once I have two electrons in this 1s orbital -- so let's say the electron configuration of helium is 1s2 -- the third electron I can't put there anymore, because there's only room for two electrons. The way I think about it is these two electrons are now going to repel the third one I want to add. So then I have to go to the 2s orbital. it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves." + }, + { + "Q": "He stated at 12:10 that the electron configuration for Nitrogen is 1s^2 2s^2 2p^3. However, how can 2p have 3 electrons when each subshell can only have 2 electrons?\n", + "A": "It is orbitals that can only have 2 electrons, not subshells. The p subshell contains 3 orbitals, so it can hold at most 6 electrons. The d subshell contains 5 orbitals, so it can hold at most 10 electrons. The f subshell contains 7 orbitals, so it can hold at most 14 electrons. And, of course, the s subshell has only 1 orbital so it can only hold 2 electrons.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 730 + ], + "3min_transcript": "Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2." + }, + { + "Q": "\nat 1:27 he talks about the s, why are the orbitals called s,p,f,etc?", + "A": "Those letters represent terms that are now outdated and no longer used. The terms were sharp, principal, diffuse, and one I can t remember.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 87 + ], + "3min_transcript": "In the last few videos we learned that the configuration of electrons in an atom aren't in a simple, classical, Newtonian orbit configuration. And that's the Bohr model of the electron. And I'll keep reviewing it, just because I think it's an important point. If that's the nucleus, remember, it's just a tiny, tiny, tiny dot if you think about the entire volume of the actual atom. And instead of the electron being in orbits around it, which would be how a planet orbits the sun. Instead of being in orbits around it, it's described by orbitals, which are these probability density functions. So an orbital-- let's say that's the nucleus it would describe, if you took any point in space around the nucleus, the probability of finding the electron. So actually, in any volume of space around the nucleus, it would tell you the probability of finding the electron within that volume. And so if you were to just take a bunch of snapshots of electrons -- let's say in the 1s orbital. You can barely see it there, but it's a sphere around the nucleus, and that's the lowest energy state that an electron can be in. If you were to just take a number of snapshots of electrons. Let's say you were to take a number of snapshots of helium, which has two electrons. Both of them are in the 1s orbital. It would look like this. If you took one snapshot, maybe it'll be there, the next snapshot, maybe the electron is there. Then the electron is there. Then the electron is there. Then it's there. And if you kept doing the snapshots, you would have a bunch of them really close. And then it gets a little bit sparser as you get out, as you get further and further out away from the electron. But as you see, you're much more likely to find the electron close to the center of the atom than further out. Although you might have had an observation with the electron sitting all the way out there, or sitting over here. So it really could have been anywhere, but if you take multiple observations, It's saying look, there's a much lower probability of finding the electron out in this little cube of volume space than it is in this little cube of volume space. And when you see these diagrams that draw this orbital like this. Let's say they draw it like a shell, like a sphere. And I'll try to make it look three-dimensional. So let's say this is the outside of it, and the nucleus is sitting some place on the inside. They're just saying -- they just draw a cut-off -- where can I find the electron 90% of the time? So they're saying, OK, I can find the electron 90% of the time within this circle, if I were to do the cross-section. But every now and then the electron can show up outside of that, right? Because it's all probabilistic. So this can still happen. You can still find the electron if this is the orbital we're talking about out here. Right? And then we, in the last video, we said, OK, the electrons fill up the orbitals from lowest energy state to high energy state." + }, + { + "Q": "Hi\u00ef\u00bc\u008cI am confused about P orbital. I can understand s orbital and it seems like it follows orders of electrons numbers in each circle. (it seems like follow 2,8,8,18,18,32,32..). However, in the video 7:56, it suddenly jumps to P orbitals without explanations. What does P orbitals refer to\n", + "A": "The orbitals each have a given amount of electrons, s can have 2, and then it must move up to p. The p orbital can have up to 6 electrons, and when you run out, you must move on to the d orbital, which can have 10 electrons. Lastly, you have the f orbital, and that orbital can have up to 14 electrons.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 476 + ], + "3min_transcript": "that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon. the first two electrons go into, so, 1s1, 1s2. So then it fills-- sorry, you can't see everything. So it fills the 1s2, so carbon's configuration. It fills 1s1 then 1s2. And this is just the configuration for helium. And then it goes to the second shell, which is the second period, right? That's why it's called the periodic table. We'll talk about periods and groups in the future. And then you go here. So this is filling the 2s. We're in the second period right here. That's the second period. One, two. Have to go off, so you can see everything. So it fills these two. So 2s2. And then it starts filling up the p orbitals. So then it starts filling 1p and then 2p. And we're still on the second shell, so 2s2, 2p2. if we just wanted to visualize this orbital right here, the p orbitals? So we have two electrons. So one electron is going to be in a-- Let's say if this is, I'll try to draw some axes. That's too thin. So if I draw a three-dimensional volume kind of axes. If I were to make a bunch of observations of, say, one of the electrons in the p orbitals, let's say in the pz dimension, sometimes it might be here, sometimes it might be there, sometimes it might be there. And then if you keep taking a bunch of observations, you're going to have something that looks like this bell shape, this barbell shape right there. And then for the other electron that's maybe in the x direction, you make a bunch of observations. Let me do it in a different," + }, + { + "Q": "\nAt 14:00, could someone explain why all the electron configurations look so strange?", + "A": "They are based on complex math equations that calculate the likelihood of an electron being within the configurations with 90% accuracy.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 840 + ], + "3min_transcript": "So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus, to fill in more of these bizarro-shaped orbitals. But these are kind of the balance -- I will talk about standing waves in the future -- but these are kind of a balance between trying to get close to the nucleus and the proton and those positive charges, because the electron charges are attracted to them, while at the same time avoiding the other electron charges, or at least their mass distribution functions. Anyway, see you in the next video." + }, + { + "Q": "At 5:13 Sal mentions that the water molecules on the skin surface are linked by hydrogen bonds. Would the cooling effect of a liquid be different if there was no hydrogen bond? I would assume that evaporation of other liquids without hydrogen bonds would take away less energy (=heat), because not having to break up a hydrogen bond would need less energy? Am I correct?\n", + "A": "Yes, hydrogen bonds take energy to form, and actually take a larger amount of heat to disrupt a hydrogen bond putting hydrogen bonds. So in theory other liquids with the same evaporating temperatures would take less energy.", + "video_name": "_eEONOJHnEs", + "timestamps": [ + 313 + ], + "3min_transcript": "So average kinetic energy is going to go down. Or another way of saying it, is that your temperature is going to go down. Your temperature is going to go down because as these molecules turn into water vapor, they're going to be the highest kinetic energy, energy is transferred to them, and then they escape. And so what's left over is going to have a lower average kinetic energy. And you're saying, \"Well, how does that \"actually cool down my hand?\" Well, your hand is made up of molecules as well. So let's say this is the surface of your hand, those are the molecules, they have some average kinetic energy, they are kind of vibrating in place, especially if we're talking about they're... they're a solid. And so maybe I'll draw the more, you know, they're vibrating like this, they're bonded to each other in some way. I won't go into the details of what types of molecules these are, but then if you have your water molecules here, water molecules that are sitting on the surface, and I'm drawing this is kind of a cross-section. I'll draw them as blue molecules. So this is an H2O right over here. H2O. This is an H2O. And this is an H2O. And they have some hydrogen bonding, so there is some hydrogen bonding going on. Well, as the high kinetic energy water molecules escape, I'll say this one right over here escapes, and so the average kinetic energy of what's left over is lower, so then the temperature has gone down, and now your body molecules, the ones that are all warmed up, and because of whatever's going on inside of your body, well, those can now bump into, they can vibrate and bump into these water molecules and increase their kinetic energy more than the ones that have the most kinetic energy. Those might escape again. And so it's a--one way of thinking about it is that all that heat is being used to allow these individual water molecules to escape in order to vaporize. And so that heat is leaving your body, so it allows you to cool down. Cooling down happens by heat actually leaving. I wrote evaporative cool... That's how evaporative cooling... That's how evaporative cooling actually works." + }, + { + "Q": "\nAt the beginning of the video (around 0:38) he said that life is carbon based. What is that supposed to mean?", + "A": "What Sal means at 0:38 is that all known lifeforms contain the element carbon. Of course, there are other elements that are found in a majority of life-forms as well. But Carbon is the most prominent. Scientists have yet to find a living organism that does not contain carbon in it s Chemical Make-Up. Therefore, as far as we can tell, all life is carbon-based.", + "video_name": "JgYlogdtJDo", + "timestamps": [ + 38 + ], + "3min_transcript": "You're probably already familiar with some forms of carbon. For example this graph I write over here, this is one form carbon takes. Very important when you're writing with a pencil, otherwise you would not see any writing. If you didn't have the graphite scraping onto your, scraping onto your paper, which you and your paper is also. It's not pure carbon, but it has a lot of carbon in it. This right over here is a raw diamond, another form the carbon can take under intense time or after a long period of time under intense pressure. But what you may or may not realize is that carbon is actually essential for life. In fact life as we know it is carbon-based, so carbon-based, based life. When we look for signs of life, at least life as we know on other planets,we are looking for signs of carbon-based life. And there might be other forms, other other elements that form the backbone of life. But carbon is the only one that we have been able to observe. Now why is carbon so valuable for life? Why does it form the backbone of the molecules that make life possible ? What all comes down to where it and its atomic number and how it tends to bond with things? So this is why Chemistry is important. So carbon we see over here has an atomic number six, which by definition means it has 6 protons. So far we draw its nucleus it would have 1,2 , 3, 4, 5, 6 protons. And the most common isotope of carbon on earth is carbon 12, which also has six neutrons. So let me draw that in this nucleus 1, 2, 3, 4, 5, 6 neutrons. And then neutral carbon is going to have six electrons. And so two of them are going to be in their inner most in the first shell .So that's two of them right over there. These are the inner shell. I guess you could say or the so that's the first two electrons ,and then you have four remaining in its outermost shell. And these four considered valence electrons, these are the electrons that the reacting. And if any of this seems unfamiliar to you,I encourage you to watch the videos on Khan Academy on things like valence electrons, but this is a little bit overview right over here. Carbon has four valence electrons. Valence electrons are the ones that do or that tend to do the reacting. And so I could if I wanted to simplify this drawng over here,I could say ,\"Okay carbon.\"And if I register as valence electrons which is a typical thing to do. I could say carbon has 1, 2, 3 ,4 valence electrons. Now you might remember the octet rule that that atoms tend to be more stable when they at least pretend, like they with that they're sharing or that they have eight electrons in there outermost shell. So carbon can do that by forming 4 covalent bonds. For example it could do that with hydrogen.This hydrogen over here has one valence electron. It actually has one electron. Now the hydrogen feels good. It feels like its sharing two-electrons filling its its first shell .Hydrogen is just trying to fill out the first" + }, + { + "Q": "At 13:10 isn't the height 6m?\n", + "A": "No, if you think about it, if that ball has a radius of 2m. So when the ball is touching the ground, it s center of mass will actually still be 2m from the ground. It s true that the center of mass is initially 6m from the ground, but when the ball falls and touches the ground the center of mass is again still 2m from the ground. That means the height will be 4m. If you work the problem where the height is 6m, the ball would have to fall halfway through the floor for the center of mass to be at 0 height.", + "video_name": "5eX5WnPDnvs", + "timestamps": [ + 790 + ], + "3min_transcript": "and so, now it's looking much better. We just have one variable in here that we don't know, V of the center of mass. This I might be freaking you out, this is the moment of inertia, what do we do with that? With a moment of inertia of a cylinder, you often just have to look these up. The moment of inertia of a cylinder turns out to be 1/2 m, the mass of the cylinder, times the radius of the cylinder squared. So we can take this, plug that in for I, and what are we gonna get? If I just copy this, paste that again. If we substitute in for our I, our moment of inertia, and I'm gonna scoot this over just a little bit, our moment of inertia was 1/2 mr squared. So I'm gonna have 1/2, and this is in addition to this 1/2, so this 1/2 was already here. There's another 1/2, from the moment of inertia term, 1/2mr squared, but this r is the same as that r, so look it, I've got a, I've got a r squared and a one over r squared, these end up canceling, and this is really strange, it doesn't matter what the radius of the cylinder was, and here's something else that's weird, not only does the radius cancel, So no matter what the mass of the cylinder was, they will all get to the ground with the same center of mass speed. In other words, all yo-yo's of the same shape are gonna tie when they get to the ground as long as all else is equal when we're ignoring air resistance. No matter how big the yo-yo, or have massive or what the radius is, they should all tie at the ground with the same speed, which is kinda weird. So now, finally we can solve for the center of mass. We've got this right hand side. The left hand side is just gh, that's gonna equal, so we end up with 1/2, V of the center of mass squared, plus 1/4, V of the center of mass squared. That's just equal to 3/4 speed of the center of mass squared. If you take a half plus a fourth, you get 3/4. So if I solve this for the speed of the center of mass, I'm gonna get, if I multiply gh by four over three, and we take a square root, we're gonna get the square root of 4gh over 3, and so now, I can just plug in numbers. If I wanted to, I could just say that this is gonna equal times 9.8 meters per second squared, times four meters, that's where we started from, that was our height, divided by three, is gonna give us a speed of the center of mass of 7.23 meters per second. Now, here's something to keep in mind, other problems might look different from this, but the way you solve them might be identical. For instance, we could just take this whole solution here, I'm gonna copy that. Let's try a new problem, it's gonna be easy. It's not gonna take long. Let's say we take the same cylinder and we release it from rest at the top of an incline that's four meters tall and we let it roll without slipping to the bottom of the incline, and again, we ask the question, \"How fast is the center of mass of this cylinder \"gonna be going when it reaches the bottom of the incline?\" Well, it's the same problem. It looks different from the other problem, but conceptually and mathematically, it's the same calculation. This thing started off with potential energy, mgh, and it turned into conservation of energy says" + }, + { + "Q": "\nAt 1:48, why isn't the concentration of Cl- taken into consideration? Although it's an extremely weak base, I thought it would still make some difference in the pH.", + "A": "HCl is a strong acid, so that means Cl- has to be an extremely weak base as you stated. Essentially the reverse reaction of Cl- + H+ -> HCl does not happen, so no need to account for it.", + "video_name": "lsHq5aqz4uQ", + "timestamps": [ + 108 + ], + "3min_transcript": "- [Voiceover] Let's do some buffer solution calculations using the Henderson-Hasselbalch equation. So in the last video I showed you how to derive the Henderson-Hasselbalch equation, and it is pH is equal to the pKa plus the log of the concentration of A minus over the concentration of HA. So we're talking about a conjugate acid-base pair here. HA and A minus. And for our problem HA, the acid, would be NH four plus and the base, A minus, would be NH three or ammonia. So the first thing we need to do, if we're gonna calculate the pH of our buffer solution, is to find the pKa, all right, and our acid is NH four plus. So let's say we already know the Ka value for NH four plus and that's 5.6 times 10 to the negative 10. To find the pKa, all we have to do is take the negative log of that. of 5.6 times 10 to the negative 10. So let's get out the calculator and let's do that math. So the negative log of 5.6 times 10 to the negative 10. Is going to give us a pKa value of 9.25 when we round. So pKa is equal to 9.25. So we're gonna plug that into our Henderson-Hasselbalch equation right here. So the pH of our buffer solution is equal to 9.25 plus the log of the concentration of A minus, our base. Our base is ammonia, NH three, and our concentration in our buffer solution is .24 molars. We're gonna write .24 here. And that's over the concentration of our acid, that's NH four plus, and our concentration is .20. So let's find the log, the log of .24 divided by .20. And so that is .080. So 9.25 plus .08 is 9.33. So the final pH, or the pH of our buffer solution, I should say, is equal to 9.33. So remember this number for the pH, because we're going to compare what happens to the pH when you add some acid and when you add some base. And so our next problem is adding base to our buffer solution. And we're gonna see what that does to the pH. So now we've added .005 moles of a strong base to our buffer solution. Let's say the total volume is .50 liters. So what is the resulting pH?" + }, + { + "Q": "\nThis may seem trivial, but at 3:40, why is the hydroxide ion written with the charge on the left-hand side, instead of the right? I've seen it that way consistently in these chemistry videos, but never anywhere else.", + "A": "It is preferable to put the charge on the atom that has the charge, so we should write \u00e2\u0081\u00bbOH or HO\u00e2\u0081\u00bb. However, many people still write the formula as OH\u00e2\u0081\u00bb. It s OK, as long as you remember that the O atom has the charge.", + "video_name": "lsHq5aqz4uQ", + "timestamps": [ + 220 + ], + "3min_transcript": "So let's find the log, the log of .24 divided by .20. And so that is .080. So 9.25 plus .08 is 9.33. So the final pH, or the pH of our buffer solution, I should say, is equal to 9.33. So remember this number for the pH, because we're going to compare what happens to the pH when you add some acid and when you add some base. And so our next problem is adding base to our buffer solution. And we're gonna see what that does to the pH. So now we've added .005 moles of a strong base to our buffer solution. Let's say the total volume is .50 liters. So what is the resulting pH? moles of sodium hydroxide, and our total volume is .50. So if we divide moles by liters, that will give us the concentration of sodium hydroxide. .005 divided by .50 is 0.01 molar. So that's our concentration of sodium hydroxide. And since sodium hydroxide is a strong base, that's also our concentration of hydroxide ions in solution. So this is our concentration of hydroxide ions, .01 molar. So we're adding a base and think about what that's going to react with in our buffer solution. So our buffer solution has NH three and NH four plus. The base is going to react with the acids. So hydroxide is going to react with NH four plus. Let's go ahead and write out the buffer reaction here. ammonium is going to react with hydroxide and this is going to go to completion here. So if NH four plus donates a proton to OH minus, OH minus turns into H 2 O. So we're gonna make water here. And if NH four plus donates a proton, we're left with NH three, so ammonia. Alright, let's think about our concentrations. So we just calculated that we have now .01 molar concentration of sodium hydroxide. For ammonium, that would be .20 molars. So 0.20 molar for our concentration. And for ammonia it was .24. So let's go ahead and write 0.24 over here. So if .01, if we have a concentration of hydroxide ions of .01 molar, all of that is going to react with the ammonium. So we're gonna lose all of this concentration here for hydroxide." + }, + { + "Q": "\nAt 1:00 how do you know if the Ka value is 5.6 x10^10 ?", + "A": "So in this problem, it is just given to you. Even though it doesn t appear to be written as part of the original question, during exams you would be given a Kb value or a Ka value to work with. Also, ammonia s Kb value is 1.8 x 10^-5 (memorized from constant use). We know that (Ka x Kb = Kw) so all we have to do is take 1.0 x 10^-14 / 1.8 x 10^-5 (Kw / Kb) which gives us 5.6 x 10^-10 (Ka). Hope that helped!", + "video_name": "lsHq5aqz4uQ", + "timestamps": [ + 60 + ], + "3min_transcript": "- [Voiceover] Let's do some buffer solution calculations using the Henderson-Hasselbalch equation. So in the last video I showed you how to derive the Henderson-Hasselbalch equation, and it is pH is equal to the pKa plus the log of the concentration of A minus over the concentration of HA. So we're talking about a conjugate acid-base pair here. HA and A minus. And for our problem HA, the acid, would be NH four plus and the base, A minus, would be NH three or ammonia. So the first thing we need to do, if we're gonna calculate the pH of our buffer solution, is to find the pKa, all right, and our acid is NH four plus. So let's say we already know the Ka value for NH four plus and that's 5.6 times 10 to the negative 10. To find the pKa, all we have to do is take the negative log of that. of 5.6 times 10 to the negative 10. So let's get out the calculator and let's do that math. So the negative log of 5.6 times 10 to the negative 10. Is going to give us a pKa value of 9.25 when we round. So pKa is equal to 9.25. So we're gonna plug that into our Henderson-Hasselbalch equation right here. So the pH of our buffer solution is equal to 9.25 plus the log of the concentration of A minus, our base. Our base is ammonia, NH three, and our concentration in our buffer solution is .24 molars. We're gonna write .24 here. And that's over the concentration of our acid, that's NH four plus, and our concentration is .20. So let's find the log, the log of .24 divided by .20. And so that is .080. So 9.25 plus .08 is 9.33. So the final pH, or the pH of our buffer solution, I should say, is equal to 9.33. So remember this number for the pH, because we're going to compare what happens to the pH when you add some acid and when you add some base. And so our next problem is adding base to our buffer solution. And we're gonna see what that does to the pH. So now we've added .005 moles of a strong base to our buffer solution. Let's say the total volume is .50 liters. So what is the resulting pH?" + }, + { + "Q": "at 0:01 the picture are real or not\n", + "A": "Yes, those pictures are real", + "video_name": "N6IAzlugWw0", + "timestamps": [ + 1 + ], + "3min_transcript": "Let's explore the scientific method. Which at first might seem a bit intimidating, but when we walk through it, you'll see that it's actually almost a common-sense way of looking at the world and making progress in our understanding of the world and feeling good about that progress of our understanding of the world. So, let's just use a tangible example here, and we'll walk through what we could consider the steps of the scientific method, and you'll see different steps articulated in different ways, but they all boil down to the same thing. You observe something about reality, and you say, well, let me try to come up with a reason for why that observation happens, and then you try to test that explanation. It's very important that you come up with explanations that you can test, and then you can see if they're true, and then based on whether they're true, you keep iterating. If it's not true, you come up with another explanation. If it is true, but it doesn't explain everything, well once again, you try to explain more of it. in I don't know, northern Canada or something, and let's say that you live near the beach, but there's also a pond near your house, and you notice that the pond, it tends to freeze over sooner in the Winter than the ocean does. It does that faster and even does it at higher temperatures than when the ocean seems to freeze over. So, you could view that as your observation. So, the first step is you're making an observation. Observation. In our particular case is that the pond freezes over at higher temperatures than the ocean does, and it freezes over sooner in the Winter. Well, the next question that you might wanna, or the next step you could view as a scientific method. It doesn't have to be this regimented, but this is a structured way of thinking about it. Well, ask yourself a question. Ask a question. Why does, so in this particular question, or in this particular scenario, why does the pond tend than the ocean does? Well, you then try to answer that question, and this is a key part of the scientific method, is what you do in this third step, is that you try to create an explanation, but what's key is that it is a testable explanation. So, you try to create a testable explanation. Testable explanation, and this is kind of the core, one of the core pillars of the scientific method, and this testable explanation is called your hypotypothesis. Your hypothesis. And so, in this particular case, a testable explanation could be that, well the ocean is made up of salt water, and this pond is fresh water, so your testable explanation could be salt water, salt water has lower freezing point." + }, + { + "Q": "at about 9:00, you show the chlorine in AlCl4 giving its' electron to the hydrogen, then the hydrogen giving it's electron back to the benzene ring so it can form a double bond and become aromatic again. You don't mention what happens to that chlorine atom and hydrogen after? do they then form HCl? thank you!\n", + "A": "I just watched the next video, Friedel Crafts Acylation Addendum, that answers my question!", + "video_name": "vFfriC55fFw", + "timestamps": [ + 540 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:40, what is the general formula for an exponential decay and wat does each letter in the formula represent?\n", + "A": "The formula for an exponential decay is (1-r)^t. T is time, r is the rate or percent,", + "video_name": "Hqzakjo_dYg", + "timestamps": [ + 280 + ], + "3min_transcript": "It makes the calculator math a little bit easier. The same thing. So if I do 2 natural log, divided by 0.05, it is equal to 13.86. So when t is equal to 13.86. And I'm assuming that we're dealing with time in years. That tends to be the convention, although sometimes it could be something else and you'd always have to convert to years. But assuming that this original formula, where they gave this k value 0.05, that was with the assumption that t is in years, and I've just solved its half-life. I just solved that after 13.86 years, you can expect to have 1/2 of the substance left. We started with 100, we ended up with 50. I could have started with x and ended up with x over 2. Let's do one more of these problems, just so that we're Let's say that I have something with a half-life of, I don't know, let's say I have it as one month. Half-life of one month. And after, well let's say that I-- well let me just for the sake of time, let me make it a little bit simpler. Let's say I just have my k value is equal to-- I mean you can go from half-life to a k value, we did that in the Let's say my k value is equal to 0.001. So my general formula is the amount of product I have, is equal to the amount that I started with times e to the minus 0.001 times t. And I gave you this, if you have to figure it out from half-life, I did that in the previous video with carbon-14. But let's say this is the formula. And let's say that after, I don't know, let's say after The decay formula for whatever element is described by this formula. How much did I start off with? So essentially I need to figure out N sub 0, right? I'm saying that after 1000 years, so N of 1000, which is equal to N sub naught times e to the minus 0.001, times 1000. Right? That's the N of 1000. And I'm saying that that's equal to 500 grams. That equals 500 grams. So I just have to solve for N sub naught. So what's the e value? So if I have 0.0001 times 1000, so this is N sub naught." + }, + { + "Q": "\n6:26 What is \"e\" for? why the alphabet becomes 2.71?", + "A": "Sal has done a few videos on where e comes from. Look for the Introduction to compound interest and e video in the Precalculus playlist.", + "video_name": "Hqzakjo_dYg", + "timestamps": [ + 386 + ], + "3min_transcript": "Let's say that I have something with a half-life of, I don't know, let's say I have it as one month. Half-life of one month. And after, well let's say that I-- well let me just for the sake of time, let me make it a little bit simpler. Let's say I just have my k value is equal to-- I mean you can go from half-life to a k value, we did that in the Let's say my k value is equal to 0.001. So my general formula is the amount of product I have, is equal to the amount that I started with times e to the minus 0.001 times t. And I gave you this, if you have to figure it out from half-life, I did that in the previous video with carbon-14. But let's say this is the formula. And let's say that after, I don't know, let's say after The decay formula for whatever element is described by this formula. How much did I start off with? So essentially I need to figure out N sub 0, right? I'm saying that after 1000 years, so N of 1000, which is equal to N sub naught times e to the minus 0.001, times 1000. Right? That's the N of 1000. And I'm saying that that's equal to 500 grams. That equals 500 grams. So I just have to solve for N sub naught. So what's the e value? So if I have 0.0001 times 1000, so this is N sub naught. equal to 500 grams. Or I could multiply both sides by e, and I have N sub naught is equal to 500e, which is about 2.71. So it's 500 times 2.71. I don't actually have e on this calculator or at least I don't see it. So we'll have 1,355 grams. So it's equal to 1,355 grams, or 1.355 kilograms. That's what I started with. So hopefully you see now. I mean, I think we've approached this pretty much at almost any direction that a chemistry test or teacher could throw the problem at you. But you really just need to remember this formula. And this applies to a lot of things. Later you'll learn, you know, when you do compound interest in finance, the k will just be a positive value, but it's essentially the same formula." + }, + { + "Q": "6:26 What is \"e\" for? why the alphabet becomes 2.71?\n", + "A": "e is an important mathematical constant that is roughly equal to 2.71.", + "video_name": "Hqzakjo_dYg", + "timestamps": [ + 386 + ], + "3min_transcript": "Let's say that I have something with a half-life of, I don't know, let's say I have it as one month. Half-life of one month. And after, well let's say that I-- well let me just for the sake of time, let me make it a little bit simpler. Let's say I just have my k value is equal to-- I mean you can go from half-life to a k value, we did that in the Let's say my k value is equal to 0.001. So my general formula is the amount of product I have, is equal to the amount that I started with times e to the minus 0.001 times t. And I gave you this, if you have to figure it out from half-life, I did that in the previous video with carbon-14. But let's say this is the formula. And let's say that after, I don't know, let's say after The decay formula for whatever element is described by this formula. How much did I start off with? So essentially I need to figure out N sub 0, right? I'm saying that after 1000 years, so N of 1000, which is equal to N sub naught times e to the minus 0.001, times 1000. Right? That's the N of 1000. And I'm saying that that's equal to 500 grams. That equals 500 grams. So I just have to solve for N sub naught. So what's the e value? So if I have 0.0001 times 1000, so this is N sub naught. equal to 500 grams. Or I could multiply both sides by e, and I have N sub naught is equal to 500e, which is about 2.71. So it's 500 times 2.71. I don't actually have e on this calculator or at least I don't see it. So we'll have 1,355 grams. So it's equal to 1,355 grams, or 1.355 kilograms. That's what I started with. So hopefully you see now. I mean, I think we've approached this pretty much at almost any direction that a chemistry test or teacher could throw the problem at you. But you really just need to remember this formula. And this applies to a lot of things. Later you'll learn, you know, when you do compound interest in finance, the k will just be a positive value, but it's essentially the same formula." + }, + { + "Q": "At 1:11 Sal says light travels fastest in a vacuum, I thought the speed of light was constant.\n", + "A": "light slows down more in more densely packed things since the photon crashes into more particle in a time period. however, the speed of light in a vacuum is constant.", + "video_name": "y55tzg_jW9I", + "timestamps": [ + 71 + ], + "3min_transcript": "In the last couple of videos we talked about reflection. And that's just the idea of the light rays bouncing off of a surface. And if the surface is smooth, the incident angle is going to be the same thing as the reflected angle. We saw that before, and those angles are measured relative to a perpendicular. So that angle right there is going to be the same as that angle right there. That's essentially what we learned the last couple of videos. What we want to cover in this video is when the light actually doesn't just bounce off of a surface but starts going through a different medium. So in this situation, we will be dealing with refraction. Refraction. Refraction, you still have the light coming in to the interface between the two surfaces. So let's say--so that's the perpendicular right there, actually let me continue the perpendicular all the way down like that. And let's say we have the incident light ray coming in at some, at some angle theta 1, just like that...what will happen--and so let's say that this up here, this is a vacuum. In a vacuum. There's nothing there, no air, no water, no nothing, that's where the light travels the fastest. And let's say that this medium down here, I don't know, let's say it's water. Let's say that this is water. All of this. This was all water over here. This was all vacuum right up here. So what will happen, and actually, that's kind of an unrealistic-- well, just for the sake of argument, let's say we have water going right up against a vacuum. This isn't something you would normally just see in nature but let's just think about it a little bit. Normally, the water, since there's no pressure, it would evaporate and all the rest. But for the sake of argument, let's just say that this is a medium where light will travel slower. What you're going to have is is this ray is actually going to switch direction, it's actually going to bend. Instead of continuing to go in that same direction, it's going to bend a little bit. It's going to go down, in that direction is the refraction. That's the refraction angle. Refraction angle. Or angle of refraction. This is the incident angle, or angle of incidence, and this is the refraction angle. Once again, against that perpendicular. And before I give you the actual equation of how these two things relate and how they're related to the speed of light in these two media-- and just remember, once again, you're never going to have vacuum against water, the water would evaporate because there's no pressure on it and all of that type of thing. But just to--before I go into the math of actually how to figure out these angles relative to the velocities of light in the different media I want to give you an intuitive understanding of not why it bends, 'cause I'm not telling you actually how light works this is really more of an observed property and light, as we'll learn, as we do more and more videos about it, can get pretty confusing. Sometimes you want to treat it as a ray, sometimes you want to treat it as a wave, sometimes you want to treat it as a photon." + }, + { + "Q": "At 8:19, what does axial and equatorial mean?\n", + "A": "It refers to which way the groups of a cycloalkane are pointing in the chair conformation. Axial = up or down Equatorial = sideways", + "video_name": "FGq9-R6Yw18", + "timestamps": [ + 499 + ], + "3min_transcript": "I've shown the sigma bonds here rotated a little bit differently. So all these sigma bonds in here have free rotation. So they're going to rotate to allow our necleophile to attack our electrophile a little bit easier. So I can think about lone pairs of electrons on our oxygen, and now it's a little bit easier to see the nucleophilic attack. It's going to attack right here at this carbon, and push these electrons in here off onto this oxygen. So let's go ahead and show the result of that. So now I have my oxygen, so it is bonded to this carbon now, and I've formed my ring. So let's follow those electrons, these electrons right here in magenta now form this bond right here. And I'm going to say that this oxygen right here in red is actually going to go up in the plane, so we'll come back to that in a minute. It had two lone pairs of electrons around it. It picked up another one, so that's where our negative one formal charge is. The oxygen in our ring still has the hydrogen bonded to it, and a lone pair of electrons that gets a plus one formal charge like that. And then I'm saying this hydrogen here is down, so this is our intermediate. And without showing all of the acid-base steps here, let's just think about what happens. We know that we next deprotonate, so a base comes along and takes this proton, and then these electrons in here move off onto our oxygen. And then we know after that we protonate our negative charge right here. So let's go ahead and draw one of the possible products. So we have our oxygen, that's part of our ring here, drawn in the chair conformation. We now have an OH right here, and then, so we have an OH equitorial as one of our possible products. And then two lone pairs of electrons on this oxygen. Hopefully you can see how this is one of the possible products and that here we drew it with like a flat plane, and here we've drawn it in more of a chair conformation. So the OH equatorial for this one. Let's go back to our original situation over here on the left, where the nucleophile attacked the carbonyl carbon. We know that the geometry at our carbonyl carbon is trigonal planar, so it's possible the nucelophile could attack from the opposite side and if that happened, then the oxygen would go down relative to the plane. So let's go ahead and show that now. We have another possibility here. So let's go ahead and draw it out. So we have our oxygen is now going to be part of our ring, so we've formed our ring. This time instead of putting" + }, + { + "Q": "\nat 3:08 Sal says that \"w\" (omega) is the angular velocity but souldn't it be the angular frequency", + "A": "Omega is in rad/s. I do believe that angular frequency would be in rev/s. They are essentially the same thing just their units are different.", + "video_name": "xoUppFlif04", + "timestamps": [ + 188 + ], + "3min_transcript": "" + }, + { + "Q": "\nAround 5:30 the Hank talks about the centrosomes moving away from the nucleas and leaving behind a trail of microtubles. But the animation shows the tubules growing from the centrosomes towards the chromosomes. I am confused on how the microtubles form. Anyone got a good tubular breakdown for meh?", + "A": "To my understanding, the centrosomes are building the mitotic spindle/microtubules in the prophase, whilst they are moving.", + "video_name": "X1bmedVziGw", + "timestamps": [ + 330 + ], + "3min_transcript": "And part of what's really amazing about this whole process is that while we know what these stages are, we don't always know the underlying mechanisms that make all of them happen and this is part of science. Science isn't like all the stuff we know. It's how we're trying to figure all this stuff out. Consider job security if you ever want to be a biologist. There is a lot of stuff that future biologists have to still figure out and this is one of them. All right, let's get our clone on. So, most of their lives, cells hang out in this limbo period called interphase, which means they're in between episodes of mitosis, mostly growing and working and doing all the stuff that makes them useful to us. During interphase, the long strings of DNA are loosely coiled and messy, like that dust bunny of dog fur and laundry lint under your bed. That mess of DNA is called chromatin, but as the mitosis process begins to gear up lots of things start happening in the cell to get ready for the big division. One of the more important things that happens is that this little set of protein cylinders next to the nucleus, called the centrosome, duplicates itself. We're going to have to move a lot of stuff around by these centrosomes. The other thing that happens is that all of the DNA begins to replicate itself too, giving the cell two copies of every strand of DNA. To brush up on how DNA replicates itself like this check out this episode and then come on back. Now the cell enters the first phase, or the prophase, when that mess of chromatin condenses and coils up on itself to produce thick strands of DNA wrapped around proteins. Those, my friends, are your chromosomes. Instead of dust bunnies, the DNA is starting to look a little bit more like dread locks and the duplicates that have been made don't just float around freely. They stay attached to the original and together they look like little x's. These are called the chromatids and one copy is the left leg and arm of the x and the other copy is the right leg and arm. Where they meet in the middle is called the centromere. Just so you know, these x's are also called chromosomes. Sometimes double chromosomes or double stranded chromosomes and when the chromatids separate, they're considered individual chromosomes too. Now, while the chromosomes are forming by like, completely disintegrating and the centrosomes then peel away from the nucleus, start heading to the opposite ends of the cell. As they go, they leave behind a wide trail of protein ropes called microtubules running from one centrosome to the other. You might recall from our anatomy of the animal cell the microtubules help provide a kind of structure to the cell and this is exactly what they're doing here. Now we reach the metaphase, which literally means after phase and it's the longest phase of mitosis. It could take up to 20 minutes. During the metaphase, the chromosomes attach to those ropey microtubules right in the middle at their centromeres. The chromosomes then begin to be moved around and this seems to be being done by molecules called motor proteins. And while we don't know too much about how these motors work, we do know, for instance that there are two of them on each side of the centromere. These are called centromere associated protein E. So these motor proteins attached to the microtubule ropes basically serve to spool up the tubule slack. Now at the same time, another protein called dynein" + }, + { + "Q": "At 4:47 , is Hank saying that a double chromosome is still considered one chromosome? His use of the plural is a bit confusing.\n", + "A": "A chromosome that has reproduced and is still attached at the centromere is still considered one chromosome.", + "video_name": "X1bmedVziGw", + "timestamps": [ + 287 + ], + "3min_transcript": "And part of what's really amazing about this whole process is that while we know what these stages are, we don't always know the underlying mechanisms that make all of them happen and this is part of science. Science isn't like all the stuff we know. It's how we're trying to figure all this stuff out. Consider job security if you ever want to be a biologist. There is a lot of stuff that future biologists have to still figure out and this is one of them. All right, let's get our clone on. So, most of their lives, cells hang out in this limbo period called interphase, which means they're in between episodes of mitosis, mostly growing and working and doing all the stuff that makes them useful to us. During interphase, the long strings of DNA are loosely coiled and messy, like that dust bunny of dog fur and laundry lint under your bed. That mess of DNA is called chromatin, but as the mitosis process begins to gear up lots of things start happening in the cell to get ready for the big division. One of the more important things that happens is that this little set of protein cylinders next to the nucleus, called the centrosome, duplicates itself. We're going to have to move a lot of stuff around by these centrosomes. The other thing that happens is that all of the DNA begins to replicate itself too, giving the cell two copies of every strand of DNA. To brush up on how DNA replicates itself like this check out this episode and then come on back. Now the cell enters the first phase, or the prophase, when that mess of chromatin condenses and coils up on itself to produce thick strands of DNA wrapped around proteins. Those, my friends, are your chromosomes. Instead of dust bunnies, the DNA is starting to look a little bit more like dread locks and the duplicates that have been made don't just float around freely. They stay attached to the original and together they look like little x's. These are called the chromatids and one copy is the left leg and arm of the x and the other copy is the right leg and arm. Where they meet in the middle is called the centromere. Just so you know, these x's are also called chromosomes. Sometimes double chromosomes or double stranded chromosomes and when the chromatids separate, they're considered individual chromosomes too. Now, while the chromosomes are forming by like, completely disintegrating and the centrosomes then peel away from the nucleus, start heading to the opposite ends of the cell. As they go, they leave behind a wide trail of protein ropes called microtubules running from one centrosome to the other. You might recall from our anatomy of the animal cell the microtubules help provide a kind of structure to the cell and this is exactly what they're doing here. Now we reach the metaphase, which literally means after phase and it's the longest phase of mitosis. It could take up to 20 minutes. During the metaphase, the chromosomes attach to those ropey microtubules right in the middle at their centromeres. The chromosomes then begin to be moved around and this seems to be being done by molecules called motor proteins. And while we don't know too much about how these motors work, we do know, for instance that there are two of them on each side of the centromere. These are called centromere associated protein E. So these motor proteins attached to the microtubule ropes basically serve to spool up the tubule slack. Now at the same time, another protein called dynein" + }, + { + "Q": "At 8:23, why use the volume of room rather than the volume of the container? Shouldn't the ideal gas law produced in the video be designated exclusively to the container of the water since the water is exclusively within the two liters, rather than the 42500 liters of room?\n", + "A": "The 2.00 L is the volume of the liquid water in the open container. When the water evaporates, the water vapour expands to fill the whole room, so that\u00e2\u0080\u0099s the volume you use.", + "video_name": "-QpkmwIoMaY", + "timestamps": [ + 503 + ], + "3min_transcript": "Now, the hardest thing about this is just making sure you have your units right and you're using the right ideal gas constant for the right units, and we'll do that right here. So what I want to do, because the universal gas constant that I have is in terms of atmospheres, we need to figure out this vapor pressuree- this equilibrium pressure between vapor and liquid-- we need to write this down in terms of atmospheres. So let me write this down. So the vapor pressure is equal to 23.8 millimeters of mercury. And you can look it up at a table if you don't have this One atmosphere is equivalent to 760 millimeters of mercury. atmospheres-- let me get my calculator out, get the calculator out, put it right over there-- so it's going to be 23.8 times 1 over 760, or just divided by 760. And we have three significant digits, so it looks like 0.0313. So this is equal to 0.0313 atmospheres. That is our vapor pressure. So let's just deal with this right here. So the number of molecules of water that are going to be in the air in the gaseous state, in the vapor state, is going to be equal to our vapor pressure. That's our equilibrium pressure. If more water molecules evaporate after that point, then we're going to have a higher pressure, which will state, so we'll go kind of past the equilibrium, which is not likely. Or another way to think about it-- more water molecules are not going to of evaporate at a faster rate than they are going to condense beyond that pressure. Anyway, the pressure here is 0.0313 atmospheres. The volume here-- they told us right over here-- so that's the volume-- 4.25. 4.25 times 10 to the fourth liters. And then we want to divide that by-- and you want to make sure that your universal gas constant has the right units, I just looked mine up on Wikipedia-- 0.08-- see everything has three significant digits. So let me just allow that more significant digits and we'll just round at the end. 0.082057, and the units here are liters atmospheres per" + }, + { + "Q": "\nAt 1:23, WHAT keeps water in its liquid state ?", + "A": "The attractive forces among water molecules keep water in its liquid state.", + "video_name": "-QpkmwIoMaY", + "timestamps": [ + 83 + ], + "3min_transcript": "This exercises is from chapter 12 of the Kotz, Treichel in Townsend and Chemistry and Chemical Reactivity book, and I'm doing it with their permission. So they tell us you place 2 liters of water in an open container in your dormitory room. The room has a volume of 4.25 times 10 to the fourth liters. You seal the room and wait for the water to evaporate. Will all of the water evaporate at 25 degrees Celsius? And then they tell us at 25 degrees Celsius, the density of water is 0.997 grams per milliliter. And its vapor pressure is 23.8 millimeters of mercury. And this is actually the key clue to tell you how to solve And just as a bit of review, lets just think about what vapor pressure is. Let's say it's some temperature, and in this case we're dealing at 25 degrees Celsius. I have a bunch of water, and let me do that in a water color. I have a bunch of water molecules At 25 degrees Celsius, they're all bouncing around in every which way. And every now and then one of them is going to have enough kinetic energy to kind of escape the hydrogen bonds and all the things that keep liquid water in its liquid state, and it will escape. It'll go off in that direction, and then another one will. And this'll just keep happening. The water will naturally vaporize in a room. But at some point, enough of these molecules have vaporized over here that they're also bumping back into the water. And maybe some of them can be captured back into the liquid state. Now, the pressure at which this happens is the vapor pressure. As you can imagine, as more and more these water molecules vaporize and go into the gaseous state, more and more will also create pressure, downward pressure. surface of the water. And the pressure at which the liquid and the vapor states are in equilibrium is the vapor pressure. And they're telling us right now. It is 23.8 millimeters of mercury. Now, what we need to do to figure out this problem is say, OK, if we could figure out how many molecules need to evaporate, how many molecules of water need to evaporate to give us this vapor pressure, we can then use the density of water to figure out how many liters of water that is. So how do we figure out how many molecules-- let me write this down-- how many molecules of water need to evaporate to give us the vapor pressure of 23.8 millimeters of mercury?" + }, + { + "Q": "\nAt 10:50, the video notes that the two chair forms of cyclohexane are in equilibrium. So, does a single molecule spontaneously flip back and forth between the two forms? Or, must a single molecule, say, collide with another molecule for the flip to occur?", + "A": "When cyclohexane is synthesized, it is in the two forms, but remember that equilibrium s can lean towards one side. The chair formation is a little bit more stable so the atoms rearrange (this is kind of your question, not much is known about how they do this, but it would help to understand hybridization) to be different than the boat conformation. Also some of the chair configuration can go back into the boat configuration.", + "video_name": "YUEkOBvJSNg", + "timestamps": [ + 650 + ], + "3min_transcript": "chair configurations are equally stable. And let me just touch on that a second. So you have, well, I don't have to-- actually, let me see. I won't copy and paste. I'll just redraw the other chair configuration for this guy. Actually let me just do it separately over here because I've made the colors here so confusing. Let me draw two, the same cyclohexane, but in two different chair configurations that it could be equilibrium in. So you could have this one, you could have this one, so this could be one chair configuration, and I'll draw it like this. And then the same hydrocarbon could be in-- or the same cyclohexane could be in equilibrium with the this other chair configuration that looks like this. Let me have a little more space here. So it looks like this. Let me do the pink. It goes up like that, like that. This pink guy goes like this. And then the blue guy is going to be just like this. So notice, in this situation this carbon appears kind of at the top of the chair, and this carbon is at the bottom, and then they've flipped. But these are equally stable configurations. But one way to think about is all of the axial guys on this carbon here turned into equatorial on this carbon and Let me show it to you. Let me just draw the hydrogens on this carbon. This carbon's hydrogens has an axial hydrogen, and has an equatorial hydrogen, whose bond would be parallel to that just like that. And this guy would have an equatorial hydrogen whose bond is parallel to actually both of these guys. And an axial hydrogen. hydrogens, but when this structure flips like that, what happens? Well, this hydrogen over here goes into this position, and this yellow hydrogen over here goes into this position. So over here, it was equatorial, and now it becomes axial. The same argument can be made over here. This equatorial hydrogen, when it flips-- when this whole blue part flips down-- now becomes axial. And this axial hydrogen, when you flip it down, becomes equatorial. And you can actually do that for all of the hydrogens. Over here you have an axial hydrogen. Once you flip it, you have an axial hydrogen, and then you have an equatorial hydrogen. When you flip it, these two equatorial hydrogens become axial." + }, + { + "Q": "Around 4:00, where is the boundary between non-polar and polar?\n", + "A": "The boundary is approximately a difference of 1.7 between elements, but there are exceptions.", + "video_name": "126N4hox9YA", + "timestamps": [ + 240 + ], + "3min_transcript": "Carbon is losing a little bit of negative charge. So carbon used to be neutral, but since it's losing a little bit of negative charge, this carbon will end up being partially positive, like that. So the carbon is partially positive. And the oxygen is partially negative. That's a polarized situation. You have a little bit of negative charge on one side, a little bit of positive charge on the other side. So let's say it's still a covalent bond, but it's a polarized covalent bond due to the differences in electronegativities between those two atoms. Let's do a few more examples here where we show the differences in electronegativity. So if I were thinking about a molecule that has two carbons in it, and I'm thinking about what happens to the electrons in red. Well, for this example, each carbon has the same value for electronegativity. So the carbon on the left has a value of 2.5. The carbon on the right has a value of 2.5. That's a difference in electronegativity of zero. aren't going to move towards one carbon or towards the other carbon. They're going to stay in the middle. They're going to be shared between those two atoms. So this is a covalent bond, and there's no polarity situation created here since there's no difference in electronegativity. So we call this a non-polar covalent bond. This is a non-polar covalent bond, like that. Let's do another example. Let's compare carbon to hydrogen. So if I had a molecule and I have a bond between carbon and hydrogen, and I want to know what happens to the electrons in red between the carbon and hydrogen. We've seen that. Carbon has an electronegativity value of 2.5. And we go up here to hydrogen, which has a value of 2.1. So that's a difference of 0.4. So there is the difference in electronegativity between those two atoms, but it's a very small difference. And so most textbooks would consider the bond between carbon and hydrogen Let's go ahead and put in the example we did above, where we compared the electronegativities of carbon and oxygen, like that. When we looked up the values, we saw that carbon had an electronegativity value of 2.5 and oxygen had a value of 3.5, for difference of 1. And that's enough to have a polar covalent bond. Right? This is a polar covalent bond between the carbon and the oxygen. So when we think about the electrons in red, electrons in red are pulled closer to the oxygen, giving the oxygen a partial negative charge. And since electron density is moving away from the carbon, the carbon gets a partial positive charge. And so we can see that if your difference in electronegativity is 1, it's considered to be a polar covalent bond. And if your difference in electronegativity is 0.4, that's considered to be a non-polar covalent bond. So somewhere in between there must" + }, + { + "Q": "\n09:55, it is said that if the electronegativity is greater than 1.7, the compound is usually considered ionic. In the previous video (\"Electronegativity\"), it is said that the compound is considered ionic if the electronegativity is greater than 2.0.\nWhich one is more reliabe? How can I tell in an exam if a compound is ionic or covalent or nonpolar?", + "A": "Different books use different cutoffs, though 1.7 is the most common, as Ryan mentioned. Ask you teacher which cutoff they want you to use for an exam.", + "video_name": "126N4hox9YA", + "timestamps": [ + 595 + ], + "3min_transcript": "And we'll put in our electrons. And we know that this bond consists of two electrons, like that. Let's look at the differences in electronegativity between sodium and chlorine. All right. So I'm going to go back up here. I'm going to find sodium, which has a value of 0.9, and chlorine which has a value of 3. So 0.9 for sodium and 3 for chlorine. So sodium's value is 0.9. Chlorine's is 3. That's a large difference in electronegativity. That's a difference of 2.1. And so chlorine is much more electronegative than sodium. And it turns out, it's so much more electronegative that it's no longer going to share electrons with sodium. It's going to steal those electrons. So when I redraw it here, I'm going to show chlorine being surrounded by eight electrons. So these two electrons in red-- let me go ahead and show them-- these two electrons in red here between the sodium and the chlorine, since chlorine is so much more in red so strongly that it completely steals them. So those two electrons in red are going to be stolen by the chlorine, like that. And so the sodium is left over here. And so chlorine has an extra electron, which gives it a negative 1 formal charge. So we're no longer talking about partial charges here. Chlorine gets a full negative 1 formal charge. Sodium lost an electron, so it ends up with a positive formal charge, like that. And so we know this is an ionic bond between these two ions. So this represents an ionic bond. So the difference in electronegativity is somewhere between 1.5 and 2.1, between a polar covalent bond and an ionic bond. So most textbooks we'll see approximately somewhere around 1.7. So if you're higher than 1.7, it's generally considered to be mostly an ionic bond. Lower than 1.7, in the polar covalent range. So we'll come back now to the example between carbon and lithium. So if we go back up here to carbon and lithium, here we treat it like a polar covalent bond. But sometimes you might want to treat the bond in red as being an ionic bond. So let's go ahead and draw a picture of carbon and lithium where we're treating it as an ionic bond. So if carbon is more electronegative than lithium, carbon's going to steal the two electrons in red. So I'll go ahead and show the electrons in red have now moved on to the carbon atom. So it's no longer sharing it with the lithium. Carbon has stolen those electrons. And lithium is over here. So lithium lost one of its electrons, giving it a plus 1 formal charge. Carbon gained an electron, giving it a negative 1 formal charge. And so here, we're treating it like an ionic bond. Full formal charges here. And this is useful for some organic chemistry reactions. And so what I'm trying to point out here is these divisions, 1.7, it's not absolute." + }, + { + "Q": "\nAt about 4:30, in discussing microstates, Sal says that we can know the position and the momentum. I thought that, because of the uncertainty principle, we can never know both the position and the momentum of any given particle. Is he just simplifying things so that we understand the difference between microstates and macrostates, or am I missing something?\nThanks.", + "A": "According to the Heisenberg Uncertainty Principle, you cant know both the location and momentum of a particle so you are correct. I believe it is a simplification.", + "video_name": "5EU-y1VF7g4", + "timestamps": [ + 270 + ], + "3min_transcript": "Now we know that that pressure is due to things like, you have a bunch of atoms bumping around. And let's say that this is a gas-- it's a balloon- it's going to be a gas. And we know that the pressure is actually caused-- and I've done several, I think I did the same video in both the chemistry and the physics playlist. I did them a year apart, so you can see if my thinking has evolved at all. But we know that the pressure's really due by the bumps of these particles as they bump into the walls and the side of the balloon. And we have so many particles at any given point of time, some of them are bumping into the wall the balloon, and that's what's essentially keeping the balloon pushed outward, giving it its pressure and its volume. We've talked about temperature, as essentially the average kinetic energy of these-- which is a function of these particles, which could be either the molecules of gas, or if it's an ideal gas, it could be just the atoms of the gas. Maybe it's atoms of helium or neon, or something like that. So for example, I could describe what's going on with I could say, hey, you know, there are-- I could just make up some numbers. The pressure is five newtons per meters squared, or some The units aren't what's important. In this video I really just want to make the differentiation between these two ways of describing what's going on. I could say the temperature is 300 kelvin. I could say that the volume is, I don't know, maybe it's one liter. And I've described a system, but I've described in on a macro level. Now I could get a lot more precise, especially now that we know that things like atoms and molecules exist. What I could do, is I could essentially label every one of contained in the balloon. And I could say, at exactly this moment in time, I could say at time equals 0, atom 1 has-- its momentum is equal to x, and its position, in three-dimensional coordinates, is x, y, and z. And then I could say, atom number 2-- its momentum-- I'm just using rho for momentum-- it's equal to y. And its position is a, b, c. And I could list every atom in this molecule. Obviously we're dealing with a huge number of atoms, on the order of 10 to the 20 something. So it's a massive list I would have to give you, but I could literally give you the state of every atom in this balloon." + }, + { + "Q": "At 3:47, don't chloroplasts have a double membrane instead of just 1?\n", + "A": "Yes , the chloroplast technically has 2 membranes ; the outer and the inner one", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 227 + ], + "3min_transcript": "a little bit rigid. So let's say that these are plant cells right here. Each of these squares, each of these quadrilaterals is a plant cell. And then in these plant cells you have these organelles called chloroplasts. Remember organelles are like organs of a cell. They are subunits, membrane-bound subunits of cells. And of course, these cells have nucleuses and DNA and all of the other things you normally associate with cells. But I'm not going to draw them here. I'm just going to draw the chloroplasts. And your average plant cell-- and there are other types of living organisms that perform photosynthesis, but we'll focus on plants. Because that's what we tend to associate it with. Each plant cell will contain 10 to 50 chloroplasts. I make them green on purpose because the chloroplasts contain chlorophyll. Which to our eyes, appear green. But remember, they're green because they reflect green wavelengths of light. Because it's reflecting. But it's absorbing all the other wavelengths. But anyway, we'll talk more about that in detail. But you'll have 10 to 50 of these chloroplasts right here. And then let's zoom in on one chloroplast. So if we zoom in on one chloroplast. So let me be very clear. This thing right here is a plant cell. That is a plant cell. And then each of these green things right here is an organelle called the chloroplast. And let's zoom in on the chloroplast itself. If we zoom in on one chloroplast, it has a membrane like that. And then the fluid inside of the chloroplast, inside of its membrane, so this fluid right here. All of this fluid. That's called the stroma. chloroplast itself, you have these little stacks of these folded membranes, These little folded stacks. Let me see if I can do justice here. So maybe that's one, two, doing these stacks. Each of these membrane-bound-- you can almost view them as pancakes-- let me draw a couple more. Maybe we have some over here, just so you-- maybe you have some over here, maybe some over here. So each of these flattish looking pancakes right here, these are called thylakoids. So this right here is a thylakoid. That is a thylakoid. The thylakoid has a membrane. And this membrane is especially important. We're going to zoom in on that in a second. So it has a membrane, I'll color that in a little bit. The inside of the thylakoid, so the space, the fluid inside" + }, + { + "Q": "At 3:00, Sal said that the chloroplasts reflect green light, so does that mean that if \"green\" photons were to hit the chloroplast, photosynthesis wouldn't occur?\n", + "A": "Well, green photons can t be absorbed because they are the wrong wavelength. The energy comes from photons that are absorbed. So the green photons can t provide energy for photosynthesis, and if you shone only green light on the chloroplast it would have no energy to photosynthesize. So, you are correct.", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 180 + ], + "3min_transcript": "It just doesn't need the photons from the sun. But let's focus first on the light-dependent reactions. The part that actually uses photons from the sun. Or actually, I guess, even photons from the heat lamp that you might have in your greenhouse. And uses those photons in conjunction with water to produce ATP and reduce NADP plus to NADPH. Remember, reduction is gaining electrons or hydrogen atoms. And it's the same thing, because when you gain a hydrogen atom, including its electron, since hydrogen is not too electronegative, you get to hog its electron. So this is both gaining a hydrogen and gaining electron. But let's study it a little bit more. So before we dig a little deeper, I think it's good to know a little bit about the anatomy of a plant. So let me draw some plant cells. a little bit rigid. So let's say that these are plant cells right here. Each of these squares, each of these quadrilaterals is a plant cell. And then in these plant cells you have these organelles called chloroplasts. Remember organelles are like organs of a cell. They are subunits, membrane-bound subunits of cells. And of course, these cells have nucleuses and DNA and all of the other things you normally associate with cells. But I'm not going to draw them here. I'm just going to draw the chloroplasts. And your average plant cell-- and there are other types of living organisms that perform photosynthesis, but we'll focus on plants. Because that's what we tend to associate it with. Each plant cell will contain 10 to 50 chloroplasts. I make them green on purpose because the chloroplasts contain chlorophyll. Which to our eyes, appear green. But remember, they're green because they reflect green wavelengths of light. Because it's reflecting. But it's absorbing all the other wavelengths. But anyway, we'll talk more about that in detail. But you'll have 10 to 50 of these chloroplasts right here. And then let's zoom in on one chloroplast. So if we zoom in on one chloroplast. So let me be very clear. This thing right here is a plant cell. That is a plant cell. And then each of these green things right here is an organelle called the chloroplast. And let's zoom in on the chloroplast itself. If we zoom in on one chloroplast, it has a membrane like that. And then the fluid inside of the chloroplast, inside of its membrane, so this fluid right here. All of this fluid. That's called the stroma." + }, + { + "Q": "At 12:00, When H+ goes through the ATP synthase, it \"puts\" ADP and Pi to form ATP?\n", + "A": "Well, assume the synthase is a wind turbine. Wind turbines generate electricity, but they cannot do so without any wind. Now, pretend that the flow of protons (H+) is that wind. Once the wind flows, the turbine is able to spin and generate ATP.", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 720 + ], + "3min_transcript": "the electrons. So you have all of these hydrogen protons. Hydrogen protons get pumped into the lumen. They get pumped into the lumen and so you might remember this from the electron transport chain. In the electron transport chain, as electrons went from a high potential, a high energy state, to a low energy state, that energy was used to pump hydrogens through a membrane. And in that case it was in the mitochondria, here the membrane is the thylakoid membrane. But either case, you're creating this gradient where-- because of the energy from, essentially the photons-- the electrons enter a high energy state, they keep going into a lower energy state. And then they actually go to photosystem I and they get hit by another photon. Well, that's a simplification, but that's how you can think of it. Enter another high energy state, then they go to a lower, lower and lower energy state. But the whole time, that energy from the electrons going from a high energy state to a low energy state is used So you have this huge concentration of hydrogen protons. And just like what we saw in the electron transport chain, that concentration is then-- of hydrogen protons-- is then used to drive ATP synthase. So the exact same-- let me see if I can draw that ATP synthase here. You might remember ATP synthase looks something like this. Where literally, so here you have a huge concentration of hydrogen protons. So they'll want to go back into the stroma from the lumen. And they go through the ATP synthase. Let me do it in a new color. So these hydrogen protons are going to make their way back. Go back down the gradient. And as they go down the gradient, they literally-- it's like an engine. And I go into detail on this when I talk about respiration. And that turns, literally mechanically turns, this top And it puts ADP and phosphate groups together. It puts ADP plus phosphate groups together to produce ATP. So that's the general, very high overview. And I'm going to go into more detail in a second. But this process that I just described is called photophosphorylation. Let me do it in a nice color. Why is it called that? Well, because we're using photons. That's the photo part. We're using light. We're using photons to excite electrons in chlorophyll. As those electrons get passed from one molecule, from one electron acceptor to another, they enter into lower and lower energy states. As they go into lower energy states, that's used to drive," + }, + { + "Q": "At 5:30, he mentions that organelles were independent organisms that at some point evolved together (a process called \"endosymbiosis\"). This is true of mitochondria and chloroplasts, but I believe it is inaccurate to say of all organelles, correct?\n", + "A": "this is possible because they both have DNA", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 330 + ], + "3min_transcript": "wavelengths of light. Because it's reflecting. But it's absorbing all the other wavelengths. But anyway, we'll talk more about that in detail. But you'll have 10 to 50 of these chloroplasts right here. And then let's zoom in on one chloroplast. So if we zoom in on one chloroplast. So let me be very clear. This thing right here is a plant cell. That is a plant cell. And then each of these green things right here is an organelle called the chloroplast. And let's zoom in on the chloroplast itself. If we zoom in on one chloroplast, it has a membrane like that. And then the fluid inside of the chloroplast, inside of its membrane, so this fluid right here. All of this fluid. That's called the stroma. chloroplast itself, you have these little stacks of these folded membranes, These little folded stacks. Let me see if I can do justice here. So maybe that's one, two, doing these stacks. Each of these membrane-bound-- you can almost view them as pancakes-- let me draw a couple more. Maybe we have some over here, just so you-- maybe you have some over here, maybe some over here. So each of these flattish looking pancakes right here, these are called thylakoids. So this right here is a thylakoid. That is a thylakoid. The thylakoid has a membrane. And this membrane is especially important. We're going to zoom in on that in a second. So it has a membrane, I'll color that in a little bit. The inside of the thylakoid, so the space, the fluid inside This light green color right there. That's called the thylakoid space or the thylakoid lumen. And just to get all of our terminology out of the way, a stack of several thylakoids just like that, that is called a grana. That's a stack of thylakoids. That is a grana. And this is an organelle. And evolutionary biologists, they believe that organelles were once independent organisms that then, essentially, teamed up with other organisms and started living inside of their cells. So there's actually, they have their own DNA. So mitochondria is another example of an organelle that people believe that one time mitochondria, or the ancestors of mitochondria, were independent organisms. That then teamed up with other cells and said, hey, if I produce your energy maybe you'll give me some food or whatnot. And so they started evolving together." + }, + { + "Q": "Did he mean hydrogen ions instead of hydrogen \"protons\"? (11:07)\n", + "A": "A hydrogen ion (H+) is a proton. So, the terms are used interchangeably.", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 667 + ], + "3min_transcript": "But the general idea-- I'll tell you the general idea and then we'll go into the specifics-- of what happens during the light reaction, or the light dependent reaction, is you have some photons. Photons from the sun. They've traveled 93 million miles. so you have some photons that go here and they excite electrons in a chlorophyll molecule, in a chlorophyll A molecule. And actually in photosystem II-- well, I won't go into the details just yet-- but they excite a chlorophyll molecule so those electrons enter into a high energy state. Maybe I shouldn't draw it like that. They enter into a high energy state. And then as they go from molecule to molecule they keep going down in energy state. But as they go down in energy state, you have hydrogen the electrons. So you have all of these hydrogen protons. Hydrogen protons get pumped into the lumen. They get pumped into the lumen and so you might remember this from the electron transport chain. In the electron transport chain, as electrons went from a high potential, a high energy state, to a low energy state, that energy was used to pump hydrogens through a membrane. And in that case it was in the mitochondria, here the membrane is the thylakoid membrane. But either case, you're creating this gradient where-- because of the energy from, essentially the photons-- the electrons enter a high energy state, they keep going into a lower energy state. And then they actually go to photosystem I and they get hit by another photon. Well, that's a simplification, but that's how you can think of it. Enter another high energy state, then they go to a lower, lower and lower energy state. But the whole time, that energy from the electrons going from a high energy state to a low energy state is used So you have this huge concentration of hydrogen protons. And just like what we saw in the electron transport chain, that concentration is then-- of hydrogen protons-- is then used to drive ATP synthase. So the exact same-- let me see if I can draw that ATP synthase here. You might remember ATP synthase looks something like this. Where literally, so here you have a huge concentration of hydrogen protons. So they'll want to go back into the stroma from the lumen. And they go through the ATP synthase. Let me do it in a new color. So these hydrogen protons are going to make their way back. Go back down the gradient. And as they go down the gradient, they literally-- it's like an engine. And I go into detail on this when I talk about respiration. And that turns, literally mechanically turns, this top" + }, + { + "Q": "At 1:00 I don't why there is NH3. Is it solvent?\n", + "A": "NH3 is the solvent. If you are talking about -NH2, that is from NaNH2 and is a very strong base.", + "video_name": "HbDWBeRJboE", + "timestamps": [ + 60 + ], + "3min_transcript": "Let's look at two ways to prepare alkynes from alkyl halides. So here I have an alkyl halide. So this is a dihalide, and my two halogens are attached to one carbon. We call this a geminal dihalide. So this is going to be a geminal dihalide reacting with a very strong base, sodium amide. So this is going to give us an E2 elimination reaction. So we're going to get an E2 elimination reaction, and this E2 elimination reaction is actually going to occur twice. And we're going to end up with an alkyne as our final product. So let's take a look at the mechanism of our double E2 elimination of a geminal dihalide. So let's start with our dihalide over here. And this time we're going to put in all of our lone pairs of electrons on our halogen, like that. So let me go ahead and put all of those in there, and then I have two hydrogens on this carbon. Sodium amide is a source of amide anions, which So a strong base means that a lone pair of electrons here on our nitrogen is going to take this proton. And these electrons, in here, are going to kick in to form a double bond at the same time these electrons kick off onto our halogen. So an E2 elimination mechanism. You can watch the previous videos on E2 elimination reactions for more details. So we're going to form ammonia as one of our products. And our other product is going to be carbon double-bonded to another carbon. And then we're going to still have our halogen down here. And over here, in the carbon on the right, we're still going to have a hydrogen, like that. So we're not quite to our alkyne yet. So we've done one E2 elimination reaction, and we're going to do one more. So we get another-- another amide anion comes along, and it's negatively charged. It's going to function as a base. It's going to take this proton this time. in here to form our triple bond. And these electrons are going to kick off onto our halogen, like that. So that is going to finally form our alkyne here. So you always have to have your base in excess, if you're trying to do this. Let's look at a very similar reaction, a double E2 elimination. This time the halogens are not on the same carbon. So let's go ahead and draw the general reaction for this. We have two carbons right here, and we have two halogens right here. And then hydrogen, and then hydrogen. This time we have two halogens on adjacent carbons. So this is called vicinal dihalides. So let's go ahead and write that. So this is vicinal, and the one we did before was geminal. So a vicinal dihalide will react in a very similar way if you add a strong base like sodium amide and you use ammonia for your solvent." + }, + { + "Q": "\nAt about 4:20, you mention that we cannot decipher the design of the rim of a wheel of a moving car because the parvo pathway has poor temporal resolution. I agree that this is true. However, then why can we not use the Magno Pathway to determine the design of the rim of the wheel since it has better temporal resolution?", + "A": "Because the magno pathway is not good at detecting fine detail, like the the rims of the wheels. (I think this is the right answer).", + "video_name": "0ugcw7wOZBg", + "timestamps": [ + 260 + ], + "3min_transcript": "we also need to figure out, OK, what are the boundaries of the rose, so the boundaries of the stem, the boundaries of the leaf, the boundaries of the petals, from the background. And this is also really important, because not only do we need to distinguish the boundaries, but we also need to figure out, OK, what shape are the leaves, what shape are the petals. And these are all very important things that your brain ultimately is able to break down. So in order for us to figure out what the form of an object is, we use a very specialized pathway that exists in our brain, which is known as the parvo pathway. So the parvo pathway is responsible for figuring out what the shape of an object is. So another way to say this is that the parvo pathway is really good at spatial resolution. Let me write that down. So spatial resolution. is that it's really good at figuring out what the boundaries of an object are, what the little details that make up the object are. So if something isn't moving, such as when you're looking at a picture or when you're looking at a rose, you're able to break down and look at the little tiny details. So you're able to look at the little veins that make up the rose leaf. You're able to see all the little nuances of the rose. And that's because you're using the parvo pathway, which has a really high level of spatial resolution. One negative aspect of the parvo pathway is that it has really poor temporal resolution. And what I mean by this is that temporal resolution is motion. So if a rose is in motion, if I throw it across the room, I can't really use the parvo pathway to track the rose. The parvo pathway is used for stationary objects to acquire high levels of detail of the object. But as soon as that object starts moving, in the object. And you've probably noticed this. So when you're in a car, you're driving along, and there's a little Volkswagen Beetle driving right by you, it's going pretty slow. You can acquire a good number of details about the car, about the driver. But if you look down at the wheels, the wheels are spinning really, really quickly. And so if you try and look at the rims and figure out what design are the rims, it's really hard to figure that out. That's because the wheels are spinning so fast that it's really difficult to acquire any type of detail about the shape of the rims, about what they look like, and things like that. And finally, the parvo pathway also allows us to see things in color. So the parvo pathway not only allows us to acquire fine details about what we're looking at, but it also allows us to see in color. So if something is moving, we can't use the parvo pathway. But what we do use is the magno pathway. So we use the magno pathway in our brain. And the magno pathway is basically a set of specialized cells-- just" + }, + { + "Q": "At 8:10, you say that the force of gravity (Fg) that acts on A is equal to the force that the table acts on A (Ft; normal force) because of Newton's second law of motion. Is it also correct to say that these two forces are equal because of Newton's first law of motion?\n", + "A": "If the object is not accelerating, we know that the net force must be zero. The first law is sort of just a special case of the second law.", + "video_name": "VfpKzwrhmqQ", + "timestamps": [ + 490 + ], + "3min_transcript": "So even if I came in all guns a blazing, Chuck Norris style, trying to dropkick some wall. That does not look like the correct form for a drop kick. But even if I came in, flying at this wall, as soon as I start to make contact with the wall, I'm gonna exert a force on the wall, and the wall has to exert a force back. So I'd exert a force on the wall to the right. And this would be the force on the wall, by my foot. There'd have to be an equal and opposite force instantly transmitted backwards, on my foot. So this would be the force on my foot, by the wall. This happens instantaneously, there is no delay. You can't kick this wall fast enough, for this other force to not be generated instantaneously. As soon as your foot starts to exert any force on the wall what so ever, the wall is gonna start exerting that same force back on your foot. So Newtons's Third Law is universal, but people still have trouble identifying these third law partner forces. So one of the best ways to do it, as soon as you list both objects, well to figure out where the partner force is, you can just reverse these labels. So I know over here, if one of my forces is the force on the wall by my foot, to find the partner force to this force, I can just reverse the labels and say it's gotta be the force on my foot, by the wall, which I drew over here. So this is a great way to identify the third law partner forces, cause it's not always obvious what force is the partner force. So to show you how this can be tricky, consider this example. Say we got the ground and a table. So this example drives people crazy for some reason. If I've got a box sitting on a table, we'll call it box A. Box A is gonna have forces exerted on it. One of those forces is gonna be the gravitational force. So the force of gravity is gonna pull straight down on box A, and if I were to ask you, what force is the third law partner force to this force of gravity, I'm willing to bet a lot of people might say, well there's an upwards force on box A, exerted by the table. And if this box A is just sitting here, not accelerating, these two forces are going to be equal and opposite. So it's even more tempting to say that these two forces are equal and opposite because of the third law, but that's not true. These two forces are equal and opposite because of the second law. The second law says if there's no acceleration, then the net force has to be zero, the forces have to cancel. And that's what's happening here. These forces are equal and opposite, they're canceling on box A. Which is a way to know that they are not third law partner forces, cause third law partner forces are always exerted on different objects. They can never cancel if they're third law partner forces. So what's going on over here? We've got two forces that are canceling, that are equal and opposite, but they're not third law partner forces, they're partner forces are somewhere else. I haven't drawn their partner forces yet. So let's try to figure out what they're partner forces are. So let's get rid of this, let's come back to here, let's slow it down to figure out what the partner force is, name the two objects interacting. So this force of gravity, I shouldn't be vague," + }, + { + "Q": "\nAt 0:25, David says opposite and writes a negative. Is there such a thing as a negative force?", + "A": "Yes, it s a force that is in the opposite direction of whatever direction you decided is positive.", + "video_name": "VfpKzwrhmqQ", + "timestamps": [ + 25 + ], + "3min_transcript": "- [Voiceover] We should talk a little more about Newtons's Third Law, because there are some deep misconceptions that many people have about this law. It seems simple, but it's not nearly as simple as you might think. So people often phrase it as, for every action there's an equal and opposite reaction. But that's just way too vague to be useful. So a version that's a little better, says that for every force, there's an equal and opposite force. So this is a little better. The equal sign means that these forces are equal in magnitude. And this negative sign means they're just different by the direction of the vector. So these are vectors, so this says that this pink vector F, has the opposite direction, but equal in magnitude to this green vector F. But to show you why this is still a little bit too vague, consider this, if this is all you knew about Newtons's Third Law, that for every force, there's an equal and opposite force, you might wonder, if you were clever, you might be like, wait a minute, if for every force F, right, there's got to be a force that's equal and opposite. that every force in the universe cancels? Shouldn't every force just cancel then, at that point? Doesn't that just mean that there's no acceleration that's even possible? Because if I go and exert a force F on something, if there's gonna be a force negative F, doesn't that mean that no matter what force I put forward, it's just gonna get cancelled? And the answer no, and the reason it's no is because these two forces are exerted on different objects. So you have to be careful. So the reason I say that this statement of Newtons's Third Law is still a little bit too vague, is because this is really on different objects. So if this is the force on object A, exerted by object B, then this force over here has to be the force on object B, exerted by object A. In other words, these forces down here are exerted on different objects. I'm gonna move this over to this side. I'm gonna move this over to here. Let's draw two different objects So if there was some object A, so I put some object A in here. Just wanna make sure there's an object A. Let's say this is object A, and it had this green force exerted on it, F. So this object right here is A. Well, there's gonna be another object, object B. We'll just make it another circle. So we'll make it look like this. So here's object B. And it's gonna have this pink force, F, negative F exerted on it. So I'm gonna call this object B. Now we're okay, now we know these forces can't cancel, and the reason these forces can't cancel, is cause they're on two different objects. But when you just say that Newtons's Third Law, is that every force has an equal and opposite force, it's not clear that it has to be on different objects. But it does have to be on different objects. So these Newtons force law pairs, often times is called force pairs, or Newton's third law partner forces, are always on different objects." + }, + { + "Q": "\nOn 3:54, why does Sal say that helium's configuration is 1s2, when helium is in the p block?", + "A": "Helium is not part of the p block. It appears on the far right of the periodic table because it is a Nobel gas, so has similar reactivity to the other elements on the far right of the periodic table. However, helium has no electrons in p orbitals so does not count as part of the p block.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 234 + ], + "3min_transcript": "well it's base configuration is the same as neon. It has a base configuration of neon. Neon is one S two, two S two, two P six, that's what this represents and then to get to sodium, you would then have three S one. How many Valence electrons does sodium have? Well its highest energy, furthest out electron or I say the electron that's in a non-stable shell. That's in a shell that hasn't been stabilized. It hasn't gotten to its fully stable state. There's only one electron in that situation right over here, the three S one electron. Sodium as well, you could depict like that. It only has one Valence electron that's the electron that could be swiped away from it or that somehow could be involved in a covalent bond somehow. Now let's do things with more, more Valence electrons but the important thing to realize and actually for the example of hydrogen sodium is that all of these group one elements are going to have one Valence electron. They're going to have one electron that they tend to use when they are either getting lost to form an ion or that they might be able to use to form a covalent bond. Now let's think about helium and helium's an interesting character because all of the rest of the noble gases have eight Valence electrons which makes them very stable but helium only has two Valence electrons. The reason why it's included here is because helium is also very stable because for that first shell, you only need two electrons to fill full, to fill stable. Helium has two Valence electrons, its electron configuration is one S two. Once again the reason why it's out here with the noble gases is because it's very stable and very inert like the noble gases that's why we now use those helium for balloons It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus" + }, + { + "Q": "\nIn 5:19, Sal said Carbon has 2 + 2 valence electrons. Meaning the electrons from the s2 shell and the p2 shell. Is the s2 shell not filled? Why not use only the p2, which is the outermost shell? Help please.", + "A": "2s and 2p are both subshells of shell number 2. So you have to consider both the s and p.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 319 + ], + "3min_transcript": "but the important thing to realize and actually for the example of hydrogen sodium is that all of these group one elements are going to have one Valence electron. They're going to have one electron that they tend to use when they are either getting lost to form an ion or that they might be able to use to form a covalent bond. Now let's think about helium and helium's an interesting character because all of the rest of the noble gases have eight Valence electrons which makes them very stable but helium only has two Valence electrons. The reason why it's included here is because helium is also very stable because for that first shell, you only need two electrons to fill full, to fill stable. Helium has two Valence electrons, its electron configuration is one S two. Once again the reason why it's out here with the noble gases is because it's very stable and very inert like the noble gases that's why we now use those helium for balloons It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange." + }, + { + "Q": "Apologies if this has already been answered, but at 9:20 Sal tells us that from the n=4 S subshell, the element \"backfills\" electrons into the n=3 D subshell. Why does it do this, rather than going directly to 4d^6? Is this a special case, or does it happen for all elements in the D subshell and/or beyond? Thanks!\n", + "A": "This happens for all elements in the d block and beyond. In the fourth level and beyond, the energy levels get all mixed up. For example, in cerium (Ce, element 58), the outermost electron configuration is 6s\u00c2\u00b2 4f\u00c2\u00b9 5d\u00c2\u00b9 6s2 4f1 5d1", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 560 + ], + "3min_transcript": "This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy." + }, + { + "Q": "\n\"They would say, \"Okay,\" \"these are the two Valence electrons\" \"for all of these transition metals.\" Well that doesn't hold up for all of them because you even have special cases like copper and chromium that only go four S one and then start filling three D depending on the circumstances.\" Could someone explain to me how can copper and chromium only go 4s^1 not 4s^2 (they have 2 groups in s subshells in their period)? 10:15", + "A": "You would expect Cu to be 4s\u00c2\u00b23d\u00e2\u0081\u00b4. But a half-filled d subshell is more stable than one with only 4 d electrons. It takes a little energy to promote a 4s electron to the 3d level, but you get back more than this by getting a 4s3d\u00e2\u0081\u00b5 configuration. In the same way, a filled d subshell is more stable than one with only 9 d electrons. It takes a little energy to promote a 4s electron to the 3d level, but you get back more than this by getting a 4s3d\u00c2\u00b9\u00e2\u0081\u00b0 configuration.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 615 + ], + "3min_transcript": "could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy. figuring out the Valence electrons, the electrons that are most slightly right becomes a little bit hard to predict. Some people, some even textbooks will say, \"Oh, all the transition metals\" \"have two Valence electrons\" \"because they all get the four S two\" \"and then they're back filling.\u201d They would say, \"Okay,\" \"these are the two Valence electrons\" \"for all of these transition metals.\" Well that doesn't hold up for all of them because you even have special cases like copper and chromium that only go four S one and then start filling three D depending on the circumstances. Sometimes it does it otherwise but even for the other transition elements like say iron is not necessarily the case so these are the one, the only two electrons that are going to react. You might have some of your D electrons, your three D electrons which are high energy might also be involved in reaction. Might be taken away or might form a bond somehow." + }, + { + "Q": "at 10:16, why does the d group belong to the number 3 not four. Since it is in the fourth group why is it written at 3d6\n", + "A": "Well, when you get to the D sub-level you have to take into account that the 4s orbital is filled BEFORE the 3d! The 5s orbital fills before the 4d, so, it even though the 5s seems like it would be farther out than the 4d, the 4d has more energy, so to speak.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 616 + ], + "3min_transcript": "could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy. figuring out the Valence electrons, the electrons that are most slightly right becomes a little bit hard to predict. Some people, some even textbooks will say, \"Oh, all the transition metals\" \"have two Valence electrons\" \"because they all get the four S two\" \"and then they're back filling.\u201d They would say, \"Okay,\" \"these are the two Valence electrons\" \"for all of these transition metals.\" Well that doesn't hold up for all of them because you even have special cases like copper and chromium that only go four S one and then start filling three D depending on the circumstances. Sometimes it does it otherwise but even for the other transition elements like say iron is not necessarily the case so these are the one, the only two electrons that are going to react. You might have some of your D electrons, your three D electrons which are high energy might also be involved in reaction. Might be taken away or might form a bond somehow." + }, + { + "Q": "At 6:50, Sal talks of Methane having a logical structure of CH^4, with 8 valence electrons. How then, is methane so reactive and explosive? Hydrogen is flammable, yet saying that Methane is flammable only because it has Hydrogen atoms would be like saying water is flammable because it is H^2O.\n", + "A": "A full octet of electrons doesn t mean it will completely unreactive to anything. It s not simply hydrogen atoms that are flammable, methane can be turned into more stable products (CO2 and H2O) when it is burned.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 410 + ], + "3min_transcript": "How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange. Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways" + }, + { + "Q": "\nAt 5:14, he says \"that has not been completed yet\" what does he mean by that? Doesn't the 2s^2 mean that it's 2nd s orbital has been full so it has to move onto the p orbital? A little confused on Valence electrons still.", + "A": "The 2s orbital is full, but those electrons are still part of the outermost shell. The 2s and the three 2p orbitals make up the second shell.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 314 + ], + "3min_transcript": "but the important thing to realize and actually for the example of hydrogen sodium is that all of these group one elements are going to have one Valence electron. They're going to have one electron that they tend to use when they are either getting lost to form an ion or that they might be able to use to form a covalent bond. Now let's think about helium and helium's an interesting character because all of the rest of the noble gases have eight Valence electrons which makes them very stable but helium only has two Valence electrons. The reason why it's included here is because helium is also very stable because for that first shell, you only need two electrons to fill full, to fill stable. Helium has two Valence electrons, its electron configuration is one S two. Once again the reason why it's out here with the noble gases is because it's very stable and very inert like the noble gases that's why we now use those helium for balloons It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange." + }, + { + "Q": "At 7:15, Sal mentions covalent bonds between hydrogen and carbon. What are covalent bonds?\n", + "A": "As opposed to ionic bonds, which donate and receive electrons to get charged, get full shells and stick together, covalent bonds share electrons to stick together and get full shells. Covalent bonds happen between elements with a smaller electronegativity difference than the amount required for an ionic bond, so generally nonmetals.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 435 + ], + "3min_transcript": "How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange. Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways" + }, + { + "Q": "at 5:38, i don't really understand why 2s\u00c2\u00b22p\u00c2\u00b3......also, for school purpose, should i memories the whole table like Sal did and how to make it easier...someone help me\n", + "A": "just see and by heart the name of elements minimum till 30 and some important like gold or zinc fe etc.. and to explain the first question its not possible in this box i would suggest you to go through all the previous topics of periodic table or read a source like your text and then go through this. it will explain everything as Mr khan always does:)", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 338 + ], + "3min_transcript": "It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange. Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair" + }, + { + "Q": "\nAt 9:32 Sal asks : \"what are the highest energy electrons ?\" and then at 9:36 he asks \"what are the furthest out ?\".\n\nHow do we determine which one are the furthest out and the one which have highest energy ? Is it a specificity of blocks ?", + "A": "You can t determine it unless you have prior knowledge of it. But Sal is not correct here, because the 4s electrons are both the furtherest out and the highest energy for the transition metals.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 572, + 576 + ], + "3min_transcript": "could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy. figuring out the Valence electrons, the electrons that are most slightly right becomes a little bit hard to predict. Some people, some even textbooks will say, \"Oh, all the transition metals\" \"have two Valence electrons\" \"because they all get the four S two\" \"and then they're back filling.\u201d They would say, \"Okay,\" \"these are the two Valence electrons\" \"for all of these transition metals.\" Well that doesn't hold up for all of them because you even have special cases like copper and chromium that only go four S one and then start filling three D depending on the circumstances. Sometimes it does it otherwise but even for the other transition elements like say iron is not necessarily the case so these are the one, the only two electrons that are going to react. You might have some of your D electrons, your three D electrons which are high energy might also be involved in reaction. Might be taken away or might form a bond somehow." + }, + { + "Q": "In 5:43, Why is it written as [He]2s2, 2p2 instead of 1s2, 2s2, 2p2? And why does Carbon only have four valence?\n", + "A": "Because Helium s electronic configuration is 1s2, which is a complete electron shell, so 1s2 is replaced with [He] instead. For example, potassium(K) s electronic configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 4s1. Since the last complete electron shell is 1s2, 2s2, 2p6, 3s2, 3p6, which is Argon, it could be substituted into K s electronic configuration. Thus, it could be written as [Ar] 4s1. Carbon has four valency electrons and needs to gain or lose FOUR more electrons to obtain a complete electron shell.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 343 + ], + "3min_transcript": "It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange. Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair" + }, + { + "Q": "Why does Sal, at 7:39 state that any element in Carbon's group will have four valence electrons?\n", + "A": "Because of the group it is in. All group 1 elements have one valence electron, group 2 elements have two, skip groups 3-12 for now (they are the exceptions to the rule), group 13 has three, group 14 has four (carbon being part of this group), and so on.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 459 + ], + "3min_transcript": "Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base" + }, + { + "Q": "\n8:22 so? H2Te or Li20 could both be similar to H20? or even Li2Te? or did I just give a completely different molecule", + "A": "Yeah H2Te and Li2O are similar to H2O and also to Li2Te. Slightly tricky since Hydrogen is NOT an alkali metal(group I metal) It s position is analogous.That s why Hydrogen s box is green. So, Hydrogen won t have the properties of Lithium. But Oxygen and Tellurium on the other hand are both Chalcogens and exhibit similar properties.So, H2Te is similar to H2O Li2O is similar to Li2Te", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 502 + ], + "3min_transcript": "Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base" + }, + { + "Q": "At 1:56 How did we know that Carbon has sp3 hybridized orbitals and that it forms a tetrahedral shape?\n", + "A": "Because each carbon has four other atoms directly attached to it. You have already learned that this requires sp3 hybridization and the sp3 bond angles are 109.5\u00c2\u00b0.", + "video_name": "IkmM4CPnqF0", + "timestamps": [ + 116 + ], + "3min_transcript": "In the video on sp3 hybridized orbitals, we went in pretty good detail about how a methane molecule looks. But just as a bit of a review, it's the tetrahedral shape. You have a carbon in the middle, and then you would have a hydrogen-- you can imagine I'm drawing it like this because this hydrogen is poking out of the page. Then maybe you have another hydrogen that's in the page. You have one above the carbon, and you have one that's behind the page. You could imagine it's like a tripod with a pole sticking out of the top of the tripod. Or if you were to imagine the shape another way, if you were to connect the hydrogens, you would have a four-sided pyramid with a triangle as each of the sides. So it would to look something like this. I'm trying my best to draw-- the pyramid would look something like this if you could see through it. So this would be one side, another side would be over here, and then the backside would be over here, and then transparent out front. So the fourth side would be the actual kind of thing that we're looking through when we look at this pyramid. It would be this front side right over here. So you can imagine it different ways, this was the case with methane. Now let's extend this into a slightly more complex molecule, and that's ethane. So the way we've been drawing it so far-- I guess the simplest way to draw ethane, is just like that. By implication this is ethane. By implication you have a carbon there, and a carbon there, and they'll each have three hydrogens bonded to it. And we've drawn it something like this. Three hydrogens bonded to each of these guys. But now we know that carbon has these sp3 hybridized orbitals, that it likes to form more of a tetrahedral shape when it bonds. So an ethane molecule would actually look more like this. So I'll do the carbons in orange. So if that's the carbon and that's the carbon. So you can imagine you have a carbon molecule here. I'll draw it as this little circle. And then if we have some perspective, so the carbon-carbon bond is going to look like that. And then you have another carbon molecule right over there. So that's that bond over here. And we want both the carbons, all of their bonds to be kind of in a tetrahedral shape. So then you could imagine this bond over here going like this. This bond going like that. And you have your hydrogen at the end. Let's make the green circles the hydrogens. So you have that hydrogen, and then-- or actually just the circles, we'll call them hydrogen-- and then you can imagine this one, maybe it's coming out of the page a little bit. That is that hydrogen. Let me label the hydrogens, actually. I'm doing it in all different colors so you can see what I'm And then this hydrogen is going right below it, maybe pointed back a little bit." + }, + { + "Q": "\nAround 2:50 Sal started talking about prokaryotes. I know about how organisms and species have evolved over time but how is this possible for single celled organisms to evolve exactly? Did these organisms gather in massses and shift into the next cellular organism? Or did they latch on to soil thus creating plants?", + "A": "This isn t known at all, but we do know that they evolved somehow. It s kind of like the apes evolving into humans, it s just natural I guess.", + "video_name": "nYFuxTXDj90", + "timestamps": [ + 170 + ], + "3min_transcript": "Because this was a time where so many things from outer space were hitting Earth, that it was so violent, that it might have killed off any kind of primitive, self-replicating organisms or molecules that might have existed before it. And I won't go into the physics of the Late Heavy Bombardment. But we believe that it happened, because Uranus and Neptune-- so if this is the sun right here-- that is the sun. This is the asteroid belt. That's outside the orbits of the inner, rocky planets. That Uranus and Neptune, their orbits moved outward. And I'm not going to go into the physics. But what that caused is, gravitationally, it caused a lot of the asteroids in the asteroid belt to move inward and start impacting the inner planets. And of course, Earth was one of the inner planets. And I should make the sun like orange or something, not blue. I don't want you to think that's Earth. And it's more obvious on the moon, because the moon does not have an atmosphere to kind of smooth over the impact. So the consensus is that only after the Late Heavy Bombardment was Earth kind of ready for life. And we believe that the first life formed 3.824 billion years ago. Remember, g for giga, for billion years ago. And when we talk about life at this period, we're not talking about squirrels or panda bears. We're talking about extremely simple life forms. We're talking about prokaryotes. And let me give you a little primer on that right now, though we go into much more detail in the biology playlist. We're talking about prokaryotes. And I'll compare them to eukaryotes. Prokaryotes are, for the most part, unicellular organisms that have no nucleuses. They also don't have any other membrane-bound that perform specific functions, like mitochondria. So their DNA is just kind of floating around. So let me draw this character's DNA. So it's just floating around, just like that. And prokaryote literally means before kernel or before a nucleus. Eukaryotes do have a nucleus, where all of their DNA is. So this is the nuclear membrane. And then all of its DNA is floating inside of the nucleus. And then it also has other membrane-bound organelles. Mitochondria is kind of the most famous of them. So it also has things like mitochondria. We'll learn more about that in future videos. Mitochondria, we believe, is essentially one prokaryote crawling inside of another prokaryote and kind of starting to become a symbiotic organism with each other. But I won't go into that right now. But when we talk about life at this period, we're talking about prokaryotes. And we still have prokaryotes on the planet." + }, + { + "Q": "\nAt 9:08 Sal said that all life was happening in the ocean and the ocean was still iron rich. Doesn't too much iron kill you?", + "A": "It wouldn t be enough to kill them, in fact it might help (hemoglobin?).", + "video_name": "nYFuxTXDj90", + "timestamps": [ + 548 + ], + "3min_transcript": "And so we have rocks from, that are roughly 3.8 billion years So we kind of put that as the beginning of the Archean Eon. And so there's two things there. One, rocks have survived from the beginning of the Archean And also, that's roughly when we think that the first life existed. And so we're now in the Archean Eon. And you might say, oh, maybe Earth is a more pleasant place now. But it would not be. It still has no to little oxygen in the environment. If you were to go to Earth at that time, it might have looked something like this. It would have a been a reddish sky. You would have had nitrogen and methane and carbon dioxide in the atmosphere. There would have been nothing for you to breathe. There still would have been a lot of volcanic activity. This right here, these are pictures of stromatolites. And these are formed from bacteria And over time, these things get built up. But the most significant event in the Archean period, at least in my humble opinion, was what we believe started to happen about 3.5 billion years ago. And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun, to actually do photosynthesis. And the real fascinating byproduct of that, other than the fact that they can now use energy directly from the sun, is that it started to produce oxygen, so starts to produce oxygen. And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere. Because you had all of this iron that was dissolved in the oceans. And let me be clear. the next several billion years, it all occurred in the ocean. We had no ozone layer now. The land was being irradiated. The land was just a completely inhospitable environment for life. So all of this was occurring in the ocean. And so the first oxygen that actually got produced, it actually, instead of just being released into the atmosphere, it ended up bonding with the iron that was dissolved in the ocean at that time. So it actually didn't have a chance to accumulate in the atmosphere. And when we fast forward past the Archean period, we're going to see, that once a lot of that iron was oxidized and the oxygen really did start to get released in the atmosphere, it actually had-- it's funny to say-- a cataclysmic effect or a catastrophic effect on the other anaerobic life on the planet at the time. And it's funny to say that because it was a catastrophe for them. But it was kind of a necessary thing that had to happen for us to happen. So for us, it was a blessing that this cyanobacteria" + }, + { + "Q": "9:00 on the video there is suddenly an ocean. What happened? Where did it come from?\n", + "A": "actually, when a meteriote formed, it had some ice on it. that ice melted and started the water cycle. this happened over thousands of years", + "video_name": "nYFuxTXDj90", + "timestamps": [ + 540 + ], + "3min_transcript": "And so we have rocks from, that are roughly 3.8 billion years So we kind of put that as the beginning of the Archean Eon. And so there's two things there. One, rocks have survived from the beginning of the Archean And also, that's roughly when we think that the first life existed. And so we're now in the Archean Eon. And you might say, oh, maybe Earth is a more pleasant place now. But it would not be. It still has no to little oxygen in the environment. If you were to go to Earth at that time, it might have looked something like this. It would have a been a reddish sky. You would have had nitrogen and methane and carbon dioxide in the atmosphere. There would have been nothing for you to breathe. There still would have been a lot of volcanic activity. This right here, these are pictures of stromatolites. And these are formed from bacteria And over time, these things get built up. But the most significant event in the Archean period, at least in my humble opinion, was what we believe started to happen about 3.5 billion years ago. And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun, to actually do photosynthesis. And the real fascinating byproduct of that, other than the fact that they can now use energy directly from the sun, is that it started to produce oxygen, so starts to produce oxygen. And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere. Because you had all of this iron that was dissolved in the oceans. And let me be clear. the next several billion years, it all occurred in the ocean. We had no ozone layer now. The land was being irradiated. The land was just a completely inhospitable environment for life. So all of this was occurring in the ocean. And so the first oxygen that actually got produced, it actually, instead of just being released into the atmosphere, it ended up bonding with the iron that was dissolved in the ocean at that time. So it actually didn't have a chance to accumulate in the atmosphere. And when we fast forward past the Archean period, we're going to see, that once a lot of that iron was oxidized and the oxygen really did start to get released in the atmosphere, it actually had-- it's funny to say-- a cataclysmic effect or a catastrophic effect on the other anaerobic life on the planet at the time. And it's funny to say that because it was a catastrophe for them. But it was kind of a necessary thing that had to happen for us to happen. So for us, it was a blessing that this cyanobacteria" + }, + { + "Q": "\ni have another question.My teacher mentioned that nuclues in unicellular organisms help in reproduction.ie-the organism breaks itself into 2 and so,there is a new organism.The video mentioned that there were so many in number(i am talking about the procaryotes).So did many procaryotes come into being in a flash because it cant reproduce?The time could be approximately 3:00", + "A": "Nothing comes into being in a flash.", + "video_name": "nYFuxTXDj90", + "timestamps": [ + 180 + ], + "3min_transcript": "Because this was a time where so many things from outer space were hitting Earth, that it was so violent, that it might have killed off any kind of primitive, self-replicating organisms or molecules that might have existed before it. And I won't go into the physics of the Late Heavy Bombardment. But we believe that it happened, because Uranus and Neptune-- so if this is the sun right here-- that is the sun. This is the asteroid belt. That's outside the orbits of the inner, rocky planets. That Uranus and Neptune, their orbits moved outward. And I'm not going to go into the physics. But what that caused is, gravitationally, it caused a lot of the asteroids in the asteroid belt to move inward and start impacting the inner planets. And of course, Earth was one of the inner planets. And I should make the sun like orange or something, not blue. I don't want you to think that's Earth. And it's more obvious on the moon, because the moon does not have an atmosphere to kind of smooth over the impact. So the consensus is that only after the Late Heavy Bombardment was Earth kind of ready for life. And we believe that the first life formed 3.824 billion years ago. Remember, g for giga, for billion years ago. And when we talk about life at this period, we're not talking about squirrels or panda bears. We're talking about extremely simple life forms. We're talking about prokaryotes. And let me give you a little primer on that right now, though we go into much more detail in the biology playlist. We're talking about prokaryotes. And I'll compare them to eukaryotes. Prokaryotes are, for the most part, unicellular organisms that have no nucleuses. They also don't have any other membrane-bound that perform specific functions, like mitochondria. So their DNA is just kind of floating around. So let me draw this character's DNA. So it's just floating around, just like that. And prokaryote literally means before kernel or before a nucleus. Eukaryotes do have a nucleus, where all of their DNA is. So this is the nuclear membrane. And then all of its DNA is floating inside of the nucleus. And then it also has other membrane-bound organelles. Mitochondria is kind of the most famous of them. So it also has things like mitochondria. We'll learn more about that in future videos. Mitochondria, we believe, is essentially one prokaryote crawling inside of another prokaryote and kind of starting to become a symbiotic organism with each other. But I won't go into that right now. But when we talk about life at this period, we're talking about prokaryotes. And we still have prokaryotes on the planet." + }, + { + "Q": "At 8:10, he mentions stromatalites, something about sediment and bacteria. What are they exactly?\n", + "A": "Stromatolites are layered mounds, columns, and sedimentary rocks. They were formed upon layer by layer of cyanobacteria, a single celled organism that lives in many different habitats. I hope that helped you.", + "video_name": "nYFuxTXDj90", + "timestamps": [ + 490 + ], + "3min_transcript": "oh, 4.6 billion years ago to 3.8 billion. Oh, that's just 800 million years. Remember-- and I'll talk about this. Grass has only existed for 50 million years. This is 800 million years. Humans and chimpanzees only diverged 5 million years ago. This is 800 million years we're talking about, from ancient Greece to now, we're only talking about 2,500 years. You multiply that times 1,000, you get 2 and 1/2 a million years. And this is 800 million years we're talking about. So these are extremely huge periods of time. And that's why we call them eons. Eons are 500 million to a billion years. Now, the dividing line between the Hadean Eon and the Archean Eon-- and it's kind of a fuzzy dividing line, but most people place it about 3.8 billion years ago. And so we have rocks from, that are roughly 3.8 billion years So we kind of put that as the beginning of the Archean Eon. And so there's two things there. One, rocks have survived from the beginning of the Archean And also, that's roughly when we think that the first life existed. And so we're now in the Archean Eon. And you might say, oh, maybe Earth is a more pleasant place now. But it would not be. It still has no to little oxygen in the environment. If you were to go to Earth at that time, it might have looked something like this. It would have a been a reddish sky. You would have had nitrogen and methane and carbon dioxide in the atmosphere. There would have been nothing for you to breathe. There still would have been a lot of volcanic activity. This right here, these are pictures of stromatolites. And these are formed from bacteria And over time, these things get built up. But the most significant event in the Archean period, at least in my humble opinion, was what we believe started to happen about 3.5 billion years ago. And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun, to actually do photosynthesis. And the real fascinating byproduct of that, other than the fact that they can now use energy directly from the sun, is that it started to produce oxygen, so starts to produce oxygen. And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere. Because you had all of this iron that was dissolved in the oceans. And let me be clear." + }, + { + "Q": "at the end of the video< im kind of confused on how sal got the ratio 1:2. can you explain?\n", + "A": "We discovered, in this video, that the bottle of molecular substance we were dealing with had 1.02 moles of Sulfur, 2.04 moles of Hydrogen, and 4.08 moles of Oxygen. The goal of the Empirical Formula is to represent the elements in simplest form. To simplify, we can divide each element by the least common factor (in this case, 1.02 happens to be the lcf) and we come up with 1, 2 and 4. We cannot reduce these any more.", + "video_name": "sXOIIEZh6qg", + "timestamps": [ + 62 + ], + "3min_transcript": "I said I would get you a more interesting mass composition to empirical formula problem, one that doesn't just have a straight-up 2:1 ratio. And so here it is. I have a bag of stuff. Or let's call this a bottle of stuff. Maybe it's in its liquid form. And it happens to be 2.04% hydrogen, 65.3% oxygen, and 32.65% sulfur. What is the empirical formula, what's our best stab at the empirical formula, of this substance? So what we would do, like we do in all these problems, let's just assume we've got 100 grams of the stuff. So we have 100 grams of the stuff. So we assume 100 grams. Let me do that in a good yellow. So let's say, assume I have 100 grams. How many grams of hydrogen do I have? If I have 100 grams total, 2.04% of that is hydrogen, so I have 2.04 grams of hydrogen. I have 65.3 grams of oxygen. Now, what we need to do now is figure out how many moles of hydrogen is this. How many moles of oxygen. And how many moles of sulfur. Then we can compare the ratios and we should be able to know the empirical formula. So how much 1 mole of hydrogen? What is the mass of 1 mole of hydrogen? Let me write that. So 1 mole of hydrogen. Well we know what the mass number for hydrogen is. It's 1. And especially, the atomic weight, also for hydrogen, if we were to take it on Earth. The composition, you pretty much just find hydrogen nucleuses. If it's neutral, it has an electron, but it has no neutrons. So it has an atomic mass of one atomic mass unit. So one mole of hydrogen. If you have a ton of hydrogens together, or a mole of them, not a ton, I shouldn't say, you have 6.02 times 10 to the 23 hydrogens. Then you take hydrogen's atomic mass number in atomic mass units. And you say, well, it'll be that many grams of hydrogen, right? So if you immediately look up here, if we have 2.04 grams of hydrogen, how many moles of hydrogen do we have? Well, one mole is one gram, so we have 2.04 moles of hydrogen. Notice, this said what the mass of the hydrogen is. This tells us how many hydrogen molecules we have. Remember, this is 2.04 times 6.02 times 10 to the 23 hydrogen atoms. Moles of hydrogen. Maybe I should write that down. So one mole of hydrogen. There you go. And then oxygen. One mole of oxygen. Oxygen's mass number, in case you forgot, is 16." + }, + { + "Q": "My question is: In the case of Fe2O3 that are in the ratio of 70% iron and 30% oxygen.\nThe atomic weight of iron is 56 and I get 70/56 = 1.25 moles.\nThe atomic weight of oxygen is 16 and I get 30/16 = 1.87 moles.\nHow do I know that their ratio is 2:3?\nI know that 1.25 / 1.87 = 0.666666667, so I have to find which numbers divided among themselves make 0.666666667 and discover that is 2/3.\nThere is no way to know immediately without having to do a lot of divisions? Thank you.\n", + "A": "Another way to write 0.6 repeating is 6/9 or 2/3. If you think of the fraction as the ratio of the number of Iron Atoms (numerator) to the number of Oxygen Atoms (denominator) then you get Fe2O3 pretty quickly. Conversely, if you were to multiply the numerator and denominator by 1000 you get 1250/1875 (you rounded 30/16 to 1.87, but it is 1.875). when you reduce that you get 2/3. As the numerator represented Fe, you have 2 atoms of that and the denominator represented O, you had 3 atoms of that", + "video_name": "sXOIIEZh6qg", + "timestamps": [ + 123 + ], + "3min_transcript": "I said I would get you a more interesting mass composition to empirical formula problem, one that doesn't just have a straight-up 2:1 ratio. And so here it is. I have a bag of stuff. Or let's call this a bottle of stuff. Maybe it's in its liquid form. And it happens to be 2.04% hydrogen, 65.3% oxygen, and 32.65% sulfur. What is the empirical formula, what's our best stab at the empirical formula, of this substance? So what we would do, like we do in all these problems, let's just assume we've got 100 grams of the stuff. So we have 100 grams of the stuff. So we assume 100 grams. Let me do that in a good yellow. So let's say, assume I have 100 grams. How many grams of hydrogen do I have? If I have 100 grams total, 2.04% of that is hydrogen, so I have 2.04 grams of hydrogen. I have 65.3 grams of oxygen. Now, what we need to do now is figure out how many moles of hydrogen is this. How many moles of oxygen. And how many moles of sulfur. Then we can compare the ratios and we should be able to know the empirical formula. So how much 1 mole of hydrogen? What is the mass of 1 mole of hydrogen? Let me write that. So 1 mole of hydrogen. Well we know what the mass number for hydrogen is. It's 1. And especially, the atomic weight, also for hydrogen, if we were to take it on Earth. The composition, you pretty much just find hydrogen nucleuses. If it's neutral, it has an electron, but it has no neutrons. So it has an atomic mass of one atomic mass unit. So one mole of hydrogen. If you have a ton of hydrogens together, or a mole of them, not a ton, I shouldn't say, you have 6.02 times 10 to the 23 hydrogens. Then you take hydrogen's atomic mass number in atomic mass units. And you say, well, it'll be that many grams of hydrogen, right? So if you immediately look up here, if we have 2.04 grams of hydrogen, how many moles of hydrogen do we have? Well, one mole is one gram, so we have 2.04 moles of hydrogen. Notice, this said what the mass of the hydrogen is. This tells us how many hydrogen molecules we have. Remember, this is 2.04 times 6.02 times 10 to the 23 hydrogen atoms. Moles of hydrogen. Maybe I should write that down. So one mole of hydrogen. There you go. And then oxygen. One mole of oxygen. Oxygen's mass number, in case you forgot, is 16." + }, + { + "Q": "\nAt 0:07 why is hydrogen at 2.04 %", + "A": "That is just the way the problem is set up.", + "video_name": "sXOIIEZh6qg", + "timestamps": [ + 7 + ], + "3min_transcript": "I said I would get you a more interesting mass composition to empirical formula problem, one that doesn't just have a straight-up 2:1 ratio. And so here it is. I have a bag of stuff. Or let's call this a bottle of stuff. Maybe it's in its liquid form. And it happens to be 2.04% hydrogen, 65.3% oxygen, and 32.65% sulfur. What is the empirical formula, what's our best stab at the empirical formula, of this substance? So what we would do, like we do in all these problems, let's just assume we've got 100 grams of the stuff. So we have 100 grams of the stuff. So we assume 100 grams. Let me do that in a good yellow. So let's say, assume I have 100 grams. How many grams of hydrogen do I have? If I have 100 grams total, 2.04% of that is hydrogen, so I have 2.04 grams of hydrogen. I have 65.3 grams of oxygen. Now, what we need to do now is figure out how many moles of hydrogen is this. How many moles of oxygen. And how many moles of sulfur. Then we can compare the ratios and we should be able to know the empirical formula. So how much 1 mole of hydrogen? What is the mass of 1 mole of hydrogen? Let me write that. So 1 mole of hydrogen. Well we know what the mass number for hydrogen is. It's 1. And especially, the atomic weight, also for hydrogen, if we were to take it on Earth. The composition, you pretty much just find hydrogen nucleuses. If it's neutral, it has an electron, but it has no neutrons. So it has an atomic mass of one atomic mass unit. So one mole of hydrogen. If you have a ton of hydrogens together, or a mole of them, not a ton, I shouldn't say, you have 6.02 times 10 to the 23 hydrogens. Then you take hydrogen's atomic mass number in atomic mass units. And you say, well, it'll be that many grams of hydrogen, right? So if you immediately look up here, if we have 2.04 grams of hydrogen, how many moles of hydrogen do we have? Well, one mole is one gram, so we have 2.04 moles of hydrogen. Notice, this said what the mass of the hydrogen is. This tells us how many hydrogen molecules we have. Remember, this is 2.04 times 6.02 times 10 to the 23 hydrogen atoms. Moles of hydrogen. Maybe I should write that down. So one mole of hydrogen. There you go. And then oxygen. One mole of oxygen. Oxygen's mass number, in case you forgot, is 16." + }, + { + "Q": "\nAt 3:30, doesn't P1=V2 and P2=V1", + "A": "No. Because its not P1/Vi=0 and P2/V2=0. If that was the case then you would have been correct. But as the temperature, no. of moles and constant does not change, both P1*V1 and P2*V2 are equal to nRT. Hope this helps :)", + "video_name": "GwoX_BemwHs", + "timestamps": [ + 210 + ], + "3min_transcript": "and they're exerting a certain pressure on their container, and if we were to make the container smaller, we have the same number of particles. n doesn't change. The average kinetic energy doesn't change, so they're just going to bump into the walls more. So that when we make the volume smaller, when the volume goes down, the pressure should go up. So let's see if we can calculate the exact number. So we can take our ideal gas equation: pressure times volume is equal to nRT. Now, do the number of particles change when I did this situation when I shrunk the volume? No! We have the same number of particles. I'm just shrinking the container, so n is n, R doesn't change, that's a constant, and then the temperature doesn't change. So my old pressure times volume is going to be equal to nRT, -- so let me call this P1 and V1. That's V2. V2 is this, and we're trying to figure out P2. P2 is what? Well, we know that P1 times V1 is equal to nRT, and we also know that since temperature and the number of moles of our gas stay constant, that P2 times V2 is equal to nRT. And since they both equal the same thing, we can say that the pressure times the volume, as long as the temperature is held constant, will be a constant. So P1 times V1 is going to equal P2 times V2. So what was P1? P1, our initial pressure, was 3 atmospheres. is equal to our new pressure times 3 liters. And if we divide both sides of the equation by 3, we get 3 liters cancel out, we're left with 9 atmospheres. And that should make sense. When you decrease the volume by 2/3 or when you make the volume 1/3 of your original volume, then your pressure increases by a factor of three. So this went by times 3, and this went by times 1/3. That's a useful thing to know in general. If temperature is held constant, then pressure times volume are going to be a constant. Now, you can take that even further." + }, + { + "Q": "\nAt 2:30, how can V1 and V2 be equal to each other when their volume is clearly decreasing?", + "A": "The volume in each IS different V1=9L and V2=3Liters. When setting up the equation both are equal to nRT since he states these are constants. The next step he makes them equal to each other divides which eliminates 3 and the L leaving 9 atm.", + "video_name": "GwoX_BemwHs", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's do some more problems that involve the ideal gas equation. Let's say I have a gas in a container and the current pressure is 3 atmospheres. And let's say that the volume of the container is 9 liters. Now, what will the pressure become if my volume goes from 9 liters to 3 liters? So from the first ideal gas equation video, you can kind of have the intuition, that you have a bunch of -- and we're holding-- and this is important. We're holding the temperature constant, and that's an important thing to realize. So in our very original intuition behind the ideal gas equation we said, look, if we have a certain number of particles and they're exerting a certain pressure on their container, and if we were to make the container smaller, we have the same number of particles. n doesn't change. The average kinetic energy doesn't change, so they're just going to bump into the walls more. So that when we make the volume smaller, when the volume goes down, the pressure should go up. So let's see if we can calculate the exact number. So we can take our ideal gas equation: pressure times volume is equal to nRT. Now, do the number of particles change when I did this situation when I shrunk the volume? No! We have the same number of particles. I'm just shrinking the container, so n is n, R doesn't change, that's a constant, and then the temperature doesn't change. So my old pressure times volume is going to be equal to nRT, -- so let me call this P1 and V1. That's V2. V2 is this, and we're trying to figure out P2. P2 is what? Well, we know that P1 times V1 is equal to nRT, and we also know that since temperature and the number of moles of our gas stay constant, that P2 times V2 is equal to nRT. And since they both equal the same thing, we can say that the pressure times the volume, as long as the temperature is held constant, will be a constant. So P1 times V1 is going to equal P2 times V2. So what was P1? P1, our initial pressure, was 3 atmospheres." + }, + { + "Q": "At 3:30 in the video the answer is revealed to be 9 atm. When you divide the 3 over, why isn't the 9 divided as well, because should ((3atm x 9L)/(3L))= 3atm?\n", + "A": "Sorry, but no. If you divide a sum by a number, then you divide each member of the sum. But if the numbers are all multiplied, you only divide one time.", + "video_name": "GwoX_BemwHs", + "timestamps": [ + 210 + ], + "3min_transcript": "and they're exerting a certain pressure on their container, and if we were to make the container smaller, we have the same number of particles. n doesn't change. The average kinetic energy doesn't change, so they're just going to bump into the walls more. So that when we make the volume smaller, when the volume goes down, the pressure should go up. So let's see if we can calculate the exact number. So we can take our ideal gas equation: pressure times volume is equal to nRT. Now, do the number of particles change when I did this situation when I shrunk the volume? No! We have the same number of particles. I'm just shrinking the container, so n is n, R doesn't change, that's a constant, and then the temperature doesn't change. So my old pressure times volume is going to be equal to nRT, -- so let me call this P1 and V1. That's V2. V2 is this, and we're trying to figure out P2. P2 is what? Well, we know that P1 times V1 is equal to nRT, and we also know that since temperature and the number of moles of our gas stay constant, that P2 times V2 is equal to nRT. And since they both equal the same thing, we can say that the pressure times the volume, as long as the temperature is held constant, will be a constant. So P1 times V1 is going to equal P2 times V2. So what was P1? P1, our initial pressure, was 3 atmospheres. is equal to our new pressure times 3 liters. And if we divide both sides of the equation by 3, we get 3 liters cancel out, we're left with 9 atmospheres. And that should make sense. When you decrease the volume by 2/3 or when you make the volume 1/3 of your original volume, then your pressure increases by a factor of three. So this went by times 3, and this went by times 1/3. That's a useful thing to know in general. If temperature is held constant, then pressure times volume are going to be a constant. Now, you can take that even further." + }, + { + "Q": "\nSo, before Sal changed the pressure in the initial problem at 6:10, the proportion was set up as (1atm x 2m^3)/300K = (2atm x 1m^3)/T. If the proportion was solved as such, wouldn't a temperature of 300K have been achieved even though the pressure of the second container was increased and the volume of the second container was decreased?", + "A": "I think I understand the algebra behind the problem, I was just a little confused conceptually. If the pressure of a container is increased and the volume is decreased, wouldn t the result be a smaller, more pressurized space? And would this scenario not lead to a higher temperature?", + "video_name": "GwoX_BemwHs", + "timestamps": [ + 370 + ], + "3min_transcript": "the two things that we know don't change in the vast majority of exercises we do is the number of molecules we're dealing with, and obviously, R isn't going to change. So if we divide both sides of this by T, we get PV over T is equal to nR, or you could say it's equal to a constant. This is going to be a constant number for any system where we're not changing the number of molecules in the container. So if initially we start with pressure one, volume one, and some temperature one that's going to be equal to this constant. And if we change any of them, if we go back to pressure two, volume two, temperature two, they should still be equal to this constant, so they equal each other. So for example, let's say I start off with a pressure of 1 atmosphere. and I have a volume of-- I'll switch units here And let's say our temperature is 27 degrees Celsius. Well, and I just wrote Celsius because I want you to always remember you have to convert to Kelvin, so 27 degrees plus 273 will get us exactly to 300 Kelvin. Let's figure out what the new temperature is going to be. Let's say our new pressure is 2 atmospheres. The pressure has increased. Let's say we make the container smaller, so 1 meter cubed. So the container has been decreased by half and the pressure is doubled by half. Actually, no. Let me make the pressure even larger. Let me make the pressure into 5 atmospheres. Now we want to know what the second temperature is, and we set up our equation. And so we have 2/300 atmosphere meters cubed per Kelvin is equal to 5/T2, our new temperature, and then we have 1,500 is equal to 2 T2. Divide both sides by 2. You have T2 is equal to 750 degrees Kelvin, which makes sense, right? We increased the pressure so much and we decreased the volume at the same time that the temperature just had to go up. Or if you thought of it the other way, maybe we increased the temperature and that's what drove the pressure to be so much higher, especially since we decreased the volume." + }, + { + "Q": "At 6:45 sal says that 2 over 300 = 5 over t2. I get that, but then he says that 1500=2t2. how did he get 1500 and how did he get 2t2?wouldn't it be t2?\n", + "A": "If you have 2/300=5/t2, you cross multiply on both sides. So basically, first you multiply both sides by t2, which results in 2t2/300=5. Then you multiply both sides by 300, resulting in 2t2=5x300 which is 2t2=1500.", + "video_name": "GwoX_BemwHs", + "timestamps": [ + 405 + ], + "3min_transcript": "And let's say our temperature is 27 degrees Celsius. Well, and I just wrote Celsius because I want you to always remember you have to convert to Kelvin, so 27 degrees plus 273 will get us exactly to 300 Kelvin. Let's figure out what the new temperature is going to be. Let's say our new pressure is 2 atmospheres. The pressure has increased. Let's say we make the container smaller, so 1 meter cubed. So the container has been decreased by half and the pressure is doubled by half. Actually, no. Let me make the pressure even larger. Let me make the pressure into 5 atmospheres. Now we want to know what the second temperature is, and we set up our equation. And so we have 2/300 atmosphere meters cubed per Kelvin is equal to 5/T2, our new temperature, and then we have 1,500 is equal to 2 T2. Divide both sides by 2. You have T2 is equal to 750 degrees Kelvin, which makes sense, right? We increased the pressure so much and we decreased the volume at the same time that the temperature just had to go up. Or if you thought of it the other way, maybe we increased the temperature and that's what drove the pressure to be so much higher, especially since we decreased the volume. is this pressure went up so much, it went up by factor of five, it went from 1 atmosphere to 5 atmospheres, because on one level we shrunk the volume by a factor of 1/2, so that should have doubled the pressure, so that should have gotten us to two atmospheres. And then we made the temperature a lot higher, so we were also bouncing into the container. We made the temperature 750 degrees Kelvin, so more than double the temperature, and then that's what got us to 5 atmospheres. Now, one other thing that you'll probably hear about is the notion of what happens at standard temperature and pressure. Let me delete all of the stuff over here. Standard temperature and pressure. Let me delete all this stuff that I don't need. Standard temperature and pressure. And I'm bringing it up because even though it's called standard temperature and pressure, and sometimes called STP," + }, + { + "Q": "at 4:41 What is quantum mechanics ?\n", + "A": "Quantum mechanics is basically what chemists and physicists use to describes how subatomic particles behave. You ll learn more about it as you learn more about chemistry :)", + "video_name": "Rd4a1X3B61w", + "timestamps": [ + 281 + ], + "3min_transcript": "that have all of these different properties. So when you think about chemistry, yes, it might visually look something like this. These are obviously much older pictures. But at its essence, it's how do we create models and understand the models that describe a lot of the complexity in the universe around us? And just to put chemistry in, I guess you could say, in context with some of the other sciences, many people would say at the purest level, you would have mathematics. That math, you're studying ideas, which could even be independent, you're seeing logical ideas that could be even independent of anything that you've ever observed or experienced. And a lot of folks that say if we ever communicate with another intelligent species that could be completely different than us, math might be that common language. Because even if we perceive the world differently math might be that common language. But on top of math, we start to say, well how is our reality actually structured? At the most basic level, what are the constituents of matter and what are the mathematical properties that describe how they react together? And then, or interact with each other? Then you go one level above that, you get to the topic of this video, which is chemistry. Which is very closely related to physics. When we talk about these chemical equations and we create these molecular structures, the interactions between these atoms, these are quantum mechanical interactions which we do not fully understand at the deepest level yet. But with chemistry, we can start to make use of the math and they physics to start to think about how all of these different building blocks can interact to explain all sorts of different phenomena. This chemical equation you see right here, This is hydrogen combusting with oxygen to produce a lot of energy. To produce energy. When we imagine combustion, we think of fire. But what even is fire at its most fundamental level? How do we get, why do we perceive this thing here? And chemistry is super important because on top of that, we build biology. We build biology. And as you'll see as you study all of these things, there's points where these things start to bleed together. But the biology in, say, a human being, or really in any species, it's based on molecular interactions. Interactions between molecules, between atoms, which, at the end of the day, is all about chemistry. As I speak, the only reason why I'm able to speak is because of really, hard to imagine the number of chemical interactions happening in me right now to create this soundness. To create this thing that thinks it exists that wants to make a video about how awesome" + }, + { + "Q": "at 2:30 the carbon is more electronegative than hydrogen has been said and it atract the shared pair of electrons\nbut why cant oxygen atom do this inspite o is more electronegative than c and h.\n", + "A": "Oxygen is more electronegative than carbon and hydrogen. It does attract the bonding electrons significantly more than those two.", + "video_name": "rhuYuerbhIE", + "timestamps": [ + 150 + ], + "3min_transcript": "- [Voiceover] In this video, we're going to find the oxidation state of carbon in several different molecules. In earlier video, we've already seen the definition for oxidation state, and also how to calculate it. So let's start with methane, and let's find the oxidation state of carbon and methane. One approach is more of a general chemistry approach where we know that hydrogen usually has an oxidation state of plus one, and we have four hydrogens for a total of plus four. The sum has to be equal to zero, so we know that carbon's oxidation state must be minus four immediately, since we only have one carbon here. So let's go ahead and verify that with our dot structure. So remember, when we're calculating the oxidation state using dot structures, we're thinking about bonding electrons, and we know that each bond consists of two electrons, so we need to put in the bonding electrons for all of our bonds. Next, we think about the oxidation state for carbon, and we start with a number or the number of valence electrons that carbon is supposed to have, and we know carbon is supposed to have four valence electrons, so from that number, we subtract the number of valence electrons in the bonded atom, or the number of valence electrons carbon has in our drawing. But now we need to think about these covalent bonds as being ionic, and so the more electronegative atom is going to take all of the electrons in the bond, so we need to think about electronegativity differences, and we're comparing carbon to hydrogen. So which is more electronegative? We know that carbon is more electronegative than hydrogen, so the two electrons in this bond here, carbon is going to take both of them, so it's winner takes all. Carbon's going to hog those electrons in this bond. All right, same for this next carbon-hydrogen bond. Carbon is more electronegative, so it takes those electrons, and all the way around. by eight electrons right, let's count them up here, one, two, three, four, five, six, seven, eight, so four minus eight is equal to minus four, so we already knew that minus four was going to be the oxidation state for carbon. Let's move on to another molecule here, so C2H4, this is ethene, or ethylene. What's the oxidation state of carbon in this molecule? Well hydrogen should be plus one, and we have four of them for a total of plus four. So the total for carbon should be minus four, because that total has to sum to zero, but this time we have two carbons, so minus four divided by two gives us minus two, each carbon should have an oxidation state of minus two. And let's verify that, let's put in our bonding electrons," + }, + { + "Q": "\nAt 3:58 Sal says a \"slightly negative pH\" i think what he means is a pH slightly below 7, but can you actually have a negative pH?", + "A": "Yes, you can. You can easily have a 10 mol/L solution of a strong acid, so [H+] = 10 mol/L. pH = -log[H+] = -log10 = -1. When acids are very acidic, though, we must use measures of acidity other than pH. You have not yet learned about these, though.", + "video_name": "BBIGR0RAMtY", + "timestamps": [ + 238 + ], + "3min_transcript": "But since it was a strong acid, those conjugate bases don't do anything. They don't add anything to the pH. They're not really basic. The chlorine in hydrogen chloride, the chlorine ion, doesn't change the pH. So this is a strong acid. And this one, when we got to the equivalence point-- when we had used up all of the acid in a solution, and then we hit this in inflection point, where any OH we added was significantly increasing the pH-- when we hit that equivalence point, our pH was already basic. And that's because we had all of the conjugate base of the weak acid, which does make the solution more basic. So this is a weak acid. And in both of these situations, we were increasing the concentration of OH minus. Maybe by adding sodium hydroxide to the solution, a strong base. Now, In these situations, we start with a base, and we add a strong acid to it. Maybe whatever base. We're adding hydrogen chloride, something that will Here, we want to sop up the OH and bring its concentration down, until some point that we have sopped up all of the OH. All of the base is gone. Or most of it is gone. In this situation, we're in a completely neutral situation. So when we sopped up all of the base, we're completely neutral. No basic conjugate bases left. So this is a strong base. And here, the titration, we're increasing the hydrogen solution, or the hydrogen concentration, to sop up all the base. Same thing here. We're sopping up all of the base. We start over here. But over here, the inflection point happens right over here. So we've sopped up all of its base, but some of its conjugate acid is still left over, even after we've sopped up all of its base. So we end up with a slightly negative pH at the equivalence point. So this is a weak base. Let me actually draw that reaction for you. Maybe its A minus is in equilibrium-- that second equilibrium arrow is a little too wild for my blood-- is equilibrium with AH. It grabs hydrogen ions from the surrounding water. Everything is in an aqueous solution. So after you add hydrochloric acid to this-- remember, HcL disassociates completely into hydrogen ions plus chlorine anions. If you add hydrochloric acid to this, these things are going to just completely sop up these things. So we keep sopping up those things. Our concentration of OH goes down and down and down. And as we sop up this, our reaction goes in that direction because Le Chatelier's Principle. More and more of this is going to get formed into this and that. Until some point, we're out of that, and we have a ton of this left." + }, + { + "Q": "at 2:16 what does he mean by PH\n", + "A": "I am struggling with my online Chemistry class and was wondering if someone could help me understand the group numbers. For example what are the group numbers for X and Y?", + "video_name": "BBIGR0RAMtY", + "timestamps": [ + 136 + ], + "3min_transcript": "I've drawn a bunch of titration curves here. So let's see if we can review everything we've learned to kind of have a more holistic understanding of interpreting these things. So the first thing to look at is which of these are the titration of acids versus bases? And everything I've done now is acids, but the logic for base titration is the exact same thing as acid. So for example, these are acid titrations. We start with low pH's. In all of these, this axis is pH. I should have drawn that ahead of time before I asked you the question, but I think you knew that already. So before we add any of the titrator or the reagent, in this reaction, we're starting with a low pH. So this is kind of our starting point. So we have a low pH there. We have a low pH there. So these are both clearly acids. Here, our starting point before we start titrating at all, it's a high pH. So both of these are bases. Let me write that down. These are clearly both bases. Now, we haven't covered bases. But it's the same exact idea. In an acid titration, you start with an acid and you add a strong base to it to sop up all of the acid until all of the acid is sopped up and you hit the equivalence point. You hit the point that all of the acid is sopped up. And now, as you add more and more strong base, you're making it superbasic. So in this acid, our equivalence point is over here. And in this acid, our equivalence point is over here. This is how much solution we had to add to sop up all of the acid. Right there. So given what we already know, which one's a strong acid, which one's a weak acid? Well, this one, when sopped up all of the acid, we have a completely neutral solution. So this must have been a strong acid. Everything has been converted to water in its natural state. pH of 7. And we might have had some neutral leftover conjugate But since it was a strong acid, those conjugate bases don't do anything. They don't add anything to the pH. They're not really basic. The chlorine in hydrogen chloride, the chlorine ion, doesn't change the pH. So this is a strong acid. And this one, when we got to the equivalence point-- when we had used up all of the acid in a solution, and then we hit this in inflection point, where any OH we added was significantly increasing the pH-- when we hit that equivalence point, our pH was already basic. And that's because we had all of the conjugate base of the weak acid, which does make the solution more basic. So this is a weak acid. And in both of these situations, we were increasing the concentration of OH minus. Maybe by adding sodium hydroxide to the solution, a strong base. Now, In these situations, we start with a base, and we add a strong acid to it. Maybe whatever base. We're adding hydrogen chloride, something that will" + }, + { + "Q": "1:21 Sal says that Sulphur's mass number is 32, but on the periodic table it says that the mass number is 32.07. Why?\n", + "A": "We calculate the mass number of one specific sulphur atom, which has 16 neutrons. So, it s mass number is exactly 32. In this big wide world, there exist other isotopes of sulphur too, having different mass numbers. The average mass of all sulphur atoms is 32.07, as calculated by scientists.", + "video_name": "koAFBScR41A", + "timestamps": [ + 81 + ], + "3min_transcript": "- [Narrator] An isotope contains 16 protons, 18 electrons, and 16 neutrons. What is the identity of the isotope? And I encourage you to pause the video and see if you can figure it out and I'll give you a hint, you might want to use this periodic table here. All right, so I'm assuming you've had a go at it. So, an element is defined by the number of protons it has. So if someone tells you the number of protons, you should be able to look at a periodic table and figure out what element they are talking about. So, because it is 16 protons, well we can go right over here to the atomic number, what has 16 protons, well anything that has 16 protons by definition is going to be sulfur right over here. So I could write a big S. Now, the next thing we might want to think about is the mass number of this particular isotope. Remember, an isotope, all sulfur atoms are going numbers of neutrons. So, the sulfurs that have different number of neutrons, those would be different isotopes. So, this case we have 16 protons and we have 16 neutrons, so if you add the protons plus the neutrons together, you're going to get your mass number. So 16 plus 16 is 32. Now let's figure out if there's going to be any charge here. Well, the protons have a positive charge. The electrons have a negative charge. If you have an equal amount of protons and electrons, then you would have no charge. But in this case, we have a surplus of electrons. We have two more electrons than protons and since we have a surplus of the negative charged particles we, and we have two more, we're going to have a negative two charge and we write that as two minus. So this is actually an ion, it has a charge. So this is the isotope of sulfur that has a mass and it has two more electrons than protons which gives it this negative charge. Let's do another example where we go the other way. Where we are told, we are given some information about what isotope and really what ion we're dealing with because this has a negative charge and we need to figure out the protons, electrons, and neutrons. Well, the first thing that I would say is, well look, they tell us that this is fluorine. As soon as you know what element we're dealing with, you know what it's atomic number is when you look at the periodic table and you can figure out the number of protons. Remember, your atomic number is the number of protons and that's what defines the element. That's what makes this one fluorine. So let's go up to the, our periodic table and we see fluorine right over here has an atomic number of nine. That means any fluorine has nine protons." + }, + { + "Q": "\nAt 0:58, Sal mentions that every isotope of Sulfur will have 16 protons.\nAre there exceptions to this rule? Can an isotope have a number of protons different from its' Atomic Number?", + "A": "No, because then it would not be an isotope of sulfur. Every sulfur atom in the universe has 16 protons. If it had say 15 protons then it would be phosphorus instead.", + "video_name": "koAFBScR41A", + "timestamps": [ + 58 + ], + "3min_transcript": "- [Narrator] An isotope contains 16 protons, 18 electrons, and 16 neutrons. What is the identity of the isotope? And I encourage you to pause the video and see if you can figure it out and I'll give you a hint, you might want to use this periodic table here. All right, so I'm assuming you've had a go at it. So, an element is defined by the number of protons it has. So if someone tells you the number of protons, you should be able to look at a periodic table and figure out what element they are talking about. So, because it is 16 protons, well we can go right over here to the atomic number, what has 16 protons, well anything that has 16 protons by definition is going to be sulfur right over here. So I could write a big S. Now, the next thing we might want to think about is the mass number of this particular isotope. Remember, an isotope, all sulfur atoms are going numbers of neutrons. So, the sulfurs that have different number of neutrons, those would be different isotopes. So, this case we have 16 protons and we have 16 neutrons, so if you add the protons plus the neutrons together, you're going to get your mass number. So 16 plus 16 is 32. Now let's figure out if there's going to be any charge here. Well, the protons have a positive charge. The electrons have a negative charge. If you have an equal amount of protons and electrons, then you would have no charge. But in this case, we have a surplus of electrons. We have two more electrons than protons and since we have a surplus of the negative charged particles we, and we have two more, we're going to have a negative two charge and we write that as two minus. So this is actually an ion, it has a charge. So this is the isotope of sulfur that has a mass and it has two more electrons than protons which gives it this negative charge. Let's do another example where we go the other way. Where we are told, we are given some information about what isotope and really what ion we're dealing with because this has a negative charge and we need to figure out the protons, electrons, and neutrons. Well, the first thing that I would say is, well look, they tell us that this is fluorine. As soon as you know what element we're dealing with, you know what it's atomic number is when you look at the periodic table and you can figure out the number of protons. Remember, your atomic number is the number of protons and that's what defines the element. That's what makes this one fluorine. So let's go up to the, our periodic table and we see fluorine right over here has an atomic number of nine. That means any fluorine has nine protons." + }, + { + "Q": "\nAt 0:56, Sal used the word reproduce. What does that mean? I forgot.", + "A": "Reproduction is the process by which a living organism is able to create their own offsprings.", + "video_name": "dQCsA2cCdvA", + "timestamps": [ + 56 + ], + "3min_transcript": "- [Voiceover] I would like to welcome you to Biology at Khan Academy. And biology, as you might now, is the study of life. And I can't really imagine anything more interesting than the study of life. And when I say \"life,\" I'm not just talking about us, human beings. I'm talking about all animals. I'm talking about plants. I'm talking about bacteria. And it really is fascinating. How do we start off with inanimate molecules and atoms? You know, this right here is a molecule of DNA. How do we start with things like that, and we get the complexity of living things? And you might be saying, well, what makes something living? Well, living things convert energy from one form to another. They use that energy to grow. They use that energy to change. And I guess growth is a form of change. They use that energy to reproduce. And these are all, in and of themselves, How do they do this? You know, we look around us. How do we, you know, eat a muffin? And how does that allow us to move around and think and do all the things we do? Where did the energy from that muffin come from? How are we similar to a plant or an insect? And we are eerily or strangely similar to these things. We actually have a lot more in common with, you know, that tree outside your window, or that insect, that bee, that might be buzzing around, than you realize. Even with the bacteria that you can only even see at a microscopic level. In fact, we have so much bacteria as part of what makes us, us. So these are fascinating questions. How did life even emerge? And so over the course of what you see in Biology on Khan Academy, we're going to answer these fundamental, fascinating questions. We're going to think about things like energy and the role of energy in life. We're going to thing about important molecules in biology. DNA and its role in reproduction and containing information. And we're going to study cells, which are the basic building block of life. And as we'll see, even though we view cells as these super, super small, small things, cells in and of themselves are incredibly complex. And if you compare them to an atomic scale, they're quite large. In fact, this entire blue background that I have there, that's the surface of an immune cell. And what you see here emerging from it, these little yellow things. These are HIV viruses, emerging from an immune cell. So even though you imagine cells as these very, very small microscopic things, this incredible complexity. Even viruses. Viruses are one of these fascinating things that kind of are right on the edge between life and nonlife. They definitely reproduce, and they definitely evolve. But they don't necessarily have a metabolism." + }, + { + "Q": "\nAt 5:41pmvideo Biology overview why is biology so important to us humans.", + "A": "It s important because we need to know how life around us works. How the living matters that surround us came to be. It s good to have a basic knowledge down about living matter.", + "video_name": "dQCsA2cCdvA", + "timestamps": [ + 341 + ], + "3min_transcript": "They don't necessarily use energy and growth in the same way that we would associate with life. And then perhaps one of the biggest questions of all is how did life come about? And we will study that as we look at evolution and natural selection. So welcome to Khan Academy's Biology section. I think you're going to find it fascinating. You're going to realize that biology, in some way, is the most complex of the sciences. And in a lot of ways, the one that we understand the least. It's going to be built on top of chemistry, which in turn is built on top of physics, which in turn is built on top of mathematics. And biology is one of our Frankly, even in the last hundred years, we're just starting to scratch the surface of understanding it. But what's really exciting is where the field of biology is going. at a molecular level, we're going to start thinking about how can we even do things like engineer biology, or affect the world around us? It's going to raise all sorts of fascinating and deep and ethical questions. So, hopefully you enjoy this. Biology is one of the most, arguably, maybe the most fascinating subject of all. I don't want to offend the chemists and the physicists out there. I actually find those quite fascinating as well. But we're going to answer, or attempt to start to answer, some of the most fundamental questions of our existence." + }, + { + "Q": "At about 3:00, bacteria is between life and non-life, what would you call non-life?\n", + "A": "Viruses are non-living.", + "video_name": "dQCsA2cCdvA", + "timestamps": [ + 180 + ], + "3min_transcript": "How do they do this? You know, we look around us. How do we, you know, eat a muffin? And how does that allow us to move around and think and do all the things we do? Where did the energy from that muffin come from? How are we similar to a plant or an insect? And we are eerily or strangely similar to these things. We actually have a lot more in common with, you know, that tree outside your window, or that insect, that bee, that might be buzzing around, than you realize. Even with the bacteria that you can only even see at a microscopic level. In fact, we have so much bacteria as part of what makes us, us. So these are fascinating questions. How did life even emerge? And so over the course of what you see in Biology on Khan Academy, we're going to answer these fundamental, fascinating questions. We're going to think about things like energy and the role of energy in life. We're going to thing about important molecules in biology. DNA and its role in reproduction and containing information. And we're going to study cells, which are the basic building block of life. And as we'll see, even though we view cells as these super, super small, small things, cells in and of themselves are incredibly complex. And if you compare them to an atomic scale, they're quite large. In fact, this entire blue background that I have there, that's the surface of an immune cell. And what you see here emerging from it, these little yellow things. These are HIV viruses, emerging from an immune cell. So even though you imagine cells as these very, very small microscopic things, this incredible complexity. Even viruses. Viruses are one of these fascinating things that kind of are right on the edge between life and nonlife. They definitely reproduce, and they definitely evolve. But they don't necessarily have a metabolism. They don't necessarily use energy and growth in the same way that we would associate with life. And then perhaps one of the biggest questions of all is how did life come about? And we will study that as we look at evolution and natural selection. So welcome to Khan Academy's Biology section. I think you're going to find it fascinating. You're going to realize that biology, in some way, is the most complex of the sciences. And in a lot of ways, the one that we understand the least. It's going to be built on top of chemistry, which in turn is built on top of physics, which in turn is built on top of mathematics. And biology is one of our Frankly, even in the last hundred years, we're just starting to scratch the surface of understanding it. But what's really exciting is where the field of biology is going." + }, + { + "Q": "at 6:00\nthe alcohol used is connected to a tertiary carbon. Why does an Sn2 reaction occur and not an Sn1 ?\n\nThank you\n", + "A": "The alcohol used is not connected to a tertiary carbon, it is connected to a secondary carbon. Tertiary carbons are required to be bonded to three other carbon atoms (do not confuse these with the number of bonds they make with atoms that aren t carbon). In the video, the alpha carbon is connected to two other carbons and one oxygen atom. This makes it a secondary carbon. Therefore, SN2 can occur.", + "video_name": "j-rBgs_p-bg", + "timestamps": [ + 360 + ], + "3min_transcript": "So 4 minus 1 gives us an oxidation state of plus 3. So once again, an increase in the oxidation state means oxidation. If you oxidize an aldehyde, you will get a carboxylic acid. Let's look at a secondary alcohol now. All right. So we'll go down here to our secondary alcohol. And once again, identify the alpha carbon-- the one attached to your OH group. We need to have at least one hydrogen on that alpha carbon. We have one right here. If we were to oxidize our secondary alcohol-- so we're going to oxidize our secondary alcohol. Once again, a simple way of doing is thinking-- my alpha carbon has one bond to oxygen. So I could increase that to two bonds, and that should be an oxidation reaction. In the process, I'm going to lose a bond it to my alpha hydrogen. So I'm now going to have two bonds of carbon to oxygen, with the hydrogen there. So that leaves my two alkyl groups. So now I have two alkyl groups. And of course, this would be a ketone functional group. If you oxidize a secondary alcohol, you're going to end up with a ketone. I can assign oxidation states. So once again, let's show that this really is an oxidation reaction here. And I go ahead and put in my electrons on my alpha carbon and think about electronegativity differences. Once again, oxygen beats carbon. Carbon versus carbon is a tie. Carbon versus hydrogen, carbon wins. And carbon versus carbon, of course, is a tie again. Normally, four valence electrons. In this example, it's surrounded by four. 4 minus 4 gives us an oxidation state of 0 for our secondary alcohol. And when I oxidize it, I'm going to get this ketone over here So let's take a look at the oxidation state of the carbon that used to be is now our carbonyl carbon. Once again, we put in our electrons. And we think about electronegativity difference. Right? So oxygen is going to beat carbon. So we go like that. Carbon versus carbon is a tie. Carbon versus carbon is a tie. Once again, it's normally 4. Minus 2 this time around that carbon, giving us an oxidation state of plus 2. So to go from a secondary alcohol to a ketone, we see there's an increase in the oxidation state. So this is definitely an oxidation reaction. Let's look now at a tertiary alcohol. So here is my tertiary alcohol. And when I find my alpha carbon, I see that this time there are no hydrogen bonded to my alpha carbon. According to the mechanism-- which we'll see in a minute-- there's no way we can oxidize this tertiary alcohol under normal conditions anyway. If we attempted to oxidize this, we would say there's no reaction here" + }, + { + "Q": "How are you going to know which six month period to choose in order to have an isosceles triangle?At 1:49 he mentions finding the maximum parallax angle that the star makes during the year.Doesn't that require daily observation for a year?\n", + "A": "isn t it just the 2 different 6 month periods? aren t there only 2?", + "video_name": "6FP-hLuAlr4", + "timestamps": [ + 109 + ], + "3min_transcript": "I got a comment on the video where we first introduced parallax, especially relative to stars. Essentially asking, how do we know that this angle and this angle is always the same? Or how do we know that we're always looking at an isosceles triangle, where this side is equal to this side? It worked out for this example that I drew right here. But what if the star was over here? What if the star was over here. Then if you just look at it this way. If you take at this point, the triangle is no longer, it's clearly no longer, an isosceles triangle. It looks more like a scalene triangle, I guess, where all of the sides are different. And so a lot of that trigonometry won't apply. Because we won't be able to assume that this is a right triangle over here. And what I want to make clear is that that is true. You would not be able to pick these two points during the year. These two points in our orbit six months apart, in order to do the same math that we did in the last video. In order to calculate this and still is pick two different points six months apart. So you want to do is if this is the sun, you want to pick two different points six months apart, where it does form an isosceles triangle. So if this is the distance from the sun to this other star right over here, you want to pick a point in Earth's orbit around the sun here. And then another point in the orbit six months later, which would put us right over here. And if you do that, then we are, now all of a sudden, we are looking at two right triangles, if we pick those periods correctly. And the best way to think about whether this is a perpendicular angle, is you're going to try to find the maximum parallax from center in each of these time periods. Here it's going to be maximally shifted in one direction. And then when you go to this six months later, it's going to be maximally shifted in the other direction. So to answer that question, the observation is right. all stars will not form an isosceles triangle with the sun and the earth. But you could pick other points in time around the year six months apart where any star will form an isosceles triangle. Hopefully you found that helpful." + }, + { + "Q": "\nAt 4:55, you mention prothrombin becoming thrombin, but this isn't mentioned again as you go through the cascade. Is prothrombin one of the Roman Numerals in the cascade or how does it fit in?", + "A": "Yes, it is one of the numerals in the cascade. Prothrombin is Clotting Factor II, and Thrombin is IIa (activated factor 2). For whatever reason, most people use prothrombin/thrombin rather than calling it Factor 2 / Factor 2a.", + "video_name": "FNVvQ788wzk", + "timestamps": [ + 295 + ], + "3min_transcript": "how does your body know to convert fibrinogen into fibrin at the site of injury? The answer is that when you injure your endothelium here, you're going to expose your blood to new proteins. And maybe your actual endothelial cells will release some proteins because they're damaged. So basically, you have new proteins that weren't seen before and that are seen now. Those proteins will eventually cause fibrinogen to turn into fibrin. So while evolution was designing us, it could've said, let's use these little yellow guys to convert fibrinogen to fibrin. That might've worked, but it's actually not the most efficient way to do things. The reason is, imagine if you and a couple of friends have a huge amount of work to do. Let's say you need to convert a million fibrinogen to fibrin. Is the best way to do it, to actually sit down and crank it out? Or would it be more efficient to have each of you And ask those friends to each call five friends. And ask those friends to each call five friends. Well, obviously that would get the job done much faster assuming you had those friends. That's also what your body does. So actually, it doesn't use these yellow guys to convert fibrinogen to fibrin. There's another player which does that, and it's an important one, so we'll give it a little drawing like that, and it is called thrombin. Thrombin, just like fibrinogen, is activated from an inactive form, which we call prothrombin. The prothrombin has a little piece on the end that prevents it from working, so this is prothrombin. That piece is removed when you want to get to work. Well, actually, he's not either because the chain of amplification is much longer than that. To draw the actual amplification cascade, you just need to see it and practice. But there is an easier way to draw it than it usually is drawn, so we'll do that now. So let's say you were counting down from XII. Normally you would start at XII. You'd go to XI. You'd go to X, and then you'd go to IX. But let's say that you weren't very good at counting. You would start with XII. You'd go to XI. Then you'd make a little mistake - you'd go to IX. Then you'd realize you forgot X, so you'd go to X. Now it's good that you remembered X because X is a big deal, and he's going to help bring us thrombin. Turns out that thrombin is also known as II. We know very well that thrombin helps give us fibrin," + }, + { + "Q": "if this sn2 mechanism only works in 1* or 2* alcohols, why is he starting with tert butanol? at 0:00?\n", + "A": "It s not impossible for SN2 to happen to tertiary alcohols, it s just slow. There is a strong nucleophile and we have formed a good leaving group so it is possible.", + "video_name": "LccmkSz-Y-w", + "timestamps": [ + 0 + ], + "3min_transcript": "In this video we're going to see how to prepare alkyl halides from alcohols. And so if we start with this alcohol over here on the left, and we add SOCl2, which is called thionyl chloride, and pyridine to it. We're going to substitute a chlorine atom for the OH group. And this mechanism occurs via an SN2 type mechanism, which means that it's only going to work with primary or secondary alcohols. And you will get inversion of configuration if you have a chirality center present in your final product. So let's take a look at the mechanism. And we'll start with our alcohol. And so the oxygen is going to have to leave somehow. But by itself OH is not the best leaving group. And so we're going to react this alcohol with thionyl chloride to convert it into a better leaving group. And so if we draw the dot structure for thionyl chloride, we would have to sulfur double bonded to an oxygen here. And then the sulfur is also bonded to chlorine. So I'll go ahead and put in those lone pairs of electrons on the chlorines, like this. it turns out you need two more. And those go on the sulfur. It's OK for sulfur to violate the octet rule, since it is in the third period now. So a lone pair of electrons on oxygen is going to form a bond with our sulfur atom, which would therefore kick these electrons in here off onto the top oxygen. So if we go ahead and draw what we get from that first step of our mechanism, now our oxygen is bonded to our sulfur. The oxygen is also a bonded to a hydrogen. One lone pair of electrons formed that new bond, so one pair of electrons is left behind. Which would give this oxygen a plus 1 formal charge. Connected to the sulfur, this top oxygen here had two lone pairs of electrons. Picked up one more lone pair, which gives it a negative 1 formal charge. And this sulfur is still bonded to chlorine. So we can go ahead and draw those in. And we can go ahead and draw that lone pair of electrons on that sulfur like that. And so in the next step of the mechanism, we're going to reform the double bond between oxygen and sulfur. And these electrons would kick off on to that chlorine. So when we draw the next intermediate here, we would now have our oxygen, still bonded to a hydrogen, still with a plus 1 formal charge like that. And now our sulfur is double bonded to our oxygen again with two lone pairs of electrons on the oxygen. The sulfur is also bonded to one chlorine now, so one of the chlorines left. And we can go ahead and draw in that chlorine. So one of the chlorines left here. It's a negatively charged chloride anion. And then still there's a lone pair of electrons on our sulfur like that. So at this part of the mechanism, the pyridine comes along. So if we go ahead and draw the dot structure for pyridine. It's a base, and so it looks like a benzene ring, except we have a nitrogen here instead. And there'd be a lone pair of electrons on this nitrogen. And so that lone pair of electrons" + }, + { + "Q": "Why does the formation of an alkyl chloride (at 3:00) require a base , while the formation of an alkyl bromide (at 8:00) doesn't?\n", + "A": "Pyridine is required in chlorination of alcohols by SOCl2 to nuetralise the HCl produced during the reaction. By the way, preparation of alkyl chloride by SOCl2 can also take place in abscence of pyridine via sNi mechanism{look up in wikipedia}.", + "video_name": "LccmkSz-Y-w", + "timestamps": [ + 180, + 480 + ], + "3min_transcript": "it turns out you need two more. And those go on the sulfur. It's OK for sulfur to violate the octet rule, since it is in the third period now. So a lone pair of electrons on oxygen is going to form a bond with our sulfur atom, which would therefore kick these electrons in here off onto the top oxygen. So if we go ahead and draw what we get from that first step of our mechanism, now our oxygen is bonded to our sulfur. The oxygen is also a bonded to a hydrogen. One lone pair of electrons formed that new bond, so one pair of electrons is left behind. Which would give this oxygen a plus 1 formal charge. Connected to the sulfur, this top oxygen here had two lone pairs of electrons. Picked up one more lone pair, which gives it a negative 1 formal charge. And this sulfur is still bonded to chlorine. So we can go ahead and draw those in. And we can go ahead and draw that lone pair of electrons on that sulfur like that. And so in the next step of the mechanism, we're going to reform the double bond between oxygen and sulfur. And these electrons would kick off on to that chlorine. So when we draw the next intermediate here, we would now have our oxygen, still bonded to a hydrogen, still with a plus 1 formal charge like that. And now our sulfur is double bonded to our oxygen again with two lone pairs of electrons on the oxygen. The sulfur is also bonded to one chlorine now, so one of the chlorines left. And we can go ahead and draw in that chlorine. So one of the chlorines left here. It's a negatively charged chloride anion. And then still there's a lone pair of electrons on our sulfur like that. So at this part of the mechanism, the pyridine comes along. So if we go ahead and draw the dot structure for pyridine. It's a base, and so it looks like a benzene ring, except we have a nitrogen here instead. And there'd be a lone pair of electrons on this nitrogen. And so that lone pair of electrons and take this proton here on the oxygen. And that would kick these electrons back off onto this oxygen. So when we go ahead and draw that-- let's go ahead and get some more room here-- so what would we get? We would now have our carbon bonded to our oxygen. Our oxygen now has two lone pairs of electrons around it. And we have our sulfur, and our chlorine, and our lone pair of electrons on the sulfur. And now we've made a better leaving group. So this is a better leaving group than the OH was in the beginning. And if we think about an SN2 type mechanism now, we know that the bond between carbon and oxygen is polarized, right? Oxygen being more electronegative, it will be partially negative. And this carbon here be partially positive. And so now we can think about our SN2 type mechanism. Our nucleophile will be this chloride anion up here that we formed in the mechanism." + }, + { + "Q": "From 6:33 to 6:46, Sal talks about how light takes 8 minutes to hit the earth from the sun. At 6:42 he says that if the sun were to disappear, it would take 8 minutes for the people on Earth to know that light disappeared. Say the sun did disappear. What would happen in terms of gravity? Would we feel the absence of its gravitational effects right away? If not, why? And if so, wouldn't that mean the effects were faster than the speed of light?\n", + "A": "gravity travels at the speed of light", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 393, + 406, + 402 + ], + "3min_transcript": "than a raindrop. If I were to draw it on this scale, where the sun is even smaller, the earth would be about that big. Now, what isn't obvious, because we've all done our science projects in third and fourth grade--or we always see these diagrams of the solar system that look something like this-- is that these planets are way further away. Even though these are depicted to scale, they're way further away from the sun than this makes it look. So the earth is 150 million kilometers from the sun. So if this is the sun right here, at this scale you wouldn't even be able to see the earth. It wouldn't even be a pixel. But it would be 150 million kilometers from the earth. and we'll be using that term in the next few videos just because it's an easier way to think about distance-- sometimes abbreviated AU, astronomical unit. And just to give a sense of how far this is, light, which is something that we think is almost infinitely fast and that is something that looks instantaneous, that takes eight minutes to travel from the sun to the earth. If the sun were to disappear, it would take eight minutes for us to know that it disappeared on earth. Or another way, just to put it in the sense of this jet airplane-- let's get the calculator back out. So we're talking about 150 million kilometers. it would take us 150,000 hours at the speed of a bullet or at the speed of a jet plane to get to the sun. And just to put that in perspective, if we want it in days, there's 24 hours per day. So this would be 6,250 days. Or, if we divided by 365, roughly 17 years. If you were to shoot a bullet straight at the sun it would take 17 years to get there, if it could maintain its velocity somehow. So this would take a bullet or a jet plane 17 years to get to the sun. Or another way to visualize it-- this sun right over here, on my screen it has about a five- or six-inch diameter. If I were to actually do it at scale, this little dot right here, which is the earth, this" + }, + { + "Q": "6:35 Sal says if the sun disappeared it would take 8 minutes for us to notice it. Would the sun's gravitational effects travel as quickly were it to somehow disappear? Does gravity propagate as quickly as light?\n", + "A": "Yes the effect of gravity propagates through space at the speed of light.", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 395 + ], + "3min_transcript": "than a raindrop. If I were to draw it on this scale, where the sun is even smaller, the earth would be about that big. Now, what isn't obvious, because we've all done our science projects in third and fourth grade--or we always see these diagrams of the solar system that look something like this-- is that these planets are way further away. Even though these are depicted to scale, they're way further away from the sun than this makes it look. So the earth is 150 million kilometers from the sun. So if this is the sun right here, at this scale you wouldn't even be able to see the earth. It wouldn't even be a pixel. But it would be 150 million kilometers from the earth. and we'll be using that term in the next few videos just because it's an easier way to think about distance-- sometimes abbreviated AU, astronomical unit. And just to give a sense of how far this is, light, which is something that we think is almost infinitely fast and that is something that looks instantaneous, that takes eight minutes to travel from the sun to the earth. If the sun were to disappear, it would take eight minutes for us to know that it disappeared on earth. Or another way, just to put it in the sense of this jet airplane-- let's get the calculator back out. So we're talking about 150 million kilometers. it would take us 150,000 hours at the speed of a bullet or at the speed of a jet plane to get to the sun. And just to put that in perspective, if we want it in days, there's 24 hours per day. So this would be 6,250 days. Or, if we divided by 365, roughly 17 years. If you were to shoot a bullet straight at the sun it would take 17 years to get there, if it could maintain its velocity somehow. So this would take a bullet or a jet plane 17 years to get to the sun. Or another way to visualize it-- this sun right over here, on my screen it has about a five- or six-inch diameter. If I were to actually do it at scale, this little dot right here, which is the earth, this" + }, + { + "Q": "3:13 109 times the circumfrence. Is that 109 times bigger/wider/taller (Pretty much is the sun 109 x, y, z or all of the above then the earth)\n", + "A": "Probably bigger.", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 193 + ], + "3min_transcript": "depending on the type of gun and all of that-- about 280 meters per second, which is about 1,000 kilometers per hour. And this is also roughly the speed of a jet. So just to give a sense of scale here, the earth's circumference-- so if you were to go around the planet-- is about 40,000 kilometers. So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the earth. You might have taken a 12- or 15-hour flight that gets you-- not all the way around the earth-- but gets you pretty far. San Francisco to Australia, or something like that. So right now these aren't scales that are too crazy. Although, even for me, the earth itself is a pretty mind-blowingly large object. Now, with that out of the way let's think about the sun. Because the sun starts to approach something far huger. So this obviously here is the sun. And I think most people appreciate that the sun is much larger than the earth, and that it's pretty far away from the earth. But I don't think most people, including myself, fully appreciate how large the sun is or how far it is away from the earth. So just to give you a sense, the sun is 109 times the circumference of the earth. if we said, OK, if I'm traveling at the speed of a bullet or the speed of a jetliner, it would take me 40 hours to go around the earth. Well, how long would it take to go around the sun? So if you were to get on a jet plane and try to go around the sun, or if you were to somehow ride a bullet and try to go around the sun-- do a complete circumnavigation of the sun-- it's going to take you 109 times as long as it would have taken you to do the earth. So it would be 100 times-- I could do 109, but just for approximate-- it's roughly 100 times the circumference of the earth. So 109 times 40 is equal to 4,000 hours. And just to get a sense of what 4,000 is-- actually, since I have the calculator out, let's do the exact calculation. It's 109 times the circumference of the earth times 40 hours." + }, + { + "Q": "At 1:42, how did we find the circumference of the Earth?\n", + "A": "I think I read somewhere in the physics textbook that you can approximate the radius of Earth by using two point of surface to the center of earth and use basic phytagoras to solve it. And yeah, after finding the radius you can easily find the circumference since Earth is nearly round. So, it s possible to find it with math, especially with geometry.", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 102 + ], + "3min_transcript": "My goal in this video and the next video is to start giving a sense of the scale of the earth and the solar system. And as we see, as we start getting into to the galaxy and the universe, it just becomes almost impossible to imagine. But we'll at least give our best shot. So I think most of us watching this video know that this right here is earth. And just to get a sense of scale here, I think probably the largest distance that we can somehow relate to is about 100 miles. You can get into a car for an hour, hour and a half, and go about 100 miles. And on the earth that would be about this far. It would be a speck that would look something like that. That is 100 miles. And also to get us a bit of scale, let's think about a speed that at least we can kind of comprehend. And that would be, maybe, the speed of a bullet. Maybe we can't comprehend it, but I'll say this is the fastest thing that we could maybe comprehend. depending on the type of gun and all of that-- about 280 meters per second, which is about 1,000 kilometers per hour. And this is also roughly the speed of a jet. So just to give a sense of scale here, the earth's circumference-- so if you were to go around the planet-- is about 40,000 kilometers. So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the earth. You might have taken a 12- or 15-hour flight that gets you-- not all the way around the earth-- but gets you pretty far. San Francisco to Australia, or something like that. So right now these aren't scales that are too crazy. Although, even for me, the earth itself is a pretty mind-blowingly large object. Now, with that out of the way let's think about the sun. Because the sun starts to approach something far huger. So this obviously here is the sun. And I think most people appreciate that the sun is much larger than the earth, and that it's pretty far away from the earth. But I don't think most people, including myself, fully appreciate how large the sun is or how far it is away from the earth. So just to give you a sense, the sun is 109 times the circumference of the earth." + }, + { + "Q": "\nabout 10:20, route 19.6h m^2/s^2 = route velocity(f)^2\nbut it turns out velocity(f) = - route 19.6h m/s. Where - < negative sign come from?", + "A": "he explains it a about 8:55. It s because you took a square root (which has positive and negative solutions) and you want the one that goes in the downward direction, which you ve defined as negative", + "video_name": "2ZgBJxT9pbU", + "timestamps": [ + 620 + ], + "3min_transcript": "Our convention is very important here. So our displacement over here is going to be negative h meters. So this is the variable, and this is the shorthand for meters. So when you multiply these two things out, lucky for us these negatives cancel out, and you get 19.6h meters squared per second squared is equal to our final velocity squared. And notice, when you square something you lose the sign information. If our final velocity was positive, you square it, you still get a positive value. If it was negative and you square it, you still get a positive value. But remember, in this example, we're going to be moving downward. So we want the negative version of this. So to really figure out our final velocity, of both sides of this equation. So if we were to take the square root of both sides of this, you take the square root of that side, you take the square root of that side, you will get-- and I'll flip them around-- your final velocity, we could say, is equal to the square root of 19.6h. And you can even take the square root of the meter squared per second squared, treat them almost like variables, even though they're units. And then outside of the radical sign, you will get a meters per second. And the thing I want to be careful here is if we just take the principal root here, the principal root here is the positive square root. But we know that our velocity is going to be downwards here, because that is our convention. So we want to make sure we get the negative square root. So let's try it out with some numbers. We've essentially solved what we set out to solve at the beginning of this video, how fast would we be falling, as a function of the height. Well, let's try it out with something. let's say the height is 5 meters, which would be probably jumping off of a or throwing a rock off of a one-story, maybe a commercial one-story building. That's about 5 meters, would be about 15 feet. So yeah, about the roof of a commercial building, give or take. So let's turn it on. And so what do we get? If we put 5 meters in here, we get 19.6 times 5 gives us 98. So almost 100. And then, we want to take the square root of that, so it's going to be almost 10. So the square root of 98 gives us roughly 9.9. And we want the negative square root of that. in that situation, when the height is 5 meters-- So if you jump off of a one-story commercial building, right at the bottom, or if you throw a rock off that, right" + }, + { + "Q": "\nAt 9:45 Sal says that we need to take the negaitve square root of 19.6*h m/s. I don\u00c2\u00b4t quite understand why he does this. Is it because our convention was that downward vectors have to have a negative algebraic sign? Thank you for your answers!", + "A": "You can define direction however you want. If you make up positive, then down is negative, and you just have to stay consistent throughout the problem. So if he defined up as positive earlier in the problem, then down has to be negative.", + "video_name": "2ZgBJxT9pbU", + "timestamps": [ + 585 + ], + "3min_transcript": "Our convention is very important here. So our displacement over here is going to be negative h meters. So this is the variable, and this is the shorthand for meters. So when you multiply these two things out, lucky for us these negatives cancel out, and you get 19.6h meters squared per second squared is equal to our final velocity squared. And notice, when you square something you lose the sign information. If our final velocity was positive, you square it, you still get a positive value. If it was negative and you square it, you still get a positive value. But remember, in this example, we're going to be moving downward. So we want the negative version of this. So to really figure out our final velocity, of both sides of this equation. So if we were to take the square root of both sides of this, you take the square root of that side, you take the square root of that side, you will get-- and I'll flip them around-- your final velocity, we could say, is equal to the square root of 19.6h. And you can even take the square root of the meter squared per second squared, treat them almost like variables, even though they're units. And then outside of the radical sign, you will get a meters per second. And the thing I want to be careful here is if we just take the principal root here, the principal root here is the positive square root. But we know that our velocity is going to be downwards here, because that is our convention. So we want to make sure we get the negative square root. So let's try it out with some numbers. We've essentially solved what we set out to solve at the beginning of this video, how fast would we be falling, as a function of the height. Well, let's try it out with something. let's say the height is 5 meters, which would be probably jumping off of a or throwing a rock off of a one-story, maybe a commercial one-story building. That's about 5 meters, would be about 15 feet. So yeah, about the roof of a commercial building, give or take. So let's turn it on. And so what do we get? If we put 5 meters in here, we get 19.6 times 5 gives us 98. So almost 100. And then, we want to take the square root of that, so it's going to be almost 10. So the square root of 98 gives us roughly 9.9. And we want the negative square root of that. in that situation, when the height is 5 meters-- So if you jump off of a one-story commercial building, right at the bottom, or if you throw a rock off that, right" + }, + { + "Q": "Is the formula shown at 9:00 applicable for every free falling projectile dropped with zero initial velocity? Vf=(-)sqrt(19.6h)\n", + "A": "Yes. As long as the air resistance is neglected and your g = 9.8m/s^2, you can use the kinematic equations for uniform motion. Here for calculating final velocity: v^2 - u^2 = 2*g*h if u = 0 and downward direction is taken as positive, then v = sqrt(2*9.8*h). Note that the final velocity doesn t depend on the mass and dimensions of the body. As long as your projectile can be approximated as a point mass particle, you are good to go.", + "video_name": "2ZgBJxT9pbU", + "timestamps": [ + 540 + ], + "3min_transcript": "So let me write this over here. So this is negative 9.8. So we have 2 times negative 9.8-- let me just multiply that out. So that's negative 19.6 meters per second squared. And then what's our displacement going to be? What's the displacement over the course of dropping this rock off of this ledge or off of this roof? So you might be tempted to say that our displacement is h. But remember, these are vector quantities, so you want to make sure you get the direction right. From where the rock started to where it ends, what's it doing? It's going to go a distance of h, but it's going to go a distance of h downwards. And our convention is down is negative. So in this example, our displacement from when it leaves your hand to when it hits the ground, the displacement is going to be equal to negative h. It's going to travel a distance of h, but it's going to travel that distance downwards. Our convention is very important here. So our displacement over here is going to be negative h meters. So this is the variable, and this is the shorthand for meters. So when you multiply these two things out, lucky for us these negatives cancel out, and you get 19.6h meters squared per second squared is equal to our final velocity squared. And notice, when you square something you lose the sign information. If our final velocity was positive, you square it, you still get a positive value. If it was negative and you square it, you still get a positive value. But remember, in this example, we're going to be moving downward. So we want the negative version of this. So to really figure out our final velocity, of both sides of this equation. So if we were to take the square root of both sides of this, you take the square root of that side, you take the square root of that side, you will get-- and I'll flip them around-- your final velocity, we could say, is equal to the square root of 19.6h. And you can even take the square root of the meter squared per second squared, treat them almost like variables, even though they're units. And then outside of the radical sign, you will get a meters per second. And the thing I want to be careful here is if we just take the principal root here, the principal root here is the positive square root. But we know that our velocity is going to be downwards here, because that is our convention. So we want to make sure we get the negative square root. So let's try it out with some numbers. We've essentially solved what we set out to solve at the beginning of this video, how fast would we be falling, as a function of the height. Well, let's try it out with something." + }, + { + "Q": "at 0:29 sal says cellular membrane is semi permeable. But actually it is selectively permeable. *why does he say so?*\n", + "A": "Both semi permeable and selectively permeable refer to the same thing that the cellular membrane only allows certain substances (molecules or ions) to pass through it. Usually, the solvent molecules can pass through however certain solute molecules can not. That s why both terms can be used.", + "video_name": "afWnU10ZNfg", + "timestamps": [ + 29 + ], + "3min_transcript": "- [Voiceover] I have three different scenarios here of a cell being immersed in a solution, and the cell is this magenta circle, that's the cellular membrane. I have the water molecules depicted by these blue circles, and then, I have the solute inside of the solution, inside of the water solution that we depict with these yellow circles. I've clearly exaggerated the size of the water molecules and the solute particles relative to the size of the cell, but I did that so that we can visualize what's actually going on. We're going to assume that the cellular membrane, this phospholipid bilayer, is semipermeable, that it will allow water molecules to pass in and out, so a water molecule could go from the inside to the outside, or from the outside to the inside, but we're gonna assume that it does not allow the passage of the solute particles, so that's why it's semipermeable. It's permeable to certain things, or we could say, selectively permeable. Now, what do we think is going to happen? Well, the first thing that you might observe is we have a lower concentration of solute on the outside than we have on the inside, some water molecules moving in just the right direction to go from the outside to the inside, and you will also have some water molecules that might be in just the right place to go from the inside to the outside, but what's more likely to happen, and what's going to happen more over a certain period of time? The water molecules that are on the outside, and we talk about this in the osmosis video, they're going to be less obstructed by solute particles. If this one happens to be moving in that direction, well, it's gonna make its way to the membrane, and then, maybe get through the membrane, while something, maybe, if this water molecule was moving in this direction, well, gee, it's gonna be obstructed now, maybe this is bouncing back, and it's gonna ricochet off of it, so the water molecules on the inside are more obstructed. They're less likely to be able to fully interact with the membrane or move in the right direction. They're being obstructed by these solute particles. Even though you're going to have water molecules going back and forth, in a given period of time, you have a higher probability so you're going to have a net inflow. Net inflow of H2O, of water molecules. Now, a situation like this, where we're talking about a cell and it's in a solution that has a lower concentration of solute, it's important that we're talking about a solute that is not allowed to go to the membrane, the membrane is not permeable to that solute. We call this type of situation, this type of solution that the cell is immersed in, we call this a hypotonic solution. Hypotonic solution. Anytime we're talking about hypotonic, or as we'll see, isotonic and hypertonic, we're talking about relative concentrations of solute that cannot get through some type of a membrane. The word hypo, you might've seen it in other things. It's a prefix that means less of something, so in this case, we have a lower concentration of solute in the solution" + }, + { + "Q": "\nAt 4:49, how do you know which ones are Adenine?", + "A": "He assigned random bases as Adenine (not accurate in the sense that those were actually Adenine) then assigned Thymine to complement Adenine.", + "video_name": "AmOO4j0E408", + "timestamps": [ + 289 + ], + "3min_transcript": "Well, the word deoxyribonucleic acid comes from the fact that this backbone is made up of a combination of sugar and phosphate. And the sugar that makes up the backbone is deoxyribose. So that's essentially the D in DNA. And then the phosphate group is acidic and that's now where you get the acid part of it. And nucleic is, hey this was found in nuclei of cells. It is nucleic acid. Deoxyribonucleic acid. It is actually mildly acidic all in total but for every acid it actually also has a base, and those bases form the rung of the ladders. And actually each rung is a pair of bases and as I said, that's where the information is actually stored. Well what am I talking about? Well let me talk about the four different bases that make up the rungs of a DNA molecule. So, you have adenine. And so for example, this part right over here. This section of that rung might be adenine. Maybe this right over here is adenine. This right over here. Remember, each of these rungs are made up by it's a pair of bases. And that might be adenine. Maybe this is adenine and I could stop there, I mean I'll do a little more adenine. Maybe that's adenine right over there. And adenine always pairs with the base thymine. So let me write that down. So adenine pairs with thymine. Thymine. So, if that's an adenine there then this is going to be a thymine. If this is an adenine then this is going to be a thymine. Or if I drew the thymine first, well say, okay it's gonna pair with the adenine. So this is going to be a thymine right over here. This is going to be a thymine. If I were to draw this, this would be a thymine right over here. Now the other two bases, you have cytosine which pairs with guanine So guanine and we're not gonna go into the molecular structure of these bases just yet, although these are good names to know because they show up a lot and they really form kind of the code, your genetic code. Guanine. Guanine pairs with cytosine. Guanine and cytosine. Cytosine. So actually if this is, let's say there's some cytosine there, let's say cytosine right over here. Maybe this is a cytosine, maybe this is cytosine, maybe this is cytosine, this is cytosine and maybe this is cytosine. Then it always pairs with the guanine. So, let's see, this is guanine then and this will be guanine. This is guanine, this is guanine. I actually didn't draw stuff here. This is guanine, I didn't say what these could be but these would be maybe the pairs of they could be adenine-thymine pairs and it could be adenine on either side" + }, + { + "Q": "at 6:59 what if your characteristics are nothing like you parents and does this mean we all come from two people and if so why are we all so different?\n", + "A": "Characteristics don t always have to be like your parents, sometimes you find that people have eyes like their grandmother or their uncle s nose. Remember that mutations do occur, and the process of crossing over contributes to genetic variety. These characteristics are recessive. The idea that we all come 2 people is biblical and not necessarily scientific.", + "video_name": "AmOO4j0E408", + "timestamps": [ + 419 + ], + "3min_transcript": "So guanine and we're not gonna go into the molecular structure of these bases just yet, although these are good names to know because they show up a lot and they really form kind of the code, your genetic code. Guanine. Guanine pairs with cytosine. Guanine and cytosine. Cytosine. So actually if this is, let's say there's some cytosine there, let's say cytosine right over here. Maybe this is a cytosine, maybe this is cytosine, maybe this is cytosine, this is cytosine and maybe this is cytosine. Then it always pairs with the guanine. So, let's see, this is guanine then and this will be guanine. This is guanine, this is guanine. I actually didn't draw stuff here. This is guanine, I didn't say what these could be but these would be maybe the pairs of they could be adenine-thymine pairs and it could be adenine on either side and they could be made of guanine-cytosine pairs where the guanine or the cytosine is on the other side. Actually just to make it a little bit more complete let me just color in the rungs here as best as I can. So those are guanines so they're gonna pair with cytosine. Pair with cytosine, pair with cytosine. When you straw in this way you might start to see how this is essentially a code, the order of which the bases are... I guess the order in which we have these or the sequence of these bases essentially in code the information that make you, you, and you could be. Well how much of it is nature versus nurture and when people say nature, you know, it's literally genetic, and that's an ongoing debate, an ongoing debate but it does code for things like your hair color. When you see that your smile is similar to your parents it is because that information to a large degree is encoded genetically. It affects a lot of what makes you you and actually not even just within a species Humans have more genetic material in common with other humans than they do with say a plant. But all living creatures as we know them have genetic information. This is the basis by which they are passing down their actual traits. Now you might be saying well, how much genetic information does a human being have? And the number will either disappoint you or you might find it mind-boggling. The human genome and every species has a different number of base pairs to large degree correlated with how complex they are although not always. But the human genome has 6,000,000. Sorry, not 6,000,000, 6,000,000,000. 6,000,000 would be disappointing, even billion might be disappointing. 6,000,000,000 base pairs. 6,000,000,000. 6,000,000,000 base pairs. And when you have your full complement of chromosomes and this is in most of the cells in your body and outside of your sex cells," + }, + { + "Q": "12:00 Why are Adenine pairing with Thymine and Guanine with Cytosine ??\n", + "A": "It is a direct result of the molecular structure of the bases, A/T only have 2 side groups which can form hydrogen bonds whereas C/G have 3.", + "video_name": "AmOO4j0E408", + "timestamps": [ + 720 + ], + "3min_transcript": "kind of the scale of this thing. This is a very dense way to actually store information. But just to have an appreciation of and you might have seen it when I was coloring in on why the structure lends itself to being able to replicate the information or even to be able to translate or express the information. Let's think about if you were to take this ladder and you were to just kind of split all the base pairs. So, you just have 1/2 of them. So you essentially have half of the ladder. And so if you only have half of the ladder, you're able to construct the other half of the ladder. Let's take an example, let's say and I'll just use the first letter to abbreviate for each of these bases. Let's say you have some... So let's say this is one of the, this is the sugar phosphate backbone right over here. So this could be one of the sides. Let's say there's some adenine. Actually we do in the right color. So you got some adenine, adenine. Maybe some adenine right over here and maybe there's an adenine there. And maybe you have some thymine, thymine, and then you have some guanine, guanine, guanine. And then let's say you have some cytosine and you have some cytosine. So with just half of this ladder I guess you could say, you're able to construct the other half, and this is actually how DNA replicates. This ladder splits and then each of those two halves of that ladder are able to construct versions of the other half, or versions of the other half are able to constructed on top of that, on top of that half. So how does that happen? Well, it's based on how these bases pair. Adenine always pairs with thymine if we're talking about DNA. So if you have an A there, you're gonna have a T on this end, T on this end. T's right all over here, T right over there. If you have a T on that end you're gonna have an A right over there. A, A. If you have a G, a guanine on this side, you're gonna have a cytosine on the other side. And if you have a cytosine you're gonna have a guanine on the other side. Hopefully that gives you an appreciation of how DNA can replicate itself. And as we'll see also how this information can be translated to other forms of either related molecules but eventually to proteins. And just to kind of round out this video, to get a real visual sense what the DNA molecule looks like or I guess a different visual depiction from this. I found this animated gif that, you know, if you haven't fully digested what a double helix looks like, this is it. And you see here, you see your sugar phosphate bases here. You see kind of the sugars and phosphate, the sugars and the phosphates alternating along this backbone, and then the rungs of the ladder are these base pairs. So this is one of the bases, that's the corresponding, that's this corresponding, I guess you can say partner. And you can see that along all the way up and down in this molecule. Very exciting." + }, + { + "Q": "at 9:50 you said that the carbon dioxide will diffuse across and into the plasma, and that some of the carbon dioxide can make its way across the membrane into the red blood cell itself where it will be converted into carbonic acid, so does only some of the carbon dioxide get converted into carbonic acid and the rest remain in the plasma? because i thought that all the carbon dioxide gets converted into carbonic acid?\n", + "A": "thank you, but does it travel in the blood as carbonic acid or as carbon dioxide?", + "video_name": "LWtXthfG9_M", + "timestamps": [ + 590 + ], + "3min_transcript": "the small capillaries, which a lot of people believe helps them release their contents and maybe some of the oxygen that they have in them. So you have a red blood cell that's coming in here. It's being squeezed through this capillary right here. It has a bunch of hemoglobin-- and when I say a bunch, you might as well know right now, each red blood cell has 270 million hemoglobin proteins. And if you total up the hemoglobin in the entire body, it's huge because we have 20 to 30 trillion red blood cells. And each of those 20 to 30 trillion red blood cells have 270 million hemoglobin proteins in them. So we have a lot of hemoglobin. So anyway, that was a little bit of a-- so actually, red blood cells make up roughly 25% of all of the We have about 100 trillion or a little bit more, give or take. I've never sat down and counted them. But anyway, we have 270 million hemoglobin particles or proteins in each red blood cell-- explains why the red blood cells had to shed their nucleuses to make space for all those hemoglobins. They're carrying oxygen. So right here we're dealing with-- this is an artery, right? It's coming from the heart. The red blood cell is going in that direction and then it's going to shed its oxygen and then it's going to become a vein. Now what's going to happen is you have this carbon dioxide. You have a high concentration of carbon dioxide in the It eventually, just by diffusion gradient, ends up-- let me do that same color-- ends up in the blood plasma just like that and some of it can make its way across the membrane into the actual red blood cell. In the red blood cell, you have this carbonic anhydrase essentially become carbonic acid, which then can release protons. Well, those protons, we just learned, can allosterically inhibit the uptake of oxygen by hemoglobin. So those protons start bonding to different parts and even the carbon dioxide that hasn't been reacted with-- that can also allosterically inhibit the hemoglobin. So it also bonds to other parts. And that changes the shape of the hemoglobin protein just enough that it can't hold onto its oxygens that well and it starts letting go. And just as we said we had cooperative binding, the more oxygens you have on, the better it is at accepting more-- the opposite happens. When you start letting go of oxygen, it becomes harder to retain the other ones. So then all of the oxygens let go. So this, at least in my mind, it's a brilliant, brilliant mechanism because the oxygen gets let go just where it needs to let go. It doesn't just say, I've left an artery and" + }, + { + "Q": "(2:35) Sal says that cooperative binding is when \"one binding makes other bindings more likely.\" What causes binding to stop in the first place?\n", + "A": "So when one oxygen binds (perhaps in the lungs), that encourages additional oxygen to bind (up to 3 more) as hemoglobin can carry 4 O2 molecules. When the red blood cell carrying hemoglobin gets to other tissues outside the lung, the increased carbon dioxide diffuses into the red blood cell and through carbonic anhydrase and the production of H+, induces hemoglobin to let go of the oxygen - where the binding stops. The oxygen then diffuses out of the RBC to the tissues.", + "video_name": "LWtXthfG9_M", + "timestamps": [ + 155 + ], + "3min_transcript": "You have four heme groups and the globins are essentially describing the rest of it-- the protein structures, the four peptide chains Now, this heme group-- this is pretty interesting. It actually is a porphyrin structure. And if you watch the video on chlorophyll, you'd remember a porphyrin structure, but at the very center of it, in chlorophyll, we had a magnesium ion, but at the very center of hemoglobin, we have an iron ion and this is where the oxygen binds. So on this hemoglobin, you have four major binding sites for oxygen. You have right there, maybe right there, a little bit behind, right there, and right there. Now why is hemoglobin-- oxygen will bind very well here, but hemoglobin has a several properties that one, make it really good at binding oxygen and then also really good at dumping oxygen when it needs to dump oxygen. And this is just the principle that once it binds to one oxygen molecule-- let's say one oxygen molecule binds right there-- it changes the shape in such a way that the other sites are more likely to bind oxygen. So it just makes it-- one binding makes the other bindings more likely. Now you say, OK, that's fine. That makes it a very good oxygen acceptor, when it's traveling through the pulmonary capillaries and oxygen is diffusing from the alveoli. That makes it really good at picking up the oxygen, but how does it know when to dump the oxygen? This is an interesting question. this guy's running right now and so he's generating a lot of carbon dioxide right now in these capillaries and he needs a lot of oxygen in these capillaries surrounding his quadriceps. I need to deliver oxygen. It doesn't know it's in the quadraceps. How does the hemoglobin know to let go of the oxygen there? And that's a byproduct of what we call allosteric inhibition, which is a very fancy word, but the concept's actually pretty straightforward. When you talk about allosteric anything-- it's often using the context of enzymes-- you're talking about the idea that things bind to other parts. Allo means other. So you're binding to other parts of the protein or the enzyme-- and enzymes are just proteins-- and it affects the ability of the protein or the enzyme to do" + }, + { + "Q": "At 11:40 the carbon that takes the 1st H ends up having 5 bonds towards the end. How would that be correct? The last H that he adds in there shouldn't be there. It was a double bond that \"broke\" and an H is added, or did I misunderstand?\n", + "A": "I can t see where any carbon has 5 bonds? Are you sure you counted correctly? Is it the front of back one that originally had the double bond? Note that there looks to be one very long bond from the front carbon to a hydrogen, but that s actually a bond between the front and back carbons, and then a bond between the back carbon and the hydrogen.", + "video_name": "fSk1Crn3R2E", + "timestamps": [ + 700 + ], + "3min_transcript": "off like that. And my double bond is going to go right here. And then this is going to be a methyl group. And then up here there are going to be two methyl groups, like that. So this is alpha-pinene, found in turpentine. And you can see there's an alkene on this. So if I took this alpha-pinene molecule and I wanted to hydrogenate it, I could use palladium and charcoal, palladium and carbon. And if I think about what happens in this mechanism, I know that my metal catalyst there, my palladium, is going to be flat, like that. And so, when it has those hydrogens, when the palladium adsorbs those hydrogens, it's going to add those two hydrogens to my double bond, think about this guy over here, think about the alpha-pinene as molecules like a spaceship, And the spaceship is approaching the docking station. The spaceship is going to approach the docking station. And there's only one way the spaceship can approach the docking station. And that is the way in which we have drawn it right here. It could not flip upside down and approach it from the top, because of the steric hindrance of these methyl groups. Right? So this is the way that it approaches. In this part of the molecule, your alkene, is the flat part, right? So it's easiest for the molecule to approach in this way. The spaceship analogy always helps my students. So there's only one product for this reaction. And let's see if we can draw it here. And let's see what it would look like. It would look something like this. So we have our two methyl groups right here. So the hydrogens are going to add from below, right? So this hydrogen, let's say it adds right here. That's going to push this methyl group up. when that hydrogen adds right down here. And then this other hydrogen is going to add to the opposite side. And so we can show the addition of that hydrogen. So there's my syn addition of these two hydrogens. And there was something else in that carbon. It was another hydrogen. So another hydrogen got pushed up right here as well. So that is your only product, the only product of this reaction. The hydrogenation reaction is very sensitive to steric conditions." + }, + { + "Q": "\nDoes adsorb mean anything (0:34) or is it just a different pronunciation of absorb?", + "A": "They are different words. Adsorb means to hold molecules of a gas as a thin film on the outside surface of a solid. Absorb means to suck up or soak up. For example, a sponge absorbs water.", + "video_name": "fSk1Crn3R2E", + "timestamps": [ + 34 + ], + "3min_transcript": "To hydrogenate an alkene, you need hydrogen gas and a metal catalyst, something like platinum or palladium or nickel. And there are many others, but these are the ones most commonly used. So what happens is those two hydrogens from the hydrogen gas are added across their double bond. And they're added on the same side of where the double bond used to be. So it's a syn addition. Let's take a look at why this is a syn addition of hydrogens. So we have our metal catalyst over here. So let's go ahead and draw our flat metal catalyst. And these metals are chosen, because they adsorb hydrogen really well, which means that if you bubble hydrogen gas through, the hydrogen is going to be adsorbed to the surface of that metal catalyst, like that. And then your alkene comes along. And your alkene is also flat, right? The portion of the molecule that contains the double bond, right? So these two carbons, this carbon and this carbon, these sp2 hybridized, which means that the stereo chemistry around those two carbons is going to be flat. So you have one thing that's flat approaching something else that's flat. So the only way those hydrogens can add are to add them on to the same side, right? So if this carbon and this carbon, if you add this hydrogen to the carbon on the left and add this hydrogen to the carbon on the right, and then you go ahead and you draw the rest of the bonds, right? This would now be a wedge and then a dash, and then this would be a wedge and then a dash. You can see those two hydrogens have added on to the same side. So these two hydrogens or these two hydrogens for our syn addition. Notice we're also changing from sp2 hybridization to sp3 hybridization over here on the right. So we have to think about stereochemistry for this reaction for your products as well. So let's take a look at an actual reaction here. And let's see if we can follow along. So if this was my reaction, I want to hydrogenate this alkene. So I would add some hydrogen gas and I could choose whichever metal catalyst I wanted to. So I could add two hydrogens on the same side. So just like I did up there. So we would get now everything changes from sp2 hybridization So we have wedges and dashes to worry about. And usually you wouldn't see it drawn like this. That's too much work, quite frankly. It would be much easier just to say, well, all I have to do is take away the double bond, and there's my product. So for some of these reactions, they're very, very, very simple. Just take away the double bond and you'll end up with your alkene-like product. Let's take a look at oxidation states for this reaction. So I'm going to redraw this reaction. And this time I'm going to draw in my atoms. And I'm also going to draw in my electrons here in a second. So I'm just drawing out all the atoms here. So I have all these methyl groups to worry about. And then I have electrons in these bonds, right? Each one of these bonds consists of two electrons." + }, + { + "Q": "I'm confused.... at 6:30 where jay says that Carbon has been reduced as it gained an electron from Hydrogen, I thought C-H interactions were non-polar so they share electrons as oppose to carbon gaining one.... :(\n", + "A": "In reality it does not mean that C-H is non-polar. Because of the small difference in electronegativities, the C\u00e2\u0088\u0092H bond is generally regarded as being non-polar. But theres still a difference in electronegativities. But it doesnt really matter in this case because concerning the oxidation state you simply look at the atom which is more electronegative, which gain more electron.", + "video_name": "fSk1Crn3R2E", + "timestamps": [ + 390 + ], + "3min_transcript": "we just drew around it for our dot structure. So that would be four. Each one of those carbons has four. So each of these carbons has an oxidation state of zero. Let's look at the product, and let's see if we can assign some oxidation states for the product. So our product over here on the right, we had a carbon and we had some methyl groups bonded to that carbon. We added on a hydrogen. And so each one of these carbons got a hydrogen added onto it. And let's go ahead and fill in our electrons in these bonds. So once again, each bond consists of two electrons, like that. And now we have a single bond between our carbons. And let's assign some oxidation states. So once again we know that the two carbons have the same electronegativity, right? So the tug of war for these two electrons right here, it's a tie. So it's a tie, it's a tie. Carbon is actually more electronegative than hydrogen. So in the war over the two electrons in the carbon-hydrogen bond, carbon wins, because it's a little bit more electronegative. So we're going to assign this extra electron here to carbon. And then again, carbon versus carbon. So that carbon gets that electron as well. Same thing down here, right? So it's a tie, it's a tie. Carbon beats hydrogen. And over here, it's a tie. So in the dot structure on the right, the oxidation states that the normal number of valence electrons would be four. From that we subtract the number of electrons in our picture here, which would be five electrons. Each one of these carbons has five electrons around it. So it gained electron. And it's a 4 minus 5 will give us a negative 1. So the oxidation states of these two carbons is negative 1. And we can look at our original oxidation states of being zero, went from zero to negative 1. That's a decrease in the oxidation state, right? A decrease in the oxidation state means a reduction. So the alkene is reduced by the addition of these two hydrogens. And you'll see other definitions for oxidation states. You'll see a gain in hydrogens is reduction. That's another definition that's often found in organic chemistry textbooks. And while that's true, to me it makes more sense to go ahead and assign your oxidation states and watch the oxidation states change as you add those hydrogens, as your molecule gains hydrogens. So this is a reduction. Let's look at the stereochemistry of the hydrogenation reaction. So let's do an example involving stereochemistry. So let's say your alkene-- let's do that ring again, it wasn't a very good one-- so let's say your alkene looked something like this. And you're going to react that with hydrogen and with platinum. All right, well, your first thought" + }, + { + "Q": "At 2:10, Sal says that speed is the magnitude of velocity but speed may not always be the magnitude of velocity (as in the case of average speed and average velocity). I'm still a little confused by what Sal meant here\n", + "A": "THe magnitude of velocity is a speed It may or may not be the speed you are looking for in a particular problem", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 130 + ], + "3min_transcript": "Now that we know a little bit about Newton's First Law, let's give ourselves a little quiz. And what I want you to do is figure out which of these statements are actually true. And our first statement is, \"If the net force on a body is zero, its velocity will not change.\" Interesting. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" Also an interesting statement. Statement number three, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" And statement four, \"An unbalanced force on an object will always change the object's direction.\" So I'll let you think about that. So let's think about these statement by statement. So our first statement right over here, its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope," + }, + { + "Q": "\nAt 1:56, Sal said that if he'd written the word velocity rather than speed then the second statement would be true, but if we consider an unbalanced force on a body, so maybe the force will \"only\" change it's direction or the other way around, so i didn't understood that how can we say the unbalanced force will Always impact it's \"velocity\" (speed+direction) when it maybe changing it's direction \"only\" and not the speed or vise versa.\nI hope my question is clear.", + "A": "If the force changes the direction of motion, then it changed the velocity, even if the speed is the same.", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 116 + ], + "3min_transcript": "Now that we know a little bit about Newton's First Law, let's give ourselves a little quiz. And what I want you to do is figure out which of these statements are actually true. And our first statement is, \"If the net force on a body is zero, its velocity will not change.\" Interesting. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" Also an interesting statement. Statement number three, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" And statement four, \"An unbalanced force on an object will always change the object's direction.\" So I'll let you think about that. So let's think about these statement by statement. So our first statement right over here, its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope," + }, + { + "Q": "@7:14 i m bit confuse here like centripetal force balances centrifugal force so here will the ice scater have balanced force............\nnd the speed example was really good\nwant to learn through such examples from daily life\n", + "A": "The forces are not balanced, if they were the skater would have uniform motion and not be accelerating in a circle. Centripetal force is a real force which accelerates an object to travel in a circle. Centrifugal force is fictitious force which is felt from being in a reference frame which is rotating. The perceived centrifugal force is equal and opposite to the centripetal force.", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 434 + ], + "3min_transcript": "It could be its speed, its direction, or both, but it doesn't have to be both. It could be just the speed or just the direction. So this is an incorrect statement. Now the third statement, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" This is absolutely true. And this is the example we gave. If I take an object, if I take my book and I try to slide it across the desk, the reason why it eventually comes to stop is because we have the unbalanced force of friction-- the grinding of the surface of the book with the grinding of the table. If I'm inside of a pool or even if there's absolutely no current in the pool, and if I were to try to push some type of object inside the water, it eventually comes to stop because of all of the resistance of the water itself. It's providing an unbalanced force in a direction opposite it's motion. That is what's slowing it down. why we don't see these things go on and on forever is that we have these frictions, these air resistants, or the friction with actual surfaces. And then the last statement, \"An unbalanced force on an object will always change the object's direction.\" Well, this one actually is maybe the most intuitive. We always have this situation. Let's say I have a block right over here, and it's traveling with some velocity in that direction-- five meters per second. If I apply an unbalanced force in that same direction-- so that's my force right over there. If I apply it in that same direction, I'm just going to accelerate it in that same direction. So I won't necessarily change it. Even if I were to act against it, I might decelerate it, but I won't necessarily change its direction. I could change its direction by doing something like this, but I don't necessarily. I'm not always necessarily changing the object's direction. So this is not true. An unbalanced force on an object will not always It can, like these circumstances, but not always. So \"always\" is what makes this very, very, very wrong." + }, + { + "Q": "at 1:30, sal says the is a minuscule amount of gravity in space, what causes gravity in space? is their any forces apart from the small about of gravity in space?\n", + "A": "If you look at the equation for the gravitational acceleration A = G * M/r^2 you can see that A goes to 0 at an infinite distance so even deep space in not infinity far from any mass so there still be a small amount of gravity out there.", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 90 + ], + "3min_transcript": "Now that we know a little bit about Newton's First Law, let's give ourselves a little quiz. And what I want you to do is figure out which of these statements are actually true. And our first statement is, \"If the net force on a body is zero, its velocity will not change.\" Interesting. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" Also an interesting statement. Statement number three, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" And statement four, \"An unbalanced force on an object will always change the object's direction.\" So I'll let you think about that. So let's think about these statement by statement. So our first statement right over here, its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope," + }, + { + "Q": "At 6:46 in the video, Mr. Kahn mentions 'deceleration' but my science book tells me there is no such thing as \"deceleration.\" It says that there is acceleration in the direction of velocity or in the opposite direction of velocity, but in physics, it is always called acceleration. Which is accurate?\n", + "A": "It s better to talk about negative acceleration but of course people do use the term deceleration and scientists do need to be able to talk to regular people. It s certainly not right to say there s no such thing as deceleration, because everyone knows there is. But that doesn t make the term or concept something we want to use when we are trying to do physics.", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 406 + ], + "3min_transcript": "It could be its speed, its direction, or both, but it doesn't have to be both. It could be just the speed or just the direction. So this is an incorrect statement. Now the third statement, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" This is absolutely true. And this is the example we gave. If I take an object, if I take my book and I try to slide it across the desk, the reason why it eventually comes to stop is because we have the unbalanced force of friction-- the grinding of the surface of the book with the grinding of the table. If I'm inside of a pool or even if there's absolutely no current in the pool, and if I were to try to push some type of object inside the water, it eventually comes to stop because of all of the resistance of the water itself. It's providing an unbalanced force in a direction opposite it's motion. That is what's slowing it down. why we don't see these things go on and on forever is that we have these frictions, these air resistants, or the friction with actual surfaces. And then the last statement, \"An unbalanced force on an object will always change the object's direction.\" Well, this one actually is maybe the most intuitive. We always have this situation. Let's say I have a block right over here, and it's traveling with some velocity in that direction-- five meters per second. If I apply an unbalanced force in that same direction-- so that's my force right over there. If I apply it in that same direction, I'm just going to accelerate it in that same direction. So I won't necessarily change it. Even if I were to act against it, I might decelerate it, but I won't necessarily change its direction. I could change its direction by doing something like this, but I don't necessarily. I'm not always necessarily changing the object's direction. So this is not true. An unbalanced force on an object will not always It can, like these circumstances, but not always. So \"always\" is what makes this very, very, very wrong." + }, + { + "Q": "\nAt 1:01, If the net force on a body is zero, would it have any velocity or would it be a stationary object?", + "A": "it can either be in motion or can also be in rest. force is not required to keep the object which is already in constant velocity in motion thats why they say a object in constant velocity to have a net force of 0", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 61 + ], + "3min_transcript": "Now that we know a little bit about Newton's First Law, let's give ourselves a little quiz. And what I want you to do is figure out which of these statements are actually true. And our first statement is, \"If the net force on a body is zero, its velocity will not change.\" Interesting. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" Also an interesting statement. Statement number three, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" And statement four, \"An unbalanced force on an object will always change the object's direction.\" So I'll let you think about that. So let's think about these statement by statement. So our first statement right over here, its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope," + }, + { + "Q": "At 10:45 Sal said that a hot spot could have created the Mid-Atlantic Ridge as well as the African Rift Valley. The Mid-Atlantic Ridge is much bigger than the African rift valley, so how could the same type of hot spot have created them both?\n", + "A": "A hot spot cannot create a ridge, they create volcanic island chains, Hawaii and its islands are prime examples. The Mid-Atlantic Ridge was created by diverging tectonic plates, the magma which streamed up from the mantel solidified in the ocean and created the ridge.", + "video_name": "FK1s1-OJ5BE", + "timestamps": [ + 645 + ], + "3min_transcript": "You essentially have the Indian Ocean flowing into this rift that formed from this hot spot. And then if you fast forward a bunch so that finally the magma can kind of surface. So let's fast forward from even this point even more. So let's fast forward even more, and let's say now the land has been pushed a good bit apart. Now the hot spot has actually surfaced. Now the crust might look something like this. So it's been pushed apart a good bit at this point. Now we're talking about on the order of hundreds of thousands of years or tens of thousands of years. So the land, for example, the land that was here, this part of the land might now be out here. And this part of the land might now be out here. What's going to happen is that this hot spot is going to continue to fuel, and we're assuming everything's underwater at this point. Since this depression that was created is now so low the crust was stretched thin. The hot spot is essentially going to come out of underwater volcanoes and start creating what's now-- this body of water's gotten large enough that we can call it a mid-oceanic ridge. And so it'll actually start creating an actual ridge with volcanoes in the center. So that's why one, we see things like the Rift Valley in Africa, we see things like the Red Sea. And maybe even more importantly, that's why we see something like the mid-Atlantic rift in the middle of the Atlantic Ocean, where you have all of this depressed land that was essentially analogous to that Rift Valley but it's at a much later stage. And that's why it's able to collect water, because when the land was pushed out and stretched thin water was able to flow into it, going when this bread was baking and this part of the crust pushed outwards, you had this rift form, and then if there was some water on the bread, or if it was raining, or if it was connected to a body of water, water would've eventually flowed in here. And if that bread kept growing this rift would have kept growing, eventually to the size of the Atlantic Ocean in our theoretical bread. And so that's why you have this huge depressed area where the ocean can form, but in the middle of it you kind of have this you have this submersed you have this actual submersed mountain chain, this submersed chain of volcanoes, this submersed ridge where the land actually does go up a little bit because of all that magma flowing directly out of it. So hopefully that clears up a little bit. That was always confusing to me why you see uplifted land but then everything around the uplifted land is much lower, and why the whole thing is submersed as it's moving away. So hopefully that clears things up a little bit." + }, + { + "Q": "\nAround 7:40, Sal says the time taken for the momentum to change is 2x/v, bt isn't the actual time taken that tiny instant it JUST strikes the wall and reverses its velocity?", + "A": "| _.->_| Starting |<-. __| Back at the same point, but not with the same velocity. | _.->_| Back at the same point with the same velocity. The distance traveled therefore must be twice the length. This distance over the velocity component in this direction will be the time required for the average pressure equation.", + "video_name": "qSFY7GKhSRs", + "timestamps": [ + 460 + ], + "3min_transcript": "Now, if I come in with a momentum of mv, and I ricochet off with a momentum of minus mv, what's my change in momentum? My change in momentum, off of that ricochet, is equal to-- well, it's the difference between these two, which is just 2mv. Now, that doesn't give me the force. I need to know the change in momentum per unit of time. So how often does this happen? How frequently? Well, it's going to happen every time we come here. We're going to hit this wall. Then the particle is going to have to travel here, bounce off of that wall, and then come back here and hit it again. So that's how frequently it's going to happen. So how long of an interval do we have to wait between the collisions? Well, the particle has to travel x going back. It's going to collide. It's going to have to travel x to the left. This distance is x. Let me do that in a different color. It's going to have to travel x to go back. Then it's going to have to travel x back. So it's going to have to travel 2x distance. And how long will it take it to travel 2x distance? Well, the time, delta T, is equal to, we know this. Distance is equal to rate times time. Or if we do distance divided by rate, we'll get the amount of time we took. This is just our basic motion formula. Our delta T, the distance we have to travel is back and forth. So it's 2 x's, divided by-- what's our rate? Well, our rate is our velocity. Divided by v. There you go. So this is our delta T right here. So our change in momentum per time is equal to 2 times our Because we ricocheted back with the same magnitude, but negative momentum. So that's our change in momentum. And then our change in time is this value over here. It's the total distance we have to travel between collisions of this wall, divided by our velocity. So it is, 2x divided by v, which is equal to 2mv times the reciprocal of this-- so this is just fraction math-- v over 2x. And what is this equal to? The 2's cancel out. So that is equal to mv squared, over x. Interesting. We're getting someplace interesting already. And if it doesn't seem too interesting, just hang on with me for a second. Now, this is the force being applied by one particle, is this-- force from one particle on this wall." + }, + { + "Q": "\nAround 6:31, why is the change in momentum a positive number? if P2 = -mv due to the velocity being in the opposite direction, then should change in momentum not be = P2 - P1, which is -2mv ?? Thanks for helping me out of my misery:-(", + "A": "I have this exact same question, and no one seems to have answered it yet. The only explanation I can come up with is that the negative is just an indication of the direction, and when looking at the force that results from this pressure, a positive value should be used.", + "video_name": "qSFY7GKhSRs", + "timestamps": [ + 391 + ], + "3min_transcript": "And, of course, we know that this could be rewritten as this is equal to-- mass is a constant and shouldn't change for the physics we deal with-- so it's delta. We could put that inside of the change. So it's delta mv over change in time. And this is just change in momentum, right? So this is equal to change in momentum over change in time. So that's another way to write force. So what's the change in momentum going to be for this particle? Well, it's going to bump into this wall. In this direction, right now, it has some momentum. Its momentum is equal to mv. And it's going to bump into this wall, and then going to ricochet straight back. And what's its momentum going to be? Well, it's going to have the same mass and the same velocity. We'll assume it's a completely elastic collision. Nothing is lost to heat or whatever else. But the velocity is in the other direction. So the new momentum is going to be minus mv, because the Now, if I come in with a momentum of mv, and I ricochet off with a momentum of minus mv, what's my change in momentum? My change in momentum, off of that ricochet, is equal to-- well, it's the difference between these two, which is just 2mv. Now, that doesn't give me the force. I need to know the change in momentum per unit of time. So how often does this happen? How frequently? Well, it's going to happen every time we come here. We're going to hit this wall. Then the particle is going to have to travel here, bounce off of that wall, and then come back here and hit it again. So that's how frequently it's going to happen. So how long of an interval do we have to wait between the collisions? Well, the particle has to travel x going back. It's going to collide. It's going to have to travel x to the left. This distance is x. Let me do that in a different color. It's going to have to travel x to go back. Then it's going to have to travel x back. So it's going to have to travel 2x distance. And how long will it take it to travel 2x distance? Well, the time, delta T, is equal to, we know this. Distance is equal to rate times time. Or if we do distance divided by rate, we'll get the amount of time we took. This is just our basic motion formula. Our delta T, the distance we have to travel is back and forth. So it's 2 x's, divided by-- what's our rate? Well, our rate is our velocity. Divided by v. There you go. So this is our delta T right here. So our change in momentum per time is equal to 2 times our" + }, + { + "Q": "isnt it sn=3+2=5 in ~5:40?\n", + "A": "No. Before the electrons move, there are 3 \u00cf\u0083 bonds and 1 lone pair, so SN = 3 + 1 = 4", + "video_name": "kQCS1AhAnMI", + "timestamps": [ + 340 + ], + "3min_transcript": "they can't participate in resonance; that lone pair of electrons in blue is localized to that nitrogen. And so, this is why you can think about an amide being different from an amine, in terms of functional group, and in terms of how they react and how they behave. If we look at another example, so this molecule right here, and we assume the lone pair of electrons on that nitrogen is localized to that nitrogen, let's go ahead and calculate the steric number. So the steric number'd be equal to sigma bonds, so that's one, two, and three; so three sigma bonds. Plus lone pairs of electrons, there's one lone pair of electrons on that nitrogen, so three plus one is four; so four hybrid orbitals, which implies SP three hybridization on that nitrogen. But we know that, that lone pair of electrons is not localized to that nitrogen; that lone pair of electrons because we have this pattern here. All right, so this pattern of a lone pair of electrons, in blue, next to a pi bond, which I will make magenta, and so we could draw a resonance structure. So I could take the electrons in blue, move them into here, too many bonds to this carbon, so I push the electrons in magenta off, onto this carbon. So we draw the resonance structure, so I have my ring here, the nitrogen's bonded to a hydrogen, the electrons in blue moved in to form a pi bond, and the electrons in magenta, moved off, onto this carbon right here, to give that carbon a negative one formal charge. All right, let's go ahead and calculate a steric number for the nitrogen here, which gets a plus one formal charge. All right, so the steric number will be equal to number of sigma bonds: So there's one sigma bond, in our double-bond, one of them is a sigma, and one of them is a pi; so I'm saying that's our sigma bond. So our steric number is equal to three, plus the number of lone pairs of electrons, now zero. So that's a steric number of three, which implies three hybrid orbitals, which says SP two hybridization. And since we know that, that lone pair is de-localized, it's going to occupy a P orbital, and so therefore this nitrogen is SP two hybridized, because we know SP two hybridization has three SP two hybrid orbitals, and one P orbital. So that lone pair in blue is actually de-localized; it's occupying a P orbital, and so let's go ahead and draw that, down here, so let's say this is the nitrogen, and you're looking at it, at a bit of an angle. If that nitrogen is SP two hybridized, that nitrogen has a P orbital, so we can go ahead and draw in a P orbital, on that nitrogen. And so, the electrons in blue, since those electrons" + }, + { + "Q": "\n@0:45 is the structure on the left trigonal pyramidal because of the lone pair? is it still a resonance structure with the right given the different pyramidal/planar structures of Nitrogen?", + "A": "If it is not conjugated, the N atom with a lone pair is trigonal pyramidal. If the N atom is conjugated, it is sp\u00c2\u00b2 hybridized and therefore trigonal planar.", + "video_name": "kQCS1AhAnMI", + "timestamps": [ + 45 + ], + "3min_transcript": "Voiceover: Let's look at the amide, or \"amid\" functional group, and let's start by assigning a steric number to this nitrogen. So, the steric number is equal to the number of sigma bonds; so here's a sigma bond, here's a sigma bond, and here's a sigma bond; so three sigma bonds. Plus number of lone pairs of electrons, so there's one lone pair of electrons on that nitrogen, so I'll go ahead and highlight them there. So three plus one gives us a total of four for the steric number, which means four hybrid orbitals, which implies SP three hybridization for that nitrogen, and from earlier videos, you know that SP three hybridization means, a trigonal, pyramidal, geometry for that nitrogen. And so, that's one way of looking at this function group, and that lone pair of electrons being localized to that nitrogen; however, now that we know resonance structures, we know that, that lone pair of electrons is not localized to that nitrogen; it's de-localized in resonance. So we could take that lone pair of electrons, force us to push some pi electrons off, onto this oxygen, so let's go ahead and draw the other resonance structure. So this top oxygen would now have three lone pairs of electrons around it, a negative one formal charge, and there'd be a double-bond between this carbon and this nitrogen, so let's go ahead and draw in everything. This nitrogen how has a plus one formal charge, and we can go ahead and complete our resonance bracket here. So let's follow those electrons along, the electrons in magenta, the lone pair of electrons moved in here to form our pi bond, and the pi electrons over here in blue, came off onto the oxygen, to give the oxygen a negative one formal charge. All right, let's now calculate a steric number for this nitrogen, in our second resonance structure. So, steric number is equal to number of sigma bonds: So here's a sigma bond, here's a sigma bond, and our and one of them is a pi bond. So we have a total of three sigma bonds around our nitrogen, zero lone pairs of electrons, so three plus zero gives us a steric number of three; that implies three hybrid orbitals, which means SP two hybridization, and a trigonal planar geometry around that nitrogen, so a planar geometry. And, here we've shown our electrons being de-localized, so the lone pair of electrons are de-localized due to resonance, and so, experimental studies have shown that the amide function group is planar, so these atoms actually are planar here, which means that the electrons in magenta are not localized to that nitrogen; they are actually de-localized. And so, that implies this nitrogen is SP two hybridized, and has a P orbital, and that allow that lone pair of electrons in magenta to be de-localized. And so, here's a situation where drawing" + }, + { + "Q": "\nHow can p orbital from N (4:44) create pi bond if it already has 2 electrons? Confused...", + "A": "What makes you think that would stop it from being able to form a pi bond..? Something to keep in mind in aromatic molecules is that they technically are not double bonds at all, every atom in the ring is sharing the 6 pi electrons because that makes it aromatic, and being aromatic is VERY favourable energetically.", + "video_name": "kQCS1AhAnMI", + "timestamps": [ + 284 + ], + "3min_transcript": "actually happening: That lone pair is participating in resonance, which makes this nitrogen SP two hybridized, so it has a P orbital. All right, let's look at this example down here, and let's look first at this left side of the molecule, and so we can see our amide functional group, and if I look at the lone pair of electrons on this nitrogen, we've just talked about the fact that this lone pair of electrons is actually de-localized, so this lone pair is participating in resonance. And so, that affects the geometry, and how you think about the hybridization of this nitrogen, here. So, the electrons in magenta are de-localized because they participate in resonance, and if I think about, let's make this a different color here; let's make these electrons in here blue, so the electrons in blue they can't participate in resonance; that lone pair of electrons in blue is localized to that nitrogen. And so, this is why you can think about an amide being different from an amine, in terms of functional group, and in terms of how they react and how they behave. If we look at another example, so this molecule right here, and we assume the lone pair of electrons on that nitrogen is localized to that nitrogen, let's go ahead and calculate the steric number. So the steric number'd be equal to sigma bonds, so that's one, two, and three; so three sigma bonds. Plus lone pairs of electrons, there's one lone pair of electrons on that nitrogen, so three plus one is four; so four hybrid orbitals, which implies SP three hybridization on that nitrogen. But we know that, that lone pair of electrons is not localized to that nitrogen; that lone pair of electrons because we have this pattern here. All right, so this pattern of a lone pair of electrons, in blue, next to a pi bond, which I will make magenta, and so we could draw a resonance structure. So I could take the electrons in blue, move them into here, too many bonds to this carbon, so I push the electrons in magenta off, onto this carbon. So we draw the resonance structure, so I have my ring here, the nitrogen's bonded to a hydrogen, the electrons in blue moved in to form a pi bond, and the electrons in magenta, moved off, onto this carbon right here, to give that carbon a negative one formal charge. All right, let's go ahead and calculate a steric number for the nitrogen here, which gets a plus one formal charge. All right, so the steric number will be equal to number of sigma bonds: So there's one sigma bond," + }, + { + "Q": "12:25 a willing mother could be one going for an abortion? i mean....instead of killing the embryo,doctors could extract it and utilise the stem cells?? just saying....: )\n", + "A": "Actually no, embryonic stem cells are derived from the surplus embryos not used or not suitable for use from in-vitro fertilization. While a religious person might still consider this immoral, it is not technically the same thing as an abortion.", + "video_name": "-yCIMk1x0Pk", + "timestamps": [ + 745 + ], + "3min_transcript": "That's another word that you might hear. Let me write that down, too: plasticity. And the word essentially comes from, you know, like a plastic can turn into anything else. When we say that something has plasticity, we're talking about its potential to turn into a lot of different things. So the theory is, and there's already some trials that seem to substantiate this, especially in some lower organisms, that, look, if you have some damage at some point in your body-- let me draw a nerve cell. Let me say I have a-- I won't go into the actual mechanics of a nerve cell, but let's say that we have some damage at some point on a nerve cell right there, and because of that, someone is paralyzed or there's some nerve dysfunction. We're dealing with multiple sclerosis or who knows what. The idea is, look, we have these cell here that could turn into anything, and we're just really understanding how It really has to look at its environment and say, hey, what are the guys around me doing, and maybe that's what helps dictate what it does. But the idea is you take these things that could turn to anything and you put them where the damage is, you layer them where the damage is, and then they can turn into the cell that they need to turn into. So in this case, they would turn into nerve cells. They would turn to nerve cells and repair the damage and maybe cure the paralysis for that individual. So it's a huge, exciting area of research, and you could even, in theory, grow new organs. If someone needs a kidney transplant or a heart transplant, maybe in the future, we could take a colony of these embryonic stem cells. Maybe we can put them in some type of other creature, or who knows what, and we can turn it into a replacement heart or a So there's a huge amount of excitement about what these can do. I mean, they could cure a lot of formerly uncurable diseases might otherwise die. But obviously, there's a debate here. And the debate all revolves around the issue of if you were to go in here and try to extract one of these cells, you're going to kill this embryo. You're going to kill this developing embryo, and that developing embryo had the potential to become a human being. It's a potential that obviously has to be in the right environment, and it has to have a willing mother and all of the rest, but it does have the potential. And so for those, especially, I think, in the pro-life camp, who say, hey, anything that has a potential to be a human being, that is life and it should not be killed. So people on that side of the camp, they're against the destroying of this embryo. I'm not making this video to take either side to that argument, but it's a potential to turn to a human being." + }, + { + "Q": "at 00:27 seconds, you say that only one sperm can get into the egg. If so, how would a couple get twins?\n", + "A": "Sumangal is right concerning monozygotic or identical twins. In the case of fraternal (dizygotic) twins, the mother produces 2 egg cells that are then fertilized by 2 separate sperm. Each develops in its own amniotic sac with its own placenta (identical twins share a placenta but usually have separate sacs depending on when they divided) and is no more or less identical than any other pair of siblings born at different times.", + "video_name": "-yCIMk1x0Pk", + "timestamps": [ + 27 + ], + "3min_transcript": "Where we left off after the meiosis videos is that we had two gametes. We had a sperm and an egg. Let me draw the sperm. So you had the sperm and then you had an egg. Maybe I'll do the egg in a different color. That's the egg, and we all know how this story goes. The sperm fertilizes the egg. And a whole cascade of events start occurring. The walls of the egg then become impervious to other sperm so that only one sperm can get in, but that's not the focus of this video. The focus of this video is how this fertilized egg develops once it has become a zygote. So after it's fertilized, you remember from the meiosis videos that each of these were haploid, or that they had-- oh, I added an extra i there-- that they had half the contingency of the DNA. As soon as the sperm fertilizes this egg, now, all of a sudden, you have a diploid zygote. So now let me pick a nice color. So now you're going to have a diploid zygote that's going to have a 2N complement of the DNA material or kind of the full complement of what a normal cell in our human body would have. So this is diploid, and it's a zygote, which is just a fancy way of saying the fertilized egg. And it's now ready to essentially turn into an organism. So immediately after fertilization, this zygote starts experiencing cleavage. It's experiencing mitosis, that's the mechanism, but it doesn't increase a lot in size. So this one right here will then turn into-- it'll just split up via mitosis into two like that. And, of course, these are each 2N, and then those are going to split into four like that. as that first zygote, and it keeps splitting. And this mass of cells, we can start calling it, this right here, this is referred to as the morula. And actually, it comes from the word for mulberry because it looks like a mulberry. So actually, let me just kind of simplify things a little bit because we don't have to start here. So we start with a zygote. This is a fertilized egg. It just starts duplicating via mitosis, and you end up with a ball of cells. It's often going to be a power of two, because these cells, at least in the initial stages are all duplicating all at once, and then you have this morula. Now, once the morula gets to about 16 cells or so-- and we're talking about four or five days. This isn't an exact process-- they started differentiating a" + }, + { + "Q": "\nStarting at 6:34, why did he not use pythagorean theorem to describe V?", + "A": "Good question. The reason he said it this way is because he was referring to vectors and not the magnitude (length) of the vectors. If you want the magnitude, then you are correct in saying that you would need the Pythagorean theorem.", + "video_name": "2QjdcVTgTTA", + "timestamps": [ + 394 + ], + "3min_transcript": "or sometimes it's called a caret character-- that tells us that it is a vector, but it is a unit vector. It has a magnitude of 1. And by definition, the vector j goes and has a magnitude of 1 in the positive y direction. So the y component of this vector, instead of saying it's 5 meters per second in the upwards direction or instead of saying that it's implicitly upwards because it's a vertical vector or it's a vertical component and it's positive, we can now be a little bit more specific about it. We can say that it is equal to 5 times j. Because you see, this magenta vector, it's going the exact same direction as j, it's just 5 times longer. I don't know if it's exactly 5 times. I'm just trying to estimate it right now. It's just 5 times longer. Now what's really cool about this is besides just being able to express the components as now able to do that-- which we did do, we're representing the components as explicit vectors-- we also know that the vector v is the sum of its components. If you start with this green vector right here and you add this vertical component right over here, you have head to tails. You get the blue vector. And so we can actually use the components to represent the vector itself. We don't always have to draw it like this. So we can write that vector v is equal to-- let me write it this way-- it's equal to its x component vector plus the y component vector. And we can write that, the x component vector is 5 square roots of 3 times i. And then it's going to be plus the y component, the vertical component, which is 5j, which is 5 times j. vector in two dimensions by some combination of i's and j's or some scaled up combinations of i's and j's. And if you want to go into three dimensions, and you often will, especially as the physics class moves on through the year, you can introduce a vector in the positive z direction, depending on how you want to do it. Although z is normally up and down. But whatever the next dimension is, you can define a vector k that goes into that third dimension. Here I'll do it in a kind of unconventional way. I'll make k go in that direction. Although the standard convention when you do it in three dimensions is that k is the up and down dimension. But this by itself is already pretty neat because we can now represent any vector through its components and it's also going to make the math much easier." + }, + { + "Q": "\nso are we sapouse to add up all the small areas to get the total work amount? 7:14\nthanks again", + "A": "Yep, on 7:14 you do.", + "video_name": "M5uOIy-JTmo", + "timestamps": [ + 434 + ], + "3min_transcript": "This is after removing each of the pebbles, so that our pressure and volume macro states are always well defined. But in state 2, we now have a pressure low and volume is high. The volume is high, you can just see that, because we kept pushing the piston up slowly, slowly, trying to maintain ourselves in equilibrium so our macrostates are always defined. And our pressure is lower just because we could have the same number of particles, but they're just going to bump into the walls a little bit less, because they have a little bit more room to move around. And that's all fair and dandy. So this describes the path of our system as it transitioned or as it experienced this process, which was a quasi-static process. Everything was defined at every point. Now we said that the work done at any given point by the system is the pressure times the change in volume. Now, how does that relate to here? Change in volume is just a certain distance along this x-axis. This is a change in volume. We started off at this volume, and let's say when we removed one pebble we got to this volume. Now, we want to multiply that times our pressure. Since we did it over such a small increment, and we're so close to equilibrium, we could assume that our pressure's is roughly constant over that period of time. So we could say that this is the pressure over that period of time. And so how much work we did, it's this pressure over here, times this volume, which is the area of this rectangle right there. And for any of you all who've seen my calculus videos, this should start looking a little bit familiar. And then what about when we could take our next pebble? Well now our pressure is a little bit lower. This is our new pressure. Our pressure is a little bit lower. And we multiply that times our new change in volume-- times this change in volume-- and we have that increment of work. And if you keep doing that, the amount of work we do is essentially the area of all of these rectangles as we remove each pebble. And now you might say, especially those of you who haven't watched my calculus videos, gee, you know, this might be getting close, but the area of these rectangles isn't exactly the area of this curve. It's a little inexact. And what I would say is, well if you're worried about that, what you should do is use smaller increments of volume. And if you want to have smaller changes in volume along each step, what you do is you remove even smaller pebbles. And this goes back to trying to get to that ideal quasi-static process. So if you did that of, eventually the delta V's would get smaller and smaller and smaller, and the rectangles would get thinner and thinner and thinner. You'd have to do it over more and more steps. But eventually you'll get to a point, if you assume really small changes in our delta V." + }, + { + "Q": "\nAt 0:59, aren't 3 sulfur atoms called sulfate instead of sulfide?", + "A": "No, the sulfate ion has the formula SO4^2-", + "video_name": "vVTwzjvWySs", + "timestamps": [ + 59 + ], + "3min_transcript": "- [Instructor] So we have the formula for an ionic compound here, and the goal of this video is what do we call this thing? It clearly involves some cobalt and some sulfur, but how do we name it? Well, the convention is, is the first element to be listed is going to be our cation, and if we look at cobalt over here, we see that it is a D-block element and D-block elements are tricky because you don't know exactly how it will ionize. So we know that this is going to be our cation, it's going to be our positive ion, but we don't know what the charge on each of those cobalt is actually going to be. So now let's look at the anion, let's look at the sulfur, or as an anion, the sulfide. So let me underline that. And on the periodic table, we see sulfur is out here that in its group, it would want to gain two electrons in order to have a complete outer shell. So the sulfide anion will look like this. So it will have sulfur when it ionizes will have a two minus charge, just like oxygen, just like everything else in this group. It would want to gain one, two electrons so that its outer shell looks like that of a normal gas, looks like that of argon. We can use this as a clue to figure out what must be the charge on the cobalts because we have three of the sulfides. Each of the sulfides has a two minus charge, and we have three of them, so that's going to give us a six minus charge all in. And then the cobalt, we have two of them. And so these two cobalt have to offset They have to have a six plus charge. Well that means that each of them need to have a three plus charge. If each of these have a three plus charge and you have two of them, then you're gonna have six plus on the positive side and you're gonna have six minus from the sulfides. And the reason why this is useful for us is now we can name this. We would call this ionic compound Cobalt III, cobalt and you would write it with Roman numerals here, Cobalt III Sulfide, Cobalt III Sulfide. Now I know what you might be thinking. Hey, when we looked at other ionic compounds, I didn't have to write the charge of the cation there and the reason why the convention is to do it here, and I don't have to write in upper case there, so let me rewrite it as" + }, + { + "Q": "On the last example at 5:00, what is the difference between 5.60x10^4 and 5.6x10^4\nthat trailing 0 seems pretty important and I don't understand why.\n", + "A": "I understand it simply as how detailed you want your number to be. 5.6 could actually be 5.61 or 5.64, depends on how many decimals you want, and that means that 5.60 is more accurate.", + "video_name": "eMl2z3ezlrQ", + "timestamps": [ + 300 + ], + "3min_transcript": "your best assumption is probably that they just measured to the nearest thousand. That they didn't measure exactly the one and just happened to get exactly on 37,000. So if there's no decimal, let me write it this way-- it's ambiguous, which means that you're not sure what it means, it's not clear what it means. And you're probably safer assuming to not count it. If someone really does measure, if you were to really measure something to the exact one, then you should put a decimal at the end like that. And there is a notation for specifying. Let's say you do measure-- and let me do a different number. Let's say you do measure 56,000. And there is a notation for specifying that 6 definitely is the last significant digit. And sometimes you'll see a bar put over the 6, And that could be useful because maybe your last significant digit is this zero over here. Maybe you were able to measure to the hundreds with a reasonable level of precision. And so then you would write something like 56,000, but then you would put the bar above that zero, or the bar below that zero to say that that was the last significant digit. So if you saw something like this, you would say three significant digits. This isn't used so frequently. A better way to show that you've measured to three significant digits would be to write it in scientific notation. There's a whole video on that. But to write this in scientific notation, you could write this as 5.60 times 10 to the fourth power. Because if you multiply this times to the fourth, you would move this decimal over four spaces and get us to 56,000. So 5.60 times 10 to the fourth. on scientific notation. It will hopefully clarify things a little bit. But when you write a number in scientific notation, it makes it very clear about your precision and how many significant digits you're dealing with. So instead of doing this notation that's a little bit outdated-- I haven't seen it used much with these bars below or above the high significant digit, instead you could represent it with a decimal in scientific notation. And there it's very clear that you have three significant digits. So hopefully that helps you out. In the next couple of videos, we'll explore a little bit more why significant digits are important, especially when you do calculations with multiple measurements." + }, + { + "Q": "\nWhat does trans specify at 5:20", + "A": "Trans is a type of isomerism. All it specifies is that it is a type of isomer. Just so you know - Isomers have the same molecular formulae but different structural formulae. Hope that helps.", + "video_name": "z8h7QgevqjM", + "timestamps": [ + 320 + ], + "3min_transcript": "This has a double bond right here. So it's hepten. If this was just an alkene, we would just called heptene, but we're not going to put this last e here, because this is the carboxylic acid. And to specify where that double bond is, we need to start numbering, and we start numbering at the carbonyl carbon. One, two, three, four, five, six, seven. So you could either name this 3 hepten, and I haven't finished it yet, I haven't put this final e over here. Or you could name it hept 3 ene, just like that. This is the more typical one that you would see, because it tells you we have a double bond, and it starts at the number three carbon, goes from the three to the four carbon. But this isn't just a regular alkene, this is a carboxylic acid. So instead of writing that final e, for an alkene, we write it as we have a carboxyl group right here, so this is 3 And we are done. Actually if you wanted to get really fancy on this one right over here, you could see that these two carbons that are on the double bond, so this carbon and this carbon, it's kind of a range like this. Let me draw it like this. They both have other hydrogens off there that we didn't draw, they're implicitly there. But if you wanted to rewrite or redraw this molecule, you could draw it like this. You have two carbons, just like this. This one has a hydrogen popping up like that; that one has a hydrogen popping down like that. And then this carbon over here has this big functional group over here. We'll call that R. And then this one over here-- I'll do it in green-- has this other functional group, has these three carbons. We can call that R prime. And if you look at it this way, the functional groups are on opposite sides of the double bond. So if you wanted to, you could also call this trans 3 heptenoic acid. And this will specify that these guys are on opposite ends. But this is only if you're assuming that I drew it in the actual three dimensional configuration in some way. Anyway, hopefully you found that useful." + }, + { + "Q": "At 4:37, Why did we put the hydrogen molecules opposite to each other? Why not in the same direction?\n", + "A": "NO. This is an equilibrium reaction. Therefore, the most stable molecules will form. Trans is much better then cis....steric repulsion (eclipsed atoms).", + "video_name": "z8h7QgevqjM", + "timestamps": [ + 277 + ], + "3min_transcript": "This has a double bond right here. So it's hepten. If this was just an alkene, we would just called heptene, but we're not going to put this last e here, because this is the carboxylic acid. And to specify where that double bond is, we need to start numbering, and we start numbering at the carbonyl carbon. One, two, three, four, five, six, seven. So you could either name this 3 hepten, and I haven't finished it yet, I haven't put this final e over here. Or you could name it hept 3 ene, just like that. This is the more typical one that you would see, because it tells you we have a double bond, and it starts at the number three carbon, goes from the three to the four carbon. But this isn't just a regular alkene, this is a carboxylic acid. So instead of writing that final e, for an alkene, we write it as we have a carboxyl group right here, so this is 3 And we are done. Actually if you wanted to get really fancy on this one right over here, you could see that these two carbons that are on the double bond, so this carbon and this carbon, it's kind of a range like this. Let me draw it like this. They both have other hydrogens off there that we didn't draw, they're implicitly there. But if you wanted to rewrite or redraw this molecule, you could draw it like this. You have two carbons, just like this. This one has a hydrogen popping up like that; that one has a hydrogen popping down like that. And then this carbon over here has this big functional group over here. We'll call that R. And then this one over here-- I'll do it in green-- has this other functional group, has these three carbons. We can call that R prime. And if you look at it this way, the functional groups are on opposite sides of the double bond. So if you wanted to, you could also call this trans 3 heptenoic acid. And this will specify that these guys are on opposite ends. But this is only if you're assuming that I drew it in the actual three dimensional configuration in some way. Anyway, hopefully you found that useful." + }, + { + "Q": "\nAt 6:28 Sal writes ||a|| = 5. What does '|| ||' mean?", + "A": "It means the magnitude, or length, of the vector. This is the hypotenuse length formed by the vector components.", + "video_name": "xp6ibuI8UuQ", + "timestamps": [ + 388 + ], + "3min_transcript": "this vector right here in green and this vector right here in red. Notice, X starts at the tail of the green vector and goes all the way to the head of the magenta vector. And the magenta vector starts at the head of the green vector and then finishes, I guess, well where it finishes is where vector X finishes. And the reason why I do this... And, you know, hopefully from this comparable explanation right here, says, okay, look, the green vector plus the magenta vector gives us this X vector. That should make sense. I put the head of the green vector to the tail of this magenta vector right over here. But the whole reason why I did this is, if I can express X as a sum of these two vectors, it then breaks down X into its vertical component and its horizontal component. So I could call this or I should say the vertical component. X vertical. And then I could call this over here the X horizontal. Or another way I could draw it, I could shift this X vertical over. Remember, it doesn't matter where I draw it, as long as it has the same magnitude and direction. And I could draw it like this. X vertical. And so what you see is is that you could express this vector X... Let me do it in the same colors. You can express this vector X as the sum of its horizontal and its vertical components. As the sum of its horizontal and its vertical components. Now we're gonna see over and over again that this is super powerful because what it can do is it can turn a two-dimensional problem into two separate one-dimensional problems, one acting in a horizontal direction, one acting in a vertical direction. Now let's do it a little bit more mathematical. I've just been telling you about length and all of that. Let me just show you what this means, to break down the components of a vector. So let's say that I have a vector that looks like this. Let me do my best to... Let's say I have a vector that looks like this. It's length is five. So let me call this vector A. So vector A's length is equal to five. And let's say that its direction... We're gonna give its direction by the angle between the direction its pointing in and the positive X axis. So maybe I'll draw an axis over here. So let's say that this right over here is the positive Y axis going in the vertical direction. This right over here is the positive X axis going in the horizontal direction. And to specify this vector's direction I will give this angle right over here. And I'm gonna give a very peculiar angle, but I picked this for a specific reason, just so things work out neatly in the end." + }, + { + "Q": "At 3:40, Sal mentions that the formula for vectors was a + b = c\nwhere 'a' and 'b' are the horizontal and vertical vectors. But when you add up vector 'a' (3) and 'b' (4) in this problem the don't equal 'c' (5).\n", + "A": "and ply if it forms a right triangle else you use the formula (a^2+b^2+2ab cos{theta})^1/2 which will be equal to the sum of given two vectors", + "video_name": "xp6ibuI8UuQ", + "timestamps": [ + 220 + ], + "3min_transcript": "As long as it has the same magnitude, the same length, and the same direction. And the whole reason I'm doing that is because the way to visually add vectors... If I wanted to add vector A plus vector B... And I'll show you how to do it more analytically in a future video. I can literally draw vector A. I draw vector A. So that's vector A, right over there. And then I can draw vector B, but I put the tail of vector B to the head of vector A. So I shift vector B over so its tail is right at the head of vector A. And then vector B would look something like this. It would look something like this. And then if you go from the tail of A all the way to the head of B, all the way to the head of B, and you call that vector C, that is the sum of A and B. Let's say these were displacement vectors. So A shows that you're being displaced this much in this direction. B shows that you're being displaced this much in this direction. So the length of B in that direction. And if I were to say you have a displacement of A, and then you have a displacement of B, what is your total displacement? So you would have had to be, I guess, shifted this far in this direction, and then you would be shifted this far in this direction. So the net amount that you've been shifted is this far in that direction. So that's why this would be the sum of those. Now we can use that same idea to break down any vector in two dimensions into, we could say, into its components. And I'll give you a better sense of what that means in a second. So if I have vector A. Let me pick a new letter. Let's call this vector \"vector X.\" Let's call this \"vector X.\" this vector right here in green and this vector right here in red. Notice, X starts at the tail of the green vector and goes all the way to the head of the magenta vector. And the magenta vector starts at the head of the green vector and then finishes, I guess, well where it finishes is where vector X finishes. And the reason why I do this... And, you know, hopefully from this comparable explanation right here, says, okay, look, the green vector plus the magenta vector gives us this X vector. That should make sense. I put the head of the green vector to the tail of this magenta vector right over here. But the whole reason why I did this is, if I can express X as a sum of these two vectors, it then breaks down X into its vertical component and its horizontal component. So I could call this" + }, + { + "Q": "\nat 3:00 cant we draw vector A from the tail of vector of vector B\nwill there be any difference", + "A": "No. There will not be any difference. If you find the resultant vector with any head-tail arrangement, you ll still get the same resultant vector.", + "video_name": "xp6ibuI8UuQ", + "timestamps": [ + 180 + ], + "3min_transcript": "by the direction of the arrow. So it's going in that direction. Now let's say I have another vector. Let's call it vector B. Let's call it vector B. It looks like this. Now what I wanna do in this video is think about what happens when I add vector A to vector B. So there's a couple things to think about when you visually depict vectors. The important thing is, for example, for vector A, that you get the length right and you get the direction right. Where you actually draw it doesn't matter. So this could be vector A. This could also be vector A. Notice, it has the same length and it has the same direction. This is also vector A. I could draw vector A up here. It does not matter. I could draw vector A up there. I could draw vector B. I could draw vector B over here. It's still vector B. It still has the same magnitude and direction. Notice, we're not saying that its tail has to start at the same place that vector A's tail starts at. I could draw vector B over here. So I can always have the same vector but I can shift it around. As long as it has the same magnitude, the same length, and the same direction. And the whole reason I'm doing that is because the way to visually add vectors... If I wanted to add vector A plus vector B... And I'll show you how to do it more analytically in a future video. I can literally draw vector A. I draw vector A. So that's vector A, right over there. And then I can draw vector B, but I put the tail of vector B to the head of vector A. So I shift vector B over so its tail is right at the head of vector A. And then vector B would look something like this. It would look something like this. And then if you go from the tail of A all the way to the head of B, all the way to the head of B, and you call that vector C, that is the sum of A and B. Let's say these were displacement vectors. So A shows that you're being displaced this much in this direction. B shows that you're being displaced this much in this direction. So the length of B in that direction. And if I were to say you have a displacement of A, and then you have a displacement of B, what is your total displacement? So you would have had to be, I guess, shifted this far in this direction, and then you would be shifted this far in this direction. So the net amount that you've been shifted is this far in that direction. So that's why this would be the sum of those. Now we can use that same idea to break down any vector in two dimensions into, we could say, into its components. And I'll give you a better sense of what that means in a second. So if I have vector A. Let me pick a new letter. Let's call this vector \"vector X.\" Let's call this \"vector X.\"" + }, + { + "Q": "Starting after 6:00 with the new problem, why were the numbers used so specific?\n", + "A": "Well, for an example, if you take out your calculator and take the sin of a random number, it ll most likely be a non-terminating number and will look like something just vomited random numbers out. I think he used such a specific angle on purpose to get his answer close to 3 in this problem. A nice, whole number is better than something like 3.58912385423168.", + "video_name": "xp6ibuI8UuQ", + "timestamps": [ + 360 + ], + "3min_transcript": "this vector right here in green and this vector right here in red. Notice, X starts at the tail of the green vector and goes all the way to the head of the magenta vector. And the magenta vector starts at the head of the green vector and then finishes, I guess, well where it finishes is where vector X finishes. And the reason why I do this... And, you know, hopefully from this comparable explanation right here, says, okay, look, the green vector plus the magenta vector gives us this X vector. That should make sense. I put the head of the green vector to the tail of this magenta vector right over here. But the whole reason why I did this is, if I can express X as a sum of these two vectors, it then breaks down X into its vertical component and its horizontal component. So I could call this or I should say the vertical component. X vertical. And then I could call this over here the X horizontal. Or another way I could draw it, I could shift this X vertical over. Remember, it doesn't matter where I draw it, as long as it has the same magnitude and direction. And I could draw it like this. X vertical. And so what you see is is that you could express this vector X... Let me do it in the same colors. You can express this vector X as the sum of its horizontal and its vertical components. As the sum of its horizontal and its vertical components. Now we're gonna see over and over again that this is super powerful because what it can do is it can turn a two-dimensional problem into two separate one-dimensional problems, one acting in a horizontal direction, one acting in a vertical direction. Now let's do it a little bit more mathematical. I've just been telling you about length and all of that. Let me just show you what this means, to break down the components of a vector. So let's say that I have a vector that looks like this. Let me do my best to... Let's say I have a vector that looks like this. It's length is five. So let me call this vector A. So vector A's length is equal to five. And let's say that its direction... We're gonna give its direction by the angle between the direction its pointing in and the positive X axis. So maybe I'll draw an axis over here. So let's say that this right over here is the positive Y axis going in the vertical direction. This right over here is the positive X axis going in the horizontal direction. And to specify this vector's direction I will give this angle right over here. And I'm gonna give a very peculiar angle, but I picked this for a specific reason, just so things work out neatly in the end." + }, + { + "Q": "\nAt 5:25, Can the same fusion process that causes Hydrogen to turn into Deuterium and then Helium continue past Helium and make all the elements given enough gravity?", + "A": "Yes, the next few videos explain it pretty well.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 325 + ], + "3min_transcript": "to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original. results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling. This is still hydrogen because it has one proton and one neutron now. It is not helium yet. This does not have two-- it does not have two protons. But then the deuterium keeps fusing. And then we eventually end up with helium. And we can even see that on the periodic table. Oh, I lost my periodic table. Well, I'll show you the next video. But we know hydrogen in its atomic state has an atomic number of 1. And it also has a mass of 1. It only has one nucleon in its nucleus. But it's being fused. It goes to hydrogen-2, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused-- and I'm not going into the detail of the reaction-- into helium. And by definition, helium has two protons and two neutrons. So it has-- or we're talking about helium-4, in particular, that isotope of helium-- it has an atomic mass of 4." + }, + { + "Q": "\nIn 1:14 you have said that the atoms condense. But condensation happens only when temperature reduces. But you have said temperature increases. Aren't they contradicting?", + "A": "Not condensation as the moisture that appears on cool objects. Condensing as in the atoms fill a smaller volume, getting denser. This builds up pressure, which builds temperature.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 74 + ], + "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring" + }, + { + "Q": "At around 5:37, what is a nucleon? Sorry if I missed it in another video...\n", + "A": "A nucleon is a particle that exists in the center of an atom, like a proton or neutron.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 337 + ], + "3min_transcript": "results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling. This is still hydrogen because it has one proton and one neutron now. It is not helium yet. This does not have two-- it does not have two protons. But then the deuterium keeps fusing. And then we eventually end up with helium. And we can even see that on the periodic table. Oh, I lost my periodic table. Well, I'll show you the next video. But we know hydrogen in its atomic state has an atomic number of 1. And it also has a mass of 1. It only has one nucleon in its nucleus. But it's being fused. It goes to hydrogen-2, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused-- and I'm not going into the detail of the reaction-- into helium. And by definition, helium has two protons and two neutrons. So it has-- or we're talking about helium-4, in particular, that isotope of helium-- it has an atomic mass of 4. Because the atomic mass of the helium that gets produced is slightly lower than four times the atomic mass of each of the constituent hydrogens. So all of this energy, all this energy from the fusion-- but it needs super high pressure, super high temperatures to happen-- keeps the star from collapsing. And once a star is in this stage, once it is using hydrogen-- it is fusing hydrogen in its core, where the pressure and the temperature is the most, to form helium-- it is now in its main sequence. This is now a main sequence star. And that's actually where the sun is right now. Now there's questions of, well, what if there just wasn't enough mass to get to this level over here? And there actually are things that never get to quite that threshold to fuse all the way into helium. There are a few things that don't quite" + }, + { + "Q": "At 3:52, Sal says that one of the protons degrades into a neutron - how does that happen?\n", + "A": "The instability essentially forces one of the constituent up quarks to decay into a down quark, resulting in a proton becoming a neutron.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 232 + ], + "3min_transcript": "Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original. results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling." + }, + { + "Q": "At 5:56, why it wouldn't form tritium from deuterium?\n", + "A": "It can if a free neutron fuses with deuterium. But free neutrons aren t readily available at the early stages of stellar fusion, and tritium is unstable and will decay to Helium-3.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 356 + ], + "3min_transcript": "results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling. This is still hydrogen because it has one proton and one neutron now. It is not helium yet. This does not have two-- it does not have two protons. But then the deuterium keeps fusing. And then we eventually end up with helium. And we can even see that on the periodic table. Oh, I lost my periodic table. Well, I'll show you the next video. But we know hydrogen in its atomic state has an atomic number of 1. And it also has a mass of 1. It only has one nucleon in its nucleus. But it's being fused. It goes to hydrogen-2, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused-- and I'm not going into the detail of the reaction-- into helium. And by definition, helium has two protons and two neutrons. So it has-- or we're talking about helium-4, in particular, that isotope of helium-- it has an atomic mass of 4. Because the atomic mass of the helium that gets produced is slightly lower than four times the atomic mass of each of the constituent hydrogens. So all of this energy, all this energy from the fusion-- but it needs super high pressure, super high temperatures to happen-- keeps the star from collapsing. And once a star is in this stage, once it is using hydrogen-- it is fusing hydrogen in its core, where the pressure and the temperature is the most, to form helium-- it is now in its main sequence. This is now a main sequence star. And that's actually where the sun is right now. Now there's questions of, well, what if there just wasn't enough mass to get to this level over here? And there actually are things that never get to quite that threshold to fuse all the way into helium. There are a few things that don't quite" + }, + { + "Q": "@ 0:18 sal is talking about gravity...well, what type of gravity is present in the stars...?\nand how these hydrogen atoms are attracted to each other due to gravity...?\n", + "A": "There is only one type of gravity Hydrogen atoms are attracted to each other because they have mass All mass attracts other mass. That s what gravity is.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 18 + ], + "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring" + }, + { + "Q": "\nAt 2:20, why don't the positivily charged nucleuses want to go any where near each other?", + "A": "The positively charged nuclei don t want to go anywhere near each other because both of them a positively charged, and just like a magnet, positives don t attract each other.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 140 + ], + "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring" + }, + { + "Q": "\nat 3:40 sal says when the fusion occurs, the atoms that fuse weigh a little bit less. One of the atoms turns into a neutron. Where does that extra weight go? Is it turned into energy or another substance? If it's energy it needs to have weight and that's not possible.", + "A": "Well, that doesn t exactly happen, You must have got confused between Beta decay and fusion, where the proton emits a positron and turns into a neutron. But that isn t exactly the case over here. The fusion occurs when two hydrogen atoms fuse together, and no weight is lost either.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 220 + ], + "3min_transcript": "Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original. results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling." + }, + { + "Q": "\nat 0:45 why does cloud get dense", + "A": "because of the forces of attraction acting over forces of repulsion in between the two nucleus and this works because the size is massive.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 45 + ], + "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring" + }, + { + "Q": "\nAt 0:03, it is said that we consider that two things have the same charge. What if they are not of the same charge? Will like charges not repel each other in that case? And opposite charges attract each other?", + "A": "Like charges always repel. Opposites always attract.", + "video_name": "2GQTfpDE9DQ", + "timestamps": [ + 3 + ], + "3min_transcript": "- [Voiceover] So we've already started to familiarize ourselves with the notion of charge. We've seen that if two things have the same charge, so they're either both positive, or they are both negative, then they are going to repel each other. So in either of these cases these things are going to repel each other. But if they have different charges, they are going to attract each other. So if I have a positive and I have a negative they are going to attract each other. This charge is a property of matter that we've started to observe. We've started to observe of how these different charges, this framework that we've created, how these things start to interact with each other. So these things are going to, these two things are going to attract each other. But the question is, what causes, how can we predict how strong the force of attraction or repulsion is going to be between charged particles? And this was a question people have noticed, I guess what you could call electrostatics, for a large swathe of recorded human history. But it wasn't until the 16 hundreds and especially the 17 hundreds, as something that they could manipulate and even start to predict in a kind of serious, mathematical, scientific way. And it wasn't until 1785, and there were many that came before Coulomb, but in 1785 Coulomb formally published what is known as Coulomb's law. And the purpose of Coulomb's law, Coulomb's law, is to predict what is going to be the force of the electrostatic force of attraction or repulsion between two forces. And so in Coulomb's law, what it states is is if I have two charges, so let me, let's say this charge right over here, and I'm gonna make it in white, because it could be positive or negative, but I'll just make it q one, it has some charge. And then I have in Coulombs. and then another charge q two right over here. Another charge, q two. And then I have the distance between them being r. So the distance between these two charges is going to be r. that the magnitude of the force, so it could be a repulsive force or it could be an attractive force, which would tell us the direction of the force between the two charges, but the magnitude of the force, which I'll just write it as F, the magnitude of the electrostatic force, I'll write this sub e here, this subscript e for electrostatic. Coulomb stated, well this is going to be, and he tested this, he didn't just kind of guess this. People actually were assuming that it had something to do with the products of the magnitude of the charges and that as the particles got further and further away the electrostatic force dissipated. But he was able to actually measure this and feel really good about stating this law. Saying that the magnitude of the electrostatic force is proportional, is proportional, to the product of the magnitudes of the charges. So I could write this as q one times q two, and I could take the absolute value of each, which is the same thing as just" + }, + { + "Q": "At 11:13, why does the final answer have two significant figures when the measurements only have 1 significant figure? I thought that, when multiplying or dividing significant figures, the result can only have as much significant figures as the measurement with the least amount of significant figures (which in this case is 1).\n", + "A": "sal often makes mistakes with sig figs", + "video_name": "2GQTfpDE9DQ", + "timestamps": [ + 673 + ], + "3min_transcript": "so that negative is going to go away. All of that over, all of that over and we're in kind of the home stretch right over here, 0.5 meters squared. 0.5 meters squared. And so, let's just do a little bit of the math here. So first of all, let's look at the units. So we have Coulomb squared here, then we're going to have Coulombs times Coulombs there that's Coulombs squared divided by Coulombs squared that's going to cancel with that and that. You have meters squared here, and actually let me just write it out, so the numerator, in the numerator, we are going to have so if we just say nine times five times, when we take the absolute value, it's just going to be one. So nine times five is going to be, nine times five times negative... five times negative one is negative five, but the absolute value there, so it's just going to be five times nine. So it's going to be 45 times 10 to the nine, minus three, minus one. so that's going to be 10 to the fifth, 10 to the fifth, the Coulombs already cancelled out, and we're going to have Newton meter squared over, over 0.25 meters squared. These cancel. And so we are left with, well if you divide by 0.25, that's the same thing as dividing by 1/4, which is the same thing as multiplying by four. So if you multiply this times four, 45 times four is 160 plus 20 is equal to 180 times 10 to the fifth Newtons. And if we wanted to write it in scientific notation, well we could divide this by, we could divide this by 100 and then multiply this by 100 and so you could write this as 1.80 times one point... and actually I don't wanna make it look like I have more significant digits than I really have. 1.8 times times 10 to the seventh units, I just divided this by 100 and I multiplied this by 100. And we're done. This is the magnitude of the electrostatic force between those two particles. And it looks like it's fairly significant, and this is actually a good amount, and that's because this is actually a good amount of charge, a lot of charge. Especially at this distance right over here. And the next thing we have to think about, well if we want not just the magnitude, we also want the direction, well, they're different charges. So this is going to be an attractive force. This is going to be an attractive force on each of them acting at 1.8 times ten to the seventh Newtons. If they were the same charge, it would be a repulsive force, or they would repel each other with this force. But we're done." + }, + { + "Q": "In 11:04s why the [10^-4*10^-1] do not have to take off the minus sign? Isn't it a absolute value?\n", + "A": "There is a modulus sign there so -tive ans will not be considered\u00f0\u009f\u0098\u008a", + "video_name": "2GQTfpDE9DQ", + "timestamps": [ + 664 + ], + "3min_transcript": "so that negative is going to go away. All of that over, all of that over and we're in kind of the home stretch right over here, 0.5 meters squared. 0.5 meters squared. And so, let's just do a little bit of the math here. So first of all, let's look at the units. So we have Coulomb squared here, then we're going to have Coulombs times Coulombs there that's Coulombs squared divided by Coulombs squared that's going to cancel with that and that. You have meters squared here, and actually let me just write it out, so the numerator, in the numerator, we are going to have so if we just say nine times five times, when we take the absolute value, it's just going to be one. So nine times five is going to be, nine times five times negative... five times negative one is negative five, but the absolute value there, so it's just going to be five times nine. So it's going to be 45 times 10 to the nine, minus three, minus one. so that's going to be 10 to the fifth, 10 to the fifth, the Coulombs already cancelled out, and we're going to have Newton meter squared over, over 0.25 meters squared. These cancel. And so we are left with, well if you divide by 0.25, that's the same thing as dividing by 1/4, which is the same thing as multiplying by four. So if you multiply this times four, 45 times four is 160 plus 20 is equal to 180 times 10 to the fifth Newtons. And if we wanted to write it in scientific notation, well we could divide this by, we could divide this by 100 and then multiply this by 100 and so you could write this as 1.80 times one point... and actually I don't wanna make it look like I have more significant digits than I really have. 1.8 times times 10 to the seventh units, I just divided this by 100 and I multiplied this by 100. And we're done. This is the magnitude of the electrostatic force between those two particles. And it looks like it's fairly significant, and this is actually a good amount, and that's because this is actually a good amount of charge, a lot of charge. Especially at this distance right over here. And the next thing we have to think about, well if we want not just the magnitude, we also want the direction, well, they're different charges. So this is going to be an attractive force. This is going to be an attractive force on each of them acting at 1.8 times ten to the seventh Newtons. If they were the same charge, it would be a repulsive force, or they would repel each other with this force. But we're done." + }, + { + "Q": "\nAt 8:28: The Coulomb's Law equation is used to calculate the electrostatic force (in Newtons) between two charges. But is there an independent way to measure this force (and thus verify that Coulomb's equation is correct)?", + "A": "You could think about lot of ways of measuring this, you could set the two charges (known charge) at a known distance and use a spring to keep them in place, you could extrapolate the force through the hook law, by knowing the constant or the spring. Or,for not depending on hooke law you could put your guessed Force(given by the Coulomb\u00c2\u00b4s equation) in the way of weight, and see if a spring , identical to the ones in the beginning , elongated the same, if it does the formula is correct.", + "video_name": "2GQTfpDE9DQ", + "timestamps": [ + 508 + ], + "3min_transcript": "let's that this distance right here is 0.5 meters. So given that, let's figure out what the what the electrostatic force between these two are going to be. it's going to be an attractive force because they have different signs. And that was actually part of Coulomb's law. This is the magnitude of the force, if these have different signs, it's attractive, if they have the same sign then they are going to repel each other. And I know what you're saying, \"Well in order to actually calculate it, \"I need to know what K is.\" What is this electrostatic constant? What is this electrostatic constant going to actually be? And so you can measure that with a lot of precision, and we have kind of modern numbers on it, but the electrostatic constant, especially for the sake of this problem, I mean if we were to get really precise it's 8.987551, we could keep gone on and on times 10 to the ninth. But for the sake of our little example here, where we really only have one significant digit for each of these. it'll make the math a little bit easier, I won't have to get a calculator out, let's just say it's approximately nine times 10 to the ninth. Nine times 10 to the ninth. Nine times, actually let me make sure it says approximately, because I am approximating here, nine times 10 to the ninth. And what are the units going to be? Well in the numerator here, where I multiply Coulombs times Coulombs, I'm going to get Coulombs squared. This right over here is going to give me, that's gonna give me Coulombs squared. And this down over here is going to give me meters squared. This is going to give me meters squared. And what I want is to get rid of the Coulombs and the meters and end up with just the Newtons. And so the units here are actually, the units here are Newtons. Newton and then meters squared, and that cancels out with the meters squared in the denominator. Newton meter squared over Coulomb squared. Over, over Coulomb squared. Over, over Coulomb squared. So, these meter squared will cancel those. Those Coulomb squared in the denomin... over here will cancel with those, and you'll be just left with Newtons. But let's actually do that. Let's apply it to this example. I encourage you to pause the video and apply this information to Coulomb's law and figure out what the electrostatic force between these two particles is going to be. So I'm assuming you've had your go at it. So it is going to be, and this is really just applying the formula. It's going to be nine times 10 to the ninth, nine times 10 to the ninth, and I'll write the units here, Newtons meter squared over Coulomb squared. And then q one times q two, so this is going to be, let's see, this is going to be, actually let me just write it all out for this first this first time. So it's going to be times five times ten to the negative three Coulombs. Times, times negative one. Time ten to the negative one Coulombs" + }, + { + "Q": "is there any mistakes on this video?\nI guess after meiosis I, you duplicate one diploid cell to two diploid cell; so that for meiosis II, each of two diploid cell will be \"simplified\" to four haploid cell? (4:14)\n", + "A": "Yes, two division events occur. You start with a replication of DNA, then divide the cells so you have 2 diploid cells, then you divide each of those cells to get 4 haploid cells.", + "video_name": "IQJ4DBkCnco", + "timestamps": [ + 254 + ], + "3min_transcript": "that has a diploid number of chromosomes. And in it's interphase, it also replicates its DNA. And then it goes through something called Meiosis One. And in Meiosis One, what you end up with is two cells that now have haploid number of chromosomes. So you end up with two cells, You now have two cells that each have a haploid number of chromosomes. So you have n and you have n. So if we're talking about human beings, you have 46 chromosomes here, and now you have 23 chromosomes in this nucleus. And now you have 23 in this nucleus. But you're still not done. Then each of these will go through a phase, which I'll talk about in a second, which is very similar to mitosis, which will duplicate this entire cell into two. So actually, let me do it like this. So now, this one, that each have the haploid number that each have the haploid number of chromosomes. And they don't all necessarily have the same genetic informatioin anymore. Because as we go through this first phase, right over here of meiosis, and this first phase here you go from diploid to haploid, right over here, this is called Meiosis One. Meiosis One, you're essentially splitting the homologous pairs and so this one might get some of the ones that you originally got from your father, and some that you originally got from your mother, some that you originally got from your father, some that you originally got from your mother, they split randomly, but each homogolous pair gets split up. And then in this phase, Meiosis Two, so this phase right over here is called Meiosis Two, it's very similar to mitosis, except your now dealing with cells that start off with the haploid number. These cells that you have over here, these are gametes. This are sex cells. These are gametes. This can now be used in fertilization. If we're talking about, if you're male, this is happening in your testes, and these are going to be sperm cells If you are female, this is happening in your ovaries and these are going to be egg cells. If you a tree, this could be pollen or it could be an ovul. But these are used for fertilization. These will fuse together in sexual reproduction to get to a fertilized egg, which then can undergo mitosis to create an entirely new organism. So not a cycle here, although these will find sex cells from another organism and fuse with them and those can turn into another organism. And I guess the whole circle of life starts again." + }, + { + "Q": "\nAt 4:22, it's noted that the cells at the end of meiosis II become gametes. Why can't the cells produced at the end of meiosis I become gametes, if they're also haploid cells? And if meiosis isn't a cycle, how can so many gametes be produced--are there just a ton of diploid cells undergoing meiosis?", + "A": "Prior to a cell division of a diploid cell, chromosomes are duplicated and condensed. In meiosis I homologous chromosomes swap chunks of genes with each other (called crossing over), then they are randomly assorted into two cells. Meioses II is responsible for separating the sister chromatids of a chromosome (of the two cells formed in meiosis I) yielding two more cells that are also haploid, for a total of four haploid cells.", + "video_name": "IQJ4DBkCnco", + "timestamps": [ + 262 + ], + "3min_transcript": "that has a diploid number of chromosomes. And in it's interphase, it also replicates its DNA. And then it goes through something called Meiosis One. And in Meiosis One, what you end up with is two cells that now have haploid number of chromosomes. So you end up with two cells, You now have two cells that each have a haploid number of chromosomes. So you have n and you have n. So if we're talking about human beings, you have 46 chromosomes here, and now you have 23 chromosomes in this nucleus. And now you have 23 in this nucleus. But you're still not done. Then each of these will go through a phase, which I'll talk about in a second, which is very similar to mitosis, which will duplicate this entire cell into two. So actually, let me do it like this. So now, this one, that each have the haploid number that each have the haploid number of chromosomes. And they don't all necessarily have the same genetic informatioin anymore. Because as we go through this first phase, right over here of meiosis, and this first phase here you go from diploid to haploid, right over here, this is called Meiosis One. Meiosis One, you're essentially splitting the homologous pairs and so this one might get some of the ones that you originally got from your father, and some that you originally got from your mother, some that you originally got from your father, some that you originally got from your mother, they split randomly, but each homogolous pair gets split up. And then in this phase, Meiosis Two, so this phase right over here is called Meiosis Two, it's very similar to mitosis, except your now dealing with cells that start off with the haploid number. These cells that you have over here, these are gametes. This are sex cells. These are gametes. This can now be used in fertilization. If we're talking about, if you're male, this is happening in your testes, and these are going to be sperm cells If you are female, this is happening in your ovaries and these are going to be egg cells. If you a tree, this could be pollen or it could be an ovul. But these are used for fertilization. These will fuse together in sexual reproduction to get to a fertilized egg, which then can undergo mitosis to create an entirely new organism. So not a cycle here, although these will find sex cells from another organism and fuse with them and those can turn into another organism. And I guess the whole circle of life starts again." + }, + { + "Q": "\naround 2:40, sal says that each of the two cells after the first division has 23 chromosomes. I thought that during S1 the parent cell duplicates its chromosomes to 46 pairs, 92 total, while still being 2n, and then the homologous pairs split and the two cells then each has 46 chromosomes 2n. Then after meiosis 2 each of the four daughter cells has 23 chromosomes. Is this correct?", + "A": "The chromosomes are still considered 1 because they are connected. The sister chromatids are just split apart.", + "video_name": "IQJ4DBkCnco", + "timestamps": [ + 160 + ], + "3min_transcript": "are just like this cell was, it can go through interphase again. It grows and it can replicate its DNA and centrosomes and grow some more then each of these can go through mitosis again. And this is actually how most of the cells in your body grow. This is how you turn from a single cell organism into you, or for the most part, into you. So that is mitosis. It's a cycle. After each of these things go through mitosis, they can then go through the entire cell cycle again. Let me write this a little bit neater. Mitosis, that s was a little bit hard to read. Now what happens in meiosis? What happens in meiosis? I'll do that over here. In meiosis, something slightly different happens and it happens in two phases. You will start with a cell that has a diploid number of chromosomes. that has a diploid number of chromosomes. And in it's interphase, it also replicates its DNA. And then it goes through something called Meiosis One. And in Meiosis One, what you end up with is two cells that now have haploid number of chromosomes. So you end up with two cells, You now have two cells that each have a haploid number of chromosomes. So you have n and you have n. So if we're talking about human beings, you have 46 chromosomes here, and now you have 23 chromosomes in this nucleus. And now you have 23 in this nucleus. But you're still not done. Then each of these will go through a phase, which I'll talk about in a second, which is very similar to mitosis, which will duplicate this entire cell into two. So actually, let me do it like this. So now, this one, that each have the haploid number that each have the haploid number of chromosomes. And they don't all necessarily have the same genetic informatioin anymore. Because as we go through this first phase, right over here of meiosis, and this first phase here you go from diploid to haploid, right over here, this is called Meiosis One. Meiosis One, you're essentially splitting the homologous pairs and so this one might get some of the ones that you originally got from your father, and some that you originally got from your mother, some that you originally got from your father, some that you originally got from your mother, they split randomly, but each homogolous pair gets split up. And then in this phase, Meiosis Two, so this phase right over here is called Meiosis Two, it's very similar to mitosis, except your now dealing with cells that start off with the haploid number." + }, + { + "Q": "at 0:02 sal sayes arbitrary amino acids. what does that mean?\n", + "A": "Any of a class of organic compounds that contains at least one amino group, \u00e2\u0080\u0093NH 2, and one carboxyl group, \u00e2\u0080\u0093COOH: the alpha-amino acids, RCH(NH 2)COOH, are the building blocks from which proteins are constructed. Your welcome!", + "video_name": "nv2kfBFkv4s", + "timestamps": [ + 2 + ], + "3min_transcript": "- [Voiceover] So I've got two arbitrary amino acids here. We recognize the telltale signs of an amino acid. We have an amino group right over here that gives us the amino and amino acid. We have a carboxyl group right over here. This is the acid part of an amino acid. And in between we have a carbon, and we call that the alpha carbon. And that alpha carbon is gonna be bonded to a hydrogen and some type of a side chain, and we're just gonna call this side chain R1, and then we're gonna call this side chain R2. And what we're gonna concern ourselves with in this video is, how do you take two amino acids and form a peptide out of them? And just as a reminder, a peptide is nothing more... than a chain of amino acids. And so, how do you take these two amino acids and form a dipeptide like this? That would be the smallest possible peptide, but then you could keep adding amino acids and form polypeptides. And a very high-level overview of this reaction is that this nitrogen uses its lone pair to form a bond with this carbonyl carbon right over here. So this lone pair goes to this carbonyl carbon, forms a bond, and then this hydrogen, this hydrogen, and this oxygen could be used net net to form a water molecule... that's let go from both of these amino acids. So this reaction, you end up with the nitrogen being attached to this carbon, and a release of a water molecule. And because you have the release of this water molecule, this type of reaction, and we've seen it many other times with other types of molecules, we call this a condensation reaction, or a dehydration synthesis. So condensation... or dehydration synthesis. We saw this type of reaction when we were putting glucoses together, when we were forming carbohydrates. Dehydration synthesis. But whenever I see a reaction like this, it's somewhat satisfying to just be able to do the counting and say, \"All right, this is gonna bond \"with that, we see the bond right over there, \"and I'm gonna let go of an oxygen and two hydrogens, \"which net net equals H2O, equals a water molecule.\" But how can we actually imagine this happening? Can we push the electrons around? Can we do a little bit of high-level organic chemistry to think about how this happens? And that's what I wanna do here. I'm not gonna do a formal reaction mechanism, but really get a sense of what's going on. Well, nitrogen, as we said, has got this lone pair, it's electronegative. And this carbon right over here, it's attached to two oxygens, oxygens are more electronegative. The oxygens might hog those electrons." + }, + { + "Q": "\nAt about 13:35 when he says it's intuitive for Heat to be represented as Q, because heat does not start with Q, was he being sarcastic, or am I missing some reason as to why Q is heat?", + "A": "Sal is being sarcastic about Q being an intuitive representation of heat, since Q is completely unrelated to the word heat. However, he is serious about the use of Q to represent heat, as an accepted method.", + "video_name": "Xb05CaG7TsQ", + "timestamps": [ + 815 + ], + "3min_transcript": "We're going to assume we're dealing with an ideal gas. And even better, we're going to assume we're dealing with a monoatomic ideal gas. And maybe this is just helium, or neon. One of the ideal gases. They don't want to bond with each other. They don't form molecules with each other. Let's just assume that they're not. They're just individual atoms. And in that case, the internal energy, we really can simplify to it being the kinetic energy, if we ignore all of these other things. But it's important to realize, internal energy is everything. It's all of the energy inside of a system. If you said, what's the energy of the system? Its internal energy. So the first law of thermodynamics says that energy cannot be created or destroyed, only transformed. So let's say that internal energy is changing. So I have this system, and someone tells me, look, the So delta U, that's just a capital delta that says, what is the change an internal energy? It's saying, look, if your internal energy is changing, your system is either having something done to it, or it's doing something to someone else. Some energy is being transferred to it or away from it. So, how do we write that? Well the first law of thermodynamics, or even the definition of internal energy, says that a change in internal energy is equal to heat added to the system-- and once again a very intuitive letter for heat, because heat does not start with Q, but the convention is to use Q for heat. The letter h is reserved for enthalpy, which is a very, very, very similar concept to heat. We'll talk about that maybe in the next video. It's equal to the heat added to the system, minus the work done by the system. Sometimes it's written like this. Sometimes it's written that the change in internal energy is equal to the heat added to the system, plus the work done on the system. And this might be very confusing, but you should just always-- and we'll really kind of look at this 100 different And actually this is a capital U. Let me make sure that I write that as a capital U. But we're going to do it 100 different ways. But if you think about it, if I'm doing work I lose energy. I've transferred the energy to someone else. So this is doing work. Likewise, if someone is giving me heat that is increasing my energy, at least to me these are reasonably intuitive Now if you see this, you say, OK, if my energy is going up, if this is a positive thing, I either have to have this go up, or work is being done to me." + }, + { + "Q": "At 2:50 why does he round the mass of the elements? Isn't it more accurate to keep it unrounded?\n", + "A": "He does that to make the examples easier to follow. In real calculations you cannot do that. It is standard practice to use 4 significant digits for your atomic masses, unless your problem requires using more.", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 170 + ], + "3min_transcript": "" + }, + { + "Q": "\n2:15 , 4:40 , 8:42 ! Why is the Charge on each of the individual Capacitors in series is SAME as the charge on the Equivalent Capacitor ?\nFull video is on this Idea!!\nBut, Why ?", + "A": "Picture each conductor with one end at a capacitor plate, and the other end at another plate. Since this conductor needs to be with total charge 0, then if the plate at the right has charge +Q, the other must have charge -Q. Thus, the charge of each pair of plates is the same.", + "video_name": "-MaD9Ycy3a4", + "timestamps": [ + 135, + 280, + 522 + ], + "3min_transcript": "Having to deal with a single capacitor hooked up to a battery isn't all that difficult, but when you have multiple capacitors, people typically get much, much more confused. There's all kinds of different ways to hook up multiple capacitors. But if capacitors are connected one after the other in this way, we call them capacitors hooked up in series. So say you were taking a test, and on the test it asked you to find the charge on the leftmost capacitor. What some people might try to do is this. Since capacitance is the charge divided by the voltage, they might plug in the capacitance of the leftmost capacitor, which is 4 farads, plug in the voltage of the battery, which is 9 volts. Solving for the charge, they'd get that the leftmost capacitor stores 36 coulombs, which is totally the wrong answer. To try and figure out why and to figure out how to properly deal with this type of scenario, let's look at what's actually going on in this example. will start to flow from the right side of capacitor 3, which makes a negative charge get deposited on the left side of capacitor 1. This makes a negative charge flow from the right side of capacitor 1 on to the left side of capacitor 2. And that makes a negative charge flow from the right side of capacitor 2 on to the left side of capacitor 3. Charges will continue doing this. And it's important to note something here. Because of the way the charging process works, all of the capacitors here must have the same amount of charge stored on them. It's got to be that way. Looking at how these capacitors charge up, there's just nowhere else for the charge to go but on to the next capacitor in the line. This is actually good news. This means that for capacitors in series, the charge stored on every capacitor is going to be the same. So if you find the charge on one of the capacitors, you've found the charge on all of the capacitors. is going to be? Well, there's a trick we can use when dealing with situations like this. We can imagine replacing our three capacitors with just a single equivalent capacitor. If we choose the right value for this single capacitor, then it will store the same amount of charge as each of the three capacitors in series will. The reason this is useful is because we know how to deal with a single capacitor. We call this imaginary single capacitor that's replacing multiple capacitors the \"equivalent capacitor.\" It's called the equivalent capacitor because its effect on the circuit is, well, equivalent to the sum total effect that the individual capacitors have on the circuit. And it turns out that there's a handy formula that lets you determine the equivalent capacitance. The formula to find the equivalent capacitance of capacitors hooked up in series looks like this." + }, + { + "Q": "\nAt 6:32, why is the charge stored on each of the individual capacitors equal to the charge stored on the equivalent capacitor? Why aren't the charges divided between the four- like each one has 192/4 C of charge?", + "A": "NO, remember that the Capacitance unit is F, not C, So basically you messed up, you should NOT sum like this, they have the same amount of Charge NOT Capacitance. So you add (1/48F) + (1/16F) + (1/96F) + (1/32F) = 0.125F, Then taking the reciprocal you get 8F which is the equivalent of CAPACITANCE. When you try to find the Voltage you do this ( 192/48 ) + ( 192/16 ) + ( 192/96 ) + ( 192/32 ) = 24v which is the same voltage of the battery. I Hope that helped!", + "video_name": "-MaD9Ycy3a4", + "timestamps": [ + 392 + ], + "3min_transcript": "This time, let's say you had four capacitors hooked up in series to a 24-volt battery. The arrangement of these capacitors looks a little different from the last example, but all of these capacitors are still in series because they're hooked up one right after the other. In other words, the charge has no choice but to flow directly from one capacitor straight to the next capacitor. So these capacitors are still considered to be in series. Let's try to figure out the charge that's going to be stored on the 16-farad capacitor. We'll use the same process as before. First we imagine replacing the four capacitors with a single equivalent capacitor. We'll use the formula to find the equivalent capacitance of capacitors in series. Plugging in our values, we find that 1 over the equivalent capacitance is going to equal 0.125. Be careful. We still have to take 1 over this value to get that the equivalent capacitance for this circuit Now that we know the equivalent capacitance, we can use the formula capacitance equals charge per voltage. We can plug in the value of the equivalent capacitance, 8 farads. And since we have a single capacitor now, the voltage across that capacitor is going to be the same as the voltage of the battery, which is 24 volts. So we find that our imaginary equivalent capacitor would store a charge of 192 coulombs. This means that the charge on each of the individual capacitors is also going to be 192 coulombs. And this gives us our answer, that the charge on the 16-farad capacitor is going to be 192 coulombs. In fact, we can go even further. Now that we know the charge on each capacitor, we can solve for the voltage that's going to exist across each of the individual capacitors. We'll again use the fact that capacitance is the charge per voltage. If we plug in the values for capacitor one, we'll plug in a capacitance of 32 farads. So we can solve for the voltage across capacitor 1, and we get 6 volts. If we were to do the same calculation for each of the other three capacitors, always being careful that we use their particular values, we'll get that the voltages across the capacitors are 2 volts across the 96-farad capacitor, 12 volts across the 16-fard capacitor, and 4 volts across the 48-farad capacitor. Now, the real reason I had us go through this is because I wanted to show you something neat. If you add up the voltages that exist across each of the capacitors, you'll get 24 volts, the same as the value of the battery. This is no coincidence. If you add up the voltages across the components in any single-loop circuit like this, the sum of the voltages is always going to equal the voltage of the battery. And this principle will actually let" + }, + { + "Q": "\nAt 6:19 why does Sal not put 3d10 instead of 4s2 and then 3d8? Does it make a difference which way you do it?", + "A": "Yes, it makes a difference. 3d\u00c2\u00b9\u00e2\u0081\u00b0 means 10 electrons in the 3d orbitals. 4s\u00c2\u00b2 3d\u00e2\u0081\u00b8 means two electrons in a 4s orbital and 8 electrons in the 3d orbitals. Since the 4s orbital is lower in energy than the 3d orbitals, the 4s orbital gets filled up first. 3d\u00c2\u00b9\u00e2\u0081\u00b0 would represent an excited state of the atom.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 379 + ], + "3min_transcript": "Well we're not going to get to f. But you could write f and g and h and keep going. What's going to happen is you're going to fill this one first, then you're going to fill this one, then that one, then this one, then this one. Let me actually draw it. So what you do is, these are the shells that exist, period. These are the shells that exist, in green. What I'm drawing now isn't the order that you fill them. This is just, they exist. So there is a 3d subshell. There's not a 3f subshell. There is a 4f subshell. Let me draw a line here, just so it becomes a little bit neater. And the way you fill them is you make these diagonals. So first you fill this s shell like that, then you fill this one like that. Then you do this diagonal down like that. Then you do this diagonal down like that. And then this diagonal down like that. six in p, in this case, 10 in d. And we can worry about f in the future, but if you look at the f-block on a periodic table, you know how many there are in f. So you fill it like that. So first you just say, OK. For nickel, 28 electrons. So first I fill this one out. So that's 1s2. 1s2. Then I go, there's no 1p, so then I go to 2s2. Let me do this in a different color. So then I go right here, 2s2. That's that right there. Then I go up to this diagonal, and I come back down. And then there's 2p6. And you have to keep track of how many electrons you're dealing with, in this case. So we're up to 10 now. So we used that one up. Then the arrow tells us to go down here, so now we do the third energy shell. So 3s2. And then where do we go next? Then we follow the arrow. We start there, there's nothing there, there's something here. So we go to 3p6. And then the next thing we fill out is 4s2. So then we go to 4s2. And then what's the very next thing we fill out? We have to go back to the top. We come here and then we fill out 3d. And then how many electrons do we have left to fill out? So we're going to be in 3d. So 3d. And how many have we used so far? 2 plus 2 is 4. 4 plus 6 is 10. 10 plus 2 is 12. 18. 20. We've used 20, so we have 8 more electrons to configure. And the 3d subshell can fit the 8 we need, so we have 3d8. And there you go, you've got the exact same answer that we had when we used the first method. Now I like the first method because you're looking at the periodic table the whole time, so you kind of understand an intuition" + }, + { + "Q": "what does he mean by \"go back to fill the previous shell, so subtract 1\" at 2:27\n", + "A": "when the shells are filled, there are certain spots, where are still no electrons, the d and f blocks fill up those spots(because thats closer to the nucleus).", + "video_name": "YURReI6OJsg", + "timestamps": [ + 147 + ], + "3min_transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that" + }, + { + "Q": "For the element Zr, discussion starting around 7:50, energy level 5s (5s2) is filled before level 4d (4d2). Will the electrons \"peel\" in the opposite order that they're \"put on\"? or will 5s electrons \"peel\" before 4d electrons? for this element, which would be considered the valence electrons? Thanks.\n", + "A": "The electrons fill in that order, but when they are used in bonding ( peeling? ), the electrons with the highest quantum number (period 5 before period 4 electrons) are used first. So the 5s2 will be used before the 4d electrons.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 470 + ], + "3min_transcript": "Then we follow the arrow. We start there, there's nothing there, there's something here. So we go to 3p6. And then the next thing we fill out is 4s2. So then we go to 4s2. And then what's the very next thing we fill out? We have to go back to the top. We come here and then we fill out 3d. And then how many electrons do we have left to fill out? So we're going to be in 3d. So 3d. And how many have we used so far? 2 plus 2 is 4. 4 plus 6 is 10. 10 plus 2 is 12. 18. 20. We've used 20, so we have 8 more electrons to configure. And the 3d subshell can fit the 8 we need, so we have 3d8. And there you go, you've got the exact same answer that we had when we used the first method. Now I like the first method because you're looking at the periodic table the whole time, so you kind of understand an intuition And you also don't have to keep remembering, OK, how many have I used up as I filled the shells? Right? Here you have to say, i used two, then I used two more. And you have to draw this kind of elaborate diagram. Here you can just use the periodic table. And the important thing is you can work backwards. Here there's no way of just eyeballing this and saying, OK, our most energetic electrons are going to be and our highest energy shell is going to be 4s2. There's no way you could get that out of this without going through this fairly involved process. But when do you use this method, you can immediately say, OK, if I'm worried about element Zr, right here. If I'm worried about element Zr. I could go through the whole exercise of filling out the entire electron configuration. But usually the highest shell, or the highest energy electrons, are the ones that matter the most. So you immediately say, OK, I'm filling in 2 d there, but remember, d, you go one period below. Right? Because the period is five. So you say, 4d2. 4d2. And then, before that, you filled out the 5s2 electrons. The 5s2 electrons. And then you could keep going backwards. And you filled out the 4p6. 4p6. And then, before you filled out the 4p6. then you had 10 in the d here. But what is that? It's in the fourth period, but d you subtract one from it, so this is 3d10. So 3d10. And then you had 4s2. This is getting messy. Let me just write that. So you have 4d2. That's those two there. Then you have 5s2. 5s2. Then we had 4p6. That's over here. Then we had 3d10. Remember, 4 minus 1, so 3d10." + }, + { + "Q": "at 2:07 why do we fill 4s orbital 1st than 3d ?\nand if we do so why do we consider 4s as the valence shell and not 3d??\n", + "A": "We put them in that order because that is the order they actually fill up in. The 4s is very much valence. A partially filled 3d is counted as valence electrons by some scientists but not by others. The reason for the disagreement has to do with the way that d electrons get involved in chemical reactions -- it is not as straightforward as what happens with the outermost s and p orbitals.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 127 + ], + "3min_transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that" + }, + { + "Q": "How did Sal at 2:00 know the third shell was not full? Why does the formula 2(n)^2 determine the shells max capacity?\n", + "A": "Sal was using the relative energy levels of the different orbitals to explain the order in which they are filled. The empty 4s orbital is of slightly lower energy than the empty 3d orbitals so is filled first. The formula you quote works simply because of the algebraic relationship between n and the magnetic quantum number.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 120 + ], + "3min_transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that" + }, + { + "Q": "\nAt 8:01 Sal says you have to minus 1 from the period for d. Why is this?", + "A": "It s just how it goes. The 3d orbitals happen to be about the same energy as the 4s orbital. Similar story for the 4d and 5s etc.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 481 + ], + "3min_transcript": "Then we follow the arrow. We start there, there's nothing there, there's something here. So we go to 3p6. And then the next thing we fill out is 4s2. So then we go to 4s2. And then what's the very next thing we fill out? We have to go back to the top. We come here and then we fill out 3d. And then how many electrons do we have left to fill out? So we're going to be in 3d. So 3d. And how many have we used so far? 2 plus 2 is 4. 4 plus 6 is 10. 10 plus 2 is 12. 18. 20. We've used 20, so we have 8 more electrons to configure. And the 3d subshell can fit the 8 we need, so we have 3d8. And there you go, you've got the exact same answer that we had when we used the first method. Now I like the first method because you're looking at the periodic table the whole time, so you kind of understand an intuition And you also don't have to keep remembering, OK, how many have I used up as I filled the shells? Right? Here you have to say, i used two, then I used two more. And you have to draw this kind of elaborate diagram. Here you can just use the periodic table. And the important thing is you can work backwards. Here there's no way of just eyeballing this and saying, OK, our most energetic electrons are going to be and our highest energy shell is going to be 4s2. There's no way you could get that out of this without going through this fairly involved process. But when do you use this method, you can immediately say, OK, if I'm worried about element Zr, right here. If I'm worried about element Zr. I could go through the whole exercise of filling out the entire electron configuration. But usually the highest shell, or the highest energy electrons, are the ones that matter the most. So you immediately say, OK, I'm filling in 2 d there, but remember, d, you go one period below. Right? Because the period is five. So you say, 4d2. 4d2. And then, before that, you filled out the 5s2 electrons. The 5s2 electrons. And then you could keep going backwards. And you filled out the 4p6. 4p6. And then, before you filled out the 4p6. then you had 10 in the d here. But what is that? It's in the fourth period, but d you subtract one from it, so this is 3d10. So 3d10. And then you had 4s2. This is getting messy. Let me just write that. So you have 4d2. That's those two there. Then you have 5s2. 5s2. Then we had 4p6. That's over here. Then we had 3d10. Remember, 4 minus 1, so 3d10." + }, + { + "Q": "At 2:30, Sal says that in the d-block, the period is subtracted by 1. Why is this the case and how was it calculated?\n", + "A": "Because the 4s2 orbital electrons are in a higher energy state than the 3d8 orbital electrons, so they are further from the nucleus. So instead of 4d8, it s 3d8 (-1, as in lower energy). The 4s2 orbital only fills with electrons before the 3d8 orbital", + "video_name": "YURReI6OJsg", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that" + }, + { + "Q": "\nIs there any difference between \"3-phosphoglycerate\" (3:17), and Glycerate 3-phosphate (which is what I have in my textbook)?", + "A": "no they are theoretically the same thing with just a slightly different structure (isomerization)", + "video_name": "DnNqe8o0ehc", + "timestamps": [ + 197 + ], + "3min_transcript": "but because it's a mouthful, the shorthand notation is R-U-B-P. Sometimes people might say Roo-B-P, or I guess you could even say Rube-P somehow, but each of these six Rube-P, or RuBPs, can then react with a carbon dioxide. So if I have six RuBPs, well, they're gonna react with six carbon dioxides, and so one way to think about it is, it's fixing the carbon in that carbon dioxide. It's taking this carbon that's part of this gaseous carbon dioxide, and fixing it as part of an organic molecule. Now, you might be tempted to say, well, it's gonna create six carbon molecules, but then those will immediately become 12 three carbon molecules. And notice, and it's important to keep doing this. Pause the video if you need to. You can make sure that the carbons are all accounted for. Well, we have six times five, so that's 30 carbons right over here, and here we have six times one carbon, so that's six carbons right over here. So if we wanna account for all of our carbons, we should have 36 carbons right over here, and we do. We have 12 three carbon molecules. This three carbon molecules, when we go into some detail here in the video on the Calvin cycle, it's called three phosphoglycerate, but that's not what the focus is on this video. The focus of this video is the enzyme that actually does the fixing of the carbon along with the RuBP. And that enzyme, that character, the character with the quirks that we're going to talk about, the shorthand, its name, you could call it ribulose one, five, bisphosphate oxygenase-carboxylase, but that's even more of a mouthful than RuBP, so people call it the nice friendly name rubisco, rubisco for short. But you can learn a lot about what rubisco does and you can even learn a little bit about its quirk that we're about to talk about. So it obviously involves ribulose one, five bisphosphate, and it does indeed involve that, and then you see oxygenase, dash, carboxylase. Well, the carboxylase is what tells us that it can deal with the carbon dioxide right over here. The carbon dioxide can be one of the substrates in a reaction with the ribulose one, five, bisphosphate. And so that's exactly what it's doing in this reaction. In a normal Calvin cycle, it's acting as a carboxylase. It is fixing that carbon. It's making it part of, if you view, you know, if you view that carbon-- Actually, I won't do it that way because here we have 12 as many. But it's taking these carbon molecules, and it's fixing them into organic molecules, some of which can eventually be used to create glucose. And that's what happens in a typical Calvin cycle." + }, + { + "Q": "At 4:41\nhow can three different orientation of the same sub shell exist at the same time?\nIn other words how can the sub shell be aligned at the x-axis, the y-axis and z-axis at the same time.\n", + "A": "Because there are exactly three p orbitals per shell (except the first)", + "video_name": "qLU0X154wlE", + "timestamps": [ + 281 + ], + "3min_transcript": "that would be two minus one, that would be one. So we go from zero and then we go to one and then we have to stop. So we have only two allowed values for the angular momentum quantum number. So if you have n equal to two, you get two allowed values here. We already talked about what l is equal to zero means, l is equal to zero refers to an s orbital. And there's one s orbital. So in the second energy level, there's another s orbital. This is different from the s orbital in the first energy level that we just talked about. So there's another s orbital here. It too is shaped liked a sphere. What I drew here is misleading. I drew this as being a little bit smaller then the one before. Remember, when n is equal to two, you're further away. You're electron is on average further away from your nucleus here, l is equal to one so if l is equal to one, what is the allowed values for the magnetic quantum number? So remember, the magnetic quantum number So negative l would be negative one, and then we include zero and then we go to positive one. So there are three possible values for the magnetic quantum number. One, two, three, the magnetic quantum number told us the orientation so there are three different orientations. And we talked about l is equal to one is referring to a p orbital which is shaped like a dumbbell. So we have three different orientations, we have three different p orbitals in the second energy level. One of them goes along the x-axis, one of them the y and one of them the z. So we talked about this in the previous video. So a total of three p orbitals here. So how many orbitals are there in the second energy level? Well we have one s orbital and three p orbitals. So one plus three gives us four. We could have done this math, n squared, so two is n squared which gives us four. Alright, let's do electrons now. So four, let's go back to the s orbital here. fit a maximum of two electrons in one orbital. For the p orbitals, we have three p orbitals. If each p orbital's holding a maximum of two electrons, three times two gives us six, so we have a total of eight electrons in the second energy level. So eight electrons and we can get that from two n squared again right? Because if n is equal to two, square that and you get four, multiply that by two and you get eight. Alright, let's go to the third energy level or the third shell here. So when n is equal to three what are the allowed values for the angular momentum quantum number l? So remember l goes from zero all the way up to n minus one. So l goes from zero all the way to n minus one. So l is equal to zero, l is equal to one and l is equal to two because three minus one is equal to two, so if we have n is equal to three," + }, + { + "Q": "\n@ 1:07 Sal says that when silk is rubbed with glass for long, one of the objects might discharge on touching another object ? What does that mean ?", + "A": "That means that the charges are transferred to the earth through that object.", + "video_name": "IDQYakHRAG8", + "timestamps": [ + 67 + ], + "3min_transcript": "- I'm guessing that you've had the experience of rubbing a balloon against your hair and then when you take the balloon away from your hair, your hair sticks up. And if you haven't had that experience, you might think about trying to lead a more rich and fun life, but I'm guessing most of you all have done that. And you had a sense that it had something to do with the balloon or your hair, somehow exchanging charge or now one is going to be more positive or negative than the other, and so now they are somehow attracted. And if you were thinking of those things, you are generally right. What you just experienced after you rubbed the balloon on your head, and then your hair is now attracted to the balloon, that's actually called the triboelectric effect, let me write that down, tribo, triboelectric, electric effect. And human beings have been observing this for a long long time, and it wasn't necessarily with balloons at birthday parties or whatever, it's with other things, they rub a silk cloth on a piece of attraction, or they might see that if they do that enough, one of the objects might discharge when it touches another object. People have observed things like lightning, where it looks like there's some type of a buildup and some type of a potential and then all of a sudden it discharges and you have this lightning and then this thunder blast sound that happens too. So this is something that humans have observed for a long long time, and scientists or people with a, I guess you could say a scientific mind have been trying to understand it for a long long time, and trying to come up with a framework for what exactly is happening. Well lucky for us, we now have a framework for it that explains it quite well. And that framework for what is going on with that triboelectric effect, is a framework around charge. Is a framework that we now have around charge. And this tells us, this way of looking at the world, says look, there's some things that just have a property called charge. Some things have a positive charge, and it's somewhat of an arbitrary name, we just happen to call it positive. And some things have what we say is an opposite charge, or a negative charge, a negative charge. We could have called this the magenta charge, and this the green charge, we could have called this the hippopotamus charge and this the ostrich charge. And we could have said that hippopotami, I believe plural for hippopotamus, they're always attracted to ostriches, but they always repel other hippopotami, and likewise. The like charges repel or like hippo... You get the general idea. But I'll stick to the words that people are used to using. And so if we say something has a charge, say a positive charge, and something else has a negative charge, then in our framework that we're setting up, these two things are going to attract. So opposite charges are going to attract, while like charges are going to repel. So if you have a positive charge, and you have a positive charge," + }, + { + "Q": "\nAt 6:35... when the hair losing electrons , does that affect on the structure of the hair because the atoms of the hair lost electrons ?", + "A": "great question! My answer is this. I dont think it would; with these reasons a) the electrons come from surface of the hair not from inside. So it is less likely to effect structure. b) although the effect is strong , the actual number of electrons exchanged is relatively small. c) your hair is possibly being charged up many times a day as objects (including the wind) rub against it and no harm seems to come of it. ok??", + "video_name": "IDQYakHRAG8", + "timestamps": [ + 395 + ], + "3min_transcript": "We associate mass as just, oh this is just something that we get, we understand it in our everyday life, but even mass, this is just a property of objects, it's just a property of matter, and we feel like we understand it because on our scales we understand notions of things like weight and volume, but even mass can get quite exotic. But anyway, the whole point of this video is not to talk about mass, it's to talk about charge. But all of these things that we talk about in physics, these are just properties that will help us deal with these notions, these behaviors in different frameworks. But anyway, let's get back to this little atom that I was constructing. So this atom, let's say it has two electrons, and obviously this is not drawn to scale, and each of these electrons have a negative charge, and they're kind of jumping around here, buzzing around this nucleus of this atom. And the reason why, this model, even going down to the atomic scale and thinking in protons and electrons is interesting, is that it allows us to start explaining what is happening in the triboelectric effect. rub that balloon on your hair, because of the property of the balloon, the material of the balloon, and the materials of your hair, when they come in contact and they rub, the balloon is grabbing electrons from your hair. So the balloon is grabbing electrons from your hair, and so it is getting more negatively charged, it is getting more negatively charged, and your hair is getting more positively charged, or essentially it's lost these electrons. And so when you put the balloon now close to your hair, remember like charges repel each other, so the electrons in your hair try to move away from these other electrons, the negative charge tries to move away from the negative charge, and I guess you could say that the tips of your hair will then become more positive. Are more positive and they will be attracted, and they will be attracted to the balloon. of transfer of electrons, that's exactly what's happening. And so when you think that way, it's like ok, we are scientists, this is a nice model, we can start to think about what's happening here. This model actually explains a whole ton of behavior that we've observed in the universe, including things like, lightning and whatever else, you know the static shock that you get when you might touch a doorknob after rubbing your shoes along the carpet. But we like to start, we like to quantify things, so we can start seeing how much they repel or how much they attract each other. And so the fundamental unit of charge, or one of the fundamental units of charge, or I guess you could say the elementary unit of charge is defined in terms of the charge of a proton or an electron. So the fundamental, or I guess you could say the elementary unit of charge is denoted by the letter e, and this is the charge of a proton, this is e for elementary, charge of proton." + }, + { + "Q": "\nAt 2:57 why did Sal said like charges are going to repel (It is just a convention). Why can't they attract each other.", + "A": "They just don t. It s like asking why does gravity pull mass together instead of repel them. That s just not how it works. Like charges repel, and unlike charges attract. Why they do that, we don t know.", + "video_name": "IDQYakHRAG8", + "timestamps": [ + 177 + ], + "3min_transcript": "of attraction, or they might see that if they do that enough, one of the objects might discharge when it touches another object. People have observed things like lightning, where it looks like there's some type of a buildup and some type of a potential and then all of a sudden it discharges and you have this lightning and then this thunder blast sound that happens too. So this is something that humans have observed for a long long time, and scientists or people with a, I guess you could say a scientific mind have been trying to understand it for a long long time, and trying to come up with a framework for what exactly is happening. Well lucky for us, we now have a framework for it that explains it quite well. And that framework for what is going on with that triboelectric effect, is a framework around charge. Is a framework that we now have around charge. And this tells us, this way of looking at the world, says look, there's some things that just have a property called charge. Some things have a positive charge, and it's somewhat of an arbitrary name, we just happen to call it positive. And some things have what we say is an opposite charge, or a negative charge, a negative charge. We could have called this the magenta charge, and this the green charge, we could have called this the hippopotamus charge and this the ostrich charge. And we could have said that hippopotami, I believe plural for hippopotamus, they're always attracted to ostriches, but they always repel other hippopotami, and likewise. The like charges repel or like hippo... You get the general idea. But I'll stick to the words that people are used to using. And so if we say something has a charge, say a positive charge, and something else has a negative charge, then in our framework that we're setting up, these two things are going to attract. So opposite charges are going to attract, while like charges are going to repel. So if you have a positive charge, and you have a positive charge, are going to accelerate away from each other. And that's not just true for positive positive, that's also true for negative and negative, these two things are going to repel because they are like charges. Now it's very interesting to think about this because we are so used to thinking in terms of charge, even you know if, especially in kind of the world of electricity you have the positive and negative terminal. You think of charging up your phone or whatever else. That it seems like, we completely, charge is just something that is fundamental about the universe, and that's true to some, that's true, but you'd have to appreciate that these are arbitrary words and they're really just to describe a property that we have observed in the world. And if you go down to the atomic level, we can get to a fundamental level of where the charge is happening. But once again, these are really models for our brain to describe, these are frameworks and models for our brain to be able to predict and describe" + }, + { + "Q": "How does top part of the hair have positive charge and bottom part have negative charge and doesn't the hair build up tension at 6:51?\n", + "A": "By rubbing the balloon electrons move from your hair to the balloon. The top of the hair becomes positively charged because they are attracted to the negative charge of the balloon. The negative ions still in the hair are repelled from the negative charge of the balloon. Yes, it will a little bit!", + "video_name": "IDQYakHRAG8", + "timestamps": [ + 411 + ], + "3min_transcript": "We associate mass as just, oh this is just something that we get, we understand it in our everyday life, but even mass, this is just a property of objects, it's just a property of matter, and we feel like we understand it because on our scales we understand notions of things like weight and volume, but even mass can get quite exotic. But anyway, the whole point of this video is not to talk about mass, it's to talk about charge. But all of these things that we talk about in physics, these are just properties that will help us deal with these notions, these behaviors in different frameworks. But anyway, let's get back to this little atom that I was constructing. So this atom, let's say it has two electrons, and obviously this is not drawn to scale, and each of these electrons have a negative charge, and they're kind of jumping around here, buzzing around this nucleus of this atom. And the reason why, this model, even going down to the atomic scale and thinking in protons and electrons is interesting, is that it allows us to start explaining what is happening in the triboelectric effect. rub that balloon on your hair, because of the property of the balloon, the material of the balloon, and the materials of your hair, when they come in contact and they rub, the balloon is grabbing electrons from your hair. So the balloon is grabbing electrons from your hair, and so it is getting more negatively charged, it is getting more negatively charged, and your hair is getting more positively charged, or essentially it's lost these electrons. And so when you put the balloon now close to your hair, remember like charges repel each other, so the electrons in your hair try to move away from these other electrons, the negative charge tries to move away from the negative charge, and I guess you could say that the tips of your hair will then become more positive. Are more positive and they will be attracted, and they will be attracted to the balloon. of transfer of electrons, that's exactly what's happening. And so when you think that way, it's like ok, we are scientists, this is a nice model, we can start to think about what's happening here. This model actually explains a whole ton of behavior that we've observed in the universe, including things like, lightning and whatever else, you know the static shock that you get when you might touch a doorknob after rubbing your shoes along the carpet. But we like to start, we like to quantify things, so we can start seeing how much they repel or how much they attract each other. And so the fundamental unit of charge, or one of the fundamental units of charge, or I guess you could say the elementary unit of charge is defined in terms of the charge of a proton or an electron. So the fundamental, or I guess you could say the elementary unit of charge is denoted by the letter e, and this is the charge of a proton, this is e for elementary, charge of proton." + }, + { + "Q": "At 6:38, why we got different masses for different points? Aren't they the same? because they are all on the same ball.\n", + "A": "Great question. If you sliced up the baseball into lots of little chunks, then, if you were careful, you might be able to make every chunk have the same mass. But dividing a sphere into many, many little pieces that all have the same volume (and thus mass if the density is constant) is very difficult. So it would probably be the case that each small mass element would have a different mass.", + "video_name": "o7_zmuBweHI", + "timestamps": [ + 398 + ], + "3min_transcript": "times the angular velocity, or the angular speed gives you the regular speed. This formula is really handy, so we're gonna replace V with R omega, and this is gonna give us R omega and you still have to square it and at this point you're probably thinking like this is even worse, what do we do this for. Well watch, if we add this is up I'll have one half M. I'm gonna get an R squared and an omega squared, and the reason this is better is that even though every point on this baseball has a different speed V, they all have the same angular speed omega, that was what was good about these angular quantities is that they're the same for every point on the baseball no matter how far away you are from the axis, and since they're the same for every point I can bring that out of the summation so I can rewrite this summation and bring everything that's constant for all of the masses out of the summation so I can write this as one half times the summation of M times R squared and end that quantity, because it's the same for each term. I'm basically factoring this out of all of these terms in the summation, it's like up here, all of these have a one half. You could imagine factoring out a one half and just writing this whole quantity as one half times M one V one squared plus M two V two squared and so on. That's what I'm doing down here for the one half and for the omega squared, so that's what was good about replacing V with R omega. The omega's the same for all of them, you can bring that out. You might still be concerned, you might be like, we're still stuck with the M in here cause you've got different Ms at different points. We're stuck with all these R squareds in here, all these points at the baseball are different Rs, they're all different points from the axis, different distances from the axis, we can't bring those out so now what do we do, well if you're clever you recognize this term. This summation term is nothing but the total moment of inertia of the object. Remember that the moment of inertia of an object, so the moment of inertia of a point mass is M R squared and the moment of inertia of a bunch of point masses is the sum of all the M R squareds and that's what we've got right here, this is just the moment of inertia of this baseball or whatever the object is, it doesn't even have to be of a particular shape, we're gonna add all the M R squareds, that's always going to be the total moment of inertia. So what we've found is that the K rotational is equal to one half times this quantity, which is I, the moment of inertia, times omega squared and that's the formula we got up here just by guessing. But it actually works and this is why it works, because you always get this quantity down here, which is one half I omega squared, no matter what the shape of the object is. So what this is telling you, what this quantity gives us is the total rotational kinetic energy of all the points on that mass about the center of the mass but here's what it doesn't give you. This term right here does not include" + }, + { + "Q": "dose that dash preforms away must be three lines? or it can be random dashes?\nat 2:40\n", + "A": "It doesn t have to be three lines, just dash it :-)", + "video_name": "ZAgQH2azx3w", + "timestamps": [ + 160 + ], + "3min_transcript": "tetrahedral geometry around that carbon. And if you look at that carbon on the picture here, you can see that this bond and this bond are in the same plane. So if you had a flat sheet of paper, you could say those bonds are in the same plane. So a line represents a bond in the plane of the paper, let me go ahead and draw that, so this is the carbon in magenta, and then we have these two bonds here, and those bonds are in the plane of the paper. Next, let's look at what else is connected to the carbon in magenta. Well, obviously, there is an OH, so let me go ahead and circle that. There is an OH we can see there is an OH here, and the OH, the OH in our picture, is coming out at us in space, so hopefully you can visualize that this bond in here is coming towards you in space, which is why this oxygen, this red oxygen atom, looks so big. So this is coming towards you, so let me go ahead and draw a wedge in here, and a wedge means that the bond is in front of your paper, so this means the OH is coming out at you in space, let me draw in the OH like that. Now let's look at what else is connected to that carbon in magenta, we know there's a hydrogen. We didn't draw it over here, but we know there's a hydrogen connected to that carbon, and we can see that this hydrogen, this hydrogen right here, let me go ahead and switch colors, this hydrogen is going away from us in space, so this bond is going away from us in space, or into the paper, or the bond is behind the paper. And we represent that with a dash, so I'm going to draw a dash here, showing that this hydrogen is going away from us. So we're imagining our flat sheet of paper and the OH coming out at us, and that hydrogen going away from us. All right, next, let's look at the carbon on the left here, so this carbon in blue, so that's this carbon, and I'll say that's this carbon over here on the left. we can see that this bond, and this bond are in the same plane, so let's go ahead and draw in the carbon, so the carbon that I just put in is the carbon in blue, and this hydrogen over here on the left, right, this bond is in the same plane, so I'm going to draw a line representing the bond is in the plane of the paper, and so we have a hydrogen right here. What about the other two hydrogens? Well, let me highlight those, so this hydrogen, hopefully you can see that this is coming out at us in space. So we represent that with a wedge, so we draw a wedge right here, and then we draw in a hydrogen, so the bond is in front of the paper, the bond is coming towards us in space. And there's another hydrogen bonded to the carbon in blue, and my thumb here is blocking it a little bit but hopefully you can see that's going away from us in space, so this hydrogen is going away from us in space. So the carbon in blue, is also SP3 hybridized." + }, + { + "Q": "at around 3:30 he calls carbon-12 an isotope, but i thought that isotopes were defined as atoms that have more neutrons than protons in their nucleus. carbon-12 has 6 protons and 6 neutrons in the nucleus, why is it an isotope?\n", + "A": "All atoms are isotopes. Isotope simply refers to the specific number of nucleons for a given atom. It doesn t matter if the number of neutrons is more, less or equal to the number of protons, or if there are no neutrons, the atom is still a specific isotope.", + "video_name": "NG-rrorZcM8", + "timestamps": [ + 210 + ], + "3min_transcript": "Now weight, on the other hand, is not ... it's different than mass. Weight is a force, it's how much the Earth, or whatever planet you happen to be on, is pulling on you. This right over here is a force. And, in the metric system, you measure weight, not with grams or kilograms, but with Newtons. Newtons. Really, when you ask someone their weight in Europe, they should give it to you in Newtons. If you ask them their mass, what they're telling you is actually their mass. They should say, \"My mass is 60 kilograms,\" or, \"70 kilograms,\" or whatever they might be. It's a very important difference in physics. If I go from Earth to the Moon, my mass does not change, but my weight does change because the force with which the Moon is pulling on me, or that we're pulling on each other, is less than it would be on Earth. In fact, even on the surface of the Earth, if you were to even go to the top of a building, Yeah, it would be very hard to measure it, but you're just slightly further from the center of the Earth, so there's a different gravitational force. Your weight will change ever so slightly, but your mass does not change. You go to deep space, and there's very little gravitational influence, you have pretty much, or close to, zero weight. But you're in deep space, and if there's no planets nearby, but your mass is still going to be whatever your mass happens to be. That's a primer on mass and weight. Now, with that out of the way, I might confuse you because, as we go into a chemistry context, it starts getting a little bit more muddled again. Let me go to chemistry, chemistry. And in any science, if people just talk generally about mass or weight, this is what they're talking about. They're talking about a measure of inertia for mass, and they're talking about a force when they're talking about weight. But in chemistry, we start thinking about things on an atomic scale. You'll hear ... \"Atomic mass.\" Atomic mass is, literally, a measure of mass. It is measured in atomic mass units. Atomic mass units, which is, and we'll talk in the future videos, a very, very, very, very, small fraction. One atomic mass unit is a very, very, very, very, very, very small fraction of a gram. It is actually defined using the most common isotope of carbon. It's defined using carbon-12. The current definition is carbon-12. Carbon-12 has a mass, has a mass, has a mass of exactly, exactly, exactly 12 atomic mass units. So they can then," + }, + { + "Q": "\nAt 3:29 in this video Sal says that an Atomic Mass Unit is a very, very small fraction of a gram; but in an earlier video he said that an Atomic Mass Unit was a very, very small fraction of a kilogram. So what exactly is an Atomic Mass Unit a very small fraction of?", + "A": "A gram is a small fraction of a kilogram (1/1000) So both of those statements you mentioned are true, aren t they?", + "video_name": "NG-rrorZcM8", + "timestamps": [ + 209 + ], + "3min_transcript": "Now weight, on the other hand, is not ... it's different than mass. Weight is a force, it's how much the Earth, or whatever planet you happen to be on, is pulling on you. This right over here is a force. And, in the metric system, you measure weight, not with grams or kilograms, but with Newtons. Newtons. Really, when you ask someone their weight in Europe, they should give it to you in Newtons. If you ask them their mass, what they're telling you is actually their mass. They should say, \"My mass is 60 kilograms,\" or, \"70 kilograms,\" or whatever they might be. It's a very important difference in physics. If I go from Earth to the Moon, my mass does not change, but my weight does change because the force with which the Moon is pulling on me, or that we're pulling on each other, is less than it would be on Earth. In fact, even on the surface of the Earth, if you were to even go to the top of a building, Yeah, it would be very hard to measure it, but you're just slightly further from the center of the Earth, so there's a different gravitational force. Your weight will change ever so slightly, but your mass does not change. You go to deep space, and there's very little gravitational influence, you have pretty much, or close to, zero weight. But you're in deep space, and if there's no planets nearby, but your mass is still going to be whatever your mass happens to be. That's a primer on mass and weight. Now, with that out of the way, I might confuse you because, as we go into a chemistry context, it starts getting a little bit more muddled again. Let me go to chemistry, chemistry. And in any science, if people just talk generally about mass or weight, this is what they're talking about. They're talking about a measure of inertia for mass, and they're talking about a force when they're talking about weight. But in chemistry, we start thinking about things on an atomic scale. You'll hear ... \"Atomic mass.\" Atomic mass is, literally, a measure of mass. It is measured in atomic mass units. Atomic mass units, which is, and we'll talk in the future videos, a very, very, very, very, small fraction. One atomic mass unit is a very, very, very, very, very, very small fraction of a gram. It is actually defined using the most common isotope of carbon. It's defined using carbon-12. The current definition is carbon-12. Carbon-12 has a mass, has a mass, has a mass of exactly, exactly, exactly 12 atomic mass units. So they can then," + }, + { + "Q": "At 3:35 Sal references carbon-12 as being an isotope. Why is carbon-12 considered an isotope? The number of protons matches the number of neutrons in the nucleus.\n", + "A": "Every atom is an isotope. It has nothing at all to do with the number of protons matching the number of neutrons. There is nothing special at all about having equal protons and neutrons. There are quite a few elements whose most common isotopes happen to have that, but if you look at the periodic table you will see that as the elements get heavier you tend to have more neutrons than protons.", + "video_name": "NG-rrorZcM8", + "timestamps": [ + 215 + ], + "3min_transcript": "Yeah, it would be very hard to measure it, but you're just slightly further from the center of the Earth, so there's a different gravitational force. Your weight will change ever so slightly, but your mass does not change. You go to deep space, and there's very little gravitational influence, you have pretty much, or close to, zero weight. But you're in deep space, and if there's no planets nearby, but your mass is still going to be whatever your mass happens to be. That's a primer on mass and weight. Now, with that out of the way, I might confuse you because, as we go into a chemistry context, it starts getting a little bit more muddled again. Let me go to chemistry, chemistry. And in any science, if people just talk generally about mass or weight, this is what they're talking about. They're talking about a measure of inertia for mass, and they're talking about a force when they're talking about weight. But in chemistry, we start thinking about things on an atomic scale. You'll hear ... \"Atomic mass.\" Atomic mass is, literally, a measure of mass. It is measured in atomic mass units. Atomic mass units, which is, and we'll talk in the future videos, a very, very, very, very, small fraction. One atomic mass unit is a very, very, very, very, very, very small fraction of a gram. It is actually defined using the most common isotope of carbon. It's defined using carbon-12. The current definition is carbon-12. Carbon-12 has a mass, has a mass, has a mass of exactly, exactly, exactly 12 atomic mass units. So they can then, to figure out what the atomic mass, or the mass of any other atom. And you might say, \"Oh, why didn't they just do a hydrogen, \"and just say that's one atomic mass unit, and all that,\" and actually, they had started there. They had been there at an earlier stage, but for a whole set of reasons, carbon-12 is kind of being the benchmark, as having 12 atomic mass units, is what people went with. Now, what is atomic weight, then? Atomic weight. Let me write this in a different color. I'll do it in blue. Atomic weight. So if you draw the same analogy that we did up here, you might say, \"Okay, this must be a ... \"This must be a force. \"It should maybe, you know, \"an atomic weight unit would be a small fraction of, \"very small fraction of a unit.\" But it turns out in chemistry, when we talk about atomic weight, we're still measuring in atomic mass units." + }, + { + "Q": "I thought atomic weight is determined using the relative abundances of only stable isotopes of elements. At 7:16, Mr. Khan says that the weighted average of C-12 and C-14 is used to obtain the atomic weight of carbon. Is this a mistake? The only stable isotopes of C are C-12 and C-13. C-14 is radioactive.\n", + "A": "Atomic weight is determined using the relative abundance of the element as it occurs in nature. Highly unstable isotopes will only be found in very small amounts in nature, because they will have decayed. C-14 has a half-life over 5000 years and it is constantly produced in the atmosphere, so a small amount of the natural carbon is C14", + "video_name": "NG-rrorZcM8", + "timestamps": [ + 436 + ], + "3min_transcript": "Atomic mass units. But it's not the mass of just one atom or just one molecule. It's a weighted average across many, many ... of how typically, what you would see, or the makeup of what you would see on Earth. What do I mean by that? Well, on Earth, there are two ... The primary isotope of carbon is carbon-12. Carbon-12, which is defined as having a mass of exactly 12 atomic mass units. But there's also some carbon-14. Carbon-14. What do these numbers mean, just as a reminder? Well, carbon-12 has six protons, and the six protons are what make it carbon. Carbon-14 is also going to have six protons. But carbon-12, carbon-12 also has six neutrons. Six neutrons. I know what you're already thinking. You're, like, \"Well, wait. \"Why don't we say that a proton or a neutron \"weighs one atomic mass unit? \"Because it looks like this is 12, \"and I'm guessing that this, \"that this, the mass of this is going to be \"pretty close to 14.\" If you're thinking that way, that's not an unreasonable way to think. In fact, when I'm kind of just working through chemistry, that is how I think about it. But they don't weigh exactly one atomic mass unit by this definition. Remember, the electron is ever so small, it has very small mass, but it is contributing, or the electrons are contributing, something to the mass. So, a proton or a neutron have very, very, very close ... They are close to one atomic mass unit. Let me write this down. One proton, one proton, or one neutron, one neutron, very close to one atomic mass unit, but not exactly. But anyway, going back to what atomic weight is, the most common isotope of carbon ... Remember, when we're saying \"isotopes,\" we're saying the same element, we have the same number of protons, but we have different number of neutrons. The most common isotope on Earth is carbon-12, but there's also some carbon-14. If you were to take a weighted average, as found on the Earth, of all the carbon-12 and all of the carbon-14, the weighted average of the atomic masses is the atomic weight. And the atomic weight of carbon ... And you'll see this on a periodic table. In fact, I have one right over here. Notice, the six protons, this is what defines it to be carbon. But then they write 12.011, which is the weighted average of the masses of all of the carbons. Now, it's very close to 12, as opposed to being closer to 14, because most of the carbon on Earth is carbon-12. We could write this down. This is the atomic weight. This is the atomic weight of carbon on Earth. This is 12.011." + }, + { + "Q": "\nat 6:28 you explained that carbon 12 has 6/ neutrons and protons each. so doesnt it make a balanced atom but you are saying its an isotope??", + "A": "All atoms are isotopes. There is no such concept of a balanced atom as far as protons and neutrons are concerned. There are stable isotopes, but stability can apply to multiple isotopes, and the ratio of protons to neutrons varies as the atomic size increases.", + "video_name": "NG-rrorZcM8", + "timestamps": [ + 388 + ], + "3min_transcript": "to figure out what the atomic mass, or the mass of any other atom. And you might say, \"Oh, why didn't they just do a hydrogen, \"and just say that's one atomic mass unit, and all that,\" and actually, they had started there. They had been there at an earlier stage, but for a whole set of reasons, carbon-12 is kind of being the benchmark, as having 12 atomic mass units, is what people went with. Now, what is atomic weight, then? Atomic weight. Let me write this in a different color. I'll do it in blue. Atomic weight. So if you draw the same analogy that we did up here, you might say, \"Okay, this must be a ... \"This must be a force. \"It should maybe, you know, \"an atomic weight unit would be a small fraction of, \"very small fraction of a unit.\" But it turns out in chemistry, when we talk about atomic weight, we're still measuring in atomic mass units. Atomic mass units. But it's not the mass of just one atom or just one molecule. It's a weighted average across many, many ... of how typically, what you would see, or the makeup of what you would see on Earth. What do I mean by that? Well, on Earth, there are two ... The primary isotope of carbon is carbon-12. Carbon-12, which is defined as having a mass of exactly 12 atomic mass units. But there's also some carbon-14. Carbon-14. What do these numbers mean, just as a reminder? Well, carbon-12 has six protons, and the six protons are what make it carbon. Carbon-14 is also going to have six protons. But carbon-12, carbon-12 also has six neutrons. Six neutrons. I know what you're already thinking. You're, like, \"Well, wait. \"Why don't we say that a proton or a neutron \"weighs one atomic mass unit? \"Because it looks like this is 12, \"and I'm guessing that this, \"that this, the mass of this is going to be \"pretty close to 14.\" If you're thinking that way, that's not an unreasonable way to think. In fact, when I'm kind of just working through chemistry, that is how I think about it. But they don't weigh exactly one atomic mass unit by this definition. Remember, the electron is ever so small, it has very small mass, but it is contributing, or the electrons are contributing, something to the mass. So, a proton or a neutron have very, very, very close ... They are close to one atomic mass unit. Let me write this down. One proton, one proton, or one neutron, one neutron, very close to one atomic mass unit, but not exactly. But anyway, going back to what atomic weight is," + }, + { + "Q": "At 0:55, he says it's confusing when you go to Europe, where they use the metric system to give you their weight. Why is it different if you were say in the USA where they use pounds? Are pounds a form of mass or weight??\n", + "A": "I suppose it s because Sal is based in the USA so therefore he sees weight from the US pound perspective.", + "video_name": "NG-rrorZcM8", + "timestamps": [ + 55 + ], + "3min_transcript": "Let's have a little bit of a primer on weight and mass, especially if we start talking about atomic weight and atomic mass. If we're sitting in a physics class, weight and mass mean something very, very ... well, they mean different things. It might be a discovery, or a new learning, for some of you, because in everyday life, when we say something's mass, we think, \"Well, the more mass it has, the more weight it has.\" Or, if we think something has more or less weight, we think, \"Okay, that relates to its mass.\" But in physics class, we see that these actually represent two different ideas, albeit related ideas. Mass is a notion of how much of something there is, or you could say, how hard is it to accelerate or decelerate it. Or you could view it as a measure of an object's inertia. We typically, it, kind of a human scale, might measure mass in terms of grams or kilograms. What's confusing is, if you go to Europe, and you ask someone their weight, they'll often give you their weight in terms of kilogram, Now weight, on the other hand, is not ... it's different than mass. Weight is a force, it's how much the Earth, or whatever planet you happen to be on, is pulling on you. This right over here is a force. And, in the metric system, you measure weight, not with grams or kilograms, but with Newtons. Newtons. Really, when you ask someone their weight in Europe, they should give it to you in Newtons. If you ask them their mass, what they're telling you is actually their mass. They should say, \"My mass is 60 kilograms,\" or, \"70 kilograms,\" or whatever they might be. It's a very important difference in physics. If I go from Earth to the Moon, my mass does not change, but my weight does change because the force with which the Moon is pulling on me, or that we're pulling on each other, is less than it would be on Earth. In fact, even on the surface of the Earth, if you were to even go to the top of a building, Yeah, it would be very hard to measure it, but you're just slightly further from the center of the Earth, so there's a different gravitational force. Your weight will change ever so slightly, but your mass does not change. You go to deep space, and there's very little gravitational influence, you have pretty much, or close to, zero weight. But you're in deep space, and if there's no planets nearby, but your mass is still going to be whatever your mass happens to be. That's a primer on mass and weight. Now, with that out of the way, I might confuse you because, as we go into a chemistry context, it starts getting a little bit more muddled again. Let me go to chemistry, chemistry. And in any science, if people just talk generally about mass or weight, this is what they're talking about. They're talking about a measure of inertia for mass, and they're talking about a force when they're talking about weight. But in chemistry, we start thinking about things on an atomic scale. You'll hear ..." + }, + { + "Q": "\nat 2:58 , can the (CH3)2CHOH also be (CH3)2 CH2O?", + "A": "The -O- is connected to the -C-. One -H- is connected to the -C- , while the other -H- is connected to the -O-. In short : No.", + "video_name": "XEPdMvZqCHQ", + "timestamps": [ + 178 + ], + "3min_transcript": "And the carbon in magenta is bonded to three other hydrogens. So we could represent that as a CH three. So I could write CH three here, and the carbon in red is this one and the carbon in magenta is this one. On the left side, the carbon in red is bonded to another carbon in blue and the carbon in blue is bonded to three hydrogens, so there's another CH three on the left side, so let me draw that in, so we have a CH three on the left and the carbon in blue is directly bonded to the carbon in red. So this is called a partially condensed structure so this is a partially condensed, partially condensed structure. We haven't shown all of the bonds here but this structure has the same information as the Lewis structure on the left. it's just a different way to represent that molecule. We could keep going. We could go for a fully condensed structure. So let's do that. Focus in on the carbon in red. So this one right here. So let me draw in that carbon over here. So that's that carbon. That carbon is bonded to two CH three groups. There's a CH three group on the right, so there's the CH three group on the right. And there's a CH three group on the left. So I could write CH three and then I could write a two here which indicates there are two CH three groups bonded to directly bonded to the carbon in red. What else is bonded to the carbon in red? There's a hydrogen, so I'll put that in. So the carbon is bonded to a hydrogen. The carbon is also bonded to an OH, so I'll write in here an OH. This is the fully condensed version, so this is completely condensed and notice there are no bonds shown. you have to infer you have to infer the bonding from the condensed. All right, let's start with the condensed and go all the way to a Lewis structure, so we'll start with a condensed and then we'll do partially condensed structure, and then we'll go to a full Lewis structure. Just to get some more practice here. So I'll draw in a condensed one, so we have CH three, three and then COCH three. All right let's turn that into a partially condensed structure. So this carbon in red right here we're gonna start with that carbon, so I'll start drawing in that carbon right here. What is bonded to that carbon? Well, we have CH three groups and we have three of them. So there are three CH three groups directly bonded to that carbon. So let me draw them in. So here's one CH three group here is another CH three group, and then finally here's the third CH three group." + }, + { + "Q": "At 5:28, Jay says that the lone pairs for Oxygen can form a pi bond between Oxygen and Carbon. Why can't they form pi electrons inside the Benzene ring?\n", + "A": "If the electrons in the C=O pi bond (from the oxygen lone pair) were pushed into the ring, that would leave oxygen with a +2 charge and an incomplete octet. This would be very unstable so does not happen.", + "video_name": "i9rfWOAEplk", + "timestamps": [ + 328 + ], + "3min_transcript": "we need to show the nitro group adding onto the ortho position. So we need to show the nitro group adding onto this carbon. And so if the nitro group is going to add onto this carbon, then these are the pi electrons that can function as a nucleophile in our mechanism. So we have a nucleophile electrophile reaction for the first step of our mechanism. So the nucleophile, these pi electrons are going to attack that positively charged nitrogen, which kicks these electrons off onto the oxygen. So if we draw the result of that nucleophilic attack, we still have our methoxy substituent up here. I'm showing the nitro group adding onto the ortho position. And remember there's still a hydrogen attached to that carbon. So I have pi electrons over here, pi electrons over here. And I'm saying that these pi electrons are the ones that formed a bond with this nitrogen like that. That takes away a bond from this carbon. We can show some resonance structures. So we can show some resonance stabilization of this cation here. So I could show these pi electrons moving over to here. And we could draw another resonance structure. So let's go ahead and show the movement of those pi electrons over to this position. So let me go ahead and draw in the rest of the ion here. So we have a hydrogen here. We have an NO2 here. And we took these pi electrons right here, moved them over to this position, took a bond away from that carbon. So we get a +1 formal charge on this carbon. And that's another resonance structure. We can draw another one. We can show the movement of these pi electrons into here. So let's go ahead and show that. We have our ring. We have our methoxy group. We have, once again, the nitro group in the ortho position. And now we show the movement of those pi electrons over to here. So let me go ahead and highlight those. These electrons in red move down to here. I took a bond away from this carbon. So that carbon is the one that gets a plus 1 formal charge now. Since the oxygen is right next to this carbon-- the oxygen has a lone pair of electrons. And so that lone pair of electrons can give us yet another resonance structure. So these electrons could move into here to draw a fourth resonance structure. So the presence of that methoxy substituent with the lone pair of electrons on that oxygen allows you to draw a fourth resonance structure. So this will give this oxygen a +1 formal charge. We have these pi electrons over here. We have our nitro group, once again, in the ortho position. And let me just go ahead and show the movement of those electrons. So these electrons-- I'll make them green-- these electrons right here are going" + }, + { + "Q": "At 2:40, would not natural selection cause organisms that were attracted to useless traits to drop out of the gene pool? Organisms attracted to useful traits would survive and sexual selection would no longer be a factor as being attracted to useful traits would actually make the organism fitter, thereby making it natural, and not sexual, selection.\n", + "A": "No. Other ones with okay mutations wouldn t drop, so the gene pool gets rid of REALLY useless organisms.", + "video_name": "tzqZsPjHFVQ", + "timestamps": [ + 160 + ], + "3min_transcript": "it's the study of how populations of a species change genetically overtime leading to species evolving. Let's start up by defining what a population is. It's simply a group of individuals of a species that can interbreed. Because we have a whole bunch of fancy genetic testing gadgets and because unlike Darwin we know a whole lot about heredity. We can now study the genetic change in populations over just a couple of generations. This is really exciting and really fun because it's basically like scientific instant gratification. I can now observe evolution happening within my lifetime. You know, just cross that off the bucket list. Now, part of population genetics or pop-gen and now we've got fancy abbreviations for everything now, involves the study of factors that cause changes and what's called allele frequency. Which is just how often certain alleles turn up within a population. Those changes are at the heart of how and why evolution happens. There are several factors that change allele frequency Just like Fast and Furious movies, there are five of them. Unlike Fast and Furious movies, they're actually very, very important and are the basic reason why all complex life on earth exists. The main selective pressure is simply natural selection itself, Darwin's sweet little baby which he spent a lot of his career defending from haters. Obviously we know these natural selection makes the alleles that make animals the strongest and most virile and least likely to die more frequent in the population. Most selective pressures are environmental ones like food supply, predators or parasites. At the population level, one of the most important evolutionary forces is sexual selection. Population genetics gets its special attention particularly when it comes to what's called non-random mating which is a lifestyle that I encourage in all of my students, do not mate randomly. Sexual selection is the idea that certain individuals will be more attractive mates than others because of specific traits. This means they'll be chosen to have more sex and therefore offspring. The pop-gen spend on things if that sexual selection There are specific traits that are preferred even though they may not make the animals technically more fit for survival. Sexual selection changes a genetic make up of a population because the alleles of the most successful maters are going to show up more often in the gene-pool. Maters are going to mate. Another important factor here, and another thing that Darwin wished he had understood is mutation. Sometimes when eggs and sperm are formed through meiosis, a mistake happens in the copying process of DNA, that errors in the DNA could result in the death or deformation of offspring. But not all mutations are harmful. Sometimes these mistakes can create new alleles that benefit the individual by making it better at finding food or avoiding predators or finding a mate. These good errors and the alleles they made are then passed to the next generation and into the population. Fourth, we have genetic drift which is when an alleles frequency changes due to random chance. A chance that's greater if the population is small. Those happens much more quickly if the population" + }, + { + "Q": "\nIs the cyclical form of ribose a furanose ? And where does the Oxygen attached to Carbon 2 goes ? I guess the Hydrogen attached to Carbon 2 in deoxyribose form is not the same from the O-H. 08:56. So, where does the Hydrogen comes from? Thanks.", + "A": "I think it is because the phosphate groups are negative and hydrogen is a positive ion, so it attracts it.", + "video_name": "L677-Fl0joY", + "timestamps": [ + 536 + ], + "3min_transcript": "that hydrogen proton right over there and this green bond that gets formed between the four prime carbon and or between the oxygen that's attached to the four prime carbon and the one prime carbon, that's this. That's this bond right over here. This oxygen is that oxygen right there. Notice, this oxygen is bound to the four prime carbon and now it's also bound to the one prime carbon. It was also attached to a hydrogen. It was also attached to a hydrogen so that hydrogen is there but then that can get nabbed up by another passing water molecule to become hydronium so it can get lost. It grabs up a hydrogen proton right over here and so it can lose a hydrogen proton right there. It's not adding or losing in that net. You form this cyclical form and the cyclical form right over here is very close to what we see in a DNA molecule. It's actually what we would see in an RNA molecule, in a ribonucleic acid. when we say deoxyribonucleic acid. Well, you can start with you have a ribose here but if we got rid of one of the oxygen groups and in particular one of... Well, actually if we just got rid of one of the oxygens we replace a hydroxyl with just a hydrogen, well then you're gonna have deoxyribose and you see that over here. This five-member ring, you have four carbons right over here. it looks just like this. The hydrogens are implicit to the carbons, we've seen this multiple time. The carbons are at where these lines intersect or I guess at the edges or maybe and also where these lines end right over there. But you see this does not have an... This molecule if we compare these two molecules, if we compare these two molecules over here, we see that this guy has an OH, and this guy implicitly just has... This has an OH and an H. This guy implicitly has just two hydrogens over here. He's missing an oxygen. This is deoxyribose. Deoxyribose doesn't have this oxygen. It does not have the oxygen on the two prime carbon. So this if you get rid of that, this is deoxyribose. So let me circle that. This thing right over here, this thing right over here, that is deoxyribose. Deoxy or it's based on deoxyribose I guess before it bonded to these other constituents. You could consider this deoxyribose. That's where the deoxyribo comes from and then the last piece of it, the last piece of it is this chunk right over here. These we call nitrogenous bases. Nitrogenous. Nitrogenous. Nitrogenous bases. You could see we have different types of nitrogenous bases. This is a nitrogenous base. This right over here is a different nitrogenous base. This right over here is another different nitrogenous base." + }, + { + "Q": "At around 0:26 in the video, why does Sal do \"deoxy\" and \"ribo\" in two different colors? Isn't deoxyribose one word and one type of sugar?\n", + "A": "he is separating the roots of the word. deoxy means lack of oxygen and ribo, I think means with sugar, and you know acid. So, therefore, it is a lack of acid, with sugar and acid.", + "video_name": "L677-Fl0joY", + "timestamps": [ + 26 + ], + "3min_transcript": "- [Voiceover] We already have an overview video of DNA and I encourage you to watch that first. What I want to do in this video is dig a little bit deeper. Actually get into the molecular structure of DNA. This is a starting point. Let's just remind ourselves what DNA stands for. I'm gonna write the different parts of the word in different colors. It stands for deoxy. Deoxyribonucleic. Ribonucleic. Ribonucleic acid. Ribonucleic acid. So I'm just gonna put this on the side and now let's actually look at the molecular structure and how it relates to this actual name, deoxyribonucleic acid. DNA is just a junction for nucleic acid and it's the term nucleic that comes from the fact that it's found in the nucleus. It's found in the nucleus of eukaryotes. That's where the nucleic comes from and we'll talk about in a second why it's called an acid but I'll wait on that. Now each DNA molecule is made up of a chain What we call nucleotides. It's made up of nucleo, nucleo, nucleotides. What does a nucleotide look like? Well, what I have right over here is I have two strands, I've zoomed two strands of DNA or I've zoomed in two strands of DNA. You could view this side right over here as one of the, I guess you can say the backbones of one side of the ladder. This is the other side of the ladder and then each of these bridges, and I will talk about what molecules these are. These are kind of the rungs of the ladder. A nucleotide, let me separate off the nucleotide. A nucleotide would... What I am cordoning off, what I am cordoning off right over here could be considered, could be considered a nucleotide. That's one nucleotide and then it's connected to another. Another nucleotide right over here. On the right hand side we have a nucleotide, we have a nucleotide right over there and then, actually I want to do it, let me do it slightly different. We have a nucleotide right over here on the right side and then right below that we have another. We have another nucleotide. We have another nucleotide. Depicted here, we essentially have four nucleotides. These two are on this left side of the ladder, these two are on the right side of the ladder. Now let's think about the different pieces of that nucleotide. The one thing that might jump out at you is we have these phosphate groups. This is a phosphate group right over here. This is a phosphate group right over here. Each of these nucleotides have a phosphate group. This is a phosphate group over here and this is a phosphate group over here." + }, + { + "Q": "\nAt 11:34, can we classify Adenine and Guanine as Purines and Cytosine and Thymine as Pyrimidines", + "A": "Yes, that s how they are classified.", + "video_name": "L677-Fl0joY", + "timestamps": [ + 694 + ], + "3min_transcript": "this one has two rings. This one over here has two rings and we have different names for these nitrogenous bases. The ones with two rings, the general categorization we call them purines. Nitrogenous bases if you have two rings, if you have two rings we call them purines. That's a general classification term. Let me make sure, purines. If you have one ring. Anyway, I'll just write this way. One ring. One ring, we call these pyrimidines. Pyrimidine. Pyrimidines. We call these pyrimidines. These particular, these two on the right, these two purines, this one up here this is adenine, and we talk about how they pair in the overview video on DNA. This one right over here is adenine, this nitrogenous base. This one over here is guanine. That is guanine. And then over here, over here, which makes it a pyrimidine, this is thymine. This right over here is thymine. This is thymine and then last but not least if we're talking about DNA, when we go into RNA, we're also gonna talk about uracil. But when we talk about DNA this one over here is cytosine. Cytosine. You could see the way it's structured. The thymine is attracted to adenine. It bonds with adenine and cytosine bonds with guanine. How are they bonding? Well, the way that these nitrogenous bases form the rungs of the ladder, how they want they're drawn to each other, this is our good old friend hydrogen bonds. This all comes out of the fact, that nitrogen is quite electronegative. When nitrogen is bound to a hydrogen you're going to have a partially negative charge at the nitrogen. Let me do this in green. You're going to have a partial negative charge at the nitrogen and a partially positive charge at the hydrogen. as being electronegative so it has a partial negative charge. The partial negative charge of this oxygen is going to be attracted to the partial positive charge of this hydrogen, and so you're going to have a hydrogen bond. That's then going to happen between this hydrogen which is going... Its electrons are being hogged by this nitrogen and this nitrogen with who, which itself hogs electrons. That forms a hydrogen bond. And then down here you have a hydrogen that has a partially positive charge because its electrons are being hogged. And then you have this oxygen with a partially negative charge, they're going to be attracted to each other. That's a hydrogen bond. Same thing between this nitrogen and that hydrogen, and same thing between this oxygen and that hydrogen. That's why cytosine and guanine pair up and that's why thymine and adenine pair up, and we talk about that as well in the overview video of DNA." + }, + { + "Q": "at 0:50 he dose not go deeply into what reactants are\n", + "A": "Reactants are the things that react together in a reaction. They are on the left side of a chemical equation.", + "video_name": "TStjgUmL1RQ", + "timestamps": [ + 50 + ], + "3min_transcript": "- [Voiceover] Let's talk a little bit about chemical reactions. And chemical reactions are a very big deal. Without chemical reactions, you or I would not exist. In your body right now, there are countless chemical reactions going on every second. Without chemical reactions, we would have no life, we would not even have the universe as we know it. So what are chemical reactions. Well, they're any time that you have bonds being formed or broken between atoms or molecules. So what are we talking about there? Well this is maybe one of the most fundamental chemical reactions. Once again if one never occurred, we'd be in trouble, we would not have, we would not have any water. But let's think about what it is actually describing. So over here on the left-hand side we have the reactants. Let me write that down. So here we have the reactants. These are the molecules that are going to reaction. And then we have an arrow that moves us to the product. So let me do that in a different color. or we could say the products. And so what are the reactants here? Well we have molecular hydrogen and we have molecular oxygen. Now why did I say molecular hydrogen? Because molecular hydrogen, which is the state that you would typically find hydrogen in if you just have it by itself, it is actually made up of two hydrogen atoms. You see it right over here, one, two hydrogen atoms. And what we have in order to have this reaction, you don't just need one molecular hydrogen and one, or one molecule of hydrogen and one molecule of oxygen. For every, for this reaction to happen, you actually have two molecules of molecular hydrogen. So this is actually made up of four hydrogen atoms. So let me make this clear. So this right over here, this is two molecules of molecular hydrogen. And that's why we have the two right out front of the H sub-two. This little subscript two tells us there's two of the hydrogen atoms in this molecule. that we have right over here, that tells us that we're dealing with two of those molecules for this reaction to happen, that we need two of these molecules for every, for every molecule of molecular oxygen. And molecular oxygen, once again, this is composed of two oxygen atoms. One, two. So under the right conditions, so you need a little bit of energy to make this happen. If under the right conditions these two things are going to react. And actually it's very, very reactive, molecular hydrogen and molecular oxygen. So much so that it's actually used for rocket fuel. You are going to produce two molecules of water. We see that right over here. And look, I did not create or destroy any atoms. I had one, I had one, I had one oxygen atom here. It was part of the oxygen molecule right here, then I have the second one right over here now. Now they are part of separate molecules. I had, I had a, I had one two, three, four hydrogens." + }, + { + "Q": "At 6:35 what are the reactants and products in reversible reaction?\n", + "A": "Typically, you name the ones on the left the reactants and the ones on the right the products, although they are technically both reactants and products. See also which chemicals you start with, and these will be your reactants. Alternatively, those being reduced in concentration to suit the equilibrium are the reactants.", + "video_name": "TStjgUmL1RQ", + "timestamps": [ + 395 + ], + "3min_transcript": "And just to get an appreciation of how much energy this produces, let me just show you this picture right over here. That's the space shuttle and this, this big tank right over here, let me... This big tank contains a bunch of liquid oxygen and hydrogen. And to create this incredible amount of energy, it actually just... You mix the two together with a little bit of energy and then you produce a ton of energy that makes the rocket, that makes the space shuttle. Well, space shuttle's been discontinued now, but back when they did it, to make it get it's necessary, it's necessary velocity. Now let's talk about the idea. So, you know, this reaction, strongly goes in this, in the direction of going to water. But it can actually go the other way, but it's very, very hard for it to go the other way. So in general we would consider this to be an irreversible reaction, even though it is. You know irreversible sounds like, It just really means that it's very unlikely to go the other way. You have to supply a lot of energy to go the other way. To make this reaction go the other way, you would have to do something called electrolysis, you provide energy, etcetera, etcetera. But in general, the way that this is written, because the arrow is only pointing in one direction, this is implying that it is irreversible. Irreversible. Irreversible. Which probably makes you think, well what about reversible reactions? And I have an example of a reversible reaction, right over here. I have a one bicarbonate ion. And the word ion, that's just used to describe any molecule or atom that has either, has an imbalance of electrons or protons that cause it to have a net charge. So this makes this an ion, and actually right over here, this is a hydrogen, this is a hydrogen ion right over here. Both of these are charged. One has a positive charge, one has a negative charge. And this reactions right over here, you have the bicarbonate ion that looks something like this. This is just my hand-drawing of it. Reacting with a hydrogen ion, it's really a hydrogen atom that has lost it's electron, so some people would even say this is a proton right over here. This is an equilibrium reaction, where it can form carbonic acid. And notice all that's happening is this hydrogen is attaching to one of the oxygens over here. And this is an equilibrium because if in an actual, in an actual solution, it's going back and forth. If you actually provide more reactants, you're gonna go more in that direction. If you provide more of the products over here, then you're gonna go in that direction. And so in an actual, in an actual environment, in an actual system, it's constantly going back and forth between these two things. And different reversible reactions might tend to one side or the other. If you provide more of the stuff on one side," + }, + { + "Q": "\nat 3:30 Sal said energy is provided but from where does it come from?", + "A": "Humans e.g. a spark of fire (Heat) can overcome the activation energy to kickstart the reaction and allow it to occur spontaneously from that point forward", + "video_name": "TStjgUmL1RQ", + "timestamps": [ + 210 + ], + "3min_transcript": "or we could say the products. And so what are the reactants here? Well we have molecular hydrogen and we have molecular oxygen. Now why did I say molecular hydrogen? Because molecular hydrogen, which is the state that you would typically find hydrogen in if you just have it by itself, it is actually made up of two hydrogen atoms. You see it right over here, one, two hydrogen atoms. And what we have in order to have this reaction, you don't just need one molecular hydrogen and one, or one molecule of hydrogen and one molecule of oxygen. For every, for this reaction to happen, you actually have two molecules of molecular hydrogen. So this is actually made up of four hydrogen atoms. So let me make this clear. So this right over here, this is two molecules of molecular hydrogen. And that's why we have the two right out front of the H sub-two. This little subscript two tells us there's two of the hydrogen atoms in this molecule. that we have right over here, that tells us that we're dealing with two of those molecules for this reaction to happen, that we need two of these molecules for every, for every molecule of molecular oxygen. And molecular oxygen, once again, this is composed of two oxygen atoms. One, two. So under the right conditions, so you need a little bit of energy to make this happen. If under the right conditions these two things are going to react. And actually it's very, very reactive, molecular hydrogen and molecular oxygen. So much so that it's actually used for rocket fuel. You are going to produce two molecules of water. We see that right over here. And look, I did not create or destroy any atoms. I had one, I had one, I had one oxygen atom here. It was part of the oxygen molecule right here, then I have the second one right over here now. Now they are part of separate molecules. I had, I had a, I had one two, three, four hydrogens. four hydrogens, just like that. And actually this produces a... So we could say some energy, and I'm being inexact right over here. Some energy and then we could say a lot of energy. A lot of energy. So this is a reaction that you just give it a little bit of a kick-start and it really wants to happen. A lot, a lot of energy. So one thing that you might wonder, and this is something that I first wondered when I learned about reactions, well how do, how does this happen? You know, is this a very organized thing? You know, do these molecules somehow know to react with each other? And the answer's no. Chemistry is a incredibly messy thing. You have these things bouncing around, they have energy. They're bouncing around all over the place and actually when you provide energy, they're gonna bounce around even more rigorously, enough so that they collide in the right ways so that they break their old bonds" + }, + { + "Q": "At 2:37 he calls the water molecules, but later calls them compounds. What is the difference between a molecule and a compound, and why is he able to call water both names?\n", + "A": "A molecule made of more than one element is a compound. Water is a compound of hydrogen and oxygen.", + "video_name": "TStjgUmL1RQ", + "timestamps": [ + 157 + ], + "3min_transcript": "or we could say the products. And so what are the reactants here? Well we have molecular hydrogen and we have molecular oxygen. Now why did I say molecular hydrogen? Because molecular hydrogen, which is the state that you would typically find hydrogen in if you just have it by itself, it is actually made up of two hydrogen atoms. You see it right over here, one, two hydrogen atoms. And what we have in order to have this reaction, you don't just need one molecular hydrogen and one, or one molecule of hydrogen and one molecule of oxygen. For every, for this reaction to happen, you actually have two molecules of molecular hydrogen. So this is actually made up of four hydrogen atoms. So let me make this clear. So this right over here, this is two molecules of molecular hydrogen. And that's why we have the two right out front of the H sub-two. This little subscript two tells us there's two of the hydrogen atoms in this molecule. that we have right over here, that tells us that we're dealing with two of those molecules for this reaction to happen, that we need two of these molecules for every, for every molecule of molecular oxygen. And molecular oxygen, once again, this is composed of two oxygen atoms. One, two. So under the right conditions, so you need a little bit of energy to make this happen. If under the right conditions these two things are going to react. And actually it's very, very reactive, molecular hydrogen and molecular oxygen. So much so that it's actually used for rocket fuel. You are going to produce two molecules of water. We see that right over here. And look, I did not create or destroy any atoms. I had one, I had one, I had one oxygen atom here. It was part of the oxygen molecule right here, then I have the second one right over here now. Now they are part of separate molecules. I had, I had a, I had one two, three, four hydrogens. four hydrogens, just like that. And actually this produces a... So we could say some energy, and I'm being inexact right over here. Some energy and then we could say a lot of energy. A lot of energy. So this is a reaction that you just give it a little bit of a kick-start and it really wants to happen. A lot, a lot of energy. So one thing that you might wonder, and this is something that I first wondered when I learned about reactions, well how do, how does this happen? You know, is this a very organized thing? You know, do these molecules somehow know to react with each other? And the answer's no. Chemistry is a incredibly messy thing. You have these things bouncing around, they have energy. They're bouncing around all over the place and actually when you provide energy, they're gonna bounce around even more rigorously, enough so that they collide in the right ways so that they break their old bonds" + }, + { + "Q": "\nAround 8:30, he mentions that carbolic acid is good for your body. But how does it help our bodies?", + "A": "For digestion", + "video_name": "TStjgUmL1RQ", + "timestamps": [ + 510 + ], + "3min_transcript": "It just really means that it's very unlikely to go the other way. You have to supply a lot of energy to go the other way. To make this reaction go the other way, you would have to do something called electrolysis, you provide energy, etcetera, etcetera. But in general, the way that this is written, because the arrow is only pointing in one direction, this is implying that it is irreversible. Irreversible. Irreversible. Which probably makes you think, well what about reversible reactions? And I have an example of a reversible reaction, right over here. I have a one bicarbonate ion. And the word ion, that's just used to describe any molecule or atom that has either, has an imbalance of electrons or protons that cause it to have a net charge. So this makes this an ion, and actually right over here, this is a hydrogen, this is a hydrogen ion right over here. Both of these are charged. One has a positive charge, one has a negative charge. And this reactions right over here, you have the bicarbonate ion that looks something like this. This is just my hand-drawing of it. Reacting with a hydrogen ion, it's really a hydrogen atom that has lost it's electron, so some people would even say this is a proton right over here. This is an equilibrium reaction, where it can form carbonic acid. And notice all that's happening is this hydrogen is attaching to one of the oxygens over here. And this is an equilibrium because if in an actual, in an actual solution, it's going back and forth. If you actually provide more reactants, you're gonna go more in that direction. If you provide more of the products over here, then you're gonna go in that direction. And so in an actual, in an actual environment, in an actual system, it's constantly going back and forth between these two things. And different reversible reactions might tend to one side or the other. If you provide more of the stuff on one side, are gonna, they're gonna be more likely to interact, Or if you provide more of this, it might go in the other direction because these might more likely react with their surroundings or disassociate in some way. Now just to get a sense of, you know, it's nice to kind of, you know, are these just some random letters that I wrote here? Carbonic acid is actually an incredibly important molecule, or we could call it a compound because it's made up of two or, two or more elements, in living systems and in fact, you know, even in the environment. And even when you go out to get some fast food. When you have carbonated drinks, it has carbonic acid in it that disassociates into carbon dioxide and that carbon dioxide is what you see bubbling up. Carbonic acid is incredibly important in how your body deals with excess carbon dioxide in its bloodstream. Carbonic acid is involved in the oceans taking up carbon dioxide from the atmosphere. So when you're studying chemistry, especially in the context of biology, these aren't just," + }, + { + "Q": "what happens if at 8:15, you didn't have the valves? what would happen to you blood\n", + "A": "If you didn t have the valves, your blood might flow backwards. Your blood might not get the oxygen it needs from the respiratory system. Also, the muscles in your body might not receive the oxygen it needs to expand and contract if the oxygenated blood is flowing backwards and away from the muscle tissue that needs oxygen to move. This will also cause the pressure in the circulatory system to be unstable. Hank talks about this at 8:16.", + "video_name": "L1qpKn2hNF0", + "timestamps": [ + 495 + ], + "3min_transcript": "I'm going to explain. We're used to talking about the heart as the head honcho of the circulatory system, and yeah, you would be in serious trouble if you didn't have a heart. But the heart's job is to basically power the circulatory system, move the blood all around your body, and get it back to the lungs so that it can pick up more oxygen and get rid of the CO2. As a result, the circulatory system of mammals essentially makes a figure 8. Oxygenated blood is pumped from the heart to the rest of the body, and then when it makes its way back to the heart again, it's then pumped on a shorter circuit to the lungs to pick up more oxygen and unload CO2 before it goes back to the heart and starts the whole cycle over again. So even though the heart does all the heavy lifting in the circulatory system, the lungs are the home base for the red blood cells, the postal workers that carry the oxygen and the CO2. Now the way that your circulatory system moves the blood around is pretty nifty. Remember when I was talking about air moving from high pressure to low pressure? Well, so does blood. A four-chambered heart, which is just one big honking beast of muscle, has very high pressure. In fact, the reason it seems like the heart is situated a little bit to the left of center is because the left ventricle is so freaking enormous and muscle-y. It has to be that way in order to keep the pressure high enough that the oxygenated blood will get out of there. From the left ventricle, the blood moves through the aorta, a giant tube, and then through the arteries and blood vessels that carry the blood away from the heart to the rest of the body. Arteries are muscular and thick-walled to maintain high pressure as the blood travels along. As arteries branch off to go to different places, they form smaller arterioles, and finally, the very little capillary beds, which, through their huge surface area, facilitate the delivery of oxygen to all of the cells in the body that need it. The capillary beds are also where the blood picks up CO2. From there, the blood keeps moving down the pressure gradient through a series of veins. These do the opposite of what the arteries did. Instead of splitting off from each other, Little ones flow together to make bigger and bigger veins to carry the deoxygenated blood back to the heart. The big difference between most veins and most arteries is that instead of being thick-walled and squeezy, veins have thinner walls and have valves that keep the blood from flowing backwards, which would be bad. This is necessary because the pressure in the circulatory system keeps dropping lower and lower until the blood flows in to two major veins. The first is the inferior vena cava, which runs pretty much down the center of the body and handles blood coming from the lower part of your body. The second is the superior vena cava, which sits on top of the heart and collects the blood from the upper body. Together, they run into the right atrium of the heart, which is the point of the lowest pressure in the circulatory system. All this deoxygenated blood is now back in the heart, and it needs to sop up some more oxygen. So, it flows into the right ventricle and then into the pulmonary artery. Now arteries, remember, flow away from the heart, even though, in this case, it contains deoxygenated blood. And pulmonary means \"of the lungs,\"" + }, + { + "Q": "\nwhat does the sub exponent thing do? (at 1:24)", + "A": "The subscripts on a and b allow you to have lots of different coefficients without using up all the letters of the alphabet. All the cosine terms have coefficients named a_sub_something and all the sine terms use b_sub_something. This way of naming is handy for talking about the terms in order: the 0th term, the 1st term, etc.", + "video_name": "UKHBWzoOKsY", + "timestamps": [ + 84 + ], + "3min_transcript": "- [Voiceover] So I have the graph of y is equal to f of t here, our horizontal axis is in terms of time, in terms of seconds. And this type of function is often described as a square wave, and we see that it is a periodic function, that it completes one cycle every two pi seconds. And so we could say its period is equal to two pi, if we wanna put the units we could say two pi, two pi seconds per cycle, we could write it like that. We could also just write s for seconds. And its frequency is gonna be one over that. So we could write its frequency is equal to one over two pi cycles per second, cycles per second, it could also be described as hertz. And what we're gonna explore in this video, is can we take a periodic function like this and represent it as an infinite sum of sines and cosines of different periods or different frequencies? So to write that out a little bit more clearly, and write it as the sum of sines and cosines? So can we write it, so it's going to be sum, let's say baseline constant, that'll shift it up or down, and as we'll see, that's going to be based on the average value of the function over one period. So a sub zero, and then, let's start adding some periodic functions here. And so let's take a sub one times cosine of t. Now, why am I starting with cosine of t? And I could also add a sine of t, so plus b sub one, of sine of t. Why am I starting with cosine of t and sine of t? Well, if our original function has a period of two pi, and I just set up this one so it does have a period of two pi, well it would make sense that it would involve some functions that have periods of two pi. If a one is much larger than b one, well it says, okay, this has a lot more of cosine of t in it, than it has of sine of t in it. And that by itself isn't going to describe this function, because we know what this would look like. This would look like a very clean sinusoid, not like a square wave. And so what we're gonna do is we're gonna add sinusoids of frequencies that are multiples of these frequencies. So let's add a sub two, so another waiting coefficient, times cosine of two t. This has a frequency of one over two pi, this has twice the frequency, this has a frequency of one over pi. And then a sub three times cosine of three t. And I'm gonna keep going on and on and on forever. And I'm gonna do the same thing with the sines. So plus b two sine of two t plus b three sine of three t." + }, + { + "Q": "\nat 7:55 the second molecule is determined to be di-substituted. How come the methyl group on the second carbon is not counted in to be tri-substituted?", + "A": "It is the number of carbon atoms directly attached to the double-bonded carbons that determines the degree of substitution. C-1 has no C atoms attached to it. C-2 has two C atoms attached to it ( a methyl and an ethyl group). Total attached atoms = 0 + 2 = 2. So the alkene is disubstituted.", + "video_name": "MDh_5n0OO2M", + "timestamps": [ + 475 + ], + "3min_transcript": "So, let me use red for this. If we think about the degree of substitution for the alkene on the right, by drawing my hydrogen right here, it makes it a little bit easier to see we have three alkyl groups, so this one, this one, and this one. So this would be a trisubstituted alkene. So the one on the right is a trisubstituted alkene, and the one on the left, so this one right here, would be a disubstituted alkene. These are the two carbons across our double bond. We have two hydrogens on this carbon, and the carbon on the right has two alkyl groups bonded to it. So this one is a disubstituted alkene. Now we've gone through the whole E1 mechanism, and we've seen that we get a disubstituted product, and a trisubstituted. Now let's think about regiochemistry. For this reaction, it's the region of the molecule where the double bond forms. the double bond formed in this region of the molecule, and for the trisubstituted product, the double bond formed in this region. The trisubstituted product is the major product, and it's also the more stable alkene. So remember, from the video in alkene stability, the more substituted your alkene is, the more stable it is, so this product is more stable, and that's why we form more of it. And the more stable products or the more substituted product is called the Zaitsev product. So we say that this E1 reaction is regioselective because it has a preference to form the more stable product, the more substituted product, which we call the Zaitsev product." + }, + { + "Q": "\nIn the equation shown at 1:52, why is H2O formed and not H3O+? How would I know which to write as a product?", + "A": "Acid + base -> salt + water This is something you should memorise", + "video_name": "aj34f2Bg9Vw", + "timestamps": [ + 112 + ], + "3min_transcript": "- [Voiceover] Let's do another titration problem, and once again, our goal is to find the concentration of an acidic solution. So we have 20.0 milliliters of HCl, and this time, instead of using sodium hydroxide, we're going to use barium hydroxide, and it takes 27.4 milliliters of a 0.0154 molar solution of barium hydroxide to completely neutralize the acid that's present. All right, so let's start with what we know. We know the concentration of barium hydroxide. It's 0.0154 molar, and we also know that molarity is equal to moles over liters. All right, so we have 0.0154 as equal to, let's make moles X, over liters. 27.4 milliliters is 0.0274 liters, right? So that's 0.0274 liters. the moles of barium hydroxide. So let's get out the calculator here, and let's do that. So let's get some room over here. So we take 0.0154 and we multiply that by 0.0274, and that gives us, this will be 4.22 times 10 to the negative fourth, right? So that's equal to 0.000422 moles of barium hydroxide. All right, next, let's write the neutralization reaction. So we have barium hydroxide reacts with HCl. So barium hydroxide plus HCl gives us, for our products, we have H plus and OH minus, so that's H20. And then our other product, this is barium two plus, right? So we have BA2 plus and CL minus 1, so you could cross those over. So BACl2, right? So BACl2, barium chloride, as our other product here. All right, next we need to balance our equation, right? We need to balance the neutralization reaction here. So let's start by looking at the chlorines. So over here on the left, we have one chlorine. On the right, we have two. So we need to put a two right here, and now we have two chlorines on both sides. Next, let's look at hydrogens. So on the left side, we have two hydrogens here, and then we have two over here. So we have four hydrogens on the left. On the right, we have only two hydrogens. So we need to put a two here for this coefficient to give us four hydrogens on the right. So now we have four, and we should be balanced, right? Everything else should be balanced. Let's look at the mole ratio for barium hydroxide to HCl." + }, + { + "Q": "At 0:32 he said lesser and great apes. What is the difference between lesser and greater apes?\n", + "A": "The great apes are chimpanzees, gorillas, humans, and orangutans. The lesser apes are gibbons.", + "video_name": "oFGkYA_diDA", + "timestamps": [ + 32 + ], + "3min_transcript": "In the first video on evolution, I drew something that I called an ape, and then I drew a tail on it. And what I want to do in this video is clarify that that was absolutely wrong. Apes - Apes have no tails. And it's actually one of the main distinguishing characteristics of apes. There's obviously other primates that also have no tails, but apes definitely have no tails. And just to clarify, there's kind of two families within apes and their common names are \"the lesser apes\" and \"the great apes.\" And the lesser apes - the lesser apes are things like gibbons, And the great apes are things like chimpanzees and gorillas, and, me, or human beings! So these right here, these are the great apes. These are the great apes, And clearly this great ape right here did not have a great sense of style in 1994. And really didn't feel the need to have a haircut!" + }, + { + "Q": "\nAt 4:00sec why did you take the force(10n) as kg.m/s2", + "A": "That s what a Newton is. It is 1 kg*m/s^2. It s the force you need to apply to 1 kg to get it to accelerate at 1 m/s^2.", + "video_name": "ou9YMWlJgkE", + "timestamps": [ + 240 + ], + "3min_transcript": "going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff so weight is a force mass is telling you how much stuff there is. And this is really neat, that this formula is so simple because maybe we could have lived in a universe where force is equal to mass squared times acceleration times the square root of acceleration which would have made all of our math much more complicated. But it's nice that it's just this constant of proportionality right over here. It's just this nice simple expression. And just to get our feet wet a little bit with computations involving force mass and acceleration, let's say that I have a force and the unit of force is appropriately called the Newton. So let's say I have a force of 10Newtons - and just to be clear, a Newton is the same thing - so this is the same thing as 10kilogram.metre per seconds squared as kilograms.metres per second square because that's exactly what you get on this side of the formula. So let's say I have a force of 10 Newtons and it is acting on -it is acting on a mass, let's say that the mass is 2 kilograms and I wanna know the acceleration. And once again in this video, these are vector quantities. If I have a positive value here I'm going to--we're going to make the assumption that it's going to the right. If I had a negative value then it would be going to the left. So implicitly I'm giving you not only the magnitude of the force but I'm also giving you the direction. I'm saying it is to the right because it is positive. So what will be the acceleration? Well we just use F=ma You have-on the left hand side 10 - I could write 10 Newtons here or I could write 10kilograms.metres per second squared and that is going to be equal to the mass which is 2 kilograms times the acceleration." + }, + { + "Q": "AT 3:23 HOW CAN WE WRITE F=M^2*A^1/2\n", + "A": "We can t, that was just an example of a more complicated equation that, in another universe, could have been the relationship between mass, force, and acceleration. He mentioned it just to point out how simple the real equation, F=ma, was.", + "video_name": "ou9YMWlJgkE", + "timestamps": [ + 203 + ], + "3min_transcript": "-actually I won't pick favorites here- but this one gives us the famous formula; Force is equal to mass times acceleration And acceleration is a vector quantity and force is a vector quantity. And what it tells us- 'cause we're saying ok if you apply force it might change that constant velocity but how does it change that constant velocity? Well say I have a brick right here and it is floating in space Newton's second law tells us that it's pretty nice for us that the laws of the universe or at least in the classical sense before Einstein showed up The laws of the universe actually dealt with pretty simple mathematics. What it tells us is if you apply a net force let's say on this side of the object and we talk about net force because if you apply two forces that cancel out and that have zero net force then the object won't change it's constant velocity. If you have a net force applied to one side of this object going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff so weight is a force mass is telling you how much stuff there is. And this is really neat, that this formula is so simple because maybe we could have lived in a universe where force is equal to mass squared times acceleration times the square root of acceleration which would have made all of our math much more complicated. But it's nice that it's just this constant of proportionality right over here. It's just this nice simple expression. And just to get our feet wet a little bit with computations involving force mass and acceleration, let's say that I have a force and the unit of force is appropriately called the Newton. So let's say I have a force of 10Newtons - and just to be clear, a Newton is the same thing - so this is the same thing as 10kilogram.metre per seconds squared" + }, + { + "Q": "\nAt 1:58 what does netforce mean?", + "A": "net force means total force. remember that forces can cancel each other out as well if opposing forces act upon an object. for example, if your bank account is overdrawn by $100 and you deposit $100, your net amount will be $0 (-100+100=0), forces behave in the same way depending on their magnitude (size) and direction (which direction they act on an object).", + "video_name": "ou9YMWlJgkE", + "timestamps": [ + 118 + ], + "3min_transcript": "Newton's first law tells us that an object at rest will stay at rest, and an object with a constant velocity will keep having that constant velocity unless it's affected by some type of net force or you actually can say that an object with constant velocity will stay having a constant velocity unless it's affected by net force because really this takes into consideration the situation where an object is at rest. You could just have a situation where the constant velocity is zero. So Newton's first law-you're gonna have your constant velocity it could be zero, it's going to stay being that constant velocity unless it's affected, unless there's some net force that acts on it. So that leads to the natural question. How does a net force affect the constant velocity or how does it affect the state of an object? And that's what Newton's second law gives us- Newton's Second Law of Motion And this one is maybe the most famous -actually I won't pick favorites here- but this one gives us the famous formula; Force is equal to mass times acceleration And acceleration is a vector quantity and force is a vector quantity. And what it tells us- 'cause we're saying ok if you apply force it might change that constant velocity but how does it change that constant velocity? Well say I have a brick right here and it is floating in space Newton's second law tells us that it's pretty nice for us that the laws of the universe or at least in the classical sense before Einstein showed up The laws of the universe actually dealt with pretty simple mathematics. What it tells us is if you apply a net force let's say on this side of the object and we talk about net force because if you apply two forces that cancel out and that have zero net force then the object won't change it's constant velocity. If you have a net force applied to one side of this object going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff" + }, + { + "Q": "\nIn many Khan academy videos, Sal does what I belive is called a dimensional analasis (5:10 in this video), and while I understand how this works, I cannot find anything on this site about where to use them, and the rules for doing so. Help?", + "A": "The rules of it are just the rules of algebra. You should use it pretty much all the time when you are doing physics problems. It provides a good check on your answers. If the units don t come out right, you did something wrong.", + "video_name": "ou9YMWlJgkE", + "timestamps": [ + 310 + ], + "3min_transcript": "so weight is a force mass is telling you how much stuff there is. And this is really neat, that this formula is so simple because maybe we could have lived in a universe where force is equal to mass squared times acceleration times the square root of acceleration which would have made all of our math much more complicated. But it's nice that it's just this constant of proportionality right over here. It's just this nice simple expression. And just to get our feet wet a little bit with computations involving force mass and acceleration, let's say that I have a force and the unit of force is appropriately called the Newton. So let's say I have a force of 10Newtons - and just to be clear, a Newton is the same thing - so this is the same thing as 10kilogram.metre per seconds squared as kilograms.metres per second square because that's exactly what you get on this side of the formula. So let's say I have a force of 10 Newtons and it is acting on -it is acting on a mass, let's say that the mass is 2 kilograms and I wanna know the acceleration. And once again in this video, these are vector quantities. If I have a positive value here I'm going to--we're going to make the assumption that it's going to the right. If I had a negative value then it would be going to the left. So implicitly I'm giving you not only the magnitude of the force but I'm also giving you the direction. I'm saying it is to the right because it is positive. So what will be the acceleration? Well we just use F=ma You have-on the left hand side 10 - I could write 10 Newtons here or I could write 10kilograms.metres per second squared and that is going to be equal to the mass which is 2 kilograms times the acceleration. divide both sides by 2 kilograms So let's divide the left by 2 kilograms let's divide the right by 2 kilograms that cancels out. The 10 and the 2-- 10 divided by 2 is 5 and then you have kilograms cancelling kilograms. Your left hand side you get 5 metres per second squared and then that's equal to your acceleration. Now just for fun, what happens if I double that force? Well then I have 20Newtons--I'll actually work it out-- 20 kilograms.metres per second squared is equal to --I'll actually color-code this-- 2 kilograms times the acceleration" + }, + { + "Q": "At 4:00 why are seconds squared?\nAnd what does Seconds^2 mean? Other then Second * Seconds.\n", + "A": "F = ma. Force is mass times acceleration. Acceleration is change in velocity over time. Velocity is distance over time. So acceleration is change in distance over time over time, or distance over time squared.", + "video_name": "ou9YMWlJgkE", + "timestamps": [ + 240 + ], + "3min_transcript": "going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff so weight is a force mass is telling you how much stuff there is. And this is really neat, that this formula is so simple because maybe we could have lived in a universe where force is equal to mass squared times acceleration times the square root of acceleration which would have made all of our math much more complicated. But it's nice that it's just this constant of proportionality right over here. It's just this nice simple expression. And just to get our feet wet a little bit with computations involving force mass and acceleration, let's say that I have a force and the unit of force is appropriately called the Newton. So let's say I have a force of 10Newtons - and just to be clear, a Newton is the same thing - so this is the same thing as 10kilogram.metre per seconds squared as kilograms.metres per second square because that's exactly what you get on this side of the formula. So let's say I have a force of 10 Newtons and it is acting on -it is acting on a mass, let's say that the mass is 2 kilograms and I wanna know the acceleration. And once again in this video, these are vector quantities. If I have a positive value here I'm going to--we're going to make the assumption that it's going to the right. If I had a negative value then it would be going to the left. So implicitly I'm giving you not only the magnitude of the force but I'm also giving you the direction. I'm saying it is to the right because it is positive. So what will be the acceleration? Well we just use F=ma You have-on the left hand side 10 - I could write 10 Newtons here or I could write 10kilograms.metres per second squared and that is going to be equal to the mass which is 2 kilograms times the acceleration." + }, + { + "Q": "At 1:15, the second law of motion is stated. How is this equation derived?\n", + "A": "It s not derived, it s an empirical observation.", + "video_name": "ou9YMWlJgkE", + "timestamps": [ + 75 + ], + "3min_transcript": "Newton's first law tells us that an object at rest will stay at rest, and an object with a constant velocity will keep having that constant velocity unless it's affected by some type of net force or you actually can say that an object with constant velocity will stay having a constant velocity unless it's affected by net force because really this takes into consideration the situation where an object is at rest. You could just have a situation where the constant velocity is zero. So Newton's first law-you're gonna have your constant velocity it could be zero, it's going to stay being that constant velocity unless it's affected, unless there's some net force that acts on it. So that leads to the natural question. How does a net force affect the constant velocity or how does it affect the state of an object? And that's what Newton's second law gives us- Newton's Second Law of Motion And this one is maybe the most famous -actually I won't pick favorites here- but this one gives us the famous formula; Force is equal to mass times acceleration And acceleration is a vector quantity and force is a vector quantity. And what it tells us- 'cause we're saying ok if you apply force it might change that constant velocity but how does it change that constant velocity? Well say I have a brick right here and it is floating in space Newton's second law tells us that it's pretty nice for us that the laws of the universe or at least in the classical sense before Einstein showed up The laws of the universe actually dealt with pretty simple mathematics. What it tells us is if you apply a net force let's say on this side of the object and we talk about net force because if you apply two forces that cancel out and that have zero net force then the object won't change it's constant velocity. If you have a net force applied to one side of this object going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff" + }, + { + "Q": "\nat 14:12 , to get the value of T2 cant we just assume that mg=T1+T2 i.e. 30N=60N+T2?\nThat would be T2=-30N..... would that be wrong?", + "A": "Yes that would be wrong because you have to separate forces and tensions acting on the x-axis from those acting on the y-axis. That has to be done because forces and tensions acting on the x-axis have no effect on those acting on the y-axis (at least as long as they act on the same point of the object). T2 is horizontal so it acts on the x-axis, mg is vertical so it acts on the y-axis and T1 actually acts on both the x-axis and the y-axis which is why David had to break it into its components.", + "video_name": "aHlOp5nYs28", + "timestamps": [ + 852 + ], + "3min_transcript": "I'll get T one times sine theta, and then I divide both sides by sine theta. I'll end up with T one equals T one in the y direction divided by sine theta. I know T one in the y direction. That was 30 degree, or sorry, not 30 degrees. That was 30 Newtons. So I've got 30 Newtons, that's my force, upward. This vertical component right here had to be 30 Newtons 'cause it had to balance gravity, divided by sine of the angle. But what is this angle? We know this angle's 30. And you could probably convince yourself, if I draw a triangle this way, let's try to figure out, we wanna figure out what this angle is right here 'cause that's what this angle is here. So if this is 30 and that's 90, then this has to be 60. And if that's 60, and this is 90, then this has to be 30. So this angle is 30 degrees right here. So that's 30 degrees. So this angle right here, So when I'm taking my sine, I'm taking my sine of 30 degrees. And I get 30 Newtons divided by sine of 30, and sine of 30 is 1/2. So .5, so I get that this is 60 Newtons. And that might seem crazy. You might be like, \"Wait a minute. \"T one is 60 Newtons? \"60 Newtons? \"The weight of this chalkboard is only 30 Newtons. \"How in the world can the tension in this rope \"be 60 Newtons?\" I mean, if we just hung it by a single string, if we just hung this chalkboard by a single string over the center of mass, you'd just get a tension of 30 Newtons. How can this be 60 Newtons? And the reason is, this part's gotta be 30 Newtons. We know that 'cause it has to balance gravity. But that's only part of the total tension. So if the total tension, if part of the total tension is 30, all of the tension's gotta be more than 30. And in this case, it's 60 Newtons. 'cause it's at an angle. So this component has to equal gravity, and this total amount has to be bigger than that so that its component is equal to gravity. Right, how do we figure out T two? Well, you don't invent a new strategy. We keep going, we're just gonna say that the acceleration in the horizontal direction is the net force in the horizontal direction divided by the mass, so we still stick with Newton's Second Law even when we wanna find this other force. This force is horizontal, so it makes sense that we're gonna use Newton's Second Law for the horizontal direction. Again, if this chalkboard is not accelerating, the acceleration is zero, so I'll draw a line here to keep my calculations separate. Equals net force in the x direction. Okay, now I'm gonna have T one in the x direction. So this is gonna be T one in the x. So I'll have T one in the x direction. That's positive 'cause it points right, and I'm gonna consider rightward positive. Minus T two, all of T two. I don't have to break T two up. T two points completely in the horizontal direction." + }, + { + "Q": "\nAt 11:31 mvr = Iw. What is the Iw stand for?", + "A": "moment of Inertia*angular velocity", + "video_name": "viLpmZtQYzE", + "timestamps": [ + 691 + ], + "3min_transcript": "if r is zero, there's gonna be no torque exerted by that axis. And if there's no torque exerted externally, there's no change in angular momentum of the system. So this system of ball and rod is gonna have no external torque on it. That means the angular momentum has to stay the same. This is a classic conservation of angular momentum problems. So we're gonna say that L initial, the initial angular momentum, has to equal the final angular momentum. And we'll just say, for our entire system, what had angular momentum initially? Well, it was this mass. So this mass had the angular momentum. And how do we find that? Remember it's m times v times R, and the r is that distance of closest approach. So we're gonna use this here for the whole four meters as this R. Yes, you can consider this hypotenuse R and a sine of the angle, but that's harder than it needs to be. You can find angular momentum, mvR, that's gonna equal the final angular momentum. So since this ball comes to rest, and it's only the bar that has angular momentum afterward, we only have to worry about the angular momentum of the bar on the final side. And to find the angular momentum of an extended object, a rigid object, you can use I omega. And this would let us solve for what is the final angular velocity of this rod after the collision. So that's what we wanna figure out. What is the final angular velocity of the rod after the collision? Now we can figure it out. We know the mass the of the ball, m. We know the speed of the ball initially. We know the R, line of closest approach. That's four meters. What's the moment of inertia here? Well, it's just gonna be 1/3 m L squared. Let me clean this up a little bit. Let me take this. I'll just copy that. Put that right down over here, and we could say that the moment of inertia of a mass of a rod, it's rotating around its end, is always gonna be 1/3 m L squared. So 1/3 times the mass of the rod, times the length of the rod squared, because this ball's line of closest approach was jus equal to the entire length of the rod, since it struck it at the very end, and then times omega. So we can solve this for omega now. We can say that omega. I'm gonna bring this down around here, so we go some room. Omega final of the rod is just gonna be, what? It's gonna be mass of the ball times the initial speed of the ball, times the line of closet approach. And then I'm gonna divide by 1/3 the mass of the rod times the length of the rod. I can just call that R, it's the same variable, length of the rod squared, So I can cancel off one of these Rs, and then I can plug in numbers if I wanted to actually get a number. I could say that the final angular velocity of this rod is gonna be five kilograms, that was the mass of the ball, times eight meters per second, that was the initial speed of the ball, and then I'm gonna divide by 1/3 of the mass of the rod was 10 kilograms, and then the length of the rod, which is this line of closest approach," + }, + { + "Q": "At 10:10, David says that there is no external torque exerted on the system. But, isn't the mass exerting torque on the system by hitting the moment arm?\n", + "A": "The system includes the mass, so the mass is not external to the system, although it is external to the rod. This is why the angular momentum of the rod can change without violating conservation of momentum, but the combined angular momentum of rod plus mass cannot.", + "video_name": "viLpmZtQYzE", + "timestamps": [ + 610 + ], + "3min_transcript": "is always just equal to that, so you could make your life easy. Just imagine, when this ball comes in, at what point is it closest to the axis? That would be this point. And then how far is it when it is closest? That gives you this R value. You can take mvR. That gives you the angular momentum of this point mass. It tells you the total amount of angular momentum that thing could transfer to something else if it lost all of its angular momentum. That's how much angular momentum something, like that rod, could get. So let's try an example. Let's do this example. It's actually a classic, a ball hitting a rod. Man, I'm telling you, physics teachers and professors, You should know how to do this. Let's get you prepared here. So let's say this ball comes in. It hits a rod, right? And so the ball is gonna come in. Ball is gonna hit a rod, and let's put some numbers on this thing, so we can actually solve this example. let's say the ball had a mass of five kilograms. It was going eight meters per second, hits the end of the rod, and the rod is 10 kilograms, four meters long. Let's assume this rod has uniform density, evenly throughout it, and it can rotate around the end. So when the ball gets in here, strikes the end of the rod, the rod is gonna rotate around its axis. And let's make another assumption. Let's assume when this ball does hit the rod, the ball stops. So after hitting the rod, the ball has stopped, and the rod moves on with all the angular momentum that the ball had. That will just make it a little easier. We'll talk about what to do if that doesn't happen. It's not that much harder. Let's just say that's the case initially or so. I'll move the ball back over to here. How do we solve this problem? Well, we're gonna try to use conservation of angular momentum. We're gonna say that even though there's an axis his exerting a force, the force that that axis is gonna exert on our system is gonna exert zero torque, because the r value. Torque is equal to r F sine theta. And if the r is zero, if r is zero, there's gonna be no torque exerted by that axis. And if there's no torque exerted externally, there's no change in angular momentum of the system. So this system of ball and rod is gonna have no external torque on it. That means the angular momentum has to stay the same. This is a classic conservation of angular momentum problems. So we're gonna say that L initial, the initial angular momentum, has to equal the final angular momentum. And we'll just say, for our entire system, what had angular momentum initially? Well, it was this mass. So this mass had the angular momentum. And how do we find that? Remember it's m times v times R, and the r is that distance of closest approach. So we're gonna use this here for the whole four meters as this R. Yes, you can consider this hypotenuse R and a sine of the angle, but that's harder than it needs to be. You can find angular momentum, mvR, that's gonna equal the final angular momentum." + }, + { + "Q": "\nAt around 09:00 , the second step of the Sn1 reaction, I was just wondering if instead of a new metanol, an Iodide could come and take the proton away? Isn't that even more electronegative than the metanol? And if so, which of the two happens more often? Thank you", + "A": "Yes, an iodide ion could certainly remove the proton. but you must remember that the solvent is methanol, so there are many more methanol molecules than iodide ions present in the solution. Also, the cation is probably solvated by a shell of methanol molecules. The probability of the H being attacked by a methanol molecule is therefore much greater than the probability of being attacked by an iodide ion.", + "video_name": "MtwvLru62Qw", + "timestamps": [ + 540 + ], + "3min_transcript": "We're going to get substituted with the weak base, and the weak base here is actually the methanol. The weak base here is the methanol. So let me draw some methanol here. It's got two unbonded pairs of electrons and one of them, it's a weak base. It was willing to give an electron. It has a partial negative charge over here because oxygen is electronegative, but it doesn't have a full negative charge, so it's not a strong nucleophile. But it can donate an electron to this carbocation, and that's what is going to happen. It will donate an electron to this carbocation. And then after that happens, it will look like this. That's our original molecule. Now this magenta electron has been donated to the carbocation. The other end of it is this blue electron right here on the oxygen. That is our oxygen. Here's that other pair of electrons on that oxygen, and it is bonded to a hydrogen and a methyl group. And then the last step of this is another weak base might be able to come and nab off the hydrogen proton right there. Oh, I want to be very clear here. The oxygen was neutral. The methanol here is neutral. It is giving away an electron to the carbocation. The carbocation had a positive charge because it had lost it originally. Now it gets an electron back. It becomes neutral. The methanol, on the other hand, was neutral, gives away an electron, so now it becomes-- it now is positive. So now you might have another methanol. You might have another methanol molecule sitting out here someplace that might also nab the proton off of this positive ion. So this one right here, it would nab it or it It would give the electron to the hydrogen proton, really. then that becomes neutral. So in the final step, it'll all look like this. We have that over here. The methanol that had originally bonded has lost its hydrogen, so it looks like this. We just have the oxygen and the CH3 there. It is now neutral because it gained an electron when that hydrogen proton was nabbed. So if you wanted to draw it, it has actually those two extra electrons, just like that. And if you want to draw this last methanol, it's now a positive cation, so it looks like this. So it's OH, CH3, H, and then it has unbonded pair right there, and now this has a positive charge. So that was the Sn1 reaction. Now, the other reaction that's going to occur is the E1. Once again, our first step-- nope, I" + }, + { + "Q": "\nAt 13:20 Sal showns that the change in entropy after the cycle is (2Q_f)/T. The left side of the equation is a subtraction, not an addition like in the last video. In the last video, he showed that the change in entropy is equal to Q_1/T_1 + Q_2/T_2. What's going on here? These are clearly not the same. This doesn't make any sense.", + "A": "In this video Sal talks about irreversible system. That s where (2Q_f)/T comes from: it s a heat generated by friction. About the signs of Q: in the previous video Sal puts a plus sign in front of the Q, but actual value of Q is negative (he actually mentioned that we ll see its actually negative somewhere in the video). In this video Sal assumes the value of Q to be positive (as he mentions at about 10:20, so he puts a negative in front.", + "video_name": "PFcGiMLwjeY", + "timestamps": [ + 800 + ], + "3min_transcript": "This is still positive. In either direction, when we move upwards or downwards, the system is generating friction. Now, we always said, we went all the way here, we went all the way back. So the sum of these has to be equal to 0, because this is a state variable. So if the sum of all of this has to be equal to 0, let's So this gets us to Qa minus Qr. So the heat accepted minus the heat released. The W's cancel out. Plus-- let me see right here-- plus 2 times the heat of friction in either direction. All of that has to be equal to 0. Let's see. What we can do is, we can rewrite this as the heat accepted minus the heat released is equal to minus 2 times the amount of heat generated from friction. And then if we just switch these around, we'll get the just wanted to get all positive numbers-- 2 times the heat of friction. Now why did I do all of this? Because I wanted to do an experiment with an irreversible system, and this was a very simple experiment with an irreversible system. Now, we said that delta S, which a long time ago I defined as Q divided by T-- and in this video, I said it had to be reversible. And I wanted to show you right now that what if I didn't make the constraint that this has to be reversible? Because if this doesn't have to be reversible, and I just use this definition right here, you'll see that your delta S here would be-- you just divide everything by T-- because our temperature was constant the entire time, we were just on a reservoir-- you'll see that this is going to be your delta s. This is your total change in the, I guess you could say, So this is, let me say, this is the heat added to the system. Let me do it this way. Heat added to the system, divided by the temperature at which it was added. Which is a positive number. Even though we got to the exact same place on this date diagram. So in an reversible system, this wouldn't be a valid state variable. So it's only a valid state variable if it's reversible. Now, does that mean that you can only talk about entropy for reversible reactions? No. You can talk about entropy for anything. But what you do is-- and this is another important point. So let's say that I have some irreversible reaction that goes from here to here. And I want to figure out its change in entropy." + }, + { + "Q": "\nAt 1:29, it's said that heat is released (Q2), and right after he says that the change is adiabatic, and no heat is being transferred to and from the system. Isn't that contradictory?", + "A": "He was pointing to the horizontal paths when talking about heat transfer, but was pointing to the vertical paths when talking about the adiabatic paths.", + "video_name": "PFcGiMLwjeY", + "timestamps": [ + 89 + ], + "3min_transcript": "In the video where I first introduced the concept of entropy, I just tried something out. I defined my change in entropy as being equal to the heat added to a system, divided by the temperature at which it was added to the system. And then I tested to see if this was a valid state variable. And when I did that, I looked at the Carnot cycle. And this is a bit of a review. Never hurts to review. Let me draw the PV diagram here. We saw that we start at this state here, and then we proceed isothermically. We removed little pebbles off the piston. So we increased the volume and lowered the pressure. Then we proceed adiabatically, where we isolated things and we moved like that. That was adiabatically. Then at this other isotherm, we added the pebbles back. And then we isolated the system again. got back to our original state. And I did a couple of videos where I show that if you take the heat added here-- so this is all being done at some high temperature, T1. This is being done at some low temperature, T2. There's some heat being added here, Q1, and that there's some heat being released here, Q2. And since these are adiabatic, there's no transfer of heat to and from the system. And when I looked at this, and when I looked at the Carnot cycle, and I used this definition of entropy, I saw that the total change in S, when I go from this point all the way around and got back, the change in S, was equal to Q1 over T1 plus Q2 over T2. And then I actually showed you that this was equal to 0, which is exactly the result that I wanted to see. Because in order for this to be a state variable, in order how I got there. It should only be dependent on my state variables. So even if I go on some crazy path, at the end of the day, it should get back to 0. But I did something, I guess, a little bit-- what I did wasn't a proof that this is always a valid state variable. It was only a proof that it's a valid state variable if we look at the Carnot cycle. But it turns out that it was only valid because the Carnot cycle was reversible. And this is a subtle but super important point, and I really should've clarified this on the first video. I guess I was too caught up showing the proof of the Carnot cycle to put the reversibility there. And before I even show you why it has to be reversable, let me just review what reversibility means. Now, we know that in order to even define a path here, the system has to be pretty close to equilibrium the whole time. That's the whole reason why throughout these videos, I've" + }, + { + "Q": "at 6:27, how 2sp2 and why not 2sp3?\n", + "A": "sp3 is in the case of a single bond sp2 is in the case of the double bond sp is in the case of a triple bond", + "video_name": "lJX8DxoPRfk", + "timestamps": [ + 387 + ], + "3min_transcript": "" + }, + { + "Q": "From 11:20 to 12:06, why do pi bonds pull the nuclei of the two carbons in ethylene closer together? Wouldn't the electrons be repelling each other?\n", + "A": "pi bonds occurs when there is a parallel bond between two p orbitals.We know that an orbital is nothing but a probability cloud representing possible positions of electrons.Now, because in the pi bonds the two p orbitals overlap, this would mean that in that particular region there is a higher possibility of finding an electron.More electrons (-ve charge) would attract the nuclei of the two atoms more strongly, thus bringing them closer.This is also the reason of why these molecules are smaller. Hope this helped...!", + "video_name": "lJX8DxoPRfk", + "timestamps": [ + 680, + 726 + ], + "3min_transcript": "" + }, + { + "Q": "At 6:57, how do you find the probability, and if you use probability would the answer be accurate?\n", + "A": "You don t! You measure the concentrations experimentally. Sal s use of probabilities here was just a way of trying to get you to intuitively understand why the expression for the equilibrium constant looks the way it does.", + "video_name": "ONBJo7dXJm8", + "timestamps": [ + 417 + ], + "3min_transcript": "what's the probability of having five heads? Well you would multiply the probability of one head five times. So the forward reaction probability is going to be the concentration of V to the a power, and, of course, that's not enough to have the reaction happen. You also need to have b of the X molecules there. So you have the concentration of X to the b power. And I want to make sure you understand this. My claim is that this is approximation -- or actually it's a pretty good way of calculating-- the probability. So let me write it this way. The rate is equal to some constant that takes into account the temperature and the molecular configurations times the probability of having a V molecules and b X molecules And the best way to approximate that is with their concentration. Obviously, the higher the concentration, the higher the moles per liter, the more likely you're going to find that many of molecules in kind of that little small space that you care about, and the temperature and the configuration are going to matter more. But if you use the concentration as the probability of a -- let me switch colors. If the probability of having a V molecule in some volume -- if we assume that the solution is homogeneous, that the V molecules are roughly evenly distributed, it's going to be -- this isn't even an approximation. It's going to be the concentration of the V molecules If we want the probability of a, where a is a number, it could be five V molecules, a V's in some volume, it's the probability of finding this a times. So it's going to be equal to -- and this is just from the probability concepts that we learned in the whole probability playlist. So if you want to have five heads in a row, it's 1/2 to the fifth power. If you want to have V molecules there, five of them at the same time in some volume, or a of them, it's going to be V to the a power times the volume. If you also care about the probability so you want all of that, so a V's and b X's in some volume," + }, + { + "Q": "\nIf at around 11:20, Sal divided by [Y]^c [Z]^d, wouldn't that change the answer? But then again a constant divided by a constant is another constant. So couldnt the equilibrium constant be the reciprocal of what it actually is?", + "A": "If he had divided by [Y]^c[Z]^d instead, he d end up with the reciprocal, yes. This, however, gives the equilibrium constant for the reverse direction of the reaction, i.e, the products reacting to form the reactants. As a general rule, the equlibrium constant for the reverse reaction is equal to the reciprocal of the equlibrium constant of the forward reaction.", + "video_name": "ONBJo7dXJm8", + "timestamps": [ + 680 + ], + "3min_transcript": "-- let's call that K-minus-- the same exact logic holds. We're just going in this direction now. If we look at our original one, we're going in that direction. So for this reaction, we do the same thing. We literally just do different letters, so the reverse reaction is just going to be the concentration of the Y molecule to the c power, because we need c of them there roughly at the same time, times the concentration of the Z molecule to the d power. Now, just at the beginning of the video, we said that equilibrium is when these rates equal each other. I wrote it down right here. So if the reverse rate is equal to some constant times this, and the forward rate is equal to some constant times that, then we reach equilibrium when these two are equal to each other. Let me clear up somespace here. Let me clear this up, too. So when are they going to be equal to each other? -- the forward rate is this. That's our forward constant, which took into account a whole bunch of temperature and molecular structure and all of that-- times the concentration of our V molecule to the a power. You can kind of view that as what's the probability of finding in a certain volume -- and that certain volume can be factored into that K factor as well-- but what's the probability of finding V things, a V molecules in some volume. And it's the concentration of V to the a power times concentration of X to the b power -- that's the forward reaction-- and that has to equal the reverse reactions. So K-minus times the concentration of Y to the c power times the concentration of Z to the d power. Now, if we divide both sides by -- let me erase more space. Nope, not with that. All right. So let's divide both sides by K-minus so you get K-plus over K-minus is equal to that, is equal to Y to the c times Z to the d. All of that over that-- V to the a times the concentration of X to the b. Let me put this in magenta just so you know that this was this K-minus right here. And then, these are just two arbitrary constants, so we could just replace them and call them the equilibrium constant. And we're there where we need to be. We're at the formula for the equilibrium constant. Now, I know this was really hand wavy, but I want you to at least get the sense that this doesn't come from out of the blue, and there is -- at least I think there is-- there's an intuition here. These are really calculating the probabilities of finding -- this is the forward reaction rate probabilities" + }, + { + "Q": "\nAt 9:00, what does the Keq stand for?", + "A": "The Equilibrium Constant K", + "video_name": "ONBJo7dXJm8", + "timestamps": [ + 540 + ], + "3min_transcript": "If we want the probability of a, where a is a number, it could be five V molecules, a V's in some volume, it's the probability of finding this a times. So it's going to be equal to -- and this is just from the probability concepts that we learned in the whole probability playlist. So if you want to have five heads in a row, it's 1/2 to the fifth power. If you want to have V molecules there, five of them at the same time in some volume, or a of them, it's going to be V to the a power times the volume. If you also care about the probability so you want all of that, so a V's and b X's in some volume, So it's going to be equal to the concentration of V to the a power times the concentration of X to the b power times the volume. So the probability of finding the right number of V particles and X particles in the right place in some volume is going to be proportional to exactly this. And we're saying that the reaction rate, the forward reaction rate, is also proportional to this thing. So that's where we get the forward reaction rate. So the rate forward is equal to the concentration of our V molecules to the a power times the concentration of our X molecules to the b power. Now, if we want to find the reverse rate, so this is the rate forward. If we want to find the rate of the reverse reaction, -- let's call that K-minus-- the same exact logic holds. We're just going in this direction now. If we look at our original one, we're going in that direction. So for this reaction, we do the same thing. We literally just do different letters, so the reverse reaction is just going to be the concentration of the Y molecule to the c power, because we need c of them there roughly at the same time, times the concentration of the Z molecule to the d power. Now, just at the beginning of the video, we said that equilibrium is when these rates equal each other. I wrote it down right here. So if the reverse rate is equal to some constant times this, and the forward rate is equal to some constant times that, then we reach equilibrium when these two are equal to each other. Let me clear up somespace here. Let me clear this up, too. So when are they going to be equal to each other?" + }, + { + "Q": "@ 5:00 sal says that the concentration is a rough approximation of the probability. But what if Concentration is greater than one? we know that probability is always less than or equal to one. Then how can concentration be an approximation for probability in that case?\n", + "A": "If you think of concentration as the number of molecules of the substance of interest divided by the number of all other molecules in the mixture then concentration will have the same limits as probability, that is, between 0 and 1.", + "video_name": "ONBJo7dXJm8", + "timestamps": [ + 300 + ], + "3min_transcript": "and in the right place and kind of close enough in order for the reaction to happen. So the reaction is really going to be driven by, if you think about it, the probability of finding a V molecules and b molecules all within close enough confines that they can actually react. So you could say that the rate is going to be driven by -- maybe it's going to be proportional. Let's say it's just equal to-- let's say some constant that takes into account things like temperature and how the molecules are actually configured. Because it's not dependent just on them being there. You have to have worry about their kinetic energies. You have to worry about their shape, because some shapes are going to be more conducive to reaction than others. So let's just let that be taken into account with a K. And we're talking about the forward reaction, right? So in order for the forward reaction to happen, let's call that K plus for the forward reaction. We have to have So what's the probability of having a molecules of X? Or what's a rough approximation of the probability? Well, the concentration. Let's think about this a second. When we write the concentration of the molecule V, which I think when I did this was the blue one right here, what is that given in? That is given in moles per liter. Moles is just a number, so this tells us, look, in any given volume, roughly how many of the molecules do you expect to find? That's what concentration is. So if I wanted to figure out the probability of finding a of these molecules, because that's how many I need, I need to multiply this by itself a times, because I need a of them. The probability of having just one molecule in just some small fraction, you would just use the concentration once. But you're going to use it a times, because you want a of those molecules there, right? what's the probability of having five heads? Well you would multiply the probability of one head five times. So the forward reaction probability is going to be the concentration of V to the a power, and, of course, that's not enough to have the reaction happen. You also need to have b of the X molecules there. So you have the concentration of X to the b power. And I want to make sure you understand this. My claim is that this is approximation -- or actually it's a pretty good way of calculating-- the probability. So let me write it this way. The rate is equal to some constant that takes into account the temperature and the molecular configurations times the probability of having a V molecules and b X molecules" + }, + { + "Q": "\nat 4:40,sal says when sun is just setting,why not observe it as sun is just rising; like in the previous case?", + "A": "If you looked at the star as the sun was rising exactly 6 months later, while on the other side of the sun, you would be facing the wrong direction. At 5:21, Sal says that at sunset 6 months later straight up is the same direction .", + "video_name": "ETzUpoqZIHY", + "timestamps": [ + 280 + ], + "3min_transcript": "up is going to be at some angle to the left of straight up. It's going to be right over there. And obviously the star won't be that big relative to your entire field of vision, but you get the idea. Maybe I'll draw it a little bit smaller, just like that. So there's going to be some angle here. And this angle, whatever it is, let's just call it Theta, that's going to be the same angle as this. And when I talk about the angle, I'm talking about if you measure from one side of the horizon to the other side of the horizon, you're essentially looking halfway around the earth. That would be 180 degrees. So you could literally measure what this angle is right over here. Now let's say we waited six months. What's going to happen? Six months, we're going to be on this side of the sun. We're assuming that our distance is relatively constant at one astronomical unit. Remember, the earth is rotating like this. So if we wait, right at sunset, right when the last glimpse of the sun has just gone away-- because you can remember, right now, the sun is illuminating this side of the earth. The sun is going to be illuminating that side of the earth. So if we're sitting right at the equator right over there, right when the sun is just setting, we look straight up. Let me do that in the same color. We look straight up. So six months later when we look straight up, where is the star relative to straight up? Well now the star will be to the right. It'll be in the direction. So if this is our field of vision six months later, now the sun is setting all the way to the right, on the right horizon. And if we look straight up, this star now is going to be to the right of straight up. Well, it looks like relative to straight up-- and we're looking at the exact kind of position of the earth. We're making sure that we're picking times of year and times of day where straight up is the same direction. We're looking in the same direction of the universe. It looks like the position of that star has actually shifted. And let's say that this is the middle of summer, and that this is the middle of winter. Doesn't have to be. It could be any other two points in time six months apart. Then when we look at this star in the summer, it's going to be over here. Summer, it's going to be right over there. And when we look at the star in the winter, it is going to be over here. And, in general, for any star, especially stars that are in the same plane as the solar system, you can find two points in the year where that star is at a kind of a maximum distance from center." + }, + { + "Q": "\nAt 3:46, the switching on and off is called pulse-width modulation, right?", + "A": "That is correct.", + "video_name": "a16uKH2K7gM", + "timestamps": [ + 226 + ], + "3min_transcript": "And that's an AC wave, sign wave. And we want the power to flow in a straight line, which is DC. So the AC, basically if you look at it it means the current's flowing in one direction, then it switches and flows in the other. So it's going like this. So we want to convert it to DC. So those four diodes, these four guys right here, convert it to DC by basically flipping the wave over like this. And then the power only flows in one direction. Which is a key aspect there. Let's look at some of the other components. We have this resistor right here. This resistor helps the board in current sensing. Then we have these two capacitors here. They are basically part of an electromagnetic interference filter. This is the transformer. The transformer is a high frequency transformer. In order for a transformer to function, there has to be a change in the voltage. The way we make change in the voltage is we do what's called pulse width modulation. The voltage is going to be coming out like this. And then it's creating a square wave, what's called a square wave. This is direct current, DC. But it's being... So this is 100% power and this is zero... So it's being switched on and off very fast. That turning on and off allows for the induction to happen, which occurs inside of this transformer here. The transformer steps the voltage down from 120 volts to both 12 volts and five volts. That pulse width modulation, or that square wave This is the IC chip, that controls the PWM or pulse width modulation. You can see there are a number of resistors here. They're to protect the circuitry on the IC chip most likely. We've got some diodes as well, those control the flow of current. The opto-isolator is used to sample output voltage and regulate it during different, basically different load conditions. And this guy right here, it looks like a transistor but it is actually a shunt regulator, which basically functions like a zener diode, or a variable zener diode. A zener diode allows current to flow in one direction to a certain point and then above a certain point, it allows current to flow in both directions. So, this sort of functions like a variable zener diode. And I wanted to say thank you to Toom-Too-Pro who provided some awesome feedback on the first video," + }, + { + "Q": "@7:30 ish, if it was a high frequency, why would we hear? I thought we couldn't hear high frequencies and only certain animals could or is that just a myth?\n", + "A": "He is saying that the high frequency affects other electronic devices, such as radios, it would have the same effect as when you are getting a phone call or message when you are near the radio.", + "video_name": "a16uKH2K7gM", + "timestamps": [ + 450 + ], + "3min_transcript": "This is the IC chip, that controls the PWM or pulse width modulation. You can see there are a number of resistors here. They're to protect the circuitry on the IC chip most likely. We've got some diodes as well, those control the flow of current. The opto-isolator is used to sample output voltage and regulate it during different, basically different load conditions. And this guy right here, it looks like a transistor but it is actually a shunt regulator, which basically functions like a zener diode, or a variable zener diode. A zener diode allows current to flow in one direction to a certain point and then above a certain point, it allows current to flow in both directions. So, this sort of functions like a variable zener diode. And I wanted to say thank you to Toom-Too-Pro who provided some awesome feedback on the first video, or switch-mode power supplies can work." + }, + { + "Q": "At 1:13 how does a radio uses electricity and how much ?\n", + "A": "it uses 5watt-hours, or after 200 days of such use", + "video_name": "a16uKH2K7gM", + "timestamps": [ + 73 + ], + "3min_transcript": "- [Instructor] So this is our switch mode power supply for the DVD player. Now, the switch mode power supply operates at a high frequency. Linear power supplies operate at lower frequencies. The key differences are that switch mode power supplies tend to be smaller in form factor. They produce more electromagnetic interference, so they have to have a lot more filtering, but they also operate with much more efficiency. They can change the amount of power they supply and they therefore are able to... They don't use near as much power as a linear power supply. So those are the main differences there. Now, if you look right here, you can see this is where the power comes in from the house. There's a fuse here and this fuse is to protect the house from any failure on the board. So if there's a failure, the fuse will fail. This safety capacitor is also designed to fail in the case of a short. And it's got all these markings on it for regulation. It's basically meant to fail, and protect the house, again, if there's a problem. The induction coils here, and this capacitor here, they're basically used to filter out noise from the power supply. The power supply, since it operates at a high frequency produces noise. And these prevent the noise from going back through to the house and causing other problems in the line. Now, if you look here, you can see these four diodes. Diodes prevent current flow in one direction. So they act like an electrical valve. So, the four diodes are set up there to act as a bridge rectifier. And they take the current that comes in, which is DC current or AC current, and they convert it to DC current. So I can show you really quickly right here. And that's an AC wave, sign wave. And we want the power to flow in a straight line, which is DC. So the AC, basically if you look at it it means the current's flowing in one direction, then it switches and flows in the other. So it's going like this. So we want to convert it to DC. So those four diodes, these four guys right here, convert it to DC by basically flipping the wave over like this. And then the power only flows in one direction. Which is a key aspect there. Let's look at some of the other components. We have this resistor right here. This resistor helps the board in current sensing. Then we have these two capacitors here. They are basically part of an electromagnetic interference filter." + }, + { + "Q": "\nAt 0:24, the hydrogen atom is represented by the Bohr model but isn't the Bohr model incorrect because electrons are actually not orbiting the nucleus in a circle and actually can behave as a wave as well? Then why is this model used?", + "A": "Because the Bohr model will help you explain basic things in Chemistry. You could use a quantum mechanical model of an atom or even think of electrons as cloud charges instead of having a particle-wave duality --- but why would you if a simpler model can suffice? Models are (almost) always wrong, not just in Chemistry either, and that s OK because they are supposed to help you understand and a Bohr model will do this.", + "video_name": "AznXSVx2xX0", + "timestamps": [ + 24 + ], + "3min_transcript": "- We've been talking about the Bohr model for the hydrogen atom, and we know the hydrogen atom has one positive charge in the nucleus, so here's our positively charged nucleus of the hydrogen atom and a negatively charged electron. If you're going by the Bohr model, the negatively charged electron is orbiting the nucleus at a certain distance. So, here I put the negatively charged electron a distance of r1, and so this electron is in the lowest energy level, the ground state. This is the first energy level, e1. We saw in the previous video that if you apply the right amount of energy, you can promote that electron. The electron can jump up to a higher energy level. If we add the right amount of energy, this electron can jump up to a higher energy level. So now this electron is a distance of r3, so we're talking about the third energy level here. This is the process of absorption. The electron absorbs energy and jumps up to a higher energy level. This is only temporary though, It's eventually going to fall back down to the ground state. Let's go ahead and put that on the diagram on the right. Here's our electron, it's at the third energy level. It's eventually going to fall back down to the ground state, the first energy level. Here's the electron going back to the first energy level here. When it does that, it's going to emit a photon. It's going to emit light. When the electron drops from a higher energy level to a lower energy level, it emits light. This is the process of emission. I could represent that photon here. This is how you usually see it in textbooks. We emit a photon, which is going to have a certain wavelength. Lambda is the symbol for wavelength. We need to figure out how to relate lambda to those different energy levels. The energy of the photon is, the energy of the emitted photon is those two energy levels. We have energy with the third energy level and the first energy level. The difference between those... So, the energy of the third energy level minus the energy of the first energy level. That's equal to the energy of the photon. This is equal to the energy of that photon here. We know the energy of a photon is equal to h nu. Let me go ahead and write that over here. Energy of a photon is equal to h nu. H is Planck's constant, this is Planck's constant. Nu is the frequency. We want to think about wavelength. We need to relate the frequency to the wavelength. The equation that does that is of course, C is equal to lambda nu. So, C is the speed of light, lambda is the wavelength, and nu is the frequency." + }, + { + "Q": "5:25 what is brass?\n", + "A": "Brass is basically an alloy of copper and zinc. If you do not know what an alloy is, it is just a metal formed by mixing two or more other metals to give it a special property. Steel for example is an alloy as well and its purpose is as you might know, is to make it super hard! Hope this was informative", + "video_name": "qLMsZKx_a8s", + "timestamps": [ + 325 + ], + "3min_transcript": "So it's really a handy way to do it. It used to be you'd have to look it up or just memorize what the different color band codes meant. But this particular resistor, it's got a green band. So we'll put it on green. There we go. And it's got a-- it looks like a navy blue band, and a gold band, green-- oh, wait. Actually, there's a black one. Sorry-- and then the gold one. There we go. So this is a 56 ohm resistor. And that's the amount of resistance that that resistor provides. And the switch right here is just a momentary switch. All right. It's not a momentary switch. It's a continuous switch. So that means when you push it down, it stays down. So the light will stay on after you push it. And the circuit is extremely simple. Basically, you've got the power from the batteries. It comes in through the loop here and the switch basically opens and closes and stops the power flow. Or when you push down on it, it closes and allows the power to flow in the continuous loop there. And so that's what's inside of a tap light. Let's take the batteries out really quick so. These are double A's. And now, it looks like this back panel here was injection molded. And you can see the ejector pins there. Those are the pins that push it out of the mold. And so it looks like it was injection molded. And I would say-- it doesn't have the plastic designation marking on it, but I would guess that it's probably either polypropylene or ABS plastic. are probably made out of-- I initially thought that they were made out of steel. But let's take a look. We've got some magnets here, so we'll take one of our magnets and-- I don't think they are. So they're probably brass contacts, because the magnets are not attracted to them. So it's not a ferrous metal. So then we've got this loop here. And this plastic loop prevents the positive terminal on the battery from slipping below the contacts. So it stays in constant contact and keeps everything together. And you can see there's another one on this side. And that's pretty much it. Oh, and there's also a feature right here so that if you have a screw or a nail on your wall, you can put the tap light in and just hang it like that. But that's pretty much the tap light." + }, + { + "Q": "at 9:08, when you say force, are you talking about the net force?\n", + "A": "yes, as a body is constantly acted upon by a number of forces at any instant, it is rather easier to calculate net fore than the individual forces", + "video_name": "Mz2nDXElcoM", + "timestamps": [ + 548 + ], + "3min_transcript": "There will be no acceleration because this wedge is here. So the wedge is exerting a force that completely counteracts the force, the perpendicular component of gravity. You might guess what it's called. So the wedge is exerting a force, just like that, that's going to be 98 newtons upward. The wedge is going to be exerting a force that is 49 square roots of 3, because this right here is 49 square roots of 3 newtons into. And so this is 49 square roots of 3 newtons out of the surface, out of the surface. And this is the normal force. It is the force perpendicular to the surface that essentially, you could kind of view as the contact force that the, in this case, that the surface is exerting We're not talking about accelerating straight towards the center of the earth. We're talking about accelerating in that direction. We broke up the force into kind of the perpendicular direction and the parallel direction. So you have this counteracting normal force. And that's why you don't have the block plummeting or accelerating into the plane. Now what other forces do we have? Well, we have the force that's parallel to the surface. And if we assume that there's no friction-- and I can assume that there's no friction in this video because we are assuming that it is ice on ice-- what is going to happen? There's no counteracting force to this 49 newtons. 49 newtons parallel downwards, I should say parallel downwards, to the surface of the plane. So what's going to happen? Well, it's going to accelerate in that direction. You have force is equal to mass times acceleration. Or you divide both sides by mass, you get force over mass is equal to acceleration. Over here, our force is 49 newtons in that direction, parallel downwards to the surface of the plane. And so if you divide both by mass, if you divide both of these by mass. So that's the same thing as dividing it by 10 kilograms, dividing by 10 kilograms, that will give you acceleration. That will give you our acceleration. So acceleration is 49 newtons divided by 10 kilograms in that direction, in this direction right over there. And 49 divided by 10 is 4.9, and then newtons divided by kilograms is meters per second squared. So then you get your acceleration. Your acceleration is going to be 4.9 meters per second squared." + }, + { + "Q": "what is happening at 4:50?\n", + "A": "At 4:50 Sal was finishing talking about the supposed formation of the Earth, and started talking about the supposed extinction of the dinosaurs.", + "video_name": "DRtLXagrMHw", + "timestamps": [ + 290 + ], + "3min_transcript": "Actually even on a worldwide basis, it was the first secular democracy based on a kind of a constitutional democracy that showed up on the planet. They said we don't want the king of England anymore. And this was about 234 years ago. And I always remembered because I was born almost on the 200th anniversary. So you just have to add my age to 200. So this is 234 years ago. So these are all events or periods of time that we've heard about and we've talked about. And people throw around these type of years. But what I want to do in this video is relate it to time scales that we can comprehend. So instead of the Big Bang occurring 13.7 billion years ago, let's pretend like it occurred 10 years ago. Because most of us, especially if you're over the age of 10, can kind of understand what 10 years is. It's a very, very long period of time. But something that's well within our lifetimes, well within our experience. Bang occurred 13.7 billion years ago, let's pretend like it occurred 10 years ago. And if we pretend that it occurred 10 years ago, let's think about how many years, or minutes, or hours ago each of these events would have occurred. So if Big Bang, which is really 13.7 billion years, if it really had occurred 10 years ago, and we scaled everything down, if we had scaled everything down, then the Earth would have been created about 3.3 years ago. So this would have been 3.3 years ago. So there's nothing kind of amazing about this. This is a significant fraction of the age of the universe. So not that mind blowing just yet. But if we go all the way to when the dinosaurs were extinct, the last land dinosaurs, now the 65 million years-- and this will give you an appreciation of the difference between million then the dinosaurs would have been extinct 17 days ago. Not even a month ago, the dinosaurs would have been extinct. So if the universe was created when I was just graduated-- well, I'm in my '30s now, so when I was 24-- just last month, the dinosaurs would have gone extinct. And it gets even crazier. 17 days ago, the dinosaurs would have extinct. Australopithecus afarensis would have walked on the Earth 19 hours ago, yesterday. 19 hours ago, she would have been walking around And modern humans wouldn't have shown up until 80 minutes ago, 80 minutes, a little over an hour. There wasn't even a modern human. Then the universe was 10 years, it didn't take until just very recently, the last hour, for us to see someone that looks something like us, looks and thinks something like us." + }, + { + "Q": "\n2:10 How can we know at what time humans had evolved to the point that they were \"modern humans\" who appeared and thought in similar ways to the way we did today? How can we really project an estimate for something like that? Evolutions is really a gradual process, and people say that we are still evolving now.", + "A": "Highly sophisticated, beautiful paintings of animals found in caves in Europe have been dated back 40,000 years. That says to me that modern humans have been around for at least 40,000 years.", + "video_name": "DRtLXagrMHw", + "timestamps": [ + 130 + ], + "3min_transcript": "What I've done here is I've copied and pasted a bunch of pictures that signify events in our history, when you think about history on a grander scale, that most of us have some relation to or we kind of have heard it talked about a little bit. And the whole point of this is to try to understand, or try to begin to understand, how long 13.7 billion years is. So just to start off, I have here-- this is the best depiction I could find where it didn't have copyrights. This is from NASA-- of the Big Bang. And I've talked about it several times. The Big Bang occurred 13.7 billion years ago. And then if we go a little bit forward, actually a lot forward, we get to the formation of our actual solar system and the Earth. This is kind of the protoplanetary disk or a depiction of a protoplanetary disk forming around our young Sun. And so this right here is 4.5 billion years ago. Now this over here-- once again, these aren't pictures of them. These are just depictions because no one was there This is what we think the asteroid that killed the dinosaurs looked like when it was impacting Earth. And it killed the dinosaurs 65 million years ago. So until then, we had land dinosaurs. And then this, as far as the current theories go, got rid of them. Now, we'll fast forward a little bit more. At about 3 million years ago-- let me do this in a color that you can see-- about 3 million, so three million years ago, our ancestors look like this. This is Australopithecus afarensis. This is I think a depiction of-- this is Lucy. I believe the theory is that all of us have some DNA from her. But this was 3 million years ago. And you fast forward some more and you actually that looked and thought like you and me. This is 200,000 years ago. That's right over here. Obviously, this drawing was done much later. But this is a depiction of a modern human, so 200,000 years ago. And then you fast forward even more. And I don't want to keep picking on Jesus. I did that with him getting on the jet liner. And I genuinely don't mean any offense to anyone. I just keep picking Jesus because frankly our calendar is kind of-- he's a good person that most people know about, 2,000 years ago. And so when we associate kind of a lot of modern history occurring after his birth. So this right here is obviously a painting of the birth of Jesus. And this is 2,000 years ago. And then this might be a little bit American-centric. But the Declaration of Independence, it" + }, + { + "Q": "\nAt 6:40,how is velocity 19.6m/s if time is zero ?", + "A": "You can start the stopwatch whenever you want, right?", + "video_name": "T0zpF_j7Mvo", + "timestamps": [ + 400 + ], + "3min_transcript": "so times zero, what is our velocity? well if we use this expression right here time zero or delta t equal zero this expression right here is gonna be zero, and it's just going to be initial velocity, in the last video we gave our initial velocity is going to be 19.6m/s, so it is going to be 19.6m/s I will plot that over here, time zero, it's going to be 19.6 m/s what is our initial displacement at time zero? Our change in time zero, so you look at this up here delta t is zero so this expression here is going to be zero so we haven't done any displacement yet when no time has gone by So we have done no displacement, we are right over there what is now our velocity? well our initial velocity right here is 19.6m/s that was a given, and our acceleration is negative 9.8m/s*s so it's negative right over there and you multiply that by delat t in every situation, so in this situation we are gonna multiply it by 1 delta t is 1, so you have 19.6 minus 9.8 that gives exactly 9.8m/s and the unit we got cause we multiply here is second so its give us meters per second 19.6m/s minus 9.8m/s one of these seconds go away multiply by second give you 9.8m/s so after 1s our velocity is now half of what it was before so we are now going 9.8m/s, let me draw a line here Let me rewrite this displacement formular here with all the infomation So we know that displacement is going to be equal to initial veloctiy which is 19.6m/s now I won't write the units here just for the sake of space times change in time, times our, use the same color to see what is what times our change in time, plus one half let me be clear one half times negative 9.8m/s*s so one half times a is going to be I rewrite this right over here cause this is gonna be negative 9.8m/s*s times one half, so this is going to be negative 4.9 All I did is one half times negative 9.8 over here It is important that is why the vector quantity is start to matter because if you put a positive here you wouldn't have the obeject slowing" + }, + { + "Q": "At 14:06 in the video, Sal says the velocity is decreasing at a constant pace. Isn't it really decreasing gradually, and then increasing gradually in the opposite direction?\n", + "A": "isnt that the same thing? Think about the acceleration. What is its value throughout?", + "video_name": "T0zpF_j7Mvo", + "timestamps": [ + 846 + ], + "3min_transcript": "we are talking about delat t, our change in time is 3s so that's square So times 9 and that gives us 14.7 meters, so 14.7 meters So after 3s we are 14.7m again, the same position with 1s but the difference is now we are moving downard over here we were moving upwards and finally what happen after 4s? Or what's our velocity? Let me just get the calculator out or you might be figure this out in your head Our velocity is going to be 19.6 - 9.8 times 4s just minus 19.6 m/s So we are going a magnitude of velocity which is the same initially threw the ball except now it's going at opposite direction and what is our displacement? get the calculator out So we have out displacement is 19.6 times 4, 4s has gone by minus 4.9 times 4 square which is 16 times, which is equal to zero! the displacement here is zero! We are back on the ground So if you plot the displacement, you will actually got a parabola a downward opening parabola that looks something like this I best draw it relatively neatly So my check to do it better than that Dotted line, dotted line is always easier to adjust than its streamed so if you plot displacement verses time it look something like this it's velocity just downward sloping line, and the acceleration is constant that velocity the whole time is decreasing at constant rate and that make sense because that's the rate which the velocity increases and decreases as the acceleration, and the acceleration base on our convention is downward so that why it's decreasing, we have a negative slope here we have a negative slope of negative 9.8 m/s*s and just to think about what's happening for this ball I know this video is getting long as it goes I'm gonna draw the vectors for velocity So I'm gonna do that in orange, or maybe I will do that in blue So velocity in blue, so right when we start, it has a positive velocity of 19.6m/s, so I will draw a big vector like this 19.6m/s that's velocity, but after 1s is 9.8m/s so half of that so then its maybe would look something like this" + }, + { + "Q": "Around near 2:40,if both the vectors point toward the centre of circle i,e gravity,what will be the force on the Wooden floor then?\n", + "A": "Newton s Third law says that if the Wooden floor is exerting a force on the ball then the floor is exerting the same amount of force on the ball.", + "video_name": "Xpgsg-fY4DY", + "timestamps": [ + 160 + ], + "3min_transcript": "the force between the two surfaces? This is what we'd have to know in order to figure out if our structure is strong enough to contain this bowling ball as it goes around in a circle. And it's also a classic centripetal force problem, so let's do this. What do we do first? We should always draw a force diagram. If we're looking for a force, you draw a force diagram. So what are the forces on this ball? You're gonna have a force of gravity downward, and the magnitude of the force of gravity is always given by M times G, where G represents the magnitude of the acceleration due to gravity. And we're gonna have a normal force as well. Now which way does this normal force point? A common misconception, people wanna say that that normal force points up because in a lot of other situations, the normal force points up. If you're just standing on the ground over here, the normal force on you is upward because it keeps you from falling through the ground, but that's not what this loop structure's doing up here. The loop structure isn't keeping you up. The loop structure's keeping you from flying out of the loop have to point downward. So this is weird for a lot of people to think about, but because the surface is above this ball, the surface pushes down. Surfaces can only push. If the surface is below you, the surface has to push up. If the surface was to the side of you, the surface would have to push right. And if the surface was to the right of you, the surface would have to push left. Normal forces in other words, always push. So the force on the ball from the track is gonna be downward but vice versa. The force on the track from the ball is gonna be upward. So if this ball were going a little too fast and this were made out of wood, you might see this thing splinter because there's too much force pushing on the track this way. But if we're analyzing the ball, the force on the ball from the track is downward. And after you draw a force diagram, the next step is usually, if you wanna find a force, to use Newton's Second Law. And to keep the calculation simple, we typically use Newton's Second Law for a single dimension at at time, i.e. vertical, horizontal, centripetal. because the normal force is pointing toward the center of the circular path and the normal force is the force we wanna find, we're gonna use Newton's Second Law for the centripetal direction and remember centripetal is just a fancy word for pointing toward the center of the circle. So, let's do it. Let's write down that the centripetal acceleration should equal the net centripetal force divided by the mass that's going in the circle. So if we choose this, we know that the centripetal acceleration can always be re-written as the speed squared divided by the radius of the circular path that the object is taking, and this should equal the net centripetal force divided by the mass of the object that's going in the circle and you gotta remember how we deal with signs here because we put a positive sign over here because we have a positive sign for our centripetal acceleration and our centripetal acceleration points toward the center of the circle always. Then in toward the center of the circle is going to be our positive direction," + }, + { + "Q": "\nat 11:44 when Hank is describing the electron configuration chain I got lost what exactly is happening and where is the ATP being produced?", + "A": "ATP is produced in the mitochondria inside the cell.", + "video_name": "CIyAs0bxeoI", + "timestamps": [ + 704 + ], + "3min_transcript": "that are related to B vitamins. Derivatives of Niacin and Riboflavin, which you might have seen in the vitamin aisle. These B vitamins are good at holding onto high energy electrons and keeping that energy until it can get released later in the electron transport chain. In fact, they're so good at it, that they show up in a lot of those high-energy vitamin powders that the kids are taking these days. NAD+s and FADs are like batteries, big awkward batteries that pick up hydrogen and energized electrons from each pyruvate, which in effect charges them up. The addition of hydrogen turns them into NADH and FADH2, respectively. Each pyruvate yields three NADHs and one FADH2 per cycle, and since each glucose has been broken down into two pyruvates that means each glucose molecule can produce six NADHs and two FADH2s. The main purpose of the Krebs Cycle is to make these powerhouses And now comes the time when your saying, \"Sweet pyruvate sandwiches, Hank, \"aren't we supposed to be making ATP here? \"Let's make it happen, Capt'n! What's the holdup?\" Well friends, your patience is finally paying off because when it comes to ATPs, the electron transport chain is the real moneymaker. In a very efficient cell, it can net a whopping 34 ATPs. So, remember all those NADHs and FADH2s we made in the Krebs Cycle? Well, their electrons are going to provide the energy that will work as a pump along a chain of channel proteins across the inner membrane of the mitochondria where the Krebs Cycle occurred. These proteins will swap these electrons to send hydrogen protons from inside the very center of the mitochondria, across its inner membrane to the outer compartment of the mitochondria. But once they're out, the protons want to get back to the other side of the inner membrane, because there's a lot of other protons out there and as we've learned, nature always tends to seek a nice, peaceful balance So all of these anxious protons are allowed back in through a special protein called ATP synthase. And the energy of this proton flow drives this crazy spinning mechanism that squeezes some ADP and some phosphates together to form ATP. So, the electrons from the 10 NADHs that come out of the Krebs Cycle, have just enough energy to produce roughly three ATPs each. And we can't forget our friends the FADH2s. We have two of them and they make two ATPs each. And voila! That is how animal cells, the world over, make ATP through cellular respiration. Now just to check, let's reset our ATP counter and do the math for a single glucose molecule once again. We made two ATPs for each pyruvate during glycolysis. We made two during the Krebs Cycle, and then during the electron transport chain we made about 34. And that is just for one molecule of glucose. Imagine how much your body makes and uses every single day." + }, + { + "Q": "\nat 3:57 Hank is saying that usually 29-30 molecules of ATP produced for 1 glucose, but if it was a best scenario it would be 38 ATPs, so where do the rest ATPs are going, why there is not always 38?", + "A": "People are still studying this topic.", + "video_name": "CIyAs0bxeoI", + "timestamps": [ + 237 + ], + "3min_transcript": "Same goes with energy, in order to be able to use it, our cells need energy to be transferred into adenosine triphosphate to be able to grow, move, create electrical impulses in our nerves and brains, everything. A while back, for instance, we talked about how cells use ATP to transport some kinds of materials in and out of its membranes. To jog your memory about that, you can watch that episode right here. Now before we see how ATP is actually put together, let's look at how cells can cash in on the energy that's stashed in there. Well, adenosine triphosphate is made up of a nitrogenous base called adenine with a sugar called ribose and three phosphate groups attached to it. Now one thing you need to know about these three phosphate groups, is that they are super uncomfortable sitting together in a row like that, like three kids on a bus who hate each other all sharing the same seat. So because the phosphate groups are such terrible company for each other, ATP is able to do this nifty trick where it shoots one of the phosphate groups off the end of the seat, creating ADP, or adenosine diphosphate, because now, And this reaction when the third jerk kid is kicked off the seat, energy is released. And since there are a lot of water molecules just floating around nearby, an OH pairing, that's called a hydroxide, from one of the H2Os comes over and takes the place of that third phosphate group, and everybody is much happier. By the way, when you use water to break down a compound like this, it's called hydrolysis, \"hydro\" from water and \"lysis\" from the Greek word \"for separate\". So now that you know how ATP is spent, let's see how it is minted, nice and new, by cellular respiration. Like I said, it all starts with oxygen and glucose. In fact, textbooks make a point of saying that through cellular respiration, one molecule of glucose can yield a bit of heat and 38 molecules of ATP. Now, it's worth noting that this number is kind of a best-case scenario. Usually it's more like 29 or 30 ATPs, but whatever. People are still studying this stuff, so let's stick with that number, 38. Now, cellular respiration isn't something that just happens all at once. Glucose is transformed into ATPs over three separate stages. the Krebs Cycle, and the electron transport chain. Traditionally, these stages are described as coming one after the other but really everything in the cell, is kind of happening all at the same time. But let's start with the first step, glycolysis, or the breaking down of the glucose. Glucose, of course, is a sugar, you know this because it's got an \"ose\" at the end of it. And glycolysis is just the breaking up of glucose's six-carbon ring into two three-carbon molecules called pyruvic acids or pyruvate molecules. Now in order to explain how exactly glycolysis works, I'd need about an hour of your time, and a giant cast of finger puppets each playing a different enzyme, and though it would pain me to do it, I would have to use words like phosphoglucoisomerase, but a simple way of explaining it, is like this, if you wanna make some money, you gotta spend some money. Glycolysis needs the investment of two ATPs in order to work and in the end, it generates four ATPs, for a net profit if you will of two ATPs." + }, + { + "Q": "\nHow do you know when to do a hydride shift? Couldn't you of just moved the Cl where the positive charge is in the second hexane? 6:28", + "A": "You want the most stable carbocation - meaning you want that positive charge to be on a carbon connected to as many other carbons as possible, or to be resonance stabilized. That hydride shift happens to move the positive charge onto the most stable carbocation (from the secondary to the tertiary carbon).... Tertiary > Secondary > Primary being the order of preference there.", + "video_name": "iEKA0jUstPs", + "timestamps": [ + 388 + ], + "3min_transcript": "So let's go ahead and write that. See if we can spell Markovnikov. The halogen adds the more substituted carbon. And the reason it does that is because the more substituted carbon is the one that was the more stable carbocation in the mechanism. So let's do another mechanism here. Whenever you have a carbocation present, you could have rearrangement. So let's do one where there's some rearrangement. So let's start out with this as our alkene and react that with hydrochloric acid once again. First steps-- pi electrons function as a base. These electrons kick off onto your chlorine. So which side do we add the proton to? Right? We could add the proton to the left side of the double bond. We could add the proton to the right side of the double bond. the most stable carbocation that we can. So it makes sense to add the proton to the right side of the double bond right here because that's going to give us this as a carbocation. What kind of carbocation is that? So let's identify this carbon as the one that has our positive charge. That carbon is bonded to two other carbons. So it is a secondary carbocation. If we had added on the proton to the left side of the double bond, we would have a primary carbocation here. So a secondary carbocation is more stable. Can we form a tertiary carbocation? Because we know tertiary carbocations are even more stable than secondary carbocations. And of course, we can. There's a hydrogen attached to this carbon. And we saw-- in our earlier video on carbocations and rearrangements-- we could get a hydride shift here. All right. So the proton and these two electrons here are hydride anion. to move over here, shift over one carbon, and form a new covalent bond. So what would we get if we get a hydride shift in our mechanism? Well, now our hydride has shifted over here to that carbon. This carbon no longer has a positive charge on it. We took a bond away from this carbon. So now, this is where our positive charge is. So we have a carbocation. How would we classify this carbocation? Well, one, two, three other carbons. So it's tertiary. It's more stable than our secondary carbocation. So in the final step of our mechanism, we had our chloride anion over here from the first step of our mechanism. So a chloride anion, negatively charged nucleophile file. So a nucleophilic attack on our carbocation. So right there. And we're going to form a bond between that halogen and that carbon. So our final product is going to end up" + }, + { + "Q": "At 12:00 carbon at 1 is categorized as tertiary because it is connected to 3 other carbons (implies connection to ring (connection beyond zigzag) is not taken into account. However, later at 12:42 carbon 1 is said to be connected to 3 other carbons and hence tertiary while according to earlier logic the connection to the ring should not be counted and so carbon #1 is not tertiary it is secondary. What am I missing?\n", + "A": "That carbon 1 at 12:00 is quaternary, but he seems to be talking about where the prefixes might come from not the actual classification of those carbons.", + "video_name": "joQd0qVnX4M", + "timestamps": [ + 720, + 762 + ], + "3min_transcript": "And I'm going to once again take off one of the hydrogens. So I'll make it an R double prime group. And now if I wanted to classify my central carbon, now this is connected to 1, 2, 3 other carbons, so it is said to be tertiary. So that is a tertiary carbon like that. And finally, I have one more example, of course. I take off the last hydrogen. So now I have R, R prime, R double prime, and R triple prime. So what is the classification of this carbon now connected to four other carbons? So it is said to be quaternary. So that is a quaternary carbon right here. All right, so if I'm trying to think about where some of these common names come from I can see, oh, well, right here I have-- well, that would be secondary. So S-E-C for my prefix. So let's go back up here, and let's see if we can find those examples. So here I have this carbon bonded to two other carbons. So this carbon was said to be secondary, so I think that's where this comes from. I've never seen that explained in a textbook or anywhere, but it just makes sense. So it's ignoring the fact that this carbon is actually attached to a ring. It's saying this carbon on my complex substituent is bonded to two other carbons, so it is secondary on that complex substituent. What about tertiary? So carbon bonded to three other carbons is said to be tertiary. So if I go back up here again I can say, well, that would make sense, because if I look at this carbon, it's bonded to three other carbons, right? And once again I'm ignoring the fact that this carbon is actually bonded to another carbon on the ring. So if you just look at the complex substituent, that carbon is said to be tertiary, which I think is where the name comes from. Let's do one more example of assigning classification of carbons to this molecule. So let's look at this carbon right here. This carbon is bonded to one other carbon and three hydrogens. So this carbon is said to be primary. This carbon right here is bonded to two other carbons, so it is said to be a secondary. This carbon right here is bonded to three other carbons so it is tertiary. This carbon is bonded to one other carbon so it is primary. This carbon is bonded to three other carbons, so it is tertiary. And all of the carbons on the ring right here are bonded to two other carbons, so they are all said to be secondary. So that's a very important skill to develop classifying your carbons." + }, + { + "Q": "\nAt 1:05, why is the longest carbon chain consisting of only 2 carbons? Why cant it be 3 carbons?", + "A": "You number the continuously-attached carbons of the side-chain from the carbon that is attached to the main chain. If you start at C-1, you can go out only one carbon with your pencil. To get to the third carbon without removing your pencil, you would have to re-trace your steps back to C-1, and that is not permitted.", + "video_name": "joQd0qVnX4M", + "timestamps": [ + 65 + ], + "3min_transcript": "So how do we name this molecule? Well, we start with the longest carbon chain. So there are seven carbons in my longest carbon chain. So I would call this heptane. And I number it to give the substituent the lowest number possible. So in this example, it doesn't really matter if I start from the left or from the right. In both examples, you would end up with a 4 for your substituent there. Now, this substituent looks different from ones we've seen before. There are three carbons in it, but those carbons are not in a straight-chain alkyl group. So if I look at it, right there are three carbons, but they're not going in a straight chain. They're branching of branching here. So this is kind of weird. How do we name this substituent? Well, down here, I have the same substituent, and I'm going to draw this little zigzag line to indicate that that substituent is coming off of some straight-chain alkane. And when you're naming a complex substituent like this, you actually use the same rules that you would use for a straight-chain alkane. which in this case is only two carbons. So that would be an ethyl group coming off of my carbon chain. So I'm going to go ahead and name that as an ethyl group. I'm going to go ahead number it to give my branching group there the lowest number possible. So I go 1 and 2. So what is my substituent coming off of my ethyl group? Well, that's a methyl group coming off of carbon 1. So I name it as 1-methylethyl. OK, so now, that complex substituent is named as 1-methylethyl. So I could go ahead and put that into my name. So coming off of carbon 4, I have 1-methylethyl. And I'm going to put that in parentheses. And all of that is coming off of carbon 4 for my molecule. way of naming that molecule. So if your naming your complex substituent as 1-methylethyl, that's the official IUPAC way, but there are also common names for these complex substituents. So the common name for 1-methylethyl is isopropyl. So isopropyl is the common name. And isopropyl is used so frequently that it's perfectly acceptable to use isopropyl for the name of this molecule as well. So you could have said, oh, this is 4-isopropylheptane, and you would have been absolutely correct. So that's yet another IUPAC name. So iso means same, and it probably comes from the fact that you have these two methyl groups giving you this Y shape that are the same. So that's one complex substituent, one that has three carbons on it. Let's look at a bunch of complex substituents" + }, + { + "Q": "At 12:48, isn't that carbon bonded to 4 carbons\n", + "A": "no it bonded to 3 carbons and the extra one available bond is attached to hydrogen , and this is why its called Primary", + "video_name": "joQd0qVnX4M", + "timestamps": [ + 768 + ], + "3min_transcript": "well, that would be secondary. So S-E-C for my prefix. So let's go back up here, and let's see if we can find those examples. So here I have this carbon bonded to two other carbons. So this carbon was said to be secondary, so I think that's where this comes from. I've never seen that explained in a textbook or anywhere, but it just makes sense. So it's ignoring the fact that this carbon is actually attached to a ring. It's saying this carbon on my complex substituent is bonded to two other carbons, so it is secondary on that complex substituent. What about tertiary? So carbon bonded to three other carbons is said to be tertiary. So if I go back up here again I can say, well, that would make sense, because if I look at this carbon, it's bonded to three other carbons, right? And once again I'm ignoring the fact that this carbon is actually bonded to another carbon on the ring. So if you just look at the complex substituent, that carbon is said to be tertiary, which I think is where the name comes from. Let's do one more example of assigning classification of carbons to this molecule. So let's look at this carbon right here. This carbon is bonded to one other carbon and three hydrogens. So this carbon is said to be primary. This carbon right here is bonded to two other carbons, so it is said to be a secondary. This carbon right here is bonded to three other carbons so it is tertiary. This carbon is bonded to one other carbon so it is primary. This carbon is bonded to three other carbons, so it is tertiary. And all of the carbons on the ring right here are bonded to two other carbons, so they are all said to be secondary. So that's a very important skill to develop classifying your carbons. with different functional groups." + }, + { + "Q": "\nAt about 8:45 in the video i just wanted to verify that FADH is being created and not FADH2. Is the 2 supposed to be out in front of the FADH?", + "A": "It s easy to get these confused, because two things are created: NADH and FADH2. The 2 should appear as a subscript (as in, there are 2 hydrogens). In addition, plants don t use NAD+/NADH, they use NADP+/NADPH. Just remember, at the end of the day, these molecules are very similar in their function, as they all contribute to the electron transport chain (next video).", + "video_name": "juM2ROSLWfw", + "timestamps": [ + 525 + ], + "3min_transcript": "six carbons. When you do this whole process once, you are generating three molecules of carbon dioxide. But you're going to do it twice. You're going to have six carbon dioxides produced. Which accounts for all of the carbons. You get rid of three carbons for every turn of this. Well, two for every turn. But really, for the steps after glycolysis you get rid of three carbons. But you're going to do it for each of the pyruvates. You're going to get rid of all six carbons, which will have to exhale eventually. But this cycle, it doesn't just generate carbons. The whole idea is to generate NADHs and FADH2s and ATPs. So we'll write that here. And this is a huge simplification. I'll show you the detailed picture in a second. We'll reduce some NAD plus into NADH. We'll do it again. And of course, these are in separate steps. There's intermediate compounds. I'll show you those in a second. It will produce some ATP. Some ADP will turn into ATP. Maybe we have some-- and not maybe, this is what happens-- some FAD gets-- let me write it this way-- some FAD gets oxidized into FADH2. And the whole reason why we even pay attention to these, you might think, hey cellular respiration is all about ATP. Why do we even pay attention to these NADHs and these FADH2s that get produced as part of the process? The reason why we care is that these are the inputs into the electron transport chain. These get oxidized, or they lose their hydrogens in the electron transport chain, and that's where the bulk of the ATP is actually produced. And then maybe we'll have another NAD get reduced, or gain in hydrogen. Or gaining a hydrogen whose electron you can hog. NADH. And then we end up back at oxaloacetic acid. And we can perform the whole citric acid cycle over again. So now that we've written it all out, let's account for what we have. So depending on-- let me draw some dividing lines so we know what's what. So this right here, everything to the left of that line right there is glycolysis. We learned that already. And then most-- especially introductory-- textbooks will give the Krebs cycle credit for this pyruvate oxidation, but that's really a preparatory stage. The Krebs cycle is really formally this part where you start with acetyl-CoA, you merge it with oxaloacetic acid. And then you go and you form citric acid, which essentially gets oxidized and produces all of these things that will need to either directly produce ATP or will do it indirectly in the electron transport chain. But let's account for everything that we have. Let's" + }, + { + "Q": "At 8:32 , Sal says that FAD is oxidised to FADH2 when before he mentioned that the NAD+ molecules are reduced to NADH2. This seems contradictory to me since they both involve the gain of hydrogen, therefore wouldn't that mean that both the NAD+ and FAD are reduced? If so then how do my notes say that the CAC includes oxidations which generate high-energy electrons that will be used to power the synthesis of ATP ?\n", + "A": "Oh I think I get it now,,, they are produced by reduction making them high electron carriers and they are then oxidized in the ETC for synthesis of ATP. Thanks :)", + "video_name": "juM2ROSLWfw", + "timestamps": [ + 512 + ], + "3min_transcript": "six carbons. When you do this whole process once, you are generating three molecules of carbon dioxide. But you're going to do it twice. You're going to have six carbon dioxides produced. Which accounts for all of the carbons. You get rid of three carbons for every turn of this. Well, two for every turn. But really, for the steps after glycolysis you get rid of three carbons. But you're going to do it for each of the pyruvates. You're going to get rid of all six carbons, which will have to exhale eventually. But this cycle, it doesn't just generate carbons. The whole idea is to generate NADHs and FADH2s and ATPs. So we'll write that here. And this is a huge simplification. I'll show you the detailed picture in a second. We'll reduce some NAD plus into NADH. We'll do it again. And of course, these are in separate steps. There's intermediate compounds. I'll show you those in a second. It will produce some ATP. Some ADP will turn into ATP. Maybe we have some-- and not maybe, this is what happens-- some FAD gets-- let me write it this way-- some FAD gets oxidized into FADH2. And the whole reason why we even pay attention to these, you might think, hey cellular respiration is all about ATP. Why do we even pay attention to these NADHs and these FADH2s that get produced as part of the process? The reason why we care is that these are the inputs into the electron transport chain. These get oxidized, or they lose their hydrogens in the electron transport chain, and that's where the bulk of the ATP is actually produced. And then maybe we'll have another NAD get reduced, or gain in hydrogen. Or gaining a hydrogen whose electron you can hog. NADH. And then we end up back at oxaloacetic acid. And we can perform the whole citric acid cycle over again. So now that we've written it all out, let's account for what we have. So depending on-- let me draw some dividing lines so we know what's what. So this right here, everything to the left of that line right there is glycolysis. We learned that already. And then most-- especially introductory-- textbooks will give the Krebs cycle credit for this pyruvate oxidation, but that's really a preparatory stage. The Krebs cycle is really formally this part where you start with acetyl-CoA, you merge it with oxaloacetic acid. And then you go and you form citric acid, which essentially gets oxidized and produces all of these things that will need to either directly produce ATP or will do it indirectly in the electron transport chain. But let's account for everything that we have. Let's" + }, + { + "Q": "\nat 08:37 sal says FADis oxideised......but FAD is gaining proton...so as per the rule(OIL RIG) reduction is gaining proton so FAD will be reduced not oxidised.", + "A": "Oxidation and reduction involve the lose or gain or electrons, not protons. What Sal is saying is that in the Krebs cycle, FAD is first reduced and gains electrons becoming FADH2, which it later looses (is oxidized) for the electron transport chain.", + "video_name": "juM2ROSLWfw", + "timestamps": [ + 517 + ], + "3min_transcript": "six carbons. When you do this whole process once, you are generating three molecules of carbon dioxide. But you're going to do it twice. You're going to have six carbon dioxides produced. Which accounts for all of the carbons. You get rid of three carbons for every turn of this. Well, two for every turn. But really, for the steps after glycolysis you get rid of three carbons. But you're going to do it for each of the pyruvates. You're going to get rid of all six carbons, which will have to exhale eventually. But this cycle, it doesn't just generate carbons. The whole idea is to generate NADHs and FADH2s and ATPs. So we'll write that here. And this is a huge simplification. I'll show you the detailed picture in a second. We'll reduce some NAD plus into NADH. We'll do it again. And of course, these are in separate steps. There's intermediate compounds. I'll show you those in a second. It will produce some ATP. Some ADP will turn into ATP. Maybe we have some-- and not maybe, this is what happens-- some FAD gets-- let me write it this way-- some FAD gets oxidized into FADH2. And the whole reason why we even pay attention to these, you might think, hey cellular respiration is all about ATP. Why do we even pay attention to these NADHs and these FADH2s that get produced as part of the process? The reason why we care is that these are the inputs into the electron transport chain. These get oxidized, or they lose their hydrogens in the electron transport chain, and that's where the bulk of the ATP is actually produced. And then maybe we'll have another NAD get reduced, or gain in hydrogen. Or gaining a hydrogen whose electron you can hog. NADH. And then we end up back at oxaloacetic acid. And we can perform the whole citric acid cycle over again. So now that we've written it all out, let's account for what we have. So depending on-- let me draw some dividing lines so we know what's what. So this right here, everything to the left of that line right there is glycolysis. We learned that already. And then most-- especially introductory-- textbooks will give the Krebs cycle credit for this pyruvate oxidation, but that's really a preparatory stage. The Krebs cycle is really formally this part where you start with acetyl-CoA, you merge it with oxaloacetic acid. And then you go and you form citric acid, which essentially gets oxidized and produces all of these things that will need to either directly produce ATP or will do it indirectly in the electron transport chain. But let's account for everything that we have. Let's" + }, + { + "Q": "\nAt 5:00 you say the force to move the particle towards the charge needs to be equal and opposite to the force being pushed out...but shouldnt it be greater? Shouldnt the force needed to move the particle towards the charge be greater since equal and opposite forces would just cancel each other out?", + "A": "Equal and opposite force pairs do not cancel out because they don t act on the same object. The moon is attracted to earth with the same force that the earth attracts the moon. If you draw those forces, you will see that only one of them is acting on the moon, and one of them is acting on the earth, so there s no canceling going on. This is a common hurdle people face in understanding Newton s 3rd law.", + "video_name": "CqsYCIjSm9A", + "timestamps": [ + 300 + ], + "3min_transcript": "Because the field is pushing it outward. It takes work to push it inward. So let's say we want to push it in. Let's say it's at 10 meters. Let's say that this distance right here-- let me draw a radial line-- let's say that this distance right here is 10 meters, and I want to push this particle in 5 meters, so it eventually gets right here. This is where I'm eventually going to get it so then it's going to be 5 meters away. So how much work does it take to move it 5 meters towards this charge? Well, the way you think about it is the field keeps changing, right? But we can assume over a very, very, very, very infinitely small distance, and let's call that infinitely small distance dr, change in radius, and as you can see, we're about to If you don't understand what any of this is, you might want to review or learn the calculus in the calculus playlist, but how much work does it require to move this particle a very, very small distance? Well, let's just assume over this very, very, very small distance, that the electric field is roughly constant, and so we can say that the very, very small amount of work to move over that very, very small distance is equal to Coulomb's constant q1 q2 over r squared times dr. Now before we move on, let's think about something for a second. Coulomb's Law tells us that this is the outward force that this charge is exerting on this particle or that the field is exerting on this particle. The force that we have to apply to move the particle from here to here has to be an inward force. to be a negative. And why is that? Because we have to completely offset the force of the field. Maybe if the particle was already moving a little bit, then our force will keep it from decelerating from the field, and if it wasn't already moving, we would have to nudge it just an infinitely small amount just to get it moving, and then our force would completely offset the force of the field, and the particle would neither accelerate nor decelerate. So this is the amount of work, and I just want to explain that we want to put that negative sign there because we going in the opposite direction of the field. So how do we figure out the total amount of work? We figured out the amount of work to get it from here to here, and I even drew it much bigger than it would be. These dr's, this is an infinitely small change in radius. If we want to figure out the total work, then we keep adding them up. We say, OK, what's the work to go from here to here, then the work to go from there to there, then the work to go from there to there, all the way until we get to 5 meters" + }, + { + "Q": "I have a question regarding the integral setup at 6:44. Mr. Khan said that the lower limit should be x=10 and upper limit x=5. However, would we still need the negative sign (attached to the force equation) since the integral takes care of direction? Or does the sign stuff actually matter (I'm not sure if work is considered a vector or scalar quantity)? Thank you and great video!\n", + "A": "Yes,we would still require the negative sign as it specifies that we are moving in the opposite direction of the force..and integral has negative sign because we are going from 10 to 5.", + "video_name": "CqsYCIjSm9A", + "timestamps": [ + 404 + ], + "3min_transcript": "to be a negative. And why is that? Because we have to completely offset the force of the field. Maybe if the particle was already moving a little bit, then our force will keep it from decelerating from the field, and if it wasn't already moving, we would have to nudge it just an infinitely small amount just to get it moving, and then our force would completely offset the force of the field, and the particle would neither accelerate nor decelerate. So this is the amount of work, and I just want to explain that we want to put that negative sign there because we going in the opposite direction of the field. So how do we figure out the total amount of work? We figured out the amount of work to get it from here to here, and I even drew it much bigger than it would be. These dr's, this is an infinitely small change in radius. If we want to figure out the total work, then we keep adding them up. We say, OK, what's the work to go from here to here, then the work to go from there to there, then the work to go from there to there, all the way until we get to 5 meters And what we do when we take the sum of these, we assume that it's an infinite sum of infinitely small increments. And as you learned, that is nothing but the integral, and so that is the total work is equal to the integral. That's going to be a definite integral because we're starting at this point. We're summing from-- our radius is equal to 10 meters-- that's our starting point-- to radius equals 5 meters. That might be a little unintuitive that we're starting at the higher value and ending at the lower value, We're pushing it inwards. And then we're taking the integral of minus k q1 q2 over r squared dr. All of these are constant terms up here, right? So we could take them out. So this is the same thing-- I don't want to run out of over r squared-- or to the negative 2-- dr. And that equals minus k-- I'm running out of space-- q1 q2. We take the antiderivative. We don't have to worry about plus here because it's a definite integral. r to the negative 2, what's the antiderivative? It's minus r to the negative 1. Well, that minus r, the minus on the minus r will just cancel with this. That becomes a plus r to the negative 1, And you evaluate it at 5 and then subtract it and evaluate it at 10. And then-- let me just go up here. Actually, let me erase some of this. Let me erase this up here." + }, + { + "Q": "At around 7:17 wasn't Sal meant to write 1/-k*q1*q2* the integral? This doubt arised due to Sal getting rid of the -k*q1*q2 by diving the integral by the term.\n", + "A": "Here -k*q1*q2 is just a constant. Because of the linearity of the integration the integral of c*f is equal to c times the integral of f, where f is some function and c is a constant.", + "video_name": "CqsYCIjSm9A", + "timestamps": [ + 437 + ], + "3min_transcript": "to be a negative. And why is that? Because we have to completely offset the force of the field. Maybe if the particle was already moving a little bit, then our force will keep it from decelerating from the field, and if it wasn't already moving, we would have to nudge it just an infinitely small amount just to get it moving, and then our force would completely offset the force of the field, and the particle would neither accelerate nor decelerate. So this is the amount of work, and I just want to explain that we want to put that negative sign there because we going in the opposite direction of the field. So how do we figure out the total amount of work? We figured out the amount of work to get it from here to here, and I even drew it much bigger than it would be. These dr's, this is an infinitely small change in radius. If we want to figure out the total work, then we keep adding them up. We say, OK, what's the work to go from here to here, then the work to go from there to there, then the work to go from there to there, all the way until we get to 5 meters And what we do when we take the sum of these, we assume that it's an infinite sum of infinitely small increments. And as you learned, that is nothing but the integral, and so that is the total work is equal to the integral. That's going to be a definite integral because we're starting at this point. We're summing from-- our radius is equal to 10 meters-- that's our starting point-- to radius equals 5 meters. That might be a little unintuitive that we're starting at the higher value and ending at the lower value, We're pushing it inwards. And then we're taking the integral of minus k q1 q2 over r squared dr. All of these are constant terms up here, right? So we could take them out. So this is the same thing-- I don't want to run out of over r squared-- or to the negative 2-- dr. And that equals minus k-- I'm running out of space-- q1 q2. We take the antiderivative. We don't have to worry about plus here because it's a definite integral. r to the negative 2, what's the antiderivative? It's minus r to the negative 1. Well, that minus r, the minus on the minus r will just cancel with this. That becomes a plus r to the negative 1, And you evaluate it at 5 and then subtract it and evaluate it at 10. And then-- let me just go up here. Actually, let me erase some of this. Let me erase this up here." + }, + { + "Q": "At the minute 2:34 I cannot understand why the middle C of the propyl group is a secondary carbon: ok, it's a secondary carbon if I don't consider its bond with the main chain (DECANE)... So I don't have to consider the bond with the main chain? Always?\n", + "A": "Is it a secondary carbon because it is the second-to-last carbon in the group?", + "video_name": "O9RPGJcAfJk", + "timestamps": [ + 154 + ], + "3min_transcript": "So we have this group right over here. This has two carbons in it, one carbon, two carbons. And so, because it has two carbons, we would use the prefix eth. Remember, meth is one carbon, eth is two carbons. And since it's a group and we're not talking about the backbone, this is an ethyl group. And we have another ethyl group right over here. Two carbons attached right over here. This is also an ethyl group. And now, this group right over here is interesting. We can count the carbons in it. So it has 1, 2, 3 carbons. So you could think about, well, this has three carbons. Our prefix for three carbons is prop. So you could say, hey, maybe this is a propyl group. This right over here, you could say maybe this is propyl group. And you wouldn't be completely off base by saying that. But we have to be a little bit more careful when we name it. assume that you're attaching to one end of the propyl group. But we're not attaching to one end of the propyl group. We're attaching essentially to the second carbon, to the middle carbon. And this is a secondary carbon. The reason why it's called a secondary carbon is because it's attached to two other carbons. If it was attached to three other carbons, it would be a tertiary carbon. If it was attached only one carbon, it would be a primary carbon. Since we're attached to the secondary carbon right over here, this is sometimes called a sec-propyl group. And it's also sometimes called isopropyl, an isopropyl group. And you'll actually see isopropyl a little bit more frequently. And these would both be referred to as common names for this group. Now, if you wanted to name this systematically, then you would do it very similar to the way that you You would look for the longest chain here. And the longest chain in this molecule, starting with where you are attached, is a chain of two carbons. And so the backbone right over here is ethyl. Let me write this right over here. It's an ethyl backbone here. And then you could view this carbon as a group attached to that ethyl backbone. And we would start counting right where we are attached to the main chain. So this is the one carbon, this is the two carbon. So this right over here, this is just one carbon group. This right over here is a methyl group. So you have a methyl group attached to the one carbon of an ethyl group. So the systematic name for this, and this is a little bit less typical for a group as small as a propyl group, but you could call this 1-methylethyl." + }, + { + "Q": "\nAt 4:15, Mr. Sal Khan mentioned the systematic name 1-methylethyl. Would it be wrong to name this compound in alphabetical order( i.e.. 1-ethylmethyl)?", + "A": "Yes, it would be wrong. The longest chain in the group is 2 carbons long, so the base name is ethyl. There is a methyl group on C1, so the name of the group is 1-methylethyl.", + "video_name": "O9RPGJcAfJk", + "timestamps": [ + 255 + ], + "3min_transcript": "assume that you're attaching to one end of the propyl group. But we're not attaching to one end of the propyl group. We're attaching essentially to the second carbon, to the middle carbon. And this is a secondary carbon. The reason why it's called a secondary carbon is because it's attached to two other carbons. If it was attached to three other carbons, it would be a tertiary carbon. If it was attached only one carbon, it would be a primary carbon. Since we're attached to the secondary carbon right over here, this is sometimes called a sec-propyl group. And it's also sometimes called isopropyl, an isopropyl group. And you'll actually see isopropyl a little bit more frequently. And these would both be referred to as common names for this group. Now, if you wanted to name this systematically, then you would do it very similar to the way that you You would look for the longest chain here. And the longest chain in this molecule, starting with where you are attached, is a chain of two carbons. And so the backbone right over here is ethyl. Let me write this right over here. It's an ethyl backbone here. And then you could view this carbon as a group attached to that ethyl backbone. And we would start counting right where we are attached to the main chain. So this is the one carbon, this is the two carbon. So this right over here, this is just one carbon group. This right over here is a methyl group. So you have a methyl group attached to the one carbon of an ethyl group. So the systematic name for this, and this is a little bit less typical for a group as small as a propyl group, but you could call this 1-methylethyl. Now, the systematic name, you might say, hey, why go through the pain of doing this for something so simple that we could just call isopropyl? This is useful if this was a much larger or a much more complex group that was attached to this main chain. But more typically, and this is why it's called the common name, you'll see this thing right over here just called isopropyl. And sometimes you would see it called sec-propyl even s-propyl. Now that we've named all of the groups, let's think about what carbons they are attached to and where we can start counting from. And the way that this is done is that you would start counting from the end of your carbon chain, this decane backbone, and you'd count from the end that bumps into the most groups faster. So, for example, if you count from this end, this would be the 1 carbon, 2 carbon, 3 carbon, 4 carbon," + }, + { + "Q": "In the \"work example problems\" video, they say that in lifting an object at constant velocity, the net work is 0, since the change in kinetic energy is 0 (we give energy into the system to push it up, and the gravity does negative work...in total, no work is done). They don't even consider PEgravitational.\n\nBut here, they say work is the same as change in energy, so lifting an object changes its gravitational potential energy and thus wok is done (2:35).\n\nWas the first video wrong?\nThanks\n", + "A": "The NET work includes the work you do to lift AND the work done by gravity, which is opposite of the work you do. So the KE doesn t change. But you did work against gravity so you added PE to the system.", + "video_name": "sZG-zHkGR4U", + "timestamps": [ + 155 + ], + "3min_transcript": "One way to find the amount of work done is by using the formula Fd cosine theta. But this number for the amount of work done represents the amount of energy transferred to an object. For instance, if you solve for the work done and you get positive 200 joules, it means that the force gave something 200 joules of energy. So if you have a way of determining the amount of energy that something gains or loses, then you have an alternate way of finding the work done, since the work done on an object is the amount of energy it gains or loses. For instance, imagine a 50-kilogram skateboarder that starts at rest. If a force starts the skateboarder moving at 10 meters per second, that force did work on the skateboarder since it gave the skateboarder energy. The amount of kinetic energy gained by the skateboarder is 2,500 joules. That means that the work done by the force on the skateboarder was positive 2,500 joules. It's positive because the force on the skateboarder If a force gives energy to an object, then the force is doing positive work on that object. And if a force takes away energy from an object, the force is doing negative work on that object. Now imagine that the skateboarder, who's moving with 10 meters per second, gets stopped because he crashes into a stack of bricks. The stack of bricks does negative work on the skateboarder because it takes away energy from the skateboarder. To find the work done by the stack of bricks, we just need to figure out how much energy it took away Since the skateboarder started with 2,500 joules of kinetic energy and ends with zero joules of kinetic energy, it means that the work done by the bricks on the skateboarder was negative 2,500 joules. It's negative because the bricks took away energy from the skateboarder. Let's say we instead lift the bricks, which are 500 kilograms, upwards a distance of four meters. To find the work that we've done on the bricks, we could use Fd cosine theta. We could just figure out the amount of energy that we've given to the bricks. The bricks gain energy here. And they're gaining gravitational potential energy, which is given by the formula mgh. If we solve, we get that the bricks gained 19,600 joules of gravitational potential energy. That means that the work we did on the bricks was positive 19,600 joules. It's positive because our force gave the bricks energy. This idea doesn't just work with gravitational potential energy and kinetic energy. It works for every kind of energy. You can always find the work done by a force on an object if you could determine the energy that that force gives or takes away from that object. [MUSIC PLAYING]" + }, + { + "Q": "where do you get the yellow thing that you used at 5:00?\n", + "A": "those are alligator clips, they are wires with little alligator like clips on them.", + "video_name": "Kq0Er6JBMmc", + "timestamps": [ + 300 + ], + "3min_transcript": "for us to run on and allow us to run our high-current, high-voltage motors and control them with our low-power, low-current, low-voltage Arduino. So that's why we need the motor controller, and we go into more detail on that in the motor controller video. So let's get started taking our hair dryer motor apart, or taking it out of the hair dryer, I should say. And since that's going to be what we're using to move our craft around, we want to start to experiment with it and see how much power it's going to require and how much torque it's going to have and things like that. All right. so we're going to experiment with our motor and see what it's going to take to power it. We've got our alligator clips connected. And we're going to use a 1.5-volt AA battery to see if we can make the motor turn and to kind of get a sense for how much air it will pull through it at 1.5 volts. So we're running it, and we're holding the plastic up, and you can see it's not moving the plastic at all. So that's not going to work. We're going to need more volts than that. And we can increase the voltage by combining the cells in a battery holder, and that allows us to wire the cells in series so we go from 1.5 volts to over 12 volts because those cells are new. So let's see what impact that has on the motor. OK, so we're connecting our battery to our hair dryer motor. And whoa, you can see it's moving much more quickly now. Now it'll push the plastic completely out of the way. And we're getting a fair amount of air coming out of it. But I don't think we're going to use this method for moving our craft because, even though it's blowing a fair amount of air, it only works really efficiently in one direction. In the other direction, it doesn't work as well because it's only meant to blow air in one direction, out of the hair dryer. so we know exactly what we can run it on and how much voltage it needs. To do that, we're going to need to remove the propeller and the outer housing around the motor. So we're just going to trim that off with our hack saw there. And we're time lapsing this so you don't have to sit through all of it. But in any case, we're going to trim the propeller off. And then it's a really tough thing to get off because it's friction fitted on to this brass fitting on the end of the motor. And so it's really hard. They definitely did a good job of press fitting that onto the brass fitting so that it won't come off as the hair dryer moves around. So we're taking our nipper pliers here, and we're just going to trim the rest of the propeller off so we can get to the motor. And we'll move the end of it off there and then unscrew the last two screws and slide the ends off." + }, + { + "Q": "At 10:19 why does he say \"Then it goes to the heart, rubs up against some alveoli \"?\n", + "A": "Well, rubbing up against some alveoli means that the oxygen diffuses from any alveolus (singular for alveoli) to enter in the blood stream as oxygenated blood.", + "video_name": "QhiVnFvshZg", + "timestamps": [ + 619 + ], + "3min_transcript": "And in the left atrium, the blood is entering-- and remember, the left atrium is on the right-hand side from our point of view-- on the left atrium, the blood is entering from above from the lungs, from the pulmonary veins. Veins go to the heart. Then it goes into-- and I'll go into more detail-- into the left ventricle and then the left ventricle pumps that oxygenated blood to the rest of the body via the non-pulmonary arteries. So everything pumps out. Let me make it a nice dark, non-blue color. So it pumps it out through there. You don't see it right here, the way it's drawn. It's a little bit of a strange drawing. It's hard to visualize, but I'll show it in more detail and then it goes to the rest of the body. Let me show you that detail right now. So we said, we have de-oxygenated blood. Let's label it right here. This is the superior vena cava. our arms and heads. This is the inferior vena vaca. This is veins from our abdomen and from our legs and the rest of our body. So it it first enters the right atrium. Remember, we call the right atrium because this is someone's heart facing us, even though this is on the left-hand side. It enters through here. It's de-oxygenated blood. It's coming from veins. the body used the oxygen. Then it shows up in the right ventricle, right? These are valves in our heart. And it passively, once the right ventricle pumps and then releases, it has a vacuum and it pulls more blood from the It pumps again and then it pushes it through here. Now this blood right here-- remember, this one still is de-oxygenated blood. De-oxygenated blood goes to the lungs to become oxygenated. So this right here is the pulmonary-- I'm using the word pulmonary because it's going to or from the lungs. And it's going away from the heart. It's the pulmonary artery and it is de-oxygenated. Then it goes to the heart, rubs up against some alveoli and then gets oxygenated and then it comes right back. Now this right here, we're going to the heart. So that's a vein. It's in the loop with the lungs so it's a pulmonary vein and it rubbed up against the alveoli and got the oxygen diffused into it so it is oxygenated. And then it flows into your left atrium. Now, the left atrium, once again, from our point of view, is on the right-hand side, but from the dude looking at it, it's his left-hand side." + }, + { + "Q": "At 3:56 he say blue one is an artery but at 7:15 Bro Sal says that blue one is a vein??\nEm confused! :/\n", + "A": "Remember, blue simply represents deoxygenated blood. So, the pulmonary artery carries deoxygenated blood from the heart to the lungs. The vena cava veins carry deoxygenated blood from the rest of the body back to the heart.", + "video_name": "QhiVnFvshZg", + "timestamps": [ + 236, + 435 + ], + "3min_transcript": "So I've been all zoomed in here on the alveolus and these capillaries, these pulmonary capillaries-- let's zoom out a little bit-- or zoom out a lot-- just to understand, how is the blood flowing? And get a better understanding of pulmonary arteries and veins relative to the other arteries and veins that are in the body. So here-- I copied this from Wikipedia, this diagram of the human circulatory system-- and here in the back you can see the lungs. Let me do it in a nice dark color. So we have our lungs here. You can see the heart is sitting right in the middle. And what we learned in the last few videos is that we have our little alveoli and our lungs. Remember, we get to them from our bronchioles, which are branching off of the bronchi, which branch off of the trachea, which connects to our larynx, which connects to our pharynx, which connects to our mouth and nose. we have the capillaries. So when we go away from the heart-- and we're going to delve a little bit into the heart in this video as well-- so when blood travels away from the heart, it's de-oxygenated. It's this blue color. So this right here is blood. This right here is blood traveling away from the heart. It's going behind these two tubes right there. So this is the blood going away from the heart. So this blue that I've been highlighting just now, these are the pulmonary arteries and then they keep splitting into arterials and all of that and eventually we're in capillaries-- super, super small tubes. They run right past the alveoli and then they become oxygenated and now we're going back to the heart. So we're talking about pulmonary veins. So we go back to the heart. Now we're going to go back to the heart. Hope you can see what I'm doing. And we're going to enter the heart on this side. You actually can't even see where we're entering the heart. We're going to enter the heart right over here-- and I'm going to go into more detail on that. Now we have oxygenated blood. And then that gets pumped out to the rest of the body. Now this is the interesting thing. When we're talking about pulmonary arteries and veins-- remember, the pulmonary artery was blue. As we go away from the heart, we have de-oxygenated blood, but it's still an artery. Then as we go towards the heart from the lungs, we have a vein, but it's oxygenated." + }, + { + "Q": "\nI thought the veins carried oxygenated blood, 9:08 says the vein carried deoxygenated blood ?? I'm confused", + "A": "Arteries take blood away from the heart. Most arteries carry oxygenated blood away from the heart to other parts of the body. The veins take blood back to the heart. Veins carry oxygenated blood is carried in veins. The pulmonary artery takes blood away from the right ventricle to the lungs where it is oxygenated. The pulmonary vein takes oxygenated blood from the lungs to the left atrium - back to the heart.", + "video_name": "QhiVnFvshZg", + "timestamps": [ + 548 + ], + "3min_transcript": "So you don't see it. I'm going to do a detailed diagram in a second-- into the pulmonary artery. We're going away from the heart. This was a vein, right? This is a vein going to the heart. This is a vein, inferior vena cava vein. This is superior vena cava. They're de-oxygenated. Then I'm pumping this de-oxygenated blood away from the heart to the lungs. Now this de-oxygenated blood, this is in an artery, right? This is in the pulmonary artery. It gets oxygenated and now it's a pulmonary vein. And once it's oxygenated, it shows up here in the left-- let me do a better color than that-- it shows up right here in the left atrium. Atrium, you can imagine-- it's kind of a room with a skylight or that's open to the outside and in both of these cases, things are entering from above-- not sunlight, but blood is entering from above. And in the left atrium, the blood is entering-- and remember, the left atrium is on the right-hand side from our point of view-- on the left atrium, the blood is entering from above from the lungs, from the pulmonary veins. Veins go to the heart. Then it goes into-- and I'll go into more detail-- into the left ventricle and then the left ventricle pumps that oxygenated blood to the rest of the body via the non-pulmonary arteries. So everything pumps out. Let me make it a nice dark, non-blue color. So it pumps it out through there. You don't see it right here, the way it's drawn. It's a little bit of a strange drawing. It's hard to visualize, but I'll show it in more detail and then it goes to the rest of the body. Let me show you that detail right now. So we said, we have de-oxygenated blood. Let's label it right here. This is the superior vena cava. our arms and heads. This is the inferior vena vaca. This is veins from our abdomen and from our legs and the rest of our body. So it it first enters the right atrium. Remember, we call the right atrium because this is someone's heart facing us, even though this is on the left-hand side. It enters through here. It's de-oxygenated blood. It's coming from veins. the body used the oxygen. Then it shows up in the right ventricle, right? These are valves in our heart. And it passively, once the right ventricle pumps and then releases, it has a vacuum and it pulls more blood from the It pumps again and then it pushes it through here. Now this blood right here-- remember, this one still is de-oxygenated blood. De-oxygenated blood goes to the lungs to become oxygenated. So this right here is the pulmonary-- I'm using the word pulmonary because it's going to or from the lungs." + }, + { + "Q": "\nhow long dos it take to get your blood circulating 0:20 minutes", + "A": "round about 1 sec actually time taken for one breath", + "video_name": "QhiVnFvshZg", + "timestamps": [ + 20 + ], + "3min_transcript": "Where I left off in the last video, we talked about how the hemoglobin in red blood cells is what sops up all of the oxygen so that it increases the diffusion gradient-- or it increases the incentive, we could say, for the oxygen to go across the membrane. We know that the oxygen molecules don't know that there's less oxygen here, but if you watch the video on diffusion you know how that process happens. If there's less concentration here than there, the oxygen will diffuse across the membrane and there's less inside the plasma because the hemoglobin is sucking it all up like a sponge. Now, one interesting question is, why does the hemoglobin even have to reside within the red blood cells? Why aren't hemoglobin proteins just freely floating in the blood plasma? That seems more efficient. You don't have to have things crossing through, in and out of, these red blood cell membranes. You wouldn't have to make red blood cells. What's the use of having these containers of hemoglobin? It's actually a very interesting idea. If you had all of the hemoglobin sitting in your the flow of the blood. The blood would become more viscous or more thick. I don't want to say like syrup, but it would become thicker than blood is right now-- and by packaging the hemoglobin inside these containers, inside the red blood cells, what it allows the blood to do is flow a lot better. Imagine if you wanted to put syrup in water. If you just put syrup straight into water, what's going to happen? The water's going to become a little syrupy, a little bit more viscous and not flow as well. So what's the solution if you wanted to transport syrup in water? Well, you could put the syrup inside little containers or inside little beads and then let the beads flow in the water and then the water wouldn't be all gooey-- and that's exactly what's happening inside of our blood. Instead of having the hemoglobin sit in the plasma and make it gooey, it sits inside these beads that we call red blood cells that allows the flow to still be So I've been all zoomed in here on the alveolus and these capillaries, these pulmonary capillaries-- let's zoom out a little bit-- or zoom out a lot-- just to understand, how is the blood flowing? And get a better understanding of pulmonary arteries and veins relative to the other arteries and veins that are in the body. So here-- I copied this from Wikipedia, this diagram of the human circulatory system-- and here in the back you can see the lungs. Let me do it in a nice dark color. So we have our lungs here. You can see the heart is sitting right in the middle. And what we learned in the last few videos is that we have our little alveoli and our lungs. Remember, we get to them from our bronchioles, which are branching off of the bronchi, which branch off of the trachea, which connects to our larynx, which connects to our pharynx, which connects to our mouth and nose." + }, + { + "Q": "at \"0:42\" I could not hear it with my volume at full, what did he say?\n", + "A": "He didn t really say anything important. He said I don t know and came up with (1/2)m^2 for the second area.", + "video_name": "xlJYYM5TWoA", + "timestamps": [ + 42 + ], + "3min_transcript": "Let's say I have a horizontal pipe that at the left end of the pipe, the cross-sectional area, area 1, which is equal to 2 meters squared. Let's say it tapers off so that the cross-sectional area at this end of the pipe, area 2, is equal to half a square meter. We have some velocity at this point in the pipe, which is v1, and the velocity exiting the pipe is v2. The external pressure at this point is essentially being applied rightwards into the pipe. The pressure at this end, the pressure 2-- that's the external pressure at that point in the pipe-- that is equal to 6,000 pascals. Given this information, let's say we have water in this pipe. We're assuming that it's laminar flow, so there's no friction within the pipe, and there's no turbulence. Using that, what I want to do is, I want to figure out what is the flow or the flux of the water in this pipe-- how much volume goes either into the pipe per second, or out of the We know that those are the going to be the same numbers, because of the equation of continuity. We know that the flow, which is R, which is volume per amount of time, is the same thing as the input velocity times the input area. The input area is 2, so it's 2v1, and that also equals the output area times output velocity, so it equals 1/2 v2. We could rewrite this, that v1 is equal to 1/2 R, and that v2 is equal to 2R. This immediately tells us that v2 is coming out at a faster rate, and this is based on the size of the openings. We know, because V2 is coming out at a faster rate, but we also know because we have much higher pressure at this end" + }, + { + "Q": "At around 5:15, why are the 3 branches drawn off of the monocyte cell branch? Could it have been drawn just coming from the myeloid (Is it just to save space)?\n", + "A": "There is a reason to it :) The Common Myeloid Progenitor cell (the first red one) grows into 4 different cells; RBC, Mast cell, Megakaryocyte and MYELOBLAST. Myeloblast then later grows into; Neutrophils, Eosinphils, Basophils, and monocyte. He kinda skipped this step because it makes it a little bit more complicated.", + "video_name": "ddifthdMNVc", + "timestamps": [ + 315 + ], + "3min_transcript": "And those two lineages are the myeloid lineage and the lymphoid lineage. And each of these lineages gives rise to many different cells. The myeloid lineage gives rise to red blood cells, which are biconcave in shape. They are the most common of all blood cells. Now, the myeloid lineage also gives rise to a big cell called a megakaryocyte. Now, you might have never heard of this before, but the megakaryocytes themselves produce platelets, which I think that you've probably have heard of. They're little fragments of cells, They kind of squeeze out little pieces of cytoplasm that become platelets. And now I have a challenge for you. Do you think that a macrophage, which is an immune cell that likes to eat up invaders like bacteria, do you think that macrophages come from the myeloid lineage or the lymphoid lineage? So, I was surprised to find out that they actually come from the myeloid lineage. I was surprised because macrophages are immune cells, but they actually come from the same lineage as red blood cells and platelets. So here is a... This is actually not yet a macrophage, this is a monocyte. A lot of crazy words here, but this is a monocyte. Monocytes actually become macrophages once they settle down in the tissues. But, before that, while they're still circulating, they are monocytes. the myeloid lineage gives rise to three guys, one of whom you have heard of probably, two of whom you may not have heard of. I'll just draw them here. So, the one you might have heard of is the, I'm running out of space here, but it's the neutrophil. Neutrophils are the most common immune cell in the blood. The other two are called eosinophils, which are significantly more rare than neutrophils. And, even more rare than eosinophils, are something called the basophils. So, it's the three phils. So, now let's go over to the lymphoid lineage. There's three important cells that come from this one. Two of them you've probably heard of." + }, + { + "Q": "\nAt approximately 6:00, what was Sal implying by the term, \"superimposed\" when he was referring to enantiomers?", + "A": "To superimpose means to put on top of eachother, so putting the image of one on top of the other and only seeing one since they re the same (or not since they re different.", + "video_name": "z8M4EciPpYI", + "timestamps": [ + 360 + ], + "3min_transcript": "in three dimensions. We don't just care about what's bonded to what or the constituents and actually this one is, as we'll see, is also a stereoisomer because this carbon is bonded to the same things in either case. So these are both, these are both situations, there are both stereoisomers, stereoisomers, and this particular variation of stereoisomer is called a cis trans isomer. Cis is when you have the two groups on the same side, cis, and trans is when you have the two groups on the opposite sides of the double bond. Cis trans isomers. Cis trans isomers. Isomers, and these are often called geometric isomers. Geometric, geometric isomers. So that's a subset, so when I'm talking about cis trans or geometric, I'm talking about these two characters over here. They are a subset of the stereoisomers. I have no double bond, I'm not talking about cis and trans. The carbon, as I've just said, is bonded to fluorine, chlorine, bromine, and a hydrogen, fluorine, chlorine, bromine, and a hydrogen. How are these two things different? And the way that they're different is if you were to actually try to superimpose them on each other. You will see that it is impossible. There are mirror images of each other and because there's four different constituents here, you can actually not superimpose this molecule onto this molecule over here and actually because of that, they actually have different chemical properties, and so this over here, these two characters, which is a subset of stereoisomers. Stereoisomers are concerned with how things are positioned in three dimensions, not just how their bonding is different, but this subset where you have these mirror images that cannot be superimposed, we call these enantiomers. So these two characters, these are enantiomers. Enantiomers, and enantio comes from Greek, So these are opposites of each other, they cannot be superimposed, they're mirror, they're mirror images. So all of these are different variations of isomers and once again, you might say, okay theses are clearly two different molecules that have different bonding, but even cis trans isomer will have different chemical properties. These two in particular, they aren't that different but they do have different chemical properties, but sometimes they're so different that one might be able to exist in a biological system while the other is not. One might be okay for your health, and the other might not be okay for your health. Same thing for enantiomers. One might be biologically active in a certain way and the other one might not be biologically active in that same way." + }, + { + "Q": "Perhaps I missed it in the video, but at 3:13 Sal is talking about the Helium absorbing light energy. Is this light coming from the Cepheid itself, or does he mean light from an outside source?\n", + "A": "I think he meant from the inside of the cepheid.", + "video_name": "X_3QAB3o4Vw", + "timestamps": [ + 193 + ], + "3min_transcript": "is doubly and singly ionized helium. And just to review, helium, so neutral helium, let me draw a neutral helium, neutral helium's got two protons, it's got two protons, two neutrons, two neutrons, and then two electrons and obviously this is not drawn to scale. So this is neutral helium right over here. Now, if you singly ionize helium you knock off one of these electrons. And these type of things happen in stars when you have a lot of heat, easier to ionize things. So singly ionized helium will look like this. It'll have the same nucleus, two protons, two neutrons. One of the electrons gets knocked off so now you only have one electron. And now you have a net positive charge. So here, let me do this in a different color, this helium now has a net charge, we could write one plus here, but if you just write a plus you implicitly mean a positive charge of one. Now you can also double the ionized helium if the environment is hot enough. and doubly ionizing helium is essentially knocking off both of the electrons. So then it's really just a helium nucleus. It's really just a helium nucleus like this. This right here is doubly, doubly ionized helium. Now I just said in order to do this you have to have a hotter environment. There has to be a hotter environment in order to be able to knock off both these, this electron really doesn't want to leave, to take an electron off of something that's already positive is difficult. You have to have a lot of really pressure and temperature. This is cooler. And this is all relative, we're talking about the insides of stars. So, you know, it's hot, this is a hotter part of the star versus a cooler part of the star I guess is a way you think about this, it's still a very hot environment by our traditional, every day standards. Now the other thing about the doubly ionized helium is that it is more opaque. It is more opaque, it doesn't allow light to go through it, it actually absorbs light. It is more opaque, it absorbs light. It absorbs light. it absorbs that light energy that energy will make it even hotter. So that's just something to think about. Now, the singly ionized helium is more transparent. This is more transparent. More transparent, it allows the light to pass through it. So it doesn't get heated as much by photons that are kind of going near it, or through it, or whatever. It allows them to go through it here the photons are going to actually heat up, heat up the ion. So let's think about how this might cause cepheid variable to pulsate. So assuming that cepheid variables have a large enough quantity, I should say, of these ions, we can imagine that when a cepheid variable is dim, so let me draw a dim cepheid variable, so I'll draw that like, I'll draw this in a dim color" + }, + { + "Q": "\nAt approximately 3:17 he states the peak at 1100cm^-1 corresponds to a C-O single bond. However, in prior videos he described the area below 1500 as the \"fingerprint\" area, and the area above 1500 as the diagnostics area. Can you explain the significance of this?", + "A": "Some peaks are so strong and so characteristic that you can identify them even they are in the fingerprint region.", + "video_name": "ALLSsIDhFdU", + "timestamps": [ + 197 + ], + "3min_transcript": "are going to have different amounts of hydrogen bonding. Some molecules might have a little bit of hydrogen bonding, so k decreases a little bit, and the wavenumber decreases a little bit. But other molecules might have a lot of hydrogen bonding, and so we can decrease k even more, therefore we are going to decrease the wavenumber even more. You get a range of wavenumbers, and since you get a range of wavenumbers for the OH bond, when hydrogen bonding is present, you get a very broad signal on your IR spectrum. So, if we go over here in this region, so we're talking about the IR spectrum for 1-hexanol, this is the region for bonds to hydrogen. So we draw a line at 3000, and we know that just below 3000, we're talking about a carbon-hydrogen bond stretch, where the carbon is sp3-hybridized. But, this over here, this very broad signal right here, So let me go ahead and highlight that. This bond right here, this oxygen-hydrogen bond, gives us a very broad signal on our IR spectrum because of hydrogen bonding. So we get this very broad signal because of the different wavenumbers. And usually you're going to see this somewhere around 3500 to 2900. So if I find this is 31, 32, 33, 34, 35... so usually in this range, maybe even a little bit higher than that, you're going to find this very broad signal. In this case, the oxygen-hydrogen bond. And so you know immediately to think about the possibility of an alcohol functional group in your molecule. Also, we can draw a line at 1500 here, and this signal actually, so somewhere around 1100 wavenumbers, this is actually the carbon-oxygen single bond. Let me go ahead and highlight that. So we have a carbon-oxygen single bond, And that's where-- that's this stretch right here. Not always going to be super useful to you, but it's just thinking about what we talked about in the earlier video, I think we calculated the approximate wavenumber for a carbon-oxygen single bond. So that's what the typical spectrum for an alcohol is going to look like. Look for that broad signal there. Alright, let's compare this alcohol to another one here. So, this molecule is butylated hydroxytoluene, or BHT, and I drew two BHT molecules in there for a reason. Let's think about why. So, you might think at first, \"OK, I have another opportunity for hydrogen bonding.\" So, here's an opportunity for hydrogen bonding, so we're going to get a broad signal for this OH bond. So I'm going to highlight it here. I might expect, since I have hydrogen bonding," + }, + { + "Q": "\n10:31 Why is keto form more stable than enol form ?", + "A": "Good Question! Enols are less stable because the C=C (double bond) is weakened by the electronegativity of oxygen.", + "video_name": "NdRl1C6Jr5o", + "timestamps": [ + 631 + ], + "3min_transcript": "hybridized with tetrahedral geometry. Now, this carbon is SP two hybridized with trigonal planar geometry. Whatever stereochemical information we had over here on the left, whether it was the R or the S enantiomer, it's been lost now that we've formed the enol. The enol is achiral, it's flat, it's planar. When we reform the keto form, so one of the possibilities is to form the enantiomer that we started with but the other possibility is to form the other enantiomer. You can see that's what I've shown here. I've shown the hydrogen now going away from us and our R double prime group coming out at us. This is the enantiomer. Because we formed the enol we can get a mixture of enantiomers. Enolization can lead to racemization. We can get a mixture of enantiomers and if we wait long enough, we can get an equal mixture of these guys. with our enol form. That's something to think about if you have a chiral center at your alpha carbon. Let's look at two quick examples of keto and enol forms. Over here on the left we have cyclohexanone and on the right would be the enol version of it. You could think about one of these as being your alpha carbon, right, and you could move these electrons in here and push those electrons off. You could see that would give you this enol form. It turns out that the keto form is favored. The equilibrium is actually far to the left favoring formation of the keto form. Even under just normal conditions, so not acid or base-catalyzed. There's only a trace amount of the enol presence however, there are some cases where the enol is extra-stabilized and that's the case for this example down here. We have the keto form and we have the enol form. Once again, you could think about pushing those electrons off giving you your enol form. This is a specially-stabilized enol, right? This is phenol right here. We know that phenol has an aromatic ring. The formation of the enol form is extra-stabilize because of this aromatic ring. This time the equilibrium is actually to the right and much more of it is in the enol form than in the keto form. In this case, we have some special stabilization." + }, + { + "Q": "Would the reaction still be complete if we didn't \"open up the structure\"? (Meaning leave it at the epoxide Jay drew at 4:16)\n", + "A": "Yes, if you simply want the epoxide, the reaction would be complete.", + "video_name": "KfTosrMs5W0", + "timestamps": [ + 256 + ], + "3min_transcript": "And then we show the bond between those like that. And then up at the top here, here's my carbonyl carbon. So now there's only one bond between that carbon and this oxygen. There is a new bond that formed between that oxygen and that hydrogen, and there is an R group over here. And then there used to be only one bond to this oxygen, but another lone pair of electrons moved in to form a carbonyl here. So this is our other product, which you can see is a carboxylic acid. Let's color code these electrons so we can follow them a little bit better. So let's make these electrons in here, those electrons are going to form the bond on the left side between the carbon and the oxygen like that. Let's follow these electrons next. So now let's look at these electrons in here, the electrons in this pi bond. Those are the ones that are going And let's make our oxygen-oxygen bond blue here. So the electrons in this bond, those are the ones that moved in here to form our carbonyl like that. And then let's go ahead and make these green right here, the electrons in this bond right here. These are the ones that moved out here to form the bond between our oxygen and our hydrogen. So our end result is to form a carboxylic acid and our epoxide. Let's look at a reaction, an actual reaction for the formation of epoxide, and then we'll talk about how to form a diol from that. So if we start with cyclohexene-- let's go ahead and draw cyclohexene in here. Let's do another one. That one wasn't very good. So we draw our cyclohexene ring like that. And to cyclohexene, we're going to add peroxyacetic acid. So what does peroxyacetic acid look like? But it has one extra oxygen in there, so it looks like that. So that's our peroxyacetic acid. So we add cyclohexene to peroxyacetic acid, we're going to form an epoxide. So we're going to form a three-membered ring, including oxygen. I'm going to say the oxygen adds to the top face of our ring. It doesn't really matter for this example, but we'll go ahead and put in our epoxide using wedges here. And that must mean going away from us, those are hydrogens in space. So that's the epoxide that would form using the mechanism that we put above there. Let's go ahead and open this up epoxide using acid. So just to refresh everyone's memory, go back up here. Now we're going to look at this second part where we add H3O plus to form our diol. So let's take a look at that now. So we're going to add H3O plus to this epoxide. And I'm going to redraw our epoxide" + }, + { + "Q": "1:30 How come the electrons between the O and H move twards the alkene's C? What triggers it? Thank you.\n", + "A": "i think C is more electronegative than H so that due to inductive effect electrons move towards C rather than H", + "video_name": "KfTosrMs5W0", + "timestamps": [ + 90 + ], + "3min_transcript": "If you start with an alkene and add to that alkene a percarboxylic acid, you will get epoxide. So this is an epoxide right here, which is where you have oxygen in a three-membered ring with those two carbons there. You can open up this ring using either acid or base catalyzed, and we're going to talk about an acid catalyzed reaction in this video. And what ends up happening is you get two OH groups that add on anti, so anti to each other across from your double bond. So the net result is you end up oxidizing your alkene. So you could assign some oxidation numbers on an actual problem and find out that this is an oxidation reaction. All right. Let's look at the mechanism to form our epoxide. So we start with our percarboxylic acid here, which looks a lot like a carboxylic acid except it has an extra oxygen. And the bond between these two oxygen atoms is weak, so this bond is going to break in the mechanism. The other important thing to note about the structure of our percarboxylic acid is the particular confirmation that it's in. So this hydrogen ends up being very a source of attraction between those atoms. There's some intramolecular hydrogen bonding that keeps it in this conformation. When the percarboxylic acid approaches the alkene, when it gets close enough in this confirmation, the mechanism will begin. This is a concerted eight electron mechanism, which means that eight electrons are going to move at the same time. So the electrons in this bond between oxygen and hydrogen are going to move down here to form a bond with this carbon. The electrons in this pi bond here are going to move out and grab this oxygen. That's going to break this weak oxygen-oxygen bond, and those electrons move into here. And then finally, the electrons in this pi bond are going to move to here to form an actual bond between that oxygen and that hydrogen. So let's see if we can draw the results of this concerted eight electron mechanism. So, of course, at the bottom here we're going to form our epoxide. So we draw in our carbons, and then we can put in our oxygen And then we show the bond between those like that. And then up at the top here, here's my carbonyl carbon. So now there's only one bond between that carbon and this oxygen. There is a new bond that formed between that oxygen and that hydrogen, and there is an R group over here. And then there used to be only one bond to this oxygen, but another lone pair of electrons moved in to form a carbonyl here. So this is our other product, which you can see is a carboxylic acid. Let's color code these electrons so we can follow them a little bit better. So let's make these electrons in here, those electrons are going to form the bond on the left side between the carbon and the oxygen like that. Let's follow these electrons next. So now let's look at these electrons in here, the electrons in this pi bond. Those are the ones that are going" + }, + { + "Q": "\nAt 7:46, why does Jay say that there is only 1 pi bond? Are pi bonds made up of more than 2 valence electrons?", + "A": "There s only one pi bond. It contains two electrons - one electron comes from the p-orbital on the left carbon and the other from the p-orbital on the right carbon.", + "video_name": "ROzkyTgscGg", + "timestamps": [ + 466 + ], + "3min_transcript": "here's another head on overlap of orbitals. The carbon carbon bond, here's also a head on overlap of orbitals and then we have these two over here. We have a total of five sigma bonds in our molecules. Let me go ahead and write that over here. There are five sigma bonds. If I would try to find those on my dot structure this would be a sigma bond. This would be a sigma bond. One of these two is a sigma bond and then these over here. A total of five sigma bonds and then we have a new type of bonding. These unhybridized P orbitals can overlap side by side. Up here and down here. We get side by side overlap of our P orbitals and this creates a pi bond. A pi bond, let me go ahead and write that here. A pi bond is side by side overlap. There is overlap above and below this sigma bond here When we're looking at the example of ethane, we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here, this pi bond is going to prevent rotations so we don't get different confirmations of the ethylene molecules. No free rotation due to the pi bonds. When you're looking at the dot structure, one of these bonds is the pi bonds, I'm just gonna say it's this one right here. If you have a double bond, one of those bonds, the sigma bond and one of those bonds is a pi bond. We have a total of one pi bond in the ethylene molecule. If you're thinking about the distance between the two carbons, let me go ahead and use a different color for that. The distance between this carbon and this carbon. It turns out to be approximately 1.34 angstroms, between the two carbons in the ethane molecule. Remember for ethane, the distance was approximately 1.54 angstroms. A double bond is shorter than a single bond. One way to think about that is the increased S character. This increased S character means electron density is closer to the nucleus and that's going to make this lobe a little bit shorter than before and that's going to decrease the distance between these two carbon atoms here. 1.34 angstroms. Let's look at the dot structure again and see how we can analyze this using the concept of steric number. Let me go ahead and redraw the dot structure. We have our carbon carbon double bond here and our hydrogens like that. If you're approaching this situation using steric number remember to find the hybridization." + }, + { + "Q": "@12:17 there's a bubble that pops up which says 'trigonal planar', as if in correction of what Jay says in the video. (He calls it planar). However, I think that what he says, planar, is correct, and not the box: the empty p orbital exists on BOTH sides of the molecule, and so the sp2 orbitals don't bend out of shape\n", + "A": "No the word trigonal planar is correct as it lies in plane and it is not a 3D structure!", + "video_name": "ROzkyTgscGg", + "timestamps": [ + 737 + ], + "3min_transcript": "If you wanna draw the dot structure of BF3, you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine. I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here. What is the hybridization stage of this boron? Let's use the concept of steric number. Once again, let's use steric number. Find the hybridization of this boron. Steric number is equal to number of sigma bonds. That's one, two, three. Three sigma bonds plus lone pairs of electrons. That's zero. Steric number of three tells us this boron is SP2 hybridized. This boron is gonna have three SP2 hybrid orbitals and one P orbital. One unhybridized P orbital. Let's go ahead and draw that. and also it's going to have an unhybrized P orbital. Now, remember when you are dealing with Boron, it has one last valance electron and carbon. Carbon have four valance electrons. Boron has only three. When you're thinking about the SP2 hybrid orbitals that you create. SP2 hybrid orbital, SP2, SP2 and then one unhybridized P orbital right here. Boron only has three valance electrons. Let's go ahead and put in those valance electrons. One, two and three. It doesn't have any electrons in its unhybridized P orbital. Over here when we look at the picture, this has an empty orbital and so boron can accept a pair of electrons. We're thinking about its chemical behavior, one of the things that BF3 can do, the Boron can accept an electron pair and function as a lewis acid. That's one way in thinking about how hybridizational and how something might react. This boron turns out to be SP2 hybridized. This boron here is SP2 hybridized and so we can also talk about the geometry of the molecule. It's planar. Around this boron, it's planar and so therefore, your bond angles are 120 degrees. If you have boron right here and you're thinking about a circle. A circle is 360 degrees. If you divide a 360 by 3, you get 120 degrees for all of these bond angles. In the next video, we'll look at SP hybridization." + }, + { + "Q": "at 2:00 aren't there 4 bonds on every carbon? so why don't we get sp3\n", + "A": "To determine the hybridization, look at the # of sigma bonds + # of lone pairs, rather than the total number of bonds. Here, each carbon does have 4 bonds, but only 3 sigma bonds, so it is sp2.", + "video_name": "ROzkyTgscGg", + "timestamps": [ + 120 + ], + "3min_transcript": "Voiceover: In an earlier video, we saw that when carbon is bonded to four atoms, we have an SP3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees. If you look at one of the carbons in ethenes, let's say this carbon right here, we don't see the same geometry. The geometry of the atoms around this carbon happens to be planar. Actually, this entire molecule is planar. You could think about all this in a plane here. And the bond angles are close to 120 degrees. Approximately, 120 degree bond angles and this carbon that I've underlined here is bonded to only three atoms. A hydrogen, a hydrogen and a carbon and so we must need a different hybridization for each of the carbon's presence in the ethylene molecule. We're gonna start with our electron configurations over here, the excited stage. We have carbons four, valence electron represented. One, two, three and four. In the video on SP3 hybridization, to make four SP3 hybrid orbitals. In this case, we only have a carbon bonded to three atoms. We only need three of our orbitals. We're going to promote the S orbital. We're gonna promote the S orbital up and this time, we only need two of the P orbitals. We're gonna take one of the P's and then another one of the P's here. That is gonna leave one of the our P orbitals unhybridized. Each one of these orbitals has one electron and it's like that. This is no longer an S orbital. This is an SP2 hybrid orbital. This is no longer a P orbital. This is an SP2 hybrid orbital and same with this one, an SP2 hybrid orbital. We call this SP2 hybridization. Let me go and write this up here. and use a different color here. This is SP2 hybridization because we're using one S Orbital and two P orbitals to form our new hybrid orbitals. and same with this carbon. Notice that we left a P orbital untouched. We have a P orbital unhybridized like that. In terms of the shape of our new hybrid orbital, let's go ahead and get some more space down here. We're taking one S orbital. We know S orbitals are shaped like spheres. We're taking two P orbitals. We know that a P orbital is shaped like a dumbbell. We're gonna take these orbitals and hybridized them to form three SP2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that. Once again, when we draw the pictures, we're going to ignore the smaller back lobe. This gives us our SP2 hybrid orbitals. In terms of what percentage character, we have three orbitals that we're taking here and one of them is an S orbital. One out of three, gives us 33% S character" + }, + { + "Q": "At 8:21 shouldn't it be A with a circle on the top? Not on the side? Or are both forms of the angstrom acceptable?\n", + "A": "Yes. The symbol for angstrom is \u00c3\u0085. In cursive writing, we don t always get the little circle directly on top of the A. Of course, it would be preferable to use the SI units of picometres (pm) instead of angstroms.", + "video_name": "ROzkyTgscGg", + "timestamps": [ + 501 + ], + "3min_transcript": "here's another head on overlap of orbitals. The carbon carbon bond, here's also a head on overlap of orbitals and then we have these two over here. We have a total of five sigma bonds in our molecules. Let me go ahead and write that over here. There are five sigma bonds. If I would try to find those on my dot structure this would be a sigma bond. This would be a sigma bond. One of these two is a sigma bond and then these over here. A total of five sigma bonds and then we have a new type of bonding. These unhybridized P orbitals can overlap side by side. Up here and down here. We get side by side overlap of our P orbitals and this creates a pi bond. A pi bond, let me go ahead and write that here. A pi bond is side by side overlap. There is overlap above and below this sigma bond here When we're looking at the example of ethane, we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here, this pi bond is going to prevent rotations so we don't get different confirmations of the ethylene molecules. No free rotation due to the pi bonds. When you're looking at the dot structure, one of these bonds is the pi bonds, I'm just gonna say it's this one right here. If you have a double bond, one of those bonds, the sigma bond and one of those bonds is a pi bond. We have a total of one pi bond in the ethylene molecule. If you're thinking about the distance between the two carbons, let me go ahead and use a different color for that. The distance between this carbon and this carbon. It turns out to be approximately 1.34 angstroms, between the two carbons in the ethane molecule. Remember for ethane, the distance was approximately 1.54 angstroms. A double bond is shorter than a single bond. One way to think about that is the increased S character. This increased S character means electron density is closer to the nucleus and that's going to make this lobe a little bit shorter than before and that's going to decrease the distance between these two carbon atoms here. 1.34 angstroms. Let's look at the dot structure again and see how we can analyze this using the concept of steric number. Let me go ahead and redraw the dot structure. We have our carbon carbon double bond here and our hydrogens like that. If you're approaching this situation using steric number remember to find the hybridization." + }, + { + "Q": "At 11:45, jay tells us that boron acts a lewis acid because it has no electrons in its unhybridised porbital and hence can gain two electrons.\nbut the sp2 hybrid orbital have only 3 electrons and hence will they be able to gain 3 MORE electrons?\n", + "A": "Each sp\u00c2\u00b2 orbital has one electron, and they each gain another electron when they form bonds to the fluorine atoms. That makes six electrons in the valence shell of boron. Boron needs two more electrons to complete its octet.", + "video_name": "ROzkyTgscGg", + "timestamps": [ + 705 + ], + "3min_transcript": "If you wanna draw the dot structure of BF3, you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine. I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here. What is the hybridization stage of this boron? Let's use the concept of steric number. Once again, let's use steric number. Find the hybridization of this boron. Steric number is equal to number of sigma bonds. That's one, two, three. Three sigma bonds plus lone pairs of electrons. That's zero. Steric number of three tells us this boron is SP2 hybridized. This boron is gonna have three SP2 hybrid orbitals and one P orbital. One unhybridized P orbital. Let's go ahead and draw that. and also it's going to have an unhybrized P orbital. Now, remember when you are dealing with Boron, it has one last valance electron and carbon. Carbon have four valance electrons. Boron has only three. When you're thinking about the SP2 hybrid orbitals that you create. SP2 hybrid orbital, SP2, SP2 and then one unhybridized P orbital right here. Boron only has three valance electrons. Let's go ahead and put in those valance electrons. One, two and three. It doesn't have any electrons in its unhybridized P orbital. Over here when we look at the picture, this has an empty orbital and so boron can accept a pair of electrons. We're thinking about its chemical behavior, one of the things that BF3 can do, the Boron can accept an electron pair and function as a lewis acid. That's one way in thinking about how hybridizational and how something might react. This boron turns out to be SP2 hybridized. This boron here is SP2 hybridized and so we can also talk about the geometry of the molecule. It's planar. Around this boron, it's planar and so therefore, your bond angles are 120 degrees. If you have boron right here and you're thinking about a circle. A circle is 360 degrees. If you divide a 360 by 3, you get 120 degrees for all of these bond angles. In the next video, we'll look at SP hybridization." + }, + { + "Q": "\nAt 2:37, why has he written 36?", + "A": "This is a known error in the video. A box pops up and tells you it should be 81 and the final answer is sqrt(97)", + "video_name": "gluN2wxqES0", + "timestamps": [ + 157 + ], + "3min_transcript": "" + }, + { + "Q": "When you squared that negative number in 2:37, does it become non negative?\n", + "A": "Yep. Squares of negative numbers are positive(non negative) since we are multiplying a negative with a negative which gives a positive result.", + "video_name": "gluN2wxqES0", + "timestamps": [ + 157 + ], + "3min_transcript": "" + }, + { + "Q": "\nWhen you squared that negative number in 2:37, does it become non negative?", + "A": "yes...for example squaring is the same thing as multiplying by the same number ..so here -4 X -4=16 ..the negatives cancel out", + "video_name": "gluN2wxqES0", + "timestamps": [ + 157 + ], + "3min_transcript": "" + }, + { + "Q": "8:26 would even a metal chain stretch?\n", + "A": "sure, it would stretch a bit. You might not be able to notice it.", + "video_name": "QKXeZFwFPS0", + "timestamps": [ + 506 + ], + "3min_transcript": "We divide by 3 kg, 'cause that's the mass. But I've still got a problem. I don't know this acceleration or this tension. So what do I do? You might notice, if you're clever you'll say wait, I've got my unknown on this side is acceleration and tension. My unknown on this side is acceleration and tension. It seems like I've got two equations, two unknowns, maybe we should combine them. And that's exactly how you do these. So I've got tension in both of these equations. Let me solve for tension over here, where it's kind of simple. And I'll just get the tension equals 5 kg, times the acceleration of the 5 kg mass in the X direction. So now I know what tension is. Tension is equal to this. And that tension over on this side is the same as the tension on this side. So I can take this and I can plug it in for this tension right here. And let's see what we get. We get that the acceleration of the 3 kg mass vertically, what am I gonna get? I'm gonna get, so T is the same as 5AX. So I'll plug in 5 kg times the acceleration of the 5 kg mass in the X direction. And then I get all of this stuff over here. So I'll get the rest of this right here. I'll just bring that down right there. Alright, now what do I have? I've got 3 kg on the bottom still, so I have to put that here. Are we any better off? Yeah, we're better, because now my only unknowns are acceleration. But these are not the same acceleration. Look, this acceleration here is the acceleration of the 3 kg mass, vertically. This acceleration here is the acceleration of the 5kg mass horizontally. Now here's where I'm gonna have to make an argument, and some people don't like this. But, it's crucial to figuring out this problem. And the key idea is this, if this 3 kg mass moves down, Well then this 5 kg mass had better move forward one meter. Because if it doesn't, then it didn't provide the one meter of rope that this 3 kg mass needed to go downward. Which means either the rope broke, or the rope stretched. And we're gonna assume that our rope does not break or stretch. All ropes are gonna stretch a little bit under tension. We're gonna assume that stretch is negligible. So the argument is that if this 3 kg mass moves downward a certain amount, this 5 kg mass has to move forward by that same amount in order to feed that amount of rope for this 3 kg mass to go downward by that amount. Otherwise, think about it. If this 5 kg mass just sat here and the 3 kg moved, or the 3 kg moved farther than the 5 kg mass, then this rope is stretching or breaking. So if you believe that, if you don't believe it, pause it and think about it. 'Cause you've gotta convince yourself of that. If you believe that then you can also convince yourself" + }, + { + "Q": "\nAcceleration of 3Kg box is downwards so it shotld be negative but you wrote it as positive in the equation above at 10:06", + "A": "Typically the acceleration of the 3kg block would be considered to be negative, because downwards and to the left are typically considered to be the negative direction and upwards and to the right are typically considered to be positive directions. However in this case he just used different signs instead of following the normal convention. He chose downwards and leftwards to be positive and upwards and rightwards to be negative.", + "video_name": "QKXeZFwFPS0", + "timestamps": [ + 606 + ], + "3min_transcript": "Well then this 5 kg mass had better move forward one meter. Because if it doesn't, then it didn't provide the one meter of rope that this 3 kg mass needed to go downward. Which means either the rope broke, or the rope stretched. And we're gonna assume that our rope does not break or stretch. All ropes are gonna stretch a little bit under tension. We're gonna assume that stretch is negligible. So the argument is that if this 3 kg mass moves downward a certain amount, this 5 kg mass has to move forward by that same amount in order to feed that amount of rope for this 3 kg mass to go downward by that amount. Otherwise, think about it. If this 5 kg mass just sat here and the 3 kg moved, or the 3 kg moved farther than the 5 kg mass, then this rope is stretching or breaking. So if you believe that, if you don't believe it, pause it and think about it. 'Cause you've gotta convince yourself of that. If you believe that then you can also convince yourself at a certain speed, let's say two meters per second. Then the 5 kg mass had better also be moving forward two meters per second because otherwise it wouldn't be feeding rope at a rate that this 3 kg needs to move downward at that rate. And finally, if you believe all that, it's not too much harder to convince yourself that this 3 kg mass, no matter what its acceleration downward must be, this 5 kg mass had better have the same magnitude of acceleration forward so that it's again, feeding the rope so this rope doesn't break, or snap, or stretch. 'Cause we're gonna assume the rope doesn't do that. So what I'm saying is that the acceleration of the 3 kg mass in the Y direction had better equal the magnitude. So these magnitudes have to be the same. The sign doesn't have to be the same. So this 3 kg mass has a negative acceleration just 'cause it points down, and we're assuming up is positive, down is negative. This 5 kg mass has a positive acceleration 'cause is the positive horizontal direction. So, they can have different signs, but the magnitudes had better be the same so that your feeding this rope at a rate that the other one needs in order to move. And so we can say that the magnitudes are the same. In this case, since one is negative of the other, I can say that the acceleration of the 3 kg mass vertically downward is gonna be equal to, let's say negative of the acceleration of the 5 kg mass in the X direction. I could have written it the other way. I could have wrote that A of the 5 kg mass in the X direction is a negative A of the 3 kg mass in the Y direction. They're just different by a negative sign is all that's important here. Okay, so this is the link we need. This is it. So this allows us to put this final equation here in terms of only one variable. 'Cause I know I've got A3Y on this left hand side. I know A3Y should always be -A5X. If I take this and just plug it in for A3Y right here," + }, + { + "Q": "why cant the outer shell electrons repel the inner shell electrons towards the nucleus? (5:57) the inner shell atoms could easily get sucked into the nucleus with the pushing( (repelling) force from the outer shell and the pulling force from positive charge of the nucleus, Right?\n", + "A": "The thing about atoms is they have this property where they only can occupy certain states, so electrons can only jump between orbits if they are empty and energy levels. There is no orbit in the nucleus and thus they can t be there. I d suggest looking at the lesson in chemistry about the electronic structure of atoms.", + "video_name": "rKoIcgBM4Vg", + "timestamps": [ + 357 + ], + "3min_transcript": "charged inner shell electrons are going to repel it. So let me go ahead and highlight these guys right here. These are our inner shell electrons. Like charges repel. And so you could think about this electron right here wanting to push this outer electron that way, and this electron wanting to push this electron that way. And so the nucleus attracts a negative charge, and the inner shell electrons repel the outer electron. And then we call this shielding, because the inner shell electrons are shielding that magenta electron from the pole of the nucleus. So this is called electronic shielding or electron screening. Now, it's going to be important concepts. So now let's go ahead and draw the atom for beryllium, so atomic number 4. And so here's our nucleus for beryllium. With an atomic number of 4, that means there are four protons in the nucleus, so a charge of four plus in our nucleus. And we have four electrons to worry about this time, in my inner orbital in our first energy level. And then we have two electrons in our outer orbital, or our second energy level. And so again, this is just a rough approximation for an idea of what beryllium might look like. And so when we think about what's happening, we're moving from a charge of 3 plus with lithium to a charge of 4 plus with beryllium. And the more positive your charges, the more it's going to attract those outer electrons. And when you think about the idea of electron screening, so once again we have these electrons in green here shielding our outer shell electrons from the effect of that positively charged nucleus. Now, you might think that outer shell electrons could shield, too. So you might think that oh, this electron right here in magenta could shield the other electron in magenta. much the same distance from the nucleus, so outer shell electrons don't really shield each other. It's more of these inner shell electrons. And because you have the same number of inner shell electrons shielding as in the lithium example-- so let me go ahead and highlight those again. So we have two inner shell electrons shielding a beryllium. We also have two inner shell electrons shielding in lithium. Because you have the same number of shielding but you have a higher positive charge, those outer electrons are going to feel more of a pull from the nucleus. And they're going to be pulled in even tighter than you might imagine, or at least tighter than our previous example. So these electrons are pulled in even more. And because of that, you're going to get the beryllium atom as being smaller than the lithium atom, hence the trend. Hence as you go across the period, you're always going to increase in the number of protons and that increased whole is going to pull those outer electrons in closer, therefore decreasing the size of the atom." + }, + { + "Q": "at 9:07, wouldn't it be that the neutral atom is smaller than the cation? If not, how come?\n", + "A": "Cations are positively charged because they lost an electron, so they have fewer electrons than protons and therefore the pull toward to nucleus is stronger per electron. The cation is therefore smaller than the neutral atom.", + "video_name": "rKoIcgBM4Vg", + "timestamps": [ + 547 + ], + "3min_transcript": "Let's look at ionic radius now. And ionic radius can be kind of complicated depending on what chemistry you are involved in. So this is going to be just a real simple version. If I took a neutral lithium atom again, so lithium-- so we've drawn this several times. Let me go ahead and draw it once more. So we have our lithium nucleus, which we have three electrons. So once again I'll go ahead and sketch in our three electrons real fast. Two electrons in the inner shell, and one electron in the outer shell like that. And let's say you were going to form a cation, so we are going to take away an electron from our neutral atom. So we have-- let me go ahead and draw this in here-- we had a three protons in the nucleus and three electrons those cancel each other out to be a neutral atom. And if we were to take away one of those electrons, so let's go ahead and show lithium losing an electron. So if lithium loses an electron, it's So the nucleus still has a plus 3 charge, because it has three protons in it. And we still have our two inner shell electrons like that, but we took away that outer shell electron. So we took away this electron in magenta, so let me go ahead and label this. So we lost an electron, so that's this electron right here, and so you could just show it over here like that. And by doing so, now we have three positive charges in our nucleus and only two electrons. And so therefore our lithium gets a plus 1 charge. So it's Li plus, it's a cation. And so we formed a cation, which is smaller than the neutral atom itself. And that just makes intuitive sense. If you take away this outer electron, now you have three positive charges in the nucleus and only two electrons here. So it's pulling those electrons in, you lost that outer electron, it's getting smaller. And so we've seen that neutral atoms will shrink when you convert them to cations, so it kind of makes sense that if you take a neutral atom and add an electron, it's going to get larger. And so that's our next concept here. So if we took something like chlorine, so a neutral chlorine atom, and we added an electron to chlorine, that would give it a negative charge. So we would get chlorine with a negative charge, or the chloride anion, I should say. And so in terms of sizes, let's go ahead and draw a representative atom here. So if this is our neutral chlorine atom and we add an electron to it, it actually gets a lot bigger. So the anion is bigger than the neutral atom. And let's see if we can think about why here. So if we were to draw an electron configuration, or to write a noble gas electron configuration for the neutral chlorine-- so you should already know how to do this-- you would just" + }, + { + "Q": "at 4:40 he says insulator is charged, how can an insulator get charged , i wont get an electric shock from it if i touch it\n", + "A": "Yes you can get an electric shock from an non-conducting material. It is commonly known that if you walk on a carpet with socks on, and then tough a doorknob you feel a shock. Basically you are being shocked by the charge accumulated by your socks travelling through the point of contact. This is called electrostatic discharge or ESD. On a grander scale, lightning is also an example of ESD caused by charge build up.", + "video_name": "ZgDIX2GOaxQ", + "timestamps": [ + 280 + ], + "3min_transcript": "and this side of the atom would be more positive. Even though the electron doesn't move, and the electrons don't move, now because this is set up where the positive is shifted from the negative, this material, if you get all of them to do this or a lot of them, this can create an overall electrical effect where this insulator can interact with other charges nearby and exert forces on them. Even though the charges can't flow through an insulator, they can still interact electrically. Now, let's see what happens if we add extra charge to these insulators or conductors. I mean, the way they started off right here we had just as many positives in the nucleus as there are negatives surrounding them and that's true for the conductors and insulators. What happens if we add extra charge? Maybe we add extra negatives into here. Then what happens? Well, it'll get really messy if we try to draw it with all the atoms, so since these all cancel out their overall charge, I am not going to draw every atom and nucleus. I'm just going to pretend like those are there I'm just going to draw the actual extra charge. Let's say we added extra negative charges to this insulator. What would happen? Let's say I just add a negative charge here and a negative charge there, and here and there, I have added a bunch of negative charges to this insulator. What would happen? Well, we know these negatives can't move throughout and insulator. Charges can't flow through an insulator so they're stuck which means for an insulator, I could charge the whole thing uniformly if I wanted to where the charge is spread out throughout the whole thing or I could make them bunch up on one side if I wanted to and they'd be stuck there. The point is that they're stuck. For a conductor, what would happen if I tried to put a negative here and a negative there, some extra negative charge on a conductor? They don't have to stay here if they don't want to. If you put extra negatives in here, they are not going to want to because negatives repel each other So what are they going to do? Well, this negative is going to try to get as far away from this other negative as it can so go over here. This negative is going to try to get as far away as it can. It repels it. Now, it can't jump off the conductor. That takes a lot more energy, but it can go to the very edge. That's what charges do for conductors. You've got a solid conducting material, you put extra charge on it, it's all... All that charge is going to reside on the outside edge whether you've added extra negative or positive, always on the outside edge. You can only add charge to the outside edge for a conductor, because if it wasn't on the outside edge it will quickly find its way to the outside edge because all these negatives repel each other. I said this is true for positives or negative. You might wonder, \"How do we add a positive?\" Well, the way you add a positive is by taking away a negative. If you started off with a material that had just as many positives as negatives and you took away a negative," + }, + { + "Q": "\nIn the example at 10:0, what would happen if the can wasnt connected through a wire to a metal but they were just touching. Would a transfer of electrons still occur?", + "A": "yes but there would be virtually no resistance", + "video_name": "ZgDIX2GOaxQ", + "timestamps": [ + 600 + ], + "3min_transcript": "Charge by induction says alright, first imagine I just take this and I bring it nearby but don't touch it. Just bring it near by this other piece of metal and I don't touch it. What would happen? There is negatives in here, I haven't drawn them. There's positives in here. The negatives can move if they wanted to. Do they want to? Yeah, they want to! These negatives are coming nearby, they want to get as far away from them as possible. Even though there are already some negatives here, a net amount of negatives are going to get moved over to this side. They were located with their atom on this side, but they want to get away from this big negative charge so they can move over here, which leaves a total amount of positive charge over here. I.E. There is a deficit of electrons over here, so this side ends up positively charged. You might think, \"Okay, well that's weird. \"Does anything else happen?\" Yeah because now these positives are closer to the negatives and these positives in this charge rod are attracting these positives. These negatives in this conducting rod are attracting these positive charges because like charges repel and opposites attract but they are also repelling. These negatives in this rod are repelling these negatives. Do those forces cancel? They actually don't because the closer you are to the charge the bigger the force. This would cause this rod to get attracted to the other rod. That's kind of cool. If you took a charged rod, brought it to an empty soda can, let that can sit on the table in this orientation so it could roll, if you bring the rod close the can will start moving towards the rod. It's kind of cool, you should try it if you can. But, that's not charge by induction. Charge by induction is something more. It says alright, take this piece of metal and conduct it to ground. What's ground? If you took a big metal pipe and stuck it in the ground that would count, or any other huge supply of electron, a place where you can gain, steal, basically take infinitely many electrons or deposit infinitely many electrons and this ground would not care. So the frame of your car, the actual metal, is a good ground because it can supply a ton of electrons or take them. Or a metal pipe in the earth. Some place you can deposit electrons or take them and that thing won't really notice or care. Now what would happen? If I bring this negative rod close to this rod that was originally had no net charge? Now instead of going to the other side of this, they say \"Hey, I can just leave. \"Let me get the heck out of here.\" These negatives can leave. A whole bunch of negatives can start leaving and what happens when that happens is that your rod is no longer uncharged. It has a net amount of charge now. They won't all leave. You're not going to get left with no electrons in here." + }, + { + "Q": "\nat 6:30 he says most plastics are insulator, can you give me an example of a plastic which is not an insulator?", + "A": "exposure of few plastics in ionic beam shows conductance like metals, and some are made semi conductors.", + "video_name": "ZgDIX2GOaxQ", + "timestamps": [ + 390 + ], + "3min_transcript": "I'm just going to draw the actual extra charge. Let's say we added extra negative charges to this insulator. What would happen? Let's say I just add a negative charge here and a negative charge there, and here and there, I have added a bunch of negative charges to this insulator. What would happen? Well, we know these negatives can't move throughout and insulator. Charges can't flow through an insulator so they're stuck which means for an insulator, I could charge the whole thing uniformly if I wanted to where the charge is spread out throughout the whole thing or I could make them bunch up on one side if I wanted to and they'd be stuck there. The point is that they're stuck. For a conductor, what would happen if I tried to put a negative here and a negative there, some extra negative charge on a conductor? They don't have to stay here if they don't want to. If you put extra negatives in here, they are not going to want to because negatives repel each other So what are they going to do? Well, this negative is going to try to get as far away from this other negative as it can so go over here. This negative is going to try to get as far away as it can. It repels it. Now, it can't jump off the conductor. That takes a lot more energy, but it can go to the very edge. That's what charges do for conductors. You've got a solid conducting material, you put extra charge on it, it's all... All that charge is going to reside on the outside edge whether you've added extra negative or positive, always on the outside edge. You can only add charge to the outside edge for a conductor, because if it wasn't on the outside edge it will quickly find its way to the outside edge because all these negatives repel each other. I said this is true for positives or negative. You might wonder, \"How do we add a positive?\" Well, the way you add a positive is by taking away a negative. If you started off with a material that had just as many positives as negatives and you took away a negative, But again, the net positive charge, the net negative charge always resides on the outside edge of the conductor because charges try to get as far away from each other as possible. So what physical materials actually do this? What physical materials are insulators? These are things like glass is an insulator. Wood is an insulator. Most plastics are insulators. All of these display this kind of behavior where you can distribute charge and the charge can't flow through it. You can stick charge on it. In fact, you can stick charge on the outside edge and it will stay there. There's conductors. These are things like metals, like gold or copper is typically used because it's kind of cheap. Cheaper than gold, certainly. Or any other metal. Silver works very well. These are materials where charges can flow freely through them. Now that we see how conductors and insulators work," + }, + { + "Q": "\nAt 8:26, why did you add the area of the triangle and the rectangle together? Wouldn't the displacement only be the area of the triangle because the base of the triangle is where the initial velocity is?", + "A": "The initial velocity causes displacement, too, doesn t it? If the velocity didn t change, wouldn t there still be displacement?", + "video_name": "MAS6mBRZZXA", + "timestamps": [ + 506 + ], + "3min_transcript": "and we can use a little symbol of geometry to break it down into two different areas, it's very easy to calculate their areas two simple shapes, you can break it down to two, blue part is the rectangle right over here, easy to figure out the area of a rectangle and we can break it down to this purple part, this triangle right here easy to figure out the area of a triangle and that will be the total distance we travel even this will hopefully make some intuition because this blue area is how far we would have travel if we are not accelerated, we just want 5m/s for 4s so you goes 5m/s 1s 2s 3s 4s so you are going from 0 to 4 you change in time is 4s so if you go 5m/s for 4s you are going to go 20 m this right here is 20m this purple or magentic area tells you how furthur than this are you going because you are accelerating because kept going faster and faster and faster it's pretty easy to calculate this area the base here is still 5(4) because that's 5(4) second that's gone by what's the height here? The height here is my final velocity minus my initial velocity minus my initial velocity or it's the change in velocity due to the accleration 13 minus 5 is 8 or this 8 right over here it is 8m/s so this height right over here is 8m/s the base over here is 4s that's the time that past what's this area of the triangle? the area of this triangle is one half times the base which is 4s times 8m/s second cancel out one half time 4 is 2 times 8 is equal to 16m So the total distance we travel is 20 plus 16 is 36m that is the total I could say the total displacement and once again is to the right, since it's positive so that is our displacement What I wanna do is to do the exact the same calculation keep it in variable form, that will give another formula many people often memorize You might understand this is completely intuitive formula and that just come out of the logical flow of reasoning that we went through this video what is the area once again if we just think about the variables? well the area of this rectangle right here is our initial velocity" + }, + { + "Q": "at 8:05, it says that the area of the triangle is 1/2 x 4 x 8. in the other videos it says that the area of the triangle is one half times base times height. why?\n", + "A": "that s exactly the SAME 1/2x 4 x 8 = HALF= 1/2 BASE= 4 HIEGHT= 8", + "video_name": "MAS6mBRZZXA", + "timestamps": [ + 485 + ], + "3min_transcript": "and we can use a little symbol of geometry to break it down into two different areas, it's very easy to calculate their areas two simple shapes, you can break it down to two, blue part is the rectangle right over here, easy to figure out the area of a rectangle and we can break it down to this purple part, this triangle right here easy to figure out the area of a triangle and that will be the total distance we travel even this will hopefully make some intuition because this blue area is how far we would have travel if we are not accelerated, we just want 5m/s for 4s so you goes 5m/s 1s 2s 3s 4s so you are going from 0 to 4 you change in time is 4s so if you go 5m/s for 4s you are going to go 20 m this right here is 20m this purple or magentic area tells you how furthur than this are you going because you are accelerating because kept going faster and faster and faster it's pretty easy to calculate this area the base here is still 5(4) because that's 5(4) second that's gone by what's the height here? The height here is my final velocity minus my initial velocity minus my initial velocity or it's the change in velocity due to the accleration 13 minus 5 is 8 or this 8 right over here it is 8m/s so this height right over here is 8m/s the base over here is 4s that's the time that past what's this area of the triangle? the area of this triangle is one half times the base which is 4s times 8m/s second cancel out one half time 4 is 2 times 8 is equal to 16m So the total distance we travel is 20 plus 16 is 36m that is the total I could say the total displacement and once again is to the right, since it's positive so that is our displacement What I wanna do is to do the exact the same calculation keep it in variable form, that will give another formula many people often memorize You might understand this is completely intuitive formula and that just come out of the logical flow of reasoning that we went through this video what is the area once again if we just think about the variables? well the area of this rectangle right here is our initial velocity" + }, + { + "Q": "@4:27 why did Sal cancel the \u00e2\u0080\u009cseconds\u00e2\u0080\u009d and the \u00e2\u0080\u009cseconds squared\u00e2\u0080\u009d to make it just \u00e2\u0080\u009cseconds\u00e2\u0080\u009d? I thought we were multiplying. Wouldn\u00e2\u0080\u0099t it be \u00e2\u0080\u009cseconds cubed\u00e2\u0080\u009d? PLEASE HELP!\n", + "A": "If you have a number, let s say X^2 and you have that X^2 divided by X, you end up with just having X as the answer because X^2 is the same as X*X, So, if you divide X*X by X, you cancel out one X in the numerator, and the X in the denominator. Hope that helps!", + "video_name": "MAS6mBRZZXA", + "timestamps": [ + 267 + ], + "3min_transcript": "So that is my time axis, time this is velocity This is my velocity right over there and I'm starting off with 5m/s, so this is 5m/s right over here So vi is equal to 5m/s And every second goes by it goes 2m/s faster that's 2m/s*s every second that goes by So after 1 second when it goes 2m/s faster it will be at 7 another way to think about it is the slope of this velocity line is my constant accleration, my constant slope here so it might look something like that So what has happend after 4s? So 1 2 3 4 this is my delta t So my final velocity is going to be right over there so this is v this is my final velocity what would it be? Well I'm starting at 5m/s So we are doing this both using the variable and concretes Some starting with some initial velocity I'm starting with some initial velocity Subscript i said i for initial and then each second that goes by I'm getting this much faster so if I gonna see how much faster have I gone I multiply the number of second, I will just multiply the number second it goes by times my acceleration, times my acceleration and once again, this right here, subscript c saying that is a constant acceleration, so that will tell my how fast I have gone If I started at this point and multiply the duration time with slope I will get this high, I will get to my final velocity I'm just taking this to make it concrete in your mind you have 5m/s plus 4s plus, I wanna do it in yellow plus 4s times our acceleration with 2m per second square and what is this going to be equal to? you have a second that is cancelling out one of the second down here You have 4 times, so you have 5m/s plus 4 times 2 is 8 this second gone, we just have 8m/s or this is the same thing as 13m/s which is going to be our final velocity and I wanna take a pause here, you can pause and think about it yourself this whole should be intuitive, we are starting by going with 5m/s every second goes by we are gonna going 2m/s faster" + }, + { + "Q": "\nat 8:15 with the answer equaling 77664439914.3 my calculator gets 7.77*10^-52 is there a fix around this?", + "A": "With an answer of that magnitude different, you must have typed something into your calculator wrong. If you type the following into Google it gets pretty close to Jay s answer (not exact because he hasn t rounded the first number) 3.53E-20 * 2 / 9.11E-31", + "video_name": "vuGpUFjLaYE", + "timestamps": [ + 495 + ], + "3min_transcript": "to free the electron and so we've exceeded that minimum amount of energy, and so we will produce a photoelectron. So, this photon is high-energy enough to produce a photoelectron. So let's go ahead and find the kinetic energy of the photoelectron that's produced. So we're gonna use this equation right up here. So let me just go and get some more room, and I will rewrite that equation. So we have the kinetic energy of the photoelectron, kinetic energy of the photoelectron, is equal to the energy of the photon, energy of the photon, minus the work function. So let's plug in our numbers. The energy of the photon was 3.78 times 10 the negative 19 joules, and then the work function is right up here again, it's 3.43, so minus 3.43 times 10 to the negative 19 joules. So let's get out the calculator again. 3.43 times 10 to the negative 19 and we get 3.5 times 10 to the negative 20. So let's go ahead and write that. This is equal to 3.5 times 10 to the negative 20 joules. This is equal to the kinetic energy of the photoelectron, and we know that kinetic energy is equal to one half mv squared. The problem asked us to solve for the velocity of the photoelectron. So all we have to do is plug in the mass of an electron, which is 9.11 times 10 to the negative 31st kilograms, times v squared. This is equal to 3.5 times 10 to the negative 20. So, let's do that math. So we take 3.5 times 10 to the negative 20, we multiply that by 2, 9.11 times 10 to the negative 31st, and this gives us that number, which we need to take the square root of. So, square root of our answer gives us the velocity of the electron, 2.8 times 10 to the 5th. So if you look at your decimal place here, this'll be one, two, three, four, five, so 2.8 times 10 to the 5th meters per second. So here's the velocity of the photoelectron produced, 2.8 times 10 to the 5th meters per second, and if you increased the intensity of this light, so you had more photons, they would produce more photoelectrons. So one photon knocks out one photoelectron if it has enough energy to do so. So let's think about this same problem," + }, + { + "Q": "(about 0:30): If shining a light of the right frequency and wavelength on a substance can knock electrons loose, therefore creating ions, wouldn't that change the properties of the substance? Is the requirement for light to free electron a minimum amount of energy, or is it a range? Does it have anything to do with why things bleach in the sun, or why, in some museums, there is no flash-photography allowed?\n", + "A": "When an electron is knocked loose, the remaining material is positively charged, which means you can t knock too many electrons loose because it will get harder and harder for them to move away. The positively charged object will attract electrons from the surroundings to neutralize itself. So the photoelectric effect is not going to make any significant change to the material. It has nothing to do with bleaching, which is a process of chemical breakdown due to the incoming energy from the sun.", + "video_name": "vuGpUFjLaYE", + "timestamps": [ + 30 + ], + "3min_transcript": "- Sometimes light seems to act as a wave, and sometimes light seems to act as a particle. And, an example of this, would be the Photoelectric effect, as described by Einstein. So let's say you had a piece of metal, and we know the metal has electrons. I'm gonna go ahead and draw one electron in here, and this electron is bound to the metal because it's attracted to the positive charges in the nucleus. If you shine a light on the metal, so the right kind of light with the right kind of frequency, you can actually knock some of those electrons loose, which causes a current of electrons to flow. So this is kind of like a collision between two particles, if we think about light as being a particle. So I'm gonna draw in a particle of light which we call a photon, so this is massless, and the photon is going to hit this electron, and if the photon has enough energy, it can free the electron, right? So we can knock it loose, and so let me go ahead and show that. So here, we're showing the electron being knocked loose let's just say, this direction, with some velocity, v, and if the electron has mass, m, we know that there's a kinetic energy. The kinetic energy of the electron would be equal to one half mv squared. This freed electron is usually referred to now as a photoelectron. So one photon creates one photoelectron. So one particle hits another particle. And, if you think about this in terms of classical physics, you could think about energy being conserved. So the energy of the photon, the energy that went in, so let me go ahead and write this here, so the energy of the photon, the energy that went in, what happened to that energy? Some of that energy was needed to free the electron. So the electron was bound, and some of the energy freed the electron. I'm gonna call that E naught, the energy that freed the electron, and then the rest of that energy must have gone into the kinetic energy of the electron, kinetic energy of the photoelectron that was produced. So, kinetic energy of the photoelectron. So let's say you wanted to solve for the kinetic energy of that photoelectron. So that would be very simple, it would just be kinetic energy would be equal to the energy of the photon, energy of the photon, minus the energy that was necessary to free the electron from the metallic surface. And this E naught, here I'm calling it E naught, you might see it written differently, a different symbol, but this is the work function. Let me go ahead and write work function here, and the work function is different for every kind of metal. So, it's the minimum amount of energy that's necessary to free the electron, and so obviously that's going to be different depending on what metal you're talking about. All right, let's do a problem. Now that we understand the general idea of the Photoelectric effect, let's look at what this problem asks us." + }, + { + "Q": "\nAt 9:30, Why do non bonding electrons take up more space than bonding electrons?", + "A": "Bonding electrons have to spend most of their tine between two nuclei. Nonbonding electrons are attracted to a nucleus only from one side, so they are free to wander further away.", + "video_name": "0na0xtIHkXA", + "timestamps": [ + 570 + ], + "3min_transcript": "We put sulfur in the center here. We know sulfur is bonded to 4 fluorines. So we put our fluorines around like that. And let's see how many valence electrons we've shown so far- 2, 4, 6, and 8. So 34 minus 8 gives us 26 valence electrons we still need to account for on our dot structure." + }, + { + "Q": "\nAt 9:25 are these resonant structures? Is the second structure possible as later at 11:58 it is said that VSEPR theory predicts first one more?", + "A": "They aren t resonance structures, they re two possible ways to arrange the electron pairs. The molecule will do whichever is more stable.", + "video_name": "0na0xtIHkXA", + "timestamps": [ + 565, + 718 + ], + "3min_transcript": "We put sulfur in the center here. We know sulfur is bonded to 4 fluorines. So we put our fluorines around like that. And let's see how many valence electrons we've shown so far- 2, 4, 6, and 8. So 34 minus 8 gives us 26 valence electrons we still need to account for on our dot structure." + }, + { + "Q": "\nAt 5:49, Sal explains just out of the blue how we reduced 2NAD+'s into 2NADH's when we went from the 2 pyruvates to Acetil-CoA, but nowhere in the previous minutes he explains how that came to happen. Any help? :(", + "A": "Glucose has 12 hydrogens. Pyruvate has 3 hydrogens (X2=6 for two molecules). 4H from the glucose molecule go to produce the 2 NADH in the first step (2H are needed to produce one NADH). The remaining 2H are used with 2H from two coenzymeA molecules (H-S-CoA) to produce 2 NADH when pyruvate is converted to acetyl CoA.", + "video_name": "9zoS5WGsmpc", + "timestamps": [ + 349 + ], + "3min_transcript": "then the acetyl group, bonding with that sulfur, and by doing that, you form acetyl-CoA. And acetyl-CoA, just so you know, you only see three letters here, but this is actually a fairly involved molecule. This is actually a picture of acetyl-CoA, I know it's really small, but hopefully you'll appreciate that it's a more involved molecule. That, the acetyl group that we're talking about is just this part, right over here, and it's a coenzyme. It's really acting to transfer that acetyl group, and we'll see that in a second. But it's also fun to look at these molecules, because once again, we see these patterns over and over again in biology or biochemistry. Acetyl-CoA, you have an adenine right over here. It's hard to see, but you have a ribose, and you also have two phosphate groups. So this end of the acetyl-CoA is essentially, is essentially an ADP. But it's used as a coenzyme. Everything that I'm talking about, this is all going to be facilitated by enzymes, and the enzymes will have cofactors, coenzymes, if we're talking about organic And as we see, the acetyl group joins on to the coenzyme A, forming acetyl-CoA, but that's just a temporary attachment. The acetyl-CoA is, essentially, gonna transfer the acetyl group over to, and now we're going to enter into the citric acid cycle. It's gonna transfer these two carbons over to oxaloacetic acid, to form citric acid. So it's gonna transfer these two carbons to this one, two, three, four carbon molecule, to form a one, two, three, four, five, six carbon molecule. But before we go into the depths of the citric acid cycle, I wanna make sure that I don't lose track of my accounting, because, even that step right over here, where we decarboxylated the pyruvate, we went from pyruvate to acetyl-CoA, that also reduced some NAD to NADH. Now, this is gonna happen once for each pyruvate, but we're gonna- all the accounting we're gonna say, is for one glucose molecule. gonna happen for each of the pyruvates. So this is going to be times- This is going to be times two. So we're gonna produce two, two NADH's in this step, going from pyruvate to acetyl-CoA. Now, the bulk of, I guess you could say, the catabolism, of the carbons, or the things that are eventually going to produce our ATP's, are going to happen in what we call the citric acid, or the Kreb cycle. It's called the citric acid cycle because, when we transferred the acetyl group from the coenzyme A to the oxaloacetic acid, we formed citric acid. And citric acid, this is the thing that you have in lemons, or orange juice. It is this molecule right over here. And the citric acid cycle, it's also called the Kreb cycle, when you first learn it, seems very, very complex, and some could argue that it is quite complex. But I'm just gonna give you an overview of what's going on. The citric acid, once again, six-carboned, it keeps getting broken down, through multiple steps," + }, + { + "Q": "At 3:05 Sal says that the most compact state of water is liquid. Why?\n", + "A": "Because when water turns into a solid, it forms crystals, and those crystals take up a little more volume than when the water molecules are allowed to mix freely without forming crystals.", + "video_name": "zjIVJh4JLNo", + "timestamps": [ + 185 + ], + "3min_transcript": "" + }, + { + "Q": "\n~4:00 he says the CL is smaller than the Na. Why is it a smaller molecule? Doesn't it have more electrons?", + "A": "as we go across the periodic table nuclear charge as well as no. of electrons increase BUT increase in nuclear charge dominates over increase in no. of electrons so overall atomic radius decreases and this is the case with Cl and Na.", + "video_name": "zjIVJh4JLNo", + "timestamps": [ + 240 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 3:21, Sal said that some molecules will be hitting the wall and would have a change in momentum. Does the net energy of the molecules change with the change in momentum?", + "A": "Of course energy changes if momentum changes.", + "video_name": "tQcB9BLUoVI", + "timestamps": [ + 201 + ], + "3min_transcript": "in that direction. They all have their own little velocity vectors, and they're always constantly bumping into each other, and bumping into the sides of the container, and ricocheting here and there and changing velocity. In general, especially at this level of physics, we assume that this is an ideal gas, that all of the bumps that occur, there's no loss of energy. Or essentially that they're all elastic bumps between the different molecules. There's no loss of momentum. Let's keep that in mind, and everything you're going to see in high school and on the AP test is going to deal with ideal gases. Let's think about what pressure means in this context. A lot of what we think about pressure is something pushing on an area. If we think about pressure here-- let's pick an arbitrary area. Let's take this side. Let's take this surface of its container. generated onto this surface? It's going to be generated by just the millions and billions and trillions of little bumps every time-- let me draw a side view. If this is the side view of the container, that same side, every second there's always these little molecules of gas moving around. If we pick an arbitrary period of time, they're always ricocheting off of the side. We're looking at time over a super-small fraction of time. And over that period of time, this one might end up here, this one maybe bumped into it right after it ricocheted and came here, this one changes momentum and goes like that. This one might have already been going in that direction, But what's happening is, at any given moment, since there's so many molecules, there's always going to be some molecules that are bumping into the side of the wall. When they bump, they have a change in momentum. All force is change in momentum over time. What I'm saying is that in any interval of time, over any period or any change in time, there's just going to be a bunch of particles that are changing their momentum on the side of this wall. That is going to generate force, and so if we think about how many on average-- because it's hard to keep track of each particle individually, and when we did kinematics and stuff, we'd keep track of the individual object at play. But when we're dealing with gases and things on a macro" + }, + { + "Q": "At 7:21 he says \"they have the same kinetic energy.\" But in order to \"squeeze\" the box, at least one wall must move inward, and while moving it will hit some of the particles and make them move faster. So wouldn't the kinetic energy increase (both the energy of some individual particles and the average energy)? Maybe this is negligible (especially if we squeeze the box very slowly)?\n", + "A": "We don t squeeze the box it s more like transferring the particles from a larger box to a smaller box keeping the kinetic energy same.", + "video_name": "tQcB9BLUoVI", + "timestamps": [ + 441 + ], + "3min_transcript": "With that out of the way, let me give you a formula. I hope by the end of this video you have the intuition for why this formula works. In general, if I have an ideal gas in a container, the pressure exerted on the gas-- on the side of the container, or actually even at any point within the gas, because it will all become homogeneous at some point-- and we'll talk about entropy in future videos-- but the pressure in the container and on its surface, times the volume of the container, is equal to some constant. We'll see in future videos that that constant is actually proportional to the average kinetic energy of the molecules bouncing around. That should make sense to you. If the molecules were moving around a lot faster, then you would have more kinetic energy, and then they would be changing momentum on the sides of the surface a lot more, so you would have more pressure. why pressure times volume is a constant. Let's say I have a container now, and it's got a bunch of molecules of gas in it. Just like I showed you in that last bit right before I erased, these are bouncing off of the sides at a certain rate. Each of the molecules might have a different kinetic energy-- it's always changing, because they're always transferring momentum to each other. But on average, they all have a given kinetic energy, they keep bumping at a certain rate into the wall, and that determines the pressure. What happens if I were able to squeeze the box, and if I were able to decrease the volume of the box? in it, but I squeeze. I make the volume of the box smaller-- what's going to happen? I have the same number of molecules in there, with the same kinetic energy, and on average, they're moving with the same velocities. So now what's going to happen? They're going to be hitting the sides more often-- at the same time here that this particle went bam, bam, now it could go bam, bam, bam. They're going to be hitting the sides more often, so you're going to have more changes in momentum, and so you're actually going to have each particle exert more force on each surface. Because it's going to be hitting them more often in a given amount of time. The surfaces themselves are smaller. You have more force on a surface, and on a smaller surface, you're going to have higher pressure." + }, + { + "Q": "at 8:51 they have written cos60 equals 3^1/2/2but actually cos60 equals 1/2\n", + "A": "Yup he took the value wrong.", + "video_name": "KDHuWxy53uM", + "timestamps": [ + 531 + ], + "3min_transcript": "product-- taking the dot product, to the force and the distance factor. And we know that the definition is the magnitude of the force vector, which is 100 newtons, times the magnitude of the distance vector, which is 10 meters, times the cosine of the angle between them. Cosine of the angle is 60 degrees. So that's equal to 1,000 newton meters times cosine of 60. Cosine of 60 is what? It's square root of 3 over 2. Square root of 3 over 2, if I remember correctly. So times the square root of 3 over 2. So the 2 becomes 500. So it becomes 500 square roots of 3 joules, whatever that is. I don't know 700 something, I'm guessing. Maybe it's 800 something. I'm not quite sure. But the important thing to realize is that the dot It applies to work. It actually calculates what component of what vector goes in the other direction. Now you could interpret it the other way. You could say this is the magnitude of a times b cosine of theta. And that's completely valid. And what's b cosine of theta? Well, if you took b cosine of theta, and you could work this out as an exercise for yourself, that's the amount of the magnitude of the b vector that's going in the a direction. So it doesn't matter what order you go. So when you take the cross product, it matters whether you do a cross b, or b cross a. But when you're doing the dot product, it doesn't matter what order. So b cosine theta would be the magnitude of vector b that goes in the direction of a. So if you were to draw a perpendicular line here, b cosine theta would be this vector. That would be b cosine theta. The magnitude of b cosine theta. So you could say how much of vector b goes in the same Or you could say how much of vector a goes in the same direction is vector b? And then multiply the two magnitudes. And now, this is, I think, a good time to just make sure you understand the difference between the dot product and the cross product. The dot product ends up with just a number. You multiply two vectors and all you have is a number. You end up with just a scalar quantity. And why is that interesting? Well, it tells you how much do these-- you could almost say-- these vectors reinforce each other. Because you're taking the parts of their magnitudes that go in the same direction and multiplying them. The cross product is actually almost the opposite. You're taking their orthogonal components, right? The difference was, this was a a sine of theta. I don't want to mess you up this picture too much. But you should review the cross product videos. And I'll do another video where I actually compare and But the cross product is, you're saying, let's multiply the magnitudes of the vectors that are perpendicular to each other, that aren't going in the same direction, that are" + }, + { + "Q": "At 6:38, why is that proton acidic?\n", + "A": "The carbon atom is sp hybridized (50% s character). Since s electrons are held more tightly to the carbon nucleus, they are further from the hydrogen nucleus. The H atom is not tightly held, so it is more easily removed. The proton is acidic.", + "video_name": "_-I3HdmyYfE", + "timestamps": [ + 398 + ], + "3min_transcript": "is going to interact with that negatively charged carbanion like that. So that's so that's the first reaction, Formation of your alkynide anion. And then if you want to do an alkylation, it's a separate reaction. You take this, and let's react it with the ethyl bromide. So CH3CH2Br. If you think about what's going to happen, the lone pair of electrons on the carbon is going to attack this carbon, the one that's bonded to your halogen, like that. The halogen is going to leave, and you're going to put this alkyl group onto your alkyne. So you're going to end up with an ethyl group on your alkyne. So let's go ahead and draw that. So we have hydrogen and then carbon triple bonded to another carbon, and then we have to put our alkyl group on there. So a CH2CH3, like that. So we've alkylated our alkyne. This is a very useful reaction for organic synthesis. So let's take the molecule we just made, So if I took this-- let me go ahead and redraw it over here. So if I took this alkyne, so we just formed this. And let's react it. Let's first react it with our base again. So let's use sodium amide right here. And in our second step, we'll react it with a primary alkyl halide. So let's go ahead and draw a primary alkyl halide here, so that is our molecule. So we think to ourselves, what happens? I have a strong base. I still have an acidic proton left on my alkyne, right? So the proton over here on the left. So that's what the base is going to do. The base is going to take that proton forming a negatively charged carbanion, an alkynide anion. And then that anion is going to be our nucleophile for an SN2 reaction. So when you're thinking about it, these electrons in here that are going to be on that carbon giving a negative 1 formal charge are going to come all the way over to our carbon. So let's go ahead and draw the products of that. We're going to have our benzene ring. So let's go ahead and draw our benzene ring here, so let's put in our electrons going around my benzene ring. And then on that benzene ring is a CH2. So that CH2 is the red one that we marked right here, and this is the alkyl group that gets put onto your alkyne. So let's just go ahead and finish drawing our alkyne here. So we have now our triple bonds, right? Carbon triple bonded to another carbon. And then our ethyl group. So CH2CH3. So you'll see in later videos how we use the acidity of terminal alkynes to alkylate when we do a few different synthesis problems." + }, + { + "Q": "8:37 Howcome theres not actual recorded history we learn about, on humans from 200,000 years ago?\n", + "A": "We have not found any records that go that far back. The earliest writing we have discovered is from around 3200 BCE (about 5,200 years ago).", + "video_name": "MS7x2hDEhrw", + "timestamps": [ + 517 + ], + "3min_transcript": "believe, a huge rock, a six-mile in diameter rock, colliding with what is now the Yucatan Peninsula in Mexico, or right off the coast of the Yucatan Peninsula. And it destroyed all of the large land life forms, especially the dinosaurs. And to put all of this in perspective-- and actually the thing that really was an aha moment for me-- it's, OK, plants are 450 million years ago. Grass, I kind of view as this fundamental thing in nature. But grass has only been around for about-- I've seen multiple estimates-- 40 to 70 million years. Grass is a relatively new thing on the planet. Flowers have only been around for 130 million years. So there was a time where you had dinosaurs, but you did not have flowers and you did not have grass. And so you fast forward all the way. And so when you look at this scale, it's kind of funny to look at this. This is the time period where the dinosaurs showed up. This whole brown line is where the mammals showed up. And then, of course, the dinosaurs died out here. Our ancestors, when the giant rock hit the Earth, must have been boroughed in holes and were able to stash some food away, or who knows what, and didn't get fully affected. I'm sure most of the large mammals were destroyed. But what's almost-- it's humbling, or almost humorous, or almost ridiculous, when you look at this chart is they put a little dot-- you can't even see it here, They say 2 million years ago, the first humans-- and even this is being pretty generous when they say first humans. These are really the first prehumans. The first humans that are the same as us, if you took one of those babies and your brought them up in the suburbs and gave them haircuts and stuff, they would be the same thing as we are, those didn't exist until 200,000 years ago, give or take. 200,000 to 400,000 years ago, I've seen estimates. So this is actually a very generous period of time to say first humans. It's actually 200,000 years ago. we are and how new evolution is, it was only 5 million years ago-- and I mentioned this in a previous video-- it was only 5 million years ago-- so this is just to get a sense. This is 0 years. Homo sapien sapien, only around for 200,000 years. The Neanderthals, they were cousin species. They weren't our ancestors. Many people think they were. They were a cousin species. We come from the same root. Although there are now theories that they might have remixed in with Homo sapiens. So maybe some of us have some Neanderthal DNA. And it shouldn't be viewed as an insult. They had big brains. Well, they didn't necessarily have big brains. They had big heads. But that seems to imply a big brain. But who knows? We always tend to portray them as somehow inferior. But I don't want to get into the political correctness of how to portray Neanderthals. But anyway, this is a very small period of time, 200,000. If you go 2 million years, then you" + }, + { + "Q": "\nAt 3:39, Sal says there is only one Snowball Earth. However, the chart says there are two.", + "A": "He probably only mentioned one.", + "video_name": "MS7x2hDEhrw", + "timestamps": [ + 219 + ], + "3min_transcript": "layer to build up. Ozone is just three oxygen atoms. It is O3. And by the end of the Proterozoic Eon-- so we're talking, I don't know, maybe 550 million years ago, give or take tens of, or hundreds, or maybe 100 million years-- these are all moving targets-- the ozone layer was dense enough to protect the land from UV rays. We talked about that in the last video, that the Earth is being bombarded with UV rays. And the ozone layer is the only thing that really keeps us from being seriously irradiated by the Sun and allows land animals to actually live. And so coinciding with that time period, around 550 million years ago, you start to have life colonizing, especially significant life, colonizing land. So life colonizes land, colonizes the land. And this was kind of an interesting-- when I first You always assume that kind of trees and grasses are kind of part of the background. They come part and parcel with land. But it turns out that animals colonized land before plants did. Plants didn't come into the picture until about 450 million years ago, give or take a few tens of millions of years. And so we're now entering the end of the Proterozoic Eon. Life has started to colonize land. We now have an ozone layer. And what happens-- and actually there's another snowball glaciation or a snowball Earth near the end of the Proterozoic Era, Eon, I should say. And there's a bunch of theories about why it came about. And then why disappeared. Maybe there were volcanoes, greenhouse gases, who knows. But as we enter the end of that, we start seeing life began to flourish. And it starts to really flourish as we enter the Phaner-- I always have trouble And it's not even labeled here. The Phanerozoic Eon is this chunk of time right over here. And let me write it out. So this right over here is the Phanerozoic, the Phanerozoic Eon. And so this chart, these divisions right here are eons. And then they jump into, instead of doing eons here, they then break into eras. Eras are subsets of eons. They are hundreds of millions of years. So this is the Paleozoic Era, the Mesozoic Era, and the Cenozoic Era. And that's actually our current era. But perhaps the most interesting-- well, I don't want to pick favorites here. But it's one of the most interesting times in the geologic era-- is the first period in the Paleozoic Era, which is the first era in the Phanerozoic Eon." + }, + { + "Q": "\nAt 2:00, when Sal explains, why is there in the velocity formula, an \"s\" instead of a \"d\" can the letter \"s\" stand for space to be covered? in other words moving from one point to another?", + "A": "the letters don t matter. You can define any variable to be anything you want. Focus on the concepts.", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 120 + ], + "3min_transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time." + }, + { + "Q": "Isn't there any other way to convert it? 7:26 Maths seems to be easier here in Brazil...\n", + "A": "The conversion method is certified by the system international units so if we use it it would be better and can be applied on almost every sum used in maths and physics. by the way no math is difficult or easy in any country provided the way u think lol,", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 446 + ], + "3min_transcript": "and you can treat the units the same way you would treat the quantities in a fraction. 5/1 kilometers per hour, and then to the north. Or you could say this is the same thing as 5 kilometers per hour north. So this is 5 kilometers per hour to the north. So that's his average velocity, 5 kilometers per hour. And you have to be careful, you have to say \"to the north\" if you want velocity. If someone just said \"5 kilometers per hour,\" they're giving you a speed, or rate, or a scalar quantity. You have to give the direction for it to be a vector quantity. You could do the same thing if someone just said, You could have said, well, his average speed, or his rate, would be the distance he travels. The distance, we don't care about the direction now, is 5 kilometers, and he does it in 1 hour. His change in time is 1 hour. So this is the same thing as 5 kilometers per hour. So once again, we're only giving the magnitude here. This is a scalar quantity. If you want the vector, you have to do the north as well. Now, you might be saying, hey, in the previous video, we talked about things in terms of meters per second. Here, I give you kilometers, or \"kil-om-eters,\" depending on how you want to pronounce it, kilometers per hour. What if someone wanted it in meters per second, or what if I just wanted to understand how many meters he travels in a second? And there, it just becomes a unit conversion problem. And I figure it doesn't hurt to work on that right now. So if we wanted to do this to meters per second, Well, the first step is to think about how many meters we are traveling in an hour. So let's take that 5 kilometers per hour, and we want to convert it to meters. So I put meters in the numerator, and I put kilometers in the denominator. And the reason why I do that is because the kilometers are going to cancel out with the kilometers. And how many meters are there per kilometer? Well, there's 1,000 meters for every 1 kilometer. And I set this up right here so that the kilometers cancel out. So these two characters cancel out. And if you multiply, you get 5,000. So you have 5 times 1,000. So let me write this-- I'll do it in the same color-- 5 times 1,000. So I just multiplied the numbers. When you multiply something, you can switch around the order. Multiplication is commutative-- I always" + }, + { + "Q": "\nAt 11:16 in the video , sal calculates 5000/3600 which is somewhere 1.38888888888889. so my question is how did that 1.888888888889 become 1.39 ?", + "A": "when we estimate 1.38888888 ,we round it off to 1.39 because it is better to write it shorter.if in case you had to multiply it with another number,it would be difficult .For example, 1.38888888888888.....*19 will be never ending ,so we do it.And by the way its 1.3888 not 1.88888", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 676 + ], + "3min_transcript": "are in an hour. So that's your gut check. We should get a smaller number than this when we want to say meters per second. But let's actually do it with the dimensional analysis. So we want to cancel out the hours, and we want to be left with seconds in the denominator. So the best way to cancel this hours in the denominator is by having hours in the numerator. So you have hours per second. So how many hours are there per second? Or another way to think about it, 1 hour, think about the larger unit, 1 hour is how many seconds? Well, you have 60 seconds per minute times 60 minutes per hour. The minutes cancel out. 60 times 60 is 3,600 seconds per hour. or if you flip them, you would get 1/3,600 hour per second, or hours per second, depending on how you want to do it. So 1 hour is the same thing as 3,600 seconds. And so now this hour cancels out with that hour, and then you multiply, or appropriately divide, the numbers right here. And you get this is equal to 5,000 over 3,600 meters per-- all you have left in the denominator here is second. Meters per second. And if we divide both the numerator and the denominator-- I could do this by hand, but just because this video's already getting a little bit long, let me get my trusty calculator out. I get my trusty calculator out just for the sake of time. as 50 divided by 36, that is 1.3-- I'll just round it over here-- 1.39. So this is equal to 1.39 meters per second. So Shantanu was traveling quite slow in his car. Well, we knew that just by looking at this. 5 kilometers per hour, that's pretty much just letting the car roll pretty slowly." + }, + { + "Q": "at 2:20 he says you use d for calculus, but don't you put an arrow on top for displacement? Do you use an arrow for the dirrivative operation too?\n", + "A": "Haha no, it just means that our alphabet ran out of letters, and it is generally a bad idea to use one letter for 2 purposes.", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 140 + ], + "3min_transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time." + }, + { + "Q": "\nAt 7:23 Sal says km is going to cancel out with the denominators but the numerator is meters.\nCould someone pls. help", + "A": "What Sal meant was at the denominator, the unit was also km. in direct variation, you cancel out the units to change a value of a particular unit into a different value with a different unit. In other words, Sal was saying that in 5 km X 1000 m/ 1 km, the kilometers will cancel out, and you will be left with just the meters. He then multiplied it and came out with 5000 meters.", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 443 + ], + "3min_transcript": "and you can treat the units the same way you would treat the quantities in a fraction. 5/1 kilometers per hour, and then to the north. Or you could say this is the same thing as 5 kilometers per hour north. So this is 5 kilometers per hour to the north. So that's his average velocity, 5 kilometers per hour. And you have to be careful, you have to say \"to the north\" if you want velocity. If someone just said \"5 kilometers per hour,\" they're giving you a speed, or rate, or a scalar quantity. You have to give the direction for it to be a vector quantity. You could do the same thing if someone just said, You could have said, well, his average speed, or his rate, would be the distance he travels. The distance, we don't care about the direction now, is 5 kilometers, and he does it in 1 hour. His change in time is 1 hour. So this is the same thing as 5 kilometers per hour. So once again, we're only giving the magnitude here. This is a scalar quantity. If you want the vector, you have to do the north as well. Now, you might be saying, hey, in the previous video, we talked about things in terms of meters per second. Here, I give you kilometers, or \"kil-om-eters,\" depending on how you want to pronounce it, kilometers per hour. What if someone wanted it in meters per second, or what if I just wanted to understand how many meters he travels in a second? And there, it just becomes a unit conversion problem. And I figure it doesn't hurt to work on that right now. So if we wanted to do this to meters per second, Well, the first step is to think about how many meters we are traveling in an hour. So let's take that 5 kilometers per hour, and we want to convert it to meters. So I put meters in the numerator, and I put kilometers in the denominator. And the reason why I do that is because the kilometers are going to cancel out with the kilometers. And how many meters are there per kilometer? Well, there's 1,000 meters for every 1 kilometer. And I set this up right here so that the kilometers cancel out. So these two characters cancel out. And if you multiply, you get 5,000. So you have 5 times 1,000. So let me write this-- I'll do it in the same color-- 5 times 1,000. So I just multiplied the numbers. When you multiply something, you can switch around the order. Multiplication is commutative-- I always" + }, + { + "Q": "Can you simplify velocity? At about 7:20, Sal stated 5 km per hour is the velocity. Usually, in physics, do you simplify final answers or leave it in their original measurements, the question asked with?\n", + "A": "we should probably simplify them because we usually count them per 1 hour.", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 440 + ], + "3min_transcript": "and you can treat the units the same way you would treat the quantities in a fraction. 5/1 kilometers per hour, and then to the north. Or you could say this is the same thing as 5 kilometers per hour north. So this is 5 kilometers per hour to the north. So that's his average velocity, 5 kilometers per hour. And you have to be careful, you have to say \"to the north\" if you want velocity. If someone just said \"5 kilometers per hour,\" they're giving you a speed, or rate, or a scalar quantity. You have to give the direction for it to be a vector quantity. You could do the same thing if someone just said, You could have said, well, his average speed, or his rate, would be the distance he travels. The distance, we don't care about the direction now, is 5 kilometers, and he does it in 1 hour. His change in time is 1 hour. So this is the same thing as 5 kilometers per hour. So once again, we're only giving the magnitude here. This is a scalar quantity. If you want the vector, you have to do the north as well. Now, you might be saying, hey, in the previous video, we talked about things in terms of meters per second. Here, I give you kilometers, or \"kil-om-eters,\" depending on how you want to pronounce it, kilometers per hour. What if someone wanted it in meters per second, or what if I just wanted to understand how many meters he travels in a second? And there, it just becomes a unit conversion problem. And I figure it doesn't hurt to work on that right now. So if we wanted to do this to meters per second, Well, the first step is to think about how many meters we are traveling in an hour. So let's take that 5 kilometers per hour, and we want to convert it to meters. So I put meters in the numerator, and I put kilometers in the denominator. And the reason why I do that is because the kilometers are going to cancel out with the kilometers. And how many meters are there per kilometer? Well, there's 1,000 meters for every 1 kilometer. And I set this up right here so that the kilometers cancel out. So these two characters cancel out. And if you multiply, you get 5,000. So you have 5 times 1,000. So let me write this-- I'll do it in the same color-- 5 times 1,000. So I just multiplied the numbers. When you multiply something, you can switch around the order. Multiplication is commutative-- I always" + }, + { + "Q": "Can someone please explAin the h2o part which sal talks about from 3:00. Thanks\n", + "A": "What he s saying is that in each H-O bond two electrons are being shared. However, oxygen, being more electronegative than hydrogen, grabs more than its fair share of the electron density in the bond. (You can think of it as a tug-of-war between oxygen and hydrogen, with oxygen winning by dragging more electron density its way.) This means that the bonds are polarised, with oxygen having a partial negative charge and each of the hydrogens a partial positive charge.", + "video_name": "Rr7LhdSKMxY", + "timestamps": [ + 180 + ], + "3min_transcript": "So how badly wants to hog, and this is an informal definition clearly, hog electrons, keep the electrons, to spend more of their time closer to them then to the other party in the covalent bond. And this is how, how much they like electrons, or how much affinity they have towards electrons. So how much they want electrons. And you can see that these are very, these are very related notions. This is within the context of a covalent bond, how much electron affinity is there? Well this, you can think of it as a slightly broader notion, but these two trends go absolutely in line with each other. And to think about, to just think about electronegativity makes it a little bit more tangible. Let's think about one of the most famous sets of covalent bonds, and that's what you see in a water molecule. you have an oxygen atom, and you have two hydrogens. Each of the hydrogen's have one valence electron, and the oxygen has, we see here, at it's outermost shell, it has one, two, three, four, five, six valence electrons. One, two, three, four, five, six valence electrons. And so you can imagine, hydrogen would be happy if it was able to somehow pretend like it had another electron then it would have an electron configuration a stable, first shell that only requires two electrons, the rest of them require eight, hydrogen would feel, hey I'm stable like helium if it could get another electron. And oxygen would feel, hey I'm stable like neon if I could get two more electrons. And so what happens is they share each other's electrons. This, this electron can be shared in conjunction with this electron for this hydrogen. So that hydrogen can kind of feel like it's using it stabilizes the outer shell, or it stabilizes the hydrogen. And likewise, that electron could be, can be shared with the hydrogen, and the hydrogen can kind of feel more like helium. And then this oxygen can feel like it's a quid pro quo, it's getting something in exchange for something else. It's getting the electron, an electron, it's sharing an electron from each of these hydrogens, and so it can feel like it's, that it stabilizes it, similar to a, similar to a neon. But when you have these covalent bonds, only in the case where they are equally electronegative would you have a case where maybe they're sharing, and even there what happens in the rest of the molecule might matter, but when you have something like this, where you have oxygen and hydrogen, they don't have the same electronegativity. Oxygen likes to hog electrons more than hydrogen does. And so these electrons are not gonna spend an even amount of time. Here I did it kind of just drawing these, you know, these valence electrons as these dots. But as we know, the electrons are in this kind of blur around, around the," + }, + { + "Q": "9:05 Is nuclear envelope and nuclear membrane the same thing?\n", + "A": "Yes, they are just different terms. You could call my shirt clothing or a textile envelope, although most people say the first I prefer the latter.", + "video_name": "mMCcBsSAlF4", + "timestamps": [ + 545 + ], + "3min_transcript": "and now going down over here. Over here. And let me draw the microtubules that are really... well, I've said it multiple times, super involved in actual the movement going on. They're elongating, they're these motor proteins that are moving the chromosomes along, once again, they're connected at the kinetochores right over here. Connected at the kinetochores. Right over there. And now we're almost done, we're ready to move into telephase II. So we're now going to go into telephase II. Telephase II. Where my two cells are now becoming four cells, so telephase II, I'm gonna show the cytokinesis starting to happen. So telephase II. So turning into four cells, In this cell up here, I have this character, and has a little bit of magenta right over here. That's this right over there and then you have the shorter magenta one. And actually, they are starting to, they're starting to unravel into their chromatid form, so maybe I'll draw that a little bit, and then this one, right over here, is starting to unravel into its chromatid form. And so it this that one. Whoops, wanna do that in that magenta color. Starting to unravel into its chromatid form, I wanna do it over here, this one is starting to unravel. And so is this one. So is... (laughs) I'm having trouble changing colors. And so is that one, and then up here, this one's starting to unravel, this one over here, and... this longer, mostly magenta one Also starting to unravel. You start having your nuclear envelope formed again, so your nuclear envelope is forming again. Nuclear envelope is forming. Your microtubules are dissolving. Let me draw the centrosomes, they're outside of the nuclear envelope. Outside of the nuclear envelope. And of course, you're finally dividing the cells, your cytokinesis happens, so now you have your four, your four cells, each have a haploid number. They each have two chromosomes. Remember, you diploid number was four, the germ cell had four chromosomes, two pairs of homologous chromosomes. Now each of your resulting gametes, these are now gametes now, these are gametes, they have a haploid number. But we started with a haploid number at the beginning of meiosis II, so that's why meiosis II" + }, + { + "Q": "\nAt 7:04, how exactly do the glycolipids help the cell to be recognized or tagged? Do they form specific patterns across the cell membrane? Do they have specific components that signal different things?", + "A": "One example is that immune cells have receptors that can attach to them. So a cell with a type of glycolipid that isn t normally present in your body (the glycolipid is then an antigen) will attach to the immune cell, and so be identified as an invader.", + "video_name": "cP8iQu57dQo", + "timestamps": [ + 424 + ], + "3min_transcript": "this is a protein, this is a protein, and I just drew some blobs to be indicative of the variety of proteins. But the important thing to realize is, if we think of cells, there's all of this diversity. There's all of this complexity that is on, or embedded, inside of its membrane. So instead of just thinking of it as just kind of as a uniform phospholipid bilayer, there's all sorts of stuff, maybe if we view this as a cross-section, there's all sorts of stuff embedded in it and we see it right over here in this diagram. You could say there's a mosaic of things embedded in it. A mosaic is a picture made up of a bunch of different components of all different colors, and you can see that you have all different components here, different types of proteins. You have proteins like this, that go across the membrane. We call these transmembrane proteins, they're a special class of integral protein. You have integral proteins like this, that might only interact with one part of the bilayer while these kind of go across it. So this right over here, this is a glycolipid, which is fascinating. It lodges itself in the membrane because it has this lipid end, so that's going to be hydrophobic. It's going to get along with all of the other hydrophobic things, but then it has an end that's really a chain of sugars and that part is going to be hydrophilic, it's going to sit outside of the cell. And these chains of sugars, these are actually key for cell-cell recognition. Your immune system uses these to differentiate between which cells are the ones that are actually from my body, the ones I don't want to mess with, the ones I want to protect and which cells are the ones that are foreign, the ones that I might want to attack. When people talk about blood type, they're talking about, well, what type of specific glycolipids do you have on cells. And there's all sorts of, that's not all we're talking about when we talk about glycolipids as a way or to be tagged in different ways. So it's a fascinating thing that these chains of sugars can lead to such complex behavior, and frankly, such useful behavior, from our point of view. But you don't just want to have sugar chains on lipids, you also have sugar chains on proteins. This, right over here, is an example of a glycoprotein. And as you can see, when you put all this stuff together, you get a mosaic, and I'm actually not even done. You have things like cholesterol embedded. Cholesterol is a lipid, so it's going to sit in the hydrophobic part of the membrane and that actually helps with the fluidity of the membrane, making sure it's not too fluid or not too stiff. So this is cholesterol, right over there. So you see this mosaic of stuff, but what about the fluid part? And I just talked about cholesterol's value in making sure that it's just the right amount of fluidity. What's neat about this, is this isn't a rigid structure." + }, + { + "Q": "\nAt 4:49, he states the concentration of the Hydronium ions is 5.0 x 10^-14\nWhere did he get that figure from? Thanks", + "A": "Well, (1.0 * 10^-14) / (0.2) is the equal to 5.0 * 10^-14. All he did was divide both sides by 0.2 in order to isolate for the variable x . Hope this helps!", + "video_name": "gsu4gjrFApA", + "timestamps": [ + 289 + ], + "3min_transcript": "and magnesium hydroxide. Let's do this problem. Calculate the pH of a 0.20 molar solution of sodium hydroxide. Let's go ahead and write sodium hydroxide here is a strong base. You get 100% dissociation in water, so 100% dissociation. Sodium hydroxide consist of Na plus and OH minus so you get dissociation and you get the sodium cation and the hydroxide anion, so Na plus and OH minus in water. If we have .2 molar of NaOH, we're also going to get .20 molar of hydroxide ions in solutions and sodium hydroxide is a strong base and we have 100% dissociation. If the concentration of hydroxide ion is 0.2, and so one way to solve this problem is to use this equation. The concentration of hydronium times the concentration of hydroxide is equal to 1.0 times 10 to the negative 14 from an earlier video. We can plug in our concentration for hydroxide and we can say the concentration of hydronium is x. We have x times 0.20 is equal to 1.0 times 10 to the negative 14. If you do that math, all right, so if you do that math, x is equal to the concentration of hydronium ions. This would be five, right. This would be 5.0 times 10 to the negative 14. Now that we have the concentration of hydronium ions, we can now calculate the pH of our solution because pH is equal to negative log We can do this on our calculator here. Let's get out the calculator to find the pH. The pH is equal to the negative log of, this is 5.0 times 10 to the negative 14. We get a pH of 13.30. All right, so let's go ahead and write that down here. pH is equal to 13.30. There's another way to do this problem, all right, and that is when you find, let me use a different color for this. When you find the concentration of hydroxides, you can immediately find the pOH, right. The pOH is equal to the negative log of the concentration of hydroxide ions. All right, so that would be .2, so we could plug that into here to find the pOH. pOH is equal to the negative log of the concentration of our hydroxide and ion." + }, + { + "Q": "At 1:30, he says that the concentration of HNO3 andH3O are the same. Would the NO3- also have that same concentration of .03M? Why or why not?\n", + "A": "It would, we just don t care about it", + "video_name": "gsu4gjrFApA", + "timestamps": [ + 90 + ], + "3min_transcript": "- [Voiceover] Here we have some strong acids which ionize 100% in solution. HClO4 is perchloric acid. HCl is hydrochloric acid. HBr is hydrobromic acid. HI is hydriodic acid. H2SO4 is sulfuric acid and HNO3 is nitric acid. Let's do a calculation using nitric acid here. Calculate the pH of a 0.030 molar solution of nitric acid. Well nitric acid is a strong acid which means it ionizes 100% in solution. If we have HNO3 and H2O here, the nitric acid is gonna donate a proton to water and if water accepts a proton, water turns into H3O plus, the hydronium ion. If HNO3 loses a proton, we're left with NO3 minus or nitrate. Since nitric acid is strong, and everything turns into our products. If this is our concentration of HNO3, this would also be our concentration of hydronium ions. The concentration of hydronium ions is .030 molar since we're dealing with a strong acid. To calculate the pH, all we have to do is plug in to our definition of pH. The pH is equal to the negative log of the concentration of hydronium ions. All we have to do is plug this number into here so the pH is equal to the negative log of 0.030. Let's get out the calculator and do that. All right, so we have the negative log of .030. We get 1.52. so the pH is equal to 1.52. For our significant figures, we have two significant figures here so we have two to the right of our decimal point. Instead of writing it this way, all right, so instead of writing this, you could have written a shortened version. You could have just written HNO3, nitric acid ionizes 100% and so it turns into H plus and NO3 minus, so that's just a shortened version of the same thing we have above. If you're working with this, you could have said that the pH is equal to the negative log of the concentration of H+. That's the same thing as we did above. All right, next let's look at strong bases. We're only gonna talk about metal hydroxides as being strong bases in this video. Some common metal hydroxides, sodium hydroxide," + }, + { + "Q": "At 11:40, why do we not square the concentration of OH? From previous lessons, I remember we have to raise the concentration to the power of the molar ratio. So shouldn't it be 1.0*10^-14/(0.012)^2 for OH? Thanks\n", + "A": "That rule is only for finding the equilibrium constant. But here, when we are finding the pH, we are supposed to multiply it by 2 as we need to know the number of moles.", + "video_name": "gsu4gjrFApA", + "timestamps": [ + 700 + ], + "3min_transcript": "and let's do that calculation here. Let's get some room, we have .0030 divided by .250 and we get .012 molar. All right, so this is .012 molar for the concentration of hydroxide. Now that you have the concentration of hydroxide ions and solution, you can find the pH using one of the two ways that we talked about above. All right, so you could go for this equation, hydronium ion concentration times hydroxide ion concentration is equal to 1.0 times 10 to the negative 14. All right, and you could take this and plug it into here. You get x times 0.012 is equal to 1.0 times 10 to the negative 14 and let's get a little bit more room here We have 1.0 times 10 to the negative 14. We're gonna divide that by .0. Actually let me do that again. We're going to divide that by .012 and this is going to give us 8.3 times 10 to the negative 13. All right, so let's write this down, x is equal to 8.3 times 10 to the negative 13. Remember what x referred to, x referred to the concentration of hydronium ions so this is the concentration of hydronium ions which means we can now calculate the pH. All right, the pH be equal to the negative log of that concentration. The negative log of 8.3 times 10 to the negative 13. Let's do one more final calculation here. We're gonna do negative log 10 to the negative 13, and we're going to get 12.08. All right, so the pH is equal to 12.08 and we're finally done with our problem here. Notice the pH is in the base range, so calcium hydroxide is a base." + }, + { + "Q": "At 3:35 in the video, I attached my 2 hydrogens to the nitrogen at different places. I put one hydrogen on the top and one on the bottom. Is that incorrect? Would my way of drawing it be acceptable?\n", + "A": "Your way is equally correct to Sal s way. These dot structures show connectivity, but not shape, so it doesn t matter where you drawn in your H s attached to the nitrogen.", + "video_name": "BIZNBfBuu1w", + "timestamps": [ + 215 + ], + "3min_transcript": "that each hydrogen is surrounded by two electrons. And so if I find hydrogen here, hydrogen is in the first energy level. And so here's one electron and here's two electrons. So in the first energy level, there is only an s orbital. And so that s orbital holds a maximum of two electrons. And we get to the electron configuration of a noble gas. And so hydrogen is stable with having only two electrons around it. Let's look at another dot structure. And let's do one that has nitrogen in it. So if I look at the molecular formula CH3 NH2, I'm going to once again start with carbon in the center with its four valence electrons around it, like that. And I know that there are three hydrogens on that carbon. So I can go ahead and put in those three hydrogens. Each hydrogen has one valence electron, like that. And then on the right side, I'm going to think about nitrogen. So I need to find nitrogen on my periodic table. Nitrogen is in group V. Therefore, I can represent those valence electrons as one, two, three, four, and five, like that. And I still have two hydrogens to worry about, right? So I have still have these two hydrogens here. And I can see there's a place for them on the nitrogen. So I can go ahead and put a hydrogen in here and a hydrogen in here, and connect the dots. And I have my dot structure. And I can also check on my octet rule. So carbon has an octet. And nitrogen has an octet as well. So let's go ahead and verify that. So there's two electrons here, four, six, and eight. So nitrogen is in the second period. And so nitrogen is also going to follow the octet rule when you're drawing your dot structures. Let's do one with oxygen next. So if I wanted to draw the dot structure for methanol, methanol is CH3 OH. with the four valence electrons. And I have three hydrogens, each one with one valence electron, like that. And so I can go ahead and put in those three hydrogens. Next I have oxygen. So I need to find oxygen on my organic periodic table. And I can see that oxygen is in group VI right here. So oxygen is going to have six valence electrons around it. So I can go ahead and draw in oxygen. And I can put its six valence electrons in-- one, two, three, four, five, and six, like that. And then I'm going to put in the hydrogen, right? So now I have a hydrogen to worry about. And I know that hydrogen has one valence electron. So I can see there's a place for it over here. And once again, I can connect the dots and see all of the single covalent bonds in this molecule. So that's one bond. That's another bond." + }, + { + "Q": "How will I know which part will I put the hydrogen? For example @4:54. How did you know that the hydrogen is to be put at the right side of oxygen?\n", + "A": "It doesn t matter exactly where as long as it s bonded to the oxygen. It would be just as valid to put it on the top or bottom of the oxygen instead of right.", + "video_name": "BIZNBfBuu1w", + "timestamps": [ + 294 + ], + "3min_transcript": "I can represent those valence electrons as one, two, three, four, and five, like that. And I still have two hydrogens to worry about, right? So I have still have these two hydrogens here. And I can see there's a place for them on the nitrogen. So I can go ahead and put a hydrogen in here and a hydrogen in here, and connect the dots. And I have my dot structure. And I can also check on my octet rule. So carbon has an octet. And nitrogen has an octet as well. So let's go ahead and verify that. So there's two electrons here, four, six, and eight. So nitrogen is in the second period. And so nitrogen is also going to follow the octet rule when you're drawing your dot structures. Let's do one with oxygen next. So if I wanted to draw the dot structure for methanol, methanol is CH3 OH. with the four valence electrons. And I have three hydrogens, each one with one valence electron, like that. And so I can go ahead and put in those three hydrogens. Next I have oxygen. So I need to find oxygen on my organic periodic table. And I can see that oxygen is in group VI right here. So oxygen is going to have six valence electrons around it. So I can go ahead and draw in oxygen. And I can put its six valence electrons in-- one, two, three, four, five, and six, like that. And then I'm going to put in the hydrogen, right? So now I have a hydrogen to worry about. And I know that hydrogen has one valence electron. So I can see there's a place for it over here. And once again, I can connect the dots and see all of the single covalent bonds in this molecule. So that's one bond. That's another bond. And the oxygen has bonded to this hydrogen as well. Again, we can check our octet rule. So the carbon has eight electrons around it. And so does the oxygen. So this would be two right here, and then four, and then six, and then eight. So oxygen is going to follow the octet rule. Now when you're drawing dot structures, you don't always have to do this step where you're drawing each individual atom and summing all of your valence electrons that way. You can just start drawing it. So for an example, if I gave you C2 H6, which is ethane, another way to do it would just be starting to draw some bonds here. And so I have two carbons. And it's a pretty good bet those two carbons are going to be connected to each other. And then I have six hydrogens. And if I look at what's possible around those carbons, I could put those six hydrogens around those two carbon atoms, like that. And if I do that, I'll have an octet around each carbon atom." + }, + { + "Q": "\nI don't understand the part after 1:34 . The part about carbon with 8 electrons and hydrogen with 2", + "A": "Hydrogen can only have two electrons total but Carbon can have 8 valence electrons total.", + "video_name": "BIZNBfBuu1w", + "timestamps": [ + 94 + ], + "3min_transcript": "In this video, we're going to look at how to draw dot structures of simple organic molecules that have single bonds. So if I look at the molecular formula CH4, which is methane, and I want to draw a dot structure for the methane molecule, I would go over here to my organic periodic table and find carbon. And I can see carbon is in group IV. Therefore, carbon will have four valence electrons. So I can draw a carbon with its four valence electrons around it like that. Remember from general chemistry, valence electrons are the electrons in the outermost energy level. So carbon has four valence electrons in its outermost energy level. Next, I have to think about hydrogen. And hydrogen is in group I on the periodic table. Therefore, hydrogen will have one valence electron. And so I can go ahead and put a hydrogen in there with one valence electron. And I know I have to do that three more times. So I keep putting in hydrogens, each with one valence electron, so a total of four hydrogens. And now I can start connecting my dots. I know that two valence electrons equals one single covalent bond. There is a single covalent bond. And then I have two more here. So this is my complete dot structure for methane. Now I can see that carbon is surrounded by eight electrons So we can go ahead and highlight those. So if I'm counting the electrons around carbon, it would be two, four, six, and eight, like that. And eight electrons around carbon makes carbon very stable. And if we look at the periodic table, we can see why. So if I look at the second period, I can see that the valence electrons for carbon would be one, two, three, and four. And to get to eight electrons, we would go five, six, seven, eight. So if carbon is surrounded by eight electrons, it's like it has the electron configuration of a noble gas, which makes it very stable, because all of the orbitals in that energy level are now full. So an octet of electrons is the maximum number of electrons for carbon. that each hydrogen is surrounded by two electrons. And so if I find hydrogen here, hydrogen is in the first energy level. And so here's one electron and here's two electrons. So in the first energy level, there is only an s orbital. And so that s orbital holds a maximum of two electrons. And we get to the electron configuration of a noble gas. And so hydrogen is stable with having only two electrons around it. Let's look at another dot structure. And let's do one that has nitrogen in it. So if I look at the molecular formula CH3 NH2, I'm going to once again start with carbon in the center with its four valence electrons around it, like that. And I know that there are three hydrogens on that carbon. So I can go ahead and put in those three hydrogens. Each hydrogen has one valence electron, like that. And then on the right side, I'm going to think about nitrogen. So I need to find nitrogen on my periodic table. Nitrogen is in group V. Therefore," + }, + { + "Q": "At 5:49, Sal forgot to right hemoglobin!\n", + "A": "Everyone makes mistakes. :)", + "video_name": "xKJ3txXIuQk", + "timestamps": [ + 349 + ], + "3min_transcript": "to reproduce itself. But you immediately see on this picture how small the HIV virus is compared to the actual T-cell. Each of these small little things each of these small things, is an HIV virus, which we already saw is a lot bigger than something like a hemoglobin protein. And so a hemoglobin protein you wouldn't even be able to on this scale, maybe it would be a pixel, if that. And on a similar scale, is this T-cell, you have a, you have things like red blood cells. This is actually a comparison this side by side. This is using an electron, this is using an electron microscope you see a red blood cell right over here and they're roughly roughly on the same size, or at least the same order of magnitude size. And a red blood cell is going to be six to eight micrometers, micrometers wide. So this is six to eight millionths of a meter. So if we were to as the average. Seven millionths, seven seven millionths of a meter. Over here, we're talking about a millionth of a millimeter. Now we're talking about seven millionths of a meter. And just to get an appreciation for size, we already compared the virus, the HIV virus, to this cell. We're seeing it directly as a emerge from the cell. But each of these red blood cells are gonna contain roughly 280 million hemoglobin molecules. So there's gonna be 200 each of these there's gonna be 280 million of these. So 280 million, that's a million million hemoglobins in each one of these. So hopefully this starts to give you an appreciation for even though we categorize cells as these unimaginably small things, they're actually far larger they're ginormous compared to things And especially when you think of things on the molecular, or the atomic scale. And that's why cells are so interesting. They actually have a lot of complexity to them. But just to have an appreciation also for how small cells are even though we've just described these red blood cells and these T-cells there's these kind of worlds unto themselves. They are these incredibly complex things. If I were to draw the width of a human hair on this screen right now relative to the scale of these red blood cells, it would be about as wide as this video. So from if I were to draw a human hair it would go from there roughly to there and there's actually a lot of variance in the width of a human hair. But the width of a human hair would be just about like that. If you looked at the scale of if you looked at the scale of this picture right over here. If you looked at these scales it would be much much much bigger. And I encourage you the width of a human hair. Look at it. Put it on a piece of paper. It's hard to even discern the width." + }, + { + "Q": "\nAt 10:23, David talks about how voltage is the difference in Electric Potential. So is the analogy Volts:Electrical Potential::Impulse:Momentum correct in the sense that one is the measurement of the difference of the other?", + "A": "eh, sort of, but that s not a very useful way to think of it. Impulse is a transfer of momentum. Volts are units that we use to measure electrical potential difference. Voltage is the electrical engineers terms for what physicists call electric potential difference . You can subtract any two quantitites you want from one another. That doesn t mean that any time you choose two pairs of quantities to subtract that there s some interesting comparison to make between the differences.", + "video_name": "ks1B1_umFk8", + "timestamps": [ + 623 + ], + "3min_transcript": "per Coulomb of charge that you put there. And it works for any point, if I picked a point twice as close, it's half as far away, let's say some point over here, let's say this r value here was only 4.5 centimeters, well I'm dividing this by r, so if the r is half as big this point over here will have a V value of 200 Joules per Coulomb and the closer I get, if I went even closer, if I went to a point that was three centimeters away, well this is a third as much as this other distance, so if I'm only dividing by a third as much distance as you get three times the result 'cause r is not squared, it's just r. So at this point, we'll have a V value of 300 Joules per Coulomb. This tells me, if I wanted to get a charge that have a whole bunch of Potential energy, I should stick it over here, this will give me a lot of Potential energy. Not quite as much, even less, the further I put my charge the less Potential energy it will have. There will be no Potential energy until there is a charge, there'll just be Electric Potential. But once you place another charge in that region to go with the first one, then you'll have Electric Potential energy and this will be a way to find it, Q times the V that you get out of this calculation. You gotta be careful though, sometimes people get sloppy, and V looks, you know, we use V for Electric Potential and we use V for Voltage, what's the difference? Are they the same? Hmmm, not quite. Sometimes you can treat them as the same but sometimes you do and messes you up. Voltage is a, technically a change in Electric Potential between two points, between two points in space, so it's got the same units 'cause the change in Electric Potential still gonna have units of Joule per Coulomb, it's just, when it's a change in we give this a new title, we call the Joule per Coulomb unit a Volt. So Joules per Coulomb are Volts, but the word Voltage specifically refers to a difference in Electric Potential, what am I talking about? Well, look at, this point is 300 Joules per Coulomb, this point over here 100 Joules per Coulomb, so the delta V, if I were to take delta V between these two points right here and I ask, what's the difference in V? Well the difference in V is 200, 200 Joules per Coulomb, that means the Voltage between those two points in space is 200 Volts, that's what it means. So, when you're talking about a difference" + }, + { + "Q": "\nWhat differentiates the blue star 3:55 lifecycle from the red giant or main sequence stars? What makes them burn hotter and faster?", + "A": "Its all about mass. A massive star will be hotter and burn brighter. Every star was once main sequence. It does not matter how massive, as long as it is stably burning hydrogen, it is main sequence. Once a star runs out of hydrogen or leaves the main sequence, it becomes a red giant or supergiant.", + "video_name": "w3IKEa_GOYs", + "timestamps": [ + 235 + ], + "3min_transcript": "in fact is, just so you have the number, this thing is 7000 light years away 7000 light years away which means that what we are seeing now, the photons that are reaching our eyes or telescopes right now left this region of space 7000 years ago so we're seeing it as it was 7000 years ago so a lot of this gas, a lot of this hydrogen, may have already condensed into many many more stars so the structure might not be the way it looks right now and actually there was another super nova that happened that might have blown away a lot of this stuff and we won't even be able to see the effects of this super nova for another thousand years but anyway, this is just a pretty amazing photograph in my opinion especially, and its beautiful at any scale and it's even more mindblowing when you think that this is 7, this is a structure that is 7 light years tall one of the pillars of creation this right here is a star field, and this is as we're looking towards the center of our galaxy this is the Sagittarius star field the neat thing here you see is such a diversity in stars this is also kind of mind numbing because every one of these stars, are inside of our galaxy this is looking towards the center of our galaxy this isn't one of those where we're looking beyond our galaxy or looking at clusters of galaxies this is just stars here but the thing here is that you see a huge variety, you see some stars that are shining red, right over here and obviously, the apparent size, you cannot completely tell because the different stars are at different distances and at difference intensities but the redder stars, these are stars in their red giant phase or they're probably at their red giant phase i haven't done specific research on these stars but that's what we suspect those are in their red giant phase the ones that are kind of in the yellowish white part of the spectrum probably not too different than own sun the ones that are in the yellowish white, closer to orange-yellowish-white part of the spectrum and the ones that look a little more bluish, or a little bit more greenish these are burning super fast let me see if i can find, this one looks a little big bluish to me, these are burning super super fast, and so the super massive stars, they burn kind of fast and furious and then just die out but the smaller stars, the ones with less mass they burn slower over a much much longer period of time so the ones that are burning fast are emitting a lot of energy at the smaller wavelength part of the light spectrum that's why they look bluer or greener and these are going to be more massive stars the ones that look white or bluer or greener" + }, + { + "Q": "At 2:40 Sal mentions a supernova that may have blown away a portion of the Eagle Nebula, and that we wont be able to see the effects of it for another 1000 years. Why only 1000 years and not 7,000? Do we know that said supernova occured 6,000 years ago?\n", + "A": "Yes, from what we can see right now, which is the Eagle Nebula 7,000 years ago, a burst of energy from a nearby supernova is heading towards the Eagle Nebula. At such a rate, it has destroyed the Eagle Nebula 6,000 years ago.", + "video_name": "w3IKEa_GOYs", + "timestamps": [ + 160 + ], + "3min_transcript": "so this is an enormous amount of distance remember, the distance from earth to the nearest star was about 4 light years it would take voyager, if it were pointed in the right direction moving at 60 thousand kilometers per hour it would take Voyager 80 thousand years to go 4 light years just this pillar is 7 light years but i wanted to show you this because these type of nebulae, the plural of nebula are where stars can form. so this right here, you actually see, is actually a breeding ground for the birth of new stars this gas is condensing, just like we talked about a couple of videos ago. Until it gets to that critical temperature, the critical density, where you can actually get fusion of hydrogen so this is just a huge interstellar cloud of hydrogen gas and over here you can see its just this breeding ground for stars and we don't even, we think that this structure doesn't even exist anymore in fact is, just so you have the number, this thing is 7000 light years away 7000 light years away which means that what we are seeing now, the photons that are reaching our eyes or telescopes right now left this region of space 7000 years ago so we're seeing it as it was 7000 years ago so a lot of this gas, a lot of this hydrogen, may have already condensed into many many more stars so the structure might not be the way it looks right now and actually there was another super nova that happened that might have blown away a lot of this stuff and we won't even be able to see the effects of this super nova for another thousand years but anyway, this is just a pretty amazing photograph in my opinion especially, and its beautiful at any scale and it's even more mindblowing when you think that this is 7, this is a structure that is 7 light years tall one of the pillars of creation this right here is a star field, and this is as we're looking towards the center of our galaxy this is the Sagittarius star field the neat thing here you see is such a diversity in stars this is also kind of mind numbing because every one of these stars, are inside of our galaxy this is looking towards the center of our galaxy this isn't one of those where we're looking beyond our galaxy or looking at clusters of galaxies this is just stars here but the thing here is that you see a huge variety, you see some stars that are shining red, right over here and obviously, the apparent size, you cannot completely tell because the different stars are at different distances and at difference intensities but the redder stars, these are stars in their red giant phase or they're probably at their red giant phase i haven't done specific research on these stars but that's what we suspect those are in their red giant phase the ones that are kind of in the yellowish white part of the spectrum" + }, + { + "Q": "\n2:40 \"There was another supernova that happened that could have blown away some of this dust\".\nHow do you know it happened if you can't see it yet?", + "A": "Supernovae happen all the time, its very likely one has happened.", + "video_name": "w3IKEa_GOYs", + "timestamps": [ + 160 + ], + "3min_transcript": "so this is an enormous amount of distance remember, the distance from earth to the nearest star was about 4 light years it would take voyager, if it were pointed in the right direction moving at 60 thousand kilometers per hour it would take Voyager 80 thousand years to go 4 light years just this pillar is 7 light years but i wanted to show you this because these type of nebulae, the plural of nebula are where stars can form. so this right here, you actually see, is actually a breeding ground for the birth of new stars this gas is condensing, just like we talked about a couple of videos ago. Until it gets to that critical temperature, the critical density, where you can actually get fusion of hydrogen so this is just a huge interstellar cloud of hydrogen gas and over here you can see its just this breeding ground for stars and we don't even, we think that this structure doesn't even exist anymore in fact is, just so you have the number, this thing is 7000 light years away 7000 light years away which means that what we are seeing now, the photons that are reaching our eyes or telescopes right now left this region of space 7000 years ago so we're seeing it as it was 7000 years ago so a lot of this gas, a lot of this hydrogen, may have already condensed into many many more stars so the structure might not be the way it looks right now and actually there was another super nova that happened that might have blown away a lot of this stuff and we won't even be able to see the effects of this super nova for another thousand years but anyway, this is just a pretty amazing photograph in my opinion especially, and its beautiful at any scale and it's even more mindblowing when you think that this is 7, this is a structure that is 7 light years tall one of the pillars of creation this right here is a star field, and this is as we're looking towards the center of our galaxy this is the Sagittarius star field the neat thing here you see is such a diversity in stars this is also kind of mind numbing because every one of these stars, are inside of our galaxy this is looking towards the center of our galaxy this isn't one of those where we're looking beyond our galaxy or looking at clusters of galaxies this is just stars here but the thing here is that you see a huge variety, you see some stars that are shining red, right over here and obviously, the apparent size, you cannot completely tell because the different stars are at different distances and at difference intensities but the redder stars, these are stars in their red giant phase or they're probably at their red giant phase i haven't done specific research on these stars but that's what we suspect those are in their red giant phase the ones that are kind of in the yellowish white part of the spectrum" + }, + { + "Q": "\nAt 5:46, Isn't (10^6) ^2 (ten to the sixth power squared) supposed to be 36?", + "A": "Nope. (10^6)^2 is 10^12. Remeber, (a^m)^n = a^mn", + "video_name": "391txUI76gM", + "timestamps": [ + 346 + ], + "3min_transcript": "How many meters is that? It's 6 million meters, right? And then, you know, the extra meter to get to my center of mass, we can ignore for now, because it would be .001, so So it's 6-- and soon. I'll write it in scientific notation since everything else is in scientific notation-- 6.371 times 10 to the sixth 6,000 kilometers is 6 million meters. So let's write that down. So the distance is going to be 6.37 times 10 to the sixth meters. We have to square that. Remember, it's distance squared. So let's see if we can simplify this a little bit. Let's just multiply those top numbers first. Force is equal to-- let's bring the variable out. Mass of Sal times-- let's do this top part. So we have 6.67 times 5.97 is equal to 39.82. multiply the 10's. So 10 to the negative 11th times 10 to the negative 24th. We can just add the exponents. They have the same base. So what's 24 minus 11? It's 10 to the 13th, right? And then what does the denominator look like? It's going to be the 6.37 squared times 10 to the sixth squared. So it's going to be-- whatever this is is going to be like 37 or something-- times-- what's 10 to the sixth squared? It's 10 to the 12th, right? 10 to the 12th. So let's figure out what 6.37 squared is. This little calculator I have doesn't have squared, so I have to-- so it's 40.58. Sal times-- let's divide, 39.82 divided by 40.58 is equal to 9.81. That's just this divided by this. And then 10 to the 13th divided by 10 to the 12th. Actually no, this isn't 9.81. Sorry, it's 0.981. 0.981, and then 10 to the 13th divided by 10 to the 12th is just 10, right? 10 to the first, times 10, so what's 0.981 times 10? Well, the force is equal to 9.81 times the mass of Sal." + }, + { + "Q": "Dear Sal,\nat 3:09 of the video above, you said that you found out the Earth's mass on Wikipedia. My teachers always say not to trust wikipedia because it is based off of other people's information/opinions. After watching the whole video (which by the way was really helpful) I found out that the Earth's mass is 5.972 multiplied by 10 to the 24th with the unit as kilograms. On your video, did you just round up the decimal? I just want to make sure so I don't get it wrong on my test next week.\nSincerely,\nAva F.\n", + "A": "Teachers should stop telling students not to trust wikipedia. It is quite reliable for scientific information.", + "video_name": "391txUI76gM", + "timestamps": [ + 189 + ], + "3min_transcript": "So that's simple enough. So let's play around with this, and see if we can get some results that look reasonably familiar to us. So let's use this formula to figure out what the acceleration, the gravitational acceleration, is at the surface of the Earth. So let's draw the Earth, just so we know what we're talking about. So that's my Earth. And let's say we want to figure out the gravitational acceleration on Sal. That's me. And so how do we apply this equation to figure out how much I'm accelerating down towards the center of Earth or the Earth's center of mass? The force is equal to-- so what's this big G thing? The G is the universal gravitational constant. Although, as far as I know, and I'm not an expert on this, It's not truly, truly a constant, or I guess when on different scales, it can be a little bit different. But for our purposes, it is a constant, and the constant in most physics classes, is this: 6.67 times 10 to the negative 11th meters cubed per kilogram seconds squared. I know these units are crazy, but all you have to realize is these are just the units needed, that when you multiply it times a mass and a mass divided by a distance squared, you get Newtons, or kilogram meters per second squared. So we won't worry so much about the units right now. Just realize that you're going to have to work with meters in kilograms seconds. So let's just write that number down. I'll change colors to keep it interesting. 6.67 times 10 to the negative 11th, and we want to know the acceleration on Sal, so m1 is the mass of Sal. And I don't feel like revealing my mass in this And then what's the mass 2? It's the mass of Earth. And I wrote that here. I looked it up on Wikipedia. This is the mass of Earth. So I multiply it times the mass of Earth, times 5.97 times 10 to the 24th kilograms-- weighs a little bit, not weighs, is a little bit more massive than Sal-- divided by the distance squared. Now, you might say, well, what's the distance between someone standing on the Earth and the Earth? Well, it's zero because they're touching the Earth. But it's important to realize that the distance between the two objects, especially when we're talking about the universal law of gravitation, is the distance between their center of masses. For all general purposes, my center of mass, maybe it's like three feet above the ground, because I'm not that tall. It's probably a little bit lower than that, actually. Anyway, my center of mass might be three feet above the ground, and where's Earth's center of mass? Well, it's at the center of Earth, so we have to know the radius of Earth, right? So the radius of Earth is-- I also looked it up on" + }, + { + "Q": "\nAt 0:52, Rishi draws a rectangular human cell. I thought that human cells are rounder and plant cells are the rectangular ones. Am I wrong?", + "A": "Human cells have no specific shapes. It is just for diagramming.", + "video_name": "MNKXq7c3eQU", + "timestamps": [ + 52 + ], + "3min_transcript": "So let's talk about exactly how flu causes so much damage to ourselves and why it makes us feel so lousy whenever we get the flu. I'm going to start out by drawing the flu virus here. This is our influenza virus. And we have on influenza a couple of features we have to remember. So on the outside there's this little envelope, and what's on the inside of this envelope are eight bits of RNA. Eight pieces of RNA. And so this RNA is important to remember, because in the human cell, in our cells-- I'm going to draw one of our cells right here-- we have, instead of RNA, we have DNA. Remember. And so this is our nucleus, and on the inside of our nucleus is our DNA. So this is our DNA over here. So the virus has RNA, and we have DNA. And the outside of the human cell-- actually let me label this over here. This is human cell. something called sialic acid. They're these little strands over here that are coming off. I'm drawing them far larger than they are in real life. They're not nearly this big, but they're these little tiny little things called sialic acid. And this sialic acid becomes very important in understanding how the influenza virus gets into and out of our cells. So on the outside, remember, of the influenza virus, there were a couple of proteins. And I'm going to draw one of these proteins here, and I'm going to make it look like a hand. So this is a little hand, and this protein is called hemagglutinin. In fact, previously I had called it the H protein, and you can call it that if you want. But the full name is hemagglutinin. And what hemagglutinin does is that it actually holds onto sialic acid. In fact, that's an easy way to remember it, right? Because H and H go together. It holds sialic acid. allows it the first step towards getting into the cell. Now there's another protein on the outside here-- I'm going to make it look like a pair of scissors, because that will kind of remind us what this one does. And this is called neuraminidase. And I'm going to-- neuraminidase. And I'm going to pass on explaining what it does, just for the moment. I'll tell you in a little bit what it does. So then the first step to get into the cell is for hemagglutinin to hold on to sialic acid. And then there are a few other small molecular steps that happen, important ones. But I'm going to suffice to say it gets inside. And once the influenza virus gets inside, these RNA segments, they are let loose. So these segments are going to start making their way towards the nucleus. And so once they get into the nucleus, they're in that same kind of area that the DNA is," + }, + { + "Q": "\nAt 1:28 why did Sal say that Force/Mass=Acceleration?", + "A": "Because it does. F = m*a. Solve for a.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 88 + ], + "3min_transcript": "What I want to do with this video is think about what happens to some type of projectile, maybe a ball or rock, if I were to throw it straight up into the air. To do that I want to plot distance relative to time. There are a few things I am going to tell you about my throwing the rock into the air. The rock will have an initial velocity (Vi) of 19.6 meters per second (19.6m/s) I picked this initial velocity because it will make the math a little bit easier. We also know the acceleration near the surface of the earth. We know the force of gravity near the surface of the earth is the mass of the object times the acceleration. (let me write this down) The force of gravity is going to be the mass of the object times little g. little g is gravity near the surface of the earth g is 9.8 meters per second squared (9.8m/s^2) you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is" + }, + { + "Q": "\nWhere did the formula at 2:32 come from?", + "A": "its the formula to calculate gravity between 2 objects. m1 = mass object 1 (kg) m2 = mass object 2 (kg) r = distance between objects (m) G = 0.0000000000667 Your result is in m/s^2", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 152 + ], + "3min_transcript": "you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall" + }, + { + "Q": "\nAt 2:23, Sal said over the square of the distance between the two things but wrote : (r^2) but isn't the sign for distance d?", + "A": "Either is acceptable. It s just a different letter that represents the same thing.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 143 + ], + "3min_transcript": "What I want to do with this video is think about what happens to some type of projectile, maybe a ball or rock, if I were to throw it straight up into the air. To do that I want to plot distance relative to time. There are a few things I am going to tell you about my throwing the rock into the air. The rock will have an initial velocity (Vi) of 19.6 meters per second (19.6m/s) I picked this initial velocity because it will make the math a little bit easier. We also know the acceleration near the surface of the earth. We know the force of gravity near the surface of the earth is the mass of the object times the acceleration. (let me write this down) The force of gravity is going to be the mass of the object times little g. little g is gravity near the surface of the earth g is 9.8 meters per second squared (9.8m/s^2) you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is" + }, + { + "Q": "I'm just curious, in the equation at 3:17 is m2 referring to the mass of the projectile?\n", + "A": "Indeed it is. Since m1 is the mass of the Earth, m2 must be the mass of the other body interacting with the Earth i.e the mass of the projectile.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 197 + ], + "3min_transcript": "you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall" + }, + { + "Q": "\n4:00 so the force of gravity is accelerating the object downwards (makes sense) then if you were to be nosediving the force of gravity is pushing against you, or in other words upwards, causing \"g\" forces. Why is this? Or if I'm wrong can someone explain g forces, like in the previous video when the pilot of the fighter jet was pushed back by gravity?", + "A": "Gravity does not push against you, it pulls you down. When a pilot feels g forces, that s not really because of gravity, it s because of acceleration of his plane. It s called g force because it FEELS like he is getting heavier, but that s not because of gravity, it s because the seat of the plane is pushing on his backside.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 240 + ], + "3min_transcript": "The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall the force of gravity is downwards. So, little g over here, if you want to give it its direction, is negative. Little g is -9.8m/s2. So, we have the acceleration due to gravity. The acceleration due to gravity (ag) is negative 9.8 meters per second squared (9.8m/s^2). Now I want to plot distance relative to time. Let's think about how we can set up a formula, derive a formula that, if we input time as a variable, we can get distance. We can assume these values right over here. Well actually I want to plot displacement over time because that will be more interesting. We know that displacement is the same thing as average velocity times change in time (displacement=Vavg*(t1-t2)). Right now we have" + }, + { + "Q": "Why did Sal use average velocity as opposed to just velocity in the equation: displacement = average velocity multiplied by change in time in 4:54?\n", + "A": "Because the velocity is not constant through out the process. Acceleration is constant. If you plot a graph of Velocity vs. Time, you ll see more clearly why using avg. Velocity is pretty much the same as finding the area below the curve.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 294 + ], + "3min_transcript": "If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall the force of gravity is downwards. So, little g over here, if you want to give it its direction, is negative. Little g is -9.8m/s2. So, we have the acceleration due to gravity. The acceleration due to gravity (ag) is negative 9.8 meters per second squared (9.8m/s^2). Now I want to plot distance relative to time. Let's think about how we can set up a formula, derive a formula that, if we input time as a variable, we can get distance. We can assume these values right over here. Well actually I want to plot displacement over time because that will be more interesting. We know that displacement is the same thing as average velocity times change in time (displacement=Vavg*(t1-t2)). Right now we have but not in terms of initial velocity and acceleration. We know that average velocity is the same thing as initial velocity (vi) plus final velocity (vf) over 2. (Vavg=(vi+vf)/2) If we assume constant acceleration. We can only calculate Vavg this way assuming constant acceleration. Once again when were are dealing with objects not too far from the center of the earth we can make that assumption. Assuming that we have a constant acceleration Once again we don't have what our final velocity is. So, we need to think about this a little more. We can express our final velocity in terms of our initial velocity and time. Just dealing with this part, the average velocity. So we can rewrite this expression as the initial velocity plus something over 2. and what is final velocity?" + }, + { + "Q": "\nAt 3:00, Sal says, \"The little g, is really all of the business over here........\". Is it because m2's mass is really too small compared to the Earth's mass, we say: g = [ G X (Earth's mass) ]/r\u00c2\u00b2?\nI'm getting confused of to why we exclude the 'm2' in the equation?\n\nThanks,\nRamana", + "A": "We don t exclude it. We have F = GMm/r^2. We don t want to write all of that every time. GM/r^2 doesn t change as long as you stay on the surface of the earth. So we give it a new name, g, and we re-write that equation as F = mg. The little m is still there.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 180 + ], + "3min_transcript": "you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall" + }, + { + "Q": "\n9:00 what is a delta t", + "A": "delta means change in . Delta t means change in t.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 540 + ], + "3min_transcript": "All of this was another way to write average velocity. the whole reason why I did this is because we don't have final velocity but we have acceleration and we are going to use change in time as our independent variable. We still have to multiply this by this green change in time here. multiply all of this times the green change in time. All of this is what displacement is going to be. This is displacement, and lets see... we can multiply the change in time times all this actually these 2s cancel out and we get (continued over here) We get: displacement is equal to initial velocity times change in time change in time is a little more accurate plus 1/2 (which is the same as dividing by 2) plus one half times the acceleration times the acceleration times (we have a delta t times delta t) change in time times change in time the triangle is delta and it just means \"change in\" so change in time times change in time is just change in times squared. In some classes you will see this written as d is equal to vi times t plus 1/2 a t squared this is the same exact thing they are just using d for displacement and t in place of delta t. The one thing I want you to realize with this video Maybe if you were under time pressure you would want to be able to whip this out, but the important thing, so you remember how to do this when you are 30 or 40 or 50 or when you are an engineer and you are trying to send a rocket into space and you don't have a physics book to look it up, is that it comes from the simple displacement is equal to average velocity times change in time and we assume constant acceleration, and you can just derive the rest of this. I am going to leave you there in this video. Let me erase this part right over here. We are going to leave it right over here. In the next video we are going to use this formula we just derived. We are going to use this to actually plot the displacement vs time because that is interesting and we are going to be thinking about what happens to the velocity and the acceleration as we move further and further in time." + }, + { + "Q": "\nAt 3:15, what distance should one no longer assume that the radius (the earth in this case) is constant (say, assuming 2 or 3 significant digits)?", + "A": "g = GM/r^2 If you plug in a few numbers for r, you can decide for yourself when g varies enough from 9.8 so that you have to worry about changing r Note that even on the surface of earth, g varies in the second decimal place, due primarily to density variations but also due to changes in altitude", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 195 + ], + "3min_transcript": "you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall" + }, + { + "Q": "\nAt 9:15 shouldn't the systematic name be 1,3-bis(2-methylpropyl)cyclopentane ?", + "A": "but in that case the longest chain will be of 2 carbon atoms. whereas it is of 3 carbon atoms.", + "video_name": "6BR0Q5e74bs", + "timestamps": [ + 555 + ], + "3min_transcript": "with the core: cyclopentane. That's just a simple five-carbon ring. A five carbon ring that looks like a pentagon: one, two, three, four, five. There you go. That is a five-carbon ring. We can number it however we want, so one, two, three, four, and five. This is telling us at the one and the three position we have-- and the bis- is kind of redundant. This is saying we have two of these things. Obviously, we have two. We have one at the one and one at the three. So you can kind of ignore the bis-. That's just the convention and we've seen that multiple times. But at each of those positions, we have a 1,1-dimethylethyl. So what's a dimethylethyl look like? So let's think about it a little bit. Let's think about it and let me do it orange. They obviously named it using systematic naming and what we have here, we have an ethyl as kind of the core So if an ethyl is equal to two carbons, so this is two carbons right there. So let me draw a two carbon: one, two. That is two carbons right over there. I'm just drawing it at the three spot. I'll draw it also at the one spot, actually. So that is two carbons right there. That's the ethyl part. And then on 1,1, so if we number them, we number where it's connected, so it's one, two. This is saying 1,1-dimethyl. So on this ethyl chain, you have two methyls. Remember, methyl is equal to one, so this is one carbon. You have one carbon. That's what methyl is, but you have two of them. You have dimethyl. You have it twice at the one spot. So you have one methyl here and then you have another methyl there. You have 1-methyl on the one spot and then you have another 1-methyl on the one spot. And then you are connected at positions one and positions three, so you're connected there and you are connected And you're done, That's it. That is our structure. Now, if you did this with common naming, instead of this group being a 1,1-dimethylethyl, you might see that we're connected to a group that has one, two, three, four carbons in it. The carbon that we're connected to branches off to three other carbons. It is a tert-butyl. So you can also call this a 1,3-- let me just write it down. So another name for this would be 1,3-tert-- or sometimes people just write a t there-- t-butylcyclo-- no, actually I should say di-t-butyl, because we have two of them. 1,3-di-t-butylcyclopentane." + }, + { + "Q": "Wouldn't the structure at minute 5:20 be named 4,4,7,10-tetramethyldodecane? With this numbering we get a total of 25 vs. 27 with the 3,6,9,9-tetramethyldodecane.\n", + "A": "There is no rule that says we sum the numbers together, the rule is we are looking for the lowest number at the first point of difference. How you determine this is by making a list of the numbers and compare them one at a time until we find a point of difference: yours: 4,4,7,10 the video s: 3,6,9,9 As 3 is lower than 4, the numbering in the video is correct.", + "video_name": "6BR0Q5e74bs", + "timestamps": [ + 320 + ], + "3min_transcript": "It's three carbons, so it's going to be one, two, three, and the connection point to the main ring in this case is going to be in the middle carbon, so it kind of forms a Y. All of the isos, the isopropyl, isobutyl, they all look like Y's, so it's going to be linked right over here. That's also going to happen at the ninth carbon, so at the ninth carbon we're going to have another isopropyl. We're going to have another isopropyl at the ninth carbon. All right, we've taken care of the 2,9-isopropyl. Then we have the 6-ethyl, which is just a two carbon. Remember, meth- is one, eth- is two, prop- is three. Let me write this down. So this is going to be prop- is equal to three. Isoprop- is equal to that type of shape right over there. So at six we have an ethyl group, so one, two, carbons, and it's connected at the six carbon on the main ring. And then finally we have a cyclopentyl. So if we look at-- let me find a color I haven't used yet-- cyclopentyl. so pent- is five, but it's five in a cycle, so this is a five-carbon ring that's branching off of the main ring. It's at the first spot. Let me draw a five-carbon rings, so pent- is equal to five, so it would look like this, one: two, three, four, five. It looks just like a pentagon. That's a cyclopentyl group and it's attached to the one carbon on my cyclohexadecane, so it is attached just like that. ropylcyclohexadecane. Let's do another one. I think we're getting the hang of it. So here, maybe we can do this one a little bit faster. Let's see, we have a tetramethyldodecane, so the main root here is the dodecane, do- for two, dec- for ten. This is a 12-carbon chain. It's not in a cycle, so let me just draw it out. We have one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, and so we can just number them arbitrarily, just because I could have drawn this any which way. So it's one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. That's the dodecane, all single bonds. Then we have a 2,6,9,9-tetramethyl. All this is telling us-- remember, meth- is one carbon," + }, + { + "Q": "At 5:56 Sal says that you won't see meniscus in plastic because it doesn't have the same polarity as the glass. Does this mean there is no capillary action in a plastic straw?\n", + "A": "There is capillary action in plastic straw. Ever seen a cold drink with a straw? When you insert the straw into the drink..the liquid rises over a height. But capillary action in glass is more than that observed in plastic straws. All plastics have different abilities for adhesion.", + "video_name": "eQXGpturk3A", + "timestamps": [ + 356 + ], + "3min_transcript": "Because its partially positive end, its partially positive end at the hydrogens. Let me do it in that green color. The partially positive end at the hydrogens would be attracted to the partially negative ends of the oxygens in the glass. And so it'll stick to it. This is actually a stronger partial charge than what you would actually see in the water because there's a bigger electronegativity difference between the silicon and the oxygen in the glass than the oxygen and the hydrogen in the water. So these things just keep bumping around. Maybe there's another water molecule that just get knocked in the right way. All of a sudden for, you know, a very brief moment it gets knocked up here. And then it's going to stick to the glass. And this phenomenon of something sticking to its container, we would call that adhesion. So what you see going on here, that is called adhesion, adhesion. And adhesion is the reason why you also see the water a little bit higher there. we call that cohesion. And that's what the hydrogen bonds are doing inside the water. So this right over here, that over there, that is co-, that is cohesion. So that's why we have things, why we observe a meniscus like this. But there's even more fascinating properties of adhesion. If I were to take, if I were to take a container of water. If I were to take a container of water. And just to be clear what's going on here with the mercury, the mercury is more attracted to itself than it is to the glass container, so it bulges right over there. But let's go back to water. So let's say that this is a big tub of water. I fill it. So, I fill the water right over here. And let's say I take a glass tube, and the material matters. It has to be a polar material. That's why you'll see the meniscus in glass, but you might not see it or you won't see it if you were dealing with a plastic tube because the plastic does not have that polarity. But let's say you were to take a glass tube, So much thinner than even a beaker. So you take a thin glass tube and you stick it in the water, you will observe something very cool. And I encourage you to do this if you can get your hands on a very thin glass tube. You will notice that the water is actually going to defy gravity and start climbing up this thin glass tube. And so that's interesting. Why is that happening? Well this phenomenon which we call capillary action. Capillary, capillary action. The word capillary, it'll refer to anything from you know, a very, very narrow tube and we also have capillaries in our circulation system. Capillaries are our thinnest blood vessels, those are very, very, very, very thin. And there's actually capillary action inside of our capillaries. But what we're seeing here, this is called capillary, capillary action. And it's really just this adhesion occurring more intensely because more of the water molecules are able to come in touch with the polar glass lattice. And so you can imagine we have glass here. If you also had glass over here." + }, + { + "Q": "\nAt 5:56 Sal says that you won't see meniscus in plastic because it doesn't have the same polarity as the glass. Is this true for every kind of plastic?", + "A": "Not all plastics are exactly the same but they all have the same type of carbon-polymer-structure. As far as meniscus is concerned, none of them will cause a meniscus.", + "video_name": "eQXGpturk3A", + "timestamps": [ + 356 + ], + "3min_transcript": "Because its partially positive end, its partially positive end at the hydrogens. Let me do it in that green color. The partially positive end at the hydrogens would be attracted to the partially negative ends of the oxygens in the glass. And so it'll stick to it. This is actually a stronger partial charge than what you would actually see in the water because there's a bigger electronegativity difference between the silicon and the oxygen in the glass than the oxygen and the hydrogen in the water. So these things just keep bumping around. Maybe there's another water molecule that just get knocked in the right way. All of a sudden for, you know, a very brief moment it gets knocked up here. And then it's going to stick to the glass. And this phenomenon of something sticking to its container, we would call that adhesion. So what you see going on here, that is called adhesion, adhesion. And adhesion is the reason why you also see the water a little bit higher there. we call that cohesion. And that's what the hydrogen bonds are doing inside the water. So this right over here, that over there, that is co-, that is cohesion. So that's why we have things, why we observe a meniscus like this. But there's even more fascinating properties of adhesion. If I were to take, if I were to take a container of water. If I were to take a container of water. And just to be clear what's going on here with the mercury, the mercury is more attracted to itself than it is to the glass container, so it bulges right over there. But let's go back to water. So let's say that this is a big tub of water. I fill it. So, I fill the water right over here. And let's say I take a glass tube, and the material matters. It has to be a polar material. That's why you'll see the meniscus in glass, but you might not see it or you won't see it if you were dealing with a plastic tube because the plastic does not have that polarity. But let's say you were to take a glass tube, So much thinner than even a beaker. So you take a thin glass tube and you stick it in the water, you will observe something very cool. And I encourage you to do this if you can get your hands on a very thin glass tube. You will notice that the water is actually going to defy gravity and start climbing up this thin glass tube. And so that's interesting. Why is that happening? Well this phenomenon which we call capillary action. Capillary, capillary action. The word capillary, it'll refer to anything from you know, a very, very narrow tube and we also have capillaries in our circulation system. Capillaries are our thinnest blood vessels, those are very, very, very, very thin. And there's actually capillary action inside of our capillaries. But what we're seeing here, this is called capillary, capillary action. And it's really just this adhesion occurring more intensely because more of the water molecules are able to come in touch with the polar glass lattice. And so you can imagine we have glass here. If you also had glass over here." + }, + { + "Q": "Am I missing something? At 2:27 Sal says that each Silicon is paired with two Oxygen but the video shows four Oxygens for every Silicon (though two of the Oxygen are shared in a covalent bond).\n", + "A": "No, well Sal did not draw the complete picture here, Silicon dioxide is the complete molecule here and what you see above is a silicon dioxide lattice which is just many silicon dioxide forming bonds with each other. You saw silicon bonded to 4 oxygen bonds but actually that silicon is bonded to 2 of them and the other 2 oxygen is bonded to some other silicon atom.", + "video_name": "eQXGpturk3A", + "timestamps": [ + 147 + ], + "3min_transcript": "- If you were to take a glass beaker, so let me draw it right over here. If you were to take a glass beaker and you were to fill it up with water, you might expect that the surface of the water would be flat. But it's actually not the case and I encourage you to try it. You might have even observed this before. The surface of the water will not be flat. The surface of the water will actually be higher near the glass than it is when it's away from the glass. It forms a shape that looks something like that. And so the first thing we might ask is what'll we call this thing. And this right over here is called a meniscus. Meniscus. And in particular this meniscus, because the fluid is higher near the container than it is when you're away from the container, we would call this a concave, concave meniscus. And you might say, \"Well if this is a concave meniscus, \"are there any situations where might have \"a convex meniscus?\" Well sure, you can have a convex meniscus. If you were take that same glass beaker, instead of filling it with water If you filled it with mercury, you would get a meniscus that looks like this where there's a bulge near the center when you're further away from the container than when you're at the container. And so let me just label this. This is a convex, convex meniscus. But it's one thing to just observe this and to name them. To say, \"Hey this is a meniscus.\" So this is a concave meniscus. But a more interesting question is why does it actually happen. And so you might imagine this concave meniscus is because the fluid is more attracted to the container than it is to itself. And you might be saying, \"Wait, wait. \"Hold on, hold on a second here. \"We've been talking about how water \"has this polarity, it has partial negative end. \"Each water molecule has a partially negative \"and has partially positive ends at the hydrogens.\" So let me write this down. Partial positive charges at the hydrogens. And that causes this hydrogen bonding to form all of these special properties. \"You're telling me that it's more attracted to the glass than it is to itself?\" And I would say, \"Yes, I am telling you that.\" And you could imagine why it is going to be more attracted to the glass than itself, because glass actually has, the molecules in glass actually are quite polar. Glass, typically made up of silicon oxide lattice. For every one silicon atom, you have two oxygen atoms. You see that right over here. For every one silicon, you have two oxygen atoms. And it turns out that the electronegativity difference between oxygen and silicon is even higher than the electronegativity difference between oxygen and hydrogen. Silicon is even less electronegative than hydrogen. So the oxygens are really able to hog silicon's electrons. Especially the ones that are involved in the bonding. So you have partial charges, partial positive charges form at the silicon and then you still have" + }, + { + "Q": "At 9:30 and onward, it is stated that the accelerations for each of the block would have a magnitude of 1.84 m/s^2. But doesn't that contradict from the previous video, which stated that the accelerations would have a magnitude of 3.68 m/s^2? Could someone please explain why this is? Thanks! :)\n", + "A": "No contradiction. In the previous examples, friction was ignored. In the last example, where the 1.84 m/s^2 was calculated, friction was taken into consideration. And this makes sense - if we have friction, then there s an opposing force to a moving object.", + "video_name": "UrfLAlk2b_8", + "timestamps": [ + 570 + ], + "3min_transcript": "that's just causing this mass to sit on the table and for two, it's cancelled by that normal force. So those cancel anyway, even though they're external forces. This is it, this is the only one that drives the system. So I put that in here and I divide by my total mass 'cause that tells me how much my system resists through inertia, changes in velocity, and this is what I get. I get the same thing I got before, I get back my three point six eight meters per second squared, and I get in one line. I mean, this trick is amazing and it works, and it works in every example where two masses or more masses are forced to move with the same acceleration. So this is great. This'll save you a ton of time. This is supposed to be a three here. And to show you how useful it is, let say there was friction, let's say there was a coefficient of friction of zero point three. Well, now I'd have a frictional force so there'd be an external frictional force here. It'd be applied this five kilogram mass. So if I get rid of this--- So it's not gonna be three point six eight anymore. I'm gonna have a force of friction that I have to subtract. So minus mu K so the force of friction--- I'll just put force of friction. And so to solve for the force of friction, the force of friction is gonna be equal to--- Well, I know three times nine point eight is--- Let me just write this in here, 29.4 Newtons minus the force of friction it's given by. So there's a formula for force of friction. The force of friction is always mu K FN. So the force of friction on this five kilogram mass is gonna be mu K which is point three. So it's gonna be zero point three times the normal force, not the normal force on our entire system. I don't include this three kilogram mass. It's only the normal force on this five kilogram mass that's contributing to this force of friction here. So even though we're treating this system as a whole, we still have to find individual forces So it won't be the entire mass that goes here. The normal force on the five kilogram mass is just gonna be five kilograms times nine point eight meters per second squared. I divide by my total mass down here because the entire mass is resisting motion through inertia. And if I solve this from my acceleration of the system, I get one point eight four meters per seconds squared. So this is less, less than our three point six eight and that makes sense. Now, there's a resistive force, a resistive external force, tryna prevent the system from moving. But you have to be careful. What I'm really finding here, I'm really finding the magnitude of the acceleration. This is just giving me the magnitude. If I'm playing this game where positive forces are ones that make it go and negative forces are ones that resist motion, external forces that is, I'm just getting the magnitude of the acceleration. Individual boxes will have that magnitude" + }, + { + "Q": "At around 6:30, when he starts solving the problem, is he assuming the penny immediately just falls down or something? Because the penny will move in some parabolic motion and have its own change in distance, which would then be added to the height of the hill.\n", + "A": "First of all, Sal does not assume that the penny falls down immediately, because he said that it has an initial velocity of positive 30 m/s, which means that it has an upward direction. Second, Sal assumed that the object was thrown straight upward and not at an angle, and therefore, it would not have a parabolic path.", + "video_name": "emdHj6WodLw", + "timestamps": [ + 390 + ], + "3min_transcript": "here, and that might simplify things. If we multiply both sides by 2a, we get-- and I'm just going to switch this to distance, if we assume that we always start at distances equal to 0. di, or initial distance, is always at point 0. We could right 2ad-- I'm just multiplying both sides by 2a-- is equal to vf squared minus vi squared, or you could write it as vf squared is equal to vi squared plus 2ad. I don't know what your physics teacher might show you or written in your physics book, but of these variations will show up in your physics book. The reason why I wanted to show you that previous problem first is that I wanted to show you that you could actually figure out these problems without having to always memorize formulas and resort to the formula. With that said, it's probably not bad idea to memorize some form of this formula, although you should understand how it Now that you have memorized it, or I showed you that maybe you don't have to memorize it, let's use this. Let's say I have the same cliff, and it has now turned purple. It was 500 meters high-- it's a 500 meter high cliff. This time, with the penny, instead of just dropping it straight down, I'm going to throw it straight up at positive 30 meters per second. The positive matters, because remember, we said negative is down, positive is up-- that's just the convention we use. Let's use this formula, or any version of this formula, to figure out what our final velocity was when we hit the bottom of the ground. This is probably the easiest formula to use, because it We can say the final velocity vf squared is equal to the initial velocity squared-- so what's our initial velocity? It's plus 30 meters per second, so it's 30 meters per second squared plus 2ad. So, 2a is the acceleration of gravity, which is minus 10, because it's going down, so it's 2a times minus 10-- I'm going to give up the units for a second, just so I don't run out of space-- 2 times minus 10, and what's the height? What's the change in distance? Actually, I should be correct about using change in distance, because it matters for this problem. In this case, the final distance is equal to minus 500, and the initial distance is equal to 0. The change in distance is minus 500." + }, + { + "Q": "\nAt 0:28 Sal said that the sun orbits around the Milky Way Galaxy. So does the sun also move and orbit just like the planets do?", + "A": "Yep! Our whole solar system orbits around the center of the Milky Way Galaxy, just like how our planets orbit the Sun. Though at a much higher velocity of 828,000 km/hr", + "video_name": "FEF6PxWOvsk", + "timestamps": [ + 28 + ], + "3min_transcript": "What I want to do in this video is give a very high-level overview of the four fundamental forces of the universe. And I'm going to start with gravity. And it might surprise some of you that gravity is actually the weakest of the four fundamental forces. And that's surprising because you say, wow, that's what keeps us glued-- not glued-- but it keeps us from jumping off the planet. It's what keeps the Moon in orbit around the Earth, the Earth in orbit around the Sun, the Sun in orbit around the center of the Milky Way galaxy. So if it's a little bit surprising that it's actually the weakest of the forces. And that starts to make sense when you actually think about things on maybe more of a human scale, or a molecular scale, or even atomic scale. Even on a human scale, your computer monitor and you, have some type of gravitational attraction. But you don't notice it. Or your cell phone and your wallet, there's gravitational attraction. But you don't see them being drawn to each other the way you might see two magnets drawn to each other And if you go to even a smaller scale, you'll see the it matters even less. We never even talk about gravity in chemistry, although the gravity is there. But at those scales, the other forces really, really, really start to dominate. So gravity is our weakest. So if we move up a little bit from that, we get-- and this is maybe the hardest force for us to visualize. Or it's, at least, the least intuitive force for me-- is actually the weak force, sometimes called the weak interaction. And it's what's responsible for radioactive decay, in particular beta minus and beta plus decay. And just to give you an example of the actual weak interaction, if I had some cesium-137-- 137 means it has 137 nucleons. A nucleon is either a proton or a neutron. You add up the protons and neutrons of cesium, you get 137. Now, the weak interaction is what's responsible for one of the neutrons-- essentially one of its quarks flipping and turning into a proton. And I'm not going to go into detail of what a quark is and all of that. And the math can get pretty hairy. But I just want to give you an example of what the weak interaction does. So if one of these neutrons turns into a proton, then we're going to have one extra proton. But we're going to have the same number of nucleons. Instead of an extra neutron here, you now have an extra proton here. And so now this is a different atom. It is now barium. And in that flipping, it will actually emit an electron and an anti-electron neutrino. And I'm not going to go into the details of what an anti-electron neutrino is. These are fundamental particles. But this is just what the weak interaction is. It's not something that's completely obvious to us. It's not the kind of this traditional things pulling" + }, + { + "Q": "At 5:05, why is scandium's electron configuration the same as argon's?\n", + "A": "It isn t, but each electron configuration builds on from the last. We use the noble gas from the previous row in square brackets to represent all of its electron configuration which saves us time and space. Argon has the electron configuration: 1s2 2s2 2p6 3s2 3p6 Scandium has the electron configuration: 1s2 2s2 2p6 3s2 3p6 3d1 4s2 When we use the shorthand scandium s electron configuration can be written as: [Ar] 3d1 4s2 [Ar] just means all of argon s electron configuration from earlier", + "video_name": "UXOcWAfBdZg", + "timestamps": [ + 305 + ], + "3min_transcript": "Well, sodium is going to have the same electron configuration as neon. Then it's going to go 3s1. Once again, it has one valence electron, one electron in its outermost shell. All of these elements in orange right over here, they have one valence electron and they're trying to get to the octet rule, this kind of stable nirvana for atoms. You could imagine is that they're very reactive and when they react they tend to lose this electron in their outermost shell. That is the case. These alkali metals are very, very reactive. Actually they have very similar properties. They're shiny and soft. Because they're so reactive it's hard to find them where they haven't reacted with other things. Let's keep looking at the other groups. If we move one over to the right this group two right over here, these are called the alkaline earth metals. alkaline earth metals. Once again, they have very similar ... They have very similar properties and that's because they have two valence electrons, two electrons in their outermost shell. Also for them, not as quite as reactive as the alkaline metals. Let me write this out, alkaline earth metals. But for them it's easier to lose two electrons than to try to gain six to get to eight. And so these tend to also be reasonably reactive and they react by losing those two outer electrons. Now something interesting happens as you go to the D block. We studied this when we looked at electron configurations, but if you look at the electron configuration for say scandium right over here, the electron, let me do it in magenta, the electron configuration for scandium, so scandium, scandium's electron configuration It's going to be argon. Then you're going to fill it in we're in the one, two, three, fourth period. It's going to be 4s2. Then we start filling the D block. These are the D block elements here. You have to remember, the D block you backfill. In the D block, this is going to be now 3s1. How many electrons does it have in its outermost shell? Once again its outermost shell is its fourth shell, is its fourth shell. These are, you could argue, higher energy electrons that fills this ... These are filled before that, and there are exceptions to this especially that we see a lot in the D block. This is what's, I guess you could say to some degree, is defining its reactivity. Although in the transition metals, the D block elements, I'm sorry, I made a little mistake there. This is 4s2 3d1." + }, + { + "Q": "\n3:05 what is a valence electron?", + "A": "A valence electron is electrons in the outermost shell of the atom. They determine properties of the atom and bonding behavior. The Octet Rule states that the outermost energy shell has 8 electrons.", + "video_name": "UXOcWAfBdZg", + "timestamps": [ + 185 + ], + "3min_transcript": "but for the most part the elements in the column have very, very, very similar properties. That's because the elements in a column, or the elements in a group tend to have the same number of electrons in their outermost shell. They tend to have the same number of valence electrons. And valence electrons are electrons in the outermost shell they tend to coincide, although there's a slightly different variation. The valence electrons, these are the electrons that are going to react, which tend to be the outermost shell electrons, but there are exceptions to that. There's actually a lot of interesting exceptions that happen in the transition metals in the D block. But we're not gonna go into those details. Let's just think a little about some of the groups that you will hear about and why they react in very similar ways. If we go with group one, group one ... And hydrogen is a little bit of a strange character because hydrogen isn't trying to get to eight valence electrons. Hydrogen in that first shell just wants to get to two Hydrogen is kind of ... It doesn't share as much in common with everything else in group one as you might expect for, say, all of the things in group two. Group one, if you put hydrogen aside, these are referred to as the alkali metals. And hydrogen is not considered an alkali metal. These right over here are the alkali. Alkali metals. Now why do all of these have very similar reactions? Why do they have very similar properties? Well, to think about that you just have to think about their electron configurations. For example, the electron configuration for lithium is going to be the same as the electron configuration of helium, of helium. Then you're going to go to your second shell, 2s1. It has one valence electron. It has one electron in its outermost shell. Well, sodium is going to have the same electron configuration as neon. Then it's going to go 3s1. Once again, it has one valence electron, one electron in its outermost shell. All of these elements in orange right over here, they have one valence electron and they're trying to get to the octet rule, this kind of stable nirvana for atoms. You could imagine is that they're very reactive and when they react they tend to lose this electron in their outermost shell. That is the case. These alkali metals are very, very reactive. Actually they have very similar properties. They're shiny and soft. Because they're so reactive it's hard to find them where they haven't reacted with other things. Let's keep looking at the other groups. If we move one over to the right this group two right over here, these are called the alkaline earth metals." + }, + { + "Q": "\nat 6:42, carbon is described to have same configuration to helium, but then he says that carbon has 4 valence electrons or valency 4. how can it be similar to helium? is there a difference between configuration and valence electrons? I just couldn't understand..", + "A": "He didn t mean carbon has EXACTLY the same electron configuration as helium, but what he is showing is a shorthand way of writing out electron configurations. You take the noble gas from the previous row and put its symbol inside square brackets, this represents the electron configuration for that noble gas, eg [He] means the same thing as 1s^2, [Ne] means the same thing as 1s^2 2s^2 2p^6. This is for our convenience later on when full electron configurations get VERY long.", + "video_name": "UXOcWAfBdZg", + "timestamps": [ + 402 + ], + "3min_transcript": "It's going to be argon. Then you're going to fill it in we're in the one, two, three, fourth period. It's going to be 4s2. Then we start filling the D block. These are the D block elements here. You have to remember, the D block you backfill. In the D block, this is going to be now 3s1. How many electrons does it have in its outermost shell? Once again its outermost shell is its fourth shell, is its fourth shell. These are, you could argue, higher energy electrons that fills this ... These are filled before that, and there are exceptions to this especially that we see a lot in the D block. This is what's, I guess you could say to some degree, is defining its reactivity. Although in the transition metals, the D block elements, I'm sorry, I made a little mistake there. This is 4s2 3d1. We're backfilling the D block. But these, their outermost electrons are in ... They still have two of those outermost electrons. There, once again, are exceptions in these transition metals right here that for the most part are going in backfilling that D block. Once you've kind of backfilled those D blocks then you come over here and you start filling the P block. For example, if you look at the electron configuration for, let's say carbon, carbon is going to have the same electron configuration as helium, as helium. Then you're going to fill your S block, 2s2, and then 2p one two. So 2p2. How many valence electrons does it have? Well, in its second shell, its outermost shell, it has two plus two. It has four valence electrons. That's going to be true for the things in this group. bonding behavior to silicone, to the other things in its group. We could keep going on, for example, oxygen and sulfur. These would both want to take two electrons from someone else because they have six valence electrons and they want to get to eight. They have similar bonding behavior. You go to this yellow group right over here. These are the halogens. There's special name for them. These are the halogens. These are highly reactive because they have seven valence electrons. They would love nothing more than to get one more valence electron. They love to react. In fact, they especially love to react with the alkali metals over here. Then finally you get to kind of your atomic nirvana in the noble gases here. The noble gases, that's the other name for the group, 18 elements, noble gases." + }, + { + "Q": "\nAt 08:30, are there lots of different types of bacteria or just one?", + "A": "There are billions of species, some are harmful to us, some are beneficial.", + "video_name": "TDoGrbpJJ14", + "timestamps": [ + 510 + ], + "3min_transcript": "In bacteria,which are what people originally just classify it on whether or not you have a nucleus, in bacteria,there is no membrane surrounding the DNA. So what they have is just a big bundle of DNA. They just have this big bundle of DNA. It's sometimes in a loop all in one circle called a nucleoid. Now,whenever we look at something,and we say,oh, we have this thing;it doesn't;there's this assumption that somehow we're superior or we're more advanced beings. But the reality is that bacteria have infiltrated far more ecosystems in every part of the planet than Eukarya have, and there's far more diversity in bacteria than there is in Eukarya. these are the more successful organisms. If a comet were to hit the Earth--God forbid-- the organisms more likely to survive are going to be the bacteria than the Eukarya,than the ones with the larger--not always larger, but the organisms that do have this nucleus and have membrane-bound organelles like mitochondria and all that. We'll talk more about it in the future. Bacteria,for the most part,are just big bags of cytoplasm. They have their DNA there. They do have ribosomes because they have to code for proteins just like the rest of us do. Some of those proteins,they'll make some from-- bacteria,they'll make these flagella, which are tails that allow them to move around. They also have these things called pili. Pili is plural for pilus or pee-lus,so these pili. And we'll see in a second that the pili are kind of introducing genetic variation into their populations. Actually,I'll take a little side note here. I'm pointing out bacteria as not having a cell wall. There's actually another class that used to be categorized as type of a bacteria,and they're called Archaea. I should give them a little bit of justice. They're always kind of the stepchild. They used to be called Archaea bacteria, but now people realize,they've actually looked at the DNA, because when they originally looked at these,they said,OK, these guys also have no nucleus and a bunch of DNA running around. These must be a form of bacteria. But now that we've actually been able to look into the DNA of the things,we've seen that they're actually quite different. But all of these,both bacteria and Archaea,are considered prokaryotes. And this just means no nucleus." + }, + { + "Q": "At 1:45 sal says that if we put bacteria in milk it becomes yogurt. But then if you eat the yogurt wont the bacteria get in your body and make you sick.How is that a good thing.\n", + "A": "There are predominately two bacteria used in making yogurt. One is non-probotic so it does not survive the stomach. The other, which is probotic, in non pathogenic, but aids in fermentation (common in the intestines). It is important to remember that not all bacteria make you sick, and to make it more complicated different sub-specie of a bacteria CAN make you sick, while a different subspecie of the same bacteria is ESSENTIAL to stay alive, such as E. Coli.", + "video_name": "TDoGrbpJJ14", + "timestamps": [ + 105 + ], + "3min_transcript": "I think we've all heard of the word bacteria. And we normally associate it with negative things. You say bacteria,those are germs. So we normally associate those with germs,and they indeed are germs, and they cause a whole set of negative things. Or at least from the standard point of view, people believe that they cause a whole bunch of negative things. So let's just list them all just to make sure we know about them, we're all on the same page.So the bad things they do, they cause a lot of diseases:tuberculosis,Lyme disease. I mean,I could go on and on. You know,pretty much any time--well,I'll be careful here. Whenever people talk about an infection, it's often caused by a bacteria.It can also be caused by a virus. An infection is,in general,anything entering you and taking advantage of your body to kind of replicate itself, and in the process,making you sick. And this whole perception of bacteria being a bad thing is probably a good reason why almost any soap you see now will say antibacterial on it. Because the makers of the soap know that in conventional thinking, bacteria are viewed as a negative thing. And you're like,OK,Sal,I know where you're going with this. Bacteria isn't all bad.There are some good traits of bacteria. For example,I could stick some yogurt in some-- or I could stick some bacteria in some milk and it'll help produce some yogurt,sometimes spelled yoghurt. And that's obviously a good thing.It's a delicious thing to eat. And these are all true,but you're like,look,you know, on balance,I still think bacteria is a bad thing. I'm not going to take sides on that debate, as I tend to avoid taking sides on debates in these science videos. Maybe I'll do a whole playlist where I do nothing but take sides on debates,but here I won't take any sides on that. But I'll just point out that you are to a large degree made up of bacteria.It's not just your gut. It's not just the gut or the yogurt you might eat or the plaque on your teeth,which is caused by bacteria. It's this kind of film that's created by bacteria Bacteria actually represents a majority of the cells on your body. And it's not just the pimples on your face. Bacteria actually represents a majority of the cells on your body. So for every--and this is kind of an astounding fact." + }, + { + "Q": "\nAt the start off the video, and at 2:06, why does Sal use the word \"rate\"? Is't distance divided by time just \"speed\" ?", + "A": "Rate is used here as shorthand for rate of change. An object s speed is the rate of change of its position over time, so he could have used speed here. Using rate helps make it clear that the same mathematical tools can be used for non-speed rates of change (e.g. water filling a tank, population growth, and so on).", + "video_name": "6FTiHeius1c", + "timestamps": [ + 126 + ], + "3min_transcript": "- [Voiceover] Let's say that something is traveling at a constant rate of five meters per second. That's its velocity in one dimension. If it was negative, we'd be moving to the left. If it's positive, it's moving to the right. Let's say that we care about what is our change in distance over, the delta symbol represents change, over a change in time of four seconds. Over four seconds. I could say from t equal zero to t is equal to four. That's our change in time. That's our four-second interval that we care about. Well, one way to think about it is what a rate by definition is nothing but a change in some quantity. In this case, it's distance over a change in some other quantity. In this case, we're thinking about time. Or another way to think about it, if we multiply both sides times change in time, you get your change in distance is equal to your rate times change in time. from pre-algebra, distance is equal to rate times time. Time, that just comes from the definition of what a rate is. It's a change in one quantity with respect to another quantity. If you just apply this, you can say, \"Okay, my rate is a constant five meters \"per second and my delta t is four seconds.\" So times four seconds. Well, that's just going to give you 20. That's just going to get you 20. Let me do that in same color that I had for the change in distance. That's going to be 20. Then the seconds, cancel the seconds, 20 meters. So my total change in distance over those four seconds is going to be 20 meters. Nothing new here. Nothing too fancy. But I want to do now is connect this to the area under the rate function over this time period. So let's graph that. That's my rate axis. This is my time axis. This is going to be in seconds. This is going to be in meters per second. Let's see, one, two, three, four, five. Let's see if it's about enough, and then I go one, two, three, four, five. Our rate, at least in this example, is a constant, is a constant five meters per second. It's a constant five meters per second. That is my r of t in this example. What did we just do here? We just multiplied our change in time times our constant rate. We just multiplied our change in time. So from time equal zero seconds to four seconds. It's this length here, if we think on that axis. Now we multiply it times our constant rate. We multiply it times this right over here. If I multiply this base times this height, what am I going to get? I'm going to get this area under the rate function." + }, + { + "Q": "At 1:55, what is a mole of a molecule?\n", + "A": "The first one is easy, and I hope this explains it for you. A mole consists of an Avogadro s Number of atoms, molecules, or whatever. That number is 6.02 x 10\u00c2\u00b2\u00c2\u00b3 So if you have 6.02 x 10\u00c2\u00b2\u00c2\u00b3 molecules of water, then you have EXACTLY one mole of it.", + "video_name": "LJmFbcaxDPE", + "timestamps": [ + 115 + ], + "3min_transcript": "Let's say I have some weak acid. I'll call it HA. A is a place holder for really a whole set of elements that I could put there. It could be fluorine, it could be an ammonia molecule. If you add H it becomes ammonium. So this isn't any particular element I'm talking about. This is just kind of a general way of writing an acid. And let's say it's in equilibrium with, of course, and you've seen this multiple times, a proton. And all of this is in an aqueous solution. Between this proton jumping off of this and its conjugate base. And we also could have written a base equilibrium, where we say the conjugate base could disassociate, or it could essentially grab a hydrogen from the water and create OH. And we've done that multiple times. But that's not the point of this video. So let's just think a little bit about if we were to stress it in some way. And you can already imagine that I'm about to touch on Le Chatelier's Principal, which essentially just says, look, if you stress an equilibrium in any way, the equilibrium moves in such a way to relieve that stress. So let's say that the stress that I apply to the system --Let me do a different color. I'm going to add some strong base. That's too dark. I'm going to add some NaOH. And we know this is a strong base when you put it in a aqueous solultion, the sodium part just kind of disassociates, but the more important thing, you have all this OH in the solution, which wants to grab hydrogens away. So when you add this OH to the solution, what's going to happen for every mole that you add, not even just mole, for every molecule you add of this into the solution, Right? So for example, if you had 1 mole oh hydrogen molecules in your solution right when you do that, all this is going to react with all of that. And the OHs are going to react with the Hs and form water, and they'll both just kind of disappear into the solution. They didn't disappear, they all turned into water. And so all of this hydrogen will go away. Or at least the hydrogen that was initially there. That 1 mole of hydrogens will disappear. So what should happen to this reaction? Well, know this is an equilibrium reaction. So as these hydrogen disappear, because this is an equilibrium reaction or because this is a weak base, more of this is going to be converted into these two products to kind of make up for that loss of hydrogen." + }, + { + "Q": "\nat 4:29 Sal says when you increase the OH, you decrease the pOH and it increases the pH? I'm a bit confused by this, can someone please explain? Thanks", + "A": "The same way that you you increase the concentration of H+ the pH goes down, the pOH would also go down. Don t forget that pOH=-log10([OH-]). Looking at [OH-]=.01 and .001 respectively, the pOH is 2 and 3.", + "video_name": "LJmFbcaxDPE", + "timestamps": [ + 269 + ], + "3min_transcript": "Right? So for example, if you had 1 mole oh hydrogen molecules in your solution right when you do that, all this is going to react with all of that. And the OHs are going to react with the Hs and form water, and they'll both just kind of disappear into the solution. They didn't disappear, they all turned into water. And so all of this hydrogen will go away. Or at least the hydrogen that was initially there. That 1 mole of hydrogens will disappear. So what should happen to this reaction? Well, know this is an equilibrium reaction. So as these hydrogen disappear, because this is an equilibrium reaction or because this is a weak base, more of this is going to be converted into these two products to kind of make up for that loss of hydrogen. So this hydrogen goes down initially, and then it starts getting to equilibrium very fast. But this is going to go down. This is going to go up. And then this is going to go down less. Because sure, when you put the sodium hydroxide there, it just ate up all of the hydrogens. But then you have this -- you can kind of view as the spare hydrogen capacity here to produce hydrogens. And when these disappear, this weak base will disassociate more. The equilibrium we'll move more in this direction. So immediately, this will eat all of that. But then when the equilibrium moves in that direction, a lot of the hydrogen will be replaced. So if you think about what's happening, if I just threw this sodium hydroxide in water. So if I just did NaOH in an aqueous solution so that's just throwing it in water -- that disassociates completely into the sodium cation and hydroxide anion. increase the quantity of OHs by essentially the number of moles of sodium hydroxide you're adding, and you'd immediately increase the pH, right? Remember. When you increase the amount of OH, you would decrease the pOH, right? And that's just because it's the negative log. So if you increase OH, you're decreasing pOH, and you're increasing pH. And just think OH-- you're making it more basic. And a high pH is also very basic. If you have a mole of this, you end up with a pH of 14. And if you had a strong acid, not a strong base, you would end up with a pH of 0. Hopefully you're getting a little bit familiar with that concept right now, but if it confuses you, just play around with the logs a little bit" + }, + { + "Q": "\nat 0:54 Sal says that this oxygen could grab a hydrogen from the water essentially creating OH. What is OH?", + "A": "OH is a hydroxyl radical. But when A\u00e2\u0081\u00bb grabs a proton from the water it forms OH\u00e2\u0081\u00bb, which is hydroxide ion.", + "video_name": "LJmFbcaxDPE", + "timestamps": [ + 54 + ], + "3min_transcript": "Let's say I have some weak acid. I'll call it HA. A is a place holder for really a whole set of elements that I could put there. It could be fluorine, it could be an ammonia molecule. If you add H it becomes ammonium. So this isn't any particular element I'm talking about. This is just kind of a general way of writing an acid. And let's say it's in equilibrium with, of course, and you've seen this multiple times, a proton. And all of this is in an aqueous solution. Between this proton jumping off of this and its conjugate base. And we also could have written a base equilibrium, where we say the conjugate base could disassociate, or it could essentially grab a hydrogen from the water and create OH. And we've done that multiple times. But that's not the point of this video. So let's just think a little bit about if we were to stress it in some way. And you can already imagine that I'm about to touch on Le Chatelier's Principal, which essentially just says, look, if you stress an equilibrium in any way, the equilibrium moves in such a way to relieve that stress. So let's say that the stress that I apply to the system --Let me do a different color. I'm going to add some strong base. That's too dark. I'm going to add some NaOH. And we know this is a strong base when you put it in a aqueous solultion, the sodium part just kind of disassociates, but the more important thing, you have all this OH in the solution, which wants to grab hydrogens away. So when you add this OH to the solution, what's going to happen for every mole that you add, not even just mole, for every molecule you add of this into the solution, Right? So for example, if you had 1 mole oh hydrogen molecules in your solution right when you do that, all this is going to react with all of that. And the OHs are going to react with the Hs and form water, and they'll both just kind of disappear into the solution. They didn't disappear, they all turned into water. And so all of this hydrogen will go away. Or at least the hydrogen that was initially there. That 1 mole of hydrogens will disappear. So what should happen to this reaction? Well, know this is an equilibrium reaction. So as these hydrogen disappear, because this is an equilibrium reaction or because this is a weak base, more of this is going to be converted into these two products to kind of make up for that loss of hydrogen." + }, + { + "Q": "@13:15 how does -log ([HA]/[A-]) turn into log ([HA/A-])^-1 ?\n", + "A": "That is an application of a basic identity of logarithms. log (a/b) = log(a) - log (b) = -log (b) + log(a) = - [log(b) - log (a)] = - log (b/a) Thus, log (a/b) = - log(b/a)", + "video_name": "LJmFbcaxDPE", + "timestamps": [ + 795 + ], + "3min_transcript": "we can multiply both sides by the reciprocal of this right here. And you get hydrogen concentration. Ka times --I'm multiplying both sides times a reciprocal of that. So times the concentration of our weak acid divided by the concentration of our weak base is equal to our concentration of our hydrogen. Fair enough. Now. Let's take the negative log of both sides. So the negative log of all of that stuff, of your acidic equilibrium constant, times HA, our weak acid divided by our weak base, our hydrogen concentration. Which is just our pH, right? Negative log of hydrogen concentration is --that's the definition of pH. I'll write the p and the H in different colors. You know a p just means negative log. Minus log. That's all. Base 10. Let's see if we can simplify this any more. So our logarithmic properties. We know that when you take the log of something and you multiply it, that's the same thing as taking the log of this plus the log of that. So this can to be simplified to minus log of our Ka minus the log of our weak acid concentration divided by its conjugate base concentration. Is equal to the pH. which is just the negative log of its equilibrium constant. So this is just the pKa. And the minus log of HA over A. What we can do is we could make this a plus, and just take this to the minus 1 power. Right? That's just another logarithm property, and you can review the logarithm videos if that confused you. And this to the minus 1 power just means invert this. So we could say, plus the logarithm of our conjugate base concentration divided by the weak acid concentration is equal to the pH. And this right here, this is called the Hendersen-Hasselbalch Equation. And I really encourage you not to memorize it. Because if you do attempt to memorize it, within a few hours, you're going to forget whether this was a plus over here. You're going to forget this, and you're going to forget whether you" + }, + { + "Q": "\nAt 3:53 Sal says that the electric field has a 3 n/c electric field. However, doesn't the electric field depend on the distance between the two particles? Isn't the equation K*Q/D^2?", + "A": "It does except in this example Sal gave where it is a constant electric field. If it were not constant, you would use the equation you have defined provided the charge on the plate is defined (Q). Then you would solve in same manner by multiplying E times the test charge to get Force then multiply force times distance", + "video_name": "zqGvUbvVQXg", + "timestamps": [ + 233 + ], + "3min_transcript": "in magnitude it's pushing out, because we assume when we draw field lines that we're using a test charge with a positive charge so it's pushing outward. Let's say I have a 1-coulomb charge. Actually, let me make it 2 coulombs just to hit a point home. Say I have a 2-coulomb charge right here, and it's positive. A positive 2-coulomb charge, and it starts off at 3 meters away, and I want to bring it in 2 meters. I want to bring it in 2 meters, so it's 1 meter away. So what is the electric-- or electrical-- potential energy difference between the particle at this point and at this point? amount of work, as we've learned in the previous two videos, we need to apply to this particle to take it from here to here. So how much work do we have to apply? We have to apply a force that directly-- that exactly-- we assume that maybe this is already moving with a constant velocity, or maybe we have to start with a slightly higher force just to get it moving, but we have to apply a force that's exactly opposite the force provided by Coulomb's Law, the electrostatic force. And so what is that force we're going to have to apply? Well, we actually have to know what the electric field is, which I have not told you yet. I just realized that, as you can tell. So let's say all of these electric field lines are 3 newtons per coulomb. So at any point, what is the force being exerted from this Well, the electrostatic force on this particle is equal to the electric field times the charge, which is equal to-- I just defined the electric field as being 3 newtons per coulomb times 2 coulombs. It equals 6 newtons. So at any point, the electric field is pushing this way 6 newtons, so in order to push the particle this way, I have to completely offset that, and actually, I have to get it moving initially, and I'll keep saying that. I just want to hit that point home. So I have to apply a force of 6 newtons in the leftward direction and I have to apply it for 2 meters to get the point here. So the total work is equal to 6 newtons times 2 meters, which is equal to 12 newton-meters or 12 joules. So we could say that the electrical potential energy--" + }, + { + "Q": "\nAt 8:18 Sal said voltage is regardless of how small or big the charge is\nbut to find the potential we divide the work done by the charge\nso isn't the potential dependent upon the size of charge?", + "A": "The potential is not dependent on the size of charge, but the potential energy is. The potential is a property of the field. It s similar to gravity. g does not depend on the mass of an object on the surface of the earth. But the gravitational potential energy does depend on the mass (mgh)", + "video_name": "zqGvUbvVQXg", + "timestamps": [ + 498 + ], + "3min_transcript": "does it take to move any charge per unit charge from here to here? Well, in our example we just did, the total work to move it from here to here was 12 joules. But how much work did it take to move it from there to there per charge? Well, work per charge is equal to 12 joules for what? What was the charge that we moved? Well, it was 2 coulombs. It equals 6 joules per coulomb. That is the electric potential difference between this point and this point. So what is the distinction? Electric potential energy was associated with a particle. How much more energy did the particle have here than here? When we say electric potential, because we essentially divide by the size of the particle, it It actually just depends on our position. So electric potential, we're just saying how much more potential, irrespective of the charge we're using, does this position have relative to this position? And this electric potential, that's just another way of saying voltage, and the unit for voltage is volts. So 6 joules per coulomb, that's the same thing as 6 volts. And so if we think of the analogy to gravitation, we said gravitational potential energy was mgh, right? This was distance, right? Electric potential is essentially the amount of gravitational-- if we extend the analogy, the amount of gravitational potential energy per mass, right? So if we wanted a quick way of knowing what the gravitational the mass, we divide by the mass, and it would be the acceleration of gravity times height. Ignore that if it confused you. So what is useful about voltage? It tells us regardless of how small or big or actually positive or negative a charge is, what the difference in potential energy would be if we're at two different points. So electric potential, we're comparing points in space. Electric potential energy, we're comparing charges at points in space. Hopefully, I didn't confuse you. In the next video, we'll actually do a couple of problems where we figure out the electric potential difference or the voltage difference between two points in space as opposed to a charge at two different points in space. I will see you in the next video." + }, + { + "Q": "\n5:29\nIs that to say the electric potential energy of electrons is dependent on its position? In an electric circuit, how do we change the electric energy per charge? When electrons pass through a load in a circuit, such as a lamp, how is electrical energy lost (converted to light energy)? How does it affect current and amperes?", + "A": "Potential energy ALWAYS relates to position. That s what PE is. The battery provides chemical energy to separate charges and give them PE. The PE is converted to KE when the circuit is complete and the charges flow as current. When they pass through the load, they bump into the atoms of the load and transfer some of their KE to the load.", + "video_name": "zqGvUbvVQXg", + "timestamps": [ + 329 + ], + "3min_transcript": "amount of work, as we've learned in the previous two videos, we need to apply to this particle to take it from here to here. So how much work do we have to apply? We have to apply a force that directly-- that exactly-- we assume that maybe this is already moving with a constant velocity, or maybe we have to start with a slightly higher force just to get it moving, but we have to apply a force that's exactly opposite the force provided by Coulomb's Law, the electrostatic force. And so what is that force we're going to have to apply? Well, we actually have to know what the electric field is, which I have not told you yet. I just realized that, as you can tell. So let's say all of these electric field lines are 3 newtons per coulomb. So at any point, what is the force being exerted from this Well, the electrostatic force on this particle is equal to the electric field times the charge, which is equal to-- I just defined the electric field as being 3 newtons per coulomb times 2 coulombs. It equals 6 newtons. So at any point, the electric field is pushing this way 6 newtons, so in order to push the particle this way, I have to completely offset that, and actually, I have to get it moving initially, and I'll keep saying that. I just want to hit that point home. So I have to apply a force of 6 newtons in the leftward direction and I have to apply it for 2 meters to get the point here. So the total work is equal to 6 newtons times 2 meters, which is equal to 12 newton-meters or 12 joules. So we could say that the electrical potential energy-- The electrical potential energy difference between this point and this point is 12 joules. Or another way to say it is-- and which one has a higher potential? Well, this one does, right? Because at this point, we're closer to the thing that's trying to repel it, so if we were to just let go, it would start accelerating in this direction, and a lot of that energy would be converted to kinetic energy by the time we get to this point, right? So we could also say that the electric potential energy at this point right here is 12 joules higher than the electric potential energy at this point. Now that's potential energy. What is electric potential? Well, electric potential tells us essentially how much work is necessary per unit of charge, right? Electric potential energy was just how much total work is needed to move it from here to here." + }, + { + "Q": "\nAt 5:12, Sal mentioned other species in the Homo genus, Neanderthals. What are Neanderthals?", + "A": "Neanderthals were an extinct species of human, or possibly a subspecies. They are very closely related to modern humans, differing in DNA by only about 0.1%. Neanderthals were larger and heavier than modern humans. They went extinct sometime between 30,000 to 45,000 years ago, possibly due to inability to adapt to the changing climate of the era, or they simply interbred with humans.", + "video_name": "oHvLlS_Sc54", + "timestamps": [ + 312 + ], + "3min_transcript": "of deciding how close two animals are. But to a large degree, a lot of these categories-- deciding where to divide along kingdom, phylum, class, order, family, tribe-- these are somewhat arbitrary. These are just picked based on early taxonomists, including Carl Linnaeus, and saying, well, this looks like a grouping right over here. But they could have grouped at a broader level or a deeper So these things right over here are somewhat arbitrary. A more analytical way is just to see how much DNA you have in common and then use that as a measure of how far apart two animals are. Or really, I should say, two species are, because this taxonomy doesn't only apply just to animals. It applies to plants and bacteria and Archaea and all sorts of things, so it's actually a broader thing than just animals. Now, with that out of the way, what I thought would be fun-- just so that we could really get a sense of where modern taxonomy is, where the field that was essentially fathered by Carl Linnaeus, where it is now, how we-- and use that to figure out where we And obviously, I'm drawing just a small fraction of the universe of the organisms that we even know about right now. But at least it frames the picture in terms of something we understand-- in particular, us. In particular, humans. Now, our species, we call ourselves humans. But we're really Homo sapiens. And the sapiens is the species part, and then Homo is the genus. And what I'm doing right over here is I'm saying, well, if Homo is the genus, what other species were inside of Homo? And the reality is-- or at least as far as we know-- there are no other living species inside of Homo. We probably killed them all off. Or maybe we interbreeded with them somehow, which might have argued that maybe they weren't different species. But more likely, they were competing in the same ecosystems, and they became endangered species very quickly when they competed with our ancestors. But the most recent other species within the genus that we know about are the neanderthals, Now, if we go further up the tree of life, further up the taxonomy-- and you'll sometimes see tribe mentioned. Sometimes you won't. And we tend to get a little bit more granular the closer we get to humans. When we go further away in the tree of life, we get a little bit less granular sometimes. But that's not always the case as well. You go a little bit further up, then you get Hominini. And I'm sure I'm mispronouncing some of this as well. But another species that's in Homonini that is not in Homo-- and I'm definitely not listing all of them, and that's why I'm showing all of these other branches over here-- is what we call the common chimpanzee. And their species name is-- their genus is Pan, and their species is troglodytes. So you would refer to them as Pan troglodytes. And that's also another convention that Carl Linnaeus came up with, is that you refer to a particular species by its genus and then its species. And you capitalize the genus, and you lowercase a species." + }, + { + "Q": "Why are sharks fish and dolphins and whales mammals? (brought up at 10:02)\n", + "A": "This is due to very notable differences between the two. Sharks have cartilaginous bones, gills, and a swim bladder, all of which are associated with fish. Whales and Dolphins have dense bones, lungs, and are warm blooded, things associated with mammals. Ocean mammals also have vestigial hip bones even though they lack legs and they are incapable of drinking salt water. Like many desert mammals, ocean mammals get their water from the food that they eat.", + "video_name": "oHvLlS_Sc54", + "timestamps": [ + 602 + ], + "3min_transcript": "from things that first lived in trees, and that's why their hands and their feet look the way they do. Now you get to even a broader level of classification. You get to the mammals. And once again, probably something you're used to thinking about. Mammals are air-breathing animals, and they tend to have fur or hair. They tend to provide some form of milk for their young. They have active mammary glands. There's other things we can talk about, what makes a mammal. I'm not going to go into the rigorous definition. But just to give you an example of a mammal that is not a primate, I could show you this polar bear right over here. This is a mammal that is not a primate. And I could do other things. I could show you a tiger, or I could show you a giraffe or a horse. And so by no stretch of the imagination am I being comprehensive. But let's keep getting broader. Now let's go to the class-- we're already at the class of Mammalia. Now let's go to the phylum. we are in the phylum chordates. And chordates, we're actually in the subphylum, which I didn't write here, vertebrates, which means we have a vertebra. We have a spinal column with a spinal cord in it. Chordates are a little bit more general. Chordates is a phylum where-- kind of the arrangement of where the mouth is, where are the digestive organs, where the anus is, where the spinal column is, where are the brains, where are the eyes, where are the mouth. They're kind of all in the same place. And if you think about it, everything I've listed here kind of has the same general structure. You have a spinal column. You have a brain. You have a mouth. Then the mouth leads to some type digestive column. And at the end of it, you have an anus over there. And you have eyes in front of the brain. And so this is a general way-- and I'm not being very rigorous here, is how you describe a chordate. And to show a chordate that is not a mammal, you would just have to think of a fish or sharks. So this right over here is a non-mammal chordate Now, let's go even broader. As you'll see, now we're going to things that are very, very not human-like. So you go one step broader. Now we're in Animalia, the kingdom of animals. And this is the broadest category that Carl Linnaeus thought about. Well actually, he did go into trees as well. But when you think of kingdom animals and you think of things that aren't chordates, you start going into things like insects. And you start going into things like jellyfish. If you go even broader, now we're talking about the domain. You go to Eukarya. So these are all organisms that have cells. And inside those cells, they have complex structures. So if you're a Eukarya, you have cells with complex structures. If you're a Prokarya, you don't have complex structures inside your cell. But other Eukarya that are not animals include things like plants. And obviously, I'm giving no justice" + }, + { + "Q": "At 1:16 , The force with which we pull will be stored as potential energy and it will be converted to kinetic energy once we leave the object.... But Sal said that the kinetic energy is converted into potential energy.. If I'm wrong, please someone correct me..\n", + "A": "The energy bounces back and forth between KE and PE.", + "video_name": "Nk2q-_jkJVs", + "timestamps": [ + 76 + ], + "3min_transcript": "Let's see if we can use what we know about springs now to get a little intuition about how the spring moves over time. And hopefully we'll learn a little bit about harmonic motion. We'll actually even step into the world of differential And don't get daunted when we get there. Or just close your eyes when it happens. Anyway, so I've drawn a spring, like I've done in the last couple of videos. And 0, this point in the x-axis, that's where the spring's natural resting state is. And in this example I have a mass, mass m, attached to the spring. And I've stretched the string. I've essentially pulled it. So the mass is now sitting at point A. So what's going to happen to this? Well, as we know, the force, the restorative force of the spring, is equal to minus some constant, times the x position. The x position starting at A. So initially the spring is going to pull back this way, right? The spring is going to pull back this way. It's going to get faster and faster and faster and faster. And we learned that at this point, it has a lot of potential energy. At this point, when it kind of gets back to its resting energy, but very little potential energy. But then its momentum is going to keep it going, and it's going to compress the spring all the way, until all of that kinetic energy is turned back into potential energy. Then the process will start over again. So let's see if we can just get an intuition for what x will look like as a function of time. So our goal is to figure out x of t, x as a function of time. That's going to be our goal on this video and probably the next few. So let's just get an intuition for what's happening here. So let me try to graph x as a function of time. So time is the independent variable. And I'll start at time is equal to 0. So this is the time axis. Let me draw the x-axis. This might be a little unusual for you, for me to draw the x-axis in the vertical, but that's because x is the dependent variable in this situation. Or we could say x of t, just so you know x is a function of time, x of t. And this state, that I've drawn here, this is at time equals 0, right? So this is at 0. Let me switch colors. So at time equals 0, what is the x position of the mass? Well the x position is A, right? So if I draw this, this is A. Actually, let me draw a line there. That might come in useful. This is A. And then this is going to be-- let me try to make it relatively-- that is negative A. That's minus A. So at time t equals 0, where is it? So this is where the graph is, right? Actually, let's do something interesting. Let's define the period." + }, + { + "Q": "\nAt 1:09, shouldn't the carbonyl O be protonated first because H2O is a relatively weak nucleophile?", + "A": "Not necessarily, assuming you don t have other stronger electrophiles in your reaction vessel. H2O is actually a pretty decent nucleophile (which is why for most SN2 reactions where it is NOT the intended substituent, we take great care to remove it from our system).", + "video_name": "632MAqIB14E", + "timestamps": [ + 69 + ], + "3min_transcript": "Voiceover: Here's an example of a nucleophilic addition reaction to an aldehyde or a ketone. So over here on the left, it could be an aldehyde, or we could change that to form a ketone. And if you add water to an aldehyde or ketone, you form this product over here on the right, which is called a hydrate, or also called a gem-diol, or geminal diol because these two OHs here are on the same carbon, so like they're twins. And this reaction is at equilibrium. So let's think about the aldehyde, or the ketone. We know the carbonyl on the aldehyde or ketone is polarized, so we know that the oxygen has more electronegatives than carbons, so it withdraws some electron densities. So this oxygen here is partially negative, and this carbonyl carbon is partially positive, like that. And therefore the carbonyl carbon, since it's partially positive, is electrophillic, so it wants electrons. And it can get electrons from water. So let's go ahead and draw the water molecule right here. Water can function as a nucleophile. It has two lone pairs of electrons, this oxygen here is partially negative, and so we're going to get a nucleophile So a lone pair of electrons on the oxygen is going to attack our carbonyl carbon like that, So the nucleophile attacks the electrophillic portion of the molecule, and these pi electrons here kick off onto the oxygen. So let's go ahead and draw the results of our nucleophilic attack here, and so we now have our oxygen bonded to this carbon, and this oxygen still has two hydrogens bonded to it, so I'm gonna go ahead and draw in those two hydrogens. There's still a lone pair of electrons on that oxygen, which gives that oxygen a +1 formal charge. And then this carbon here is bonded to another oxygen, which had two lone pairs of electrons around it, and now it picked up another one, so a -1 formal charge on this oxygen, and there's still an R group bonded to it, and a hydrogen over here, like that. And so let's try to follow some electrons here. So one of the lone pairs of electrons on the oxygen formed a bond with our carbon, so I'm And then we can think about our pi electrons. So, our pi electrons in here, as kicking off onto our oxygen, so it doesn't really matter which one of these three it is, right, let's say it's that one, and we get this as our intermediate. And so next, we can think about an acid-base reaction. So, another water molecule comes along right here, and so we know water can function as an acid or a base, and so this lone pair of electrons could take, let's say it takes this proton right here, and leaves these electrons behind on our oxygen. So let's go ahead and draw the result of that acid-base reaction, and so we would have our oxygen here, would now be bonded to only one hydrogen, and, now let's see, we still have our negatively charged oxygen over here on the right, and then we have our R group, and our hydrogen like that, and we still have this lone pair of electrons," + }, + { + "Q": "at 6:40, you talk about how adding Cl groups causes it to be more reactive by withdrawing electronegativity, why is it then that Ketones are more stable than Aldehydes as stated in the reactivity video? Wouldn't the extra R group also withdraw electronegativity and cause it to be more reactive?\n", + "A": "R groups mean they re carbon substituents and assuming they have C-H bonds they re electron donating through hyperconjugation. Obviously if you replaced all the C-H bonds with something highly electron withdrawing like C-Cl bonds you will make a very reactive ketone.", + "video_name": "632MAqIB14E", + "timestamps": [ + 400 + ], + "3min_transcript": "this as our product for our hydrate. Except, we know that ketones are not as reactive as aldehydes, and so this time the equilibrium is to the left. It favors the formation of the ketone. Let's look at another one. So this is acetaldehyde. And so if we add water to acetaldehyde over here, we form this as our hydrate product. And once again, we know that these reactions occur because of our carbonyl carbon right here being partially positive, so the oxygen withdraws some electron density like that. So, we could make aldehydes or ketones more reactive by adding something else that withdraws electron density from that carbonyl carbon, and one thing you could do, is add an electronegative atom like halogen, so let's go ahead and do that. Let's add three halogens here. Let's add three chlorines to this carbon, the one adjacent to our carbonyl carbon. And those electron withdrawing groups, withdraw some electron density, so they're gonna withdraw electron density this way, once again, away from our carbonyl carbon, and so this carbonyl carbon gets even more partially positive by the addition of these electronegative atoms. And the more positive you make that carbonyl carbon, the more electrophilic you make it, and therefore, the more the nucleophile, which is water, is going to attack, and so you make it even more reactive by adding these, and so you can push the equilibrium even more to the right. You can form more of your product. And so let's go ahead and draw the product of that reaction, so we would put three chlorines on here like that, and so you could also do this with ketones, and you could make ketones much more reactive by doing that. And so this particular reaction is a little bit famous. Over here on the left is trichloroacetaldehyde, and then, once you've formed the hydrate and this is famous for being knock-out drops. And so some of the old references to it are, \"slip someone a Mickey Finn.\" So you could slip chloral hydrate into someone's drink, and it was a knock-out drop situation. And so just a little bit of an interesting reaction here for formation of hydrates." + }, + { + "Q": "\nAt 0:26, Sal is naming the alkene. He notes the methyl group at Carbon-2, but not Carbon-1 or Carbon-5-- so I'm a bit confused. We don't have to name those?\nAs in: wouldn't it be 1,2,5-methylpent-2-ene?\n\nThanks for any help!", + "A": "They re not methyl groups, they are part of the main chain. They ve already been accounted for by calling it pentene.", + "video_name": "O_yeKo6-qIg", + "timestamps": [ + 26 + ], + "3min_transcript": "- [Voiceover] Anytime you're trying to come up with a mechanism for a reaction, it's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different. So what we're starting with, we could call this one, two, three, four, five, so this is, let's see, we have methyl group on the number two carbon, it is a pentene, and that is double-bond between the number two and number three carbons, so this is two-methyl-pent-2-ene. So that's what we start with, we're in the presence, we're in an acidic environment, we've got what's gonna be catalyzed by our hydronium here, and we end up with this, and how is our product different from what we started with? Well the double bond is now gone, the number three carbon gains this hydrogen, and now the number two carbon gains a hydroxyl group. in the presence of an acid, it's acid-catalyzed, we have gained two hydrogens and an oxygen, which is what we've gained, what could be used to make a water. And this is actually called an acid-catalyzed addition of water. The water isn't sitting on one part of the molecule, but if you take the hydrogen we added, and the hydroxyl we added, if you combine them, that's what you need to make a water. So let's think about how we can, how this actually happens in the presence of our hydronium. So let me redraw this molecule right over here. So we copy and paste it, so that's not exactly it yet, that is just with the single bond. So let me draw, woops, wrong tool. Let me draw the double bond there. And now let me put it in the presence of some hydronium. Alright, so we have an oxygen bonded, two. but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton, thus making the entire molecule positive, because the hydrogen proton is positive. So there you go, this now has a positive charge. And this can be pretty reactive, 'cause we know that oxygen is quite electronegative, it lives to keep its electrons. So what is there was a way, what if there's a way for the oxygen to take back the electrons in this bond right over here, the two electrons in this bond. Well what if one of these carbons, especially the ones that have the double bonds, what if some of the electrons from this double bond could be used to snab, to take that hydrogen proton, and then oxygen can hog its electrons again. And you might say, \"oh that's reasonable, but which of these carbons would actually do it?\" And to think about which of those carbons would do it," + }, + { + "Q": "\nAt 7:40, if two oxygen electrons are shared, then why isn't a double bond formed??", + "A": "2 electrons is a single bond 4 electrons is a double bond", + "video_name": "O_yeKo6-qIg", + "timestamps": [ + 460 + ], + "3min_transcript": "And so this is now just neutral water, and we see that we have a conservation of charge here, this was positive in charge, now our original molecule is positively charged. And what feels good about this is we're getting, we're getting close to our end product, at least on our number three carbon, we now have, we now have this hydrogen. Now we need to think about, well how do we get a hydroxyl group added right over here? Well we have all this water, we have all this water floating around, let me, I could use this water molecule but the odds of it being the exact same water molecule, we don't know. But there's all sorts of water molecules, we're in an aqueous solution, so let me draw another water molecule here. So the water molecules are all equivalent, but let me draw another water molecule here. And you can imagine, if they just pump into each other in just the right way, this is, water is a polar molecule, it has a because the oxygen likes to hog the electrons, and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged, so you can imagine the oxygen end might be attracted to this tertiary carbocation, and so just bumping it in just the right way, it might form a bond. So let me say these two electrons right over here, let's say they form a bond with this, with that number two carbon, and then what is going to result? So let me draw, so what is, what is going to result, let me scroll down a little bit, and let me paste, whoops, let me copy and paste our original molecule again. So, here we go. This is the one we constructed actually, so we have the hydrogen there. We have the hydrogen, now this character, so we have the water molecule, so oxygen bonded to two hydrogens, you have this one lone pair that isn't reacting, but then you have the lone pair that does do the reacting. And so it now forms a bond. Woops, let me do it in that orange color. It now, it now forms an actual bond. And we're really close to our final product, we have our hydrogen on the number three carbon, we have more than we want on our number two carbon, we just want a hydroxyl group, now we have a whole water bonded to the carbon. So somehow we have to get one of these other hydrogens swiped off of it, well that could happen with just another water molecule. So let's draw that. So another water molecule someplace, I'll do the different color just to differentiate, although as we know water, well it's hard to see what color is water if you're looking at the molecular scale." + }, + { + "Q": "\nAt 3:22, shouldn't it be -ke^2/r instead of ke^2/r?", + "A": "the thing is, the final result of this calculation doesn t result in a vector, so we really don t need the energy, we only need the magnitude. In the second case however, it is a vector, so sign is required for vector algebra", + "video_name": "7Zin8hG9Nhw", + "timestamps": [ + 202 + ], + "3min_transcript": "with where that electron is. So we know the electron is also attracted to the nucleus. There's an electric force, alright, so this electron is pulled to the nucleus, this is an attractive force. This is the electric force, this is a centripetal force, the force that's holding that electron in a circular orbit around the nucleus here. And, once again, we talked about the magnitude of this electric force in an earlier video, and we need it for this video, too. We're gonna use it to come up with the kinetic energy for that electron. So the electric force is given by Coulomb's Law, the magnitude of the electric force is equal to K, which is a constant, \"q1\", which is, let's say it's the charge on the proton, times \"q2\", charge on the electron, divided by \"r squared\", where \"r\" is the distance between our two charges. We know that Newton's Second Law: force is equal to the mass times the acceleration. so the mass of the electron times the acceleration of the electron. The electric force is a centripetal force, keeping it in circular motion, so we can say this is the \"centripetal acceleration\". Alright, let's go ahead and write down what we know. \"K\" is a constant, we'll write that in here, \"q1\", \"q1\" is the charge on a proton, which we know is elemental charge, so it would be positive \"e\"... \"q2\" is the charge on the electron. The charge on the electron is the same magnitude as the charge on the proton, but it's a negative value. So we have negative \"e\", is the charge on the electron, divided by \"r squared\", is equal to the mass of the electron times the centripetal acceleration. So, centripetal acceleration is equal to \"v squared\" over \"r\". So, we did this in a previous video. We're gonna do the exact same thing we did before. because we already know the direction is always going to be towards the center, and therefore, we only care... we don't care about this negative sign here. We can also cancel one of the \"r\"s. So if we don't care about... if we only care about the magnitude, on the left side, we get: Ke squared over r is equal to mv squared, on the right side. And you can see, we're almost to what we want. Our goal was to try to find the expression for the kinetic energy, that's 1/2 mv squared. Here, we have mv squared, so if we multiply both sides by 1/2, right, multiply both sides by 1/2, now we have an expression for the kinetic energy of the electron. So: 1/2 mv squared is equal to the kinetic energy. So we know the kinetic energy is equal to: 1/2 Ke squared over r Alright, so we will come back to the kinetic energy. Next, we're gonna find the potential energy." + }, + { + "Q": "Whoa, hold on\u00e2\u0080\u0094that value of 0.15 at 10:40 is not a probability! Sal picked the box size arbitrarily. If he had picked a box that were ten times greater in volume, then the \"probability\" would be 1.5; that is, there would be a 150% \"chance\" of there being a hydrogen molecule in that box. That's obviously not right. Instead, it represents the expected, or average, number of hydrogen molecules per box of that volume. To go from an average value to a probability is quite hand-wavy indeed.\n", + "A": "The volume of that box is 1 Liter; because SI units.", + "video_name": "psLX080RQR8", + "timestamps": [ + 640 + ], + "3min_transcript": "So let's see, if I take the twenty-third -- so let me write it out here. So my hydrogen per box -- So my concentration of hydrogen per dV, is equal to 12 times 10 -- whoops! That's not helping when my pen malfunctions. Let me get that right. 12 times 10 to the twenty-third power times 0.125 times 10 to the minus twenty-seventh power. All of that divided by 10 to the minus 3, right? That's 1 times 10 to the minus 3. So let's cancel out some exponents. If we get rid of the minus 3 here, you divide by minus 3, then this becomes minus 24. And then the minus 24 and the minus what's 12 times 1.25? So times 12 is equal to 1.5. So the 12 times the 1.25 is equal to 1.5 times -- and then 10 to the twenty-third times 10 to the minus twenty-fourth is equal to 10 to the minus 1, right? So it's just divided by 10. So on average, your concentration of hydrogen in a little cube that's half a nanometer in each direction is equal to 0.15 molecules -- not Moles anymore -- of hydrogen molecule per my little dV, my little box. And so this is a probability, right? This is a probability, because obviously I can't have 0.15 molecules in every box. there's a 0.15 chance that I have a hydrogen molecule in my box. So if I want to go back here to this, this is 0.15, this is 0.15, this is 0.15. But how did we get this 0.15? We multiplied the concentration of hydrogen, which was this right here. That's the concentration of the hydrogen -- I should've written it in a more vibrant color -- times just a bunch of scaling factors, right? We could just say that, well, this was just equal to the concentration of hydrogen times, based on how I picked my dV, I had to do all of this scaling. But it was times some constant of scaling, scaling to my appropriate factor. So if we want to figure out each of these, this is just the concentration of hydrogen times" + }, + { + "Q": "at 7:15, what did sal mean by scaling factor?\n", + "A": "He scales the units so they are all the same. It is basically converting a unit to another. For example 0.01 l is equal to 1 ml, so the scaling here would be 1 ml * 10^-3 = 0.01 l", + "video_name": "psLX080RQR8", + "timestamps": [ + 435 + ], + "3min_transcript": "Let's pick my dV to be -- I looked up the diameter of an ammonia molecule. It was about 1/10 of a nanometer. If this was a nanometer box, you could put 10 in each direction, so you can almost fit 1,000 if you packed them really tightly. So let's make this half a nanometer in each direction. So if I pick my dV -- and remember, I don't know if this is the right distance. I'm just trying to give you the intuition behind the equilibrium formula. But if I pick this as being 0.5 nanometers by 0.5 nanometers by 0.5 nanometers, what is my volume? So my little volume is going to be 0.5 times 10 to the minus 1/9 meters-- that's a nanometer because we're dealing with cubic meters. So this is equal to 0.5 to the 1/3 power. 0.5 times 0.5 is 0.25 times 0.5 is 0.125. I want to do the math right, so let me just make sure I got that right. 0.5 to the 1/3 power. Right, 0.125 times -- negative 9 to the 1/3 power is minus 27 -- 10 to the minus 27 meters cubed. So that's my volume. Now, we know the concentration. Let's figure out what's the probability. So this is the probability in the box, right? That's what we're concerned with, the probability in the box. Well, the probability in the box, that's the probability that I have one hydrogen in the box, times the probability that I have another hydrogen in the box, times the probability that I have another hydrogen in the box --these are all in-the-box probabilities -- times the probability that I'll do the nitrogen in a different color just to ease -- oh, I should've done these in the orange because those are the color of the molecules up there. And I'll do this one in purple. What's the probability of having hydrogen in the box? Well, we know hydrogen's concentration at equilibrium is 2 Molar. So concentration of hydrogen, we know hydrogen's concentration is equal to 2 Molar, which is 2 Moles per liter, which is equal to -- 2 Moles is just 2 times 6 times 10 to the 23rd power -- Moles is just a number -- divided by liters. So 1 liter is-- we could write it in meters cubed, or we could just make the conversion. Actually, let me just do this for you. 1 liter is equal to 1 times 10 to the minus 3 meters cubed." + }, + { + "Q": "\nAt 8:13 Jay says that the Bohr model is incorrect. Is this because the Uncertainty Principle showed that the position of the electron would be at a distance greater than 2 times the radius. Whereas the Bohr model had the position of the electron at r1?", + "A": "The Bohr model could not be correct because it only worked for H and even for H it did not properly explain all the observations about the emission spectrum of H. That s why physicists continued to search for a better model, and through the work of DeBroglie, Schrodinger, Heisenberg, Born and others, the modern model was developed.", + "video_name": "PZIoFD_Z73M", + "timestamps": [ + 493 + ], + "3min_transcript": "we had the uncertainty in the position of the electron, times the uncertainty in the momentum of the electron must be greater than or equal to Planck's Constant divided by four pi. So we can take that uncertainty in the momentum and we can plug it in here. So now we have the uncertainty in the position of the electron in the ground state of the hydrogen atom times 2.0 times 10 to the negative 25. This product must be greater than or equal to, Planck's Constant is 6.626 times 10 to the negative 34. Alright, divide that by four pi. So we could solve for the uncertainty in the position. So, Delta X must be greater than or equal to, let's go ahead and do that math. So we have Planck's Constant, 6.626 times 10 to the negative 34, and then we need to divide by the uncertainty in the momentum. So we also need to divide by the uncertainty in momentum, that's 2.0 times 10 to the negative 25, and that gives us 2.6 times 10 to the negative 10. So the uncertainty in the position must be greater than or equal to 2.6 times 10 to the negative 10 and if you worked our your units, you would get meters for this. So the uncertainty in the position must be 2.6 times 10 to the negative 10 meters. Let's go back up here to the picture of the hydrogen atom. 2.6 times 10 to the negative 10 meters, that's greater than the diameter of our hydrogen atom, so the uncertainty would be greater than this diameter. the diameter of the hydrogen atom, using the Bohr model. So the Bohr model is wrong. It's telling us we know the electron is orbiting the nucleus at a certain radius, and it's moving at a certain velocity. The uncertainty principle says this isn't true. If we know the velocity fairly accurately, we don't know the position of the electron, the position of the electron is greater than the diameter, according to the Bohr model. So this just one reason why the Bohr model is wrong. But again, we keep the Bohr model around because it is useful as a simple model when you're just starting to get into chemistry. But this concept of the uncertainty principle goes against our natural intuitions. So our every day life doesn't really give us any experience with the uncertainty principle. For example, if we had a particle, let's make it a much bigger particle here, so a much bigger particle than an electron, so something that we can actually see in our real life, and so this has a much bigger mass," + }, + { + "Q": "\nAt 3:46, what does he mean by the electron being in the ground state?", + "A": "Ground state means the lowest energy state", + "video_name": "PZIoFD_Z73M", + "timestamps": [ + 226 + ], + "3min_transcript": "so the uncertainty in the momentum must increase to four, because one times four is equal to four. If I decrease the uncertainty in the position even more, so if I lower that to point five, I increase the uncertainty in the momentum, that must go up to eight. So point five times eight gives us four. And so, what I'm trying to show you here, is as you decrease the uncertainty in the position, you increase the uncertainty in the momentum. So another way of saying that is, the more accurately you know the position of a particle, the less accurately you know the momentum of that particle. And that's the idea of the uncertainty principle. And so let's apply this uncertainty principle to the Bohr model of the hydrogen atom. So let's look at a picture of the Bohr model of the hydrogen atom. Alright, we know our negatively charged electron orbits the nucleus, like a planet around the sun. And, let's say the electron is going this direction, so there is velocity going in that direction. Alright, the reason why the Bohr model is useful, is because it allows us to understand things like quantized energy levels. And we talked about the radius for the electron, so if there's a circle here, there's a radius for an electron in the ground state, this would be the radius of the first energy level, is equal to 5.3 times 10 to the negative 11 meters. So if we wanted to know the diameter of that circle, we could just multiply the radius by two. So two times that number would be equal to 1.06 times 10 to the negative 10 meters. And this is just a rough estimate of the size of the hydrogen atom using the Bohr model, with an electron in the ground state. Alright, we also did some calculations to figure out the velocity. So the velocity of an electron in the ground state we calculated that to be 2.2 times 10 to the six meters per second. And since we know the mass of an electron, we can actually calculate the linear momentum. So the linear momentum P is equal to the mass times the velocity. Let's say we knew the velocity with a 10% uncertainty associated with that number. So a 10% uncertainty. If we convert that to a decimal, we just divide 10 by 100, so we get 10% is equal to point one. So we have point one here. If I want to know the uncertainty of the momentum of that electron, so the uncertainty in the momentum of that particle, momentum is equal to mass times velocity. If there's a 10% uncertainty associated with the velocity, we need to multiply this by point one. So let's go ahead and do that. So we would have the mass of the electron is 9.11 times 10 to the negative 31st." + }, + { + "Q": "at 7:02 jay took away 2 pi electrons, one from each corresponding carbon atom. how is it possible to take 2 pi electrons from different carbon atoms. i do know about breaking a pi bond where one carbon attains a positive charge where other attains a negative charge for showing resonance. i am finding it hard to believe that two corresponding carbon atoms attains a + charge. pls do explain why.\n", + "A": "You need a reagent that really wants those electrons, like SbF\u00e2\u0082\u0085 in SO\u00e2\u0082\u0082Cl\u00e2\u0082\u0082", + "video_name": "I6wzan4hNc4", + "timestamps": [ + 422 + ], + "3min_transcript": "have three molecular orbitals. But these are my antibonding molecular orbitals. Those are higher energy. And my two points of intersection that are right on the center line here, represent two non-bonding molecular orbitals like that. So when I fill my molecular orbitals, again it's analogous to electron configurations. I have a total of eight pi electrons that I need to worry about for a planar cyclooctatetraene molecule. And so I can go ahead and start to fill in my pi electrons So that takes care of six of them and I have two more. And since this is analogous to electron configurations, I'm going to follow Hund's rule and not pair up my electrons in an orbital here. So that represents my eight pi electrons. And since I have unpaired electrons, I have two unpaired electrons, that predicts a very unstable molecule if it were to adopt a planar confirmation. And if I think about in terms of Huckel's rule, I know it doesn't follow Huckel's rule. Huckel's rule is 4n plus 2, where n is an integer. have this orbital down here. And I do have 4n right here. But I don't have 4n, where n is an integer for these two electrons up here. And so this is where it breaks down. And so I have a total of eight pi electrons, which does not follow Huckel's rule. And so because the number of electrons is incorrect, this molecule is definitely not going to adopt a planar confirmation. And so cyclooctatetraene has a tub confirmation, and not planar. It is not aromatic. It is considered to be non-aromatic because of the violation of the first criteria. But it is possible to react cyclooctatetraene. It's possible to oxidize it. And so let's see what happens when we do that. So if we take cyclooctatetraene and we oxidize it. So it's going to lose some electrons. So I'm going to say that these pi electrons are going to stay. And we're going to lose the pi electrons on the left. So if I take away a bond from these two carbons that they're going to be positively charged. And so I could draw a resonance structure for this. I could move these electrons over here. And so if I go ahead and show that resonance structure, then this carbon still has a plus 1 charge. And the other positive charge moves over here to this carbon, And you could continue drawing resonance structures for this molecule. I am not going to do that. I just want to show you that the positive charges are spread out throughout the entire ion here. And so one way to represent that would be to just show the electrons are spread out throughout this entire ion. And the whole thing has a 2 plus charge like that. And so when I analyze this dication that I got from cyclooctatetraene, I realize that all the carbons are sp2 hybridized. So if I look at these, my carbocations those are sp2 Everything with a double bond on it is sp2 hybridized. And so I have eight sp2 hybridized carbons." + }, + { + "Q": "''5:38\"\n\nno ones dug that far?\ncool!\nhow deep is the earths core?\ndoes anyone at khan academy know?\n", + "A": "did u guy ever see ice age continental drift. if so do u remember when scrat the little weird animal that loves acorns. well he went to the earths core and he didnt burn up and die.", + "video_name": "KL0i1RSnpfI", + "timestamps": [ + 338 + ], + "3min_transcript": "We don't know yet whether the inner part is liquid or solid. Now, the next point of evidence is how do we know that there's an inner core? And we can use P-waves for that. A P-wave can travel through anything, but remember, in general for the same type of material if you get denser material it's going to move faster, so it's going to refract outwards like we've seen over here. But if it goes into a liquid, in general, sound waves, or I should say P-waves, seismic waves move slower in liquids. And so the refraction patterns we get when we do measure from seismograph stations around the world is that it looks like the P-waves are kind of doing what you would expect in the mantle, but then they're getting refracted as if they're going to a slower medium as they go through the outer core. And we see that right over here. And then they get refracted again Now, that is just what you would expect if it was all liquid, but if you go to stations that are even further out it looks like, if you just look at the refraction patterns, and you can now model this with fancy computers and get all the data points, but you could say, well, the only way that reality can fit the data that we get based on when things reach here is if the P-waves are being first refracted through the outer core, but then they're refracted in a way that they're going through denser material, significantly denser material than the inner core. And then they're just continuing to refract the way you would expect. So it's really the refraction pattern of the P-waves. And frankly, the fact that there's this what you call a P-wave shadow. The P-wave shadow by itself, all that tells you is that kind of roughly crazy things are happening someplace in the core. But the real way to know that we have an inner core that's solid, as opposed to the whole thing being liquid, is that the P-waves is the pattern of when and how the P-waves reach And then you can kind of, based on modeling how waves would travel through different densities and different types of mediums, you could say, well, there's got to be an inner core right over here. And obviously, it's a lot more math than I'm going into. But if you do the math based on the shadow, and you know the speed of the material, and all of that type of thing, then you can figure out the depth at which these transitions occur. We know that we have a transition from mantle to outer core here. And then a transition from outer core to core there. So hopefully that satiates your questions about how do we know what the composition of the earth is without ever having to dig down there, because we've never even gotten below our crust." + }, + { + "Q": "At 5:36, why did he move the hydrogen to the other side of the periodic table?\n", + "A": "He showed that the hydrogen could belong to the alkali metals (because it could lose one electron like them and have a complete outer shell) OR it could gain one electron and still have a full outer shell (like the halogens which are group 7 elements). Remember the first electron shell only holds 2 electrons, then each shell after that holds 8.", + "video_name": "CCsNJFsYSGs", + "timestamps": [ + 336 + ], + "3min_transcript": "it just needs one electron. So in theory, hydrogen could have been put there. So hydrogen actually could typically could have a positive or a negative 1 oxidation state. And just to see an example of that, let's think about a situation where hydrogen is the oxidizing agent. And an example of that would be lithium hydride right Now, in lithium hydride, you have a situation where hydrogen is more electronegative. A lithium is not too electronegative. It would happily give away an electron. And so in this situation, hydrogen is the one that's oxidizing the lithium. Lithium is reducing the hydrogen. Hydrogen is the one that is hogging the electron. So the oxidation state on the lithium here is a positive 1. And the oxidation state on the hydrogen here is a negative. to make sure we get the notation. Lithium has been oxidized by the hydrogen. Hydrogen has been reduced by the lithium. Now, let's give an example where hydrogen plays the other role. Let's imagine hydroxide. So the hydroxide anion-- so you have a hydrogen and an oxygen. And so essentially, you could think of a water molecule that loses a hydrogen proton but keeps that hydrogen's electron. And this has a negative charge. This has a negative 1 charge. But what's going on right over here? And actually, let me just draw that, because it's fun to think about it. So this is a situation where oxygen typically has-- 1, 2, 3, 4, 5, 6 electrons. And when it's water, you have 2 hydrogens like that. And then you share. over there sharing that pair, covalent bond sharing that right over there. To get to hydroxide, the oxygen essentially nabs both of these electrons to become-- so you get-- that pair, that pair. Now you have-- let me do this in a new color. Now, you have this pair as well. And then you have that other covalent bond to the other hydrogen. And now this hydrogen is now just a hydrogen proton. This one now has a negative charge. So this is hydroxide. And so the whole thing has a negative charge. And oxygen, as we have already talked about, is more electronegative than the hydrogen. So it's hogging the electrons. So when you look at it right over here, you would say, well, look, hydrogen, if we had to, if we were forced to-- remember, oxidation states is just an intellectual tool which we'll find useful. If you had to pretend this wasn't a covalent bond, but an ionic bond, you'd say, OK, then maybe this hydrogen would fully lose an electron," + }, + { + "Q": "\nAt 14:03, how did the answer become -6.37 m/s^2? I substituted the same values in the exact same kinematics equation and I got -62.83 m/s^2. Did I possibly enter something wrong with my calculator (it was in radians mode).", + "A": "I know the answer to your question! Ok, so when you put them into the calculator, ALWAYS PUT PARENTHESIS FOR PI! It is so important! If you just divide -(40rad/s)^2 by 80pi (2*40pi), then you will get -62.8319... If you divide -(40rad/s)^2 by (80pi), you will get the right answer -6.3662...", + "video_name": "TBlDBaUGqNc", + "timestamps": [ + 843 + ], + "3min_transcript": "we got how fast in meters per second. It was going 160 meters per second. And the next part asks, what was the angular acceleration of the bar? All right, this one we're gonna have to actually use a kinematic formula for. We'll bring these back, put 'em over here. Again the way you use these, you identify what you know. We know the initial angular velocity was 40. So this time we know omega initial 40 radians per second. Set it revolved 20 revolutions. That's delta theta, but again, we can't just write 20. We've gotta right this in terms of radians if we're gonna use these radians per second. They have to all be in the same unit. So it's gonna be 20 revolution times two pi radians per revolution. So that's 40 pi radians. What's our third known? You always need a third known to use a kinematic formula. It's this. It says it decelerates to a stop, which means it stops. That means omega final, the final angular velocity is zero. And we want the angular acceleration, that's alpha. We wanna know alpha. We know the rest of these variables. Again to figure out which equation to use, I figure out which one got left out. And that's the time. I was neither give the time nor was I asked to find the time. Since this was left out, I'm gonna look for the formula that doesn't use time at all. And that's not the first one. That's not the second or the third, it's actually the fourth. So I'm gonna use this fourth equation. So what do we know? We know omega final was zero. So I'm gonna put a zero squared. But zero squared is still zero, equals omega initial squared. That's 40 radians per second squared. And then it's gonna be plus two times alpha. We don't know alpha, but that's what we wanna find, so I'm gonna leave that as a variable. And then delta theta we know. Delta theta was 40 pi radians since it was 20 revolutions. And if you solve this algebraically for alpha, you move the 40 over to the other side. So you'll subtract it. You get a negative 40 radians per second squared. And then you gotta divide by this two as well as the 40 pi radians, which gives me negative 6.37 radians per second squared. Well this thing slowed down to a stop. So this angular acceleration has gotta have the opposite sign to the initial angular velocity. We called this positive 40, that means our alpha's gonna be negative. So recapping, these are the rotational kinematic formulas that relate the rotational kinematic variables. They're only true if the angular acceleration is constant. But when it's constant, you can identify the three known variables and the one unknown that you're trying to find and then use the variable that got left out of the mix to identify which kinematic formula to use, since you would use the formula that does not involve that variable that was neither given nor asked for." + }, + { + "Q": "\nAfter the third resonant structure at 6:58,what happens??", + "A": "The third resonance structure can become the second resonance structure and then the first resonance structure, so on and so forth. The blue pi bond is more than one carbon away from the positive charge so it can t move to the carbocation to form a fourth resonance structure.", + "video_name": "fpq0eICjuSI", + "timestamps": [ + 418 + ], + "3min_transcript": "If this electron goes there, then it would look like this. Let me redraw it. I'll draw the resonance structures quickly. You have your hydrogen. You have your electrophile. That's not an electrophile anymore, but you have that E that's now been added. You have that hydrogen. You have a double bond here. Let me draw a little bit neater. You have this hydrogen. You have this hydrogen, this hydrogen and this hydrogen. What I said is, this is stabilized. So an electron here can actually jump over here. So if this electron jumps over here, the double bond is now over there If that goes over there like that, the double bond is now over here. Now this guy lost his electron and it would have a positive charge. And then that is resonance stabilized. It can either go back to this guy, or this electron over here can jump over there. Let me redraw the whole thing over again. This right here, you have the E and the hydrogen. You have a hydrogen here, hydrogen here, hydrogen here, hydrogen here. And normally you don't worry about the hydrogens, but one of the hydrogens is going to be nabbed later on in this mechanism, so I want to draw all the hydrogens just so you know that they are there. But as I said, this is resonance stabilized. If this electron right here jumps over there, then this double bond is now this double bond. And now this guy over here lost an electron, so it would have a positive charge. And again, once you had this double bond up here, this double bond up there is that double bond. So we can go back and forth between these. The electrons are just swishing around the ring. So it's not going to be maybe as great as the situation that we had when we had a nice benzene ring that was completely aromatic. and around the ring, stabilize the structure, but this is still a relatively stable carbocation, because the electrons can move around. You can kind of view it as a positive charge that gets dispersed between this carbon, this carbon, and that carbon over there. As I said, it's still not a great situation. The molecule wants to go back to being aromatic, wants to go to that really stable state. And the way it can go back to that really stable state is somehow an electron can be added to this thing. And the way that an electron can be added to this thing is, if we have some base flying around, and that base nabs this proton, this proton right here that's on the same carbon as where the electrophile is attached. So if this base nabs a proton, so it just nabs the hydrogen nucleus, then that electron that the hydrogen had, that" + }, + { + "Q": "\n2:28 Why are Ischemic Strokes are more common then Hemorrhagic strokes?", + "A": "It is simply more common for blood clots to form and plaques to break off and cause ischemic strokes. The rupture of a vessel is just less likely.", + "video_name": "xbyfeEW56Nc", + "timestamps": [ + 148 + ], + "3min_transcript": "You've probably heard of people having a stroke, and you're probably familiar with the notion that it has something to do with the brain, and you'd be right. In particular, it's a rapid loss of brain function because of something strange happening with the blood flow to the brain. And let me show you that in a little bit more detail. And to do that, the 2 major types of strokes. There's the ischemic strokes, and the other type of stroke is hemorrhagic, and these can kind of be sub-categorized, but I won't go into all of the details there. And if I really just define ischemia and hemorrhaging to you, I think you'll have an idea of how these strokes are different and how they interrupt the blood flow to different parts of the brain. You know from the videos on stenosis and ischemia and the videos on heart attacks that ischemia is to certain body tissues. So an ischemic stroke is actually very, very similar to what we saw in a heart attack, except it's not occurring in a coronary blood vessel, it's occurring in a blood vessel in the brain. So let me draw that right over here. Let's say that this is a blood vessel in the brain. And let's say that blood is flowing in that direction (this is an artery). And so you could imagine that maybe there is a big blood clot that forms in some part of the brain. Let me do the blood clot in magenta. This blood clot might form because -- no, that's not magenta -- the blood clot might form because maybe there's a plaque there, maybe the plaque got ruptured, either way, this clot is restricting the flow of blood. And we know that this blood clot -- we can call this a thrombus, or we could say that thrombosis has occurred over here-- either way, the blood flow is restricted, is not going to get its oxygen, and it might die; it might experience infarction. And that's why ischemic strokes are also sometimes called cerebral infarctions. These are all very fancy words, but I think, hopefully, they're becoming a little bit more common in our vocabulary, they keep showing up over and over again. And I also want to be clear: most strokes are actually ischemic strokes. The numbers I looked up, they say, 87% of strokes are ischemic. Now, the other type of way that you could have ischemia in one of these blood vessels, and this is completely analogous to what we saw in the heart, when we had heart attacks, is: you could have thrombosis, or you could also have an embolism. Whenever someone says thrombosis, or a thrombus, or thrombi, they're talking about blood clots. Whenever someone talks about an embolus, or emboli, or embolism," + }, + { + "Q": "7:17- Are the valves in the veins tethered to the walls; or do they not need to be because of the low pressure?\n", + "A": "The valves are attached to the walls, as in the are not free moving. But there are not any other attachments like there are in the heart i.e. the chordae tendineae stoping the valves from flipping back.", + "video_name": "iqRTd1NY-pU", + "timestamps": [ + 437 + ], + "3min_transcript": "because the blood that's coming into our arteries is under, let's not forget, high pressure. So the arterial system we know is a high-pressure system. So this makes perfect sense that the first few arteries, those large arteries and even those medium-sized arteries, are going to be able to deal with the pressure really well. Now, let me draw a little line here just to keep it straight. The small artery and the arteriole, these two are actually sometimes called the muscular arteries. And the reason, again, if you just want to look at the wall of the artery, you'll get the answer. The wall of the artery is actually very muscular. In fact, specifically, it's smooth muscle. So not the kind of muscle you have in your heart or in your biceps, but this is smooth muscle that's in the wall of the artery. And there's lots of it. So again, if you have a little blood vessel like this, if you imagine tons and tons of smooth muscle on the outside-- If those bands decide that they want to contract down, that they want to squeeze down, you're going to get something that looks like a little straw, because those muscles are now tight. They're tightly wound, so you're going to create like a little straw. And this process is called vasoconstriction. Vaso just means blood vessel. And constriction is kind of tightening down. So vasoconstriction, tightening down of the blood vessel. And what that does is it increases resistance. Just like if you're trying to blow through a tiny, tiny little straw, there's a lot of resistance. Well, it's the same idea here. And actually, a lot of that resistance and change in the vasoconstriction is happening at the arteriole level. So that's why they're very special and I want you to remember them. From there, blood is going to go through the capillaries. I didn't actually label them the first time, but let me just write that here. Some, as they call them, capillary beds. And then it's going to go and get collected in the venules and eventually into the veins. And the important thing about the veins-- I'm going to stop right here and just talk about it very briefly-- is that they have these little valves. And these valves make sure that the blood continues to flow in one direction. So one important thing here is the valves. And remember, the other important thing is that they are able to deal with large volumes. So unlike the arterial side where it was all about large pressure, down here with the vein side, we have to think about large volumes. Remember about 2/3 of your blood at any point in time is sitting in some vein or venule somewhere." + }, + { + "Q": "At 1:32 it says that the very small arteries are called arterioles. When does an artery turn into an arteriole, and likewise for veins and venules? Does it depend on the size or the structure of the blood vessel or something else?\n", + "A": "Size does define the classification of artery versus arterioles. When there is two or less medial layers of smooth muscle. This is typically less than 0.1 mm diameters. Structurally, arteries contain more elastic tissue and the arterioles contain more smooth muscle. Functionally, arterioles contribute more to restriction of blood flow and consequently control total peripheral resistance. There is also some gas exchange in arterioles.", + "video_name": "iqRTd1NY-pU", + "timestamps": [ + 92 + ], + "3min_transcript": "I want to figure out how blood gets from my heart, which I'm going to draw here, all the way to my toe. And I'm going to draw my foot over here and show you which toe I'm talking about. Let's say this toe right here. Now, to start the journey, it's going to have to go out of the left ventricle and into the largest artery of the body. This is going to be the aorta. And the aorta is very, very wide across. And that's why I say it's a large artery. And from the aorta-- I'm actually not drawing all the branches of the aorta. But from the aorta, it's going to go down into my belly. And it's going to branch towards my left leg and my right leg. So let's say we follow just the left leg. So this artery over here on the top, it's going to get a little bit smaller. And maybe I'd call this a medium-sized artery by this point. This is actually now getting down towards my ankle. Let's say we've gone quite a distance down in my ankle. And let's just follow the branch that goes towards my foot, which is this top one. Let's say this one goes towards my foot, and this is going to be now an even smaller artery. Let's call it small artery. From there, we're actually going to get into what we call arterioles, so it's going to get even tinier. It's going to branch. Now, these are very, very tiny branches coming off my small artery. And let's follow this one right here, and this one is my arteriole. So these are all the different branches I have to go through. And finally, I'm going to get into tiny little branches. I'm going to have to draw them very, very skinny just to convince you that we're getting smaller and smaller. Let me draw three of them. Let's draw four just for fun. And this is actually going to now get towards my little toe cells. to convince you that I actually have gotten there. Let's say one, two over here, and maybe one over here. These are my toes cells. And after the toe cells have kind of taken out whatever they need-- maybe they need glucose or maybe they need some oxygen. Whatever they've taken out, they're also going to put in their waste. So they have, of course, some carbon dioxide waste that we need to drag back. This is now going to dump into what we call a venule. And this venule is going to basically then feed into many, many other venules. Maybe there's a venule down here coming in, and maybe a venule up here coming in maybe from the second toe. And it's going to basically all kind of gather together, and again, to a giant, giant set of veins. Maybe veins are dumping in here now, maybe another vein dumping in here. And these veins are all going to dump into an enormous vein that we call the inferior vena cava." + }, + { + "Q": "At 9:04, where does the joules sign go?\n", + "A": "Sal forgot to write it in, but it should still be there!", + "video_name": "lsXcKgjg8Hs", + "timestamps": [ + 544 + ], + "3min_transcript": "Once again, that's x grams. They cancel out. So the ice will absorb 333.55 joules as it goes from zero degree ice to zero degree water. Or 333.55x joules. Let me put the x there, that's key. So the total amount of heat that the ice can absorb without going above zero degrees... Because once it's at zero degree water, as you put more heat into it, it's going to start getting warmer again. If the ice gets above zero degrees, there's no way it's going to bring the water down to zero degrees. The water can't get above zero degrees. So how much total heat can our ice absorb? So heat absorbed is equal to the heat it can absorb when it goes from minus 10 to zero degrees ice. And that's 20.5x. Plus the amount of heat we can absorb as we go from zero degree ice to zero degree water. And that's 333.55x. And of course, all of this is joules. So this is the total amount of heat that the ice can absorb without going above zero degrees. Now, how much real energy does it have to absorb? Well it has to absorb all of this 125,340 joules of energy out of the water. Because that's the amount of energy we have to extract from the water to bring it down to zero degrees. So the amount of energy the ice absorbs has to be this 125,340. So that has to be equal to 125,340 joules. We can do a little bit of algebra here. Add these two things. 20.5x plus 333.55x is 354.05x. Yeah, 330 plus 30 is 350. Then you have a 3 with a 0.5 there. 354.05x and that is equal to the amount of energy we take out of the water. You divide both sides. So x is equal to 125,340 divided by 354.05. I'll take out the calculator for this. The calculator tells me 125,340, the amount of energy that has to be absorbed by the ice, divided by 354.05 is equal to 354 grams. Roughly, there's a little bit extra. So actually, just to be careful maybe I'll take 355 grams of ice." + }, + { + "Q": "\nAt 1:32, what is spin (in terms of chemistry) ?", + "A": "If you want to think of it like a ball rotating clockwise or anticlockwise that is more than likely going to be fine for this level of understanding. The major thing it means for chemistry is that each orbital can only have at most 2 electrons, each with opposite spins (ie one spin up and one spin down per orbital)", + "video_name": "u1eGSL6J6Fo", + "timestamps": [ + 92 + ], + "3min_transcript": "Let's remind ourselves a little bit of what we already know about orbitals and I've gone over this early on in the regular chemistry playlist. Let's say that this is the nucleus of our atom, super small, and around that we have our first orbital, the 1s orbital. The 1s orbital, you can kind of just view it as a cloud around the nucleus. So you have your 1s orbital and it can fit two electrons, so the first electron will go into the 1s orbital and then the second electron will also go into the 1s orbital. For example, hydrogen has only one electron, so it would go into 1s. Helium has one more, so that will also go into the 1s orbital. After that is filled, then you move onto the 2s orbital. The 2s orbital, you can view it as a shell around the 1s orbital, and all of these, you can't really view it in our conventional way of thinking. You can kind of view it as a probability cloud of where you might find the electrons. kind of a shell cloud around the 1s orbital. So imagine that it's kind of a fuzzy shell around the 1s orbital, so it's around the 1s orbital, and your next electron will go there. Then the fourth electron will also go there, and I drew these arrows upward and downward because the first electron that goes into the 1s orbital has one spin and then the next electron to go into 1s orbital will have the opposite spin, and so they keep pairing up in that way. They have opposite spins. Now, if we keep adding electrons, now we move to the 2p orbitals. Actually, you can view it as there are three 2p orbitals and each of them can hold two electrons, so it can hold a total of six electrons in the 2p orbitals. Let me draw them for you just so you can visualize it. So if we were to label our axis here, so think in three So imagine that that right there is the x-axis. Let me do this in different colors. Let's say that this right here is our y-axis and then we have a z-axis. I'll do that in blue. Let's say we have a z-axis just like that. You actually have a p orbital that goes along each of those axes. So you could have your two-- let me do it in the same color. So you have your 2p sub x orbital, and so what that'll look like is a dumbbell shape that's going in the x-direction. So let me try my best attempt at drawing this. It's a dumbbell shape that goes in the x-direction, in kind of both directions, and it's actually symmetric. I'm drawing this end bigger than that end so it looks like it's coming out at you a little bit, but let me draw it a little bit better than that. I can do a better job." + }, + { + "Q": "At 12:08 he says that the Carbon has 1s and 3sp3 orbital, so my question is that ; that each of those 3 sp3 orbitals would contains a single electron ? Am i right?\n", + "A": "Instead of having one s and three p orbitals, the carbon atom has four sp\u00c2\u00b3 orbitals. Each of those four sp\u00c2\u00b3 orbitals contains a single electron.", + "video_name": "u1eGSL6J6Fo", + "timestamps": [ + 728 + ], + "3min_transcript": "just-- the first one doesn't look like it's just in the s orbital and then the p and y and z for the other three. They all look like they're a little bit in the s and a little bit in the p orbitals. Let me make that clear. So instead of this being a 2s, what it really looks like for carbon is that this looks like a 2sp3 orbital. This looks like a 2sp3 orbital, that looks like a 2sp3 orbital, that looks like a 2sp3 orbital. They all look like they're kind of in the same orbital. This special type of-- it sounds very fancy. This sp3 hybridized orbital, what it actually looks like is something that's in between an s and a p orbital. It has a 25% s nature and a 75% p nature. You can imagine it as being a mixture of these four things. That's the behavior that carbon has. So when you mix them all, instead of having an s cross-section, an s orbital looks like that and the p orbital looks something like that in cross-section. So this is a an s and that is a p. When they get mixed up, the orbital looks like this. An sp3 orbital looks something like this. This is a hybridized sp3 orbital. Hybrid just means a combination of two things. A hybrid car is a combination of gas and electric. A hybridized orbital is a combination of s and p. Hybridized sp3 orbitals are the orbitals when carbon bonds with things like hydrogen or really when it bonds with anything. So if you looked at a molecule of methane, and people talk about sp3 hybridized orbitals, all they're saying is that you have a carbon in the center. And instead of having one s and three p orbitals, it has four sp3 orbitals. So let me try my best at drawing the four sp3 orbitals. Let's say this is the big lobe that is kind of pointing near us, and then it has a small lobe in the back. Then you have another one that has a big lobe like that and a small lobe in the back. Then you have one that's going back behind the page, so let me draw that. You can kind of imagine a three-legged stool, and then its small lobe will come out like that. And then you have one where the big lobe is pointing straight up, and it has a small lobe going down. You can imagine it as kind of a three-legged stool. One of them is behind like that and it's pointing straight up, So a three-legged stool with something-- it's kind of like a tripod, I guess is the best way to think about it." + }, + { + "Q": "At 5:57 when you is not dissolved, do you mean it is not dissolved in water or in gas.\n", + "A": "In this particular reaction, all the other components are gases, so it means that the C is not a gas. In other words, the carbon is not dissolved in the gaseous phase.", + "video_name": "TsXlTWgyItw", + "timestamps": [ + 357 + ], + "3min_transcript": "rate forward, is going to be dependent on some forward constant times just the concentration of the boron trifluoride. The water's everywhere, so you don't have to multiply it times the concentration of water, whatever that means, because the water's everywhere. So the denominator here, you do not put the solvent. So the correct answer for this one is you only put whatever is actually dissolved in the solution. Because frankly, the concentration doesn't actually makes sense for everything else, and if you think about it from the probability point of view, that also makes sense, because there's always water around. If you said, OK, what's the probability of finding water at any small volume of fluid so you could just multiply it by a 1 there, but that doesn't make a difference. Now, what about the following reaction? Any equilibrium where you have different states of matter is called a heterogeneous equilibrium. So let's say I have H2O in the gaseous state and that's essentially steam -- so it's not going to be the solvent this time-- plus carbon in the solid state. And let's say that that's an equilibrium with hydrogen in the gas state plus carbon dioxide in the gaseous state. This is a heterogeneous equilibrium because you have things in the gaseous and the solid state. And solid state, by definition, it can't be dissolved either into the gas or into the -- when we talk about solutions, we talked about colloids and suspensions and mixtures before, but we're talking about solutions. By definition, if this is in the solid state, it's not dissolved. If this was dissolved, we would write an aq here. So if you talk about the forward reaction, what's the forward reaction going to be dependent on? So the rate forward, well, the solid, there's a big block of carbon sitting there. There's a big cube of carbon there, and there's steam, there's water gas all around it. So if you pick any volume, especially if you pick some volume near the boundary of the carbon, you're always going to have carbon around. It's just what matters is the concentration of the water gas. That's what's going to drive the forward rate, so the forward rate is going to be dependent on some constant times the concentration of the water gas. And, of course, the backwards rate, so you need to get some h2, some molecules of -- let me draw it like that, because it has 2 hydrogen molecules plus a carbon dioxide," + }, + { + "Q": "At 7:12 Sal mentioned about buoyancy effect. What do you mean by that ?\n", + "A": "buoyancy is upthrust acting on a body", + "video_name": "R5CRZONOHCU", + "timestamps": [ + 432 + ], + "3min_transcript": "And then I want to multiply that times the mass of Earth, which is right over here. That is 5.9722 times 10 to the 24th. So times 10 to the 24th power. And we want to divide that by the radius of Earth squared. So divided by the radius of Earth is-- so this is in kilometers. And I just want to make sure that everything is the same units. So 6,371 kilometers-- actually, let me scroll over. Well, you can't see the kilometers right now. But this is kilometers. It is the same thing as 6,371,000 meters. If you just multiply this by 1,000. Or you could even write this as 6.371. 6.371 times 10 to the sixth meters. That's the radius of the Earth. The distance between the center of mass of Earth and the center of mass of this object, which is sitting at the surface of the Earth. And so let's get our drum roll. And we get 9.8. And if we round, we actually get something a little bit higher than what the textbooks give us. We get 9.82. Let's just round. So we get 9.82-- 9.82 meters per second squared. And so you might say, well, what's going on here? Why do we have this discrepancy between what the universal law of gravitation gives us and what the average measured acceleration due to the force of gravity at the surface of the Earth. And the discrepancy here, the discrepancy between these two numbers, is really because Earth is not a uniform sphere of uniform density. And that's what we have to assume over here when we use the universal law of gravitation. It's actually a little bit flatter than a perfect sphere. The different layers of the Earth have different densities. You have all sorts of different interactions. And then you also, if you measure effective gravity, there's also a little bit of a buoyancy effect from the air. Very, very, very, very negligible, I don't know if it would have been enough to change this. But there's other minor, minor effects, irregularities. Earth is not a perfect sphere. It is not of uniform density. And that's what accounts for the bulk of this. Now, with that out of the way, what I'm curious about is what is the acceleration due to gravity if we go up 400 kilometers? So now, the main difference here, g will stay the same. The mass of Earth will stay the same, but the radius is now going to be different. Because now we're placing the center of mass of our object-- whether it's a space station or someone sitting in the space station, they're going to be 400 kilometers higher. And I'm going to exaggerate what 400 kilometers looks like. This is not drawn to scale. But now the radius is going to be the radius of the Earth plus 400 kilometers." + }, + { + "Q": "\n8:50 I did not understand the equation. It sounds logical that if we have parallel flow the resistance is lower, but how do we get to the equation?", + "A": "We have less resistance when there is less flow. When we double the amount of identical paths, we halve the amount of flow through each individual path. By doubling the amount of paths, we halve the resistance. Opening more doors makes it easier to exit the theater. Even if only one person gets out through the new door, everyone else experiences less competition for all the other doors.", + "video_name": "E-q9JpkGc-8", + "timestamps": [ + 530 + ], + "3min_transcript": "Here it stays at 2. But here in the middle, it goes from 2 to 32 because it's 16 times greater. So you end up increasing the resistance in the middle section by a lot. So let me just write that out for you. So 2 times 16 gets us to 32. So here the resistance is 32. And so if I wanted to calculate the total resistance, I'd get something like this-- 32 plus 2 plus 2 is 36. So I actually went from 6 to 36 when this blood clot came and clogged up part of that vessel. So just keep that in mind. We'll talk about that a little bit more, but I just wanted to use this example and also kind of cement the idea of what you do with resistance in a series. Let's contrast that to a different situation. And this is when you have resistance in parallel. through all of my vessels, I could also do something like this-- I could say, well, let's say, I have three vessels again. And this time, I'm going to change the length and the radius. And let's say this one's really big. And the resistance here, let's say, is 5, here is 10, and here is 6. So you've got three different resistances. And the blood now can choose to go through any one of these paths. It doesn't have to go through all three. So how do I figure out now what the total resistance is? So what is the total resistance? Well, the total resistance this time is going to be 1 over 1 R1, plus 1 over R2, plus 1 over R3. And you can go on and on just as before. So let's just put that there, that there, and that there. And I can figure this out pretty easily. So I can say 1 over 1 over 6 plus 1 over 10 plus 1/5. And the common denominator there is 30. So I could say 5/30. This is 3/30, and this would be 6/30. And adding that up together, I get 1 over 14/30 or 30 over 14, which is 2 and let's say 0.1. So 2.1. So the total resistance here is 2.1. Putting all three of these together is pretty interesting. And I want you to realize that the resistance in total is actually less than any component part." + }, + { + "Q": "\nAt 2:53 why we only use centripetal acceleration on the left side of the equation? Shouldn't it be centripetal acceleration plus acceleration due to gravity? When the ball is in the air isn't it always accelerating downwards due to Earth's gravity?", + "A": "I believe, it is because the direction chosen in the free body/force diagram is towards the centre of the circular path. The effect of gravity is included in the weight or gravity force( m x g ).", + "video_name": "2lcaBPLLoLo", + "timestamps": [ + 173 + ], + "3min_transcript": "going four meters per second? And if it's a force you want to find, the first step always is to draw a quality force diagram. So let's do that here. Let's ask what forces are on this yo-yo. Well, if we're near Earth and we're assuming we're going to be near the surface of the Earth playing with our yo-yo, there's gonna be a force of gravity and that force of gravity is gonna point straight downward. So the magnitude of that force of gravity is gonna be m times g, where g is positive 9.8. g represents the magnitude of the acceleration due to gravity, and this expression here represents the magnitude of the force of gravity. But there's another force. The string is tied to the mass, so this string can pull on the mass. Strings pull, they exert a force of tension. Which way does that tension go? A lot of people want to draw that tension going upward, and that's not good. Ropes can't push. If you don't believe me, go get a rope, try to push on something. You'll realize, oh yeah, it can't push, but it can pull. So that's what this rope's gonna do. This rope's gonna pull. How much? I don't know. This is gonna be the force of tension right here, and we'll label it with a capital T. We could have used F with the sub T. There's different ways to label the tension, but no matter how you label it, that tension points in towards the center of the circle 'cause this rope is pulling on the mass. So, after you draw a force diagram, if you want to find a force, typically, you're just gonna use Newton's second law. And we're gonna use this formula as always in one dimension at a time so vertically, horizontally, centripetally, one dimension at a time to make the calculations as simple as possible. And since we have a centripetal motion problem, we have an object going in a circle, and we want to find one of those forces that are directed into the circle, we're gonna use Newton's second law for the centripetal direction. So we'll use centripetal acceleration here and net force centripetally here. So in other words, we're gonna write down that the centripetal acceleration is gonna be equal to the net centripetal force exerted on the mass that's going around in that circle. So because we chose the centripetal direction, the centripetal acceleration with the formula for centripetal acceleration. The centripetal acceleration's always equivalent to v squared over r, the speed of the object squared divided by the radius of the circle that the object is traveling in. So we set that equal to the net centripetal force over the mass, and the trickiest part here, the part where the failure's probably gonna happen is trying to figure out what do you plug in for the centripetal force. And now we gotta decide what is acting as our centripetal force and plug those in here with the correct signs. So, let's just see what forces we have on our object. There's a force of tension and a force of gravity. So, when you go try to figure out what to plug in here, people start thinking, they start looking all over. No, you drew your force diagram. Look right there. Our force diagram holds all the information about all the forces that we've got as long as we drew it well. And we did draw it well. We included all the forces, so we'll just go one by one. Should we include, should we even include the force of gravity in this centripetal force calculation?" + }, + { + "Q": "\nin 1:42 the example, does that actually work?", + "A": "Try it! Put a pot of water on a stove and only heat one side. See what happens!", + "video_name": "f8GK2oEN-uI", + "timestamps": [ + 102 + ], + "3min_transcript": "What I want to do in this video is talk a little bit about maybe why the plates are actually moving in the first place. And nothing I'm talking about in this video has been definitively proved. This is just kind of the current thinking, the leading thinking, on why plates are actually moving. Although we haven't seen the definitive evidence yet, and it's probably a combination of a bunch of things. Now, before we even talk about plates, let's just talk about convection. And you might already be familiar with the term, but just in case you're not, let's do a little bit of review of convection. So let's say I have a pot over here. So that is my pot. And it contains some water. So I have water in my pot. And let's say I only heat one end of the pot. So I put a flame right over at that end of the pot. So what's going to happen? Well, the water that's right over the flame is going to be warmed up more than any of the other water. So this water is going to get warm. But when it gets warm, it also becomes less dense. the molecules are vibrating more. They have more kinetic energy. They're going to a bounce further distances away from each other. They will become less dense. And if you have something that's less dense, and it's surrounded by things that are more dense, and we're dealing in kind of a fluid state right here, that warm, less dense water is going to move upwards. It's going to move upwards. Well, when it moves upwards something has to replace it. So you're going to have cooler water from this side of the container kind of replacing where that water was. Now, this water, as it rises, what's going to happen to it? Well, it's going to cool down. It's going to get further from the flame. It's going to mix with maybe some of the other water, or transfer some of its kinetic energy to the neighboring water. So it'll cool down. But once it cools down, what's it going to want to do? Remember, in general, the closer you are to the flame. is going to be warmer. So all of this stuff is going to be warmer, and all of this stuff up here, like the coldest water, is always going to be furthest from the flame. And so the coldest water is going to be over here. But remember, the coldest water is also the densest water. So this water over here is dense. And so it will sink. It's denser than the water around it. And it also helps replace the water that's going here to get warmed up again. And so what you do is you have this cycle here. Warm water rises, moves over to the right down here, and then goes back down as it cools down, and it's dense, and then it gets warmed up again. And so this process, essentially what it's doing is it's transferring the heat. It's allowing the heat to be transferred from this one spot throughout the fluid. And so we call this process, this is convection. Now, the reason why we think the plates are moving is because we think that there are similar types of convection" + }, + { + "Q": "\nYou know, that is something I wondered. At 4:54, Sal says \"degrees Kelvin\", even though it's supposed to be just plain \"Kelvin\". I know that. I'm cool with that. But why is that? Why are the other temperature readings in degrees but Kelvin is not?", + "A": "Because the other scales don t actually measure a physical quantity, just relative warmth. 0 degrees F is not zero somethings . Same for Celsius. But zero kelvin represents zero thermal energy, and 100 K is twice as hot as 50 K.", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 294 + ], + "3min_transcript": "So for example, the lowest possible temperature that can be achieved in the universe, when you think about it in Celsius, let me draw a little temperature scale here. So if that's the temperature scale. I'll draw two, one for Celsius and one for Kelvin. So the lowest possible temperature that can be achieved in the universe, and when we say the lowest possible temperature that means that the average kinetic energy of the molecules or the atoms are zero. They're just not moving. They're just stationary. So in Celsius, it's minus 273.15 degrees Celsius. So zero might be some place over here. Zero, that's where water freezes. And then 100 degrees, that's where water boils. And you can immediately see, the whole Celsius scale and the boiling point of water. Now, Kelvin. So look at this and you say, if I have something that's 5 degrees and I have another thing that's 10 degrees, when you look at the Celsius scale, you're like, oh, maybe the 10 degree thing it has twice as much energy as the 5 degree thing. It has twice the temperature. But when you look at it from the absolute distance to zero. Let me see if I can draw this. So the 10 degree is all the way over here and the 5 degree is almost as far, that far. So the 10 degrees Celsius is only a slight increment over 5 degrees Celsius, if you were to divide the two. It's not twice as hot. And that's why they came up with the Kelvin scale. Because in the Kelvin scale, absolute zero is defined as 0. Zero Kelvin. So this right here is zero degrees Kelvin. And so zero degrees Kelvin is absolute zero. So what is zero degrees Celsius? And the increments are the same. is one degree change in Kelvin. So at least they keep it, it's just a shift. So this is going to be plus 273 degrees Kelvin. And then 5 degrees would be plus 278 10 degrees would be plus 283 Kelvin. And then you see that 5 and 10 degrees really aren't that different from each other. But in general, if you want to convert from Celsius to Kelvin you just add 273 degrees. So 30 degrees Celsius is what? Well, this 5 and 10 I drew too close to 100. But let's say it's sitting here. It would be 303 degrees Kelvin. So this is equal to 303 degrees Kelvin. All right, so now for our temperature, that's what we were worried about. We wanted to put in the temperature there. So now we can put in our 303 degrees Kelvin. Now we have to figure out what constant to use here." + }, + { + "Q": "At 3:24 Sal says the lowest possible temperature in C is -273.15 Degrees C, and I was wondering, what would happen if we could quickly freeze a person to -273.15 Degrees C? Then if we could return their temperature back to a normal temperature, would they be ok?\nI know it sounds kind of odd, but I was considering forensic science, and just wondering....\nThanks!- MP\n", + "A": "Technically speaking, you cannot actually reach absolute zero, because the would be a state of no thermal energy at all. But we can get very close to it. No, the person would die.", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 204 + ], + "3min_transcript": "So each molecule has two hydrogens in it. And let's say I'm measuring it at 30 degrees Celsius. Use different color. 30 degrees Celsius. My brain is really malfunctioning. 30 degrees, not 30 percent, 30 degrees Celsius. And let's say that the pressure on the outside of the balloon, we've measuredat two atmospheres. So my question to you is how many moles of hydrogen do we have? How many moles... So let's apply our ideal gas equation. And since we're dealing with liters and atmospheres, But in general, if we keep pressure. So our pressure is given in atmospheres. Let me write down all the units, actually. So we have 2 atmospheres times our volume is 2 liters, is equal to n. n is the number of particles we care about, and we care about it in moles, but let's just write n there for now. Is equal to n times R. I'll do R in a second times. R times T. Now you might be atempted to just put 30 degrees in there. But in all of these problems-- in fact in general, whenever you're doing any of these gas problems or thermodynamics problems, or any time you're doing math with temperature-- you should always convert into Kelvin. And just as a bit of review as to what Kelvin is, So for example, the lowest possible temperature that can be achieved in the universe, when you think about it in Celsius, let me draw a little temperature scale here. So if that's the temperature scale. I'll draw two, one for Celsius and one for Kelvin. So the lowest possible temperature that can be achieved in the universe, and when we say the lowest possible temperature that means that the average kinetic energy of the molecules or the atoms are zero. They're just not moving. They're just stationary. So in Celsius, it's minus 273.15 degrees Celsius. So zero might be some place over here. Zero, that's where water freezes. And then 100 degrees, that's where water boils. And you can immediately see, the whole Celsius scale" + }, + { + "Q": "At 6:12 I understand why you would pick a certain R value, but how are they derived? On my chemistry worksheet they aren't given or anything.\n", + "A": "R is nothing more than the Boltzmann constant multiplied by Avogadro s constant with a few unit conversion factors thrown in to adjust for whatever units you are using.", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 372 + ], + "3min_transcript": "and the boiling point of water. Now, Kelvin. So look at this and you say, if I have something that's 5 degrees and I have another thing that's 10 degrees, when you look at the Celsius scale, you're like, oh, maybe the 10 degree thing it has twice as much energy as the 5 degree thing. It has twice the temperature. But when you look at it from the absolute distance to zero. Let me see if I can draw this. So the 10 degree is all the way over here and the 5 degree is almost as far, that far. So the 10 degrees Celsius is only a slight increment over 5 degrees Celsius, if you were to divide the two. It's not twice as hot. And that's why they came up with the Kelvin scale. Because in the Kelvin scale, absolute zero is defined as 0. Zero Kelvin. So this right here is zero degrees Kelvin. And so zero degrees Kelvin is absolute zero. So what is zero degrees Celsius? And the increments are the same. is one degree change in Kelvin. So at least they keep it, it's just a shift. So this is going to be plus 273 degrees Kelvin. And then 5 degrees would be plus 278 10 degrees would be plus 283 Kelvin. And then you see that 5 and 10 degrees really aren't that different from each other. But in general, if you want to convert from Celsius to Kelvin you just add 273 degrees. So 30 degrees Celsius is what? Well, this 5 and 10 I drew too close to 100. But let's say it's sitting here. It would be 303 degrees Kelvin. So this is equal to 303 degrees Kelvin. All right, so now for our temperature, that's what we were worried about. We wanted to put in the temperature there. So now we can put in our 303 degrees Kelvin. Now we have to figure out what constant to use here. Remember, we're dealing with atmospheres and liters. So I wrote down a couple of versions of R right here. Let's see we're dealing with atmospheres and liters. And in the denominator we're always dealing with mole and Kelvin no matter what. So those are always going to be there. So we should use this proportionality constant. R is equal to 0.082 liter atmospheres per mole Kelvin. Let me write that down. So let me rewrite our whole equation actually. So I have 2 atmospheres times 2 liters is equal to n times, I have a bad memory, 0.082 liter atmospheres per mole Kelvin, times 303 degrees Kelvin. So let's see what we can do. Let's see if all of the units work out." + }, + { + "Q": "At 6:11, where did Sal come up with the numbers for R?\n", + "A": "R is the universal or ideal gas constant in PV = nRT. These problems are called ideal gas equations because it is assumed that the gas behaves in an ideal manner, which would allow us to use the constant that Sal shows. It is a constant that makes the equation work... you ll either have a table or have to memorize this number, unfortunately.", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 371 + ], + "3min_transcript": "and the boiling point of water. Now, Kelvin. So look at this and you say, if I have something that's 5 degrees and I have another thing that's 10 degrees, when you look at the Celsius scale, you're like, oh, maybe the 10 degree thing it has twice as much energy as the 5 degree thing. It has twice the temperature. But when you look at it from the absolute distance to zero. Let me see if I can draw this. So the 10 degree is all the way over here and the 5 degree is almost as far, that far. So the 10 degrees Celsius is only a slight increment over 5 degrees Celsius, if you were to divide the two. It's not twice as hot. And that's why they came up with the Kelvin scale. Because in the Kelvin scale, absolute zero is defined as 0. Zero Kelvin. So this right here is zero degrees Kelvin. And so zero degrees Kelvin is absolute zero. So what is zero degrees Celsius? And the increments are the same. is one degree change in Kelvin. So at least they keep it, it's just a shift. So this is going to be plus 273 degrees Kelvin. And then 5 degrees would be plus 278 10 degrees would be plus 283 Kelvin. And then you see that 5 and 10 degrees really aren't that different from each other. But in general, if you want to convert from Celsius to Kelvin you just add 273 degrees. So 30 degrees Celsius is what? Well, this 5 and 10 I drew too close to 100. But let's say it's sitting here. It would be 303 degrees Kelvin. So this is equal to 303 degrees Kelvin. All right, so now for our temperature, that's what we were worried about. We wanted to put in the temperature there. So now we can put in our 303 degrees Kelvin. Now we have to figure out what constant to use here. Remember, we're dealing with atmospheres and liters. So I wrote down a couple of versions of R right here. Let's see we're dealing with atmospheres and liters. And in the denominator we're always dealing with mole and Kelvin no matter what. So those are always going to be there. So we should use this proportionality constant. R is equal to 0.082 liter atmospheres per mole Kelvin. Let me write that down. So let me rewrite our whole equation actually. So I have 2 atmospheres times 2 liters is equal to n times, I have a bad memory, 0.082 liter atmospheres per mole Kelvin, times 303 degrees Kelvin. So let's see what we can do. Let's see if all of the units work out." + }, + { + "Q": "At 1:37, how big is one atm?\n", + "A": "One atm is the amount of air pressure that is normally on you if you are near sea level.", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 97 + ], + "3min_transcript": "In the last video we hopefully learned the intuition behind the ideal gas equation, that pressure times volume is equal to the number of molecules we have times some constant times the temperature. And that's all nice and it hopefully it makes sense to you how all of these fit together. That pressure should be inverse to volume and that's why you're multiplying both sides by each other. You could take volume and put it on this side of the equation. Or that pressure should be proportional to the number of particles and the temperature. But now let's apply it and actually do some problems. Because just knowing this isn't good enough. So let's say that I have a two liter container, or let's say a two liter balloon, containing hydrogen gas. So each molecule has two hydrogens in it. And let's say I'm measuring it at 30 degrees Celsius. Use different color. 30 degrees Celsius. My brain is really malfunctioning. 30 degrees, not 30 percent, 30 degrees Celsius. And let's say that the pressure on the outside of the balloon, we've measuredat two atmospheres. So my question to you is how many moles of hydrogen do we have? How many moles... So let's apply our ideal gas equation. And since we're dealing with liters and atmospheres, But in general, if we keep pressure. So our pressure is given in atmospheres. Let me write down all the units, actually. So we have 2 atmospheres times our volume is 2 liters, is equal to n. n is the number of particles we care about, and we care about it in moles, but let's just write n there for now. Is equal to n times R. I'll do R in a second times. R times T. Now you might be atempted to just put 30 degrees in there. But in all of these problems-- in fact in general, whenever you're doing any of these gas problems or thermodynamics problems, or any time you're doing math with temperature-- you should always convert into Kelvin. And just as a bit of review as to what Kelvin is," + }, + { + "Q": "\nThis might be a silly question but at 7:42, how come the (.082 L . atm/mole . K) becomes 1/mole??\n\nCheers", + "A": "Just look at the units. If you ve used consistent units then RT/PV has the units of 1/mol because the pressure and volume units have cancelled out. Or, a more useful way of looking at it is : PV = nRT divide by RT PV/RT = n Thus, PV/RT has units of mol, with all the other units cancelling out.", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 462 + ], + "3min_transcript": "Remember, we're dealing with atmospheres and liters. So I wrote down a couple of versions of R right here. Let's see we're dealing with atmospheres and liters. And in the denominator we're always dealing with mole and Kelvin no matter what. So those are always going to be there. So we should use this proportionality constant. R is equal to 0.082 liter atmospheres per mole Kelvin. Let me write that down. So let me rewrite our whole equation actually. So I have 2 atmospheres times 2 liters is equal to n times, I have a bad memory, 0.082 liter atmospheres per mole Kelvin, times 303 degrees Kelvin. So let's see what we can do. Let's see if all of the units work out. you can treat units like numbers. So if you divide both sides of this equation by atmospheres, the atmospheres cancel out. Divide both sides of this equation by liters, liters cancel out. You have a Kelvin in the numerator, Kelvin in the denominator, that cancels out. And so we have 2 times 2 is equal to n times 0.082 times 303. And then we have just a per mole and a 1 over the mole. So to solve for n, or the number of moles, what we do is we divide both sides of this equation by all of this stuff. So we get 2 times 2 is 4. 4 divided by 0.082 divided by 303. dividing both sides by it. And when you divide by a per mole, if you put a 1 over a mole here, that's the same thing as multiplying by a mole. So it's good, the units all worked out. We're getting n in terms of moles. And so we just have to get the calculator out and figure out how many moles we're dealing with. So we have 4 divided by 0.082 divided by 303 is equal to 0.16. If we wanted to go more digits, 0.161, but we'll just round. So this is equal to 0.16 moles of H2. I am telling you actually here, the exact number of hydrogen molecules. But if you wanted a number, you'd just multiply this times 6.02 times 10 to the 23 and then you would have a number in kind of the traditional sense. And of course, if you wanted to know" + }, + { + "Q": "at 4:04 why does it say peed in orange?\n", + "A": "I think it reads PE = 0. That is, the potential energy is equal to zero.", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 244 + ], + "3min_transcript": "It cannot be created or destroyed. It can just be converted from one form to another. But I'm just showing you, this object had 100 joules of energy or, in this case, gravitational potential energy. And down here, it has no energy. Or at least it has no gravitational potential energy, and that's the key. That gravitational potential energy was converted into something else. And that something else it was converted into is kinetic energy. And in this case, since it has no potential energy, all of that previous potential energy, all of this 100 joules that it has up here is now going to be converted into kinetic energy. And we can use that information to figure out its velocity right before it hits the ground. So how do we do that? Well what's the formula for kinetic energy? And we solved it two videos ago, and hopefully it shouldn't be too much of a mystery to you. It's something good to memorize, but it's also good to know how we got it and go back two videos if you forgot. into kinetic energy. We had 100 joules of potential energy, so we're still going to have 100 joules, but now all of it's going to be kinetic energy. And kinetic energy is 1/2 mv squared. So we know that 1/2 mv squared, or the kinetic energy, is now going to equal 100 joules. What's the mass? The mass is 1. And we can solve for v now. 1/2 v squared equals 100 joules, and v squared is equal to 200. And then we get v is equal to square root of 200, which is something over 14. We can get the exact number. Let's see, 200 square root, 14.1 roughly. The velocity is going to be 14.1 meters per second squared downwards. Right before the object touches the ground. Right before it touches the ground. And you might say, well Sal that's nice and everything. We learned a little bit about energy. solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it." + }, + { + "Q": "\nAt 4:04, Sal said you could solve this problem using kinematic formulas but how could you? Don't you only know distance and acceleration?", + "A": "s = 1/2*a*t^2", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 244 + ], + "3min_transcript": "It cannot be created or destroyed. It can just be converted from one form to another. But I'm just showing you, this object had 100 joules of energy or, in this case, gravitational potential energy. And down here, it has no energy. Or at least it has no gravitational potential energy, and that's the key. That gravitational potential energy was converted into something else. And that something else it was converted into is kinetic energy. And in this case, since it has no potential energy, all of that previous potential energy, all of this 100 joules that it has up here is now going to be converted into kinetic energy. And we can use that information to figure out its velocity right before it hits the ground. So how do we do that? Well what's the formula for kinetic energy? And we solved it two videos ago, and hopefully it shouldn't be too much of a mystery to you. It's something good to memorize, but it's also good to know how we got it and go back two videos if you forgot. into kinetic energy. We had 100 joules of potential energy, so we're still going to have 100 joules, but now all of it's going to be kinetic energy. And kinetic energy is 1/2 mv squared. So we know that 1/2 mv squared, or the kinetic energy, is now going to equal 100 joules. What's the mass? The mass is 1. And we can solve for v now. 1/2 v squared equals 100 joules, and v squared is equal to 200. And then we get v is equal to square root of 200, which is something over 14. We can get the exact number. Let's see, 200 square root, 14.1 roughly. The velocity is going to be 14.1 meters per second squared downwards. Right before the object touches the ground. Right before it touches the ground. And you might say, well Sal that's nice and everything. We learned a little bit about energy. solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it." + }, + { + "Q": "At 2:15 how does the object have no energy?I thought that everything always has energy\n", + "A": "everything has some combination of potential energy, kinetic energy, and internal energy...the internal energy of the block (related to the material at the atomic level) doesn t change during this example so it got left out of the explanation...but, yes, everything does always have some energy in the real world", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 135 + ], + "3min_transcript": "Welcome back. At the end of the last video, I left you with a bit of a question. We had a situation where we had a 1 kilogram object. This is the 1 kilogram object, which I've drawn neater in this video. That is 1 kilogram. And we're on earth, and I need to mention that because gravity is different from planet to planet. But as I mentioned, I'm holding it. Let's say I'm holding it 10 meters above the ground. So this distance or this height is 10 meters. And we're assuming the acceleration of gravity, which we also write as just g, let's assume it's just 10 meters per second squared just for the simplicity of the math instead of the 9.8. So what we learned in the last video is that the potential energy in this situation, the potential energy, which equals m times g times h is equal to the mass is 1 kilogram times the acceleration of gravity, which is 10 I'm not going to write the units down just to save space, although you should do this when you do it on your test. And then the height is 10 meters. And the units, if you work them all out, it's in newton meters or joules and so it's equal to 100 joules. That's the potential energy when I'm holding it up there. And I asked you, well when I let go, what happens? Well the block obviously will start falling. And not only falling, it will start accelerating to the ground at 10 meters per second squared roughly. And right before it hits the ground-- let me draw that in brown for ground-- right before the object hits the ground or actually right when it hits the ground, what will be the potential energy of the object? Well it has no height, right? Potential energy is mgh. The mass and the acceleration of gravity stay the same, but the height is 0. So they're all multiplied by each other. So down here, the potential energy is going to be equal to 0. And I told you in the last video that we have the law of conservation of energy. It cannot be created or destroyed. It can just be converted from one form to another. But I'm just showing you, this object had 100 joules of energy or, in this case, gravitational potential energy. And down here, it has no energy. Or at least it has no gravitational potential energy, and that's the key. That gravitational potential energy was converted into something else. And that something else it was converted into is kinetic energy. And in this case, since it has no potential energy, all of that previous potential energy, all of this 100 joules that it has up here is now going to be converted into kinetic energy. And we can use that information to figure out its velocity right before it hits the ground. So how do we do that? Well what's the formula for kinetic energy? And we solved it two videos ago, and hopefully it shouldn't be too much of a mystery to you. It's something good to memorize, but it's also good to know how we got it and go back two videos if you forgot." + }, + { + "Q": "\nAt 9:00- 9:30 Sal talks about how the velocity can be determined, but wouldn't that be affected by the shape of the slope? Which we haven't taken into account? An upslope at that particular height would cause a different velocity than a downslope surely?\nPlease clarify", + "A": "Your argument is totally correct ,but according to the question we have to only find the velocity of the body on the surface and not on the iceberg.", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 540, + 570 + ], + "3min_transcript": "But what else can this do for me? And this is where it's really cool. Not only can I figure out the velocity when all of the potential energy has disappeared, but I can figure out the velocity of any point-- and this is fascinating-- along this slide. So let's say when the box is sliding down here, so let's say the box is at this point. It changes colors too as it falls. So this is the 1 kilogram box, right? It falls and it slides down here. And let's say at this point it's height above the ground is 5 meters. So what's its potential energy here? So let's just write something. All of the energy is conserved, right? So the initial potential energy plus the initial kinetic energy is equal to the final potential energy plus the final kinetic energy. I'm just saying energy is conserved here. Well the potential energy is 100 and the kinetic energy is 0 because it's stationary. I haven't dropped it. I haven't let go of it yet. It's just stationary. So the initial energy is going to be equal to 100 joules. That's cause this is 0 and this is 100. So the initial energy is 100 joules. At this point right here, what's the potential energy? Well we're 5 meters up, so mass times Mass is 1, times gravity, 10 meters per second squared. Times height, times 5. So it's 50 joules. That's our potential energy at this point. And then we must have some kinetic energy with the velocity going roughly in that direction. Plus our kinetic energy at this point. And we know that no energy was destroyed. It's just converted. So we know the total energy still has to be 100 joules. So essentially what happened, and if we solve for this-- the kinetic energy is now also going to be equal to 50 joules. Halfway down, essentially half of the potential energy got converted to kinetic energy. And we can use this information that the kinetic energy is 50 joules to figure out the velocity at this point. 1/2 mv squared is equal to 50. The mass is 1. Multiply both sides by 2. You get v squared is equal to 100. The velocity is 10 meters per second along this crazy, icy slide. And that is something that I would have challenged you to solve using traditional kinematics formulas, especially considering that we don't know really much about the surface of this slide. And even if we did, that would have been a million times harder than just using the law of conservation of energy and realizing that at this point, half the potential energy is now kinetic energy and it's going along the direction of the slide." + }, + { + "Q": "At 6:15 , all the potential energy gets converted to K.E.\nLet us say it fell straight down. At the last instant all the P.E. would be converted into K.E.\nBut then it hits the ground and will come to a rest. Then its height = 0 and so is its velocity. So both P.E. and K.E. are zero, then where did that energy go?\n", + "A": "That energy went into impacting the ground - making noise, deforming it, heating it.", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 375 + ], + "3min_transcript": "solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it. This is the ground down here. This is the ground. So what's going to happen this time? I'm still 10 meters in the air, so let me draw that. That's still 10 meters. I should switch colors just so not everything is ice. So that's still 10 meters, but instead of the object going straight down now, it's going to go down here and then start It's going to go sliding along this hill. And then at this point it's going to be going really fast in the horizontal direction. And right now we don't know how fast. And just using our kinematics formula, this would have been a really tough formula. This would have been difficult. I mean you could have attempted it and it actually would have taken calculus because the angle of the slope changes continuously. We don't even know the formula for the angle of the slope. You would have had to break it out into vectors. You would have to do all sorts of complicated things. This would have been a nearly impossible problem. But using energy, we can actually figure out what the velocity of this object is at this point. And we use the same idea. Here we have 100 joules of potential energy. Down here, what's the height above the ground? Well the height is 0. So all the potential energy has disappeared. And just like in the previous situation, all of the potential energy is now converted into kinetic energy. And so what is that kinetic energy going to equal? It's going to be equal to the initial potential energy. So here the kinetic energy is equal to 100 joules. And that equals 1/2 mv squared, just like we just solved. And if you solve for v, the mass is 1 kilogram. So the velocity in the horizontal direction will be, if you solve for it, 14.1 meters per second. Instead of going straight down, now it's going to be going in the horizontal to the right. And the reason why I said it was ice is because I wanted this to be frictionless and I didn't want any energy lost to heat or anything like that. And you might say OK Sal, that's kind of interesting. And you kind of got the same number for the velocity than if I just dropped the object straight down." + }, + { + "Q": "\nAt 6:10 when the object is sliding , some of the energy will be converted to heat energy due to fiction right?? So, all the energy will not be converted to kinetic energy right?", + "A": "That s correct", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 370 + ], + "3min_transcript": "solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it. This is the ground down here. This is the ground. So what's going to happen this time? I'm still 10 meters in the air, so let me draw that. That's still 10 meters. I should switch colors just so not everything is ice. So that's still 10 meters, but instead of the object going straight down now, it's going to go down here and then start It's going to go sliding along this hill. And then at this point it's going to be going really fast in the horizontal direction. And right now we don't know how fast. And just using our kinematics formula, this would have been a really tough formula. This would have been difficult. I mean you could have attempted it and it actually would have taken calculus because the angle of the slope changes continuously. We don't even know the formula for the angle of the slope. You would have had to break it out into vectors. You would have to do all sorts of complicated things. This would have been a nearly impossible problem. But using energy, we can actually figure out what the velocity of this object is at this point. And we use the same idea. Here we have 100 joules of potential energy. Down here, what's the height above the ground? Well the height is 0. So all the potential energy has disappeared. And just like in the previous situation, all of the potential energy is now converted into kinetic energy. And so what is that kinetic energy going to equal? It's going to be equal to the initial potential energy. So here the kinetic energy is equal to 100 joules. And that equals 1/2 mv squared, just like we just solved. And if you solve for v, the mass is 1 kilogram. So the velocity in the horizontal direction will be, if you solve for it, 14.1 meters per second. Instead of going straight down, now it's going to be going in the horizontal to the right. And the reason why I said it was ice is because I wanted this to be frictionless and I didn't want any energy lost to heat or anything like that. And you might say OK Sal, that's kind of interesting. And you kind of got the same number for the velocity than if I just dropped the object straight down." + }, + { + "Q": "\nIn example at 9:00 we evaluate velocity knowing PE, height and mass. But what about a slope? If you would drown the slope differently, let's say an infinite straight line at height 5 m. , then an object would have stopped after some amount of time due to friction or air resistance. Does Sal assumes that there are no other forces but gravity?", + "A": "yes, he s ignoring air resistance and friction so that you can focus on the concepts of PE and KE.", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 540 + ], + "3min_transcript": "But what else can this do for me? And this is where it's really cool. Not only can I figure out the velocity when all of the potential energy has disappeared, but I can figure out the velocity of any point-- and this is fascinating-- along this slide. So let's say when the box is sliding down here, so let's say the box is at this point. It changes colors too as it falls. So this is the 1 kilogram box, right? It falls and it slides down here. And let's say at this point it's height above the ground is 5 meters. So what's its potential energy here? So let's just write something. All of the energy is conserved, right? So the initial potential energy plus the initial kinetic energy is equal to the final potential energy plus the final kinetic energy. I'm just saying energy is conserved here. Well the potential energy is 100 and the kinetic energy is 0 because it's stationary. I haven't dropped it. I haven't let go of it yet. It's just stationary. So the initial energy is going to be equal to 100 joules. That's cause this is 0 and this is 100. So the initial energy is 100 joules. At this point right here, what's the potential energy? Well we're 5 meters up, so mass times Mass is 1, times gravity, 10 meters per second squared. Times height, times 5. So it's 50 joules. That's our potential energy at this point. And then we must have some kinetic energy with the velocity going roughly in that direction. Plus our kinetic energy at this point. And we know that no energy was destroyed. It's just converted. So we know the total energy still has to be 100 joules. So essentially what happened, and if we solve for this-- the kinetic energy is now also going to be equal to 50 joules. Halfway down, essentially half of the potential energy got converted to kinetic energy. And we can use this information that the kinetic energy is 50 joules to figure out the velocity at this point. 1/2 mv squared is equal to 50. The mass is 1. Multiply both sides by 2. You get v squared is equal to 100. The velocity is 10 meters per second along this crazy, icy slide. And that is something that I would have challenged you to solve using traditional kinematics formulas, especially considering that we don't know really much about the surface of this slide. And even if we did, that would have been a million times harder than just using the law of conservation of energy and realizing that at this point, half the potential energy is now kinetic energy and it's going along the direction of the slide." + }, + { + "Q": "\nAround 9:05 Sal says that glycolysis is aerobic but , my understanding is that a cell without enough oxygen will go through fermentation instead of glycolysis. So, how is glycolysis still anaerobic?", + "A": "fermentation happens instead of the TCA cycle and oxidative phosphorylation and thus after glycolysis. Glycolysis does not require oxygen, and as such it is considered anaerobic; however, unlike fermentation it still occurs in the presence of oxygen.", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 545 + ], + "3min_transcript": "that's added on to that. You know, these things are all bonded to other things, with oxygens and hydrogens and whatever. But each of these 3-carbon backbone molecules are called pyruvate. We'll go into a lot more detail on that. But glycolysis, it by itself generates-- well, it needs two ATPs. And it generates four ATPs. So on a net basis, it generates two-- let me write this in a different color-- it generates two net ATPs. So that's the first stage. And this can occur completely in the absence of oxygen. I'll do a whole video on glycolysis in the future. Then these byproducts, they get re-engineered a little bit. And then they enter into what's called the Krebs cycle. And then, and this is kind of the interesting point, there's another process that you can say happens after the Krebs cycle. But we're in a cell and everything's bumping into everything all of the time. But it's normally viewed to be after glycolysis And this requires oxygen. So let me be clear, glycolysis, this first step, no oxygen required. Doesn't need oxygen. It can occur with oxygen or without it. Oxygen not needed. Or you could say this is called an anaerobic process. This is the anaerobic part of the respiration. Let me write that down too. Anaerobic. Maybe I'll write that down here. Glycolysis, since it doesn't need oxygen, we You might be familiar with the idea of aerobic exercise. The whole idea of aerobic exercise is to make you breathe hard because you need a lot of oxygen to do aerobic exercise. So anaerobic means you don't need oxygen. Aerobic means it needs oxygen. Anaerobic means the opposite. You don't need oxygen. So, glycolysis anaerobic. And it produces two ATPs net. And then you go to the Krebs cycle, there's a little bit of setup involved here. And we'll do the detail of that in the future. But then you move over to the Krebs cycle, which is aerobic. It is aerobic. It requires oxygen to be around. And then this produces two ATPs. And then this is the part that, frankly, when I first learned it, confused me a lot. But I'll just write it in order the way it's Then you have something called-- we're using the same colors too much-- you have something called the electron" + }, + { + "Q": "At 13:05, it is said that each NADH produces 3 ATPs in the electron transport chain. So, glycolysis and Kreb's cycle give us 10 NADHs, which amounts to 10 \u00c3\u0097 3 = 30 ATPs in the electron transport chain. However, the electron transport chain produces 34 ATPs as said. May anyone please tell me from where the other 4 ATPs come from (All mechanisms proceed ideally, of course)? Thank you for any help.\n", + "A": "The Krebs cycle also produces 2 FADH2, which produce 1.5-2 ATP each during the electron transport chain = ~4 more ATP", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 785 + ], + "3min_transcript": "That's called alcohol fermentation. And we, as human beings, I guess fortunately or unfortunately, our muscles do not directly produce alcohol. They produce lactic acid. So we do lactic acid fermentation. Let me write that down. Lactic acid. That's humans and probably other mammals. But other things like yeast will do alcohol fermentation. So this is when you don't have oxygen. It's actually this lactic acid that if I were to sprint really hard and not be able to get enough oxygen, that my muscles start to ache because this lactic acid starts to build up. But that's just a side thing. If we have oxygen we can move to the Krebs cycle, get our two ATPs, and then go on to the electron transport chain and produce 34 ATPs, which is really the bulk of what happens in respiration. Now I said this as an aside, that to some degree this isn't fair. producing these other molecules. They're not producing them entirely, but what they're doing is, they're taking-- and I know this gets complicated here, but I think over the course of the next few videos we'll get an intuition for it-- in these two parts of the reaction, glycolysis and the Krebs cycle, we're constantly taking NAD-- I'll write it as NAD plus-- and we're adding hydrogens to it to form NADH. And this actually happens for one molecule of glucose, this happens to 10 NADs. Or 10 NAD plusses to become NADHs. And those are actually what drive the electron transport chain. And I'll talk a lot more about it and kind of how that happens and why is energy being derived and how is this an oxidative reaction and all of that. And what's getting oxidized and what's being reduced. But I just wanted to give due credit. These guys aren't just producing two ATPs in each of They're also producing, actually combined, 10 NADHs, which each produce three ATPs in an ideal situation, the And they're also doing it to this other molecule, FAD, which is very similar. But they're producing FADH. Now I know all of this is very complicated. I'll make videos on this in the future. But the important thing to remember is cellular respiration, all it is is taking glucose and kind of repackaging the energy in glucose, and repackaging it in the form of, your textbooks will tell you, 38 ATPs. If you're doing an exam, that's a good number to write. It tends to, in reality be a smaller number. It's also going to produce heat. Actually most of it is going to be heat. But 38 ATPs, and it does it through three stages. The first stage is glycolysis, where you're just literally splitting the glucose into two. You're generating some ATPs. But the more important thing is, you're generating some NADHs that are going to be used later in the electron transport chain. Then those byproducts are split even more in the Krebs" + }, + { + "Q": "\nAt 13:11, Sal says that FAD is being reduced to FADH during these processes. Isn't the reduced form of FAD actually FADH2?", + "A": "You are correct. I think if you go to that spot in the video, a small box should come up with the correction on the bottom right corner.", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 791 + ], + "3min_transcript": "That's called alcohol fermentation. And we, as human beings, I guess fortunately or unfortunately, our muscles do not directly produce alcohol. They produce lactic acid. So we do lactic acid fermentation. Let me write that down. Lactic acid. That's humans and probably other mammals. But other things like yeast will do alcohol fermentation. So this is when you don't have oxygen. It's actually this lactic acid that if I were to sprint really hard and not be able to get enough oxygen, that my muscles start to ache because this lactic acid starts to build up. But that's just a side thing. If we have oxygen we can move to the Krebs cycle, get our two ATPs, and then go on to the electron transport chain and produce 34 ATPs, which is really the bulk of what happens in respiration. Now I said this as an aside, that to some degree this isn't fair. producing these other molecules. They're not producing them entirely, but what they're doing is, they're taking-- and I know this gets complicated here, but I think over the course of the next few videos we'll get an intuition for it-- in these two parts of the reaction, glycolysis and the Krebs cycle, we're constantly taking NAD-- I'll write it as NAD plus-- and we're adding hydrogens to it to form NADH. And this actually happens for one molecule of glucose, this happens to 10 NADs. Or 10 NAD plusses to become NADHs. And those are actually what drive the electron transport chain. And I'll talk a lot more about it and kind of how that happens and why is energy being derived and how is this an oxidative reaction and all of that. And what's getting oxidized and what's being reduced. But I just wanted to give due credit. These guys aren't just producing two ATPs in each of They're also producing, actually combined, 10 NADHs, which each produce three ATPs in an ideal situation, the And they're also doing it to this other molecule, FAD, which is very similar. But they're producing FADH. Now I know all of this is very complicated. I'll make videos on this in the future. But the important thing to remember is cellular respiration, all it is is taking glucose and kind of repackaging the energy in glucose, and repackaging it in the form of, your textbooks will tell you, 38 ATPs. If you're doing an exam, that's a good number to write. It tends to, in reality be a smaller number. It's also going to produce heat. Actually most of it is going to be heat. But 38 ATPs, and it does it through three stages. The first stage is glycolysis, where you're just literally splitting the glucose into two. You're generating some ATPs. But the more important thing is, you're generating some NADHs that are going to be used later in the electron transport chain. Then those byproducts are split even more in the Krebs" + }, + { + "Q": "at 7:35 it needs 2 ATPs for cellular respiration?\n", + "A": "2ATP is needed to start cellular repiration (in glycollosis). In short, the 2ATP breaks into 2ADP and the remaining phosphate bonds to the gluclose molecule (and the other remianing phophate later in the cycle). This gives the energy required for glycollosis to occur and produce energy which results in a gain in at (38 in the full cycle for one glucose)", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 455 + ], + "3min_transcript": "And it's actually a cycle. Let me show you what glucose actually looks like. This is glucose right here. And notice you have one, two, three, four, five, six carbons. I got this off of Wikipedia. Just look up glucose and you can see this diagram if you want to kind of see the details. You can see you have six carbons, six oxygens. That's one, two, three, four, five, six. And then all these little small blue things are my hydrogens. So that's what glucose actually looks like. But the process of glycolysis, you're essentially just taking-- I'm writing it out as a string, but you could imagine it as a chain-- and it has oxygens and hydrogens added to each of these carbons. But it has a carbon backbone. And it breaks that carbon backbone in two. That's what glycolysis does, right there. So you've kind of lysed the glucose and each of these things. that's added on to that. You know, these things are all bonded to other things, with oxygens and hydrogens and whatever. But each of these 3-carbon backbone molecules are called pyruvate. We'll go into a lot more detail on that. But glycolysis, it by itself generates-- well, it needs two ATPs. And it generates four ATPs. So on a net basis, it generates two-- let me write this in a different color-- it generates two net ATPs. So that's the first stage. And this can occur completely in the absence of oxygen. I'll do a whole video on glycolysis in the future. Then these byproducts, they get re-engineered a little bit. And then they enter into what's called the Krebs cycle. And then, and this is kind of the interesting point, there's another process that you can say happens after the Krebs cycle. But we're in a cell and everything's bumping into everything all of the time. But it's normally viewed to be after glycolysis And this requires oxygen. So let me be clear, glycolysis, this first step, no oxygen required. Doesn't need oxygen. It can occur with oxygen or without it. Oxygen not needed. Or you could say this is called an anaerobic process. This is the anaerobic part of the respiration. Let me write that down too. Anaerobic. Maybe I'll write that down here. Glycolysis, since it doesn't need oxygen, we" + }, + { + "Q": "I don't understand what Sal means by \"on a net basis\" at 7:37. Someone please explain this concept to me...\n", + "A": "He summarized the glycolyse reaction - in some stage of this reaction you need to use 2ATPs [-2] and in some you produce 4ATPs [+4]. And, summing up, you generate ( produce ) 2 ATPs [-2 +4 = +2].", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 457 + ], + "3min_transcript": "And it's actually a cycle. Let me show you what glucose actually looks like. This is glucose right here. And notice you have one, two, three, four, five, six carbons. I got this off of Wikipedia. Just look up glucose and you can see this diagram if you want to kind of see the details. You can see you have six carbons, six oxygens. That's one, two, three, four, five, six. And then all these little small blue things are my hydrogens. So that's what glucose actually looks like. But the process of glycolysis, you're essentially just taking-- I'm writing it out as a string, but you could imagine it as a chain-- and it has oxygens and hydrogens added to each of these carbons. But it has a carbon backbone. And it breaks that carbon backbone in two. That's what glycolysis does, right there. So you've kind of lysed the glucose and each of these things. that's added on to that. You know, these things are all bonded to other things, with oxygens and hydrogens and whatever. But each of these 3-carbon backbone molecules are called pyruvate. We'll go into a lot more detail on that. But glycolysis, it by itself generates-- well, it needs two ATPs. And it generates four ATPs. So on a net basis, it generates two-- let me write this in a different color-- it generates two net ATPs. So that's the first stage. And this can occur completely in the absence of oxygen. I'll do a whole video on glycolysis in the future. Then these byproducts, they get re-engineered a little bit. And then they enter into what's called the Krebs cycle. And then, and this is kind of the interesting point, there's another process that you can say happens after the Krebs cycle. But we're in a cell and everything's bumping into everything all of the time. But it's normally viewed to be after glycolysis And this requires oxygen. So let me be clear, glycolysis, this first step, no oxygen required. Doesn't need oxygen. It can occur with oxygen or without it. Oxygen not needed. Or you could say this is called an anaerobic process. This is the anaerobic part of the respiration. Let me write that down too. Anaerobic. Maybe I'll write that down here. Glycolysis, since it doesn't need oxygen, we" + }, + { + "Q": "\nAt 7:34, what does it mean by net ATP?", + "A": "Some ATP are used in the production of ATP. Net is the produced - the used", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 454 + ], + "3min_transcript": "And it's actually a cycle. Let me show you what glucose actually looks like. This is glucose right here. And notice you have one, two, three, four, five, six carbons. I got this off of Wikipedia. Just look up glucose and you can see this diagram if you want to kind of see the details. You can see you have six carbons, six oxygens. That's one, two, three, four, five, six. And then all these little small blue things are my hydrogens. So that's what glucose actually looks like. But the process of glycolysis, you're essentially just taking-- I'm writing it out as a string, but you could imagine it as a chain-- and it has oxygens and hydrogens added to each of these carbons. But it has a carbon backbone. And it breaks that carbon backbone in two. That's what glycolysis does, right there. So you've kind of lysed the glucose and each of these things. that's added on to that. You know, these things are all bonded to other things, with oxygens and hydrogens and whatever. But each of these 3-carbon backbone molecules are called pyruvate. We'll go into a lot more detail on that. But glycolysis, it by itself generates-- well, it needs two ATPs. And it generates four ATPs. So on a net basis, it generates two-- let me write this in a different color-- it generates two net ATPs. So that's the first stage. And this can occur completely in the absence of oxygen. I'll do a whole video on glycolysis in the future. Then these byproducts, they get re-engineered a little bit. And then they enter into what's called the Krebs cycle. And then, and this is kind of the interesting point, there's another process that you can say happens after the Krebs cycle. But we're in a cell and everything's bumping into everything all of the time. But it's normally viewed to be after glycolysis And this requires oxygen. So let me be clear, glycolysis, this first step, no oxygen required. Doesn't need oxygen. It can occur with oxygen or without it. Oxygen not needed. Or you could say this is called an anaerobic process. This is the anaerobic part of the respiration. Let me write that down too. Anaerobic. Maybe I'll write that down here. Glycolysis, since it doesn't need oxygen, we" + }, + { + "Q": "\nIn my textbook, it is said that in cellular respiration, energy is produced in the form of ATP. But, in this video, around 3:19, Sal said that energy is used to produce ATP. Which one is correct ? I am confused. :(", + "A": "ATP is the energy currency in the cell and can be thought of as chemical form of energy. One way to think about is to think about it like a rechargeable battery. ATP can be used, giving up its stored energy in the phosphate bond and then recharged with the addition of energy to reform the phosphate bond.", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 199 + ], + "3min_transcript": "mole of glucose, if you had six moles of molecular oxygen running around the cell, then-- and this is kind of a gross simplification for cellular respiration. I think you're going to appreciate over the course of the next few videos, that one can get as involved into this mechanism as possible. But I think it's nice to get the big picture. But if you give me some glucose, if you have one mole of glucose and six moles of oxygen, through the process of cellular respiration-- and so I'm just writing it as kind of a big black box right now, let me pick a nice color. So this is cellular respiration. Which we'll see is quite involved. But I guess anything can be, if you want to be particular enough about it. Through cellular respiration we're going to produce six moles of carbon dioxide. Six moles of water. going to produce energy. We're going to produce energy. And this is the energy that can be used to do useful work, to heat our bodies, to provide electrical impulses in our brains. Whatever energy, especially a human body needs, but it's not just humans, is provided by this cellular respiration mechanism. And when you say energy, you might say, hey Sal, on the last video didn't you just-- well, if that was the last video you watched, you probably saw that I said ATP is the energy currency for biological systems. And so you might say, hey, well it looks like glucose is the energy currency for biological systems. And to some degree, both answers would be correct. But to just see how it fits together is that the process of cellular respiration, it does produce energy directly. So if I were to break down this energy portion of cellular respiration right there, some of it would just be heat. You know, it just warms up the cell. And then some of it is used-- and this is what the textbooks will tell you. The textbooks will say it produces 38 ATPs. It can be more readily used by cells to contract muscles or to generate nerve impulses or do whatever else-- grow, or divide, or whatever else the cell might need. So really, cellular respiration, to say it produces energy, a little disingenuous. It's really the process of taking glucose and producing ATPs, with maybe heat as a byproduct. But it's probably nice to have that heat around. We need to be reasonably warm in order for our cells to operate correctly. So the whole point is really to go from glucose, from one mole of glucose-- and the textbooks will tell you-- to 38 ATPs." + }, + { + "Q": "At 8:54, he said that the structure on right is not going to contribute to the hybrid as much, what does he mean by that? If the structure is supposed to be a hybrid of both, does that mean the resonance structure is closer to the structure on the left?\n", + "A": "It means the true structure of the molecule is closer to the left one than the right. The bond lengths in acetone are consistent with a C-O double bond which gives a bit of weight to what he s saying.", + "video_name": "UHZHkZ6_H5o", + "timestamps": [ + 534 + ], + "3min_transcript": "so I put that in, and so when you're doing this for cations, you're not gonna move a positive charge, so when you're drawing your arrows, you're showing the movement of electrons, so the arrow that I drew over here, let me go ahead a mark it in magenta. So this arrow in magenta is showing the movement of those electrons in blue, and when those electrons in blue move, that creates a plus-one formal charge on this carbon, and so don't try to move positive charges: Remember, you're always pushing electrons around. Then finally, let's do one more. So, for this situation, this is for acetone, so we have a carbon right here, double-bonded to an oxygen, and we know that there are differences in electronegativity between carbon and oxygen: Oxygen is more electronegative. So what would happen if we took those pi electrons? blue for pi electrons, so these pi electrons right here, and we move those pi electrons off, onto the more electronegative atom, like that, so let's go ahead and draw our resonance structure. So this top oxygen would have three lone pairs of electrons: one of those lone pairs are the ones in blue, those pi electrons; that's gonna give the oxygen a negative-one formal charge, and we took a bond away from this carbon, so we took a bond away from this carbon, and that's going to give that carbon a plus-one formal charge. And so, when you think about your resonance structures, first if all, I should point out that one negative charge and one positive charge give you an overall charge of zero, so charge is conserved, and over here, of course, the charge is zero. So if you're thinking about the resonance hybrid, we know that both structures contribute to the overall hybrid, but the one on the right isn't going to contribute as much, so this one you have a positive and a negative charge, and the goal, of course, is to get to overall neutral. But, what's nice about drawing this resonance structure, and thinking about this resonance structure, is it's emphasizing the difference in electronegativity, so, for this one, you could just say oxygen get a partial negative, and this carbon right here, gets a partial positive. So that's one way of thinking about it, which is very helpful for reactions. But drawing this resonance structure is just another way of thinking about, emphasizing the fact that when you're thinking about the hybrid, you're thinking about a little more electron density on that oxygen. All right, so once again, do lots of practice; the more you do, the better you get at drawing resonance structures, and the more the patterns, the easier the patterns become." + }, + { + "Q": "at 9:40 , why is the resonance structure on the right a minor contributor compared to the one on the left when the right one has both negative and positive charges assigned corresponding to the electro negativity of the atoms in the compound?\n", + "A": "Because in the right structure those full formal charges have been introduced where the left structure has none Less formal charges are generally better structures", + "video_name": "UHZHkZ6_H5o", + "timestamps": [ + 580 + ], + "3min_transcript": "so I put that in, and so when you're doing this for cations, you're not gonna move a positive charge, so when you're drawing your arrows, you're showing the movement of electrons, so the arrow that I drew over here, let me go ahead a mark it in magenta. So this arrow in magenta is showing the movement of those electrons in blue, and when those electrons in blue move, that creates a plus-one formal charge on this carbon, and so don't try to move positive charges: Remember, you're always pushing electrons around. Then finally, let's do one more. So, for this situation, this is for acetone, so we have a carbon right here, double-bonded to an oxygen, and we know that there are differences in electronegativity between carbon and oxygen: Oxygen is more electronegative. So what would happen if we took those pi electrons? blue for pi electrons, so these pi electrons right here, and we move those pi electrons off, onto the more electronegative atom, like that, so let's go ahead and draw our resonance structure. So this top oxygen would have three lone pairs of electrons: one of those lone pairs are the ones in blue, those pi electrons; that's gonna give the oxygen a negative-one formal charge, and we took a bond away from this carbon, so we took a bond away from this carbon, and that's going to give that carbon a plus-one formal charge. And so, when you think about your resonance structures, first if all, I should point out that one negative charge and one positive charge give you an overall charge of zero, so charge is conserved, and over here, of course, the charge is zero. So if you're thinking about the resonance hybrid, we know that both structures contribute to the overall hybrid, but the one on the right isn't going to contribute as much, so this one you have a positive and a negative charge, and the goal, of course, is to get to overall neutral. But, what's nice about drawing this resonance structure, and thinking about this resonance structure, is it's emphasizing the difference in electronegativity, so, for this one, you could just say oxygen get a partial negative, and this carbon right here, gets a partial positive. So that's one way of thinking about it, which is very helpful for reactions. But drawing this resonance structure is just another way of thinking about, emphasizing the fact that when you're thinking about the hybrid, you're thinking about a little more electron density on that oxygen. All right, so once again, do lots of practice; the more you do, the better you get at drawing resonance structures, and the more the patterns, the easier the patterns become." + }, + { + "Q": "\nAt 4:00 min when sal names the 7,7, dibromo oct - 5 - yn - 4 -ol isn't there a chiral center on the number 4 carbon?", + "A": "You re right, there is a chiral center on C4. Since it s drawn flat (no wedges or dashes) there is no way to tell whether it s R or S and we d just assume that there is a racemic mixture (which you could indicate explicitely with +/- or R/S in a more complete name). If, for example, the OH was shown on a wedge or a dash, then we would indicate whether it was R or S in the name.", + "video_name": "nQ7QSV4JRSs", + "timestamps": [ + 240 + ], + "3min_transcript": "We have 1, 2, 3, 4, 5 carbons. So it's going to be pent. And there's no double bonds. So I'll just write pentane right then. And we're not going to just write a pentane because actually, the fact that makes it an alcohol, that takes precedence over the fact that it is an alkane. So it actually, the suffix of the word will involve the alcohol part. So it is pentanol. That tells us that's an alcohol. And to know where the OH is grouped, we'll start numbering closest to the OH. So 1, 2, 3, 4, 5. Sometimes it'll be called 2-pentanol. And this is pretty clear because we only have one group here, only one OH. So we know that that is what the 2 applies to. But a lot of times, if people want to be a little bit more particular, they might write pentan-2-ol. multiple functional groups. So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try the name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. We want to call it an octyne. But because we have an alcohol there, we want to call this an octyne-- let me make it very clear. Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that So it is oct-5-yn. That tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-al. And now we have these two bromo groups here on the 7 carbon. So it's 7,7-dibromo oct-5-yn-4-al. And this would all be one word. Let me make sure that you realize that I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex." + }, + { + "Q": "at 5:20, is the informal name phenol?\n", + "A": "its not a benzene either way because he doesnt identify the conjugated double bonds. so it is only cyclic but not aromatic", + "video_name": "nQ7QSV4JRSs", + "timestamps": [ + 320 + ], + "3min_transcript": "multiple functional groups. So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try the name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. We want to call it an octyne. But because we have an alcohol there, we want to call this an octyne-- let me make it very clear. Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that So it is oct-5-yn. That tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-al. And now we have these two bromo groups here on the 7 carbon. So it's 7,7-dibromo oct-5-yn-4-al. And this would all be one word. Let me make sure that you realize that I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex. It's a cyclohexane. But then of course, the hydroxide or the hydroxy group I should call it, takes dominance. It's a hexanol. So this is a cyclohexanol. And once again, that comes from the OH right there. And you don't have to number it. Because no matter what carbon it's on, it's on the same one. If you had more than one of these OH groups, then we would have to worry about numbering them. Let's just do this one right over here. So once again, what is our carbon chain? We have 1, 2, 3 carbons. And we have the hydroxy group attached to the 1 and the 3 carbon. Prop is our prefix. It is an alkane. So we would call this-- and there's a couple of ways to do this. We could call this 1 comma 3 propanediol." + }, + { + "Q": "at 4:22 why is it not OCT-3-yn-5-ol since the double bond ought to take precedence over the -OH group? thanks!\n", + "A": "The double bond does not take precedence over the alcohol. Carboxylic acids > esters > amides > ketones > aldehydes > alcohols > amines > alkenes > alkynes > R = OR = X = Ph (those of equal precedence go in alphabetical order.", + "video_name": "nQ7QSV4JRSs", + "timestamps": [ + 262 + ], + "3min_transcript": "We have 1, 2, 3, 4, 5 carbons. So it's going to be pent. And there's no double bonds. So I'll just write pentane right then. And we're not going to just write a pentane because actually, the fact that makes it an alcohol, that takes precedence over the fact that it is an alkane. So it actually, the suffix of the word will involve the alcohol part. So it is pentanol. That tells us that's an alcohol. And to know where the OH is grouped, we'll start numbering closest to the OH. So 1, 2, 3, 4, 5. Sometimes it'll be called 2-pentanol. And this is pretty clear because we only have one group here, only one OH. So we know that that is what the 2 applies to. But a lot of times, if people want to be a little bit more particular, they might write pentan-2-ol. multiple functional groups. So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try the name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. We want to call it an octyne. But because we have an alcohol there, we want to call this an octyne-- let me make it very clear. Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that So it is oct-5-yn. That tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-al. And now we have these two bromo groups here on the 7 carbon. So it's 7,7-dibromo oct-5-yn-4-al. And this would all be one word. Let me make sure that you realize that I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex." + }, + { + "Q": "At 8:55, I don't understand clearly why i1 equals (V1-V2)/R1.\n", + "A": "V1 and V2 are node voltages . That means they are measured with respect to the reference node (ground). This is suggested by the two orange arrows that start at the ground node and curve up to each node. Resistor R1 is connected between node 1 and node 2. The element voltage that appears across R1 is the difference of the two node voltages, V1-V2. Using Ohm s Law you compute the current through R1 as i = Voltage across resistor / R i = (V1 - V2) / R", + "video_name": "2lY757QaaKs", + "timestamps": [ + 535 + ], + "3min_transcript": "let me move up a little bit. Step Four is write the Kirchoff's Current Law equations directly from the circuit. We're going to do this in a special way, We're going to perform at this node here, at node two. We're going to write the current law for this. That means we got to identify the currents. There's a current, we'll call that a current, and that's a current. Let me give some names to these currents just to be clear. We'll call this one I1 because it goes through R1. We'll call this one here, I2 because it goes through R2. This one is already Is. Now let's write Kirchoff's Current Law just in terms of I, and we'll say all the currents flowing into the node so they're going to get negative signs when we write Kirchoff's Current Law. Let's do that right here. And we write I1 minus I2 minus Is equals to zero. So right now we're working on Step Four. This is the essence of the Node Voltage Method. This is where we do something new that we haven't done before. We're going to write these currents in terms of the node voltages. So we can write I1, I1 is current flowing this way through this current. I1 equals V1 minus V2 over R1. in terms of node voltages. The current flowing down through I2, now we have to subtract I2, so we just apply Ohm's Law directly, which means that the current in I2 is equal to V2 divided by R2. The last current is Is, minus Is. We'll write that in terms of Is, like that, and that equals zero. This means we have now completed Step Four. That is KCL written using the terminology of node voltages. We could check off that we've done Step Four." + }, + { + "Q": "at 5:09, can plasma be considered to be the fourth state of matter?\n", + "A": "Yes, it is the fourth state of matter.", + "video_name": "WenwtcuqOj8", + "timestamps": [ + 309 + ], + "3min_transcript": "where does it exist? Well, probably closest to home, it exists in lightning. And that's worthy of an entire video. But the idea is that you start having a huge potential difference between the clouds and the ground. And then because you have this huge voltage difference between the two, you have electrons that are essentially wanting to go into the ground. You have a build-up of electrons up here that want to go into the ground. They can't because air is normally a fairly bad conductor. It's an insulator. But what happens is because there's so much electropotential here, the electrons that are close in the molecules up here, at least how I visualize it, So their electrons start to want to move away in the air molecules. Whether you're talking about the air is a mixture of oxygen, and nitrogen, and carbon dioxide. They start wanting to get away from the clouds. So they start disassociating and start forming this ionized air. And eventually, at some point, this happens to such a degree that you can actually get conduction from the cloud to the ground. And that conduction is when the air is in a plasma state. The conduction allows extremely high temperatures and the electrons to flow all the way to the ground. The other common example, you might see something like this, well actually not like this, but at least a plasma state, is in stars. And that's because you have extremely strong electromagnetic fields, extremely high pressure, and in that type of environment, once again, I'm old super over simplifying it, you can get to a state where the electrons can get disassociated from things that otherwise wouldn't I thought I would touch on that because it's an interesting subject. And it exists in the universe. On the universal level, because stars are pretty much all plasma, it is actually the most common state of matter in the universe. Although in our everyday life, we probably encounter solids, liquids and gases a lot more. And one other thing I want to maybe clarify from the last video is, I talk about the bonding between water molecules. And let's say we're talking in the solid state. So I have an oxygen, a hydrogen, a hydrogen. And I have some electrons here, some electrons here. Let's say there's another hydrogen here, an oxygen, and a hydrogen. Maybe there's an oxygen here. That has hydrogen. And then this has a hydrogen. And it has two electrons, two electron pairs." + }, + { + "Q": "7:46: Sal mentions NH3 then at 8:09 Sal mentions HF and water is H2O. Is there a certain way to label a compound? Because the H appears in different spots in the examples provided.\n", + "A": "In general it is convention to have carbon first then hydrogen then followed by other compounds (C6H12O6). However, there is also a convention that in binary acidic compounds the hydrogen is at the beginning (HCl, HF, HBr etc.), and at the end of organic acids (R-COOH). Long story short there are several different formula conventions and naming conventions and they use separate rules, so you wind up with some variations in writing classic formulas.", + "video_name": "WenwtcuqOj8", + "timestamps": [ + 466, + 489 + ], + "3min_transcript": "That oxygen is so much more electronegative that it hogs the electrons. And so the oxygen side starts to have a partial negative charge. While the hydrogen side starts to have a partial positive side. Because with the hydrogen, essentially all of its electrons are hanging out close to the oxygen, hydrogen ends up just becoming like this proton that's floating out there. Because we said it doesn't even have neutrons in most cases. So this has a slightly positive charge. This one has a positive charge. And the positive polar end of the water molecule is attracted to the negative polar end. And I called it polar bonds, and it shows you my memory from high school chemistry is not ideal, I really should have called it hydrogen bonds. So this is a hydrogen bond, and this is a hydrogen bond. It's just a matter of the name I used. I just want to clarify that because that is what's typically used in your chemistry class. And that is just the bond that exists from a partially positive hydrogen atom. Because its electrons are hanging out near the oxygen. And a partially negative oxygen atom in the water molecule. Because it has stolen all of these electrons from the hydrogen. You draw it like that, it's called a hydrogen bond. And hydrogen bonds tend to form between hydrogen or really only a handful of super electronegative atoms. And that's nitrogen, fluorine, and oxygen. And these are actually the three most electronegative atoms. So the nitrogen, NH3, when it bonds with hydrogen, is essentially so electronegative that you have the same situation. All the electrons hang out here, so you have a partial negative charge, partial positive on the hydrogen ends. You get the same HF. You get the same type of hydrogen bonds. And so in this case, these guys would be attracted to the nitrogen part of other molecules and would form I just want to get that out of the way. And with that done, I think we can return to some of the ideas of the last video and actually do some problems. Let's take the case with water. Actually, let me just state the problem first. So let's say that we have a" + }, + { + "Q": "at 0:58, I don't understand what a plasma is\n", + "A": "Plasma is when an atom is vibrating so fast that the electrons are shaken off. This can be accomplished with high heat, low pressure, or an electromagnetic field.", + "video_name": "WenwtcuqOj8", + "timestamps": [ + 58 + ], + "3min_transcript": "In the last video we touched on the three states of matter that are really most familiar to our everyday experience. The solid, the liquid, and the gas. And I kind of hinted that there is a fourth state, which I don't cover, because it's usually not the domain of an introductory chemistry course. But a little bit of a discussion ensued on the message board for that video. So I thought I would at least touch on that fourth stage. And that's plasma. I'll do it in a suitably bright color. Plasma. And people consider it a fourth state because it has some properties of gases. In some ways it's almost a subset of gases. But it also has properties of conductivity that you normally wouldn't associate with a gas. And just so you know, when you first hear it you think, oh that's a fairly exotic thing, plasma. And in the first video, I said it's only something that occurs at high temperatures, which isn't exactly 100% right. It doesn't have to be at high temperatures. I really should have said that under extenuating electromagnetic field. Or something has to happen to essentially bump the the electrons, or move the electrons off of gases that would've otherwise have kept their electrons. So it's kind of analogous to what happens in metal. When we talk about metal bonds, we talk about this notion of a sea of electrons. Let's say if we talked about iron. What happens with most metals is that they have so many electrons, and they are so willing to give them, that the electrons just kind of float outside of the atoms themselves and create this kind of big sea of electrons. And then the atoms themselves become positively charged ions. Because they essentially donated some So they're attracted to the sea and that's what makes them malleable and even more importantly what allows them to conduct electricity. But they're all really packed closely together and it's a very dense structure. Plasma is a situation where if you take gases, and remember, So you take a bunch of gases and they have high kinetic energy. Although, they don't have to be, that could be under very But they're moving around and bumping into each other. But they're not close to each other. They don't have a fixed structure with each other. Or they're not rubbing against each other like in the case of a liquid. But what happens in a plasma, or one situation is, that you apply such a strong electromagnetic field that the electrons want to disassociate. So let's say these electrons start bumping off of the plasma. And so a solid has its own shape. A plasma will take the shape of its container like a gas. And sometimes it is described as an ionized gas. And it's described as ionized because electrons are bumped off. And when the electrons are bumped off, the otherwise neutral atoms now have positive charges. And what this allows is, essentially a conduction of electricity. Because now these electrons are free to move." + }, + { + "Q": "at 4:45 hank mentions that the hagfish doesn't have a vertebrae yet it is a vertebrate. how?\n", + "A": "He actually doesn t classify it as a vertebrate, just a chordate. Chordates are known to have spinal chords be their main attraction. But to be classified you also need a skull, a post-anus tail at one point in one s life cycle, the ability to make mucus, and other minor details. The hagfish, while not containing most of these has at least a skull and a post-anus tail so it can be grouped into this section because there nowhere better to put it.", + "video_name": "c7Yy9v8dH8s", + "timestamps": [ + 285 + ], + "3min_transcript": "These four traits all began to appear during the Cambrian explosion, more than 500 million years ago. Today, they're shared by members of all three chordate sub-phyla, even if the animals in those sub-phyla look pretty much nothing like each other. For instance, our new friends here in cephalochordata are the oldest living sub-phyla, but you can't forget the other invertebrate group of chordates, the urochordata - literally, tail cords. There are over 2,000 species here, including sea squirts. If you're confused about why this ended up in a phylum with us, it's because they have tadpole-like larva with all four chordate characteristics. The adults, which actually have a highly-developed internal structure, with a heart and other organs, retain the pharyngeal slits, but all the other chordate features disappear or reform into other structures. The third and last, and most complex sub-phylum, is the vertebrata, and has the most species in it, because its members have a hard backbone, which has allowed for an explosion in diversity, You can see how fantastic this diversity really is when you break down vertebrata into its many, many classes, from slimy sea snake-y things to us warm and fuzzy mammals. As these classes become more complex, you can identify the traits they each develop that gave them an evolutionary edge over the ones that came before. For example, how's this for an awesome trait - a brain. Vertebrates with a head that contains sensory organs and a brain are called craniates. They also always have a heart with at least two chambers. Since this is science, you're gonna have to know that there's an exception for every rule that you're gonna have to remember. The exception in this case is the myxini, or hagfish, the only vertebrate class that has no vertebra, but is classified with us because it has a skull. This snake-like creature swims by using segmented muscles to exert force against its notochord. Whatever, hagfish. Closely related to it is the class petromyzontida, otherwise known as lampreys, the oldest-living lineage of vertebrates. Now, these have a backbone made of cartilage, With the advent of a backbone, we see vertebrates getting larger, developing more complex skeletons, and becoming more effective at catching food and avoiding predators. But, did you notice anything missing? Lampreys, and other early vertebrates, are agnafins, literally no jaws, and if you want to be able to chew food, it really helps to have a jaw and teeth. Most scientists think that the jaw evolved from structures that supported the first two pharyngeal slits near the mouth. And teeth? Well, the current theory is that they evolved from sharp scales on the face. Gnathostomes, or jaw mouths, arrived on the scene 470 million years ago, and one of the oldest, most successful groups of gnathostomes that have survived to present day are the class chondrichthyes - the cartilage fish. You know them as the sharks, and skates and rays, and as their name says, their skeleton is made up mostly of cartilage, but they show the beginnings of a calcified skeleton. Chondrichthyes haven't changed much in the past 300 million years or so, and their success stems from the paired fins that allow for fish and swimming," + }, + { + "Q": "\nhow can I find the plane of symmetry in the meso compound in 9:30....it is impossible to find it?!", + "A": "Carbons on each end, and a OH group attached on both. Try to imagine drawing a line of symmetry DIAGONALLY.", + "video_name": "zNAL1R-hZr0", + "timestamps": [ + 570 + ], + "3min_transcript": "our chiral centers here, so this one has an OH coming out at us, and then that one has it going away from us, this one has an OH coming out at us and this one has it going away from us, so we have opposite configurations at both chirality centers. Let's look at three and four next. So what is the relationship between these two? Well, first, we might think these could be enantiomers because at this carbon, we have OH on a wedge, and then here we have OH on a dash. And then here we have OH on a dash, and here we have it on a wedge. So that might be your first guess. But let's look at the video and let's look at the model sets to help us out. Remember, I'm leaving the hydrogens off the methyl groups and the hydrogens off the oxygens in the video just to help us see the molecule more clearly. On the left, we have a model of drawing three. So here's our carbon chain with an OH going away from us in space, and an OH coming out at us in space. Here's our carbon chain with an OH coming out at us in space and an OH going away from us in space. So I'll hold the two models and we'll compare them. First let's see if they are mirror images of each other. So I'll take the one on the right and I'll rotate it and I'll hold it up next to the one on the left. And now we can see that these two are mirror images of each other. So next, let's see if one is superimposable on the other. So I'll go back to the starting point and I'll rotate it around like that, and let's see if we can superimpose the one on the right, the mirror image, on the molecule on the left. And notice that we can. All of the atoms line up. So all of the hydrogens, carbons and oxygens are in the same place. So this is a compound that has chirality centers, but it is achiral, the mirror image is superimposable on itself. So we should be able to find a plane of symmetry. So I'll just pick one of these models, doesn't matter which one because they represent so we can see a plane of symmetry. So right there is our plane of symmetry. This is a meso compound. So three and four actually represent the same compound. So this is one meso compound. So these two are the same, and we have one meso compound. It's not really obvious looking at these bond line structures that these represent the same molecule. So definitely get a model set and try this out for yourself. So we thought there might be four stereoisomers, but actually there are only three. We have a pair of enantiomers, and we have one meso compound." + }, + { + "Q": "In the last section of the video with the astronaut glove won't the reaction force of the heavy box push you even further away from the space arm contraption? slightly puzzled , Thankyou in advance\n(Sorry the timing is 1:09 and onwards)\n", + "A": "no. newton s third law says that every action has an equal and opposite reaction. So, pushing the box away from the space arm results in a force applied in the opposite direction, which is towards the space arm.", + "video_name": "By-ggTfeuJU", + "timestamps": [ + 69 + ], + "3min_transcript": "" + }, + { + "Q": "At 7:16, when the astronaut is pushing an object away from their body, will the force increase if the astronaut increases the amount of time it takes to throw the object? (i.e. will the object have a greater equal and opposite force if the astronaut takes a long time to throw the object than if it was thrown quickly)\n", + "A": "No, the force is determined by how hard the astronaut pushes, not how long. The total momentum for the astronaut is a combination of how hard he pushes and how long", + "video_name": "By-ggTfeuJU", + "timestamps": [ + 436 + ], + "3min_transcript": "" + }, + { + "Q": "At, 7:40, will the astronaut have to let go of the object to accelerate in the opposite direction or not?\n", + "A": "Of course she has to let go.", + "video_name": "By-ggTfeuJU", + "timestamps": [ + 460 + ], + "3min_transcript": "" + }, + { + "Q": "3:15 - 4:30. Would it not mean if you took your foot away (presuming that your foot is not part of you), the sand would accelerate upwards. In the same way as: two persons pushing against each other and one of them moves out of the way. It's obvious that the person that didn't move out the way would fly forwards, unbalanced. Is this not the same thing as what Newton is saying. I'm not disregarding Newton's Law, I just don't understand it.\n", + "A": "The sand pushes up on you foot and your foot pushes down on the sand. If you foot does not push on the sand, the sand does not push on your foot. For the sand to accelerate upwards, something would have to be pushing upward on the sand. The sand cannot push on itself.", + "video_name": "By-ggTfeuJU", + "timestamps": [ + 195, + 270 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 4:09 , Sal said that the molecules of the rock would align themselves to the poles , but how do people track molecular alignment?", + "A": "In this instance, the alignment is evident because the rock has become magnetized.", + "video_name": "6EdsBabSZ4g", + "timestamps": [ + 249 + ], + "3min_transcript": "but it is kind of a curious thing to look at. And not only is there this ridge. There's lot of underwater volcanic activity. You have magma flowing out and lava flowing out into the water, and it's kind of forming this ridge that really goes across the whole Atlantic Ocean. There are other ridges in the world like that, underwater ridges. You have one over here in the Pacific Ocean. You have these here in the Indian Ocean. That's just a little clue, but that by itself doesn't tell you that these plates are actually moving apart at the ridge. The more conclusive-- this is just the beginning of the clue-- but what made this conclusive is one, the separate discovery. And this is what's interesting is that you have these separate discoveries in different domains that eventually let you come to a pretty neat conclusion. So you've had a separate discovery that if you look at different eras of magnetic rock, or maybe I should say magnetic rock from different periods And you can tell where they are in geologic time by how they're layered. So this would be newer rock. And then this would be a little bit older. And then this would be even, even older. Geologists noticed something interesting. If I were to take magnetic rock, and if it was molten lava, and if it were to harden, remember it's magnetic rock so it would want to align with the poles the same way a compass would. So if I had a bunch of magnetic-- so let's say this is some lava right here. And so the molecules can align themselves. Since it's a liquid and they can align themselves, they are going to naturally want to align with the poles. So they'll naturally all want to align in one direction because of Earth's magnetic field. that alignment will kind of be frozen. Now, if Earth's magnetic field was constant over time, then when you look at magnetic rocks from any period, you would expect them all to be aligned in the same direction. So since we're taking a cross section of rock here let's say an alignment towards the North pole looks like this. And I draw it like that. That's kind of an arrow pointing into our screen. And let's say an alignment pointing to the South pole would look like this. This would be an arrow pointing out of our screen. So what you would expect is the newer rock that kind of the alignment, the field, the alignment of the rock, would go into the screen, and then the older rock, it would still go older into the screen. So if I were to draw a top view-- Let me draw it like this just so I make sure that everyone is on the same page. So let me just draw a cross section like this so that we know what we're talking about." + }, + { + "Q": "\ni have studied in my text books that index finger should point towards the magnetic field.so i am a bit confused cz in this video at approx 6:40 and 10:00 ,sal has said that index finger should be pointing in the direction of the current....", + "A": "There are many different versions of these rules. Some use the right hand, some use the left, some define fingers one way, some do it another. They all lead to the same answer. Use whatever one works for you, and stick with that.", + "video_name": "l3hw0twZSCc", + "timestamps": [ + 400, + 600 + ], + "3min_transcript": "It's just a notation. I've seen professors do it either way, I've seen it written either way, as well. Cross the magnetic field that it's in. What's the magnetic field that it's in? The magnetic field-- I'll do it in magenta, because it's the magnetic field created by current 1. So it's magnetic field 1, which is this magnetic field. So before going into the math, let's just figure out what direction is this net force going to be in? So here we say, well, the current is a scalar, so that's not going to affect the direction. What's the direction of L2? This is L2. I didn't label it L2 on the diagram. What's the direction of L2? Well, it's up. And then the direction of B1, the magnetic field created by current 1, is going into the page here. So here we just do the standard cross product. Let me see if I can pull this off. This is actually an easy one to draw. And then I put my middle finger in the direction of the field. So my middle finger's going to point straight down into this page. My other fingers just do what they would naturally do. And then my thumb would go in the direction of the net force. This is just the cross product. You'll see teachers teach the cross product other ways, where they tell you to put your thumb in the direction of the field, and this and that, your palm-- those are all valid. They're just different variations of the same thing. I find this one easier to remember. Because when I take the cross product, index finger is the first term of the cross product. Middle finger is the second term of the cross product. Thumb is the direction of the cross product. So anyway, this is the direction of L2. The magnetic field, we already know, goes into the page. So my middle finger is going into the page. And my thumb is in the direction of the force on the magnetic field. So that's the direction of the force. So there you have it. is-- we know from this wrap around rule that pops out here and it goes in here-- the effect that it has on this other wire is that where the current is going in the same direction, is that it will be attracted. So the net force you is going in that direction. We could say the force from 1 on 2. That's just my convention. Maybe other people would have written it the force given to 2 by 1. That's the force given by 1 to 2. That's how I'm writing it. Now what's going to be the force on current 1 from I2? Well, it's going to be the current-- well, it's going to be the force there. Well it's going to be the same thing. Let me draw I2's magnetic field. You do the wrap around rule, it's going to look the same. So I2, sure, on this side its field is going to be going into the page. But what's I2's field going to be doing here?" + }, + { + "Q": "\nAt 8:40, you talked about voltage drop. So, does the voltage differ from resistor to resistor? what does voltage drop mean?", + "A": "The voltage drop refers to the difference in electric potential at opposite ends of the resistor. If the resistor is carrying current there must be a potential difference, otherwise why are the charges moving. The PD is given by V = IR, where I is the current through the resistor.", + "video_name": "7vHh1sfZ5KE", + "timestamps": [ + 520 + ], + "3min_transcript": "That's a 1, not an I. I1 times R1, right? And similarly, if I measured the voltage from here to here, that voltage is going to be equal to I2 times R2. Let's say this is where I3 is. So the voltage, if I were to measure it from here to here-- But anyway, if we look at the voltage from here to here, it's going to be I3 times R3. So what we see is that the voltage across the entire circuit, which I can write as V-total, is going to be equal to the potential drops, the total potential drop across each of these devices. So the way to think about it is that-- well, let's think about the electrons. The electrons here, they really want to get here. get here, they've experienced some potential drop. So the electrons here actually are a little bit less eager to get here. And then once they've gone through here, maybe they're just tired of bumping around so much. And once they're here, they're a little bit less eager to get here. So there's a voltage drop across each device, right? So the total voltage is equal to the voltage drop across each of the devices. And now let's go back to the convention, and we'll say that the current is going in that direction. The total voltage drop is equal to V1 plus V2 plus V3, so the total voltage drop is equal to I1 R1 plus I2 R2 plus I3 R3. And what's the total voltage drop? Well, that's equal to the total current through the whole system. I-total, or we just call it I, times the total resistance is Well, we know that all the I's are the same. Hopefully, you can take it as, just conceptually it makes sense to you that the current through the entire circuit will be the same. So all these I's are the same, so we can just cancel them out. Divide both sides by that I. We assume it's non-zero, so I, I, I, I, and then we have that the total resistance of the circuit is equal to R1 plus R2 plus R3. So when you have resistors in series like this, the total resistance, their combined resistance, is just equal to their sum. And that was just a very long-winded way of explaining something very simple, and I'll do an example. Let's say that this voltage is-- I don't know. Let's say it's 20 volts. Let's say resistor 1 is 2 ohms. Let's say resistor 2 is" + }, + { + "Q": "\nAt around 10:50 in the video sal goes from 20= current * 10 ohms to current = 2 AMPERES. How did he get to amperes from ohms?", + "A": "Voltage = Current x Resistance in terms of units: (volts)=(amperes) x (ohms) (ohms)=(volts) / (amperes) with a little rearranging, you get: (amperes)=(volts) / (ohms)", + "video_name": "7vHh1sfZ5KE", + "timestamps": [ + 650 + ], + "3min_transcript": "Well, we know that all the I's are the same. Hopefully, you can take it as, just conceptually it makes sense to you that the current through the entire circuit will be the same. So all these I's are the same, so we can just cancel them out. Divide both sides by that I. We assume it's non-zero, so I, I, I, I, and then we have that the total resistance of the circuit is equal to R1 plus R2 plus R3. So when you have resistors in series like this, the total resistance, their combined resistance, is just equal to their sum. And that was just a very long-winded way of explaining something very simple, and I'll do an example. Let's say that this voltage is-- I don't know. Let's say it's 20 volts. Let's say resistor 1 is 2 ohms. Let's say resistor 2 is total resistance through this circuit? Well, the total resistance is 2 ohms plus 3 ohms plus 5 ohms, so it's equal to 10 ohms. So total resistance is equal to 10 ohms. So if I were to ask you what is the current going through this circuit? Well, the total resistance is 10 ohms. We know Ohm's law: voltage is equal to current times resistance. The voltage is just equal to 20. 20 is equal to the current times 10 ohms, right? We just added the resistances. Divide both sides by 10. You get the current is equal to 2 amps or 2 coulombs per second. So what seemed like a very long-winded explanation actually results in something that's very, very, very easy to apply. I will see you in the next video." + }, + { + "Q": "At 13:00, why did you choose this as the longest chain? they're all the same.. can we choose any of the three lines as our main group? And the rest is methyl ? or is there a specific rule to follow?\nThank you so much.\n", + "A": "Right. They are all the same length. You can choose any one of them. Whichever one you choose, it will be an ethyl group with two methyl groups on C-1 (1,1-dimethylethyl).", + "video_name": "TJUm860AjNw", + "timestamps": [ + 780 + ], + "3min_transcript": "Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached They're both attached to the one. We have two of them. That's why we wrote di- over there. So it's 1,1-dimethylethyl- and then finally, cyclopentane. So hopefully, that doesn't confuse you too much. I think if you watch the video over and over and try to practice it with your own problems, you'll see that the systematic name way is actually pretty, pretty logical. And actually, if you have more than five or six carbons in the group, they always or they tend to always use the systematic naming." + }, + { + "Q": "\nAt 12:42, why is it necessary to say \"dimethyl\" instead of simply \"methyl\" if it's already indicated by the \"1,1\" that there are two methyls branching out?", + "A": "It works the other way. You count the methyl groups first, then you tell where each of them is. The name says, You have two methyl groups, and they are each on carbon-1 .", + "video_name": "TJUm860AjNw", + "timestamps": [ + 762 + ], + "3min_transcript": "Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached They're both attached to the one. We have two of them. That's why we wrote di- over there. So it's 1,1-dimethylethyl- and then finally, cyclopentane. So hopefully, that doesn't confuse you too much. I think if you watch the video over and over and try to practice it with your own problems, you'll see that the systematic name way is actually pretty, pretty logical. And actually, if you have more than five or six carbons in the group, they always or they tend to always use the systematic naming." + }, + { + "Q": "I am not sure but at 11:50 wont the compound name be (2,2-dimethylpropyl)cyclopentane ?\n", + "A": "In the older systematic name you use the point of connection to the parent chain as carbon #1. So from that carbon the longest chain is 2 carbons long which is why it is ethyl. There s also two other carbons coming off that first carbon so that is why it is 1,1-dimethyl. All together that is (1,1-dimethylethyl)", + "video_name": "TJUm860AjNw", + "timestamps": [ + 710 + ], + "3min_transcript": "Now, that describes just the group. 1-methylpropyl describes just this part right here. That describes just that right over there. And then to have the whole compound, to describe the whole compound, you put this in parentheses, so this is the systematic naming. So 1-methyl-- I put an L there. Let me do it in the same color. 1-methyl, because you're starting where you're attaching. So 1-methyl, you have a methyl group right there on that first carbon. It's a propyl chain. One, two, three, propyl, and then you would say cyclopentane. That's the systematic name for that. Now if you look at this one right here, and the common name is iso-butyl, what you do is you Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached" + }, + { + "Q": "\nIs there any reason that Isobutylcyclopentane can't be written as i-butylcyclopentane?\nOr is it just one of those, \"we don't do this because they tell us not to,\" kind of science situations? (8:48 in video)", + "A": "No. There s no reason. isobutylcyclopentane, i-butylcyclopentane, and 3-methylpropylcyclopentane are all IUPAC names. You can write any of these.", + "video_name": "TJUm860AjNw", + "timestamps": [ + 528 + ], + "3min_transcript": "the difference here. The common naming, it's easier to say and easier to spell, but it's sometimes a little confusing. This is just straight up butyl so you would call this butylcyclopentane. This is sec-butyl, because you have this guy connected to two carbons. That's where the sec- comes from. Sometimes it'll be s-butyl. So this could be called sec-butylcyclopentane or s-butylcyclopentane. This, because we're attached to the end away from this branching off, is still a butyl group, since we have four carbons. But since we're attached here, this is iso-butyl, so this is iso-butylcyclopentane, And then finally, since the carbon we're attaching to is attached to one, two, three other carbons, it is a tert-butyl or a t-butyl group. So this is t-butylcyclopentane. That's the common naming. So maybe I should clear out systematic here just so it's clear to you that everything we've done So let me write it down. It won't hurt to write them down again because the more familiar you are with these, the better. So this is just butylcyclopentane. This is s- or sec- butylcyclopentane. And this is iso-butylcyclopentane. I'm going off the screen here. And then finally, this is tert-butyl, or t-butylcyclopentane. Now, I said these are the common naming. What are the systematic naming? Well, in the systematic, this is still butylcyclopentane. So let me write this down. Systematic, this is still This is very clearly a cyclopentane. This is very clearly a butyl group. But in the systematic naming, what we try to do is we try to name this group right here just as we would name a traditional chain, but we ended it with an -yl. So if you look at this right here, what we do is we just consider the chain where we attach. We attached over here, so the longest chain from that point is there and there. So if you look at it like that, it looks like you have one, two, three carbons, and you have one carbon attached on the beginning. So this little group right here in the systematic naming, this looks like a one, two, three. Three carbons, that's the prop- prefix, so we're dealing with a prop-, and it's all going to be one group, so it's a propyl group. This is a propyl group, but it has a methyl-- remember, meth- is one carbon. It has a methyl group attached on the first carbon." + }, + { + "Q": "\nIN 4:30 WHAT DOES 'sec' actually means?", + "A": "It means that the bond coming from the rest is attached to 2 carbons.", + "video_name": "TJUm860AjNw", + "timestamps": [ + 270 + ], + "3min_transcript": "this first carbon on the butyl right there. I could have just as easily done it like this. I could have just as easily had it like this, where-- let me draw my butyl again, so I have one, two, three, four. So, once again, this is a butyl, but instead of being bonded to the cyclopentane on my first carbon, maybe it's bonded right here. Let me do it with that yellow color. Maybe it's bonded right here. This seems like maybe this could also be butylcyclopentane. It looks like we have a butyl group. This is a butyl right here. I drew a butyl group right over here, and I also drew a butyl group right over here. But these are fundamentally two different molecular structures. I'm touching the first carbon here. Now, there's two ways to differentiate this. One is the common naming and one is the systematic naming. So let me differentiate between the two. So in the common naming, and this can get a little bit involved, and this frankly is probably the most complicated part of naming organic compounds. Systematic is often more complicated, but it's easier to systematically come up with it. So there's a common and then there's a systematic. So the common way of doing it is, if you just say butylcyclopentane, that implies that you are bonding to the first or, depending on how you view it, the last carbon in the chain. So this right here is butylcyclopentane. This right here is not just butylcyclopentane. What you would do is you definitely have a cyclopentane ring, so this would definitely be a cyclopentane. Let me put some space here. And you do have a butyl group on it, so we do have a butyl group, but because we are bonded-- we aren't bonded to the first carbon. We're bonded to a carbon that is bonded to two other carbons. We call this sec-butylcyclopentane, so this is sec-. And everything I'm doing is obviously free-hand. If you were to see this in a book, the sec- would be italicized, or sometimes it would be written as s-butylcyclopentane. And this sec- means that we have attached to a carbon that is touching two other carbons. So you look at the butyl group, and say, well, which of these carbons is attached to two others? It's either that one or that one. And regardless of whether you're attached to this or this, if you think about it, it's fundamentally the same" + }, + { + "Q": "at 4:22, why do you call it secbutyl cyclopentane instead of 2-butyl-cyclopentane ?\n", + "A": "It is called sec-butyl cyclopentane because the sec shows how many carbons are attached to the carbon in the butyl chain that attaches to the cyclopentane, in this case, 2 carbons (on either side of the attached carbon). If it were 2-butyl cyclopentane, it would mean that you had a butane substituent in the second position of the cyclopentane..", + "video_name": "TJUm860AjNw", + "timestamps": [ + 262 + ], + "3min_transcript": "this first carbon on the butyl right there. I could have just as easily done it like this. I could have just as easily had it like this, where-- let me draw my butyl again, so I have one, two, three, four. So, once again, this is a butyl, but instead of being bonded to the cyclopentane on my first carbon, maybe it's bonded right here. Let me do it with that yellow color. Maybe it's bonded right here. This seems like maybe this could also be butylcyclopentane. It looks like we have a butyl group. This is a butyl right here. I drew a butyl group right over here, and I also drew a butyl group right over here. But these are fundamentally two different molecular structures. I'm touching the first carbon here. Now, there's two ways to differentiate this. One is the common naming and one is the systematic naming. So let me differentiate between the two. So in the common naming, and this can get a little bit involved, and this frankly is probably the most complicated part of naming organic compounds. Systematic is often more complicated, but it's easier to systematically come up with it. So there's a common and then there's a systematic. So the common way of doing it is, if you just say butylcyclopentane, that implies that you are bonding to the first or, depending on how you view it, the last carbon in the chain. So this right here is butylcyclopentane. This right here is not just butylcyclopentane. What you would do is you definitely have a cyclopentane ring, so this would definitely be a cyclopentane. Let me put some space here. And you do have a butyl group on it, so we do have a butyl group, but because we are bonded-- we aren't bonded to the first carbon. We're bonded to a carbon that is bonded to two other carbons. We call this sec-butylcyclopentane, so this is sec-. And everything I'm doing is obviously free-hand. If you were to see this in a book, the sec- would be italicized, or sometimes it would be written as s-butylcyclopentane. And this sec- means that we have attached to a carbon that is touching two other carbons. So you look at the butyl group, and say, well, which of these carbons is attached to two others? It's either that one or that one. And regardless of whether you're attached to this or this, if you think about it, it's fundamentally the same" + }, + { + "Q": "At 13:05 near the end of the video, when describing the molecule. Is it not important to label the molecule in alphabetical order? For example. the end result is (1,1 dimethylethyl)cyclopentane. Would it not be more correct to have it as (ethyl - 1,1 -dimethyl)cyclopentane? as E is before M in the alphabet.\n", + "A": "(1,1-dimethylethyl) does not need to go alphabetically. Ethyl is the parent chain of this group, the 2 methyl groups are coming off of that ethyl, so the name needs to show that. It would be similar to naming 4-ethyldecane as decane-4-ethyl because d comes before e.", + "video_name": "TJUm860AjNw", + "timestamps": [ + 785 + ], + "3min_transcript": "Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached They're both attached to the one. We have two of them. That's why we wrote di- over there. So it's 1,1-dimethylethyl- and then finally, cyclopentane. So hopefully, that doesn't confuse you too much. I think if you watch the video over and over and try to practice it with your own problems, you'll see that the systematic name way is actually pretty, pretty logical. And actually, if you have more than five or six carbons in the group, they always or they tend to always use the systematic naming." + }, + { + "Q": "I had learned that while naming, the functional groups should be added in alphabetic order, . In 13:01, you named t-butyl cyclopentane as (1,1-dimethyl ethyl) cyclopentane. shouldn't it be (ethyl 1,1, dimethyl)cyclopentane because 'e' comes before 'm'?\nThanks!\n", + "A": "Hu hu. they re right", + "video_name": "TJUm860AjNw", + "timestamps": [ + 781 + ], + "3min_transcript": "Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached They're both attached to the one. We have two of them. That's why we wrote di- over there. So it's 1,1-dimethylethyl- and then finally, cyclopentane. So hopefully, that doesn't confuse you too much. I think if you watch the video over and over and try to practice it with your own problems, you'll see that the systematic name way is actually pretty, pretty logical. And actually, if you have more than five or six carbons in the group, they always or they tend to always use the systematic naming." + }, + { + "Q": "\nAt 9:55 where did he get the square root of 3/2 from?", + "A": "The cosine of 30 degrees is equal to the square root of three over two.", + "video_name": "_UrfHFEBIpU", + "timestamps": [ + 595 + ], + "3min_transcript": "a little confusing to you. We just said, this point is stationery. It's not moving up or down. It's not accelerating up or down. And so we know that there's a downward force of 100 Newtons, so there must be an upward force that's being provided by these two wires. This wire is providing no upward force. So all of the upward force must be the y component or the upward component of this force vector on the first wire. So given that, we can now solve for the tension in this first wire because we have T1-- what's sine of 30? Sine of 30 degrees, in case you haven't memorized it, sine of 30 degrees is 1/2. So T1 times 1/2 is equal to 100 Newtons. Divide both sides by 1/2 and you get T1 is equal to 200 Newtons. second wire is. And we also, there's another clue here. This point isn't moving left or right, it's stationary. So we know that whatever the tension in this wire must be, it must be being offset by a tension or some other force in the opposite direction. And that force in the opposite direction is the x component of the first wire's tension. So it's this. So T2 is equal to the x component of the first wire's tension. And what's the x component? Well, it's going to be the tension in the first wire, 200 Newtons times the cosine of 30 degrees. It's adjacent over hypotenuse. And that's square root of 3 over 2. So it's 200 times the square root of 3 over 2, which equals 100 square root of 3. completely offsets to the left and the x component of this wire is 100 square root of 3 Newtons to the right. Hopefully I didn't confuse you. See you in the next video." + }, + { + "Q": "\nAt 16:00, Sal talks about all the different variations, 2 to the 46th. how does this work in twins? My cousins are identical but one has slightly curlier hair than the other, how can this work if they have exactly the same DNA, there both very different from each other in personality too. I know there monozigotic, but I don't know what that means.", + "A": "Even identical twins are not completely identical. There were slight differences in the womb, there continue to be differences when they are babies. Genes can be expressed more or expressed less depending on what other genes do and depending on environmental factors.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 960 + ], + "3min_transcript": "to any son or daughter that I might have. We'll talk about how that happens when we talk about meiosis or mitosis, that when I generate my sperm cells, sperm cells essentially takes one-- instead of having 23 pairs of chromosomes in sperm, you only have 23 chromosomes. So, for example, I'll take one from each of those, and through the process of meiosis, which we'll go into, I'll generate a bunch of sperm cells. And each sperm cell will have one from each of these pairs, one version from each of those pairs. So maybe for this chromosome I get it from my dad, from the next chromosome, I get it from my mom. Then I donate a couple more from-- I should've drawn them next to each other. I donate a couple more from my mom. Then for chromosome number 5, it comes from my dad, and so But there's 2 to the twenty-third combinations here, because there are 23 pairs that I'm collecting from. Now, my wife's egg is going to have the same situation. There are 2 to the 23 different combinations of DNA that she can contribute just based on which of the homologous pairs she will contribute. So the possible combinations that just one couple can produce, and I'm using my life as an example, but this applies to everything. This applies to every species that experiences sexual reproduction. So if I can give 2 to twenty-third combinations of DNA and my wife can give 2 to the 23 combinations of DNA, then we can produce 2 to the forty-sixth combinations. Now, just to give an idea of how large of a number this is, this is roughly 12,000 times the number of human beings on So there's a huge amount of variation that even one couple can produce. And if you thought that even that isn't enough, it turns out that amongst these homologous pairs, and we'll talk about when this happens in meiosis, you can actually have DNA recombination. And all that means is when these homologous pairs during meiosis line up near each other, you can have this thing called crossover, where all of this DNA here crosses over and touches over here, and all this DNA crosses over and touches over there. So all of this goes there and all of this goes there. What you end up with after the crossover is that one DNA, the one that came from my mom, or that I thought came from my mom, now has a chunk that came from my dad, and the chunk that came from my dad, now has a chunk that came from my mom. Let me do that in the right color. It came from my mom like that. And so that even increases the amount of variety even more." + }, + { + "Q": "\nAt 18:26 why did Professor Sal think that Sexual Reproduction is variation of population?", + "A": "Sexual reproduction is a huge contributor to genetic variation in a population. Because each parent only donates half of their genome, each indivudual offspring is guaranteed to have a unique genome of its own (Except in the case of identical twins). Mutation also contributes to variation in an important but less immediately noticeable way.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 1106 + ], + "3min_transcript": "So there's a huge amount of variation that even one couple can produce. And if you thought that even that isn't enough, it turns out that amongst these homologous pairs, and we'll talk about when this happens in meiosis, you can actually have DNA recombination. And all that means is when these homologous pairs during meiosis line up near each other, you can have this thing called crossover, where all of this DNA here crosses over and touches over here, and all this DNA crosses over and touches over there. So all of this goes there and all of this goes there. What you end up with after the crossover is that one DNA, the one that came from my mom, or that I thought came from my mom, now has a chunk that came from my dad, and the chunk that came from my dad, now has a chunk that came from my mom. Let me do that in the right color. It came from my mom like that. And so that even increases the amount of variety even more. chromosomes that you're contributing where the chromosomes are each of these collections of DNA, you're now talking about-- you can almost go to the different combinations at the gene level, and now you can think about it in almost infinite form of variation. You can think about all of the variation that might emerge when you start mixing and matching different versions of the same gene in a population. And you don't just look at one gene. I mean, the reality is that genes by themselves very seldom code for a specific-- you can very seldom look for one gene and say, oh, that is brown hair, or look for one gene and say, oh, that's intelligence, or that is how likable someone is. It's usually a whole set of genes interacting in an incredibly complicated way. You know, hair might be coded for by this whole set of genes on multiple chromosomes and this might be coded for a whole set of genes on multiple chromosomes. And so then you can start thinking about all of the different combinations. And then all of a sudden, maybe some combination that never existed before all of a sudden emerges, and that's But I'll leave you to think about it because maybe that combination might be passed on, or it may not be passed on because of this recombination. But we'll talk more about that in the future. But I wanted to introduce this idea of sexual reproduction to you, because this really is the main source of variation within a population. To me, it's kind of a philosophical idea, because we almost take the idea of having males and females for granted because it's this universal idea. But I did a little reading on it, and it turns out that this actually only emerged about 1.4 billion years ago, that this is almost a useful trait, because once you introduce this level of variation, the natural selection can start-- you can kind of say that when you have this more powerful form of variation than just pure mutations, and maybe you might have some primitive form of crossover before, but now that you have this sexual reproduction and you have this variation, natural selection can occur in a" + }, + { + "Q": "\n14:01. Why is it 2 to the power 23 of possible combinations?", + "A": "Statistics: you have 23 [blanks] to be filled in a new chromosome. In each [blank] you can have genetic material from one or the other type (2 options). Therefore, the total combination is 2^23.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 841 + ], + "3min_transcript": "another word, and I'm overwhelming you with words here. So my genotype is exactly what alleles I have, what versions of the gene. So I got like the fifth version of the curly allele. There could be multiple versions of the curly allele in our gene pool. And maybe I got some version of the straight allele. That is my genotype. My phenotype is what my hair really looks like. So, for example, two people could have different genotypes with the same-- they might code for hair that looks pretty much the same, so it might have a very similar phenotype. So one phenotype can be represented by multiple genotypes. So that's just one thing to think about, and we'll talk a lot about that in the future, but I just wanted to introduce you to that there. Now, I entered this whole discussion because I wanted to So how does variation happen? Well, what's going to happen when I-- well, let What's going to happen when I reproduce? And I have. I have a son. Well, my contribution to my son is going to be a random collection of half of these genes. For each homologous pair, I'm either going to contribute the one that I got from my mother or the one that I got from my father, right? So let's say that the sperm cell that went on to fertilize my wife's egg, let's say it happened to have that one, that one, or I could just pick one from each of these 23 sets. And you say, well, how many combinations are there? Well, for every set, I could pick one of the two homologous chromosomes, and I'm going to do that 23 times. 2 times 2 times 2, so that's 2 to the twenty third. to any son or daughter that I might have. We'll talk about how that happens when we talk about meiosis or mitosis, that when I generate my sperm cells, sperm cells essentially takes one-- instead of having 23 pairs of chromosomes in sperm, you only have 23 chromosomes. So, for example, I'll take one from each of those, and through the process of meiosis, which we'll go into, I'll generate a bunch of sperm cells. And each sperm cell will have one from each of these pairs, one version from each of those pairs. So maybe for this chromosome I get it from my dad, from the next chromosome, I get it from my mom. Then I donate a couple more from-- I should've drawn them next to each other. I donate a couple more from my mom. Then for chromosome number 5, it comes from my dad, and so" + }, + { + "Q": "At 12:22, it mentions different versions of allele genes. How many are there?\n", + "A": "We don t know. A gene can have any number of different alleles. Some alleles may have no phenotype at all, others may have a very significant phenotype.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 742 + ], + "3min_transcript": "reproduction have this complete set of chromosomes in it, which I find amazing. But only certain chromosomes-- for example, these genes will be completely useless in my fingernails, because all of a sudden, the straight and the curly don't matter that much. And I'm simplifying. Maybe they will on some other dimension. But let's say for simplicity, they won't matter in certain places. So certain genes are expressed in certain parts of the body, but every one of your body cells, and we call those somatic cells, and we'll separate those from the sex sells or the germs that we'll talk about later. So this is my body cells. So this is the great majority of your cells, and this is opposed to your germ cells. And the germ cells-- I'll just write it here, just so you get a clear-- for a male, that's the sperm cells, and for female that's the egg cells, or the ova. and what I want to give you the idea is that for every trait, I essentially have two versions: one from my mother and one from my father. Now, these right here are called homologous chromosomes. What that means is every time you see this prefix homologous or if you see like Homo sapiens or even the word homosexual or homogeneous, it means same, right? You see that all the time. So homologous means that they're almost the same. They're coding for the most part the same set of genes, but they're not identical. They actually might code for slightly different versions of the same gene. So depending on what versions I get, what is actually another word, and I'm overwhelming you with words here. So my genotype is exactly what alleles I have, what versions of the gene. So I got like the fifth version of the curly allele. There could be multiple versions of the curly allele in our gene pool. And maybe I got some version of the straight allele. That is my genotype. My phenotype is what my hair really looks like. So, for example, two people could have different genotypes with the same-- they might code for hair that looks pretty much the same, so it might have a very similar phenotype. So one phenotype can be represented by multiple genotypes. So that's just one thing to think about, and we'll talk a lot about that in the future, but I just wanted to introduce you to that there. Now, I entered this whole discussion because I wanted to" + }, + { + "Q": "How can an organism be both a male and a female? @4:29\n", + "A": "Many organisms in nature are hermaphordites, possessing both male and female reproductive organs. The worm C. elegans is a common example of a hermaphrodite.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 269 + ], + "3min_transcript": "these code for proteins or they code for things that control other proteins, but maybe you have a change in one of them. Maybe this cytosine for whatever reason becomes a guanine randomly, or maybe these got deleted, and that would change the DNA. But you could imagine, if I went to someone's computer code and just randomly started changing letters and randomly started inserting letters without really knowing what I'm doing, most of the time, I'm going to break the computer program. Most of the time, the great majority of the time, this is going to go nowhere. It'll either do nothing, for example, if I go into someone's computer program and if I just add a couple of spaces or something, that might not change the computer program, but if I start getting rid of semicolons and start changing numbers and all that, it'll probably make the computer program break. So it'll either do nothing or it'll actually kill the organisms most of the time. Mutations: sometimes, they might make the actual cell kind of run amok, and we'll do a whole maybe series of videos organism as well as a whole, although if it occurs after the organism has reproduced, it might not be something that selects against the organism and it also wouldn't be passed on. But anyway, I won't go detailed into that. But the whole point is that mutations don't seem to be a satisfying source of variation. They could be a source or kind of contribute on the margin, but there must be something more profound than mutations that's creating the diversity even within, or maybe I should call it the variation, even within a population. And the answer here is really it's kind of right in front of us. It really addresses kind of one of the most fundamental things about biology, and it's so fundamental that a lot of people never even question why it is the way it is. And that is sexual reproduction. And when I mean sexual reproduction, it's this notion that have nucleuses-- and we call those eukaroytes. Maybe I'll do a whole video on eukaryotes versus prokaryotes, but it's the notion that if you look universally all the way from plants-- not universally, but if you look at cells that have nucleuses, they almost universally have this phenomenon that you have males and you have females. In some organisms, an organism can be both a male and a female, but the common idea here is that all organisms kind of produce versions of their genetic material that mix with other organisms' version of their genetic material. If mutations were the only source of variation, then I could just bud off other Sals. Maybe just other Sals would just bud off from me, and then randomly one Sal might be a little bit different and" + }, + { + "Q": "\nsal u just told(15:07)that we get the chromosome from our parents and that means our parent too follow this then this will mean that the chromosome that the first human ever had are shill carried.am i right?", + "A": "Unless the first human didn t have children.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 907 + ], + "3min_transcript": "So how does variation happen? Well, what's going to happen when I-- well, let What's going to happen when I reproduce? And I have. I have a son. Well, my contribution to my son is going to be a random collection of half of these genes. For each homologous pair, I'm either going to contribute the one that I got from my mother or the one that I got from my father, right? So let's say that the sperm cell that went on to fertilize my wife's egg, let's say it happened to have that one, that one, or I could just pick one from each of these 23 sets. And you say, well, how many combinations are there? Well, for every set, I could pick one of the two homologous chromosomes, and I'm going to do that 23 times. 2 times 2 times 2, so that's 2 to the twenty third. to any son or daughter that I might have. We'll talk about how that happens when we talk about meiosis or mitosis, that when I generate my sperm cells, sperm cells essentially takes one-- instead of having 23 pairs of chromosomes in sperm, you only have 23 chromosomes. So, for example, I'll take one from each of those, and through the process of meiosis, which we'll go into, I'll generate a bunch of sperm cells. And each sperm cell will have one from each of these pairs, one version from each of those pairs. So maybe for this chromosome I get it from my dad, from the next chromosome, I get it from my mom. Then I donate a couple more from-- I should've drawn them next to each other. I donate a couple more from my mom. Then for chromosome number 5, it comes from my dad, and so But there's 2 to the twenty-third combinations here, because there are 23 pairs that I'm collecting from. Now, my wife's egg is going to have the same situation. There are 2 to the 23 different combinations of DNA that she can contribute just based on which of the homologous pairs she will contribute. So the possible combinations that just one couple can produce, and I'm using my life as an example, but this applies to everything. This applies to every species that experiences sexual reproduction. So if I can give 2 to twenty-third combinations of DNA and my wife can give 2 to the 23 combinations of DNA, then we can produce 2 to the forty-sixth combinations. Now, just to give an idea of how large of a number this is, this is roughly 12,000 times the number of human beings on" + }, + { + "Q": "2:35\ncan we modify DNA accroding to our wish.\ncan we design the codons on our own.\n", + "A": "Yes, technically in I think it was late 2007 a group a microbiologists created a new species of bacteria that was blue. The blue bit was just so that they could see them easier if it worked, but it s true. However the technology, like CarlBiologist said, is in its infancy and when they created the bacteria it was impossible for them to create anything much more complex then that. Medium-complexity bacteria was their limit, just because of how much trouble it is to assemble long strands of DNA.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 155 + ], + "3min_transcript": "And if they are drastic enough, maybe these guys start becoming dominant and start not liking these guys, because they're so different or whatever else. We could see a lot of different reasons. This could eventually turn into a different species. Now, the obvious question is what leads to this variation? In a population what leads to this-- in fact, even in our population, what leads to one person having dirty blonde hair, one person having brown hair, one person having black hair, and we have the spectrum of skin complexions and heights is pretty much infinite. What causes that? And then one thing that I kind of point to, we talked about this a little bit in the DNA video, is this notion of mutations. DNA, we learned, is just a sequence of these bases. So adenine, guanine, let's say I've got some thymine going. I have some more adenine, some cytosine. And that these code, if you have enough of these in a row, these code for proteins or they code for things that control other proteins, but maybe you have a change in one of them. Maybe this cytosine for whatever reason becomes a guanine randomly, or maybe these got deleted, and that would change the DNA. But you could imagine, if I went to someone's computer code and just randomly started changing letters and randomly started inserting letters without really knowing what I'm doing, most of the time, I'm going to break the computer program. Most of the time, the great majority of the time, this is going to go nowhere. It'll either do nothing, for example, if I go into someone's computer program and if I just add a couple of spaces or something, that might not change the computer program, but if I start getting rid of semicolons and start changing numbers and all that, it'll probably make the computer program break. So it'll either do nothing or it'll actually kill the organisms most of the time. Mutations: sometimes, they might make the actual cell kind of run amok, and we'll do a whole maybe series of videos organism as well as a whole, although if it occurs after the organism has reproduced, it might not be something that selects against the organism and it also wouldn't be passed on. But anyway, I won't go detailed into that. But the whole point is that mutations don't seem to be a satisfying source of variation. They could be a source or kind of contribute on the margin, but there must be something more profound than mutations that's creating the diversity even within, or maybe I should call it the variation, even within a population. And the answer here is really it's kind of right in front of us. It really addresses kind of one of the most fundamental things about biology, and it's so fundamental that a lot of people never even question why it is the way it is. And that is sexual reproduction. And when I mean sexual reproduction, it's this notion" + }, + { + "Q": "\nAround 14:10, Sal says he has 2^23 forms of contributing to his \"son or daughter\" . Wouldn't it be 2^22 since one chromosome indicates the sex?", + "A": "No, he will contribute a chromosome that will help decide what gender the child could be.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 850 + ], + "3min_transcript": "another word, and I'm overwhelming you with words here. So my genotype is exactly what alleles I have, what versions of the gene. So I got like the fifth version of the curly allele. There could be multiple versions of the curly allele in our gene pool. And maybe I got some version of the straight allele. That is my genotype. My phenotype is what my hair really looks like. So, for example, two people could have different genotypes with the same-- they might code for hair that looks pretty much the same, so it might have a very similar phenotype. So one phenotype can be represented by multiple genotypes. So that's just one thing to think about, and we'll talk a lot about that in the future, but I just wanted to introduce you to that there. Now, I entered this whole discussion because I wanted to So how does variation happen? Well, what's going to happen when I-- well, let What's going to happen when I reproduce? And I have. I have a son. Well, my contribution to my son is going to be a random collection of half of these genes. For each homologous pair, I'm either going to contribute the one that I got from my mother or the one that I got from my father, right? So let's say that the sperm cell that went on to fertilize my wife's egg, let's say it happened to have that one, that one, or I could just pick one from each of these 23 sets. And you say, well, how many combinations are there? Well, for every set, I could pick one of the two homologous chromosomes, and I'm going to do that 23 times. 2 times 2 times 2, so that's 2 to the twenty third. to any son or daughter that I might have. We'll talk about how that happens when we talk about meiosis or mitosis, that when I generate my sperm cells, sperm cells essentially takes one-- instead of having 23 pairs of chromosomes in sperm, you only have 23 chromosomes. So, for example, I'll take one from each of those, and through the process of meiosis, which we'll go into, I'll generate a bunch of sperm cells. And each sperm cell will have one from each of these pairs, one version from each of those pairs. So maybe for this chromosome I get it from my dad, from the next chromosome, I get it from my mom. Then I donate a couple more from-- I should've drawn them next to each other. I donate a couple more from my mom. Then for chromosome number 5, it comes from my dad, and so" + }, + { + "Q": "At 2:45 what does delta T mean?\n", + "A": "Say we heat something from 20 Celsius to 50 Celsius, the delta T would be 50 - 20 = 30, i.e. 30 is the change in temperature. Now say we cool something from 60 Celsius to 10 Celsius, the change in temp is 60 - 10 = 50 celsius. It doesnt matter if we heat or cool something, we still have to use energy to change the state of the substance; the only difference is when we heat something, it takes in energy, when it cools it releases energy, but the calculation is the same. I hope u understand", + "video_name": "H7nrVDV8ahc", + "timestamps": [ + 165 + ], + "3min_transcript": "And it's going to be a little mathy, but I think the math is pretty straightforward, especially if you've taken a first-year course in calculus. And this is actually a pretty neat application of it. So let's just think a little bit about the rate of change, or the probability, or the number particles that are changing at any given time. So if we say, the difference or change in our number of particles, or the amount of particles, in any very small period of time, what's this going to be dependent on? This is the number particles we have in a given period time. This is our rate of change. So one thing, we know that our rate of change is going down. We know it's a negative number. We know that, in the case of radioactive decay, I could do the same exercise with compounding growth, where I would say, oh no, it's not a negative number, that our growth is dependent on how much we have. In this case the amount we're decaying is proportional, but it's going already have. Let me explain that. So what I'm saying is, look, our amount of decay is proportional to the amount of the substance that we already are dealing with. And just to maybe make that a little bit more intuitive, imagine a situation here where you have 1 times 10 to the 9th. You have a billion carbon atoms. And let's say over here you have 1 times 10 to the 6th carbon atoms. And if you look at it at over some small period of time, let's say, if you look at it over one second, let's say our dt. dt as an infinitesimally small time, but let's say it's a change in time. It's a delta t. And let's say over one second, you observe that this sample had, I don't know, let's say you saw 1000 carbon particles. You really wouldn't see that with carbon-14, but this is just for the sake of our intuition. Let's say over one second you saw 1000 carbon particles per Well here you have 1000th of the number particles in this sample as this one. So, for every thousand particles you saw decaying here, you'd really expect to see one carbon particle per second here. Just because you have a smaller amount. Now I don't know what the actual constant is. But we know that no matter what substance we're talking about, this constant is dependent on the substance. Carbon's going to be different from uranium, is going to be different from, you know, we looked at radon. They're all going to have different quantities right here. And we can see that. We'll actually do it in the next video, you can actually calculate this from the half-life. But the rate of change is always going to be dependent on the number of particles you have, right? I mean, we saw that here with half-life. When you have 1/2 the number of particles, you lose 1/2 as much. Here, if we start with 100 particles here, we went to 50 particles, then we went to 25. When you start with 50, in a period of time you lose 25." + }, + { + "Q": "At 1:00 where do the 2 hydrogen cations come from as showed on the products side of glycolysis?\n", + "A": "The H+ comes from NADH", + "video_name": "ArmlWtDnuys", + "timestamps": [ + 60 + ], + "3min_transcript": "- [Voiceover] So let's give ourselves an overview of glycolysis. and glycolysis is an incredibly important biochemical pathway. It occures in practically all life as we know it and it's all about taking glucose as a fuel and, in the process of breaking it up, lycing the glucose, glycolysis, breaking it up into two pyruvate molecules. Glucose is a six carbon molecule. Each of the pyruvates are three carbon molecules. In the process of doing that, you produce two ATPs net. It actually turns out that you need to use two ATPs and then you produce four. So you use two ATPs. That's often called the investment phase and we'll talk about that in a second. And then you produce four ATPs for a net of plus two ATPs and that's what we see right over here. You see a net of two ATPs being produced directly the reduction of NAD to NADH. Remember, reduction is all about gaining electrons, and over here, NAD, that's nicotinamide adinine dinucleotide, we have other videos on that, it's an interesting molecule, it's actually a fairly decent-sized molecule, we see this positive charge, but then we see that not only does it gain a hydrogen, but it loses its positive charge. It gains a hydrogen and an electron. You can think on a net basis it's gaining a hydride. Now a hydride anion's not going to typically be all by itself, but on a net basis, you can think about that's what's happening. And so it's gaining a hydrogen and an extra electron and so this, the NAD+, this is going to get reduced. That is going to get reduced to NADH. So this is getting reduced to NADH. be oxidized in the electron transport chain. We'll study that later on when we think about oxidative phosphorylation, to produce even more ATPs. But on a very high-level, simple basis. Glucose being broken down in pyruvate, six carbons, three carbons each of these pyruvates, now there's other things attached to the carbons, and we'll see that in a little bit. Two ATPs net generated, and you have the reduction of two NADs to two NADHs, and those can be used later on to produce more ATPs. Now, glycolysis is typically just the beginning of cellular respiration. If oxygen is around, then you have these products, some of these moving into the mitochondria where you can have the citric acid cycle, Krebs cycle, and the oxidative phosphorylation occur. If you don't have oxygen around, then you're going to do anaerobic respiration, or you're" + }, + { + "Q": "Around 9:00 Hank says \"Haeckel influenced from Darwin and Darwin disagreed with him\".\n\nWhat does that mean? Is it a mistake?\n", + "A": "It is not a mistake, it just means that Haeckel was inspired by Darwin, but Darwin didnt agree with where Haeckel was taking his (Darwins) ideas.", + "video_name": "cstic6WHr2E", + "timestamps": [ + 540 + ], + "3min_transcript": "the esophagus and stomach and colon and stuff. And in addition, some of the cells start breaking off between the endoderm and the ectoderm and form another layer called the mesoderm. These cells will eventually end up as the muscles and the circulatory system and the reproductive systems and in the case of vertebrates, most of the bone. So, what's our embryo looking like now? Awesome! From here, this little guy is gonna go on to fulfill his destiny as a ladybug or a walrus or whatever. And now this seems to me like a great time to take a look at a completely dis-proven theory that biologists hold in the highest contempt, but which is actually a kind of useful way to think about the way that an animal embryo develops into a fully formed animal. Plus, it makes for a great biolo-graphy! (music) Back in the mid-1800's a German zoologist named Ernst Haeckel tried to prove what we now refer to as recapitulation theory. Basically, and this is not basic at all, \"Ontogeny recapitulates phylogeny.\" Uhh! In other words, ontogeny, or the growth and development of an embryo recapitulates or sums up phylogeny, which is the evolutionary history of a species. So this means, for instance, that a human embryo over the course of it's development will go through all of the hundreds of millions of years worth of evolutionary steps that it took for a single celled organism to evolve into a fully tricked out person. Haeckel was a contemporary of Darwin and on the origin of species, made a giant impression on him, especially a section of it that notes how cool it is that all vertebrate embryos look pretty similar to one another, regardless of whether they're a mammal or bird or reptile. Darwin, however, cautioned that this probably wasn't a very good way of reconstructing the history of evolution. He just thought it meant that the embryological similarities were evidence of common ancestry between species. Well, Haeckel was kind of a spaz and he definitely heard the first part of Darwin's idea, and very quickly wrote a couple of books about how the development of an embryo mirrors the evolutionary development of adults of a species, which is exactly what Darwin said was not happening. Anyway, Haeckel did spend a lot of time looking at embryos and observed that the slits in the neck of a human embryo resembled the gill slits of fish, which he took to mean that we must have, at one point, had a fish-like ancestor. He drew tons of figures of different animal embryos in different stages of development to prove his theory and his illustrations of embryos started to make their way into textbooks all over the world. Haeckel is exactly the sort of person who really ticks other scientists off because real science loving scientists like to sit and think about stuff and find out all the problems with an idea before they start publishing books about it. And here, Haeckel was firing off volume after volume and before long all the \"data\" he had \"collected\" convinced a bunch of other people," + }, + { + "Q": "\n09:09 PE=m*g*h right? why he did not multiply it to the gravity?", + "A": "The weight of 10 N is already the result of Fg = m * g.", + "video_name": "vSsK7Rfa3yA", + "timestamps": [ + 549 + ], + "3min_transcript": "of this machine is equal to 10 Newtons. Mechanical advantage is the output over the input, so the mechanical advantage is equal to the force output by the force input, which equals 10/5, which equals 2. And that makes sense, because I have to pull twice as much for this thing to move up half of that distance. Let's see if we can do another mechanical advantage problem. Actually, let's do a really simple one that we've really been working with a long time. Let's say that I have a wedge. A wedge is actually considered a machine, which it took me a little while to get my mind around that, but a wedge is a machine. And why is a wedge a machine? Because it gives you mechanical advantage. So if I have this wedge here. And this is a 30-degree angle, if this distance up here, distance going to be? Well, it's going to be D sine of 30. And we know that the sine of 30 degrees, hopefully by this point, is 1/2, so this is going to be 1/2D. You might want to review the trigonometry a little bit if that doesn't completely ring a bell for you. So if I take an object, if I take a box-- and let's assume it has no friction. We're not going to go into the whole normal force and all that. If I take a box, and I push it with some force all the way up here, what is the mechanical advantage of this system? Well, when the box is up here, we know what its potential energy is. Its potential energy is going to be the weight of the box. So let's say this is a 10-Newton box. The potential energy at this point is going to be 10 Newtons times its height. So potential energy at this point has to equal 10 Newtons And that's also the amount of work one has to put into the system in order to get it into this state, in order to get it this high in the air. So we know that we would have to put 5 joules of work in order to get the box up to this point. So what is the force that we had to apply? Well, it's that force, that input force, times this distance has to equal 5 joules. So this input force-- oh, sorry, this is going to be-- sorry, this isn't 5 joules. It's 10 times 1/2 times the distance. It's 5D joules. This isn't some kind of units. It's 10 Newtons times the distance that we're up, and that's 1/2D, so it's 5D joules. Sorry for confusing you. And so the force I'm pushing here times this distance has to also equal to 5D joules." + }, + { + "Q": "\nAround 9:22, Sal mentioned that PE is 10kg times height. What about gravity? I thought PE equals Mxgxheight?? Please help. Thanks.", + "A": "I think you misheard. He didn t say 10kg he said 10N. N would already include any relevant gravitational force.", + "video_name": "vSsK7Rfa3yA", + "timestamps": [ + 562 + ], + "3min_transcript": "of this machine is equal to 10 Newtons. Mechanical advantage is the output over the input, so the mechanical advantage is equal to the force output by the force input, which equals 10/5, which equals 2. And that makes sense, because I have to pull twice as much for this thing to move up half of that distance. Let's see if we can do another mechanical advantage problem. Actually, let's do a really simple one that we've really been working with a long time. Let's say that I have a wedge. A wedge is actually considered a machine, which it took me a little while to get my mind around that, but a wedge is a machine. And why is a wedge a machine? Because it gives you mechanical advantage. So if I have this wedge here. And this is a 30-degree angle, if this distance up here, distance going to be? Well, it's going to be D sine of 30. And we know that the sine of 30 degrees, hopefully by this point, is 1/2, so this is going to be 1/2D. You might want to review the trigonometry a little bit if that doesn't completely ring a bell for you. So if I take an object, if I take a box-- and let's assume it has no friction. We're not going to go into the whole normal force and all that. If I take a box, and I push it with some force all the way up here, what is the mechanical advantage of this system? Well, when the box is up here, we know what its potential energy is. Its potential energy is going to be the weight of the box. So let's say this is a 10-Newton box. The potential energy at this point is going to be 10 Newtons times its height. So potential energy at this point has to equal 10 Newtons And that's also the amount of work one has to put into the system in order to get it into this state, in order to get it this high in the air. So we know that we would have to put 5 joules of work in order to get the box up to this point. So what is the force that we had to apply? Well, it's that force, that input force, times this distance has to equal 5 joules. So this input force-- oh, sorry, this is going to be-- sorry, this isn't 5 joules. It's 10 times 1/2 times the distance. It's 5D joules. This isn't some kind of units. It's 10 Newtons times the distance that we're up, and that's 1/2D, so it's 5D joules. Sorry for confusing you. And so the force I'm pushing here times this distance has to also equal to 5D joules." + }, + { + "Q": "At min 9:00 you say that the heater is switched off by the bimetallic strip if it gets too hot. That means that the fan switches off too, right? Because as you said, the heater works as a resistor to the fan motor....\nThis implies that the dryer stops from working if it gets hot, which obviously cant be true!\nI am loving this video btw :D\n", + "A": "If the circuit is broken then both the heater and the motor would stop functioning, however metal takes time to cool down. I do not know the schematic of the circuit", + "video_name": "Vq7EOmvU1eQ", + "timestamps": [ + 540 + ], + "3min_transcript": "So a lot of times when you heat up other metals, they'll oxidize, they'll rust. And then you get problems with that. This doesn't do that. It just heats up, and then it cools back down and there's no oxidization. So it works really well as a heater. And on the front, here, you can see there's this 90 degree opposed bracket that holds everything 90 degrees apart, which keeps everything working as far as air flow and stuff like that. It also holds this mica sheet separated. And then you can see on the inside, we have these brass contacts. And they're little brass rivets. And they distribute the power around to different parts of the heater. And right here, this part is a bimetallic strip. So that bimetallic strip is made out of two pieces of metal. And what it does is when the metal is heated to a certain point, it causes the bimetallic strip to bend. than the other piece of metal. And it causes the strip to bend. So maybe, this one on the outside-- if this was the bimetallic strip. This one expands faster than this one, and it causes it to bend. So you could have two different types of metal like, say, an alloy called invar and another piece of metal called copper. And the copper is going to respond to heat at a certain rate, and it's going to expand. And the invar is going to not expand nearly as fast. And it's going to cause that switch to open up. And that will shut down the power to the heater. So if it gets too hot, the bimetallic strip will expand, and it will pull itself away from the contact and shut down the electricity to the heater. You can see, we also have a diode, right here, in line. That just controls the flow of electricity. . It's like a little electrical valve. And then we have a thermal resistor, here. This is called a thermal fuse. And the thermal fuse, basically, is another safety precaution. If temperatures get too high, the fuse will blow. And it will shut down the hair dryer and prevent the housing from melting, or the hair dryer from getting too hot and potentially blowing air out that could burn you. And so, remember, the heater has functioned as a resistor. And it has dropped the voltage down to 12 volts. But it's still AC power, which means alternating current. Alternating current, basically, functions as a sine wave, It flows back and forth. And the motor that we have here is designed to run on direct current, so just flowing in a continuous loop. And the way that this unit, this hair dryer, deals with that is it has these four diodes. You can see them-- one, two, three, four. And so what those four diodes do is they function like a bridge rectifier." + }, + { + "Q": "\nAt 5:20, why does the lone pair repel the other bonds more strongly? Didn't it act just like a regular bond in the previous videos?", + "A": "When you are drawing dot structures, you are not dealing directly with geometries. When you start talking about bond angles, the difference between the repulsion from two electrons belonging to a single atom (lone pair) vs the repulsion from two electrons being shared by two atoms (sigma bond) becomes important.", + "video_name": "BM-My1AheLw", + "timestamps": [ + 320 + ], + "3min_transcript": "and then we also have a bond angle, let me go ahead and draw that in, so a bond angle, this hydrogen-carbon-hydrogen bond angle in here, is approximately 109 point five degrees. All right, let's go ahead and do the same type of analysis for a different molecule, here. So let's do it for ammonia, next. So we have NH three, if I want to find the steric number, the steric number is equal to the number of sigma bonds, so that's one, two, three; so three sigma bonds. Plus number of lone pairs of electrons, so I have one lone pair of electrons here, so three plus one gives me a steric number of four. So I need four hybridized orbitals, and once again, when I need four hybridized orbitals, I know that this nitrogen must be SP three hybridized, because SP three hybridization gives us four hybrid orbitals, and so let's go ahead and draw those four hybrid orbitals. So we would have nitrogen, and let's go ahead and draw in all four of those. those are the four hybrid orbitals. When you're drawing the dot structure for nitrogen, you would have one electron, another electron, another electron, and then you'd have two in this one, like that. And then you'd go ahead, and put in your hydrogens, so, once again, each hydrogen has one electron, in a hybridized S orbital, so we go ahead and draw in those hydrogens, so our overlap of orbitals, so here's a sigma bond, here's a sigma bond, and here's a sigma bond; so three sigma bonds in ammonia, and then we have this lone pair up here. So the arrangement of these electron pairs, is just what we talked about before: So we have this tetrahedral arrangement of electron pairs, or electron groups, so the VSEPR theory tells us that's how they're going to repel. However, that's not the shape of the molecule, so if I go ahead and draw in another picture over here, to talk about the molecular geometry, and go ahead and draw in the bonding electrons, like that, and then I'll put in my non-bonding housed in an SP three hybridized orbital. So, the arrangement of the atoms turns out not to be tetrahedral, and that has to do with this lone pair of electrons up here, at the top. So, this lone pair of electrons is going to repel these bonding electrons more strongly than in our previous example, and because it's going to repel those electrons a little bit more strongly, you're not gonna get a bond angle of 109 point five; it's going to decrease the bond angle. So let me go ahead, and use the same color we used before, so this bond angle is not 109 point five; it goes down a bit, because of the extra repulsion, so it turns out to be approximately 107 degrees. And in terms of the shape of the molecule, we don't say \"tetrahedral\"; we say \"trigonal-pyramidal.\" So let me go ahead, and write that here, so the geometry of the ammonia molecule is trigonal-pyramidal, and let's analyze that a little bit." + }, + { + "Q": "at around 4:50 minutes sal said the volume of 1 pound lead is more than volume of 1 pound lead .shouldn't it be the reverse of it?\n", + "A": "he says that the volume of one pound of feathers will be a lot more than the volume of 1 pound of lead because feathers have a smaller density than lead", + "video_name": "5EWjlpc0S00", + "timestamps": [ + 290 + ], + "3min_transcript": "Actually, I shouldn't say water-- let me change this, because I said that this is going to be some random liquid, and the mass is a liquid. The force down is going to be equal to the mass of the liquid times gravity. What is that mass of the liquid? Well, now I'll introduce you to a concept called density, and I think you understand what density is-- it's how much there is of something in a given amount of volume, or how much mass per volume. That's the definition of density. The letter people use for density is rho-- let me do that in a different color down here. rho, which looks like a p to me, equals mass per volume, The units are kilograms per meter cubed-- that is density. I think you might have an intuition that if I have a cubic meter of lead-- lead is more dense than marshmallows. Because of that, if I have a cubic meter of lead, it will have a lot more mass, and in a gravitational field, weigh a lot more than a cubic meter of marshmallows. Of course, there's always that trick people say, what weighs more-- a pound of feathers, or a pound of lead? Those, obviously, weigh the same-- the key is the volume. A cubic meter of lead is going to weigh a lot more than a cubic meter of feathers. Making sure that we now know what the density is, let's go back to what we were doing before. We said that the downward force is equal to the mass of the liquid times the gravitational force, and so We could use this formula right here-- density is equal to mass times volume, so we could also say that mass is equal to density times volume. I just multiply both sides of this equation times volume. In this situation, force down is equal to-- let's substitute this with this. The mass of the liquid is equal to the density of the liquid times the volume of the liquid-- I could get rid of these l's-- times gravity. What's the volume of the liquid? The volume of the liquid is going to be the cross-sectional area of the cylinder times the height. So let's call this cross-sectional area A. A for area-- that's the area of the cylinder or the foil that's floating within the water." + }, + { + "Q": "at 7:13 howcome ligers arnt species\n", + "A": "Organisms are considered part of a species if they can breed amongst themselves. Since ligers are sterile and unable to produce offspring, then that s your explanation.", + "video_name": "Tmt4zrDK3dA", + "timestamps": [ + 433 + ], + "3min_transcript": "And that is a different hybrid that has slightly different properties than a liger. I encourage you to look up what a tiglon is. Similarly, you give me a male donkey. And donkeys are clearly a species by themselves, because if you give me a male donkey and a female donkey they can reproduce, produce another donkey, and then that donkey can mate with other donkeys to produce more and more donkeys. So not only can a donkey interbreed with another donkey, but that product, that child donkey, can keep interbreeding with other donkeys. Similarly, horses, they can interbreed and produce fertile offspring. But if you give me a male donkey and a female horse they can mate and they can produce a mule. But once again, like the ligers, mules are not, at least as far as I know, mules they're not fertile. They cannot interbreed with each other. And because, even though donkeys and horses can breed and produce mules, their offspring aren't fertile. We don't consider donkeys and horses part of the same species. And we would consider mules, like a liger or a tiglon, we would consider them a hybrid. So these are all hybrids, or we would call cross. In general, the word hybrid is used when you have two things, two different types that are somehow coming together, somehow having a combination. And once again, like the case with the tiglon, you might say, well, what if I had a female donkey and a male horse? And then you would actually produce something called a hinny, which isn't as common as a mule. And people like to use mules, they're actually very good work animals because they have some of the good properties of both donkeys and horses. Hinnies are less common, but once again, it is possible. And they have different properties than mules. And I do want to emphasize this idea. to think about, well, how can we classify things? And we said, hey, maybe things that look and act similar, we can call a species. And maybe things that look and act different, we shouldn't call them species. But I want to show you a very typical case, one that's really all around us all the time, where this definition-- animals that can interbreed and the offspring are fertile-- really does seem to become much, much more important than just some notion of animals that look alike or animals that act the same. And the best example of that is with dogs. As I said, this is a very typical species here, because dogs-- and I just took a sample of some of the different types of breeds of dogs-- they can look very, very different. It's obvious, look at the difference between these dogs. For example, this little chihuahua here and this dog right over here. Obviously, they're size-wise, their look, and even how they act are much, much more different than maybe how this donkey would act relative to this horse," + }, + { + "Q": "At,08:23,why is the upper limit zero ,as we are putting r^infinity which will be zero so what happened to the h^2?\n", + "A": "As r gets larger, theta gets larger. As theta gets larger, the vertical force component gets shorter. At point infinity, theta becomes a right angle and the force is effectively flat (0 vertical).", + "video_name": "TxwE4_dXo8s", + "timestamps": [ + 503 + ], + "3min_transcript": "H squared plus r squared is u. We do that by definition. So u to the 3/2, which is equal to the antiderivative of-- we could write this as u to the minus 3/2 du. This is just kind of reverse the exponent rule. So that equals minus 2u to the minus 1/2, and we can confirm, right? If we take the derivative of this, minus 1/2 times minus 2 is 1, and then subtract 1 from here, we get minus 3/2. And then we could add plus c, but since we're eventually going to do a definite integral, the c's all cancel out. Or we could say that this is equal to-- since we made that substitution-- minus 2 over-- minus 1/2, that's the same thing as over the square root of h squared So all of the stuff I did in magenta was just to figure out the antiderivative of this, and we figured it out to be this: minus 2 over the square root of h squared plus r squared. So with that out of the way, let's continue evaluating our definite integral. So this expression simplifies to-- this is a marathon problem, but satisfying-- K-- let's get all the constants-- Kh pi sigma-- we can even take this minus 2 out-- times minus 2, and all of that, and we're going to evaluate the definite integral at the two boundaries-- 1 over the square root of h squared plus r squared evaluated at infinity minus it evaluated at 0, right? What is 1 over the square root of h squared plus infinity, right? What happens when we evaluate r at infinity? Well, the square root of infinity is still infinity, and 1 over infinity is 0, so this expression right here just becomes 0. When you evaluate it at infinity, this becomes 0 minus this expression evaluated at 0. So what happens when it's at 0? When r squared is 0, we get 1 over the square root of h squared, right? So let's write it all out. This becomes minus 2Kh pi sigma times 0 minus 1 over the square root of h squared. Well this equals minus 2Kh pi sigma times-- well, 1 over the" + }, + { + "Q": "6:14\nHe integrates u^(-3/2) to -2u^(-1/2). Shouldn't it instead be (-2^(-1/2))/3 ? (since you have to divide by -3/2)\n", + "A": "Try taking the derivative of your answer and see if you get the original equation", + "video_name": "TxwE4_dXo8s", + "timestamps": [ + 374 + ], + "3min_transcript": "You'll see why in a second, but I'm going to take all the other constants out that we're not integrating across. So it's equal to Kh pi sigma times the integral from zero to infinity of what is this? So what did I leave in there? I left a 2r, so we could rewrite this as-- well, actually, I'm running out of space. 2r dr over h squared plus r squared to the 3/2, or we could think of it as the negative 3/2, right? So what is the antiderivative of here? Well, this is essentially the reverse chain rule, right? I could make a substitution here, if you're more comfortable using the substitution rule, but you We could make the substitution that u is equal to-- if we just want to figure out the antiderivative of this-- if u is equal to h squared plus r squared-- h is just a constant, right-- then du is just equal to-- I mean, the du dr-- this is a constant, so it equals 2r, or we could say du is equal to 2r dr. And so if we're trying to take the antiderivative of 2r dr over h squared plus r squared to the 3/2, this is the exact same thing as taking the antiderivative with this substitution. 2r dr, we just showed right here, that's the same thing as du, right? H squared plus r squared is u. We do that by definition. So u to the 3/2, which is equal to the antiderivative of-- we could write this as u to the minus 3/2 du. This is just kind of reverse the exponent rule. So that equals minus 2u to the minus 1/2, and we can confirm, right? If we take the derivative of this, minus 1/2 times minus 2 is 1, and then subtract 1 from here, we get minus 3/2. And then we could add plus c, but since we're eventually going to do a definite integral, the c's all cancel out. Or we could say that this is equal to-- since we made that substitution-- minus 2 over-- minus 1/2, that's the same thing as over the square root of h squared" + }, + { + "Q": "At 5:12, David says 'from 3 to 5'. Wouldn't that be -5?\n", + "A": "Yes, that s correct. Since the turtle never makes it to +5m, you can assume that whenever he says five in this video, he means -5m.", + "video_name": "GtoamALPOP0", + "timestamps": [ + 312 + ], + "3min_transcript": "Sorry. Now what happens? Turtle at some later time, four seconds, is at negative five meters. That's all the way back here. So between two seconds and four seconds, this turtle rocketed back this way. That's also awkward. Turn down the reverse booster. What a noob, ah turtle. Here we go. Made it all the way back to here. Then what does the turtle do? After that point, turtle rockets forward. Makes it back to zero at this point. And then all the way back to three meters, so this turtle rockets forward back to three meters. That's what the turtle did. That's what this graph is representing, and that's how you read it. But there's more than that in here. I told you there was a lot of information and there is. So one piece of information you can get is the displacement of the turtle. And the displacement I'm gonna represent this with a delta x. And remember the displacement is the final position. Minus the initial position. You can find the displacement between any two times here, for the total time shown on the graph. But I could've found it between zero and like four seconds. Let's just do zero to 10, the whole thing. So what's the final position? The final position would be the position the turtle has. At 10 seconds she was at three meters. At 10 seconds 'cause I read the graph right there. Minus initially, 'cause we're considering the total time, at zero seconds, the turtle was also at three, that means the total displacement was zero. And that makes sense 'cause this turtle started at three. Rocketed back to five, well actually started at three, stood there for a second or two, rocketed back to five, rocketed back to three, ended at the same place she started, no total displacement. What else can we find? We can figure out the total distance. For the total distance traveled remember distance is the sum of all the path links traveled. So for this first path link, there was no distance traveled. That was the awkward part. We're not gonna talk about that. Then, so this is zero meters, plus between two seconds and four seconds, the turtle went from three to five. That's a distance traveled of eight meters. And should we make that negative? Nope. Distance is always positive. We make all these path links positive, we round them all up. So eight meters. Because the turtle went from three all the way back to five. That's the total distance of eight meters traveled. Plus between four seconds and 10 seconds, the turtle made it from negative five meters all the way back to three meters. That means she traveled another eight meters. That means the total distance traveled was 16 meters for the whole trip. Again you could have found this for two points any two points on here. Alright what else can you figure out? You can figure out the say average velocity, sometimes people represent that with a bar. Sometimes they just say the AVG." + }, + { + "Q": "\nWhy at 4:04 does the .0100 turn into acetate?", + "A": "As you use up the reactants, you are making products. So, as CH3COOH and OH are being used up( hence the (-) sign), CH3COO is being made ( hence the (+) sign).", + "video_name": "WbDL7xN-Pn0", + "timestamps": [ + 244 + ], + "3min_transcript": "So notice we have the same number of moles of acid as we do of base. And the base is going to neutralize the acid. Let's go ahead and write the reaction. Let's write the neutralization reaction. If we have, we start with some acetic acid. And to our acidic solution we add our sodium hydroxide. So, we're adding some sodium hydroxide here. The hydroxide ions are going to take the acidic proton. So, a hydroxide ion takes this acidic proton right here. H plus and OH minus give us H2O. If you take away the acidic proton from acetic acid, you're left with acetate, CH 3 COO minus. And we're starting with .01 moles of acetic acid. So, let's color coordinate here. So, we're starting with 0.0100 moles. So, 0.0100 moles of acetic acid. And that's the same number of moles of base. 0.0100, so we have 0.0100 moles of base. Notice our mol ratio is one to one, so we have one to one here. So, all of the base is going to react. It's going to completely neutralize the acid that we originally had present. So, all of our base reacts, and we end up with zero here, and all of our acid has been completely neutralized. We lose all of this. So, we lose all of that, and so we've neutralized all of our acid too. So, this is our equivalence point for this titration. And if we're losing acetic acid, we're converting acetic acid into acetate. So, if we think about starting with zero moles of acetate, and we lose that turns into acetate. So, we have to write plus 0.0100 over here, so we're making that. So, we end up with 0.0100 moles of our acetate anion. All right, next. If we have moles of our acetate, and we have a concentration. So, if we find the total volume of our solution, we can find the concentration of acetate anions. So, let's do that next. We'll go back up here. What is the total volume now? We started with 50 mL, and we added 200 more. So, 50.0 and 200.0 give us 250.0 mL. So, that's our total volume now. And 250.0 mL would be, move our decimal place three, 0.25 liters. So, next we find our concentration of acetate. So, what is our concentration of acetate now?" + }, + { + "Q": "at 00:27, whats velosity\n", + "A": "first of all it is velocity not velosity velocity is the rate of change of the position of an object ,equivalent to a specification of its speed and direction of motion or Or it is speed in a specific direction", + "video_name": "zAx61CO5mDw", + "timestamps": [ + 27 + ], + "3min_transcript": "Let's say we have some object that's moving in a circular path Let's say this is the center of the object path, the center of the circle So the object is moving in a circular path that looks something like that counterclockwise circular path--you could do that with clockwise as well I want to think about how fast it is spinning or orbiting around this center how that relates to its velocity? So let's say that this thing right over here is making five revolutions every second So in 1 second, 1 2 3 4 5. Every second it's making 5 revolutions So how could we relate that to how many radians it is doing per second? Remember radians is just one way to measure angles You could do with how degrees per second If we do it with radians, we know that each revolution is 2 pi radians which is really just you say you've gone 2 pi radii, whatever the radius of the circle is and that's where actually the definition of the radian comes from So if you go 5 revolutions per second and they're 2 pi per revolution then you can do a little bit of dimensional analysis. These cancel out and you get 5 times 2 pi which gets us to 5 times 2 pi gets us 10 pi radians per second And it works out the dimensional analysis and hopefully it also makes sense to you intuitively If you're doing five revolution a second, each of those revolutions is 2 pi radians so you're doing 10 pi radians per second. You're going 1 2 3 4 5, so that gives us 10, or 2 pi 2 pi 2 pi 2 pi 2 pi radians So this right here, either five revs per second or 10 pi radians per second they're both essentially measuring the same thing how fast are you orbiting around this central point? And this measure of how fast you're orbiting around a central point is called angular velocity It's called angular velocity because if you think about it this is telling us how fast is our angle changing, or speed of angle changing When you're dealing with it in two dimensions and this is typically when in a recent early physics course how we do deal with it Even though it's called the angular velocity it tends to be treated as angular speed It actually is a vector quantity and it's a little unintuitive that the vector's actually popping out of the page for this. It's actually a pseudo-vector and we'll talk more about that in the future So it is a vector quantity and the direction of the vector" + }, + { + "Q": "\nIs angular velocity a pseudo scalar quantity or a pseudo vector quantity? He mentions it as a (pseudo) scalar at 3:47 but as a pseudo vector at 2:57. Which one is correct?", + "A": "Angular velocity is a vector. Sal mentions at 2:57 that angular velocity is a vector quantity and at the 3:47 he says it is often treated as a scaler not that it is a scaler.", + "video_name": "zAx61CO5mDw", + "timestamps": [ + 227, + 177 + ], + "3min_transcript": "So this right here, either five revs per second or 10 pi radians per second they're both essentially measuring the same thing how fast are you orbiting around this central point? And this measure of how fast you're orbiting around a central point is called angular velocity It's called angular velocity because if you think about it this is telling us how fast is our angle changing, or speed of angle changing When you're dealing with it in two dimensions and this is typically when in a recent early physics course how we do deal with it Even though it's called the angular velocity it tends to be treated as angular speed It actually is a vector quantity and it's a little unintuitive that the vector's actually popping out of the page for this. It's actually a pseudo-vector and we'll talk more about that in the future So it is a vector quantity and the direction of the vector when it's spinning in a counterclockwise direction there is a vector, the real angular vector does pop out of the page We start thinking about operating in three dimensions And if it's going clockwise, the angular velocity vector would pop into the page The way you think about that, right-hand rule Curl your fingers of your right hand in the direction that it's spinning and then your thumb is essentially pointing in the direction that the actual vector or the pseudo-vector's gonna going We'll not think too much about that For our purposes, when we're just thinking about two-dimensional plane right over here we can really think of an angular velocity as a--the official term is a pseudo-scaler but we can include that as a scaler quantity, as long as we do specify which way it is rotating So this right over here, this 10 pi radians per second we could call this its angular velocity a lower case omega right there Upper case omega looks like this So there's a couple of ways you could think about it You could say angular velocity is equal to change in angle over a change in time So for example, this is telling us 10 pi radians per second Or if you want to do in the calculus sense and take instantaneous angular velocity it would be the derivative of your angle with respect to time How the angle is changing with respect to time With that out of the way, I want to see if we can see how this relates to speed How does this relate to the actual speed of the object? So to get the speed of the object, we just have to think about how far is this object traveling" + }, + { + "Q": "\nAt 2:04 you were talking about net force. I do not know what the net force is. So please tell me.", + "A": "Net force is a resultant force or the vector sum of all the forces onto the body.", + "video_name": "1E3Z_R5AHdg", + "timestamps": [ + 124 + ], + "3min_transcript": "What I want to do in this video is think about the two different ways of interpreting lowercase g. Which as we've talked about before, many textbooks will give you as either 9.81 meters per second squared downward or towards the Earth's center. Or sometimes it's given with a negative quantity that signifies the direction, which is essentially downwards, negative 9.81 meters per second squared. And probably the most typical way to interpret this value, as the acceleration due to gravity near Earth's surface for an object in free fall. And this is what we're going to focus on this video. And the reason why I'm stressing this last part is because we know of many objects that are not in free fall. For example, I am near the surface of the Earth right now, and I am not in free fall. What's happening to me right now is I'm sitting in a chair. And so this is my chair-- draw a little stick drawing on my chair, and this is me. And let's just say that the chair is supporting all my weight. So I have-- my legs are flying in the air. So this is me. And so what's happening right now? If I were in free fall, I would be accelerating towards the center of the Earth at 9.81 meters per second squared. But what's happening is, all of the force due to gravity is being completely offset by the normal force from the surface of the chair onto my pants, and so this is normal force. And now I'll make them both as vectors. is equal to 0, especially in this vertical direction. And because the net force is equal to 0, I am not accelerating towards the center of the Earth. I am not in free fall. And because this 9.81 meters per second squared still seems relevant to my situation-- I'll talk about that in a second. But I'm not an object in free fall. Another way to interpret this is not as the acceleration due to gravity near Earth's surface for an object in free fall, although it is that-- a maybe more general way to interpret this is the gravitational-- or Earth's gravitational field. Or it's really the average acceleration, or the average, because it actually changes slightly throughout the surface of the Earth. But another way to view this, as the average gravitational field at Earth's surface. Let me write it that way in pink. So the average gravitational field-- and we'll" + }, + { + "Q": "At 1:56,what is the normal force?\n", + "A": "The normal force is the force from the surface perpendicular to the object. In this case, it is the force of the chair pushing up on the person.", + "video_name": "1E3Z_R5AHdg", + "timestamps": [ + 116 + ], + "3min_transcript": "What I want to do in this video is think about the two different ways of interpreting lowercase g. Which as we've talked about before, many textbooks will give you as either 9.81 meters per second squared downward or towards the Earth's center. Or sometimes it's given with a negative quantity that signifies the direction, which is essentially downwards, negative 9.81 meters per second squared. And probably the most typical way to interpret this value, as the acceleration due to gravity near Earth's surface for an object in free fall. And this is what we're going to focus on this video. And the reason why I'm stressing this last part is because we know of many objects that are not in free fall. For example, I am near the surface of the Earth right now, and I am not in free fall. What's happening to me right now is I'm sitting in a chair. And so this is my chair-- draw a little stick drawing on my chair, and this is me. And let's just say that the chair is supporting all my weight. So I have-- my legs are flying in the air. So this is me. And so what's happening right now? If I were in free fall, I would be accelerating towards the center of the Earth at 9.81 meters per second squared. But what's happening is, all of the force due to gravity is being completely offset by the normal force from the surface of the chair onto my pants, and so this is normal force. And now I'll make them both as vectors. is equal to 0, especially in this vertical direction. And because the net force is equal to 0, I am not accelerating towards the center of the Earth. I am not in free fall. And because this 9.81 meters per second squared still seems relevant to my situation-- I'll talk about that in a second. But I'm not an object in free fall. Another way to interpret this is not as the acceleration due to gravity near Earth's surface for an object in free fall, although it is that-- a maybe more general way to interpret this is the gravitational-- or Earth's gravitational field. Or it's really the average acceleration, or the average, because it actually changes slightly throughout the surface of the Earth. But another way to view this, as the average gravitational field at Earth's surface. Let me write it that way in pink. So the average gravitational field-- and we'll" + }, + { + "Q": "At 1:05 , if the air resistance isn't assumed negligible,is vertical velocity going to slow down,too?\n\n(sorry if grammar mistakes... not very good at English...)\n", + "A": "yes. Both vertical and horizontal velocities will be affected by air resistance", + "video_name": "sTp4cI9VyCU", + "timestamps": [ + 65 + ], + "3min_transcript": "In the last video, I told you that we would figure out the final velocity of when this thing lands. So let's do that. I forgot to do it in the last video. So let's figure out the final velocity-- the vertical and the horizontal components of that final velocity. And then we can reconstruct the total final velocity. So the horizontal component is easy, because we already know that the horizontal component of its velocity is this value right over here, which we-- this 30 cosine of 80 degrees. And that's not going to change at any point in time. So this is going to be the horizontal component of the projectile's velocity when it lands. But what we need to do is figure out the vertical component of its velocity. Well, one thing we did figure out in the last video, we figured out what the time in the air is going to be. And we know a way of figuring out our final velocity from an initial velocity given our time in the air. We know that a change in velocity -- and we're only dealing in the vertical now-- we're only dealing with the vertical, We've assumed that air resistance is negligible. So we're only dealing with the vertical component right over here. We know that the change in velocity-- or, we could say the horizontal-- the vertical component of the change in velocity, is equal to the vertical component of the acceleration times time. Now, we know what the change in time is, we know it is-- I'll just write down times our time. And what is our change in velocity? Well, our change in velocity is our final vertical velocity minus our initial vertical velocity. And we know what our initial vertical velocity is, we solved for it. Our initial vertical velocity, we figured out, was 29.54 meters per second. That's 30 sine of 80 degrees, 29.54 meters per second. is equal to-- our acceleration in the vertical direction is negative, because it's accelerating us downwards, negative 9.8 meters per second squared. And our time in the air is 5.67 seconds. Times 5.67 seconds. And so we can solve for the vertical component of our final velocity. So once again, this is the vertical component. This isn't the total one. So, the vertical component. Let me-- well I wrote vertical up here. So there's the vertical component. So let's solve for this. So if you add 29.54 to both sides, you get the vertical component of your final velocity. Well, this is a vertical component, I didn't mark it up here properly-- is equal to 29.54 meters per second plus 9.8 plus -- or I should say minus-- meters per second. Minus 9.8 meters per second squared, times 5.67 seconds." + }, + { + "Q": "@3:40 What don't we use velocity as a vector quantity?\n", + "A": "Sal does. It is showing the direction ( displacement ). At least this is my understanding. You can try to ask Sal.", + "video_name": "vZOk8NnjILg", + "timestamps": [ + 220 + ], + "3min_transcript": "I'm just looking at it from the object's point of view how does the velocity vector change from each of these points in time to the next? Let me get all of these in there This green one That. Copy and paste it That. I could keep going, keep drawing velocity vectors around the circle but let me do this orange one right over here Copy and paste So between this magenta time and this purple time what was the change in velocity? Well, we could look at that purely from these vectors right here That is our change in velocity So I take this vector and say in what direction was the velocity changing when this vector was going on this part of the arc It's roughly--if I just translate that vector right over here it's roughly going in that direction So that is the direction of our change in velocity This triangle is delta; delta is for change Now think about the next time period between this blue or purple period and this green period Our change in velocity would look like that So while it's traveling along this part of the arc roughly it's the change in velocity if we draw the vector starting at the object It would look something like this I'm just translating this vector right over here I'll do it one more time From this green point in time to this orange point in time and the change in velocity actually continues changing but hopefully you're going to see the pattern here So between those two points in time, this is our change in velocity And let me translate that vector right over there It would look something like that change in velocity So what do you see, if I were to keep drawing more of these change in velocity vectors you would see at this point, the change in velocity would have to be going generally in that direction At this point, the change in velocity would have to be going generally in that direction So what do you see? What's the pattern for any point along this circular curve? Well, the change in velocity first of all, is perpendicular to the direction of the velocity itself And we haven't proved it, but it at least looks like it Looks like this is perpendicular And even more interesting, it looks like it's seeking the center" + }, + { + "Q": "at 3:00 sal starts drawing change in velocity vectors. they are essentially the difference of the 2 velocity vectors, right?!\n", + "A": "Essentially, yes. The 2nd velocity vector minus the 1st velocity vector.", + "video_name": "vZOk8NnjILg", + "timestamps": [ + 180 + ], + "3min_transcript": "in particular the direction of the force would have to act on this object in order for the velocity vector to change like that? This remind ourselves if there was no force acting on this body this comes straight from Newton's 1st Law of motion then the velocity would not change neither the magnitude nor the direction of the velocity will change If there were no force acting on the subject it would just continue going on in the direction it was going it wouldn't curve; it wouldn't turn; the direction of its velocity wasn't changing Let's think about what the direction of that force would have to be and to do that, I'm gonna copy and paste these velocity vectors and keep track of what the direction of the change in velocity has to be Copy and paste that So that is our first velocity vector Copy all of these. This is our second one right over here I'm just looking at it from the object's point of view how does the velocity vector change from each of these points in time to the next? Let me get all of these in there This green one That. Copy and paste it That. I could keep going, keep drawing velocity vectors around the circle but let me do this orange one right over here Copy and paste So between this magenta time and this purple time what was the change in velocity? Well, we could look at that purely from these vectors right here That is our change in velocity So I take this vector and say in what direction was the velocity changing when this vector was going on this part of the arc It's roughly--if I just translate that vector right over here it's roughly going in that direction So that is the direction of our change in velocity This triangle is delta; delta is for change Now think about the next time period between this blue or purple period and this green period Our change in velocity would look like that So while it's traveling along this part of the arc roughly it's the change in velocity if we draw the vector starting at the object It would look something like this I'm just translating this vector right over here I'll do it one more time From this green point in time to this orange point in time" + }, + { + "Q": "\n@3:04 when the tail of the two vectors are joined together, shouldn't the resultant vector be the diagnol (parallelogram law of addition) ?", + "A": "Sal is calculating change in velocity, not overall velocity. If it was overall velocity, then yes, you would draw the head of one of the arrows adjoining the tail of the other, and calculate the diagonal.", + "video_name": "vZOk8NnjILg", + "timestamps": [ + 184 + ], + "3min_transcript": "in particular the direction of the force would have to act on this object in order for the velocity vector to change like that? This remind ourselves if there was no force acting on this body this comes straight from Newton's 1st Law of motion then the velocity would not change neither the magnitude nor the direction of the velocity will change If there were no force acting on the subject it would just continue going on in the direction it was going it wouldn't curve; it wouldn't turn; the direction of its velocity wasn't changing Let's think about what the direction of that force would have to be and to do that, I'm gonna copy and paste these velocity vectors and keep track of what the direction of the change in velocity has to be Copy and paste that So that is our first velocity vector Copy all of these. This is our second one right over here I'm just looking at it from the object's point of view how does the velocity vector change from each of these points in time to the next? Let me get all of these in there This green one That. Copy and paste it That. I could keep going, keep drawing velocity vectors around the circle but let me do this orange one right over here Copy and paste So between this magenta time and this purple time what was the change in velocity? Well, we could look at that purely from these vectors right here That is our change in velocity So I take this vector and say in what direction was the velocity changing when this vector was going on this part of the arc It's roughly--if I just translate that vector right over here it's roughly going in that direction So that is the direction of our change in velocity This triangle is delta; delta is for change Now think about the next time period between this blue or purple period and this green period Our change in velocity would look like that So while it's traveling along this part of the arc roughly it's the change in velocity if we draw the vector starting at the object It would look something like this I'm just translating this vector right over here I'll do it one more time From this green point in time to this orange point in time" + }, + { + "Q": "\nat 5:00 why did you preform SN1 even though the nucleophile carries a negative charge and it is supposed to be a strong nucleophile so it must go SN2 ??", + "A": "The answer is: steric hindrance. There isn\u00e2\u0080\u0099t enough room for the nucleophile to attack the back side of the carbon bearing the leaving group.", + "video_name": "sDZDgctzbkI", + "timestamps": [ + 300 + ], + "3min_transcript": "when the nucleophile attacked from the left, and on the right I'm holding when the nucleophile attacks from the right. So what's the relationship between these two? Well they're mirror images of each other, but if I try to superimpose one on top of the other, you can see I can't do it. So these are non-superimposable mirror images of each other. These are enantiomers. Now let's look at our products from a different viewpoint. So I'm going to take the product on the left, and I'm going to turn it so that the S-H is coming out at me in space. So here we can see the S-H coming out at us in space, the methyl group going away from us, the ethyl group on the right, and the propyl group on the left. So now let's look at our other product, and this time if we're going to keep the same carbon chain, the methyl group's coming out at us in space, the S-H is going away from us, the ethyl group is on the right, and the propyl group is still on the left. Here are the two products that we got from the video, let's go back to the drawings over here and pretend like we don't have a model set. We know that our nucleophile attacks our electrophile and a bond forms between the sulfur and that carbon. So if I draw in my carbon chain here I know a bond formed between the sulfur and the carbon. Let me highlight the electrons. So let's say a lone pair of electrons in magenta on the sulfur form this bond, and here's our product. But if you look at our product, notice that this carbon is a chiral center. There are four different groups attached to that carbon. So if you think about the stereochemistry of this mechanism, with the nucleophile approaching the electrophile from either side of that plane, you should get a mixture of enantiomers as your product. So if I draw in my carbon chain here, I could represent one enantiomer by putting the S-H on a wedge, so let me just draw that in here. So here's our S-H on a wedge, which means the methyl group must be going away from us in space. I would have to show the S-H going away from us in space, which means the methyl group is coming out at us. And notice that these two products match the model sets that we drew here. Since there's an equal likelihood that the nucleophile could attack from one side or the other, we would expect to see an equal mixture of our products. So I'm going to say here approximately 50% is this enantiomer, and approximately 50% of our product is this enantiomer. Finally let's go through the hybridization states of this carbon in red one more time. So for our starting alkyl halide, the carbon in red is tetrahedral, right? It's sp3 hybridized so it has tetrahedral geometry. When we formed our carbocation, the carbon in red is now sp2 hybridized, so it has planar geometry. But for our products we're back to an sp3 hybridized carbon" + }, + { + "Q": "\nAt 1:40, where is the blood pressure taken?", + "A": "on the upper arm where the brachial artery is. this is because the artery is close to the skin here and it is more convenient than taking it from the neck or the leg", + "video_name": "J97G6BeYW0I", + "timestamps": [ + 100 + ], + "3min_transcript": "So recently I went into my doctor's office, and I was told that my blood pressure was 115/75. So I thought we would talk about exactly what this means and try to figure out how to think about blood pressure in general, using these numbers and this experience as kind of a launching point. So the way I think about blood pressure is I always imagine kind of a tube and I imagine blood going through that tube. And this tube is like a blood vessel. So there's blood trying to get its way from one side to the other. And on its way, the neat thing that it's doing is as it flows, it's pushing out. So it's forcing against these walls, and specifically what I mean by that is there are cells and there's plasma, and all that stuff is pushing out against the walls of the blood vessels. So you've got a force, and that force So it's force over a surface area. And any time you see force over an area, you know that equals a pressure. And in this case, it's a blood pressure because it's the blood that's doing that work. So this is how I think of blood pressure, specifically as those little blue arrows. And the two questions that kind of pop into my mind anytime I'm thinking about blood pressure are where is the blood pressure being taken, and when are you taking it? So let's start with the first question, where? And by that I mean where in the circulatory system. So you've got the heart-- and this is my Valentine's Day heart-- and you've got the aorta coming off of the heart. And it's got lots of branches, but I'm going to just draw one branch, which is the artery leading off to my arms. This is called the brachial artery going off to my arm. maybe even more, the blood pressure that you're getting recorded, or the number that's being told to you, is being checked at this point. I marked it with a little x because that's kind of the upper arm. So that's usually where they're checking the blood pressure. And again, they're checking the force that the blood is putting on the vessel walls. So these little blue arrows. So that answers the where question. And certainly, you can imagine if I checked blood pressure let's say at some other spot, let's say over here or over here, you might get a different blood pressure reading than if I checked it at the yellow x. So really were just talking about that one spot. Now the other question is when are you checking it? So for this, let me show you a little table or a figure, So imagine that over time, time is this way, you have different recordings for blood pressure. So this will be blood pressure." + }, + { + "Q": "\nAt 02:06, why blood pressure is measured on brachial artery", + "A": "It s easy to get to, in most people it s a very reliable measurement, and most people don t mind having the cuff around their arm during the measurement process.", + "video_name": "J97G6BeYW0I", + "timestamps": [ + 126 + ], + "3min_transcript": "So recently I went into my doctor's office, and I was told that my blood pressure was 115/75. So I thought we would talk about exactly what this means and try to figure out how to think about blood pressure in general, using these numbers and this experience as kind of a launching point. So the way I think about blood pressure is I always imagine kind of a tube and I imagine blood going through that tube. And this tube is like a blood vessel. So there's blood trying to get its way from one side to the other. And on its way, the neat thing that it's doing is as it flows, it's pushing out. So it's forcing against these walls, and specifically what I mean by that is there are cells and there's plasma, and all that stuff is pushing out against the walls of the blood vessels. So you've got a force, and that force So it's force over a surface area. And any time you see force over an area, you know that equals a pressure. And in this case, it's a blood pressure because it's the blood that's doing that work. So this is how I think of blood pressure, specifically as those little blue arrows. And the two questions that kind of pop into my mind anytime I'm thinking about blood pressure are where is the blood pressure being taken, and when are you taking it? So let's start with the first question, where? And by that I mean where in the circulatory system. So you've got the heart-- and this is my Valentine's Day heart-- and you've got the aorta coming off of the heart. And it's got lots of branches, but I'm going to just draw one branch, which is the artery leading off to my arms. This is called the brachial artery going off to my arm. maybe even more, the blood pressure that you're getting recorded, or the number that's being told to you, is being checked at this point. I marked it with a little x because that's kind of the upper arm. So that's usually where they're checking the blood pressure. And again, they're checking the force that the blood is putting on the vessel walls. So these little blue arrows. So that answers the where question. And certainly, you can imagine if I checked blood pressure let's say at some other spot, let's say over here or over here, you might get a different blood pressure reading than if I checked it at the yellow x. So really were just talking about that one spot. Now the other question is when are you checking it? So for this, let me show you a little table or a figure, So imagine that over time, time is this way, you have different recordings for blood pressure. So this will be blood pressure." + }, + { + "Q": "At 8:24, why is there a layer of non-fusing helium? isn't all helium supposed to fuse and form carbon and oxygen? Help!\n", + "A": "That layer isn t at a temperature/pressure where it can fuse.", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 504 + ], + "3min_transcript": "But anyway, the core is now experiencing helium fusion. It has a shell around it of helium that is not quite there, does not quite have the pressures and temperatures to fuse yet. So just regular helium. But then outside of that, we do have the pressures and temperatures for hydrogen to continue to fuse. So out here, you do have hydrogen fusion. And then outside over here, you just have the regular hydrogen plasma. So what just happened here? When you have helium fusion all of a sudden-- now this is, once again, providing some type of energetic outward support for the core. So it's going to counteract the ever-increasing contraction of the core as it gets more and more dense, because now we have energy going outward, energy pushing things outward. But at the same time that that is happening, is fusing into helium. So it's making this inert part of the helium core even larger and larger and denser, even larger and larger, and putting even more pressure on this inside part. And so what's actually going to happen within a few moments, I guess, especially from a cosmological point of view, this helium fusion is going to be burning super-- I shouldn't use-- igniting or fusing at a super-hot level. But it's contained due to all of this pressure. But at some point, the pressure won't be able to contain it, and the core is going to explode. But it's not going to be one of these catastrophic explosions where the star is going to be destroyed. It's just going to release a lot of energy all of a sudden into the star. And that's called a helium flash. But once that happens, all of a sudden, then now the star is going to be more stable. And I'll use that in quotes without writing it down getting to be less stable than a main sequence star. But once that happens, you now will have a slightly larger volume. So it's not being contained in as small of a tight volume. That helium flash kind of took care of that. So now you have helium fusing into carbon and oxygen. And there's all sorts of other combinations of things. Obviously, there's many elements in between helium and carbon and oxygen. But these are the ones that dominate. And then outside of that, you have helium forming. You have helium that is not fusing. And then outside of that, you have your fusing hydrogen. Over here, you have hydrogen fusing into helium. And then out here in the rest of the radius of our super-huge red giant, you just have your hydrogen plasma out here. Now what's going to happen as this star ages? Well, if we fast forward this a bunch-- and remember," + }, + { + "Q": "\nAt 3:15 is are sun a \"main sequence\" star?", + "A": "Yes, for now. In 5 billion years, it will run out of hydrogen and move away from the main sequence line and into the red giant phase.", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 195 + ], + "3min_transcript": "So core becomes more dense. And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster. Because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing, so it starts to fuse hotter. So let me write this, so the fusion, so hydrogen fuses faster. And actually, we even see this in our sun. Our sun today is brighter and hotter. It's fusing faster than it was when it was born 4.5 or 4.6 billion years ago. But eventually you're going to get to the point so that the core, you only have helium. So there's going to be some point where the entire core is all helium. And it's going to be way denser than this core over here. All of that mass over there has now been turned into helium. But most of it is now in helium, and it's going to be at a much, much smaller volume. And the whole time, the temperature is increasing, the fusion is getting faster and faster. And now there's this dense volume of helium that's not fusing. You do have, and we saw this in this video, a shell around it of hydrogen that is fusing. So this right here is hydrogen fusion going on. And then this over here is just hydrogen plasma. Now the unintuitive thing, or at least this was unintuitive to me at first, is what's going on the core is that the core is getting more and more dense. It's fusing at a faster rate. And so it's getting hotter and hotter. So the core is hotter, fusing faster, getting more and more dense. I kind of imagine it's starting to collapse. Every time it collapses, it's getting hotter and more dense. But at the same time that's happening, the star itself is getting bigger. Red giants are much, much larger than main sequence stars. But the whole time that this is getting more dense, the rest of the star is, you could kind of view it as getting less dense. And that's because this is generating so much energy that it's able to more than offset, or better offset the gravitational pull into it. So even though this is hotter, it's able to disperse the rest of the material in the sun over a larger volume. And so that volume is so big that the surface, and we saw this in the last video, the surface of the red giant is actually cooler-- let me write that a little neater-- is actually cooler than the surface of a main sequence star. This right here is hotter. And just to put things in perspective, when the sun becomes a red giant, and it will become a red giant, its diameter will be 100 times the diameter that it is today." + }, + { + "Q": "At 1:41, it is mentioned that the sun today is 'brighter and hotter' as it is fusing faster than it was when it was born 4.5 to 4.6 billion years ago. Is this an antithesis for global warming?\n", + "A": "No, the Sun warming process is much slower than global warming appears to be. The Sun will get 1% brighter in the next 100 million years.", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 101 + ], + "3min_transcript": "In the last video, we started with a star in its main sequence, like the sun. And inside the core of that star, you have hydrogen fusion going on. So that is hydrogen fusion, and then outside of the core, you just had hydrogen. You just hydrogen plasma. And when we say plasma, it's the electrons and protons of the individual atoms have been disassociated because the temperatures and pressures are so high. So they're really just kind of like this soup of electrons and protons, as opposed to proper atoms that we associate with at lower temperatures. So this is a main sequence star right over here. And we saw in the last video that this hydrogen is fusing into helium. So we start having more and more helium here. And as we have more and more helium, the core becomes more and more dense, because helium is a more massive atom. It is able to pack more mass in a smaller volume. So core becomes more dense. And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster. Because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing, so it starts to fuse hotter. So let me write this, so the fusion, so hydrogen fuses faster. And actually, we even see this in our sun. Our sun today is brighter and hotter. It's fusing faster than it was when it was born 4.5 or 4.6 billion years ago. But eventually you're going to get to the point so that the core, you only have helium. So there's going to be some point where the entire core is all helium. And it's going to be way denser than this core over here. All of that mass over there has now been turned into helium. But most of it is now in helium, and it's going to be at a much, much smaller volume. And the whole time, the temperature is increasing, the fusion is getting faster and faster. And now there's this dense volume of helium that's not fusing. You do have, and we saw this in this video, a shell around it of hydrogen that is fusing. So this right here is hydrogen fusion going on. And then this over here is just hydrogen plasma. Now the unintuitive thing, or at least this was unintuitive to me at first, is what's going on the core is that the core is getting more and more dense. It's fusing at a faster rate. And so it's getting hotter and hotter. So the core is hotter, fusing faster, getting more and more dense. I kind of imagine it's starting to collapse. Every time it collapses, it's getting hotter and more dense. But at the same time that's happening, the star itself is getting bigger." + }, + { + "Q": "At 11:58 Sal says it would take a long time for a white dwarf to turn into a black dwarf, but do we have any estimation for the time it would take? And how long does it take for a planetary nebula to disappear?\n", + "A": "If we define a black dwarf as a white dwarf cooled to 5 K, then it would take 10\u00c2\u00b9\u00e2\u0081\u00b5 years to form (which is considerable larger than the current age of the universe). Planetary nebulae, however, are very short-lived. They only last for tens of thousands of years on average.", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 718 + ], + "3min_transcript": "All of this hydrogen, all of this fusing hydrogen will run out. All of this fusion helium will run out. This is the fusing hydrogen. This is the inert helium, which will run out. It'll be used in kind of this core, being fused into the carbon and oxygen, until you get to a point where you literally just have a really hot core of carbon and oxygen. And it's super-dense. This whole time, it will be getting more and more dense as heavier and heavier elements show up in the course. So it gets denser and denser and denser. But the super dense thing will not, in the case of the sun-- and if it was a more massive star, it would get there-- but in the case of the sun, it will not get hot enough for the carbon and the oxygen to form. So it really will just be this super-dense ball of carbon and oxygen and all of the other material in the sun. Remember, it was superenergetic. The more that we progressed down this, the more energy was releasing outward, and the larger the radius of the star became, and the cooler the outside of the star became, until the outside just becomes this kind of cloud, this huge cloud of gas around what once was the star. And in the center-- so I could just draw it as this huge-- this is now way far away from the star, much even bigger than the radius or the diameter of a red giant. And all we'll have left is a mass, a superdense mass of, I would call it, inert carbon or oxygen. This is in the case of the sun. And at first, when it's hot, and it will be releasing radiation because it's so hot. We'll call this a white dwarf. This right here is called a white dwarf. And it'll cool down over many, many, many, many, many, many, many, years, until it becomes, when it's completely it'll just be this superdense ball of carbon and oxygen, at which point, we would call it a black dwarf. And these are obviously very hard to observe because they're not emitting light. And they don't have quite the mass of something like a black hole that isn't even emitting light, but you can see how it's affecting things around it. So that's what's going to happen to the sun. In the next few videos, we're going to talk about what would happen to things less massive than the sun and what would happen to things more massive can imagine the more massive. There would be so much pressure on these things, because you have so much mass around it, that these would begin to fuse into heavier and heavier elements until we get to iron." + }, + { + "Q": "\nAt 8:38 why did Sal say 'you have helium that is not fusing'- whatever helium is there outside the core is continuously fusing due to the pressure right?", + "A": "No, that s what he just explained to you. The helium outside the core may not be fusing because the pressure may not be high enough.", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 518 + ], + "3min_transcript": "is fusing into helium. So it's making this inert part of the helium core even larger and larger and denser, even larger and larger, and putting even more pressure on this inside part. And so what's actually going to happen within a few moments, I guess, especially from a cosmological point of view, this helium fusion is going to be burning super-- I shouldn't use-- igniting or fusing at a super-hot level. But it's contained due to all of this pressure. But at some point, the pressure won't be able to contain it, and the core is going to explode. But it's not going to be one of these catastrophic explosions where the star is going to be destroyed. It's just going to release a lot of energy all of a sudden into the star. And that's called a helium flash. But once that happens, all of a sudden, then now the star is going to be more stable. And I'll use that in quotes without writing it down getting to be less stable than a main sequence star. But once that happens, you now will have a slightly larger volume. So it's not being contained in as small of a tight volume. That helium flash kind of took care of that. So now you have helium fusing into carbon and oxygen. And there's all sorts of other combinations of things. Obviously, there's many elements in between helium and carbon and oxygen. But these are the ones that dominate. And then outside of that, you have helium forming. You have helium that is not fusing. And then outside of that, you have your fusing hydrogen. Over here, you have hydrogen fusing into helium. And then out here in the rest of the radius of our super-huge red giant, you just have your hydrogen plasma out here. Now what's going to happen as this star ages? Well, if we fast forward this a bunch-- and remember, and the reactions happen faster and faster, and this core is expelling more and more energy outward, the star keeps growing. And the surface gets cooler and cooler. So if we fast forward a bunch, and this is what's going to happen to something the mass of our sun, if it's more massive, then at some point, the core of carbon and oxygen that's forming can start to fuse into even heavier elements. But in the case of the sun, it will never get to that 600 million Kelvin to actually fuse the carbon and the oxygen. And so eventually you will have a core of carbon and oxygen, or mainly carbon and oxygen surrounded by fusing helium surrounded by non-fusing helium surrounded by fusing hydrogen, which is surrounded by non-fusing hydrogen, or just the hydrogen plasma of the sun. But eventually all of this fuel will run out." + }, + { + "Q": "What part(s) of a star are luminous? Does the outside gas layer of a star produce light? (Such as at 11:11).\n", + "A": "The core of the star produces light from fusion. The rest the gas just puts pressure on the core and scatter the light emitted by fusion.", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 671 + ], + "3min_transcript": "and the reactions happen faster and faster, and this core is expelling more and more energy outward, the star keeps growing. And the surface gets cooler and cooler. So if we fast forward a bunch, and this is what's going to happen to something the mass of our sun, if it's more massive, then at some point, the core of carbon and oxygen that's forming can start to fuse into even heavier elements. But in the case of the sun, it will never get to that 600 million Kelvin to actually fuse the carbon and the oxygen. And so eventually you will have a core of carbon and oxygen, or mainly carbon and oxygen surrounded by fusing helium surrounded by non-fusing helium surrounded by fusing hydrogen, which is surrounded by non-fusing hydrogen, or just the hydrogen plasma of the sun. But eventually all of this fuel will run out. All of this hydrogen, all of this fusing hydrogen will run out. All of this fusion helium will run out. This is the fusing hydrogen. This is the inert helium, which will run out. It'll be used in kind of this core, being fused into the carbon and oxygen, until you get to a point where you literally just have a really hot core of carbon and oxygen. And it's super-dense. This whole time, it will be getting more and more dense as heavier and heavier elements show up in the course. So it gets denser and denser and denser. But the super dense thing will not, in the case of the sun-- and if it was a more massive star, it would get there-- but in the case of the sun, it will not get hot enough for the carbon and the oxygen to form. So it really will just be this super-dense ball of carbon and oxygen and all of the other material in the sun. Remember, it was superenergetic. The more that we progressed down this, the more energy was releasing outward, and the larger the radius of the star became, and the cooler the outside of the star became, until the outside just becomes this kind of cloud, this huge cloud of gas around what once was the star. And in the center-- so I could just draw it as this huge-- this is now way far away from the star, much even bigger than the radius or the diameter of a red giant. And all we'll have left is a mass, a superdense mass of, I would call it, inert carbon or oxygen. This is in the case of the sun. And at first, when it's hot, and it will be releasing radiation because it's so hot. We'll call this a white dwarf. This right here is called a white dwarf. And it'll cool down over many, many, many, many, many, many, many, years, until it becomes, when it's completely" + }, + { + "Q": "at 2:30 why does the pulse turns upside down after reflection?\n", + "A": "on reflection, there is a change in phase. (If the medium is fixed... if it is free to move, then there is no phase change)", + "video_name": "gT0IqL1dyyk", + "timestamps": [ + 150 + ], + "3min_transcript": "- [Instructor] If you've got a medium and you disturb it, you can create a wave. And if you create a wave in a medium that has no boundaries, in other words, a medium that's so big, this wave basically never meets the boundary, then there's nothing really stopping you from making a wave of any wave length or frequency whatsoever. In order words, there's not really any naturally preferred wave lengths, they're all pretty much as good as any other wave length. However, if you confine this wave into a medium that has boundaries, this wave is gonna reflect when it meets the boundary and that means it's gonna overlap with itself. And when this happens, you can create something that are called standing waves. And we'll talk about what these mean in a minute but the reason we care about them, is because when standing waves happen, they select preferred wave lengths and frequencies. Only particular wave lengths and frequencies are gonna set up these standing waves and what ends up happening is that these often become dominant and that's why these standing waves are important to study. So, let's study some standing waves. Let's take a particular example. Let's say you've got a string, whoa, not that many strings. this string down at both ends. So, you're gonna prevent any motion from happening at the end of this string. This string can wiggle in the middle but it can't wiggle at the end points. And this isn't that crazy. A guitar string is basically a string fixed at both ends. Piano strings are strings fixed at both ends. So, the physics behind standing waves determines the types of notes you're gonna get on all of these instruments. And by the way, this point over here, we're basically making sure that it has no motion. So, by nailing it down, what I really mean is that there's gong to be no motion at this end point and no motion at this end point. And instead of calling those no motion points, physicists came up with a name for that. They call these nodes. So node is really just a fancy word for not moving at that point. So, for this string, there's gonna be nodes at each end. And we'll see, when you set up a standing wave, it's possible that there's nodes in the middle as well but there don't have to be. For this string though, we're making sure that there have to be nodes at each end. Well, let's say, you give the end of the string here a little pluck and you cause a disturbance, that disturbance is gonna move down the line because that's what waves do. It's gonna come over here. Once it meets a boundary, it's gonna reflect back to the left. Now, it turns out, when a string hits a boundary where it's fixed, when it hits a node, in other words, it gets flipped over. So, you might have tried this before with the hose. If you send a pulse down the line and you try to see how it reflects, it gets reflected upside down. It doesn't matter too much for our purposes, but every time it's gonna reflect, it flips its direction and it keeps bouncing. Now, let's say instead of sending in a single pulse, we send in a whole bunch of pulses, right. We send in like a simple harmonic wave. Now when this thing reflects, it's gonna reflect back on top of itself because this leading edge will get reflected upside down this way and it's gonna meet all the rest of the wave behind it and overlap with it, creating some total wave" + }, + { + "Q": "\nat 1:47 does anyone know what galaxies the circled ones are?", + "A": "I don t think anyone would bother to name all the billions of galaxys that Hubble can see. If they had names, they would be numbers.", + "video_name": "Wl4re38deh0", + "timestamps": [ + 107 + ], + "3min_transcript": "The whole point of this video is really just to look at, what in my mind is one of the coolest pictures ever taken by anything, and this was actually taken by the Hubble telescope and what they did is, they pointed the telescope at this area of our night sky and obviously the Hubble telescope, it's out in orbit, so it doesn't have to worry about all the interference from our actual atmosphere, so it gets a nice good look at things, but it's right over here, relative to the moon, and obviously the moon is moving around, but on that day, it was here, relative to the moon and they picked this location. They picked this location right over here because there weren't that many nearby stars there so it really allowed the telescope, because if there were nearby stars that light would've kind of outshone things that are behind it further away maybe perhaps galaxies. So just keep in mind everything you're going to see is in this little patch of the night sky, and I think the main point for showing the moon here. Obviously the moon is moving around so I'm not going to tell you which exact patch of sky this is. But to really give you an idea of how small of a patch of sky that really is. You really could've blocked things, but the galaxies are there, beyond that. The clusters of galaxies, and the super clusters of galaxies. So with that said out of the way, just remember everything we're talking about in this video is in this little patch of sky right over here. The whole point of this, once again like all of these videos is really to kind of just blow your mind. So this right here is what the Hubble telescope saw in that patch. Everything, everything that I'm showing you right here. I just want to be clear, some of these things are nearby stars, but most of these things are galaxies. That's a galaxy. That is a galaxy. That is a galaxy. That's a galaxy, and that's a galaxy. The reason I want to draw you this, obviously in the last video I showed you our local group, I showed you the Virgo super cluster, I showed you the kind of clusters of clusters and I even showed you the depiction of the observable universe. But what is amazing galaxy. Oh look there's another galaxy up here. That you can keep doing that forever and this is just in that little patch of sky, and this is obviously not all of the galaxies of the universe. These are just the ones that we could see. There ones that might be even further, or there definitely are ones that are further back and their light is just even more. We could probably even focus on a patch of sky like that and see that many galaxies again. You can kind of keep, keep zooming in, but I just this thing, and I encourage you I mean there's so many unbelievable images you can look up. They're all on NASA's website, a lot of these are on Wikipedia. But these images are just unbelievable. You can see a galaxy, another galaxy, another galaxy, another galaxy. I suspect some of this stuff might actually be clusters of galaxies. Galaxy, another galaxy. And remember I'm just circling these galaxies, just left and right, and you kind of lose sight of what each galaxy actually is. These galaxies have hundred of billions" + }, + { + "Q": "\nAt 0:36, she says that there is a dynamic equilibrium. I thought we were discussing chemical equilibrium. How do chemical and dynamic equilibriums relate?", + "A": "They are the same thing - the chemical equilibrium is dynamic, as per Le Chatelier s Principle.", + "video_name": "5gujU2QcGcY", + "timestamps": [ + 36 + ], + "3min_transcript": "- [Voiceover] In this video we're going to go through an example reaction that uses Le Chatelier's principle. So what we're gonna do is, we're gonna apply Le Chatelier's principle to look at various changes to this reaction when we perturb our reaction from equilibrium. Just as a reminder, what do I mean when I say, This reaction is at equilibrium.? So what that means is we have a reversible reaction. We have the forward reaction which has some rate K forward. The reverse reaction has some rate K backward and a dynamic equilibrium. These rate are equal to each other, so all of the concentrations are going to stay constant, and then what we do is we decide to see what happens when we add some carbon dioxide gas. If we add carbon dioxide gas, the concentration of carbon dioxide gas will go up, or you can think about it as a partial presh going up. that if we had a reaction at equilibrium and then we perturbed it by adding more CO2, it will shift to try to reduce the effect of that change. It will favor the reverse reaction, so if we add CO2, what happens is, we favor our reactants. Another way we can see this is by looking at the equilibrium constant for this reaction. We can write our equilibrium constant K, where this is a capital K. It's kind of confusing, but I will try to make this look like a capital K. (laughs) We can write it in two ways, we can write it in terms of the molar concentration, and if we write Kc, the expression will be the product concentration, our CO2 gas concentration. And that's it, because when we write out Kc, we write out concentrations of gases but we don't include solids. Kc is just the concentration of CO2 at equilibrium. I'm going to write an eq there just to show that's the equilibrium concentration. I said that you can also write it in terms of partial pressures, so there's our fancy capital K with a p subscript, which means that instead of concentrations, we're writing everything for gases in terms of partial pressures. We have the partial pressure of CO2 and again, that's it, because everything else is a solid, so we don't include those in our equilibrium expression. Writing these expressions out will be really helpful for our second condition. We're going to think about what happens, when you increase the volume of our container. We can rewrite the partial pressure, actually, in terms of the volume, so if you use the ideal gas law, the partial pressure of CO2 is equal to" + }, + { + "Q": "At 5:36 there is a mention of when you add calcium carbonate (solid), you don't interrupt equilibrium. However thinking about that makes me wonder... For example, you add a scoop of calcium carbonate, which has a certain volume into this 'fixed' volume, isn't it so that it will change the pressure and/or volume slightly? And by doing so change the equilibrium a bit? Hope I'm clear. Thanks\n", + "A": "That s a great question Sander Sloof! The assumption made in the video, which I unfortunately failed to state, is that changing the amount of solid did not change the total volume of the container significantly. I apologize for any confusion! Your thinking is entirely correct-if enough solid was added to change the volume of the container, that would also change gas pressures and perturb the reaction from equilibrium. Thanks for commenting!", + "video_name": "5gujU2QcGcY", + "timestamps": [ + 336 + ], + "3min_transcript": "Argon gas is an inert gas. We don't expect it to react to anything. One thing that will happen when you add the argon gas is it will increase the overall pressure of your container, so it will increase the total pressure. But that actually doesn't tell us what it does to our equilibrium. Let's look back at our equilibrium expressions Kc and Kp. We can see that the partial pressure for Kp only depends on the moles of our CO2 and our volume. Since we didn't change the moles of CO2 and we also didn't change the volume. Even though we increased P-total, the partial pressure of CO2 stayed the same. That means we didn't perturb our reaction from equilibrium and since we didn't perturb it from equilibrium, We are still at equilibrium. The concentrations will still stay the same. What happens when we add more calcium carbonate? That's our starting material and it is a solid. Our equilibrium expressions are determined by our CO2 concentrations, so adding more calcium carbonate, which is a solid, isn't actually going to perturb our reaction from equilibrium. Our reaction is still going to be at equilibrium and we will get no shift in concentrations. We're actually gonna look at one more thing. We're going to think about what happens when you add a catalyst. Let's say we want to speed up this reaction. We can envision what's going on here when we add the catalyst by using an energy diagram. If we have an energy diagram. We have energy on the Y axis between our reactants, or starting materials, with our products and I just sort of made up these relative energies. The way that I have this drawn here, we can see that our product is lower energy than our starting material and our forward rate, Kf, which is is up here, is determined by the size of this activation barrier between our starting material and our transition state. Our backward rate, Kb, is determined by the size of this energy barrier, so the difference in energy between the product and our transition state. If we add a catalyst to our reaction, we can think about it as lowering the activation energy for our reaction and that means we have a lower energy barrier for our forward reaction, so our forward reaction is gonna get faster," + }, + { + "Q": "\nat 0:22 my beloved Mr.Khan said if there is g the situation will be little bit different but from equation T = 2pi sqrt(m/k) , must not g have any effect on this case?", + "A": "Good observation :-). He said that because the figure that he made wouldn t be as simple as it is if we were dealing with gravity. There would be the weight of the object acting downwards (which would affect its position) and a whole lot other things. But eventually the formula would turn out to be the same.(T=2pi sqrt(m/k)). So I think he said that so that we don t get confused by the diagram. That s it. :-)", + "video_name": "oqBHBO8cqLI", + "timestamps": [ + 22 + ], + "3min_transcript": "And if you were covering your eyes because you didn't want to see calculus, I think you can open your eyes again. There shouldn't be any significant displays of calculus in this video. But just to review what we went over, we just said, OK if we have a spring-- and I drew it vertically this time-- but pretend like there's no gravity, or maybe pretend like we're viewing-- we're looking at the top of a table, because we don't want to look at the effect of We just want to look at a spring by itself. So this could be in deep space, or something else. But we're not thinking about gravity. But I drew it vertically just so that we can get more intuition for this curve. Well, we started off saying is if I have a spring and 0-- x equals 0 is kind of the natural resting point of the spring, if I just let this mass-- if I didn't pull on the spring at all. But I have a mass attached to the spring, and if I were to stretch the spring to point A, we said, well what happens? Well, it starts with very little velocity, but there's a restorative force, that's going to be pulling it back towards this position. So that force will accelerate the mass, accelerate the mass, accelerate the mass, until it gets right here. And then it'll have a lot of velocity here, but then it'll And then it'll decelerate, decelerate, decelerate. Its velocity will stop, and it'll come back up. And if we drew this as a function of time, this is what happens. It starts moving very slowly, accelerates. At this point, at x equals 0, it has its maximum speed. So the rate of change of velocity-- or the rate of change of position is fastest. And we can see the slope is very fast right here. And then, we start slowing down again, slowing down, until we get back to the spot of A. And then we keep going up and down, up and down, like that. And we showed that actually, the equation for the mass's position as a function of time is x of t-- and we used a little bit of differential equations to prove it. But this equation-- not that I recommend that you memorize anything-- but this is a pretty useful equation to memorize. Because you can use it to pretty much figure out anything-- about the position, or of the mass at any given anything else. Even the velocity, if you know a little bit of calculus, you can figure out the velocity at anytime, of the object. And that's pretty neat. So what can we do now? Well, let's try to figure out the period of this oscillating system. And just so you know-- I know I put the label harmonic motion on all of these-- this is simple harmonic motion. Simple harmonic motion is something that can be described by a trigonometric function like this. And it just oscillates back and forth, back and forth. And so, what we're doing is harmonic motion. And now, let's figure out what this period is. Remember we said that after T seconds, it gets back to its original position, and then after another T seconds, it gets back to its original position. Let's figure out with this T is. And that's essentially its period, right? What's the period of a function? It's how long it takes to get back to your starting point. Or how long it takes for the whole cycle to happen once." + }, + { + "Q": "\nat 9:28, Sal used the formular Qr=2*pi*r*dr*sigma (i'm sorry i can't write the sigma symbol on my laptop) to calculate the area of the ring but i don't quite get it. Can anyone show me how to understand it? Thank you very much.", + "A": "2*pi*r shows the circumference of the ring which the width can be ignored. cut the ring into infinite tiny rectangles, each rectangle has a width of dr. so the total area can be expressed as 2*pi*r*dr sigma. Hope this would help.", + "video_name": "prLfVucoxpw", + "timestamps": [ + 568 + ], + "3min_transcript": "So now let's see if we can figure out what the magnitude of the electric field is, and then we can put it back into this and we'll figure out the y-component from this point. And actually, we're not just going to figure out the electric field just from that point, we're going to figure out the electric field from a ring that's surrounding this. So let me give you a little bit of perspective or draw it with a little bit of perspective. So this is my infinite plate again. I'll draw it in yellow again since I originally drew it in yellow. This is my infinite plate. It goes in every direction. And then I have my charge floating above this plate someplace at height of h. And this point here, this could have been right here maybe, but what I'm going to do is I'm going to draw a ring that's of an equal radius around this point right here. Let's draw a ring, because all of these points are going to be the same distance from our test charge, right? They all are exactly like this one point that I drew here. You could almost view this as a cross-section of this ring that I'm drawing. So let's figure out what the y-component of the electric force from this ring is on our point charge. So to do that, we just have to figure out the area of this ring, multiply it times our charge density, and we'll have the total charge from that ring, and then we can use Coulomb's Law to figure out its force or the field at that point, and then we could use this formula, which we just figured out, to figure out the y-component. I know it's involved, but it'll all be worth it, because you'll know that we have a constant electric field. So let's do that. So first of all, Coulomb's Law tells us-- well, first of all, let's figure out the charge from this ring. So Q of the ring, it equals what? width of the ring. So let's say the circumference is 2 pi r, and let's say it's a really skinny ring. It's really skinny. It's dr. Infinitesimally skinny. So it's width is dr. So that's the area of the ring, and so what's its charge going to be? It's area times the charge density, so times sigma. That is the charge of the ring. And then what is the electric field generated by the ring at this point here where our test charge is? Well, Coulomb's Law tells us that the force generated by the ring is going to be equal to Coulomb's constant times the charge of the ring times our test charge divided by the distance squared, right? Well, what's the distance between really any point on" + }, + { + "Q": "at 9:52 sal mentioned test charge what does that mean?\n", + "A": "A test charge is a small charge that we imagine placing somewhere in a field to help us think about the direction and magnitude of the field at that point.", + "video_name": "prLfVucoxpw", + "timestamps": [ + 592 + ], + "3min_transcript": "Let's draw a ring, because all of these points are going to be the same distance from our test charge, right? They all are exactly like this one point that I drew here. You could almost view this as a cross-section of this ring that I'm drawing. So let's figure out what the y-component of the electric force from this ring is on our point charge. So to do that, we just have to figure out the area of this ring, multiply it times our charge density, and we'll have the total charge from that ring, and then we can use Coulomb's Law to figure out its force or the field at that point, and then we could use this formula, which we just figured out, to figure out the y-component. I know it's involved, but it'll all be worth it, because you'll know that we have a constant electric field. So let's do that. So first of all, Coulomb's Law tells us-- well, first of all, let's figure out the charge from this ring. So Q of the ring, it equals what? width of the ring. So let's say the circumference is 2 pi r, and let's say it's a really skinny ring. It's really skinny. It's dr. Infinitesimally skinny. So it's width is dr. So that's the area of the ring, and so what's its charge going to be? It's area times the charge density, so times sigma. That is the charge of the ring. And then what is the electric field generated by the ring at this point here where our test charge is? Well, Coulomb's Law tells us that the force generated by the ring is going to be equal to Coulomb's constant times the charge of the ring times our test charge divided by the distance squared, right? Well, what's the distance between really any point on Well, this could be one of the points on the ring and this could be another one, right? And this is like a cross-section. So the distance at any point, this distance right here, is once again by the Pythagorean theorem because this is also r. This distance is the square root of h squared plus r squared. It's the same thing as that. So it's the distance squared and that's equal to k times the charge in the ring times our test charge divided by Well, distance is the square root of h squared plus r squared, so if we square that, it just becomes h squared plus r squared. And if we want to know the electric field created by that ring, the electric field is just the force per test charge, so if we divide both sides by Q, we learned that the electric field of the ring is equal to Coulomb's constant" + }, + { + "Q": "\nAt 8:50, \"Acyl\" groups are introduced. But what is the difference between an acyl group and a carbonyl group? Don't they have the same structures?", + "A": "Great question, very tricky to answer too! A carbonyl has the structure of C=O only. An acyl has the structure of R-C=O, where R is any hydrocarbon. A carbonyl is present in a acyl group. Example: (CH3)2-C=O is has an acyl group, which is CH3-C=O. It also has a carbonyl group, C=O.", + "video_name": "OpyTJbzA7Fk", + "timestamps": [ + 530 + ], + "3min_transcript": "is this green bond. And this second green bond is this green bond. And this third green bond is that green bond. The way I've drawn it right now, each of these oxygens haven't let go of its hydrogens. And that could actually happen before or after or all at the same time. Chemistry actually is not a clean thing. But I could, if I want, I could draw the hydrogens here. I could draw the hydrogens. I could draw the hydrogens over here and then these would have a positive charge. These would have a positive charge. But then you could imagine another water molecule coming by and kind of taking one of these hydrogen protons, taking the hydrogen protons away from each of those oxygens. this molecule right over here. And remember we produced three water molecules. So that's one, two and three water molecules. And now this molecule, if you ignore the water molecules out there, this is a triglyceride. Let me write it again. Actually, let me write the slightly more technical term. Sometimes referred to as triacylglycerol. Well we know where the glycerol comes from. It has this glycerine or this glycerol backbone right over here. Now what is \"acyl\" mean? Well acyl is a functional group where you have a carbon that's part of a carbonyl group. It can be bound to a kind of an organic chain And then it's bound to something else. And so we have three acyl groups so \"triacyl\". So this right over here, this right here is an acyl group. This right over here is an acyl group. This right over here is an acyl group. Each of them, they're bound to an oxygen right over here. And actually that gives us more practice with another functional group. When you have a situation-- Let me get another color out. When you have a carbonyl group and let's say you have just an organic chain right over here and then you have an oxygen and then you have another organic chain right over there. This thing is called an ester. And you actually see three esters right over here. So you see this ester. This ester right over here." + }, + { + "Q": "\nat 7:48 why would Oxygen have a positive charge?", + "A": "2 electrons in the S orbital (-2), and 1 lone pair of electrons (-2), and 3 covalent bonds (3 x-1), give a total electron charge of -7 and oxygen has 8 protons so it ends up with a net positive charge.", + "video_name": "OpyTJbzA7Fk", + "timestamps": [ + 468 + ], + "3min_transcript": "and then just as he's leaving, this guy comes back. So there's different ways that all of these could happen but this is the general idea. And then you have it happening a third time down here. One of these lone pairs come and form a bond with this carbon. This carbon in the carbonyl group, part of this carboxyl group. And so once again, this guy can take those two electrons away and maybe share that pair with a hydrogen proton. Once again, this is forming a water molecule, this is forming a water molecule. So three water molecules are going to be produced. This is why we call it dehydration synthesis. We're losing three water molecules in order to form these bonds. So what's it going to look like after this has happened? Well, let me scroll down here. And actually let me just zoom. Actually just let me scroll down here. So this green bond over here is going to now-- is this green bond. And this second green bond is this green bond. And this third green bond is that green bond. The way I've drawn it right now, each of these oxygens haven't let go of its hydrogens. And that could actually happen before or after or all at the same time. Chemistry actually is not a clean thing. But I could, if I want, I could draw the hydrogens here. I could draw the hydrogens. I could draw the hydrogens over here and then these would have a positive charge. These would have a positive charge. But then you could imagine another water molecule coming by and kind of taking one of these hydrogen protons, taking the hydrogen protons away from each of those oxygens. this molecule right over here. And remember we produced three water molecules. So that's one, two and three water molecules. And now this molecule, if you ignore the water molecules out there, this is a triglyceride. Let me write it again. Actually, let me write the slightly more technical term. Sometimes referred to as triacylglycerol. Well we know where the glycerol comes from. It has this glycerine or this glycerol backbone right over here. Now what is \"acyl\" mean? Well acyl is a functional group where you have a carbon that's part of a carbonyl group. It can be bound to a kind of an organic chain" + }, + { + "Q": "I noticed how you said, at 3:10, \"The period does not change this way.\" This statement was referring to the horizontal period I presume. Is there a vertical period I am unaware of?\n", + "A": "The period is the time for one oscillation. There s no horizontal or vertical to it.", + "video_name": "6M_bjRzyUn0", + "timestamps": [ + 190 + ], + "3min_transcript": "Some of you might say, yes, it should increase the period because look, now it has farther to travel, right? Instead of just traveling through this amount, whoa that looked horrible, instead of just traveling through this amount back and forth, it's gotta travel through this amount back and forth. Since it has farther to travel, the period should increase. But some of you might also say, wait a minute. If we pull this mass farther, we know Hooke's law says that the force is proportional, the force from the spring, proportional to the amount that the spring is stretched. So, if I pulled this mass back farther, there's gonna be a larger force that's gonna cause this mass to have a larger velocity when it gets to you, a larger speed when it gets to the equilibrium position, so it's gonna be moving faster than it would have. So, since it moves faster, maybe it takes less time for this to go through a cycle. But it turns out those two effects offset exactly. In other words, the fact that this mass and the fact that it will now be traveling faster offset perfectly and it doesn't affect the period at all. This is kinda crazy but something you need to remember. The amplitude, changes in the amplitude do not affect the period at all. So pull this mass back a little bit, just a little bit of an amplitude, it'll oscillate with a certain period, let's say, three seconds, just to make it not abstract. And let's say we pull it back much farther. It should oscillate still with three seconds. So it has farther to travel, but it's gonna be traveling faster and the amplitude does not affect the period for a mass oscillating on a spring. This is kinda crazy, but it's true and it's important to remember. This amplitude does not affect the period. In other words, if you were to look at this on a graph, let's say you graphed this, put this thing on a graph, if we increase the amplitude, what would happen to this graph? Well, it would just stretch this way, right? We'd have a bigger amplitude, but you can do that and there would not necessarily If you leave everything else the same and all you do is change the amplitude, the period would remain the same. The period this way would not change. So, changes in amplitude do not affect the period. So, what does affect the period? I'd be like, alright, so the amplitude doesn't affect it, what does affect the period? Well, let me just give you the formula for it. So the formula for the period of a mass on a spring is the period here is gonna be equal to, this is for the period of a mass on a spring, turns out it's equal to two pi times the square root of the mass that's connected to the spring divided by the spring constant. That is the same spring constant that you have in Hooke's law, so it's that spring constant there. It's also the one you see in the energy formula for a spring, same spring constant all the way. This is the formula for the period of a mass on a spring. Now, I'm not gonna derive this because the derivations typically involve calculus. If you know some calculus and you want to see how this is derived," + }, + { + "Q": "At 1:58 why is it that although there are two sets of genetic code, the sister chromatids are still considered as only 1 chromosome?\n", + "A": "The sister chromatids are still attached to each other by a centromere. Only when separated are they considered 2 chromosomes. You usually visualize a chromosome as X-shaped, do you not? That is a chromosome that has undergone DNA replication therefore it has two sets of genetic code.", + "video_name": "TKGcfbyFXsw", + "timestamps": [ + 118 + ], + "3min_transcript": "- In the previous video, we talked about interphase which is the bulk of a cell's life cycle as it grows and its DNA replicates, and it grows some more. And now, we're gonna talk about the actual cell division. We're gonna talk about mitosis. And if you wanna be precise, mitosis is the process by which this one nucleus will turn into two nuclei that each have the original genetic information. Now, as we exit mitosis, we get into cytokineses which will then split each of the nuclei into a separate cell when we split the cytoplasm right over here. We split or the cell as it turns into to cells. Well, let's see how all of this happens. So the first phase, and I'll leave the end of interphase right over here. We have this big cell, our DNA has been replicated. We have two centrosomes right over here. The first phase of mitosis involves the cell enough space here. So, involve. So, this is the cell right over here. So we're gonna go to this phase right over here. And a few things start happening. One, the DNA, the chromosomes go from being in their chromatin form where they're all spread out to kind of a more condensed form that you can actually see from a light microscope. So for example, that magenta chromosome which is now made up of two sister chromatids after replication, we talked about that in the interphase video. It might look something like this in a ... It might look something like this if you were to look in a microscope. It's unlikely to be magenta but it's gonna have kind of that classic chromosome shape that you're used to seeing in textbooks. And it has the centromere that connects these two sister chromatids. Right now, both of these two sister chromatids combined are considered to be one chromosome. the magenta stuff was still considered to be one chromosome. And we can draw the blue chromosome. Once again, it's now in the condensed form. That's one sister chromatid right over there. That's another sister chromatid. They are connected at the centromere. So they're condensing now as we enter into mitosis. And the nuclear membrane starts to go away. So the nuclear membrane is starting to go away. And these two centrosomes are starting to migrate to opposite sides of the cell. So one of them is going over here and one of them is maybe going to go over here. So they're migrating to opposite sides of the cell. And it's pretty incredible. It's easy to say, \"Oh, this happens and then that happens.\" Remember, this cell doesn't have a brain. This is all happening through chemical and thermodynamic" + }, + { + "Q": "\nat 8:36 what is happening here how the two events are experienced at different times\nin A and B frame of reference", + "A": "Time is relative.", + "video_name": "2BVGig1LXLs", + "timestamps": [ + 516 + ], + "3min_transcript": "if you think about the time axis to the left and the right. And since they're equally skewed, the time dilation relative to this rest frame is the scaling is going to be the same. So we could put both of these on, both of these on the same scale. And you could see that this neutral frame of reference, this ct prime prime frame of reference, I drew them over here, that if you look at either of A or B's frame of reference, they're going to be in between A and B's frame of reference. But here it's the neutral one. It's the one where we're drawing the time or we're drawing the two axes being perpendicular to each other. And now if you look at these two events, so if you look at this first event, where you have our delta ct, so you look at this first event where you have a delta in ct between right when the spaceships pass and right over there. You see that if you were to look at that, if you were to look at the ct prime coordinate for that event, when you go parallel to the x prime axis, So this is, that is your change in ct prime. And likewise, if you have that other event that I drew in that blue-green color, right over here. And this is your change in ct prime. This is your change in ct prime, well then if you think of it from A's frame of reference, well we just follow the x-axis not the x prime. We go parallel to it. We end up right over there. Now what's really interesting about this, what's really interesting about this is from A's frame of reference, the yellow event happens before the blue event. But in B's frame of reference, the blue event happens before the yellow event. So really, really, really fascinating things going on but what I really like about this diagram is that A and B's frame of reference are going to have the same scale since they're both equally skewed to the left and the right if we're thinking about the time axis. And this type of diagram is called a Loedel diagram. a variation of a Minkowski diagram but it lets us really appreciate the symmetry between these frames of reference." + }, + { + "Q": "\nOK, at 2:20, why did you divide acceleration by 2?", + "A": "mm. alight thnx", + "video_name": "P7LKEkcNibo", + "timestamps": [ + 140 + ], + "3min_transcript": "Just want to follow up on the last video, where we threw balls in the air, and saw how long they stayed up And we used that to figure out how fast we initially threw the ball and how high they went in the air. And in the last video, we did it with specific numbers. In this video, I just want to see if we can derive some interesting formulas so that we can do the computations really fast in our brains, while we're playing this game out on some type of a field, and we don't necessarily have any paper around. So let's say that the ball is in the air for delta t. Delta t is equal to time in the air. Then we know that the time up is going to be half that, which is the same thing as the time down. The time up is going to be equal to delta t-- I'm going to do that in the same color-- is going to be equal to the time in the air divided by 2. So what was our initial velocity? Well, all we have to do is remind ourselves that the change in velocity, which is the same thing as the final velocity minus the initial velocity. we're just talking about half of the path of this ball. So the time that it gets released, and it's going at kind of its maximum upward velocity, and it goes slower, and slower, slower, all the way until it's stationary for just a moment, and then it starts going down again. Now remember, the acceleration is constant downwards this entire time. So what is the final velocity if we just consider half of the time? Well, it's the time. It's 0. So it's going to be 0 minus our initial velocity, when it was taking off. That's our change in velocity. This is our change in velocity, is going to be equal to the acceleration of gravity, negative 9.8 meters per second squared-- or the acceleration due to gravity when an object is in free fall, to be technically correct-- times the time that we are going up. So times delta t up, which is the same thing. Delta t up is the same thing as our total time in the air divided by 2. And so we get negative the initial velocity is equal to-- this thing, when you divide it by 2, is going to be 4.9 meters per second squared-- we still have our negative out front-- times our delta t. Remember, this is our total time in the air, not just the time up. This is our total time in the air. And then we multiply both sides times a negative. We get our initial velocity is just going to be equal to 4.9 meters per second squared times the total time that we are in the air. Or you could say it's going to be 9.8 meters per second squared times half of the time that we're in the air. Either of those would get you the same calculation." + }, + { + "Q": "I don't understand Sal's \"Oil Rig\" reference at 8:57, it is something to do with NADPH's function in cellular respiration?\n", + "A": "It refers to Redox reactions! Its a little acronym to remember the difference between oxidation and reduction. O-oxidation i-is l-lose and R-reductuion i-is g-gain. I was taught Leo the lion says Ger. Which is the same principle! (Loss electrons= oxidation (LEO) and Gain electrons=reduction (GER))", + "video_name": "-rsYk4eCKnA", + "timestamps": [ + 537 + ], + "3min_transcript": "although they do occur when the sun is out. They don't need those photons, but they need the byproducts from the light reaction to occur, so that's why it's called the light-independent reaction. They occur while the sun is out, but they don't need the sun. This needs the sun, so let me make it very clear. So this requires sunlight. This requires photons. And let me just make a very brief overview of this. This'll maybe let us start building a scaffold from which we can dig deeper. So the light reactions need photons, and then it needs water. So water goes into the light reactions and out of the other side of the light reactions. We end up with some molecular oxygen. So that's what happens in the light reactions, and I'm going to go much deeper into what actually occurs. And what the light the actions produce is ATP, which we know It produces ATP and it produces NADPH. Now, when we studied cellular respiration, we saw the molecule NADH. NADPH is very similar. You just have this P there. You just have this phosphate group there, but they really perform similar mechanisms. That this agent right here, this molecule right here, is able to give away-- now let's think about what this means-- it's able to give away this hydrogen and the electron associated with this hydrogen. So if you give away an electron to someone else or someone else gains an electron, that something else is being reduced. Let me write that down. This is a good reminder. OIL RIG. Oxidation is losing an electron. Your charge is reduced when you gain an electron. It has a negative charge. So this is a reducing agent. It gets oxidized by losing the hydrogen and the electron with it. I have a whole discussion on the biological versus chemistry view of oxidation, but it's the same idea. When I lose a hydrogen, I also lose the ability to hog that hydrogen's electron. So this right here, when it reacts with other things, it's a reducing agent. It gives away this hydrogen and the electron associated with it, and so the other thing gets reduced. So this thing is a reducing agent. And what's useful about it is when this hydrogen, and especially the electron associated with that hydrogen, goes from the NADPH to, say, another molecule and goes to a lower energy state, that energy can also be used in the" + }, + { + "Q": "At 10:35, Sal says that the dark reactions are called the Calvin Cycle, aren't light reactions called the Calvin cycle and dark reactions, the Hill reactions?\n", + "A": "The Light Reactions are called the Light Reactions and the Calvin Cycle is sometimes called the dark reactions, but it is more accurate to describe the Calvin Cycle as the light independent reactions because the Calvin Cycle occurs whether the sun is shining or not.", + "video_name": "-rsYk4eCKnA", + "timestamps": [ + 635 + ], + "3min_transcript": "Your charge is reduced when you gain an electron. It has a negative charge. So this is a reducing agent. It gets oxidized by losing the hydrogen and the electron with it. I have a whole discussion on the biological versus chemistry view of oxidation, but it's the same idea. When I lose a hydrogen, I also lose the ability to hog that hydrogen's electron. So this right here, when it reacts with other things, it's a reducing agent. It gives away this hydrogen and the electron associated with it, and so the other thing gets reduced. So this thing is a reducing agent. And what's useful about it is when this hydrogen, and especially the electron associated with that hydrogen, goes from the NADPH to, say, another molecule and goes to a lower energy state, that energy can also be used in the And we saw in cellular respiration the very similar molecule, NADH, that through the Kreb Cycle, or actually more importantly, that through the electron transport chain, was able to help produce ATP as it gave away its electrons and they went to lower energy states. But I don't want to confuse you too much. So the light reactions, you take in photons, you take in water, it spits out oxygen, and it spits out ATP and NADPH that can then be used in the dark reactions. And the dark reactions, for most plants we talk about, it's called the Calvin Cycle. And I'll go into a lot more detail of what actually occurs in the Calvin Cycle, but it takes in the ATP, the NADPH, and it produces-- it doesn't directly produce glucose. It produces-- oh, you probably saw this. You could call it PGAL. These all stand for-- let me write these down-- this is phosphoglyceraldehyde. My handwriting broke down. Or you could call it glyceraldehyde 3-phosphate. Same exact molecule. You can almost imagine it as-- this is a very gross oversimplification-- as three carbons with a phosphate group attached to it. But this can then be used to produce other carbohydrates, including glucose. If you have two of these, you can use those two to produce glucose. So let's just take a quick overview again because this is super important. I'm going to make videos on the light reactions and the" + }, + { + "Q": "at 9:20, the molecule has 3 carbons instead of 4, right? because the double bond is between two carbon molecules\n", + "A": "Yes that molecule only has three carbons", + "video_name": "8x8tA4YPhJw", + "timestamps": [ + 560 + ], + "3min_transcript": "So let me draw in some electrons here and I'll use red, so this bond is two electrons, and then we have a bunch of electrons in here, so how many electrons around oxygen do we have so far? Well, we would have three, just these three, and we need five, which means we need two more, which means we need a lone pair of electrons on the oxygen. So that's a little bit too complicated, I think, for figuring it out, but you could use that method, or you could just learn this pattern, right, and eventually, you'll have to have this pattern down pretty well. Let's look at another example for assigning formal charge to oxygen. So our goal is to find the formal charge on oxygen in this example, and we put in our electrons in this bond. Each bond represents two electrons, so the formal charge on oxygen, is equal to the number of valence electrons that oxygen is supposed to have which is six, minus the number of valence electrons and in this example, right, we would take one of these electrons from this bond, and how many electrons is that total around oxygen? This would be one, two, three, four, five, six, seven, so six minus seven gives us a formal charge of negative one. So I could re-draw that over here, I could say oxygen has three lone pairs of electrons, and a negative one formal charge, so our pattern, this time, our pattern is one bond. Here's the one bond, and then three lone pairs of electrons. So let me write that down. So the pattern is one bond, when oxygen has one bond, and three lone pairs of electrons the formal charge is negative one, just like we saw up here with the calculation. So we could leave those electrons off if you wanted to save some time, we could just say, oh, this is oxygen with a negative one formal charge, three lone pairs of electrons on that oxygen. Let's look at one more example, where formal charge is negative one. So right here, this oxygen has a negative one formal charge, and we can see it already has one bond to it. And so the pattern, of course, is one bond plus three lone pairs of electrons. So we already have the one bond. In order for that oxygen to have a negative one formal charge, we need three lone pairs of electrons. So we could re-draw this, so that is one way to represent that ion, and we could also represent it like this, with putting three lone pairs of electrons on that oxygen with a negative one formal charge. So again, become familiar with these patterns." + }, + { + "Q": "\nat 8:22, why doesnt the n\n-14 also have a half life", + "A": "Some atoms are stable, others aren t. C-14 is not, so it decays, but N is stable, so it doesn t decay.", + "video_name": "9REPnibO4IQ", + "timestamps": [ + 502 + ], + "3min_transcript": "chance that any of the guys that are carbon will turn to nitrogen. So if you go back after a half-life, half of the atoms will now be nitrogen. So now you have, after one half-life-- So let's ignore this. So we started with this. All 10 grams were carbon. 10 grams of c-14. This is after one half-life. And now we have five grams of c-14. And we have five grams of nitrogen-14. Fair enough. Let's think about what happens after another half-life. Well we said that during a half-life, 5,740 years in the case of carbon-14-- all different elements have a different half-life, if they're radioactive-- over atom-- there's a 50% chance it'll decay. So if we go to another half-life, if we go another half-life from there, I had five grams of carbon-14. So let me actually copy and paste this one. This is what I started with. Now after another half-life-- you can ignore all my little, actually let me erase some of this up here. Let me clean it up a little bit. After one one half-life, what happens? Well I now am left with five grams of carbon-14. Those five grams of carbon-14, every one of those atoms still has, over the next-- whatever that number was, 5,740 years-- after 5,740 years, all of those once again have a 50% chance. And by the law of large numbers, half of them will So we'll have even more conversion into nitrogen-14. So now half of that five grams. So now we're only left with 2.5 grams of c-14. And how much nitrogen-14? Well we have another two and a half went to nitrogen. So now we have seven and a half grams of nitrogen-14. And we could keep going further into the future, and after every half-life, 5,740 years, we will have half of the carbon that we started. But we'll always have an infinitesimal amount of carbon. But let me ask you a question. Let's say I'm just staring at one carbon atom. Let's say I just have this one carbon atom. You know, I've got its nucleus, with its c-14. So it's got its six protons. 1, 2, 3, 4, 5, 6. It's got its eight neutrons. It's got its six electrons. 1, 2, 3, 4, 5, 6, whatever." + }, + { + "Q": "At about 7:25, why didn't he just add T2 to the zero in the blue equation so he would get sqrt(3)T1 = T2 and plug that into any T2s as opposed to him multiplying the second equation from the bottom by the sqrt(3) and making it more complicated? I got the same answer for both. Is there a mathematical reason he didn't do that?\n", + "A": "That s just the strategy that he used to solve this calculus , your strategy is correct too.I think he did this way because it makes more easy to solve the system of equations by elimination.", + "video_name": "zwDJ1wVr7Is", + "timestamps": [ + 445 + ], + "3min_transcript": "to be 10 Newtons. Because it's offsetting this force of gravity. So what's this y component? Well, this was T1 of cosine of 30. This should start to become a little second nature to you that this is T1 sine of 30, this y component right here. So T1-- Let me write it here. T1 sine of 30 degrees plus this vector, which is T2 sine of 60 degrees. You could review your trigonometry and your SOH-CAH-TOA. Frankly, I think, just seeing what people get confused on is the trigonometry. But you can review the trig modules and maybe some of the earlier force vector modules that we did. And hopefully, these will make sense. I'm skipping a few steps. And these will equal 10 Newtons. And let's rewrite this up here where I substitute the values. Actually, let me do it right here. What's the sine of 30 degrees? The sine of 30 degrees is 1/2 so we get 1/2 T1 plus the sine of 60 degrees, which is square root of 3 over 2. Square root of 3 over 2 T2 is equal to 10. And then I don't like this, all these 2's and this 1/2 here. So let's multiply this whole equation by 2. So 2 times 1/2, that's 1. So you get T1 plus the square root of 3 T2 is equal to, 2 times 10 , is 20. Similarly, let's take this equation up here and let's multiply this equation by 2 and bring it down here. So this is the original one that we got. So if we multiply this whole thing by 2-- I'll do it in this color so that you know that it's a different equation. So if you multiply square root of 3 over 2 times 2-- I'm just doing this to get rid of the 2's in the denominator. So you get square root of 3 T1 minus T2 is equal to 0 because And let's see what we could do. What if we take this top equation because we want to start canceling out some terms. Let's take this top equation and let's multiply it by-- oh, I don't know. Let's multiply it by the square root of 3. So you get the square root of 3 T1. I'm taking this top equation multiplied by the square root of 3. This is just a system of equations that I'm solving for. And the square root of 3 times this right here. Square root of 3 times square root of 3 is 3. So plus 3 T2 is equal to 20 square root of 3. And now what I want to do is let's-- I know I'm doing a lot of equation manipulation here." + }, + { + "Q": "\n7:25 \u00c2\u00bfand according with that graph how many electrons will lose the element with tahat energy 1 or all the shell?", + "A": "That graph is for the first ionisation energy only, so the energy required for an atom to lose one electron only.", + "video_name": "5CBs36jtZxY", + "timestamps": [ + 445 + ], + "3min_transcript": "they don't want their electron configurations messed with. So, it would be very hard... Neon on down has their eight electrons that (mumbling) Octet Rule. Helium has two which is full for the first shell, and so it's very hard to remove an electron from here, and so it has a very high ionization energy. Low energy, easy to remove electrons. Or especially the first electron, and then here you have a high ionization energy. I know you have trouble seeing that H. So, this is high, high ionization energy, and that's the general trend across the periodic table. As you go from left to right, you go from low ionization energy to high ionization energy. Now, what about trends up and down the periodic table? Well, within any group, if we, even if we look at the Alkali, if we look at the Alkali Metals right over here, if we're down at the bottom, if we're looking at, if we're looking at, say, Cesium right over here, that electron in the, one, two, three, four, five, six, further from that one electron that Lithium has and its second shell. So, it's going to be, it's going to be further away. It's not going to be as closely bound to the nucleus, I guess you could say. So, this is going to be even, that one electron's gonna even easier to remove than the one electron in the outermost shell of Lithium. So, this one has even lower, even lower, even lower... And that's even going to be true of the Noble Gases out here that Xenon, that it's electrons in its outermost shell, even though it has eight valence electrons, they're further away from the nucleus, and so they're a little, the energy required to remove them is still going to be high but it's going to be lower than the energy from, from say Neon or Helium. So, this is low. So, once again, ionization energy low to high as we go from left to right, and low to high as we go from bottom to top. that if we go from the bottom left to the top right we go from low ionization energy, very easy to remove an electron from these characters right over here to high ionization energy, very hard to move, remove an electron from these characters over here. And you can see it if, you could see in a trend of actual measured ionization energies and I like to see charts like this because it kind of show you where the periodic table came from when people noticed these kind of periodic trends. It's like, hey, it looks like there's some common patterns here. But on this one in particular we see on this axis we have ionization energy and electron volts, that's actually, it's literally a, this is units of energy. You could convert it to Joules if you like. Then over here, we're increasing the atomic numbers. So, we're (mumbling), we're starting with Hydrogen then we go to Helium, and we keep, and then we go, go from Hydrogen to Helium to Lithium and let me show you what's happening right over here." + }, + { + "Q": "At 2:26, how do you know what to dram for the ion size?\n", + "A": "These are just schematics (pictures not intended to realistically depict something or indicate actual values)!", + "video_name": "HBi8xjMchZc", + "timestamps": [ + 146 + ], + "3min_transcript": "In the video on solubility, I draw little pictures of sodium and chloride ions when sodium chloride dissolves or disassociates into water. This is sodium and this is chloride. And my simple brain, when I looked at it, I said, OK, how should I draw these things? I said, well, they're in the same period, and sodium is a Group 1 element. It's an alkali metal, while chlorine is a halogen, so chlorine's going to have a smaller atomic radius. And the logic there, just to review from the atomic table trends, is that both of their valence electrons are in the third shell. Sodium only has 11 protons pulling in the center. It has 11 in the center, and it has only one electron out there in its valence shell. So the attraction isn't as strong as the case of chlorine, which has 17 protons in the center. Although it has more valence electrons -- it has 7 of them-- these protons are going to have a stronger attraction on them. you'd expect the sodium neutral atom to be bigger than the chlorine neutral atom. Because this guy has more protons pulling everything in. And that's how I drew the ions in that video. I said, oh, when I disassociate in water, I'll have a big sodium ion and a smaller chlorine ion, which is incorrect. Because think about -- and this was pointed out to me by one of the viewers, and they're correct, and I should have realized it. What happens when you ionize these things? This guy will lose an electron, right? He gives the electron to this guy. So his electron configuration is actually going to look a lot more like neon. He now will have no electrons in that third shell, in the third energy state. So now he's going to have an atomic radius that's actually much more similar to neon here, right? Because he's going to have filled up the second shelf. So actually, the sodium ion, this is completely incorrect. The sodium ion is going to have Actually, it will be even a little smaller than neon because it has the same electron configuration, but it has one more proton. So the sodium ion is actually going to be smaller. Because it gets rid of the electron in that third shell, and the chlorine cation, gained an electron, so it has completely completed its third shell. So here you have where the chlorine ion is going to be bigger. So in that solubility video, I should've actually switched the places between the sodium and the chlorine, at least in size-wise. And, of course, i showed how they disassociate in water, and this would be attracted to the oxygen end of the water, and you have the hydrogen end and all that. But you can watch the solubility video for that. It doesn't change the real takeaway from the video. But I think this is a really interesting point that it brings up, that when you ionize these neutral atoms, it can significantly change, especially significantly change their relative atomic sizes. Anyway, hopefully, you found that interesting." + }, + { + "Q": "\nSal says that fog is an example of a colloid. And he says at 5:38 that fog is water molecules(H2O) that are inside air. Isn't H2O's molecular mass much less than 2nm. I think it has to be because Cs is 0.26 nm. If the H2O molecule is less than 2nm, shouldn't it also be a solution ?", + "A": "Eventhough, the size of H2O particles is less than 2nm, fog is a colloid as it exhibits tyndall effect(scattering of light), a distinguishing traits among colloids.", + "video_name": "3ROWXs3jtQU", + "timestamps": [ + 338 + ], + "3min_transcript": "where we're at 2 to 500 nanometers, we're dealing with a colloid. That word, I remember in seventh grade, I think you learned it in science class: the colloid. And a friend and I, we thought it was a more appropriate word for some type of gastrointestinal problem. But it's not a gastrointestinal problem. It's a type of homogeneous mixture. And it's a homogeneous mixture where the particles are small enough that they stay suspended. So maybe they could call it a better suspension or a permanent suspension. So here the molecules are -- so let's say that's my mixture. So water, maybe it's water. It doesn't have to be water. It could be air or whatever. Now the molecules are small enough that they stay suspended. So the forces, either their buoyancy or the force -- actually, more important, the forces between the particles and the intermolecular forces kind of outweigh to exit the solution in either direction. And so common examples of these -- well, the one I alway think of, for me, the colloid is Jell-O. but gelatin is a colloid. but gelatin is a colloid. The gelatin molecules stay suspended in the -- the gelatin powder stays suspended in the water that you add to it, and you can leave it in the fridge forever and it just won't ever deposit out of it. Other examples, fog. Fog, you have water molecules inside of an air mixture. And then you have smoke. Fog and smoke, these are examples of aerosols. This is an aerosol where you have a liquid in the air. This is an aerosol where you have a solid in the air. Smoke just comes from little dark particles and they'll never come out of the air. They're small enough that they'll always just float around with the air. Now, if you get below 2 nanometers -- maybe I should eliminate my homogenized milk. If you get below 2 nanometers -- I'm trying to draw in black. If you're less than 2 nanometers, you're now in the realm of the solution. And although this is very interesting in the everyday world, a lot of things that we-- and this is a fun thing to think about in your house, or when you encounter things, is this a suspension? Well, first, you should just think is it homogeneous? And then think is it a suspension? Is it eventually going to not be in the state it's in and then I'll have to shake it? Is it a colloid where it will stay in this kind of nice, thick state in the case of Jell-O or fog or smoke where it will really just stay in the state Or is it a solution? And solution is probably the most important in chemistry." + }, + { + "Q": "\nAt around 4:45 Sal draws a cup with liquid in it and says \"this could be air, whatever...\" . Funny cause AIR isn't liquid.", + "A": "Gases are quite able to make solutions.", + "video_name": "3ROWXs3jtQU", + "timestamps": [ + 285 + ], + "3min_transcript": "you've got to make sure that the can is well shaken. Otherwise, you're going to get an inconsistent coat. The other, that's close to my heart, is chocolate milk. Because when you mix it up, it's nice and it seems homogeneous, right? It's nice. And I already have milk here. So right at first when you stir it nice, you have all the little chocolate clumps in there, at least the chocolate when I make it is like that. But then if you let it sit around for a long time, eventually all the chocolate is going to collect at the bottom of the glass. Actually, different parts of it. I've seen situations where the sugar all collects at the bottom and then you have these little clumps at the top. But you get the idea, that the mixture separates. And that's because the particles in either the paint or the chocolate milk are greater than 500 nanometers. Now, if we get to a range that's a little bit smaller than that, where we're at 2 to 500 nanometers, we're dealing with a colloid. That word, I remember in seventh grade, I think you learned it in science class: the colloid. And a friend and I, we thought it was a more appropriate word for some type of gastrointestinal problem. But it's not a gastrointestinal problem. It's a type of homogeneous mixture. And it's a homogeneous mixture where the particles are small enough that they stay suspended. So maybe they could call it a better suspension or a permanent suspension. So here the molecules are -- so let's say that's my mixture. So water, maybe it's water. It doesn't have to be water. It could be air or whatever. Now the molecules are small enough that they stay suspended. So the forces, either their buoyancy or the force -- actually, more important, the forces between the particles and the intermolecular forces kind of outweigh to exit the solution in either direction. And so common examples of these -- well, the one I alway think of, for me, the colloid is Jell-O. but gelatin is a colloid. but gelatin is a colloid. The gelatin molecules stay suspended in the -- the gelatin powder stays suspended in the water that you add to it, and you can leave it in the fridge forever and it just won't ever deposit out of it. Other examples, fog. Fog, you have water molecules inside of an air mixture. And then you have smoke. Fog and smoke, these are examples of aerosols. This is an aerosol where you have a liquid in the air. This is an aerosol where you have a solid in the air. Smoke just comes from little dark particles" + }, + { + "Q": "at 7:20 what does aq stand for?\n", + "A": "aq is the state description for an aqueous solution.", + "video_name": "3ROWXs3jtQU", + "timestamps": [ + 440 + ], + "3min_transcript": "to exit the solution in either direction. And so common examples of these -- well, the one I alway think of, for me, the colloid is Jell-O. but gelatin is a colloid. but gelatin is a colloid. The gelatin molecules stay suspended in the -- the gelatin powder stays suspended in the water that you add to it, and you can leave it in the fridge forever and it just won't ever deposit out of it. Other examples, fog. Fog, you have water molecules inside of an air mixture. And then you have smoke. Fog and smoke, these are examples of aerosols. This is an aerosol where you have a liquid in the air. This is an aerosol where you have a solid in the air. Smoke just comes from little dark particles and they'll never come out of the air. They're small enough that they'll always just float around with the air. Now, if you get below 2 nanometers -- maybe I should eliminate my homogenized milk. If you get below 2 nanometers -- I'm trying to draw in black. If you're less than 2 nanometers, you're now in the realm of the solution. And although this is very interesting in the everyday world, a lot of things that we-- and this is a fun thing to think about in your house, or when you encounter things, is this a suspension? Well, first, you should just think is it homogeneous? And then think is it a suspension? Is it eventually going to not be in the state it's in and then I'll have to shake it? Is it a colloid where it will stay in this kind of nice, thick state in the case of Jell-O or fog or smoke where it will really just stay in the state Or is it a solution? And solution is probably the most important in chemistry. 99% of everything we'll talk about in chemistry involves solutions. And in general, it's an aqueous solution, when you stick something in water. So sometimes you'll see something like this. You'll see some compound x in a reaction and right next to it they'll write this aq. They mean that x is dissolved in water. It's a solute with water as the solvent. So actually, let me put that terminology here, just because I used it just now. So you have a solute. This is the thing that's usually whatever you have a smaller amount of, so thing dissolved. And then you have the solvent. This is often water or it's the thing that's in larger quantity. Or you can think of it as the thing that's all around or the thing that's doing the dissolving. Thing dissolving." + }, + { + "Q": "\nAt 2:38 you talked about the amplitude, but if I have a higher amplitude does it make the frequency low or high?", + "A": "Amplitude is independent of frequency.", + "video_name": "tJW_a6JeXD8", + "timestamps": [ + 158 + ], + "3min_transcript": "This is a pulse wave because we only have, essentially, one perturbation of the string. Now if I kept doing that-- if I kept going up and down, and up and down, essentially, if I periodically did it at regular intervals, then my string would look something like this. Doing my best to draw it neatly. It might look something like this, where once again, the perturbations are going towards-- the disturbances are going to move to-- the right. They're going to move to the right with some velocity. And what I want to do in this video is really focus on this type of wave. This type of wave right here, which you can imagine, since I'm periodically moving this left side up and down, up and down, and creating these periodic movements in the wave, we call this a periodic wave. This is So what I want to talk about is some of the properties of a periodic wave. Now, the first thing you might say is, hey, how far are you jerking it up and down? How far are these movements from rest? So if this is the resting position right there, how far are these movements above the resting position and below the resting position? And we call that the amplitude of the wave. So that distance right there- I'll do it in magenta-- that distance right there is the amplitude. Sometimes mariners will have an idea of wave height. Wave height, they normally refer to from the bottom-- from the trough-- of a wave to its peak. Amplitude, we're talking about from the resting position to it's peak. So let me label peak. I think you know what peak means. Peak is the highest point on the wave. That's the peak. If you're in a fishing boat and you wanted to see how big a wave is, you'd probably care about the wave height-- not so much if your boat's sitting down here, you have to care about this whole distance. But anyway, we won't talk too much about that. So that's the first interesting idea behind a wave. And not all waves are being generated by Sal jiggling a string on the left-hand side. But I think you get the idea that these waves can represent many different-- this graph can represent many different types of wave forms. And this, essentially, displacement, from the resting position, or from the zero position, that is your amplitude. Now the next question you might ask is, OK, I know how far you're jiggling this string up and down, but how quickly are you doing it? So how long does it take for you to go all the way up, all the way down, and back again? So how long for each cycle?" + }, + { + "Q": "\nat 6:48 in the video i still don't get the 10 cycles over second what does that equal", + "A": "A cycle is the same as a period. That is, one full wave from peak to peak. The number of cycles per second is the frequency of the wave. So 10 cycles per second means there are 10 full waves every second, and this means the frequency is 10 Hz (Hertz is a unit which just means cycles per second ). It also means that the period is T = .1 s. The frequency and the period are always reciprocals of each other.", + "video_name": "tJW_a6JeXD8", + "timestamps": [ + 408 + ], + "3min_transcript": "seconds per cycle. Or maybe it's 2 seconds per cycle. But what if we're asked how many cycles per second? So we're asking the opposite question. It's not how long, how many seconds does it take for me to go up, down, and back again. We're saying in each second, how many times am I going up, down, back again? So how many cycles per second? That's the inverse of period. So period, the notation is normally a big capital T for period. This is frequency. It's normally denoted by an F. And this, you're going to say cycles per second. So if you're going 5 seconds per cycle, that means you're doing 1/5 of a cycle or, 1/5 of a cycle per second. And that make sense. Because the period and the frequency are inversions of This is how many seconds per cycle. How long does one up, down, back again take? And this is how many up, down, back agains are there in a second? So they are inverses of each other. So we could say that frequency is equal to 1 over the period. Or you could say that period is equal to 1 over the frequency. So if I told you that I'm vibrating the left end of this rope at 10 cycles per second-- and, by the way, the unit of cycles per second, this is a hertz, so I could have also written this down as 10 hertz, which you've probably heard before. 10 hertz just means 10 cycles per second. If my frequency is 10 cycles per second, my period is going So 1 over 10 seconds per cycle, which makes sense. If in 10 times, I can go up and down, a whole up, down, back again, if I can do that 10 times in a second, it's going to take me 1/10 of a second to do it each time. Now another question we might ask ourselves is, how quickly is this wave moving, in this case, to the right? Since I'm jiggling the left end of the string. How quickly is it moving to the right? So the velocity. So to do that, we need to figure out how far did the wave go after one cycle? Or after one period? So after I jiggled this once, how far did the wave go? What is this distance from this resting point to this resting point there? And we call that a wavelength." + }, + { + "Q": "\nAt 8:54, Doesn't the wave length get reduced or shortened as time progresses?", + "A": "actually the amplitude of the wave decreases which slowly brings it to end the wavelength remains constant.", + "video_name": "tJW_a6JeXD8", + "timestamps": [ + 534 + ], + "3min_transcript": "So 1 over 10 seconds per cycle, which makes sense. If in 10 times, I can go up and down, a whole up, down, back again, if I can do that 10 times in a second, it's going to take me 1/10 of a second to do it each time. Now another question we might ask ourselves is, how quickly is this wave moving, in this case, to the right? Since I'm jiggling the left end of the string. How quickly is it moving to the right? So the velocity. So to do that, we need to figure out how far did the wave go after one cycle? Or after one period? So after I jiggled this once, how far did the wave go? What is this distance from this resting point to this resting point there? And we call that a wavelength. wavelength. You could view a wavelength as how far the initial pulse went after completing exactly one cycle. Or you could view it as the distance from one peak to another peak. That is also going to be the wavelength. Or you could view it as a distance from one trough to the other trough. That's also the wavelength. Or in general, you could view the wavelength as one exactly equal point on the wave. From that distance to that distance. That is also one wavelength. Where you're completing, between that point and that point, you're completing one entire cycle to get exactly back to that same point. And when I say exactly back to that same point, this point doesn't count. Because this point, although we're in the same position, we're now going down. We want to go to the point where we're in the same position. And notice over here, we're going up. We want to be going up again. So distance is not one wavelength. back to the same position. And we're moving in the same direction. So this is also one wavelength. So if we know how far we've travelled after one period-- let me write it this way; wavelength is equal to how far the wave has traveled after one period. Or you could say after one cycle. Because remember, a period is how long does it take to complete one cycle. One to complete up, down, and back again notion. So if we know how far we've traveled, and we know how long it took us, it took us one period, how can we figure out the velocity? Well, the velocity is equal to distance divided by time." + }, + { + "Q": "at 1:04 you said that waves don't really need a medium to travel in. Can you give an example of a type of wave that doesn't need a medium(other than light)\n", + "A": "Any other electromagnetic wave. Microwaves, for example. Mechanic waves always require a medium.", + "video_name": "tJW_a6JeXD8", + "timestamps": [ + 64 + ], + "3min_transcript": "In the last video we talked about the idea that if I start with some type of a string there, and if I were to take the left end of the string-- I could just have equally have done the right, but if I take the left end of the string and jerk it up, then all the way down, and then back to its resting position, it'll generate this disturbance in the string. And the disturbance might initially look like this after I've done that jerking up and down once. And that disturbance is going to propagate down the string. It's going to move down the string like that. Let me color this in black. So this is right after I do that first cycle-- that first jerking up and down. The string might look something like that. And if we wait a little while, the string might look something like this, assuming that I only did that once. The string might look something like this, where that pulse has actually propagated down the string. That pulse has propagated down the string. And in the last video, we said, hey, this disturbance that's propagating down the string, or propagating down this medium-- although it doesn't necessarily have to have a medium- we called this a wave. And in particular, This is a pulse wave because we only have, essentially, one perturbation of the string. Now if I kept doing that-- if I kept going up and down, and up and down, essentially, if I periodically did it at regular intervals, then my string would look something like this. Doing my best to draw it neatly. It might look something like this, where once again, the perturbations are going towards-- the disturbances are going to move to-- the right. They're going to move to the right with some velocity. And what I want to do in this video is really focus on this type of wave. This type of wave right here, which you can imagine, since I'm periodically moving this left side up and down, up and down, and creating these periodic movements in the wave, we call this a periodic wave. This is So what I want to talk about is some of the properties of a periodic wave. Now, the first thing you might say is, hey, how far are you jerking it up and down? How far are these movements from rest? So if this is the resting position right there, how far are these movements above the resting position and below the resting position? And we call that the amplitude of the wave. So that distance right there- I'll do it in magenta-- that distance right there is the amplitude. Sometimes mariners will have an idea of wave height. Wave height, they normally refer to from the bottom-- from the trough-- of a wave to its peak. Amplitude, we're talking about from the resting position to it's peak. So let me label peak. I think you know what peak means. Peak is the highest point on the wave. That's the peak." + }, + { + "Q": "Is Sal saying 'Lamda' at 12:55 or is he saying 'Landa'?\n", + "A": "Lambda (the Greek letter that looks like an upside down Y ).", + "video_name": "tJW_a6JeXD8", + "timestamps": [ + 775 + ], + "3min_transcript": "wavelength over period. Which is the same thing as wavelength times 1 over my period. And we just said that 1 over the period, this is the same thing is your frequency. So velocity is equal to wavelength times your frequency. And if you know this, you can pretty much solve all of the basic problems that you might encounter in waves. So for example, if someone tells you that I have a velocity of-- I don't know-- 100 meters per second to the right, so in that direction-- velocity you have to give a direction-- and they were to tell you that my frequency is equal to-- let's say my frequency is 20 cycles per second, which is the same thing as 20 hertz. only able to observe this part of your wave, you'd only observe that part of my string. If we're talking about 20 hertz, then in 1 second, you would see this go up and down twenty times. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. In exactly one second, you would see this go up and down 20 times. That's what we mean by the frequency being 20 hertz, or 20 cycles per second. So, they gave you the velocity. They gave you the frequency. What's the wavelength here? So the wavelength, in this situation-- you would say the velocity-- your velocity is equal to your wavelength times your frequency. Divide both sides by 20. And actually, let me make sure I get the units right. So this is meters per second, is equal to lambda times 20 cycles per second. 100 meters per second times 1/20 seconds per cycle. And then this becomes a 5. This becomes a 1. So you get 5, and then the seconds cancel out. So you get 5 meters per cycle. So this is equal to 5 meters per cycle, which would be your wavelength in this situation. So it's 5 meters. You could say 5 meters per cycle, but wavelength implies that you're talking about the distance per cycle. So in this situation, if this is moving to the right at 100 meters per second and this frequency-- I see this moving up and down 20 times in a second-- then this distance, right here, must be 5 meters. Likewise, we can figure out the period very easily. The period here is just going to be 1 over the frequency. It's going to be 1/20 seconds per cycle." + }, + { + "Q": "At 10:15 when Sal says that he could write it as a Vector, What does he mean? What is a vector?\n", + "A": "A vector is a value that also has a direction. Since velocity has a direction, Sal is saying he could write down not only the value of the velocity (the magnitude ), but also the direction. Doing math with vectors is little more complicated and the subject of later material. Sal didn t bother to do that in this video.", + "video_name": "tJW_a6JeXD8", + "timestamps": [ + 615 + ], + "3min_transcript": "wavelength. You could view a wavelength as how far the initial pulse went after completing exactly one cycle. Or you could view it as the distance from one peak to another peak. That is also going to be the wavelength. Or you could view it as a distance from one trough to the other trough. That's also the wavelength. Or in general, you could view the wavelength as one exactly equal point on the wave. From that distance to that distance. That is also one wavelength. Where you're completing, between that point and that point, you're completing one entire cycle to get exactly back to that same point. And when I say exactly back to that same point, this point doesn't count. Because this point, although we're in the same position, we're now going down. We want to go to the point where we're in the same position. And notice over here, we're going up. We want to be going up again. So distance is not one wavelength. back to the same position. And we're moving in the same direction. So this is also one wavelength. So if we know how far we've travelled after one period-- let me write it this way; wavelength is equal to how far the wave has traveled after one period. Or you could say after one cycle. Because remember, a period is how long does it take to complete one cycle. One to complete up, down, and back again notion. So if we know how far we've traveled, and we know how long it took us, it took us one period, how can we figure out the velocity? Well, the velocity is equal to distance divided by time. vector, but I think you get the general idea. Your velocity-- what's the distance you travel in a period? Well, the distance you travel in a period is your wavelength after one up, down, back again. The wave pulse would have traveled exactly that far. That would be my wavelength. So I've traveled the distance of a wavelength, and how long did it take me to travel that distance? Well, it took me a period to travel that distance. So it's wavelength divided by period. Now I just said that 1 over the period is the same thing as the frequency. So I could rewrite this as wavelength. And actually, I should be clear here. The notation for wavelength tends to be the Greek letter lambda." + }, + { + "Q": "why does nitrogen not 'like to react' with other chemicals? 8:25\n", + "A": "In its gaseous form (N2) it is triple bonded which is quite strong and thus doesn t want to react.", + "video_name": "lzWUG4H5QBo", + "timestamps": [ + 505 + ], + "3min_transcript": "to go both to us, or to Jack and to be used by itself. So it actually has an excess of oxygen that it's making. So this is actually kind of interesting and good to know. Now if you think about it, if I was to, let's say sketch out a planet. Let's draw a little planet over here, and ask you the question. If this was your planet Earth, and you've got thousands, instead of just one Jack, let's say now you have thousands of Jacks and thousands of beanstalks, in fact, not even thousands. Let's say billions, because really, that's what we have. We have a planet full of humans, and full of other animals, and full of plants. What would the atmosphere look like? This is the atmosphere. What would the atmosphere look like? Well you'd guess that the atmosphere is gas. And so what would those gases be? Well, the way I've drawn things again, of carbon dioxide. So I would say well, I guess there must be, I don't know, maybe 50-50 carbon dioxide and oxygen based on what we know so far. And the truth is, that's actually not true. That if you actually look at air, if you actually break down the atmosphere or air-- I'm just going to write \"air\" here-- if you actually break it down, turns out that the ratios are actually a little different. So for example, oxygen makes up about 21% of our air. This is our air breakdown. And carbon dioxide makes up about less than 1%. So, that leaves you wondering, what the heck is making up all that other parts of air? What is it made of? And in truth, it's about 78% nitrogen. Now you know, you've got nitrogen in your proteins, we've got nitrogen in our DNA. things. But nitrogen gas, specifically, is actually N2. And N2, this nitrogen gas, really is not too reactive. It kind of just hangs out by itself, does not like to react with things. So, looking at our little atmosphere graph, if you want to now think about it, knowing that we've got very little carbon dioxide and about 21% oxygen-- you could think of oxygen being, let's say, something like that-- well then relative to that nitrogen would be much more. You have much more nitrogen hanging out. And so this is really what our atmosphere looks like. It looks more filled with nitrogen than anything else. And in terms of carbon dioxide, it's just got a little smidge of carbon dioxide. Maybe right there. That could be carbon dioxide, maybe even less than that." + }, + { + "Q": "at 4:18 what is ATP?\n", + "A": "Adenosine Tri-Phosphate (ATP) is a small molecule that cells use as a coenzyme. It s one of the most basic signal transporting molecules in living things. It does many biochemical processes: From activating membrane transporters to regulating transcription.", + "video_name": "lzWUG4H5QBo", + "timestamps": [ + 258 + ], + "3min_transcript": "kind of relying on each other to really work. So you kind of need both of them to work well. And so let's actually take a moment to write out these processes that are happening between Jack and the beanstalk. So let's start with the process of photosynthesis, the beanstalk. So on the one hand, you've got what? You've got water because, of course, the beanstalk needs water, and you've got carbon dioxide. And I'm going to do carbon oxide in orange. So it's taking in water and carbon dioxide. And it's going to put out, it's going to actually take these ingredients if you want to think of it as kind of cooking, it's going to take these ingredients and it's going to put out. It's going to put out what? Oxygen and glucose. So I'll put glucose up top and oxygen down below. So these are the inputs and outputs of photosynthesis. You've got inputs. You've got glucose and oxygen going in. You're going to start seeing some serious similarities here. You've got glucose and oxygen going in. So Jack is taking in those two things. And he is again, of course, processing them. And he's putting out water and carbon dioxide. So this looks really, really nice, right? Looks perfect actually. Because everything is nice and balanced. And you can see how it makes perfect sense that, not only did Jack need the beanstalk, but actually it sounds like the beanstalk needed Jack, based on how I've drawn it. Now remember, none of this would even happen if there was no sunlight. So we actually need light energy. In fact, that's the whole purpose of this, right? Getting energy. So you have to have some light energy. I'm going to put a big plus sign, and I might even circle it because it's so important. And on the other side, of course, Jack is getting something as well. He's getting chemical energy. In fact, he's using the chemical energy to help him climb the beanstalk. And so the chemical energy comes in the form of what we call ATP, which is just a molecule of high energy. And so Jack and the beanstalk are basically going from light energy to chemical energy using these two Now here's the part that people don't always appreciate. And I'm actually going to take just a moment to show you that this isn't the full story. There's actually something else going on as well. And that is that there's actually some cellular respiration happening on the plant's side. So remember, not only does the human, or the Jack, need energy, but so does the plant. The plant needs energy as well. And in fact, if it takes in light energy right here," + }, + { + "Q": "\nAt 3:53, can someone clarify what temperature has to go with the energy in the molecule?", + "A": "Temperature IS kinetic energy of molecules.", + "video_name": "WScwPIPqZa0", + "timestamps": [ + 233 + ], + "3min_transcript": "Sal is about to introduce us to a new concept. It's a new way of looking at probably a very familiar concept to you. And that's temperature. Temperature can and should be viewed as the average energy of the particles in the system. So I'll put a little squiggly line, because there's a lot of ways to think about it. Average energy. And mostly kinetic energy, because these particles are moving and bouncing. The higher the temperature, the faster that these particles move. And the more that they're going to bounce into the side of the container. But temperature is average energy. It tells us energy per particle. So obviously, if we only had one particle in there with super high temperature, that's going to have less pressure than if we have a million particles in there. If I have, let's take two cases right here. One is, I have a bunch of particles with a certain temperature, moving in their different directions. And the other example, I have one particle. And maybe they have the same temperature. That on average, they have the same kinetic energy. The kinetic energy per particle is the same. Clearly, this one is going to be applying more pressure to its container, because at any given moment more of these particles are going to be bouncing off the side than in this example. This guy's going to bounce, bam, then going to go and move, bounce, bam. So he's going to be applying less pressure, even though his temperature might be the same. Because temperature is kinetic energy, or you can view it as kinetic energy per particles. Or it's a way of looking at kinetic energy per particle. So if we wanted to look at the total energy in the system, we would want to multiply the temperature times the number And just since we're dealing on the molecular scale, the number of particles can often be represented as moles. Remember, moles is just a number of particles. So we're saying that that pressure-- well, I'll say it's proportional, so it's equal to some constant, let's call that R. Because we've got to make all the units work out in the end. I mean temperature is in Kelvin but we eventually want to get back to joules. So let's just say it's equal to some constant, or it's proportional to temperature times the number of particles. And we can do that a bunch of ways. But let's think of that in moles. If I say there are 5 mole particles there, you know that's 5 times 6 times 10 to the 23 particles. So, this is the number of particles. This is the temperature. And this is just some constant." + }, + { + "Q": "\nWhen the equation P=nRT is formed at 4:56 I tried to find the units for R. I got it as m(superscript) -3 i.e. volume. So if the equation is correct then why do we need the equation PV=nRT?", + "A": "R is a constant for accuracy. R s units should be atm*L/moles*K. That allows all the units to cancel. The reason why you need Volume in the equation is because volume also affects pressure.", + "video_name": "WScwPIPqZa0", + "timestamps": [ + 296 + ], + "3min_transcript": "If I have, let's take two cases right here. One is, I have a bunch of particles with a certain temperature, moving in their different directions. And the other example, I have one particle. And maybe they have the same temperature. That on average, they have the same kinetic energy. The kinetic energy per particle is the same. Clearly, this one is going to be applying more pressure to its container, because at any given moment more of these particles are going to be bouncing off the side than in this example. This guy's going to bounce, bam, then going to go and move, bounce, bam. So he's going to be applying less pressure, even though his temperature might be the same. Because temperature is kinetic energy, or you can view it as kinetic energy per particles. Or it's a way of looking at kinetic energy per particle. So if we wanted to look at the total energy in the system, we would want to multiply the temperature times the number And just since we're dealing on the molecular scale, the number of particles can often be represented as moles. Remember, moles is just a number of particles. So we're saying that that pressure-- well, I'll say it's proportional, so it's equal to some constant, let's call that R. Because we've got to make all the units work out in the end. I mean temperature is in Kelvin but we eventually want to get back to joules. So let's just say it's equal to some constant, or it's proportional to temperature times the number of particles. And we can do that a bunch of ways. But let's think of that in moles. If I say there are 5 mole particles there, you know that's 5 times 6 times 10 to the 23 particles. So, this is the number of particles. This is the temperature. And this is just some constant. We gave these two examples. Obviously, it is dependent on the temperature; the faster each of these particles move, the higher pressure we'll have. It's also dependent on the number of particles, the more particles we have, the more pressure we'll have. What about the size of the container? The volume of the container. If we took this example, but we shrunk the container somehow, maybe by pressing on the outside. So if this container looked like this, but we still had the same four particles in it, with the same average kinetic energy, or the same temperature. So the number of particles stays the same, the temperature is the same, but the volume has gone down. Now, these guys are going to bump into the sides of the container more frequently and there's less area. So at any given moment, you have more force and less area. So when you have more force and less area, your pressure is going to go up. So when the volume went down, your pressure went up." + }, + { + "Q": "\nAt 2:20, can you please explain how is temperature related to average energy?\nCan you also tell me which video explains pressure,density,volume for a substance are explained ?", + "A": "It comes from the video on the Kinetic Molecular Theory of gases. The result at the end of the video is that KE(avg) = \u00c2\u00bdkT. So KE(avg) \u00e2\u0088\u009d T and T \u00e2\u0088\u009d KE(avg). I m don t think there is a video explaining how P, V, and \u00cf\u0081 are related for substances in general, but the video on Boyle s Law explains the relationship between P and V for an ideal gas.", + "video_name": "WScwPIPqZa0", + "timestamps": [ + 140 + ], + "3min_transcript": "Let's say I have a balloon. And in that balloon I have a bunch of particles bouncing around. They're gas particles, so they're floating freely. And they each have some velocity, some kinetic energy. And what I care about, let me just draw a few more, what I care about is the pressure that is exerted on the surface of the balloon. So I care about the pressure. And what's pressure? It's force per area. So the area here, you can think of it as the inside surface of the balloon. And what's going to apply force to that? Well any given moment-- I only drew six particles here, but in a real balloon you would have gazillions of particles, and we could talk about how large, but more particles than you can really probably imagine-- but at any given moment, some of those particles are bouncing off the wall of the container. bouncing there, this guy's bouncing like that. And when they bounce, they apply force to the container. An outward force, that's what keeps the balloon blown up. So think about what the pressure is going to be dependent on. So first of all, the faster these particles move, the higher the pressure. Slower particles, you're going to be bouncing into the container less, and when you do bounce into the container, it's going to be less of a ricochet, or less of a change in momentum. So slower particles, you're going to have pressure go down. Now, it's practically impossible to measure the kinetic energy, or the velocity, or the direction of each individual particle. Especially when you have gazillions of them in a balloon. So we do is we think of the average energy of the particles. Sal is about to introduce us to a new concept. It's a new way of looking at probably a very familiar concept to you. And that's temperature. Temperature can and should be viewed as the average energy of the particles in the system. So I'll put a little squiggly line, because there's a lot of ways to think about it. Average energy. And mostly kinetic energy, because these particles are moving and bouncing. The higher the temperature, the faster that these particles move. And the more that they're going to bounce into the side of the container. But temperature is average energy. It tells us energy per particle. So obviously, if we only had one particle in there with super high temperature, that's going to have less pressure than if we have a million particles in there." + }, + { + "Q": "1:45 does that mean that helium particles \"whiz around\" faster then air particles?\n", + "A": "yes. its for the same reason when you inhale helium gas and try to speak your voice seems to be at a very high pitch", + "video_name": "WScwPIPqZa0", + "timestamps": [ + 105 + ], + "3min_transcript": "Let's say I have a balloon. And in that balloon I have a bunch of particles bouncing around. They're gas particles, so they're floating freely. And they each have some velocity, some kinetic energy. And what I care about, let me just draw a few more, what I care about is the pressure that is exerted on the surface of the balloon. So I care about the pressure. And what's pressure? It's force per area. So the area here, you can think of it as the inside surface of the balloon. And what's going to apply force to that? Well any given moment-- I only drew six particles here, but in a real balloon you would have gazillions of particles, and we could talk about how large, but more particles than you can really probably imagine-- but at any given moment, some of those particles are bouncing off the wall of the container. bouncing there, this guy's bouncing like that. And when they bounce, they apply force to the container. An outward force, that's what keeps the balloon blown up. So think about what the pressure is going to be dependent on. So first of all, the faster these particles move, the higher the pressure. Slower particles, you're going to be bouncing into the container less, and when you do bounce into the container, it's going to be less of a ricochet, or less of a change in momentum. So slower particles, you're going to have pressure go down. Now, it's practically impossible to measure the kinetic energy, or the velocity, or the direction of each individual particle. Especially when you have gazillions of them in a balloon. So we do is we think of the average energy of the particles. Sal is about to introduce us to a new concept. It's a new way of looking at probably a very familiar concept to you. And that's temperature. Temperature can and should be viewed as the average energy of the particles in the system. So I'll put a little squiggly line, because there's a lot of ways to think about it. Average energy. And mostly kinetic energy, because these particles are moving and bouncing. The higher the temperature, the faster that these particles move. And the more that they're going to bounce into the side of the container. But temperature is average energy. It tells us energy per particle. So obviously, if we only had one particle in there with super high temperature, that's going to have less pressure than if we have a million particles in there." + }, + { + "Q": "At 3:42, the example compares two balloons, one with many particles and one with only one particle. The particles are the gas particles, how can we have only 1 gas particle in a balloon?\n", + "A": "The example is a simplification of the real world. Each particle in the example represents many billions of trillions of particles in a real balloon.", + "video_name": "WScwPIPqZa0", + "timestamps": [ + 222 + ], + "3min_transcript": "Sal is about to introduce us to a new concept. It's a new way of looking at probably a very familiar concept to you. And that's temperature. Temperature can and should be viewed as the average energy of the particles in the system. So I'll put a little squiggly line, because there's a lot of ways to think about it. Average energy. And mostly kinetic energy, because these particles are moving and bouncing. The higher the temperature, the faster that these particles move. And the more that they're going to bounce into the side of the container. But temperature is average energy. It tells us energy per particle. So obviously, if we only had one particle in there with super high temperature, that's going to have less pressure than if we have a million particles in there. If I have, let's take two cases right here. One is, I have a bunch of particles with a certain temperature, moving in their different directions. And the other example, I have one particle. And maybe they have the same temperature. That on average, they have the same kinetic energy. The kinetic energy per particle is the same. Clearly, this one is going to be applying more pressure to its container, because at any given moment more of these particles are going to be bouncing off the side than in this example. This guy's going to bounce, bam, then going to go and move, bounce, bam. So he's going to be applying less pressure, even though his temperature might be the same. Because temperature is kinetic energy, or you can view it as kinetic energy per particles. Or it's a way of looking at kinetic energy per particle. So if we wanted to look at the total energy in the system, we would want to multiply the temperature times the number And just since we're dealing on the molecular scale, the number of particles can often be represented as moles. Remember, moles is just a number of particles. So we're saying that that pressure-- well, I'll say it's proportional, so it's equal to some constant, let's call that R. Because we've got to make all the units work out in the end. I mean temperature is in Kelvin but we eventually want to get back to joules. So let's just say it's equal to some constant, or it's proportional to temperature times the number of particles. And we can do that a bunch of ways. But let's think of that in moles. If I say there are 5 mole particles there, you know that's 5 times 6 times 10 to the 23 particles. So, this is the number of particles. This is the temperature. And this is just some constant." + }, + { + "Q": "at 6:50 or any other equations why are there constant?\nlike PV=nRT <- what is that \"R\"?\nand they don't explain what's it for.\nare those things so complicated that we can't understand ?\n", + "A": "The R is just a conversion factor that takes account of the types of units you chose to use for P, V, and T. Nature does not arrange herself to match Pascals, cubic meters, and Kelvin, so we have to adjust to her.", + "video_name": "WScwPIPqZa0", + "timestamps": [ + 410 + ], + "3min_transcript": "We gave these two examples. Obviously, it is dependent on the temperature; the faster each of these particles move, the higher pressure we'll have. It's also dependent on the number of particles, the more particles we have, the more pressure we'll have. What about the size of the container? The volume of the container. If we took this example, but we shrunk the container somehow, maybe by pressing on the outside. So if this container looked like this, but we still had the same four particles in it, with the same average kinetic energy, or the same temperature. So the number of particles stays the same, the temperature is the same, but the volume has gone down. Now, these guys are going to bump into the sides of the container more frequently and there's less area. So at any given moment, you have more force and less area. So when you have more force and less area, your pressure is going to go up. So when the volume went down, your pressure went up. proportional to volume. So let's think about that. Let's put that into our equation. We said that pressure is proportional-- and I'm just saying some proportionality constant, let's call that R, to the number of particles times the temperature, this gives us the total energy. And it's inversely proportional to the volume. And if we multiply both sides of this times the volume, we get the pressure times the volume is proportional to the number of particles times the temperature. So PV is equal to RnT. And just to switch this around a little bit, so it's in a form that you're more likely to see in your chemistry book, if we just switch the n and the R term. You get pressure times volume is equal to n, the number of And this right here is the ideal gas equation. Hopefully, it makes some sense to you. When they say ideal gas, it's based on this little mental exercise I did to come up with this. I made some implicit assumptions when I did this. One is I assumed that we're dealing with an ideal gas. And so you say what, Sal, is an ideal gas? An ideal gas is one where the molecules are not too concerned with each other. They're just concerned with their own kinetic energy and bouncing off the wall. So they don't attract or repel each other. Let's say they attracted each other, then as you increased the number of particles maybe they'd want to" + }, + { + "Q": "\nAt 4:39, he changes one of the hydrogens to an R group for the H3O+. Why was this necessary? I understood that that could stay H3O+.", + "A": "I believe the reaction will be taking place with alcohol as the solvent (not in water), so there really wouldn t be any H3O+ ...", + "video_name": "rL0lDESwwv8", + "timestamps": [ + 279 + ], + "3min_transcript": "those pi electrons in here, in our carbonyl, had kicked off onto our oxygen, like that. So, that gives us our intermediate, and we're almost to the formation of a hydrate, the last step would be to deprotonate: So a molecule of water comes along, this time, it's gonna function as a base, and take this proton away, leaving these two electrons behind, on this oxygen. So let's go ahead, and show our product, formation of a hydrate, so we have a carbon bonded to two OH groups now, and so we form our hydrate, or our gem-diol. Let's follow those electrons now. So, let's say that these electrons right in here, moved off onto this top oxygen, doesn't really matter which lone pair you say they are, and that's formation of our hydrate. We could have followed through as a ketone, so instead of an aldehyde, we could have started off here as a ketone, and so, this would've been our prime, and then, this would've been our prime, and then, finally, this would've been our prime; so, that's still formation of a hydrate. If we wanted to form a hemiacetal, we wouldn't have started with water; we would have started with an alcohol, so let's go ahead and show that. So let's make this R double prime OH, and, of course, an acid-catalyzed version, so R double prime OH and H plus, instead of forming a hydrate, this is gonna form a hemiacetal, so instead of H three O plus, which is what we had in our acid-catalyzed hydration, we're gonna protonate the alcohol, so let me go ahead and change this, this would no longer gonna be an \"H,\" here, this would be \"R double prime,\" like that, and so, when we protonate our carbonyl oxygen here, and make our carbon more electrophilic, this wouldn't be water as our nucleophile; this would be alcohol. So our alcohol molecule would be a nucleophile, and attack our carbonyl carbon, So this wouldn't be an \"H,\" this would be \"R double prime,\" and therefore, in our final product, this wouldn't be an \"H,\" this would be \"R double prime,\" and hopefully you can see, that this is now a hemiacetal. It's actually difficult to isolate hemiacetals most of the time, and if you're doing a reaction in acid, usually, the hemiacetal is going to move on to form a full acetal. So let's go ahead, and show the structure of a full acetal. And, this isn't just one step, and I won't go through the steps in this video, that'll be in the next video, but I just wanted to show you the structure of a full acetal here. So we have R double prime, and lone pairs of electrons on our oxygen; and we have, over here, R double prime and lone pairs of electrons on our oxygen, so this would be an acetal, and let's, once again, compare structures. So, with an acetal, we have this OR double prime group, and another OR double prime group" + }, + { + "Q": "\nAt 9:12, we learn that the oxygen was reduced by the carbon. Wasn't it also reduced by the hydrogen, as 2 out of the 4 oxygens gained their negative reduction because of the addition of 2 positive hydrogens each?", + "A": "The hydrogens don t change oxidation states through the reaction though so they didn t take part in the reduction/oxidation part of this reaction", + "video_name": "OE0MMIyMTNU", + "timestamps": [ + 552 + ], + "3min_transcript": "On the left-hand side, you have 2O2's neutral oxidation number. And on the right-hand side, what do you have? You have 4 total oxygens. I'll combine these together. I'll just write this as 4 total oxygens. And what's each of their oxidation numbers? Well, we see it's a negative 2. So what happened to each of these 4 oxygens? I could have written 4O here instead of 2O2. Either way, I'm just really trying to account for the oxygens. Here I have 4 oxygens with a neutral oxidation number-- with an oxidation number of 0. And here I have 4 oxygens with a negative oxidation number. How do you go from 0 to negative? Each of them must have gained 2 electrons. If you have 4 oxygens, each of them gained 2 electrons, Actually, let me write it like this. Let me move this part. So cut and paste. Let me move it over to the right a little bit because what I want to show is the gaining of the electrons. So plus 8 electrons. So what happened to oxygen? Well, oxygen gained electrons. What is gaining electrons? Reduction is gaining-- RIG. GER-- gaining electrons is reduction. Oxygen has been reduced. Now, what oxidized carbon? Well, carbon lost electrons to the oxygen. So carbon oxidized by the oxygen, which is part of the motivation for calling it \"oxidation.\" And what reduced the oxygen? So oxygen reduced by the carbon. And this type of reaction-- where you have both oxidation and reduction taking place, and really they're two sides of the same coin. One thing is going to be oxidized if another thing is being reduced, and vice versa. We call these oxidation reduction reactions. Or sometimes \"redox\" for short. Take the \"red\" from \"reduction\" and the \"ox\" from \"oxidation,\" and you've got \"redox.\" This is a redox reaction. Something is being oxidized. Something else is being reduced. Not everything is being oxidized or reduced, and we can see that very clearly when we depict it in these half reactions. And one way to check that your half reactions actually makes sense is you can actually sum up the two sides." + }, + { + "Q": "Hello! I'm a beginner in chemistry and at 8:20 I lost it. Why does the oxygen gain 8 electrons and not 2? It's oxidation number is 2- Can someone explain? What happens to the other 6 gained electrons?\n", + "A": "There s four oxygen atoms because of the coefficients. Four of them each gaining 2 electrons = 8", + "video_name": "OE0MMIyMTNU", + "timestamps": [ + 500 + ], + "3min_transcript": "Or LEO-- losing electrons is oxidation-- the lion says GER-- gaining electrons is reduction. So it's clear right over here that carbon is being oxidized. It is losing electrons. Oxidation is losing electrons. Carbon is oxidized. Let's think about the hydrogen. On the left-hand side, you have 4 hydrogens that each have an oxidation number of plus 1. On the right-hand side, you have 4 hydrogens. We're writing it a slightly different way. We could write it like this-- 2H2's that each have an oxidation number of plus 1. The oxidation numbers for the hydrogens has not changed. The hydrogens has neither been oxidized nor reduced. On the left-hand side, you have 2O2's neutral oxidation number. And on the right-hand side, what do you have? You have 4 total oxygens. I'll combine these together. I'll just write this as 4 total oxygens. And what's each of their oxidation numbers? Well, we see it's a negative 2. So what happened to each of these 4 oxygens? I could have written 4O here instead of 2O2. Either way, I'm just really trying to account for the oxygens. Here I have 4 oxygens with a neutral oxidation number-- with an oxidation number of 0. And here I have 4 oxygens with a negative oxidation number. How do you go from 0 to negative? Each of them must have gained 2 electrons. If you have 4 oxygens, each of them gained 2 electrons, Actually, let me write it like this. Let me move this part. So cut and paste. Let me move it over to the right a little bit because what I want to show is the gaining of the electrons. So plus 8 electrons. So what happened to oxygen? Well, oxygen gained electrons. What is gaining electrons? Reduction is gaining-- RIG. GER-- gaining electrons is reduction. Oxygen has been reduced. Now, what oxidized carbon? Well, carbon lost electrons to the oxygen. So carbon oxidized by the oxygen, which is part of the motivation for calling it \"oxidation.\" And what reduced the oxygen?" + }, + { + "Q": "\nAt around 8:00, is hydrogen also reducing oxygen by giving oxygen electrons?", + "A": "No. Hydrogen doesn t change its oxidation state in this reaction. Thus, it cant oxidise or reduce anything. Hope this helps :)", + "video_name": "OE0MMIyMTNU", + "timestamps": [ + 480 + ], + "3min_transcript": "Or LEO-- losing electrons is oxidation-- the lion says GER-- gaining electrons is reduction. So it's clear right over here that carbon is being oxidized. It is losing electrons. Oxidation is losing electrons. Carbon is oxidized. Let's think about the hydrogen. On the left-hand side, you have 4 hydrogens that each have an oxidation number of plus 1. On the right-hand side, you have 4 hydrogens. We're writing it a slightly different way. We could write it like this-- 2H2's that each have an oxidation number of plus 1. The oxidation numbers for the hydrogens has not changed. The hydrogens has neither been oxidized nor reduced. On the left-hand side, you have 2O2's neutral oxidation number. And on the right-hand side, what do you have? You have 4 total oxygens. I'll combine these together. I'll just write this as 4 total oxygens. And what's each of their oxidation numbers? Well, we see it's a negative 2. So what happened to each of these 4 oxygens? I could have written 4O here instead of 2O2. Either way, I'm just really trying to account for the oxygens. Here I have 4 oxygens with a neutral oxidation number-- with an oxidation number of 0. And here I have 4 oxygens with a negative oxidation number. How do you go from 0 to negative? Each of them must have gained 2 electrons. If you have 4 oxygens, each of them gained 2 electrons, Actually, let me write it like this. Let me move this part. So cut and paste. Let me move it over to the right a little bit because what I want to show is the gaining of the electrons. So plus 8 electrons. So what happened to oxygen? Well, oxygen gained electrons. What is gaining electrons? Reduction is gaining-- RIG. GER-- gaining electrons is reduction. Oxygen has been reduced. Now, what oxidized carbon? Well, carbon lost electrons to the oxygen. So carbon oxidized by the oxygen, which is part of the motivation for calling it \"oxidation.\" And what reduced the oxygen?" + }, + { + "Q": "at 2:28, does it matter if the equation isn't balanced?\n", + "A": "In most cases yes! If you have an unbalanced redox equation involving charged ions, then if the equation is not balanced, the oxidation numbers will be wrong and therefore your half equations will be wrong. In the case of the video, balancing would not matter, but it s safest to balance first and then work out oxidation numbers. Hope that helps! :)", + "video_name": "OE0MMIyMTNU", + "timestamps": [ + 148 + ], + "3min_transcript": "What we have depicted right over here is a combustion reaction. We have the hydrocarbon methane right over here. You take that. You take some molecular oxygen. You give them enough heat. And then, they are going to combust. And they're going to produce carbon dioxide, water, and then more energy than you put in. This is an exothermic reaction. More energy comes out than you put in. This is why fires keep spreading. This is why combustion is used to power things. But that's not what we're going to focus on in this video. In this video, we want to think about-- which of these components, of these molecules, what's being oxidized and what is being reduced? And to do that, let's first think about the oxidation states of the input atoms and the oxidation states of the outputs-- of the different constituents of these molecules. So I encourage you to now pause this video and try to figure that out on your own. So I'm assuming you've given a go at it. Let's first think about the methane. And I have a bunch of electronegativities here based on the Pauling scale listed out here, but let's just visualize methane. Methane is a carbon bonded to 4 hydrogens. In our oxidation state world-- even though, this, in reality, is a covalent bond-- we pretend like they're hypothetically ionic bonds. So we have to give the electron pair to one of the parties to the bond. If we look between carbon and hydrogen, carbon is more electronegative than hydrogen. So we will assume that carbon will be hogging the electrons and that hydrogen will be giving away the electrons. So carbon's going to take-- in this hypothetical world-- hypothetically take an electron from each of these hydrogens. It is going to have an oxidation state of negative 4, an electron from each of 4 hydrogens. Negative 4. And once again, we write the sign after the number probably so that we don't get these confused with exponents. Each of those hydrogens is having-- in this hypothetical world-- an electron taken away from it. We could say it has an oxidation state of plus 1, which we could write as 1 plus. Or we could just write a positive right over here-- a plus sign. Now, we have molecular oxygen-- oxygen bonded to oxygen. Well, all oxygens are created equal. Or we'll assume that these oxygens are created equal, that they're not different isotopes or anything like that. In this reality, there's no reason why one oxygen would hog any electrons from the other oxygen. In this world, the oxygen has an oxidation state-- when it's in this molecular oxygen form-- it has an oxidation state of 0 or an oxidation number of 0. Now, let's think about this side-- the products. Now, what's happening here with carbon dioxide? Carbon dioxide is a carbon double" + }, + { + "Q": "\nAt 9:15 , why don't both the oxygen's make double bonds with sulfur? I know this breaks the octet rule, but sulfur is in the 3rd period, so isn't it more important to have no formal charge than obey the octet rule?", + "A": "It is because Oxygen is more electronegative, and therefore will consume more of the electrons. Sulfur, being less electronegative, cannot have more electrons around it than the more electronegative oxygen", + "video_name": "3RDytvJYehY", + "timestamps": [ + 555 + ], + "3min_transcript": "So I could take, let's say-- let me go ahead and make these blue here. So I could take a lone pair of electrons from either oxygen. I'm just going to say that we take a lone pair of electrons from that oxygen and move them in here to form a double bond between the sulfur and the oxygen. So if I do that, now I have a double bond between the sulfur and the oxygen. The oxygen on the right now has only two lone pairs of electrons around it. The oxygen on the left still has three lone pairs of electrons around it like that. And the sulfur still has a lone pair of electrons here in the center. All right, so if we assign formal charges now-- let's go ahead and do that really fast. So we know that we have electrons in these bonds here. And so if we assign a formal charge to the oxygen on the left-- let's do that one first, all right, so this oxygen on the left here. All right, oxygen normally has six valence electrons And in our dot structure, we give to the sulfur. So you can see the oxygen is surrounded by seven valence electrons. This one's a little bit harder to see, so let me go ahead and mark it there. So 6 minus 7 gives us a formal charge of negative 1 on this oxygen. And when we do the same formal charge for the sulfur here, we can see that sulfur is surrounded by five valence electrons. Sulfur's in group six. So in the free atom, there's six. 6 minus 5 gives us a formal charge of plus 1. So we have a formal charge of plus 1 in the sulfur, formal charge of negative 1 on this oxygen. And even though we don't have a formal charge of zero on these atoms, this is about as good as we're going to get in terms of this representation of the molecule here. And another thing to think about is the fact that I didn't have to take the lone pair of electrons from this oxygen, right? I could have taken the lone pair of electrons from over here, on that oxygen. And that would just be another resonance structure of this. So I don't want to go too in detail We talked about those in earlier videos. So make sure that you watch them. But this is our final dot structure here. So let me go ahead and redraw it here. This is one of the possible dot structures you can draw sulfur dioxide that fulfills our rules for drawing dot structures. So let me go ahead and put in these valence electrons here so we can see it a little bit better. All right, so I'll leave out formal charges now because we're just focusing in on geometry. We're concerned about VSEPR theory. So we have our dot structure. We go back up to check our steps for predicting the shape. And now we are going to count the number of electron clouds that surround our central atoms, the regions of electron density, which could be valence electrons in bonds, so bonding electrons, or non-bonding electrons, like lone pairs of electrons. So if we look at our central atom, which is sulfur, and let's see if we can count up our electron clouds. So this is an electron cloud over here." + }, + { + "Q": "Isn't boron trifluoride tetrahedral, not trigonal planar as the video states at 4:25?\n", + "A": "No, BF\u00e2\u0082\u0083 is trigonal planar.", + "video_name": "3RDytvJYehY", + "timestamps": [ + 265 + ], + "3min_transcript": "of electrons in the bonded atom. And so when you look at the bonds between boron and fluorine, one of those electrons goes to fluorine. And one of those electrons goes to boron here. So we can see that's the same for all three of these bonds. And so now boron is surrounded by three valence electrons in the bonded atom. 3 minus 3 gives us a formal charge of 0. So, remember, the goal is to minimize a formal charge when you're drawing your dot structures. And so this is a completely acceptable dot structure here, even though boron isn't following an octet. Now, boron can be surrounded by eight electrons. And so you'll even see some textbooks say, well, one of these lone pairs electrons on one of these fluorines could actually move in here to surround the boron with eight electrons, giving it an octet. And that's fine. That would give the boron a formal charge. And that might actually contribute to the overall structure of this molecule. But for us, for our purposes we're, just going to stick with this as being our dot structure here. So let me go ahead and redraw that. And I'm going to draw it in a slightly different way when So let me go ahead and put in our lone pairs of electrons here on the fluorine. And let's think about step two. We're going to count the number of electron clouds that surround the central atom here. So, remember, electron clouds are either the bonding electrons or non-bonding electrons-- the valence electrons in bonds or the lone pairs-- just regions of electron density that can repel each other. All right, so if I'm looking at my central atom, which is my boron, I can see that here are some electrons. All right, so that's an electron cloud. These electrons right here occupy an electron cloud, as well. And then I have another electron cloud here. So I have three electron clouds that are going to repel each other. And that allows us to predict the geometry of those electron clouds around that central atom. They're going to try to get as far away from each other as they possibly can. And it turns out, that happens when those electron So I'm going to draw sheet of paper here, a plane. And we're going to put our boron atom in the center. And here's one of our electron clouds. And then here's another one, and here's our third one here. So they're going to get as far away from each other as they possibly can. We call this shape trigonal planar. So let me go ahead and write that here. So this is a trigonal planar geometry of my electron clouds surrounding my central atom. And since we don't have any lone pairs of electrons to worry about on our central atom, we can go ahead and predict the geometry of the molecule as being the same here as the geometry of the electron cloud. So the molecule has a trigonal planar shape, as well. Now, when we think about bond angles, right? So the easiest way to think about what's the bond angle for trigonal planar or what would we predict the bond angle to be, you think about a circle. And since the electrons repel each other equally, you want to divide your circle into three equivalent angles." + }, + { + "Q": "At 10:15, why wouldn't the 2 non-bonded electrons be in different electron clouds, or different orbitals? It seems that Hund's rule would cause the 2 electrons to join different oribitals.\n", + "A": "Which orbital do you suppose the second one would go in to? Something to keep in mind: VSEPR theory predicts a bond angle of 120 degrees for SO2, and that s very close to what we actually observe (119 degrees)", + "video_name": "3RDytvJYehY", + "timestamps": [ + 615 + ], + "3min_transcript": "to the sulfur. So you can see the oxygen is surrounded by seven valence electrons. This one's a little bit harder to see, so let me go ahead and mark it there. So 6 minus 7 gives us a formal charge of negative 1 on this oxygen. And when we do the same formal charge for the sulfur here, we can see that sulfur is surrounded by five valence electrons. Sulfur's in group six. So in the free atom, there's six. 6 minus 5 gives us a formal charge of plus 1. So we have a formal charge of plus 1 in the sulfur, formal charge of negative 1 on this oxygen. And even though we don't have a formal charge of zero on these atoms, this is about as good as we're going to get in terms of this representation of the molecule here. And another thing to think about is the fact that I didn't have to take the lone pair of electrons from this oxygen, right? I could have taken the lone pair of electrons from over here, on that oxygen. And that would just be another resonance structure of this. So I don't want to go too in detail We talked about those in earlier videos. So make sure that you watch them. But this is our final dot structure here. So let me go ahead and redraw it here. This is one of the possible dot structures you can draw sulfur dioxide that fulfills our rules for drawing dot structures. So let me go ahead and put in these valence electrons here so we can see it a little bit better. All right, so I'll leave out formal charges now because we're just focusing in on geometry. We're concerned about VSEPR theory. So we have our dot structure. We go back up to check our steps for predicting the shape. And now we are going to count the number of electron clouds that surround our central atoms, the regions of electron density, which could be valence electrons in bonds, so bonding electrons, or non-bonding electrons, like lone pairs of electrons. So if we look at our central atom, which is sulfur, and let's see if we can count up our electron clouds. So this is an electron cloud over here. So we have one electron cloud. Over here on the right, this double bond, we can consider it as an electron cloud. We're not worried about numbers of electrons, just regions of them. And then for the first time, we now have a lone pair of electrons, right? And this, we can also think about as occupying an electron cloud. So we have three electron clouds. And we saw in the previous example that when you have three electron clouds, the electron clouds are going to try to adopt a trigonal planar shape. So I could redraw this dot structure and attempt to show it in more of a trigonal planar shape here. So let's go ahead and show it looking like this-- once again, not worried about drawing informal charges here-- so something like this for the structure. Let me go ahead and put those electrons in our orbital here so we can see that electron cloud a little bit better. And so, once again, our electron clouds are in a trigonal planar geometry." + }, + { + "Q": "At 2:50 he said that the completed-octet version of the dot structure and the dot structure with minimized charge are acceptable representations of the molecule. But which ones do molecules generally \"pick\" in nature?\n", + "A": "The actual structure is a hybrid of all possible structures, but the ones with minimized charge separation are closest to the actual structure.", + "video_name": "3RDytvJYehY", + "timestamps": [ + 170 + ], + "3min_transcript": "We're going to put those leftover electrons on a terminal atoms, which are our fluorines in this case. Fluorine follows the octet rule. So each fluorine is now surrounded by two valence electrons. So each fluorine needs six more to have an octet of electrons. So I'll go ahead and put six more valence electrons on each of the three fluorines. And 6 times 3 is 18. So we just represented 18 more valence electrons. And so now we are all set. We've represented all 24 valence electrons in our dot structure. And some of you might think, well, boron is not following the octet rule here. And that is true. It's OK for boron not to follow the octet rule. And to think about why, let's assign a formal charge to our boron atom here. And so, remember, each covalent bond consists of two electrons. Let me go ahead and draw in those electrons in blue here. And when you're assigning formal charge, remember how to do that. You take the number of valence electrons in the free atoms, which is three. of electrons in the bonded atom. And so when you look at the bonds between boron and fluorine, one of those electrons goes to fluorine. And one of those electrons goes to boron here. So we can see that's the same for all three of these bonds. And so now boron is surrounded by three valence electrons in the bonded atom. 3 minus 3 gives us a formal charge of 0. So, remember, the goal is to minimize a formal charge when you're drawing your dot structures. And so this is a completely acceptable dot structure here, even though boron isn't following an octet. Now, boron can be surrounded by eight electrons. And so you'll even see some textbooks say, well, one of these lone pairs electrons on one of these fluorines could actually move in here to surround the boron with eight electrons, giving it an octet. And that's fine. That would give the boron a formal charge. And that might actually contribute to the overall structure of this molecule. But for us, for our purposes we're, just going to stick with this as being our dot structure here. So let me go ahead and redraw that. And I'm going to draw it in a slightly different way when So let me go ahead and put in our lone pairs of electrons here on the fluorine. And let's think about step two. We're going to count the number of electron clouds that surround the central atom here. So, remember, electron clouds are either the bonding electrons or non-bonding electrons-- the valence electrons in bonds or the lone pairs-- just regions of electron density that can repel each other. All right, so if I'm looking at my central atom, which is my boron, I can see that here are some electrons. All right, so that's an electron cloud. These electrons right here occupy an electron cloud, as well. And then I have another electron cloud here. So I have three electron clouds that are going to repel each other. And that allows us to predict the geometry of those electron clouds around that central atom. They're going to try to get as far away from each other as they possibly can. And it turns out, that happens when those electron" + }, + { + "Q": "\nAt 2:00, Khan resembles a ray to a marching band, however, I don't see how one part of a single ray would leave the water before its other part. Also, if (in any way), a piece of a ray does leave he water first, why doesn't the speed differentiation cause the photon (ray= photon?) to start spinning instead of bending??", + "A": "A ray is just a way to draw light. The light is really a wave, right? Not an arrow. To understand refraction, you really have to use the wave behavior of light, not think of it as a photon.", + "video_name": "jxptCXHLxKQ", + "timestamps": [ + 120 + ], + "3min_transcript": "Before doing more examples with Snell's Law which essentially amounts to math problems what I do is give you an intuitive understanding for why this straw looks bent in this picture right over here To do that, let me just do a simplified version of that picture This is the side profile of the cup, or glass right over here The best I can draw it And then let me draw the actual straw. I'll first draw the straw where it actually is coming in off the side of the cup and the straw is actually not bending goes to the bottom of the cup just like that and then it goes up like that and then it goes slightly above. Then it actually does bent up here It's irrelevant to what we want to talk about What I want to do in this video is talk about when we look over here why does it look like the straw got bent? As the light from the straw down here changes as it go from one medium to another Now we know from refraction indices or just in general the light moves slower in water than it does in air slower in water; faster in air Let's think about what's going to happen Let me draw 2 rays that are coming from this point on the straw right over here I draw one ray right over here. I'm gonna take an arbitrary direction. Like that Now when it goes from the slower medium to the faster medium, what's going to happen to it? Until this light go here so the left side of the ray is going to end up in the air before the right side and I'm using the car example to think about which way this light's going to bend The left side of the marching band is gonna get out before the right side and start moving faster So this is going to turn to the right Let me do another ray Let the ray come from the same point Right along the straw, so another ray just like that It will also turn to the right Now if someone's eye is right over here-- Draw their nose and all the rest If they're looking down where does it look like this 2 light rays? Let's say his eye's big enough to capture both of these rays Where does it look like this 2 light rays are coming from? So if you trace both of these rays back if you just assume that there's a line here--that's what our eyes and brains do-- if you assume whatever direction this ray is currently going it's the direction it came from" + }, + { + "Q": "\nat 1:02, why is refractive index useful and what is it used to measure", + "A": "Refractive index is used to measure how much the speed of light travels faster in a medium than that of air. For example, refractive index of diamond is 2.41. So, in air, the light travels 2.41 times faster than in diamond. I hope this helped!", + "video_name": "jxptCXHLxKQ", + "timestamps": [ + 62 + ], + "3min_transcript": "Before doing more examples with Snell's Law which essentially amounts to math problems what I do is give you an intuitive understanding for why this straw looks bent in this picture right over here To do that, let me just do a simplified version of that picture This is the side profile of the cup, or glass right over here The best I can draw it And then let me draw the actual straw. I'll first draw the straw where it actually is coming in off the side of the cup and the straw is actually not bending goes to the bottom of the cup just like that and then it goes up like that and then it goes slightly above. Then it actually does bent up here It's irrelevant to what we want to talk about What I want to do in this video is talk about when we look over here why does it look like the straw got bent? As the light from the straw down here changes as it go from one medium to another Now we know from refraction indices or just in general the light moves slower in water than it does in air slower in water; faster in air Let's think about what's going to happen Let me draw 2 rays that are coming from this point on the straw right over here I draw one ray right over here. I'm gonna take an arbitrary direction. Like that Now when it goes from the slower medium to the faster medium, what's going to happen to it? Until this light go here so the left side of the ray is going to end up in the air before the right side and I'm using the car example to think about which way this light's going to bend The left side of the marching band is gonna get out before the right side and start moving faster So this is going to turn to the right Let me do another ray Let the ray come from the same point Right along the straw, so another ray just like that It will also turn to the right Now if someone's eye is right over here-- Draw their nose and all the rest If they're looking down where does it look like this 2 light rays? Let's say his eye's big enough to capture both of these rays Where does it look like this 2 light rays are coming from? So if you trace both of these rays back if you just assume that there's a line here--that's what our eyes and brains do-- if you assume whatever direction this ray is currently going it's the direction it came from" + }, + { + "Q": "4:08 I think this is hept-2,4-diene.\n", + "A": "ya it is", + "video_name": "KWv5PaoHwPA", + "timestamps": [ + 248 + ], + "3min_transcript": "So this tells us that we have a seven carbon chain that has a double bond starting-- the ene tells us a double bond. Let me write that down. So this double bond right there, that's what the ene tells us. Double bond between two carbons, it's an alkene. The double bond starts-- if you start at this point-- the double bond starts at number two carbon, and then it will go to the number three carbon. Now you might be asking, well, what if I had more than one double bond here? So let me draw a quick example of that. Let's say I have something like, one, two, three, four, five, six, seven. So this is the same molecule again. One, two, three, four, five, six, seven. The way we drew it up here, it would look something like this. What if I had another double bond sitting right here? How would we specify this? One, two, three, four, five, six, seven. So we're still going to have a hept here. It's still going to be an alkene, so we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three, so it would be hept-2. And we also have another double bond starting from four and going to five, so hept-2,4-ene. That's what this molecule right there is. Sometimes, this is the-- I guess-- proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene, they'll write that as 2-heptene, probably because it's easier to say 2-heptene. And from this, you would be able to draw this thing over here, so it's giving you the same amount of information. Similarly over here, they might say 2,4-heptene. But this is the specific, this is the correct It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon." + }, + { + "Q": "\nAt 3:30 why is the answer hepta-2,4-diene and not hept-2,4-diene. Where does the letter a come from? Does the letter a indicate something or is it just spelling?", + "A": "It s just how you name molecules with multiple double bonds. Also makes it flow a bit better off the tongue.", + "video_name": "KWv5PaoHwPA", + "timestamps": [ + 210 + ], + "3min_transcript": "So this tells us right here that we're dealing with an alkene, not an alkane. If you have a double bond, it's an alkene. Triple bond, alkyne. We'll talk about that in future videos. This is hept, and we'll put an ene here, but we haven't specified where the double bond is and we haven't numbered our carbons. When you see an alkene like this, you start numbering closest to the double bond, just like as if it was a alkyl group, as if it was a side chain of carbons. So this side is closest to the double bond, so let's start numbering there. One, two, three, four, five, six, seven. The double bond is between two and three, and to specify its location, you start at the lowest of these numbers. So this double bond is at two. So this tells us that we have a seven carbon chain that has a double bond starting-- the ene tells us a double bond. Let me write that down. So this double bond right there, that's what the ene tells us. Double bond between two carbons, it's an alkene. The double bond starts-- if you start at this point-- the double bond starts at number two carbon, and then it will go to the number three carbon. Now you might be asking, well, what if I had more than one double bond here? So let me draw a quick example of that. Let's say I have something like, one, two, three, four, five, six, seven. So this is the same molecule again. One, two, three, four, five, six, seven. The way we drew it up here, it would look something like this. What if I had another double bond sitting right here? How would we specify this? One, two, three, four, five, six, seven. So we're still going to have a hept here. It's still going to be an alkene, so we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three, so it would be hept-2. And we also have another double bond starting from four and going to five, so hept-2,4-ene. That's what this molecule right there is. Sometimes, this is the-- I guess-- proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene, they'll write that as 2-heptene, probably because it's easier to say 2-heptene. And from this, you would be able to draw this thing over here, so it's giving you the same amount of information. Similarly over here, they might say 2,4-heptene. But this is the specific, this is the correct" + }, + { + "Q": "at \"4:17\", why did the first C of the chain named H3C and why not CH3?\n", + "A": "It was written as H3C to highlight that it is the carbon of the methyl group (CH3) that bonds the the neighboring carbon in the chain, not a hydrogen. The methyl group can be written as either H3C or CH3, it does not matter; it s just clearer if written as H3C in this case. H3C is usually used at the ends of molecules.", + "video_name": "KWv5PaoHwPA", + "timestamps": [ + 257 + ], + "3min_transcript": "So this tells us that we have a seven carbon chain that has a double bond starting-- the ene tells us a double bond. Let me write that down. So this double bond right there, that's what the ene tells us. Double bond between two carbons, it's an alkene. The double bond starts-- if you start at this point-- the double bond starts at number two carbon, and then it will go to the number three carbon. Now you might be asking, well, what if I had more than one double bond here? So let me draw a quick example of that. Let's say I have something like, one, two, three, four, five, six, seven. So this is the same molecule again. One, two, three, four, five, six, seven. The way we drew it up here, it would look something like this. What if I had another double bond sitting right here? How would we specify this? One, two, three, four, five, six, seven. So we're still going to have a hept here. It's still going to be an alkene, so we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three, so it would be hept-2. And we also have another double bond starting from four and going to five, so hept-2,4-ene. That's what this molecule right there is. Sometimes, this is the-- I guess-- proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene, they'll write that as 2-heptene, probably because it's easier to say 2-heptene. And from this, you would be able to draw this thing over here, so it's giving you the same amount of information. Similarly over here, they might say 2,4-heptene. But this is the specific, this is the correct It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon." + }, + { + "Q": "Brother Sal, I don't understand this> At 4:35 u said that \"double bond will take presidence....\". What does it mean?\n", + "A": "well answered =]", + "video_name": "KWv5PaoHwPA", + "timestamps": [ + 275 + ], + "3min_transcript": "One, two, three, four, five, six, seven. So we're still going to have a hept here. It's still going to be an alkene, so we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three, so it would be hept-2. And we also have another double bond starting from four and going to five, so hept-2,4-ene. That's what this molecule right there is. Sometimes, this is the-- I guess-- proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene, they'll write that as 2-heptene, probably because it's easier to say 2-heptene. And from this, you would be able to draw this thing over here, so it's giving you the same amount of information. Similarly over here, they might say 2,4-heptene. But this is the specific, this is the correct It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon. We have a double bond starting from the second carbon to the third carbon. So this thing right here, this double bond from the second carbon to the third carbon. So it's hept-2,3-ene-- sorry, not 2,3, 2-ene. You don't write both endpoints. If there was a three, then there would have been another double bond there. It's hept-2-ene. And then we have this methyl group here, which is also sitting on the second carbon. So this methyl group right there on the second carbon. So we would say 2-methyl-hept-2-ene. It's a hept-2-ene, that's all of this part over here, double bonds starting on the two if we're numbering from the right. And then the methyl group is also attached to that second carbon." + }, + { + "Q": "at 6:22, if you are trying to find the largest carbon chain, shouldn't it be 8 if you start counting from the carbon on the far left methyl group? Is it different if its a cycle?\n", + "A": "If there is a ring structure that has more carbons than any chains connected to the ring, then use the number of carbons in the ring as the longest chain. In this case, 6 carbons = hex , and since it is in a ring use the prefix cyclo- . The methyl group then becomes a substituent.", + "video_name": "KWv5PaoHwPA", + "timestamps": [ + 382 + ], + "3min_transcript": "It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon. We have a double bond starting from the second carbon to the third carbon. So this thing right here, this double bond from the second carbon to the third carbon. So it's hept-2,3-ene-- sorry, not 2,3, 2-ene. You don't write both endpoints. If there was a three, then there would have been another double bond there. It's hept-2-ene. And then we have this methyl group here, which is also sitting on the second carbon. So this methyl group right there on the second carbon. So we would say 2-methyl-hept-2-ene. It's a hept-2-ene, that's all of this part over here, double bonds starting on the two if we're numbering from the right. And then the methyl group is also attached to that second carbon. So we have a cycle here, and once again the root is going to be the largest chain or the largest ring here. Our main ring is the largest one, and we have one, two, three, four, five, six, carbon. So we are dealing with hex as our root for kind of the core of our structure. It's in a cycle, so it's going to be cyclohex. So let me write that. So it's going to be cyclohex. But it has a double bond in it. So it's cyclohex ene, cyclohexene. Let me do this in a different color. So we have this double bond here, and that's why we know it is an ene. Now you're probably saying, Hey Sal, how come we didn't have to number where the ene is? So if you only have one double bond in a ring, it's assumed that one end point of the double bond is your 1-carbon." + }, + { + "Q": "\nAt 2:47 Sal says 'if we are stationary relative to the ether right now'. How can earth be stationary relative to the ether at a point in its orbit, if the earth is always moving at 30 km/s.", + "A": "Exactly! That s Sal s point.", + "video_name": "88hs5LcCoX4", + "timestamps": [ + 167 + ], + "3min_transcript": "the Earth is rotating, but not only is it rotating on its own axis, but it's rotating around the sun. So if this is the sun right over here, this is the Earth. The Earth is rotating, and these are all rough figures, the Earth is moving around the Sun at approximately 30 kilometers per second. 30 kilometers per second! By our everyday standards, that's quite fast, but we're not done yet. 'Cuz the Sun is also moving around the center of the galaxy. And this isn't an actual picture of the Milky Way; obviously we haven't gotten this far from our own galaxy to actually get this type of a vantage point, but if the Sun were right over there, the Sun, estimates are, are moving with a speed of 200, roughly, 200, let me write that in a better color so you can actually see it, 200 kilometers per second. around the center of the Milky Way, and then the Milky Way itself could be moving. So we don't know our actual, kind of, our orientation relative to the Ether, but we are, we're constantly changing our orientation, we're moving in these orbital patterns. If there is some type of luminiferous Ether, if there is some type of luminiferous Ether, and I'm just gonna draw these lines over here to kind of show our luminiferous Ether, we must be moving relative to it if we orient ourselves just the right way. In fact, the odds of us being stationary relative to the Ether are pretty close to zero, especially if we wait a little. If we're stationary relative to the Ether right now, let's say at this point, since we're changing our direction, we're not going to be stationary relative to the Ether at that point. And that's just when you consider the Earth's orbit around the Sun. It's even more true when you think about the solar system's orbit around the center of the galaxy, So, we should be moving relative to the Ether, or the Ether should be moving relative to us. So we should be able to detect some type of, some type of what's called an \"Ether wind\". 'Cuz it should be moving relative... Ether wind. Now how would you detect an Ether wind? Well, let's think about some other type of medium moving relative to us. Let's say that we are sitting on an island, let me do this in a better color for an island. So let's say that we're on an island that's in the middle of a stream. So these are the shores of the stream. These are the shores of the stream. And there is some type of a current. So the water is moving in that direction. So that's the medium. And now let's start a wave propagating through this. So if I were to just take a pebble and drop it" + }, + { + "Q": "\n1:30 why Na+ cant react with water ?", + "A": "Na+ is a cation, having a positive charge. Needing a negative charge from water, it could potentially react with either H+ or OH-. H+ doesn t work since it has a positive charge. OH- does not work because the supposedly formed substance would be NaOH, a strong base. Strong bases are not formed since they dissociate to near completion.", + "video_name": "HwkEQfsJenk", + "timestamps": [ + 90 + ], + "3min_transcript": "- Let's say we have some hydrochloric acid, and a solution of sodium hydroxide. We know that hydrochloric acid is a strong acid so we can think about it as consisting of H+ and Cl-. Sodium hydroxide is a strong base, so in solution we're going to have sodium ions and hydroxide anions. Let's think about the products for this reaction. One product would be H+, and OH-. If you put H+ and OH- together, you form H20. So water is one of our products, and the other product would be what's leftover. We have Na+ and Cl-, so that gives us NaCl which is sodium chloride. This is an example of an acid base neutralization reaction where an acid reacts with the base to give you water and a salt. In this case, our salt is sodium chloride. Let's think about an aqueous solution of sodium chloride. and you dissolve your sodium chloride in water to make your solution. In solution, you're gonna have sodium cations and chloride anions. Let's think about what those would do with water. We know that the pH of water is seven. The pH of water is equal to seven. Sodium ions don't react with water so they're not going to affect the pH of our solution. You might think that the chloride anion could function as a weak base, and take a proton from water. However, that's not really gonna happen very well because the chloride anion, Cl-, this is the conjugate base to HCl. We know that HCl is a strong acid, and the stronger the acid, the weaker the conjugate base. With a very strong acid, we're gonna get a very weak conjugate base from water very well so the pH is unaffected. The pH of our solution of sodium chloride is equal to seven. When you have a salt that was formed from a strong acid and a strong base, so sodium chloride was formed from a strong acid and a strong base, these salts form neutral solutions. So your pH should be equal to seven. Let's compare this situation to the salt that's formed from a weak acid, and a strong base. Over here we have acetic acid which we know is a weak acid. Then we have sodium hydroxide which is our strong base. In solution, we would have Na+ and OH-. Hydroxide is going to take the acidic proton on acetic acid, and this is the acidic proton on acetic acid. Once again, H+ and OH- give us H20." + }, + { + "Q": "\nAt 1:18 , how is pH related to the formation of salts?", + "A": "Salt of strong acid + strong base \u00e2\u0086\u0092 pH = 7. Salt of strong acid + weak base \u00e2\u0086\u0092 pH < 7. Salt of weak acid + strong base \u00e2\u0086\u0092 pH > 7. Salt of weak acid + weak base \u00e2\u0086\u0092 pH depends on which is stronger acid or base.", + "video_name": "HwkEQfsJenk", + "timestamps": [ + 78 + ], + "3min_transcript": "- Let's say we have some hydrochloric acid, and a solution of sodium hydroxide. We know that hydrochloric acid is a strong acid so we can think about it as consisting of H+ and Cl-. Sodium hydroxide is a strong base, so in solution we're going to have sodium ions and hydroxide anions. Let's think about the products for this reaction. One product would be H+, and OH-. If you put H+ and OH- together, you form H20. So water is one of our products, and the other product would be what's leftover. We have Na+ and Cl-, so that gives us NaCl which is sodium chloride. This is an example of an acid base neutralization reaction where an acid reacts with the base to give you water and a salt. In this case, our salt is sodium chloride. Let's think about an aqueous solution of sodium chloride. and you dissolve your sodium chloride in water to make your solution. In solution, you're gonna have sodium cations and chloride anions. Let's think about what those would do with water. We know that the pH of water is seven. The pH of water is equal to seven. Sodium ions don't react with water so they're not going to affect the pH of our solution. You might think that the chloride anion could function as a weak base, and take a proton from water. However, that's not really gonna happen very well because the chloride anion, Cl-, this is the conjugate base to HCl. We know that HCl is a strong acid, and the stronger the acid, the weaker the conjugate base. With a very strong acid, we're gonna get a very weak conjugate base from water very well so the pH is unaffected. The pH of our solution of sodium chloride is equal to seven. When you have a salt that was formed from a strong acid and a strong base, so sodium chloride was formed from a strong acid and a strong base, these salts form neutral solutions. So your pH should be equal to seven. Let's compare this situation to the salt that's formed from a weak acid, and a strong base. Over here we have acetic acid which we know is a weak acid. Then we have sodium hydroxide which is our strong base. In solution, we would have Na+ and OH-. Hydroxide is going to take the acidic proton on acetic acid, and this is the acidic proton on acetic acid. Once again, H+ and OH- give us H20." + }, + { + "Q": "at 1:25 what is an oxidation state?\n", + "A": "An oxidation state is a tool to write the electronic charge(s) of a molecule without resorting to partial charges (e.g. 1/2-).", + "video_name": "M7PnxSQedkM", + "timestamps": [ + 85 + ], + "3min_transcript": "Now that we know a little bit about oxidation and reduction, what I want to do is really just do an exercise to just make sure that we can at least give our best shot at figuring out the oxidation states for the constituent atoms that make up a compound. So, for example, here I have magnesium oxide, which is used in cement. It has other applications. And this is magnesium hydroxide, which is actually used in antacids. It's used in deodorant. And what I want you to think about, and I encourage you to pause this video right now, is given these two molecules, these two compounds, and what we know about the periodic table, try to come up with the oxidation states for the different elements in each of these compounds. So I'm assuming that you've given a go at it. Now let's try to work through this or think through this together. So first of all, magnesium. Magnesium right over here. We see it's group two. It's an alkaline earth metal. It's not that electronegative. We've already seen that something in this group right over here with two valence electrons, it's likely to give them away. So if it were to form ionic bonds, or if it were to be ionized, it's likely to lose two electrons. If you lose two electrons, you would have a plus 2 charge. So magnesium would typically have a plus 2 oxidation state. On the other side of the periodic table, oxygen, group seven. It has six valence electrons. It's very electronegative, so electronegative that oxidation is named for it. It likes to take electrons from other elements. And oxygen in particular likes to take two electrons. So it's not unusual to see, actually anything in this group, but especially oxygen, taking two electrons from something else. If you take two electrons, and you started off neutrally, or you started in a neutral state, it's not unusual to see oxygen at a negative 2 oxidation state. So given that, it seems like this could work out. And actually when you write it as a superscript here, the convention is to write the positive after the 2. And oxygen would have or could have a negative 2 oxidation state. And this makes sense relative to the overall charge of the molecule. Positive 2 plus negative 2 is going to be 0. And that makes sense. This thing overall is a neutral molecule. And not only in this case is the oxidation state a hypothetical ionic charge, if these were to be ionic bonds, this actually is an ionic compound. Oxygen actually does take two electrons. And magnesium actually does give away two electrons. So in this case the oxidation state is actually describing what is happening ionically. Now let's think about this one right over here, magnesium hydroxide. Well, just like before, magnesium typically has" + }, + { + "Q": "\nAt 2:27 what is the reducing agent?", + "A": "a substance that tends to bring about reduction by being oxidized and losing electrons.", + "video_name": "M7PnxSQedkM", + "timestamps": [ + 147 + ], + "3min_transcript": "Now that we know a little bit about oxidation and reduction, what I want to do is really just do an exercise to just make sure that we can at least give our best shot at figuring out the oxidation states for the constituent atoms that make up a compound. So, for example, here I have magnesium oxide, which is used in cement. It has other applications. And this is magnesium hydroxide, which is actually used in antacids. It's used in deodorant. And what I want you to think about, and I encourage you to pause this video right now, is given these two molecules, these two compounds, and what we know about the periodic table, try to come up with the oxidation states for the different elements in each of these compounds. So I'm assuming that you've given a go at it. Now let's try to work through this or think through this together. So first of all, magnesium. Magnesium right over here. We see it's group two. It's an alkaline earth metal. It's not that electronegative. We've already seen that something in this group right over here with two valence electrons, it's likely to give them away. So if it were to form ionic bonds, or if it were to be ionized, it's likely to lose two electrons. If you lose two electrons, you would have a plus 2 charge. So magnesium would typically have a plus 2 oxidation state. On the other side of the periodic table, oxygen, group seven. It has six valence electrons. It's very electronegative, so electronegative that oxidation is named for it. It likes to take electrons from other elements. And oxygen in particular likes to take two electrons. So it's not unusual to see, actually anything in this group, but especially oxygen, taking two electrons from something else. If you take two electrons, and you started off neutrally, or you started in a neutral state, it's not unusual to see oxygen at a negative 2 oxidation state. So given that, it seems like this could work out. And actually when you write it as a superscript here, the convention is to write the positive after the 2. And oxygen would have or could have a negative 2 oxidation state. And this makes sense relative to the overall charge of the molecule. Positive 2 plus negative 2 is going to be 0. And that makes sense. This thing overall is a neutral molecule. And not only in this case is the oxidation state a hypothetical ionic charge, if these were to be ionic bonds, this actually is an ionic compound. Oxygen actually does take two electrons. And magnesium actually does give away two electrons. So in this case the oxidation state is actually describing what is happening ionically. Now let's think about this one right over here, magnesium hydroxide. Well, just like before, magnesium typically has" + }, + { + "Q": "3:36 so if the 2 was positive it would not be neutral ?\n", + "A": "Right it would be positive, and vice versa. But in both of these compounds the positive and negative charges cancel each other out.", + "video_name": "M7PnxSQedkM", + "timestamps": [ + 216 + ], + "3min_transcript": "And actually when you write it as a superscript here, the convention is to write the positive after the 2. And oxygen would have or could have a negative 2 oxidation state. And this makes sense relative to the overall charge of the molecule. Positive 2 plus negative 2 is going to be 0. And that makes sense. This thing overall is a neutral molecule. And not only in this case is the oxidation state a hypothetical ionic charge, if these were to be ionic bonds, this actually is an ionic compound. Oxygen actually does take two electrons. And magnesium actually does give away two electrons. So in this case the oxidation state is actually describing what is happening ionically. Now let's think about this one right over here, magnesium hydroxide. Well, just like before, magnesium typically has So it could have an oxidation state of positive 2, which would imply that the entire hydroxide anion-- And let's just say hydroxide for now. Well I'll say hydroxide anion. I kind of gave it away a little bit-- that this hydroxide, or this part of the molecule, the right-hand part of what I've written here, for this whole thing to be neutral, it should have a negative 2 oxidation state. Now how does that make sense? Well we have two hydroxides here. Notice this subscript right over here. So if each of those hydroxides has a negative 1 charge, or a negative 1, I guess you could say, total oxidation state, then when you take two of them together, they would net out against the magnesium. And that does seem to make sense. If oxygen has a negative 2 oxidation state, hydrogen has a positive 1 oxidation state. is going to have a net oxidation state of negative 1. But then you have two of them. So the net oxidation for this part of the molecule or the compound is going to be negative 2 nets out with the positive 2 from magnesium. So once again, it makes sense." + }, + { + "Q": "At 1:19,why does magnesium loose two electrons?\n", + "A": "It loses two electrons because it strives as any element to become stable, which is to say to have either 8 or 2 electrons on its outer shell. So to become stable, it only needs to lose the 2 electrons on its outer shell for the shell under the initial shell to become the outer shell.", + "video_name": "M7PnxSQedkM", + "timestamps": [ + 79 + ], + "3min_transcript": "Now that we know a little bit about oxidation and reduction, what I want to do is really just do an exercise to just make sure that we can at least give our best shot at figuring out the oxidation states for the constituent atoms that make up a compound. So, for example, here I have magnesium oxide, which is used in cement. It has other applications. And this is magnesium hydroxide, which is actually used in antacids. It's used in deodorant. And what I want you to think about, and I encourage you to pause this video right now, is given these two molecules, these two compounds, and what we know about the periodic table, try to come up with the oxidation states for the different elements in each of these compounds. So I'm assuming that you've given a go at it. Now let's try to work through this or think through this together. So first of all, magnesium. Magnesium right over here. We see it's group two. It's an alkaline earth metal. It's not that electronegative. We've already seen that something in this group right over here with two valence electrons, it's likely to give them away. So if it were to form ionic bonds, or if it were to be ionized, it's likely to lose two electrons. If you lose two electrons, you would have a plus 2 charge. So magnesium would typically have a plus 2 oxidation state. On the other side of the periodic table, oxygen, group seven. It has six valence electrons. It's very electronegative, so electronegative that oxidation is named for it. It likes to take electrons from other elements. And oxygen in particular likes to take two electrons. So it's not unusual to see, actually anything in this group, but especially oxygen, taking two electrons from something else. If you take two electrons, and you started off neutrally, or you started in a neutral state, it's not unusual to see oxygen at a negative 2 oxidation state. So given that, it seems like this could work out. And actually when you write it as a superscript here, the convention is to write the positive after the 2. And oxygen would have or could have a negative 2 oxidation state. And this makes sense relative to the overall charge of the molecule. Positive 2 plus negative 2 is going to be 0. And that makes sense. This thing overall is a neutral molecule. And not only in this case is the oxidation state a hypothetical ionic charge, if these were to be ionic bonds, this actually is an ionic compound. Oxygen actually does take two electrons. And magnesium actually does give away two electrons. So in this case the oxidation state is actually describing what is happening ionically. Now let's think about this one right over here, magnesium hydroxide. Well, just like before, magnesium typically has" + }, + { + "Q": "9:21 Why aren't children vaccinated against meningococcus earlier than the age o two?\n\nEdit: I found an answer to this question in an Infectious diseases textbook. Apparently the immune system of children less then two years old is not mature enough to respond to a polysacharide antigen of N. meningitidis, thus vaccination before the age of two would be futile.\n", + "A": "I disagree with Amelia on one point. Babies are given a Heb B vaccine at birth. I d assume that there would be just as much of a stigma regarding that. Also, babies are now given a whole slew of vaccines at age 2 months old. If it were safe and effective in neonates, it would be on the schedule starting at 2 months.", + "video_name": "CnXuSCaCNBo", + "timestamps": [ + 561 + ], + "3min_transcript": "And then, most importantly, the fluid is examined with something called a Gram stain, a special kind of stain. It's Gram with a capital G, named after Dr. Gram. And a Gram stain can determine whether or not there are bacteria present. MORGAN THEIS: OK. So you're actually staining the bacteria. DR. CHARLES PROBER: Exactly. And if there are sufficient bacteria present, the Gram stain will reveal those bacteria. And so with bacterial meningitis, the second prudent principle is to know the usual pathogens. So if a spinal fluid is obtained, there are lots of white cells, the glucose is low. Even with a negative Gram stain, one can guess the usual pathogens, because the list is short in normal children. And those bacteria, the short list includes a bacteria called Haemophilus influenzae, type B. I mean, an E at the end? DR. CHARLES PROBER: There is an E on the end. A second bacteria is the new pnuemococcus. MORGAN THEIS: Now, that's funny. It sounds like it causes pneumonia. DR. CHARLES PROBER: And it does indeed cause pneumonia as well. But it also causes bacterial meningitis. And a third bacteria is called meningicoccus. And those are the prominent bacteria in normal children with bacterial meningitis. The reason though, we're not seen as much bacterial meningitis in 2011 as we were seeing 10 and 20 years ago is, we now have vaccination against each of those three different pathogens. MORGAN THEIS: All of them? DR. CHARLES PROBER: We do. We vaccinate against Haemophilus influenzae, type B, starting at two months of age. And by the time the child is about a year and a half, they're completely protected against The pneumococcus, we also vaccinate against. And it's very successful. The vaccine is very successful at reducing the frequency of pneumococcal meningitis. It also starts at two months of age. And meningicoccus, the vaccine is relatively new and is used in children who are a bit older. They're over two years of age, under special circumstances. So that means that we still can see, and do see, cases of meningicoccal meningitis, because it occurs in children under two years of age. MORGAN THEIS: I see. DR. CHARLES PROBER: But those are the usual pathogens. And when you go to other parts of the world who don't use vaccines, those are the pathogens that will be prominent in causing bacterial meningitis. And knowing those pathogens, we go to the third principle of antibiotic prescribing, which is knowing what antibiotics typically kill those bacteria. MORGAN THEIS: OK. So what should I call that category, like matching?" + }, + { + "Q": "I've noted that Jay writes ionic compounds like this: Na+-OH (7:32)\nIs this the standard way of writing it? I'm used to seeing Na+OH-, with the sign after the anion... (both of course in superscript).\n", + "A": "see the things are same........but specifically in the hydroxide anion , due to the more electronegative character of the oxygen atom the bonded pair of electrons in the molecule rest near the oxygen atom.....hence jay has made the negative charge over the oxygen atom...", + "video_name": "My5SpT9E37c", + "timestamps": [ + 452 + ], + "3min_transcript": "So we are gonna use DMSO. And we know in an SN2 mechanism the nucleophile attacks our alkyl halide at the same time our leaving group leaves. So our nucleophile is the hydroxide ion. It is going to attack this carbon and these electrons are gonna come off on to the bromide to form our bromide anion. So our OH replaces our bromine and we can see that over here in our product. In an SN2 mechanism we need a strong nucleophile to attack our alkyl halide. And DMSO is gonna help us increase the effectiveness of our nucleophile which is our hydroxide ion. So let's look at some pictures of how it helps us. So we have sodium hydroxide here. So first let's focus in on the sodium, our cation. So here is the sodium cation. DMSO is a good solvator of cations and that's because oxygen has a partial negative charge. The sulfur has a partial positive charge help to stabilize the positive charge on our sodium. So same thing over here. Partial negative, partial positive and again we are able to solvate our cation. So the fact that our polar aprotic solvent is a good solvator of a cation means we can separate this ion from our nucleophile. That increases the effectiveness of the hydroxide ion. The hydroxide ion itself is not solvated by a polar aprotic solvent. So you might think, okay well if the oxygen is partially negative and the sulfur is partially positive. The partially positive sulfur could interact with our negatively charged nucleophile. But remember we have these bulky methyl groups here. And because of steric hindrance that prevents our hydroxide ion from interacting with DMSO. So the hydroxide ion is all by itself which of course increases its effectiveness as a nucleophile. If we had used something like water, we know that water is a polar protic solvent with the oxygen being partially negative and the hydrogens being partially positive and a polar protic solvent would interact with our nucleophile solvating it and essentially decreasing the effectiveness of our nucleophile. So that's why polar protic solvents don't work as well if you want an SN2 mechanism. A polar aprotic solvent increases the effectiveness of our nucleophile therefore favoring our SN2 mechanism." + }, + { + "Q": "\nat 10:06 what does he mean when he says increases the strength of the nucleophile?", + "A": "In a polar protic solvent, the nucleophile is solvated by the solvent molecules. The nucleophile has to push these solvating molecules out of the way in order to attack the \u00ce\u00b1 carbon. A polar aprotic solvent preferentially solvates the metal cation. That means that the nucleophile is relatively unsolvated or \u00e2\u0080\u009cbare\u00e2\u0080\u009d. It can then much more easily attack the \u00ce\u00b1 carbon. This makes it a stronger nucleophile.", + "video_name": "My5SpT9E37c", + "timestamps": [ + 606 + ], + "3min_transcript": "If we had used something like water, we know that water is a polar protic solvent with the oxygen being partially negative and the hydrogens being partially positive and a polar protic solvent would interact with our nucleophile solvating it and essentially decreasing the effectiveness of our nucleophile. So that's why polar protic solvents don't work as well if you want an SN2 mechanism. A polar aprotic solvent increases the effectiveness of our nucleophile therefore favoring our SN2 mechanism." + }, + { + "Q": "At 4:43 it shows water molecules moving through protein channels. I am aware that the cell membrane does not allow watersoluble/hydrophillic/lipidphobic substances to pass straight through but I thought small molecules such as oxygen gas, carbon dioxide and water were able to pass through easily because of their size, without needing to move through channel proteins. Is this incorrect? Thanks\n", + "A": "Channel proteins allow passive transport of the small molecules like oxygen gas and water, meaning these molecules can flow through from the higher concentration to the lower one without using energy. This is what the video means by easily passing through the membrane, since no ATP is required. The water and oxygen cannot pass through anywhere they want, otherwise the membrane would be so full of holes that it would be useless! Hope this helps!", + "video_name": "TyZODv-UqvU", + "timestamps": [ + 283 + ], + "3min_transcript": "We want the in-the-middle Charlie Sheen who can just make us laugh and be happy; and that is the state that water concentrations are constantly seeking, it's called isotonic when the concentration is the same on both sides, outside and in. And this works in real life, we can actually show it to you. This vase if full of fresh water, and we also have a sausage casing, which is actually made out of cellulose, and inside of that we have salt water. We've died it so that you can see it move through the casing, which is acting as our membrane. This time lapse shows how over a few hours, the salt water diffuses into the pure water. It'll keep diffusing until the concentration of salt in the water is the same inside the membrane as outside. When water does this, attempting to become isotonic, it's called moving across its concentration gradient. Most of my cells right now are bathed in a solution that has the same concentration as inside of them, and this is important. For example, if you took one of my red blood cells and put in a glass of pure water, it would be so hypertonic, so much stuff would be in the cell compared to outside the cell, that water would rush into the red blood cell, then it would literally explode. But if the concentration of my blood plasma were too high, all the water would rush out of my cell and it would shrivel up and be useless. And that's why your kidneys are constantly on the job regulating the concentration of your blood plasma to keep it isotonic. No, water can permeate a cell membrane without any help, but it's not actually particularly easy. As we discussed in the last episode, cell membranes are made out of phospholipids; and the phospholipid bilayer is hydrophilic, or water-loving on the outside, and hydrophobic, or water-hating, on the inside. So, water molecules have a hard time passing through these layers because they get stuck at the non-polar hydrophobic core. That is where the channel proteins come in. They allow passage of stuff like water and ions without using any energy. They straddle the width of the membrane, and inside they have channels that are hydrophilic, which draws the water through. The proteins that are specifically for channeling water are called aquaporins. Each one can pass 3 billion water molecules a second. Make me have to pee just thinking about it. Things like oxygen and water that cells need constantly, but most chemicals, they use what's called active transport. This is especially useful if you want to move something in the opposite direction of it concentration gradient from low concentration to a high concentration. So, say we're back at that show, and I'm keeping company with John, who's being all anti-social in his polite and charming way, but after half a beer and an argument about who's the best Doctor Who, I want to get back to my friends across the crowded bar. So, I transport myself against the concentration gradient of humans, spending a lot of energy dodging stomping feet, throwing an elbow to get to them. That is high energy transport. In a cell, getting the energy necessary to do pretty much anything, including moving something the wrong direction across its concentration gradient, requires ATP. ATP, adenosine tri-phosphate. You just wanna replay that over and over again, until it just rolls off the tongue because it's one of the most important chemicals that you will ever, ever, ever hear about." + }, + { + "Q": "Sorry I'm confused because at 4:00 it was mentioned, I thought Hypotonic was when water would rush in and explode while hypertonic was when it would shrivel up.\n", + "A": "That is correct - however, this terminology is very tricky. See the example below: You have a situation with less water in cell, more in envirnoment The cell is hypertonic to the solution, and the solution is hypotonic to the cell. The cell is in a hypotonic solution.", + "video_name": "TyZODv-UqvU", + "timestamps": [ + 240 + ], + "3min_transcript": "until it was a uniform mass of John Greens throughout the club. When oxygen gets crowded, it finds places that are less crowded and moves into those spaces. When water gets crowded, it does the same thing. It moves to where there is less water. When water goes across a membrane, it's a kind of diffusion called osmosis. This is how your cells regulate their water content. Not only does this apply to water itself, which as we've discussed is the world's best solvent, you're gonna learn more about water in our water episode, it also works with water that contains dissolved materials or solutions, like solutions of salt water or solutions of sugar water, or booze, which is just a solution of ethanol and water. If the concentration of a solution is higher inside of a cell than it is outside of the cell, then that solution is called hypyertonic, like power thirst, it's got everything packed into it. And if the concentration inside of the cell is lower than outside of the cell, it's called hypotonic. Which is sort of a sad version of hypertonic. So, like with Charlie Sheen, we don't want the crazy, manic Charlie Sheen, We want the in-the-middle Charlie Sheen who can just make us laugh and be happy; and that is the state that water concentrations are constantly seeking, it's called isotonic when the concentration is the same on both sides, outside and in. And this works in real life, we can actually show it to you. This vase if full of fresh water, and we also have a sausage casing, which is actually made out of cellulose, and inside of that we have salt water. We've died it so that you can see it move through the casing, which is acting as our membrane. This time lapse shows how over a few hours, the salt water diffuses into the pure water. It'll keep diffusing until the concentration of salt in the water is the same inside the membrane as outside. When water does this, attempting to become isotonic, it's called moving across its concentration gradient. Most of my cells right now are bathed in a solution that has the same concentration as inside of them, and this is important. For example, if you took one of my red blood cells and put in a glass of pure water, it would be so hypertonic, so much stuff would be in the cell compared to outside the cell, that water would rush into the red blood cell, then it would literally explode. But if the concentration of my blood plasma were too high, all the water would rush out of my cell and it would shrivel up and be useless. And that's why your kidneys are constantly on the job regulating the concentration of your blood plasma to keep it isotonic. No, water can permeate a cell membrane without any help, but it's not actually particularly easy. As we discussed in the last episode, cell membranes are made out of phospholipids; and the phospholipid bilayer is hydrophilic, or water-loving on the outside, and hydrophobic, or water-hating, on the inside. So, water molecules have a hard time passing through these layers because they get stuck at the non-polar hydrophobic core. That is where the channel proteins come in. They allow passage of stuff like water and ions without using any energy. They straddle the width of the membrane, and inside they have channels that are hydrophilic, which draws the water through. The proteins that are specifically for channeling water are called aquaporins. Each one can pass 3 billion water molecules a second. Make me have to pee just thinking about it. Things like oxygen and water that cells need constantly," + }, + { + "Q": "\nat 8:34 Sal said that if we cut a magnet into half then two more magnets will appear. But notice that the left side (from the viewer's point of view) of the 1st magnet is north pole but the left side of the second magnet is south. But why?", + "A": "because a magnet is made of a bunch of smaller magnets that stick together, north to south and south to north", + "video_name": "8Y4JSp5U82I", + "timestamps": [ + 514 + ], + "3min_transcript": "something in that field that can be affected by it, it'll be some net force acting on it. So actually, before I go into magnetic field, I actually want to make one huge distinction between magnetism and electrostatics. Magnetism always comes in the form of a dipole. It means that we have two poles. A north and a south. In electrostatics, you do have two charges. You have a positive charge and a negative charge. So you do have two charges. But they could be by themselves. You could just have a proton. You don't have to have an electron there right next to it. You could just have a proton and it would create a positive electrostatic field. And our field lines are what a positive point charge would do. And it would be repelled. So you don't always have to have a negative charge there. And you don't have to have a proton there. So you could have monopoles. These are called monopoles, when you just have one charge when you're talking about electrostatics. But with magnetism you always have a dipole. If I were to take this magnet, this one right here, and if I were to cut it in half, somehow miraculously each of those halves of that magnet will turn into two more magnets. Where this will be the south, this'll be the north, this'll be the south, this will be the north. And actually, theoretically, I've read-- my own abilities don't go this far-- there could be such a thing as a magnetic monopole, although it has not been observed yet in nature. So everything we've seen in nature has been a dipole. So you could just keep cutting this up, all the way down to if it's just one electron left. And it actually turns out that even one electron is still a magnetic dipole. It still is generating, it still has a north pole and a And actually it turns out, all magnets, the magnetic field is actually generated by the electrons within it. By the spin of electrons and that-- you know, when we talk about electron spin we imagine some little ball of charge spinning. But electrons are-- you know, it's hard to-- they do have mass. But it starts to get fuzzy whether they are energy or mass. And then how does a ball of energy spin? Et cetera, et cetera. So it gets very almost metaphysical. So I don't want to go too far into it. And frankly, I don't think you really can get an intuition. It is almost-- it is a realm that we don't normally operate in. But even these large magnets you deal with, the magnetic field is generated by the electron spins inside of it and by the actual magnetic fields generated by the electron motion around the protons. Well, I hope I'm not overwhelming you. And you might say, well, how come sometimes a metal bar can be magnetized and sometimes it won't be? Well, when all of the electrons are doing random" + }, + { + "Q": "In 6:39, what is the nearest star? I just want to know.\n", + "A": "Proxima Centauri", + "video_name": "5FEjrStgcF8", + "timestamps": [ + 399 + ], + "3min_transcript": "So if this is the Sun-- and if I were to draw Jupiter, it would look something like-- I'll do Jupiter in pink-- Jupiter would be around that big. And then the Earth would be around that big if you were to put them all next to each other. So the Sun, once again, is huge. Even though we see it almost every day, it is unimaginably huge. Even the Earth is unimaginably huge. And the Sun is 100 times more unimaginably bigger. Now we're going to start getting really, really, really wacky. You multiply the diameter of the Sun, which is already 100 times the diameter of the Earth-- you multiply that times 100. And that is the distance from the Earth to the Sun. So I've drawn the Sun here as a little pixel. And I didn't even draw the Earth as a pixel. Because a pixel would be way too large. It would have to be a hundredth of a pixel in order to draw the Earth properly. So this is a unbelievable distance It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star-- so remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times-- and once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball-- in our part of the galaxy in a volume the size of the Earth-- so if you had a big volume the size of the Earth, if the stars were the sizes of basketballs, in our part of the galaxy, you would only have a handful of basketballs per that volume. So unbelievably sparse. Even though, when you look at the galaxy-- and this is just an artist's depiction of it-- it looks like something that has the spray of stars, and it looks reasonably dense, there is actually a huge amount of space that the great, great, great, great, great majority of the volume in the galaxy is just empty, empty space. There's no stars, no planets, no nothing. I mean, this is a huge jump that I'm talking about. And then if you really want to realize how large a galaxy, itself, can be, you take this distance between the Sun, or between our solar system and the nearest star--" + }, + { + "Q": "I'm not sure where I heard this (probably early elementary school), but I remember learning that the red spot on Jupiter is 3x as big as Earth. If that's true, I feel like Jupiter can't be 10x as big as Earth, like Sal says at 4:12, because the red spot isn't 30 percent of the planet. Am I crazy in remembering that the red spot is the size of 3 Earths?\n", + "A": "Jupiter s circumference is 11 times that of Earths. But that gives it a surface area more than 100x larger. The red spot is about 2 or 3 earth diameters wide.", + "video_name": "5FEjrStgcF8", + "timestamps": [ + 252 + ], + "3min_transcript": "it's starting to get larger than what we're used to processing on a daily basis. A bridge-- we've been on a bridge. We know what a bridge looks like. We know that a bridge is huge. But it doesn't feel like something that we can't comprehend. Already, a city is something that we can't comprehend all at once. We can drive across a city. We can look at satellite imagery. But if I were to show a human on this, it would be unbelievably, unbelievably small. You wouldn't actually be able to see it. It would be less than a pixel on this image. A house is less than a pixel on this image. But let's keep multiplying by 10. If you multiply by 10 again, you get to something roughly the size of the San Francisco Bay Area. This whole square over here is roughly that square right over there. Let's multiply by 10 again. So this square is about 100 miles by 100 miles. So this one would be about 1,000 miles by 1,000 miles. And now you're including a big part of the Western United States. You Nevada here. You have Arizona and New Mexico-- so a big chunk of a big continent we're already including. And frankly, this is beyond the scale that we're used to operating. We've seen maps, so maybe we're a little used to it. But if you ever had to walk across this type of distance, it would take you a while. To some degree, the fact that planes goes so fast-- almost unimaginably fast for us-- that it's made it feel like things like continents aren't as big. Because you can fly across them in five or six hours. But these are already huge, huge, huge distances. But once again, you take this square that's about 1,000 miles by 1,000 miles, and you multiply that by 10. And you get pretty close-- a little bit over-- the diameter of the Earth-- a little bit over the diameter of the Earth. But once again, we're on the Earth. We kind of relate to the Earth. If you look carefully at the horizon, you might see a little bit of a curvature, especially if you were to get into the plane. So even though this is, frankly, larger can kind of relate to the Earth. Now you multiply the diameter of Earth times 10. And you get to the diameter of Jupiter. And so if you were to sit Earth right next to Jupiter-- obviously, they're nowhere near that close. That would destroy both of the planets. Actually, it would definitely destroy Earth. It would probably just be merged into Jupiter. So if you put Earth next to Jupiter, it would look something like that right over there. So I would say that Jupiter is definitely-- on this diagram that I'm drawing here-- is definitely the first thing that I have I can't comprehend. The Earth, itself, is so vastly huge. Jupiter is-- it's 10 times bigger in diameter. It's much larger in terms of mass, and volume, and all the rest. But just in terms of diameter, it is 10 times bigger. But let's keep going. 10 times Jupiter gets us to the sun." + }, + { + "Q": "at 0:15 Sal says that we can't comprehend thing that are small compared to the size of the universe, is he referring to the galaxies?\n", + "A": "An example would be galaxies, other examples also include black holes, superclusters, clusters, et.c", + "video_name": "5FEjrStgcF8", + "timestamps": [ + 15 + ], + "3min_transcript": "The purpose of this video is to just begin to appreciate how vast and enormous the universe is. And frankly, our brains really can't grasp it. What we'll see in this video is that we can't even grasp things that are actually super small compared to the size of the universe. And we actually don't even know what the entire size of the universe is. But with that said, let's actually just try to appreciate how small we are. So this is me right over here. I am 5 foot 9 inches, depending on whether I'm wearing shoes-- maybe 5 foot 10 with shoes. But for the sake of this video, let's just roughly approximate around 6 feet, or around roughly-- I'm not to go into the details of the math-- around 2 meters. Now, if I were to lie down 10 times in a row, you'd get about the length of an 18-wheeler. That's about 60 feet long. So this is times 10. Now, if you were to put an 18-wheeler-- if you were to make it tall, as opposed to long-- somehow stand it up-- and you were to do that 10 times in a row, you'll get to the height of roughly a 60-story skyscraper. you'll get about a 60-story skyscraper. Now, if you took that skyscraper and if you were to lie it down 10 times in a row, you'd get something of the length of the Golden Gate Bridge. And once again, I'm not giving you the exact numbers. It's not always going to be exactly 10. But we're now getting to about something that's a little on the order of a mile long. So the Golden Gate Bridge is actually longer than a mile. But if you go within the twin spans, it's roughly about It's actually a little longer than that. But that gives you a sense of a mile. Now, if you multiply that by 10, you get to the size of a large city. And this right here is a satellite photograph of San Francisco. This is the actual Golden Gate Bridge here. And when I copy and pasted this picture, I tried to make it roughly 10 miles by 10 miles just so you appreciate the scale. And what's interesting here-- and this picture's interesting. Because this is the first time we can relate to cities. it's starting to get larger than what we're used to processing on a daily basis. A bridge-- we've been on a bridge. We know what a bridge looks like. We know that a bridge is huge. But it doesn't feel like something that we can't comprehend. Already, a city is something that we can't comprehend all at once. We can drive across a city. We can look at satellite imagery. But if I were to show a human on this, it would be unbelievably, unbelievably small. You wouldn't actually be able to see it. It would be less than a pixel on this image. A house is less than a pixel on this image. But let's keep multiplying by 10. If you multiply by 10 again, you get to something roughly the size of the San Francisco Bay Area. This whole square over here is roughly that square right over there. Let's multiply by 10 again. So this square is about 100 miles by 100 miles. So this one would be about 1,000 miles by 1,000 miles. And now you're including a big part of the Western United States." + }, + { + "Q": "At 6:56 Sal said that the nearest star is 200,000 times the distance from the earth to the sun; but isn't the sun considered a star?\n", + "A": "Yes. Obviously he meant the nearest star other than the sun, right?", + "video_name": "5FEjrStgcF8", + "timestamps": [ + 416 + ], + "3min_transcript": "So if this is the Sun-- and if I were to draw Jupiter, it would look something like-- I'll do Jupiter in pink-- Jupiter would be around that big. And then the Earth would be around that big if you were to put them all next to each other. So the Sun, once again, is huge. Even though we see it almost every day, it is unimaginably huge. Even the Earth is unimaginably huge. And the Sun is 100 times more unimaginably bigger. Now we're going to start getting really, really, really wacky. You multiply the diameter of the Sun, which is already 100 times the diameter of the Earth-- you multiply that times 100. And that is the distance from the Earth to the Sun. So I've drawn the Sun here as a little pixel. And I didn't even draw the Earth as a pixel. Because a pixel would be way too large. It would have to be a hundredth of a pixel in order to draw the Earth properly. So this is a unbelievable distance It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star-- so remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times-- and once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball-- in our part of the galaxy in a volume the size of the Earth-- so if you had a big volume the size of the Earth, if the stars were the sizes of basketballs, in our part of the galaxy, you would only have a handful of basketballs per that volume. So unbelievably sparse. Even though, when you look at the galaxy-- and this is just an artist's depiction of it-- it looks like something that has the spray of stars, and it looks reasonably dense, there is actually a huge amount of space that the great, great, great, great, great majority of the volume in the galaxy is just empty, empty space. There's no stars, no planets, no nothing. I mean, this is a huge jump that I'm talking about. And then if you really want to realize how large a galaxy, itself, can be, you take this distance between the Sun, or between our solar system and the nearest star--" + }, + { + "Q": "5:37 What does AU stand for?\n", + "A": "It stands for Astronomical Unit, which is the average distance between the Earth and the Sun, about 93 million miles, or 1.5*10^11 meters.", + "video_name": "5FEjrStgcF8", + "timestamps": [ + 337 + ], + "3min_transcript": "can kind of relate to the Earth. Now you multiply the diameter of Earth times 10. And you get to the diameter of Jupiter. And so if you were to sit Earth right next to Jupiter-- obviously, they're nowhere near that close. That would destroy both of the planets. Actually, it would definitely destroy Earth. It would probably just be merged into Jupiter. So if you put Earth next to Jupiter, it would look something like that right over there. So I would say that Jupiter is definitely-- on this diagram that I'm drawing here-- is definitely the first thing that I have I can't comprehend. The Earth, itself, is so vastly huge. Jupiter is-- it's 10 times bigger in diameter. It's much larger in terms of mass, and volume, and all the rest. But just in terms of diameter, it is 10 times bigger. But let's keep going. 10 times Jupiter gets us to the sun. So if this is the Sun-- and if I were to draw Jupiter, it would look something like-- I'll do Jupiter in pink-- Jupiter would be around that big. And then the Earth would be around that big if you were to put them all next to each other. So the Sun, once again, is huge. Even though we see it almost every day, it is unimaginably huge. Even the Earth is unimaginably huge. And the Sun is 100 times more unimaginably bigger. Now we're going to start getting really, really, really wacky. You multiply the diameter of the Sun, which is already 100 times the diameter of the Earth-- you multiply that times 100. And that is the distance from the Earth to the Sun. So I've drawn the Sun here as a little pixel. And I didn't even draw the Earth as a pixel. Because a pixel would be way too large. It would have to be a hundredth of a pixel in order to draw the Earth properly. So this is a unbelievable distance It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star-- so remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times-- and once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball--" + }, + { + "Q": "\nAt around 7:45, the second equation of CO2 plus HbO2 does not seem to be a balanced equation. Where does an extra H+ on the right hand side of the equation come from? According to the reactants, shouldn't it be CO2+HbO2 = HbCOO- + O2 ??", + "A": "Hemoglobin is a protein so it s formed of amino acids , carboxylic acid having amide group -NH2, so in this equation CO2 binds to the terminal amide group forming amide linkage and -NH2 loses H+ _-NH2 + CO2 ------> _-NHCOO- + H+", + "video_name": "QP8ImP6NCk8", + "timestamps": [ + 465 + ], + "3min_transcript": "Now, if it's an acid, try to keep in mind what acids do. Acids are going to kick off a proton. So this becomes HCO3 minus. And it kicks off a proton. And notice that now you've got bicarb and proton on this side. And this bicarb is actually going to just make its way outside. So the bicarb goes outside the cell. And the proton, what it does is, it meets up with one of these oxyhemoglobins. It kind of finds an oxyhemoglobin. Remember, there are millions of them around. And it literally binds to hemoglobin. And it boots off the oxygen. So it binds to hemoglobin and oxygen falls away. So this is interesting because now this is a second reason for why oxygen gets delivered to the tissues. And that is that, protons compete with hemoglobin. So they're competing for hemoglobin. Now I said there is another thing that happens to the carbon dioxide. So what's the other thing? Turns out that carbon dioxide actually sometimes independently seeks out oxyhemoglobin. Remember, again, there are millions of them. So it'll find one. And it'll do the same thing. It'll say, well, hey, hemoglobin, why don't you just come bind with me and get rid of that oxygen? So it also competes with oxygen. So you've got some competition from protons, some competition from carbon dioxide. And when carbon dioxide actually binds, interesting thing is that it makes a proton. So guess what happens? That proton can go and compete again by itself. It can compete with oxyhemoglobin and try to kick off another one, kick off another oxygen. because now you've got a few reasons why you have oxygen delivery. You've got protons competing. You've got now CO2 competing with oxygen. So you've got a couple of sources of competition. And you've got, of course, just simply the fact that there's just not too much oxygen around. So these are reasons for oxygen delivery. So at this point, you've got oxygen that's delivered to the cells. And these hemoglobin molecules, they're still our cell, of course, inside of a red blood cell. And these hemoglobin molecules have now been bound by different things. So they're no longer bound by oxygen. So you can't really call them oxyhemoglobin anymore. Instead they have protons on them like this. And they might have some COO minus on them. So they might have-- actually, let me do that in the original kind of orangey color. So they basically have different things binding to them." + }, + { + "Q": "At 8:01, we see an equation where CO2 bonds with Hb and creates H+. Where does the hydrogen come from? Has it already created carbonic acid by reacting with H2O?\n", + "A": "Yes you re right CO2 already has reacted with water and then formed carbonic acid, from which the hydrogen dissociates..", + "video_name": "QP8ImP6NCk8", + "timestamps": [ + 481 + ], + "3min_transcript": "Now, if it's an acid, try to keep in mind what acids do. Acids are going to kick off a proton. So this becomes HCO3 minus. And it kicks off a proton. And notice that now you've got bicarb and proton on this side. And this bicarb is actually going to just make its way outside. So the bicarb goes outside the cell. And the proton, what it does is, it meets up with one of these oxyhemoglobins. It kind of finds an oxyhemoglobin. Remember, there are millions of them around. And it literally binds to hemoglobin. And it boots off the oxygen. So it binds to hemoglobin and oxygen falls away. So this is interesting because now this is a second reason for why oxygen gets delivered to the tissues. And that is that, protons compete with hemoglobin. So they're competing for hemoglobin. Now I said there is another thing that happens to the carbon dioxide. So what's the other thing? Turns out that carbon dioxide actually sometimes independently seeks out oxyhemoglobin. Remember, again, there are millions of them. So it'll find one. And it'll do the same thing. It'll say, well, hey, hemoglobin, why don't you just come bind with me and get rid of that oxygen? So it also competes with oxygen. So you've got some competition from protons, some competition from carbon dioxide. And when carbon dioxide actually binds, interesting thing is that it makes a proton. So guess what happens? That proton can go and compete again by itself. It can compete with oxyhemoglobin and try to kick off another one, kick off another oxygen. because now you've got a few reasons why you have oxygen delivery. You've got protons competing. You've got now CO2 competing with oxygen. So you've got a couple of sources of competition. And you've got, of course, just simply the fact that there's just not too much oxygen around. So these are reasons for oxygen delivery. So at this point, you've got oxygen that's delivered to the cells. And these hemoglobin molecules, they're still our cell, of course, inside of a red blood cell. And these hemoglobin molecules have now been bound by different things. So they're no longer bound by oxygen. So you can't really call them oxyhemoglobin anymore. Instead they have protons on them like this. And they might have some COO minus on them. So they might have-- actually, let me do that in the original kind of orangey color. So they basically have different things binding to them." + }, + { + "Q": "\nAt 1:10, is diffusion kind of like osmosis?", + "A": "Yes, osmosis is specifically for water, while diffusion is for other solutes :)", + "video_name": "QP8ImP6NCk8", + "timestamps": [ + 70 + ], + "3min_transcript": "Let's talk about exactly how oxygen and carbon dioxide come into and out of the lungs. So you know this is our alveolus in the lungs. This is the last little chamber of air where the lungs are going to interface with blood vessels. So this is our blood vessel down here. And oxygen is going to make its way from this alveolus. It's going to go into the blood vessel. And it's going to go from the blood vessel into a little red blood cell. This is my red blood cell here. He's headed out for the first delivery of oxygen that day. And he's going to pick up some oxygen. And it's going to get inside of the red blood cell through diffusion. That's how it gets inside. So the oxygen has made its way into the red blood cell. And where do you think it goes first? Well, this red blood cell is, we sometimes think of it as a bag of hemoglobin. It's got millions and millions and millions of hemoglobin proteins. So this is our hemoglobin protein. It's got four parts to it. So hemoglobin, I can shorten this to Hb. Now, oxygen is going to bump into, quite literally bump into one of these hemoglobins. And it's going to bind, let's say, right here. And initially, it's kind of tricky because oxygen doesn't feel very comfortable sitting on the hemoglobin or binding to hemoglobin. But once a single oxygen is bound, a second one will come and bind as well. And then a third will find it much easier. Because what's happening is that as each oxygen binds, it actually changes the conformation or shape of hemoglobin. And so each subsequent oxygen has an easier time binding. We call that cooperativity. Has the word, almost like cooperation in it. And an easy way to think of cooperativity, the way I think of it, is that if you're at a dinner party, you are much more likely to sit where two or three of your friends are already sitting, if you think of this as a table with four chairs, rather than just sitting at a table by yourself being the first one to sit there. is kind of a friendly molecule. And so it also likes to sit where or bind where other oxygens have already bound. What are the two, then, major ways, based on this diagram, how I've drawn it. What are the two major ways that oxygen is going to be transported in the blood? One is hemoglobin binding oxygen. And we call that HbO2. Just Hb for hemoglobin, O2 for oxygen. And this molecule, or this enzyme, then, is not really called hemoglobin anymore. Technically, it's called oxyhemoglobin. That's the name for it. And another way that you can actually transport oxygen around is, that some of this oxygen-- I actually underlined it there-- is dissolved, O2 is dissolved in plasma. So some of the oxygen actually just gets dissolved right into the plasma. And that's how it gets moved around. Now, the majority, the vast majority of it is actually going to be moved through binding to hemoglobin. So just a little bit is dissolved in the plasma." + }, + { + "Q": "At 0:29 when Sal mentions that c3 mixes with RuBis what does that mean.\n", + "A": "C3 means the number of c in an triose phosphate molecule and in the calvin cycle the after some ofthe triose phosphate is sent to make stuff like glucose and lipids, the remaining triose phosphate does not mix but is converted to Rubp with the help of ATP.", + "video_name": "xp6Zj24h8uA", + "timestamps": [ + 29 + ], + "3min_transcript": "A couple of videos ago, we saw that in classic C3 photosynthesis And once again, it's called C3 because the first time that carbon dioxide (CO2) is fixed, it's fixed into a 3-carbon molecule. But we saw the problem with C3 photosynthesis is that the enzyme that does the carbon fixation, it can also react with oxygen. And when oxygen essentially reacts with ribulose biphosphate instead of your carbon, you get an unproductive reaction. Not only is it unproductive, it'll actually suck up your ATP and your NADPH and you'll go nowhere. So every now and then, when oxygen bonds here instead of a carbon dioxide, you get nothing produced, net. Everything becomes less efficient. And so in the last video, we saw that some plants have evolved a way to get around this. And what they do is, they fix their carbon on the outside, on cells that are actually exposed to the air. And then once they fix the carbon they actually fix it that gets turned into malate. Then they pump the malate deeper within the leaf, where you aren't exposed to oxygen. And then they take the carbon dioxide off the malate, and this is where they actually perform the Calvin cycle. And even though you do have your RuBisCO still there, your RuBisCO isn't going to have - the photorespiration is not going to occur. Because it only has access to carbon dioxide. It does not have access to this oxygen out here. Now that's a very efficient way of producing sugars. And that's why some of the plants that we associate with being very strong sugar, or even ethanol producers, all perform C4 photosynthesis. Corn, sugarcane and crab grass. And these are all very, very efficient sugar producers. Because they don't have to worry too much about photorespiration. Now some plants have a slightly different problem. They're not so worried about the efficiency of the process. They're more worried about losing water. These are plants that are in the desert. Because these stomata, these pores that are on the leaves, they let in air, but they can also let out water. I mean, if I'm in the rainforest, I don't care about that. But if I'm in the middle of the desert, I don't want to let out water vapour through my stomata. So the ideal situation is, I would want my stomata closed during the daytime. This is what I want. So I want, if I'm in the desert, let me make this clear. If I'm in the desert, I want stomata closed during the day. For obvious reasons. I don't want all my water to vaporise out of these holes in my leaves. But at the same time, the problem is photosynthesis can only occur during the daytime. And that includes the dark reactions. Remember, I've said multiple times, the dark reactions are" + }, + { + "Q": "\n7:00 Wouldn't getting the dam out of the way release the rest of the toxic copper sediments?", + "A": "They would use bioremediation to remove toxic elements from the water before getting rid of the dam.", + "video_name": "3BBqL_F9fxQ", + "timestamps": [ + 420 + ], + "3min_transcript": "extinction vortex. The next step is to figure out how small a population is too small. Ecologists do this by calculating what's called the minimum viable population, which is the smallest size at which a population can survive and sustain itself. To get at this number, you have to know the real breeding population of, say, grizzly bears in Yellowstone National Park. And then you figure out everything you can about a grizzly's life history, how long they live, who gets to breed the most, how often they can have babies, that kind of thing. After all that information is collected, ecologists can run the numbers and figure out that, for the grizzlies in Yellowstone, a population of, say hypothetically, 90 bears would have about a 95% chance of surviving for 100 years. But if there were a population of 100 bears, the population would likely be able to survive for 200 years. Something to note, ecology involves a lot of math. So if you're interested in this, that's just the way it is. So that's the small population approach to conservation. Another way of preserving biodiversity focuses on populations whose numbers are in decline, no matter how large the original population was. and it involves answering a series of related questions that get at the root of what's causing an organism's numbers to nosedive. First, you have to determine whether the population's actually declining. Then you have to figure out how big the population historically was and what its requirements were. And finally, you have to get at what's causing the decline and figure out how to address it. Milltown Dam actually gives us a good example of this process. In the winter of 1996, authorities had to release some of the water behind the dam as an emergency measure because of a big ice floe in the river that was threatening to break the dam. But when they released the water, a bunch of toxic sediment went with it, which raised the copper concentrations downriver to almost 43 times what state standards allowed. As a result, it's estimated that about half of the fish downstream died. Half of the fish, dead. And researchers have been monitoring the decline in populations ever since. This information was really helpful in determining what to do with the dam because we knew what the fish population was like before and after the release It was decided that it would be best to get the dam out as soon as possible rather than risk another 1996 scenario. Which brings me to the place where conservation biology and restoration ecology intersect. Restoration ecology is kind of where the rubber meets the road in conservation biology. It comes up with possible solutions for ecological problems. Now, short of a time machine, which I'm working on, you can't really get a natural environment exactly the way that it used to be. But you can at least get rid of whatever is causing the problem, and help recreate some of the elements that the ecosystem needs to function properly. All of this involves a whole suite of strategies. For instance, what's happening in Milltown is an example of structural restoration, basically the removal and cleanup of whatever human impact was causing the problem, in this case, the dam and the toxic sediments behind it. And then the rebuilding of the historical natural structure, here the meanders of the river channel and the vegetation. Another strategy is bioremediation, which recruits organisms temporarily to help remove toxins, like bacteria that eat wastes or plants that leach out metals from tainted soils. Some kinds of fungi and bacteria are even" + }, + { + "Q": "At 1:06, It is great to learn about the restoration of the river system in Montana. Is there other examples (positive or negative) of Restoration Elology? Is there a way to restore Chernobyl nuclear site, since we know exactly what went wrong?\n", + "A": "Well, one example is the Chesepeake Bay Watershed. The watershed was very polluted at one time but now, laws, regulations, and ordinary efforts from people are making it better! This applies to most polluted places in the US because of the EPA. Oh, and about Chernobyl, we have to wait until the radiation goes away", + "video_name": "3BBqL_F9fxQ", + "timestamps": [ + 66 + ], + "3min_transcript": "For the past 12 weeks, we've been investigating our living planet together, learning how it works on many levels, how populations of organisms interact, how communities thrive and ecosystems change, and how humans are wrecking the nice, perfectly functioning systems Earth has been using for hundreds of thousands of years. And now it's graduation day. This here is like the commencement speech where I talk to you about the future and our role in it and how what we're doing to the planet is totally awful, but we're taking steps to undo some of the damage So what better way to wrap up our series on ecology than by taking a look at the growing fields of conservation biology and restoration ecology. These disciplines use all the Kung Fu moves that we've learned about in the past 11 weeks and apply them to protecting ecosystems and cleaning up the messes that we've already made. And one of the main things they teach us is that doing these things is difficult, like in the way that uncooking bacon is difficult. So let's look at what we're doing and try to uncook this unbelievably large pile of bacon we've made. we've got a Superfund site. Not super-fun. Superfund, a hazardous waste site that the government is in charge of cleaning up. The mess here was made more than 100 years ago, when there was a dam in the Clark Fork River behind me called the Milltown Dam. This part of Montana has a long history of copper mining, and back in 1908, there was a humongous flood that washed about 4.5 million cubic meters of mine tailings chock full of arsenic and toxic heavy metals into the Clark Fork River. And most of it washed into the reservoir created by the Milltown Dam. I mean, actually it was lucky that the dam was there-- it had only been completed six months before-- or the whole river system all the way to the Pacific Ocean would have been a toxic mess. As it happened, though, only about 160 kilometers of the river was all toxic, messed up. A lot of it recuperated over time. But all that nasty hazardous waste was still sitting behind Milltown Dam. that started polluting nearby residents' wells. So scientists spent decades studying the extent of the damage caused by the waste and coming up with ways to fix it. And from 2006 to 2010, engineers carefully removed all the toxic sediment as well as the dam itself. Now this stretch of the Clark Fork River runs unimpeded for the first time in over a century, and the restored area where the dam used to be is being turned into a state park. Efforts like this show us conservation biology and restoration ecology in action. Conservation biology involves measuring the biodiversity of an ecosystem and determining how to protect it. In this case, it was used to size up the health of fish populations in the Clark Fork River, which were severely affected by the waste behind the dam, and the damn blocking their access to spawning grounds upstream, and figuring out how to protect them during the dam's removal. Restoration ecology, meanwhile, is the science of restoring broken ecosystems, like taking an interrupted, polluted river and turning it into what you see taking shape here." + }, + { + "Q": "At 2:11, Sal says that \"almost nothing\" is completely smooth on an atomic level. Is there any known substance that is completely smooth?\n", + "A": "A substance that is completely smooth will be frictionless. As far as our knowledge is concerned, there is no substance that is completely frictionless. Thus, till date, we do not know of any substance that is completely smooth, that is, frictionless. This is because the atoms of all substances interact and nudge each other when they are close to each other. Hope this helps :)", + "video_name": "J9BWNiOSGlc", + "timestamps": [ + 131 + ], + "3min_transcript": "I mentioned in the last several videos that the coefficient of kinetic friction tends to be less, sometimes it'll be roughly equal to, the coefficient of static friction. But this might lead you to-- at least, a question that I've had in my mind, and I still have to some degree-- is why? Why is the coefficient of kinetic friction lower? Or why can it be lower? And the current best theory-- one I can visualize in my head, and based on the reading that I've done-- is the difference between-- so let's think about it this way. So if we look at it at a kind of a regular human level, maybe we have a block. So this is the static case. So let's think about the static case. Let me draw it like this. So I'll draw the static case over here. So I have a block that is stationary on top of-- let me do the surface in a different color-- on top of some type of surface right over here. And over here, I'm going to have a block moving at a constant velocity relative to some surface, And so let me draw it out. So this is moving at some constant velocity. And so the interesting thing here is, assuming that these are the same masses, that these are the same surfaces, is, why should the coefficient of friction here-- why should the coefficient of static friction-- so here, since this is stationary, what's under play is the coefficient of static friction. Why should that be larger than the coefficient of kinetic friction over here? Why should that be large than the coefficient of kinetic friction? Or another way to think about it is, you would need to apply more force to overcome the static friction here, and start to get this accelerating, than you would need to apply to get this already moving body to accelerate. Because there would be kind of a less of a responsive friction force. So what I'm going to do is zoom in into the atomic level. And so when you zoom in to the atomic level, almost nothing is completely smooth. So the surface over here might look something like this. So I'm going to draw the molecules that make up the surface, the best to my ability. So the molecules, when you zoom up really close for the surface, might look something like this. So we're really zooming into the atomic level, unimaginably small level. Much smaller than that box I just drew. But I'm just trying to look at what's happening with the atoms where they contact, or the molecules where they contact? And the box's molecules might look something like this. They aren't completely smooth. And hopefully this video also emphasizes that all of these forces and all of this contact that we're talking about in these videos-- and it's actually" + }, + { + "Q": "At 7:24, wouldn't the liquid just stay inside because the hole is the only way for air to get in?\n", + "A": "not only that, but a vacuum would have negative pressure wouldn t it? unless outside is a vacuum too. the liquid would just turn into a vapor and fill the container.", + "video_name": "QX2YLR09Q78", + "timestamps": [ + 444 + ], + "3min_transcript": "Similarly, if pressure increases, then velocity is going to decrease. That might be a little unintuitive, but the other way, it makes a lot of sense. When velocity increases, this pressure is going to decrease, and that's actually what makes planes fly and all sorts of neat things happen, but we'll get more into that in a second. Let's see if we can use Bernoulli's equation to do something useful. You should memorize this, and it shouldn't be too hard to memorize. It's pressure, and then you have this potential energy term, but instead of mass, you have density. You have this kinetic energy term. It's not kinetic energy anymore, because we manipulated it some, but instead of mass, you have density. With that said, let's do a problem. I'll keep this down here, since you probably haven't memorized it as yet. Let me erase everything else. That's not how I wanted to erase it. That's how I wanted to erase it. I wanted to erase it like that without getting rid of OK, that's good enough. And then let me clean up. Clean up all this stuff. Let's say that I have a cup. I'll just draw a cup. It's easier to draw sometimes then to draw straight lines and all of that. No, that's too dark. Do purple. I'm using a super-wide tool. I have to switch the length. OK, so that's my cup. It has some fluid. Actually, let's say it has a top to it, and I have some fluid in it. Maybe it happens to be red. We haven't been dealing with red fluids as yet,. Let me-- oh, I didn't want to do that. So you know there's a fluid there. Let's say that h-- we don't know what units are, but let's say h meters below the surface of the fluid. This is all fluid here. I poke a hole right there, and fluid starts spurting out. My question to you is, what is this output velocity of the fluid as a function of this height? Let me tell you something else. Let's say that this hole is so small, let's call the area of that hole A2, and let's say that the surface area of the water is A1. Let's say that hole is so small that the surface area the water-- let's say that A2 if equal to 1/1,000 of A1. This is a small hole relative to the surface" + }, + { + "Q": "\nAt 2:00, the carbon gaining the negative charge is sp3 hybridized right? The only reason the compound is still aromatic is that the negative charge(lone pair) is delocalised even when that carbon does not have a p orbital. Am I correct? If not, please correct me. Thanks :)", + "A": "No it s sp2, it doesn t rehybridise itself. When you have a lone pair next to a pi system, that lone pair will be in a p orbital because there is a large energy benefit to the aromatic system.", + "video_name": "wvVdgGTrh-o", + "timestamps": [ + 120 + ], + "3min_transcript": "Voiceover: In the last video, we used the criteria for aromaticity to see that heterocycles can be aromatic too. In this video, we're going to look at more aromatic heterocycles, specifically five-membered rings. We'll start with pyrrole right down here. The pyrrole molecule, as you can see, five atoms in the ring, and if we take a look at the carbons in the ring, we can see that those carbons all have a double bond to them. Therefore, each of those carbons is sp two hybridized, meaning each of the carbons has a free p orbital. So I can go ahead and draw a free p orbital on each of those four carbons like that. In terms of the nitrogen on the ring, I need to know the hybridization state of this nitrogen. The best way to do that is to calculate the steric number. We know the steric number is equal to the number of sigma bonds plus number of lone pairs of elections. I can see that here is a sigma bond, here is a sigma bond, and here is a sigma bond, so three sigma bonds plus lone pairs of electrons. There's one lone pair of electrons on that nitrogen there, which would imply an sp three hybridization state for pyrrole. We know that's not the case because pyrrole is an aromatic molecule, and sp three hybridized nitrogen would mean no p orbitals at that nitrogen, which would violate the first criterion for this compound to be aromatic. And so there must be some way to get that nitrogen to be sp two hybridized, and of course, we saw how to do that in the end to the last video. This lone pair of electrons on this nitrogen is actually not localized to this nitrogen. We can take this lone pair of electrons and move them in here so that lone pair of electrons can participate in resonance. If those lone pairs of electrons move into there to form a pi bond, that would kick these electrons off onto this carbon, so the resonance structure will have nitrogen with a pi bond here now and a lone pair of electrons on this carbon, which would give this carbon a negative one formal charge. We still have a pi bond over here like that, Now when we analyze the hybridization state of this nitrogen, we can see that once again, we're going for sigma bonds, so one sigma bond, two sigma bonds, three sigma bonds, so three sigma bonds, this time no lone pairs of electrons because that lone pair of electrons is now de-localized in resonance, and so three plus zero is of course three, meaning that this nitrogen is now sp two hybridized. Since that nitrogen is sp two hybridized, it has a free p orbital, so we can go ahead and draw the p orbital on that nitrogen. You could think about in terms of dot structure, these two electrons over here, these two electrons are actually de-localized and participate in resonance, so that lone pair of electrons you could think about as occupying a p orbital here and they're actually de-localized. We have all these pi electrons de-localized throughout our ring, and so let's go ahead and check the criteria" + }, + { + "Q": "at 2:42 although the nitrogen in pyrrole is sp2 hybridised once the lone pair of electrons participate in resonance does the positive charge on nitrogen not mean anything when said that it is sp2 hybridised.\n", + "A": "the fact that nitrogen has a positive formal charge in some of the resonance structures is simply a consequence of the way formal charge is calculated. it has no bearing on the hybridization of nitrogen. Geometry specifies hybridization.", + "video_name": "wvVdgGTrh-o", + "timestamps": [ + 162 + ], + "3min_transcript": "which would imply an sp three hybridization state for pyrrole. We know that's not the case because pyrrole is an aromatic molecule, and sp three hybridized nitrogen would mean no p orbitals at that nitrogen, which would violate the first criterion for this compound to be aromatic. And so there must be some way to get that nitrogen to be sp two hybridized, and of course, we saw how to do that in the end to the last video. This lone pair of electrons on this nitrogen is actually not localized to this nitrogen. We can take this lone pair of electrons and move them in here so that lone pair of electrons can participate in resonance. If those lone pairs of electrons move into there to form a pi bond, that would kick these electrons off onto this carbon, so the resonance structure will have nitrogen with a pi bond here now and a lone pair of electrons on this carbon, which would give this carbon a negative one formal charge. We still have a pi bond over here like that, Now when we analyze the hybridization state of this nitrogen, we can see that once again, we're going for sigma bonds, so one sigma bond, two sigma bonds, three sigma bonds, so three sigma bonds, this time no lone pairs of electrons because that lone pair of electrons is now de-localized in resonance, and so three plus zero is of course three, meaning that this nitrogen is now sp two hybridized. Since that nitrogen is sp two hybridized, it has a free p orbital, so we can go ahead and draw the p orbital on that nitrogen. You could think about in terms of dot structure, these two electrons over here, these two electrons are actually de-localized and participate in resonance, so that lone pair of electrons you could think about as occupying a p orbital here and they're actually de-localized. We have all these pi electrons de-localized throughout our ring, and so let's go ahead and check the criteria Pyrrole does contain a ring of continuously overlapping p orbitals, and it does have four n plus two pi electrons in that ring, so let's go ahead and highlight those. We had these pi electrons, so that's two, these pi electrons, so that's four, and then these pi electrons here in magenta are actually de-localized in the ring, so that gives us six pi electrons. So if n is equal to one, four times one plus two gives me six pi electrons. Pyrrole has six pi electrons and also has a ring of continuously overlapping p orbitals, so we can say that it is aromatic. Let's go ahead and look at another molecule here so similar to it. This is imidazole. For imidazole, once again, we have the same sort of situation that we had for pyrrole with this nitrogen right here, so at first, it looks like that nitrogen might be sp three hybridized, but we can draw a resonance structure for it." + }, + { + "Q": "At 2:57, Sal uses the resistance formula to combine the 4 parallel resistors. But doesn't the current just go through the path of smallest resistance? If that is the case, then wouldn't all the current go through 3 ohms resistor and then the 1 ohm resistor? Combining these, since they are in series, gives us a 4 ohm resistor. Then by V=IR, we have 5 amps.\n", + "A": "No, current goes through any path it can go through. There is no such law as current goes through the path of smallest resistance . If you put 5 volts across a 1 ohm resistor, 5 amps will go through the resistor. The same 5 volts across a 10 ohm resistor will give you 0.5 amps through that resistor. Since parallel resistors have the same potential difference across them, if you put the 1 ohm and the 10 ohm both connected to the 5 volt source, that source is going to send out 1.5 amps.", + "video_name": "3NcIK0s3IwU", + "timestamps": [ + 177 + ], + "3min_transcript": "that is 3 ohms. And let's say I have a resistor here. Let's just make it simple: 1 ohm. And just to make the numbers reasonably easy-- I am doing this on the fly now-- that's the positive terminal, negative terminal. Let's say that the voltage difference is 20 volts. So what I want us to do is, figure out what is the current flowing through the wire at that point? Obviously, that's going to be different than the current at that point, that point, that point, that point, all of these different points, but it's going to be the same as the current flowing at this point. So the easiest way to do this is try to figure out the equivalent resistance. Because once we know the equivalent resistance of this big hairball, then we can just use Ohm's law and be done. So first of all, let's just start at, I could argue, the simplest part. Let's see if we could figure out the equivalent resistance of these four resistors in parallel. Well, we know that that resistance is going to be equal to 1/4 plus 1/8 plus 1/16 plus 1/16. So that resistance-- and now it's just adding fractions-- over 16. 1/4 is 4/16 plus 2/16 plus 1 plus 1, so 1/R is equal to 4 plus 2 is equal to 8/16-- the numbers are working out-- is equal to 1/2, so that equivalent resistance is 2. So that, quickly, we just said, well, all of these resistors combined is equal to 2 ohms. So let me erase that Simplify it. So that whole thing could now be simplified as 2 ohms. I lost some wire here. I want to make sure that circuit can still flow. So that easily, I turned that big, hairy mess into something that is a lot less hairy. Well, what is the equivalent resistance of this resistor and this resistor? Well, they're in series, and series resistors, they just add up together, right? So the combined resistance of this 2-ohm resistor and this 1-ohm resistor is just a 3-ohm resistor. So let's erase and simplify. So then we get that combined resistor, right?" + }, + { + "Q": "At 4:20 when there are 4 H is it impossible there would be a lone pair or lone single e- making FC -1 or 0 respectively? Is steric hindrance at play?\n", + "A": "Yes it is not possible. Nitrogen can only have up to 8 valence electrons around it and in this case there are 4 bonds which is 8 electrons. It can fit no more.", + "video_name": "5-MM39VCwc0", + "timestamps": [ + 260 + ], + "3min_transcript": "So three bonds and one lone pair of electrons, the formal charge is equal to zero. So when nitrogen has three bonds and one lone pair of electrons, the formal charge is equal to zero. And sometimes you don't want to draw in lone pairs of electrons, so you could just leave those off. You could just say alright, well if I just draw this and you know the formal charge of nitrogen is zero, then it's assumed you also know there's a lone pair of electrons on that nitrogen. So this is just another way of representing the same molecule, leaving off the lone pair, because you should know it's there. Let's look at other examples where nitrogen has a formal charge of zero. So we'll start with the example on the left here and if we look at this nitrogen and we know it has a formal charge of zero, let's see how many bonds it has. Let's use red here. So here's one bond, two bonds, and then three bonds. we know there should be a lone pair of electrons on that nitrogen. So you could leave it off and just know it's there, or you could draw them in. So I'll go ahead and draw in the lone pair of electrons on the nitrogen. So formal charge of zero. Let's look at the one on the right. So if we assume that nitrogen has a formal charge of zero, let's see how many bonds we have here. So here's one, two, and three. So we have three bonds, so we'd still need one lone pair of electrons. So if you wanted to show the lone pair of electrons you could put them in there like that. Notice this gives nitrogen an octet of electrons around it. So count those up, here's two, four, six, and eight. So nitrogen would have an octet. And remember, you could just leave off that lone pair of electrons and it's assumed if we know nitrogen has a formal charge of zero that there is a lone pair and we just didn't want to take the time to draw them in. Let's assign formal charge to another nitrogen, So what is the formal charge of nitrogen now? Let's draw in our electrons. So each bond is two electrons, so I draw those in there. And the formal charge on nitrogen is equal to the number of valence electrons that nitrogen is supposed to have, which we already know is five, so we put a five in here, and from that we subtract the number of valence electrons that nitrogen actually has in our drawing. So for these bonds, hydrogen gets one electron and nitrogen gets one for each of these bonds. So that allows us to see there are four electrons around nitrogen. So here's one, two, three, and four. So in our drawing, nitrogen only has four electrons around it, so this would be five minus four, which gives us a formal charge of plus one. So it's like nitrogen lost a valence electron. It's supposed to have five and here we see only four around it, so it's as if it lost" + }, + { + "Q": "\n0:05 Sal mentioned that \"parsec\" is science fiction...is this true?", + "A": "It is often used in science fiction shows like Star Trek and Star Wars (Han Solo misuses it in episode IV). But a parsec is a real unit that astronomers use.", + "video_name": "6zV3JEjLoyE", + "timestamps": [ + 5 + ], + "3min_transcript": "You've probably heard the word parsec before in science fiction movies or maybe even some things dealing with astronomy. And what I want to do in this video is really just tell you where the word and the definition of the word really come from. And just to kind of cut to the chase, it's just a unit of distance. It's just about 3.26 light years. But what I want to do is just think about where did this weird distance come from, this distance that is roughly 3.26 light years? It comes from the distance of something, probably a star. But let me say \"something\" because there are no stars exactly this far away from us. The distance of something that has a parallax angle of one And the word comes from the \"par\" in parallax and the \"second\" in arc seconds. So it's literally par-- let me do this in a different color. It's literally parsec. You could think of it as kind of the parallax arc second. How far would this thing be? It turns out it's 3.26 light years. So we can actually calculate that. And that's actually what I'm going to do in this video. So let's say there is something. So this is the sun. This is the Earth at some point in time. This is the Earth six months later at the opposite end of the orbit. And we are looking at some distance. We are looking at some object some distance away. We know that this distance right here is one astronomical unit. And what we want to do is figure out the distance of this object. And all we know is that it has parallax angle of one arc If we're looking right at-- remember, we're looking from above the solar system. So the Earth is rotating in this direction in either case. And so in this point in the year-- we don't know when this is. Depends on what star that is. At this point in the year right at sunrise, right when we first catch the first glimpses of the sun's light, if we look straight up, the angle between that object in the night sky and straight up is going to be the parallax angle. So this is going to be one arc second. And just to make it consistent with the last few videos we did on parallax, let's just visualize how that would look in the night sky. So let me draw the night sky over here. Let me do that in purple maybe. Let me draw the night sky over here." + }, + { + "Q": "\nat 8:52 its been said that DMSO i.e a polar aprotic solvent increases the nucleophilicity of the ethoxide aninon therefore increasing SN2.... but i dont understand how DMSO increases the nucleophilic strengh of ethoxide anion?? plz help", + "A": "I don t think he said it s favoring substitution over elimination in the way that you re thinking, which is that the SN2 products are the major products. I think he s just trying to say that in order to increase the % of SN2 products, you can change the solvent to make it happen.", + "video_name": "vFSZ5PU0dIY", + "timestamps": [ + 532 + ], + "3min_transcript": "with phosphoric acid and heat. And we saw a lot of these types of problems in the videos on elimination reactions. So, it's not gonna be SN1 or SN2 and we don't have a strong base, so don't think E2, think E1. And our first step would be to protonate our alcohol to form a better leaving group. So phosphoric acid is a source of protons and we're going to protonate this oxygen for our first step. So, let's draw in our ring and we protonate our oxygen, so now our oxygen has two bonds to hydrogen, one lone pair of electrons and a plus one formal charge on the oxygen. So, this lone pair of electrons on the oxygen picked up a proton from phosphoric acid to form this bond. And now we have a better leaving group than the hydroxide ion. These electrons come off onto the oxygen and we remove a bond from this carbon in red which would give us a secondary carbocation. and the carbon in red is this one and that carbon would have a plus one formal charge. So, let me draw in a plus one formal charge here. And now we have water which can function as a weak base in our E1 reaction and take a proton from a carbon next to our carbon with a positive charge. So, let's say this carbon right here. It has two hydrogens on it. I'll just draw one hydrogen in and water functions as a base, takes this proton and these electrons move in to form a double bond. So, let's draw our final product here. We would have a ring, we would have a double bond between these two carbons, so our electrons in, let's use magenta, electrons in magenta moved in to form our double bond. So, our product is cyclohexane. So, a secondary alcohol undergoes an E1 reaction if you use something like sulfuric acid For this reaction we have this secondary alkyl halide reacting with an aqueous solution of formic acid. Formic acid is a weak nucleophile and water is a polar protic solvent. A weak nucleophile and a polar protic solvent should make us think about an SN1 type mechanism because water as a polar protic solvent can stabilize the formation of a carbocation. So, let's draw the carbocation that would result. These electrons would come off onto our bromine and we're taking a bond away from this carbon in red. So, the carbon in red gets a plus one formal charge and let's draw our carbocation. So, we have our benzine ring here. I'll put in my pi electrons and the carbon in red is this one, so that carbon gets a plus one formal charge. This is a secondary carbocation" + }, + { + "Q": "\nAt 2:10, why was the E1 mechanism excluded ?", + "A": "The strong base favors E2. I think this is because carbocations are energetically unfavorable. So, if you can do E2 or E1, then E2 will win and in the presence of a strong base you can do E2 ...", + "video_name": "vFSZ5PU0dIY", + "timestamps": [ + 130 + ], + "3min_transcript": "- [Instructor] Let's look at elimination versus substitution for a secondary substrate. And these are harder than for a primary or tertiary substrate because all four of these are possible to start with. So, if we look at the structure of our substrate and we say it's secondary, we next need to look at the reagent. So, we have NaCl which we know is Na plus and Cl minus and the chloride anion functions only as a nucleophile. So, we would expect a substitution reaction, nucleophilic substitution. So, E1 and E2 are out. Between SN1 and SN2 with the secondary substrate, we're not sure until we look at the solvent and DMSO is a polar aprotic solvent, which we saw in an earlier video, favors an SN2 mechanism. So, SN1 is out and we're gonna think about our chloride anion functioning as a nucleophile. So, let me draw it in over here. So, this is with a negative one formal charge. And an SN2 mechanism are nucleophile attacks and our nucleophile is going to attack this carbon in red. So, we're gonna form a bond between the chlorine and this carbon in red and when the nucleophile attacks, we also get loss of our leaving group. So, these electrons come off onto the oxygen and we know that tosylate is a good leaving group. So, when we draw our product, let's draw this in here, and the carbon in red is this one, we know an SN2 mechanism means inversion of configuration. The nucleophile has to attack from the side opposite of the leaving group. So, we had a wedge here for our leaving groups, so that means we're gonna have a dash for our chlorines. We're gonna put the chlorine right here and that's the product of our SN2 reaction. For our next problem, we have a secondary alkyl halide. So, just looking at our reactions, we can't really rule any out here. So, all four are possible, until we look at our reagent. Now, we saw in an earlier video, it does not act like a nucleophile. So SN1 and SN2 are out. And a strong base means an E2 reaction. So, E1 is out. Now that we know we're doing an E2 mechanism, let's analyze the structure of our alkyl halide. The carbon that's directly bonded to our halogen is our alpha carbon and the carbons directly bonded to the alpha carbon are the beta carbons. So, I'll just do the beta carbon on the right since they are the same essentially. And we know that our base is gonna take a proton from that beta carbon. So, let me just draw in a hydrogen here. And DBN is a neutral base, so I'll just draw a generic base here. Our base is going to take this proton at the same time these electrons move in to form a double bond and these electrons come off to form our bromide anion. So, our final product is an alkyne and our electrons in magenta in here" + }, + { + "Q": "at 7:08 as T=>Ta ,then how T can be HOTTER if T=Ta\n", + "A": "Sal should have written T > Ta. In that case, T will always be hotter than Ta.", + "video_name": "IICR-w1jYcA", + "timestamps": [ + 428 + ], + "3min_transcript": "and once again I could put a constant here, but I'm going to end up with a constant on the right hand side too so I'm just going to merge them into the constant on the right hand side. So that is going to be equal to, now here, this is going to be negative kt, and once again we have plus C. And now we can raise e to both of these powers, or another way of interpreting this is if e to this thing is going to be the same as that. So we can write this as, the absolute value, let me do that in that same blue color. We can write this as the absolute value of T minus T sub a is equal to e, something about e I always think of the color green. e to the negative kt plus C. This of course is the same thing as, we've done this multiple times before. Negative kt times e to the C power. And we could just call this another arbitrary constant. If we called this C1, then we could just call this whole thing C. So this we could say is Ce to the negative kt. So at least it's starting to resemble what we did when we were modelling population. We'll see it's a little bit different. Instead of just temperature on this left hand side, we have temperature minus our ambient temperature. And so, we can do a couple of things. If, in a world, say we were dealing with a hot cup of tea, something that's hotter than the ambient temperature. So we could imagine a world where T is So that means this is hot, or it's hotter, I guess we could say. So if we're dealing with something hotter than the ambient temperature, then this absolute value is going to be positive or the thing inside the absolute value So we don't need the absolute value. Or the absolute value of it is going to be the same thing as it. And then we can just add T sub a to both sides, and then we would have our temperature, and I can even write this as a function of time, is going to be equal to this business, is going to be equal to Ce, let me do that in that same color. Ce to the negative kt plus T sub a. All I did is I'm assuming that this inside the absolute value is going to be positive, so the absolute value is not going to change the value. And I added T sub a to both sides to get this. So this right over here is going to be our general solution," + }, + { + "Q": "\nA quick, slightly off topic question. At \"11:45\" Sal says g = 9.8 m/s^2, and that the acceleration is 33m/s^2. Why exactly do we square the seconds? Couldn't we do a few more steps and express it as m/s? That would seem to be easier for people to process.", + "A": "The m/s^2 just means meters per second every second...so if gravity is 9.8 meters per second every second.", + "video_name": "VYgSXBjEA8I", + "timestamps": [ + 705 + ], + "3min_transcript": "And what we're gonna get is, I'll just write this in one color, it's going to be 72 divided by 160, times, we have in the numerator, meters squared over seconds squared, we're squaring the units, and then we're going to be dividing by meters. So times, I'll do this in blue, times one over meters. Right? Because we have a meters in the denominator. And so what we're going to get is this meters squared divided by meters, that's going to cancel out, we're going to get meters per second squared. Which is cool because that's what acceleration should be in. And so let's just get the calculator out, to calculate this exact acceleration. So we have to take, oh sorry, this is 72 squared, let me write that down. So this is, this is going to be 72 squared, don't want to forget about this part right over here. 72 squared divided by 160. right over here that we calculated, so let's just square that, and then divide that by 160, divided by 160. And if we go to 2 significant digits, we get 33, we get our acceleration is, our acceleration is equal to 33 meters per second squared. And just to give you an idea of how much acceleration that is, is if you are in free fall over Earth, the force of gravity will be accelerating you, so g is going to be equal to 9.8 meters per second squared. So this is accelerating you 3 times more than what Earth is making you accelerate if you were to jump off of a cliff or something. So another way to think about this is that the force, and we haven't done a lot on force yet, we'll talk about this in more depth, more than 3 times the force of gravity, more than 3 g's. 3 g's would be about 30 meters per second squared, this is more than that. So an analogy for how the pilot would feel is when he's, you know, if this is the chair right here, his pilot's chair, that he's in, so this is the chair, and he's sitting on the chair, let me do my best to draw him sitting on the chair, so this is him sitting on the chair, flying the plane, and this is the pilot, the force he would feel, or while this thing is accelerating him forward at 33 meters per second squared, it would feel very much to him like if he was lying down on the surface of the planet, but he was 3 times heavier, or more than 3 times heavier. Or if he was lying down, or if you were lying down, like this, let's say this is you, this is your feet, and this is your face, this is your hands," + }, + { + "Q": "At 3:57, Sal starts to take displacement for the solution. Why did he not use any other formula except for the displacement or distance?\n", + "A": "Because the problem asks for ddisplacement/", + "video_name": "VYgSXBjEA8I", + "timestamps": [ + 237 + ], + "3min_transcript": "So if we want to convert this into seconds, we have, we'll put hours in the numerator, 1 hour, so it cancels out with this hour, is equal to 3600 seconds. I'll just write 3600 s. And then if we want to convert it to meters, we have 1000 meters is equal to 1 km, and this 1 km will cancel out with those kms right over there. And whenever you're doing any type of this dimensional analysis, you really should see whether it makes sense. If I'm going 260 km in an hour, I should go much fewer km in a second because a second is so much shorter amount of time, and that's why we're dividing by 3600. If I can go a certain number of km in an hour a second, I should be able to go a lot, many many more meters in that same amount of time, and that's why we're multiplying by 1000. When you multiply these out, the hours cancel out, you have km canceling out, and you have 260 times 1000 So let me get my trusty TI-85 out, and actually calculate that. So I have 260 times 1000 divided by 3600 gets me, I'll just round it to 72, because that's about how many significant digits I can assume here. 72 meters per second. So all I did here is I converted the take-off velocity, so this is 72 m/s, this has to be the final velocity after accelerating. So let's think about what that acceleration could be, given that we know the length of the runway, and we're going to assume constant acceleration here, just to simplify things a little bit. But what does that constant acceleration have to be? So let's think a little bit about it. The total displacement, I'll do that in purple, the total displacement is going to be times the difference in time, or the amount of time it takes us to accelerate. Now, what is the average velocity here? It's going to be our final velocity, plus our initial velocity, over 2. It's just the average of the initial and final. And we can only do that because we are dealing with a constant acceleration. And what is our change in time over here? What is our change in time? Well our change in time is how long does it take us to get to that velocity? Or another way to think about it is: it is our change in velocity divided by our acceleration. If we're trying to get to 10 m/s, or we're trying to get 10 m/s faster, and we're accelerating at 2 m/s squared, it'll take us 5 seconds. Or if you want to see that explicitly written in a formula, we know that acceleration is equal to" + }, + { + "Q": "\nAt 7:34, where is 72 coming from?", + "A": "He took 260 km/h and changed the units to m/s seconds, which means: 260 km/h, 1km =1000m, 1h = 3600 seconds. Hence: (260 * 1000 m) / 36000 s = 72 m/s.", + "video_name": "VYgSXBjEA8I", + "timestamps": [ + 454 + ], + "3min_transcript": "If we just work through this math, and I'll try to write a little bigger, I see my writing is getting smaller, our displacement can be expressed as the product of these two things. And what's cool about this, well let me just write it this way: so this is our final velocity plus our initial velocity, times our final velocity minus our initial velocity, all of that over 2 times our acceleration. Our assumed constant acceleration. And you probably remember from algebra class this takes the form: a plus b times a minus b. And so this equal to -- and you can multiply it out and you can review in our algebra playlist how to multiply out two binomials like this, but this numerator right over here, I'll write it in blue, is going to be equal to our final velocity squared minus our initial velocity squared. into the sum of the two terms times the difference of the two terms, so that when you multiply these two out you just get that over there, over 2 times the acceleration. Now what's really cool here is we were able to derive a formula that just deals with the displacement, our final velocity, our initial velocity, and the acceleration. And we know all of those things except for the acceleration. We know that our displacement is 80 meters. We know that this is 80 meters. We know that our final velocity, just before we square it, we know that our final velocity is 72 meters per second. And we know that our initial velocity is 0 meters per second. And so we can use all of this information to solve for our acceleration. And you might see this formula, displacement, the scalar version, and really we are thinking only in the scalar, we're thinking about the magnitudes of all of these things for the sake of this video. We're only dealing in one dimension. But sometimes you'll see it written like this, sometimes you'll multiply both sides times the 2 a, and you'll get something like this, where you have 2 times, really the magnitude of the acceleration, times the magnitude of the displacement, which is the same thing as the distance, is equal to the final velocity, the magnitude of the final velocity, squared, minus the initial velocity squared. Or sometimes, in some books, it'll be written as 2 a d is equal to v f squared minus v i squared. And it seems like a super mysterious thing, but it's not that mysterious. We just very simply derived it from displacement, or if you want to say distance, if you're just thinking about the scalar quantity, is equal to average velocity times the change in time. So, so far we've just derived ourselves a kind of a" + }, + { + "Q": "\nAt 14:00 Sal says that after 2 seconds, the plane would go 66 meters. Shouldn't it be 99 meters because of acceleration?", + "A": "aceleration is at the rate of 33m/s^2 .so, after 2 seconds its speed will be 33m/s x 2 = 66m/s", + "video_name": "VYgSXBjEA8I", + "timestamps": [ + 840 + ], + "3min_transcript": "more than 3 times the force of gravity, more than 3 g's. 3 g's would be about 30 meters per second squared, this is more than that. So an analogy for how the pilot would feel is when he's, you know, if this is the chair right here, his pilot's chair, that he's in, so this is the chair, and he's sitting on the chair, let me do my best to draw him sitting on the chair, so this is him sitting on the chair, flying the plane, and this is the pilot, the force he would feel, or while this thing is accelerating him forward at 33 meters per second squared, it would feel very much to him like if he was lying down on the surface of the planet, but he was 3 times heavier, or more than 3 times heavier. Or if he was lying down, or if you were lying down, like this, let's say this is you, this is your feet, and this is your face, this is your hands, essentially two more people stacked above you, roughly, I'm just giving you the general sense of it, that's how it would feel, a little bit more than two people, that squeezing sensation. So his entire body is going to feel 3 times heavier than it would if he was just laying down on the beach or something like that. So it's very very very interesting, I guess, idea, at least to me. Now the other question that we can ask ourselves is how long will it take to get catapulted off of this carrier? And if he's accelerating at 33 meters per second squared, how long would it take him to get from 0 to 72 meters per second? So after 1 second, he'll be going 33 meters per second, after 2 seconds, he'll be going 66 meters per second, so it's going to take, and so it's a little bit more than 2 seconds. So it's going to take him a little bit more than 2 seconds. 72 meters per second, and you divide it by 33, it'll take him 2.18 seconds, roughly, to be catapulted off of that carrier." + }, + { + "Q": "\nAt 9:34, the amino acid is being called a \"Zwitter ion.\" Sal explains that \"zwitter\" means \"hybrid\" in German. Why can't it be called \"polar?\"", + "A": "It cannot be explained as polar because polar substance is that which has a partial negative charge inside the molecule. While this has a charge which can be explained as a full charge. Polar substances are not either anion or cation.", + "video_name": "Pk4d9lY48GI", + "timestamps": [ + 574 + ], + "3min_transcript": "is the term for two or more amino acids connected together, so this would be a dipeptide, and the bond isn't this big, I just, actually let me just, let me draw it a little bit smaller. So... That's serine. This is valine. They can form a peptide bond, and this would be the smallest peptide, this would be a dipeptide right over here. \"Peptide,\" \"peptide bond,\" or sometimes called a peptide linkage. And as this chain forms, that polypeptide, as you add more and more things to it, as you add more and more amino acids, this is going to be, this can be a protein or can be part of a protein that does all of these things. Now one last thing I wanna talk about, this is the way, the way these amino acids have been drawn is a way you'll often see them in a textbook, but at physiological pH's, the pH's inside of your body, which is in that, you know, that low sevens range, What you have is this, the carboxyl group right over here, is likely to be deprotonated, it's likely to have given away its hydrogen, you're gonna find that more likely than when you have... It's gonna be higher concentrations having been deprotonated than being protonated. So, at physiological conditions, it's more likely that this oxygen has taken both of those electrons, and now has a negative charge, so it's given, it's just given away the hydrogen proton but took that hydrogen's electron. So it might be like this, and then the amino group, the amino group at physiological pH's, it's likely to actually grab a proton. So nitrogen has an extra loan pair, so it might use that loan pair to grab a proton, in fact it's physiological pH's, you'll find a higher concentration of it having grabbed a proton than not grabbing a proton. use its loan pairs to grab a proton, and so it is going to have... So it is going to have a... It is going to have a positive charge. And so sometimes you will see amino acids described this way, and this is actually more accurate for what you're likely to find at physiological conditions, and these molecules have an interesting name, a molecule that is neutral even though parts of it have charge, like this, this is called a zwitterion. That's a fun, fun word. Zwitterion. And \"zwitter\" in German means \"hybrid,\" and \"ion\" obviously means that it's going to have charge, and so this has hybrid charge, even though it has charges at these ends, the charges net out to be neutral." + }, + { + "Q": "\nhi, at \"4:00\" when mr khan was naming the compound, i was wondering if he made a mistake by saying 2,3-dimethyl instead of saying 2,3-trimethyl coz there are three CH3's.....or maybe someone correct me and explain it to me please? =) thank you x", + "A": "There are 3 CH3s but only two of them are groups, one of them is part of the main chain", + "video_name": "peQsBg9P4ms", + "timestamps": [ + 240 + ], + "3min_transcript": "to use should have as many simple groups attached to it as possible, as opposed to as few complex groups. So if we used this carbon as part of our longest chain, then this will be a group that's attached to it, which would be a bromomethyl group, which is not as simple as maybe it could be. But if we use this carbon in our longest chain, we'll have We'll have a bromo attached, and we'll also have a methyl group. And that's what we want. We want more simple groups attached to the longest chain. So what we're going to do is we're going to use this carbon, this carbon, this carbon, and that carbon as our longest chain. And we want to start from the end that is closest to something being attached to it, and that bromine is right there. So there's going to be our number one carbon, our number two carbon, our number three carbon, and our number four carbon. figure out what order they should be listed in. So this is a 1-bromo and then this will be a 2-methyl right here. And then just a hydrogen. Then three we have a fluoro, so on a carbon three, we have a fluoro, and then on carbon three, we also have a methyl group right here, so we also have a 3-methyl. So when we name it, we put in alphabetical order. Bromo comes first, so this thing right here is 1-bromo. Then alphabetically, fluoro comes next, 1-bromo-3-fluoro. We have two methyls, so it's going to be 2 comma 3-dimethyl. And remember, the D doesn't count in alphabetical order. chain is four carbons. Dimethylbutane. So that's just the standard nomenclature rules. We still haven't used the R-S system. Now we can do that. Now to think about that, we already said that this is our chiral center, so we just have to essentially rank the groups attached to it in order of atomic number and then use the Cahn-Ingold-Prelog rules, and we'll do all that in this example. So let's look at the different groups attached to it. So when you look at it, this guy has three carbons and a hydrogen. Carbon is definitely higher in atomic number on It has an atomic number 6. Hydrogen is 1. You probably know that already. So hydrogen is definitely going to be number four. So let me put number four there next to the hydrogen. And let me find a nice color. I'll do it in white. So hydrogen is definitely the number four group. We have to differentiate between this carbon group," + }, + { + "Q": "@6:10 so even if the carbon next to the chiral center has a F plus 2 CH3 groups attached to it, we still consider the C attached to the Br as the higher one? Thanks\n", + "A": "oh, I see. Now what if there were more CH3 groups attached to the other C ? And thanks on your answer!", + "video_name": "peQsBg9P4ms", + "timestamps": [ + 370 + ], + "3min_transcript": "chain is four carbons. Dimethylbutane. So that's just the standard nomenclature rules. We still haven't used the R-S system. Now we can do that. Now to think about that, we already said that this is our chiral center, so we just have to essentially rank the groups attached to it in order of atomic number and then use the Cahn-Ingold-Prelog rules, and we'll do all that in this example. So let's look at the different groups attached to it. So when you look at it, this guy has three carbons and a hydrogen. Carbon is definitely higher in atomic number on It has an atomic number 6. Hydrogen is 1. You probably know that already. So hydrogen is definitely going to be number four. So let me put number four there next to the hydrogen. And let me find a nice color. I'll do it in white. So hydrogen is definitely the number four group. We have to differentiate between this carbon group, And the way you do it, if there's a tie on the three carbons, you then look at what is attached to those carbons, and you compare the highest thing attached to each of those carbons to the highest things attached to the other carbons, and then you do the same ranking. And if that's a tie, then you keep going on and on and on. So on this carbon right here, we have a bromine. Bromine has an atomic number of 35, which is higher than carbon. So this guy has a bromine attached to it. This guy only has hydrogen attached to it. This guy has a fluorine attached to it. That's the highest thing. So this is going to be the third lowest, or I should say the second to lowest, because it only has hydrogens attached to it, so that is number three. The one has the bromine attached to it is going to be number one, and the one that has the fluorine attached to it is number two. have to look at the next highest constituent, and even if this had three fluorines attached to it, the bromine would still trump it. You compare the highest to the highest. So now that we've done that, let me redraw this molecule so it's a little bit easier to visualize. So I'll draw our chiral carbon in the middle. And I'm just doing this for visualization purposes. And right here we have our number one group. I'll literally just call that our number one group. So right there that is our number one group. It's in the plane of the screen. So I'll just call that our number one group. Over here, also in the plane of the screen, I have our number two group. So let me do it like that. So then you have your number two group, just like that." + }, + { + "Q": "at 9:15 why does the # 1 group stay in place and the rest of the groups rotate?\n", + "A": "Group 1 stays in the same place because the rest of the groups need to rotate in order to put the hydrogen behind all other atoms. You need to think about the molecule in three dimension. The hydrogen in this situation is pointing towards you and therefore in order to put it away from you, you would need to keep group 1 in the same spot and rotate the rest of the groups in order to make hydrogen face away from you.", + "video_name": "peQsBg9P4ms", + "timestamps": [ + 555 + ], + "3min_transcript": "molecule right now the way it's drawn. I'll do that in magenta. So then you have your number three group. It's behind the molecule, so I'll draw it like this. This is our number three group. And then we have our number four group, which is the hydrogen pointing out right now. And I'll just do that in a yellow. We have our number four group pointing out in front right now. So that is number four, just like that. Actually, let me draw it a little bit clearer, so it looks a little bit more like the tripod structure that it's So let me redraw the number three group. The number three group should look like-- so this is our number three group. Let me draw it a little bit more like this. The number three group is behind us. And then finally, you have your number four group in straight out. So that is coming straight out of-- well, not straight out, but at an angle out of the page. So that's our number four group, I'll just label it It really is just a hydrogen, so I really didn't have to simplify it much there. Now by the R-S system, or by the Cahn-Ingold-Prelog system, we want our number four group to be the one furthest back. So we really want it where the number three position is. And so the easiest way I can think of doing that is you can imagine this is a tripod that's leaning upside down. Or another way to view it is you can view it as an umbrella, where this is the handle of the umbrella and that's the top of the umbrella that would block the rain, I guess. But the easiest way to get the number four group that's actually a hydrogen in the number three position would be to rotate it. You could imagine, rotate it around the axis defined by the number one group. So the number one group is just going to stay where it is. The number four is going to rotate to the Number three is going to rotate around to the number two group, and then the number two group is going to rotate to where the number four group is right now. So if we were to redraw that, let's redraw our chiral carbon. So let me scroll over a little bit. So we have our chiral carbon. I put the little asterisk there to say that that's our chiral carbon. The number four group is now behind. I'll do it with the circles. It makes it look a little bit more like atoms. So the number four group is now behind where the number three group used to be, so number four is now there. Number one hasn't changed. That's kind of the axis that we rotated around. So the number one group has not changed. Number one is still there. Number two is now where number four used to be, so number two is now jutting out of the page." + }, + { + "Q": "\nAt 3:10 What is mole ratio?", + "A": "A mole ratio is a ratio between the amount of mols of each substance. In this case, it s the ratio of the amount of mols of Fe2O3, which is 1, and the amount of mols of Al, which is 2. So, the mole ratio is 1:2", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 190 + ], + "3min_transcript": "we know how to get to the balanced point. A balanced chemical equation. So let's do some stoichiometry. Just so we get practice balancing equations, I'm always going to start with unbalanced equations. Let's say we have iron three oxide. Two iron atoms with three oxygen atoms. Plus aluminum, Al. And it yields Al2 O3 plus iron. So remember when we're doing stoichiometry, first of all we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation. But I think it's good practice to actually balance the equations ourselves. So let's try to balance this one. We have two iron atoms here in this iron three oxide. How many iron atoms do we have on the right hand side? We only have one. So let's multiply this by 2 right here. We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams. how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here." + }, + { + "Q": "i didnt get the part at 8:20 that why do you multiply to get the mass of aluminum required.\ncan u pls help me?\n", + "A": "You multiply avogadro s number by the number of grams in aluminium to get the number of amus in aluminum. I recommend watching the video The mole and Avogadro s number on khan academy. Hope this helps! Excellent question. If you require further clarification, please don t hesitate to comment on my answer.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 500 + ], + "3min_transcript": "is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units. Or 6.02 times 10 to 23 aluminium atoms is going to be 27 grams. So if we need 1.06 moles, how many is that going to be? So 1.06 moles of aluminium is equal to 1.06 times 27 grams. And what is that? What is that? Equals 28.62. So we need 28.62 grams of aluminium, I won't write the whole thing there, in order to essentially use up our 85 grams of the iron three oxide. And if we had more than 28.62 grams of aluminium," + }, + { + "Q": "at 7:12 how does Sal get how many moles of Al are there\n", + "A": "he just multiplyed the Fe2O3 s mole(0.53) by 2. so you get 1.06. in the chemical equation. we have 2 Al for every Fe2O3. it s that easy", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 432 + ], + "3min_transcript": "So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units." + }, + { + "Q": "I have a question that says \"How many liters of oxygen are required to form 6.0 L of water\" and it gives me this balanced equation: 2H2+1O2-->2H20\nWhat I don't understand is what I'm supposed to to with the 6 L and the 2 mol of H20 to get to O. Do I divide the 6 by two so I have a 1:1 ratio or what?\n", + "A": "You shouldn t divide the 6, because the equation isn t dealing with liters, it s dealing with moles. 2 moles of H(2) and 1 mole of O(2) form 2 moles of H(2)O. 6 L is 6000 grams, so your next step is to find how many moles are in 6000 grams of water. THEN, you find the number of moles of oxygen you need using the 1:2 ratio, and then find the number of grams(then liters) of oxygen that you need, which should be your response.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 61 + ], + "3min_transcript": "We know what a chemical equation is and we've learned how to balance it. Now, we're ready to learn about stoichiometry. And this is an ultra fancy word that often makes people think it's difficult. But it's really just the study or the calculation of the relationships between the different molecules in a reaction. This is the actual definition that Wikipedia gives, stoichiometry is the calculation of quantitative, or measurable, relationships of the reactants and the products. And you're going to see in chemistry, sometimes people use the word reagents. For most of our purposes you can use the word reagents and reactants interchangeably. They're both the reactants in a reaction. The reagents are sometimes for special types of reactions where you want to throw a reagent in and see if something happens. And see if your belief about that substance is true or things like that. But for our purposes a reagent and reactant is the same thing. So it's a relationship between the reactants and the products in a balanced chemical equation. we know how to get to the balanced point. A balanced chemical equation. So let's do some stoichiometry. Just so we get practice balancing equations, I'm always going to start with unbalanced equations. Let's say we have iron three oxide. Two iron atoms with three oxygen atoms. Plus aluminum, Al. And it yields Al2 O3 plus iron. So remember when we're doing stoichiometry, first of all we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation. But I think it's good practice to actually balance the equations ourselves. So let's try to balance this one. We have two iron atoms here in this iron three oxide. How many iron atoms do we have on the right hand side? We only have one. So let's multiply this by 2 right here. We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams." + }, + { + "Q": "\nAt 4:34, does it make more sense to use the average mass number?", + "A": "Yes ideally he should not be rounding these. He should have used the number on the periodic table and round to the appropriate significant figures at the end of the calculation.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 274 + ], + "3min_transcript": "how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that?" + }, + { + "Q": "At 6:00, what is the relationship between moles and grams?\n", + "A": "For atoms 1 mole equals the relative atomic mass with units of grams (g). e.g. 1 mole of oxygen atoms has a mass of 16 g. For molecules, 1 mole equals the relative formula mass with units of grams (g). e.g. 1 mole of water (H2O) has a mass of (2(1) + 16) = 18 g.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 360 + ], + "3min_transcript": "So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum." + }, + { + "Q": "\nat 8:04, in the balanced equation, the coefficient is 2 for Al. So why wouldnt you multiply it's atomic mass by two, since 2 Al atoms are necessary?", + "A": "When balancing, you are initially not looking for mass ratios, but atom ratios. For instance magnesium oxide is MgO, or a 1 to 1 ratio between Mg and O. So the balanced chemical equation would be: 2 Mg + 1 O2 --> 2 MgO Next you could use the masses to find out what mass of Mg would react with what mass of O2 and thus give the product mass formed IF both reactants get used up completely.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 484 + ], + "3min_transcript": "is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units. Or 6.02 times 10 to 23 aluminium atoms is going to be 27 grams. So if we need 1.06 moles, how many is that going to be? So 1.06 moles of aluminium is equal to 1.06 times 27 grams. And what is that? What is that? Equals 28.62. So we need 28.62 grams of aluminium, I won't write the whole thing there, in order to essentially use up our 85 grams of the iron three oxide. And if we had more than 28.62 grams of aluminium," + }, + { + "Q": "At 6:05, you said that there was Fe3O3 but before that you said it was Fe2O3, so wouldn't the final answer be different\n", + "A": "He meant Fe203. If you have the pop ups enabled, a pop up shows up saying Sal wrote Fe303 but meant 1 mole of Fe203", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 365 + ], + "3min_transcript": "So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum." + }, + { + "Q": "\nAt 8:40 Sal said that the Black hole at the center of our galaxy is 4 million times the mass of our sun. How is this possible aren't black holes an object of infinite density", + "A": "The mass is in an infinitesimally small volume which would make any finite amount of mass have an infinite density.", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 520 + ], + "3min_transcript": "when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough, so that you could kind of have this snowballing effect, so that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the Sun. And, this is maybe even the more interesting part, those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going pass it fast enough, it'll just start going in orbit around the black hole. And so you could imagine that this is how the early galaxies or even our galaxy formed. what about the black hole at the center of the Milky Way? And we think there is one. We think there is one because we've observed stars orbiting very quickly around something at the center of the universe-- sorry, at the center of our Milky Way. I want to be very clear, not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the Sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are and somehow they grow into supermassive black holes and that everything in between we just can't observe. Or that they really are a different class of black holes. Maybe they formed near the beginning of the actual universe. When the density of things was a little uniform, things condensed into each other. And what we're going to talk about in the next video is how these supermassive black holes can help generate unbelievable sources of radiation, even though the black holes themselves aren't emitting them. And those are going to be quasars." + }, + { + "Q": "\nAt 6:15, Sal mentioned quantum fluctuations. What does that mean?", + "A": "It is the temporary change in the amount of energy in a point in space.", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 375 + ], + "3min_transcript": "So another possible explanation-- my inclinations lean towards this one because it kind of explains the gap-- is that these supermassive black holes actually formed shortly after the Big Bang, that these are primordial black holes. These started near the beginning of our universe, primordial black holes. Now remember, what do you need to have a black hole? You need to have an amazingly dense amount of matter or a dense amount of mass. If you have a lot of mass in a very small volume, then their gravitational pull will pull them closer, and closer, and closer together. And they'll be able to overcome all of the electron degeneracy pressures, and the neutron degeneracy pressures, and the quark degeneracy pressures, to really collapse into what we think is a single point. I want to be clear here, too. We don't know it's a single point. We've never gone into the center of a black hole. Just the mathematics of the black holes, or at least into a single point where the math starts to break down. So we're really not sure what happens at that very small center point. But needless to say, it will be an unbelievably, maybe infinite, maybe almost infinitely, dense point in space, or dense amount of matter. And the reason why I kind of favor this primordial black hole and why this would make sense is right after the formation of the universe, all of the matter in the universe was in a much denser space because the universe was smaller. So let's say that this is right after the Big Bang, some period of time after the Big Bang. Now what we've talked about before when we talked about cosmic background is that at that point, the universe was relatively uniform. It was super, super dense but it was relatively uniform. So a universe like this, there's no reason why anything would collapse into black holes. Because if you look at a point here, sure, there's a ton of mass very close to it. So the gravitational force would be the same in every direction if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly nonuniform. So let's say it becomes slightly nonuniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly nonuniform, but extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you could imagine in that primordial universe," + }, + { + "Q": "Wait, at 0:42, he talks about the event horizon saying that if anything gets past it, it cannot go back, but wouldn't gravity be like a constant, no just one line that separates things?\n", + "A": "The farther away you move from an object, the weaker its gravity gets. This is why the astronauts in orbit don t feel Earth s gravity and can float about. Well the event horizon is the distance where the gravity becomes strong enough to draw in light.", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 42 + ], + "3min_transcript": "In the videos on massive stars and on black holes, we learned that if the remnant of a star, of a massive star, is massive enough, the gravitational contraction, the gravitational force, will be stronger than even the electron degeneracy pressure, even stronger than the neutron degeneracy pressure, even stronger than the quark degeneracy pressure. And everything would collapse into a point. And we called these points black holes. And we learned there's an event horizon around these black holes. And if anything gets closer or goes within the boundary of that event horizon, there's no way that it can never escape from the black hole. All it can do is get closer and closer to the black hole. And that includes light. And that's why it's called a black hole. So even though all of the mass is at the central point, this entire area, or the entire surface of the event horizon, this entire surface of the event horizon-- I'll do it in purple because it's It will emit no light. Now these type of black holes that we described, we call those stellar black holes. And that's because they're formed from collapsing massive stars. And the largest stellar black holes that we have observed are on the order of 33 solar masses, give or take. So very massive to begin with, let's just be clear. And this is what the remnant of the star has to be. So a lot more of the original star's mass might have been pushed off in supernovae. That's plural of supernova. Now there's another class of black holes here and these are somewhat mysterious. And they're called supermassive black holes. And to some degree, the word \"super\" isn't big enough, supermassive black holes, than stellar black holes. They're are a lot more massive. They're on the order of hundreds of thousands to billions of solar masses, hundred thousands to billions times the mass of our Sun, solar masses. And what's interesting about these, other than the fact that there are super huge, is that there doesn't seem to be black holes in between or at least we haven't observed black holes in between. The largest stellar black hole is 33 solar masses. And then there are these supermassive black holes that we think exist. And we think they mainly exist in the centers of galaxies. And we think most, if not all, centers of galaxies actually have one of these supermassive black holes. But it's kind of an interesting question, if all black holes were formed from collapsing stars, wouldn't we see things in between? So one theory of how these really massive black holes form" + }, + { + "Q": "At 5:06, when Sal says that the math starts to break down, what does he mean? Is it that different laws of physics apply inside a black hole as opposed to outside of it? How is that possible?\n", + "A": "The math starts to break down means that you are starting to try to do things like divide by zero, which you generally can t do.", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 306 + ], + "3min_transcript": "in an area that has a lot of matter that it can accrete around it. So I'll draw the-- this is the event horizon around it. The actual black hole is going to be in the center of it, or rather the mass of the black hole will be in the center of it. And then over time, you have just more and more mass just falling into this black hole. Just more and more stuff just keeps falling into this black hole. And then it just keeps growing. And so this could be a plausible reason, or at least the mass in the center keeps growing and so the event horizon will also keep growing in radius. Now this is a plausible explanation based on our current understanding. But the reason why this one doesn't gel that well is if this was the explanation for supermassive black holes, you expect to see more black holes in between, maybe black holes with 100 solar masses, or a 1,000 solar masses, or 10,000 solar masses. But we're not seeing those right now. We just see the stellar black holes, So another possible explanation-- my inclinations lean towards this one because it kind of explains the gap-- is that these supermassive black holes actually formed shortly after the Big Bang, that these are primordial black holes. These started near the beginning of our universe, primordial black holes. Now remember, what do you need to have a black hole? You need to have an amazingly dense amount of matter or a dense amount of mass. If you have a lot of mass in a very small volume, then their gravitational pull will pull them closer, and closer, and closer together. And they'll be able to overcome all of the electron degeneracy pressures, and the neutron degeneracy pressures, and the quark degeneracy pressures, to really collapse into what we think is a single point. I want to be clear here, too. We don't know it's a single point. We've never gone into the center of a black hole. Just the mathematics of the black holes, or at least into a single point where the math starts to break down. So we're really not sure what happens at that very small center point. But needless to say, it will be an unbelievably, maybe infinite, maybe almost infinitely, dense point in space, or dense amount of matter. And the reason why I kind of favor this primordial black hole and why this would make sense is right after the formation of the universe, all of the matter in the universe was in a much denser space because the universe was smaller. So let's say that this is right after the Big Bang, some period of time after the Big Bang. Now what we've talked about before when we talked about cosmic background is that at that point, the universe was relatively uniform. It was super, super dense but it was relatively uniform. So a universe like this, there's no reason why anything would collapse into black holes. Because if you look at a point here, sure, there's a ton of mass very close to it." + }, + { + "Q": "At 1:12 Sal said there are Stellar Blackholes and at 4:18 he said there are Primorial (super massive) Blackholes. Is there any other type of Blackhole?\n", + "A": "There could be, but those are the only two types we have discovered.", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 72, + 258 + ], + "3min_transcript": "In the videos on massive stars and on black holes, we learned that if the remnant of a star, of a massive star, is massive enough, the gravitational contraction, the gravitational force, will be stronger than even the electron degeneracy pressure, even stronger than the neutron degeneracy pressure, even stronger than the quark degeneracy pressure. And everything would collapse into a point. And we called these points black holes. And we learned there's an event horizon around these black holes. And if anything gets closer or goes within the boundary of that event horizon, there's no way that it can never escape from the black hole. All it can do is get closer and closer to the black hole. And that includes light. And that's why it's called a black hole. So even though all of the mass is at the central point, this entire area, or the entire surface of the event horizon, this entire surface of the event horizon-- I'll do it in purple because it's It will emit no light. Now these type of black holes that we described, we call those stellar black holes. And that's because they're formed from collapsing massive stars. And the largest stellar black holes that we have observed are on the order of 33 solar masses, give or take. So very massive to begin with, let's just be clear. And this is what the remnant of the star has to be. So a lot more of the original star's mass might have been pushed off in supernovae. That's plural of supernova. Now there's another class of black holes here and these are somewhat mysterious. And they're called supermassive black holes. And to some degree, the word \"super\" isn't big enough, supermassive black holes, than stellar black holes. They're are a lot more massive. They're on the order of hundreds of thousands to billions of solar masses, hundred thousands to billions times the mass of our Sun, solar masses. And what's interesting about these, other than the fact that there are super huge, is that there doesn't seem to be black holes in between or at least we haven't observed black holes in between. The largest stellar black hole is 33 solar masses. And then there are these supermassive black holes that we think exist. And we think they mainly exist in the centers of galaxies. And we think most, if not all, centers of galaxies actually have one of these supermassive black holes. But it's kind of an interesting question, if all black holes were formed from collapsing stars, wouldn't we see things in between? So one theory of how these really massive black holes form" + }, + { + "Q": "at 8:18 he says the center of the universe (unintentionally, and then fixes it) but it got me thinking .Is there a center of the universe?\n", + "A": "There is no center of the universe, so far as we can tell. There is a center of portion of the universe that we can observe, which we call the observable universe. By definition, we are at the center of that, because we can observe the same distance in every direction.", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 498 + ], + "3min_transcript": "So the gravitational force would be the same in every direction if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly nonuniform. So let's say it becomes slightly nonuniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly nonuniform, but extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you could imagine in that primordial universe, when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough, so that you could kind of have this snowballing effect, so that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the Sun. And, this is maybe even the more interesting part, those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going pass it fast enough, it'll just start going in orbit around the black hole. And so you could imagine that this is how the early galaxies or even our galaxy formed. what about the black hole at the center of the Milky Way? And we think there is one. We think there is one because we've observed stars orbiting very quickly around something at the center of the universe-- sorry, at the center of our Milky Way. I want to be very clear, not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the Sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are and somehow they grow into supermassive black holes and that everything in between we just can't observe. Or that they really are a different class of black holes." + }, + { + "Q": "\nat 7:31, he said that there were might be black holes between stellar mass black holes and supermassive black holes. are there?", + "A": "Yes but not very many", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 451 + ], + "3min_transcript": "So the gravitational force would be the same in every direction if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly nonuniform. So let's say it becomes slightly nonuniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly nonuniform, but extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you could imagine in that primordial universe, when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough, so that you could kind of have this snowballing effect, so that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the Sun. And, this is maybe even the more interesting part, those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going pass it fast enough, it'll just start going in orbit around the black hole. And so you could imagine that this is how the early galaxies or even our galaxy formed. what about the black hole at the center of the Milky Way? And we think there is one. We think there is one because we've observed stars orbiting very quickly around something at the center of the universe-- sorry, at the center of our Milky Way. I want to be very clear, not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the Sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are and somehow they grow into supermassive black holes and that everything in between we just can't observe. Or that they really are a different class of black holes." + }, + { + "Q": "\nAt 8:14 the stars could be orbitting something besides a black hole that scientists haven't discovered yet, right?", + "A": "We aren t just orbiting the black hole (its much to far away to affect us greatly), we are orbiting every other thing in our galaxy that is closer to the center. We orbit the common center of mass for our galaxy. Our galaxy began just orbiting that one black hole but as it got bigger, it needed the gravity of every other thing in it to keep it together.", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 494 + ], + "3min_transcript": "So the gravitational force would be the same in every direction if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly nonuniform. So let's say it becomes slightly nonuniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly nonuniform, but extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you could imagine in that primordial universe, when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough, so that you could kind of have this snowballing effect, so that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the Sun. And, this is maybe even the more interesting part, those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going pass it fast enough, it'll just start going in orbit around the black hole. And so you could imagine that this is how the early galaxies or even our galaxy formed. what about the black hole at the center of the Milky Way? And we think there is one. We think there is one because we've observed stars orbiting very quickly around something at the center of the universe-- sorry, at the center of our Milky Way. I want to be very clear, not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the Sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are and somehow they grow into supermassive black holes and that everything in between we just can't observe. Or that they really are a different class of black holes." + }, + { + "Q": "4:31... Why would the two groups attached to a carbon be named in order of how big each of the R-groups are, and not alphabetically?\n\nI would have thought that butyl should come before propyl, not the other way round. (and I do realize that this is a common name, not the IUPAC name)\n", + "A": "Even in common names, the rule is to list the groups alphabetically. The name methyl ethyl ketone is traditional and was assigned long before anyone thought of a system for common names. That name is still used by many people. But the systematic common name puts the names of the groups in alphabetical order.", + "video_name": "wD15pD5pCt4", + "timestamps": [ + 271 + ], + "3min_transcript": "Propa-, and instead of calling it propane, we get rid of that \"e\" over there and we would call it propanone. That tells us that this right here is a ketone. And you have to know where this double bond is. And actually, for propanone, you don't have to specify it, because if you know it's a ketone, you know that it has to have a carbon on either side of the carbonyl group, so you actually don't even have to specify where the carbonyl group is. But if you wanted to, you could say, OK, that's going to be on the two carbon. No matter what direction you start counting from, it's going to be on the two carbon. But the two is kind of optional for propanone. Let's do a couple of other ones. So let's say we had a molecule that looks like this. So the traditional way of naming it, you'd say, OK, on So on that end, I have three carbons. That is a propyl group. And on this other side of the ketone right over here, I have only one carbon. That is a methyl group. So then you would just name them. And you name then in order of increasing chain size, molecule size, or group size. So this one you'd write methyl first. Methyl, because it's only one carbon. So this is methyl propyl ketone. This is kind of the traditional or the common way, often kind of the most used way, of naming this molecule. But the systematic way of naming it, you look at the longest carbon chain and you say, OK, I have one, two, three, four, five carbons. So it's going to be pent-. And then you want to start numbering it so that the So you want to start numbering on the right side: one, two, three, four and five. So this right here, so we said the prefix would be pent-. So it's penton, and instead of saying it's pentane, you say it's pentanone. And to specify where the carbonyl group is, you say it's 2. This is 2-pentanone. And you might also see it written like this: pentan-2-one. Either one of these right here would be acceptable. Let's do a slightly more complicated example. Let's say we had something that looked like this. So we have something, a molecule, that looks like this. And let me stick some chlorines over here. So what would this be? Well, our longest chain, once again, is this cyclohexane:" + }, + { + "Q": "\nFrom 6:05 to 6:30, Jay mentioned that exactly 10.2 eV is needed to promote the level 1 electron to energy level 2. Does that mean that if x eV of energy is given to the level 1 electron, such that 10.2 < x < 12.9, the electron will not be promoted to level 2?", + "A": "Yes, photons whose energies don t match the possible jumps are not absorbed.", + "video_name": "nJ-PtF14EFw", + "timestamps": [ + 365, + 390 + ], + "3min_transcript": "When N is equal to one, that was negative 13.6 electron volts, that's the energy associated with that electron as it orbits the nucleus. And so if we go over here on the right, and we say this top line here represents energy is equal to zero, then this would be negative 13.6 electron volts. So none of this is drawn perfectly to scale, but this is just to give you an idea about what's happening. So this is when N is equal to one, the electron is at a distance R one away from the nucleus, we're talking about the first energy level, and there's an energy of negative 13.6 electron volts associated with that electron. Alright, let's say the electron was located a distance R two from the nucleus. Alright, so that's N is equal to two, and we just calculated that energy is equal to negative 3.4 electron volts, alright? And then let's say the electron was at R three and once again we just calculated that energy to be equal to negative 1.51 electron volts. And so it's useful to compare these two diagrams together, because we understand this concept of energy much better. For example, let's say we wanted to promote the electron that I drew, so this electron right here I just marked. Let's say we wanted to promote that electron from the lower energy level to a higher energy level. So let's say we wanted to add enough energy to cause that electron to go from the first energy level to the second energy level, so that electron is jumping up here to the second energy level. We would have to give that electron this much energy, so the difference in energy between our two energy levels, so the difference between these two numbers. And if you're thinking about just in terms of magnitude, alright, 13.6 minus 3.4, alright? And so if you gave that electron 10.2 electron volts of energy, right, you could cause that electron to jump from the first energy level all the way here to the second energy level. But you would have to provide the exact right amounts of energy in order to get it to do that. Alright, let's say you wanted to cause the electron to jump, let's say, from the first energy level all the way to the third energy level, alright? So from the first energy level to the third energy level. So that would be, here's our electron in the first energy level, let's say we wanted to cause it to jump all the way up to here, alright? So once again, you would have to provide enough energy in order to do that. So, just thinking about the magnitudes, right? This was negative 1.51, this was negative 13.6, if we just take 13.6 minus 1.51, alright, we would get how much energy we need to put in" + }, + { + "Q": "At 1:11, what hormone is being referred to? Also, at 3:43, what neurotransmitter is she talking about?\n", + "A": "Ghrelin is the hormone that signals hunger. Orexin/hypocretin is missing in narcolepsy patients (in most cases).", + "video_name": "VBcEz8bVbL0", + "timestamps": [ + 71, + 223 + ], + "3min_transcript": "Voiceover: I\u2019m sure we\u2019ve all had trouble sleeping at one point or another, maybe trouble falling asleep, staying asleep or waking up or maybe you're forcing yourself to sleep less because you have too much to do to lie in bed. But sleep deprivation can be a serious issue. People who don't get enough sleep are more irritable and perform worse on memory and detention tasks than people who do. So all this can be just a minor annoyance in everyday life, imagine the long-term implications for let's say, airline pilots, firefighters, security officers or the person driving next to you on the freeway. For example, one study in Canada showed that the Monday after the Spring time change, so when people lose an hour of sleep, the number of traffic accidents increases sharply compared to the Monday after the Fall time change when people get an extra hour of sleep, the number of accidents decreases sharply. So that's just one example, but sleep deprivation also makes people more susceptible to obesity. which is a hormone that tells your body to make more fat. You also produce more of the hormone that tells your body you're hungry, so you end up eating more and turning more of what you eat into fat which can contribute to weight gain. And finally sleep deprivation can also increase your risk for depression and one theory about this link is that REM sleep helps your brain process emotional experiences, which in turn helps protect against depression though we're still not entirely sure about this link. Most people, now most people experience sleep deprivation at some points in their lives, but the good news is that most people can get back on track by getting a few nights of good sleep, sort of paying back your sleep debt. Your next question might be then, \"How much sleep is enough sleep?\" That's kind of a hard question to answer, but most adults need about 7-8 hours of sleep, Babies need a lot more sleep, for example, than older adults often sleep less than 10 or 8 hours without severe detriments. Again everyone has trouble falling asleep at some point, but people who have persistent problems in falling or staying asleep have a more serous sleep disorder called insomnia. There are various medications that can help people fall asleep, but taking them for too long can result in dependence and tolerance, which is if a person continues to rely on the medication then their body will get used to it and they'll eventually need more and more to get the same affects. Now, that can often be a bad thing because there are side effects to drugs, but so treatment for insomnia often involves psychological training as well as or sometimes instead of medication and some lifestyle changes might also be necessary. For example, exercise regularly but not right before bed or spend some time just relaxing before bed" + }, + { + "Q": "\nWhy Momentum of Ball A and B in y direction at initial position is zero ? 3:07", + "A": "okay thanks i got it :)", + "video_name": "CFygKiTB-4A", + "timestamps": [ + 187 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt around 4:52, Sal calculated that the X-component is the square root of 3. I thought since the velocity was 3, you would just move the vector and the x-component would be 3. Could someone explain please?", + "A": "Actually v=3 is the initial velocity of the ball before the collision.After collision it reduces to \u00e2\u0088\u009a3.Just think of a collision between two carrom pieces or two snooker balls if you still don t understand this. Hope it helps", + "video_name": "CFygKiTB-4A", + "timestamps": [ + 292 + ], + "3min_transcript": "" + }, + { + "Q": "At 7:41 you write the name of the second molecule and you say\n\"which alphabet comes after b ,whcih is h (for heptyl)\"\nbut must it not be which chain comes is nearer to the first C Atom?\n", + "A": "No, when naming molecules, the order in which you name the chains is alphabetically. To indicate where each branch goes, you use numbers to put in front of the name of the branch", + "video_name": "Se-ekDNhCDk", + "timestamps": [ + 461 + ], + "3min_transcript": "This over here, let's see what we're dealing with. We have one, two, three carbons, so that is just a standard propyl group. Then here, we have one, two, three, four carbons, so we could say this is a butyl group. But this isn't just any butyl group. If we use the common naming, the carbon we immediately touch on or that we immediately get to when we go off of our main ring, that branches off into three other carbons, so this is tert-, tert- for three. The tert- is usually written in italics. It's hard to differentiate that when you see it. I'll write it in cursive. This is a tertbutyl group. Now, the next question is how do we specify where these different groups sit on this main ring? If you just had one group, you wouldn't have to specify it, but when you have more than one, what you actually do is you figure out which one would be alphabetically first, and that would be number one. Now, we have an H in heptyl, a P in propyl, and tert-butyl, you might say, well, do I use the T or do I use the B? And this is just the convention, you use the B. If you have sec-butyl or tert-butyl, you ignore the sec- or the tert-. If this was an isobutyl or an isopropyl, you actually would use the I, so it's a little bit-- I guess the best way to think about it is there's a dash here so you can kind of ignore it, but if this was isobutyl it would all be one word, so you would consider the I. So in this situation, we would consider the B, and B comes before P or an H, so that is where we will start numbering one. Then to figure out which direction to keep numbering to encounter the first or where we will encounter the first side chain, so we'll go in this direction because we get right to the propyl group. One, two, three, four, five, six, seven, eight, nine. So this compound, we're going to start with the alphabetically first side chain, so it's 1-tert-butyl. I'll write this in cursive. 1-tert-butyl. Then the next one alphabetically is the heptyl group. That's H for heptyl, so then it is 5-heptyl. And then we have the propyl, and then it is 2-propyl, and then finally cyclononane, and we're done!" + }, + { + "Q": "\nAt around 0:48, Sal says to always choose the chain that has the most alkyl groups on it. Why?", + "A": "It s a rule made to make naming strict and universal.", + "video_name": "Se-ekDNhCDk", + "timestamps": [ + 48 + ], + "3min_transcript": "We've got a few more molecular structures to name, so let's look at this first one right here. The first thing you always want to do is identify the longest chain. If we start over here, we have one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen carbons, looks pretty long. Now what if we start over here? This looks like it could also be a long chain: one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen. So we have two different chains, depending on whether we want to go up here or whether we go over here, that have a length of thirteen. So you're probably asking, which chain do you choose? You should always, if we have two chains of equal lengths and they're the longest chains, you pick the one that will have more branches or more alkyl groups on it. So this group right here, if we pick this from here to here as our chain, we only have one group on it, that group up there. If we pick this chain, starting over here and then We would have this group over here and then we would have this group over here. This is the better chain to use because it has more groups on it. It has more groups, but the groups are smaller and simpler. So let's start counting. And the direction we want to count, we always want to start on the side of the chain where we're going to encounter something first, and everything is closer this end of the chain so we'll start counting here. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, so we have thirteen carbons on our main chain. Let me draw our main chain. So our main chain is this thing in orange I'm drawing right here. That is our main chain. The prefix is tridec- and it's tridecane because we have all single bonds here, so it's tridecane. Tridecane is the main chain. Then we have two groups over here. This one in green, this only has one the carbon branching off of the main chain, so it's prefix will be meth- and it'll be a methyl group, so that is methyl. That is a methyl group right there. Then this one down here, we have one, two, three carbons. the prefix is prop-, so this is a propyl group. And the methyl is sitting on the three carbon of our main chain and the propyl group is sitting on the four carbon: one, two, three, four. Now when we figure out what order to list them in when we" + }, + { + "Q": "did you mean displacement near 3:35 or are you talking about distance?\n", + "A": "Silver Night, That s a good question considering how displacement and distance can be easily mixed up despite having subtle differences. In his example, he is referring to displacement considering he uses it in an equation involving velocity instead of speed, and denoting it has a directional component.", + "video_name": "Y5cSGxdDHz4", + "timestamps": [ + 215 + ], + "3min_transcript": "write this in a different color, so we have some variety-- velocity times time, or d equals rate times time. The reason why I have change in distance here, or change in time, is that I'm not assuming necessarily that we're starting off at the point 0 or at time 0. If we do, then it just turns out to the final distance is equal to the average velocity times the final time, but let's stick to this. We want to figure out time given this set of inputs. Let's go from this equation. If we want to solve for time, or the change in time, we could just could divide both sides by the average velocity-- actually, no, let's not do that. Let's just stay in terms of change in distance. this and start again. We're given change in distance, initial velocity, and acceleration, and we want to figure out what the time is-- it's really the change in time, but let's just assume that we start time 0, so it's kind of the final time. Let's just start with the simple formula: distance, or change in distance-- I'll use them interchangeably, with a lower case d this time-- is equal to the average velocity times time. What's the average velocity? The average velocity is just the initial velocity plus the The only reason why we can just average the initial and the final is because we're assuming constant acceleration, and that's very important, but in most projectile problems, we do have constant acceleration-- downwards-- and that's gravity. We can assume, and we can do this-- we can say that the average of the initial and the final velocity is the average velocity, and then we multiply that times time. Can we use this equation directly? No. we know acceleration, but don't know final velocity. If we can write this final velocity in terms of the other things in this equation, then maybe we can solve for time. Let's try to do that: distance is equal to-- let me take a little side here. What do we know about final velocity? We know that the change in velocity is equal to acceleration times time, assuming that time" + }, + { + "Q": "At 0:52, (specifically at the definition you gave) could you also say \"Work is the amount of Kinetic Energy that is required for an object to get from its resting position to its current position\"?\n", + "A": "No. Kinetic energy is energy associated with motion, not position. You could replace KE in your sentence with PE, and add that at the current position the object is also at rest, and then it would be right.", + "video_name": "3mier94pbnU", + "timestamps": [ + 52 + ], + "3min_transcript": "Welcome back. In the last video, I showed you or hopefully, I did show you that if I apply a force of F to a stationary, an initially stationary object with mass m, and I apply that force for distance d, that that force times distance, the force times the distance that I'm pushing the object is equal to 1/2 mv squared, where m is the mass of the object, and v is the velocity of the object after pushing it for a distance of d. And we defined in that last video, we just said this is work. Force times distance by definition, is work. And 1/2 mv squared, I said this is called kinetic energy. And so, by definition, kinetic energy is the amount of work-- and I mean this is the definition right here. It's the amount of work you need to put into an object or apply to an object to get it from rest So its velocity over here. So let's just say I looked at an object here with mass m and it was moving with the velocity v. I would say well, this has a kinetic energy of 1/2 mv squared. And if the numbers are confusing you, let's say the mass was-- I don't know. Let's say this was a 5 kilogram object and it's moving at 7 meters per second. So I would say the kinetic energy of this object is going to be 5-- 1/2 times the mass times 5 times 7 squared, times It's times 49. So let's see. 1/2 times 49, that's a little under 25. So it'll be approximately 125 Newton meters, which is approximately-- and Newton meter is just a joule-- 125 joules. So this is if we actually put numbers in. And so when we immediately know this, even if we didn't Let's say we didn't know that someone else had applied a force of m for a distance of d to this object, just by calculating its kinetic energy as 125 joules, we immediately know that that's the amount of work that was necessary. And we don't know if this is exactly how this object got to this velocity, but we know that that is the amount of work that was necessary to accelerate the object to this velocity of 7 meters per second. So let's give another example. And instead of this time just pushing something in a horizontal direction and accelerating it, I'm going to show you an example we're going to push something up, but its velocity really isn't going to change. Invert. Let's say I have a different situation, and we're on this planet, we're not in deep space. And I have a mass of m and I were to apply a force." + }, + { + "Q": "At 2:05, Is it ok if we make the same approximations or should we work things out accurately?\n", + "A": "It depends on how the answer is expected. if the answer is expected in whole nos. then you can do approximations to make your given values into whole nos. if answer is needed upto 2 or 3 decimal places then you can t make approximations.", + "video_name": "3mier94pbnU", + "timestamps": [ + 125 + ], + "3min_transcript": "Welcome back. In the last video, I showed you or hopefully, I did show you that if I apply a force of F to a stationary, an initially stationary object with mass m, and I apply that force for distance d, that that force times distance, the force times the distance that I'm pushing the object is equal to 1/2 mv squared, where m is the mass of the object, and v is the velocity of the object after pushing it for a distance of d. And we defined in that last video, we just said this is work. Force times distance by definition, is work. And 1/2 mv squared, I said this is called kinetic energy. And so, by definition, kinetic energy is the amount of work-- and I mean this is the definition right here. It's the amount of work you need to put into an object or apply to an object to get it from rest So its velocity over here. So let's just say I looked at an object here with mass m and it was moving with the velocity v. I would say well, this has a kinetic energy of 1/2 mv squared. And if the numbers are confusing you, let's say the mass was-- I don't know. Let's say this was a 5 kilogram object and it's moving at 7 meters per second. So I would say the kinetic energy of this object is going to be 5-- 1/2 times the mass times 5 times 7 squared, times It's times 49. So let's see. 1/2 times 49, that's a little under 25. So it'll be approximately 125 Newton meters, which is approximately-- and Newton meter is just a joule-- 125 joules. So this is if we actually put numbers in. And so when we immediately know this, even if we didn't Let's say we didn't know that someone else had applied a force of m for a distance of d to this object, just by calculating its kinetic energy as 125 joules, we immediately know that that's the amount of work that was necessary. And we don't know if this is exactly how this object got to this velocity, but we know that that is the amount of work that was necessary to accelerate the object to this velocity of 7 meters per second. So let's give another example. And instead of this time just pushing something in a horizontal direction and accelerating it, I'm going to show you an example we're going to push something up, but its velocity really isn't going to change. Invert. Let's say I have a different situation, and we're on this planet, we're not in deep space. And I have a mass of m and I were to apply a force." + }, + { + "Q": "Hi Sal, in the example \"5:30\", how could the distance between two points in the universe be 30 million light years if the whole universe has only been expanding for 300K light years? Could the universe be expanding faster than the speed of light?\n", + "A": "Actually, there s nothing preventing the universe from expanding faster than the speed of light.", + "video_name": "6nVysrZQnOQ", + "timestamps": [ + 330 + ], + "3min_transcript": "If we were talking about smaller time scales or I guess smaller distances, you could say approximately that. The expansion of the universe itself would not make as much of a difference. And let me make it even more clear. I'm talking about an object over there. But we could even talk about that coordinate in space. And actually, I should say that coordinate in space-time, because we're viewing it at a certain instant as well. But that coordinate is not 13.7 billion light years away from our current coordinate. And there's a couple of reasons to think about it. First of all, think about it, that light was emitted 13.7 billion years ago. When that light was emitted, we were much closer to that coordinate. This coordinate was much closer to that. Where we are in the universe right now was much closer to that point in the universe. as this-- let me actually draw it. So let's go 300,000 years after that initial expansion of that singularity. So we're just 300,000 years into the universe's history right now. So this is roughly 300,000 years into the universe's life. I guess we could view it that way. And first of all, at that point things haven't differentiated in a meaningful way yet right now. And we'll talk more about this when we talk about the cosmic microwave background radiation. But at this point in the universe, it was kind of this almost uniform white-hot plasma of hydrogen. And then we're going to talk about it. It was emitting microwave radiation. And we'll talk more about that in a future video. But let's just think about two points in this early universe. And let's say you have the coordinate where we are right now. You have the coordinate where we are right now. And in fact, I'll just make that roughly-- I won't make it the center just because I think it makes it easier to visualize if it's not the center. And let's say at that very early stage in the universe, if you were able to just take some rulers instantaneously and measure that, you would measure this distance to be 30 million light years. And let's just say right at that point, this object over here-- I'll do it in magenta-- this object over here emits a photon, maybe in the microwave frequency range. And we'll see that that was the range that it was emitting in. But it emits a photon. And that photon is traveling at the speed of light. It is light. And so that photon, says, you know what, I only got 30 million light years to travel. That's not too bad. I'm going to get there in 30 million years. And I'm going to do it discrete." + }, + { + "Q": "At 2:26, arc length can also be found using a degree.\nso why do we use radians if we can find arc length by using\n(Theta/360)*2pi*r? I mean why do we have to use radians?\n", + "A": "radians have no units. This makes them much more convenient than degrees.", + "video_name": "dVYYh8C80zo", + "timestamps": [ + 146 + ], + "3min_transcript": "- [Instructor] So in the previous video, we defined all the new angular motion variables and we made an argument that those are more useful in many cases to use than the regular motion variables for things that are rotating in a circle. Since every point on the string in tennis ball, let's say this is a tennis ball you tied a string to and you're whirling around in a circle. Every point on the string including the tennis ball will have the same angular displacement, angular velocity and angular acceleration. But even though using angular motion variables is more convenient for these rotational motion problems, it's also really important to know how to translate those angular motion variables back into the regular motion variables. So that's what I wanna show you in this video how to translate angular motion variables back into regular motion variables. So let's do this. The simplest possible angular motion variable was the angular displacement because that just represented how much angle an object has rotated through. So let's say it rotated through this much. We represented the angular displacement with a delta theta In Physics, we typically choose to measure this in radians for a reason and I'll show you in just a second. Now, how would we convert this into a regular motion variable? What regular motion variable would that be? If I were to come at this for the first time, I'd be like all right, this is the angular displacement. Let's figure out how to relate it to the regular displacement but that would be weird. Because just think about it, the regular displacement for the ball that started over here and made it over here would be from this point to that point, that would be the regular displacement of the ball, the regular linear displacement of the ball. That's a little weird. I don't wanna show you how to find that for one, you have to use the law of cosines. That's a little more in depth than I'd wanna get to in this video. For two, the better reason this isn't all that useful in turns out. There's a much more useful quantity that would tell you how far the ball went. That's the arc length of the ball. So the ball traced out of path through space We call this the arc length that it turns out this is much more useful in a variety of problems. Good news is it's much easier to find than that regular displacement. So this is the arc length. People vary on what letter to use here. I've seen an l but most math books use s so we'll just use s as well. You might think this is hard to find but it's not. In fact, if we use radians and this is why we use radians, it's extremely easy to find. If we wanted to find the arc length of this tennis ball, we're just gonna take the radius of the circular path that tennis ball is tracing out. So in this case it'd be the length of the string. We take that radius. If we're in radians, we just multiply by the angular displacement. This is why the radians are so convenient. We just take that measurement in radians multiplied by the radius of the circular path the object is tracing out. You get the arc length which is the number of meters along this path that the object has traveled." + }, + { + "Q": "at 7:13, Sal says \"...solar systems...\", then something says Sal meant \"planetary systems\". way is that? way aren't they called solar systems?\n", + "A": "Sol is the old name for our sun specifically. so our planetary system is called The Solar system. Other planetary systems would be named after the star they orbit, like the Wolf system.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 433 + ], + "3min_transcript": "looking at an object that's sitting-- let me do this in a darker color-- if we're sitting here on Earth and we're looking at an object out here that's 500 light years away, we're looking at it as it was 500 years ago because the light that is reaching our eyeballs right now, or our telescopes right now, left this guy over here 500 years ago. In fact, he's not going to even be there anymore. He probably has moved around a little bit. So just even on this scale, we're talking about these unimaginably huge distances. And then when we zoom out, this is kind of our local part of the galaxy right over here. This piece right here, this is called the Orion Spur. And people are still trying to work out exactly the details of the actual shape the Milky Way Galaxy, the galaxy that we're in. But we're pretty sure-- actually, we're very sure-- we have these spiral arms But it's actually very hard to come up with the actual shape, especially because you can't see a lot of the galaxy, because it's kind of on the other side, on the other side of the center. But really just to get a sense of something that at least-- I mean it blows my mind if you really think about what it's saying-- these unbelievable distances show up as a little dot here. This whole drawing shows up as a dot here. Now when we zoom out, over here that dot would no longer even show up. It wouldn't even register a pixel on this drawing right over here. And then this whole drawing, this whole thing right over here, this whole picture is this grid right over here. It is this right over here. So hopefully, that gives you a sense of how small even our local neighborhood is relative to the galaxy as a whole. And the galaxy as a whole, just to give you a sense, has 200 to 400 billion stars. to give you a sense that when we saw the solar system, it's not just the Sun. There's all this neat, dynamic stuff. And there are planets, and asteroids, and solar winds. And so there's 200 to 400 billion stars and for the most part, 200 and 400 billion solar systems. So it's an unimaginably, I guess, complex or huge place. And just to make it clear, even when we zoom in to this picture right here, and I think it was obvious based on telling you about this, that these little white pockets right here, This isn't two stars. These are thousands of stars here. So when you go over here, each little blotch of white that you see, that's not a star, that's not a thousand stars. We're starting to talk in the millions of stars when you look at certain blotches here and there. I mean, maybe it might be one star that's closer to you or might be a million stars that are far apart" + }, + { + "Q": "\nat the picture in 12:23 what is that shiny bright light in the lower left corner", + "A": "It is a planet. Difficult to say which one without knowing when this picture was taken. Although my guess would most likely be Jupiter.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 743 + ], + "3min_transcript": "it's hard to say the edge of the galaxy, because there's always going to be a few more stars and other things orbiting around the galaxy as you go further and further out, but it gets less dense with stars. But the main density, the main disk, is about 100,000 light years. 100,000 light years is the diameter roughly of the main part of the galaxy. And it's about 1,000 light years thick. So you can kind of imagine it as this disk, this thing that's fairly flat. But it's 1,000 light years thick. It's 1,000 light years thick. You would have to do this distance 250 times just to go from the top part of the galaxy to the bottom part, much less going across the galaxy. So it might seem relatively flat. But it's still immensely, immensely thick. And just as another way to visualize it, if this thing right over here that includes the Oort Cloud, roughly a light year in diameter, is a grain of sand, then the universe as a whole is going to be the diameter of a football field. And that might tell you, OK, those are two tractable things. I can imagine a grain of sand, a millimeter wide grain of sand in a football field. But remember, that grain of sand is still 50,000 or 60,000 times the diameter of Earth's orbit. And Earth's orbit, it would still take a bullet or something traveling as fast as a jet plane 15 hours to just go half of that-- or sorry, not-- 15 years or 17 years, I forgot the exact number. But it was 15, 16, 17 years to even cover half of that distance. So 30 years just to cover the diameter of Earth's orbit. That's 1/60,000 of our little grain of sand in the football And just to kind of really, I don't know, have an appreciation for how mind-blowing this really is, this is actually a picture of the Milky Way Galaxy, our galaxy, from our vantage point. As you can see, we're in the galaxy and this is looking towards the center. And even this picture, you start to appreciate the complexity of what 100 billion stars are. But what I really want to point out is even in this picture, when you're looking at these things, some of these things that look like stars, those aren't stars. those are thousands of stars or millions of stars. Maybe it could be one star closer up. But when we're starting to approach the center of the galaxy, these are thousands and thousands and millions of stars or solar systems that we're actually looking at. So really, it starts to boggle the mind to imagine what might actually be going on over there." + }, + { + "Q": "At 1:27 he says that the Voyager space craft was the fastest thing that has been made by man and sent out in space. I thaght it was the New Horizens space craft on the way to Pluto that was the fastest. Wich is the faster one?\n", + "A": "While New Horizons may have started off faster, Voyager 1 is the fastest as it got more of a speed boost from its gravitational assists from Jupiter and Saturn.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 87 + ], + "3min_transcript": "In the last video, I hinted that things were about to get wacky. And they are. So if we start where we left off in the last video, we started right over here, looking at the distance to the nearest star. And just as a reminder, in this drawing right here, this depiction right here, this circle right here, this solar system circle, it's not the size of the Sun. It's not the size of the orbits of the Earth, or Pluto, or the Kuiper Belt. This is close to the size of the Oort Cloud. And the actual orbit of Earth is about-- well, the diameter of the orbit of Earth is about 1/50,000 of this. So you wouldn't even see it on this. It would not even make up a pixel on this screen right here, much less the actual size of the Sun or something much smaller. And just remember that orbit of the Earth, that was at a huge distance. It takes eight minutes for light to get from the Sun to the Earth, this super long distance. If you shot a bullet at the Sun from Earth, So one thing, this huge distance wouldn't even show up on this picture. Now, what we saw in the last video is that if you travel at unimaginably fast speeds, if you travel at 60,000 kilometers per hour-- and I picked that speed because that's how fast Voyager 1 actually is traveling. That's I think the fastest object we have out there in space right here. And it's actually kind of leaving the solar system as we speak. But even if you were able to get that fast, it would still take 80,000 years, 75,000 or 80,000 years, to travel the 4.2 light years to the Alpha Centauri cluster of stars. To the nearest star, it would take 80,000 years. And that scale of time is already an amount of time that I have trouble comprehending. As you can imagine, all of modern civilization But most of recorded history is in the last 4,000 or 5,000 years. So this is 80,000 years to travel to the nearest star. So it's a huge distance. Another way to think about it is if the Sun were the size of a basketball and you put that basketball in London, if you wanted to do it in scale, the next closest star, which is actually a smaller basketball, right over here, Proxima Centauri, that smaller basketball you would have to put in Kiev, Ukraine in order to have a similar scale. So these are basketballs sitting in these cities. And you would have to travel about 1,200 miles to place the next basketball. And these basketballs are representing these super huge things that we saw in the first video. The Sun, if you actually made the Earth relative to these basketballs, these would be little grains of sand. So there are any little small planets over here, they would have to be grains of sand" + }, + { + "Q": "At 8:54 in the video, Sal mentioned the word supermassive black holes.What is a supermassive black hole and what is it's purpose?\n", + "A": "Supermassive black holes are giant black holes that are in the center of galaxies (well most of them). The Milky Way has one called Sagittarius A*", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 534 + ], + "3min_transcript": "to give you a sense that when we saw the solar system, it's not just the Sun. There's all this neat, dynamic stuff. And there are planets, and asteroids, and solar winds. And so there's 200 to 400 billion stars and for the most part, 200 and 400 billion solar systems. So it's an unimaginably, I guess, complex or huge place. And just to make it clear, even when we zoom in to this picture right here, and I think it was obvious based on telling you about this, that these little white pockets right here, This isn't two stars. These are thousands of stars here. So when you go over here, each little blotch of white that you see, that's not a star, that's not a thousand stars. We're starting to talk in the millions of stars when you look at certain blotches here and there. I mean, maybe it might be one star that's closer to you or might be a million stars that are far apart And everything has to be used in kind of loose terms here. And we'll talk more about other galaxies. But even this isn't the upper bound of galaxies. People believe the Andromeda Galaxy has a trillion stars in it, a trillion solar systems. We're talking about these huge, huge, immense distances. And so just to give you a sense of where we fit in the picture, this is a rough location of our Sun. And remember, that little dot I drew just now is including millions of stars, millions of solar systems, already unimaginable the distances. But if you really want to get at the sense relative to the whole galaxy, this is an artist's depiction. Once again, we could never obviously get this perspective on the galaxy. It would take us forever to travel this far so that you could see the galaxy from above. But this is our best guess looking at things from our vantage point. And we actually can't even see this whole area over here because it's on the other side of the center of the galaxy, which is super, super dense and super bright. We think-- or actually there's a super massive black hole at the center of the galaxy. And we think that they're at the center of all or most galaxies. But you know the whole point of this video, actually this whole series of videos, this is just kind of-- I don't know-- to put you in awe a little bit of just how huge Because when you really think about the scale-- I don't know-- no words can really describe it. But just to give you a sense, we're about 25,000 light years from the center of the galaxy. So even when we look at things in the center of the galaxy, that's as they were 25,000 years ago. It took 25,000 years for that light to get to us. I mean when that light left the center of the galaxy, I won't even guess to think what humanity was like at that point in time. So it's these huge distances in the whole galaxy over here." + }, + { + "Q": "\nAt 12:14, Khan shows a picture of the Milky Way galaxy. In the bottom (and a little to the left) there is an IMMENSLEY COLOSSAL star or billions of stars that make a huge, bright and perfect star-shaped figure or formation. Please explain all about this form so I will understand.\n-Thanks, Easton", + "A": "But it is an artist hypothesized sketch?", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 734 + ], + "3min_transcript": "it's hard to say the edge of the galaxy, because there's always going to be a few more stars and other things orbiting around the galaxy as you go further and further out, but it gets less dense with stars. But the main density, the main disk, is about 100,000 light years. 100,000 light years is the diameter roughly of the main part of the galaxy. And it's about 1,000 light years thick. So you can kind of imagine it as this disk, this thing that's fairly flat. But it's 1,000 light years thick. It's 1,000 light years thick. You would have to do this distance 250 times just to go from the top part of the galaxy to the bottom part, much less going across the galaxy. So it might seem relatively flat. But it's still immensely, immensely thick. And just as another way to visualize it, if this thing right over here that includes the Oort Cloud, roughly a light year in diameter, is a grain of sand, then the universe as a whole is going to be the diameter of a football field. And that might tell you, OK, those are two tractable things. I can imagine a grain of sand, a millimeter wide grain of sand in a football field. But remember, that grain of sand is still 50,000 or 60,000 times the diameter of Earth's orbit. And Earth's orbit, it would still take a bullet or something traveling as fast as a jet plane 15 hours to just go half of that-- or sorry, not-- 15 years or 17 years, I forgot the exact number. But it was 15, 16, 17 years to even cover half of that distance. So 30 years just to cover the diameter of Earth's orbit. That's 1/60,000 of our little grain of sand in the football And just to kind of really, I don't know, have an appreciation for how mind-blowing this really is, this is actually a picture of the Milky Way Galaxy, our galaxy, from our vantage point. As you can see, we're in the galaxy and this is looking towards the center. And even this picture, you start to appreciate the complexity of what 100 billion stars are. But what I really want to point out is even in this picture, when you're looking at these things, some of these things that look like stars, those aren't stars. those are thousands of stars or millions of stars. Maybe it could be one star closer up. But when we're starting to approach the center of the galaxy, these are thousands and thousands and millions of stars or solar systems that we're actually looking at. So really, it starts to boggle the mind to imagine what might actually be going on over there." + }, + { + "Q": "At 9:40, he talks about how we see things in space as they were. How do they come up with these numbers? If we see something 5 billion light years away, how do we know if it's not 3, or 7 billion light years away?\n", + "A": "Astronomers have several techniques for measuring distance. These include stellar parallax, spectroscopic parallax, the use of Cepheid variable stars as standard candles , and measurement of the spectral red shift of distant galaxies. Sal has videos about parallax and Cepheid variables. He also has vids about the redshift, although I am not sure he explains how it can be used to estimate distance.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 580 + ], + "3min_transcript": "And everything has to be used in kind of loose terms here. And we'll talk more about other galaxies. But even this isn't the upper bound of galaxies. People believe the Andromeda Galaxy has a trillion stars in it, a trillion solar systems. We're talking about these huge, huge, immense distances. And so just to give you a sense of where we fit in the picture, this is a rough location of our Sun. And remember, that little dot I drew just now is including millions of stars, millions of solar systems, already unimaginable the distances. But if you really want to get at the sense relative to the whole galaxy, this is an artist's depiction. Once again, we could never obviously get this perspective on the galaxy. It would take us forever to travel this far so that you could see the galaxy from above. But this is our best guess looking at things from our vantage point. And we actually can't even see this whole area over here because it's on the other side of the center of the galaxy, which is super, super dense and super bright. We think-- or actually there's a super massive black hole at the center of the galaxy. And we think that they're at the center of all or most galaxies. But you know the whole point of this video, actually this whole series of videos, this is just kind of-- I don't know-- to put you in awe a little bit of just how huge Because when you really think about the scale-- I don't know-- no words can really describe it. But just to give you a sense, we're about 25,000 light years from the center of the galaxy. So even when we look at things in the center of the galaxy, that's as they were 25,000 years ago. It took 25,000 years for that light to get to us. I mean when that light left the center of the galaxy, I won't even guess to think what humanity was like at that point in time. So it's these huge distances in the whole galaxy over here. it's hard to say the edge of the galaxy, because there's always going to be a few more stars and other things orbiting around the galaxy as you go further and further out, but it gets less dense with stars. But the main density, the main disk, is about 100,000 light years. 100,000 light years is the diameter roughly of the main part of the galaxy. And it's about 1,000 light years thick. So you can kind of imagine it as this disk, this thing that's fairly flat. But it's 1,000 light years thick. It's 1,000 light years thick. You would have to do this distance 250 times just to go from the top part of the galaxy to the bottom part, much less going across the galaxy. So it might seem relatively flat. But it's still immensely, immensely thick. And just as another way to visualize it, if this thing right over here that includes the Oort Cloud, roughly a light year in diameter, is a grain of sand," + }, + { + "Q": "This may sound like a stupid question, but if you look in the middle of the galaxy at around 5:30 you see that there is a very bright yellow spot that is immensely long. What is it?\n", + "A": "That is a very large cluster of densely packed yellow stars. There are probably several clusters like that in the center of the galaxy.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 330 + ], + "3min_transcript": "So this is a massive, massive distance, already, at least in my mind, unimaginable. And when it gets really wacky is when you start really looking at this. Even this is a super, super small distance relative to the galactic scale. So this whole depiction of kind of our neighborhood of stars, this thing over here is about, give or take-- and we're doing rough estimates right here-- it's about 30 light years. I'll just do LY for short. So that's about 30 light years, And once again, you can take pictures of our galaxy from our point of view. But you actually can't take a picture of the whole galaxy from above it. So these are going to be artists' depictions. But if this is 30 light years, this drawing right here of kind of our local neighborhood of the galaxy, this right here is roughly-- and these are all approximations. This is about 1,000 light years. And this is the 1,000 light years of our Sun's neighborhood, if you can even call it a neighborhood anymore. Even this isn't really a neighborhood if it takes you 80,000 years to get to your nearest neighbor. But this whole drawing over here-- now, it would take forever to get anywhere over here-- it would be 1/30 of this. So it would be about that big, this whole drawing. And what's really going to blow your mind is this would be roughly a little bit more than a pixel on this drawing right here, that spans a 1,000 light years. But then when you start to really put it into perspective-- so now, let's zoom out a little bit-- so this drawing right here, this 1,000 light years is now this 1,000 light years over here. So this is the local vicinity of the Sun. And once again, the word \"local\" is used in a very liberal way at this point. So this right here is 1,000 light years. looking at an object that's sitting-- let me do this in a darker color-- if we're sitting here on Earth and we're looking at an object out here that's 500 light years away, we're looking at it as it was 500 years ago because the light that is reaching our eyeballs right now, or our telescopes right now, left this guy over here 500 years ago. In fact, he's not going to even be there anymore. He probably has moved around a little bit. So just even on this scale, we're talking about these unimaginably huge distances. And then when we zoom out, this is kind of our local part of the galaxy right over here. This piece right here, this is called the Orion Spur. And people are still trying to work out exactly the details of the actual shape the Milky Way Galaxy, the galaxy that we're in. But we're pretty sure-- actually, we're very sure-- we have these spiral arms" + }, + { + "Q": "@ 6:55, during the discussion of London dispersion forces, it is assumed that the molecules produce opposite, and therefore attractive, momentary charges. Why wouldn't the molecules produce just as many same, and therefore repulsive, momentary charges, thus canceling any net effect?\n", + "A": "That is very possible!", + "video_name": "pBZ-RiT5nEE", + "timestamps": [ + 415 + ], + "3min_transcript": "that you should remember for hydrogen bonding are fluorine, oxygen, and nitrogen. And so the mnemonics that students use is FON. So if you remember FON as the electronegative atoms that can participate in hydrogen bonding, you should be able to remember this intermolecular force. The boiling point of water is, of course, about 100 degrees Celsius, so higher than what we saw for acetone. And this just is due to the fact that hydrogen bonding is a stronger version of dipole-dipole interaction, and therefore, it takes more energy or more heat to pull these water molecules apart in order to turn them into a gas. And so, of course, water is a liquid at room temperature. Let's look at another intermolecular force. And this one is called London dispersion forces. So these are the weakest intermolecular forces, and they have to do with the electrons that are always And even though the methane molecule here, if we look at it, we have a carbon surrounded by four hydrogens for methane. And it's hard to tell in how I've drawn the structure here, but if you go back and you look at the video for the tetrahedral bond angle proof, you can see that in three dimensions, these hydrogens are coming off of the carbon, and they're equivalent in all directions. And there's a very small difference in electronegativity between the carbon and the hydrogen. And that small difference is canceled out in three dimensions. So the methane molecule becomes nonpolar as a result of that. So this one's nonpolar, and, of course, this one's nonpolar. And so there's no dipole-dipole interaction. There's no hydrogen bonding. The only intermolecular force that's holding two methane molecules together would be London dispersion forces. And so once again, you could think about the electrons that are in these bonds moving in those orbitals. And let's say for the molecule on the left, if for a brief transient moment in time you get a little bit of negative charge to be those electrons have a net negative charge on this side. And then for this molecule, the electrons could be moving the opposite direction, giving this a partial positive. And so there could be a very, very small bit of attraction between these two methane molecules. It's very weak, which is why London dispersion forces are the weakest intermolecular forces. But it is there. And that's the only thing that's holding together these methane molecules. And since it's weak, we would expect the boiling point for methane to be extremely low. And, of course, it is. So the boiling point for methane is somewhere around negative 164 degrees Celsius. And so since room temperature is somewhere around 20 to 25, obviously methane has already boiled, if you will, and turned into a gas. So methane is obviously a gas at room temperature and pressure. Now, if you increase the number of carbons, you're going to increase the number of attractive forces" + }, + { + "Q": "\nAt 7:38 is there an omitted hydrogen on the first carbon? I understand that hydrogens aren't always drawn but is the hydrogen that was added bonded to that firs carbon?", + "A": "At about 5:33 you ll see pi bond picked up the H and it was added to the far left carbon, which formed a secondary carbocation on the carbon next to it. The secondary carbocation is more stable than if the H was added to the second carbon from the left which would result in a primary carbocation.", + "video_name": "dJhxphep_gY", + "timestamps": [ + 458 + ], + "3min_transcript": "We can form a tertiary carbocation if we think about the possibility of a hydride shift. So right here there is a hydrogen attached to that carbon. And if the proton and these two electrons are going to move over here and form a new bond with our positively charged carbon, so we get a hydride shift at this point. So let's draw what would result from that hydride shift. We moved a hydrogen over here. That took a bond away from this carbon. So that is the carbon that's going to end up with the positive charge now. We added a bond to what used to be our secondary carbocation carbon. And so that formal charge goes away. The formal charge moves to this carbon right here, which is now a tertiary carbocation. If you look at the carbons connected to that carbon, this is a tertiary carbocation. So we know tertiary carbocations are more stable than secondary. So now we're at the step of the mechanism where a water molecule is going to come along. is going to function as a nucleophile and attack our positively charged carbon, like that. So let's go ahead and draw what the result of that nucleophilic attack would look like. So we have our carbon skeleton, and we have an oxygen atom now bonded to that carbon. So two hydrogens here, and once again, one lone pair of electrons now participate in that reaction, giving this oxygen a plus 1 formal charge. And then finally, instead of showing the last step, a water molecule comes along, takes one of the protons off of our positively charged oxygen and gives us our major product with the OH adding on to this carbon right here. So this is a major product. This is our major product. And we would get some of the alcohol that forms from the secondary carbocation. So a minor product, that's what we would get if this oxygen had attacked And you will get some of that. But if your test asks for the major product, you should show the product of this rearrangement. Now, we're lucky in this instance because our product here, this carbon right here, ends up not being a chirality center. Because I have two methyl groups attached to that carbon, so I don't have to worry about my stereochemistry here. Let's do a reaction where we do have to worry about stereochemistry. OK. So let's look at this reaction right here. So we take this as our starting alkene. So I'll put the double bond right there. And let's make that a little bit more clear here. So the double bond is between these two carbons right here. And once again, we're going to add water and sulfuric acid. So H2SO4. And when we think about the mechanism," + }, + { + "Q": "\nAt 8:30 so there is no cis or trans in this molecule? And why do?", + "A": "Because one side has a methyl group and the other side does not. Both carbons of the double bond would have to have the same substituents in order for there to be cis/trans isomerism.", + "video_name": "dJhxphep_gY", + "timestamps": [ + 510 + ], + "3min_transcript": "We can form a tertiary carbocation if we think about the possibility of a hydride shift. So right here there is a hydrogen attached to that carbon. And if the proton and these two electrons are going to move over here and form a new bond with our positively charged carbon, so we get a hydride shift at this point. So let's draw what would result from that hydride shift. We moved a hydrogen over here. That took a bond away from this carbon. So that is the carbon that's going to end up with the positive charge now. We added a bond to what used to be our secondary carbocation carbon. And so that formal charge goes away. The formal charge moves to this carbon right here, which is now a tertiary carbocation. If you look at the carbons connected to that carbon, this is a tertiary carbocation. So we know tertiary carbocations are more stable than secondary. So now we're at the step of the mechanism where a water molecule is going to come along. is going to function as a nucleophile and attack our positively charged carbon, like that. So let's go ahead and draw what the result of that nucleophilic attack would look like. So we have our carbon skeleton, and we have an oxygen atom now bonded to that carbon. So two hydrogens here, and once again, one lone pair of electrons now participate in that reaction, giving this oxygen a plus 1 formal charge. And then finally, instead of showing the last step, a water molecule comes along, takes one of the protons off of our positively charged oxygen and gives us our major product with the OH adding on to this carbon right here. So this is a major product. This is our major product. And we would get some of the alcohol that forms from the secondary carbocation. So a minor product, that's what we would get if this oxygen had attacked And you will get some of that. But if your test asks for the major product, you should show the product of this rearrangement. Now, we're lucky in this instance because our product here, this carbon right here, ends up not being a chirality center. Because I have two methyl groups attached to that carbon, so I don't have to worry about my stereochemistry here. Let's do a reaction where we do have to worry about stereochemistry. OK. So let's look at this reaction right here. So we take this as our starting alkene. So I'll put the double bond right there. And let's make that a little bit more clear here. So the double bond is between these two carbons right here. And once again, we're going to add water and sulfuric acid. So H2SO4. And when we think about the mechanism," + }, + { + "Q": "At 3:42, Sal gets the average speed. How does he do that? What is the 12pi?\n", + "A": "12pi is the circumference of the circle. The formula for circumference is C = 2(pi)(r), and the radius of the loop is 6 meters, so the circumference of the loop is 12pi meters. He finds the speed of the car by doing distance divided by time. The time for the loop is 2 and 14/30 seconds.", + "video_name": "zcnnZz2pCSg", + "timestamps": [ + 222 + ], + "3min_transcript": "and in the previous video, we figured out the radius was 6 m So it's 2 pi times 6 m which is equal to 12 pi meters If you wanted to figure out its average speed-- the velocity is constantly changing because the direction is changing but the magnitude of the velocity-- if we wanted to figure out the average magnitude of the velocity or the average speed the total distance traveled is 12 pi meters divided by the time required to travel the 12 pi meters so that is 2 and 14/30 seconds Now let's get our calculator out to actual calculate that value So we're going to have the distance 12 pi m divided by And then this gives us in meters per second 15.3 m/s So the average speed is approximately 15.3 m/s which is almost twice as fast as the minimum speed we figured out because you want that margin of safety and you want to be able to have some traction with the road Although you don't want to go too fast, because then the G force is going to be too big then this--maybe we'll talk about that in the future video I'll just relate this into kilometers per hour. Let's figure out what that is I want to use the calculator here So that's in the meters per second So that's how many meters per hour, and divide it by 1000 which you can see right over there that is 55 km/h If you want to do it in miles, it's rough approximation, divide it by 1.6 It's about 35 mph give or take or 55 km/h So this is approximately 55 km/h So the driver here luckily they did the physics ahead of time and he had the margin of safety He was well in excess of the minimum velocity just to maintain the circular motion So he probably has some nice traction with the track up here" + }, + { + "Q": "\nat 2:05, aren't there other pieces of evidence besides fossils, such as mountain ranges?", + "A": "I think there is.", + "video_name": "axB6uhEx628", + "timestamps": [ + 125 + ], + "3min_transcript": "We know that new plate material is being formed, and these lithosphere plates on the surface of the Earth are moving around. And that might raise the question in your brain-- what happens if we kind of reverse things? We know the direction they're moving in. What does that tell us about where they came from? So let's just do the thought experiment. Right now, South America and Africa are moving away from each other, because of new plate material being created at the mid-Atlantic rift. Let's rewind it. Let's bring them back together. We know that India is jamming into the Eurasian Plate right now, causing the Himalayas to get higher and higher. What if we rewind that? Let's bring India back down towards Antarctica. Same thing with Australia. We have new plate material being formed between Australia and Antarctica that's making the continents move apart. Let's bring them back together. Let's rewind the clock. but if you actually look at the GPS data, it becomes pretty obvious that North America, right now is moving in a counterclockwise rotation. So let's rewind it into a-- let's go back, moving it in a clockwise direction. Let's, instead of Eurasia going further away from North America, let's bring it back together. And so what you could imagine is a reality where India, Australia are jammed down into South America-- sorry, into Antarctica. South America and Africa are jammed together. North America is jammed in there. And essentially, Eurasia is also jammed in there. So it looks like they all would clump together if you go back a few hundred million years. And based on, literally-- based on just that thought experiment, you could imagine at one point, all of the continents on the world were merged into one supercontinent. And that supercontinent is called Pangaea-- pan for the world. And it turns out that all of the evidence we've seen actually does make us believe that there was a supercontinent called-- well, we call it Pangaea, now. Obviously, there probably weren't things on the planet calling it anything back then. Or, there were things back then, but not things that would actually go and try to label continents that we know of. But all of the evidence tells us that Pangaea existed about 200 to 300 million years ago, roughly maybe 250 million, give or take, years ago. And I want to be clear. This was not the first supercontinent. To a large degree, it's kind of the most recent supercontinent. And it's easiest for us to construct because it was the most recent one. But we believe that there were other supercontinents before this. That if you rewind even more that you would have to break up Pangaea and it would reform. But we're now going back in time. Or that there were several supercontinents" + }, + { + "Q": "at 6:07 it is easy to understand how the unpaired electrons repel the hydrogens attached to nitrogen so that a trigonal pyramidal shape is achieved. But what forces drive the formation of a tetrahedral shape for methane? Why doesn't it assume a flat plane? Wouldn't that create the greatest distance among the hydrogens?\n", + "A": "If it assumed a flat plane, then the angles between the Hs would be 360\u00c2\u00b0/4 = 90\u00c2\u00b0. A tetrahedral shape allows the angles between the Hs to be 109.5\u00c2\u00b0, greater than the 90\u00c2\u00b0 that a flat plane would allow.", + "video_name": "ka8Yt4bTODs", + "timestamps": [ + 367 + ], + "3min_transcript": "surrounded the central atom. So regions of electron density. And let's go ahead and find those. So I can see that I have these bonding electrons. That's a region of electron density, so that's an electron cloud. Here's another one. Here's another one. So that's three. And this lone pair of electrons, this non-bonding pair of electrons is also going to be counted as an electron cloud. It's a region of electron density too. And so once again we have four regions of electron density. When you're thinking about the geometry of those electron clouds, those four electron clouds are going to, once again, try to point towards the corners of a tetrahedron. So we can kind of sketch out the ammonia molecule. And we can draw the base the same way we did before, with our three hydrogens right here. And then we're going to go ahead and put our lone pair of electrons right up here. And so again, it's an attempt to show the electron clouds Let's go back up here and look at our steps again. So in step three, we predicted the geometry electron clouds are going to attempt to be in a tetrahedron shape around our central atom. But when we're actually talking about the geometry or shape of the molecule, we're going to ignore any lone pairs when we predict the geometry of the molecule. So when we look at the ammonia molecule, we're going to ignore that lone pair of electrons on top of the nitrogen and we're just going to focus in on the bottom part for the shape here. And so when we do that, we get something that looks like a little squat pyramid here. So if I'm [? ignoring ?] that lone pair of electrons up there at the top, it's going to look something like that for the shape. And we call this trigonal pyramidal. So this is a trigonal pyramidal shape. So even though the electron clouds are attempting to be in a tetrahedron fashion, because we ignore any lone pairs of electrons. In terms of a bond angle, this lone pair of electrons on the nitrogen actually occupies more space. These non-bonding electrons occupy a little more space than bonding electrons. And because of that, those non-bonding electrons are going to repel these bonding electrons. I'm going to go ahead and put them in blue here just as an example. Repel these a little bit more than in the previous example that we saw. And that's actually going to make the bond angle a little bit smaller than the ideal bonding angle we saw before for 109.5 for a tetrahedral arrangement of electron clouds. And so it turns out that this bond angle between the atoms, the hydrogen nitrogen hydrogen bond angle gets a little bit smaller than 109.5. So it actually gets smaller to approximately 107 degrees here for a trigonal pyramidal situation." + }, + { + "Q": "At 3:50, how is it that 109.5 separates all the clouds better on a single plane compared to 90?\n", + "A": "But it isn t on a single plane, it s in 3D space. 109.5 degrees just happens to be the maximum angle when you have 4 equal groups. If you need proof then Jay did do a mathy video that proves this on here, might be easier just accept it for what it is.", + "video_name": "ka8Yt4bTODs", + "timestamps": [ + 230 + ], + "3min_transcript": "of that tetrahedron, that four sided figure here. And so we've created the geometry of the electron clouds around our central atom. And in step four, we ignore any lone pairs around our central atom, which we have none this time. And so therefore, the geometry the molecule is the same as the geometry of our electron pairs. So we can say that methane is a tetrahedral molecule like that. All right, in terms of bond angles. So our goal now is to figure out what the bond angles are in a tetrahedral molecule. Turns out to be 109.5 degrees in space. So that's having those bonding electrons as far away from each other as they possibly can using VSEPR theory. So 109.5 degrees turns out to be the ideal bond angle for a tetrahedral molecule. Let's go ahead and do another one. Let's look at ammonia. So we have NH3. So we start by finding our valence electrons, nitrogen in group five. So 5 valence electrons. Hydrogen in group one, and I have three of them. So 1 times 3 plus 5 is 8. So once again, we have 8 valence electrons to worry about. We put nitrogen in the center and we know nitrogen has bonded to 3 hydrogens, so we go ahead and put our 3 hydrogens in there like that. Let's see how many valence electrons we've used up so far. 2, 4, and 6. So 8 minus 6 is 2 valence electrons left. We can't put them on our terminal atoms, because the hydrogens are already surrounded by two electrons. So we go ahead and put those two valence electrons on our central atom, which is our nitrogen like that. And so now we've gone ahead and represented those two valence electrons. So we have all eight valence electrons shown for our dot structure. All right, we go back up here to our steps to remind us what to do after we've drawn our dot structure. And we can see that now we're going surrounded the central atom. So regions of electron density. And let's go ahead and find those. So I can see that I have these bonding electrons. That's a region of electron density, so that's an electron cloud. Here's another one. Here's another one. So that's three. And this lone pair of electrons, this non-bonding pair of electrons is also going to be counted as an electron cloud. It's a region of electron density too. And so once again we have four regions of electron density. When you're thinking about the geometry of those electron clouds, those four electron clouds are going to, once again, try to point towards the corners of a tetrahedron. So we can kind of sketch out the ammonia molecule. And we can draw the base the same way we did before, with our three hydrogens right here. And then we're going to go ahead and put our lone pair of electrons right up here. And so again, it's an attempt to show the electron clouds" + }, + { + "Q": "\n5:30 how do you apply bond breaking energy? just heat?", + "A": "Heat is certainly an option because it is a form of energy, in fact, any form of energy will work as long as there is enough quantity of energy to break the bond. In fact, some experiments are so spontaneous that yelling at the reactants (sound energy) will start the reaction", + "video_name": "Ce4BGV1DVVg", + "timestamps": [ + 330 + ], + "3min_transcript": "Notice this is different from the previous video where we talked about electronegativity. There we were comparing elements in the same period on the periodic table. So we were moving horizontally across our periodic table this way. And in that video, the fluoride anion was the most stable one because fluorine's our most electronegative element, and therefore, best able to stabilize a negative charge. But as you go down a group on the periodic table, your electronegativity decreases. So that can't be the dominant trend because if your electronegativity decreases as you go down, just thinking about electronegativity, that would predict HF to be the strongest acid, and that's not what we observe. So as you go down a group on the periodic table, it's the size of the anion that determines the stability of the conjugate base. So the larger the anion, and therefore, the more stable the conjugate base. The more stable the conjugate base, the more likely HX is to donate a proton, and therefore, the stronger the acid. Another very important factor to think about is the strength of the bond. We've already said that hydroiodic acid is our strongest acid with the lowest pKa value. So this bond right here must be the easiest to break. If it's easy to break this bond, that makes it easy to donate this proton. So we can get an idea of the bond strengths for our binary acids by looking at bond association energy. So we could also call these bond energies or bond enthalpies. So remember bond dissociation energy measures the amount of energy that's needed to break a bond in the gaseous state. So if we look at our hydrogen halides and we think about our bonds, notice what happens to the bond energy. It's the hardest to break this bond, This takes the most energy to break this bond, and as you go down, we see we decrease in bond dissociation energy. So it only takes 299 kilojoules per mole to break this bond between H and I. I should say these are approximate bond energies and you'll see several different values in different textbooks. So if the HI bond is the easiest to break, that means when you're thinking about the acids, this bond is the easiest to break, therefore, it's the most likely to donate a proton, and therefore, it has the lowest value for the pKa. Hydroiodic acid is the strongest out of these four binary acids." + }, + { + "Q": "\nAt 0:45, how can a reaction have a zero order while the molecularity can never be zero? Because the powers raised over the concentration in an elementary reaction while using rate law is equal to the stoichiometric coefficients of reactants and these coefficients can also never be zero.", + "A": "I think you are gravely mistaken the powers are not the stoichiometric coefficients, they are the order of reaction with respect to the given atom or molecule, and order as we discussed is experimental and molecularity is very different from order they are two different things..i hope you got your error...", + "video_name": "49LcF9Zf9TI", + "timestamps": [ + 45 + ], + "3min_transcript": "- Let's see we have a zero order reaction where A turns into our products and when time is equal to zero we're starting with our initial concentration of A and after some time period T, we would have the concentration of A at that time. So we could express the rate of our reaction, one way to do it would be to say the rate of the reaction is equal to the negative change in the concentration of A over the change in time. And another way to do this would be to right the rate law. So the rate of our reaction is equal to the rate constant K times the concentration of A and since I said this is a zero order reaction, this would be A to the zero power. And any number to the zero power is equal to one. Therefore, the rate of the reaction would be equal to K times one or the rate is just equal to the rate constant K. So the rate of a zero order reaction is a constant. Next, we can set these equal. Right we can say that K is equal to the negative change in the concentration of A over the change in time. So we have the negative change in the concentration of A over the change in time is equal to the rate constant K. And next we could think about our calculus, alright, instead of writing change in A over change in time, we're going to write the rate of change of the concentration of A with respect to time. We have our negative sign in here and then we have our K on the side. So we're ready to think about our differential equation and we're gonna multiple by both sides by DT. Let's go ahead and multiple both sides by negative DT, so then we would have DA on the left side, right, and then we would have negative KDT on the right side. And we're ready to integrate. Right, so we're gonna integrate on the left, K is a constant so we can pull it out of our integral on the right. about what would be integrating from. So we'd be going from time is equal to zero to time is equal to T and from our initial concentration to our concentration at time T. So you plug those in we're going from time is equal to zero to time is equal to T and then for our concentration we're going from our initial concentration to our concentration at time T. So we have some easy integrals here, right? What's the integral of DA? That would be of course A or the concentration of A. So we have the concentration of A, right. We are evaluating this from our initial concentration to our concentration of A at time T. On the right side we have another easy integral, integral of DT. That's just T. So we have negative KT from zero to T. Next fundamental theorem of calculus, right, so we would get the concentration of A" + }, + { + "Q": "At 6:46 in the diagram of the 1.0 M NaCl solution what are the two blue and 1 green circles representing? and the red ones?\n", + "A": "The red and white molecule (which forms Mickey Mouse s head) would be H2O, water. The blue and green molecule is NaCl, salt. As he says, And here they visualise Sodium Chloride at the surface. while pointing to the green/blue molecule.", + "video_name": "z9LxdqYntlU", + "timestamps": [ + 406 + ], + "3min_transcript": "It probably doesn't have much of an effect down here, but some of it's going to be bouncing on the surface, so they're going to be taking up some of the surface area. And because, and this is at least how I think of it, since they're going to be taking up some of the surface area, you're going to have less surface area exposed to the solvent particle or to the solution or the stuff that'll actually vaporize. You're going to have a lower vapor pressure. And remember, your boiling point is when the vapor pressure, when you have enough particles with enough kinetic energy out here to start pushing against the atmospheric pressure, when the vapor pressure is equal to the atmospheric pressure, you start boiling. But because of these guys, I have a lower vapor pressure. So I'm going to have to add even more kinetic energy, more heat to the system in order to get enough vapor pressure up here to start pushing back the atmospheric pressure. So solute also raises the boiling point. solute, when you add something to a solution, it's going to make it want to be in the liquid state more. Whether you lower the temperature, it's going to want to stay in liquid as opposed to ice, and if you raise the temperature, it's going to want to stay in liquid as opposed to gas. I found this neat -- hopefully, it shows up well on this video. I have to give due credit, this is from chem.purdue.edu/gchelp/solutions/eboil.html, but I thought it was a pretty neat graphic, or at least a visualization. This is just the surface of water molecules, and it gives you a sense of just how things vaporize as well. There's some things on the surface that just bounce off. And here's an example where they visualized sodium chloride at the surface. And because the sodium chloride is kind of bouncing around on the surface with the water molecules, fewer of those water molecules kind of have the room to escape, so the boiling point gets elevated. And this is one of the neat things in life is that the answer is actually quite simple. The change in boiling or freezing point, so the change in temperature of vaporization, is equal to some constant times the number of moles, or at least the mole concentration, the molality, times the molality of the solute that you're putting into your solution. So, for example, let's say I have 1 kilogram of -- so let's say my solvent is water. I'll switch colors. And I have 1 kilogram of water, and let's say we're just at atmospheric pressure." + }, + { + "Q": "@3:12, did he mean \"lower the freezing point\"?\n", + "A": "I believe he meant lower the melting point.", + "video_name": "z9LxdqYntlU", + "timestamps": [ + 192 + ], + "3min_transcript": "they're just vibrating in place. So you have to get a little bit orderly right there, right? And then, obviously, this lattice structure goes on and on with a gazillion water molecules. But the interesting thing is that this somehow has to get organized. And what happens if we start introducing molecules into this water? Let's say the example of sodium-- actually, I won't do any example. Let's just say some arbitrary molecule, if I were to introduce it there, if I were to put something-- let me draw it again. So now I'll just use that same -- I'll introduce some molecules, and let's say they're pretty large, so they push all of these water molecules out of the way. So the water molecules are now on the outside of that, and let's have another one that's over here, some relatively large molecules of solute relative to water, and this is because a water molecule really isn't that big. Now, do you think it's going to be easier Are you going to have to remove more or less energy to get to a frozen state? Well, because these molecules, they're not going to be part of this lattice structure because frankly, they wouldn't even fit into it. They're actually going to make it harder for these water molecules to get organized because to get organized, they have to get at the right distance for the hydrogen bonds to form. But in this case, even as you start removing heat from the system, maybe the ones that aren't near the solute particles, they'll start to organize with each other. But then when you introduce a solute particle, let's say a solute particle is sitting right here. It's going to be very hard for someone to organize with this guy, to get near enough for the hydrogen bond to start taking hold. This distance would make it very difficult. And so the way I think about it is that these solute particles make the structure irregular, or they add more disorder, and we'll eventually talk about entropy and all of that. and it's making it harder to get into a regular form. And so the intuition is that this should lower the boiling point or make it -- oh, sorry, lower the melting point. So solute particles make you have a lower boiling point. Let's say if we're talking about water at standard temperature and pressure or at one atmosphere then instead of going to 0 degrees, you might have to go to negative 1 or negative 2 degrees, and we're going to talk a little bit about what that is. Now, what's the intuition of what this will do when you want to go into a gaseous state, when you want to boil it? So my initial gut was, hey, I'm already in a disordered state, which is closer to what a gas is, so wouldn't that make it easier to boil? But it turns out it also makes it harder to boil, and this is how I think about it. Remember, everything with boiling deals with what's happening at the surface," + }, + { + "Q": "(1:50) since the dark reactions are independent, why do they require ATP, NADPH, AND CO2, but not the protons, can anyone help me better understand?\n", + "A": "ATP and NADPH in non-cyclic photophosphorylation are created in the light reactions using the photons. This byproducts from the Light dependent reactions are used in the Calvin cycle. So, ATP and NADPH are created in light reactions because this are needed in the calvin cycle to produce glucose or any other carbohydrate which can be used by the plants later. Hope this helps.", + "video_name": "slm6D2VEXYs", + "timestamps": [ + 110 + ], + "3min_transcript": "" + }, + { + "Q": "9:30 How do we get glucose (C6H12O6) from 2 G3Ps which is 3 carbon with phosphate? Don't we need Hydrogen as well as more Oxygen?\n", + "A": "It s because there are materials from the Light Dependent Reactions (Photosynthesis) that go into making glucose", + "video_name": "slm6D2VEXYs", + "timestamps": [ + 570 + ], + "3min_transcript": "" + }, + { + "Q": "\nCan someone refresh my memory here? At 2:10, Sal says the Oxygen wants to give away an electron. I thought Oxygen's goal was to take electrons in order to complete it's valence shell.", + "A": "Remember that all molecules want a neutral charge. The oxygen atom here has a negative charge. When he says it wants to give away an electron he really means that the oxygen really wants to share its electron with a proton, forming a covalent bond, which in turn will keep its octet, and make the molecule neutral (no charge).", + "video_name": "J0gXdEAaSiA", + "timestamps": [ + 130 + ], + "3min_transcript": "Let's try to come up with the reaction of when we have this molecule right here reacting with sodium methoxide in a methanol solution, or with methanol as the solvent. Just so we get a little practice with naming, let's see, this is one, two, three, four carbons. So it has but- as a prefix and no double bond or triple bonds, so it's butane. And we have a chloro group here. So if we start numbering at the side closest to it, one, two. So it's 2-chlorobutane. Let's think about what might happen here. The first thing-- let me just redraw the molecule right there. The first thing you need to realize is this sodium methoxide is a salt. When it's not dissolved, it's made up of a positive sodium cation. Let me draw them right here. negative methoxide anion. Let me draw the methoxide part right here. It normally would have just two pairs, but now we have three pairs here. The oxygen has an extra electron. Actually, let me draw another electron as the extra electron. This'll be useful for our mechanism. It could be any of them. The oxygen has an extra electron. It has a negative charge. In solid form, when they're not dissolved, they had formed an ionic bond and they form a crystal-like structure. It's a salt, but when you dissolve it in something, in a solution, in this case, we have methanol as the solution, they will disassociate from each other. And what you have right here is this methoxide right here. This is a very, very, very strong base. And if you use the Lewis definition of a base, that means it really, really, really wants to give away this electron to something else. If you use the Bronsted-Lowry definition, this means that it really, really, really wants to take a proton off of something else. In this situation, that is exactly what it will do. I'll actually give you the most likely reaction to occur here, and we'll talk about other reactions, and why this is the most likely reaction in future videos. So it wants to nab a proton. It is a strong base. It wants to give away this electron. Let's say that it gives away this electron to this hydrogen right over here. Now, this hydrogen already had an electron." + }, + { + "Q": "At around 7:15, Sal says that the speed is independent of the inertial frame of reference. But if Sal and Sally are traveling at different speeds, how can they observe the speed of light as being the same? Isn't Sally closer to the speed of light, so she has greater relative velocity compared to it?\n", + "A": "What you are saying makes common sense but that is not how velocities that are a significant fraction of the speed of light work. Any observer that is not accelerating will always measure light in a vacuum traveling at the same speed.", + "video_name": "OIwp8m3S30c", + "timestamps": [ + 435 + ], + "3min_transcript": "or it looks like the velocity of that photon is one and a half times 10 to the 8th meters per second in the positive x direction. And this should hopefully makes sense from a Newtonian point of view, or a Galilean point of view. These are called Galilean transformations because if I'm in a car and there's another car and you see this on the highway all the time, if I'm in a car going 60 miles per hour, there's another car going 65 miles per hour, from my point of view, it looks like it's only moving forward at five miles per hour. So that photon will look slower to Sally. Similarly, if we assume this Newtonian, this Galilean world, if she had a flashlight, if she had a flashlight right over here and right at time equals zero she turned it on, and that first photon we were to plot it on her frame of reference, well, it should go the speed of light So it starts here at the origin. And then after one second, in the s prime, in the s prime coordinates, it should have gone three times 10 to the 8th meters. After two seconds, it should've gone six times 10 to the 8th meters. And so it's path on her space-time diagram should look like that. That's her photon, that first photon that was emitted from it. So you might be noticing something interesting. That photon from my point of view is going faster than the speed of light. After one second, its x coordinate is 4.5 times 10 to the 8th meters. It's going 4.5 times 10 to the 8th meters per second. It's going faster than the speed of light. It's going faster than my photon, and that might make intuitive sense except it's not what we actually observe in nature. And anytime we try to make a prediction that's not what's observed in nature, it means that our understanding of the universe is not complete inertial reference frame we are in, the speed of light, regardless of the speed or the relative velocity of the source of that light, is always going three times 10 to the 8th meters per second. So we know from observations of the universe that Sally, when she looked at my photon, she wouldn't see it going half the speed of light, she would see it going three times 10 to the 8th meters per second. And we know from observations of the universe that Sally's photon, I would not observe it as moving at 4.5 times 10 to the 8th meters per second, that it would actually still be moving at three times 10 to the 8th meters per second. So something has got to give. This is breaking down our classical, our Newtonian, our Galilean views of the world. It's very exciting. We need to think of some other way to conceptualize things, some other way to visualize these space-time diagrams for the different frames of reference." + }, + { + "Q": "At 3:00, why does the carbonyl reform? I thought oxygen can hold the negative charge?\n", + "A": "Yes, it can, but a C=O double bond with no charge is more stable than a C-O\u00e2\u0081\u00bb single bond with a charge.", + "video_name": "rNJPNlgmhbk", + "timestamps": [ + 180 + ], + "3min_transcript": "will react with that faster. And so once again, our product will depend on what our starting material is. So the mechanism for the reduction of aldehydes or ketones with lithium aluminum hydride is just like the one for sodium borohydride. So we'll move on to a mechanism for the reduction of an ester. So let's go ahead and do that. So let's start with an ester down here. So we have our carbonyl. Like that. So we'll put in our lone pairs. And down here, we have our R prime group. Like that. So there's our ester. And we add lithium aluminum hydride in excess. So in terms of molar equivalence, let's go ahead and put lithium aluminum hydride down here, Li+. And then, we have Al bonded to 4 hydrogens. And that's going to give the aluminum a negative 1 formal charge. Like that. So the first step in the mechanism is just like the one we did in the previous video. All right. in electronegativity between this carbon and this oxygen-- oxygen being more electronegative, pulling these electrons closer towards it and the double bond, giving it a partial negative charge. Whereas the carbon down here is losing some electron density, becoming partially positive. So the carbonyl carbon is an electrophile. It wants electrons. And of course, it's going to get electrons from these 2 electrons in here. So these 2 electrons are going to attack this carbon. So a nucleophilic attack. And kick these electrons off onto our oxygen. So that's our first step-- the nucleophilic attack portion. So now, we have R. And it used to have 2 bonds to carbon and oxygen. Now, it has only 1 bond because those electrons moved off onto the oxygen, giving the oxygen a negative 1 formal charge. Like that. So we added on our hydrogen. Like that. And then, we still have our O and our R prime group In the next step of this reaction, the carbonyl's going to reform. So the electrons in here are going to kick back into here to reform our carbonyl. That would mean 5 bonds to carbon-- which we know never happens-- so that these electrons are going to have to break and come off onto the oxygen. So let's go ahead and draw the results of that. So now, we have R bonded to carbon. Now, we reformed our carbonyls. Now, we have only two lone pairs of electrons on that oxygen now. And then, we still have a hydrogen. Right here. So we lost an oxygen with an R prime group. And that oxygen is going to have three lone pairs around it with a negative 1 formal charge. Oxygen being relatively electronegative, it can handle that formal charge fairly well and be relatively stable. So now we have an aldehyde. And we know that this reaction can occur again with an aldehyde. So since you have extra lithium aluminum hydride floating" + }, + { + "Q": "\nat 10:28 ,isn't h2 and Pd going to reduce the aldehyde ester AND the benzene ring also.", + "A": "no, reducing the benzene would need H2 over a Rhodium/Carbon catalyst. H2 over Palladium catalyst is selective to alkenes and alkynes than aldehydes and ketones if it s in the same molecule.", + "video_name": "rNJPNlgmhbk", + "timestamps": [ + 628 + ], + "3min_transcript": "forms your primary alcohol as your product. And the rest of the molecule's going to stay the same. So this ester is going to remain untouched down here. So it's the chemoselectivity of that reaction. Let's say we start with the same starting material. And the first step-- this time we add lithium aluminum hydride. Like that. And the second step-- we add some water. Well, lithium aluminum hydride will reduce aldehydes and ketones, and it will also reduce esters. So lithium aluminum hydride in excess-- so let's just assume this is at an excess here-- it's going to react with this aldehyde portion of the molecule. It's also going to react with this ester portion of the molecule. So it's going to reduce both of those and form alcohol. So let's go ahead and try to draw the product here. So we have our benzene ring, which is untouched. And up here, we know that lithium aluminum hydride Like that. And then, down here, what used to be our ester functional group, we're going to add two molar equivalents of hydrogen to that carbonyl. And we're going to end up breaking that bond between the carbonyl carbon and that oxygen. All of this over here is going to go away as a leaving group in our mechanism. And we're going to add on two hydrogens to that carbon. And then, that's going to form our alcohol. So we're going to add two hydrogens onto that carbon, forming a primary alcohol down here as well-- just like in the mechanism that we just discussed. So reduction of esters using lithium aluminum hydride. What about if we were to add a hydrogen gas and palladium as our metal catalyst here? Well, this is also a reduction reaction that we talked about earlier. Hydrogenation is an example of a reduction reaction. And it's going to be chemoselective. reaction, the only thing the hydrogenation reaction is going to touch is this double bond. It's going to reduce this double bond. So let's go ahead and draw the product. It's not going to touch the aldehyde. It's not going to touch the ester. And it's not going to touch the benzene ring. So let's go ahead and draw the product. The benzene ring is not hydrogenated under normal conditions, but we're going to add on two hydrogens across that double bond. And the aldehyde is untouched. And down here, the ester is going to be untouched as well. So that would be our product from a hydrogenation. So three different reductions, three different products. Now, hydrogen will reduce carbonyls under the right conditions. Usually, if you have increased pressure and increased temperatures, you actually can reduce those carbonyls. But again, you can control those conditions. So you can control what part of the molecule is reduced." + }, + { + "Q": "Near 8:01, Jay said that NaBH4,reduces the aldehyde to alcohol.\n\nIt's correct. I have no objections in that. But, my teacher taught us that NaBH4 also reduces the alpha-beta-conjugated double bonds. So,by my teacher's side, the product should be a cyclic ring with two double bonds and no double bond in the right side... Was my teacher correct or is Jay correct?\n", + "A": "No, NaBH4 is a mild reducing agent and will only react with more reactive carbonyls like ketones and aldehydes. Alkenes require more reactive metal catalysts like Pd or Pt. There are special cases where the nucleophilicity of the alkene can be adjusted to react with NaBH4, but these are not addressed in an undergraduate class.", + "video_name": "rNJPNlgmhbk", + "timestamps": [ + 481 + ], + "3min_transcript": "So this one and this one. So the reaction happened twice. So if you're doing this reaction with a carboxylic acid, it's a similar mechanism. We don't have time to go through it. But you're going to end up with the same product. You're going to add on two hydrogens on to that original carbonyl carbon. Like that. So let's look at the chemoselectivity of this reaction. So now that we've covered sodium borohydride and lithium aluminum hydride, let's see how you can choose which one of those reagents is the best to use. So if I start here with our reactants-- so let's make it a benzene ring. Like that. And let's put stuff on the benzene ring. So let's go ahead and put a double bond here. And then, we'll make this an aldehyde functional group on one end. And then over here on this end, I'm going to put an ester. Like that. transform different parts of this molecule using different reagents. So let's say we were to do a reaction wherein we add on sodium borohydride. And then, the proton source in the second step. So we need to think about what's going to happen. Sodium borohydride is selective for aldehydes and ketones only. It will not reduce carboxylic acids or esters. So it's only going to react with the aldehyde at the top right portion of this molecule. So let's see if we can draw this in here. So it's going to react with the aldehyde in the top right portion. So we are still going to have our double bond here. And the aldehyde's going to go away to form a primary alcohol. So we're going to get primary alcohol where the aldehyde used to be. Sodium borohydride has reduced that carbonyl. forms your primary alcohol as your product. And the rest of the molecule's going to stay the same. So this ester is going to remain untouched down here. So it's the chemoselectivity of that reaction. Let's say we start with the same starting material. And the first step-- this time we add lithium aluminum hydride. Like that. And the second step-- we add some water. Well, lithium aluminum hydride will reduce aldehydes and ketones, and it will also reduce esters. So lithium aluminum hydride in excess-- so let's just assume this is at an excess here-- it's going to react with this aldehyde portion of the molecule. It's also going to react with this ester portion of the molecule. So it's going to reduce both of those and form alcohol. So let's go ahead and try to draw the product here. So we have our benzene ring, which is untouched. And up here, we know that lithium aluminum hydride" + }, + { + "Q": "at 1:18 Sal circles one set of arrows on the Nazca plate and just a little to the right and above there is another set of arrows. These arrows are pointing in a directions almost perpendicular to the set Sal circled. How can a plate be moving in two directions at once? Or is this map from wikipedia inaccurate?\n", + "A": "plates move in different directions, so in one part a plate will slide under another and in another part of the same plate might be seperating away from another or they can spread or converge at the same time", + "video_name": "f2BWsPVN7c4", + "timestamps": [ + 78 + ], + "3min_transcript": "What I want to do in this video is talk a little bit about plate tectonics. And you've probably heard the word before, and are probably, or you might be somewhat familiar with what it discusses. And it's really just the idea that the surface of the Earth is made up of a bunch of these rigid plates. So it's broken up into a bunch of rigid plates, and these rigid plates move relative to each other. They move relative to each other and take everything that's on them for a ride. And the things that are on them include the continents. So it literally is talking about the movement of these plates. And over here I have a picture I got off of Wikipedia of the actual plates. And over here you have the Pacific Plate. Let me do that in a darker color. You have the Pacific Plate. You have a Nazca Plate. You have a South American Plate. I could keep going on. You have an Antarctic Plate. It's actually, obviously whenever you do a projection onto two dimensions of a surface of a sphere, the stuff at the bottom and the top look much bigger than they actually are. Antarctica isn't this big relative It's just that we've had to stretch it out to fill up the rectangle. But that's the Antarctic Plate, North American Plate. And you can see that they're actually moving relative to each other. And that's what these arrows are depicting. You see right over here the Nazca Plate and the Pacific Plate are moving away from each other. New land is forming here. We'll talk more about that in other videos. You see right over here in the middle of the Atlantic Ocean the African Plate and the South American Plate meet each other, and they're moving away from each other, which means that new land, more plate material I guess you could say, is somehow being created right here-- we'll talk about that in future videos-- and pushing these two plates apart. Now, before we go into the evidence for plate tectonics or even some of the more details about how plates are created and some theories as to why the plates might move, what I want to do is get a little bit of the terminology of plate tectonics out of the way. and that's not exactly right. And to show you the difference, what I want to do is show you two different ways of classifying the different layers of the Earth and then think about how they might relate to each other. So what you traditionally see, and actually I've made a video that goes into a lot more detail of this, is a breakdown of the chemical layers of the Earth. And when I talk about chemical layers, I'm talking about what are the constituents of the different layers? So when you talk of it in this term, the top most layer, which is the thinnest layer, is the crust. Then below that is the mantle. Actually, let me show you the whole Earth, although I'm not going to draw it to scale. So if I were to draw the crust, the crust is the thinnest outer layer of the Earth. You can imagine the blue line itself is the crust. Then below that, you have the mantle. So everything between the blue and the orange line," + }, + { + "Q": "\nHey :)\nAt 13:38 , I dont really understand how can NH3 give NH4 and hydroxide ions?\nand isnt it NOT reversible reaction? :O\n\n-Sahil, Mumbai", + "A": "Put in the first term of reation water and you can understand better. Amonia is a weak base and water, in this case, is a weak acid. There is a Ka in this case.", + "video_name": "3Gm4nAAc3zc", + "timestamps": [ + 818 + ], + "3min_transcript": "Well, if you know the pKa for a weak acid-- For example, let's say we have NH4 plus. This is a weak acid, right? It can donate an H, but it's not an irreversible reaction. That H can be gained back. So this is a weak acid. If you look it up on Wikipedia, it'll say, hey, the pKa of NH4 is equal to 9.25. Right? So this is 9.25 for NH4. For ammonium. So what is going to be the pKb for ammonia? Right? Let me write that reaction down. So this is NH4. Is in equilibrium. This is plus. It can get rid of one of its hydrogen protons, and you're just left with ammonia. So this is what the pKa is associated with. So the equilibrium constant for this reaction is is equal to 9.25. And if I had the reverse reaction, the conjugate base reaction, so ammonia converts to ammonium. Plus it grabbed that hydrogen proton from a water molecule. If I wanted to figure out the pKb, or the equilibrium constant, or the negative log of the equilibrium constant for this reaction, what is it? Well, this one's 9.25. And 9.25 plus this pKb have to be equal to 14. So what's 14 minus 9.25? It's what, 4.75. So we immediately know the equilibrium constant for the conjugate base reaction. So it's a useful thing to know. And always remember, you see these pKa and pKb, and you say, what is that? Well, if you see a p, it's a negative log of something. And in this case, it's the negative log of the equilibrium constant for an acidic reaction. Plus the negative log of the equilibrium basic reaction, where this is the conjugate base of this acid. It's always going to be equal to 14 if we're dealing with an aqueous solution at 25 degrees Celsius, which is essentially room temperature, which is usually going to be the case." + }, + { + "Q": "At 1:40, Sal says that is a conjugate-base. How do we know that is not a conjugate-acid?\n", + "A": "Watch the conjugate acids and bases video. Because HA is an acid in the first place, A- has to be the conjugate base.", + "video_name": "3Gm4nAAc3zc", + "timestamps": [ + 100 + ], + "3min_transcript": "In the last video we learned that if I had some -- let's say I have some weak acid. So it's hydrogen plus some-- the rest of whatever the molecule was called. We'll just call it A. And I think this tends to be the standard convention for the rest of the acid. They can disassociate, or it's in equilibrium because it's a weak acid. So it can be in equilibrium with -- since it's a weak acid, it's going to produce some hydrogen proton. And then the rest of the molecule is going to keep the electrons. So it's going to be plus -- oh, this is in an aqueous solution. Let me do that. Aqueous solution. And then you're going to have the rest of the acid, whatever it might be. A minus, and that's also going to be in an aqueous solution. And that's the general pattern. We've seen the case where A could be an NH3, right? If A is an NH3, then when you have this, you have an NH4 plus, and this would be ammonia. A could be a fluorine molecule right there, because then this would be hydrogen fluoride or hydrofluoric acid. And this would just be the negative ion of fluorine. Or a fluorine with an extra electron. So it could be a bunch of stuff. You can just throw in anything there, and it'll work. Especially for the weak acids. So we learned a last video that if this is the acid, then this is the conjugate base. And we could write the same reactions, essentially, as kind of more of a basic reaction. So we could say, if I start with A minus -- it's in an aqueous solution -- that's in equilibrium with-- This thing could grab a hydrogen It's still in an aqueous solution. And then one of those water molecules that it plucked that hydrogen off of is now going to be a hydroxide molecule. Right? Because it's hydrogen. Remember, whenever I say pluck the hydrogen, just the proton, not the electron for the hydrogen. So the electron stays on that water molecule, so it has a negative charge. It's in an aqueous solution. So we could write the same reaction both ways. And we can write equilibrium constants for both of these reactions. So let's do that. Let me erase this, just because I can erase this stuff right there, and then use that space. So an equilibrium reaction for this first one. I could call this the K sub a, because the equilibrium reaction for an acid. And so this is going to be equal to its products. So the concentration of my hydrogen times" + }, + { + "Q": "\nAt 4:13, why do we know that Si has 4 bonds to F atoms?", + "A": "Si has 4 valence electrons so it needs to form 4 bonds to complete its octet.", + "video_name": "p7Fsb21B2Xg", + "timestamps": [ + 253 + ], + "3min_transcript": "But it is not possible for elements to have more than eight electrons. Always check your dot structures, and make sure that if you have an element in the second period, you do not exceed eight electrons. Once you get to the third period, you have even more orbitals available to you now. So in the first energy level, you have only one s orbital. In the second energy level you have s and p orbitals, and in the third energy level you have s, p, and d orbitals. So you can fit more than eight electrons. And so therefore it's possible to exceed the octet rule for elements in the third period and beyond. And we will see a few examples of that in this video, and some of the ones to come here. So getting back to our molecule, silicon tetrafluoride, if I wanted to find out how many total valence electrons are in this molecule, I need to find these elements on my periodic table. So I go over here and I find silicon, and I see it's in group four. So therefore one atom of silicon has four valence electrons. Fluorine is over here in a group of seven, will have seven valence electrons. And I have four of them. So 7 times 4 gives me 28 valence electrons for my fluorine. The total number valence electrons for my molecule will be 28 plus 4. So I have to account for 32 valence electrons when I draw this dot structure. So let's go ahead and move on to the next step. Let's go back up here and look at our guidelines. So we figured out how many valence electrons we need to account for for our dot structure. We don't have any kind of charges, so we don't need to worry about the rest of step one here. We move on to step two, where we decide on the central atom of our dot structure. And the way to do this is to pick the least electronegative element that we have here, and then draw the bonds. And so for our example, we're working with silicon and fluorine. And so we can go ahead and find those again on our periodic table. Here's fluorine. Fluorine is the most electronegative element, and so therefore, for silicon tetrafluoride, we're going to put the silicon atom at the center of our dot So I'm going to start with silicon here. And I know that silicon has four bonds to fluorine atoms. I'm going to go ahead and put in some fluorines right here. So here's some fluorines like that. So I just drew four covalent bonds, and we know that each covalent bond represents two valence electrons, right? So here's two valence electrons, here's two, so that's a total of four, six, and eight. So we've represented eight valence electrons so far in our dot structure. So we originally had to represent 32. So I'm going to go ahead and subtract 8 from 32. So 32 minus 8 gives me 24. So now I only have to account for 24 valence electrons. Let's go back up and look at our steps again. So let's find out where we are. So we've decided the central atom, and we've drawn the bonds, and we just subtracted the electrons that we used to draw those bonds from the total that we got in step one. So we're on to step three, where we assign the leftover" + }, + { + "Q": "\n5:05 You mention the term terminal atoms. What is that ?", + "A": "The word Terminal Means End so a terminal atom is just the outermost atom of the molecule.", + "video_name": "p7Fsb21B2Xg", + "timestamps": [ + 305 + ], + "3min_transcript": "will have seven valence electrons. And I have four of them. So 7 times 4 gives me 28 valence electrons for my fluorine. The total number valence electrons for my molecule will be 28 plus 4. So I have to account for 32 valence electrons when I draw this dot structure. So let's go ahead and move on to the next step. Let's go back up here and look at our guidelines. So we figured out how many valence electrons we need to account for for our dot structure. We don't have any kind of charges, so we don't need to worry about the rest of step one here. We move on to step two, where we decide on the central atom of our dot structure. And the way to do this is to pick the least electronegative element that we have here, and then draw the bonds. And so for our example, we're working with silicon and fluorine. And so we can go ahead and find those again on our periodic table. Here's fluorine. Fluorine is the most electronegative element, and so therefore, for silicon tetrafluoride, we're going to put the silicon atom at the center of our dot So I'm going to start with silicon here. And I know that silicon has four bonds to fluorine atoms. I'm going to go ahead and put in some fluorines right here. So here's some fluorines like that. So I just drew four covalent bonds, and we know that each covalent bond represents two valence electrons, right? So here's two valence electrons, here's two, so that's a total of four, six, and eight. So we've represented eight valence electrons so far in our dot structure. So we originally had to represent 32. So I'm going to go ahead and subtract 8 from 32. So 32 minus 8 gives me 24. So now I only have to account for 24 valence electrons. Let's go back up and look at our steps again. So let's find out where we are. So we've decided the central atom, and we've drawn the bonds, and we just subtracted the electrons that we used to draw those bonds from the total that we got in step one. So we're on to step three, where we assign the leftover So in this case, the terminal atoms would be the fluorines. So let's go back down here and look at our dot structure. So fluorine would be the terminal atoms, and we're going to assign electrons to those fluorines. But how many do we need to assign? Well, going back to our periodic table over here, so fluorine is in the second period. So pretty good bet it's going to follow the octet rule here. So we need to surround each fluorine atom with eight electrons. Each fluorine already has two electrons around it, so I'm going to go ahead and put six more around each fluorine, like that. So each fluorine get six more valence electrons. And since I'm assigning six valence electrons to four fluorines, 6 times 4 gives me 24. And so therefore we've now accounted for all of the valence electrons. And so this should be the should be the final dot structure here. And so we don't even need to go on to step four for this molecule. This is a very simple molecule to draw." + }, + { + "Q": "\nWhat is \"lumen\" mentioned at 7:08 ?", + "A": "Lumen is the inside space of a tubular structure, such as an artery or intestine. Source: Wikipedia", + "video_name": "6UqtgH_Zy1Y", + "timestamps": [ + 428 + ], + "3min_transcript": "a separate organelle. So you get this thing that looks like this, and I'll just do it the best that I can draw it. And this right over here is called the endoplasmic reticulum. So this right here is endoplasmic reticulum, which I've always thought would be a good name for a band. And the endoplasmic reticulum is key for starting to produce and then later on package proteins that are either embedded in the cellular membrane or used outside of the cell itself. So how does that happen? Well, the endoplasmic reticulum really has two regions. It has the rough endoplasmic reticulum. And the rough endoplasmic reticulum has a bunch of ribosomes. So that's a free ribosome right over here. These are ribosomes that are attached to the membrane of the endoplasmic reticulum. So this region where you have attached ribosomes right over here, that is the rough endoplasmic reticulum. I'll call it the rough ER for short. Perhaps an even better name for a band. And then there's another region, which is the smooth endoplasmic reticulum. And the role that this plays in protein synthesis, or at least getting proteins ready for the outside of the cell, is you can have messenger RNA-- let me do that in that lighter green color-- you can have messenger RNA find one of these ribosomes associated with the rough endoplasmic reticulum. And as the protein is translated, it won't be translated inside the cytosol. It'll be translated on the other side of the rough endoplasmic reticulum. in the lumen of the rough endoplasmic reticulum. Let me make that a little bit-- let me draw that a little bit better. So let's say that this right over here, that right over here is the membrane of the endoplasmic reticulum. And then as a protein, or as a mRNA is being translated into protein, the ribosome can attach. And let's say that this right over here is the mRNA that is being translated. Let's say it's going in that direction right over here. Here is the membrane of the ER. So ER membrane. This right over here-- and actually, the way I've drawn it right over here, this is just one bilipid layer. So let me just draw it like this. I could do it like this. And this is actually, this bilipid layer is continuous." + }, + { + "Q": "\nAt 2:51, wouldn't the magnetic field be flowing out of the page instead of into on the right side of wire 2? Sal has it going into the page, but with my right thumb down my fingers go out at that position?", + "A": "The field he has drawn there is the field from wire 1, that is, the field that produces a magnetic force on wire 2. There will also be a (larger) contribution to the total magnetic field at that point from wire 2 pointing into the page, but since the field from wire 2 is symmetric around wire 2, it will produce no net force on wire 2.", + "video_name": "4tctB1wZNiI", + "timestamps": [ + 171 + ], + "3min_transcript": "And of course, it's going into the page, into the video screen, all the way out to infinity. It gets weaker and weaker. It's inversely proportional to the radius away from the wire, But even here, this magnetic field is going into the page. Now we know, just as a little bit of review, the force created by current 1 on current 2-- that's just the convention I'm using, you wouldn't always put the 1 first-- is equal to current 2 times some length-- let's call that length 2-- along the wire. This is going to be a vector because it's a magnitude of length and a direction. And it goes in the same direction as the current. So let's say that that is L2. So we're talking about from here to here. That's the current. Cross product that with the magnetic field. I'll switch back to that. Now it all seems pretty complicated, but you can just take your right hand rule and figure out the direction. So we put our index finger-- I'm doing it right now, you can't see it-- you put your index finger in the direction of L2. You can write the 2 down here, instead of writing a big 2 up there. Put your index finger in the direction of L2. I keep redoing it just to make sure I'm drawing it right. Put your middle finger in the direction of-- so this is L2. This is this. Goes in the direction of the middle finger. Sorry, the index finger. Your middle finger is going to go in the direction of the field. So it's going to be pointing downwards, because the field is going into the page, on this side of this wire. And then your other hands are going to do what they will. And then your thumb is going to go in the direction of the net force. So your thumb is going to go like that. So there you have it. You have your little veins or tendons, whatever those are, that's your nail. So in this situation, when the current is going in opposite direction, the net force is actually going to be outward on this wire. The net force is outward. And then if you don't believe me, you might want to try it yourself, but the force on current 1 or on wire 1, or some length of wire 1, caused by the magnetic field due to current 2, is also going to be outwards. So here, if you want to think about it little bit, or have a little bit of intuition, if the current's going in the same direction they will attract, and if currents are going in opposite directions they will repel each other. So anyway, let's apply some numbers. Let's apply some numbers to this problem. Let's do it with the opposite current direction. So let's say that current 1-- I'm just going to make up some numbers-- is 2 amperes. Current 2 is, I don't know, 3 amperes." + }, + { + "Q": "\nAt 1:18, why is the magnetic field of the right wire going into the page? When I use the right hand rule, I flip my thumb upside down which makes my fingers go clockwise, as opposed to counter-clockwise like the left wire.", + "A": "Yup, that s wrong :).", + "video_name": "4tctB1wZNiI", + "timestamps": [ + 78 + ], + "3min_transcript": "In the last video, we saw that if we have two currents, or two wires carrying current, and the current is going in the same direction, that they'll attract each other. Now what would happen-- before we break into the numbers-- what would happen if the two currents are going in opposite directions? Would they attract or repel each other? And you can probably guess that, but let's go through the exercise. Because I realize that last time I did it, I got And I'll do it a little bit cleaner this-- I don't have to draw as many magnetic field lines. So let's say that's wire 1. That's wire 2. And I'll just make the currents go in opposite So this is I1. And this is I2. So what would the magnetic field created by current 1 look like? Well, let's do the wrap around rule. Put our thumb in the direction of the current, and then the magnetic field will wrap around. It'll go into the page here and it'll go out of the page here. If you put your thumb up like that. Your right hand, always use your right hand. And of course, it's going into the page, into the video screen, all the way out to infinity. It gets weaker and weaker. It's inversely proportional to the radius away from the wire, But even here, this magnetic field is going into the page. Now we know, just as a little bit of review, the force created by current 1 on current 2-- that's just the convention I'm using, you wouldn't always put the 1 first-- is equal to current 2 times some length-- let's call that length 2-- along the wire. This is going to be a vector because it's a magnitude of length and a direction. And it goes in the same direction as the current. So let's say that that is L2. So we're talking about from here to here. That's the current. Cross product that with the magnetic field. I'll switch back to that. Now it all seems pretty complicated, but you can just take your right hand rule and figure out the direction. So we put our index finger-- I'm doing it right now, you can't see it-- you put your index finger in the direction of L2. You can write the 2 down here, instead of writing a big 2 up there. Put your index finger in the direction of L2. I keep redoing it just to make sure I'm drawing it right. Put your middle finger in the direction of-- so this is L2. This is this. Goes in the direction of the middle finger. Sorry, the index finger. Your middle finger is going to go in the direction of the field. So it's going to be pointing downwards, because the field is going into the page, on this side of this wire. And then your other hands are going to do what they will. And then your thumb is going to go in the direction of the net force. So your thumb is going to go like that. So there you have it." + }, + { + "Q": "Why did you multiply by area 4:18?\n", + "A": "He multiplied for 1 teorically, [1 = a/a = b/b =A/A] so you can multiply and divide by the area whenever you want, he did that so that it lets him take [F/A = P]", + "video_name": "uqyLOuAzbvo", + "timestamps": [ + 258 + ], + "3min_transcript": "that point of the system, plus the kinetic energy at that point of the system. Then we know from the conservation of energy that that has to equal the output work plus the output potential energy plus the output kinetic energy. A lot of times in the past, we've just said that the potential energy input plus the kinetic energy input is equal to the potential energy output plus the kinetic energy output, but the initial energy in the system can also be done by work. So we just added work to this equation that says that the energy in is equal to the energy out. With that information, let's see if we can do anything interesting with this pipe that I've drawn. So what's the work that's being put into this system? It's the force in times the distance in, and so over a period of time, t, what has been done? We learned in the last video that over a period of time, t, the fluid here might have moved this far. What is this distance? This distance is the input velocity times whatever amount of time we're dealing with, so T-- so that's the distance. What's the force? The force is just pressure times area, and we can figure that out by just dividing force by, area and then multiply by area, so we get input force divided by area input, times area input. It's divided and multiplied by the same number-- that's It's equal to the input distance over that amount of time, and that's velocity times time, so the work input is equal to the input pressure times the input area times input velocity times time. What is this area times velocity times time, times this distance? That's the volume of fluid that flowed in over that amount of time. So that equals the volume of fluid over that period of time, so we could call that volume in, or volume i--" + }, + { + "Q": "\nAt 6:33 when Sal is substituting for the potential and kinetic energies for the energy equation he substitutes with mass times gravity times height and mass times velocity squared divided by 2 respectively. In subsequent videos he uses rho instead of mass. I was wondering which version is correct.", + "A": "rho is density or mass per unit volume in this case", + "video_name": "uqyLOuAzbvo", + "timestamps": [ + 393 + ], + "3min_transcript": "We know that density is just mass per volume, or that volume times density is equal to mass, or we know that volume is equal to mass divided by density. The work that I'm putting into the system-- I know I'm doing a lot of crazy things, but it'll make sense so far-- is equal to the input pressure times the amount of volume of fluid that moved over that period of time. That volume of fluid is equal to the mass of the fluid that went in at that period of time, and we'll call that the input mass, divided by the density. Hopefully, that makes a little bit of sense. As we know, the input volume is going to be equal to the doesn't change-- is equal to the output mass, so we don't have to write an input and output for the mass. The mass is going to be constant; in any given amount of time, the mass that enters the system will be equivalent to the mass that exits the system. There we go: we have an expression, an interesting expression, for the work being put into the system. What is the potential energy of the system on the left-hand side? The potential energy of the system is going to be equal to that same mass of fluid that I talked about times gravity times this input height-- the initial height-- times h1. The initial kinetic energy of the fluid equals the mass of the fluid-- this mass right here, of that same cylinder volume that I keep pointing to-- times the velocity of the We remember this from kinetic energy divided by 2. So what's the total energy at this point in the system over this period of time? How much energy has gone into the system? It's going to be the work done, which is the input pressure-- I'm running out of space, so let me erase all of this. I'll probably have to run out of time, too, but that's OK-- it's better than being confused. Back to what we were doing. So, the total energy going into the system is the work being done into the system, and I rewrote it in this" + }, + { + "Q": "At \"3:38\" SAL says that Vi*t= D in the formula work=fd. what does distance have to do with time?\n", + "A": "distance is the amount something travels in a given time. So, if distance is unknown, you would need time to solve for it- requiring either solving mathematically or measuring. If it needs to be solved mathematically you would need the time to figure out how far something traveled in a certain amount of time. Additionally, Force= mass * acceleration- where again time could be important to know as acceleration is the change in velocity over a given time.", + "video_name": "uqyLOuAzbvo", + "timestamps": [ + 218 + ], + "3min_transcript": "that point of the system, plus the kinetic energy at that point of the system. Then we know from the conservation of energy that that has to equal the output work plus the output potential energy plus the output kinetic energy. A lot of times in the past, we've just said that the potential energy input plus the kinetic energy input is equal to the potential energy output plus the kinetic energy output, but the initial energy in the system can also be done by work. So we just added work to this equation that says that the energy in is equal to the energy out. With that information, let's see if we can do anything interesting with this pipe that I've drawn. So what's the work that's being put into this system? It's the force in times the distance in, and so over a period of time, t, what has been done? We learned in the last video that over a period of time, t, the fluid here might have moved this far. What is this distance? This distance is the input velocity times whatever amount of time we're dealing with, so T-- so that's the distance. What's the force? The force is just pressure times area, and we can figure that out by just dividing force by, area and then multiply by area, so we get input force divided by area input, times area input. It's divided and multiplied by the same number-- that's It's equal to the input distance over that amount of time, and that's velocity times time, so the work input is equal to the input pressure times the input area times input velocity times time. What is this area times velocity times time, times this distance? That's the volume of fluid that flowed in over that amount of time. So that equals the volume of fluid over that period of time, so we could call that volume in, or volume i--" + }, + { + "Q": "\nAt 1:50, why is there 1% of carbon dioxide in the alveoli? I thought we breathe all the carbon dioxide out.", + "A": "because we are not perfect human beings and 1% will remain", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 110 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 14:12, how can red blood cells have NO nucleus or DNA. How do blood tests work if they don't?", + "A": "When they blood type they are looking for the presence of cell surface antigens. A antigens, B antigens, both of them together (AB), or no antigens at all (O type).", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 852 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 3:02 you said that Artery's go away from the heart but don't they come from the heart?", + "A": "Going away from the heart and coming from the heart are the same thing. Going away from the heart and coming to the heart are opposites. Arteries carry blood away from the heart This can also be said as blood comes from the heart through arteries.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 182 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 9:08, Sal says that the N2 does diffuse in the blood, so what do we do with it? And can we make amino acids out of nitrogen?", + "A": "simply no. Nitrogen is removed as a waste product. Nitrogen is either expelled out of our body when we breathe, or removed from our waste products.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 548 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 12:35, Sal stated that from the outside of the body, veins appear a blue-green color. I looked more closely and saw that little purple colored vessels. Is this a capalary or an artery and if a capalary then what is contained in side of it?", + "A": "No, the technical term for them is veins. Veins are the vessels that carry blood towards the lungs so this blood is not oxygenated which makes it blue. When oxygen binds to the iron in the blood, the iron rusts so the blood is red. When the iron is not bonded to oxygen, then it is gray and the other colors in the cell are more prominent making it blue.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 755 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 15:16 you said \"They get rid of their nucleus.\" Does the dent in the center have anything to do with that, or is that just a structure created by evolution?", + "A": "The dent increases surface area allowing for more oxygen to be absorbed, yes, you could say it s evolution that imposed its structural development.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 916 + ], + "3min_transcript": "" + }, + { + "Q": "At around 3:00, when he's talking about arteries, he said earlier in the video that it was capillaries. Are arteries and capillaries the same thing?\n", + "A": "arteries are blood vessels that take blood aay from the heart and lungs. Capillaries are tiny, very tiny blood vessels that help with diffusion etc.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 180 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 1:40 he says there is less than 1 percent of Carbon dioxide in the atmosphere but their are trillions of humans breathing it every second how is that possible?", + "A": "Photosynthetic organisms (plants, bacteria etc.) take CO2 and transform it to O2 in the process of photosynthesis. That s why there is more O2 than CO2 in the atmosphere. Interestingly, before photosynthetic organisms appeared there was less O2 in the atmosphere. About 300 million years ago, the amount of Oxygen peaked at about 30%. Also, (as of 2012.) there are about 7 billion humans on Earth not trillions.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 100 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 2:06 sal says pulmonary capillaries.\nwhat are pulmonary capillaries?", + "A": "Pulmonary capillaries surround the alveoli and are where gas exchange takes place between the pulmonary arteries and pulmonary veins.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 126 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 3:56, why is this membrane so thin, and why is it there at all?", + "A": "The membrane is thin because you need to get O2 and CO2 across it. The thicker a membrane is, the harder it would be for small molecules like this to diffuse through. And it is there in the first place to keep everything else in the right place. Without a membrane there, your blood would pour into the alveoli and you would cough it up. Not very good for oxygen delivery to your tissues! :)", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 236 + ], + "3min_transcript": "" + }, + { + "Q": "5:16 to 5:30 does that mean time/2 * acceleration of gravity = initial velocity. I want to now if in this equation does it have to be acceleration of gravity or can it be just acceration.\n", + "A": "I think it can just be acceleration.", + "video_name": "IYS4Bd9F3LA", + "timestamps": [ + 316, + 330 + ], + "3min_transcript": "We know that the acceleration We know the acceleration of gravity here, we are assuming it's constant or slightly not constant but we are gonna assume it's constant We are just dealing close to the surface of the earth is negative 9.8 m/s*s, so let's think about it This change in velocity, are change in velocity Their change in velocity is the final velocity minus the initial velocity which is the same thing as zero minus the initial velocity which is the negative of the initial velocity That's another way to think about change in velocity We just shown the definition of acceleration change in velocity is equal to acceleration, is equal to acceleration negative 9.8 m/s*s times time or times change in time, our change in time, So the change in time is 2.5 s, times 2.5 s So what is our change in velocity which is also the same thing as negative of our initial velocity Get the calculator out, let me get my calculator, bring it on to the screen, so it is negative 9.8 m/s times 2.5 s Times 2.5 s, it gives us negative 24.5, so this gives us I will write it in new color This gives us negative 24.5 m/s, this seconds cancels out With one of these seconds in the denominator we only have one of the denominator out m/s, and this is the same thing as the negative, as the negative initial velocity Negative initial velocity that's the same thing as change in velocity So that simply we are able to figure out what our velocity is So literally you take the time, the total time in the air divide by two And multiply that by acceleration of gravity and if you take I guess you can take the absolute value of that or take the positive version of that And that gives you your initial velocity So your initial velocity here is literally 24.5 m/s Since it's a positive quantity it is upwards in this example So that's my initial velocity, so we already figure out part of this game The initial velocity that threw upward That's also going to be, we gonna have the same magnitude of velocity The balls about to hit the ground although is gonna be in the other direction So what is the distance or let me make it clear what is the displacement of the ball from its lowest point right when it leaves your hand" + }, + { + "Q": "Why at around 7:10 Sal divides by 2?\n", + "A": "He finding the average velocity. To get the average of 2 number you add them and devide by 2. Since one of the numbers is 0 you end up dividing the other by two. For example the average of 1 and 3 is (1 + 3)/2 = 4/2 = 2 but the average of 0 and 2 is (0 + 2)/2 = 2/2 = 1", + "video_name": "IYS4Bd9F3LA", + "timestamps": [ + 430 + ], + "3min_transcript": "So that simply we are able to figure out what our velocity is So literally you take the time, the total time in the air divide by two And multiply that by acceleration of gravity and if you take I guess you can take the absolute value of that or take the positive version of that And that gives you your initial velocity So your initial velocity here is literally 24.5 m/s Since it's a positive quantity it is upwards in this example So that's my initial velocity, so we already figure out part of this game The initial velocity that threw upward That's also going to be, we gonna have the same magnitude of velocity The balls about to hit the ground although is gonna be in the other direction So what is the distance or let me make it clear what is the displacement of the ball from its lowest point right when it leaves your hand We just have to remember, all of these come from very straight forward ideas Change in velocity is equal to acceleration times change in time And the other simple idea is that displacement is equal to average velocity, average velocity times change in time Now what is our average velocity? Our average velocity is your initial velocity plus your final velocity Divided by 2, or we assume that acceleration is constant So literally just the arithmetic mean of your initial and final velocity So what is that? That's gonna be 24.5 m/s plus our final velocity In this situation we are just going over to the first 2.5 s So our final velocity is once again 0 m/s We are just talking about when we get to his point over here And we just gonna divide that by 2 This will give us the average velocity And we wanna multiply that by 2.5 s, times 2.5 s So we get this part right over here 24.5 divided by 2 When you go with the 0, it is still 24.5 It gives us 12.25 times 2.5, and remember this right over here is in seconds let me write the units down So this is 12.25 m/s times 2.5 seconds And just to remind ourselves We are calculating the displacement over the first 2.5 seconds So this gives us, I get the calculator out once again We have 12.25 times 2.5 seconds gives us 30.625" + }, + { + "Q": "\n@8:00 Sal is talking about dividing by 2 then multiplying by 2.5, why cant you multiply by 1.5 instead of complicating it? I did the math and it doesn't add up the same. Why is that?", + "A": "I am not sure how you got the factor of 1.5. Sal is talking about taking the average of the initial velocity which is 24.5 m/s and the final velocity in the upward trajectory of the ball, which is 0 m/s. In constant acceleration motion, to find the average velocity, you just take the average of final and initial velocities: (u+v)/2. Displacement \u00ce\u0094S = average velocity * \u00ce\u0094t. = {(u+v)/2} * \u00ce\u0094t", + "video_name": "IYS4Bd9F3LA", + "timestamps": [ + 480 + ], + "3min_transcript": "We just have to remember, all of these come from very straight forward ideas Change in velocity is equal to acceleration times change in time And the other simple idea is that displacement is equal to average velocity, average velocity times change in time Now what is our average velocity? Our average velocity is your initial velocity plus your final velocity Divided by 2, or we assume that acceleration is constant So literally just the arithmetic mean of your initial and final velocity So what is that? That's gonna be 24.5 m/s plus our final velocity In this situation we are just going over to the first 2.5 s So our final velocity is once again 0 m/s We are just talking about when we get to his point over here And we just gonna divide that by 2 This will give us the average velocity And we wanna multiply that by 2.5 s, times 2.5 s So we get this part right over here 24.5 divided by 2 When you go with the 0, it is still 24.5 It gives us 12.25 times 2.5, and remember this right over here is in seconds let me write the units down So this is 12.25 m/s times 2.5 seconds And just to remind ourselves We are calculating the displacement over the first 2.5 seconds So this gives us, I get the calculator out once again We have 12.25 times 2.5 seconds gives us 30.625 The seconds cancel out This is actually a ton, you know, roughly give or take about 90 feet throw into the air, this looks like a nine stories building And I frankly do not have the arm for that But if someone is able to throw the ball for 5 seconds in the air They have thrown 30 meters in the air Hopefully you will find that entertaining In my next video I'll generalize this maybe we can get a little bit of formula so maybe you can generalize it So regardless of the measurement of time you can get the displacement in the air Or even better, try to derive it yourself And you will see how, at least how I tackle it in the next video" + }, + { + "Q": "At 3:24, Sal showed that -9.8 was the velocity. I am confused about how he got that number. Was it from the previous video? Because i remember seeing it there.\n", + "A": "-9.8 is the acceleration not the velocity maybe if you ask again i can clarify more", + "video_name": "IYS4Bd9F3LA", + "timestamps": [ + 204 + ], + "3min_transcript": "We gonna assume air resistance is negligible And what that does for us is we can assume that the time up That the time for the ball to go up to its peak height is the same thing as the time that takes it to go down If you look at this previous video, we've plot it displacement verse time You see after 2 seconds the ball went from being on the ground or I guess the thrower's hand all the way to its peak height And then the next 2 seconds it took the same amount of time to go back down to the ground which makes sense whatever the initial velocity is, it take half the time to go to zero and it takes the same amount of time to now be accelerated into downward direction back to that same magnitude of velocity but now in the downward direction So let's play around with some numbers here Just so you get a little bit more of concrete sense So let's say I throw a ball in the air And you measure using the stop watch and the ball is in the air for 5s Well the first thing we could do is we could say look at the total time in the air was 5 seconds that mean the time, let me write it, that means the change in time to go up during the first half, I guess the ball time in the air is going to be 2.5 seconds and which tells us that over this 2.5 seconds we went from our initial velocity, whatever it was We went from our initial velocity to our final velocity which is a velocity of 0 m/s in the 2.5 seconds And this is a graph for that example, This is the graph for the previous one, The previous example we knew the initial velocity but in whatever the time is you are going from you initial velocity to be stationery at the top, right with the ball being stationery and then start getting increasing velocity in the downward direction So it takes 2.5 seconds to go from some initial velocity to 0 seconds We know that the acceleration We know the acceleration of gravity here, we are assuming it's constant or slightly not constant but we are gonna assume it's constant We are just dealing close to the surface of the earth is negative 9.8 m/s*s, so let's think about it This change in velocity, are change in velocity Their change in velocity is the final velocity minus the initial velocity which is the same thing as zero minus the initial velocity which is the negative of the initial velocity That's another way to think about change in velocity We just shown the definition of acceleration change in velocity is equal to acceleration, is equal to acceleration negative 9.8 m/s*s times time or times change in time, our change in time," + }, + { + "Q": "\nHi! At 1:13, Sal says that the time it takes the object to go up is the same as the time it takes for the object to come down. I understand that we are not taking air resistance into account, but wouldn't gravity make the ball come down faster? The ball is working against gravity on it's way up and falling with gravity on the way down so it seems like the ball's decent would be faster than its ascent. Thanks for the help!", + "A": "The acceleration due to gravity is the same throughout the flight. Acceleration is rate of change in velocity. Won t it take exactly the same amount of time to go from 20 m/s to 0 as it will take to go from 0 to -20 m/s?", + "video_name": "IYS4Bd9F3LA", + "timestamps": [ + 73 + ], + "3min_transcript": "Let's say you and I are playing a game or I'm trying to figure out how high a ball is being thrown in the air How fast would we throwing that ball in the air? And what we do is one of us has a ball and the other one has a stop watch over here So this is my best attempt to make it more like a cat than a stop watch but I think you get the idea And what we do is one of us throw the ball the other one times how long the ball is in the air And what we do is gonna use that time in the air to figure out how fast the ball was thrown straight up and how long it was in the air or how high it got And there is going to be one assumption I make here frankly that's an assumption we are gonna make in all of these projectile motion type problem is that air resistance is negligible And for something like it, this is a baseball or something like that That's a pretty good approximation So when can I get the exact answer, I encourage you experiment it on your own or even to see what air resistance does to your calculations We gonna assume for this projectile motion in future one We gonna assume air resistance is negligible And what that does for us is we can assume that the time up That the time for the ball to go up to its peak height is the same thing as the time that takes it to go down If you look at this previous video, we've plot it displacement verse time You see after 2 seconds the ball went from being on the ground or I guess the thrower's hand all the way to its peak height And then the next 2 seconds it took the same amount of time to go back down to the ground which makes sense whatever the initial velocity is, it take half the time to go to zero and it takes the same amount of time to now be accelerated into downward direction back to that same magnitude of velocity but now in the downward direction So let's play around with some numbers here Just so you get a little bit more of concrete sense So let's say I throw a ball in the air And you measure using the stop watch and the ball is in the air for 5s Well the first thing we could do is we could say look at the total time in the air was 5 seconds that mean the time, let me write it, that means the change in time to go up during the first half, I guess the ball time in the air is going to be 2.5 seconds and which tells us that over this 2.5 seconds we went from our initial velocity, whatever it was We went from our initial velocity to our final velocity which is a velocity of 0 m/s in the 2.5 seconds And this is a graph for that example, This is the graph for the previous one, The previous example we knew the initial velocity but in whatever the time is you are going from you initial velocity to be stationery at the top, right with the ball being stationery and then start getting increasing velocity in the downward direction So it takes 2.5 seconds to go from some initial velocity to 0 seconds" + }, + { + "Q": "how does he know (19:34) that there is more heat going in the systm than heat going out?\n", + "A": "Since work=heat, and from step C to A (that the system is receiving heat) the work is less than from step A to C. And this is because when the temperature of a system is at lower state you need less force to do work to the system.", + "video_name": "aAfBSJObd6Y", + "timestamps": [ + 1174 + ], + "3min_transcript": "And in the process, it was also doing some work. And what was the work that it did? Well, it's the area under this curve, or the area inside of this cycle. So this is the work done by our Carnot engine. And the way you think about it is, when you're going in the rightward direction with increasing volume, it's the area under the curve is the work done by the system. And then when you move in the leftward direction with decreasing volume, you subtract out the work done to the system, and then you're left with just the area in the curve. So we can write this Carnot engine like this. It's taking, it's starting-- so you have a reservoir at T1. And then you have your engine, right here. And then it's connected-- so this takes Q1 in from this reservoir. The work is represented by the amount of-- the work right here is the area inside of our cycle. And then it transfers Q2, or essentially the remainder from Q1, into our cold reservoir. So T2. So it transfers Q2 there. So the work we did is really the difference between Q1 and Q2, right? You say, hey. If I have more heat coming in than I'm letting out, where did the rest of that heat go? It went to work. Literally. So Q1 minus Q2 is equal to the amount of work we did. And actually, this is a good time to emphasize again that heat and work are not a state variable. A state variable has to be the exact same value when we complete a cycle. Now, we see here that we completed a cycle, and we had a net amount of work done, or a net amount of heat added to the system. So we could just keep going around the cycle, and keep having heat added to the system. So there is no inherent heat state variable right here. this point in time. All you could say is what amount of heat was added or taken away from the system, or you can only say the amount of work that was done to, or done by, the system. Anyway, I want to leave you there right now. We're going to study this a lot more. But the real important thing is, and if you never want to get confused in a thermodynamics class, I encourage you to even go off on your own, and do this yourself. Kind of-- you can almost take a pencil and paper, and redo this video that I just did. Because it's essential that you understand the Carnot engine, understand this adiabatic process, understand what isotherms are. Because if you understand that, then a lot of what we're about to do in the next few videos with regard to entropy will be a little bit more intuitive, and not too confusing." + }, + { + "Q": "At 4:30 you said you want the triple bond to have the lowest # of Carbon placement possible for its naming. But what if there is a double bond within the structure too, and they happen to tie for this \"lowest # of Carbon placement possible.\" Which bond would get priority for naming?\n", + "A": "To answer your question, alkenes have priority over alkynes. Thus the molecule C-C\u00e2\u0089\u00a1C-C=C-C would be named hex-2-en-4-yne by IUPAC.", + "video_name": "qZTeyhR1akA", + "timestamps": [ + 270 + ], + "3min_transcript": "So when we talked about two carbons for nomenclature, our root was eth-, and since we're dealing with alkynes, our ending is going to be -yne here. So I stick the root together with the ending, so I take eth- and I add the -yne to it, so it's called ethyne. So ethyne would be the IUPAC name for this molecule. That's probably not the name that's used most frequently. This is the most famous alkyne, also called acetylene. So you've heard of acetylene torches before, and you can do a very cool demonstration called underwater fireworks where you use acetylene in there. So this is acetylene, which is the common name or ethyne, which is more the IUPAC name, and this is the simplest alkyne. So let's name another one here. Let's get an actual chain, a carbon chain in here, and let's see how to name this one like that. Well, the first thing to notice is the straight line And remember, that's because that portion of the molecule is linear, because those carbons are sp hybridized. So when you're drawing Lewis Dot Structures, you'll notice that almost everyone will draw that portion of the molecule straight to better reflect the actual geometry of the molecule. All right, so I need to figure out how many carbons are in this alkyne. So I find my longest carbon chain including my triple bond. So let's see, here is one carbon, two carbons, three carbons, four, and five. So a five carbon alkyne. So remember your root for five carbons would be pent-. So it would be pentane if it was an alkane, but since it's an alkyne, it would be pentyne. So this is pentyne. So I have that much so far. Next, I need to figure out how to number my chain. So I could number it from the left, My goal is to give my triple bond the lowest number possible, so it's a lot like double bonds here. So I want to give my triple bond the lowest number possible, and that means, of course, I need to start from the right here. So if I start from the right, and I say this is carbon one, then my triple bond starts at carbon two. And then this would be carbon three, carbon four, and carbon five, like that. So the triple bond starts at carbon two, so all I have to do is put a two in here and say 2-pentyne. So in that respect, it's just like naming alkenes here. So that's 2-pentyne. Let's do another one, one that's a little bit more complicated than that. So let's look at this molecule here. So there's my triple bond, and then I have-- let's see, I'll put in a chain and some methyl groups there-- and so this is my molecule. All right, so once again find the longest carbon chain that includes your triple bond." + }, + { + "Q": "At 2:24 sal says halogens are group 7 elements but they are group 17\n", + "A": "In the new system halogens occupy group 17... Older system puts halogens into SevenA group. Also keep in mind that halogens have seven valence electrons. he may have been referring to that...", + "video_name": "__zy-oOLPug", + "timestamps": [ + 144 + ], + "3min_transcript": "- [Voiceover] When you're studying chemistry you'll often see reactions, in fact you always see reactions. For example if you have hydrogen gas it's a diatomic molecule, 'cause hydrogen bonds with itself in the gassy state, plus iodine gas, I2, that's also in the gassy state, it's very easy to just sort of, oh you know, if you put 'em together they're going to react and form the product, if you have two moles of, hydrogen, two moles of iodine, so it's gonna form two moles of hydrogen iodide. That's all nice and neat and it makes it seem like it's a very clean thing that happens without much fuss. But we know that that isn't the reality and we also know that this doesn't happen just instantly, it's not like you can just take some hydrogen, put it with some iodine, and it just magically turns into hydrogen iodide. That there's some process going on, that these gaseous state particles are bouncing around, and somehow they must bounce into each other and break bonds that they were in before, and that's what we're going to study now. This whole study of how the reaction progresses, and the rates of the reactions is called kinetics. Which is a very fancy word, but you're probably familiar with it because we've talked a lot about kinetic energy. Kinetics. Which is just the study of the rate of reactions. How fast do they happen, and how do they happen? So let's just in our minds, come up with a intuitive way that hydrogen and iodine can combine. So let's think about what hydrogen looks like. So if we get our periodic table out, hydrogen's got one valence electron so if they have two hydrogen atoms they can share them with each other. And then iodine, iodine has seven valence electrons, so if they each share one they get complete as well. So let's just review that right now. So hydrogen this hydrogen might have one, well, will have one electron out there. And then you can have another hydrogen that has another electron out there, this hydrogen can pretend like he has this electron, this hydrogen can pretend like she has that electron, and then they're happy. They both feel like they've completed their 1S shell. Same thing on the iodine side. Where you have two iodines, they both have seven valence electrons. They're halogens, you know that already. Halogens are the group seven elements, so they have seven electrons this guy's got one here, this guy's got one here, if this guy can pretend like he's got that electron, he's happy, he has eight valence electrons. If this guy can pretend like he's got that one, same thing. So there's a bond right here, and this is why hydrogen is a diatomic molecular gas, and this is why iodine is the same. Now, when they're in the gaseous state, you have a bunch of these things that are moving around bumping into each other, I'll do it like this. So the hydrogen might look something like this, the hydrogen is these two atomic spheres that are bonded together," + }, + { + "Q": "\nAt 0:44, Why do sal put O2 instead of just O?", + "A": "Because oxygen doesn t hang around as single atoms on the surface of earth, it comes as a diatomic gas with 2 oxygen atoms bonded together, so O2", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 44 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 10:14, why is he converting moles to moles?", + "A": "You can also think about it this way grams A -> moles A -> moles B -> grams B", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 614 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 12:00 why did Sal multiply 0.833 with 32? Where did the 32 come from?", + "A": "For every mol of O2 there are 32g O2, so you can multiply something by 32g O2/1mol O2. It s like multiplying by 1/1. He did that so that you can cancel out the mol O2 on top and bottom and solve for g of O2.", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 720 + ], + "3min_transcript": "" + }, + { + "Q": "at 0:25 why does glucose react with oxygen?\n", + "A": "It s the equation for respiration and is how the body acquires energy.", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 25 + ], + "3min_transcript": "" + }, + { + "Q": "\n@10:50, is it 0.833 of O2 or O? Does it make a difference?", + "A": "Its moles of O2. This makes a difference in calculating grams as there are 2 oxygen atoms per mole. Hope this helps :)", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 650 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 10:11, so the coefficient symbolizes the amount of moles?", + "A": "Mostly, but not exact. The coefficients represent the ratios of the numbers of molecules of the chemicals involved in the reaction. As such, something like 2H\u00e2\u0082\u0082 (g) + O\u00e2\u0082\u0082(g) \u00e2\u0086\u0092 2H\u00e2\u0082\u0082O (g) The coefficients could mean 2 molecules of H\u00e2\u0082\u0082 and 1 molecule of O\u00e2\u0082\u0082 reacting. Or it could be 2 moles of H\u00e2\u0082\u0082 and 1 molecule of O\u00e2\u0082\u0082 reacting. Or it could be 0.0150 moles of H\u00e2\u0082\u0082 and 0.0075 moles of O\u00e2\u0082\u0082 reacting. So, it doesn t actually specify the quantity you have, it is just the ratio of the numbers of molecules,", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 611 + ], + "3min_transcript": "" + }, + { + "Q": "\n4:35 oxygen is spelled incorrectly", + "A": "Yes, Sal has missed the y in the spelling of oxygen.", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 275 + ], + "3min_transcript": "" + }, + { + "Q": "at 12:51, you used every one mole of glucose we produce 6 mole of carbon dioxide\" and found the moles of co2 first, does it make any difference if i take the moles of glucose and find out the number of h2o moles first, or i cant do that because there is no carbon in h2o so i have to find the moles of carbons first?\n", + "A": "He didn t even have to write the balanced equation. He knew that i mol glucose contains 6 mol f C, and 6 mol of C must produce 6 mol of CO\u00e2\u0082\u0082. You can start anywhere you want when balancing an equation, but usually the procedure that works best is to start with the most complicated formula (glucose); balance all atoms other than O and H (i.e., balance C); balance O; balance H last.", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 771 + ], + "3min_transcript": "" + }, + { + "Q": "at 0:56, is there a way to check your stoichiometry problem to make sure its right?\n", + "A": "You can google the reaction after trying it out for yourself to see if you are right :) read the language in the question carefully to work out what is a reactant (left side) and what is a product (right side). Most gases like Oxygen and Hydrogen will exist as O2 and H2, not just a single atom O Fun fact! This reaction is known as respiration.", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 56 + ], + "3min_transcript": "" + }, + { + "Q": "at 1:42 sal uses the term notional charge\nwhat does it mean?\n", + "A": "Physicists do not use the term notional charge. It is probably because Sal has a finance background and is bringing terminology from that world into play. In this case, he is using the term notional charge to imply a test charge.", + "video_name": "0YOGrTNgGhE", + "timestamps": [ + 102 + ], + "3min_transcript": "Let's imagine that instead of having two charges, we just have one charge by itself, sitting in a vacuum, sitting in space. So that's this charge here, and let's say its charge is Q. That's some number, whatever it is. That's it's charge. And I want to know, if I were to place another charge close to this Q, within its sphere of influence, what's going to happen to that other charge? What's going to be the net impact on it? And we know if this has some charge, if we put another charge here, if this is 1 coulomb and we put another charge here that's 1 coulomb, that they're both positive, they're going to repel each other, so there will be some force that pushes the next charge away. If it's a negative charge and I put it here, it'll be even a stronger force that pulls it in because it'll be closer. So in general, there's this notion of what we can call an electric field around this charge. And what's an electric field? We can debate whether it really exists, but what it affecting the space around it in some way that whenever I put-- it's creating a field that whenever I put another charge in that field, I can predict how the field will affect that charge. So let's put it in a little more quantitative term so I stop confusing you. So Coulomb's Law told us that the force between two charges is going to be equal to Coulomb's constant times-- and in this case, the first charge is big Q. And let's say that the second notional charge that I eventually put in this field is small q, and then you divide by the distance between them. Sometimes it's called r because you can kind of view the distance as the radial distance between the two charges. So sometimes it says r squared, but it's the distance between them. So what we want to do if we want to calculate the field, we want to figure out how much force is there placed per charge at any point around this Q, so, say, at a given At this distance, we want to know, for a given Q, what is the force going to be? So what we can do is we could take this equation up here and divide both sides by this small 1, and say, OK, the force-- and I will arbitrarily switch colors. The force per charge at this point-- let's call that d1-- is equal to Coulomb's constant times the charge of the particle that's creating the field divided by-- well, in this case, it's d1-- d1 squared, right? Or we could say, in general-- and this is the definition of the electric field, right? Well, this is the electric field at the point d1, and if we wanted a more general definition of the electric field, we'll just make this a general variable, so instead of having a particular distance, we'll define the field for all distances away from the point Q." + }, + { + "Q": "\nat 8:22 , he said positive test charge . what is a test charge ?", + "A": "A test charge is a charge that we imagine is so small that it does not affect the field we are trying to test.", + "video_name": "0YOGrTNgGhE", + "timestamps": [ + 502 + ], + "3min_transcript": "that circle around it, this is generating a field that if I were to put-- let's say I were to place a 1 coulomb charge here, the force exerted on that 1 coulomb charge is going to be equal to 1 coulomb times the electric fields, times 2,500 newtons per coulomb. So the coulombs cancel out, and you'll have 2,500 newtons, which is a lot, and that's because 1 coulomb is a very, very large charge. And then a question you should ask yourself: If this is 1 times 10 to the negative 6 coulombs and this is 1 coulomb, in which direction will the force be? Well, they're both positive, so the force is going to be outwards, right? So let's take this notion and see if we can somehow draw an electric field around a particle, just to get an intuition of what happens when we later put a charge anywhere So there's a couple of ways to visualize an electric field. One way to visualize it is if I have a-- let's say I have a point charge here Q. What would be the path of a positive charge if I placed it someplace on this Q? Well, if I put a positive charge here and this Q is positive, that positive charge is just going to accelerate outward, right? It's just going to go straight out, but it's going to accelerate at an ever-slowing rate, right? Because here, when you're really close, the outward force is very strong, and then as you get further and further away, the electrostatic force from this charge becomes weaker and weaker, or you could say the field becomes weaker and weaker. But that's the path of a-- it'll just be radially outward-- of a positive test charge. And then if I put it here, well, it would be radially outward that way. It wouldn't curve the way I drew it. It would be a straight line. If I did it here, it would be like that, but then I can't draw the arrows. If I was here, it would out like that. I think you get the picture. At any point, a positive test charge would just go straight out away from our charge Q. And to some degree, one measure of-- and these are called electric field lines. And one measure of how strong the field is, is if you actually took a unit area and you saw how dense So here, they're relatively sparse, while if I did that same area up here-- I know it's not that obvious. I'm getting more field lines in. But actually, that's not a good way to view it because I'm covering so much area. Let me undo both of them. You can imagine if I had a lot more lines, if I did this area, for example, in that area, I'm capturing two of these field lines. Well, if I did that exact same area out here, I'm only" + }, + { + "Q": "What is static electric force at 12:10?\n", + "A": "static electric force is the force acting between two charges Static charge is charge that is not moving (unlike current)", + "video_name": "0YOGrTNgGhE", + "timestamps": [ + 730 + ], + "3min_transcript": "shorter and shorter the electric field vectors get. And so, in general, there's all sorts of things you can draw the electric fields for. Let's say that this is a positive charge and that this is a negative charge. Let me switch colors so I don't have to erase things. If I have to draw the path of a positive test charge, it would go out radially from this charge, right? But then as it goes out, it'll start being attracted to this one the closer it gets to the negative, and then it'll curve in to the negative charge and these arrows go like this. And if I went from here, the positive one will be repelled really strong, really strong, it'll accelerate fast and it's rate of acceleration will slow down, but then as it gets closer to the negative one, it'll speed up again, and then that would be its path. Similarly, if there was a positive test charge here, its path would be like that, right? If it was here, its path would be like that. If it was there, maybe its path is like that, and at some point, its path might never get to that-- this out here might just go straight out that way. That one would just go straight out, and here, the field lines would just come in, right? A positive test charge would just be naturally attracted to that negative charge. So that's, in general, what electric field lines show, and we could use our little area method and see that over here, if we picked a given area, the electric field is much weaker than if we picked that same area right here. We're getting more field lines in than we do right there. So that hopefully gives you a little sense for what an electric field is. It's really just a way of visualizing what the impact would be on a test charge if you bring it close to another charge. And hopefully, you know a little bit about Coulomb's constant. And let's just do a very simple-- I'm getting this out of the AP Physics book, but they say-- let's do a little simple problem: Calculate the static electric force between So 6 times-- oh, no, that's not on an electric field. Oh, here it says: What is the force acting on an electron placed in an external electric field where the electric field is-- they're saying it is 100 newtons per coulomb at that point, wherever the electron is. So the force on that, the force in general, is just going to be the charge times the electric field, and they say it's an electron, so what's the Well, we know it's negative, and then in the first video, we learned that its charge is 1.6 times 10 to the negative nineteenth coulombs times 100 newtons per coulomb. The coulombs cancel out. And this is 10 squared, right? This is 10 to the positive 2, so it'll be 10 to the minus 19" + }, + { + "Q": "\nWhat is silican? As said in 3:23", + "A": "Silicon is an element. It has the symbol Si.", + "video_name": "T2DaaGuKOTo", + "timestamps": [ + 203 + ], + "3min_transcript": "(lively intro music) I'm a collection of organic compounds called Hank Green. An organic compound is more or less any chemical that contains carbon, and carbon is awesome. Why? Lots of reasons. I'm gonna give you three. First, carbon is small. It doesn't have that many protons and neutrons. Almost always 12, rarely it has some extra neutrons making it C-13 or C-14. Because of that, carbon does not take up a lot of space and can form itself into elegant shapes. It can form rings. It can form double or even triple bonds. It can form spirals and sheets and all kinds of really awesome things that bigger molecules would never manage to do. Basically, carbon is like an olympic gymnast. It can only do the remarkable and beautiful things it can do because it's petite. Second, carbon is kind. It's not like other elements that desperately want to gain or lose or share electrons No, carbon knows what it's like to be lonely, so it's not all, \"I can't live without your electrons.\" Needy, like chlorine or sodium is. This is why chlorine tears apart your insides if you breathe it in gaseous form, and why sodium metal, if ingested, will explode. Carbon, though, eh. It wants more electrons, but it's not going to kill for them. It's easy to work with. It makes and breaks bonds like a 13-year-old mall rat, but it doesn't ever really hold a grudge. Third, carbon loves to bond because it needs 4 extra electrons, so it will bond with whoever happens to be nearby. Usually, it will bond with 2 or 3 or 4 of them at the same time. Carbon can bond with lots of different elements. Hydrogen, oxygen, phosphorus, nitrogen, and other atoms of carbon. It can do this in infinite configurations, allowing it to be the core element of the complicated structures that make living things like ourselves. Because carbon is small, kind, and loves to bond, life is pretty much built around it. Carbon is the foundation of biology. even conceiving of life that is not carbon-based. Silicon, which is analogous to carbon in many ways, is often cited as a potential element for alien life to be based on, but it's bulkier, so it doesn't form the same elegant shapes as carbon. It's also not found in any gases, meaning that life would have to be formed by eating solid silicon, whereas life here on earth is only possible because carbon is constantly floating around in the air in the form of carbon dioxide. Carbon, on its own, is an atom with 6 protons, 6 electrons, and 6 neutrons. Atoms have electron shells, and they need or want to have these shells filled, in order to be happy, fulfilled atoms. The first electron shell called the S-orbital needs 2 electrons to be full. Then there's the 2nd S-orbital, which also needs 2, carbon has this filled as well. Then we have the first P-orbital, which needs 6 to be full. Carbon only has 2 left over, so it wants 4 more. Carbon forms a lot of bonds that we call \"covalent\". These are bonds where the atoms actually share electrons," + }, + { + "Q": "\nAt 8:35, he says that since water has positive and negatively charged sides, the molecules will stick together, hydrogen side to oxygen side, and I am wondering, does that explain the hexagonal patterns that are formed at a molecular level when water freezes?", + "A": "That is exactly why water freezes the way it does.", + "video_name": "T2DaaGuKOTo", + "timestamps": [ + 515 + ], + "3min_transcript": "around the oxygen than around the hydrogens. This creates a slight positive charge around the hydrogens and a slight negative charge around the oxygen. When something has a charge, we say that it's polar. It has a positive and negative pole. This is a polar covalent bond. Ionic bonds occur when instead of sharing electrons, atoms just donate or accept an electron from another atom completely and then live happily as a charged atom or ion. Atoms would, in general, prefer to be neutral but compared with having the full electron shells is not that big of a deal. The most common ionic compound in our daily lives? that would be good old table salt, NaCl, sodium chloride, but don't be fooled by its deliciousness. Sodium chloride, as I previously mentioned, is made of 2 very nasty elements. Chlorine is a halogen, or an element that only needs one proton to fill its octet, while sodium is an alkali metal, an element that only has one electron in its octet. They will happily tear apart any chemical compound searching to satisfy the octet rule. No better outcome could occur than sodium meeting chlorine. They immediately transfer electrons so sodium doesn't have its extra and the chlorine fills its octet. They become Na+ and Cl-, and are so charged that they stick together, and that stickiness is what we call an ionic bond. These chemical changes are a big deal, remember? Sodium and chlorine just went from being deadly to being delicious. They're also hydrogen bonds, which aren't really bonds, so much. So, you remember water? I hope you didn't forget about water. Water is important. Since water is stuck together with a polar covalent bond, the hydrogen bit of it is a little bit positively-charged and the oxygen is a little negatively-charged. When water molecules move around, they actually stick together a little bit, hydrogen side to oxygen side. This kind of bonding happens in all sorts of molecules, particularly in proteins. It plays an extremely important role in how proteins fold up to do their jobs. bonds, even when they're written with dashes or solid lines, Sometimes ionic bonds are stronger than covalent bonds, though that's the exception rather than the rule, and covalent bond strength varies hugely. The way that those bonds get made and broken is intensely important to how life and our lives operate. Making and breaking bonds is the key to life itself. It's also like if you were to swallow some sodium metal, the key to death. Keep all of this in mind as you move forward in biology. Even the hottest person you have ever met is just a bunch of chemicals rambling around in a bag of water. That, among many other things, is what we're gonna talk about next time." + }, + { + "Q": "At 0:01 you started out with the static friction and the kinetic friction. What is the formula to be able to figure those amounts out? I need help for my physics class,.. Please help.\n", + "A": "He gives these equations at 1:24 and 2:01. The equation for static friction is: ||Fb|| / ||Fn|| = \u00ce\u00bcs Where ||Fb|| is the budging force, or the amount of force required to make an object start moving, ||Fn|| is the normal force, and \u00ce\u00bcs is the coefficient of static friction. The equation for kinetic friction is: ||Ff|| / ||Fn|| = \u00ce\u00bck Where ||Ff|| is the force of friction, ||Fn|| is the normal force, and \u00ce\u00bck is the coefficient of kinetic friction.", + "video_name": "ZA_D4O6l1lo", + "timestamps": [ + 1 + ], + "3min_transcript": "So I have got this block of wood here that has a mass of 5 kilograms and it is sitting on some dirt and we are near the surface of the earth and the coefficient of static friction between this type of wood and this type of dirt is 0.60 and the coefficient of kinetic friction between this type of wood and this type of dirt is 0.55 This was measured by someone else long ago or you found it in some type of a book someplace And let's say we push on this side of the block with a force of a 100 N What is going to happen? So the first thing you might realize is if there is no friction if this was a completely frictionless boundary and there is no air resistance, we are assuming that there is no air resistance in this example That in this dimension, in the horizontal dimension there would only be one force here, this 100 N force It would be completely unbalanced and that would be the net force and so you would have a force going in that direction of a 100 N on a mass of 5 kilograms Force = Mass times acceleration acceleration and force are vector quantities would give you 20 meters per second of acceleration in the rightward direction That is if there were no friction but there is friction in this situation So let's think about how we'll deal with it So the coefficient of friction tells us So this right here is the ratio between the magnitude of the force that I have called the budging force The amount of force you need to apply to get this thing to budge to get this thing to start moving. So we can start using the coefficient of kinetic friction It's the ratio between that and the magnitude of the force of contact between this block and the floor or ground here And the magnitude of that force of contact is the same thing as the normal force that the ground is applying on the block the magnitude of the normal force the ground is applying on the block Then once its moving then we can say that this is going to be--this will then be equal to this over here will be equal to the force of friction and this over here will be equal to the force of friction The magnitude of the force of friction over the force of contact the contact force between those two, so over the normal force and it makes sense that the larger the contact force the more that these are being pressed together the little at the atomic level, they kind of really get into each others grooves the more budging force you would need or the more friction force would go against your motion And in either situation the force of friction is going against your motion So even if you push it in that way sounds like force of friction is all of a sudden going to help you So let's think about what the necessary force will we need to overcome the force of friction right here in the static situation So the force of gravity on this block is going to be the gravitational field which is 9.8 m/s^2 times 5 kilograms 9.8 m/s times 5 kilograms gives 49 kilogram meters per second or 49 newtons down" + }, + { + "Q": "1:22 I don't understand why he says that coefficient of static friction deals with the force to move the object (budging force). Is it not rather maximum friction force that can be applied BEFORE it moves. Meaning any more force and then it will move (becoming kinetic friction)?\n", + "A": "The coefficient of static friction tells you how much frictional force you get for a given normal force. Friction force = coeff of sf * normal force. Friction is a reactive force, which means that it only responds to applied forces. When you multiply the coefficient of static friction times the normal force, that tells you the maximum budging force that the friction will resist. Apply more force than that, and the object will start to move.", + "video_name": "ZA_D4O6l1lo", + "timestamps": [ + 82 + ], + "3min_transcript": "So I have got this block of wood here that has a mass of 5 kilograms and it is sitting on some dirt and we are near the surface of the earth and the coefficient of static friction between this type of wood and this type of dirt is 0.60 and the coefficient of kinetic friction between this type of wood and this type of dirt is 0.55 This was measured by someone else long ago or you found it in some type of a book someplace And let's say we push on this side of the block with a force of a 100 N What is going to happen? So the first thing you might realize is if there is no friction if this was a completely frictionless boundary and there is no air resistance, we are assuming that there is no air resistance in this example That in this dimension, in the horizontal dimension there would only be one force here, this 100 N force It would be completely unbalanced and that would be the net force and so you would have a force going in that direction of a 100 N on a mass of 5 kilograms Force = Mass times acceleration acceleration and force are vector quantities would give you 20 meters per second of acceleration in the rightward direction That is if there were no friction but there is friction in this situation So let's think about how we'll deal with it So the coefficient of friction tells us So this right here is the ratio between the magnitude of the force that I have called the budging force The amount of force you need to apply to get this thing to budge to get this thing to start moving. So we can start using the coefficient of kinetic friction It's the ratio between that and the magnitude of the force of contact between this block and the floor or ground here And the magnitude of that force of contact is the same thing as the normal force that the ground is applying on the block the magnitude of the normal force the ground is applying on the block Then once its moving then we can say that this is going to be--this will then be equal to this over here will be equal to the force of friction and this over here will be equal to the force of friction The magnitude of the force of friction over the force of contact the contact force between those two, so over the normal force and it makes sense that the larger the contact force the more that these are being pressed together the little at the atomic level, they kind of really get into each others grooves the more budging force you would need or the more friction force would go against your motion And in either situation the force of friction is going against your motion So even if you push it in that way sounds like force of friction is all of a sudden going to help you So let's think about what the necessary force will we need to overcome the force of friction right here in the static situation So the force of gravity on this block is going to be the gravitational field which is 9.8 m/s^2 times 5 kilograms 9.8 m/s times 5 kilograms gives 49 kilogram meters per second or 49 newtons down" + }, + { + "Q": "\nAt 11:40 Sal says that he should have looked up the exact frequency of hemophilia in men. He estimates that is in roughly 1/7000. Does anyone know a more exact frequency?", + "A": "Hemophilia occurs in about one out of every 7,500 live male births. There are approximately 17,000 people in the United States who have hemophilia. Factor VIII deficiency accounts for about 80% (1 in every 5,000 male births) of the hemophilia population and Factor IX deficiency accounts for about 20% (1 in every 30,000 male births) of the hemophilia population", + "video_name": "-ROhfKyxgCo", + "timestamps": [ + 700 + ], + "3min_transcript": "and let's say this is her genotype. She has one regular X chromosome and then she has one X chromosome that has the-- I'll put a little superscript there for hemophilia-- she has the hemophilia mutation.She's just going to be a carrier. Her phenotype right here is going to be no hemophilia. She'll have no problem clotting her blood. The only way that a woman could be a hemophiliac is if she gets two versions of this, because this is a recessive mutation. Now this individual will have hemophilia. Now men,they only have one X chromosome. So for a man to exhibit hemophilia to have this phenotype, he just needs it only on the one X chromosome he has. So this man will have hemophilia. So a natural question should be arising is, hey, you know this guy-- let's just say that this is a relatively infrequent mutation that arises on an X chromosome-- the question is who's more likely to have hemophilia? A male or a female? All else equal,who's more likely to have it? Well if this is a relatively infrequent allele,a female, in order to display it,has to get two versions of it. So let's say that the frequency of it-- and I looked it up before this video-- roughly they say between 1 in 5,000 to 10,000 men exhibit hemophilia. So let's say that the allele frequency of this is 1 in 7,000, the frequency of Xh,the hemophilia version of the X chromosome. because it's completely determined whether-- there's a 1 in 7,000 chance that this X chromosome they get is the hemophilia version. Who cares what the Y chromosome they get is, cause that essentially doesn't code at all for the blood clotting factors and all of the things that drive hemophilia. Now,for a woman to get hemophilia,what has to happen? She has to have two X chromosomes with the mutation. Well the probability of each of them having the mutation is 1 in 7,000 So the probability of her having hemophilia is 1 in 7,000 times 1 in 7,000,or that's 1 in what,49 million. So as you can imagine,the incidence of hemophilia in women is much lower than the incidence of hemophilia in men. And in general for any sex-linked trait, if it's recessive, if it's a recessive sex-linked trait,which means men, if they have it,they're going to show it," + }, + { + "Q": "12:25 If haemophilia is recessive, why would a man who carries only one of the alleles for haemophilia (because he is XY, and so it can only ever be on one gene) have the condition present? If he has two chromosome (the X and Y) and only one can / does carry the allele for haemophilia, and then the condition is present, wouldn't that make it dominant?\n", + "A": "No, in order for an allele to be dominant, it has to show its phenotype even in the presence of another recessive allele of that gene. When only one gene is present an allele cannot be determined to be dominant or recessive. Haemophilia is recessive as in females (who are XX), there are indeed two copies of the gene and one haemophilia-causing allele is not enough for the disease to be present.", + "video_name": "-ROhfKyxgCo", + "timestamps": [ + 745 + ], + "3min_transcript": "and let's say this is her genotype. She has one regular X chromosome and then she has one X chromosome that has the-- I'll put a little superscript there for hemophilia-- she has the hemophilia mutation.She's just going to be a carrier. Her phenotype right here is going to be no hemophilia. She'll have no problem clotting her blood. The only way that a woman could be a hemophiliac is if she gets two versions of this, because this is a recessive mutation. Now this individual will have hemophilia. Now men,they only have one X chromosome. So for a man to exhibit hemophilia to have this phenotype, he just needs it only on the one X chromosome he has. So this man will have hemophilia. So a natural question should be arising is, hey, you know this guy-- let's just say that this is a relatively infrequent mutation that arises on an X chromosome-- the question is who's more likely to have hemophilia? A male or a female? All else equal,who's more likely to have it? Well if this is a relatively infrequent allele,a female, in order to display it,has to get two versions of it. So let's say that the frequency of it-- and I looked it up before this video-- roughly they say between 1 in 5,000 to 10,000 men exhibit hemophilia. So let's say that the allele frequency of this is 1 in 7,000, the frequency of Xh,the hemophilia version of the X chromosome. because it's completely determined whether-- there's a 1 in 7,000 chance that this X chromosome they get is the hemophilia version. Who cares what the Y chromosome they get is, cause that essentially doesn't code at all for the blood clotting factors and all of the things that drive hemophilia. Now,for a woman to get hemophilia,what has to happen? She has to have two X chromosomes with the mutation. Well the probability of each of them having the mutation is 1 in 7,000 So the probability of her having hemophilia is 1 in 7,000 times 1 in 7,000,or that's 1 in what,49 million. So as you can imagine,the incidence of hemophilia in women is much lower than the incidence of hemophilia in men. And in general for any sex-linked trait, if it's recessive, if it's a recessive sex-linked trait,which means men, if they have it,they're going to show it," + }, + { + "Q": "\nIn the video ,10:28 you said a man only one hemophelia cromosome to past it to the off spring. so does that mean that most cancer comes from men?", + "A": "Cancer is a mutation of genes, not genetically inherited from parents (unless the cancer was present in gametes).", + "video_name": "-ROhfKyxgCo", + "timestamps": [ + 628 + ], + "3min_transcript": "But what it tells you is it does very little other than determining what the gender is. And the way it determines that, it does have one gene on it called the SRY gene. You don't have to know that. SRY,that plays a role in the development of testes or the male sexual organ.So if you have this around, this gene right here can start coding for things that will eventually lead to the development of the testicles. And if you don't have that around,that won't happen, so you'll end up with a female. And I'm making gross oversimplifications here. But everything I've dealt with so far,OK, this clearly plays a role in determining sex. But you do have other traits on these genes. And the famous cases all deal with specific disorders. The genes,or the mutations I should say. So the mutations that cause color blindness. Red-green color blindness,which I did in green, which is maybe a little bit inappropriate. Color blindness and also hemophilia. This is an inability of your blood to clot. Actually, there's several types of hemophilia. But hemophilia is an inability for your blood to clot properly. And both of these are mutations on the X chromosome. And they're recessive mutations.So what does that mean? It means both of your X chromosomes have to have-- let's take the case for hemophilia-- both of your X chromosomes have to have the hemophilia mutation in order for you to show the phenotype of having hemophilia. and let's say this is her genotype. She has one regular X chromosome and then she has one X chromosome that has the-- I'll put a little superscript there for hemophilia-- she has the hemophilia mutation.She's just going to be a carrier. Her phenotype right here is going to be no hemophilia. She'll have no problem clotting her blood. The only way that a woman could be a hemophiliac is if she gets two versions of this, because this is a recessive mutation. Now this individual will have hemophilia. Now men,they only have one X chromosome. So for a man to exhibit hemophilia to have this phenotype, he just needs it only on the one X chromosome he has." + }, + { + "Q": "is the X chromosome always longer compared to the Y?\njust curious... 07:43\n", + "A": "Yes it is a huge giant by comparison.", + "video_name": "-ROhfKyxgCo", + "timestamps": [ + 463 + ], + "3min_transcript": "I mean it's not just the case with kings. It's probably true,because most of our civilization is male dominated, that you've had these men who are obsessed with producing a male heir to kind of take over the family name. And,in the case of Henry the VIII,take over a country. And they become very disappointed and they tend to blame their wives when the wives keep producing females,but it's all their fault. Henry the VIII,I mean the most famous case was with Ann Boleyn. I'm not an expert here,but the general notion is that he became upset with her that she wasn't producing a male heir. And then he found a reason to get her essentially decapitated, even though it was all his fault. He was maybe producing a lot more sperm He eventually does produce a male heir so he was-- and if we assume that it was his child-- then obviously he was producing some of these,but for the most part, it was all Henry the VIII's fault. So that's why I say there's a little bit of irony here. Is that the people doing the blame are the people to blame for the lack of a male heir? Now one question that might immediately pop up in your head is, Sal,is everything on these chromosomes related to just our sex-determining traits or are there other stuff on them? So let me draw some chromosomes. So let's say that's an X chromosome and this is a Y chromosome. Now the X chromosome,it does code for a lot more things, although it is kind of famously gene poor. It codes for on the order of 1,500 genes. And the Y chromosome,it's the most gene poor of all the chromosomes. It only codes for on the order of 78 genes. But what it tells you is it does very little other than determining what the gender is. And the way it determines that, it does have one gene on it called the SRY gene. You don't have to know that. SRY,that plays a role in the development of testes or the male sexual organ.So if you have this around, this gene right here can start coding for things that will eventually lead to the development of the testicles. And if you don't have that around,that won't happen, so you'll end up with a female. And I'm making gross oversimplifications here. But everything I've dealt with so far,OK, this clearly plays a role in determining sex. But you do have other traits on these genes. And the famous cases all deal with specific disorders." + }, + { + "Q": "At around 5:48 what would the hybridization of the transition state be?\n", + "A": "The hybridization of the carbon atom in the transition state would be sp\u00c2\u00b2.", + "video_name": "3LiyCxCTrqo", + "timestamps": [ + 348 + ], + "3min_transcript": "We can see in our final products here, the nucleophile has substituted for the leaving group. The N stands for nucleophilic because of course it is our nucleophile that is doing the substituting. And finally the two here refers to the fact that this is bimolecular which means that the rate depends on the concentration of two things. The substrate and the nucleophile. That's different from an SN1 mechanism where the rate is dependent only on the concentration of one thing. The rate of the reaction also depends on the structure of the alkyl halide, on the structure of the substrate. On the left we have a methyl halide followed by a primary alkyl halide. The carbon bonded to our bromine is directly attached to one alkyl group followed by a secondary alkyl halide, the carbon bonded to the bromine is bonded to two alkyl groups, followed by a tertiary alkyl halide. Turns out that the methyl halides and the primary alkyl halide react the fastest in an SN2 mechanism. Secondary alkyl halides react very slowly and tertiary alkyl halides react so, so slowly that we say they are unreactive toward an SN2 mechanism. And this makes sense when we think about the mechanism because remember, the nucleophile has to attack the electrophile. The nucleophile needs to get close enough to the electrophilic carbon to actually form a bond and steric hindrance would prevent that from happening. Something like a tertiary alkyl halide has this big bulky methyl groups which prevent the nucleophile for attacking. Let's look at a video so we can see this a little bit more clearly. Here's our methyl halide with our carbon directly bonded to a halogen which I'm seeing as yellow. And here's our nucleophile which could be the hydroxide ion. for the side opposite of the leaving group and you can see with the methyl halide there's no steric hindrance. When we move to a primary alkyl halide, the carbon bonded to the halogen has only one alkyl group bonded to it, it's still easy for the nucleophile to approach. When we move to a secondary alkyl halide, so for a secondary you can see that the carbon bonded to the halogen has two methyl groups attached to it now. It gets a little harder for the nucleophile to approach in the proper orientation. These bulky methyl groups make it more difficult for the nucleophile to get close enough to that electrophilic carbon. When we go to a tertiary alkyl halide, so three alkyl groups. There's one, there's two and there's three. There's a lot more steric hindrance and it's even more difficult for our nucleophile to approach. As we saw on the video, for an SN2 reaction we need decreased steric hindrance. So, if we look at this alkyl halide," + }, + { + "Q": "8:40 what represnt those arrows?\n", + "A": "The arrows represent the electrons. You can have up to two electrons in an orbital but, if there are two electrons, they must have opposite spin. (Spin is defined by the fourth quantum number that you will have seen in earlier videos.) Opposite spin is represented by drawing one arrow pointing up and the other arrow pointing down.", + "video_name": "649ZlWMp0LE", + "timestamps": [ + 520 + ], + "3min_transcript": "of the two p electron from the full attraction of the nucleus, right? So, even though we have five protons in the nucleus, and a positive five charge for boron, the fact that these two s electrons add a little bit of extra shielding means it's easier to pull this electron away. So, it turns out to be a little bit easier to pull this electron in the two p orbital away due to these two s electrons. And that's the reason for this slight decrease in ionization energy. As we go from boron to carbon, we see an increase in ionization energy, from carbon to nitrogen, an increase in ionization energy. Again, we attribute that to increased effective nuclear charge, but when we go from nitrogen to oxygen, we see a slight decrease again. From about 1400 kilojoules per mole, down to about 1300 kilojoules per mole for oxygen. So, let's see if we can explain that by writing out Nitrogen has seven electrons to think about. So it's electron configuration is one s two, two s two, and two p three. So that takes care of all seven electrons. For oxygen, we have another electron, so one s two, two s two, two p four is the electron configuration for oxygen. Let's just draw using orbital notation the two s orbital and the two p orbital. So for nitrogen, here's our two s orbital. We have two electrons in there, so let's draw in our two electrons. And for our two p orbitals, we have three electrons. So here are the two p orbitals, and let's draw in our three electrons using orbital notation. Let's do the same thing for oxygen. So there's the two s orbital for oxygen, which is full, so we'll sketch in those two electrons, and we have four electrons in the two p orbitals. There's one electron, there's two, there's three, and notice what happens when we add the fourth electron. We're adding it to an orbital that already has an electron in it, so when I add that fourth electron to the two p orbital, it's repelled by the electron that's already there, which means it's easier to remove one of those electrons, so electrons have like charges, and like charges repel. And so that's the reason for this slight decrease in ionization energy. So, it turns out to be a little bit easier to remove an electron from an oxygen atom, than nitrogen, due to this repulsion in this two p orbital. From there on, we see our general trend again. The ionization energy for fluorine is up to 1681, and then again for neon, we see an increase in the ionization energy due to the increased effective nuclear charge." + }, + { + "Q": "1.At 5:24, Sal said isostatic process. What does it mean?\n2. Why did he use natural log instead of common log? Either way is fine?\nThanks!\n", + "A": "1. By isostatic, he means any one of the isobaric, isochoric, adiabatic, isothermal, etc (if any) processes. 2. Since integral of 1/x = ln x, he used natural log so that the answers match.", + "video_name": "WLKEVfLFau4", + "timestamps": [ + 324 + ], + "3min_transcript": "Or maybe if I had a smaller container, I would also have fewer potential states. There would be fewer potential places for our little particles to exist. So I want to create some type of state variable that tells me, well, how many states can my system be in? So this is kind of a macrostate variable. It tells me, how many states can my system be in? And let's call it s for states. For the first time in thermodynamics, we're actually using a letter that in some way is related to what we're actually trying to measure. s for states. And since the states, they can grow really large, let's say I like to take the logarithm of the number of states. Now this is just how I'm defining my state variable. I get to define it. So I get to put a logarithm out front. So let me just put a logarithm. So in this case, it would be the logarithm of my number of states-- so it would be x to the n, where this is number of And you know, we need some kind of scaling factor. Maybe I'll change the units eventually. So let me put a little constant out front. Every good formula needs a constant to get our units right. I'll make that a lowercase k. So that's my definition. I call this my state variable. If you give me a system, I should, in theory, be able to tell you how many states the system can take on. So let me close that box right there. Now let's say that I were to take my box that I had-- let me copy and paste it. I take that box. And it just so happens that there was an adjacent box next to it. They share this wall. They're identical in size, although what I just drew But they're close enough. They're identical in size. And what I do, is I blow away this wall. I just evaporate it, all of a sudden. It just disappears. So this wall just disappears. Now, what's going to happen? Well, as soon as I blow away this wall, this is very much not an isostatic process. All hell's going to break loose. I'm going to blow away this wall, and you know, the particles that were about to bounce off the wall are just going to keep going. They're going to keep going until they can maybe bounce off of that wall. So right when I blow away this wall, there's no pressure here, because these guys have nothing to bounce off to. While these guys don't know anything. They don't know anything until they come over here and say, oh, no wall. So the pressure is in flux. Even the volume is in flux, as these guys make their way across the entire expanse of the new volume. So everything is in flux. Right? And so what's our new volume? If we call this volume, what's this?" + }, + { + "Q": "\nat around 10:06 you say that the temperature doesnt change because everything is in isolation thus adiabatic. In previous videos, with the cilinder/pebbles, when everything was adiabatic, the temperature DID change. The only time the temperature didnt change back then was when there was a reservoir added (Isotherm).\n\nSo why is it different here? Just because the system doesnt have to do any work?", + "A": "I guess i found my answer at ~6:55, sorry", + "video_name": "WLKEVfLFau4", + "timestamps": [ + 606 + ], + "3min_transcript": "Now, each particle could be in 2x different states. Why do I say 2x? Because I have twice the area to be in. Now, the states aren't just, you know, position in space. But everything else-- so, you know, before here, maybe I had a positions in space times b positions, or b momentums, you know, where those are all the different momentums, and that was equal to x. Now I have 2a positions in volume that I could be in. I have twice the volume to deal with. So I have 2a positions in volume I can be at, but my momentum states are going to still be-- I just have b momentum states-- so this is equal to 2x. I now can be in 2x different states, just because I have 2 times the volume to travel around in, right? Well, each particle can be in 2x states. So this is 2x times 2x times 2x. And I'm going to do that n times. So my new s-- so this is, you know, let's call this s initial-- so my s final, my new way of measuring my states, is going to be equal to that little constant that I threw in there, times the natural log of the new number of states. It's 2x to the n power. So my question to you is, what is my change in s when I blew away the wall? You know, there was this room here the entire time, although these particles really didn't care because this wall was there. So what is the change in s when I blew away this wall? The temperature didn't change, because no kinetic energy was expended. I should have said. It's adiabatic. There's no transfer of heat. So that's also why the temperature didn't change. So what is our change in s? Our change in s is equal to our s final minus our s initial, which is equal to-- what's our s final? It's this expression, right here. It is k times the natural log--and we can write this as 2 to the n, x to the n. That's just exponent rules. And from that, we're going to subtract out our initial s value, which was this. k natural log of x to the n. Now we can use our logarithm properties to say, well, you know, you take the logarithm of a minus the logarithm of b, you can just divide them. So this is equal to k-- you could factor that out-- times the logarithm of 2 to the N-- it's uppercase N, This is uppercase N. I don't want to get confused with Moles." + }, + { + "Q": "@02:13,\nshouldn't it be x(x-1)(x-2)(x-3).....\nbecause a already ocupied 1 state so b has only x-1 states to go to?\n", + "A": "But the states are changing as the molecules have velocity. If they wouldn t have had a velocity then your proposition would have been correct", + "video_name": "WLKEVfLFau4", + "timestamps": [ + 133 + ], + "3min_transcript": "If you followed some of the mathematics, and some of the thermodynamic principles in the last several videos, what occurs in this video might just blow your mind. So not to set expectations too high, let's just start off with it. So let's say I have a container. And in that container, I have gas particles. Inside of that container, they're bouncing around like gas particles tend to do, creating some pressure on the container of a certain volume. And let's say I have n particles. Now, each of these particles could be in x different states. Let me write that down. What do I mean by state? Well, let's say I take particle A. Let me make particle A a different color. have some velocity like that. It could also be in that corner and have a Those would be two different states. It could be up here, and have a velocity like that. It could be there and have a velocity like that. If you were to add up all the different states, and there would be a gazillion, of them, you would get x. That blue particle could have x different states. You don't know. We're just saying, look. I have this container. It's got n particles. So we just know that each of them could be in x different states. Now, if each of them can be in x different states, how many total configurations are there for the system as a whole? Well, particle A could be in x different states, and then particle B could be in x different places. So times x. If we just had two particles, then you would multiply all the different places where X could be times all the different places where the red particle could be, then you'd get all the different configurations for the system. But we don't have just two particles. We have n particles. So for every particle, you'd multiply it times the number a total of n times. And this is really just combinatorics here. You do it n times. This system would have n configurations. For example, if I had two particles, each particle had three different potential states, how many different configurations could there be? Well, for every three that one particle could have, the other one could have three different states, so you'd have nine different states. If you had another particle with three different states, you'd multiply that by three, so you have 27 different states. Here we have n particles. Each of them could be in x different states. So the total number of configurations we have for our system-- x times itself n times is just x to the n. So we have x to the n states in our system. Now, let's say that we like thinking about how many states a system can have. Certain states have less-- for example, if I had fewer particles, I would have fewer" + }, + { + "Q": "\nThe 1,3-Diflurocyclopentane.....at 8:58; wouldnt it have an axis of symmetry straight down the center as if folding a sheet of paper hot dog style?", + "A": "Yes. Sal was just using the same axis of symmetry to compare it to the first example. You would, in fact, have an axis of symmetry if you chose to do it that way; however, it still would not be chiral, because if you started counting from either fluorine there would still be an unequal number of carbon/fluorine groups on each side.", + "video_name": "QQMZ1ljepWg", + "timestamps": [ + 538 + ], + "3min_transcript": "If you were to go clockwise, you'd have CH2, then a CH, which happens to be connected to a fluorine. So you're actually going to see something different, depending whether you're going down into that group or into And then it's also bonded to a hydrogen and a fluorine, four different groups. This is also a chiral center. Another way to think about it, and it's actually interesting to compare it to this molecule up here, which was not chiral and did not have a chiral center, this molecule up here-- let me draw it a little different to make it a little bit more clear. So this one, I could draw it like this. If you have the chlorine like that, over here, we thought about this as a potential chiral center, and it's kind of playing the same role as in that example down here, but you see over here, this is not a chiral center because that goes through that carbon. So you can actually just draw an axis of symmetry that goes exactly through that carbon. The way I drew it, it's not completely neat, but you can see that that is the reflection of that, if I were to draw the bonds actually a little bit more symmetric. Over here, if we try to do the exact same thing, if we try to draw an axis of symmetry over here, if we try to draw an axis of symmetry, we can make that bond to the fluorine go through our axis of symmetry,, we'll see that that still is not the reflection of this because we have a fluorine up here. We don't have a fluorine over here. Now, we can do the same thing with this end. If you try to do an axis of symmetry, fluorine up there, no fluorine over here. So each of these are definitely chiral centers, while this carbon up here was not a chiral center. Now, the next question is, well, this thing's got two chiral centers, two chiral carbons. It's probably a chiral molecule. center, you had a chiral molecule. But let's take its mirror image. To take its mirror image, let me clear out some real estate over here, So let me clear out this. Let me clear it out. So what's the mirror image going to look like? Let me draw first the mirror. So the mirror image, you're going to have a fluorine over there. Then you're going to bond to a carbon, which is also bonded to a hydrogen. That's going to bond to a CH2. That's going to bond to a CH. That's the mirror image of that, which bonds to a fluorine. That's the mirror image of that. And then you go down. This is the mirror image of CH2 here. This is the mirror image of this. You connect them. Now, these are mirror images of each other. But they are also the exact same molecule. I could just literally move this guy over to the right, and it would be superimposed. They are exactly the same. So even though we have two chiral atoms, two chiral" + }, + { + "Q": "\nhow can a structure have a chiral atom and be not chiral? does it matter? 9:33.", + "A": "Well a Meso compound would have a chiral atom but not actually be chiral, wouldn t it?", + "video_name": "QQMZ1ljepWg", + "timestamps": [ + 573 + ], + "3min_transcript": "that goes through that carbon. So you can actually just draw an axis of symmetry that goes exactly through that carbon. The way I drew it, it's not completely neat, but you can see that that is the reflection of that, if I were to draw the bonds actually a little bit more symmetric. Over here, if we try to do the exact same thing, if we try to draw an axis of symmetry over here, if we try to draw an axis of symmetry, we can make that bond to the fluorine go through our axis of symmetry,, we'll see that that still is not the reflection of this because we have a fluorine up here. We don't have a fluorine over here. Now, we can do the same thing with this end. If you try to do an axis of symmetry, fluorine up there, no fluorine over here. So each of these are definitely chiral centers, while this carbon up here was not a chiral center. Now, the next question is, well, this thing's got two chiral centers, two chiral carbons. It's probably a chiral molecule. center, you had a chiral molecule. But let's take its mirror image. To take its mirror image, let me clear out some real estate over here, So let me clear out this. Let me clear it out. So what's the mirror image going to look like? Let me draw first the mirror. So the mirror image, you're going to have a fluorine over there. Then you're going to bond to a carbon, which is also bonded to a hydrogen. That's going to bond to a CH2. That's going to bond to a CH. That's the mirror image of that, which bonds to a fluorine. That's the mirror image of that. And then you go down. This is the mirror image of CH2 here. This is the mirror image of this. You connect them. Now, these are mirror images of each other. But they are also the exact same molecule. I could just literally move this guy over to the right, and it would be superimposed. They are exactly the same. So even though we have two chiral atoms, two chiral It is a non-chiral molecule." + }, + { + "Q": "At around 5:30, the video is talking about chromate being CrO4^2-, but when it is dichromate it is Cr2O7^2-. Why are there 7 oxygen atoms? I thought there would be 8, as 4*2 is 8. Sorry for the notation.\n", + "A": "The di means that there are two chromium atoms in the ion, not that it consists of two chromate ions. The missing O atom ends up as water. For example: 2K\u00e2\u0082\u0082CrO\u00e2\u0082\u0084 + H\u00e2\u0082\u0082SO\u00e2\u0082\u0084 \u00e2\u0086\u0092 K\u00e2\u0082\u0082Cr\u00e2\u0082\u0082O\u00e2\u0082\u0087 + H\u00e2\u0082\u0082O + K\u00e2\u0082\u0082SO\u00e2\u0082\u0084", + "video_name": "DpnUrVXSLaQ", + "timestamps": [ + 330 + ], + "3min_transcript": "And then we have four Oxygens, so if we go to three Oxygens, SO three two minus, this is Sulfite, cause ite means fewer Oxygens. What about if we took Sulfate, SO four two minus, and we added on an H plus. So H plus and SO four two minus should give us HSO four and then, instead of a negative two here, instead of a two minus, we would just have a one minus, because we added on a positive charge. So one positive charge and two negative charges, give us one negative charge. So HSO four minus is called the Hydrogen Sulfate ion. You might also hear Bisulfate for this one. Next CO three two minus is called Carbonate, so if we add on an H plus to CO three two minus, we'd get HCO three and then we go from minus two or two minus, to minus one, 'cause we're adding on a positive charge here. and you'll also hear Bicarbonate a lot. Next we have PO four three minus, which is called Phosphate. If we add on an H plus to Phosphate, think about what we would get. We would get HPO four and then instead of three minus, we're adding on positive charge, so we get two minus. So we call this Hydrogen Phosphate. Alright, let's add on a proton to Hydrogen Phosphate. So we're adding an H plus onto Hydrogen Phosphate. That would give us two H's. PO four and we'd go from two minus down to one minus. So H two PO four minus is called Dihydrogen Phosphate. Alright, let's continue on. One more set of polyatomic ions to know. So we have CrO four two minus, which is called Chromate. And if we have two Chromiums, so Cr two O seven two minus Next, C two O four two minus is called the Oxalate ion. and we have O two, two minus is called Peroxide. And here we have SCN minus, which we call Thiocyanate. So thio, think about sulfur if you see thio there. So for our next one, we have sulfur present again, is S two O three two minus and this one's called Thiosulfate. So you might see a few additional polyatomic ions in your class, but these are the ones that you see most frequently. So make sure to memorize your polyatomic ions." + }, + { + "Q": "At around 4:45, he did not explain where the second hydrogen atom came from to make hydrogen phosphate become dihydrogen phosphate. Can someone explain that to me?\n", + "A": "It doesn t really matter where it came from, he s just showing you how those phosphate molecules are related to one another. It probably came from water if you have to have an answer to understand.", + "video_name": "DpnUrVXSLaQ", + "timestamps": [ + 285 + ], + "3min_transcript": "And then we have four Oxygens, so if we go to three Oxygens, SO three two minus, this is Sulfite, cause ite means fewer Oxygens. What about if we took Sulfate, SO four two minus, and we added on an H plus. So H plus and SO four two minus should give us HSO four and then, instead of a negative two here, instead of a two minus, we would just have a one minus, because we added on a positive charge. So one positive charge and two negative charges, give us one negative charge. So HSO four minus is called the Hydrogen Sulfate ion. You might also hear Bisulfate for this one. Next CO three two minus is called Carbonate, so if we add on an H plus to CO three two minus, we'd get HCO three and then we go from minus two or two minus, to minus one, 'cause we're adding on a positive charge here. and you'll also hear Bicarbonate a lot. Next we have PO four three minus, which is called Phosphate. If we add on an H plus to Phosphate, think about what we would get. We would get HPO four and then instead of three minus, we're adding on positive charge, so we get two minus. So we call this Hydrogen Phosphate. Alright, let's add on a proton to Hydrogen Phosphate. So we're adding an H plus onto Hydrogen Phosphate. That would give us two H's. PO four and we'd go from two minus down to one minus. So H two PO four minus is called Dihydrogen Phosphate. Alright, let's continue on. One more set of polyatomic ions to know. So we have CrO four two minus, which is called Chromate. And if we have two Chromiums, so Cr two O seven two minus Next, C two O four two minus is called the Oxalate ion. and we have O two, two minus is called Peroxide. And here we have SCN minus, which we call Thiocyanate. So thio, think about sulfur if you see thio there. So for our next one, we have sulfur present again, is S two O three two minus and this one's called Thiosulfate. So you might see a few additional polyatomic ions in your class, but these are the ones that you see most frequently. So make sure to memorize your polyatomic ions." + }, + { + "Q": "\nBasic question... Why at 3:45, the H is coming out, instead of stay in the sheet planar?", + "A": "We have to consider what the molecule looks like in three dimensions, in reality the molecules are not flat like the paper we draw them on. There really is no good way to prove this to yourself other than playing around with a molecular modelling kit. They are very handy for learning about stereoisomers. In the example in this video there is a methyl group going away from us, and we can only see 3 bonds to that specific carbon, so there is an implied hydrogen atom coming out of the page towards us.", + "video_name": "9IYTqlVk_ZI", + "timestamps": [ + 225 + ], + "3min_transcript": "The carbon on the left is directly bonded to three hydrogens, so we write down here three hydrogens and the carbon on the right is directly bonded to two carbons and one hydrogen, so we write carbon, carbon, hydrogen. Next, we look for the first point of difference and that's the first atom here, so carbon versus hydrogen. Carbon has the higher atomic number, so this group wins, so this group on the right is higher priority which means this must be number two for our groups, and the methyl group gets a number three. Now that we've assigned priority to our four groups, the next step was to orient the molecules so that the lowest priority group is projecting away from us, and that's what we have here because the hydrogen is the lowest priority group so we can ignore this and we can focus in on one, two, and three, so let me label this on the right, the OH was the highest highest priority and the methyl group was the third highest priority. Next we draw a circle and determine whether we're going around clockwise or counterclockwise, so we draw a circle from one to two to three, and obviously, we're going around in a clockwise direction and clockwise is R, so we are R at carbon two, so I write here 2R. Now, let's focus on carbon three, so we know that carbon three is a chiral center, so what is directly bonded to this carbon? Well, there's a carbon attached to three hydrogens, so there's a methyl group, there's a hydrogen that must be coming out at us in space and on the right, there's a carbon bonded to two hydrogens and directly bonded to another carbon. On the left, there's a carbon directly bonded to an oxygen. There must be a hydrogen going away from us, and this carbon is directly bonded to another carbon, so here is our chiral center. We look at the atoms directly bonded to that carbon. There are three carbons, one, two, three, and a hydrogen. We know the hydrogen has the lowest atomic number, so hydrogen is the lowest priority and we call that group four. Now we have three carbons. We have a tie because carbon, of course, has the same atomic number. To break a tie, we look at the atoms that are bonded to those carbons, so let's start with the carbon on the right here, so the carbon on the right is directly bonded to one carbon and then two hydrogens, so we write carbon, hydrogen, hydrogen, so in decreasing atomic number, in order of decreasing atomic number. This carbon down here is directly bonded to three hydrogens so one, two, and three. The carbon on the left is directly bonded to an oxygen, a carbon and a hydrogen, so we put these in order of decreasing atomic number, so we put oxygen first, then carbon, then hydrogen." + }, + { + "Q": "Starting around 7:30ish, Sal is using Hess's Law. Correct?\n", + "A": "Sorry, I think you aren t right Hess s law states that the overall enthalpy change in a reaction is the sum of all the reactions for the process and is independent to the route taken this might look like Hess law at the first glance...", + "video_name": "fYUwEAPejbY", + "timestamps": [ + 450 + ], + "3min_transcript": "That's what makes the electrolyte paste alkaline. And you are going to then go to, they're going to react and you're going to form zinc oxide. Zinc oxide in the solid state, plus water in the liquid state, plus two, plus two electrons. And so this reaction that I just described, this is going to be happening right over here. So you could think of it as the hydroxide anions get formed at the cathode, and then they move their way over to the left to the anode where they react with the zinc and they turn, or the zinc reacts and forms zinc oxide, so you have more and more zinc oxide being formed, water, which can then seep its way back into the electrolyte paste and then it can eventually react again at the cathode, and then you have these two electrons. So these electrons, this turns into an electron source right over here, and then the electrons would migrate to react again. And so you can start to see how this will be an energy source that you can tap into this current that'll form to do some useful work. That's why you have an energy source. All right, so let's read that questions now that we have a decent understanding of what's going on. Early forms of metal air cells use zinc as the anode. Well, that's the example we just thought about. Zinc oxide is produced as the cell operates according to the overall equation below. For every two molecules of zinc and one molecule of molecular oxygen, you produce two molecules of zinc oxide. Use the data in the table above, calculate the cell potential for the zinc air cell. So let's think about, let's break down this reaction. Actually, we can just break it down into these two steps here. Because notice, this is, we can't see them both at the same time but this, the things that are reacting, Let me underlie the things that are reacting. You have zinc, you have zinc. You have oxygen, you have oxygen. And so if you return this, if you return the second reaction around like we did before, and let me rewrite it, let me rewrite both of them actually a little bit lower right over here. So you have this top reaction. So you have O2 gaseous state plus two H2O liquid, plus four electrons. Whoops, four electrons, yields four hydroxide anions in aqueous solution. And then this one, we said we're gonna go in the other direction. So you're going to have zinc, you're going to have zinc in the solid state, plus two hydroxide anions. And then that's going to yield zinc," + }, + { + "Q": "at \"8:52\", if the crab nebula is 65,00 light years away, couldn't a star or more than one star formed in the 65,00 years it took for that image to reach earth?\n", + "A": "It takes more than 6500 (that was the number in the video) years to form a star. It actually takes more than 65,000 years to form a star.", + "video_name": "qOwCpnQsDLM", + "timestamps": [ + 532 + ], + "3min_transcript": "it'll be something about maybe two times the mass of the sun, give or take one and a half to three times the mass of the sun. So this is one and a half to three times the mass of the sun in a volume that has a diameter of about 10-- on the order of tens of kilometers. So it's roughly the size of a city, in a diameter of a city. So this is unbelievably dense, diameter of a city. I mean, we know how much larger the sun is relative to the Earth. And we know how much larger the Earth is relative to a city. But this is something large-- more mass than the sun being squeezed into the density, or into the size of a city, so unbelievably dense. Now if the original star is even more massive, if it's more than 20 times the sun-- so let me write it over here. Let me scroll up. If it's greater than 20 times the sun, even the neutrons' inability to squeeze further will give up. And it'll turn into a black hole. And that's-- and I could do many videos on that. And that's actually an open area of research, still, on exactly what's going on inside of a black hole. But then you turn into a black hole, where essentially all of the mass gets condensed into an infinitely small and dense point, so something unbelievably hard to imagine. And just to give you a sense of it, so this will be more mass then even three times the mass So we're talking about an incredibly high amount of mass. Just to kind of visualize things, here is actually a remnant of a supernova. This is the Crab Nebula. This is, right here, is the Crab Nebula. And it's about 6,500 light years away. So it's still, from a galactic scale-- if you think of our galaxy as being 100,000 light years But it's an enormous distance. The closest star to us is four light years away. And it would take Voyager travelling at 60,000 kilometers an hour, 80,000 years to get there. So this is a very, very-- that's only four light years. Now this is 6,500 light years. But this supernova, it's believed happened 1,000 years ago, right at the center. And so at the center here, we should have a neutron star. And this cloud, the shock wave that you see here, this is still the material traveling outward from that supernova over 1,000 years. This shock wave, or the diameter of this sphere of material, is six light years. So we could say this distance right here is six light years. So this is an enormously big shock wave cloud. And actually, we believe that our solar system started to form, started to condense because of a shock wave created" + }, + { + "Q": "\nWhat does exothermic mean? At around 00:50", + "A": "Exothermic reactions are one of the two types of chemical reactions. An exothermic reaction releases energy, while endothermic reactions absorb energy.", + "video_name": "qOwCpnQsDLM", + "timestamps": [ + 50 + ], + "3min_transcript": "Where we left off in the last video, we had a mature massive star, a star that had started forming a core of iron. It has enormous pressure, enormous inward pressure on this core. Because as we form heavier and heavier elements in the core, the core gets denser and denser and denser. And so we keep fusing more and more elements into iron. This iron core becomes more and more massive, more and more dense. It's squeezing in on itself. And it's not fusing. That is not exothermic anymore. If iron were to fuse, it would not even be an exothermic process. It would require energy. So it wouldn't be even something that could be helped to fend off this squeezing, to fend off this increasing density of the core. So we have this iron here, and it just gets more and more massive, more and more dense. And so at some mass, already a reasonably high mass, the only thing that's keeping this from just completely So let me write this here, electron degeneracy pressure. And all this means is we have all of these iron atoms getting really, really, really close to each other. And the only thing that keeps it from collapsing at this earlier stage, the only thing that keeps it from collapsing altogether, is that they have these electrons. You have these electrons, and these are being squeezed together, now. I mean, we're talking about unbelievably dense states of matter. And electron degeneracy pressure is, essentially-- it's saying these electrons don't want to be in the same place at the same time. I won't go into the quantum mechanics of it. But they cannot be squeezed into each other any more. So that, at least temporarily, holds this thing from collapsing even further. in the case of a white dwarf, that's how a white dwarf actually maintains its shape, because of the electron degeneracy pressure. But as this iron core gets even more massive, more dense, and we get more and more gravitational pressure-- so this is our core, now-- even more gravitational pressure, eventually even this electron degeneracy-- I guess we could call it force, or pressure, this outward pressure, this thing that keeps it from collapsing-- even that gives in. And then we have something called electron capture, which is essentially the electrons get captured by protons in the nucleus. They start collapsing into the nucleuses. It's kind of the opposite of beta negative decay, where you have the electrons get captured, protons get turned into neutrons." + }, + { + "Q": "\nAt 2:51 he talks about \"Beta negative decay\". What is it?", + "A": "It s the conversion of a neutron into a proton in the nucleus of an atom. When that happens, an electron is emitted. That electron is called a beta particle.", + "video_name": "qOwCpnQsDLM", + "timestamps": [ + 171 + ], + "3min_transcript": "So let me write this here, electron degeneracy pressure. And all this means is we have all of these iron atoms getting really, really, really close to each other. And the only thing that keeps it from collapsing at this earlier stage, the only thing that keeps it from collapsing altogether, is that they have these electrons. You have these electrons, and these are being squeezed together, now. I mean, we're talking about unbelievably dense states of matter. And electron degeneracy pressure is, essentially-- it's saying these electrons don't want to be in the same place at the same time. I won't go into the quantum mechanics of it. But they cannot be squeezed into each other any more. So that, at least temporarily, holds this thing from collapsing even further. in the case of a white dwarf, that's how a white dwarf actually maintains its shape, because of the electron degeneracy pressure. But as this iron core gets even more massive, more dense, and we get more and more gravitational pressure-- so this is our core, now-- even more gravitational pressure, eventually even this electron degeneracy-- I guess we could call it force, or pressure, this outward pressure, this thing that keeps it from collapsing-- even that gives in. And then we have something called electron capture, which is essentially the electrons get captured by protons in the nucleus. They start collapsing into the nucleuses. It's kind of the opposite of beta negative decay, where you have the electrons get captured, protons get turned into neutrons. But you can imagine an enormous amount of energy is also being released. So this is kind of a temporary-- and then all of a sudden, this collapses. This collapses even more until all you have-- and all the protons are turning into neutrons. Because they're capturing electrons. So what you eventually have is this entire core is collapsing into a dense ball of neutrons. You can kind of view them as just one really, really, really, really, really massive atom because it's just a dense ball of neutrons. At the same time, when this collapse happens, you have an enormous amount of energy being released in the form of neutrinos. Did I say that neutrons are being released? No, no, no, the electrons are being captured by the protons, protons turning into neutrons-- this dense ball of neutrons right here-- and in the process, neutrinos get released, these fundamental particles." + }, + { + "Q": "\nHey Sal, at 9:56 you mention that the current model for the formation of our solar system is that it was created by a supernova. That should have left behind a neutron star. Do we have a good candidate for the particular neutron star that is the remnant of the supernova that created our solar system?", + "A": "The objects in our galaxy orbit around the center at different speed. After 4.5 billion years, it could be anywhere. Currently, any neutron star could be from the supernova that stimulated the nebula collapse.", + "video_name": "qOwCpnQsDLM", + "timestamps": [ + 596 + ], + "3min_transcript": "even the neutrons' inability to squeeze further will give up. And it'll turn into a black hole. And that's-- and I could do many videos on that. And that's actually an open area of research, still, on exactly what's going on inside of a black hole. But then you turn into a black hole, where essentially all of the mass gets condensed into an infinitely small and dense point, so something unbelievably hard to imagine. And just to give you a sense of it, so this will be more mass then even three times the mass So we're talking about an incredibly high amount of mass. Just to kind of visualize things, here is actually a remnant of a supernova. This is the Crab Nebula. This is, right here, is the Crab Nebula. And it's about 6,500 light years away. So it's still, from a galactic scale-- if you think of our galaxy as being 100,000 light years But it's an enormous distance. The closest star to us is four light years away. And it would take Voyager travelling at 60,000 kilometers an hour, 80,000 years to get there. So this is a very, very-- that's only four light years. Now this is 6,500 light years. But this supernova, it's believed happened 1,000 years ago, right at the center. And so at the center here, we should have a neutron star. And this cloud, the shock wave that you see here, this is still the material traveling outward from that supernova over 1,000 years. This shock wave, or the diameter of this sphere of material, is six light years. So we could say this distance right here is six light years. So this is an enormously big shock wave cloud. And actually, we believe that our solar system started to form, started to condense because of a shock wave created And just to answer another question that was kind of jumping up, probably, in the last video-- and this is still not really, really well understood. We talk about how elements up to iron, or maybe nickel, can be formed inside of the cores of massive stars. So you could imagine when the star explodes, a lot of that material is released into the universe. And so that's why we have a lot of these materials in our own bodies. In fact, we could not exist if these heavier elements were not formed inside of the cores of primitive stars, stars that have supernova-ed a long time ago. Now the question is, how do these heavier elements form? How do we get all of this other stuff on the periodic table? How do we get all these other heavier elements? And they're formed during the supernova itself. It's so energetic. You have all sorts of particles streaming out and streaming in, streaming out because of the force of the shock wave, streaming in because of the gravity. But you have all sorts of kind of a mishmash of elements" + }, + { + "Q": "\n\"During the eclipse Sal talks about at 5:28, why is the moon the exact size of the sun, offering complete 1 for 1 coverage?\"", + "A": "Happy coincidence.", + "video_name": "qOwCpnQsDLM", + "timestamps": [ + 328 + ], + "3min_transcript": "But you can imagine an enormous amount of energy is also being released. So this is kind of a temporary-- and then all of a sudden, this collapses. This collapses even more until all you have-- and all the protons are turning into neutrons. Because they're capturing electrons. So what you eventually have is this entire core is collapsing into a dense ball of neutrons. You can kind of view them as just one really, really, really, really, really massive atom because it's just a dense ball of neutrons. At the same time, when this collapse happens, you have an enormous amount of energy being released in the form of neutrinos. Did I say that neutrons are being released? No, no, no, the electrons are being captured by the protons, protons turning into neutrons-- this dense ball of neutrons right here-- and in the process, neutrinos get released, these fundamental particles. But it's an enormous amount of energy. And this actually is not really, really well understood, of all of the dynamics here. Because at the same time that this iron core is undergoing through this-- at first it kind of pauses due to the electron degeneracy pressure. And then it finally gives in because it's so massive. And then it collapses into this dense ball of neutrons. But when it does it, all of this energy's released. And it's not clear how-- because it has to be a lot of energy. Because remember, this is a massive star. So you have a lot of mass in this area over here. But it's so much energy that it causes the rest of the star to explode outward in an unbelievable, I guess, unbelievably bright or energetic explosion. And that's called a supernova. it comes from, I believe-- I'm not an expert here-- Latin for \"new.\" And the first time people observed a nova, they thought it was a new star. Because all of a sudden, something they didn't see before, all of a sudden, it looks like a star appeared. Because maybe it wasn't bright enough for us to observe it before. But then when the nova occurred, it did become bright enough. So it comes from the idea of new. But a supernova is when you have a pretty massive star's core collapsing. And that energy is being released to explode the rest of the star out at unbelievable velocities. And just to kind of fathom the amount of energy that's being released in a supernova, it can temporarily outshine an entire galaxy. And in a galaxy, we're talking about hundreds of billions of stars. Or another way to think about it, in that very short period of time, it can release as much energy as the sun will in its entire lifetime. So these are unbelievably energetic events. And so you actually have the material that's not in the core being shot out of the star" + }, + { + "Q": "\nAt 7:45, is killing itself the process of apoptosis? or is it another mechanism specific to the immune system?", + "A": "Apoptosis is where a cell receives signals in order for it to kill itself. In the immune system, the cytotoxic T cells release granzymes and other chemicals that kill the cell. For example, spores may be punctured in the cell causing an increase in water intake and eventual lysing (bursting).", + "video_name": "YdBXHm3edL8", + "timestamps": [ + 465 + ], + "3min_transcript": "so it's presenting that same antigen on this MHC one complex. Remember, the variable portions need to match up. Let's say that this is an effector cytotoxic T cell and actually let me draw in a little bit different. Let me draw it like this. This is effector cytotoxic T cell. Its receptor, its variable portion is the one that's compatible with this antigen that's being presented right over here. Let me just label this again. This is the MHC one complex. This is an effector cytotoxic T cell. We'll put the C there for cytotoxic and what it does is essentially kind of latches on to the cell that needs to die and it does it, not only have this receptor interfacing with the MHC one complex but actually has a whole series of proteins This would be much smaller or relative to the scale cell. It essentially latches on between the two and I'm not going to go in detail but essentially forms what you can call an immuno synapse which is kind of where the two things are interacting with each other. When it identifies this, it says, okay, I need to kill this thing or essentially I need to make this thing kill itself. It starts releasing all of these molecules. It can release molecules like preference so to release this preference which will essentially cause gaps or holes to form in the membrane of the cell that needs to die and it could release other things like granzymes that can go in and essentially cause this thing to kill itself. The whole point of this video is to appreciate I guess what we haven't talked about yet. We had already talked about what happens when you identify shady things outside of the cells and then how you can kind of bring them in and then present them and then use that to further activate Now we're talking about identifying shady things inside the cells. Those get presented by MHC one complexes and then the cytotoxic T cells recognize them and then force the cell to kill themselves. This wouldn't just be cancer cells, this could also apply to a cell that has already been affected by a virus. For example a cell like this all ready, so that's its nucleus. It's already been infected by some virus so the virus has hijacked the cell's replication machinery in order to replicate itself. The proper immune response is hey look, I'm a virus making machine. I should kill myself. It will wreck some of the antigens that are being produced inside by the viruses. They're going to bind to MHC one complexes. Pieces of the virus are going to bind to MHC one complexes" + }, + { + "Q": "at 8:00 why is the o positive , didn't it make a covalent bond ? so shouldn't it be partially positive ?\n", + "A": "For oxidation number purposes, the O lost one of its lone pair electrons when it bonded with the H, so it gained a full positive charge.", + "video_name": "U9dGHwsewNk", + "timestamps": [ + 480 + ], + "3min_transcript": "In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs? What's our final product? Let me draw it here. This part of the reaction is going to happen fast. The rate-determining step happened slow. The leaving group had to leave. The carbocation had to form. That's not going to happen super fast but once that forms, it's not that stable and then this thing will happen. This is fast. Let me paste everything again. So now we already had the bromide. It had left. Now the hydrogen is gone. The hydrogen from that carbon right there is gone. This electron is still on this carbon but the electron that That electron right here is now over here, and now this bond right over here, is this bond. We formed an alkene and now, what was an ethanol took a hydrogen proton and now becomes a positive cation. Let me draw that. So this electron ends up being given. It's no longer with the ethanol. It gets given to this hydrogen right here. That hydrogen right there. And now they have formed a new bond and since this oxygen gave away an electron, it now has a positive charge. And all along, the bromide anion had left in The bromide anion is floating around with its eight valence electrons, one, two, three, four, five, six, seven, and then it has this one right over here. That makes it negative. Then our reaction is done. We had a weak base and a good leaving group, a tertiary carbon, and the leaving group left. We only had one of the reactants involved. It was eliminated. And then once it was eliminated, then the weak base was then able to take a hydrogen off of this molecule, and that allowed this molecule to become an alkene, formed a double bond. This is called, and I already told you, an E1 reaction. E for elimination, in this case of the halide. One, because the rate-determining step only involved one of the molecules. It did not involve the weak base. We'll talk more about this, and especially different circumstances where you might have the different types of E1" + }, + { + "Q": "\nAt 5:53 why will the O of ethanol donate its electron to the bonded H atom ?? Why not directly to the positively charged carbocation ? Well I think O can donate an electron to C (+) ......", + "A": "This is because of the steric hindrance of the ethanol. Due to its larger/bulkier size it is more apt to react with the hydrogen that is further away and less crowded by the overall molecule . This is ultimately why an E1 reaction occurred and why and S1 reaction didn t occur.", + "video_name": "U9dGHwsewNk", + "timestamps": [ + 353 + ], + "3min_transcript": "doesn't occur nothing else will. But now that this does occur everything else will happen quickly. In our rate-determining step, we only had one of the reactants involved. It's analogous to the SN1 reaction but what we're going to see here is that we're actually eliminating. We're going to call this an E1 reaction. We're going to see that in a second. Actually, elimination is already occurred. The bromide has already left so hopefully you see why this is called an E1 reaction. It's elimination. E for elimination and the rate-determining step only involves one of the reactants right here. It didn't involve in this case the weak base. Now that the bromide has left, let's think about whether this weak base, this ethanol, can actually do anything. It does have a partial negative charge over here. It does have a partial negative charge and on these ends it has partial positive charges, so it is somewhat attracted to hydrogen, or to protons I should say, to positive charges. But not so much that it can swipe it off of things that Now that this guy's a carbocation, this entire molecule actually now becomes pretty acidic, which means it wants to give away protons. Another way you could view it is it wants to take electrons, depending on whether you want to use the Bronsted-Lowry definition of acid, or the Lewis definition. Either way, it wants to give away a proton. It could be that one. It has excess positive charge. It wants to get rid of its excess positive charge. So it's reasonably acidic, enough so that it can react with this weak base. What you have now is the situation, where on this partial negative charge of this oxygen-- let me pick a nice color here-- let's say this purple electron right here, it can be donated, or it will swipe the hydrogen proton. Then hydrogen's electron will be taken In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs? What's our final product? Let me draw it here. This part of the reaction is going to happen fast. The rate-determining step happened slow. The leaving group had to leave. The carbocation had to form. That's not going to happen super fast but once that forms, it's not that stable and then this thing will happen. This is fast. Let me paste everything again. So now we already had the bromide. It had left. Now the hydrogen is gone. The hydrogen from that carbon right there is gone. This electron is still on this carbon but the electron that" + }, + { + "Q": "6:20 - Would EtOH take the H if the C were not a cation? In other words, is if necessary for the carbocation to exist to enable EtOH to take the H? Does the presence of the carbocation enable EtOH to take the H by reducing the strength of the H-C bond by pulling electrons towards itself? Do the electrons in the H-C bond move towards the carbocation before the H is taken, or only afterwards?\n", + "A": "No. The EtOH would not take the H if the C were not a cation. The cation must be there to make the H atom acidic enough to be removed by the relatively weak base EtOH. The cation does this by pulling the electrons in the C-H bond towards itself. The electrons in the C-H bond have already moved towards the carbon before the EtOH attacks, But they move entirely to C atom after the EtOH has attacked.", + "video_name": "U9dGHwsewNk", + "timestamps": [ + 380 + ], + "3min_transcript": "doesn't occur nothing else will. But now that this does occur everything else will happen quickly. In our rate-determining step, we only had one of the reactants involved. It's analogous to the SN1 reaction but what we're going to see here is that we're actually eliminating. We're going to call this an E1 reaction. We're going to see that in a second. Actually, elimination is already occurred. The bromide has already left so hopefully you see why this is called an E1 reaction. It's elimination. E for elimination and the rate-determining step only involves one of the reactants right here. It didn't involve in this case the weak base. Now that the bromide has left, let's think about whether this weak base, this ethanol, can actually do anything. It does have a partial negative charge over here. It does have a partial negative charge and on these ends it has partial positive charges, so it is somewhat attracted to hydrogen, or to protons I should say, to positive charges. But not so much that it can swipe it off of things that Now that this guy's a carbocation, this entire molecule actually now becomes pretty acidic, which means it wants to give away protons. Another way you could view it is it wants to take electrons, depending on whether you want to use the Bronsted-Lowry definition of acid, or the Lewis definition. Either way, it wants to give away a proton. It could be that one. It has excess positive charge. It wants to get rid of its excess positive charge. So it's reasonably acidic, enough so that it can react with this weak base. What you have now is the situation, where on this partial negative charge of this oxygen-- let me pick a nice color here-- let's say this purple electron right here, it can be donated, or it will swipe the hydrogen proton. Then hydrogen's electron will be taken In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs? What's our final product? Let me draw it here. This part of the reaction is going to happen fast. The rate-determining step happened slow. The leaving group had to leave. The carbocation had to form. That's not going to happen super fast but once that forms, it's not that stable and then this thing will happen. This is fast. Let me paste everything again. So now we already had the bromide. It had left. Now the hydrogen is gone. The hydrogen from that carbon right there is gone. This electron is still on this carbon but the electron that" + }, + { + "Q": "\nat \"7:40\" could you also write 350. feet (with a decimal at the end) to show that you meausred it to the nearest one? Or does nobody do that? Because I remember you doing that in the first significant figures video.", + "A": "You could also write 350. feet, although that has three significant didgits when the building was 350 feet with only two significant didgits, meaning that the measurement of the building may have been a guess or may have been exact. We just cannot tell without the decimal point, which is what the guy was trying to express.", + "video_name": "xHgPtFUbAeU", + "timestamps": [ + 460 + ], + "3min_transcript": "and so we can only legitimately say, if we want to represent what we did properly that the tower is 3.99 meters. And I also want to make it clear that this doesn't just apply to when there is a decimal point. If I were tell you that... Let's say that I were to measure... I want to measure a building. I was only able to measure the building to the nearest 10 feet. So I tell you that that building is 350 feet tall. So this is the building. This is a building. And let's say there is a manufacturer of radio antennas, so... or radio towers. And the manufacturers has measured their tower to the nearest foot. And they say, their tower is 8 feet tall. So notice: here they measure to the nearest 10 feet, here they measure to the nearest foot. And actually to make it clear, because once again, as I said, this is ambiguous, Maybe it was exactly 350 feet or maybe they just rounded it to the nearest 10 feet. So a better way to represent this, they... would be to say instead of writing it 350, a better way to write it would be 3.5 times 10 to the second feet tall. And when you are writing in scientific notation, that makes it very clear that there is only 2 significant digits here, you are only measuring to the nearest 10 feet. Other way to represent it: you could write 350 this notation has done less, but sometimes the last significant digit has a line on the top of it, or the last significant digit has a line below it. Either of those are ways to specify it, this is probably the least ambiguous, but assuming that they only make measure to the nearest 10 feet, If someone were ask you: \"How tall is the building plus the tower?\" Well, your first reaction were, let's just add the 350 plus 8, you get 358. You'd get 358 feet. So this is the building plus the tower. 358 feet. We are making it look like we were able to measure the combination to the nearest foot. But we were able to measure only the tower to the nearest foot. So in order to represent our measurement at the level of precision at we really did, we really have to round this to the nearest 10 feet. Because that was our least precise measurement. So we would really have to round this up to, 8 is greater-than-or-equal to 5, so we round this up to 360 feet. So once again, whatever is... Just to make it clear, even this ambiguous, maybe we put a line over to show, that is our level of precision, that we have 2 significant digits. Or we could write this as 3.6 times 10 to the second. Which is times 100. 3.6 times 10 to the second feet in scientific notation. And this makes it very clear that we only have 2 significant digits here." + }, + { + "Q": "\nat 7:23, 350 ft is 2 significant figures but when you multiply out 3.5 e^2, that would make it 350. which is 3 significant figures, not 2, right?", + "A": "No, the significant digits are still 2. You are not adding precision by multiplying. If you want to represent 3 significant digits, you will right 3.50*10^2", + "video_name": "xHgPtFUbAeU", + "timestamps": [ + 443 + ], + "3min_transcript": "0 plus 9 is 9, 9 plus 0 is 9, you get the decimal point, 1 plus 2 is 3. So you get 3.991. And the problem with this, the reason why this is a little bit... it's kind of misrepresenting how precise you measurement is. You don't know, if I told you that the tower is 3.991 meters tall, I'm implying that I somehow was able to measure the entire tower to the nearest millimeter. The reality is that I was only be able to measure the part of the tower to the millimeter. This part of the tower I was able to measure to the nearest centimeter. So to make it clear the our measurement is only good to the nearest centimeter, because there is more error here, then... it might overwhelm or whatever the precision we had on the millimeters there. To make that clear, we have to make this only as precise as the least precise thing that we are adding up. So over here, the least precise thing was, we went to the hundredths, so over here we have to round to the hundredths. and so we can only legitimately say, if we want to represent what we did properly that the tower is 3.99 meters. And I also want to make it clear that this doesn't just apply to when there is a decimal point. If I were tell you that... Let's say that I were to measure... I want to measure a building. I was only able to measure the building to the nearest 10 feet. So I tell you that that building is 350 feet tall. So this is the building. This is a building. And let's say there is a manufacturer of radio antennas, so... or radio towers. And the manufacturers has measured their tower to the nearest foot. And they say, their tower is 8 feet tall. So notice: here they measure to the nearest 10 feet, here they measure to the nearest foot. And actually to make it clear, because once again, as I said, this is ambiguous, Maybe it was exactly 350 feet or maybe they just rounded it to the nearest 10 feet. So a better way to represent this, they... would be to say instead of writing it 350, a better way to write it would be 3.5 times 10 to the second feet tall. And when you are writing in scientific notation, that makes it very clear that there is only 2 significant digits here, you are only measuring to the nearest 10 feet. Other way to represent it: you could write 350 this notation has done less, but sometimes the last significant digit has a line on the top of it, or the last significant digit has a line below it. Either of those are ways to specify it, this is probably the least ambiguous, but assuming that they only make measure to the nearest 10 feet, If someone were ask you: \"How tall is the building plus the tower?\" Well, your first reaction were, let's just add the 350 plus 8, you get 358. You'd get 358 feet. So this is the building plus the tower. 358 feet." + }, + { + "Q": "\n5:30 What would drive the reformation of the carbonyl?", + "A": "Delocalization of electrons leads to increased stability. Before 5:30, the negative charge is localized on the O atom. After 5:30. there is no charge and a lone pair on O is now delocalized in the \u00cf\u0080 orbital of the C=O bond.", + "video_name": "CafRuKs7EfE", + "timestamps": [ + 330 + ], + "3min_transcript": "have formed this bond now, and this carbon is bonded to this nitrogen, bonded to this, and our R double prime group, and if we showed, let's make these electrons in here green, these electrons move off onto our nitrogen, like that. And so our carbon is also double-bonded to this nitrogen, with R double prime group. So in the next step our amine is gonna function as a nucleophile. So let's go ahead and draw in our amine. So we have our nitrogen with two hydrogens, and an R prime group, lone pair of electrons on our nitrogen, makes this amine able to function as a nucleophile, so let me go ahead and put those electrons in magenta. So right here, this oxygen is partially negative, It's gonna withdraw some electron density from this carbon, so partially positive. going to attack our electrophile. So our amine is going to attack this carbon, push these electrons off onto our oxygen. So let's go ahead and show the result of this nucleophilic attack. Alright, so let's get some room down here. We would have our carbon bonded to this oxygen. So let's go ahead and draw in all of those electrons, so, -1 formal charge on this oxygen, so if these electrons in here in green move off onto our oxygen we get a -1 formal charge. This carbon is bonded to an R group, it's also bonded to this nitrogen, this nitrogen now has a +1 formal charge. So +1 formal charge on our nitrogen, after the electrons in magenta move in here to form this bond. So we still have our carbon bonded to this oxygen, and draw in our lone pairs of electrons, and then we have this carbon, we have a nitrogen, we have our R double prime group, we have a hydrogen, lone pair of electrons double-bonded to group, so a lot of stuff going on here. So, the use of DCC gives you a good leaving group. So if we think about all this stuff over here, this is an excellent leaving group. So if we reform our carbonyl, let's go ahead and show that, so if these electrons in here move in to reform our carbonyl, these electrons could come off onto our oxygen, and we could even show them moving over to here, to save some time, and if there's a proton out here, these electrons could pick up that proton, and we have an excellent leaving group, so this actually forms dicyclohexylurea over here on the right. So if I circle all of this stuff, we're gonna get dicyclohexylurea, and let's go ahead and show what would happen. So if we reform our carbonyl, let's use those electrons in here in red, so if those electrons in red move in, now we would have our R group," + }, + { + "Q": "\n10:00 I'm not sure I understand why it's important to know what components of the vectors in the 2-d plane are perpendicular to the other... I mean as long as your 3rd vector is sticking in or out of the page it will be perpendicular to both a & b so how does finding their components help with this ?", + "A": "This will become important with torque, which will most likely be seen in the next video. If they don t cover it I would google it.", + "video_name": "o_puKe_lTKk", + "timestamps": [ + 600 + ], + "3min_transcript": "magnitude of vector a that is perpendicular to vector b, and those are the two numbers that I want to multiply and then give it that direction as specified by the right hand rule. And I'll show you some applications. This is especially important-- well, we'll use it in torque and we'll also use it in magnetic fields, but it's important in both of those applications to figure out the components of the vector that are perpendicular to either a force or a radius in question. So that's why this cross product has the sine theta because we're taking-- so in this, if you view it as magnitude of a sine theta times b, this is kind of saying this is the magnitude of the component of a perpendicular to b, or you could interpret it the other way. You could interpret it as a times b sine theta, right? Put a parentheses here. And then you could view it the other way. that is perpendicular to a. Let me draw that, just to hit the point home. So that's my a, that's my b. This is a, this is b. So b has some component of it that is perpendicular to a, and that is going to look something like-- well, I've run out of space. Let me draw it here. If that's a, that's b, the component of b that is perpendicular to a is going to look like this. It's going to be perpendicular to a, and it's going to go that far, right? And then you could go back to SOH CAH TOA and you could prove to yourself that the magnitude of this vector is b sine theta. So that is where the sine theta comes from. It makes sure we're multiplying the components of the vectors that are perpendicular to each other to get a third vector that is perpendicular to both of them. And then the people who invented the cross product said, well, it's still ambiguous because it doesn't tell us-- there's always two vectors that are perpendicular to these two. One goes in, one goes out. They're in opposite directions. And that's where the right hand rule comes in. They'll say, OK, well, we're just going to say a convention that you use your right hand, point it like a gun, make all your fingers perpendicular, and then you know what direction that vector points in. Anyway, hopefully, you're not confused. Now I want you to watch the next video. This is actually going to be some physics on electricity, magnetism and torque, and that's essentially the applications of the cross product, and it'll give you a little bit more intuition of how to use it. See you soon." + }, + { + "Q": "Can you please explain that what does Sal means by distances are proportional? (at 5:35)\n", + "A": "It means that if you increase the distance to the edge, the distance the edge travels down also increases. When he says that the distance increases proportionally, he means that if the distance to the 7N force increases by 10%, the travel distance will also increase by 10%.", + "video_name": "DiBXxWBrV24", + "timestamps": [ + 335 + ], + "3min_transcript": "compound it further and et cetera, et cetera, using some of the other concepts we've learned. But I won't worry about that right now. So let's say that I'm going to push up here. Well no let me see what I want to do. I want to push down here with a force of-- let's say that this distance right here is 35 meters, this distance is 5 meters-- and let's say I'm going to push down with the force of 7 newtons, and what I want to figure out is how heavy of an object can I lift here. How heavy of an object. Well, all we have to do is use the same formula. But the moments-- and I know I used that word once before, so you might not know what it is-- but the moments on both sides of the fulcrum have to be the same. So what's the moment again? Well, the moment is just the force times the distance from the force to the fulcrum. So the input moment is 7 newtons times 35 meters. And realize that that does not work, because the distance this force is traveling is not 35 meters. The distance this force is traveling is something like, here. But this 35 meters is going to be proportional to the distance that this is traveling when you compare it to this other side. So this quantity, 7 newtons times 35 meters, is the moment. And that is going to be equal to the moment on this side, the output moment. So that is equal to 5 meters times the force that I'm lifting, or the lifting force of the machine, times let's say the force out. So we can figure out the force out by just dividing both sides by 5. So let's see, 35 divided by 5 is 7, so you get 7 times 7 And you can see that, because you can see that the length of this side of the lever is 7 times the length of this side of the lever. So when you input a force of 7, you output a force of 7 times that. And of course, in order to move the block 1 meter up in this direction, you're going to have to push down for 7 meters. And that's where we know that the input work is equal to the output work. Well anyway hopefully I didn't confuse you and you have a reasonable sense of how levers work. In the next couple of videos, I'll introduce you to other machines, simple machines like a wedge-- I've always had trouble calling a wedge a machine, but it is one-- and pulleys. I'll see you in the next video." + }, + { + "Q": "So basically the Na-K pump exchanges Na+ for K+ ions? (1:50)\nWhy would the cell exchange one ion for the other?\nIf the answer is concentration gradients then another question.\nSuppose one side of the wall contains a higher concentration of both Na+ and K+ than the other side. In that case what will happen?\n", + "A": "Yes the answer is concentration gradient. If there is hihger conc at one side then a potential difference will be established that causes polarization of membrane that propagates. This happens in the case of impulse conduction by nerves.", + "video_name": "vh166DKxYiM", + "timestamps": [ + 110 + ], + "3min_transcript": "- [Voiceover] What I hope to do in this video is give ourselves an appreciation for the sodium-potassium pump, and as the name implies, it pumps sodium and potassium, but it does it in different directions. So this little depiction right over here, this is my drawing, my rendition of the sodium-potassium pump, it's a trans-membrane, I guess you could say protein complex right over here. And in this resting state, it is open to the inside of the cell and it has an affinity for sodium ions. And so the sodium ions, you see three sodium ions depicted here in blue. They're going to bind to the pump. And once they bind to it, then it's going to want to be phosphorylated by an ATP, and we see that right over here. This is ATP, adenosine triphosphate. And when it gets phosphorylated, it's a release of energy and it allows the confirmation of the actual protein to change. So the new confirmation of the protein, it's now going to open up to the outside, close off to the inside, have an affinity for sodium ions, but an affinity for potassium ions, and this is fascinating, that release of energy, change of confirmation, that these proteins really are these molecular machines, these fascinating molecular machines. But once that happens, this change of confirmation, the sodium ions are going to be released outside of the cell. And then you're going to have potassium ions that are going to bind from the outside. And then once that happens, the change in confirmation, it's going to get dephosphorylated and then you're going to go back to your original confirmation, your original confirmation right over here. Where you no longer have an affinity for potassium ions, they're going to be released, and then you're going to be back in the original phase. So this is fascinating. By using ATP, by using energy, this is active transport, it takes energy to do this. Let me write this down. This is active, this is active transport that we are talking about right over here. We're able to pump, using an ATP, we're able to pump three sodium ions out, so let me write that down. Three sodium ions out. And in the process, we pump two potassium ions in. So we pump two potassium ions in. Now you might say, okay, the outside, since these both have positive charge, but I have three sodium going out, two potassium going in. That must make the outside more positive than the inside, and that actually is true. But that by itself isn't fully responsible. It's actually only partially responsible for the electric potential difference between the inside of the membrane and the outside of the membrane. What really sets that up is that you actually have channel proteins that allow potassium ions to move down, to diffuse down their concentration gradient. So let's think about what happens before I even talk about these channel proteins right over here. Because of the sodium-potassium pump, what is sodium's concentration gradient?" + }, + { + "Q": "At 4:00, aren't the three dot structures the same, just rotated? Why are they considered different?\n", + "A": "I wondered the same thing initially. I think that an important thing to consider is that the diagrams are only the-same-but-rotated if you don t care which oxygen atom is which. If you labeled the oxygen atoms, then it wouldn t be the same. So in real life, if you were somehow able to hold the molecule still and look at just one oxygen atom, the three structures would not be the same (it could have either a single or double bond to the nitrogen).", + "video_name": "bUCu7bPkZeI", + "timestamps": [ + 240 + ], + "3min_transcript": "And there are a couple of different ways that we could give nitrogen an octet. For example, we could take a lone pair of electrons from this top oxygen here and move them into here to share those electrons between that top oxygen and that nitrogen. So let's go ahead and draw that resulting dot structure. So we would have our nitrogen now with a double bond to our top oxygen. Our top oxygen had three lone pairs of electrons. But now it has only two, because electrons in green moved in to form a double bond. This nitrogen is bonded to an oxygen on the bottom left and an oxygen on the bottom right here. So this is a valid dot structure. We followed our steps. And we'll go ahead and put this in brackets and put a negative charge outside of our brackets like that. So that's one possible dot structure. But we didn't have to take a lone pair of electrons from the top oxygen. We could've taken a lone pair of electrons from the oxygen on the bottom left here. So if those electrons in blue moved in here, would have been equally valid. We could have shown this oxygen on the bottom left now bonded to this nitrogen, and it used to have three lone pairs. Now it has only two. And now this top oxygen is still a single bond with three lone pairs around it. And this bottom right oxygen is still a single bond with three lone pairs around it. So this is a valid dot structure as well. So let's go ahead and put our brackets with a negative charge. And then, of course, we could have taken a lone pair of electrons from the oxygen on the bottom right. So I could have moved these in here to form a double bond. And so now, we would have our nitrogen double bonded to an oxygen on the bottom right. The oxygen on the bottom right now has only two lone pairs of electrons. The oxygen at the top, single bond with three lone pairs. And then the same situation for this oxygen on the bottom left. And so this is, once again, another possible dot structure. And so these are considered to be And the way to represent that would be this double-headed resonance arrow here. And I think when students first see resonance structures, the name implies that, in this case, the ion is resonating back and forth between these three different possible, equally valid dot structures. And that's not quite what's going on here. Each of these dot structures is an attempt to represent the structure of the ion. But they're really not the best way of doing that. You need to think about combining these three dot structures in a resonance hybrid of each other. And so let's go ahead and draw just a simple representation of a way of thinking about a resonance hybrid. So if I combined all three of my dot structures here into one picture, I had a double bond to one oxygen in each of my three resonance structures And so the top oxygen had a double bond in one of them, the bottom left in the middle one, and then the bottom right" + }, + { + "Q": "At 1:41 doesn't each O atom should have a negative charge and at 2:49 N atom should have a possitive charge?\n", + "A": "Yes to both of these", + "video_name": "bUCu7bPkZeI", + "timestamps": [ + 101, + 169 + ], + "3min_transcript": "Now that we know how to draw dot structures, let's apply our rules to the nitrate anion. And we're going to see that we can draw a few different dot structures for this anion. And we're going to call those resonance structures of each other. But first, we need to calculate the total number of valence electrons. And so nitrogen is in Group 5 in the period table, therefore, five valence electrons. Oxygen is in Group 6, therefore, six valence electrons for each oxygen. I have three of them. So 6 times 3 is 18 valence electrons, plus the 5 from the nitrogen gives me 23. And I have a negative charge. This is an anion here. So we have to add one electron to that. So 23 plus 1 gives us a total of 24 valence electrons that we need to represent in our dot structure. So we know that nitrogen is going to go in the center, because oxygen is more electronegative. So nitrogen goes in the center. Nitrogen is bonded to three oxygens. So I can go ahead and put them in there like that. And let's see. How many valence electrons have we represented so far? Therefore, 24 minus 6 gives us 18 valence electrons left over. We're going to put those leftover valence electrons on our terminal atoms, which are our oxygens. And oxygen's going to follow the octet role. Currently, each oxygen has two valence electrons around it, the ones in magenta. So if each oxygen has two, each oxygen needs six more to complete the octet. And so I go ahead and put six more valence electrons on each one of my oxygens. Now each oxygen is surrounded by eight electrons. So the oxygens are happy. We added a total of six valence electrons to three oxygens. So 6 times 3 is 18. So we've used up all of the electrons that we need to represent. And so this dot structure, so far, it has all of our valence electrons here. Oxygen has an octet. So oxygen is happy. But nitrogen does not have an octet. If you look at the electrons in magenta, there are only six electrons around the nitrogen. And there are a couple of different ways that we could give nitrogen an octet. For example, we could take a lone pair of electrons from this top oxygen here and move them into here to share those electrons between that top oxygen and that nitrogen. So let's go ahead and draw that resulting dot structure. So we would have our nitrogen now with a double bond to our top oxygen. Our top oxygen had three lone pairs of electrons. But now it has only two, because electrons in green moved in to form a double bond. This nitrogen is bonded to an oxygen on the bottom left and an oxygen on the bottom right here. So this is a valid dot structure. We followed our steps. And we'll go ahead and put this in brackets and put a negative charge outside of our brackets like that. So that's one possible dot structure. But we didn't have to take a lone pair of electrons from the top oxygen. We could've taken a lone pair of electrons from the oxygen on the bottom left here. So if those electrons in blue moved in here," + }, + { + "Q": "\nI don't quite understand when at 6:28, Sal says that two particles \"buzz past each other.\" Can two particles collide too quickly? I thought that as long as they had energy greater than that of the reaction's activation energy, they would react.", + "A": "Water and carbon dioxide can form carbonic acid . But in high temperature they can t form a lot because it will breakdown (They did not just buzz past each other , they did react ) (positive delta S & H, high T , negative delta G , not spontaneously reactions ). In low temperature they can form a lot of carbonic acid because it is stable. (positive delta S & H, low T , positive delta G). It will not spontaneously breaking down (not enough reaction s activation energy)", + "video_name": "CHHu-iTwHjg", + "timestamps": [ + 388 + ], + "3min_transcript": "and whether they're spontaneous depends on the temperature. So, over here, if we are dealing, our Delta H is less than zero. So, we're going to have a release of energy here, but our entropy decreases. What's gonna happen? Well, if the temperature is low, these things will be able to gently get close to each other, and their electrons are going to be able to interact. Maybe they get to a lower energy state, and they can release energy. They're releasing energy, and the electrons will spontaneously do this. But, the entropy has gone down. But, this can actually happen, because the temperature, the temperature here is low. \"Wait, doesn't that violate \"The Second Law of Thermodynamics?\" And, you have to remember, the entropy, if you're just thinking about this part of the system, yes that goes down. But, you have heat being released. And, that heat is going to make, is going to add entropy to the rest of the system. So, still, The Second Law of Thermodynamics holds because of this released heat. But, if you just look at the constituents here, the entropy went down. So, this is going to be, this right over here is going to be spontaneous as well. And, we're always wanting to back to the formula. If this is negative and this is negative, well, this is going to be a positive term. But, if 'T' low enough, this term isn't going to matter. 'T' is, you confuse it as the weighing factor on entropy. So, if 'T' is low, the entropy doesn't matter as much. Then, enthalpy really takes over. So, in this situation, Delta G, we're assuming 'T' is low enough to make Delta G negative. And, this is going to be spontaneous. Now, if you took that same scenario, but you had a high temperature, well now, you have these same two molecules. Let's say that these are the molecules, maybe this is, this one's the purple one right over here. You have the same two molecules here. Hey, they could get to a more kind of a, they could release energy. But over here, you're saying, The change in enthalpy is negative. But, they're buzzing past each other so fast that they're not gonna have a chance. Their electrons aren't gonna have a chance to actually interact in the right way for the reaction to actually go on. And so, this is a situation where it won't be spontaneous, because they're just gonna buzz past each other. They're not gonna have a chance to interact properly. And so, you can imagine if 'T' is high, if 'T' is high, this term's going to matter a lot. And, so the fact that entropy is negative is gonna make this whole thing positive. And, this is gonna be more positive than this is going to be negative. So, this is a situation where our Delta G is greater than zero. So, once again, not spontaneous. And, everything I'm doing is just to get an intuition for why this formula for Gibbs Free Energy makes sense. And, remember, this is true under constant pressure and temperature. But, those are reasonable assumptions if we're dealing with, you know, things in a test tube, or if we're dealing with a lot of biological systems. Now, let's go over here. So, our enthalpy," + }, + { + "Q": "\nAt 3:03 mitosis goes in to a bunch of what?", + "A": "Into a bunch of me . In other words, the bunch of cells turns into a human s organs.", + "video_name": "PvoigrzODdE", + "timestamps": [ + 183 + ], + "3min_transcript": "is going to keep replicating. So it will, you know, after one, so after one, and we're going to go into the details of the mechanics of mitosis, but after one round of mitosis, it is now two cells. It is now two cells. And I'm going to draw it, once again, I'm not going to draw it at scale. It's now two cells. I want to make sure I have enough space on my little chalkboard here. It has two cells. Instead of drawing all of the chromosomes, let me just say that each of these, in my nucleus, I still have 2N. I have the diploid number. So each of these two cells that it has differentiated to still have the full contingency. That's what mitosis does. It essentially replicates the entire cells. You have the same number of chromosomes. And then this process is just going to keep happening. These two characters are going to replicate, are going to replicate, and so then you're going to have, through mitosis, Mitsosis. So they just keep duplicating themselves. And each of these cells have the full contingency. 2N, the diploid number of chromosomes for, well, in this case it's going to be 46 And then this process is just going to keep happening. So this process is going to keep happening. I'll do dot dot dot to show that, you know, a lot of this has been going on. So mitosis is just going to keep happening. And so eventually you're going to have thousands of these cells, and eventually as we'll see, you're going to have millions and ten millions of them. So let me draw them really really really small. There's a bunch of them there. And each of them, each of them are going to have the diploid number of chromosomes. They're going to have 46 chromosomes. 23 pair of homologous chromosomes. So we now have a big ball of these here. they're going to differentiate into me. They're going to differentiate into the different parts of my body. So for example, these cells right over here might eventually, they'll keep replicating, but then it's them and their offspring might eventually differentiate into my brain cells. These cells here will keep replicating, and them and their offspring, I guess you could say, or the things that they replicate into, might differentiate into my heart. These right over here might differentiate into my lungs, and of course all of these eventually will differentiate into all the different, and they and their offspring will differentiate into all of the things that make me me. And so you have a lot more of this mitosis. You're eventually going to have a human being. So let me just say this is more mitosis going on. Mitosis." + }, + { + "Q": "\nOk so just to clarify, can mutations happen in both the female and male cells? Sal only sued sperm cells in his example at 8:45 so I just wanted to make sure. Also when Sal is talking about how the somatic cells differentiate into all the different body parts at 2:31 aren't they called stem cells?", + "A": "the mutation occurs in both cells they somatic cells can be called as stem cells", + "video_name": "PvoigrzODdE", + "timestamps": [ + 525, + 151 + ], + "3min_transcript": "So if you're female you're going to produce eggs. If you're male you're going to produce sperm. But this is through the process of meiosis. Meiosis you're going to produce sperm in the case of a man, and you're going to produce ova in the case of a female. And this brings up a really interesting thing, because throughout biology we talk about mutations and natural selection and whatever else. And it's important to realize how mutations may affect you and your offspring. So if you have a mutation in one of the somatic cells here, let's say in a skin cell, or in you brain, or in the heart, that may affect your ability to, you know, especially if God forbid it's a really dangerous thing like cancer, and it happens when you're young, before you've had a chance to reproduce and you're not able to survive, But if this is happening in a somatic cell, it's not going to affect the DNA make-up of what you pass on. The DNA make-up of what you pass on, that's determined by what's going on in the gametes. So a mutation, if on the way to differentiating into gametes a mutation happens, so if one of these mutate and then keep replicating, so let's say there's a mutation here, and they keep replicating and they differentiate into the germ cells the mutation is right over there, then through meiosis that produces some mutated sperm. Then that would pass on to your, well, that has a chance of passing on to your children. Because once again, it might not be all of the sperm cells that have that mutation. It could be only a handful of the two to three hundred million of the sperm cells, and so if that mutation somehow makes it harder to kind of function, either fuse with an egg or even potentailly develop and kind of swim through fluid, then it still might not be the thing that makes it. So mutations only affect your offspring in the situation where those, the cells in which they happen are eventually differentiated into things, into gametes that you will pass on to your children." + }, + { + "Q": "\nIt talks about Theia colliding with Earth in 4:51 , how come the Earth is a perfect sphere? Did the collision do anything?", + "A": "The collision would have released so much energy that the surface of the Earth would have been turned into molten rock. Gravity acts on the molten rock, and the molten rock fills the crater and leaves a relatively smooth surface.", + "video_name": "VbNXh0GaLYo", + "timestamps": [ + 291 + ], + "3min_transcript": "because you can kinda view this as kind of a big cloud of gas, so they're always bumping into each other, but for the most part it was their angular velocity, and over the next tens of millions of years they'll slowly bump into each other and clump into each other. Even small particles have gravity, and they're gonna slowly become rocks and asteroids and, eventually, what we would call \"planetesimals,\" which are, kinda view them as seeds of planets or early planets, and then those would have a reasonable amount of gravity and other things would be attracted to them and slowly clump up to them. This wasn't like a simple process, you know, you could imagine you might have one planetesimal form, and then there's another planetesimal formed, and instead of having a nice, gentle those two guys accreting into each other, they might have huge relative velocities and ram into each other, and then just, you know, shatter, so this wasn't just a nice, gentle process of constant accretion. actually happened early in Earth's history, and we actually think this is why the Moon formed, so at some point you fast-forward a little bit from this, Earth would have formed, I should say, the mass that eventually becomes our modern Earth would have been forming. Let me draw it over here. So, let's say that that is our modern Earth, and what we think happened is that another proto-planet or another, it was actually a planet because it was roughly the size of Mars, ran into our, what it is eventually going to become our Earth. This is actually a picture of it. This is an artist's depiction of that collision, where this planet right here is the size of Mars, and it ran into what would eventually become Earth. This we call Theia. This is Theia, and what we believe happened, and if you look up, if you go onto the Internet, you'll see some simulations It wasn't a direct hit that would've just kinda shattered each of them and turned into one big molten ball. We think it was a glancing blow, something like this. This was essentially Earth. Obviously, Earth got changed dramatically once Theia ran into it, but Theia is right over here, and we think it was a glancing blow. It came and it hit Earth at kind of an angle, and then it obviously the combined energies from that interaction would've made both of them molten, and frankly they probably already were molten because you had a bunch of smaller collisions and accretion events and little things hitting the surface, so probably both of them during this entire period, but this would've had a glancing blow on Earth and essentially splashed a bunch of molten material out into orbit. It would've just come in, had a glancing blow on Earth, and then splashed a bunch of molten material, some of it would've been captured by Earth, so this is" + }, + { + "Q": "\nAt 8:18, Sal mentions we don't have any rocks or artifacts of any sort from the Hadean Eon. So what makes us sure we had a Hadeon Eon?", + "A": "Actually, in the last few decades of the 20th century, some rocks from this period were found in several countries. They were found in Greenland, Canada and west Australia. They were apparently altered due to volcanic dike. But, you see, there is proof that the Hadean Eon existed.", + "video_name": "VbNXh0GaLYo", + "timestamps": [ + 498 + ], + "3min_transcript": "kind of this molten, super hot ball, and some of it just gets splashed into orbit from the collision. Let me just see if I can draw Theia here, so Theia has collided, and it is also molten now because huge energies, and it splashes some of it into orbit. If we fast-forward a little bit, this stuff that got splashed into orbit, it's going in that direction, that becomes our Moon, and then the rest of this material eventually kind of condenses back into a spherical shape and is what we now call our Earth. So that's how we actually think right now that the Moon actually formed. Even after this happened, the Earth still had a lot more, I guess, violence to experience. Just to get a sense of where we are in the history of Earth, this time clock starts right here at the formation of our solar system, 4.6 billion years ago, probably coinciding with some type of supernova, and as we go clockwise on this diagram, we're moving forward in time, and we're gonna go all the way forward to the present period, and just so you understand some of the terminology, \"Ga\" means \"billions of years ago\" 'G' for \"Giga-\" \"Ma\" means \"millions of years ago\" 'M' for \"Mega-\" So where we are right now, the Moon has formed, and we're in what we call the Hadean period or actually I shouldn't say \"period.\" It's the Hadean eon of Earth. \"Period\" is actually another time period, so let me make this very clear. It's the Hadean, we are in the Hadean eon, and an eon is kind of the largest period of time that we talk about, especially relative to Earth, and it's roughly 500 million to a billion years well, from a geological point of view what makes it distinctive is really we don't have any rocks from the Hadean period. We don't have any kind of macroscopic-scale rocks from the Hadean period, and that's because at that time, we believe, the Earth was just this molten ball of kind of magma and lava, and it was molten because it was a product of all of these accretion events and all of these collisions and all this kinetic energy turning into heat. If you were to look at the surface of the Earth, if you were to be on the surface of the Earth during the Hadean eon, which you probably wouldn't want to be because you might get hit by a falling meteorite or probably burned by some magma, whatever, it would look like this, and you wouldn't be able to breathe anyway; this is what the surface of the Earth would look like. It would look like a big magma pool, and that's why we don't have any rocks from there because the rocks were just constantly being recycled, being dissolved and churned" + }, + { + "Q": "\nAt 9:50, during the late Hadean, Do these asteroid impacts contribute to the elemental makeup of our planet? I understand some metals are so heavy that they can only form in supernovae. But do these impacts provide enough energy to form new elements?", + "A": "Yes, it added compounds (like most of our water) and elements to the planet. But it DID NOT create new elements.. Elements are only created inside a sun or a supernova, An asteroid impact will never create a new element.", + "video_name": "VbNXh0GaLYo", + "timestamps": [ + 590 + ], + "3min_transcript": "well, from a geological point of view what makes it distinctive is really we don't have any rocks from the Hadean period. We don't have any kind of macroscopic-scale rocks from the Hadean period, and that's because at that time, we believe, the Earth was just this molten ball of kind of magma and lava, and it was molten because it was a product of all of these accretion events and all of these collisions and all this kinetic energy turning into heat. If you were to look at the surface of the Earth, if you were to be on the surface of the Earth during the Hadean eon, which you probably wouldn't want to be because you might get hit by a falling meteorite or probably burned by some magma, whatever, it would look like this, and you wouldn't be able to breathe anyway; this is what the surface of the Earth would look like. It would look like a big magma pool, and that's why we don't have any rocks from there because the rocks were just constantly being recycled, being dissolved and churned the Earth still is a giant molten ball, it's just we live on the super-thin, cooled crust of that molten ball. If you go right below that crust, and we'll talk a little bit more about that in future videos, you will get magma, and if you go dig deeper, you'll have liquid iron. I mean, it still is a molten ball. And this whole period is just a violent, not only was Earth itself a volcanic, molten ball, it began to harden as you get into the late Hadean eon, but we also had stuff falling from the sky and constantly colliding with Earth, and really just continuing to add to the heat of this molten ball. Anyway, I'll leave you there, and, as you can imagine, at this point there was no, as far as we can tell, there was no life on Earth. Some people believe that maybe some life could've formed in the late Hadean eon, but for the most part this was just completely inhospitable for any life forming. we'll talk a little bit about the Archean eon." + }, + { + "Q": "At 5:03 What was the evidence that was used to come up with the idea that collision occurred between Earth and Theia? How strong is the evidence?\n", + "A": "We have noticed that the Moon has a lot less heavy elements than the Earth and that the composition of the surface of the Moon is very similar to that of the Earth except that sometime in the past, it experienced extreme heat. This could mean a few things but it would seem to suggest that the Moon used to be part of the Earths mantle or crust and was launched into orbit by a powerful event.", + "video_name": "VbNXh0GaLYo", + "timestamps": [ + 303 + ], + "3min_transcript": "because you can kinda view this as kind of a big cloud of gas, so they're always bumping into each other, but for the most part it was their angular velocity, and over the next tens of millions of years they'll slowly bump into each other and clump into each other. Even small particles have gravity, and they're gonna slowly become rocks and asteroids and, eventually, what we would call \"planetesimals,\" which are, kinda view them as seeds of planets or early planets, and then those would have a reasonable amount of gravity and other things would be attracted to them and slowly clump up to them. This wasn't like a simple process, you know, you could imagine you might have one planetesimal form, and then there's another planetesimal formed, and instead of having a nice, gentle those two guys accreting into each other, they might have huge relative velocities and ram into each other, and then just, you know, shatter, so this wasn't just a nice, gentle process of constant accretion. actually happened early in Earth's history, and we actually think this is why the Moon formed, so at some point you fast-forward a little bit from this, Earth would have formed, I should say, the mass that eventually becomes our modern Earth would have been forming. Let me draw it over here. So, let's say that that is our modern Earth, and what we think happened is that another proto-planet or another, it was actually a planet because it was roughly the size of Mars, ran into our, what it is eventually going to become our Earth. This is actually a picture of it. This is an artist's depiction of that collision, where this planet right here is the size of Mars, and it ran into what would eventually become Earth. This we call Theia. This is Theia, and what we believe happened, and if you look up, if you go onto the Internet, you'll see some simulations It wasn't a direct hit that would've just kinda shattered each of them and turned into one big molten ball. We think it was a glancing blow, something like this. This was essentially Earth. Obviously, Earth got changed dramatically once Theia ran into it, but Theia is right over here, and we think it was a glancing blow. It came and it hit Earth at kind of an angle, and then it obviously the combined energies from that interaction would've made both of them molten, and frankly they probably already were molten because you had a bunch of smaller collisions and accretion events and little things hitting the surface, so probably both of them during this entire period, but this would've had a glancing blow on Earth and essentially splashed a bunch of molten material out into orbit. It would've just come in, had a glancing blow on Earth, and then splashed a bunch of molten material, some of it would've been captured by Earth, so this is" + }, + { + "Q": "at 06:43 so Thea formed into a ball fusing with earth, is that why earth is egg shaped?\n", + "A": "Earth is not egg shaped. It is very close to a sphere, with a small bulge near the equator. The bulge is the result of the rotation of the earth.", + "video_name": "VbNXh0GaLYo", + "timestamps": [ + 403 + ], + "3min_transcript": "It wasn't a direct hit that would've just kinda shattered each of them and turned into one big molten ball. We think it was a glancing blow, something like this. This was essentially Earth. Obviously, Earth got changed dramatically once Theia ran into it, but Theia is right over here, and we think it was a glancing blow. It came and it hit Earth at kind of an angle, and then it obviously the combined energies from that interaction would've made both of them molten, and frankly they probably already were molten because you had a bunch of smaller collisions and accretion events and little things hitting the surface, so probably both of them during this entire period, but this would've had a glancing blow on Earth and essentially splashed a bunch of molten material out into orbit. It would've just come in, had a glancing blow on Earth, and then splashed a bunch of molten material, some of it would've been captured by Earth, so this is kind of this molten, super hot ball, and some of it just gets splashed into orbit from the collision. Let me just see if I can draw Theia here, so Theia has collided, and it is also molten now because huge energies, and it splashes some of it into orbit. If we fast-forward a little bit, this stuff that got splashed into orbit, it's going in that direction, that becomes our Moon, and then the rest of this material eventually kind of condenses back into a spherical shape and is what we now call our Earth. So that's how we actually think right now that the Moon actually formed. Even after this happened, the Earth still had a lot more, I guess, violence to experience. Just to get a sense of where we are in the history of Earth, this time clock starts right here at the formation of our solar system, 4.6 billion years ago, probably coinciding with some type of supernova, and as we go clockwise on this diagram, we're moving forward in time, and we're gonna go all the way forward to the present period, and just so you understand some of the terminology, \"Ga\" means \"billions of years ago\" 'G' for \"Giga-\" \"Ma\" means \"millions of years ago\" 'M' for \"Mega-\" So where we are right now, the Moon has formed, and we're in what we call the Hadean period or actually I shouldn't say \"period.\" It's the Hadean eon of Earth. \"Period\" is actually another time period, so let me make this very clear. It's the Hadean, we are in the Hadean eon, and an eon is kind of the largest period of time that we talk about, especially relative to Earth, and it's roughly 500 million to a billion years" + }, + { + "Q": "\n@3:47 Sal draws a double bond? what is this? is this just showing they are all attached together? If they weren't drawn would this be incorrect?", + "A": "It s complicated. Benzene doesn t really have alternating single and double bonds as depicted but there isn t a very good way to show its true structure. Another common way is the hexagon but with a circle in the middle.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 227 + ], + "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have" + }, + { + "Q": "At 6:29, Sal draws a water molecule. I've always seen water molecules drawn wit the hydrogen atoms at the same angle. what is that angle and why is it at that angle\n", + "A": "this is to do with bonding angles. as the water molecules have two lone pairs and two bonding pairs, the electrons are trying to get far away as possible which is 107.5 degrees which is why Sal had drawn them that way. hope this helps", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 389 + ], + "3min_transcript": "but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have Well, if it's not drawn, then it must be a hydrogen. That's actually the convention that people use in organic chemistry. So there's multiple ways to do a structural formula, but this is a very typical one right over here. As you see, I'm just getting more and more and more information as I go from empirical to molecular to structural formula. Now, I want to make clear, that empirical formulas and molecular formulas aren't always different if the ratios are actually, also show the actual number of each of those elements that you have in a molecule. A good example of that would be water. Let me do water. Let me do this in a different color that I, well, I've pretty much already used every color. Water. So water we all know, for every two hydrogens, for every two hydrogens, and since I already decided to use blue for hydrogen let me use blue again for hydrogen, for every two hydrogens you have an oxygen. It just so happens to be, what I just wrote down I kind of thought of in terms of empirical formula, in terms of ratios, but that's actually the case. A molecule of hydrogen, sorry, a molecule of water has exactly two hydrogens and, and one oxygen. If you want to see the structural formula, you're probably familiar with it or you might be familiar with it. Each of those oxygens in a water molecule are bonded to two hydrogens, are bonded to two hydrogens. So hopefully this at least begins to appreciate different ways of referring to or representing a molecule." + }, + { + "Q": "At 3:45, Sal said that you have a double bond every other bond. Could someone explain why that is and what do double bonds mean?\n", + "A": "A double bond is just 2 bonds to the same atom, it s shown by 2 lines between the atoms instead of 1. With benzene it s a bit more complicated than this but it isn t worth worrying about yet.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 225 + ], + "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have" + }, + { + "Q": "\nAt 3:44 why we have a double bond in every other carbon atom", + "A": "That is a way of drawing the structure of benzene, which has what we call an aromatic ring. You will learn more about this when you start to study organic chemistry.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 224 + ], + "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have" + }, + { + "Q": "At 3:46, why are there three double bonds in the carbon hexagon ? What do they signify ?\n", + "A": "That is the structural formula for benzene. In a double bond, four electrons are shared between the carbon two atoms, compared with only two in a single bond. In fact, the structure of benzene is a little more complicated than shown and you will learn more about single and double bonds, and the benzene structure, in later videos. Benzene is not the best example that Sal could have chosen to explain the different types of formulae.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 226 + ], + "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have" + }, + { + "Q": "at 3:47, he says that every other bond is a double bond, but he doesn't explain why that is. How do we know when something has a double bond and in this case would be every other one?\n", + "A": "Yeah I think using this isn t the best example as this exact question seems to come up often. You know what a single covalent bond is right? 1 bond (2 shared electrons) between 2 atoms. Well a double bond is 2 covalent bonds (4 shared electrons) between atoms. You can t really know where to place those bonds simply from the formula once the number of carbon atoms gets this high, but if you continue with organic chemistry you are going to become very familiar with that benzene molecule.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 227 + ], + "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have" + }, + { + "Q": "\nAt 4:40 ; What is organic chemistry?", + "A": "It also involves the shape!", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 280 + ], + "3min_transcript": "That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have Well, if it's not drawn, then it must be a hydrogen. That's actually the convention that people use in organic chemistry. So there's multiple ways to do a structural formula, but this is a very typical one right over here. As you see, I'm just getting more and more and more information as I go from empirical to molecular to structural formula. Now, I want to make clear, that empirical formulas and molecular formulas aren't always different if the ratios are actually, also show the actual number of each of those elements that you have in a molecule. A good example of that would be water. Let me do water. Let me do this in a different color that I, well, I've pretty much already used every color. Water. So water we all know, for every two hydrogens, for every two hydrogens, and since I already decided to use blue for hydrogen let me use blue again for hydrogen, for every two hydrogens you have an oxygen." + }, + { + "Q": "2:35, How much evidence have scientists found for this theory? If that theory turns out true, that would be amazing!\n", + "A": "First thing I will say is that a theory is never proven. We can t test all conceivable circumstances that could happen and with historical events like the incorporation of ancient ancestor of the mitochondria into a cell we were not there to observe it. As for the mitochondria having been a separate bacteria like entity it is very well supported by what we know about bacteria and the mitochondria.", + "video_name": "i1dAnpSFbyI", + "timestamps": [ + 155 + ], + "3min_transcript": "and this is actually a little bit more of a textbook visualization, as we'll learn in a few minutes or seconds that we now have more sophisticated visualizations of what's actually going on inside of a mitochondria, but we haven't actually answered all of our questions, but you might have already learned that, so let me make it clear, these are mitochondria. That's the plural. If we're just talking about one of them, we're talking about a mitochondrion. That's the singular of mitochondria. But you might have already learned, some time in your past or in another Khan Academy video, that these are viewed as the ATP factories for cells. So let me right it this way. So ATP factories. A-T-P factories and if you watched the videos on ATP or cellular respiration or other videos, I'd repeatedly talk about how ATP is really the currency for energy in the cell that when it's in its ATP form you have adenosine triphosphate. If you pop one of the phosphate groups off, you pop one of the P's off, it release energy from movement to thinking to all sorts of things that actually go on in your bodies, so you can imagine mitochondria are really important for energy, for when the cell has to do things. And that's why you'll find more mitochondria in things like muscle cells, things that have to use a lot of energy. Now before I get into the structure of mitochondria, I wanna talk a little bit about its fascinating past because we think of cells as the most basic unit of life and that is true, that comes straight out of cell theory, but it turns out the most prevalent theory of how mitochondria got into our cells is that at one time the predecessors, the ancestors to our mitochondria, were free, independent organisms, microorganisms. So they're descendent from bacterial-like microorganisms that might have been living on their own and they were maybe really good at processing energy or maybe they were even good at other things, they got ingested by what the ancestors of our cells and instead of just being engulfed and being torn to shreds and kind of being digested and eaten, it was like, \"Hey, wait, if these things stick around, \"those cells are more likely to survive \"because they're able to help process glucose \"or help generate more energy out of things.\" And so the cells that were able to kind of live in symbiosis have them kind of give a place for the mitochondria to live or the pre-mitochondria, the ancestor mitochondria, those survived and then through kind of the processes of natural selection, this is what we now associate, we now associate eukaryotic cells as having mitochondria, so I find this whole idea of one organism being inside of another organism in symbiosis even at the cellular level, that's kind of mind-boggling, but anyway, I'll stop talking about that and now let's just talk about the present, let's talk about what the actual structure of mitochondria are." + }, + { + "Q": "\nIn 9:14, Sal mentions Glycolysis. What is that?", + "A": "Glycolysis is the process by which glucose or any other food substance is broken down for use in the creation of ATP", + "video_name": "i1dAnpSFbyI", + "timestamps": [ + 554 + ], + "3min_transcript": "if you're talking about one of these folds, you're talking about a crista, but if you're talking about more than one of them, you would call that a cristae, cristae. Sometimes I've seen the pronunciation of this as cristae, cristae or cristae, that's plural for crista. These are just folds in the inner membrane and once again the inner membrane is also a phospholipid bilayer. Now inside of the inner membranes, so between the outer membrane and the inner membrane you could imagine what this is gonna be called. That space is called the intermembrane space, not too creative of a name, intermembrane space and because of the porins, the small molecule concentration of the intermembrane space and then outside of the mitochondria, those concentrations are gonna be similar, but then the inner membrane does not have the porins in it and so you can actually have a different concentration on either side and that is essential for the electron transport chain. The electron transport chain really culminates with hydrogen, a hydrogen ion gradient being built between the two sides and then they flow down that gradient through a protein called ATP synthase which helps us synthesize ATP, but we'll talk more about that maybe in this video or in a future video, but let's finish talking about the different parts of a mitochondrion. So inside the inner membrane you have this area right over here is called the matrix. It's called, let me use this in a different color, this is the matrix and it's called the matrix 'cause it actually has a much higher protein concentration, it's actually more viscous than the cytosol that would be outside of the So this right over here is the matrix. When we we talk about cellular respiration, cellular respiration has many phases in it. We talk about glycolysis. Glycolysis is actually occurring in the cytosol. So glycolysis can occur in the cytosol. Glycolysis. But the other major phases of cellular respiration. Remember we talk about the citric acid cycle also known as the Krebs cycle, that is occuring in the matrix. So Krebs cycle is occuring in the matrix and then I said the electron transport chain which is really what's responsible for producing the bulk of the ATP, that is happening through proteins that are straddling the inner membrane or straddling the cristae right over here. Now we're just done. Probably one of the most fascinating parts of mitochondria, we said that we think that they are descendent" + }, + { + "Q": "at around 1:55 he says that there are 29 protons and electrons. are the amount of electrons and protons equal in every atom?\n", + "A": "The electrons and protons are equal in every neutral atom. If an atom loses or gains an electron, we call them ions .", + "video_name": "ZRLXDiiUv8Q", + "timestamps": [ + 115 + ], + "3min_transcript": "- [Voiceover] All right, now we're gonna talk about the idea of an electric current. The story about currents starts with the idea of charge. We've learned that we have two kinds of charges, positive and negative charge. We'll just make up two little charges like that. And we know if they're the opposite sign, that there'll be a force of attraction between them. And if they have two like signs, here's two charges that are both positive, and these charges are gonna repel each other. So this is the basic electrostatics idea, and the same thing for two minus charges. They also repel. So like charges repel, and unlike charges attract. That's one idea. We have the idea of charge. And now we need a place to get some charge. One of the places we like to get charge from is copper, copper wires. A copper atom looks like this. Copper atom has a nucleus and it also has electrons flying around the outside, electrons in orbits around the outside. So we'll draw the electrons like this. There'll be orbits around this nucleus. Pretty good circles. And there'll be electrons in these. Little minus signs. There's electrons stacked up in this. And even farther out, there's electrons. So there's kind of a interesting looking copper atom. Copper, the symbol for copper is Cu, and its atomic number is 29. That means there's 29 protons inside here, and there's 29 electrons outside. It turns out, just as a coincidence for copper, that the last orbital out here that guy right there. And that's the one that is the easiest to pull away from copper and have it go participate in conduction, in electric current. If I have a chunk of copper, every copper atom will have the opportunity to contribute one, this one lonely electron out here. If we look at another element, like for instance silver, silver has this same kind of electron configuration, where there's just one out here. And that's why silver and copper are such good, good conductors. Now we're gonna build, let's build a copper wire. Here's sort of a copper wire. It's just made of solid copper. It's all full of copper atoms." + }, + { + "Q": "At 1:10 he talks about electrons in orbits; I thought electrons are believed to flow around the nucleus in \"shells\" (delineated by mathematical probability) as opposed to defined orbits...\n", + "A": "Hello Chaba, You are correct about the mathematical probability. However, for a first introduction it is preferable to use the simplified model that defines orderly fixed orbits. It is easier to explain and visualize. Regards, APD", + "video_name": "ZRLXDiiUv8Q", + "timestamps": [ + 70 + ], + "3min_transcript": "- [Voiceover] All right, now we're gonna talk about the idea of an electric current. The story about currents starts with the idea of charge. We've learned that we have two kinds of charges, positive and negative charge. We'll just make up two little charges like that. And we know if they're the opposite sign, that there'll be a force of attraction between them. And if they have two like signs, here's two charges that are both positive, and these charges are gonna repel each other. So this is the basic electrostatics idea, and the same thing for two minus charges. They also repel. So like charges repel, and unlike charges attract. That's one idea. We have the idea of charge. And now we need a place to get some charge. One of the places we like to get charge from is copper, copper wires. A copper atom looks like this. Copper atom has a nucleus and it also has electrons flying around the outside, electrons in orbits around the outside. So we'll draw the electrons like this. There'll be orbits around this nucleus. Pretty good circles. And there'll be electrons in these. Little minus signs. There's electrons stacked up in this. And even farther out, there's electrons. So there's kind of a interesting looking copper atom. Copper, the symbol for copper is Cu, and its atomic number is 29. That means there's 29 protons inside here, and there's 29 electrons outside. It turns out, just as a coincidence for copper, that the last orbital out here that guy right there. And that's the one that is the easiest to pull away from copper and have it go participate in conduction, in electric current. If I have a chunk of copper, every copper atom will have the opportunity to contribute one, this one lonely electron out here. If we look at another element, like for instance silver, silver has this same kind of electron configuration, where there's just one out here. And that's why silver and copper are such good, good conductors. Now we're gonna build, let's build a copper wire. Here's sort of a copper wire. It's just made of solid copper. It's all full of copper atoms." + }, + { + "Q": "at 3:30 Hank says that all organelles are suspended in cytoplasm, but isn't cytosol the correct term for the fluid inside of the cell?\n", + "A": "No cytoplasm is the correct term.", + "video_name": "d9GkH4vpK3w", + "timestamps": [ + 210 + ], + "3min_transcript": "to more than 400 million years ago. These plants were lycophytes, which are still around today and which reproduce through making a bunch of spores, shedding them, saying a couple of Hail Marys, and hoping for the best. Some of these lycophytes went on to evolve into scale trees, which are now extinct, but huge, swampy forests of them used to cover the earth. Some people call these scale tree forests coal forest because there were so many of them, they were so dense and they covered the whole earth, that they eventually fossilized into giant seams of coal, which are very important to our lifestyles today. This is now called the Carboniferous Period. See what we did there? Because coal is made of carbon, so they named the epoch of geological history over how face-meltingly intense and productive these forests were. I would give my left eyeball, three fingers on my left hand, the middle ones so that I could hang loose, and my pinky toe, if I were able to go back and see these scale forests, because they would be freaking awesome. Anyway, angiosperms, or plants that used flowers the Cretaceous Period, about 65 million years ago, just as the dinosaurs were dying out. Which makes you wonder if, in fact, the first angiosperms assassinated all the dinosaurs. I'm not saying that's definitely what happened. I'm just, it's a little bit suspicious. Anyway, on a cellular level, plant and animal cells are actually pretty similar. They're called eukaryotic cells, which means they have a good kernel, and that kernel is the nucleus. Not \"nuculus\". The nucleus can be found in all sorts of cells: animal cells, plant cells, algae cells. Basically, all the popular kids. Eukaryotic cells are way more advanced than prokaryotic cells. We have the eukaryotic cell and we have the prokaryotic cell. Prokaryotic basically means before the kernel, pro-kernel. Then we have the eukaryotic, which means good kernel. The prokaryotes include your bacteria and your archea, which you've probably met before in your lifetime. Every time you've had strep throat, for example. Or if you've ever been in a hot spring, or an oil well or something, they're everywhere: Like I said, eukaryotes have that separately enclosed nucleus, that is, that all-important nucleus that contains its DNA and is enclosed by a separate membrane, because a eukaryotic cell is a busy place. There's chemical reactions going on in all different parts of the cell. It's important to keep those places divided up. Eukaryotic cells also have these little stuff-doing factories called organelles, because we decided that we'd name everything something weird. But organelles, and they're suspended in cytoplasm, continuing with the really esoteric terminology that you're going to have to know. Cytoplasm is mostly just water, but it's some other stuff, too. Well, basically, if you want to know about the structure of the eukaryotic cell, you should watch my video on animal cells, which let's just link to it right here. Plant and animal cells are very similar environments. They control themselves in very similar ways. But obviously plants and animals are very different things, so what are the differences in a plant cell that makes it so different from an animal? That's what we're going to go over now. First, plants are thought to have evolved from green algae," + }, + { + "Q": "At 5:59, Hank says that animals who can digest cellulose and lignin have a certain type of bacteria in their stomachs that break down these complex molecules into glucose molecules. What exactly are these bacteria, and how are they able to do this?\n", + "A": "Fibrobacter succinogenes, Ruminococci, etc They have the enzyme cellulase that can break down cellulose to get the nutrients they need.", + "video_name": "d9GkH4vpK3w", + "timestamps": [ + 359 + ], + "3min_transcript": "Something plants inherited from their ancestors was a rigid cell wall surrounding the plasma membrane of each cell. This cell wall of plants is mainly made out of cellulose and lignin, which are two really tough compounds. Cellulose is by far the most common and easy to find complex carbohydrate in nature, though if you were to include simple carbohydrates as well, glucose would win that one. This is because, fascinating fact, cellulose is in fact just a chain of glucose molecules. You're welcome. If you want to jog your memory about carbohydrates and other organic molecules, you can watch this episode right here. Anyway, as it happens, you know who needs carbohydrates to live? Animals. But you know what's a real pain in the ass to digest? Cellulose. Plants weren't born yesterday. Cellulose is a far more complex structure than you'll generally find in a prokaryotic cell, and it's also one of the main things that differentiates a plant cell from an animal cell. Animals do not have this rigid cell wall. They have a flexible membrane that frees them up to move around and eat plants and stuff. However, the cell wall gives structure and also protects it, to a degree. Which is why trees aren't squishy, and they don't giggle when you poke them. The combination of lignin and cellulose is what makes trees, for example, able to grow really, really freaking tall. Both of these compounds are extremely strong and resistant to deterioration. When we eat food, lignin and cellulose is what we call roughage because we can't digest it. It's still useful for us on certain aspects of our digestive system, but it's not nutritious. Which is why eating a stick is really unappetizing and like your shirt. This is a 100% plant shirt, but it doesn't taste good. We can't go around eating wood like a beaver, or grass like a cow, because our digestive systems just aren't set up for that. A lot of other animals that don't have access to delicious donut burgers have either developed gigantic stomachs, like sloths, or multipe stomachs, like goats, in order to make a living eating cellulose. These animals have a kind of bacteria in their stomachs that actually does the digestion of the cellulose for it. glucose molecules which can then be used for food. Other animals, like humans, mostly carnivores, don't have any of that kind of bacteria, which is why it's so difficult for us to digest sticks. But there is another reason why cellulose and lignin are very, very useful to us as humans. It burns, my friends. This is basically what would happen in our stomachs. It's oxidizing. It's producing the energy that we would get out of it if we were able to, except it's doing it very, very quickly. This is the kind of energy, like this energy that's coming out of it right now, is the energy that would be useful to us if we were cows. But we're not, so instead, we just use it to keep ourselves warm on the cold winter nights. (blows air) Ow; it's on me; ow. Anyway, while we animals are walking around, spending our lives searching for ever more digestable plant materials, plants don't have to do any of that. They just sit there and they make their own food. We know how they do that." + }, + { + "Q": "At 4:08 why are zeros that are in between other digits significant? Aren't those in-between zeros just placeholders?\n", + "A": "because there not at the very beginning they represent numbers in 705.001 the two zeros represent tenths and hundredths.", + "video_name": "eCJ76hz7jPM", + "timestamps": [ + 248 + ], + "3min_transcript": "look, I measured this far. If they didn't measure this far, they would have just left these 0's off. And they would have just told you 7 meters, not 7.00. Let's do the next one. So based on the same idea, we have the 5 and the 2. The non-zero digits are going to be significant figures. You don't include this leading 0, by the same logic that if this was 0.052 kilometers, this would be the same thing as 52 meters, which clearly only has two significant figures. So you don't want to count leading 0's before the first non-zero digit, I guess we could say. You don't want to include those. You just want to include all the non-zero digits and everything in between, and trailing 0's if a decimal point is involved. So over here, the person did 370. And then they wrote the decimal point. If they didn't write the decimal point, it would be a little unclear on how precise this was. But because they wrote the decimal point, it means that they measured it exactly to be 370. They didn't get 372 and then round down. Or they didn't have kind of a roughness only to the nearest tens place. This decimal tells you that all three of these are significant. So this is three significant figures over here. Then on this next one, once again, this decimal tells us that not only did we get to the nearest one, but then we put another trailing 0 here, which means we got to the nearest tenth. So in this situation, once again, we have three significant figures. Over here, the 7 is in the hundreds. But the measurement went all the way down to the thousandths place. And even though there are 0's in between, those 0's are part of our measurement, because they are in between non-zero digits. the way it's written, is a significant digit. So you have six significant digits. Now, this last one is ambiguous. The 37,000-- it's not clear whether you measured exactly 37,000. Maybe you measured to the nearest one, and you got an exact number. You got exactly 37,000. Or maybe you only measured to the nearest thousand. So there's a little bit of ambiguity here. If you just see something written exactly like this, you would probably say, if you had to guess-- or not guess. If there wasn't any more information, you would say that there's just two significant figures or significant digits. For this person to be less ambiguous, they would want to put a decimal point right over there. And that lets you know that this is actually five digits of precision, that we actually go to five significant figures. So if you don't see that decimal point, I would go with two." + }, + { + "Q": "At 3:51 I don't get how 10.0 is 3 significant digits because I thought zeros after the decimal or \"leading zeros\" is consider non-signifcant\n", + "A": "The 1 is a significant digit because it is not a zero. The 3rd zero is a significant digit because it is after a decimal point AND after a significant digit (the 1). Also, the 2nd zero is significant because this zero is between two other significant digits.", + "video_name": "eCJ76hz7jPM", + "timestamps": [ + 231 + ], + "3min_transcript": "look, I measured this far. If they didn't measure this far, they would have just left these 0's off. And they would have just told you 7 meters, not 7.00. Let's do the next one. So based on the same idea, we have the 5 and the 2. The non-zero digits are going to be significant figures. You don't include this leading 0, by the same logic that if this was 0.052 kilometers, this would be the same thing as 52 meters, which clearly only has two significant figures. So you don't want to count leading 0's before the first non-zero digit, I guess we could say. You don't want to include those. You just want to include all the non-zero digits and everything in between, and trailing 0's if a decimal point is involved. So over here, the person did 370. And then they wrote the decimal point. If they didn't write the decimal point, it would be a little unclear on how precise this was. But because they wrote the decimal point, it means that they measured it exactly to be 370. They didn't get 372 and then round down. Or they didn't have kind of a roughness only to the nearest tens place. This decimal tells you that all three of these are significant. So this is three significant figures over here. Then on this next one, once again, this decimal tells us that not only did we get to the nearest one, but then we put another trailing 0 here, which means we got to the nearest tenth. So in this situation, once again, we have three significant figures. Over here, the 7 is in the hundreds. But the measurement went all the way down to the thousandths place. And even though there are 0's in between, those 0's are part of our measurement, because they are in between non-zero digits. the way it's written, is a significant digit. So you have six significant digits. Now, this last one is ambiguous. The 37,000-- it's not clear whether you measured exactly 37,000. Maybe you measured to the nearest one, and you got an exact number. You got exactly 37,000. Or maybe you only measured to the nearest thousand. So there's a little bit of ambiguity here. If you just see something written exactly like this, you would probably say, if you had to guess-- or not guess. If there wasn't any more information, you would say that there's just two significant figures or significant digits. For this person to be less ambiguous, they would want to put a decimal point right over there. And that lets you know that this is actually five digits of precision, that we actually go to five significant figures. So if you don't see that decimal point, I would go with two." + }, + { + "Q": "\n0:50. He shows that there are only three significant digits in that number. How do you discern significant digits from insignificant digits?", + "A": "Significant digits are the numbers that actually give you the precision. The 0 s after the 7 tell you that the measurement is more precise, but the 0 s before are just basically placeholders. It is the first non-zero number then all the numbers, including zeros, after that which are significant figures.", + "video_name": "eCJ76hz7jPM", + "timestamps": [ + 50 + ], + "3min_transcript": "Let's see if we can learn a thing or two about significant figures, sometimes called significant digits. And the idea behind significant figures is just to make sure that when you do a big computation and you have a bunch of digits there, that you're not over-representing the amount of precision that you had, that the result isn't more precise than the things that you actually measured, that you used to get that result. Before we go into the depths of it and how you use it with computation, let's just do a bunch of examples of identifying significant figures. Then we'll try to come up with some rules of thumb. But the general way to think about it is, which digits are really giving me information about how precise my measurement is? So on this first thing right over here, the significant figures are this 7, 0, 0. So over here, you have three significant figures. And it might make you a little uncomfortable that we're not including these 0's that are after the decimal point and before this 7, that we're not including those. Because you're just like, that does help define the number. how precise our measurement is. And to try to understand this a little bit better, imagine if this right over here was a measurement of kilometers, so if we measured 0.00700 kilometers. This would be the exact same thing as 7.00 meters. Maybe, in fact, we just used a meter stick. And we said it's exactly 7.00 meters. So we measured to the nearest centimeter. And we just felt like writing it in kilometers. These two numbers are the exact same thing. They're just different units. But I think when you look over here, it makes a lot more sense why you only have three significant figures. These 0's are just shifting it based on the units of measurement that you're using. But the numbers that are really giving you the precision are the 7, the 0, and the 0. And the reason why we're counting these trailing 0's is that whoever wrote this number didn't have to write them down. look, I measured this far. If they didn't measure this far, they would have just left these 0's off. And they would have just told you 7 meters, not 7.00. Let's do the next one. So based on the same idea, we have the 5 and the 2. The non-zero digits are going to be significant figures. You don't include this leading 0, by the same logic that if this was 0.052 kilometers, this would be the same thing as 52 meters, which clearly only has two significant figures. So you don't want to count leading 0's before the first non-zero digit, I guess we could say. You don't want to include those. You just want to include all the non-zero digits and everything in between, and trailing 0's if a decimal point is involved." + }, + { + "Q": "\nAt 3:40, why was all three numbers in the number 10.0 significant figures when the number was rounded to the nearest tenth?", + "A": "When counting significant numbers, you count all the numbers, counting from the right of the first non-zero number. so as 1 is the first non-zero number and there are two zeros to the right of it you have 3 significant numbers. As opposed to the first example, which has 3 leading zeros and seven being the first non-zero number you would count 3 significant numbers in the number 0.00700", + "video_name": "eCJ76hz7jPM", + "timestamps": [ + 220 + ], + "3min_transcript": "look, I measured this far. If they didn't measure this far, they would have just left these 0's off. And they would have just told you 7 meters, not 7.00. Let's do the next one. So based on the same idea, we have the 5 and the 2. The non-zero digits are going to be significant figures. You don't include this leading 0, by the same logic that if this was 0.052 kilometers, this would be the same thing as 52 meters, which clearly only has two significant figures. So you don't want to count leading 0's before the first non-zero digit, I guess we could say. You don't want to include those. You just want to include all the non-zero digits and everything in between, and trailing 0's if a decimal point is involved. So over here, the person did 370. And then they wrote the decimal point. If they didn't write the decimal point, it would be a little unclear on how precise this was. But because they wrote the decimal point, it means that they measured it exactly to be 370. They didn't get 372 and then round down. Or they didn't have kind of a roughness only to the nearest tens place. This decimal tells you that all three of these are significant. So this is three significant figures over here. Then on this next one, once again, this decimal tells us that not only did we get to the nearest one, but then we put another trailing 0 here, which means we got to the nearest tenth. So in this situation, once again, we have three significant figures. Over here, the 7 is in the hundreds. But the measurement went all the way down to the thousandths place. And even though there are 0's in between, those 0's are part of our measurement, because they are in between non-zero digits. the way it's written, is a significant digit. So you have six significant digits. Now, this last one is ambiguous. The 37,000-- it's not clear whether you measured exactly 37,000. Maybe you measured to the nearest one, and you got an exact number. You got exactly 37,000. Or maybe you only measured to the nearest thousand. So there's a little bit of ambiguity here. If you just see something written exactly like this, you would probably say, if you had to guess-- or not guess. If there wasn't any more information, you would say that there's just two significant figures or significant digits. For this person to be less ambiguous, they would want to put a decimal point right over there. And that lets you know that this is actually five digits of precision, that we actually go to five significant figures. So if you don't see that decimal point, I would go with two." + }, + { + "Q": "\n2:20--How is it that there can be any kind of \"wind\" in space? Wind is movement between atoms, right? In space there's nothing to move....", + "A": "Why thank you, Andrew M.", + "video_name": "jEeJkkMXt6c", + "timestamps": [ + 140 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 7:27, he says voyager 1 has a speed about 17 km/s. So how can be New Horizons the fastest spacecraft with a speed of about 15 km/s?", + "A": "New Horizons has the fastest launch velocity of any spacecraft. However, Voyager 1 got several gravity assist boosts from the gas giant planets to its velocity, giving it a faster overall velocity.", + "video_name": "jEeJkkMXt6c", + "timestamps": [ + 447 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 1:25, it was stated high energy electrons were spewed out at 400km/s. isnt that faster than light? i thought nothing travels faster than light?", + "A": "That is a lot slower than light. Light speed in a vacuum is 299,792,458 m/s or 299,792.458 km/s. Empty space can expand faster than light, in the first few Planck times after the Big Bang, space was expanding at around 100-1000 light years per Planck time (light travels one Planck length per Planck Time, there are 5.85356655 \u00c3\u0097 10^50 planck lengths in a light year). However, space expanding faster than light does not violate Einstein s Special Relativity.", + "video_name": "jEeJkkMXt6c", + "timestamps": [ + 85 + ], + "3min_transcript": "" + }, + { + "Q": "6:42 If Alpha Centauri is closest, what's that really close light ball diagonal from the sun.\n", + "A": "This is a 2D picture of a 3D mapping of stars. That star would simply be within the line of sight of the Sun, but closer or further from the viewer than the Sun.", + "video_name": "jEeJkkMXt6c", + "timestamps": [ + 402 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 3:17, Sal says that light is able to pass through the half-silvered mirror. How can light pass through a mirror?", + "A": "Same way it passes through other things it passes through, like windows.", + "video_name": "3G_Q6AggQF8", + "timestamps": [ + 197 + ], + "3min_transcript": "and then the whole solar system is orbiting around the center of the galaxy at a nice clip. The galaxy itself might be moving, so if you have some absolute frame of reference that's defined by the ether, well we are going to be moving relative to it. And if we're moving relative to it well maybe you just measure the speed of light in different directions and see whether the speed of light is faster or slower in a certain direction and then that might help you identify-- well, one, validate that the ether exists-- but also think about what our velocity is relative to the ether, relative to that absolute frame of reference. But the problem in the 19th century is that we didn't have any precise way of actually measuring--or a precise enough way of measuring--the speed of light where we could detect the relative difference due to the light going for or against, or into or away from, the actual direction of the ether wind. with first kind of breaking things open, starting to really make a dent in this whole idea of a luminiferous ether, is the Michelson-Morley Experiment. Michelson-Morley Experiment. They recognized, okay, we can't measure the speed of light with enough precision to detect has it gotten slowed down by the ether wind or sped up by the ether wind, but what we could do, and this is what Michelson and Morley did do, and I'm gonna do an oversimplification of the experiment, is that, okay, you have a light source, you have a light source right over here. So, you have a light source. And so that's going to send light in this direction. It's going to send light just like that. And what you do is you have a half-silvered mirror that allows half the light to pass directly through it and half of it to be reflected. So let's put a half-silvered mirror right over here. So, there's a half-silvered mirror. and this is just a simplification of it. Let me do it a little neater than that. So half will bounce off like that. And then the other half will be able to go through it. Will be able to go through it. It's a half-silvered mirror. And then we make each of those light rays-- we've essentially taken our original light ray and split it into two-- well then we'll then bounce those off mirrors. Bounce those off mirrors that are equidistant. And there are some adjustments when you actually have to factor in everything, but just as a simple notion, these things are just now going to bounce back. So, this one is now going to bounce back. It's half-silvered, it can go through, or part of it can go through, that mirror. So that's that ray. And then this one is going to bounce back. This one's going to go bounce back. And part of it is going to bounce into this direction. And then you can detect what you see." + }, + { + "Q": "At 1:57, why is the information of carbon-13 being added to carbon-12's information?\n", + "A": "Because that s how the atomic weight of an element is calculated. (mass of isotope 1 x abundance of isotope 1) + (mass of isotope 2 x abundance of isotope 2) + ...", + "video_name": "EPvd-3712U8", + "timestamps": [ + 117 + ], + "3min_transcript": "- [Instructor] We have, listed here... We know that carbon 12 is the most common isotope of carbon on Earth. 98.89% of the carbon on Earth in carbon 12. And we know that, by definition, its mass is exactly 12 atomic mass units. Now that's not the only isotope of carbon on Earth. There are other isotopes. The next most frequent one is carbon 13. 1.11% of the carbon on Earth is carbon 13. And we can experimentally find that its mass is 13.0034 atomic mass units. So, these numbers that we have here, just as a review, these are atomic mass. These are atomic mass. And so, what we're gonna think about, in this video, is how do they come up with the atomic weight number that they'll give you on a periodic table like that? So, atomic weight. Well, in the video on atomic weight and on atomic mass, we see that the atomic weight is the weighted average of the atomic masses of the various isotopes of that element. So, to find this roughly 12.01, we take the weighted average of these two things. And what do we weight it by? We weight it by how common that isotope actually is. So, what we wanna do is, we could take 98.89% and multiply it by 12. And I'll rewrite this percentage as a decimal. So it'll be 0.9889 times 12. And, to that, we are going to add... We are going to add 1.11% times 13.0034. So, as a decimal, that's going to be 0.011. That's 1.11% is 0.011, oh, 111. So, what does that give us? Let's get our calculator out here. So, we are going to have 0.9889 times 12 is equal to 11.8668. And, to that, we are going to add... We are going to add 0.0111 times 13.0034. And I know it's going to do this multiplication first because it's a calculator knows about order of operations. And so, that's all going to be, as you can see, 12.01113774, which, if you were to round to the hundredths place, is how this atomic weight was gotten." + }, + { + "Q": "\nAt 9:00 and shortly after in the video, he refers to the snowball period in which the earth kind of iced over. My question is how did life survive this? Wouldn't that cold have negatively affected (rather, killed) both the prokaryotic and the eukaryotic organisms?", + "A": "could be but some could survive underground where it is always 64 degrees Fahrenheit so they would be able to survive the whole snowball timeline and when everything thaws out they can come on the surface again", + "video_name": "E1P79uFLCMc", + "timestamps": [ + 540 + ], + "3min_transcript": "by with ultraviolet radiation from the Sun, which is very inhospitable to DNA and to life. And so the only life at this point could occur in the ocean, where it was protected to some degree from the ultraviolet radiation. The land was just open to it. Anything on the land would have just gotten irradiated. It's DNA would get mutated. It just would not be able to live. So what happened, and what I guess has to happen, and the reason why we are able to live on land now is that we have an ozone layer. We have an ozone layer up in the upper atmosphere that helps absorb, that blocks most of the UV radiation from the Sun. And now that oxygen began to accumulate, we have the Oxygen Catastrophe. Oxygen accumulates in the atmosphere. Some of that oxygen goes into the upper atmosphere. So we're now in this time period right over here. It goes into the upper atmosphere. to turn into ozone, which then can help actually block the UV light. And I'll do another video maybe on the ozone/oxygen cycle. So this oxygen production, it's crucial, one, to having an ozone layer so that eventually life can exist on the land. And it's also crucial because eukaryotic organisms need that oxygen. Now, the third thing that happened, and this is also pretty significant event, we believe that that oxygen that started to accumulate in the atmosphere, reacted with methane in the atmosphere. So it reacted with methane. And methane is an ozone-- not an ozone. It's a greenhouse gas. It helps retain heat in the atmosphere. And once it reacts with the oxygen and starts dropping out of the atmosphere as methane, we believe the Earth cooled down. And it entered its first, and some people believe it's longest, snowball period. So that's what they talk about right here It's sometimes called the Huronic glaciation. And that happened because we weren't able to retain our heat, if that theory is correct. And so the whole-- as the theory goes-- the whole Earth essentially just iced over. So as we go through the Proterozoic Eon, I guess the big markers of it is it's the first time that we now have an oxygen-rich atmosphere. It's the first time that eukaryotes can now come into existence because they now have oxygen to, I guess we could say, breathe. And the other big thing is now this is where the ozone forms. So this kind of sets the stage for in the next eon, for animals or living things, to eventually get on to the land. And we'll talk about that in the next video." + }, + { + "Q": "At 6:11, why is tension considered an internal force? What is the difference between internal and external forces?\n", + "A": "My teacher taught me to just draw a big circle around the whole system you re trying to deal with. Anything outside of that circle is external, and anything inside is internal. If you drew a circle around both of the boxes and the string attaching them, the tension force is inside of the circle and thus internal.", + "video_name": "_0nDUXO0k7o", + "timestamps": [ + 371 + ], + "3min_transcript": "it's equal to the force of kinetic friction \"mu\" \"k\" times \"Fn\" and the \"mu\" \"k\" is going to be 0.2, you have to be careful because the \"Fn\" is not just equal to \"m\" \"g\" the reason is that on an incline the normal force points this way so the normal force doesn't have to counteract all of gravity on an incline it just has to counteract that component of gravity that's directed perpendicular to the incline and that happens to be \"m\" \"g\" \"cos(theta)\" for an object on an incline and if that makes no sense go back and look at the video on inclines or look at the article on inclines and you'll see that this component of gravity pointing into the surface is \"m\" \"g\" cosine that means that normal force is \"m\" \"g\" cosine. Because there's no acceleration in this perpendicular direction and I have to multiply by 0.2 because I'm not really plugging in the in this perpendicular direction. I'm plugging in the kinetic frictional force this 0.2 turns this perpendicular force into this parallel force, so I'm plugging in the force of kinetic friction and it just so happens that it depends on the normal force. That's why I'm plugging that in, I'm gonna need a negative 0.2 times 4 kg times 9.8 meters per second squared. And then I need to multiply by cosine of the angle in this case the angle is 30 degrees. Alright, now finally I divide by my total mass because I have no other forces trying to propel this system or to make it stop and my total mass is going to be 13 kg. You might object and think wait a minute, there's other forces here like this tension going this way, why don't we include that? benefit and appeal of treating this two-mass system as if it were a single mass is that we don't have to worry about these internal forces, it's there but that tension is also over here and on this side it's resisting the motion because it's pointing opposite the directional motion. On this side it's helping the motion, it's an internal force the internal force is canceled that's why we don't care about them, that's what this trick allows us to do by treating this two-mass system as a single object we get to neglect any internal forces because internal forces always cancel on that object. So if we just solve this now and calculate, we get 4.75 meters per second squared is the acceleration of this system. So this 4 kg mass will accelerate up the incline parallel to it with an acceleration of 4.75 meters per second squared. This 9 kg mass will accelerate downward" + }, + { + "Q": "At 1:42 he says hat you shouldn't open a plug without a professional, why not? what might happen?\n", + "A": "I would say only worry about that if you plan to plug it back in. if certain parts are messed up inside the plug it can be a fire hazard, parts may explode, or it may ruin some of the wiring in your house. If you make sure it is unplugged and you won t be plugging it back in you wouldn t need a professional.", + "video_name": "gFFvaLzhYew", + "timestamps": [ + 102 + ], + "3min_transcript": "Alright today we're going to take a look at the Conair 1875 hair dryer.We're going to look at the different systems and functions inside of it, how it was made and how it works. And we're also going to take a look at how they were able to produce a hair dryer for less than $8.00 and still make a profit and still stay in business as a company because that's a very low price and the way low price. And the way they've done that is they've reduced a lot of cost and complexity and we'll take a look at how they've done that. So the first thing I want to take a look at is the plug here. This is called a \"ground fault interruptor circuit plug\" and it has two different sized prongs right here. There's a larger prong and a smaller prong. And that's very important. The larger prong is the neutral prong and that means you can't plug this in incorrectly, it has to go in in only one way. And that means that the power is grounded properly. So the power always goes to ground and that's a critical thing in a circuit like this. So what this plug does it's actually pretty smart it can tell if there's a power difference might occur when the hair dryer was, say, dropped in water or there was some sort of short that happened. Inside the hair dryer there are open electrical contacts that if they're put into water or some other conductive fluid they'll short out. And it will cause the, you know, it'll electrify the fluid. And in the past that was a huge problem because people would get schocked or electrocuted and now it's not as big a deal because we have these ground fault interruptor circuits. So let's take a look at what's inside of that. And I've already popped this apart to some degree. I'm going to see if I can get it the rest of the way here. Now I want to say one thing really quick here from a safety stand point: It's absolutely critical that you DON'T take apart any plugs ever without a professionnal! And if you do have a professional and you do end up taking apart a plug like this make sure that you never ever moulded housing it was injection moulded. There were two pieces of steel that came together and the molten plastic was injected and you can see there are little pin marks right here. And pins came in inside the mold and pushed this part out. And then there's a little plastic piece here with a spring and that's for the test switch. So the test switch pushes on this part right here on the printed circuit board and the reset switch is right here. So you push on the reset switch and it will reset it so if it gets triggered you can still use your hair dryer again later. So one thing I want to take a look at here is the printed circuit board here. So we've got a lot of really cool things happening on this printed circuit board. It is made out of fiberglass. It's got a thin layer of copper applied to it. And then on top of the copper is a layer of lacquer. (The copper) Before they put the copper layer down they actually etch away parts of the copper. So" + }, + { + "Q": "\nHow would you name tert-butyl chloride in the UPAC system please? (7:41)", + "A": "First, find the longest carbon chain: 3 carbons, so it is a propane. Next, find the substituents: one methyl group at carbon 2 and one chloro group also at carbon 2. We name the substituents in alphabetical order, so the overall name is: 2-chloro-2-methylpropane.", + "video_name": "aaZ-isZs4ko", + "timestamps": [ + 461 + ], + "3min_transcript": "So let me write that in here, so dimethyl. This time we have a methyl group at two and five. So I'll write in here two and five. And then a bromine at five, so it'd be 5-bromo. So our full name would be 5-bromo-2,5-dimethylheptane. Now let's name this compound. So it's the same approach so let's count up our carbons here in our longest carbon chain. One, two, three, four, and five. So a five carbon alkane is pentane so I write in here pentane. Next I think about how do I number my carbon chain to give the lowest number possible to my first substituent? If I number it from the left, I give the bromine here a one. So that's the best way to number it. So that'd be one, two, three, four, and five. So I have a methyl group at four, a chlorine at three, and a bromine at one. So it's gonna be the bromine first, then the chlorine and then the methyl. So I'll go ahead and put the methyl in here. So this is at carbon four so 4-methyl. Next I have my chlorine at three so that would be 3-chloro. And finally my bromine at one, so 1-bromo. So the full name is 1-bromo-3-chloro-4-methylpentane. Now let's look at this compound. So we have one, two, three, four, five and six. So six carbons in our chain, a six carbon alkane is hexane. So I write in here hexane. Now I have two substituents, I have a bromine and I have a methyl group. So let's number from the left and see what happens. One, two, three, four, five and six. That gives me a bromine at two and a methyl at five. So this would be one, two, three, four, five, six. This gives me a methyl at two and a bromine at five. So just looking at numbers, we can't decide who wins here. So we have two versus two which is a tie, five versus five so that's a tie. So the way to break the tie is to think about the alphabet for your substituents. So we have bromine versus methyl so b versus m. Obviously b comes before m in the alphabet so the bromine's gonna win. We're gonna give the bromine the lower number. And that of course is the example on the left where the bromine is coming off of carbon two. So we're gonna choose the system on the left or the way of numbering the carbon chain from the left. Which means we have a methyl group at five. So 5-methylhexane. And then bromine at two, so 2-bromo." + }, + { + "Q": "The compound at 2:05 is a haloalkane so shouldn't it be named 3-Bromo-4,6-dimethylheptane? I was taught that the functional group is given first preference?\n", + "A": "You were taught wrong. A halogen substituent and an alkyl substituent have equal priority, so you number from the end that is closer to either one. IN A TIE, halogens take priority over alkyl groups. Hence, 2-bromo-4-methylhexane is correct, while 4-bromo-2-methylhexane is not.", + "video_name": "aaZ-isZs4ko", + "timestamps": [ + 125 + ], + "3min_transcript": "- [Lecturer] You often see two different ways to name alkele halides. And so we'll start with the common way first. So think about alkele halides. First you wanna think about an alkele group, and this alkele group is an ethyl group, there are two carbons on it. So we write in here ethyl. And then since it's alkele halides, you wanna think about the halogen you have So this is chlorine so it's gonna end in i, so chloride. So ethyl chloride would be the name for this compound. Now let's name the same molecule using IUPAC nomenclature. In this case it's gonna be named as a halo alkane. So for a two carbon alkane that would be ethane. So I write in here ethane. And of course our halogen is chlorine, so this would be chloro. So chloroethane is the name of this molecule. If I had fluorine instead of chlorine, it would be fluoroethane. So let me write in here fluoro, notice the spelling on that. If I had a bromine instead of the chlorine, And finally if I had an iodine instead of the chlorine, it would be iodoethane. So let me write in here ioto. Let's name this compound using our common system. So again think about the alkyl group that is present. So we saw in earlier videos this alkyl group is isopropyl. So I write in here isopropyl. And again we have chlorine attached to that. So it would be isopropyl chloride using the common system. If I'm naming this using the IUPAC system, I look for my longest carbon chain, so that'd be one, two and three. I know that is propane so I write in here propane. And we have a chlorine attached to carbon two. So that would be 2-chloro, 2-chloropropane. Let's look at how to classify alkyl halides. We find the carbon that's directly bonded to our halogen and we see how many alkyl groups There's only one alkyl group, this methyl group here, attached to this carbon so that's called primary. So ethyl chloride is an example of a primary alkyl halide. If you look at isopropyl chloride down here. This is the carbon that's bonded to our halogen and that carbon is bonded to two alkyl groups. So that's said to be a secondary alkyl halide. And let me draw in an example of another one here really fast. So for this compound the carbon that is bonded to our halogen is bonded to three alkyl groups. So three methyl groups here. So that's called a tertiary alkyl halide. And the name of this compound is tert-butyl chloride. So that's the common name for it. And that's the one that you see used most of the time. For larger molecules it's usually easier to use" + }, + { + "Q": "At 7:22 I got lost at the finding the ration of H2C2O4 AND NaOH how did he get this result of 1;2\n", + "A": "H2C2O4 + 2NaOH ----> Na2C2O4+2H20 The chemical equation.", + "video_name": "XjFNmfLv9_Q", + "timestamps": [ + 442 + ], + "3min_transcript": "And we know its concentration, 0.485 molar-- so let me do that in a different color-- 0.485 molar, this information allows us to figure out the actual molecules of sodium hydroxide. So we want to multiply this by-- we have 0.485 moles of sodium hydroxide for every 1 liter of this solution. That's what the molarity tells us. We have 0.485 moles per liter. So the liters cancel out, and then now we're going to actually have to get a calculator out. And this'll tell us how many moles of sodium hydroxide we have in this solution. So let me get my calculator. There we go. All right, let me just multiply these two numbers. So we have 0.03447 times 0.485 is equal to-- let me put this And we only have three significant digits here, so we're going to round to three significant digits. So we'll just go with 0.0167. So let me move that over off the screen. So this is going to be equal to 0.0167, and all we have left here are moles of sodium hydroxide. Now we know that this many moles of sodium hydroxide are going to completely react with however many moles of oxalic acid we have. Now we know that we need two moles of this for Or for every mole of oxalic acid that completely reacts, we need two moles of this. So let's write that down. And then you color. So we need two moles of sodium hydroxide, we got that from our balanced equation right there, and it's obvious it needs one mole, or one molecule will take this proton, and then you need another molecule to take that proton. So we need two moles of sodium hydroxide for every one mole of oxalic acid. For every one mole of H2C2O4. So essentially, we are just going to divide this number by 2. Let me get the calculator back. So we're just going to divide 0.0167 divided by 2. Once again, three significant digits 0.00835." + }, + { + "Q": "\nat 2:09, why do they pump blood into the vein not the artery?", + "A": "It because the Artery is under very high pressure, and the vein has very low. Does this help?", + "video_name": "Nnqp_3HMlDU", + "timestamps": [ + 129 + ], + "3min_transcript": "- [Voiceover] Let's say that this is a red blood cell. What makes up the outer layer of this red blood cell? Since it's a cell, it has a cell membrane, and that's made up of lipids. But embedded in those lipids, there's all kinds of proteins and molecules, some of which I'm drawing here, that have all kinds of different functions. There are two of them that are sort of more important than the others, at least for this topic that we're going to talk about, and those are the A molecule and the B molecule. I think you'll be pretty happy with those names, not too hard to remember. I'm actually calling them molecules and not proteins because they're actually not proteins. They are actually something called glycolipids, which I actually didn't realize at first. Glycolipids, glyco meaning a sugar group, lipid meaning a fat group, so it's some kind of mix of a sugar and fat. You can look it up if you're interested. have both of these molecules, these glycolipids, on their red blood cells. Some people do have both, but some have only one. For example, some people might have only the A, or, as you can imagine, some people would have only the B, and some people, can you figure out the last possibility? Some people have neither A nor B. Of course, all these people have all kinds of other proteins and molecules embedded in their red blood cells. The reason why we care about this and why I'm talking about these As and these Bs, is that in medicine, we often have to give blood transfusions. Let's say you got in a car accident and you lost a lot of blood. You're rushed to the hospital. If you've lost enough blood, they'll give you a transfusion of blood. Transfusion just means they'll put a needle in your vein and pump blood into your veins. and it has to do with these A and B groups. For example, it turns out, and we'll explain this, but for example, it turns out that if you are the kind of person who has this kind of blood that only has As on your blood cells, then it turns out that you can't get a blood transfusion from someone who has this kind of blood, As and Bs. Let's learn why. If you remember from the immune system, there's something called an antibody. We usually draw it in this shape here. If you remember from the immune system, your body has something called antibodies, and it uses these antibodies to fight things that it doesn't want in the body. For example, if you have a bacterium, I'm drawing one here, you'll have an antibody that will bind to that bacterium. The purpose of that is that now your body knows that is should destroy this bacterium. This antibody is kind of like a tag" + }, + { + "Q": "\non 4:23, shouldn't there also be a H2O at the left side of the equation?", + "A": "No, because the H20 (2 Hydrogens + 1 Oxygen) bonded with one more Hydrogen. Thus it s now H3O+ (positive balance since H is a proton). An equation features the same molecules but rearranged.", + "video_name": "Y4HzGldIAss", + "timestamps": [ + 263 + ], + "3min_transcript": "so the hydrogen is just going to be left as a hydrogen proton. And then the chlorine, the chlorine has just nabbed that electron. It had the electrons it had before, and then it just nabbed an electron from the hydrogen, and so it now has a negative charge, and these are both in aqueous solution still. It's still, they're still both dissolved in the water. And so you see very clearly here, you put this in an aqueous solution, you're going to increase the amount of, you're going to increase the amount of hydrogen ions, the amount of protons in the solution. And we've talked about this before, you'll often see a reaction written like this, but the hydrogen protons, they just don't sit there by themselves in the water. They are going to bond with the water molecules to actually form hydronium. So another way that you'll often see this is like this. You have the hydrochloric acid, hydrochloric acid. It's in an aqueous solution, and then you have the H2O. You have the water molecules, H2O, and you'll sometimes see written, okay, it's in its liquid form, and it's going to yield. Instead of just saying that you have a hydrogen ion right over here, you'll say, \"Okay, that thing, \"the hydrogen is actually gonna get bonded \"to a water molecule.\" And so what you're gonna be left with is actually H3O. Now this thing, this was a water molecule, and all it got was a hydrogen ion. All that is is a proton. It didn't come with any electrons, so now this is going to have a positive charge. It's going to have a positive charge, and we could now say that this is going to be in an aqueous solution, hydronium is going to be in an aqueous solution, and you're going to have plus, now you're going to still have the chloride ion, Chloride, chloride anion, and this is still in an aqueous solution. It is dissolved in water, and remember all that happened here is that the chlorine here took all of the electrons, leaving hydrogen with none. Then that hydrogen proton gets nabbed by a water molecule and becomes hydronium. So even by this definition you might say it increases the concentration of hydrogen protons. You could say it increase the concentration of hydronium, of hydronium right over here. Hydronium ions. So that makes, by the Arrhenius definition, that makes hydrochloric acid a strong acid. That makes it a strong acid. Now what would be a strong base by the Arrhenius definition of acids and bases? Well one would be sodium hydroxide. So let me write that down, so if I have sodium hydroxide," + }, + { + "Q": "\nAt 6:40 if the water splits the sodium hydroxide to OH- and Na +\nthen why the sodium ion dont react with the water? its an alkaline metal", + "A": "The polar nature of water causes water to form intermolecular (between molecule) bonds with the sodium, which is what causes it to split from hydroxide ion (OH-). The sodium ion doesn t react with water to form a new chemical substance because it is happy with eight outer shell electrons.", + "video_name": "Y4HzGldIAss", + "timestamps": [ + 400 + ], + "3min_transcript": "and then I have the hydroxide. That's an oxygen bonded to a hydrogen. So that's sodium hydroxide, and actually if you wanted to see what this molecule looked like you have a oxygen having a covalent bond to a hydrogen. Let me do these in different colors. Oxygen has a covalent bond to a hydrogen. to a hydrogen right over here. And it actually has three alone pairs. It actually has three alone pairs right over here. It's actually nabbed the electron from, from somebody some place, and so it's going to have a negative charge. It is going to have a negative charge. Actually I could write it both, It has a negative charge, and then you have a sodium ion that has lost its electron somehow. So you have a sodium ion that has lost an electron somehow, so it has a positive charge, to the oxygen right over here, making the oxygen negative and making the sodium positive, and so this is now positive, this is negative, they're going to be attracted to each other, and they form an ionic bond, so sodium hydroxide, they have an ionic bond because the sodium is actually positive, and the hydroxide part right over here. That is negative, and that's what draws them together, but anyway, you put this in an aqueous solution. You throw some sodium hydroxide into an aqueous solution, it will disassociate into, into sodium with its positive charge, the sodium ions, and actually you know the sodium ion is still part of this. That's what makes it attracted to the hydroxide anion, but it's still going to be in an aqueous solution, and then you're going to have the hydroxide. You're going to have the hydroxide anion, This has a negative charge, and it's still going to be dissolved in the water, so aqueous solution. So you throw sodium hydroxide in water, it's going to increase the concentration. It's going to increase the concentration of hydroxide in the water. It's going to increase the hydroxide concentration, and so by the Arrhenius definition of acids and bases, this would be a strong Arrhenius base. This would be a strong, a strong base by the Arrhenius definition. Now, and I encourage you to look at that relative to the other definitions, the Bronsted-Lowry definition of acids and bases and the Lewis definition of acids and bases, and see how you would think about categorizing things." + }, + { + "Q": "how does sal say in 5:03 that hcl is a strong acid?\n", + "A": "Arrhenius acids are proton donors, they completely ionize to give H+ and conjugate base. HCl - H+ Cl- H2SO4 - 2H+ SO4-2 HNO3- H+ - NO3-1 The key word is completely, they do not form an equilibrium.", + "video_name": "Y4HzGldIAss", + "timestamps": [ + 303 + ], + "3min_transcript": "and then you have the H2O. You have the water molecules, H2O, and you'll sometimes see written, okay, it's in its liquid form, and it's going to yield. Instead of just saying that you have a hydrogen ion right over here, you'll say, \"Okay, that thing, \"the hydrogen is actually gonna get bonded \"to a water molecule.\" And so what you're gonna be left with is actually H3O. Now this thing, this was a water molecule, and all it got was a hydrogen ion. All that is is a proton. It didn't come with any electrons, so now this is going to have a positive charge. It's going to have a positive charge, and we could now say that this is going to be in an aqueous solution, hydronium is going to be in an aqueous solution, and you're going to have plus, now you're going to still have the chloride ion, Chloride, chloride anion, and this is still in an aqueous solution. It is dissolved in water, and remember all that happened here is that the chlorine here took all of the electrons, leaving hydrogen with none. Then that hydrogen proton gets nabbed by a water molecule and becomes hydronium. So even by this definition you might say it increases the concentration of hydrogen protons. You could say it increase the concentration of hydronium, of hydronium right over here. Hydronium ions. So that makes, by the Arrhenius definition, that makes hydrochloric acid a strong acid. That makes it a strong acid. Now what would be a strong base by the Arrhenius definition of acids and bases? Well one would be sodium hydroxide. So let me write that down, so if I have sodium hydroxide, and then I have the hydroxide. That's an oxygen bonded to a hydrogen. So that's sodium hydroxide, and actually if you wanted to see what this molecule looked like you have a oxygen having a covalent bond to a hydrogen. Let me do these in different colors. Oxygen has a covalent bond to a hydrogen. to a hydrogen right over here. And it actually has three alone pairs. It actually has three alone pairs right over here. It's actually nabbed the electron from, from somebody some place, and so it's going to have a negative charge. It is going to have a negative charge. Actually I could write it both, It has a negative charge, and then you have a sodium ion that has lost its electron somehow. So you have a sodium ion that has lost an electron somehow, so it has a positive charge," + }, + { + "Q": "at 2:41 Shouldn't the 2-propanol be named as 1-methyl-1-ethanol since it would give the hydroxyl group the lowest number ie. 1?\n", + "A": "No, you pick the longest chain that has the OH attached and then give the lowest possible number to the C bearing the OH group. The longest chain has three C atoms, and the OH group is on C-2.", + "video_name": "kFpLDQfEg1E", + "timestamps": [ + 161 + ], + "3min_transcript": "Now, if that O-H weren't there, then we'd have just a three carbon alkane, which we would call propane. But since we have our O-H there, this is actually an alcohol. Alcohol is going to have the -ol ending. So this is called propanol. So let's go ahead and write propanol here. And the O-H group is coming off of carbon one. So we're going to say that's one propanol like that. How would we classify this alcohol? Well, the carbon right here that is bonded to the O-H, that carbon is bonded to one other carbon right here. So this would be a primary alcohol. So one propanol is a primary alcohol in terms of its classification. Let's look at a similar-looking molecule. Still three carbons, but this time we put the O-H on the carbon in the middle there. So once again, you're going to go ahead and number it. This is carbon one, this carbon two, this is carbon three. So it's also called propanol. The difference is the hydroxl group is on a different carbon, It's now on carbon two. So we're going to write two-propanol here, which is the IUPAC name. This is also called isopropanol, rubbing alcohol, it's all the same stuff. But two-propanol would be the proper IUPAC nomenclature. How would you classify two-propanol? So once again, we find the carbon attached to the O-H. That's this one. How many carbons is that carbon attached to? It's attached to one and two other carbons. So therefore, this is a secondary alcohol. So we have an example of a primary alcohol, and an example of a secondary alcohol here. Let's do a little bit more complicated nomenclature question. And so let's go ahead and draw out a larger molecule with more substituents. So let's put an O-H here. And then let's go ahead and do that as well. So give the full IUPAC name for this molecule. So you want to find the longest carbon chain that includes the O-H. OK so you have to find the longest carbon chain that includes the O-H, and you want to give the O-H the lowest number possible. So that's going to mean that you're going to start over here. And make this carbon number one like that. So if that's carbon number one, this must be carbon number two, three, four, five, six, and seven. So we have a seven-carbon alcohol. So seven-carbon alcohol would be heptanol. So we can go and start naming this. Make sure to give us plenty of space here. So we have heptanol. And we know that the O-H is coming off of carbon two. So we can go ahead and write two-heptanol like that. Let's look at the other substituents that we have. Well, what do we have right here coming off of our ring?" + }, + { + "Q": "At 4:52 as by the longest chain rule i think the name would be 6-chloro-6-methyl-4-octane-1-ethyl-ol or maybe 6-chloro-6-methyl-1-ethyl-ol-4-octane instead of what you said in the video : 5-chloro-5-methyl-3-propyl-2-heptanol ?? I am not sure if i am saying right or wrong but i have learned from previous videos that the chain should be longest.\n", + "A": "You took into consideration only the longest carbon chain rule and forgot about other rules. we can never have ethyl-ol it should be ethan- ol and according to you, that should be octan-1-ol. that s true that the chain should be longest but it also has been mentioned in the video that we have to provide the lowest locant rule to the -OH and other groups.", + "video_name": "kFpLDQfEg1E", + "timestamps": [ + 292 + ], + "3min_transcript": "And then let's go ahead and do that as well. So give the full IUPAC name for this molecule. So you want to find the longest carbon chain that includes the O-H. OK so you have to find the longest carbon chain that includes the O-H, and you want to give the O-H the lowest number possible. So that's going to mean that you're going to start over here. And make this carbon number one like that. So if that's carbon number one, this must be carbon number two, three, four, five, six, and seven. So we have a seven-carbon alcohol. So seven-carbon alcohol would be heptanol. So we can go and start naming this. Make sure to give us plenty of space here. So we have heptanol. And we know that the O-H is coming off of carbon two. So we can go ahead and write two-heptanol like that. Let's look at the other substituents that we have. Well, what do we have right here coming off of our ring? So that would be propyl. So we have three-propyl. So go ahead and write three-propyl in here. And what else do we have? At carbon five, we have two substituents. So we have a chloro group right here. And we have a methyl group right here. And remember your alphabet. Right, so C comes before M. So we can go ahead and put our methyl in there. All right, coming off of carbon five, so that would be five-methyl, like that. And then also coming off five, we have chloro. So five-chloro. Right in here. And that should do it. Everything follows the alphabet rule. So we have five-chloro, five-methyl, three-propyl, two-heptanol for this molecule. What about a problem that includes some stereochemistry? So let's say they give us one where we have to worry about stereochemistry. So let's see, something like that. And let's make an O-H group going away from us. And then let's go ahead and make this one coming out at us like that. So give the full IUPAC name for this molecule, and you have to include stereochemistry. So once again, find your longest carbon chain that includes your O-H group. And you want to give that O-H the lowest number possible so it takes precedence over things like alkyl groups, and halogens, and double bonds. So we're going to start from the left. So one, two, three, four, five, six, seven, eight, nine like that. So we have a nine-carbon alcohol. So that would be nonanol. And the alcohol is coming off of carbon three. The O-H is coming off of carbon three. So we have three-nonanol. Like that. So three-nonanol." + }, + { + "Q": "\nWhat greek character is Sal using at 7:30 when he's explaining the polarity of water and what does it mean? For instance, a theta usually means angle and delta means change.", + "A": "It is another form of the Greek letter delta and in this case means slightly or partially .", + "video_name": "Rw_pDVbnfQk", + "timestamps": [ + 450 + ], + "3min_transcript": "their electrons to roam, they all become slightly positive. And so they're kind of embedded in this mesh or this sea of electrons. And so the metallic crystals, depending on what cases you look at, sometimes they're harder than the ionic crystals, sometimes not. Obviously, we could list a lot of very hard metals, but we could list a lot of very soft metals. Gold, for example. If you take a screwdriver and a hammer, you know, pure gold, 24-carat gold, if you take a screwdriver and hit it onto the gold, it'll dent it, right? So this one isn't as brittle as the ionic crystal. It'll often mold to what you want to do with it. It's a little bit softer. Even if you talk about very hard metals, they tend to not be as brittle, because the sea of electrons kind of gives you a little give when you're moving around the metal. But that's not to say that it's not hard. In fact, sometimes that give that a metal has, or that ability to bend or flex, is what actually gives it its strength So the strength, and I've touched on this, it also goes into the boiling point. So because these bonds are pretty strong, it has a higher boiling point. If you just took salt crystal and tried to boil it, you'd have to add a lot of heat into the system. So this has a higher boiling point than say-- I mean, definitely things that have just van der Waals forces like the noble gases, but it'll also have a higher boiling point than, say, hydrogen fluoride. Hydrogen fluoride, if you remember from the last video, just had dipole-dipole forces. But what's interesting about this is they have a very high boiling point unless they're dissolved in water. So these are very hard, high boiling point, but the ionic crystals can actually be dissolved in water. And when they are dissolved in water, they form ionic dipole bonds. What does that mean? Ionic dipole or ionic polar bonds. -- and this is actually why it dissolves in water. Because the water molecule, we've gone over this tons of times, it has a negative end, because oxygen is hoarding the electrons, and then the hydrogen ends are positive because the electron's pretty stripped of it. So when you put these sodium and chloride ions in the room, or in the water solution, the positive sodiums want to get attracted to the negative side of this dipole, and then the negative chlorides, Cl minus, want to go near the hydrogens. So they kind of get dissolved in this. They don't necessarily want to be-- they still want to be attracted to each other, but they're still also attracted to different sides of the water, so it allows them to get dissolved and go with the flow of the water. So in this case, when you actually dissolve an ionic crystal into water," + }, + { + "Q": "Sal you mention at 7:53 that the Na+ is attracted to the o- of the water and Cl- is attracted towards the H+ side of water. My question is that why they get attracted to the o- and h+ poles of the water? Is this due to the more positive charge on the h+ than Na+ and cause Na have remaining 10 electrons in its outer shell? and the same case in Cl-.\n", + "A": "The Na+ is much more positive than the partial H+ in water, BUT many H2O molecules work together to seperate each Ion from the crystal lattice. So the attraction between 1 Na+ and 1 Cl- is stronger than 1 Cl- and one partial H+, many partial H+ polar molecules can pull the ions apart. This works the same for the Na+ and partial O- attraction as well.", + "video_name": "Rw_pDVbnfQk", + "timestamps": [ + 473 + ], + "3min_transcript": "So the strength, and I've touched on this, it also goes into the boiling point. So because these bonds are pretty strong, it has a higher boiling point. If you just took salt crystal and tried to boil it, you'd have to add a lot of heat into the system. So this has a higher boiling point than say-- I mean, definitely things that have just van der Waals forces like the noble gases, but it'll also have a higher boiling point than, say, hydrogen fluoride. Hydrogen fluoride, if you remember from the last video, just had dipole-dipole forces. But what's interesting about this is they have a very high boiling point unless they're dissolved in water. So these are very hard, high boiling point, but the ionic crystals can actually be dissolved in water. And when they are dissolved in water, they form ionic dipole bonds. What does that mean? Ionic dipole or ionic polar bonds. -- and this is actually why it dissolves in water. Because the water molecule, we've gone over this tons of times, it has a negative end, because oxygen is hoarding the electrons, and then the hydrogen ends are positive because the electron's pretty stripped of it. So when you put these sodium and chloride ions in the room, or in the water solution, the positive sodiums want to get attracted to the negative side of this dipole, and then the negative chlorides, Cl minus, want to go near the hydrogens. So they kind of get dissolved in this. They don't necessarily want to be-- they still want to be attracted to each other, but they're still also attracted to different sides of the water, so it allows them to get dissolved and go with the flow of the water. So in this case, when you actually dissolve an ionic crystal into water, not a lot of charge that is really movable in this state. But here, all of a sudden, we have these charged particles that can move. And because they can move, all of a sudden, when you put salt, sodium chloride, in water, that does become conductive. So anyway, I wanted you to be at least exposed to all of these different forms of matter. And now, you should at least get a sense when you look at something and you should at least be able to give a pretty good guess at how likely it is to have a high boiling point, a low boiling point, or is it strong or not. And the general way to look at it is just how strong are the intermolecular bonds. Obviously, if the entire structure is all one molecule, it's going to be super-duper strong. And on the other hand, if you're just talking about neon, a bunch of neon molecules, and all they have are the London dispersion forces, this thing's going to have ultra-weak bonds." + }, + { + "Q": "\nAt 1:37, what does covalent network mean?", + "A": "It means the atoms form a rigid network as Sal explains.", + "video_name": "Rw_pDVbnfQk", + "timestamps": [ + 97 + ], + "3min_transcript": "In the last video, I talked about some of the weaker intermolecular forces or structures of elements. The weakest, of course, was the London dispersion force. In this video, I'll start with the strongest structure, and that's the covalent network. So if you have a covalent network crystal and let me actually define the word crystal. Crystal is just when you have a solid, where the molecules that make up the solid are in a regular, relatively consistent pattern, and this is versus an amorphous solid, where everything is kind of just a hodge-podge and there's different concentrations of different things, of different ions, and different molecules, and different parts of the solid. So crystal is just a very regular structure. Ice is a crystal, because once you get the temperature low enough in water, the hydrogen bonds form a crystal, a regular structure. And we've talked about that a bunch. But the strongest of all crystal structures And the biggest, or the prime, example of that is carbon when it forms a diamond. So in the covalent network, carbon has four valence electrons, so it always wants four more. So when carbon shares with itself, it's very happy. So what it can do is it can form four bonds to four more carbons, and then each of those carbons can form four more bonds to four more carbons. And this one, 1, 2, 3, and it just keeps going on. This is the structure of a diamond. And the reason why this is such a strong structure is because you can almost view the entire -- in fact, you should view the entire diamond as one molecule, because they all have covalent bonds. These are actual sharing of electrons, and these are actually the strongest of all molecular bonds. So you can imagine if the entire solid you're going to have an extremely strong, extremely high boiling point substance, and that's why a diamond is so strong, and that's why it's so hard to boil a diamond. Now, the next two, and it depends on your special cases of the next most solid version of a solid, and it depends which case you're talking about, one are the ionic crystals, and I'll do them both here, because one isn't necessarily -- ionic crystal-- and the next is the metal. Well, it's not the next. They're kind of the metallic crystal. And these bonds, I mean, let's say the most common ionic molecule or -- that's not exactly the right word, because to some degree, let's say if I had some sodium and some chloride -- and just remember, what happens with sodium chloride is sodium here really has one extra electron" + }, + { + "Q": "At 2:15, what is the melting point of carbon?\n", + "A": "That depends on what allotrope of carbon it is and what the atmospheric pressure is. Graphite normally sublimes rather than boils. The temperature at which this happens depends on the pressure but it is usually above 3900 K. Diamond can be made to boil at very, very extreme pressures and temperatures. If you can get the pressure high enough, diamond will boil at over 5100 K, which is close to the temperature of the surface of the sun.", + "video_name": "Rw_pDVbnfQk", + "timestamps": [ + 135 + ], + "3min_transcript": "In the last video, I talked about some of the weaker intermolecular forces or structures of elements. The weakest, of course, was the London dispersion force. In this video, I'll start with the strongest structure, and that's the covalent network. So if you have a covalent network crystal and let me actually define the word crystal. Crystal is just when you have a solid, where the molecules that make up the solid are in a regular, relatively consistent pattern, and this is versus an amorphous solid, where everything is kind of just a hodge-podge and there's different concentrations of different things, of different ions, and different molecules, and different parts of the solid. So crystal is just a very regular structure. Ice is a crystal, because once you get the temperature low enough in water, the hydrogen bonds form a crystal, a regular structure. And we've talked about that a bunch. But the strongest of all crystal structures And the biggest, or the prime, example of that is carbon when it forms a diamond. So in the covalent network, carbon has four valence electrons, so it always wants four more. So when carbon shares with itself, it's very happy. So what it can do is it can form four bonds to four more carbons, and then each of those carbons can form four more bonds to four more carbons. And this one, 1, 2, 3, and it just keeps going on. This is the structure of a diamond. And the reason why this is such a strong structure is because you can almost view the entire -- in fact, you should view the entire diamond as one molecule, because they all have covalent bonds. These are actual sharing of electrons, and these are actually the strongest of all molecular bonds. So you can imagine if the entire solid you're going to have an extremely strong, extremely high boiling point substance, and that's why a diamond is so strong, and that's why it's so hard to boil a diamond. Now, the next two, and it depends on your special cases of the next most solid version of a solid, and it depends which case you're talking about, one are the ionic crystals, and I'll do them both here, because one isn't necessarily -- ionic crystal-- and the next is the metal. Well, it's not the next. They're kind of the metallic crystal. And these bonds, I mean, let's say the most common ionic molecule or -- that's not exactly the right word, because to some degree, let's say if I had some sodium and some chloride -- and just remember, what happens with sodium chloride is sodium here really has one extra electron" + }, + { + "Q": "Since the element would change if the amount of protons in an atom changed (7:31), then why is the subscript necessary? If all carbon atoms have 6 protons, why do we need to make a subscript that says it has 6 protons, if by definition carbon has 6 protons? Why doesn't just noting the element suffice?\n", + "A": "Yes it s redundant when you know the identity of the element. The only time you re likely to use atomic number as a subscript is in nuclear chemistry reactions.", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 451 + ], + "3min_transcript": "electrons and neutrons there are. So let's first think about protons. Well we know that the subscript is the atomic number and the atomic number is equal to the number of protons. So there are six protons in this atom of carbon. And if it's a neutral atom of carbon, the number of electrons must be equal to the number of protons. So if there are six protons, there must also be six electrons. And finally, how do we figure out the number of neutrons? Well let's go ahead and write down the formula we discussed. The mass number is equal to the atomic number plus the number of neutrons. So the mass number was right here, that's 12. So we can put in a 12. The atomic number was six, right here. So we put in a six. Plus the number of neutrons. So the number of neutrons is just equal to 12 minus six, which is, of course, six. So there are six neutrons. So just subtract the atomic number from the mass number and you'll get the number of neutrons in your atom. Let's do another one. This is carbon and this time we have a superscript of 13. The atomic number doesn't change when you're talking about an isotope. If you change the atomic number, you change the element. So there's still six protons in the nucleus of this atom and in a neutral atom, there must be the equal number of electrons. So six electrons and then finally, how many neutrons are there? Well just like we did before, we subtract the atomic number from the mass number. So we just have to 13 minus six So 13 minus six is, of course, seven. So there are seven neutrons in this atom. Another way to represent isotopes, let's say we wanted to represent this isotope in a different way, sometimes you'll see it where you write the name of the element. So this is carbon. And then you put a hyphen here and then you put the mass number. So carbon hyphen 13 refers to this isotope of carbon and this is called hyphen notation. So let me go ahead and write this hyphen notation. Alright, let's do one more example here. Let's do one that looks a little bit scarier. So let's do uranium. So U is uranium. The atomic number of uranium is 92. The mass number for this isotope is 235. So how many protons, electrons, and neutrons in this atom of uranium?" + }, + { + "Q": "\nAre there non-neutral atoms?At 6:21, Jay said,\"if it is a neutral atom.... It got me confused", + "A": "Of course. Atoms that aren t neutral are called ions. They are still atoms, they just do not have the same number of electrons and protons.", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 381 + ], + "3min_transcript": "And for protium, let's look at protium here. So in the nucleus there's only one proton and zero neutrons, so one plus zero gives us a mass number of one. And I'll use red here for mass number so we can distinguish. Alright, so mass number is red and let me use a different color here for the atomic number. Let me use magenta here. So the subscript is the atomic number and that's Z, and the superscript is the mass number and that's A. So this symbol represents the protium isotope. Let's draw one for deuterium. So it's hydrogen so we put an H here. There is still one proton in the nucleus, right one proton in the nucleus, so we put an atomic number of one. The mass number is the superscript, So we look in the nucleus here. There's one proton and one neutron. So one plus one is equal to two. So we put a two here for the superscript. And finally for tritium, it's still hydrogen. So we put hydrogen here. There's one proton in the nucleus, atomic number of one, so we put a one here. And then the combined numbers of protons and neutrons, that would be three. So one proton plus two neutrons gives us three. So there's the symbol for tritium. So here are the isotopes of hydrogen and using these symbols allows us to differentiate between them. So let's take what we've learned and do a few more practice problems here. So let's look at a symbol for carbon. So here we have carbon with subscript six, superscript 12. electrons and neutrons there are. So let's first think about protons. Well we know that the subscript is the atomic number and the atomic number is equal to the number of protons. So there are six protons in this atom of carbon. And if it's a neutral atom of carbon, the number of electrons must be equal to the number of protons. So if there are six protons, there must also be six electrons. And finally, how do we figure out the number of neutrons? Well let's go ahead and write down the formula we discussed. The mass number is equal to the atomic number plus the number of neutrons. So the mass number was right here, that's 12. So we can put in a 12. The atomic number was six, right here. So we put in a six. Plus the number of neutrons." + }, + { + "Q": "At 2:41 the third isotope is called tritium, I'm wondering if there are specialized names for each element's isotopes or if all elements use the same names. For carbon to have more/less neutrons are its isotopes also called deuterium and tritium?\n", + "A": "Only hydrogen isotopes had the privilege to be called with their own names. All other elements isotopes are simply called by the element s name and the atomic mass (for instance carbon fourteen, uranium twohundred and thirtyfive, and so on).", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 161 + ], + "3min_transcript": "So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen. Well we're going to write little symbols to represent these isotopes. And so the symbol that we'll draw here for protium is going to have the element symbol, which is, of course, hydrogen, and then down here we're going to write the atomic number. So the subscript is the atomic number which is one, because there's one proton in the nucleus, and then for the superscript, we're going to write in the mass number. So let me move down here so we can look at the definition for the mass number. The mass number is the combined number of protons and neutrons in a nucleus, so it's protons and neutrons, and it's symbolized by A. So A is the mass number, which is equal to the number of protons, that's the atomic number which we symbolized by Z, plus the number of neutrons." + }, + { + "Q": "At 2:10, how do a neutrons have mass? Isn't neutrons part of an atom?\n", + "A": "Yes, neutrons have mass. An isolated neutron is about 1.008665 u. And neutrons are found in all atoms beyond Hydrogen-1.", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 130 + ], + "3min_transcript": "So the atomic number is symbolized by Z and it refers to the number of protons in a nucleus. And you can find the atomic number on the periodic table. So we're going to talk about hydrogen in this video. So for hydrogen, hydrogen's atomic number is one. So it's right here, so there's one proton in the nucleus of a hydrogen atom. In a neutral atom, the number of protons is equal to the number of electrons, because in a neutral atom there's no overall charge and the positive charges of the protons completely balance with the negative charges of the electrons. So let's go ahead and draw an atom of hydrogen. We know the atomic number of hydrogen is one, so there's one proton in the nucleus. So there's my one proton in the nucleus, and we're talking about a neutral hydrogen atom, so there's one electron. I'm going to draw that one electron somewhere outside the nucleus and I'm going to use the oversimplified Bohr model. So this isn't actually what an atom looks like, So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen." + }, + { + "Q": "At 1:17, why do the isotopes of hydrogen all have different names (protium, deuterium, tritium), and where do they come from?\n", + "A": "They have names for convenience. Pro means 1 Deu means 2 Tri means 3 See?", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 77 + ], + "3min_transcript": "So the atomic number is symbolized by Z and it refers to the number of protons in a nucleus. And you can find the atomic number on the periodic table. So we're going to talk about hydrogen in this video. So for hydrogen, hydrogen's atomic number is one. So it's right here, so there's one proton in the nucleus of a hydrogen atom. In a neutral atom, the number of protons is equal to the number of electrons, because in a neutral atom there's no overall charge and the positive charges of the protons completely balance with the negative charges of the electrons. So let's go ahead and draw an atom of hydrogen. We know the atomic number of hydrogen is one, so there's one proton in the nucleus. So there's my one proton in the nucleus, and we're talking about a neutral hydrogen atom, so there's one electron. I'm going to draw that one electron somewhere outside the nucleus and I'm going to use the oversimplified Bohr model. So this isn't actually what an atom looks like, So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen." + }, + { + "Q": "\nAt 3:20, does he mean that protium and deuterium, and tritium are specific words used only for NITROGEN atoms, or does he mean you can use these words on any element?\nThank you!", + "A": "They are specific terminology for Hydrogen only. Hydrogen with 1 proton and 0 neutrons= Protium, Hydrogen with 1 proton and 1 neutron = Deuterium, Hydrogen with 1 proton and 2 neutrons =tritium.", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 200 + ], + "3min_transcript": "So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen. Well we're going to write little symbols to represent these isotopes. And so the symbol that we'll draw here for protium is going to have the element symbol, which is, of course, hydrogen, and then down here we're going to write the atomic number. So the subscript is the atomic number which is one, because there's one proton in the nucleus, and then for the superscript, we're going to write in the mass number. So let me move down here so we can look at the definition for the mass number. The mass number is the combined number of protons and neutrons in a nucleus, so it's protons and neutrons, and it's symbolized by A. So A is the mass number, which is equal to the number of protons, that's the atomic number which we symbolized by Z, plus the number of neutrons." + }, + { + "Q": "\nAt 8:57, he says the atomic mass is 235, but isn't the atomic mass 238.03?", + "A": "He says 235 is the mass number of this isotope The relative atomic mass or atomic weight of uranium is 238.03. These are not the same thing!", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 537 + ], + "3min_transcript": "So the number of neutrons is just equal to 12 minus six, which is, of course, six. So there are six neutrons. So just subtract the atomic number from the mass number and you'll get the number of neutrons in your atom. Let's do another one. This is carbon and this time we have a superscript of 13. The atomic number doesn't change when you're talking about an isotope. If you change the atomic number, you change the element. So there's still six protons in the nucleus of this atom and in a neutral atom, there must be the equal number of electrons. So six electrons and then finally, how many neutrons are there? Well just like we did before, we subtract the atomic number from the mass number. So we just have to 13 minus six So 13 minus six is, of course, seven. So there are seven neutrons in this atom. Another way to represent isotopes, let's say we wanted to represent this isotope in a different way, sometimes you'll see it where you write the name of the element. So this is carbon. And then you put a hyphen here and then you put the mass number. So carbon hyphen 13 refers to this isotope of carbon and this is called hyphen notation. So let me go ahead and write this hyphen notation. Alright, let's do one more example here. Let's do one that looks a little bit scarier. So let's do uranium. So U is uranium. The atomic number of uranium is 92. The mass number for this isotope is 235. So how many protons, electrons, and neutrons in this atom of uranium? at the atomic number, that's 92. So there must be 92 protons. In a neutral atom, the number of electrons is equal to the number of protons. So there are 92 electrons and then finally, to figure out the number of neutrons, we subtract this number from the mass number. So we just need to do 235 minus 92. And that gives us 143. So there are 143 neutrons." + }, + { + "Q": "What is meant by superscript ? said at 3:27\n", + "A": "A superscript is just text set above the line and smaller than usual, like the 2 in x\u00c2\u00b2. Similarly, a subscript is like the 0 in x\u00e2\u0082\u0080. I used special characters to make this work, so you can copy the \u00c2\u00b2 if you want to use it elsewhere.", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 207 + ], + "3min_transcript": "So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen. Well we're going to write little symbols to represent these isotopes. And so the symbol that we'll draw here for protium is going to have the element symbol, which is, of course, hydrogen, and then down here we're going to write the atomic number. So the subscript is the atomic number which is one, because there's one proton in the nucleus, and then for the superscript, we're going to write in the mass number. So let me move down here so we can look at the definition for the mass number. The mass number is the combined number of protons and neutrons in a nucleus, so it's protons and neutrons, and it's symbolized by A. So A is the mass number, which is equal to the number of protons, that's the atomic number which we symbolized by Z, plus the number of neutrons." + }, + { + "Q": "At 0:07, he talks about the Atomic Number (z), why do scientists use letters to symbolize numbers or elements? Just curious.\n", + "A": "i think its because it helps in writing chemical formula and equation ..its easier to use them symbolically coz there are so many elements and compounds formed from them(elements)", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 7 + ], + "3min_transcript": "So the atomic number is symbolized by Z and it refers to the number of protons in a nucleus. And you can find the atomic number on the periodic table. So we're going to talk about hydrogen in this video. So for hydrogen, hydrogen's atomic number is one. So it's right here, so there's one proton in the nucleus of a hydrogen atom. In a neutral atom, the number of protons is equal to the number of electrons, because in a neutral atom there's no overall charge and the positive charges of the protons completely balance with the negative charges of the electrons. So let's go ahead and draw an atom of hydrogen. We know the atomic number of hydrogen is one, so there's one proton in the nucleus. So there's my one proton in the nucleus, and we're talking about a neutral hydrogen atom, so there's one electron. I'm going to draw that one electron somewhere outside the nucleus and I'm going to use the oversimplified Bohr model. So this isn't actually what an atom looks like, So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen." + }, + { + "Q": "At 5:41 why can you not have more than three isotopes or can you and you just did not show it?\n", + "A": "There is no set number of isotopes per element. Each element has whatever number of naturally occurring isotopes it just happens to have. With hydrogen, there are three naturally occurring isotopes. There are a few more that have been made artificially, but they are so radioactive they only exist for a tiny fraction of a second. On the other hand, mercury has seven stable, naturally occurring isotopes and dozens of radioactive isotopes (mostly artificially made isotopes).", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 341 + ], + "3min_transcript": "And for protium, let's look at protium here. So in the nucleus there's only one proton and zero neutrons, so one plus zero gives us a mass number of one. And I'll use red here for mass number so we can distinguish. Alright, so mass number is red and let me use a different color here for the atomic number. Let me use magenta here. So the subscript is the atomic number and that's Z, and the superscript is the mass number and that's A. So this symbol represents the protium isotope. Let's draw one for deuterium. So it's hydrogen so we put an H here. There is still one proton in the nucleus, right one proton in the nucleus, so we put an atomic number of one. The mass number is the superscript, So we look in the nucleus here. There's one proton and one neutron. So one plus one is equal to two. So we put a two here for the superscript. And finally for tritium, it's still hydrogen. So we put hydrogen here. There's one proton in the nucleus, atomic number of one, so we put a one here. And then the combined numbers of protons and neutrons, that would be three. So one proton plus two neutrons gives us three. So there's the symbol for tritium. So here are the isotopes of hydrogen and using these symbols allows us to differentiate between them. So let's take what we've learned and do a few more practice problems here. So let's look at a symbol for carbon. So here we have carbon with subscript six, superscript 12. electrons and neutrons there are. So let's first think about protons. Well we know that the subscript is the atomic number and the atomic number is equal to the number of protons. So there are six protons in this atom of carbon. And if it's a neutral atom of carbon, the number of electrons must be equal to the number of protons. So if there are six protons, there must also be six electrons. And finally, how do we figure out the number of neutrons? Well let's go ahead and write down the formula we discussed. The mass number is equal to the atomic number plus the number of neutrons. So the mass number was right here, that's 12. So we can put in a 12. The atomic number was six, right here. So we put in a six. Plus the number of neutrons." + }, + { + "Q": "\nAt 8:15 we are shown another way to depict an isotope is by writing it like, carbon-14, for instance. Would it not be simplest to just use the atomic symbol and keep the mass number in the superscript since the atomic number in subscript will never change?", + "A": "You can use whatever method is most convenient. It is simpler to avoid superscripts in typing, but you can t avoid them when writing nuclear equations.", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 495 + ], + "3min_transcript": "electrons and neutrons there are. So let's first think about protons. Well we know that the subscript is the atomic number and the atomic number is equal to the number of protons. So there are six protons in this atom of carbon. And if it's a neutral atom of carbon, the number of electrons must be equal to the number of protons. So if there are six protons, there must also be six electrons. And finally, how do we figure out the number of neutrons? Well let's go ahead and write down the formula we discussed. The mass number is equal to the atomic number plus the number of neutrons. So the mass number was right here, that's 12. So we can put in a 12. The atomic number was six, right here. So we put in a six. Plus the number of neutrons. So the number of neutrons is just equal to 12 minus six, which is, of course, six. So there are six neutrons. So just subtract the atomic number from the mass number and you'll get the number of neutrons in your atom. Let's do another one. This is carbon and this time we have a superscript of 13. The atomic number doesn't change when you're talking about an isotope. If you change the atomic number, you change the element. So there's still six protons in the nucleus of this atom and in a neutral atom, there must be the equal number of electrons. So six electrons and then finally, how many neutrons are there? Well just like we did before, we subtract the atomic number from the mass number. So we just have to 13 minus six So 13 minus six is, of course, seven. So there are seven neutrons in this atom. Another way to represent isotopes, let's say we wanted to represent this isotope in a different way, sometimes you'll see it where you write the name of the element. So this is carbon. And then you put a hyphen here and then you put the mass number. So carbon hyphen 13 refers to this isotope of carbon and this is called hyphen notation. So let me go ahead and write this hyphen notation. Alright, let's do one more example here. Let's do one that looks a little bit scarier. So let's do uranium. So U is uranium. The atomic number of uranium is 92. The mass number for this isotope is 235. So how many protons, electrons, and neutrons in this atom of uranium?" + }, + { + "Q": "Just to make sure I did not misunderstood this: At 1:19, we have four covalent bond for ammonium but calculated that we need 9 valence electrons.\nIs the reason that there are only 8 electrons (covalent bonds) because we have a cation and therefore took one electron \"away\", as in 9-1?\nI know it is explained pretty clearly but I want to make sure that I understood the reasoning behind it.\n", + "A": "Yes that is it. It is a +1 cation so there is 1 less electron than protons, so there s 8 valence electrons.", + "video_name": "dNPs-cr_6Bk", + "timestamps": [ + 79 + ], + "3min_transcript": "The previous video, we saw some steps for drawing dot structures. In this video we're going to use the same steps to draw a few more structures. But we're also going to talk about how formal charge relates to dot structure. So we'll get back to this definition in a minute. Right now, let's draw a quick dot structure for the ammonium cation. So NH4 plus. The first thing you do is find the total number of valence electrons. And so to do that you look at a periodic table and find nitrogen, which is in group five. Therefore, nitrogen has five valence electrons. Right, each hydrogen has one. And we have four of them. So we have 5 plus 4, giving us 9 electrons. However, there's a plus 1 charge. Meaning this is a cation, meaning we're going to lose an electron here. So instead of representing nine in our dot structure, going to represent eight electrons. And so let's go ahead and put the nitrogen at the center. Remember you put the least electronegative atom at the center, except for hydrogen. So nitrogen is going to go in the center here. And we know it's going to have bonds to four hydrogens, so we go ahead and put in those hydrogens right here. we've used up in our dot structure. Two, four, six, and eight. So that takes care of all eight valence electrons that we were supposed to represent. So this is the structure. And I can go ahead and put some brackets around it here. And also a plus 1 charge to indicate that this is an ion. And so that's the dot structure for the ammonium cation here. Let's see if we can assign formal charges to the nitrogen and the hydrogen. So I'm going to go ahead and redraw our dot structure here. And I'm also going to draw in the electrons, right? We know that each of those covalent bonds consists of two electrons. I'm going to go ahead and put in those two electrons right here. And if I want to find a formal charge for, let's say, the central nitrogen. What I would do is think about the number of valence electrons in the free atoms. So if you had a nitrogen all by itself, right? You look at the periodic table, it's in group five. And so therefore we're talking about five valence electrons From that number we're going to subtract the number of valence electrons in the bonded atom. And the way to approach that is to look at your dot structure here, and think about those two electrons in those covalent bonds. One of them we're going to assign to the hydrogen, and one of them are going to assign to the nitrogen. And so we go around, we do that for each one of our covalent bonds like that. And so now, we can see that nitrogen is surrounded by four valence electrons in the bonded atom. So let me go ahead and write that. So it's 5 minus 4. And so 5 minus 4 is of course plus 1. So we have a plus 1 of formal charge on the nitrogen. So this nitrogen as a plus 1 formal charge. Now let's do it for hydrogen here. So hydrogen's in group one on the periodic table. So let me just point this out. This is for . Nitrogen and then for hydrogen. It's in group one. So one valence electron in the free atom. And from that we're going to subtract a number of valence electrons in the bonded atom." + }, + { + "Q": "\nAt 9:49, he mentions significant figures. What are significant figures? My science teacher didn't exactly explain them properly. And she explained them differently.", + "A": "At 9:45, when he is trying to round the pKa to 2 significant figures, he actually rounds to 4 significant figures (10.57 has 4 significant figures). The pKa correctly rounded to 2 significant figures would be 11 (I rounded 10.5686362358 up to 11).", + "video_name": "DGMs81-Rp1o", + "timestamps": [ + 589 + ], + "3min_transcript": "as the log of Ka plus the log of Kb equal to the log of Kw. If we take the negative of everything, let's go ahead and do that, the negative of everything, negative log of Ka. I'll put that in parenthesis, plus the negative log of Kb is equal to the negative log of Kw. The negative log of Ka, we know that this is equal to the pKa. The negative log of Ka was our definition for our pKa, and the negative log of Kb was our definition for pKb. pKa plus pKb is equal to finally the negative log of Kw. That would give you 14. 14.00. The negative log of 1.0 times 10 to the negative 14 Now we have something else that we can work with, so let me go ahead and box this right here. Let's take the Ka value that we just found. Let's find the pKa. The pKa would be equal to the negative log of 2.7 times 10 to the negative 11. Let's do that on our calculator here. Let's get some room. The negative log of 2.7 times 10 to the negative 11 gives us 10.57. We have to round that. 10.57. These are our two significant figures because we have two significant figures here. Let's go back up to our problem here, so that's the pKa for the methylammonium ion. Let's say you're given the pKa for the methylammonium ion What is the pKb for methylamine? All we have to do is plug in to our equation. 10.57 was our pKa value. Let's go ahead and write that in here. We have 10.57 plus pkb is equal to 14. When we solve for the pKb that would give us 3.43. So the pKb is equal to 3.43. We could double-check that. Let's go back up here and we could double-check that, because if we took the negative log of this number that's what we should get. Let's go ahead and do that. Let's take the negative log of 3.7 times 10 to the negative 4." + }, + { + "Q": "At 2:26, what does he mean that we multiply them?\n", + "A": "He means that we multiply the given values of Ka and Kb respectively. We do it in order to equate it to 1 x 10^-14 which is actually Kw..... Hope this helps.... : )", + "video_name": "DGMs81-Rp1o", + "timestamps": [ + 146 + ], + "3min_transcript": "- We've already seen that NH4 plus and NH3 are a conjugate acid-base pair. Let's look at NH4 plus. The ammonium ion would function as an acid and donate a proton to water to form H3O plus. If NH4 plus donates a proton you're left with NH3. The Ka for this reaction is 5.6 times 10 to the negative 10. Now let's look at NH3 which we know is a weak base, and it's going to take a proton from water, therefore forming NH4 plus. If we take a proton from water we're left with OH minus. Since we talked about a base here we're gonna use Kb, and Kb for this reaction is 1.8 times 10 to the negative 5. What would happen if we add these two reactions together? We have two water molecules for our reactants, so let me go ahead and write H2O What about ammonium? We have ammonium on the left side for reactant, we also have ammonium over here for our product. That cancels out. Same thing happens with ammonia, NH3. We have NH3 on the left. We have NH3 on the right. We have NH3 as a reactant, NH3 as a product. We can cancel those out too. Our only reactants would be two H2O. For our products we would get H3O plus and OH minus, so H3O plus, hydronium, and hydroxide. This reaction should sound familiar to you. This net reaction is the auto-ionization of water where one water molecule acts as an acid, one water molecule acts as a base. We get H3O plus and OH minus. The equilibrium constant for the auto-ionization of water you've already seen that Kw is equal to 1.0 We added these two reactions together and we got this for our net reaction. What would we do with Ka and Kb to get Kw? It turns out that you multiply them, Ka times Kb for a conjugate acid-base pair is equal to Kw. Let's do that math. Ka is 5.6 times 10 to the negative 10. So 5.6 times 10 to the negative 10. Kb is equal to 1.8 times 10 to the negative 5, 1.8 times 10 to the negative 5, and let's get out the calculator and let's go ahead and do that math. We have 5.6 times 10 to the negative 10. We're going to multiply that by 1.8 times 10 to the negative 5." + }, + { + "Q": "\nWhat did he mean by ''overcome the heat of fusion'' at 0:37? there's no fusion occurring there.", + "A": "by definition heat of fusion is: The energy required to change a gram of a substance from the solid to the liquid state without changing its temperature is commonly called it s heat of fusion . So this is the energy to break the bonds in the solid state to be able to flow in the liquid state. In this case fusion is not about two atoms fusing to form a heavier atom as in a physics way, but that the atoms are in a giant solid lattice that must be broken to change the state.", + "video_name": "hA5jddDYcyg", + "timestamps": [ + 37 + ], + "3min_transcript": "We know that when we have some substance in a liquid state, it has enough kinetic energy for the molecules to move past each other, but still not enough energy for the molecules to completely move away from each other. So, for example, this is a liquid. Maybe they're moving in that direction. These guys are moving a little bit slower in that direction so there's a bit of this flow going on, but still there are bonds between them. They kind of switch between different molecules, but they want to stay close to each other. There are these little bonds between them and they want to If you increase the average kinetic energy enough, or essentially increase the temperature enough and then overcome the heat of fusion, we know that, all of a sudden, even these bonds aren't strong enough to even keep them close, and the molecules separate and they get into a gaseous phase. And there they have a lot of kinetic energy, and they're bouncing around, and they take the shape of their container. But there's an interesting thing to think about. Which implies, and it's true, that all of the molecules do not have the same kinetic energy. Let's say even they did. Then these guys would bump into this guy, and you could think of them as billiard balls, and they transfer all of the momentum to this guy. Now this guy has a ton of kinetic energy. These guys have a lot less. This guy has a ton. These guys have a lot less. There's a huge distribution of kinetic energy. If you look at the surface atoms or the surface molecules, and I care about the surface molecules because those are the first ones to vaporize or-- I shouldn't jump the gun. They're the ones capable of leaving if they had enough kinetic energy. If I were to draw a distribution of the surface molecules-- let me draw a little graph here. So in this dimension, I have kinetic energy, and on this And this is just my best estimate, but it should give you the idea. So there's some average kinetic energy at some temperature, right? This is the average kinetic energy. And then the kinetic energy of all the parts, it's going to be a distribution around that, so maybe it looks something like this: a bell curve. You could watch the statistics videos to learn more about the normal distribution, but I think the normal distribution-- this is supposed to be a normal, so it just gets smaller and smaller as you go there. And so at any given time, although the average is here, there's some molecules that have a very low kinetic energy. They're moving slowly or maybe they have-- well, let's just say they're moving slowly. And at any given time, you have some molecules that have a very high kinetic energy, maybe just because of the random bumps that it gets from other molecules. It's accrued a lot of velocity or at least a lot of momentum." + }, + { + "Q": "\nAt 12:08, Is the atmosphere part of the closed container as well? Otherwise, I am not sure how the molecules that comprise the atmosphere are interacting with the gaseous water molecules.\nAlso, when we speak of partial pressure does that indicate that gaseous water molecules within the container experience different amount of pressure than the liquid water molecules?", + "A": "The closed container is assumed to contain air which will exert pressure on the surface of the liquid water. Both the molecules in the air and the molecules in the gaseous water are exerting pressure on the surface of the liquid water.", + "video_name": "hA5jddDYcyg", + "timestamps": [ + 728 + ], + "3min_transcript": "state in order for the equilibrium to be reached. Let me do it all in the same color. So the pressure created by its evaporated molecules is going to be higher for it to get to that equilibrium state, so it has high vapor pressure. And on the other side, if you're at a low temperature or you have strong intermolecular forces or you have a heavy molecule, then you're going to have a low vapor pressure. For example, iron has a very low vapor pressure because it's not vaporizing while-- let me think of something. Carbon dioxide has a relatively much higher vapor pressure. Much more of carbon dioxide is going to evaporate when you have it. straight from the liquid to the solid state, but I think you get the idea. And something that has a high vapor pressure, that wants to evaporate really bad, we say it has a high volatility. You've probably heard that word before. So, for example, gasoline has a higher-- it's more volatile than water, and that's why it evaporates, and it also has a higher vapor pressure. Because if you were to put it in a closed container, more gasoline at the same temperature and the same atmospheric pressure, will enter into the vapor state. And so that vapor state will generate more pressure to offset the natural inclination of the gasoline to want to escape than in the case with water. Now, an interesting thing happens when this vapor pressure is equal to the atmospheric pressure. the atmosphere here at a certain pressure. Let's say until now, we've assumed that the atmosphere was at a higher pressure, for the most part keeping these molecules contained. Maybe some atmosphere molecules are coming in here, and maybe some of the vapor molecules are escaping a bit, but it's keeping it contained because this is at a higher pressure out here than this vapor pressure. And of course the pressure right here, at the surface of the molecule, is going to be the combination of the partial pressure due to the few atmospheric molecules that come in, plus the vapor pressure. But once that vapor pressure becomes equal to that atmospheric pressure, so it can press out with the same amount of force-- you can kind of view it as force per area-- so then the molecules can start to escape. It can push the atmosphere back. And so you start having a gap here. You start having a vacuum. I don't want to use exactly a vacuum, but since the" + }, + { + "Q": "at 7:50 ,does given temperature mean that the temperature has to be kept constant while this process goes on?\n", + "A": "Yes. The vapor pressure depends on the temperature of the liquid. As you raise the temperature, you raise the average kinetic energy of the molecules. More of these energetic molecules will be able to escape from the liquid, and the vapor pressure will increase. This means that you must always state the temperature at which you measured the vapor pressure. For example, the vapor pressure of water is 0.6 kPa at 0\u00c2\u00b0C, 2.3 kPa at 20\u00c2\u00b0C, 12.3 kPa at 50\u00c2\u00b0C, and 101.3 kPa at 100\u00c2\u00b0C.", + "video_name": "hA5jddDYcyg", + "timestamps": [ + 470 + ], + "3min_transcript": "Now something interesting happens. This is the distribution of the molecules in the liquid state. Well, there's also a distribution of the kinetic energies of the molecules in the gaseous state. Just like different things are bumping into each other and gaining and losing kinetic energy down here, the same thing is happening up here. So maybe this guy has a lot of kinetic energy, but he bumps into stuff and he loses it. And then he'll come back down. So there's some set of molecules. I'll do it in another set of blue. These are still the water-- or whatever the fluid we're talking about-- that come back from the vapor state back into the liquid state. And so what happens is, there's always a bit of evaporation and there's always a bit of condensation because you always have this distribution of kinetic energies. At any given moment in time, out of the vapor above the liquid, some of the vapor loses its kinetic energy and then it goes back into the liquid state. Some of the surface liquid gains kinetic energy by random And the vapor state will continue to happen until you get to some type of equilibrium. And when you get that equilibrium, we're at some pressure up here. So let me see, some pressure. And the pressure is caused by these vapor particles over here, and that pressure is called the vapor pressure. I want to make sure you understand this. So the vapor pressure is the pressure created, and this is at a given temperature for a given molecule, right? Every molecule or every type of substance will have a different vapor pressure at different temperatures, and obviously every different type of substance will also have different vapor pressures. For a given temperature and a given molecule, it's the pressure at which you have a pressure created by the vapor molecules where you have an equilibrium. going back into the liquid state. And we learned before that the more pressure you have, the harder it is to vaporize even more, right? We learned in the phase state things that if you are at 100 degrees at ultra-high pressure, and you were dealing with water, you would still be in the liquid state. So the vapor creates some pressure and it'll keep happening, depending on how badly this liquid wants to evaporate. But it keeps vaporizing until the point that you have just as much-- I guess you could kind of view it as density up here, but I don't want to think-- you have just as many molecules here converting into this state as molecules here converting into this state. So just to get an intuition of what vapor pressure is or how it goes with different molecules, molecules that really want to evaporate-- and so why would a molecule want to evaporate? It could have high kinetic energy, so this would be at a" + }, + { + "Q": "Can anyone define equilibrium for me to understand? Sal says it at 10:11.\n", + "A": "The word equilibrium in general means balance. The situation Sal is talking about in the video is balanced because in a certain amount of time the -same number- of particles will evaporate and leave the liquid to become gas as -the same number -condense and return to the liquid state. So the number of particles in each state does not change. It is in equilibrium.", + "video_name": "hA5jddDYcyg", + "timestamps": [ + 611 + ], + "3min_transcript": "going back into the liquid state. And we learned before that the more pressure you have, the harder it is to vaporize even more, right? We learned in the phase state things that if you are at 100 degrees at ultra-high pressure, and you were dealing with water, you would still be in the liquid state. So the vapor creates some pressure and it'll keep happening, depending on how badly this liquid wants to evaporate. But it keeps vaporizing until the point that you have just as much-- I guess you could kind of view it as density up here, but I don't want to think-- you have just as many molecules here converting into this state as molecules here converting into this state. So just to get an intuition of what vapor pressure is or how it goes with different molecules, molecules that really want to evaporate-- and so why would a molecule want to evaporate? It could have high kinetic energy, so this would be at a It could have low intermolecular forces, right? It could be molecular. Obviously, the noble gases have very low molecular forces, but in general, most hydrocarbons or gasoline or methane or all of these things, they really want to evaporate because they have much lower intermolecular forces than, say, water. Or they could just be light molecules. You could look at the physics lectures, but kinetic energy it's a function of mass and velocity. So you could have a pretty respectable kinetic energy because you have a high mass and a low velocity. So if you have a light mass and the same kinetic energy, you're more likely to have a higher velocity. You could watch the kinetic energy videos for that. But something that wants to evaporate, a lot of its molecules-- let me do it in a different color. Something that wants to evaporate really bad, a lot state in order for the equilibrium to be reached. Let me do it all in the same color. So the pressure created by its evaporated molecules is going to be higher for it to get to that equilibrium state, so it has high vapor pressure. And on the other side, if you're at a low temperature or you have strong intermolecular forces or you have a heavy molecule, then you're going to have a low vapor pressure. For example, iron has a very low vapor pressure because it's not vaporizing while-- let me think of something. Carbon dioxide has a relatively much higher vapor pressure. Much more of carbon dioxide is going to evaporate when you have it." + }, + { + "Q": "\nAt 4:47, we calculate the total moles of NH4 given off by the reaction assuming it starts with 0 moles of NH4. Although, wouldn't we start with more than 0 moles because the NH3 molecule would have been disassociating before we even added the HCl?", + "A": "You start with NH3 and you have 0.004mol of NH3. Afterward, it reacts with HCl to form NH4. Before the reaction, NH3 remains as NH3 and does not start dissociating.", + "video_name": "kWucfgOkCIQ", + "timestamps": [ + 287 + ], + "3min_transcript": "NH three plus HCl. So let's write it that way too. So ammonia plus HCl, HCl donates a proton to NH three giving us NH four plus and once HCl donates a proton, you have Cl minus, the chloride anion, the conjugate base. So this would be NH four plus Cl minus. Alright, let's write down how many moles of acid that we have. So we added 0.004 moles of our acid. So let's write down 0.004 here, for moles of acid that we added. And for moles of base that we started with, that's also equal to 0.004. Let's get some more room down here. So we started with 0.004 moles of our base. If you look at your molar ratios there, one to one mole ratio. to completely react with the base that we originally had present. So this represents the equivalence point of our titration. This is the equivalence point for our titration, because we've now added enough acid to completely react with our base. So all of this acid is going to react, right? So we use up all of that and it uses up all of the base that we had present. So all this base is going to react with the acid we're gonna lose, all 0.004 moles of NH three. So we're left with zero of that. Ammonia turns into ammonium, NH four plus. And so, if we start with zero for NH four plus, whatever we lose for ammonia, was what we gain for NH four plus. So for losing 0.004 moles of ammonia, that's how many moles of ammonium we are gaining. So we gain 0.004 moles of NH four plus. we're left with 0.004 moles of NH four plus. So this is at our equivalence point. So we have some ammonium present. What's the concentration of ammonium that we have present? Well, the concentration, once again, is moles over liters. So we have 0.004 moles of NH four plus, so let's write 0.004 moles of NH four plus. What is the volume? What's the total volume now? Well let's go back up to the problem. We started with 40 milliliters of our... We started with 40 milliliters and we added another 40. So 40 plus 40 is 80 milliliters. Or one, two, three, 0.08 liters. So our total volume now is 0.08 liters. So let's go ahead and write that in here." + }, + { + "Q": "At 8:23 when you start the last reaction mechanism, I know it is E1 elimination, but the first two were dehydration of alcohol, is there a specific name for this third reaction?\n", + "A": "Dehydrohalogenation - loss of a hydro(gen) and a halogen.", + "video_name": "l-g2xEV-z7o", + "timestamps": [ + 503 + ], + "3min_transcript": "would be ethanol, so let me go ahead and draw in lone pairs of electrons on the oxygen, so notice we're also heating this reaction, so the ethanol is gonna function as a base, so ethanol's not a strong base, but it can take a proton, so let me go ahead and draw in a proton right here, and a lone pair of electrons on the oxygen is going to take this proton, and the electrons would move into here to form our alkene, so let me go ahead and draw our product, let me put that in here, and let me highlight some electrons, so the electrons in blue moved in here to form our double bond. So a couple of points about this reaction, one point is, when you're looking at SN1 mechanisms, the first step is loss of a leaving group to form your carbocation, so when you get to this carbocation, you might think, well, why is ethanol acting as a base here? Why couldn't it act as a nucleophile? act as a nucleophile, and it would attack the positively-charged carbon, and you would definitely get a substitution product for this reaction as well, so if ethanol acts as a nucleophile, you're gonna get a substitution reaction, an SN1 mechanism. If the ethanol acts as a base, you're gonna get an E1 elimination mechanism, so here, we're just gonna focus on the elimination product, and we won't worry about the substitution product, but we will talk about this stuff in a later video, 'cause that would definitely happen. Alright, something else I wanna talk about is we had three beta carbons over here, and if I look at these three beta carbons, and I just picked one of them, I just said that this carbon right here, let me highlight it, I just took a proton from this carbon, but it doesn't matter which of those carbons that we take a proton off because of symmetry, let me go ahead and draw this in over here, so this is my carbocation, let's say, so our weak base comes along, and takes a proton from here, and these electrons have moved into here, that would give us the same product, right? So this would be, let me go and highlight those electrons, so these electrons in dark blue would move in to form our double bond, but this is the same as that product. Alcohols can also react via an E1 mechanism. The carbon that's bonded to the OH would be the alpha carbon, and the carbon next to that would be the beta carbon, so reacting an alcohol with sulfuric acid and heating up your reaction mixture will give you an alkene, and sometimes, phosphoric acid is used instead of sulfuric acid. So we saw the first step of an E1 mechanism was loss of a leaving group, but if that happens here, if these electrons come off onto the oxygen, that would form hydroxide as your leaving group, and the hydroxide anion is a poor leaving group, and we know that by looking at pKa values." + }, + { + "Q": "\nAt 4:00 the guy says that the water will take the the proton from hydrogen attached to the beta carbon...\n\nwhy would the water behave in such a way? we just said that oxygen is very electro negative and does not like to have a positive formal charge.\n\nso what's the deal?", + "A": "In solution there will be a ton of water and a small quantity of acid......since water is amphoteric (it is both an acid and a base) it can act as a Bronsted base....", + "video_name": "l-g2xEV-z7o", + "timestamps": [ + 240 + ], + "3min_transcript": "of only your substrate, this over here on the left, so it's first order with respect to the substrate. And that's because of this rate determining step. The loss of the leaving group is the rate determining step, and so the concentration of your substrate, your starting material, that's what matters. Your base can't do anything until you lose your leaving group. And so, since the base does not participate in the rate determining step, it participates in the second step, the concentration of the base has no effect on the rate of the reaction, so it's the concentration of the substrate only, and since it's only dependent on the concentration of the substrate, that's where the one comes from in E1, so I'm gonna go ahead and write this out here, so in E1 mechanism, the one comes from the fact this is a unimolecular, a unimolecular rate law here, and the E comes from the fact that this is an elimination reaction, so when you see E1, it's an elimination reaction, and it's unimolecular, the overall rate of the reaction only depends on the concentration of your substrate, so if you increase, let's say you have, let's say this was your substrate right here, and you increase the concentration of your substrate, let me just write this down, so if you increase the concentration of your substrate by a factor of two, you would also increase the rate of reaction by a factor of two, so it's first order with respect to the substrate, so this is some general chemistry here. If you increase the concentration of your base by a factor of two, you would have no effect on the overall rate of the reaction. So let's talk about one more point here in the mechanism, and that is the formation of this carbocation. Since we have a carbocation in this mechanism, we need to think about the possibility of rearrangements in the mechanism, and you need to think what would form, what substrate would form a stable carbocation, forming a tertiary carbocation would be favorable for an E1 mechanism. Here we have a tertiary alkyl halides, and let's say this tertiary alkyl halide undergoes an E1 elimination reaction. So the carbon that's bonded to the iodine must be our alpha carbon, and then we would have three beta carbons, so that's a beta carbon, that's a beta carbon, and that's a beta carbon. So the first step in an E1 mechanism is loss of our leaving group, so if I draw the lone pairs of electrons in here on iodine, I know that these electrons in this bond would come off onto iodine to form the iodide anion, so let me draw that in here, so we would make the iodide anion, and let me highlight our electron, so the electrons in this bond come off onto the iodine to form the iodide anion. And this is an excellent leaving group." + }, + { + "Q": "At 1:06 I Noticed that elements 58-71 are missing? Was this intentional or a bad Periodic Table?\n", + "A": "At 2:10 it shows why.", + "video_name": "t_f8bB1kf6M", + "timestamps": [ + 66 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 1:09 when you're numbering the groups with A's after them, why do you skip groups 3 through 12?", + "A": "Because some chemists decided a long time ago that that is how they are to be numbered. It is an old system and both the European and US ways to number groups used the same numbers and letters to refer to different groups. That is why it has been depreciated and have officially been replaced with 1-18. Still some people continue to use it today.", + "video_name": "t_f8bB1kf6M", + "timestamps": [ + 69 + ], + "3min_transcript": "" + }, + { + "Q": "What are valence electrons? (Mentioned at 1:29.)\n", + "A": "With sal s vids about electron configurations", + "video_name": "t_f8bB1kf6M", + "timestamps": [ + 89 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 00:12, the Speaker explains the definition of acceleration as \"the change in velocity over time\" and that got me to rethink about its meaning. Does his definition mean the velocity increases and decreases? Before, my thoughts were acceleration only meant the increase in velocity.", + "A": "Acceleration does not only mean increase in velocity, but also decrease.", + "video_name": "FOkQszg1-j8", + "timestamps": [ + 12 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 4:23 if acceleration is 20 miles per hour per second then how can we multiply seconds with hour ?", + "A": "You aren t multiplying seconds and hours. by saying 20 miles per hour per second you are saying that every second you are going to go 20 miles per hour faster. if you start at 0 miles per hour then 1 second later you would be going 20 miles per hour and the next second you would be going 40 miles per hour. so it is the change in speed over the amount of time it takes to change the speed.", + "video_name": "FOkQszg1-j8", + "timestamps": [ + 263 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 0:20 Sal said that \"Acceleration is the change in velocity over time.\" but shouldn't the definition be 'The rate of the velocity INCREASING (over time' because if the velocity decreases it is DECELERATION.", + "A": "We define acceleration as increasing or decreasing. All we do is change the direction of the acceleration. For example, you are moving east at 10m/s but the acceleration is 5m/s^2 to the west. So you are still accelerating. Deceleration can be used at that point but the scientific and more correct term would be what I described above.", + "video_name": "FOkQszg1-j8", + "timestamps": [ + 20 + ], + "3min_transcript": "" + }, + { + "Q": "At 5:20 Sal says he is going to use a logarithmic scale. What is that and how does it proportionally work?\n", + "A": "A logarithmic scale is a nonlinear scale used when there is a large range of quantities. Common uses include the earthquake strength, sound loudness, light intensity, and pH of solutions.", + "video_name": "BWs-ONRDDG4", + "timestamps": [ + 320 + ], + "3min_transcript": "But you could see these stars in other galaxies. And what's even more interesting about them is that their intensity is variable. That they become brighter and dimmer with a well-defined period. So if you're looking at a Cepheid variable star-- and this is just kind of a simulation, a very cheap simulation-- it might look like this. And then over the course of the next three, four days, it might reduce in intensity to something like this. And then after three, four days again, it will look like this. And then it'll look like this again. So it's actual intensity is going up and down with a well-defined period. So if this takes three days and then this is another three days, then the period, one entire cycle of its going from low intensity back to high intensity, is going to be six days. So this is a six-day period. And what Henrietta Leavitt saw, and this wasn't an obvious thing to do, she are roughly the same distance away. Everything in the Large Magellanic Cloud is roughly the same distance away. And it's obviously not exact. This is an entire galaxy. So you have obviously things further away in that galaxy and things closer up. You have stars here and here. And their distance isn't going to be exactly the same to us, even though we're sitting maybe over here someplace. But it's going to be close. It wasn't a bad approximation. And by making that assumption, she saw something pretty neat. If she plotted-- so let me plot this right over here. So she plotted on the horizontal axis, if she plotted the relative luminosity. So really, the only way that she can measure this is just how bright did they look to her? And she's assuming that they're same distance. So obviously, if you have a brighter star, but it's much, much further away, it's going to look dimmer. So if you assume that they're all roughly the same distance, how bright it is at the actual star. So she plotted relative luminosity of the star on one axis. And on the other axis, she plotted the period of these variable stars. She plotted the period. And what I'm going to do is I'm going to do this on a logarithmic scale. So let's say this is in days. So this is one day. This is 10 days. This is 100 days, right over here. It's a logarithmic scale because I'm going up in powers of 10. If we take the log of these, this would be 0, this would be 1, this would be 2. And so that's what I'm using as a scale. So I'm using the log of the period or I'm just marking them as 1, 10, 100. But I'm giving each of these factors of 10 an equal spacing. But when you plot it on this scale, the relative luminosity versus the period, she got a plot that looked something like this. And this is obviously not exact. She got a plot that looks something like this." + }, + { + "Q": "\nAt around 7:09 Sal says that the heat added to the system is equal to the work we did. but it is the system (the piston and pebbles) that do the work by expanding right?", + "A": "Work is the transfer of energy. There are two instances where energy is transferred. First from the thermal reservoir to the thermodynamic system (gas enclosed by cylinder walls and piston). This energy is then transferred from this volume of gas to the piston and pebbles. The piston and pebbles gain (kinetic) energy as their speed increases.", + "video_name": "M_5KYncYNyc", + "timestamps": [ + 429 + ], + "3min_transcript": "This is the heat we put into it. So it's the net heat we applied to the system divided by the heat we put into it. So this is equal to-- Q1 divided by Q1 is 1, minus Q2 over Q1. So once again, this is another interesting definition of They're all algebraic manipulations using the definition of internal energy, and whatever else. Now let's see if we can somehow relate efficiency to our temperatures. Now, let me-- this is Q1 right there. So what were Q2 and Q1? What were they? What were their absolute values, not looking at the signs of them? I mean, we know that Q2 was transferred out of the system, so if we said Q2 in terms of the energy applied to the system, it would be a negative number. But if we just wanted to know the magnitude of Q2, what would it be, and the magnitude of Q1? Well, the magnitude of Q1-- let me draw a new Carnot cycle, just for cleanliness. That's my volume axis. That's my pressure axis. PV. I start here at some state, and then I go isothermally. What's a good color for an isothermal expansion? Maybe purple. It's kind of-- let's see. An isothermal expansion. I'm on an isotherm here. So I go down there, and I go to state, and I went down to state B. So this is A to B. And we know we are an isotherm. This is when Q1 was added. This is an isotherm. If your temperature didn't change, your internal energy didn't change. And like I said before, if your internal energy is 0, then the heat applied to the system is the same as the work you did. They cancel each other out. That's why you got to 0. So this Q1 that we apply to the system, it must be equal to the work we did. And the work we did is just the area under this curve. And why is it the area under the curve? Because it's a bunch of rectangles of pressure times volume. And then you just add up all the rectangles, an infinite number of infinitely narrow rectangles, and you get the area. And what is that? Just a review-- pressure times volume was the work, right? Because we're expanding the cylinder. We're moving up that piston. We're doing force times distance. So the amount Q1 is equal to that integral-- the amount of work we did as well-- is equal to the integral from V final-- I shouldn't say V final. from VB, from our volume at B-- oh sorry. From our volume at A to our volume at B. We're starting here, and we're going to our volume at B. And we're taking the integral of pressure-- and I've done this multiple times, but I'll do it again. So the pressure, our height, times our change in volume, dv. We go back to our ideal gas formula, PV is equal to nRT," + }, + { + "Q": "At 3:00, would the kinetic energy in the same gas be transferred?\n", + "A": "Yes of course, just like there are some parts of a thing that are colder or hotter, I suppose that every molecule has a different escalar value of kinetic energy, but due to the collisions in a very small area they are approximately equal.", + "video_name": "PA-T6lMxCBI", + "timestamps": [ + 180 + ], + "3min_transcript": "They're gonna have, each of them are gonna have kinetic energy and if you average them, that's gonna be proportional to temperature. So let me depict each of this individual molecule's kinetic energy. Maybe this one is doing that. Maybe this one is doing that. Maybe this one is going in this direction. Maybe that one is going that direction. That one is going in that direction. This is going in that direction. That is going in that direction. So, notice, they all have different directions. And the magnitude of their velocity can be different. They all have different speeds. They all might have different speeds. So they have different speeds right over here. And they're all bumping into each other. Transferring their kinetic energy, transferring their momentum to from one particle to another. But when we talk about temperature we talk about the average kinetic energy or what's proportional to the average kinetic energy of the system. Well this one, each of these molecules are also gonna have some kinetic energy. But on average it's gonna be lower. Maybe this one is doing something like this. This is doing something like this. This is doing something like this. So they're different, but on average they're gonna be lower. So, hopefully you see that this magenta arrows are bigger than these blue arrows that I'm doing. And they don't all have to be, for example this one might have a lot of kinetic energy. But when you average it out, the average here is gonna be lower than the average here. So, just like that. Now, if this is our initial state, what do we think is going to start happening? Well, before our different groups of gases were colliding with itself or the magenta was colliding with the magenta. The blue is colliding with the blue. But now they're gonna start colliding with each other. And so you can imagine when this molecule right over here, collides with this molecule, it's gonna transfer some kinetic energy to it. So after the collision, this one might be going. After the collision. So lets just say they just bounced in to. So this is right before, So right after they finish bouncing into each other, this one might ricochet off. So this one is going to go this way. Let me do this in a different color. So it might hit this one, bounce off, and then transfer some of its kinetic energy and then it bounces off in this direction. While this one, after the collision, after the collision is going to is maybe going to move much faster in this direction. And so notice, you have a transfer of energy. Just with that one collision you had a transfer of kinetic energy from this molecule to that molecule. And this is going to happen throughout the system. That the faster molecules, the ones with more kinetic energy as they collide you're gonna have transfer of energy. So you're gonna have a transfer of energy from the higher temperature to the lower temperature. Transfer, transfer of energy. And this transfer of energy." + }, + { + "Q": "At 2:10, why is there a kelvin scale? where did it come from?\n", + "A": "The kelvin is a measure for temperature. It is an absolute scale for thermodynamics as it has 0 in the middle instead of at the beginning like the celcius scale ie. The scale of Celsius scale starts with 0 degrees but in the Kelvin scale, 0 is at the centre. To convert some temperature from Celsius to Kelvin, we add 273.15 from the temperature. For example, 45 degrees Celsius= 45 + 273.15 = 318.15 kelvin. Hope this helps :)", + "video_name": "tvO0358YUYM", + "timestamps": [ + 130 + ], + "3min_transcript": "I have this problem here from chapter five of the Kotz, Treichel, and Townsend Chemistry and Chemical Reactivity book, and I'm doing this with their permission. So they tell us that ethanol, C2H5OH, boils at-- let me do this in orange-- it boils at 78.29 degrees Celsius. How much energy, in joules, is required to raise the temperature of 1 kilogram of ethanol from 20 degrees Celsius to the boiling point and then change the liquid to vapor at that temperature? So there's really two parts of this problem. How much energy, in joules, to take the ethanol from 20 degrees to 78.29 degrees Celsius? That's the first part. And then once we're there, we're going to have 78.29 degrees Celsius liquid ethanol. But then we also need the energy to turn it into vapor. So those are going to be the two parts. So let's just think about just raising the liquid Let's figure out how we're going to do that. Just the liquid temperature. So the first thing I looked at is well how many degrees are we raising the temperature? Well we're going from 20 degrees Celsius-- let me write Celsius there just so it's clear-- 20 degrees Celsius to 78.29 degrees Celsius. So how much did we raise it? Well, 78.29 minus 20 is 58.29. So our change in temperature is equal to 58.29 degrees Celsius or this could even be 58.29 kelvin. the Celsius scale and the kelvin scale are the same thing. The kelvin scale is just a shifted version of the Celsius scale. If you added 273 to each of these numbers you would have the kelvin temperature, but then if you take the difference, it's going to be the exact same difference. Either way you do it, 78.29 minus 20. So that's how much we have to raise the temperature. So let's figure out how much energy is required to raise that temperature. So we want a delta T. We want to raise the temperature 58.29. I'll stay in Celsius. Actually let me just change it to kelvin because that looks like what our units are given in terms of specific heat. So let me write that down. 58.29 kelvin is our change in temperature. I could have converted either of these to kelvin first, then found the difference, and gotten the exact same number. Because the Celsius scale and the kelvin scale, the increments are the same amount. Now, that's our change in temperature." + }, + { + "Q": "at 6:43 why didn't sal multiply it with \u00ce\u0094T?\n", + "A": "Because there was no change in temperature at that point (liquid turning into vapour). Heat of vaporisation is just mass times latent heat.", + "video_name": "tvO0358YUYM", + "timestamps": [ + 403 + ], + "3min_transcript": "So this is going to be 142-- we'll just round down-- 142,000 kelvin. So this is 142,000. Sorry 142,000 joules. Joules is our units. We want energy. So this right here is the amount of energy to take our ethanol, our 1 kilogram of ethanol, from 20 degrees Celsius to 78.29 degrees Celsius. Or you could view this as from 293 kelvin to whatever this number is plus 273, that temperature in kelvin. Either way, we've raised its temperature by 58.29 kelvin. Now, the next step is, it's just a lot warmer ethanol, liquid ethanol. We now have to vaporize it. It has to become vapor at that temperature. So now we have to add the heat of vaporization. So that's right here. We should call it the enthalpy of vaporization. The enthalpy of vaporization, they tell us, is And this is how much energy you have to do to vaporize a certain amount per gram of ethanol. Assuming that it's already at the temperature of vaporization, assuming that it's already at its boiling point, how much extra energy per gram do you have to add to actually make it vaporize? So we have this much. And we know we have 1,000 grams of enthanol. The grams cancel out. 855 times 1,000 is 855,000 joules. So it actually took a lot less energy to make the ethanol go from 20 degrees Celsius to 78.29 degrees Celsius than it took it to stay at 78.29, but go from the liquid form to the This took the bulk of the energy. But if we want to know the total amount of energy, let's 855,000 plus 842,000. 800 plus 100 is 900. That's 900,000. 50 plus 40 is 90. 5 plus 2 is 7. So it's 997,000 joules or 997 kilojoules. Or we could say it's almost 1 megajoule, if we wanted to speak in those terms. But that's what it will take for us to vaporize that 1 kilogram of ethanol." + }, + { + "Q": "\nAt 3:25 Ronald says \"It's called rhodopsin because it's in a rod\" but I feel like that's not actually true. Wikipedia says it's derived from the greek word \"rhodos\" which sounds much more likely. And the Opsines in cones are called Iodopsines.", + "A": "True. But Ron s reasoning serves as at least a mnemonic to remember the protein name in rods. :)", + "video_name": "CqN-XIPhMpo", + "timestamps": [ + 205 + ], + "3min_transcript": "So normally, he is turned on. When there's no light, the rod is turned on. But when light is present, it actually turns him off. So with light, he's turned off. So how does this happen? So this occurs through this thing called the phototransduction cascade. So the phototransduction cascade is the set of steps that occurs at the molecular level that basically takes this rod and turns him off. And in turning him off, he's actually able to turn on a bunch of other cells that eventually lets the brain know, hey, there's light here. So it allows the brain to realize that there's light and allows it to comprehend what's going on make sense of the world. So let's go ahead and examine this phototransduction cascade. So let's give ourselves a little bit of space here, just look at it in a little bit more detail. So I'm going to redraw the rod over here. And let's just look at this part of the rod, just this very top little bit of the rod. And I'm going to draw it a lot bigger, so that we can really make sense of what's going on here. So inside the rod, which we just made much bigger, there are a bunch of these little disks. So they are really thin little disks, and they're just stacked on top of one another, and there are hundreds of them. And basically they are like this. They just go stacked on top of one another, and they fill up the entire rod. Same thing with cones but we're just going to look at the rods here. So inside of these little disks, there are a whole bunch of different proteins all interspersed throughout the disks. So this protein that I'm drawing in red-- let me just go ahead and draw it a lot bigger, so we're going to go ahead and blow this protein up. so it consists of a bunch of different subunits. There are actually seven subunits, so there are 5, 6, and 7. So there are all these subunits that make up this little protein right here. So this protein as a whole is called rhodopsin. And it's called rhodopsin because it's in a rod. If it were in a cone, it would be called cone opsin, but it's basically the same protein. And sitting inside of this protein is a small molecule, and it sits just inside. Let's go ahead and just zoom in a little bit, so we can make a little bit more sense of what we're looking at here. Let's go ahead and zoom in over here. So this little molecule is just sitting inside of rhodopsin, and it's kind of bent. You can see here how I drew it just a little bit bent. So this little molecule is called retinal." + }, + { + "Q": "At 8:24, it is said \"cells are hyperpolarized and turn off\". why is it OFF, not ON? hyperpolarization activates the cell and ON seems to be better to name the event.\n", + "A": "Hyperpolarization is inhibitory. It opposes the depolarization necessary to fire an action potential that allows glutamate to be released into the synapse. This glutamate is inhibitory in ON bipolar cells and excitatory in OFF bipolar cells. When the the rod is hyperpolarized, it is no longer able to release glutamate and it loses the inhibition of the ON bipolar cells, thus turning them on.", + "video_name": "CqN-XIPhMpo", + "timestamps": [ + 504 + ], + "3min_transcript": "shape and causes rhodopsin to change shape, transducin breaks away from rhodopsin. And the alpha subunit actually comes over here to another part of the disk and binds to a protein called phosphodiesterase, which I'll just draw as this little box over here. So this little protein is called cyclic GMP phosphodiesterase, or PDE for short. So PDE, basically what it does-- let's go ahead and zoom out a little bit. What PDE does when it's activated is it takes cyclic GMP, which is floating all and it basically takes the cyclic GMP and converts it into just regular GMP. So this basically reduces the concentration of cyclic GMP and increases the concentration of GMP. And the reason that this is important is because there's another channel over here, so there's a whole bunch of these sodium channels and they're all over the cell. So they're just a whole bunch of them, and basically what they let the cell do is they allow the cell to take in sodium from the outside. So let's just say there's a little sodium ion, and it allows it to come inside the cell. So in order for this sodium channel to be open, it actually needs cyclic GMP to be bound to it. So as long as cyclic GMP is bound, the channel is open. But as the concentration of cyclic GMP decreases because causes sodium channels to close. So now we have a closing of sodium channels, and now we basically have less sodium entering the cell. And as less sodium enters the cell, it actually causes the cell to hyperpolarize and turn off. So as the sodium channels close, it actually causes the rods to turn off. So basically, without light, the rods are on because these sodium channels are open. Sodium is flowing through, and the rods are turned on. They can actually produce an action potential and activate the next cell and so on. But as soon as they're turned off, what happens is very interesting because-- let's just look at this rod over here. So what happens is really interesting, because there's this other cell over here that is called the bipolar cell. And we'll just give it a kind of a neutral base, because it's bipolar." + }, + { + "Q": "\nat 12:30,one carbon is oxidised but one carbon is reduced.This implies that there must have been a reducing as well as an oxidising agent simultaneously.How is that possible?", + "A": "To answer that, you really need to see the chemical reaction ( redox reaction) which created the change in oxidation states. Remember, individual atoms within a molecule are not oxidizing or reducing agents, but the entire molecule ITSELF is the agent.", + "video_name": "bJMUKNbAsTY", + "timestamps": [ + 750 + ], + "3min_transcript": "directly bonded to two more carbons. On the right, those carbons now only have a single bond between them and this carbon on the left is bonded to a hydrogen. This carbon on the right is directly bonded to an oxygen and let's put in those other carbons, right? We have another carbon, carbon bonds. I mean, let's put in those carbon, carbon bonds like that. Let's assign our oxidation states. So, we draw in our bonding electrons and we think about electronegativity differences but we're going assume once again that our carbons have the same electronegativity on the left and so when we're assigning electrons let's just pick one of those carbons. So, let's pick the carbon on the left, right? We divide up those two electrons in that bond. Give one electron to one carbon. One electron to the other. We do the same thing here. And for the double bond with four electrons we divide up those four electrons. We give two electrons to each carbon. and around it, this carbon has one, two, three, and four. So, four minus four is an oxidation state of zero. So, this carbon has an oxidation state of zero. Same thing for the carbon on the right side of the double bond, right? So, the same situation. So, it also has an oxidation state of zero. What about for our product, right? Let's examine those two carbons. Let's put in our bonding electrons. So, we draw in our bonding electrons here. So, we put those in. A lot more bonding electrons to draw. And we think about electronegativity differences, right? Things have changed. So, now, now let's focus in on the carbon on the left. So, the carbon on the left is now bonded to a hydrogen and carbon is a little bit more electronegative. So, we give both of those electrons to carbon and then we have carbon bonded to carbon, carbon bonded to carbon, and carbon bonded to carbon. So, what's the oxidation state Carbon is supposed to have four valence electrons and how many do have around it? Let's see, one, two, three, four, five. So, four minus five gives us an oxidation state of minus one. So, this carbon on the left now has an oxidation state of minus one. What about the carbon on the right? Well, we have a tie here, right? So, we divide up those electrons, a tie here, but oxygen is more electronegative than carbon. So, oxygen gets both of those electrons. So, the carbon on the right should have an oxidation state of for minus three. Here are the three electrons around carbon once we've accounted for electronegativity. So, that's an oxidation state of plus one. So, this carbon has an oxidation state of plus one. So, overall, what happened here? Well, let me use red. So, the carbon on the left went from an oxidation state of zero to an oxidation state of minus one. That is a decrease in the oxidation state." + }, + { + "Q": "\nIs there a difference between amplitude as described in :47 (which is on the displacement vs time graph) and the green arrows at 4:20 (on the displacement vs 'position x' graph)?", + "A": "No the amplitude is same", + "video_name": "-_xZZt99MzY", + "timestamps": [ + 260 + ], + "3min_transcript": "Typical sounds have frequencies in the 100s or even 1000s of hertz. For instance, this note, which is an A note, is causing air to oscillate back and forth 440 times per second. So, the frequency of this A note is 440 hertz. Higher notes have higher frequencies, and lower notes have lower frequencies. Humans can hear frequencies as low as about 20 hertz and as high as about 20,000 hertz, but if a speaker were to oscillate air back and forth more than about 20,000 times per second, it would create sound waves, but we wouldn't be able to hear them. (sound starts, then stops) For instance, this speaker is still playing a note, but we can't hear it right now. Dogs could hear this note, though. Dogs can hear frequencies up to at least 40,000 hertz. Another key idea in sound waves is the wavelength of the sound wave. The idea of a wavelength is that when this sound the air molecules will be compressed close together in some regions and spread far apart from each other in other regions. If you find the distance between two compressed regions, that would be the wavelength of that sound wave. Since the wavelength is a distance, we measure it in meters. People get wavelength and period mixed up all the time. The period of a sound wave is the time it takes for an air molecule to oscillate back and forth one time. The wavelength of a sound wave is the distance between two compressed regions of air. People get these mixed up because there's an alternate way to create a graph of this sound wave. Consider this. Before the wave moves through the air, each air molecule has some undisturbed position from the speaker that we can measure in meters. This number represents the equilibrium undisturbed position of that air molecule. Then as the sound wave passes by, the air molecules get displaced slightly from that position. would be the displacement of the air molecule versus the undisturbed position or equilibrium position of that air molecule. This graph would let us know for a particular moment in time how displaced is that air molecule at that particular position in space. This graph shows us that in some regions the air is displaced a lot from its equilibrium position, and in other regions, the air is not displaced much at all from its equilibrium position. For this kind of graph, the distance between peaks represents the wavelength of the sound wave, not the period, because it would be measuring the distance between compressed regions in space. So, be careful. For a sound wave, a displacement versus time graph represents what that particular air molecule is doing as a function of time, and on this type of graph, the interval between peaks represents the period of the wave, but a displacement versus position graph represents a snapshot of the displacement of all the air molecules along that wave" + }, + { + "Q": "\nI just realized something. At 7:30, Sal says that the length of the vial is 1 centimeter, while the problem says that it is 1.0 centimeter. When he gets to his final answer at 12:15, he reports the answer to three significant digits. Since the length of the vial only had 2 significant digits, shouldn't the answer be 0.010, instead of 0.0998? Thanks for the input.", + "A": "Yes, Sal should only keep 2 significant figures if the length of the vial is to two significant figures. You are correct in your understanding of this.", + "video_name": "VqAa_cmZ7OY", + "timestamps": [ + 450, + 735 + ], + "3min_transcript": "So you see the linear relationship? Let me draw the line. I don't have a line tool here, so I'm just going to try to freehand it. I'll draw a dotted line. Dotted lines are a little bit easier to adjust. I'm doing it in a slight green color, but I think you see this linear This is the Beer-Lambert law in effect. Now let's go back to our problem. We know that a solution, some mystery solution, has an absorbance of 0.539-- let me do our mystery solution in-- well, I've pretty much run out of colors. I'll do it in pink-- of 0.539. So our absorbance is 0.5-- this is 0.55, so 0.539 is going to be right over there. And we want to know the concentration of potassium permanganate. Well, if we just follow the Beer-Lambert law, it's got to sit on that line. this line right over here. And this over here looks like 0.10 molar. So this right here is 0, or at least just estimating it, looking at this, that looks like 0.10 molar, or 0.10 molarity for that solution. So that's the answer to our question just eyeballing it off of this chart. Let's try to get a little bit more exact. We know the Beer-Lambert law, and we can even figure out the constant. The Beer-Lambert law tells us that the absorbance is equal to some constant, times the length, times the concentration, where the length is measured in centimeters. So that is measured in centimeters. And the concentration is measured in moles per liter, So we can figure out-- just based on one of these data points because we know that it's 0-- at 0 concentration the absorbance is going to be 0. So that's our other one. We can figure out what exactly this constant is right here. So we know all of these were measured at the same length, or at least that's what I'm assuming. They're all in a 1 centimeter cell. That's how far the light had to go through the solution. So in this example, our absorbance, our length, is equal to 1 centimeter. So let's see if we can figure out this constant right here for potassium permanganate at-- I guess this is probably standard temperature and pressure right here-- for this frequency of light. Which they told us up here it was 540 nanometers." + }, + { + "Q": "At 1:15 he said 'spectrometer', we he soon corrected. I wonder, what is the difference between spectrometer and a spectrophotometer?\n", + "A": "A spectrometer is An apparatus used for recording and measuring spectra, esp. as a method of analysis. . However, a spectrophotometer is ;An apparatus for measuring the intensity of light in a part of the spectrum, esp. as transmitted or emitted by particular substances.;, Hope that helped!", + "video_name": "VqAa_cmZ7OY", + "timestamps": [ + 75 + ], + "3min_transcript": "Let's see if we can tackle this spectrophotometry example. I took this from the Kotz, Treichel and Townsend Chemistry & Chemical Reactivity book and did it with their permission. So let's see what the problem is. It says a solution of potassium permanganate-- let me underline that in a darker color-- potassium permanganate has an absorbance of 0.539 when measured at 540 nanometer in a 1 centimeter cell. So this 540 nanometers is the wavelength of light that we're measuring the absorbance of. And so this is probably a special wavelength of light for potassium permanganate, one that it tends to be good at absorbing. So it'll be pretty sensitive to how much solute we have in the solution. OK, and the beaker is 1 centimeter. So that's just the length right there. What is the concentration of potassium permanganate? solution, the following calibration data were collected for the spectrophotometer. The absorbances of these known concentrations were already measured. So what we're going to do is we're going to plot these. And then, essentially, this absorbance is going to sit on the line. We learned from the Beer-Lambert law, that is a linear relationship between absorbance and concentration. So this absorbance is going to sit some place on this line. And we're just going to have to read off where that concentration is. And that will be our unknown concentration. So let's plot this first. Let's plot our concentrations first. So this axis, the horizontal axis, will be our concentration axis. I'll draw the axis in blue right there. I just need to make sure I have all this data here. So this is concentration in molarity. And let's see, it goes from 0.03 all the way to 0.15. So let's make this 0.03, then go three more. This over here is 0.06. One, two, three, then this over here is 0.09. This over here is 0.12. And then this over here is 0.15. And then the absorbances go-- well it's close to 0, or close to 0.1-- all the way up to close to 1. So let's make this right here 0.1. Let's make this 0.2, 0.3, 0.4, 0.5-- almost done-- 0.6, 0.7," + }, + { + "Q": "At 0:08, What is a Spectrophotometer?\n", + "A": "A spectrophotometer is a photometer that can measure intensity as a function of the light source wavelength. Important features of spectrophotometers are spectral bandwidth and linear range of absorption or reflectance measurement. Therefore it is an apparatus that uses light, that is a particular colour (wavelength), and measures the absorption/transmittance or reflected light strength/intensity.", + "video_name": "VqAa_cmZ7OY", + "timestamps": [ + 8 + ], + "3min_transcript": "Let's see if we can tackle this spectrophotometry example. I took this from the Kotz, Treichel and Townsend Chemistry & Chemical Reactivity book and did it with their permission. So let's see what the problem is. It says a solution of potassium permanganate-- let me underline that in a darker color-- potassium permanganate has an absorbance of 0.539 when measured at 540 nanometer in a 1 centimeter cell. So this 540 nanometers is the wavelength of light that we're measuring the absorbance of. And so this is probably a special wavelength of light for potassium permanganate, one that it tends to be good at absorbing. So it'll be pretty sensitive to how much solute we have in the solution. OK, and the beaker is 1 centimeter. So that's just the length right there. What is the concentration of potassium permanganate? solution, the following calibration data were collected for the spectrophotometer. The absorbances of these known concentrations were already measured. So what we're going to do is we're going to plot these. And then, essentially, this absorbance is going to sit on the line. We learned from the Beer-Lambert law, that is a linear relationship between absorbance and concentration. So this absorbance is going to sit some place on this line. And we're just going to have to read off where that concentration is. And that will be our unknown concentration. So let's plot this first. Let's plot our concentrations first. So this axis, the horizontal axis, will be our concentration axis. I'll draw the axis in blue right there. I just need to make sure I have all this data here. So this is concentration in molarity. And let's see, it goes from 0.03 all the way to 0.15. So let's make this 0.03, then go three more. This over here is 0.06. One, two, three, then this over here is 0.09. This over here is 0.12. And then this over here is 0.15. And then the absorbances go-- well it's close to 0, or close to 0.1-- all the way up to close to 1. So let's make this right here 0.1. Let's make this 0.2, 0.3, 0.4, 0.5-- almost done-- 0.6, 0.7," + }, + { + "Q": "In 5:57, does 'visible universe' mean how much space we can see with really powerful telescopes or even more than that? If it is more than why is it 'visible'?\n", + "A": "It is that part of the universe which is within range so that light from there would have had time to reach us since the time of the big bang. Light that originated too far away has not had time to reach us, and objects at that distance are outside the visible (or observable ) universe, regardless of how good our telescopes are.", + "video_name": "JiE_kNk3ucI", + "timestamps": [ + 357 + ], + "3min_transcript": "the distance from the Milky Way to Andromeda, which was 2 and 1/2 million light years, which would be just a little dot just like that, that would be the distance between the Milky Way and Andromeda. And now, we're looking at the Virgo Super Cluster that is 150 million light years. But we're not done yet. We can zoom out even more. We can zoom out even more, and over here. So you had your Virgo Super Cluster, 150 million light years was that last diagram, this diagram right over here. I want to keep both of them on the screen if I can. This diagram right here, 150 million light years across. That would fit right about here on this diagram. So this is all of the super clusters that are near us. And once again, \"near\" has to be used very, very, very loosely. A billion light years is-- two, three, four, five-- a billion light years is about from here to there. So we're starting to talk on a fairly massive-- I guess we've always been talking on a massive scale. But now, it's an even more massive scale. But we're still not done. Because this whole diagram-- now these dots that you're seeing now, I want to make it very clear. These aren't stars. These aren't even clusters of stars. These aren't even clusters of millions or even billions of stars. Each of these dots are clusters of galaxies, each of those galaxies having hundreds of billions to trillions of stars. So we're just on an unbelievably massive scale at this point. But we're still not done. We're still not done. This is roughly about a billion light years across. But right here is actually the best estimate of the visible universe. And in future videos, we're going to talk a lot more about what the visible universe means. this entire diagram right here, about a billion light years, would fit just like that. So we're talking about a super small amount of this part right here. And this is just the visible universe. I want to make it clear. This is not the entire universe. And we say it's the visible universe because think about what's happening. When we think about the a point out here, and we're observing it, and that's let's say 13 billion light years away. Let's say that point 13 billion. We're going to talk more about this in future videos, 13 billion light years. And I feel it's almost a sacrilege to be writing on this because this complexity that we're seeing here is just mind boggling. But this 13 billion light year away object, the light is just getting to us. This light left some point 13 billion light years ago." + }, + { + "Q": "\nWhy do some people think that the universe is flat? At 5:55, the visible universe looks like a sphere. So how can it be flat?", + "A": "The flatness refers to the curvature of space itself, not the geometry of the boundary of the visible universe, which obviously would have to be spherical.", + "video_name": "JiE_kNk3ucI", + "timestamps": [ + 355 + ], + "3min_transcript": "the distance from the Milky Way to Andromeda, which was 2 and 1/2 million light years, which would be just a little dot just like that, that would be the distance between the Milky Way and Andromeda. And now, we're looking at the Virgo Super Cluster that is 150 million light years. But we're not done yet. We can zoom out even more. We can zoom out even more, and over here. So you had your Virgo Super Cluster, 150 million light years was that last diagram, this diagram right over here. I want to keep both of them on the screen if I can. This diagram right here, 150 million light years across. That would fit right about here on this diagram. So this is all of the super clusters that are near us. And once again, \"near\" has to be used very, very, very loosely. A billion light years is-- two, three, four, five-- a billion light years is about from here to there. So we're starting to talk on a fairly massive-- I guess we've always been talking on a massive scale. But now, it's an even more massive scale. But we're still not done. Because this whole diagram-- now these dots that you're seeing now, I want to make it very clear. These aren't stars. These aren't even clusters of stars. These aren't even clusters of millions or even billions of stars. Each of these dots are clusters of galaxies, each of those galaxies having hundreds of billions to trillions of stars. So we're just on an unbelievably massive scale at this point. But we're still not done. We're still not done. This is roughly about a billion light years across. But right here is actually the best estimate of the visible universe. And in future videos, we're going to talk a lot more about what the visible universe means. this entire diagram right here, about a billion light years, would fit just like that. So we're talking about a super small amount of this part right here. And this is just the visible universe. I want to make it clear. This is not the entire universe. And we say it's the visible universe because think about what's happening. When we think about the a point out here, and we're observing it, and that's let's say 13 billion light years away. Let's say that point 13 billion. We're going to talk more about this in future videos, 13 billion light years. And I feel it's almost a sacrilege to be writing on this because this complexity that we're seeing here is just mind boggling. But this 13 billion light year away object, the light is just getting to us. This light left some point 13 billion light years ago." + }, + { + "Q": "\n5:55 I thought that the universe was much \"darker\" compared to this picture. Or white is just used to mark superclusters and not actual visible electromagnetic radiation? If we could actually have a look at this universe from somewhere else (impossible, since what we see here is essentially a map of time, the periphery of the circle being the first radiation reaching the Earth that we could detect) I think we would not see much light...", + "A": "I believe that the white dots represent superclusters, and not electromagnetic radiation. I agree with your idea that if we were to look at the universe in a real time picture of the periphery of the circle we would see a much darker image, new born stars reduced to black dwarfs, etc.", + "video_name": "JiE_kNk3ucI", + "timestamps": [ + 355 + ], + "3min_transcript": "the distance from the Milky Way to Andromeda, which was 2 and 1/2 million light years, which would be just a little dot just like that, that would be the distance between the Milky Way and Andromeda. And now, we're looking at the Virgo Super Cluster that is 150 million light years. But we're not done yet. We can zoom out even more. We can zoom out even more, and over here. So you had your Virgo Super Cluster, 150 million light years was that last diagram, this diagram right over here. I want to keep both of them on the screen if I can. This diagram right here, 150 million light years across. That would fit right about here on this diagram. So this is all of the super clusters that are near us. And once again, \"near\" has to be used very, very, very loosely. A billion light years is-- two, three, four, five-- a billion light years is about from here to there. So we're starting to talk on a fairly massive-- I guess we've always been talking on a massive scale. But now, it's an even more massive scale. But we're still not done. Because this whole diagram-- now these dots that you're seeing now, I want to make it very clear. These aren't stars. These aren't even clusters of stars. These aren't even clusters of millions or even billions of stars. Each of these dots are clusters of galaxies, each of those galaxies having hundreds of billions to trillions of stars. So we're just on an unbelievably massive scale at this point. But we're still not done. We're still not done. This is roughly about a billion light years across. But right here is actually the best estimate of the visible universe. And in future videos, we're going to talk a lot more about what the visible universe means. this entire diagram right here, about a billion light years, would fit just like that. So we're talking about a super small amount of this part right here. And this is just the visible universe. I want to make it clear. This is not the entire universe. And we say it's the visible universe because think about what's happening. When we think about the a point out here, and we're observing it, and that's let's say 13 billion light years away. Let's say that point 13 billion. We're going to talk more about this in future videos, 13 billion light years. And I feel it's almost a sacrilege to be writing on this because this complexity that we're seeing here is just mind boggling. But this 13 billion light year away object, the light is just getting to us. This light left some point 13 billion light years ago." + }, + { + "Q": "\nCa. at 9:40 we see the Virgo Supercluster oriented as center. Is there an actual center to the visible universe and where is the Milky Way located in proximity to that center?", + "A": "The center of the visible universe is Earth. This is because it is from our reference point that we determine if something is visible.", + "video_name": "JiE_kNk3ucI", + "timestamps": [ + 580 + ], + "3min_transcript": "where we are in the Virgo Super Cluster, inside of the Milky Way Galaxy, where we are was much closer to that point. It was on the order of-- and I want to make sure I get this right-- 36 million light years. So we were super close by, I guess, astronomical scales. We were super close, only 36 million light years, to this object, when that light was released. But that light was coming to us and the whole time the universe So we were also moving away from it, if you just think about all of the space, that everything is expanding away from each other, And only 13 billion years later did it finally catch up with us. But the whole time that that was happening, this object has also been moving. This object has also been moving away from us. And so our best estimate of where this object is now, based on how space is expanding, is on the order a 40 or 45 billion light years away. We're just observing where that light And I want to be very clear. What we are observing, this light is coming from something very, very, very primitive. That object or that area of space where that light was emitted from has now condensed into way more, I guess, mature astronomical structures. If you take it from the other point of view, people sitting where in this point of space now, and they've now moved 46 billion light years out, when they observe our region of space, they're not going to see us. They're not going to see Earth as it is now. They're going to see the region of space where Earth is at a super primitive stage, shortly after the Big Bang. And when I use words like \"shortly,\" I use that also loosely. We're still talking about hundreds of thousands or even millions of years. So we'll talk more about that in a future video. But the whole point of this video is it's beyond mind numbing. I would say the last video, about the Milky Way, that alone was mind numbing. But now, we're going in a reality where just the Milky Way becomes something that's about this picture right here. And the really mind numbing thing is, if someone told me that this is the entire universe, this by itself would certainly put things in perspective. But it's unknown what's beyond it. There's some estimates that this might be only be 1 times 10 to the 23rd of the entire universe. And it might even be the reality that the entire universe is smaller than this. And that's an interesting thing to think about. But I'll leave you there because I think no matter how you think about it, it's just-- I don't know. I actually, before doing this video, I stared at some of these photos for half an hour. This is my least productive day just because it's just so awe inspiring to think about what these dots and dots of the dots really are." + }, + { + "Q": "12:57 Can you give examples of specific protein receptors on the postsynaptic membrane that the neurotransmitters would bind to? Are they all channel coupled receptors? Thanks:)\n", + "A": "Most of them are channel coupled receptors. One example would be a ligand-gated sodium channel ( ligand-gated because the presence of a ligand, such as a neurotransmitter, is what opens the gate, and sodium channel because it causes an influx of sodium into the post-synaptic neuron)", + "video_name": "Tbq-KZaXiL4", + "timestamps": [ + 777 + ], + "3min_transcript": "Let me do a zoom in of that just to make it clear what's going on. So after they've bonded-- this is kind of before the calcium comes in, bonds to those SNARE proteins, then the SNARE protein will bring the vesicle ultra-close to the presynaptic membrane. So that's the vesicle and then the presynaptic membrane will look like this and then you have your SNARE proteins. And I'm not obviously drawing it exactly how it looks in the cell, but it'll give you the idea of what's going on. Your SNARE proteins have essentially pulled the things together and have pulled them apart so that these two membranes merge. And then the main side effect-- the reason why all this is happening-- is it allows those neurotransmitters to be dumped into the synaptic cleft. So those neurotransmitters that were inside of our vesicle then get dumped into the synaptic cleft. It's exiting the cytoplasm, you could say, of the presynaptic neuron. These neurotransmitters-- and you've probably heard the specific names of many of these-- serotonin, dopamine, epinephrine-- which is also adrenaline, but that's also a hormone, but it also acts as a neurotransmitter. Norepinephrine, also both a hormone and a So these are words that you've probably heard before. But anyway, these enter into the synaptic cleft and then they bond on the surface of the membrane of the post-synaptic neuron or this dendrite. Let's say they bond here, they bond here, and they bond here. So they bond on special proteins on this membrane surface, but the main effect of that is, that will trigger So let's say that this neuron is exciting this dendrite. So when these neurotransmitters bond on this membrane, maybe sodium channels open up. So maybe that will cause a sodium channel to open up. So instead of being voltage gated, it's neurotransmitter gated. So this will cause a sodium channel to open up and then sodium will flow in and then, just like we said before, if we go to the original one, that's like this getting excited, it'll become a little bit positive and then if it's enough positive, it'll electrotonically increase the potential at this point on the axon hillock and then we'll have another neuron-- in this case, this neuron being stimulated. So that's essentially how it happens. It actually could be inhibitory. You could imagine if this, instead of triggering a sodium ion channel, if it triggered a potassium ion channel. If it triggered a potassium ion channel, potassium ion's" + }, + { + "Q": "\n13:50 What is the deciding factor in neurotransmitters being excitatory (Na channel) vs inhibitors (K channel). Does one neuron hold multiple types of neurotransmitters or are they specific?", + "A": "First Question: I m not really sure either. Sorry :( Second Question: Neurons often release more than one transmitter. Until relatively recently, it was believed that a given neuron produced only a single type of neurotransmitter. There is now convincing evidence, however, that many types of neurons contain and release two or more different neurotransmitters. Hope that helps! =D", + "video_name": "Tbq-KZaXiL4", + "timestamps": [ + 830 + ], + "3min_transcript": "It's exiting the cytoplasm, you could say, of the presynaptic neuron. These neurotransmitters-- and you've probably heard the specific names of many of these-- serotonin, dopamine, epinephrine-- which is also adrenaline, but that's also a hormone, but it also acts as a neurotransmitter. Norepinephrine, also both a hormone and a So these are words that you've probably heard before. But anyway, these enter into the synaptic cleft and then they bond on the surface of the membrane of the post-synaptic neuron or this dendrite. Let's say they bond here, they bond here, and they bond here. So they bond on special proteins on this membrane surface, but the main effect of that is, that will trigger So let's say that this neuron is exciting this dendrite. So when these neurotransmitters bond on this membrane, maybe sodium channels open up. So maybe that will cause a sodium channel to open up. So instead of being voltage gated, it's neurotransmitter gated. So this will cause a sodium channel to open up and then sodium will flow in and then, just like we said before, if we go to the original one, that's like this getting excited, it'll become a little bit positive and then if it's enough positive, it'll electrotonically increase the potential at this point on the axon hillock and then we'll have another neuron-- in this case, this neuron being stimulated. So that's essentially how it happens. It actually could be inhibitory. You could imagine if this, instead of triggering a sodium ion channel, if it triggered a potassium ion channel. If it triggered a potassium ion channel, potassium ion's outside of the cell. So positive things are going to leave the cell if it's potassium. Remember, I used triangles for potassium. And so if positive things leave the cell, then if you go further down the neuron, it'll become less positive and so it'll be even harder for the action potential to start up because it'll need even more positive someplace else to make the threshold gradient. I hope I'm not confusing you when I say that. So this connection, the way I first described it, it's exciting. When this guy gets excited from an action potential, calcium floods in. It makes these vesicles dump their contents in the synaptic cleft and then that will make other sodium gates open up and then that will stimulate this neuron, but if it makes potassium gates open up, then it will inhibit it-- and that's how, frankly, these synapses work. I was about to say there's millions of synapses, but" + }, + { + "Q": "At around 14:07, Sal points out that if a K+ pump is triggered, K+ ions will flow out. Why would it go out and not in?\n", + "A": "Sal mentions they move out according to the concentration gradient meaning it moves from the area of high concentration K+ to low concentration K+", + "video_name": "Tbq-KZaXiL4", + "timestamps": [ + 847 + ], + "3min_transcript": "It's exiting the cytoplasm, you could say, of the presynaptic neuron. These neurotransmitters-- and you've probably heard the specific names of many of these-- serotonin, dopamine, epinephrine-- which is also adrenaline, but that's also a hormone, but it also acts as a neurotransmitter. Norepinephrine, also both a hormone and a So these are words that you've probably heard before. But anyway, these enter into the synaptic cleft and then they bond on the surface of the membrane of the post-synaptic neuron or this dendrite. Let's say they bond here, they bond here, and they bond here. So they bond on special proteins on this membrane surface, but the main effect of that is, that will trigger So let's say that this neuron is exciting this dendrite. So when these neurotransmitters bond on this membrane, maybe sodium channels open up. So maybe that will cause a sodium channel to open up. So instead of being voltage gated, it's neurotransmitter gated. So this will cause a sodium channel to open up and then sodium will flow in and then, just like we said before, if we go to the original one, that's like this getting excited, it'll become a little bit positive and then if it's enough positive, it'll electrotonically increase the potential at this point on the axon hillock and then we'll have another neuron-- in this case, this neuron being stimulated. So that's essentially how it happens. It actually could be inhibitory. You could imagine if this, instead of triggering a sodium ion channel, if it triggered a potassium ion channel. If it triggered a potassium ion channel, potassium ion's outside of the cell. So positive things are going to leave the cell if it's potassium. Remember, I used triangles for potassium. And so if positive things leave the cell, then if you go further down the neuron, it'll become less positive and so it'll be even harder for the action potential to start up because it'll need even more positive someplace else to make the threshold gradient. I hope I'm not confusing you when I say that. So this connection, the way I first described it, it's exciting. When this guy gets excited from an action potential, calcium floods in. It makes these vesicles dump their contents in the synaptic cleft and then that will make other sodium gates open up and then that will stimulate this neuron, but if it makes potassium gates open up, then it will inhibit it-- and that's how, frankly, these synapses work. I was about to say there's millions of synapses, but" + }, + { + "Q": "At 4:41 he said water can dissolve more molecules than any liquid. What makes it better than lava?\n", + "A": "It has to do with the polarity of water, it causes ionic compounds to dissociate because there s an attraction between the molucules. Lava isn t dissolving things, it s burning or melting them because of the intense heat instead of interacting with them on a molecular level like a solvent would. Melting something is not the same as dissolving something.", + "video_name": "QymONNa5C6s", + "timestamps": [ + 281 + ], + "3min_transcript": "wax paper or some teflon or something where the water beads up like this. Some leaves of plants do it really well; it's quite cool. Since water adheres weakly to the wax paper or to the plant, but strongly to itself, the water molecules are holding those droplets together in a configuration that creates the least amount of surface area. This is high surface tension that allows some bugs and even I think one lizard and also one Jesus to be able to walk on water. The cohesive force of water does have its limits, of course. There are other substances that water quite likes to stick to. Take glass, for example. This is called adhesion. The water is spreading out here instead of beading up because the adhesive forces between the water and the glass are stronger than the cohesive forces of the individual water molecules in the bead of water. Adhesion is attraction between two different substances, so in this case the water molecules and the glass molecules. These properties lead to one of my favorite things about water; the fact that it can defy gravity. That really cool thing that just happened is called capillary action. Explaining it can be easily done Thanks to adhesion, the water molecules are attracted to the molecules in the straw. As the water molecules adhere to the straw, other molecules are drawn in by cohesion following those fellow water molecules. Thank you cohesion. The surface tension created here causes the water to climb up the straw. It will continue to climb until eventually gravity pulling down on the weight of the water in the straw overpowers the surface tension. The fact that water's a polar molecule also makes it really good at dissolving things, which we call it's a good solvent then. Scratch that. Water isn't a good solvent. It's an amazing solvent. There are more substances that can be dissolved in water than in any other liquid on earth. Yes, that includes the strongest acid that we have ever created. These substances that dissolve in water, sugar or salt being ones that we're familiar with, are called hydrophilic, and they are hydrophilic because they are polar. Their polarity is stronger than the cohesive forces of the water. When you get one of these polar substances in water, it's strong enough that it breaks all the little Instead of hydrogen bonding to each other, the water will hydrogen bond around these polar substances. Table salt is ionic, and right now it's being separated into ions as the poles of our water molecules interact with it. What happens when there is a molecule that cannot break the cohesive forces of water? It can't penetrate and come into it. Basically, what happens when that substance can't overcome the strong cohesive forces of water, can't get inside of the water? That's when we get what we call a hydrophobic substance, or something that is fearful of water. These molecules lack charged poles. They are non-polar and are not dissolving in water because essentially they're being pushed out of the water by water's cohesive forces. Water, we may call it the universal solvent, but that does not mean that it dissolves everything. (boppy music) There have been a lot of eccentric scientists throughout history, but all this talk about water" + }, + { + "Q": "\nAt 5:38, since we started with a ketone, shouldn't the compound be called a hemiketal?", + "A": "It could (and maybe should) be called a hemiketal. Some people use hemiacetal for both types of intermediates. SInce this reaction type works for both aldehydes and ketones, I guess they just used the more general term hemiacetal .", + "video_name": "8-ccnvn9DxI", + "timestamps": [ + 338 + ], + "3min_transcript": "we would have, this oxygen over here, would now have two lone pairs of electrons around it, so let's show those, so let's make 'em blue here. So these electrons moved out onto our oxygen, like that. And we just formed a bond between the oxygen on our ethanol, and this carbon, so we have a bond here, like that. And this still had a hydrogen attached to it, an ethyl group, and a plus one formal charge, like that. Alright, so next, let's get a little bit of room down here. The third step would be deprotonation, so let me go ahead and write that. So, step three, we deprotonate. So, another molecule of ethanol could come along and function as a base, and a lone pair of electrons on ethanol could take this proton, which leaves these electrons behind on our oxygen. So let's go ahead, and show that. we would have an OH over here, on the left, let's go ahead and put in those electrons, and then over here, on the right, we would have, this time, two lone pairs of electrons on our oxygen. So let me go ahead, and use green for those. These electrons right in here moved off, onto our oxygen, and so, if you look at that structure closely, that's a hemiacetal. So deprotonation yields our hemiacetal here, which is an intermediate in our reaction. Alright, so next step, next step here is protonations; let me go ahead, and mark this as being step four. We're going to protonate this OH over here, on the left. And so, one of the possibilities would be a protonated ethanol over here, functioning as an acid, so let's go ahead, and draw that. So a plus one formal charge on this oxygen, and a lone pair of electrons picks up a proton, and so let's go ahead and show that. So we protonate the OH, and the reason why protonating the OH would be good, is that would give us water as a leaving group. So, let's once gain show those electrons; let's use magenta again. So these electrons, right here, picked up a proton, and let's show these electrons as being that bond now. And then over here, on the right, we have, once again, our oxygen, and ethyl, and then we have two lone pairs of electrons, and then, let's keep this lone pair green right here. And then, since we protonated the OH, we get a plus one formal charge on this oxygen here, and, if you look closely, let me use red for this, if you look closely over here, you can kinda see water hiding, right? We know water's an excellent leaving group, so, if these electrons in green moved in here," + }, + { + "Q": "I haven't understood why the final velocity is negative (1:24) and neither why the gravitational acceleration is negative too... (2:02)\n", + "A": "Because we said that up is positive.", + "video_name": "15zliAL4llE", + "timestamps": [ + 84, + 122 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 1:36, Sal says velocity is a scalar quantity but isn't velocity a vector quantity since it requires both magnitude and direction?", + "A": "Yeah,they corrected it later,you see?", + "video_name": "15zliAL4llE", + "timestamps": [ + 96 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:23, I don't get why acceleration due to gravity is negative because acceleration is a scalar, so it does not have direction?!\n", + "A": "Acceleration is a vector, not a scalar.", + "video_name": "15zliAL4llE", + "timestamps": [ + 143 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 6:28, I am confused on how you found the momentum of the car.", + "A": "The momentum of the car is it s mass (1,000kg) multiplied by it s velocity (9m/s eastward). The units of momentum are kg-m/s The momentum of any object is it s mass multiplied by it s velocity vector.", + "video_name": "XFhntPxow0U", + "timestamps": [ + 388 + ], + "3min_transcript": "momentum problems. Whoops. Invert colors. OK. So let's say we have a car. Say it's a car. Let me do some more interesting colors. A car with a magenta bottom. And it is, let's see, what does this problem say? It's 1,000 kilograms. So a little over a ton. And it's moving at 9 meters per second east. So its velocity is equal to 9 meters per second east, or to the right in this example. And it strikes a stationary 2, 000 kilogram truck. So here's my truck. Here's my truck and this is a 2,000 kilogram truck. And it's stationary, so the velocity is 0. somehow gets stuck in the truck and they just both keep moving together. So they get stuck together. The question is, what is the resulting speed of the combination truck and car after the collision? Well, all we have to do is think about what is the combined momentum before the collision? The momentum of the car is going to be the mass times the car-- mass of the car. Well the total momentum is going to the mass of the car times the velocity of the car plus the mass of the truck times the velocity of the truck. And this is before they hit each other. So what's the mass of the car? That's 1,000. What's the velocity of the car? It's 9 meters per second. So as you can imagine, a unit of momentum would be kilogram meters per second. So it's 1,000 times 9 kilogram meters per second, but I won't save space. And then the mass of the truck is 2,000. And what's its velocity? Well, it's 0. It's stationary initially. So the initial momentum of the system-- this is 2,000 times 0-- is 9,000 plus 0, which equals 9,000 kilogram meters per second. That's the momentum before the car hits the back of the truck. Now what happens after the car hits the back of the truck? So let's go to that situation. So we have the truck. I'll draw it a little less neatly. And then you have the car and it's probably a little bit-- well, I won't go into whether it's banged up and whether it released heat and all of that. Let's assume that there was nothing-- if this is a simple problem that we can do. So if we assume that, there would be no change in momentum. Because we're saying that there's no net forces acting on the system. And when I say system, I mean the combination of the car and the truck." + }, + { + "Q": "In Right hand rule, doesn't the index finger indicate the direction of magnetic field and the middle finger the direction of current? Then why does the video say otherwise? (at 9:45)\n", + "A": "Try it your way and try it the way it s described in the video and see what happens.", + "video_name": "jQ2nD8ZGeEw", + "timestamps": [ + 585 + ], + "3min_transcript": "magnitude of l. Actually, let me write it. Well, 2 meters times the magnetic field, 1 tesla. And so when you take a cross product of something, this is just a reminder. l cross B. That's equal to the magnitude of l times the magnitude of B times the sine of the angle between them times some unit directional vector that we figure out with the right hand rule. So we already did the magnitude of the distance vector. That was 2 meters. We did the magnitude of the magnetic field. And what's the sine of the angle between them? Well, if the magnetic field is going into the screen, if it's going straight into the screen, you could imagine a bunch of arrows shooting into the screen. Those are the vectors. While our distance vector, or this l is in the screen, they So this angle is 90 degrees. So this actually just becomes 1. So in terms of the magnitude, we're done. The l cross B magnitude is 2 times 1 tesla. And then we multiply that times the current. And then we actually have the magnitude of the force. The magnitude of this force is going to be equal to 5 amperes times 2 meters times 1 tesla. Which is equal to 10 newtons. And then the only question left is, what is the direction of the force that the magnetic field is exerting? And this is where we break out the right hand rule. And it's no different. You could just imagine one of the positive particles moving in that direction, and just use the right hand rule. So let's take our hand out. And if we-- let me draw a hand. A right hand. So this is my right hand. If I have my thumb sticking out like that. So the l is going to be my index finger. And then the B is the magnetic field. That's going into the screen. So you can't see it. All you can take my word for it is that my middle finger is pointed downwards into the screen and then my other fingers are just doing something else. And there you have it. Your thumb is actually the direction of the force. Your index finger is the direction of-- we'll say l for these purposes. And then the magnetic field is going into it, so you can't see my middle finger but it's pointing downwards. I could draw a little x there, to show it's going downwards. And then the force is what my thumb is doing. So the force on this wire, or at least on that section of wire, is going to be perpendicular to the direction of the current. And that direction is going to be a 10 newton force. Anyway, I've run out of time." + }, + { + "Q": "\nThe particles in the experiment, shown in 7:51, seems to have an increased radius. How come? The particle is experiencing a constant acceleration due to the electrical force - is it really true that the absolute value of v stays the same? Should the radius in our example really be constant?", + "A": "If the speed is constant, the radius is constant.", + "video_name": "b1QFKLZC11U", + "timestamps": [ + 471 + ], + "3min_transcript": "can actually visualize. And so this whole business of magnetic fields making charged particles go into circles, this is one of the few times that I can actually say has a direct application into things that you've seen. Namely, your TV. Or at least the old-school TVs. The non-plasma or LCD TVs, your cathode ray TVs, take advantage of this. Where you essentially have a beam of not protons but electrons. And a magnet-- if you take apart a TV, which I don't think you should do, because you're more likely to hurt yourself because there's a vacuum in there that can implode, and all that-- but essentially, you have a magnet that deflects this electron beam and does it really fast so it scans your entire screen of different intensities, and that's what forms the image. I won't go into that detail. Maybe one day I'll do a whole video on how TVs work. So that's one application of a magnetic field causing a beam of charged particles to curve. And then the other application, and this is actually one where it's actually useful to make the particle go in a circle, is these cyclotrons that you read in circles really, really fast, and then they smash them together. Well, have you ever wondered, how do they even make a proton go in a circle? It's not like you could hold it and guide it around in a circle. They pass it through an appropriate strength magnetic field, and it curves the path of the proton so that it can keep going through the same field over and over again. And then they can actually use those electric fields. I don't claim to have any expertise in this, but then they can keep speeding it up using the same devices, because it keeps passing through the same part of the collider. And then once it collides, you've probably seen those pictures. You know, that you spend billions of dollars on supercolliders, and you end up with these pictures. And somehow these physicists are able to take these pictures and say, oh, this is some new particle because of the way it moved. Well, what they're actually talking about is these are moving at relativistic speeds. move at different velocities, their mass is changing, and all that. But the basic idea is what we just learned. They move in circles. They move in circles because they're going through a magnetic field. But their radiuses are different because their charges and their velocities are going to be different. And actually some will move to the left and some will move to the right. And that might be because they're positive or negative, and then the radius will be dependent on their masses. Anyway, I don't want to confuse you. But I just wanted to show you that we actually are touching on some physics that a physicist would actually care about. Now with that said, what would have happened if this wasn't a proton but if this was an electron moving at this velocity at 6 times 10 to the seventh meters per second through a 0.5 tesla magnetic field popping out of this video. What would have happened? Well, this formula would have still been safe. The magnitude of the force is the charge-- but it wouldn't" + }, + { + "Q": "\nAt 2:13, what's an arduino?", + "A": "A programmable computer chip.", + "video_name": "VnfpSf6YxuU", + "timestamps": [ + 133 + ], + "3min_transcript": "This is all our parts all laid out for the Bit-zee bot, the things that you'll need to make one. Now you can make yours out of a broad variety of things, and we highly recommend that you do that. The only thing that you really have to have is the Arduino. Everything else you can switch out for other things. You can use different types of batteries. You can use different motors, et cetera. I'm going to go through what I've got here and where the products came from, or where the parts came from, I should say. And then we're going to start to put a Bit-zee together on this board so you can see how it's all wired up. But if you don't happen to have two hair dryers that you can take apart, you can either go and buy two electric motors and get some wheels for them. Or there's a variety of things you can do to solve that problem. So again, these are two motors from our hair dryer. You can see the hair dryer blower fan there. And underneath that is a sheet of Lexan. It's a stiff plastic that's really resilient. And that sheet of Lexan, it's easy to machine. So it's going to be used for mounting some of our devices. And you can get that at a hardware store for a few dollars. And this is a universal remote, and it can be gotten at Target for around $8. And we're going to use that to control our Bit-zee bot. And then we've got some electrical tape and different 22-gauge wire. And then we have some solder here. We'll use that to make our solder connections. Just like if you saw the video for the motor controller, it was used to solder that together. And this is a motor controller, which will allow us to control the speed and direction of our motors. And this is our Arduino. It's our microprocessor that we can plug into-- I should say a microcontroller that we can plug into our computer and download code to it to get the motor controller and other things to function the way we want them to. So this is a breadboard, and it's used for prototyping. And we're going to show you how to wire it up And this is our digital recording module, and it's for basically recording sounds and playing them back. And we're going to use the Arduino to trigger that so that when the little bot drives around it can make some sounds. Of course, these are just double-A batteries, and they're going to go in this battery holder. The double-As are 1.5 volts. But when we connect them in series together, they're going to be 12 volts. So that'll be great for powering our motors, because they want to run on a higher voltage than 1.5. And so we have our different transistors here that we're going to use to do some switching in our circuits. And we've got some three-color LED and some screws and nuts and then a bunch of resistors. And these are 330 ohm, 10K ohm, 220 ohm. We'll go into the details on that kind of stuff later. And then I have some of the board of our alarm clock radio, so we're going to use some components off of that board." + }, + { + "Q": "At 2:08, I am not getting why he divided the concept into chiral molecules and chiral atoms, while in the end they are chiral molecules and not atoms and have the same concept?\n", + "A": "Chiral atom is a potentially confusing term - chiral centre is a better expression. It s important to be able to identify a chiral centre because that leads to the identification of a chiral molecule. Understanding the exact configuration of groups around the chiral centre is also necessary for naming the molecule.", + "video_name": "tk-SNvCPLCE", + "timestamps": [ + 128 + ], + "3min_transcript": "If I were to draw a hand, and let me just draw a hand really fast, so I'll draw a left hand. It looks something like that. That is a left hand. Now, if I were to take its mirror image, let's say that this is a mirror right there, and I want to take its mirror image, and I'll draw the mirror image in green. So its mirror image would look something like this. Not exact, but you get the idea. The mirror image of a left hand looks a lot like a right hand. Now, no matter how I try to shift or rotate this hand like this, I might try to maybe rotate it 180 degrees, so that the thumb is on the other side like this image right here. But no matter what I do, I will never be able to make this thing look like that thing. I can shift it and rotate it, it'll just never happen. I will never be able to superimpose the blue hand on top of this green hand. When I say superimpose, literally put it exactly on top of the green hand. So whenever something is not superimposable on its mirror So this hand drawing right here is an example of a chiral object. Or I guess the hand is an example of a chiral object. This is not superimposable on its mirror image. And it makes sense that it's called chiral because the word chiral comes from the Greek word for hand. And this definition of not being able to be superimposable on its mirror image, this applies whether you're dealing with chemistry, or mathematics, or I guess, So if we extend this definition to chemistry, because that's what we're talking about, there's two concepts here. There are chiral molecules, and then there are chiral centers or chiral-- well, I call them chiral atoms. They tend to be carbon atoms, so sometimes they call them chiral carbons. So you have these chiral atoms. Now, chiral molecules are literally molecules that are not superimposable on their mirror image. I'm not going to write the whole thing. You know, not superimposable-- I'll just write the whole thing. Not superimposable on mirror image. Now, for chiral atoms, this is essentially true, but when you look for chiral atoms within a molecule, the best way to spot" + }, + { + "Q": "\nAt 9:57, what eactly does it mean to exert a repelling chage of Fe=5N/C in this case? Does the charge that's being exerted this force move away from the other positive charge at a rate that we can compare to F=ma?", + "A": "If the charge is 1 C, it will experience a force of 5N. if there are no other forces on it, it will accelerate accordingly. If it is 2 C it will experience a force of 10 N", + "video_name": "elJUghWSVh4", + "timestamps": [ + 597 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 2:46, Sal said that we need an equal and opposite force to lift the mass but wouldn't the force we've just applied and the gravitational force cancel each other?", + "A": "Yes, and that would enable the mass to maintain its velocity, whatever that velocity is.", + "video_name": "elJUghWSVh4", + "timestamps": [ + 166 + ], + "3min_transcript": "" + }, + { + "Q": "around 9:10 sal says that to get the charge moving downwards, we have to exert a force of 10N. But if we exert that force in the downward direction, seeing that the metal plate is ALSO exerting a force by the same amount, won't the charge just stay stationary over there (like suspended in the electric field)???\nPLEASE HELP ME ON THIS ONE :(\nVERY IMPORTANT\n", + "A": "The force you actually have to exert is 10.000000000(etc)1 N, which is basically 10, so that s the answer people give. Theoretically, yes it would float if they were both exactly 10N.", + "video_name": "elJUghWSVh4", + "timestamps": [ + 550 + ], + "3min_transcript": "" + }, + { + "Q": "At 11:06 David says that the height is -3m\\s, using the area formula for the second triangle, i-e Area of Triangle = 1/2(Base)(height) . So how is this possible that height can be measured using a negative sign, after all Height ain't no Vector quantity\n", + "A": "Above the axis is positive, below is negative. Scalars can be positive or negative But in this case the height represents displacement, which is a vector", + "video_name": "DD58B2siDv0", + "timestamps": [ + 666 + ], + "3min_transcript": "so we're gonna end with positive 11 meters per second. You might object. You might say, \"Wait a minute, hold on now. \"If we want delta v, \"right, and that's positive two, \"shouldn't delta v be the whole thing \"from like zero to six seconds? \"Shouldn't I say v at six seconds minus v at zero \"is positive two meters per second?\" I can't do that. The reason I can't do that is because look at what I did on the left hand side, my time interval goes from zero to six but on the right hand side, I only included the area from four to six. That's the area, there's a yellow triangle right here. If I wanted to put six and zero on this left hand side, I could do that but from my total area, I wouldn't use that. I have to use the total area. In other words, the total are from zero all the way to six These sides have to agree with each other. So from zero to six, my total area would be, this area here was eight, right? We found that rectangle was eight. This area here was two. So my total area would be 10. I can do that if I want. I could say v at six minus v at zero was, well v at zero we said was one because I just gave you that, equals 10 meters per second. I get that the v at six would be 11 meters per second just like we got it before. So you can still do it mathematically like this but make sure your time intervals agree on those sides. Now let's do the last part here. So we can find this area. This area and the area always represents the area from the curve to the horizontal axis. So in this case it's below the horizontal axis. That means it can negative area. The reason is it's a triangle again. So 1/2 base times height. So 1/2, the base is one, two, three seconds. negative now, negative three meters per second squared. I get that the total area is gonna be negative 4.5 meters per second. All right, now Daisy's gonna have a change in velocity of negative 4.5. If we want to get the velocity at nine, there's a few ways we can do it. Right, just conceptually, we can say that Daisy started at six with a velocity of 11. Her change during this period was negative 4.5. If you just add the two, you add the change to the value she started with. Well you're gonna get positive 6.5 if I add 11 and negative 4.5 meters per second or, if that sounded like mathematical witchcraft, you can say that, all right, delta v equals, what, negative 4.5 meters per second. Delta v would be, all right, you gotta be careful," + }, + { + "Q": "at 6:24. how is it possible to have infinitely small rectangles?\n", + "A": "It s not, but we can imagine that we can get as many rectangles as we want, as small as we want. And then we can imagine what the limit of that is as the number of rectangles goes up and up and up.", + "video_name": "DD58B2siDv0", + "timestamps": [ + 384 + ], + "3min_transcript": "has to just be the change in velocity. So area and change in velocity are representing the exact same thing on this graph. Area is the change in velocity. That's gonna be really useful because when you come over to here the area is still gonna be the change in velocity. That's useful because I know how to easily find the area of a triangle. The area of a triangle is just 1/2 base times height. I don't easily know how to deal with an acceleration that's varying within this formula but I do know how to find the area. For instance the area here, though I have 1/2, the base is two seconds, the height is gonna be positive two meters per second squared. What are we gonna get? One of the halves, cancel. Well, the half cancels one of the twos and I'm gonna get that this is gonna be equal to two meters per second. That's gonna be the area that represents the change in velocity. by two meters per second during this time. Now you might object. You might say, \"Wait a minute. \"I'll buy this over here because height times width \"is just a times delta t, \"but triangle, that has an extra factor of a half in it, \"and there's no half up here. \"How does this, I mean, how can we still make this claim?\" We can make this claim because we'll do the same thing we always do. We can imagine, all right, imagine a rectangle here. We're gonna estimate the area with a bunch of rectangles. Then this rectangle, and this rectangle in your line like that looks horrible. That doesn't look like the area of a triangle at all. It's got all these extra pieces right here, right? You don't want all of that. And okay, I agree. That didn't work so well. Let's make them even smaller, right? Smaller width. So we'll do a rectangle like that. We'll do this one. You see we're getting better. This is definitely closer. This is not as bad as the other one but it's still not exact. so we'll make it even smaller rectangle and an even smaller rectangle here all of these at the same width but they're even smaller than the ones before. Now we're getting really close. This area is really gonna get close to the area of the triangle. The point is if you make them infinite testable small, they'll exactly represent the area of a triangle. Each one of them can be found with this formula. The delta v for each one will be the area, or sorry, the acceleration of the height of that rectangle times the small infinite testable width and you'll get the total delta v which is so gonna be the total area. Long story short, area on a, acceleration versus time graphs represents the change in velocity. This is one you got to remember. this is the most important aspect of an acceleration graph, oftentimes the most useful aspect of it, the way you analyze it. So why do we care about change in velocity? Because it will allow us to find the velocity." + }, + { + "Q": "\nAt 9:00 to 10:40 he keeps subtracting the initial (positive) velocity from the change in velocity, yet ends up with a greater number. Why or how is this?", + "A": "He adds 1m/s to 10m/s on the right side of the equation so that it cancels out on the left side so you are left with the velocity at 6 seconds equals 11 m/s.", + "video_name": "DD58B2siDv0", + "timestamps": [ + 540, + 640 + ], + "3min_transcript": "then we can find the velocity at any other point. For instance, let's say I gave you the velocity Daisy had. For some reason I'm gonna stopwatch. I start my stopwatch at right at that moment. At t equals zero, Daisy had a velocity of, let's say positive one meter per second. So Daisy was traveling that fast at t equals zero. That was her velocity at t equals zero seconds. Now I can get the velocity wherever I want. If I want the velocity at four, let's figure this out. To get the velocity at four, I can say that the delta v during this time period right here, this four seconds. I know what that delta v was. That delta v was positive eight. We found that area, height times width. So positive eight is what the delta v is gotta equal. What's delta v? That's v at four seconds minus v at zero seconds. I know what v at zero second was. That was one. So we can get that v at four minus one meter per second is equal to positive eight meters per second. So I get the velocity at four was positive nine meters per second. And you're like, phew, that was hard. I don't wanna do that every time. Yeah, I wouldn't wanna do that every time either so there's a quick way to do it. We can just do this. What's the velocity we had to start with? That was one. What was our change in velocity? That was positive eight. So what's our final velocity? Well, one plus eight gives us our final velocity. It's positive nine. Well it's just gonna take this change in velocity of this area which represents the change in velocity which is gonna add our initial velocity to it when we solve for this final velocity. for instance, if I didn't make sense, for instance, if we want to find the velocity at six, well, we can just say we started at t equals four seconds with a velocity of positive nine. We start here with positive nine. so we're gonna end with positive 11 meters per second. You might object. You might say, \"Wait a minute, hold on now. \"If we want delta v, \"right, and that's positive two, \"shouldn't delta v be the whole thing \"from like zero to six seconds? \"Shouldn't I say v at six seconds minus v at zero \"is positive two meters per second?\" I can't do that. The reason I can't do that is because look at what I did on the left hand side, my time interval goes from zero to six but on the right hand side, I only included the area from four to six. That's the area, there's a yellow triangle right here. If I wanted to put six and zero on this left hand side, I could do that but from my total area, I wouldn't use that. I have to use the total area. In other words, the total are from zero all the way to six" + }, + { + "Q": "what is meant by \"6 followed by roughly 230's of molecules\" at 1:50 ?\n", + "A": "Well, a mole of any element is equal to 6.022*10^23 atoms. So, if we ignore .022, it means roughly 6*10^23 atoms. This means that there are roughly 6 followed by 23 zeroes(6000.....) atoms. That s what Sal meant at 1:50. Hope this helps :)", + "video_name": "5B1i26dUwME", + "timestamps": [ + 110 + ], + "3min_transcript": "Before I continue, I want to introduce you to what might be an unfamiliar concept, although if you've taken chemistry, you might know a little bit about it-- it's called a mole. This isn't the thing that grows on your face with the hair in it, or the animal that digs in your backyard, although those are also called moles. We're talking about the SI unit called a mole. A mole is just a number. It's like saying a mole of something means a certain number of something, just like a dozen-- it's like saying that a dozen eggs is 12 eggs. Just like that, a mole would be 6-- I always forget the exact number, but it's 6.023, or something of that nature. You could look it up-- I think it's 6.023 times 10 to the twentieth. Let me look up the exact number, just because I think The mole is 6.022 times 10 to the 23 of something, so it's a very number of something. Normally, we don't deal in moles of eggs-- I don't think there has been a mole of eggs ever produced in the history of the universe. 10 to the 23 is a very, very, very large number. Where is it useful? A mole is useful for counting atoms, and so what is a mole of atoms or molecules? What's that many molecules? It's 6 followed by roughly 23 0's of molecules-- a very, very big number. What's interesting about a mole is that when I have a mole of something, its mass-- let's say its mass in grams. A to-- so if I have this many carbon molecules, its mass in grams is x grams. It'll have a mass of x grams, where x is the atomic mass number of an atom of carbon, although if I was talking about a mole of a molecule, I would figure out the atomic mass of the entire molecule. What's an atomic mass number? Let me see if I can do a Web search on a periodic table. I'm showing you what I do here, and it's not fancy. Let me go to Google, and let's look up periodic table, and" + }, + { + "Q": "at 3:45 why didn't Sal just write C6H8O2 in stead of HC6H7O2 for sorbic acid?\n", + "A": "Both formulas are correct for sorbic acid! Sal wrote it with one of the hydrogen atoms at the beginning to make it a little more clear that the underlying structure of sorbic acid has one proton (or hydrogen ion) that is reactive, since it can be donated to a base. By writing it as HC6H7O2 and knowing that is an acid, one can figure out that once sorbic acid donates a proton, you are left with H+ and C6H7O2-.", + "video_name": "jzcB3faNdq0", + "timestamps": [ + 225 + ], + "3min_transcript": "is we think about well, if these are dissolved in water, it's an aqueous solution, these are going to disassociate into their, into ions. And so we would write that out on both of the, on both the reactant and the product side. So the potassium sorbate, we can write that as, it's gonna be a potassium ion dissolved in an aqueous solution, plus the C six H seven O two, this is also going to be an ion dissolved in the aqueous solution, plus the hydrochloric acid will dissolve, so you have the hydrogen proton dissolved in the aqueous solution plus the chloride ion, or anion I guess we could say it. so that's going to be in our aqueous solution, What happens? Well, you're going to have the C six H seven O two react with the hydrogen proton to get to sorbic acid. So you're gonna have sorbic acid, H C six H seven O two, that's the sorbic acid. It's going to be in an aqueous solution. So I took care so far of that and that, and then you're going to have, and then you're going to have your potassium ions, your potassium ions and your chloride ions. It's going to be just like that. So this right over here is the ionic equation, not the net ionic equation. I have the ions on the reaction, on the reactant side, and then on the product side right over here. And did I, yep, I included everything. Now you do the net ionic reaction, you can imagine what's going to happen here. I have potassium ions on the right. I could net them out on both sides. So let's net them out on both sides. I have chloride ions on the left. I have chloride ions on the right. I could net them out on both sides. And then I can write the net ionic equation. So the net ionic equation, net ionic equation, is going to be well, I have my C six H seven O two ion dissolved in an aqueous solution. You combine that with the hydrogen proton dissolved in the aqueous solution, and it's going to, it's going to give us sorbic acid: H C six H seven O two in our aqueous solution. So there you have it. That is the net ionic equation." + }, + { + "Q": "\nat 8:18 sal says that helium is smaller than hydrogen because it has 2 electrons oin the valence shell but in the case of helium , how can the nucleus attract the valence electrons ,it is a noble gas\nplease explain ?", + "A": "Valence electrons are the electron in the outer shell of an atom. Helium and hydrogen both have valence electrons: hydrogen has 1, helium has 2. Why does the nucleus attract electrons? Because the nucleus is made up from protons (positively charged) that are attracting electrons (negatively charged) and opposite charges attract. Whether or not it is a noble gas is irrelevant. Note that the valance electrons are simply those in the outer most layer and nothing more.", + "video_name": "q--2WP8wXtk", + "timestamps": [ + 498 + ], + "3min_transcript": "So, you have some, I guess you could say Coulom force that is attracting it, that is keeping it there. But if you go to krypton, all of a sudden you have much more positive charge in the nucleus. So you have 1, 2, 3, 4, 5, 6, 7, 8- I don't have to do them all. You have 36. You have a positive charge of 36. Let me write that, you have plus 36. Here you have plus 19. And you have 36 electrons, you have 36 electrons- I don't know, I've lost track of it, but in your outermost shell, in your fourth, you're going to have the 2S and then you're going to have the 6P. So you have 8 in your outermost shell. So that'd be 1, 2, 3, 4, 5, 6, 7, 8. So one way to think about it, if you have more positive charge in the center, and you have more negative charge on that outer shell, so that's going to bring that outer shell inward. It's going to have more I guess you could imagine one way, And because of that, that outer most shell is going to drawn in. Krypton is going to be smaller, is going to have a smaller atomic radius than potassium. So the trend, as you go to the right is that you are getting, and the general trend I would say, is that you are getting smaller as you go to the right in a period. That's the reason why the smallest atom of all, the element with the smallest atom is not hydrogen, it's helium. Helium is actually smaller than hydrogen, depending on how you, depending on what technique you use to measure it. That's because, if we take the simplest case, hydrogen, you have 1 proton in the nucleus and then you have 1 electron in that 1S shell, and in helium you have 2, 2 protons in the nucleus and I'm not drawing the neutrons and obviously there's different isotopes, but you have 2 electrons now in your outer most shell. So you have more, I guess you could say, you could have more Coulomb attraction here. This is plus 2 and then these 2 combined are negative 2. They're going to be drawn inward. So, that's the trend as we go to the right, as we go from the left to the right of the periodic table, we're getting smaller. Now what do you think is going to happen as we go down the period table? As we go down the periodic table in a given group? Well, as we go down a group, each new element down the group, we're adding, we're in a new period. We're adding a new shell. So you're adding more and more and more shells. Here you have just the first shell, now the second shell and each shell is getting further and further and further away. So as you go down the periodic table, you are getting, you are getting larger. You're having a larger atomic radius depending on how you are measuring it. So what's the general trend? Well if you get larger as you go down, that means you're getting smaller as you go up." + }, + { + "Q": "At 8:10 Sal mentions Coulomb attraction; Exactly what is that?\n", + "A": "It s the attraction between two opposite charges due to Coulombs law. Law of electrostatics: The force between two point charges (e.g. proton & electron) is proportional to the product of their charges, and inversely proportional to the square of the distance between them. So the more charged particles are there, the higher is the charge pulling the electrons to the nucleus, the smaller becomes the distance between them", + "video_name": "q--2WP8wXtk", + "timestamps": [ + 490 + ], + "3min_transcript": "So, you have some, I guess you could say Coulom force that is attracting it, that is keeping it there. But if you go to krypton, all of a sudden you have much more positive charge in the nucleus. So you have 1, 2, 3, 4, 5, 6, 7, 8- I don't have to do them all. You have 36. You have a positive charge of 36. Let me write that, you have plus 36. Here you have plus 19. And you have 36 electrons, you have 36 electrons- I don't know, I've lost track of it, but in your outermost shell, in your fourth, you're going to have the 2S and then you're going to have the 6P. So you have 8 in your outermost shell. So that'd be 1, 2, 3, 4, 5, 6, 7, 8. So one way to think about it, if you have more positive charge in the center, and you have more negative charge on that outer shell, so that's going to bring that outer shell inward. It's going to have more I guess you could imagine one way, And because of that, that outer most shell is going to drawn in. Krypton is going to be smaller, is going to have a smaller atomic radius than potassium. So the trend, as you go to the right is that you are getting, and the general trend I would say, is that you are getting smaller as you go to the right in a period. That's the reason why the smallest atom of all, the element with the smallest atom is not hydrogen, it's helium. Helium is actually smaller than hydrogen, depending on how you, depending on what technique you use to measure it. That's because, if we take the simplest case, hydrogen, you have 1 proton in the nucleus and then you have 1 electron in that 1S shell, and in helium you have 2, 2 protons in the nucleus and I'm not drawing the neutrons and obviously there's different isotopes, but you have 2 electrons now in your outer most shell. So you have more, I guess you could say, you could have more Coulomb attraction here. This is plus 2 and then these 2 combined are negative 2. They're going to be drawn inward. So, that's the trend as we go to the right, as we go from the left to the right of the periodic table, we're getting smaller. Now what do you think is going to happen as we go down the period table? As we go down the periodic table in a given group? Well, as we go down a group, each new element down the group, we're adding, we're in a new period. We're adding a new shell. So you're adding more and more and more shells. Here you have just the first shell, now the second shell and each shell is getting further and further and further away. So as you go down the periodic table, you are getting, you are getting larger. You're having a larger atomic radius depending on how you are measuring it. So what's the general trend? Well if you get larger as you go down, that means you're getting smaller as you go up." + }, + { + "Q": "at 9:35 why does sal make a diagonal line? I understand the horizontal line but why is there anything vertical involved?\n", + "A": "Each element in the table has one more proton than the last and that means one more electron if the element is not ionized. So as you go down the table, you get more and more electrons, and more importantly, more orbitals. The more orbitals, an element has, the larger it will be.", + "video_name": "q--2WP8wXtk", + "timestamps": [ + 575 + ], + "3min_transcript": "And because of that, that outer most shell is going to drawn in. Krypton is going to be smaller, is going to have a smaller atomic radius than potassium. So the trend, as you go to the right is that you are getting, and the general trend I would say, is that you are getting smaller as you go to the right in a period. That's the reason why the smallest atom of all, the element with the smallest atom is not hydrogen, it's helium. Helium is actually smaller than hydrogen, depending on how you, depending on what technique you use to measure it. That's because, if we take the simplest case, hydrogen, you have 1 proton in the nucleus and then you have 1 electron in that 1S shell, and in helium you have 2, 2 protons in the nucleus and I'm not drawing the neutrons and obviously there's different isotopes, but you have 2 electrons now in your outer most shell. So you have more, I guess you could say, you could have more Coulomb attraction here. This is plus 2 and then these 2 combined are negative 2. They're going to be drawn inward. So, that's the trend as we go to the right, as we go from the left to the right of the periodic table, we're getting smaller. Now what do you think is going to happen as we go down the period table? As we go down the periodic table in a given group? Well, as we go down a group, each new element down the group, we're adding, we're in a new period. We're adding a new shell. So you're adding more and more and more shells. Here you have just the first shell, now the second shell and each shell is getting further and further and further away. So as you go down the periodic table, you are getting, you are getting larger. You're having a larger atomic radius depending on how you are measuring it. So what's the general trend? Well if you get larger as you go down, that means you're getting smaller as you go up. So, what are going to be, what's going to be the smallest ones? Well, we've already said helium is the smallest. So what are going to be some of the largest? What are going to be some of the largest atoms? Well that's going to be the atoms down here at the bottom left. So, these are going to be large, these are going to be small. So, large over here, small over here and the general trend, as you go from the bottom left to the top right you are getting, you are getting smaller." + }, + { + "Q": "\nAt 2:40 Sal draws this putting two atoms together WITHOUT bonding / sharing electrons. Then he draws the other example WITH bondings(electrons share the same atom) ... At the first example you can take half of it and the second you even though can take half of it. So it seems to be that there isn't a difference in Bonding or Not Bonding?!", + "A": "No, these are different ways of estimating an atomic radius. They give different results because atoms do not have a fixed or definite boundary.", + "video_name": "q--2WP8wXtk", + "timestamps": [ + 160 + ], + "3min_transcript": "That would work except for the fact that this is not the right way to conceptualize how electrons or how they move or how they are distributed around a nucleus. Electrons are not in orbits the way that planets are in orbit around the sun and we've talked about this in previous videos. They are in orbitals which are really just probability distributions of where the electrons can be, but they're not that well defined. So, you might have an orbital, and I'm just showing you in 2 dimensions. It would actually be in 3 dimensions, where maybe there's a high probability that the electrons where I'm drawing it in kind of this more shaded in green. But there's some probability that the electrons are in this area right over here and some probability that the electrons are in this area over here, and let's say even a lower probability that the electrons are over this, like this over here. And so you might say, well at a moment the electron's there. You might say well that's the radius. But in the next moment, there's some probability it might be likely that it ends up here. But there's some probability that it's going to be over there. Then the radius could be there. So electrons, these orbitals, these diffuse probability distributions, they don't have a hard edge, so how can you say what the size of an atom actually is? There's several techniques for thinking about this. One technique for thinking about this is saying, okay, if you have 2 of the same atom, that are- 2 atoms of the same element that are not connected to each other, that are not bonded to each other, that are not part of the same molecule, and you were able to determine somehow the closest that you could get them to each other without them bonding. So, you would kind of see, what's the closest that they can, they can kind of get to each other? So let's say that's one of them and then this is the other one right over here. that closest, that minimum distance, without some type of, you know, really, I guess, strong influence happening here, but just the minimum distance that you might see between these 2 and then you could take half of that. So that's one notion. That's actually called the Van der Waals radius. Another way is well what about if you have 2 atoms, 2 atoms of the same element that are bonded to each other? They're bonded to each other through a covalent bond. So a covalent bond, we've already- we've seen this in the past. The most famous of covalent bonds is well, a covalent bond you essentially have 2 atoms. So that's the nucleus of one. That's the nucleus of the other. And they're sharing electrons. So their electron clouds actually, their electron clouds actually overlap with each other, actually overlap with each other" + }, + { + "Q": "\nDo we have a line diagram for methane as Sal explained for Propane at 6:42", + "A": "There s no way to represent it using a line structure.As each line represents a carbon-carbon bond, and methane only has carbon-hydrogen bonds, there is no line structure.", + "video_name": "pMoA65Dj-zk", + "timestamps": [ + 402 + ], + "3min_transcript": "And the important thing is, no matter what the notation, as long as you can figure out the exact molecular structure, as long as you can-- so there's this last CH3. Whether you have this, this, or this, you know what the molecular structure is. You could draw any one of these given any of the others. Now, there's an even simpler way to write this. You could write it just like this. Let me do it in a different color. You literally could write it so we have three carbons. So one, two, three. Now, this seems ridiculously simple and you're like, how can this thing right here give you the same information as all of these more complicated ways to draw it? Well, in chemistry, and in organic chemistry in particular, any of these-- let me call it a line diagram or a line angle diagram. It's the simplest way and it's actually probably the most useful way to show chains of carbons or to show organic molecules. Once they start to get really, really complicated, because something like this, you assume that the end points of any lines have a carbon on it. So if you see something like that, you assume that there's a carbon at that end point, a carbon at that end point, and a carbon at that end point. And then you know that carbon makes four bonds. There are no charges here. All the carbons are going to make four bonds, and each of the carbons here, this carbon has two bonds, so the other two bonds are implicitly going to be with hydrogens. If they don't draw them, you assume that they're going to be with hydrogens. This guy has one bond, so the other three must be with hydrogen. This guy has one bond, so the other three must be hydrogens. So just drawing that little line angle thing right there, I actually did convey the exact same information as this depiction, this depiction, or this depiction. So you're going to see a lot of this. This really simplifies things. And sometimes you see things that are in between. You might see someone draw it like this, where they'll write CH3, and then they'll draw it like that. molecule where you write the CH3's for the end points, but then you implicitly have the CH2 on the inside. You assume that this end point right here is a C and it's bonded to two hydrogens. So these are all completely valid ways of drawing the molecular structures of these carbon chains or of these organic compounds." + }, + { + "Q": "At 4:30, do you need to show all electrons for propane.\n", + "A": "Unless you are drawing a dot structure no it s a waste of time. That s why we almost always draw structures like he does at 5:35, it tells us exactly the same information as the other structures it just saves time.", + "video_name": "pMoA65Dj-zk", + "timestamps": [ + 270 + ], + "3min_transcript": "Now let's explore slightly larger chains. So let's say I have a two-carbon chain. Well, let me do a three-carbon chain so it really looks like a chain. So if I were to draw everything explicitly it might So I have a carbon. It has one, two, three, four electrons. Maybe I have another carbon here that has-- let me do the carbons in slightly different shades of yellow. I have another carbon here that has one, two, three, four electrons. And then let me do the other carbon in that first yellow. And then I have another carbon so we're going to have a three-carbon chain. It has one, two, three, four valence electrons. Now, these other guys are unpaired, and if you don't specify it, it's normally going to be hydrogen, so let me draw some hydrogens over here. So you're going to have a hydrogen there, a hydrogen over there, a hydrogen over here, a hydrogen over here, a done, a hydrogen there, and then a hydrogen there. Now notice, in this molecular structure that I've drawn, I have three carbons. They were each able to form four bonds. This guy has bonds with three hydrogens and another carbon. This guy has a bond with two hydrogens and two carbons. This guy has a bond with three hydrogens and then this carbon And so this is a completely valid molecular structure, but it was kind of a pain to draw all of these valence electrons here. So what we typically would want to do is, at least in this structure, and we're going to see later in this video there's even simpler ways to write it, so if we want at least do it with these lines, we can draw it like this. So you have a carbon, carbon, carbon, and then they are bonded to the hydrogens. So you'll almost never see it written like this because this is just kind of crazy. Hyrdrogen, hydrogen-- at least crazy to write. It takes forever. And it might be messy, like it might not be clear where these I didn't write it as clearly as I could. So they have two electrons there. They share with these two guys. Hopefully, that was reasonably clear. But if we were to draw it with the lines, it looks just like that. So it's a little bit neater, faster to draw, same exact idea here and here. And in general, and we'll go in more detail on it, this three-carbon chain, where everything is a single bond, is propane. Let me write these words down because it's helpful to get. This is methane. And you're going to see the rhyme-- you're going to see the reason to this naming soon enough. This is methane; this is propane. And there's an even simpler way to write propane. You could write it like this. Instead of explicitly drawing these bonds, you could say that this part right here, you could write that that part right there, that is CH3, so you have a CH3, connected to a-- this is a CH2, that is CH2 which is then connected to" + }, + { + "Q": "\nAt 11:35, why does the horizontal velocity stay the same? Doesn't the object horizontally slow down as it goes further in space (by which the velocity would decrease)?", + "A": "In the real world the horizontal velocity would decrease because of air resistance but usually in problems like this air resistance is ignored to illustrate the concept without complicating it with extra details.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 695 + ], + "3min_transcript": "cosine of 30 degrees is equal to the adjacent side. Is equal to the adjacent side, which is the magnitude of our horizontal component, is equal to the adjacent side over the hypotenuse. Over 10 meters per second. multiply both sides by 10 meters per second, you get the magnitude of our adjacent side, color transitioning is difficult, the magnitude of our adjacent side is equal to 10 meters per second. Is equal to 10 meters per second. Times the cosine, times the cosine of 30 degrees. And you might not remember the cosine of 30 degrees, you can use a calculator for this. Or you can just, if you do remember it, you know that it's the square root of three over two. Square root of three over two. So to figure out the actual component, I'll stop to get a calculator out if I want, well I don't have to use it, do it just yet, because I have 10 times Which is going to be 10 divided by two is five. So it's going to be five times the square root of three meters per second. So if I wanna figure out the entire horizontal displacement, so let's think about it this way, the horizontal displacement, we're trying to figure out, the horizontal displacement, a S for displacement, is going to be equal to the average velocity in the x direction, or the horizontal direction. And that's just going to be this five square root of three meters per second because it doesn't change. So it's gonna be five, I don't want to do that same color, is going to be the five square roots of 3 meters per second times the change in time, times how long it is in the air. And we figure that out! Its 1.02 seconds. Times 1.02 seconds. The seconds cancel out with seconds, and we'll get that answers in meters, and now we get our calculator out to figure it out. times 1.02. It gives us 8.83 meters, So this is going to be equal to, this is going to be equal to, this is going to be oh, sorry. this is going to be equal to 8.8, is that the number I got? 8.83, 8.83 meters. And we're done. And the next video, I'm gonna try to, I'll show you another way of solving for this delta t. To show you, really, that there's multiple ways to solve this. It's a little bit more complicated but it's also a little bit more powerful if we don't start and end at the same elevation." + }, + { + "Q": "\nAt 6:44, why is the delta velocity be -10?", + "A": "What are the values that determine the change in velocity, and what numbers represent them in this video?", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 404 + ], + "3min_transcript": "and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air," + }, + { + "Q": "\n6:45 why is velocity -5 m/s and not 0 m/s?", + "A": "Why would it be 0? Because the projectile hits the ground and stops? What we are trying to find out is how fast the projectile is going when it hits the ground. We wouldn t need to do any work to know that after it hits the ground it just sits there.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 405 + ], + "3min_transcript": "and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air," + }, + { + "Q": "\nAt 4:05, Sal takes only the MAGNITUDE of the vert component of the velocity vector. What happens to its Direction ?", + "A": "The direction is up which is what has been chosen as the positive vertical direction", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 245 + ], + "3min_transcript": "We want to break it down it with x- and y-components, or its horizontal and vertical components. so that's its horizontal, let me draw a little bit better, that's its horizontal component, and that its vertical component looks like this. This is its vertical component. So let's do the vertical component first. So how do we figure out the vertical component given that we know the hypotenuse of this right triangle and we know this angle right over here. And the angle, and the side, this vertical component, or the length of that vertical component, or the magnitude of it, is opposite the angle. So we want to figure out the opposite. We have to hypotenuse, so once again we write down so-cah, so-ca-toh-ah. Sin is opposite over hypotenuse. So we know that the sin, the sin of 30 degrees, the sin of 30 degrees, is going to be equal to the magnitude So this is the magnitude of velocity, I'll say the velocity in the y direction. That's the vertical direction, y is the upwards direction. Is equal to the magnitude of our velocity of the velocity in the y direction. Divided by the magnitude of the hypotenuse, or the magnitude of our original vector. Divided by ten meters per second. Ten meters per second. And then, to solve for this quantity right over here, we multiply both sides by 10. And you get 10, sin of 30. 10, sin of 30 degrees. 10 sin of 30 degrees is going to be equal to the magnitude of our, the magnitude of our vertical component. And so what is the sin of 30 degrees? And this, you might have memorized this from your basic trigonometry class. You can get the calculator out if you want, It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation" + }, + { + "Q": "At around 6:48, why is the final vertical velocity -5 m/s? Wouldn't the final velocity be 0, since the ball momentarily comes to a complete stop when hitting the ground?\n", + "A": "You re right, but at the very instant that the ball hits the ground, before the ground stops it, it will be moving at -5 m/s. Just like how the ball isn t moving before it is thrown, but its initial velocity is still 5 m/s.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 408 + ], + "3min_transcript": "and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air," + }, + { + "Q": "\nAt 06:26, why does Sal say the final velocity is -5 m/s? Surely it would be 0 m/s when it hits the ground as it inevitably will?", + "A": "We are interested in its velocity right before it touches the ground. It is pretty obvious that the body will be at rest when it touches the ground, so calculating the velocity at that time is pretty useless. Since we d rather take into account the velocity with which the body strikes the ground, we consider -5 m/s (which is the velocity RIGHT before the body touches the ground).", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 386 + ], + "3min_transcript": "It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information" + }, + { + "Q": "\ni am confused about the initial velocity and the final velocity when you talked about it in 5:50. how do you figure it out?", + "A": "The initial velocity is the velocity you start with (in this case, in the vertical direction). The final velocity is the velocity you finish with, which is exactly the same one as you started with but in the opposite direction (going down, instead of going up).", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 350 + ], + "3min_transcript": "It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information" + }, + { + "Q": "At 6:28 Sal says that the final velocity is -5m/s. I thought it would be 0 since it hits the ground and at that moment in time, it will have no velocity.\n", + "A": "We re studying projectile motion. It s not a projectile when it s lying on the ground,.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 388 + ], + "3min_transcript": "It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information" + }, + { + "Q": "At 7:34 I am still confused on how the acceleration in the vertical direction became -9.8 meter per second squared...?\n", + "A": "Downward direction is always taken a negative acceleration, a deceleration and is thus represented by a -ve.!", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 454 + ], + "3min_transcript": "And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air, negative 9.8 meters per second squared. So we get, lets just do that, I wanna do that in the same color. So I do it in, that's not, well, that close enough. So we get negative 9.8 meters per second squared. Negative 9.8 meters per second squared. That cancels out, and I get my change in time. And I'll just get the calculator. I have a negative divided by a negative so that's a positive, which is good, because we want to go in positive time. We assume that the elapsed time is a positive one. And so what we get? If I get my calculator out, I get my calculator out. I have, this is the same thing as positive 10 divided by 9.8. 10, divided by 9.8. Gives me 1.02. I'll just round to two digits right over there. So that gives me 1.02 seconds So our change in time, so this right over here is 1.02." + }, + { + "Q": "\nAt about 8:14 in video, both sides are divided by -9.8m/s/s. The left side has a -10m/s unit, but when the division is done the units are ignored. Can you explain why it is ok to divide m/s by m/s/s?", + "A": "It s just algebra. (m/s) / ((m/s)/s) = 1/ (1/s) = s", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 494 + ], + "3min_transcript": "And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air, negative 9.8 meters per second squared. So we get, lets just do that, I wanna do that in the same color. So I do it in, that's not, well, that close enough. So we get negative 9.8 meters per second squared. Negative 9.8 meters per second squared. That cancels out, and I get my change in time. And I'll just get the calculator. I have a negative divided by a negative so that's a positive, which is good, because we want to go in positive time. We assume that the elapsed time is a positive one. And so what we get? If I get my calculator out, I get my calculator out. I have, this is the same thing as positive 10 divided by 9.8. 10, divided by 9.8. Gives me 1.02. I'll just round to two digits right over there. So that gives me 1.02 seconds So our change in time, so this right over here is 1.02." + }, + { + "Q": "\nat 10:20 how is it that the velocity of the object in the horizontal direction remains the same. i can't seem to put my head around this one.", + "A": "Why would the velocity in the horizontal direction change? Is there a force in the horizontal direction? When you are in an airplane and you drop your peanuts, they land right at your feet, don t they? Their horizontal velocity remained the same.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 620 + ], + "3min_transcript": "negative 9.8 meters per second squared. So we get, lets just do that, I wanna do that in the same color. So I do it in, that's not, well, that close enough. So we get negative 9.8 meters per second squared. Negative 9.8 meters per second squared. That cancels out, and I get my change in time. And I'll just get the calculator. I have a negative divided by a negative so that's a positive, which is good, because we want to go in positive time. We assume that the elapsed time is a positive one. And so what we get? If I get my calculator out, I get my calculator out. I have, this is the same thing as positive 10 divided by 9.8. 10, divided by 9.8. Gives me 1.02. I'll just round to two digits right over there. So that gives me 1.02 seconds So our change in time, so this right over here is 1.02. delta t, I'm using lowercase now but I can make this all lower case. Is equal to 1.02 1.02 seconds. Now how do we use this information to figure out how far this thing travels? Well if we assume that it retains its horizontal component of its velocity the whole time, we just assume we can this multiply that times our change in time and we'll get the total displacement in the horizontal direction. So to do that, we need to figure out this horizontal component, So this is the component of our velocity in the x direction, or the horizontal direction. Once again, we break out a little bit of trigonometry. This side is adjacent to the angle, so the adjacent over hypotenuse is the cosine of the angle. Cosine of an angle is adjacent over hypotenuse. So we get cosine. Cosine of 30 degrees, cosine of 30 degrees is equal to the adjacent side. Is equal to the adjacent side, which is the magnitude of our horizontal component, is equal to the adjacent side over the hypotenuse. Over 10 meters per second. multiply both sides by 10 meters per second, you get the magnitude of our adjacent side, color transitioning is difficult, the magnitude of our adjacent side is equal to 10 meters per second. Is equal to 10 meters per second. Times the cosine, times the cosine of 30 degrees. And you might not remember the cosine of 30 degrees, you can use a calculator for this. Or you can just, if you do remember it, you know that it's the square root of three over two. Square root of three over two. So to figure out the actual component, I'll stop to get a calculator out if I want, well I don't have to use it, do it just yet, because I have 10 times" + }, + { + "Q": "\n11:00\n\nSal simplified 10*cos30 to 10*((squareRoot of 3)/2) to 5*squareroot of 3... wouldn't it still be 5*((squareRoot of 3)/2)?\n\nSorry, it's late and maybe I need to take a break...", + "A": "No it would not. 10/2 is 5 so 10(square root of 3)/2 = 5(square root of 3).", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 660 + ], + "3min_transcript": "delta t, I'm using lowercase now but I can make this all lower case. Is equal to 1.02 1.02 seconds. Now how do we use this information to figure out how far this thing travels? Well if we assume that it retains its horizontal component of its velocity the whole time, we just assume we can this multiply that times our change in time and we'll get the total displacement in the horizontal direction. So to do that, we need to figure out this horizontal component, So this is the component of our velocity in the x direction, or the horizontal direction. Once again, we break out a little bit of trigonometry. This side is adjacent to the angle, so the adjacent over hypotenuse is the cosine of the angle. Cosine of an angle is adjacent over hypotenuse. So we get cosine. Cosine of 30 degrees, cosine of 30 degrees is equal to the adjacent side. Is equal to the adjacent side, which is the magnitude of our horizontal component, is equal to the adjacent side over the hypotenuse. Over 10 meters per second. multiply both sides by 10 meters per second, you get the magnitude of our adjacent side, color transitioning is difficult, the magnitude of our adjacent side is equal to 10 meters per second. Is equal to 10 meters per second. Times the cosine, times the cosine of 30 degrees. And you might not remember the cosine of 30 degrees, you can use a calculator for this. Or you can just, if you do remember it, you know that it's the square root of three over two. Square root of three over two. So to figure out the actual component, I'll stop to get a calculator out if I want, well I don't have to use it, do it just yet, because I have 10 times Which is going to be 10 divided by two is five. So it's going to be five times the square root of three meters per second. So if I wanna figure out the entire horizontal displacement, so let's think about it this way, the horizontal displacement, we're trying to figure out, the horizontal displacement, a S for displacement, is going to be equal to the average velocity in the x direction, or the horizontal direction. And that's just going to be this five square root of three meters per second because it doesn't change. So it's gonna be five, I don't want to do that same color, is going to be the five square roots of 3 meters per second times the change in time, times how long it is in the air. And we figure that out! Its 1.02 seconds. Times 1.02 seconds. The seconds cancel out with seconds, and we'll get that answers in meters, and now we get our calculator out to figure it out." + }, + { + "Q": "\nAt 3:21 why does Sal put two lines in the front and back of the unknown magnitude?", + "A": "That is a way to indicate take the magnitude of the vector in between these bars . It s sort of like the vector version of an absolute value sign.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 201 + ], + "3min_transcript": "we can use that to figure out how long will this rock stay in the air. Because it doesn't matter what its horizontal component is. Its vertical component is gonna determine how quickly it decelerates due to gravity and then re-accelerated, and essentially how long it's going to be the air. And once we figure out how long it's in the air, we can multiply it by, we can multiply it by the horizontal component of the velocity, and that will tell us how far it travels. And, once again, the assumption that were making this videos is that air resistance is negligible. Obviously, if there was significant air resistance, this horizontal velocity would not stay constant while it's traveling through the air. But we're going to assume that it does, that this does not change, that it is negligible. We can assume that were doing this experiment on the moon if we wanted to have a, if we wanted to view it in purer terms. But let's solve the problem. So the first that we want to do is we wanna break down this velocity vector. We want to break down this velocity vector that has a magnitude of ten meters per second. We want to break it down it with x- and y-components, or its horizontal and vertical components. so that's its horizontal, let me draw a little bit better, that's its horizontal component, and that its vertical component looks like this. This is its vertical component. So let's do the vertical component first. So how do we figure out the vertical component given that we know the hypotenuse of this right triangle and we know this angle right over here. And the angle, and the side, this vertical component, or the length of that vertical component, or the magnitude of it, is opposite the angle. So we want to figure out the opposite. We have to hypotenuse, so once again we write down so-cah, so-ca-toh-ah. Sin is opposite over hypotenuse. So we know that the sin, the sin of 30 degrees, the sin of 30 degrees, is going to be equal to the magnitude So this is the magnitude of velocity, I'll say the velocity in the y direction. That's the vertical direction, y is the upwards direction. Is equal to the magnitude of our velocity of the velocity in the y direction. Divided by the magnitude of the hypotenuse, or the magnitude of our original vector. Divided by ten meters per second. Ten meters per second. And then, to solve for this quantity right over here, we multiply both sides by 10. And you get 10, sin of 30. 10, sin of 30 degrees. 10 sin of 30 degrees is going to be equal to the magnitude of our, the magnitude of our vertical component. And so what is the sin of 30 degrees? And this, you might have memorized this from your basic trigonometry class. You can get the calculator out if you want," + }, + { + "Q": "\nAt 6:25 why is the final velocity equal to -5m/s and not 0 m/s ?", + "A": "He is talking about the vertical velocity at which it falls back down to the ground. You know the old saying, what goes up must come down . Well, in order to come back down, it must do so at some speed. Assuming no loss of energy due to air resistance, it will come back down at the same speed at which it went up, except in the opposite direction, hence negative rather than positive.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 385 + ], + "3min_transcript": "It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information" + }, + { + "Q": "\nAt about 7:42 Sal said that vertical velocity = acceleration x change in time. Then he said that 10m/s = -9.8m/s2 x t I do not understand why the acceleration equals -9.8m/s2. I know that -9.8m/s2 is the acceleration of an object falling, but couldn't a projectile be thrown up at ANY acceleration? (For example, if the projectile is thrown up at an acceleration of 7.3m/s2 then 10m/s2 = 7.3m/s2 x t and t = 1.37 seconds, not 1.02 seconds)", + "A": "No, a projectile will accelerate at 9.8 m/s^2 downward as soon as it is released. We can give it whatever initial upward velocity we want, but the moment it is released, there s no upward force on the object, so gravity takes over.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 462 + ], + "3min_transcript": "And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air, negative 9.8 meters per second squared. So we get, lets just do that, I wanna do that in the same color. So I do it in, that's not, well, that close enough. So we get negative 9.8 meters per second squared. Negative 9.8 meters per second squared. That cancels out, and I get my change in time. And I'll just get the calculator. I have a negative divided by a negative so that's a positive, which is good, because we want to go in positive time. We assume that the elapsed time is a positive one. And so what we get? If I get my calculator out, I get my calculator out. I have, this is the same thing as positive 10 divided by 9.8. 10, divided by 9.8. Gives me 1.02. I'll just round to two digits right over there. So that gives me 1.02 seconds So our change in time, so this right over here is 1.02." + }, + { + "Q": "\nThroughout this lecture series, (at this video time 0:37) you state that a chiral carbon is \"usually\" a carbon bonded to four different groups. This begs the question, when is a carbon chiral, that is not bonded to four different groups? Thank you, Mike Johnston", + "A": "A chiral carbon always has four different groups. But you can have a chiral molecule that contains no chiral atoms.", + "video_name": "0XSSPow5oAc", + "timestamps": [ + 37 + ], + "3min_transcript": "In the last video we learned a little bit about what a chiral molecule or what a chiral carbon or a chiral atom is. What I want to do in this video is go through a bunch of examples and see if we can identify if there are any chiral atoms and to also see if we're dealing with a chiral molecule. So let's look at our examples here. So here I have, what is this? This is chlorocyclopentane. So the first question is do we have any chiral atoms? And when we look at our definition that we thought of chiral atoms, it all comes from this notion of handedness and not being able to be superimposable on your mirror image, but we said that they're usually carbons bonded to four different groups. Let's see, do we have any carbons here bonded to four different groups? Well, all the CH2's, they're bonded to another CH2 and then two H'2. I could draw it like this: H and H. So they're bonded to two of the same group, so none of these CH2's are good candidates for being a chiral They're both bonded to-- or all of them are bonded to two hydrogens and two other very similar-looking CH2 groups, although you have to look at the entire group that it's bonded to. But they're all definitely bonded to two hydrogens, so it's not four different groups. If we look at this CH right here, we could separate it out like this. We could separate the H out like this, and so since it's bonded to a hydrogen. This carbon is bonded to a chlorine, and then it's bonded to-- well, it's not clear when you look at it right from the get-go whether this group is different than this group. But if you go around, if you were to split it half-way like this, or maybe another better way to think about is if you were to go around this molecule in that direction, the counterclockwise direction, you would encounter a CH2 group, and then you encounter a CH2 group, and then you would encounter a third, and then you would encounter a fourth CH2 group, then you would come back to So you would encounter four CH2's and then you'd come back to where you were before. If you go in this direction, what happens? You encounter one, two, three, four CH2's and you come back to where you were before. So all of this, this bottom group, depending on how far you want to extend it, and this top group, are really the same group. So this is not a chiral center or not a chiral carbon. It's not bonded to four different groups. And this is also not a chiral molecule, because it does not have a chiral center. And to see that it's not a chiral molecule-- let me see if I can backtrack this back to the way I wrote it right before. So you see that it's not a chiral molecule. There's a couple of ways you could think about it. The easiest way, or the way my brain likes to think about it, is just to think about its mirror image. Its mirror image will look like this." + }, + { + "Q": "\nI love the part when Sal is looking for a spot to draw the flipped molecule at 3:55", + "A": "I like that part, too. :)", + "video_name": "0XSSPow5oAc", + "timestamps": [ + 235 + ], + "3min_transcript": "So you would encounter four CH2's and then you'd come back to where you were before. If you go in this direction, what happens? You encounter one, two, three, four CH2's and you come back to where you were before. So all of this, this bottom group, depending on how far you want to extend it, and this top group, are really the same group. So this is not a chiral center or not a chiral carbon. It's not bonded to four different groups. And this is also not a chiral molecule, because it does not have a chiral center. And to see that it's not a chiral molecule-- let me see if I can backtrack this back to the way I wrote it right before. So you see that it's not a chiral molecule. There's a couple of ways you could think about it. The easiest way, or the way my brain likes to think about it, is just to think about its mirror image. Its mirror image will look like this. Then you have a CH, CH2, CH2, and you have a CH2, CH2, and then you complete your cyclopentane. Now, in this situation, is there any way to rotate this to get this over there? Well, if you just took this molecule right here and you just rotated it 180 degrees, what would it look like? Well, maybe a little over-- yeah, well, not quite 180 degrees, but if you were to rotate it so that the chlorine goes about that far, you would get this exact molecule. You would get something. It would look a little bit different. It would look like this. Let me see if I can do it justice. It would look like this. You would have a CH2. little bit more space. If I were to rotate this about that far, I would get a CH. You get the chlorine and then you have your CH2, and then you have another CH2, CH2, and then you would have your CH2 up there. If you were to rotate this all the way around, or actually this is almost exactly 180 degrees, it would look like this. And the only difference between this and this is just how we drew this bond here. I could have easily, instead of drawing that bond like that, I could draw it facing up like that, and these are the exact same molecule. So this molecule is also not chiral. So let's go to this one over here. So what is this? This is a bromochlorofluoromethane, just" + }, + { + "Q": "\nat 5:14 why does it produce 2 halid groups and not a alcohol group?", + "A": "Answered starting at 2:15.", + "video_name": "1k6MUeM-pEo", + "timestamps": [ + 314 + ], + "3min_transcript": "And our halogen now has an extra lone pair of electrons, giving it a negative charge, making it a halide anion. So in the next step of the mechanism, the halide anion will function as our nucleophile and attack the carbon bonded to the oxygen, kicks these electrons off onto the oxygen. And that's how we make our second alkyl halide. So we'd form R-X as our other product. And also water would be produced in this as well-- so H2O, like that. Now, I drew this second part of the mechanism like it's an SN2 mechanism. And it would be an SN2 mechanism if we were starting with a primary alcohol. So if this guy over here is a primary alcohol, and after it gets protonated, a primary alcohol would work the best for an SN2 mechanism because that would be decreased steric hindrance. However, if we were dealing with something like a tertiary alcohol at this point, things would likely proceed via an SN1 type mechanism. Let's do an example of the acidic cleavage of ethers. And we'll start with an ether that looks like this. So we're going to react this ether with excess hydrobromic acid. And we're going to heat things up. And when we think about our products, we know that the ether's going to go away. And we know that we're going to get two alkyl halides out So we just need to find our alkyl groups. So if I look over here, here's one of my alkyl groups. And if I look over here, here's my other alkyl group. So all I have to do is turn those alkyl groups into alkyl halides. And they're going to be alkyl bromides, since we're using hydrobromic acid here. So I'm going to draw one of my alkyl halides like that. So it would be bromine attached to the ring. And then my other alkyl halide will be this methyl group over here. So I take a methyl group and I attach it to bromine. And we'd also produce water, as well. But your major organic products would be these two alkyl halides. Let's do another one. This one will make it a little bit tricky. So in three dimensions-- so we're going to have to think about acidic cleavage of ethers in three dimensions. So it makes it much harder. So if I look at an ether, which looks like this, and I add, once again, excess hydrobromic acid, and I heat things up, I'm going to get acidic cleavage of that ether. And when I'm trying to figure out the products, I know that oxygen's going to go away. And I know that the carbons that are bonded to that oxygen are the ones that are going to form my alkyl halides. So I look at this carbon that's bonded to my oxygen. That's going to be bonded to a halogen. And if I look at this carbon on the other side of my ether," + }, + { + "Q": "At 7:16 what are the names of those molecule?\n", + "A": "It can be named 3-aminopropanal or 3-aminopropionaldehyde", + "video_name": "GuaozMpFS3w", + "timestamps": [ + 436 + ], + "3min_transcript": "that there are only single-bonds around that carbon, only sigma bonds, and so, therefore we know that carbon is SP three hybridized, with tetrahedral geometry, so SP three hybridized, tetrahedral geometry. All right, let's look at this carbon right here; it's the exact same situation, right, only sigma, or single bonds around it, so this carbon is also SP three hybridized, and so, therefore tetrahedral geometry. Let's next look at the oxygen here, so if I wanted to figure out the hybridization and the geometry of this oxygen, steric number is useful here, so let's go ahead and calculate the steric number of this oxygen. So that's number of sigma bonds, so here's a single-bond, so that's a sigma bond, and then here's another one; so I have two sigma bonds, so two plus number of lone pairs of electrons around the atom, so here's a lone pair of electrons, and here's a lone pair of electrons. So, I have two lone pairs of electrons, so I need four hybridized orbitals for this oxygen, and we know that occurs when you have SP three hybridization, so therefore, this oxygen is SP three hybridized: There are four SP three hybrid orbitals around that oxygen. All right, let's do geometry of this oxygen. So, the electron groups, there are four electron groups around that oxygen, so each electron group is in an SP three hydbridized orbital. The geometry of those electron groups might be tetrahedral, but not the geometry around the oxygen here, so the geometry around the oxygen, if you ignore the lone pairs of electrons, you can see that it is bent, so even though that oxygen is SP three hybridized, it's geometry is not tetrahedral; the geometry of that oxygen there is bent or angual. All right, and because of symmetry, this carbon right here is the same as this carbon, so it's also is the same as this carbon, so it's also SP three hybridized, so symmetry made our lives easy on this one. All right, let's do one more example. So, once again, our goal is to find the hybridization states, and the geometries for all the atoms, except for hydrogen, and so, once again, let's start with carbon; let's start with this carbon, right here. All right, so once again, our goal is to find the hybridization state, so the fast way of doing it, is to notice there's one double-bond to that carbon, so it must be SP two hybridized, and therefore the geometry is trigonal planar, so trigonal planar geometry. Let's do the steric number way, so if I were to calculate the steric number: Steric number is equal to the number of sigma bonds. So here's a sigma bond, here's a sigma bond; I have a double-bond between the carbon and the oxygen, so one of those is a sigma bond," + }, + { + "Q": "\nI also wanted to ask that though in the eg at 7:00, we figured out that the S.N. of oxygen and I have clearly understood the concept of finding the steric no. but I wanted to know that how can oxygen with its elec. confi. 1s2 2s2 2px2 2py1 2pz1 excite its electron(or by any other means) form 4 sp3 hybridised just as we have seen so clearly in the case of carbon.", + "A": "You don t excite the electrons. You hybridize the orbitals first and then puts the electrons in. So you hybridize the s orbital and two 2pz orbitals. This gives 2sp\u00c2\u00b2, 2sp\u00c2\u00b2, 2sp\u00c2\u00b2 and an unhybridized 2pz. The 6 electrons of O give the configuration (2sp\u00c2\u00b2)\u00c2\u00b2, (2sp\u00c2\u00b2)\u00c2\u00b2, (2sp\u00c2\u00b2)\u00c2\u00b9, (2pz)\u00c2\u00b9. The filled sp\u00c2\u00b2 orbitals are the lone pairs, the half-filled sp\u00c2\u00b2 orbital forms the \u00cf\u0083 bond to C, and the half-filled pz orbital forms the \u00cf\u0080 bond to C.", + "video_name": "GuaozMpFS3w", + "timestamps": [ + 420 + ], + "3min_transcript": "that there are only single-bonds around that carbon, only sigma bonds, and so, therefore we know that carbon is SP three hybridized, with tetrahedral geometry, so SP three hybridized, tetrahedral geometry. All right, let's look at this carbon right here; it's the exact same situation, right, only sigma, or single bonds around it, so this carbon is also SP three hybridized, and so, therefore tetrahedral geometry. Let's next look at the oxygen here, so if I wanted to figure out the hybridization and the geometry of this oxygen, steric number is useful here, so let's go ahead and calculate the steric number of this oxygen. So that's number of sigma bonds, so here's a single-bond, so that's a sigma bond, and then here's another one; so I have two sigma bonds, so two plus number of lone pairs of electrons around the atom, so here's a lone pair of electrons, and here's a lone pair of electrons. So, I have two lone pairs of electrons, so I need four hybridized orbitals for this oxygen, and we know that occurs when you have SP three hybridization, so therefore, this oxygen is SP three hybridized: There are four SP three hybrid orbitals around that oxygen. All right, let's do geometry of this oxygen. So, the electron groups, there are four electron groups around that oxygen, so each electron group is in an SP three hydbridized orbital. The geometry of those electron groups might be tetrahedral, but not the geometry around the oxygen here, so the geometry around the oxygen, if you ignore the lone pairs of electrons, you can see that it is bent, so even though that oxygen is SP three hybridized, it's geometry is not tetrahedral; the geometry of that oxygen there is bent or angual. All right, and because of symmetry, this carbon right here is the same as this carbon, so it's also is the same as this carbon, so it's also SP three hybridized, so symmetry made our lives easy on this one. All right, let's do one more example. So, once again, our goal is to find the hybridization states, and the geometries for all the atoms, except for hydrogen, and so, once again, let's start with carbon; let's start with this carbon, right here. All right, so once again, our goal is to find the hybridization state, so the fast way of doing it, is to notice there's one double-bond to that carbon, so it must be SP two hybridized, and therefore the geometry is trigonal planar, so trigonal planar geometry. Let's do the steric number way, so if I were to calculate the steric number: Steric number is equal to the number of sigma bonds. So here's a sigma bond, here's a sigma bond; I have a double-bond between the carbon and the oxygen, so one of those is a sigma bond," + }, + { + "Q": "\nThis might be a silly question, but at 8:45 when he draws in the H - does it matter that he draws it down? Could it drawn up?", + "A": "The carbon bonding with the H you mentioned is sp3, so it tetrahedral geometric. It is a common way to draw the H like that", + "video_name": "7p2qfyqiXHc", + "timestamps": [ + 525 + ], + "3min_transcript": "we have carbon with three bonds to hydrogen and one loan pair of electrons on this carbon, a negative one formal charge, so we can represent that here with our negative sign next to that carbon. A carbon with a negative charge is called a carbanion, so this is a carbanion, and let's analyze the pattern that we have for our carbanion. We have one, two, three bonds, so let me write that down, we have three bonds, and this time we have one loan pair of electrons, so we have one loan pair, so three single bonds plus one loan pair of electrons for a carbon will give us a negative one formal charge on that carbon, we will have a carbanion. These also come up in mechanisms in organic chemistry, so let's analyze some carbanions, so down here, let's start with the carbanion on the left, and the negative one formal charge is on So we should have three bonds and one loan pair of electrons on that carbon, well, let's analyze it, the carbon in red is bonded to a CH3, a CH3, and a CH3, so that takes care of our three bonds, and of course, here's the one loan pair of electrons. Let's move on to the next example, so the carbon with the negative one formal charge is this carbon, that I just marked in red, the carbon in red is directly bonded to a carbon here, and directly bonded to a carbon here, so that's two bonds, I need a total of three bonds. 'Cause I already have a loan pair of electrons, right here's my loan pair of electrons on that carbon, so I need one more bond, and that bond, of course, must be to a hydrogen, so I can draw in a hydrogen here, again, that hydrogen is usually left off when you're drawing dot structures, but it's important to realize that hydrogen is actually there. Finally, one more example. The negative one formal charge is on this carbon, it already has a loan pair of electrons, so so far, we have one bond and one loan pair, we need a total of three bonds, so we need two more bonds on that carbon in red, and so those last two bonds, of course, must be to two hydrogens. So it's important to be able to assign formal charge and to do the math, it's important to be able to do these calculations, but eventually you won't need to do the math anymore, eventually you'll be able to look at a carbon and come with a formal charge after you've done enough problems." + }, + { + "Q": "At 2:53 it is mentioned that in this example carbon only has 6 electrons around it, but what about the single electron in P orbital that is not bonding?\n", + "A": "With a carbocation there is no electron in that p orbital. Another atom has taken it when the carbocation was formed.", + "video_name": "7p2qfyqiXHc", + "timestamps": [ + 173 + ], + "3min_transcript": "so I could give that one valence electron back to hydrogen, and one valence electron to carbon, and so we're going to divide up all of our bonds that way, alright, give one valence electron to hydrogen, and the other valence electron to carbon, because that makes it easier for us to see that carbon has four valence electrons in our drawing, so let me highlight them here, one, two, three, four. So we're going to subtract four from four, so four minus four is equal to zero. So carbon has a formal charge of zero in methane. Let's do another example, this one down here. You can see it's different because this time we have three bonds. So let me draw in the electrons in those bonds, and let's find the formal charge on carbon. The formal charge on carbon is equal to the number of valence electrons that carbon is supposed to have, which we know is four, and from that we subtract the number of valence electrons We divide up the electrons in our bonds, just like we did before, and we can see that carbon has only three electrons around it this time, so I'll highlight those, one, two, and three. So four minus three is equal to plus one, so carbon has a formal charge of plus one. So carbon's supposed to have four valence electrons, it has only three around it, so it lost one of its electrons, which gives it a formal charge of plus one. Let me go ahead and redraw that, so over here on the right, we have carbon with three bonds to hydrogen, and this carbon has a plus one formal charge, so we can represent that by putting a + charge here next to the carbon. Notice that carbon does not have an octet of electrons around it, it has only six electrons around it, and that's actually okay, carbon can never exceed an octet, but it's okay for carbon to have less than eight electrons. it's a positively charged carbon, we call those carbocations, so let me write down here, this is a carbocation. And carbocations come up a lot in organic chemistry mechanisms, so it's really important to understand them. Alright, let's think about the pattern that we see here, we have three single bonds around this carbon, let me go ahead and highlight them here, so here's one, two, and three, so we have three single bonds around our carbon, and we have zero lone pairs of electrons around that carbon, so three bonds plus zero loan pairs of electrons will give you a positively charged carbon, will give you a carbocation. What is the hybridization of this positively charged carbon? Well, there's one, two, three single bonds and zero loan pairs of electrons, and so from the videos on hybridization, you should know that" + }, + { + "Q": "When he was talking in the pic about 1:25 your mouse shifted over long waves what are they?\n", + "A": "long waves are the lowest part of the electromagnetic radiation scale even less than your tv and radio produce", + "video_name": "PX_XSnVWlNc", + "timestamps": [ + 85 + ], + "3min_transcript": "I want to make a quick correction to the video on quasars. In that video I said, and I mistakenly said that the creation disk that's really releasing the energy of the quasar, that it's releasing energy predominately in the x-ray part of the electromagnetic spectrum. And that was incorrect. Most quasars are actually emitting electromagnetic radiation across the spectrum, all the way from x-rays, as high frequency as x-rays, all the way down to infrared. And some quasars even release super high frequency gamma rays, and they'll release low frequency electromagnetic waves all the way down to radio waves. So I just wanted to make that correction. It's not predominately in the x-ray part of the spectrum. It's across the spectrum right over here. It's this entire range of the spectrum, and sometimes even a wider range. Now, the other thing I want to clarify is this is the range of the spectrum that's being emitted. But we have to remember that most, or actually all of these quasars are quite far away. Many of them are many, many billions of light years away. And so they're moving away from us at a very fast speed, or they're getting redshifted because the universe is expanding so fast relative to us at that point, or that coordinate is moving so fast away from our coordinate. And so even though this is the spectrum that's being emitted, it's all going to be redshifted. It's all going to be redshifted down, and so we are going to observe things at a much lower frequency, maybe around the radio part of the frequency. So everything will be redshifted down. And that's why these were originally called quasi stellar radio sources. Anyway, hopefully that clears things up a little bit." + }, + { + "Q": "\n5:38 What does cohesion mean?", + "A": "Cohesion is the intermolecular attraction between like-molecules.", + "video_name": "6G1evL7ELwE", + "timestamps": [ + 338 + ], + "3min_transcript": "This is going to form a partial negative charge at the, I guess you could say, the non-hydrogen end that is the end that has, that's well I guess this top end, the way I've drawn it right over here. And this Greek letter delta, this is to signify a partial charge, and it's a partial negative charge. Because electrons are negative. And then over here, since you have a slight deficiency of electrons, because they're spending so much time around the oxygen, it forms a partial positive charge right over there. So right when you just look at one water molecule, that doesn't seem so interesting. But it becomes really interesting when you look at many water molecules interacting together. So let me draw another water molecule right over here. So it's oxygen, you have two hydrogens, and then you have the bonds between them. You have a partially negative charge there. Partially positive charge on that end. And so you can imagine the partial, the side that has a partially negative charge is going to be And that attraction between these two, this is called a hydrogen bond. So that right over there is called a hydrogen bond. And this is key to the behavior of water. And we're going to see that in future videos. All the different ways that hydrogen bonds give water its unique characteristics. Hydrogen bonds are weaker than covalent bonds, but they're strong enough to give water that kind of nice fluid nature when we're thinking about kind of normal, or you could say, normal temperatures and pressures. This nice fluid nature, it allows these things to be attracted to each other, to have some cohesion, but also to break and reform and flow past each other. So you can imagine another hydrogen bond with another water molecule right over here. So put my hydrogens over there. Put my hydrogens, your bonds, partial negative, partial positive right over there. And so we'll see in future videos, hydrogen bonds, Key for its properties to its ability to take in heat. Key for its ability to regulate temperature. The key for its ability is why lakes don't freeze over. It's key for some of it's properties around evaporative cooling and surface tension and adhesion and cohesion, and we'll see that. And probably most important, and it's hard to rank of these things, if we're thinking about biological systems, this polarity that we have in water molecules and these hydrogen bonding, it's key for its ability to be a solvent, for it to be able to have polar molecules be dissolved inside of water. And we'll see that in future videos." + }, + { + "Q": "\nwhy is oxygen a electronegative element at 3:16- 3:18?? and why is it way more electronegative that hydrogen at 3:23-3:25??", + "A": "See the electronegativity values for oxygen and hydrogen and you ll see that the difference is great that s why oxygen is more electronegative than hydrogen. Also, as you go through the periodic table, from right to left, electronegativity increases. See the periodic trends for electronegativity for better understanding.", + "video_name": "6G1evL7ELwE", + "timestamps": [ + 196, + 198, + 203, + 205 + ], + "3min_transcript": "And they are bonded with covalent bonds. And covalent bonds, each of these bonds is this pair of electrons that both of these atoms get to pretend like they have. And so you have these two pairs. And you might be saying, \"Well, why did I draw \"the two hydrogens on this end? \"Why didn't I draw them on opposite sides of the oxygen?\" Well that's because oxygen also has two lone electron pairs. Two lone electron pairs. And these things are always repelling each other. The electrons are repelling from each other, and so, in reality if we were looking at it in three dimensions, the oxygen molecule is kind of a tetrahedral shape. I could try to, let me try to draw it a little bit. So if this is the oxygen right over here then you would have, you could have maybe one lone pair of electrons. I'll draw it as a little green circle there. Another lone pair of electrons back here. Then you have the covalent bond. You have the covalent bond to And then you have the covalent bond to the other hydrogen atom. And so you see it forms this tetrahedral shape, It's pretty close to a tetrahedron. Just like this, but the key is that the hydrogens are on one end of the molecule. And this is, we're going to see, very very important to the unique properties, or to the, what gives water its special properties. Now, one thing to realize is, it's very, in chemistry we draw these electrons very neatly, these dots up here. We draw these covalent bonds very neatly. But that's not the way that it actually works. Electrons are jumping around constantly. They're buzzing around, it's actually much more of a, even when you think about electrons, it's more of a probability of where you might find them. And so instead of thinking of these electrons as definitely here or definitely in these bonds, They're actually more of in this cloud around the different atoms. They're in this cloud that kind of describes a probability and they jump around. And what's interesting about water is oxygen is extremely electronegative. So oxygen, that's oxygen and that's oxygen, it is extremely electronegative, it's one of the more electronegative elements we know of. It's definitely way more electronegative than hydrogen. And you might be saying, \"Well, Sal, \"what does it mean to be electronegative?\" Well, electronegative is just a fancy way of saying that it hogs electrons. It likes to keep electrons for itself. Hogs electrons, so that's what's going on. Oxygen like to keep the electrons more around itself than the partners that it's bonding with. So even in these covalent bonds, you say, \"Hey, we're supposed to be sharing these electrons.\" Oxygen says, \"Well I still want them to \"spend a little bit more time with me.\" And so they actually do spend more time on the side without the hydrogens than they do around the hydrogens." + }, + { + "Q": "\n2:18 Why are only 4 electrons shown? There should be 8 electrons shown.", + "A": "Each bubble represents either a lone pair of electrons or a covalent bond in both cases they represent 2 electrons therefore 8 in total", + "video_name": "6G1evL7ELwE", + "timestamps": [ + 138 + ], + "3min_transcript": "- [Voiceover] I don't think it's any secret to anyone that water is essential to life. Most of the biological, or actually in fact all of the significant biological processes in your body are dependent on water and are probably occurring inside of water. When you think of cells in your body, the cytoplasm inside of your cells, that is mainly water. In fact, me, who is talking to you right now, I am 60% to 70% water. You could think of me as kind of this big bag of water making a video right now. And it's not just human beings that need water. Life as we know it is dependent on water. That why when we have the search for signs of life on other planets we're always looking for signs of water. Maybe life can occur in other types of substances, but water is essential to life as we know it. And to understand why water is so special let's start to understand the structure of water and how it interacts with itself. And so water, as you probably already know, is made up of one oxygen atom and two hydrogen atoms. And they are bonded with covalent bonds. And covalent bonds, each of these bonds is this pair of electrons that both of these atoms get to pretend like they have. And so you have these two pairs. And you might be saying, \"Well, why did I draw \"the two hydrogens on this end? \"Why didn't I draw them on opposite sides of the oxygen?\" Well that's because oxygen also has two lone electron pairs. Two lone electron pairs. And these things are always repelling each other. The electrons are repelling from each other, and so, in reality if we were looking at it in three dimensions, the oxygen molecule is kind of a tetrahedral shape. I could try to, let me try to draw it a little bit. So if this is the oxygen right over here then you would have, you could have maybe one lone pair of electrons. I'll draw it as a little green circle there. Another lone pair of electrons back here. Then you have the covalent bond. You have the covalent bond to And then you have the covalent bond to the other hydrogen atom. And so you see it forms this tetrahedral shape, It's pretty close to a tetrahedron. Just like this, but the key is that the hydrogens are on one end of the molecule. And this is, we're going to see, very very important to the unique properties, or to the, what gives water its special properties. Now, one thing to realize is, it's very, in chemistry we draw these electrons very neatly, these dots up here. We draw these covalent bonds very neatly. But that's not the way that it actually works. Electrons are jumping around constantly. They're buzzing around, it's actually much more of a, even when you think about electrons, it's more of a probability of where you might find them. And so instead of thinking of these electrons as definitely here or definitely in these bonds, They're actually more of in this cloud around the different atoms. They're in this cloud that kind of describes a probability" + }, + { + "Q": "\nat 5:31 sal mentions \"action potential\". what is it?", + "A": "An action potential is a short-lasting event in which the electrical membrane potential of a cell rapidly rises and falls, following a consistent trajectory. You will see this often in physiology, particularly is neurons because this is how signals travel through our bodies", + "video_name": "ob5U8zPbAX4", + "timestamps": [ + 331 + ], + "3min_transcript": "actually work, but it's good just to have the anatomical structure first. So these are called Schwann cells and they're covering-- they make up the myelin sheath. So this covering, this insulation, at different intervals around the axon, this is called the myelin sheath. So Schwann cells make up the myelin sheath. I'll do one more just like that. And then these little spaces between the myelin sheath-- just so we have all of the terminology from-- so we know the entire anatomy of the neuron-- these are called the nodes of Ranvier. I guess they're named after Ranvier. Maybe he was the guy who looked and saw they had these little slots here where you don't have myelin sheath. So the general idea, as I mentioned, is that you get a signal here. We're going to talk more about what the signal means-- and then that signal gets-- actually, the signals can be summed, so you might have one little signal right there, another signal right there, and then you'll have maybe a larger signal there and there-- and that the combined effects of these signals get summed up and they travel to the hillock and if they're a large enough, they're going to trigger an action potential on the axon, which will cause a signal to travel down the balance of the axon and then over here it might be connected via synapses to other dendrites or muscles. And we'll talk more about synapses and those might help trigger other things. So you're saying, what's triggering these things here? Well, this could be the terminal end of other neurons' axons, like in the brain. This could be some type of sensory neuron. This could be on a taste bud someplace, so a salt molecule be some type of sensor. It could be a whole bunch of different things and we'll talk more about the different types of neurons." + }, + { + "Q": "at 4:22, sam says the axon is covered by schwann cells which form the myelin sheath. but as far as i know myelin sheath have cells called schwann cells. i am confused. please help.\n", + "A": "The myelin sheath is made by Schwann cells. The Schwann cells wrap around the axon and produce myelin around it, creating the myelin sheath", + "video_name": "ob5U8zPbAX4", + "timestamps": [ + 262 + ], + "3min_transcript": "probably in the next few. So this is where it receives the signal. So this is the dendrite. This right here is the soma. Soma means body. This is the body of the neuron. And then we have kind of a-- you can almost view it as a tail of the neuron. It's called the axon. A neuron can be a reasonably normal sized cell, although there is a huge range, but the axons can be quite long. They could be short. Sometimes in the brain you might have very small axons, but you might have axons that go down the spinal column or that go along one of your limbs-- or if you're talking about one of a dinosaur's limbs. So the axon can actually stretch several feet. Not all neurons' axons are several feet, but they could be. And this is really where a lot of the distance of the signal gets traveled. Let me draw the axon. So the axon will look something like this. connect to other dendrites or maybe to other types of tissue or muscle if the point of this neuron is to tell a muscle to do something. So at the end of the axon, you have the axon terminal right there. I'll do my best to draw it like that. So this is the axon. This is the axon terminal. And you'll sometimes hear the word-- the point at which the soma or the body of the neuron connects to the axon is as often referred to as the axon hillock-- maybe you can kind of view it as kind of a lump. It starts to form the axon. And then we're going to talk about how the impulses travel. And a huge part in what allows them to travel efficiently are these insulating cells around the axon. actually work, but it's good just to have the anatomical structure first. So these are called Schwann cells and they're covering-- they make up the myelin sheath. So this covering, this insulation, at different intervals around the axon, this is called the myelin sheath. So Schwann cells make up the myelin sheath. I'll do one more just like that. And then these little spaces between the myelin sheath-- just so we have all of the terminology from-- so we know the entire anatomy of the neuron-- these are called the nodes of Ranvier. I guess they're named after Ranvier. Maybe he was the guy who looked and saw they had these little slots here where you don't have myelin sheath." + }, + { + "Q": "why does sal say temperature changes at 1:50, if PV is the only thing determining it? i mean, if PV is constant throughout the process, and there is no loss of kinetic energy, how can temperature become lower?\n", + "A": "Temperature will change because PV is determining it. So you logic is right, and so is Sal s. I think you just misinterpreted the way Sal said it.", + "video_name": "lKq-10ysDb4", + "timestamps": [ + 110 + ], + "3min_transcript": "SAL: In the last video, where we talked about macrostates, we set up this situation where I had this canister, or the cylinder, and had this movable ceiling. I call that a piston. And the piston is being kept up by the pressure from the And it's being kept down by, in the last example I had, a rock or a weight on top. And above that I had a vacuum. So essentially there's some force per area, or pressure, being applied by the bumps of the particles into this piston. And if this weight wasn't here-- let's assume that the piston itself or this movable ceiling itself, it has no mass-- if that weight wasn't there it would just be pushed indefinitely far, because there'd be no pressure from But this weight is applying some force on that same area downwards. So we're at some equilibrium point, some stability. And we plotted that on this PV diagram right here. I'll do it in magenta. So that's our state 1 that we were in right there. And then what I did in the last video, I just blew away And as soon as I blew away half of this block, obviously the force that's being applied by the block will immediately go down by half, and so the gas will push up on it. And it happened so fast that, al of a sudden the gas is pushing up. Right when it happens, the gas near the top of the canister is going to have lower pressure, because it has less pushing up against it. The molecules that are down here don't even know that I blew away this block yet. It's going to take some time. And essentially the gas is going to push it up, and then maybe it'll oscillate down, and then push it up, and oscillate down a little bit. It'll take some time eventually until we get to another equilibrium state, where we have a new, probably, or definitely lower pressure. We definitely have a higher volume. I won't talk too much about it yet, but we probably have a lower temperature as well. And this is our new state. And our macrostate's pressure and volume are defined once we're at the new equilibrium, so we're right here. how did we get here? Is there any way to have defined a path to get from our first state-- where pressure and volume were well defined, because the system wasn't thermodynamic equilibrium-- to get to our second state? And the answer was no. Because between this state and this state all hell broke loose. I had different temperatures at different points in the system. I could have had a different pressure here than I had up here. The volume might have been fluctuating from moment to moment. So when you're outside of equilibrium-- and I had written it down over there-- you cannot define, or you can't say that those macro variables are well defined. So there was no path that you could say how we got from-- erase this-- how we got from state 1 to state 2. You could just say, OK, we were in some type of equilibrium. So we were in state 1." + }, + { + "Q": "At 0:26 can we determine the value of this pressure? If so, how?\n", + "A": "i think it is P = t/v (temp/volume) as volume is inversely proportional to pressure i.e if p increases v decreases and vice-versa", + "video_name": "lKq-10ysDb4", + "timestamps": [ + 26 + ], + "3min_transcript": "SAL: In the last video, where we talked about macrostates, we set up this situation where I had this canister, or the cylinder, and had this movable ceiling. I call that a piston. And the piston is being kept up by the pressure from the And it's being kept down by, in the last example I had, a rock or a weight on top. And above that I had a vacuum. So essentially there's some force per area, or pressure, being applied by the bumps of the particles into this piston. And if this weight wasn't here-- let's assume that the piston itself or this movable ceiling itself, it has no mass-- if that weight wasn't there it would just be pushed indefinitely far, because there'd be no pressure from But this weight is applying some force on that same area downwards. So we're at some equilibrium point, some stability. And we plotted that on this PV diagram right here. I'll do it in magenta. So that's our state 1 that we were in right there. And then what I did in the last video, I just blew away And as soon as I blew away half of this block, obviously the force that's being applied by the block will immediately go down by half, and so the gas will push up on it. And it happened so fast that, al of a sudden the gas is pushing up. Right when it happens, the gas near the top of the canister is going to have lower pressure, because it has less pushing up against it. The molecules that are down here don't even know that I blew away this block yet. It's going to take some time. And essentially the gas is going to push it up, and then maybe it'll oscillate down, and then push it up, and oscillate down a little bit. It'll take some time eventually until we get to another equilibrium state, where we have a new, probably, or definitely lower pressure. We definitely have a higher volume. I won't talk too much about it yet, but we probably have a lower temperature as well. And this is our new state. And our macrostate's pressure and volume are defined once we're at the new equilibrium, so we're right here. how did we get here? Is there any way to have defined a path to get from our first state-- where pressure and volume were well defined, because the system wasn't thermodynamic equilibrium-- to get to our second state? And the answer was no. Because between this state and this state all hell broke loose. I had different temperatures at different points in the system. I could have had a different pressure here than I had up here. The volume might have been fluctuating from moment to moment. So when you're outside of equilibrium-- and I had written it down over there-- you cannot define, or you can't say that those macro variables are well defined. So there was no path that you could say how we got from-- erase this-- how we got from state 1 to state 2. You could just say, OK, we were in some type of equilibrium. So we were in state 1." + }, + { + "Q": "\nAt 03:06 Sal says\"the temperature also probably went down\" Why would that be?", + "A": "When the volume increases, the pressure and temperature usually both go down. Both pressure and temperature are products of molecular motion. When the same amount of molecules occupy a larger space, then they bounce off the walls less frequently. Therefore, the pressure and temperature are lower.", + "video_name": "lKq-10ysDb4", + "timestamps": [ + 186 + ], + "3min_transcript": "And as soon as I blew away half of this block, obviously the force that's being applied by the block will immediately go down by half, and so the gas will push up on it. And it happened so fast that, al of a sudden the gas is pushing up. Right when it happens, the gas near the top of the canister is going to have lower pressure, because it has less pushing up against it. The molecules that are down here don't even know that I blew away this block yet. It's going to take some time. And essentially the gas is going to push it up, and then maybe it'll oscillate down, and then push it up, and oscillate down a little bit. It'll take some time eventually until we get to another equilibrium state, where we have a new, probably, or definitely lower pressure. We definitely have a higher volume. I won't talk too much about it yet, but we probably have a lower temperature as well. And this is our new state. And our macrostate's pressure and volume are defined once we're at the new equilibrium, so we're right here. how did we get here? Is there any way to have defined a path to get from our first state-- where pressure and volume were well defined, because the system wasn't thermodynamic equilibrium-- to get to our second state? And the answer was no. Because between this state and this state all hell broke loose. I had different temperatures at different points in the system. I could have had a different pressure here than I had up here. The volume might have been fluctuating from moment to moment. So when you're outside of equilibrium-- and I had written it down over there-- you cannot define, or you can't say that those macro variables are well defined. So there was no path that you could say how we got from-- erase this-- how we got from state 1 to state 2. You could just say, OK, we were in some type of equilibrium. So we were in state 1. The pressure went down, the volume went up. The temperature also probably went down. And so I ended up in this other state once I reached equilibrium. And that's all fair and good, but wouldn't it have been nice if there was some way? If we could have said, look, you know, there's some way that we got from this point to this point? If we could perform my little rock experiment in a slightly different manner, so that all this hell didn't break loose, so that maybe at every point in between my macro variables are actually defined? So how could I do that? Remember, I said that the macro variables, the macrostates, whether it's pressure, temperature, volume, and there are others, but I said these are only defined when we are in a thermodynamic equilibrium. And that just means that things have reached a stability point. That, for example, the temperature is consistent throughout the system. If it's not consistent throughout the system, I shouldn't be talking about it. If the temperature is different here than it is up here, I shouldn't say that the temperature of" + }, + { + "Q": "If calories, or kilocalories, (3:00) are heat or energy in your body, why is it unhealthy to have to many calories in your body?\n", + "A": "The energy is stored in fat, and evidence suggests that excess fat is not good for your health.", + "video_name": "h-31O7CaF2o", + "timestamps": [ + 180 + ], + "3min_transcript": "I have to do a better job of drawing sand. So this is sand right over here. If I wanna raise that one degree celsius, I would need a different amount. It actually turns out that I need less heat, I need less heat to raise the sand one degree celsius than I need to raise the temperature of the water or one gram of the water one degree celsius. So let's say this is a gram of water and this is a gram of sand. I'm going to need more heat here to raise this one degree celsius than to raise that. That's because water has a higher specific heat. So higher, higher, relatively higher specific heat. Specific heat. And sand, or at least relative to water, has a lower specific heat. Lower specific heat. So two ways you could think about it. Let me write this. Lower, lower specific, specific heat. Two ways to think about it. If you wanna raise one gram of each of them into the water than you're going to have to put into the sand. Or the other way around, if you put the same amount of energy into both, you're gonna raise the temperature of the sand a lot more than you would raise the temperature of the water. And water, actually, its specific heat has a special name and this is a name that you have seen before. The specific heat of water is called the calorie. Specific, specific heat of water is called the calorie. And you have seen this word before. When you've wanted to cut calories, when you've looked at the back of nutritional labels on food. There's one clarification. The calorie that people talk about when they're talking about nutritional labels or how many calories are actually in food, that's actually kilocalories. So if you see, if someone hands you a, let's say a bowl of ice cream. Let me draw a bowl of ice cream here. So if someone hands you a bowl of ice cream right over here and they tell you that this is 500 calories. This is 500 calories. If we're thinking of it in terms of specific heat, it's actually 500 kilocalories. 500 kilocalories. So there's a couple of ways that you could think about 500 kilocalories. You could think about it as, this is, this ice cream has enough energy to raise, to raise 500 kilograms of water one degree celsius. You could also view it as the amount, well actually if you wanna think of it in more human terms, most humans are roughly grown people are between 50 kilograms or 60, 70 kilograms roughly over there. You could say 50 kilograms of water. Actually a grown male might be composed of about 50 kilograms of water and then there's obviously other things that make up their weight, I'm just approximating. 50 kilograms of water" + }, + { + "Q": "\nAt 6:00 you said that sand heats up faster than water due to less specific heat. could that also be because the sun heats up the top inch or so of sand but penetrates deeper into the water? So the sun is heating up a greater volume of water than sand per area? Just curious.", + "A": "If you take the same amount of sand and water (and place it with the same shape), still sand is going to heat faster. You should see the last part of the video when Sal talks about the friction of the hydrogen bonds in water.", + "video_name": "h-31O7CaF2o", + "timestamps": [ + 360 + ], + "3min_transcript": "in the top case, but in this case you would raise it 10. You would raise it 10 degrees celsius. And that's actually happening in our body. Your body heat is actually caused, partially, the energy from food, some of it is to process your movement and the different functions of your muscles and the brain and all the things you body does, and some of it is just producing heat. Sometimes as a byproduct of that movement and sometimes, frankly, just for the sake of producing heat. So, 500 calories you see on a food label, that's really 500 kilocalories, and that's enough energy to raise 500,000 grams of water, remember 500 kilograms, 500,000 grams of water, one degree celsius, or 50,000 grams of water 10 degrees celsius. But anyway, water has a special name, it's calorie. It's neat to be able to connect it to, well, what we're eating and to think about what a calorie actually means, or what kilocalorie actually means. But this notion that water has a higher relative specific heat It's actually one of the reasons why it's often nice to live near the coast. Because, let me draw, let me draw a coastline. Right over here. This is, actually I'll draw it from, I'll draw it from above. So that's the coastline. This is land. Let's just say it's made up of sand, for the sake of argument, and other things. This is water. Let's think about it, first let's think about it in the summer. Let's think about in the summer, when it is hot. Think about just a sunny day. So in the summer, when things are hot, so you have the sun, you have the sun right over here and it's radiating energy and obviously the area also has, is also warming up the things that it comes in touch with. But you're having, at least at the coastline, the air is roughly the same temperature over both and, although there will be some variation, and they're getting the same amount of sunlight. But since water has a higher specific heat, So this is going to get less hot. Let me write it over here. This is going to get less hot. Less hot. And the land is going to get more hot. More hot. More hot. And that's why, when you're on the coastline, the air is also affected by what's in touch with it, it's gonna be in tough with this less hot water, and so as the air, especially if it's coming from the ocean, if the air is going in this direction, then if you're at the coast, you're gonna get probably a cooler breeze than if you're more inland over here. If you're more inland over here, the air is going to be hotter. So this is going to be hotter air if you're inland. And it's less hot air if you're at the coastline because the air is being cooled down to some degree, it's not being warmed up as much by the water. Now you have the opposite effect if you think about the coldest times of the year. If you think about the middle of the night in the winter. So let me draw that. If you think about nighttime in the winter," + }, + { + "Q": "\nAt 7:40 Sal shifts the parallel component down and joins it up to the Fg component. My question is, how can you know that this line will meet up exactly with the Fg component and therefore become a right angled triangle? If you don't know the length(force component) of the perpendicular line, how can you assume that it will be at a 90deg angle to the Fg line?", + "A": "The projected force is skewed by the angle theta of the ramp. Meaning if the ramp wasn t inclined then Fg1 would be a line perpendicular with the ground. Since the ramp is inclined we know that the force is displaced by the same angle, as Sal proved in the video. Thus, Fg1 forms a right angle with the ramp whether the ramp in inclined or not.", + "video_name": "TC23wD34C7k", + "timestamps": [ + 460 + ], + "3min_transcript": "This too will be 90 minus theta degrees. 90 minus theta degrees. Now, given that, can we figure out this angle? Well one thing, we're assuming that this yellow force vector right here is perpendicular to the surface of this plane or perpendicular to the surface of this ramp. So that's perpendicular. This right here is 90 minus theta. So what is this angle up here going to be equal to? This angle, let me do it in green. What is this angle up here going to be equal to? So this angle plus 90 minus theta plus 90 must be equal to 180, or this angle plus 90 minus theta must be equal to-- let me write this down. I don't want to do too much in your head. So let's call it x. So x plus 90 minus theta. Plus this 90 degrees right over here, plus this 90 degrees, needs to be equal to 180 degrees. So we subtract 90 twice, you subtract 180 degrees and you get x minus theta is equal to 0, or x is equal to theta. So whatever the inclination of the plane is or of this ramp, that is also going to be this angle right over here. And the value to that is that now we can use our basic trigonometry to figure out this component and this component of the force of gravity. And to see that a little bit clearer, let me shift this force vector down over here. The parallel component, let me shift it over here. And you can see the perpendicular component plus the parallel component is equal to the total force due to gravity. And you should also see that this is a right triangle that I have set up over here. This is parallel to the plane. This is perpendicular to the plane. And so we can use basic trigonometry to figure out the magnitudes of the perpendicular force due to gravity and the parallel force due to gravity. I'll do it over here. The magnitude of the perpendicular force due to gravity. Or I should say the component of gravity that's perpendicular to the ramp, the magnitude of that vector-- a lot of fancy notation but it's really just the length of this vector right over here. So the magnitude of this over the hypotenuse of this right triangle. Well, what the hypotenuse of this right triangle? Well, it's going to be the magnitude of the total gravitational force. I guess you could say that. And so you could say that is mg. We could write it like this. But that's really-- well, I could write it like that. And so this is going to be equal to what? We have the, if we're looking at this angle right here, we have the adjacent over the hypotenuse. Remember." + }, + { + "Q": "\nAt about 2:30, he draws the force due to gravity and the force of the gravity of the perpendicular. I'm confused on which one of those is the force of gravity normal.", + "A": "I don t get it", + "video_name": "TC23wD34C7k", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's say I have some type of a block here. And let's say this block has a mass of m. So the mass of this block is equal to m. And it's sitting on this-- you could view this is an inclined plane, or a ramp, or some type of wedge. And we want to think about what might happen to this block. And we'll start thinking about the different forces that might keep it in place or not keep it in place and all of the rest. So the one thing we do know is if this whole set up is near the surface of the Earth-- and we'll assume that it is for the sake of this video-- that there will be the force of gravity trying to bring or attract this mass towards the center of the Earth, and vice versa, the center of the earth towards this mass. So we're going to have some force of gravity. Let me start right at the center of this mass right over here. And so you're going to have the force of gravity. The force due to gravity is going to be equal to the gravitational field And so we'll call that g. We'll call that g times the mass. Let me just write it. The mass times the gravitational field near the surface of the Earth. And it's going to be downwards, we know that, or at least towards the surface of the Earth. Now, what else is going to be happening here? Well, it gets a little bit confusing, because you can't really say that normal force is acting directly against this force right over here. Because remember, the normal force acts perpendicular to a surface. So over here, the surface is not perpendicular to the force So we have to think about it a little bit differently than we do if this was sitting on level ground. Well, the one thing we can do, and frankly, that we should do, is maybe we can break up this force, the force due to gravity. We can break it up into components that are either perpendicular to the surface or that are parallel to the surface. what's likely to happen. What are potentially the netting forces, or balancing forces, over here? So let's see if we can do that. Let's see if we can break this force vector, the force due to gravity, into a component that is perpendicular to the surface of this ramp. And also another component that is parallel to the surface of this ramp. Let me do that in a different color. That is parallel to the surface of this ramp. And this is a little bit unconventional notation, but I'll call this one over here the force due to gravity that is perpendicular to the ramp. That little upside down t, I'm saying that's perpendicular. Because it shows a line that's perpendicular to, I guess, this bottom line, this horizontal line over there. And this blue thing over here, I'm going to call this the part of force due to gravity that is parallel." + }, + { + "Q": "\nIs the sign of parallel. || . and magnetude same?? Becuz in 7:56 sal tells the parallel lines as magnitude??", + "A": "parallel: || magnitude: ||x||", + "video_name": "TC23wD34C7k", + "timestamps": [ + 476 + ], + "3min_transcript": "This too will be 90 minus theta degrees. 90 minus theta degrees. Now, given that, can we figure out this angle? Well one thing, we're assuming that this yellow force vector right here is perpendicular to the surface of this plane or perpendicular to the surface of this ramp. So that's perpendicular. This right here is 90 minus theta. So what is this angle up here going to be equal to? This angle, let me do it in green. What is this angle up here going to be equal to? So this angle plus 90 minus theta plus 90 must be equal to 180, or this angle plus 90 minus theta must be equal to-- let me write this down. I don't want to do too much in your head. So let's call it x. So x plus 90 minus theta. Plus this 90 degrees right over here, plus this 90 degrees, needs to be equal to 180 degrees. So we subtract 90 twice, you subtract 180 degrees and you get x minus theta is equal to 0, or x is equal to theta. So whatever the inclination of the plane is or of this ramp, that is also going to be this angle right over here. And the value to that is that now we can use our basic trigonometry to figure out this component and this component of the force of gravity. And to see that a little bit clearer, let me shift this force vector down over here. The parallel component, let me shift it over here. And you can see the perpendicular component plus the parallel component is equal to the total force due to gravity. And you should also see that this is a right triangle that I have set up over here. This is parallel to the plane. This is perpendicular to the plane. And so we can use basic trigonometry to figure out the magnitudes of the perpendicular force due to gravity and the parallel force due to gravity. I'll do it over here. The magnitude of the perpendicular force due to gravity. Or I should say the component of gravity that's perpendicular to the ramp, the magnitude of that vector-- a lot of fancy notation but it's really just the length of this vector right over here. So the magnitude of this over the hypotenuse of this right triangle. Well, what the hypotenuse of this right triangle? Well, it's going to be the magnitude of the total gravitational force. I guess you could say that. And so you could say that is mg. We could write it like this. But that's really-- well, I could write it like that. And so this is going to be equal to what? We have the, if we're looking at this angle right here, we have the adjacent over the hypotenuse. Remember." + }, + { + "Q": "At 8:56 Sal refers to the perpendicular component Fg as the adjacent but I thought that was supposed to be the opposite component which kinda throws off all the calculations after this. Did I miss something because I am really confused.\n", + "A": "A side is termed as Adjacent or Opposite in reference to the the Angle you are considering. Here, Sal refers \u00d1\u00b2 (Theta). So the Fg\u00e2\u008a\u00a5 is the Adjacent side here. Hypotenuse is constant in position. But Adjacent and Opposite are variable, Because they change with respect to your preferred Angle.", + "video_name": "TC23wD34C7k", + "timestamps": [ + 536 + ], + "3min_transcript": "So we subtract 90 twice, you subtract 180 degrees and you get x minus theta is equal to 0, or x is equal to theta. So whatever the inclination of the plane is or of this ramp, that is also going to be this angle right over here. And the value to that is that now we can use our basic trigonometry to figure out this component and this component of the force of gravity. And to see that a little bit clearer, let me shift this force vector down over here. The parallel component, let me shift it over here. And you can see the perpendicular component plus the parallel component is equal to the total force due to gravity. And you should also see that this is a right triangle that I have set up over here. This is parallel to the plane. This is perpendicular to the plane. And so we can use basic trigonometry to figure out the magnitudes of the perpendicular force due to gravity and the parallel force due to gravity. I'll do it over here. The magnitude of the perpendicular force due to gravity. Or I should say the component of gravity that's perpendicular to the ramp, the magnitude of that vector-- a lot of fancy notation but it's really just the length of this vector right over here. So the magnitude of this over the hypotenuse of this right triangle. Well, what the hypotenuse of this right triangle? Well, it's going to be the magnitude of the total gravitational force. I guess you could say that. And so you could say that is mg. We could write it like this. But that's really-- well, I could write it like that. And so this is going to be equal to what? We have the, if we're looking at this angle right here, we have the adjacent over the hypotenuse. Remember. We can do this in a new color. SOH CAH TOA. Cosine is adjacent over hypotenuse. So this is equal to cosine of the angle. So cosine of theta is equal to the adjacent over the hypotenuse. So if you multiply both sides by the magnitude of the hypotenuse, you get the component of our vector that is perpendicular to the surface of the plane is equal to the magnitude of the force due to gravity times the cosine of theta. Times the cosine of theta. We'll apply this in the next video just so you can make the numbers a lot more concrete. Sometimes just the notation makes it confusing. You'll see it's really actually pretty straightforward. And then this second thing, we can use the same logic. If we think about the parallel vector right over here, the magnitude of the component of the force" + }, + { + "Q": "At about 2:30, he draws the force due to gravity and the force of the gravity of the perpendicular. I'm confused on which one of those is the force of gravity normal.\n", + "A": "The normal force is always perpendicular to the surface of contact.", + "video_name": "TC23wD34C7k", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's say I have some type of a block here. And let's say this block has a mass of m. So the mass of this block is equal to m. And it's sitting on this-- you could view this is an inclined plane, or a ramp, or some type of wedge. And we want to think about what might happen to this block. And we'll start thinking about the different forces that might keep it in place or not keep it in place and all of the rest. So the one thing we do know is if this whole set up is near the surface of the Earth-- and we'll assume that it is for the sake of this video-- that there will be the force of gravity trying to bring or attract this mass towards the center of the Earth, and vice versa, the center of the earth towards this mass. So we're going to have some force of gravity. Let me start right at the center of this mass right over here. And so you're going to have the force of gravity. The force due to gravity is going to be equal to the gravitational field And so we'll call that g. We'll call that g times the mass. Let me just write it. The mass times the gravitational field near the surface of the Earth. And it's going to be downwards, we know that, or at least towards the surface of the Earth. Now, what else is going to be happening here? Well, it gets a little bit confusing, because you can't really say that normal force is acting directly against this force right over here. Because remember, the normal force acts perpendicular to a surface. So over here, the surface is not perpendicular to the force So we have to think about it a little bit differently than we do if this was sitting on level ground. Well, the one thing we can do, and frankly, that we should do, is maybe we can break up this force, the force due to gravity. We can break it up into components that are either perpendicular to the surface or that are parallel to the surface. what's likely to happen. What are potentially the netting forces, or balancing forces, over here? So let's see if we can do that. Let's see if we can break this force vector, the force due to gravity, into a component that is perpendicular to the surface of this ramp. And also another component that is parallel to the surface of this ramp. Let me do that in a different color. That is parallel to the surface of this ramp. And this is a little bit unconventional notation, but I'll call this one over here the force due to gravity that is perpendicular to the ramp. That little upside down t, I'm saying that's perpendicular. Because it shows a line that's perpendicular to, I guess, this bottom line, this horizontal line over there. And this blue thing over here, I'm going to call this the part of force due to gravity that is parallel." + }, + { + "Q": "At 7:52, it is stated that v+ = v_in. But isn't v_in = v+ - v-? How are these two statements consistent with each other?\n", + "A": "In this video, v_in is the name of the voltage created by the voltage source on the left side of the circuit. It is connected directly to the v+ input of the opamp. In other opamp videos, the same variable name might be assigned a different meaning. like v_in = v+ - v- as you suggest. There are no standard rules or conventions for which signal gets to be called v_in. It all depends on the person who draws the schematic.", + "video_name": "_Ut-nQ535iE", + "timestamps": [ + 472 + ], + "3min_transcript": "Kay, so what I'm going to do next is I'm going to take this expression and stuff it right in there. Let's do that. See if we got enough room, okay now let's go over here. Now I can say that V out equals A times V plus minus V out times R2 over R1 plus R2, alright so far so good. Let's keep going, let's keep working on this. V not equals A times V plus minus A V not, R2 over R1 plus R2. terms over on the left hand side. Let's try that. So that gives me, V not plus A V not, times R2 over R1 plus R2 and that equals A times V plus, and actually I can change that now V plus is what? V plus is V in. Okay let's keep going I can factor out the V not. V not is one plus A R2 over R1 plus R2 and that equals AV in. and our original goal, we want to find V out in terms of V in. So I'm going to take this whole expression here and divide it over to the other side, so then I have just V not on this side, and V in on the other side. Make some more room. I can do that, I can say V not equals A V in divided by this big old expression, one plus A R2 over R1 plus R2. Alright so that's our answer. That's the answer. That's V out equals some function of V in. Now I want to make a really important observation here." + }, + { + "Q": "At 3:05 he said \"the water isn't exerting force on anything.\" but isn't it exerting force on the molecules around it?\n", + "A": "Everything has gravity, so yes. However, the gravity of a molecule of water is so minuscule it can largely be ignored.", + "video_name": "NGpJPz44JYc", + "timestamps": [ + 185 + ], + "3min_transcript": "You can get rid of the area 1 on both sides, and then you're saying that the velocity up here is equal to the rate at which the top of the surface moves down and is equal to 1/1,000 of the velocity of the liquid spurting out of this little hole. With that, we actually have the three variables for the left-hand side of Bernoulli's equation. What are the variables on the left-hand side? What is the pressure at this point where we have a hole? This is an important thin. When we talk about Bernoulli's-- let me rewrite Bernoulli's equation. It's P1 plus rho gh1 plus rho V1 squared over 2 is equal to We figured out all of these terms. Now let's figure out the things that we have to input here. What is the pressure at point two? You might want to say, and this was my initial reaction, too, and that why I made a mistake, is that what 's the pressure at this depth in the fluid? That's not what Bernoulli's equation is telling us. Bernoulli's equation is telling us actually what is the external pressure at that hole. When we did the derivation, we were saying how much work-- this was kind of the work term, although we played around with it a little bit. But if we look at the water that's spurting out of the exerting force against anything so it's not actually doing work. When we think about the pressure, the output pressure, it's not the pressure at that depth of the fluid. You should think of it as the external pressure at the hole. In this case, there is no external pressure at the hole. Let's say that if we closed the hole, then at that point, sure. The pressure would be the pressure that's being exerted by the outside of the canister to contain the water, in which case, we would end up with no velocity. The water wouldn't spurt anywhere. But now we're seeing the external pressure is zero. That's what the hole essentially creates. We're going to say that P2 is zero, so this pressure was zero, because we're in a vacuum. P2 is also zero, so both of these are zero. Remember, that's the external pressure." + }, + { + "Q": "At 10:50, when do you use the kb value or the ka value?\n", + "A": "When you set up the equation for the equilibrium constant, the products are on the top line and the reactants are on the bottom line (for example, K = [NH4+][[OH-]/[NH3]). If the reactant is a weak base (ie, NH3 in this example) then K is Kb. But if the reactant is a weak acid (eg, CH3COOH), then K is Ka. (For example, K = [H3O+][CH3COO-]/[CH3COOH].)", + "video_name": "223KLPnJCBI", + "timestamps": [ + 650 + ], + "3min_transcript": "And we're saying that we have zero for our initial concentration of hydroxide. So when the reaction comes to equilibrium here, for ammonia we would have 0.15 minus x. For ammonium, we have two sources right? So this is a common ion here. So we have 0.35 plus x. And then for hydroxide we would have just x. So since ammonia is acting as a weak base here, let's go ahead and write our equilibrium expression. And we would write Kb. And Kb for ammonia is 1.8 times 10 to the negative five. So this is equal to concentration of our products over reactants. So we have concentration of NH4+ times the concentration of OH- all over the concentration of our reactants, leaving out water. So we just have ammonia here, So let's go ahead and plug in what we have. For the concentration of ammonium, we have 0.35 plus x. So we put 0.35 plus x. For the concentration of hydroxide, we have x. So we go ahead and put an x in here. And then that's all over the concentration of ammonia at equilibrium, and we go over here, and for ammonia at equilibrium it's 0.15 minus x. So we write over here 0.15 minus x. And we can plug in the Kb value, 1.8 times 10 to the negative five. So let's go ahead and plug in the Kb. So we have 1.8 times 10 to the negative five is equal to, okay once again, we're going to make the assumption. So if we say that x is extremely small number, then we don't have to worry about it And so we just say this is equal to 0.35. So 0.35 plus x is pretty close to 0.35. So this is times x. And make sure you understand this x is this x. And then once again, 0.15 minus x, if x is a very small number, that's approximately equal to 0.15. So now, we would have this. And we need to solve for x. So let's go ahead and do that. Let's get out the calculator here. We need to solve for x. So 1.8 times 10 to the negative five times 0.15, and then we need to divide that by 0.35. And that gives us what x is equal to. And so x is equal to 7.7 times 10 to the negative six. So x is equal to 7.7 times 10 to the negative six. x represents the concentration" + }, + { + "Q": "\nAt 9:50 we have 0.35 M NH4NO3, but how is that the same as having 0.35 M NH4? How come we just completely ignore the NO3 part of the compound?", + "A": "because nitrate ions are mainly found as free ions on both sides of the equation, i.e, they are not bonded to any other ion while they are in solution, and thus are called spectator ions and can be neglected while writing a net ionic equation.", + "video_name": "223KLPnJCBI", + "timestamps": [ + 590 + ], + "3min_transcript": "take an H+ away from H2O, we form OH- or the hydroxide ion. So once again we start with our initial concentration. And we're going to pretend like it's one of the problems that we've been doing in earlier videos. So if we have 0.15 molar concentration of ammonia, we go ahead and put 0.15 here. We think about the change. Whatever concentration we lose for ammonia is the same concentration that we gain for ammonium since ammonia turns into ammonium. And therefore that's also the same concentration we gain for hydroxide. So one source for the ammonium ion it would be the protonation of ammonia. So that's one source. But we have an additional source because we also have 0.35 molar ammonium nitrate. So there's another source for ammonium ions. So we have 0.35 molar, so we go ahead and put 0.35 molar in here And we're saying that we have zero for our initial concentration of hydroxide. So when the reaction comes to equilibrium here, for ammonia we would have 0.15 minus x. For ammonium, we have two sources right? So this is a common ion here. So we have 0.35 plus x. And then for hydroxide we would have just x. So since ammonia is acting as a weak base here, let's go ahead and write our equilibrium expression. And we would write Kb. And Kb for ammonia is 1.8 times 10 to the negative five. So this is equal to concentration of our products over reactants. So we have concentration of NH4+ times the concentration of OH- all over the concentration of our reactants, leaving out water. So we just have ammonia here, So let's go ahead and plug in what we have. For the concentration of ammonium, we have 0.35 plus x. So we put 0.35 plus x. For the concentration of hydroxide, we have x. So we go ahead and put an x in here. And then that's all over the concentration of ammonia at equilibrium, and we go over here, and for ammonia at equilibrium it's 0.15 minus x. So we write over here 0.15 minus x. And we can plug in the Kb value, 1.8 times 10 to the negative five. So let's go ahead and plug in the Kb. So we have 1.8 times 10 to the negative five is equal to, okay once again, we're going to make the assumption. So if we say that x is extremely small number, then we don't have to worry about it" + }, + { + "Q": "\nAt 11:14, would I be correct if I said that increasing the resistance decreases the volume (I'm guessing that more resistance means less current to the speaker, and less current means less volume). Am I correct?", + "A": "you are correct it like water, a small rock would have low resistance but a big rock would have high resistance and so would let less water through", + "video_name": "xuQcB-oo-4U", + "timestamps": [ + 674 + ], + "3min_transcript": "When you press the button, the pin moves and the switch gets triggered. When you wanna snooze in the morning, you push this. It shifts the pin and it causes the sleep button to be triggered. That's kinda how that works. This is kind of ingenious in another way too because it holds a bunch of different things together. It's got the pins. It holds the speaker and it also holds the ferrite rod with the copper coil around it which functions as an antenna. That's an antenna for AM/FM radio. The signals come from here. We got a wire broken there. Signals come from here and they go to this thing which is a setup of four variable capacitors and they help to tune out frequencies we don't want. When we turn our dial, we can go right to 101.1 FM or 538 AMR, whichever station we want. This helps us to select those things. Those variable capacitors help us to filter out unwanted frequencies. and they can be used to oscillate at a particular frequency if they're coupled with a capacitor. That can be useful in performing radio functions as well. This guy right here is a, it's a radio chip. It's an IC chip that helps to demodulate or to separate the music or the signal that you want from the actual wave. AM is amplitude modulation so that means that the wave is changed in its height. FM is frequency modulation so that means that the wave is changed in how often it occurs in order to embed the signal that we get to listen to as radio sound. This chip basically decodes that and says, this is the original wave and then this is embedded signal. That's able to be then sent to our speaker Before it gets to the speaker, it goes past this variable resistor right here which is also called a potentiometer. When we turn that, it changes the resistance in the circuit and it either increases or decreases the volume, and increases or decreases the amount of power running to these wires. This one actually has come undone. The wires come here and there's a copper coil and a magnet. When the powers run to the copper coil and the magnet, it causes the paper cone to vibrate and that produces a pressure wave and we interpret that as sound. That's how that works. Then right here you can see there's two different switches here. We've got a switch that controls whether we're an AM or FM and then another switch that's just sort of let's us select different functions like turn the alarm clock off or have it set to buzzer instead of a radio, and things like that." + }, + { + "Q": "\nAt 13:10, what is the material they use to etch away the area that isn't protected?", + "A": "It depends. There are many, and I mean VERY many materials on the market that can do this, but I personally use a non-corrosive drying paste that protects certain areas from the acid, although this is time-taking and is not favored by most industries -The Weekend tinkerer", + "video_name": "xuQcB-oo-4U", + "timestamps": [ + 790 + ], + "3min_transcript": "Before it gets to the speaker, it goes past this variable resistor right here which is also called a potentiometer. When we turn that, it changes the resistance in the circuit and it either increases or decreases the volume, and increases or decreases the amount of power running to these wires. This one actually has come undone. The wires come here and there's a copper coil and a magnet. When the powers run to the copper coil and the magnet, it causes the paper cone to vibrate and that produces a pressure wave and we interpret that as sound. That's how that works. Then right here you can see there's two different switches here. We've got a switch that controls whether we're an AM or FM and then another switch that's just sort of let's us select different functions like turn the alarm clock off or have it set to buzzer instead of a radio, and things like that. That resist electric, current flow. That can be useful because it helps to prevent too much power from flowing to certain components on the board and things like that. This is a transistor. These things are transistors. They can function as switches. These guys right here are filters and they can help to reduce noise or electromagnetic interference, They're probably used in the radio circuit here to help to clean up the signal. On the back here, you can see again, these prongs connect to the battery. This is the printed circuit board here on the back. Basically, it's a thin layer of copper that's been applied to this fiber glass board. Then a chemical was used. They used basically a photomotion process which is like a similar to remove ... They shine a light on it and they use certain areas of the copper and keep other areas. They'll use an asset or a material to etch away the areas that aren't protected. Those result in copper traces. Those copper traces are basically very well ordered, little tiny wires that are very flat and they allow us to connect all these different components in a very small space very efficiently so we can just push the components. These are called through-hole components. Through-holes and then solder them on the back and then they're all wired up together. We don't have to worry about a lot of messy wires and things not being connected correctly. You can see there's different components, small components on the back. The little surface mount resistors and things like that. That's our alarm clock radio and those are the insides. Hope you've enjoyed it." + }, + { + "Q": "\nAt 12:56, he mentions this photo emulsion process, but I couldn't tell what he meant. What is it?", + "A": "please do check the information and confirm: when a photograph (NON DIGITAL) is taken, photo-chemical compounds react and produce different colours and shades. i guess it is quite similar...", + "video_name": "xuQcB-oo-4U", + "timestamps": [ + 776 + ], + "3min_transcript": "Before it gets to the speaker, it goes past this variable resistor right here which is also called a potentiometer. When we turn that, it changes the resistance in the circuit and it either increases or decreases the volume, and increases or decreases the amount of power running to these wires. This one actually has come undone. The wires come here and there's a copper coil and a magnet. When the powers run to the copper coil and the magnet, it causes the paper cone to vibrate and that produces a pressure wave and we interpret that as sound. That's how that works. Then right here you can see there's two different switches here. We've got a switch that controls whether we're an AM or FM and then another switch that's just sort of let's us select different functions like turn the alarm clock off or have it set to buzzer instead of a radio, and things like that. That resist electric, current flow. That can be useful because it helps to prevent too much power from flowing to certain components on the board and things like that. This is a transistor. These things are transistors. They can function as switches. These guys right here are filters and they can help to reduce noise or electromagnetic interference, They're probably used in the radio circuit here to help to clean up the signal. On the back here, you can see again, these prongs connect to the battery. This is the printed circuit board here on the back. Basically, it's a thin layer of copper that's been applied to this fiber glass board. Then a chemical was used. They used basically a photomotion process which is like a similar to remove ... They shine a light on it and they use certain areas of the copper and keep other areas. They'll use an asset or a material to etch away the areas that aren't protected. Those result in copper traces. Those copper traces are basically very well ordered, little tiny wires that are very flat and they allow us to connect all these different components in a very small space very efficiently so we can just push the components. These are called through-hole components. Through-holes and then solder them on the back and then they're all wired up together. We don't have to worry about a lot of messy wires and things not being connected correctly. You can see there's different components, small components on the back. The little surface mount resistors and things like that. That's our alarm clock radio and those are the insides. Hope you've enjoyed it." + }, + { + "Q": "\nIn the diagram at 10:13 , u say that the top ones are correct for IUPAC naming system .I would request u to please state that whether both are applicable or only one ? If one please state which one and also why ?", + "A": "The top 2 would be named exactly the same way: 3,4-diethyl-2-methylheptane", + "video_name": "F8RCR_1jIAk", + "timestamps": [ + 613 + ], + "3min_transcript": "So let's see what we have. Let's see how many carbons we have if we said this is our longest carbon chain. So let's number them. Let's call this carbon one. Let's call this carbon two, three, four, five, six, So once again this would be called heptane. What sort of substituents do we have coming off this molecule? We have a methyl group coming off of carbon two. We have an ethyl group coming off of carbon three. And we have another ethyl group coming off of carbon four. So that's the exact same situation we had for the first example here. So these are the same thing. So it doesn't really matter which one of those you chose, you'd be naming it the exact same name. Let's compare those two to the molecule to the example down here. Again it's the same molecule, but let's say you chose a different path. Let's say you chose down here. So you said, oh, this looks like it's So you go like this, and you say, all right, that's my longest carbon chain. How many carbons are in that? Well, this would be one, two, three, four, five, six, and seven. So what sort of substituents do we have in this molecule? Well coming off of carbon four we can see there is an ethyl group. Coming off of carbon three we can see this looks kind of complicated. It's not really a straight chain. This is much more complex substituent, which we'll get to naming in a future video. So for this molecule we have a total of two substituents. For the top molecule we have an example of three substituents. So the question is which one of these will be the correct way to name my molecule according to IUPAC nomenclature? So I have two chains of equal length. Both of these chains are seven carbons. So how do I break that tie? IUPAC rules state you choose the parent chain So the top one has three subtituents, the bottom example has two substituents. So if you were to name this molecule using IUPAC packed nomenclature you would choose the top way of naming it, which again we will get to in more detail in the next few videos here. So let's look at cyclo alkanes now. So we've just done straight chain alkanes. We looked at branched chain alkanes. Let's look at cyclo alkanes. So this is a pretty funny dot structure here. Let's see how many carbons are in this triangle. Well, of course, there's one, two, and three carbons. So if I were to draw what this molecule looks like, if I were to draw all the atoms involved there'd be three carbons like that. And to complete the octet around carbon there'd have to be two hydrogens on each carbon like that. So that's a cyclo alkane." + }, + { + "Q": "At 0:26 why is the oxygen said to get the plus one charge instead of the hydrogen. If oxygen is more electronegative than the hydrogen, wouldn't it make more sense to say the hydrogen is positively charged. Is this because when you draw these molecules, you usually leave out the hydrogens?\n", + "A": "The oxygen will have a +1 formal charge. To find formal charge, you take the number of valence electrons of a free atom, subtract 1/2 # of shared e-, and subtract #of lone e-. In this case for oxygen in H3O+: Oxygen has 6 valence e- , has three bonds, and has 2 electrons that fill its octet but aren t involved in bonding. Therefore the formal charge is 6-3-2=+1.", + "video_name": "BeHOvYchtBg", + "timestamps": [ + 26 + ], + "3min_transcript": "- [Voiceover] Let's look at this acid base reaction. So water is gonna function as a base that's gonna take a proton off of a generic acid HA. So lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the A. Oxygen, oxygen is now bonded to three hydrogens. So it picked up a proton. That's gonna give this oxygen a plus one formal charge and we can follow those electrons. So these two electrons in red here are gonna pick up this proton forming this bond. So we make hydronium H30 plus and these electrons in green right here are going to come off onto the A to make A minus. Let's go ahead and draw that in. So we're gonna make A minus. Let me draw these electrons in green and give this a negative charge like that. Let's analyze what happened. HA donated a proton so this is our Bronsted-Lowry acid. Once HA donates a proton, we're left with the conjugate base Water, H2O accepted a proton, so this is our Bronsted-Lowry base and then once H2O accepts a proton, we turn into hydronium H3O plus. So this is the conjugate acid. So H3O plus, the conjugate acid and then A minus would be a base. If you think about the reverse reaction, H3O plus donating a proton to A minus then you would get back H2O and HA. Once this reaction reaches equilibrium, we can write an equilibrium expression and we're gonna consider the stuff on the left to be the reactants. We're gonna think about the fourth reaction and the stuff on the right to be the products. Let's write our equilibrium expression. And so we write our equilibrium constant and now we're gonna write KA which we call the acid, the acid ionization constant. So this is the acid ionization constant so acid dissociation. So either one is fine. All right and we know when we're writing an equilibrium expression, we're gonna put the concentration of products over the concentration of reactants. Over here for our products we have H3O plus, so let's write the concentration of hydronium H3O plus times the concentration of A minus, so times the concentration of A minus. All over the concentration of our reactant, so we have HA over here, so we have HA. So we could write that in and then for water, we leave water out of our equilibrium expression. It's a pure liquid. Its concentration doesn't change and so we leave, we leave H2O out of our equilibrium expression. All right, so let's use this idea of writing an ionization constant and let's apply this to a strong acid." + }, + { + "Q": "\nat 4:35, the equation had a value d. what is it?", + "A": "The d in the capacitor and inductor equations is an expression used in calculus. You can read it as a tiny change in ... something . So dv means a tiny change in voltage , and dt means a tiny change in time . The quotient dv/dt represents the rate of change (slope) of voltage with respect to time.", + "video_name": "l-h72j2-X0o", + "timestamps": [ + 275 + ], + "3min_transcript": "and we're gonna be very consistent about this and that's gonna keep us from making mistakes. All right, so let's go back to our resistor and we're gonna do the equation for a resistor. What is the I-V equation for a resistor? I-V equation means what relates current to voltage and for a resistor, it's V equals i times R so the voltage across the resistor is equal to the current through the resistor times this constant of proportionality that we call the resistance. This has a very important name. This is called Ohm's law and you're gonna use this a lot so that's Ohm's law right there. This is Ohm's law. Now for the IV relationship for the capacitor, the capacitor has that property that the current of the voltage, not to the voltage but to the rate of change of the voltage and the way we write that is current equals, C is the proportionality constant, and we write dv, dt so this is the rate of change of voltage with respect to time. We multiply that by this property of this device called capacitance and that gives us the current. This doesn't have a special name but I'm gonna refer to it as the capacitor equation so now we have two equations. Let's do the third equation which is for the inductor. The inductor has the property very similar to the capacitor. It has the property that the voltage across is proportional to the time rate of change of the current flowing through the inductor so this is a similar but opposite of how a capacitor works. The voltage is proportional to the time rate of change is voltage equals L, di, dt. The voltage is proportional. The proportionality constant is the inductants. The inductance of the inductor and this is the time rate of change of voltage, OH sorry, the time rate of change of current flowing through the inductor so this gives us our three equations. Here they are. These are three element equations and we're gonna use these all the time, right there, those three equations. One final point I wanna make is for both these equations of components, these are ideal, ideal components." + }, + { + "Q": "\nWhen, at ~ 2:41 Sal is talking about the specific heats of water is it assuming atmospheric pressure? Just wondering because I've been taught that it'll have lower boiling and evaporating points at lower pressures and higher ones at higher pressures.\nOn a separate note, should I do the section on the gas laws before this section?\n\nI'm not trying to be a smart Alec.\nThanks.", + "A": "Specific heat has nothing to do with boiling. It is the amount of heat you need to add to 1 kg of the material to get the temperature to go up by 1 K.", + "video_name": "zz4KbvF_X-0", + "timestamps": [ + 161 + ], + "3min_transcript": "that heat energy is being used to kind of break the lattice structure. To add potential energy to the ice. Or essentially melt it. So for it here, right here, we're ice. Right at this point, we're zero degree ice. And then as we add more and more heat we get to zero degree water. So at zero degrees, you can either have water or ice. And if you have water, to turn it into ice you have to take heat out of it. And if you have ice, and you want to turn it into water you have to put heat into it. And then the heat is used again to warm up the water at some rate. And then at 100 degrees, which is the boiling point of water, right here, a similar phase change happens. Where the increased heat is not used to increase the temperature of the water, it's used to put potential energy into the system. So the water molecules are forced away from each other. The same way that if I'm forced away from the planet Earth. I have potential energy because I can fall back to the earth. of falling back to each other. But this energy right here is the energy necessary to vaporize the water. Right here you have 100 degree water. 100 degree liquid. And here you have 100 degree vapor, water vapor And then as you add more and more heat once again it increases the temperature But you, Sal, I already learned this a few videos ago, I have the intuition. But I want to deal with real numbers. I want to know exactly how much heat is required for these different things to happen. And for that, we can get these numbers. And these are specific to the different states of water. If you looked up any other element or molecule, you would have different values for these numbers we're going to be dealing with right now. But this first number right here is the heat of fusion And this is the amount of heat that's required to fuse 100 degree water into 100 degree ice. Or the amount of energy you have to take out of the water. So this distance right here or along this axis, If you're going in the the leftward direction you have to take that much out of the system to turn into ice. If you're going in the rightward direction you have to add that much to turn into water. So heat of fusion. It's called the heat of fusion because when you fuse something together you make it solid. So it could also be considered the heat of melting. Just two different words for the same thing depending on what direction you going. The important thing is the number, 333. Similarly, you have the heat of vaporization 2257 joules per gram. That's this distance along this axis right here. So if you had one gram of 100 degree liquid water and you wanted to turn it into one gram of 100 degree liquid vapor. And in all of this we're assuming that nothing silly is happening to the pressure that we're under constant pressure. You would have to put 2257 joules into the system." + }, + { + "Q": "What is potential energy? I'm confused since you mentioned it at 1:48.\n", + "A": "Any form of stored energy is potential energy. Most commonly in physics you have gravitational potential energy and electric potential energy. You can also have elastic potential energy (the energy stored in a spring) and chemical potential energy (stored in chemical bonds), A physicist might tell you those are ultimately forms of electric potential energy but don t worry about that for now.", + "video_name": "zz4KbvF_X-0", + "timestamps": [ + 108 + ], + "3min_transcript": "A couple of videos ago, we learned that --if we started with solid water or ice at a reasonably low temperature --maybe this temperature right here is minus 10 degrees Celsius. And we can deal with Celsius when we're dealing with these phase changes because we really just care about the difference in temperature and not necessarily the absolute temperature. So one degree in Celsius is the same thing as one degree in Kelvin. So the differences are the same whether you're dealing with Celsius or Kelvin. So we're starting with minus 10 degree Celsius ice or solid water. And we learned that as you heat it up as you add heat energy to the water, the temperature goes up. The molecules, at least while they're in that ice lattice network, they just start vibrating. And their average kinetic energy goes up until we get to zero degrees. Which is the melting point of water. And at zero degrees, we already learned something interesting happens. The added heat in the system does not increase the temperature of the ice anymore. At least over that little period right here. that heat energy is being used to kind of break the lattice structure. To add potential energy to the ice. Or essentially melt it. So for it here, right here, we're ice. Right at this point, we're zero degree ice. And then as we add more and more heat we get to zero degree water. So at zero degrees, you can either have water or ice. And if you have water, to turn it into ice you have to take heat out of it. And if you have ice, and you want to turn it into water you have to put heat into it. And then the heat is used again to warm up the water at some rate. And then at 100 degrees, which is the boiling point of water, right here, a similar phase change happens. Where the increased heat is not used to increase the temperature of the water, it's used to put potential energy into the system. So the water molecules are forced away from each other. The same way that if I'm forced away from the planet Earth. I have potential energy because I can fall back to the earth. of falling back to each other. But this energy right here is the energy necessary to vaporize the water. Right here you have 100 degree water. 100 degree liquid. And here you have 100 degree vapor, water vapor And then as you add more and more heat once again it increases the temperature But you, Sal, I already learned this a few videos ago, I have the intuition. But I want to deal with real numbers. I want to know exactly how much heat is required for these different things to happen. And for that, we can get these numbers. And these are specific to the different states of water. If you looked up any other element or molecule, you would have different values for these numbers we're going to be dealing with right now. But this first number right here is the heat of fusion And this is the amount of heat that's required to fuse 100 degree water into 100 degree ice. Or the amount of energy you have to take out of the water. So this distance right here or along this axis," + }, + { + "Q": "\nShouldn't we always calculate with K in the equation not C? Especially @6:11 because that's how the K's cancel out.", + "A": "When you re talking about change in temperature, C is actually the same as K. The difference between -10 C (263 K) and 0 C (273 K) is the same in both cases: 10 degrees. It s important to note the difference when you re talking in absolute values (0 C vs. 273 K) but when you re talking in relative values, the amounts will be the same. (Note that this does NOT apply to F. Each degree F is about half a degree C or K.)", + "video_name": "zz4KbvF_X-0", + "timestamps": [ + 371 + ], + "3min_transcript": "and you wanted to condense it, you would have to take that much energy out of the system. OK, fine, you know how much energy is required for the phase changes. But what about these parts right here? How much energy is required to warm up a gram of ice by one degree Celsius or Kelvin? And for that we look at the specific heat. It takes 2 joules of energy to warm up 1 gram 1 degree Kelvin. When water is in the solid state When it's in the liquid state, it takes about double that. It takes about 4 joules per gram to raise it 1 degree Kelvin. And when you're in the vapor state, it's actually more similar to the solid state. So given what we know now, we can actually figure out how much energy it would take to go from minus 10 degree ice to 110 degree vapor. So the first thing we're going to be doing is we're going to be going from minus 10 degree ice to zero degree ice. So we're going to go 10 degrees. We have to figure out how much heat does it take to warm up ice by 10 degrees. So the heat is going to be equal to the change in temperature. So actually let me write the specific heat first. So 2.05 joules per gram Kelvin. Oh, and I should tell you we can't different values for the amount of ice we're warming up to vapor, so let's say we're dealing with 200 grams. So it'll be the specific heat times the number of grams we're warming up of ice times the change in temperature So, times 10 degrees Kelvin Let's just say it's a 10 degrees Kelvin change, it doesn't matter if we're using Kelvin or Celsius. I could have written a Celsius here. Let's put Celsius right there. So what is that equal to? Get the calculator. Clear it out. 2.05 times 200 times 10 is equal to 4,100 joules. Let me do this in a different color This is 4,100 joules. Fair enough. Now, so what we've done is just this part right here. This distance right here is 4,100 joules. Now we have to turn that zero degree ice into zero degree water. So now we have to go from 0 degree ice to 0 degree water." + }, + { + "Q": "\nat 4:19 shouldnt x approach zero\nthen and only,'I think ' we might get closer to zero\nPLEASE correct if i am wrong\nI am talking about the slope at point 'a' on the second graph", + "A": "Since x = a+h, when h approaches 0, x approaches a.", + "video_name": "Df2escG-Vu0", + "timestamps": [ + 259 + ], + "3min_transcript": "you would just substitute a into your function definition. And you would say, well, that's going to be the limit as h approaches 0 of-- every place you see an x, replace it with an a. f of-- I'll stay in this color for now-- blank plus h minus f of blank, all of that over h. And I left those blanks so I could write the a in red. Notice, every place where I had an x before, it's now an a. So this is the derivative evaluated at a. So this is one way to find the slope of the tangent line when x equals a. Another way-- and this is often used as the alternate form of the derivative-- would be to do it directly. So this is the point a comma f of a. So let's say this is the value x. This point right over here on the function would be x comma f of x. And so what's the slope of the secant line between these two points? Well, it would be change in the vertical, which would be f of x minus f of a, over change in the horizontal, over x minus a. Actually, let me do that in that purple color. Over x minus a. Now, how could we get a better and better approximation for the slope of the tangent line here? Well, we could take the limit as x approaches a. As x gets closer and closer and closer to a, the secant line slope is going to better and better and better approximate the slope of the tangent line, this tangent line that I have in red here. So we would want to take the limit as x approaches a here. We have an expression for the slope of a secant line. And then we're bringing those x values of those points closer and closer together. So the slopes of those secant lines better and better and better approximate that slope of the tangent line. And at the limit, it does become the slope of the tangent line. That is the definition of the derivative. So this is the more standard definition of a derivative. It would give you your derivative as a function of x. And then you can then input your particular value of x. Or you could use the alternate form of the derivative. If you know that, hey, look, I'm just looking to find the derivative exactly at a. I don't need a general function of f. Then you could do this. But they're doing the same thing." + }, + { + "Q": "So, for resistors in parallel, the circuit experiences no drop in voltage when the current reaches the resistors (11:19)? And for resistors in series, there is a voltage drop? Also, are the formulas for total resistance for resistors in parallel and for resistors in series different?\n", + "A": "There is a voltage drop, but it is identical for each resistor. The equations are very different. Series: Rtotal = R1 + R2 +...+Rn Parallel: 1/Rtotal = (1/R1 +1/R2 + ... +1/R3)", + "video_name": "ZrMw7P6P2Gw", + "timestamps": [ + 679 + ], + "3min_transcript": "That's 4/20, right? So 1 over our total resistance is equal to 5/20, which is 1/4. So if 1/R is equal to 1/4, R must be equal to 4. R is equal to 4 ohms. So we could redraw this crazy circuit as this. I'll try to draw it small down here. We could redraw this where this resistance is 4 ohms and this is 16 volts. We could say that this whole thing combined is really just a resistor that is 4 ohms. Well, if we have a 16-volt potential difference, current is flowing that way, even though that's not what the electrons are doing. And that's what our resistance is, 4 ohms. What is the current? V equals IR, Ohms law. It equals the current times 4 ohms. So current is equal to 16 divided by 4, is equal to 4 amps. So let's do something interesting. Let's figure out what the current is flowing through. What's this? What's the current I1 and what's this current I2? Well, we know that the potential difference from here to here is also 16 volts, right? Because this whole thing is essentially at the same potential and this whole thing is essentially at the same potential, so you have 16 volts across there. 16 volts divided by 20 ohms, so let's call this I1. So I1 is equal to 16 volts divided by 20 ohms, which is equal to what? 4/5. So it equals 4/5 of an ampere, or 0.8 amperes. through here? I2? I'm going to do this in a different color. It's getting confusing. I'll do it in the vibrant yellow. So the current flowing through here, once again, the potential difference from here-- that's not different enough-- the potential difference from here to here is also 16 volts, right? So the current is going to be I2, is going to be equal to 16/5, which is equal to 3 1/5 amps. So most of the current is actually flowing through this, and that makes sense because the resistance is less, right? So that should hopefully give you a little bit of intuition And less current is flowing through here, so I2 through the 20-ohm resistor is 0.8 amps is I1, and I2 through the 5-ohm resistor is equal to 3.2 amps. And it makes sense that when you add these two currents together, the 3.2 amperes flowing through here and the" + }, + { + "Q": "9:37 - Could you not also use the formula R(total) = 1 / ((1 / R1) + (1 / R2) + ... (1 / Rn)) , where \"Rn\" is the last resistance that is compiled into the equation?\n", + "A": "Hello Swaggy, Yes! Personally I prefer this the equation you mentioned as it is very fast to solve using a calculator and the 1/x key. Regards, APD", + "video_name": "ZrMw7P6P2Gw", + "timestamps": [ + 577 + ], + "3min_transcript": "drew in the previous diagram, although now I will assign numbers to it. Let's say that this resistance is 20 ohms and let's say that this resistance is 5 ohms. What I want to know is, what is the current through the system? First, we'll have to figure out what the equivalent resistance is, and then we could just use Ohm's law to figure out the current in the system. So we want to know what the current is, and we know that the convention is that current flows from the positive terminal to the negative terminal. So how do we figure out the equivalent resistance? Well, we know that we just hopefully proved to you that the total resistance is equal to 1 over this resistor plus 1 over this resistor. So 1 over-- I won't keep writing it. What's 1 over 20? Well, actually, let's just make it a fraction. That's 4/20, right? So 1 over our total resistance is equal to 5/20, which is 1/4. So if 1/R is equal to 1/4, R must be equal to 4. R is equal to 4 ohms. So we could redraw this crazy circuit as this. I'll try to draw it small down here. We could redraw this where this resistance is 4 ohms and this is 16 volts. We could say that this whole thing combined is really just a resistor that is 4 ohms. Well, if we have a 16-volt potential difference, current is flowing that way, even though that's not what the electrons are doing. And that's what our resistance is, 4 ohms. What is the current? V equals IR, Ohms law. It equals the current times 4 ohms. So current is equal to 16 divided by 4, is equal to 4 amps. So let's do something interesting. Let's figure out what the current is flowing through. What's this? What's the current I1 and what's this current I2? Well, we know that the potential difference from here to here is also 16 volts, right? Because this whole thing is essentially at the same potential and this whole thing is essentially at the same potential, so you have 16 volts across there. 16 volts divided by 20 ohms, so let's call this I1. So I1 is equal to 16 volts divided by 20 ohms, which is equal to what? 4/5. So it equals 4/5 of an ampere, or 0.8 amperes." + }, + { + "Q": "\nIf the voltage difference between two points in the circuit is always the same as long as the resistence is between these two points,is there no voltage difference between two points which are both in the blue region, for example (04:43)? So, in a circuit without a resistence, wouldn't there be a voltage difference? But then there would no current at all... I don't understand this at all. Also, isn't the voltage difference which provokes the movement of the electrons?\n\nThanks!", + "A": "Virtually all conductors have resistance, albeit very small. If you short a battery with a wire, you will get a very large amount of current flowing (likely the largest that battery can output). There will be a voltage drop across that wire that is equal to the voltage increase across the battery. Also batteries do have some built in internal resistance which also limits the maximum output current and results in a small voltage drop within the battery when current is flowing.", + "video_name": "ZrMw7P6P2Gw", + "timestamps": [ + 283 + ], + "3min_transcript": "Or you could even think of it that the current entering-- when the currents I2 and I1 merge, that they combine and they become Current 1, right? I mean, think about it. In a given second, if this is 5 coulombs per second-- I'm just making up numbers-- and this is 6 coulombs per second, in a given second right here, you're going to have 5 coulombs coming from this branch and 6 coulombs coming from this branch, so you're going to have 11 coulombs per second coming out once they've merged, so this would be 11 So I think hopefully that makes sense to you that this current is equal to the combination of this current and that current. Now, what do we also know? We also know the voltage along this entire ideal wire is color, in blue. So, for example, the voltage anywhere along this blue that I'm filling in is going to be the same, because this wire is an ideal conductor, and you can almost view this blue part as an extension of the positive terminal of the battery. And very similarly-- I'll do it in yellow-- we could draw this wire as an extension of the negative terminal of the battery. This is an extension of the negative terminal of the battery. So the voltage difference between here and here-- so let's call that the total voltage, or let's just call it the voltage, right? The voltage difference between that point and that point is the exact same thing as the voltage difference between this point and this point, which is the exact same thing point and this point. So what can we say? What is the total current in the system? If we just viewed this as a black box, that this is some type of total resistance, well, the total current in the system would be the total voltage, the voltage divided by-- let's call this our total resistance, right? Let's say we couldn't see this and we just said, oh, that's just some total resistance, and that is equal to the current going through R1. This is I1. This is a 1 right here. This is current I1. What's current I1? Well, it's going to be the voltage across this resistor divided by the resistance, right? That's what Ohm's law tells us: V is equal to IR, or another way we could say it is V over R is equal to I, right? So I1 is equal to the voltage across this resistor, but we" + }, + { + "Q": "\nAt 9:25 in the video the resistance comes out to be 4 ohms. While the mathematics behind this makes perfect sense, the physics doesn't. The smallest resistor on the entire circuit is 5 ohms. I understand that most of the electrons will want to pass through this, but wouldn't that mean that the total resistance should never come out to be less than the smallest resistor?", + "A": "No, the resistance of two resistors in parallel will always be less than the resistance of either of the two. Think about it this way. Take the bigger resistor out. Now the resistance is just 4 ohms, right. Now put it back. There s another path for current to go through that wasn t there before. Therefore more current is going to flow, therefore the total resistance is less than it was before.", + "video_name": "ZrMw7P6P2Gw", + "timestamps": [ + 565 + ], + "3min_transcript": "So let's see if we can use this information we have learned to actually solve a problem, and I actually find that it's always easier to solve a problem than to explain the theory behind a problem. You'll see that with most of these circuit problems, it's actually very basic mathematics. So let's say I have a 16 volt battery plus, minus, it's 16 volts. And just to hit the point home that you always don't have to draw circuits the same, although it is nice if you're actually drawing complicated circuits, I could draw it like this. I could draw the circuit like this, and let's say that there's a resistor here. And then let's say there's a wire and then there's another resistor here, and that this decides to do some random loopy thing here and that they connect here, and that they come back here. This strange thing that I have drawn, which you will never see in any textbook, because most people are more reasonable than me, is the exact same-- you can almost drew in the previous diagram, although now I will assign numbers to it. Let's say that this resistance is 20 ohms and let's say that this resistance is 5 ohms. What I want to know is, what is the current through the system? First, we'll have to figure out what the equivalent resistance is, and then we could just use Ohm's law to figure out the current in the system. So we want to know what the current is, and we know that the convention is that current flows from the positive terminal to the negative terminal. So how do we figure out the equivalent resistance? Well, we know that we just hopefully proved to you that the total resistance is equal to 1 over this resistor plus 1 over this resistor. So 1 over-- I won't keep writing it. What's 1 over 20? Well, actually, let's just make it a fraction. That's 4/20, right? So 1 over our total resistance is equal to 5/20, which is 1/4. So if 1/R is equal to 1/4, R must be equal to 4. R is equal to 4 ohms. So we could redraw this crazy circuit as this. I'll try to draw it small down here. We could redraw this where this resistance is 4 ohms and this is 16 volts. We could say that this whole thing combined is really just a resistor that is 4 ohms. Well, if we have a 16-volt potential difference, current is flowing that way, even though that's not what the electrons are doing. And that's what our resistance is, 4 ohms. What is the current? V equals IR, Ohms law." + }, + { + "Q": "\nAt 0:07, Sal mentioned that at higher temperatures you get a 4th state(plasma). But doesn't our blood contain plasma, and isn't our average temperature is 37C?", + "A": "good question. blood plasma has nothing to do with the fourth state of matter. It s just has the same (inaccurate, of course) name.", + "video_name": "pKvo0XWZtjo", + "timestamps": [ + 7 + ], + "3min_transcript": "I think we're all reasonably familiar with the three states of matter in our everyday world. At very high temperatures you get a fourth. But the three ones that we normally deal with are, things could be a solid, a liquid, or it could be a gas. And we have this general notion, and I think water is the example that always comes to at least my mind. Is that solid happens when things are colder, relatively colder. And then as you warm up, you go into a liquid state. And as your warm up even more you go into a gaseous state. So you go from colder to hotter. And in the case of water, when you're a solid, you're ice. When you're a liquid, some people would call ice water, but let's call it liquid water. I think we know what that is. vapor or steam. So let's think a little bit about what, at least in the case of water, and the analogy will extend to other types of molecules. But what is it about water that makes it solid, and when it's colder, what allows it to be liquid. And I'll be frank, liquids are kind of fascinating because you can never nail them down, I guess is the best way to view them. Or a gas. So let's just draw a water molecule. So you have oxygen there. You have some bonds to hydrogen. And then you have two extra pairs of valence electrons in the oxygen. And a couple of videos ago, we said oxygen is a lot more electronegative than the hydrogen. It likes to hog the electrons. So even though this shows that they're sharing At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules." + }, + { + "Q": "At 0:06, what is the fourth state of matter Sal refers to?\n", + "A": "Plasma would be the fourth state of matter that Sal is referring to.", + "video_name": "pKvo0XWZtjo", + "timestamps": [ + 6 + ], + "3min_transcript": "I think we're all reasonably familiar with the three states of matter in our everyday world. At very high temperatures you get a fourth. But the three ones that we normally deal with are, things could be a solid, a liquid, or it could be a gas. And we have this general notion, and I think water is the example that always comes to at least my mind. Is that solid happens when things are colder, relatively colder. And then as you warm up, you go into a liquid state. And as your warm up even more you go into a gaseous state. So you go from colder to hotter. And in the case of water, when you're a solid, you're ice. When you're a liquid, some people would call ice water, but let's call it liquid water. I think we know what that is. vapor or steam. So let's think a little bit about what, at least in the case of water, and the analogy will extend to other types of molecules. But what is it about water that makes it solid, and when it's colder, what allows it to be liquid. And I'll be frank, liquids are kind of fascinating because you can never nail them down, I guess is the best way to view them. Or a gas. So let's just draw a water molecule. So you have oxygen there. You have some bonds to hydrogen. And then you have two extra pairs of valence electrons in the oxygen. And a couple of videos ago, we said oxygen is a lot more electronegative than the hydrogen. It likes to hog the electrons. So even though this shows that they're sharing At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules." + }, + { + "Q": "\nAround 14:00 minutes, the temperature goes up but the state does not change (yet). So does that mean that there is some ice that can be colder or warmer that other ice?", + "A": "Sure, ice can be -10 C, or -20C, or -100C, or 0C.", + "video_name": "pKvo0XWZtjo", + "timestamps": [ + 840 + ], + "3min_transcript": "So what happens at zero degrees? Which is also 273.15 Kelvin. Let's say that's that line. What happens to a solid? Well, it turns into a liquid. Ice melts. Not all solids, we're talking in particular about water, about H2O. So this is ice in our example. All solids aren't ice. Although, you could think of a rock as solid magma. Because that's what it is. I could take that analogy a bunch of different ways. But the interesting thing that happens at zero degrees. Depending on what direction you're going, either the freezing point of water or the melting point of ice, something interesting happens. As I add more heat, the temperature does not to go up. As I add more heat, the temperature does not go up for Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or But something happens. And they're really getting kinetically active. But just like when you went from solid to liquid, there's a certain amount of energy that you have to contribute to the system. And actually, it's a good amount at this point. Where the water is turning into vapor, but it's not getting any hotter. So we have to keep adding heat, but notice that the temperature didn't go up. We'll talk about it in a second what was happening then. And then finally, after that point, we're completely vaporized, or we're completely steam. Then we can start getting hot, the steam can then get hotter as we add more and more heat to the system. So the interesting question, I think it's intuitive, that as you add heat here, our temperature is going to go up." + }, + { + "Q": "At about14:50, Sal said that the steam will get hotter. However, is there a limit to that, or does it just heat up forever?\n", + "A": "No, you cannot heat steam forever . Like all molecules, when sufficiently hot it will decompose. There is not a set temperature at which it decomposes, it is more of a gradual process as the temperature increases. You start getting significant quantities of water decomposing at around 2500 K. At about 3000 K you get around 50% of water molecules decomposing. Of course, this is a rather hot temperature and it is rather impractical to generate, much less create a container to house the reaction.", + "video_name": "pKvo0XWZtjo", + "timestamps": [ + 890 + ], + "3min_transcript": "Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or But something happens. And they're really getting kinetically active. But just like when you went from solid to liquid, there's a certain amount of energy that you have to contribute to the system. And actually, it's a good amount at this point. Where the water is turning into vapor, but it's not getting any hotter. So we have to keep adding heat, but notice that the temperature didn't go up. We'll talk about it in a second what was happening then. And then finally, after that point, we're completely vaporized, or we're completely steam. Then we can start getting hot, the steam can then get hotter as we add more and more heat to the system. So the interesting question, I think it's intuitive, that as you add heat here, our temperature is going to go up. We were adding heat. So over here we were turning our heat into kinetic energy. Temperature is average kinetic energy. But over here, what was our heat doing? Well, our heat was was not adding kinetic energy to the system. The temperature was not increasing. But the ice was going from ice to water. So what was happening at that state, is that the kinetic energy, the heat, was being used to essentially break these bonds. And essentially bring the molecules into a higher energy state. So you're saying, Sal, what does that mean, higher energy state? Well, if there wasn't all of this heat and all this kinetic energy, these molecules want to be very close to each other. For example, I want to be close to the surface of the earth. When you put me in a plane you have put me in a higher energy state. I have a lot more potential energy. I have the potential to fall towards the earth. Likewise, when you move these molecules apart, and you go" + }, + { + "Q": "\nat 10:07 he said that the months stay the same. Is that what causes leap years?", + "A": "No AegonTargaryen is right, a year is not completly 365 days. If it were, there wouldn t be February 29th...which BTW is my birthday, but we usually celebrate it on February 28th", + "video_name": "2o-Sef6wllg", + "timestamps": [ + 607 + ], + "3min_transcript": "important to see what happens to our calendar-- if we wait 1,800 years this arrow, it will still have a tilt of 23.4 degrees, but instead of pointing in this direction, it might be pointing in this direction. Or in fact, it will be pointing in this direction. I'm obviously not drawing it that exact. And then the bottom of the arrow will come out over here. So if you think about that, if you wait 1,800 years, and once again, the tilt hasn't changed or it's changed a little bit, but what the precession has done, tracing out this circle has changed the direction of this arrow, it changed the direction of our axis of rotation. And if you wait 1,800 years when will the Northern Hemisphere be pointed most away from the sun. Well, now, it won't be pointed most away from the sun at this point in space relative to the sun anymore because now its axis of rotation looks something like this. it will be most pointed away, or the Northern Hemisphere will be most pointed away from the sun, about a month earlier. So about a month earlier. It'll be most pointed away from the sun about a month earlier. So this is when it will be most pointed away from the sun. To today's time, we would say, no, that's still not the most pointed away, but since we have this precession, since the direction of the tilt I guess we could say, or the direction of our rotational axis is changing, we are now at a different point in our orbit where we are most pointed away from the sun. So this is 1,800 years later, approximately. So now based on this, and I think this is what Vicksoma might have been hinting at, you say, look, OK, it's earlier in our orbit. Wouldn't this now be November? It will still be December 22nd. This will still be December 21st or 22nd Depending on the year. Still be the same date, and that's because our calendar is based on when we are most tilted away or when we are most tilted towards the sun. So by definition, this is when we are most tilted away so this will be the winter solstice. So what happens is every year-- so the way I drew it right over here and, actually, this perihelion actually changes over time as well. There's a precession of the perihelion as well, but I'm not going to go into that right now. So if you fast forward 1,800 years, all that's going to happen is that what we consider by our calendar to be December 22nd in an absolute point in our orbit we'll be earlier in our orbit, but we're still going to call it December 22nd. And so the perihelion is going to be further away from that December 22nd, it's actually going to be a month further" + }, + { + "Q": "\nAt 1:40, why must those electrons also attack? Why can't we just be left with a halogen ion and 1-halogen-ethane?", + "A": "We could. If the molecule is ethane, the attack of the \u00cf\u0080 electrons would give a halide ion and a 1\u00c2\u00b0 2-haloethyl carbocation. But the 1\u00c2\u00b0 cation is less stable than the cyclic halonium ion, so the reaction does not use that pathway.", + "video_name": "Yiy84xYQ3es", + "timestamps": [ + 100 + ], + "3min_transcript": "Let's take a look at the halogenation of alkenes. So on the left, I have my alkene. And I'm going to add a halogen to it, so something like bromine or chlorine. And I can see that those 2 halogen atoms are going to add anti to each other-- so they'll add on opposite sides of where the double bond used Let's take a look at the mechanism so we can figure out why we get an anti addition. So I start with my alkene down here. And I'm going to show the halogen approaching that alkene. So it's going to approach this way. And I put in my lone pairs of electrons. Like that. If I think about that halogen molecule, I know that it's nonpolar. Because if I think about the electrons in the bond between my two halogens here, both halogen atoms, of course, have the exact same electronegativity. So neither one is pulling more strongly. And so overall, the molecule is nonpolar. However, if the pi electrons in my alkene-- so I'm going to say that these electrons right here are my pi electrons-- if those electrons get too close to the electrons in blue, they would, of course, since they're like charge. And if the electrons in magenta repel the electrons in blue, the electrons in blue would be forced closer to the top halogen, like that, giving the top halogen a partial negative charge and leaving the bottom halogen with a partial positive charge because it's losing a little bit of electron density. Now, I could think about that bottom halogen as acting like an electrophile because it wants electrons. And so in this mechanism, the pi electrons are going to function as a nucleophile. And the pi electrons are going to attack my electrophile, like that. At the same time, these electrons over here-- this electron pair on the left side of the halogen-- is going to attack this carbon. And the electrons in blue are also going to kick off onto the top halogen, like that. So let's go ahead and draw the result of all those electrons moving around. So now, I have carbon singly-bonded to another carbon, like that. on the right and my halogen, like that. And these electrons over here, I'm going to mark in red. So this lone pair of electrons on my halogen are going to form a bond with the carbon on the left, And that halogen still has two lone pairs of electrons on it. So I'm going to put in those two lone pairs of electrons. And that halogen has a plus 1 formal charge. This is called a cyclic halonium ion, and it's been proven to occur in this mechanism. If I think about that positively charged halogen, halogens are very electronegative, and they want electrons. So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here losing a little bit of electron density, giving it a partial positive charge. And so that's going to function as our electrophile in the next step." + }, + { + "Q": "At 3:00 shouldn't both carbon atoms have a partial + charge?\n", + "A": "Yes, they should. He just put the \u00ce\u00b4\u00e2\u0081\u00ba on one of the carbons to illustrate the mechanism.", + "video_name": "Yiy84xYQ3es", + "timestamps": [ + 180 + ], + "3min_transcript": "since they're like charge. And if the electrons in magenta repel the electrons in blue, the electrons in blue would be forced closer to the top halogen, like that, giving the top halogen a partial negative charge and leaving the bottom halogen with a partial positive charge because it's losing a little bit of electron density. Now, I could think about that bottom halogen as acting like an electrophile because it wants electrons. And so in this mechanism, the pi electrons are going to function as a nucleophile. And the pi electrons are going to attack my electrophile, like that. At the same time, these electrons over here-- this electron pair on the left side of the halogen-- is going to attack this carbon. And the electrons in blue are also going to kick off onto the top halogen, like that. So let's go ahead and draw the result of all those electrons moving around. So now, I have carbon singly-bonded to another carbon, like that. on the right and my halogen, like that. And these electrons over here, I'm going to mark in red. So this lone pair of electrons on my halogen are going to form a bond with the carbon on the left, And that halogen still has two lone pairs of electrons on it. So I'm going to put in those two lone pairs of electrons. And that halogen has a plus 1 formal charge. This is called a cyclic halonium ion, and it's been proven to occur in this mechanism. If I think about that positively charged halogen, halogens are very electronegative, and they want electrons. So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here losing a little bit of electron density, giving it a partial positive charge. And so that's going to function as our electrophile in the next step. in the previous step. So we had a halogen that had 3 lone pairs of electrons around it. It picked up the electrons in blue. Right? So now, it has 4 lone pairs of electrons-- 8 total electrons-- giving it a negative 1 formal charge, meaning it can now function as a nucleophile. So if I think about this cyclic halonium ion here, the halogen on top is going to prevent the nucleophile from attacking from the top. It's going to have to attack from below here. So this negatively charged halide anion is going to nucleophilic attack this electrophile here-- this carbon. And that's going to kick these electrons in magenta off onto this halogen here. So let's go ahead and draw the results of that nucleophilic attack. All right. So now, I'm going to have my 2 carbons still bonded to each other like that. And the top halogen has swung over here to the carbon" + }, + { + "Q": "At 3:42, does the halide ion attack the carbon that is most substituted or can it attack any carbon equally? I know in the halohydrin reaction the H2O attacks the most substituted carbon and the halogen goes on the least substituted.\n", + "A": "The halide ion preferentially attacks the more substituted carbon, just like water does in the halohydrin reaction.", + "video_name": "Yiy84xYQ3es", + "timestamps": [ + 222 + ], + "3min_transcript": "on the right and my halogen, like that. And these electrons over here, I'm going to mark in red. So this lone pair of electrons on my halogen are going to form a bond with the carbon on the left, And that halogen still has two lone pairs of electrons on it. So I'm going to put in those two lone pairs of electrons. And that halogen has a plus 1 formal charge. This is called a cyclic halonium ion, and it's been proven to occur in this mechanism. If I think about that positively charged halogen, halogens are very electronegative, and they want electrons. So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here losing a little bit of electron density, giving it a partial positive charge. And so that's going to function as our electrophile in the next step. in the previous step. So we had a halogen that had 3 lone pairs of electrons around it. It picked up the electrons in blue. Right? So now, it has 4 lone pairs of electrons-- 8 total electrons-- giving it a negative 1 formal charge, meaning it can now function as a nucleophile. So if I think about this cyclic halonium ion here, the halogen on top is going to prevent the nucleophile from attacking from the top. It's going to have to attack from below here. So this negatively charged halide anion is going to nucleophilic attack this electrophile here-- this carbon. And that's going to kick these electrons in magenta off onto this halogen here. So let's go ahead and draw the results of that nucleophilic attack. All right. So now, I'm going to have my 2 carbons still bonded to each other like that. And the top halogen has swung over here to the carbon It picked up the electrons in magenta. So that's what the carbon on the left will look like. The carbon on the right is still bonded to 2 other things. And the halide anion had to add from below. So now we're going to have this halogen down here. Like that. And so now we understand why it's an anti addition of my 2 halogen atoms. Let's go ahead and do a reaction. So we're going to start with cyclohexane as our reactant And we're going to react cyclohexane with bromine-- so Br2. Now, if I think about the first step of the mechanism, I know I'm going to form a cyclic halonium ion. So I'm going to draw that ring. And I'm going to show the formation of my cyclic halonium ion. It's called a bromonium ion. So I'm going to form a ring like this." + }, + { + "Q": "\nAt 3:08 Hank says Ground tissue does photosynthesis but how can it get light if there is Dermal Tissue on the outside layer? Is the dermal tissue transparent?", + "A": "Very good question. I might not be the right person to answer this, and I could be wrong, but I think photosynthesis does not require all of the visible light, but only some energy from the light. There are wavelengths of the light that will continue through the dermal tissue. I suppose the energy from those light waves contain enough energy to keep the photosynthesis going.", + "video_name": "VFtOcdXeP0Y", + "timestamps": [ + 188 + ], + "3min_transcript": "But these things alone can't explain a vascular plants extraordinary evolutionary success. I mean, algae was photosynthesizing long before plants made it fashionable and as we learned last week, non-vascular plants have reproductive strategies that are tricked out six ways from Sunday. So, what gives? The secret to vascular plants success is their defining trait, conductive tissues that can take food and water from one part of the plant to another part of a plant. This may sound simple enough, but the ability to move stuff from one part of an organism to another was a huge evolutionary break through for vascular plants. It allowed them to grow exponentially larger, store food for lean times and develop some fancy features that allowed them to spread farther and faster. It was one of the biggest revolutions in the history of life on earth. The result? Plants dominated earth long before animals even showed up and even today, they hold most of the world records. The largest organism in the world is a redwood in northern California, 115 meters tall. Bigger than three blue whales laid end-to-end. of quaking aspen in Utah, all connected by the roots. Weighing a total of 13 million pounds and the oldest living thing, a patch of sea grass in the Mediterranean dating back 200,000 years. We spend a lot of time congratulating ourselves on how awesomely magnificent and complex the human animal is, but you guys, I got to hand it to you. (music) So you know by now, the more specialized tissues an organism has, the more complex they are and the better they typically do, but you also know that these changes don't take place overnight. The tissues that define vascular plants didn't evolve all at once, but today we recognize three types that make these plants what they are. Dermal tissues make up their outermost layers and help prevent damage and water loss. Vascular tissues do all of that conducting of materials I just mentioned and the most abundant tissue type, ground tissues, carry out some of the most important and the storage of left over food. Now, some plants never go beyond these basics. They sprout from a germinated seed, develop these tissues and then stop. This is called primary growth and plants that are limited to this stage are herbaceous. As the name says, they are like herbs; small, soft and flexible and typically they die down to the root or die completely after one growing season. Pretty much everything you see growing in a backyard garden; herbs and flowers and broccoli and that kind of stuff. Those are herbaceous. But a lot of vascular plants go on to secondary growth, which allows them to grow, not just taller, but wider. This is made possible by the development of additional tissues, particularly woody tissues. These are your woody plants, which include; shrubs and bark covered vines, called lianas and, of course, your trees, but no matter how big they may or may not grow all vascular plants are organized into three main organs. All of which you are intimately familiar with, not just because you knew what they were when you were in second grade, but also because you probably eat them everyday. First, the root. It absorbs water and nutrients" + }, + { + "Q": "At about 10:18, the sap Hank is talking about gives the Ponderosa (spelling??) its delicious smell. In a maple tree, is that sap we turn into maple syrup produced in the same way the Ponderosa-scent-sap is? Thanks! :D\n", + "A": "Maple sap is very watery and needs to be boiled down into what you can buy in the store.", + "video_name": "VFtOcdXeP0Y", + "timestamps": [ + 618 + ], + "3min_transcript": "while in cold, dry years they're light and thin. These woody remains form tree rings, which scientists can use, not only to track the age of a tree, but also the history of the climate that it lived in. Now, at the top of the xylem, water arrives at it's final destination, the leaf. Here water travels through an increasingly minuscule network of vein-like structures until it's dumped into a new kind of tissue called the mesophyll. As you can tell from it's name, meso meaning middle and phyll meaning leaf, this layer sits between the top and the bottom epidermis of the leaf, forming the bacon in the BLT that is the leaf structure. This, my friends, marks our entry into the ground tissue. I'm sure you're as excited about that as I am. Despite it's name, ground tissue isn't just in the ground and it's actually just defined as any tissue that's either not dermal or vascular. Regardless of this low billing though, it's where the money is and by money, I mean food. And the mesophyll is chalk full of parenchyma cells of various shapes and sizes and many of them are arranged loosely to let CO2 and other materials These cells contain the photosynthetic organelles, chloroplasts, which as you know, host the process of photosynthesis, but where is this CO2 coming from? Well, some of the neatest features on the leaf are these tiny openings in the epidermis called stomata. Around each stoma are two guard cells connected at both ends that regulate it's size and shape. When conditions are dry and the guard cells are limp, they stick together, closing the stoma, but when the leaf is flush with water the guard cells plump up and bow out from each other opening the stoma to allow water to evaporate and let carbon dioxide in. This is what allows evapotranspiration to take place, as well as photosynthesis. And you remember photosynthesis through a series of brain-rackingly complicated reactions sparked by the energy from the sun, the CO2 combines with hydrogen from the water to create glucose. The left over oxygen is released through the stomata and the glucose is ready for shipping. Now, if you've been paying attention, you'll notice that earlier I said that there are two kinds of vascular tissue as the sugar exits the leaf through the phloem. The phloem is mostly made of cells stacked in tubes with perforated plates at either end. After the glucose is loaded into these cells called sieve cells or sieve-tube elements, they then absorb water from the nearby xylem to form a rich, sugary sap to transport the sugar. This sweet sap, by the way, is what gives the ponderosa it's delicious smell. By way of internal pressure and diffusion, the sap travels wherever it's needed to parts of the plant experiencing growth during the growing season or down to the root if it's dormant like during winter, where it's stored until spring. So, now that you understand everything that it takes for vascular plants to succeed, I hope you see why plants equals winning and I'm not just talking about them sweeping the contest for biggest, heaviest, oldest living things, though again, congrats on that guys. Plants are not only responsible for like making rain happen, they're also the first and most important link in our food chain and that's why the world's most plant rich habitats like rain forests and grasslands are so crucial to our survival." + }, + { + "Q": "At 0:44 sec in the video how is the slope coming from positive infinity at negative four?\n", + "A": "At x = -4, the slope is vertical and has a positive infinity slope, thus, the graph is going from that positive infinity amount to 0 at y=4.", + "video_name": "NFzma7NsHtI", + "timestamps": [ + 44 + ], + "3min_transcript": "I have a function f of x here, and I want to think about which of these curves could represent f prime of x, could represent the derivative of f of x. Well, to think about that, we just have to think about, well, what is a slope of the tangent line doing at each point of f of x and see if this corresponds to that slope, if the value of these functions correspond to that slope. So we can see when x is equal to negative 4, the slope of the tangent line is essentially vertical. So you could say it's not really defined there. But as we go slightly to the right of x equals negative 4, we just have a very, very, very positive slope. So you could kind of view it as our slope is going from infinity to very, very positive to a little bit less positive to a little bit less positive, to a little bit less positive, to a little bit less positive. So which of these graphs here have that property? Remember, this is trying to graph the slope. So which of these functions down here, which of these graphs, have a value that is essentially kind of approaching infinity when x is equal to negative 4, and then it So this one, it looks like it's coming from negative infinity, and it's getting less and less and less negative. So that doesn't seem to meet our constraints. This one looks like it is coming from positive infinity, and it's getting less and less and less positive, so that seems to be OK. This has the same property. It's getting less and less and less positive. This one right over here starts very negative and gets less and less and less negative. So we can rule that out. Now let's think about what happens when x gets to 0. When x gets to 0, the tangent line is horizontal. We're at a maximum point of this curve right over here. The slope of a horizontal line is 0. Remember, we're trying to look for which one of these curves represent the value of that slope. So which one of these curves hit 0 when x is equal to 0? Well, this one doesn't. So the only candidate that we have left is this one, And let's see if it keeps satisfying what we need for f prime of x. So after that point, it should start getting more and more negative. The slope should get more and more and more negative, essentially approaching negative infinity as x approaches 4. And we see that here. The value of this function is getting more and more negative, and it's approaching negative infinity as x approaches 4. So we'll go with this one. This looks like a pretty good candidate for f prime of x." + }, + { + "Q": "Shouldn't Hydrogen be the one with the partially positive charge and Fluorine with partially negative charge? Then why in the video did Sal do oppositely at 10:09 mins?\n", + "A": "You are correct, the H would be partially positive and the F should be partially negative due to the differences in electronegativities. Also, Fluorine has chemical symbol of F, not Fl. Fl is now the symbol for Flerovium, element number 114.", + "video_name": "8qfzpJvsp04", + "timestamps": [ + 609 + ], + "3min_transcript": "And because this van der Waals force, this dipole-dipole interaction is stronger than a London dispersion force. And just to be clear, London dispersion forces occur in all molecular interactions. It's just that it's very weak when you compare it to pretty much anything else. It only becomes relevant when you talk about things with noble gases. Even here, they're also London dispersion forces when the electron distribution just happens to go one way or the other for a single instant of time. But this dipole-dipole interaction is much stronger. And because it's much stronger, hydrogen chloride is going to take more energy to, get into the liquid state, or even more, get into the gaseous state than, say, just a sample of helium gas. Now, when you get even more electronegative, when this guy's even more electronegative when you're dealing with nitrogen, oxygen or fluorine, you get into a special case of dipole-dipole interactions, and that's the hydrogen bond. So it's really the same thing if a bunch of hydrogen fluorides around the place. Maybe I could write fluoride, and I'll write hydrogen fluoride here. Fluoride its ultra-electronegative. It's one of the three most electronegative atoms on the Periodic Table, and so it pretty much hogs all of the electrons. So this is a super-strong case of the dipole-dipole interaction, where here, all of the electrons are going to be hogged around the fluorine side. So you're going to have a partial positive charge, positive, partial negative, partial positive, partial negative and so on. So you're going to have this, which is really a dipole interaction. But it's a very strong dipole interaction, so people call it a hydrogen bond because it's dealing with hydrogen and a very electronegative atom, is pretty much hogging all of hydrogen's one electron. So hydrogen is sitting out here with just a proton, so it's going to be pretty positive, and it's really attracted to the negative side of these molecules. But hydrogen, all of these are van der Waals. So van der Waals, the weakest is London dispersion. Then if you have a molecule with a more electronegative atom, then you start having a dipole where you have one side where molecule becomes polar and you have the interaction 404 00:10:31,330 --> 00:10:32,470 between the positive and the negative side of the pole. It gets a dipole-dipole interaction. And then an even stronger type of bond is a hydrogen bond because the super-electronegative atom is essentially stripping off the electron of the hydrogen, or almost stripping it off. It's still shared, but it's all on that side of the molecule. Since this is even a stronger bond between molecules, it will have even a higher boiling point. So London dispersion, and you have dipole or polar bonds, and then you have hydrogen bonds." + }, + { + "Q": "\nAt 11:00, why is the partial positive at fluorine? If the fluorine is more electronegative, why does it have a partial positive?", + "A": "I believe that is the video s mistake, because generally in hydrogen bonds it is the hydrogen side that is slightly positive and the more electronegative element is slightly negative. You are correct, probably just a small mistake on the video s end.", + "video_name": "8qfzpJvsp04", + "timestamps": [ + 660 + ], + "3min_transcript": "a bunch of hydrogen fluorides around the place. Maybe I could write fluoride, and I'll write hydrogen fluoride here. Fluoride its ultra-electronegative. It's one of the three most electronegative atoms on the Periodic Table, and so it pretty much hogs all of the electrons. So this is a super-strong case of the dipole-dipole interaction, where here, all of the electrons are going to be hogged around the fluorine side. So you're going to have a partial positive charge, positive, partial negative, partial positive, partial negative and so on. So you're going to have this, which is really a dipole interaction. But it's a very strong dipole interaction, so people call it a hydrogen bond because it's dealing with hydrogen and a very electronegative atom, is pretty much hogging all of hydrogen's one electron. So hydrogen is sitting out here with just a proton, so it's going to be pretty positive, and it's really attracted to the negative side of these molecules. But hydrogen, all of these are van der Waals. So van der Waals, the weakest is London dispersion. Then if you have a molecule with a more electronegative atom, then you start having a dipole where you have one side where molecule becomes polar and you have the interaction 404 00:10:31,330 --> 00:10:32,470 between the positive and the negative side of the pole. It gets a dipole-dipole interaction. And then an even stronger type of bond is a hydrogen bond because the super-electronegative atom is essentially stripping off the electron of the hydrogen, or almost stripping it off. It's still shared, but it's all on that side of the molecule. Since this is even a stronger bond between molecules, it will have even a higher boiling point. So London dispersion, and you have dipole or polar bonds, and then you have hydrogen bonds. but because the strength of the intermolecular bond gets stronger, boiling point goes up because it takes more and more energy to separate these from each other. In the next video-- i realize I'm out of time. So this is a good survey, I think, of just the different types of intermolecular interactions that aren't necessarily covalent or ionic. In the next video, I'll talk about some of the covalent and ionic types of structures that can be formed and how that might affect the different boiling points." + }, + { + "Q": "\nIf the sides of atoms become negative and positive (4:26), then that means at one moment in time, one side of the neon atom will be attracted to another neon atom. As the atoms change charges again, the neon atom will be attracted in the other direction. But that would mean, even at absolute zero, the particles would be moving. How can this be? This would mean the presence of kinetic energy even though the gas has no heat.", + "A": "It is not correct to say that atoms are not moving at 0 K. If that were the case. we would then know both the position and the momentum of the atom. That would violate the Heisenberg Uncertainty Principle. Just as an electron in a hydrogen atom has a certain minimum allowed energy (a 1s orbital), an atom at 0 K must have a certain minimum energy. For example, liquid helium does not freeze under atmospheric pressure at any temperature because of its zero-point energy.", + "video_name": "8qfzpJvsp04", + "timestamps": [ + 266 + ], + "3min_transcript": "-- it's a probability cloud and it's what neon's atomic configuration is. 1s2 and it's outer orbital is 2s2 2p6, right? So it's highest energy electron, so, you know, it'll look-- I don't know. It has the 2s shell. The 1s shell is inside of that and it has the p-orbitals. The p-orbitals look like that in different dimensions. That's not the point. And then you have another neon atom and these are-- and I'm just drawing the probability distribution. I'm not trying to draw a rabbit. But I think you get the point. Watch the electron configuration videos if you want more on this, but the idea behind these probability distributions is that the electrons could be anywhere. There could be a moment in time when all the electrons out over here. There could be a moment in time where all the electrons are over here. Same thing for this neon atom. If you think about it, out of all of the possible configurations, there's actually a very low likelihood that they're going to be completely evenly distributed. There's many more scenarios where the electron distribution is a little uneven in one neon atom or another. So if in this neon atom, temporarily its eight valence electrons just happen to be like, you know, one, two, three, four, five, six, seven,eight, then what does this neon atom look like? It temporarily has a slight charge in this direction, right? It'll feel like this side is more negative than this side or this side is more positive than that side. Similarly, if at that very same moment I had another neon that has 1 2 3 4 5 6 7 8... that had a similar-- actually, let me do that differently. Let's say that this neon atom is like this: one, two, three,four, five, six, seven, eight. So here, and I'll do it in a dark color So this would be a little negative. Temporarly, just for that single moment in time, this will be kind of negative. That'll be positive. This side will be negative. This side will be positive. So you're going to have a little bit of an attraction for that very small moment of time between this neon and this neon, and then it'll disappear, because the electrons will reconfigure. But the important thing to realize is that almost at no point is neon's electrons going to be completely distributed. So as long as there's always going to be this haphazar distribution, there's always going to be a little bit of a-- I don't want to say polar behavior, because that's almost too strong of a word. But there will always be a little bit of an extra charge on one side or the other side of an atom, which will allow it to attract it to the opposite side charges of other similarly imbalanced molecules. And this is a very, very, very weak force. It's called the London dispersion force. I think the guy who came up with this, Fritz London, who was neither--" + }, + { + "Q": "At 7:24, why isn't the hydrogen-chlorine bond a hydrogen bond?\n", + "A": "Because a hydrogen bond is a weak intermolecular bond i.e between two or more molecules due to dipole-moment. The H-Cl bond is simply a bond between two atoms to form a molecule. Hope that helps.", + "video_name": "8qfzpJvsp04", + "timestamps": [ + 444 + ], + "3min_transcript": "I think he was German-American. London dispersion force, and it's the weakest of the van der Waals forces. I'm sure I'm not pronouncing it correctly. And the van der Waals forces are the class of all of the intermolecular, and in this case, neon-- the molecule, is an atom . It's just a one-atom molecule, I guess you could say. The van der Waals forces are the class of all of the intermolecular forces that are not covalent bonds and that aren't ionic bonds like we have in salts, and we'll touch on those in a second. And the weakest of them are the London dispersion forces. So neon, these noble gases, actually, all of these noble gases right here, the only thing that they experience are London dispersion forces, which are the weakest of all of the intermolecular forces. And because of that, it takes very little energy to get them into a gaseous state. So at a very, very low temperature, That's why they're called noble gases, first of all. And they're the most likely to behave like ideal gases because they have very, very small attraction to each other. Fair enough. Now, what happens when we go to situations when we go to molecules that have better attractions or that are a little bit more polar? Let's say I had hydrogen chloride, right? Hydrogen, it's a little bit ambivalent about whether or not it keeps its electrons. Chloride wants to keep the electrons. Chloride's quite electronegative. It's less electronegative than these guys right here. These are kind of the super-duper electron hogs, nitrogen, oxygen, and fluorine, but chlorine is pretty electronegative. So if I have hydrogen chloride, so I have the chlorine atom right here, it has seven electrons and then it shares an electron with the hydrogen. It shares an electron with the hydrogen, Because this is a good bit more electronegative than hydrogen, the electrons spend a lot of time out here. So what you end up having is a partial negative charge on the side, where the electron hog is, and a partial positive side. And this is actually very analogous to the hydrogen bonds. Hydrogen bonds are actually a class of this type of bond, which is called a dipole bond, or dipole-dipole interaction. So if I have one chlorine atom like that and if I have another chlorine atom, the other chlorin eatoms looks like this. If I have the other chlorine atom-- let me copy and paste it-- right there, then you'll have this attraction between them. You'll have this attraction between these two chlorine atoms-- oh, sorry, between these two hydrogen chloride molecules. And the positive side, the positive pole of this dipole is the hydrogen side, because the electrons have kind of left it, will be attracted to the chlorine side" + }, + { + "Q": "\nAt 7:33, if the bond is between two atoms that are the same such as a C-C bond, which way would the dipole moment go towards?", + "A": "when equally electronegative atoms form a bond the electrons do not prefer to go more towards either of the atoms i.e., they are shared equally between the atoms giving a zero dipole moment", + "video_name": "q3g3jsmCOEQ", + "timestamps": [ + 453 + ], + "3min_transcript": "" + }, + { + "Q": "What's the relation with center of mass, at 2:51?\n", + "A": "The molecule has different centers of charge and mass, meaning that the position where charge is zero is not the same as the average mass position.", + "video_name": "q3g3jsmCOEQ", + "timestamps": [ + 171 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 5:18, why do you care about the lone paris? Bc then wouldnt Sf2 be non-olar as well?", + "A": "By Sf2 do you mean SF2 (sulfur difluoride)? If so it s polar. The lone pairs around S mean that the molecule is NOT linear, it looks sort of like a V instead. This means that the polar bonds between S and each F do not cancel one another out and the molecule is polar.", + "video_name": "q3g3jsmCOEQ", + "timestamps": [ + 318 + ], + "3min_transcript": "" + }, + { + "Q": "At around 7:20, I don't understand how you can replace V with 4 m^3.\n\nmeters^3 is not a unit for volume so it is confusing.\n", + "A": "meters^3 IS a unit for volume. cubic meters. any length cubed is a measure of volume.", + "video_name": "d4bqNf37mBY", + "timestamps": [ + 440 + ], + "3min_transcript": "So let's see, if we do 2100 grams, remember I want to do everything in grams, so I just want to do a unit conversion there. 2100 grams times 2.86% is equal to about 60 grams. So hydrogen, this 2% of that 2100 grams is 60 grams. And then what's the molar mass of one hydrogen. That's H2. So we know that the hydrogen atom by itself has a mass of 1, doesn't have a neutron in most cases. So the atomic mass of this is 2. Or the molar mass of this is 2 grams. So one mole of H2 is equal to two grams. We have 60 grams. So we clearly have 60 divided by 2, we have 30 moles. was a super small fraction of the total mass of the gas that we have inside of the container, we actually have more actual particles, more actual molecules of hydrogen than we do of oxygen. That's because each molecule of hydrogen only has an atomic mass of 2 atomic mass units, while each molecule of oxygen has 32 because there's two oxygen atoms. So already we're seeing we actually have more particles do the hydrogen than do the oxygen. And the particles are what matter, not the mass, when we talk about part pressure and partial pressure. So the first thing we can think about is how many total moles of gas, how many total particles do we have bouncing around? 20 moles of oxygen, 30 moles of hydrogen, 50 moles of nitrogen gas. Add them up. We have 100 moles of gas. So if we want to figure out the total pressure first, we can just apply this 100 moles. Let me erase this. There you go. And I can erase some stuff that you're not seeing off the screen. And now I'm ready. So we have 100 moles. So we just do our PV is equal to nRT. We're trying to solve for P. P times 4 meters cubed is equal to n. n is the number of moles. We have 100 moles. Is equal to 100 moles times R. I'll put a blank there for R, because we have to figure out which R we want to use. Times temperature, remember we have to do it in Kelvin. So 0 degrees Celsius is 273 Kelvin. And then which R do we use? I always like to write my R's down here. So we're dealing with meters cubed, we're not dealing with liters, so let's use this one. 8.3145 meters cubed pascals per mole Kelvin." + }, + { + "Q": "\nat 9:50 sal said 101,325Pa=1atm , I thought it was 101.3Pa=1atm, thats what i remember learning in chemistry, or am I wrong.", + "A": "1 atm = 101.3 kPa, that is kiloPascals, or 101 325 Pa", + "video_name": "d4bqNf37mBY", + "timestamps": [ + 590 + ], + "3min_transcript": "these are in meters cubed pascals divided by moles Kelvin. And then our temperature was 273 Kelvin. Now let's do a little dimensional analysis to make sure that we're doing things right. These meters cancel out with those meters. We divide both sides of the equation by meters. These moles cancel with these moles. The moles of the numerator, the moles of the denominator. Kelvin in the numerator, Kelvin in the denominator. And all we're left with is pascals. Which is good because that is a unit of pressure. So if we divide both sides of this equation by 4. I'll just divide the 100 by 4. 25 times 8.3145 times 273. Which is nice, because that's a unit of pressure. So, let's do the math, 25 times 8.3145 times 273 is equal to 56,746 pascals. And that might seem like a crazy number. But the pascal is actually a very small amount of pressure. It actually turns out that 101,325 pascals is equal to one atmosphere. So if we want to figure out how many atmospheres this is, we could just divide that. Let me look it up on this table. Yes, 101,325. That's 56.746 kilopascals. Or if we wanted it in atmospheres we just take 56,746 divided by 101,325. It equals 0.56 atmospheres. So that's the total pressure being exerted from all of the gases. I deleted that picture. So this is the total pressure. So our question is, what's the partial pressure? We could use either of these numbers, they're just in different units. What's the partial pressure of just the oxygen by itself? Well, you look at the moles, because we don't care about the actual mass. Because we're assuming that they're ideal gases. We want to look at the number of particles. Because remember, we said pressure times volume is proportional to the number of particles times temperature. And they're all at the same temperature. So the number of particles is what matters. So oxygen represents 20% of the particles." + }, + { + "Q": "At 8:04 Sal mentions multiple values or R in PV=nRT, but my science book only has one. How do you know which to use?\n", + "A": "You use the version of R that contains the same units that you have for volume and pressure.", + "video_name": "d4bqNf37mBY", + "timestamps": [ + 484 + ], + "3min_transcript": "was a super small fraction of the total mass of the gas that we have inside of the container, we actually have more actual particles, more actual molecules of hydrogen than we do of oxygen. That's because each molecule of hydrogen only has an atomic mass of 2 atomic mass units, while each molecule of oxygen has 32 because there's two oxygen atoms. So already we're seeing we actually have more particles do the hydrogen than do the oxygen. And the particles are what matter, not the mass, when we talk about part pressure and partial pressure. So the first thing we can think about is how many total moles of gas, how many total particles do we have bouncing around? 20 moles of oxygen, 30 moles of hydrogen, 50 moles of nitrogen gas. Add them up. We have 100 moles of gas. So if we want to figure out the total pressure first, we can just apply this 100 moles. Let me erase this. There you go. And I can erase some stuff that you're not seeing off the screen. And now I'm ready. So we have 100 moles. So we just do our PV is equal to nRT. We're trying to solve for P. P times 4 meters cubed is equal to n. n is the number of moles. We have 100 moles. Is equal to 100 moles times R. I'll put a blank there for R, because we have to figure out which R we want to use. Times temperature, remember we have to do it in Kelvin. So 0 degrees Celsius is 273 Kelvin. And then which R do we use? I always like to write my R's down here. So we're dealing with meters cubed, we're not dealing with liters, so let's use this one. 8.3145 meters cubed pascals per mole Kelvin. these are in meters cubed pascals divided by moles Kelvin. And then our temperature was 273 Kelvin. Now let's do a little dimensional analysis to make sure that we're doing things right. These meters cancel out with those meters. We divide both sides of the equation by meters. These moles cancel with these moles. The moles of the numerator, the moles of the denominator. Kelvin in the numerator, Kelvin in the denominator. And all we're left with is pascals. Which is good because that is a unit of pressure. So if we divide both sides of this equation by 4. I'll just divide the 100 by 4. 25 times 8.3145 times 273." + }, + { + "Q": "\nAt 4:33 he says no net work due to no change in kinetic energy. Is there not a change in potential energy that would account for work being done?", + "A": "When PE increases, no NET work is done. For example, when you lift a 1 kg book 1 meter in the air, you did 1 Joule of work on it, but gravity did -1 Joule of work, so the net work is zero, and thats why the book is not moving. This is called the work-KE theorem.", + "video_name": "udgMh3Y-dTk", + "timestamps": [ + 273 + ], + "3min_transcript": "But the angle between the gravitational force and the direction of the displacement is 90 degrees in this case. And since cosine of 90 is 0, the gravitational force does no work on this trashcan. Similarly, if we were to find the work done by the normal force, the angle between the direction of the displacement and the normal force is 90 degrees. So the normal force also does no work on the trashcan. This makes sense because forces that are perpendicular to the motion can never do any work on that object. So that's how you can find the work done by individual forces. And if we wanted to know the net work done on this trashcan, we could just add up the work done by each individual force. So the net work is going to be 200 joules. Now that we know the net work done on the trashcan, we can use the work-energy principle to figure out the speed of the trashcan after it's slid the 10 meters. The work-energy principle says that the net work done in kinetic energy of that object. So 200 joules is going to equal the difference in kinetic energy. If we assume the trashcan started at rest, which seems reasonable, the initial velocity is 0. So we can solve for the final speed of the trashcan, which comes out to be 10 meters per second. This time, let's say you take the trashcan and lift it upwards with a constant velocity for a distance of 2 meters. In order to lift the trashcan up with a constant velocity, you need to push with a force equal to the weight of the trashcan, which means you have to push upwards with a force of 39.2 newtons. So to find the work done by the force that you exert, the force is going to be 39.2 newtons. The displacement is going to be 2 meters. And the angle between the force and the displacement is going to be 0 degrees because the direction of the force that you exert is in the same direction as the displacement of the trashcan. So the work that you've done in lifting up this 4 kilogram To find the work done by the force of gravity, we can use the force of gravity, which is again 39.2 newtons. The displacement is again 2 meters. But the angle between the direction of the displacement and the gravitational force is 180 degrees because the displacement points up and the gravitational force points down. So the work done by the gravitational force is negative 78.4 joules, which means the net work done on the trashcan is 0. And that makes sense. Because since the trashcan moved upwards with constant velocity, there was no change in the kinetic energy of this object." + }, + { + "Q": "\nAt 7:32 to 7:45 If an object gets to the center of some massive object will it be moving back and forth because gravity is pulling and pushing it or will it stay at center?", + "A": "It will stay at the center as the gravity is 0 at that point.", + "video_name": "RpOHZc6cDIw", + "timestamps": [ + 452, + 465 + ], + "3min_transcript": "If this distance right here is r/1,000 wouldn't some photon here, or atom here, or molecule, or whatever it's over here, wouldn't that experience the same force, this million times the force as this thing? And you've got to remember, all of a sudden when this thing is inside of this larger mass, what's happening? The entire mass is no longer pulling on it in that direction. It's no longer pulling it in that inward direction. You now have all of this mass over here. Let me think of the best way that's doing it. So you can think of it all of this mass over here is pulling it in an outward direction. It's not telling. What that mass out there is doing, since that mass itself is being pulled inward, It is exerting pressure on that point. But the actual gravitational force that that point is experiencing is actually going to be less. It's actually going to be mitigated by the fact that there's so much mass over here pulling in the other direction. And so you could imagine if you were in the center of a really massive object-- so that's a really massive object. If you were in the center, there would be no net gravitational force being pulled on you, because you're at its center of mass. The rest of the mass is outward. So at every point it will be pulling you outward. And so that's why if you were to enter the core of a star, if you were to get a lot closer to its center of mass, it's not going to be pulling on you with this type of force. And the only way you can get these types of forces is if the entire mass is contained in a very dense region, in a very small region. such strong gravity that not even light can escape. Hopefully that clarifies things a little bit." + }, + { + "Q": "\nAt around 3:20, Sal is talking about how, the faster the projectile is thrown, the further it goes before falling back down. How is this so? I thought that objects accelerated downward at the same rate regardless of initial horizontal velocity. Does this only apply to objects projected perfectly horizontally?", + "A": "The downward acceleration is the same, so the fall time is the same, but the horizontal velocity is faster, so it goes further in the same amount time.", + "video_name": "oIZV-ixRTcY", + "timestamps": [ + 200 + ], + "3min_transcript": "this r isn't that different. It's a little bit further than if you were at the surface of the Earth. Remember that r is measured from wherever you are to the center, from the center of the Earth, or really the center of the object to the center of the Earth. The center of the Earth represents most of the distance here. So if I'm at the surface of the Earth or if I'm just a few hundred miles above the surface of the Earth, it's not going to change r that dramatically, especially in terms of percentage. So when you look at it this way, it seems pretty clear that the force of gravity for someone who is in space only a few hundred miles above the Earth should not be that different than the force of gravity for someone who is on the surface of the Earth. So my question to you is, what gives? If there should be gravity in space, how can we see all of these pictures of people floating around like this? And the answer is that there is gravity in space, They're just moving fast enough relative to the Earth that they keep missing it. And let me show you what I'm talking about there. Let's say I'm sitting here in Africa, and I were to shoot something, if maybe I have a really good sling shot, and I were to sling something super fast and maybe at a 45 degree angle, it might take off a little bit and eventually hit another point. And this would actually already be a super duper slingshot. I just made it travel a couple of thousand miles or at least over 1,000 miles. If I make it go a little bit faster, if I put a little bit more force on, if I just propelled the projectile a little bit faster, it might go a little bit further, but it will eventually fall back to the Earth. Let's try to propel it a little bit faster than that. Then it'll still fall to the Earth. Let's propel it even faster than that. Well then, it's still eventually going to fall to the Earth. I think you might see where this is going. Let's go even faster than that. eventually it'll fall to the Earth. Even faster than that, so if you were to throw an object even faster than that, it would then go really far and then fall to the Earth. I think you see what's happening. Every time you go faster and faster, you throw this projectile faster and faster, it gets further and further, up to some velocity that you release this projectile, and whenever it's trying to fall to the Earth, it's going so fast that it keeps missing the Earth. So it'll keep going around and around and around the Earth, and a projectile like that would essentially be in orbit. So what's happening is if there was no gravity for that projectile, if there was no gravity, the projectile would just go straight away into space. But because there's gravity, it's constantly pulling it towards the center of the Earth, or the center of that projectile and the center of the Earth are being pulled towards each other," + }, + { + "Q": "At 1:56, Sal said, \"I'm making these numbers up on the fly, so bear with me.\" What did he mean? Did he mean that he's inventing these numbers temporarily and without preparing carefully before recording this video?\n", + "A": "yes, that is what it means.", + "video_name": "_k3aWF6_b4w", + "timestamps": [ + 116 + ], + "3min_transcript": "Welcome to Level 4 multiplication. Let's do some problems. Let's see, we had 235 times-- I'm going to use two different colors here, so bear with me a second. Let's say times 47. So you start a Level 4 problem just like you would normally do a Level 3 problem. We'll take that 7, and we'll multiply it by 235. So 7 times 5 is 35. 7 times 3 is 21, plus the 3 we just carried is 24. 7 times 2 is 14, plus the 2 we just carried. This is 16. So we're done with the 7. Now we have to deal with this 4. Well, since that 4 is in the tens place, we add a 0 here. You could almost view it as we're multiplying 235, not by we put that 0 there. But once you put the 0 there, you can treat it just like a 4. So you say 4 times 5, well, that's 20. Let's ignore what we had from before. 4 times 3 is 12, plus the 2 we just carried, which is 14. 4 times 2 is 8, plus the 1 we just carried, so that's 9. And now we just add up everything. 5 plus 0 is 5, 4 plus 0 is 4, 6 plus 4 is 10, carry the 1, and 1 plus 9, well, that's 11. So the answer's 11,045. Let's do another problem. Let's say I had 873 times-- and I'm making these numbers up on the fly, so bear with me-- 873 times-- some high numbers-- hopefully get a better understanding of what I'm trying to explain. Let's say 97. No, I just used a 7. Let's make it 98. So just like we did before, we go to the ones place first, and that's where that 8 is, and we multiply that 8 times 873. So 8 times 3 is 24, carry the 2. 8 times 7 is 56, plus 2 is 58, carry the 5. 8 times 8 is 64, plus the 5 we just carried. That's 69. We're done with the 8. Now we have to multiply the 9, or we could just do it as we're multiplying 873 by 90. But multiplying something by 90 is just the same thing as multiplying something by 9 and then adding a 0 at the end, so that's why I put a 0 here." + }, + { + "Q": "\nAt 5:15, how did he get the excluded values?", + "A": "he just found what values of x would make the denominator equal zero.", + "video_name": "dstNU7It-Ro", + "timestamps": [ + 315 + ], + "3min_transcript": "When I multiply them, I get 2 times 30, which is 60. And an a plus a b, when I add them, I get 17. Once again, 5 and 12 seem to work. So let's split this up. Let's split this up into 2x squared. We're going to split up the 17x into a 12x plus a 5x and that adds up to 17x. When you multiply 12 times 5, you get 60, and then plus 30. Then on this first group right here, we can factor out a 2x, so if you factor out a 2x, you get 2x times x plus 6. In that second group, we can factor out a 5, so you get Now, we can factor out an x plus 6, and we get we get x plus 6 times 2x plus 5. We've now factored the numerator and the denominator. Let's rewrite both of these expressions or write this entire rational expression with the numerator and the denominator factored. The numerator is going to be equal to x plus 4 times 2x plus 5. We figured that out right there. And then the denominator is x plus 6 times 2x plus 5. It might already jump out at you that you have 2x plus 5 in the numerator and the denominator, and we can cancel them out. We will cancel them out. But before we do that, let's work on the second part of this question. State the domain. A more interesting question is what are the x values that will make this rational expression undefined? It's the x values that will make the denominator equal to 0, and when will the denominator equal to 0? Well, either when x plus 6 is equal to 0, or when 2x plus 5 is equal to 0. We could just solve for x here. Subtract 6 from both sides, and you get x is equal to negative 6. If you subtract 5 from both sides, you get 2x is equal to negative 5. Divide both sides by 2. You get x is equal to negative 5/2. We could say the domain-- let me write this over here. The domain is all real numbers other than or except x is equal to negative 6 and x is equal to negative 5/2." + }, + { + "Q": "\nI think that I am on the wrong topic, but I really don't understand what Sal was doing at 1:16-1:28. Can someone please explain what he just did?", + "A": "Adding on to what Jerry said, you would get the factors of 5 and -1. You could then plug it into the equation like this. x^2 + 5x - x - 5. Form here, you can solve it to this. x(x + 5) - 1(x + 5) Finally you would get (x + 5) and (x - 1) as your final answer", + "video_name": "DRpdoZQtvOM", + "timestamps": [ + 76, + 88 + ], + "3min_transcript": "We already have the tools in our tool kit to factor something like x squared plus 4x minus 5. And the way that we've already thought about it is we've said, hey, let's think of two numbers that if we were to take their product, we'd get negative 5. And if we were to add the two numbers, we'd get positive 4. And the fact that their product is negative tells you one of them is going to be positive and one of them is going to be negative. And so there's a couple of ways you could think about it. Well, you could say, well, maybe one of the numbers is negative 1 and then the other one is positive 5. Actually, this one seems to work. Negative 1 times 5 is negative 5. Negative 1 plus 5 is positive 4. So this one actually seems to work. The other option would have been-- since we're just going to deal with the factors of 5, and 5's a prime number, the other option would have been something like 1 and negative 5. There's only two factors for 5. So 1 and negative 5-- their product would have been negative 5. But if you were to add these two numbers, you would have gotten a negative 4 right over here. So we're going to go with this right over here. to factor this using the tools that we already know about, we will get-- And let me write these numbers in a different color so we can keep track of them. So negative 1 and 5. We know that this would factor out to be x minus 1 times x plus 5. And you can verify this for yourself that if you were to multiply this out, you will get x squared plus 4x minus 5. You can even see this here. x times x is x squared. Negative x plus 5x is going to be 4x. And then negative 1 times 5 is negative 5. This is all review for us at this point. Now I want to tackle something a little bit more interesting. Let's say we wanted to factor x squared plus 4xy minus 5y squared. All of a sudden, I've introduced a y and a y squared here. I have two variables. How would I tackle it? But the important thing is to just take a deep breath and realize that we're not fundamentally doing something different. Now, the one little tricky thing I've done when I've written it this way-- and I encourage you to pause this and try this on your own before I explain any further. But the one tricky thing I did right over here is I wrote the x before the y. And that tends to be the convention. You just write them kind of in alphabetical order. But if we wanted it in a form that's a little bit closer to this and something that would fit this mold a little bit more is if we swapped these two. Because then we could write it as x squared plus 4yx minus 5y squared. And now it becomes pretty clear that this 4y term right over here-- this right over here is the coefficient on the x term, the same way that 4 was the coefficient on x right here. And this negative 5y squared corresponds" + }, + { + "Q": "So we are supposed to guess at 2:16?\n", + "A": "No, that side definitely goes there. Could wasn t the best choice of word.", + "video_name": "m1ZTnl4CNQg", + "timestamps": [ + 136 + ], + "3min_transcript": "Teddy knows that a figure has a surface area of 40 square centimeters. The net below has 5 centimeter and 2 centimeter edges. Could the net below represent the figure? So let's just make sure we understand what this here represents. So it tells us that it has 5 centimeter edges. So this is one of the 5 centimeter edges right over here. And we know that it has several other 5 centimeter edges because any edge that has this double hash mark right over here is also going to be 5 centimeters. So this edge is also 5 centimeters, this is also 5 centimeters, this is also 5 centimeters, and then these two over here are also 5 centimeters. So that's 5 centimeters, and that's 5 centimeters. And then we have several 2 centimeter edges. So this one has 2 centimeters. And any other edge that has the same number of hash marks, in this case one, is also going to be 2 centimeters. So all of these other edges, pretty much all the rest of the edges, are going to be 2 centimeters. Now, they don't ask us to do this in the problem, but it's always fun to start with a net like this It looks pretty clear this is going to be a rectangular prism. But let's actually draw it. So if we were to-- we're going to fold this in. We're going to fold this that way. You could view this as our base right over here. We're going to fold this in. We're going to fold that up. And then this is going to be our top. This is the top right over here. This polyhedron is going to look something like this. So you're going to have your base that has a length of 5 centimeters. So this is our base. Let me do that in a new color. So this is our base right over here. I'll do it in the same color. So that's our base, this dimension right over here. I could put the double hash marks if I want. 5 centimeters, and that's of course the same as that dimension up there. Now, when we fold up this side-- we'll do this in orange, could be this side right over here, along this 2 centimeter edge. So that's that side right over here. When you fold this side in right over here, that could be that. That's that side right over there. And then when of course we fold this side in-- that's the same color. Let me do a different color. When we fold this side in, that's the side that's kind of facing us a little bit. So that's that right over there. That's that right over there. Color that in a little bit better. And then we can fold this side in, and that would be that side. And then, of course, we have the top that's connected right over here. So the top would go-- this would be the top, and then the top would, of course, go on top of our rectangular prism. So that's the figure that we're talking about. It's 5 centimeters in this dimension. It is 2 centimeters tall, and it is 2 centimeters wide." + }, + { + "Q": "At 3:38, how does Sal get (1 - 4x^2)? I understand the -4x^2 part, but not how he gets the 1 preceding it when he factors out the e^-2x^2...\n", + "A": "All he did is factor out the e^(-2x^2). e^(-2x^2) - 4x^2 * e^(-2x^2) = e^(-2x^2) * (1-4x^2) It is the same as factoring out the x term in this expression: x -2x^2 = x * (1-2x) When factoring it out, there is still the e^(-2x^2) * 1 term that you need to take into account.", + "video_name": "MUQfl385Yug", + "timestamps": [ + 218 + ], + "3min_transcript": "What is this going to be? Well all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to one. This first part is going to be equal to e to the negative two x squared. Now the derivative of e to the negative two x squared over here. I'll do this in this pink color. This part right over here, that is going to be equal to- We'll just apply the chain rule. Derivative of e to the negative two x squared with respect to negative two x squared, well that's just going to be e to the negative two x squared. We're going to multiply that times the derivative of negative two x squared with respect to x. That's going to be what, negative four x. Times negative four x, and of course we have this x over here. We have that x over there and let's see, Well obviously both of these terms have an e to the negative two x squared. I'm going to try to figure out where this is either undefined or where this is equal to zero. Let's think about this a little bit. If we factor out e to the negative two x squared, I'll do that in green. We're going to have, this is equal to e to the negative two x squared times, we have here, one minus four x squared. One minus four x squared. This is the derivative of f. Where would this be undefined or equal to zero? e to the negative two x squared, this is going to be defined for any value of x, and this part is also going to be defined for any value of x. There's no point where this is undefined. Let's think about when this is going to be equal to zero. The product of these two expressions equalling zero, e to the negative two x squared, that will never be equal to zero. If you get this exponent to be a really, I guess you could say very negative number, you will approach zero but you will never get it to be zero. This part here can't be zero. If the product of two things are zero at least one of them has to be zero, so the only way we can get f prime of x to be equal to zero is when one minus four x squared is equal to zero. One minus four x squared is equal to zero, let me rewrite that. One minus four x squared is equal to zero, when does that happen? This one we can just solve. Add four x squared to both sides, you get one is equal to four x squared. Divide both sides by four, you get" + }, + { + "Q": "at 4:28 why the e^-2x^2 cant be zero ?\n", + "A": "The exponential function (when considered a function of real numbers) is always positive. That is, for every real number u, we have exp(u) > 0. Now let u = -2x\u00c2\u00b2, which is a real number whenever x is a real number. It follows that exp(u) = exp(-2x\u00c2\u00b2) > 0. Hence never zero. The fact that exp(u) > 0 for every real number u requires a proof in itself, but I will not provide one here (unless you really want me to).", + "video_name": "MUQfl385Yug", + "timestamps": [ + 268 + ], + "3min_transcript": "What is this going to be? Well all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to one. This first part is going to be equal to e to the negative two x squared. Now the derivative of e to the negative two x squared over here. I'll do this in this pink color. This part right over here, that is going to be equal to- We'll just apply the chain rule. Derivative of e to the negative two x squared with respect to negative two x squared, well that's just going to be e to the negative two x squared. We're going to multiply that times the derivative of negative two x squared with respect to x. That's going to be what, negative four x. Times negative four x, and of course we have this x over here. We have that x over there and let's see, Well obviously both of these terms have an e to the negative two x squared. I'm going to try to figure out where this is either undefined or where this is equal to zero. Let's think about this a little bit. If we factor out e to the negative two x squared, I'll do that in green. We're going to have, this is equal to e to the negative two x squared times, we have here, one minus four x squared. One minus four x squared. This is the derivative of f. Where would this be undefined or equal to zero? e to the negative two x squared, this is going to be defined for any value of x, and this part is also going to be defined for any value of x. There's no point where this is undefined. Let's think about when this is going to be equal to zero. The product of these two expressions equalling zero, e to the negative two x squared, that will never be equal to zero. If you get this exponent to be a really, I guess you could say very negative number, you will approach zero but you will never get it to be zero. This part here can't be zero. If the product of two things are zero at least one of them has to be zero, so the only way we can get f prime of x to be equal to zero is when one minus four x squared is equal to zero. One minus four x squared is equal to zero, let me rewrite that. One minus four x squared is equal to zero, when does that happen? This one we can just solve. Add four x squared to both sides, you get one is equal to four x squared. Divide both sides by four, you get" + }, + { + "Q": "\nAt 2:50, shouldn't d/dx e^-2x^2 = -2e^-2x^2*(-4x)?", + "A": "f(x) = e^(-2x^2) f (x) = d/dx(e^(-2x^2)) Apply Chain Rule: f (x) = e^(-2x^2)*d/dx(-2x^2) Remove constant: f (x) = e^(-2x^2)*-2*d/dx(x^2) Apply Power Rule d/dx(x^n) = n*x^(n - 1): f (x) = e^(-2x^2)*-2*2*x^(2 - 1)*d/dx(x) f (x) = e^(-2x^2)*-2*2*x^(2 - 1)*dx/dx f (x) = e^(-2x^2)*-2*2*x^(2 - 1) f (x) = e^(-2x^2)*-2*2*x^(1) [ f (x) = -4x*e^(-2x^2) ]", + "video_name": "MUQfl385Yug", + "timestamps": [ + 170 + ], + "3min_transcript": "Let's think about how we can find the derivative of this. f prime of x is going to be, well let's see. We're going to have to apply some combination of the product rule and the chain rule. It's going to be the derivative with respect to x of x, so it's going to be that, times e to the negative two x squared plus the derivative with respect to x of e to the negative two x squared times x. This is just the product rule right over here. Derivative of the x times e to the negative of two x squared plus the derivative of e to the What is this going to be? Well all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to one. This first part is going to be equal to e to the negative two x squared. Now the derivative of e to the negative two x squared over here. I'll do this in this pink color. This part right over here, that is going to be equal to- We'll just apply the chain rule. Derivative of e to the negative two x squared with respect to negative two x squared, well that's just going to be e to the negative two x squared. We're going to multiply that times the derivative of negative two x squared with respect to x. That's going to be what, negative four x. Times negative four x, and of course we have this x over here. We have that x over there and let's see, Well obviously both of these terms have an e to the negative two x squared. I'm going to try to figure out where this is either undefined or where this is equal to zero. Let's think about this a little bit. If we factor out e to the negative two x squared, I'll do that in green. We're going to have, this is equal to e to the negative two x squared times, we have here, one minus four x squared. One minus four x squared. This is the derivative of f. Where would this be undefined or equal to zero? e to the negative two x squared, this is going to be defined for any value of x," + }, + { + "Q": "\nwhat is the associauive property again? Sal says it on 2:05 to 2:10", + "A": "In the context of the video, Sal was talking about 3*a^1*a^7. The Associative Property refers to grouping numbers in a particular way to make solving the problem easier. It means that 3*(a^1*a^7) = a^7*(3*a^1) = a^1*(3*a^7). In the video at 2:10, Sal is choosing to make 3*a^1*a^7 into 3*(a^1*a^7) so he can simplify it into 3*(a^8) and into 3a^8.", + "video_name": "-TpiL4J_yUA", + "timestamps": [ + 125, + 130 + ], + "3min_transcript": "Simplify 3a times a to the fifth times a squared. So the exponent property we can use here is if we have the same base, in this case, it's a. If we have it raised to the x power, we're multiplying it by a to the y power, then this is just going to be equal to a to the x plus y power. And we'll think about why that works in a second. So let's just apply it here. Let's start with the a to the fifth times a squared. So if we just apply this property over here, this will result in a to the fifth plus two-th power. So that's what those guys reduce to, or simplify to. And of course, we still have the 3a out front. Now what I want to do is take a little bit of an aside and realize why this worked. Let's think about what a to the fifth times a squared means. A to the fifth literally means a times a times a times a times a. Now, a squared literally means a times a. So we're multiplying these five a's times these two a's. And what have we just done? We're multiplying a times itself five times, and then another two times. We are multiplying a times itself. So let me make it clear. This over here is a to the fifth. This over here is a squared. When you multiply the two, you're multiplying a by itself itself seven times. 5 plus 2. So this is a to the seventh power. a to the 5 plus 2 power. So this simplifies to 3a times a to the seventh power. Now you might say, how do I apply the property over here? What is the exponent on the a? And remember, if I just have an a over here, this is equivalent to a to the first power. So I can rewrite 3a is 3 times a to the first power. A to the first power-- and the association property of multiplication, I can do the multiplication of the a's before I worry about the 3's. So I can multiply these two guys first. So a to the first times a to the seventh-- I just have to add the exponents because I have the same base and I'm taking the product-- that's going to be a to the eighth power. And I still have this 3 out front. So 3a times a to the fifth times a squared simplifies to 3a to the eighth power." + }, + { + "Q": "\nAt 2:53 why is it that at most 1 Sals out of 100 may die in order for the insurance company not to loose money? What's the explanation? Sal didn't give any here.", + "A": "Because each Sal only pays 1% of the insurance payout over the life of the policy. So in this case 100 Sals each pay $10,000 for a total of $1,000,000. For each Sal that dies, the insurance company needs to pay that Sal s family $1,000,000. So if 2 Sals die the Insurance pay out $2,000,000 but have only collected $1,000,000 so they are losing money.", + "video_name": "NSSoMafbBqQ", + "timestamps": [ + 173 + ], + "3min_transcript": "have a higher chance of dying at that point, then it's probably going to be more expensive for me to get insurance. But I really am just worried about the next 20 years. But what I want to do in this video is think about given these numbers that have been quoted to me by the insurance company, what do they think that my odds of dying are over the next 20 years? So what I want to think about is the probability of Sal's death in 20 years, based on what the people at the insurance company are telling me. Or at least, what's the maximum probability of my death in order for them to make money? And the way to think about it, or one way to think about it, kind of a back-of-the-envelope way, is to think about what's the total premiums they're getting over the life of this policy divided by how much they're insuring me for. So they're getting $500 times 20 years is equal to, that's $10,000 over the life of this policy. So they're getting-- let's see those 0s cancel out, this 0 cancels out-- they're getting, over the life of the policy, $1 in premiums for every $100 in insurance. Or another way to think about it. Let's say that there were 100 Sals, 100 34-year-olds looking to get 20-year term life insurance. And they insured all of them. So if you multiplied this times 100, they would get $100 in premiums. This is the case where you have 100 Sals, or 100 people who are pretty similar to me. 100 Sals. And the only way that they could make money is if, at most, one of those Sals-- or really just break even-- if, at most, 1 of those Sals were to die. So break even if only 1 Sal dies. I don't like talking about this. It's a little bit morbid. So one way to think about it, they're getting $1 premium for $100 insurance. Or if they had 100 Sals, they would get $100 in premium, and the only way they would break even, if only 1 of those Sal dies. So what they're really saying is that the only way they can break even is if the probability of Sal dying in the next 20 years is less than or equal to 1 in 100. And this is an insurance company. They're trying to make money. So they're probably giving these numbers because they think the probability of me dying is a good-- maybe it's 1 in 200 or it's 1 in 300." + }, + { + "Q": "what does denote mean at 0:08?\n", + "A": "Denote means: to show, mark, or be a sign of (something). In his case, he is denoting or marking that the two lines are parallel.", + "video_name": "Ld7Vxb5XV6A", + "timestamps": [ + 8 + ], + "3min_transcript": "So I've got two parallel lines. So that's the first line right over there and then the second line right over here. Let me denote that these are parallel. These are parallel lines. Actually, I can do that a little bit neater. And let me draw a transversal, so a line that intersects both of these parallel lines, so something like that. And now let's say that we are told that this angle right over here is 9x plus 88. And this is in degrees. And we're also told that this angle right over here is 6x plus 182, once again, in degrees. So my goal here and my question for you is, can we figure out what these angles actually are, given that these are parallel lines and this is a transversal line? And I encourage you to pause this video to try this on your own. are related by the fact that they're formed from a transversal intersecting parallel lines. And we know, for example, that this angle corresponds to this angle right over here. They're going to be congruent angles. And so this is 6x plus 182. This is also going to be 6x plus 182. And then that helps us realize that this blue angle and this orange angle are actually going to be supplementary. They're going to add up to 180 degrees, because put together, when you make them adjacent, their outer rays form a line right over here. So we know that 6x plus 182 plus 9x plus 88 is going to be equal to 180 degrees. And now we just have to simplify this thing. So 6x plus 9x is going to give us 15x. Let's see, 182 plus 8, would get us to 190. And then we add another 80. It gets us to 270-- plus 270-- is equal to 180. If we subtract 270 from both sides, we get 15x is equal to negative 90. And now we can divide both sides by 15. And we get x is equal to-- what is this? Let's see, 6 times 15 is 60 plus 30 is 90. So x is going to be equal to negative 6. So far, we've made a lot of progress. We figured out what x is equal to. x is equal to negative 6, but we still haven't figured out what these angles are equal to. So this angle right over here, 9x plus 88, this is going to be equal to 9 times negative 6 plus 88." + }, + { + "Q": "Why would Sal not just subtract 34 degrees from one-hundred eighty and get the angle for the other side at 3:58?\n", + "A": "Good eye. When I come up with an alternate correct answer to a problem, as I m learning, I consider it a good sign -- you re thinking.", + "video_name": "Ld7Vxb5XV6A", + "timestamps": [ + 238 + ], + "3min_transcript": "Let's see, 182 plus 8, would get us to 190. And then we add another 80. It gets us to 270-- plus 270-- is equal to 180. If we subtract 270 from both sides, we get 15x is equal to negative 90. And now we can divide both sides by 15. And we get x is equal to-- what is this? Let's see, 6 times 15 is 60 plus 30 is 90. So x is going to be equal to negative 6. So far, we've made a lot of progress. We figured out what x is equal to. x is equal to negative 6, but we still haven't figured out what these angles are equal to. So this angle right over here, 9x plus 88, this is going to be equal to 9 times negative 6 plus 88. Let me write this down before I make a mistake. Negative 54 plus 88 is going to be-- let's see, to go from 88 minus 54 will give us 34 degrees. So this is equal to 34, and it's in degrees. So this orange angle right here is 34 degrees. The blue angle is going to be 180 minus that. But we can verify that by actually evaluating 6x plus 182. So this is going to be equal to 6 times negative 6 is negative 36 plus 182. So this is going to be equal to-- let's see, if I subtract the 6 first, I get to 176. So this gets us to 146 degrees. And you can verify-- 146 plus 34 is equal to 180 degrees. as well. We know that if this is 34 degrees, then this must be 34 degrees as well. Those are opposite angles. This angle also corresponds to this angle so it must also be 34 degrees, which is opposite to this angle, which is going to be 34 degrees. Similarly, if this one right over here is 146 degrees, we already know that this one is going to be 146. This one's going to be 146 since it's opposite. And that's going to be 146 degrees as well." + }, + { + "Q": "\nAt 5:13, couldn't you have just found the total area of all the triangles and the square and have it equal to c^2?\n\n(1/2 ab)*4= area of 4 triangles\n(b-a)^2=area of square\n\n2ab+(b-a)^2=c^2\n2ab+b^2-2ab+a^2=c^2\nb^2+a^2=c^2\nWhat's wrong with this why do we have to shift all the triangles and stuff", + "A": "I think its essentially the same thing. While you ve penned down an equation and played about with the variables algebraically, Bhaskara has played about geometrically and gotten the same answer. Also (correct me if I m wrong), you wrote down two equations before you arrived at the theorem, and Bhaskara shifted two triangles before he arrived at the theorem! Pretty neat I think.", + "video_name": "1ul8g55dYA4", + "timestamps": [ + 313 + ], + "3min_transcript": "the three angles are theta, 90 minus theta, and 90 degrees. So they all have the same exact angle, so at minimum, they are similar, and their hypotenuses are the same. So we know that all four of these triangles are completely congruent triangles. So with that assumption, let's just assume that the longer side of these triangles, that these are of length, b. So the longer side of these triangles I'm So this length right over here, I'll call that lowercase b. And let's assume that the shorter side, so this distance right over here, this distance right over here, this distance right over here, that these are all-- this distance right over here, that these are of length, a. So if I were to say this height right over here, Now we will do something interesting. Well, first, let's think about the area of the entire square. What's the area of the entire square in terms of c? Well, that's pretty straightforward. It's a c by c square. So the area here is equal to c squared. Now, what I'm going to do is rearrange two of these triangles and then come up with the area of that other figure in terms of a's and b's, and hopefully it gets us to the Pythagorean theorem. And to do that, just so we don't lose our starting point because our starting point is interesting, let me just copy and paste this entire thing. So I don't want it to clip off. So let me just copy and paste this. Copy and paste. So this is our original diagram. And what I will now do-- and actually, let me clear that out. I'm now going to shift. This is the fun part. I'm going to shift this triangle here in the top left. I'm going to shift it below this triangle on the bottom right. And I'm going to attempt to do that by copying and pasting. So let's see how much-- well, the way I drew it, it's not that-- well, that might do the trick. I want to retain a little bit of the-- so let me copy, or let me actually cut it, and then let me paste it. So that triangle I'm going to stick right over there. And let me draw in the lines that I just erased. So just to be clear, we had a line over there, and we also had this right over here. And this was straight up and down, and these were straight side to side. Now, so I moved this part over down here. So I moved that over down there. And now I'm going to move this top right triangle down" + }, + { + "Q": "\nI REALLY don't understand why we rewrite (7x - 5) for a second time on the right hand side (2:59) sec in to the video?? why?", + "A": "What sal is doing is the distribution property. He is distributing the value (7x-5) to g(x) s components. Recall: (a)(b+c) = ab+ac", + "video_name": "JKvmAexeMgY", + "timestamps": [ + 179 + ], + "3min_transcript": "f of x is equal to 7x minus 5. g of x is equal to x to the third power plus 4x. And then they ask us to find f times g of x So the first thing to realize is that this notation f times g of x is just referring to a function that is a product of f of x and g of x. So by definition, this notation just means f of x times g of x. And then we just have to substitute f of x with this definition, g of x with this definition, and then multiply out these algebraic expressions. f of x is right over there. And g of x, is right over there. So let's do it. So this is going to be equal to-- switch back to the orange color. It's going to be equal to f of x, which is 7x minus 5 times g of x, and g of x is x to the third power plus 4x. that each have two terms. You could use FOIL if you like. I don't like using FOIL because you might forget what it's even about. Foil is really just using the distributive property twice. So for example, you take this expression. Whatever you have out here, if you had a 9 out here, or an a, or an x, or anything. Now you have 7x minus 5. If you're multiplying it times this expression, you would multiply this times each term over here. So when you multiply 7x minus 5 times x to the third, you get-- I'll write it this way. You get x to the third times-- actually, let me write it the other way. You get 7x minus 5 times x to the third. And then you have plus 7x minus 5 times 4x. And now we can do the distributive property again. the things we distribute on the right hand side. It's the same exact idea. We could put the x to the third here as well. And when we distribute, you multiply x to the third times 7x and times negative 5. x to the third times 7x is 7x to the fourth power. X to the third times negative 5 is minus 5x to the third. And then you do it over here. You distribute the 4x over the 7x. 4x times 7x is plus 28x squared. 4x times negative 5 is minus 20x. And let's see if we can simplify this. We only have one fourth degree term, one third degree term, one second degree term, and one first degree term. Actually, we can't simplify this anymore. And we're done. This is the product of those two function definitions. This is f times g of x. It is a new function created by multiplying the other two functions." + }, + { + "Q": "At 0:25 He say (f*g)(x). then says f of x and g of x is the same thing, why is it like that? Also why doesn't he use y?\n", + "A": "Because you are doing what he also says to do in the video, distributing. You are distributing the x in (f*g)(x) to f and g, thus getting f(x)+g(x)", + "video_name": "JKvmAexeMgY", + "timestamps": [ + 25 + ], + "3min_transcript": "f of x is equal to 7x minus 5. g of x is equal to x to the third power plus 4x. And then they ask us to find f times g of x So the first thing to realize is that this notation f times g of x is just referring to a function that is a product of f of x and g of x. So by definition, this notation just means f of x times g of x. And then we just have to substitute f of x with this definition, g of x with this definition, and then multiply out these algebraic expressions. f of x is right over there. And g of x, is right over there. So let's do it. So this is going to be equal to-- switch back to the orange color. It's going to be equal to f of x, which is 7x minus 5 times g of x, and g of x is x to the third power plus 4x. that each have two terms. You could use FOIL if you like. I don't like using FOIL because you might forget what it's even about. Foil is really just using the distributive property twice. So for example, you take this expression. Whatever you have out here, if you had a 9 out here, or an a, or an x, or anything. Now you have 7x minus 5. If you're multiplying it times this expression, you would multiply this times each term over here. So when you multiply 7x minus 5 times x to the third, you get-- I'll write it this way. You get x to the third times-- actually, let me write it the other way. You get 7x minus 5 times x to the third. And then you have plus 7x minus 5 times 4x. And now we can do the distributive property again. the things we distribute on the right hand side. It's the same exact idea. We could put the x to the third here as well. And when we distribute, you multiply x to the third times 7x and times negative 5. x to the third times 7x is 7x to the fourth power. X to the third times negative 5 is minus 5x to the third. And then you do it over here. You distribute the 4x over the 7x. 4x times 7x is plus 28x squared. 4x times negative 5 is minus 20x. And let's see if we can simplify this. We only have one fourth degree term, one third degree term, one second degree term, and one first degree term. Actually, we can't simplify this anymore. And we're done. This is the product of those two function definitions. This is f times g of x. It is a new function created by multiplying the other two functions." + }, + { + "Q": "At 2:28 why is it that 7x * x^3 equals to x^4 ?\n", + "A": "Diego, 7x^3 means 7*x*x*x 7x^3 * x = 7*x*x*x * x 7x^4 means 7*x*x*x*x That is why 7x^3 * x = 7x^4 I hope that helps", + "video_name": "JKvmAexeMgY", + "timestamps": [ + 148 + ], + "3min_transcript": "f of x is equal to 7x minus 5. g of x is equal to x to the third power plus 4x. And then they ask us to find f times g of x So the first thing to realize is that this notation f times g of x is just referring to a function that is a product of f of x and g of x. So by definition, this notation just means f of x times g of x. And then we just have to substitute f of x with this definition, g of x with this definition, and then multiply out these algebraic expressions. f of x is right over there. And g of x, is right over there. So let's do it. So this is going to be equal to-- switch back to the orange color. It's going to be equal to f of x, which is 7x minus 5 times g of x, and g of x is x to the third power plus 4x. that each have two terms. You could use FOIL if you like. I don't like using FOIL because you might forget what it's even about. Foil is really just using the distributive property twice. So for example, you take this expression. Whatever you have out here, if you had a 9 out here, or an a, or an x, or anything. Now you have 7x minus 5. If you're multiplying it times this expression, you would multiply this times each term over here. So when you multiply 7x minus 5 times x to the third, you get-- I'll write it this way. You get x to the third times-- actually, let me write it the other way. You get 7x minus 5 times x to the third. And then you have plus 7x minus 5 times 4x. And now we can do the distributive property again. the things we distribute on the right hand side. It's the same exact idea. We could put the x to the third here as well. And when we distribute, you multiply x to the third times 7x and times negative 5. x to the third times 7x is 7x to the fourth power. X to the third times negative 5 is minus 5x to the third. And then you do it over here. You distribute the 4x over the 7x. 4x times 7x is plus 28x squared. 4x times negative 5 is minus 20x. And let's see if we can simplify this. We only have one fourth degree term, one third degree term, one second degree term, and one first degree term. Actually, we can't simplify this anymore. And we're done. This is the product of those two function definitions. This is f times g of x. It is a new function created by multiplying the other two functions." + }, + { + "Q": "at 1:17 how do you determine that x =2?\n", + "A": "Because u need it to equal zero so u must replace x with what is necessary witch was 2 in that case.", + "video_name": "7QMoNY6FzvM", + "timestamps": [ + 77 + ], + "3min_transcript": "We're asked to graph the equation y is equal to negative 2 times x minus 2 squared plus 5. So let me get by scratch pad out so we could think about this. So y is equal to negative 2 times x minus 2 squared plus 5. So one thing, when you see a quadratic or a parabola graph expressed in this way, the thing that might jump out at you is that this term right over here is always going to be positive because it's some quantity squared. Or I should say, it's always going to be non-negative. It could be equal to 0. So it's always going to be some quantity squared. And then we're multiplying it by a negative. So this whole quantity right over here is going to be non positive. It's always going to be less than or equal to 0. So this thing is always less than or equal to 0, the maximum value that y will take on is when this thing actually does equal 0. So the maximum value for y is at 5. And when does that happen? Well, y hits 5 when this whole thing is 0. And when does this thing equal 0? Well, this whole thing equals 0 when x minus 2 is equal to 0. And x minus 2 is equal to 0 when x is equal to 2. So the point 2 comma 5 is the maximum point for this parabola. And it is actually going to be the vertex. So if we were to graph this, so the point 2 comma 5. So that's my y-axis. This is my x-axis. So this is 1, 2, 1, 2, 3, 4, 5. So this right here is the point 2 comma 5. This is a maximum point, it's a maximum point for this parabola. And now I want to find two more points so that I can really determine the parabola. Three points completely determine a parabola. So that's 1, the vertex, that's interesting. Now, what I'd like to do is just get two points that are equidistant from the vertex. what happens when x is equal to 1 and when x is equal to 3. So I could make a table here actually, let me do that. So I care about x being equal to 1, 2, and 3, and what the corresponding y is. We already know that when x is equal to 2, y is equal to 5. 2 comma 5 is our vertex. When x is equal to 1, 1 minus 2 is negative 1, squared is just 1. So this thing is going to be negative 2 plus 5, so it's going to be 3. And when x is equal to 3, this is 3 minus 2, which is 1 squared is 1 times negative 2 is negative 2 plus 5 is 3 as well. So we have three points. We have the point 1 comma 3, the point 2 comma 5, and the point 3 comma 3 for this parabola. So let me go back to the exercise and actually put those three points in. And so we have the point 1 comma 3, we have the point 2 comma 5," + }, + { + "Q": "\n4:52 He labeled the postulate as AAS but once he made it clear BE was congruent to EC doesn't it become SAS after that is marked?", + "A": "He used AAS to prove that the two triangles are congruent, which by definition means that all corresponding sides/angles are congruent, including \u00f0\u009d\u0090\u00b5\u00f0\u009d\u0090\u00b8 \u00e2\u0089\u008c \u00f0\u009d\u0090\u00b6\u00f0\u009d\u0090\u00b8 (which means that \u00f0\u009d\u0090\u00b5\u00f0\u009d\u0090\u00b8 and \u00f0\u009d\u0090\u00b6\u00f0\u009d\u0090\u00b8 are of equal length, which is what we wanted to prove).", + "video_name": "RFesGHsuFZw", + "timestamps": [ + 292 + ], + "3min_transcript": "I'll just write a little code here. So Alt interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle. And then the next side is congruent to the next side So pink, green, side. Pink, green, side. So we can employ AAS, angle-angle-side. And it's in the right order. So now, we know that triangle-- we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB-- actually, let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle, that we haven't labeled. So angle BEA, we can say, is congruent to angle-- we start with the magenta vertices-- C, go to the center, E, and then go the unlabeled one, D. And we know this because of angle-angle-side. And they correspond to each other-- magenta-green-side, magenta-green-side. They're all congruent. So this is from AAS. And then, if we know that they are congruent, then that means corresponding sides are congruent. So then we know these two triangles are congruent. So that means that their corresponding sides are congruent. So then we know that length of BE is going to be equal-- and that's the segment that's between the magenta and the green angles. The corresponding side is side CE between the magenta and the green angles-- is equal to CE. If we number them, that's 1, that's 2, and that's 3. And so that comes out of statement 3. And so we have proven this. E is the midpoint of BC. It comes straight out of the fact that BE is equal to CE. So I can mark this off with hash. This line segment right over here is congruent to this line segment right over here, because we know that those two triangles are congruent. And I've inadvertently, right here, done a little two-column proof. This over here on the left-hand side is my statement. And then on the right-hand side, I gave my reason. And we're done." + }, + { + "Q": "\nat 4:21 Did you mean angle side angle?", + "A": "NO because Angle Side Angle mean the side is in between the 2 angles. Since the side is not between them so it SSA. Hope this help.", + "video_name": "RFesGHsuFZw", + "timestamps": [ + 261 + ], + "3min_transcript": "So this line right over here, this is a transversal. And there's actually several ways that we can do this problem. But we know that this is a transversal. And there's a couple of ways to think about it right over here. So let me just continue the transversal, so we get to see all of the different angles. You could say that this angle right here, angle ABE-- so this is its measure right over here-- you could say that it is the alternate interior angle to angle ECD, to this angle right over there. And if that didn't jump out at you, you would say that the corresponding angle to this one right over here is this angle right up here. If you were to continue this line off a little bit, these are the corresponding angles. And then this one is vertical. But either way, angle ABE-- let me be careful. Angle ABE is going to be congruent to angle DCE. I'll just write a little code here. So Alt interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle. And then the next side is congruent to the next side So pink, green, side. Pink, green, side. So we can employ AAS, angle-angle-side. And it's in the right order. So now, we know that triangle-- we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB-- actually, let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle, that we haven't labeled. So angle BEA, we can say, is congruent to angle-- we start with the magenta vertices-- C, go to the center, E, and then go the unlabeled one, D. And we know this because of angle-angle-side. And they correspond to each other-- magenta-green-side, magenta-green-side. They're all congruent. So this is from AAS. And then, if we know that they are congruent, then that means corresponding sides are congruent. So then we know these two triangles are congruent. So that means that their corresponding sides are congruent. So then we know that length of BE is going to be equal-- and that's the segment that's between the magenta and the green angles. The corresponding side is side CE between the magenta and the green angles-- is equal to CE." + }, + { + "Q": "\nI'm a little confused on how he got 3 at 5:12... how did he get that?", + "A": "Statement #3 - triangle-BEA is congruent to triangle-CED, comes from angle-angle-side. The first angle is from statement #2 (ABE = DCE via alternate interior angles), the second angle is from statement #1 (AEB = DEC via vertical angles), and the side is given in the original drawing (AE=DE).", + "video_name": "RFesGHsuFZw", + "timestamps": [ + 312 + ], + "3min_transcript": "I'll just write a little code here. So Alt interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle. And then the next side is congruent to the next side So pink, green, side. Pink, green, side. So we can employ AAS, angle-angle-side. And it's in the right order. So now, we know that triangle-- we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB-- actually, let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle, that we haven't labeled. So angle BEA, we can say, is congruent to angle-- we start with the magenta vertices-- C, go to the center, E, and then go the unlabeled one, D. And we know this because of angle-angle-side. And they correspond to each other-- magenta-green-side, magenta-green-side. They're all congruent. So this is from AAS. And then, if we know that they are congruent, then that means corresponding sides are congruent. So then we know these two triangles are congruent. So that means that their corresponding sides are congruent. So then we know that length of BE is going to be equal-- and that's the segment that's between the magenta and the green angles. The corresponding side is side CE between the magenta and the green angles-- is equal to CE. If we number them, that's 1, that's 2, and that's 3. And so that comes out of statement 3. And so we have proven this. E is the midpoint of BC. It comes straight out of the fact that BE is equal to CE. So I can mark this off with hash. This line segment right over here is congruent to this line segment right over here, because we know that those two triangles are congruent. And I've inadvertently, right here, done a little two-column proof. This over here on the left-hand side is my statement. And then on the right-hand side, I gave my reason. And we're done." + }, + { + "Q": "\nAt 6:50, so for every eigenvector there is also a corresponding eigenvalue?", + "A": "Actually, that s a good question. Some matrices transform vectors so that some of the vectors don t rotate (they re eigenvectors ). If a transformed vector x isn t rotated, what is x , the transformed vector? x is a scalar (called lambda - the eigenvalue ) times the original vector x - that is, Ax = x = lambda*x. So: Does every eigenvector have an eigenvalue?", + "video_name": "PhfbEr2btGQ", + "timestamps": [ + 410 + ], + "3min_transcript": "So in the example I just gave where the transformation is flipping around this line, v1, the vector 1, 2 is an eigenvector of our transformation. So 1, 2 is an eigenvector. And it's corresponding eigenvalue is 1. This guy is also an eigenvector-- the vector 2, minus 1. He's also an eigenvector. A very fancy word, but all it means is a vector that's just scaled up by a transformation. It doesn't get changed in any more meaningful way than just the scaling factor. And it's corresponding eigenvalue is minus 1. transformation matrix is. I forgot what it was. We actually figured it out a while ago. If this transformation matrix can be represented as a matrix vector product-- and it should be; it's a linear transformation-- then any v that satisfies the transformation of-- I'll say transformation of v is equal to lambda v, which also would be-- you know, the transformation of [? v ?] would just be A times v. These are also called eigenvectors of A, because A is just really the matrix representation of the transformation. So in this case, this would be an eigenvector of A, and this would be the eigenvalue associated with the eigenvector. So if you give me a matrix that represents some linear transformation. Now the next video we're actually going to figure out a way to figure these things out. But what I want you to appreciate in this video is that it's easy to say, oh, the vectors that don't get changed much. But I want you to understand what that means. It literally just gets scaled up or maybe they get reversed. Their direction or the lines they span fundamentally don't change. And the reason why they're interesting for us is, well, one of the reasons why they're interesting for us is that they make for interesting basis vectors-- basis vectors whose transformation matrices are maybe computationally more simpler, or ones that make for better coordinate systems." + }, + { + "Q": "At 4:11, Sal assumes he can draw a line through B, C, and D, making BCD a transversal of the two parallel lines, an assumption which allows him to solve the whole problem. I understand how that all works out. But unless I'm mistaken, it's never stated that points B, C, and D all lie on the same line. Sure, they look like they do, but without that being explicitly stated, isn't it possibly an erroneous assumption?\n", + "A": "Let s say that BC and CD are subtly off kilter. The technique will still work because you can extend line segment BC along (infinite) line BC, and there will be a new point (call it Q) to use instead of D. From there, the technique will be the same. The important element is that AB and CE are parallel, which means that BC (not necessarily BC and CD) is the transversal.", + "video_name": "0gzSreH8nUI", + "timestamps": [ + 251 + ], + "3min_transcript": "Bring down the 0. 5 goes into 20 four times, and then you're not going to have a remainder. 4 times 5 is 20. No remainder. So it's 34 times. So x is equal to 34. So the second largest angle has a measure of 34 degrees. This angle up here is going to be 4 times that. So 4 times 34-- let's see, that's going to be 120 degrees plus 16 degrees. This is going to be 136 degrees. Is that right? 4 times 4 is 16, 4 times 3 is 120, 16 plus 120 is 136 degrees. So we're done. The three measures, or the sizes of the three angles, are 10 degrees, 34 degrees, and 136 degrees. Let's do another one. So let's see. We have a little bit of a drawing here. And what I want to do is-- and we could think about different things. We could say, let's solve for x. I'm assuming that 4x is the measure of this angle. 2x is the measure of that angle right over there. And then if we know x, we can figure out what the actual measures of these angles are, assuming that we can figure out x. And the other thing that they tell us is that this line over here is parallel to this line over here. And it was very craftily drawn. Because it's parallel, but one stops here, and then one starts up there. So the first thing I want to do-- if they're telling us that these two lines are parallel, there's probably going to be something involving transversals or something. It might be something involving-- the other option is something involving triangles. And at first, you might say, wait, is this angle and that angle vertical angles? But you have to be very careful. This is not the same line. This line is parallel to that line. This line, it's bending right over there, so we can't make any type of assumption like that. So the interesting thing-- and I'm not sure if this will lead in the right direction-- is to just make it clear that these two are part of parallel lines. So I could continue this line down like this. And then I can continue this line up like that. And then that starts to look a little bit more like we're used to when we're dealing with parallel lines. could even say line BC, if we were to continue it on. If we were to continue it on and on, even pass D, then this is clearly a transversal of those two parallel lines. This is clearly a transversal. And so if this angle right over here is 4x, it has a corresponding angle. Half of the-- or maybe most of the work on all of these is to try to see the parallel lines and see the transversal and see the things that might be useful for you. So that right there is the transversal. These are the parallel lines. That's one parallel line. That is the other parallel line. You can almost try to zone out all of the other stuff in the diagram. And so if this angle right over here is 4x, it has a corresponding angle where the transversal intersects the other parallel line. This right here is its corresponding angle. So let me draw it in that same yellow. This right over here is a corresponding angle. So this will also be 4x." + }, + { + "Q": "\nAt 5:04, instead of finding the corresponding angle for 4x, couldn't we find the corresponding angle for 2x and solve? I know you get the same answer, but I'm just clarifying if you can't.", + "A": "there is more than one way to solve most problems. You can do it both ways in this problem.", + "video_name": "0gzSreH8nUI", + "timestamps": [ + 304 + ], + "3min_transcript": "And then if we know x, we can figure out what the actual measures of these angles are, assuming that we can figure out x. And the other thing that they tell us is that this line over here is parallel to this line over here. And it was very craftily drawn. Because it's parallel, but one stops here, and then one starts up there. So the first thing I want to do-- if they're telling us that these two lines are parallel, there's probably going to be something involving transversals or something. It might be something involving-- the other option is something involving triangles. And at first, you might say, wait, is this angle and that angle vertical angles? But you have to be very careful. This is not the same line. This line is parallel to that line. This line, it's bending right over there, so we can't make any type of assumption like that. So the interesting thing-- and I'm not sure if this will lead in the right direction-- is to just make it clear that these two are part of parallel lines. So I could continue this line down like this. And then I can continue this line up like that. And then that starts to look a little bit more like we're used to when we're dealing with parallel lines. could even say line BC, if we were to continue it on. If we were to continue it on and on, even pass D, then this is clearly a transversal of those two parallel lines. This is clearly a transversal. And so if this angle right over here is 4x, it has a corresponding angle. Half of the-- or maybe most of the work on all of these is to try to see the parallel lines and see the transversal and see the things that might be useful for you. So that right there is the transversal. These are the parallel lines. That's one parallel line. That is the other parallel line. You can almost try to zone out all of the other stuff in the diagram. And so if this angle right over here is 4x, it has a corresponding angle where the transversal intersects the other parallel line. This right here is its corresponding angle. So let me draw it in that same yellow. This right over here is a corresponding angle. So this will also be 4x. this angle that has measure 4x and this angle that measures 2x-- we see that they're supplementary. They're adjacent to each other. Their outer sides form a straight angle. So they're supplementary, which means that their measures add up to 180 degrees. They kind of form-- they go all the way around like that if you add the two adjacent angles together. So we know that 4x plus 2x needs to be equal to 180 degrees, or we get 6x is equal to 180 degrees. Divide both sides by 6. You get x is equal to 30, or x is equal to-- well, I shouldn't say-- well, x could be 30. And then this angle right over here is 2 times x. So it's going to be 60 degrees. So this angle right over here is going to be 60 degrees. And this angle right over here is 4 times x. So it is 120 degrees, and we're done." + }, + { + "Q": "Also, can you get a definite answer for part 2:30 where the area is going to be x?\n", + "A": "Yes and No, it depends on what you mean. No if you do not know anything like in thew video. What Sal is showing is like a format. (you can fill out the sides and the inside) Yes if you know the side L*W. What he is trying to show you is that for a square each side, if it is squared will get you the area. Hope this helped!", + "video_name": "87_qIofPwhg", + "timestamps": [ + 150 + ], + "3min_transcript": "- [Voiceover] We already know a little bit about square roots. For example, if I were to tell you that seven squared is equal to 49, that's equivalent to saying that seven is equal to the square root of 49. The square root essentially unwinds taking the square of something. In fact, we could write it like this. We could write the square root of 49, so this is whatever number times itself is equal to 49. If I multiply that number times itself, if I square it, well I'm going to get 49. And that's going to be true for any number, not just 49. If I write the square root of X and if I were to square it, that's going to be equal to X and that's going to be true for any X for which we can evaluate the square root, evaluate the principle root. Now typically and as you advance in math you're going to see that this will change, but typically you say, okay if I'm going to take X has to be non-negative. This is going to change once we start thinking about imaginary and complex numbers, but typically for the principle square root, we assume that whatever's under the radical, whatever's under here, is going to be non-negative because it's hard to square a number at least the numbers that we know about, it's hard to square them and get a negative number. So for this thing to be defined, for it to make sense, it's typical to say that, okay we need to put a non-negative number in here. But anyway, the focus of this video is not on the square root, it's really just to review things so we can start thinking about the cube root. And as you can imagine, where does the whole notion of taking a square of something or a square root come from? Well it comes from the notion of finding the area of a square. If I have a square like this and if this side is seven, well if it's a square, all the sides are going to be seven. it would be seven times seven or seven squared. That would be the area of this. Or if I were to say, well what is if I have a square, if I have, and that doesn't look like a perfect square, but you get the idea, all the sides are the same length. If I have a square with area X. If the area here is X, what are the lengths of the sides going to be? Well it's going to be square root of X. All of the sides are going to be the square root of X, so it's going to be the square root of X by the square root of X and this side is going to be the square root of X as well and that's going to be the square root of X as well. So that's where the term square root comes from, where the square comes from. Now what do you think cube root? Well same idea. If I have a cube. If I have a cube. Let me do my best attempt at drawing a cube really fast. If I have a cube and a cube, all of it's dimensions have the same length so this is a two, by two, by two cube," + }, + { + "Q": "At 1:14, he mentions imaginary and complex numbers.\n\nWhat does he mean by \"imaginary\" numbers?\n", + "A": "Imaginary numbers are complex numbers in the form a + bi, in which a = 0. So, you d simply have i and its coefficient.", + "video_name": "87_qIofPwhg", + "timestamps": [ + 74 + ], + "3min_transcript": "- [Voiceover] We already know a little bit about square roots. For example, if I were to tell you that seven squared is equal to 49, that's equivalent to saying that seven is equal to the square root of 49. The square root essentially unwinds taking the square of something. In fact, we could write it like this. We could write the square root of 49, so this is whatever number times itself is equal to 49. If I multiply that number times itself, if I square it, well I'm going to get 49. And that's going to be true for any number, not just 49. If I write the square root of X and if I were to square it, that's going to be equal to X and that's going to be true for any X for which we can evaluate the square root, evaluate the principle root. Now typically and as you advance in math you're going to see that this will change, but typically you say, okay if I'm going to take X has to be non-negative. This is going to change once we start thinking about imaginary and complex numbers, but typically for the principle square root, we assume that whatever's under the radical, whatever's under here, is going to be non-negative because it's hard to square a number at least the numbers that we know about, it's hard to square them and get a negative number. So for this thing to be defined, for it to make sense, it's typical to say that, okay we need to put a non-negative number in here. But anyway, the focus of this video is not on the square root, it's really just to review things so we can start thinking about the cube root. And as you can imagine, where does the whole notion of taking a square of something or a square root come from? Well it comes from the notion of finding the area of a square. If I have a square like this and if this side is seven, well if it's a square, all the sides are going to be seven. it would be seven times seven or seven squared. That would be the area of this. Or if I were to say, well what is if I have a square, if I have, and that doesn't look like a perfect square, but you get the idea, all the sides are the same length. If I have a square with area X. If the area here is X, what are the lengths of the sides going to be? Well it's going to be square root of X. All of the sides are going to be the square root of X, so it's going to be the square root of X by the square root of X and this side is going to be the square root of X as well and that's going to be the square root of X as well. So that's where the term square root comes from, where the square comes from. Now what do you think cube root? Well same idea. If I have a cube. If I have a cube. Let me do my best attempt at drawing a cube really fast. If I have a cube and a cube, all of it's dimensions have the same length so this is a two, by two, by two cube," + }, + { + "Q": "at 5:08 where does the tan function come from?\n", + "A": "tan theta = opp/adj opp = sin theta adj = cos theta A simple proof: I dont know how to do theta symbol, so just pretend the no. 0 is theta show that: tan0 = sin0/cos0 RHS = opp/hyp divided by adj/hyp = opp/hyp times hyp/adj (when dividing fractions, invert and multiply) = opp/adj = tan0 = LHS", + "video_name": "8RasCV_Lggg", + "timestamps": [ + 308 + ], + "3min_transcript": "Well, this is forming an angle of theta with a positive real axis and so the horizontal coordinate over here by definition is going to be cosine of theta. Cosine of theta. That's the unit circle definition of cosine of theta, and the vertical coordinate is going to be sine of theta. Sine of theta. And so what would the horizontal and vertical coordinates of this point be? We obviously know they're negative three and two, but what would they be in terms of cosine theta and sine theta? Well look, this point right over here is a radius of one away from the origin. So this distance right over here is one, but now we are r away from the origin. We're r times as far. So if we're r times as far in that direction, then we're going to be r times as far in the vertical direction and r times We're going to scale everything by r. So the horizontal coordinate of this point right here instead of being cosine of theta is going to be r times cosine of theta. So this point right over here, which we know is negative three, is going to be equal to r cosine of theta, and by the same logic, this point over here, the vertical coordinate, we're going to scale up sine theta by r, we're r times as far. So this point right over here is going to be r sine theta and we already know that that's equal to two. r sine theta. So given that, can we now figure out what r and theta are? So let's first think about figuring out what theta is. So to do that, let's think about some of our trig functions. So one trig function that involves sine theta and cosine theta is tangent theta. tangent of our angle, tangent of theta, is equal to sine of theta over cosine of theta. We could also multiply the numerator and denominator here by r. That won't change the value. So that's the same thing as r sine theta over r cosine theta and we know r sine theta is going to be equal to two and we know that r cosine theta is negative three. So this whole thing is going to be negative 2/3. Another way of thinking about it is the tangent of theta is going to be the same thing as the slope of this line right over here. And what is the slope of that line? Well if you start at z and you want to go to the origin you're going to go positive three in the x direction" + }, + { + "Q": "sal why are you dividing when you could be subtracting at 00:55\n", + "A": "Because you wouldn t get the correct answer. Whenever dealing with decimals and percents, in situations like Sal had in 00:55, divide.", + "video_name": "gKywkLHV6Ko", + "timestamps": [ + 55 + ], + "3min_transcript": "A zoo has 15 emperor penguins who make up 30% of the total number of penguins at the zoo. How many penguins live at the zoo? So let's let x equal the total number of penguins who live at the zoo. So they're telling us that 30% of that are the emperor penguins, which are 15. So they're saying if we take 30%, so 30% of x is equal to 15. Or another way of saying this is we could say, Instead of writing 30%, we could write that as a decimal, as 0.30 of x. So it's 0.30 times x is equal to 15. And now to solve this, we just have to divide both sides by 0.30. So let's do that. 0.30, 0.30, we get x is equal to 15 divided by 0.30. this one. So if we take 0-- it's 15 divided by 0.30. So just as a refresher of dividing decimals, we could essentially multiply both of these by 100. And when you multiply both of them by 100, you can't just multiply one of them. Then you would get a different answer. But if you multiply both of them by 100, you move this decimal point over two spots, so it becomes a 30. And you move this decimal over two spots, so it becomes 1,500. So it really boils down to how many times-- the same number of times that 30 goes into 1,500 is the number of times that 0.30 will go into 15. So let's think about this. Let me rewrite it here, just so it's neat. 30 goes into 1,500, all we did is we moved both decimals over to the right twice. Or you could say we just multiplied here by 100, which wouldn't change the value of the fraction. So 30 goes into-- let's see, it doesn't go into 1. It does go into 150. It goes into 150 five times. 5 times 30 is 150. Subtract, you get 0. And then 30 goes into 0 zero times. So this 30 goes into 1,500 fifty times. So this right over here is equal to 50. And you can verify it. Multiply 0.3 times 50, and you will get 15. Now, there is another way of doing it. You could say the fraction of emperor penguins over the total number of penguins at the zoo. So you could say 15. You could say 15 over the total number of penguins at the zoo is equal to 30%. Percent literally means 30 hundreds. So it's equal to 30 for every 100." + }, + { + "Q": "Hi Sal, at 4:44 it says a mean of 7.5 is impossible for a sample of size n=2. But, wouldn't a mean of 7.5 be achieved by a random sample with values 6 and 9, which was a mean of (6+9)/2 = 7.5? Thanks, Chris Broski\n", + "A": "Hi I just saw the answer to my questions below. I understand now, yes, it is possible to get 7.5 as the mean for a sample of size n =2. Thanks, Chris Broski", + "video_name": "NYd6wzYkQIM", + "timestamps": [ + 284 + ], + "3min_transcript": "When n is pretty small, it doesn't approach a normal distribution that well. So when n is small-- let's take the extreme case. What happens when n is equal to 1? And that literally just means I take one instance of this random variable and average it. Well, it's just going to be that thing. So if I just take a bunch of trials from this thing and plot it over time, what's it look like? Well, it's definitely not going to look like a normal distribution. You're going to have a couple of 1's. You're going to have a couple of 2's. You're going to have more 3's like that. You're going to have no 4's. You're going to have a bunch of 5's. You're going to have some 6's that look like that. And you're going to have a bunch of 9's. So there, your sampling distribution of the sample mean for an n of 1 is going to look-- I don't care how many trials you do, it's not going to look like a normal distribution. So the central limit theorem, although I said you do a bunch of trials, it'll look like a normal distribution, definitely doesn't work for n equals 1. As n gets larger, though, it starts to make sense. Let's see, if we've got n equals 2-- and I'm all just doing this in my head. I don't know what the actual distributions would look like. for it to become an exact normal distribution. But then you could get more instances, you could get more-- you might get things from all of the above. But in each of your baskets that you're averaging, you're only going to get two numbers. For example, you're never going to get a 7 and 1/2 in your sampling distribution of the sample mean for n is equal to 2, because it's impossible to get a 7, and it's impossible to get an 8. So you're never going to get 7 and 1/2 as-- so maybe when you plot it, maybe it looks like this. But there'll be a gap at 7 and 1/2 because that's impossible. And maybe it looks something like that. So it still won't be a normal distribution when n is equal to 2. So there's a couple of interesting things here. So one thing-- and I didn't mention this the first time, just because I really wanted you to get the gut sense of what the central limit theorem is. The central limit theorem says as n approaches, really as it approaches infinity, then is when you get the real normal distribution. you don't have to get that much beyond n equals 2. If you get to n equals 10 or n equals 15, you're getting very close to a normal distribution. So this converges to a normal distribution very quickly. Now, the other thing is you obviously want many, many trials. So this is your sample size. That is your sample size. That's the size of each of your baskets. In the very first video I did on this, I took a sample size of 4. In the simulation I did in the last video, we did sample sizes of 4 and 10 and whatever else. This is a sample size of 1. So that's our sample size. So as that approaches infinity, your actual sampling distribution of the sample mean will approach a normal distribution. Now, in order to actually see that normal distribution, and actually to prove it to yourself, you would have to do this many, many-- remember the normal distribution happens-- this is kind of the population, or this is the random variable. That tells you all of the possibilities. In real life, we seldom know all of the possibilities." + }, + { + "Q": "\nat around 4:25, Sal says we are never going to get 7.5 when n=2. But if 6 and 9 are randomly selected, then 7.5 would be the average, so I'm not quite understanding his reasoning here?", + "A": "You re totally correct, Mike G! Your logic is solid; Sal made a mistake.", + "video_name": "NYd6wzYkQIM", + "timestamps": [ + 265 + ], + "3min_transcript": "So these boxes are really small. So we just do a bunch of these trials. At some point, it might look a lot like a normal distribution. Obviously, there are some average values. It won't be a perfect normal distribution, because you can never get anything less than 0, or anything less than 1, really, as an average. You can't get 0 as an average. And you can't get anything more than 9. So it's not going to have infinitely long tails but, at least for the middle part of it, a normal distribution might be good approximation. In this video, what I want to think about is what happens as we change n. So in this case, n was 4. n is our sample size. Every time we do a trial, we took four and we took their average, and we plotted it. We could have had n equal 10. We could've taken 10 samples from this population, you could say, or from this random variable, averaged them, and then plotted them here. And in the last video, we ran the simulation. I'm going to go back to that simulation in a second. We saw a couple of things. And I'll show it to you in a little bit more depth When n is pretty small, it doesn't approach a normal distribution that well. So when n is small-- let's take the extreme case. What happens when n is equal to 1? And that literally just means I take one instance of this random variable and average it. Well, it's just going to be that thing. So if I just take a bunch of trials from this thing and plot it over time, what's it look like? Well, it's definitely not going to look like a normal distribution. You're going to have a couple of 1's. You're going to have a couple of 2's. You're going to have more 3's like that. You're going to have no 4's. You're going to have a bunch of 5's. You're going to have some 6's that look like that. And you're going to have a bunch of 9's. So there, your sampling distribution of the sample mean for an n of 1 is going to look-- I don't care how many trials you do, it's not going to look like a normal distribution. So the central limit theorem, although I said you do a bunch of trials, it'll look like a normal distribution, definitely doesn't work for n equals 1. As n gets larger, though, it starts to make sense. Let's see, if we've got n equals 2-- and I'm all just doing this in my head. I don't know what the actual distributions would look like. for it to become an exact normal distribution. But then you could get more instances, you could get more-- you might get things from all of the above. But in each of your baskets that you're averaging, you're only going to get two numbers. For example, you're never going to get a 7 and 1/2 in your sampling distribution of the sample mean for n is equal to 2, because it's impossible to get a 7, and it's impossible to get an 8. So you're never going to get 7 and 1/2 as-- so maybe when you plot it, maybe it looks like this. But there'll be a gap at 7 and 1/2 because that's impossible. And maybe it looks something like that. So it still won't be a normal distribution when n is equal to 2. So there's a couple of interesting things here. So one thing-- and I didn't mention this the first time, just because I really wanted you to get the gut sense of what the central limit theorem is. The central limit theorem says as n approaches, really as it approaches infinity, then is when you get the real normal distribution." + }, + { + "Q": "\nAt 1:46, shouldn't Sal have noticed he wrote \"Armaan\" when it should be Arman?", + "A": "This is a known problem with the video. A box pops up with the correction.", + "video_name": "W-5liMGKgHA", + "timestamps": [ + 106 + ], + "3min_transcript": "Let's say that Arman today is 18 years old. And let's say that Diya today is 2 years old. And what I am curious about in this video is how many years will it take-- and let me write this down-- how many years will it take for Arman to be three times as old as Diya? So that's the question right there, and I encourage you to try to take a shot at this yourself. So let's think about this a little bit. We're asking how many years will it take. That's what we don't know. That's what we're curious about. How many years will it take for Arman So let's set some variable-- let's say, y for years. Let's say y is equal to years it will take. So given that, can we now set up an equation, given this information, to figure out how many years it will take for Arman to be three times as old as Diya? Well, let's think about how old Arman will be in y years. How old will he be? Let me write here. In y years, Arman is going to be how old? Arman is going to be-- well, he's 18 right now-- and in y years, he's going to be y years older. So in y years, Arman is going to be 18 plus y. And what about Diya? How old will she be in y years? she will just be 2 plus y. So what we're curious about, now that we know this, is how many years will it take for this quantity, for this expression, to be three times this quantity? So we're really curious. We want to solve for y such that 18 plus y is going to be equal to 3 times 2 plus y. Notice, this is Arman in y years. This is Diya in y years. And we're saying that what Arman's going to be in y years is three times what Diya is going to be in y years. So we've set up our equation. Now we can just solve it. So let's take this step by step. So the left hand side-- and maybe I'll do this in a new color, just so I don't have to keep switching-- so on the left hand side," + }, + { + "Q": "The third problem, The \"Bizarre looking shape\" at 5:09, would it be possible to find the area of that dodecagon?\n", + "A": "Yes. Since the dodecagon is made up of all right angles, as Sal said, finding the area would be easy. All you need to do is cut it up into rectangles, find the rectangles areas, and add them up.", + "video_name": "vWXMDIazHjA", + "timestamps": [ + 309 + ], + "3min_transcript": "at 90 degrees and we could call this point E. And what is interesting here is we can split this up into something we recognize a rectangle and a right triangle. But you might say how do, how do we figure out what these you know we have this side and that side, so we can figure out the area of this rectangle pretty straight forwardly. But how would we, how would we figure out the area of this triangle? Well if this side is 6 then that means that this that EC is also going to be 6. If AB is 6, notice we have a rectangle right over her, opposite side of a rectangle are equal. So if AB equals 6, implies that EC is equal to 6, EC is equal to 6, so EC is equal to 6 and if EC is equal to 6 then that tells us that DE is going to be 3. DE is going to be 3, this distance right over here is going to be 3. 9 was the length of this entire, of the entire base of this figure right over here. 9 was this entire distance so 9 minus 6 gives us 3, and now we have all the information that we need to figure out the area. The area of this part right over here of this rectangle is just going to be 6 times 7, so is going to be equal to 42 plus the area of this triangle right over here. Plus the area of this triangle right over here, and that is one half base times height one half. The base over her is 3, one half times 3 and the height over here is once again going to be 7 this is a rectangle, opposite sides are equal, so if this is 7, this is also going to be 7 one half times 3 times 7, so it is going be 42, lets see. 3 times 7 is 21, 21 divided by 2 is 10.5, 10.5 so this is going to be equal to 52.5 Lets do one more. So here I have a bizarre looking, a bizarre looking shape, and we need to figure out its perimeter. And it it first seems very daunting because they have only given us this side and this side and they have only given us this side right over here. And one thing that we are allowed to assume in this and you don't always have to make you can't always make that assumption and I just didn't draw it here I had time because it would had really crowded out this this diagram. Is it all of the angles in this diagrams are right angles,so i could have drawn a right angle here a right angle here, a right angle there, right angle there, but as you can see it kind of makes things a little bit, it makes things a little bit messy. But how do we figure out the perimeter if we don't know these little distances, if we don't know these little distances here. And the secret here is to kind of shift the sides because all we want to care about is the sum of the sides of the sides. So what I will do is a little exercise in shifting the sides. So this side over here I am going to shift" + }, + { + "Q": "\nI'm really confused about why the top equation was multiplied by -2 at 17:20. Surely it's not an arbitrary number, right?", + "A": "Sal was setting up the elimination step. The next thing he does is add the two equations and the C_1 variable is eliminated allowing us to solve for C_2. Multiplying by -2 was the easiest way to get the C_1 term to cancel. Another question is why he chooses to use elimination. The first equation is already solved for C_1 so it would be very easy to use substitution. He may have chosen elimination because that is how we work with matrices.", + "video_name": "Qm_OS-8COwU", + "timestamps": [ + 1040 + ], + "3min_transcript": "It was 1, 2, and b was 0, 3. Let me remember that. So my vector a is 1, 2, and my vector b was 0, 3. Now my claim was that I can represent any point. Let's say I want to represent some arbitrary point x in R2, so its coordinates are x1 and x2. I need to be able to prove to you that I can get to any x1 and any x2 with some combination of these guys. So let's say that my combination, I say c1 times a plus c2 times b has to be equal to my vector x. Let me show you that I can always find a c1 or c2 given that you give me some x's. So let's just write this right here with the actual vectors So we have c1 times this vector plus c2 times the b vector 0, 3 should be able to be equal to my x vector, should be able to be equal to my x1 and x2, where these are just arbitrary. So let's see if I can set that to be true. So if this is true, then the following must be true. c1 times 1 plus 0 times c2 must be equal to x1. We just get that from our definition of multiplying vectors times scalars and adding vectors. And then we also know that 2 times c2-- sorry. c1 times 2 plus c2 times 3, 3c2, should be equal to x2. Now, if I can show you that I can always find c1's and c2's any point in R2 using just these two vectors. So let me see if I can do that. So this is just a system of two unknowns. This is just 0. We can ignore it. So let's multiply this equation up here by minus 2 and put it here. So we get minus 2, c1-- I'm just multiplying this times minus 2. We get a 0 here, plus 0 is equal to minus 2x1. And then you add these two. You get 3c2, right? These cancel out. You get 3-- let me write it in a different color. You get 3c2 is equal to x2 minus 2x1. Or divide both sides by 3, you get c2 is equal to 1/3 x2 minus x1. Now we'd have to go substitute back in for c1." + }, + { + "Q": "\nAround 13:50 when Sal gives a generalized mathematical definition of \"span\" he defines \"i\" as having to be greater than one and less than \"n\". Is this because \"i\" is indicating the instances of the variable \"c\" or is there something in the definition I'm missing?", + "A": "Its because we are looking at the Span of the Vectors v1 to vn, so for every vi there is a ci. Thats why i has to be between 1 and n.", + "video_name": "Qm_OS-8COwU", + "timestamps": [ + 830 + ], + "3min_transcript": "And in our notation, i, the unit vector i that you learned in physics class, would be the vector 1, 0. So this is i, that's the vector i, and then the vector j is the unit vector 0, 1. This is what you learned in physics class. Let me do it in a different color. This is j. j is that. And you learned that they're orthogonal, and we're going to talk a lot more about what orthogonality means, but in our traditional sense that we learned in high school, it means that they're 90 degrees. But you can clearly represent any angle, or any vector, in R2, by these two vectors. And the fact that they're orthogonal makes them extra nice, and that's why these form-- and I'm going to throw out a word here that I haven't defined yet. These form the basis. These form a basis for R2. In fact, you can represent anything in R2 by these two I'm not going to even define what basis is. But let me just write the formal math-y definition of span, just so you're satisfied. So if I were to write the span of a set of vectors, v1, v2, all the way to vn, that just means the set of all of the vectors, where I have c1 times v1 plus c2 times v2 all the way to cn-- let me scroll over-- all the way to cn vn. So this is a set of vectors because I can pick my ci's to be any member of the real numbers, and that's true for i-- so I should write for i to be anywhere between 1 and n. All I'm saying is that look, I can multiply each of these vectors by any value, any arbitrary value, real value, And now the set of all of the combinations, scaled-up combinations I can get, that's the span of these vectors. You can kind of view it as the space of all of the vectors that can be represented by a combination of these vectors right there. And so the word span, I think it does have an intuitive sense. I mean, if I say that, you know, in my first example, I showed you those two vectors span, or a and b spans R2. I wrote it right here. That tells me that any vector in R2 can be represented by a linear combination of a and b. And actually, just in case that visual kind of pseudo-proof doesn't do you justice, let me prove it to you algebraically. I'm telling you that I can take-- let's say I want to represent, you know, I have some-- let me rewrite my a's and b's again." + }, + { + "Q": "At 1:10 why does he ask that question?\n", + "A": "he was adding the number thats it", + "video_name": "Oe1PKI_6-38", + "timestamps": [ + 70 + ], + "3min_transcript": "Add, and they give us 5x squared plus 8x minus 3 plus 2x squared minus 7x plus 13x. So we can really view this as just adding two polynomials. And actually, the second polynomial here can be simplified right off the bat. We have two like terms here. It's not the 2x squared. There's no other 2x squared term in this polynomial. But you have a negative 7x term. And then you have a plus 13x. So we could actually add these two terms. What is the negative 7 of something plus 13 of that something? Or another way to view it, what's 13x minus 7x? 13 of something minus 7 of that something? Well, that's just going to be 6 of that something. So that is just going to be 6x. And then you have your 2x squared. And you have your 2x squared right over here. Let me write it. So 2x squared. So this polynomial right over here simplifies to 2x squared plus 6x. And then we want to add it to this polynomial right And what we can do is, what I like to do is just rewrite this polynomial under this polynomial. or the like terms, the terms that have the same variable raised to the same power. So we have x raised to the second power. So let's put the 5x squared over here. We have an x raised to the first power. We have an x raised to the first power. So let's put the 8x over here. And then we don't have a constant term in this yellow polynomial. We don't have anything raised to the zeroth power, any constant terms. We do here. So let's just put a minus 3. And then we could add these two things up. We have 2x squared plus 5x squared. That is 2 of something plus 5 of something is 7 of that something. So it's going to be 7x squared. And then to that, if I 6 of something and I add 8 of that something to it, well, I'm going to have 14 of that something. If I have 6x's and I give you 8 more x's, you're going to 14x's. And then we have nothing here. And so if we just add nothing to negative 3, you're just going to get a negative 3. we get 7x squared plus 14x minus 3." + }, + { + "Q": "at time 9:10 when he puts the boundary conditions, he replaces each of the y variables with -1. i do not get this point as boundary condition means the whole function should evaluate to -1 as x=0. some one please help me with that.\n", + "A": "thanks a lot i got it now....!!!....", + "video_name": "C5-lz0hcqsE", + "timestamps": [ + 550 + ], + "3min_transcript": "can use this initial condition, when x is 0, y is 1, to figure out the constant. So let's first separate this equation. So let's multiply both sides by 2 times y minus 1. And you get 2 times y minus 1 times dy dx is equal to 3x squared plus 4x plus 2. Multiply both sides times dx. This is really just an exercise in algebra. And I can multiply this one out, too, you get 2y minus 2, that's just this, dy. I multiplied both sides times dx, so that equals 3x squared plus 4x plus 2 dx. I have separated the equations. I've separated the independent from the dependent variable, and their relative differentials, and so now I can integrate. What's the antiderivative of this expression with respect to y? Well, let's just see. It's y squared minus 2y. I won't write the plus c, I'll just do it on the right hand side. That is equal to 3x squared. Well, the antiderivative is x to the third, plus 2x squared, plus 2x plus c. And that c kind of takes care of the constant for both sides of the equation, and hopefully you understand why from the last example. But we can solve for c using the initial condition y of 0 is equal to negative 1. So let's see. When x is 0, y is negative 1. So let's put y as negative 1, so we get negative 1 squared minus 2 times negative 1, that's the value of y, is equal to when x is equal to 0. So when x is equal to 0, that's 0 to the third plus 2 times 0 squared plus 2 times 0 plus c. All of these, this is all 0. This is, let's see, negative 1 squared, that's 1. Minus 2 times minus 1, that's plus 2, is equal to c. And we get c is equal to 3. So, the implicit exact solution, the solution of our differential equation-- remember now, it's not a class, because they gave us an initial condition-- is y squared minus 2y is equal to x to the third plus 2x squared plus 2x plus 3. We figured out that's what c was. And actually, if you want, you could write this in an explicit form by completing the square. This is just algebra this point. This is an implicit form. If you wanted to make it explicit, you could add 1 to I'm just completing the square here. So y squared minus 2y plus 1." + }, + { + "Q": "\n1:10 I don't understand Sal's explanation of the vertical tangent as having infinite dy. I know that a vertical line has an undefined slope because there's no change in x so dy is being divided by zero, but what does infinity have to do with that?\nAlso, how does a small change in x give an undefined slope when it's still a defined number that returns another real number when divided by a defined change in y?", + "A": "But we don t have a defined change in y for a vertical line. The change could be anything. If we have a vertical line, then y is not a function of x. Would you be happier if we said dy was undefined for a vertical line, rather than infinite ? To be honest I m not sure if you think a vertical line is differentiable with respect to x or something else is troubling you.", + "video_name": "pwh1dK3vTkM", + "timestamps": [ + 70 + ], + "3min_transcript": "- [Voiceover] The graph of function f is given below. It has a vertical tangent at the point three comma zero. Three comma zero has a vertical tangent, let me draw that. It has a vertical tangent right over there, and a horizontal tangent at the point zero comma negative three. Zero comma negative three, so it has a horizontal tangent right over there, and also has a horizontal tangent at six comma three. Six comma three, let me draw the horizontal tangent, just like that. Select all the x-values for which f is not differentiable. Select all that apply. F prime, f prime, I'll write it in short hand. We say no f prime under it's going to happen under three conditions. The first condition you could say well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative? Well, remember, our derivative is with respect to x, but when you have a vertical tangent, you change your x a very small amount, you have an infinite change in y, either in the positive or the negative direction. That's one situation where you have no derivative. They tell us where we have a vertical tangent in here, where x is equal to three. We have no ... F is not differentiable at x equals three because of the vertical tangent. You might say what about horizontal tangents? No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero. F prime of six is equal to zero. F prime of zero is equal to zero. What are other scenarios? Well another scenario where you're not gonna have a defined derivative is where the graph is not continuous. Not continuous. We see right over here at x equals negative three, our graph is not continuous. Those are thee only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right. These there I guess would be interesting cases. They haven't given us those choices here. We already said, at x equals 0, the derivative is zero. It's defined. At x equals six, the derivative is zero. We have a flat tangent. Once again it's defined there as well. Let's do another one of these. Actually, I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn. This isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that, or like, well no, that doesn't look too sharp, or like this. The reason why where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that." + }, + { + "Q": "\nat 0:53 it is said that a point having a vertical tangent is not differentiable because for a change in x there is an infinite change in y at that particular point . how is it so?", + "A": "Sal is a tad off with his wording there. The change in y is finite but it doesn t get smaller as x gets smaller so you have a limit of dy/dx where dy decreases much slower than dx as dx --> 0 so the limit is unbounded.", + "video_name": "pwh1dK3vTkM", + "timestamps": [ + 53 + ], + "3min_transcript": "- [Voiceover] The graph of function f is given below. It has a vertical tangent at the point three comma zero. Three comma zero has a vertical tangent, let me draw that. It has a vertical tangent right over there, and a horizontal tangent at the point zero comma negative three. Zero comma negative three, so it has a horizontal tangent right over there, and also has a horizontal tangent at six comma three. Six comma three, let me draw the horizontal tangent, just like that. Select all the x-values for which f is not differentiable. Select all that apply. F prime, f prime, I'll write it in short hand. We say no f prime under it's going to happen under three conditions. The first condition you could say well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative? Well, remember, our derivative is with respect to x, but when you have a vertical tangent, you change your x a very small amount, you have an infinite change in y, either in the positive or the negative direction. That's one situation where you have no derivative. They tell us where we have a vertical tangent in here, where x is equal to three. We have no ... F is not differentiable at x equals three because of the vertical tangent. You might say what about horizontal tangents? No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero. F prime of six is equal to zero. F prime of zero is equal to zero. What are other scenarios? Well another scenario where you're not gonna have a defined derivative is where the graph is not continuous. Not continuous. We see right over here at x equals negative three, our graph is not continuous. Those are thee only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right. These there I guess would be interesting cases. They haven't given us those choices here. We already said, at x equals 0, the derivative is zero. It's defined. At x equals six, the derivative is zero. We have a flat tangent. Once again it's defined there as well. Let's do another one of these. Actually, I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn. This isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that, or like, well no, that doesn't look too sharp, or like this. The reason why where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that." + }, + { + "Q": "At 2:52 how can we be sure that the three angles equal 180 degrees? It may appear to be a straight line, but without any notation ensuring that it is, how can we be sure?\n", + "A": "Hi J, Right at the very beginning of the video, at 0:01, Sal uses the words in this larger triangle here as he outlines triangle ABE. If this was a straight word question however, the question would start out with something like, Given triangle ABE, prove that ... Hope that helps!", + "video_name": "aDCXPdzyS0s", + "timestamps": [ + 172 + ], + "3min_transcript": "corresponds to the E vertex in ECD. So all-- everything that I've done in magenta, all of these angles are congruent, and then we also know that the C angle. So in BCA-- sorry, BCD, this angle right over here, is congruent to the C angle in BCA. BCA, the C angle is right over here, or C is the vertex for that angle in BCA. And that is also the C angle, I guess we could call it, in ECD. But in ECD, we're talking about this angle right over here. So these three angles are going to be congruent. And I think you could already guess a way to come up with the values of those three angles. But let's keep looking at everything else that they're telling us. Finally, we have vertex D over here. So angle-- so this is the last one in where we listed-- so corresponds to the A vertex angle in BCA. So BCA, that's going to correspond to this angle right over here. It's really the only one that we haven't labeled yet. And that corresponds to this angle, this vertex right over here, that angle right over there. And just to make it consistent, this C should also be circled in yellow. And so we have all these congruences, and now we can come up with some interesting things about them. First of all, here, angle BCA, angle BCD, and angle DCE, they're all congruent, and when you add them up together, you get to 180 degrees. If you put them all adjacent, as they all are right here, they end up with a straight angle, if you look at their outer sides. So you have, if these are each x, you have three of them added together have to be 180 degrees, which tells us that each of these have to be 60 degrees. That's the only way you have three of the same thing adding up to 180 degrees. Fair enough. Well, we have these two characters up here. They are both equal and they add up to 180 degrees. They are supplementary, the only way you can have two equal things that add up to 180 is if they're both 90 degrees. So these two characters are both 90 degrees. Or we could say this is a right angle, that's a right angle. And this is congruent to both of those, so that is also 90 degrees, and then we're left with these magenta parts of the angle. And here, we could just say, well, 90 plus 60 plus something is going to add up to 180. 90 plus 60 is 150. So this has to be 30 degrees to add up to 180. And if that's 30 degrees, then this is 30 degrees. And then this thing right over here is 30 degrees. And then the last thing-- we've actually done what we said we would do, we found out all of the angles. We could also think about these outer angles. So this-- or not the outer angles, or these combined angles. So angle say AC-- or say, angle ABE, so this whole angle we see is 60 degrees." + }, + { + "Q": "At \"3:12\" I do not exactly understand how Sal pinpoints point C in the designated location? Please explain how he knew where to find all of the points of the hexagon given the special set of values for y, thank you.\n", + "A": "because the hexagon had to be convex and with all sides equal.", + "video_name": "Ec-BKdC8vOo", + "timestamps": [ + 192 + ], + "3min_transcript": "So let me draw my x-axis. That is my x-axis. Right over there. And then my y-axis. My y-axis would look like that. Y-axis. We know that the vertex A sits at the point 0,0. That is the vertex A. Now, we know that all of the vertices have y-coordinates that are either 0, 2, 4, 6, 8, 10. And they are distinct members of the set. Which means no 2 of the vertices share the same y-coordinate. So they're not going to be on the same horizontal line. So let me draw these horizontal lines-- the x-axis is 0. Then you have y is equal to 2. Then you have 4, then you have 6, and then you have 8, and then you have 10 up here. Now, B we already know. So first of all, we've already used up to the 0 for A. A Is already using up the 0. is 2. So we use that as well. Let me see if I can draw B over here-- It sits on this horizontal someplace. And the hexagon has side length S, where we don't know what that length is, but they're all the same. So let's just call this S it's going to help me think about it. Now that I know that this is equilateral equaliateral hexagon, all the sides are going to be the same length. We're going to go out here the coordinate B comma 2. We don't know what B is, but that is our vertex B. Now, F is the other vertex that is connected to A. F cannot sit on this horizontal-- cannot sit on y is equal to 2. It can't sit on y is equal to 6, because then this distance would be super far. Clearly much further than this distance over here. Or, actually, you could have that. But then you would you wouldn't be able to draw really a convex hexagon. So it's going to be S away. Maybe it will be something like that. So let me draw it-- so that is the next vertex. That is vertex F. Because we're going A, B,C,D, E, F, and then back to A. Fair enough. Now what about vertex C? Well, vertex C can't be on the 4 horizontal. So it's going to have to be on the 6 horizontal. So vertex C is going to have to be someplace like that. That's vertex C. And once again, that length is S this length is S. Now what about vertex E? Can't be on the 6 horizontal, already taken up by vertex C. So the 4 the 6 are taken up. So it has to be at the 8 horizontal. And so this is length S. And we also know that we're going back to the origin now. So this is the vertex E right here." + }, + { + "Q": "At 3:07 why does he add 16?\n", + "A": "I ve just looked at this video and cannot find any 16 mentioned. Did you reference the correct video?", + "video_name": "PupNgv49_WY", + "timestamps": [ + 187 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 7:11 why do you all of a sudden flip 81 into 1/81", + "A": "Sal was substituting into A / C. A was 1 / 9 and C was 81. So, A / C = ( 1 / 9 ) / 81 which is the same as ( 1 / 9 ) * ( 1 / 81 ). In words, dividing by a number is the same as multiplying by its reciprocal.", + "video_name": "PupNgv49_WY", + "timestamps": [ + 431 + ], + "3min_transcript": "" + }, + { + "Q": "In 2:14, we have the same bases on one side of the equation but what if there was a value on the other side of the equation without the same base. For example, logbase3 of 5 +logbase3of 2= 8 How would you approach the problem?\n", + "A": "First you would combine the bases. Then you will exponentiate both sides of the equation. What this means is you make all of those numbers into a power. For example, x = y. When you exponentiate it, you will get something like 3^x = 3^y. You can use this to your advantage to eliminate logs.", + "video_name": "PupNgv49_WY", + "timestamps": [ + 134 + ], + "3min_transcript": "" + }, + { + "Q": "is any number in the sequence of numbers 1 2 4 8 16 32 64 128 256 512 etc... able to some how convert into a higher number in the sequence? this question came from 3:33 in the video, it might have been before, when you converted the number 8 in the sequence into 25, because i have noticed that many numbers are multiples of those higher numbers.\n", + "A": "I m not exactly sure what you are asking but the sequence you wrote down is the sequence you get by multiplying by 2 to get the next number. Another way to write that sequences is 2^0,2^1,2^2,2^3,2^4,2^5,2^6,.... I m not sure what you mean by converting 8 into 25.", + "video_name": "PupNgv49_WY", + "timestamps": [ + 213 + ], + "3min_transcript": "" + }, + { + "Q": "Why does Sal change colors, like at 4:37?\n", + "A": "It lets the viewer understand things more easily. He also does it so that it isn t boring and he color codes some things. Try to learn things the KHAN way!!", + "video_name": "PupNgv49_WY", + "timestamps": [ + 277 + ], + "3min_transcript": "" + }, + { + "Q": "at 1:58 why did u divide the number ab=-3 by a , why not multiplied it by a?\n", + "A": "Sal is solving the b . To move the a , we look at the relationship between the a and b in ab . It is multiplication. Then we always use the opposite operation to move the a . The opposite operation to multiplication is division. If you multiplied both sides by a , you would get a*ab = -3a or a^2b = -3a . The b is not by itself. Hope this helps.", + "video_name": "3_DxJwDTbyQ", + "timestamps": [ + 118 + ], + "3min_transcript": "I've written some example relationships between two variables-- in this case between m and n, between a and b, between x and y. And what I want to do in this video is see if we can identify whether the relationships are a direct relationship, whether they vary directly, or maybe they vary inversely, or maybe it is neither. So let's explore it a little bit. So over here, we have m/n is equal to 1/7. So let's see how we can manipulate this. If we multiply both sides by n. What are we going to get? And in general, you want to separate them so that the two variables are on different sides of the equation, so you can see is it going to meet, is it going to be the pattern-- let me write it this way. m is equal to k times n. This would be direct variation. Or is it going to be the pattern m is equal to k times 1/n? This is inverse variation. And you see in either one of these, they're on different sides of the equal sign. So let's take this first relationship right now. Let's multiply both sides by n. And you get m, because these cancel out, So this actually meets the direct variation pattern. It's some constant times n. m is equal to some constant times n. So this right over here is direct. They vary directly. This is direct variation. Let's see, ab is equal to negative 3. So if we want to separate them-- and we could do it with either variable, we could divide both sides. I don't know, let's divide both sides by a. We could have done it by b. If we divide both sides by a, we get b is equal to negative 3/a. Or we could also write this as b is equal to negative 3 times 1/a. And once again, this is this pattern right here. One variable is equal to a constant times 1 over the other variable. In this case, our constant is negative 3. So over here, they vary inversely. This is an inverse relationship. Let's try this one over here. I'll try to do it in that same color. xy is equal to 1/10. isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m." + }, + { + "Q": "at around 4:25 he says that you can manipulate it to make one variable seem more DEPENDENT then another. What does he mean? If anyone could help me that would be nice, thanks.\n", + "A": "m=n/7, m/n = 1/7, n = 7m, n/m = 7. The only thing I can come up with is that whichever variable has the number 7 on the same side with it might seem to be the independent variable, as in y = f(x) , y = 3x .", + "video_name": "3_DxJwDTbyQ", + "timestamps": [ + 265 + ], + "3min_transcript": "isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m. And remember, if I say that n varies inversely with m, that also means that m varies inversely with n. Those two things imply each other. Now let's try it with this expression over here. And this one's a little bit of a trickier one, because we've already separated the variables on both hands. And then we have this kind of-- if this was b is equal to 1/3 times a, then we would have direct variation, then b would vary directly with a. But in this case, we have 1/3 minus a. And you say, hey, maybe they're opposites, or whatever. And it actually turns out that this is neither. This is neither. And to make that point 100% clear, let's look at two of these examples. In direct variation, if you scale up one variable in one direction, you would scale up the other variable by the same amount. So if we have x going-- if x doubles from 1 to 2, So m and n. So when-- and the way I've written it here, although you could algebraically manipulate it so that one looks more dependent than the other. But in this situation where n is 1, m is 1/7. And when n is 7, m is going to be 1. So you have the situation that if n is scaled up by 7, then m is also scaled up by 7, or vice versus. So it's more of a relationship. I could have expressed n in terms of m, but when you scale one variable up by 7, you also have to scale up the other variable by 7. When you scale it up by some amount, you have to scale the other variable by the same amount. So this is direct variation. Let's take the inverse, or when two variables vary inversely, this situation right over here. Let's take a and b." + }, + { + "Q": "\ncan anyone hrlp me. at 2:15 he says 1/10/x = 1/10x. I can't see how he got there..", + "A": "(1/10)/x is the same as multiplying 1/10 by the reciprocal of x i.e. 1/x. So, (1/10)/x = 1/10 * 1/x = 1/10x. HOPE THAT HELPED YOU!", + "video_name": "3_DxJwDTbyQ", + "timestamps": [ + 135 + ], + "3min_transcript": "I've written some example relationships between two variables-- in this case between m and n, between a and b, between x and y. And what I want to do in this video is see if we can identify whether the relationships are a direct relationship, whether they vary directly, or maybe they vary inversely, or maybe it is neither. So let's explore it a little bit. So over here, we have m/n is equal to 1/7. So let's see how we can manipulate this. If we multiply both sides by n. What are we going to get? And in general, you want to separate them so that the two variables are on different sides of the equation, so you can see is it going to meet, is it going to be the pattern-- let me write it this way. m is equal to k times n. This would be direct variation. Or is it going to be the pattern m is equal to k times 1/n? This is inverse variation. And you see in either one of these, they're on different sides of the equal sign. So let's take this first relationship right now. Let's multiply both sides by n. And you get m, because these cancel out, So this actually meets the direct variation pattern. It's some constant times n. m is equal to some constant times n. So this right over here is direct. They vary directly. This is direct variation. Let's see, ab is equal to negative 3. So if we want to separate them-- and we could do it with either variable, we could divide both sides. I don't know, let's divide both sides by a. We could have done it by b. If we divide both sides by a, we get b is equal to negative 3/a. Or we could also write this as b is equal to negative 3 times 1/a. And once again, this is this pattern right here. One variable is equal to a constant times 1 over the other variable. In this case, our constant is negative 3. So over here, they vary inversely. This is an inverse relationship. Let's try this one over here. I'll try to do it in that same color. xy is equal to 1/10. isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m." + }, + { + "Q": "At 3:41 I didn't understand how it is neither\n", + "A": "Cause the funcion has a sign minus (-) and is not clear if a is negative or is a sustraction. If you see the direct/inverse variation has no addtions or sustractions, just multiplication and divisions. I hope being helpfully", + "video_name": "3_DxJwDTbyQ", + "timestamps": [ + 221 + ], + "3min_transcript": "isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m. And remember, if I say that n varies inversely with m, that also means that m varies inversely with n. Those two things imply each other. Now let's try it with this expression over here. And this one's a little bit of a trickier one, because we've already separated the variables on both hands. And then we have this kind of-- if this was b is equal to 1/3 times a, then we would have direct variation, then b would vary directly with a. But in this case, we have 1/3 minus a. And you say, hey, maybe they're opposites, or whatever. And it actually turns out that this is neither. This is neither. And to make that point 100% clear, let's look at two of these examples. In direct variation, if you scale up one variable in one direction, you would scale up the other variable by the same amount. So if we have x going-- if x doubles from 1 to 2, So m and n. So when-- and the way I've written it here, although you could algebraically manipulate it so that one looks more dependent than the other. But in this situation where n is 1, m is 1/7. And when n is 7, m is going to be 1. So you have the situation that if n is scaled up by 7, then m is also scaled up by 7, or vice versus. So it's more of a relationship. I could have expressed n in terms of m, but when you scale one variable up by 7, you also have to scale up the other variable by 7. When you scale it up by some amount, you have to scale the other variable by the same amount. So this is direct variation. Let's take the inverse, or when two variables vary inversely, this situation right over here. Let's take a and b." + }, + { + "Q": "\nwhat does he mean at 1:11 when he says constant?", + "A": "Well, what he means by constant is that when one side of a function goes up or down, multiplying or dividing, the other side will go up by the same amount, since it is a direct variation. If the function was a inverse variation, it wouldn t be a constant, because when one side of the function goes up, the other side of the function goes down, so therefore the variable isn t constant, because it doesn t do the same thing for both sides, it does the opposite. HOPE THIS HELPS!!", + "video_name": "3_DxJwDTbyQ", + "timestamps": [ + 71 + ], + "3min_transcript": "I've written some example relationships between two variables-- in this case between m and n, between a and b, between x and y. And what I want to do in this video is see if we can identify whether the relationships are a direct relationship, whether they vary directly, or maybe they vary inversely, or maybe it is neither. So let's explore it a little bit. So over here, we have m/n is equal to 1/7. So let's see how we can manipulate this. If we multiply both sides by n. What are we going to get? And in general, you want to separate them so that the two variables are on different sides of the equation, so you can see is it going to meet, is it going to be the pattern-- let me write it this way. m is equal to k times n. This would be direct variation. Or is it going to be the pattern m is equal to k times 1/n? This is inverse variation. And you see in either one of these, they're on different sides of the equal sign. So let's take this first relationship right now. Let's multiply both sides by n. And you get m, because these cancel out, So this actually meets the direct variation pattern. It's some constant times n. m is equal to some constant times n. So this right over here is direct. They vary directly. This is direct variation. Let's see, ab is equal to negative 3. So if we want to separate them-- and we could do it with either variable, we could divide both sides. I don't know, let's divide both sides by a. We could have done it by b. If we divide both sides by a, we get b is equal to negative 3/a. Or we could also write this as b is equal to negative 3 times 1/a. And once again, this is this pattern right here. One variable is equal to a constant times 1 over the other variable. In this case, our constant is negative 3. So over here, they vary inversely. This is an inverse relationship. Let's try this one over here. I'll try to do it in that same color. xy is equal to 1/10. isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m." + }, + { + "Q": "At 2:53, why doesn't y = -2x count as inverse variation? The values move in opposite directions; as x increases y decreases. Thanks for any help.\n", + "A": "Inverse variation is when the letter is under, so divides the coefficient and doesn t multiply it.", + "video_name": "3_DxJwDTbyQ", + "timestamps": [ + 173 + ], + "3min_transcript": "So this actually meets the direct variation pattern. It's some constant times n. m is equal to some constant times n. So this right over here is direct. They vary directly. This is direct variation. Let's see, ab is equal to negative 3. So if we want to separate them-- and we could do it with either variable, we could divide both sides. I don't know, let's divide both sides by a. We could have done it by b. If we divide both sides by a, we get b is equal to negative 3/a. Or we could also write this as b is equal to negative 3 times 1/a. And once again, this is this pattern right here. One variable is equal to a constant times 1 over the other variable. In this case, our constant is negative 3. So over here, they vary inversely. This is an inverse relationship. Let's try this one over here. I'll try to do it in that same color. xy is equal to 1/10. isolate them on either side of the equation. Let's divide both sides by x. You could divide by y, because you're really just trying to find an inverse or direct relationship. So divide both sides by x. You get y is equal to 1/10 over x, which is the same thing as 1/10x, which is the same thing as 1/10 times 1/x. So y is equal to some constant times 1/x. Once again, this is an inverse, y and x vary inversely. Let's do this one over here. 9 times m-- I'll go to that same orange color-- 9 times 1/m is equal to n. So this one's actually already done for us. And it might be a little bit clearer if we just flip this around. If we just flip the left and the right hand side, we get n is equal to 9 times 1/m. n is equal to some constant times 1/m. And remember, if I say that n varies inversely with m, that also means that m varies inversely with n. Those two things imply each other. Now let's try it with this expression over here. And this one's a little bit of a trickier one, because we've already separated the variables on both hands. And then we have this kind of-- if this was b is equal to 1/3 times a, then we would have direct variation, then b would vary directly with a. But in this case, we have 1/3 minus a. And you say, hey, maybe they're opposites, or whatever. And it actually turns out that this is neither. This is neither. And to make that point 100% clear, let's look at two of these examples. In direct variation, if you scale up one variable in one direction, you would scale up the other variable by the same amount. So if we have x going-- if x doubles from 1 to 2," + }, + { + "Q": "\nDid you know that the reason things float to the edge of a overfilled cup if you put something in it (say, a penny) is because that the middle of the water is actually just a teeny bit higher then the sides, so the penny automatically goes to the lowest point in the cup.\nVice Versa, if you were to underfill a cup, and put a penny on the side, it would automatically go towards the middle because that is the lowest point (3:02)", + "A": "Yep. Surface tension is an amazing thing, isn t it? In science, when you re measuring liquid in a beaker or graduated cylinder, you would call the lowest point the meniscus", + "video_name": "lOIP_Z_-0Hs", + "timestamps": [ + 182 + ], + "3min_transcript": "would be directly above it. So that's no good, because it blocks all the light or something. You can go 180 degrees, to have the next leaf directly opposite, which seems ideal. Only once you go 180 again, the third leaf is right over the first. In fact, any fraction of a circle with a whole number as a base is going to have complete overlap after that number of turns. And unlike when you're doodling, as a plant you're not smart enough to see you've gone all the way around and now should switch to adding things in between. If you try and postpone the overlap by making the fraction really small, you just get a ton of overlap in the beginning, and waste all this space, which is completely disastrous. Or maybe other fractions are good. The kind that position leaves in a star like pattern. It will be a while before it overlaps, and the leaves will be more evenly spaced in the meantime. But what if there were a fraction that never completely overlapped? For any rational fraction, eventually the star will close. But what if you used an irrational number? The kind of number that can't be expressed as a whole number What if you used the most irrational number? If you think it sounds weird to say one irrational number is might want to become a number theorist. If you are a number theorist, you might tell us that phi is the most irrational number. Or you might say, that's like saying, of all the integers, 1 is the integer-iest. Or you might disagree completely. But anyway, phi. It's more than 1, but less than 2, more than 3/2, less than 5/3. Greater than 8/5, but 13/8 is too big. 21/13 is just a little too small, and 34/21 is even closer, but too big, and so on. Each pair of adjacent Fibonacci numbers creates a ratio that gets closer and closer to Phi as the numbers increase. Those are the same numbers on the sides of these squares. Now stop being a number theorist, and start being a plant again. You put your first leaf somewhere, and the second leaf at an angle, which is one Phi-th of a circle. Which depending on whether you're going one way or the other, could be about 222.5 degrees, or about 137.5. Great, your second leaf is pretty far from the first, gets lots of space in the sun. And now let's add the next one a Phi-th of the circle away. You can see how new leaves tend to pop up in the spaces left between old leaves, but it never quite fills things evenly. So there's always room for one more leaf, without having to do a whole new layer. It's very practical and as a plant you probably like this. It would also be a good way to give lots of room to seed pods and petals and stuff. As a plant that follows this scheme, you'd be at an advantage. Where do spirals come in? Let's doodle a pinecone using this same method. By the way, you can make your own phi angle-a-tron by dog-earing a corner of your notebook. If you folded it so the edges of the line-- You have 45 degrees plus 90, which is 135. Pretty close to 137.5. If you're careful, you can slip in a couple more degrees. Detach your angle-a-tron and you're good to go. Add each new pine cone-y thing a phi angle around, and make them a little farther out each time. Which you can keep track of by marking the distance on your angle-a-tron. The spirals form by themselves. And if we count the number of arms, look it's five and eight." + }, + { + "Q": "\nCan you explain the sentence written at 2:12 to me: \"Of all the integers, 1 is the integeriest.\"?", + "A": "She is saying that some people say that saying that phi is the most irrational number is like saying that of all the integers 1 is the most integer-y.", + "video_name": "lOIP_Z_-0Hs", + "timestamps": [ + 132 + ], + "3min_transcript": "Say you're me and you're in math class, and you're doodling flowery petally things. If you want something with lots of overlapping petals, you're probably following a loose sort of rule that goes something like this. Add new petals where there's gaps between old petals. You can try doing this precisely. Start with some number of petals, say five, then add another layer in between. But the next layer, you have to add 10, then the next has 20. The inconvenient part of this is that you have to finish a layer before everything is even. Ideally, you'd have a rule that just lets you add petals until you get bored. Now imagine you're a plant, and you want to grow in a way that spreads out your leaves to catch the most possible sunlight. Unfortunately, and I hope I'm not presuming too much in thinking that, as a plant, you're not very smart. You don't know how to add number to create a series, you don't know geometry and proportions, and can't draw spirals, or rectangles, or slug cats. But maybe you could follow one simple rule. Botanists have noticed that plants seem to be fairly consistent when it comes to the angle between one leaf and the next. So let's see what you could do with that. So you grow your first leaf, and if you didn't change angle would be directly above it. So that's no good, because it blocks all the light or something. You can go 180 degrees, to have the next leaf directly opposite, which seems ideal. Only once you go 180 again, the third leaf is right over the first. In fact, any fraction of a circle with a whole number as a base is going to have complete overlap after that number of turns. And unlike when you're doodling, as a plant you're not smart enough to see you've gone all the way around and now should switch to adding things in between. If you try and postpone the overlap by making the fraction really small, you just get a ton of overlap in the beginning, and waste all this space, which is completely disastrous. Or maybe other fractions are good. The kind that position leaves in a star like pattern. It will be a while before it overlaps, and the leaves will be more evenly spaced in the meantime. But what if there were a fraction that never completely overlapped? For any rational fraction, eventually the star will close. But what if you used an irrational number? The kind of number that can't be expressed as a whole number What if you used the most irrational number? If you think it sounds weird to say one irrational number is might want to become a number theorist. If you are a number theorist, you might tell us that phi is the most irrational number. Or you might say, that's like saying, of all the integers, 1 is the integer-iest. Or you might disagree completely. But anyway, phi. It's more than 1, but less than 2, more than 3/2, less than 5/3. Greater than 8/5, but 13/8 is too big. 21/13 is just a little too small, and 34/21 is even closer, but too big, and so on. Each pair of adjacent Fibonacci numbers creates a ratio that gets closer and closer to Phi as the numbers increase. Those are the same numbers on the sides of these squares. Now stop being a number theorist, and start being a plant again. You put your first leaf somewhere, and the second leaf at an angle, which is one Phi-th of a circle. Which depending on whether you're going one way or the other, could be about 222.5 degrees, or about 137.5. Great, your second leaf is pretty far from the first, gets lots of space in the sun. And now let's add the next one a Phi-th of the circle away." + }, + { + "Q": "\nAt 9:10 to 9:33 Sal says that a rational can be a repeating decimal but it can't go backwards. But his answer was backwards. Is he wrong or am I not getting something.", + "A": "Sal is completely right! What he meant was that you can t say (by that statement) that a number is rational because it can be written as a repeating decimal. To conclude that a number that can t be written as a repeating decimal is not rational, he didn t goes backwards. He did use a strait foward reasoning to deduce. If all rationals can be expressed that way, a number that can t be written as a repeting decimal is not rational.", + "video_name": "GluohfOedQE", + "timestamps": [ + 550, + 573 + ], + "3min_transcript": "All right. And then, the last one, 2 is congruent to 3. 2 is congruent to 3. Well, by the same logic, if 1 is congruent to 4, and since 1 and 2 are opposite, it's also the same as 2. And 4, because it's opposite of 3, it's congruent to that, all these angles have to be the same thing. So 2 and 3 would also be congruent angles. So all of the other ones must be true, B, C and D. So A is definitely our choice. Next problem. Let me copy and paste it. OK. Consider the arguments below. Every multiple of 4 is even, 376 is a multiple of 4. Fair enough. A number can be written as a repeating decimal if it is rational. Pi cannot be written as a repeating decimal. Therefore, pi is not rational. Which ones, if any, use deductive reasoning? OK, so statement one, every multiple of 4 is even. 376 is a multiple of 4. So this is deductive reasoning. Because you know that every multiple of 4 is even. So you pick any multiple of 4, it's going to be even. 376 is a multiple of 4. Therefore, it has to be even. So this is correct logic. So statement one is definitely deductive reasoning. Let's see, statement number two. A number can be written as a repeating decimal if it is rational. So if you're rational, that means that you can write it as a repeating decimal. That's like 0.33333. That's 1/3. That's all they mean by a repeating decimal. written as a repeating decimal if it is rational. That doesn't say that a repeating decimal means that It just means that a rational number can be written as a repeating decimal. This statement doesn't let us go the other way. It doesn't say that a repeating decimal can definitely be written as a rational number. It just says that if it is rational, a number can be written as a repeating decimal. Fair enough. And then it says pi cannot be written as a repeating decimal. Pi cannot be written as a repeating decimal. So if pi cannot be written as a repeating decimal, pi be rational? Well if pi was rational, if pi was in this set, if pi were rational, then you could write it as a repeating decimal. But it said that you cannot write it as a repeating decimal. So therefore, pi cannot be rational." + }, + { + "Q": "\nSal converted the fractions into improper fractions in 1:38. Is that the easiest way?", + "A": "converting mixed fractions into improper fractions helps in comparing mixed fractions", + "video_name": "QS1LMomm0Gk", + "timestamps": [ + 98 + ], + "3min_transcript": "Find the sum 3 and 1/8 plus 3/4 plus negative 2 and 1/6. Let's just do the first part first. It's pretty straightforward. We have two positive numbers. Let me draw a number line. So let me draw a number line. And I'll try to focus in. So we're going to start at 3 and 1/8. So let's make this 0. So you have 1, 2, 3, and then you have 4. 3 and 1/8 is going to be right about there. So let me just draw its absolute value. So this 3 and 1/8 is going to be 3 and 1/8 to the right of 0. So it's going to be exactly that distance from 0 to the right. So this right here, the length of this arrow, you could view it as 3 and 1/8. Now whenever I like to deal with fractions, especially when they have different denominators and all of that, I like to deal with them as improper fractions. It makes the addition and the subtraction, and, actually, the multiplication and the division plus 1 is 25 over 8. So this is 25 over 8, which is the same thing as 3 and 1/8. Another way to think about it, 3 is 24 over 8. And then you add 1/8 to that, so you get 25 over 8. So this is our starting point. Now to that, we are going to add 3/4. We are going to add 3/4. So we're going to move another 3/4. We are going to move another 3/4. It's hard drawing these arrows. We're going to move another 3/4 to the right. So this right here, the length of this that we're moving to the right is 3/4. So plus 3/4. Now where does this put us? Well, both of these are positive integers. So we can just add them. We just have to find a like denominator. So we have 25 over 8. We have 25 over 8 plus 3/4. That's the same thing as we need to find a common denominator of 4 and 8 is 8. So it's going to be something over 8. To get from 4 to 8, we multiply by 2. So we have to multiply 3 by 2 as well. So you get 6. So 3/4 is the same thing as 6/8. If we have 25/8 and we're adding 6/8 to that, that gives us 25 plus 6 is 31/8. So this number right over here, this number right over here, is 31/8. And it makes sense because 32/8 would be 4. So it should be a little bit less than four. So this number right over here is 31/8. Or the length of this arrow, the absolute value of that number, is 31/8, a little bit less than 4. If you wanted to write that as a mixed number, it would be what? It would be 3 and 7/8. So that's that right over here. This is 31/8." + }, + { + "Q": "At 1:57, Why do we need to find a common denominator? Can't we just add 25/8 + 3/4 directly?\n", + "A": "Cause a common denominators solves all of life s problems", + "video_name": "QS1LMomm0Gk", + "timestamps": [ + 117 + ], + "3min_transcript": "Find the sum 3 and 1/8 plus 3/4 plus negative 2 and 1/6. Let's just do the first part first. It's pretty straightforward. We have two positive numbers. Let me draw a number line. So let me draw a number line. And I'll try to focus in. So we're going to start at 3 and 1/8. So let's make this 0. So you have 1, 2, 3, and then you have 4. 3 and 1/8 is going to be right about there. So let me just draw its absolute value. So this 3 and 1/8 is going to be 3 and 1/8 to the right of 0. So it's going to be exactly that distance from 0 to the right. So this right here, the length of this arrow, you could view it as 3 and 1/8. Now whenever I like to deal with fractions, especially when they have different denominators and all of that, I like to deal with them as improper fractions. It makes the addition and the subtraction, and, actually, the multiplication and the division plus 1 is 25 over 8. So this is 25 over 8, which is the same thing as 3 and 1/8. Another way to think about it, 3 is 24 over 8. And then you add 1/8 to that, so you get 25 over 8. So this is our starting point. Now to that, we are going to add 3/4. We are going to add 3/4. So we're going to move another 3/4. We are going to move another 3/4. It's hard drawing these arrows. We're going to move another 3/4 to the right. So this right here, the length of this that we're moving to the right is 3/4. So plus 3/4. Now where does this put us? Well, both of these are positive integers. So we can just add them. We just have to find a like denominator. So we have 25 over 8. We have 25 over 8 plus 3/4. That's the same thing as we need to find a common denominator of 4 and 8 is 8. So it's going to be something over 8. To get from 4 to 8, we multiply by 2. So we have to multiply 3 by 2 as well. So you get 6. So 3/4 is the same thing as 6/8. If we have 25/8 and we're adding 6/8 to that, that gives us 25 plus 6 is 31/8. So this number right over here, this number right over here, is 31/8. And it makes sense because 32/8 would be 4. So it should be a little bit less than four. So this number right over here is 31/8. Or the length of this arrow, the absolute value of that number, is 31/8, a little bit less than 4. If you wanted to write that as a mixed number, it would be what? It would be 3 and 7/8. So that's that right over here. This is 31/8." + }, + { + "Q": "\nReferencing the Banach-Tarski Paradox made my day. (2:24)\nIf only physics wouldn't get in the way of my math!", + "A": "I know, right? Physics rocks UNTIL it gets in the way of doing things like the Banach-Tarski paradox!", + "video_name": "F5RyVWI4Onk", + "timestamps": [ + 144 + ], + "3min_transcript": "At my house, no Thanksgiving dinner is complete without mathed potatoes. To make mathed potatoes, start by boiling the potatoes until they're soft, which will take about 15 to 20 minutes. After you drain them and let them cool slightly, you're ready for the math. Take one potato and divide evenly to get half a potato, plus half a potato. Then divide the halves into fourths and the fourths into eights and so on. Eventually, you will have a completely mathed potato that looks like this. Once you have proven this result for one potato, you can apply it to other potatoes without going through the entire process. That's how math works. While I prefer refined and precise methods for mathing a potato, many people just apply brute force algorithms. You can also add other variables like butter, cream, garlic, salt, and pepper. Place in a hemisphere and garnish with an organic hyperbolic plane, and your math potatoes are ready for the table. Together with a cranberry cylinder and a nice basket of bread spheres with butter prism, you'll be well on your way to creating a delicious and engaging Thanksgiving meal. Here's a serving tip. When arranging mathed potatoes on your plate, If you just make a mound, the gravy will fall off. It's best to create some kind of trough or pool. But what shape will maximize the amount of gravy it can hold? Due to the structural properties of mathed potatoes, this can essentially be reduced to a two-dimensional gravy pool problem, where you want the most gravy area given a certain potato perimeter. When I think of this question, I like to think about inflating shapes. Say you inflated a triangle. It would add more area and round out into a circle. And then, if the perimeter can't change, it would pop. In fact, all 2D shapes inflate into circles. And in 3D, it's spheres, which is my bubbles like to be round. And turkeys are spheroid, because that optimizes for maximum stuffing. The limited three-dimensional capacity of mathed potatoes may confuse things a little. But since a mathed potato sphere can't support itself, you're really stuck with extrusions of 2D shapes. So what's better? A deep mathed potato cylinder or a shallow but wider one? Well, think of it like this. If you slice the deep version in half, you'll see it has equivalent gravy-holding capacity to two And the perimeter of two circles would be more efficient if combined into one bigger circle. So the solution is to create the biggest, roundest, shallowest gravy pool you can. In fact, maybe you should just skip the mathed potatoes and get a bowl. Anyway, I hope this simple recipe helps you have an optimal Thanksgiving experience. Advanced chef-amaticians may wish to try Banach-Tarski potatoes, wherein after you cut a potato in a particular way, you put the pieces back together and get two potatoes. Stay tuned for more delicious and extremely practical Thanksgiving recipes this week." + }, + { + "Q": "\n2:27 If I had this sort of problem on a test, what are other ways of stating that there's no solution? I'm assuming the symbol for the empty set would be one of them? Or is that basically it?", + "A": "Yes, the symbol for the empty set would be an acceptable way of saying there is no solution.", + "video_name": "ZF_cZ-GX9PI", + "timestamps": [ + 147 + ], + "3min_transcript": "Solve for x, 5x - 3 is less than 12 \"and\" 4x plus 1 is greater than 25. So let's just solve for X in each of these constraints and keep in mind that any x has to satisfy both of them because it's an \"and\" over here so first we have this 5 x minus 3 is less than 12 so if we want to isolate the x we can get rid of this negative 3 here by adding 3 to both sides so let's add 3 to both sides of this inequality. The left-hand side, we're just left with a 5x, the minus 3 and the plus 3 cancel out. 5x is less than 12 plus 3 is 15. Now we can divide both sides by positive 5, that won't swap the inequality since 5 is positive. So we divide both sides by positive 5 and we are left with just from this constraint that x is less than 15 over 5, which is 3. So that constraint over here. But we have the second constraint as well. is greater than 25. So very similarly we can subtract one from both sides to get rid of that one on the left-hand side. And we get 4x, the ones cancel out. is greater than 25 minus one is 24. Divide both sides by positive 4 Don't have to do anything to the inequality since it's a positive number. And we get x is greater than 24 over 4 is 6. And remember there was that \"and\" over here. We have this \"and\". So x has to be less than 3 \"and\" x has to be greater than 6. So already your brain might be realizing that this is a little bit strange. This first constraint says that x needs to be less than 3 so this is 3 on the number line. We're saying x has to be less than 3 so it has to be in this shaded area right over there. greater than 6. So if this is 6 over here, it says that x has to greater than 6. It can't even include 6. And since we have this \"and\" here. The only x-es that are a solution for this compound inequality are the ones that satisfy both. The ones that are in the overlap of their solution set. But when you look at it right over here it's clear that there is no overlap. There is no x that is both greater than 6 \"and\" less than 3. So in this situation we have no solution." + }, + { + "Q": "At 8:54, why is tan theta described as sine theta/ Cosine theta?\n", + "A": "You will need to know these definitions: sin\u00ce\u00b8 = y/r cos\u00ce\u00b8 = x/r tan\u00ce\u00b8 = y/x Since we re in a unit circle, r = 1, so sin\u00ce\u00b8 = y and cos\u00ce\u00b8 = x. Since tan\u00ce\u00b8 = y/x, y = sin\u00ce\u00b8, and x = cos\u00ce\u00b8, we can make the substitutions. y = sin\u00ce\u00b8 x = cos\u00ce\u00b8 tan\u00ce\u00b8 = y/x = sin\u00ce\u00b8/cos\u00ce\u00b8", + "video_name": "1m9p9iubMLU", + "timestamps": [ + 534 + ], + "3min_transcript": "is the same thing as cosine of theta. And b is the same thing as sine of theta. Well, that's interesting. We just used our soh cah toa definition. Now, can we in some way use this to extend soh cah toa? Because soh cah toa has a problem. It works out fine if our angle is greater than 0 degrees, if we're dealing with degrees, and if it's less than 90 degrees. We can always make it part of a right triangle. But soh cah toa starts to break down as our angle is either 0 or maybe even becomes negative, or as our angle is 90 degrees or more. You can't have a right triangle with two 90-degree angles in it. It starts to break down. Let me make this clear. So sure, this is a right triangle, so the angle is pretty large. I can make the angle even larger and still have a right triangle. Even larger-- but I can never get quite to 90 degrees. have a right triangle any more. It all seems to break down. And especially the case, what happens when I go beyond 90 degrees. So let's see if we can use what we said up here. Let's set up a new definition of our trig functions which is really an extension of soh cah toa and is consistent with soh cah toa. Instead of defining cosine as if I have a right triangle, and saying, OK, it's the adjacent over the hypotenuse. Sine is the opposite over the hypotenuse. Tangent is opposite over adjacent. Why don't I just say, for any angle, I can draw it in the unit circle using this convention that I just set up? And let's just say that the cosine of our angle is equal to the x-coordinate where we intersect, where the terminal side of our angle intersects the unit circle. to be equal to the y-coordinate where the terminal side of the angle intersects the unit circle? So essentially, for any angle, this point is going to define cosine of theta and sine of theta. And so what would be a reasonable definition for tangent of theta? Well, tangent of theta-- even with soh cah toa-- could be defined as sine of theta over cosine of theta, which in this case is just going to be the y-coordinate where we intersect the unit circle over the x-coordinate. In the next few videos, I'll show some examples where we use the unit circle definition to start evaluating some trig ratios." + }, + { + "Q": "At 2:34, shouldn't the point on the circle be (x,y) and not (a,b)? [Since horizontal goes across 'x' units and vertical goes up 'y' units--- A full explanation will be greatly appreciated]\n", + "A": "It would be x and y, but he uses the letters a and b in the example because a and b are the letters we use in the Pythagorean Theorem a\u00c2\u00b2+b\u00c2\u00b2 = c\u00c2\u00b2 and they re the letters we commonly use for the sides of triangles in general. It doesn t matter which letters you use so long as the equation of the circle is still in the form a\u00c2\u00b2+b\u00c2\u00b2 = 1.", + "video_name": "1m9p9iubMLU", + "timestamps": [ + 154 + ], + "3min_transcript": "And the way I'm going to draw this angle-- I'm going to define a convention for positive angles. I'm going to say a positive angle-- well, the initial side of the angle we're always going to do along the positive x-axis. So you can kind of view it as the starting side, the initial side of an angle. And then to draw a positive angle, the terminal side, we're going to move in a counterclockwise direction. So positive angle means we're going counterclockwise. And this is just the convention I'm going to use, and it's also the convention that is typically used. And so you can imagine a negative angle would move in a clockwise direction. So let me draw a positive angle. So a positive angle might look something like this. This is the initial side. And then from that, I go in a counterclockwise direction And then this is the terminal side. So this is a positive angle theta. And what I want to do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates a comma b. The x value where it intersects is a. The y value where it intersects is b. And the whole point of what I'm doing here is I'm going to see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I want to do is I want to make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here. And let me make it clear that this is a 90-degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. you is, what is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of 1. So the hypotenuse has length 1. Now, what is the length of this blue side right over here? You could view this as the opposite side to the angle. Well, this height is the exact same thing as the y-coordinate of this point of intersection. So this height right over here is going to be equal to b. The y-coordinate right over here is b. This height is equal to b. Now, exact same logic-- what is the length of this base going to be? The base just of the right triangle? Well, this is going to be the x-coordinate" + }, + { + "Q": "\nat 4:29, why do we find cosine?", + "A": "Sal is just demonstrating how all of the trig functions work as they are defined on a unit circle.", + "video_name": "1m9p9iubMLU", + "timestamps": [ + 269 + ], + "3min_transcript": "And then this is the terminal side. So this is a positive angle theta. And what I want to do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates a comma b. The x value where it intersects is a. The y value where it intersects is b. And the whole point of what I'm doing here is I'm going to see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I want to do is I want to make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here. And let me make it clear that this is a 90-degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. you is, what is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of 1. So the hypotenuse has length 1. Now, what is the length of this blue side right over here? You could view this as the opposite side to the angle. Well, this height is the exact same thing as the y-coordinate of this point of intersection. So this height right over here is going to be equal to b. The y-coordinate right over here is b. This height is equal to b. Now, exact same logic-- what is the length of this base going to be? The base just of the right triangle? Well, this is going to be the x-coordinate If you were to drop this down, this is the point x is equal to a. Or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine-- let me use the same green-- what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our soh cah toa definition. That's the only one we have now. We are actually in the process of extending it-- soh cah toa definition of trig functions. And the cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side-- for this angle, the adjacent side has length a. So it's going to be equal to a over-- what's" + }, + { + "Q": "At 3:48, how do we know that side b of the triangle is the same as the chord \"b\", like Sal says?\n", + "A": "He never says that the length of the chord you are creating is equivalent to the length of side b. He s simply saying that the length of side b, or the height as he says, is the y-value, b, which is what he marks on the y-axis. He s not creating a chord b, it s just a dashed line to show that the y-value corresponds to the height of side b of the triangle. Hope this helps!", + "video_name": "1m9p9iubMLU", + "timestamps": [ + 228 + ], + "3min_transcript": "And then this is the terminal side. So this is a positive angle theta. And what I want to do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates a comma b. The x value where it intersects is a. The y value where it intersects is b. And the whole point of what I'm doing here is I'm going to see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I want to do is I want to make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here. And let me make it clear that this is a 90-degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. you is, what is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of 1. So the hypotenuse has length 1. Now, what is the length of this blue side right over here? You could view this as the opposite side to the angle. Well, this height is the exact same thing as the y-coordinate of this point of intersection. So this height right over here is going to be equal to b. The y-coordinate right over here is b. This height is equal to b. Now, exact same logic-- what is the length of this base going to be? The base just of the right triangle? Well, this is going to be the x-coordinate If you were to drop this down, this is the point x is equal to a. Or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine-- let me use the same green-- what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our soh cah toa definition. That's the only one we have now. We are actually in the process of extending it-- soh cah toa definition of trig functions. And the cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side-- for this angle, the adjacent side has length a. So it's going to be equal to a over-- what's" + }, + { + "Q": "At 7:05,Sal said \"bottom boundary\" as z=2-2/3*x-y/3 but from graph it seems\nz=0 to z=z=2-2/3*x-y/3. I think the bottom limit for dz is zero? i am confused?\n", + "A": "Here is the clarification: Review from 1:13 where Sal defines the plane (via x,y,z intercepts). Then at 2:11 he says he cares about the volume above the plane (to make the problem more complicated). So he is asking for the volume between this plane and the plane z=2.", + "video_name": "ZN2PfqZ4ihM", + "timestamps": [ + 425 + ], + "3min_transcript": "And that is equal to dx -- no, sorry, this is dy. Let me do this in yellow, or green even better. So dy, which is this. dy times dx, dx times dz. That's the volume of that little cube. And if we wanted to know the mass of that cube, we would multiply the density function at that point times this dv. So the mass, you could call it d -- I don't know, dm. The mass differential is going to be equal to that times that. So x squared y z times this. dy, dx, and dz. And we normally switch this order around, depending on what we're going to integrate with respect to first so we don't get confused. So let's try to do this. Let's try to set up this integral. So let's do it traditionally. The last couple of triple integrals we did we integrated So let's do that. So we're going to integrate with respect to z first. so we're going to take this cube and we're going to sum up all of the cubes in the z-axis. So going up and down first, right? So if we do that, what is the bottom boundary? So when you sum up up and down, these cubes are going to turn to columns, right? So what is the bottom of the column, the bottom bound? What's the surface? It's the surface defined right here. So, if we want that bottom bound defined in terms of z, we just have to solve this in terms of z. So let's subtract. So what do we get. If we want this defined in terms of z, we get 3z is equal to 6 minus 2x minus y. Or z is equal 2 minus 2/3x minus y over 3. This is the same thing as that. But when we're talking about z, explicitly defining a So the bottom boundary -- and you can visualize it, right? The bottom of these columns are going to go up and down. We're going to add up all the columns in up and down direction, right? You can imagine summing them. The bottom boundary is going to be this surface. z is equal to 2 minus 2/3x minus y over 3. And then what's the upper bound? Well, the top of the column is going to be this green plane, and what did we say the green plane was? It was z is equal to 2. That's this plane, this surface right here. Z is equal to 2. And, of course, what is the volume of that column? Well, it's going to be the density function, x squared yz times the volume differential, but we're integrating with respect to z first. Let me write dz there. I don't know, let's say we want to integrate with respect to -- I don't know, we want to integrate with respect" + }, + { + "Q": "\nAt 10:10, I don't get it", + "A": "There is an exact same question at the back, with the same point of time reference which has some good answers. You should have a look at them.", + "video_name": "ZN2PfqZ4ihM", + "timestamps": [ + 610 + ], + "3min_transcript": "In the last couple of videos, I integrated So let's do x just to show you it really doesn't matter. So we're going to integrate with respect to x. So, now we have these columns, right? When we integrate with respect to z, we get the volume of each of these columns wher the top boundary is that plane. Let's see if I can draw it decently. The top boundary is that plane. The bottom boundary is this surface. Now we want to integrate with respect to x. So we're going to add up all of the dx's. So what is the bottom boundary for the x's? Well, this surface is defined all the way to -- the volume under question is defined all the way until x is equal to 0. And if you get confused, and it's not that difficult to get confused when you're imagining these three-dimensional things, say you know what, we already integrated with respect to z. The two variables I have left are x and y. Let me draw the projection of our volume onto the xy plane, and what does that look like? Because that actually does help simplify things. So if we twist it, if we take this y and flip it out like that, and x like that we'll get in kind of the traditional way that we learned when we first learned algebra. The xy-axis. So this is x, this is y. And this point is what? Or this point? That's x is equal to 3. So it's 1, 2, 3. That's x is equal to 3. And this point right here is y is equal to 6. So 1, 2, 3, 4, 5, 6. So on the xy-axis, kind of the domain -- you can view it that -- looks something like that. So one way to think about it is we've figured out if these columns -- we've integrated up/down or along the z-axis. But when you view it looking straight down onto it, you're looking on the xy plane, each of our columns are going to of your screen in the z direction. But the base of each column is going to dx like that, and then dy up and down, right? So we decided to integrate with respect to x next. So we're going to add up each of those columns in the x direction, in the horizontal direction. So the question was what is the bottom boundary? What is the lower bound in the x direction? Well, it's x is equal to 0. If there was a line here, then it would be that line probably as a function of y, or definitely as a function of y. So our bottom bound here is x is equal to 0. What's our top bound? I realize I'm already pushing. Well, our top bound is this relation, but it has to be in terms of x, right? So, you could view it as kind of saying well, if z is equal to 0, what is this line? What is this line right here? So z is equal to 0. We have 2x plus y is equal to 6. We want the relationship in terms of x. So we get 2x is equal to 6 minus y where x is equal" + }, + { + "Q": "at 4:35 cant the x and y axi, axises, axis(multiple axis) be a line and go on forever in any direction and seems how forever is a relative term go on in both directions in so far it is impossible to reach the end what so ever.\n", + "A": "The axes do go on forever, so to speak, in both directions, and it is indeed impossible to reach the end (if we progress by finite distances). Did it seem to you that Sal was implying something different? Is your question related to the restriction in the domain of arctangent?", + "video_name": "QGfdhqbilY8", + "timestamps": [ + 275 + ], + "3min_transcript": "so inverse tangent of x minus six plus three pi over two so let me write that down. Let me type this. G inverse of x is going to be the inverse tangent so I can write it like this, the inverse tangent of x minus six and yes it interpreted it correctly. Inverse tangent you can do that as arctangent of x minus six plus three pi over two and it did interpret it correctly but then we have to think about what is the domain of g inverse? What is the domain of g inverse of x? Let's think about this a little bit more. The domain of g inverse of x, so let's just think about what tangent is doing. The tangent function if we imagine a unit circle, so that's a unit circle right over there. Guess we can imagine to be a unit circle. My pen tool is acting up a little bit it's putting this little gaps and things Let's just say for the sake of argument that that's a unit circle, that's the x axis and that's the y axis. If you form an angle theta. If you form some angle theta right over here, the tangent of theta is essentially the slope of this terminal ray of the angle or the ... Or I guess we can call it the terminal ray of the angle. The angles form by that ray and this ray along the positive x axis. The tangent of theta is the slope right over there and you can get a tangent of any theta except for a few. You can find the tangent of that, you could find the slope there, you could find the slope there, you could also find the slope there, you could find the slope there but the place where you can't find the slope is when this ray goes straight up, or this ray goes straight down. Those were the cases where you can't find the slope. They are the slope you could say is approaching The domain of tangent, so tangent domain so the domain is essentially all real numbers, all reals except multiples of pi over \u2026 I guess you can say pi over two plus multiples of pi, except pi over two plus multiples of pi where k could be any integer so you could also be subtracting pi because if you have pi over two, if you add pi, you go straight down here. You add another pi you go up there, if you subtract pi you go down here, add, subtract another pi you go over there. This is the domain but given this domain you can get any real number. The range here is all reals because you can get any slope here," + }, + { + "Q": "\nAt 0:43 what did mean by 90 degrees. Do you mean that all sides are the same length for a square?", + "A": "It is true that all sides of a square have equal length but when he said the square had 90 degrees he meant that all the angles of the square were equal to 90 degrees. Have an awesome day!\u00f0\u009f\u0098\u008a", + "video_name": "1pHhMX0_4Bw", + "timestamps": [ + 43 + ], + "3min_transcript": "A parallelogram is a blank with two sets of parallel lines. So let's see what the options are. So one option is a quadrilateral. And a parallelogram is definitely a quadrilateral. A quadrilateral is a four-sided figure, and it is definitely a four-sided figure. A parallelogram is not always a rhombus. A rhombus is a special case of a parallelogram where not only do you have to sets of parallel lines as your sides, two sets of parallel sides, but all of the sides are the same length in a rhombus. And a square is a special case of a rhombus where all of the angles are 90 degrees. So here, all we can say is that a parallelogram is a quadrilateral. And so let's check our answer. And it's always a good idea to look at hints. And so it'll kind of say the same thing that we just said, but it would say it for the particular problem that you're actually looking at. Let's do a few more of these. Suzanne is on an expedition to save the universe. Sounds like a reasonable expedition to go on. a game called Find the Rhombuses. A wizard tells her that she has a square, a quadrilateral, and a parallelogram, and she must identify which of the shapes are also rhombuses. Which of these shapes should she pick to save the universe? So a square is a special case of a rhombus. Just to remind ourselves, a rhombus, the opposite sides are parallel to each other. You have two sets of parallel sides. A square has two sets of parallel sides, and it has the extra condition that all of the angles are right angles. So a square is definitely going to be a rhombus. Now, all rhombuses have four sides. So all rhombuses are quadrilaterals. But not all quadrilaterals are rhombuses. You could have a quadrilateral where none of the sides are parallel to each other. So we won't click this one. Once again, a parallelogram. So all rhombuses are parallelograms. two sets of parallel line segments representing their sides. But all parallelograms are not rhombuses. So I would say that if someone gives you square, you can say, look, a square is always going to be a rhombus. A quadrilateral isn't always going to be a rhombus, nor is a parallelogram always going to be a rhombus. We got it right." + }, + { + "Q": "At 0:34-0:39, did he also mean to say that all sides are of equal length? A rectangle is also a quadrilateral with only 90-degree angles, but all the sides aren't always the same length.\n", + "A": "Yes, rhombus is a quadrilateral with 4 sides of equal length. So rectangles are not always rhombus.", + "video_name": "1pHhMX0_4Bw", + "timestamps": [ + 34, + 39 + ], + "3min_transcript": "A parallelogram is a blank with two sets of parallel lines. So let's see what the options are. So one option is a quadrilateral. And a parallelogram is definitely a quadrilateral. A quadrilateral is a four-sided figure, and it is definitely a four-sided figure. A parallelogram is not always a rhombus. A rhombus is a special case of a parallelogram where not only do you have to sets of parallel lines as your sides, two sets of parallel sides, but all of the sides are the same length in a rhombus. And a square is a special case of a rhombus where all of the angles are 90 degrees. So here, all we can say is that a parallelogram is a quadrilateral. And so let's check our answer. And it's always a good idea to look at hints. And so it'll kind of say the same thing that we just said, but it would say it for the particular problem that you're actually looking at. Let's do a few more of these. Suzanne is on an expedition to save the universe. Sounds like a reasonable expedition to go on. a game called Find the Rhombuses. A wizard tells her that she has a square, a quadrilateral, and a parallelogram, and she must identify which of the shapes are also rhombuses. Which of these shapes should she pick to save the universe? So a square is a special case of a rhombus. Just to remind ourselves, a rhombus, the opposite sides are parallel to each other. You have two sets of parallel sides. A square has two sets of parallel sides, and it has the extra condition that all of the angles are right angles. So a square is definitely going to be a rhombus. Now, all rhombuses have four sides. So all rhombuses are quadrilaterals. But not all quadrilaterals are rhombuses. You could have a quadrilateral where none of the sides are parallel to each other. So we won't click this one. Once again, a parallelogram. So all rhombuses are parallelograms. two sets of parallel line segments representing their sides. But all parallelograms are not rhombuses. So I would say that if someone gives you square, you can say, look, a square is always going to be a rhombus. A quadrilateral isn't always going to be a rhombus, nor is a parallelogram always going to be a rhombus. We got it right." + }, + { + "Q": "\nShouldn't it be (2x-2y) - (2x-2y)(dy/dx) instead of (2x-2y)+(2x-2y)(dy/dx) at 2:15", + "A": "He has switched the order of the variables, so that it is (2x-2y)+(2 y-2 x)(dy/dx). -(2x-2y) = 2y-2x", + "video_name": "9uxvm-USEYE", + "timestamps": [ + 135 + ], + "3min_transcript": "Let's get some more practice doing implicit differentiation. So let's find the derivative of y with respect to x. We're going to assume that y is a function of x. So let's apply our derivative operator to both sides of this equation. So let's apply our derivative operator. And so first, on the left hand side, we essentially are just going to apply the chain rule. First we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1, and the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x. We're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do. Let's make it clear. This right over here, I can rewrite as 2x minus 2y. So let me do that so I can save some space. This is 2x minus 2y If I just distribute the 2. onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides." + }, + { + "Q": "At 2:12, why does Sal simply multiply (2x-2y) *1 and then by the (dy/dx), instead of FOIL'ing it out and getting something like 2x-2x(dy/dx)-2y+2y(dy/dx) ?\n", + "A": "He could have done that. Then he would have had to put the dy/dx s back together: 2x - 2y + 2y(dy/dx) - 2x(dy/dx) 2x - 2y + (2y - 2x)(dy/dx) He wants to get all the dy/dx s together such that he can manipulate them into one instance of dy/dx multiplied by a factor (in this case that factor ends up being (2y - 2x - 1)), such that he can isolated dy/dx on the left and have his answer.", + "video_name": "9uxvm-USEYE", + "timestamps": [ + 132 + ], + "3min_transcript": "Let's get some more practice doing implicit differentiation. So let's find the derivative of y with respect to x. We're going to assume that y is a function of x. So let's apply our derivative operator to both sides of this equation. So let's apply our derivative operator. And so first, on the left hand side, we essentially are just going to apply the chain rule. First we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1, and the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x. We're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do. Let's make it clear. This right over here, I can rewrite as 2x minus 2y. So let me do that so I can save some space. This is 2x minus 2y If I just distribute the 2. onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides." + }, + { + "Q": "at 3:05 shouldnt it be (2y-2x) time negative dy/dx but why is there no negative\n", + "A": "at 2:15 - 2:25 he explains that instead of doing -(2x - 2y) he chose to do (2y - 2x) (they are equal).", + "video_name": "9uxvm-USEYE", + "timestamps": [ + 185 + ], + "3min_transcript": "of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x. We're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do. Let's make it clear. This right over here, I can rewrite as 2x minus 2y. So let me do that so I can save some space. This is 2x minus 2y If I just distribute the 2. onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides. 2y from that side. And then we could also subtract a dy dx from both sides, so that all of our dy dx's are on the left hand side, and all of our non dy dx's are on the right hand side. So let's do that. So we're going to subtract a dy dx on the right and a dy dx here on the left. And so what are we left with? Well, on the left hand side, these cancel out. And we're left with 2y minus 2x dy dx minus 1 dy dx, or just minus a dy dx. Let me make it clear. We could write this as a minus 1 dy dx. So this is we can essentially just add these two coefficients. So this simplifies to 2y minus 2x minus 1 times the derivative of y with respect to x, which is going" + }, + { + "Q": "At 0:52, why does he say she wants to divide the blueberries into 6 groups? That number doesn't take Kali into account, and we don't know how many berries she may want. She could want 1 blueberry, or 0, or 12. Why then, does he not at least say \"we can assume that Kali doesn't want any berries, if she is giving them all too her friends.\"?\n", + "A": "That s right. I agree", + "video_name": "QXNg_u5Tv8Q", + "timestamps": [ + 52 + ], + "3min_transcript": "- Kali is having a picnic for her six friends. The oldest friend, Vikram, is 10 years old. The youngest, Diya, is six years old. She has a total of 48 blueberries and wants to split them evenly between her friends. How many blueberries does each friend get? And I encourage you to now, pause this video and try to figure it out on your own. How many blueberries does each friend get? Let's think about this a little bit. So she has six friends. She has six friends and she wants them all to be able to get the same amount. So, she wants to split the 48 blueberries evenly amongst her six friends. The ages of her friends don't matter. So, she's going to take the 48 blueberries. She's going to take the 48 blueberries and divide it, and divide it by six. She wants to divide it into six groups. So she wants to divide it into six groups. 48 divided by 6. And so this, is going to be equal to the number of blueberries that each friend gets. So 48 divided by six is equal to question mark, is the same thing as saying that 48, 48 is equal to, is equal to question mark times six, times six. So if we could figure out what number we can multiply by six to get 48, then we know what 48 divided by six actually is. For example, this question mark, this is the number of blueberries per friend. The number of blueberries per friend times six friends, well that should tell us the total number of blueberries, which is 48. So what is this number? Well, let's think about, let's just think about all of the multiples of six. So, six times one is equal to six. Six times two is equal to 12. And really we're just increasing by six each time. Six times three is equal to 18. Six times four is equal to 24. Six times five is equal to 30. Six times six is equal to 36. Six times seven is equal to 42. Notice we're just adding six every time. Six times eight is equal to 48. Is equal to 48. So we now know that question mark, we now know that the question mark must be eight. Six times eight and eight times six is the same thing. So this is going to be equal to six times question mark. Six times question mark and now we learned that question mark is equal to 48. Sorry, question mark is equal to eight. So, each of her friends are going to get eight blueberries. So this is, right over here, 48 divided by six is equal to eight blueberries." + }, + { + "Q": "so it ends at 4:42 ?\n", + "A": "Yes. Why wouldn t it its the end of the video.", + "video_name": "vAlazPPFlyY", + "timestamps": [ + 282 + ], + "3min_transcript": "it's a vertical angle, it's the one on the opposite side of the intersection. It's one of these angles that it is not adjacent to. So it would be this angle right over here. So going back to the question, a vertical angle to angle EGA, well if you imagine the intersection of line EB and line DA, then the non-adjacent angle formed to angle EGA is angle DGB. Actually, what we already highlighted in magenta right over here. So this is angle DGB. Which could also be called angle BGD. These are obviously both referring to this angle up here. Name an angle that forms a linear pair with the angle DFG. So we'll put this in a new color. Angle DFG. Sorry, DGF, all of these should have G in the middle. So linear pair with angle DGF, so that's this angle right over here. So an angle that forms a linear pair will be an angle that is adjacent, where the two outer rays combined will form a line. So for example, if you combine angle DGF, which is this angle, and angle DGC, then their two outer rays form this entire line right over here. So we could say angle DGC. Or, if you look at angle DFG, you could form a line this way. If you take angle AGF, so if you take this one, then the outer rays will form this line. So angle AGF would also work. Angle AGF. Let's do one more. Name a vertical angle to angle FGB. So this is FGB right over here. when CF-- let me highlight this, that's hard to see. This is the last one, so I can make a mess out of this. That angle is formed when CF and EB intersect with each other. And four angles are formed. The one question, FGB, these two angles that are adjacent to it, it shares a common ray. And then the vertical angle, the one that sits on the opposite side. So this angle, this angle right over here, which is angle EGC. Or you could also call it angle CGE. So angle CGE." + }, + { + "Q": "at 0:22, what is a better under standing of what adjacent is?\n", + "A": "It just means next to", + "video_name": "vAlazPPFlyY", + "timestamps": [ + 22 + ], + "3min_transcript": "We're asked to name an angle adjacent to angle BGD. So angle BGD, let's see if we can pick it out. So here is B, here is G, and here is D, right over here. So angle BGD is this entire angle right over here. So when we talk about adjacent angles, we're talking about an angle that has one of its rays in common. So for example, angle AGB has one of the rays in common, it has GB in common with angle BGD. So we could say angle AGB, which could obviously also be called angle BGA, BGA and AGB are both this angle right over here. You could also go with angle FGB, because that also has GB in common. So you go angle FGB, which could also be written as angle BGF. So you could do this angle right over here, angle EGD. Or you could go all the way out here, angle FGD. These last two sharing ray GD in common. So any one of these responses would satisfy the question of just naming an angle, just naming one. Let's do this next one. Name an angle vertical to angle EGA. So this is this angle right over here. And the way you think about vertical angles is, imagine two lines crossing. So imagine two lines crossing, just like this. And they could literally be lines, and they're intersecting at a point. This is forming four angles, or you could imagine it's forming two sets of vertical angles. it's a vertical angle, it's the one on the opposite side of the intersection. It's one of these angles that it is not adjacent to. So it would be this angle right over here. So going back to the question, a vertical angle to angle EGA, well if you imagine the intersection of line EB and line DA, then the non-adjacent angle formed to angle EGA is angle DGB. Actually, what we already highlighted in magenta right over here. So this is angle DGB. Which could also be called angle BGD. These are obviously both referring to this angle up here. Name an angle that forms a linear pair with the angle DFG. So we'll put this in a new color. Angle DFG. Sorry, DGF, all of these should have G in the middle." + }, + { + "Q": "\nAt 4:00 Sal cancels out the terms that approach zero as if they dont exist and states that the limit is 3/6 or 1/2.\n\nI dont understand how he can be sure the function isn't defined for 1/2.\n\nAs it approaches infinity it is (3 minus a tiny fraction) over (6 minus a tiny fraction)\n\nAs I see it, unless those tiny fractions are exactly the same, the limit will be a tiny amount greater or less than 1/2, making the function defined for the point Sal claims it never reaches.\n\nIs there something I'm misunderstanding?", + "A": "I think you just answered your own question without realizing it. The value will always be a tiny amount greater or less than 1/2. It will never be exactly equal to 1/2.", + "video_name": "P0ZgqB44Do4", + "timestamps": [ + 240 + ], + "3min_transcript": "These other terms are going to matter less obviously minus 54 isn't going to grow at all and minus 18X is going to grow much slower than the three X squared, the highest degree terms are going to be what dominates. If we look at just those terms then you could think of simplifying it in this way. F of X is going to get closer and closer to 3/6 or 1/2. You could say that there's a horizontal asymptote at Y is equal to 1/2. Another way we could have thought about this if you don't like this whole little bit of hand wavy argument that these two terms dominate is that we can divide the numerator and the denominator by the highest degree or X raised to the highest power in the numerator and the denominator. The highest degree term is X squared in the numerator. Let's divide the numerator and the denominator or I should say the highest degree term in the numerator and the denominator is X squared. and denominator by that. If you multiply the numerator times one over X squared and the denominator times one over X squared. Notice we're not changing the value of the entire expression, we're just multiplying it times one if we assume X is not equal zero. We get two. In our numerator, let's see three X squared divided by X squared is going to be three minus 18 over X minus 81 over X squared and then all of that over six X squared times one over X squared, this is going to be six and then minus 54 over X squared. What's going to happen? If you want to think in terms of if you want to think of limits as something approaches infinity. If you want to say the limit as X approaches infinity here. What's going to happen? Well this, this and that are going to approach zero Now, if you say this X approaches negative infinity, it would be the same thing. This, this and this approach zero and once again you approach 1/2. That's the horizontal asymptote. Y is equal to 1/2. Let's think about the vertical asymptotes. Let me write that down right over here. Let me scroll over a little bit. Vertical asymptote or possibly asymptotes. Vertical maybe there is more than one. Now it might be very tempting to say, \"Okay, you hit a vertical asymptote\" \"whenever the denominator equals to zero\" \"which would make this rational expression undefined\" and as we'll see for this case that is not exactly right. Just making the denominator equal to zero by itself will not make a vertical asymptote. It will definitely be a place" + }, + { + "Q": "\nAt 7:55, there was a reference to the function g:R^2-->R. How did that function morph into g(x1,x2)=2.", + "A": "so you have a vector space that is R^2, and a vector space that is R. the R^2 vector space simply has more data than the vector space R. the transformation G fits vectors from the R^2 vector space into vectors in R, or at least attempts to.", + "video_name": "BQMyeQOLvpg", + "timestamps": [ + 475 + ], + "3min_transcript": "It could be equal to the codomain. It's some subset. A set is a subset of itself, every member of a set is also a member of itself, so it's a subset of itself. So range is a subset of the codomain which the function actually maps to. So let me give you an example. Let's say I define the function g, and it is a mapping from the set of real numbers. Let me say it's a mapping from R2 to R. mapping it to R. And I will define g, I'll write it a couple of different ways. So now I'm going to take g of two values, I could say xy or I could say x1, x2. Let me do it that way. g of x1, x2 is always equal to 2. It's a mapping from R2 to R, but this always equals 2. And let me actually write the other notation just because you probably haven't seen this much. g maps any points x1 and x2 to the point 2. This makes the mapping a little bit clearer. But just to get the notation right, what is our domain? That was part of my function definition, I said we're mapping from R2, so my domain is R2. Now what is my codomain? My codomain is the set that I'm potentially mapping to, and is part of the function definition. This by definition is the codomain. So my codomain is R. Now what is the range of my function? The range is the set of values that the function actually maps to. In this case, we always map to the value 2, so the range is actually just the value 2. And if we were to visualize this-- R2 is actually-- I wouldn't draw it as a blurb, I would draw it as the entire Cartesian space, but I'm just giving you an abstract notion. If I really have to draw R, I'd draw it as some type of a number line." + }, + { + "Q": "\n@ 4:00 why is the exponent -10?", + "A": "At around 2:10 you will see that it is counting the number of places from the standard decimal place, and it is negative because you re moving it to the right, towards the negative side of the number line.", + "video_name": "6phoVfGKKec", + "timestamps": [ + 240 + ], + "3min_transcript": "to 0-- let me actually-- I skipped a step right there. Let me add 1 times 10 to the 0, so we have something natural. So this is one times 10 to the first. One times 10 to the 0 is equal to 1 times 1, which is equal to 1. 1 times 10 to the negative 1 is equal to 1/10, which is equal to 0.1. If I do 1 times 10 to the negative 2, 10 to the negative 2 is 1 over 10 squared or 1/100. So this is going to be 1/100, which is 0.01. What's happening here? When I raise it to a negative 1 power, I've essentially moved the decimal from to the right of the 1 to the left of the 1. I've moved it from there to there. When I raise it to the negative 2, I moved it two over to the left. So how many times are we going to have to move it over to the left to get this number right over here? So we have to move it one time just to get in front of the 3. And then we have to move it that many more times to get all of the zeroes in there so that we have to move it one time to get the 3. So if we started here, we're going to move 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 times. So this is going to be 3.457 times 10 to the negative 10 power. Let me just rewrite it. So 3.457 times 10 to the negative 10 power. So in general, what you want to do is you want to find the first non-zero number here. Remember, you want a number here that's between 1 and 10. And it can be equal to 1, but it has to be less than 10. 3.457 definitely fits that bill. It's between 1 and 10. And then you just want to count the leading zeroes between the decimal and that number and include the number because that tells you to actually get this number up here. And so we have to shift this decimal 10 times to the left to get this thing up here." + }, + { + "Q": "At 9:45, how does \u00e2\u0088\u009a10 . \u00e2\u0088\u009a10 = 10 in the denominator?\n", + "A": "That equals 10 because at 9:04, he says that 1/radical 10 is irrational. If you wanted to get rid of the radical, you would have to multiply square root 10 over square root 10. 1x square root 10= 10. Square root 10x square root of 10=10 because 10 times 10 is 100. 100 is a perfect square so it simplifys back to 10 again. If my explaination dosen t help much, watch other videos on this topic.", + "video_name": "BpBh8gvMifs", + "timestamps": [ + 585 + ], + "3min_transcript": "recognize that, gee, if I multiply 0.4 times 0.4, I'll get this. But I'll show you a more systematic way of doing it, if that wasn't obvious to you. So this is the same thing as the square root of 16/100, right? That's what 0.16 is. So this is equal to the square root of 16 over the square root of 100, which is equal to 4/10, which is equal to 0.4. Let's do a couple more like that. Part I was the square root of 0.1, which is equal to the square root of 1/10, which is equal to the square root of 1 over the square root of 10, which is equal to 1 over-- now, the square root of 10-- 10 is just 2 times 5. So that doesn't really help us much. A lot of math teachers don't like you leaving that radical But I can already tell you that this is irrational. You'll just keep getting numbers. You can try it on your calculator, and it will never repeat. Your calculator will just give you an approximation. Because in order to give the exact value, you'd have to have an infinite number of digits. But if you wanted to rationalize this, just to show you. If you want to get rid of the radical in the denominator, you can multiply this times the square root of 10 over the square root of 10, right? This is just 1. So you get the square root of 10/10. These are equivalent statements, but both of them are irrational. You take an irrational number, divide it by 10, you still have an irrational number. Let's do J. We have the square root of 0.01. This is the same thing as the square root of 1/100. Which is equal to the square root of 1 over the square root of 100, which is equal to 1/10, or 0.1. It's being written as a fraction. This one up here was also rational. It can be written expressed as a fraction." + }, + { + "Q": "\nWhy in example D @ 4:37 the square of 2*2=2 and in example E @ 6:24 the square of 2*2=4? What makes the two identically appearing values different?", + "A": "Because in D There was only one 2*2 and in E there are two: (2*2)*(2*2)*5*5*5 And we can simplify that as 4.", + "video_name": "BpBh8gvMifs", + "timestamps": [ + 277, + 384 + ], + "3min_transcript": "Square root of 20. Once again, 20 is 2 times 10, which is 2 times 5. So this is the same thing as the square root of 2 times 2, right, times 5. Now, the square root of 2 times 2, that's clearly just going to be 2. It's going to be the square root of this times square root of that. 2 times the square root of 5. And once again, you could probably do that in your head The square root of the 20 is 4 times 5. The square root of 4 is 2. You leave the 5 in the radical. So let's do part D. We have to do the square root of 200. Same process. Let's take the prime factors of it. So it's 2 times 100, which is 2 times 50, which is 2 times 25, which is 5 times 5. Let me scroll to the right a little bit. This is equal to the square root of 2 times 2 times 2 times 5 times 5. Well we have one perfect square there, and we have another perfect square there. So if I just want to write out all the steps, this would be the square root of 2 times 2 times the square root of 2 times the square root of 5 times 5. The square root of 2 times 2 is 2. The square root of 2 is just the square root of 2. The square root of 5 times 5, that's the square root of 25, that's just going to be 5. So you can rearrange these. 2 times 5 is 10. 10 square roots of 2. And once again, this it is irrational. You can't express it as a fraction with an integer and a numerator and the denominator. And if you were to actually try to express this number, it Well let's do part E. The square root of 2000. I'll do it down here. Part E, the square root of 2000. Same exact process that we've been doing so far. Let's do the prime factorization. That is 2 times 1000, which is 2 times 500, which is 2 times 250, which is 2 times 125, which is 5 times 25, which is 5 times 5. And we're done. So this is going to be equal to the square root of 2 times 2-- I'll put it in parentheses-- 2 times 2, times 2 times 2, times 2 times 2, times 5 times 5," + }, + { + "Q": "At 8:36, how come the fraction isn't reduced?\n", + "A": "Zoe.daniele, you can either reduce the fraction, or you can put it into decimal form. Either way, your answer will be correct (make sure to check with your teacher whether he or she wants you to use fractions or decimals in your answer). Hope that helps!", + "video_name": "BpBh8gvMifs", + "timestamps": [ + 516 + ], + "3min_transcript": "which is equal to 1/2. Which is clearly rational. It can be expressed as a fraction. So that's clearly rational. Part G is the square root of 9/4. Same logic. This is equal to the square root of 9 over the square root of 4, which is equal to 3/2. Let's do part H. The square root of 0.16. recognize that, gee, if I multiply 0.4 times 0.4, I'll get this. But I'll show you a more systematic way of doing it, if that wasn't obvious to you. So this is the same thing as the square root of 16/100, right? That's what 0.16 is. So this is equal to the square root of 16 over the square root of 100, which is equal to 4/10, which is equal to 0.4. Let's do a couple more like that. Part I was the square root of 0.1, which is equal to the square root of 1/10, which is equal to the square root of 1 over the square root of 10, which is equal to 1 over-- now, the square root of 10-- 10 is just 2 times 5. So that doesn't really help us much. A lot of math teachers don't like you leaving that radical But I can already tell you that this is irrational. You'll just keep getting numbers. You can try it on your calculator, and it will never repeat. Your calculator will just give you an approximation. Because in order to give the exact value, you'd have to have an infinite number of digits. But if you wanted to rationalize this, just to show you. If you want to get rid of the radical in the denominator, you can multiply this times the square root of 10 over the square root of 10, right? This is just 1. So you get the square root of 10/10. These are equivalent statements, but both of them are irrational. You take an irrational number, divide it by 10, you still have an irrational number. Let's do J. We have the square root of 0.01. This is the same thing as the square root of 1/100. Which is equal to the square root of 1 over the square root of 100, which is equal to 1/10, or 0.1." + }, + { + "Q": "\nat 6:30 why we can't write \"36 35 34 33 32 1 1 1 1\" instead of \"1 1 1 1 32 31 30 29 28\"", + "A": "Because knowing you have all the 1s, you can t have the whole 36 options. The 1s are part of the 36 unique cards, so you can t pick one of them and then add them again, that would make it five 1s. So once you locked the 1s to be in your hand, you only have 32 cards left as the first choice for the non-1s card.", + "video_name": "ccrYD6iX_SY", + "timestamps": [ + 390 + ], + "3min_transcript": "So that's the total number of hands. Now a little bit more of a nuanced thought process is, how do we figure out the number of ways in which the event can happen, in which we can have all four 1's. So let's figure that out. So number of ways-- or maybe we should say this-- number of hands with four 1's. And just as a little bit of a thought experiment, imagine if we were only taking four cards, if a hand only had four cards in it. Well if a hand only had four cards in it, then the number of ways to get a hand with four 1's, there'd only be one way, one combination. You'd just have four 1's. That's the only combination with four 1's, if we were only picking four cards. But here, we're not only picking four cards. One, two, three, four. But the other five cards are going to be different. So one, two, three, four, five. So for the other five cards-- if you imagine this slot-- considering that of the 36 we would have to pick four of them already in order for us to have four 1's. Well, we've used up four of them, so there's 32 possible cards over in that position of the hand. And then there'd be 31 in that position of the hand. And then there'd be 30 because every time we're picking a card, were using it up. And now we only have 30 to pick from. Then we only have 29 to pick from. And then we have 28 to pick from. And just like we did before, we don't care about order. We don't care if we pick the 5 of clubs first or whether we pick the 5 of clubs last. So we shouldn't double count it. five cards can be arranged. So we have to divide this by the different ways that five cards can be arranged. The first card or the first position can be any one of five cards, then four cards, then three cards, then two cards, then one cards. So the number of hands with four 1's is actually just this number. You're actually looking at all of the different ways you can fill up the remaining cards. These four 1's are just going to be four 1's. There's only one way to get that if the remaining cards that's going to give all of the different combinations of having four 1's. So this will be a count of all of the different combinations because all of the different extra stuff that you have will be all of the different hands. Now we know the total number of hands with four 1's is this number. And now we can divide it by the total number of possible hands. And I didn't multiply them out on purpose so that we can cancel things out. So let's do that. Let's take this and divide by that. So let me just copy and paste it." + }, + { + "Q": "In 0 4:04 Sal tells about a factorial. What is the definition of a factorial\n", + "A": "In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example, 5! is 5*4*3*2*1. 76! would be 76*75*74*73*72*71*...*3*2*1. They are essential in probability.", + "video_name": "ccrYD6iX_SY", + "timestamps": [ + 244 + ], + "3min_transcript": "number of hands. Now let's figure out the total number of hands first, because on some level this might be more intuitive and we've actually done this before. Now, the total number of hands, we're picking nine cards. And we're picking them from a set of 36 unique cards. And we've done this many, many times. Let me write this, total number of hands, or total number of possible hands. That's equal to-- you can imagine, you have nine cards to pick from. The first card you pick, it's going to be 1 of 36 cards. Then the next one is going to be 1 of 35. Then the next one is going to be 1 of 34, 33, 32, 31. We're going to do this nine times, one, two, three, four, five, six, seven, eight, and nine. So that would be the total number of hands if order mattered. But we know-- and we've gone over this before-- that we don't care about the order. So we're overcounting here. We're overcounting for all of the different rearrangements that these cards could have. It doesn't matter whether the Ace of diamonds is the first card I pick or the last card I pick. The way I've counted them right now, we are counting those as two separate hands. But they aren't two separate hands, so order doesn't matter. So we have to do is, we have to divide this by the number of ways you can arrange nine things. So you could put nine of the things in the first position, then eight in the second, seven in the third, so forth and so on. It essentially becomes 9 factorial times 2 times 1. And we've seen this multiple times. This is essentially 36 choose 9. This expression right here is the same thing-- just you can relate it to the combinatorics formulas that you might be familiar with-- this is the same thing as 36 factorial over 36 minus 9 factorial-- that's what this orange part is over here-- divided by 9 factorial or over 9 factorial. So that's the total number of hands. Now a little bit more of a nuanced thought process is, how do we figure out the number of ways in which the event can happen, in which we can have all four 1's. So let's figure that out. So number of ways-- or maybe we should say this-- number of hands with four 1's. And just as a little bit of a thought experiment, imagine if we were only taking four cards, if a hand only had four cards in it. Well if a hand only had four cards in it, then the number of ways to get a hand with four 1's, there'd only be one way, one combination. You'd just have four 1's. That's the only combination with four 1's, if we were only picking four cards. But here, we're not only picking four cards." + }, + { + "Q": "\nat 03:55 , why does it equal to 36!/(36-9)!9! ??", + "A": "Before that in the numerator we have 36*35*...*28, and 9! in the denominator. 36! is 36*35*....28*27...*2*1. So in writing 36! as the new numerator, we have to multiply by 27*26...*2*1 both the numerator and denominator. (36-9)! = 27!, which is exactly that factor.", + "video_name": "ccrYD6iX_SY", + "timestamps": [ + 235 + ], + "3min_transcript": "number of hands. Now let's figure out the total number of hands first, because on some level this might be more intuitive and we've actually done this before. Now, the total number of hands, we're picking nine cards. And we're picking them from a set of 36 unique cards. And we've done this many, many times. Let me write this, total number of hands, or total number of possible hands. That's equal to-- you can imagine, you have nine cards to pick from. The first card you pick, it's going to be 1 of 36 cards. Then the next one is going to be 1 of 35. Then the next one is going to be 1 of 34, 33, 32, 31. We're going to do this nine times, one, two, three, four, five, six, seven, eight, and nine. So that would be the total number of hands if order mattered. But we know-- and we've gone over this before-- that we don't care about the order. So we're overcounting here. We're overcounting for all of the different rearrangements that these cards could have. It doesn't matter whether the Ace of diamonds is the first card I pick or the last card I pick. The way I've counted them right now, we are counting those as two separate hands. But they aren't two separate hands, so order doesn't matter. So we have to do is, we have to divide this by the number of ways you can arrange nine things. So you could put nine of the things in the first position, then eight in the second, seven in the third, so forth and so on. It essentially becomes 9 factorial times 2 times 1. And we've seen this multiple times. This is essentially 36 choose 9. This expression right here is the same thing-- just you can relate it to the combinatorics formulas that you might be familiar with-- this is the same thing as 36 factorial over 36 minus 9 factorial-- that's what this orange part is over here-- divided by 9 factorial or over 9 factorial. So that's the total number of hands. Now a little bit more of a nuanced thought process is, how do we figure out the number of ways in which the event can happen, in which we can have all four 1's. So let's figure that out. So number of ways-- or maybe we should say this-- number of hands with four 1's. And just as a little bit of a thought experiment, imagine if we were only taking four cards, if a hand only had four cards in it. Well if a hand only had four cards in it, then the number of ways to get a hand with four 1's, there'd only be one way, one combination. You'd just have four 1's. That's the only combination with four 1's, if we were only picking four cards. But here, we're not only picking four cards." + }, + { + "Q": "From minute 5:07 to 5:53, why does it have to be 1x1x1x1x32x31x30x29x28 instead of 36x35x34x33x32x1x1x1x1? Obviously the end result will be different but I'm asking why or what is the rule, idea, concept or whatever that says to go with the former rather than the latter.\n", + "A": "Even if you used your logic, you are still saying that the 4 ones are drawn at the end, so the first 5 cards could not be ones, so it is still 32*31*30*29*28*1*1*1*1.", + "video_name": "ccrYD6iX_SY", + "timestamps": [ + 307, + 353 + ], + "3min_transcript": "So we're overcounting here. We're overcounting for all of the different rearrangements that these cards could have. It doesn't matter whether the Ace of diamonds is the first card I pick or the last card I pick. The way I've counted them right now, we are counting those as two separate hands. But they aren't two separate hands, so order doesn't matter. So we have to do is, we have to divide this by the number of ways you can arrange nine things. So you could put nine of the things in the first position, then eight in the second, seven in the third, so forth and so on. It essentially becomes 9 factorial times 2 times 1. And we've seen this multiple times. This is essentially 36 choose 9. This expression right here is the same thing-- just you can relate it to the combinatorics formulas that you might be familiar with-- this is the same thing as 36 factorial over 36 minus 9 factorial-- that's what this orange part is over here-- divided by 9 factorial or over 9 factorial. So that's the total number of hands. Now a little bit more of a nuanced thought process is, how do we figure out the number of ways in which the event can happen, in which we can have all four 1's. So let's figure that out. So number of ways-- or maybe we should say this-- number of hands with four 1's. And just as a little bit of a thought experiment, imagine if we were only taking four cards, if a hand only had four cards in it. Well if a hand only had four cards in it, then the number of ways to get a hand with four 1's, there'd only be one way, one combination. You'd just have four 1's. That's the only combination with four 1's, if we were only picking four cards. But here, we're not only picking four cards. One, two, three, four. But the other five cards are going to be different. So one, two, three, four, five. So for the other five cards-- if you imagine this slot-- considering that of the 36 we would have to pick four of them already in order for us to have four 1's. Well, we've used up four of them, so there's 32 possible cards over in that position of the hand. And then there'd be 31 in that position of the hand. And then there'd be 30 because every time we're picking a card, were using it up. And now we only have 30 to pick from. Then we only have 29 to pick from. And then we have 28 to pick from. And just like we did before, we don't care about order. We don't care if we pick the 5 of clubs first or whether we pick the 5 of clubs last. So we shouldn't double count it." + }, + { + "Q": "\nat 1:26, i think you meant to write DCE = BAE. (You wrote DEC = BAE)", + "A": "Indeed, it should be DCE = BAE.", + "video_name": "TErJ-Yr67BI", + "timestamps": [ + 86 + ], + "3min_transcript": "So we have a parallelogram right over here. And what I want to prove is that its diagonals bisect each other. So the first thing that we can think about-- these aren't just diagonals. These are lines that are intersecting, parallel lines. So you can also view them as transversals. And if we focus on DB right over here, we see that it intersects DC and AB. And since we know that they're parallel-- this is a parallelogram-- we know the alternate interior angles must be congruent. So that angle must be equal to that angle there. And let me make a label here. Let me call that middle point E. So we know that angle ABE must be congruent to angle CDE by alternate interior angles of a transversal intersecting parallel lines. Now, if we look at diagonal AC-- or we should call it transversal AC-- we can make the same argument. These two lines are parallel. So alternate interior angles must be congruent. So angle DEC must be-- so let me write this down-- angle DEC must be congruent to angle BAE, for the exact same reason. Now we have something interesting, if we look at this top triangle over here and this bottom triangle. We have one set of corresponding angles that are congruent. We have a side in between that's going to be congruent. Actually, let me write that down explicitly. We know-- and we proved this to ourselves in the previous video-- that parallelograms-- not only are opposite sides parallel, they are also congruent. So we know from the previous video that that side is equal to that side. So let me go back to what I was saying. We have two sets of corresponding angles that are congruent, we have a side in between of corresponding angles that are congruent. So we know that this triangle is congruent to that triangle by angle-side-angle. So we know that triangle-- I'm going to go from the blue to the orange to the last one-- triangle ABE is congruent to triangle-- blue, orange, then the last one-- CDE, by angle-side-angle congruency. Now, what does that do for us? Well, we know if two triangles are congruent, all of their corresponding features, especially all of their corresponding sides, are congruent. So we know that side EC corresponds to side EA. Or I could say side AE corresponds to side CE. They're corresponding sides of congruent triangles," + }, + { + "Q": "I really don't get it. At 1:42, why do you have to do f(x) times g(x)^-1?\n", + "A": "Dividing by a number is the same as multiplying by the inverse of that number, which you can get by setting the exponent equal to -1 : Thus, the inverse of B is B^-1 since B^-1 = 1/B^1 = 1/B. Therefore, A/B = A*B^-1. That s why f(x)/g(x) = f(x)*g(x)^-1.", + "video_name": "ho87DN9wO70", + "timestamps": [ + 102 + ], + "3min_transcript": "We already know that the product rule tells us that if we have the product of two functions-- so let's say f of x and g of x-- and we want to take the derivative of this business, that this is just going to be equal to the derivative of the first function, f prime of x, times the second function, times g of x, plus the first function, so not even taking its derivative, so plus f of x times the derivative of the second function. So two terms, in each term we take the derivative of one of the functions and not the other, and then we switch. So over here is the derivative of f, not of g. Here it's the derivative of g, not of f. This is hopefully a little bit of review. This is the product rule. Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule. I have mixed feelings about the quotient rule. If you know it, it might make some operations a little bit rule. And I frankly always forget the quotient rule, and I just rederive it from the product rule. So let's see what we're talking about. So let's imagine if we had an expression that could be written as f of x divided by g of x. And we want to take the derivative of this business, the derivative of f of x over g of x. The key realization is to just recognize that this is the same thing as the derivative of-- instead of writing f of x over g of x, we could write this as f of x times g of x to the negative 1 power. And now we can use the product rule with a little bit of the chain rule. What is this going to be equal to? Well, we just use the product rule. It's the derivative of the first function right over here-- so it's going to be f prime of x-- times just the second function, which is just which is just f of x, times the derivative of the second function. And here we're going to have to use a little bit of the chain rule. The derivative of the outside, which we could kind of view as something to the negative 1 power with respect to that something, is going to be negative 1 times that something, which in this case is g of x to the negative 2 power. And then we have to take the derivative of the inside function with respect to x, which is just g prime of x. And there you have it. We have found the derivative of this using the product rule and the chain rule. Now, this is not the form that you might see when people are talking about the quotient rule in your math book. So let's see if we can simplify this a little bit. All of this is going to be equal to-- we can write this term right over here as f prime of x over g of x." + }, + { + "Q": "\nAt 3:54 Sal says f(x)=1/(2+2cos(theta)), but it should be f(theta)=\u00e2\u0080\u00a6.\nIt's not a function of x, but it is a function of theta.", + "A": "Yup You got it right Sal Made A Mistake, it should be f(theta).", + "video_name": "d8qtbGMB2gI", + "timestamps": [ + 234 + ], + "3min_transcript": "zero over zero. Well, we can, we got some trig functions here, so maybe we can use some of our trig identities to simplify this. And the one that jumps out at me is that we have the sine squared of theta and we know from the Pythagorean, Pythagorean Identity in Trigonometry, it comes straight out of the unit circle definition of sine and cosine. We know that, we know that sine squared theta plus cosine squared theta is equal to one or, we know that sine squared theta is one minus cosine squared theta. One minus cosine squared theta. So we could rewrite this. This is equal to one minus cosine theta over two times one minus cosine squared theta. Now, this is one minus cosine theta. This is a one minus cosine squared theta, so it's not completely obvious yet of how you can simplify it, as a difference of squares. If you view this as, if you view this as A squared minus B squared, we know that this can be factored as A plus B times A minus B. So I could rewrite this. This is equal to one minus cosine theta over two times, I could write this as one plus cosine theta times one minus cosine theta. One plus cosine theta times one minus, one minus cosine theta. And now this is interesting. I have one minus cosine theta in the numerator and I have a one minus cosine theta in the denominator. Now we might be tempted to say, \"Oh, let's just cross that out with that \"and we would get, we would simplify it \"and get F of X is equal to one over \"and we could distribute this two now.\" We could say, \"Two plus two cosine theta.\" \"Well, aren't these the same thing?\" And we would be almost right, because F of X, this one right over here, this, this is defined this right over here is defined when theta is equal to zero, while this one is not defined when theta is equal to zero. When theta is equal to zero, you have a zero in the denominator. And so what we need to do in order for this F of X or in order to be, for this to be the same thing, we have to say, theta cannot be equal to zero. But now let's think about the limit again. Essentially, what we want to do is we want to find the limit as theta approaches zero of F of X. And we can't just do direct substitution into, if we do, if we really take this seriously, 'cause we're gonna like, \"Oh well, if I try to put zero here, \"it says theta cannot be equal to zero \"F of X is not defined at zero.\" This expression is defined at zero but this tells me, \"Well, I really shouldn't apply zero to this function.\" But we know that if we can find another function" + }, + { + "Q": "2:12 to 2:23 is so unclear, why do they take out the 2s and 3s? How did they get the numbers? I took the test and did it another way...\n", + "A": "that is how it goes", + "video_name": "-UagBvxCReA", + "timestamps": [ + 132, + 143 + ], + "3min_transcript": "All right. What we've got here are 12 pirates. They're going to divide out a treasure chest of gold. And here's how they're going to do it. First pirate's going to come along, take 1/12 of the gold that's in the chest. Second pirate's going to come along, take 2/12 of whatever's left after the first pirate is Third pirate's going to take 3/12 of whatever's left after the second pirate finished, and on, and on, and on. Let's see what happens here. Each pirate gets a positive whole number of coins. And the number of coins that was in the chest is the smallest number of coins for which it's possible for each pirate to get, a positive number of coins, a positive whole number of coins using this process. We're going to start with x because x marks the spot. x is the number of coins that was in the chest at the beginning. And the first pirate comes along, takes 1/12. That leaves 11/12 remaining for the next pirate who And the next pirate, second pirate takes 2/12 and leaves 10/12 of what was there So she gets there and there's this much. She's going to leave 10/12 of this amount for the next pirate. The next pirate comes along, takes 3/12, leaves 9/12 of this for the following pirate, and on, and on, and on we go until we get to the last few pirates. The 11th pirate takes 11/12, leaves 1/12 of what was there for the last pirate, who comes along and takes everything that's left. Well, that's what we want to figure out. How much does the last pirate receive? So we want to figure out what the value of this expression We can write this a lot shorter as x times 11 factorial over 12 to the 11th. And we want to figure out what this is. Now, x is the smallest value that makes this an integer. Actually, x is the smallest value that makes sure each pirate gets an integer number of coins. I'm not going to worry about that right now. I'm just going to worry about the last pirate and figure what the last pirate gets. well, this will just come out to be an integer. But 11 factorial is not any of these choices. So we can simplify this fraction. We can take out all the factors of 2 and 3 from this 11 factorial and see what's left. We're going to be left with a factor of 11. And then we're going to have two 5's from the 5 and the 10. And we're going to have a 7 sitting in there. And then we need to figure out well, we're going to simplify this fraction, take out factors of 2. We could stop right here, just compute this and call that the answer. But I'm a little bothered by that whole every pirate has to get an integer number of coins thing. But let's go ahead and simplify this fraction. The number of 2's in 11 factorial, the number of factors of 2, you get 2, 4, 6, 8, 10. That's 5. You get an extra one from the 4, two extra ones from the 8. 8 factors of 2 up here are 22 down there. That leaves us 2 to the 14th. And then the factors of 3, you have 3, 6, and 9 up there. You get an extra factor of 3 in the 9." + }, + { + "Q": "\nAt 1:30, how did he get x((4x)2", + "A": "You mean, how he got to x( (4x)^2 + 2*4*3 + 3^2) ?", + "video_name": "BFW2lHobO4E", + "timestamps": [ + 90 + ], + "3min_transcript": "- [Voiceover] So let's say that we've got the polynomial 16x to the third plus 24x squared plus nine x. Now what I'd like you to do is pause the video and see if you could factor this polynomial completely. Now let's work through it together. So the first thing that you might notice is that all of the terms are divisible by x so we can actually factor out an x. So let's do that, and actually, if we look at these coefficients, it looks like, let's see, it looks like they don't have any common factors other than one. So it looks like the largest monomial that we can factor out is just going to be an x. So let's do that. Let's factor out an x. So then this is gonna be x times. When you factor out an x from 16x to the third you're gonna be left with 16x squared, and then plus 24x and then plus nine. Now this is starting to look interesting so let me just rewrite it. It's gonna be x times. This part over here looks interesting this looks like a perfect square. Let me write it out. 16x squared. That's the same thing as four x squared, and then we have a nine over there, which is clearly a perfect square. That is three squared. Three squared. And when we look at this 24x, we see that it is four times three times two, and so we can write it as, let me write it this way. So this is going to be plus two times four times three x. So let me make it, so two times four times three times three x. Now why did I take the trouble, why did I take the trouble of writing everything like this? Because we see that it fits the pattern for a perfect square. What do I mean by that? Well in previous videos, we saw that Ax plus B, and you were to square it, you're going to get Ax squared plus two ABx plus B squared, and we have that form right over here. This is the Ax squared. Let me do the same color. The Ax squared. Ax squared. We have the B squared. You have the B squared. And then you have the two ABx. Two ABx right over there. So this section, this entire section, we can rewrite as being we know it, what A and B are. A is four and B is three so this is going to be Ax, so four x plus four x plus b, which we know to be three. That whole thing, that whole thing squared, and now we" + }, + { + "Q": "\nWould using a different trig identity work @4:00. For instance, we could use the \"Product-to-Sum\" trig I.D of Sin(t-Tau)Cos(Tau). Then we could just split the integrals and it would make life a bit nicer right?\nTrig I.D.: sin(u)cos(v) = sin(u + v) + sin(u \u00e2\u0088\u0092 v)", + "A": "Yes, you can use that identity, and the integration does indeed is simpler. But that identity is not so easy to memorise, Sal decided to use only the most basic of identities.", + "video_name": "IW4Reburjpc", + "timestamps": [ + 240 + ], + "3min_transcript": "one of these things? And to kind of give you that comfort, let's actually compute a convolution. Actually, it was hard to find some functions that are very easy to analytically compute, and you're going to find that we're going to go into a lot of trig identities to actually compute this. But if I say that f of t, if I define f of t to be equal to the sine of t, and I define cosine of t-- let me do it in orange-- or I define g of t to be equal to the cosine of t. Now let's convolute the two functions. So the convolution of f with g, and this is going to be a function of t, it equals this. I'm just going to show you how to apply this integral. integral from 0 to t of f of t minus tau. This is my f of t. So it's is going to be sine of t minus tau times g of tau. Well, this is my g of t, so g of tau is cosine of tau, cosine of tau d tau. So that's the integral, and now to evaluate it, we're going to have to break out some trigonometry. So let's do that. This almost is just a very good trigonometry and integration review. So let's evaluate this. But I wanted to evaluate this in this video because I want to show you that this isn't some abstract thing, that you can actually evaluate these functions. So the first thing I want to do-- I mean, I don't know what the antiderivative of this is. It's tempting, you see a sine and a cosine, maybe they're the derivatives of each other, but this is the sine of t minus tau. So let me rewrite that sine of t minus tau, and we'll just use the trig identity, that the sine of t minus tau is the sine of tau times the cosine of t. And actually, I just made a video where I go through all of these trig identities really just to review them for myself and actually to make a video in better quality on them as well. So if we make this subsitution, this you'll find on the inside cover of any trigonometry or calculus book, you get the convolution of f and g is equal to-- I'll just write that f-star g; I'll just write it with that-- is equal to the integral from 0 to t of, instead of sine of t minus tau, I'm going to write this thing right there. So I'm going to write the sine of t times the cosine of tau minus the sine of tau times the cosine of t, and then all" + }, + { + "Q": "in the first example at 0:42, what where the m's and the b's?\n", + "A": "Here, m and b are constants; in a linear equation, m is traditionally used to represent slope, and b is used to represent the y-intercept. In other words, they re just the good old slope-intercept form pieces from algebra.", + "video_name": "zid7J4EhZN8", + "timestamps": [ + 42 + ], + "3min_transcript": "- So let's get a little bit more comfort in our understanding of what a differential equation even is. So here we have a differential equation. We haven't started exploring how we find the solutions for a differential equations yet. But let's just say you saw this, and someone just walked up to you on the street and says, \"Hey, I will give you a clue, \"that there's a solution to this differential equation \"that is essentially a linear function, \"where y is equal to mx plus b, \"and you just need to figure out \"the m's and the b's, or maybe the m and the b \"that makes this linear function \"satisfy this differential equation.\" What I now encourage you to do, is pause the video and see if you can do it. So I'm assuming you have paused it, and had a go at it. So let's think this through together. If we know that this kind of a solution can be described in this way, we need to figure out some m's and b's here. This is telling us that if we were to take the derivative of this with respect to x, if we take the derivative of mx plus b with respect to x, that that should be equal to negative 2 times x well, we know y is this thing, minus 5. And that should be true for all x's, in order for this to be a solution to this differential equation. Remember, the solution to a differential equation is not a value or a set of values. It is a function or a set of functions. So in order for this to satisfy this differential equation, it needs to be true for all of these x's here. So let's work through it. Let's figure out first what our dy dx is. So dy dx. We'll just take the derivative here with respect to x dy dx is derivative of mx with respect to x, is just going to be m. And of course derivative of b with respect to x, just a constant, so it's just going to be zero. So dy dx is m. So we could write m is equal to negative 2x, is equal to negative 2x, plus 3 times, and instead of putting y there I could write mx plus b. Remember y is equal to mx plus b. this has to be true for all x's. mx plus b, and then of course we have the minus 5. So if you weren't able to solve it the first time, I encourage you to start from here, and now figure out what m and b needs to be in order for this equation right over here, in order for this to be true for all x's. In order for this to be true for all x's. So assuming you have paused again and had a go at it, let's just keep algebraically manipulating this. I'll just switch to one color here. So we have m. m is equal to negative 2x plus, if we distribute this 3 we're going to have 3mx plus 3b, and then of course we're going to have minus 5. And now we can group the x terms. So if we were to group, if we were to group ... Let me find a new color here, maybe this blue. So if we were to take these two and add them together" + }, + { + "Q": "\nat 3:14, can you explain the use of the chain rule that results in 2y * y'? thank you.", + "A": "The derivative of y^2 is 2y * (dy/dx). (dy/dx) = y so then 2y * (dy/dx) = 2y * y", + "video_name": "ZtI94pI4Uzc", + "timestamps": [ + 194 + ], + "3min_transcript": "with different notation. So let's take the derivative of this thing right over here. Well we're going to apply the chain rule. Actually, we're going to apply the chain rule multiple times here. The derivative of e to the something with respect to that something is going to be e to the something times the derivative of that something with respect to x. So times the derivative of xy squared. So that's our left-hand side. We aren't done taking the derivative yet. And on our right-hand side, the derivative of x is just 1. And the derivative with respect to x of y is just going to be minus-- or I could write-- negative dy dx. But instead of writing dy dx, I'm going to write y prime. As you can tell, I like this notation and this notation more because it makes it explicit Here, we just have to assume that we're taking the derivative with respect to x. Here, we have to assume that's the derivative of y with respect to x. But anyway let's stick with this notation right over here. Actually, let me make all of my y primes, all my derivatives of y with respect to x, let me make them pink so I keep track of them. So once again, this is going to be equal to e to the xy squared times the derivative of this. Well the derivative of this, we can just use the product and actually a little bit of the chain rule So the derivative of x is just 1 times the second function. So it's going to be times y squared. And then to that, we're going to add the product of the first function which is this x times the derivative of y squared with respect to x. Well that's going to be the derivative of y squared times the derivative of y with respect to x, which we are now writing as y prime. And then that's going to be equal to 1 minus y prime. And like we've been doing, we now have to just solve for y prime. So let's distribute this exponential, this e to the xy squared. And we get e, or maybe I should say y squared times e to the xy squared. So that's that. Plus 2xye to the xy squared. y prime, the derivative of y with respect to x, is equal to 1 minus the derivative of y with respect" + }, + { + "Q": "At 6:09, Sal describes taking the integral at a specific value c. What would happen if you took the derivate of the Dirac delta function combined with a certain function at value c and then applied the Laplace transform?\n", + "A": "The derivative of the Dirac delta function is undefined at c, so it would not work very well.", + "video_name": "vhfjEpQWWeE", + "timestamps": [ + 369 + ], + "3min_transcript": "Let me draw this, what we're trying to do. So let me draw what we're trying to take the integral of. And we only care from zero to infinity, so I'll only do it from zero to infinity. And I'll assume that c is greater than zero, that the delta function pops up someplace in the positive t-axis. So what is this first part going to look like? What is that going to look like? e to the minus st times f of t? I don't know. It's going to be some function. e to the minus st starts at 1 and drops down, but we're multiplying it times some arbitrary function, so I'll just draw it like this. Maybe it looks something like this. This right here is e to the minus st times f of t. And the f of t is what kind of gives it its arbitrary shape. Fair enough. Now, let's graph our Dirac delta function. With zero everywhere except right at c, right at c right there, it pops up infinitely high, but we only draw an I mean, normally when you graph things you don't draw arrows, but this arrow shows that the area under this infinitely high thing is 1. So we do a 1 there. So if we multiply this, we care about the area under this whole thing. When we multiply these two functions, when we multiply this times this times the delta function, this is-- let This is the delta function shifted to c. If I multiply that times that, what do I get? This is kind of the key intuition here. Let me redraw my axes. Let me see if I can do it a little bit straighter. Don't judge me by the straightness of my axes. So that's t. So what happens when I multiply these two? Everywhere, when t equals anything other than c, the Dirac delta function is zero. So it's zero times anything. I don't care what this function is going to do, it's going to be zero. So it's going to be zero everywhere, except something At t equals c, what's the value of the function? Well, it's going to be the value of the Dirac delta function. It's going to be the Dirac delta function times whatever height this is. This is going to be this point right here or this right there, that point. This is going to be this function evaluated at c. I'll mark it right here on the y-axis, or on the f of t, whatever you want to call it. This is going to be e to the minus sc times f of c. All I'm doing is I'm just evaluating this function at c, so that's the point right there. So if you take this point, which is just some number, it could be 5, 5 times this, you're just getting 5 times the Dirac delta function. Or in this case, it's not 5. It's this little more abstract thing. I could just draw it like this. When I multiply this thing times my little delta function" + }, + { + "Q": "In the first equation, Sal simplifies the equation to: (7x - 2) / (15 - 5/x). He states at 2:32 that 7x will approach negative infinity, however he simplified this value from 7x^2 which by definition will always turn negative numbers positive. Shouldn't he have simplified the equation to state: 7|x|?\n", + "A": "Sal simplified the expression down to it s dominate terms: 7x/15. If you are going to use the 7x^2, then you also need to use the 15x in the denominator. Yes, 7x^2 would be positive, but 15x would be negative. A positive / a negative = a negative. Hope this helps.", + "video_name": "Vtcmyr5IGYY", + "timestamps": [ + 152 + ], + "3min_transcript": "by one over x, or another way of thinking about it is we're dividing both the numerator and the denominator by x. And if we're doing the same thing to the numerator and the denominator, if we're multiplying or dividing them by the same value, I should say, well then, I'm just really just multiplying it by one. So, I'm not changing its value. This will make it a little bit more interesting, and a little bit easier for us to think about what happens when x becomes very, very, very negative. So, 7x-squared divided by x, or being multiplied by one over x, is going to be 7x. 2x times one over x, or 2x divided by x, is just two. And then all of that over 15x divided by x, or 15x over x, is just going to be 15. And then you have five over x. Five times one over x is equal to five over x. Minus five over x. Now, this is equivalent, for our purposes, to think about what happens when x gets very, very, very, very negative. Well, when x gets very, very, very, very, very, very, very negative, this is going to become a very large negative number. You subtract two from it, it really won't matter much. You divide that by 15, well, that's not gonna matter much. And this is just going to become very, very, very small. You're taking five and you're dividing it by ever-larger negative numbers, or more and more negative numbers. So, this right over here is gonna go to zero. This thing over here is gonna go towards infinity. Or, I should say, it's gonna go towards negative infinity. Seven times a negative trillion, seven times a negative googol, seven times a negative googolplex, we're getting more and more negative numbers, this is gonna get, this is going to approach negative infinity. Doesn't matter that you're subtracting two from that. In fact, that'll get even more negative. And it doesn't matter if you then divide that by 15, you're still approaching negative infinity. If you had a arbitrarily negative number, an arbitrarily negative number. And, so, you could say that this is going to go to negative infinity. Now, another way that you could've thought about it. This is actually how I do think about it when I'm trying to, when I see these types of problems. I say, well which terms in the numerator and the denominator are going to dominate? And what do I mean by \"dominate\"? Well, as x gets very positive or x gets very negative, another way to think about it is the magnitude of x gets large, the absolute value of x gets large. The higher degree terms are going to grow much faster than the lesser degree terms. And so, we could say that for large x, for large x, and when I say \"large\" I mean high absolute value. High absolute value. And if we're going to negative infinity, that's high absolute value. So, f of x is going to be approximately equal to" + }, + { + "Q": "\nAt 2:00, why would I need to see if the sides are parallel to each other?", + "A": "To know if any of the shapes are the right answer. And to know for sure that what you think is right", + "video_name": "vsgrWDLEzcQ", + "timestamps": [ + 120 + ], + "3min_transcript": "Classify quadrilateral ABCD. Choose the option that best suits the quadrilateral. We're going to pick whether it's a square, rhombus, rectangle, parallelogram, trapezoid, none of the above. And I'm assuming we're going to pick the most specific one possible, because obviously all squares are rhombuses, or rhombi, I guess you'd say. Not all rhombi are squares. All squares are also rectangles. All squares, rhombi, and rectangles are parallelograms. So we want to be as specific as possible in picking this. So let's see, point A is at 1 comma 6. I encourage you to pause this video and actually try this on your own before seeing how I do it, but I'll just proceed. So that's point A right over there. Point B is at negative 5 comma 2. That's point B. Point C is at-- just had some carbonated water, so some air is coming up. Point C is at negative 7 comma 8. And then finally, point D is at 2 comma 11. And actually that kind of goes off the screen. This is 10, 11 would be right like this. So that would be 2 comma 11. If we were to extend this, this is 10 and this is 11 right up here. 2 comma 11. So let's see what this quadrilateral looks like. You have this line right over here, this line right over there, that line right over there, and then you have this line like this, and then you have this like this. So right off the bat, well it's definitely a quadrilateral. I have four sides. But the key question, are any of these sides parallel to any of the other sides? So just looking at it, side CB is clearly not parallel to AD. Now it also looks like CD is not parallel to BA, but maybe I just drew it badly. Maybe they actually are parallel. So let's see if we can verify that. So the way to tell whether two things are parallel is to actually figure out their slopes. So let's first figure out the slope of AB, or BA. So let's figure out our slope here. Your slope is going to be the change in y over your change in x. And in this case, you could think of it as we're starting at the point negative 5 comma 2, and we're ending at the point 1 comma 6. So what's our change in y? Our change in y, we go from 2-- we're going from 2 all the way to 6." + }, + { + "Q": "\nAt 7:17, how does Sal choose the starting point? Does it matter? Could it be the other point?", + "A": "The starting/ending points could certainly be any as long as they are on the line c. I think these were chosen because they both had integer values for x and y and thus made the computation a little easier.", + "video_name": "Iqws-qzyZwc", + "timestamps": [ + 437 + ], + "3min_transcript": "So our change in y was 3 when our change in x was 6. Now, one of the things that confuses a lot of people is how do I know what order to-- how did I know to do the 0 first and the negative 6 second and then the 1 first and then the negative 2 second. And the answer is you could've done it in either order as long as you keep them straight. So you could have also have done change in y over change in x. We could have said, it's equal to negative 2 minus 1. So we're using this coordinate first. Negative 2 minus 1 for the y over negative 6 minus 0. Notice this is a negative of that. That is the negative of that. But since we have a negative over negative, they're going to cancel out. So this is going to be equal to negative 3 over negative 6. The negatives cancel out. This is also equal to 1/2. So the important thing is if you use this y-coordinate x-coordinate first as well. If you use this y-coordinate first, as we did here, then you have to use this x-coordinate first, as you did there. You just have to make sure that your change in x and change in y are-- you're using the same final and starting points. Just to interpret this, this is saying that for every minus 6 we go in x. So if we go minus 6 in x, so that's going backwards, we're going to go minus 3 in y. But they're essentially saying the same thing. The slope of this line is 1/2. Which tells us for every 2 we travel in x, we go up 1 in y. Or if we go back 2 in x, we go down 1 in y. That's what 1/2 slope tells us. Notice, the line with the 1/2 slope, it is less steep than the line with a slope of 3. Let's do line c right here. I'll do it in pink. Let's say that the starting point-- I'm just picking this arbitrarily. Well, I'm using these points that they've drawn here. The starting point is at the coordinate negative 1, 6 and that my finishing point is at the point 5, negative 6. Our slope is going to be-- let me write this-- slope is going to be equal to change in x-- sorry, change in y. I'll never forget that. Change in y over change in x. Sometimes it's said rise over run. Run is how much you're moving in the horizontal direction. Rise is how much you're moving in the vertical direction. Then we could say our change in y is our finishing y-point minus our starting y-point. This is our finishing y-point." + }, + { + "Q": "At 2:27,Can it still be -12 over -4? (-12/-4)\n", + "A": "yes it can be (-12/-4)", + "video_name": "Iqws-qzyZwc", + "timestamps": [ + 147 + ], + "3min_transcript": "In this video I'm going to do a bunch of example slope problems. Just as a bit of review, slope is just a way of measuring the inclination of a line. And the definition-- we're going to hopefully get a good working knowledge of it in this video-- the definition of it is a change in y divided by change in x. This may or may not make some sense to you right now, but as we do more and more examples, I think it'll make a good amount of sense. Let's do this first line right here. Line a. Let's figure out its slope. They've actually drawn two points here that we can use as the reference points. So first of all, let's look at the coordinates of those points. So you have this point right here. What's its coordinates? Its x-coordinate is 3. Its y-coordinate is 6. And then down here, this point's x-coordinate is negative 1 and its y-coordinate is negative 6. So there's a couple of ways we can think about slope. We could say change in y-- so slope is change in y over change in x. We can figure it out numerically. I'll in a second draw it graphically. So what's our change in y? Our change in y is literally how much did our y values change going from this point to that point? So how much did our y values change? Our y went from here, y is at negative 6 and it went all the way up to positive 6. So what's this distance right here? It's going to be your end point y value. It's going to be 6 minus your starting point y value. Minus negative 6 or 6 plus 6, which is equal to 12. You say one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. So when we changed our y value by 12, we had to change our x same change in y? Well we went from x is equal to negative 1 to x is equal to 3. Right? x went from negative 1 to 3. So we do the end point, which is 3 minus the starting point, which is negative 1, which is equal to 4. So our change in y over change in x is equal to 12/4 or if we want to write this in simplest form, this is the same thing as 3. Now the interpretation of this means that for every 1 we move over-- we could view this, let me write it this way. Change in y over change in x is equal to-- we could say it's 3 or we could say it's 3/1. Which tells us that for every 1 we move in the positive x-direction, we're going to move up 3 because this is a positive 3 in the y-direction. You can see that. When we moved 1 in the x, we moved up 3 in the y. When we moved 1 in the x, we moved up 3 in the y." + }, + { + "Q": "At 2:29 isn't it suppose to be y2-y1 over x2-x1?\n", + "A": "Yes . . . and that s what Sal did. y2 is 6, and y1 is -6, so we get y2 - y1 = 6 - (-6) = 12 For the change in x Sal basically just counted the number of squares horizontally between the two points, but that s the same as finding x2 - x1.", + "video_name": "Iqws-qzyZwc", + "timestamps": [ + 149 + ], + "3min_transcript": "In this video I'm going to do a bunch of example slope problems. Just as a bit of review, slope is just a way of measuring the inclination of a line. And the definition-- we're going to hopefully get a good working knowledge of it in this video-- the definition of it is a change in y divided by change in x. This may or may not make some sense to you right now, but as we do more and more examples, I think it'll make a good amount of sense. Let's do this first line right here. Line a. Let's figure out its slope. They've actually drawn two points here that we can use as the reference points. So first of all, let's look at the coordinates of those points. So you have this point right here. What's its coordinates? Its x-coordinate is 3. Its y-coordinate is 6. And then down here, this point's x-coordinate is negative 1 and its y-coordinate is negative 6. So there's a couple of ways we can think about slope. We could say change in y-- so slope is change in y over change in x. We can figure it out numerically. I'll in a second draw it graphically. So what's our change in y? Our change in y is literally how much did our y values change going from this point to that point? So how much did our y values change? Our y went from here, y is at negative 6 and it went all the way up to positive 6. So what's this distance right here? It's going to be your end point y value. It's going to be 6 minus your starting point y value. Minus negative 6 or 6 plus 6, which is equal to 12. You say one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. So when we changed our y value by 12, we had to change our x same change in y? Well we went from x is equal to negative 1 to x is equal to 3. Right? x went from negative 1 to 3. So we do the end point, which is 3 minus the starting point, which is negative 1, which is equal to 4. So our change in y over change in x is equal to 12/4 or if we want to write this in simplest form, this is the same thing as 3. Now the interpretation of this means that for every 1 we move over-- we could view this, let me write it this way. Change in y over change in x is equal to-- we could say it's 3 or we could say it's 3/1. Which tells us that for every 1 we move in the positive x-direction, we're going to move up 3 because this is a positive 3 in the y-direction. You can see that. When we moved 1 in the x, we moved up 3 in the y. When we moved 1 in the x, we moved up 3 in the y." + }, + { + "Q": "At 4:39, Sal didn't label the x values as x1 and x2, and the y-values as y1 and y2. Why is that?\n", + "A": "You don t need to label the points as x1, x2, y1 and y2. You just need to know which is which.", + "video_name": "Iqws-qzyZwc", + "timestamps": [ + 279 + ], + "3min_transcript": "move 6 in the y. 6/2 is the same thing as 3. So this 3 tells us how quickly do we go up as we increase x. Let's do the same thing for the second line on this graph. Graph b. Same idea. I'm going to use the points that they gave us. But really you could use any points on that line. So let's see, we have one point here, which is the point 0, 1. You have 0, 1. And then the starting point-- we could call this the finish point-- the starting point right here, we could view it as x is negative 6 and y is negative 2. So same idea. What is the change in y given some change in x? So let's do the change in x first. So what is our change in x? So in this situation, what is our change in x? delta x. It's one, two, three, four, five, six. It's going to be 6. But if you didn't have a graph to count from, you could literally take your finishing x-position, so it's 0, and subtract from that your starting x-position. 0 minus negative 6. So when your change in x is equal to-- so this will be 6-- what is our change in y? Remember we're taking this as our finishing position. This is our starting position. So we took 0 minus negative 6. So then on the y, we have to do 1 minus negative 2. What's 1 minus negative 2? That's the same thing as 1 plus 2. That is equal to 3. So it is 3/6 or 1/2. So notice, when we moved in the x-direction by 6, we moved So our change in y was 3 when our change in x was 6. Now, one of the things that confuses a lot of people is how do I know what order to-- how did I know to do the 0 first and the negative 6 second and then the 1 first and then the negative 2 second. And the answer is you could've done it in either order as long as you keep them straight. So you could have also have done change in y over change in x. We could have said, it's equal to negative 2 minus 1. So we're using this coordinate first. Negative 2 minus 1 for the y over negative 6 minus 0. Notice this is a negative of that. That is the negative of that. But since we have a negative over negative, they're going to cancel out. So this is going to be equal to negative 3 over negative 6. The negatives cancel out. This is also equal to 1/2. So the important thing is if you use this y-coordinate" + }, + { + "Q": "In the end of 10:00 , we can see, slope of one line is -2 (negative two) and another is +1. So which one's rate of change will be higher ?\n", + "A": "The higher rate of change should be -2 as the rate of change deals with the magnitude of the slope.", + "video_name": "Iqws-qzyZwc", + "timestamps": [ + 600 + ], + "3min_transcript": "minus our starting x-point. If that confuses you, all I'm saying is, it's going to be equal to our finishing y-point is negative 6 minus our starting y-point, which is 6, over our finishing x-point, which is 5, minus our starting x-point, which is negative 1. So this is equal to negative 6 minus 6 is negative 12. 5 minus negative 1. That is 6. So negative 12/6. That's the same thing as negative 2. Notice we have a negative slope here. That's because every time we increase x by 1, we go down in the y-direction. So this is a downward sloping line. It's going from the top left to the bottom right. As x increases, the y decreases. And that's why we got a negative slope. This line over here should have a positive slope. So I'll use the same points that they use right over there. So this is line d. Slope is equal to rise over run. How much do we rise when we go from that point to that point? We could do it this way. We are rising-- I could just count it out. We are rising one, two, three, four, five, six. We are rising 6. How much are we running? We are running-- I'll do it in a different color. We're running one, two, three, four, five, six. We're running 6. So our slope is 6/6, which is 1. Which tells us that every time we move 1 in the x-direction-- positive 1 in the x-direction-- we go positive 1 in the y-direction. For every x, if we go negative 2 in the x-direction, we're going to go negative 2 in the y-direction. y in this slope. Notice, that was pretty easy. If we wanted to do it mathematically, we could figure out this coordinate right there. That we could view as our starting position. Our starting position is negative 2, negative 4. Our finishing position is 4, 2. So our slope, change in y over change in x. I'll take this point 2 minus negative 4 over 4 minus negative 2. 2 minus negative 4 is 6. Remember that was just this distance right there. Then 4 minus negative 2, that's also 6. That's that distance right there. We get a slope of 1. Let's do another one." + }, + { + "Q": "\nAt 2:11, isn't it a change of -4 because y=-4?", + "A": "y=-4 is like y= 0x -4. In order for the x to be gone you would need a zero at its coefficient. since the coefficient of the x is the slope then the slope is zero.", + "video_name": "J43CIbKpdWc", + "timestamps": [ + 131 + ], + "3min_transcript": "- [Instructor] What is the equation of the horizontal line through the point negative four comma six? So let's just visualize this. Once you get the hang of it, you might not have to draw a graph, but for explanatory purposes, it might be useful. So negative four comma six, that's going to be in the second quadrant. So if this is my That is my y-axis. I'm going to go negative four in the x direction. So one, two, three, four. Negative four. And then one, three, four, five, six, in the y direction. So the point that we care about is going to be right over there. Negative four comma six. And what is the equation of the horizontal line? It is a horizontal line. So it's just going to go straight left, right like this. That is what the line would actually look like. So what is that equation? Well, for any x, y is going to be equal to six. y is equal to six. Doesn't matter what x you input here, you're gonna get y equals six. It just stays constant right over there. So the equation is y is equal to six. Let's do another one of these. So here we are asked what is the slope of the line y is equal to negative four? So let's visualize it and then in the future, you might not have to draw it like this. But let's just draw our axis again. X-axis y-axis and the slope of line y equals negative four. So for whatever x you have, y is going to be negative four. Let's say that's negative four right over there. And so, the line is y the line is y equals negative four. So I can draw it like this. So what's the slope of that? Well, slope is change in y for given change in x. y doesn't change. It stays at negative four. My change in y over change in x. Doesn't matter what my change in x is. My change in y is always going to be zero. It's constant. So the slope here is going to be equal to zero. Y doesn't change, no matter how much you change x. Let's do another one of these. This is fun. So now they are asking us, what is the slope of the line x equals negative three? Let me graph that one. So, I'm just going to draw my axis real fast. X-axis y-axis X is equal to negative three. So negative one, negative two, negative three. And so, this line is going to look let me, it's going to look like this. No matter what y or you can say no matter what y is. X is going to be equal to negative three. So," + }, + { + "Q": "\nAt 3:45 he points out that there's a new slowest time, and uses that to determine that the 4th bullet point is wrong. Does that new slowest time even matter though? Even if that new slowest dot wasn't there (ie, it had been at the 52 mark in the final round), does that change anything regarding the 4th bullet point question?", + "A": "It does matter because, as pointed out, one swimmer did swim slower so the statement all swam faster cannot be true.", + "video_name": "KXDOOmquZag", + "timestamps": [ + 225 + ], + "3min_transcript": "So his time definitely got worse. And this is at 53.8 seconds. So let's look at the statements and see which of these apply. The swimmers had faster times, on average, in the finals. Is this true? Faster times on average in the finals? So if we look at the finals right over here, we could take each of these times, add them up, and then divide by 8, the number of times we have. But let's see if we can get an intuition for where this is, because we're really just comparing these two plots, or these two distributions, we could say. And so let's see, if all the data was these three points and these three points, we could intuit that the mean would be right around there. It would be around 53.2 or 53.3 seconds, right around there. And then we have this point and this point, if you just found the mean of that point and that point, would get you right around there. So the mean of those two points would bring down the mean a little bit. And once again, I'm not figuring out the exact number. But maybe it would be around 53.2, 53.1, or 53.2 seconds. So that's my intuition for the mean of the final round. And now let's think about the mean of the semifinal round. Let's just look at these bottom five dots. If you find their mean, you could intuit it would be maybe someplace around here, pretty close to around 53.3 seconds. And then you have all these other ones that are at 53.5 and 53.3, which will bring the mean even higher. So I think it's fair to say that the mean in the final around and the time is less than the mean up here. And you could calculate it yourself, but I'm just trying to look at the distributions and get an intuition here. And at least in this case, it looks pretty clear that the swimmers had faster times, on average, It took them less time. One of the swimmers was disqualified from the finals. Well, that's not true. We have 8 swimmers in the semifinal round. And we have 8 swimmers in the final round. So that one's not true. The times in the finals vary noticeably more than the times in the semifinals. That does look to be true. We see in the semifinals, a lot of the times were clumped up right around here at 53.3 seconds and 53.5 seconds. The high time isn't as high as this time. The low time isn't as low there. So the final round is definitely-- they vary noticeably more. Individually, the swimmers all swam faster in the finals than they did in the semifinals. Well, that's not true. Whoever this was, clearly they were one of these data points up here. This data point took more time than all of these data points. So this represents someone who took more time in the finals than they did in the semifinals. And we got it right." + }, + { + "Q": "At 4:35, I still don't get why -(-e)/f is equal to e/f.\n", + "A": "When there are 2 negatives, it equals to a positive. When someone says Jump , it s a positive. When someone says Don t eat , it is negative. Meanwhile, if someone says Don t not eat , that s back to saying Eat which is a positive.", + "video_name": "9eSPhvhuInw", + "timestamps": [ + 275 + ], + "3min_transcript": "So this is going to be right. Now this one, negative divided by a negative, well that's just going to be positive. So that's the same thing as five over b. One way to think about it is that well the negatives kind of cancel each other out. So five over b, that looks good too. And of course I won't select none of the above because I found two choices that worked. All right, let's do one more. Which of the following expressions are equal to negative e over negative f? And remember we just have to take this step by step here. Actually let's try to just simplify this directly. So negative e over negative f. Well we just need to remind ourselves that this part right over here. Negative e over negative f. Let me write an equal sign. Negative e over, and I'm gonna put this negative. Let me do this in a different color. Let me do this in purple. So we have this purple. So we have that purple negative right over there. And negative e over negative f. That's the same thing as negative's divided by a negative is a positive. That's the same thing as e over f, as positive e over f. So this whole thing will simplify to negative e over f. So let's see which of these choices are that. Well this right here is positive e over f. So that's not the choice. This one over here. This one we could write it several ways actually. We could write it negative negative e over f. Which of course is equal to positive e over f. We could also write this. We could put the negative in the denominator. We could say that this thing. Actually let me write it over here as negative e over negative f. This is also a legitimate thing to do. You could take this negative and multiply it times the denominator. Right over here. But either way it's going to be equal to positive e over f. These two are actually evaluate to the same expression. So here, I would select. Finally, I would select. All right, hopefully that helps." + }, + { + "Q": "at 1:08 he says that the sin of 7pi/12 is the length of the magenta line however isn't that the sin of pi/2 and shouldn't that length be one considering it is a unit circle?\n", + "A": "Sal has drawn a right triangle. The hypotenuse (the diagonal line, or the green line furthest to the left) is = 1 because it goes from the center of the circle out to the circle. The angle 7pi/12 goes up to the hypotenuse. You are trying to make the angle into 90 degrees (pi/2)rather than the 7pi/12. Next, it is hard to see, but the magenta line is slightly shorter than 1. This is why it doesn t = 1. Hope this helps.", + "video_name": "2RbKfRfzD-M", + "timestamps": [ + 68 + ], + "3min_transcript": "Voiceover:What I want to attempt to do in this video is figure out what the sine of seven pi over 12 is without using a calculator. And so let's just visualize seven pi over 12 in the unit circle. One side of the angle is going along the positive x-axis if we go straight up, that's pi over two, which is the same thing as six pi over 12, so then we essentially just have another pi over 12 to get right over there. This is the angle that we're talking about that is seven pi over 12 radians, by the unit circle definition of sine, it's the y-coordinate of where this ray intersects the unit circle. This is the unit circle, has radius one where it intersects the unit. The y-coordinate is the sine. Another way to think about it, it's the length of this line right over here. I encourage you to pause the video right now and try to think about it on your own. See if you can use your powers of trigonometry to figure out what sine of seven pi over 12 is I'm assuming you've given a go at it, and if you're like me, your first temptation might have been just to focus on this triangle right over here that I drew for you. The triangle looks like this. It looks like this, where that's what you're trying to figure out, this length right over here, sine of seven pi over 12. We know the length of the hypotenuse is one. It's a radius of the unit circle. It's a right triangle right over there. We also know this angle right over here, which is this angle right over here, this gets us six pi over 12, and then we have another pi over 12, so we know that that is pi over 12, not pi over 16. We know that this angle right over here is pi over 12. Given this information, we can figure out this, or we can at least relate this side to this other side using a trig function relative to this angle. This is the adjacent side. this magenta side over one, or you could just say it's equal to this magenta side. You could say that this is cosine of pi over 12. We just figured out that sine of seven pi over 12 is the same thing as cosine of pi over 12, but that still doesn't help me. I don't know offhand what the cosine of pi over 12 radians is without using a calculator. Instead of thinking about it this way, let's see if we can compose this angle or if we can decompose it into some angles for which we do know the sine and cosine. What angles are those? Those are the angles in special right triangles. For example, we are very familiar with 30-60-90 triangles. 30-60-90 triangles look something like this. This is my best attempt at hand drawing it. Instead of writing 30-degree side, since we're thinking in radians, I'll write that as pi over six radians. The 60-degree side, I'm going to write that as pi over three radians, and of course, this is the right angle." + }, + { + "Q": "At 6:16, why do you need to collect multiple data points to determine if a question is a statistical question?\n", + "A": "Statistics is the comparison of data at different points. If you have a single data point then there is nothing to compare it to. So you need multiple points so you can compare the data, an thus be using statistics.", + "video_name": "OjzfQDFf7Uk", + "timestamps": [ + 376 + ], + "3min_transcript": "variability. So you're saying, okay that kinda makes sense, but I need to see some tangible questions or tangible examples of things that are statistical questions and things that are not statistical questions. I would say fair enough. Let's look at some examples. So here I have six questions, and I encourage you to pause this video right now. Before I work through it, think about it. Based on this definition of a statistical question, which of these questions are statistical, would require your statistical toolkit, and which of these are not statistical? So assuming you had a go at it, let's go through these one by one. So the first question, how much does my pet grapefruit weigh? You know, it's bizarre to begin with to have a pet grapefruit, but is this a statistical question? What do I need to do to answer it? I have to take my pet grapefruit out. I have to weigh it. Then I have to just write that down. Just doing that I am collecting data, to mess with statistics a little bit, but I'm just getting one data point. So I might weigh it and I might see my grapefruit weighs one pound, but that's not data with variability. That's just one data point. In order to have variability you have to have multiple data points and should be at least possible that they could vary. So, for example, all of these folks ate zero bricks but maybe it was possible that someone actually ate a brick. But here I have just one data point. With one data point, you can't have variability, so this is not a statistical question. I just collect a data point. Next question, what is the average number of cars in a parking lot on Monday mornings? To think about whether it is a statistical question, we just have to think about what do I have to do to answer that question? I would have to go out to the parking lot on multiple Monday mornings, and measure the number of cars. So on the first Monday morning I might see there are 50 cars. The next Monday morning I might go out there and count there's 49 cars. The next Monday morning I might see 63 cars. So I'm collecting multiple data points to answer this question. Then I'm going to take the average of all these, but I'm collecting multiple data points to answer this question. It's definitely possible that there could be variation here, that there could be variability, so this is a statistical question. Next question, am I hungry? It's an important question. We ask it to ourselves multiple times. In fact, sometimes our bodies just tell it to us. But I am definitely not collecting ... I guess you could say I'm collecting some type of feelings from my stomach or how weak I feel or not, but it's definitely not data with variability. I'm either hungry or not hungry on a given day. I mean if you said broader, how does my hunger change from day to day and you came up some type of a scale for rating your hunger, all right maybe that's more statistical. But just am I hungry, a yes-no question. This is not ..." + }, + { + "Q": "\nAt 3:31, Why does Sal multiply 6x5x4x3?", + "A": "nPr is used because nPr counts all answer sets (including duplicate unordered sets, which is what we want since the duplicates have different meaning than the ordered version) theres 6 total items so n=6 we re picking 4 items so r=4 the formula is n!/(n-r)!=6!/2!=6*5*4*3", + "video_name": "oQpKtm5TtxU", + "timestamps": [ + 211 + ], + "3min_transcript": "different than green, red, yellow and blue. We're going to assume that these are not the same code. Even though we've picked the same 4 colors, we're going to assume that these are 2 different codes, and that makes sense because we're dealing with codes. So these are different codes. So this would count as 2 different codes right here, even though we've picked the same actual colors. The same 4 colors, we've picked them in different orders. Now, with that out of the way, let's think about how many different ways we can pick 4 colors. So let's say we have 4 slots here. 1 slot, 2 slot, 3 slot and 4 slots. And at first, we care only about, how many ways can we We haven't picked any colors yet. Well, we have 6 possible colors, 1, 2, 3, 4, 5, 6. So there's going to be 6 different possibilities for this slot right there. So let's put a 6 right there. Now, they told us that the colors cannot be repeated, so whatever color is in this slot, we're going to take it out of the possible colors. So now that we've taken that color out, how many possibilities are when we go to this slot, when we go to the next slot? How many possibilities when we go to the next slot right here? Well, we took 1 of the 6 out for the first slot, so there's only 5 possibilities here. And by the same logic when we go to the third slot, we've used up 2 of the slots-- 2 of the colors already, so there would only 4 possible colors left. And then for the last slot, we would've used up 3 of the colors, so there's only 3 possibilities left. So if we think about all of the possibilities, all of the permutations-- and permutations are when you think about all the possibilities and you do care about order; where you say that this is different than So all of the different permutations here, when you pick 4 colors out of a possible of 6 colors, it's going to be 6 possibilities for the first 1, times 5 for the second bucket, times 4 for the third or the third bucket of the third position, times 3. So 6 times 5 is 30, times 4 is times 3. So 30 times 12. So this is 30 times 12, which is equal to their 360 possible 4-color codes." + }, + { + "Q": "\nhow is 1 m = to 100 cm 02:57", + "A": "Centi refers to 100. The root is the same as in century (100 years). A centimeter is one one-hundredth of a meter (1/100). So 100 centimeters are equal to 1 meter. Note that a decimeter is 1/10 of a meter. That word looks a little like decade, doesn t it?", + "video_name": "byjmR7JBXKc", + "timestamps": [ + 177 + ], + "3min_transcript": "that would be represented by 1 centimeter in the floor plan. And it goes the other way around. 1 centimeter on the floor plan would represent 80 centimeters in the house. And it's always important to do-- if this confuses you, just always do a reality check that the house should be bigger than the floor plan. So if the floor plan for this dimension of our living room is 4 centimeters, the actual house will be 80 times that. And 80 times 4 is 320-- let me do that in a blue color-- is equal to 320 centimeters. And we can do the same thing for the length of the living room. So 80 times 5 centimeters is going to get us to-- is going to be-- 80 times 5 is 400 centimeters. So we could figure out the area of this room in centimeters, if we like, and I guess, why not? It might be easier to convert it to meters later. Let me write this down. 400 times 320. Let's think about it. 4 times 32 is going to be 120, plus 8, 128. And I have 1, 2, 3 zeroes. 1, 2, 3. So it's going to be 128,000 centimeters squared. Now that's a lot of square centimeters. What would we do if we wanted to convert it into meters? Well, we just have to figure out how many square centimeters are there in a square meter. So let's think about it this way. A meter is equal to-- 1 meter is equal to 100 centimeters. So a square meter, so that's right over there. is the same thing as 100 centimeters by 100 centimeters. And so if you were to calculate this area in centimeters, 100 times 100 is 10,000, is equal to 10,000 centimeters squared. So you have 10,000 square centimeters for every square meter. And so, if you want to convert 128,000 centimeters squared to meters squared, you would divide by 10,000. So dividing that by 10,000 would give us 12.8 square meters. Now, another way you could've done it, and maybe this would have been easier, is to convert it up here. Instead of saying 400 centimeters times 320 centimeters, you would say, well," + }, + { + "Q": "\nwhy does it say the problem reads 80:1 scale but should read 1:80 scale", + "A": "They corrected it", + "video_name": "byjmR7JBXKc", + "timestamps": [ + 4801, + 140 + ], + "3min_transcript": "" + }, + { + "Q": "i am confused, is the 80 in 80:1 centimeters as well?\n", + "A": "It is saying that 80 centimeters in the real world is 1 centimeter on the blueprint", + "video_name": "byjmR7JBXKc", + "timestamps": [ + 4801 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 1:21 i never understand how he got y/x 3/1 6/2=27/9 im soooooooooooooooooooooooooooooooo lost right now", + "A": "A proportional relationship just means that the 2 fractions are equal. Your 3 fractions are all equal fractions. Start with 3/1 If you multiply 3/1 * 2/2 = 6/2. If you reduce it, you get back 3/1 If you multiply 3/1 * 9/9 = 27/9. If you reduce it, you get back 3/1. Hope this helps.", + "video_name": "qYjiVWwefto", + "timestamps": [ + 81 + ], + "3min_transcript": "What I want to introduce you to in this video is the notion of a proportional relationship. And a proportional relationship between two variables is just a relationship where the ratio between the two variables is always going to be the same thing. So let's look at an example of that. So let's just say that we want to think about the relationship between x and y. And let's say that when x is one, y is three, and then when x is two, y is six. And when x is nine, y is 27. Now this is a proportional relationship. Why is that? Because the ratio between y and x is always the same thing. And actually the ratio between y and x or, you could say the ratio between x and y, is always the same thing. So, for example-- if we say the ratio y over x-- this is always equal to-- it could be three over one, which is just three. It could be six over two, It could be 27 over nine, which is also just three. So you see that y over x is always going to be equal to three, or at least in this table right over here. And so, or at least based on the data points we have just seen. So based on this, it looks like that we have a proportional relationship between y and x. So this one right over here is proportional. So given that, what's an example of relationships that are not proportional. Well those are fairly easy to construct. So let's say we had-- I'll do it with two different variables. So let's say we have a and b. And let's say when a is one, b is three. And when a is two, b is six. And when a is 10, b is 35. when a is one, b is three so the ratio b to a-- you could say b to a-- you could say well when b is three, a is one. Or when a is one, b is three. So three to one. And that's also the case when b is six, a is two. Or when a is two, b is six. So it's six to two. So these ratios seem to be the same. They're both three. But then all of sudden the ratio is different right over here. This is not equal to 35 over 10. So this is not a proportional relationship. In order to be proportional the ratio between the two variables always has to be the same. So this right over here-- This is not proportional. Not proportional. So the key in identifying a proportional relationship is look at the different values that the variables take on when one variable is one value," + }, + { + "Q": "\nIn the charts that Sal makest at 3:36, is the y side always going to be numerator in fraction form?", + "A": "It doesn t matter which side you use as the numerator, as long as the proportions are consistent. So if you have x/y = 2x/2y = 3x/3y, that will work for sure. But if you have x/y = 2y/2x = 3x/3y, that won t work. Sure you can place the y on the top but make sure it stays on the top.", + "video_name": "qYjiVWwefto", + "timestamps": [ + 216 + ], + "3min_transcript": "It could be 27 over nine, which is also just three. So you see that y over x is always going to be equal to three, or at least in this table right over here. And so, or at least based on the data points we have just seen. So based on this, it looks like that we have a proportional relationship between y and x. So this one right over here is proportional. So given that, what's an example of relationships that are not proportional. Well those are fairly easy to construct. So let's say we had-- I'll do it with two different variables. So let's say we have a and b. And let's say when a is one, b is three. And when a is two, b is six. And when a is 10, b is 35. when a is one, b is three so the ratio b to a-- you could say b to a-- you could say well when b is three, a is one. Or when a is one, b is three. So three to one. And that's also the case when b is six, a is two. Or when a is two, b is six. So it's six to two. So these ratios seem to be the same. They're both three. But then all of sudden the ratio is different right over here. This is not equal to 35 over 10. So this is not a proportional relationship. In order to be proportional the ratio between the two variables always has to be the same. So this right over here-- This is not proportional. Not proportional. So the key in identifying a proportional relationship is look at the different values that the variables take on when one variable is one value, And then take the ratio between them. Here we took the ratio y to x, and you see y to x, or y divided by x-- the ratio of y to x is always going to be the same here so this is proportional. And you could actually gone the other way. You could have said, well what's the ratio of x to y? Well over here it would be one to three, which is the same thing as two to six, which is the same thing as nine to 27. When you take this ratio-- if you say the ratio of x to y instead of y to x, you see that it is always one third. But any way you look at it-- the ratio between these two variables-- if you say y to x, it's always going to be three. Or x to y is always going to be one third. So this is proportional while this one is not." + }, + { + "Q": "At 7:34, why are all the sides equal to the square root of a^2 + b^2?\n", + "A": "If one of the sides is equal to the square root of a^2 + b^2 (using Pythagoras theorem), then all of the sides are equal to it (since all the sides of a rhombus are equal).", + "video_name": "jpKjXtywTlQ", + "timestamps": [ + 454 + ], + "3min_transcript": "so all four of these triangles are congruent. So you could take the area of this triangle, multiply it by 4-- you have the area of the rhombus. Let me write this down. The rhombus area is equal to 4 times 1/2ab. 1/2ab gives us just this triangle right over here. 4 times that, so 4 times 1/2ab is 2ab, is going to be the area of the rhombus. If we can somehow figure out a and b, we can figure out the rhombus' area. So let's focus on this first piece of information. Let's focus on triangle AB and D. Now they tell us that its circumradius is 12.5. So let's just use this formula right over here. We get 12.5. It's circumradius is 12.5 is equal to the product So what's the length of the sides here? So we have this side right over here, side bd. That's just going to be 2a, right? That's an a plus another a, so it's going to be 2a, times this side right over here. What's this side, which is just one of the sides of the rhombus. Well, this is the hypotenuse of this right triangle right over here, right? This is a right angle. So it's going to be the square root of a squared plus b squared. But all of the sides are going to be that. It's a rhombus. All the sides are the same-- a squared plus b squared. They're all going to have that exact same length. So the product of the sides-- you have 2a-- that's the length of bd-- times the length of ba, which is going to be the square root of a squared plus b squared, times the length of ad, which is the square root of a squared plus b squared. 4 times the area of ABD. Now, what's the area of a, b, and d? Well, ABD is just two of these triangles right over here. This guy right over here is 1/2ab. This guy over here is also 1/2ab. So the entire area is going to be two of these guys, so it's just going to be a times b. It gives you the area of both of these triangles. Each of them are 1/2ab. So instead of writing area right here, I could write ab. Now let's see. This simplifies to 12.5, is equal to. Divide the numerator and the denominator by 2. So that becomes a 1. That becomes a 2, divided by a. That becomes a 1. That becomes a 1. Square root of a squared plus b squared times square root of a squared plus b squared is just a squared plus b squared." + }, + { + "Q": "\nAt 12:19, Sal simplified (2a)^2 to 4a^2. Wouldn't it just be 4a?", + "A": "(2a)^2 = (2a)*(2a) = 2*2*a*a = (2*2)*(a*a) = 2^2*a^2.", + "video_name": "jpKjXtywTlQ", + "timestamps": [ + 739 + ], + "3min_transcript": "You add them together, you just get ab. You just get ab, 2 divided by 2, you get a 1 there. You get a 2 here. Divide by b, get a 1. That just becomes an a. And so you get 25 is equal to the numerator, square root of a squared b squared times itself is just going to be a squared plus b squared over 2a. So that second triangle, its circumradius being 25 gives us this equation right over here. Now we can use both of this. We have two equations with two unknowns. Let's solve for a and b. If we know a and b, we can then go back here and figure out the rhombus's area. So over here, we get-- let's multiply both sides by 2b. We get 25b is equal to a squared plus b squared. Over here, if we multiply both sides by 2a, we get 50a is equal to a squared plus b squared. 25b is equal to a squared plus b squared. So 25b must be the same thing as 50a. They're both a squared plus b squared. So we get 25b must be equal to 50a. They're both equal to a squared plus b squared. Now, divide both sides by 25, you get b is equal to 2a. b is equal to-- actually I wanted to do that in the magenta. b is equal to 2a. So we can take this information and then now substitute back into either one of these equations to solve for b, and then we can solve for a. So let's go back into this one. So we get 50a-- actually we'll solve for a first. 50a is equal to a squared plus b squared. so let's write 2a squared. So we get 50a is equal to a squared plus 4a squared. Or we get 50a is equal to 5a squared. We could divide both sides by 5a. If we divide this side by 5a, we get 10. And if we divide this side by 5a, we get a. So a is equal to 10. And then we could just substitute back here to figure out b. 2 times a is equal to b. b is equal to 2 times 10, which is equal to 20. So we know a is 10. b is 20. We just have to go right back here to figure out the area of the rhombus. The area of the rhombus is equal to 2 times-- a was 10-- 2 times 10 times 20. This is 20 times 20." + }, + { + "Q": "at 12:23 , he said 50=5a^2 , then he divided by 5 and got 10=a. what about that squared symbol?\n", + "A": "The equation was 50a=5a^2. He then took each side divided by 5a, which came out to be 10=a, or a=10.", + "video_name": "jpKjXtywTlQ", + "timestamps": [ + 743 + ], + "3min_transcript": "You add them together, you just get ab. You just get ab, 2 divided by 2, you get a 1 there. You get a 2 here. Divide by b, get a 1. That just becomes an a. And so you get 25 is equal to the numerator, square root of a squared b squared times itself is just going to be a squared plus b squared over 2a. So that second triangle, its circumradius being 25 gives us this equation right over here. Now we can use both of this. We have two equations with two unknowns. Let's solve for a and b. If we know a and b, we can then go back here and figure out the rhombus's area. So over here, we get-- let's multiply both sides by 2b. We get 25b is equal to a squared plus b squared. Over here, if we multiply both sides by 2a, we get 50a is equal to a squared plus b squared. 25b is equal to a squared plus b squared. So 25b must be the same thing as 50a. They're both a squared plus b squared. So we get 25b must be equal to 50a. They're both equal to a squared plus b squared. Now, divide both sides by 25, you get b is equal to 2a. b is equal to-- actually I wanted to do that in the magenta. b is equal to 2a. So we can take this information and then now substitute back into either one of these equations to solve for b, and then we can solve for a. So let's go back into this one. So we get 50a-- actually we'll solve for a first. 50a is equal to a squared plus b squared. so let's write 2a squared. So we get 50a is equal to a squared plus 4a squared. Or we get 50a is equal to 5a squared. We could divide both sides by 5a. If we divide this side by 5a, we get 10. And if we divide this side by 5a, we get a. So a is equal to 10. And then we could just substitute back here to figure out b. 2 times a is equal to b. b is equal to 2 times 10, which is equal to 20. So we know a is 10. b is 20. We just have to go right back here to figure out the area of the rhombus. The area of the rhombus is equal to 2 times-- a was 10-- 2 times 10 times 20. This is 20 times 20." + }, + { + "Q": "at 0:42 , how do we get to know if an equation is homogeneous equation. i mean what is the working rule for homogeneous equation?\n", + "A": "For a first order equation you can determine it s not separable because an x term is added to a y term. Homogeneous is probably the next thing to try. It works because the exponent on the x and y terms are the same. I don t know of any perfect test to determine if it s homogeneous. I would say it s trial and error and, of course, practice.", + "video_name": "6YRGEsQWZzY", + "timestamps": [ + 42 + ], + "3min_transcript": "Let's do one more homogeneous differential equation, or first order homogeneous differential equation, to differentiate it from the homogeneous linear differential equations we'll do later. But anyway, the problem we have here. It's the derivative of y with respect to x is equal to-- that x looks like a y-- is equal to x squared plus 3y squared. I'm writing it a little bit small, so that I don't run out of space. Divided by 2xy. So in all of these homogeneous equations-- and obviously, we don't know it's homogeneous yet. So we have to try to write it as a function of y divided by x. So it looks like we could do that, if we divide the top and the bottom by x squared. So if we just multiply-- let me do it in a different color-- 1 over x squared, or x to the negative 2, over 1 over x squared. We're essentially just multiplying by 1. And then, that is equal to what? 1 plus 3y squared over x squared divided by 2-- if you times y over x. Or we could just rewrite the whole thing, and this is just equal to 1 plus 3y over x squared divided by 2 times y over x. So yes, this is a homogeneous equation. Because we were able to write it as a function of y divided by x. So now, we can do the substitution with v. And hopefully, this is starting to become a little bit of second nature to you. So we can make the substitution that v is equal to y over x, or another way of writing that is that y is equal to xv. And then, of course, the derivative of y with respect to x, or if we take the derivative with respect to x of both sides of this, that's equal to the derivative of x is 1 times v, this is just the product rule, plus x times the And now, we can substitute the derivative of y with respect to x is just this. And then the right hand side of the equation is this. But we can substitute v for y over x. So let's do that. And so we get v plus x. Instead of dv [? dv ?] x, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared, we're making the substitution v is equal to y over x. All of that over 2v. Now, let's see what we can do. This is where we just get our algebra hat on, and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. So we'll get 2v squared plus 2xv v prime-- 2v times x, yep," + }, + { + "Q": "\nI have an dummy question, but at minute 1:03, i dont get why the multiplication (1/x\u00c2\u00b2)/1/x\u00c2\u00b2 makes the after result.\nThanks", + "A": "Distribute the 1/x^2 on top to both terms, then multiply the bottom term by 1/x^2. You ll get the same result as the video.", + "video_name": "6YRGEsQWZzY", + "timestamps": [ + 63 + ], + "3min_transcript": "Let's do one more homogeneous differential equation, or first order homogeneous differential equation, to differentiate it from the homogeneous linear differential equations we'll do later. But anyway, the problem we have here. It's the derivative of y with respect to x is equal to-- that x looks like a y-- is equal to x squared plus 3y squared. I'm writing it a little bit small, so that I don't run out of space. Divided by 2xy. So in all of these homogeneous equations-- and obviously, we don't know it's homogeneous yet. So we have to try to write it as a function of y divided by x. So it looks like we could do that, if we divide the top and the bottom by x squared. So if we just multiply-- let me do it in a different color-- 1 over x squared, or x to the negative 2, over 1 over x squared. We're essentially just multiplying by 1. And then, that is equal to what? 1 plus 3y squared over x squared divided by 2-- if you times y over x. Or we could just rewrite the whole thing, and this is just equal to 1 plus 3y over x squared divided by 2 times y over x. So yes, this is a homogeneous equation. Because we were able to write it as a function of y divided by x. So now, we can do the substitution with v. And hopefully, this is starting to become a little bit of second nature to you. So we can make the substitution that v is equal to y over x, or another way of writing that is that y is equal to xv. And then, of course, the derivative of y with respect to x, or if we take the derivative with respect to x of both sides of this, that's equal to the derivative of x is 1 times v, this is just the product rule, plus x times the And now, we can substitute the derivative of y with respect to x is just this. And then the right hand side of the equation is this. But we can substitute v for y over x. So let's do that. And so we get v plus x. Instead of dv [? dv ?] x, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared, we're making the substitution v is equal to y over x. All of that over 2v. Now, let's see what we can do. This is where we just get our algebra hat on, and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. So we'll get 2v squared plus 2xv v prime-- 2v times x, yep," + }, + { + "Q": "I believe at 2:23 and other instances during previous videos he makes the mistake of putting the 'approx. equal to' sign. It is exactly equal to e^x.\n", + "A": "Well, technically, yes and no, since we are summing something to infinity, it can never be exact exact, because it has to go on forever(e is transcendental anyway), but he probably could have used equals, or approx. equals, the definitions get quite similiar while summing things to infinity.", + "video_name": "JYQqml4-4q4", + "timestamps": [ + 143 + ], + "3min_transcript": "Now let's do something pretty interesting. And this will, to some degree, be one of the easiest functions to find the Maclaurin series representation of. But let's try to approximate e to the x. f of x is equal to e to the x. And what makes this really simple is, when you take the derivative-- and this is, frankly, one of the amazing things about the number e-- is that when you take the derivative of e to the x, you get e to the x. So this is equal to f prime of x. This is equal to f, the second derivative of x. This is equal to the third derivative of x. This is equal to the n-th derivative of x. It's always equal to e to the x. That's kind of the first mind blowing thing about the number e. It's just, you could keep taking its derivative. The slope at any point on that curve is the same as the value of that point on that curve. That's kind of crazy. Anyway, with that said, let's take its Maclaurin representation. So we have to find f of 0, f prime of 0, the second derivative at 0. Well, when you take e to the 0, e to the 0 is just equal to 1. This is going to be equal to f prime of 0. It's going to be equal to any of the derivatives evaluated at 0. The n-th derivative evaluated at 0. And that's why it makes applying the Maclaurin series formula fairly straightforward. If I wanted to approximate e to the x using a Maclaurin series-- so e to the x-- and I'll put a little approximately over here. And we'll get closer and closer to the real e to the x as we keep adding more and more terms. And especially if we had an infinite number of terms, it would look like this. f of 0-- let me do it in-- what colors did I use for cosine and sine? So I used pink and I used green. So let me use a non-pink, non-green. I'll use the yellow here. So f of 0 is 1 plus f prime of 0 times x. f prime of 0 is also 1. So plus x plus, this is also 1, so it's going to be x squared over 2 factorial. All of these things are going to be 1. This is 1, this is 1, when we're talking about e to the x. So you go to the third term. This is 1. You just have x to the third over 3 factorial. Plus x to the third over 3 factorial. And I think you see the pattern here. We just keep adding terms. x to the fourth over 4 factorial plus x to the fifth over 5 factorial plus x to the sixth over 6 factorial. And something pretty neat is starting to emerge. Is that e to x, 1-- this is just really cool-- that e to the x can be approximated by 1 plus x plus x squared over 2 factorial plus x to the third over 3 factorial. Once again, e to the x is starting to look like a pretty cool thing. This also leads to other interesting results. That if you wanted to approximate e, you just evaluate this at x is equal to 1. So if you wanted to approximate e, you'd say" + }, + { + "Q": "at 0:24 cant you just do -31+(1-1)-7 ?\n", + "A": "yes, you can but the way Sal did it is simpler.", + "video_name": "GA_yxxeFYBU", + "timestamps": [ + 24 + ], + "3min_transcript": "- So I have a function here, h of n, and let's say that it explicitly defines the terms of a sequence. Let me make a little... Let me make a quick table here. We have n, and then we have h of n. When n is equal to one, h of n is negative 31, minus seven times one minus one, which is going to be... This is just going to be zero, so it's going to be negative 31. When n is equal to two, it's going to be negative 31 minus seven times two minus one, so two minus one. This is just going to be one, so it's negative 31 minus seven, which is equal to negative 38. When n is equal to three, it's gonna be negative 31 minus seven times three minus one, which is just two, so we're gonna subtract seven twice. It's gonna be negative 31 minus 14, which is equal negative 45. We're starting at negative 31, and then we keep subtracting, we keep subtracting negative seven. We keep subtracting negative seven from that. In fact we subtract negative seven one less than the term... We subtract negative seven one less times than the term we're dealing with. If we're dealing with the third term we subtract negative seven twice. If we're dealing with the second term we subtract negative seven once. This is all nice, but what I want you to do now is pause the video and see if you can define this exact same sequence. The sequence here is you start at negative 31, and you keep subtracting negative seven, so negative 38, negative 45. The next one is gonna be negative 52, and you go on and on and on. You keep subtracting negative seven. Can we define this sequence in terms of a recursive function? Why don't you have a go at that. Let's try to define it in terms of a recursive function. Let's just call that g of n, In some ways a recursive function is easier, because you can say okay look. The first term when n is equal to one, if n is equal to one, let me just write it, If n is equal to one, if n is equal to one, what's g of n gonna be? It's gonna be negative 31, negative 31. And if n, if n is greater than one and a whole number, so this is gonna be defined for all positive integers, and whole, and whole number, it's just going to be the previous term, so g of n minus one minus seven, minus seven. We're saying hey if we're just picking an arbitrary term we just have to look at the previous term and then subtract, and then subtract seven. It all works out nice and easy, because you keep looking at previous, previous, previous terms all the way until you get to the base case, which is when n is equal to one, and you can build up back from that. You get this exact same sequence." + }, + { + "Q": "During 0:21, he screen became blurred and it was tough to concentrate. Still, love your videos( Your ScratchPad makes me jealous)!!\u00f0\u009f\u0098\u0083\n", + "A": "HaCk(ERROR<1>) X>0 Y<0{heading 0 Ipx} =hAck", + "video_name": "qW-Ce44ll0Q", + "timestamps": [ + 21 + ], + "3min_transcript": "- We have three different number lines here, and on each number line they mark off a couple of these marks, this is negative two this is negative 10, negative five, negative 11, and then we need to figure out what the blue dot represents. And as a little bit of a hint, each mark here is not necessarily incrementing by one, it could be by more than one. So I encourage you to pause this video, and to try it on your own. Try to figure out what number does this blue dot represent on these different number lines? So let's tackle this first one. So we're gonna go where they gave us this mark is negative two, this is negative 10, and we need to figure out this blue one that's further to the left of negative 10. Well just going from negative two to negative 10, what has to happen to negative two to get to negative 10? Well we'll have to subtract eight. And so if we move two to the left, that's the equivalent of subtracting eight. So if we move two to the left again, that's subtracting eight again. So this must be negative 18 and so if we move two to the left again, that also must be subtracting eight. that gets us to negative 26. Now another way we could have thought about it is, if we jump two to the left and that's negative eight, then one jump to the left is gonna be negative four. So you could say \"Well this is negative four\", that would get you to negative six, then negative four again gets you to negative 10, then this would be negative 14, negative 18, negative 22, negative 26. Now let's tackle this one. So here we're gonna, we have to figure out what this blue dot here is on the right. So if we started at 11, you make two jumps to go to negative five. So what do I have to add to negative 11 to get to negative five? Well I have to add six. So if I add six over two jumps, that means that each of these jumps, that must be plus three, and this one must be plus three. So negative five plus three... So plus three is gonna get us to negative two. Plus three gets us to positive one, negative two plus three is positive one, plus three gets us to four. So here we got to negative 26, here we get to four. Now let's tackle this one. So we're at negative seven, or this is negative seven this is one and then we have to figure out this mark right over here. So if you take two jumps to the right, from negative seven to one, how much did we have to add? Well to get from negative seven to one, you have to add eight. You have to add eight. So then if we make another two jumps to the mystery number that means we added eight again. So one plus eight is nine. Another way we could have thought about it, if we added eight over two jumps that means we added four on each jump. So this one must be negative three, add four you get to one," + }, + { + "Q": "\nAt 2:09 he basically tells us we can't assume and we don't know the angels.\nWell.. We know the angle bisector is equal (no duh...) But doesn't that mean that the 2 angels at D are 90 degrees. And if we know that we also know that the Angle A and Angle C are equal am I right?", + "A": "We actually don t know that the 2 angles at D are 90\u00c2\u00b0. They look like they could be, but that information wasn t given in the question. If they were, then you re correct that we could find the other angles and prove that the triangles are congruent by ASA (or AAAS).", + "video_name": "TpIBLnRAslI", + "timestamps": [ + 129 + ], + "3min_transcript": "What I want to do first is just show you what the angle bisector theorem is and then we'll actually prove it for ourselves. So I just have an arbitrary triangle right over here, triangle ABC. And what I'm going to do is I'm going to draw an angle bisector for this angle up here. And we could have done it with any of the three angles, but I'll just do this one. I'll make our proof a little bit easier. So I'm just going to bisect this angle, angle ABC. So let's just say that's the angle bisector of angle ABC, and so this angle right over here is equal to this angle right over here. And let me call this point down here-- let me call it point D. The angle bisector theorem tells us that the ratio between the sides that aren't this bisector-- so when I put this angle bisector here, it created two smaller triangles out of that larger one. The angle bisector theorem tells us the ratios between the other sides of these two triangles So it tells us that the ratio of AB to AD is going to be equal to the ratio of BC to, you could say, CD. So the ratio of-- I'll color code it. The ratio of that, which is this, to this is going to be equal to the ratio of this, which is that, to this right over here-- to CD, which is that over here. So once you see the ratio of that to that, it's going to be the same as the ratio of that to that. So that's kind of a cool result, but you can't just accept it on faith because it's a cool result. You want to prove it to ourselves. And so you can imagine right over here, we have some ratios set up. So we're going to prove it using similar triangles. And unfortunate for us, these two triangles right here We know that these two angles are congruent to each other, but we don't know whether this angle is equal to that angle or that angle. We don't know. We can't make any statements like that. So in order to actually set up this type of a statement, we'll have to construct maybe another triangle that will be similar to one of these right over here. And one way to do it would be to draw another line. And this proof wasn't obvious to me the first time that I thought about it, so don't worry if it's not obvious to you. What happens is if we can continue this bisector-- this angle bisector right over here, so let's just continue it. It just keeps going on and on and on. And let's also-- maybe we can construct a similar triangle to this triangle over here if we draw a line that's parallel to AB down here. So let's try to do that. So I'm just going to say, well, if C is not on AB, you could always find a point or a line that goes through C that is parallel to AB." + }, + { + "Q": "How did Sal know to divide by two at 2:04. Thats part of taking the average... right?\n", + "A": "Sal was trying to find the midpoint of thouse two points. He averaged the coordinates to find a new set of coordinates exactly in the middle. The midpoint formula is... (x+x/2, y+y/2)", + "video_name": "63mWxNXQQAk", + "timestamps": [ + 124 + ], + "3min_transcript": "- [Voiceover] We're asked to use the \"Reflection\" tool to define a reflection that will map line segment ME, line segment ME, onto the other line segment below. So we want to map ME to this segment over here and we want to use a Reflection. Let's see what they expect from us if we want to add a Reflection. So if I click on this it says Reflection over the line from, and then we have two coordinate pairs. So they want us to define the line that we're going to reflect over with two points on that line. So let's see if we can do that. To do that I think I need to write something down so let me get my scratch pad out and I copied and pasted the same diagram. And the line of reflection, one way to think about it, we want to map point E, we want to map point E, to this point right over here. We want to map point M to this point over here. And so between any point and its corresponding point on the image after the reflection, these should be equidistant from the line of reflection. This and this should be equidistant from the This and this should be E, and this point should be equidistant from the line of reflection. Or another way of thinking about it, that line of reflection should contain the midpoint between these two magenta points and it should contain the midpoint between these two deep navy blue points. So let's just calculate the midpoints. So we could do that with a little bit of mathematics. The coordinates for E right over here, that is, let's see that is x equals negative four, y is equal to negative four, and the coordinates for the corresponding point to E in the image. This is x is equal to two, x is equal to two, and y is equal to negative six. So what's the midpoint between negative four, negative four, and two, comma, negative six? Well you just have to take the average of the x's and take the average of the y's. Let me do that, actually I'll do it over here. So if I take the average of the x's it's going to be negative four, negative four, plus two, plus two, And then the average of the y's, it's going to be negative four plus negative six over two. Negative four plus negative six, over two and then close the parentheses. Let's see, negative four plus two is negative two, divided by two is negative one. So it's going to be negative one, comma. Negative four plus negative six, that's the same thing as negative four minus six which is going to be negative 10. Divided by two is negative five. Let me do that in a blue color so you see where it came from. Is going to be negative five. So there you have it. That's going to be the midpoint between E and the corresponding point on its image. So let's see if I can plot that. So this is going to be, this point right over here is going to be negative one, comma, negative five. So x is negative one, y is negative five. So it's this point right over here and it does indeed look like the midpoint." + }, + { + "Q": "\nI still don't understand. e.g. Sharing $20 in the ratio of 2:3. How would you do it? As well as both numbers wouldn't be equal if you separated. Any advice?", + "A": "Well ratios don t have to be equal that is the thing.2+3=5 so 20\u00c3\u00b75=4 so the first person who got 2 would get 4x2=8 so that would be 8 for the first person then 4x3=12 and so to check do 8+12=20 tada", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 123 + ], + "3min_transcript": "Voiceover:We've got some apples here and we've got some oranges and what I want to think about is, what is the ratio, what is the ratio of apples ... Of apples, to oranges? To oranges. To clarify what we're even talking about, a ratio is giving us the relationship between quantities of 2 different things. So there's a couple of ways that we can specify this. We can literally count the number of apples. 1, 2, 3, 4, 5, 6. So we have 6 apples. And we can say the ratio is going to be 6 to, 6 to ... And then how many oranges do we have? 1, 2, 3, 4, 5, 6, 7, 8, 9. It is 6 to 9. The ratio of apples to oranges is 6 to 9. And you could use a different notation. You could also write it this way. 6 to ... You would still read the ratio as being 6 to 9. because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges" + }, + { + "Q": "\nInstead of using \"to\" for the ratio, we could also tell it as \"is to\"....For example 6:9 could be told as 6 is to 9 too...", + "A": "Yes, I think using is to is the same as using to", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 369 + ], + "3min_transcript": "And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples." + }, + { + "Q": "\nIn ratio when you write something like 6:10, why do you have to write a colon??", + "A": "The colons are used to seperate two different amounts of something/item.", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 370 + ], + "3min_transcript": "And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples." + }, + { + "Q": "If the problem is like this one where it says find the ratio to apples and oranges 6:9 does it matter which one goes first could it be 9:6\n", + "A": "It matters which goes first and it is very important. In your example, we have the ratio of apples to oranges so the number of apples will be first, 6:9 (I assume that 6 is the number of apples). If you write it as 9:6, you also need to switch your sentence into the ratio of oranges to apples or it won t be the same.", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 369, + 546 + ], + "3min_transcript": "And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples." + }, + { + "Q": "\nat 3:13 what does he say", + "A": "This is what he said 3:1= Here we 9 oranges for every 6 apples. This can also be shown as 6 apples for every 9 oranges Hope this helps.", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 193 + ], + "3min_transcript": "because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples." + }, + { + "Q": "\nAt 3:54,is 9:6 the same as 6:9,is it only switched around?", + "A": "Uh, no. For example, you d have 9 red cars for every 6 blue cars, while if it was 6:9, it would be 6 red cars for every 9 blue cars.", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 234, + 546, + 369 + ], + "3min_transcript": "because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples." + }, + { + "Q": "At 3:09, Sal said that 0^0th power = 1. But I used my calculator and saw that 0^0 = undefined.\nIn the previous videos, he showed that decreasing the index/exponent would mean dividing by the number whose power is taken. In other words, 0^1 to 0^0= 1*0/0=0/0=undefined.\nBut then there is the case of other powers. what is actually going on?\n", + "A": "Well, what Sal is saying is that 0^0th power could equal one, depending on what theory you think is correct. It comes out undefined, because the true answer to 0^0 power could be 0 or 1: we just don t know. Because theorists can t agree, they settled with undefined, which pretty much means, we don t know for sure .", + "video_name": "PwDnpb_ZJvc", + "timestamps": [ + 189 + ], + "3min_transcript": "once again, you are going to get 0. And I think you see a pattern here. If I take 0 to any non-zero number-- so to the power of any non-zero, so this is some non-zero number, then this is going to be equal to 0. Now, this raises a very interesting question. What happens at 0 to the 0-th power? So here, 0 to the millionth power is going to be 0. 0 to the trillionth power is going to be 0. Even negative or fractional exponents, which we haven't talked about yet, as long as they're non-zero, this is just going to be equal to 0, kind of makes sense. But now, let's think about what 0 to the 0-th power is, because this is actually a fairly deep question. Well, actually, why don't you pause the video and think a little bit about what 0 to the 0-th power should be. Well, there's two trains of thought here. You could say, look, 0 to some non-zero number is 0. So why don't we just extend this to all numbers and say 0 to any number should be 0. And so maybe you should say that 0 to the 0-th power is 0. But then, there was another train of logic that we've already learned, that any non-zero number, if you take any non-zero number, and you raise it to the 0-th power. We've already established that you start with a 1, and you multiply it times that non-zero number 0 times. So this is always going to be equal to 1 for non-zero numbers. So maybe say, hey, maybe we should extend this to all numbers, including 0. So maybe 0 to the 0-th power should be 1. should be equal to 1. So you see a conundrum here, and there's actually really good cases, and you can get actually fairly sophisticated with your mathematics. And there's really good cases for both of these, for 0 to 0-th being 0, and 0 to the 0-th power being 1. And so when mathematicians get into this situation, where they say, well, there's good cases for either. There's not a completely natural one. Either of these definitions would lead to difficulties in mathematics. And so what mathematicians have decided to do is, for the most part-- and you'll find people who will dispute this; people will say, no, I like one more than the other-- but for the most part, this is left undefined. 0 to the 0-th is not defined by at least just kind of more conventional mathematics. In some use cases, it might be defined to be one of these two things. So 0 to any non-zero number, you're going to get 0." + }, + { + "Q": "At 0:12, what is the answer\n", + "A": "Zero, because multiplying ANYTHING by zero is going to be zero.", + "video_name": "PwDnpb_ZJvc", + "timestamps": [ + 12 + ], + "3min_transcript": "So let's think a little bit about powers of 0. So what do you think 0 to the first power is going to be? And I encourage you to pause this video. Well, let's just think about it. One definition of exponentiation is that you start with a 1, and then, you multiply this number times a 1 one time. So this is literally going to be 1 times-- let me do it in the right color-- it's 1 times 0. You're multiplying the 1 by 0 one time. 1 times 0, well, that's just going to be equal to 0. Now, what do you think 0 squared or 0 to the second power is going to be equal to? Well, once again, one way of thinking about this is that you start with a 1, and we're going to multiply it by 0 two times. So times 0 times 0. Well, what's that going to be? once again, you are going to get 0. And I think you see a pattern here. If I take 0 to any non-zero number-- so to the power of any non-zero, so this is some non-zero number, then this is going to be equal to 0. Now, this raises a very interesting question. What happens at 0 to the 0-th power? So here, 0 to the millionth power is going to be 0. 0 to the trillionth power is going to be 0. Even negative or fractional exponents, which we haven't talked about yet, as long as they're non-zero, this is just going to be equal to 0, kind of makes sense. But now, let's think about what 0 to the 0-th power is, because this is actually a fairly deep question. Well, actually, why don't you pause the video and think a little bit about what 0 to the 0-th power should be. Well, there's two trains of thought here. You could say, look, 0 to some non-zero number is 0. So why don't we just extend this to all numbers and say 0 to any number should be 0. And so maybe you should say that 0 to the 0-th power is 0. But then, there was another train of logic that we've already learned, that any non-zero number, if you take any non-zero number, and you raise it to the 0-th power. We've already established that you start with a 1, and you multiply it times that non-zero number 0 times. So this is always going to be equal to 1 for non-zero numbers. So maybe say, hey, maybe we should extend this to all numbers, including 0. So maybe 0 to the 0-th power should be 1." + }, + { + "Q": "At about the time frame 5:20 to 5:40 you told us that the dot plot would probably be the easiest to use. did you mean for everything as it was stressed that way in \" and this is once again where maybe the dot plot is the most (pause), the most, it jumps out at you.\" So do mean that it is the easiest for all the questions or just those few? -thanks\n", + "A": "No, he was talking about those that were shown there. So, the dot plot was the best for that in comparison with the list and the table. Hope this helps!", + "video_name": "gdE46YSedvE", + "timestamps": [ + 320, + 340 + ], + "3min_transcript": "A dot plot. And a dot plot, we essentially just take the same information, and even think about it the same way. But we just show it visually. In a dot plot, what we would have ... In a dot plot, what we would have ... Actually, let me just not draw an even arrow there. We have the different age groups, so five, six, seven, eight, nine, 10, 11, and 12, and we have a dot to represent, or we use a dot for each student at that age. So there's two five-year-olds, so I'll do two dots. One, and two. There's one six-year-old, so that's gonna be one dot, right over here. There's four seven-year-olds, so one, two, three, four dots. There's four nine-year-olds, so one, two, three, and four. There's one 10-year-old. So let's put a dot, one dot, right over there for that one 10-year-old. There's no 11-year-olds. I'm not gonna put any dots there. And then there's two 12-year-olds. So one 12-year-old, and another 12-year-old. So there you go, we have a frequency table, dot plot, list of numbers. These are all showing the same data, just in different ways. Once you have it represented in any of these ways, we can start to ask questions about it. So we could say, \"What is the most frequent age?\" Well, the most frequent age, when you look at it visually, or the easiest thing might be just to look at the dot plot because you see visually, the most frequent age are the two highest stacks. There's actually seven and nine are tied for the most frequent age. You'd have also seen it here, where seven and nine are tied at four. you would actually, you'd have to count all of them to kind of come up with this again and say, \"Okay, there's four sevens, four nines. \"That's the largest number.\" So this is, if you're looking for, what's the most frequent age? When you just visually inspect here, probably pops out at you the fastest. But there's other questions we can ask ourselves. We can ask ourselves, \"What is the range? \"What is the range of ages in the classroom?\" And this is once again where maybe the dot plot jumps out at you the most, because the range is just the maximum age ... or, the maximum data point minus the minimum data point. So what's the maximum age here? Well, the maximum age here, we see it from the dot plot, is 12. And the minimum age here, you see, is five. So there's a range of seven. The difference between the maximum and the minimum is seven. But you could have also done that over here. You could say, \"The maximum age here is 12. \"Minimum age here is five. \"And so let's subtract ...\" You find the difference between 12 and five, which is seven." + }, + { + "Q": "I thought I understood this but at 6:18 I stopped understanding it.:(\n", + "A": "What exactly did you not understand? The range is a measure that helps you figure how large the difference between the smallest and largest values of your data set is, and to calculate it you take the smallest value and subtract it from the largest one. It s really just a way to figure out how spread out your data is. Hope this helped, if not don t hesitate to ask away :)", + "video_name": "gdE46YSedvE", + "timestamps": [ + 378 + ], + "3min_transcript": "There's four nine-year-olds, so one, two, three, and four. There's one 10-year-old. So let's put a dot, one dot, right over there for that one 10-year-old. There's no 11-year-olds. I'm not gonna put any dots there. And then there's two 12-year-olds. So one 12-year-old, and another 12-year-old. So there you go, we have a frequency table, dot plot, list of numbers. These are all showing the same data, just in different ways. Once you have it represented in any of these ways, we can start to ask questions about it. So we could say, \"What is the most frequent age?\" Well, the most frequent age, when you look at it visually, or the easiest thing might be just to look at the dot plot because you see visually, the most frequent age are the two highest stacks. There's actually seven and nine are tied for the most frequent age. You'd have also seen it here, where seven and nine are tied at four. you would actually, you'd have to count all of them to kind of come up with this again and say, \"Okay, there's four sevens, four nines. \"That's the largest number.\" So this is, if you're looking for, what's the most frequent age? When you just visually inspect here, probably pops out at you the fastest. But there's other questions we can ask ourselves. We can ask ourselves, \"What is the range? \"What is the range of ages in the classroom?\" And this is once again where maybe the dot plot jumps out at you the most, because the range is just the maximum age ... or, the maximum data point minus the minimum data point. So what's the maximum age here? Well, the maximum age here, we see it from the dot plot, is 12. And the minimum age here, you see, is five. So there's a range of seven. The difference between the maximum and the minimum is seven. But you could have also done that over here. You could say, \"The maximum age here is 12. \"Minimum age here is five. \"And so let's subtract ...\" You find the difference between 12 and five, which is seven. You still could have done it. You'd say, \"Okay, what's the lowest? \"Let's see, five. Are there any fours here? \"Nope, there's no fours. \"So five's the minimum age. \"And what's the largest? \"Is it seven? No. \"Is it nine? Nine, maybe 10. \"Oh, 12. 12. \"Are there any 13s? No. \"12 is the maximum.\" So you say, \"12 minus five is seven\" to get the range. But then we could ask ourselves other questions. We could say, \"How many ... \"How many older ... \"older than nine?\" is a question we could ask ourselves. And then, if we were to look at the dot plot, we'd say, \"Okay, this is nine.\" And we'd care about how many are older than nine. So that would be this one, two and three. Or you could look over here. How many are older than nine? Well, it's the one person who's 10 and then the two who are 12. So there are three. And over here, if you said, \"How many are older than nine?\" Well, then you'd just have to go through the list and say, \"Okay, no, no, no, no, no, no, no," + }, + { + "Q": "I am wondering if it's possible to raise a number to a power to result in 0. x^x = 0 I've been graphing functions with the variable in the exponent (i.e. y = 4^x) and was wondering if it ever actually reaches zero. @ 4:13 Sal says it never quite reaches zero, so is it impossible? The line keeps going on forever in the X direction?\n", + "A": "It s my firm belief that infinity doesn t exist. As you can see by the graphs of y = a^x, y is always positive, and can only get close to zero where x is negative. No matter how large a negative number -x you get, you can always find another so large that the first one is negligible in comparison. And a^(-x) = 1/(a^x) is never zero. (Kind of crazy, insn t it?)", + "video_name": "9SOSfRNCQZQ", + "timestamps": [ + 253 + ], + "3min_transcript": "That could be my x-axis. And then let's make this my y-axis. I'll draw it as neatly as I can. So let's make that my y-axis. And my x values, this could be negative 2. Actually, make my y-axis keep going. So that's y. This is x. That's a negative 2. That's negative 1. That's 0. That is 1. And that is positive 2. And let's plot the points. x is negative 2. y is 1/25. Actually, let me make the scale on the y-axis. So let's make this. So we're going to go all the way to 25. So let's say that this is 5. Actually, I have to do it a little bit smaller than that, too. So this is going to be 5, 10, 15, 20. And then 25 would be right where I wrote the y, give or take. So now let's plot them. Negative 2, 1/25. So 1/25 is going to be really, really close to the x-axis. That's about 1/25. So that is negative 2, 1/25. It's not going to be on the x-axis. 1/25 is obviously greater than 0. It's going to be really, really, really, really, close. Now let's do this point here in orange, negative 1, 1/5. Negative 1/5-- 1/5 on this scale is still pretty close. It's pretty close. So that right over there is negative 1, 1/5. And now in blue, we have 0 comma 1. 0 comma 1 is going to be right about there. If this is 2 and 1/2, that looks about right for 1. And then we have 1 comma 5. 1 comma 5 puts us right over there. And then finally, we have 2 comma 25. When x is 2, y is 25. 2 comma 25 puts us right about there. And so I think you see what happens with this function, The further in the negative direction we go, 5 to ever-increasing negative powers gets closer and closer to 0, but never quite. So we're leaving 0, getting slightly further, further, further from 0. Right at the y-axis, we have y equal 1. Right at x is equal to 0, we have y is equal to 1. And then once x starts increasing beyond 0, then we start seeing what the exponential is good at, which is just this very rapid increase. Some people would call it an exponential increase, which is obviously the case right over here. So then if I just keep this curve going, you see it's just going on this sometimes called a hockey stick. It just keeps on going up like this at a super fast rate, ever-increasing rate. So you could keep going forever to the left, and you'd get closer and closer and closer to 0 without quite getting to 0. So 5 to the negative billionth power" + }, + { + "Q": "\n1:30 how can 5^0=1?", + "A": "You have to think of it in groups a bit, but also considering the increase/decrease. Ex: 2^3, or 2x2x2=8 2^2, or 2x2=4, with a decrease/increase of 4. 2^1 or 2 = 2, with a decrease/increase of 2. So the number is basically halved each time, specifically only for 2s. For 3s it would be divided in to thirds, 4s would be fourths and so forth. So what s one half of 2? 2 x 1/2 cross multiply: 1 x 1/1 = 1. for 3s: 3 x 1/3 cross multiply: 1 x 1/1= 1. and so forth. Posted this above, hope it helps :D", + "video_name": "9SOSfRNCQZQ", + "timestamps": [ + 90 + ], + "3min_transcript": "We're asked to graph y is equal to 5 to the x-th power. And we'll just do this the most basic way. We'll just try out some values for x and see what we get for y. And then we'll plot those coordinates. So let's try some negative and some positive values. And I'll try to center them around 0. So this will be my x values. This will be my y values. Let's start first with something reasonably negative but not too negative. So let's say we start with x is equal to negative 2. Then y is equal to 5 to the x power, or 5 to the negative 2 power, which we know is the same thing as 1 over 5 to the positive 2 power, which is just 1/25. Now let's try another value. What happens when x is equal to negative 1? Then y is 5 to the negative 1 power, which is the same thing as 1 over 5 to the first power, or just 1/5. Now let's think about when x is equal to 0. Then y is going to be equal to 5 to the 0-th power, which is going to be equal to 1. So this is going to be equal to 1. And then finally, we have-- well, actually, let's try a couple of more points here. Let me extend this table a little bit further. Let's try out x is equal to 1. Then y is 5 to the first power, which is just equal to 5. And let's do one last value over here. Let's see what happens when x is equal to 2. Then y is 5 squared, 5 to the second power, which is just equal to 25. And now we can plot it to see how this actually looks. So let me get some graph paper going here. My x's go as low as negative 2, as high as positive 2. And then my y's go all the way from 1/25 all the way to 25. So I have positive values over here. So let me draw it like this. That could be my x-axis. And then let's make this my y-axis. I'll draw it as neatly as I can. So let's make that my y-axis. And my x values, this could be negative 2. Actually, make my y-axis keep going. So that's y. This is x. That's a negative 2. That's negative 1. That's 0. That is 1. And that is positive 2. And let's plot the points. x is negative 2. y is 1/25. Actually, let me make the scale on the y-axis. So let's make this. So we're going to go all the way to 25. So let's say that this is 5. Actually, I have to do it a little bit smaller than that, too. So this is going to be 5, 10, 15, 20. And then 25 would be right where I wrote the y, give or take. So now let's plot them. Negative 2, 1/25." + }, + { + "Q": "on 4:11 it says to the 3rd power but how because in the formula it says to the 2nd power so I don't know how he got 3rd power\n", + "A": "The units of volume is to the third power because it is related to multiplying length \u00e2\u0080\u00a2 width \u00e2\u0080\u00a2 height. Radius squared provides two, height is the third direction.", + "video_name": "hC6zx9WAiC4", + "timestamps": [ + 251 + ], + "3min_transcript": "of this entire can, this entire cylinder. But if you just want the cone, it's 1/3 of that. It is 1/3 of that. And that's what I mean when I say it's surprisingly clean that this cone right over here is 1/3 the volume of this cylinder that is essentially-- you could think of this cylinder as bounding around it. Or if you wanted to rewrite this, you could write this as 1/3 times pi or pi/3 times hr squared. However you want to view it. The easy way I remember it? For me, the volume of a cylinder is very intuitive. You take the area of the base. And then you multiply that times the height. And so the volume of a cone is just 1/3 of that. It's just 1/3 the volume of the bounding cylinder is one way to think about it. But let's just apply these numbers, just to make sure that it makes sense to us. So let's say that this is some type of a conical glass, And let's say that we're told that it holds 131 cubic centimeters of water. And let's say that we're told that its height right over here-- I want to do that in a different color. We're told that the height of this cone is 5 centimeters. And so given that, what is roughly the radius of the top of this glass? Let's just say to the nearest 10th of a centimeter. Well, we just once again have to apply the formula. The volume, which is 131 cubic centimeters, is going to be equal to 1/3 times pi times the height, which is 5 centimeters, times the radius squared. If we wanted to solve for the radius squared, we could just divide both sides by all of this business. equal to 131 centimeters to the third-- or 131 cubic centimeters, I should say. You divide by 1/3. That's the same thing as multiplying by 3. And then, of course, you're going to divide by pi. And you're going to divide by 5 centimeters. Now let's see if we can clean this up. Centimeters will cancel out with one of these centimeters. So you'll just be left with square centimeters only in the numerator. And then to solve for r, we could take the square root of both sides. So we could say that r is going to be equal to the square root of-- 3 times 131 is 393 over 5 pi. So that's this part right over here. Once again, remember we can treat units just" + }, + { + "Q": "\nI know at 3:04 he gave the formula for finding the radius, but how do you change it to find the height?", + "A": "If you already know the volume and the radius, to find the height, the formula wound be the following: h = 3 time v divided by pi times r squared.", + "video_name": "hC6zx9WAiC4", + "timestamps": [ + 184 + ], + "3min_transcript": "You could call this distance right over here h. And the formula for the volume of a cone-- and it's interesting, because it's close to the formula for the volume of a cylinder in a very clean way, which is somewhat surprising. And that's what's neat about a lot of this three-dimensional geometry is that it's not as messy as you would think it would be. It is the area of the base. Well, what's the area of the base? The area of the base is going to be pi r squared. It's going to be pi r squared times the height. And if you just multiplied the height times pi r squared, that would give you the volume of an entire cylinder that looks something like that. So this would give you this entire volume of the figure that looks like this, where its center of the top is the tip right over here. So if I just left it as pi r squared of this entire can, this entire cylinder. But if you just want the cone, it's 1/3 of that. It is 1/3 of that. And that's what I mean when I say it's surprisingly clean that this cone right over here is 1/3 the volume of this cylinder that is essentially-- you could think of this cylinder as bounding around it. Or if you wanted to rewrite this, you could write this as 1/3 times pi or pi/3 times hr squared. However you want to view it. The easy way I remember it? For me, the volume of a cylinder is very intuitive. You take the area of the base. And then you multiply that times the height. And so the volume of a cone is just 1/3 of that. It's just 1/3 the volume of the bounding cylinder is one way to think about it. But let's just apply these numbers, just to make sure that it makes sense to us. So let's say that this is some type of a conical glass, And let's say that we're told that it holds 131 cubic centimeters of water. And let's say that we're told that its height right over here-- I want to do that in a different color. We're told that the height of this cone is 5 centimeters. And so given that, what is roughly the radius of the top of this glass? Let's just say to the nearest 10th of a centimeter. Well, we just once again have to apply the formula. The volume, which is 131 cubic centimeters, is going to be equal to 1/3 times pi times the height, which is 5 centimeters, times the radius squared. If we wanted to solve for the radius squared, we could just divide both sides by all of this business." + }, + { + "Q": "\nAt 15:20 Would we be wrong if we write \u00e2\u0088\u00ab f cos\u00ce\u00b8 dt?", + "A": "You would want \u00e2\u0088\u00ab ||f||.||r|| cos\u00ce\u00b8 dt, with the magnitude of f and r, and \u00ce\u00b8 all defined as functions of t, which given the way Sal has set up the problem (with f and r defined as rectangular functions of t) would be difficult.", + "video_name": "t3cJYNdQLYg", + "timestamps": [ + 920 + ], + "3min_transcript": "How do we actually calculate something like this? Especially because we have everything parameterized in terms of t. How do we get this in terms of t? And if you just think about it, what is f dot r? Or what is f dot dr? Well, actually, to answer that, let's remember what dr looked like. If you remember, dr/dt is equal to x prime of t, I'm writing it like, I could have written dx dt if I wanted to do, times the i-unit vector, plus y prime of t, times the j-unit vector. And if we just wanted to dr, we could multiply both sides, if we're being a little bit more hand-wavy with the differentials, not too rigorous. We'll get dr is equal to x prime of t dt times the unit vector i plus y prime of t times the differential dt So this is our dr right here. And remember what our vector field was. It was this thing up here. Let me copy and paste it. And we'll see that the dot product is actually not so crazy. So copy, and let me paste it down here. So what's this integral going to look like? This integral right here, that gives the total work done by the field, on the particle, as it moves along that path. Just super fundamental to pretty much any serious physics that you might eventually find yourself doing. So you could say, well gee. It's going to be the integral, let's just say from t is equal to a, to t is equal to b. Right? a is where we started off on the path, t is equal to a to t is equal to b. You can imagine it as being timed, as a particle And then what is f dot dr? Well, if you remember from just what the dot product is, you can essentially just take the product of the corresponding components of your of vector, and add them up. So this is going to be the integral from t equals a to t equals b, of p of p of x, really, instead of writing x, y, it's x of t, right? x as a function of t, y as a function of t. So that's that. Times this thing right here, times this component, right? We're multiplying the i-components. So times x prime of t d t, and then that plus, we're going to do the same thing with the q function. So this is q plus, I'll go to another line. Hopefully you realize I could have just kept writing, but I'm running out of space. Plus q of x of t, y of t, times the component of our dr. Times" + }, + { + "Q": "\nHolla. At 06:00, \"add one to itself and times.\" I know, sounds silly, but cannot get it, since adding one to itself is two. What am I missing here?", + "A": "He actually said (or at least he meant) add one to itself n times . If n = 4 then it would be 1 + 1 + 1 + 1 = 4 = n.", + "video_name": "sRVGcYGjUk8", + "timestamps": [ + 360 + ], + "3min_transcript": "this is just a constant term. This is just a constant term. So you can take it out. Times mu squared times the sum from i equals 1 to n. And what's going to be here? It's going to be a 1. We just divided a 1. We just divided this by 1. And took it out of the sigma sign, out of the sum. And you're just left with a 1 there. And actually, we could have just left the mu squared there. But either way, let's just keep simplifying it. So this we can't really do-- well, actually we could. Well, no, we don't know what the x sub i's are. So we just have to leave that the same. So that's the sum. Oh sorry, and this is just the numerator. This whole simplification, we're just simplifying the numerator. And later, we're just going to divide by n. So that is equal to that divided by n, which is equal to this thing divided by n. Because it's the numerator that's the confusing part. We just want to simplify this term up here. So let's keep doing this. So this equals the sum from i equals 1 to n of x sub i squared. And let's see, minus 2 times mu-- sorry, that mu doesn't look good. Edit, Undo, minus 2 times mu times the sum from i is equal to 1 to n of xi. And then, what is this? What is another way to write this? Essentially, we're going to add 1 to itself n times. This is saying, just look, whatever you have here, just iterate through it n times. If you had an x sub i here, you would use the first x term, then the second x term. When you have a 1 here, this is just essentially saying, add one to itself n times, which is the same thing as n. And then see if there's anything else we can do here. Remember, this was just the numerator. So this looks fine. We add up each of those terms. So we just have minus 2 mu from i equals 1 to-- oh well, think about this. What is this? What is this thing right here? Well actually, let's bring back that n. So this simplified to that divided by n, which simplifies to that whole thing, which is simplified to this whole thing, divided by n, which simplifies to this whole thing divided by n, which is the same thing as each of the terms divided by n, which" + }, + { + "Q": "\nIsn't the \"proof\" for the equivalence of the angles formed by a transversal across two parallel lines (4:09) simply that if the angles were not equivalent the lines would meet and not be parallel? Is that not considered a proof? If not, why not?", + "A": "Yes it is. People think it isn t but Khan knows better\u00f0\u009f\u0098\u009c", + "video_name": "H-E5rlpCVu4", + "timestamps": [ + 249 + ], + "3min_transcript": "This is a transversal line. It is transversing both of these parallel lines. This is a transversal. And what I want to think about is the angles that are formed, and how they relate to each other. The angles that are formed at the intersection between this transversal line and the two parallel lines. So we could, first of all, start off with this angle right over here. And we could call that angle-- well, if we made some labels here, that would be D, this point, and then something else. But I'll just call it this angle right over here. We know that that's going to be equal to its vertical angles. So this angle is vertical with that one. So it's going to be equal to that angle right over there. We also know that this angle, right over here, is going to be equal to its vertical angle, or the angle that is opposite the intersection. So it's going to be equal to that. And sometimes you'll see it specified like this, where you'll see a double angle mark like that. Or sometimes you'll see someone write and these two are equal right over here. Now the other thing we know is we could do the exact same exercise up here, that these two are going to be equal to each other and these two are going to be equal to each other. They're all vertical angles. What's interesting here is thinking about the relationship between that angle right over there, and this angle right up over here. And if you just look at it, it is actually obvious what that relationship is-- that they are going to be the same exact angle, that if you put a protractor here and measured it, you would get the exact same measure up here. And if I drew parallel lines-- maybe I'll draw it straight left and right, it might be a little bit more obvious. So if I assume that these two lines are parallel, and I have a transversal here, what I'm saying is that this angle is going to be the exact same measure as that angle there. And to visualize that, just imagine tilting this line. And as you take different-- so it If you take the line like this and you look at it over here, it's clear that this is equal to this. And there's actually no proof for this. This is one of those things that a mathematician would say is intuitively obvious, that if you look at it, as you tilt this line, you would say that these angles are the same. Or think about putting a protractor here to actually measure these angles. If you put a protractor here, you'd have one side of the angle at the zero degree, and the other side would specify that point. And if you put the protractor over here, the exact same thing would happen. One side would be on this parallel line, and the other side would point at the exact same point. So given that, we know that not only is this side equivalent to this side, it is also equivalent to this side over here. And that tells us that that's also equivalent to that side over there. So all of these things in green are equivalent. And by the same exact argument, this angle" + }, + { + "Q": "At 6:49, what does \"deduce\" mean?\n", + "A": "Deduce means to make a conclusion given the evidence that you have,", + "video_name": "H-E5rlpCVu4", + "timestamps": [ + 409 + ], + "3min_transcript": "If you take the line like this and you look at it over here, it's clear that this is equal to this. And there's actually no proof for this. This is one of those things that a mathematician would say is intuitively obvious, that if you look at it, as you tilt this line, you would say that these angles are the same. Or think about putting a protractor here to actually measure these angles. If you put a protractor here, you'd have one side of the angle at the zero degree, and the other side would specify that point. And if you put the protractor over here, the exact same thing would happen. One side would be on this parallel line, and the other side would point at the exact same point. So given that, we know that not only is this side equivalent to this side, it is also equivalent to this side over here. And that tells us that that's also equivalent to that side over there. So all of these things in green are equivalent. And by the same exact argument, this angle And that's going to be the same as this angle, because they are opposite, or they're vertical angles. Now the important thing to realize is just what we've deduced here. The vertical angles are equal and the corresponding angles at the same points of intersection are also equal. And so that's a new word that I'm introducing right over here. This angle and this angle are corresponding. They represent kind of the top right corner, in this example, of where we intersected. Here they represent still, I guess, the top or the top right corner of the intersection. This would be the top left corner. They're always going to be equal, corresponding angles. And once again, really, it's, I guess, for lack of a better word, it is a bit obvious. Now on top of that, there are other words that people will see. We've essentially just proven that not only is this angle equivalent to this angle, but it's also equivalent to this angle right over here. And these two angles-- let me label them so that we can make some headway here. for the angles themselves. So let's call this lowercase a, lowercase b, lowercase c. So lowercase c for the angle, lowercase d, and then let me call this e, f, g, h. So we know from vertical angles that b is equal to c. But we also know that b is equal to f because they are corresponding angles. And that f is equal to g. So vertical angles are equivalent, corresponding angles are equivalent, and so we also know, obviously, that b is equal to g. And so we say that alternate interior angles are equivalent. So you see that they're kind of on the interior of the intersection. They're between the two lines, but they're on all opposite sides of the transversal. Now you don't have to know that fancy word, alternate interior angles, you really just have to deduce what we just saw over here. Know that vertical angles are going to be equal and corresponding angles are going to be equal. And you see it with the other ones, too. We know that a is going to be equal to d, which is going to be equal to h, which is going to be equal to e." + }, + { + "Q": "\nAt 1:46, do the lines have to be parallel for the line to be a transversal?", + "A": "No, not necessarily. A transversal is a line/line segment/ ray intersecting two other lines/line segments/ rays.", + "video_name": "H-E5rlpCVu4", + "timestamps": [ + 106 + ], + "3min_transcript": "Let's say we have two lines over here. Let's call this line right over here line AB. So A and B both sit on this line. And let's say we have this other line over here. We'll call this line CD. So it goes through point C and it goes through point D. And it just keeps on going forever. And let's say that these lines both sit on the same plane. And in this case, the plane is our screen, or this little piece of paper that we're looking at right over here. And they never intersect. So they're on the same plane, but they never intersect each other. If those two things are true, and when they're not the same line, they never intersect and they can be on the same plane, then we say that these lines are parallel. They're moving in the same general direction, in fact, the exact same general direction. If we were looking at it from an algebraic point of view, we would say that they have the same slope, but they have different y-intercepts. They involve different points. they would intersect that at a different point, but they would have the same exact slope. And what I want to do is think about how angles relate to parallel lines. So right over here, we have these two parallel lines. We can say that line AB is parallel to line CD. Sometimes you'll see it specified on geometric drawings like this. They'll put a little arrow here to show that these two lines are parallel. And if you've already used the single arrow, they might put a double arrow to show that this line is parallel to that line right over there. Now with that out of the way, what I want to do is draw a line that intersects both of these parallel lines. So here's a line that intersects both of them. Let me draw a little bit neater than that. So let me draw that line right over there. Well, actually, I'll do some points over here. Well, I'll just call that line l. And this line that intersects both of these parallel lines, This is a transversal line. It is transversing both of these parallel lines. This is a transversal. And what I want to think about is the angles that are formed, and how they relate to each other. The angles that are formed at the intersection between this transversal line and the two parallel lines. So we could, first of all, start off with this angle right over here. And we could call that angle-- well, if we made some labels here, that would be D, this point, and then something else. But I'll just call it this angle right over here. We know that that's going to be equal to its vertical angles. So this angle is vertical with that one. So it's going to be equal to that angle right over there. We also know that this angle, right over here, is going to be equal to its vertical angle, or the angle that is opposite the intersection. So it's going to be equal to that. And sometimes you'll see it specified like this, where you'll see a double angle mark like that. Or sometimes you'll see someone write" + }, + { + "Q": "at 3:30 when Sal is discussing lim x--> c for f(x)/g(x), you can only calculate the limit if the bottom is not zero!\n", + "A": "That is true until you learn L Hospital s rule and then it will be possible to calculate the limit of a function when g(x) is zero or when f(x) and g(x) are both infinity", + "video_name": "lSwsAFgWqR8", + "timestamps": [ + 210 + ], + "3min_transcript": "plus the limit of g of x as x approaches c. Which is equal to, well this right over here is-- let me do that in that same color-- this right here is just equal to L. It's going to be equal to L plus M. This right over here is equal to M. Not too difficult. This is often called the sum rule, or the sum property, of limits. And we could come up with a very similar one with differences. The limit as x approaches c of f of x minus g of x, is just going to be L minus M. It's just the limit of f of x as x approaches c, minus the limit of g of x as x approaches c. So it's just going to be L minus M. And we also often call it the difference rule, or the difference property, of limits. intuitive. Now what happens if you take the product of the functions? The limit of f of x times g of x as x approaches c. Well lucky for us, this is going to be equal to the limit of f of x as x approaches c, times the limit of g of x, as x approaches c. Lucky for us, this is kind of a fairly intuitive property of limits. So in this case, this is just going to be equal to, this is L times M. This is just going to be L times M. Same thing, if instead of having a function here, we had a constant. So if we just had the limit-- let me do it in that same color-- the limit of k times f of x, as x approaches c, where k is just some constant. This is going to be the same thing as k times the limit And that is just equal to L. So this whole thing simplifies to k times L. And we can do the same thing with difference. This is often called the constant multiple property. We can do the same thing with differences. So if we have the limit as x approaches c of f of x divided by g of x. This is the exact same thing as the limit of f of x as x approaches c, divided by the limit of g of x as x approaches c. Which is going to be equal to-- I think you get it now-- this is going to be equal to L over M. And finally-- this is sometimes called the quotient property-- finally we'll look at the exponent property." + }, + { + "Q": "At 2:52, Sal said, \"anything divided by anything is just one.\" I think what he meant to say was \"anything divided by itself is just one.\" :)\n", + "A": "Yeah, Sal was referring to the first anything when he said anything the second time : )", + "video_name": "UquFdMg6Z_U", + "timestamps": [ + 172 + ], + "3min_transcript": "The other way to think about it is that we're multiplying this entire expression. So this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you-- they're are all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here, we could just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. And then you have x to the fourth divided by x to the-- well, they don't tell us. But if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the 4 minus 1 power, or x Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1/2. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just-- well, I'll write it. I could write a 1 here. And let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. You could view it two ways-- anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1, Either way, you knew how to do this before you even learned that exponent property. Because x divided by x is 1, and then assuming x does not equal to 0. And we kind of have to assume x doesn't equal 0 in this whole thing. Otherwise, we would be dividing by 0. And then finally, we have 4 over 6x. And there's a couple of ways to think about it. So the simplest way is negative 4 over 6 is the same thing as negative 2/3. Just simplified that fraction. And we're multiplying that times 1/x. So we can view this 4 times 1/x. Another way to think about it is you could have viewed this 4 as being multiplied by x to the 0 power, and this being x to the first power. And then when you tried to simplify it using your exponent properties, you would have-- well, that would be x to the 0 minus 1 power, which is x to the negative 1 power." + }, + { + "Q": "\nAt 0:30, why couldn't long division work?", + "A": "That is another way to do basically the same thing. Whichever way is easiest for you!", + "video_name": "UquFdMg6Z_U", + "timestamps": [ + 30 + ], + "3min_transcript": "Simplify the expression 18x to the fourth minus 3x squared plus 6x minus 4, all of that over 6x. So there's a couple ways to think about them. They're all really equivalent. You can really just view this up here as being the exact same thing as 18x to the fourth over 6x plus negative 3x squared over 6x, or you could say minus 3x squared over 6x, plus 6x over 6x, minus 4 over 6x. Now, there's a couple of ways to think about it. One is I just kind of decomposed this numerator up here. If I just had a bunch of stuff, a plus b plus c over d, that's clearly equal to a/d plus b/d plus c/d. Or maybe not so clearly, but hopefully that helps clarify it. Another way to think about it is kind of like you're distributing the division. If I divide a whole expression by something, that's equivalent to dividing each of the terms The other way to think about it is that we're multiplying this entire expression. So this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you-- they're are all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here, we could just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. And then you have x to the fourth divided by x to the-- well, they don't tell us. But if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the 4 minus 1 power, or x Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1/2. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just-- well, I'll write it. I could write a 1 here. And let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. You could view it two ways-- anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1," + }, + { + "Q": "At 2:14 to 2:32, could the -\u00c2\u00bdx also be written as -x/2?\n", + "A": "Yes, they are the same.", + "video_name": "UquFdMg6Z_U", + "timestamps": [ + 134, + 152 + ], + "3min_transcript": "Simplify the expression 18x to the fourth minus 3x squared plus 6x minus 4, all of that over 6x. So there's a couple ways to think about them. They're all really equivalent. You can really just view this up here as being the exact same thing as 18x to the fourth over 6x plus negative 3x squared over 6x, or you could say minus 3x squared over 6x, plus 6x over 6x, minus 4 over 6x. Now, there's a couple of ways to think about it. One is I just kind of decomposed this numerator up here. If I just had a bunch of stuff, a plus b plus c over d, that's clearly equal to a/d plus b/d plus c/d. Or maybe not so clearly, but hopefully that helps clarify it. Another way to think about it is kind of like you're distributing the division. If I divide a whole expression by something, that's equivalent to dividing each of the terms The other way to think about it is that we're multiplying this entire expression. So this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you-- they're are all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here, we could just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. And then you have x to the fourth divided by x to the-- well, they don't tell us. But if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the 4 minus 1 power, or x Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1/2. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just-- well, I'll write it. I could write a 1 here. And let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. You could view it two ways-- anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1," + }, + { + "Q": "at around 1:15 you say that you subtract 4x from each side so wouldn't it be 2y = 8-4x not 2y=4x-8\n", + "A": "The equation was 4x + 2y = - 8 You simply missed out the negative signs. Subtracting 4x from the equation above will give us: 2y = -8 -4x which is the same as 2y = -4x -8 If you cannot understand why changing orders will still be the same, try taking -2 -4 and -4 -2. Either way, you will get -6! Same analogy here :) Hope this helped", + "video_name": "V6Xynlqc_tc", + "timestamps": [ + 75 + ], + "3min_transcript": "We're asked to convert these linear equations into slope-intercept form and then graph them on a single coordinate plane. We have our coordinate plane over here. And just as a bit of a review, slope-intercept form is a form y is equal to mx plus b, where m is the slope and b is the intercept. That's why it's called slope-intercept form. So we just have to algebraically manipulate these equations into this form. So let's start with line A, so start with a line A. So line A, it's in standard form right now, it's 4x plus 2y is equal to negative 8. The first thing I'd like to do is get rid of this 4x from the left-hand side, and the best way to do that is to subtract 4x from both sides of this equation. So let me subtract 4x from both sides. The left hand side of the equation, these two 4x's cancel out, and I'm just left with 2y is equal to. or negative 8 minus 4, however you want to do it. Now we're almost at slope-intercept form. We just have to get rid of this 2, and the best way to do that that I can think of is divide both sides of this equation by 2. So let's divide both sides by 2. So we divide the left-hand side by 2 and then divide the right-hand side by 2. You have to divide every term by 2. And then we are left with y is equal to negative 4 divided by 2 is negative 2x. Negative 8 divided by 2 is negative 4, negative 2x minus 4. So this is line A, let me graph it right now. So line A, its y-intercept is negative 4. So the point 0, negative 4 on this graph. If x is equal to 0, y is going to be equal to negative 4, you So 0, 1, 2, 3, 4. That's the point 0, negative 4. That's the y-intercept for line A. And then the slope is negative 2x. So that means that if I change x by positive 1 that y goes down by negative 2. So let's do that. So if I go over one in the positive direction, I have to go down 2, that's what a negative slope's going to do, negative 2 slope. If I go over 2, I'm going to have to go down 4. If I go back negative 1, so if I go in the x direction negative 1, that means in the y direction I go positive two, because two divided by negative one is still negative two, so I go over here. If I go back 2, I'm going to go up 4. Let me just do that. Back 2 and then up 4. So this line is going to look like this." + }, + { + "Q": "10:40. How can you define a function S such that S(y) is the unique solution in X to f(x) = y? How can you know such a function even exists?\n", + "A": "Sal assumed that for any point in Y there is a unique solution to f(x) = y. That means that there must be a function. Remember, a function is anything that maps each value of y to a unique x.", + "video_name": "7GEUgRcnfVE", + "timestamps": [ + 640 + ], + "3min_transcript": "you have to be a little bit precise about it in order to get to the point you want. Let's see if the opposite is true. Let's see if we assume-- let's see if we start from the assumption, that for all y that is a member of our set Y, that the solution, that the equation f of x, is equal to y has a unique solution. Let's assume this and see if it can get us the other way. If given this, we can prove invertibility. So let's think about the first way. So we're saying that for any y-- let me draw my sets again. So this is my set X and this is my set Y right there. Now we're working for the assumption that you can pick here has a unique solution. Let's call that unique solution. Well, we could call it whatever. But a unique solution x. So you can pick any point here, and I've given you, we're assuming now, that, look, you pick a point in Y, I can find you some point in X such that f of x is equal to y. And not only can I find that for you, that is a unique solution. So given that, let me define a new function. Let me define the function s. The function s is a mapping from y to x. It's a mapping from y to x and s of, let's say s of y, where, of course, y is a member of our set capital-Y. s of y is equal to the unique solution in x to f of x is Now, you're saying, hey, Sal, it looks a little convoluted. But think about it. This is a completely valid function definition. Right? We're starting with the idea that you give me any y here. You give me any member of this set, and I can always find you a unique solution to this equation. Well OK, so that means that any guy here can be associated with a unique solution in the set X, where the unique solution is the unique solution to this equation here. So, why don't I just define a function that says, look I'm going to associate every member y with its unique solution to f of x is equal to y. That's how I'm defining this function right here. And, of course, this is a completely valid mapping from y to x. And we know that this only has one legitimate value because this, any value y, any lower-case value y, in this set has a unique solution to f of x is equal to y." + }, + { + "Q": "\n@ 3:16 I didnt get how' 2 is the same thing as e to the natural log of 2 ' ?can someone please explain?", + "A": "e^x and lnx are opposite functions. It would be akin to multiplying by some number and then dividing by that same number. The operations cancel one another out. e to the natural log of elephant = elephant.", + "video_name": "C5Lbjbyr1t4", + "timestamps": [ + 196 + ], + "3min_transcript": "log of x is equal to 1/x. So we have some expression, and we have its derivative, which tells us that we can use substitution. Sometimes you can do in your head, but this problem, it's still not trivial to do in your head. So let's make the substitution. Let's substitute this right here with a u. So let's do that. So if you define u, and it doesn't have to be u, it's just, that's the convention, it's called u-substitution, it could have been s-substitution for all we care. Let's say u is equal to the natural log of x, and then du dx, the derivative of u with respect to x, of course is equal to 1/x. Or, just the differential du, if we just multiply both sides by dx is equal to 1 over x dx. So let's make our substitution. So this will be equal to the indefinite integral, or the antiderivative, of 2 to the now u, so 2 to the u, Now what is 1 over x dx? That's just du. So this term times that term is just our du. Let me do it in a different color. 1 over x times dx is just equal to du. That's just equal to that thing, right there. Now, this still doesn't look like an easy integral, although it's gotten simplified a good bit. And to solve, you know, whenever I see the variable that I'm integrating against in the exponent, you know, we don't have any easy exponent rules here. The only thing that I'm familiar with, where I have my x or my variable that I'm integrating against in my exponent, is the case of e to the x. We know that the integral of e to the x, dx, is equal to e to the x plus c. So if I could somehow turn this into some variation of e to the x, maybe, or e to the u, maybe I can make this integral a So let's see. How can we redefine this right here? Well, 2, 2 is equal to what? 2 is the same thing as e to the natural log of 2, right? The natural log of 2 is the power you have to raise e to to get 2. So if you raise e to that power, you're, of course going to get 2. This is actually the definition of really, the natural log. You raise e to the natural log of 2, you're going to get 2. So let's rewrite this, using this-- I guess we could call this this rewrite or-- I don't want to call it quite a substitution. It's just a different way of writing the number 2. So this will be equal to, instead of writing the number 2, I could write e to the natural log of 2. And all of that to the u du. And now what is this equal to? Well, if I take something to an exponent, and then to another" + }, + { + "Q": "At 2:42 Sal says he wants to multiply the length of vector v by the vector itself (||v|| v), but then immediately after he defines the unit vector as \"1/||v|| v\". Where did he get the \"1/\" from?\n", + "A": "If a vector v has a length of e.g. 5, then v/5 has length 1.", + "video_name": "lQn7fksaDq0", + "timestamps": [ + 162 + ], + "3min_transcript": "u, it's just equal to the square root of the squared sums of all of its components. And if you think about it, this is just an extension of the Pythagorean theorem, to some degree. So it's u1 squared plus u2 squared all the way to un squared. And it's the square root of that. If this is a unit vector, if this is a unit vector, so this is a unit vector, that implies that the length of u will be equal to 1. And that doesn't matter in what dimension space we are. This could be R100 this could be R2. For it to have a unit vector in any of those spaces, their length is 1. The next obvious question is, how do you construct a unit vector? Let's say that I have some vector, v. And let's say it's not a unit vector. And I want to turn it into some vector u that is a unit vector, that just goes in the same direction. So u will go in the same direction, as v, but the length of u is going to be equal to 1. How do I construct this vector u here? What I could do is, I could take the length of v. I could find out what the length of v is, and we know how to do that. We just apply this definition of vector length. And what happens if I figure out the length of v, and then I multiply the vector v times that? What if I make my u, what if I say u is equal to, 1 over the length of v times v itself? What happens here? here, what do I get? The length of u is equal to the length of this scalar. Remember this is just some number, right? It's equal to this scalar, and I'm assuming v is a non-zero vector. The length of whatever this scalar number is times v. And we know that we can take this scalar out of the formula, we can show that -- I think I've shown it in a previous video -- that the length of c times v is equal to c times the length of v. Let me write that down. And that's essentially what I'm assuming right here. That if I take the length of c times some vector, v, that is equal to c times the length of v. I think we showed this when we first were introduced to the idea of length. So we know that this is going to be equal to 1 over the" + }, + { + "Q": "At 0:50,why is it that dividing is the same as multiplying the fractions?\n", + "A": "Dividing by a fraction is the same as multiplying by that fraction with it s denominator and numerator swapped.", + "video_name": "yb7lVnY_VCY", + "timestamps": [ + 50 + ], + "3min_transcript": "Tommy is studying for final exams this weekend. He will spend 1/5 of the weekend studying. What fraction of the weekend will he spend studying for each of his 4 subjects if he spends the same amount of time studying for each subject? So the total amount of time he's going to spend studying this weekend is 1/5 of the weekend. And he has to divide that into 4 equal sections. And he's going to spend that much time on each subject. So he's going to divide this by 4. Now, we've already seen that dividing by a number is the same thing as multiplying by its reciprocal. You might say, hey, well, what's the reciprocal of 4? You just have to remind yourself that 4 is the same thing as 4/1. So 1/5 divided by 4/1 is the same thing as 1/5 times 1/4. And you could also view this as 1/4 of 1/5 or 1/5 But here we multiply our numerators to get 1. And then we multiply our denominators, 4 times 5 is 20. So you get 1/20 of the weekend will be spent studying for each subject. Now, let's also try to think about this visually. Let's imagine that this is his entire weekend. And I've divided it into 5 equal sections. And so we already know that the total amount of his weekend spent studying is 1/5. So that's the total amount studying for the weekend is 1/5. Now, he has to divide this into 4 equals section. So let's do that. He's got four subjects, and he's going to spend the same amount of time on each of the 4 subjects. So he's going to divide this into 4 equal sections. So how much time does he spend on one subject? Well, in each subject, that would be this little area that I'm doing in yellow right over here. And what is that? Well, that's 1 over-- how many equal sections are there Well, I've just drawn out the grid. You had 5 rows, and now you have 4 columns. So 5 rows times 4 columns, you have 20 equal sections. So once again, looking at it visually, he's spending 1/20 of his weekend on each of the 4 subjects. And then if you do this for 4 subjects, that means that in this whole weekend, 1/5 will be spent studying. But the question that they're asking, he's spending 1/20 of the weekend on each subject." + }, + { + "Q": "\nAt 2:32 you say that 5 + 5 cannot represent a vector. (5 represents a possible starting point on the number line and the additional 5 can represent a distance in the positive direction on the number line.)", + "A": "A vector has to have BOTH a magnitude and a direction. As is, it is just a magnitude. Had there been a direction attached then it would be a vector.", + "video_name": "n8Ic2Oj-zvA", + "timestamps": [ + 152 + ], + "3min_transcript": "Voiceover:Which of the following can represent a vector? A vector is something that has both a magnitude and a direction. Let's see which one of these could represent something that has a magnitude and a direction. The first choice right over here is the number 5. 5 could represent a magnitude, you could say, essentially, how large something is. That's all the information has. It doesn't say 5 in a certain direction. This one by itself would not be a or it couldn't represent a vector. You need to specify a direction as well. The angle measure 5 degrees. Well the angle measure 5 degrees could represent a direction. If you say it's 5 degrees. Let's say that's the positive X axis that's the positive Y axis. If you said it's 5 degrees relative to the positive X axis. If you said something like this, so if you had 5 degrees over there, that would be specifying a direction, but it's not giving us a magnitude. It's not saying how far in that direction or how large in that direction. you're giving us a direction, but no magnitude. Let's just cross these out. The point (5,5). Now this is interesting. If we say that the point is the end point of a vector that starts at the origin, so lets draw that out. That's the positive X axis, positive Y axis. This is 1,2,3,4,5 in the horizontal direction. 1,2,3,4,5 in the positive vertical direction. The .5,5 is going to be right over there. If you said that this represents the head of a vector who's tail is at the origin. The vector would look something like this. Now, you're giving both a magnitude and a direction. What's the magnitude? What's the distance between the origin and this point, right over here. and the direction, it's this general direction. This could specify a magnitude and a direction. This could represent a vector. The outcome of 5+5, obviously that's just 10. That's just a number. For the same reason, this first choice, if I just have a number, that could be a magnitude, but we don't have a direction as well. This could not represent and that one could not represent a vector." + }, + { + "Q": "I haven't had the time to find the video to explain this type of equation so I'll just ask it here. If I had an equation such as 6t3rd + 9t2nd \u00e2\u0080\u0093 15t(that's 6t to the third plus 9t to the second minus 15t)how would I factor this to a form similar to the one shown at 3:35?\n", + "A": "first factor out 3t so you get 3t(2t2nd+3t-5). Then factor (2t2nd+3t-5). to factor it, you need to find 2 numbers that multiply to make -10 (2*-5) and add together to make 3 (the middle number). those two numbers are 5 and -2. Split the middle term into those 2 numbers which gives you 3t(2t2nd-2t+5t-5). Then factor by grouping. that gives you 3t(2t(t-1)+5(t-1)). Simplify to 3t(2t+5)(t-1). The next video is very helpful in understanding this. Hope that helps!", + "video_name": "u1SAo2GiX8A", + "timestamps": [ + 215 + ], + "3min_transcript": "So we could try out things like 5 and 12, 5 and negative 12, because one has to be negative. If you add these two you get negative 7, if you did negative 5 and 12 you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10? Then you get a negative 4, if you added these two. But we want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our two numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So this 4y can be rewritten as negative 6y plus 10y, right? Because if you add those you get 4y. And then the other sides of it, you have your 4y squared, your 4y squared and then you have your minus 15. coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the term. Let me do it in a different color. So if I take these two guys, what can I factor out of those two guys? Well, there's a common factor, it looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared, divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. So this group gets factored into 2y times 2y, minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos I've explained why this works. Now here, the greatest common factor is a 5. So we can factor out a 5, so this is equal to plus 5 times 10y, divided by 5 is 2y. And so we have 2y times 2y minus 3, plus 5 times 2y minus 3. So now you have two terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3, so this is equal to 2y minus 3, times 2y, times that 2y, plus that 5. There's no magic happening here, all I did is undistribute the 2y minus 3. I factored it out of both of these guys. I took it out of the parentheses. If I distribute it in, you'd get back to this expression. But we're done, we factored it. We factored it into two binomial expressions. 4y squared plus 4y, minus 15 is 2y minus 3, times 2y plus 5." + }, + { + "Q": "\nAt 3:25, I'm confused about where the 5 came from. Can someone shed some light on this?", + "A": "They divided 10y-15 by two and got 5(2y-3).", + "video_name": "u1SAo2GiX8A", + "timestamps": [ + 205 + ], + "3min_transcript": "So we could try out things like 5 and 12, 5 and negative 12, because one has to be negative. If you add these two you get negative 7, if you did negative 5 and 12 you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10? Then you get a negative 4, if you added these two. But we want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our two numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So this 4y can be rewritten as negative 6y plus 10y, right? Because if you add those you get 4y. And then the other sides of it, you have your 4y squared, your 4y squared and then you have your minus 15. coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the term. Let me do it in a different color. So if I take these two guys, what can I factor out of those two guys? Well, there's a common factor, it looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared, divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. So this group gets factored into 2y times 2y, minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos I've explained why this works. Now here, the greatest common factor is a 5. So we can factor out a 5, so this is equal to plus 5 times 10y, divided by 5 is 2y. And so we have 2y times 2y minus 3, plus 5 times 2y minus 3. So now you have two terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3, so this is equal to 2y minus 3, times 2y, times that 2y, plus that 5. There's no magic happening here, all I did is undistribute the 2y minus 3. I factored it out of both of these guys. I took it out of the parentheses. If I distribute it in, you'd get back to this expression. But we're done, we factored it. We factored it into two binomial expressions. 4y squared plus 4y, minus 15 is 2y minus 3, times 2y plus 5." + }, + { + "Q": "In 3:49, how does he know that the triangle is equilateral?\n", + "A": "EC = EB as he has proven in 3:17 , and EB and CB are radiuses of the same circle, therefore EB = CB. And, if you think about it you will find out that you can say EC=EB=CB by combining the two relationships above. That s why it is equilateral.", + "video_name": "3n0LvI99-KM", + "timestamps": [ + 229 + ], + "3min_transcript": "So let me draw segment EC. I'll draw that as straight as possible. I can draw a better job of that, so segment EC. Now something becomes interesting. Because what is the relationship between triangle EBG and triangle ECG? Well, they both definitely share this side right over here. They both share side EG. And then BG is equal to GC, and they both have 90-degree angles. You have a 90-degree angle here. You have a 90-degree angle there. So you see by side-angle-side that these two triangles are going to be congruent. So we know that triangle EBG is congruent to triangle ECG-- I should emphasize the C, not the E-- And that also tells us that all of the corresponding angles and sides are going to be the same. So that tells us right there that EC is equal to EB. So we know that EB is equal to EC. And what also is equal to that length? Well, once again, this is the radius of the circle. BE is one radius of the circle going from the center to the arc. But so is BC. It is also a radius of the circle going from the center to the arc. So this is also equal to BC. So I could draw the three things right here. I'm referring to the entire thing, not just one of the segments, all of BC. So what kind of a triangle is this right over here, triangle BEC? And we know that because all three sides are equal, so that tells us that all of its angles are equal. So that tells us that the measure of angle BEC-- we're not done yet, but it gets us close-- is 60 degrees. So the measure of angle BEC right over there is 60 degrees. So that gives us part of the problem. BEC is part of the angle BED. If we can just figure out the measure of angle CED now, if we can figure out this angle right over here, we just add that to 60 degrees, and we're done. We've figured out the entire BED. Now, let's think about how we can do this right over here. So there's a couple of interesting things that we already do know. We know that this right over here is equal to the radius of the circle. And we also know that this length down here-- this is a square-- we know that this length down here is the same as this length up here, that these" + }, + { + "Q": "\nat 5:18 were did he get 9/3?????i understand the 3 but not the 9 plz help me", + "A": "i am the helper", + "video_name": "Zn-GbH2S0Dk", + "timestamps": [ + 318 + ], + "3min_transcript": "And in general if you wanna work this out before I give you how I do it that now's a good time to actually pause the video. And you could, you could try to work it out and then, play it again and, and see what I have to say about it. But assuming you wanna hear it, let me go and do it. So let's do the same thing. We, first of all, we can merge these two Xs on the left-hand side. Remember, you can't add the 5 and the 3 because the 3 is just a constant term while the 5 is 5 times x. But the 5 times x and the negative 7 times actually can merge. So 5, you just add the coefficient. So, it's 5 and negative 7. So, that becomes negative 2x minus 3 is equal to x plus 8. Now, if we wanna take this x that's on the right-hand side and put it over the left-hand side, we can just subtract x from both sides. to, these two Xs cancel out, is equal to 8. Now, we can just add 3 to both sides to get rid of that constant term 3 on left hand-side. These two 3's will cancel out. And you get minus 3x is equal to 11. Now, you just multiply both sides by negative one-third. And once again, this is just the same thing as dividing both sides by negative 3. And you get x equals negative 11 over 3. Actually let's, let's, just for fun, let's check this just to see. And the cool thing about algebra is if you have enough time, you can always make sure you got the right answer. So we have 5x, so we have 5 times negative 11 over 3. So that's, I'm just, I'm just gonna take this and substitute it back into the original equation. So you have minus 55 over 3, that's just 5 times negative 11 over 3, that's a 3, minus 3. And what's 3? Three could also be written as, minus 9 over 3. I'm skipping some steps, but I think you, you know your fractions pretty good by this point. So that's minus 9 over 3. And then, minus 7x is the same thing, as plus 77 over 3. Because we have the minus 7 times minus 11, so it's plus 77. And, and the equation is saying that should equal minus 11 over 3, that's what x is, plus an 8 is nothing more than 24 over 3. Let's add this up. Minus 55 minus 9, that's minus 64, if I'm right, yeah, that's minus 64." + }, + { + "Q": "\nAt 3:00 i didnt get how he got 8", + "A": "Sal doesn t seem to arriving at an 8 at 3:00. Maybe you mean how he gets -8 at 2:00? At that point he is adding -3 to both sides of the equation so he can isolate the term with x (i.e. the variable) on the left side of the = sign and the contants on the right side. When adding -3 to -5 you get -8.", + "video_name": "Zn-GbH2S0Dk", + "timestamps": [ + 180 + ], + "3min_transcript": "And we could do that by adding 7x to both sides, 7x. This is a review. We, we're adding the opposite. So, it's negative 7x, so we add 7x so that's why. And we do that, become the right side, these two will cancel. And the left side, we get 10x plus 3 equals, and on the right side, all we have left is the negative 5. Almost there, now we're at a level, what is this, a level two problem. And now we just have to take this 3 and move it to the other side. And we can do that by subtracting 3 from both sides. That's a 3 minus 3. The left-hand side, the 3s cancel out, that's why we subtract it in the first place. And you have 10x equals and then minus 5 minus 3, well that equals minus 8. the reciprocal of 10, which is the coefficient on x, times 1 over 10. You could also, some people would say, well, we're just dividing both side by 10 which is essential what we're doing. If you divide by 10, that's the same thing as multiplying by 1 over 10. Well, anyway, the left-hand side, 1 over 10 times 10. Well, that equals 1, so we're just left with x equals negative 8 over 10. And that can be reduced further. They both share the common factor 2. So you divide by 2. So it's minus 4 over 5. I think that's right, assuming that I didn't make any careless mistakes. Let's do another problem. Let's say I had 5, that's a 5x minus 3 minus And in general if you wanna work this out before I give you how I do it that now's a good time to actually pause the video. And you could, you could try to work it out and then, play it again and, and see what I have to say about it. But assuming you wanna hear it, let me go and do it. So let's do the same thing. We, first of all, we can merge these two Xs on the left-hand side. Remember, you can't add the 5 and the 3 because the 3 is just a constant term while the 5 is 5 times x. But the 5 times x and the negative 7 times actually can merge. So 5, you just add the coefficient. So, it's 5 and negative 7. So, that becomes negative 2x minus 3 is equal to x plus 8. Now, if we wanna take this x that's on the right-hand side and put it over the left-hand side, we can just subtract x from both sides." + }, + { + "Q": "At 5:41, Sal says that he could define the f(x) as -x^2-3x+28 or (x+7)(x-4), as they are the same thing.\nWhen you graph -x^2-3x+28 you get a parabola opening up\nWhen you graph (x+7)(x-4) you get a graph opening down\n\nHow can they be the same thing if their graphs are not the same?\n", + "A": "Ok, so I think he misspoke when he said he could use either one, yes? Or is the sign for some reason not important in this case?", + "video_name": "xdiBjypYFRQ", + "timestamps": [ + 341 + ], + "3min_transcript": "try out the number 0 and make sure that 0 doesn't work, right, because 0 is between the two roots. It actually turns out that when x is equal to 0, f of x is minus 6, which is definitely less than 0. So I think this will give you a visual intuition of what this quadratic inequality means. Now with that visual intuition in the back of your mind, let's do some more problems and maybe we won't have to go through the exercise of drawing it, but maybe I will draw it just to make sure that the point hits home. Let me give you a slightly trickier problem. Let's say I had minus x squared minus 3x plus 28, let me say, is greater than 0. Well I want to get rid of this negative sign in front of the x squared. I just don't like it there because it makes it look more confusing to factor. I'm going to multiply everything by negative 1. I get x squared plus 3x minus 28, and when you multiply or to swap the sign. So this is now going to be less than 0. And if we were to factor this, we get x plus 7 times x minus 4 is less than 0. So if this was equal to 0, we would know that the two roots of this function -- let's define the function f of x -- let's define the function as f of x is equal to -- well we can define it as this or this because they're the same thing. But for simplicity let's define it as x plus 7 times x minus 4. That's f of x, right? Well, after factoring it, we know that the roots of this, the roots are x is equal to minus 7, and x is equal to 4. this inequality true? If this was any equality we'd be done. But we want to know what makes this inequality true. I'll give you a little bit of a trick, it's always going to be the numbers in between the two roots or outside of the two roots. So what I do whenever I'm doing this on a test or something, I just test numbers that are either between the roots or outside of the two roots. So let's pick a number that's between x equals minus 7 and x equals 4. Well let's try x equals 0. Well, f of 0 is equal to -- we could do it right here -- f of 0 is 0 plus 7 times 0 minus 4 is just 7 times minus 4, which is minus 28. So f of 0 is minus 28." + }, + { + "Q": "\n7:47 Is there a video where Sal proves this? Thanks", + "A": "using the same diagram as in the video, imagine the following: draw segment AD and segment BC, angle AED is congruent to angle BEC by vertical angles theorem, angle DAE is congruent to angle BCE because angles that subtends the same arc are congruent (arc DB) so, triangle ADE is similar to triangle CBE by aa similarity postulate. corresponding sides of similar triangles are proportional, so AE/CE = DE/BE, cross multiply and you get the CHORD-CHORD PRODUCT THEOREM, AE*BE = CE*DE.", + "video_name": "FJIZPvE3O1A", + "timestamps": [ + 467 + ], + "3min_transcript": "This is its radius. Let's call that r. Well what's the area of the square going to be? If that's the radius, this is also the radius. So one side of this square up here, is going to be 2r. So this side is also going to be 2r. It's a square, all the sides are the same. So they want to know the ratio of the area of the circle to the area of the square. The area of the square is just 2r times 2r. Which is 4r squared. Area of the circle is just pi r squared. You hopefully learned the formula for area of a circle. Divide the numerator and the denominator by r squared. You're left with pi/4. That's choice D. In the circle below, AB and CD are chords intersecting at E. Fair enough. If AE is equal to 5, BE is equal to 12, what is the value of DE? CE is equal to 6. What is the value of DE. Let's call that x. Now, I'm not going to prove it here, just for saving time. But there's a neat property of chords within a circle. That if I have two chords intersecting a circle, it turns out that the two segments when you multiply them times each other, are always going to be equal to the same thing. So in this case, 5 times 12. That's going to be equal to these two segments multiplied by each other. It's going to be equal to x times 6. So you get 60 is equal to 6x. Divide both sides by 6, you get x is equal to 10. And that is choice C. That might be a fun thing for you to think about after this video of why that is. And maybe you want to play around with chords and prove to yourself that that's always the case. At least that it makes intuition for you, makes sense. RB is tangent to a circle. Tangent means that it just touches the outside of the circle right there at only one point. And it's actually perpendicular to the radius at that point. So this is the radius of that point. The center is at A. This is a radius. And it's tangent at point B, so it's perpendicular to the" + }, + { + "Q": "at 4:18, couldnt you go over 6 THEN up 4? if not why? please help! i just joined the pre-algebra class, and im way behind.\n", + "A": "no because 6 would be a negative number (moving to the left) and 4 would be a positive number. If you did that,then you would have a negative slope. The answer, is a positive slope. Hope that helps :)", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 258 + ], + "3min_transcript": "Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6? So it's the same value, you just have to be consistent. If this is my start point, I went down 4, and then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4, so it would be a change in y would be 4, and then my change in x would be 6. And either way, once again, change in y over change in x is going to be 4 over 6, 2/3. So no matter which point you choose, as long as you kind of think about it in a consistent way, you're going to get the same value for slope." + }, + { + "Q": "\nOn 3:36 he goes to the right but it still comes out negative. Why? If anyone knows the answer to this, please reply.", + "A": "At 3:36 he goes 6 units to the left (-6) - he doesn t go to the right.", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 216 + ], + "3min_transcript": "Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6? So it's the same value, you just have to be consistent. If this is my start point, I went down 4, and then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4, so it would be a change in y would be 4, and then my change in x would be 6. And either way, once again, change in y over change in x is going to be 4 over 6, 2/3. So no matter which point you choose, as long as you kind of think about it in a consistent way, you're going to get the same value for slope." + }, + { + "Q": "can you go up first? So like at 3:24 he is going down first then over six. But from the point at -3 could you go up 4 and then over 6 to get to your other point?\n", + "A": "Yes you can. Sal likely was more comfortable going down then over, but it will work either way.", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 204 + ], + "3min_transcript": "that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6?" + }, + { + "Q": "\nAt 2:57 Sal said that the change was -4. Why can't that number be 4? Does that change the outcome of the answer? Thanks!", + "A": "You basically switched one of the values in y2-y1/x2-x1 so you made it y1-y2 as long as you switch the x value as well to x1-x2, then nothing changes but if you only switch the y, then it changes the slope.", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 177 + ], + "3min_transcript": "that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6?" + }, + { + "Q": "at 1:21, Sal draws a line going across, but I thought it was rise (which is going up), then run, (go across). Am I misunderstanding something?\n", + "A": "Slope is rise/run, but it doesn t matter what order you calculate them in. If you find the run (x) first, just make sure you are putting that value in the denominator of your slope.", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 81 + ], + "3min_transcript": "Find the slope of the line in the graph. And just as a bit of a review, slope is just telling us how steep a line is. And the best way to view it, slope is equal to change in y over change in x. And for a line, this will always be constant. And sometimes you might see it written like this: you might see this triangle, that's a capital delta, that means change in, change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here-- let me do it in a more vibrant color-- so let's say we start at that point right there. And we want to go to another point that's pretty that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is" + }, + { + "Q": "At 0:25, why is it y over x and not x over y?\n", + "A": "Slope is always rise over run.", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 25 + ], + "3min_transcript": "Find the slope of the line in the graph. And just as a bit of a review, slope is just telling us how steep a line is. And the best way to view it, slope is equal to change in y over change in x. And for a line, this will always be constant. And sometimes you might see it written like this: you might see this triangle, that's a capital delta, that means change in, change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here-- let me do it in a more vibrant color-- so let's say we start at that point right there. And we want to go to another point that's pretty that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is" + }, + { + "Q": "\nAt 1:04 how can a numerator be equal to 0 when there is only a value such 4/x-2", + "A": "Well... When does 4 = 0? Let s keep an eye out for pigs on the wing.", + "video_name": "fvC0dm2wzIo", + "timestamps": [ + 64 + ], + "3min_transcript": "Voiceover:Right over here, I have the graph of f of x, and what I want to think about in this video is whether we could have sketched this graph just by looking at the definition of our function, which is defined as a rational expression. We have 2x plus 10 over 5x minus 15. There is a couple of ways to do this. First, you might just want to pick out any numbers that are really easy to calculate. For example, what happens when X is equal to 0? We could say f of 0 is going to be equal to, well, all the x term is going to be 0, so you're going to be left with 10 over negative 15, which is negative 10/15, which is negative 2/3. You can plot that one. x equals 0. f of x or y equals f of x is negative 2/3, and we see that point, let me do that in a darker color, you see that point right over there, so we could have plotted that point. We could also say, \"When does this function equal 0?\" Well, the function is equal to 0 when ... is if you get this numerator equal to 0, so you could try to solve 2x plus 10 is equal to 0. That's going to happen when 2x is equal to negative 10. I just subtracted 10 from both sides. If I divide both sides by 2, that's going to happen when x is equal to negative 5. You see that, you see this right over here. When x is equal to negative 5, the function intersects the x-axis. but that still doesn't give us enough to really form this interesting shape over here. You could think about what other functions have this type of shape. Now what I want to think about is the behavior of the function at different points. First, I want to think about when this function is undefined and what type of behavior we might expect for that function when it's undefined. This function is going to be undefined. The only way I can think of to make this undefined is if I make the denominator equal to 0. We don't know what it means to divide by 0. That is undefined. The function is going to be undefined let me do this in blue, when 5x minus 15 is equal to 0, or adding 15 to both sides, when 5x is equal to 15, or dividing both sides by 5, when x is equal to 3, f is undefined. Now, there is a couple of ways for a function to be undefined at a point. You could have something like this. Let me draw some axes right over here. Let's say that this is 3. You could have your function, it could look like this. It could be defined. It might approach something but just not be defined right at 3 and then just keep on going like that, or the other possibility is it might have a vertical asymptote there. If it has a vertical asymptote, it's going to look something like this. It might be approach, it might just pop up to infinity, and it might pop down from infinity on this side, or it might go from negative infinity right over here." + }, + { + "Q": "3:19 Shouldn't it be commutative instead of associative, since commutative is about switching orders? Thanks.\n", + "A": "Idk! I m just a fifth grader, so... -Brianna-", + "video_name": "vr0sTKbV7lI", + "timestamps": [ + 199 + ], + "3min_transcript": "This is x is equal to-- let's see, 1, 2, 3. y is between 0 and 4. 1, 2, 3, 4. So the x-y plane will look something like this. The kind of base of our cube will look something like this. And then z is between 0 and 2. So 0 is the x-y plane, and then 1, 2. So this would be the top part. And maybe I'll do that in a slightly different color. So this is along the x-z axis. You'd have a boundary here, and then it would come in like this. You have a boundary here, come in like that. A boundary there. So we want to figure out the volume of this cube. And you could do it. You could say, well, the depth is 3, the base, the width is 4, so this area is 12 times the height. 12 times 2 is 24. You could say it's 24 cubic units, whatever units we're doing. But let's do it as a triple integral. Well, what we could do is we could take the volume of a very small-- I don't want to say area-- of a very small volume. So let's say I wanted to take the volume of a small cube. Some place in this-- in the volume under question. And it'll start to make more sense, or it starts to become a lot more useful, when we have variable boundaries and surfaces and curves as boundaries. But let's say we want to figure out the volume of this little, small cube here. That's my cube. It's some place in this larger cube, this larger rectangle, cubic rectangle, whatever you want to call it. So what's the volume of that cube? Let's say that its width is dy. So that length right there is dy. It's height is dx. Sorry, no, it's height is dz, right? The way I drew it, z is up and down. And it's depth is dx. This is dx. This is dz. This is dy. So you can say that a small volume within this larger volume differential. And that would be equal to, you could say, it's just the width times the length times the height. dx times dy times dz. And you could switch the orders of these, right? Because multiplication is associative, and order doesn't matter and all that. But anyway, what can you do with it in here? Well, we can take the integral. All integrals help us do is help us take infinite sums of infinitely small distances, like a dz or a dx or a dy, et cetera. So, what we could do is we could take this cube and first, add it up in, let's say, the z direction. So we could take that cube and then add it along the up and down axis-- the z-axis-- so that we get the volume of a column. So what would that look like? Well, since we're going up and down, we're adding-- we're taking the sum in the z direction. We'd have an integral." + }, + { + "Q": "At 5:54 is the trapezoid the same thing as a trapezium?\n", + "A": "This is a very good question. If you are american then the accepted answer would be no. By american definition a trapezium has no parralel sides while a trapezoid has 1 set of sides that are parallel to each other. The British definition states that a trapezium has 2 sides that are parallel which would be the same thing as a trapezoid.", + "video_name": "10dTx1Zy_4w", + "timestamps": [ + 354 + ], + "3min_transcript": "now you might recognize this purple one on the left is a special type a not only are all the corners square not all do they have these right angles but all the sides are the same length at least it looks like it from the picture this side is the same length as this side is the same length as that side same length as that side. So if the sides are of same length and all the corners are square where they have right angles, then we call it a square so this right over here is a, this right over here is a square special type a rectangle this one not all the side to the same this one all the sides are the same out if we go down here looks like all the sides are the same one two three four but it doesn't have square corners. Notice if I tried to put a little square here it wouldn't fit perfectly, if I tried square here doesn't fit perfectly these So if all four sides are the same and that's all we know about it we call it a rhombus. We call it a rhombus so you might be saying, the square all four sides are the same, is that a rhombus? yes it is. So the square is a rectangle it's a rectangle and is a rhombus so this last shape here this last quadrilateral what's interesting about it is you have two of the side's going in the same direction so this one right over here knows what I mean by the same direction it sometimes is called parallel if I draw two lines here these two lines are never going to cross each other so these two lines are parallel if you have opposite one pair of opposite sides that are parallel we call them trapezoids we call them trapezoids. Notice the other blinds the other opposite sides these will cross, these will cross" + }, + { + "Q": "\nI dont understand 3:17?", + "A": "You can t have a domain that maps to more than 1 number in the range.", + "video_name": "Uz0MtFlLD-k", + "timestamps": [ + 197 + ], + "3min_transcript": "maybe that is associated with 2 as well. So 2 is also associated with the number 2. And so notice, I'm just building a bunch of associations. I've visually drawn them over here. Here I'm just doing them as ordered pairs. We could say that we have the number 3. 3 is in our domain. Our relation is defined for number 3, and 3 is associated with, let's say, negative 7. So this is 3 and negative 7. Now this type of relation right over here, where if you give me any member of the domain, and I'm able to tell you exactly which member of the range is associated with it, this is also referred to as a function. And in a few seconds, I'll show you a relation that is not a function. Because over here, you pick any member of the domain, and the function really is just a relation. It's really just an association, sometimes called and particular members of the range. So you give me any member of the domain, I'll tell you exactly which member of the range it maps to. You give me 1, I say, hey, it definitely maps it to 2. You give me 2, it definitely maps to 2 as well. You give me 3, it's definitely associated with negative 7 as well. So this relation is both a-- it's obviously a relation-- but it is also a function. Now to show you a relation that is not a function, imagine something like this. So once again, I'll draw a domain over here, and I do this big, fuzzy cloud-looking thing to show you that I'm not showing you all of the things in the domain. I'm just picking specific examples. And let's say that this big, fuzzy cloud-looking thing is the range. And let's say in this relation-- and I'll build it the same way that we built it over here-- let's say in this relation, 1 is associated with 2. So let's build the set of ordered pairs. So 1 is associated with 2. Let's say that 2 is associated with, let's say that 2 is associated with negative 3. And let's say on top of that, we also associate, we also associate 1 with the number 4. So we also created an association with 1 with the number 4. So we have the ordered pair 1 comma 4. Now this is a relationship. We have, it's defined for a certain-- if this was a whole relationship, then the entire domain is just the numbers 1, 2-- actually just the numbers 1 and 2. It's definitely a relation, but this is no longer a function. And the reason why it's no longer a function is, if you tell me, OK I'm giving you 1 in the domain, what member of the range is 1 associated with? Over here, you say, well I don't know, is 1 associated with 2, or is it associated with 4? It could be either one. So you don't have a clear association. If I give you 1 here, you're like, I don't know, do I hand you a 2 or 4? That's not what a function does. A function says, oh, if you give me a 1, I know I'm giving you a 2. If you give me 2, I know I'm giving you 2. Now with that out of the way, let's actually" + }, + { + "Q": "At 1:05, I was wondering if there was an actual name for the type of function which maps 2 numbers in the domain to just 1 in the range, e.g. f(x)=x^2\n", + "A": "Good question! We say that functions are not one-to-one. :) Therefore, what we actually say is that one-to-one functions are functions for which each element in the domain is paired with its own unique element of the range (or, that each element of the range is paired with exactly one element from the domain). Hope that helps!", + "video_name": "Uz0MtFlLD-k", + "timestamps": [ + 65 + ], + "3min_transcript": "Is the relation given by the set of ordered pairs shown below a function? So before we even attempt to do this problem, right here, let's just remind ourselves what a relation is and what type of relations can be functions. So in a relation, you have a set of numbers that you can kind of view as the input into the relation. We call that the domain. You can view them as the set of numbers over which that relation is defined. And then you have a set of numbers that you can view as the output of the relation, or what the numbers that can be associated with anything in domain, and we call that the range. And it's a fairly straightforward idea. So for example, let's say that the number 1 is in the domain, and that we associate the number 1 with the number 2 in the range. So in this type of notation, you would say that the relation has 1 comma 2 in its set of ordered pairs. These are two ways of saying the same thing. maybe that is associated with 2 as well. So 2 is also associated with the number 2. And so notice, I'm just building a bunch of associations. I've visually drawn them over here. Here I'm just doing them as ordered pairs. We could say that we have the number 3. 3 is in our domain. Our relation is defined for number 3, and 3 is associated with, let's say, negative 7. So this is 3 and negative 7. Now this type of relation right over here, where if you give me any member of the domain, and I'm able to tell you exactly which member of the range is associated with it, this is also referred to as a function. And in a few seconds, I'll show you a relation that is not a function. Because over here, you pick any member of the domain, and the function really is just a relation. It's really just an association, sometimes called and particular members of the range. So you give me any member of the domain, I'll tell you exactly which member of the range it maps to. You give me 1, I say, hey, it definitely maps it to 2. You give me 2, it definitely maps to 2 as well. You give me 3, it's definitely associated with negative 7 as well. So this relation is both a-- it's obviously a relation-- but it is also a function. Now to show you a relation that is not a function, imagine something like this. So once again, I'll draw a domain over here, and I do this big, fuzzy cloud-looking thing to show you that I'm not showing you all of the things in the domain. I'm just picking specific examples. And let's say that this big, fuzzy cloud-looking thing is the range. And let's say in this relation-- and I'll build it the same way that we built it over here-- let's say in this relation, 1 is associated with 2. So let's build the set of ordered pairs. So 1 is associated with 2. Let's say that 2 is associated with, let's say that 2 is associated with negative 3." + }, + { + "Q": "\nat 1:37 why did he multiply 5x5 ?", + "A": "He evaluated what was inside the parentheses first: (9-4) = 5 So, 5(9-4) = 5(5) = 25", + "video_name": "Badvask-UDU", + "timestamps": [ + 97 + ], + "3min_transcript": "Rewrite the expression five times 9 minus 4-- that's in parentheses-- using the distributive law of multiplication over subtraction. Then simplify. So let me just rewrite it. This is going to be 5 times 9 minus 4, just like that. Now, if we want to use the distributive property, well, You could just evaluate 9 minus 4 and then multiply that times 5. But if you want to use the distributive property, you distribute the 5. You multiply the 5 times the 9 and the 4, so you end up with 5 times 9 minus 5 times 4. Notice, we distributed the 5. We multiplied it times both the 9 and the 4. In the first distributive property video, we gave you an idea of why you have to distribute the 5, why it makes And we're going to verify that it gives us the same answer as if we just evaluated the 9 minus 4 first. But anyway, what are these things? So 5 times 9, that is 45. So we have 45 minus-- what's 5 times 4? Well, that's 20. 45 minus 20, and that is equal to 25, so this is using the distributive property right here. If we didn't want to use the distributive property, if we just wanted to evaluate what's in the parentheses first, we would have gotten-- let's go in this direction-- 5 times-- what's 9 minus 4? 9 minus 4 is 5. Let me do that in a different color. 5 times 9 minus 4. So it's 5 times 5. 5 times 5 is just 25, so we get the same answer either way. This is using the distributive law of multiplication over subtraction, usually just referred to as the This is evaluating the inside of the parentheses first and then multiplying by 5." + }, + { + "Q": "I dont get what he says at 2:05\n", + "A": "That 74.7 is one sigma away from the mean. What he should have said maybe would be like this. Where does that get us?Well, 81-6.3 is 74.7 which is one standard deviation from the mid", + "video_name": "Wp2nVIzBsE8", + "timestamps": [ + 125 + ], + "3min_transcript": "Here's the second problem from CK12.org's AP statistics FlexBook. It's an open source textbook, essentially. I'm using it essentially to get some practice on some statistics problems. So here, number 2. The grades on a statistics midterm for a high school are normally distributed with a mean of 81 and a standard deviation of 6.3. All right. Calculate the z-scores for each of the following exam grades. Draw and label a sketch for each example. We can probably do it all on the same example. But the first thing we'd have to do is just remember what is a z-score. What is a z-score? A z-score is literally just measuring how many standard deviations away from the mean? Just like that. So we literally just have to calculate how many standard deviations each of these guys are from the mean, and So let me do part a. So we have 65. So first we can just figure out how far is 65 from the mean. Let me just draw one chart here that we can use the entire time. So it's just our distribution. Let's see. We have a mean of 81. That's our mean. And then a standard deviation of 6.3. So our distribution, they're telling us that it's normally distributed. So I can draw a nice bell curve here. They're saying it's normally distributed, so that's as good of a bell curve as I'm capable of drawing. This is the mean right there at 81. And the standard deviation is 6.3. So one standard deviation above and below is going to be 6.3 away from that mean. So if we go 6.3 in the positive direction, that value right there is going to be 87.3. where does that get us? What, 74.7? Right, if we add 6, it'll get us to 80.7, and then 0.3 will get us to 81. So that's one standard deviation below and above the mean, and then you'd add another 6.3 to go 2 standard deviations, so on and so forth. So that's a drawing of the distribution itself. So let's figure out the z-scores for each of these grades. 65 is how far? 65 is maybe going to be here someplace. So we first want to say, well how far is it just from our mean? So the distance is, you just want to positive number here. Well actually, you want a negative number. Because you want your z-score to be positive or negative. Negative would mean to the left of the mean and positive would mean to the right of the mean. So we say 65 minus 81. So that's literally how far away we are." + }, + { + "Q": "second q is 1/4??? Paused @ o2:42\n", + "A": "there are 13 hearts in a 52 card standard ddeck of playing cards...1/4.....13/52", + "video_name": "obZzOq_wSCg", + "timestamps": [ + 162 + ], + "3min_transcript": "a Jack of spades, or a Jack of hearts. So if you just multiply these two things-- you could take a deck of playing cards, take out the jokers and count them-- but if you just multiply this you have four suits, each of those suits have 13 types. So you're going to have 4 times 13 cards, or you're going to have 52 cards in a standard playing deck. Another way you could have said, look, there's 13 of these ranks, or types, and each of those come in four different suits-- 13 times 4. Once again, you would have gotten 52 cards. Now, with that of the way, let's think about the probabilities of different events. So let's say I shuffle that deck. I shuffle it really, really well and then I randomly pick a card from that deck. And I want to think about what is the probability that I pick a Jack. Well, how many equally likely events are there? Well, I could pick any one of those 52 cards. So there's 52 possibilities for when I pick that card. And how many of those 52 possibilities are Jacks? the Jack of clubs, and the Jack of hearts. There's four Jacks in that deck. So it is 4 over 52-- these are both divisible by 4-- 4 divided by 4 is 1, 52 divided by 4 is 13. Now, let's think about the probability. So I'll start over. I'm going to put that Jack back and I'm going to reshuffle the deck. So once again, I still have 52 cards. So what's the probability that I get a hearts? What's the probability that I just randomly pick a card from a shuffled deck and it is a heart? Well, once again, there's 52 possible cards I could pick from. 52 possible, equally likely events that we're dealing with. And how many of those have our hearts? Well, essentially 13 of them are hearts. For each of those suits you have 13 types. So there are 13 hearts in that deck. There are 13 spades in that deck. There are 13 clubs in that deck. So 13 of the 52 would result in hearts, and both of these are divisible by 13. This is the same thing as 1/4. One in four times I will pick it out, or I have a one in four probability of getting a hearts when I randomly pick a card from that shuffled deck. Now, let's do something that's a little bit more interesting, or maybe it's a little obvious. What's the probability that I pick something that is a Jack-- I'll just write J-- and it is a hearts? Well, if you are reasonably familiar with cards you'll know that there's actually only one card that is both a Jack and a heart. It is literally the Jack of hearts. So we're saying, what is the probability that we pick the exact card, the Jack of hearts? Well, there's only one event, one card, that meets this criteria right over here, and there's 52 possible cards." + }, + { + "Q": "\nin 00:20, what did Sal mean by Suits??", + "A": "thanks a lot!", + "video_name": "obZzOq_wSCg", + "timestamps": [ + 20 + ], + "3min_transcript": "Let's do a little bit of probability with playing cards. And for the sake of this video, we're going to assume that our deck has no jokers in it. You could do the same problems with the joker, you'll just get slightly different numbers. So with that out of the way, let's first just think about how many cards we have in a standard playing deck. So you have four suits, and the suits are the spades, the diamonds, the clubs, and the hearts. You have four suits and then in each of those suits you have 13 different types of cards-- and sometimes it's called the rank. You have the ace, then you have the two, the three, the four, the five, the six, seven, eight, nine, ten, and then you have the Jack, the King, and the Queen. And that is 13 cards. So for each suit you can have any of these-- you can have any of the suits. a Jack of spades, or a Jack of hearts. So if you just multiply these two things-- you could take a deck of playing cards, take out the jokers and count them-- but if you just multiply this you have four suits, each of those suits have 13 types. So you're going to have 4 times 13 cards, or you're going to have 52 cards in a standard playing deck. Another way you could have said, look, there's 13 of these ranks, or types, and each of those come in four different suits-- 13 times 4. Once again, you would have gotten 52 cards. Now, with that of the way, let's think about the probabilities of different events. So let's say I shuffle that deck. I shuffle it really, really well and then I randomly pick a card from that deck. And I want to think about what is the probability that I pick a Jack. Well, how many equally likely events are there? Well, I could pick any one of those 52 cards. So there's 52 possibilities for when I pick that card. And how many of those 52 possibilities are Jacks? the Jack of clubs, and the Jack of hearts. There's four Jacks in that deck. So it is 4 over 52-- these are both divisible by 4-- 4 divided by 4 is 1, 52 divided by 4 is 13. Now, let's think about the probability. So I'll start over. I'm going to put that Jack back and I'm going to reshuffle the deck. So once again, I still have 52 cards. So what's the probability that I get a hearts? What's the probability that I just randomly pick a card from a shuffled deck and it is a heart? Well, once again, there's 52 possible cards I could pick from. 52 possible, equally likely events that we're dealing with. And how many of those have our hearts? Well, essentially 13 of them are hearts. For each of those suits you have 13 types. So there are 13 hearts in that deck." + }, + { + "Q": "\nGreat video! but just a word not clear... at 5:48 what is being said..four fifty what?", + "A": "4/52=1/13 is the probability of a jack being drawn.", + "video_name": "obZzOq_wSCg", + "timestamps": [ + 348 + ], + "3min_transcript": "pick the Jack of hearts-- something that is both a Jack and it's a heart. Now, let's do something a little bit more interesting. What is the probability-- you might want to pause this and think about this a little bit before I give you the answer. What is the probability of-- so I once again, I have a deck of 52 cards, I shuffled it, randomly pick a card from that deck-- what is the probability that that card that I pick from that deck is a Jack or a heart? So it could be the Jack of hearts, or it could be the Jack of diamonds, or it could be the Jack of spades, or it could be the Queen of hearts, or it could be the two of hearts. So what is the probability of this? And this is a little bit more of an interesting thing, because we know, first of all, that there are 52 possibilities. But how many of those possibilities meet these conditions that it is a Jack or a heart. And to understand that, I'll draw a Venn diagram. Sounds kind of fancy, but nothing fancy here. represents all of the outcomes. So if you want, you could imagine it has an area of 52. So this is 52 possible outcomes. Now, how many of those outcomes result in a Jack? So we already learned, one out of 13 of those outcomes result in a Jack. So I could draw a little circle here, where that area-- and I'm approximating-- represents the probability of a Jack. So it should be roughly 1/13, or 4/52, of this area right over here. So I'll just draw it like this. So this right over here is the probability of a Jack. There's four possible cards out of the 52. So that is 4/52, or one out of 13. Now, what's the probability of getting a hearts? Well, I'll draw another little circle here that represents that. 13 out of 52 cards represent a heart. and a Jack. So I'm actually going to overlap them, and hopefully this will make sense in a second. So there's actually 13 cards that are a heart. So this is the number of hearts. And actually, let me write this top thing that way as well. It makes it a little bit clearer that we're actually looking at the number of Jacks. And of course, this overlap right here is the number of Jacks and hearts-- the number of items out of this 52 that are both a Jack and a heart-- it is in both sets here. It is in this green circle and it is in this orange circle. So this right over here-- let me do that in yellow since I did that problem in yellow-- this right over here is a number of Jacks and hearts. So let me draw a little arrow there. It's getting a little cluttered, maybe" + }, + { + "Q": "This is where the ray terminates.\n1:40\nIt's an endpoint.\nLOOK at the above statement In regarding to the ray, he messed up as there is not the end point but rather the commencement point or the source of the ray ,that is not where it terminates but starts were it a line transfoming into a ray then we would say thats where it terminates and an end point but since its genesis is where there is a point it is wrong to refer it as an end or use terminate what do you think?\n", + "A": "It starts there, endlessly going in whatever direction. But it ends there, because when the ray reaches that part, it ends.", + "video_name": "DkZnevdbf0A", + "timestamps": [ + 100 + ], + "3min_transcript": "Any pair of points can be connected by a line segment. That's right. Connect two pairs of black points in a way that creates two parallel line segments. So let's see if we can do that. So I could create one segment that connects this point to this point and then another one that connects this point to this point. And they look pretty parallel. In fact, I think this is the right answer. If we did it another way, if we had connected that point to that point and this point to this point, then it wouldn't look so parallel. These clearly, if they were to keep going, they would intersect at some point. So let me set it back up the way I did it the first time. Let me make these two points parallel. And these are line segments because they have two end points. They each have two end points. And they continue forever. Well they don't continue forever. They continue forever in no directions, in zero directions. If it was a ray, it would continue forever If it's a line, it continues forever In fact, it wouldn't have end points because it would just continue forever in both of these directions. Let's do one more. Drag the ray so it has an endpoint at A, so we want to make its endpoint at A where the ray terminates and goes through one of the other black points. The ray should also be parallel to the pink line. So I have two options. I could make it go through this black point, but it's clearly not parallel. In fact, it looks perpendicular here. So let's try to make it go through this point. Well, yes, when I do that, it does indeed look like my ray is parallel to the pink line. And this is a ray because it has one endpoint. This is where the ray terminates. It's an endpoint. It literally ends there. And it continues forever in one direction. In this case, the direction is to the right. It continues forever to the right. So it continues forever in one-- continues forever in one direction." + }, + { + "Q": "At 0:40, he puts a zero as a place holder. my teachers at school say to just drop the number. Wouldn't that change the answer? Could you tell me which one is correct and why?\n", + "A": "i think you have gotten confused. you can only do that while you are adding decimals. but i guess you can also do that if the number is at the top and it is the larger one over zero hope this helps!", + "video_name": "MufbvU4tGh8", + "timestamps": [ + 40 + ], + "3min_transcript": "Let's try to calculate 10.1 minus 3.93. And I encourage you to pause this video and try it on your own first, and then we can think about whether we did it the same way. So let's just rewrite it, aligning the decimal and the different place values. So 10.1 minus-- the 3 is in the ones place, so I'll put it right under the 0-- 3.93. Now, let's just try to calculate this. Now, before we subtract, we want all the numbers on top to be larger than the numbers on the bottom. And we don't even have a number here. We could stick a 0 here. Let me do that in a different color here. We could stick a 0 here. 10.1 is the same thing as 10.10, but we still face an issue 0 is less than 3. 1 is less than 9. 0 is less than 3. So we're going to do a little bit of regrouping. So let's do that regrouping. So we could take a 10 away, one 10 away, and then one 10 is the same thing as 10 ones. So I could write a 10 in the ones place. away so I'm going to have nine ones, and give that one to the tenths place. Well, one is 10 tenths. 10 tenths plus 1 tenth is going to be 11 tenths. Now, I could take one of those tenths away and give it to the hundredths place. 1 tenth is 10 hundredths. And now I have a higher digit in the numerator, or at least as equal in the numerator as I have in the denominator. So 10 minus 3 is 7. 10 minus 9 is 1. I have the decimal. 9 minus 3 is 6. And then I have nothing over here. So 10.1 minus 3.93, 6.17." + }, + { + "Q": "\nWait I do not get it do you have to do what he said at 0:55", + "A": "You sort of do. Since 3 is larger than 0, you need to borrow. The 1 ten is the same as 10 ones. So if you add them together, you d get ten. Now you can subtract with 10 and 3.", + "video_name": "MufbvU4tGh8", + "timestamps": [ + 55 + ], + "3min_transcript": "Let's try to calculate 10.1 minus 3.93. And I encourage you to pause this video and try it on your own first, and then we can think about whether we did it the same way. So let's just rewrite it, aligning the decimal and the different place values. So 10.1 minus-- the 3 is in the ones place, so I'll put it right under the 0-- 3.93. Now, let's just try to calculate this. Now, before we subtract, we want all the numbers on top to be larger than the numbers on the bottom. And we don't even have a number here. We could stick a 0 here. Let me do that in a different color here. We could stick a 0 here. 10.1 is the same thing as 10.10, but we still face an issue 0 is less than 3. 1 is less than 9. 0 is less than 3. So we're going to do a little bit of regrouping. So let's do that regrouping. So we could take a 10 away, one 10 away, and then one 10 is the same thing as 10 ones. So I could write a 10 in the ones place. away so I'm going to have nine ones, and give that one to the tenths place. Well, one is 10 tenths. 10 tenths plus 1 tenth is going to be 11 tenths. Now, I could take one of those tenths away and give it to the hundredths place. 1 tenth is 10 hundredths. And now I have a higher digit in the numerator, or at least as equal in the numerator as I have in the denominator. So 10 minus 3 is 7. 10 minus 9 is 1. I have the decimal. 9 minus 3 is 6. And then I have nothing over here. So 10.1 minus 3.93, 6.17." + }, + { + "Q": "\nAt 5:00, all I have to do is graph is x=5 and x=-1 correct?", + "A": "I don t believe you would graph x=5 and x=-1 because the solutions are points, not lines. What you have are vertical lines, and we don t really want that, because the solutions to the quadratic are one x with one y only. When asked for the answers, you are correct to say that x=5 and -1. But do not actually graph that.", + "video_name": "VTlvg4wJ1X0", + "timestamps": [ + 300 + ], + "3min_transcript": "that looks something like that. Let's say that the y is equal to some other function, not necessarily this f of x. Y is equal to g of x. The x-values where you intersect, where you intersect the x-axis. Well, in order to intersect the x-axis, y must be equal to zero. So, y is equal to zero there. Notice our y-coordinate at either of those points are going to be equal to zero. And that means that our function is equal to zero. So, figuring out the x-values where the graph of y equals f of x intersects the x-axis, this is equivalent to saying, \"For what x-values does f of x equal zero?\" So we could just say, \"For what x-values does this thing right over here \"equal zero?\" So, let me just write that down. So we could rewrite this as x, x minus two squared minus nine equals zero. x minus two squared is equal to nine. And just like we saw before, that means that x minus two is equal to the positive or negative square root of nine. So, we could say x minus two is equal to positive three or x minus two is equal to negative three. Well, you add two to both sides of this, you get x is equal to five, or x is equal to, if we add two to both sides of this equation, you'll get x is equal to negative one. And you can verify that. If x is equal to five, five minus two is three, squared is nine, minus nine is zero. So, the point five comma zero is going to be on this graph. And also, if x is equal to negative one, negative one minus two, negative three. Squared is positive nine, minus nine is zero. So, also the point negative one comma zero So those are the points where, those are the x-values where the function intersects the x-axis." + }, + { + "Q": "At about 7:56 Sal simplifies 2x^2 - 3x^2 into -1x^2... How come the outcome isn't -1x^4?\n", + "A": "If we let x\u00c2\u00b2 = y Then our problem becomes: 2y - 3y Which is obviously equal to: -1y But since y = x\u00c2\u00b2 We have: 2x\u00c2\u00b2 - 3x\u00c2\u00b2 = -1x\u00c2\u00b2 = -x\u00c2\u00b2", + "video_name": "aoXUWSwiDzE", + "timestamps": [ + 476 + ], + "3min_transcript": "negative 4 is negative 8, or 15 plus negative 8 is 15 minus 8. And now, the numerator becomes negative 3 minus 8, which is negative 11. And the denominator is 15 minus 8, which is 7. So problem 10, we simplified it to negative 11 over 7. Right there. Let's do a couple of these over here. OK, we see some exponents. I'll pick one of the harder ones. Let's do this one over here, problem 18. So 2x squared minus 3x squared, plus 5x minus 4. OK, well, this wasn't that hard. But what we could do here-- let me write this down. 2x squared minus 3x squared, plus 5x minus 4. And they tell us that x is equal to negative 1. One thing we could do is simplify this before we even substitute for negative 1. So what's 2 of something minus 3 of something? This is 2x squareds minus 3x squareds. So 2 of something minus 3 of something, that's going to be minus 1 of that something. So that right there-- or negative 1 of that something-- that would be negative 1x squared plus 5x minus 4. And they tell us x is equal to negative 1. So this is negative 1 times x squared, negative 1 squared, plus 5 times x, which is negative 1, minus 4. So what is this? Negative 1 squared is just 1. So this whole expression simplifies to negative 1 plus 5 times negative 1-- we do the multiplication first, of course. So that's minus 5, or negative 5 minus 4. So negative 1 minus 5 is negative 6, minus 4 is equal to negative 10. And I'll do these last two just to get a sample of all of the types of problems in this variable expression section. The weekly cost, c, of manufacturing x remote controls-- so the cost is c, x is the remote controls-- is given by this formula. The cost is equal to 2,000 plus 3 times the number of remote controls, where cost is given in dollars. Question a, what is the cost of producing 1,000 remote controls?" + }, + { + "Q": "\n@ 4:43 #6 Why is \"\" + -4 \"' The same as \"\" - 4 \"\" ?", + "A": "The plus and the negative sign before that four immediately cancel each other out. Same as -(-4) : the minus negative x become a plus x.", + "video_name": "aoXUWSwiDzE", + "timestamps": [ + 283 + ], + "3min_transcript": "And every time we see a d, we'll put a minus 4 there. And I'll do a couple of these. I won't do all of them, just for the sake of time. So let's say problem number 5. They gave us 2 times a plus 3 times b. Well, this is the same thing as 2 times-- instead of an a, we know that a is going to be equal to negative 3. So 2 times minus 3, plus 3 times b-- what's b? They're telling us that b is equal to 2-- so 3 times 2. And what is this equal to? 2 times minus 3-- let me do it in a different color-- 2 times negative 3 is negative 6, plus 3 times 2. 3 times 2 is 6. That's positive 6. So that is equal to 0. And notice the order of operations. before we added the two numbers. Multiplication and division takes precedence over addition and subtraction. Let's do problem 6. I'll do that right here. So you have 4 times c. 4 times-- now what's c equal to? They tell us c is equal to 5. So 4 times 5, that's our c, plus d. d is minus or negative 4. So we have 4 times 5 is 20, plus negative 4-- that's the same thing as minus 4, so that is equal to 16. Problem 6. Now, let's do one of the harder ones down here. This problem 10 looks a little bit more daunting. Problem 10 right there. So we have a minus 4b in the numerator, if you can read it, it's kind of small. a is minus 3. b is 2. So 4 times 2. Remember, this right here is a, that right there is b. They're telling me up here. And then all of that over-- all of that is over 3c plus 2d. So 3 times-- what was c? c is 5 plus 2 times d. What is d? d is negative 4. So let's figure this out. So we have to do order of operations. Multiplication comes first before addition and subtraction. So this is going to be equal to minus 3 minus 4 times 2," + }, + { + "Q": "\nat 00:41, where did you get the x and it shows 2 x11r?", + "A": "it is actually an x it looks like an r", + "video_name": "aoXUWSwiDzE", + "timestamps": [ + 41 + ], + "3min_transcript": "Let's do some practice problems dealing with variable expressions. So these first problems say write the following in a more condensed form by leaving out the multiplication symbol or leaving out a multiplication symbol. So here we have 2 times 11x, so if we have 11 x's and then we're going to have 2 times those 11 x's, we're going to have 22 x's. So another way you could view this, 2 times 11x, you could view this as being equal to 2 times 11, and all of that times x, and that's going to be equal to 22 x's. You had 11 x's, you're going to have 2 times as many x's, so you're going to have 22 x's. Let's see, you have 1.35 times y. Now here we're just going to do a straight simplifying how we write it. So 1.35 times y-- I'll do it in a different color-- 1.35 times y-- that's a little dot there. If we have a variable following a number, we know that means 1.35 times that variable. So that, we could rewrite as just being equal to 1.35y. We've condensed it by getting rid of the multiplication sign. Let's see, here we have 3 times 1/4. Well, this is just straight up multiplying a fraction. So in problem 3-- this was problem 1, this is problem 2, problem 3-- 3 times 1/4, that's the same thing as 3 over 1 times 1/4. Multiply the numerators, you get 3. Multiply the denominators, 1 times 4, you get 4. So number 3, I got 3/4. And then finally, you have 1/4 times z. We could do the exact same thing we did up here in problem number 2. This was the same thing as 1.35y. That's the same thing as 1.35 times y. 1/4z, or we could view this as being equal to 1 over 4 times z over 1, which is the same thing as z times 1, over 4 times 1, or the same thing as z over 4. So all of these are equivalent. Now, what do they want us to do down here? Evaluate the following expressions for a is equal to 3, b is equal to 2, c is equal to 5, and d is equal to minus 4-- or, actually, I should say negative 4 is the correct terminology. Negative 4. So we just substitute. Every time we see an a, we're going to put a minus 3 there, or a negative 3 there. Every time we see a b, we'll put a positive 2 there." + }, + { + "Q": "\nAt 18:30 al finishes a problem as 40x10^-7 could that be simplified?", + "A": "Yes it can, the number is really 0.000040, he simply wrote it in scientific notation (which should be explained in the video). You could think of scientific notation like a kind of shorthand. 0.000040 equals 40 \u00c3\u0097 10^-7, so you can write is either way, it s simply easier, and less cumbersome, to write it in scientific notation.", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 1110 + ], + "3min_transcript": "at this small a dose. Or maybe you would, I don't want to get into that. But how would I write this in scientific notation? So I start off with the first non-zero number, if I'm starting from the left. So it's going to be 8.192. I just put a decimal and write 0.192 times-- times 10 to what? Well, I just count. Times 10 to the 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I have to include that number, 10 to the minus 10. And I think you'll find it reasonably satisfactory that this number is easier to write than that number over there. Now, and this is another powerful thing about scientific notation. Let's say I have these two numbers and I want to multiply them. Let's say I want to multiply the number 0.005 times the number 0.0008. This is actually a fairly straightforward one to do, And especially if you're dealing with twenty or thirty 0's on either sides of the decimal point. Put a 0 here to make my wife happy. But when you do it in scientific notation, it will actually simplify it. This guy can be rewritten as 5 times 10 to the what? I have 1, 2, 3 spaces behind the decimal. 10 to the third. And then this is 8, so this is times 8 times 10 to the-- sorry, this is 5 times 10 to the minus 3. That's very important. 5 times 10 to the 3 would have been 5,000. Be very careful about that. Now, what is this guy equal to? This is 1, 2, 3, 4 places behind the decimal. So it's 8 times 10 to the minus 4. If we're multiplying these two things, this is the same thing as 5 times 10 to the minus 3 times 8 times 10 to the minus 4. There's nothing special about the scientific notation. It literally means what it's saying. So for multiplying, you could write it out like this. And multiplication, order doesn't matter. 10 to the minus 3 times 10 to the minus 4. And then, what is 5 times 8? 5 times 8 we know is 40. So it's 40 times 10 to the minus 3 times 10 to the minus 4. And if you know your exponent rules, you know that when you multiply two numbers that have the same base, you can just add their exponents. So you just add the minus 3 and the minus 4. So it's equal to 40 times 10 to the minus 7. Let's do another example. Let's say we were to multiply Avogadro's number. So we know that's 6.022 times 10 to the 23rd. Now, let's say we multiply that times some really small number. So times, say, 7.23 times 10 to the minus 22. So this is some really small number. You're going to have a decimal, and then you're going to have" + }, + { + "Q": "\n20:26 wouldnt 6.022x7.23 be att least 42 because 6x7=42", + "A": "Yes, you are correct. Well done for noticing Sal s mistake :)", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 1226 + ], + "3min_transcript": "10 to the minus 3 times 10 to the minus 4. And then, what is 5 times 8? 5 times 8 we know is 40. So it's 40 times 10 to the minus 3 times 10 to the minus 4. And if you know your exponent rules, you know that when you multiply two numbers that have the same base, you can just add their exponents. So you just add the minus 3 and the minus 4. So it's equal to 40 times 10 to the minus 7. Let's do another example. Let's say we were to multiply Avogadro's number. So we know that's 6.022 times 10 to the 23rd. Now, let's say we multiply that times some really small number. So times, say, 7.23 times 10 to the minus 22. So this is some really small number. You're going to have a decimal, and then you're going to have So this is a really small number. But the multiplication, when you do it in scientific notation, is actually fairly straightforward. This is going to be equal to 6.0-- let me write it properly. 6.022 times 10 to the 23rd times 7.23 times 10 to the minus 22. We can change the order, so it's equal to 6.022 times 7.23. That's that part. So you can view it as these first parts of our scientific notation times 10 to the 23rd times 10 to the minus 22. And now, this is-- you're going to do some little decimal multiplication right here. It's going to be-- some number-- 40 something, I think. I can't do this one in my head. But this part is pretty easy to calculate. I'll just leave this the way it is. But this part right here, this will be times. 10 to the 23rd times 10 to the minus 22. You get times 10 to the first power. And then this number, whatever it's going to equal, I'll just leave it out here since I don't have a calculator. 0.23. Let's see, it will be 7.2. Let's see, 0.2 times-- it's like a fifth. It'll be like 41-something. So this is approximately 41 times 10 to the 1. Or, another way is approximately-- it's going to be 410-something. And to get it right, you just have to actually perform this multiplication. So hopefully you see that scientific notation is, one, really useful for super large and super small numbers. And not only is it more useful to kind of understand the numbers and to write the numbers, but it also simplifies operating on the numbers." + }, + { + "Q": "\nAt 6:01, what is avogadro's number?", + "A": "it s the number of atoms in one mole of a substance, and it s equal to 6.02214179 x 10^23", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 361 + ], + "3min_transcript": "And actually, this might be interesting, just as an aside. You may or may not know what this number is called. This is called a googol. A googol. In the early '90s if someone said, hey, that's a googol, you wouldn't have thought of a search engine. You would have thought of the number 10 to the 100th power, which is a huge number. It's more than the number of atoms, or the estimated number of atoms, in the known universe. In the known universe. It raises the question of what else is there out there. But I was reading up on this not too long ago. And if I remember correctly, the known universe has the order of 10 to the 79th to 10 to the 81 atoms. And this is, of course, rough. No one can really count this. People are just kind of estimating it. Or even better, guesstimating this. But this is a huge number. What may be even more interesting to you a very popular search engine-- Google. Google is essentially just a misspelling of the word \"googol\" with the O-L. And I don't know why they called it Google. Maybe they got the domain name. Maybe they want to hold this much information. Maybe that many bytes of information. Or, it's just a cool word. Whatever it is-- maybe it was the founder's favorite number. But it's a cool thing to know. But anyway, I'm digressing. This is a googol. It's just 1 with a hundred 0's. But I could equivalently have just written that as 10 to 100, which is clearly an easier way. This is an easier way to write this. This is easier. In fact, this is so hard to write that I didn't even take the trouble to write it. It would have taken me forever. This was just twenty 0's right here. A hundred 0's I would have filled up this screen and you have found it boring. So I didn't even write it. So clearly, this is easier to write. But how can we write something that isn't a direct power of 10? How can we use the power of this simplicity? How can we use the power of the simplicity somehow? And to do that, you just need to make the realization. This number, we can write it as-- so this has how many digits in it? It has 1, 2, 3, and then twenty 0's. So it has 23 digits after the 6. 23 digits after the 6. So what happens if I use this-- if I try to get close to it with a power of 10? So what if I were to say 10 to the 23? Do it in this magenta. 10 to the 23rd power. That's equal to what? That equals 1 with 23 0's. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23. You get the idea, that's 10 to the 23rd." + }, + { + "Q": "at 12:44 Sal says with negative exponents with a base of 10 (unlike positive exponents) it cannot be thought of as the number of zeroes. (i.e. 10^5 = 10 000, but 10^-5 = .00001) That only has 4 zeroes. :(\nBut couldn`t you write the answer to 10^-5 as 0.00001? That has 5 zeroes. (Sorry this has been really bugging me for a while. :) )\n", + "A": "No, the decimal is supposed to move 5 times. Think about it. 10^-5 can also be written as 1/10^5. 10^5=100000, and you do 1/100000. When you divide, you find the quotient to be 0.00001. Hope this helps!", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 764 + ], + "3min_transcript": "Well, it's just 6. So 6 is equal to 6 times 10 to the 0. You wouldn't actually have to write it this way. This is much simpler, but it shows you that you really can express any number in scientific notation. Now, what if we wanted to represent something like this? I had started off the video saying in science you deal with very large and very small numbers. So let's say you had the number-- do it in this color. And you had 1, 2, 3, 4. And then, let's say five 0's. And then you have followed by a 7. Well, once again, this is not an easy number to deal with. But how can we deal with it as a power of 10? As a power of 10? So what's the largest power of 10 that fits into this number, that this number is divisible by? So let's think about it. All the powers of 10 we did before were going to positive or going to-- well, yeah, positive powers of 10. We could also do negative powers of 10. We know that 10 to the 0 is 1. 10 to the minus 1 is equal to 1/10, which is equal to 0.1. Let me switch colors. I'll do pink. 10 to the minus 2 is equal to 1 over 10 squared, which is equal to 1/100, which is equal to 0.01. And you I think you get the idea that the--, well, let me just do one more so that you can get the idea. 10 to the minus 3. 10 to the minus 3 is equal to 1 over 10 to the third, which is equal to 1/1,000, which is equal to the 0.001. So the general pattern here is 10 to the whatever negative power is however many places you're going to have behind the decimal point. So here, it's not the number of 0's. In here, 10 to the minus 3, you only have two 0's but you have three places behind the decimal point. So what is the largest power of 10 that goes into this? Well, how many places behind the decimal point do I have? I have 1, 2, 3, 4, 5, 6. and we're going to have six places behind the decimal point. And the last place is going to be a 1. So you're going to have Five 0's and a 1. That's 10 to the minus 6. Now, this number right here is 7 times this number. If we multiply this times 7, we get 7 times 1. And then we have 1, 2, 3, 4, 5, 6 numbers behind the decimal point. So 1, 2, 3, 4, 5, 6. So this number times 7 is clearly equal to the number that we started off with. So we can rewrite this number. Instead of writing this number every time, we can write it as being equal to this number. Or, we could write it as 7. This is equal to 7 times this number. But this number is no better than that number. But this number is the same thing as 10 to the minus 6. 7 times 10 to the minus 6." + }, + { + "Q": "\nWhat is the difference between estimating and guestimating? 4:57", + "A": "An estimate is more certain than a guestimate", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 297 + ], + "3min_transcript": "That's equal to 10. What's 10 squared? That's 10 times 10. That's 100. What is 10 to the third? 10 to the third is 10 times 10 times 10, which is equal to 1,000. I think you see a general pattern here. 10 to the 0 has no 0's. No 0's in it. 10 to the 1 has one 0. 10 to the second power-- I was going to say the two-th power. 10 to the second power has two 0's. Finally, 10 to the third has three 0's. Don't want to beat a dead horse here, but I think you get the idea. Three 0's. If I were to do 10 to the 100th power, what would that look like? I don't feel like writing it all out here, but it would be 1 followed by-- you could guess it-- a hundred 0's. So it would just be a bunch of 0's. And if we were to count up all of those 0's, you And actually, this might be interesting, just as an aside. You may or may not know what this number is called. This is called a googol. A googol. In the early '90s if someone said, hey, that's a googol, you wouldn't have thought of a search engine. You would have thought of the number 10 to the 100th power, which is a huge number. It's more than the number of atoms, or the estimated number of atoms, in the known universe. In the known universe. It raises the question of what else is there out there. But I was reading up on this not too long ago. And if I remember correctly, the known universe has the order of 10 to the 79th to 10 to the 81 atoms. And this is, of course, rough. No one can really count this. People are just kind of estimating it. Or even better, guesstimating this. But this is a huge number. What may be even more interesting to you a very popular search engine-- Google. Google is essentially just a misspelling of the word \"googol\" with the O-L. And I don't know why they called it Google. Maybe they got the domain name. Maybe they want to hold this much information. Maybe that many bytes of information. Or, it's just a cool word. Whatever it is-- maybe it was the founder's favorite number. But it's a cool thing to know. But anyway, I'm digressing. This is a googol. It's just 1 with a hundred 0's. But I could equivalently have just written that as 10 to 100, which is clearly an easier way. This is an easier way to write this. This is easier. In fact, this is so hard to write that I didn't even take the trouble to write it. It would have taken me forever. This was just twenty 0's right here. A hundred 0's I would have filled up this screen and you have found it boring. So I didn't even write it. So clearly, this is easier to write." + }, + { + "Q": "At 0:18 how do you know the difference between a large and a very large number?\n", + "A": "One is BIG and the other is SMALL", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 18 + ], + "3min_transcript": "I don't think it's any secret that if one were to do any kind of science, they're going to be dealing with a lot of numbers. It doesn't matter whether you do biology, or chemistry, or physics, numbers are involved. And in many cases, the numbers are very large. They are very, very large numbers. Very large numbers. Or, they're very, small, very small numbers. Very small numbers. You could imagine some very large numbers. If I were to ask you, how many atoms are there in the human body? Or how cells are in the human body? Or the mass of the Earth, in kilograms, those are very large numbers. If I were to ask you the mass of an electron, that would be a very, very small number. So any kind of science, you're going to be dealing with these. And just as an example, let me show you one of the most common numbers you're going to see, in especially chemistry. It's called Avogadro's number. Avogadro's number. And if I were write it in just the standard way of writing as-- do it in a new color. It would be 6022-- and then another 20 zeroes. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. And even I were to throw some commas in here, it's not going to really help the situation to make it more readable. Let me throw some commas in here. This is still a huge number. If I have to write this on a piece of paper or if I were to publish some paper on using Avogadro's number, it would take me forever to write this. And even more, it's hard to tell if I forgot to write a zero or if I maybe wrote too many zeroes. So there's a problem here. Is there a better way to write this? So is there a better way to write this than to write it all out like this? To write literally the 6 followed by the 23 digits, or the 6022 followed by the 20 zeroes there? you're curious, Avogadro's number, if you had 12 grams of carbon, especially 12 grams of carbon-12, this is how many atoms you would have in that. And just so you know, 12 grams is like a 50th of a pound. So that just gives you an idea of how many atoms are laying around at any point in time. This is a huge number. The point of here isn't to teach you some chemistry. The point of here is to talk about an easier way to write this. And the easier way to write this we call scientific notation. Scientific notation. And take my word for it, although it might be a little unnatural for you at this video. It really is an easier way to write things like things like that. Before I show you how to do it, let me show you the underlying theory behind scientific notation. If I were to tell you, what is 10 to the 0 power? We know that's equal to 1." + }, + { + "Q": "\nat 18:36 sal added two powers 10^-3 and 10^-4 = 10^-7. should it not be 10^7 ?", + "A": "No, Sal is correct. -3 + -4 = -7. A negative plus a negative equals a bigger negative. (You may be getting confused with multiplication; a negative TIMES a negative is a positive.)", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 1116 + ], + "3min_transcript": "And especially if you're dealing with twenty or thirty 0's on either sides of the decimal point. Put a 0 here to make my wife happy. But when you do it in scientific notation, it will actually simplify it. This guy can be rewritten as 5 times 10 to the what? I have 1, 2, 3 spaces behind the decimal. 10 to the third. And then this is 8, so this is times 8 times 10 to the-- sorry, this is 5 times 10 to the minus 3. That's very important. 5 times 10 to the 3 would have been 5,000. Be very careful about that. Now, what is this guy equal to? This is 1, 2, 3, 4 places behind the decimal. So it's 8 times 10 to the minus 4. If we're multiplying these two things, this is the same thing as 5 times 10 to the minus 3 times 8 times 10 to the minus 4. There's nothing special about the scientific notation. It literally means what it's saying. So for multiplying, you could write it out like this. And multiplication, order doesn't matter. 10 to the minus 3 times 10 to the minus 4. And then, what is 5 times 8? 5 times 8 we know is 40. So it's 40 times 10 to the minus 3 times 10 to the minus 4. And if you know your exponent rules, you know that when you multiply two numbers that have the same base, you can just add their exponents. So you just add the minus 3 and the minus 4. So it's equal to 40 times 10 to the minus 7. Let's do another example. Let's say we were to multiply Avogadro's number. So we know that's 6.022 times 10 to the 23rd. Now, let's say we multiply that times some really small number. So times, say, 7.23 times 10 to the minus 22. So this is some really small number. You're going to have a decimal, and then you're going to have So this is a really small number. But the multiplication, when you do it in scientific notation, is actually fairly straightforward. This is going to be equal to 6.0-- let me write it properly. 6.022 times 10 to the 23rd times 7.23 times 10 to the minus 22. We can change the order, so it's equal to 6.022 times 7.23. That's that part. So you can view it as these first parts of our scientific notation times 10 to the 23rd times 10 to the minus 22. And now, this is-- you're going to do some little decimal multiplication right here. It's going to be-- some number-- 40 something, I think. I can't do this one in my head. But this part is pretty easy to calculate. I'll just leave this the way it is. But this part right here, this will be times. 10 to the 23rd times 10 to the minus 22." + }, + { + "Q": "\nIn problem #10, minute 1:56, wouldn't the triangles be called RTP or PTR or just T for the one in purple, and the other to be called APT or TPA or just P. I thought that the major/main angle of the triangle needed to be in-between the other two letters when naming it?", + "A": "I m not aware of defining a triangle based solely on the reference to the major angle.", + "video_name": "bWTtHKSEcdI", + "timestamps": [ + 116 + ], + "3min_transcript": "Actually, just right after stopping that video, I realized a very simple way of showing you that RP is congruent to TA, a little bit more of a rigorous definition. If we can show that this triangle right there, that one I drew in purple, and this triangle right here are congruent, then we could make a fairly reasonable argument that RP is going to be congruent to TA, because they're essentially the corresponding sides of the two congruent triangles. This congruent triangles would be kind of flipping each other. So how can we make that argument? Well, on the purple triangle, this angle is going to be equal to this angle on the yellow triangle. Actually, we got that from the fact that this is an isosceles trapezoid, so the base angles are going to be the same. know that this side, right there, is going to be congruent to this side. And then finally, they both share this side right here. So we could use the argument-- once again, side, angle, side-- that the side, angle, and side are congruent to this side, angle, and side. So you could say by SAS, triangle TRP is congruent to triangle TAP. And if they're congruent, then all of the corresponding sides are equal, so then TA is congruent to RP. Once again, you didn't have to do all that. It's a multiple choice test. But I wanted to give you that. I felt bad I wasn't giving you a more rigorous definition. A more rigorous proof. So anyway, problem number 11. If a quadrilateral has perpendicular diagonals, then it is a rhombus. Fair enough. Which of the following is a counterexample to the statement above? So they're saying, if it's perpendicular diagonals, then it's a rhombus. So if we could find something that has perpendicular diagonals that is not a rhombus, then we have a counterexample. Then this would not be true. So let's find something with perpendicular diagonals that is not a rhombus. Well, this one has perpendicular diagonals. The diagonals are perpendicular to each other, all 90-degree angles. And this is clearly not a rhombus. This is like a kite. This is not parallel to this and that is not parallel, so this is not a rhombus. So this is definitely a counterexample." + }, + { + "Q": "at about 3:30, the instructor states that (a^b)^d=a^bd. How is it possible for the exponent to shift down a degree?\n", + "A": "This is how exponents work. It doesn t shift it down a degree. Here are a few cases and examples: (2^3)^2=8^2=64 (2^3)^2=2^(3*2)=2^6=64 (2^1=2,2^2=2,2^3=8,2^4=16,2^5=32,2^6=64) (3^2)^3=9^3=729 (3^2)^3=3^(2*3)=3^6=729 (3,9,27,81,243,729) This is not unlike what happens when you multiply two numbers with the same base raise to different powers. In that case you add the exponents: 2^2*2^4=2^(2+4)=2^6=64 2^2*2^4=4*16=64 Good question. It is confusing at first, but when you get the hang of it, it s not too hard.", + "video_name": "Pb9V374iOas", + "timestamps": [ + 210 + ], + "3min_transcript": "Well, this is the exponent right over here. That's the same thing as z. So that's going to be the same thing as-- let me do this in a different color-- that 3 is the same thing-- we could put it out front-- that's the same thing as 3 times the logarithm base 5 of x. This is just another way of writing it using this property. And so you could argue that this is a what-- maybe this is a simplification because you took the exponent outside of the logarithm, and you're multiplying the logarithm by that number now. Now with that out of the way, let's think about why that actually makes sense. So let's say that we know that-- and I'll just pick some arbitrary letters here-- let's say that we know that a to the b power is equal to c. And so if we know that-- that's written as an exponential equation. If we wanted to write the same truth as a logarithmic equation, we would say logarithm base a of c To what power do I have to raise a to get c? I raise it to the bth power. a to the b power is equal to c. Fair enough. Now let's take both sides of this equation right over here, and raise it to the dth power. So let me make it-- so let's raise-- take both sides of this equation and raise it to the dth power. Instead of doing it in place, I'm just going to rewrite it over here. So I wrote the original equation, a to the b is equal to c, which is just rewriting this statement. But let me take both sides of this to the dth power. And I should be consistent. I'll use all capital letters. So this should be a B. Actually, let's say I'm using all lowercase letters. This is a lowercase c. So let me write it this way, a to the-- so I'm going to raise this to the dth power, and I'm going to raise this to the dth power. Obviously, if these two things are equal to each other, if I raise both sides to the same power, the equality is still going to hold. Now, what's interesting over here is we can use what we know about exponent properties. Say, look, if I have a to the b power, and then I raise that to the d power, our exponent properties say that this is the same thing. This is equal to a to the bd power. This is equal to a to the bd. Let me write it here. This is-- let me do that in a different color. I've already used that green. This right over here, using what we know about exponent properties, this is the same thing as a to the bd power. So we have a to the bd power is equal to c to the dth power. And now this exponential equation, if we would write it as a logarithmic equation, we would say log base a of c to the dth power is equal to bd. What power do I have to raise a to get to c to the dth power?" + }, + { + "Q": "I understood the video, but one part confused me. When he's talking about B = -B at 1:12 through 1:53, that confused me. Wouldn't B equal -B because it is after the zero? Thanks in advance!\n", + "A": "No. If you assume the scale is one then B = -1. -B would then be - -1. Two negatives is a positive so B= -1 and -B = 1. 1 is not equal to -1 so B=-B is not true. Just because there is a negative in front of the variable does not mean it is negative. It just means that it is multiplied by -1. I hope this clears up your confusion.", + "video_name": "vRa6XxykfbY", + "timestamps": [ + 72, + 113 + ], + "3min_transcript": "- [Voiceover] So let's do a couple of examples of the number opposites exercise on Khan Academy. Just to make sure that we've really digested what it means to be the opposite of a number. So they have a number line here, and then they've marked some of the numbers with letters. And this is A, B, C, D, E. And then they say, \"What can we say about point B?\" And point B is one hash mark to the left of zero, all right. \"Select all that applies.\" So the first thing, they say is, \"B is equal to negative D.\" So that means that B needs to be the opposite of D. Let's see if that's true. D is here. D is one hash mark to the right of zero. The opposite of D, or negative D, should be one hash mark to the left of zero. Which is B. B is the opposite of D. B is negative D, so this is true. The next statement is, \"B is the opposite of E.\" Let's see, E is one, two, three hash marks to the right of zero. So the opposite of E should be one, two, three hash marks to the left of zero. B is not the opposite of E. A is the opposite of E, but that isn't what they write here so we're not going to check that. And then they say, \"B is equal to negative B.\" That's saying that B is the same thing as the opposite of B. That is not true. The opposite of B is going to go on the other side of the number line. That's D, which is not B. So this is also not true. This would have been true if they said, \"C is equal to negative C.\" Because C is at zero, it's zero away from zero. C is zero. The opposite of zero is going to be zero away on the other side of the number line. Well, that's still zero away. So zero is negative zero. Or we could have said C is equal to negative C. But the only number for which that is true is zero. It's not going to be true for B, because B is clearly not zero. So let's do a couple of here. \"What can we say about negative C?\" All right, so now C is zero. \"Select all that apply.\" Yeah, the negative of zero is going to still be zero. The negative of zero, the opposite of zero, is still going to be zero. The opposite of zero, that's not going to be this number. This is some non-zero number. So that's that. All right. \"What can we say about point T?\" T once again is at zero. So we say, \"T is zero.\" \"Zero is the opposite of zero.\" It is equal to negative zero. Both of them -- one is zero, you could say zero to the left, one is zero to the right. Well, that still means that they're still on zero. And they're right, they are zero. \"T is equal to the opposite of the opposite of T.\" Well, this would actually be true for any number. For any number, the opposite of the opposite is going to be that number. We could have put the Q here, the R here. Really for any number the opposite of the opposite is going to be that number again. And so it's definitely going to be true for zero as well. So that's definitely going to be true. So hopefully that kind of gives you a little bit" + }, + { + "Q": "\nAt 5:38, does Sal mean Law of Sines?", + "A": "yes you are definitely correct that is a funny mess up on Sal part", + "video_name": "VjmFKle7xIw", + "timestamps": [ + 338 + ], + "3min_transcript": "that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A. so four times the sine of 105 gives us, it's approximately equal to, let's just round to the nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square root of two over two. Come to think of it, B is four times the sine of 45 degrees. If we wanted actual numerical value, we could just write this as two square roots of two. But let's actually figure out what that is. Two square roots of two is equal to 2.83. So B is approximately equal to 2.83. So [I'm] be clear, this four divided by two is two square roots of two, which is 2.8. Which is approximately equal to 2.83 if we round to the nearest 100th, 2.83, which also seems pretty reasonable here. So the key of the Law of Cosines is if you have two angles and a side, you're able to figure out everything else about it. Or if you actually had two sides and an angle, you also would be able to figure out everything else about the triangle." + }, + { + "Q": "At 3:36, you said that we should take a reciprocal to the both sides of the equation. But can't we just cross multiply the equation to get a=4xsin105 degrees\n", + "A": "Either way works. The answer is the same.", + "video_name": "VjmFKle7xIw", + "timestamps": [ + 216 + ], + "3min_transcript": "So sine of 45 degrees over B. And so if we wanted to figure out A, we could solve this equation right over here. And if we wanted to solve for B, we could just set this equal to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you might just remember it from your unit circles or from even 30, 60, 90 triangles and that's 1/2. And if you don't remember it, you can use a calculator to verify that. I have already verified that this is in degree mode, so it's 0.5. So this is going to be equal to 1/2 over two. So another way of thinking about it, that's going to be equal to 1/4, this piece is equal to 1/4 is equal to sine of a 105 degrees over A. Let me write this, this is equal to sine of 105 degrees over A. And actually, we could also say, that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A. so four times the sine of 105 gives us, it's approximately equal to, let's just round to the nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square root of two over two. Come to think of it, B is four times the sine of 45 degrees." + }, + { + "Q": "\nI calculated the sides using the Pythagorean Theorem, and I came up with a = 3.47, which is .4 off the answer they said at 4:14. When I calculated the side of b from the given answer (3.86) I got 3.3, also about .4 off the listed answer at 5:14 (2.83). Is this just because of rounding, or was something wrong?", + "A": "Hello GoMcLucky, The Pythagorean Theorem doesn t apply to this problem as we do not have right triangle. Instead we want to set up a ratio: SIN(105) / a = sin(30)/2 a = 3.86 Regards, APD", + "video_name": "VjmFKle7xIw", + "timestamps": [ + 254, + 314 + ], + "3min_transcript": "So sine of 45 degrees over B. And so if we wanted to figure out A, we could solve this equation right over here. And if we wanted to solve for B, we could just set this equal to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you might just remember it from your unit circles or from even 30, 60, 90 triangles and that's 1/2. And if you don't remember it, you can use a calculator to verify that. I have already verified that this is in degree mode, so it's 0.5. So this is going to be equal to 1/2 over two. So another way of thinking about it, that's going to be equal to 1/4, this piece is equal to 1/4 is equal to sine of a 105 degrees over A. Let me write this, this is equal to sine of 105 degrees over A. And actually, we could also say, that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A. so four times the sine of 105 gives us, it's approximately equal to, let's just round to the nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square root of two over two. Come to think of it, B is four times the sine of 45 degrees." + }, + { + "Q": "At 1:04 why is Sal Subtracting those numbers?\n", + "A": "Remember that the sum of a triangle s three angles is 180 degrees. You can see that we know two of the angles already, 45 and 30. To find the final unknown angle, Sal simply subtracts the other two angles from 180, the sum of a triangle s angles.", + "video_name": "VjmFKle7xIw", + "timestamps": [ + 64 + ], + "3min_transcript": "Voiceover:We've got a triangle here where we know two of the angles and one of the sides. And what I claim, is that I can figure out everything else about this triangle just with this information. You give me two angles and a side, and I can figure out what the other two sides are going to be. And I can, of course, figure out the third angle. So, let's try to figure that out. And the way that we're going to do it, we're going to use something called the Law of Sines. In a future video, I will prove the Law of Sines. But here, I am just going to show you how we can actually apply it. And it's a fairly straightforward idea. The Law of Sines just tells us that the ratio between the sine of an angle, and the side opposite to it, is going to be constant for any of the angles in a triangle. So for example, for this triangle right over here. This is a 30 degree angle, This is a 45 degree angle. They have to add up to 180. So this right over here has to be a, let's see, it's going to be 180 minus 45 minus 30. That's 180 minus 75, so this is going to equal And so applying the Law of Sines, actually let me label the different sides. Let's call this side right over here, side A or has length A. And let's call this side, right over here, has length B. So the Law of Sines tells us that the ratio between the sine of an angle, and that the opposite side is going to be constant through this triangle. So it tells us that sine of this angle, sine of 30 degrees over the length of the side opposite, is going to be equal to sine of a 105 degrees, over the length of the side opposite to it. Which is going to be equal to sine of 45 degrees. So sine of 45 degrees over B. And so if we wanted to figure out A, we could solve this equation right over here. And if we wanted to solve for B, we could just set this equal to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you might just remember it from your unit circles or from even 30, 60, 90 triangles and that's 1/2. And if you don't remember it, you can use a calculator to verify that. I have already verified that this is in degree mode, so it's 0.5. So this is going to be equal to 1/2 over two. So another way of thinking about it, that's going to be equal to 1/4, this piece is equal to 1/4 is equal to sine of a 105 degrees over A. Let me write this, this is equal to sine of 105 degrees over A. And actually, we could also say," + }, + { + "Q": "At around 5:03 how does he go from 4*\u00e2\u0088\u009a2/2 to 2\u00e2\u0088\u009a2?\nSolving both problems on the calculator give the same answer so they are both the same thing, but how did he simplify it from 4*\u00e2\u0088\u009a2/2 to 2\u00e2\u0088\u009a2?\n\nEDIT: Ok... its pretty obvious It just didn't click for some reason. He just simplified the 2 in 4*\u00e2\u0088\u009a2/2 to convert it to 2\u00e2\u0088\u009a2.. idk how I didn't see that.\n", + "A": "that got me as well", + "video_name": "VjmFKle7xIw", + "timestamps": [ + 303 + ], + "3min_transcript": "that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A. so four times the sine of 105 gives us, it's approximately equal to, let's just round to the nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square root of two over two. Come to think of it, B is four times the sine of 45 degrees. If we wanted actual numerical value, we could just write this as two square roots of two. But let's actually figure out what that is. Two square roots of two is equal to 2.83. So B is approximately equal to 2.83. So [I'm] be clear, this four divided by two is two square roots of two, which is 2.8. Which is approximately equal to 2.83 if we round to the nearest 100th, 2.83, which also seems pretty reasonable here. So the key of the Law of Cosines is if you have two angles and a side, you're able to figure out everything else about it. Or if you actually had two sides and an angle, you also would be able to figure out everything else about the triangle." + }, + { + "Q": "At 8:34, does it matter where you place your coordinate letter points? Like for example, can you put point T at the top left of the isosceles trapezoid instead of the bottom left?\n", + "A": "Yes, as long the the diagonals correspond and the letters are in order going around the trapezoid. The answer would still be the same!", + "video_name": "4PPMnI8-Zsc", + "timestamps": [ + 514 + ], + "3min_transcript": "Given, TRAP, that already makes me worried. Given TRAP is an isosceles trapezoid with diagonals RP and TA, which of the following must be true? OK, let's see what we can do here. So an isosceles trapezoid means that the two sides that lead up from the base to the top side are equal. Kind of like an isosceles triangle. So let me draw that. Actually, I'm kind of guessing that. I haven't seen the definition of an isosceles triangle anytime in the recent past. An isosceles trapezoid. But it sounds right. So I'll go with it. And that's a good skill in life. So I think what they say when they say an isosceles trapezoid, they are essentially saying that this to be equal to that. They're saying that this side is equal to that side. An isosceles trapezoid. And they say RP and TA are diagonals of it. Let me draw that. So let me actually write the whole TRAP. So this is T R A P is a trapezoid. Let me draw the diagonals. RP is that diagonal. And TA is this diagonal right here. OK. All right, let's see what we can do. Which of the following must be true? RP is perpendicular to TA. Well, I can already tell you that that's not going to be true. And you don't even have to prove it. Because you can even visualize it. If you squeezed the top part down. If you were to squeeze the top down, they didn't tell us how high it is. Then these angles, let me see if I can draw it. That angle and that angle, which are opposite or vertical angles, which we know is the U.S. word for it. Those are going to get smaller and smaller if we squeeze it down. And in order for both of these to be perpendicular those would have to be 90 degree angles. And we already can see that that's definitely not the case. All right. RP is parallel to TA. Well that's clearly not the case, they intersect. All right, they're the diagonals. RP is congruent to TA. Well, that looks pretty good to me. Because it's an isosceles trapezoid. If we drew a line of symmetry here, everything you see on this side is going to be kind of congruent to its mirror" + }, + { + "Q": "At 0:13, what does 'hone in' mean?\n", + "A": "To hone in means to sharpen or make something more precise", + "video_name": "ojFuf9RYmzI", + "timestamps": [ + 13 + ], + "3min_transcript": "Voiceover:So we've got 3 Y squared plus 6 X to the third and we're raising this whole to the fifth power and we could clearly use a binomial theorem or pascal's triangle in order to find the expansion of that. But what I want to do is really as an exercise is to try to hone in on just one of the terms and in particular I want to hone in on the term that has some coefficient times X to the sixth, Y to the sixth. So in this expansion some term is going to have X to the sixth, Y to sixth and I want to figure out what the coefficient on that term is and I encourage you to pause this video and try to figure it out on your own. So I'm assuming you've had a go at it and you might have at first found this to be a little bit confusing. I'm only raising it to the fifth power, how do I get X to the sixth, Y to the sixth? But then when you look at the actual terms of the binomial it starts to jump out at you. Okay, I have a Y squared term, I have an X to the third term, so when I raise these to powers I'm going to get, I could have powers higher than the fifth power. But to actually think about which of these terms has the X to the sixth, Y to the sixth, in I guess the actual expansion without even thinking about its coefficients. And we've seen this multiple times before where you could take your first term in your binomial and you could start it off it's going to start of at a, at the power we're taking the whole binomial to and then in each term it's going to have a lower and lower power. So let me actually just copy and paste this. So this is going to be, so copy and so that's first term, second term, let me make sure I have enough space here. Second term, third term, fourth term, fourth term, fifth term, and sixth term it's going to have 6 terms to it, you always have one more term than the exponent. And then, actually before I throw the exponents on it, let's focus on the second term. So the second term, actually I'll write it like this. let me copy and paste that, whoops. So let me copy and paste that. So we're going to put that there. That there. And that there. And that there. And then over to off your screen. I wrote it over there. We'll see if we have to go there. And then let's put the exponents. So this exponent, this is going to be the fifth power, fourth power, third power, second power, first power and zeroeth power. And for the blue expression, for 6 X to the third, this is going to be the zeroeth power, first power, first power, second power, third power, fourth power, and then we're going to have the fifth power right over here." + }, + { + "Q": "\nAt 1:34, why are the points 2,0 and 0,2?", + "A": "how does that help", + "video_name": "wyTjyQMVvc4", + "timestamps": [ + 94 + ], + "3min_transcript": "So let's say I have a function of x and y; f of x and y is equal to x plus y squared. If I try to draw that, let's see if I can have a good attempt at it. That is my y axis-- I'm going to do a little perspective here --this is my x axis-- I make do the negative x and y axis, could do it in that direction --this is my x axis here. And if I were to graph this when y is 0, it's going to be just a-- let me draw it in yellow --is going to be just a straight line that looks something like that. And then for any given actually, we're going to have a parabola in y. y is going to look something like that. I'm just going to it in the positive quadrant. It's going to look something like that. It'll actually, when you go into the negative y, you're going to see the other half of the parabola, but I'm not going to worry about it too much. So you're going to have this surface. it looks something like that. Maybe I'll do to another attempt at drawing it. But this is our ceiling we're going to deal with again. And then I'm going to have a path in the xy plane. equal to 2, y is 0. And I'm going to travel, just like we did in the last video, I'm going to travel along a circle, but this time the circle's going to have of radius 2. Move counter clockwise in that circle. This is on the xy plane, just to be able to visualize it properly. So this right here's a point 0, 2. And I'm going to come back along the y axis. This is my path; I'm going to come back along the y access and then so I look a left here, and then I'm going to take another left here in and come back along the x axis. I drew it in these two shades of green. That is my contour. And what I want to do is I want to evaluate the surface area of essentially this little building that has the roof of f of xy is equal to x plus y squared, and I want to find the surface area of its walls. Then you're going to have this wall, which is along the curve; it's going to look something like kind of funky wall on that curved side right there. I'll try my best effort to try to-- it's going to be curving way up like that and then along the y axis. It's going to have like a half a parabolic wall right there. I'll do that back wall along the y axis. I'll do that in orange, I'll use magenta. That is the back wall along the y axis. Then you have this front wall along the x axis. And then you have this weird curvy curtain or wall-- do that maybe in blue --that goes along this curve right here, this part of a circle of radius 2. So hopefully you get that visualization. It's a little harder; I'm not using any graphic program at this time." + }, + { + "Q": "\nAt 8:14, when Sal says this curve actually has another point where the slope is equal to the average slope, I was wondering whether there are as many such points as the number of inflection points in the graph, or atleast in the closed interval we are concerned about.\nIs this true? Are there really as many such points as the number of inflection points?", + "A": "You re almost right. If n = the number of inflection points, then there will be n + 1 points where the slope is equal to the average slope. Except straight lines, of course, where the slope of every point is equal to the average slope.", + "video_name": "bGNMXfaNR5Q", + "timestamps": [ + 494 + ], + "3min_transcript": "Let me do it in a blue. So that's the average slope between those two points, right? So what does the mean value theorem tell us? It says, if f of x is defined over this closed interval from a to b, and f of x is continuous, and it's differentiable, that you could take the derivative at any point, that there must be some points c f prime of c is equal to this thing. So is equal to f prime of c. I shouldn't have written it here. So what is that telling us? So all that's telling us, is if we're continuous, differentiable, defined over the closed interval, that there's some point c, oh, and c has to be between a and b, of the points, but there's some point c where the derivative at c, or the slope at c, the instantaneous slope at c, is exactly equal to the average slope over that interval. So what does that mean? So we can look at it visually. Is there any point along this curve where the slope looks very similar to this average slope that we calculated? Well, sure, let's see. It looks like, maybe, this point, right here? This is very inexact. But that point looks like the slope, you know, I could say the slope is something like that, right there. So we don't know what, analytically, this function is, but visually, you could see that at this point c, the derivative, so I just picked that point. So this could be our point c. And how do we just say that? to the average slope. So f prime of c is this thing, and it's going to be equal to the average slope over the whole thing. And this curve actually probably has another point where the slope is equal to the average slope. Let's see. This one looks, like, right around there. Just the way I drew it. Looks like the slope there could look something like, could be parallel as well. These lines should be parallel. The tangent lines should be parallel. So hopefully that makes a little sense to you. Another way to think about it is that your average, actually, let me draw a graph just to make sure that we hit the point home. Let's draw my position as a function of time. So this is something, this'll make it applicable to the real world. So that's my x-axis, or the time axis, that's my position axis. This is going back to our original intuition of what" + }, + { + "Q": "why was it that one corner was 30* and one was 49* at 8:44? I still don't understand :(\n", + "A": "It was because when you add up 101 (the measure of corner 1) and 30 (measure of corner 2), you get 131. In a triangle, when you add up all 3 of the angles, you get 180*, no matter what. So, 180-131=49. Therefore, the measure of the other angle would be 49*. Hope that helped! ^_^", + "video_name": "kqU_ymV581c", + "timestamps": [ + 524 + ], + "3min_transcript": "What is this angle? This is a good time to pause because I will now show you the solution. So what can we do here? So this angle, well jeez, I just like to just mess around and see what I can figure out. So, if this angle here is 101 degrees, what other angles can we figure out? We could figure out -- well, we could figure out this angle. We could figure out a bunch of angles. We could figure out that -- let me switch the color, these are my figure out angles. So that's 101, then this is supplementary, that's 79 degrees, right? That's also 79 degrees because this is also supplementary. This angle right here is opposite to it, so this angle right here is going to be 101 degrees. What else can figure out? We could figure out this angle because it's supplementary, we could figure out this angle. We could also figure out this angle because we see this triangle right here. So let's call this angle b, b for blue. So b plus 75 plus 75 is going to equal 180. And I'm just using this triangle right here. So b plus 150 is equal to 180, or b is equal to 30 degrees. So we're able to figure this out. Now, what will you do if I told you that we are now ready to figure out this yellow angle? It might not be obvious to you. You kind of have to look at the triangle in the right way, and the SAT will do this to you all the time, all the time. That's why I'm testing you this way. Well, let me give you a little hint. Look at this triangle. Non-ideal color, let me do it in red so it really stands out. Look at this triangle. just looking at the right triangle and kind of seeing that oh wow, I actually can figure out something. Look at this triangle right here. We know this angle of it, 101 degrees. We know this angle, we just figured it out, it was 30 degrees. So all we have left is to figure out this yellow angle, call it x. So x plus 101 plus 30 is equal to 180 degrees because the angles in a triangle add up to 180 degrees. So x plus 131 is equal to 180. x is equal to what? 49 degrees. There you go. We've done the second problem in the angle game. I think that's all of the time I have now in this video. In the next video maybe I'll do a couple more of these angle game problems. See you soon." + }, + { + "Q": "At 9:55 shouldn't the sqrt of 1 be +1 and -1 and then say that its because its in the first quadrant we can drop -1?\n", + "A": "Sal took the positive square root of 1 because ds(the length of a small segment of the curve) is always positive because it is a length.", + "video_name": "uXjQ8yc9Pdg", + "timestamps": [ + 595 + ], + "3min_transcript": "curtain that has our curve here as kind of its base, and has this function, this surface as it's ceiling. So we go back down here, and let me rewrite this whole thing. So this becomes the integral from t is equal to o to t is equal to pi over 2-- I don't like this color --of cosine of t, sine of t, cosine times sine-- that's just the xy --times ds, which is this expression right here. And now we can write this as-- I'll go switch back to that color I don't like --the derivative of x with respect to t is minus sine of t, and we're going to square it, plus the derivative of y with respect to t, that's cosine of t, and we're going to square it-- let me make my radical a little bit bigger --and then all of that times dt. you realize that this right here, and when you take a negative number and you squared it, this is the same thing. Let me rewrite, do this in the side right here. Minus sine of t squared plus the cosine of t squared, this is equivalent to sine of t squared plus cosine of t squared. You lose the sign information when you square something; it just becomes a positive. So these two things are equivalent. And this is the most basic trig identity. This comes straight out of the unit circle definition: sine squared plus cosine squared, this is just equal to 1. So all this stuff under the radical sign is just equal to 1. And we're taking the square root of 1 which is just 1. So all of this stuff right here will just become 1. And so this whole crazy integral simplifies a good bit pi over 2 of-- and I'm going to switch these around just because it will make it a little easier in the next step --of sine of t times cosine of t, dt. All I did, this whole thing equals 1, got rid of it, and I just switched the order of that. It'll make the next up a little bit easier to explain. Now this integral-- You say sine times cosine, what's the antiderivative of that? And the first thing you should recognize is, hey, I have a function or an expression here, and I have its derivative. The derivative of sine is cosine of t. So you might be able to a u substitution in your head; it's a good skill to be able to do in your head. But I'll do it very explicitly here. So if you have something that's derivative, you define that something as u. So you say u is equal to sine of t and then du, dt, the derivative of u with respect to t is equal to cosine of t." + }, + { + "Q": "\nAt 1:55 isn't it supposed to be 3.14159", + "A": "Sal used 3.14 as an approximate value for \u00cf\u0080. (3.14159 is also an approximate, but a little more precise.)", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 115 + ], + "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi." + }, + { + "Q": "why did you divide by 2 at 1:10\n", + "A": "because the diameter is twice as big as radius. so you divide 2 from diameter to find the radius", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 70 + ], + "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi." + }, + { + "Q": "\nat 1:29, why is it 64 pi not 16 pi because dont you multiply pi and the diameter to get the answer", + "A": "pi * diameter (same as 2*pi*radius) would be the circumference of the wafer which is 16 pi. The question wants us to compute the area so we need to use pi*r^2.", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 89 + ], + "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi." + }, + { + "Q": "How do you get 64 at 1:27? Do you just multiply the number that's the radius by itself or something? I'm confused with this.\n", + "A": "Yes. pi*r^2 means radius in second power. What s power? How many times the number needs to be multiplied by itself. If power is 2 than the number needs to be multiplied twice. If radius is 8 than r^2 is 8^2 which is 8*8 (multiply by itself twice) which equals 64.", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 87 + ], + "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi." + }, + { + "Q": "\nAt 3:50, How does 8*8*8 simplify to just 8?", + "A": "good question. this simplifies to 8 because all those 8s are inside of a cube root symbol. so we had \u00e2\u0088\u009a8*8*8. this simplifies to just 8. sry if this doesn t help :( its kinda hard to explain in words.", + "video_name": "DKh16Th8x6o", + "timestamps": [ + 230 + ], + "3min_transcript": "And 4 is 2 times 2. So we got a lot of twos. So essentially, if you multiply 2 one, two, three, four, five, six, seven, eight, nine times, you're going to get 512, or 2 to the ninth power is 512. And that by itself should give you a clue of what the cube root is. But another way to think about it is, can we find-- there's definitely three twos here. But can we find three groups of twos, or we could also find-- let me look at it this way. We can find three groups of two twos over here. So that's 2 times 2 is 4. 2 times 2 is 4. So definitely 4 multiplied by itself three times is divisible into this. But even better, it looks like we can get three groups of three twos. So one group, two groups, and three groups. That is 8. This is 2 times 2 times 2. That's 8. And this is also 2 times 2 times 2. So that's 8. So we could write 512 as being equal to 8 times 8 times 8. And so we can rewrite this expression right over here as the cube root of 8 times 8 times 8. So this is equal to negative 1, or I could just put a negative sign here, negative 1 times the cube root of 8 times 8 times 8. So we're asking our question. What number can we multiply by itself three times, or to the third power, to get 512, which is the same thing as 8 times 8 times 8? Well, clearly this is 8. So the answer, this part right over here, is just going to simplify to 8. And so our answer to this, the cube root of negative 512, is negative 8. And we are done. And you could verify this. Multiply negative 8 times itself three times. Negative 8 times negative 8 is positive 64. You multiply that times negative 8, you get negative 512." + }, + { + "Q": "0:32 You need to use prime factorization to help you with finding cube roots?\n", + "A": "The prime factorization of the base is rather important in finding any type of root (whether square root, cube root, 5th root, or whatever). Do you need instruction on how to do this?", + "video_name": "DKh16Th8x6o", + "timestamps": [ + 32 + ], + "3min_transcript": "We are asked to find the cube root of negative 512. Or another way to think about it is if I have some number, and it is equal to the cube root of negative 512, this just means that if I take that number and I raise it to the third power, then I get negative 512. And if it doesn't jump out at you immediately what this is the cube of, or what we have to raise to the third power to get negative 512, the best thing to do is to just do a prime factorization of it. And before we do a prime factorization of it to see which of these factors show up at least three times, let's at least think about the negative part a little bit. So negative 512, that's the same thing-- so let me rewrite the expression-- this is the same thing as the cube root of negative 1 times 512, which is the same thing as the cube root of negative 1 times the cube root of 512. What number, when I raise it to the third power, do I get negative 1? Well, I get negative 1. This right here is negative 1. Negative 1 to the third power is equal to negative 1 times negative 1 times negative 1, which is equal to negative 1. So the cube root of negative 1 is negative 1. So it becomes negative 1 times this business right here, times the cube root of 512. And let's think what this might be. So let's do the prime factorization. So 512 is 2 times 256. 256 is 2 times 128. 128 is 2 times 64. We already see a 2 three times. 64 is 2 times 32. 32 is 2 times 16. We're getting a lot of twos here. 16 is 2 times 8. And 4 is 2 times 2. So we got a lot of twos. So essentially, if you multiply 2 one, two, three, four, five, six, seven, eight, nine times, you're going to get 512, or 2 to the ninth power is 512. And that by itself should give you a clue of what the cube root is. But another way to think about it is, can we find-- there's definitely three twos here. But can we find three groups of twos, or we could also find-- let me look at it this way. We can find three groups of two twos over here. So that's 2 times 2 is 4. 2 times 2 is 4. So definitely 4 multiplied by itself three times is divisible into this. But even better, it looks like we can get three groups of three twos. So one group, two groups, and three groups." + }, + { + "Q": "At 3:48, I don't get it how Sal got the square root of 8*8*8 to get the answer 8.....Can someone explain to me what happened there?\n", + "A": "well he was answering the cubed root of -512 thats how he got 8 so -8*-8*-8 is -512", + "video_name": "DKh16Th8x6o", + "timestamps": [ + 228 + ], + "3min_transcript": "And 4 is 2 times 2. So we got a lot of twos. So essentially, if you multiply 2 one, two, three, four, five, six, seven, eight, nine times, you're going to get 512, or 2 to the ninth power is 512. And that by itself should give you a clue of what the cube root is. But another way to think about it is, can we find-- there's definitely three twos here. But can we find three groups of twos, or we could also find-- let me look at it this way. We can find three groups of two twos over here. So that's 2 times 2 is 4. 2 times 2 is 4. So definitely 4 multiplied by itself three times is divisible into this. But even better, it looks like we can get three groups of three twos. So one group, two groups, and three groups. That is 8. This is 2 times 2 times 2. That's 8. And this is also 2 times 2 times 2. So that's 8. So we could write 512 as being equal to 8 times 8 times 8. And so we can rewrite this expression right over here as the cube root of 8 times 8 times 8. So this is equal to negative 1, or I could just put a negative sign here, negative 1 times the cube root of 8 times 8 times 8. So we're asking our question. What number can we multiply by itself three times, or to the third power, to get 512, which is the same thing as 8 times 8 times 8? Well, clearly this is 8. So the answer, this part right over here, is just going to simplify to 8. And so our answer to this, the cube root of negative 512, is negative 8. And we are done. And you could verify this. Multiply negative 8 times itself three times. Negative 8 times negative 8 is positive 64. You multiply that times negative 8, you get negative 512." + }, + { + "Q": "\nAt 4:33 was Sal meant to write \"X2 +2\" or am I confused ?", + "A": "You re right, he should have written X2 +2 instead of X1 +2", + "video_name": "r4bH66vYjss", + "timestamps": [ + 273 + ], + "3min_transcript": "It's just the way that we do it. The way we visualize things. If I wanted to plot the point 1, 1, I go to my coordinate axes. The first point I go along the horizontal, what we traditionally call our x-axis. And I go 1 in that direction. And then convention is, the second point I go 1 in the vertical direction. So the point 1, 1. Oh, sorry, let me be very clear. This is 2 and 2, so one is right here, and one is right there. So the point 1, 1 would be right there. That's just the standard convention. Now our convention for representing vectors are, you might be tempted to say, oh, maybe I just represent this vector at the point minus 1, 2. And on some level you can do that. I'll show you in a second. But the convention for vectors is that you can start at any point. Let's say we're dealing with two dimensional vectors. You can start at any point in R2. So let's say that you're starting at the point x1, and x2. To represent the vector, what we do is we draw a line from that point to the point x1. And let me call this, let's say that we wanted to draw a. So x1 minus 1. So this is, I'm representing a. So this is, I want to represent the vector a. x1 minus 1, and then x1 plus 2. Now if that seems confusing to you, when I draw it, it'll be very obvious. So let's say I just want to start at the point, let's just say for quirky reasons, I just pick a random point here. I just pick a point. That one right there. That's my starting point. So minus 4, 4. Now if I want to represent my vector a, what I just said is So x1 plus minus 1 or x1 minus 1. So my new one is going to be, so this is my x1 minus 4. So now it's going to be, let's see, I'm starting at the point minus 4 comma 4. If I want to represent a, what I do is, I draw an arrow to minus 4 plus this first term, minus 1. And then 4 plus the second term. 4 plus 2. And so this is what? This is minus 5 comma 6. So I go to minus 5 comma 6. So I go to that point right there and I just draw a line. So my vector will look like this. I draw a line from there to there. And I draw an arrow at the end point. So that's one representation of the vector minus 1, 2. Actually let me do it a little bit better. Because minus 5 is actually more, a little closer to right here. Minus 5 comma 6 Is right there, so I draw" + }, + { + "Q": "At 5:09, shouldn't it be (x1-1, x2+2)?\n", + "A": "Yes it should", + "video_name": "r4bH66vYjss", + "timestamps": [ + 309 + ], + "3min_transcript": "It's just the way that we do it. The way we visualize things. If I wanted to plot the point 1, 1, I go to my coordinate axes. The first point I go along the horizontal, what we traditionally call our x-axis. And I go 1 in that direction. And then convention is, the second point I go 1 in the vertical direction. So the point 1, 1. Oh, sorry, let me be very clear. This is 2 and 2, so one is right here, and one is right there. So the point 1, 1 would be right there. That's just the standard convention. Now our convention for representing vectors are, you might be tempted to say, oh, maybe I just represent this vector at the point minus 1, 2. And on some level you can do that. I'll show you in a second. But the convention for vectors is that you can start at any point. Let's say we're dealing with two dimensional vectors. You can start at any point in R2. So let's say that you're starting at the point x1, and x2. To represent the vector, what we do is we draw a line from that point to the point x1. And let me call this, let's say that we wanted to draw a. So x1 minus 1. So this is, I'm representing a. So this is, I want to represent the vector a. x1 minus 1, and then x1 plus 2. Now if that seems confusing to you, when I draw it, it'll be very obvious. So let's say I just want to start at the point, let's just say for quirky reasons, I just pick a random point here. I just pick a point. That one right there. That's my starting point. So minus 4, 4. Now if I want to represent my vector a, what I just said is So x1 plus minus 1 or x1 minus 1. So my new one is going to be, so this is my x1 minus 4. So now it's going to be, let's see, I'm starting at the point minus 4 comma 4. If I want to represent a, what I do is, I draw an arrow to minus 4 plus this first term, minus 1. And then 4 plus the second term. 4 plus 2. And so this is what? This is minus 5 comma 6. So I go to minus 5 comma 6. So I go to that point right there and I just draw a line. So my vector will look like this. I draw a line from there to there. And I draw an arrow at the end point. So that's one representation of the vector minus 1, 2. Actually let me do it a little bit better. Because minus 5 is actually more, a little closer to right here. Minus 5 comma 6 Is right there, so I draw" + }, + { + "Q": "\nAt 21:32, Khan says that the vector \"y-x\" is equal to (-6, -5) and graphs it by going six points down and five points to the left. Shouldn't he graph it by going six points to the left and five points down?", + "A": "Good catch. You are correct.", + "video_name": "r4bH66vYjss", + "timestamps": [ + 1292 + ], + "3min_transcript": "x minus y. That's the difference between the two vectors. You can view the difference as, how do you get from one vector to another vector, right? Like if, you know, let's go back to our kind of second grade world of just scalars. If I say what 7 minus 5 is, and you say it's equal to 2, well that just tells you that 5 plus 2 is equal to 7. Or the difference between 5 and 7 is 2. And here you're saying, look the difference between x and y is this vector right there. It's equal to that vector right there. Or you could say look, if I take 5 and add 2 I get 7. Or you could say, look, if I take vector y, and I add vector x minus y, then I get vector x. Now let's do something else that's interesting. Let's do what y minus x is equal to. y minus x. What is that equal to? Do it in another color right here. Well we'll take minus 4, minus 2 which is minus 6. It's minus 5. So y minus x is going to be, let's see, if we start here we're going to go down 6. 1, 2, 3, 4, 5, 6. And then back 5. So back 2, 4, 5. So y minus x looks like this. It's really the exact same vector. Remember, it doesn't matter where we start. It's just pointing in the opposite direction. So if we shifted it here. I could draw it right on top of this. It would be the exact as x minus y, but just in the opposite direction. Which is just a general good thing to know. So you can kind of do them as the negatives of each other. And actually let me make that point very clear. You know we drew y. Actually let me draw x, x we could draw as 2, 3. So you go to the right 2 and then up 3. I've done this before. This is x in non standard position. What is negative x? Negative x is minus 2 minus 3. So if I were to start here, I'd go to minus 2, then I'd go minus 3. So minus x would look just like this. Minus x. It looks just like x. It's parallel. It has the same magnitude. It's just pointing in the exact opposite direction. And this is just a good thing to kind of really get seared into your brain is to have an intuition for these things. Now just to kind of finish up this kind of idea of adding and subtracting vectors. Everything I did so far was in R2. But I want to show you that we can generalize them. And we can even generalize them to vector spaces that aren't normally intuitive for us to actually visualize. So let me define a couple of vectors. Let me define vector a to be equal to 0, minus 1, 2, and 3." + }, + { + "Q": "At 25:07, it looks like he accidentally enabled the smudge tool on the tablet. Do vectors like these have anything to do with matrix algebra?\n", + "A": "Yes, you could actually see the columns of matrices as vectors (if I am not mistaken).", + "video_name": "r4bH66vYjss", + "timestamps": [ + 1507 + ], + "3min_transcript": "We can do the same addition and subtraction operations with them. It's just it'll be hard to visualize. We can keep them in just vector form. So that it's still useful to think in four dimensions. So if I were to say 4 times a. This is the vector a minus 2 times b. What is this going to be equal to? This is a vector. What is this going to be equal to? Well we could rewrite this as 4 times this whole column vector, 0, minus 1, 2, and 3. Minus 2 times b. Minus 2 times 4, minus 2, 0, 5. And what is this going to be equal to? This term right here, 4 times this, you're going to get, the 4 times this, you're going to get 4 times 0, 0, minus 4, 8. 4 times 3 is 12. And then minus, I'll do it in yellow, minus 2 times 4 is 8. 2 times minus 2 is minus 4. 2 times 0 is 0. 2 times 5 is 10. This isn't a good part of my board, so let me just. It doesn't write well right over there. I haven't figured out the problem, but if I were just right it over here, what do we get? With 0 minus 8? Minus 8. Minus 4, minus 4. Minus negative 4. So that's minus 4 plus 4, so that's 0. 8 minus 0 is 8. 12 minus, what was this? Oh, this is a 10. Now you can see it again. Something is very bizarre. 2 times 5 is 10. So it's 12 minus 10, so it's 2. So when we take this vector and multiply it by 4, and subtract 2 times this vector, we just get this vector. And even though you can't represent this in kind of an easy kind of graph-able format, this is a useful concept. And we're going to see this later when we apply some of these vectors to multi-dimensional spaces." + }, + { + "Q": "At 5:35 how did Sal know that the two areas are equal? The problem doesn't say that they function has x intercept at 1.5, and picture isn't necessarily to scale\n", + "A": "looking at the graph and where the coordinates are given and marked, we can clearly see that the two areas are equal", + "video_name": "OvMBNVi5bLY", + "timestamps": [ + 335 + ], + "3min_transcript": "So it's going to be pi, and then we're going to divide it by four, because it's only one fourth of that circle. And we're going to subtract that. So we have negative pi and we were multiplying it times 1/4 out here, because it's just a quarter circle in either case. And we're subtracting it, because the area is below the x-axis. And so this simplifies to-- this is equal to 1/4 times 8 pi, which is the same thing as 2 pi. Did I do that right? 1/4 times 8 pi. So this all simplifies to 2 pi. So g of negative 4 is equal to negative 8 minus 2 pi. So clearly it is a negative number here. More negative than negative 8. So let's try the other bounds. So let's see what g of positive 3 is. So g of positive 3, when x is equal to 3, that-- we go back to our definition-- that is 2 times 3 plus the integral from 0 to 3 f of t dt. And this is going to be equal to 2 times 3 is 6. And the integral from 0 to 3 f of t dt, that's this entire area. So we have positive area over here. And then we have an equal negative area right over here because it's below the x-axis. So the integral from 0 to 3 is just going to be 0, you're going to have this positive area, and then this negative area right over here is going to completely cancel it out. Because it's symmetric right over here. So this thing is going to evaluate to 0. So g of 3 is 6. So we already know that our starting point, g of negative 4-- that when x is equal to negative 4-- that is not where g hits a global maximum. Because that's a negative number. And we already found the end point, where g hits a positive value. So negative 4 is definitely not a candidate. where g has a global absolute maximum. Now what we have to do is figure out any critical points that g has in between. So points in there where it's either undifferentiable or its derivative is equal to 0. So let's look at this derivative. So g prime of x-- we just take the derivative of this business up here. Derivative of 2x is 2. Derivative of this definition going from 0 to x of f of t dt-- we did that in part a, this is just the fundamental theorem of calculus-- this is just going to be plus f of x right over there. So it actually turns out that g is differentiable over the entire interval. You give any x value over this interval, we have a value for f of x. f of x isn't differentiable everywhere, but definitely f of x is defined everywhere, over the interval. So you'll get a number here, and obviously two is just two," + }, + { + "Q": "\nAt 1:38 ,does he have to write out all of the possibility's like that . i tried that and it took a long time (not to mention how tricky it was). Can't he use a possibility diagram/probability space diagram?", + "A": "It is just a way to show every possible outcome, like with truth tables. A shorter way to represent this is to determine the rule governing the outcomes and simply calculate mathematically. I think Sal is trying to show all possibilities so that even students witj no prior exposure to probabiliy can understand.", + "video_name": "xSc4oLA9e8o", + "timestamps": [ + 98 + ], + "3min_transcript": "Let's think about the situation where we have a completely fair coin here. So let me draw it. I'll assume it's a quarter or something. So this is a quarter. Let me draw my best attempt at a profile of George Washington. Well, that'll do. So it's a fair coin. And we're going to flip it a bunch of times and figure out the different probabilities. So let's start with a straightforward one. Let's just flip it once. So with one flip of the coin, what's the probability of getting heads? Well, there's two equally likely possibilities. And the one with heads is one of those two equally likely possibilities, so there's a 1/2 chance. Same thing if we were to ask what is the probability of getting tails? There are two equally likely possibilities, and one of those gives us tails, so 1/2. And this is one thing to realize. If you take the probabilities of heads plus the probabilities of tails, you get 1/2 plus 1/2, which is 1. And this is generally true. The sum of the probabilities of all of the possible events should be equal to 1. all of these fractions, and the numerator will then add up to all of the possible events. The denominator is always all the possible events. So you have all the possible events over all the possible events when you add all of these things up. Now let's take it up a notch. Let's figure out the probability of-- I'm going to take this coin, and I'm going to flip it twice-- the probability of getting heads and then getting another heads. There's two ways to think about it. One way is to just think about all of the different possibilities. I could get a head on the first flip and a head on the second flip, head on the first flip, tail on the second flip. I could get tails on the first flip, heads on the second flip. Or I could get tails on both flips. So there's four distinct, equally likely possibilities. And one way to think about is on the first flip, I have two possibilities. On the second flip, I have another two possibilities. And so I have four possibilities. For each of these possibilities, for each of these two, I have two possibilities here. So either way, I have four equally likely possibilities. And how many of those meet our constraints? Well, we have it right over here, this one right over here-- having two heads meets our constraints. And there's only one of those possibilities. I've only circled one of the four scenarios. So there's a 1/4 chance of that happening. Another way you could think about this-- and this is because these are independent events. And this is a very important idea to understand in probability, and we'll also study scenarios that are not independent. But these are independent events. What happens in the first flip in no way affects what happens in the second flip. And this is actually one thing that many people don't realize. There's something called the gambler's fallacy, where someone thinks if I got a bunch of heads in a row, then all of a sudden, it becomes more likely on the next flip to get a tails. That is not the case." + }, + { + "Q": "At 2:32 you said 240x^2. Why and How is it squared?\n", + "A": "It s squared because you are multiplying 30x and 8x. When multiplying variables, you write the variable, (x), and add what powers that they are each to together, (x^1 times x^1).You will end up with x^2. Here is an example. x^3 times x^2. First, write the variable, (x), and then, add their powers together, (^3+^2)=^5. Your answer will be x^5. :)", + "video_name": "bamcYQDzVTw", + "timestamps": [ + 152 + ], + "3min_transcript": "So the volume is going to be the area of the base, which is 3x squared plus 30x plus 5, times the height, times 8x minus 5. Now, to multiply something like this out, it might seem really complicated, but we once again, we just have to do the distributive property. If you viewed this big thing in pink here as just a number, if this was like the number 7, you'd just say, well, this is going to be 7 times 8x minus 7 times 5 or minus 5 times 7. You would just distribute it out. You would just multiply this entire thing times each of the terms. That's what the distributive property tells us when you first learned it. So let's do that. So it's going to be this entire thing times 8x, or we can view it as 8x times this entire thing. So 8x times this entire thing: 3x squared plus 30x plus 5, times minus 5. So once again you get 3x squared plus 30x plus 5. And now we just multiply these out. We just distribute out the 8x over this whole thing and we distribute out the negative 5 over that whole thing. So then we get 8x times 3x squared is 24x to the third power. 8x times 30x is what? That's 240x squared, so plus 240x squared. 8x times 5 is plus 40x. And then we multiply this negative 5 out. Negative 5 times 3x squared is negative 15x squared. Negative 5 times 30x is negative 150x. And then we just have to simplify it from here. So we only have one third-degree term, one thing that has an x to the third in it. We have this term right here, so we'll write that as 24x to the third. And then what are our x squared terms? We have 240x squared minus 15x squared. So what's two 240 minus 15? It's 225x squared. So plus 225x squared. That's adding that term to that term right over there. And then we have 40x minus 150x. That would be negative 110x. And then finally we have just this negative 25 out here. That's the only constant term. And we're done! We found the volume of the tank." + }, + { + "Q": "I think he said that 3-dimensional object volumes were the products of the base and the height, somewhere around 1:00. That probably is a mistake, or I heard incorrectly, because pyramid volume has a different formula. So, are we just supposed to assume that the tank is not a pyramid? I've never heard of a pyramid-shaped tank, so that might be it.\n", + "A": "He said a three-dimensional object like this referring to a cylinder, and in this case, and with similar 3D objects (like prisms), base x height applies.", + "video_name": "bamcYQDzVTw", + "timestamps": [ + 60 + ], + "3min_transcript": "Find the volume of a tank whose base has an area of 3x squared plus 30x plus 5 square feet and whose height is 8x minus 5. So let me draw a potential tank here. Maybe it's a cylinder, so let me draw a cylindrical tank here, just like that. It's supposed to be a cylinder, and maybe if it was transparent, you would see this back side over there. And these two are the same. These are supposed to be straight lines. And the volume of a-- let me just label it first. So the area of the base, which is the same as the area of the top here or of the base here, the area of that is 3x squared plus 30x plus 5. And they tell us the height is 8x minus 5. The height of this tank is 8x minus 5. And if you want to find the volume of a three-dimensional object like this, you just multiply the area of the base So the volume is going to be the area of the base, which is 3x squared plus 30x plus 5, times the height, times 8x minus 5. Now, to multiply something like this out, it might seem really complicated, but we once again, we just have to do the distributive property. If you viewed this big thing in pink here as just a number, if this was like the number 7, you'd just say, well, this is going to be 7 times 8x minus 7 times 5 or minus 5 times 7. You would just distribute it out. You would just multiply this entire thing times each of the terms. That's what the distributive property tells us when you first learned it. So let's do that. So it's going to be this entire thing times 8x, or we can view it as 8x times this entire thing. So 8x times this entire thing: 3x squared plus 30x plus 5, times minus 5. So once again you get 3x squared plus 30x plus 5. And now we just multiply these out. We just distribute out the 8x over this whole thing and we distribute out the negative 5 over that whole thing. So then we get 8x times 3x squared is 24x to the third power. 8x times 30x is what? That's 240x squared, so plus 240x squared. 8x times 5 is plus 40x. And then we multiply this negative 5 out. Negative 5 times 3x squared is negative 15x squared. Negative 5 times 30x is negative 150x." + }, + { + "Q": "(3:33)\nThe C(A) is (1, 2, 3) and (1, 1, 4). We know this because X1 and X2 are the pivot variables?\nand columns 3 and 4 are not part of the column space because X3 and X4 are free variables?\n", + "A": "C(A) is the span of v1 = (1, 2, 3) and v2 = (1, 1, 4). Why? Because by setting x3 and x4 in turn to 0 we can show that v3 and v4 are each in the span of v1 and v2. Then clearly all 4 column vectors are in the span of the 1st two, and by closure the span of any vectors within a span is in the span.", + "video_name": "EGNlXtjYABw", + "timestamps": [ + 213 + ], + "3min_transcript": "So this is x1. Let me scroll up a little bit. This is associated with x1, right? When you multiply this times x1, you get this column times x1, this column times x2, this column times x3, this column times x4 like that. When you look at just regular A, when you look at just your matrix A, it's the same thing. If you were to write Ax equal to 0, this column would be associated with x1, this column would be associated with x2, x3 and x4 like that. What you can do is you put it in reduced row echelon form. You say which columns have my pivot entries or are associated with pivot variables? You say, OK, x1 and x2 are associated with pivot variables, or they are the pivot variables, and they're associated with these first two columns, and so those first two columns would be a basis for the column space. How did I get this? Am I just making up things on the fly? Well, no! a situation where the vectors associated with the free variables you can write them as linear combinations of the vectors associated with the pivot variables, and we did a special case of that last time. But a very quick and dirty way of doing it, and I don't know if it's actually dirty, is just take your matrix, put it in it reduced row echelon form, and you say, look, this column and this column are associated with my free variables. Therefore, this column and this column must be associated with my free variables. The solution sets are the same to Ax equal to 0 or the reduced row echelon form of Ax is equal to 0. So they're the same. So if this column and this column are associated with free variables, so are this column and this column, which means that they can be expressed just by judiciously selecting your values of your free variables as linear variables, with the pivot entries, which are that column and that column. So this column and this column would be a basis for A, and we see that. We see that all the way down here. 1, 2, 3 and 1, 1, 4, We did a lot of work and we got all the way there, and we said this is a basis for the column span of A. Now, doing all of that work, let's see if we can actually visualize what the column space of A looks like. I have a strange feeling I might have said column span a couple of times, but the column space, what does it look like? Well, there's a couple of ways to think about what it looks like. One way is we can say, look, this span this is 2-- that's a member of R3. That's a vector in R3 and this is a vector in R3. Let me draw my x, z and-- well, normally it's drawn--" + }, + { + "Q": "\nAt 2:25 Why isn't y = 2sin(t)?\nThen again at 7:12 Why isn't x = 2cos(t)?\nWe were given these values in the beginning of the question (previous video), and yet Sal made these up? How?", + "A": "It could be, if you wanted it to be. The point is that on the c2 path y needs to go from 2 down to 0. On the c3 path x needs to go from 0 up to 2. How it gets there isn t that important, and doesn t have to be the same or even related to the c1 path parametrization, so Sal just picked the simplest possible way to do what he wanted.", + "video_name": "Qqanbd3gLhw", + "timestamps": [ + 145, + 432 + ], + "3min_transcript": "In the last video, we set out to figure out the surface area of the walls of this weird-looking building, where the ceiling of the walls was defined by the function f of xy is equal to x plus y squared, and then the base of this building, or the contour of its walls, was defined by the path where we have a circle of radius 2 along here, then we go down along the y-axis, and then we take another left, and we go along the x-axis, and that was our building. And in the last video, we figured out this first wall's surface area. In fact, you can think of it, our original problem is, we wanted to figure out the line integral along the closed path, so it was a closed line integral, along the closed path c of f of xy, and we're always multiplying f of xy times a little bit, a little, small distance of our path, ds. We're writing this in the most abstract way possible. And what we saw in the last video is, the easiest way to do this is to break this up into multiple paths, or into So you can imagine, this whole contour, this whole path we call c, but we could call this part, we figured out in the last video, c1. This part we can call, let me make a point, c2, and this point right here is c3. So we could redefine, or we can break up, this line integral, this closed-line integral, into 3 non-closed line integrals. This will be equal to the line integral along the path c1 of f of xy ds, plus the line integral along c2 of f of x y ds plus the line integral, you might have guessed it, along c3 of f of xy ds, and in the last video, we got as far as figuring out this first part, this first curvy wall all right here. Its surface area, we figured out, was 4 plus 2 pi. Now we've got to figure out the other 2 parts. And in order to do it, we need to do another parameterization of x and y. It's going to be different than what we did for this part. We're no longer along this circle, we're just along the y-axis. So as long as we're there, x is definitely going to be equal to 0. So that's my parameterization, x is equal to 0. If we're along the y-axis, x is definitely equal to 0. And then y, we could say it starts off at y is equal to 2. Maybe we'll say y is equal to 2 minus t, for t is between 0, t is greater than or equal to 0, less than or equal to 2. And that should work. When t is equal to 0, we're at this point right there, and then as t increases towards 2, we move down the y-axis, until eventually when t is equal to 2, we're at that point right there. So that's our parameterization. And so let's evaluate this line, and we could do our derivatives, too, if we like. What's the derivative, I'll write it over here. What's dx dt?" + }, + { + "Q": "\nAt 12:18, when Sal is simplifying x^4 = 81, he straightaway puts x^2 = 9. It could have been -9 as well, but wouldn't have worked in the end, since it would yield complex values for x.\nWhat do you think - was that unintentional (a mistake), or he knew it would not be helpful in the end?", + "A": "Could have been either of those, you can t really know for sure. You re right that it wouldn t have been useful though, as at 6:30 Sal even mentions that we re assuming that we re dealing with real numbers. Had he put x^2 = +- 9, the only thing that would have changed would be the length of the video, as he d have probably explained what you ve just covered in your post. :)", + "video_name": "zC_dTaEY2AY", + "timestamps": [ + 738 + ], + "3min_transcript": "if it's negative before x, has to be positive after x,or if it's positive before x, has to be negative after x. So let's test that out. So the first thing we need to do is find these candidate points. Remember, the candidate points are where the second derivative is equal to 0. We're going to find those points, and then see if this is true, that the sign actually changes. We want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where this our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12x squared minus 4x to the sixth is equal to 0. Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So 4x squared. Now we'll have 27 times, if we factor 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the fourth is equal to 0. So the x's that will make this equal to 0 will satisfy either, I'll switch colors, either 4x squared is equal to 0, or, now 27 times 3, I can do that in my head. That's 81. 20 times 3 is 60, 7 times 3 is 21, 60 plus 21is 81. Or 81 minus x to the fourth is equal to 0. Any x that satisfies either of these will make this entire expression equal 0. Because if this thing is 0, the whole thing is If this thing is 0, the whole thing is going to be equal to 0. Let me be clear, this is 81 right there. So let's solve this. This is going to be 0 when x is equal to 0, itself. If we add x to the fourth to both sides, you get x to the fourth is equal to 81. If we take the square root of both sides of this, you get x squared is equal to 9, or so you get x is plus or minus 3. x is equal to plus or minus three. So these are our candidate inflection points, x is equal to 0, x is equal to plus 3, or x is equal to minus 3. So what we have to do now, is to see whether the second derivative changes signs around these points in order to be able to label them inflection points. So what happens when x is slightly below 0? So let's take the situation, let's do all the scenarios. What happens when x is slightly below 0? Not all of them, necessarily, but if x is like 0.1. What is the second derivative going to be doing?" + }, + { + "Q": "\nat 12:00 did he factor out 4x^2 correctly? it doesn't look right for some reason.", + "A": "It s correct. 4x\u00c2\u00b2*81 - 4x\u00c2\u00b2*x^4 = 27 * 12 * x\u00c2\u00b2 - 4x^6", + "video_name": "zC_dTaEY2AY", + "timestamps": [ + 720 + ], + "3min_transcript": "if it's negative before x, has to be positive after x,or if it's positive before x, has to be negative after x. So let's test that out. So the first thing we need to do is find these candidate points. Remember, the candidate points are where the second derivative is equal to 0. We're going to find those points, and then see if this is true, that the sign actually changes. We want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where this our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12x squared minus 4x to the sixth is equal to 0. Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So 4x squared. Now we'll have 27 times, if we factor 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the fourth is equal to 0. So the x's that will make this equal to 0 will satisfy either, I'll switch colors, either 4x squared is equal to 0, or, now 27 times 3, I can do that in my head. That's 81. 20 times 3 is 60, 7 times 3 is 21, 60 plus 21is 81. Or 81 minus x to the fourth is equal to 0. Any x that satisfies either of these will make this entire expression equal 0. Because if this thing is 0, the whole thing is If this thing is 0, the whole thing is going to be equal to 0. Let me be clear, this is 81 right there. So let's solve this. This is going to be 0 when x is equal to 0, itself. If we add x to the fourth to both sides, you get x to the fourth is equal to 81. If we take the square root of both sides of this, you get x squared is equal to 9, or so you get x is plus or minus 3. x is equal to plus or minus three. So these are our candidate inflection points, x is equal to 0, x is equal to plus 3, or x is equal to minus 3. So what we have to do now, is to see whether the second derivative changes signs around these points in order to be able to label them inflection points. So what happens when x is slightly below 0? So let's take the situation, let's do all the scenarios. What happens when x is slightly below 0? Not all of them, necessarily, but if x is like 0.1. What is the second derivative going to be doing?" + }, + { + "Q": "\nAt 2:36, Sal said that the function of the graph is y = | x - 3 |. Can someone please explain this?", + "A": "Two things can make the graph of functions shift, a change in the x or a change in the y. For absolute value functions, the shifts are defined by y = | x - h | + k, so a y shift is outside the absolute value and follows the sign. Inside for the x shift, however, the graph shifts in the opposite direction, so a -3 shifts it in the positive x direction. The other way to think about is that if x = 3, then the value inside the absolute value is 0.", + "video_name": "Wri26sPEBoI", + "timestamps": [ + 156 + ], + "3min_transcript": "and think about how that might change the equation. So let's just visualize what we're even talking about. So if we're gonna shift three to the right, it would look like, it would look like this. So that's what we first wanna figure out the equation for and so how would we think about it? Well, one way to think about it is, well, something interesting is happening right over here at x equals three. Before, that interesting thing was happening at x equals zero. Now, it's happening at x equals three. And the interesting thing that happens here is that you switch signs inside the absolute value. Instead of taking an absolute value of a negative, you're now taking the absolute value as you cross this point of a positive and that's why we see a switch in direction here of this line and so you see the same thing happening right over here. So at this point right over here, we know that our equation needs to evaluate out to zero and this is where it's going to switch signs and so we say, okay, well, this looks like an absolute value so it's going to have the form, y is equal to the absolute value of something and so you say, okay, if x is three, how do I make that equal to zero? Well, I can subtract three from it. If I say y is equal to the absolute value of x minus three, well, let's try it out. Let's see if it makes sense. So when x is equal to three, three minus three is zero, absolute value of that is zero. That works out. When x is equal to four, four minus three is one, absolute value of one is indeed, is indeed one. And if x is equal to two, well, two minus three is negative one So once again, I'm showing you this by really trying out numbers, trying to give you a little bit of an intuition because that wasn't obvious to me when I first learned this that if I'm shifting to the right which it looks like I'm increasing an x value but what I would really do is replace my x with an x minus the amount that I'm shifting to the right but I encourage you to try numbers and think about what's happening here. At this vertex right over here, whatever was in the absolute value sign was equaling zero. It's when whatever was in the absolute value sign is switching from negative signs to positive signs. So once again, if you shift three to the right, that has to happen at x equals three. So whatever is inside the absolute value sign has to be equal to zero at x equals three and this, pause this video and really think about this if it isn't making sense and even as you get more and more familiar with this," + }, + { + "Q": "At around 2:30 in the video Sal invokes the chain rule: d/dx(ln(e^x))=d/dx(e^x)*d/d e^x(ln(e^x))=d/dx(e^x)*(1/e^x)=1. The derivative of the composite function ln(e^x) is the derivative of the inner function w.r.t. x multiplied by the derivative of the outer function w.r.t. e^x. I don't recognize this chain rule. I always thought the chain rule was: d/dx(e^x)*ln(e^x)+d/d e^x(ln(e^x))*e^x. Furthermore, is there a more direct proof; without starting from the composite function d/dx(ln(e^x))?\n", + "A": "Well, you could define e^x such that d/dx e^x = d/dx e^x, and then show that the e in question actually is the e. However, in rigorous calculus they first define a function ln x to be \u00e2\u0088\u00ab{0,x}(dt/t), such that its derivative clearly is 1/x (and d/dx ln(f(x)) = f\u00c2\u00b4(x)/f(x)) and then define e^x to be its inverse function.", + "video_name": "sSE6_fK3mu0", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's prove with the derivative of e to the x's, and I think that this is one of the most amazing things, depending on how you view it about either calculus or math or the universe. Well we're essentially going to prove-- I've already told you before that the derivative of e to the x is equal to e to the x, which is amazing. The slope at any point of that line is equal to the x value-- is equal to the function at that point, not the x value. The slope at any point is equal to e. That is mind boggling. And that also means that the second derivative at any point is equal to the function of that value or the third derivative, or the infinite derivative, and that never ceases to amaze me. But anyway back to work. So how are we going to prove this? Well we already proved-- I actually just did it right before starting this video-- that the derivative-- and some people actually call this the definition of e. They go the other way around. They say there is some number for which this is true, So it could almost be viewed as a little bit circular, but be we said that e is equal to the limit as n approaches infinity of 1 over 1 plus n to the end. And then using this we actually proved that derivative of ln of x is equal to 1/x. The derivative of log base e of x is equal to 1/x. So now that we prove this out, let's use this to prove this. Let me keep switching colors to keep it interesting. Let's take the derivative of ln of e to the x. This is almost trivial. This is equal to the logarithm of a to the b is equal to b times the logarithm of a, so this is equal to the derivative of x ln of e. Well, to the first power, right? So this just equals the derivative of x, which we have shown as equal to 1. I think we have shown it, hopefully we've shown it. If we haven't, that's actually a very easy one to prove. OK fair enough. We did that. But let's do this another way. Let's use the chain rule. So what doe the chain rule say? If we have f of g of x, where we have one function embedded in another one, the chain rule say we take the derivative of the inside function, so d/dx of e to the x. And then we take the derivative of the outside function or the derivative of the outside function with respect to the inner function. You can almost view it that way. So the derivative ln of e to the x with" + }, + { + "Q": "At 0:56 I don't know why he added the four there, but I see how it works with the equation. Where does the four come from?\nThanks for your help!\n", + "A": "Hi, I would be happy to help you; is the 0:56 time reference that you made correct? (That part of the video is long division)", + "video_name": "Y2-tz27nKoQ", + "timestamps": [ + 56 + ], + "3min_transcript": "PROBLEM: \"Express the rational number 19/27 (or 19 27ths) as a terminating decimal or a decimal that eventually repeats. Include only the first six digits of the decimal in your answer.\" Let me give this a shot. So we want to express 19/27 \u2013 which is the same thing as 19 \u00f7 27 \u2013 as a decimal. So let's divide 27 into 19. So 27 going into 19. And we know it's going to involve some decimals over here, because 27 is larger than 19, and it doesn't divide perfectly. So let's get into this. So 27 doesn't go into 1. It doesn't go into 19. It does go into 190. And it looks like 27 is roughly 30. It's a little less than 30. 30 times 6 would be 180. So let's go with it going 6 times. Let's see if that works out. Well, 6 \u00d7 7 is 42. 6 \u00d7 2 is 12, + 4 is 16. Actually, we could've had another 27 in there. Because when we subtract \u2013 So we get a 10 from the 10's place. So that becomes 8 10's. This became 28. So we could have put one more 27 in there. So let's do that. So let's put one more 27 in there. So 7 27's. 7 \u00d7 7 is 49. 7 \u00d7 2 is 14, + 4 is 18. And now our remainder is 1. We can bring down another 0. 27 goes into 10 0 times. 0 \u00d7 27 is 0. [Not \"10,\" as Sal states by mistake.] Subtract \u2013 we have a remainder of 10. But now, we have to bring down another 0. So let's bring down this 0 right over here. So now, 27 goes into 100 3 times. 3 \u00d7 27 is 60 + 21, is 81. Well, we could take 100 from the 100's place, and make it 10 10's. And then we could take 1 of those 10's from the 10's place and turn it into 10 1's. And so 9 10's minus 8 10's is equal to 1 10. And then 10 -1 is 9. So it's equal to 19. You probably \u2013 You might have been able to do that in your head. And then we have \u2013 And I see something interesting here \u2013 because when we bring down our next 0, we see 190 again. We saw 190 up here. But let's just keep going. So 27 goes into 190 \u2013 And we already played this game. It goes into it 7 times. 7 \u00d7 27 \u2013 we already figured out \u2013 was 189. We subtracted. We had a remainder of 1. Then we brought down another 0. We said 27 goes into 10 0 times. 0 \u00d7 27 is 0. Subtract. Then you have \u2013" + }, + { + "Q": "What does Sal mean at 2:01 when he says \"Any of those - well that's just going to give you 3.\"? I also don't understand how he got 15 either. I've been stuck on this lesson re-watching it trying to figure out what he is saying, but it doesn't make sense!\n", + "A": "The square root of 3*3 is 3. 3*3 = 9, and the square root of nine is three.", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 121 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "in 2:06 to 2:14 -- why wasn't the square root of 3*3 simplified to the square root of 9 to get 81\n", + "A": "You are taking the square root of 9 not squaring 9. When you do 9 square or 9^2 you get 81. When your taking square root , you finding which number you have to multiply twice to get to 9. By which your answer will be 3, since 3*3 = 9", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 126, + 134 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "At 1:55 why did he write 13 separately?\n", + "A": "The \u00e2\u0088\u009a(117) is split into 2 square roots: \u00e2\u0088\u009a(9) * \u00e2\u0088\u009a(13). The reason we do this is we can simplify \u00e2\u0088\u009a(9). It = 3. So, the radical will go away on that piece. We can not simplify \u00e2\u0088\u009a(13), so it stays in that form. Hope this helps.", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 115 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "\nKahn takes the derivative at 2:17, but what is the purpose of that? Does that signify the rate of change of the area or is that the area from (x,x+\u00e2\u0088\u0086x)?\n\nPlease answer both questions.", + "A": "It would indeed signify the rate of change of the area, but that is not why he is doing it. Sal is demonstrating that the integral (in terms of area under a curve) happens to coincide with a sort-of inverse operation of taking a derivative. This concept is at the core of the Fundamental Theorem of Calculus.", + "video_name": "pWtt0AvU0KA", + "timestamps": [ + 137 + ], + "3min_transcript": "Let's say that we've got some function f that is continuous on the interval a to b. So let's try to see if we can visualize that. So this is my y-axis. This right over here, I want to make it my t-axis. We'll use x little bit later. So I'll call this my t-axis. And then let's say that this right over here is the graph of y is equal to f of t. And we're saying it's continuous on the interval from a to b. So this is t is equal to a. This is t is equal to b. So we're saying that it is continuous over this whole interval. Now, for fun, let's define a function capital F of x. And I will do it in blue. Let's define capital F of x as equal to the definite integral from a is a lower bound to x of f of t-- x is in this interval, where a is less than or equal to x is less than or equal to b. Or that's just another way of saying that x is in this interval right over here. Now, when you see this, you might say, oh, the definite integral, this has to do with differentiation an antiderivatives and all that. But we don't know that yet. All we know right now is that this is the area under the curve f between a and x, so between a, and let's say this right over here is x. So f of x is just this area right over here. That's all we know about it. We don't know it has anything to do with antiderivatives just yet. That's what we're going to try to prove in this video. And we're going to do it just using the definition of derivatives and see what we get when we take the derivative using the definition of derivatives. So the derivative, f prime of x-- well this definition of derivatives, it's the limit as delta x approaches 0 of capital F of x plus delta x minus f of x, all of that over delta x. This is just the definition of the derivative. Now, what is this equal to? Well, let me rewrite it using these integrals right up here. This is going to be equal to the limit as delta x approaches 0 of-- what's f of x plus delta x? Well, put x in right over here. You're going to get the definite integral from a to x plus delta" + }, + { + "Q": "I don't understand the factoring. At 1:09 when Sal starts simplifying and factoring everything out, how did he get that (x2+3x+2) was the same thing as (x+2) (x+1) and that x2-4 was the same as (x+2)(x-2)?\n", + "A": "He noticed that the factors of 2 (1 and 2) add up to 3, and that (x^2-4) is a difference of squares, which has a simple factoring rule. If you re having trouble with factoring, you may want to go back and review your algebra before going further. If your algebra is rusty, calculus will be very difficult.", + "video_name": "oUgDaEwMbiU", + "timestamps": [ + 69 + ], + "3min_transcript": "The function, f of x is equal to 6x squared plus 18x plus 12 over x squared minus 4, is not defined at x is equal to positive or negative 2. And we see why that is, if x is equal to positive or negative 2 then x squared is going to be equal to positive 4, and 4 minus 4 is 0, and then we're going to have a 0 in the denominator. And that's not defined. We don't know what that happens when you divide-- well we've never defined what happens when you divide by 0. So they say, what value should be assigned to f of negative 2 to make f of x continuous at that point? So to think about that, let's try to actually simplify f of x. So f of x-- I'll just rewrite it-- is equal to-- Actually let me just start simplifying right from the get go. So in the numerator I can factor out a 6 out of every one of those terms. So it's 6 times x squared plus 3x plus 2 over-- and the denominator, this is a difference of squares. This is x plus 2 times x minus 2. So this is going to be equal to 6 times-- let me do it a different color. So we think of two numbers and if I take their product If I take their sum I get 3. The most obvious one is 2 and 1. So this is 6 times x plus 2 times x plus 1. When you take the product there you'll get x squared plus 3x plus 2, and then all of that over x plus 2 times x minus 2. Now, if we know that x does not equal negative 2. Then we can divide both the numerator and the denominator by x plus 2. The reason why I'm making that constraint is that if x were to be equal to negative 2 then x plus 2 is going to be equal to 0. And you won't be able to do that. We don't know what it means divide something by 0. So we could say that this is going to be equal to-- so we can divide the numerator equal to negative 2. So this is equal to 6 times-- we're going divided by x plus 2 in the numerator, x plus 2 in the denominator-- so it's going to be 6 times x plus 1 over x minus 2. And we have to put the constraint here because now we've changed it. Now this expression over here is actually defined at x equals negative 2. But in order to be equivalent to the original function we have to constrain it. So we will say for x not equal to negative 2. And it's also obvious that x can't be equal to 2 here. This one also isn't defined at positive 2 because you're dividing by 0. So you could say, for x does not equal to positive or negative 2 if you want to make it very explicit. But they ask us, what could we assign f of negative 2 to make the function continuous at the point?" + }, + { + "Q": "\nAt 1:58. isn't the question mark supposed to be on top of the equals sign, to indicate that we are not sure that the equation is equal to -11, instead of next to -11?", + "A": "You could do either the way he does it in this is more informal", + "video_name": "SkMNREAMNvc", + "timestamps": [ + 118 + ], + "3min_transcript": "Is negative 1 comma 7 a solution for the system of linear equations below? And they give us the first equation is x plus 2y is equal to 13. Second equation is 3x minus y is equal to negative 11. In order for negative 1 comma 7 to be a solution for the system, it needs to satisfy both equations. Or another way of thinking about it, x equals 7, and y-- sorry, x is equal to negative 1. This is the x coordinate. X equals negative 1, and y is equal to 7, need to satisfy both of these equations in order for it to be a Solution. So let's try it out. Let's try it out with the first equation. So we have x plus 2y is equal to 13. So if we're thinking about that, we're testing to see if when x is equal to negative 1, and y is equal to 7, will x plus 2y equals 13? So we have negative 1 plus 2 times 7-- y should be 7-- this needs to be equal to 13. because we don't know whether it does. So this is the same thing as negative 1 plus 2 times 7 plus 14. That does, indeed, equal 13. Negative 1 plus 14, this is 13. So 13 does definitely equal 13. So this point it does, at least, satisfy this first equation. This point does sit on the graph of this first equation, or on the line of this first equation. Now let's look at the second equation. I'll do that one in blue. We have 3 times negative 1 minus y, so minus 7, needs to be equal to negative 11. I'll put a question mark here because we don't know whether it's true or not. So let's see, we have 3 times negative 1 is negative 3. And then we have minus 7 needs to be equal to negative 11-- I put the question mark there. Negative 3 minus 7, that's negative 10. No, negative 10 does not equal a negative 11. So x equaling negative 1, and y equaling 7 does not satisfy the second equation. So it does not sit on its graph. So this over here is not a solution for the system. So the answer is no. It satisfies the first equation, but it doesn't satisfy the second. In order to be a solution for the system, it has to satisfy both equations." + }, + { + "Q": "\nAt 0:33, how did he know that i squared equals -1?", + "A": "Hi Anushka, Recall that i is defined as sqrt(-1). if we square this we get liberate radicand yielding -1. Here is a challenge for you. Calculate i squared, i cubed, and i squared. Look for the pattern. It will help you in the future. Regards, APD", + "video_name": "SP-YJe7Vldo", + "timestamps": [ + 33 + ], + "3min_transcript": "Voiceover:Most of your mathematical lives you've been studying real numbers. Real numbers include things like zero, and one, and zero point three repeating, and pi, and e, and I could keep listing real numbers. These are the numbers that you're kind of familiar with. Then we explored something interesting. We explored the notion of what if there was a number that if I squared it I would get negative one. We defined that thing that if we squared it we got negative one, we defined that thing as i. So we defined a whole new class of numbers which you could really view as multiples of the imaginary unit. So imaginary numbers would be i and negative i, and pi times i, and e times i. This might raise another interesting question. What if I combined imaginary and real numbers? What if I had numbers that were essentially sums or differences of real or imaginary numbers? Let's say I call it z, and z tends to be the most used variable when we're talking about what I'm about to talk about, complex numbers. Let's say that z is equal to, is equal to the real number five plus the imaginary number three times i. So this thing right over here we have a real number plus an imaginary number. You might be tempted to add these two things, but you can't. They won't make any sense. These are kind of going in different, we'll think about it visually in a second, but you can't simplify this anymore. You can't add this real number to this imaginary number. A number like this, let me make it clear, that's real and this is imaginary, imaginary. A number like this we call a complex number, a complex number. It has a real part and an imaginary part. or someone will say what's the real part? What's the real part of our complex number, z? Well, that would be the five right over there. Then they might say, \"Well, what's the imaginary part? \"What's the imaginary part of our complex number, z? And then typically the way that this function is defined they really want to know what multiple of i is this imaginary part right over here. In this case it is going to be, it is going to be three. We can visualize this. We can visualize this in two dimensions. Instead of having the traditional two-dimensional Cartesian plane with real numbers on the horizontal and the vertical axis, what we do to plot complex numbers is we on the vertical axis we plot the imaginary part, so that's the imaginary part. On the horizontal axis we plot the real part. We plot the real part just like that." + }, + { + "Q": "\nat 0:51 y would u cross out the division and turn it into 2/3\nPLEASE ANSWER THIS QUESTION I WANT TO STUDY THIS TOPIC", + "A": "what i dont understand what that means can u be more specific", + "video_name": "Io9i1JkKgN4", + "timestamps": [ + 51 + ], + "3min_transcript": "Determine whether 30/45 and 54/81 are equivalent fractions. Well, the easiest way I can think of doing this is to put both of these fractions into lowest possible terms, and then if they're the same fraction, then they're equivalent. So 30/45, what's the largest factor of both 30 and 45? 15 will go into 30. It'll also go into 45. So this is the same thing. 30 is 2 times 15 and 45 is 3 times 15. So we can divide both the numerator and the denominator by 15. So if we divide both the numerator and the denominator by 15, what happens? Well, this 15 divided by 15, they cancel out, this 15 divided by 15 cancel out, and we'll just be left with 2/3. So 30/45 is the same thing as 2/3. It's equivalent to 2/3. 2/3 is in lowest possible terms, or simplified form, Now, let's try to do 54/81. Now, let's see. Nothing really jumps out at me. Let's see, 9 is divisible into both of these. We could write 54 as being 6 times 9, and 81 is the same thing as 9 times 9. You can divide the numerator and the denominator by 9. So we could divide both of them by 9. 9 divided by 9 is 1, 9 divided by 9 is 1, so we get this as being equal to 6/9. Now, let's see. 6 is the same thing as 2 times 3. 9 is the same thing as 3 times 3. We could just cancel these 3's out, or you could imagine this is the same thing as dividing both the numerator and the denominator by 3, or multiplying both the numerator and the denominator by 1/3. These are all equivalent. I could write divide by 3 or multiply by 1/3. Let me write divide by 3 for now. I don't want to assume you know how to multiply fractions, because we're going to learn that in the future. So we're going to divide by 3. 3 divided by 3 is just 1. 3 divided by 3 is 1, and you're left with 2/3. So both of these fractions, when you simplify them, when you put them in simplified form, both end up being 2/3, so they are equivalent fractions." + }, + { + "Q": "could somebody explain to me which formula is Sal using at 6:00? is it the Heron's formula?\n", + "A": "it is the distance formula", + "video_name": "GiGLhXFBtRg", + "timestamps": [ + 360 + ], + "3min_transcript": "The centroid of a triangle is just going to be the average of the coordinates of the vertices. Or the coordinate of the centroid here is just going to be the average of the coordinates of the vertices. So this coordinate right over here is going to be-- so for the x-coordinate, we have 0 plus 0 plus a. So we have three coordinates. They add up to a, and we have to divide by 3. So it's a over 3. The y-coordinate is going to b plus 0 plus 0. They add up to b, but we have three of them, so the average is b over 3. And then same thing-- we do it for the z-coordinate. The average is going to be c, is c over 3. And I'm not proving it to you right here. You could verify it for yourself. But it's going to be the average, that if you were to figure out what this line is, this line is, and this line is, this centroid, or this center is just the average of these coordinates. Now, what we want to do is use this information. Let's just use this coordinate right here and then compare just using the distance formula. Let's compare this distance up here in orange to this distance down here in yellow. And remember, this point right over here-- this is the median of this bottom side right over here. It's just going to be the average of these two points. And so the x-coordinate-- 0 plus a over 2 is going to be a over 2. b plus 0 over 2 is going to be b over 2. And then it has no z-coordinates, so it's just going to be 0. 0 plus 0 over 2 is 0. So we know the coordinates for this point that point and that point. So we can calculate the yellow distance and we can calculate the orange distance. So let's calculate the orange distance. So that is going to be equal to the square root of-- of these points squared. So it's a over 3 minus 0 squared. So that's going to be a squared over 9, plus b over 3 minus 0 squared. So that's b squared over 9. Plus c over 3 minus c, which is negative 2/3. And we want to square that. So we're going to have positive 4 over 9c squared. Did I do that right? c over 3, so 1/3 minus 1 is negative 2/3. So this is negative 2/3 c. You're going to get 4/9 c squared. So that's the orange distance. Now, let's calculate-- and if we want to do it, we can express this-- let me express it a little bit simpler than this. This is the same thing as the square root of a squared plus b" + }, + { + "Q": "At 1:41, why are angles C and B 90 degrees? I know it was explained in the video, but I didn't really understand it.\n", + "A": "AB and AC are tangents, meaning that they intersect the circle at one and only one point. One property of tangents are that if you draw a radius to the tangent point, the tangent will be perpendicular to that radius.", + "video_name": "ZiqHJwzv_HI", + "timestamps": [ + 101 + ], + "3min_transcript": "Angle A is a circumscribed angle on circle O. So this is angle A right over here. Then when they say it's a circumscribed angle, that means that the two sides of the angle are tangent to the circle. So AC is tangent to the circle at point C. AB is tangent to the circle at point B. What is the measure of angle A? Now, I encourage you to pause the video now and to try this out on your own. And I'll give you a hint. It will leverage the fact that this is a circumscribed angle as you could imagine. So I'm assuming you've given a go at it. So the other piece of information they give us is that angle D, which is an inscribed angle, is 48 degrees and it intercepts the same arc-- so this is the arc that it intercepts, arc CB I guess you could call it-- it intercepts this arc right over here. It's the inscribed angle. The central angle that intersects that same arc So this is going to be 96 degrees. I could put three markers here just because we've already used the double marker. Notice, they both intercept arc CB so some people would say the measure of arc CB is 96 degrees, the central angle is 96 degrees, the inscribed angle is going to be half of that, 48 degrees. So how does this help us? Well, a key clue is that angle is a circumscribed angle. So that means AC and AB are each tangent to the circle. Well, a line that is tangent to the circle is going to be perpendicular to the radius of the circle that intersects the circle at the same point. So this right over here is going to be a 90-degree angle, and this right over here is going to be a 90-degree angle. OC is perpendicular to CA. OB, which is a radius, is perpendicular to BA, which is a tangent line, and they both intersect right We have a quadrilateral going on here. ABOC is a quadrilateral, so its sides are going to add up to 360 degrees. So we could know, we could write it this way. We could write the measure of angle A plus 90 degrees plus another 90 degrees plus 96 degrees is going to be equal to 360 degrees. Or another way of thinking about it, if we subtract 180 from both sides, if we subtract that from both sides, we get the measure of angle A plus 96 degrees is going to be equal to 180 degrees." + }, + { + "Q": "\nJust a quick question, at 9:38 you cannot cancel the top vector v and the bottom vector v right? Is this because they are dot products and not multiplication signs?", + "A": "You get a different answer (a vector divided by a vector, not a scalar), and the answer you get isn t defined.", + "video_name": "27vT-NWuw0M", + "timestamps": [ + 578 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 2:08, he says \"sin^2(theta)\" why didn't he say \"sin(theta)^2\"?", + "A": "sin ^2 theta is the more formal way to write out the equation, as it avoids the possibility of mistaking the theta as theta^2 .", + "video_name": "HnDvUaVjQ1I", + "timestamps": [ + 128 + ], + "3min_transcript": "So we've got a right triangle drawn over here where this base's length is a, the height here is b, and the length of the hypotenuse is c. And we already know when we see something like this, we know from the Pythagorean theorem, the relationship between a, b, and c, we know there's a squared plus b squared is going to be equal to the hypotenuse squared, is going to be equal to c squared. What I want to do in this video is explore how we can relate trig functions to, essentially, the Pythagorean theorem. And to do that, let's pick one of these non-right angles. So let's pick this angle right over here as theta, and let's just think about this what the sine of theta is and what the cosine of theta is, and see if we can mess with them a little bit to somehow leverage the Pythagorean theorem. So before we do that, let's just write down sohcahtoa just so we remember the definitions of these trig functions. So sine is opposite over hypotenuse. Cah, cosine is adjacent over hypotenuse. And toa, tan is opposite over adjacent, So let's think about sine of theta. I will do it, I'll do it in this blue color. So sine of theta is what? It is opposite over hypotenuse, so it is equal to the length of b or it is equal to b-- b is the length-- b over the length of the hypotenuse, which is c. Now what is cosine of theta? Well, the adjacent side, the side of this angle that is not the hypotenuse, it has length a. So it's the length of the adjacent side over the length of the hypotenuse. Now how could I relate these things? Well it seems like, if I square sine of theta, then I'm going to have sine squared theta is equal to b squared over c squared, and cosine squared theta is going to be a squared over c squared. Seems like I might be able to add them to get something that's pretty close to the Pythagorean theorem here. So let's try that out. So sine squared theta is equal to b squared over c squared. Cosine squared theta is equal to a squared over c squared. So what's this sum? What's sine squared theta plus cosine squared theta? Is going to be equal to what? Sine squared theta is b squared over c squared, plus a squared over c squared, which is going to be equal to-- Well we have a common denominator of c squared. And the numerator, we have b squared plus a squared. Now, what is b squared plus a squared?" + }, + { + "Q": "At 0:35, Sal said a greatest common factor and a greatest common divisor are \"kinda\" the same thing. Does he mean exactly the same?\n", + "A": "Yes GCF & GCD are the same thing", + "video_name": "jFd-6EPfnec", + "timestamps": [ + 35 + ], + "3min_transcript": "" + }, + { + "Q": "\n4:59.", + "A": "i get it", + "video_name": "jFd-6EPfnec", + "timestamps": [ + 299 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 2:37 arent the base and height souposed to be switched? or they can stay the same?", + "A": "well, at 2:37, the base is supposed to be one of the sides, and the height should be inside the triangle.", + "video_name": "ukPjc3Oyad4", + "timestamps": [ + 157 + ], + "3min_transcript": "And now the height, I guess you could say this, if you were to drop a penny from here, it's sitting outside the triangle. So it looks different from this one. But this would still be the height. This would still be the height. Right over here, ya do the same thing. One half times base times height would give you the area of this triangle. So how can we apply that over here? Well, this triangle is on this grid, but it's kind of at an angle. With this grid it's hard to pick out the base and the height. for this triangle as a whole. But what we could do, there's actually several ways that we can approach this. Is we can break this triangle up into two or more triangles where we can figure out the base and the height for each of them. So for example, I can break this one. Let's see I could, I'm picking this point and this point because it breaks it up into two triangles where I can figure out the base and the height. Well what am I talking about? Well this triangle over here, that I am shading in blue. I've rotated 90 degrees. But if you view this yellow. If you view this yellow as the base of this triangle, you see that the base is three. So, let me write the base is equal to three units. And what's the height here? Well the height here is going to be The height here is going to be this distance right over here, which is four. Height is equal to four. So the area, the area of that triangle right over there, is going to be one half times three times four, which is equal to six. So this part right over here, the area is six. And now we can do a similar thing with this other triangle. 'Cause once again, we can view this yellow line, or now I have this yellow and blue line, as the base. The base is equal to three. So I could write that base is equal to three. And once again, I've rotated. So now the base is on the side. And then the height here, the height of this triangle is two. If this is the base, remember if this is the base here we've just rotated it. Then this right over here is the height Height would be equal to two. So what's the area of this one? The area of this one is going to be one half times the base, three, times the height which is two. One half times two is one times three. This is going to be equal to three. So the area of the whole thing is gonna be this area of three plus this area of six. It's gonna be the area of nine. Area is equal to nine. Now that's one way you could do it. Is you could just break it up into triangles where you could figure out the base and the height. Another way, and this is you can kinda view it as a maybe a trickier way. Or ya kinda have to think a little bit outside of the box, or maybe outside of the triangle to do it this way. Is to, instead of doing it this way, visualize this triangle. And actually, let me, let me clean this up a little bit. Let me undo all this work that I just did. Let me undo this to show you the other approach." + }, + { + "Q": "\nWhat is happening at 4:27 when Sal states that Cos(theta) = -1 can be viewed as (2n + 1)pi? Why add the one to that function, when the previous function \"2n(pi)\" does the same thing?", + "A": "(2n+1) where n is an integer, is the standard way of specifying any odd integer. Whereas 2n is even. So, cos \u00ce\u00b8 = \u00e2\u0088\u00921 only when \u00ce\u00b8 is an odd integer multiplied by \u00cf\u0080.", + "video_name": "SdHwokUU8xI", + "timestamps": [ + 267 + ], + "3min_transcript": "on and on, and it makes sense. Theta equaled, or sorry, cosine of theta, the x-coordinate on this unit circle equaled one right when we were at zero angle, and we had to go all the way around the circle to get back to that point, two pi radians. But then it'll be again when we get to four pi radians, and then six pi radians, so two pi, four pi, six pi, and I guess you could see the pattern here. We're gonna keep hitting cosine of theta equals one every two pi, so you could really kind of view this as every multiple of two pi. Two pi n, where n is an integer, n is integer... is an integer. And that applies also for negative values. If you're going the other way around, if we're going the other way around, we don't get back until we get to negative two pi. Notice we were at zero, and then the next time we're at one again is at negative two pi, and then negative four pi, and then over and over and over again. But this applies, if n is an integer, n can be to all of the negative values of theta where cosine of theta is equal to one. Now let's think about when cosine of theta is equal to negative one. So cosine of theta is equal to negative one at theta is equal to, well we can just look at this graph right over here. Well when theta is equal to pi, when theta is equal to pi, and let's see, well, it kinda goes off this graph, but this graph would keep going like this, would keep going like this, and you'd see it would also be at three pi. And you can visualize it over here. Theta, cosine of theta is equal to negative one when we're at this point on the unit circle. So that happens when we get to pi radians, and then it won't happen again until we get to two pi, three pi radians, three pi radians. And it won't happen again until we go to two pi, until we add another two pi, until we make one entire revolution, And you can keep going on and on and on, and that's also true in the negative direction, so if we take two pi away from this, so if we were here, and if we go all the way around back to negative pi, it should also be the case, and you actually see it right over here on the graph. So you could think about this as two pi, two pi n plus pi, or you could view it as two n plus one, or two n plus one times pi, where pi is, sorry, where n is an integer. Let me right that a little bit neater, n is integer. At every one of those points, cosine, or for every one of these thetas, cosine of theta is going to keep hitting negative one over and over again. And you see it, it goes it goes from one bottom, where you can kind of valley to the next valley, it takes two pi to get to the next valley," + }, + { + "Q": "at 3:50, why is it that it goes pi, 3 pi, 5 pi for cos (-1) rather than o, 2, 4 pi...?\n", + "A": "The cosine is just the x-value of a point on the unit circle. At 0, 2pi, 4pi, etc, the x-value is 1. But at pi, 3pi, 5pi, etc, the x-value is -1.", + "video_name": "SdHwokUU8xI", + "timestamps": [ + 230 + ], + "3min_transcript": "on and on, and it makes sense. Theta equaled, or sorry, cosine of theta, the x-coordinate on this unit circle equaled one right when we were at zero angle, and we had to go all the way around the circle to get back to that point, two pi radians. But then it'll be again when we get to four pi radians, and then six pi radians, so two pi, four pi, six pi, and I guess you could see the pattern here. We're gonna keep hitting cosine of theta equals one every two pi, so you could really kind of view this as every multiple of two pi. Two pi n, where n is an integer, n is integer... is an integer. And that applies also for negative values. If you're going the other way around, if we're going the other way around, we don't get back until we get to negative two pi. Notice we were at zero, and then the next time we're at one again is at negative two pi, and then negative four pi, and then over and over and over again. But this applies, if n is an integer, n can be to all of the negative values of theta where cosine of theta is equal to one. Now let's think about when cosine of theta is equal to negative one. So cosine of theta is equal to negative one at theta is equal to, well we can just look at this graph right over here. Well when theta is equal to pi, when theta is equal to pi, and let's see, well, it kinda goes off this graph, but this graph would keep going like this, would keep going like this, and you'd see it would also be at three pi. And you can visualize it over here. Theta, cosine of theta is equal to negative one when we're at this point on the unit circle. So that happens when we get to pi radians, and then it won't happen again until we get to two pi, three pi radians, three pi radians. And it won't happen again until we go to two pi, until we add another two pi, until we make one entire revolution, And you can keep going on and on and on, and that's also true in the negative direction, so if we take two pi away from this, so if we were here, and if we go all the way around back to negative pi, it should also be the case, and you actually see it right over here on the graph. So you could think about this as two pi, two pi n plus pi, or you could view it as two n plus one, or two n plus one times pi, where pi is, sorry, where n is an integer. Let me right that a little bit neater, n is integer. At every one of those points, cosine, or for every one of these thetas, cosine of theta is going to keep hitting negative one over and over again. And you see it, it goes it goes from one bottom, where you can kind of valley to the next valley, it takes two pi to get to the next valley," + }, + { + "Q": "\nI don't get it... At 4:08 5/9 was given as the answer for cleaning the entire bathroom. I just cannot wrap my head around it because I feel like 5/9 is for cleaning 3/5 of the bathroom still? Please help. Thanks", + "A": "Anon is right... just break it down on a scratchpad", + "video_name": "2DBBKArGfus", + "timestamps": [ + 248 + ], + "3min_transcript": "He's able to take one third of a bottle so I could write it here 13 of a bottle to clean to clean three-fifths of a bathroom three-fifths of a bathroom three-fifths of a bathroom So hopefully it's clear now. Why this was helpful? We say okay? We want to take how many bottles it takes to clean a certain number of bathrooms one third of a bottle take can clean? three-fifths of a bathroom and makes it clear that we need to take the one-third and Divide it by three-fifths because then we're going to get bottles per bathroom. We're going to get the rate We're not going to get bathrooms per bottle we're going to get bottles per bathroom Which is what we care about what fraction of the bottle well? It will it take to clean entire bathroom to clean one entire bathroom? So now we just have to take one-third and divide it by three-fifths 13 three-fifths Divided by three fifths and the units are going to be bottles per bathroom. I'll write it like that Bottles or since if we know that that's going to be a fraction. We could just say rate bottle bathroom bottle Bathroom and so this is going to be the same thing as well we're going to have one third divided by three fifths is the same thing as One-Third times the reciprocal of three fifths so one-third times five thirds one-Third times five thirds Bottle bathroom, so let me write that down bottle bathroom Bottle bathroom and what is this going to be well? We multiply the numerators 1 times 5 is going to give us 5 and the denominator three times three is nights? bathroom Bathroom I wrote these a little bit further apart than I would want to let me write that a little bit closer together 59 59 of a bottle bathroom So just as a reminder of what we did here It could see him a little bit daunting But we said okay look he we want how much I'm what fraction of the bottle to clean his entire bathroom So we care about bottles per bathroom or since we're talking about a fraction I guess we could just say bottle some fraction of a bottle per bathroom so it's bottle Bathrooms that's going to give us the correct rate So it took one third of a bottle to clean Three-fifths of a bathroom so you divide one third by three-fifths? You get five ninths of a bottle per bathroom, so this? Right over here that is 59 it takes five ninths of a bottle of cleaning solution to clean his entire bathroom" + }, + { + "Q": "at 1:56 what do you mean by bottles an bathroom?\n", + "A": "at 1:56 he is just using this ratio as an example nothing too important", + "video_name": "2DBBKArGfus", + "timestamps": [ + 116 + ], + "3min_transcript": "It's a little daunting because it has fractions, but then when we work through it step-by-step Hopefully, it'll feel a little bit more intuitive so it says Calvin cleans three-fifths of his bathroom with one third of a bottle of cleaning solution at this rate What fraction of the bottle of cleaning Solution will Calvin used to clean his entire bathroom? And like always encourage you to pause the video and try to take a stab at this yourself So as [I] mentioned it's a little bit You know it's a three-fifths of his bathroom with one third of a bottle How do we [think] about this and what my brain does is I'd like to say well How would we like to answer the question what fraction of the bottle of cleaning solution will Calvin used to clean his entire bathroom? So we want to figure out is we want to figure out how many bottles so?? bottles bottles Per let me write it this way So we want to figure out he uses a certain number up with that? bottles per Bathroom if we knew this then we have the answer to the question if this would this might be I don't know [two-fifths] Bottles two-fifths of a bottle bathroom it might be two Bottles per bathroom, but if we know whatever this? Is and we know the answer How many bottles doesn't need to take to clean a bathroom or what fraction of a bottle? We don't know that they're hinting that it's a fraction of a bottle but how much of a bottle or how many bottles per bathroom or another way to think about this if we want to know if We wanted to express it as a as a rate more mathematically. We could say? bottles bottles per bathroom Bathroom and the Reason why this is helpful It makes it clear that look we want to figure out we want to take our Units tell us that we want to divide the number of bottles or the fraction of a bottle it takes? To clean the number of bathrooms or certain fraction of the bathrooms and they tell us over here they tell us that He's able to take one third of a bottle so I could write it here 13 of a bottle to clean to clean three-fifths of a bathroom three-fifths of a bathroom three-fifths of a bathroom So hopefully it's clear now. Why this was helpful? We say okay? We want to take how many bottles it takes to clean a certain number of bathrooms one third of a bottle take can clean? three-fifths of a bathroom and makes it clear that we need to take the one-third and Divide it by three-fifths because then we're going to get bottles per bathroom. We're going to get the rate We're not going to get bathrooms per bottle we're going to get bottles per bathroom Which is what we care about what fraction of the bottle well? It will it take to clean entire bathroom to clean one entire bathroom? So now we just have to take one-third and divide it by three-fifths" + }, + { + "Q": "\nAt 2:13 Sal said lets see how they look an a ___ diagram. Does anyone know what he said? \"Argan\"\nThanks!\nI am assuming it is named after a mathematician.", + "A": "It s called an Argand Diagram.", + "video_name": "BZxZ_eEuJBM", + "timestamps": [ + 133 + ], + "3min_transcript": "I want to make a quick clarification and then add more tools in our complex number toolkit. In the first video, I said that if I had a complex number z, and it's equal to a plus bi, I used a word. And I have to be careful about that word, because I used in the everyday sense. But it also has a formal reality to it. So clearly, the real part of this complex number is a. Clearly, that is the real part. And clearly, this complex number is made up of a real number plus an imaginary number. So just kind of talking in everyday terms, I called this the imaginary part. I called this imaginary number the imaginary part. But I want to just be careful there. I did make it clear that if you were to see the function the real part of z, this would spit out the a. And the function the imaginary part of z, this would spit out-- and we talked the number that's scaling the i. So it would spit out the b. So if someone is talking in the formal sense about the imaginary part, they're really talking about the number that is scaling the i. But in my brain, when I think of a complex number, I think of it having a real number and an imaginary number. And if someone were to say, well, what part of that is the imaginary number? I would have given this whole thing. But if someone says just what's the imaginary part, where they give you this function, just give them the b. Hopefully, that clarifies things. Frankly, I think the word \"imaginary part\" is badly named because clearly, this whole thing is an imaginary number. This right here is not an imaginary number. It's just a real number. It's the real number scaling the i. So they should call this the number scaling the imaginary part of z. Anyway, with that said, what I want to introduce you to is the idea of a complex number's conjugate. So if this is z, the conjugate of z-- it'd Sometimes it's z with a little asterisk right over there. That would just be equal to a minus bi. So let's see how they look on an Argand diagram. So that's my real axis. And then that is my imaginary axis. And then if I have z-- this is z over here-- this height over here is b. This base, or this length, right here is a. That's z. The conjugate of z is a minus bi. So it comes out a on the real axis, but it has minus b as its imaginary part, so just like this. So this is the conjugate of z. So just to visualize it, a conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis. You can imagine if this was a pool of water, we're seeing its reflection over here." + }, + { + "Q": "At 2:34 Why does Sal draw the number Z with an arrow like a vector?\n", + "A": "It is not an arrow, but simply a line on top of Z, it stands for conjugate of Z in this case.", + "video_name": "BZxZ_eEuJBM", + "timestamps": [ + 154 + ], + "3min_transcript": "the number that's scaling the i. So it would spit out the b. So if someone is talking in the formal sense about the imaginary part, they're really talking about the number that is scaling the i. But in my brain, when I think of a complex number, I think of it having a real number and an imaginary number. And if someone were to say, well, what part of that is the imaginary number? I would have given this whole thing. But if someone says just what's the imaginary part, where they give you this function, just give them the b. Hopefully, that clarifies things. Frankly, I think the word \"imaginary part\" is badly named because clearly, this whole thing is an imaginary number. This right here is not an imaginary number. It's just a real number. It's the real number scaling the i. So they should call this the number scaling the imaginary part of z. Anyway, with that said, what I want to introduce you to is the idea of a complex number's conjugate. So if this is z, the conjugate of z-- it'd Sometimes it's z with a little asterisk right over there. That would just be equal to a minus bi. So let's see how they look on an Argand diagram. So that's my real axis. And then that is my imaginary axis. And then if I have z-- this is z over here-- this height over here is b. This base, or this length, right here is a. That's z. The conjugate of z is a minus bi. So it comes out a on the real axis, but it has minus b as its imaginary part, so just like this. So this is the conjugate of z. So just to visualize it, a conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis. You can imagine if this was a pool of water, we're seeing its reflection over here. to visually add the complex number and its conjugate. So we said these are just like position vectors. So if we were to add z and its conjugate, we could essentially just take this vector, shift it up here, do heads to tails. So this right here, we are adding z to its conjugate. And so this point right here, or the vector that specifies that point, is z plus z's conjugate. And you can see right here, just visually, this is going to be 2a. And to do that algebraically. If we were to add z-- that's a plus bi-- and add that to its conjugate, so plus a minus bi, what are we going to get? These two guys cancel out. We're just going to have 2a. Or another way to think about it-- and really, we're just playing around with math-- if I take any complex number, and to it I add its conjugate," + }, + { + "Q": "\nAt 2:14, he referenced a z*. What does that mean? Additionally, is an Argand diagram just an imaginary coordinate plane?", + "A": "z is the conventional variable for a complex number. a z with a bar over it, or a z with an asterisk means the conjugate of z, which is what the video is all about.", + "video_name": "BZxZ_eEuJBM", + "timestamps": [ + 134 + ], + "3min_transcript": "I want to make a quick clarification and then add more tools in our complex number toolkit. In the first video, I said that if I had a complex number z, and it's equal to a plus bi, I used a word. And I have to be careful about that word, because I used in the everyday sense. But it also has a formal reality to it. So clearly, the real part of this complex number is a. Clearly, that is the real part. And clearly, this complex number is made up of a real number plus an imaginary number. So just kind of talking in everyday terms, I called this the imaginary part. I called this imaginary number the imaginary part. But I want to just be careful there. I did make it clear that if you were to see the function the real part of z, this would spit out the a. And the function the imaginary part of z, this would spit out-- and we talked the number that's scaling the i. So it would spit out the b. So if someone is talking in the formal sense about the imaginary part, they're really talking about the number that is scaling the i. But in my brain, when I think of a complex number, I think of it having a real number and an imaginary number. And if someone were to say, well, what part of that is the imaginary number? I would have given this whole thing. But if someone says just what's the imaginary part, where they give you this function, just give them the b. Hopefully, that clarifies things. Frankly, I think the word \"imaginary part\" is badly named because clearly, this whole thing is an imaginary number. This right here is not an imaginary number. It's just a real number. It's the real number scaling the i. So they should call this the number scaling the imaginary part of z. Anyway, with that said, what I want to introduce you to is the idea of a complex number's conjugate. So if this is z, the conjugate of z-- it'd Sometimes it's z with a little asterisk right over there. That would just be equal to a minus bi. So let's see how they look on an Argand diagram. So that's my real axis. And then that is my imaginary axis. And then if I have z-- this is z over here-- this height over here is b. This base, or this length, right here is a. That's z. The conjugate of z is a minus bi. So it comes out a on the real axis, but it has minus b as its imaginary part, so just like this. So this is the conjugate of z. So just to visualize it, a conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis. You can imagine if this was a pool of water, we're seeing its reflection over here." + }, + { + "Q": "\nAt 8:15 is the reason that ||a-b||=||b-a|| because the length is always a positive value?", + "A": "Yes. Remember that when you subtract two things (x - y) it is equal to -1 * (y - x). Since the magnitude only deals with the absolute value, the magnitudes will be the same, but the direction will be exactly opposite.", + "video_name": "5AWob_z74Ks", + "timestamps": [ + 495 + ], + "3min_transcript": "But here now, if I put little parentheses here, now I can apply the triangle inequality. And I say, well, you know what? This is going to be, by the triangle inequality, which we've proved, it's going to be less than or equal to the lengths of each of these vectors. Vector b plus the length of vector a minus b. So we know that the length of a is less than the sum of that one and that one. So we don't have to worry about this being our problem. We know that that is not true. Now let's look at b. So is there any way that I can rewrite b as a sum of two Well sure. I can write it as a sum of a plus, let me put it this way. If that vector right there is a minus b, the same vector in the reverse direction is going to be the vector b minus a. That's the same thing as b. And you can see it right here. The a's would cancel out and you're just left with the b there. Now by the triangle inequality, we know that this is less than or equal to the length of vector a plus the length of vector b minus a. Now you're saying hey, Sal, you're dealing with b minus a. This is the length of a minus b. And I can leave this for you to prove it based on our definition of vector lengths, but the length of b minus a is equal to minus 1 times a minus b. And I'll leave it to you to say that look, these lengths are equal. Because essentially-- I could leave that, but I think you can take that based on just the visual depiction of them And I have to be careful with length because it's not just in two dimensions. But I think you get the idea and I'll leave that for you to prove that these lengths are the same thing. So we know that b is less than the length of those two things. So we don't have to worry about that one right there. Finally, a minus b. The magnitude or the length of vector a minus b. Well I can write that as the length of-- or I can write that as vector a plus vector minus b. If we just put a minus b right there and go in the other directions, we could say minus b, which would be in that direction plus a would give us our vector a minus b." + }, + { + "Q": "\nThere is one but. In the video (2:43) it said that after the price for topping became 2$, the equation must be 8+2(p+8)/p+8. While I was doing the exercises I got the same problem, but in the hints the answer was that the second part should be 24+2p/p+8. Where did they get 24 and how?", + "A": "The answer is (24+2p)/(p+8), because when you simplify [8+2(p+8)]/(p+8), that s what you get. Here s the process: [8+2(p+8)]/(p+8) (8+2p+16)/(p+8) -- used distributive property (24+2p)/(p+8) -- added together; 8+16=24", + "video_name": "jQ15tkoXZoA", + "timestamps": [ + 163 + ], + "3min_transcript": "and then her total cost per pizza after if she bakes 8 more pizzas. So before, we're going to use p to say that's the number of pizzas she baked per day before the change in price. So before the change in price, on a given day, she would spend $8 on the oven and then $1.50 on ingredients So 1.5, or $1.50, times the number of pizzas. This would be her total cost on all the pizzas in that day. It's the oven cost plus it's the ingredients cost. So if you wanted this on a per pizza basis, you would just divide by the number of pizzas. Now let's think about what happens after the change in price. After the change in price, her cost per day for the oven is still $8. But now she has to spend $2 per pizza on ingredients. So $2 per pizza. And instead of saying that she's baking p pizzas, So it's going to be p plus 8. And so this is going to be her total cost for all of the pizzas she's now baking. And so if you want it on a per pizza basis, well, she's now making p plus 8 pizzas, you would divide by p plus 8. And the problem tells us that these two things are equivalent. Here you had a higher cost in ingredients per pizzas, but since you are now baking more pizzas, you're spreading the oven cost amongst more and more pizzas. So let's think about what p has to be. p has to be some number, some number of pizzas, so that these two expressions are equal. Her total cost per pizza before, when she only made p, is going to be the same as her total cost per pizza when she's making p plus 8 pizzas. So these two things need to be equal. So we did that first part, or we did what they asked us. We wrote an equation to find out how many pizzas Dominique baked And we used p to represent the number of pizzas. But now for fun, let's actually just solve for p. So let's just simplify things a little bit. So this part right over here. Actually, let's just cross multiply this on both sides. Or another way of thinking is multiply both sides times p plus 8 and multiply both sides times p. So if we multiply by p plus 8, and we multiply by p, we multiply by p plus 8, and we multiply by p, that cancels with that. That cancels with that. On the left-hand side-- so let's see. We have to just do the distributive property twice right over here. What is p times 8 plus 1.5p? Well, that's going to be 8p. I'm just multiplying the p times this stuff first. Plus 1.5p squared. And now let's multiply the 8 times both of these terms." + }, + { + "Q": "At 1:14, why is it the square root of x+2? I know that the square root of x looks like that, but how can you determine if it is x + 2?\n", + "A": "If x+2 is substituted for x, then the graph is centered at x= -2. That s because the general format is x-c, where c is where the graph is centered. For example, If x-2 was substituted in, then the graph would be centered at x=2.", + "video_name": "IJWDyPFXGyM", + "timestamps": [ + 74 + ], + "3min_transcript": "Select the piecewise function whose graph is shown below. Or I guess we should say to the right. I copied and pasted it so it's on the right now. So we have this piecewise continuous function. So it's not defined for x being negative 2 or lower. But then starting at x greater than negative 2, it starts being defined. It's continuous all the way until we get to the point x equals 2 and then we have a discontinuity. And then it starts getting it defined again down here. And then it is continuous for a little while all the way. And then when x is greater than 6, it's once again undefined. So let's think about which of these functions describe this one over here. So this one looks like a radical function shifted. So square root of x would look like this. Let me do it in a color that you could see. Square root of x would look like this. And this just looks like square root of x shifted 2 to the left. So this looks like square root of x plus 2. This one right over here looks like square root of x plus 2. And you could verify that. When x is negative 2, negative 2 plus 2, square root of that is going to be 0. And it's not defined there, but we see that if we were to continue it would have been defined there. Let's try some other points. When x is negative 1, negative 1 plus 2 is 1. Principal root of 1 is 1. Let's try 2. When x is equal to 2, 2 plus 2 is 4. Principal root of 4 is positive 2. So this just looks like a pretty good candidate. So it looks like our function. So let's see, it looks like our function would be-- and I'm not going to call it anything because it could be p, h, g, or f. But our function, if I were to write it out, So it looks like it's the square root of x plus 2 for negative 2 being less than x-- it's not defined at negative 2, but as long as x is greater than negative 2 and x is less than or equal to 2. And x is less than or equal to 2. So that's this part of the function. And then it jumps down. Now, this looks like x to the third. x to the third looks something like this. So x to the third power looks something like that. So let's see, negative 2 to the third power is negative 8. So it looks something like that. That's what x to the third looks like. So 2 to the third power is 8. So x to the third looks something like that. This looks like x to the third shifted over 4. So I'm guessing that this is x minus 4 to the third power." + }, + { + "Q": "\nisnt the slope of the first graph on the left corner negative..and how at 2:17, does SAL say that the slope is close to 1?.?", + "A": "it s positive because as x increases y increases, too", + "video_name": "nZrxs-U9d8o", + "timestamps": [ + 137 + ], + "3min_transcript": "Let's say that this right over here is the graph of lowercase f of x. That's lowercase f of x there. And let's say that we have some other function capital F of x. And if you were to take its derivative, so, capital F prime of x, that's equal to lowercase f of x, lower case f of x. So given that, which of these, which of these, could be the graph of capital, of capital F of x? And I encourage you to pause this video, and try and think about it on your own before we work through it. Well if, if this curve is going to be the derivative of one of them. That means that any, for any x value it's describing what the instantaneous rate of change. Or what the slope of the tangent line is, of which ever one of these is the possible capital F of x. So, let's just look at a couple of things right here so what, what do we know about lower case f of x? Well, one thing we know is it's always positive. It, it has as we go to negative infinity, it asymptotes towards 0. But it's always positive. So since this is describing the slope of one of these. That means that the slope of one of these always, or out of the candidates, has to always be positive. And, if we look at this, the slope of the tangent line here is, indeed, always positive. The slope of the tangent line here does look like it's positive. Every time we increase an x, we're increasing by y. Here it's positive. But here it's negative. When we increase by x, we decrease by y. So, we can rule, we can rule this one out. Now, what else, what else do we know? Well, this is the derivative. This is telling us the slope of the tangent line. So, for example, when x is equal to, when x is equal to negative 4, f of, f of negative 4 is pretty close to 0. It's pretty close to 0. So, it's slightly, slightly more than 0. of x has to be pretty close to 0, when x equals negative 4. So, let's see, when x equals negative 4, the slope of tangent line, here, isn't close to 0, this actually looks closer to 1. So, we could rule this one out. Over here, when x is equal to negative 4, the slope of the tangent line, yeah, that actually does look pretty close to 0. So I won't rule that one out. And over here, the slope of the tangent line, when x is equal to negative 4, that also looks pretty close to 0. So these are still both in the running. So let's see how we can think of it different. So let's just pick another point. When x is equal to, when x is equal to 0, f of 0 looks like it's pretty close to 1. I don't know if it's exactly to 1. Actually it looks almost exactly. Almost exactly equal to 1. So when capital F of, so at capital F of 0, the" + }, + { + "Q": "\nAt 11:26 he forgets to write a line above the x2 (the standard deviation of the difference of the sample 'means'). Right?", + "A": "Yes, he does, as commented in the next video and as you can see from his calculation of the confidence intervals, he gets a little confused sometimes in videos that are longer than 10 minutes ;D After all, its not that easy to talk about so serious stuff for quiet long, without making any small mistakes :)", + "video_name": "hxZ6uooEJOk", + "timestamps": [ + 686 + ], + "3min_transcript": "variances of each of those distributions. Let me write it this way. So the variance, I'll kind of re-prove it. The variance of our distribution is going to be equal to the sum of the variances of each of these sampling distributions. And we know that the variance of each of these sampling distributions is equal to the variance of this sampling distribution, is equal to the variance of the population distribution, divided by our sample size. And the variance of this sampling distribution, for our control, is going to be equal to the variance of the population distribution for the control divided by its sample size. And since we don't know what these are, we can approximate them. Especially, because our n is greater than 30 for both circumstances. We can approximate these with our sample variances for each of these distributions. So let me make this clear. Our sample variances for each of these distributions. So this is going to be our sample standard deviation one squared, which is the sample variance for that distribution, over 100. Plus my sample standard deviation for the control squared, which is the sample variance. variance divided by 100. And this will give us the variance for this distribution. And if we want the standard deviation, we just take the square roots of both sides. If we want the standard deviation of this distribution right here, this is the variance right now, so we just need to take the square roots. Let's calculate this. We actually know these values. S1, our sample standard deviation for group one is 4.67. We wrote it right here, as well. It's 4.76 and 4.04. The S is 4.67, we're going to have to square it. And the S2 is 4.04, we're going to have to square it. So let's calculate that. So we're going to take the square root of 4.67 squared" + }, + { + "Q": "7:45 Sal says there's a 95% chance that 1.91 is within 1.96 standard deviations. Shouldn't the interpretation be that \"we are 95% confident...etc.?\" Because this isn't really about probability, right?\n", + "A": "Yes, because he is talking about the specific value that was observed (1.91). If we re still talking theoretically, we can talk about probability. So we can say there is a 95% chance that xbar will be within 1.96\u00cf\u0083 of \u00ce\u00bc. When we start plugging in numbers and get, say, that xbar is 9.31, then we no longer have a random variable, and the observed mean either is or is not (100% or 0%) within 1.96\u00cf\u0083 of \u00ce\u00bc.", + "video_name": "hxZ6uooEJOk", + "timestamps": [ + 465 + ], + "3min_transcript": "So this critical Z value right here is 1.96 standard deviations. This is 1.96 times the standard deviation of x1 minus x2. And then this right here is going to be negative 1.96 times the same thing. Let me write that. So this right here, it's symmetric. This distance is going to be the same as that distance. So this is negative 1.96 times the standard deviation of this distribution. So let's put it this way, there's a 95% chance that our sample as a difference of these other samples. There's a 95% chance that 1.91 lies within 1.96 times the standard deviation of that distribution. So you could view it as a standard error of this statistic. So x1 minus x2. Let me finish that sentence. There's a 95% chance that 1.91, which is the sample statistic, or the statistic that we got, is within 1.96 times the standard deviation of this distribution of the true mean of of the distribution. There's a 95% chance that the true mean of the distribution is within 1.96 times the standard deviation of the distribution of 1.91. These are equivalent statements. If I say I'm within three feet of you, that's equivalent to saying you're within three feet of me. That's all that's saying. But when we construct it this way, it becomes pretty clear, how do we actually construct the confidence interval? We just have to figure out what this distance right over here is. And to figure out what that distance is, we're going to have to figure out what the standard deviation of this distribution is. Well the standard deviation of the differences of the sample means is going to be equal to, and we saw this in the last video-- in fact, I think I have it right at the bottom" + }, + { + "Q": "\nAt 3:34, Sal drew a distribution of the difference of sample means. This looks like it's a normal distribution? In the last video, the distribution of the difference of sample means he drew looked normal, too. How do we know this is normally distributed? If you add two normal r.v.s together, is the resulting distribution normal, and why?\n\nThanks! Beth", + "A": "Central Limit Theorem. Sal has vids on it.", + "video_name": "hxZ6uooEJOk", + "timestamps": [ + 214 + ], + "3min_transcript": "So just based on what we see, maybe you lose an incremental 1.91 pounds every four months if you are on this diet. And what we want to do in this video is to get a 95% confidence interval around this number. To see that in that 95% confidence interval, maybe, do we always lose weight? Or is there a chance that we can actually go the other way with the low-fat diet? So in this video, 95% confidence interval. In the next video, we'll actually do a hypothesis test using this same data. And now to do a 95% confidence interval, let's think about the distribution that we're thinking about. So let's look at the distribution. Of course we're going to think about the distribution that we're thinking about. We want to think about the distribution of the difference of the means. So it's going to have some true mean here. of the sample means. Let me write that. It's not a y, it's an x1 and x2. So it's the sample mean of x1 minus the sample mean of x2. And then this distribution right here is going to have some standard deviation. So it's the standard deviation of the distribution of the mean of x1 minus the sample mean of x2. It's going to have some standard deviation here. And we want to make an inference about this. Or I guess, the best way to think about it, we want to get a 95% confidence interval. Based on our sample, we want to create an interval around this, where we're confident that there's a 95% chance that this true mean, the true mean of the differences, lies within that interval. And to do that let's just think of it the other way. are 95% sure that any sample from this distribution, and this is one of those samples, that there is a 95% chance that we will select from this region right over here. So we care about a 95% region right over here. So how many standard deviations do we have to go in each direction? And to do that we just have to look at a Z table. And just remember, if we have 95% in the middle right over here, we're going to have 2.5% over here and we're going to have 2.5% over here. We have to have 5% split between these two symmetric tails. So when we look at a Z table, we want the critical Z value that they give right over here. And we have to be careful here. We're not going to look up 95%, because a Z table gives us the cumulative probability up to that critical Z value." + }, + { + "Q": "at 4:23, how do you find the height and acquaint as srt2 over 2? I cannot understand.\n", + "A": "What do you mean by acquaint ? Numbers don t get acquainted with each other, people do. Regarding how did Sal find the sine and cosine of -45 degrees... 45 degree angles are a very common occurrence. Sal has memorized these values from the unit circle. You might want to search the internet and find a good picture of the unit circle to use as reference.", + "video_name": "Idxeo49szW0", + "timestamps": [ + 263 + ], + "3min_transcript": "And so if you draw a line-- Let me draw a little unit circle here. So if I have a unit circle like that. And let's say I'm at some angle. Let's say that's my angle theta. And this is my y-- my coordinates x, y. We know already that the y-value, this is the sine of theta. Let me scroll over here. Sine of theta. And we already know that this x-value is the cosine of theta. So what's the tangent going to be? It's going to be this distance divided by this distance. Or from your algebra I, this might ring a bell, because we're starting at the origin from the point 0, 0. This is our change in y over our change in x. Or it's our rise over run. Or you can kind of view the tangent of theta, or it really The slope. So you could write slope is equal to the tangent of theta. So let's just bear that in mind when we go to our example. If I'm asking you-- and I'll rewrite it here --what is the inverse tangent of minus 1? And I'll keep rewriting it. Or the arctangent of minus 1? I'm saying what angle gives me a slope of minus 1 on the unit circle? So let's draw the unit circle. Let's draw the unit circle like that. Then I have my axes like that. And I want a slope of minus 1. A slope of minus 1 looks like this. If it was like that, it would be slope of plus 1. So what angle is this? So in order to have a slope of minus 1, this distance is the same as this distance. And you might already recognize that this is a right angle. So this has to be a 45 45 90 triangle. This is an isosceles triangle. These two have to add up to 90 and they have to be the same. So this is 45 45 90. And if you know your 45 45 90-- Actually, you don't even have to know the sides of it. In the previous video, we saw that this is going to be-- Right here. This distance is going to be square root of 2 over 2. So this coordinate in the y-direction is minus square root of 2 over 2. And then this coordinate right here on the x-direction is square root of 2 over 2 because this length right there is that. So the square root of 2 over 2 squared plus the square root of 2 over 2 squared is equal to 1 squared. But the important thing to realize is this is a 45 45 90 triangle. So this angle right here is-- Well if you're just looking at the triangle by itself, you would say that this is a 45 degree angle. But since we're going clockwise below the x-axis, we'll call" + }, + { + "Q": "\nat \"9:00\", Why do you include the first quadrant in your restriction of Theta? Why wouldn't you restrict your range of Theta to only the fourth quadrant?", + "A": "You need to have all the positive and all the negative possibilities. If you exclude Q1 you can t have any positive answers. Q1 covers the positive angles and Q4 covers the negative angles.", + "video_name": "Idxeo49szW0", + "timestamps": [ + 540 + ], + "3min_transcript": "It would be-- I'll just say 2 pi minus pi over 4. Or 4 pi minus pi. It can't map to all of these different things. So I have to constrict the range on the inverse tan function. And we'll restrict it very similarly to the way we restricted the sine-- the inverse sine range. We're going to restrict it to the first and fourth quadrants. So the answer to your inverse tangent is always going to be something in these quadrants. But it can't be this point and that point. Because a tangent function becomes undefined at pi over 2 and at minus pi ever 2. Because your slope goes vertical. You start dividing-- Your change in x is 0. You're dividing-- Your cosine of theta goes to 0. So if you divide by that, it's undefined. So your range-- So if I-- Let me write this down. So if I have an inverse tangent of x, I'm going to-- Well, So if I have the tangent of theta is equal to x, what are all the different values that x could take on? These are all the possible values for the slope. And that slope can take on anything. So x could be anywhere between minus infinity and positive infinity. x could pretty much take on any value. But what about theta? Theta, you can only go from minus pi over 2 all the way to pi over 2. And you can't even include pi over 2 or minus pi over 2 because then you'd be vertical. So then you say your-- So if I'm just dealing with vanilla tangent. Not the inverse. The domain-- Well the domain of tangent can go multiple times around, so let me not make that statement. But if I want to do inverse tangent so I don't have a 1 to many mapping. I want to cross out all of these. I'm going to restrict theta, or my range, to be greater than And so if I restrict my range to this right here and I exclude that point and that point. Then I can only get one answer. When I say tangent of what gives me a slope of minus 1? And that's the question I'm asking right there. There's only one answer. Because if I keep-- This one falls out of it. And obviously as I go around and around, those fall out of that valid range for theta that I was giving you. And then just to kind of make sure we did it right. Our answer was pi over 4. Let's see if we get that when we use our calculator. So the inverse tangent of minus 1 is equal to that. Let's see if that's the same thing as minus pi over 4. Minus pi over 4 is equal to that. So it is minus pi over 4. But it was good that we solved it without a calculator because" + }, + { + "Q": "\nThe restrictions on this test seem redundant.\n\nfrom 0:53 to 1:33, Sal gives three restrictions on the series:\n\n1) Bn \u00e2\u0089\u00a5 0 for all relevant n (namely positive integers n).\n2) lim as Bn\u00e2\u0086\u0092\u00e2\u0088\u009e = 0\n3) {Bn} is a decreasing sequence.\n\nDon't the first two rules imply the third?", + "A": "Consider the function f(n) = x*e^(-n). This will be our Bn. What do you get at n= 0, n=1, n=2? f(0) = 0, f(1) = e^-1 = 0.37, f(2) = 2/e^-2 = 0.27 So the first 3 terms of the sequence are: [0, 0.37, 0.27]. Notice the b2> b1 this function is not always decreasing! Yet all terms are greater than zero. And the limit as n\u00e2\u0086\u0092\u00e2\u0088\u009e = 0. Some functions like this rise a little bit, then fall back down as they go on. So the third rule is necessary.", + "video_name": "91qVGeyTl44", + "timestamps": [ + 53, + 93 + ], + "3min_transcript": "- [Voiceover] Let's now expose ourselves to another test of conversions, and that's the alternating series test. And I'll explain the alternating series test, and I'll apply it to an actual series while I do it to make the explanation of the alternating series test a little bit more concrete. So let's say that I have some series, some infinite series. Let's say it goes from n equals k to infinity of a sub n. Let's say I can write it as, or I can rewrite a sub n. So a sub n is equal to negative one to the n, times b sub n, or a sub n is equal to negative one to the n plus one, times b sub n, where b sub n is greater than or equal to zero for all the n's we care about. So for all of these integer n's greater than or equal to k. and we know two more things. And we know number one, the limit as n approaches infinity of b sub n is equal to zero. And number two, b sub n is a decreasing sequence. Then that lets us know that the original infinite series is going to converge. So this might seem a little bit abstract right now. Let's use this with an actual series to make it a little bit more concrete. So let's say that I had the series from n equals one to infinity And we can write it out just to make this series a little bit more concrete. When n is equal to one, this is going to be negative one to the one power. Actually, let's just make this a little bit, let's make this a little bit more interesting. Let's make this negative one to the n plus one. So when n is equal to one, this is going to be negative one squared over one, which is going to be one. And then when n is two, it's going to be negative one to the third power, which is going to be negative one half. So it's minus one half, plus one third, minus one fourth, plus, minus, and it just keeps going on and on forever. Now, can we rewrite this a sub n like this? Well sure, the negative one to the n plus one is actually explicitly called out. We can rewrite our a sub n, so let me do that. So a sub n, which is equal to negative one," + }, + { + "Q": "At 0:38-0:48: Why is it necessary that the alternating sign be expressed by either (-1)^n or (-1)^(n+1). Aren't there a whole slew of other ways to produced an alternating sign? What about (-1)^(n+2), (-1)^(n^2), or any polynomial exponent with odd integer coefficients?\n", + "A": "We are only talking about the form the series takes on. We know that it alternates, so the question is, is a negative term first, or a positive term. Given n goes from 1 to infinity, the first term of the (-1)^n series will be negative, and the first term of the (-1)^(n+1) series will be positive. That is all that is meant by the form of the series. Why make it any more complicated? It is an alternating series, either the first term is positive, or the first term is negative.", + "video_name": "91qVGeyTl44", + "timestamps": [ + 38, + 48 + ], + "3min_transcript": "- [Voiceover] Let's now expose ourselves to another test of conversions, and that's the alternating series test. And I'll explain the alternating series test, and I'll apply it to an actual series while I do it to make the explanation of the alternating series test a little bit more concrete. So let's say that I have some series, some infinite series. Let's say it goes from n equals k to infinity of a sub n. Let's say I can write it as, or I can rewrite a sub n. So a sub n is equal to negative one to the n, times b sub n, or a sub n is equal to negative one to the n plus one, times b sub n, where b sub n is greater than or equal to zero for all the n's we care about. So for all of these integer n's greater than or equal to k. and we know two more things. And we know number one, the limit as n approaches infinity of b sub n is equal to zero. And number two, b sub n is a decreasing sequence. Then that lets us know that the original infinite series is going to converge. So this might seem a little bit abstract right now. Let's use this with an actual series to make it a little bit more concrete. So let's say that I had the series from n equals one to infinity And we can write it out just to make this series a little bit more concrete. When n is equal to one, this is going to be negative one to the one power. Actually, let's just make this a little bit, let's make this a little bit more interesting. Let's make this negative one to the n plus one. So when n is equal to one, this is going to be negative one squared over one, which is going to be one. And then when n is two, it's going to be negative one to the third power, which is going to be negative one half. So it's minus one half, plus one third, minus one fourth, plus, minus, and it just keeps going on and on forever. Now, can we rewrite this a sub n like this? Well sure, the negative one to the n plus one is actually explicitly called out. We can rewrite our a sub n, so let me do that. So a sub n, which is equal to negative one," + }, + { + "Q": "\nat 3:25 he said twice,but ment two times", + "A": "Twice is the same as two times. Once, twice, thrice are still used occasionally.", + "video_name": "N1X0vf5PUz4", + "timestamps": [ + 205 + ], + "3min_transcript": "How many of these 4 equal pieces would I have to eat. I encourage you to pause the video and think about that. Well, I would eat this piece. You could imagine me eating this piece and this piece right over here. I've eaten the same amount of the pizza. Each of these pieces you could imagine got cut into 2 pieces when I cut the whole pizza this way. And so now I have to eat 2 slices of the 4, as opposed to 1 slice of the 2. So I just ate 2 out of the 4 slices. I'm using different numbers here. Here I'm using a 1 in the numerator and 2 in the denominator. Here, I'm using a 2 in the numerator and a 4 in the denominator. These two fractions represent the same quantity. I ate the same amount of pizza. If I eat 2/4 of a pizza, if I eat 2 out of 4 equal pieces, that's the same fraction of the pizza So we would say that these two things are equivalent fractions. Now let's do another one like this. Instead of just dividing it into 4 equal pieces, let's divide it into 8 equal pieces. So now we could cut once like this. So now we have 2 equal pieces. Cut once like this. Now we have 4 equal pieces. And then divide each of those 4 pieces into 2 pieces. So I'll cut those in-- So let's see. I want to make them equal pieces. Those don't look as equal as I would like. So that looks more equal, and that looks reasonably equal. So now how many equal pieces do I have? I have 8 equal pieces. But let's say I wanted to eat the same fraction of the pizza. Well, how many of those 8 equal pieces have I eaten? Well, I've eaten 1, 2, 3, 4 of those 8 equal pieces. And so once again, this fraction, 4 of 8, or 4/8, is equivalent to 2/4, which is equivalent to 1/2. And you might see a little bit of a pattern here. Going from this scenario to this scenario, I got twice as many equal slices. Because I had twice as many equal slices, I needed to eat two times the number of slices. So I multiply the denominator by 2, and I multiply the numerator by 2. If I multiply the numerator and the denominator by the same number, then I'm not changing what that fraction represents. And you see that over here. Going from 4 slices to 8 slices, I cut every slice, I turned every slice into 2 more slices, so I had twice as many slices. And then if I want to eat the same amount," + }, + { + "Q": "2:41 alligator eats the bigger number\n", + "A": "Im finally done this thing!", + "video_name": "nFsQA2Zvy1o", + "timestamps": [ + 161 + ], + "3min_transcript": "These are both equal quantities. What I have here on the left hand side, this is 1 plus 1 minus 1 is 1 and this right over here is 1. These are both equal quantities. Now I will introduce you to other ways of comparing numbers. The equal sign is when I have the exact same quantity on both sides. Now we'll think about what we can do when we have different quantities on both sides. So let's say I have the number 3 and I have the number 1 and I want to compare them. So clearly 3 and 1 are not equal. In fact, I could make that statement with a not equal sign. So I could say 3 does not equal 1. But let's say I want to figure out which one is a larger So if I want to have some symbol where I can compare them, where I can tell, where I can state which of these is larger. And the symbol for doing that is the greater than symbol. This literally would be read as 3 is greater than 1. 3 is a larger quantity. And if you have trouble remembering what this means-- greater than-- the larger quantity is on the opening. I guess if you could view this as some type of an arrow, or some type of symbol, but this is the bigger side. Here, you have this little teeny, tiny point and here you have the big side, so the larger quantity is on the big side. This would literally be read as 3 is greater than-- so let me write that down-- greater than, 3 is greater than 1. I could write an expression. I could write 1 plus 1 plus 1 is greater than, let's say, well, just one 1 right over there. This is making a comparison. But what if we had things the other way around. What if I wanted to make a comparison between 5 and, let's say, 19. So now the greater than symbol wouldn't apply. It's not true that 5 is greater than 19. I could say that 5 is not equal to 19. So I could still make this statement. But what if I wanted to make a statement about which one is larger and which one is smaller? Well, as in plain English, I would want to say 5 is less than 19. So I would want to say-- let me write that down-- I want to write 5 is less than 19." + }, + { + "Q": "\nAt 3:41 can you explain how you went from 4(y^2-4y+4) to 4(y-2)^2?", + "A": "Sal factored the trinomial. Find 2 numbers that multiply to +4 but also add to -4. The numbers are -2 and -2 This creates the factors 4(y -2)(y - 2). If you write it using exponents, you get: 4 (y-2)^2 Hope this helps.", + "video_name": "cvA4VN1dpuY", + "timestamps": [ + 221 + ], + "3min_transcript": "Is equal to 109. And the things we're going to add, those are what complete the square. Make these things a perfect square. So, if I take this, you have a minus 4 here. I take half of that number. This is just completing the square, I encourage you to watch the video on completing the square where I explain why this works. But I think I have a minus 4. I take half of that, it's minus 2. And then minus 2 squared is plus 4. Now, I can't do one thing to one side of the equation without doing it to the other. And I didn't add a 4 to the left-hand side of the equation. I actually added a 4 times 4, right? Because you have this 4 multiplying it out front. So I added at 16 to the left side of the equation, so I have to also add it to the right-hand side of the equation. This is equivalent to also having a plus 16 here. That might make a little bit clearer, right? When you factor it out, and it becomes a 4. And we would have added a 16 up here as well. Likewise, if we take half of this number here. 1 squared is 1. We didn't add a 1 to the left-hand side of the equation, we added a 1 times minus 25. So we want to put a minus 25 here. And, likewise, this would have been the same thing as adding a minus 25 up here. And you do a minus 25 over here. And now, what does this become? The y terms become 4 times y minus 2 squared. y minus 2 squared. Might want to review factoring a polynomial, if you found that a little confusing, that step. Minus 25 times x plus 1 squared. That's that, right there. x plus 1 squared, is equal to, let's see, 109 plus 16 is 25 minus 25, it equals 100. We're almost there. So we want a 1 here, so let's divide both sides So, you will get y minus 2 squared. 4 divided by 100 is the same thing as 1/25, so this becomes over 25. Minus, let's see, 25/100 is the same thing as 1/4, so this becomes x plus 1 squared over 4 is equal to 1. And there you have it. We have it in standard form and, yes, indeed, we do have a hyperbola. Now, let's graph this hyperbola. So the first thing we know is where the center of this hyperbola is. Is the center of this hyperbola is at the point x is equal to minus one. So it's an x is equal to minus 1. y is equal to 2. And let's figure out the asymptotes of this hyperbola. So if this was -- this is the way I always do it, because I always forget the actual formula. If this was centered at 0 and it looked something like this. y squared over 25 minus x squared over 4 is equal to 1." + }, + { + "Q": "\nAt about 1:14 he starts writing the \"number tree\". Do you always put the larger number on the right?", + "A": "The order of the numbers don t change the actual math. Usually people will use one side or the other for the prime numbers which makes a clear picture where it is easy to find all the primes. However, it is not fixed and when working with students I ll break it up by whatever factors they can think of and circle the primes as we get them. (For example, when factoring 48 a student may say 6*8 initially)", + "video_name": "znmPfDfsir8", + "timestamps": [ + 74 + ], + "3min_transcript": "What is the least common multiple of 36 and 12? So another way to say this is LCM, in parentheses, 36 to 12. And this is literally saying what's the least common multiple of 36 and 12? Well, this one might pop out at you, because 36 itself is a multiple of 12. And 36 is also a multiple of 36. It's 1 times 36. So the smallest number that is both a multiple of 36 and 12-- because 36 is a multiple of 12-- is actually 36. There we go. Let's do a couple more of these. That one was too easy. What is the least common multiple of 18 and 12? And they just state this with a different notation. The least common multiple of 18 and 12 is equal to question mark. So let's think about this a little bit. So there's a couple of ways you can think about-- so let's just write down our numbers that we care about. We care about 18, and we care about 12. So there's two ways that we could approach this. One is the prime factorization approach. of these numbers and then construct the smallest number whose prime factorization has all of the ingredients of both of these numbers, and that will be the least common multiple. So let's do that. 18 is 2 times 9, which is the same thing as 2 times 3 times 3, or 18 is 2 times 9. 9 is 3 times 3. So we could write 18 is equal to 2 times 3 times 3. That's its prime factorization. 12 is 2 times 6. 6 is 2 times 3. So 12 is equal to 2 times 2 times 3. Now, the least common multiple of 18 and 12-- let me write this down-- so the least common multiple of 18 and 12 is going to have to have enough prime factors to cover because we want the least common multiple or the smallest common multiple. So let's think about it. Well, it needs to have at least 1, 2, a 3 and a 3 in order to be divisible by 18. So let's write that down. So we have to have a 2 times 3 times 3. This makes it divisible by 18. If you multiply this out, you actually get 18. And now let's look at the 12. So this part right over here-- let me make it clear. This part right over here is the part that makes up 18, makes it divisible by 18. And then let's see. 12, we need two 2's and a 3. Well, we already have one 3, so our 3 is taken care of. We have one 2, so this 2 is taken care of. But we don't have two 2s's. So we need another 2 here. So, notice, now this number right over here has a 2 times 2 times 3 in it, or it has a 12 in it, and it has a 2 times 3 times 3, or an 18 in it. So this right over here is the least common multiple" + }, + { + "Q": "\n@ 7:43 why and how did you guess 'y' to be = Ae^(2x) ?", + "A": "Because we knew that after some algebra with both the second derivative, the first derivative and the original function, the result must be 3e\u00c2\u00b2\u00cb\u00a3, so a good candidate that will maintain that exponent but with enough room to play with the coefficients was Ae\u00c2\u00b2\u00cb\u00a3", + "video_name": "znE4Nq9NJCQ", + "timestamps": [ + 463 + ], + "3min_transcript": "And so our general solution-- I'll call that h. Well, let's call that y general. y sub g. So our general solution is equal to-- and we've done this many times-- C1 e to the 4x plus C2 e to the minus 1x, or minus x. Fair enough. So we solved the homogeneous equation. So how do we get, in that last example, a j of x that will give us a particular solution, so on the right-hand side we get this. Well here we just have to think a little bit. And this method is called The Method of Undetermined And you have to say, well, if I want some function where I take a second derivative and add that or subtracted some multiple of its first derivative minus some multiple of the function, I get e to the 2x. That function and its derivatives and its second times e to the 2x. So essentially we take a guess. We say well what does it look like when we take the various derivatives and the functions and we multiply multiples of it plus each other? And all of that. We would get e the to 2x or some multiple of e to the 2x. Well, a good guess could just be that j-- well I'll call it y particular. Our particular solution here could be that-- and particular solution I'm using a little different than the particular solution when we had initial conditions. Here we can view this as a particular solution. A solution that gives us this on the right-hand side. So let's say that the one I pick is some constant A times e to the 2x. If that's my guess, then the derivative of that is equal to 2Ae the to 2x. And the second derivative of that, of my particular And now I can substitute in here, and let's see if I can solve for A, and then I'll have my particular solution. So the second derivitive, that's this. So I get 4Ae to the 2x minus 3 times the first derivitive. So minus 3 times this. So that's minus 6Ae to the 2x minus 4 times the function. So minus 4Ae to the 2x, and all of that is going to be equal to 3e to the 2x. Well we know e to the 2x equal 0, so we can divide both sides by that. Just factor it out, really. Get rid of all of the e's to the 2x. On the left-hand side, we have 4A and a minus 4A. Well, those cancel out. And then lo and behold, we have minus 6A is equal to 3. Divide both sides by 6 and get A is equal to minus 1/2." + }, + { + "Q": "@ 8:38 ish, what happens when the left side all adds up to 0?\nex: y''(x)-6y'(x)+9y(x)=5e^3x is the equation and my guess would be Ae^3x\ni'd end up with coefficients of 9-18+9=5e^3x\n", + "A": "In that case, you ve proved your guess wrong by contradiction. That means you must change your guess accordingly. For example, I would change y(x) = Ae^3x to y(x) = Ae^Bx instead. This gives y(x) = Ae^Bx, y (x) = ABe^Bx and y (x) = AB^2e^3x y (x) - 6y (x) + 9y(x) = AB^2e^3x - 6ABe^3x + 9Ae^3x = 5e^3x Therefore, AB^2 - 6AB + 9A = 5. Solving for A, we have A = 5/(B - 3)^2; B =/= 3 That s it. Hope that helps!", + "video_name": "znE4Nq9NJCQ", + "timestamps": [ + 518 + ], + "3min_transcript": "times e to the 2x. So essentially we take a guess. We say well what does it look like when we take the various derivatives and the functions and we multiply multiples of it plus each other? And all of that. We would get e the to 2x or some multiple of e to the 2x. Well, a good guess could just be that j-- well I'll call it y particular. Our particular solution here could be that-- and particular solution I'm using a little different than the particular solution when we had initial conditions. Here we can view this as a particular solution. A solution that gives us this on the right-hand side. So let's say that the one I pick is some constant A times e to the 2x. If that's my guess, then the derivative of that is equal to 2Ae the to 2x. And the second derivative of that, of my particular And now I can substitute in here, and let's see if I can solve for A, and then I'll have my particular solution. So the second derivitive, that's this. So I get 4Ae to the 2x minus 3 times the first derivitive. So minus 3 times this. So that's minus 6Ae to the 2x minus 4 times the function. So minus 4Ae to the 2x, and all of that is going to be equal to 3e to the 2x. Well we know e to the 2x equal 0, so we can divide both sides by that. Just factor it out, really. Get rid of all of the e's to the 2x. On the left-hand side, we have 4A and a minus 4A. Well, those cancel out. And then lo and behold, we have minus 6A is equal to 3. Divide both sides by 6 and get A is equal to minus 1/2. It is equal to minus 1/2 e to the 2x. And now, like I just showed you before I cleared the screen, our general solution of this non-homogeneous equation is going to be our particular solution plus the general solution to the homogeneous equation. So we can call this the most general solution-- I don't know. I'll just call it y. It is our general solution C1e to the 4x plus C2e to the minus x plus our particular solution we found. So that's minus 1/2e to the 2x. Pretty neat. Anyway, I'll do a couple more examples of this. And I think you'll get the hang of it. In the next examples, we'll do something other than an e to the 2x or an e function here." + }, + { + "Q": "\nAt 2:34, could you replace the outermost parentheses with brackets so it can be 4+[2(7- 2x)] instead of 4+(2(7-2x)?\nAre both expressions valid or is one expression more correct than the other?", + "A": "Yes, they both mean the same thing. The brackets would be understood to be parentheses; they are usually used to set the groups of things you multiply first apart.", + "video_name": "xLYVo_k0_us", + "timestamps": [ + 154 + ], + "3min_transcript": "First consider the expression for negative 5 plus the quantity of 4 times x. Now, take the product of negative 8 and that expression and then add 6. So let's do it step by step. First, we're going to have this expression-- negative 5 plus something. So it's going to be negative 5 plus the quantity of 4 times x. Well, that's just going to be 4x. So it's going to be negative 5 plus 4x. So that's this expression up here. Now, take the product of negative 8, so were going to just take negative 8, and we're going to multiply the product of negative 8 and that expression. So we're going to take negative 8 and multiply it so that expression is this thing right over here. So if we say the product of negative 8 in that expression is going to be negative 8 times that expression, that expression is negative 5 plus 4x, so that's negative 8. The product of the two, so we could put a multiplication sign there, or we could just leave that out and implicitly it would mean multiplication, take the product of negative 8 and that expression and then add 6. So that would be then adding 6 right over here. So we could write it as negative 8 open parentheses negative 5 plus 4x and then add 6. Let's do one more. First, consider the expression the sum of 7 and-- so that's going to be 7 plus something-- and the product of negative 2 and x. The product of negative 2 and x is negative 2x. So it's 7 plus negative 2x. We could write that as 7 minus 2x. So this is equal to 7 minus 2x. These are the same expression. So now we're saying 4 plus the quantity of 2 times that expression? So it's going to be 4 plus some quantity. I'll put that in parentheses. The quantity of 2 times-- I'll do this in magenta or in yellow. 2 times that expression-- let me do this in blue-- that expression is this thing right over here. So 4 plus the quantity of 2 times that expression, 2 times 7 minus 2x. And we are done." + }, + { + "Q": "I don't understand at 1:25 what falls out of the pythagorean theorem\n", + "A": "The video states that the distance formula falls out of the the Pythagorean Theorem. The distance formula is A squared plus B squared equals C squared. C represents the length of the hypotenuse of a right triangle (The longest side). A or B can represent the length of one of the sides that make a right angle in a right triangle.", + "video_name": "9ASWQi14FlE", + "timestamps": [ + 85 + ], + "3min_transcript": "Point A is at negative 5 comma 5. So this is negative 5 right over here. This is 1, 2, 3, 4, 5. That's 5 right over there. So point A is right about there. So that is point A, just like that, at negative 5 comma 5. And then, it's a center of circle A, which I won't draw just yet, because I don't know the radius of circle A. Point B is at-- let me underline these in the appropriate color-- point B is at 3 comma 1. So 1, 2, 3 comma 1. So that's point B right over there, so center of circle B. Point P is at 0, 0, so it's right over there at the origin. And it is on circles A and B. Well, that's a big piece of information. Because that tells us, if this is on both circles, then that means that this is B's radius away from the point B from the center. And this tells us that it is circle A's radius away from its center, which is at point A. And so we can imagine-- let me draw a radius or the radius for circle A-- we now know since P sits on it, that this could be considered the radius for circle A. And you could use a distance formula, but what we'll see is that the distance formula is really just falling out of the Pythagorean theorem. So the distance formula tells us the radius right over here, this is just the distance between those two points. So the radius or the distance between those two points squared is going to be equal to our change in x values between A and P. So our change in x values, we could write it as negative 5 minus 0 squared. That's our change in x, negative 5 minus 0 squared, plus our change in y, 5 minus 0 squared, which is the length of the radius squared, is equal to negative 5 squared plus 5 squared. Or we could say that the radius is equal to the square root of, this is 25, this is 25, 50. 50, we can write as 25 times 2. So this is equal to the square root of 25 times the square root of 2, which is 5 times the square root of 2. So this distance right over here is 5 times the square root Now, I said this is just the same thing as the Pythagorean theorem. Why? Well, if we were to construct a right triangle right over here, then we can look at this distance. This distance would be the absolute value of negative 5 Or you could say, this is 0 minus negative 5. This distance right over here is 5." + }, + { + "Q": "At 8:15, when he proves T(V) is a subspace, is he proving simultaneously that R^m is a subspace as well??\n", + "A": "If R^n is a vector space (which it is, as it is easy to show), then yes, because T(x) = Ax maps R^n to R^m.", + "video_name": "hZ827mfh1Jo", + "timestamps": [ + 495 + ], + "3min_transcript": "So what is this equal to? Well, we know from our properties, our definition of linear transformations, the sum of the transformations of two vectors is equal to the transformation of the sum of their of vectors. Now, is the transformation of a plus b, is this a member of TV? Is it a member of our image? Well, a plus b is a member of V, and the image contains the transformation of all of the members of V. So the image contains the transformation of this guy. This guy, a plus b is a member of V. So you're taking a transformation of a member of V which, by definition, is in your image of V under T. So this is definitely true. Now, let's ask the next question. If I take a scalar multiple of some member of my image of V If I take the sum scalar, what is this equal to? By definition for linear transformation, this is the same thing as a transformation of the scalar times the vector. Now is this going to be a member of our image of V under T? Well we know that ca is definitely in V, right? That's from the definition of a subspace. This is definitely in V. And so, if this is in V, the transformation of this has to be in V's image under T. So this is in -- this is also a member of V. And obviously, you can set this equal to 0. The zero vector is a member of V, so any transformation of -- if you just put a 0 here, you'll get the zero vector. So the zero vector is definitely -- I don't care what this is, if you multiply it times 0, you are going to get the zero vector. also a member of TV. So we come on the result that T -- the image of V under T, is a subspace. Which is a useful result which we will be able to use later on. But this, I guess, might naturally lead to the question, what if we go -- everything we have been dealing with so far have been subsets, with the case of this triangle, or subspaces, in the case of V. But what if I were to take the image Rn under T, right? This is the image of Rn under T. Let's think about what this means. This means, what do we get when we take any member of Rn, what is the set of all of the vectors? Then when we take the transformation of all of the members of Rn, let me write this." + }, + { + "Q": "AT 8:00 shouldn't it be PLUS zero not MINUS?\n", + "A": "Actually during 5:43 and 5:53", + "video_name": "fyJkXBvcA2Q", + "timestamps": [ + 480 + ], + "3min_transcript": "Let's figure out what the Laplace transform of t squared is. And I'll do this one in green. And maybe we'll see a pattern emerge. The Laplace transform of t squared. Well, it equals 1/s times the Laplace transform of it's So what's it's derivative? Times the Laplace transform of 2t plus this evaluated at 0. Well, that's just 0. So this is equal to-- well we can just take this constant out. This is equal to 2/s times the Laplace transform of t. Well, what does that equal? We just solved it. So it's 2/s times 1/s squared. So it's equal to 2/s to the third. Well, let me just do t the third. And I think then you'll see the pattern. The pattern will emerge. The Laplace transform. And this is actually kind of fun. I recommend you do it. It's somehow satisfying. It's much more satisfying than integration by parts. So the Laplace transform of t to the third is 1/s times the Laplace transform of it's derivative, which is 3t squared. Which is-- take the constant out because it's a linear operator. 3/s times the Laplace transform of t squared. So it equals what? What's the Laplace transform of t squared? It's 2/s to the third. So this equals 3 times 2 over what? s to the fourth. And you can put a t/n here and use an inductive argument to And that general formula is-- I think you see the pattern here. Whatever my exponent is, the Laplace transform has an s in the denominator with one larger exponent. And then the numerator is the factorial of my exponent. So in general, and this is one more entry in our Laplace transform table. The Laplace transform of t to the nth power is equal to n factorial over s to the n plus 1. That's a parenthesis. I guess I didn't have to write those parenthesis. That just confuses it. But anyway, when you see this in a Laplace transform table, it seems intimidating. Oh boy, I have n's and I have n factorials and all of that. But it's just saying with this pattern we showed, t to the" + }, + { + "Q": "At 7:24, Sal said g(x) = 3 (1/3)^x\nCould this be simplified down to 1^x?\n", + "A": "No, you need to retain the base of 3, but you could simplify it to 3^(1-x) Like this: 3*(1/3)^x = (3^1)*(3^(-1))^x = (3^1)*(3^(-x)) = 3^(1-x) But it s harder to see what this will do than in the earlier form.", + "video_name": "gFdh_rE2XgU", + "timestamps": [ + 444 + ], + "3min_transcript": "would be a times r to the negative one. They tell us that g of negative one is going to be equal to nine. G of negative one is equal to nine. And so we could write this a times r to the negative one. That's the same thing as a over r is equal to nine, or we could multiply both sides by r, and we could say a is equal to 9r. Now let's use this other point. This other point, they tell us. They tell us that g of one, which would be the same thing as a times r to the first power or just a times r, that that is equal to one, or a times r is equal to one. So how can we use this information right here, a is equal to 9r and a times r is equal to one, to solve for a and r? Well, I have a little system here. We could just take this a and substitute it in right over here for a, and so we would get 9r for a. This first constraint tells us a must be equal to 9r. So 9r, instead of writing an a here, I'll write 9r, times r is equal to one. Or we could write, let me scroll down a little bit. We could write 9r squared is equal to one. Divide both sides by nine. R squared is equal to one over nine. And now to figure out r, you might wanna take the positive and negative square root of both sides, but they tell us that r is greater than zero, so we can just take the principal root of both sides and we get r is equal to 1/3. And then we could substitute this back into either one of these other two to figure out what a is. We know that a is equal to nine times r. So our exponential function could be written as g of x is equal to a, which is three, times r, which is 1/3, 1/3 to the x power." + }, + { + "Q": "\nAt 5:07 in the equation ar^-1, does -1 apply to r only or both a and r. I'm not sure because Sal didn't put any parentheses in.", + "A": "if no parentheses => applies only to one symbol which it is written after, (r)", + "video_name": "gFdh_rE2XgU", + "timestamps": [ + 307 + ], + "3min_transcript": "I gotta be careful here, I got a little bit. Every time you increase your x by one, you're decreasing your y. And here on the x axis, we're marking off every half. So every time you increase your x by one, you are decreasing your y. You are decreasing your y by four there, so that makes sense that the slope is negative four. So now let's think about what b is. So to figure out b, we could use either one of these points to figure out, given an x, what f of x is, and then we can solve for by. Let's try one, 'cause one is a nice, simple number. So we could write f of one, which would be negative four times one plus b. And they tell us that f of one is one, is equal to one. And so this part right over here, we could write that as negative four plus b is equal to one, and then we get b is equal to five. So we get f of x is equal to negative 4x plus five. Now, does that make sense that the y-intercept here is five? Well, you see that right over here. By inspection, you could have guessed, actually, that the y-intercept here is five, but now we've solved it. Maybe this was 5.00001 or something, but now we know for sure it's negative 4x plus five. Or another way you could've said it, if the slope is negative four, if this right over here is nine, you increase one in the x direction, you're gonna decrease four in the y direction, and that will get you to y is equal to five, so that is the y-intercept. But either way, we have figured out the linear function. Now let's figure out the exponential function. So here we could just use the two points to figure out these two unknowns. So, for example, let's try this first point. So g of negative one, which if we look would be a times r to the negative one. They tell us that g of negative one is going to be equal to nine. G of negative one is equal to nine. And so we could write this a times r to the negative one. That's the same thing as a over r is equal to nine, or we could multiply both sides by r, and we could say a is equal to 9r. Now let's use this other point. This other point, they tell us. They tell us that g of one, which would be the same thing as a times r to the first power or just a times r, that that is equal to one, or a times r is equal to one. So how can we use this information right here, a is equal to 9r and a times r is equal to one, to solve for a and r? Well, I have a little system here." + }, + { + "Q": "At 6:40 how to get from r^2=1/9 to r=1/3. What was done to get this answer. I don't get it. I'm confused. Could someone explain please?\n", + "A": "You need to take the square root of both sides. The square root of r^2 = r. The square root of 1/9 = 1/3.", + "video_name": "gFdh_rE2XgU", + "timestamps": [ + 400 + ], + "3min_transcript": "would be a times r to the negative one. They tell us that g of negative one is going to be equal to nine. G of negative one is equal to nine. And so we could write this a times r to the negative one. That's the same thing as a over r is equal to nine, or we could multiply both sides by r, and we could say a is equal to 9r. Now let's use this other point. This other point, they tell us. They tell us that g of one, which would be the same thing as a times r to the first power or just a times r, that that is equal to one, or a times r is equal to one. So how can we use this information right here, a is equal to 9r and a times r is equal to one, to solve for a and r? Well, I have a little system here. We could just take this a and substitute it in right over here for a, and so we would get 9r for a. This first constraint tells us a must be equal to 9r. So 9r, instead of writing an a here, I'll write 9r, times r is equal to one. Or we could write, let me scroll down a little bit. We could write 9r squared is equal to one. Divide both sides by nine. R squared is equal to one over nine. And now to figure out r, you might wanna take the positive and negative square root of both sides, but they tell us that r is greater than zero, so we can just take the principal root of both sides and we get r is equal to 1/3. And then we could substitute this back into either one of these other two to figure out what a is. We know that a is equal to nine times r. So our exponential function could be written as g of x is equal to a, which is three, times r, which is 1/3, 1/3 to the x power." + }, + { + "Q": "\nAt 0:23 Sal says \"literally\". Is that a math term I am not aware of or something?", + "A": "He means that it s condensed to basic numbers. Like a regular equation is 6743-2340 but a literal equation is (6000+700+40+3) - (2000+300+40)", + "video_name": "RRk5qLd__Ro", + "timestamps": [ + 23 + ], + "3min_transcript": "We've got 6,798 minus 3,359. So let's see how far we can get with the subtraction. So immediately, when we go into the ones place, we're going to try to subtract a 9 from an 8. So we immediately reach a little bit of a stumbling block. And to see what our options are here in terms of regrouping, I'm going to rewrite both of these numbers. So I'm going to rewrite 6,798 literally. So this is equal to 6,000. That's this right over here. That's 6,000, plus 700, plus 90, plus 8. Minus all of this. So I could subtract each of the places. So I could say this is going to be minus 3,000 minus 300 minus 50-- a 5 in a tens place So here, we're just explicitly showing what those place values represent. A 6 in the thousands place is 6,000. A 3 in the hundreds place is 300. Let's go back to our problem. We wanted to subtract a 9 from an 8. Well, that's a little bit of a stumbling block. But what if we could take some of the value from some of these and give it to the 8? In particular, we could go one place value up to the 90. And what if we were to take 10 from the 90? So let's do that. If we were to take 10 from the 90, then 90 becomes 80. But we don't want to change the value of the entire number. So we're going to give that 10 to this 8. We're essentially regrouping right over here. And then that 8 can become an 18. Notice, I did not change the value of the number. Instead of saying it's 6,000 plus 700 plus 90 plus 8, I'm just saying that it's 6,000 plus 700 plus 80 plus 18. Those are both going to give you 6,798. But now it becomes a little bit easier for us to actually subtract. Now, if we subtract, I have 18 minus 9, which is 9. I have 80-- not 90, now. I have 80 minus 50, which is 30. And these are all positive, so this is plus 9. This is a positive 30. 80 minus 50 is 30. I have 700 minus 300, which is 400. And I have 6,000 minus 3,000, which is 3,000. So this is literally going to be 3,000 plus 400 plus 30 plus 9, or 3,439. Now, how would you do it if you didn't want to write it out like this? And this is where you'll see a kind of shorthand notation." + }, + { + "Q": "In 0:32, How did he get 200cm squared for the height of the cereal box?\n", + "A": "It was the area of the front of the cereal box, so he did 20cm(height of box) times 10cm (width of front of box) which equals 200 cm which is the area of the front of the box.", + "video_name": "1iSBNSYhvIU", + "timestamps": [ + 32 + ], + "3min_transcript": "- [Voiceover] Let's see if we can figure out the surface area of this cereal box. And there's a couple of ways to tackle it. The first way is, well let's figure out the surface area of the sides that we can see, and then think about what the surface area of the sides that we can't see are and how they might relate, and then add them all together. So let's do that. So the front of the box is 20 centimeters tall and 10 centimeters wide. It's a rectangle, so to figure out its area we can just multiply 20 centimeters times 10 centimeters, and that's going to give us 200 centimeters. 200 centimeters, or 200 square centimeters, I should say. 200 square centimeters, that's the area of the front. And let me write it right over here as well, 200. Now we also know there's another side that has the exact same area as the front of the box, and that's the back of the box. And so let's write another 200 square centimeters for the back of the box. Now let's figure out the area of the top of the box. The top of the box, we see the box is three centimeters deep so this right over here is three centimeters. It's three centimeters deep and it's 10 centimeters wide. So the top of the box is gonna be three centimeters times 10 centimeters, which is 30 square centimeters of area. So that's the top of the box, 30 square centimeters. Well, the bottom of the box is gonna have the exact same area, we just can't see it right now, so that's gonna be another 30. Then we have two more sides, cuz this box has six sides, We have this side panel that is 20 centimeters tall, we see that the height of the box is 20 centimeters and three centimeters deep. So three times 20, let me write that a little bit neater. Three times 20, that's 20 centimeters right there. Three centimeters times 20 centimeters is going to give us 60 square centimeters. Now that's this side panel, but there's another side panel that has the exact same area that's on the other side of the box, so it's 60 centimeters squared for this side, and then another 60 for the corresponding side And now we can just add up all of these together. And so we get zero, this is going to be carry the one, or regroup the one, it's a 100, and then we have 500. So we get 580 square centimeters is the surface area of this box." + }, + { + "Q": "At 0:00, is the bead counter actually an 'Abacus'? Thank you if you answer.\n\n-TheAmericanBerserker\n", + "A": "Yes , at 0:00 , the bead counter is an abacus.-facepalm- Stop doing this just for badges.", + "video_name": "v3gdX07Q6qE", + "timestamps": [ + 0 + ], + "3min_transcript": "Hey, Sal Hello, Brit I picked this up at a garage sale and I know you like colours. I love colours! You wear colourful shirts everyday and I thought you might like this. I do. This is very kind. I like it, yeah. So what are you hoping to do with this thing? Well at minimum maybe just represent numbers. Like counts? Keep track of numbers? Yeah you know counting the number of days or? 1, 2, 3. Yeah I could imagine doing that. Oh ok. So moving the bead down is a number, great. And that there's, there's 10. 10 there, so maybe that'd be 20, 30, 40, 50, 60, 70, 80, 90, 100. Yeah you could count 100 beads to keep track of things. What if I need to count to 105 or 6? Do I need to buy another one? I don't even know where to get more than these. Um, that would be an option. I guess maybe that's not an option if you don't know where to get it. Let's see. Um. Well, you know the different colours, you know, just like we have different forms of currency. Maybe we can have each of these, each of these colours or maybe the columns they represent a different amount. But what if we had one of these red beads represent 10 of these blue beads on this first column. So then you could go, this would be 10. Or you could say that is 10? So there's two representations of 10 here. On, yeah the way we just worked it out here, yeah. Okay. And then 11 would be that. Okay, and this is 21. Yeah, you have two 10s and a 1. The wooden colour or whatever colour. The skintone bead is going to represent all of the red beads. Yeah if each of one of these. Yeah that's a good system. If each column represents 10 of the beads to the column to its right that could be interesting. Because if this represents 10 red beads. That's ten 10s That'd be equivalent, this is equivalent to 100 blue beads. Which would be 100 of the red beads. Which would be 1,000 of the blue beads. And this would be 10,000 of the blue beads. And this would be 100,000 of the other blue beads. This would be 1,000,000 of the blue beads. And we're going to be able to repesent all the numbers in between? 1 and say 1,000,000? I think we can. Let me just give you a number and see how you do. What about 15,003? So let's think a little bit about this. So let's try the big numbers first. So see each of these is 1 and each of these is 10. Each of these is 100. Each of these is 1,000. I don't have 15 of these. But these are 10,000. Each of these are 10,000 So this is one 10,000 And then I could do 5 thousands. 1, 2, 3, 4, 5. So this is 15 thousand. So 1 ten thousand, 5 thousands. Then 0 hundreds, 0 tens, and then throw a 3 there. So 15,003." + }, + { + "Q": "In discussing reflections at 4:38, why does not Sal place the reflection line through the origin? This approach would better help novice learners of the concept.\n", + "A": "I agree especially if you do not have a computer program to do these reflections - along the line y=x makes easy reflections, the line on the graph does not appear to be precisely defined, but it sort of looks like y = x + 1/2.", + "video_name": "XiAoUDfrar0", + "timestamps": [ + 278 + ], + "3min_transcript": "The point of rotation, actually, since D is actually the point of rotation that one actually has not shifted, and just 'til you get some terminology, the set of points after you apply the transformation this is called the image of the transformation. So, I had quadrilateral BCDE, I applied a 90-degree counterclockwise rotation around the point D, and so this new set of points this is the image of our original quadrilateral after the transformation. I don't have to just, let me undo this, I don't have to rotate around just one of the points that are on the original set that are on our quadrilateral, I could rotate around, I could rotate around the origin. I could do something like that. Notice it's a different rotation now. It's a different rotation. I could rotate around any point. Now let's look at another transformation, and that would be the notion of a reflection, You imagine the reflection of an image in a mirror or on the water, and that's exactly what we're going to do over here. If we reflect, we reflect across a line, so let me do that. This, what is this one, two, three, four, five, this not-irregular pentagon, let's reflect it. To reflect it, let me actually, let me actually make a line like this. I could reflect it across a whole series of lines. Woops, let me see if I can, so let's reflect it across this. Now, what does it mean to reflect across something? One way I imagine is if this was, we're going to get its mirror image, and you imagine this as the line of symmetry that the image and the original shape they should be mirror images across this line we could see that. Let's do the reflection. There you go, and you see we have a mirror image. This is this far away from the line. This, its corresponding point in the image This point over here is this distance from the line, and this point over here is the same distance but on the other side. Now, all of the transformations that I've just showed you, the translation, the reflection, the rotation, these are called rigid transformations. Once again you could just think about what does rigid mean in everyday life? It means something that's not flexible. It means something that you can't stretch or scale up or scale down it kind of maintains its shape, and that's what rigid transformations are fundamentally about. If you want to think a little bit more mathematically, a rigid transformation is one in which lengths and angles are preserved. You can see in this transformation right over here the distance between this point and this point, between points T and R, and the difference between their corresponding image points, that distance is the same. The angle here, angle R, T, Y, the measure of this angle over here," + }, + { + "Q": "Do units in this case kilometers (km) cancel each other out? At 3:06 Sal mentions km cancel km to just get hours.\n", + "A": "Yes, the km units cancel creating a final unit in hours.", + "video_name": "Q0tTfe2lKIc", + "timestamps": [ + 186 + ], + "3min_transcript": "I'll draw a bigger building, that's his school, and we know that this distance is 35, 35 kilometers and we also know that it took one and a half hours. One and a half hours, now he traveled at different rates for different distances. So he traveled some distance to the school bus so this is, or to the bus stop, so that's the bus stop, right over there, and we're seeing this distance to the bus stop, that's how much he covered by walking. So this right over here, this distance, right over here, that is W and the rest of the distance, he covered by the bus, so the rest of this distance, all of this distance right over there, that is going to be B. So what do we know? We know the distance covered by walking plus the distance covered by bus is going to be 35 kilometers, 35 kilometers here, this is the entire, that is the entire distance from home to school, so we know that W plus B, is equal to 35 kilometers and just with one equation, we're not going to be able to figure out what W and B are but we have another constraint. We know the total amount of time. So the total amount of time is going to be one and half hours, so we'll just write that over here. This is going to be 1.5, so what's the time traveled by, what's the time he walks? Let me write this over here, time time walking, we'll that's going to be the distance walking divided by the rate walking. So the distance walking is W kilometers W kilometers divided by his rate, the distance divided by your rate is gonna give you your time, so let's see, his rate is five kilometers per hour, five kilometers per hour and so you're gonna and if you divide by or if you have one over hours in the denominator, that's going to be the same thing, this is gonna be W over five hours, so the units work out. So his time walking is W over five, W over five and by that same logic, his time on the bus is going to be the distance on the bus divided by, divided by the average speed of the school bus, so this is going to be 60. This is all going to be in hours and now we can solve this system of equations. We have two linear equations with two unknowns. We should be able to find W and B that satisfy both of these. Now what's an easy thing to do? Let's see, if I can multiply this second equation by negative five, and I'm gonna, this is going to be a negative W here so it'll cancel out with this W up there. So let's do that, let's multiply the second equation by, I'm just gonna switch to one color here," + }, + { + "Q": "\nMr Sal At 3:43\nhow can square root of X-3 be greater or equal to 0\nif i plug 0 in place of x i would -3 which is wrong", + "A": "If x=0, then it would be sqrt(-3). Then would be an imaginary number. That s why domain comes in handy.", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 223 + ], + "3min_transcript": "And what's 1 over 0? I don't know what it is, so this is undefined. No one ever took the trouble to define what 1 over 0 should be. And they probably didn't do, so some people probably thought about what should be, but they probably couldn't find out with a good definition for 1 over 0 that's consistent with the rest of mathematics. So 1 over 0 stays undefined. So f of 0 is undefined. So we can't put 0 in and get a valid answer for f of 0. So here we say the domain is equal to -- do little brackets, that shows kind of the set of what x's apply. That's those little curly brackets, I didn't draw it that well. x is a member of the real numbers still, such that So here I just made a slight variation on what I had before. Before we said when f of x is equal to x squared that x is just any real number. Now we're saying that x is any real number except for 0. This is just a fancy way of saying it, and then these curly brackets just mean a set. Let's do a couple more ones. Let's say f of x is equal to the square root of x minus 3. So up here we said, well this function isn't defined when we get a 0 in the denominator. But what's interesting about this function? Can we take a square root of a negative number? Well until we learn about imaginary and complex numbers we can't. So here we say well, any x is valid here except for the x's So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the" + }, + { + "Q": "\nHey there! I'm new to Khan Academy and I've got a doubt regarding inequalities involving modulo, i.e, |x|.\n\nAt 5:58, Sal says that when |x|>=3 ---> x<=(-3) or x>=3, But what if x was positive, how then would x<=(-3)?\n\nThanks!", + "A": "First |x| represent the absolute value not modulo (modulo is a very different thing). As for your question, the solution is that x<=3 OR x>=3 so if x is positive, it satisfies the right half of the inequality and since it is an OR it doesn t need to satisfy the other side, likewise for a negative number <-3. If the solution was instead x<=-3 AND x>=3 then we would have a problem since anything that satisfied one half could not satisfy the other, thus resulting in no valid solution.", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 358 + ], + "3min_transcript": "So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive So, once again, x has to be less than negative 3 or x has to be greater than 3, so we have our domain. So we have it as x is a member of the reals -- I always forget. Is that the line? I forget, it's either a colon or a line. I'm rusty, it's been years since I've done this kind of stuff. But anyway, I think you get the point. It could be any real number here, as long as x is less than negative 3, less than or equal to negative 3, or x is greater than or equal to 3. Let me ask a question now. What if instead of this it was -- that was the denominator, this is all a separate problem up here. So now we have 1 over the square root of the absolute value of x minus 3. So now how does this change the situation? So not only does this expression in the denominator," + }, + { + "Q": "At 6:04 when hes discussing that x can < or = to -3, im lost. If x was -4 and then you subtracted 3 youd get -7. If x has to be = or > than 0 then i dont understand how it can be -3 or less. Some body please explain this.\n", + "A": "i understand now thank you", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 364 + ], + "3min_transcript": "So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive So, once again, x has to be less than negative 3 or x has to be greater than 3, so we have our domain. So we have it as x is a member of the reals -- I always forget. Is that the line? I forget, it's either a colon or a line. I'm rusty, it's been years since I've done this kind of stuff. But anyway, I think you get the point. It could be any real number here, as long as x is less than negative 3, less than or equal to negative 3, or x is greater than or equal to 3. Let me ask a question now. What if instead of this it was -- that was the denominator, this is all a separate problem up here. So now we have 1 over the square root of the absolute value of x minus 3. So now how does this change the situation? So not only does this expression in the denominator," + }, + { + "Q": "\nAt 6:21, Sal asks, \"Is 'such that' the colon or the line?\" I've always used those interchangeably. Is there actually a difference?", + "A": "Well I ve always used the pipe ( | ) symbol there (e.g { x in R | x \u00e2\u0089\u00a0 -2 } ) out of laziness, but Sal s comment in this video made me question that practice...", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 381 + ], + "3min_transcript": "So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive So, once again, x has to be less than negative 3 or x has to be greater than 3, so we have our domain. So we have it as x is a member of the reals -- I always forget. Is that the line? I forget, it's either a colon or a line. I'm rusty, it's been years since I've done this kind of stuff. But anyway, I think you get the point. It could be any real number here, as long as x is less than negative 3, less than or equal to negative 3, or x is greater than or equal to 3. Let me ask a question now. What if instead of this it was -- that was the denominator, this is all a separate problem up here. So now we have 1 over the square root of the absolute value of x minus 3. So now how does this change the situation? So not only does this expression in the denominator," + }, + { + "Q": "\n4:28 It is to my understanding that if f(x)=radical x-3 then the x cannot be any less than three, which means x greater or EQUAL TO 0* cannot be so it is just x is ONLY greater than or equal to zero. If I am missing something please let me know.\n*just putting an emphasis", + "A": "Sal said it was greater than or equal to zero, so I don t know what you mean", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 268 + ], + "3min_transcript": "And what's 1 over 0? I don't know what it is, so this is undefined. No one ever took the trouble to define what 1 over 0 should be. And they probably didn't do, so some people probably thought about what should be, but they probably couldn't find out with a good definition for 1 over 0 that's consistent with the rest of mathematics. So 1 over 0 stays undefined. So f of 0 is undefined. So we can't put 0 in and get a valid answer for f of 0. So here we say the domain is equal to -- do little brackets, that shows kind of the set of what x's apply. That's those little curly brackets, I didn't draw it that well. x is a member of the real numbers still, such that So here I just made a slight variation on what I had before. Before we said when f of x is equal to x squared that x is just any real number. Now we're saying that x is any real number except for 0. This is just a fancy way of saying it, and then these curly brackets just mean a set. Let's do a couple more ones. Let's say f of x is equal to the square root of x minus 3. So up here we said, well this function isn't defined when we get a 0 in the denominator. But what's interesting about this function? Can we take a square root of a negative number? Well until we learn about imaginary and complex numbers we can't. So here we say well, any x is valid here except for the x's So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the" + }, + { + "Q": "\nAround 6:10, I don't undersand how x could be smaller than -3.\nSay x= -4, than we get square root of -4 -3 = square root of -7, which isn't possible, or at least doesn't make sense to me..", + "A": "It is because you are taking the absolute value of x. So, for x = -4, we would get: \u00e2\u0088\u009a( |x| - 3) = \u00e2\u0088\u009a( |-4| - 3) = \u00e2\u0088\u009a( 4 - 3) = \u00e2\u0088\u009a( 1) = \u00c2\u00b1 1 It is only when x is between -3 and 3 that we would get a negative number under the square root.", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 370 + ], + "3min_transcript": "So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive So, once again, x has to be less than negative 3 or x has to be greater than 3, so we have our domain. So we have it as x is a member of the reals -- I always forget. Is that the line? I forget, it's either a colon or a line. I'm rusty, it's been years since I've done this kind of stuff. But anyway, I think you get the point. It could be any real number here, as long as x is less than negative 3, less than or equal to negative 3, or x is greater than or equal to 3. Let me ask a question now. What if instead of this it was -- that was the denominator, this is all a separate problem up here. So now we have 1 over the square root of the absolute value of x minus 3. So now how does this change the situation? So not only does this expression in the denominator," + }, + { + "Q": "\nAt 9:12 why can't we make the trasformation R2+R3 and then R3+R2 to recover two zero rows as in the case lambda=3? This way at the end I would get an eigenspace that is equal to the span of the vectors (-1,1,0) and (-1,0,1). And if you sum them up you actually get the same vector Sal recovered in the video! Are they simply two \"equivalent\" solutions or is there something wrong about mine?\nThanks in advance for any answer :D", + "A": "You can replace any row with a scalar multiple of itself, or with itself minus another row (or equivalently, another row minus itself ). There are different paths to the same result. If you don t interchange any columns, you should end up with the same equation Sal got, otherwise you made an error.", + "video_name": "3Md5KCCQX-0", + "timestamps": [ + 552 + ], + "3min_transcript": "And all the other things don't change. Minus 2, minus 2, 1. Minus 2, minus 2 and 1. And then that times vectors in the eigenspace that corresponds to lambda is equal to minus 3, is going to be equal to 0. I'm just applying this equation right here which we just derived from that one over there. So, the eigenspace that corresponds to lambda is equal to minus 3, is the null space, this matrix right here, are all the vectors that satisfy this equation. So what is-- the null space of this is the same thing as the null space of this in reduced row echelon form So let's put it in reduced row echelon form. So the first thing I want to do, I'm going to keep my first row the same. I'm going to write a little bit smaller than I normally do because I think I'm going to run out of space. So minus 2, minus 2, minus 2. I will skip some steps. Let's just divide the first row by minus 2. So we get 1, 1, 1. And then let's replace this second row with the second row plus this version of the first row. So this guy plus that guy is 0 minus 5 plus minus-- or let me Let me replace it with the first row minus the second row. So minus 2 minus minus 2 is 0. Minus 2 minus minus 5 is plus 3. And then minus 2 minus 1 is minus 3. And then let me do the last row in a different color for fun. And I'll do the same thing. I'll do this row minus this row. So minus 2 minus minus 2 is a 0. Minus 2 plus 2. Minus 2 minus 1 is minus 3. And then we have minus 2 minus minus 5. So that is 3. Now let me replace-- and I'll do it in two steps. So this is 1, 1, 1. I'll just keep it like that. And actually, well let me just keep it like that. And then let me replace my third row with my third row plus my second row. It'll just zero out. If you add these terms, these all just become 0. That guy got zeroed out. And let me take my second row and divide it by 3. So this becomes 0, 1, minus 1. And I'm almost there. I'll do it in orange. So let me replace my first row with my first row minus my second row. So this becomes 1, 0, and then 1 minus minus 1 is 2. 1 minus minus 1 is 2. And then in the second row is 0, 1, minus 1." + }, + { + "Q": "At 2:56 while dividing 54 by 4, why did 4 went to twenty while it is written as 2.0? I see that it is number two because the decimal is just next to it on the right hand side. In the other words 2.0 is equals to 2 x 10^0. While the number twenty is equals to 2 x 10^1. Also, the number twenty is equals to 20.0, which looks different than the twenty that is written when we added a zero at 2:54.\n", + "A": "Ah, I believe that Sal was carrying the decimal point down so you can see where it would be.", + "video_name": "jTCZfMMcHBo", + "timestamps": [ + 176, + 174 + ], + "3min_transcript": "So how much does the amount of money I make change when I work a certain number of hours, when my hours worked change by a certain amount. So let's just take some data points here. We could take really any of these data points, I'll take some of the smaller numbers. So let's say if when I go from four to eight hours, so my change in x is going to be what? If I go from four to eight, might change in x is going to be eight minus four, four hours, right? So this is going to be my change in x. I'm just picking these two points, I could have picked four and forty if I wanted, but the math would become more complicated. But how much does the amount of money earn change if I go from four hours to eight hours? Well, I go from $54 to $108, so the difference in the amount of money I make is $108 minus $54. Well, that's going to be $108 minus $54, that's just $54. And then what was the change in the amount of hours I worked? Well, the change in the hours I worked was four hours. So, if I work four more hours, I make 54 more dollars. Let me put a little equal sign there. So what is 54 divided by four? So four goes into 54-- looks like there's going to be decimal here-- four goes into five one time, one times four is four. Subtract, you get five minus four is one, bring down this four you get 14. Four goes into 14 three times, three times four is 12. Fourteen minus 12 is two, bring down a 0 right here, four goes into 20 five times. And of course you have this decimal right here. Five times four is 20. So this is equal to 13.5, but since we're talking in terms of dollars, maybe say $13.50, because that's our numerator, right? This is money earned, dollars per hour, because that's our denominator, dollars per hour. So that essentially answers our question. What does the slope represent in this situation? It represents the hourly wage for working at wherever this might be. Frankly, for this problem, you didn't even have to take two We could have said hey, if you work four hours and make $54, 54 divided by four is 13.50. Or we could have said hey, if we work eight hours, we get $108, 108 divided by eight is 13.50. So you didn't even have to take two data points here, you could have just taken any of these numbers divided by any of these numbers. But hopefully we also learned a little bit about what slope is." + }, + { + "Q": "\nAt 4:24, 3=2 has no solutions but what about if x =0? Then 0=0 would be one solution.", + "A": "If you are asking if X=0 can be a solution to the middle equation, the answer is no. It really has no solution. To verify whether or not X=0 is solution, substitute into the equation. -7x + 3 = 2x + 2 - 9x -7(0) + 3 = 2(0) + 2 - 9(0) 0 + 3 = 0 + 2 + 0 3 = 2 This is false, so X=0 is not a solution. Hope this helps.", + "video_name": "qsL_5Y8uWPU", + "timestamps": [ + 264 + ], + "3min_transcript": "If we subtract 2 from both sides, we are going to be left with-- on the left hand side we're going to be left with negative 7x. And on the right hand side, you're going to be left with 2x. This is going to cancel minus 9x. 2x minus 9x, If we simplify that, that's negative 7x. You get negative 7x is equal to negative 7x. And you probably see where this is going. This is already true for any x that you pick. Negative 7 times that x is going to be equal to negative 7 times that x. So we already are going into this scenario. But you're like hey, so I don't see 13 equals 13. Well, what if you did something like you divide both sides by negative 7. At this point, what I'm doing is kind of unnecessary. You already understand that negative 7 times some number is always going to be negative 7 times that number. But if we were to do this, we would get x is equal to x, and then we could subtract x from both sides. zero, which is true for any x that you pick. Zero is always going to be equal to zero. So any of these statements are going to be true for any x you pick. So for this equation right over here, we have an infinite number of solutions. Let's think about this one right over here in the middle. So once again, let's try it. I'll do it a little bit different. I'll add this 2x and this negative 9x right over there. So we will get negative 7x plus 3 is equal to negative 7x. So 2x plus 9x is negative 7x plus 2. Well, let's add-- why don't we do that in that green color. Let's do that in that green color. Plus 2, this is 2. Now let's add 7x to both sides. Well if you add 7x to the left hand side, you're just going to be left with a 3 there. And if you add 7x to the right hand side, going to be left with a 2 there. So all I did is I added 7x. I added 7x to both sides of that equation. And now we've got something nonsensical. I don't care what x you pick, how magical that x might be. There's no way that that x is going to make 3 equal to 2. So in this scenario right over here, we have no solutions. There's no x in the universe that can satisfy this equation. Now let's try this third scenario. So once again, maybe we'll subtract 3 from both sides, just to get rid of this constant term. So we're going to get negative 7x on the left hand side. On the right hand side, we're going to have 2x minus 1. And now we can subtract 2x from both sides. To subtract 2x from both sides, you're going to get-- so subtracting 2x," + }, + { + "Q": "At 3:20 the symbol Sal used, is that the sign of infinite?\n", + "A": "Yes \u00e2\u0088\u009e is the infinite sign. Great educated guess!", + "video_name": "qsL_5Y8uWPU", + "timestamps": [ + 200 + ], + "3min_transcript": "and you were to get something like 3 equals 5, and you were to ask yourself the question is there any x that can somehow magically make 3 equal 5, no. No x can magically make 3 equal 5, so there's no way that you could make this thing be actually true, no matter which x you pick. So if you get something very strange like this, this means there's no solution. On the other hand, if you get something like 5 equals 5-- and I'm just over using the number 5. It didn't have to be the number 5. It could be 7 or 10 or 113, whatever. And actually let me just not use 5, just to make sure that you don't think it's only for 5. If I just get something, that something is equal to itself, which is just going to be true no matter what x you pick, any x you pick, this would be true for. Well, then you have an infinite solutions. So with that as a little bit of a primer, let's try to tackle these three equations. So over here, let's see. Maybe we could subtract. If we want to get rid of this 2 here on the left hand side, If we subtract 2 from both sides, we are going to be left with-- on the left hand side we're going to be left with negative 7x. And on the right hand side, you're going to be left with 2x. This is going to cancel minus 9x. 2x minus 9x, If we simplify that, that's negative 7x. You get negative 7x is equal to negative 7x. And you probably see where this is going. This is already true for any x that you pick. Negative 7 times that x is going to be equal to negative 7 times that x. So we already are going into this scenario. But you're like hey, so I don't see 13 equals 13. Well, what if you did something like you divide both sides by negative 7. At this point, what I'm doing is kind of unnecessary. You already understand that negative 7 times some number is always going to be negative 7 times that number. But if we were to do this, we would get x is equal to x, and then we could subtract x from both sides. zero, which is true for any x that you pick. Zero is always going to be equal to zero. So any of these statements are going to be true for any x you pick. So for this equation right over here, we have an infinite number of solutions. Let's think about this one right over here in the middle. So once again, let's try it. I'll do it a little bit different. I'll add this 2x and this negative 9x right over there. So we will get negative 7x plus 3 is equal to negative 7x. So 2x plus 9x is negative 7x plus 2. Well, let's add-- why don't we do that in that green color. Let's do that in that green color. Plus 2, this is 2. Now let's add 7x to both sides. Well if you add 7x to the left hand side, you're just going to be left with a 3 there. And if you add 7x to the right hand side," + }, + { + "Q": "5:40 Why that line is called secant line?\n", + "A": "A secant line is a line that intersects a curve of some sort, at two points. A secant line is what we use to find average rates of change.", + "video_name": "f4MYCepzLyQ", + "timestamps": [ + 340 + ], + "3min_transcript": "it still is positive, but it seems a lot less positive. So it looks like the rate of change. It looks like the rate of change is actually changing here. In this case it looks like the line is getting steeper and steeper and steeper as time goes by. Or another way to think about it is, our rate of change of distance with respect to time is Increasing its not just constant. It's not a constant slope. The slope is increasing as time increases. So for something like this, how do we think about rate of change? How do we think about rate of change of one variable with respect to another, in particular distance with respect to time? Well later on in your math careers you'll find out that this is actually what most of differential calculus is all about and you will get to differential calculus. But for our purposes, we have another tool at our disposal and this is actually a good foundation for the calculus that you will learn in the future. This is the notion of average rate of change. Let me write this down average rate of change I already hopefully gave you an argument Why it's very hard you'll need actually calculus to figure out the Instantaneous rate of change the rate of change Right at that point the slope of the tangent line that you're going to need a calculus But to figure out the average rate of change between two points We can use very similar tools that we use to figure out the slope of a line. So for example: We could figure out the average rate of change between any two points on this curve. So for example: We could say the average rate of change from when we go from t equals 0 to t equals 3 That's going to be the slope of this secant line So let me draw that. So let me do this in a more fun color. I'll do it in this color All right So as I said the actual Instantaneous rate of change is constantly changing. In this case it's increasing, and we'll need calculus for that. But now we could think about average rate of change which would be the slope of the line that connects these two points. an average rate of change. As we see, the actual curve its rate of change is lower earlier on, and then it's rate of change is higher as we get closer and closer to 3 seconds. But the average rate of change is going to be the slope of this line right over here. And we could think about that. This is going to be our change. So the slope of this line, is going to be a change in distance over change in time. An average rate of change especially when you're talking about a curve like this it depends on what starting and ending point This is the average rate of change for the first 3 seconds, is going to be- Well what is our distance right at the third second? Well, it's going to be d of 3. Now what was our initial distance? It's going to be d of zero. So this expression right over here. It's going to give me my change in distance. It's going to give me my change in distance right over here Let me write. It actually maybe I could do it this way So I could do it over here. So our change in distance is this." + }, + { + "Q": "\nAt 0:50 what does sal mean by with respect to time?", + "A": "Sal meant . what will be the change in distance with respect to time . Example :- you car moves with a speed of 80/kmph , so what do you mean by this, your car moves with a speed of 80 km, per hour and that is your speed . if you want to write it in meter/ sec it would be 22.2/sec . =)", + "video_name": "f4MYCepzLyQ", + "timestamps": [ + 50 + ], + "3min_transcript": "definitions for the function D over here over here d of t is equal to 3 t plus 1. We could imagine the d could represent distance as a function of time. Distance measured in meters and time measured in seconds. So over here when time is 0 right when we're starting our distance is going to be 1 meter. After 1 second has gone by our distance is now going to be, so 3 times 1 plus 1 is going to be 4. Our distance now is going to be 4 is going to be 4 meters. After 2 seconds our distance is going to be 3 times two is 6 plus 1 it's going to be 7 it's going to be 7 meters. So given this definition of D of T this function definition. What is the rate of change of distance with respect to time and let me write it this way. What is the change in distance the rate of change of distance with respect to time which people sometimes called speed. Well, what is this going to be? Let's just take two points Let's just say the change in distance over change in time when time goes from time equals 0 to time equals 1 So over here our change in time is equal to 1 our change in time is 1 and what's our change in distance? Well our change in distance when our time increased by 1 our distance increases by 3 it goes from 1 meters to 4 meters. So our change in distance is equal to 3. So it's going to be equal to 3 over 1 or just 3. If we wanted the units it would be 3 meters, every 1 second. Now let's think about it, does that change if we pick any other two points? what if we were to say between 1 second and 2 seconds so between 1 second and 2 seconds our change in time is 1 second and then our change in distance is So once again our change in distance over change in time is 3 meters per second. This is all review and you might recognize, we pick any two points on this line here and we're going to have the same rate of change of distance with respect to time. In fact that's what defines a line. Or one of the ways to think about a line or a linear function is that the rate of change of one variable with respect to the other one, is constant. In this particular one we're talking about the rate of change of the vertical variable with respect to the horizontal one. We're talking about the slope of the line. This is the slope. This line has a slope of 3. That's what defines a line or one of the things that defines a line is the slope between any two points is going to be exactly 3. Just as a little bit of review from other Algebra you've seen before. You can even pick it out in the function definition." + }, + { + "Q": "\nI still don't understand why at 1:36 you have to multiply both sides by Pi to simplify the equation.", + "A": "He multiplied both sides by the reciprocal.", + "video_name": "tVcasOt55Lc", + "timestamps": [ + 96 + ], + "3min_transcript": "I have a circle here whose circumference is 18 pi. So if we were to measure all the way around the circle, we would get 18 pi. And we also have a central angle here. So this is the center of the circle. And this central angle that I'm about to draw has a measure of 10 degrees. So this angle right over here is 10 degrees. And what I'm curious about is the length of the arc that subtends that central angle. So what is the length of what I just did in magenta? And one way to think about it, or actually maybe the way to think about it, is that the ratio of this arc length to the entire circumference-- let me write this down-- should be the same as the ratio of the central angle to the total number of angles if you were So let's just think about that. We know the circumference is 18 pi. We're looking for the arc length. I'm just going to call that a. a for arc length. That's what we're going to try to solve for. We know that the central angle is 10 degrees. So you have 10 degrees over 360 degrees. So we could simplify this by multiplying both sides by 18 pi. And we get that our arc length is equal to-- well, 10/360 is the same thing as 1/36. So it's equal to 1/36 times 18 pi, so it's 18 pi over 36, which is the same thing as pi/2. to be pi/2, whatever units we're talking about, long. Now let's think about another scenario. Let's imagine the same circle. So it's the same circle here. Our circumference is still 18 pi. There are people having a conference behind me or something. That's why you might hear those mumbling voices. But this circumference is also 18 pi. But now I'm going to make the central angle an obtuse angle. So let's say we were to start right over here. This is one side of the angle. I'm going to go and make a 350 degree angle. So I'm going to go all the way around like that. So this right over here is a 350 degree angle. And now I'm curious about this arc that subtends this really huge angle. So now I want to figure out this arc length-- so all of this." + }, + { + "Q": "\nAt 0:51 he said 9/5 is equals to 1 4/5, but why?", + "A": "9/5 = 5/5 + 4/5 = 1 + 4/5 = 1 4/5. Have a blessed, wonderful new year!", + "video_name": "-lUEWEEpmIo", + "timestamps": [ + 51 + ], + "3min_transcript": "Let's see if we can figure out what 30% of 6 is. So one way of thinking about 30%-- this literally means 30 per 100. So you could view this as 30/100 times 6 is the same thing as 30% of 6. Or you could view this as 30 hundredths times 6, so 0.30 times 6. Now we could solve both of these, and you'll see that we'll get the same answer. If you do this multiplication right over here, 30/100-- and you could view this times 6/1-- this is equal to 180/100. We can simplify. We can divide the numerator and the denominator by 10. And then we can divide the numerator and the denominator by 2. And we will get 9/5, which is the same thing as 1 and 4/5. And then if we wanted to write this as a decimal, 4/5 is 0.8. And if you want to verify that, you and there's going to be a decimal. So let's throw some decimals in there. It goes into 4 zero times. So we don't have to worry about that. It goes into 40 eight times. 8 times 5 is 40. Subtract. You have no remainder, and you just have 0's left here. So 4/5 is 0.8. You've got the 1 there. This is the same thing as 1.8, which you would have gotten if you divided 5 into 9. You would've gotten 1.8. So 30% of 6 is equal to 1.8. And we can verify it doing this way as well. So if we were to multiply 0.30 times 6-- let's do that. And I could just write that literally as 0.3 times 6. Well, 3 times 6 is 18. I have only one digit behind the decimal amongst both of these numbers that I'm multiplying. I only have the 3 to the right of the decimal. So I'm only going to have one number to the right of the decimal here. So I just count one number. It's going to be 1.8. So either way you think about it or calculate it, 30% of 6" + }, + { + "Q": "\nAt 0:19 why does Sal say that we can get a 1 coefficient? Why do we need to get a 1 coefficient?", + "A": "Your 1st step in factoring should always be to look for and remove the greatest common factor. The reason this should be your 1st step is that it makes doing any other factoring technique easier to do. Think of it has cleaning out the clutter, all the other numbers get smaller and easier to work with. And, you have to factor it out sometime, so do it as your 1st step and make the rest of the work easier.", + "video_name": "R-rhSQzFJL0", + "timestamps": [ + 19 + ], + "3min_transcript": "We're asked to factor 35k squared plus 100k, minus 15. And because we have a non-1 coefficient out here, the best thing to do is probably to factor this by grouping. But before we even do that, let's see if there's a common factor across all of these terms, and maybe we can get a 1 coefficient, out there. If we can't get a 1 coefficient, we'll at least have a lower coefficient here. And if we look at all of these numbers, they all look divisible by 5. In fact their greatest common factor is 5. So let's at least factor out a 5. So this is equal to 5 times-- 35k squared divided by 5 is 7k squared. 100k divided by 5 is 20k. And then negative 15 divided by 5 is negative 3. So we were able to factor out a 5, but we still don't have a 1 coefficient here, so we're still going to have to factor by grouping. But at least the numbers here are smaller so it'll be easier to think about it in terms of finding numbers whose product sum is equal to 20. So let's think about that. Let's figure out two numbers that if I were to add them, or even better if I were to take their product, I get 7 times negative 3, which is equal to negative 21. And if I were to take their sum, if I add those two numbers, it needs to be equal to 20. Now, once again, because their product is a negative number, that means they have to be of different signs, so when you add numbers of different signs, you could view it as you're taking the difference of the positive versions. So the difference between the positive versions of the number has to be 20. So the number that immediately jumps out is we're probably going to be dealing with 20 and 21, and 1 will be the negative, because we want to get to a positive 20. So let's think about it. So if we think of 20 and negative 1, their product is Sorry. If we take 21 and negative 1, their product is negative 21. 21 times negative 1 is negative 21. and if you take their sum, 21 plus negative 1, that is equal to 20. So these two numbers right there fit the bill. Now, let's break up this 20k right here into a 21k and a negative 1k. So let's rewrite the whole thing. We have 5 times 7k squared, and I'm going to break this 20k into a-- let me do it in this color right here-- I'm going to break that 20k into a plus 21k, minus k. Or you could say minus 1k if you want. I'm using those two factors to break it up. And then we finally have the minus 3 right there. Now, the whole point of doing that is so that we can now factor each of the two groups." + }, + { + "Q": "\nI'm really confused. Where does Sal get the 21 from @1:18? O_O", + "A": "He multiplies 7 by -3. He factored the polynomial, then applied the product/sum theorem for quadratic polynomials to get the answers for their roots.", + "video_name": "R-rhSQzFJL0", + "timestamps": [ + 78 + ], + "3min_transcript": "We're asked to factor 35k squared plus 100k, minus 15. And because we have a non-1 coefficient out here, the best thing to do is probably to factor this by grouping. But before we even do that, let's see if there's a common factor across all of these terms, and maybe we can get a 1 coefficient, out there. If we can't get a 1 coefficient, we'll at least have a lower coefficient here. And if we look at all of these numbers, they all look divisible by 5. In fact their greatest common factor is 5. So let's at least factor out a 5. So this is equal to 5 times-- 35k squared divided by 5 is 7k squared. 100k divided by 5 is 20k. And then negative 15 divided by 5 is negative 3. So we were able to factor out a 5, but we still don't have a 1 coefficient here, so we're still going to have to factor by grouping. But at least the numbers here are smaller so it'll be easier to think about it in terms of finding numbers whose product sum is equal to 20. So let's think about that. Let's figure out two numbers that if I were to add them, or even better if I were to take their product, I get 7 times negative 3, which is equal to negative 21. And if I were to take their sum, if I add those two numbers, it needs to be equal to 20. Now, once again, because their product is a negative number, that means they have to be of different signs, so when you add numbers of different signs, you could view it as you're taking the difference of the positive versions. So the difference between the positive versions of the number has to be 20. So the number that immediately jumps out is we're probably going to be dealing with 20 and 21, and 1 will be the negative, because we want to get to a positive 20. So let's think about it. So if we think of 20 and negative 1, their product is Sorry. If we take 21 and negative 1, their product is negative 21. 21 times negative 1 is negative 21. and if you take their sum, 21 plus negative 1, that is equal to 20. So these two numbers right there fit the bill. Now, let's break up this 20k right here into a 21k and a negative 1k. So let's rewrite the whole thing. We have 5 times 7k squared, and I'm going to break this 20k into a-- let me do it in this color right here-- I'm going to break that 20k into a plus 21k, minus k. Or you could say minus 1k if you want. I'm using those two factors to break it up. And then we finally have the minus 3 right there. Now, the whole point of doing that is so that we can now factor each of the two groups." + }, + { + "Q": "What is the difference between a square root and a principle square root? 0:49 3:00\n", + "A": "A square root can be negative or positive, but the principle square root is always positive.", + "video_name": "s03qez-6JMA", + "timestamps": [ + 49, + 180 + ], + "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i." + }, + { + "Q": "\nAt 1:50 to 1:57, is the reason why that following that logic you would get i^infinity 2\u00e2\u0088\u009a13 or am I missing something?", + "A": "I m confused where you got the idea of i^infinity from. Sal is explaining why sqrt(+52) cannot be sqrt(-1) \u00c3\u0097 sqrt(-52). When you simplify this, you end up with i \u00c3\u0097 i\u00c3\u0097sqrt(52), which is -sqrt(52). This is not the correct answer; therefore it doesn t work.", + "video_name": "s03qez-6JMA", + "timestamps": [ + 110, + 117 + ], + "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i." + }, + { + "Q": "At, 2:12 Sal says the property does not work if both nos. are negative. I wanted to know why?\n", + "A": "If both numbers are negative, you get something like what he showed in the video. To recap: if sqrt(1) = sqrt(-1) * sqrt(-1) <-- both numbers are negative then 1 = -1 <-- you already know sqrt(1) = 1, and sqrt(x) * sqrt(x) = x As you can see, it just doesn t work. Because of this, both numbers can t be negative.", + "video_name": "s03qez-6JMA", + "timestamps": [ + 132 + ], + "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i." + }, + { + "Q": "\nAt 2:03 , Sal says u cant separate a principal square root number into the product of two numbers with the principal square root of a negative number. But you can, if you take the only the square root (not principal square root ) of the number ,say 4 , right? Cuz, u'll end up with i^2 sq.root(4) which\nalso can be written as [ -sq.root(4) ] which will give you the same two answers. Jst need a confirmation. Thanks.", + "A": "I can understand what you are trying to say, but -2 x -2 = 4 not -4, likewise 2 x 2 is not = -4 hopefully this helps if you still need the question answered after 2 years.", + "video_name": "s03qez-6JMA", + "timestamps": [ + 123 + ], + "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i." + }, + { + "Q": "When you are answering the question for credit on a test or something, would you write the constraints explained at 3:12 along with your answer?\n", + "A": "yes, you would its part of the answer", + "video_name": "XChok8XlF90", + "timestamps": [ + 192 + ], + "3min_transcript": "factor this top expression over here\u2014which was the expression for the width\u2014this is of the form a-squared minus b-squared, where b-squared is 9. So this is going to be the same thing as a plus the square root of 9 times a minus the square root of 9. So this is a plus 3 times a minus 3 and I just recognized that from just the pattern; if you ever see something a-squared minus b-squared, it's a plus b times a minus b and you can verify that for yourself; multiply this out, you'll get a-squared minus b-squared. So, the width can be factored into a plus 3 times a minus 3. Let's see if we can do something for the denominator. So here, if we want to factor this out, we have to think of two numbers that when we add them, I get positive 6, and when I take their product, I get negative 27. Let's see, if I have positive 9 and negative 3, that would work. 9 times a is 9a, a times negative 3 is negative 3a, when you add those two middle terms together, you'll get 6a, just like that and then, 9 times negative 3 is negative 27\u2014of course the a times a is a squared. So I've factored the two expressions and let's see if we can simplify it. And before we simplify it, because when we simplify it we lose information, let's just remember what are allowable a's here, so we don't lose that information. Are there any a values here that will make this expression undefined? Well, any a value that makes the denominator zero will make this undefined. So a cannot be equal to negative 9 or 3, 'cause if a was negative 9 or 3, then the denominator would be zero; this expression would be undefined. we don't want to change this domain; we don't want to allow things that weren't allowable to begin with, so let's just remember this right over here. Now with that said, now that we've made this constraint, we can simplify it more; we can say, look we have an a minus 3 in the numerator and we have an a minus 3 in the denominator and we're assuming that a is not going to be equal to 3, so it's not like we're dividing zero over zero. So a will not be equal to 3, any other number; this will be an actual number, you divide the numerator and denominator by that same value, and we are left with a plus 3 over a plus 9 and the constraint here\u2014we don't want to forget the constraints on our domain\u2014 a cannot equal negative 9 or 3. And it's important that we write this here, because over here we lost the information that a could not be equal to 3, but in order for this to really be the same thing as this thing over here, when a was equal to 3 it wasn't defined" + }, + { + "Q": "\nIs the constraint a \u00e2\u0089\u00a0 -9 at 3:47 really necessary? The simplified expression is already undefined at a = -9, so it looks as if it doesn't have to be restated.\n\nAdditionally, the original context of the word problem provides a rectangle whose width and length are rational expressions. Can't you not have a negative number in the simplified expression anyway?", + "A": "1) Yes, it s obvious. 2) Set the area > 0. You get a^4+6a^3-36a^2-54a+261 > 0", + "video_name": "XChok8XlF90", + "timestamps": [ + 227 + ], + "3min_transcript": "9 times a is 9a, a times negative 3 is negative 3a, when you add those two middle terms together, you'll get 6a, just like that and then, 9 times negative 3 is negative 27\u2014of course the a times a is a squared. So I've factored the two expressions and let's see if we can simplify it. And before we simplify it, because when we simplify it we lose information, let's just remember what are allowable a's here, so we don't lose that information. Are there any a values here that will make this expression undefined? Well, any a value that makes the denominator zero will make this undefined. So a cannot be equal to negative 9 or 3, 'cause if a was negative 9 or 3, then the denominator would be zero; this expression would be undefined. we don't want to change this domain; we don't want to allow things that weren't allowable to begin with, so let's just remember this right over here. Now with that said, now that we've made this constraint, we can simplify it more; we can say, look we have an a minus 3 in the numerator and we have an a minus 3 in the denominator and we're assuming that a is not going to be equal to 3, so it's not like we're dividing zero over zero. So a will not be equal to 3, any other number; this will be an actual number, you divide the numerator and denominator by that same value, and we are left with a plus 3 over a plus 9 and the constraint here\u2014we don't want to forget the constraints on our domain\u2014 a cannot equal negative 9 or 3. And it's important that we write this here, because over here we lost the information that a could not be equal to 3, but in order for this to really be the same thing as this thing over here, when a was equal to 3 it wasn't defined right over there; a cannot be equal to 3. Hopefully you found that useful." + }, + { + "Q": "what is that sound at 3:40 ?\n", + "A": "some one laughing", + "video_name": "X2jVap1YgwI", + "timestamps": [ + 220 + ], + "3min_transcript": "of space-- 95 times 0.15. 5 times 5 is 25, 9 times 5 is 45 plus 2 is 47, 1 times 95 is 95, bring down the 5, 12, carry the 1, 15. And how many decimals do we have? 1, 2. 15.25. Actually, is that right? I think I made a mistake here. See 5 times 5 is 25. 5 times 9 is 45, plus 2 is 47. And we bring the 0 here, it's 95, 1 times 5, 1 times 9, then we add 5 plus 0 is 5, 7 plus 5 is 12-- oh. I made a mistake. So I'll ask you an interesting question? How did I know that 15.25 was a mistake? Well, I did a reality check. I said, well, I know in my head that 15% of 100 is 15, so if 15% of 100 is 15, how can 15% of 95 be more than 15? I think that might have made sense. The bottom line is 95 is less than 100. So 15% of 95 had to be less than 15, so I knew my answer of 15.25 was wrong. And so it turns out that I actually made an addition error, and the answer is 14.25. So the answer is going to be 95 plus 15% of 95, which is the same thing as 95 plus 14.25, well, that equals what? 109.25. especially this 2 here. 109.25. So if I start off with $95.00 and my portfolio grows-- or the amount of money I have-- grows by 15%, I'll end up with $109.25. Let's do another problem. Let's say I start off with some amount of money, and after a year, let's says my portfolio grows 25%, and after growing 25%, I now have $100. How much did I originally have? Notice I'm not saying that the $100 is growing by 25%." + }, + { + "Q": "At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?\n", + "A": "It s because from Earth the moon is closer than the sun therefore it looks like it is blocking it.", + "video_name": "X2jVap1YgwI", + "timestamps": [ + 331 + ], + "3min_transcript": "especially this 2 here. 109.25. So if I start off with $95.00 and my portfolio grows-- or the amount of money I have-- grows by 15%, I'll end up with $109.25. Let's do another problem. Let's say I start off with some amount of money, and after a year, let's says my portfolio grows 25%, and after growing 25%, I now have $100. How much did I originally have? Notice I'm not saying that the $100 is growing by 25%. by 25%, and I end up with $100 after it grew by 25%. To solve this one, we might have to break out a little bit of algebra. So let x equal what I start with. So just like the last problem, I start with x and it grows by 25%, so x plus 25% of x is equal to 100, and we know this 25% of x we can just rewrite as x plus 0.25 of x is equal to 100, and now actually we have a level-- actually this might be level 3 system, level 3 linear equation-- but the bottom x is the same thing as 1x, right? So 1x plus 0.25x, well that's just the same thing as 1 plus 0.25, plus x-- we're just doing the distributive property in reverse-- equals 100. And what's 1 plus 0.25? That's easy, it's 1.25. So we say 1.25x is equal to 100. Not too hard. And after you do a lot of these problems, you're going to intuitively say, oh, if some number grows by 25%, and it becomes 100, that means that 1.25 times that number is equal to 100. And if this doesn't make sense, sit and think about it a little bit, maybe rewatch the video, and hopefully it'll, over time, start to make a lot of sense to you. This type of math is very very useful. I actually work at a hedge fund, and I'm doing this type of math in my head day and night. So 1.25 times x is equal to 100, so x would equal" + }, + { + "Q": "\nAny one knows the answer for this it would be so helpful\nA Squate and a rectangle have the same area.\nThe sides of the rectangle are in the ratio 4:1. It's perimeter is 200cm.\nWhat is the length of the sides of the Square?", + "A": "For problems like this, it is useful to write one of the unknown in terms of a symbol or letter. (i ll go with x) Here, length of shorter side of the rectangle is x. (Since the longer side is 4 times longer, i ll call it 4x) perimeter of rectangle= length+width+length+width = 200 Therefore, x+4x+x+4x =200 10x=200 x=20 Now to find the area of the rectangle, area= lengtth * width =20*80 =1600 area of square = 1600 = length*length length= sqrt(1600) =40 (fill in the units everywhere when you work on paper)", + "video_name": "4ywTWCaLmXE", + "timestamps": [ + 241 + ], + "3min_transcript": "is equal to question mark. And instead of writing question mark, I'll put a variable in there. Actually, let me put a question mark there just so you really understand it is equal to a question mark in a box number cups of flour over 9 cups of oatmeal. And so I like this first way we did it because it's really just common sense. If we're tripling the oatmeal, then we're going to have to triple the flour to make the recipe in the same proportion. Another way, once you set up an equation like this, is actually to do a little bit of algebra. Some people might call it cross-multiplying, but that cross-multiplying is still using a little bit of algebra. In cross-multiplication, whenever you have a proportion set up like this, people will multiply the diagonals. So when you use cross-multiplication, you'll say that 2 times 9 must be equal to question mark times 3, must be equal to whatever is in this question mark, the number of cups of flour times 3. Or we get 18 is equal to whatever our question mark was times 3. So the number of cups of flour we need to use times 3 needs to be equal to 18. What times 3 is equal 18? You might be able to do that in your head. That is 6. Or you could divide both sides by 3, and you will get 6. So we get question mark in a box needs to be equal to 6 cups of flour. Same answer we got through kind of common sense. this cross-multiplying doesn't make any intuitive sense. Why does that work? If I have something set up like this proportion set up, why does it work that if I take the denominator here and multiply it by the numerator there that that needs to be equal to the numerator here times the denominator there? And that comes from straight up algebra. And to do that, I'm just going to rewrite this part as x just to simplify the writing a little bit. So we have 2/3 is equal to-- instead of that question mark, I'll write x over 9. And in algebra, all you're saying is that this quantity over here is equal to this quantity over here. So if you do anything to what's on the left, if you want it to still be equal, if the thing on the right still needs to be equal, you have to do the same thing to it. Now, what we want to do is we want to simplify this so all we have on the right-hand side is an x. So what can we multiply this by so that we're just left with an x? So that we've solved for x? Well, if we multiply this times 9, the 9's are going to cancel out. So let's multiply the right by 9. But of course, if we multiply the right by 9," + }, + { + "Q": "\nAt 4:55 , how is that multiplying 9 times 9 times cancels each other?\n9*9=81....am so confuse", + "A": "Ah I see whats going on. Alright, theres 2/3 = x/9. Lets change this just a tiny bit to 2/3 = x * 1/9, now multiplying both sides by 9. we get 9 * 2/3 = x * 1/9 * 9. On the right side it s One Divided by 9 that gets multiplied by 9. Or if I just swapped it around. 9 * 2/3 = x * 9 * 1/9 = x * 9/9. So now multiplying and dividing it all out it is. 6 = x", + "video_name": "4ywTWCaLmXE", + "timestamps": [ + 295 + ], + "3min_transcript": "In cross-multiplication, whenever you have a proportion set up like this, people will multiply the diagonals. So when you use cross-multiplication, you'll say that 2 times 9 must be equal to question mark times 3, must be equal to whatever is in this question mark, the number of cups of flour times 3. Or we get 18 is equal to whatever our question mark was times 3. So the number of cups of flour we need to use times 3 needs to be equal to 18. What times 3 is equal 18? You might be able to do that in your head. That is 6. Or you could divide both sides by 3, and you will get 6. So we get question mark in a box needs to be equal to 6 cups of flour. Same answer we got through kind of common sense. this cross-multiplying doesn't make any intuitive sense. Why does that work? If I have something set up like this proportion set up, why does it work that if I take the denominator here and multiply it by the numerator there that that needs to be equal to the numerator here times the denominator there? And that comes from straight up algebra. And to do that, I'm just going to rewrite this part as x just to simplify the writing a little bit. So we have 2/3 is equal to-- instead of that question mark, I'll write x over 9. And in algebra, all you're saying is that this quantity over here is equal to this quantity over here. So if you do anything to what's on the left, if you want it to still be equal, if the thing on the right still needs to be equal, you have to do the same thing to it. Now, what we want to do is we want to simplify this so all we have on the right-hand side is an x. So what can we multiply this by so that we're just left with an x? So that we've solved for x? Well, if we multiply this times 9, the 9's are going to cancel out. So let's multiply the right by 9. But of course, if we multiply the right by 9, Otherwise they still wouldn't be equal. If they were equal before being multiplied by 9, for them to still be equal, you have to multiply 9 times both sides. On the right-hand side, the 9's cancel out, so you're just left with an x. On the left-hand side, you have 9 times 2/3, or 9/1 times 2/3. Or this is equal to 18/3. And we know that 18/3 is the same thing as 6. So these are all legitimate ways to do it. I wanted you to understand that what I'm doing right here That's actually the reasoning why cross-multiplication works. But for a really simple problem like this, you could really just use common sense. If you're increasing the cups of oatmeal by a factor of 3, then increase the cups of flour by a factor of 3." + }, + { + "Q": "At 2:010 in the video above, is the pink loop twice the size of the smaller loop?\n", + "A": "It should be, since the pink loop was going around the original loop twice.", + "video_name": "Am-a5x9DGjg", + "timestamps": [ + 121 + ], + "3min_transcript": "So I saw these Fruit by the Foots, or Fruit by the Feet. Maybe Fruits by the Foot. Anyway, I figured they had mathematical potential. So I decided to just record myself playing with them. The first thing that comes to mind when there's a strip of paper, even paper covered in fruit flavored sugar, is to make a Mobius strip. So I did. Thing about Mobius strips is they have one side, while Fruits by the Foot are, by nature, two-sided. A normal loop would have a paper side and a sugar side. Putting a half twist in would make a sudden transition from one to the other. But you can also do this. Wrap it around twice so the sugary part sticks to itself and the entire outside is covered in paper. Here is our Mobius strip. After confirming that Sharpie flip chart markers won't bleed through the paper, I drew a line along a single side of the Mobius strip. But that's not telling us anything that we didn't already know. Just like one line can cover both sides, except really one side, the paper does, too. So unlike a normal loop of Fruit by the Foot covered in paper, which would need two pieces, one for each side, the Mobius strip can be unwrapped by pulling off a single strip of paper. and then came back to see what else I could do. The strip of fruit-flavored gunk has these two lines going down it, perforations, dividing it into thirds. One of the go-to Mobius strip fun things is to cut it in half down the middle. But after that, cutting it into thirds is the next thing to do. And it's as if Fruit by the Foot is designed for this purpose. So I started separating it along this line. You might want to pause here and think about what you think would happen. So I continue around. And when I loop around once, now I'm on the other side. And it turns out to not have been two lines, but one line that goes around twice. Let's see what we've got. It's completely whoa. I mean, this is totally magical to me, that there's these two loops and they're linked together, and they're not even the same size. And they're made of flavored sugar gunk. But let's understand what's going on. You can do it with paper, too. I cut out a strip, twist, and tape. Now I'm coloring the edge, both to demonstrate that the Mobius Because when we cut a strip into thirds, you could also think of it like this. You're cutting the edge off. This leaves a thinner Mobius strip and a long looping edge. Because of the twist, the edge loops around the body of a Mobius strip so the two are linked. Really, though, you should just try it yourself. After making a Mobius strip and ripping it into thirds, I poked at the leftover bit of Fruit by the Foot, waiting to be inspired. It's spirally? I opened up another package so I'd have more to play with. Whoa. It's got this pattern on it. I mean, I don't remember this at all about Fruit by the Foot. But then again, it's not like I've had one in the past 15 years. And last time I had a Fruit by the Foot, I wouldn't have recognized that this is a frieze pattern, a symmetric pattern that repeats in one dimension. I mean, repeating patterns are good because they have a rolling stamp that just kind of presses this pattern into it. But it's got other symmetry, too. The two halves of this pattern are exactly the same, but not" + }, + { + "Q": "At 1:50, why did it divide like, into two loops? How come they aren't the same? Help me!\n", + "A": "Because there is only one edge on a M\u00c3\u00b6bius strip, when she cut the edge off, it completely separated. Because of the twist in the M\u00c3\u00b6bius strip, the longer edge curled around the thicker center.", + "video_name": "Am-a5x9DGjg", + "timestamps": [ + 110 + ], + "3min_transcript": "So I saw these Fruit by the Foots, or Fruit by the Feet. Maybe Fruits by the Foot. Anyway, I figured they had mathematical potential. So I decided to just record myself playing with them. The first thing that comes to mind when there's a strip of paper, even paper covered in fruit flavored sugar, is to make a Mobius strip. So I did. Thing about Mobius strips is they have one side, while Fruits by the Foot are, by nature, two-sided. A normal loop would have a paper side and a sugar side. Putting a half twist in would make a sudden transition from one to the other. But you can also do this. Wrap it around twice so the sugary part sticks to itself and the entire outside is covered in paper. Here is our Mobius strip. After confirming that Sharpie flip chart markers won't bleed through the paper, I drew a line along a single side of the Mobius strip. But that's not telling us anything that we didn't already know. Just like one line can cover both sides, except really one side, the paper does, too. So unlike a normal loop of Fruit by the Foot covered in paper, which would need two pieces, one for each side, the Mobius strip can be unwrapped by pulling off a single strip of paper. and then came back to see what else I could do. The strip of fruit-flavored gunk has these two lines going down it, perforations, dividing it into thirds. One of the go-to Mobius strip fun things is to cut it in half down the middle. But after that, cutting it into thirds is the next thing to do. And it's as if Fruit by the Foot is designed for this purpose. So I started separating it along this line. You might want to pause here and think about what you think would happen. So I continue around. And when I loop around once, now I'm on the other side. And it turns out to not have been two lines, but one line that goes around twice. Let's see what we've got. It's completely whoa. I mean, this is totally magical to me, that there's these two loops and they're linked together, and they're not even the same size. And they're made of flavored sugar gunk. But let's understand what's going on. You can do it with paper, too. I cut out a strip, twist, and tape. Now I'm coloring the edge, both to demonstrate that the Mobius Because when we cut a strip into thirds, you could also think of it like this. You're cutting the edge off. This leaves a thinner Mobius strip and a long looping edge. Because of the twist, the edge loops around the body of a Mobius strip so the two are linked. Really, though, you should just try it yourself. After making a Mobius strip and ripping it into thirds, I poked at the leftover bit of Fruit by the Foot, waiting to be inspired. It's spirally? I opened up another package so I'd have more to play with. Whoa. It's got this pattern on it. I mean, I don't remember this at all about Fruit by the Foot. But then again, it's not like I've had one in the past 15 years. And last time I had a Fruit by the Foot, I wouldn't have recognized that this is a frieze pattern, a symmetric pattern that repeats in one dimension. I mean, repeating patterns are good because they have a rolling stamp that just kind of presses this pattern into it. But it's got other symmetry, too. The two halves of this pattern are exactly the same, but not" + }, + { + "Q": "At 2:04. Why does Sal say, \" third of the way around the triangle \". What does it mean?\nThanks.\n", + "A": "A triangle s three angles measure up to 180. Since it is a equilateral triangle Sal is constructing, each angle is 60 degrees. Many people become confused when Sal says 120 degrees is one third of the triangle, but he is talking about the arc of the circle. To find the arc or the angle formed by the arc, use this equation: arcX = 2 angleX. Thus, the angle is 60 degrees and one third of a triangle, due to what I said earlier. (60 x 3 = 180)", + "video_name": "gWMTTP58_J0", + "timestamps": [ + 124 + ], + "3min_transcript": "Construct an equilateral triangle inscribed inside the circle. So let me construct a circle that has the exact same dimensions as our original circle. Looks pretty good. And now, let me move this center, so it sits on our original circle. So they now sit on each other. Or their centers now sit on each other. So I can make it, and that looks pretty good. And now, let's think about something. If I were to draw this segment right over here, this, of course, has the length of the radius. Now, let's do another one. And that's either of their radii, because they have the same length. Now, let's just center this at our new circle and take it out here. Now, this is equal to the radius of the new circle, which is the same as the radius of the old circle. It's going to be the same as this length here. So these two segments have the same length. Now, if I were to connect that point to that point, this is a radius of our original circle. So this right over here, I have constructed an equilateral triangle. Now, why is this at all useful? Well, we know that the angles in an equilateral triangle are 60 degrees. So we know that this angle right over here is 60 degrees. Now, why is this being 60 degrees interesting? Well, imagine if we constructed another triangle out here, just symmetrically, but kind of flipped down just like this. Well, the same argument, this angle right over here between these two edges, this is also going to be 60 degrees. So this entire interior angle, if we add those two up, are going to be 120 degrees. Now, why is that interesting? Well, if this interior angle is 120 degrees, then that means that this arc right over here is 120 degrees. Or it's a third of the way around the triangle. Since that's a third of the way around the triangle, if I were to connect these two dots, that is going to be, this right over here is going to be a side of our equilateral triangle. This right over here, it's secant to an arc that is 1/3 of the entire circle. And now, I can keep doing this. Let's move-- I'll reuse these-- let's move our circle around. And so now, I'm going to move my circle along the circle. And once again, I just want to intersect these two points. And so now, let's see, I could take one of these, take it there, take it there, same exact argument. This angle that I haven't fully drawn, or this arc you could say, is 120 degrees. So this is going to be one side of our equilateral triangle." + }, + { + "Q": "\nAt 1:02, what does Sal mean with the fact that the sum of 1/n is unbounded, but that the sum of 1/n^2 is bounded?", + "A": "i think an unbound sum is basically like a divergent integral and a bound sum is like a convergent integral one goes to infinity and the other has a fixed value", + "video_name": "u1UKIljUWuc", + "timestamps": [ + 62 + ], + "3min_transcript": "- [Voiceover] Let's explore a bit the infinite series from n equals one to infinity of one over n squared. Which of course is equal to one plus one fourth, that's one over two squared, plus one over three squared, which is one ninth, plus one sixteenth and it goes on and on and on forever. So there's a couple of things that we know about it. The first thing is that all of the terms here are positive. So all of the terms here are positive. So they're all positive and that they're decreasing. It looks like they're decreasing quite quickly here from one to one fourth to one ninth to one sixteenth, and so they're quickly approaching zero, which makes us feel pretty good that this thing has a chance of converging. And because they're all positive we know that this sum right over here, if it does converge is going to be greater than zero. So the only reason why it wouldn't converge is if which we know if this was one over n it would be unbounded towards infinity. So this says that's a possibility here. So if we could show that this is bounded, then that will be a pretty good argument for why this thing right over here converges because the only reason why you could diverge is if you went to either positive infinity or negative infinity. We already know that this thing isn't going to go to negative infinity because it's all positive terms. Or you could diverge if this thing oscillates, but it's not going to oscillate because all of these terms are just adding to the sum, none of them are taking away because none of these terms are negative. So let's see if we can make a good argument for why this sum right over here is bounded, especially if we can come up with the bound, then that's a pretty good argument that this infinite series should converge. And the way that we're going to do that is we're going to explore a related function. So what I wanna do is I wanna explore f of x is equal to one over x squared. one over n squared as f of n if I were to write it this way. So why is this interesting? Well let's graph it. So that's the graph of y is equal to f of x. And notice this is a continuous, positive, decreasing function, especially over the interval that I care about right over here. I guess we could say for positive values of x, it is a continuous, positive, decreasing function. And what's interesting is we can use this as really an underestimate for this area right over here. What do I mean by that? Well one, this first term right over here, you could view that as the area of this block right over here." + }, + { + "Q": "At 6:33 -- Does Sal mean\n(1 + def-integral from 2 to inf of 1/(x^2)) instead of (1 + def-integral from 1 to inf of 1/(x^2))?\n", + "A": "No, it is the indefinite integral from 1 to infinity. The series does start at n=2 and goes to infinity, because we took out the 1. But both the integral and the series, that starts at n = 2, start at x = 1. The only difference is that the series is an understimate (a right Reimann Sum) of the actual area of the curve (the indefinite integral).", + "video_name": "u1UKIljUWuc", + "timestamps": [ + 393 + ], + "3min_transcript": "So our original series from n equals one to infinity of one over n squared. It's going to be equal to this first block, the area of this first block plus the area of all the rest of the blocks, the one fourth plus one ninth plus one sixteenth, let me do this in a new color. Which we could write as the sum from n equals two to infinity of one over n squared. So I just kind of expressed this as a sum of this plus all of that stuff. Now what's interesting is that this, what I just wrote in this blue notation that's this block plus this block plus the next block, which is going to be less than this definite integral right over here. This definite integral, notice it's an underestimate it's always below the curve, so it's going to So we can write that this thing is going to be less than one plus instead of writing this I'm gonna write the definite integral. One plus the definite integral from one to infinity of one over x squared dx. Now why is that useful? Well we know how to evaluate this and I encourage you to review the section on Khan Academy on improper integrals if this looks unfamiliar, but I'll evaluate this down here. We know that this is the same thing as the limit as, I'm going to introduce a variable here, t approaches infinity of the definite integral from one to t of and I'll just write this as x to the negative two dx. Which is equal to the limit as t approaches infinity of negative x to the negative one, or actually I And we're going to evaluate that at t and at one, which is equal to the limit as t approaches infinity of negative one over t and then minus negative one over one so that would just be plus one. And as t approaches infinity this term right over here is going to be zero, so this is just going to simplify to one. So this whole thing evaluates to one. So just like that we were able to place an upper-bound on this series. We're able to say that the series under question or in question, so the infinite sum from n equals one to infinity of one over n squared is going to be less than one plus one or it's going to be less than two. Or another way to think about it, it's going to be the two" + }, + { + "Q": "\nAt 5:35, why does Sal define the range of the Volume function according to the height (x) and width (20 - 2x) of the box? Why that specific combination of dimensions and not say height (x) & length (30 - 2x) or width & length instead?", + "A": "At that point he isn t defining the range of the volume, he s defining the realistic range for x, where x will ultimately represent the height of the box. Eventually some value of x will result in an optimum volume relative to its height (where the height is a direct function of x), width and length, all of which have been defined in terms of x.", + "video_name": "MC0tq6fNRwU", + "timestamps": [ + 335 + ], + "3min_transcript": "So what would the volume be as a function of x? Well, the volume as a function of x is going to be equal to the height, which is x, times the width, which is 20 minus x-- sorry, 20 minus 2x times the depth, which is 30 minus 2x. Now, what are possible values of x that give us a valid volume? Well, x can't be less than 0. You can't make a negative cut here. Somehow we would have to add cardboard or something there. So we know that x is going to be greater than or equal to 0. So let me write this down. x is going to be greater than or equal to 0. And what does it have to be less than? Well, I can cut at most-- we can see here the length right over here, this pink color, this mauve color, So this has got to be greater than 0. This is always going to be shorter than the 30 minus 2x, but the 20 minus 2x has to be greater than or equal to 0. You can't cut more cardboard than there is. Or you could say that 20 has to be greater than or equal to 2x, or you could say that 10 is going to be greater than or equal to x, which is another way of saying that x is going to be less than or equal to 10. That's a different shade of yellow. x is going to be less than or equal to 10. So x has got to be between 0 and 10. Otherwise we've cut too much, or we're somehow adding cardboard or something. So first let's think about the volume at the endpoints of our-- essentially of our domain, of what x can be for our volume. Well, our volume when x is equal to 0 is equal to what? We can have 0 times all of this stuff. You're not going to have any height here. So you're not going to have any volume, so our volume would be 0. What is our volume when x is equal to 10? Well, if x equaled 10, then the width here that I've drawn in pink would be 0. So once again, we would have no volume. And this term right over here, if we just look at it algebraically would also be, equal to 0, so this whole thing would be equal to 0. So someplace in between x equals 0 and x equals 10 we should achieve our maximum volume. And before we do it analytically with a bit of calculus, let's do it graphically. So I'll get my handy TI-85 out. So let me get my TI-85 out. And so first let me set my range appropriately before I attempt to graph it. So I'll put my graph function. Let me set my range. So my minimum x-value, let me make that 0. We know that x cannot be less than 0. My maximum x-value, well, 10 seems pretty good. My minimum y-value, this is essentially" + }, + { + "Q": "Towards the end of the video at around 19:00, didn't Sal mean to write rank(A) = n?\n", + "A": "I believe he did", + "video_name": "M3FuL9qKTBs", + "timestamps": [ + 1140 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 4:22, Sal said that there is variability in the data. can you please explain this?", + "A": "Variability means that something changes. So, if your set of data has values that differ from each other or change, then there is variability in the data.", + "video_name": "qyYSQDcSNlY", + "timestamps": [ + 262 + ], + "3min_transcript": "Do dogs run faster than cats? Once again, there are many dogs and many cats, and they all run at different speeds. Some dogs run faster than some cats, and some cats run faster than some dogs. So we would need some statistics to get a sense of in general or on average how fast do dogs run and then maybe on average how fast do cats run. Then we could compare those averages, or we could compare the medians in some way. So this is definitely a statistical question. Once again, we're talking about, in general, a whole population of dogs, the whole species of dogs, versus cats. And there's variation in how fast dogs run and how fast cats run. If we were talking about a particular dog and a particular cat, well, then there would just be an answer. Does dog A run faster than cat B? Well, sure. That's not going to be a statistical question. You don't have to use the tools of statistics. Actually, no, this fits the pattern of the previous one. Do wolves weigh more than dogs? Once again, there are some very light dogs and some very heavy So those wolves definitely weigh more than those dogs. But there's some very, very, very heavy dogs. And so what you want to do here, because we have variability in each of these, is you might want to come with some central tendency. On average, what's the median wolf weight? What's the average, the mean wolf weight? Compare that to the mean dog's weight. So once again, since we're speaking in general about wolves, not a particular wolf, and in general about dogs, and there's variation in the data, and we're trying to glean some numbers from that to compare, this is definitely a statistical question. Does your dog weigh more than that wolf? And we're assuming that we're pointing at a particular wolf. Now this is the particular. We're comparing a particular dog to a particular wolf. and come up with an absolute answer. There's no variability in this dog's weight, at least at the moment that we weigh it, no variability in this wolf's weight at the moment that we weigh it. This is not a statistical question. I'll put an x next to the ones that are not statistical questions. Does it rain more in Seattle than Singapore? Once again, there is variation here. And we would also probably want to know, does it rain more in Seattle than Singapore in a given year, over a decade, or whatever? But regardless of those questions, however we ask it, in some years, it might rain more in Seattle. In other years, it might rain more in Singapore. Or if we just picked Seattle, it rains a different amount from year to year. In Singapore, it rains a different amount from year So how do we compare? Well, that's where the statistics could be valuable. There's variability in the data. So we can look at the data set for Seattle and come up with some type of an average, some type of a central tendency, and compare that to the average, the mean, the mode-- the mode probably wouldn't be that useful here-- to Singapore." + }, + { + "Q": "7:54... Shouldn't the new vector have \"m\" components? One for each row?\n", + "A": "Yes (mistake).", + "video_name": "7Mo4S2wyMg4", + "timestamps": [ + 474 + ], + "3min_transcript": "m-th row, and then the m-th row will be am1. This is the m-th row first column. am1 times x1 plus -- it's hard to keep switching colors -- plus am2 times x2, all the way until we get to amn times xn. So what is this vector going to look like? It's essentially going to have -- let's say we call this vector-- Let's say it's equal to vector b. What does vector b look like? How many entries is it going to have? Well it has an entry for each row of this, right? We're taking each row and we're essentially taking the dot product of this row vector with this column vector. notation in a second. But I think you understand that this is a dot product. The first component times the first component plus the second component times the second component plus the third component times the third component, all the way to the n-th component plus the n-th component times the n-th component. So this is essentially the dot product of this row vector. We've been writing all of our vectors as columns, so we could call them column vectors, you're just writing them as rows. And we can be a little bit more specific with the notation in a second, but what's this going look like? Well we're doing this m times, so we're going to have m entries. You're going to b1 b2 all the way to bn. If you viewed these all as matrices, you can kind of view it as -- and this will eventually work for the matrix math we're going to learn -- this is an m by n matrix and we're multiplying it by -- how many rows does this guy have? He has n rows. So m by n times an n by 1, you essentially can ignore these middle two terms, and they'll result with -- how many rows does this guy have? He has m rows, and 1 column. These middle two terms have to be equal to each other just for the multiplication to be defined, and then you're left with an m by 1 matrix. So this was all abstract, let me actually apply it to some actual numbers. But it's important to actually set the definition. Now that we have the definition we can apply it to some actual matrices and vectors. So let's say we have the matrix. Let's say I want to multiply the matrix minus 3, 0, 3, 2. Now I'll do this one in yellow. 1, 7, minus 1, 9. And I want to multiply that by the vector." + }, + { + "Q": "At 0:36 could the sequence be written instead as the sum from n=0 to infinity of (1/2)^n? and if so which one is used most commonly?\n", + "A": "Sure - both versions are widely used.", + "video_name": "rcRg_gO7-7E", + "timestamps": [ + 36 + ], + "3min_transcript": "Let's say that I have a geometric series. A geometric sequence, I should say. We'll talk about series in a second. So a geometric series, let's say it starts at 1, and then our common ratio is 1/2. So the common ratio is the number that we keep multiplying by. So 1 times 1/2 is 1/2, 1/2 times 1/2 is 1/4, 1/4 times 1/2 is 1/8, and we can keep going on and on and on forever. This is an infinite geometric sequence. And we can denote this. We can say that this is equal to the sequence of a sub n from n equals 1 to infinity, with a sub n equaling 1 times our common ratio to the n minus 1. So it's going to be our first term, which is just 1, times our common ratio, which is 1/2. 1/2 to the n minus 1. And you can verify it. This right over here you can view as 1/2 to the 0 power. This is 1/2 half to the first power, this is 1/2 squared. So the first term is 1/2 to the 0. The second term is 1/2 to the 1. The third term is 1/2 squared. So the nth term is going to be 1/2 to the n minus 1. So this is just really 1/2 to the n minus 1 power. Fair enough. Now, let's say we don't just care about looking at the sequence. We actually care about the sum of the sequence. So we actually care about not just looking at each of these terms, see what happens as I keep multiplying by 1/2, but I actually care about summing 1 plus 1/2 plus 1/4 plus 1/8, and keep going on and on and on forever. So this we would now call a geometric series. And because I keep adding an infinite number of terms, this is an infinite geometric series. So this right over here would be the infinite geometric series. A series you can just view as the sum of a sequence. Now, how would we denote this? Well, we can use summing notation. We could say that this is equal to the sum. Let me make sure I'm not falling off the page. Let me just scroll over to the left a bit. The sum from n equals 1 to infinity of a sub n. And a sub n is just 1/2 to the n minus 1. 1/2 to the n minus 1 power. So you just say OK, when n equals 1, it's 1/2 to the 0, which is 1. Then I'm going to sum that to when n equals 2, which is 1/2, when n equals 3, it's 1/4. On and on, and on, and on. So all I want to do in this video is to really clarify differences between sequences and series, and make you a little bit comfortable with the notation. In the next few videos, we'll actually try to take sums of geometric series and see if we actually get a finite value." + }, + { + "Q": "\n3:04 she says that circles can be defined by three points, can't a circle be defined by more than three?", + "A": "No. A circle is only defined by three points. If you have three points, then there is only one circle that can pass through all three (unless they are collinear in which case no circle can). A fourth point would either be redundant or make the circle previously defined invalid.", + "video_name": "DK5Z709J2eo", + "timestamps": [ + 184 + ], + "3min_transcript": "and it's cool that that's possible. But I'm not really interested in doing calculations. So we'll come back to camels. Here's a fractal. You start with these circles in a circle, and then keep drawing the biggest circle that fits in the spaces between. This is called an Apollonian Gasket. And you can choose a different starting set of circles, and it still works nicely. It's well known in some circles because it has some very interesting properties involving the relative curvature of the circles, which is neat, But it also looks cool and suggests an awesome doodle Step 1, draw any shape. Step 2, draw the biggest circle you can within this shape. Step 3, draw the biggest circle you can in the space left. Step 4, see step 3. As long as there is space left over after the first circle, meaning don't start with a circle, this method turns any shape into a fractal. You can do this with triangles. You can do this with stars. And don't forget to embellish. You can do this with elephants, or snakes, or a profile of one I choose Abraham Lincoln. Awesome. OK, but what about other shapes besides circles? For example, equilateral triangles, say, filling this other triangle, which works because the filler triangles, and orientation matters. This yields our friend, Sierpinski's triangle, which, by the way, you can also make out of Abraham Lincoln. But triangles seem to work beautifully in this case. But that's a special case. And the problem with triangles is that they don't always fit snugly. For example, with this blobby shape, the biggest equilateral triangle has this lonely hanging corner. And sure, you don't have to let that stop you, and it's a fun doodle game. But I think it lacks some of the beauty of the circle game. Or what if you could change the orientation of the triangle to get the biggest possible one? What if you didn't have to keep it equilateral? Well, for polygonal shapes, the game runs out pretty quickly, so that's no good. But in curvy, complicated shapes, the process itself becomes difficult. How do you find the biggest triangle? It's not always obvious which triangle has more area, especially when you're starting shape is not very well defined. This is an interesting sort of question because there is a correct answer, but if you were going to write a computer program that filled a given shape with another shape, following even the simpler version of the rules, you might need to learn some computational geometry. And certainly, we can move beyond triangles to squares, or even elephants. But the circle is great because it's just so fantastically round. A circle can be defined by three points. So draw three, arbitrary points, and then try and find the circle they belong to. So one of the things that intrigues me about the circle game is that, whenever you have one of these sorts of corners, you know there's going to be an infinite number of circles Thing is, for every one of those infinite circles, you create a few more little corners that are going to need an infinite number of circles. And for every one of those, and so on. You just get an incredible number of circles breeding more circles. And you can see just how dense infinity can be. Though the astounding thing is that this kind of infinity is still the smallest, countable kind of infinity. And there are kinds of infinity that are just mind bogglingly infiniter. But wait, here's an interesting thing. If you call this distance 1 arbitrary length unit, then this distance plus this, dot, dot, dot, is an infinite series that approaches 1. And this is another, different, series that still approaches 1. And here's another, and another. And as long as the outside shape is well defined, so will the series be. But if you want the simple kind of series, where each circle's diameter is a certain percentage of the one before it," + }, + { + "Q": "\nat 0:56, vi says a camel is only a third of the page, but if so, couldn't you just mark 1/6 of the page, and 1/12, 1/24, & so on?", + "A": "Yeah, she could ve just acted like the 2/3 left was 1 page, then made the camels half of that, then half of that, and so on. She could ve also made the camels bigger or only used 2/3 of the page for camels, making the 1/2 of 1/2 of 1/2\u00e2\u0080\u00a6 again and used the extra for a circle game shape.", + "video_name": "DK5Z709J2eo", + "timestamps": [ + 56 + ], + "3min_transcript": "So you mean you're in math class, yet again, because they make you go every single day. And you're learning about, I don't know, the sums of infinite series. That's a high school topic, right? Which is odd, because it's a cool topic. But they somehow manage to ruin it anyway. So I guess that's why they allow infinite serieses in the curriculum. So, in a quite understandable need for distraction, you're doodling and thinking more about what the plural of series should be than about the topic at hand. Serieses, serises, seriesen, seri? Or is it that the singular should be changed? One serie, or seris, or serum? Just like the singular of sheep should be shoop. But the whole concept of things like 1/2 plus 1/4 plus 1/8 plus 1/16, and so on approaches 1 is useful if, say, you want to draw a line of elephants, each holding the tail of the next one. Normal elephant, young elephant, baby elephant, dog-sized elephant, puppy-sized elephant, all the way down to Mr. Tusks, and beyond. Which is at least a tiny bit awesome because you can get an infinite number of elephants in a line and still have it fit across a single notebook page. But there's questions, like what if you started with a camel, which, being smaller than an elephant, only goes across a third of the page. How big should the next camel be in order to properly approach the end of the page? and it's cool that that's possible. But I'm not really interested in doing calculations. So we'll come back to camels. Here's a fractal. You start with these circles in a circle, and then keep drawing the biggest circle that fits in the spaces between. This is called an Apollonian Gasket. And you can choose a different starting set of circles, and it still works nicely. It's well known in some circles because it has some very interesting properties involving the relative curvature of the circles, which is neat, But it also looks cool and suggests an awesome doodle Step 1, draw any shape. Step 2, draw the biggest circle you can within this shape. Step 3, draw the biggest circle you can in the space left. Step 4, see step 3. As long as there is space left over after the first circle, meaning don't start with a circle, this method turns any shape into a fractal. You can do this with triangles. You can do this with stars. And don't forget to embellish. You can do this with elephants, or snakes, or a profile of one I choose Abraham Lincoln. Awesome. OK, but what about other shapes besides circles? For example, equilateral triangles, say, filling this other triangle, which works because the filler triangles, and orientation matters. This yields our friend, Sierpinski's triangle, which, by the way, you can also make out of Abraham Lincoln. But triangles seem to work beautifully in this case. But that's a special case. And the problem with triangles is that they don't always fit snugly. For example, with this blobby shape, the biggest equilateral triangle has this lonely hanging corner. And sure, you don't have to let that stop you, and it's a fun doodle game. But I think it lacks some of the beauty of the circle game. Or what if you could change the orientation of the triangle to get the biggest possible one? What if you didn't have to keep it equilateral? Well, for polygonal shapes, the game runs out pretty quickly, so that's no good. But in curvy, complicated shapes, the process itself becomes difficult. How do you find the biggest triangle? It's not always obvious which triangle has more area, especially when you're starting shape is not very well defined. This is an interesting sort of question because there is a correct answer, but if you were going to write a computer program that filled a given shape with another shape, following even the simpler version of the rules, you might need to learn some computational geometry. And certainly, we can move beyond triangles to squares, or even elephants. But the circle is great because it's just so fantastically round." + }, + { + "Q": "At 3:57, couldn't you just make the intervals on 0 and 3 and not include the numbers -1 and 4?\n", + "A": "Remember, there are lots of numbers (fractions and decimals) that are located between the integers. If you make the interval 0 to 3, you lose all the real numbers from -1 to 0 and 3 to 4 (numbers like -0.5, -1/4, -0.999999, 3.6; 3.999, etc.). So, you have a completely different set of numbers and a much smaller set of numbers. Thus, the interval needs to be (-1, 4)", + "video_name": "UJQkqV2zGv0", + "timestamps": [ + 237 + ], + "3min_transcript": "that we include, this bracket on the left says that we include negative three, and this bracket on the right says that we include positive two in our interval. Sometimes you might see things written a little bit more math-y. You might see x is a member of the real numbers such that... And I could put these curly brackets around like this. These curly brackets say that we're talking about a set of values, and we're saying that the set of all x's that are a member of the real number, so this is just fancy math notation, it's a member of the real numbers. I'm using the Greek letter epsilon right over here. It's a member of the real numbers such that. This vertical line here means \"such that,\" negative three is less x is less than-- negative three is less than or equal to x, is less than or equal to two. I could also write it this way. I could write x is a member of the real numbers such that x is a member, such that x is a member So these are all different ways of denoting or depicting the same interval. Let's do some more examples here. So let's-- Let me draw a number line again. So, a number line. And now let me do-- Let me just do an open interval. An open interval just so that we clearly can see the difference. Let's say that I want to talk about the values between negative one and four. Let me use a different color. So the values between negative one and four, but I don't want to include negative one and four. So this is going to be an open interval. So I'm not going to include four, and I'm not going to include negative one. Notice I have open circles here. Over here had closed circles, the closed circles told me that I included negative three and two. it's all the values in between negative one and four. Negative .999999 is going to be included, but negative one is not going to be included. And 3.9999999 is going to be included, but four is not going to be included. So how would we-- What would be the notation for this? Well, here we could say x is going to be a member of the real numbers such that negative one-- I'm not going to say less than or equal to because x can't be equal to negative one, so negative one is strictly less than x, is strictly less than four. Notice not less than or equal, because I can't be equal to four, four is not included. So that's one way to say it. Another way I could write it like this. x is a member of the real numbers such that x is a member of... Now the interval is from negative one to four but I'm not gonna use these brackets." + }, + { + "Q": "\nAt 8:13 could it also be expressed as {X \u00cf\u00b5 R | 1 < X < 1 } ?", + "A": "Sorry, but no. 1 < X < 1 is the notation for and , not or . 1 < X < 1 is the null set since both 1 < X and x < 1 must be true. There is no number that is both greater than one and less than one.", + "video_name": "UJQkqV2zGv0", + "timestamps": [ + 493 + ], + "3min_transcript": "So we're not going to include negative four. Negative four is strictly less than, not less than or equal to, so x can't be equal to negative four, open circle there. But x could be equal to negative one. It has to be less than or equal to negative one. It could be equal to negative one so I'm going to fill that in right over there. And it's everything in between. If I want to write it with this notation I could write x is a member of the real numbers such that x is a member of the interval, so it's going to go between negative four and negative one, but we're not including negative four. We have an open circle here so I'm gonna put a parentheses on that side, but we are including negative one. We are including negative one. So we put a bracket on that side. That right over there would be the notation. Now there's other things that you could do You could say, well hey, everything except for some values. Let me give another example. Let's get another example here. Let's say that we wanna talk about all the real numbers except for one. We want to include all of the real numbers. All of the real numbers except for one. Except for one, so we're gonna exclude one right over here, open circle, but it can be any other real number. So how would we denote this? Well, we could write x is a member of the real numbers such that x does not equal one. So here I'm saying x can be a member of the real numbers but x cannot be equal to one. It can be anything else, but it cannot be equal to one. You could say x is a member of the real numbers such that x is less than one, or x is greater than one. So you could write it just like that. Or you could do something interesting. This is the one that I would use, this is the shortest and it makes it very clear. You say hey, everything except for one. But you could even do something fancy, like you could say x is a member of the real numbers such that x is a member of the set going from negative infinity to one, not including one, or x is a member of the set going from-- or a member of the interval going from one, not including one, all the way to positive, all the way to positive infinity. And when we're talking about negative infinity or positive infinity, you always put a parentheses." + }, + { + "Q": "When you work out the problem at 3:55, wouldn't it be f(x+1)/f(x)=1/4*2^x+1/1/4*2^x=1? Because 1/4 and x's cancel and you are left with 2/2 which is equal to 1.\n", + "A": "First off, be careful how you write it, you need extra sets of parentheses or your equation is totally different so f(x+1)/f(x) = 1/4 * 2^ (x+1) / (1/4*2^x). The 1/4 cancel so you are left with 2^(x+1)/2^x x s do not cancel, we have to use exponent rules where dividing same bases require us to subtract the exponents, so as Sal noted x + 1 - x is just 1, so we have 2^1 or just 2", + "video_name": "G2WybA4Hf7Y", + "timestamps": [ + 235 + ], + "3min_transcript": "Five times one is just five. So the initial value is once again, that. So if you have exponential functions of this form, it makes sense. Your initial value, well if you put a zero in for the exponent, then the number raised to the exponent is just going to be one, and you're just going to be left with that thing that you're multiplying by that. Hopefully that makes sense, but since you're looking at it, hopefully it does make a little bit. Now, you might be saying, well what do we call this number? What do we call that number there? Or that number there? And that's called the common ratio. The common common ratio. And in my brain, we say well why is it called a common ratio? Well, if you thought about integer inputs into this, especially sequential integer inputs into it, you would see a pattern. For example, h of, let me do this in that green color, h of zero is equal to, we already established one-fourth. Now, what is h of one going to be equal to? two to the first power. So it's going to be one-fourth two. What is h of two going to be equal to? Well, it's going to be one-fourth times two squared, so it's going to be times two times two. Or, we could just view this as this is going to be two times h of one. And actually I should have done this when I wrote this one out, but this we can write as two times h of zero. So notice, if we were to take the ratio between h of two and h of one, it would be two. If we were to take the ratio between h of one it would be two. That is the common ratio between successive whole number inputs into our function. So, h of I could say plus one over h of n is going to be equal to is going to be equal to actually I can work it out mathematically. One-fourth times two to the n plus one one-fourth two to the n. Two to the n plus one, divided by two to the n is just going to be equal to two. That is your common ratio. So for the function h. For the function f, our common ratio is three. If we were to go the other way around, if someone said, hey, I have some function whose initial value, so let's say, I have some function, I'll do this in a new color, I have some function, g, and we know that its initial initial value is five. And someone were to say its common ratio its common ratio is six, what would this exponential function look like? And they're telling you this is an exponential function. Well, g of let's say x is the input, is going to be equal to our initial value, which is five. That's not a negative sign there, Our initial value is five." + }, + { + "Q": "\nAt 2:43, what does the word arithmetic mean in general? I've heard people use the word a lot.", + "A": "The worldwide definition of arithmetic is: the branch of mathematics relating to the manipulation of numbers. Fractions, addition and subtraction, division and multiplication fall into this category.", + "video_name": "uhxtUt_-GyM", + "timestamps": [ + 163 + ], + "3min_transcript": "and we'll talk a lot about samples versus populations, but I think just a basic sense of what that is. If I survey three people who are going to vote for president, I clearly haven't surveyed the entire population. I've surveyed a sample. But what inferential statistics are all about are if we can do some math on the samples, maybe we can make inferences or conclusions about the population as a whole. Anyway, that's just the big picture of what statistics is all about. So let's just get into the meat of it. We'll start with the descriptive. So the first thing that I would want to do it, or I think most people would want to do when they're given a whole set of numbers and they're told to describe it. It's like well, maybe I can come up with some number that is most indicative of all of the numbers in that set, or some number that represents the central tendency. This is a word you'll see a lot in statistics books. The central tendency of a set of numbers. And I'll be a little bit more exact here than I normally am with the word average. When I talk about in this context, it just means that the average is a number that somehow is giving us a sense of the central tendency, or maybe a number that is most representative of a set. I know that sounds all very abstract, but let's do a couple of examples. So there's a bunch of ways you can actually measure of the central tendency, or the average, of a set of numbers, and you've probably seen these before. They are the mean-- actually, there are types of means, but we'll stick with the arithmetic mean. Later when we talk about stock returns and things, we'll do geometric means. And maybe we'll cover the harmonic mean one day. There's a mean, the median, and the mode. can kind of be representative of a data set, or a population's central tendency, or a sample central tendency. And they all are collectively-- they can all be forms of an average. And I think when we see examples, it'll make a little bit more sense. In every day speak when people talk about an average-- I think you've already computed averages in your life-- they're usually talking about the arithmetic mean. So normally when someone says, let's take the average of these numbers and they expect you to do something, they want you to figure out the arithmetic mean. They don't want you to figure out the median or the mode. But before we go any further, let's figure out what these things are. So let me make up a set of numbers. Let's say I have the number 1, let's say I have another 1, 2, 3, let's say I have a 4. That's good enough. We just wanted a simple example. So the mean, or the arithmetic mean," + }, + { + "Q": "In this video about Statistics in 8:04 and 10:42 when Sal was talking about The median does this type of rule apply When there is odd numbers there is a middle number but when there is an even number you have to divde it by the two middle numbers that is your median.Does that rule apply in Statistics? Just want to know.\n", + "A": "Thanks Andrew for letting me know!", + "video_name": "uhxtUt_-GyM", + "timestamps": [ + 484, + 642 + ], + "3min_transcript": "And I know what you might be thinking-- well, that was easy enough when we had five numbers. But what if we had six numbers? What if it was like this? What if this was our set of numbers? 1, 1, 2, 3, let's add another 4 there. So now there's no middle number. 2 is not the middle number because there's two less than it and three larger than it. And then 3 is not the middle number because there's two larger and three smaller than it. So there's no middle number. So when you have a set with even numbers and someone tells you to figure out the median, what you do is you take the middle two numbers. And then you take the arithmetic mean of those two numbers. So in this case, of this set, the median would be 2.5. Fair enough. But let's put this aside, because I want to compare the median and the means and the modes for the same set of numbers. But that's a good thing to know, because sometimes it can be a little confusing. These are all mathematical tools for getting our heads around numbers. It's not like one day someone saw one of these formulas on the face of the sun and says, oh, that's part of the universe. That is how the average should be calculated. These are human constructs to just get our heads around large sets of data. I mean, this isn't a large set of data. But instead of five numbers, if we had five million numbers, you could imagine that you don't like thinking about every number individually. Anyway, before I talk more about that, let me tell you what the mode is. And the mode, to some degree, it's the one that I think most people forget or never learn. And when they see it on an exam, it confuses them because they're like, oh, that sounds very advanced. But in some ways, it is the easiest of all of the measures of central tendency or of average. The mode is essentially what number is most common in a set. So in this example, there's two ones and then there's one of everything else. So the mode here is 1. So mode you can say is the most common number. 1, 1, 2, 3, 4, 4. Here, I have two 1s and I have two 4s. And this is where the mode gets a little bit tricky, because either of these would have been a decent answer for the mode. You could have actually said the mode of this is 1, or the mode of this is 4. And it gets a little bit ambiguous, and you probably want a little clarity from the person asking you. Most times on a test when they ask you, there's not going to be this ambiguity. There will be a most common number in the set. Now you're saying, oh, why wasn't just one of these good enough? Why did we learn averages? Why don't we just use arithmetic mean all the time? What's median and mode good for? Well, I'll try to do one example of that and see if it rings true with you. And then you could think a little bit more. Let's say I had this set of numbers-- 3, 3, 3, 3, 3, and 100." + }, + { + "Q": "\nAt 11:10 sal, u wrote 115 /6 what is that supposed to mean were talking about arithmatic not fractions and also your example for the mode did not make sense may you clarify that for me", + "A": "He was calculating the mean of the data set {3,3,3,3,3,100}, which is found by taking the sum of all those numbers (=115) and dividing by the number of elements in the set (6). Hence 115/6, which is, as he says, 19 1/6. You could also write it as 19.7, but 115/6 is more precise, as you don t have to round. For that same data set, the mode is pretty straightforward. The number 3 is clearly the most common number in that set (thee are five 3 s but only one 100), so the mode is 3.", + "video_name": "uhxtUt_-GyM", + "timestamps": [ + 670 + ], + "3min_transcript": "What's the mean here? So one, two, three, four, five 3s and 100. So it would be 115 divided by 6. Because I have one, two, three, four, five, six numbers. 115 is just the sum of all of these. So that's equal to-- how many times does 6 go into 115? 6 goes into it one time. 1 times 6 is 6. 55 goes into it 9 times. 9 times 6 is 54. So it's equal to 19 and 1/6. Fair enough. I just added all the numbers and divided by how many there are. But my question is, is this really representative of this set? I mean, I have a ton of 3s and then I have 100 all of a sudden. And we're saying that the central tendency is 19 and 1/6. 19 and 1/6 doesn't really seem indicative of this set. I mean, maybe it does, depending on your application. But it just seems a little bit off. I mean, my intuition would be that central tendency is something closer to 3. So what the median tell us? We already put these numbers in order. If I had given you out of order, you'd And you'd say, what's the middle number? Let's see. The middle two numbers, since I have an even number, are 3 and 3. So if I take the average of 3 and 3-- I should be particular with my language-- if I take the arithmetic mean of 3 and 3, I get 3. And this is maybe a better measurement of the central tendency or of the average of this set of numbers. Essentially, what it does is by taking the median, I wasn't so much affected by this really large number that's very different than the others. In statistics, they call that an outlier. If you talked about average home prices, maybe every house in the city is $100,000 and then there's one house that costs a trillion dollars. And then if someone told you the average house price was a million dollars, you might have a very wrong perception of that city. But the median house price would be $100,000, are like. So similarly, this median gives you a better sense of what the numbers in this set are like. The arithmetic mean was skewed by what they'd call an outlier. Being able to tell what an outlier is is one of those things that a statistician will say, well, I know it when I see it. There isn't really a formal definition for it, but it tends to be a number that really sticks out. And sometimes it's due to a measurement error, or whatever. And then finally, the mode. What is the most common number in this set? Well, there's five 3s and there's one 100. So the most common number is, once again, it's a 3. So in this case, when you had this outlier, the median and the mode tend to be maybe a little bit better about giving you an indication of what these numbers represent. Maybe this was just a measurement error. We don't actually know what these represent. If these are house prices, then I would argue that these are probably more indicative measures of what the houses in an area cost." + }, + { + "Q": "At 18:10 when Sal says/writes \"set X spans subspace V and X has 5 elements you now know that no set than spans the subspace V can have fewer than 5 elements\", isn't this statement incorrect? The set X didn't claim to be a BASIS for V, just span V. (I know Sal continues with \"even better if X is a basis...\" but it sounds like he's going on to make a separate statement).\n", + "A": "Indeed. This threw me for a loop. Sal needs to correct this with a pop-up.", + "video_name": "Zn2K8UIT8r4", + "timestamps": [ + 1090 + ], + "3min_transcript": "Remember we said that n is greater than m. Or when we defined B, we said that m is less than n. Same thing, that this was a smaller set. Now we're saying that this spans V, but at the same time we said this was a basis. This is just our starting fact, that this is a basis for V. Basis means two things: it means it spans V and it means it's linearly independent. Now we just got this result by assuming that we had some set B that's smaller than this set here that spans V. We were able to construct this by saying that a1 through am also spans V. The result we got is that this spans V. But if this subset of A spans V, then A becomes linearly Because if this subset spans V, that means that an can be represented as some linear combination of these guys. contradiction with our original statement that set A is a basis for V, because that means it's linearly independent. If you're able to do this, then this means that there was some smaller spanning set, you get the result that A has to be linearly dependant, even though we said it was linearly independent. So we now know, we get our contradiction, we say that there cannot be [typing] a spanning set B that has fewer elements than A. And this is a pretty neat outcome, because now, if I the subspace V again. Then you know that X has five elements. You now know that no set that spans the subspace V can have fewer than five elements. Even better, if I told you that X is a basis for V, and I told you it has five elements, and Y is a basis for V. [typing] You know that Y also has to have exactly five elements." + }, + { + "Q": "At 4:41, why are you using the last equation?\n", + "A": "It doesn t matter what equation of the 3 is used to find out the value z. Any of the three equations can be used.", + "video_name": "f7cX-Ar2cEM", + "timestamps": [ + 281 + ], + "3min_transcript": "and add the two equations. So let's do that. So let's multiply this times 7. 7 times 8 is 56, so it's 56x minus 7y is equal to 7 times negative 10, is equal to negative 70. And now we can add these two equations. I'm now trying to eliminate the y's. So we have 16x plus 56x. That is 72x. So we have 72x, these guys eliminate, equal to negative 72, negative 2 plus negative 70. Divide both sides by 72, and we get x is equal to negative 1. And now we just have to substitute back to figure out what y and z are equal to. So we have 8x minus y is equal to negative 10. If x is equal to negative 1, that means 8 times negative 1, or negative 8 minus y is equal to negative 10. We can add 8 to both sides. And so we have negative y is equal to negative 2, or multiplying both sides by negative 1, y is equal to 2. let me square that off. So x is equal to negative 1, y is equal to 2. We now just have to worry about z, and we can go back to any of these up here. So I'll just use this last one. The numbers seem lower. So if we substitute back into this last equation right over here, we have 3 times x, which is 3 times negative 1 plus y, which is 2, minus z is equal to 3. So it's negative 3 plus 2 minus z is equal to 3. And this is negative 1 minus z is equal to 3. negative z is equal to 4. Multiplying both sides by negative 1, you get z is equal to negative 4. So we're done. Let's verify that these solutions, or the solution of x is negative 1, y is equal to 2, z is equal to negative 4, actually satisfies all three of these constraints. So let's substitute into this first one. So you have x plus 2y plus 5z. So x is negative 1 plus 2 times y, so plus 4, plus 5z. So minus 20 has to be equal to negative 17. And this is negative. This right here is positive 3 minus 20 is indeed equal to negative 17. So it satisfies the first constraint. Let's look at the second one. 2 times x, 2 times negative 1. That's negative 2 minus 3 times y." + }, + { + "Q": "at 2:30 why is he adding all the numbers when he could be subtracting?\n", + "A": "so that he can eliminate the z variable.", + "video_name": "f7cX-Ar2cEM", + "timestamps": [ + 150 + ], + "3min_transcript": "Solve this system. And once again, we have three equations with three unknowns. So this is essentially trying to figure out where three different planes would intersect in three dimensions. And to do this, if we want to do it by elimination, if we want to be able to eliminate variables, it looks like, well, it looks like we have a negative z here. We have a plus 2z. We have a 5z over here. If we were to scale up this third equation by positive 2, then you would have a negative 2z here, and it would cancel out with this 2z there. And then if you were to scale it up by 5, you'd have a negative 5z here, and then that could cancel out with that 5z over there. So let's try to cancel out. Let's try to eliminate the z's first. So let me start with this equation up here. I'll just rewrite it. So we have-- I'll draw an arrow over here-- we have x plus 2y plus 5z is equal to negative 17. And then to cancel out or to eliminate the z's, I'll multiply this equation here times 5. So I'm going to multiply this equation times 5. So 3x times 5 is 15x, y times 5 is plus 5y, and then negative z times 5 is negative 5z-- that's the whole point and why we're multiplying it by 5-- is equal to 3 times 5, which is equal to 15. And so if we add these two equations, we get x plus 15x is 16x, 2y plus 5y is 7y, and 5z minus 5z or plus negative 5z, those are going to cancel out. And that is going to be equal to negative 17 plus 15 is negative 2. So we were able to use the constraints in that equation and that equation, and now we have an equation in just x and y. So let's try to do the same thing. Let's trying to eliminate the z's. But now I'll use this equation and this equation. So this equation-- let me just rewrite it over here. We have 2x minus 3y plus 2z is equal to negative 16. And now, so that this 2z gets eliminated, let's multiply this equation times 2. So let's multiply it times 2, so we'll have a negative 2z here to eliminate with the positive 2z. So 2 times 3x is 6x, 2 times y is plus 2y, and then 2 times negative z is negative 2z is equal to 2 times 3 is equal to 6. And now we can add these two equations. 2x plus 6x is 8x, negative 3y plus 2y is negative y, and then these two guys get canceled out. And then that is equal to negative 16 plus 6 is negative 10. So now we have two equations with two unknowns. We've eliminated the z's. And let's see, if we want to eliminate again, we have a negative y over here. We have a positive 7y." + }, + { + "Q": "at around 15:40. Are you allowed to just say dx*(1/dt) = dx/dt and have it be the derivative of x with respect to t??\n", + "A": "I m not sure what you are asking, but remember that since dx and dt were on different sides of the radical, you d have to deal with them slightly differently.", + "video_name": "_60sKaoRmhU", + "timestamps": [ + 940 + ], + "3min_transcript": "algebraically manipulate differentials, what we can do is let us multiply and divide by dt. So one way to think about it, you could rewrite, so let me just do this orange part right here. Let's do a little side right here. So if you take this orange part, and write it in pink, and you have dx squared, and then you have plus dy squared, and let's say you just multiply it times dt over dt, right? That's a small change in t, divided by a small change in t. That's 1, so of course you can multiply it by that. If we're to bring in this part inside of the square root sign, right, so let me rewrite this. This is the same thing as 1 over dt times the square root of dx squared plus dy squared, and then times that dt. I just wanted to write it this way to show you I'm just multiplying by 1. And here, I'm just taking this dt, writing it there, and And now if I wanted to bring this into the square root sign, this is the same thing, this is equal to, and I'll do it very slowly, just to make sure, I'll allow you to believe that I'm not doing anything shady with the algebra. This is the same thing as the square root of 1 over dt squared, let me make the radical a little bit bigger, times dx squared plus dy squared, and all of that times dt, right? I didn't do anything, you could just take the square root of this and you'd get 1 over dt. And if I just distribute this, this is equal to the square root, and we have our dt at the end, of dx squared, or we could even write, dx over dt squared, plus dy over dt squared. Right? dx squared over dt squared is just dx over dt squared, same thing with the y's. And now all of a sudden, this starts to look We said that these are equivalent. And I'll switch colors, just for the sake of it. So we have the integral. From t is equal to a. Let me get our drawing back, if I-- from t is equal to a to t is equal to b of f of x of t times, or f of x of t and f of, or and y of t, they're both functions of t, and now instead of this expression, we can write the square root of, well, what's dx, what's the change in x with respect to, whatever this parameter is? What is dx dt? dx dt is the same thing as g prime of t. Right? x is a function of t. The function I wrote is g prime of t. And then dy dt is same thing as h prime of t." + }, + { + "Q": "Okay, I guess Sal is using Synthetic Division at 10:20, but I don't see how he's doing it.\nFor example, when he uses the 1:\n\u00e2\u0080\u00a2 We write down the 1\n1, ...\n\u00e2\u0080\u00a2 1 x 1 + (-3) = -2\n1, -3, ...\n\u00e2\u0080\u00a2 1 x (-2) + (-9) = -11\n1, -3, -11, ...\n\u00e2\u0080\u00a2 1x (-11) + 27 = 16\n1, -3, -11, 16\nNow this doesn't add up to zero, but it still doesn't make any sense to me.\nUsing 3 I get:\n1, 0, -9, 0\nWhat am I doing wrong?\n", + "A": "I don t use synthetic division. It s quicker when you get used to it, but harder to understand. Long division (what he used in this video) always works. You don t have to relearn it whenever you haven t used it for a while.", + "video_name": "11dNghWC4HI", + "timestamps": [ + 620 + ], + "3min_transcript": "it times that guy. So plus lambda squared. Minus 4 lambda plus 4. And now of course, we have these terms over here. So we're going to have to simplify it again. So what are all of our constant terms? We have a 23 and we have a plus 4. So we have a 27. Plus 27. And then, what are all of our lambda terms? We have a minus 9 lambda and then we have a-- let's see. We have a minus 9 lambda, we have a plus 4 lambda, and then we have a minus 4 lambda. So these two cancel out. So I just have a minus 9 lambda. And then, what are my lambda squared terms? I have a plus lambda squared and I have a minus 4 lambda squared. So if you add those two that's going to be minus 3 lambda squared. And then finally, I have only one lambda cubed term, that So this is the characteristic polynomial for our matrix. So this is the characteristic polynomial and this represents the determinant for any lambda. The determinant of this matrix for any lambda. And we said that this has to be equal to 0 if any only if lambda is truly an eigenvalue. So we're going to set this equal to 0. And unlucky or lucky for us, there is no real trivial-- there is no quadratic. Well there is, actually, but it's very complicated. And so it's usually a waste of time. So we're going to have to do kind of the art of factoring a quadratic polynomial. I got this problem out of a book and I think it's fair to say that if you ever do run into this in an actual linear algebra class or really, in an algebra class generally-- it doesn't even have to be in the context of eigenvalues, you probably will be dealing with integer solutions. And if you are dealing with integer solutions, then your Especially if you have a 1 coefficient out here. So your potential roots-- in this case, what are the factors of 27? So 1, 3, 9 and 27. So all these are potential roots. So we can just try them out. 1 cubed is 1 minus 3. So let me try 1. So if we try a 1, it's 1 minus 3 minus 9 plus 27. That does not equal 0. It's minus 2 minus 9 is minus 11. Plus 16. That does not equal 0. So 1 is not a root. If we try 3 we get 3 cubed, which is 27. Minus 3 times 3 squared is minus 3 times 3, which is minus 27. Minus 9 times 3, which is minus 27. Plus 27. That does equal 0. So lucky for us, on our second try we were able to find one 0 for this." + }, + { + "Q": "At 2:18, I didn't get what the speaker meant by \"that multiplies the rest of the term\". Can anyone explain?\n", + "A": "A number is a coefficient, which defines how much there is. In maths , especially algebra, numbers get an x or another letter. When you put a 5 before a x you get 5x. Whereby 5 is the coefficient and x the rest of the term . It s get tricky when you add multiple variables to one coefficient. Like this 4xy or 57abc.", + "video_name": "9_VCk9tWT0Y", + "timestamps": [ + 138 + ], + "3min_transcript": "What I want to do in this video is think about how expressions are formed and the words we use to describe the different parts of an expression. And the reason why this is useful is when you hear other people refer to some expression and say, oh, I don't agree with the second term, or the third term has four factors, or why is the coefficient on that term 6, you'll know what they're talking about, and you can communicate in the same way. So let's think about what those words actually mean. So we have an expression here. And the first thing I want to think about are the terms of an expression or what a term is. And one way to think about it is the terms are the things that are getting added and subtracted. So, for example, in this expression right over here, you have three things that are getting added and subtracted. The first thing, you're taking 2 times 3. You're adding that to 4. And then from that, you're subtracting 7y. So in this example, you have three terms. The first term is 2 times 3. The second term is just the number 4. Now, let's think about the term \"factor.\" And when people are talking about a factor, especially in terms of an expression, they're talking about the things that are getting multiplied in each term. So, for example, if you said, what are the factors of the first term? The first term refers to this one right over here-- 2 times And there's two factors. There's a 2 and a 3, and they are being multiplied by each other. So here you have two factors in the first term. What about the second term? This was the first term. The second term here has only one factor, just the 4. It's not being multiplied by anything. And the third term here, once again, has two factors. It's the product of 7 times y. So we have two factors here. We have a 7 and a y. And this constant factor here, this number 7 also has a special name. It is called the coefficient of this term-- coefficient. And the coefficient is the nonvariable that multiplies the rest of the term. That's one way of thinking about it. So here's 7y. Even if it was 7xy or 7xyz or 7xyz squared, that nonvariable that's multiplying everything else, we would consider to be the coefficient. Now, let's do a few more examples. And in each of these-- I encourage you actually right now to pause the video-- think about what the terms are. How many terms are there in each expression, how many factors in each term, and what are the coefficients? So let's look at this first one. It's clear that we have three things being added together. This is the first term. This is the second term. And this is the third term. So this is the first term. This is the second term. This is the third term. And they each have two factors. This first one has the factors 3 and x." + }, + { + "Q": "\nAt 1:03, Sal says that the term is 7y, but isn't it -7y ?", + "A": "Whether it s negative (7y) or not, it really depends on what the value of y is. : If y is negative, then 7y will be a negative number too : However if y is positive, then this term will be positive", + "video_name": "9_VCk9tWT0Y", + "timestamps": [ + 63 + ], + "3min_transcript": "What I want to do in this video is think about how expressions are formed and the words we use to describe the different parts of an expression. And the reason why this is useful is when you hear other people refer to some expression and say, oh, I don't agree with the second term, or the third term has four factors, or why is the coefficient on that term 6, you'll know what they're talking about, and you can communicate in the same way. So let's think about what those words actually mean. So we have an expression here. And the first thing I want to think about are the terms of an expression or what a term is. And one way to think about it is the terms are the things that are getting added and subtracted. So, for example, in this expression right over here, you have three things that are getting added and subtracted. The first thing, you're taking 2 times 3. You're adding that to 4. And then from that, you're subtracting 7y. So in this example, you have three terms. The first term is 2 times 3. The second term is just the number 4. Now, let's think about the term \"factor.\" And when people are talking about a factor, especially in terms of an expression, they're talking about the things that are getting multiplied in each term. So, for example, if you said, what are the factors of the first term? The first term refers to this one right over here-- 2 times And there's two factors. There's a 2 and a 3, and they are being multiplied by each other. So here you have two factors in the first term. What about the second term? This was the first term. The second term here has only one factor, just the 4. It's not being multiplied by anything. And the third term here, once again, has two factors. It's the product of 7 times y. So we have two factors here. We have a 7 and a y. And this constant factor here, this number 7 also has a special name. It is called the coefficient of this term-- coefficient. And the coefficient is the nonvariable that multiplies the rest of the term. That's one way of thinking about it. So here's 7y. Even if it was 7xy or 7xyz or 7xyz squared, that nonvariable that's multiplying everything else, we would consider to be the coefficient. Now, let's do a few more examples. And in each of these-- I encourage you actually right now to pause the video-- think about what the terms are. How many terms are there in each expression, how many factors in each term, and what are the coefficients? So let's look at this first one. It's clear that we have three things being added together. This is the first term. This is the second term. And this is the third term. So this is the first term. This is the second term. This is the third term. And they each have two factors. This first one has the factors 3 and x." + }, + { + "Q": "\non 5:34 that expression the last 3 factors are x, y, z what would be the coefficients of that term?", + "A": "The coefficient would be 1. It doesn t have to be stated that 1*x*y^2*z^5 for 1 to be the coefficient as the 1 multiplies way to x*y^2*z^5. Also the 1 is not a factor of that term.", + "video_name": "9_VCk9tWT0Y", + "timestamps": [ + 334 + ], + "3min_transcript": "And if you look at that level, if you look at the first term, and you say, well, how many factors does that have? Well, you would say that it has three factors-- x, y, and z. How many factors does the second term have? Well, you could say, well, it has two factors. One factor is x plus y, and then the other factor is y. The first factor is x plus 1. And the second one is y. It's multiplying this expression. This smaller expression itself is one of the factors. And the other one is y. And then this third one also has two factors, a 4 and an x. And if someone said, hey, what's the coefficient on this term? You would say, hey, look, the coefficient is the 4. Now, let's look at this one over here. Actually, before I look at that one, what was interesting about this is that here you had a little smaller expression itself acting as one of the factors. So then you can go and then zoom in on this expression right over here. And you can ask the same question. On this smaller expression, how many terms does it have? Well, it has two terms-- an x and a 1. And each of them have exactly one factor. So when we're giving these, you can keep nesting these expressions to think about when you talk about terms or factors or factors of terms, you have to really specify what part of the nesting you're thinking about. If you're talking about the terms of this whole expression, there's one, two, three. But then you could look at this subexpression, which itself is a factor of a term, and say, oh, well, there's only two terms in this one. Now, let's look at this one. How many terms? Well, once again, there's clearly three. Actually, let me add one more, because I'm tired of expressions with three terms. So I'm just going to add a 1 here. So now, we clearly have four terms. This is the first term, second term, third term, fourth term. And how many factors are in each of them? Well, this is interesting. You might say, well, the factors are the things that are being multiplied. But here I'm dividing by y. Remember, dividing by y is the same thing as multiplying by its reciprocal. to have three factors here, where the factors are 3x and 1/y. If you multiplied 3 times x times 1/y, you're going to get exactly what you have right over here. So you would say this has three factors. If someone asks, what's the coefficient here? Well, you'd say, well, that 3 is the coefficient. Here how many factors do you have? And this is a little bit tricky, because you might say, well, isn't 5x squared times y, isn't that equal to 5 times x times x times y? And you'd be right. So it would be very tempting to say that you have four factors. But the convention, the tradition that most people use, is that they consider the exponent with x as a base as just one factor, this as just one factor. So traditionally, people will say this has three factors. It has a 5x squared and a y. x squared is just considered a factor. And once again, what's the coefficient? It's the 5." + }, + { + "Q": "At 4:08, Sal's final expression (2rp - 4\u00cf\u0080r^2) had three terms : r, p, and r^2. But didn't the question ask for a binomial, which has two terms? I'm confused....\n", + "A": "This expression has two terms. I think you counted the types of variables in the expression instead of the terms. A term is an expression that doesn t have the operations of addition and subtraction. 2rp<---One term 4\u00cf\u0080r^2<---Second term There are two terms in this expression; therefore, it s a binomial. I hope this helped!", + "video_name": "EvvxBdNIUeQ", + "timestamps": [ + 248 + ], + "3min_transcript": "If p is greater than 7r, then 2-- let me write it this way. We know that p is greater than 7r. So if we're going to multiply both sides of this equation by 2rr-- and 2r is positive, we're dealing with positive distances, positive lengths-- so if we multiply both sides of this equation by 2r, it shouldn't change the equation. So multiply that by 2r, and then multiply this by 2r. And then our equation becomes 2rp is greater than 14r squared. Now, why is this interesting? Actually, why did I even multiply this by 2r? Well, that's so that this becomes the same as the area of the rectangle. So this is the area of the rectangle. And what's 14r squared? Well, 4 times pi, is going to get us something less than 14. So this is 4 pi is less than 14. 14 is 4 times 3 and 2-- let me put it this way. 4 times 3.5 is equal to 14. So 4 times pi, which is less than 3.5, is going to be less than 14. So we know that this over here is larger than this quantity over here. It's larger than 4 pi r squared. And so we know that this rectangle has a larger area than the circle. So we can just subtract the circle's area from the rectangle's area to find the difference. So the difference is going to be the area of the rectangle, which we already figured out is 2rp. And we're going to subtract from that the area of the circle. The area of the circle is 4 pi r squared. And one point I want to clarify. I gave the equation of the area of a circle to be pi r squared. And then we said that the radius is actually 2r in this case. So I substituted 2r for r. Hopefully that doesn't confuse you. This r is the general term for any radius. They later told us that the actual radius is 2 times some letter r. So I substitute that into the formula. Anyway, hopefully you found that useful." + }, + { + "Q": "5:42: How did he get 2x squared?\n", + "A": "during the Pythagorean theorem he had x^2 + X^2 = c^2 right? Well when u put x^2 and x^2 together u get 2x^2", + "video_name": "McINBOFCGH8", + "timestamps": [ + 342 + ], + "3min_transcript": "Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x. And then we can use the Pythagorean Theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the square of length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. We can take the principal root of both sides of that. I wanted to just change it to yellow. Last, take the principal root of both sides of that. The left-hand side you get, principal root of 2 is just square root of 2, and then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2 So if you have a right isosceles triangle, whatever the two legs are, they're going to have the same length. That's why it's isosceles. The hypotenuse is going to be square root of 2 times that. So c is equal to x times the square root of 2. So for example, if you have a triangle that looks like this. Let me draw it a slightly different way. It's good to have to orient ourselves in different ways every time. So if we see a triangle that's 90 degrees, 45 and 45 like this, and you really just have to know two of these angles to know what the other one is going to be, and if I tell you that this side right over here is 3-- I actually don't even have to tell you that this other side's going to be 3. This is an isosceles triangle, so those two legs are going to be the same. And you won't even have to apply the Pythagorean theorem if you know this-- and this is a good one to know-- that the hypotenuse here, the side opposite the 90 degree side, is just going to be square root of 2 times the length of either of the legs. So it's going to be 3 times the square root of 2. in a 45-45-90 triangle or a right isosceles triangle, the ratio of the sides are one of the legs can be 1. Then the other leg is going to have the same measure, the same length, and then the hypotenuse is going to be square root of 2 times either of those. 1 to 1, 2 square root of 2. So this is 45-45-90. That's the ratios. And just as a review, if you have a 30-60-90, the ratios were 1 to square root of 3 to 2. And now we'll apply this in a bunch of problems." + }, + { + "Q": "Why do you not express the ratio of the 30,60,90 as 3:4:5? I seems a lot less confusing.\n", + "A": "Just because 3:4:5 immediately relates to pytha. thereom... which isn t always the case with 30.60.90 s.", + "video_name": "McINBOFCGH8", + "timestamps": [ + 184 + ], + "3min_transcript": "on this ratio right over here. Just an example, if you see a triangle that looks like this, where the sides are 2, 2 square root of 3, and 4. Once again, the ratio of 2 to 2 square root of 3 is 1 to square root of 3. The ratio of 2 to 4 is the same thing as 1 to 2. This right here must be a 30-60-90 triangle. What I want to introduce you to in this video is another important type of triangle that shows up a lot in geometry and a lot in trigonometry. And this is a 45-45-90 triangle. Or another way to think about is if I have a right triangle that is also isosceles. You obviously can't have a right triangle that is equilateral, because an equilateral triangle has all of their angles have to be 60 degrees. But you can have a right angle, you can have a right triangle, that is isosceles. And isosceles-- let me write this-- this is a right isosceles triangle. are equal. So these are the two sides that are equal. And then if the two sides are equal, we have proved to ourselves that the base angles are equal. And if we called the measure of these base angles x, then we know that x plus x plus 90 have to be equal to 180. Or if we subtract 90 from both sides, you get x plus x is equal to 90 or 2x is equal to 90. Or if you divide both sides by 2, you get x is equal to 45 degrees. So a right isosceles triangle can also be called-- and this is the more typical name for it-- it can also be called a 45-45-90 triangle. And what I want to do this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle. Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x. And then we can use the Pythagorean Theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the square of length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. We can take the principal root of both sides of that. I wanted to just change it to yellow. Last, take the principal root of both sides of that. The left-hand side you get, principal root of 2 is just square root of 2, and then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2" + }, + { + "Q": "What does delta mean like at 5:08?\n", + "A": "Delta means the change in a quantity. It is a Greek letter that looks like a triangle. It is used in Math, Science, and others to show that you re looking at how something is changing without having to write out: the change in... . Kinda like when you re laughing out loud and you text someone lol .", + "video_name": "9wOalujeZf4", + "timestamps": [ + 308 + ], + "3min_transcript": "One, two, three. Our delta y-- and I'm just doing it because I want to hit an even number here-- our delta y is equal to-- we go down by 2-- it's equal to negative 2. So for A, change in y for change in x. When our change in x is 3, our change in y is negative 2. So our slope is negative 2/3. When we go over by 3, we're going to go down by 2. Or if we go over by 1, we're going to go down by 2/3. You can't exactly see it there, but you definitely see it when you go over by 3. So that's our slope. We've essentially done half of that problem. Now we have to figure out the y-intercept. So that right there is our m. Now what is our b? Our y-intercept. Well where does this intersect the y-axis? Well we already said the slope is 2/3. So this is the point y is equal to 2. gone down by 2/3. So this right here must be the point 1 1/3. Or another way to say it, we could say it's 4/3. That's the point y is equal to 4/3. Right there. A little bit more than 1. About 1 1/3. So we could say b is equal to 4/3. So we'll know that the equation is y is equal to m, negative 2/3, x plus b, plus 4/3. That's equation A. Let's do equation B. Hopefully we won't have to deal with as many fractions here. Equation B. Let's figure out its slope first. Let's start at some reasonable point. We could start at that point. Let me do it right here. B. Equation B. When our delta x is equal to-- let me write it this way, delta x. So our delta x could be 1. When we move over 1 to the right, what happens We go up by 3. delta x. delta y. Our change in y is 3. So delta y over delta x, When we go to the right, our change in x is 1. Our change in y is positive 3. So our slope is equal to 3. What is our y-intercept? Well, when x is equal to 0, y is equal to 1. So b is equal to 1. So this was a lot easier. Here the equation is y is equal to 3x plus 1. Let's do that last line there. Line C Let's do the y-intercept first. You see immediately the y-intercept-- when x is equal to 0, y is negative 2. So b is equal to negative 2. And then what is the slope? m is equal to change in y over change in x." + }, + { + "Q": "\nI dont get it. Any equation on how 0.2 becomes 1/5? at 8:03?", + "A": "0.2 is equal to 1/5. A simple way to explain it is how many 0.2 s in 1.0. There are 5. 0.2 is just 1 of those 5, so 1/5.", + "video_name": "9wOalujeZf4", + "timestamps": [ + 483 + ], + "3min_transcript": "If we go over to the right by one, two, three, four. So our change in x is equal to 4. What is our change in y? Our change in y is positive 2. So change in y is 2 when change in x is 4. So the slope is equal to 1/2, 2/4. So the equation here is y is equal to 1/2 x, that's our slope, minus 2. And we're done. Now let's go the other way. Let's look at some equations of lines knowing that this is the slope and this is the y-intercept-- that's the m, that's the b-- and actually graph them. Let's do this first line. I already started circling it in orange. The y-intercept is 5. When x is equal to 0, y is equal to 5. You can verify that on the equation. So when x is equal to 0, y is equal to one, two, three, four, five. That's the y-intercept and the slope is 2. in the y-direction. If I move 1 in the x-direction, I move up 2 in If I move 1 in the x-direction, I move up 2 in the y-direction. If I move back 1 in the x-direction, I move down 2 in If I move back 1 in the x-direction, I move down 2 in I keep doing that. So this line is going to look-- I can't draw lines too neatly, but this is going to be my best shot. It's going to look something like that. It'll just keep going on, on and on and on. So that's our first line. I can just keep going down like that. Let's do this second line. y is equal to negative 0.2x plus 7. Let me write that. y is equal to negative 0.2x plus 7. It's always easier to think in fractions. So 0.2 is the same thing as 1/5. We could write y is equal to negative 1/5 x plus 7. So it's one, two, three, four, five, six. That's our y-intercept when x is equal to 0. This tells us that for every 5 we move to the right, we move down 1. We can view this as negative 1/5. The delta y over delta x is equal to negative 1/5. For every 5 we move to the right, we move down 1. So every 5. One, two, three, four, five. We moved 5 to the right. That means we must move down 1. We move 5 to the right. One, two, three, four, five. We must move down 1. If you go backwards, if you move 5 backwards-- instead of this, if you view this as 1 over negative 5. These are obviously equivalent numbers. If you go back 5-- that's negative 5. One, two, three, four, five. Then you move up 1. If you go back 5-- one, two, three, four, five-- you move up 1." + }, + { + "Q": "I still don't get how to find x and how you know that it is automatically zero based on something else at 9:12 and why is it that when y=-x that equeals M but when y=3.75 that equeals b. is x always zero?\n", + "A": "i think you subtract,add, divide,or multiply 3.75 to both sides of the equation depending on what the sign is you do the opposite :)", + "video_name": "9wOalujeZf4", + "timestamps": [ + 552 + ], + "3min_transcript": "in the y-direction. If I move 1 in the x-direction, I move up 2 in If I move 1 in the x-direction, I move up 2 in the y-direction. If I move back 1 in the x-direction, I move down 2 in If I move back 1 in the x-direction, I move down 2 in I keep doing that. So this line is going to look-- I can't draw lines too neatly, but this is going to be my best shot. It's going to look something like that. It'll just keep going on, on and on and on. So that's our first line. I can just keep going down like that. Let's do this second line. y is equal to negative 0.2x plus 7. Let me write that. y is equal to negative 0.2x plus 7. It's always easier to think in fractions. So 0.2 is the same thing as 1/5. We could write y is equal to negative 1/5 x plus 7. So it's one, two, three, four, five, six. That's our y-intercept when x is equal to 0. This tells us that for every 5 we move to the right, we move down 1. We can view this as negative 1/5. The delta y over delta x is equal to negative 1/5. For every 5 we move to the right, we move down 1. So every 5. One, two, three, four, five. We moved 5 to the right. That means we must move down 1. We move 5 to the right. One, two, three, four, five. We must move down 1. If you go backwards, if you move 5 backwards-- instead of this, if you view this as 1 over negative 5. These are obviously equivalent numbers. If you go back 5-- that's negative 5. One, two, three, four, five. Then you move up 1. If you go back 5-- one, two, three, four, five-- you move up 1. I have to just connect the dots. I think you get the idea. I just have to connect those dots. I could've drawn it a little bit straighter. Now let's do this one, y is equal to negative x. Where's the b term? I don't see any b term. You remember we're saying y is equal to mx plus b. Where is the b? Well, the b is 0. You could view this as plus 0. Here is b is 0. When x is 0, y is 0. That's our y-intercept, right there at the origin. And then the slope-- once again you see a negative sign. You could view that as negative 1x plus 0. So slope is negative 1. When you move to the right by 1, when change in x is 1, change in y is negative 1. When you move up by 1 in x, you go down by 1 in y. Or if you go down by 1 in x, you're going to go up by 1 in y. x and y are going to have opposite signs. They go in opposite directions. So the line is going to look like that." + }, + { + "Q": "\n5:20 Why is it written n greater than or equal to 2?", + "A": "In the formula being used, the nth term = the (n-1)th term + a common difference. In other words, the value of the nth term depends on the value of the term before it If n were 1, the term before it would be the 0th term --- which doesn t exist. So n is restricted to 2 and above in order that the term before will be no less than the first term.", + "video_name": "_cooC3yG_p0", + "timestamps": [ + 320 + ], + "3min_transcript": "And we could write that this is the sequence a sub n, n going from 1 to infinity of-- and we could just say a sub n, if we want to define it explicitly, is equal to 100 plus we're adding 7 every time. And then each term-- the second term we added 7 once. Third term-- we add 7 twice. So for the nth term, we're going to add 7 n minus 1 times. So this is an explicit definition of it, but we could also do it recursively. So just to be clear, this is one definition where we write it like this, or we could write a sub n, from n equals 1 to infinity. And in either case I should write with. And if I want to define it recursively, I could say a sub 1 is equal to 100. And then, for anything larger than 1, for any index above 1, And so we're done. This is another way of defining it. So in general, if you wanted a generalizable way to spot or define an arithmetic sequence, you could say an arithmetic sequence is going to be of the form a sub n-- if we're talking about an infinite one-- from n equals 1 to infinity. If you want to define it explicitly, you could say a sub n is equal to some constant, which would essentially the first term. It would be some constant plus some number that your incrementing-- or I guess this could be a negative number, or decrementing by-- times n minus 1. So this is one way to define an arithmetic sequence. In this case, d was 2. In this case, d is 7. That's how much you're adding by each time. And in this case, k is negative 5, and in this case, k is 100. the recursive way of defining an arithmetic sequence generally, you could say a sub 1 is equal to k, and then a sub n is equal to a sub n minus 1. A given term is equal to the previous term plus d for n greater than or equal to 2. So once again, this is explicit. This is the recursive way of defining it. And we would just write with there. Now the last question I have is is this one right over here an arithmetic sequence? Well, let's check it out. We start at 1. Then we add 2. Then we add 3. So this is an immediate giveaway that this is not an arithmetic sequence. Now we are adding 4. We're adding a different amount every time. So this, first of all, this is not arithmetic. This is not an arithmetic sequence. But how could we define this, since we're trying to define our sequences?" + }, + { + "Q": "At 4:19, Sal pronounces \"arithmetic\" wrong.\n", + "A": "That was defiantly Worth pointing out", + "video_name": "_cooC3yG_p0", + "timestamps": [ + 259 + ], + "3min_transcript": "I could say a sub 1 is equal to negative 5. And then each successive term, for a sub 2 and greater-- so I could say a sub n is equal to a sub n minus 1 plus 3. Each term is equal to the previous term-- oh, not 3-- plus 2. So this is for n is greater than or equal to 2. So either of these are completely legitimate ways of defining the arithmetic sequence that we have here. We can either define it explicitly, or we could define it recursively. Now let's look at this sequence. Is this one arithmetic? Well, we're going from 100. We add 7. 107 to 114, we're adding 7. 114 to 121, we are adding 7. So this is indeed an arithmetic sequence. So just to be clear, this is one, And we could write that this is the sequence a sub n, n going from 1 to infinity of-- and we could just say a sub n, if we want to define it explicitly, is equal to 100 plus we're adding 7 every time. And then each term-- the second term we added 7 once. Third term-- we add 7 twice. So for the nth term, we're going to add 7 n minus 1 times. So this is an explicit definition of it, but we could also do it recursively. So just to be clear, this is one definition where we write it like this, or we could write a sub n, from n equals 1 to infinity. And in either case I should write with. And if I want to define it recursively, I could say a sub 1 is equal to 100. And then, for anything larger than 1, for any index above 1, And so we're done. This is another way of defining it. So in general, if you wanted a generalizable way to spot or define an arithmetic sequence, you could say an arithmetic sequence is going to be of the form a sub n-- if we're talking about an infinite one-- from n equals 1 to infinity. If you want to define it explicitly, you could say a sub n is equal to some constant, which would essentially the first term. It would be some constant plus some number that your incrementing-- or I guess this could be a negative number, or decrementing by-- times n minus 1. So this is one way to define an arithmetic sequence. In this case, d was 2. In this case, d is 7. That's how much you're adding by each time. And in this case, k is negative 5, and in this case, k is 100." + }, + { + "Q": "\nat 6:41, is there not a way to write the third question in an explicit from? If there is, can someone show me how?", + "A": "Actually you re asking for the common formula that is taught in the next lessons: a_n = (n * (n + 1))/2", + "video_name": "_cooC3yG_p0", + "timestamps": [ + 401 + ], + "3min_transcript": "the recursive way of defining an arithmetic sequence generally, you could say a sub 1 is equal to k, and then a sub n is equal to a sub n minus 1. A given term is equal to the previous term plus d for n greater than or equal to 2. So once again, this is explicit. This is the recursive way of defining it. And we would just write with there. Now the last question I have is is this one right over here an arithmetic sequence? Well, let's check it out. We start at 1. Then we add 2. Then we add 3. So this is an immediate giveaway that this is not an arithmetic sequence. Now we are adding 4. We're adding a different amount every time. So this, first of all, this is not arithmetic. This is not an arithmetic sequence. But how could we define this, since we're trying to define our sequences? So we could say, this is equal to a sub n, where n is starting at 1 and it's going to infinity, with-- we'll say our base case-- a sub 1 is equal to 1. And then for n is 2 or greater, a sub n is going to be equal to what? So a sub 2 is the previous term plus 2. a sub 3 is the previous term plus 3. a sub 4 is the previous term plus 4. So it's going to be the previous term plus whatever your index is. So this looks close, but notice here we're changing the amount that we're adding based on what our index is. We're adding the amount of index to the previous term. And so this is for n is greater than or equal to 2. Well for an arithmetic sequence, we're adding the same amount regardless of what our index is. Here we're adding the index itself. an interesting sequence nonetheless." + }, + { + "Q": "On 0:08, is it first first quadratic formula? Yes or no?\n\nAnd on 2:27-2:28, what's binomial? Anyone tell me??\n", + "A": "The equation shown at 0:08 is the standard form of a quadratic expression or equation, not the quadratic formula. A binomial is a polynomial having only two terms. Hope that helped.", + "video_name": "r3SEkdtpobo", + "timestamps": [ + 8, + 147, + 148 + ], + "3min_transcript": "In the last video, I told you that if you had a quadratic equation of the form ax squared plus bx, plus c is equal to zero, you could use the quadratic formula to find the solutions to this equation. And the quadratic formula was x. The solutions would be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And we learned how to use it. You literally just substitute the numbers a for a, b for b, c for c, and then it gives you two answers, because you have a plus or a minus right there. What I want to do in this video is actually prove it to you. Prove that using, essentially completing the square, I can get from that to that right over there. So the first thing I want to do, so that I can start completing the square from this point right here, is-- let me rewrite the equation right here-- so we have ax-- bx, plus c is equal to 0. So the first I want to do is divide everything by a, so I just have a 1 out here as a coefficient. So you divide everything by a, you get x squared plus b over ax, plus c over a, is equal to 0 over a, which is still just 0. Now we want to-- well, let me get the c over a term on to the right-hand side, so let's subtract c over a from both sides. And we get x squared plus b over a x, plus-- well, I'll just leave it blank there, because this is gone now; we subtracted it from both sides-- is equal to negative c over a I left a space there so that we can complete the square. And you saw in the completing the square video, you and you square it. So what is b over a divided by 2? Or what is 1/2 times b over a? Well, that is just b over 2a, and, of course, we are going to square it. You take 1/2 of this and you square it. That's what we do in completing a square, so that we can turn this into the perfect square of a binomial. Now, of course, we cannot just add the b over 2a squared to the left-hand side. We have to add it to both sides. So you have a plus b over 2a squared there as well. Now what happens? Well, this over here, this expression right over here, this is the exact same thing as x plus b over 2a squared. And if you don't believe me, I'm going to multiply it out." + }, + { + "Q": "At 1:17 Sal divides \"everything by a\" but writes ... x(b/a) instead of (b * x) / a. Why does x not get divided by a?\n", + "A": "x(b/a) is exactly the same as (b * x) / a Here are the steps: x (b / a) = x/1 * (b / a) = (b * x) / (a * 1) = (b * x) / a", + "video_name": "r3SEkdtpobo", + "timestamps": [ + 77 + ], + "3min_transcript": "In the last video, I told you that if you had a quadratic equation of the form ax squared plus bx, plus c is equal to zero, you could use the quadratic formula to find the solutions to this equation. And the quadratic formula was x. The solutions would be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And we learned how to use it. You literally just substitute the numbers a for a, b for b, c for c, and then it gives you two answers, because you have a plus or a minus right there. What I want to do in this video is actually prove it to you. Prove that using, essentially completing the square, I can get from that to that right over there. So the first thing I want to do, so that I can start completing the square from this point right here, is-- let me rewrite the equation right here-- so we have ax-- bx, plus c is equal to 0. So the first I want to do is divide everything by a, so I just have a 1 out here as a coefficient. So you divide everything by a, you get x squared plus b over ax, plus c over a, is equal to 0 over a, which is still just 0. Now we want to-- well, let me get the c over a term on to the right-hand side, so let's subtract c over a from both sides. And we get x squared plus b over a x, plus-- well, I'll just leave it blank there, because this is gone now; we subtracted it from both sides-- is equal to negative c over a I left a space there so that we can complete the square. And you saw in the completing the square video, you and you square it. So what is b over a divided by 2? Or what is 1/2 times b over a? Well, that is just b over 2a, and, of course, we are going to square it. You take 1/2 of this and you square it. That's what we do in completing a square, so that we can turn this into the perfect square of a binomial. Now, of course, we cannot just add the b over 2a squared to the left-hand side. We have to add it to both sides. So you have a plus b over 2a squared there as well. Now what happens? Well, this over here, this expression right over here, this is the exact same thing as x plus b over 2a squared. And if you don't believe me, I'm going to multiply it out." + }, + { + "Q": "At 6:30, why couldn't you take \"b^2\" out of the radical? Because the square root of b^2 is just plus/minus b ? Or am I missing something?\n", + "A": "You can only take factors out of root signs if they re factors, but in this case it s being added not multiplied so you can t split it up.", + "video_name": "r3SEkdtpobo", + "timestamps": [ + 390 + ], + "3min_transcript": "So the left-hand side simplifies to this. The right-hand side, maybe not quite as simple. Maybe we'll leave it the way it is right now. Actually, let's simplify it a little bit. So the right-hand side, we can rewrite it. This is going to be equal to-- well, this is going to be b squared. I'll write that term first. This is b-- let me do it in green so we can follow along. So that term right there can be written as b squared over 4a square. And what's this term? What would that become? This would become-- in order to have 4a squared as the denominator, we have to multiply the numerator and the denominator by 4a. So this term right here will become minus 4ac over 4a squared. And you can verify for yourself that that is the same thing as that. denominator by 4a. In fact, the 4's cancel out and then this a cancels out and you just have a c over a. So these, this and that are equivalent. I just switched which I write first. And you might already be seeing the beginnings of the quadratic formula here. So this I can rewrite. This I can rewrite. The right-hand side, right here, I can rewrite as b squared minus 4ac, all of that over 4a squared. This is looking very close. Notice, b squared minus 4ac, it's already appearing. We don't have a square root yet, but we haven't taken the square root of both sides of this equation, so let's do that. So if you take the square root of both sides, the left-hand side will just become x plus-- let me scroll down a little bit-- x plus b over 2a is going to be equal to the plus or minus square root of this thing. numerator over the square root of the denominator. So it's going to be the plus or minus the square root of b squared minus 4ac over the square root of 4a squared. Now, what is the square root of 4a squared? It is 2a, right? 2a squared is 4a squared. The square root of this is that right here. So to go from here to here, I just took the square root of both sides of this equation. Now, this is looking very close to the quadratic. We have a b squared minus 4ac over 2a, now we just essentially have to subtract this b over 2a from both sides of the equation and we're done. So let's do that. So if you subtract the b over 2a from both sides of this equation, what do you get? You get x is equal to negative b over 2a, plus or minus the square root of b squared minus 4ac over 2a, common" + }, + { + "Q": "\nAt 5:00, How do you already know that the sin^2(a) plus the cos^2(a) equals one?", + "A": "let us imagine a right triangle \u00ef\u00bc\u008co=opposite\u00ef\u00bc\u008ch=hypotenuse\u00ef\u00bc\u008ca= adjacent. sin\u00ef\u00bc\u0088a\u00ef\u00bc\u0089=o/h\u00ef\u00bc\u008ccos\u00ef\u00bc\u0088a\u00ef\u00bc\u0089=a/h\u00ef\u00bc\u008cright\u00ef\u00bc\u009fso \u00ef\u00bc\u009a sin \u00c2\u00b2\u00ef\u00bc\u0088a\u00ef\u00bc\u0089=o\u00c2\u00b2/h\u00c2\u00b2 and cos\u00c2\u00b2\u00ef\u00bc\u0088a\u00ef\u00bc\u0089=a\u00c2\u00b2/h\u00c2\u00b2 we can add up these\u00ef\u00bc\u008cand get\u00ef\u00bc\u009a sin \u00c2\u00b2\u00ef\u00bc\u0088a\u00ef\u00bc\u0089+ cos\u00c2\u00b2\u00ef\u00bc\u0088a\u00ef\u00bc\u0089=\u00ef\u00bc\u0088o\u00c2\u00b2+a\u00c2\u00b2\u00ef\u00bc\u0089/h\u00c2\u00b2 \u00ef\u00bc\u008c we know that \u00ef\u00bc\u009a\u00ef\u00bc\u0088o\u00c2\u00b2+a\u00c2\u00b2\u00ef\u00bc\u0089=h\u00c2\u00b2 \u00ef\u00bc\u008cthen we can get the answer\u00ef\u00bc\u008chope that could help you\u00ef\u00bc\u0081", + "video_name": "a70-dYvDJZY", + "timestamps": [ + 300 + ], + "3min_transcript": "well, you use these same properties. Cosine of minus b, that's still going to be cosine on b. So that's going to be the cosine of a times the cosine-- cosine of minus b is the same thing as cosine of b. But here you're going to have sine of minus b, which is the same thing as the minus sine of b. And that minus will cancel that out, so it'll be plus sine of a times the sine of b. When you have a plus sign here you get a minus there. When you don't minus sign there, you get a plus sign there. But fair enough. I don't want to dwell on that too much because we have many more identities to show. So what if I wanted an identity for let's say, the cosine of 2a? So the cosine of 2a. Well that's just the same thing as the cosine of a plus a. If my second a is just my b, then this is just equal to cosine of a times the cosine of a minus the sine of a times the sine of a. My b is also an a in this situation, which I could rewrite as, this is equal to the cosine squared of a. I just wrote cosine of a times itself twice or times itself. Minus sine squared of a. This is one I guess identity already. Cosine of 2a is equal to the cosine squared of a minus the sine squared of a. Let me box off my identities that we're showing in this video. So I just showed you that one. What if I'm not satisfied? What if I just want it in terms of cosines? Well, we could break out the unit circle definition of our trig functions. This is kind of the most fundamental identity. The sine squared of a plus the cosine squared Or you could write that-- let me think of the best way to do this. You could write that the sine squared of a is equal to 1 minus the cosine sign squared of a. And then we could take this and substitute it right here. So we could rewrite this identity as being equal to the cosine squared of a minus the sine squared of a. But the sine squared of a is this right there. So minus-- I'll do it in a different color. Minus 1 minus cosine squared of a. That's what I just substituted for the sine squared of a. And so this is equal to the cosine squared of a minus 1 plus the cosine squared of a. Which is equal to-- we're just adding. I'll just continue on the right. We have 1 cosine squared of a plus another cosine squared of a, so it's 2 cosine squared of a minus 1." + }, + { + "Q": "At 12:06pm, How do i verify csc^4 (x) - cot^4 (x) = 1 + 2 cot^2 (x)\ni got to (csc^2 (x) + cot^2 (x))1 = 1 + 2cot^2 (x) (i think the right side can convert to 2cscx but im stuck)\n", + "A": "Identity to use: 1 + cot\u00c2\u00b2 x = csc\u00c2\u00b2 x We need to square this. Thus, [1 + cot\u00c2\u00b2 x]\u00c2\u00b2 = [csc\u00c2\u00b2 x ]\u00c2\u00b2 1 + 2cot\u00c2\u00b2 x + cot\u00e2\u0081\u00b4 x = csc\u00e2\u0081\u00b4 x Let us then use this to substitute for csc\u00e2\u0081\u00b4 x in your equation: csc\u00e2\u0081\u00b4 (x) - cot\u00e2\u0081\u00b4 (x) = 1 + 2 cot\u00c2\u00b2 (x) (1 + 2cot\u00c2\u00b2 x + cot\u00e2\u0081\u00b4 x) - cot\u00e2\u0081\u00b4 (x) = 1 + 2 cot\u00c2\u00b2 (x) 1 + 2cot\u00c2\u00b2 x = 1 + 2 cot\u00c2\u00b2 (x) As, both sides of the equation are now the same thing, we have proved the original.", + "video_name": "a70-dYvDJZY", + "timestamps": [ + 726 + ], + "3min_transcript": "squared of a is equal to 1/2 times 1 minus cosine of 2a. And we have our other discovery I guess we could call it. Our finding. And it's interesting. It's always interesting to look at the symmetry. Cosine squared-- they're identical except for you have a plus 2a here for the cosine squared and you have a minus cosine of 2a here for the sine squared. So we've already found a lot of interesting things. Let's see if we can do anything with the sine of 2a. Let me pick a new color here that I haven't used. Well, I've pretty much used all my colors. So if I want to figure out the sine of 2a, this is equal to the sine of a plus a. Which is equal to the sine of a times the co-- well, I don't Times the cosine of a plus-- and this cosine of a, that's the second a. Actually, you could view it that way. Plus the sine-- I'm just using the sine of a plus b. Plus the sine of the second a times the cosine of the first a. I just wrote the same thing twice, so this is just people to 2 sine of a, cosine of a. That was a little bit easier. So sine of 2a is equal to that. So that's another result. I know I'm a little bit tired by playing with all of these sine and cosines. And I was able to get all the results that I needed for my calculus problem, so hopefully this was a good review for you because it was a good review for me. You can write these things down. You can memorize them if you want, but the really important take away is to realize that you really can derive all of these formulas really from these initial formulas that we just had. And even these, I also have proofs to show you how to get trig functions." + }, + { + "Q": "\nat 2:01, vi said \"hamentashe\" or something. What are those?!", + "A": "Hamantash is a triangular cookie or pastry, often associated with the Jewish holiday Purim.", + "video_name": "o6KlpIWhbcw", + "timestamps": [ + 121 + ], + "3min_transcript": "So you're me and you're in math class and-- triangles, triangles, triangles-- I don't know. The teacher keeps saying words, and you're supposed to be doing something with trigons, whatever those are. But you're bored and-- triangles, triangles, triangles. Sure you could draw your triangles separately, They're happiest when snuggled up together into a triangle party. Everybody knows triangles love parties. Sometimes they get together and do these triangle congo lines. If you keep adding new triangles on the same side, it gets all curvy and spirally. Or you can alternate and it goes pretty straight. In fact, since all the sides of the triangle are supposed to be straight lines, and since they're all lying on top of some previous straight line, then this whole line would have to be straight if these were actually triangles. Since it's not, it's proof that these aren't quite triangles. Maybe they've been partying a little too hard, but hey, at least you're not doing math. Speaking of which, the teacher is still going on about types of trigons, and you're supposed to be taking notes. But you're more interesting in types of triangles, which you already know all about. There are fat triangles, and pointy triangles, and perfect triangles, and cheese slice triangles which are a kind of pointy triangle but are symmetric like a slice of cheese or cake. Super pointy triangles are fun to stack into triangle stacks. You can put all the points facing one direction, but the stack starts to wobble too much towards that direction. So it's good to put some facing the other direction before you go too far. You'll notice pretty quickly, that the skinnier the triangle, the less wobble it adds to the stack. can put just one not so skinny triangle that's pointing the other way. Or maybe you want to wobble, because you have to navigate your triangle stack around your notes. In which case you can even alternate back and forth as long as you make the triangles point towards where you want to go, a little less skinny. The easy part about triangle stacks, is that there's really only one part of the triangle that's important, as far as the stack is concerned. And that's the pointy point. The other two angles can be fat and skinny, or skinny and fat, or both the same if it's a cheese slice, and it doesn't change the rest of the stack. Unless the top angle is really wide, because then you'll get two skinny points, and which side should you continue to stack on? Also, instead of thinking fat and skinny, you should probably create code words that won't set off your teacher's mind reading alarms for non-math related thoughts. So you pick two words off the board, obtuse and acute, which by sheer coincidence I'm sure, just happen to mean fat angle and skinny angle. Of course, those are also kinds of triangles. Which doesn't make much sense, because the obtuser one angle of a triangle is, the more acute the other two get. Yet, if you make an acute triangle with the same perimeter, it has more area, which seems like an obtuse quality. And then, can you still call an obtuse equilateral triangle a cheese or cake-slice triangle, because these look more like [INAUDIBLE]. but at least you're not paying attention to the stuff the teacher is saying about trigonometric functions. You'd rather think about the functions of triangles. And you already know some of those. There are sines, and cosines, but enough of this tangent. The thing to pay attention to is what affects your triangle how. If you start drawing the next triangle on your triangle stack this way, by this point, you already know what the full triangle would have to be. Because you just continue this edge until it meets this invisible line, and then fill the rest in. In fact, you can make an entire triangle stack just by piling on triangle parts and adding the points later to see what happens. There's some possible problems though. If you start a triangle like this, you can see that it's never going to close, no matter how far you extend the lines. Since this triangle isn't real, let's call it a Bermuda Triangle. This happens when two angles together are already more than 180 degrees. And since all the angles in a triangle add up to 180-- which, by the way, you can test by ripping one up and putting the three points into a line-- that means that if these are two 120 degree angles of a triangle, the third angle is off somewhere being negative 60. Of course, you have no problem being a Bermuda Triangle on a sphere, were angles always add up to more than 180," + }, + { + "Q": "\nat 1:09 did she say ''beacaus you have to navigate your triangle stack around your nose''?\nor ''beacaus you have to navigate your triangle stack around your notes''?", + "A": "I heard nose but notes is probably more correct", + "video_name": "o6KlpIWhbcw", + "timestamps": [ + 69 + ], + "3min_transcript": "So you're me and you're in math class and-- triangles, triangles, triangles-- I don't know. The teacher keeps saying words, and you're supposed to be doing something with trigons, whatever those are. But you're bored and-- triangles, triangles, triangles. Sure you could draw your triangles separately, They're happiest when snuggled up together into a triangle party. Everybody knows triangles love parties. Sometimes they get together and do these triangle congo lines. If you keep adding new triangles on the same side, it gets all curvy and spirally. Or you can alternate and it goes pretty straight. In fact, since all the sides of the triangle are supposed to be straight lines, and since they're all lying on top of some previous straight line, then this whole line would have to be straight if these were actually triangles. Since it's not, it's proof that these aren't quite triangles. Maybe they've been partying a little too hard, but hey, at least you're not doing math. Speaking of which, the teacher is still going on about types of trigons, and you're supposed to be taking notes. But you're more interesting in types of triangles, which you already know all about. There are fat triangles, and pointy triangles, and perfect triangles, and cheese slice triangles which are a kind of pointy triangle but are symmetric like a slice of cheese or cake. Super pointy triangles are fun to stack into triangle stacks. You can put all the points facing one direction, but the stack starts to wobble too much towards that direction. So it's good to put some facing the other direction before you go too far. You'll notice pretty quickly, that the skinnier the triangle, the less wobble it adds to the stack. can put just one not so skinny triangle that's pointing the other way. Or maybe you want to wobble, because you have to navigate your triangle stack around your notes. In which case you can even alternate back and forth as long as you make the triangles point towards where you want to go, a little less skinny. The easy part about triangle stacks, is that there's really only one part of the triangle that's important, as far as the stack is concerned. And that's the pointy point. The other two angles can be fat and skinny, or skinny and fat, or both the same if it's a cheese slice, and it doesn't change the rest of the stack. Unless the top angle is really wide, because then you'll get two skinny points, and which side should you continue to stack on? Also, instead of thinking fat and skinny, you should probably create code words that won't set off your teacher's mind reading alarms for non-math related thoughts. So you pick two words off the board, obtuse and acute, which by sheer coincidence I'm sure, just happen to mean fat angle and skinny angle. Of course, those are also kinds of triangles. Which doesn't make much sense, because the obtuser one angle of a triangle is, the more acute the other two get. Yet, if you make an acute triangle with the same perimeter, it has more area, which seems like an obtuse quality. And then, can you still call an obtuse equilateral triangle a cheese or cake-slice triangle, because these look more like [INAUDIBLE]. but at least you're not paying attention to the stuff the teacher is saying about trigonometric functions. You'd rather think about the functions of triangles. And you already know some of those. There are sines, and cosines, but enough of this tangent. The thing to pay attention to is what affects your triangle how. If you start drawing the next triangle on your triangle stack this way, by this point, you already know what the full triangle would have to be. Because you just continue this edge until it meets this invisible line, and then fill the rest in. In fact, you can make an entire triangle stack just by piling on triangle parts and adding the points later to see what happens. There's some possible problems though. If you start a triangle like this, you can see that it's never going to close, no matter how far you extend the lines. Since this triangle isn't real, let's call it a Bermuda Triangle. This happens when two angles together are already more than 180 degrees. And since all the angles in a triangle add up to 180-- which, by the way, you can test by ripping one up and putting the three points into a line-- that means that if these are two 120 degree angles of a triangle, the third angle is off somewhere being negative 60. Of course, you have no problem being a Bermuda Triangle on a sphere, were angles always add up to more than 180," + }, + { + "Q": "At 2:49, how did she make the triangle look upset when she didn't even change sides?\n", + "A": "it was a short smile she just made a big arch and the she got a frown", + "video_name": "o6KlpIWhbcw", + "timestamps": [ + 169 + ], + "3min_transcript": "can put just one not so skinny triangle that's pointing the other way. Or maybe you want to wobble, because you have to navigate your triangle stack around your notes. In which case you can even alternate back and forth as long as you make the triangles point towards where you want to go, a little less skinny. The easy part about triangle stacks, is that there's really only one part of the triangle that's important, as far as the stack is concerned. And that's the pointy point. The other two angles can be fat and skinny, or skinny and fat, or both the same if it's a cheese slice, and it doesn't change the rest of the stack. Unless the top angle is really wide, because then you'll get two skinny points, and which side should you continue to stack on? Also, instead of thinking fat and skinny, you should probably create code words that won't set off your teacher's mind reading alarms for non-math related thoughts. So you pick two words off the board, obtuse and acute, which by sheer coincidence I'm sure, just happen to mean fat angle and skinny angle. Of course, those are also kinds of triangles. Which doesn't make much sense, because the obtuser one angle of a triangle is, the more acute the other two get. Yet, if you make an acute triangle with the same perimeter, it has more area, which seems like an obtuse quality. And then, can you still call an obtuse equilateral triangle a cheese or cake-slice triangle, because these look more like [INAUDIBLE]. but at least you're not paying attention to the stuff the teacher is saying about trigonometric functions. You'd rather think about the functions of triangles. And you already know some of those. There are sines, and cosines, but enough of this tangent. The thing to pay attention to is what affects your triangle how. If you start drawing the next triangle on your triangle stack this way, by this point, you already know what the full triangle would have to be. Because you just continue this edge until it meets this invisible line, and then fill the rest in. In fact, you can make an entire triangle stack just by piling on triangle parts and adding the points later to see what happens. There's some possible problems though. If you start a triangle like this, you can see that it's never going to close, no matter how far you extend the lines. Since this triangle isn't real, let's call it a Bermuda Triangle. This happens when two angles together are already more than 180 degrees. And since all the angles in a triangle add up to 180-- which, by the way, you can test by ripping one up and putting the three points into a line-- that means that if these are two 120 degree angles of a triangle, the third angle is off somewhere being negative 60. Of course, you have no problem being a Bermuda Triangle on a sphere, were angles always add up to more than 180, Which is fine, unless you're afraid your triangle might get eaten by sea monsters. Anyway, stacking triangles into a curve is nice, and you probably want to make a spiral. But if you're not careful, it'll crash into itself. So you'd better think about your angles. Though, if you do it just right, instead of a crash disaster, you'll get a wreath thing. Or you can get a different triangle circle by starting with a polygon, extending the sides in one direction, and then triangling around it, to get this sort of aperture shape. And then you should probably add more triangles, triangles, triangles-- One last game. Start with some sort of asterisk, then go around a triangle it up. Shade out from the obtusest angles, and it'll look pretty neat. You can then extend it with another layer of triangles, and another, and if you shade the inner parts of these triangles, it's guaranteed to be an awesome triangle party. And there's lots of other kinds of triangle parties just waiting to happen. Ah, the triangle. So simple, yet so beautiful. The essence of two-dimensionality, the fundamental object of Euclidean geometry, the three points that define a plane. You can have your fancy complex shapes, they're just made up of triangles. Dissect a square into triangles, make symmetric arrangements, some reminiscent of spherical and hyperbolic geometry. Triangles branches into binary fractal trees." + }, + { + "Q": "at 5:17 do you multiply positive numbers the same way you multiply regular numbers..? if so do the positive numbers cancel out and the awnser become negative..?\n", + "A": "ok but what I don t understand is how negative x negative =positive but positive x positive =positive how does that work..? shouldn t it be positive x positive =negative...?", + "video_name": "47wjId9k2Hs", + "timestamps": [ + 317 + ], + "3min_transcript": "This right over here is a positive six. So we have another rule of thumb here. If I have a negative times a negative, the negatives are going to cancel out. And that's going to give me a positive number. Now with these out of the way, let's just do a bunch of examples. I'm encouraging you to try them out before I do them. Pause the video, try them out, and see if you get the same answer. So let's try negative one times negative one. Well, one times one would be one. And we have a negative times a negative. They cancel out. Negative times a negative give me a positive. So this is going to be positive one. I can just write one, or I can literally write a plus sign there to emphasize. This is a positive one. What happened if I did negative one times zero? Now this might seem, this doesn't fit into any of these circumstances, zero is neither positive nor negative. And here you just have to remember anything times zero So negative one times zero is going to be zero. Or I could've said zero times negative seven hundred and eighty-three, that is also going to be zero. Now what about two--let me do some interesting ones. What about--I'm looking a new color. Twelve times negative four. Well, once again, twelve times positive four would be fourty-eight. And we are in the circumstance where one of these two numbers, right over here, is negative. This one right here. If exactly one of the two numbers is negative, then the product is going to be negative. We are in this circumstance, right over here. We have one negative, so the product is negative. You could imagine this as repeatedly adding negative four twelve times And so you will get to negative fourty-eight. Let's do another one. What is seven times three? Well, this is a bit of a trick. There are no negative numbers here. Positive seven times positive three. The first circumstance, which you already knew how to do before this video. This would just be equal to twenty-one. Let's do one more. So if I were to say negative five times negative ten-- well, once again, negative times a negative. The negatives cancel out. You are just left with a positive product. So it's going to be five times ten. It's going to be fifty. The negative and the negative cancel out. Your product is going to be positive. That's this situation right over there." + }, + { + "Q": "\nAt 2:48 why does pos times neg have to be neg? is there a reason for that?", + "A": "If you tell yourself that Multiplication is the same as repeated addition, then: 4x -9 is the same as saying four times -9 , or: -9 + -9 + -9 + -9 -> -9-9-9-9 = -36", + "video_name": "47wjId9k2Hs", + "timestamps": [ + 168 + ], + "3min_transcript": "if I had two times three, I would get six. But because one of these two numbers is negative, then my product is going to be negative. So if I multiply, a negative times a positive, I'm going to get a negative. Now what if we swap the order which we multiply? So if we were to multiply three times negative two, it shouldn't matter. The order which we multiply things don't change, or shouldn't change the product. When we multiply two times three, we get six. When we multiply three times two, we will get six. So we should have the same property here. Three times negative two should give us the same result. It's going to be equal to negative six. And once again we say, three times two would be six. One of these two numbers is negative, and so our product is going to be negative. So we could draw a positive times a negative And both of these are just the same thing with the order which we are multiplying switched around. But this is one of the two numbers are negative. Exactly one. So one negative, one positive number is being multiplied. Then you'll get a negative product. Now we'll think about the third circumstance, where both of the numbers are negative. So if I were to multiply--I'll just switch colors for fun here-- If I were to multiply negative two times negative three-- this might be the least intuitive for you of all, and here I'm going to introduce you the rule, in the future I will explore why this is, and why this makes mathematics more--all fit together. But this is going to be, you see, two times three would be six. And I have a negative times a negative, one way you can think about it is that negatives cancel out! So you'll actually end up with a positive six. Actually I don't have to draw a positive here. This right over here is a positive six. So we have another rule of thumb here. If I have a negative times a negative, the negatives are going to cancel out. And that's going to give me a positive number. Now with these out of the way, let's just do a bunch of examples. I'm encouraging you to try them out before I do them. Pause the video, try them out, and see if you get the same answer. So let's try negative one times negative one. Well, one times one would be one. And we have a negative times a negative. They cancel out. Negative times a negative give me a positive. So this is going to be positive one. I can just write one, or I can literally write a plus sign there to emphasize. This is a positive one. What happened if I did negative one times zero? Now this might seem, this doesn't fit into any of these circumstances, zero is neither positive nor negative. And here you just have to remember anything times zero" + }, + { + "Q": "\nAt 0:25 - what's the point of writing down \"a^2 + -ab...\"? Why not just \"a^2 - ab\"? The teachers I've had would have marked the way you'd used as wrong.", + "A": "You could do that, and you re correct, it is proper form. Sal just did that to keep everyone on the same page, and solve the problem step by step. Later in the video, it did get simplified down to a^2 - ab, and as long as your final answer is simplified properly, it won t really matter where you do it.", + "video_name": "YahJQvY396o", + "timestamps": [ + 25 + ], + "3min_transcript": "We need to factor 49x squared minus 49y squared. Now here there's a pattern that you might already be familiar with. But just to make sure you are, let's think about what happens if we multiply a plus b-- where these are just two terms in a binomial-- times a minus b. If you multiply this out, you have a times a, which is a squared, plus a times negative b, which is negative ab-- that's a times negative b-- plus b times a, which is the same thing as ab. And then you have b times negative b, which is negative b squared. So when you do that, you have a negative ab and a positive ab, they cancel out. And you're just going to be left with an a squared minus a b squared. Now, this thing that we have here is exactly that pattern. 49x squared is a perfect square. 49y squared is a perfect square. We can rewrite it like that. We could rewrite this over here as 7x squared minus-- and And so you see it's a pattern. It's a squared minus b squared. So if you wanted to factor this-- if you would just use this pattern that we just derived-- you would say that this is the same thing as a, 7x plus b plus 7y times 7x minus b, minus 7y. And you'd be done. Now there's one alternate way that you could factor this and it'd be completely legitimate. You could start from the beginning and say, you know what? 49 is a common factor here, so let me just factor that out. So you could say it's equivalent to 49 times x squared minus y squared. And you say, oh, this fits the pattern of-- this is a squared minus b squared. So this will be x plus y times x minus y. times x minus y. And to see that this, right here, is the exact same thing as this right over here, you could just factor 7 out of You'd factor out a 7 out of that term, a factor 7 out of that term. And when you multiply them, you'd get the 49. So these are-- this or this-- these are both ways to factor this expression." + }, + { + "Q": "\nAround 9:13, what type of function would it be if there was an x value that wasn't mapped to a y value?", + "A": "Such a scenario doesn t really exist. Say we had f: {a, b, c} -> {1, 2}, defined by f(a)=1, f(b)=2, and f(c)=7. Then either it is a typo where (whoever gave such a problem) forgot to put 7 in {1, 2}, or the object f is not a function, because a function f: A->B must be a subset of AxB. Since 7 is not in B, such an f wouldn t be a subset of AxB, so it wouldn t be a function.", + "video_name": "xKNX8BUWR0g", + "timestamps": [ + 553 + ], + "3min_transcript": "most one x that maps to it. Or another way to say it is that for any y that's a member of y-- let me write it this way --for any y that is a member y, there is at most one-- let me write most in capital --at most one x, such that f of x is equal to y. There might be no x's that map to it. So for example, you could have a little member of y right here that just never gets mapped to. Everyone else in y gets mapped to, but that guy never gets mapped to. So this would be a case where we don't have a surjective function. This is not onto because this guy, he's a member of the co-domain, but he's not a member of the He doesn't get mapped to. But this would still be an injective function as long as every x gets mapped to a unique y. Now, how can a function not be injective or one-to-one? And I think you get the idea when someone says one-to-one. Well, if two x's here get mapped to the same y, or three get mapped to the same y, this would mean that we're not dealing with an injective or a one-to-one function. So that's all it means. Let me draw another example here. Let's actually go back to this example right here. When I added this e here, we said this is not surjective anymore because every one of these guys is not being mapped to. Is this an injective function? Well, no, because I have f of 5 and f of 4 both mapped to d. injectiveness. This is what breaks it's surjectiveness. Now if I wanted to make this a surjective and an injective function, I would delete that mapping and I would change f of 5 to be e. Now everything is one-to-one. I don't have the mapping from two elements of x, going to the same element of y anymore. And everything in y now gets mapped to. So this is both onto and one-to-one." + }, + { + "Q": "\nAt 0:40, Sal starts thinking about the question. Maybe that makes sense but what I was thinking was different. What I did was, x^5 simplify with x^5, x^2 with x^2 and then I tried solving the question. I got a very different answer.....-0.03......That is way off 2/3! I know 2/3 is an approximation but having a difference of signs is a big difference! Why?", + "A": "Ack! NEVER, EVER, EVER, cancel out anything next to a plus or minus sign! This is a very bad algebra mistake. Here s an example so you can see why this is not okay: Consider (x + 1) / (x + 2). If you could just cancel out the x s, you would be left with 1/2. But, try plugging in some values for x and you will see that this can t be right. For instance, if you plug in x = 1, you will get (1+1)/(1+2) = 2/3, not 1/2. Different values of x would give you different ratios.", + "video_name": "gv9ogppphso", + "timestamps": [ + 40 + ], + "3min_transcript": "So we have f of x equaling 4x to the fifth minus 3x squared plus 3, all of that over 6x to the fifth minus 100x squared minus 10. Now, what I want to think about is-- what is the limit of f of x, as x approaches infinity? And there are several ways that you could do this. You could actually try to plug in larger and larger numbers for x and see if it seems to be approaching some value. Or you could reason through this. And when I talk about reasoning through this, it's to think about the behavior of this numerator and denominator as x gets very, very, very large. And when I'm talking about that, what I'm saying is, as x gets very, very large-- let's just focus on the numerator. As x gets very, very large, this term right over here in the numerator-- 4x to the fifth-- is going to become a much, much more significant than any of these other things. Something squaring gets large. But something being raised to the fifth power gets raised that much, much faster. Similarly, in the denominator, this term right over here-,, is going to grow much, much, much faster than any of these Even though this has 100 as a coefficient or a negative 100 as a coefficient, when you take something to the fifth power, it's going to grow so much faster than x squared. So as x gets very, very, very large, this thing is going to approximate 4x to the fifth over 6x to the fifth for a very large, large x Or we could say as x approaches infinity. Now, what could this be simplified to? Well, you have x to the fifth divided by x to the fifth. These are going to grow together. So these you can think of them as canceling out. And so you are left with 2/3. So what you could say is-- the limit of f of x, as x approaches infinity, as x gets larger and larger and larger, all of these other terms aren't going to matter that much. And so it's going to approach 2/3. if that actually makes sense. What we're actually saying is that we have a horizontal asymptote at y is equal to 2/3. So lets look at the graph. So right here is the graph. Got it from Wolfram Alpha. And we see, indeed, as x gets larger and larger and larger, f of x seems to be approaching this value that looks right at around 2/3. So it looks like we have a horizontal asymptote right over here. Let me draw that a little bit neater. We have a horizontal asymptote right at 2/3. So let me draw it as neatly as I can. So this right over here is y is equal to 2/3. The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2/3. And when we just look at the graph here, it seems like the same thing is happening from the bottom direction, when x approaches negative infinity." + }, + { + "Q": "i didn't understand at 2:40 why x squared was x. I thought this was radicals not square roots. Or is x just 1\nHelp Please :/\n", + "A": "In the video, it actually shows the square root of x squared, not just x squared. So, if you think of the equation as the square root of x*x, you get x. Hope this helps!", + "video_name": "egNq4tSfi1I", + "timestamps": [ + 160 + ], + "3min_transcript": "plus three \"a\"'s which will give you four \"a\"'s, in this case \"a\" is all of this business right over here so we added those terms, and then we wanted to think about we have four principle roots of \"a\" and we have one more principle roots of \"a\", so same idea you have four of these things I am circling in magenta and you have one more of these things that I am circling in magenta, that one co-efficient is implicit so if I have four of something plus one more of something it becomes five of that something so plus plus five times the square root, plus five times the square root of eight and now we'll see if we can simplify this anymore, we have four of something and we have five of something else, so you can't just add these two things together, but maybe we can simplify this a little bit so we know that the principle root of two x squared, this is the same thing as, so let me write the four out front, so we have the four, and the principle root of two x squared is the same thing as the principle and then we have plus five times, now eight can be written as a product of a perfect square and a not so perfect square, eight can be written as four times two, so lets write it that way so if we view this whole, this is the principle root, the square root of four times two, we can re-write this as the five times the square root of four, or the principle root of four times the principle root of two and what can we simplify here? well we know what the principle root of x squared is, it is the positive square root of x squared, so it is not just x, you might be tempted to say it is x but since we know it is the positive square root we have to say it is the absolute value of x, because what if x was negative? if was x was negative, you'd have , lets say it was negative three, you'd have negative three squared, and so it wouldn't just be x, it wouldnt be negative three, it would be positive three, so you have to take the absolute value, and the other thing that is a perfect square is the four right here, its principle root is two, its principle square root i should say is two, so now you have, if we just change the order we are multiplying right here, you have four, four times the absolute value of x, four times the absolute value of x, times the square root of two, times the square root of two, I want to do that in that same yellow color, times the square root of two, plus plus we have five times two, which is ten, right, this whole thing is simplified to two, so we have plus ten square roots of two, now we could call it a day, and say we are all done adding and simplifying or you could add a little bit more depending on how you wanna view it, because over here you have" + }, + { + "Q": "\nAt 2:47, Sal says that sqrt(x^2) is the absolute value of x, and I get that. How come at 3:14, he says that sqrt(4) would be 2? Shouldn't he also consider -2?", + "A": "He qualified his answer by saying that the principal square root of 4 is 2. The principal square root is the unique nonnegative square root of a nonnegative real number.", + "video_name": "egNq4tSfi1I", + "timestamps": [ + 167, + 194 + ], + "3min_transcript": "plus three \"a\"'s which will give you four \"a\"'s, in this case \"a\" is all of this business right over here so we added those terms, and then we wanted to think about we have four principle roots of \"a\" and we have one more principle roots of \"a\", so same idea you have four of these things I am circling in magenta and you have one more of these things that I am circling in magenta, that one co-efficient is implicit so if I have four of something plus one more of something it becomes five of that something so plus plus five times the square root, plus five times the square root of eight and now we'll see if we can simplify this anymore, we have four of something and we have five of something else, so you can't just add these two things together, but maybe we can simplify this a little bit so we know that the principle root of two x squared, this is the same thing as, so let me write the four out front, so we have the four, and the principle root of two x squared is the same thing as the principle and then we have plus five times, now eight can be written as a product of a perfect square and a not so perfect square, eight can be written as four times two, so lets write it that way so if we view this whole, this is the principle root, the square root of four times two, we can re-write this as the five times the square root of four, or the principle root of four times the principle root of two and what can we simplify here? well we know what the principle root of x squared is, it is the positive square root of x squared, so it is not just x, you might be tempted to say it is x but since we know it is the positive square root we have to say it is the absolute value of x, because what if x was negative? if was x was negative, you'd have , lets say it was negative three, you'd have negative three squared, and so it wouldn't just be x, it wouldnt be negative three, it would be positive three, so you have to take the absolute value, and the other thing that is a perfect square is the four right here, its principle root is two, its principle square root i should say is two, so now you have, if we just change the order we are multiplying right here, you have four, four times the absolute value of x, four times the absolute value of x, times the square root of two, times the square root of two, I want to do that in that same yellow color, times the square root of two, plus plus we have five times two, which is ten, right, this whole thing is simplified to two, so we have plus ten square roots of two, now we could call it a day, and say we are all done adding and simplifying or you could add a little bit more depending on how you wanna view it, because over here you have" + }, + { + "Q": "Why at 9:45 do we use n! in the denominator instead of just the n. I understood up to the second derivative why we introduce 1/2 but then on the third it I do not understand why we have 1/3! as this is 1/6 and not 1/3 which means that it will not cancel out. I hope you understand my questions.\nThank you very much for your help :)\n", + "A": "For the second derivative, it s not 1 / 2 it s really 1 / 2! It s just that 2! happens to be equal to 2. The factorial works well for repeated differentiation: d/dx x^5 / 5! = x^4 / 4! d/dx x^4 / 4! = x^3 / 3! d/dx x^3 / 3! = x^2 / 2! d/dx x^2 / 2! = x / 1! d/dx x / 1! = 1 / 0! d/dx 1 / 0! = 0", + "video_name": "epgwuzzDHsQ", + "timestamps": [ + 585 + ], + "3min_transcript": "And let's see how it does on its third derivative, or I should say the second derivative. So p prime prime of x is equal to-- this is a constant, so its derivative is 0. So you just take the coefficient on the second term is equal to f prime prime of 0. So what's the second derivative of p evaluated at 0? Well, it's just going to be this constant value. It's going to be f prime prime of 0. So notice, by adding this term, now, not only is our polynomial value the same thing as our function value at 0, its derivative at 0 is the same thing as the derivative of the function at 0. And its second derivative at 0 is the same thing as the second derivative of the function at 0. So we're getting pretty good at this. And you might guess that there's a pattern here. Every term we add, it'll allow us to set up the situation so that the n-th derivative of our approximation at 0 of our function at 0. So in general, if we wanted to keep doing this, if we had a lot of time on our hands and we wanted to just keep adding terms to our polynomial, we could-- and let me do this in a new color. Maybe I'll do it in a color I already used. We could make our polynomial approximation. So the first term, the constant term, will just be f of 0. Then the next term will be f prime of 0 times x. Then the next term will be f prime prime of 0 times 1/2 times x squared. I just rewrote that in a slightly different order. Then the next term, if we want to make their third derivative the same at 0, would be f prime prime prime of 0. The third derivative of the function at 0, times 1/2 times 1/3, so 1 over 2 times 3 times x to the third. And we can keep going. Plus, if we want to make their fourth derivatives at 0 coincide, it would be the fourth derivative of the function. I could put a 4 up there, but this is really emphasizing-- it's the fourth derivative at 0 times 1 over-- and I'll change the order. Instead of writing it in increasing order, I'll write it as 4 times 3 times 2 times x to the fourth. And you can verify it for yourself. If we just had this only, and if you were to take the fourth derivative of this, evaluate it at 0, it'll be the same thing as the fourth derivative of the function evaluated at 0. And in general, you can keep adding terms where the n-th term will look like this. The n-th derivative of your function evaluated at 0 times x to the n over n factorial. Notice this is the same thing as 4 factorial. 4 factorial is equal to 4 times 3 times 2 times 1." + }, + { + "Q": "at about 3:09, Sal adds an \"x\" after f '(0). Why?\n", + "A": "It s so p (0) still equals f (0) after he takes the derivative of his expression f(0)+f (0)x. If the x wasn t there, then the f (0) would end up equaling zero when he took the derivative, just as the f(0) did. Reviewing power rule may be helpful in clarifying this.", + "video_name": "epgwuzzDHsQ", + "timestamps": [ + 189 + ], + "3min_transcript": "So at first, maybe we just want p of 0, where p is the polynomial that we're going to construct, we want p of 0 to be equal to f of 0. So if we want to do that using a polynomial of only one term, of only one constant term, we can just set p of x is equal to f of 0. So if I were to graph it, it would look like this. It would just be a horizontal line at f of 0. And you could say, Sal, that's a horrible approximation. It only approximates the function at this point. Looks like we got lucky at a couple of other points, but it's really bad everywhere else. And now I would tell you, well, try to do any better using a horizontal line. At least we got it right at f of 0. So this is about as good as we can do with just a constant. And even though-- I just want to remind you-- this might not look like a constant, but we're assuming that given the function, will just give us a number. So whatever number that was, we would put it right over here. We'd say p of x is equal to that number. It would just be a horizontal line right there at f of 0. But that obviously is not so great. So let's add some more constraints. Beyond the fact that we want p of 0 to be equal to f of 0, let's say that we also want p prime at 0 to be the same thing as f prime at 0. Let me do this in a new color. So we also want, in the new color, we also want-- that's not a new color. We also want p prime. We want the first derivative of our polynomial, when evaluated at 0, to be the same thing as the first derivative of the function when evaluated at 0. And we don't want to lose this right over here. So what if we set p of x as being equal to f of 0? So we're taking our old p of x, but now we're Plus f prime of times x. So let's think about this a little bit. If we use this as our new polynomial, what happens? What is p is 0? p of 0 is going to be equal to-- you're going to have f of 0 plus whatever this f prime of 0 is times 0. If you put a 0 in for x, this term is just going to be 0. So you're going to be left with p of 0 is equal to f of 0. That's cool. That's just as good as our first version. Now what's the derivative over here? So the derivative is p prime of x is equal to-- you take the derivative of this. This is just a constant, so its derivative is 0. The derivative of a coefficient times x is just going to be the coefficient. So it's going to be f prime of 0. So if you evaluate it at 0-- so p prime of 0." + }, + { + "Q": "\nWhat is the function approaching as x=1? At 3:30 could anyone explain or phrase that in a different way or ways?", + "A": "What is the y-value of the line (or function) approaching as x gets closer to 1? This example is simple as y is always 1 except on the exact point of (1,1) since we cannot divide by zero. when x=0, y=1. when x=0.5, y=1. when x=0.9, y=1.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 210 + ], + "3min_transcript": "which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here. you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing." + }, + { + "Q": "At 5:06 Sal drew a curve like thing after writing g(x).What is that curve?\n", + "A": "Braces ( curly braces ), it is the same of using { .", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 306 + ], + "3min_transcript": "you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this." + }, + { + "Q": "\nAt 10:34, Sals calculator said that 1.999999999999^2 was 4. Well it is actually 3.999999996...but close enough.", + "A": "Listen again, he said his calculator rounded the result to 4 but that the result wasn t exactly 4, but it was really really really really close to 4. And that is the whole deal with limits. What is the value of a function as you get really really really really close to the limit value. Keep Studying!", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 634 + ], + "3min_transcript": "as we get closer and closer to it. And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1. The limit of g of x as x approaches 2 is equal to 4. And you could even do this numerically using a calculator, and let me do that, because I think that will be interesting. So let me get the calculator out, let me get my trusty TI-85 out. So here is my calculator, and you could numerically say, OK, what's it going to approach as you approach x equals 2. So let's try 1.94, for x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared. so 1.99, and once again, let me square that. Well now I'm at 3.96. What if I do 1.999, and I square that? I'm going to have 3.996. Notice I'm going closer, and closer, and closer to our point. And if I did, if I got really close, 1.9999999999 squared, what am I going to get to. It's not actually going to be exactly 4, this calculator just rounded things up, but going to get to a number really, really, really, really, really, really, really, really, really close to 4. And we can do something from the positive direction too. And it actually has to be the same number when we approach from the below what we're trying to approach, and above what we're trying to approach. So if we try to 2.1 squared, we get 4.4. let me go a couple of steps ahead, Now we are getting much closer to 4. So the closer we get to 2, the closer it seems like we're getting to 4. So once again, that's a numeric way of saying that the limit, as x approaches 2 from either direction of g of x, even though right at 2, the function is equal to 1, because it's discontinuous. The limit as we're approaching 2, we're getting closer, and closer, and closer to 4." + }, + { + "Q": "Hey @4:15 This might sound stupid but why is the answer 1 and not zero?\n", + "A": "I think I was thinking about slopes and derivatives. I was a little rusty on my calculus at the time I watched this video.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 255 + ], + "3min_transcript": "And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here. you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety," + }, + { + "Q": "at 5:22, Sal says \"it\" what is he referring to?\n", + "A": "His it is referring to g(x) , and yes he is trying to say g(2) = 1 , or at the x-value 2, g(x) outputs a 1.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 322 + ], + "3min_transcript": "you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this." + }, + { + "Q": "\nAt 6:39 he says the discontinuity is when X=2, why can't it also be at X=-2?", + "A": "That is the whole point of limits, x cannot equal 2 because that is your limit!", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 399 + ], + "3min_transcript": "Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that. this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function" + }, + { + "Q": "at 7:08 on the y=f(x) axis you put an open circle at 4 the top why\n", + "A": "Putting an empty circle in a graph of this kind indicates that the function does not have a value on the curve at that point. Often this is because the function is undefined at that point, but in this case it is because the function has a quirky definition stating that the function has a different value, not on this curve, at that particular point.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 428 + ], + "3min_transcript": "Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that. this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function" + }, + { + "Q": "\nAt 2:44 why did he draw an open circle and not an asymptote?", + "A": "Good question. It s because there is a finite limit as x approaches 1. He uses the open circle to show that the function is undefined at that point. If you graph something like 1/(x-1) you ll notice that it goes to +infiniti as you approach 1+ and -infiniti as you approach 1-. So there is an asymptote at 1. In the example of (x-1)/(x-1) the function is 1 at every x value except 1 so there is no asymptote.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 164 + ], + "3min_transcript": "which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here. you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing." + }, + { + "Q": "\nAt 0:10, Sal said that limits are the ideas that calculus is based upon. How important are limits for anyone more familiar with calculus? (just curious)", + "A": "Limits are very important to calculus. Derivatives and integrals use calculus and many modeling applications in calculus use limits. Limits are very important in calculus because many concepts are based on that sole aspect.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 10 + ], + "3min_transcript": "In this video, I want to familiarize you with the idea of a limit, which is a super important idea. It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really, really, really simple idea. So let me draw a function here, actually, let me define a function here, a kind of a simple function. So let's define f of x, let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal look, I have the same thing in the numerator and denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true, the difference between f of x equals 1 and this thing right over here, is that this thing can never equal-- this thing is undefined when x is equal to 1. Because if you set, let me define it. Let me write it over here, if you have f of, sorry not f of 0, if you have f of 1, what happens. which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here." + }, + { + "Q": "At 4:40 , f(x) must tend to one but why it is equal to one?\n", + "A": "It is the two-sided limit of the function as x approaches 1.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 280 + ], + "3min_transcript": "you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this." + }, + { + "Q": "At 6:27, why is he using g(x)? Can't it be something else?\n", + "A": "What do you mean? It is just used to write a function.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 387 + ], + "3min_transcript": "familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that." + }, + { + "Q": "At 0:35, Sal explains another way of finding the determinant, by taking each value of the first row terms in the original matrix and multiplying it by the corresponding terms in the cofactor matrix, and taking the sum of those products. Why does this work?\n", + "A": "This is called rule of Sarrus, only applies to the 3 x 3 matrices. It s just a coincidence of the traditional minor and cofactor method.", + "video_name": "ArcrdMkEmKo", + "timestamps": [ + 35 + ], + "3min_transcript": "We're nearing the home stretch of our quest to find the inverse of this three-by-three matrix here. And the next thing that we can do is find the determinant of it, which we already have a good bit of practice doing. So the determinant of C, of our matrix-- I'll do that same color-- C, there are several ways that you could do it. You could take this top row of the matrix and take the value of each of those terms times the cofactor-- times the corresponding cofactor-- and take the sum there. That's one technique. Or you could do the technique where you rewrite these first two columns, and then you take the product of the top to left diagonals, sum those up, and then subtract out the top right to the bottom left. I'll do the second one just so that you can see that you get the same result. The determinant is going to be equal to-- I'll 1, negative 2, 2, 2, 1, 1, 3, 4, 5. And let me now, just to make it a little bit simpler, rewrite these first two columns. So negative 1, negative 2, 2, 1, 3, 4. So the determinant is going to be equal to-- so let me write this down. So you have negative 1 times 1 times 5. Well that's just going to be negative 5, taking that product. Then you have negative 2 times 1 times 3. Well that's negative 6. So we'll have negative 6. Or you could say plus negative 6 there. And then you have 2 times 2 times 4. Well that's just 4 times 4, which is just 16. So we have plus 16. And then we do the top right to the bottom left. Well that's negative 4 times 5. So that is negative 20. But we're going to subtract negative 20. So that's negative 4 times 5, negative 20, but we're going to subtract negative 20. Obviously that's going to turn into adding positive 20. Then you have negative 1 times 1 times 4, which is negative 4. But we're going to subtract these products. We're going to subtract negative 4. And then you have 2 times 1 times 3, which is 6. But we have to subtract it. So we have subtracting 6. And so this simplifies to negative 5 minus 6 is negative 11, plus 16 gets us to positive 5. So all of this simplifies to positive 5. And then we have plus 20 plus 4." + }, + { + "Q": "At 0:57, 2i-7i is -5i. Based on this, is i a coefficient?\n", + "A": "No, the coefficients are the 2, -7 and -5. i is like a variable. But, it is unique because we know its value. i always = sqrt(-1).", + "video_name": "SfbjqVyQljk", + "timestamps": [ + 57 + ], + "3min_transcript": "We're asked to add the complex number 5 plus 2i to the other complex number 3 minus 7i. And as we'll see, when we're adding complex numbers, you can only add the real parts to each other and you can only add the imaginary parts to each other. So let's add the real parts. So we have a 5 plus a 3. And then the imaginary parts-- we have a 2i. So plus 2i. And then we have a negative 7i, or we're subtracting 7i. So minus 7i right over here. And 5 plus 3-- that's pretty straightforward. That's just going to be 8. And then if I have two of something and from that I subtract seven of that something-- and in this case, the something is the imaginary unit, the number i. If I have two i's and I take away seven i's, then I have negative five i's. 2 minus 7 is negative 5. So then I have negative 5i. So when you add these two complex numbers, You get another complex number. It has a real part and an imaginary part." + }, + { + "Q": "\n6*3 is 3 times larger than 2*3. how is this in-between as Sal says @1:30\nI can perform the computation, I just don't know why it works. Can someone help.", + "A": "Note that if we multiply 2 \u00e2\u0080\u00a2 3 by 3, we can have 3 \u00e2\u0080\u00a2 2 \u00e2\u0080\u00a2 3. Then, we can use the associativity of multiplication to find that this is equivalent to (3 \u00e2\u0080\u00a2 2) \u00e2\u0080\u00a2 3 = 6 \u00e2\u0080\u00a2 3. Therefore, 6 times 3 is thrice 2 times 3.", + "video_name": "j3-XYLnxJDY", + "timestamps": [ + 90 + ], + "3min_transcript": "So right here, we have a four-sided figure, or a quadrilateral, where two of the sides are parallel to each other. And so this, by definition, is a trapezoid. And what we want to do is, given the dimensions that they've given us, what is the area of this trapezoid. So let's just think through it. So what would we get if we multiplied this long base 6 times the height 3? So what do we get if we multiply 6 times 3? Well, that would be the area of a rectangle that is 6 units wide and 3 units high. So that would give us the area of a figure that looked like-- let me do it in this pink color. The area of a figure that looked like this would be 6 times 3. So it would give us this entire area right over there. Now, the trapezoid is clearly less than that, but let's just go with the thought experiment. Now, what would happen if we went with 2 times 3? has a width of 2 and a height of 3. So you could imagine that being this rectangle right over here. So that is this rectangle right over here. So that's the 2 times 3 rectangle. Now, it looks like the area of the trapezoid should be in between these two numbers. Maybe it should be exactly halfway in between, because when you look at the area difference between the two rectangles-- and let me color that in. So this is the area difference on the left-hand side. And this is the area difference on the right-hand side. If we focus on the trapezoid, you see that if we start with the yellow, the smaller rectangle, it reclaims half of the area, half of the difference between the smaller rectangle It gets exactly half of it on the left-hand side. And it gets half the difference between the smaller and the larger on the right-hand side. So it completely makes sense that the area of the trapezoid, this entire area right over here, should really just be the average. It should exactly be halfway between the areas of the smaller rectangle and the larger rectangle. So let's take the average of those two numbers. It's going to be 6 times 3 plus 2 times 3, all of that over 2. So when you think about an area of a trapezoid, you look at the two bases, the long base and the short base. Multiply each of those times the height, and then you could take the average of them. Or you could also think of it as this is the same thing as 6 plus 2. And I'm just factoring out a 3 here. 6 plus 2 times 3, and then all of that over 2," + }, + { + "Q": "At 1:36,why the point (1,-1) is the center?\n", + "A": "you set each numerator equal to zero. so x-1=0 ==> x=1 and y+1=0 ==> y=-1 If you want to know why setting the each numerator finds the center... I think about it like this. (x-1)^2 is symmetric about x=1. That is if you add or subtract the same number from x, say 1 (x=2, x=0), then you will get the same output. (2-1)^2=(0-1)^2. The same goes for (y+1)^2. ***this may not be the exact terminology but is a way to think about it.", + "video_name": "lGQw-W1PxBE", + "timestamps": [ + 96 + ], + "3min_transcript": "Let's see if we can tackle a slightly more difficult hyperbola graphing problem. Let's add the hyperbola. Make this up on the fly x minus 1 squared over 16 minus y plus 1 squared over 4 is equal to 1. So the first thing to recognize is that this is a hyperbola and we'll in a few videos, do a bunch of problems where the first point is just to identify what type of conic section we have and then the second step is actually graph the conic section. I already told you that we're going to be doing a hyperbola problem, so you know it's a hyperbola. But the way to recognize that is that you have this minus of the y squared term and then we actually have it shifted. The classic or the standard non-shifted form of a hyperbola or a hyperbola centered at 0 would look something like this. Especially if it has the same asymptotes just shifted, but minus y squared over 4 is equal to 1. And the difference between this hyperbola and this hyperbola the center of this hyperbola is at the point x is equal to 1 y is equal to minus 1. And the way to think about it is x equals 1 makes this whole term 0, and so that's why it's the center. And y equal to minus 1 makes this whole term 0. And on here, of course, the center is the origin. Center is 0, 0. So the easy way to graph this is to really graph this one, but you shift it so you use the center being 1 minus 1 instead of the center being 0, 0. So let's do that. So let's figure out the slope of the two asymptotes here and for this hyperbola right here. So if we go with this one, let's just solve for y. That's what I always like to do whenever I'm graphing a hyperbola. So we get minus y squared over 4. Subtracting x squared over 16 from both sides minus x squared over 16 plus 1. I'm working on this hyperbola right here, not this one, and then I'm going to just shift it later. And then let's say multiply both sides by minus 4 and you get y squared is equal to-- see the minus cancels out with that and then 4 over 16 is x squared over 4 minus 4 and so y is equal to plus or minus square root of x squared" + }, + { + "Q": "\nIn 3:35, why didn't Sal factor x^2/4-4 in (x/2+2)(x/2-2)?", + "A": "He s taking the square root of everything except the constant anyway, so it makes more sense and is easier to square root in the end.", + "video_name": "lGQw-W1PxBE", + "timestamps": [ + 215 + ], + "3min_transcript": "for this hyperbola right here. So if we go with this one, let's just solve for y. That's what I always like to do whenever I'm graphing a hyperbola. So we get minus y squared over 4. Subtracting x squared over 16 from both sides minus x squared over 16 plus 1. I'm working on this hyperbola right here, not this one, and then I'm going to just shift it later. And then let's say multiply both sides by minus 4 and you get y squared is equal to-- see the minus cancels out with that and then 4 over 16 is x squared over 4 minus 4 and so y is equal to plus or minus square root of x squared And to figure out the asymptotes you just have to think about well what happens as x approaches positive or negative infinity. As x gets really positive or x gets really negative. And we've done this a bunch of times already. This is more important than just memorizing the formula, because it gives you an intuition of where those equations for the lines of the asymptote actually come from. Because these are what this graph or this equation or this function approaches as x approaches positive or negative infinity. As x approaches positive or negative infinity, what is y approximately equal to, in this case? Well once again, this term is going to dominate. This is just a 4 right here. You could imagine when x is like a trillion or a negative trillion, this is going to be huge number and this is going to be just like you know you almost view it like is going to dominate. So as you approach positive or negative infinity, y is going to be approximately equal to the square root, the positive and negative square root, of x squared over 4. So y would be approximately equal to positive or negative x over 2, or 1/2x. Let's do that. Let's draw our asymptotes. And remember, these are the asymptotes for this situation. But now of course, we're centered at 1 negative 1. So I'm going to draw two lines with these slopes, with positive 1/2 and negative 1/2 slopes, but they're going to be centered at this point. I just got rid of the shift just so I could figure out the asymptotes but of course this is the real thing that we're trying to graph, so let me do that. This is my y-axis this is my x-axis and the center of" + }, + { + "Q": "At ~6:20, shouldn't the second multiple choice said \"If \u00e2\u0088\u00a0AOC is rotated...\", not \"If Ray OA and Ray OC are rotated...\"? Because you can't rotate points, rays, lines, or line segments due to them not having a vertex, right?\n", + "A": "Yes, you can rotate those around a point/origin, actually.", + "video_name": "uYXhga17q1g", + "timestamps": [ + 380 + ], + "3min_transcript": "but it really doesn't help us establishing that phi is equal to theta. So that one I also don't feel good about. So and it's good because we felt good about the middle choice. Let's do one more of these. So they are telling us that line AOB, and they could have just said line AB, but I guess they wanted to put the O in there to show that point O is on that line, that AOB are colinear. And COD is our straight lines, all right, fair enough. Which of these statements prove vertical angles are always equal? So vertical angles would be the angles on the opposite sides of an intersection. So in order to prove that vertical, so for example angle AOC, and angle DOB, are vertical angles. And if we wanted to prove that they are equal, we would say well their measures are gonna be equal, so theta should be equal to phi. So this one says segment OA is congruent to OD. OA is congruent to OD. We don't know that, they never even told us that. So I don't even have to read the rest of it, this is already saying, I don't know how far D is away from O, I don't know if it's the same distance A is from O. So we can just rule this first choice out. I can stop reading, this started with a statement that we don't know, based on the information they gave us. So let's look at the second choice. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to OB and OD respectively. If two rays are rotated by the same amount, the angle between them will not change, so phi must be equal to theta. So this is interesting, so let's just slow down and think about what they're saying. If ray OA and OC are each rotated 180 degrees, if you rotate it 180 degrees, it's gonna go all the way around and point in the other direction, it's going to become, it's going to map to ray OB. So I definitely believe that. OA is going to map to ray OB, and ray OC, if you rotate it 180 degrees, is going to map to ray OD. And so this first statement is true. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to ray OB and OD respectively. And when people say respectively, they're saying in the same order. That ray OA maps to ray OB, and that ray OC maps to ray OD. And we saw that, ray OA maps, if you rotate it all the way around 180 degrees, it'll map to OB, and then OC if you rotate 180 degrees, will map to OD." + }, + { + "Q": "\ni don't understand 2:20", + "A": "its just saying that 70% is equal to 100% - 30%", + "video_name": "d1oNF88SAgg", + "timestamps": [ + 140 + ], + "3min_transcript": "Let's say I go to the fruit store today and they have a sale on guavas. Everything is 30% off. This is for guavas. And it's only today. Only today. So I say, you know what, let me go buy a bunch of guavas. So I go and I buy 6 guavas. So I buy six guavas. And it ends up, when I go to the register, and we're assuming no tax, it's a grocery and I live in a state where they don't tax groceries. So for the 6 guavas, they charge me, I get the 30% off. They charge me $12.60. $12.60. So this is the 30% off sale price on 6 guavas. I go home, and then my wife tells me, you know, Sal, can you go get 2 more guavas tomorrow? So the next day I go and I want to buy 2 more guavas. So, 2 guavas. But now the sale is off. That was only that first day that I bought the 6. So how much are those two guavas going to cost me? How much are those two guavas going to cost at full price? At full price? So, a good place to start is, to think about how much would those 6 guavas have cost us at full price? This is the sale price, right here? This is the sale price. How much would those have cost me at full price? So let's do a little bit of algebra here. Pick a suitable color for the algebra. Maybe this grey color. So, let's say that x is equal to the cost of 6 guarvas. 6 guavas, at full price. So, essentially, if we take 30% off of this, we should get $12.60. So let's do that. So if we have the full price of 6 guavas, we're going So that's the same thing as 0.30. Or I could just write 0.3. I could ignore that zero if I like. Actually, let me write it like this. My wife is always bugging me to write zeroes before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. Some I'm just taking 30% off of the full price, off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We could imagine there's a 1 in front -- you know, x is the same thing as 1x. So 1x minus 0.3x is going to be equal to 0.7x. So we get 0.7x, or we could say 0.70 if you like. Same number. Point, or 0.7x, is equal to 12.60. And once you get used to these problems, you might just skip" + }, + { + "Q": "\nWhy are we subtracting 0.30 from x at 1:59?\nI don't get it\nThis algebra is a bit confusing", + "A": "x would be considered the full price of the guavas, and you are subtracting 0.30 from it, as you are finding 30% of the full price. 0.30 is 30% in decimal notation, so by subtracting x by 0.30, you are finding the price after the 30% discount", + "video_name": "d1oNF88SAgg", + "timestamps": [ + 119 + ], + "3min_transcript": "Let's say I go to the fruit store today and they have a sale on guavas. Everything is 30% off. This is for guavas. And it's only today. Only today. So I say, you know what, let me go buy a bunch of guavas. So I go and I buy 6 guavas. So I buy six guavas. And it ends up, when I go to the register, and we're assuming no tax, it's a grocery and I live in a state where they don't tax groceries. So for the 6 guavas, they charge me, I get the 30% off. They charge me $12.60. $12.60. So this is the 30% off sale price on 6 guavas. I go home, and then my wife tells me, you know, Sal, can you go get 2 more guavas tomorrow? So the next day I go and I want to buy 2 more guavas. So, 2 guavas. But now the sale is off. That was only that first day that I bought the 6. So how much are those two guavas going to cost me? How much are those two guavas going to cost at full price? At full price? So, a good place to start is, to think about how much would those 6 guavas have cost us at full price? This is the sale price, right here? This is the sale price. How much would those have cost me at full price? So let's do a little bit of algebra here. Pick a suitable color for the algebra. Maybe this grey color. So, let's say that x is equal to the cost of 6 guarvas. 6 guavas, at full price. So, essentially, if we take 30% off of this, we should get $12.60. So let's do that. So if we have the full price of 6 guavas, we're going So that's the same thing as 0.30. Or I could just write 0.3. I could ignore that zero if I like. Actually, let me write it like this. My wife is always bugging me to write zeroes before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. Some I'm just taking 30% off of the full price, off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We could imagine there's a 1 in front -- you know, x is the same thing as 1x. So 1x minus 0.3x is going to be equal to 0.7x. So we get 0.7x, or we could say 0.70 if you like. Same number. Point, or 0.7x, is equal to 12.60. And once you get used to these problems, you might just skip" + }, + { + "Q": "\nI noticed Sal uses x in his previous video, and then he uses it here at 1:40. I know x is the same as 1x, but why is 1 used instead of 2 or 3? Is 1 supposed to be the opposite of 0? Because in a number line, the numbers between 0 and 1 would be 0.01 to 0.99. Does 1 represent 100% in this case?", + "A": "Yes, x here represent 100%, that s why x is the same as 1x. Sal is just using x as a variable. You can use whatever letter you like, y, z, a, b, etc...", + "video_name": "d1oNF88SAgg", + "timestamps": [ + 100 + ], + "3min_transcript": "Let's say I go to the fruit store today and they have a sale on guavas. Everything is 30% off. This is for guavas. And it's only today. Only today. So I say, you know what, let me go buy a bunch of guavas. So I go and I buy 6 guavas. So I buy six guavas. And it ends up, when I go to the register, and we're assuming no tax, it's a grocery and I live in a state where they don't tax groceries. So for the 6 guavas, they charge me, I get the 30% off. They charge me $12.60. $12.60. So this is the 30% off sale price on 6 guavas. I go home, and then my wife tells me, you know, Sal, can you go get 2 more guavas tomorrow? So the next day I go and I want to buy 2 more guavas. So, 2 guavas. But now the sale is off. That was only that first day that I bought the 6. So how much are those two guavas going to cost me? How much are those two guavas going to cost at full price? At full price? So, a good place to start is, to think about how much would those 6 guavas have cost us at full price? This is the sale price, right here? This is the sale price. How much would those have cost me at full price? So let's do a little bit of algebra here. Pick a suitable color for the algebra. Maybe this grey color. So, let's say that x is equal to the cost of 6 guarvas. 6 guavas, at full price. So, essentially, if we take 30% off of this, we should get $12.60. So let's do that. So if we have the full price of 6 guavas, we're going So that's the same thing as 0.30. Or I could just write 0.3. I could ignore that zero if I like. Actually, let me write it like this. My wife is always bugging me to write zeroes before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. Some I'm just taking 30% off of the full price, off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We could imagine there's a 1 in front -- you know, x is the same thing as 1x. So 1x minus 0.3x is going to be equal to 0.7x. So we get 0.7x, or we could say 0.70 if you like. Same number. Point, or 0.7x, is equal to 12.60. And once you get used to these problems, you might just skip" + }, + { + "Q": "At 3:21, why can we divide 12.60 by 0.7 to get the full price?\n", + "A": "The equation has 0.7 times x . To isolate x , we use the opposite operation. The opposite of multiplication is division. This is why Sal is dividing both sides of the equation by 0.7", + "video_name": "d1oNF88SAgg", + "timestamps": [ + 201 + ], + "3min_transcript": "That was only that first day that I bought the 6. So how much are those two guavas going to cost me? How much are those two guavas going to cost at full price? At full price? So, a good place to start is, to think about how much would those 6 guavas have cost us at full price? This is the sale price, right here? This is the sale price. How much would those have cost me at full price? So let's do a little bit of algebra here. Pick a suitable color for the algebra. Maybe this grey color. So, let's say that x is equal to the cost of 6 guarvas. 6 guavas, at full price. So, essentially, if we take 30% off of this, we should get $12.60. So let's do that. So if we have the full price of 6 guavas, we're going So that's the same thing as 0.30. Or I could just write 0.3. I could ignore that zero if I like. Actually, let me write it like this. My wife is always bugging me to write zeroes before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. Some I'm just taking 30% off of the full price, off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We could imagine there's a 1 in front -- you know, x is the same thing as 1x. So 1x minus 0.3x is going to be equal to 0.7x. So we get 0.7x, or we could say 0.70 if you like. Same number. Point, or 0.7x, is equal to 12.60. And once you get used to these problems, you might just skip Where you say, 70% of the full price is equal to my sale price, right? I took 30% off. This is 70% of the full price. You might just skip to this step once you get used to these problems in a little bit. And now we just have to solve for x. Divide both sides by 0.7, so you get x is equal to 12.60 divided by 0.7. We could use a calculator, but it's always good to get a little bit of practice dividing decimals. So we get 0.7 goes into 12.60. Let's multiply both of these numbers by 10, which is what we do when we move both of their decimals one to the right. So the 0.7 becomes a 7. Ignore that right there. The 12.60 becomes 126, put the decimal right there. Decimal right there. And we're ready to just do straight up long division. So this is now a 7, not a .7. So 7 goes into 12 1 time." + }, + { + "Q": "\nIn 1:13, why is sal putting it like -400w+1,100 instead of subtracting it like 1,100-400w?", + "A": "the first looks like slope intercept form, y = mx+ b", + "video_name": "2EwPpga_XPw", + "timestamps": [ + 73 + ], + "3min_transcript": "Just as you were solving the potato chip conundrum in the last video, the king's favorite magical bird comes flying along and starts whispering into the king's ear. And this makes you a little bit self-conscious, a little bit insecure, so you tell the king, what is the bird talking about. And the king says, well, the bird says that he thinks that there's another way to do the problem. And you're not used to taking advice from birds. And so being a little bit defensive, you say, well, if the bird thinks he knows so much, let him do this problem. And so the bird whispers a little bit more in the king's ear and says, OK, well I'll have to do the writing because the bird does not have any hands, or at least can't manipulate chalk. And so the bird continues to whisper in the king's ear. And the king translates and says, well, the bird says, let's use one of these equations to solve for a variable. So let's say, let's us this blue equation right over here to solve for a variable. And that's essentially going to be a constraint of one variable in terms of another. So let's see if we can do that. So here, if we want to solve for m, we could subtract 400 w from both sides. If we subtract 400w from the left, this 400w goes away. If we subtract 400w from the right, we have is equal to negative 400w plus 1,100. So what got us from here to here is just subtracting 400w from both sides. And then if we want to solve for m, we just divide both sides by 100. So we just divide all of the terms by 100. And then we get m is equal to negative 400 divided by 100, is negative 4w. 1,100 divided by 100 is 11. Plus 11. So now we've constrained m in terms of w. This is what the bird is saying, using the king as his translator. Why don't we take this constraint and substitute it back for m in the first equation? And then we will have one equation with one unknown. 200, so he's looking at that first equation now, he says 200. Instead of putting an m there, the bird says well, by the second constraint, m is equal to negative 4w plus 11. So instead of writing an m, we substitute for m the expression negative 4w plus 11. And then we have the rest of it, plus 300w, is equal to 1,200. So just to be clear, everywhere we saw an m, we replaced it with this right over here, in that first equation. So the first thing, you start to scratch your head. And you say, is this a legitimate thing to do. Will I get the same answer as I got when I solved the same problem with elimination? And I want you to sit and think about that for a second." + }, + { + "Q": "At about 2:19, what is the word \"constraint\" used for? What exactly does that mean for this problem?\n", + "A": "A constraint is simply another word for a limitation or restriction of the possible set of solutions satisfying both equalities. While either equality will represent visually a different line on a graph or in other words a different relationship to each variable given a set of constants, the solution set is limited to the intersection of the equalities. Thus in a system of equations, additional equalities or equations represent further restriction of the possible set of all solutions.", + "video_name": "2EwPpga_XPw", + "timestamps": [ + 139 + ], + "3min_transcript": "Just as you were solving the potato chip conundrum in the last video, the king's favorite magical bird comes flying along and starts whispering into the king's ear. And this makes you a little bit self-conscious, a little bit insecure, so you tell the king, what is the bird talking about. And the king says, well, the bird says that he thinks that there's another way to do the problem. And you're not used to taking advice from birds. And so being a little bit defensive, you say, well, if the bird thinks he knows so much, let him do this problem. And so the bird whispers a little bit more in the king's ear and says, OK, well I'll have to do the writing because the bird does not have any hands, or at least can't manipulate chalk. And so the bird continues to whisper in the king's ear. And the king translates and says, well, the bird says, let's use one of these equations to solve for a variable. So let's say, let's us this blue equation right over here to solve for a variable. And that's essentially going to be a constraint of one variable in terms of another. So let's see if we can do that. So here, if we want to solve for m, we could subtract 400 w from both sides. If we subtract 400w from the left, this 400w goes away. If we subtract 400w from the right, we have is equal to negative 400w plus 1,100. So what got us from here to here is just subtracting 400w from both sides. And then if we want to solve for m, we just divide both sides by 100. So we just divide all of the terms by 100. And then we get m is equal to negative 400 divided by 100, is negative 4w. 1,100 divided by 100 is 11. Plus 11. So now we've constrained m in terms of w. This is what the bird is saying, using the king as his translator. Why don't we take this constraint and substitute it back for m in the first equation? And then we will have one equation with one unknown. 200, so he's looking at that first equation now, he says 200. Instead of putting an m there, the bird says well, by the second constraint, m is equal to negative 4w plus 11. So instead of writing an m, we substitute for m the expression negative 4w plus 11. And then we have the rest of it, plus 300w, is equal to 1,200. So just to be clear, everywhere we saw an m, we replaced it with this right over here, in that first equation. So the first thing, you start to scratch your head. And you say, is this a legitimate thing to do. Will I get the same answer as I got when I solved the same problem with elimination? And I want you to sit and think about that for a second." + }, + { + "Q": "\nAt 0:33 if he did regroup instead of leaving it, would the outcome be the same or would it be different?", + "A": "If by outcome you mean the solution to 65*78 then yes, it would be the same.", + "video_name": "p0jCw2sqZgs", + "timestamps": [ + 33 + ], + "3min_transcript": "I'm going to multiply 78 times 65 in a little less than standard way, but hopefully it'll make some sense, and you'll realize that there's multiple ways that you can multiply. And this is actually the way that I multiply numbers in my head. So 78 times 65. And then we're actually going to think about what the different parts of this process represent on this area model. So 78 times 65. So I'm going to start just the way we normally start when we multiply. I'm going to start with this 5 in this ones place, and I'm going to say 5 times 8 is 40. And instead of just writing a 0 and carrying a 4 right over here, I'm just going to write the number 40. So this was the 5 times the 8. Now, I'm going to multiply the 5 times the 7. And we have to be a little bit careful here because 5 times-- this isn't just any 7, this is a 70. So what is 5 times 70? Well, 5 times 7 would be 35, so five times 70 is 350. So I'll write that down, 350. If you add these two together, this is going to be 5 times 78. Now let's go over to the 6. So let's multiply the 6 times the 8. Now we have to be careful again. This 6 is not just a regular 6, it's in the tens place. This is a 60. 60 times 8. Well, 6 times 8 is 48, so 60 times 8 is going to be 480. So it's going to be 480. And then 6 times 7. Well, that would be 42, but we have to be careful. This is 60 times 70, so we're going to have two zeroes at the end. This is 4,200, not just 42. So 6 times 7 is 4,200. And now we can add everything together. And this is a very similar process to what we do when we do the traditional method of multiplying. I just made it a little bit more explicit But we can add everything together. In the ones place, we have a 0. In the tens place, we have 4 plus 5 is 9. 9 plus 8 is 17. And now we can carry a 1. 1 plus 3 is 4. 4 plus 4 is 8. 8 plus 2 is 10. Carry a 1, regroup of 1 even, and then you have a 5 right over there. So you get 5,070. Now, I want to think about-- I want to visualize what was going on here using this area model. So once again, we had 78. So I'm going to make this vertical length represent 78. So this distance right over here represents the 70. That's the 70, and then we'll make this distance right over here represent the 8. Let me make that a little bit cleaner so you see what I'm talking about. So this distance right over here represents the 8 and then we're going to multiply that times 65." + }, + { + "Q": "At 0:24 why is that shape a parallelogram because I thought it was scalene how come it is a paralleogram I don't get it\n", + "A": "Sal says like a parallelogram . He doesn t mean to say that it IS a parallelogram. Hope that helps you understand! :)", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 24 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 2:14 what does adjacent mean?", + "A": "Adjacent means that the two sides are directly next to each other. Not opposite in any way but connected.", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 134 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 0:10 Sal said that mathmeticians have looked at the way kites are drawn in cartoons, why would they do that? I thought that the kite that we fly and draw in cartoons was copied off the math kite.", + "A": "They tried to compare kites as a shape to something used in cartoons so they could see the new shape.", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 10 + ], + "3min_transcript": "" + }, + { + "Q": "Using the kite in the video, preferably looking around 2:00 , are the top left and bottom right sides parallel?\n", + "A": "Kites do not usually have any parallel sides. The only exception I know of is the rhombus (which, depending on how kite is defined is a special kind of kite).", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 120 + ], + "3min_transcript": "" + }, + { + "Q": "\nsorry for got time 5:01", + "A": "He said it can be an arbitrarily large number as long as it s greater", + "video_name": "UTs4uZhu5t8", + "timestamps": [ + 301 + ], + "3min_transcript": "right corner. Say we have x plus 15 is greater than or equal to negative 60. Notice, now we have greater than or equal. So let's solve this the same way we solved We can subtract 15 from both sides. And I like to switch up my notation. Here I added the 5, kind of, on the same line. You could also do your adding or subtracting below the line, like this. So if I subtract 15 from both sides, so I do a minus 15 there, and I do a minus 15 there. Then the left-hand side just becomes an x. Because obviously you have 15 minus 15. That just cancels out. And you get x is greater than or equal to negative 60 minus 15 is negative 75. If something is greater than or equal to something else, if I take 15 away from this and from that, the greater than or So our solution is x is greater than or equal to negative 75. Let's graph it on the number line. So let me draw a number line here. I'll have-- let's say that's negative 75, that's negative 74, that's negative 73, that's negative 76. And so on and so forth. I could keep plotting things. Now, x has to be greater than or equal to negative 75. So x can be equal to negative 75. So we can include the point, because we have this greater than or equal sign. Notice we're not making it hollow like we did there, we're making it filled in because it can equal negative 75, or it needs to be greater than. So greater than or equal. We'll shade in everything above negative 75 as well. So in orange is the solution set. And this obviously, we could keep going to the right. x could be a million, it could be a billion, it could be a googol. It can be an arbitrarily large number as long as it's greater Let's do a couple more. Let's do x minus 2 is less than or equal to 1. Once again, we want to get just our x on the left-hand side. Get rid of this negative 2. Let's add 2 to both sides of this equation. Plus 2. The left-hand side just becomes an x. You have a less than or equal sign. That won't change by adding or subtracting the same thing to both sides of the inequality. And then 1 plus 2 is 3. So x needs to be less than or equal to 3. Any x that is less than or equal to 3 will satisfy this equation. So let's plot it. And I'd try out any x that's less than or equal to 3 and verify for yourself that it does indeed satisfy this I shouldn't call it an equation. This inequality. So let me graph the solution set. So let's say this is 0, 1, 2, 3, 4." + }, + { + "Q": "Starting at 3:35 sal subtracts to isolate the variable, but in previous videos, any time you subtract from both sides you have to change the inequality sign. Is that only done in a multiply or divide situation or did he just not switch the sign?\n", + "A": "Correct, do not change the direction of the inequality when you add or subtract a negative. Always change the inequality when you multiply or divide by a negative. 3 < 4 -3 > -4", + "video_name": "UTs4uZhu5t8", + "timestamps": [ + 215 + ], + "3min_transcript": "And then we could go below 40. 39, 38. You can just keep going below 40. It just keeps going on in both directions. And any x that is less than 40 will satisfy this. So it can't be equal to 40, because if x is equal to 40, 40 minus 5 is 35. It's not less than 35. So x has to be less than 40. And to show this on the number line, we do a circle around 40 to show that we're not including 40. But then we can shade in everything below 40. So everything that's just exactly below 40 is included So everything I've shaded in yellow is included in our solution set. So 39, 39.999999, repeating, which is about as close as you can get to 40 as possible, that's in our solution set. But 40 is not. That's why we put that open circle around it. Let's do another one. Let me do it in another color as well. right corner. Say we have x plus 15 is greater than or equal to negative 60. Notice, now we have greater than or equal. So let's solve this the same way we solved We can subtract 15 from both sides. And I like to switch up my notation. Here I added the 5, kind of, on the same line. You could also do your adding or subtracting below the line, like this. So if I subtract 15 from both sides, so I do a minus 15 there, and I do a minus 15 there. Then the left-hand side just becomes an x. Because obviously you have 15 minus 15. That just cancels out. And you get x is greater than or equal to negative 60 minus 15 is negative 75. If something is greater than or equal to something else, if I take 15 away from this and from that, the greater than or So our solution is x is greater than or equal to negative 75. Let's graph it on the number line. So let me draw a number line here. I'll have-- let's say that's negative 75, that's negative 74, that's negative 73, that's negative 76. And so on and so forth. I could keep plotting things. Now, x has to be greater than or equal to negative 75. So x can be equal to negative 75. So we can include the point, because we have this greater than or equal sign. Notice we're not making it hollow like we did there, we're making it filled in because it can equal negative 75, or it needs to be greater than. So greater than or equal. We'll shade in everything above negative 75 as well. So in orange is the solution set. And this obviously, we could keep going to the right. x could be a million, it could be a billion, it could be a googol. It can be an arbitrarily large number as long as it's greater" + }, + { + "Q": "\nAt 4:19 Sal says that sqrt(2) * sqrt(2) is 2. How does that work?", + "A": "We can multiply square roots. sqrt(2) * sqrt(2) = sqrt(4) What is the sqrt(4)? It = 2.", + "video_name": "s9ppnjgmiyk", + "timestamps": [ + 259 + ], + "3min_transcript": "And to identify the perfect squares you would say, Alright, are there any factors where I have at least two of them? Well I have two times two here. And I also have five times five here. So I can rewrite the square root of 200 as being equal to the square root of two times two. Let me just write it all out. Actually I think I'm going to run out of space. So the square root, give myself more space under the radical, square root of two times two times five times five times two. And I wrote it in this order so you can see the perfect squares here. Well this is going to be the same thing as the square root of two times two. This second method is a little bit more monotonous, (laughing) I guess is one way to think about it. And they really, they boil down to the same method. We're still going to get to the same answer. So square root of two times two times the square root times the square root of five times five, times the square root of two. Well the square root of two times two is just going to be, this is just two. Square root of five times five, well that's just going to be five. So you have two times five times the square root of two, which is 10 times the square root of two. So this right over here, square root of 200, we can rewrite as 10 square roots of two. So this is going to be equal to one over 10 square roots of two. Now some people don't like having a radical in the denominator and if you wanted to get rid of that, you could multiply both 'Cause notice we're just multiplying by one, we're expressing one as square root of two over square root of two, and then what that does is we rewrite this as the square root of two over 10 times the square root of two times the square root of two. Well the square root of two times the square root of two is just going to be two. So it's going to be 10 times two which is 20. So it could also be written like that. So hopefully you found that helpful. In fact, even this one, you could write if you want to visualize it slightly differently, you could view it as one twentieth times the square root of two. So these are all the same thing." + }, + { + "Q": "\nAt 3:40 Why does the the square root of 2 times 2 times the square root of 5 times 5 times the the square root of 2 equal 10 times the square root of 2? Shouldn't , sense, the square root of 2 times 2 is 2 and the square root of 5 times 5 is, it be 7 times the square root of 2?", + "A": "you re multiplying the square roots not adding them.", + "video_name": "s9ppnjgmiyk", + "timestamps": [ + 220 + ], + "3min_transcript": "And to identify the perfect squares you would say, Alright, are there any factors where I have at least two of them? Well I have two times two here. And I also have five times five here. So I can rewrite the square root of 200 as being equal to the square root of two times two. Let me just write it all out. Actually I think I'm going to run out of space. So the square root, give myself more space under the radical, square root of two times two times five times five times two. And I wrote it in this order so you can see the perfect squares here. Well this is going to be the same thing as the square root of two times two. This second method is a little bit more monotonous, (laughing) I guess is one way to think about it. And they really, they boil down to the same method. We're still going to get to the same answer. So square root of two times two times the square root times the square root of five times five, times the square root of two. Well the square root of two times two is just going to be, this is just two. Square root of five times five, well that's just going to be five. So you have two times five times the square root of two, which is 10 times the square root of two. So this right over here, square root of 200, we can rewrite as 10 square roots of two. So this is going to be equal to one over 10 square roots of two. Now some people don't like having a radical in the denominator and if you wanted to get rid of that, you could multiply both 'Cause notice we're just multiplying by one, we're expressing one as square root of two over square root of two, and then what that does is we rewrite this as the square root of two over 10 times the square root of two times the square root of two. Well the square root of two times the square root of two is just going to be two. So it's going to be 10 times two which is 20. So it could also be written like that. So hopefully you found that helpful. In fact, even this one, you could write if you want to visualize it slightly differently, you could view it as one twentieth times the square root of two. So these are all the same thing." + }, + { + "Q": "\nAt 3:18 what does Sal mean when he says that each triangle has 180 degrees?", + "A": "All of the internal angles of a triangle will always add up to 180\u00c2\u00b0.", + "video_name": "_HJljJuVHLw", + "timestamps": [ + 198 + ], + "3min_transcript": "So the measure of this angle is also x plus 40. Because they're corresponding angles, and you could see that by inspection, and if you moved around the transversal, it would make sense that that's the case. So this is x plus 40 and this is minus 40, and they're clearly supplements of each other. They're supplementary angles. Then the sum of these two angles have to be equal to 180. So let's figure it out. So x minus 40 plus x plus 40 is equal to 180, because they're supplementary. The 40's cancel out. So you just minus 40, plus 40 adds up to zero. So you're left with 2x is equal to 180. x is equal to 90. So it's D. 47: The measures of the interior angles of a pentagon are 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. OK, so first of all, we have to remember what is the sum of the interior angles of a pentagon? And that's where I always draw an arbitrary pentagon. Let me see if I can do that. Actually, there's a polygon tool here. How does it work? I'm just trying to draw a pentagon. I don't know if that's any different than the line tool, So how many triangles can I draw in a pentagon? And that tells me what my total interior angles are. And there is a formula for that, but I like relying on your reasoning more than the formula, because you might forget the formula, or even worse, you might remember it, but not have the confidence to use it, or you might remember it wrong ten years in the future. So the best thing to do, if you have a polygon, is to count the triangles in it. Straightforward enough. So a pentagon has three triangles in it. So the sum of its interior angles are going to be 3 times 180, because it has 3 triangles in it. Each triangle has 180 degrees. And I know you can't see what I just wrote. So the sum of all of these angles are going to be the sum of all of the angles in all three triangles. So it's 3 times 180 is equal to 540 degrees. So that's the sum of all of these interior angles. Now, they say that each of them are 2x, 6x, et cetera. So the sum of all of these terms have to be equal 540. I'm going to write them vertically. It makes them easier to add. So if we write there 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. This is going to be the largest, right?" + }, + { + "Q": "\nAt 10:28, why isn't the exterior angle 240 degrees, like a major arc in a circle?", + "A": "all exterior angles add up to 360 in any polygon. in regular polygons all exterior angles are equal. thus one angle=360/6=60", + "video_name": "_HJljJuVHLw", + "timestamps": [ + 628 + ], + "3min_transcript": "equal to 180 minus 100 which equals 80 degrees. So this angle right here is equal to 80 degrees. And the angle they want us to figure out is the opposite of this angle, or in the U.S., I guess, they say vertical angles. And so opposite or vertical angles are equal or they're congruent, so this is going to be 80 degrees as well. And that is choice A. Problem 50: What is the measure of an exterior angle of a regular hexagon? A regular hexagon tells us that all of the sides are the same, it's equilateral, and all of the angles are the same, equiangular. So if we just knew what's the total degree measure of the interior angles, we could just divide that by 6, and then and then we could use that information to figure out the Let's just do it. So once again, I like to just draw a hexagon. Let's just draw a hexagon and count the triangles in it. Two sides, three sides, four sides, five sides and six sides. And how many triangles do I have here? One, two, three. So I have one, two, three, four triangles. The sum of the interior angles of this hexagon, of any hexagon, whether it's regular or not, are going to be 4 times 180 and that's 720 degrees. And it's a regular hexagon, so all the interior angles are going to be the same. And there's six of them. So each of them are going to be 720 divided by 6. So each of the interior angles are going to be 120 degrees. And I didn't draw it that regular, but we can assume that all of these are each 120 degrees. Fair enough. Now, if all of those are each 120 degrees, what is the measure of an exterior angle? Well, we could just extend one of these sides out a little bit. We could say, OK, if this is 120 degrees, what is its supplement? Well, these have to add up to 180, so 180 minus 120 is 60 degrees. I could do it on any side. I could extend that line out there, and I'd say, oh, that's 60 degrees. So any of the exterior angles are 60 degrees. B. All right, do I have time for one more? I'll wait for the next one in the next video. See you soon." + }, + { + "Q": "At 2:44, Sal says there is a formula, but he doesn't reveal it. What is the formula?\n", + "A": "The sum of interior angles in an n-sided convex polygon is: 180(n - 2)", + "video_name": "_HJljJuVHLw", + "timestamps": [ + 164 + ], + "3min_transcript": "So the measure of this angle is also x plus 40. Because they're corresponding angles, and you could see that by inspection, and if you moved around the transversal, it would make sense that that's the case. So this is x plus 40 and this is minus 40, and they're clearly supplements of each other. They're supplementary angles. Then the sum of these two angles have to be equal to 180. So let's figure it out. So x minus 40 plus x plus 40 is equal to 180, because they're supplementary. The 40's cancel out. So you just minus 40, plus 40 adds up to zero. So you're left with 2x is equal to 180. x is equal to 90. So it's D. 47: The measures of the interior angles of a pentagon are 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. OK, so first of all, we have to remember what is the sum of the interior angles of a pentagon? And that's where I always draw an arbitrary pentagon. Let me see if I can do that. Actually, there's a polygon tool here. How does it work? I'm just trying to draw a pentagon. I don't know if that's any different than the line tool, So how many triangles can I draw in a pentagon? And that tells me what my total interior angles are. And there is a formula for that, but I like relying on your reasoning more than the formula, because you might forget the formula, or even worse, you might remember it, but not have the confidence to use it, or you might remember it wrong ten years in the future. So the best thing to do, if you have a polygon, is to count the triangles in it. Straightforward enough. So a pentagon has three triangles in it. So the sum of its interior angles are going to be 3 times 180, because it has 3 triangles in it. Each triangle has 180 degrees. And I know you can't see what I just wrote. So the sum of all of these angles are going to be the sum of all of the angles in all three triangles. So it's 3 times 180 is equal to 540 degrees. So that's the sum of all of these interior angles. Now, they say that each of them are 2x, 6x, et cetera. So the sum of all of these terms have to be equal 540. I'm going to write them vertically. It makes them easier to add. So if we write there 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. This is going to be the largest, right?" + }, + { + "Q": "\nAt 0:37, Sal gives a negative number as common ratio, so the terms are alternatively positive and negative. But a sequence usually has terms which keep increasing or decreasing, right?", + "A": "Sequences might usually keep increasing or decreasing, but there s no rule that says they have to. As Sal s example illustrates.", + "video_name": "CecgFWTg9pQ", + "timestamps": [ + 37 + ], + "3min_transcript": "In the last video we saw that a geometric progression, or a geometric sequence, is just a sequence where each successive term is the previous term multiplied by a fixed value. And we call that fixed value the common ratio. So, for example, in this sequence right over here, each term is the previous term multiplied by 2. So 2 is our common ratio. And any non-zero value can be our common ratio. It can even be a negative value. So, for example, you could have a geometric sequence that looks like this. Maybe start at one, and maybe our common ratio, let's say it's negative 3. So 1 times negative 3 is negative 3. Negative 3 times negative 3 is positive 9. Positive 9 times negative 3 is negative 27. And then negative 27 times negative 3 is positive 81. And you could keep going on and on and on. What I now want to focus on in this video is the sum of a geometric progression or a geometric sequence, and we would call that Let's scroll down a little bit. So now we're going to talk about geometric series, which is really just the sum of a geometric sequence. So, for example, a geometric series would just be a sum of this sequence. So if we just said 1 plus negative 3, plus 9, plus negative 27, plus 81, and we were to go on, and on, and on, this would be a geometric series. And we could do it with this one up here just to really make it clear of what we're doing. So if we said 3 plus 6, plus 12, plus 24, plus 48, this once again is a geometric series, just the sum of a geometric sequence or a geometric progression. So how would we represent this in general terms and maybe Well, we'll start with whatever our first term is. And over here if we want to speak in general terms we could call that a, our first term. So we'll start with our first term, a, and then each successive term that we're going to add is going to be a times our common ratio. And we'll call that common ratio r. So the second term is a times r. Then the third term, we're just going to multiply this one times r. So it's going to be a times r squared. And then we can keep going, plus a times r to the third power. And let's say we're going to do a finite geometric series. So we're not going to just keep on going forever. Let's say we keep going all the way until we get to some a times r to the n. a times r to the n-th power." + }, + { + "Q": "3:28 is there a symbol for not existing or for a false statement ?\n", + "A": "The not-equal-to sign, =/=, means that something does not equal something else. Another one to use is the naught sign, meaning that no possible solutions exist. It is denoted by a zero with a diagonal slash, like (/), only more circular.", + "video_name": "nOnd3SiYZqM", + "timestamps": [ + 208 + ], + "3min_transcript": "1.99, 1.99999 . As x gets closer and closer from those values, what is f of x approaching? And we see here that it is approaching 5. But what if we were asked the natural other question-- What is the limit of f of x as x approaches 2 from values greater than 2? So this is a little superscript positive right over here. So now we're going to approach x equals 2, but we're going to approach it from this direction-- x equals 3, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001. And we're going to get closer and closer to 2, but we're coming from values that are larger than 2. So here, when x equals 3, f of x is here. When x equals 2.5, f of x is here. When x equals 2.01, f of x looks like it's right over here. to f of x equaling 1. It never does quite equal that. It actually then just has a jump discontinuity. This seems to be the limiting value when we approach when we approach 2 from values greater than 2. So this right over here is equal to 1. And so when we think about limits in general, the only way that a limit at 2 will actually exist is if both of these one-sided limits are actually the same thing. In this situation, they aren't. As we approach 2 from values below 2, the function seems to be approaching 5. And as we approach 2 from values above 2, the function seems to be approaching 1. So in this case, the limit-- let me write this down-- the limit of f of x, as x approaches 2 equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that." + }, + { + "Q": "3:27 If both limits must be equal for the whole limit to exist, what about in instances where there is a piecewise function like this, but with the filled-in dot continuous on one of the lines? As it is approached from either side, you will get different values.\n\nTherefore, for all piecewise functions, is the limit nonexistent at the discontinuity?\n", + "A": "Yes, if there is a discontinuity, or if the graph asymptotically approaches infinity, there is not limit. However, as you will soon learn (or maybe already have), there is a such thing as a one sided limit. For example the one sided limit of sqrt(x) approaching 0 from the right is 0, even though the whole limit does not exist at that point. I hope this helped.", + "video_name": "nOnd3SiYZqM", + "timestamps": [ + 207 + ], + "3min_transcript": "1.99, 1.99999 . As x gets closer and closer from those values, what is f of x approaching? And we see here that it is approaching 5. But what if we were asked the natural other question-- What is the limit of f of x as x approaches 2 from values greater than 2? So this is a little superscript positive right over here. So now we're going to approach x equals 2, but we're going to approach it from this direction-- x equals 3, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001. And we're going to get closer and closer to 2, but we're coming from values that are larger than 2. So here, when x equals 3, f of x is here. When x equals 2.5, f of x is here. When x equals 2.01, f of x looks like it's right over here. to f of x equaling 1. It never does quite equal that. It actually then just has a jump discontinuity. This seems to be the limiting value when we approach when we approach 2 from values greater than 2. So this right over here is equal to 1. And so when we think about limits in general, the only way that a limit at 2 will actually exist is if both of these one-sided limits are actually the same thing. In this situation, they aren't. As we approach 2 from values below 2, the function seems to be approaching 5. And as we approach 2 from values above 2, the function seems to be approaching 1. So in this case, the limit-- let me write this down-- the limit of f of x, as x approaches 2 equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that." + }, + { + "Q": "\nAt 5:30, why does he say that the limit as x approaches 4 is 5 instead of -5?", + "A": "Because he made a mistake.", + "video_name": "nOnd3SiYZqM", + "timestamps": [ + 330 + ], + "3min_transcript": "equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that. As x equals 4 from below-- So when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999-- we're getting closer and closer to our function equaling negative 5. So the limit as we approach 4 from below-- this one-sided limit from the left, we could say-- this is going to be equal to negative 5. And if we were to ask ourselves the limit of f of x, as x approaches 4 from the right, from values larger than 4, well, same exercise. f of 5 gets us here. f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of 4 is actually defined, but we're getting closer and closer to it. Even if f of 4 was not defined on either side, we would be approaching negative 5. So this is also approaching negative 5. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say-- so these two things are equal. And because these two things are equal, we know that the limit of f of x, as x approaches 4, is equal to 5. Let's look at a few more examples. So let's ask ourselves the limit of f of x-- now, this is our new f of x depicted here-- as x approaches 8. And let's approach 8 from the left. As x approaches 8 from values less than 8. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to 8." + }, + { + "Q": "At 5:30 I think Sal made a speed mistake. The limit as x approaches 4 is equal to -5, not 5. Am I right? :)\n", + "A": "Yes, of course, and the correction is on there now :)", + "video_name": "nOnd3SiYZqM", + "timestamps": [ + 330 + ], + "3min_transcript": "equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that. As x equals 4 from below-- So when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999-- we're getting closer and closer to our function equaling negative 5. So the limit as we approach 4 from below-- this one-sided limit from the left, we could say-- this is going to be equal to negative 5. And if we were to ask ourselves the limit of f of x, as x approaches 4 from the right, from values larger than 4, well, same exercise. f of 5 gets us here. f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of 4 is actually defined, but we're getting closer and closer to it. Even if f of 4 was not defined on either side, we would be approaching negative 5. So this is also approaching negative 5. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say-- so these two things are equal. And because these two things are equal, we know that the limit of f of x, as x approaches 4, is equal to 5. Let's look at a few more examples. So let's ask ourselves the limit of f of x-- now, this is our new f of x depicted here-- as x approaches 8. And let's approach 8 from the left. As x approaches 8 from values less than 8. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to 8." + }, + { + "Q": "\nAt 3:13, sal says the limit does not exist. Does it mean that the graph is not possible, or that f(X) is not defined for that graph?", + "A": "Another way to think of the limit failing to exist at x=2 is that f(2) cannot be re-defined in such a way that the graph becomes continuous at x=2. In other words, the discontinuity in the graph at x=2 cannot be removed even if the value of f(2) is changed! Have a blessed, wonderful day!", + "video_name": "nOnd3SiYZqM", + "timestamps": [ + 193 + ], + "3min_transcript": "1.99, 1.99999 . As x gets closer and closer from those values, what is f of x approaching? And we see here that it is approaching 5. But what if we were asked the natural other question-- What is the limit of f of x as x approaches 2 from values greater than 2? So this is a little superscript positive right over here. So now we're going to approach x equals 2, but we're going to approach it from this direction-- x equals 3, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001. And we're going to get closer and closer to 2, but we're coming from values that are larger than 2. So here, when x equals 3, f of x is here. When x equals 2.5, f of x is here. When x equals 2.01, f of x looks like it's right over here. to f of x equaling 1. It never does quite equal that. It actually then just has a jump discontinuity. This seems to be the limiting value when we approach when we approach 2 from values greater than 2. So this right over here is equal to 1. And so when we think about limits in general, the only way that a limit at 2 will actually exist is if both of these one-sided limits are actually the same thing. In this situation, they aren't. As we approach 2 from values below 2, the function seems to be approaching 5. And as we approach 2 from values above 2, the function seems to be approaching 1. So in this case, the limit-- let me write this down-- the limit of f of x, as x approaches 2 equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that." + }, + { + "Q": "\nwhy is it the matrix vector product and not the dot product at 13:00", + "A": "Because the dot products are for just one row and he s talking about the whole system.", + "video_name": "qvyboGryeA8", + "timestamps": [ + 780 + ], + "3min_transcript": "We represented our system of equation like this and we put it into reduced row echelon form, so this is A and this is 0. This right here is, let me make sure I have some space, let me put it right here. That right there is the reduced row echelon form of A. And so where essentially this equation, this is a linear equation that is trying to solve this problem. The reduced row echelon form of A times our vector x is equal to 0. So, all the solutions to this are also the solutions to our original problem, to our original ax is equal to 0. So what's the solution to this? All the x's that satisfy this, these are the null space of the reduced row echelon form of A. Right? So here are all of the x's, this is the null space, this problem, if we find all of the x's here, this is the null But we're saying that this problem is the same problem as this one, right? So we can write that the null space of A is equal to the null space of the reduced row echelon form of A. And that might seem a little bit confusing, hey, why are you even writing this out, but it's the actually very useful when you're trying to calculate null spaces. So we didn't even have to write a big augmented matrix here. We can say, take our matrix A, put it in reduced row echelon form and then figure out it's null space. We would have gone straight to this point right here. This is the reduced row echelon form of A, and then I could have immediately solved these equations, right? I would have just taken the dot product of the reduced row echelon form or, not the dot product, the matrix vector product of the reduced row echelon form of A with this vector, and I would've gotten these equations, and then these equations would immediately, I can just rewrite them in this form, and I would But anyway, hopefully you found that reasonably useful." + }, + { + "Q": "\nAt 3:34, how does Sal know the triangle is a 30-60-90?", + "A": "Because of the sides. The basic 30-60-90 triangle has sides 2, 1, and sqr 3 (Watch Example: Solving a 30-60-90 triangle , Intro to 30-60-90 Triangles , 30-60-90 Triangles II ...), you can use them to find out angles and points on graphs, with this question, instead of 1 the side is 1/2, so to find the rest of the sides you simply half all the sides of the basic triangle and it is still a 30-60-90 triangle but now it fits the triangle on the graph and you can solve the problem.", + "video_name": "eTDaJ4ebK28", + "timestamps": [ + 214 + ], + "3min_transcript": "OK, so that is my y-axis, that is my x-axis. Not the most neatly drawn axes ever, but it'll do. Let me draw my unit circle. Looks more like a unit ellipse, but you get the idea. And the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle. So if we have some angle, the x-value is going to be equal a minus 1/2. So we got a minus 1/2 right here. And so the angle that we have to solve for, our theta, is the angle that when we intersect the unit circle, the x-value is minus 1/2. So let me see, this is the angle that we're trying to figure out. This is theta that we need to determine. So how can we do that? So this is minus 1/2 right here. Let's figure out these different angles. out this angle right here. And if I know that angle, I can just subtract that from 180 degrees to get this light blue angle that's kind of the solution to our problem. So let me make this triangle a little bit bigger. So that triangle, let me do it like this. That triangle looks something like this. Where this distance right here is 1/2. That distance right there is 1/2. This distance right here is 1. Hopefully you recognize that this is going to be a 30, 60, 90 triangle. You could actually solve for this other side. You'll get the square root of 3 over 2. And to solve for that other side you just need to do the Pythagorean theorem. Actually, let me just do that. Let me just call this, I don't know, just call this a. So you'd get a squared, plus 1/2 squared, which is 1/4, which is equal to 1 squared, which is 1. You get a squared is equal to 3/4, or a is equal to the So you immediately know this is a 30, 60, 90 triangle. And you know that because the sides of a 30, 60, 90 triangle, if the hypotenuse is 1, are 1/2 and square root of 3 over 2. And you also know that the side opposite the square root of 3 over 2 side is 60 degrees. That's 60, this is 90. This is the right angle, and this is 30 right up there. But this is the one we care about. This angle right here we just figured out is 60 degrees. So what's this? What's the bigger angle that we care about? What is 60 degrees supplementary to? It's supplementary to 180 degrees. So the arccosine, or the inverse cosine, let me write that down. The arccosine of minus 1/2 is equal to 120 degrees." + }, + { + "Q": "can someone explain how sal gets to 3/4 ? at 4:03?\n", + "A": "He subtracts 1/4 from both sides.", + "video_name": "eTDaJ4ebK28", + "timestamps": [ + 243 + ], + "3min_transcript": "OK, so that is my y-axis, that is my x-axis. Not the most neatly drawn axes ever, but it'll do. Let me draw my unit circle. Looks more like a unit ellipse, but you get the idea. And the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle. So if we have some angle, the x-value is going to be equal a minus 1/2. So we got a minus 1/2 right here. And so the angle that we have to solve for, our theta, is the angle that when we intersect the unit circle, the x-value is minus 1/2. So let me see, this is the angle that we're trying to figure out. This is theta that we need to determine. So how can we do that? So this is minus 1/2 right here. Let's figure out these different angles. out this angle right here. And if I know that angle, I can just subtract that from 180 degrees to get this light blue angle that's kind of the solution to our problem. So let me make this triangle a little bit bigger. So that triangle, let me do it like this. That triangle looks something like this. Where this distance right here is 1/2. That distance right there is 1/2. This distance right here is 1. Hopefully you recognize that this is going to be a 30, 60, 90 triangle. You could actually solve for this other side. You'll get the square root of 3 over 2. And to solve for that other side you just need to do the Pythagorean theorem. Actually, let me just do that. Let me just call this, I don't know, just call this a. So you'd get a squared, plus 1/2 squared, which is 1/4, which is equal to 1 squared, which is 1. You get a squared is equal to 3/4, or a is equal to the So you immediately know this is a 30, 60, 90 triangle. And you know that because the sides of a 30, 60, 90 triangle, if the hypotenuse is 1, are 1/2 and square root of 3 over 2. And you also know that the side opposite the square root of 3 over 2 side is 60 degrees. That's 60, this is 90. This is the right angle, and this is 30 right up there. But this is the one we care about. This angle right here we just figured out is 60 degrees. So what's this? What's the bigger angle that we care about? What is 60 degrees supplementary to? It's supplementary to 180 degrees. So the arccosine, or the inverse cosine, let me write that down. The arccosine of minus 1/2 is equal to 120 degrees." + }, + { + "Q": "\nAt 6:38, Sal says that, \"We need to restrict it's range to the UPPER hemisphere\". Why do we actually need to restrict it only to UPPER hemisphere?", + "A": "it is not mandatory that you have to restrict only the upper hemisphere. Anyways you have to restrict it to one hemisphere as there will be another value equal to it on the other hemisphere. Also as Sal said you could go one full round and have the same trig function value which would be unacceptable", + "video_name": "eTDaJ4ebK28", + "timestamps": [ + 398 + ], + "3min_transcript": "No, it's 180 minus the 60, this whole thing is 180, so this is, right here is, 120 degrees, right? 120 plus 60 is 180. Or, if we wanted to write that in radians, you just right 120 degrees times pi radian per 180 degrees, degrees cancel out. 12 over 18 is 2/3, so it equals 2 pi over 3 radians. So this right here is equal to 2 pi over 3 radians. Now, just like we saw in the arcsine and the arctangent videos, you probably say, hey, OK, if I have 2 pi over 3 radians, that gives me a cosine of minus 1/2. And I can write that. cosine of 2 pi over 3 is equal to minus 1/2. statement up here. But I can just keep going around the unit circle. For example, I could, how about this point over here? Cosine of this angle, if I were to add, if I were to go this far, would also be minus 1/2. And then I could go 2 pi around and get back here. So there's a lot of values that if I take the cosine of those angles, I'll get this minus 1/2. So we have to restrict ourselves. We have to restrict the values that the arccosine function can take on. So we're essentially restricting it's range. We're restricting it's range. What we do is we restrict it's range to this upper hemisphere, the first and second quadrants. So if we say, if we make the statement that the arccosine of x is equal to theta, we're going to restrict our range, theta, to that top. So theta is going to be greater than or equal to 0 and less than or equal to 2 pi. Less than or equal to pi, right? Where this is also 0 degrees, or 180 degrees. We're restricting ourselves to this part of the hemisphere right there. And so you can't do this, this is the only point where the cosine of the angle is equal minus 1/2. We can't take this angle because it's outside of our range. And what are the valid values for x? Well any angle, if I take the cosine of it, it can be between minus 1 and plus 1. So x, the domain for the arccosine function, is going to be x has to be less than or equal to 1 and greater than or equal to minus 1. And once again, let's just go check our work. Let's see if the value I got here, that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the TI-85." + }, + { + "Q": "\njust got confused after 9:06 .... HELP!", + "A": "x = f(f^-1(x)) - a minor correction: f^-1 goes before f(x) Cheers", + "video_name": "eTDaJ4ebK28", + "timestamps": [ + 546 + ], + "3min_transcript": "Less than or equal to pi, right? Where this is also 0 degrees, or 180 degrees. We're restricting ourselves to this part of the hemisphere right there. And so you can't do this, this is the only point where the cosine of the angle is equal minus 1/2. We can't take this angle because it's outside of our range. And what are the valid values for x? Well any angle, if I take the cosine of it, it can be between minus 1 and plus 1. So x, the domain for the arccosine function, is going to be x has to be less than or equal to 1 and greater than or equal to minus 1. And once again, let's just go check our work. Let's see if the value I got here, that the arccosine of minus 1/2 really is 2 pi over 3 as calculated by the TI-85. So i need to figure out the inverse cosine, which is the same thing as the arccosine of minus 1/2, of minus 0.5. It gives me that decimal, that strange number. Let's see if that's the same thing as 2 pi over 3. 2 times pi divided by 3 is equal to, that exact same number. So the calculator gave me the same value I got. But this is kind of a useless, well, it's not a useless number. It's a valid, that is the answer. But it doesn't, it's not a nice clean answer. I didn't know that this is 2 pi over 3 radians. And so when we did it using the unit circle, we were able to get that answer. So hopefully, actually let me ask you, let me just finish this up with an interesting question. And this applies to all of them. If I were to ask you, you know, say I were to take the arccosine of x, and then I were to take the cosine of that, Well, this statement right here can be said, well, let's say that the arccosine of x is equal to theta, that means that the cosine of theta is equal to x, right? So if the arccosine of x is equal to theta, we can replace this with theta. And then the cosine of theta, well the cosine of theta is x. So this whole thing is going to be x. Hopefully I didn't get confuse you there, right? I'm saying look, arccosine of x, just call that theta. Now, by definition, this means that the cosine of theta is equal to x. These are equivalent statements. These are completely equivalent statements right here. So if we put a theta right there, we take the cosine of theta, it has to be equal to x. Now let me ask you a bonus, slightly trickier question." + }, + { + "Q": "\nfrom 3:30 , when sal defines the error fxn, he uses a term *'bound* how good is p(x) fitting in f(x)...\n\nBut what is meant by Bound and how is the error function bounded ?", + "A": "Being bound simply means that you know that a value is definitely between two limits. For instance, if 10 < x < 15, then x is bound between 10 and 15. You ll actually do that bounding in another video when Sal gets to questions like, how many terms do we need in order to ensure that our approximation is good to one part in a thousand? See my other answer above for an example of when you might look for error bounds in a physical world example as opposed to pure math.", + "video_name": "wgkRH5Uoavk", + "timestamps": [ + 210 + ], + "3min_transcript": "Sometimes you'll see something like N comma a to say it's an Nth degree approximation centered at a. Actually, I'll write that right now. Maybe we might lose it if we have to keep writing it over and over but you should assume that it is an Nth degree polynomial centered at a. And it's going to look like this. It is going to be f of a, plus f prime of a, times x minus a, plus f prime prime of a, times x minus a squared over-- Either you could write two or two factorial, they're the same value. I'll write two factorial. You could write a divided by one factorial over here, if you like. And then plus, you go to the third derivative of f at a times x minus a to the third power, I think you see where this is going, over three factorial. And you keep going, I'll go to this line right here, all the way to your Nth degree term which is the Nth derivative of f to the N over N factorial. And this polynomial right over here, this Nth degree polynomial centered at a, f or P of a is going to be the same thing as f of a. And you can verify that because all of these other terms have an x minus a here. So if you put an a in the polynomial, all of these other terms are going to be zero. And you'll have P of a is equal to f of a. Let me write that down. P of a is equal to f of a. And so it might look something like this. And it's going to fit the curve better the more of these terms that we actually have. So it might look something like this. I'll try my best to show what it might look like. So this is all review, I have this polynomial that's approximating this function. The more terms I have, the higher degree of this polynomial, the better that it will fit this curve But what I wanna do in this video is think about if we can bound how good it's fitting this function as we move away from a. So what I wanna do is define a remainder function. Or sometimes, I've seen some text books call it an error function. And I'm going to call this-- I'll just call it an error-- Just so you're consistent with all the different notations you might see in a book, some people will call this a remainder function and sometimes they'll write a remainder function for an Nth degree polynomial centered at a. Sometimes you'll see this as an error function. The error function is sometimes avoided because it looks like expected value from probability. But you'll see this often, this is E for error. E for error, R for remainder. And sometimes they'll also have the subscripts over there like that. And what we'll do is, we'll just define this function to be the difference between f of x and our approximation of f of x for any given x." + }, + { + "Q": "\nis 3 x = 8 i at 1:01", + "A": "Not quite, actually 3 x = 9 at 1:01", + "video_name": "kbqO0YTUyAY", + "timestamps": [ + 61 + ], + "3min_transcript": "So once again, we have three equal, or we say three identical objects. They all have the same mass, but we don't know what the mass is of each of them. But what we do know is that if you total up their mass, it's the same exact mass as these nine objects And each of these nine objects have a mass of 1 kilograms. So in total, you have 9 kilograms on this side. And over here, you have three objects. They all have the same mass. And we don't know what it is. We're just calling that mass x. And what I want to do here is try to tackle this a little bit more symbolically. In the last video, we said, hey, why don't we just multiply 1/3 of this and multiply 1/3 of this? And then, essentially, we're going to keep things balanced, because we're taking 1/3 of the same mass. This total is the same as this total. That's why the scale is balanced. Now, let's think about how we can represent this symbolically. So the first thing I want you to think about is, can we set up an equation that expresses that we have these three things of mass x, Can we express that as an equation? And I'll give you a few seconds to do it. Well, let's think about it. Over here, we have three things with mass x. So their total mass, we could write as-- we could write their total mass as x plus x plus x. And over here, we have nine things with mass of 1 kilogram. I guess we could write 1 plus 1 plus 1. That's 3. Plus 1 plus 1 plus 1 plus 1. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9. And actually, this is a mathematical representation. If we set it up as an equation, it's an algebraic representation. It's not the simplest possible way we can do it, but it is a reasonable way to do it. If we want, we can say, well, if I have an x plus another x plus another x, I have three x's. So I could rewrite this as 3x. And 3x will be equal to? Well, if I sum up all of these 1's right over here-- 1 We have 9 of them, so we get 3x is equal to 9. And let me make sure I did that. 1, 2, 3, 4, 5, 6, 7, 8, 9. So that's how we would set it up. And so the next question is, what would we do? What can we do mathematically? Actually, to either one of these equations, but we'll focus on this one right now. What can we do mathematically in order to essentially solve for the x? In order to figure out what that mystery mass actually is? And I'll give you another second or two to think about it. Well, when we did it the last time with just the scales we said, OK, we've got three of these x's here. We want to have just one x here. So we can say, whatever this x is, if the scale stays balanced, it's going to be the same as whatever we have there. There might be a temptation to subtract two of the x's maybe from this side, but that won't help us. And we can even see it mathematically over here. If we subtract two x's from both sides, on the left-hand side you're going to have 3x minus 2x. And on the right-hand side, you're going to have 9 minus 2x." + }, + { + "Q": "\nAt 0:57 why does Sal write down 6 with a 1 above it and say six squared-isn't it(the drawing of six squared) supposed to be a six with a 2(instead of a 1) above it?", + "A": "it is a 2, but as the quality is low, it looks like a squished up 2 or a 1.", + "video_name": "nMhJLn5ives", + "timestamps": [ + 57 + ], + "3min_transcript": "I promised you that I'd give you some more Pythagorean theorem problems, so I will now give you more Pythagorean theorem problems. And once again, this is all about practice. Let's say I had a triangle-- that's an ugly looking right triangle, let me draw another one --and if I were to tell you that that side is 7, the side is 6, and I want to figure out this side. Well, we learned in the last presentation: which of these sides is the hypotenuse? Well, here's the right angle, so the side opposite the right angle is the hypotenuse. So what we want to do is actually figure out the hypotenuse. So we know that 6 squared plus 7 squared is equal to the hypotenuse squared. hypotenuse, so we'll use C here as well. And 36 plus 49 is equal to C squared. 85 is equal to C squared. Or C is equal to the square root of 85. And this is the part that most people have trouble with, is actually simplifying the radical. So the square root of 85: can I factor 85 so it's a product of a perfect square and another number? 85 isn't divisible by 4. So it won't be divisible by 16 or any of the multiples of 4. 5 goes into 85 how many times? No, that's not perfect square, either. I don't think 85 can be factored further as a So you might correct me; I might be wrong. This might be good exercise for you to do later, but as far as I can tell we have gotten our answer. The answer here is the square root of 85. And if we actually wanted to estimate what that is, let's think about it: the square root of 81 is 9, and the square root of 100 is 10 , so it's some place in between 9 and 10, and it's probably a little bit closer to 9. So it's 9 point something, something, something. And that's a good reality check; that makes sense. If this side is 6, this side is 7, 9 point something, something, something makes sense for that length. Let me give you another problem. [DRAWING] Let's say that this is 10 . This is 3. What is this side? First, let's identify our hypotenuse. We have our right angle here, so the side opposite the right angle is the hypotenuse and it's also the longest side. So it's 10." + }, + { + "Q": "At 1:35 you said radical is 85 the radical?\n", + "A": "also i don t know about perfect squares but the square root of 85 can be simplified to 7 square root of 17", + "video_name": "nMhJLn5ives", + "timestamps": [ + 95 + ], + "3min_transcript": "I promised you that I'd give you some more Pythagorean theorem problems, so I will now give you more Pythagorean theorem problems. And once again, this is all about practice. Let's say I had a triangle-- that's an ugly looking right triangle, let me draw another one --and if I were to tell you that that side is 7, the side is 6, and I want to figure out this side. Well, we learned in the last presentation: which of these sides is the hypotenuse? Well, here's the right angle, so the side opposite the right angle is the hypotenuse. So what we want to do is actually figure out the hypotenuse. So we know that 6 squared plus 7 squared is equal to the hypotenuse squared. hypotenuse, so we'll use C here as well. And 36 plus 49 is equal to C squared. 85 is equal to C squared. Or C is equal to the square root of 85. And this is the part that most people have trouble with, is actually simplifying the radical. So the square root of 85: can I factor 85 so it's a product of a perfect square and another number? 85 isn't divisible by 4. So it won't be divisible by 16 or any of the multiples of 4. 5 goes into 85 how many times? No, that's not perfect square, either. I don't think 85 can be factored further as a So you might correct me; I might be wrong. This might be good exercise for you to do later, but as far as I can tell we have gotten our answer. The answer here is the square root of 85. And if we actually wanted to estimate what that is, let's think about it: the square root of 81 is 9, and the square root of 100 is 10 , so it's some place in between 9 and 10, and it's probably a little bit closer to 9. So it's 9 point something, something, something. And that's a good reality check; that makes sense. If this side is 6, this side is 7, 9 point something, something, something makes sense for that length. Let me give you another problem. [DRAWING] Let's say that this is 10 . This is 3. What is this side? First, let's identify our hypotenuse. We have our right angle here, so the side opposite the right angle is the hypotenuse and it's also the longest side. So it's 10." + }, + { + "Q": "At about 3:12, do you mean James Grime from Numberphile?\n", + "A": "Yes, that is exactly what Vi meant.", + "video_name": "lA6hE7NFIK0", + "timestamps": [ + 192 + ], + "3min_transcript": "are bigger than other infinities. This is metaphorically resonant and all but whether infinity really exists or if anything can last forever or whether a life contains infinite moments. Those aren't the kind of questions you can answer with math but if life does contain infinite moments, one for each real number time, that you can do math to. This time, we're not just going to do metaphors. We're going to prove it. Understanding different infinities starts with some really basic questions like is five bigger than four. You learned that it is but how do you know? Because this many is more than this many, they're both just one hand equal to each other except to fold it into slightly different shapes. Unless you're already abstracting out the idea of numbers and how you learn they're suppose to work just as you learned a long life is supposed to be somehow more than a short life rather than just a life equal to any other but folded into a different shape. Is five and six bigger than 12? Five and six is two things after all and twelve is just one thing and what about infinity? If I want to make up a number bigger than infinity, how would I know whether it really is bigger and not just the same infinity folded into a different shape? The way five plus five is just another shape for 10. One way to make a big number is to take a number of numbers, meta numbers. This is where a box containing five and six has two things and is actually bigger than a box with only the number 12. You could take the number of numbers from one to five and put them in a box and you'd have a box set of five or you could take the number of numbers that are five which is one or you could take the number of counting numbers or the number of real numbers. It's kind of funny that the number of counting numbers is not itself a counting number but an infinite number often referred to This size of infinity is usually called countable infinity because it's like counting infinitely but I like James Grime's way of calling it listable infinity because the usual counting numbers basically make an infinite list and many other numbers of numbers are also listable. You can put all positive whole numbers on an infinite list like this. You can put all whole numbers including negative ones by alternating. You can list all whole numbers along with all half way points between them. You can even list all the rational numbers by cleverly going through all possible combinations of one whole number divided by another whole number. All countably infinite numbers of things, all aleph null. Countable infinity is like saying if I make an infinite list of these things, I can list all the things. The weird thing is that it seems like this definition should be obvious that no matter how many things there are," + }, + { + "Q": "\n@ 3:05, she mentions a type of infinity called \"aleph null\". Could someone help me out because I understand that there are multiple infinities, but what is \"aleph null\"?", + "A": "Aleph null is the smallest infinite cardinal number. A cardinal number is the type of number we normally think of when counting things: three videos, six A s, one question, and aleph null integers. Aleph null is defined to be the number of natural numbers.", + "video_name": "lA6hE7NFIK0", + "timestamps": [ + 185 + ], + "3min_transcript": "are bigger than other infinities. This is metaphorically resonant and all but whether infinity really exists or if anything can last forever or whether a life contains infinite moments. Those aren't the kind of questions you can answer with math but if life does contain infinite moments, one for each real number time, that you can do math to. This time, we're not just going to do metaphors. We're going to prove it. Understanding different infinities starts with some really basic questions like is five bigger than four. You learned that it is but how do you know? Because this many is more than this many, they're both just one hand equal to each other except to fold it into slightly different shapes. Unless you're already abstracting out the idea of numbers and how you learn they're suppose to work just as you learned a long life is supposed to be somehow more than a short life rather than just a life equal to any other but folded into a different shape. Is five and six bigger than 12? Five and six is two things after all and twelve is just one thing and what about infinity? If I want to make up a number bigger than infinity, how would I know whether it really is bigger and not just the same infinity folded into a different shape? The way five plus five is just another shape for 10. One way to make a big number is to take a number of numbers, meta numbers. This is where a box containing five and six has two things and is actually bigger than a box with only the number 12. You could take the number of numbers from one to five and put them in a box and you'd have a box set of five or you could take the number of numbers that are five which is one or you could take the number of counting numbers or the number of real numbers. It's kind of funny that the number of counting numbers is not itself a counting number but an infinite number often referred to This size of infinity is usually called countable infinity because it's like counting infinitely but I like James Grime's way of calling it listable infinity because the usual counting numbers basically make an infinite list and many other numbers of numbers are also listable. You can put all positive whole numbers on an infinite list like this. You can put all whole numbers including negative ones by alternating. You can list all whole numbers along with all half way points between them. You can even list all the rational numbers by cleverly going through all possible combinations of one whole number divided by another whole number. All countably infinite numbers of things, all aleph null. Countable infinity is like saying if I make an infinite list of these things, I can list all the things. The weird thing is that it seems like this definition should be obvious that no matter how many things there are," + }, + { + "Q": "How come at 1:28 it sounds like her voice is echoing?\n", + "A": "beacuse it is she is in an open room", + "video_name": "lA6hE7NFIK0", + "timestamps": [ + 88 + ], + "3min_transcript": "Voiceover: Any sequence you can come up with, whatever pattern looks fun. All your favorite celebrities birthdays lead into end followed by random numbers, whatever. All of that plus every sequence you can't come up with, each of those are the decimal places of a badly named so called real number and any of those sequences with one random digit changed is another real number. That's the thing most people don't realize about the set of all real numbers. It includes every possible combination of digits extending infinitely among aleph null decimal places. There's no last digit. The number of digits is greater than any real number, any counting number which makes it an infinite number of digits. Just barely an infinite number of digits because it's only barely greater than any finite number but even though it's only the smallest possible infinity of digits. This infinity is still no joke. It's still big enough, that for example point nine repeating is exactly precisely one and not epsilon less. You don't get that kind of point nine repeating equals one action unless your infinity really is infinite. are bigger than other infinities. This is metaphorically resonant and all but whether infinity really exists or if anything can last forever or whether a life contains infinite moments. Those aren't the kind of questions you can answer with math but if life does contain infinite moments, one for each real number time, that you can do math to. This time, we're not just going to do metaphors. We're going to prove it. Understanding different infinities starts with some really basic questions like is five bigger than four. You learned that it is but how do you know? Because this many is more than this many, they're both just one hand equal to each other except to fold it into slightly different shapes. Unless you're already abstracting out the idea of numbers and how you learn they're suppose to work just as you learned a long life is supposed to be somehow more than a short life rather than just a life equal to any other but folded into a different shape. Is five and six bigger than 12? Five and six is two things after all and twelve is just one thing and what about infinity? If I want to make up a number bigger than infinity, how would I know whether it really is bigger and not just the same infinity folded into a different shape? The way five plus five is just another shape for 10. One way to make a big number is to take a number of numbers, meta numbers. This is where a box containing five and six has two things and is actually bigger than a box with only the number 12. You could take the number of numbers from one to five and put them in a box and you'd have a box set of five or you could take the number of numbers that are five which is one or you could take the number of counting numbers or the number of real numbers. It's kind of funny that the number of counting numbers is not itself a counting number but an infinite number often referred to" + }, + { + "Q": "At 3:11, what is listable infinity and who is Jason Grime?\n", + "A": "James Grime is a mathematician. He is often seen in several Numberphile videos, and his website is singingbanana.com. Listable infinity is his way of saying countable infinity, because the idea is that you can list the elements of a countable set, but you can t actually count them, because there are an infinite number.", + "video_name": "lA6hE7NFIK0", + "timestamps": [ + 191 + ], + "3min_transcript": "are bigger than other infinities. This is metaphorically resonant and all but whether infinity really exists or if anything can last forever or whether a life contains infinite moments. Those aren't the kind of questions you can answer with math but if life does contain infinite moments, one for each real number time, that you can do math to. This time, we're not just going to do metaphors. We're going to prove it. Understanding different infinities starts with some really basic questions like is five bigger than four. You learned that it is but how do you know? Because this many is more than this many, they're both just one hand equal to each other except to fold it into slightly different shapes. Unless you're already abstracting out the idea of numbers and how you learn they're suppose to work just as you learned a long life is supposed to be somehow more than a short life rather than just a life equal to any other but folded into a different shape. Is five and six bigger than 12? Five and six is two things after all and twelve is just one thing and what about infinity? If I want to make up a number bigger than infinity, how would I know whether it really is bigger and not just the same infinity folded into a different shape? The way five plus five is just another shape for 10. One way to make a big number is to take a number of numbers, meta numbers. This is where a box containing five and six has two things and is actually bigger than a box with only the number 12. You could take the number of numbers from one to five and put them in a box and you'd have a box set of five or you could take the number of numbers that are five which is one or you could take the number of counting numbers or the number of real numbers. It's kind of funny that the number of counting numbers is not itself a counting number but an infinite number often referred to This size of infinity is usually called countable infinity because it's like counting infinitely but I like James Grime's way of calling it listable infinity because the usual counting numbers basically make an infinite list and many other numbers of numbers are also listable. You can put all positive whole numbers on an infinite list like this. You can put all whole numbers including negative ones by alternating. You can list all whole numbers along with all half way points between them. You can even list all the rational numbers by cleverly going through all possible combinations of one whole number divided by another whole number. All countably infinite numbers of things, all aleph null. Countable infinity is like saying if I make an infinite list of these things, I can list all the things. The weird thing is that it seems like this definition should be obvious that no matter how many things there are," + }, + { + "Q": "\nWhat does the small hand represent in 13:11:4", + "A": "The small hand represents the hour; in this case it is pointing to nine.", + "video_name": "ftndEjAg6qs", + "timestamps": [ + 791 + ], + "3min_transcript": "" + }, + { + "Q": "What is the elapsed time in start time 2:30 endtime 6:30\n", + "A": "the answer is 4:00", + "video_name": "ftndEjAg6qs", + "timestamps": [ + 150, + 390 + ], + "3min_transcript": "We're asked, what time is it? So first, we want to look at the hour hand, which is the shorter hand, and see where it is pointing. So this right over here would have been 12 o'clock, 1 o'clock, 2 o'clock, 3 o'clock, 4 o'clock. And it looks like it's a little bit past 4 o'clock. So we are in the fourth hour. So the hour is 4. And then we have to think about the minutes. The minutes are the longer hand, and every one of these lines represent 5 minutes. We start here. This is 0 minutes past the hour, then 5 minutes past the hour, then 10 minutes past the hour. So the time is-- the minutes are 10, 10 minutes past the hour, and the hour is 4, or it's 4:10. Let's do a few more. What time is it? So first, we want to look at the hour hand. That's the shorter hand right over here. It's at-- let's see. This is 12, 1, 2, 3, 4, 5, 6, 7, 8, 9. It's just past 9. So it's still in the ninth hour. It hasn't gotten to the 10th hour yet. The ninth hour's from starting with 9 all the way until it's right almost before it gets to 10, and then it gets to the 10th hour. So the hour is 9, and then we want the minutes. Well, we can just count from 0 starting at the top of the clock. So 0, 5, 10, 15, 20, 25, 30. It's 9:30. And that also might make sense to you, because we know there are 60 minutes in an hour. And this is exactly halfway around the clock. And so half of 60 is 30. Let's do one more. What time is it? This is 12, 1-- actually, we can even count backwards. We can go 12, 11, 10. So right now we're in the 10th hour. The hour hand has passed 10, but it hasn't gotten to 11 yet. So we are in the 10th hour. So this would be 0, 5, 10, 15, 20 minutes past the hour. That's where the longer hand is pointing. It is 10:20." + }, + { + "Q": "\nat 2:57 how did you come up with x+y= 5?", + "A": "Looking at the scale to the right, you can see that one side of the scale has 5 blocks(of 1). The other side has a X block and a Y block. The scale is not tipped, therefore a X and a Y have the same weight/ equal 5. So x+y=5", + "video_name": "h9ZgZimXn2Q", + "timestamps": [ + 177 + ], + "3min_transcript": "So,it doesn't get too spread out. On the left hand side, I got 2X plus a mass of Y. That's the total mass. The total mass on the left hand side is 2X plus Y, the total mass on the right hand side is just 8. 1,2,3,4,5,6,7,8. It is equal to 8 And since we see that the scale is balance, this total mass must be equal to this total mass. So, we can write an equal sign there. Now my question to you is there anything we can do just based on the information that we have here to solve for either the mass X or for the mass Y. Is there anything that we can do. Well the simple answer is just with this information here, there's actually very little. You might say that \"Oh well, let me take the Y from both sides\" You might take this Y block up. But if you take this Y block up you have to take away Y from this side and you don't know what Y is. So, you're not gonna get rid of the Y. Same thing with the X's, you actually don't have enough information. Y depends on what X is,and X depends on what Y is. Lucky for us however,we do have some more of these blocks laying around. And what we do is we take one of these X blocks. And I stack it over here,and I also take one of the Y block and I stack it right over there. And then I keep adding all these ones until I balance these things out. So, I keep adding these ones. Obviously if I just place this, this will go down cause there's nothing on that side. But I keep adding these blocks until it all balances out and I find that my scale balances once I have 5 kg on the right hand side So, once again let me ask you this information having X and Y on the left hand side and a 5kg on the right hand side Well our total mass on the left hand side is X plus Y. And our total mass, let me right that once again a little bit closer to the center. It's X plus Y on the left hand side and the right hand side I have 5 kg. I have 5kg. I have 5 kg on the right hand side. And we know that's actually balance the scale. So these total masses must be equal to each other. And this information by itself, once again. There's nothing I can do with it. I don't know what X and Y. If Y is 4 maybe X is 1 or maybe X is 4, Y is 1. Who knows what these are. The interesting thing is we can actually use both of these information to figure out what X and Y actually is." + }, + { + "Q": "3:03 but how are you supposed to do this in the real world if you don't have a scale?\n", + "A": "x+y=5 is just a given, a tool for solving the original problem. Sal simply skips over explaining that fact. On a test or worksheet the problem might be presented as something like... Solve 2x+y=8 for x if x+y=5", + "video_name": "h9ZgZimXn2Q", + "timestamps": [ + 183 + ], + "3min_transcript": "So,it doesn't get too spread out. On the left hand side, I got 2X plus a mass of Y. That's the total mass. The total mass on the left hand side is 2X plus Y, the total mass on the right hand side is just 8. 1,2,3,4,5,6,7,8. It is equal to 8 And since we see that the scale is balance, this total mass must be equal to this total mass. So, we can write an equal sign there. Now my question to you is there anything we can do just based on the information that we have here to solve for either the mass X or for the mass Y. Is there anything that we can do. Well the simple answer is just with this information here, there's actually very little. You might say that \"Oh well, let me take the Y from both sides\" You might take this Y block up. But if you take this Y block up you have to take away Y from this side and you don't know what Y is. So, you're not gonna get rid of the Y. Same thing with the X's, you actually don't have enough information. Y depends on what X is,and X depends on what Y is. Lucky for us however,we do have some more of these blocks laying around. And what we do is we take one of these X blocks. And I stack it over here,and I also take one of the Y block and I stack it right over there. And then I keep adding all these ones until I balance these things out. So, I keep adding these ones. Obviously if I just place this, this will go down cause there's nothing on that side. But I keep adding these blocks until it all balances out and I find that my scale balances once I have 5 kg on the right hand side So, once again let me ask you this information having X and Y on the left hand side and a 5kg on the right hand side Well our total mass on the left hand side is X plus Y. And our total mass, let me right that once again a little bit closer to the center. It's X plus Y on the left hand side and the right hand side I have 5 kg. I have 5kg. I have 5 kg on the right hand side. And we know that's actually balance the scale. So these total masses must be equal to each other. And this information by itself, once again. There's nothing I can do with it. I don't know what X and Y. If Y is 4 maybe X is 1 or maybe X is 4, Y is 1. Who knows what these are. The interesting thing is we can actually use both of these information to figure out what X and Y actually is." + }, + { + "Q": "At about 3:00, where did he get the 5 from?\n", + "A": "On the left side of the scale, there is an x and a y cube. That s where he gets x + y from. On the right side of the scale, there s 5 yellow cubes. That s where he gets 5 from. x + y must equal 5, because that s how the scale is balanced.", + "video_name": "h9ZgZimXn2Q", + "timestamps": [ + 180 + ], + "3min_transcript": "So,it doesn't get too spread out. On the left hand side, I got 2X plus a mass of Y. That's the total mass. The total mass on the left hand side is 2X plus Y, the total mass on the right hand side is just 8. 1,2,3,4,5,6,7,8. It is equal to 8 And since we see that the scale is balance, this total mass must be equal to this total mass. So, we can write an equal sign there. Now my question to you is there anything we can do just based on the information that we have here to solve for either the mass X or for the mass Y. Is there anything that we can do. Well the simple answer is just with this information here, there's actually very little. You might say that \"Oh well, let me take the Y from both sides\" You might take this Y block up. But if you take this Y block up you have to take away Y from this side and you don't know what Y is. So, you're not gonna get rid of the Y. Same thing with the X's, you actually don't have enough information. Y depends on what X is,and X depends on what Y is. Lucky for us however,we do have some more of these blocks laying around. And what we do is we take one of these X blocks. And I stack it over here,and I also take one of the Y block and I stack it right over there. And then I keep adding all these ones until I balance these things out. So, I keep adding these ones. Obviously if I just place this, this will go down cause there's nothing on that side. But I keep adding these blocks until it all balances out and I find that my scale balances once I have 5 kg on the right hand side So, once again let me ask you this information having X and Y on the left hand side and a 5kg on the right hand side Well our total mass on the left hand side is X plus Y. And our total mass, let me right that once again a little bit closer to the center. It's X plus Y on the left hand side and the right hand side I have 5 kg. I have 5kg. I have 5 kg on the right hand side. And we know that's actually balance the scale. So these total masses must be equal to each other. And this information by itself, once again. There's nothing I can do with it. I don't know what X and Y. If Y is 4 maybe X is 1 or maybe X is 4, Y is 1. Who knows what these are. The interesting thing is we can actually use both of these information to figure out what X and Y actually is." + }, + { + "Q": "\nAt 1:40 why do you divide x^3-1 by x-1?\nWhy don't you divide it by x+1? Does it make a difference?", + "A": "Two reasons. First, factoring by (x + 1) doesn t get us anywhere, because our problem is how to find the limit at x = 1, which means we need to get rid of a factor of (x - 1), because that s the factor that gives us a zero in the denominator at x = 1. And second, (x + 1) isn t a factor of x^3 - 1. Try dividing it and you ll see that you get a remainder.", + "video_name": "rU222pVq520", + "timestamps": [ + 100 + ], + "3min_transcript": "Let's try to find the limit as x approaches 1 of x to the third minus 1 over x squared minus 1. And at first when you just try to substitute x equals 1, you get 0/0 1 minus 1 over 1 minus 1. So that doesn't help us. So let's see if we can try to simplify this in some way. So you might immediately recognize-- so let's rewrite this expression right over here so it's x to the third minus 1 over x squared minus 1. This on the bottom immediately jumps out as a difference of squares. So we know on the bottom that this could be factored as x minus 1 times x plus 1. And so if somehow this thing on the top also has an x minus 1 as a factor, then that x minus 1 will cancel with this, and then we're not going to have an issue of dividing by 0. The reason why I care about the x minus 1 term is that this is what's making our denominator equal 0. When you say x equals 1, you have 1 minus 1 times 1 plus 1. So if we can have an x minus 1 up here, then we can cancel these out for any x not equal to 1. And then we might have a much simpler thing to find the limit of. So let's think about whether x to the third minus 1 is the product of x minus 1 and something else. And to do that we can do a little bit of algebraic long division. Some of you guys might already recognize a pattern here, but we'll try to do-- well, let's divide x minus 1 into it to see whether it divides evenly into x to the third minus 1. So x minus 1-- we just look at the highest degree term-- x goes into x to the third x squared times. Goes x squared times. Actually, let me do it this way so that way we can keep track of the place. So this would be x-- this would be the second degree place, first degree place, and this would be the constant. So x to the third minus 1. x squared times x is x to the third. x squared times negative 1 is minus x squared. And now we're going to want to subtract this. So we are then left with x squared. x goes into x squared x times plus x. x times x is x squared. x times minus 1 is minus x. And once again we're going to subtract this. We'll swap the signs, negative and positive. And so these cancel out, and we're left with x. And then we bring down a minus 1. x minus 1 goes into x minus 1 exactly one time. 1 times x minus 1 is x minus 1. And then you subtract, and then you have no remainder. So this numerator right over here can be factored as x minus 1 times x squared plus x plus 1. And so we can say that this is the same exact thing." + }, + { + "Q": "\nat 5:43, do the answers to the \"b\" value always end up as weird fractions (i.e. in the video 13/3)?", + "A": "Not always. The y-intercept can be any real number.", + "video_name": "XMJ72mtMn4Y", + "timestamps": [ + 343 + ], + "3min_transcript": "y is equal to negative 5 thirds, that's our slope, x plus b. So we still need to solve for y-intercept to get our equation. And to do that, we can use the information that we know in fact we have several points of information We can use the fact that the line goes through the point (-1,6) you could use the other point as well. We know that when is equal to negative 1, So y is eqaul to 6. So y is equal to six when x is equal to negative 1 So negative 5 thirds times x, when x is equal to negative 1 y is equal to 6. So we literally just substitute this x and y value back into this and know we can solve for b. So let's see, this negative 1 times negative 5 thirds. So we have 6 is equal to positive five thirds plus b. And now we can subtract 5 thirds from both sides of this equation. From the left handside and subtracted from the rigth handside And then we get, what's 6 minus 5 thirds. So that's going to be, let me do it over here We take a common denominator. So 6 is the same thing as Let's do it over here. So 6 minus 5 over 3 is the same thing as 6 is the same thing as 18 over 3 minus 5 over 3 6 is 18 over 3. And this is just 13 over 3. And this is just 13 over 3. And then of course, these cancel out. So we get b is equal to 13 thirds. So we are done. We know We know the slope and we know the y-intercept. The equation of our line is y is equal to negative 5 thirds x plus our y-intercept which is 13 which is 13 over 3. And we can write these as mixed numbers. if it's easier to visualize. 13 over 3 is four and 1 thirds. this y-intercept right over here. That's 0 coma 13 over 3 or 0 coma 4 and 1 thirds. And even with my very roughly drawn diagram it those looks like this. And the slope negative 5 thirds that's the same thing as negative 1 and 2 thirds. You can see here the slope is downward because the slope is negative. It's a little bit steeper than a slope of 1. It's not quite a negative 2. It's negative 1 and 2 thirds. if you write this as a negative, as a mixed number. So, hopefully, you found that entertaining." + }, + { + "Q": "At 4:28, would the answer be the same if you used other set of points?\n", + "A": "I am not sure what you are asking because through any two points there is only one line. If you give different points, unless they are on the same line, then you would get a different line.", + "video_name": "XMJ72mtMn4Y", + "timestamps": [ + 268 + ], + "3min_transcript": "our starting point and make that our ending point. So what is our change in y? So our change in y, to go we started at y is equal to six, we started at y is equal to 6. And we go down all the way to y is equal to negative 4 So this is rigth here, that is our change in y You can look at the graph and say, oh, if I start at 6 and I go to negative 4 I went down 10. or if you just want to use this formula here it will give you the same thing We finished at negative 4, we finished at negative 4 and from that we want to subtract, we want to subtract 6. This right here is y2, our ending y and this is our beginning y This is y1. So y2, negative 4 minus y1, 6. or negative 4 minus 6. That is equal to negative 10. And all it does is tell us the change in y you go from this point to that point we have to go down 10. That's where the negative 10 comes from. Now we just have to find our change in x. So we can look at this graph over here. We started at x is equal to negative 1 and we go all the way to x is equal to 5. So we started at x is equal to negative 1, and we go all the way to x is equal to 5. So it takes us one to go to zero and then five more. So are change in x is 6. You can look at that visually there or you can use this formula same exact idea, our ending x-value, our ending x-value is 5 and our starting x-value is negative 1. 5 minus negative 1. 5 minus negative 1 is the same thing as 5 plus 1. So it is 6. So our slope here is negative 10 over 6. wich is the exact same thing as negative 5 thirds. as negative 5 over 3 I devided the numerator and the denominator by 2. y is equal to negative 5 thirds, that's our slope, x plus b. So we still need to solve for y-intercept to get our equation. And to do that, we can use the information that we know in fact we have several points of information We can use the fact that the line goes through the point (-1,6) you could use the other point as well. We know that when is equal to negative 1, So y is eqaul to 6. So y is equal to six when x is equal to negative 1 So negative 5 thirds times x, when x is equal to negative 1 y is equal to 6. So we literally just substitute this x and y value back into this and know we can solve for b. So let's see, this negative 1 times negative 5 thirds. So we have 6 is equal to positive five thirds plus b. And now we can subtract 5 thirds from both sides of this equation." + }, + { + "Q": "\nwhat did at sign that sal made mean? at 3:41?", + "A": "That the bisector was congruent to itself...i think..", + "video_name": "7UISwx2Mr4c", + "timestamps": [ + 221 + ], + "3min_transcript": "So the distance from B to D is going to be the same thing as the distance-- let me do a double slash here to show you it's not the same as that distance. So the distance from B to D is going to be the same thing as the distance from D to C. And obviously, between any two points, you have a midpoint. And so let me draw segment AD. And what's useful about that is that we have now constructed two triangles. And what's even cooler is that triangle ABD and triangle ACD, they have this side is congruent, this side is congruent, and they actually share this side right over here. So we know that triangle ABD we know that it is congruent to triangle ACD. You have two triangles that have three sides that are congruent, or they have the same length. Then the two triangles are congruent. And what's useful about that is if these two triangles are congruent, then their corresponding angles are congruent. And so we've actually now proved our result. Because the corresponding angle to ABC in this triangle is angle ACD in this triangle right over here. So that we then know that angle ABC is congruent to angle ACB. So that's a pretty neat result. If you have an isosceles triangle, a triangle where two of the sides are congruent, then their base angles, these base angles, are also going to be congruent. Now let's think about it the other way. Can we make the other statement? If the base angles are congruent, So let's try to construct a triangle and see if we can prove it the other way. So I'll do another triangle right over here. Let me draw another one just like that. That's not that pretty of a triangle, so let me draw it a little nicer. I'm going to draw it like this. Let me do that in a different color. So I'll call that A. I will call this B. I will call that C right over there. And now we're going to start off with the idea that this angle, angle ABC, is congruent to angle ACB. So they have the same exact measure. And what we want to do in this case-- we want to prove-- so let me draw a little line here to show that we're doing a different idea. Here we're saying if these two sides are the same, then the base angles are going to be the same. We've proved that. Now let's go the other way. If the base angles are the same, do we know that the two sides are the same? So we want to prove that segment AC is congruent to AB." + }, + { + "Q": "\nAt 0:57, Sal mentioned angles ABC and ACB. How do I know what order to put the letters in? For example, how do I know to write angle ABC instead of angle CBA?", + "A": "no, order matters. the middle letter has to be the vertex", + "video_name": "7UISwx2Mr4c", + "timestamps": [ + 57 + ], + "3min_transcript": "So we're starting off with triangle ABC here. And we see from the drawing that we already know that the length of AB is equal to the length of AC, or line segment AB is congruent to line segment AC. And since this is a triangle and two sides of this triangle are congruent, or they have the same length, we can say that this is an isosceles triangle. Isosceles triangle, one of the hardest words for me to spell. I think I got it right. And that just means that two of the sides are equal to each other. Now what I want to do in this video is show what I want to prove. So what I want to prove here is that these two-- and they're sometimes referred to as base angles, these angles that are between one of the sides, and the side that isn't necessarily equal to it, and the other side that is equal and the side that's not equal to it. I want to show that they're congruent. So I want to prove that angle ABC, I And so for an isosceles triangle, those two angles are often called base angles. And this might be called the vertex angle over here. And these are often called the sides or the legs of the isosceles triangle. And these are-- obviously they're sides. These are the legs of the isosceles triangle and this one down here, that isn't necessarily the same as the other two, you would call the base. So let's see if we can prove that. So there's not a lot of information here, just that these two sides are equal. But we have, in our toolkit, a lot that we know about triangle congruency. So maybe we can construct two triangles here that are congruent. And then we can use that information to figure out whether this angle is congruent to that angle there. And the first step, if we're going to use triangle congruency, is to actually construct two triangles. So one way to construct two triangles is let's set up another point right over here. Let's set up another point D. And let's just So the distance from B to D is going to be the same thing as the distance-- let me do a double slash here to show you it's not the same as that distance. So the distance from B to D is going to be the same thing as the distance from D to C. And obviously, between any two points, you have a midpoint. And so let me draw segment AD. And what's useful about that is that we have now constructed two triangles. And what's even cooler is that triangle ABD and triangle ACD, they have this side is congruent, this side is congruent, and they actually share this side right over here. So we know that triangle ABD we know that it is congruent to triangle ACD." + }, + { + "Q": "\nStarting around 2:40, I entered the same exact numbers into my calculator, 5*tan(65), and I got -7.350191288, instead of 10.7225346025. Any suggestions to what I could have done wrong?", + "A": "Your calculator is using radians, not degrees. You can probably change it in your calculator s settings (but I can t help you with that without knowing what calculator you have). You could could convert 65 degrees into 65*\u00cf\u0080/180 radians and enter 5*tan(65*\u00cf\u0080/180) to get the correct value.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 160 + ], + "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse." + }, + { + "Q": "\nAt 1:01 Sal says to use a trigonometric function. Since this is a right triangle, why can't you just use the Pythagorean Theorem?", + "A": "In order to solve for a side using the Pythagorean theorem, you would already have to know the lengths of 2 sides of a right triangle. In this case, you only know 1 length, so you must use the trig functions to solve for a side.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 61 + ], + "3min_transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7." + }, + { + "Q": "Wait a second.. what is the definition of the tangent of a number, what do you do with it? For example, at about 2:30, Mr. Khan finds the tangent of 65 degrees on his calculator. How does that work?\n", + "A": "If you have a right triangle with a 65 degree angle in it, then the tangent of 65\u00c2\u00ba is the ratio of the side opposite the 65\u00c2\u00ba angle to the side next to it.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 150 + ], + "3min_transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7." + }, + { + "Q": "At 2:50, Professor Khan used a TI-85 calculator to find the answer as 10.7. But, when I used my TI-84 calculator at home to find the answer, I got -7.4 (rounded to the nearest tenth). I also searched up on google \"5*tan 65\" and got -7.4 (rounded to the nearest tenth). Is there a reason for this?\n", + "A": "Make sure your calculator is set to use degrees. It is likely set to radians which would cause it to produce a different result.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 170 + ], + "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse." + }, + { + "Q": "\nat 2:50 he enters 5 times tangent of 65 into calculator and gets 10.72....\nwhen I enter the exact same thing.. I get -7.35....\nhave re-entered it over and over to ensure it's the same thing.\nsame result... yes it's the plain old \"tan\" function (not tan-1)", + "A": "Make sure you aren t in radian mode.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 170 + ], + "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse." + }, + { + "Q": "\nAt 5:54 , I believe Sal should have used the approximately equal sign (\u00e2\u0089\u0088) as opposed to the equal sign (=). Of course, it's a small difference so it could be my eyes, but did anyone else see it like I did?", + "A": "actually, he is correct because the problem clearly states to round to the nearest tenth. So his answer is exact when rounded to the nearest tenth.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 354 + ], + "3min_transcript": "And so what trigonometric ratios-- or we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that XY is exactly 5 and we don't have to deal with this approximation, let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well we see from SohCahToa cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees is equal to the length of the adjacent side, which is 5 over the length of the hypotenuse, which has a length of b. And then we can try to solve for b. You multiply both sides times b, you're left with b times cosine of 65 degrees is equal to 5. And then to solve for b, you could divide both sides by cosine of 65 degrees. So we're just dividing-- we have to figure it out what our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees. And we're left with b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is 5 divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that 5 squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out So I'll give you a few seconds to think about what the measure of angle W is. Well here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is-- well we can simplify the left-hand side right over here. 65 plus 90 is 155. So angle W plus 155 degrees is equal to 180 degrees. And then we get angle W-- if we subtract 155 from both sides-- angle W is equal to 25 degrees. And we are done solving the right triangle shown below." + }, + { + "Q": "Why is it, at 2:33, that the tangent was multiplied by 5?\n", + "A": "It was multiplied so the variable a would be isolated. It is just like in solving a regular algebraic equation- you isolate the variable first, then solve for it.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 153 + ], + "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse." + }, + { + "Q": "At 1:32, why didn't Sal use the Law of Sines?\n", + "A": "I don t know why. Using the Law of Sines would seem to make doing this problem quicker and easier. However, there are often several ways to do a problem, and maybe Sal just wanted to illustrate how to do this problem from basics (SOH CAH TOA).", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 92 + ], + "3min_transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7." + }, + { + "Q": "\nat 1:30 does it matter if I use I use sin,cos,tan? in which situation would I use each one? If it does NOT matter, can I be sure that it's the same way with every other trig problem?", + "A": "In this situation, we know the length of the adjacent side, we know the angle is 65 degrees, and we want to find the length of the opposite side. It matters very much which trig function we use in every trig problem. Here, the tan of 65 degrees = opposite / 5, so we can solve for the length of the opposite side with opposite = 5*tan 65 degrees. Which trig functions we use depends on which parts of the triangle we know.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 90 + ], + "3min_transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7." + }, + { + "Q": "(at 2:38) Hi, can anyone please tell me how to plug your equation eg. 5*tan 65 into the calculator? I seem to have trouble doing it on the Khan Academy virtual calculator on the practice exercise.\n", + "A": "First, you make sure the calculator is in degree mode. Then, you hit the buttons 5 , x , tan , 6 , 5 , ) , and = and you should get the answer.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 158 + ], + "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse." + }, + { + "Q": "\nAt 5:30 you used the calculator to do 5/ (Cos 65) my calculator gets 0.422 when I do that and it is set in degrees.", + "A": "Cos(x) is always between -1 and 1 so you should never get an absolute value below 5.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 330 + ], + "3min_transcript": "This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse. And so what trigonometric ratios-- or we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that XY is exactly 5 and we don't have to deal with this approximation, let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well we see from SohCahToa cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees is equal to the length of the adjacent side, which is 5 over the length of the hypotenuse, which has a length of b. And then we can try to solve for b. You multiply both sides times b, you're left with b times cosine of 65 degrees is equal to 5. And then to solve for b, you could divide both sides by cosine of 65 degrees. So we're just dividing-- we have to figure it out what our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees. And we're left with b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is 5 divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that 5 squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out" + }, + { + "Q": "During 1:34 he uses tangents but insted could you have figured out the other angle and then use law of sines\n", + "A": "He could have done that, but this lesson is using the basic trig functions. Additionally, using the Law of Sines would actually be slower.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 94 + ], + "3min_transcript": "We're asked to solve the right triangle shown below. Give the links to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7." + }, + { + "Q": "1:22 I dont understand how did sal said difference is -2 please help\n", + "A": "Sal didnt mean -2, he meant -1. He said this because the whole number intervals on the graph are every other square and therefore look like its -2 but it is -1. It is negative because the slope goes down 1 unit or looking from the other way it goes left 1 unit.", + "video_name": "hoRISaqp1Po", + "timestamps": [ + 82 + ], + "3min_transcript": "With the graph of the function f as an aid, evaluate the following limits. So the first one is the limit as x approaches 3 of f of x minus f of 3 over x minus 3. So let's think about x minus-- x equals 3 is right over here. This right over here is f of 3, or we could say f of 3 is 1 right over here. That's the point 3 comma f of 3. And they're essentially trying to find the slope between an arbitrary x and that point as that x gets closer and closer to 3. So we can imagine an x that is above 3, that is, say, right over here. Well, if we're trying to find the slope between this x comma f of x and 3 comma f of 3, we see that it gets this exact same form. Your end point is f of x. So it's f of x minus f of 3 is your change in the vertical axis. That's this distance right over here. And we would divide by your change And that's going to be x minus 3. So that's the exact expression that we have up here when I picked this as an arbitrary x. And we see that that slope, just by looking at the line between those two intervals, seems to be negative 2. And the slope was the same thing if we go the other side. If x was less than 3, then we also would have a slope of negative 2. Either way, we have a slope of negative 2. And that's important because this limit is just the limit as x approaches 3. So it can be as x approaches 3 from the positive direction or from the negative direction. But in either case, the slope, as we get closer and closer to this point right over here, is negative 2. Now let's think about what they're asking us here. So we have 8, f of 8. So let's think. We have 8. This is 8 comma f of 8. So that's 8 comma f of 8 right over there. So our temptation might be to say, hey, 8 plus h is going to be someplace out here. It's going to be something larger than 8. But notice, they have the limit as h approaches 0 from the negative direction. So approaching 0 from the negative direction means you're coming to 0 from below. You're at negative 1, negative 0.5, negative 0.1, negative 0.0001. So h is actually going to be a negative number. So 8 plus h would actually be-- we could just pick an arbitrary point. It could be something like this right over here. So this might be the value of 8 plus h. And this would be the value of f of 8 plus h. So once again, they're finding-- or this expression is the slope between these two points. And then we are taking the limit as h approaches 0 from the negative direction. So as h gets closer and closer to 0," + }, + { + "Q": "\nat 3:30 Sal says the slope for the limit from the positive direction of 8 is infinite. how does he know that?\nI thought the slop was -1.\ncan someone please help me?", + "A": "You re correct that the slope on the positive side of 8 is clearly -1. I m not sure if Sal worded that 100% correct. I would say the limit as h approaches 0+ would be -1. The limit at 8 does not exist because limit as h ---> 0- does not equal h---> 0+ So the infinite slope he is drawing applies only AT 8.", + "video_name": "hoRISaqp1Po", + "timestamps": [ + 210 + ], + "3min_transcript": "And that's going to be x minus 3. So that's the exact expression that we have up here when I picked this as an arbitrary x. And we see that that slope, just by looking at the line between those two intervals, seems to be negative 2. And the slope was the same thing if we go the other side. If x was less than 3, then we also would have a slope of negative 2. Either way, we have a slope of negative 2. And that's important because this limit is just the limit as x approaches 3. So it can be as x approaches 3 from the positive direction or from the negative direction. But in either case, the slope, as we get closer and closer to this point right over here, is negative 2. Now let's think about what they're asking us here. So we have 8, f of 8. So let's think. We have 8. This is 8 comma f of 8. So that's 8 comma f of 8 right over there. So our temptation might be to say, hey, 8 plus h is going to be someplace out here. It's going to be something larger than 8. But notice, they have the limit as h approaches 0 from the negative direction. So approaching 0 from the negative direction means you're coming to 0 from below. You're at negative 1, negative 0.5, negative 0.1, negative 0.0001. So h is actually going to be a negative number. So 8 plus h would actually be-- we could just pick an arbitrary point. It could be something like this right over here. So this might be the value of 8 plus h. And this would be the value of f of 8 plus h. So once again, they're finding-- or this expression is the slope between these two points. And then we are taking the limit as h approaches 0 from the negative direction. So as h gets closer and closer to 0, And these points move closer and closer and closer together. So this is really just an expression of the slope of the line, roughly-- and we see that it's constant. So what's the slope of the line over this interval? Well, you can just eyeball it and see, well, look. Every time x changes by 1, our f of x changes by 1. So the slope of the line there is 1. It would have been a completely different thing if this said limit as h approaches 0 from the positive direction. Then we would be looking at points over here. And we would see that we would slowly approach, essentially, a vertical slope, kind of an infinite slope." + }, + { + "Q": "\nA the time of the video, 1:13, it tells you the inverse of the slope. What i was wondering is since you have the coordinates and the inverse slope, why don't you put in into point slope form to satisfy the question? Since the equation is looking for the equation of line B, why cant you do it this way? is it still wrong to do so? Hopefully this has made sense... :)", + "A": "since the question does not say the answer has to be in slope-intercept form - then i guess that yes, simply putting the equation in point-slope form would be a correct answer", + "video_name": "TsEhZRT16LU", + "timestamps": [ + 73 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 2:20 he said 9 and 9 when it was 7 and 9", + "A": "yes, but it was actually at 2:14 hey, nobody s perfect, anyone can make a mistake", + "video_name": "LEFE1km5ROY", + "timestamps": [ + 140 + ], + "3min_transcript": "A statistician for a basketball team tracked the number of points that each of the 12 players on the team had in one game. And then made a stem-and-leaf plot to show the data. And sometimes it's called a stem-plot. How many points did the team score? And when you first look at this plot right over here, it seems a little hard to understand. Understand we have 0, 1, 2 under leaf you have all of these digits here. How does this relate to the number of points each student, or each player, actually scored? And the way to interpret a stem-and-leaf plot is the leafs contain-- at least the way that this statistician used it-- the leaf contains the smallest digit, or the ones digit, in the number of points that each player scored. And the stem contains the tens digits. And usually the leaf will contain the rightmost digit, or the ones digit, and then the stem will contain all of the other digits. And what's useful about this is it gives kind of a distribution of where the players were. You see that most of the players scored points that started with a 0. And then only one score scored points to started with a 2, and it was actually 20 points. So I'm going to actually write down all of this data in a way that maybe you're a little bit more used to understanding it. So I'm going to write the 0's in purple. So there's, let's see, 1, 2, 3, 4, 5, 6, 7 players had 0 as the first digit. So 1, 2, 3, 4, 5, 6, 7. I wrote seven 0's. And then this player also had a 0 in his ones digit. This player, I'm going to try to do all the colors, this player also had a 0 in his ones digit. This player right here had a 2 in his ones digit, so he scored a total of 2 points. This player, let me do orange, this player had 4 for his ones digit. This player had 7 for his ones digit. Then this player had 7 for his ones digit. this player had 9 for his ones digit. So the way to read this is, you had one player with 0 points. 0, 2, 4, 7, 9 and 9. But you can see, and it's kind of silly saying the zero was a tens digit, you could have even put a blank there. But the 0 lets us know that they didn't score anything in the tens place. But these are the actual scores for those seven players. Now let's go to the next row in the stem-and-leaf plot. So over here, all of the digits start with, or all of the points start with 1, for each of the players. And there's four of them. So 1, 1, 1, and 1. And then we have this player over here, his ones digit, or her ones digit, is a 1. So this player, this represents 11. 1 in the tens place, 1 in the ones place. This player over here also got 11. 1 in the tens place, 1 in the ones place." + }, + { + "Q": "on 1:50 does he keep adding them all together\n", + "A": "Sal is not adding them all together. What he is doing is using place value to determine what the tens and ones places should be. Re-watch the video from 0:32.", + "video_name": "LEFE1km5ROY", + "timestamps": [ + 110 + ], + "3min_transcript": "A statistician for a basketball team tracked the number of points that each of the 12 players on the team had in one game. And then made a stem-and-leaf plot to show the data. And sometimes it's called a stem-plot. How many points did the team score? And when you first look at this plot right over here, it seems a little hard to understand. Understand we have 0, 1, 2 under leaf you have all of these digits here. How does this relate to the number of points each student, or each player, actually scored? And the way to interpret a stem-and-leaf plot is the leafs contain-- at least the way that this statistician used it-- the leaf contains the smallest digit, or the ones digit, in the number of points that each player scored. And the stem contains the tens digits. And usually the leaf will contain the rightmost digit, or the ones digit, and then the stem will contain all of the other digits. And what's useful about this is it gives kind of a distribution of where the players were. You see that most of the players scored points that started with a 0. And then only one score scored points to started with a 2, and it was actually 20 points. So I'm going to actually write down all of this data in a way that maybe you're a little bit more used to understanding it. So I'm going to write the 0's in purple. So there's, let's see, 1, 2, 3, 4, 5, 6, 7 players had 0 as the first digit. So 1, 2, 3, 4, 5, 6, 7. I wrote seven 0's. And then this player also had a 0 in his ones digit. This player, I'm going to try to do all the colors, this player also had a 0 in his ones digit. This player right here had a 2 in his ones digit, so he scored a total of 2 points. This player, let me do orange, this player had 4 for his ones digit. This player had 7 for his ones digit. Then this player had 7 for his ones digit. this player had 9 for his ones digit. So the way to read this is, you had one player with 0 points. 0, 2, 4, 7, 9 and 9. But you can see, and it's kind of silly saying the zero was a tens digit, you could have even put a blank there. But the 0 lets us know that they didn't score anything in the tens place. But these are the actual scores for those seven players. Now let's go to the next row in the stem-and-leaf plot. So over here, all of the digits start with, or all of the points start with 1, for each of the players. And there's four of them. So 1, 1, 1, and 1. And then we have this player over here, his ones digit, or her ones digit, is a 1. So this player, this represents 11. 1 in the tens place, 1 in the ones place. This player over here also got 11. 1 in the tens place, 1 in the ones place." + }, + { + "Q": "2:14 he said 9 twice\n", + "A": "why ask me the same thing in a different way?", + "video_name": "LEFE1km5ROY", + "timestamps": [ + 134 + ], + "3min_transcript": "A statistician for a basketball team tracked the number of points that each of the 12 players on the team had in one game. And then made a stem-and-leaf plot to show the data. And sometimes it's called a stem-plot. How many points did the team score? And when you first look at this plot right over here, it seems a little hard to understand. Understand we have 0, 1, 2 under leaf you have all of these digits here. How does this relate to the number of points each student, or each player, actually scored? And the way to interpret a stem-and-leaf plot is the leafs contain-- at least the way that this statistician used it-- the leaf contains the smallest digit, or the ones digit, in the number of points that each player scored. And the stem contains the tens digits. And usually the leaf will contain the rightmost digit, or the ones digit, and then the stem will contain all of the other digits. And what's useful about this is it gives kind of a distribution of where the players were. You see that most of the players scored points that started with a 0. And then only one score scored points to started with a 2, and it was actually 20 points. So I'm going to actually write down all of this data in a way that maybe you're a little bit more used to understanding it. So I'm going to write the 0's in purple. So there's, let's see, 1, 2, 3, 4, 5, 6, 7 players had 0 as the first digit. So 1, 2, 3, 4, 5, 6, 7. I wrote seven 0's. And then this player also had a 0 in his ones digit. This player, I'm going to try to do all the colors, this player also had a 0 in his ones digit. This player right here had a 2 in his ones digit, so he scored a total of 2 points. This player, let me do orange, this player had 4 for his ones digit. This player had 7 for his ones digit. Then this player had 7 for his ones digit. this player had 9 for his ones digit. So the way to read this is, you had one player with 0 points. 0, 2, 4, 7, 9 and 9. But you can see, and it's kind of silly saying the zero was a tens digit, you could have even put a blank there. But the 0 lets us know that they didn't score anything in the tens place. But these are the actual scores for those seven players. Now let's go to the next row in the stem-and-leaf plot. So over here, all of the digits start with, or all of the points start with 1, for each of the players. And there's four of them. So 1, 1, 1, and 1. And then we have this player over here, his ones digit, or her ones digit, is a 1. So this player, this represents 11. 1 in the tens place, 1 in the ones place. This player over here also got 11. 1 in the tens place, 1 in the ones place." + }, + { + "Q": "\nAt 5:00, the whole equation is written and he is describing the point-slope form. In the original equation at 4:11, is it y - y1 and x - x1 or y - y2 and x - x2?", + "A": "Either point (x1, y1) or (x2, y2) (or any known point on the line) will work, because the slope between any 2 points on the line is the same.", + "video_name": "LtpXvUCrgrM", + "timestamps": [ + 300, + 251 + ], + "3min_transcript": "Again to get to this point? Well, your change in X is positive two. So your change in X is equal to two. And so what's your slope? Change in Y over change in X. Negative eight over two is equal to negative four. So now that we have a, now that we know the slope and we know a point, we know a, we actually know two points on the line, we can express this in point-slope form. And so let's do that. And the way I like to it is I always like to just take it straight from the definition of what slope is. We know that the slope between any two points on this line is going to be negative four. So if we take an arbitrary Y that sits on this line and if we find the difference between that Y and, let's focus on this point up here. So if we find the difference between that Y and this Y, and nine, and it's over the difference between This is going to be the slope between any XY on this line and this point right over here. And the slope between any two points on a line are going to have to be constant. So this is going to be equal to the slope of the line. It's going to be equal to negative four. And we're not in point-slope form or classic point-slope form just yet. To do that, we just multiply both sides times X minus four. So we get Y minus 9, we get Y minus nine is equal to our slope, negative four times X minus four. Time X minus four. And this right over here is our classic, this right over here is our classic point-slope form. We have the point, sometimes they even put parenthesis like this, but we could figure out the point from this point-slope form. The point that sits on this line with things So it would be X equals four, Y equals nine, which we have right up there, and then the slope is right over here, it's negative four. Now from this can we now express this linear equation in y-intercept form? And y-intercept form, just as a bit of a reminder, it's Y is equal to MX plus B. Where this coefficient is our slope and this constant right over here allows us to figure out our y-intercept. And to get this in this form we just have to simplify a little bit of this algebra. So you have Y minus nine. Y minus nine is equal to, well let's distribute this negative four. And I'll just switch some colors. Let's distribute this negative four. Negative four times X is negative four X. Negative four time negative four is plus 16. And now, if we just want to isolate the Y on the left hand side, we can add nine to both sides." + }, + { + "Q": "At 6:17 and 6:20, Sal says \"-a finite number of values,\". What does finite mean?\n", + "A": "It means that the values are countable and not infinite. The numbers 1 to 10 are finite. Or 1 to one billion are also finite. Contrast that with an infinite number of values where any number up to infinity is allowed.", + "video_name": "dOr0NKyD31Q", + "timestamps": [ + 377, + 380 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 08:32, does the fact that there will only be one value for Z make a difference for our discrete random variable?\n\nI ask because Y will have multiple values for the multiple students. I guess I can answer my own question because if the class contains one student then Y is equivalent to Z in that the random variable can only take on one value.", + "A": "A discrete random variable can have more than one value but the number of values it can be is limited. In the ant example the answer can be one of many whole numbers over a range or interval. If our interval is 0 to 1,000,000 the answer could be an integer in that range (say 52). However, there there cannot be 52.6583 ants born in the universe. Either an ant is born or it isn t. To be a continuous random variable it has be able to take on ANY value in the interval.", + "video_name": "dOr0NKyD31Q", + "timestamps": [ + 512 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:22, why does the Khan Academy guy say pre-algebra is necessary? I'm in Grade 11 summer school and I can barely pass it btw I'm 16 years old and I feel dumb :(.\n", + "A": "You are not dumb. Sal believe s in the growth mindset, a belief that no one is dumb.", + "video_name": "Zm0KaIw-35k", + "timestamps": [ + 142 + ], + "3min_transcript": "Jayda takes 3 hours to deliver 189 newspapers on her paper route. What is the rate per hour at which she delivers the newspaper? So this first sentence tells us that she delivers, or she takes, 3 hours to deliver 189 newspapers. So you have 3 hours for every 189 newspapers. That's what the first sentence told us. But we want to figure out the rate per hour, or the newspapers per hour, so we can really just flip this rate So if we were to just flip it, we would have 189 newspapers for every 3 hours, which is really the same information. We're just flipping what's in the numerator and what's in Now we want to write it in as simple as possible form, and let's see if this top number is divisible by 3. 1 plus 8 is 9, plus 9 is 18. So that is divisible by 3. So let's divide this numerator and this denominator by 3 to simplify things. So if you divide 189 by 3. Let's do it over on the side right here. 3 goes into 189. 3 goes and 18 six times. 6 times 3 is 18. Subtract. Bring down the 9. 18 minus 18 was nothing. 3 goes into 9 exactly three times. 3 times 3 is 9, no remainder. So if you divide 189 by 3, you get 63, and if you divide 3 by 3, you're going to get 1. You have to divide both the numerator and the denominator by the same number. So now we have 63 newspapers for every 1 hour. Or we could write this as 63 over 1 newspapers per hour. thing as 63 newspapers per hour." + }, + { + "Q": "\nHow long can pi go for? 30:00", + "A": "are you sure?", + "video_name": "Zm0KaIw-35k", + "timestamps": [ + 1800 + ], + "3min_transcript": "" + }, + { + "Q": "\nfigured out the answer at 0:30", + "A": "that is 1:63 newspapers per hour...... ;p", + "video_name": "Zm0KaIw-35k", + "timestamps": [ + 30 + ], + "3min_transcript": "Jayda takes 3 hours to deliver 189 newspapers on her paper route. What is the rate per hour at which she delivers the newspaper? So this first sentence tells us that she delivers, or she takes, 3 hours to deliver 189 newspapers. So you have 3 hours for every 189 newspapers. That's what the first sentence told us. But we want to figure out the rate per hour, or the newspapers per hour, so we can really just flip this rate So if we were to just flip it, we would have 189 newspapers for every 3 hours, which is really the same information. We're just flipping what's in the numerator and what's in Now we want to write it in as simple as possible form, and let's see if this top number is divisible by 3. 1 plus 8 is 9, plus 9 is 18. So that is divisible by 3. So let's divide this numerator and this denominator by 3 to simplify things. So if you divide 189 by 3. Let's do it over on the side right here. 3 goes into 189. 3 goes and 18 six times. 6 times 3 is 18. Subtract. Bring down the 9. 18 minus 18 was nothing. 3 goes into 9 exactly three times. 3 times 3 is 9, no remainder. So if you divide 189 by 3, you get 63, and if you divide 3 by 3, you're going to get 1. You have to divide both the numerator and the denominator by the same number. So now we have 63 newspapers for every 1 hour. Or we could write this as 63 over 1 newspapers per hour. thing as 63 newspapers per hour." + }, + { + "Q": "\nAt 0:21, how did you conclude that the sample standard deviation was 2.98? I understood the method you used to arrive at 17.17 for the sample mean, but in in order to find the t-statistic I must know how we got \"S\".", + "A": "The s is the sample s standard deviation.To find this take each data point subtract it by the mean(17.17) and square it. Then add it all together. finally divide that number by 10 and take the square root of that number. If you put into the calculator it would look like this: [ (15.6 - 17.17)^2 + (16.2 - 17.17)^2+.... ]/10 <-then take the square root of it and it will = 2.98.", + "video_name": "D2sMsmL0ScQ", + "timestamps": [ + 21 + ], + "3min_transcript": "The mean emission of all engines of a new design needs to be below 20 parts per million if the design is to meet new emission requirements. 10 engines are manufactured for testing purposes, and the emission level of each is determined. The emission data is, and they give us 10 data points for the 10 test engines, and I went ahead and calculated the mean of these data points. The sample mean of 17.17. And the standard deviation of these 10 data points right here is 2.98, the sample standard deviation. Does the data supply sufficient evidence to conclude that this type of engine meets the new standard? Assume we are willing to risk a type-1 error with a probability of 0.01. And we'll touch on this in a second. Before we do that, let's just define what our null hypothesis and our alternative hypothesis are going to be. Our null hypothesis can be that we don't meet the standards. That we just barely don't meet the standards. That the mean of our new engines is exactly 20 parts per million. And you essentially want the best possible value where we we still don't meet the standard. And then our alternative hypothesis says no, we do meet That the true mean for our new engines is below 20 parts per million. And to see if the data that we have is sufficient, what we're going to do is assume, we're going to assume that this is true. And given that this is true, if we assume this is true, and the probability of this occurring, and the probability of getting a sample mean of that is less than 1%, then we will reject the null hypothesis. So we are going to reject our null hypothesis if the probability of getting a sample mean of 17.17 given the null hypothesis is true, is less than 1%. 1% chance that we are making a type-1 error. A type-1 error is that we're rejecting it even though it's true. Here there's only a 1% chance, or less than a 1% chance that we will reject it if it is true. Now the next thing we have to think about is what type of distribution we should think about. And I guess the first thing that rings in my brain is we only have 10 samples here. We only have 10 samples. We have a small sample size right over here. So we're going to be dealing with a T-distribution and T-statistic. So with that said, so let's think of it this way. We can come up with a T-statistic that is based on these statistics right over here. So the T-statistic is going to be 17.17, our sample mean, minus the assumed population mean-- minus 20 parts per million over our sample standard deviation, 2.98--" + }, + { + "Q": "At 4:24, how are we able to bring the 5 to the outside of the integral?\n", + "A": "That is a basic property of limits and an integral is a limit problem. You can factor out any constant that is convenient to factor out, though you don t have to.", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 264 + ], + "3min_transcript": "is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1" + }, + { + "Q": "At around 5:10, Sal applies the power rule to x^-1, making it -x^-1. Should it not actually be -x^-2, since you are supposed to subtract 1 from the exponent. Also, would the expression not turn positive, since there is a negative numerator and a negative denominator? Thanks for any help.\n", + "A": "The power rule for integrals is the reverse of the power rule for derivatives, so you add 1 to the exponent, you don t subtract. There is no negative coefficient in the numerator, so that should be negative. The power rule for integrals, then is: \u00e2\u0088\u00ab u \u00e2\u0081\u00bf du = [ u\u00e2\u0081\u00bf\u00e2\u0081\u00ba\u00c2\u00b9 / (n+1) ] + C", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 310 + ], + "3min_transcript": "So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1 And then if we want, we can distribute the 5. So this is equal to negative 5x to the negative 1. Now, we could write plus 5 times some constant, but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could [INAUDIBLE] this. If you want it to show that it's a different constant, you could say this is c1, c1, c1. You multiply 5 times c1, you get another constant. We could just call that c, which is equal to 5 times c1. But there you have it. Negative 5x to the negative 1 plus c. And once again, all of these, try to evaluate the derivative, and you will see that you get this business, right over there." + }, + { + "Q": "\nwhat if at 4:35 i do not take 5 out of the antiderivative operator and then just operate it simply, it gives me a result -5x^(-1) +C . can anyone tell me why is this wrong?", + "A": "Why would you think it is wrong. You got the same result as Sal by a slightly different method. Whether or not to factor the coefficient outside the integral sign is purely stylistic. I think it makes for less clutter, and anything that reduces the clutter of integration is welcome.", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 275 + ], + "3min_transcript": "So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1 And then if we want, we can distribute the 5. So this is equal to negative 5x to the negative 1. Now, we could write plus 5 times some constant, but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could [INAUDIBLE] this. If you want it to show that it's a different constant, you could say this is c1, c1, c1. You multiply 5 times c1, you get another constant. We could just call that c, which is equal to 5 times c1. But there you have it. Negative 5x to the negative 1 plus c. And once again, all of these, try to evaluate the derivative, and you will see that you get this business, right over there." + }, + { + "Q": "\nAt 2:52, why is n not allowed to be equal to -1?", + "A": "Because then you would be dividing by zero.", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 172 + ], + "3min_transcript": "And then we have plus c. The derivative of a constant with respect to x-- a constant does not change as x changes, so it is just going to be 0, so plus 0. And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing-- and this is a very general terms-- is equal to x to the n. So given that, what is the antiderivative-- let me switch colors here. What is the antiderivative of x to the n? And remember, this is just the kind of strange-looking notation we use. It'll make more sense when we start doing definite integrals. But what is the antiderivative of x to the n? And we could say the antiderivative with respect to x, if we want to. And another way of calling this is the indefinite integral. is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx." + }, + { + "Q": "At 3:49, to check to see if the derivative of (x\u00e2\u0081\u00b6/6)+C is in fact x\u00e2\u0081\u00b5, why don't you use to quotient rule? Shouldn't the derivative be 6(x\u00e2\u0081\u00b6/6) * [(6*6x\u00e2\u0081\u00b5- 0*x\u00e2\u0081\u00b6)/ 6\u00c2\u00b2]? Which equals x^11, not x\u00e2\u0081\u00b5.\n", + "A": "There is no reason to use the quotient method here, but the rules will always work is used correctly (unless you end up with an undefined answer) The problem with your math is that when you multiply x\u00e2\u0081\u00b6 times zero that term goes away. So you end up with 36x\u00e2\u0081\u00b5/36", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 229 + ], + "3min_transcript": "is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1" + }, + { + "Q": "\nat 1:15, why did you not apply the derivative rule for division? It is supposed to be (f'(x)g(x) - f(x)g'(x))/(g(x))^2", + "A": "keep in mind n is a constant, so n+1 is a constant f(x)=((x^(n+1))/(n+1) by quotient rule: f (x)=(x^(n)*(n+1))-(0*x^(n+1)) f (x)=((n+1)(x^n))-0 f (x)=(n+1)(x^(n) same, see?", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 75 + ], + "3min_transcript": "Let's take the derivative with respect to x of x to the n plus 1-th power over n plus 1 plus some constant c. And we're going to assume here, because we want this expression to be defined, we're going to assume that n does not equal negative 1. If it equaled negative 1, we'd be dividing by 0, and we haven't defined what that means. So let's take the derivative here. So this is going to be equal to-- well, the derivative of x to the n plus 1 over n plus 1, we can just use the power rule over here. So our exponent is n plus 1. We can bring it out front. So it's going to be n plus 1 times x to the-- I want to use that same color. Colors are the hard part-- times x to the-- instead of n plus 1, we subtract 1 from the exponent. This is just the power rule. So n plus 1 minus 1 is going to be n. And then we can't forget that we were dividing by this n plus 1. And then we have plus c. The derivative of a constant with respect to x-- a constant does not change as x changes, so it is just going to be 0, so plus 0. And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing-- and this is a very general terms-- is equal to x to the n. So given that, what is the antiderivative-- let me switch colors here. What is the antiderivative of x to the n? And remember, this is just the kind of strange-looking notation we use. It'll make more sense when we start doing definite integrals. But what is the antiderivative of x to the n? And we could say the antiderivative with respect to x, if we want to. And another way of calling this is the indefinite integral. is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power" + }, + { + "Q": "So at around 5:18, if you ended up with -5x^-1 + 5c, would that still be correct?\n", + "A": "Yes, and I think Sal mentions this around 5:15 in the video. However, since 5 times a constant C is just another constant, we get in the habit of writing a different variable name (C_1) instead of 5C.", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 318 + ], + "3min_transcript": "So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1 And then if we want, we can distribute the 5. So this is equal to negative 5x to the negative 1. Now, we could write plus 5 times some constant, but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could [INAUDIBLE] this. If you want it to show that it's a different constant, you could say this is c1, c1, c1. You multiply 5 times c1, you get another constant. We could just call that c, which is equal to 5 times c1. But there you have it. Negative 5x to the negative 1 plus c. And once again, all of these, try to evaluate the derivative, and you will see that you get this business, right over there." + }, + { + "Q": "\nWhen he explained about the CF is equal to 9, i did not understand that well. I did not understand how CF is equal to 9 at 6:31. Please help me as my english isn't good", + "A": "CF corresponds to AB (which has a length of 9) on a similar triangle. Therefore, there is a proportional relationship between the two. CF is actually equal to 36/7, not nine, but it is proportional to nine.", + "video_name": "7aGEvpHaNJ8", + "timestamps": [ + 391 + ], + "3min_transcript": "minus x-- that's that side right there-- and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is y. So it's equal to y minus x over y. So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to x-- let me do that in a different color. It's going to be equal to x over this entire base right over here, so this entire BE, which once again, we know is y. So over y. And now this looks interesting, because it looks like we have three unknowns. Sorry, we know what DE is already. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between x and y. So we have three unknowns and only two equations, so it seems hard to solve at first, because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of x over y, and then we could do a substitution. So that's why this was a little tricky. So this one right here-- let me do it in that same green color. We can rewrite it as CF over 9 is equal to y minus x over y. or 1 minus x over y. All I did is, I essentially, I guess you could say, distributed the 1 over y times both of these terms. So y over y minus x over y, or 1 minus x minus y. And this is useful, because we already know what x over y is equal to. We already know that x over y is equal to CF over 12. So this right over here, I can replace with this, CF over 12. So then we get-- this is the home stretch here-- CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides, so you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here," + }, + { + "Q": "\nThere is a \"Acute\" angle in 5:35. A to C and E :)", + "A": "Rays EA and EC do make an acute angle. You would call it angle AEC. Always put the letter of the center point of the angle in the middle of the angle name.", + "video_name": "w9jEq6dmqPg", + "timestamps": [ + 335 + ], + "3min_transcript": "We can start at point E , go in the direction of C and go beyond C, so it is a ray, ray EC you could start in E and go in the direction of A and go beyond A, so EA is a ray, and we can start at E and go in the direction of F and go beyond F, that is ray EF ray EF and ray CF are different as the starting points are different Now lets go to point F. To the left of point F there is no other point. We can look at the right, we can have ray FE, start at F go through E and go through C and beyond C that is teh same thing as FE ray FE and ray FC are the same as the point E is on ray FC, then finally we have not focussed on point A you may think there is ray AE, but the line does not go beyond E, so it is not a ray, to the top of A there is no other point, so there is no ray there either that is all the rays based on the points specified. If they had given us a point over there, we could have had other rays," + }, + { + "Q": "at 2:53, can the rays CE and CF also be written CEF or FEC?\n", + "A": "No because you only use two points to identify a ray, using all three points means you are talking about an angle", + "video_name": "w9jEq6dmqPg", + "timestamps": [ + 173 + ], + "3min_transcript": "is JH going up, goes upto H and keeps on going in that direction beyond H, ray JH, starting from J going through H and going beyond it forever now if we go to H, there is no ray HJ as the line ends in J and does not keep going beyond J, there is no ray H as it is just one point, just usiing one point, we cannot say it as a ray. now looking at our diagram the only ray is JH. Now let us look at the other points. after C to specify it as a ray, we can have a ray CE, starts at C goes through E and goes on for ever, you can also have a ray starting at C, going through F and going on forever, CE & CF are the same rays as F sits on ray CE and E sits of ray CF, so CE & CF are the same rays, We can start at point E , go in the direction of C and go beyond C, so it is a ray, ray EC you could start in E and go in the direction of A and go beyond A, so EA is a ray, and we can start at E and go in the direction of F and go beyond F, that is ray EF ray EF and ray CF are different as the starting points are different Now lets go to point F. To the left of point F there is no other point. We can look at the right, we can have ray FE, start at F go through E and" + }, + { + "Q": "\nAt 2:38, can you express the ray as ray CEF?", + "A": "No. You only use two points to express a ray. For example ray CF or ray EF or ray EC. If you use three points you are expressing an angle, in this case a straight angle.", + "video_name": "w9jEq6dmqPg", + "timestamps": [ + 158 + ], + "3min_transcript": "is JH going up, goes upto H and keeps on going in that direction beyond H, ray JH, starting from J going through H and going beyond it forever now if we go to H, there is no ray HJ as the line ends in J and does not keep going beyond J, there is no ray H as it is just one point, just usiing one point, we cannot say it as a ray. now looking at our diagram the only ray is JH. Now let us look at the other points. after C to specify it as a ray, we can have a ray CE, starts at C goes through E and goes on for ever, you can also have a ray starting at C, going through F and going on forever, CE & CF are the same rays as F sits on ray CE and E sits of ray CF, so CE & CF are the same rays, We can start at point E , go in the direction of C and go beyond C, so it is a ray, ray EC you could start in E and go in the direction of A and go beyond A, so EA is a ray, and we can start at E and go in the direction of F and go beyond F, that is ray EF ray EF and ray CF are different as the starting points are different Now lets go to point F. To the left of point F there is no other point. We can look at the right, we can have ray FE, start at F go through E and" + }, + { + "Q": "\nFrom 0:34 and onward, he starts to talk about how you can tell if it is an axis of symmetry. Can you do this with any shape? How about with a circle??", + "A": "You don t need to worry. From 0:34, Sal started talking about that cause the topic is axis of symmetry. You always find a line of symmetry in every basic shape you find. But it is not important that there is axis of symmetry in every shape. I hope this helped!", + "video_name": "LrTn4cvsewk", + "timestamps": [ + 34 + ], + "3min_transcript": "For each of these diagrams, I want to think about whether this blue line represents an axis of symmetry. And the way we can tell is if on both sides of the blue line we essentially have mirror images. Let's take this top part of this polygon, the part that is above this blue line here, and let's reflect it across the blue line-- you could almost imagine that it's a reflection over some type of a lake or something-- and see if we get exactly what we have below. Then this would be an axis of symmetry. So this point right over here, this distance to the blue line, let's go-- the same amount on the other side would get you right there. And so you immediately see we start ending up with a point that is off what's actually here in black, the actual bottom part of the polygon. So this is a pretty good clue that this is not an axis of symmetry. But let's just continue it, just to go through the exercise. So this point, if you reflected it across this blue line, would get you here. This point-- I'll do it in a different color. This point, if you were to reflect it across this blue line, it would get I can do a straighter job than that. So if you go about that distance about it, and I want to go straight down into the blue line, and I'm going to go the same distance on the other side, it gets me to right around there. And then this point over here, if I were to drop it straight down, then if I were to go the same distance on the other side, it gets me right around there. And then finally, this point gets me right around there. So its mirror image of this top part would look something like this. My best attempt to draw it would look something like this, which is very different than the part of the polygon that's actually on the other side of this blue line. So in this case, the blue line is not an axis of symmetry. So this is no. No, this blue line is not an axis of symmetry. Now let's look at it over here. Here you can see that it looks like this blue line really divides this polygon in half. It really does look like mirror images. It really does look like, if you imagine that this is some type of a lake, a still lake, so I shouldn't actually draw waves, but this is some type of a lake, that this is the reflection. And we can even go point by point here. So this point right over here is the same distance from, if we dropped a perpendicular to this point as this one right over here; this one over here, same distance, same distance as this point right over here; and we could do that for all of the points. So in this case, the blue line does represent an axis of symmetry." + }, + { + "Q": "\nAt 3:04 why does it continue after that?", + "A": "It doesn t, it ends at 3:04.", + "video_name": "jb8mFpA1YI8", + "timestamps": [ + 184 + ], + "3min_transcript": "Let's see if we can figure out 3 times 60. Well, there's a couple of ways you could think about it. You could literally view this as 60 three times. So you could view this as 60 plus 60 plus 60. And you might be able to compute this in your head. 60 plus 60 is 120, plus another 60 is 180. And you'd be done. But another way to think about this is that 3 times 60 is the same thing as 3 times-- instead of thinking of it as 60, you could think of 60 as 6 times 10. 3 times 6 times 10. Now, when you're multiplying three numbers like this, it doesn't matter what order you multiply them in. So we could multiply the 3 times 6 first and get 18 and then multiply that times 10. And 18 times 10 is just going to be 180. It's going to be 18 with another zero. So this is going to be 180. Now, the more practice you get here, you'll realize, but I have to worry about this 0 right over here. So I'm going to put one more zero at the end. It's going to be 180. Same answer that we got right over there. Let's do another one of these. So let's say we want to multiply 50 times 7. And I encourage you to pause the video and think about it yourself, and then unpause it and see what I do. So 50, well, there's a couple of ways you could think about it. One, you could literally try to add 50 seven times. Adding 7 fifty times would take forever, but you could literally say 50 plus 50 plus 50 plus 50-- let's see, that's four-- plus 50 plus 50. Let's see, that is six. I'll do one more right over here. 50 right over here. So this is 50 seven times. If you add together 50 plus 50 is 100, 150, 200, 250, 300, 350. So you could do it that way. You just need to realize that 50 is the same thing as 10 times 5. So we could write this as 10 times 5, and then we're multiplying that times 7. Once again, the order that we multiply does not matter. So we can multiply the 5 times the 7 first. We know that that is 35, and we're going to multiply that times 10. 10 times 35, well, we're just going to stick a zero at the end of the 35. It's going to be equal to 350. Now I want to do that zero in that same color. It's going to be 350. Now, you might realize, hey, look, I could have just looked at this 5 right over here, multiplied the 5 times the 7, and have gotten the 35. And then, not forgetting that it's actually not a 5, it's a 50. So I have to multiply by 10 again, or I have to throw that 0 at the end of it to get that 0 right over here. So 50 times 7 is 350." + }, + { + "Q": "\nAt 11:15, Sal mistakenly wrote 2.5, when it was supposed to be 2.9.", + "A": "Good catch! If you look, you ll even see a little note in the lower left corner about the mistake.", + "video_name": "i6lfVUp5RW8", + "timestamps": [ + 675 + ], + "3min_transcript": "Now let's do another one, where we start with the scientific notation value and we want to go to the numeric value. Just to mix things up. So let's say you have 2.9 times 10 to the negative fifth. So one way to think about is, this leading numeral, plus all 0's to the left of the decimal spot, is going to be five digits. So you have a 2 and a 9, and then you're going to have 4 more 0's. 1, 2, 3, 4. And then you're going to have your decimal. And how did I know 4 0's? Because I'm counting,, this is 1, 2, 3, 4, 5 spaces behind the decimal, including the leading numeral. And so it's 0.000029. And just to verify, do the other technique. How do I write this in scientific notation? I count all of the digits, all of the leading 0's behind the So I have 1, 2, 3, 4, 5 digits. So it's 10 to the negative 5. And so it'll be 2.9 times 10 to the negative 5. And once again, this isn't just some type of black magic here. This actually makes a lot of sense. If I wanted to get this number to 2.9, what I would have to do is move the decimal over 1, 2, 3, 4, 5 spots, like that. And to get the decimal to move over the right by 5 spots-- let's just say with 0, 0, 0, 0, 2, 9. If I multiply it by 10 to the fifth, I'm also going to have to multiply it by 10 to the negative 5. So I don't want to change the number. This right here is just multiplying something by 1. 10 to the fifth times 10 to the negative 5 is 1. decimal 5 to the right. 1, 2, 3, 4, 5. So this will be 2.5, and then we're going to be left with times 10 to the negative 5. Anyway, hopefully, you found that scientific notation drill useful." + }, + { + "Q": "At 11:15 instead of 2.5 isn't it suppose to be 2.9\n", + "A": "That is a known error in the video. A box pops up and tells you the error and provides the correct value of 2.9", + "video_name": "i6lfVUp5RW8", + "timestamps": [ + 675 + ], + "3min_transcript": "Now let's do another one, where we start with the scientific notation value and we want to go to the numeric value. Just to mix things up. So let's say you have 2.9 times 10 to the negative fifth. So one way to think about is, this leading numeral, plus all 0's to the left of the decimal spot, is going to be five digits. So you have a 2 and a 9, and then you're going to have 4 more 0's. 1, 2, 3, 4. And then you're going to have your decimal. And how did I know 4 0's? Because I'm counting,, this is 1, 2, 3, 4, 5 spaces behind the decimal, including the leading numeral. And so it's 0.000029. And just to verify, do the other technique. How do I write this in scientific notation? I count all of the digits, all of the leading 0's behind the So I have 1, 2, 3, 4, 5 digits. So it's 10 to the negative 5. And so it'll be 2.9 times 10 to the negative 5. And once again, this isn't just some type of black magic here. This actually makes a lot of sense. If I wanted to get this number to 2.9, what I would have to do is move the decimal over 1, 2, 3, 4, 5 spots, like that. And to get the decimal to move over the right by 5 spots-- let's just say with 0, 0, 0, 0, 2, 9. If I multiply it by 10 to the fifth, I'm also going to have to multiply it by 10 to the negative 5. So I don't want to change the number. This right here is just multiplying something by 1. 10 to the fifth times 10 to the negative 5 is 1. decimal 5 to the right. 1, 2, 3, 4, 5. So this will be 2.5, and then we're going to be left with times 10 to the negative 5. Anyway, hopefully, you found that scientific notation drill useful." + }, + { + "Q": "\nAt 0:25 I wanted to know when was scientific notation even invented?", + "A": "not hepfull flaged", + "video_name": "i6lfVUp5RW8", + "timestamps": [ + 25 + ], + "3min_transcript": "There are two whole Khan Academy videos on what scientific notation is, why we even worry about it. And it also goes through a few examples. And what I want to do in this video is just use a ck12.org Algebra I book to do some more scientific notation examples. So let's take some things that are written in scientific notation. Just as a reminder, scientific notation is useful because it allows us to write really large, or really small numbers, in ways that are easy for our brains to, one, write down, and two, understand. So let's write down some numbers. So let's say I have 3.102 times 10 to the second. And I want to write it as just a numerical value. It's in scientific notation already. It's written as a product with a power of 10. So how do I write this? It's just a numeral. Well, there's a slow way and the fast way. The slow way is to say, well, this is the same thing as 3.102 times 100, which means if you multiplied 3.102 times And then we have 1, 2, 3 numbers behind the decimal point, and that'd be the right answer. This is equal to 310.2. Now, a faster way to do this is just to say, well, look, right now I have only the 3 in front of the decimal point. When I take something times 10 to the second power, I'm essentially shifting the decimal point 2 to the right. So 3.102 times 10 to the second power is the same thing as-- if I shift the decimal point 1, and then 2, because this is 10 to the second power-- it's same thing as 310.2. So this might be a faster way of doing it. the right by 1. Let's do another example. Let's say I had 7.4 times 10 to the fourth. Well, let's just do this the fast way. Let's shift the decimal 4 to the right. So 7.4 times 10 to the fourth. Times 10 to the 1, you're going to get 74. Then times 10 to the second, you're going to get 740. We're going to have to add a 0 there, because we have to shift the decimal again. 10 to the third, you're going to have 7,400. And then 10 to the fourth, you're going to have 74,000. Notice, I just took this decimal and went 1, 2, 3, 4 spaces. So this is equal to 74,000. And when I had 74, and I had to shift the decimal 1 more to the right, I had to throw in a 0 here. I'm multiplying it by 10. Another way to think about it is, I need 10 spaces between" + }, + { + "Q": "At 10:10 we have 5 digits. Is there any limit as to the total number of digits we can have? How does this affect significant figures?\n", + "A": "Nope. There are no limits. The more digits you have, the bigger the exponent of 10 needs to be. The rules for significant figures aren t affected.", + "video_name": "i6lfVUp5RW8", + "timestamps": [ + 610 + ], + "3min_transcript": "And then we're left with this one, times 10 to the negative 3. Now, a very quick way to do it is just to say, look, let me count-- including the leading numeral-- how many spaces I have behind the decimal. 1, 2, 3. So it's going to be 2.81 times 10 to the negative 1, 2, 3 power. Let's do one more like that. Let me actually scroll up here. Let's do one more like that. Let's say I have 1, 2, 3, 4, 5, 6-- how many 0's do I have Well, I'll just make up something. 0, 2, 7. And you wanted to write that in scientific notation. Well, you count all the digits up to the 2, behind the decimal. So 1, 2, 3, 4, 5, 6, 7, 8. So this is going to be 2.7 times 10 to Now let's do another one, where we start with the scientific notation value and we want to go to the numeric value. Just to mix things up. So let's say you have 2.9 times 10 to the negative fifth. So one way to think about is, this leading numeral, plus all 0's to the left of the decimal spot, is going to be five digits. So you have a 2 and a 9, and then you're going to have 4 more 0's. 1, 2, 3, 4. And then you're going to have your decimal. And how did I know 4 0's? Because I'm counting,, this is 1, 2, 3, 4, 5 spaces behind the decimal, including the leading numeral. And so it's 0.000029. And just to verify, do the other technique. How do I write this in scientific notation? I count all of the digits, all of the leading 0's behind the So I have 1, 2, 3, 4, 5 digits. So it's 10 to the negative 5. And so it'll be 2.9 times 10 to the negative 5. And once again, this isn't just some type of black magic here. This actually makes a lot of sense. If I wanted to get this number to 2.9, what I would have to do is move the decimal over 1, 2, 3, 4, 5 spots, like that. And to get the decimal to move over the right by 5 spots-- let's just say with 0, 0, 0, 0, 2, 9. If I multiply it by 10 to the fifth, I'm also going to have to multiply it by 10 to the negative 5. So I don't want to change the number. This right here is just multiplying something by 1. 10 to the fifth times 10 to the negative 5 is 1." + }, + { + "Q": "\nAt 0:47, Sal says 15 to 25.\nShouldn't he say 25 to 15 ?", + "A": "He can say 15 to 25 or 25 to 15, it depends on him, because the order in which you say ratios doesn t really matter.", + "video_name": "jNUz0P5MG9M", + "timestamps": [ + 47 + ], + "3min_transcript": "The following table describes the relationship between the number of servings of spaghetti bolognese-- I don't know if I'm pronouncing that-- or bolognese, and the number of tomatoes needed to prepare them. Test the ratios for equivalents, and determine whether the relationship is proportional. Well, you have a proportional relationship between the number of servings and the number of tomatoes is if the ratio of the number of servings to the number of tomatoes is always the same. Or if the ratio of the number of tomatoes to the number of servings is always the same. So let's just think about the ratio of the number of tomatoes to the number of servings. So it's 10 to 6, which is the same thing as 5 to 3. So here the ratio is 5 to 3. 15 to 9, if you divide both of these by 3, you get 5 to 3. So it's the same ratio. 15 to 25, if you divide both of these by 5, you get 5 to 3. So based on this data, it looks like the ratio between the number of tomatoes and the number of servings is always constant. So yes, this relationship is proportional." + }, + { + "Q": "at around 3:30, Sal siad for the second equation that why is ewual to NEGATIVE 1x +5. Why is that? Shouldn't it be positive?\n", + "A": "No, it should not, the reason for this is that the x(1x) was already negative in the equation. The original form of the equation looked like this y<5 -x(1x) , the only thing Sal did was rewrite the equation to look like the one above it. So he moved the x(1x) to the front of the equation taking the negative sign with it y< -x(1x) +5. When you rearrange an equation the signs of numbers remain to keep the problem balanced.", + "video_name": "CA4S7S-3Lg4", + "timestamps": [ + 210 + ], + "3min_transcript": "So the point 0, negative 8 is on the line. And then it has a slope of 1. You don't see it right there, but I could write it as 1x. So the slope here is going to be 1. I could just draw a line that goes straight up, or you could even say that it'll intersect if y is equal to 0, if y were equal to 0, x would be equal to 8. So 1, 2, 3, 4, 5, 6, 7, 8. And so this is x is equal to 8. If it has a slope of 1, for every time you move to the right 1, you're going to move up 1. So the line is going to look something like this. And actually, let me not draw it as a solid line. If I did it as a solid line, that would actually be this equation right here. But we're not going to include that line. We care about the y values that are greater than that line. So what we want to do is do a dotted line to show that that's just the boundary, that we're not including that in our solution set. Let me do this in a new color. So this will be the color for that line, or for that inequality, I should say. And this says y is greater than x minus 8. So you pick an x, and then x minus 8 would get us on the boundary line. And then y is greater than that. So it's all the y values above the line for any given x. So it'll be this region above the line right over here. And if that confuses you, I mean, in general I like to just think, oh, greater than, it's going to be above the line. If it's less than, it's going to be below a line. But if you want to make sure, you can just test on either side of this line. So you could try the point 0, 0, which should be in our solution set. And if you say, 0 is greater than 0 minus 8, or 0 is greater than negative 8, that works. So this definitely should be part of the solution set. And you could try something out here like 10 comma 0 and see that it doesn't work. Because you would have 10 minus 8, which would be 2, and then you'd have 0. And 0 is not greater than 2. So when you test something out here, you also see that it won't work. But in general, I like to just say, hey look, this is the boundary line, and we're greater than the boundary line for any given x. Let's do this one. The boundary line for it is going to be y is equal to 5 minus x. So the boundary line is y is equal to 5 minus x. So once again, if x is equal to 0, y is 5. So 1, 2, 3, 4, 5. And then it has a slope of negative 1. We could write this as y is equal to negative 1x plus 5. That's a little bit more traditional. So once again, y-intercept at 5. And it has a slope of negative 1. Or another way to think about it, when y is 0, x will be equal to 5. So 1, 2, 3, 4, 5. So every time we move to the right one, we go down one because we have a negative 1 slope. So it will look like this. And once again, I want to do a dotted line because we are-- so that is our dotted line. And I'm doing a dotted line because it says y is less than 5 minus x." + }, + { + "Q": "\nAt 3:14 Sal said that y is equal to 5-x even though it said y is less then 5-x?", + "A": "At 3:14 Sal is talking about the boundary line, not the area. That s why he says equal to.", + "video_name": "CA4S7S-3Lg4", + "timestamps": [ + 194 + ], + "3min_transcript": "So the point 0, negative 8 is on the line. And then it has a slope of 1. You don't see it right there, but I could write it as 1x. So the slope here is going to be 1. I could just draw a line that goes straight up, or you could even say that it'll intersect if y is equal to 0, if y were equal to 0, x would be equal to 8. So 1, 2, 3, 4, 5, 6, 7, 8. And so this is x is equal to 8. If it has a slope of 1, for every time you move to the right 1, you're going to move up 1. So the line is going to look something like this. And actually, let me not draw it as a solid line. If I did it as a solid line, that would actually be this equation right here. But we're not going to include that line. We care about the y values that are greater than that line. So what we want to do is do a dotted line to show that that's just the boundary, that we're not including that in our solution set. Let me do this in a new color. So this will be the color for that line, or for that inequality, I should say. And this says y is greater than x minus 8. So you pick an x, and then x minus 8 would get us on the boundary line. And then y is greater than that. So it's all the y values above the line for any given x. So it'll be this region above the line right over here. And if that confuses you, I mean, in general I like to just think, oh, greater than, it's going to be above the line. If it's less than, it's going to be below a line. But if you want to make sure, you can just test on either side of this line. So you could try the point 0, 0, which should be in our solution set. And if you say, 0 is greater than 0 minus 8, or 0 is greater than negative 8, that works. So this definitely should be part of the solution set. And you could try something out here like 10 comma 0 and see that it doesn't work. Because you would have 10 minus 8, which would be 2, and then you'd have 0. And 0 is not greater than 2. So when you test something out here, you also see that it won't work. But in general, I like to just say, hey look, this is the boundary line, and we're greater than the boundary line for any given x. Let's do this one. The boundary line for it is going to be y is equal to 5 minus x. So the boundary line is y is equal to 5 minus x. So once again, if x is equal to 0, y is 5. So 1, 2, 3, 4, 5. And then it has a slope of negative 1. We could write this as y is equal to negative 1x plus 5. That's a little bit more traditional. So once again, y-intercept at 5. And it has a slope of negative 1. Or another way to think about it, when y is 0, x will be equal to 5. So 1, 2, 3, 4, 5. So every time we move to the right one, we go down one because we have a negative 1 slope. So it will look like this. And once again, I want to do a dotted line because we are-- so that is our dotted line. And I'm doing a dotted line because it says y is less than 5 minus x." + }, + { + "Q": "At 4:30, when zero is said to be undefined, how come no one of the many mathematicians over the years were able to define it? Why is it so disputed over? Humans as a whole, over the many years of their existence weren't able to do so?\n", + "A": "Sal says a number divided by zero is undefined, not zero itself. The reason a number divided by zero is undefined is because no such number exists. The value of a number divided by zero is infinite.", + "video_name": "bQ-KR3clFgs", + "timestamps": [ + 270 + ], + "3min_transcript": "we're gonna wanna multiply the numerator out, and if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could've written this little \"x\" thing over here but what you're gonna see in Algebra is that the dot become much more common. Because the X becomes used for other-- People don't want to confuse it with the letter X which gets used a lot in Algebra. That's why they used the dot very often. So this just says negative seven (-7) times three (3) in the numerator, and we're gonna take that product and divide it by negative one (-1). So the numerator, negative seven (-7) times three (3), positive seven (7) times three (3) would be twenty-one (21), but since exactly one of these two are negative, this is going to be negative twenty-one (-21), that's gonna be negative twenty-one (-21) over negative one (-1). And so negative twenty-one (-21) divided by negative one (-1), negative divided by a negative is going to be a positive. So this is going to be a positive twenty-one (21). Let me write all these things down. So if I were to take a positive divided by a negative, If I had a negative divided by a positive, that's also going to be a negative. If I have a negative divided by a negative, that's going to give me a positive, and if obviously a positive divided by a positive, that's also going to give me a positive. Now let's do this last one over here. This is actually all multiplication, but it's interesting, because we're multiplying three (3) things, which we haven't done yet. And we could just go from left to right over here, and we could first think about negative two (-2) times negative seven (-7). Negative two (-2) times negative seven (-7). They are both negatives, and negatives cancel out, so this would give us, this part right over here, will give us positive fourteen (14). And so we're going to multiply positive fourteen (14) times this negative one (-1), times -1. Now we have a positive times a negative. Exactly one of them is negative, so this is going to be negative answer, Now let me give you a couple of more, I guess we could call these trick problems. What would happen if I had zero (0) divided by negative five (-5). Well this is zero negative fifths So zero divided by anything that's non-zero is just going to equal to zero. But what if it were the other way around? What happens if we said negative five divided by zero? Well, we don't know what happens when you divide things by zero. We haven't defined that. There's arguments for multiple ways to conceptualize this, so we traditionally do say that this is undefined. We haven't defined what happens when something is divided by zero. And similarly, even when we had zero divided by zero, this is still, this is still, undefined." + }, + { + "Q": "At 4:45 he says zero is undefined. Does that mean its not negative of positive?\n", + "A": "Its neither. An integer is a whole number that can be either greater than 0, called positive, or less than 0, called negative. Zero is neither positive nor negative. Two integers that are the same distance from the origin in opposite directions are called opposites.", + "video_name": "bQ-KR3clFgs", + "timestamps": [ + 285 + ], + "3min_transcript": "we're gonna wanna multiply the numerator out, and if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could've written this little \"x\" thing over here but what you're gonna see in Algebra is that the dot become much more common. Because the X becomes used for other-- People don't want to confuse it with the letter X which gets used a lot in Algebra. That's why they used the dot very often. So this just says negative seven (-7) times three (3) in the numerator, and we're gonna take that product and divide it by negative one (-1). So the numerator, negative seven (-7) times three (3), positive seven (7) times three (3) would be twenty-one (21), but since exactly one of these two are negative, this is going to be negative twenty-one (-21), that's gonna be negative twenty-one (-21) over negative one (-1). And so negative twenty-one (-21) divided by negative one (-1), negative divided by a negative is going to be a positive. So this is going to be a positive twenty-one (21). Let me write all these things down. So if I were to take a positive divided by a negative, If I had a negative divided by a positive, that's also going to be a negative. If I have a negative divided by a negative, that's going to give me a positive, and if obviously a positive divided by a positive, that's also going to give me a positive. Now let's do this last one over here. This is actually all multiplication, but it's interesting, because we're multiplying three (3) things, which we haven't done yet. And we could just go from left to right over here, and we could first think about negative two (-2) times negative seven (-7). Negative two (-2) times negative seven (-7). They are both negatives, and negatives cancel out, so this would give us, this part right over here, will give us positive fourteen (14). And so we're going to multiply positive fourteen (14) times this negative one (-1), times -1. Now we have a positive times a negative. Exactly one of them is negative, so this is going to be negative answer, Now let me give you a couple of more, I guess we could call these trick problems. What would happen if I had zero (0) divided by negative five (-5). Well this is zero negative fifths So zero divided by anything that's non-zero is just going to equal to zero. But what if it were the other way around? What happens if we said negative five divided by zero? Well, we don't know what happens when you divide things by zero. We haven't defined that. There's arguments for multiple ways to conceptualize this, so we traditionally do say that this is undefined. We haven't defined what happens when something is divided by zero. And similarly, even when we had zero divided by zero, this is still, this is still, undefined." + }, + { + "Q": "In algebra, why just not use the letter X so you can still use it for multiplication? 2:05\n", + "A": "It s just that x happens to be very common, if you don t like the use of x, you can use any other variable, like j .", + "video_name": "bQ-KR3clFgs", + "timestamps": [ + 125 + ], + "3min_transcript": "Now that we know a little bit about multiplying positive and negative numbers, Let's think about how how we can divide them. Now what you'll see is that it's actually a very similar methodology. That if both are positive, you'll get a positive answer. If one is negative, or the other, but not both, you'll get a negative answer. And if both are negative, they'll cancel out and you'll get a positive answer. But let's apply and I encourage you to pause this video and try these out yourself and then see if you get the same answer that I'm going to get. So eight (8) divided by negative two (-2). So if I just said eight (8) divided by two (2), that would be a positive four (4), but since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative. So eight (8) divided by negative two (-2) is negative four (-4). Now negative sixteen (-16) divided by positive four (4)-- now be very careful here. If I just said positive sixteen (16) divided by positive four (4), that would just be four (4). But because one of these two numbers is negative, then I'm going to get a negative answer. Now I have negative thirty (-30) divided by negative five (-5). If I just said thirty (30) divided by five (5), I'd get a positive six (6). And because I have a negative divided by a negative, the negatives cancel out, so my answer will still be positive six (6)! And I could even write a positive (+) out there, I don't have to, but this is a positive six (6). A negative divided by a negative, just like a negative times a negative, you're gonna get a positive answer. Eighteen (18) divided by two (2)! And this is a little bit of a trick question. This is what you knew how to do before we even talked about negative numbers: This is a positive divided by a positive. Which is going to be a positive. So that is going to be equal to positive nine (9). Now we start doing some interesting things, here's kind of a compound problem. We have some multiplication and some division going on. we're gonna wanna multiply the numerator out, and if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could've written this little \"x\" thing over here but what you're gonna see in Algebra is that the dot become much more common. Because the X becomes used for other-- People don't want to confuse it with the letter X which gets used a lot in Algebra. That's why they used the dot very often. So this just says negative seven (-7) times three (3) in the numerator, and we're gonna take that product and divide it by negative one (-1). So the numerator, negative seven (-7) times three (3), positive seven (7) times three (3) would be twenty-one (21), but since exactly one of these two are negative, this is going to be negative twenty-one (-21), that's gonna be negative twenty-one (-21) over negative one (-1). And so negative twenty-one (-21) divided by negative one (-1), negative divided by a negative is going to be a positive. So this is going to be a positive twenty-one (21). Let me write all these things down. So if I were to take a positive divided by a negative," + }, + { + "Q": "\nAt 0:42 why is the answer negative if the bigger number is positive?", + "A": "It doesn t matter if the bigger number is positive or not. If you re dividing (or multiplying) and exactly one of the two numbers is negative then the answer is negative.", + "video_name": "bQ-KR3clFgs", + "timestamps": [ + 42 + ], + "3min_transcript": "Now that we know a little bit about multiplying positive and negative numbers, Let's think about how how we can divide them. Now what you'll see is that it's actually a very similar methodology. That if both are positive, you'll get a positive answer. If one is negative, or the other, but not both, you'll get a negative answer. And if both are negative, they'll cancel out and you'll get a positive answer. But let's apply and I encourage you to pause this video and try these out yourself and then see if you get the same answer that I'm going to get. So eight (8) divided by negative two (-2). So if I just said eight (8) divided by two (2), that would be a positive four (4), but since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative. So eight (8) divided by negative two (-2) is negative four (-4). Now negative sixteen (-16) divided by positive four (4)-- now be very careful here. If I just said positive sixteen (16) divided by positive four (4), that would just be four (4). But because one of these two numbers is negative, then I'm going to get a negative answer. Now I have negative thirty (-30) divided by negative five (-5). If I just said thirty (30) divided by five (5), I'd get a positive six (6). And because I have a negative divided by a negative, the negatives cancel out, so my answer will still be positive six (6)! And I could even write a positive (+) out there, I don't have to, but this is a positive six (6). A negative divided by a negative, just like a negative times a negative, you're gonna get a positive answer. Eighteen (18) divided by two (2)! And this is a little bit of a trick question. This is what you knew how to do before we even talked about negative numbers: This is a positive divided by a positive. Which is going to be a positive. So that is going to be equal to positive nine (9). Now we start doing some interesting things, here's kind of a compound problem. We have some multiplication and some division going on. we're gonna wanna multiply the numerator out, and if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could've written this little \"x\" thing over here but what you're gonna see in Algebra is that the dot become much more common. Because the X becomes used for other-- People don't want to confuse it with the letter X which gets used a lot in Algebra. That's why they used the dot very often. So this just says negative seven (-7) times three (3) in the numerator, and we're gonna take that product and divide it by negative one (-1). So the numerator, negative seven (-7) times three (3), positive seven (7) times three (3) would be twenty-one (21), but since exactly one of these two are negative, this is going to be negative twenty-one (-21), that's gonna be negative twenty-one (-21) over negative one (-1). And so negative twenty-one (-21) divided by negative one (-1), negative divided by a negative is going to be a positive. So this is going to be a positive twenty-one (21). Let me write all these things down. So if I were to take a positive divided by a negative," + }, + { + "Q": "At 2:30, what does \"factoring out the negative sign\" mean? How do you factor out a negative sign? I also had a problem with this in \"Manipulating linear expressions with rational coefficients,\" in Algebra. I never understood how to do it.\n", + "A": "area of each interior angle = 180 (5-2)/5 = 108, so area of each exterior angle = 180-108 = 72, so area of five exterior angle = 72*5 = 360, Am I on the right track?", + "video_name": "95logvV8nXY", + "timestamps": [ + 150 + ], + "3min_transcript": "Now this looks like an interesting problem. We have this polygon. It looks like a pentagon right over here, has five sides. It's an irregular pentagon. Not all the sides look to be the same length. And the sides are kind of continued on. And we have these particular exterior angles of this pentagon. And what we're asked is, what is the sum of all of these exterior angles. And it's kind of daunting, because they don't give us any other information. They don't even give us any particular angles. They don't start us off anywhere. And so what we can do, let's just think about the step by step, just based on what we do know. Well, we have these exterior angles. And these exterior angles, they're each supplementary to some interior angle. So maybe if we can express them as a function of the interior angles, we can maybe write this problem in a way that seems a little bit more doable. So let's write the interior angles over here. We already got to e. So let's call this f, this interior angle f. Let's call this interior angle g. Let's call this one i. And let's call this one j. And so this sum of these particular exterior angles, a is now the same thing as 180 minus g, because a and g are supplementary. So a is 180 minus g. And then we have plus b. But b we can write in terms of this interior angle. It's going to be 180 minus h, because these two angles once again, are supplementary. We do that in a new color. So this is going to be 180 minus h. And we could do the same thing for each of them. c, we can write as 180 minus i, so plus 180 minus i. And then d, we can write as 180 minus j, so plus 180 minus j. And then finally, e, I'm running out of colors, e, we can write as 180 minus f, so plus 180 minus f. we have 180 5 times. So this is going to be equal to 5 times 180 which is what, 900. And then you have minus g, minus h, minus i, minus j, minus f. Or we could write that as minus-- I'll try to do the same colors-- g plus h-- I'm kind of factoring out this negative sign-- g plus h-- I'll do the same color as g, that's not the same color-- g plus h, plus i, plus j, plus f. And the whole reason why did this and why this is interesting now, is that we've expressed this first thing that we need to figure out. We've expressed it in terms of sum of the interior angles." + }, + { + "Q": "at the time 2:02 in the video, i don't understand what he was doing? Can anyone help me please!\n", + "A": "In this section at 2:02, he is finding the 4th root of 16 or \u00e2\u0088\u009c16 The 4th root of something is a number multiplied by itself 4 times to equal that something Well, at that time after factoring 16, he has \u00e2\u0088\u009c2\u00e2\u0088\u00992\u00e2\u0088\u00992\u00e2\u0088\u00992 written and he is talking about finding the 4th root of 2 times 2 times 2 times 2 We are looking for a factor that occurs 4 times, and there it is! We can see that 2 is a factor of 16 four different times, so \u00e2\u0088\u009c16 = \u00e2\u0088\u009c2\u00e2\u0088\u00992\u00e2\u0088\u00992\u00e2\u0088\u00992 = \u00e2\u0088\u009c2\u00e2\u0081\u00b4 = 2 And that is how he got the 2", + "video_name": "iX7ivCww2ws", + "timestamps": [ + 122 + ], + "3min_transcript": "So far, when we were dealing with radicals we've only been using the square root. We've seen that if I write a radical sign like this and put a 9 under it, this means the principal square root of 9, which is positive 3. Or you could view it as the positive square root of 9. Now, what's implicit when we write it like this is that I'm taking the square root. So I could have also written it like this. I could have also written the radical sign like this and written this index 2 here, which means the square root, the principal square root of 9. Find me something that if I square that something, I get 9. And the radical sign doesn't just have to apply to a square root. You can change the index here and then take an arbitrary root of a number. So for example, if I were to ask you, what-- You could imagine this is called the cube root, or you could call it the third root of 27. What is this? power, I'd get 27. Well, the only number that if you take it to the third power, you get 27 is equal to 3. 3 times 3 times 3 is equal to 27. 9 times 3, 27. So likewise, let me just do one more. So if I have 16-- I'll do it in a different color. If I have 16 and I want to take the fourth root of 16, what number times itself 4 times is equal to 16? And if it doesn't pop out at you immediately, you can actually just do a prime factorization of 16 to figure it out. 16 is 2 times 8. 8 is 2 times 4. 4 is 2 times 2. So this is equal to the fourth root of 2 times 2 times 2 times 2. You have these four 2's here. of this must be equal to 2. And you could also view this as kind of the fourth principal root because if these were all negative 2's, it would also work. Just like you have multiple square roots, you have multiple fourth roots. But the radical sign implies the principal root. Now, with that said, we've simplified traditional square roots before. Now we should hopefully be able to simplify radicals with higher power roots. So let's try a couple. Let's say I want to simplify this expression. The fifth root of 96. So like I said before, let's just factor this right here. So 96 is 2 times 48. Which is 2 times 24. Which is 2 times 12. Which is 2 times 6." + }, + { + "Q": "I do not entirely understand the rules in regards to a specific exponent being rewritten as a specific corresponding square root. Ex: The square root of 25= 25^1/2 power. I am absolutely lost after this point: 6:35. Are there any videos that explore this further and in greater detail? Thanks!\n", + "A": "Basically, when a number is raised to a fractional power, it is asked for that specific root. EX: 1/2 power= square root 1/3 power = cubed root", + "video_name": "iX7ivCww2ws", + "timestamps": [ + 395 + ], + "3min_transcript": "Which is 2 times 4. Which is 2 times 2. So we have 1, 2, 3, 4, 5, 6. So it's essentially 2 to the sixth power. So this is equivalent to the sixth root of 2 to the sixth-- that's what 64 is --times x to the eighth power. Now, the sixth root of 2 to the sixth, that's pretty straightforward. So this part right here is just going to be equal to 2. That's going to be 2 times the sixth root of x to the eighth power. And how can we simplify this? Well, x to the eighth power, that's the same thing as x to the sixth power times x squared. This is the same thing as x to the eighth. So this is going to be equal to 2 times the sixth root of x to the sixth times x squared. And the sixth root, this part right here, the sixth root of x to the sixth, that's just x. So this is going to be equal to 2 times x times the sixth root of x squared. Now, we can simplify this even more if you really think about. Remember, this expression right here, this is the exact same thing as x squared to the 1/6 power. And if you remember your exponent properties, when you raise something to an exponent, and then raise that to an exponent, that's equivalent to x to the 2 times 1/6 power. Or-- let me write this --2 times 1/6 power, which is the same thing-- Let me not forget to write my 2x there. And this is the same thing as 2x-- it's the same 2x there --times x to the 2/6. Or, if we want to write that in most simple form or lowest common form, you get 2x times x to the-- What do you have here? x to the 1/3. So if you want to write it in radical form, you could write this is equal to 2 times 2x times the third root of x. Or, the other way to think about it, you could just say-- So we could just go from this point right here. We could write this. We could ignore this, what we did before. And we could say, this is the same thing as 2 times x to the eighth to the 1/6 power. x to the eighth to the 1/6 power. So this is equal to 2 times x to the-- 8" + }, + { + "Q": "\nAt 6:20, why does the sixth root of X^6 equal X and not the absolute value of X?", + "A": "It depends on whether or not the radical is given in the equation or if you need to insert it. If the problem has a radical shown, then you assume it is positive (what Sal did). If you are solving the problem and you introduce a radical, then you could have both the positive and negative option (so the absolute value).", + "video_name": "iX7ivCww2ws", + "timestamps": [ + 380 + ], + "3min_transcript": "Times 3 to the 1/5. Now I have something that's multiplied. I have 2 multiplied by itself 5 times. And I'm taking that to the 1/5. Well, the 1/5 power of this is going to be 2. Or the fifth root of this is just going to be 2. So this is going to be a 2 right here. And this is going to be 3 to the 1/5 power. 2 times 3 to the 1/5, which is this simplified about as much as you can simplify it. But if we want to keep in radical form, we could write it as 2 times the fifth root 3 just like that. Let's try another one. Let me put some variables in there. Let's say we wanted to simplify the sixth root of 64 times x to the eighth. So let's do 64 first. 64 is equal to 2 times 32, which is 2 times 16. Which is 2 times 4. Which is 2 times 2. So we have 1, 2, 3, 4, 5, 6. So it's essentially 2 to the sixth power. So this is equivalent to the sixth root of 2 to the sixth-- that's what 64 is --times x to the eighth power. Now, the sixth root of 2 to the sixth, that's pretty straightforward. So this part right here is just going to be equal to 2. That's going to be 2 times the sixth root of x to the eighth power. And how can we simplify this? Well, x to the eighth power, that's the same thing as x to the sixth power times x squared. This is the same thing as x to the eighth. So this is going to be equal to 2 times the sixth root of x to the sixth times x squared. And the sixth root, this part right here, the sixth root of x to the sixth, that's just x. So this is going to be equal to 2 times x times the sixth root of x squared. Now, we can simplify this even more if you really think about. Remember, this expression right here, this is the exact same thing as x squared to the 1/6 power. And if you remember your exponent properties, when you raise something to an exponent, and then raise that to an exponent, that's equivalent to x to the 2 times 1/6 power. Or-- let me write this --2 times 1/6 power, which is the same thing-- Let me not forget to write my 2x there." + }, + { + "Q": "\nAt 3:10 Sal says that each of the objects became two groups. Exactly how are they in two groups? I'm having a hard time visualizing the two groups. I get how he divided each circle in half groups, but the 2 groups he mentions at the very end, I cannot see. Please help. Thanks.", + "A": "Oh wait, I just saw it. The four circles were split in two.", + "video_name": "tnkPY4UqJ44", + "timestamps": [ + 190 + ], + "3min_transcript": "many times it goes. That'll be the whole number part of the mixed number. And then whatever's left over will be the remaining numerator over 5. So what we'll do is take 5 into 6. 5 goes into 6 one time. 1 times 5 is 5. Subtract. You have a remainder of 1. So 6/5 is equal to one whole, or 5/5, and 1/5. This 1 comes from whatever is left over. And now we're done! 3/5 divided by 1/2 is 1 and 1/5. Now, the one thing that's not obvious is why did this work? Why is dividing by 1/2 the same thing as multiplying essentially by 2. 2/1 is the same thing as 2. And to do that, I'll do a little side-- fairly simple-- example, but hopefully, it gets the point across. So we have four objects: one, two, three, four. So I have four objects, and if I were to divide into groups of two, so I want to divide it into groups of two. So that is one group of two and then that is another group of two, how many groups do I have? Well, 4 divided by 2, I have two groups of two, so that is equal to 2. Now, what if I took those same four objects: one, two, three, four. So I'm taking those same four objects. Instead of dividing them into groups of two, I want to divide them into groups of 1/2, which means each group will have half of an object in it. So let's say that would be one group right there. That is a second group. That is a third group. I think you see each group has half of a circle in it. That is the fourth. That's the fifth. That's the sixth. You have eight groups of 1/2, so this is equal to 8. And notice, now each of the objects became two groups. So you could say how many groups do you have? Well, you have four objects and each of them became two groups. I'm looking for a different color. Each of them became two groups, and so you also have eight. So dividing by 1/2 is the same thing as multiplying by 2. And you could think about it with other numbers, but hopefully, that gives you a little bit of an intuition." + }, + { + "Q": "\nat 12:47pm. What if you are multiplying a whole number by a fraction", + "A": "Let s do 7 \u00c3\u00b7 3/4 Change the whole number into a fraction: 7/1 \u00c3\u00b7 3/4 Then change division to multiply by using the reciprocal of 3/4: 7/1 * 4/3 Multiply: 28/3 = 9 1/3 Hope this helps.", + "video_name": "tnkPY4UqJ44", + "timestamps": [ + 767 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 0:21, cant you divide by 5?", + "A": "At 0:21 you could divide =)", + "video_name": "D1cKk48kz-E", + "timestamps": [ + 21 + ], + "3min_transcript": "Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate the c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. We are multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. We would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so it's good that we did not include negative 4. So this is our solution. And this is that solution graphed. And I wanted to do that in that other green color. That's what it looks like." + }, + { + "Q": "At 0:33, couldn't you divide by -5?\n", + "A": "You could, however, some people consider it easier to multiply. Dividing by -5 is the same as dividing by -5/1. You can multiply the inverse -1/5 if you find multiplication easier.", + "video_name": "D1cKk48kz-E", + "timestamps": [ + 33 + ], + "3min_transcript": "Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate the c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. We are multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. We would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so it's good that we did not include negative 4. So this is our solution. And this is that solution graphed. And I wanted to do that in that other green color. That's what it looks like." + }, + { + "Q": "At 1:13, why does Sal flip the inequality sign? PLEASE HELP!\n", + "A": "We solve inequalities almost exactly the way we solve equations. The one exception is that if we multiply or divide by a negative, we have to reverse the inequality. Let s look at what happens if we don t. Start with: -4 < 12 This is current true. If we divide both sides by -2, what happens? -4/(-2) < 12/(-2) 2 < -6 This is no longer true. 2 is a larger number than -6. This is why we reverse the inequality. It is needed to maintain the integrity (truth) of the inequality. Hope this helps.", + "video_name": "D1cKk48kz-E", + "timestamps": [ + 73 + ], + "3min_transcript": "Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate the c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. We are multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. We would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so it's good that we did not include negative 4. So this is our solution. And this is that solution graphed. And I wanted to do that in that other green color. That's what it looks like." + }, + { + "Q": "0:25. Why would it be one fifth and not 5?\n", + "A": "-5 is the coefficient of c, not a subtraction from c. If it were -5+c or c - 5,, then the opposite would be to add 5. since the 5 and c are adjacent, it means multiply (which is what a coefficient is), so the opposite of multiplying by -5 is to divide by -5 which is the same as multiplying by (-1/5).", + "video_name": "D1cKk48kz-E", + "timestamps": [ + 25 + ], + "3min_transcript": "Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate the c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. We are multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. We would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so it's good that we did not include negative 4. So this is our solution. And this is that solution graphed. And I wanted to do that in that other green color. That's what it looks like." + }, + { + "Q": "at 1:36 why did you shade in 2 whole halves an only 1/2 on the others?\n--------------------------------------------------------------\n", + "A": "because, we have to add 1/2 (means half) 5 times", + "video_name": "4PlkCiEXBQI", + "timestamps": [ + 96 + ], + "3min_transcript": "Let's think about how or what 1/2 times 5 represents. So one way to think about it is that this could be five 1/2's added together. So you could view this as 1/2 plus 1/2 plus 1/2 plus 1/2 plus 1/2, which is the same thing as 1 plus 1 plus 1 plus 1 plus 1, over 2, which is equal to 5/2. The other way to think about this is that you start with 5 things. So let's say, that's 1 thing. Let me copy and paste that so they all look the same. So then let me paste it. So that's 2 things. That's 3 things. That's 4 things. And that's 5 things. So the other way to think about it is you start with 5 things, So what would be 1/2 of this? Well, let's see. You have 5 things, so you would get-- 5 divided by 2 would be 2 and 1/2. So you would get this far. Let me make it like this. So you would get this one. You would get this one. And you would get this one. Now, is this the same thing as 5/2? Well, what happens if we divide each of these wholes into halves? So let's do that. So if we just multiplied-- so we just divide each of these into 2. So instead of having 5 wholes, we now have 10 halves. How many of those halves have we filled in? Well, we have filled in 1, 2, 3, 4, 5. So this is also equal to 5/2. multiplication actually means. But if you said, well, how did I compute this? Well, the way you could think about it, and multiplying fractions is actually straightforward from that point of view, is as long as you can express both of them as fractions, and 5 we already know is the same thing as 5 ones, so this we can just multiply times 5/1. So now that I've expressed both of them as fractions, I can just multiply the numerator. So 1 times 5 over 2 times 1. And what's that going to be equal to? Well, 1 times 5 is 5. 2 times 1 is 2." + }, + { + "Q": "\nAt 3:10 he writes the value of pi = 3.14, but Sal uses it as a reference for radians, isn't the 3.14 value in degrees? I mean he says that 3 radians is close to 3.14. So basically I'm asking what measure are we using in normal arithmetic, degrees right? If so, how come he compared 3 radians to 3.14 ,what are presumebly, degrees? Please tolerate any ignorance you come across in my question, I'm simply trying to understand the concepts.", + "A": "Pi is simply a mathematical constant. It does not have any default units attached to it. The number 5 is not in degrees or meters, it is just the number 5. The same with pi.", + "video_name": "fYQ3GRSu4JU", + "timestamps": [ + 190 + ], + "3min_transcript": "If we go straight up, if we rotate it, essentially, if you want to think in degrees, if you rotate it counterclockwise 90 degrees, that is going to get us to pi over two. That would have been a counterclockwise rotation of pi over two radians. Now is three pi over five greater or less than that? Well, three pi over five, three pi over five is greater than, or I guess another way I can say it is, three pi over six is less than three pi over five. You make the denominator smaller, making the fraction larger. Three pi over six is the same thing as pi over two. So, let me write it this way. Pi over two is less than three pi over five. It's definitely past this. We're gonna go past this. Does that get us all the way over here? If we were to go, essentially, be pointed in the opposite direction. Instead of being pointed to the right, making a full, that would be pi radians. That would be pi radians. But this thing is less than pi. Pi would be five pi over five. This is less than pi radians. We are going to sit, we are going to sit someplace, someplace, and I'm just estimating it. We are gonna sit someplace like that. And so we are going to sit in the second quadrant. Let's think about two pi seven. Two pi over seven, do we even get past pi over two? Pi over two here would be 3.5 pi over seven. We don't even get to pi over two. We're gonna end up, we're gonna end up someplace, someplace over here. This thing is, it's greater than zero, so we're gonna definitely start moving counterclockwise, but we're not even gonna get to... This thing is less than pi over two. This is gonna throw us in the first quadrant. What about three radians? three is a little bit less than pi. Right? Three is less than pi but it's greater than pi over two. How do we know that? Well, pi is approximately 3.14159 and it just keeps going on and on forever. So, three is definitely closer to that than it is to half of that. It's going to be between pi over two, and pi. It's gonna be, if we start with this magenta ray, we rotate counterclockwise by three radians, we are gonna get... Actually, it's probably gonna be, it's gonna look something, it's gonna be something like this. But for the sake of this exercise, we have gotten ourselves, once again, into the second quadrant." + }, + { + "Q": "\nDoesn't Koch's snowflake and Mandelbrot's famous shape already address your concept of an \"infinigon/zigfinigon\" as seen around 03:09?", + "A": "She does say something fractally at 03:35 which again Koch and Mandelbrot takcle the fractal field with their well known shapes.", + "video_name": "D2xYjiL8yyE", + "timestamps": [ + 189 + ], + "3min_transcript": "is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0." + }, + { + "Q": "\nAt 3:24, your creating a pentagram, right?", + "A": "If you mean the five-sided shape she ended up with after the Zigfinite star then that s a pentagon. If not, can you explain better what you mean?", + "video_name": "D2xYjiL8yyE", + "timestamps": [ + 204 + ], + "3min_transcript": "is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0." + }, + { + "Q": "\u00e2\u0088\u009a4 is 2. But at about 2:45, it says \u00e2\u0088\u009a2 is 2. How is this possible?\n", + "A": "This is a joke, sometimes called sophistry. In fact, angular line will never be straight, so it s proof that the square root of 2 is 2 does not work =)", + "video_name": "D2xYjiL8yyE", + "timestamps": [ + 165 + ], + "3min_transcript": "is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0." + }, + { + "Q": "I did the drawing she does at about 2:50 and then divided the squares I got into right triangles and squares, and divided those squares into right triangles and squares, and so on... The top half ended up looking like Serpinsky's triangle. I encourage you to try it for yourself. Also, does anyone know why it looks like Serpinsky's triangle?\n", + "A": "She is really good at doodling :P", + "video_name": "D2xYjiL8yyE", + "timestamps": [ + 170 + ], + "3min_transcript": "is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0." + }, + { + "Q": "Why can't you just square root the whole hypotenuse formula? For example, at 1:50, instead of squaring 15, 8, and x, why can't you just use 15+8=x?\n", + "A": "That will not get us the right answer. 15+8=23, but x=17. You have to square both numbers, add them together, and only then take the square root.", + "video_name": "T971zHhZ3S4", + "timestamps": [ + 110 + ], + "3min_transcript": "What I want to do in this video is attempt to find the diameter of this circle right over here. And I encourage you to pause the video and try this out on your own. Well, let's think about what's going on right over here. AB is definitely a diameter of the circle. It's a straight line. It's going through the center of the circle. O is the center of the circle right over here. And so what do we know? Well, we could look at this angle right over here, angle C, and think about it is an inscribed angle. And think about the arc that it intercepts. It intercepts this arc right over here. This arc is exactly half of the circle. Angle C is inscribed. If you take these two sides or the two sides of the angle, it intercepts at A and B, and so it intercepts an arc, this green arc right over here. So the central angle right over here is 180 degrees, and the inscribed angle is going to be half of that. Or another way of thinking about it, it's going to be a right angle. And what that does for us is it tells us that triangle ACB is a right triangle. This is a right triangle, and the diameter is its hypotenuse. So we can just apply the Pythagorean theorem here. 15 squared plus 8 squared-- let me do this in magenta-- is going to be the length of side AB squared. So this side right over here, let me just call that x. That's going to be equal to x squared. So 15 squared, that's 225. 8 squared is 64, plus 64-- I want to do 225 plus 64 is 289 is equal to x squared. And then 289 is 17 squared. And you could try out a few numbers if you're unsure about that. So x is equal to 17. So the diameter of this circle right over here is 17." + }, + { + "Q": "\nat 4:08, why did he flip the left side?", + "A": "He did that to get t up top . He inverted both sides of the equation which is a valid thing to do. Here is why the flip is valid: start with a / b = c / d => ad = bc (multiplied both sides by b, then by d) => d / c = b / a (divided both sides by c, then by a)", + "video_name": "gD7A1LA4jO8", + "timestamps": [ + 248 + ], + "3min_transcript": "" + }, + { + "Q": "At 3:08 and 5:38, how did Sal know what the graphs of the functions would look like?\n", + "A": "Lots of practice... different types of equations create specific graphs. f(x) = x^2 is a quadratic equation. It creates a U-shaped graph. g(x) = x is a linear equation. It creates a line for a graph.", + "video_name": "96uHMcHWD2E", + "timestamps": [ + 188, + 338 + ], + "3min_transcript": "okay, we know the set of all of the valid inputs, that's called the domain, but what about all, the set of all of the outputs that the function could actually produce? And we have a name for that. That is called the range of the function. So the range. The range, and the most typical, there's actually a couple of definitions for range, but the most typical definition for range is \"the set of all possible outputs.\" So you give me, you input something from the domain, it's going to output something, and by definition, because we have outputted it from this function, that thing is going to be in the range, and if we take the set of all of the things that the function could output, that is going to make up the range. So this right over here is the set of all possible, all possible outputs. All possible outputs. So let's make that a little bit more concrete, with an example. So let's say that I have the function f(x) and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here. What this looks like. So the graph of \"f(x) is equal to x squared\" is going to look something like this. So, it's gonna look, it's going to look something like this. I'm obviously hand-drawing it, so it's not perfect. It's gonna be a parabola with a, with a vertex right here at the origin. So this is the graph, this is the graph, \"y is equal to f(x),\" this of course is the x-axis, this of course is the y-axis. So let's think about it, what is the set of all possible outputs? Well in this case, the set of all possible outputs is the set of all possible y's here. Well, we see, y can take on any non-negative value. y could be zero, y could one, y could be pi, y could be e, but y cannot be negative. So the range here is, the range... We could, well we could say it a couple of ways, we could say, \"f(x)\", let me write it this way. \"f(x) is a member of the real numbers\"" + }, + { + "Q": "What are oulpuls at 1:45- 1:55\n", + "A": "Outputs , with the crosses on the t s not drawn well.", + "video_name": "96uHMcHWD2E", + "timestamps": [ + 105, + 115 + ], + "3min_transcript": "- As a little bit of a review, we know that if we have some function, let's call it \"f\". We don't have to call it \"f\", but \"f\" is the letter most typically used for functions, that if I give it an input, a valid input, if I give it a valid input, and I use the variable \"x\" for that valid input, it is going to map that to an output. It is going to map that, or produce, given this x, it's going to product an output that we would call \"f(x).\" And we've already talked a little bit about the notion of a domain. A domain is the set of all of the inputs over which the function is defined. So if this the domain here, if this is the domain here, and I take a value here, and I put that in for x, then the function is going to output an f(x). If I take something that's outside of the domain, let me do that in a different color... If I take something that is outside of the domain and try to input it into this function, the function will say, \"hey, wait wait,\" \"I'm not defined for that thing\" \"that's outside of the domain.\" Now another interesting thing to think about, okay, we know the set of all of the valid inputs, that's called the domain, but what about all, the set of all of the outputs that the function could actually produce? And we have a name for that. That is called the range of the function. So the range. The range, and the most typical, there's actually a couple of definitions for range, but the most typical definition for range is \"the set of all possible outputs.\" So you give me, you input something from the domain, it's going to output something, and by definition, because we have outputted it from this function, that thing is going to be in the range, and if we take the set of all of the things that the function could output, that is going to make up the range. So this right over here is the set of all possible, all possible outputs. All possible outputs. So let's make that a little bit more concrete, with an example. So let's say that I have the function f(x) and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here." + }, + { + "Q": "\nAt 4:35, Is zero a \"non-negative\" number? Is it also non-positive?", + "A": "Yes, 0 is neither positive nor negative. If you were to add 0 to any number, let s say x, it would not increase nor decrease its value.", + "video_name": "96uHMcHWD2E", + "timestamps": [ + 275 + ], + "3min_transcript": "What this looks like. So the graph of \"f(x) is equal to x squared\" is going to look something like this. So, it's gonna look, it's going to look something like this. I'm obviously hand-drawing it, so it's not perfect. It's gonna be a parabola with a, with a vertex right here at the origin. So this is the graph, this is the graph, \"y is equal to f(x),\" this of course is the x-axis, this of course is the y-axis. So let's think about it, what is the set of all possible outputs? Well in this case, the set of all possible outputs is the set of all possible y's here. Well, we see, y can take on any non-negative value. y could be zero, y could one, y could be pi, y could be e, but y cannot be negative. So the range here is, the range... We could, well we could say it a couple of ways, we could say, \"f(x)\", let me write it this way. \"f(x) is a member of the real numbers\" \"or equal to zero.\" We could write it that way, if we wanted to write it in a less mathy notation, we could say that \"f(x) is going to be\" \"greater than or equal to zero.\" f(x) is not going to be negative, so any non-negative number, the set of all non-negative numbers, that is our range. Let's do another example of this, just to make it a little bit, just to make it a little bit, a little bit clearer. Let's say that I had, let's say that I had g(x), let's say I have g(x), I'll do this in white, let's say it's equal to \"x squared over x.\" So we could try to simplify g(x) a little bit, we could say, \"look, if I have x squared\" \"and I divide it by x, that's gonna,\" \"that's the same thing as g(x) being equal to x.\" \"x squared over x\" is x, but we have to be careful. in our domain, x cannot be equal to zero. If x is equal to zero, we get zero over zero, we get indeterminate form. So in order for this function to be the exact same function, we have to put that, 'cause it's not obvious now from the definition, we have to say, \"x cannot be equal to zero.\" So g(x) is equal to x for any x as long as x is not equal to zero. Now these two function definitions are equivalent. And we could even graph it. We could graph it, it's going to look, I'm gonna do a quick and dirty version of this graph. It's gonna look something like, this. It's gonna have a slope of one, but it's gonna have a hole right at zero, 'cause it's not defined at zero. So it's gonna look like this. So the domain here, the domain of g is going to be, \"x is a member of the real numbers\" \"such that x does not equal zero,\" and the range is actually going to be the same thing." + }, + { + "Q": "\nThe Formula Sal derives for m at 13:00min is not the same as the formula he derived in a previous video for m(Proof part 4 @ 3:00min) They seem to be negatives of each other.. Am I missing something?", + "A": "What would happen if you multiplied both the numerator and denominator by -1?", + "video_name": "ualmyZiPs9w", + "timestamps": [ + 780, + 180 + ], + "3min_transcript": "to what we do with regression. The expected value of X times Y, it can be approximated by the sample mean of the products of X and Y. This is going to be the sample mean of X and Y. You take each of your XY associations, take their product, and then take the mean of all of them. So that's going to be the product of X and Y. And then this thing right over here, the expected value of Y that can be approximated by the sample mean of Y, and the expected value of X can be approximated by the sample mean of X. So what can the covariance of two random variables be approximated by? What can it be approximated by? Well this right here is the mean of their product from your sample minus the mean of your sample Y's times the mean And this should start looking familiar. This should look a little bit familiar, because what is this? This was the numerator. This right here is the numerator when we were trying to figure out the slope of the regression line. So when we tried to figure out the slope of the regression line, we had the-- let me just rewrite the formula here just to remind you-- it was literally the mean of the products of each of our data points, or the XY's, minus the mean of Y's times the mean of the X's. All of that over the mean the X squareds. And you could even view it as this, over the mean of the X times the X's. But I could just write the X squareds, over here, minus the mean of X squared. This is how we figured out the slope of our regression line. if we assume in our regression line that the points that we have were a sample from an entire universe of possible points, then you could say that we are approximating the slope of our aggression line. And you might see this little hat notation in a lot of books. I don't want you to be confused. They are saying that you're approximating the population's regression line from a sample of it. Now, this right here-- so everything we've learned right now-- this right here is the covariance, or this is an estimate of the covariance of X and Y. Now what is this over here? Well, I just said, you could rewrite this very easily as-- this bottom part right here-- you could write as the mean of X times X-- that's the same thing as X squared-- minus the mean of X times the mean of X, right? That's what the mean of X squared is. Well, what's this? Well, you could view this as the covariance of X with X." + }, + { + "Q": "\nWhere exactly did that x+3 go whenever it was originally at the top of the fraction at 2:09? Was is just replaced by A and B? If so how & why...?", + "A": "The answer to your question starts at 7:20.", + "video_name": "S-XKGBesRzk", + "timestamps": [ + 129 + ], + "3min_transcript": "Let's see if we can learn a thing or two about partial fraction expansion, or sometimes it's called partial fraction decomposition. The whole idea is to take rational functions-- and a rational function is just a function or expression where it's one expression divided by another-- and to essentially expand them or decompose them into simpler parts. And the first thing you've got to do, before you can even start the actual partial fraction expansion process, is to make sure that the numerator has a lower degree than the denominator. In the situation, the problem, that I've drawn right here, I've written right here, that's not the case. The numerator has the same degree as the denominator. So the first step we want to do to simplify this and to get it to the point where the numerator has a lower degree than the denominator is to do a little bit of algebraic division. And I've done a video on this, but it never hurts to get a review here, so to do that, we divide the denominator into the squared minus 3x minus 40 into x squared minus 2x minus 37. So how many times? You look at the highest degree term, so x squared goes into x squared one time, one times this whole thing is x squared minus 3x minus 40, and now you want to subtract this from that to get the remainder. And see, if I'm subtracting, so I'm going to subtract, and then minus minus is a plus, a plus, and then you can add them. These cancel out. Minus 2x plus 3x, that's x. Minus 37 plus 40, that's plus 3. So this expression up here can be rewritten as-- let me scroll down a little bit-- as 1 plus x plus 3 over x squared This might seem like some type of magic thing I just did, but this is no different than what you did in the fourth or fifth grade, where you learned how to convert improper regular fractions into mixed numbers. Let me just do a little side example here. If I had 13 over 2, and I want to turn it into a mixed number, what you do-- you can probably do this in your head now-- but what you did is, you divide the denominator into the numerator, just like we did over here. 2 goes into 13. We see 2 goes into 13 six times, 6 times 2 is 12, you subtract that from that, you get a remainder of 1. 2 doesn't go into 1, so that's just the remainder. So if you wanted to rewrite this, it would be the number of times the denominator goes into the numerator, that's 6, plus the remainder over the denominator. Plus 6-- plus 1 over 2. And when you did it in elementary school, you would just write 6 1/2, but 6 1/2 is the same thing as 6 plus 1/2." + }, + { + "Q": "\nAt 1:20, is Sal taking the anti-derivative of the derivative of f(x)*g(x) ? So they just cancel out? Like if you had (sqrt(X))^2 and you are just left with X. Is that pretty much the same thing that is going on at that point?", + "A": "Whether they cancel out depends on what f(x) and g(x) happen to be. Integration by parts is quite useful if you need to integrate a complicated function that would be exceedingly difficult to try to integrate altogether.", + "video_name": "dh__n9FVKA0", + "timestamps": [ + 80 + ], + "3min_transcript": "What we're going to do in this video is review the product rule that you probably learned a while ago. And from that, we're going to derive the formula for integration by parts, which could really be viewed as the inverse product rule, integration by parts. So let's say that I start with some function that can be expressed as the product f of x, can be expressed as a product of two other functions, f of x times g of x. Now let's take the derivative of this function, let's apply the derivative operator right over here. And this, once again, just a review of the product rule. It's going to be the derivative of the first function times the second function. So it's going to be f-- no, I'm going to do that blue color-- it's going to be f-- that's not blue-- it's going to be f prime of x times g of x times-- that's not the same color-- times g of x plus the first function times the derivative of the second, of the second. This is all a review right over here. The derivative of the first times the second function plus the first function times the derivative of the second function. Now, let's take the antiderivative of both sides of this equation. Well if I take the antiderivative of what I have here on the left, I get f of x times g of x. We won't think about the constant for now. We can ignore that for now. And that's going to be equal to-- well what's the antiderivative of this? This is going to be the antiderivative of f prime of x times g of x dx plus the antiderivative of f of x g prime of x dx. Now, what I want to do is I'm going And to solve for that, I just have to subtract this business. I just have to subtract this business from both sides. And then if I subtract that from both sides, I'm left with f of x times g of x minus this, minus the antiderivative of f prime of x g of x-- let me do that in a pink color-- g of x dx is equal to what I wanted to solve for, is equal to the antiderivative of f of x g prime of x dx. And to make it a little bit clearer, let me swap sides here. So let me copy and paste this. So let me copy and then paste it. There you go." + }, + { + "Q": "Isnt integral of f'(x) is f(x) just like at 2:42\n", + "A": "yes it is\u00e2\u0080\u00a6as d/dx of f(x) Is f (x)", + "video_name": "dh__n9FVKA0", + "timestamps": [ + 162 + ], + "3min_transcript": "of the second. This is all a review right over here. The derivative of the first times the second function plus the first function times the derivative of the second function. Now, let's take the antiderivative of both sides of this equation. Well if I take the antiderivative of what I have here on the left, I get f of x times g of x. We won't think about the constant for now. We can ignore that for now. And that's going to be equal to-- well what's the antiderivative of this? This is going to be the antiderivative of f prime of x times g of x dx plus the antiderivative of f of x g prime of x dx. Now, what I want to do is I'm going And to solve for that, I just have to subtract this business. I just have to subtract this business from both sides. And then if I subtract that from both sides, I'm left with f of x times g of x minus this, minus the antiderivative of f prime of x g of x-- let me do that in a pink color-- g of x dx is equal to what I wanted to solve for, is equal to the antiderivative of f of x g prime of x dx. And to make it a little bit clearer, let me swap sides here. So let me copy and paste this. So let me copy and then paste it. There you go. So let me copy and paste it. So I'm just switching the sides, just to give it in a form that you might be more used to seeing in a calculus book. So this is essentially the formula for integration I will square it off. You'll often see it squared off in a traditional textbook. So I will do the same. So this right over here tells us that if we have an integral or an antiderivative of the form f of x times the derivative of some other function, we can apply this right over here. And you might say, well this doesn't seem that useful. First I have to identify a function that's like this. And then still I have an integral in it. But what we'll see in the next video is that this can actually simplify a whole bunch of things that you're trying to take the antiderivative of." + }, + { + "Q": "\nAt 0:09, Sal said that 35 doesn't goes into 6. Did he purposely choose a number with a remainder?", + "A": "He likely did. He s trying to demonstrate decimals. Since decimals are fractions, there needs to be a remainder to create a decimal value.", + "video_name": "xUDlKV8lJbM", + "timestamps": [ + 9 + ], + "3min_transcript": "Let's take 63 and divide it by 35. So the first thing that we might say is, OK, well, 35 doesn't go into 6. It does go into 63. It goes into 63 one time, because 2 times 35 is 70, so that's too big. So it goes one time. So let me write that. 1 times 35 is 35. And then if we were to subtract and we can regroup up here, we can take a 10 from the 60, so it becomes a 50, give that 10 to the 3, so it becomes a 13. 13 minus 5 is 8. 5 minus 3 is 2. So you could just say, hey, 63 divided by 35-- You could say 63 divided by 35 is equal to 1 remainder 28. But this isn't so satisfying. We know that the real answer is going to be one point something, something, something. So what I want to do is keep dividing. I want to divide this thing completely and see what type And to do that, I essentially have to add a decimal here and then just keep bringing down decimal places to the right of the decimal. So 63 is the exact same thing as 63.0, and I could add as many zeroes as I might want to add here. So what we could do is we just make sure that this decimals right over there, and we can now bring down a zero from the tenths place right over here. And you bring down that zero, and now we ask ourselves, how many times does 35 go into 280? And, as always, this is a bit of an art when you're dividing a two-digit number into a three-digit number. So let's see, it's definitely going to be-- if I were to say-- so 40 goes into 280 seven times. 30 goes into 280 about nine times. It's going to be between 7 and 9, so let's try 8. So, let's see what 35 times 8 is. 35 times 8. 5 times 8 is 40, 3 times 8 is 24, plus 4 is 28. So 35 goes into 280 exactly eight times. 8 times 5, we already figured it out. 8 times 35 is exactly 280, and we don't have any remainder now, so we don't have to bring down any more of these zeroes. So now we know exactly that 63 divided by 35 is equal to exactly 1.8." + }, + { + "Q": "02:29 Why did Sal use K instead of Z?\n", + "A": "I quickly checked the timestamp out. I assume he did so as you can use any letter that isn t already used in this situation. No special reason.", + "video_name": "lHdlHTsXbZg", + "timestamps": [ + 149 + ], + "3min_transcript": "So, once again let's think of yourself as some type of ancient philosopher/mathematician, who is trying to extend mathematics as much as possible and try to make sure that you're not being lazy and leaving things undefined, when you might be able to define them. Whenever you start extending mathematics, especially in the realm of multiplication and division, there are few things that you hold dear to. You feel that if you define some type of division operation, that needs to be undone by multiplication; this is close to your heart. So you assume... You want to assume... You would like to assume that any type of division operation, if you start with some number and if you divide with a number over which... - division by that number is defined - so when I divide by some number and then multiply by that same number that this should get me this original number right over here, this should give me x right over here. And this happens when we just multiply and divide with regular numbers. If I get 3 divided by 2 times 2, that's gonna get me 3. If I say that's going to get me 10. The other things that I want to assume... - and this is very close to my heart - I feel that any type of definitions I make have to be constant with the idea x*0 has to be 0 or any x. So these are close to my heart. I wanna extend mathematics. These two things are things that cannot be contradicted, cannot be untrue. Now, that out of the way. You wanna start exploring the divide-by-0 question. So the first thing that you say: \"Well, let me just try to define it.\" So let's start, let's assume that I have, so this is... So let's make a further assumption... that x is some non-zero number. x divided by 0 should be, how I should divide it, let's just assume there is define, and then come up with any results that there might be, there might be a resolve for. So let's say that x divided by 0 is equal to k. Well, if this is true and if we are defining what it means to divide by zero, then we are assuming that if we multiply by zero, we'll get our original number right over here. This is something that we are not willing to contradict. So let's see what happens: x divided by 0 is equal to k. On the left hand side we have a divide by zero and than multiplied by zero. Well then if two things are equal, if I do something to one thing inorder for them to stay equal, I have to do it to the other thing. This has to be equal to that. I have to multiply the left hand AND the right hand side by zero." + }, + { + "Q": "\nAt 1:59 Sal says x can not be equal to zero. Why does x have to be a non-zero number? If it could be zero, couldn't 0/0 = 0? Is there any reasoning as to why x can not equal zero, or is it just to enforce other reasoning as to why 0/0 is impossible?", + "A": "At that point it s just the boundaries of the current proof he was laying out. Later on he does another proof where 0/0=0, however that comes from multiplying the constant on the right side of the equation by 0. That constant can be any number between infinity and negative infinity, so it is indeterminate.", + "video_name": "lHdlHTsXbZg", + "timestamps": [ + 119 + ], + "3min_transcript": "So, once again let's think of yourself as some type of ancient philosopher/mathematician, who is trying to extend mathematics as much as possible and try to make sure that you're not being lazy and leaving things undefined, when you might be able to define them. Whenever you start extending mathematics, especially in the realm of multiplication and division, there are few things that you hold dear to. You feel that if you define some type of division operation, that needs to be undone by multiplication; this is close to your heart. So you assume... You want to assume... You would like to assume that any type of division operation, if you start with some number and if you divide with a number over which... - division by that number is defined - so when I divide by some number and then multiply by that same number that this should get me this original number right over here, this should give me x right over here. And this happens when we just multiply and divide with regular numbers. If I get 3 divided by 2 times 2, that's gonna get me 3. If I say that's going to get me 10. The other things that I want to assume... - and this is very close to my heart - I feel that any type of definitions I make have to be constant with the idea x*0 has to be 0 or any x. So these are close to my heart. I wanna extend mathematics. These two things are things that cannot be contradicted, cannot be untrue. Now, that out of the way. You wanna start exploring the divide-by-0 question. So the first thing that you say: \"Well, let me just try to define it.\" So let's start, let's assume that I have, so this is... So let's make a further assumption... that x is some non-zero number. x divided by 0 should be, how I should divide it, let's just assume there is define, and then come up with any results that there might be, there might be a resolve for. So let's say that x divided by 0 is equal to k. Well, if this is true and if we are defining what it means to divide by zero, then we are assuming that if we multiply by zero, we'll get our original number right over here. This is something that we are not willing to contradict. So let's see what happens: x divided by 0 is equal to k. On the left hand side we have a divide by zero and than multiplied by zero. Well then if two things are equal, if I do something to one thing inorder for them to stay equal, I have to do it to the other thing. This has to be equal to that. I have to multiply the left hand AND the right hand side by zero." + }, + { + "Q": "At @6:19 he says that he wanted to come up with a number for k, but k is already defined by 0/0 so shouldn't there be no problem since he already defined k?\n", + "A": "It s because he wants to come up with a single number for k, but he can t seem to find one. So he still technically calls it indeterminate.", + "video_name": "lHdlHTsXbZg", + "timestamps": [ + 379 + ], + "3min_transcript": "Now that out of the way... OK... This was a situation when x does not equal zero. But what about when x DOES equal zero. So let's think about that a little bit. And once again, I will try to define it. So I will assume... that 0 divided by 0 is equal to some number. Well once again, so let's say it is equal to k again. And so, once again... we are trying to do the same logic, so we'll write 0/0 is equal to k. Actually, let me colourcode these zeros. This will be a magenta zero and this is a blue zero right over here. And once again, I am not willing to give up the idea that if I start with a number x, I divide it by something over which division is defined, and then I multiply by that something, I should get my original x again. for the division. So what I am gonna do - I am gonna multiply the left-hand side times 0 and by this property that I am not willing to give up, the left-hand side should simplify to this magenta zero. It should simplify to this over here. But once again, anything I do to once side of the equation, inorder for the equation to hold true, I need to the other side of the equation. And these two were equal beforehand. Any operation I do to this inorder for it to still be equal , I need to do to that. So let me multiply the right-hand side by zero. So the left I get 0, I just get this magenta 0, and on the right I could just write the zero here, but I won't simplify it. I get k times 0. Well, this I see right over here... This actually is not a contradiction. This actually is true for any k, This is one of the core assumptions that I've made in my mathematics True for any k. It's not a contradiction. But the problem here is I wanna come up with the k, I'd like a resolve for a k. It would not be nice if this turned out to be 0. if this turned out to be one, or if this turned out to be negative one But now I see, given the assumptions right here this could be ANY... this could be absolutely any k I cannot determine what k this should be This could be a hundred thousand, this could be 75, it could be anything true for any k I cannot determine what k this should be and that's why when you get a little bit more nuance in early math people will say, well 0 divided by 0, well we don't know what that's gonna be there's no consistent answer there so we're just going to call it undefined" + }, + { + "Q": "\nAt 5:05, why isn't the antiderivative of dw just w? The dw completely disappears, which confuses me.", + "A": "The indefinite integral of only dw is just w (+C, of course). However, he didn t just take the indefinite integral of dw. He took the indefinite integral of (e^w)dw. The derivative of e^w+C is e^w, so the indefinite integral of (e^w)dw is e^w+C. Because there s a -1/5 in front of the integral, we must multiply the whole expression by -1/5. This results in -1/5(e^w)+C (-1/5 times a constant is still a constant). I hope this explains what Sal did!", + "video_name": "ShpI3gPgLBA", + "timestamps": [ + 305 + ], + "3min_transcript": "So let's try to get it in terms of the form of e to the something and not e to the negative something. So let's set. And I'm running out of colors here. Let's set w as equal to negative u. In that case, then dw, derivative of w with respect to u, is negative 1. Or if we were to write that statement in differential form, dw is equal to du times negative 1 is negative du. So this right over here would be our w. And do we have a dw here? Well, we have just a du. We don't have a negative du there. But we can create a negative du by multiplying this inside by a negative 1, but then also by multiplying the outside by a negative 1. Negative 1 times negative 1 is positive 1. We haven't changed the value. We have to do both of these in order for it to make sense. Or I could do it like this. So negative 1 over here and a negative 1 right over there. that's the same thing as negative du. This is this right over here. And so we can rewrite our integral. It's going to be equal to now it's going to be negative 1/5. Trying to use the colors as best as I can. Negative 1/5 times the indefinite integral of e to the-- well, instead of negative u, we can write w-- e to the w. And instead of du times negative 1, or negative du, we can write dw. Now this simplifies things a good bit. We know what the antiderivative of this is in terms of w. This is going to be equal to negative 1/5 e to the w. so I'll just do a plus c. And now we just have to do all of our unsubstituting. So we know that w is equal to negative u. So we could write that. So this is equal to negative 1/5. I want to stay true to my colors. Negative 1/5 e to the negative u, that's what w is equal to, plus c. But we're still not done unsubstituting. We know that u is equal to sine of 5x. u is equal to sine of 5x, so we can write this as being equal to negative 1/5 times e to the negative u, which is negative u is sine of 5x." + }, + { + "Q": "@ ~2:20-2:33, we have (1/5)\u00e2\u0088\u00ab(1/e^u)du,\n\nwhy can't we use w substitution on the e^u?\n", + "A": "I tried my, and got a different answer, or maybe just another form of the answer? Thanks a bunch for your help :)", + "video_name": "ShpI3gPgLBA", + "timestamps": [ + 140, + 153 + ], + "3min_transcript": "Let's see if we can take the integral of cosine of 5x over e to the sine of 5x dx. And there's a crow squawking outside of my window, so I'll try to stay focused. So let's think about whether u-substitution might be appropriate. Your first temptation might have said, hey, maybe we let u equal sine of 5x. And if u is equal to sine of 5x, we have something that's pretty close to du up here. Let's verify that. So du could be equal to-- so du dx, derivative of u with respect to x. Well, we just use the chain rule. Derivative of 5x is 5. Times the derivative of sine of 5x with respect to 5x, that's just going to be cosine of 5x. If we want to write this into differential form, which is useful when we do our u-substitution, we could say that du is equal to 5 cosine 5x. And we look over here, we don't have quite du there. We have just cosine of 5x dx. So when you look over here, you have a cosine of 5x dx, but we don't have a 5 cosine of 5x dx. But we know how to solve that. We can multiply by 5 and divide by 5. 1/5 times 5 is just going to be 1, so we haven't changed the value of the expression. When we do it this way, we see pretty clearly we have our u and we have our du. Our du is 5. Let me circle that. Let me do that in that blue color-- is 5 cosine of 5x dx. So we can rewrite this entire expression as-- do that 1/5 in purple. This is going to be equal to 1/5-- I hope you don't hear that crow outside. He's getting quite obnoxious. 1/5 times the integral of all the stuff in blue is my du, So how do we take the antiderivative of this? Well, you might be tempted to well, what would you do here? Well, we're still not quite ready to simply take the antiderivative here. If I were to rewrite this, I could rewrite this as this is equal to 1/5 times the integral of e to the negative u du. And so what might jump out at you is maybe we do another substitution. We've already used the letter u, so maybe we'll use w. We'll do some w-substitution. And you might be able to do this in your head, but we'll do w-substitution just to make it a little bit clearer. This would have been really useful if this was just e to the u because we know the antiderivative of e to the u" + }, + { + "Q": "6:01 What is the principle root?\n", + "A": "When you take a square root, you have two branches; a positive branch, and a negative branch. For instance, x\u00c2\u00b2 = 9 implies that x = \u00c2\u00b1 3 because (-3)\u00c2\u00b2 = 9 and 3\u00c2\u00b2 = 9 also. The principle square root is the positive branch.", + "video_name": "q7eF5Ci944U", + "timestamps": [ + 361 + ], + "3min_transcript": "We already know that OA is equal to OC. And we also know that OM is equal to itself. OM is clearly equal to OM. It's equal to itself. And we also know from the Pythagorean theorem that AM-- do this in a new color. We know that AM squared plus OM squared is equal to OA squared. The length of the two legs squared summed up is equal to the length of the hypotenuse squared. So we know that for the left triangle right over here, for AMO. And we can set up the same relationship for CMO. We know in CMO that-- and I'm going to try to do it corresponding-- that CM squared plus OM Now we know a few things. We know that OA is equal to OC. So, for example, right over here, where we have OA squared, we could replace this with OC. And then you can already see where this is going. You can already see that CM and AM are going to be the same. But if you want to do it a little bit more formally, you can subtract OM squared from both sides of this equation, and you'd get AM squared is equal to OC squared-- I've replaced this with OC squared-- minus OM squared. So that's on the left-hand side right over here. And then on the right-hand side, if we subtract OM squared from both sides, we have CM squared. CM squared is equal to OC squared minus OM squared. And then we could take the principal root care about the positive root because we don't want to have negative distances. So if you take the principal root of both sides, this becomes AM is equal to the principal root of that. And we also get that CM is equal to the principal root of that. Well, these two quantities are the same. So AM must be equal to CM. They both equal to this quantity right here. So AM is equal to CM, and it might be a bisector. And this is really common sense. If you know that two sides of two different right triangles are going to be congruent to each other, you can always use the Pythagorean theorem to get the third side. And that third side is uniquely constrained by the length of the other two sides, because it's a right triangle. And so these are all ways of really getting at the same thing. But now we can feel pretty good about the fact that if OD is perpendicular here, then it's definitely going to bisect this chord." + }, + { + "Q": "At 1:15 did anyone notice that the two smaller right triangles make a bigger right triangle\n", + "A": "Doesn t work as a proof.", + "video_name": "q7eF5Ci944U", + "timestamps": [ + 75 + ], + "3min_transcript": "In a previous video, we've already shown that if we have some circle centered at O right over here, and that if OD is a radius, and it's a radius that bisects chord AC-- so bisects means that it kind of splits it in two, that AB is equal to BC-- we've proven in a previous video that OD will be perpendicular to AC. So we've proven that we can assume that is perpendicular. And that video, if you want to look it up, and you might want to prove it yourself just out of interest, but the video-- if you do a search on Khan Academy for radius is perpendicular to a chord, you'll hopefully find the proof of that. What I want to do in this video is go the other way. If we know that OD is a radius and that it is perpendicular to chord AC, what I want to do in this video is prove that it's bisecting it. So we're not assuming that it's bisecting it in this video right over here. We're just assuming that it's perpendicular. So we're essentially going to go the other way. Here, we started with the fact that it bisected it, and we established that they were perpendicular. That was in the previous video. Now we're going to start with the assumption that they're perpendicular and then prove that they bisect. we'll set up some triangles here since we know a lot about triangles now. And we'll set up the triangles by drawing two more radii, radius OC and radius OA. And that's useful for us because we know that they're both radii for the same circle. So they have to be the same length. The radius doesn't change on a circle. So those two things are the same length. And you might already see where this is going. Let me label this point here. Let me call this M because we're hoping that ends up being the midpoint of AC. Triangle AMO is a right triangle. This is its hypotenuse. AO is its hypotenuse. Triangle OMC is a right triangle, and this is its hypotenuse right over there. And so already showed that the hypotenuses have the same length, and both of these right triangles share segment or side OM. So OM is clearly equal to itself. And in a previous video, not the same video where we explained this thing. I think the video is called \"More on why SSA is not a postulate.\" In that video, we say that SSA is not a postulate. So SSA, not a congruency postulate. But we did establish in that video that RSH is a congruency postulate. And RSH tells us that if we have a right triangle-- that's where the R comes from-- if we have a right triangle, and we have one of the sides are congruent, and the hypotenuse is congruent, then we have two congruent triangles. And if you look at this right over here, we have two right triangles. AMO is a right triangle. CMO is a right triangle. They have one leg that's congruent, right over here, MO, and then both of their hypotenuses are congruent to each other. So, by RSH, we know that triangle AMO" + }, + { + "Q": "\nAt 3:00, couldnt you also say that triangle AMO is congruent to triangle CMO by aas", + "A": "If you did that, you d either need to prove that angle A is congruent to angle C or that angle AOM is congruent to angle COM. I don t see any obvious reason to believe that either of those is true, or at least nothing that is as obvious as proving that AO is congruent to CO or that MO is congruent to itself.", + "video_name": "q7eF5Ci944U", + "timestamps": [ + 180 + ], + "3min_transcript": "we'll set up some triangles here since we know a lot about triangles now. And we'll set up the triangles by drawing two more radii, radius OC and radius OA. And that's useful for us because we know that they're both radii for the same circle. So they have to be the same length. The radius doesn't change on a circle. So those two things are the same length. And you might already see where this is going. Let me label this point here. Let me call this M because we're hoping that ends up being the midpoint of AC. Triangle AMO is a right triangle. This is its hypotenuse. AO is its hypotenuse. Triangle OMC is a right triangle, and this is its hypotenuse right over there. And so already showed that the hypotenuses have the same length, and both of these right triangles share segment or side OM. So OM is clearly equal to itself. And in a previous video, not the same video where we explained this thing. I think the video is called \"More on why SSA is not a postulate.\" In that video, we say that SSA is not a postulate. So SSA, not a congruency postulate. But we did establish in that video that RSH is a congruency postulate. And RSH tells us that if we have a right triangle-- that's where the R comes from-- if we have a right triangle, and we have one of the sides are congruent, and the hypotenuse is congruent, then we have two congruent triangles. And if you look at this right over here, we have two right triangles. AMO is a right triangle. CMO is a right triangle. They have one leg that's congruent, right over here, MO, and then both of their hypotenuses are congruent to each other. So, by RSH, we know that triangle AMO And so if we know that they're congruent, then their corresponding sides have to be congruent. So based on that, we then know that AM is a corresponding side. Let me do that in a different color. AM is a corresponding side to MC. So we know that AM must be equal to MC because they're corresponding sides. These are corresponding sides. So congruency implies that these are equal. And if those are equal, then we know that OD is bisecting AC. So we've established what we need to do. Another way that we could have proven it without RSH, is just straight up with the Pythagorean theorem. We already know, just by setting up these two radii right over here, we know that OA-- so we draw a little line here." + }, + { + "Q": "\nAt 5:00 and on, I still don't understand how you can get a pentagon by slicing it through once.", + "A": "And you would need to have a blade that is bigger than the sides of the cube so there would be parts where the blade can exit the cube before you finish cutting.", + "video_name": "aSokFEpoJFM", + "timestamps": [ + 300 + ], + "3min_transcript": "but I think you get the idea - this would be a triangle. There's different types of triangles that you can construct. You could construct an equilateral triangle, so as long as this cut is the same length as this cut right over here, is the same length as that upper length, or the length that intersects on this space of the cube, that's gonna be an equilateral triangle. If you pushed this point up more, actually I'd do that in a different color, you were going to have an isosceles triangle. If you were to bring this point really, really close, like here, you would approach having a right angle, but it wouldn't be quite a right angle: you'd still have these angles who'd still be less than 90\u00ba, you can approach 90\u00ba. So you can't quite have an exactly a right angle, and so since you can't get to 90\u00ba, sure enough you can't get near to 91\u00ba, so actually you're not gonna be able to do But you can do an equilateral, you can do an isosceles, you can do scalene triangles. I guess you could say you could do the different types of acute triangles. But now let's do some really interesting things: Can you get a pentagon by slicing a cube with a plane? And I really want you to pause the video and think about it here, because that's such a fun thing. Think about it: How can you get a pentagon by slicing a cube with a plane? All right, so here I go, this is how you can get a pentagon by slicing a cube with a plane: Imagine slicing the top - we'll do it a little bit different - so imagine slicing the top, right over there, like this, Imagine slicing this backside, like that, this back side that you can see, quite like that, now you slice this side, right over here, like this, This could be, and alike I'm gonna draw the plane - yet maybe it won't be so obvious if I try to draw the plane - but you get the actual idea: if I slice this, the right angle (not any right angle - 90\u00ba - but 'the right angle' - the proper angle. Actually I shouldn't use 'right angle', that would confuse everything.) If I slice it in the proper angle, that I'm doing with the intersection of my plane then my cube is going to be this pentagon, right over here. Now let's up the stakes something, let's up the stakes even more! What about a hexagon? Can I slice a cube in a way, with a 2D plane, to get the intersection of the plane on the cube being a hexagon? As you could imagine, I wouldn't have asked you that question unless I could. So let's see if we can do it. So if we slice this, right over there, if we slice this bottom piece," + }, + { + "Q": "at 5:35 how do you make that pentigon in one slice\n", + "A": "He uses different cuts to obtain the shape of a pentagon, the sides can be expanded to show a plane cutting into the cube. think of the solid lines as lines you can see, and think of the dotted lines as you cannot see, the front two solid lines on the pentagon you can see are lining up with the faces of the cube and the plane is tilted at an angle", + "video_name": "aSokFEpoJFM", + "timestamps": [ + 335 + ], + "3min_transcript": "But you can do an equilateral, you can do an isosceles, you can do scalene triangles. I guess you could say you could do the different types of acute triangles. But now let's do some really interesting things: Can you get a pentagon by slicing a cube with a plane? And I really want you to pause the video and think about it here, because that's such a fun thing. Think about it: How can you get a pentagon by slicing a cube with a plane? All right, so here I go, this is how you can get a pentagon by slicing a cube with a plane: Imagine slicing the top - we'll do it a little bit different - so imagine slicing the top, right over there, like this, Imagine slicing this backside, like that, this back side that you can see, quite like that, now you slice this side, right over here, like this, This could be, and alike I'm gonna draw the plane - yet maybe it won't be so obvious if I try to draw the plane - but you get the actual idea: if I slice this, the right angle (not any right angle - 90\u00ba - but 'the right angle' - the proper angle. Actually I shouldn't use 'right angle', that would confuse everything.) If I slice it in the proper angle, that I'm doing with the intersection of my plane then my cube is going to be this pentagon, right over here. Now let's up the stakes something, let's up the stakes even more! What about a hexagon? Can I slice a cube in a way, with a 2D plane, to get the intersection of the plane on the cube being a hexagon? As you could imagine, I wouldn't have asked you that question unless I could. So let's see if we can do it. So if we slice this, right over there, if we slice this bottom piece, like that, and then you slice the side that we can see right over there, (This side, I could have made it much straighter) So hopefully you get the idea! I can slice this cube so that I can actually get a hexagon. So, hopefully, this gives you a better appreciation for what you could actually do with a cube, especially if you're busy slicing it with large planar planes - or large planar blades, in some way - There's actually more to a cube that you might have imagined in the past. When we think about it, there's six sides to a cube, and so it's six surfaces to a cube, so you can cut as many as six of the surfaces when you intersect it with a plane, and every time you cut into one of those surfaces" + }, + { + "Q": "\nSal said at 2:34 that the range is limited to the first and fourth Quadrants. Why wouldn't 2.80 radians be excluded from this set?\nPerhaps he misspoke, because all positive values for y, (sin), would actually be in the first or second quadrants, correct?", + "A": "Sal didn t misspeak. If 2.80 were in the range of arcsine, then the arcsine function would be multivalued. If we allowed values in the first and second quadrants, there would be two values that every input could map to; one in the first quadrant and one in the second.", + "video_name": "NC7iWEQ9Kug", + "timestamps": [ + 154 + ], + "3min_transcript": "if we do this in a color you can see, along the positive X axis. Then the other side, so let's see, this is our angle right over here. Let's say that's some angle theta. The sine of this angle is going to be the Y value of where this ray intersects the unit circle. This right over here, that is going to be sine of theta. With that review out of the way, let's think about what X values, and we're assuming we're dealing in radians. What X values when if I take the sine of it are going to give me 1/3? When does Y equal 1/3 along the unit circle? That's 2/3, 1/3 right over here. We see it equals 1/3 exactly two places, here and here. There's two angles where, or at least two, if we just take one or two on each pass of the unit circle. Then we can keep adding multiples of two pi to get as many as we want. We see just on the unit circle we could have this angle. Or we could go all the way around to that angle right over there. Then we could add any multiple of two pi to those angles to get other angles that would also work where if I took the sine of them I would get 1/3. Now let's think about what these are. Here we can take our calculator out, and we could take the inverse sine of 1/3. Let's do that. The inverse sine of 1 over 3. We have to remember what the range of the inverse sine function is. It's going to give us a value between negative pi over 2 and pi over 2, so a value that sticks us in either the first or the fourth quadrant if we're thinking about the unit circle right over here. We see that gave us zero point, if we round to the nearest hundredth, 34. Essentially they've given us this value. They've given us 0.34. That's this angle right over here. Well, it's a positive value. It's greater than zero, but it's less than pi over two. Pi is 3.14, so pi over two is going to be 1.57 and we can go on and on and on so this right over here is 0.34 radians. But what would this thing over here be? It's going to be whatever, if we go to the negative X axis and we subtract 0.34 so we subtracted 0.34. This is 0.34 We're going to get to this angle. It's going to be if we take pi minus the previous answer, it gets us we round to the nearest hundredth, is 2.8 radians. This is 0.34 radians, and then this one, let me do it in this purple color, this one right over here if we were to go all the way around, it's pi minus 0.34 which is 2.80 radians rounding to the nearest hundredth. Now that's not all of the values. We can add multiples of two pi to each of these." + }, + { + "Q": "\nAt 0:42, when Mr. Khan says \"nesting\", is he trying to say a function within a function? This looks almost like combining, but I know it really isn't. So what exactly is composing, and what is the main difference between composing and combining?", + "A": "Composing function- applying one function to the results of another...In other words he s replacing x as the results of another function (eg. replace f(x) with f(g(2)), which is also equal to f(-3)", + "video_name": "wUNWjd4bMmw", + "timestamps": [ + 42 + ], + "3min_transcript": "Voiceover:So we have three different function definitions here. This is F of X in blue, here we map between different values of T and what G of T would be. So you could use this as a definition of G of T. And here we map from X to H of X. So for example, when X is equal to three, H of X is equal to zero. When X is equal to one, H of X is equal to two. And actually let me number this one, two, three, just like that. Now what I want to do in this video is introduce you to the idea of composing functions. Now what does it mean to compose functions? Well that means to build up a function by composing one function of other functions or I guess you could think of nesting them. What do I mean by that? Well, let's think about what it means to evaluate F of, not X, but we're going to evaluate F of, actually let's just start with a little warm-up. Let's evaluate F of G of two. and I encourage you to pause this video and think about it on your own. Well it seems kind of daunting at first, if you're not very familiar with the notation, but we just have to remember what a function is. A function is just a mapping from one set of numbers to another. So for example, when we're saying G of two, that means take the number two, input it into the function G and then you're going to get an output which we are going to call G of two. Now we're going to use that output, G of two, and then input it into the function F. So we're going to input it into the function F, and what we're going to get is F of the thing that we inputted, F of G of two. So let's just take it step by step. What is G of two? Well when T is equal to two, G of two is negative three. Well, I'm going to get negative three squared minus one, which is nine minus one which is going to be equal to eight. So this right over here is equal to eight. F of G of two is going to be equal to eight. Now, what would, using this same exact logic, what would F of H of two be? And once again, I encourage you to pause the video and think about it on your own. Well let's think about it this way, instead of doing it using this little diagram, here everywhere you see the input is X, whatever the input is" + }, + { + "Q": "\nAt 4:39\nlog\u00e2\u0082\u0083(27x) as log\u00e2\u0082\u0083(27)+log_3(x), which is simplified as 3+log\u00e2\u0082\u0083(x)\nI mean why is that? cause log\u00e2\u0082\u0083(27) suppose to be log\u00e2\u0082\u0083(3^3), then =3 log\u00e2\u0082\u00833", + "A": "Your work is ok so far, but is incomplere as 3 log\u00e2\u0082\u00833 can be simplified: `3 log\u00e2\u0082\u00833 = 3*1 = 3. Remember, 3^1 = 3. So, log\u00e2\u0082\u00833 = 1. Hope this helps.", + "video_name": "pkGrXzakRFs", + "timestamps": [ + 279 + ], + "3min_transcript": "I'm writing it as an exponential function or exponential equation, instead of a logarithmic equation. So b to the zth power is equal to c. This is the same statement or the same truth said in a different way. And this is the same truth said in a different way. Well, if we know that a is equal to this, is equal to b to the y, and c is equal to bz, then we can write b to the x power is equal to b to the y power-- that's what a is. We know that already-- times b to the z power. And we know from our exponent properties that if we take b to the y times b to the z, this is the same thing as b to the-- I'll do it in a neutral color-- b to the y plus z power. And so if b to the y plus z power is the same thing as b to the x power, that tells us that x must be equal to y plus z. If this is confusing to you, don't worry about it too much. The important thing, or at least the first important thing, is that you know how to apply it. And then you can think about this a little bit more, and you can even try it out with some numbers. You just have to realize that logarithms are really just exponents. I know when people first would tell me that, I was like, well, what does that mean? But when you evaluate a logarithm, you're getting an exponent that you would have to raise b to to get to a times c. But let's just apply this property right over here. So if we apply it to this one, we know that log base 3 of 27 times x-- I'll write it that way-- is equal to log base 3 of 27 plus log base 3 of x. This tells us, what power do I have to raise 3 to to get to 27? You could view it as this way. 3 to the ? is equal to 27. Well, 3 to the third power is equal to 27. 3 times 3 is 9 times 3 is 27. So this right over here evaluates to 3. So if we were to simplify-- or I guess I wouldn't even call it simplifying it. I would just call it expanding it out or using this property, because we now have two terms where we started off Actually, if we started with this, I'd say that this is the more simple version of it. But when we rewrite it, this first term becomes 3. And then we're left with plus log base 3 of x. So this is just an alternate way of writing this original statement, log base 3 of 27x. So once again, not clear that this is simpler than this right over here. It's just another way of writing it using logarithm properties." + }, + { + "Q": "\nAt 4:11, how does one know that the arc that subtends 'theta' is equal to r?", + "A": "Sal is just defining what a radian is. It is simply the measure of a central angle that subtends an arc equal in length to the radius.", + "video_name": "EnwWxMZVBeg", + "timestamps": [ + 251 + ], + "3min_transcript": "What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta. also the same length as the radius of the circle. So this arc is also of length r. So given that, if you were defining a new type of angle measurement, and you wanted to call it a radian, which is very close to a radius, how many radians would you define this angle to be? Well, the most obvious one, if you kind of view a radian as another way of saying radiuses, or I guess radii. Well, you say, look, this is subtended by an arc of one radius. So why don't we call this right over here one radian, which is exactly how a radian is defined. When you have a circle, and you have an angle of one radian, the arc that subtends it is exactly one radius long. Which you can imagine might be a little bit useful as we start to interpret more and more types of circles." + }, + { + "Q": "\nAt 3:13, what does he mean by \"theta\"? I can't seem to understand what it means.", + "A": "Theta is the eighth letter of the greek alphabet. Greek letters are commonly used as variables in higher math.", + "video_name": "EnwWxMZVBeg", + "timestamps": [ + 193 + ], + "3min_transcript": "And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta." + }, + { + "Q": "\nDoes theta at 3:14 mean anything it stands out in a weird way", + "A": "Not really, it s just used a lot in geometry and trigonometry to indicate angles for whatever reason. It s no different than if you d call it angle x or angle z.", + "video_name": "EnwWxMZVBeg", + "timestamps": [ + 194 + ], + "3min_transcript": "And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta." + }, + { + "Q": "\nAt 01:14 sal asks what are the theories. What I guessed was I think that since our earth almost takes 360 days around the sun to complete one revolution thats why we said one complete revolution or rotation is 360 degress. Am iCorrect?", + "A": "Woohoo I placed my guess and it was correct... It was just a guess!", + "video_name": "EnwWxMZVBeg", + "timestamps": [ + 74 + ], + "3min_transcript": "You are by now probably used to the idea of measuring angles in degrees. We use it in everyday language. We've done some examples on this playlist where if you had an angle like that, you might call that a 30-degree angle. If you have an angle like this, you could call that a 90-degree angle. And we'd often use this symbol, just like that. If you were to go 180 degrees, you'd essentially form a straight line. Let me make these proper angles. If you go 360 degrees, you have essentially done one full rotation. And if you watch figure skating on the Olympics, and someone does a rotation, they'll say, oh, they did a 360. Or especially in some skateboarding competitions, and things like that. But the one thing to realize-- and it might not be obvious right from the get-go-- but this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles. And if you think about it, you'll And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already" + }, + { + "Q": "\nat 3:22 he calls the angle theta... what does that mean? does it represent a certain angle measure, position, etc?", + "A": "That is a Greek letter named Theta. In mathematics and science disciplines, Theta is often used as the variable representing an angle measurement. You will see it very often used this way.", + "video_name": "EnwWxMZVBeg", + "timestamps": [ + 202 + ], + "3min_transcript": "And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta." + }, + { + "Q": "\nAt 3:17 what does theta mean? Is it just a name for any angle, or does it have to be a specific angle to be \"theta\"?", + "A": "Theta is just like an angle labeled as x. Mathematicians commonly use letters of the greek alphabet to label angles.", + "video_name": "EnwWxMZVBeg", + "timestamps": [ + 197 + ], + "3min_transcript": "And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta." + }, + { + "Q": "\nSo, basically every degree amount will have the same amount of radians no matter the circle? So every circle is 2pi(r) radians? Is that what this is about? 7:00", + "A": "Yes. Since radians are just a way of measuring angles, like degrees, they stay the same no matter the size of the circle. As a circle always has 360 degrees around the point, it always has 2pi radians.", + "video_name": "EnwWxMZVBeg", + "timestamps": [ + 420 + ], + "3min_transcript": "have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle. Here, the angle in radians tells you exactly the arc length that is subtending the angles. So let's do a couple of thought experiments here. So given that, what would be the angle in radians if we were to go-- so let me draw another circle here. So that's the center, and we'll start right over there. So what would happen if I had an angle-- if I wanted to measure in radians, what angle would this be in radians? And you could almost think of it as radiuses. So what would that angle be? Going one full revolution in degrees, that would be 360 degrees. Based on this definition, what would this be in radians? The arc that subtends this angle is the entire circumference of this circle. Well, what's the circumference of a circle in terms of radiuses? So if this has length r, if the radius is length r, what's the circumference of the circle in terms of r? That's going to be 2 pi r. So going back to this angle, the length of the arc that subtends this angle is how many radiuses? Well, it's 2 pi radiuses. It's 2 pi times r. So this angle right over here, I'll call this a different-- well, let's call this angle x. x in this case is going to be 2 pi radians. If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses. So given that, let's start to think about how we can convert between radians and degrees, and vice versa. If I were to have-- and we can just follow up over here. If we do one full revolution-- that is, 2 pi radians-- how many degrees is this going to be equal to? Well, we already know this. A full revolution in degrees is 360 degrees. Well, I could either write out the word degrees, or I can use this little degree notation there. Actually, let me write out the word degrees. It might make things a little bit clearer that we're kind of using units in both cases. Now, if we wanted to simplify this a little bit, we could divide both sides by 2. In which case, on the left-hand side," + }, + { + "Q": "At 4:23, what does Sal mean when he says that the ratio can be represented as a function of n?\n", + "A": "What Sal means is that your attention should be drawn to the fact that the ratio ((n+1) term over (n) term) dos not simplify to a constant ratio because it has a variable n in it. What it simplifies to is a ratio which is itself a function of the terms in question (the a sub n in question). Hope this helps!", + "video_name": "av947KCWf2U", + "timestamps": [ + 263 + ], + "3min_transcript": "we want to figure out whether a series like this, so starting at N equals five to infinity of, let's say, N to the tenth power. The numerator is growing quickly. N to the tenth power over N factorial, and factorial we know also grows very, very quickly, probably it grows much faster than even a high degree polynomial like this, or a high degree term like this, but how do we prove that it converges? We could definitely the diverges test to show that this does not diverge, but how do we prove that this actually converges? And maybe we can use a little bit of our intuition here. Well, let's see if we can come up with a common ratio. So let's see if there's a common ratio here. So let's take the N plus oneth term, which is going to be N plus one to the tenth power over N plus one factorial and divide that by the Nth term, Well, dividing by a fraction, or dividing by anything, is equivalent to multiplying by it's reciprocal, So let's just multiply by the reciprocal, so times N factorial over N to the tenth. Remember all I'm trying to do is exactly right over here. See if there is some type of a common ratio. Well, if we do a little algebra here N plus one factorial, and factorial algebra is always fun. This is the same thing as N plus one times N factorial. Times N factorial. You're not use to seeing order of operations with factorials, but this factorial only applies to this N right over here, and why is that useful? Because this N factorial cancels with this N factorial, and we're left with N plus one to the tenth power over N plus one times N to the tenth. So I know what you're thinking. \"Hey, wait, look, this isn't a fixed common ratio here.\" The ratio between consecutive terms here, when I took the N plus oneth term divided by the Nth term, it's changing depending on what my N is. This ratio, I guess you could say, is a function of N so this doesn't seem too useful, but what if I were to say, \"O.K., well look, with any of these series \"we really care about the behavior as our N's \"get really, really, really large as the limit, as our N's, go to infinity.\" So what if we were to look at the behavior here, and if this is approaching some actual values as N approaches infinity, well, it would make conceptual sense that we could kind of think of that as the limit of our common ratio. So let's do that. Let's take the limit as N approaches infinity of this thing. So remember what we're doing here." + }, + { + "Q": "\nYou can't prove that n^10/n! doesn't diverge with the divergence test as stated in 2:40 because lim an = 0 so it can still diverge ou converge. Or am I wrong?", + "A": "Absolutely correct, a series such as 1/n (harmonic series) diverges, even though the limit as n goes to infinity leads to 0 for an, so Sal made a mistake.", + "video_name": "av947KCWf2U", + "timestamps": [ + 160 + ], + "3min_transcript": "If it is not review, I encourage you to watch the videos on geometric series. And what's interesting about this is we've proven to ourselves, in the videos about geometric series, that if the common ratio, if the absolute value of the common ratio is less than one, then the series converges. And if the absolute value of R is greater than or equal to one, then the series diverges. And that makes sense, we've proven it as well, but it also makes logical sense that, look, if that absolute value of R is less than one then each term here is going to go down by that common, or it's going to be multiplied by that common ratio, and it's going to decrease on and on and on and on until it makes sense that, even though this is an infinite sum, it will converge to a finite value. Now, with that out of the way for review, let's tackle something a little bit more interesting. we want to figure out whether a series like this, so starting at N equals five to infinity of, let's say, N to the tenth power. The numerator is growing quickly. N to the tenth power over N factorial, and factorial we know also grows very, very quickly, probably it grows much faster than even a high degree polynomial like this, or a high degree term like this, but how do we prove that it converges? We could definitely the diverges test to show that this does not diverge, but how do we prove that this actually converges? And maybe we can use a little bit of our intuition here. Well, let's see if we can come up with a common ratio. So let's see if there's a common ratio here. So let's take the N plus oneth term, which is going to be N plus one to the tenth power over N plus one factorial and divide that by the Nth term, Well, dividing by a fraction, or dividing by anything, is equivalent to multiplying by it's reciprocal, So let's just multiply by the reciprocal, so times N factorial over N to the tenth. Remember all I'm trying to do is exactly right over here. See if there is some type of a common ratio. Well, if we do a little algebra here N plus one factorial, and factorial algebra is always fun. This is the same thing as N plus one times N factorial. Times N factorial. You're not use to seeing order of operations with factorials, but this factorial only applies to this N right over here, and why is that useful? Because this N factorial cancels with this N factorial, and we're left with N plus one to the tenth power over N plus one times N to the tenth." + }, + { + "Q": "At 2:20 what did Sal mean by you can rewrite this.\n", + "A": "What he actually means in 2:20 is that -15.08+526.90 is the same thing as saying 526.90-15.08. If you are careful with the numbers, you can rearrange the numbers in an equation for simplification!", + "video_name": "fFdOr8U4mnI", + "timestamps": [ + 140 + ], + "3min_transcript": "At the beginning of the week, Stewart's checking account had a balance of negative $15.08. On Monday morning, he deposited a check for $426.90. On Tuesday morning, he deposited another check for $100. How much was in Stewart's checking account after the second deposit, so after both of these deposits right over here. So he starts off with a negative balance. So a negative balance means that he's overdrawn his checking account. He actually owes the bank money now. Luckily, he's now going to put some money in his bank account. So he'll actually have a positive balance in his checking account. So he's starts off with the negative $15.08. And then to that, he adds $426.90. And then he adds another $100. So he started off with negative $15.08. And then to that, he adds $426.90 and $100. And so how much is going to have in his bank account? He started owing $15.08, and then he's going to add $526.90. So one way to visualize it is, if you think about it on a number line, if this is 0 right over here, he's going to start off at negative $15.08. But then he's going to add $526. So this right over here, this is $15.08 to the left. That's how much he owes. And to that, he's going to add $526. So I'm not drawing this to scale. But to that, he is going to add $526.90. So the amount that he's going to be in the positive is going to be $526.90 minus the $15.08. That's how much he's going to be in the positive. And that's going to be $526.90 minus $15.08. So that's going to be, and we can even just rewrite this so it actually looks exactly like that. That's exactly the same thing as $526.90 minus-- adding a negative is the same thing as subtracting a positive-- minus $15.08. And this is-- I will do this in another color-- $526.90 minus $15.08. Let's see, 0 is less than 8. Let's make that a 10 and borrow from this 9. So that becomes an 8, or I guess you could say we're regrouping. Now, everything up here is larger than everything there. So 10 minus 8 is 2. 8 minus 0 is 8. We have our decimal. 6 minus 5 is 1." + }, + { + "Q": "0:07, Sal says that Stewart has a checking account with a balance of -$15. Does this mean that Stewart owes money?\n", + "A": "Yes, he does.", + "video_name": "fFdOr8U4mnI", + "timestamps": [ + 7 + ], + "3min_transcript": "At the beginning of the week, Stewart's checking account had a balance of negative $15.08. On Monday morning, he deposited a check for $426.90. On Tuesday morning, he deposited another check for $100. How much was in Stewart's checking account after the second deposit, so after both of these deposits right over here. So he starts off with a negative balance. So a negative balance means that he's overdrawn his checking account. He actually owes the bank money now. Luckily, he's now going to put some money in his bank account. So he'll actually have a positive balance in his checking account. So he's starts off with the negative $15.08. And then to that, he adds $426.90. And then he adds another $100. So he started off with negative $15.08. And then to that, he adds $426.90 and $100. And so how much is going to have in his bank account? He started owing $15.08, and then he's going to add $526.90. So one way to visualize it is, if you think about it on a number line, if this is 0 right over here, he's going to start off at negative $15.08. But then he's going to add $526. So this right over here, this is $15.08 to the left. That's how much he owes. And to that, he's going to add $526. So I'm not drawing this to scale. But to that, he is going to add $526.90. So the amount that he's going to be in the positive is going to be $526.90 minus the $15.08. That's how much he's going to be in the positive. And that's going to be $526.90 minus $15.08. So that's going to be, and we can even just rewrite this so it actually looks exactly like that. That's exactly the same thing as $526.90 minus-- adding a negative is the same thing as subtracting a positive-- minus $15.08. And this is-- I will do this in another color-- $526.90 minus $15.08. Let's see, 0 is less than 8. Let's make that a 10 and borrow from this 9. So that becomes an 8, or I guess you could say we're regrouping. Now, everything up here is larger than everything there. So 10 minus 8 is 2. 8 minus 0 is 8. We have our decimal. 6 minus 5 is 1." + }, + { + "Q": "\nAt 3:38, I don't understand why you wouldn't just divide 9 into 2 and have the quotient of 0.22 there instead of \"borrowing\" the 10.\nI calculated 0.22*10^14 which would simplify to 2.2*10^13\nSo i would have gotten it wrong unless i \"borrowed\" the 10 during my calculation.", + "A": "0.22 x 10^14 is numerically correct, but it isn t valid scientific notation. A number which is written in scientific notation must have the mantissa (the bit before the power of ten) greater than or equal to 1, and less than 10. Here, 0.22 is not greater than or equal to 1, so it fails that test. So although 0.22 x 10^14 = 2.2 x 10^13, only the latter is correct scientific notation.", + "video_name": "0lOpqmTdtzk", + "timestamps": [ + 218 + ], + "3min_transcript": "It's clearly not less than 10. But we can convert this to scientific notation very easily. 90 is the same thing as 9 times 10, or you could even say 9 times 10 to the first. And then you multiply that times 10 to the negative 15. And then this simplifies to 9 times-- let's add these two exponents-- 10 to the negative 14. And now we can actually divide. And let's simplify this division a little bit. This is going to be the same thing as 2/9 times 1 to t over 10 to the negative 14. Well what's 1 over 10 to the negative 14? Well that's just 10 to the 14. Now you might say, OK, we just have to figure out what 2/9 is We've written this in scientific notation. But you might have already realized, look, 2/9 is not greater than or equal to 1. How can we make this greater than or equal to 1? Well we could multiply it by 10. If we multiply this by 10, then we've got to divide this by 10 to not change the value of this expression. But let's do that. So I'm going to multiply this by 10, and I'm going to divide this by 10. So I haven't changed. I've multiplied and divided by 10. So this is equal to 20/9 times 10 to the 14th divided by 10 is 10 to the 13th power. So what's 20/9? This is going to give us a number that is greater than or equal to 1 and less than 10. So let's figure it out. And I think they said round our answer to two decimal places. So 20 divided by 9-- 9 doesn't go into 2. It does go into 20 two times. 2 times 9 is 18. Subtract. Get a remainder of 2. I think you see where this show is going to go. 9 goes into 20 two times. 2 times 9 is 18. We're just going to keep getting 2's. So we get another 2, bring down a 0. Nine goes into 20 two times. So this thing right over here is really 2.2 repeating. But they said round to two decimal places, so this is going to be equal to 2.22 times 10 to the 13th power." + }, + { + "Q": "\nAt 1:28, Sal talks about tuples. This makes me think of \"quintuples,\" except the \"tu\" is pronounced differently. Is that where \"tuples\" comes from? Where does the term \"tuples\" come from?", + "A": "From Wikipedia: The term originated as an abstraction of the sequence: single, double, triple, quadruple, quintuple, sextuple, septuple, octuple, ..., n\u00e2\u0080\u0091tuple, ..., where the prefixes are taken from the Latin names of the numerals. Although this treats \u00e2\u0080\u0091tuple as the suffix, the original suffix was \u00e2\u0080\u0091ple as in triple (three-fold) or decuple (ten\u00e2\u0080\u0091fold). This originates from a medieval Latin suffix \u00e2\u0080\u0091plus (meaning more ).", + "video_name": "lCsjJbZHhHU", + "timestamps": [ + 88 + ], + "3min_transcript": "When you get into higher mathematics, you might see a professor write something like this on a board, where it's just this R with this extra backbone right over here. And maybe they write R2. Or if you're looking at it in a book, it might just be a bolded capital R with a 2 superscript like this. And if you see this, they're referring to the two-dimensional real coordinate space, which sounds very fancy. But one way to think about it, it's really just the two-dimensional space that you're used to dealing with in your coordinate plane. To go a little bit more abstract, this isn't necessarily this visual representation. This visual representation is one way to think about this real coordinate space. If we were to think about it a little bit more abstractly, the real R2, the two-dimensional real coordinate space-- let me write this down-- and the two-dimensional real coordinate space. the 2 tells us how many dimensions we're dealing with, and then the R tells us this is a real coordinate space. The two-dimensional real coordinate space is all the possible real-valued 2-tuples. Let me write that down. This is all possible real-valued 2-tuples. So what is a 2-tuple? Well, a tuple is an ordered list of numbers. And since we're talking about real values, it's going to be ordered list of real numbers, and a 2-tuple just says it's an ordered list of 2 numbers. So this is an ordered list of 2 real-valued numbers. Well, that's exactly what we did here when we thought about a two-dimensional vector. This right over here is a 2-tuple, and this is a real-valued 2-tuple. Neither of these have any imaginary parts. So you have a 3 and a 4. Order matters. And even if we were trying to represent them in our axes right over here, this vector, 4, 3, would be 4 along the horizontal axis, and then 3 along the vertical axis. And so it would look something like this. And remember, we don't have to draw it just over there. We just care about its magnitude and direction. We could draw it right over here. This would also be 4, 3, the column vector, 4 3. So when we're talking about R2, we're talking about all of the possible real-valued 2-tuples. So all the possible vectors that you can have, where each of its components-- and the components are these numbers right over here-- where each of its components are a real number. So you might have 3, 4. You could have negative 3, negative 4. So that would be 1, 2, 3, 1, 2, 3, 4," + }, + { + "Q": "\nAt 2:20 how come you do not get rid of the exponent after you square 2x^2? Is it because it still contains a variable? because after you squared (8)^2 = 64 ...no more exponent, but when you square (2x)^2, it turns into 4x^2?", + "A": "In (2x)^2, both the 2 and the X have to be squared. We can calculate 2^2 = 4. But, we don t know the numeric value for X. So, we can t calculate X*X. We just write it in exponent form: x^2. (2x)^2 = (2x) (2x) = (2*2) (x*x) = 4x^2", + "video_name": "h6HmHjkA034", + "timestamps": [ + 140 + ], + "3min_transcript": "Find the product 2x plus 8 times 2x minus 8. So we're multiplying two binomials. So you could use FOIL, you could just straight up use the distributive property here. But the whole point of this problem, I'm guessing, is to see whether you recognize a pattern here. This is of the form a plus b times a minus b, where here a is 2x and b is 8. We have 2x plus 8 and then 2x minus 8. a plus b, a minus b. What I want to do is I'm just going to multiply this out for us. And then just see what happens. Whenever you have this pattern, what the product actually looks So if you were to multiply this out, we can distribute the a plus b. We could distribute this whole thing. Distribute the whole a plus b on the a and then distribute it on the b. And I could have done this with this problem right here, and it would have taken us less time to just solve it. But I want to find out the general pattern here. So a plus b times a. So we have a times a plus b, that's this times this. that's negative b times a plus b. So I've done distributive property once, now I could do it again. I can distribute the a onto the a and this b and it gives me a squared. a times a is a squared, plus a times b, which is ab. And now I can do it with the negative b. Negative b times a is negative ab or negative ba, same thing. And negative b times b is negative b squared. Now, what does this simplify to? Well, I have an ab, and then I'm subtracting an ab. So these two guys cancel out and I am just left with a squared minus b squared. So the general pattern, and this is a good one to just kind of know super fast, is that a plus b times a minus b is always going to be a squared minus b squared. So we have an a plus b times an a minus b. So it's going to be 2x squared minus b squared minus 8 squared. 2x squared, that's the same thing as 2 squared times x squared, or 4x squared. And from that, we're subtracting 8 squared. So it's going to be 4x squared minus 64." + }, + { + "Q": "at 1:27 Sal said milli ment 1,000 shouldn`t it mean 1,000,000\n", + "A": "milli means one thousandth, micro means one millionth. :)", + "video_name": "mI84WDfhuYA", + "timestamps": [ + 87 + ], + "3min_transcript": "We're asked to arrange the following measurements in order from smallest to largest. And we have the measurements, and they're different units. This is in decameters, then we have meters, then we have millimeters, then we have centimeters. So the way I would tackle this is I would try to convert them all to the same units, maybe meters, and then compare them all in terms of meters. So let's do that. So this is just a screenshot of that exact same screen we just saw there. And so let's just convert each of these into meters. So the first one we have right over here is one decameter. So we have to remind ourselves what deca even means. Well, deca is equal to 10 meters. And actually, let me add more of these, just so that we know all the different prefixes we could have. So you have deca. You have hecto, which would be 100. You have kilo, which would be 1,000. And then, of course, you just have meters, That would just be equal 1 if you have no prefix in front of the meters. And then if you have a tenth of a meter, that is decimeter, so this is 1/10. Then you have centi, which is 1/100. And then you have milli, which is equal to 1/1,000. So let's use this information right over here to figure out how many meters each of these are. So one decameter-- we just saw deca means 10. That's 1 times 10 meters. So this is going to be equal to 10 meters. So this right over here is 10 meters. This is already written in meters. This is 13 meters. Then we have 15,000 millimeters, but milli means 1/1,000, So each millimeter is 1/1,000 of a meter. You could view this as instead of writing a milli here, I wrote 1/1,000. So 15,000 times 1/1,000, that's just going to give me 15. So this is going to be 15 meters. So this is 15 meters. Another way of thinking about it is, look, 1,000 millimeters is equal to a meter. So let's divide this into groups of 1,000. Well, this is literally 15 groups of 1,000. This is 15 groups of 1,000 millimeters, so that's going to be 15 meters. And then, finally, we have 1,900 centimeters. So 1,900, instead of writing centi, I'm going to write 1/100 of a meter." + }, + { + "Q": "At 11:40, can (0,0) on a graph be both a x-intercept and a y-intercept?\n", + "A": "Yes. (0,0) is an x-intercept and a y-intercept. We call this point the origin.", + "video_name": "_npwsLh0vws", + "timestamps": [ + 700 + ], + "3min_transcript": "" + }, + { + "Q": "\n1:20 WHAT did it mean infantine set of eyes", + "A": "There are infinite y values.", + "video_name": "_npwsLh0vws", + "timestamps": [ + 80 + ], + "3min_transcript": "In the following graph, is y a function of x? So in order for y to be a function of x, for any x that you input into the function, any x for which the function is defined. So let's say we have y is equal to f of x. So we have our little function machine. It should spit out exactly one value of y. If it spits out multiple values of y, we don't know what f of x is going to be equal to. It could be equal to any of those possible values for y. So let's see if, for this graph, whether for a given x it spits out exactly one y. Well, the function seems to be only defined so the domain of this function is x is equal to negative 2. That's the only place where we have a definition for it. And if we try to input negative 2 into this little black box, what do we get? Do we get exactly one thing? If we put in negative 2 here, we could get anything. The point negative 2, 9 is on this relation. Negative 2, 7; negative 2, 7.5; negative 2, 3.14159-- they're all on these. So if you put a negative 2 into this relation, essentially, you actually get an infinite set of values. It could be 9. It could be 3.14. It could be 8. It could be negative 8. You get an infinite number of results. So since it does not map to exactly one output of this function, in the following graph, y is not a function of x." + }, + { + "Q": "4:54 i think he showed wrong direction of resultant vector c it should be from tail of a to head of b which is shown wrong i think\n", + "A": "You might want to listen more closely to what Sal was saying. In the second example he was working, the vector C he was using was NOT the sum of vectors A and B. Once he switched directions as you suggest to create the vector -C he then set up the equation A + B = -C.", + "video_name": "BsBH8nAv5l4", + "timestamps": [ + 294 + ], + "3min_transcript": "completely perfect. So vector a, just like that. And one way of thinking about subtracting vector b is instead of adding vector b the way we did here, we could add negative b. So negative b would have the same magnitude but just the opposite direction. So that's vector a. Vector negative b will still start right over here, but will go in the opposite direction. So let's do that. So negative b is going to look like this, is going to look something like this. So that is negative b. Notice, same magnitude exactly opposite direction. We've flipped it around 180 degrees, and now the resulting vector is going to be d. So vector d is going to look like that, vector d. So c is a plus b, d is a minus b. Or you can even call this a plus, a plus negative b. and go the other way. See if we could go from the diagrams to the actual equations. So let's start with, let me draw an interesting one. So let's say that this... Let's say that's vector a. Vector a. I'll use green. Let's say that, that is vector b, and I will now use magenta. And let's say that this is vector c. Vector c. So I encourage you to pause the video and see if you can write an equation that defines this relationship. Well, this is interesting because they're all going in a circle right over here. Let's say that you started at... This is your initial point. well, if you're trying to figure out what a plus b is going to be, the resulting vector would start here and end here. But vector c is going in the opposite direction. But we could, instead of thinking about vector c like this, we could think about the opposite of vector c, which would do, so instead of calling this c, I could flip this around by calling it this negative c. So I could flip this around, and now, let me do the same color, this would be equal to negative c. Notice, before I just had vector c here and it started at this point and ended at this point. Now I just flipped it around, it has the exact opposite direction, same magnitude. Now it is negative c. And this makes it easier for us to construct an equation because negative c starts at the tip, at the initial point or the tail of vector a, and it goes to the head of vector b or the terminal point of vector b. So we can now write an equation." + }, + { + "Q": "\nIf Vector A - Vector B =Vector D in 4:08 then Vector B - Vector A= -Vector D ??", + "A": "Yes it is. Just , you are multiplying -1 to the whole equation.", + "video_name": "BsBH8nAv5l4", + "timestamps": [ + 248 + ], + "3min_transcript": "and then go to (mumbles) just draw vector a. So vector a. So once again, a vector, I can shift them around as long as I'm not changing the direction or their magnitude. So vector a looks like that. And notice, if you now go and start at the initial point of b and go to the terminal point of a, you still get vector c. So that's why a plus b and b plus a are going to give you the same thing. Now what if instead of saying a plus b I wanted to think about what a minus b is going to be? So let me write that down. Vector a minus vector b, minus vector b. And let's call that vector d. That is equal to vector d. So once again, I could start with vector a and here, order matters. So vector a looks something like this. completely perfect. So vector a, just like that. And one way of thinking about subtracting vector b is instead of adding vector b the way we did here, we could add negative b. So negative b would have the same magnitude but just the opposite direction. So that's vector a. Vector negative b will still start right over here, but will go in the opposite direction. So let's do that. So negative b is going to look like this, is going to look something like this. So that is negative b. Notice, same magnitude exactly opposite direction. We've flipped it around 180 degrees, and now the resulting vector is going to be d. So vector d is going to look like that, vector d. So c is a plus b, d is a minus b. Or you can even call this a plus, a plus negative b. and go the other way. See if we could go from the diagrams to the actual equations. So let's start with, let me draw an interesting one. So let's say that this... Let's say that's vector a. Vector a. I'll use green. Let's say that, that is vector b, and I will now use magenta. And let's say that this is vector c. Vector c. So I encourage you to pause the video and see if you can write an equation that defines this relationship. Well, this is interesting because they're all going in a circle right over here. Let's say that you started at... This is your initial point." + }, + { + "Q": "\nwhere did u get 144 on 03:54????", + "A": "it is given that the side of the square is 12 and hence the area of the square is 12*12=144.....", + "video_name": "vaOXkt7uuac", + "timestamps": [ + 234 + ], + "3min_transcript": "the area of all of the parallelograms because they are congruent. So let's see if we can figure out the area of one of the parallelograms. So there is actually a formula for the area of a parallelogram, it's actually just the base times the height. And they actually give us that. But let me show you that they give us that because it might not be obvious to you. Let me try to draw it. I'll use my line tool. Nope, that's not the line tool. One side, then go straight like that, come down like that, good enough. OK, now if I look at just this parallelgoram, they tell us that the height here is 3. And I know it's the height because they told me it's a 90 degree angle. And they tell us at the base is 5. And I'm telling you that the area of a parallelogram is But you shouldn't just take my word for it. That should make intuitive sense to you. And the way to think about it intuitively is imagine if we were to take this part of the parallelogram, and if we were to move it over here. If we were to cut that off and move it over here. Then the parallelogram would look something like this. You'd have the part that we didn't cut off. And then you move the cut-off part over here. And now the dimensions, this base would be 5, and this height would be 3. And the area of this rectangle is 15. And there's no reason why the area of this should be any different than that. We just rearranged its parts. So that's why the area of a parallelogram is just the base times the height. So the area of each of these parallelograms is 15. So the area of all of them combined is 15 times 4, which is 60. So 144 minus 60 is 84. Problem 38. What is the area in square meters of the trapezoid shown below. So to figure out the area and we could break it up into these rectangles and triangles. To figure out the area of this rectangle we need to know its height. And actually we'll need that to figure out the area of the triangles as well. So what's this height right there? Let's see, we know that this distance is going to be 6. It's a rectangle. If that distance is 6 and both of these are 5, both of these triangles here are going to be congruent. Because this length is equal to this length. This length is equal to this length. And we also we make this angle is equal to that angle. But anyway, let me do it in another color." + }, + { + "Q": "At 8:15, How does Sal get 25 roots of 3, divided by 2? I don't get where 2 came from.\n", + "A": "The 2 comes from the equation for the area of a triangle, 1/2 times the base times the height, or [1/2bh]. He divides it by 2 because that is the 1/2 in action. The base is 5 and the height is 5\u00e2\u0088\u009a3. So, the area is (25\u00e2\u0088\u009a3)/2, which could also be simplified to 12.5\u00e2\u0088\u009a3.", + "video_name": "vaOXkt7uuac", + "timestamps": [ + 495 + ], + "3min_transcript": "a squared is equal to 16. a is equal to 4. a is equal to 4. And now we're ready to figure out the area. What's the area of the rectangle? 6 times 6, it's 24. What's the area of each of these triangles? 3 times 4 times 1/2. 3 times 4 is 12 times 1/2 is 6. So the area of that triangle is 6. The area of this triangle is 6. So 6 plus 24 plus 6 is 36. B. Problem 39. What is the area in square inches of the triangle below. Interesting. OK, so this is an equilateral triangle, all the sides are equal. And so we could actually say that since these two triangles are symmetric. That's equal to that. And this comes to a general formula for the area of an equilateral triangle. But let's just figure it all out. So this side is going to be 5. If this is 5 and that's 10, what is this side right here? Let's call it x. Pythagorean theorem. This is the hypotenuse. So x squared plus 5 squared plus 25 is going to be equal to the hypotenuse squared, it's equal to 100. x squared is equal to 100 minus 25, 75. x is equal to the square root of 75. 75 is 25 times 3. So that's equal to the square root of 25 times 3. Which is equal to the square root of 25 times the square root of 3. Which is equal to 5 roots of 3. And now, what's the area of just this right triangle right here? This one on the right side. Well its base is 5, its height is 5 roots of 3. So it's going to be 1/2 times the base, 5, times the height, And that's what? 1/2 times 5 times 5. So it's 25 root 3 over 2 and that's just this triangle right there. Well this triangle's going to have the the exact same area. They are congruent triangles. So the area of the figure is this times 2. So 2 times that is equal to just the 25 root 3. And that's choice B. Next problem, problem 40. The perimeter of two squares are in a ratio of 4:9. What is the ratio between the areas of the two squares? Let me draw two squares. That's one square. Let me draw another square. That's another square. Let's say that the sides of this are x and the sides of this one are y." + }, + { + "Q": "\nAt 0:42, Sal says that the two angles are supplementary. In a real proof, wouldn't we not be able to assume that unless it was given?? Or do the opposite arrows show that it is a horizontal line?", + "A": "The arrows do indicate that it is a line. So the angles are supplementary.", + "video_name": "eTwnt4G5xE4", + "timestamps": [ + 42 + ], + "3min_transcript": "We're given a bunch of lines here that intersect in all different ways and form triangles. And what I want to do in this video, we've been given the measures of some of the angles, this angle, that angle, and that angle. And what we want to do in this video is figure out what the measure of this angle is. And we're going to call that measure x. And so I encourage you to pause the video right now and try it yourself. And then I'm going to give you the solution. So I'm assuming you've unpaused it. And you've solved it or you've given it at least a good shot of it. And what's fun about these is there's multiple ways to solve these. And you kind of just have to keep figuring out what you can figure out. So let's say you start on the left-hand side right over here. If this is 121 degrees, then you'd say, well look, this angle right over here is supplementary to this angle right over there. So this is 121 degrees plus this green angle, that has to be equal to 180 degrees. So this is going to be 180 minus 121. 80 minus 20 would be 60. So that's going to be 59 degrees. So let me write that down. That's going to be 59 degrees. Now we see that we have two angles of a triangle. If you have two angles of a triangle, you can figure out the third angle, because they need to add up to 180. Or you could say that this angle right over here-- so we'll call that question mark-- we know that 59 plus 29 plus question mark needs to be equal to 180 degrees. And if we subtract the 15 out of the 29 from both sides, we get question mark is equal to 180 minus 59 minus 29 degrees. So that is going to be 180 minus 59 minus 29, let's see, 180 minus 59, we already know, is 121. And then 121 minus 29. So if you subtract just 20, you get 101. You subtract another 9, you get 92. This is equal to 92 degrees. Well, this right here is equal to 92 degrees. This angle right here is vertical with that angle. So it is also going to be equal to 92 degrees. And now we're getting pretty close. We can zoom in on this triangle down here. And let me save some space here. So let me just say that that over there is also going to be 92 degrees. And at this triangle down here, we have two of the sides of the triangle. We just have to figure out the third. And actually, we don't even have to do much math here, because we have two of the angles of this triangle. We have to figure out the third angle. So over here, we have one angle that's 92, one angle that's 29. The other one will be 180 minus 92 minus 29. And we don't even have to do any math here, because essentially, this is the exact same angles that we have in this triangle right over here. We have a 92 degree angle, we have a 29 degree angle, and the other one is 59 degrees. So in this case, it has to be also" + }, + { + "Q": "At 3:44 Sal mentions the \"Exterior angle of 121 to be equal to the sum of the remote interior angles.\" This somehow confuses me as I don't remember him mentioning it before in previous videos. Is this true? If so does any exterior angle of a triangle mean that its equal to the triangles interior angles? Thanks in advance!\n", + "A": "Yep!! I can t remember the name of the postulate/theorem that says this, but it is definitely true. Any exterior angle of a triangle is equal to the sum of its remote interior angles (the two interior angles that are across from the exterior angle in question). Hope this helps!! :)", + "video_name": "eTwnt4G5xE4", + "timestamps": [ + 224 + ], + "3min_transcript": "This is equal to 92 degrees. Well, this right here is equal to 92 degrees. This angle right here is vertical with that angle. So it is also going to be equal to 92 degrees. And now we're getting pretty close. We can zoom in on this triangle down here. And let me save some space here. So let me just say that that over there is also going to be 92 degrees. And at this triangle down here, we have two of the sides of the triangle. We just have to figure out the third. And actually, we don't even have to do much math here, because we have two of the angles of this triangle. We have to figure out the third angle. So over here, we have one angle that's 92, one angle that's 29. The other one will be 180 minus 92 minus 29. And we don't even have to do any math here, because essentially, this is the exact same angles that we have in this triangle right over here. We have a 92 degree angle, we have a 29 degree angle, and the other one is 59 degrees. So in this case, it has to be also So over here, they'll also add up to 180. So that will also get us to 59 degrees. We could just get that by taking 180, subtracting 29, subtracting 92. And then if this is 59 degrees, then this angle is also going to be 59 degrees, because they are vertical angles. So we're done. x is equal to 59 degrees. Now there's multiple ways that you could have reasoned through this problem. You could have immediately said-- so let me start over, actually. Maybe a faster way, but you wouldn't have been able to do kind of this basic steps there, is you said, look, this is an exterior angle right over here. It is equal to the sum of the remote interior angles. So 121 is going to be 29 plus this thing right over here. And we ended up doing that when I did it step-by-step before. But here, we're just using kind of a few things that we know about triangles ahead of time Although I like to do it the other way just so we make sure we don't do anything weird. So anyway, this is going to be 129 minus 29, which is going to be 92. And if this is 92, then this is also going to be 92. And then, if this is x, then this is also going to be x. And you could say x plus 92 plus 29 is equal to 180 degrees. And then you'd say x plus 92 plus 29 is going to be 121 degrees. We already knew that before. And so that is going to equal 180 degrees. And so x is equal to 59 degrees. So there's a ton of ways that you could have thought about this problem." + }, + { + "Q": "\nat 2:46, i notised that the angel he was trying to solve enisally is about the the same as the one he is going to solve.", + "A": "Well ya, but there is an even easier way to solve for x. 180-121 is 59. So now we know the angle next to 121. There is another line with the same slope on the other side, so we know the corresponding angles are the same. Therefore we know that x is 59. So Simple.", + "video_name": "eTwnt4G5xE4", + "timestamps": [ + 166 + ], + "3min_transcript": "80 minus 20 would be 60. So that's going to be 59 degrees. So let me write that down. That's going to be 59 degrees. Now we see that we have two angles of a triangle. If you have two angles of a triangle, you can figure out the third angle, because they need to add up to 180. Or you could say that this angle right over here-- so we'll call that question mark-- we know that 59 plus 29 plus question mark needs to be equal to 180 degrees. And if we subtract the 15 out of the 29 from both sides, we get question mark is equal to 180 minus 59 minus 29 degrees. So that is going to be 180 minus 59 minus 29, let's see, 180 minus 59, we already know, is 121. And then 121 minus 29. So if you subtract just 20, you get 101. You subtract another 9, you get 92. This is equal to 92 degrees. Well, this right here is equal to 92 degrees. This angle right here is vertical with that angle. So it is also going to be equal to 92 degrees. And now we're getting pretty close. We can zoom in on this triangle down here. And let me save some space here. So let me just say that that over there is also going to be 92 degrees. And at this triangle down here, we have two of the sides of the triangle. We just have to figure out the third. And actually, we don't even have to do much math here, because we have two of the angles of this triangle. We have to figure out the third angle. So over here, we have one angle that's 92, one angle that's 29. The other one will be 180 minus 92 minus 29. And we don't even have to do any math here, because essentially, this is the exact same angles that we have in this triangle right over here. We have a 92 degree angle, we have a 29 degree angle, and the other one is 59 degrees. So in this case, it has to be also So over here, they'll also add up to 180. So that will also get us to 59 degrees. We could just get that by taking 180, subtracting 29, subtracting 92. And then if this is 59 degrees, then this angle is also going to be 59 degrees, because they are vertical angles. So we're done. x is equal to 59 degrees. Now there's multiple ways that you could have reasoned through this problem. You could have immediately said-- so let me start over, actually. Maybe a faster way, but you wouldn't have been able to do kind of this basic steps there, is you said, look, this is an exterior angle right over here. It is equal to the sum of the remote interior angles. So 121 is going to be 29 plus this thing right over here. And we ended up doing that when I did it step-by-step before. But here, we're just using kind of a few things that we know about triangles ahead of time" + }, + { + "Q": "\nWhy in the first place, would you need to find for example an inverse of a matrix like Sal said at 3:40, or a determinant, or add or subtract matrixes?", + "A": "The why is the means to find solution(s) to matrix operations.", + "video_name": "0oGJTQCy4cQ", + "timestamps": [ + 220 + ], + "3min_transcript": "And it has three columns. This is a 1 by 3 matrix. I could have a matrix-- and I think you see where all of this Figuring out the dimensions of a matrix are not too difficult. I could have a matrix that looks like this, where it's 3, 5, 0, 0, negative 1, negative 7. This right over here has three rows. So it's three rows, and it has two columns. So we would call this a 3 by 2. Let me do that in that same color. We would call it a 3 by 2 matrix, three rows and two columns. You know that a matrix is just a rectangular array of numbers. You can say what its dimensions are. You know that each of these numbers that take one of these positions-- we just call those entries. But what are matrices good for? I still might not be clear what the connection is between this and this right over here. this is just a compact representation of a bunch of numbers. It's a way of representing information. They become very valuable in computer graphics because these numbers could represent the color intensity at a certain point. They could represent whether an object is there at a certain point. And as we develop an algebra around matrices, and when we talk about developing an algebra around matrices, we're going to talk about operations that we're going to perform on matrices that we would normally perform with numbers. So we're going to essentially define how to multiply matrices, how to add matrices. We'll learn about taking an inverse of a matrix. And by coming up with an algebra of how we manipulate these things, it'll become very useful in the future when you're trying to write a computer graphics program or you're trying to do an economic simulation or a probability simulation, to say, oh, I have this matrix that represents where different particles are in space. Or I have this matrix that represents the state of some type of a game. And I know ways of doing it very efficiently so that I can multiply a bunch of them. Or I could come run a simulation, and I can actually come up with useful results. So that's all matrices are. But as you'll see through this, we can define operations on them. And then later on, when you take a linear algebra course in college, you'll learn a lot more of the depth of how they can be applied and what you can use them to represent." + }, + { + "Q": "At 1:55 in the video, why is it 72 * 7? I understand that 72 is the (9 * 8 * 7) part, but why is it multiplied by seven? Is it because there is only 7 people left after the 3 positions are filled?\n", + "A": "72 is just the 9*8 part. He multiplied it by 7 because he wasn t done with the whole multiplication of 9*8*7.", + "video_name": "l9ft9jpriNA", + "timestamps": [ + 115 + ], + "3min_transcript": "A club of nine people wants to choose a board of three officers, President, Vice President, and Secretary. Assuming the officers are chosen at random, what is the probability that the officers are Marcia for president, Sabita for Vice President, and Robert for Secretary? So to think about the probability of Marcia-- so let me write this-- President is equal to Marcia, or Vice President is equal to Sabita, and Secretary is equal to Robert. This, right here, is one possible outcome, one specific outcome. So it's one outcome out of the total number of outcomes, over the total number of possibilities. Now what is the total number of possibilities? Well to think about that, let's just think about the three positions. and you have Secretary. Now let's just assume that we're going to fill the slot of President first. We don't have to do President first, but we're just going to pick here. So if we're just picking President first, we haven't assigned anyone to any officers just yet, so we have nine people to choose from. So there are nine possibilities here. Now, when we go to selecting our Vice President, we would have already assigned one person to the President. So we only have eight people to pick from. And when we assign our Secretary, we would've already assigned our President and Vice President, so we're only going to have seven people to pick from. So the total permutations here or the total number of possibilities, or the total number of ways, to pick President, Vice President, and Secretary from nine people, is going to be 9 times 8 times 7. Which is, let's see, 9 times 8 is 72. 72 times 7, 2 times 7 is 14, 7 times 7 is 49 plus 1 is 50. So to answer the question, the probability of Marcia being President, Sabita being Vice President, and Robert being Secretary is 1 over the total number of possibilities, which is 1 over 504. That's the probability." + }, + { + "Q": "How do you tell the quadrants apart at 0:40\n", + "A": "the quadrants are usually labeled", + "video_name": "Jeh5vudjmLI", + "timestamps": [ + 40 + ], + "3min_transcript": "Plot 4 comma negative 1, and select the quadrant in which the point lies. So 4, the first number in our ordered pair, that's our x-coordinate. That says how far to move in the horizontal or the x-direction. It's a positive 4. So I'm going to go 4 to the right. And then the second coordinate says, what do we do in the vertical direction or in the y-direction? It's a negative 1. Since it's negative, we're going to go down. And it's a negative 1, so we're going to go down 1. So that right over there is the point 4 comma negative 1. So I've plotted it, but now I have to select which quadrant the point lies in. And this is just a naming convention. This is the first quadrant. This is the second quadrant. This is the third quadrant. And this is the fourth quadrant. So the point lies in the fourth quadrant, quadrant IV. And I guess you have to know your Roman numerals a little bit to know that's representing quadrant IV. Let's do a couple more of these. Plot 8 comma negative 4, and select Well, my x-coordinate is 8 so I go 8 in the positive x-direction. And then my y-coordinate is negative 4, so I go 4 down. And this is sitting, again, in not the first, not the second, not the third, but the fourth quadrant, in quadrant IV. Let's do one more of these. Hopefully we get a different quadrant. So we want to plot the point negative 5 comma 5. So now my x-coordinate is negative. It's negative 5. So I'm going to move to the left in the x-direction. So I go to negative 5. And my y-coordinate is positive so I go up 5, so negative 5 comma 5. And this is sitting not in the first quadrant, but the second quadrant. And of course, this is the third and the fourth. So this is sitting in the second quadrant. Check answer, and we got it right." + }, + { + "Q": "At 2:55, I cannot understand how does ln(67)=4.205 approximately matches between 2 and 3. plz help:(\n", + "A": "If we let ln67 = a then e^a = 67 by definition of a logarithm e by definition is about 2.71 I think If we round the 4.2 down for approx. 2^4 = 16 and 3^4 = 81 Therefore since e = 2.71 which is in between 2 and 3 we expect e^4 to be in between 16 and 81, which it is.", + "video_name": "Dpo_-GrMpNE", + "timestamps": [ + 175 + ], + "3min_transcript": "this way is log base e is referred to as the natural logarithm. And I think that's used because e shows up so many times in nature. So log base e of 67, another way of saying that-- or seeing that, and the more typical way of seeing that is the natural log. And I think this is ln, so I think it's maybe from French or something, log natural, of 67. So this is the same thing as log base e of 67. This is saying the exact same thing. To what power do I have to raise e to to get 67? When you see this ln, it literally means log base e. Now, they let us use a calculator, and that's good because I don't know off the top of my head what power I have to raise 2.71 so on and so forth-- what power I have to raise that to to get to 67. So we'll get our calculator out. So we get the TI85 out. And different calculators will have different ways of doing it. If you have a graphing calculator like this, you literally can literally type in the statement natural log So here this is the button for ln, means natural log, log natural, maybe. ln of 67, and then you press Enter, and it'll give you the answer. If you don't have a graphing calculator, you might have to press 67 and then press natural log to give you the answer, but a graphing calculator can literally type it in the way that you would write it out, and then you would press Enter. So 4.20469 and we want to round to the nearest thousandth. So this is the thousandths place right here, this 4. The digit after that is 5 or larger, it's a 6, so we're going to round up. So this is 4.205. So this is approximately equal to 4.205. And it actually makes a lot of sense, because we know that e is greater than 2, and it is less than 3. And if you think about what 2 to the fourth power And 3 to the fourth power gets you to 81. 67 is between 16 and 81 and e is between 2 and 3. So at least it feels right that's something that's like 2.71 to the little over the fourth power should get you to a number that's pretty close to 3 to the fourth power. Actually that makes sense because it's actually closer to 3. 2.71 is closer to 3 than it is to 2. So this feels right, that you take this to the fourth, little over the fourth power, you get to 67." + }, + { + "Q": "At 5:00 in the video, why does Mr. Khan place positive 2 in the x place, if the x value was -2? Was that a mistake or something else?\n", + "A": "The two aren t related, the positive two is the midpoint of the two x-intercepts and the negative two is one of the intercepts of the parabola", + "video_name": "EV57jv7JKCs", + "timestamps": [ + 300 + ], + "3min_transcript": "the x-axis at x equals negative two right over there, and x is equal to six. These are our x-intercepts. So given this, how do we figure out the vertex? Well the key idea here is to recognize that your axis of symmetry for your parabola is going to sit right between your two x-intercepts. So what is the midpoint between, or what is the average of six and negative two? Well, you could do that in your head. Six plus negative two is four divided by two is two. Let me do that. So I'm just trying to find the midpoint between the point, let's use a new color. So I'm trying to find the midpoint between the point negative two comma zero and six comma zero. Well the midpoint, those are just the average of the coordinates. The average of zero and zero is just going to be zero, it's going to sit on the x-axis. or the average negative two plus six over two. Well let's see, that's four over two, that's just going to be two, so two comma zero. And you see that there. You could have done that without even doing the math. You say okay, if I want to go right in between the two, I want to be two away from each of them. And so just like that, I could draw an axis of symmetry for my parabola. So my vertex is going to sit on that axis of symmetry. And so how do I know what the y value is? Well I can figure out, I can substitute back in my original equation, and say well what is y equal when x is equal to two? Because remember the vertex has a coordinate x equals two. It's going to be two comma something. So let's go back, let's see what y equals. So y will equal to one-half times, we're going to see when x equals two, so two minus six, times two plus two. Let's see, this is negative four, Negative four times four is negative 16. So it's equal to one-half times negative 16, which is equal to negative eight. So our vertex when x is equal to two, y is equal to negative eight. And so our vertex is going to be right over here two common negative eight. And now we can draw the general shape of our actual parabola. It's going to look something like, once again this is a hand-drawn sketch, so take it with a little bit of a grain of salt, but it's going to look something like this. And it's going to be symmetric around our axis of symmetry. That's why it's called the axis of symmetry. This art program I have, there's a symmetry tool, but I'll just use this and there you go. That's a pretty good sketch of what this parabola is or what this graph is going to look like which it is, an upward-opening parabola." + }, + { + "Q": "at 1:57, how is it equal to each other\n", + "A": "it means the answer could be that number or less", + "video_name": "ilWDSYnTEFs", + "timestamps": [ + 117 + ], + "3min_transcript": "I'm starting to take a little bit more care of my health, and I start counting my actual calories. And let's say C is equal to the number of calories I eat in a given day. And I want to lose some weight. So, in particular, I want to eat less than 1,500 calories in a day. So how can I express that as an inequality? Well, I want the number of calories in a day to be less than-- and remember, the less than symbol, I make it point to the smaller thing. So I want the calories to be less than 1,500. So this is one way of expressing it. I say, look, the number of calories that I consume in a day need to be less than 1,500. Now, one thing to keep in mind when I write that is obviously if I eat no calories in a day, or if I eat 100 calories, or if I eat 1,400 calories, Those are all less than 1,500. But what about 1,500 calories? Is it true that 1,500 is less than 1,500? 1,500 is equal to 1,500. So this is not a true statement. But what if I want to eat up to and including 1,500 calories? I want to make sure that I get every calorie in there. How can I express that? How can express that I can eat less than or equal to 1,500 calories, so I can eat up to and including 1,500 calories? Right now, this is only up to but not including 1,500. How could I express that? Well, the way I would do that is to throw this little line under the less than sign. Now, this is not just less than. This is less than or equal to. So this symbol right over here, this So now 1,500 would be a completely legitimate C, a completely legitimate number of calories to have in a day. And if we wanted to visualize this on a number line, the way we would think about it, let's say that this right over here is our number line. I'm not going to count all the way from 0 to 1,500, but let's imagine that this right over here is 0. Let's say this over here is 1,500. How would we display less than or equal to 1,500 a number line? Well, we would say, look, we could be 1,500, so we'll put a little solid circle right over there. And then we can be less than it, so then we would color in everything less than 1,500," + }, + { + "Q": "At 2:00, it is a little error in video when he writes M_y=cosx+1xe^y ( it should be M_y=cosx+2xe^y).\n", + "A": "No, he wrote it correctly. It s just a low resolution video, like many of the old KA videos are.", + "video_name": "Pb04ntcDJcQ", + "timestamps": [ + 120 + ], + "3min_transcript": "OK, I filled your brain with a bunch of partial derivatives and psi's, with respect to x's and y's. I think now it's time to actually do it with a real differential equation, and make things a little bit more concrete. So let's say I have the differential, y, the differential equation, y cosine of x, plus 2xe to the y, plus sine of x, plus-- I'm already running out of space-- x squared, e to the y, minus 1, times y prime, is equal to 0. Well, your brain is already, hopefully, in exact differential equations mode. But if you were to see this pattern in general, where you see a function of x and y, here-- this is just some function of x and y-- and then you have another function of x and y, times y prime, or times dy, d of x, your brain should immediately say if this is inseparable. because that'll take a lot of time. But if it's not separable, your brain said, oh, maybe this is an exact equation. And, you say, let me test whether this is an exact equation. So if this is an exact equation, this is our function M, which is a function of x and y. And this is our function N, which is a function of x and y. Now, the test is to see if the partial of this, with respect to y, is equal to the partial of this, with respect to x. So let's see. The partial of M, with respect to y, is equal to-- let's see, y is-- so this cosine of x is just a constant, so it's just cosine of x. Cosine of x plus-- now, what's the derivative? Well, 2x is just a constant, what's the derivative of e to the y, with respect to y? Well, it's just e to the y, right? So we have the constant on the outside, 2x times the derivative, with respect to y, so it's 2xe to the y. Fair enough. Now, what is the partial derivative of this, with So N sub x, or the partial of N, with respect to x-- so what's the derivative of sine of x, with respect to x? Well, that's easy, that's cosine of x, plus 2x times e to the y, right? e and y is just a constant, because y is constant when we're taking the partial, with respect to x. So plus 2xe to the y. And then minus 1, the derivative of a constant, with respect to anything is going to be 0. So the derivative of N-- the partial of N, with respect to x, is cosine of x, plus 2xe to the y, which, lo and behold, is the same thing as the derivative, the partial of M, with respect to y. So there we have it. We've shown that M of y is equal to-- or the partial of M, with respect to y-- is equal to the partial of N, with respect to x, which tells us that this is an exact equation. Now, given that this is an exact equation-- oh, my wife" + }, + { + "Q": "At around 2:30 Sal tell us that -4 / - 1/2 is equal to -4/1 * - 2/1. I am wondering why the 2 becomes a negative when it's going to be multiplied. I would like to know if -4/1 is the same as - 4/1? I hope I'm clear.\n\nTo be clearer: why doesn't Sal multiply -4/1 by 2/-1, and instead multiplies by -2/1\n", + "A": "He could multiply by any of those because - 2/1 = -2/1 = 2/(-1).", + "video_name": "H0q9Fqb8YT4", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's do some examples dividing fractions. Let's say that I have negative 5/6 divided by positive 3/4. Well, we've already talked about when you divide by something, it's the exact same thing as multiplying by its reciprocal. So this is going to be the exact same thing as negative 5/6 times the reciprocal of 3/4, which is 4/3. I'm just swapping the numerator and the denominator. So this is going to be 4/3. And we've already seen lots of examples multiplying fractions. This is going to be the numerators times each other. So we're going to multiply negative 5 times 4. I'll give the negative sign to the 5 there, so negative 5 times 4. Let me do 4 in that yellow color. And then the denominator is 6 times 3. You might already know that 5 times 4 is 20, and you just have to remember that we're multiplying a negative times a positive. We're essentially going to have negative 5 four times. So negative 5 plus negative 5 plus negative 5 plus negative 5 is negative 20. So the numerator here is negative 20. And the denominator here is 18. So we get 20/18, but we can simplify this. Both the numerator and the denominator, they're both divisible by 2. So let's divide them both by 2. Let me give myself a little more space. So if we divide both the numerator and the denominator by 2, just to simplify this-- and I picked 2 because that's the largest number that goes into both of these. It's the greatest common divisor of 20 and 18. 20 divided by 2 is 10, and 18 divided by 2 is 9. So negative 5/6 divided by 3/4 is-- oh, I have to be very careful here. It's negative 10/9, just how we always learned. if the signs are different, then you're going to get a negative value. Let's do another example. Let's say that I have negative 4 divided by negative 1/2. So using the exact logic that we just said, we say, hey look, dividing by something is equivalent to multiplying by its reciprocal. So this is going to be equal to negative 4. And instead of writing it as negative 4, let me just write it as a fraction so that we are clear what its numerator is and what its denominator is. So negative 4 is the exact same thing as negative 4/1. And we're going to multiply that times the reciprocal of negative 1/2. The reciprocal of negative 1/2 is negative 2/1. You could view it as negative 2/1, or you could view it as positive 2 over negative 1, or you could view it as negative 2." + }, + { + "Q": "at 1:47 it say greatest common divisor what dose that mean\n", + "A": "It is the largest number that can be divided into both 18 and 20. 18 can be divided by 2, 3, 6, and 9, while twenty can be divided by 2, 4, and 5. The only number that they have in common is 2, so two is the greatest common divisor. Notice that they can each also be divided by 1, but generally we ignore that when simplifying fractions because dividing by 1 gets us nowhere.", + "video_name": "H0q9Fqb8YT4", + "timestamps": [ + 107 + ], + "3min_transcript": "Let's do some examples dividing fractions. Let's say that I have negative 5/6 divided by positive 3/4. Well, we've already talked about when you divide by something, it's the exact same thing as multiplying by its reciprocal. So this is going to be the exact same thing as negative 5/6 times the reciprocal of 3/4, which is 4/3. I'm just swapping the numerator and the denominator. So this is going to be 4/3. And we've already seen lots of examples multiplying fractions. This is going to be the numerators times each other. So we're going to multiply negative 5 times 4. I'll give the negative sign to the 5 there, so negative 5 times 4. Let me do 4 in that yellow color. And then the denominator is 6 times 3. You might already know that 5 times 4 is 20, and you just have to remember that we're multiplying a negative times a positive. We're essentially going to have negative 5 four times. So negative 5 plus negative 5 plus negative 5 plus negative 5 is negative 20. So the numerator here is negative 20. And the denominator here is 18. So we get 20/18, but we can simplify this. Both the numerator and the denominator, they're both divisible by 2. So let's divide them both by 2. Let me give myself a little more space. So if we divide both the numerator and the denominator by 2, just to simplify this-- and I picked 2 because that's the largest number that goes into both of these. It's the greatest common divisor of 20 and 18. 20 divided by 2 is 10, and 18 divided by 2 is 9. So negative 5/6 divided by 3/4 is-- oh, I have to be very careful here. It's negative 10/9, just how we always learned. if the signs are different, then you're going to get a negative value. Let's do another example. Let's say that I have negative 4 divided by negative 1/2. So using the exact logic that we just said, we say, hey look, dividing by something is equivalent to multiplying by its reciprocal. So this is going to be equal to negative 4. And instead of writing it as negative 4, let me just write it as a fraction so that we are clear what its numerator is and what its denominator is. So negative 4 is the exact same thing as negative 4/1. And we're going to multiply that times the reciprocal of negative 1/2. The reciprocal of negative 1/2 is negative 2/1. You could view it as negative 2/1, or you could view it as positive 2 over negative 1, or you could view it as negative 2." + }, + { + "Q": "\nWhy switch the numbers at 0:34", + "A": "When dividing, you need to make the second fraction a reciprocal, or flip the fraction so that the numerator ends up as the denominator and the denominator as the numerator.", + "video_name": "H0q9Fqb8YT4", + "timestamps": [ + 34 + ], + "3min_transcript": "Let's do some examples dividing fractions. Let's say that I have negative 5/6 divided by positive 3/4. Well, we've already talked about when you divide by something, it's the exact same thing as multiplying by its reciprocal. So this is going to be the exact same thing as negative 5/6 times the reciprocal of 3/4, which is 4/3. I'm just swapping the numerator and the denominator. So this is going to be 4/3. And we've already seen lots of examples multiplying fractions. This is going to be the numerators times each other. So we're going to multiply negative 5 times 4. I'll give the negative sign to the 5 there, so negative 5 times 4. Let me do 4 in that yellow color. And then the denominator is 6 times 3. You might already know that 5 times 4 is 20, and you just have to remember that we're multiplying a negative times a positive. We're essentially going to have negative 5 four times. So negative 5 plus negative 5 plus negative 5 plus negative 5 is negative 20. So the numerator here is negative 20. And the denominator here is 18. So we get 20/18, but we can simplify this. Both the numerator and the denominator, they're both divisible by 2. So let's divide them both by 2. Let me give myself a little more space. So if we divide both the numerator and the denominator by 2, just to simplify this-- and I picked 2 because that's the largest number that goes into both of these. It's the greatest common divisor of 20 and 18. 20 divided by 2 is 10, and 18 divided by 2 is 9. So negative 5/6 divided by 3/4 is-- oh, I have to be very careful here. It's negative 10/9, just how we always learned. if the signs are different, then you're going to get a negative value. Let's do another example. Let's say that I have negative 4 divided by negative 1/2. So using the exact logic that we just said, we say, hey look, dividing by something is equivalent to multiplying by its reciprocal. So this is going to be equal to negative 4. And instead of writing it as negative 4, let me just write it as a fraction so that we are clear what its numerator is and what its denominator is. So negative 4 is the exact same thing as negative 4/1. And we're going to multiply that times the reciprocal of negative 1/2. The reciprocal of negative 1/2 is negative 2/1. You could view it as negative 2/1, or you could view it as positive 2 over negative 1, or you could view it as negative 2." + }, + { + "Q": "at 5:11 Mr Khan says that a 90 degree angle is a right angle so would an 180 degree angle be a straight angle?\n", + "A": "yes, it could be called a straight angle or just a straight line", + "video_name": "92aLiyeQj0w", + "timestamps": [ + 311 + ], + "3min_transcript": "And then I'll point AB in the-- well, assuming that I'm drawing it exactly the way that it's Normally, instead of moving the angle, you could actually move the protractor to the angle. So it looks something like that, and you could see that it's pointing to right about the 30 degree mark. So we could say that the measure of angle BAC is equal to 30 degrees. And so you can look just straight up from evaluating these numbers that 77 degrees is clearly larger than 30 degrees, and so it is a larger angle, which makes sense because it is a more open angle. And in general, there's a couple of interesting angles to think about. If you have a 0 degree angle, you actually have something that's just a closed angled. It really is just a ray at that point. As you get larger and larger or as you get more and more open, is completely straight up and down while the other one is left to right. So you could imagine an angle that looks like this where one ray goes straight up down like that and the other ray goes straight right and left. Or you could imagine something like an angle that looks like this where, at least, the way you're looking at it, one doesn't look straight up down or one does it look straight right left. But if you rotate it, it would look just like this thing right over here where one is going straight up and down and one is going straight right and left. And you can see from our measure right over here that that gives us a 90 degree angle. It's a very interesting angle. It shows up many, many times in geometry and trigonometry, and there's a special word for a 90 degree angle. It is called a \"right angle.\" So this right over here, assuming if you rotate it around, would look just like this. We would call this a \"right angle.\" You draw a little part of a box right over there, and that tells us that this is, if you were to rotate it, exactly up and down while this is going exactly right and left, if you were to rotate it properly, or vice versa. And then, as you go even wider, you get wider and wider and wider and wider until you get all the way to an angle that looks like this. So you could imagine an angle where the two rays in that angle form a line. So let's say this is point X. This is point Y, and this is point Z. You could call this angle ZXY, but it's really so open that it's formed an actual line here. Z, X, and Y are collinear. This is a 180 degree angle where we see the measure of angle ZXY" + }, + { + "Q": "\nWhat if your angle is 0 Degrees? At the end 8:18-8:21, Sal says that if your angle is all the way 180 degrees... it forms a line. But, what if the angle is 0 degrees? What does that form? Is it still an acute angle? Or is there a special vocabulary word to describe this type of angle? Appreciate the help! :)", + "A": "An angle of 0 degrees between two rays forms one ray. Think of the two rays being perfectly on top of each other, going in the same direction.", + "video_name": "92aLiyeQj0w", + "timestamps": [ + 498, + 501 + ], + "3min_transcript": "And you can actually go beyond that. So if you were to go all the way around the circle so that you would get back to 360 degrees and then you could keep going around and around and around, and you'll start to see a lot more of that when you enter a trigonometry class. Now, there's two last things that I want to introduce in this video. There are special words, and I'll talk about more types of angles in the next video. But if an angle is less than 90 degrees, so, for example, both of these angles that we started our discussion with are less than 90 degrees, we call them \"acute angles.\" So this is acute. So that is an acute angle, and that is an acute angle right over here. They are less than 90 degrees. What does a non-acute angle look like? And there's a word for it other than non-acute. So, for example-- let me do this in a color I haven't used-- an angle that looks like this, and let me draw it a little bit better than that. An angle it looks like this. So that's one side of the angle or one of the rays and then I'll put the other one on the baseline right Clearly, this is larger than 90 degrees. If I were to approximate, let's see, that's 100, 110, 120, almost 130. So let's call that maybe a 128-degree angle. We call this an \"obtuse angle.\" The way I remember it as acute, it's kind of \"a cute\" angle. It's nice and small. I believe acute in either Latin or Greek or maybe both means something like \"pin\" or \"sharp.\" So that's one way to think about it. An acute angle seems much sharper. Obtuse, I kind of imagine something that's kind of lumbering and large. Or you could think it's not acute. It's not nice and small and pointy. but this is just general terminology for different types of angles. Less than 90 degrees, you have an acute angle. At 90 degrees, you have a right angle. Larger than 90 degrees, you have an obtuse angle. And then, if you get all the way to 180 degrees, your angle actually forms a line." + }, + { + "Q": "at 6:00 he is talking about < zxy\nwouldn't it be the same if he called it:\n1 when at the top of the screen it says b<0\n", + "A": "It can t. And in the 30 seconds after 5:37, Sal explains that b>1 cannot be part of the solution because b<0 is a restriction.", + "video_name": "0_VaUYoNV7Y", + "timestamps": [ + 337 + ], + "3min_transcript": "" + }, + { + "Q": "\nWhen you divide A out at 4:20 do you always replace the empty side with 1?", + "A": "When you divide any variable out of a equation, you always replace the number or variable with 1. 1 is like the benchmark in math, it always takes the place of the number if the initial number is divided out.", + "video_name": "0_VaUYoNV7Y", + "timestamps": [ + 260 + ], + "3min_transcript": "" + }, + { + "Q": "at the end ( 5:15 - 5:20 ) when sal says b <-1 and b > 1 how does that work b cannot be no number can be b if b was 2 that would violate the first inequality if b was -2 it would violate the second someone please explain.\n", + "A": "You are trying to use b <-1 AND b > 1 . Sal is using b <-1 OR b > 1. You are correct, AND will not work as it means both conditions need to be true. But, OR is more flexible, only one condition need to be true. Hope this helps.", + "video_name": "0_VaUYoNV7Y", + "timestamps": [ + 315, + 320 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 0:49. Don't you mean 19 + 18. You said 19 over 18...", + "A": "Yeah he meant 19 + 18.. It said it in the bottom right corner..", + "video_name": "8Eb5MWwcMMY", + "timestamps": [ + 49 + ], + "3min_transcript": "Let's add 19 and 3/18 to 18 and 2/3. So I like to separate out the whole number parts from the fraction parts. So 19 and 3/18 is the same thing as 19 plus 3/18. And to that, we are going to add 18 and 2/3, which is the same thing as 18 plus 2/3. Now we can separately add the whole number parts. So we could add the 19 to the 18. So we could do 19 plus 18. And then we can add the fraction parts-- let me do this in green-- plus 3/18 plus 2/3. Now 19/18, pretty straightforward. That is what? Let's see. 19 plus 19 would be 38. So this is going to be 1 less than that. It's going to be 37. So that gives me 37. And then 3/18 plus 2/3, to add them, I need to have the same denominator. So let's convert 2/3 to something over 18. So 2/3, if I want to write it as something over 18, well, I multiplied the denominator by 6, so I'd also have to multiply the numerator by 6. So it's the same thing as 12/18. So I can rewrite 2/3 as 12/18. And now I can add these two things together. That's going to be-- so I have 37 plus-- it's going to be something over 18-- plus something over 18. 3 plus 12 is 15, plus 15/18. And so expressing this as a mixed number, I get 37 and 15/18. And that's the right number. But we can simplify it even more. We can simplify the 15/18. Both the numerator and the denominator are divisible by 3. So let's divide them both by 3. And we're not changing the value because we're And so this gives us, we still have our 37, but the numerator is now 5, and the denominator is now 6. So we get 37 and 5/6. And we're done." + }, + { + "Q": "At 1:10, where did the 6 come in?\n", + "A": "The 18 is bigger than the 3.... so you have to ask yourself what you have to multiply the 3 by to get to 18.... 3x6=18..... Whatever you do to the denominator you have to do to the numerator to arrive at an equivalent fraction... so now you also multiply the numerator by 6 to get the numerator for the equivalent fraction", + "video_name": "8Eb5MWwcMMY", + "timestamps": [ + 70 + ], + "3min_transcript": "Let's add 19 and 3/18 to 18 and 2/3. So I like to separate out the whole number parts from the fraction parts. So 19 and 3/18 is the same thing as 19 plus 3/18. And to that, we are going to add 18 and 2/3, which is the same thing as 18 plus 2/3. Now we can separately add the whole number parts. So we could add the 19 to the 18. So we could do 19 plus 18. And then we can add the fraction parts-- let me do this in green-- plus 3/18 plus 2/3. Now 19/18, pretty straightforward. That is what? Let's see. 19 plus 19 would be 38. So this is going to be 1 less than that. It's going to be 37. So that gives me 37. And then 3/18 plus 2/3, to add them, I need to have the same denominator. So let's convert 2/3 to something over 18. So 2/3, if I want to write it as something over 18, well, I multiplied the denominator by 6, so I'd also have to multiply the numerator by 6. So it's the same thing as 12/18. So I can rewrite 2/3 as 12/18. And now I can add these two things together. That's going to be-- so I have 37 plus-- it's going to be something over 18-- plus something over 18. 3 plus 12 is 15, plus 15/18. And so expressing this as a mixed number, I get 37 and 15/18. And that's the right number. But we can simplify it even more. We can simplify the 15/18. Both the numerator and the denominator are divisible by 3. So let's divide them both by 3. And we're not changing the value because we're And so this gives us, we still have our 37, but the numerator is now 5, and the denominator is now 6. So we get 37 and 5/6. And we're done." + }, + { + "Q": "What the heck is a caveat? Sal says it at 3:14. I'm guessing that it is the backwards C like object.\n", + "A": "It s the RESTRICTION or the CONDITION. You know how Sal writes p \u00e2\u0089\u00a0 -5 , that s the caveat or condition.", + "video_name": "gcnk8TnzsLc", + "timestamps": [ + 194 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 3:35, when you cancel out (p - 5), you cross out (p - 6)", + "A": "It looks like a 6, but it is a 5.", + "video_name": "gcnk8TnzsLc", + "timestamps": [ + 215 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 4:02 when Sal reached the answer, 4(p+3) over 5, providing p does not equal -5, he did not further simplify the equation. Can you not simplify it into 4p+12 over 5, providing that p does not equal -5?", + "A": "In more advanced math, you often have to take your result and do other steps. That is usually easier if the result is in factored form. Sometimes in Calculus you will want to multiply the result all the way out. Practice will help you choose...and also little hints like the instructions you receive with the assignment and choices you are given as possible answers.", + "video_name": "gcnk8TnzsLc", + "timestamps": [ + 242 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 1:02, how did you get 0?", + "A": "He is talking about the denominator: x cannot make the denominator equal to 0 or that would make the problem undefined. No matter what other solutions we come up with, we always have to exclude that value, which in this case is x.\u00e2\u0089\u00a0 -4", + "video_name": "2RnS3fSHVV8", + "timestamps": [ + 62 + ], + "3min_transcript": "Let's tackle a slightly harder problem than what we saw in the last video. I have here x minus 3 over x plus 4 is greater than or equal to 2. So the reason why this is slightly harder is I now have a greater than or equal to. And the other thing that makes it slightly harder is I don't just have a simple 0 here, I actually have a 2 here. So I'm going to do this problem the same way. I'm going to do it two ways. I'm going to do the same two ways we did last time, but I'm going to reverse the order of the two ways that I do it. So the first thing I can do, I might be tempted-- let me multiply both sides of this equation times x plus 4. And like we saw in the last video we had to be very careful because we have an inequality here. If x plus 4 is greater than 0 then we're going to keep the inequality the same. If it's less than 0 we're going to swap it. So let's look at those two situations. So one situation is-- let me make it right here. One situation is x plus 4 is greater then 0. And remember, it can ever be equal to 0 because then this So let's explore x plus 4 is greater than 0. If we subtract 4 from both sides of this equation this is equivalent to saying-- these are equivalent statements. If we subtract 4 from both sides that x is greater than minus 4. So if we assume that x is greater than minus 4 then x plus 4 is going to be greater than 0. And then when we multiply both sides of this times x plus 4-- let's do that. x minus 3 over x plus 4. You have your 2. We're going to multiply both sides of this times x plus 4. Since we're assuming x is greater than minus 4 or that x plus 4 is greater than 0, we don't have to switch the inequality sign. We're multiplying both sides by a positive. So the inequality stays the same as in our original problem. And the whole reason why we did that is because this and this will cancel out. And then we have 2 times x plus 4. Let's see what we get. x minus 3 is greater than or equal to 2 times x plus And now what can we do? We can subtract x from both sides. We get minus 3 is greater than or equal to x plus 8. I just took x from 2x. And then we can subtract 8 from both sides. So you subtract 8 from both sides. You get minus 3 minus 8 is minus 11 is greater than or equal to x. And of course, then we subtract an 8 so this guy disappears. So if we assume that x is greater than minus 4, then x has to be less than or equal to minus 11. Now that seems a little bit nonsensical. In order for this statement to be true, x has to be both greater than minus 4 and it has to be less than minus 11. Anything greater than minus 4 is going to be greater than minus 11. So there's no x that satisfies this equation." + }, + { + "Q": "In 3:26, Sal said, \"So all I did, got rid of the exponent...\", but I think he meant the negative sign\n", + "A": "Yes, he could have said it that way, but what he means is that he got rid of the negative exponent and replaced it with a positive.", + "video_name": "S34NM0Po0eA", + "timestamps": [ + 206 + ], + "3min_transcript": "to the 3 times 4 power, or 2 to the 12th power, which you could also write as raising it to the fourth power and then the third power. All this is saying is, if I raise something to a power and then raise that whole thing to a power, it's the same thing as multiplying the two exponents. This is the same thing as 2 to the 12th. So we could use that property here to say, well, 2/3 is the same thing as 1/3 times 2. So we could go in the other direction. We could say, hey look, well this is going to be the same thing as 64 to the 1/3 power and then that thing squared. Notice, I'm raising something to a power and then raising that to a power. If I were to multiply these two things, I would get 64 to the 2/3 power. Now, why did I do this? Well, we already know what 64 to the 1/3 power is. We just calculated it. That's equal to 4. So we could say that-- and I'll write it in that same yellow color-- this is equal to 4 squared, which is equal to 16. The way I think of it, let me find the cube root of 64, which is 4. And then let me square it. And that is going to get me to 16. Now I'll give you in even hairier problem. And I encourage you to try this one on your own before I work through it. So we're going to work with 8/27. And we're going to raise this thing to the-- and I'll try to color code it-- negative 2 over 3 power, to the negative 2/3 power. I encourage you to pause and try this on your own. Well the first thing I do whenever I see a negative exponent is to say, well, how can I get rid of that negative exponent? And I just remind myself, well, the negative exponent really just says, take the reciprocal of this to the positive exponent. I'm going to use that light mauve color. So this is going to be equal to 27/8. I just took the reciprocal of this right over here. It's equal to 27/8 to the positive 2/3 power. So notice, all I did, I got rid of the exponent and took the reciprocal of the base right over here. 8/27 is the base. Negative 2/3 is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over a denominator to some power-- and this is another very powerful exponent property-- this is going to be the exact same thing as raising 27 to the 2/3 power-- to the 2 over 3 power-- over 8 to the 2/3 power." + }, + { + "Q": "At 5:47, why is the answer just 9/4? Can't we just simplify it to 2/1/4?\n", + "A": "You could, but Sal must consider 9/4 a simplified fraction as well, even if it is improper, because you can t simplify it any more (although, you can change it to a mixed number). Sometimes, whatever is considered simplified depends on the standards of your teacher or professor.", + "video_name": "S34NM0Po0eA", + "timestamps": [ + 347 + ], + "3min_transcript": "I'm going to use that light mauve color. So this is going to be equal to 27/8. I just took the reciprocal of this right over here. It's equal to 27/8 to the positive 2/3 power. So notice, all I did, I got rid of the exponent and took the reciprocal of the base right over here. 8/27 is the base. Negative 2/3 is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over a denominator to some power-- and this is another very powerful exponent property-- this is going to be the exact same thing as raising 27 to the 2/3 power-- to the 2 over 3 power-- over 8 to the 2/3 power. Notice, if I have something divided by something and I'm raising the whole thing to a power, I can essentially raise the numerator to that power over the denominator raised to that power. Now, let's think about what this is. Well just like we saw before, this is going to be the same thing. This is going to be the same thing as 27 to the 1/3 power and then that squared because 1/3 times 2 is 2/3. So I'm going to raise 27 to the 1/3 power and then square whatever that is. All this color coding is making this have to switch a lot of colors. This is going to be over 8 to the 1/3 power. And then that's going to be raised to the second power. Same thing we were doing in the denominator-- we raise 8 to the 1/3 and then square that. So what's this going to be? Well, 27 to the 1/3 power is the cube root of 27. times that same number is going to be equal to 27. Well, it might jump out at you already that 3 to the third is equal to 27 or that 27 to the 1/3 is equal to 3. So the numerator, we're going to end up with 3 squared. And then in the denominator, we are going to end up with-- well, what's 8 to the 1/3 power? Well, 2 times 2 times 2 is 8. So this is 8 to the 1/3 third is 2. Let me do that same orange color. 8 to the 1/3 is 2, and then we're going to square that. So this is going to simplify to 3 squared over 2 squared, which is just going to be equal to 9/4. So if you just break it down step by step, it actually is not too daunting." + }, + { + "Q": "\ni still don't get it at here the time 0:35\nAnd like M21 how did he got 100/100\nhere the time 0:32\nAnd again help on the last second 1:48", + "A": "because percent is out of a 100,(for EG,80/100) like an exam, it is out of a 100 percent, if you get 1 wrong you get like 98% or something, and for the last one, he is just multiplying 1.501 times 100%, so you move the decimal twice because 100 has two zero. but let say it was out of 1000, so you move the zero....three times.10, that will be one time.", + "video_name": "3_caioiRu5I", + "timestamps": [ + 35, + 32, + 108 + ], + "3min_transcript": "Let's see if we can write 1.501 as a percentage. Now we really just want to write this quantity as something over 100. So what we can do is we can say that this is the same thing as 1.501 over 1. We haven't changed its value. And let's multiply it times 100/100. When we do this product-- and I haven't changed the value. This is just multiplying something times 1. But when we do it, we're going to change the way that we're representing it. The denominator now is going to be 1 times 100. So that's pretty straightforward. That's just going to be 100. And then the numerator, we're going to want to multiply 1.501 times 100. And so if we multiply this times 10, we would move the decimal one over. We want to multiply it by 100, so we want to multiply by 10 again. So we're going to move the decimal over to the right twice. We're multiplying it by 10 twice, or you're multiplying it by 100. This is multiplying by 10. This is multiplying by 100. So this is where the decimal's going to sit now. And so we've rewritten the 1.501 as 150.1 over 100, or 150.1 per 100. Let me write that. 150.1 per 100, which is the same thing as 150.1 per cent, which is the same thing as 150.1%. And we're done. And, essentially, all you're doing-- if you really want to simplify the process-- we're multiplying this by 100 to get 150.1. And then we're saying, that is the percentage. So you want to make sure that you write the percent there or the percent symbol." + }, + { + "Q": "At 1:36, it says the answer is 150.1%. Doesn't that equal 150.1/ 100? Is that even possible?\n", + "A": "Yes, it s possible. The concept is similar to improper fractions to mixed numbers.", + "video_name": "3_caioiRu5I", + "timestamps": [ + 96 + ], + "3min_transcript": "Let's see if we can write 1.501 as a percentage. Now we really just want to write this quantity as something over 100. So what we can do is we can say that this is the same thing as 1.501 over 1. We haven't changed its value. And let's multiply it times 100/100. When we do this product-- and I haven't changed the value. This is just multiplying something times 1. But when we do it, we're going to change the way that we're representing it. The denominator now is going to be 1 times 100. So that's pretty straightforward. That's just going to be 100. And then the numerator, we're going to want to multiply 1.501 times 100. And so if we multiply this times 10, we would move the decimal one over. We want to multiply it by 100, so we want to multiply by 10 again. So we're going to move the decimal over to the right twice. We're multiplying it by 10 twice, or you're multiplying it by 100. This is multiplying by 10. This is multiplying by 100. So this is where the decimal's going to sit now. And so we've rewritten the 1.501 as 150.1 over 100, or 150.1 per 100. Let me write that. 150.1 per 100, which is the same thing as 150.1 per cent, which is the same thing as 150.1%. And we're done. And, essentially, all you're doing-- if you really want to simplify the process-- we're multiplying this by 100 to get 150.1. And then we're saying, that is the percentage. So you want to make sure that you write the percent there or the percent symbol." + }, + { + "Q": "\nAt 1:03 , can I write -15+j?", + "A": "Yes, you could write it as -15+j", + "video_name": "640-86yn2wM", + "timestamps": [ + 63 + ], + "3min_transcript": "- [Voiceover] Let's do some examples of the writing expressions with variables exercise. So it says \"Write an expression to represent 11 more than a.\" Well you could just have a but if you want 11 more than a, you would wanna add 11 so you could write that as a plus 11. You could also write that as 11 plus a. Both of them would be 11 more than a. So let's check our answer here. We got it right. Let's do a few more of these. \"Write an expression to represent the sum of d and 9.\" So the sum of d and 9, that means you're gonna add d and 9. So I could write that as d plus 9 or I could write that as 9 plus d. And check our answer. Got that right. Let's do a few more of these. \"Write an expression to represent j minus 15.\" Well, I could just write it with math symbols instead of writing the word minus. Instead of writing M-I-N-U-S, And then I check my answer. Got it right. Let's do a few more of these. This is a lot of fun. \"Write an expression to represent 7 times r.\" There's a couple ways I could do it. I could use this little dot right over here, do 7 times r like that. That would be correct. I could literally just write 7r. If I just wrote 7r that would also count. Let me check my answer. That's right. Let me do a couple of other of these just so you can see that I could've just done 10 and this is not a decimal, it sits a little bit higher than a decimal. It's multiplication and the reason why once you start doing algebra, you use this symbol instead of that kind of cross for multiplication is that x-looking thing gets confused with x when you're using x as a variable so that's why this is a lot more useful. So we wanna write 10 times u, 10 times u, let's check our answer. We got it right. Let's do one more. So we could write it as 8 and then I could write a slash like that, 8 divided by d. And there you go. This is 8 divided by d. Let me check, let me check the answer. I'll do one more of these. Oh, it's 6 divided by b. Alright, same thing. So 6, I could use this tool right over here. It does the same thing as if I were to press the backslash. So 6 divided by b. Check my answer. We got it right." + }, + { + "Q": "Hi can I please get help Asap Please ?! I understood what sal was saying until he got to the slope part. To sum it up , I don't understand from 1:38 to the end\n", + "A": "If x = 1 what is y? If x = 2 what is y? How much did x change? y? What s (change in y)/(change in x)?", + "video_name": "uk7gS3cZVp4", + "timestamps": [ + 98 + ], + "3min_transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done." + }, + { + "Q": "At 0:27, how do you actually tell if -2 equals y? How is x=0 when there is 1/3 next to it?\n", + "A": "A little after your time stamp, Sal states that we know that the y intercept is -2 because the y intercept is where x=0. The idea of putting an equation in slope intercept form is so that we can quickly recognize the slope (m) and the y intercept (b). For the y intercept (where a function crosses the y axis), x will always be zero.", + "video_name": "uk7gS3cZVp4", + "timestamps": [ + 27 + ], + "3min_transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done." + }, + { + "Q": "\nAt 0:38, what does he mean by \"the y-intercept occurs when x equals to zero\"?", + "A": "The Y-intercept is the point where the graph crosses the Y-Axis and a line crosses the Y-axis when x=0, since the Y-axis can be thought of as the line X=0.", + "video_name": "uk7gS3cZVp4", + "timestamps": [ + 38 + ], + "3min_transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done." + }, + { + "Q": "At 6:22, Sal says that we are going to find the volume of the coin shaped object by multiplying pi times f(x)^2 times dx, but how is this possible because the side of the coin also slightly slopes like the rest of the cone?\n", + "A": "The point of the integral is to take the sum of an infinite number of coins, so in theory the slope is 0.", + "video_name": "btGaOTXxXs8", + "timestamps": [ + 382 + ], + "3min_transcript": "you rotate it. What do you end up with? Well, you get something that looks kind of like a coin, like a disk, like a quarter of some kind. And let me draw it out here. So that same disk out here would look something like this. And it has a depth of dx. So how can we find the volume of that disk? Let me redraw it out here, too. It's really important to visualize this stuff properly. So this is my x-axis. My disk looks something like this. My best attempt at the x-axis sits it right over there. It comes out of the center. And then this is the surface of my disk. And then this right over here is my depth dx. So that looks pretty good. to give you a little bit of the depth. So how can we find the volume of this? Well, like any disk or cylinder, you just have to think about what the area of this face is and then multiply it times the depth. So what's the area of this base? Well, we know that the area of a circle is equal to pi r squared. So if we know the radius of this face, we can figure out the area of the face. Well, what's the radius? Well, the radius is just the height of that original rectangle. And for any x, the height over here is going to be equal to f of x. And in this case, f of x is x squared. So over here, our radius is equal to x squared. So the area of the face for a particular x is going to be equal to pi times f of x squared. In this case, f of x is x squared. Well, our volume is going to be our area times the depth here. It's going to be that times the depth, times dx. So the volume of this thing right over here-- so the volume just of this coin, I guess you could call it, is going to be equal to. So my volume is going to be equal to my area times dx, which is equal to pi times x squared squared. So it's equal to pi-- x squared squared is x to the fourth-- pi x to the fourth dx. Now, this expression right over here, this gave us the volume just of one of those disks. But what we want is the volume of this entire hat, or this entire bugle or cone-looking, or I guess you could say the front-of-a-trumpet-looking" + }, + { + "Q": "\nThis seems too simple for a 10-minute video, so someone please confirm that I understand. If I fully get the concepts of span, subspace, and a matrix being defined as a set of column vectors, then it's dead simple to grasp the concept of the column space of a matrix, right? So, anyone who gets the videos in this series up to this point can watch up to 1:28, then skip to the next video, right?", + "A": "I believe the point to drive home is that any vector multiplied with the matrix must result in a vector that exists in the column space of that matrix, which in turn is the span of those column vectors. This implies that the column space is a subspace.", + "video_name": "st6D5OdFV9M", + "timestamps": [ + 88 + ], + "3min_transcript": "We spent a good deal of time on the idea of a null space. What I'm going to do in this video is introduce you to a new type of space that can be defined around a matrix, it's called a column space. And you could probably guess what it means just based on what it's called. But let's say I have some matrix A. Let's say it's an m by n matrix. So I can write my matrix A and we've seen this multiple times, I can write it as a collection of columns vectors. So this first one, second one, and I'll have n of them. How do I know that I have n of them? Because I have n columns. And each of these column vectors, we're going to have how many components? So v1, v2, all the way to vn. This matrix has m rows. So each of these guys are going to have m components. So they're all members of Rm. So the column space is defined as all of the possible linear So the column space of A, this is my matrix A, the column space of that is all the linear combinations of these column vectors. What's all of the linear combinations of a set of vectors? It's the span of those vectors. So it's the span of vector 1, vector 2, all the way to vector n. And we've done it before when we first talked about span and subspaces. But it's pretty easy to show that the span of any set of vectors is a legitimate subspace. It definitely contains the 0 vector. If you multiply all of these guys by 0, which is a valid linear combination added up, you'll see that it contains the 0 vector. If, let's say that I have some vector a that is a member of the column space of a. That means it can be represented as some linear combination. So a is equal to c1 times vector 1, plus c2 times vector Now, the question is, is this closed under multiplication? If I multiply a times some new-- let me say I multiply it times some scale or s, I'm just picking a random letter-- so s times a, is this in my span? Well s times a would be equal to s c1 v1 plus s c2 v2, all the way to s Cn Vn Which is once again just a linear combination of these column vectors. So this Sa, would clearly be a member of the column space of a. And then finally, to make sure it's a valid subspace-- and this actually doesn't apply just to column space, so this applies to any span. This is actually a review of what we've done the past. We just have to make sure it's closed under addition. So let's say a is a member of our column space." + }, + { + "Q": "\nAt 9:48 to the end of the video the lines get confusing", + "A": "Yes, They Do. Don t Worry You ll get the hang of it Good Luck ! And Happy Learning .", + "video_name": "K759mIqpvOU", + "timestamps": [ + 588 + ], + "3min_transcript": "it's gonna be negative six, and L is this X coordinate, it's going to be, it's going to be positive six. So the absolute value of J minus L is going to be our, is going to the the difference in the horizontal axis, or it's going to be the distance on the horizontal axis between this point and that point. Or you say the horizontal distance between those two points. So it would be the length of this line segment. So that is the absolute value of J minus L. Once again, the X coordinate here is negative six, the X coordinate here is positive six, and you can even figure it out. It's going to be negative six minus six, which would be negative 12, and then you take the absolute value of that, this is going to be 12. And you don't even have to figure that out here, we just know that the length of this line is the absolute value of J minus L. So that's that. And then they have the absolute value of M minus Q. the absolute value of M minus Q. So they have M over here, that's the Y coordinate here, and Q is the Y coordinate down here. So the absolute value of M minus Q is going to be the distance, the vertical distance between these two points, which is really just, 'cause the X value isn't changing, this is actually going to be the length. This is going to be the length of that side. That's going to be the absolute value of M minus Q. So yeah, if you multiply this length times this length, you're going to get the area of the rectangle. So I didn't even have to look at the other choices, I would definitely go with this one. But let's see where the other ones probably aren't correct. So this is the absolute value of J minus M. So here you're taking the difference of the X coordinate here and the Y coordinate over there. So that's kind of bizarre. This this already looks suspicious. Here you're saying the absolute value of J minus N. Absolute value of J, Well, their X coordinates are the same, so this is actually going to be, this we actually know is going to be zero. J is equal to N, they're both equal to negative six. That's not gonna give you the length of this line, because we have no change along X here. All the change is along Y. If we wanted to figure out the length of this line right over here we would have to find the absolute value of K, of the change in our Y coordinates. So absolute value of K minus O would give you the length of this line. And if you wanted the length of this right over here, you'd want your change in X, so that would be the absolute value of N minus P, or you could say the absolute value of P minus N. But they didn't use those choices. So yeah, we feel good about that." + }, + { + "Q": "\nat 5:21 How does he know it will be a right triangle?", + "A": "He knows that triangle BDM is a right triangle because point D is on a line perpendicular to line AB. As he says at 5:30, All of the points that are equidistant between A and B are going to be on a line that is perpendicular to AB. Therefore, line DM intersects line BM at a right angle, making BDM a right triangle.", + "video_name": "smtrrefmC40", + "timestamps": [ + 321 + ], + "3min_transcript": "the average of 24 and 0, is 12. That's fair enough. Now let's see what we can figure out about the sides. So we know this is a right triangle. So our gut reaction is always to use the Pythagorean theorem. We know this side and that side. So if we wanted to figure out AB, we could just say that-- we just know that 24 squared plus 7 squared is equal to AB squared. And 24 squared is 576, plus 49 is equal to AB squared. And let's see-- 576 plus 49. If it was plus 50, it would get us to 626, but it's 1 less than that. So it's 625. So 625 is equal to AB squared. So this distance, the distance of this big hypotenuse here, is 25. Or half of the distance, this is going to be 25/2, from B to M. From M to A is also going to be 25/2. Now, the other thing we know is that M right over here, the triangle CMA is an isosceles triangle. How do we know that? Well, M, if you look at its x-coordinate-- its x-coordinate is directly in between the x-coordinates for C and A. 7/2-- it's the average. This is 7. This is 0. Its M-coordinate is right over there. It's directly above the midpoint of this base over here. So this is going to be an isosceles triangle. It's symmetric. You could flip the triangle over. So this length-- and this seems useful because this is kind of the base of the triangle we care about. this is also going to be 25/2. It's an isosceles triangle. This is going to be the same as that because we're symmetric around this right over here. So let's see. We know one side of this triangle. Let's see if we could figure out this side over here. It seems pretty straightforward. This is going to be a right triangle right over here, because this line from D to M is going to be perpendicular to AB. All of the points that are equidistant between A and B are going to be on a line that is perpendicular to AB. So this is going to be a right triangle. So we can figure out DM using the Pythagorean theorem again. We get 25/2 squared plus DM squared is going to be equal to 15 squared." + }, + { + "Q": "I dont understand, how at 10:37 he got cosx=sin90-x?\n", + "A": "Its just a property of sin and cos. If you look at the unit circle you can see that the sin of say 30 is the same as the cos of 60. The sin of an angle is the same as the cos of the complementary angle.", + "video_name": "smtrrefmC40", + "timestamps": [ + 637 + ], + "3min_transcript": "And this over here is 24. This is 25/2, and this is 25/2. And so using this, we could use the law of cosines to figure out what theta actually is here. So let's do that. We're going to use a little bit of trig identities, but law of cosines. So we get the opposite to the angle squared, so we get 24 squared. 24 squared is equal to 25/2 squared, plus 25/2 squared, minus 2 times 25/2 times 25/2 times the cosine of this angle, times-- let me scroll over-- the cosine of theta plus 90 degrees. this has it in terms of cosine of theta plus 90 degrees. How can we figure out the sine of theta? That's what we actually care about, to figure out the area of this triangle, or actually to figure out the height of this triangle. And to do that, you just have to make the realization. We know the trig identity that the cosine of theta is equal-- I won't use theta because I don't want to overload theta-- cosine of x is equal to sine of 90 minus x. So the cosine of theta plus 90 degrees is going to be equal to the sine of-- let me put it in parentheses-- 90 minus whatever is here. 90 minus theta minus 90, which is equal to-- the 90's cancel out-- sine of negative theta. And we know sine of negative theta is equal to the negative sine of theta. is the negative sine of theta. So we could write the sine of theta here and then put the negative out here. And this becomes a positive. So what does this simplify to? We have 24 squared, which is 576. 576 is equal to-- let's see. I won't skip any steps here. So we have 25 squared plus 25 squared. This is 2 times 25/2 squared plus 2 times-- this is 25/2 squared again-- times sine of theta. Now we just have to solve for sine of theta. So this is going to be equal to-- well, this is 576 is equal to 2 times 25/2 squared." + }, + { + "Q": "I dont understand, how at 10:37 he got cosx=sin90-x? Also at 14:39 it is really confusing?\n", + "A": "If you take a right triangle where there are the points A,B,and C(with C being the right angle), you can see that sin(x)=cos(90-x) by looking at the sine of angle A and the cosine of angle B. (90-angle A= angle B)", + "video_name": "smtrrefmC40", + "timestamps": [ + 637, + 879 + ], + "3min_transcript": "And this over here is 24. This is 25/2, and this is 25/2. And so using this, we could use the law of cosines to figure out what theta actually is here. So let's do that. We're going to use a little bit of trig identities, but law of cosines. So we get the opposite to the angle squared, so we get 24 squared. 24 squared is equal to 25/2 squared, plus 25/2 squared, minus 2 times 25/2 times 25/2 times the cosine of this angle, times-- let me scroll over-- the cosine of theta plus 90 degrees. this has it in terms of cosine of theta plus 90 degrees. How can we figure out the sine of theta? That's what we actually care about, to figure out the area of this triangle, or actually to figure out the height of this triangle. And to do that, you just have to make the realization. We know the trig identity that the cosine of theta is equal-- I won't use theta because I don't want to overload theta-- cosine of x is equal to sine of 90 minus x. So the cosine of theta plus 90 degrees is going to be equal to the sine of-- let me put it in parentheses-- 90 minus whatever is here. 90 minus theta minus 90, which is equal to-- the 90's cancel out-- sine of negative theta. And we know sine of negative theta is equal to the negative sine of theta. is the negative sine of theta. So we could write the sine of theta here and then put the negative out here. And this becomes a positive. So what does this simplify to? We have 24 squared, which is 576. 576 is equal to-- let's see. I won't skip any steps here. So we have 25 squared plus 25 squared. This is 2 times 25/2 squared plus 2 times-- this is 25/2 squared again-- times sine of theta. Now we just have to solve for sine of theta. So this is going to be equal to-- well, this is 576 is equal to 2 times 25/2 squared." + }, + { + "Q": "\nIn 2:15 did he make a mistake?\nYou are supposed to divide with the biggest number inside the box, but he did the opposite. Is that a mistake?\nM&M", + "A": "No, it is not a mistake. The problem is 4 / 16 = 4 \u00c3\u00b7 16. 16 is the divisor, it goes outside. Maybe you need to review decimal division. This usually comes up at that time.", + "video_name": "FaDtge_vkbg", + "timestamps": [ + 135 + ], + "3min_transcript": "Let's give ourselves a little bit of practice with percentages. So let's ask ourselves, what percent of-- I don't know, let's say what percent of 16 is 4? And I encourage you to pause this video and to try it out yourself. So when you're saying what percent of 16 is 4, percent is another way of saying, what fraction of 16 is 4? And we just need to write it as a percent, as per 100. So if you said what fraction of 16 is 4, you would say, well, look, this is the same thing as 4/16, which is the same thing as 1/4. But this is saying what fraction 4 is of 16. You'd say, well, 4 is 1/4 of 16. But that still doesn't answer our question. What percent? So in order to write this as a percent, we literally have to write it as something over 100. The word \"cent\" you know from cents and century. It relates to the number 100. So it's per 100. So you could say, well, this is going to be equal to question mark over 100, the part of 100. And there's a bunch of ways that you could think about this. You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25. In the numerator to go from-- I need to also multiply by 25 in order to have an equivalent fraction. So I'm also going to multiply by 25. So 1/4 is the same thing as 25/100. And another way of saying 25/100 is this is 25 per 100, or 25%. So this is equal to 25%. Now, there's a couple of other ways you could have thought about it. You could have said well, 4/16, this is literally 4 divided by 16. and convert to a decimal, which is very easy to convert to a percentage. So let's try to actually do this division right over here. So we're going to literally divide 4 by 16. Now, 16 goes into 4 zero times. 0 times 16 is 0. You subtract, and you get a 4. And we're not satisfied just having this remainder. We want to keep adding zeroes to get a decimal answer right over here. So let's put a decimal right over here. We're going into the tenths place. And let's throw some zeroes right over here. The decimal makes sure we keep track of the fact that we are now in the tenths, and in the hundredths, and in the thousandths place if we have to go that far. But let's bring another 0 down. 16 goes into 40 two times. 2 times 16 is 32. If you subtract, you get 8. And you could bring down another 0. And we have 16 goes into 80." + }, + { + "Q": "\nisn't it 1/2 on 0:47?", + "A": "No, it s 1/4. Simplify 4/16 by going like this: 4 divided by four equals 1 16 divided by four equals 4", + "video_name": "FaDtge_vkbg", + "timestamps": [ + 47 + ], + "3min_transcript": "Let's give ourselves a little bit of practice with percentages. So let's ask ourselves, what percent of-- I don't know, let's say what percent of 16 is 4? And I encourage you to pause this video and to try it out yourself. So when you're saying what percent of 16 is 4, percent is another way of saying, what fraction of 16 is 4? And we just need to write it as a percent, as per 100. So if you said what fraction of 16 is 4, you would say, well, look, this is the same thing as 4/16, which is the same thing as 1/4. But this is saying what fraction 4 is of 16. You'd say, well, 4 is 1/4 of 16. But that still doesn't answer our question. What percent? So in order to write this as a percent, we literally have to write it as something over 100. The word \"cent\" you know from cents and century. It relates to the number 100. So it's per 100. So you could say, well, this is going to be equal to question mark over 100, the part of 100. And there's a bunch of ways that you could think about this. You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25. In the numerator to go from-- I need to also multiply by 25 in order to have an equivalent fraction. So I'm also going to multiply by 25. So 1/4 is the same thing as 25/100. And another way of saying 25/100 is this is 25 per 100, or 25%. So this is equal to 25%. Now, there's a couple of other ways you could have thought about it. You could have said well, 4/16, this is literally 4 divided by 16. and convert to a decimal, which is very easy to convert to a percentage. So let's try to actually do this division right over here. So we're going to literally divide 4 by 16. Now, 16 goes into 4 zero times. 0 times 16 is 0. You subtract, and you get a 4. And we're not satisfied just having this remainder. We want to keep adding zeroes to get a decimal answer right over here. So let's put a decimal right over here. We're going into the tenths place. And let's throw some zeroes right over here. The decimal makes sure we keep track of the fact that we are now in the tenths, and in the hundredths, and in the thousandths place if we have to go that far. But let's bring another 0 down. 16 goes into 40 two times. 2 times 16 is 32. If you subtract, you get 8. And you could bring down another 0. And we have 16 goes into 80." + }, + { + "Q": "At 2:37, how do you know the slope of the tangent is one?\n", + "A": "It is given that tan \u00ce\u00b8 = 1 in the problem.", + "video_name": "MABWdzmZFIQ", + "timestamps": [ + 157 + ], + "3min_transcript": "is going to be the inverse tangent of one. So it might be tempting to just pick this one right over here. Type inverse tangent of one into his calculator. So maybe this looks like the best choice. But remember, I said if we restrict the domain right over here. if we restrict the possible values of tangent, of theta here appropriately, then this is going to simplify to this. But there is a scenario where this does not happen. And that's if we pick thetas that are outside of the range of the inverse tangent function. What do I mean by that? Well it's really just based on the idea that there are multiple angles that have... or multiple angles whose tangent is one. And let me draw that here with a unit circle here. So we draw a unit circle, so that's my x axis, that's my y axis, let me draw my unit circle here. have to draw the unit circle, because the tangent is really much more about the slope of the ray created by the angle, than where it intersects the unit circle as would be the case with sine and cosine. So if you have.. so you could have this angle right over here. So let's say this is a candidate theta, where the tangent of this theta is the slope of this line, and this terminal angle, the terminal ray, you could say of the angle. The other side, the initial ray, is along the positive x axis. And so you could say, okay the tangent of this theta, the tangent of this theta is one. Because the slope of this line is one. Let me scroll over a little bit. Well, so let me write it this way. So tangent theta is equal to one. But I can construct another theta whose tangent is equal to one by going all the way over here and essentially going in the opposite direction but the slope of this line, so let's call this theta two, tangent of theta two is also going to be equal to one. And of course you could go another but that's functionally the same angle in terms of where it is relative to the positive x axis, or what direction it points into, but this one is fundamentally a different angle. So we do not know, we do not have enough information just given what we've been told to know exactly which theta we're talking about, whether we're talking about this orange theta or this mauve theta. So I would say the get more information, there are multiple angles which fit this description." + }, + { + "Q": "\nAt 7:43, I would have separated it into the span of [.5,0] and [0,1]. Why is that wrong?", + "A": "The span of (1/2, 1) is a line parallel to (1/2, 1). The span of {(1/2, 0), (0, 1)} is the span of {(1, 0), (0, 1)} - the xy plane.", + "video_name": "3-xfmbdzkqc", + "timestamps": [ + 463 + ], + "3min_transcript": "corresponds to this pivot column, plus or minus 1/2 times my second entry has got to be equal to that 0 right there. Or, v1 is equal to 1/2 v2. And so if I wanted to write all of the eigenvectors that satisfy this, I could write it this way. My eigenspace that corresponds to lambda equals 5. That corresponds to the eigenvalue 5 is equal to the set of all of the vectors, v1, v2, that are equal to some scaling factor. Let's say it's equal to t times what? If we say that v2 is equal to t, so v2 is going to be equal to t times 1. or 1/2 times t. Just like that. For any t is a member of the real numbers. If we wanted to, we could scale this up. We could say any real number times 1, 2. That would also be the span. Let me do that actually. It'll make it a little bit cleaner. Actually, I don't have to do that. So we could write that the eigenspace for the eigenvalue 5 is equal to the span of the vector 1/2 and 1. So it's a line in R2. Those are all of the eigenvectors that satisfy-- that work for the equation where the eigenvalue is equal to 5. Now what about when the eigenvalue is equal to minus 1? So let's do that case. When lambda is equal to minus 1, then we have-- it's going So the eigenspace for lambda is equal to minus 1 is going to be the null space of lambda times our identity matrix, which is going to be minus 1 and 0, 0, minus 1. It's going to be minus 1 times 1, 0, 0, 1, which is just minus 1 there. Minus A. So minus 1, 2, 4, 3. And this is equal to the null space of-- minus 1, minus 1 is minus 2. 0 minus 2 is minus 2. 0 minus 4 is minus 4 and minus 1 minus 3 is minus 4. And that's going to be equal to the null space of the reduced row echelon form of that guy. So we can perform some row operations right here. Let me just put it in reduced row echelon form. So if I replace my second row plus 2 times my first row." + }, + { + "Q": "at 9:40, i got 1 and 1 at the bottom row instead because I kept the bottom row the same rather than the top row. Is that also correct?\n", + "A": "Yes, that has the same meaning. You will still end up with v1 + v2 = 0.", + "video_name": "3-xfmbdzkqc", + "timestamps": [ + 580 + ], + "3min_transcript": "So the eigenspace for lambda is equal to minus 1 is going to be the null space of lambda times our identity matrix, which is going to be minus 1 and 0, 0, minus 1. It's going to be minus 1 times 1, 0, 0, 1, which is just minus 1 there. Minus A. So minus 1, 2, 4, 3. And this is equal to the null space of-- minus 1, minus 1 is minus 2. 0 minus 2 is minus 2. 0 minus 4 is minus 4 and minus 1 minus 3 is minus 4. And that's going to be equal to the null space of the reduced row echelon form of that guy. So we can perform some row operations right here. Let me just put it in reduced row echelon form. So if I replace my second row plus 2 times my first row. Minus 2, minus 2. And then my second row, I'll replace it with two times-- I'll replace it with it plus 2 times the first. Or even better, I'm going to replace it with it plus minus 2 times the first. So minus 4 plus 4 is 0. And then if I divide the top row by minus 2, the reduced row echelon form of this matrix right here or this matrix right here is going to be 1, 1, 0. So the eigenspace that corresponds to the eigenvalue minus 1 is equal to the null space of this guy right here It's the set of vectors that satisfy this equation: 1, 1, 0, 0. And then you have v1, v2 is equal to 0. Or you get v1 plus-- these aren't vectors, v1 plus v2 is equal to 0. Because 0 is just equal to that thing right there. So 1 times v1 plus 1 times v2 is going to be equal to that 0 right there. Or I could write v1 is equal to minus v2. Or if we say that v2 is equal to t, we could say v1 is equal to minus t. Or we could say that the eigenspace for the eigenvalue minus 1 is equal to all of the vectors, v1, v2 that are equal to some scalar t times v1 is minus t and v2 is plus t. Or you could say this is equal to the span of the vector minus 1 and 1. So let's just graph this a little bit just to understand what we just did. We were able to find two eigenvalues for" + }, + { + "Q": "why couldn't the five at 0:47 be plus or minus five?\n", + "A": "mj, You re correct. 25x\u00e2\u0081\u00b4-30x\u00c2\u00b2+9 can be factored to either (5x\u00c2\u00b2-3)(5x\u00c2\u00b2-3) or (-5x\u00c2\u00b2+3)(-5x\u00c2\u00b2+3)", + "video_name": "o-ZbdYVGehI", + "timestamps": [ + 47 + ], + "3min_transcript": "We need to factor 25x to the fourth minus 30x squared plus 9. And this looks really daunting because we have something to the fourth power here. And then the middle term is to the second power. But there's something about this that might pop out at you. And the thing that pops out at me at least is that 25 is a perfect square, x to the fourth is a perfect square, so 25x to the fourth is a perfect square. And 9 is also perfect square, so maybe this is the square of some binomial. And to confirm it, this center term has to be two times the product of the terms that you're squaring on either end. Let me explain that a little bit better So, 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now, what is 30x squared? So remember, this needs to be two times the product of what's inside the square, or the square root of this and the square root of that. Given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square. So we can just rewrite this as this is equal to 5x squared-- let me do it in the same color. 5x squared minus 3 times 5x squared minus 3. And if you want to verify it for yourself, multiply this out. You will get 25x to the fourth minus 30x squared plus 9." + }, + { + "Q": "Can anyone explain to me where the 2 came from at 1:26? Thanks!\n", + "A": "He s just checking to make sure that it is a perfect square", + "video_name": "o-ZbdYVGehI", + "timestamps": [ + 86 + ], + "3min_transcript": "We need to factor 25x to the fourth minus 30x squared plus 9. And this looks really daunting because we have something to the fourth power here. And then the middle term is to the second power. But there's something about this that might pop out at you. And the thing that pops out at me at least is that 25 is a perfect square, x to the fourth is a perfect square, so 25x to the fourth is a perfect square. And 9 is also perfect square, so maybe this is the square of some binomial. And to confirm it, this center term has to be two times the product of the terms that you're squaring on either end. Let me explain that a little bit better So, 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now, what is 30x squared? So remember, this needs to be two times the product of what's inside the square, or the square root of this and the square root of that. Given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square. So we can just rewrite this as this is equal to 5x squared-- let me do it in the same color. 5x squared minus 3 times 5x squared minus 3. And if you want to verify it for yourself, multiply this out. You will get 25x to the fourth minus 30x squared plus 9." + }, + { + "Q": "At 4:34, How did you turn the 370, to a 371?? Why add the 1??\n", + "A": "The solution to the problem was that she should sell 370.3 computers in order to make a profit. However, in real life you can t sell a fraction of a computer. If she sold 370 computers she would make too little money to make a profit, so you have to round up and say that she should sell 371.", + "video_name": "roHvNNFXr4k", + "timestamps": [ + 274 + ], + "3min_transcript": "And then to solve for x, we just have to divide both sides by 27. Divide both sides by 27. On our right-hand side, we have x. So let me just write this down. So we have x on our right-hand side is going to be equal to 10,000/27. I switched the right and the left-hand sides here. Now, what is this going to be? Well, we can do a little bit of long division to handle that. So 27 goes into 10,000. So 27 doesn't go into 1. Doesn't go into 10. It goes into 100 three times. 3 times 27 is what? 81. 100 minus 81 is 19. Then we can bring down a 0. 27 goes into 190? It looks like it will go into it about six times. 6 times 7 is 42. 6 times 2 is 12. Plus 4. 16. Let's see, 90 minus 62 is actually 28. Let me turn this back. So it goes 7 times. 7 times 7 is 49. 7 times 2 is 14. Plus 4 is 18. There you go. So 190 minus 189. We get 1. Let's bring down another 0. We bring down another 0. We have a 0 right there. 27 goes into 10 how many times? Well, it doesn't go into 10 at all. So we'll put a 0 right there. 0 times 27 is 0. Then we subtract. And then, we get 10 again. And now we're in the decimal range. Or we're going to start getting decimal values. We can bring down another 0. We could get 27 goes into 100 three times. And then we're going to keep going on and on and on and on. But this is enough information for us to answer our question. What is the minimum number of computers she needs to sell in a month to make a profit? Well, she can't sell a decimal number of computers, a third of a computer. She could either sell 370 computers. If she sells 370 computers, then she's not going to get to break even because that's less than the quantity she needs for break even. So she needs to sell 371. She needs to sell 371 computers in a month to make a profit." + }, + { + "Q": "\nWhy did Sal switch sides 2:28 (the variable and the number)", + "A": "He switched the sides simply because it s customary to write equations with the variable on the left and the number on the right. Switching the sides doesn t change the validity of the equation. For example, if 2 + 7 = 13 - 4, then 13 -4 = 2 + 7.", + "video_name": "roHvNNFXr4k", + "timestamps": [ + 148 + ], + "3min_transcript": "Marcia has just opened her new computer store. She makes $27 on every computer she sells and her monthly expenses are $10,000. What is the minimum number of computer she needs to sell in a month to make a profit? So I'll let you think about that for a second. Well, let's think about what we have to figure out. We have to figure out the minimum number of computers she needs to sell. So let's set that to a variable or set a variable to represent that. So let's let x equal the number of computers she sells. Number of computers sold. Now, let's think about how much net profit she will make in a month. And that's what we're thinking about. How many computers, the minimum number she needs to sell in order to make a net profit? So I'll write her profit is going to be how much money she brings in from selling the computers. And she makes $27 on every computer she sells. So her profit is going to be $27 times the number of computer she sells. But we're not done yet. She still has expenses of $10,000 per month. So we're going to have to subtract out the $10,000. What we care about is making a profit. We want this number right over here to be greater than 0. So let's just think about what number of computers would get us to 0. And then, maybe she needs to sell a little bit more than that. So let's see what gets her to break even. So break even-- that's 0 profit. Neither positive or negative-- is equal to 27 times-- and I'll do it all in one color now. 27x minus 10,000. Well, we've seen equations like this before. We can add 10,000 to both sides. Add 10,000 to both sides, so it's no longer on the right-hand side. And we are left with 10,000. And then to solve for x, we just have to divide both sides by 27. Divide both sides by 27. On our right-hand side, we have x. So let me just write this down. So we have x on our right-hand side is going to be equal to 10,000/27. I switched the right and the left-hand sides here. Now, what is this going to be? Well, we can do a little bit of long division to handle that. So 27 goes into 10,000. So 27 doesn't go into 1. Doesn't go into 10. It goes into 100 three times. 3 times 27 is what? 81. 100 minus 81 is 19. Then we can bring down a 0. 27 goes into 190? It looks like it will go into it about six times." + }, + { + "Q": "\nAt 2:03, why does Khan leave a blank space?\n\nshouldn't it be; (assume \"///\" is just a space for formatting purpose)\n///////////////////// (4/9)^1 + (4/9)^2 + (4/9)^3 + . . .\n///////////////////// (4/9)^2 + (4/9)^3 + (4/9)^4 + . . .\n//////////////////// ----------------------------------------- -\nnot\n/////////////////////////// (4/9) + (4/9)^2 + (4/9)^3 + ...\n///////////////////////////////////// (4/9)^2 + (4/9)^3 + (4/9)^4 + ...\n///////////////////// ----------------------------------------- -", + "A": "Use code blocking for monospace formatting. (Click the Formatting tips hyperreference below the entry block) (4/9)^1 + (4/9)^2 + (4/9)^3 + ... (4/9)^2 + (4/9)^3 + (4/9)^4 + ... You want the addition to be applied to equivalent exponents: (4/9)^1 + (4/9)^2 + (4/9)^3 + ... (4/9)^2 + (4/9)^3 + (4/9)^4 + ... (4/9)^1 + 2*(4/9)^2 + 2*(4/9)^3 + 2*...", + "video_name": "vBlR2xNAGmo", + "timestamps": [ + 123 + ], + "3min_transcript": "In the last video we got as far as figuring out that the area of this Koch snowflake This thing that has an infinite perimeter, can be expressed as this infinite sum over here So our job in this video is to try to simplify this, and hopefully get a finite value Let's do our best to actually simplify this thing right over here So the easiest part of this thing to simplify is this right over here So let's just focus on that Then if we can get a value for this part that I am bracketing off Then we can just place that value here and simplify the rest of it So what I've just bracketed off can be re-written as three times four ninths plus four ninths squared plus four ninths to the third power And you can go on and on and on Plus four ninths to every other power, all the way through infinity! Lucky for us, there is a way to figure out this infinite (geometric) series But I'll just do it by hand this time, just so that we don't have to resort to some magical formulas So let's say that we define some sum, this one over here (let's call it S) Let's say that S is equal to what we have in parentheses over here It's going to be equal to four ninths, plus four ninths squared, plus four ninths to the third all the way to infinity Now let's also say that we multiply S by four ninths What's four ninths S going to look like? So then, I'm just essentially multiplying every term here by four ninths So if I take this first term and multiply it by four ninths, what am I going to get? If I take the second term and multiply it by four ninths, I'm going to get four ninths to the third power And we are going to go all the way to infinity So this is interesting When I multiply four ninths times this I get all of the terms here except for this first four ninths Now, this is kind of the magic of how we can actually find the sum of an infinite geometric series We can subtract this term right over here (this pink line) from this green line If we do that, clearly this is equal to that and this is equal to that So if we subtract this from that its equivalent to subtracting the pink from the green So we get S minus four ninths S is equal to... Well, every other term, this guy minus this guy is going to cancel out" + }, + { + "Q": "Why does Sal use shorthand at 5:28?\n", + "A": "conserving space.", + "video_name": "aTjNDKlz8G4", + "timestamps": [ + 328 + ], + "3min_transcript": "Then I'll just cancel this out and I would have gotten inches per minute. So anyway, I don't want to confuse you too much with all of that unit cancellation stuff. The bottom line is you just remember, well if I'm going 28 feet per minute, I'm going to go 12 times that many inches per minute, right, because there are 12 inches per foot. So I'm going 336 inches per minute. So now I have the question, but we're not done, because the question is how many inches am I going to be traveling in 1 second. So let me erase some of the stuff here at the bottom. So 336 inches -- let's write it like that -- inches per minute, Well what do we know? We know that 1 minute -- and notice, I write it in the numerator here because I want to cancel it out with this minute here. 1 minute is equal to how many seconds? It equals 60 seconds. And this part can be confusing, but it's always good to just take a step back and think about what I'm doing. If I'm going to be going 336 inches per minute, how many inches am I going to travel in 1 second? Am I going to travel more than 336 or am I going to travel less than 336 inches per second. Well obviously less, because a second is a much shorter period of time. So if I'm in a much shorter period of time, I'm going to be traveling a much shorter distance, if I'm So I should be dividing by a number, which makes sense. I'm going to be dividing by 60. I know this can be very confusing at the beginning, but that's why I always want you to think about should I be getting a larger number or should I be getting a smaller number and that will always give you a good reality check. And if you just want to look at how it turns out in terms of units, we know from the problem that we want this minutes to cancel out with something and get into seconds. So if we have minutes in the denominator in the units here, we want the minutes in the numerator here, and the seconds in the denominator here. And 1 minute is equal to 60 seconds. So here, once again, the minutes and the minutes cancel out. And we get 336 over 60 inches per second. Now if I were to actually divide this out, actually we" + }, + { + "Q": "\n6:39 how did you get 5.6 shouldn't it be 5.3?\nSorry, I'm confused. :/", + "A": "Its basically 336 divided by 60. Your answer is 5.6. Hope that helps :)", + "video_name": "aTjNDKlz8G4", + "timestamps": [ + 399 + ], + "3min_transcript": "So I should be dividing by a number, which makes sense. I'm going to be dividing by 60. I know this can be very confusing at the beginning, but that's why I always want you to think about should I be getting a larger number or should I be getting a smaller number and that will always give you a good reality check. And if you just want to look at how it turns out in terms of units, we know from the problem that we want this minutes to cancel out with something and get into seconds. So if we have minutes in the denominator in the units here, we want the minutes in the numerator here, and the seconds in the denominator here. And 1 minute is equal to 60 seconds. So here, once again, the minutes and the minutes cancel out. And we get 336 over 60 inches per second. Now if I were to actually divide this out, actually we 6 goes into 336, what, 56 times? 56 over 10, and then we can divide that again by 2. So then that gets us 28 over 5. And 28 over 5 -- let's see, 5 goes into 28 five times, 25. 3, 5.6. So this equals 5.6. So I think we now just solved the problem. If Zack is going 28 feet in every minute, that's his speed, he's actually going 5.6 inches per second. Hopefully that kind of made sense. Let's try to see if we could do another one. per hour is that? Well, 91 feet per second. If we want to say how many miles that is, should we be dividing or should we be multiplying? We should be dividing because it's going to be a smaller number of miles. We know that 1 mile is equal to -- and you might want to just memorize this -- 5,280 feet. It's actually a pretty useful number to know. And then that will actually cancel out the feet. Then we want to go from seconds to hours, right? So, if we go from seconds to hours, if I can travel 91 feet" + }, + { + "Q": "You lost me at 6:39 >.<\n", + "A": "Its basically 336 divided by 60. Your answer is 5.6. Hope that helps :)", + "video_name": "aTjNDKlz8G4", + "timestamps": [ + 399 + ], + "3min_transcript": "So I should be dividing by a number, which makes sense. I'm going to be dividing by 60. I know this can be very confusing at the beginning, but that's why I always want you to think about should I be getting a larger number or should I be getting a smaller number and that will always give you a good reality check. And if you just want to look at how it turns out in terms of units, we know from the problem that we want this minutes to cancel out with something and get into seconds. So if we have minutes in the denominator in the units here, we want the minutes in the numerator here, and the seconds in the denominator here. And 1 minute is equal to 60 seconds. So here, once again, the minutes and the minutes cancel out. And we get 336 over 60 inches per second. Now if I were to actually divide this out, actually we 6 goes into 336, what, 56 times? 56 over 10, and then we can divide that again by 2. So then that gets us 28 over 5. And 28 over 5 -- let's see, 5 goes into 28 five times, 25. 3, 5.6. So this equals 5.6. So I think we now just solved the problem. If Zack is going 28 feet in every minute, that's his speed, he's actually going 5.6 inches per second. Hopefully that kind of made sense. Let's try to see if we could do another one. per hour is that? Well, 91 feet per second. If we want to say how many miles that is, should we be dividing or should we be multiplying? We should be dividing because it's going to be a smaller number of miles. We know that 1 mile is equal to -- and you might want to just memorize this -- 5,280 feet. It's actually a pretty useful number to know. And then that will actually cancel out the feet. Then we want to go from seconds to hours, right? So, if we go from seconds to hours, if I can travel 91 feet" + }, + { + "Q": "At 1:01, why is it called the Foil method? Does 'Foil' stand for something or is it just called that?\n", + "A": "FOIL stands for First, Outside, Inside, Last . You multiply together the first term in each binomial, the outside (leftmost and rightmost) terms, the inside terms, and the last term in each binomial. Take those four products, add them up, and you have the expanded expression.", + "video_name": "HLNSouzygw0", + "timestamps": [ + 61 + ], + "3min_transcript": "- [Instructor] We're now going to explore factoring a type of expression called a difference of squares and the reason why it's called a difference of squares is 'cause it's expressions like x squared minus nine. This is a difference. We're subtracting between two quantities that are each squares. This is literally x squared. Let me do that in a different color. This is x squared minus three squared. It's the difference between two quantities that have been squared and it turns out that this is pretty straightforward to factor. And to see how it can be factored, let me pause there for a second and get a little bit of review of multiplying binomials. So put this on the back burner a little bit. Before I give you the answer of how you factor this, let's do a little bit of an exercise. Let's multiply x plus a times x minus a where a is some number. Now, we can use that, but I like just thinking of this as a distributive property twice. We could take x plus a and distribute it onto the x and onto the a. So when we multiply it by x, we would get x times x is x squared, a times x is plus ax and then when we multiply it by the negative a, well, it'll become negative a times x minus a squared. So these middle two terms cancel out and you are left with x squared minus a squared. You're left with a difference of squares. x squared minus a squared. So we have an interesting result right over here that x squared minus a squared is equal to, is equal to x plus a, x plus a times x minus a. And so we can use, and this is for any a. Here, what is our a? Our a is three. This is x squared minus three squared or we could say minus our a squared if we say three is a and so to factor it, this is just going to be equal to x plus our a which is three times x minus our a which is three. So x plus three times x minus three. Now, let's do some examples to really reinforce this idea of factoring differences of squares. So let's say we want to factor, let me say y squared minus 25 and it has to be a difference of squares. It doesn't work with a sum of squares. Well, in this case, this is going to be y and you have to confirm, okay, yeah, 25 is five squared and y squared is well, y squared. So this gonna be y plus something times y minus something" + }, + { + "Q": "\nWhy add -11 to the other side when you could add 4 and equal them out 0:31", + "A": "Lauri, The problem was to Use the drop down menu to form a linear equation with no solution. If you chose the 4, you would have a linear equation which is valid for any x and so it has an infinite number of solutions for x.", + "video_name": "uQs100shv-A", + "timestamps": [ + 31 + ], + "3min_transcript": "We're asked to use the drop-downs to form a linear equation with no solutions. So a linear equation with no solutions is going to be one where I don't care how you manipulate it, the thing on the left can never be equal to the thing on the right. And so let's see what options they give us. One, they want us to-- we can pick the coefficient on the x term and then we can pick the constant. So if we made this negative 11x, so now we have a negative 11x on both sides. Here on the left hand side, we have negative 11x plus 4. If we do something other than 4 here, so if we did say negative 11x minus 11, then here we're not going to have any solutions. And you say, hey, Sal how did you come up with that? Well think about it right over here. We have a negative 11x here, we have a negative 11x there. If you wanted to solve it algebraically you could add 11x to both sides and both of these terms will cancel out with each other and all you would be left with is a 4 is equal to a negative 11, which is not possible for any x that you pick. Another way that you think about it is here we have negative 11 times some number taking negative 11 times that same number and we're subtracting 11 from it. So if you take a negative 11 times some number and on one side you add four, and on the other side you subtract 11, there's no way, it doesn't matter what x you pick. There's no x for which that is going to be true. But let's check our answer right over here." + }, + { + "Q": "At 1:20 Sal says that there is only one triangle, but couldn't he have pulled it to the right?\n", + "A": "If you mean, could he have pulled the bottom right point to the right more , then no. If he pulled it to the right more, then the triangle wouldn t have two sides of length 3, which is what the directions are asking for.", + "video_name": "lohMwoq3WFA", + "timestamps": [ + 80 + ], + "3min_transcript": "They're asking us to draw a right triangle. So that means it has to have a 90-degree angle. But it's also an isosceles triangle, so that means it has to have at least two sides equal and has two sides of length 3. So those two sides that are going to be equal are going to be of length 3, and it's got to be a right triangle. So let's see if we can do that. So let's try to make this right over here the right angle. And let's make this side and this side have length 3, so 3 and then 3 right over there. Let me make sure I get that right angle right. OK, there you go. So it's a right angle. It's isosceles. At least two sides are equal. And the two sides have length 3. So it seems like we've met all of our constraints. Now they say, is there a unique triangle that satisfies this condition? So another way of rephrasing that, is this the only triangle that I could have drawn that meets these conditions? I can't change this angle if I want to meet these conditions. I can't change these two lengths. And if you keep this angle constant and you keep these two lengths constant, then this point and this point are going to be So this is the only side that can connect those two points. So this is the only triangle that meets those conditions. You can't have different side lengths, or you couldn't have different angles right over here and also meet those conditions. So is there a unique triangle that satisfies the given conditions? Yes, there's only one unique triangle." + }, + { + "Q": "at 2:37.... someone tell me that was a typo on the transcript.\n", + "A": "Of course it is, it should say when is f of x increasing . A very unfortunate typo.", + "video_name": "KxOp3s9ottg", + "timestamps": [ + 157 + ], + "3min_transcript": "between a and b. At x equals a or at x equals b the value of our function is zero but it's positive when x is between a and b, a and b or if x is greater than c. X is, we could write it there, c is less than x or we could write that x is greater than c. These are the intervals when our function is positive. Let me write this, f of x, f of x positive when x is in this interval or this interval or that interval. So when is f of x negative? Let me do this in another color. F of x is going to be negative. Well, it's gonna be negative if x is less than a. So this is if x is less than a or if x is between b and c F of x is down here so this is where it's negative. So here or, or x is between b or c, x is between b and c. And I'm not saying less than or equal to because at b or c the value of the function f of b is zero, f of c is zero. That's where we are actually intersecting the x-axis. So that was reasonably straightforward. Now let's ask ourselves a different question. When is the function increasing or decreasing? So when is f of x, f of x increasing? Well increasing, one way to think about it is every time that x is increasing then y should be increasing or another way to think about it, you have a, you have a positive rate of change of y with respect to x. We could even think about it as imagine if you had a tangent line at any of these points. of that tangent line is going to be positive. But the easiest way for me to think about it is as you increase x you're going to be increasing y. So where is the function increasing? Well I'm doing it in blue. So it's increasing right until we get to this point right over here, right until we get to that point over there then it starts decreasing until we get to this point right over here and then it starts increasing again. It starts, it starts increasing again. So let me make some more labels here. So let's say that this, this is x equals d and that this right over here, actually let me do that in green color, so let's say this is x equals d. Now it's not a, d, b but you get the picture and let's say that this is x is equal to, x is equal to," + }, + { + "Q": "at about 6:45 mins into the video, sal differentiates 6tan(theta). why is 6 not differentiated but tan is?\n", + "A": "hmm.... You know what... a constant can never be differentiated...... !!!", + "video_name": "fD7MbnXbTls", + "timestamps": [ + 405 + ], + "3min_transcript": "Let me scroll over more. So that's the derivative of the outside. If the cosine theta was just an x, you would say x to the minus 1 derivative is minus 1 x to the minus 2. Now times the derivative of the inside. Of cosine of theta with respect to theta. So that's times minus sine of theta. I'm going to multiply all of that times sine of theta. The derivative of this thing, which is the stuff in green, times the first expression. So what does this equal? These cosine of theta divided by cosine of theta, that is equal to 1. And then I have a minus 1 and I have a minus sine of theta. That's plus plus. What do I have? I have sine squared, sine of theta time sine of theta over cosine squared. So plus sine squares of theta over cosine squared of theta. Which is equal to 1 plus tangent squared of theta. What's 1 plus tangent squared of theta? That's equal to secant squared of theta. secant squared of theta. All that work to get us fairly something-- it's nice when it comes out simple. So d x d theta, this is just equal to secant squared of theta. If we want to figure out what d x is equal to, d x is equal to just both sides times d theta. So it's 6 times secant squared theta d theta. That's our d x. Of course, in the future we're going to have to back substitute, so we want to solve for theta. That's fairly straightforward. Just take the arctangent of both sides of this equation. You get that the arctangent of x over 6 is equal to the theta. We'll save this for later. So what is our integral reduced to? Our integral now becomes the integral of d x? What's d x? It is 6 secant squared theta d theta. times 1 plus tangent squared of theta. We know that this right there is secant squared of theta. I've shown you that multiple times. So this is secant squared of theta in the denominator. We have a secant squared on the numerator, they cancel out. So those cancel out. So are integral reduces to, lucky for us, 6/36 which is just 1/6 d theta. Which is equal to 1/6 theta plus c. Now we back substitute using this result. Theta is equal to arctangent x over 6. The anti-derivative 1 over 36 plus x squared is equal to 1/6 times theta. Theta's just equal to the arctangent x over 6 plus c." + }, + { + "Q": "\nAt 1:05 for the slope formula of y=6 i thought the slope is 6 and the y intercept is 0? Is that right? If not, why isn't it?", + "A": "If the slope was 6, there would be a x after the 6.", + "video_name": "y5yNi08cr6I", + "timestamps": [ + 65 + ], + "3min_transcript": "We're asked which of these lines are parallel. So they give us three equations of three different lines and if they're parallel, then they have to have the So all we have to do over here is figure out the slopes of each of these lines, and if any of them are equal, they're parallel. So let's do line A. Line A, it's 2y is equal to 12x plus 10. We're almost in slope-intercept form, we can just divide both sides of this equation by 2. We get y is equal to 6x-- right, 12 divided by 2 -- 6x plus 5. So our slope in this case, we have it in slope-intercept form, our slope in this case is equal to 6. Let's try line B. Line B is y is equal to six. You might say this hey, this is a bizarre character, how do I get this into slope-intercept form, where's the x? And my answer to you is that it already is in slope-intercept form. I could just rewrite it as y is equal to 0x plus 6. slope here is 0. y is going to be equal to six no matter how much you change x. Change in y is always going to be 0, it's always going to be 6. So here, our slope is 0, so these two lines are definitely not parallel, they have different slopes. So let's try line C. Line C-- I'll do it down here. Line C, so it's y minus 2 is equal to 6 times x plus 2. And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2. So the point negative 2, 2, is being represented here because you're subtracting the points. And the slope is 6, so we already know that the slope is equal to 6. And sometimes people are more comfortable with slope-intercept form, so let's put it in slope-intercept form will still be equal to 6. So if we distribute the 6, we get y minus 2 is equal to 6 times x, 6x, plus 6 times 2 is 12. And if you add this 2 -- if you add 2 to both sides of the equation, you get y-- because these guys cancel out-- is equal to 6x plus 14. So you see, once again, the slope is 6. So line A and line C have the same the slope, so line A and line C are parallel. And they're different lines. If they had the same y-intercept, then they would just be the same line." + }, + { + "Q": "At 0:25, doesn't Sal mean \"letters,\" and not \"alphabets\"?\n", + "A": "In the general idea... letters... some people say different things.", + "video_name": "VYbqG2NuOo8", + "timestamps": [ + 25 + ], + "3min_transcript": "- [Voiceover] So let's ask ourselves some interesting questions about alphabets in the English language. And in case you don't remember and are in the mood to count, there are 26 alphabets. So if you go, \"A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, and Z,\" you'll get, you'll get 26, 26 alphabets. Now let's ask some interesting questions. So given that there are 26 alphabets in the English language, how many possible three letter words are there? And we're not going to be thinking about phonetics or how hard it is to pronounce it. So, for example, the word, the word ZGT would be a legitimate word in this example. Or the word, the word, the word SKJ would be a legitimate word in this example. So how many possible three letter words are there in the English language? I encourage you to pause the video and try to think about it. Alright, I assume you've had a go at it. So let's just think about it, for three letter words possibilities are there for the first one? Well, there's 26 possible letters for the first one. Anything from a to z would be completely fine. Now how many possibilities for the second one? And I intentionally ask this to you to be a distractor because we've seen a lot of examples. We're saying, \"Oh, there's 26 possibilities \"for the first one and maybe there's 25 for the second one, \"and then 24 for the third,\" but that's not the case right over here because we can repeat letters. I didn't say that all of the letters had to be different. So, for example, the word, the word HHH would also be a legitimate word in our example right over here. So we have 26 possibilities for the second letter and we have 26 possibilities for the third letter. So we're going to have, and I don't know what this is, 26 to the third power possibilities, or 26 times 26 time 26 and you can figure out what that is. That is how many possible three letter words we can have for the English language they are, if they meant anything and if we repeated letters. Now let's ask a different question. What if we said, \"How many possible three letter words \"are there if we want all different letters?\" So we want all different letters. So these all have to be different letters. Different, different letters and once again, pause the video and see if you can think it through. Alright, so this is where permutations start to be useful. Although, I think a lot of things like this, it's always best to reason through than try to figure out if some formula applies to it. So in this situation, well, if we went in order, we could have 26 different letters for the first one, 26 different possibilities for the first one. You know, I'm always starting with that one, but there's nothing special about the one on the left. We could say that the one on the right, there's 26 possibilities, well for each of those possibilities, for each of those 26 possibilities, there might be 25 possibilities" + }, + { + "Q": "At 4:11 when You got the 3rd row did you not make a mistake when you are converting the row, You said it was the 3rd row- 2X the second row. so for the 3rd row shouldn't it be 0 3 3 2 ? Because in the video you did the 3rd- 2x the first row.....\n", + "A": "I guess that was just misspoken, since the first row was the only row kept original. Sal meant to do the 3rd row minus 2 times the first row.", + "video_name": "QV0jsTiobU4", + "timestamps": [ + 251 + ], + "3min_transcript": "Everything above the main diagonal is 0, so this is the main diagonal right here, all the way down like that. All of these guys are going to be non-zero. All of that's going to be non-zero, and then the 0's are going to be above the diagonal, like that. We saw in the last video that the determinant of this guy is just equal to the product of the diagonal entries, which is a very simple way of finding a determinant. And you could use the same argument we made in the last video to say that the same is true of the lower triangular matrix, that its determinant is also just the product of those entries. I won't prove it here, but you can use the exact same argument you used in the video that I just did on the upper triangular. So given this, that the determinant of this is just the product of those guys, and that I can perform row operations on this guy and not change the determinant, maybe a simpler way to calculate this determinant is to get multiply the entries down the diagonal. So let's do that. So we want to find the determinant of A. It's 1, 2, 2, 1, 1, 1, 2, 4, 2, 2, 7, 5, 2, minus 1, 4, minus 6, 3. Now let's try to get this into upper triangular form. So let's replace the second row with the-- so I'm just going to keep the first row the same. 1, 2, 2, 1. And let's replace the second row with the second row minus the first row. The second row minus the first row is going to be equal to 1 minus 1 is 0. So in this case the constant is just 1. 2 minus 2 is 0. 4 minus 2 is 2. 2 minus 1 is 1. Now let's replace the third row with the third row minus 2 times the second row. So 2 minus 2 times 1 is 0. 7 minus 2 times 2 is 3. 5 minus 2 times 2 is 1. 2 minus 2 times 1 is 0. Let me get a good color here. I'll do pink. Let's replace the last row with the last row, essentially, plus the first row. You could say minus minus 1 times the first row is the same thing as the last row plus the first row. So minus 1 plus 1 is 0. 4 plus 2 is 6. Minus 6 plus 2 is minus 4." + }, + { + "Q": "\nat 2:32 why does Sal get -xe^-x^2 instead of xe^x^2?", + "A": "Those negatives do not cancel.", + "video_name": "DL-ozRGDlkY", + "timestamps": [ + 152 + ], + "3min_transcript": "If you can on one side of this equation through algebra separate out the Ys and the DYs and on the other side have all the Xs and DXs, and then integrate. Perhaps you can find the particular solution to this differential equation that contains this point. Now if you can't do it don't worry because we're about to work through it. So like I said, let's use a little bit of algebra to get all the Ys and DYs on one side and all the Xs and DXs on the other side. So one way, let's say I want to get all the Ys and DYs on the left hand side, and all the Xs and DXs on the right hand side. Well, I can multiply both sides times Y. So I can multiply both sides times Y that has the effect of putting the Ys on the left hand side and then I can multiple both sides times DX. I can multiple both sides times DX and we kind of treat, you can treat these differentials when you're manipulating it to essentially separate out the variables. And so, this will cancel with that. And so, we are left with Y, DY. Y, DY is equal to negative X. And actually let me write it this way. Let me write it as negative X, E. Actually, I might need a little more space. So negative X E to the negative X squared DX. DX. Now why is this interesting? Because we could integrate both sides. And now this also highlights why we call it the separable. You won't be able to do this with every differential equation. You won't be able to algebraically separate the Ys and DYs on one side and the Xs and DXs on the other side. But this one we were able to. a separable differential equation. Differential equation. And it's usually the first technique that you should try. Hey, can I separate the Ys and the Xs and as I said, this is not going to be true of many, if not most differential equations. But now that we did this we can integrate both sides. So let's do that. So, I'll find a nice color to integrate with. So, I'm going to integrate both sides. Now if you integrate the left hand side what do you get? You get and remember, we're integrating with respect to Y here. So this is going to be Y squared over two and we could put some constant there. I could call that plus C one. And if you're integrating that that's going to be equal to. Now the right hand side we're integrating with respect to X. And let's see, you could do U substitution or you could recognize that look, the derivative of negative X squared" + }, + { + "Q": "In 0:28 sal says 8 goes into 37 three times but can you add the remainder??\n", + "A": "8 goes into 37 4 times, ( because 8 x 4 = 32 ) and that is why he writes 4 on top of the line.. The reminder then is 5, ( because 37- 32 = 5 ) and he then uses that when he brings down another place value - 7.. So the next thing to figure out then is how many times 8 goes into 57, and so on.... Did that make any sense ?", + "video_name": "MbpmP1esh-Q", + "timestamps": [ + 28 + ], + "3min_transcript": "Let's divide 3,771 by 8. So we're going to do a little bit of long division is the way we set it up. So the first thing we want to do and think about is, well, does 8 go into 3? Well, no, 8 does not go into 3. So we'll keep moving. Does 8 go into 37? Sure. 8 times 4 is 37. 8 times 5 is 40. So that's too big. So 8 goes into 37 four times. 4 times 8 is 32. And we subtract, and we get 37 minus 32 is 5. And just so you have a little glimpse of what's actually going on here, we're really saying how many times does 8 go into 3,700? And we're saying that 8 goes into 3,700 400 times. There's 4 sitting in the hundreds place, and that 400 times 8 is 3,200. And then when you subtract 3,200 from 3,700, you're getting 500. But we'll revert back to just the traditional long division But that's what's really going on here. we're really saying 8 goes into 3,700 400 times. But let's keep going. So now, we have this 5 to deal with, which is, as we already talked about, it's in the hundreds place. It's really a 500. But now, let's bring down another place value. So let's bring down the 7. And let's think about how many times does 8 go into 57. And we're really thinking about how much 8 goes into 570. But 8 goes into 57-- let's see, 8 times 5 is 40. 8 times 6 is 48. 8 times 7 is 54. So let's go 7 times. 7 times 8. Oh, actually, I did that right. 8 times 7-- my multiplication tables are weak. It is 8 times-- so it does go in seven times. But 7 times 8 isn't 54; it's 56. I always have trouble with that one. It is 56. And so now, we subtract again. And now we have one more place value to bring down. We have-- let me do it in in a color I haven't used yet. I'll do blue. We have a 1 to bring down. And now we're literally saying, how many times does 8 go into 11? Well, 8 goes into 11 one time. 1 times 8 is 8. And we are left with-- because we have no more places to bring down-- 11 minus 8 is 3. So 8 goes into 3,771 471 times, and the remainder here is 3." + }, + { + "Q": "\nCan someone explain to me why the cube root and 3 exponent got erased? at 3:50ish", + "A": "Something raised to the power of 3 is a perfect cube, like: 2^3 = 8 You can simplify cube roots of perfect cubes: cuberroot (8) = 2. The video is just a more complicated version of the example I just did. Sal has cuberoot [(x+1)^3]. The (x+1)^3 is a perfect cube. Take the cube root, and you end up with just (x+1). Hope this helps.", + "video_name": "8GEGnSEJA2s", + "timestamps": [ + 230 + ], + "3min_transcript": "this part just simplifies to x plus one, and then I subtract one, so it all simplified out to just being equal to x. So we're just left with an x. So, f of g of x is just x. So now, let's try what g of f of x is. So, g of f of x is going to be equal to, I'll do it right over here, this is going to be equal to the cube root of actually, let me write it out, wherever I see an x, I can write f of x instead, I didn't do it that last time, I went directly and replaced with the definition of f of x but just to make it clear what I'm doing everywhere I'm seeing an x, I replace it with an f of x. So, the cube root of f of x plus one, minus seven. Well, that's going to be equal to the cube root of cube root of f of x, which is all of this business over here minus one, and then we add one and we add one, and then we subtract the seven lucky for us, subtracting one and adding one, those cancel out. Next, we're gonna take the cube root of x plus seven to the third power. Well, the cube root of x plus seven to the third power is just going to be x plus seven so, this is going to be x plus seven, for all of this business simplifies to x plus seven, and then we do subtract seven and these two cancel out, or they negate each other and we are just left with x. So, we see something very interesting. f of g of x is just x and g of f of x is x. So, in this case, if we start with an x if we start with an x, we input it into the function g and we get g of x and then we input that into the function f, then we input that into the function f, f of g of x gets us back to x. It gets us back to x. So we kind of did a round-trip. And the same thing is happening over here. If I put x into f of x... I'm sorry, if I put x into the function f, and I get f of x, the output is f of x, and then I input that into g, into the function g into the function g, once again I do this round-trip and I get back to x. Another way to think about it, these are both composite functions, one way to think about it is, if these are the set of all possible inputs into either of these composite functions, and then, these are the outputs, so you're starting with an x, I'll do this case first," + }, + { + "Q": "\nat 1:09 he says 100 times the light goes on and off. How long would that take?!?!?!", + "A": "ok... you re weird -Midnite Blaze", + "video_name": "-xYkTJFbuM0", + "timestamps": [ + 69 + ], + "3min_transcript": "So we had the hundred logicians. All of their foreheads were painted blue. And before they entered the room, they were told that at least one of you hundred logicians has your forehead painted blue. And then every time that they turned on the lights, so that they could see each other, they said OK, once you've determined that you have a blue forehead, when the lights get turned off again, we want you to leave the room. And then once that's kind of settled down, they'll turn the lights on again. And people will look at each other again. And then they'll turn them off again. And maybe people will leave the room. And so forth and so on. And they're also all told that everyone in the room is a perfect logician. They have infallible logic. So the question was, what happens? And actually maybe an even more interesting question is why does it happen? So I'll answer the first, what happens? And if just take the answer, and you don't know why, it almost seems mystical. That essentially the light gets turned on and off 100 gets turned on, and the lights get turned off again, all of them leave. They all leave. So I mean, it's kind of weird, right? Let's say I'm one of them. Or you're one of them. I go into this room. The lights get turned on. And I see 99 people with blue foreheads. And I can't see my own forehead. They see my forehead, of course. But to any other person, I'm one of the 99, right? But I see 99 blue foreheads. So essentially what happens if we were to watch the show is, the lights get turned on. You see 99 blue foreheads. Then the lights get turned off again. And then the lights get turned on again. And everyone's still sitting there. And I still see 99 blue foreheads. And that happens 100 times. everyone leaves the room. And at first glance, that seems crazy, because nothing changes. Nothing changes between every time we turn on the light. But the way you need to think about this-- and this is what makes it interesting-- is what happens instead of 100, let's say there was one person in the room. So before the show starts-- they never told me that there were going to be 100 people in the room. They just said, at least one of you, at least one of the people in the room, has your forehead painted blue. And as soon as you know that your forehead is painted blue, you leave the room. And that everyone's a perfect logician. So imagine the situation where instead of 100 there's only one perfect logician. Let's say it's me. So that's the room. I walk in. And I sit down. And maybe I should do it with blue." + }, + { + "Q": "\nso would the ratio 9.999... : .999... would be equal to 10:1", + "A": "Yup! just like 1.111\u00e2\u0080\u00a6 to 0.111\u00e2\u0080\u00a6 :) (10/9 to 1/9)", + "video_name": "TINfzxSnnIE", + "timestamps": [ + 601 + ], + "3min_transcript": "Anyway, it's 1/2 plus 1/4 plus 1/8, dot, dot, dot, dot, dot, to get 1. Each time, you fall short of 1. So how can you ever do anything? Luckily, infinity has got our backs. I mean, that's like the definition of infinity, a numbers so large, you can never get there, no matter how many steps you do, no matter how high you count. This way of writing numbers with this dot, dot, dot business, or with a bar over the repeating part, is a shorthand for an infinite series, whether it be 9/10 plus 9/100, and so on to get 1. Or 3/10 plus 3/100, and so on, to get 1/3. No matter how many 3s you write down, it will always be less than 1/3, but it will also always be less than infinity 3s. Infinity is what gets us there when no real number can. The binary equivalent of 0.9 repeating is 0.1 repeating. That's exactly 1/2, plus 1/4, plus 1/8, and so on. That's how we know a dotted, dotted, dot, dot, dot The ultimate reason that 0.9 repeating equals 1 is because it works. It's consistent, just like 1 plus 1 equals 2 is consistent, and just like 1 divided by 0 equals infinity isn't. Mathematics is about making up rules and seeing what happens. And it takes great creativity to come up with good rules. The only difference between mathematics and art is that if you don't follow your invented rules precisely in mathematics, people have a tendency to tell you you're wrong. Some rules give you elementary algebra and real numbers, and these rules can't tell the difference between 0.9 repeating and 1, just like they can't tell the difference between 0.5 and 1/2, or between 0 and negative 0. I hope you see now that the view that 9.9 repeating does not equal 10 is simply un-- 9.9 repeating --able. If you started this video thinking, I h-- 7.9 repeating that 7.9 repeating is 8, I hope now, you're thinking, oh, sweet, 4.9 repeating is 5? High 4.9 repeating! I got 98.9 repeating problems, but 0.9 repeating is 1. Here's the moral of the story. The idea of a number infinitely close to but less than 1 is not stupid or wrong, but wonderful, and beautiful, and interesting. The true mathematician takes \"you can't do that\" as a challenge. If someone tells you can't subtract a bigger number from a smaller number, just invent negative numbers. If someone tells you can't multiply a number by itself to get a negative number, then invent imaginary numbers. If someone tells you can't multiply two non-zero numbers together to get 0, or raise one non-zero number to the power of another and get 0, you should probably say, I'll do both at once, and in 8 dimensions. And if you ignore them telling you that numbers aren't 8-dimensional, and that inventing fake numbers is a useless waste of time, and then actually try to figure it out, the next thing you know, you've got split octonions, which besides being super awesome, just happen to be the perfect way to describe the wave equation of electrons and stuff." + }, + { + "Q": "\ni just still don't get why .9999..... equals 1 it doesn't make any sense.\nalso starting at 01:28 it starts to get confusing... can someone explain it to me???", + "A": "It s confusing because it deals with infinities, something that algebra and arithmetic are quite incapable of adequately describing. Our brains too, because infinities are actually quite hard for us to accurately imagine. calculus, on the other hand, is full of wonderful things like numbers infinitely close but not equal to other numbers and dividing by zero and such.", + "video_name": "TINfzxSnnIE", + "timestamps": [ + 88 + ], + "3min_transcript": "99.9 repeating percent of mathematicians agree, 0.9 repeating equals 1. If because I said so works for you, you can go ahead and do something else now. Maybe you're like, 0.9 repeating equals one, that's this 0.9 repeating-derful! Otherwise, on to reason number 2, or reason 1.9 repeating. See, it's weird, because when we think of the number 1 or 2, in most contexts we mean it as a natural number, like 1, 2, 3, 4, 5. In the sense then, the next number after 1 is 2. 9.9 repeating may equal 10, but you wouldn't say you have 9.9 repeating lords a leaping, in the same way you wouldn't say you had 9.75 lords a leaping plus 1/4 lord a leaping. Lords, leaping or otherwise, come in natural numbers. So what does this statement mean? 0.9 repeating is the same as 1? It looks pretty different, but it equals 1 in the same way that a 1/2 equals 0.5. They have the same value, You can philosophize over whether, if 1 is the loneliest number, 0.78 plus 0.22 is just as lonely, but there's no mathematical doubt that they have the same value, just as 100 years of solitude or 99.9 repeating years of solitude. So reason 2 is not a proof, but a reason to stay open minded. Numbers that look different can have the same value. Another example of this is that, in algebra, 0 equals negative 0. 0.9 repeating is a decimal number, a real number. See, if you want 0.9 repeating to be that number infinitesimally close to 1, but not 1-- and let's face it, some of you do-- then you're writing down the wrong number when you write 0.9 repeating. That number infinitely close to 1, but less than 1, is a number, but it's not 0.9 repeating or any real number. OK, let's do more 3.9 repeating-mal proof for reason 3.9 repeating. According to this 3.9 repeating-mula, 3.9 repeating is 4. First step, say 0.9 repeating equals x. Then multiply each side by 10. Third, subtract 0.9 repeating from this side, which equals x, which we subtract from the other side. And 10x minus x is 9x. Divide by 9, and you get 1 equals x, which you might notice also equals 0.9 repeating. There's no tricks here. It's simple multiplication, subtraction and division by 9, which are all allowed because they are consistent. When something is inconsisten-- 9.9 repeating --t, we just throw it out of algebra altogether. For example, in algebra if you try to divide by 0, you get this problem where anything can equal anything. I mean, if you want to say everything is equal, fine, but your algebra sucks. Normal, everyday elementary algebra, the one they shove down students throats as if it were the only algebra, doesn't allow dividing by 0. So it stays consistent and suspiciously practical. We also could have shifted the decimal point twice, multiplying by 100 to prove that if you have 99.9 repeating bottles of beer on the wall, 99.9 repeating bottles of beer. Take one down, pass it around, 99 bottles of beer on the wall." + }, + { + "Q": "At around 05:58, what did she write on the left hand side in green?\n", + "A": "Infinity times magnified", + "video_name": "TINfzxSnnIE", + "timestamps": [ + 358 + ], + "3min_transcript": "a long time ago. But elementary algebra can't deal with infinity. If you allow infinity in your algebrizations, once again, you get contradictions. Infinity may not be a real number, but it is a number, a hyperreal number. Hyperreals, like infinity and the infinitesimal, follow different rules. And while algebra can't handle them, some people thought they should be numbers, and you should be able to use them. And so they figured out how, and bam! you get something like calculus. Reason 6. Take the number 1, and subtract 0.9 repeating. It's pretty clear that it's infinite 0s, but you might be tempted to think there's some sort of final 1 beyond infinity. Let's write that down as 0.0 repeating 1. Of course, if the 0 repeats infinitely, then you never get to the 1, so you might as well leave off the number. Thus the difference between 0.9 repeating and 1 is 0. There is no difference. Here's another \"there's no difference\" proof. What's the next higher real number? The game, is for any number you claim is the next real number, I can find one that's even closer to 1. Of the many delightful things about real numbers is that for any two numbers, no matter how close, there's still an infinite amount of numbers between them, and an infinite amount of numbers between those, and so on. There is no next higher real number after 1. Likewise, there's no next lower number. If 0.9 repeating and 1 were different real numbers, there would have to be infinite other real numbers between them. If you can't name a number higher than 0.9 repeating, but lower than 1, it can only be because 0.9 repeating is 1. If you don't like it, well, go to college and learn about hyperreals, or better yet, surreals. That's a system where you can have a number that's infinitely close to 1, but not 1. But even weirder, there's infinity more numbers that are infinitely even closer. Take 0.3 repeating, a repeating decimal equal to 1/3. Multiply it by 3. Obviously, by definition, 3/3 is 1, and 0.3 repeating times 3 is 0.9 repeating, which you might have noticed is also 1. The only assumption here is that 0.3 repeating equals 1/3. Maybe you don't like decimal notation in general, which brings us to reason number 9, this sum of an infinite series thing. 9/10 plus 9/100 plus 9/1000. And we can sum this series and get 1. But I can see why you might be unhappy with this. It recalls Zeno's paradoxes. How can you get across a room, when first you have to walk halfway, and then half of that, and so on. Or, how can you shoot an arrow into a target, when first it needs to go halfway, but before it can get halfway, it needs to go half of halfway, and before that, half of half of halfway, and half of half of half of halfway, and so on." + }, + { + "Q": "my qs is two no. are in the ratio 5:6. If 8 is subtracted from each of the no. the ratio becomes 4:5. find the no.s\n", + "A": "the numbers are 40 and 48", + "video_name": "DopnmxeMt-s", + "timestamps": [ + 306, + 245 + ], + "3min_transcript": "if you want to try working this out yourself, you might want to pause the video and then play it once you're ready to see how I do it. Anyway, let me move forward assuming you haven't paused it. If we want to get rid of this 3/4, all we do is we subtract 3/4 from both sides of this equation. Minus 3/4. Well, the left hand side, the two 3/4 will just cancel. We get minus 1/2x equals, and then on the right hand side, we just have to do this fraction addition or fraction So the least common multiple of 6 and 4 is 12. So this becomes 5/6 6 is 10/12 minus 3/4 is 9/12, so we get minus 1/2x is equal to 1/12. And if that step confused you, I went a little fast, you might just want to review the adding and subtraction of fractions. So going back to where we were. So now all we have to do is, well, the coefficient on the x term is minus 1/2, and this is now a level one problem. So to solve for x, we just multiply both sides by the reciprocal of this minus 1/2x, and that's minus 2/1 times minus 1/2x on that side, and then that's times minus 2/1. The left hand side, and you're used to this by now, simplifies to x. The right hand side becomes minus 2/12, and we could simplify that further to minus 1/6. Well, let's check that just to make sure we got it right. So let's try to remember that minus 1/6. So the original problem was minus 1/2x, I just wrote only the left hand side of the original problem. So minus 1/2 times minus 1/6, well, that's positive 1/12 plus 3/4. Well, that's the same thing as 12, the 1 stays the same, plus 9. 1 plus 9 is 10 over 12. And that is equal to 5/6, which is what our original problem was. This stuff I wrote later. So it's 5/6, so the problem checks out. So hopefully, you're now ready to try some level two problems on your own. I might add some other example problems. But the only extra step here relative to level one problems is you'll have this constant term that you need to add or subtract from both sides of this equation, and you'll essentially turn it into a level one problem." + }, + { + "Q": "\nAt 4:34, I know that Sal means the twos cancel out and then it turns to one over one, which equals one, etc. But WHY do the twos cancel out?", + "A": "2/2 means 2 divided by 2, any number divided by itself = 1. To rephrase it: How many 2 s are in 2? One. Works for any number.", + "video_name": "DopnmxeMt-s", + "timestamps": [ + 274 + ], + "3min_transcript": "if you want to try working this out yourself, you might want to pause the video and then play it once you're ready to see how I do it. Anyway, let me move forward assuming you haven't paused it. If we want to get rid of this 3/4, all we do is we subtract 3/4 from both sides of this equation. Minus 3/4. Well, the left hand side, the two 3/4 will just cancel. We get minus 1/2x equals, and then on the right hand side, we just have to do this fraction addition or fraction So the least common multiple of 6 and 4 is 12. So this becomes 5/6 6 is 10/12 minus 3/4 is 9/12, so we get minus 1/2x is equal to 1/12. And if that step confused you, I went a little fast, you might just want to review the adding and subtraction of fractions. So going back to where we were. So now all we have to do is, well, the coefficient on the x term is minus 1/2, and this is now a level one problem. So to solve for x, we just multiply both sides by the reciprocal of this minus 1/2x, and that's minus 2/1 times minus 1/2x on that side, and then that's times minus 2/1. The left hand side, and you're used to this by now, simplifies to x. The right hand side becomes minus 2/12, and we could simplify that further to minus 1/6. Well, let's check that just to make sure we got it right. So let's try to remember that minus 1/6. So the original problem was minus 1/2x, I just wrote only the left hand side of the original problem. So minus 1/2 times minus 1/6, well, that's positive 1/12 plus 3/4. Well, that's the same thing as 12, the 1 stays the same, plus 9. 1 plus 9 is 10 over 12. And that is equal to 5/6, which is what our original problem was. This stuff I wrote later. So it's 5/6, so the problem checks out. So hopefully, you're now ready to try some level two problems on your own. I might add some other example problems. But the only extra step here relative to level one problems is you'll have this constant term that you need to add or subtract from both sides of this equation, and you'll essentially turn it into a level one problem." + }, + { + "Q": "\nAt 2:50, not enough to solve but we can simply i think,\n15a+15b = 3a+3a+3a+3a+3a+5b+5b+5b\ni guess you see know ...\n\n3a+5b = 2\n3a+5b = 2\n3a+5b = 2 THEN\n\n2+2+2+3a+3a = ?\n6+6a= 6 . ( 1+a)\n\nis it correct ? or am i too wrong in there =)", + "A": "You are right", + "video_name": "CLQRZ2UbQ4Q", + "timestamps": [ + 170 + ], + "3min_transcript": "Let's do a few more examples where we're evaluating expressions with unknown variables. So this first one we're told 3x plus 3y plus 3z is equal to 1, and then we're asked what's 12x plus 12y plus 12z equal to? And I'll give you a few moments to think about that. Well let's rewrite this second expression by factoring out the 12, so we get 12 times x plus y plus z. That's this second expression here, and you can verify that by distributing the 12. You'll get exactly this right up here. Now, what is 12 times x plus y plus z? Well, we don't know yet exactly what x plus y plus z is equal to, but this first equation might help us. This first equation, we can rewrite this left-hand side by factoring out the 3, so we could rewrite this as 3 times x plus y plus z is equal to 1. All I did is I factored the 3 out on the left-hand side. I just divide both sides of this equation by 3, and I'm left with x plus y plus z is equal to 1/3, and so here, instead of x plus y plus z, I can write 1/3. So this whole thing simplified to 12 times 1/3. 12 times 1/3 is the same thing as 12 divided by 3, which is equal to 4. Let's try one more. So here we are told that 3a plus 5b is equal to 2, and then we're asked what's 15a plus 15b going to be equal to? So we might-- let's see. I'll give you a few moments to try to tackle this on your own. Let's see how we might do it. We could approach it the way we've approached the last few problems, trying to rewrite the second expression. We could rewrite it as 15 times a plus b, and so we just have to figure out what a plus b is, And so, it's tempting to look up here, and say maybe we can solve for a plus b somehow, but we really can't. If we divide-- if we try to factor out a 3, we'll get 3 times a plus 5/3b, so this doesn't really simplify things in terms of a plus b. If we try to factor out a 5, we'd get 5 times 3/5a plus b is equal to 2, but neither of these gets us in a form where we can then solve for a plus b. So in this situation, we actually do not have enough information to solve this problem. So it's a little bit of a trick. Not enough info to solve. Anyway, hopefully you enjoyed that." + }, + { + "Q": "at 4:27 it shows reflection 0,1 to 1,1 and I see this as Y=1 but for the question y=x\u00e2\u0088\u00922 answered 0,2 to 2,4 and not sure why its wrong.\n", + "A": "You weren t plugging in the numbers correctly. y=x-2 for (0,2) would give you 2=0-2 which is not correct since 2 does not equal -2. Plugging in (2,4) for the second part would give you 4=2-2 which is a false statement as well.", + "video_name": "vO1Ur38PGCY", + "timestamps": [ + 267 + ], + "3min_transcript": "" + }, + { + "Q": "\nwait at 6:39 why did he simplify", + "A": "The sooner you simplify the more extra work you avoid.", + "video_name": "bAerID24QJ0", + "timestamps": [ + 399 + ], + "3min_transcript": "This 4 becomes a 1 and this becomes a 10. If you remember when you're multiplying fractions, you can simplify it like that. So it actually becomes minus-- actually, plus 30, because we have a minus times a minus and 3 times 10, over, the 4 is now 1, so all we have left is 39. And 30/39, if we divide the top and the bottom by 3, we get 10 over 13, which is the same thing as what the equation said we would get, so we know that we've got the right answer. Let's do one more problem. Minus 5/6x is equal to 7/8. yourself, now's a good time to pause, and I'm going to start doing the problem right now. So same thing. What's the reciprocal of minus 5/6? Well, it's minus 6/5. We multiply that. If you do that on the left hand side, we have to do it on the right hand side as well. Minus 6/5. The left hand side, the minus 6/5 and the minus 5/6 cancel out. We're just left with x. And the right hand side, we have, well, we can divide both the 6 and the 8 by 2, so this 6 becomes negative 3. This becomes 4. 7 times negative 3 is minus 21/20. And assuming I haven't made any careless mistakes, that should be right. Actually, let's just check that real quick. Minus 5/6 times minus 21/20. Turn this into a 4. Make this into a 2. Make this into a 7. Negative times negative is positive. So you have 7. 2 times 4 is 8. And that's what we said we would get. So we got it right. I think you're ready at this point to try some level one equations. Have fun." + }, + { + "Q": "\nat 2:10 what do u mean by recipical", + "A": "a reciprocal is the inverted number. if you have 3/4 the reciprocal is 4/3. if the number is a whole we know the denominator is 1 therefore any whole number a has the reciprocal 1/a. A number like 3 has the reciprocal of 1/3.", + "video_name": "bAerID24QJ0", + "timestamps": [ + 130 + ], + "3min_transcript": "Welcome to level one linear equations. So let's start doing some problems. So let's say I had the equation 5-- a big fat 5, 5x equals 20. So at first this might look a little unfamiliar for you, but if I were to rephrase this, I think you'll realize this is a pretty easy problem. This is the same thing as saying 5 times question mark equals 20. And the reason we do the notation a little bit-- we write the 5 next to the x, because when you write a number right next to a variable, you assume that you're multiplying them. So this is just saying 5 times x, so instead of a question mark, we're writing an x. So 5 times x is equal to 20. Now, most of you all could do that in your head. You could say, well, what number times 5 is equal to 20? Well, it equals 4. But I'll show you a way to do it systematically just in case that 5 was a more complicated number. So rewriting it, if I had 5x equals 20, we could do two things and they're essentially the same thing. We could say we just divide both sides of this equation by 5, in which case, the left hand side, those two 5's will cancel out, we'll get x. And the right hand side, 20 divided by 5 is 4, and we would have solved it. Another way to do it, and this is actually the exact same way, we're just phrasing it a little different. If you said 5x equals 20, instead of dividing by 5, we could multiply by 1/5. And if you look at that, you can realize that multiplying by 1/5 is the same thing as dividing by 5, if you know the difference between dividing and multiplying fractions. And then that gets the same thing, 1/5 times 5 is 1, so you're just left with an x equals 4. because when we start having fractions instead of a 5, it's easier just to think about multiplying by the reciprocal. Actually, let's do one of those right now. So let's say I had negative 3/4 times x equals 10/13. Now, this is a harder problem. I can't do this one in my head. We're saying negative 3/4 times some number x is equal to 10/13. If someone came up to you on the street and asked you that, I think you'd be like me, and you'd be pretty stumped. But let's work it out algebraically. Well, we do the same thing. We multiply both sides by the coefficient on x. So the coefficient, all that is, all that fancy word means, is the number that's being multiplied by x. So what's the reciprocal of minus 3/4." + }, + { + "Q": "what are conventions? 3:15\n", + "A": "A convention is an unspoken agreement to do something a certain way. Grammar, spelling, the order of operations, etc. are all systems of conventions.", + "video_name": "bAerID24QJ0", + "timestamps": [ + 195 + ], + "3min_transcript": "So rewriting it, if I had 5x equals 20, we could do two things and they're essentially the same thing. We could say we just divide both sides of this equation by 5, in which case, the left hand side, those two 5's will cancel out, we'll get x. And the right hand side, 20 divided by 5 is 4, and we would have solved it. Another way to do it, and this is actually the exact same way, we're just phrasing it a little different. If you said 5x equals 20, instead of dividing by 5, we could multiply by 1/5. And if you look at that, you can realize that multiplying by 1/5 is the same thing as dividing by 5, if you know the difference between dividing and multiplying fractions. And then that gets the same thing, 1/5 times 5 is 1, so you're just left with an x equals 4. because when we start having fractions instead of a 5, it's easier just to think about multiplying by the reciprocal. Actually, let's do one of those right now. So let's say I had negative 3/4 times x equals 10/13. Now, this is a harder problem. I can't do this one in my head. We're saying negative 3/4 times some number x is equal to 10/13. If someone came up to you on the street and asked you that, I think you'd be like me, and you'd be pretty stumped. But let's work it out algebraically. Well, we do the same thing. We multiply both sides by the coefficient on x. So the coefficient, all that is, all that fancy word means, is the number that's being multiplied by x. So what's the reciprocal of minus 3/4. to use times, and you're probably wondering why in algebra, there are all these other conventions for doing times as opposed to just the traditional multiplication sign. And the main reason is, I think, just a regular multiplication sign gets confused with the variable x, so they thought of either using a dot if you're multiplying two constants, or just writing it next to a variable to imply you're multiplying a variable. So if we multiply the left hand side by negative 4/3, we also have to do the same thing to the right hand side, minus 4/3. The left hand side, the minus 4/3 and the 3/4, they cancel out. You could work it out on your own to see that they do. They equal 1, so we're just left with x is equal to 10 times minus 4 is minus 40, 13 times 3, well, that's equal to 39." + }, + { + "Q": "I always learned the equation was A(n)=a+(n-1)*(d) where a equals the first term, n is the term number, and d is the common difference. So the equation for 3:14 is A(100)=15+(100-1)*(-6)=-579. Is this correct?\n", + "A": "Yes, that is correct (5:53). Good job!", + "video_name": "JtsyP0tnVRY", + "timestamps": [ + 194 + ], + "3min_transcript": "Then to go from 9 to 3, well, we subtracted 6 again. And then to go from 3 to negative 3, well, we subtracted 6 again. So it looks like, every term, you subtract 6. So the second term is going to be 6 less than the first term. The third term is going to be 12 from the first term, or negative 6 subtracted twice. So in the third term, you subtract negative 6 twice. In the fourth term, you subtract negative 6 three times. So whatever term you're looking at, you subtract negative 6 one less than that many times. Let me write this down just so-- Notice when your first term, you have 15, and you don't subtract negative 6 at all. Or you could say you subtract negative 6 0 times. let me write it better this way --minus 0 times negative 6. That's what that first term is right there. What's the second term? This is 15. We just subtracted negative 6 once, or you could say, minus 1 times 6. Or you could say plus 1 times negative 6. Either way, we're subtracting the 6 once. Now what's happening here? This is 15 minus 2 times negative 6-- or, sorry --minus 2 times 6. We're subtracting a 6 twice. What's the fourth term? This is 15 minus-- We're subtracting the 6 three times from the 15, so minus 3 times 6. So, if you see the pattern here, when we have our fourth The fourth term, we have a 3. The third term, we have a 2. The second term, we have a 1. So if we had the nth term, if we just had the nth term here, what's this going to be? It's going to be 15 minus-- You see it's going to be n minus 1 right here. When n is 4, n minus 1 is 3. When n is 3, n minus 1 is 2. When n is 2, n minus 1 is 1. When n is 1, n minus 1 is 0. So we're going to have this term right here is n minus 1. So minus n minus 1 times 6. So if you want to figure out the 100th term of this sequence, I didn't even have to write it in this general term, you can just look at this pattern. It's going to be-- and I'll do it in pink --the 100th term in our sequence-- I'll continue our table down --is going to be what? It's going to be 15 minus 100 minus 1, which is 99, times 6." + }, + { + "Q": "\nWhy did he put the 0 at 2:16?", + "A": "The 0 represents how many times we want to subtract 6 (or add -6, that s the same thing) from our starting value, which is 15. More generally, the value x represents how many times we want to subtract 6 from our starting value of 15. Because the very first value is 15, we don t want to subtract 6 from it, because that would give us 9, which is the second term, so we write 15 - (0)*6. Anything times 0 is 0, so that becomes 15 - 0, but subtracting or adding nothing to an equation doesn t change it s value.", + "video_name": "JtsyP0tnVRY", + "timestamps": [ + 136 + ], + "3min_transcript": "We are asked, what is the value of the 100th term in this sequence? And the first term is 15, then 9, then 3, then negative 3. So let's write it like this, in a table. So if we have the term, just so we have things straight, and then we have the value. and then we have the value of the term. I'll do a nice little table here. So our first term we saw is 15. Our second term is 9. Our third term is 3. I'm just really copying this down, but I'm making sure we associate it with the right term. And then our fourth term is negative 3. And they want us to figure out what the 100th term of this sequence is going to be. So let's see what's happening here, if we can discern some type of pattern. So when we went from the first term to the second term, what happened? 15 to 9. Looks like we went down by 6. It's always good to think about just how much the numbers changed by. That's always the simplest type of pattern. Then to go from 9 to 3, well, we subtracted 6 again. And then to go from 3 to negative 3, well, we subtracted 6 again. So it looks like, every term, you subtract 6. So the second term is going to be 6 less than the first term. The third term is going to be 12 from the first term, or negative 6 subtracted twice. So in the third term, you subtract negative 6 twice. In the fourth term, you subtract negative 6 three times. So whatever term you're looking at, you subtract negative 6 one less than that many times. Let me write this down just so-- Notice when your first term, you have 15, and you don't subtract negative 6 at all. Or you could say you subtract negative 6 0 times. let me write it better this way --minus 0 times negative 6. That's what that first term is right there. What's the second term? This is 15. We just subtracted negative 6 once, or you could say, minus 1 times 6. Or you could say plus 1 times negative 6. Either way, we're subtracting the 6 once. Now what's happening here? This is 15 minus 2 times negative 6-- or, sorry --minus 2 times 6. We're subtracting a 6 twice. What's the fourth term? This is 15 minus-- We're subtracting the 6 three times from the 15, so minus 3 times 6. So, if you see the pattern here, when we have our fourth" + }, + { + "Q": "At 5:14, Why would the sqrt of 1/2 be 1/sqrt2 and not sqrt 1/2?\n", + "A": "This is because the sqrt(1/2) is the same thing as saying sqrt(1)/sqrt(2) and since the square root of 1 is just 1 we can simplify the numerator and rewrite it as 1/sqrt(2).", + "video_name": "KoYZErFpZ5Q", + "timestamps": [ + 314 + ], + "3min_transcript": "So we know the length of segment AB, which is a radius of the circle or is the radius of the circle is length one, what else do we know about this triangle? Can we figure out the lengths of segment AD and the length of segment DB? Well sure, because we have two base angles that have the same measure, that means that the corresponding sides are also going to have the same measure. That means that this side is going to be congruent to that side. I can reorient it in a way that might make it a little easier to realize. If we were to flip it over, maybe not completely fit it over but if we were to make it look like this. So the triangle, we could make it look like, a little bit more like this. Actually I want to make it look like a right angle though. So my triangle, let me make it look like, there you go. So if this is D, this is B, this is A, this is our right angle. this is also pi over four radians. When your two base angles are the same, you know you're dealing with an isosceles triangle. Because they're not all the same, it's not equilateral. If all the angles were the same this would be equilateral. This is an isosceles non-equilateral triangle. So if your base angles are the same, then you also know that the corresponding sides are going to be the same. These two sides are the same, this is an isosceles triangle. So how does that help us figure out the lengths of the sides? Well if you say that this side has length X, that means that this side has length X. If this side has length X, then this side has length X. and now we can use the Pythagorean Theorem. We can say that this X squared plus this X squared is equal to the hypotenuse square is equal to one square. Or we could write that two X squared is equal two or taking the principal root of both sides, we get X is equal to one over the square root of two. And a lot of folks don't like having a radical denominator, they don't like having a rational number in the denominator. So we could rationalize the denominator by multiplying by the square root of two over the square root of two, which would be, let's see the numerator will have the square root of two, and the denominator we're just going to have square root of two times square root of two is just two. So we've already been able to figure out several interesting things We've been able to figure the angle of ABD in radians. We're able to figure out lengths of segment AD and the length of segment BD. Now what I want to do is figure out what are the sine, cosine, and tangent of pi over four radians, given all of the work we have done? So let's first think about, what is the sine? We'll do this in the color orange." + }, + { + "Q": "\nat 5:17 shouldn't it be sqrt 1/2 ?", + "A": "Yes, but you need to go further and simplify the radical. When your radical is simplified, 1) it will not have a fraction inside the radical. sqrt(1/2) = sqrt(1) / sqrt(2) 2) all perfect squares inside the radical will be simplified: sqrt(1) / sqrt(2) = 1 / sqrt(2) 3) there will be no radical in the denominator - rationalize the denominator: 1 / sqrt(2) = 1 / sqrt(2) * sqrt(2) / sqrt(2) = sqrt(2) / 2 hope this helps.", + "video_name": "KoYZErFpZ5Q", + "timestamps": [ + 317 + ], + "3min_transcript": "So we know the length of segment AB, which is a radius of the circle or is the radius of the circle is length one, what else do we know about this triangle? Can we figure out the lengths of segment AD and the length of segment DB? Well sure, because we have two base angles that have the same measure, that means that the corresponding sides are also going to have the same measure. That means that this side is going to be congruent to that side. I can reorient it in a way that might make it a little easier to realize. If we were to flip it over, maybe not completely fit it over but if we were to make it look like this. So the triangle, we could make it look like, a little bit more like this. Actually I want to make it look like a right angle though. So my triangle, let me make it look like, there you go. So if this is D, this is B, this is A, this is our right angle. this is also pi over four radians. When your two base angles are the same, you know you're dealing with an isosceles triangle. Because they're not all the same, it's not equilateral. If all the angles were the same this would be equilateral. This is an isosceles non-equilateral triangle. So if your base angles are the same, then you also know that the corresponding sides are going to be the same. These two sides are the same, this is an isosceles triangle. So how does that help us figure out the lengths of the sides? Well if you say that this side has length X, that means that this side has length X. If this side has length X, then this side has length X. and now we can use the Pythagorean Theorem. We can say that this X squared plus this X squared is equal to the hypotenuse square is equal to one square. Or we could write that two X squared is equal two or taking the principal root of both sides, we get X is equal to one over the square root of two. And a lot of folks don't like having a radical denominator, they don't like having a rational number in the denominator. So we could rationalize the denominator by multiplying by the square root of two over the square root of two, which would be, let's see the numerator will have the square root of two, and the denominator we're just going to have square root of two times square root of two is just two. So we've already been able to figure out several interesting things We've been able to figure the angle of ABD in radians. We're able to figure out lengths of segment AD and the length of segment BD. Now what I want to do is figure out what are the sine, cosine, and tangent of pi over four radians, given all of the work we have done? So let's first think about, what is the sine? We'll do this in the color orange." + }, + { + "Q": "\nAt around 2:06, how come Sal doesn't put (square root of y)^2 in the denominator? I thought when using the product rule, you put (g(x))^2 in the denominator.", + "A": "I suspect you re confusing the product rule with the quotient rule. Product rule: derivative of uv = u v + uv Quotient rule: derivative of u/v = (u v - uv )/v\u00c2\u00b2", + "video_name": "2CsQ_l1S2_Y", + "timestamps": [ + 126 + ], + "3min_transcript": "What I want to show you in this video is that implicit differentiation will give you the same result as, I guess we can say, explicit differentiation when you can differentiate explicitly. So let's say that I have the relationship x times the square root of y is equal to 1. This one is actually pretty straightforward to define explicitly in terms of x, to solve for y. So if we divide both sides by x, we get square root of y is equal to 1/x. And then if you square both sides, you get y is equal to 1 over x squared, which is the same thing as x to the negative 2 power. And so if you want the derivative of y with respect to x, this is pretty straightforward. This is just an application of the chain rule. So we get dy dx is equal to negative 2 x to the negative 2 minus 1-- x to the negative 3 power. So that's pretty straightforward. But what I want to see is if we get the same exact result when we differentiate implicitly. to both sides of this equation. And so let me make it clear what we're doing-- x times the square root of y and 1 right over there. When you apply the derivative operator to the expression on the left-hand side, well, actually, we're going to apply both the product rule and the chain rule. The product rule tells us-- so we have the product of two functions of x. You could view it that way. So this, the product rule tells us this is going to be the derivative with respect to x of x times the square root of y plus x, not taking its derivative, times the derivative with respect to x of the square root of y. Let me make it clear, this bracket. the derivative with respect to x of this constant, that's just going to be equal to 0. So what does this simplify to? Well, the derivative with respect to x of x is just 1. This simplifies to 1, so we're just going to be left with the square root of y right over here. So this is going to simplify to a square root of y. And what does this over here simplify to? Well the derivative with respect to x of the square root of y, here we want to apply the chain rule. So let me make it clear. So we have plus this x plus whatever business this is. And I'm going to do this in blue. Well, it's going to be the derivative of the square root of something with respect to that something. Well, the derivative of the square root of something with respect to that something," + }, + { + "Q": "His algebra is wrong at 2:37 3*1/4 is not 13/4 and 2*1/2 is not 5/2\n", + "A": "His math is not wrong. You are confusing mixed numbers: 3 1/4 with multiplication of 3 * 1/4. In a mixed number like 3 1/4 of 2 1/2 there is no multiplication symbol between the numbers. 3 1/4 is 3 whole units + 1/4 unit, or 3.25 in decimal. 2 1/2 is 2 whole units + 1/2, or 2.5 in decimal form. Seems like you need to review basic concepts of dealing with fractions and mixed numbers.", + "video_name": "x5EJG_rAtkY", + "timestamps": [ + 157 + ], + "3min_transcript": "And that's exactly what these two statements are saying. x has to be less than 2 and 1/2, and it has to be greater than negative 2 and 1/2. If this absolute value were the other way, that the absolute value of x has to be greater than 2 and 1/2, then it would be the numbers outside of this, and it would be an or. But we're dealing with the less than situation right there, so let's just do what we were able to figure out when it was just an x. The distance from this thing to 0 has to be less than 2 and 1/2, so we can write that 2r minus 3 and 1/4 has to be less than 2 and 1/2 and 2r minus 3 and 1/4 has to be greater than negative 2 and 1/2. Same exact reasoning here. Let me draw a line so we don't get confused. This quantity right here has to be between negative 2 and 1/2. It has to be greater than negative 2 and 1/2 right there. And it has to be less than 2 and 1/2, so that's all I wrote there. So let's solve each of these independently. Well, this first went over here, you've learned before that I don't like improper fractions, and I don't like fractions in general. So let's make all of these fractions. Sorry, I don't like mixed numbers. I want them to be improper fractions. So let's turn all of these into improper fractions. So if I were to rewrite it, we get 2r minus 3 and 1/4 is the same thing as 3 times 4 is 12, plus 1 is 13. 2r minus 13/4 is less than-- 2 times 2 is 4, plus 1 is five-- is less than 5/2. So that's the first equation. And then the second question-- and do the same thing here-- we have 2r minus 13 over 4 has to be a greater All right, now let's solve each of these independently. To get rid of the fractions, the easiest thing to do is to multiply both sides of this equation by 4. That'll eliminate all of the fractions, so let's do that. Let's multiply-- let me scroll to the left a little bit-- let's multiply both sides of this equation by 4. 4 times 2r is 8r, 4 times negative 13 over 4 is negative 13, is less than-- and I multiplied by a positive number so I didn't have to worry about swapping the inequality-- is less than 5/2 times 4 is 10, right? You get a 2 and a 1, it's 10. So you get 8r minus 13 is less than 10. Now we can add 13 to both sides of this equation so that we get rid of it on the left-hand side. Add 13 to both sides and we get 8r-- these guys cancel out-- is less than 23, and then we divide" + }, + { + "Q": "\n00:32 when you said 'any m&b on the surface = SE for that line', what does 'that line' exactly mean?", + "A": "That line is referring to a line drawn through your data. If you have your data plotted on a graph, you can draw a line through it to fit to the data. This line will have a gradient, m, and a y-intercept, b. It will also have a squared error - the square distance of each point of data from the line. The aim is to find the line with the minimum squared error.", + "video_name": "u1HhUB3NP8g", + "timestamps": [ + 32 + ], + "3min_transcript": "All right, so where we left off, we had simplified our algebraic expression for the squared error to the line from the n data points. We kind of visualized it. This expression right here would be a surface, I guess you could view it as a surface in three dimensions, where for any m and b is going to be a point on that surface that represents the squared error for that line. Our goal is to find the m and the b, which would define an actual line, that minimize the squared error. The way that we do that, is we find a point where the partial derivative of the squared error with respect to m is 0, and the partial derivative with respect to b is also equal to 0. So it's flat with respect to m. So that means that the slope in this direction is going to be flat. Let me do it in the same color. So the slope in this direction, that's the partial derivative with respect to m, is going to be flat. It's not going to change in that direction. The partial derivative with respect to b is going to be flat. So it will be a flat point right over there. and that is our minimum point. So let's figure out the m and b's that give us this. So if I were to take the partial derivative of this expression with respect to m. Well this first term has no m terms in it. So it's a constant from the point of view of m. Just as a reminder, partial derivatives, it's just like taking a regular derivative. You're just assuming that everything but the variable that you're doing the partial derivative with respect to, you're assuming everything else is a constant. So in this expression, all the x's, the y's, the b's, the n's, those are all constant. The only variable, when we take the partial derivative with respect to m, that matters is the m. So this is a constant. There's no m here. This term right over here, we're taking with respect to m. So the derivative of this with respect to m, it's kind of the coefficients on the m. So negative 2 times n times the mean of the xy's, that's Then this term or right here has no m's in it. So it's constant with respect to m. So its partial derivative with respect to m is 0. Then this term here, you have n times the mean of the x squared times m squared. So this is going to be-- we're talking about a partial derivative with respect to m-- so it's going to be 2 times n times the mean of the x [? squareds ?] times m. The derivative of m squared is 2m, and then you just have this coefficient there as well. Now this term, you also have an m over there. So let's see, everything else is just kind of a coefficient on this m. So the derivative with respect to m is 2bn times the mean of the x's. If I took the derivative of 3m, the derivative is just 3." + }, + { + "Q": "at around 1:40 how do we know that the top side (c or 2x) in not equal to the right side (b or x=10) ?\n", + "A": "It s because triangle ABC is an isosceles triangle. If 2x was equal to x+10, then it wouldn t be an isosceles triangle, it would be an equilateral triangle, because an equilateral triangle s angles are all equal.", + "video_name": "CVKAro3HUxQ", + "timestamps": [ + 100 + ], + "3min_transcript": "So what do we have here? We have a triangle, and we know that the length of AC is equal to the length of CB. So this is an isosceles triangle, we have two of its legs are equal to each other. And then they also tell us that this line up here, they didn't put another label there. Let me put another label there just for fun. Let's call this, you could even call this a ray because it's starting at C, that line or ray CD is parallel to this segment AB over here, and that's interesting. Then they give us these two angles right over here, these adjacent angles. They give it to us in terms of x. And what I want to do in this video is try to figure out what x is. And so given that they told us that this line and this line are parallel, and we can turn this into line CD, so it's not just a ray anymore, so it just keeps going on and on in both directions. The fact that they've given us a parallel line tells us that maybe we can use some of what we know about transversals and parallel lines to figure out some of the angles here. And you might recognize that this right over here, You might recognize that line CB is a transversal for those two parallel lines. Let's let me draw both of the parallel lines a little bit more so that you can recognize that as a transversal, and then a few things might jump out. You have this x plus 10 right over here, and its corresponding angle is right down here. This would also be x plus 10. And if this is x plus 10, then you have a vertical angle right over here that would also be x plus 10. Or you could say that you have alternate interior angles that would also be congruent. Either way, this base angle is going to be x plus 10. Well, it's an isosceles triangle. So your two base angles are going to be congruent. So if this is x plus 10, then this is going to be x plus 10 as well. And now we have the three angles of a triangle expressed as functions of-- expressed in terms of x. So when we take their sum, they need to be equal to 180, and then we can actually solve for x. We get 2x plus x plus 10 plus x plus 10 And then we can add up the x's. So we have a 2x there plus an x plus another x, that gives us 4x. 4 x's. And then we have a plus 10 and another plus 10, so that gives us a plus 20, is equal to 180. And we can subtract 20 from both sides of that, and we get 4x is equal to 160. Divide both sides by 4, and we get x is equal to 40. And we're done. We've figured out what x is, and then we could actually figure out what these angles are. If this is x plus 10 then you have 40 plus 10, this right over here is going to be a 50-degree angle. This is 2x, so 2 times 40, this is an 80-degree angle. It doesn't look at it the way I've drawn it, and that's why you should never assume anything based on how a diagram is drawn." + }, + { + "Q": "\nAt 2:55 Sal suggests that it is necessary to take the natural log of both sides but I solved it by changing it to e^(ln(x)*x) and then taking the derivative as normal and I got the right answer.", + "A": "There is always different ways to solve the same problems. Sal only mentioned one of the ways by taking the natural log (mentioned at 4:37), but using e is another way.", + "video_name": "N5kkwVoAtkc", + "timestamps": [ + 175 + ], + "3min_transcript": "natural log of x. And now we can take the derivative of both sides of this with respect to x. So the derivative with respect to x of that, and then the derivative with respect to x of that. Now we're going to apply a little bit of the chain rule. So the chain rule. What's the derivative of this with respect to x? What's the derivative of our inner expression with respect to x? It's a little implicit differentiation, so it's dy with respect to x times the derivative of this whole thing with respect to this inner function. So the derivative of natural log of x is 1/x. So the derivative of natural log of y with respect to y is 1/y. So times 1/y. And the derivative of this-- this is just the product rule, and I'll arbitrarily switch colors here-- is the derivative of the first term, which is 1, times the second term, so times which is 1/x times the first term. So times x. And so we get dy/dx times 1/y is equal to natural log of x plus-- this just turns out to be 1-- x divided by x, and then you multiply both sides of this by y. You get dy/dx is equal to y times the natural log of x plus 1. And if you don't like this y sitting here, you could just make the substitution. y is equal to x to the x. So you could say that the derivative of y with respect to x is equal to x to the x times the natural log of x plus 1. And that's a fun problem, and this is often kind of given as a trick problem, or sometimes even a bonus problem if people don't know to take the natural log of both sides of that. that's what we're going to tackle in this. But it's good to see this problem done first because it gives us the basic tools. So the more difficult problem we're going to deal with is this one. Let me write it down. So the problem is y is equal to x to the-- and here's the twist-- x to the x to the x. And we want to find out dy/dx. We want to find out the derivative of y with respect to x. So to solve this problem we essentially use the same tools. We use the natural log to essentially breakdown this exponent and get it into terms that we can deal with. So we can use the product rule. So let's take the natural log of both sides of this equation like we did last time. You get the natural log of y is equal to the natural log of x to the x to the x." + }, + { + "Q": "\nAt 4:11, why does Sal ask us how many times 16 goes into 1388 when we can see the answer on the other side of the screen?", + "A": "Sal is explaining how to do it and he is asking himself he same questions that you will ask yourself.", + "video_name": "R486L0M5cWk", + "timestamps": [ + 251 + ], + "3min_transcript": "So that's still too large, so it's going to be 6. But notice, we had to do this little side work on the side right over here to realize it wasn't seven. Now six is the largest how many times you go into 108 without going over it. So 6 times 6 is 36. Carry the 3, or regroup the 3, depending on how you think about it. 6 times 1 is 6, plus 3 is 9. Then you subtract again. 8 minus 6 is 2. And then you can just say 10 minus 9 is 1, or you could even borrow. You could make this a 10. And then that goes away. 10 minus 9 is 1. So then you have 12. And if we're not going into decimals, you're done. Because 16 does not go into 12. So 16 goes into 1,388 86 times with a remainder of 12. That right over there is your remainder. And that's all a decent way of doing it. And that's the way you traditionally But what I want to do is introduce another maybe a little more interesting way So once again, let's do our 16 goes into 1,388. And what we're going to do is give us much more leeway for approximation, or for essentially guessing. And what we want to do is just guess. We're going to make guesses for how many times 16 goes into the numbers without overestimating, without jumping too high. And now we're not just going to think about the 1 or the 13 or the 138. We're going to think about the whole number as a whole. And before we do that, I'm going to get two things out of the way, just because it will help us. I'm just going to remind ourselves what 16 times 2 and 16 times 5 are. And I'm just picking these as random numbers that we can use to approximate. You don't have to use 2 and 5. You can use any numbers. Maybe I'll show other examples there. So 16 times 2, we know, is 32. And 16 times 5 is 50 plus 30, is 80. So let's just keep these two results in mind So the first thing to think about is our best guess for how many times does 16 go into 1,388. Or another way to think about, how many times does 16 go into 1,000? Let's just do something at a very rough approximation. Well, we know it's not going to be 100, because 100 times 16 would be 1,600. You would just throw those two 0's at the end of it. And [? you'd ?] say, how many times does it go into 1,000? Well, we know if 16 times 5 is 80, we know that 16 times 50 would be 800. So let's use that. And I'm using the 5-- I'm multiplying it by another 10 to get to 50-- instead of the 2 because 800 is a lot closer than 320 to our 1,000 that we care about. So what we could say is, well, 16 times 50 will get us to 800." + }, + { + "Q": "\n@3:35 Sal Crosses out the \"hours.\" Why can he do this?", + "A": "Because the hours part can be cutout as it is both at the numerator as well as at the denominator.", + "video_name": "Uc2Tm4Lr7uI", + "timestamps": [ + 215 + ], + "3min_transcript": "at just 8 miles per hour.\" So we're given a time. And we are given a speed. We should be able to figure out a distance. So let's just do a little bit of aside here. We should be able to figure out the distance to the-- actually let me write it this way. The distance to the store will be equal to-- now we've got to make sure we have our units right. Here they gave it in minutes. Here they have 8 miles per hour. So let's convert this into hours. So 45 minutes in hours, so it's 45 minutes out of 60 minutes per hour. So that's going to give us 45/60. Divide both by 15. That's the same thing as 3/4. So it's going to be 3/4 hours is the time times an average speed So what is the distance to the store? Well, 3/4 times 8. Or you could view it as 3/4 times 8 times 1, is going to be-- well, it's going to be 24 over 4. Let me just write that. That's going to be 24 over 4 which is equal to-- did I get it? Yeah. 24 over 4, which is equal to 6. And units-wise, we're just left with miles. So the distance to the store is 6 miles. 2 times the distance to the gift store, well, this whole thing is going to be 12 miles. 12 miles is the total distance she traveled. Now, what is the time to the gift store? Well, they already told that to us. They already told us that it's 45 minutes. Now, I want to put everything in hours. I'm assuming that they want our average speed in hours. So the time to the gift store was 3/4 of an hour. And what's the time coming back from the gift store? Well, we know her speed. We know her speed coming back. We already know the distance from the gift store. It's the same as the distance to the gift store. So we can take this distance, we can take 6 miles, that's the distance to the gift store, 6 miles divided by her speed coming back, which is 24 miles per hour, so divided by 24 miles per hour. It gives us-- well, let's see. We're going to have 6 over 24 is the same thing as 1/4. It's going to be 1/4. And then miles divided by miles per hour is the same thing as miles times hours per mile. The miles cancel out. And you'll have 1/4 of an hour. So it takes her 1/4 of an hour to get back. And that fits our intuition." + }, + { + "Q": "\nAt 2:34, Sal arrives at -(sqrt of 1/7)... I tried to simplify this more got -sqrt or 1 / sqrt of 7. I squared the top and the bottom. -1^2 and sqrt 7^2 and got 1/7. Is this correct or incorrect and why?", + "A": "No, you started off ok. -(sqrt of 1/7) = - sqrt(1) / sqrt(7) But, you can not square the fraction. It isn t equivalent to the original fraction. Instead, you should 1) Simplify sqrt(1): - sqrt(1) / sqrt(7) = -1 / sqrt(7) 2) Rationalize the denominator. This means we want to convert the denominator to a rational number. We do this by multiplying top and bottom by sqrt(7) and simplify. -1 / sqrt(7) = -1 / sqrt(7) * sqrt(7) / sqrt(7) = - sqrt(7) / sqrt(49) = - sqrt(7) / 7 Hope this helps.", + "video_name": "suwJmCrSDI8", + "timestamps": [ + 154 + ], + "3min_transcript": "Now, are there any perfect squares in 35? 35 is seven times five. No, neither of those are perfect squares. So, I could just leave that as square root of 35. And, let's see, the square root of four? Well, thats going to be two. This is the principal root, so we're thinking about the positive square root. The square root of nine is three. And so, this part right over is going to be four times two, times the square root of five, so it's going to be eight square roots of five. And then, this part over here is going to be minus three, times three, times the square root of five. So, minus nine square roots of five. And all of that is going to be over the square root of 35. Square root of 35. And, so let's see, if I have eight of something, and I subtract nine of that something, I'm gonna have negative one of that something. or I could just say negative square root of five. Negative square root of five over the square root of 35. I actually think I could simplify this even more, because this is the same thing. This is equal to the negative of the square root of five over 35. Both the numerator and the denominator are divisible by five. So, we could divide them both by five, and we would get the square root of divide the numerator by five, you get one. Divide the denominator by five, you get seven. So, we can view this as the square root of 1/7th. Square root of 1/7th. And we are all done. Let's do another one of these. These are strangely, strangely fun. And once again, pause it, and see if you can work it out on your own. Perform the indicated operations. Alright, so let's first multiply. So, this essentially is doing the distributive property twice. And, actually let me just do it that way, plus the square root of six. Let's first multiply it times the square root of five. So, the square root of five times the square root of five is going to be five. Square root of five times the square root of six is the square root of 30. So, five plus the square root of 30. And then, when I take this expression, and I multiply it times the second term, times the negative square root of six. Well, negative square root of six times the square root of five is going to be the negative of the square root of 30. And then the negative of the square root of six times the square root of six is going to be, we're gonna subtract six. The square root of six times the square root of six is six, then we have the negative out there. And so, just like that, we are left with, well, let's see. Square root of 30 minus the square root of 30, well those cancel out, thats zero. And we're left with five minus six, which is going to be equal to negative one." + }, + { + "Q": "\nWith reference to 9:15 (appx) ,\n\nIs it right to say that all graphs where there is ONLY a discontinuity at one end point, are graphs with removable discontinuity ? By the looks of it, it seems so. Was just wondering if there are any special cases.", + "A": "If you have a piecewise function, and your endpoint is at the point where the function changes, then the discontinuity is not removable. It s not a requirement that a discontinuity at an endpoint must be removable or that a discontinuity should be at the endpoints.", + "video_name": "kdEQGfeC0SE", + "timestamps": [ + 555 + ], + "3min_transcript": "" + }, + { + "Q": "\n4:54 why is rationalizing important?", + "A": "Rationalizing is like simplifying. To always have to simplify your answer unless you re told not to.", + "video_name": "tSHitjFIjd8", + "timestamps": [ + 294 + ], + "3min_transcript": "We know from the Pythagorean theorem-- let's say the hypotenuse is equal to C-- the Pythagorean theorem tells us that A squared plus B squared is equal to C squared. Right? Well we know that A equals B, because this is a 45-45-90 triangle. So we could substitute A for B or B for A. But let's just substitute B for A. So we could say B squared plus B squared is equal to C squared. Or 2B squared is equal to C squared. Or B squared is equal to C squared over 2. Or B is equal to the square root of C squared over 2. the numerator and the square root of the denominator-- C over the square root of 2. And actually, even though this is a presentation on triangles, I'm going to give you a little bit of actually information on something called rationalizing denominators. So this is perfectly correct. We just arrived at B-- and we also know that A equals B-- but that B is equal to C divided by the square root of 2. It turns out that in most of mathematics, and I never understood quite exactly why this was the case, people don't like square root of 2s in the denominator. Or in general they don't like irrational numbers in the denominator. Irrational numbers are numbers that have decimal places that never repeat and never end. So the way that they get rid of irrational numbers in the denominator is that you do something called rationalizing the denominator. And the way you rationalize a denominator-- let's take our example right now. If we had C over the square root of 2, we just multiply both the numerator and the denominator by the Because when you multiply the numerator and the denominator by the same number, that's just like multiplying it by 1. The square root of 2 over the square root of 2 is 1. And as you see, the reason we're doing this is because square root of 2 times square root of 2, what's the square root of 2 times square root of 2? Right, it's 2. Right? We just said, something times something is 2, well the square root of 2 times square root of 2, that's going to be 2. And then the numerator is C times the square root of 2. So notice, C times the square root of 2 over 2 is the same thing as C over the square root of 2. And this is important to realize, because sometimes while you're taking a standardized test or you're doing a test in class, you might get an answer that looks like this, has a square root of 2, or maybe even a square root of 3 or whatever, in the denominator. And you might not see your answer if it's a multiple choice question. What you ned to do in that case is rationalize the denominator. So multiply the numerator and the denominator by square root of 2 and you'll get square root of 2 over 2." + }, + { + "Q": "\nI am confused. I think that this new formula (B=C2/square root 2) is unnecessary. Sal stated at 1:04 that in these 45-45-90 triangels, both non-hypotenuse sides are equal. Therefore, if one non-hypotenuse side of the triangle is 8, we could automatically assume its other congruent side is 8. We could then just use the Pythagorean Theorem to find the hypotenuse. Could someone please explain why we must use a different formula? I'll give a more detailed example of my point in the questions area.", + "A": "For example, in a 45-45-90 triangle where one side is labeled 2 and the hypotenuse is unknown, we can label the other non-hypotenuse side 2 also. From this point, we can simply use the Pythagorean Theorem to find the hypotenuse. We don t need the formula which Sal gives at 4:27 in the video.", + "video_name": "tSHitjFIjd8", + "timestamps": [ + 64 + ], + "3min_transcript": "Welcome to the presentation on 45-45-90 triangles. Let me write that down. How come the pen-- oh, there you go. 45-45-90 triangles. Or we could say 45-45-90 right triangles, but that might be redundant, because we know any angle that has a 90 degree measure in it is a right triangle. And as you can imagine, the 45-45-90, these are actually the degrees of the angles of the triangle. So why are these triangles special? Well, if you saw the last presentation I gave you a little theorem that told you that if two of the base angles of a triangle are equal-- and it's I guess only a base angle You could draw it like this, in which case it's maybe not so obviously a base angle, but it would still be true. If these two angles are equal, then the sides that they don't share-- so this side and this side in this example, or this are going to be equal. So what's interesting about a 45-45-90 triangle is that it is a right triangle that has this property. And how do we know that it's the only right triangle that has this property? Well, you could imagine a world where I told you that this is a right triangle. This is 90 degrees, so this is the hypotenuse. Right, it's the side opposite the 90 degree angle. And if I were to tell you that these two angles are equal to each other, what do those two angles have to be? Well if we call these two angles x, we know that the angles in a triangle add up to 180. So we'd say x plus x plus-- this is 90-- plus 90 is equal to 180. Or 2x plus 90 is equal to 180. Or 2x is equal to 90. So the only right triangle in which the other two angles are equal is a 45-45-90 triangle. So what's interesting about a 45-45-90 triangle? Well other than what I just told you-- let me redraw it. I'll redraw it like this. So we already know this is 90 degrees, this is 45 degrees, this is 45 degrees. And based on what I just told you, we also know that the sides that the 45 degree angles don't share are equal. So this side is equal to this side. And if we're viewing it from a Pythagorean theorem point of view, this tells us that the two sides that are not the hypotenuse are equal. So this is a hypotenuse." + }, + { + "Q": "At 8:30, why is 2 being multiplied by 8, but \u00e2\u0088\u009a2 isn't being multiplied by 8?\n", + "A": "Think of it as 8/1 * 2/\u00e2\u0088\u009a2. In normal multiplication, you just multiply across for example, 2/3 * 3/4 = (2*3)/(3*4) = 6/12 = 1/2", + "video_name": "tSHitjFIjd8", + "timestamps": [ + 510 + ], + "3min_transcript": "So what did we learn? This is equal to B, right? So turns out that B is equal to C times the square root of 2 over 2. So let me write that. So we know that A equals B, right? And that equals the square root of 2 over 2 times C. Now you might want to memorize this, though you can always derive it if you use the Pythagorean theorem and remember that the sides that aren't the hypotenuse in a 45-45-90 triangle are equal to each other. But this is very good to know. Because if, say, you're taking the SAT and you need to solve a problem really fast, and if you have this memorized and someone gives you the hypotenuse, you can figure out what are the sides very fast, or i8f someone gives you one of the sides, you can figure out the hypotenuse very fast. Let's try that out. I'm going to erase everything. So we learned just now that A is equal to B is equal to the So if I were to give you a right triangle, and I were to tell you that this angle is 90 and this angle is 45, and that this side is, let's say this side is 8. I want to figure out what this side is. Well first of all, let's figure out what side is the hypotenuse. Well the hypotenuse is the side opposite the right angle. So we're trying to actually figure out the hypotenuse. Let's call the hypotenuse C. And we also know this is a 45-45-90 triangle, right? Because this angle is 45, so this one also has to be 45, because 45 plus 90 plus 90 is equal to 180. So this is a 45-45-90 triangle, and we know one of the sides-- this side could be A or B-- we know that 8 is equal to the C is what we're trying to figure out. So if we multiply both sides of this equation by 2 times the square root of 2-- I'm just multiplying it by the inverse of the coefficient on C. Because the square root of 2 cancels out that square root of 2, this 2 cancels out with this 2. We get 2 times 8, 16 over the square root of 2 equals C. Which would be correct, but as I just showed you, people don't like having radicals in the denominator. So we can just say C is equal to 16 over the square root of 2 times the square root of 2 over the square root of 2. So this equals 16 square roots of 2 over 2. Which is the same thing as 8 square roots of 2." + }, + { + "Q": "At 0:54 , does Sal mean that if a triangle has less then 3 acute angles, it isn't an acute triangle? Is it true? If it is, why?\n", + "A": "Yes, it s true. You must have 3 acute angles to make an acute triangle. You need one right angle to make a right triangle. You need one obtuse angle to make an obtuse triangle.", + "video_name": "PiQxA9O7Rd8", + "timestamps": [ + 54 + ], + "3min_transcript": "Which side is perpendicular to side BC? So BC is this line segment right over here. And for another segment to be perpendicular to it, perpendicular just means that the two segments need to intersect at a right angle, or at a 90-degree angle. And we see that BC intersects AB at a 90-degree angle. This symbol right over here represents a 90-degree, or a right angle. So we just have to find side AB or BA. And that's right over here. Side AB is perpendicular to side BC. Let's do a few more of these. Put the triangles into the correct categories, so this right over here. So let's see. Let's think about our categories. Right triangles-- so that means it has a 90-degree angle in it. Obtuse triangles-- that means it has an angle larger than 90 degrees in it. Acute triangles-- that means all three angles are less than 90 degrees. So this one has a 90-degree angle. It has a right angle right over here. This one right over here, all of these angles are less than 90 degrees, just eyeballing it. So this is going to be an acute-- that's going to be an acute triangle. I'll put it under acute triangles right over there. Then this one over here, this angle up here, this is-- and we can assume that these actually are drawn to scale, this is more open than a 90-degree angle. This is an obtuse angle right over here. It's going to be more than 90 degrees. So this is an obtuse triangle. Now, this one over here, all of them seem acute. None of them even seem to be a right angle. So I would put this again into acute-- acute triangles. This one here clearly has a right angle. It's labeled as such. So we'll throw it right over here. And then this one, this angle right over here is clearly even larger. It has a larger measure than a right angle. So this angle right over here is more than 90 degrees. It's going to be an obtuse angle. So we got two in each of these. And let's check our answer. We got it right." + }, + { + "Q": "\nAt 0:30, The selected answer was side AB, but couldn't side CD also be considered a correct answer?", + "A": "Perpendicular means the two lines are forming a right angle.", + "video_name": "PiQxA9O7Rd8", + "timestamps": [ + 30 + ], + "3min_transcript": "Which side is perpendicular to side BC? So BC is this line segment right over here. And for another segment to be perpendicular to it, perpendicular just means that the two segments need to intersect at a right angle, or at a 90-degree angle. And we see that BC intersects AB at a 90-degree angle. This symbol right over here represents a 90-degree, or a right angle. So we just have to find side AB or BA. And that's right over here. Side AB is perpendicular to side BC. Let's do a few more of these. Put the triangles into the correct categories, so this right over here. So let's see. Let's think about our categories. Right triangles-- so that means it has a 90-degree angle in it. Obtuse triangles-- that means it has an angle larger than 90 degrees in it. Acute triangles-- that means all three angles are less than 90 degrees. So this one has a 90-degree angle. It has a right angle right over here. This one right over here, all of these angles are less than 90 degrees, just eyeballing it. So this is going to be an acute-- that's going to be an acute triangle. I'll put it under acute triangles right over there. Then this one over here, this angle up here, this is-- and we can assume that these actually are drawn to scale, this is more open than a 90-degree angle. This is an obtuse angle right over here. It's going to be more than 90 degrees. So this is an obtuse triangle. Now, this one over here, all of them seem acute. None of them even seem to be a right angle. So I would put this again into acute-- acute triangles. This one here clearly has a right angle. It's labeled as such. So we'll throw it right over here. And then this one, this angle right over here is clearly even larger. It has a larger measure than a right angle. So this angle right over here is more than 90 degrees. It's going to be an obtuse angle. So we got two in each of these. And let's check our answer. We got it right." + }, + { + "Q": "At 11:28, when Sal multiplies by b^2 shouldn't it be b^2 x^2/a^2 instead of b^2/ a^2 x^2 ?\n", + "A": "Actually, both are equivalent. First Case: b^2 x^2/a^2 (b^2*x^2)/a^2 Second Case: b^2/a^2 x^2 (b^2/a^2)*x^2 x^2=x^2/1. (b^2/a^2)*(x^2/1) Multiply the numerators and denominators of the fractions. (b^2*x^2)/a^2 We end up with (b^2*x^2)/a^2 in both cases, meaning that both expressions is equivalent. I hope this helps!", + "video_name": "pzSyOTkAsY4", + "timestamps": [ + 688 + ], + "3min_transcript": "never get to x equal to 0. You get to y equal 0, right here and here. But you never get to x equals 0. And actually your teacher might want you to plot these points, and there you just substitute y equals 0. And you can just look at the original equation. Actually, you could even look at this equation right here. Can x ever equal 0? If you look at this equation, if x is equal to 0, this whole term right here would cancel out, and you'd just be left with a minus b squared. Which is, you're taking b squared, and you put a negative sign in front of it. So that's a negative number. And then you're taking a square root of a negative number. So we're not dealing with imaginaries right now. So you can never have x equal to 0. But y could be equal to 0, right? You can set y equal to 0 and then you could solve for it. So in this case, actually let's do that. If y is equal to 0, you get 0 is equal to the square root of b squared over a squared x squared minus b squared. If you square both sides, you get b squared over a I know this is messy. So then you get b squared over a squared x squared is equal to b squared. You could divide both sides by b squared, I guess. You get a 1 and a 1. And then you could multiply both sides by a squared. You get x squared is equal to a squared, and then you get x is equal to the plus or minus square root of a. So this point right here is the point a comma 0, and this point right here is the point minus a comma 0. Now let's go back to the other problem. I have a feeling I might be running out of time. So notice that when the x term was positive, our hyperbola opened to the right and the left. And you could probably get from detective reasoning that when the y term is positive, which is the case in this one, we're probably going to open up and down. So let's solve for y. You get y squared over b squared. We're going to add x squared over a squared to both sides. So you get equals x squared over a squared plus 1. Multiply both sides by b squared. y squared is equal to b squared over a squared x squared plus b squared. You have to distribute the b squared. Now take the square root. I'll switch colors for that. So y is equal to the plus or minus square root of b squared over a squared x squared plus b squared. And once again-- I've run out of space-- we can make that same argument that as x approaches positive or negative infinity, this equation, this b, this little constant term right here isn't going to matter as much. You're just going to take the square root of this term right here." + }, + { + "Q": "\nAt 2:46 did Sal mean hyperbolas? It sounded like he said parabolas.", + "A": "I think you are correct. He said parabolas , but he should have said hyperbolas .", + "video_name": "pzSyOTkAsY4", + "timestamps": [ + 166 + ], + "3min_transcript": "And so this is a circle. And once again, just as review, a circle, all of the points on the circle are equidistant from the center. Or in this case, you can kind of say that the major axis and the minor axis are the same distance, that there isn't any distinction between the two. You're always an equal distance away from the center. So that was a circle. An ellipse was pretty much this, but these two numbers could be different. Because your distance from the center could change. So it's x squared over a squared plus y squared over b squared is equal to 1. That's an ellipse. And now, I'll skip parabola for now, because parabola's kind of an interesting case, and you've already touched on it. So I'll go into more depth in that in a future video. But a hyperbola is very close in formula to this. And so there's two ways that a hyperbola could be written. And I'll do those two ways. minus y squared over b squared is equal to 1. And notice the only difference between this equation and this one is that instead of a plus y squared, we have a minus y squared here. So that would be one hyperbola. The other one would be if the minus sign was the other way around. If it was y squared over b squared minus x squared over a squared is equal to 1. So now the minus is in front of the x squared term instead of the y squared term. And what I want to do now is try to figure out, how do we graph either of these parabolas? Maybe we'll do both cases. And in a lot of text books, or even if you look it up over the web, they'll give you formulas. But I don't like those formulas. One, because I'll always forget it. And you'll forget it immediately after taking the test. You might want to memorize it if you just want to be able to do the test a little bit faster. But you'll forget it. you'll probably get confused. Because sometimes they always use the a under the x and the b under the y, or sometimes they always use the a under the positive term and to b under the negative term. So if you just memorize, oh, a divided by b, that's the slope of the asymptote and all of that, you might be using the wrong a and b. So I encourage you to always re-prove it to yourself. And that's what we're going to do right here. It actually doesn't take too long. So these are both hyperbolas. And what I like to do whenever I have a hyperbola is solve for y. So in this case, if I subtract x squared over a squared from both sides, I get-- let me change the color-- I get minus y squared over b squared. That stays there. Is equal to 1 minus x squared over a squared. And then, let's see, I want to get rid of this minus, and I want to get rid of this b squared. So let's multiply both sides of this equation times minus b squared." + }, + { + "Q": "at 10:32, why is the a sqrt(a), isn't it supposed to me +- a? Because x^2=a^2 and then root both sides, x = +-a? I can't find the report a problem button though. But I might be missing something.\n", + "A": "If you keep playing the video, you should notice that in the bottom right hand corner Sal explains his mistake.", + "video_name": "pzSyOTkAsY4", + "timestamps": [ + 632 + ], + "3min_transcript": "never get to x equal to 0. You get to y equal 0, right here and here. But you never get to x equals 0. And actually your teacher might want you to plot these points, and there you just substitute y equals 0. And you can just look at the original equation. Actually, you could even look at this equation right here. Can x ever equal 0? If you look at this equation, if x is equal to 0, this whole term right here would cancel out, and you'd just be left with a minus b squared. Which is, you're taking b squared, and you put a negative sign in front of it. So that's a negative number. And then you're taking a square root of a negative number. So we're not dealing with imaginaries right now. So you can never have x equal to 0. But y could be equal to 0, right? You can set y equal to 0 and then you could solve for it. So in this case, actually let's do that. If y is equal to 0, you get 0 is equal to the square root of b squared over a squared x squared minus b squared. If you square both sides, you get b squared over a I know this is messy. So then you get b squared over a squared x squared is equal to b squared. You could divide both sides by b squared, I guess. You get a 1 and a 1. And then you could multiply both sides by a squared. You get x squared is equal to a squared, and then you get x is equal to the plus or minus square root of a. So this point right here is the point a comma 0, and this point right here is the point minus a comma 0. Now let's go back to the other problem. I have a feeling I might be running out of time. So notice that when the x term was positive, our hyperbola opened to the right and the left. And you could probably get from detective reasoning that when the y term is positive, which is the case in this one, we're probably going to open up and down. So let's solve for y. You get y squared over b squared. We're going to add x squared over a squared to both sides. So you get equals x squared over a squared plus 1. Multiply both sides by b squared. y squared is equal to b squared over a squared x squared plus b squared. You have to distribute the b squared. Now take the square root. I'll switch colors for that. So y is equal to the plus or minus square root of b squared over a squared x squared plus b squared. And once again-- I've run out of space-- we can make that same argument that as x approaches positive or negative infinity, this equation, this b, this little constant term right here isn't going to matter as much. You're just going to take the square root of this term right here." + }, + { + "Q": "At 2:35, do you have to distribute the y, could you just divide the whole equation by x?\n", + "A": "Keep in mind that we want to solve the equation for x. To do that we have to unlock the parentheses and then gather all the x terms on one side of the equation and everything else on the other side. That s why the y must be distributed. Dividing by x would severely complicate things, as you would end up with [ y ( x + 3 ) ] / x = 2 - 1/x and we would still not have only the x terms on one side of the equation.", + "video_name": "VzWxvDe8TUQ", + "timestamps": [ + 155 + ], + "3min_transcript": "You start with a member of the range and you go back to a member of the domain. That's what g inverse does. And so, there's a couple of ways to think about it. If we were to set, if we were to say, look, let's just set y is equal to g of x. So we could just call that point y, as well. So this expression here, let's just, okay, if we have an x, this is what we do. We multiply by two times and we subtract one, divided by x plus three, and then that tells us what the associated g of x, what the associated y is. But what if we're given the y, or the g of x. How do we get x? Well, we could just solve for x. So let's do that. If we said that y is equal to two x minus one over x plus three, let's solve for x, so that we know for any y we get, what the corresponding x we'll get. So how do we do that? Well, we can multiply both sides times x plus three. we're going to get y times x plus three, x plus three, is equal to, is equal to two x minus one. All I did was multiply both sides times x plus three. x plus three, and I multiplied x plus three on that side as well. The x plus three cancels out with the x plus three. And then, let's see. Let's see what we can do now. Well, we could distribute the y. We could distribute the y, and I'll switch now to a different color because that's going to get complicated if I keep trying to separate the ys in a different color. So we're going to get y times x, y x, plus three y, plus three y, is equal to two x minus one. And remember, we are now trying to solve for x. So we can collect all of the x terms on one side of this equation, So let's get all of our x terms on the left-hand side and all of our non-x terms on the right-hand side. So I want to get rid of this. This doesn't involve an x, so let me subtract three y here. And so I'll subtract three y from this side. Minus three y. And we want to get rid of this from the right-hand side, so minus two x. So we can put it right over here. Minus two x. And so, we are going to get, let me scroll down a little bit. So, whoops. Sorry for that. So this is going to be y x minus two x is going to be y minus two times x. These cancel out, which is their goal. Is equal to two x minus two x, let's cancel that, which was the goal. And then you have one minus three y. One minus three y. And now, to solve for x, you just divide both sides by y minus two. y minus two. And we're going to get x is equal to" + }, + { + "Q": "\nAt 3:52, I think Sal forgets the negative sign next to the 1. If this is true, does that mean my answer of (3y+1)/(-y+2) is correct? I took a different approach by leaving the 3y and adding the 1, instead of Sal's method, so I was hoping that it worked both ways.", + "A": "Sal does forget the sign, and a pop-up box in the lower right-hand corner mentions that. Your answer is correct. ( -1 - 3y ) / ( y - 2 ) = -1 ( 1 + 3y ) / [ -1 ( -y + 2 ) = ( 1 + 3y ) / ( -y + 2 )", + "video_name": "VzWxvDe8TUQ", + "timestamps": [ + 232 + ], + "3min_transcript": "we're going to get y times x plus three, x plus three, is equal to, is equal to two x minus one. All I did was multiply both sides times x plus three. x plus three, and I multiplied x plus three on that side as well. The x plus three cancels out with the x plus three. And then, let's see. Let's see what we can do now. Well, we could distribute the y. We could distribute the y, and I'll switch now to a different color because that's going to get complicated if I keep trying to separate the ys in a different color. So we're going to get y times x, y x, plus three y, plus three y, is equal to two x minus one. And remember, we are now trying to solve for x. So we can collect all of the x terms on one side of this equation, So let's get all of our x terms on the left-hand side and all of our non-x terms on the right-hand side. So I want to get rid of this. This doesn't involve an x, so let me subtract three y here. And so I'll subtract three y from this side. Minus three y. And we want to get rid of this from the right-hand side, so minus two x. So we can put it right over here. Minus two x. And so, we are going to get, let me scroll down a little bit. So, whoops. Sorry for that. So this is going to be y x minus two x is going to be y minus two times x. These cancel out, which is their goal. Is equal to two x minus two x, let's cancel that, which was the goal. And then you have one minus three y. One minus three y. And now, to solve for x, you just divide both sides by y minus two. y minus two. And we're going to get x is equal to over y minus two. So one way you could think about it is, you could say, well x is equal to, this expression is equal to the inverse function as a function of y. You give me a y, you give me a y that's sitting in this range, I can input it into this function definition here and I can give you the corresponding x. I can give you the corresponding x. Now, we didn't ask for g inverse of y. We asked for g inverse of x. But the important thing to remember is this variable that we use inside of, the variable that we use inside of functions right over here, they're chosen somewhat arbitrarily. They just say, okay, I'm going to call the input y, and if you're going to call the input y, well, this is how I would give you the output. But we could call the input a, and that would be one minus three a over a minus twp. Or we could even call the input, so let me make this clear." + }, + { + "Q": "\nAt 3:47, shouldn't it be -1-3y instead of 1-3y, so that we get (y-2)x = -1-3y ?? In other words, did he forget to keep the 1 negative?", + "A": "This is a known mistake in the video. There is a box that pops up at about 3:50 that states there is an error and provides the correct value.", + "video_name": "VzWxvDe8TUQ", + "timestamps": [ + 227 + ], + "3min_transcript": "we're going to get y times x plus three, x plus three, is equal to, is equal to two x minus one. All I did was multiply both sides times x plus three. x plus three, and I multiplied x plus three on that side as well. The x plus three cancels out with the x plus three. And then, let's see. Let's see what we can do now. Well, we could distribute the y. We could distribute the y, and I'll switch now to a different color because that's going to get complicated if I keep trying to separate the ys in a different color. So we're going to get y times x, y x, plus three y, plus three y, is equal to two x minus one. And remember, we are now trying to solve for x. So we can collect all of the x terms on one side of this equation, So let's get all of our x terms on the left-hand side and all of our non-x terms on the right-hand side. So I want to get rid of this. This doesn't involve an x, so let me subtract three y here. And so I'll subtract three y from this side. Minus three y. And we want to get rid of this from the right-hand side, so minus two x. So we can put it right over here. Minus two x. And so, we are going to get, let me scroll down a little bit. So, whoops. Sorry for that. So this is going to be y x minus two x is going to be y minus two times x. These cancel out, which is their goal. Is equal to two x minus two x, let's cancel that, which was the goal. And then you have one minus three y. One minus three y. And now, to solve for x, you just divide both sides by y minus two. y minus two. And we're going to get x is equal to over y minus two. So one way you could think about it is, you could say, well x is equal to, this expression is equal to the inverse function as a function of y. You give me a y, you give me a y that's sitting in this range, I can input it into this function definition here and I can give you the corresponding x. I can give you the corresponding x. Now, we didn't ask for g inverse of y. We asked for g inverse of x. But the important thing to remember is this variable that we use inside of, the variable that we use inside of functions right over here, they're chosen somewhat arbitrarily. They just say, okay, I'm going to call the input y, and if you're going to call the input y, well, this is how I would give you the output. But we could call the input a, and that would be one minus three a over a minus twp. Or we could even call the input, so let me make this clear." + }, + { + "Q": "I'm not following what Sal did at 3:18 when he got out the calculator and got a number for r. Did he just rearrange the whole logistic equation in his head and solve for r, or what?\n", + "A": "this is the equation he used: future value / present value = (1+i)^n (growth rate equation google it) i= growth rate n=number of periods. 150/100=(1+i)^20---> i=[(1.5)^(1/20)] - 1", + "video_name": "-fIsaqN-aaQ", + "timestamps": [ + 198 + ], + "3min_transcript": "This logistic function. This logistic function is a nonconstant solution, and it's the interesting one we care about if we're going to model population to the logistic differential equation. So now that we've done all that work to come up with this, let's actually apply it. That was the whole goal, was to model population growth. So let's come up with some assumptions. Let's first think about, well let's say that I have an island. So let's say that this is my island, and I start settling it with a 100 people. So I'm essentially saying N naught So I'm saying N naught is equal to 100. Let's say that this environment, given current technology of farming and agriculture, and the availability of water and whatever else, let's say it can only support 1,000 people max. So you get the idea, so we get K is equal to 1000. That's the limit to the population. So now what we have to think about is what is r going to be? So we have to come up with some assumptions. So, let's say in a generation which is about 20 years, well I'll just assume in 20 years, yeah I think it's reasonable that the population grows by, let's say that the population grows by 50%. In 20 years you have 50% growth. 50% increase, increase in the actual population. in order to after 20 years to grow by 50%. Well to think about that I'll get out my calculator. One way to think about it, growing by 50%, that means that you are at 1.5 your original population, and if I take that to the 120th power, and we'll just do 1 divided by 20th. This essentially says how much am I going to grow by or what is going to, this is telling me I'm going to grow by a factor of 1.02 every year, 1.02048. So one way to think about it is if every year I grow by 0.020 I'll just round five then over 20 years" + }, + { + "Q": "On 2:32 is scaling a type of measuring? That's all I'm asking thank you!\n", + "A": "Sort of. When you scale something up or down, you are keeping the same proportions, but making it bigger or smaller. So, if you have a 1 cm cube and scale it up by 10, then each of the measurements would increase x10. So, it keeps the same proportions relative to the other sides, but gets bigger. If one side gets bigger out of proportion, then it would lose its original shape.", + "video_name": "yUYDhmQsiXY", + "timestamps": [ + 152 + ], + "3min_transcript": "it's already written as 8/7 times 2/3. And then, this last expression, we could write it as, in the numerator, 5 times 2. And then in the denominator, it's over 5 times 3. 5 times 3, which is of course the same thing as 5/5 times 2/3. So you see, all three of these expressions involve something times 2/3. Now, looking at it this way, does it become easier to pick out which of these are the largest, which of these are the smallest, and which of these are someplace in between? I encourage you to pause it again if you haven't thought about it yet. So let's visualize each of these expressions by first trying to visualize 2/3. So let's say the height of what I am drawing right now, let's say the height of this bar right over here is 2/3. The height here is 2/3. So first, let's think about what this one on the right here represents. This is 5/5 times 2/3. Well, what's 5/5? 5/5 is the same thing as 1. This is literally just 1 times 2/3. This whole expression is the same thing as 1 times 2/3, or really, just 2/3. So this, the height here, 2/3, this is the same thing as this thing over here. This is going to be equal to-- this could also be viewed as 5 times 2 over 3 times 5, which was this first expression right over here. Now, let's think about what these would look like. So this is 7/8 times 2/3. So it's less than 8/8 times 2/3. It's less than 1 times 2/3. So we're going to scale 2/3 down. It's going to be 7/8 of 2/3. So this one right over here would look something like this. Let me see if I can draw it. Yeah, it would look something like this. If the yellow height is 2/3, then this right over here, then this height right over here-- let me make it clear. This height right over here would be 7/8 times 2/3. Likewise, let's look at this one right over here. Let's look at this one in the middle, 8/7 times 2/3. Well, 8/7 is bigger than 7/7. It's more than 1. This is more than 2/3. This is 1 and 1/7 times 2/3. So it's going to be the same height as 2/3 plus another 1/7. So it's going to look something like this. It's going to look something like this. So its height-- now we scaled the 2/3 up because 8/7" + }, + { + "Q": "how does sal know that at 2:10 the graph g(x) ill be equal to f(x)-horizontal shift + the vertical shift and not f(x) + horizontal shift + vertical shift?\n", + "A": "He explains this at 3:00 and on.", + "video_name": "MDav5OMpCto", + "timestamps": [ + 130 + ], + "3min_transcript": "- [Voiceover] So we have these two graphs that look pretty similar, y equals f of x and y is equal to g of x. And what they asked us to do is write a formula for the function g in terms of f. So let's think about how to do it and like always pause the video and see if you can work through it on your own. All right, well, what I'd like to do is I'd like to focus on this minimum point because I think that's a very easy thing to look at because of them have that minimum point right over there. And so we can think, but how do we shift f? Especially this minimum point, how do we shift it to get to overlapping with g? Well, the first thing that might jump out of this is that we would want to shift to the left. And we'd wanna shift to the left four. So let me do this in a new color. So I would wanna shift to the left by four. We have shifted to the left by four or you could say we shifted by negative four. Either way, you could think about it. And then we're gonna shift down. from y equals two to y is equal to negative five. So let me do that. So let's shift down. So we shift down by seven or you could say we have a negative seven shift. So, how do you express g of x if it's a version of f of x that shifted to the left by four and shifted down by seven. Or, you could say I have a negative four horizontal shift. I have a negative seven vertical shift. Well, one thing to think about it is g of x, g of x is going to be equal to f of, let me do it in a little darker color, it's going to be equal to f of x minus your horizontal shift, all right, horizontal shift. So x minus your horizontal shift plus your vertical shift. So plus your vertical Well, what is our horizontal shift here? Well, we're shifting to the left so it's a negative shift. So our horizontal shift is negative four. Now, what's our vertical shift? Well, we went down so our vertical shift is negative seven. So it's negative seven. So there you have it. We get g of x. Let me do that in the same color. We get g of x is equal to f of x minus negative four or x plus four and then we have plus negative seven or you can just say minus seven. And we're done. And when I look at things like this, the negative seven is somewhat, it's more intuitive to me. As I shifted it down, it made sense that I have a negative seven. But first, when you work on these, you say, \"I shifted to the left. \"Why is it a plus four?\"" + }, + { + "Q": "\n0:33 so a derivative of a function for critical points can only either equal to zero or undefined? a derivative can't have both places where it equals to zero and and places where it's undefined?\ncould you please do a video or explain how to identify the minima and maxima for a function where the derivative can be undefined instead of equalling to zero?", + "A": "Some calculus textbooks (perhaps most) refer only to points where the derivative is zero as critical points. Others may include points where the derivative is undefined. Consider this function: y = sqrt(36 - x^2) That s the equation for the top half of a circle. The derivative is zero at x=0, a maximum, and undefined at x=-6 and x=6, which are minima.", + "video_name": "pInFesXIfg8", + "timestamps": [ + 33 + ], + "3min_transcript": "We've got the function f of x is equal to x to the third power minus 12x plus 2. And what I want to do in this video is think about at what points does my function f take on minimum or maximum values? And to figure that out I have to first figure out what are the critical points for my function f. And then which of those critical points do we achieve a minimum or maximum value? And to determine the critical points we have to find the derivative of our function because our critical points are just the point at which our derivative is either equal to 0 or undefined. So the derivative of this thing right over here, we're just going to use the power rule several times, and then I guess you can call it the constant rule. But the derivative of x to the third is 3x squared. Derivative of negative 12x is negative 12. And the derivative of a constant, it doesn't change with respect to x, so it's just going to be equal to 0. So we're going to get a critical point when this thing right over here, for some value of x is either undefined or 0. Well this thing is defined for all values of x. So the only places we're going to find critical points So let's set it equal to 0. When does 3x squared minus 12 equal 0? So let's add 12 to both sides. You get 3x squared is equal to 12. Divide both sides by 3. You get x squared is equal to 4. Well this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f of 2, or let me be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime negative 2 is also, same exact reason, is also equal to 0. So we can say-- and I'll switch colors here-- that f has critical points at x equals 2 and x equals negative 2. But we still don't know whether the function takes on a minimum values at those points, maximum values of those points, or neither. To figure that out we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. I'll draw an axis right over here. I'll do it down here because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2, 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well we have when x is equal to 0 for the derivative we're at negative 12." + }, + { + "Q": "\nAt 3:08 , why did sal crossed the curve of -12 from 2 and -2 ?\nPlease explain.\nThank you", + "A": "I dont understand your question, please explain your problem.", + "video_name": "pInFesXIfg8", + "timestamps": [ + 188 + ], + "3min_transcript": "So let's set it equal to 0. When does 3x squared minus 12 equal 0? So let's add 12 to both sides. You get 3x squared is equal to 12. Divide both sides by 3. You get x squared is equal to 4. Well this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f of 2, or let me be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime negative 2 is also, same exact reason, is also equal to 0. So we can say-- and I'll switch colors here-- that f has critical points at x equals 2 and x equals negative 2. But we still don't know whether the function takes on a minimum values at those points, maximum values of those points, or neither. To figure that out we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. I'll draw an axis right over here. I'll do it down here because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2, 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well we have when x is equal to 0 for the derivative we're at negative 12. So this is, we're graphing y is equal to f prime of x. So it looks something like this. These are obviously the 0's of our derivative. So it has to move up to cross the x-axis there and over here. So what is the derivative doing at each of these critical points? Well over here our derivative is crossing from being positive, we have a positive derivative, to being a negative derivative. So we're crossing from being a positive derivative to being a negative derivative, that was our criteria for a critical point to be a maximum point. Over here we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function" + }, + { + "Q": "\n1:00\nSo sin(x)^-1 is not 1/sin(x)? Why is it written this way?", + "A": "It really isn t a well thought out syntax. If it confuses you, you can use arcsin instead of sin^-1. The difference is subtle. Btw, 1/sin(x) can be written as csc(x) (csc stands for cosecant ).", + "video_name": "57BiI_iD3-U", + "timestamps": [ + 60 + ], + "3min_transcript": "Let's see if we can remove the parameter t from a slightly more interesting example. So let's say that x is equal to 3 times the cosine of t. And y is equal to 2 times the sine of t. We can try to remove the parameter the same way we did in the previous video, where we can solve for t in terms of either x or y and then substitute back in. And I'll do that. But I want to do that first, just to show you that it kind of leads to a hairy or an unintuitive answer. So if we solve for-- let's solve for t here. We could do it either one, they're equally complex. So if we solve for t here, we would say divide both sides by 2. You'd get y over 2 is equal to sine of t. And then you would take the arcsine of both sides, or the inverse sine of both sides, and you would get-- I like writing arcsine, because inverse sine, people often confuse it with an Arcsine of y over 2 is equal to t. Actually, let me do that little aside there. I should probably do it at the trigonometry playlist, but it's a good thing to hit home. Because I think people get confused. So arcsine of anything, let's say, y. The other way of writing that is sine minus 1 of y. These two things are equivalent, when they're normally used. But I don't like using this notation most of the time, because it can be ambiguous. This could mean sine of y to the negative 1 power, which equals 1 over sine of y. And arcsine and this are definitely not the same thing. So you want to be very careful there to make sure that you don't get confused when someone writes an inverse sine like this. But they're not actually taking sine of y to On the other hand, if someone were to write sine squared of y, this is unambiguously the same thing as sine of y squared. In fact, I wish this was the more conventional notation because it wouldn't make people think, oh, 2 and minus 1 there, and of course, that's just sine of y squared. So it can be very ambiguous. And of course, if this was a negative, this would be a minus 2, and then this really would be 1 over sine of y squared. That's why, just a long-winded way of explaining why I wrote arcsine, instead of inverse sine right there. Needless to say, let's get back to the problem. So we've solved for t in terms of y. Now we can substitute back here. And we've got an expression for x in terms of y. So we get x is equal to 3 times the cosine of t. But we just solved for t. t is this thing right here. So it's the cosine of arcsine of y over 2." + }, + { + "Q": "\nAt 0:23, the equations were x=3cos(t) and y=2sin(t). Why couldn't we have written x=3sin(t) instead of cosine and y=2cos(t) instead of sine? Would it have made a different graph? Would it have been a hyperbola instead of an ellipse?", + "A": "It would have made the same graph. Try graphing it on your graphing calculator (or a free online one). Be sure that the graph is in parametric mode (or PAR ) and in radians, not degrees. The only difference is in which direction the equation goes. As Sal was mentioning at around 7:45, we don t know which direction the graph was created by simplifying the equation. If sin and cos were switched, then the graph would go in a different direction, but would look the same. Hope this helps!", + "video_name": "57BiI_iD3-U", + "timestamps": [ + 23 + ], + "3min_transcript": "Let's see if we can remove the parameter t from a slightly more interesting example. So let's say that x is equal to 3 times the cosine of t. And y is equal to 2 times the sine of t. We can try to remove the parameter the same way we did in the previous video, where we can solve for t in terms of either x or y and then substitute back in. And I'll do that. But I want to do that first, just to show you that it kind of leads to a hairy or an unintuitive answer. So if we solve for-- let's solve for t here. We could do it either one, they're equally complex. So if we solve for t here, we would say divide both sides by 2. You'd get y over 2 is equal to sine of t. And then you would take the arcsine of both sides, or the inverse sine of both sides, and you would get-- I like writing arcsine, because inverse sine, people often confuse it with an Arcsine of y over 2 is equal to t. Actually, let me do that little aside there. I should probably do it at the trigonometry playlist, but it's a good thing to hit home. Because I think people get confused. So arcsine of anything, let's say, y. The other way of writing that is sine minus 1 of y. These two things are equivalent, when they're normally used. But I don't like using this notation most of the time, because it can be ambiguous. This could mean sine of y to the negative 1 power, which equals 1 over sine of y. And arcsine and this are definitely not the same thing. So you want to be very careful there to make sure that you don't get confused when someone writes an inverse sine like this. But they're not actually taking sine of y to On the other hand, if someone were to write sine squared of y, this is unambiguously the same thing as sine of y squared. In fact, I wish this was the more conventional notation because it wouldn't make people think, oh, 2 and minus 1 there, and of course, that's just sine of y squared. So it can be very ambiguous. And of course, if this was a negative, this would be a minus 2, and then this really would be 1 over sine of y squared. That's why, just a long-winded way of explaining why I wrote arcsine, instead of inverse sine right there. Needless to say, let's get back to the problem. So we've solved for t in terms of y. Now we can substitute back here. And we've got an expression for x in terms of y. So we get x is equal to 3 times the cosine of t. But we just solved for t. t is this thing right here. So it's the cosine of arcsine of y over 2." + }, + { + "Q": "\nAt 1:57, why does Sal say that you divide x^4 by x^4 but then simply divides x^4 by 4?", + "A": "just misspoke. just taking the antiderivative of x^3", + "video_name": "cBi4a1iSaPk", + "timestamps": [ + 117 + ], + "3min_transcript": "So our goal in this video is to take the antiderivative of this fairly crazy looking expression. Or another way of saying it is to find the indefinite integral of this crazy looking expression. And the key realization right over here is that this expression is made up of a bunch of terms. And the indefinite integral of the entire expression is going to be equal to the indefinite integral of each of the term. So this is going to be equal to, we could look at this term right over here, and just take the indefinite that, 7x to the third dx. And then from that, we can subtract the indefinite integral of this thing. So we could say this is, and then minus the indefinite integral of 5 times the square root of x dx. And then we can look at this one right over here. So then we could say plus the indefinite integral of 18 square root of x. And then finally, I'm running out of colors here, finally I need more colors in my thing. We can take the antiderivative of this. So plus the antiderivative of x to the negative 40th power dx. So I've just rewritten this and color-coded things. So let's take the antiderivative of each of these. And you'll see that we'll be able to do it using our whatever we want to call it. The inverse of the power rule, or the anti-power rule, whatever you might want to call it. So let's look at the first one. So we have-- what I'm going to do is, I'm just going to find the antiderivative without the constant, and just add the constant at the end. For the sake of this one. Just to make sure we get the most general antiderivative. So here the exponent is a 3. So we can increase it by 1. So it's going to be x to the 4th. Let me do that same purple color, or pink color. It's going to be x to the 4th, or we're going to divide by x to the 4th. So it's x to the 4th over 4 is the antiderivative of x to the 3rd. And you just had this scaling quantity, the seven out front. So we get 7x to the 4th over 4. Fair enough. From that, we're going to subtract the antiderivative of this. Now at first this might not be obvious, that you could use our inverse power rule, or anti-power rule here. But then you just need to realize that 5 times the principal root of x is the same thing as 5 times x to the 1/2 power. And so once again, the exponent here is 1/2. We can increment it by 1. So it's going to be x to the 3/2. And then divide by the incremented exponent. So divide by 3/2. And of course we had this 5 out front, so we still want to have the 5 out front. Now this next expression looks even wackier. But once again, we can simplify a little bit. This is the same thing-- let me do it right over here-- this is the same thing as 18 times x to the 1/2 times" + }, + { + "Q": "At 5:00,\n5/3/2 = 5* 2/3 = 10/3\nI get 5/3/2 = 5/6\n\nWhere am I going wrong?\n", + "A": "The fraction Sal is simplifying is 5 divided by 3/2 . To write it horizontally requires parentheses (5)/(3/2). To divide 5 by 3 before dividing 3 by 2 misreads the problem leading to your wrong answer. The large fraction bar acts as a grouping symbol in common mathematical usage. Thus in order of operations you can think of there being parentheses around the numerator and denominator of any fraction written using the fraction bar.", + "video_name": "cBi4a1iSaPk", + "timestamps": [ + 300 + ], + "3min_transcript": "x to the 3rd in the denominator is the same thing as x to the negative 3. We have the same base, we could just add the exponents. So this is going to be equal to 18 times x to the 2 and 1/2 power. Or another way of thinking about it, this is the same thing as 18 times x to the 5/2 power. Did I do that right? Negative 3. Oh sorry, this is negative 2 and 1/2. Let me make this very clear. And this is going to be the negative 5/2 power. x to the negative 3 is the same thing as x to the negative 6/2. Negative 6/2 plus 1/2 is negative 5/2. So once again, we just have to increment this exponent. So negative 5/2 plus 1 is going to be negative 3/2. And then you divide by what your exponent is when you increment it. So divide it by negative 3/2. And then you had the 18 out front. And we obviously are going to have to simplify this. And then finally, our exponent in this term. Let me not use that purple anymore. The exponent in this term right over here is negative 40. If we increment it, we get x to the negative 39 power, all of that over negative 39. And now we can add our constant, c. And all we need to do is simplify all of this craziness. So the first one is fairly simplified. We can write it as 7/4 x to the 4th. Now this term right over here is essentially negative 5 divided by 3/2. So 5 over 3/2 is equal to 5 times 2/3, So this thing right over here simplifies to negative 10/3 x to the 3/2. And then we have all of this craziness. Now 18 divided by negative 3/2 is equal to 18 times negative 2/3. Which is equal to, well we can simplify this a little bit, this is the same thing as 6 times negative 2. Which is equal to negative 12. So this expression right over here is negative 12x to the negative 3/2. And then finally, this one right over here, we can just rewrite it as, if we want, we could, well, we could just write negative 1/39 x to the negative 39 plus c. And we're done. We've found the indefinite integral of all this craziness." + }, + { + "Q": "\nat 1:17 when you try to graph the slope of f(x)'s assumed parabola,how do you know that slope will decrease linearly", + "A": "though Sal didn t want to share it with you here .. but, if you want to know .. a standard parabolic function is of type a x^2 + b x + c .. now help yourself and differentiate the above function .. for x^2, we got the function to be 2x .. when you differentiate the above standard parabolic function, you will necessarily get the function 2ax + b .. this is the equation of a line or a linear relation in x irrespective of the values of a, b and c ..", + "video_name": "eVme7kuGyuo", + "timestamps": [ + 77 + ], + "3min_transcript": "So I've got this crazy discontinuous function here, which we'll call f of x. And my goal is to try to draw its derivative right over here. So what I'm going to need to think about is the slope of the tangent line, or the slope at each point in this curve, and then try my best to draw that slope. So let's try to tackle it. So right over here at this point, the slope is positive. And actually, it's a good bit positive. And then as we get larger and larger x's, the slope is still positive, but it's less positive-- and all the way up to this point right over here, where it becomes 0. So let's see how I could draw that over here. So over here we know that the slope must be equal to 0-- right over here. Remember over here, I'm going to try to draw y is equal to f prime of x. And I'm going to assume that this is some type of a parabola. But let's say that, so let's see, here the slope is quite positive. So let's say the slope is right over here. And then it gets less and less and less positive. And I'll assume it does it in a linear fashion. That's why I had to assume that it's some type of a parabola. So it gets less and less and less positive. Notice here, for example, the slope is still positive. And so when you look at the derivative, the slope is still a positive value. But as we get larger and larger x's up to this point, the slope is getting less and less positive, all the way to 0. And then the slope is getting more and more negative. And at this point, it seems like the slope is just as negative as it was positive there. So at this point right over here, the slope is just as negative as it was positive right over there. So it seems like this would be a reasonable view of the slope of the tangent line over this interval. Here the slope seems constant. Our slope is a constant positive value. So once again, our slope here is a constant positive line. Let me be careful here because at this point, our slope won't really be defined, because our slope, you could draw multiple tangent lines at this little pointy point. So let me just draw a circle right over there. But then as we get right over here, the slope seems to be positive. So let's draw that. The slope seems to be positive, although it's not as positive as it was there. So the slope looks like it is-- I'm just trying to eyeball it-- so the slope is a constant positive this entire time. We have a line with a constant positive slope. So it might look something like this." + }, + { + "Q": "Why doesn't Sal shade in the yellow circle (which marks the beginning of the red line) at 4:04?\n\nEdit: Isn't that point defined (just a jump discontinuity)?\n", + "A": "It is not defined, because the definition of derivative uses limits, and since the limit from the left diverges (meaning, f(x) is the red point, and f(x+e) where e<0 has (f(x+e)-f(x))/e going to infinity), we can t define the derivative here. It just stems from the limit definitions is all.", + "video_name": "eVme7kuGyuo", + "timestamps": [ + 244 + ], + "3min_transcript": "Here the slope seems constant. Our slope is a constant positive value. So once again, our slope here is a constant positive line. Let me be careful here because at this point, our slope won't really be defined, because our slope, you could draw multiple tangent lines at this little pointy point. So let me just draw a circle right over there. But then as we get right over here, the slope seems to be positive. So let's draw that. The slope seems to be positive, although it's not as positive as it was there. So the slope looks like it is-- I'm just trying to eyeball it-- so the slope is a constant positive this entire time. We have a line with a constant positive slope. So it might look something like this. I want these things to match up. So let me do my best. So this matches up to that. This matches up over here. And we just said we have a constant positive slope. So let's say it looks something like that over this interval. And then we look at this point right over here. So right at this point, our slope is going to be undefined. There's no way that you could find the slope over-- or this point of discontinuity. But then when we go over here, even though the value of our function has gone down, we still have a constant positive slope. In fact, the slope of this line looks identical to the slope of this line. Let me do that in a different color. The slope of this line looks identical. So we're going to continue at that same slope. It was undefined at that point, but we're going to continue at that same slope. And once again, it's undefined here So the slope will look something like that. And then we go up here. The value of the function goes up, but now the function is flat. So the slope over that interval is 0. The slope over this interval, right over here, is 0. So we could say-- let me make it clear what interval I'm talking about-- the slope over this interval is 0. And then finally, in this last section-- let me do this in orange-- the slope becomes negative. But it's a constant negative. And it seems actually a little bit more negative than these were positive. So I would draw it right over there. So it's a weird looking function. But the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point. And by doing so, we have essentially drawn the derivative over that interval." + }, + { + "Q": "\nAt 0:41 \"How many times does 4 go into 9 ?\" Why do we need to know how many times does 4 go into 9 ?", + "A": "Because that would go in the ones place for long dividing 4 / 9.", + "video_name": "KFzcwWTEDDI", + "timestamps": [ + 41 + ], + "3min_transcript": "In this video, I'll introduce you to a new way of computing division, especially for larger numbers. And then we'll think a little bit about why it works. So we're going to try to compute what 96 divided by 4 is. And I'm going to write it a little bit differently. I'm going to write 96 divided by-- so I'm going to write this strange-looking symbol right over here, this thing that covers the 96. But you could view this as 96 divided by 4. And I'll show you in a second why we write it this way. This is actually a very useful way of actually computing it. So the first thing we'll do is we'll say, well, how many times does 4 go into 9? Well, we know that 4 times 2 is equal to 8 and that 4 times 3 is equal to 12. So 3 would be too much. We would go above 9. So we want to be below 9 but not have too much left over. We want the largest number that gets us into 9 without going over 9. 4 goes into 9 two times. And then we say, what's 2 times 4? Well, 2 times 4 is 8. 4 times 2 is 8. Or 2 times 4 is 8. And now, we subtract. We subtract the 8 from the 9. And we get 1. And now we bring down the next digit, which is the 6. And then we ask ourselves, well, how many times does 4 go into 16? Well, in this case, we know that 4 goes into 16 exactly four times. 4 times 4 is 16. So we say 4 goes into 16 four times. Then we multiply 4 times 4 is 16. We subtract. And 16 minus 16, we have absolutely nothing left over. And there we have our answer. I know it seems kind of magical at this point. But in a few seconds, we're going to think about why this actually worked. We got that 96 divided by 4 is equal to-- I want to do that in a different color-- 24. and think about why did this actually work. How did we magically get the right answer here? And you can verify this. Multiply 4 times 24, and you will get 96. Well, I'm assuming you gave a go at it. And the important thing always is to keep track of the place value. And it really tells you what's going on when we do this process. When we looked at this 9 right over here, this 9 is in the tens place. This is actually representing 90. It represents 9 tens. So we're saying, well, how many times does 4 go into 90 if we're thinking about multiples of 10? Well, it goes 20 times. 4 times 20 is 80. And so we said, well, 4 times 20 is 80. But we still have 16 left over. You do 96 minus 80. You have 16 left over to divide 4 into. And then 4 goes into 16 four times." + }, + { + "Q": "\nAt 4:15 why did he flip 3/7?", + "A": "A simpler way to divide fractions is to take the first fraction and multiply it with the reciprocal of the second fraction, example: 5/7 divided by 3/2 is equal to 5/7 times 2/3.", + "video_name": "K2b8iMPY11I", + "timestamps": [ + 255 + ], + "3min_transcript": "So, this is going to be the same thing as two times, two times the reciprocal of 2/3, which is, we swap the denominator and the numerator, two times three over two. So, two times three halves, if I have three halves twice, well, that's just going to be equal to, that's going to be equal to six halves, and six halves, you can view that as, well, two halves is a whole, so this is gonna be three wholes. Or you can say six divided by two. Well, that's just gonna be equal to three. You can view it either way. And then that is equal to three. Let's do a few more of these. And let's keep making them a little bit more complicated. Just let me get some good practice. And like always, pause the video. You should get excited when you see one of these things. And pause the video and see if you can do it on your own. All right, let's do something really interesting. Let's say negative 16 over nine over, over, I'll do this in a tan color, What is this? Can you simplify this complex fraction? Can you simplify this expression over here? Well, once again, we can view this as negative 16/9 divided by 3/7. So, this could be rewritten as, and I can write it either as negative 16 over 9 or I could rewrite it as negative 16/9. So, I can put the negative in front of the whole, in front of the whole fraction like that, or I could say that it's negative 16 over nine, or I can even write this as negative 16 over negative nine. Those would all be equivalent. But right now I'm writing at negative 16/9 and I am going to now divide, we can interpret this complex fraction as dividing it by 3/7. Dividing by 3/7. And so, this is going to be equal to, this is going to be equal to negative 16/9, Actually, let me just rewrite it as negative 16 over nine. Negative 16 over nine just to show that we can do it. Times the reciprocal of this. So, times 7/3. I just swap the numerator and the denominator. Times, do that same brown color, times, times 7/3, and now what am I going to get? In the numerator, I have negative 16 times seven. Let me think about it. 10 times seven is 70. Six times seven is 42. 70 plus 42 is 112. So, this, so this part right over here, this part right over here would be negative 112. Negative 112 over nine times three. Nine times three is 20, nine times three is 27. And there you have it. You can view this as negative 112 over 27 or you can put the negative up front and you can say, \"Hey, this is hinting" + }, + { + "Q": "7:20 After the summation sign brackets are needed around the two terms following, else the summation sign only applies to the first term?\n", + "A": "You are right. A parenthesis is needed to enclose the two terms in order to include both terms in the summation. Otherwise, it would be ambiguous. Sal sometimes makes mistakes like that. However, judging from this context, even without a parenthesis, it was clear that he intended to include both terms in the sum.", + "video_name": "qUNGPqCPzMg", + "timestamps": [ + 440 + ], + "3min_transcript": "this as negative 2, you could view this as negative 2 plus 0n, plus 0 times n. That's not \"on.\" That's 0-- let me write it this way-- 0 times n. So when you look at it this way, it's clear that A plus B is the coefficient on n. That must be equal to 0. A plus B must be equal to 0. And this is kind of bread-and-butter partial fraction expansion. We have other videos on that if you need to review that. And the constant part, 2A plus B, is equal to negative 2. And so now we have two equations in two unknowns. And we could solve it a bunch of different ways. But one interesting way is let's multiply the top equation by negative 1. So then this becomes negative A minus B is equal to-- well, Now we can add these two things together. And we are left with 2A minus A is A, plus B minus B-- well, those cancel out. A is equal to negative 2. And if A is equal to negative 2, A plus B is 0, B must be equal to 2. Negative 2 plus 2 is equal to 0. We solved for A. And then I substituted it back up here. So now we can rewrite all of this right over here. We can rewrite it as the sum-- and actually, let me do a little bit instead. Let me just write it as a finite sum as opposed to an infinite sum. And then we can just take the limit as we go to infinity. So let me rewrite it like this. So this is the sum from n equals 2-- instead to infinity, I'll just say to capital N. And then later, we could take the limit as this goes to infinity of-- well, instead of writing this, I can write this right over here. So it's negative 2 over n plus 1. And then B is 2, plus B over n plus 2. So once again, I've just expressed this as a finite sum. Later, we can take the limit as capital N approaches infinity to figure out what this thing is. Oh, sorry, and B-- let me not write B anymore. We now know that B is 2 over n plus 2. Now, how does this actually go about helping us? Well, let's do what we did up here. Let's actually write out what this is going to be equal to. This is going to be equal to-- when n is 2, this is negative 2/3, so it's negative 2/3, plus 2/4. So that's n equals-- let me do it down here, because I'm" + }, + { + "Q": "\nSal loses me at about 6:00. I don't understand what he's doing with the\nA+B=0\n2A+B= -2\n-A-B=0\nA= -2\nB=2\nWhat is the thought process here? I don't understand what he did.", + "A": "When you have the same variable on both sides of the equal sign, the coefficients of that variable are going to equal each other so in this case (A + B)n = 0(n) so (A + B) = 0. Same with (2A + B). They are both constant terms and you can rewrite them as (2A + B)n^0 = -2n^0. Then set coefficients of the same variable equal to each other and you get (2A + B) = -2. Now you solve the system of equations (Sal does it here by the method of elimination) and you get A = -2 and B = 2.", + "video_name": "qUNGPqCPzMg", + "timestamps": [ + 360 + ], + "3min_transcript": "And, of course, all of that is over n plus 1 times n plus 2. So how do we solve for A and B? Well, the realization is that this thing must be equal to negative 2. These two things must be equal to each other. Remember, we're making the claim that this, which is the same thing as this, is equal to this. That's the whole reason why we started doing this. So we're making the claim that these two things are equivalent. We're making this claim. So everything in the numerator must be equal to negative 2. So how do we do that? It looks like we have two unknowns here. To figure out two unknowns, we normally need two equations. Well, the realization here is, look, we have an n term on the left-hand side here. We have no n term here. this as negative 2, you could view this as negative 2 plus 0n, plus 0 times n. That's not \"on.\" That's 0-- let me write it this way-- 0 times n. So when you look at it this way, it's clear that A plus B is the coefficient on n. That must be equal to 0. A plus B must be equal to 0. And this is kind of bread-and-butter partial fraction expansion. We have other videos on that if you need to review that. And the constant part, 2A plus B, is equal to negative 2. And so now we have two equations in two unknowns. And we could solve it a bunch of different ways. But one interesting way is let's multiply the top equation by negative 1. So then this becomes negative A minus B is equal to-- well, Now we can add these two things together. And we are left with 2A minus A is A, plus B minus B-- well, those cancel out. A is equal to negative 2. And if A is equal to negative 2, A plus B is 0, B must be equal to 2. Negative 2 plus 2 is equal to 0. We solved for A. And then I substituted it back up here. So now we can rewrite all of this right over here. We can rewrite it as the sum-- and actually, let me do a little bit instead. Let me just write it as a finite sum as opposed to an infinite sum. And then we can just take the limit as we go to infinity. So let me rewrite it like this. So this is the sum from n equals 2-- instead to infinity, I'll just say to capital N. And then later, we could take the limit as this goes to infinity of-- well, instead of writing this, I can write this right over here." + }, + { + "Q": "\nat 1:00 Sal says, \"We also have to subtract 54 from this side\" When he should have said \"We also have to subtract 5 from this side\"", + "A": "i know. it was just a typo. :D", + "video_name": "VidnbCEOGdg", + "timestamps": [ + 60 + ], + "3min_transcript": "Solve for a and check your solution. And we have a plus 5 is equal to 54. Now, all this is saying is that we have some numbers, some variable a. And if I add 5 to it, I will get 54. And you might be able to do this in your head. But we're going to do it a little bit more systematically. Because that'll be helpful for you when we do more complicated problems. So in general, whenever you have an equation like this, we want to have the variable. We want this a all by itself on one side of the equation. We want to isolate it. It's already on the left-hand side, so let's try to get rid of everything else on the left-hand side. Well, the only other thing on the left-hand side is this positive 5. Well, the best way to get rid of a plus 5, or a positive 5, is to subtract 5. So let's subtract 5. But remember, this says a plus 5 is equal to 54. If we want the equality to still hold, anything we do to the left-hand side of this equation, we have to do to the right side of the equation. So we also have to subtract 54 from the right. So we have a plus 5 minus 5. Because if you add 5 and you subtract 5, they cancel out. So a plus 0 is just a. And then 54 minus 5, that is 49. And we're done. We have solved for a. A is equal to 49. And now we can check it. And we can check it by just substituting 49 back for a in our original equation. So instead of writing a plus 5 is equal to 54, let's see if 49 plus 5 is equal to 54. So we're just substituting it back in. 49, 49 plus-- let me do that in that same shade of green. 49 plus 5 is equal to 54. We're trying to check this. 49 plus 5 is 54. And that, indeed, is equal to 54. So it all checks out." + }, + { + "Q": "At 0:38, Sal says that we need to get rid of everything on the left-hand side. Why?\n", + "A": "you do this to isolate x, or the other variable, on one side to where x equals a number. :-) hope this helps!", + "video_name": "VidnbCEOGdg", + "timestamps": [ + 38 + ], + "3min_transcript": "Solve for a and check your solution. And we have a plus 5 is equal to 54. Now, all this is saying is that we have some numbers, some variable a. And if I add 5 to it, I will get 54. And you might be able to do this in your head. But we're going to do it a little bit more systematically. Because that'll be helpful for you when we do more complicated problems. So in general, whenever you have an equation like this, we want to have the variable. We want this a all by itself on one side of the equation. We want to isolate it. It's already on the left-hand side, so let's try to get rid of everything else on the left-hand side. Well, the only other thing on the left-hand side is this positive 5. Well, the best way to get rid of a plus 5, or a positive 5, is to subtract 5. So let's subtract 5. But remember, this says a plus 5 is equal to 54. If we want the equality to still hold, anything we do to the left-hand side of this equation, we have to do to the right side of the equation. So we also have to subtract 54 from the right. So we have a plus 5 minus 5. Because if you add 5 and you subtract 5, they cancel out. So a plus 0 is just a. And then 54 minus 5, that is 49. And we're done. We have solved for a. A is equal to 49. And now we can check it. And we can check it by just substituting 49 back for a in our original equation. So instead of writing a plus 5 is equal to 54, let's see if 49 plus 5 is equal to 54. So we're just substituting it back in. 49, 49 plus-- let me do that in that same shade of green. 49 plus 5 is equal to 54. We're trying to check this. 49 plus 5 is 54. And that, indeed, is equal to 54. So it all checks out." + }, + { + "Q": "At 0:54, what are \"real numbers\"?\n", + "A": "real numbers are a value that represents a quantity along a continuous line. Both plus and minus. like 1, 2, 3, 4, 5, blah blah blah", + "video_name": "GVZUpOm3XUg", + "timestamps": [ + 54 + ], + "3min_transcript": "What I want to do in this video is introduce the idea of a universal set, or the universe that we care about, and also the idea of a complement, or an absolute complement. If we're for doing it as a Venn diagram, the universe is usually depicted as some type of a rectangle right over here. And it itself is a set. And it usually is denoted with the capital U-- U for universe-- not to be confused with the union set notation. And you could say that the universe is all possible things that could be in a set, including farm animals and kitchen utensils and emotions and types of Italian food or even types of food. But then that just becomes somewhat crazy, because you're thinking of all possible things. Normally when people talk about a universal set, they're talking about a universe of things that they care about. So the set of all people or the set of all real numbers or the set of all countries, whatever the discussion is being focused on. But we'll talk about in abstract terms right now. Now, let's say you have a subset of that universal set, set A. that I have just shaded in. What we're going to talk about now is the idea of a complement, or the absolute complement of A. And the way you could think about this is this is the set of all things in the universe that aren't in A. And we've already looked at ways of expressing this. The set of all things in the universe that aren't in A, we could also write as a universal set minus A. Once again, this is a capital U. This is not the union symbol right over here. Or we could literally write this as U, and then we write that little slash-looking thing, U slash A. So how do we represent that in the Venn diagram? Well, it would be all the stuff in U that is not in A. One way as the relative complement of A that is in U. But when you're taking the relative complement of something that is in the universal set, you're really talking about the absolute complement. Or when people just talk about the complement, that's what they're saying. What's the set of all the things in my universe that are not in A. Now, let's make things a little bit more concrete by talking about sets of numbers. Once again, our sets-- we could have been talking about sets of TV personalities or sets of animals or whatever it might be. But numbers are a nice, simple thing to deal with. And let's say that our universe that we care about right over here is the set of integers. So our universe is the set of integers. So I'll just write U-- capital U-- is equal to the set of integers. And this is a little bit of an aside, but the notation for the set of integers tends to be a bold Z. And it's Z for Zahlen, from German, for apparently integer." + }, + { + "Q": "Why is the integral of cos(x) not -sin(x) at 2:47?\n", + "A": "The fundamental theorem of calculus says that f(x) is a differentiable function, then the integral of f (x) is just f(x). Now if f(x) = sin(x), then f (x) = cos(x). This means that the integral of cos(x) is just sin(x). (Note: I wish I could use integral signs but I can t find a good way to do so).", + "video_name": "xVWCfMe97ws", + "timestamps": [ + 167 + ], + "3min_transcript": "We still haven't fully separated the y's and the x's. Let's divide both sides of this by y, and then let's see. We get 1 over y plus 2y squared divided by y, that's just 2y, times dy dx is equal to cosine of x. I can just multiply both sides by dx. 1 over y plus 2y times dy is equal to cosine of x dx. And now we can integrate both sides. So what's the integral of 1 over y with respect to y? I know your gut reaction is the natural log of y, which is correct, but there's actually a slightly broader function the natural log of the absolute value of y. And this is just a slightly broader function, because it's domain includes positive and negative numbers, it just excludes 0. While natural log of y only includes numbers larger than 0. So natural log of absolute value of y is nice, and it's actually true that at all points other than 0, its derivative is 1 over y. It's just a slightly broader function. So that's the antiderivative of 1 over y, and we proved that, or at least we proved that the derivative of natural log of y is 1 over y. Plus, what's the antiderivative of 2y with respect to y? Well, it's y squared, is equal to-- I'll do the plus c on this side. Whose derivative is cosine of x? Well, it's sine of x. And then we could add the plus c. We could add that plus c there. And what was our initial condition? y of 0 is equal to 1. So ln of the absolute value of 1 plus 1 squared is equal to sine of 0 plus c. The natural log of one, e to the what power is 1? Well, 0, plus 1 is-- sine of 0 is 0 --is equal to C. So we get c is equal to 1. So the solution to this differential equation up here is, I don't even have to rewrite it, we figured out c is equal to 1, so we can just scratch this out, and we could put a 1. The natural log of the absolute value of y plus y squared is equal to sine of x plus 1. And actually, if you were to graph this, you would see that y never actually dips below or even hits the x-axis. So you can actually get rid of that absolute value function there. But anyway, that's just a little technicality. But this is the implicit form of the solution to this" + }, + { + "Q": "\nAt 0:30 it says that f of negative 6 is 7\nHow did they get 7?", + "A": "f(x) is a function and when you put -6 in for x you get 7. He got it by finding -6 on the x-axis and going strait up to see what was the y-coordinate when the function was at -6 for x. Does that make any sense? Basically, when the function s x-coordinate is -6, the y-coordinate is 7.", + "video_name": "uaPm3Tpuxbc", + "timestamps": [ + 30 + ], + "3min_transcript": "We're asked to evaluate negative 2 times f of negative 6 plus g of 1. And they've defined, at least graphically, f of x and g of x here below. So let's see how we can evaluate this. Well, to do this, we first have to figure out what f of negative 6 is. So our input into our function is negative 6. And we'll assume that's along the horizontal axis. So our input is negative 6. And based on our function definition, f of negative 6 is 7. Let me write this down. f of negative 6 is equal to 7. And what is g of 1? Well, once again, here's our input axis. And then the function says that g of 1, which is right over there, is negative 5. g of 1 is equal to negative 5. So this statement simplifies to negative 2 times f of negative 6, which is 7. So times 7 plus g of 1, which is negative 5. Negative 2 times 7 is negative 14 plus negative 5, which is negative 19. And we are done." + }, + { + "Q": "At 7:28 , why are corrseponding altitudes at similar triangles have the same ratio as other corresponding parts such as corresponding sides? perhaps i missed the part where we proves or speaks about corresponding altitudes having the same ratio of the sides? how can we prove this?\n", + "A": "This was proven in Sal s videos on similarity.", + "video_name": "v5SAMuRanGM", + "timestamps": [ + 448 + ], + "3min_transcript": "is PR, segment PR, the length of that, times the height, which is, the height is segment AQ, so the length of segment AQ, we could just write it like that, times the length of segment AQ. So how can we simplify this a little bit? Well, we could divide the 1/2 by the 1/2. Those two cancel out. But what else do we know? Well, they gave us the ratio between AC and AQ. The ratio of AC to AQ right over here is phi plus 1 to 1. Or we could just say this is equal to phi. Or we could say this is just equal to phi plus 1. So let me rewrite this. Actually, let me write it this way. This is going to be equal to-- So we have the length of segment BD over the length of segment PR, and then this part right over here we can rewrite, this is equal to phi plus 1 over 1. So I'll just write it that way. So what's the ratio of BD to PR? So the ratio of the base of the larger triangle to the base of the smaller triangle. So let's think about it a little bit. What might jump out at you is that the larger triangle and the smaller triangle, that they are similar triangles. They both obviously have angle A in common, and since PR is parallel to BD, we know that this angle corresponds to this angle. So these are going to be congruent angles. And we know that this angle corresponds to this angle right over here. So now we have three correspondingly angles are congruent. This is congruent to itself, which is in both triangles. This is congruent to this. This is congruent to that. You have three congruent angles, you're dealing with two similar triangles. And what's useful about similar triangles are the ratio between corresponding parts. of the similar triangles are going to be the same. And they gave us one of those ratios. They gave us the ratio of the altitude of the larger triangle to the altitude of the smaller triangle. AC to AQ is phi plus 1 to phi. But since this is true for one corresponding part of the similar triangles, this is true for any corresponding parts of the similar triangle, that the ratio is going to be phi plus 1 to 1. So the ratio of BD, the ratio of the base of the larger triangle to the base of the smaller one, that's also going to be phi plus 1 over 1. Let me just write it this way. This could also be rewritten as phi plus 1 over 1. So what does this simplify to? Well, we have phi plus 1 over 1 times phi plus 1 over 1. Well, we could just divide by 1." + }, + { + "Q": "\nat 1:19 what dose sal say?", + "A": "Here s what Sal is saying between 1:17 and 1:42: Well, we can t subtract the 70 from the 20, but we have other value in the number. We have value in the hundreds place. So why don t we take a hundred from the six hundred, so that becomes five hundred, and give that hundred to the ten s place. If we give that hundred to the ten s place, what is a hundred plus twenty? Well, it s going to be \u00e2\u0080\u00a6 120. (He is explaining how regrouping or borrowing works during subtraction.)", + "video_name": "QOtam19NQcQ", + "timestamps": [ + 79 + ], + "3min_transcript": "I've written the same subtraction problem twice. Here we see we're subtracting 172 from 629. And all I did here is I expanded out the numbers. I wrote 629 as 600 plus 20 plus 9, and I rewrote 172, the one is 100. So that's there. This is 7/10. It's in the tens place, so it's 70. And then the 2 is 2 ones, so it just represents 2. And we'll see why this is useful in a second. So let's just start subtracting, and we'll start with the ones place. So we have 9 minus 2. Well, that's clearly just 7. And over here we could also say, well, 9 minus 2, we have the subtraction out front. That is going to be 7. Pretty straightforward. But then something interesting happens when we get to the tens place. We're going to try to subtract 2 minus 7, or we're going to try to subtract 7 from 2. And we haven't learned yet how to do things like negative numbers, which we'll learn in the future, so we have a problem. How do you subtract a larger number from a smaller number? regrouping, sometimes called borrowing. And that's why this is valuable. When we're trying to subtract a 7 from a 2, we're really trying to subtract this 70 from this 20. Well, we can't subtract the 70 from the 20, but we have other value in the number. We have value in the hundreds place. So why don't we take 100 from the 600, so that becomes 500, and give that 100 to the tens place? If we give that 100 to the tens place, what is 100 plus 20? Well, it's going to be 120. So all I did, I didn't change the value of 629. I took 100 from the hundreds place and I gave it to the tens place. Notice 500 plus 120 plus 9 is still 629. We haven't changed the value. So how would we do that right over here? Well, if we take 100 from the hundreds place, give that hundred to the tens place, it's going to be 10 hundreds. So this will now become a 12. This will now become a 12. But notice, this 12 in the tens place represents 12 tens, or 120. So this is just another way of representing what we've done here. There's no magic here. This is often called borrowing, where you say hey, I took a 1 from the 6, and I gave it to the 2. But wait, why did this 2 become a 12? Why was I able to add 10? Well, you've added 10 tens, or 100. You took 100 from here, so 600 became 500, and then 20 became 120. But now we're ready to subtract. 12 tens minus 7 tens is 5 tens. Or you could say 120 minus 70 is 50. And then finally, you have the hundreds place." + }, + { + "Q": "\nat 1:27, do you have to take away 100? or can you take away just 1?", + "A": "Well,it depends on what strategy you are using to subtract.If you re just doing standard algorithm then, you would just write that you are borrowing 1 but if you want to say the accurate value of the number then you would say that you re borrowing 100, which is what you are actually borrowing! Hope this helps!", + "video_name": "QOtam19NQcQ", + "timestamps": [ + 87 + ], + "3min_transcript": "I've written the same subtraction problem twice. Here we see we're subtracting 172 from 629. And all I did here is I expanded out the numbers. I wrote 629 as 600 plus 20 plus 9, and I rewrote 172, the one is 100. So that's there. This is 7/10. It's in the tens place, so it's 70. And then the 2 is 2 ones, so it just represents 2. And we'll see why this is useful in a second. So let's just start subtracting, and we'll start with the ones place. So we have 9 minus 2. Well, that's clearly just 7. And over here we could also say, well, 9 minus 2, we have the subtraction out front. That is going to be 7. Pretty straightforward. But then something interesting happens when we get to the tens place. We're going to try to subtract 2 minus 7, or we're going to try to subtract 7 from 2. And we haven't learned yet how to do things like negative numbers, which we'll learn in the future, so we have a problem. How do you subtract a larger number from a smaller number? regrouping, sometimes called borrowing. And that's why this is valuable. When we're trying to subtract a 7 from a 2, we're really trying to subtract this 70 from this 20. Well, we can't subtract the 70 from the 20, but we have other value in the number. We have value in the hundreds place. So why don't we take 100 from the 600, so that becomes 500, and give that 100 to the tens place? If we give that 100 to the tens place, what is 100 plus 20? Well, it's going to be 120. So all I did, I didn't change the value of 629. I took 100 from the hundreds place and I gave it to the tens place. Notice 500 plus 120 plus 9 is still 629. We haven't changed the value. So how would we do that right over here? Well, if we take 100 from the hundreds place, give that hundred to the tens place, it's going to be 10 hundreds. So this will now become a 12. This will now become a 12. But notice, this 12 in the tens place represents 12 tens, or 120. So this is just another way of representing what we've done here. There's no magic here. This is often called borrowing, where you say hey, I took a 1 from the 6, and I gave it to the 2. But wait, why did this 2 become a 12? Why was I able to add 10? Well, you've added 10 tens, or 100. You took 100 from here, so 600 became 500, and then 20 became 120. But now we're ready to subtract. 12 tens minus 7 tens is 5 tens. Or you could say 120 minus 70 is 50. And then finally, you have the hundreds place." + }, + { + "Q": "\nat 1:33 it just dose not make sence", + "A": "Ok. If you have a calculator on your computer (what you are using right now) I want you to multiply 10 x 32. It will equal 320. If you multiply 100 x 32, you will get 3200. Instead of multiplying by 10 you are multiplying by 100. There are two zeroes in 100, so you take 32 and add zeroes, 3200. Izzy, if this does not help comment below, and I will give you a more detailed answer.", + "video_name": "tHQOAvbyRL0", + "timestamps": [ + 93 + ], + "3min_transcript": "- [Voiceover] Let's multiply four times 80. So we can look at this a few ways. One way is to say four times, we have the number 80. So we have the number 80 one time, two times, three times, four times. Four times we have the number 80. And we could do this computation, add all of these, and get our solution. But let's look at it another way. Let's try to stick with multiplication. And one way we can do that is to break up this 80. We know a pattern for multiplying by 10, so let's try to break up this 80 to get a 10. So if we have four times, and instead of 80, let's say eight times 10. Because 80 and eight times 10 are equal; those are equivalent; so we can replace our 80 with eight times 10. which is super helpful 'cause there's a nice neat pattern in math that we can use to help us with the times 10 part. So let's start to solve this. Four times eight is 32. And then we still have 32 times 10. And then we can use our pattern for multiplying by 10, which is that anytime we multiply a whole number times 10 we take that whole number, in this case 32, and we add a zero to the end. So 32 times 10 is 320. And there's a reason that pattern works; we went into it in another video, but here just real quickly, 32 times 10 is 32 tens. And we can do a few examples. If we had, say, three times 10, that would be three tens, or a 10 plus another 10 plus another 10, which equals 30: If we had something like 12 times 10, well, that would be 12 tens. And if we listed out 10 12 times and counted 'em up, there would be 120, it would add up to 120, which again is our whole number with a zero on the end, or 12 with a zero on the end. So we can use that pattern here to see that 32 times 10 is 32 with a zero on the end. Let's try another one. Let's do something like, let's say 300 this time, we'll do hundreds instead of tens, times six. 300 we can break up, like we did with 80 in the last one, and we can say that 300 is 100 three times, or 100 times three. And then we still have our times six after that." + }, + { + "Q": "At 23:58 he mentions that parametric equations are the only way to define a line in 3D. Would parametric equations make a line in any dimension from R2 to Rn?\n", + "A": "Yes. Parametric equations will always define a line, assuming you have a parametric equation for each of the dimensions.", + "video_name": "hWhs2cIj7Cw", + "timestamps": [ + 1438 + ], + "3min_transcript": "It kind of goes into our board like this, so the y-axis comes out like that. So what you can do, and actually I probably won't graph, so the determinate for the x-coordinate, just our convention, is going to be this term right here. So we can write that x-- let me write that down. So that term is going to determine our x-coordinate. So we can write that x is equal to minus 1-- be careful with the colors-- minus 1, plus minus 1 times t. That's our x-coordinate. Now, our y-coordinate is going to be determined by this part of our vector addition because these are the y-coordinates. So we can say the y-coordinate is equal to-- I'll just write it like this-- 2 plus minus 1 times t. that there, the t shows up because t times 3-- or I could just put this t into all of this. So that the z-coordinate is equal to 7 plus t times 3, or I could say plus 3t. And just like that, we have three parametric equations. And when we did it in R2, I did a parametric equation, but we learned in Algebra 1, you can just have a regular y in terms x. You don't have to have a parametric equation. But when you're dealing in R3, the only way to define a line is to have a parametric equation. If you have just an equation with x's, y's, and z's, if I just have x plus y plus z is equal to some number, this is not a line. And we'll talk more about this in R3. This is a plane. The only way to define a line or a curve in three dimensions, if I wanted to describe the path of a fly in Or if I shoot a bullet in three dimensions and it goes in a straight line, it has to be a parametric equation. So these-- I guess you could call it-- these are the equations of a line in three dimensions. So hopefully you found that interesting. And I think this will be the first video where you have an appreciation that linear algebra can solve problems or address issues that you never saw before. And there's no reason why we have to just stop at three, three coordinates, right here. We could have done this with fifty dimensions. We could have defined a line in fifty dimensions-- or the set of vectors that define a line, that two points sit on, in fifty dimensions-- which is very hard to visualize, but we can actually deal with it mathematically." + }, + { + "Q": "At 20:24 you are making a \"general definition\" in R*3 for a parametric equations for a line.\nL= {P1 + t(P1 - P2)}, t in R but could there also be L={P2 + t(P1 - P2)} . Is that the same line and does it have the same direction? If not then the \"general definition\" is not so general(?).\n", + "A": "They re both the same line with the only difference being which point on the line you re at given a specific value for t. This definition also isn t restricted just to R3. Any line in any R^n vector space can be written in this form.", + "video_name": "hWhs2cIj7Cw", + "timestamps": [ + 1224 + ], + "3min_transcript": "can say, since this is what determines our x-coordinate, we would say that x is equal to 0 plus t times minus 2, or minus 2 times t. And then we can say that y, since this is what determines our y-coordinate, y is equal to 3 plus t times 2 plus 2t. So we could have rewritten that first equation as just x is equal to minus 2t, and y is equal to 2t plus 3. So if you watch the videos on parametric equations, this is just a traditional parametric definition of this line right there. Now, you might have still viewed this as, Sal, this was a waste of time, this was convoluted. You have to define these sets and all that. But now I'm going to show you something that you probably-- that's true of anything. But you probably haven't seen in your traditional algebra class. Let's say I have two points, and now I'm going to deal in three dimensions. So let's say I have one vector. I'll just call it point 1, because these are position vectors. We'll just call it position 1. This is in three dimensions. Just make up some numbers, negative 1, 2, 7. Let's say I have Point 2. Once again, this is in three dimensions, so you have to specify three coordinates. This could be the x, the y, and the z coordinate. Point 2, I don't know. Let's make it 0, 3, and 4. Now, what if I wanted to find the equation of the line that passes through these two points in R3? So this is in R3. Well, I just said that the equation of this line-- so I'll just call that, or the set of this line, let me just call this l. guys, it could be P1, the vector P1, these are all vectors, be careful here. The vector P1 plus some random parameter, t, this t could be time, like you learn when you first learn parametric equations, times the difference of the two vectors, times P1, and it doesn't matter what order you take it. So that's a nice thing too. P1 minus P2. It could be P2 minus P1-- because this can take on any positive or negative value-- where t is a member of the real numbers. So let's apply it to these numbers. Let's apply it right here. What is P1 minus P2? P1 minus P2 is equal to-- let me get some space here. P1 minus P2 is equal, minus 1 minus 0 is minus 1. 2 minus 3 is minus 1." + }, + { + "Q": "\nAt 24:20, the line is represented in terms of x,y and z. And x,y and z are defined in terms of t. If we assume that t is time then each value of t will give a single points on R3 at different times. Are we saying that the points on a 3-dimensional line can be represented in terms of a 4th dimension? t is a scalar, what is the interpretation of t?", + "A": "t is a free variable, getting all values in R.", + "video_name": "hWhs2cIj7Cw", + "timestamps": [ + 1460 + ], + "3min_transcript": "It kind of goes into our board like this, so the y-axis comes out like that. So what you can do, and actually I probably won't graph, so the determinate for the x-coordinate, just our convention, is going to be this term right here. So we can write that x-- let me write that down. So that term is going to determine our x-coordinate. So we can write that x is equal to minus 1-- be careful with the colors-- minus 1, plus minus 1 times t. That's our x-coordinate. Now, our y-coordinate is going to be determined by this part of our vector addition because these are the y-coordinates. So we can say the y-coordinate is equal to-- I'll just write it like this-- 2 plus minus 1 times t. that there, the t shows up because t times 3-- or I could just put this t into all of this. So that the z-coordinate is equal to 7 plus t times 3, or I could say plus 3t. And just like that, we have three parametric equations. And when we did it in R2, I did a parametric equation, but we learned in Algebra 1, you can just have a regular y in terms x. You don't have to have a parametric equation. But when you're dealing in R3, the only way to define a line is to have a parametric equation. If you have just an equation with x's, y's, and z's, if I just have x plus y plus z is equal to some number, this is not a line. And we'll talk more about this in R3. This is a plane. The only way to define a line or a curve in three dimensions, if I wanted to describe the path of a fly in Or if I shoot a bullet in three dimensions and it goes in a straight line, it has to be a parametric equation. So these-- I guess you could call it-- these are the equations of a line in three dimensions. So hopefully you found that interesting. And I think this will be the first video where you have an appreciation that linear algebra can solve problems or address issues that you never saw before. And there's no reason why we have to just stop at three, three coordinates, right here. We could have done this with fifty dimensions. We could have defined a line in fifty dimensions-- or the set of vectors that define a line, that two points sit on, in fifty dimensions-- which is very hard to visualize, but we can actually deal with it mathematically." + }, + { + "Q": "why is it vector b - vector a ? at 13:43 ?\n", + "A": "It s to get the slope of the line. The vector b\u00e2\u0083\u0097 - a\u00e2\u0083\u0097 (or a\u00e2\u0083\u0097 - b\u00e2\u0083\u0097) point in the direction of the line we want to represent.", + "video_name": "hWhs2cIj7Cw", + "timestamps": [ + 823 + ], + "3min_transcript": "all and I go up. So my vector b will look like that. Now I'm going to say that these are position vectors, that we draw them in standard form. When you draw them in standard form, their endpoints represent some position. So you can almost view these as coordinate points in R2. This is R2. All of these coordinate axes I draw are going be R2. Now what if I asked you, give me a parametrization of the line that goes through these two points. So essentially, I want the equation-- if you're thinking in Algebra 1 terms-- I want the equation for the line that goes through these two points. So the classic way, you would have figured out the slope and all of that, and then you would have substituted back in. But instead, what we can do is, we can say, hey look, this line that goes through both of those points-- you could that's a better-- Both of these vectors lie on this line. Now, what vector can be represented by that line? Or even better, what vector, if I take any arbitrary scalar-- can represent any other vector on that line? Now let me do it this way. What if I were to take-- so this is vector b here-- what if I were to take b minus a? We learned in, I think it was the previous video, that b minus a, you'll get this vector right here. You'll get the difference in the two vectors. This is the vector b minus the vector a. And you just think about it. What do I have to add to a to get to b? I have to add b minus a. So if I can get the vector b minus a-- right, we know how We just subtract the vectors, and then multiply it by any scalar, then we're going to get any point along that line. So what happens if we take t, so some scalar, times our vector, times the vectors b minus a? What will we get then? So b minus a looks like that. But if we were to draw it in standard form-- remember, in standard form b minus a would look something like this. It would start at 0, it would be parallel to this, and then from 0 we would draw its endpoint. So if we just multiplied some scalar times b minus a, we would actually just get points or vectors that lie on this line. Vectors that lie on that line right there. Now, that's not what we set out to do. We wanted to figure out an equation, or parametrization, if you will, of this line, or this set. Let's call this set l. So we want to know what that set is equal to. So in order to get there, we have to start with this, which" + }, + { + "Q": "At 0:04, why does it say \"4,5000 equals 3 thousands plus how many\" in the words when it says \"4,500 = 3 thousands + ? hundreds\" on the board?\n", + "A": "He is correct it at 0:04 it says 45,000 with subtitles on.", + "video_name": "a_mzIWvHx_Y", + "timestamps": [ + 4 + ], + "3min_transcript": "We have 4,5000 equals 3 thousands plus how many hundreds, question mark hundreds? So let's write this left-hand side, but I'm going to write it out in terms of thousands and hundreds. So I'll write the thousands in orange. So this is equal to 4 thousands, which is the same thing as just 4,000, plus 500, which you could also view as 5 hundreds. So this is the left-hand side. Now let's look at the right-hand side. We have 3 thousands. So it's 3 thousands. Now let's not even look at this right now. Let's just think about what do we have to add to this right-hand side in order to get the same thing that we have on the left-hand side? Well, if you compare the 3,000 and the 4,000, you see you have an extra 1,000 over here. So let's add an extra 1,000 on the right. So we're going to add one extra 1,000. And now we just have 3,000 plus 1,000. This makes it the 4,000. But then, of course, we also need another 500. So we're going to need plus a 500 right over here. we need to say 4,000 plus 500 is equal to 3,000 plus 1,500. Now, the way they've set this up, we need to express-- so it almost looks the same. On the left-hand side, this is 4,500. So this right over here, this is the same thing as 4,500. This is this right over here. And on the right-hand side, we have 3 thousands. So that's this right over here. That's the 3 thousands. And then we just need to express this as hundreds. So 1,500, this is the same thing as 15 hundreds. So let's rewrite everything. We can rewrite this as saying 4,500, just to get the exact same form that they wrote it over there. So we could write 4,500 is equal to 3 thousands plus-- now This is 15 hundreds. Literally, if you took 15 times 100, it's going to be equal to 1,500. So this could be viewed as 15 hundreds, so plus 15 hundreds. So in this situation, the question mark is equal to 15." + }, + { + "Q": "\nAt 6:51 Sal says that it doesn't have to defined at that point, what does he mean by that?", + "A": "He meant that the denominator does not have to be 0.", + "video_name": "igJdDN-DPgA", + "timestamps": [ + 411 + ], + "3min_transcript": "here, this is g of x. So this is my g of x. And we know that as g of x approaches-- so the g of x could look something like that, right? And we know that the limit as x approaches a of g of x is equal to L. So that's right there. So this is g of x. That's g of x. Let me do h of x in a different color. So now h of x could look something like this. Like that. So that's h of x. And we also know that the limit as x approaches a of h of x -- let's see, this is the function of x axis. So you can call it h of x, g of x, or f of x. That's just the dependent access, and this is the x-axis. So once again, the limit as x approaches a of h of x, well Or at least the limit is equal to that. And none of these functions actually have to even be defined at a, as long as these limits, this limit exists and this limit exists. And that's also an important thing to keep in mind. So what does this tell us? f of x is always greater than this green function. It's always less than h of x, right? So any f of x I draw, it would have to be in between those two, right? So no matter how I draw it, if I were to draw a function, it's bounded by those two functions just by definition. So it has to go through that point. Or at least it has to approach that point. Maybe it's not defined at that point, but the limit as we approach a of f of x also has to be at point L. And maybe f of x doesn't have to be defined right there, but the limit as we approach it is going to be L. And hopefully that makes a little bit of sense, and hopefully my calories example made a little So let's keep that in the back of our mind, And now we will use that to prove that the limit as x approaches 0 of sine of x over x is equal to 1. And I want to do that, one, because this is a super useful limit. And then the other thing is, sometimes you learn the squeeze theorem, you're like, oh, well that's obvious but when is it useful? And we'll see. Actually I'm going to do it in the next video, since we're already pushing 8 minutes. But we'll see in the next video that the squeeze theorem is tremendously useful when we're trying to prove this. I will see you in the next video." + }, + { + "Q": "7:34 Sal said that x could not equal -1. But it also cannot equal 2. Did Sal forget it?\n", + "A": "Sal did not forget the condition that x cannot be equal to 2. You know from the resulting equation not to use x = 2 but you would not know that x = -1 would produce a wrong answer. So, if you were to report to someone that she could use either equation, he would know not to use +2 in the first equation but would not know that -1 will cause an error.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 454 + ], + "3min_transcript": "multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it. So you need to think of two numbers. This is just a review of grouping. You need to think of two numbers that when we multiply them are equal to 3 times negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx. Or even better, not 3x, equal to 3. So what two numbers could there be? Let's see, our times tables. Let's see, they are three apart. One's going to have to be positive and one's negative. 9 times 6 is 54. If we make the 9 positive and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6 is negative 54. So we can rewrite this up here. We can rewrite this as 3x squared, and I'm going to say" + }, + { + "Q": "at 1:39 in the video, why do you factor out a 3 from the numerator? javascript:%20void%200\n", + "A": "When working with rational expressions (fractions), we always simplify both numerator and denominator so that each of their factors is exposed. This then allows us easily to see what can be divided out in order to further simplify the expression.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 99 + ], + "3min_transcript": "When we first started learning about fractions or rational numbers, we learned about the idea of putting things in lowest terms. So if we saw something like 3, 6, we knew that 3 and 6 share a common factor. We know that the numerator, well, 3 is just 3, but that 6 could be written as 2 times 3. And since they share a common factor, the 3 in this case, we could divide the numerator by 3 and the denominator by 3, or we could say that this is just 3/3, and they would cancel out. And in lowest terms, this fraction would be 1/2. Or just to kind of hit the point home, if we had 8/24, once again, we know that this is the same thing as 8 over 3 times 8, or this is the same thing as 1 over 3 times 8 over 8. The 8's cancel out and we get this in lowest terms as 1/3. These are rational numbers. Rational expressions are essentially the same thing, but instead of the numerator being an actual number and the denominator be an actual number, they're expressions involving variables. So let me show you what I'm talking about. Let's say that I had 9x plus 3 over 12x plus 4. Now, this numerator up here, we can factor it. We can factor out a 3. This is equal to 3 times 3x plus 1. That's what our numerator is equal to. And our denominator, we can factor out a 4. This is the same thing as 4 times 3x. 12 divided by 4 is 3. 12x over 4 is 3x. So here, just like there, the numerator and the denominator have a common factor. In this case, it's 3x plus 1. In this case, it's a variable expression. It's not an actual number, but we can do the exact same thing. They cancel out. So if we were to write this rational expression in lowest terms, we could say that this is equal to 3/4. Let's do another one. Let's say that we had x squared-- let So let's say we had x squared minus 9 over 5x plus 15. So what is this going to be equal to? So the numerator we can factor. It's a difference of squares." + }, + { + "Q": "At 10:57 why did he write (3x-6) before (x+3)? He didn't do it in previous problems and I wasn't sure if it would have made a difference if he would have written (x+3) (3x-6)\n", + "A": "nope it would not had made a difference how you wrote it cuz in the muliplication property you can switch terms around and get the same answer for example: 2*3=6 3*2=6 or: 3x*4x=12x^2 4x*3x=12x^2 hope this helps:)", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 657 + ], + "3min_transcript": "Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3. we get negative 6 times x plus 3. And now this is very clear our grouping was successful. This is the same thing as-- we can kind of undistribute this as 3x minus 6 times x plus 3. If we were to multiply this times each of these terms, you get that right there. So the top term, we can rewrite it as 3x minus 6-- let me do it in the same color. So we can rewrite it as 3x minus 6 times x plus 3. That's this term right here. I don't want to make it look like a negative sign. That's that term right there. Now let's factor this bottom part over here. Scroll to the left a little bit. to think of two numbers that when I take their product, I get 2 times 3, which is equal to 6, and they need to add up to be 5. And the two obvious numbers here are 2 and 3. I can rewrite this up here as 2x squared plus 2x plus 3x plus 3, just like that. And then if I put parentheses over here, and I decided to group the 2 with the 2 because they have a common factor of 2, and I grouped the 3 with the 3 because they have a common factor of 3. This right here is 2 and a 3. So here we can factor out a 2x. If you factor out a 2x, you get 2x times x plus 1 plus-- you factor out a 3 here-- plus 3 times x plus 1." + }, + { + "Q": "at 8:59 you did a harder way then what I was told, could you not just factor the 3 out from the beggining?\n", + "A": "Certainly! That would be easier, faster, and would give you a more thoroughly factored result. (He does some odd things with factoring throughout the videos -- certainly his approach works, but it isn t always the simplest.)", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 539 + ], + "3min_transcript": "negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it. So you need to think of two numbers. This is just a review of grouping. You need to think of two numbers that when we multiply them are equal to 3 times negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx. Or even better, not 3x, equal to 3. So what two numbers could there be? Let's see, our times tables. Let's see, they are three apart. One's going to have to be positive and one's negative. 9 times 6 is 54. If we make the 9 positive and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6 is negative 54. So we can rewrite this up here. We can rewrite this as 3x squared, and I'm going to say Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3." + }, + { + "Q": "I didn't understand any one thing he explained in this video and it's getting me frustrated!\n\nAt 4:55 you see the whole x squared - 9 = (x + 3 ) (x-3)! How does this make sence? How did u get x+3 and minus 3 in relation to x squared - 9. Help me!\n", + "A": "(x + 3)\u00e2\u0080\u00a2(x - 3) FOIL: x\u00e2\u0080\u00a2x + 3\u00e2\u0080\u00a2x - 3\u00e2\u0080\u00a2x + 3\u00e2\u0080\u00a2(-3) x^2 - 9", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 295 + ], + "3min_transcript": "And in the denominator we can factor a 5 out. This is 5 times x plus 3. So once again, a common factor in the numerator and in the denonminator, we can cancel them out. But we touched on this a couple of videos ago. We have to be very careful. We can cancel them out. We can say that this is going to be equal to x minus 3 over 5, but we have to exclude the values of x that would have made this denominator equal to 0, that would have made the entire expression undefined. So we could write this as being equal to x minus 3 over 5, but x cannot be equal to negative 3. Negative 3 would make this zero or would make this whole thing zero. So this and this whole thing are equivalent. This is not equivalent to this right here, because this is defined that x is equal to negative 3, while this isn't So to make them the same, I also have to add the extra condition that x cannot equal negative 3. So likewise, over here, if this was a function, let's say we wrote y is equal to 9x plus 3 over 12x plus 4 and we wanted to graph it, when we simplify it, the temptation is oh, well, we factored out a 3x plus 1 in the numerator and They cancel out. The temptation is to say, well, this is the same graph as y is equal to the constant 3/4, which is just a horizontal line at y is equal to 3/4. But we have to add one condition. We have to eliminate-- we have to exclude the x-values that would have made this thing right here equal to zero, and that would have been zero if x is equal to negative 1/3. If x is equal to negative 1/3, this or this denominator would be equal to zero. So even over here, we'd have to say x cannot be equal to That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So" + }, + { + "Q": "At 9:00, what numbers represent a and b?\n", + "A": "a and b represent the coefficients of what you have to split your middle term into when you factor, in order to get the quadratic to factor.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 540 + ], + "3min_transcript": "negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it. So you need to think of two numbers. This is just a review of grouping. You need to think of two numbers that when we multiply them are equal to 3 times negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx. Or even better, not 3x, equal to 3. So what two numbers could there be? Let's see, our times tables. Let's see, they are three apart. One's going to have to be positive and one's negative. 9 times 6 is 54. If we make the 9 positive and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6 is negative 54. So we can rewrite this up here. We can rewrite this as 3x squared, and I'm going to say Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3." + }, + { + "Q": "in the example of minute 10:45 wouldn't the proper way to factorize be (3x-3)(x+6) ? sal made (3x-6)(x+3)\n", + "A": "No, Sal was right. The expression 3x(x-3) -6(x-3) has a common factor of x-3. Factoring that out yields (x-3)(3x-6), which is the same pair of factors Sal had..", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 645 + ], + "3min_transcript": "Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3. we get negative 6 times x plus 3. And now this is very clear our grouping was successful. This is the same thing as-- we can kind of undistribute this as 3x minus 6 times x plus 3. If we were to multiply this times each of these terms, you get that right there. So the top term, we can rewrite it as 3x minus 6-- let me do it in the same color. So we can rewrite it as 3x minus 6 times x plus 3. That's this term right here. I don't want to make it look like a negative sign. That's that term right there. Now let's factor this bottom part over here. Scroll to the left a little bit. to think of two numbers that when I take their product, I get 2 times 3, which is equal to 6, and they need to add up to be 5. And the two obvious numbers here are 2 and 3. I can rewrite this up here as 2x squared plus 2x plus 3x plus 3, just like that. And then if I put parentheses over here, and I decided to group the 2 with the 2 because they have a common factor of 2, and I grouped the 3 with the 3 because they have a common factor of 3. This right here is 2 and a 3. So here we can factor out a 2x. If you factor out a 2x, you get 2x times x plus 1 plus-- you factor out a 3 here-- plus 3 times x plus 1." + }, + { + "Q": "\nThank you for the lesson. At 15:13, only one restriction is included. Should we also say that x cannot be 1/2 since it would make (2x-1) a zero as well?", + "A": "Yes that would be appropriate.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 913 + ], + "3min_transcript": "And a plus b needs to be equal to 5. So in this situation, it looks like if we went with 6 and negative 1, that seems to be a better situation. 6 minus 1 is 5. 6 times negative 1 is negative 6. So that would have been a horrible mistake. So we can rewrite this up here as 2x squared, and I'll group the 6 with the 2x squared because they share a common factor. So plus 6x minus x, this is the same thing as 5x minus 3. I just had to find the numbers to split this 5x into. But 6x minus x is 5x. And if I put some parentheses here, I can factor out of 2x out of this first term. I get 2x times x plus 3. times x plus 3. And then our grouping was successful. We get 2-- let me do this in a different color-- we get 2x minus 1 times x plus 3. So our denominator here is equal to 2x minus 1 times x plus 3. And once again, we have a common factor in our numerator and our denonminator, the x plus 3. But we have to add the condition that x cannot be equal to negative 3, because that would make this whole thing equal to zero. Or not equal to zero, it would make us divide by zero, which is undefined. So we have to say that x cannot be equal to negative 3. over 2x minus 1, granted that we also imposed the condition that x does not equal negative 3. Hopefully, you found that interesting." + }, + { + "Q": "I stopped understanding at around 4:00 when you started talking about conditions and why x can't equal -3, why can't it equal -3 when both of the (x+3)'s have been cancelled out??\n", + "A": "By being cancelled out, that meant that the (x+3) was divided out of the expression. But if x = - 3, then (x+3) would have equalled zero ---- and division by zero is never permitted. That s why the only way the (x+3) s are allowed to be cancelled out is if the statement that x cannot equal -3 is included with the simplified fraction.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 240 + ], + "3min_transcript": "So here, just like there, the numerator and the denominator have a common factor. In this case, it's 3x plus 1. In this case, it's a variable expression. It's not an actual number, but we can do the exact same thing. They cancel out. So if we were to write this rational expression in lowest terms, we could say that this is equal to 3/4. Let's do another one. Let's say that we had x squared-- let So let's say we had x squared minus 9 over 5x plus 15. So what is this going to be equal to? So the numerator we can factor. It's a difference of squares. And in the denominator we can factor a 5 out. This is 5 times x plus 3. So once again, a common factor in the numerator and in the denonminator, we can cancel them out. But we touched on this a couple of videos ago. We have to be very careful. We can cancel them out. We can say that this is going to be equal to x minus 3 over 5, but we have to exclude the values of x that would have made this denominator equal to 0, that would have made the entire expression undefined. So we could write this as being equal to x minus 3 over 5, but x cannot be equal to negative 3. Negative 3 would make this zero or would make this whole thing zero. So this and this whole thing are equivalent. This is not equivalent to this right here, because this is defined that x is equal to negative 3, while this isn't So to make them the same, I also have to add the extra condition that x cannot equal negative 3. So likewise, over here, if this was a function, let's say we wrote y is equal to 9x plus 3 over 12x plus 4 and we wanted to graph it, when we simplify it, the temptation is oh, well, we factored out a 3x plus 1 in the numerator and They cancel out. The temptation is to say, well, this is the same graph as y is equal to the constant 3/4, which is just a horizontal line at y is equal to 3/4. But we have to add one condition. We have to eliminate-- we have to exclude the x-values that would have made this thing right here equal to zero, and that would have been zero if x is equal to negative 1/3. If x is equal to negative 1/3, this or this denominator would be equal to zero. So even over here, we'd have to say x cannot be equal to" + }, + { + "Q": "Near 8:36 couldn't you just factor out a 3 and get 3(x^2+x-6) and then factor the rest of it using the quadratic formula or other methods?\n", + "A": "Yes, you could", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 516 + ], + "3min_transcript": "negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it. So you need to think of two numbers. This is just a review of grouping. You need to think of two numbers that when we multiply them are equal to 3 times negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx. Or even better, not 3x, equal to 3. So what two numbers could there be? Let's see, our times tables. Let's see, they are three apart. One's going to have to be positive and one's negative. 9 times 6 is 54. If we make the 9 positive and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6 is negative 54. So we can rewrite this up here. We can rewrite this as 3x squared, and I'm going to say Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3." + }, + { + "Q": "\n@ 7:17 ish you said the restriction was x cannot equal -1 but there is another restriction too it is x cannot equal 2", + "A": "The only restrictions that need to be stated separately are the ones that are fully divided out of the problem. Because ( x - 2 ) is still visible in the denominator, it is already clear that x cannot equal 2 (since division by zero is never allowed).", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 437 + ], + "3min_transcript": "That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it." + }, + { + "Q": "\nAt 7:29 don't we also have to add that x does NOT equal 2. Wouldn't 2 make the equation undefined?", + "A": "Yes, but that is obvious from the problem since x-2 is in the denominator. In this case we are reminding ourselves that x cannot be equal to -1 as well. We need the reminder because after we factored the expression, we had in the denominator (x+1) and (x-2), but the (x+1) term cancelled out with a like term in the numerator. So the original expression could not have x=2 and x=-1, but we lost the visual clue when we cancelled the (x+1) terms (see 6:44).", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 449 + ], + "3min_transcript": "That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it." + }, + { + "Q": "At 8:57, couldn't Sal have made a= -6 and b=+9 instead of the other way round as both of them are divisable by 3 and 18?\n", + "A": "Yes, he could ve had. It doesn t matter whether they are a or b as those are just place holders. As long as the numbers are correct you end up with the same answer.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 537 + ], + "3min_transcript": "negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it. So you need to think of two numbers. This is just a review of grouping. You need to think of two numbers that when we multiply them are equal to 3 times negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus b, needs to be equal to 3x because we're going to split up the 3x into an ax and a bx. Or even better, not 3x, equal to 3. So what two numbers could there be? Let's see, our times tables. Let's see, they are three apart. One's going to have to be positive and one's negative. 9 times 6 is 54. If we make the 9 positive and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6 is negative 54. So we can rewrite this up here. We can rewrite this as 3x squared, and I'm going to say Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3." + }, + { + "Q": "\nWhat did he do at 1:50 when it was 4/3 pi 27", + "A": "Alegbra. 4/3 * \u00cf\u0080 * 27 or (4 * \u00cf\u0080 * 27)/3. If you multiply it out, 108\u00cf\u0080/3=36\u00cf\u0080 or, what Sal does, is simplify through division because he recognizes 27/3 in his head, so he gets to 4 * \u00cf\u0080 * 9=36\u00cf\u0080 by skipping a bit.", + "video_name": "IXRMVcoqRRQ", + "timestamps": [ + 110 + ], + "3min_transcript": "Frank wants to fill up a spherical water balloon with as much water as possible. The balloons he bought can stretch to a radius of 3 inches-- not too big. If the volume of a sphere is-- and this is volume as a function of radius-- is equal to 4/3 pi r cubed, what volume of water in cubic inches can Frank put into the balloon? So this function definition is going-- if you give it a radius in inches, it's going to produce a volume in cubic inches. So let's rewrite it. Volume as a function of radius is equal to 4/3 pi r cubed. Now, they say the balloons he bought can stretch to a radius of 3 inches. So let's think about, if the radius gets to 3 inches, what the volume of that balloon is going to be. So we essentially would just input 3 inches into our function definition. So everywhere where we see an r, we would replace it with a 3. let me rewrite it in the same color. V of-- that's not the same color. We do it in that brownish color right over here. So V of 3 is equal to 4/3 pi-- and instead of r cubed, I would write 3 cubed-- 4/3 pi 3 cubed. This is how the function definition works. Whatever we input here, it will replace the r in the expression. So V of 3 is going to be equal to-- so this is going to be equal to 4/3 pi times-- 3 to the third power is 27. 27 divided by 3 is 9, so this is 9. 9 times 4 is 36 pi. So this is equal to 36 pi. And since this was in inches, our volume is going to be in inches cubed or cubic inches. can put in the balloon-- 36 pi cubic inches." + }, + { + "Q": "\nwhile he was rearranging the operations to solve for \"h\" ,at 1:51 why did he divide \"b\" from \"b\" instead of subtracting \"b\" from \"b\" to eliminate it?", + "A": "Yes, you would subtract it from both sides instead of dividing in order to rearrange the equation. That would result in 2A = b + h becoming 2A -b = h", + "video_name": "eTSVTTg_QZ4", + "timestamps": [ + 111 + ], + "3min_transcript": "The formula for the area of a triangle is A is equal to 1/2 b times h, where A is equal to area, b is equal to length of the base, and h is equal to the length of the height. So area is equal to 1/2 times the length of the base times the length of the height. Solve this formula for the height. So just to visualize this a little bit, let me draw a triangle here. Let me draw a triangle just so we know what b and h are. b would be the length of the base. So this distance right over here is b. And then this distance right here is our height. That is the height of the triangle-- let me do that at a lower case h because that's how we wrote it in the formula. Now, they want us to solve this formula for the height. So the formula is area is equal to 1/2 base times height. And we want to solve for h. We essentially want to isolate the h on one side of the equation. It's already on the right-hand side. So let's get rid of everything else on the right-hand side. We could kind of skip steps if we wanted to. But let's see if we can get rid of this 1/2. So the best way to get rid of a 1/2 that's being multiplied by h is if we multiply both sides of the equation by its reciprocal. If we multiply both sides of the equation by 2/1 or by 2. So let's do that. So let's multiply-- remember anything you do to one side of the equation, you also have to do to the other side of the equation. Now, what did this do? Well, the whole point behind multiplying by 2 is 2 times 1/2 is 1. So on the right-hand side of the equation, we're just going to have a bh. And on the left-hand side of the equation, we have a 2A. And we're almost there, we have a b multiplying by an h. If we want to just isolate the h, we could divide both sides of this equation by b. We're just dividing both sides. You can almost view b as the coefficient on the h. We're just dividing both sides by b. And then what do we get? Well, the right-hand side, the b's cancel out. So we get h-- and I'm just swapping the sides here. h is equal to 2A over b. And we're done. We have solved this formula for the height. And I guess this could be useful. If someone just gave you a bunch of areas and a bunch of base lengths, and they said keep giving me the height for those values, or for those triangles." + }, + { + "Q": "0:22 how should you divide if your number is a fraction? Like 11 4/6 divided by 5\n", + "A": "To divide 11 4/6 by 5, make 11 4/6 into ana improper fraction, 70/6. Then When diving a fraction like 70/6 divided by 5, Remember KEEP CHANGE FLIP. To do the strategy, first turn teh divide symbol into a multiplying symbol, then make 5 into 5/1. Then flip 5/1 into 1/5, then multiply. Then Simplify, and you SHOULD have the answer.", + "video_name": "Z_NHrwK6ALE", + "timestamps": [ + 22 + ], + "3min_transcript": "Let's see if we can divide 5.005 by-- let's divide it by 7, and see what we get. So we can rewrite this as 5.005 divided by 7. And the key here is to keep track of the decimal. But other than that, you're really treating it like a traditional long division problem. So you want to put the decimal, it's to the right of the ones place. So it's right over there. I just put the decimal right above the decimal. And now we would treat it just like a traditional long division problem. So how many times does 7 go into 5? Well, it goes into 5 zero times, and we could write 0 times 7 is 0. Actually, let's do that, just to show that it works to kind of do the process all the way through. 0 times 7 is 0. You subtract, you get 5. So you get this 5 right over here. So you get 5, and then we can bring down another 0. So now we ask ourselves, how many times does 7 go into 50? Well, 7 times 7 is 49, so it's going to go seven times. 7 times 7, 49. Subtract, you get 1. And now we can bring down this 0. Let's do that. We bring down that 0. 7 goes into 10 one time. 1 times 7 is 7. Subtract, get a 3. And now finally, we can bring down this 5. We can bring down the 5. And we ask ourselves, how many times does 7 go into 35? Well, 5 times 7 is 35. This goes 5 times. 5 times 7 is 35. Subtract, and we are done. We have divided it completely. And so 5.005 divided by 7 is equal to 0.715." + }, + { + "Q": "at 4:13pm, Solve the following system of linear equations:\n-2x + 4y - z = 8\nx + 7y + 2z = 5\n3x + 3y + 3z = -3\n\nelimination does not seem to work :(\n", + "A": "The first equation, is equal to the second minus the third. So you really have 2 equations and 3 unknowns, and there is not a unique solution.", + "video_name": "GWZKz4F9hWM", + "timestamps": [ + 253 + ], + "3min_transcript": "Now to solve for x, we'll subtract 20 from both sides to get rid of the 20 on the left hand side. On the left hand side, we're just left with the -11x and then on the right hand side we are left with -22. Now we can divide both sides by -11. And we are left with x is equal to 22 divided by 11 is 2, and the negatives cancel out. x = 2. So we are not quite done yet. We've done, I guess you can say the hard part, we have solved for x but now we have to solve for y. We could take this x value to either one of these equations and solve for y. But this second one has already explicitly solved for y and instead of x, we now know that the x value where these two intersect, you could view it that way is going to be equal to 2, so 2 * 2 - 5 let's figure out the corresponding y value. So you get y=2(2)-5 and y = 4 - 5 so y = -1. And you can verify that it'll work in this top equation If y = -1 and x=2, this top equation becomes -3(2) which is -6-4(-1) which would be plus 4. And -6+4 is indeed -2. So it satisfies both of these equations and now we can type it in to verify that we got it right, So, let's type it in... x=2 and y=-1. Excellent, now we're much less likely to be embarassed by talking birds." + }, + { + "Q": "At 14:10 why can't sqrt of 39/3 be simplified to sqrt of 13?\n", + "A": "Because the 3 is not under the radical. It is \u00e2\u0085\u0093 \u00e2\u0088\u009a39, it is not \u00e2\u0088\u009a(39\u00c3\u00b73)", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 850 + ], + "3min_transcript": "" + }, + { + "Q": "\nOn the second problem (answer at 9:16) couldn't you turn the negative square root into \u00e2\u0088\u009a(84)I?", + "A": "negative square roots are insolvable, and you cannot change intergers at free will. All we can derive is that the equation is insolvable.", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 556 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 13:36, what happened to the 2 outside of the radical?", + "A": "He simplified the fraction by dividing the top and bottom by 2.", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 816 + ], + "3min_transcript": "" + }, + { + "Q": "at 14:35, why does he divide -12 by 2 as well? Is it because it's a different number than all the stuff with the square root? I thought it would just stay the same because you already divided by two from the top.\n", + "A": "Because (a+b)/c = a/c + b/c. Concrete example (2 + 4)/2 = 6/2 = 3 Right? Now distribute: (2 + 4)/2 = 2/2 + 4/2 = 1 + 2 = 3 Same! What you are trying to do is a/c + b Using our example your way: (2 + 4)/2 = 2/2 + 4 = 1 + 4 = 5.", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 875 + ], + "3min_transcript": "" + }, + { + "Q": "at 4:37, on the final step, the problem is (4 + or - 10)/2. sal divides both terms by two. i thought he could only divide 1 term, and then the 2 would be gone. how did he do that?\n", + "A": "The whole formula must be divided by 2. Think of the quadratic formula as this. [-b \u00c2\u00b1 sqrt(b^2-4ac)] / 2 or in your case it could have been written as (-4\u00c2\u00b110)/2 doing the parentheses first would result in the same answer as dividing both the numbers by two first.", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 277 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 0:16, Sal mentioned caveat. What does caveat mean?", + "A": "It is a limitation or condition, so he says if you remember this, then you should also remember how to prove it. The condition is the then part of it.", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 16 + ], + "3min_transcript": "" + }, + { + "Q": "At 13:37, how did -12\u00c2\u00b12\u00e2\u0088\u009a39/-6 become -6\u00c2\u00b1\u00e2\u0088\u009a39/-3 ? I don't understand that division/simplification process.\n", + "A": "Think of it as (-12\u00c2\u00b12\u00e2\u0088\u009a39)/-6. If I want to factor out a 2 on the top, I would end up with 2(-6 \u00c2\u00b1 \u00e2\u0088\u009a39)/6, if I divide 2/6, I am left with 1/3, so we would have (-6 \u00c2\u00b1 \u00e2\u0088\u009a39)/3. Be careful how you write it because you do not show the -12 as being part of the numerator, and that is why I added the extra parentheses.", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 817 + ], + "3min_transcript": "" + }, + { + "Q": "At 12:38, why did he chose to stop factoring? Why didn't he continue?\n", + "A": "Because it cant be factored anymore", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 758 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 10:51 with the yellow matrix Sal just finished, could you add the pivot entries with free variables and finish the reduced row echelon form?\n\nSo something like:\n| 1 1 0 0 | 2 |\n| 0 1 0 0 | s |\n| 0 0 1 2 | 5 |\n| 0 0 0 1 | t |\nInto:\n| 1 0 0 0 | 2 - s |\n| 0 1 0 0 | s |\n| 0 0 1 0 | 5 - 2t |\n| 0 0 0 1 | t |\n\nOr wouldn't schools and universities be looking for something so detailed?", + "A": "Yes, you can reduce it all the way. Don t quote me but... the first form is called Gaussian elimination the second form is called Gauss-Jordan elimination The nice thing about fully reducing the matrix is now you have you re entire solution on the right side.", + "video_name": "JVDrlTdzxiI", + "timestamps": [ + 651 + ], + "3min_transcript": "of parallel equations, they won't intersect. And you're going to get, when you put it in reduced row echelon form, or you just do basic elimination, or you solve the systems, you're going to get a statement that zero is equal to something, and that means that there are no solutions. So the general take-away, if you have zero equals something, no solutions. If you have the same number of pivot variables, the same number of pivot entries as you do columns, so if you get the situations-- let me write this down, this is good to know. if you have zero is equal to anything, then that means no solution. If you're dealing with r3, then you probably have parallel planes, in r2 you're dealing with parallel lines. If you have the situation where you have the same number of pivot entries as columns, so it's just 1, 1, 1, 1, this I think you get the idea. That equals a, b, c, d. Then you have a unique solution. Now if, you have any free variables-- so free variables look like this, so let's say we have 1, 0, 1, 0, and then I have the entry 1, 1, let me be careful. 0, let me do it like this. 1, 0, 0, and then I have the entry 1, 2, and then I have a bunch of zeroes over here. And then this has to equal zero-- remember, if this was a bunch of zeroes equaling some variable, then I would have no solution, or equalling some constant, let's say this is equal to 5, this is equal to 2. If this is our reduced row echelon form that we eventually get to, then we have a few free variables. This is a free, or I guess we could call this column a free Because it has no pivot entries. These are the pivot entries. So this is variable x2 and that's variable x4. Then these would be free, we can set them equal to anything. So then here we have unlimited solutions, or no unique solutions. And that was actually the first example we saw. And these are really the three cases that you're going to see every time, and it's good to get familiar with them so you're never going to get stumped up when you have something like 0 equals minus 4, or 0 equals 3. Or if you have just a bunch of zeros and a bunch of rows. I want to make that very clear. Sometimes, you see a bunch of zeroes here, on the left-hand side of the augmented divide, and you might say, oh maybe I have no unique solutions, I have an infinite number of solutions. But you have to look at this entry right here. Only if this whole thing is zero and you have free variables, then you have an infinite number of solutions. If you have a statement like, 0 is equal to a, if this is equal to 7 right here, then all of the sudden you would" + }, + { + "Q": "\nAt 1:58 why is the square root of 74,74", + "A": "That square root of 74 is squared, so you have sqrt(74) and you square it, by definition you get 74.", + "video_name": "T0IOrRETWhI", + "timestamps": [ + 118 + ], + "3min_transcript": "A carpet measures 7 feet long and has a diagonal measurement of square root of 74 feet. Find the width of the carpet. So let's draw ourselves a carpet here. So let's draw a carpet. It has a length of 7 feet, so let's say that that is 7 feet, right there. And it's going to be a rectangle of some kind. So let's say that we're looking down on the carpet like that. That's our carpet. And then it has a diagonal measurement of square root of 74 feet. So that means that this distance, right here-- draw it a little bit neater than that --this distance right here, the diagonal of the carpet, is the square root of 74 feet. And what they want to know is the width of the carpet. Find the width of the carpet. So let's say that this is the width of the carpet. That is w, right there. Now, you might already realize that what I have drawn here is a right triangle. Let me make sure you realize it. And since that is a triangle that has a 90 degree angle, it's a right triangle. The side opposite the right angle, or the 90 degrees, is a hypotenuse, or the longest side. It is the square root of 74. And the shorter sides are w and 7. And the Pythagorean Theorem tells us that the sum of the squares of the shorter side will be equal to the square of the hypotenuse, so the square of the longer side. So we get w squared, this side squared. plus 7 squared, this other side squared, is going to be equal to the hypotenuse squared, square root of 74 squared. And then we get w squared plus 49 is equal to the square root of 74 squared. Well, that's just going to be 74. It is equal to 74. We can subtract 49 from both sides of this equation. Subtract 49 from both sides. The left side-- these guys are going to cancel out, we're just going to be left with a w squared --is equal to-- What's 74 minus 49? 74 minus 49, well, we can do a little bit of regrouping or borrowing here, if we don't want to do it in our head. We can make this a 14. This becomes a 6. 14 minus 9 is 5. 6 minus 4 is 2. And we have w squared is equal to 25. So w is going to be equal to the square root of 25, the positive square root. So let's take the square root of both sides, the positive square root, and we will get w is equal to 5. Because we obviously we don't want it to be negative 5. That wouldn't be a realistic distance. So the width of the carpet is 5. And we're done." + }, + { + "Q": "\nAt 00:27, wouldn't we cross off 7 and 49 and simplify them like Sal did at 00:56? I'm a bit confused.", + "A": "Yes, you can. Doing so makes the problem easier as well.", + "video_name": "pi3WWQ0q6Lc", + "timestamps": [ + 27, + 56 + ], + "3min_transcript": "Let's do a few examples multiplying fractions. So let's multiply negative 7 times 3/49. So you might say, I don't see a fraction here. This looks like an integer. But you just to remind yourself that the negative 7 can be rewritten as negative 7/1 times 3/49. Now we can multiply the numerators. So the numerator is going to be negative 7 times 3. And the denominator is going to be 1 times 49. 1 times 49. And this is going to be equal to-- 7 times 3 is 21. And one of their signs is negative, so a negative times a positive is going to be a negative. So this is going to be negative 21. You could view this as negative 7 plus negative 7 plus negative 7. And that's going to be over 49. And this is the correct value, but we can simplify it more because 21 and 49 both share 7 as a factor. So let's divide both the numerator and the denominator by 7. Divide the numerator and the denominator by 7. And so this gets us negative 3 in the numerator. And in the denominator, we have 7. So we could view it as negative 3 over 7. Or, you could even do it as negative 3/7. Let's do another one. Let's take 5/9 times-- I'll switch colors more in this one. That one's a little monotonous going all red there. 5/9 times 3/15. So this is going to be equal to-- we multiply the numerators. So it's going to be 5 times 3. 5 times 3 in the numerator. And the denominator is going to be 9 times 15. 9 times 15. you see that there is already common factors in the numerator and the denominator. Both the numerator and the denominator, they're both divisible by 5 and they're both divisible by 3, which essentially tells us that they're divisible by 15. So we can divide the numerator and denominator by 15. So divide the numerator by 15, which is just like dividing by 5 and then dividing by 3. So we'll just divide by 15. Divide by 15. And this is going to be equal to-- well, 5 times 3 is 15. Divided by 15 you get 1 in the numerator. And in the denominator, 9 times 15 divided by 15. Well, that's just going to be 9. So it's equal to 1/9. Let's do another one. What would negative 5/9 times negative 3/15 be? Well, we've already figured out what positive 5/9 times positive 3/15 would be. So now we just have to care about the sign." + }, + { + "Q": "\nin the captions of 5:23 - 5:26 it says \"maybeyou\"", + "A": "It might have just been a mistake. It s been sorted out now (it says maybe you)", + "video_name": "DPuK6ZgBGmE", + "timestamps": [ + 323, + 326 + ], + "3min_transcript": "Once again, this is two, this is three. She deviates. Her absolute deviation is three. And then we wanna take the mean of the absolute deviation. That's the M in MAD, in Mean Absolute Deviation. This is Manueala's absolute deviation, Sophia's absolute deviation, Jada's absolute deviation, Tara's absolute deviation. We want the mean of those, so we divide by the number of datapoints, and we get zero plus one, plus two, plus three, is six over four. Six over four, which is the same thing as 1 1/2. Or, lemme just write it in all the different ways. We could write it as three halves, or 1 1/2, or 1.5. Which gives us a measure of how much do these datapoints vary from the mean of four. I know what some of you are thinking. \"Wait, I thought there was a formula \"associated with the mean absolute deviation. \"It seems really complex. \"It has all of these absolute-value signs That's all we did. When we write all those absolute-value signs, that's just a fancy way of looking at each datapoint, and thinking about how much does it deviate from the mean, whether it's above or below. That's what the absolute value does. It doesn't matter, if it's three below, we just say three. If it's two above, we just say two. We don't put a positive or negative on. Just so you're comfortable seeing how this is the exact same thing you would've done with the formula, let's do it that way, as well. So the mean absolute deviation is going to be equal to. Well, we'll start with Manueala. How many bubbles did she blow? She blew four. From that you subtract the mean of four, take the absolute value. That's her absolute deviation. Of course, this does evaluate to this zero, to zero here. Then you take the absolute value. Sophia blew five bubbles, and the mean is four. Then you do that for Jada. Jada blew six bubbles; the mean is four. And then you do it for Tara. Then you divide it by the number of datapoints you have. Lemme make it very clear. This right over here, this four, is the mean. This four is the mean. You're taking each of the datapoints, and you're seeing how far it is away from the mean. You're taking the absolute value 'cause you just wanna figure out the absolute distance. Now you see, or maybe you see. Four minus four, this is. Four minus four, that is a zero. That is that zero right over there. Five minus four, absolute value of that? That's going to be. Lemme do this in a new color. This is just going to be one. This thing is the same thing as that over there. We were able to see that just by inspecting this graph, or this chart. And then, six minus four, absolute value of that, that's just going to be two. That two is that two right over here, which is the same thing as this two right over there. And then, finally, our one minus four, this negative three," + }, + { + "Q": "at 2:03, what does Sal mean by deviate?\n", + "A": "Sorry I am late but deviate basically means how far the number is from the mean. Where sal uses it, Sophia (I think that was her name) blew 5 bubbles and the mean is 4, so when you deviate, you find the distance away from the mean.", + "video_name": "DPuK6ZgBGmE", + "timestamps": [ + 123 + ], + "3min_transcript": "- This bar graph here tells us bubbles blown by each gum-chewer. We have four different gum-chewers, and they tell us how many bubbles each of them blew. What I wanna do is, I wanna figure out first the mean of the number of bubbles blown, and then also figure out how dispersed is the data, how much do these vary from the mean. I'm gonna do that by calculating the mean absolute deviation. Pause this video now. Try to calculate the mean of the number of bubbles blown. And then, after you do that, see if you can calculate the mean absolute deviation. Step one, let's figure out the mean. The mean is just going to be the sum of the number of bubbles blown divided by the number of datapoints. Manueala blew four bubbles. She blew four bubbles. Sophia blew five bubbles. Jada blew six bubbles. Tara blew one bubble. So let's divided by four. And so, this is going to be equal to four plus five is nine, plus six is 15, plus one is 16. So it's equal to 16 over four, which is 16 divided by four is equal to four. The mean number of bubbles blown is four. Lemme actually do this with a bold line right over here. This is the mean number of bubbles blown. Now what I wanna do is I wanna figure out the mean absolute deviation. Mean. MAD: Mean Absolute Deviation. What we wanna do is we wanna take the mean of how much do each of these datapoints deviate from the mean. I know I just used the word mean twice in a sentence, so it might be a little confusing, but as we work through it, hopefully, it'll make a little bit of sense. How much does Manueala's, the number of bubbles she blew, Well, Manueala actually blew four bubbles, and four is the mean. So her deviation, her absolute deviation from the mean is zero. Is zero. Actually, lemme just write this over here. Absolute deviation, that's AD, absolute deviation from the mean. Manueala didn't deviate at all from the mean. Now let's think about Sophia. Sophia deviates by one from the mean. We see that right there, she's one above. Now, we would say one whether it's one above or below, 'cause we're saying absolute deviation. Sophia deviates by one. Her absolute deviation is one. And then, we have Jada. How much does she deviate from the mean? We see it right over here. She deviates by two. She is two more than the mean. And then, how much does Tara deviate from the mean? She is at one, so that is three below the mean." + }, + { + "Q": "\nFor this problem I looked at just the equation and solved like in the \"Factoring 5th degree polynomials to find real zeroes\" (pervious) video. I took y=x^3+3x^2+x+3 and split it into two, just like in 4:20, x^3+3x^2 and x+3. When I got to the (x^2+1) (x+3) I factored out the (x^2+1) again and got (x-1) (x+1). Now I have three solutions, x= -3, -1, 1.\nWhy is this not applicable?", + "A": "Because x^2 + 1 is not equal to (x - 1)(x + 1). If you multiply that out, you will get x^2 - 1.", + "video_name": "uFZvWYPfOmw", + "timestamps": [ + 260 + ], + "3min_transcript": "see that the real function does not equal 0 at 4 or 7. Another giveaway that this is not going to be the function is that you are going to have a total of three roots. Let me write this down. So you're going to have a total of three roots. Now, , those three roots could be real or complex roots. And the big key is complex roots come in pairs. So you might have a situation with three real roots. And this is an example with three real roots, although we know this actually isn't the function right over here. Or if you have one complex root, you're going to have another complex root. So if you have any complex roots, the next possibility is one real and two complex roots. And this right over here has two real roots. That would somehow imply that you have only one complex root, which that is not a possibility. Now another way that you could have thought about this-- and this would have been the longer way. But let's say you didn't have the graphs here for you, and someone asked you to just find the roots-- well, you could have attempted to factor this. And this one actually is factorable. y is equal to x to the third plus 3x squared plus x plus 3. As mentioned in previous videos, factoring things of a degree higher than 2, there is something of an art to it. But oftentimes, if someone expects you to, you might be able to group things in interesting ways, especially when you see that several terms have some common factors. So for example, these first two terms right over here have the common factor x squared. So if you were to factor that out, you would get x squared times x plus 3, which is neat because that looks a lot like the second two terms. And then you can factor the x plus 3 out. We could factor the x plus 3 out, and we would get x plus 3 times x squared plus 1. And now, your 0's are going to happen, or this whole y-- remember this is equal to y-- y is going to equal 0 if either one of these factors is equal to 0. So when does x plus 3 equal 0? Well, subtract 3 from both sides. That happens when x is equal to negative 3. When does x squared plus 1 equal 0, I should say? Well, when x squared is equal to negative 1. Well, there's no real x's, no real valued x's." + }, + { + "Q": "\nAt around 2:20 Sal says that the polynomial function has three roots. Can someone explain how we know that ? :^)", + "A": "Because the polynomial is a 3rd degree polynomial (x^3). 2nd degree polynomials (x^2) have 2 solutions and 1st degree polynomials (x^1) have only 1 solution. Hope this helps. Good Luck.", + "video_name": "uFZvWYPfOmw", + "timestamps": [ + 140 + ], + "3min_transcript": "Use the real 0's of the polynomial function y equal to x to the third plus 3x squared plus x plus 3 to determine which of the following could be its graph. So there's several ways of trying to approach it. One, we could just look at what the 0's of these graphs are or what they appear to be and then see if this function is actually 0 when x is equal to that. So for example, in graph A-- and first of all, as always, I encourage you to pause this video and try it before I show you how to solve it. So I'm assuming you've given a go at it. So let's look at this first graph here. Its 0, it clearly has a 0 right at this point. And just by trying to inspect this graph, it looks like this is at x is equal to negative 3, if I were to estimate. So that looks like the point negative 3, 0. So let's see, if we substitute x equals negative 3 here, whether we get y equaling 0. So let's see, negative 3 to the third power plus 3 What does this give us? This gives us negative 27. This gives us positive 27. This gives, of course, negative 3. This is plus 3. These two cancel out. These two cancel out. This does indeed equal 0. So this was actually pretty straightforward. Graph A does indeed work. You could try graph B right here, and you would have to verify that we have a 0 at, this looks like negative 2. Another one, this looks like at 1, another one that looks at 3. And since we already know that A is the answer, none of these-- if you were to input x equals negative 2, x equals 1, or x equals 3 into this function definition right over here, you should not get 0. And you'll see that this doesn't work. Same thing for this one. If you tried 4 or 7 for your x's, you see that the real function does not equal 0 at 4 or 7. Another giveaway that this is not going to be the function is that you are going to have a total of three roots. Let me write this down. So you're going to have a total of three roots. Now, , those three roots could be real or complex roots. And the big key is complex roots come in pairs. So you might have a situation with three real roots. And this is an example with three real roots, although we know this actually isn't the function right over here. Or if you have one complex root, you're going to have another complex root. So if you have any complex roots, the next possibility is one real and two complex roots. And this right over here has two real roots." + }, + { + "Q": "\nAt 5:23, you say that x^3 divided by x^2 =x , why would that be?", + "A": "Note that: \u00f0\u009d\u0091\u00a5\u00c2\u00b3 = \u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5 \u00f0\u009d\u0091\u00a5\u00c2\u00b2 = \u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5 Therefore: \u00f0\u009d\u0091\u00a5\u00c2\u00b3/\u00f0\u009d\u0091\u00a5\u00c2\u00b2 = (\u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5)/(\u00f0\u009d\u0091\u00a5 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u00a5) = \u00f0\u009d\u0091\u00a5 We could also use exponent properties: \u00f0\u009d\u0091\u00a5\u00c2\u00b3/\u00f0\u009d\u0091\u00a5\u00c2\u00b2 = \u00f0\u009d\u0091\u00a5\u00c2\u00b3\u00e2\u0081\u00bb\u00c2\u00b2 = \u00f0\u009d\u0091\u00a5\u00c2\u00b9 = \u00f0\u009d\u0091\u00a5 Comment if you have questions.", + "video_name": "MZl6Mna0leQ", + "timestamps": [ + 323 + ], + "3min_transcript": "But what do these simplify to? So this first term over here, this simplifies to 2x squared times-- now you get 4 divided by 2 is 2, x to the fourth divided by x squared is x squared. And then y divided by 1 is just going to be a y. So it's 2x squared times 2x squared y, and then you have minus 2x squared times, 8 divided by 2 is 4. x to the third divided by x squared is x. And y divided by 1, you can imagine, is just y. And then finally, of course, you have minus 2x squared time-- this right here simplifies to 1-- times 1. Now, if you were to undistribute 2x squared out of the expression, you'd essentially get 2x squared Right, if you distribute this out, if you take that out of each of the terms, you're going to get 2x squared times this 2x squared y, minus 4xy, and then you have minus 1, minus 1, and we're done. We've factored the problem. Now, it looks like we did a lot of steps. And the reason why I kind of of went through great pains to show you exactly what we're doing is so you know exactly what we're doing. In the future, you might be able to do this a little bit quicker. You might be able to do many of the steps in your head. You might say OK, let me look at each of these. Well, the biggest coefficient that divides all of these is a 2, so let me put that 2, let me factor 2 out. Well, all of these are divisible by x squared. That's the largest degree of x. Let me factor an x squared out. And this guy doesn't have a y, so I can't factor a y out. this guy divided by 2x squared? Well 4 divided by 2 is 2. x to the fourth divided by x squared is x squared. y divided by 1, there's no other y degree that we factored out, so it's just going to be a y. And then you have minus 8 divided by 2 is 4. x to the third divided by x squared is x. And then you have y divided by say, 1, is just y. And then you have minus 2 divided by 2 is 1. x squared divided by x squared 1, so 2x squared divided by 2x squared is just 1. So in the future, you'll do it more like this, where you kind of just factor it out in your head, but I really want you to understand what we did here. There is no magic. And to realize that there's no magic, you could just use the distributive property to multiply this out again, to multiply it out again, and you're going to see that you get exactly this." + }, + { + "Q": "At about \"8:00\" you prove that the linear combination of the two vectors can represent any vector in R^2.And then you set to prove that they are linearly independent.\n\nIsn't it self proving that if two vectors span R^2 (and n vectors span R^n) they HAVE to be linearly independent?\nJust my intuition speaking.\n", + "A": "If n vectors spans R^n, then n vectors are linearly independant and a basis for R^n. So you are right in your intuition.", + "video_name": "zntNi3-ybfQ", + "timestamps": [ + 480 + ], + "3min_transcript": "" + }, + { + "Q": "at the end of the video 18:47 he says that in order for a set to be a basis it has to have the minimum or the most efficient set o vectors that can span R2.\nWhat does he mean for most efficient?\n", + "A": "It means the set with the smallest possible number of vectors. So the set containing <1, 0>, <0, 1>, and <1, 1> is not a basis, since <1, 0> and <1, 1> already span R\u00c2\u00b2. The <1, 0> vector is unnecessary.", + "video_name": "zntNi3-ybfQ", + "timestamps": [ + 1127 + ], + "3min_transcript": "" + }, + { + "Q": "In the example that Sal provides around 22:58, is it obvious from the start that the rank of the 3x2 matrix can be at most 2 and therefore less than 3, and therefore not \"onto\" R3?\n", + "A": "That would be a correct observation, yes! :) In his case, the function is not surjective (therefore not invertible), but injective- because rank(S) = n.", + "video_name": "eR8vEdJTvd0", + "timestamps": [ + 1378 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 3:15 on the video Sal starts by doing the subtraction, per order of operations shouldn't the division of /c be done first? If not, why not?", + "A": "Given: 1 1 \u00e2\u0094\u0080 - \u00e2\u0094\u0080 a b 1 \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00c3\u00b7 \u00e2\u0094\u0080 c d The 1/a - 1/b is treated has a parenthetical term, which gives it order precedence: \u00e2\u0094\u008c \u00e2\u0094\u0090 \u00e2\u0094\u00821 1\u00e2\u0094\u0082 \u00e2\u0094\u0082\u00e2\u0094\u0080 - \u00e2\u0094\u0080\u00e2\u0094\u0082 \u00e2\u0094\u0082a b\u00e2\u0094\u0082 \u00e2\u0094\u0094 \u00e2\u0094\u0098 1 \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00c3\u00b7 \u00e2\u0094\u0080 c d \u00e2\u0094\u008c \u00e2\u0094\u0090 \u00e2\u0094\u0082 b a \u00e2\u0094\u0082 \u00e2\u0094\u0082\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 - \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0082 \u00e2\u0094\u0082a\u00e2\u0080\u00a2b a\u00e2\u0080\u00a2b\u00e2\u0094\u0082 \u00e2\u0094\u0094 \u00e2\u0094\u0098 1 \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00c3\u00b7 \u00e2\u0094\u0080 c d b - a \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 a\u00e2\u0080\u00a2b 1 \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00c3\u00b7 \u00e2\u0094\u0080 c d b - a 1 \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00c3\u00b7 \u00e2\u0094\u0080 a\u00e2\u0080\u00a2b\u00e2\u0080\u00a2c d b - a d \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 \u00e2\u0080\u00a2 \u00e2\u0094\u0080 a\u00e2\u0080\u00a2b\u00e2\u0080\u00a2c 1 d\u00e2\u0080\u00a2(b - a) \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 a\u00e2\u0080\u00a2b\u00e2\u0080\u00a2c d\u00e2\u0080\u00a2b - d\u00e2\u0080\u00a2a \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 a\u00e2\u0080\u00a2b\u00e2\u0080\u00a2c", + "video_name": "_BFaxpf35sY", + "timestamps": [ + 195 + ], + "3min_transcript": "And once again, encourage you, encourage you to pause the video and figure it out on your own. Well, when you divide by a fraction, it is equivalent to multiplying by it's, by it's reciprocal. So this is going to be the same thing as a over b, a over b times, times the reciprocal of this. So times d over, I'm going to use the same color just so I don't confuse you, that d was purple, times d over c, times d over c and then it reduces to a problem like this. You know, and I shouldn't even use this multiplication symbol now that we're in algebra because you might confuse that with an x, so let me write that as times, times d, d over c, times d over c, Well the numerator you're going to have a times d, so it's ad, a, d, over, over bc, over b times c. Now let's do one that's maybe a little bit more involved and see if you can pull it off. So let's say that I had, let's say that I had, I don't know, let me write it as 1 over a, minus 1 over b, all of that over, all of that over c, and let's say, let's also divide that by 1 over d. So this is a more involved expression then what we've seen so far but I think we have all the tools to tackle it so I encourage you to pause the video and see if you can simplify this, if you can actually carry out these operations and come up with a one fraction that represents this. so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab," + }, + { + "Q": "At 5:40, why can't you simplify the expression (db-da)/(abc) by cancelling the a variable or the b variable to make it become (d-d)/(c)?\n", + "A": "You cannot do that because the db and the da are being subtracted. If there was a multiplication sign (ex. (db*da)/(abc) ) you would be fine to do that.", + "video_name": "_BFaxpf35sY", + "timestamps": [ + 340 + ], + "3min_transcript": "so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab, that's the same thing as multiplying by the reciprocal of c. So if I'm dividing it by c, that's the same thing as multiplying, that's the same thing as multiplying times 1 over c. And if I am, and I'll just keep going here, if I'm dividing by 1 over d, if I'm dividing, notice this is the same thing as division right over here. If I'm dividing by c, that's the same thing as multiplying by the reciprocal of c. And then finally, I'm dividing by 1 over d, that's the same thing as multiplying by the reciprocal of 1 over d. So the reciprocal of 1 over d is d over, d over 1. And so what does this result with? Well in the numerator I have b-a times 1 times d. So we can write this as d times (b-a), times (b-a) and then in the denominator I have abc, ab and c. we can distribute this d, and we're going to be left with, we deserve a minor drum roll at this point, we can write this as d times b, d times b minus d, woops, I want to do that in the same green color so you really see how it got distributed, minus d times a, all of that over ab, abc. And we are done." + }, + { + "Q": "\nat 4:10, why must we multiply by the reciprocal of c??", + "A": "Multiplication by the reciprocal of c is the same as dividing by c.", + "video_name": "_BFaxpf35sY", + "timestamps": [ + 250 + ], + "3min_transcript": "Well the numerator you're going to have a times d, so it's ad, a, d, over, over bc, over b times c. Now let's do one that's maybe a little bit more involved and see if you can pull it off. So let's say that I had, let's say that I had, I don't know, let me write it as 1 over a, minus 1 over b, all of that over, all of that over c, and let's say, let's also divide that by 1 over d. So this is a more involved expression then what we've seen so far but I think we have all the tools to tackle it so I encourage you to pause the video and see if you can simplify this, if you can actually carry out these operations and come up with a one fraction that represents this. so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab, that's the same thing as multiplying by the reciprocal of c. So if I'm dividing it by c, that's the same thing as multiplying, that's the same thing as multiplying times 1 over c. And if I am, and I'll just keep going here, if I'm dividing by 1 over d, if I'm dividing, notice this is the same thing as division right over here. If I'm dividing by c, that's the same thing as multiplying by the reciprocal of c. And then finally, I'm dividing by 1 over d, that's the same thing as multiplying by the reciprocal of 1 over d. So the reciprocal of 1 over d is d over, d over 1. And so what does this result with? Well in the numerator I have b-a times 1 times d. So we can write this as d times (b-a), times (b-a) and then in the denominator I have abc, ab and c." + }, + { + "Q": "At 3:25 Sal quickly glosses over how he transforms 1/a - 1/b into b/ba - a/ba. I think an explanation about how these two statements are equivalent would be useful!\n", + "A": "For the 1/a term, multiply it by b/b. Now you have: (1/a)(b/b) = b/ab For the 1/b term, multiply it by a/a. Now you have (1/b)(a/a) = a/ab So you have converted 1/a - 1/b into b/ab - a/ab.", + "video_name": "_BFaxpf35sY", + "timestamps": [ + 205 + ], + "3min_transcript": "And once again, encourage you, encourage you to pause the video and figure it out on your own. Well, when you divide by a fraction, it is equivalent to multiplying by it's, by it's reciprocal. So this is going to be the same thing as a over b, a over b times, times the reciprocal of this. So times d over, I'm going to use the same color just so I don't confuse you, that d was purple, times d over c, times d over c and then it reduces to a problem like this. You know, and I shouldn't even use this multiplication symbol now that we're in algebra because you might confuse that with an x, so let me write that as times, times d, d over c, times d over c, Well the numerator you're going to have a times d, so it's ad, a, d, over, over bc, over b times c. Now let's do one that's maybe a little bit more involved and see if you can pull it off. So let's say that I had, let's say that I had, I don't know, let me write it as 1 over a, minus 1 over b, all of that over, all of that over c, and let's say, let's also divide that by 1 over d. So this is a more involved expression then what we've seen so far but I think we have all the tools to tackle it so I encourage you to pause the video and see if you can simplify this, if you can actually carry out these operations and come up with a one fraction that represents this. so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab," + }, + { + "Q": "\nin the video at 5:00 d is multiplied by (b-a) and c multiplied by ab so whenc is multiplied why is the answer abc instead of acbc", + "A": "The distributive property multiplies across addition or subtraction. d (b-a) = bd - ad But, with c (ab), there is no addition or subtraction. Thus, the distributive property does not apply. You just get abc Note: think about what happens with numbers. 2 (3*5) = 2*3*5 = 6*5 = 30 You just have 3 numbers to multiply, not 4. Hope this helps.", + "video_name": "_BFaxpf35sY", + "timestamps": [ + 300 + ], + "3min_transcript": "so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab, that's the same thing as multiplying by the reciprocal of c. So if I'm dividing it by c, that's the same thing as multiplying, that's the same thing as multiplying times 1 over c. And if I am, and I'll just keep going here, if I'm dividing by 1 over d, if I'm dividing, notice this is the same thing as division right over here. If I'm dividing by c, that's the same thing as multiplying by the reciprocal of c. And then finally, I'm dividing by 1 over d, that's the same thing as multiplying by the reciprocal of 1 over d. So the reciprocal of 1 over d is d over, d over 1. And so what does this result with? Well in the numerator I have b-a times 1 times d. So we can write this as d times (b-a), times (b-a) and then in the denominator I have abc, ab and c. we can distribute this d, and we're going to be left with, we deserve a minor drum roll at this point, we can write this as d times b, d times b minus d, woops, I want to do that in the same green color so you really see how it got distributed, minus d times a, all of that over ab, abc. And we are done." + }, + { + "Q": "At about 8:00, he turns A^2=h^2-h^2/4\nto\nA^2=h^2(1-1/4)\n\nI don't get how he did that.\n", + "A": "That confused me at first too, but then I realised that since h^2/4 is 1/4 of h^2, h^2 - h^2/4 is equal to three quarters of h^2, which is the same as h^2 times 1 - 1/4.", + "video_name": "Qwet4cIpnCM", + "timestamps": [ + 480 + ], + "3min_transcript": "Because that's h over 2, and this is also h over 2. Right over here. So if we go back to our original triangle, and we said that this is 30 degrees and that this is the hypotenuse, because it's opposite the right angle, we know that the side opposite the 30 degree side is 1/2 of the hypotenuse. And just a reminder, how did we do that? Well we doubled the triangle. Turned it into an equilateral triangle. Figured out this whole side has to be the same as the hypotenuse. And this is 1/2 of that whole side. So it's 1/2 of the hypotenuse. So let's remember that. The side opposite the 30 degree side is 1/2 of the hypotenuse. Let me redraw that on another page, because I think this is getting messy. So going back to what I had originally. This is a right angle. This is the hypotenuse-- this side right here. If this is 30 degrees, we just derived that the side opposite that this is equal to 1/2 the hypotenuse. If this is equal to 1/2 the hypotenuse then what is this side equal to? Well, here we can use the Pythagorean theorem again. We know that this side squared plus this side squared-- let's call this side A-- is equal to h squared. So we have 1/2 h squared plus A squared is equal to h squared. This is equal to h squared over 4 plus A squared, is equal to h squared. Well, we subtract h squared from both sides. We get A squared is equal to h squared minus h squared over 4. This is equal to 3/4 h squared. And once going that's equal to A squared. I'm running out of space, so I'm going to go all the way over here. So take the square root of both sides, and we get A is equal to-- the square root of 3/4 is the same thing as the square root of 3 over 2. And then the square root of h squared is just h. And this A-- remember, this is an area. This is what decides the length of the side. I probably shouldn't have used A. But this is equal to the square root of 3 over 2, times h. We've derived what all the sides relative to the hypotenuse are of a 30-60-90 triangle. So if this is a 60 degree side." + }, + { + "Q": "I'm pretty sure this is true, but just want to be sure. When he make the denominators the same in the last problem around 3:30, does the five still remain? For example does 5 4/9 = 5 28/63? This makes sense to me as to why it would be right, I just want to know if actually am.\n", + "A": "Yes. The 5 stays put because you re only changing the 4/9 by multiplying it. The fraction is manipulated, but the whole number is not.", + "video_name": "R8YKuGJ0plI", + "timestamps": [ + 210 + ], + "3min_transcript": "3 and 9/10 is clearly a larger number. We have a 3 out here instead of a 1, so we will write less than. And the way I always remember it is, the opening always faces the larger number. And the point is small. It always points to the smaller number. Now let's do this next one. 4 and 7/8 versus 49/9. So let's convert this to a mixed number. 9 goes into 49 5 times, and 5 times 9 is 45. So the remainder is going to be 4. The remainder is 4, so this is 5 and 4/9. Once again, we can literally just look at the whole number parts. 5 is clearly larger than 4, so once again, less than. Point facing the smaller number, opening facing the larger number. Now 2 and 1/2 versus 11/10. 10 goes into 11 only 1 time. And if you care about the remainder, it's 1. Which is clearly smaller than 2 and 1/2. You just look at the whole number parts. 2 is clearly larger than 1. So we want our opening of our less than or greater than sign to face the larger number. So we would write it like this. And this is greater than, so 2 and 1/2 is greater than 11/10. The little point facing the smaller number. 5 and 4/9 versus 40/7. 7 goes into 40, so let me rewrite this, 7 goes into 40 5 times. And then you're going to have a remainder of 5, because 7 times 5 is 35. You have a remainder of 5 to get to 40. So it's 5 and 5/7. And if that looks like I'm doing some type of voodoo, just remember, I'm really just breaking it up. I'm just really saying that 40/7 is the same thing as 35 plus 5/7. The largest multiple of 7 that is less than this number. And this is the same thing as 35/7 plus 5/7. And 5/7 is just 5/7 there. This one is interesting because we have the same whole number out front on our mixed numbers. 5 versus 5. So now we actually do have to pay attention to the fractional part of our mixed number. We essentially have to compare 4/9 to 5/7. And there's a couple of ways to do this. You could get them to have the same denominator. That's probably the easiest way to do it. So you could rewrite-- so what's the least common multiple of 9 and 7? They share no factors, so really the least common multiple is going to be their product. So if we want to rewrite 4/9 we would write 63 in the denominator, that's 9 times 7. If we multiply the denominator by 7 we also have to multiply the numerator by 7. So that will be 28. Now 5/7, we're going to make the denominator 63. We're multiplying the denominator times 9. Then we have to multiply the numerator times 9 as well. 5 times 9 is 45." + }, + { + "Q": "\nWhen you factor it out at 1:46 what becomes of that lonely +x in the middle of the equation?", + "A": "When (x+3)(x-2) is multiplied you distribute the first and last numbers separately, meaning that you have (x^2 - 2x +3x -6) as your answer. After combining the like terms of -2x and +3x we get +1x or simply put, + x. (x^2 + x - 6)", + "video_name": "EAa3J_nDkoI", + "timestamps": [ + 106 + ], + "3min_transcript": "Let's say that f of x is equal to x squared plus x minus 6 over x minus 2. And we're curious about what the limit of f of x, as x approaches 2, is equal to. Now the first attempt that you might want to do right when you see something like this, is just see what happens what is f of 2. Now this won't always be the limit, even if it's defined, but it's a good place to start, just to see if it's something reasonable could pop out. So looking at it this way, if we just evaluate f of 2, on our numerator, we get 2 squared plus 2 minus 6. So it's going to be 4 plus 2, which is 6, minus 6, so you're going to get 0 in the numerator and you're going to get 0 in the denominator. So we don't have, the function is not defined, so not defined at x is equal 2. f not defined. So there's no simple thing there. Even if this did evaluate, if it was a continuous function, but that doesn't necessarily mean the case. But we see very clearly the function is not defined here. So let's see if we can simplify this and also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we can rewrite the top expression. And this just goes back to your algebra one, two numbers whose product is negative 6, whose sum is positive 3, well that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that over x minus 2. So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all X's except for x is equal to 2. So that's another way of looking at it. Another way we could rewrite our f of x, we could rewrite f of x, this is the exact same function, f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it. So that is, that is not anywhere near being a straight line, that is much better. So let's call this the y-axis call it y equals f of x. And then let's, over here, let me make a horizontal line, that is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3 and then the slope is 1. And it's defined for all X's except for x is equal to 2. So this is x is equal to 1, x is equal to 2." + }, + { + "Q": "at 2:55 - when it says f(x)= x +3 why did you graph on the y axis... u went up 3 and then u added the limit of 2 in there and said it would be 5 do we do that for all problems?\nso when it says f(x) its only really saying y= x+3 so we go up 2? is that the B in Y=MX +B ??\n", + "A": "In the example f(x)=x+3 the vertical axis is the f(x) axis. Generally f(x ) is the same thing as y. So f(x)=x+3 is the same thing as y=x+3.", + "video_name": "EAa3J_nDkoI", + "timestamps": [ + 175 + ], + "3min_transcript": "but that doesn't necessarily mean the case. But we see very clearly the function is not defined here. So let's see if we can simplify this and also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we can rewrite the top expression. And this just goes back to your algebra one, two numbers whose product is negative 6, whose sum is positive 3, well that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that over x minus 2. So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all X's except for x is equal to 2. So that's another way of looking at it. Another way we could rewrite our f of x, we could rewrite f of x, this is the exact same function, f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it. So that is, that is not anywhere near being a straight line, that is much better. So let's call this the y-axis call it y equals f of x. And then let's, over here, let me make a horizontal line, that is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3 and then the slope is 1. And it's defined for all X's except for x is equal to 2. So this is x is equal to 1, x is equal to 2. So let me make sure I can, so it's undefined right over there. So this is what f of x looks like. Now given this, let's try to answer our question. What is the limit of f of x as x approaches 2. Well, we can look at this graphically. As x approaches 2 from lower values in 2, so this right over here is x is equal to 2, if we get to maybe, let's say this is 1.7, we see that our f of x is right over there. If we get to 1.9, our f of x is right over there. So it seems to be approaching this value right over there. Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5 our f of x is right over there." + }, + { + "Q": "at 1:08 sal mentioned radians. what are they ??\n", + "A": "They are another unit of angle measure like degrees. In higher level math, they are much easier to use. You can learn about them in the Trigonometry section of KA.", + "video_name": "1BH2TNzAAik", + "timestamps": [ + 68 + ], + "3min_transcript": "A circle has a circumference of 20 pi. It has an arc length of 221/18 pi. What is the central angle of the arc in degrees? So they're asking for this one. So this is the arc that they're talking about, that's 221/18 pi long. And they want to know this angle that it subtends, this central angle right over here. So we just have to remind ourselves that the ratio of this arc length to the entire circumference-- let me write that down-- the ratio of this arc length, which is 221/18 pi, to the entire circumference, which is 20 pi, is going to be equal to the ratio of this central angle, which we can call theta, the ratio of theta to 360 degrees if we were to go all the way around the circle. This will give us our theta in degrees. in terms of 2 pi radians around the circle, but it was 360 degrees since we're in degrees. Now we just have to simplify. Now the easiest thing is just to multiply both sides times 360 degrees. So let's do that. So if we multiply the left-hand side by 360 degrees, we get 360 times 221 times pi over-- let's see, we have 18 times 20 times pi. And on the right-hand side, if we multiply it by 360, we are just left with theta. So we really just have to simplify this now. Pi divided by pi is going to be 1. 360 divided by 20, well, it's going to be the same thing as 36/2, which is the same thing as 18. So this all simplified to 221 degrees. Theta is 221 degrees." + }, + { + "Q": "\nAt 1:25, how can you put fraction on top of another fraction?", + "A": "It looks messy, but you can do it. When you put a fraction on top of another fraction it helps to think of it as the numerator fraction divided by the denominator fraction.", + "video_name": "1BH2TNzAAik", + "timestamps": [ + 85 + ], + "3min_transcript": "A circle has a circumference of 20 pi. It has an arc length of 221/18 pi. What is the central angle of the arc in degrees? So they're asking for this one. So this is the arc that they're talking about, that's 221/18 pi long. And they want to know this angle that it subtends, this central angle right over here. So we just have to remind ourselves that the ratio of this arc length to the entire circumference-- let me write that down-- the ratio of this arc length, which is 221/18 pi, to the entire circumference, which is 20 pi, is going to be equal to the ratio of this central angle, which we can call theta, the ratio of theta to 360 degrees if we were to go all the way around the circle. This will give us our theta in degrees. in terms of 2 pi radians around the circle, but it was 360 degrees since we're in degrees. Now we just have to simplify. Now the easiest thing is just to multiply both sides times 360 degrees. So let's do that. So if we multiply the left-hand side by 360 degrees, we get 360 times 221 times pi over-- let's see, we have 18 times 20 times pi. And on the right-hand side, if we multiply it by 360, we are just left with theta. So we really just have to simplify this now. Pi divided by pi is going to be 1. 360 divided by 20, well, it's going to be the same thing as 36/2, which is the same thing as 18. So this all simplified to 221 degrees. Theta is 221 degrees." + }, + { + "Q": "\nAt 10:14, couldn't have Sal simplified 10y and 4y by dividing it?\nSo if my reasoning is correct (you can correct me if I'm wrong, here to learn :)),\nthe answer should be 5y\u00e2\u0088\u009ay+25y divided by 2y - 25.", + "A": "No, Sal can not reduce the fraction. To reduce fractions we can only cancel common factors (items being multiplied). You are trying to cancel out parts of terms (items being added/subtracted). Since they aren t factors, you can reduce. Hope this helps.", + "video_name": "gY5TvlHg4Vk", + "timestamps": [ + 614 + ], + "3min_transcript": "So what is this going to be equal to? Well, let's just multiply the numerator and the denominator by 2 square roots of y plus 5 over 2 square roots of y plus 5. This is just 1. We are not changing the number, we're just multiplying it by 1. So let's start with the denominator. What is the denominator going to be equal to? The denominator is going to be equal to this squared. Once again, just a difference of squares. It's going to be 2 times the square root of y squared minus 5 squared. If you factor this, you would get 2 square roots of y plus 5 times 2 square roots of y minus 5. This is a difference of squares. And then our numerator is 5y times 2 square roots of y. So it would be 10. And this is y to the first power, this is y to the half power. We could write y square roots of y. Or we could write this as y to the 3/2 power or 1 and 1/2 power, however you want to view it. And then finally, 5y times 5 is plus 25y. And we can simplify this further. So what is our denominator going to be equal to? We're going to have 2 squared, which is 4. Square root of y squared is y. 4y. And then minus 25. And our numerator over here is-- We could even write this. We could keep it exactly the way we've written it here. We could factor out a y. There's all sorts of things we could do it. But just to keep things simple, we could just leave that as 10. Let me just write it different. I could write that as this is y to the first, this is y to the 1/2 power. I could write that as even a y to the 3/2 if I want. I could write that as y to the 1 and 1/2 if I want. All of those are equivalent. Plus 25y. Anyway, hopefully you found this rationalizing the denominator interesting." + }, + { + "Q": "At 10:11 why didn't he add 10y and 25y to make 35y square root y as the numerator?\n", + "A": "You can t add unlike terms. The expression in the numerator is: 10y\u00e2\u0088\u009a(y) + 25y. These are unlike because y\u00e2\u0088\u009a(y) is not the same as just y . Its like having 10yz + 25y. Hope this helps.", + "video_name": "gY5TvlHg4Vk", + "timestamps": [ + 611 + ], + "3min_transcript": "So what is this going to be equal to? Well, let's just multiply the numerator and the denominator by 2 square roots of y plus 5 over 2 square roots of y plus 5. This is just 1. We are not changing the number, we're just multiplying it by 1. So let's start with the denominator. What is the denominator going to be equal to? The denominator is going to be equal to this squared. Once again, just a difference of squares. It's going to be 2 times the square root of y squared minus 5 squared. If you factor this, you would get 2 square roots of y plus 5 times 2 square roots of y minus 5. This is a difference of squares. And then our numerator is 5y times 2 square roots of y. So it would be 10. And this is y to the first power, this is y to the half power. We could write y square roots of y. Or we could write this as y to the 3/2 power or 1 and 1/2 power, however you want to view it. And then finally, 5y times 5 is plus 25y. And we can simplify this further. So what is our denominator going to be equal to? We're going to have 2 squared, which is 4. Square root of y squared is y. 4y. And then minus 25. And our numerator over here is-- We could even write this. We could keep it exactly the way we've written it here. We could factor out a y. There's all sorts of things we could do it. But just to keep things simple, we could just leave that as 10. Let me just write it different. I could write that as this is y to the first, this is y to the 1/2 power. I could write that as even a y to the 3/2 if I want. I could write that as y to the 1 and 1/2 if I want. All of those are equivalent. Plus 25y. Anyway, hopefully you found this rationalizing the denominator interesting." + }, + { + "Q": "9:23, Why would I be multiplying as x * x, sqy * sqy , -5*5\nfor the denominator. but in the numerator its\n\n2 * 5, sqrY*y 5*5\nhow come the numerator is being applied to all terms and the denominator is being multiplied by like terms?\n", + "A": "In the numerator since we are multiplying ( 2*sqrt(y)-5 ) by its complement we are using a shortcut. If you multiply a factor by its complement like (x+a)*(x-a) you get x^2-a^2 Proof: (x+a)*(x-a)=x*x+x*(-a)+a*x-a*a=x^2-a*x+a*x-a^2=x^2-a^2 so we can use shortcut on the numerator: ( 2*sqrt(y)-5 )*(2*sqrt(y)+5)=(2*sqrt(y))^2 - (5)^2= 4y-25 or we can do it the long way like: ( 2*sqrt(y)-5 )*(2*sqrt(y)+5)=2*sqrt(y)*2*sqrt(y)+2*sqrt(y)*5-5*2*sqrt(y)-5*5 =4y-25 Both give you the same answer. Hope this makes sense.", + "video_name": "gY5TvlHg4Vk", + "timestamps": [ + 563 + ], + "3min_transcript": "It actually made it look a little bit better. And you know, I don't if I mentioned in the beginning, this is good because it's not obvious. If you and I are both trying to build a rocket and you get this as your answer and I get this as my answer, this isn't obvious, at least to me just by looking at it, that they're the same number. But if we agree to always rationalize our denominators, we're like, oh great. We got the same number. Now we're ready to send our rocket to Mars. Let's do one more of this, one more of these right here. Let's do one with variables in it. So let's say we have 5y over 2 times the square root of y minus 5. So we're going to do this exact same process. We have a binomial with an irrational denominator. It might be a rational. We don't know what y is. But y can take on any value, so at points it's going to be So what is this going to be equal to? Well, let's just multiply the numerator and the denominator by 2 square roots of y plus 5 over 2 square roots of y plus 5. This is just 1. We are not changing the number, we're just multiplying it by 1. So let's start with the denominator. What is the denominator going to be equal to? The denominator is going to be equal to this squared. Once again, just a difference of squares. It's going to be 2 times the square root of y squared minus 5 squared. If you factor this, you would get 2 square roots of y plus 5 times 2 square roots of y minus 5. This is a difference of squares. And then our numerator is 5y times 2 square roots of y. So it would be 10. And this is y to the first power, this is y to the half power. We could write y square roots of y. Or we could write this as y to the 3/2 power or 1 and 1/2 power, however you want to view it. And then finally, 5y times 5 is plus 25y. And we can simplify this further. So what is our denominator going to be equal to? We're going to have 2 squared, which is 4. Square root of y squared is y. 4y. And then minus 25. And our numerator over here is-- We could even write this. We could keep it exactly the way we've written it here. We could factor out a y. There's all sorts of things we could do it. But just to keep things simple, we could just leave that as 10. Let me just write it different. I could write that as this is y to the first, this is y to the 1/2 power. I could write that as even a y to the 3/2 if I want. I could write that as y to the 1 and 1/2 if I want." + }, + { + "Q": "\nAt @2:54 isn't the minus symbol after 8 supposed to be in the green color because of how it's a -12?", + "A": "mabey and do you want to be friends", + "video_name": "GmD7Czmol0k", + "timestamps": [ + 174 + ], + "3min_transcript": "Then I have plus five. And then I'm gonna subtract. I am subtracting eight times 0.25. 0.25, this is 1/4. I could rewrite this if I want. 0.25, that's the same thing as 1/4. Eight times 1/4, or another way to think about it is eight divided by four is gonna be equal to two. So this whole thing over here is going to be equal to two. So it's gonna be minus, we have this minus out here, so minus two. And what is this going to be? Well, let's think about it. 3.5 plus five is 8.5, minus two is going to be 6.5. So this is equal to this, is equal to 6.5. Let's do another one of these. Alright. And we'll, just like before, try to work through it on your own before we do it together. Alright, now let's do it together. plus eight minus 12 N, when M is equal to 30 and N is equal to 1/4. Alright. So everywhere I see an M I want to replace with a 30. And everywhere I see an N I want to replace with a 1/4. So this is going to be equal to 0.1 times M. M is 30. Times 30 plus eight minus 12 times N, where N is 1/4. N is 1/4. So what is, what is 1/10, This right over here, 0.1, that's the same thing as 1/10 of 30? Well 1/10 of 30, that's going to be three. So this part is three. And we have three plus eight. And then we're gonna have minus. Well what is 12 times 1/4? That's gonna be 12/4, or 12 divided by four, And now when we evaluate this, so that is equal to this, we have three plus eight minus three. Well, threes are going to, you know positive three, then you're gonna subtract three, and you're just going to be left with, you're just going to be left with an eight. And you're done. This expression when M is equal to 30 and N is equal to 1/4 is equal to eight." + }, + { + "Q": "At 3:16, can you always just cross out all the numbers that are congruent in the same equation, or is it better, more accurate, to solve the equation all the way through?\n", + "A": "He did not cross out numbers that are congruent, he crossed out additive inverses, so 3 - 3 = 0. If they were congruent, you would get 3 + 3 to get 6", + "video_name": "GmD7Czmol0k", + "timestamps": [ + 196 + ], + "3min_transcript": "Then I have plus five. And then I'm gonna subtract. I am subtracting eight times 0.25. 0.25, this is 1/4. I could rewrite this if I want. 0.25, that's the same thing as 1/4. Eight times 1/4, or another way to think about it is eight divided by four is gonna be equal to two. So this whole thing over here is going to be equal to two. So it's gonna be minus, we have this minus out here, so minus two. And what is this going to be? Well, let's think about it. 3.5 plus five is 8.5, minus two is going to be 6.5. So this is equal to this, is equal to 6.5. Let's do another one of these. Alright. And we'll, just like before, try to work through it on your own before we do it together. Alright, now let's do it together. plus eight minus 12 N, when M is equal to 30 and N is equal to 1/4. Alright. So everywhere I see an M I want to replace with a 30. And everywhere I see an N I want to replace with a 1/4. So this is going to be equal to 0.1 times M. M is 30. Times 30 plus eight minus 12 times N, where N is 1/4. N is 1/4. So what is, what is 1/10, This right over here, 0.1, that's the same thing as 1/10 of 30? Well 1/10 of 30, that's going to be three. So this part is three. And we have three plus eight. And then we're gonna have minus. Well what is 12 times 1/4? That's gonna be 12/4, or 12 divided by four, And now when we evaluate this, so that is equal to this, we have three plus eight minus three. Well, threes are going to, you know positive three, then you're gonna subtract three, and you're just going to be left with, you're just going to be left with an eight. And you're done. This expression when M is equal to 30 and N is equal to 1/4 is equal to eight." + }, + { + "Q": "\nAt 2:30 when i solved for the equation 0.7^0 multiplied 0.3^6 my answer was 0.000729 not 0.001.", + "A": "Your calculation is correct but he is rounding to the nearest 1/100. Thus, 0.000729 ~ 0.001.", + "video_name": "eL965_Lscb8", + "timestamps": [ + 150 + ], + "3min_transcript": "- [Voiceover] Now that we've spent a couple of videos exploring a scenario where I'm taking multiple free throws and figuring out the probability of making K of the scores and six attempts or in N attempts. Let's actually define a random variable using this scenario and see if we can construct it's probability distribution and we'll actually see that it's a binomial distribution. So, let's define the random variable X. So, let's say that X is equal to the number, the number of made shots, number of made free throws when taking, when taking six free throws. So, it's how many of the six do you make? And we're going to assume what we assumed in the first video in this series of this, these making free throws. So, we're gonna assume the seventy percent free throw probability right over here. So, assuming assumptions, assuming seventy percent free throw, free throw percentage. So, let's figure out the probabilities of the different values that X could actually take on. So, let's see, what is the probability, what is the probability that X is equal to zero? That even though you have a seventy percent free throw percentage that you make none of the shots and actually you could calculate this through probably some common sense without using all of these fancy things but just to make things consistent, I'm gonna write it out. So, this is going to be, this is going to be, it's going to be equal to six choose zero times zero point seven to the zeroth power times zero point three to the sixth power, and this right over here is gonna end up being one. This over here's going to end up being one. And so, you're just gonna be left with zero point three to the sixth power and I calculated it ahead of time. if we do, if we round our percentages to the nearest tenth, this is going to give you approximately, approximately, well, if we round the decimal to the nearest, to the nearest thousandth you're gonna get something like that which is approximately equal to zero point one percent chance of you missing all of them. So, roughly I'm speaking, roughly here a one in a thousand, one in a thousand chance of that happening, of missing all six free throws. Now, let's keep going, this is fun. So, what is the probability that our random variable is equal to 1? Well, this is going to be six choose one times zero point seven to the first power times zero point three to the six minus first power. So, it's a fifth power. And I calculated this and this is approximately zero point zero one" + }, + { + "Q": "\nI think you made a mistake at the end (12:18):\ny = (2/5)*x + (2/5) should be y = (2/5)*x + (4/5)", + "A": "you are correct 4/5 is b*", + "video_name": "QkepM8Vv3kw", + "timestamps": [ + 738 + ], + "3min_transcript": "We can say 6 times m star-- This is just straight Algebra 1. So 6 times our m star, so 6 times 2 over 5, plus 2 times our b star is equal to 4. Enough white, let me use yellow. So we get 12 over 5 plus 2b star is equal to 4, or we could say 2b star-- let me scroll down a little bit --2b star is equal to 4. Which is the same thing as 20 over 5, minus 12 over 5, which is equal to-- I'm just subtracting the 12 over 5 from both sides --which is equal to 8 over 5. And you divide both sides of the equation by 2, you get b star is equal to 4/5. And just like that, we got our m star and our b star. Our least squares solution is equal to 2/5 and 4/5. And remember, the whole point of this was to find an equation of the line. y is equal to mx plus b. Now we can't find a line that went through all of those points up there, but this is going to be our least squares solution. This is the one that minimizes the distance between a times our vector and b. No vector, when you multiply times that matrix a-- that's not a, that's transpose a --no other solution is going to give us a closer solution to b than when we put our newly-found x star into this equation. This is going to give us our best solution. It's going to minimize the distance to b. So let's write it out. y is equal to mx plus b. So y is equal to 2/5 x plus 2/5. y is equal to 2/5 x plus 2/5. So its y-intercept is 2/5, which is about there . This is at 1. 2/5 is right about there. And then its slope is 2/5. Let's think of it this way: for every 2 and 1/2 you go to the right, you're going to go up 1. So if you go 1, 2 and 1/2, we're going to go up 1. We're going to go up 1 like that. So our line-- and obviously this isn't precise --but our line is going to look something like this. I want to do my best shot at drawing it because this is the fun part. It's going to look something like that." + }, + { + "Q": "\nAt around 2:50, do we always have to multiply by negative one? in every system?", + "A": "Not 100% on the context, but it looks like he s eliminating a variable. The answer is no. It depends on the coefficient of the variable you re tying to eliminate. If the first equation was -3x + ... and the bottom was x + ...., you would just multiply by 3. In general terms: You want A + B = 0, where A and B are coefficients of the first term. To do that, you choose some c such that A + cB = 0. The key thing to remember is to multiply both sides of the equation by c. ex) c*(Ax + By) = 10*c", + "video_name": "z1hz8-Kri1E", + "timestamps": [ + 170 + ], + "3min_transcript": "going to be equal to 62.5 pounds. That's exactly what this first statement is telling us. The second statement, the weight of 3 televisions and 2 DVD players, so if I have 3 televisions and 2 DVD players, so the weight of 3 televisions plus the weight of 2 DVD players, they're telling us that that is 52 pounds. And so now we've set up the system of equations. We've done the first part, to create a system that represents the situation. Now we need to solve it. Now, one thing that's especially tempting when you have two systems, and both of them have something where, you know, you have a 3t here and you have a 3t here, what we can do is we can multiply one of the systems by some factor, so that if we were to add this equation to that equation, we would get one of the terms to cancel out. And that's what we're going to do right here. equations to each other, because remember, when we learned this at the beginning of algebra, anything you do to one side of an equation, if I add 5 to one side of an equation, I have to add 5 to another side of the equation. So if I add this business to this side of the equation, if I add this blue stuff to the left side of the equation, I can add this 52 to the right-hand side, because this is saying that 52 is the same thing as this thing over here. This thing is also 52. So if we're adding this to the left-hand side, we're actually adding 52 to it. We're just writing it a different way. Now, before we do that, what I want to do is multiply the second, blue equation by negative 1. And I want to multiply it by negative 1. So negative 3t plus-- I could write negative 2d is equal to negative 52. So I haven't changed the information in this equation. I just multiplied everything by negative 1. The reason why I did that is because now if I add these two equations, these 3t terms are going to cancel out. So let's do that. Let's add these two equations. And remember, all we're doing is we're adding the same thing We're adding essentially negative 52 now, now that we've multiplied everything by a negative 1. This negative 3t plus negative 2d is the same thing as negative 52. So let's add this left-hand side over here. The 3t and the negative 3t will cancel out. That was the whole point. 5d plus negative 2d is 3d. So you have a 3d is equal to 62.5 plus negative 52, or 62.5 minus 52 is 10.5. And now we can divide both sides of this equation by 3. And you get d is equal to 10.5 divided by 3. So let's figure out what that is. 3 goes into 10.5-- it goes into 10 three times. 3 times 3 is 9. Subtract. Get 1. Bring down the 5. Of course, you have your decimal point right here. 3 goes into 15 five times." + }, + { + "Q": "\nAt 0:25 does that mean you can change it around for every operation?", + "A": "No. It only works for multiplication and addition because they are commutative. Communitive means that no matter where your numbers are, you get the same answer. Helpful?", + "video_name": "SfxULALs_u8", + "timestamps": [ + 25 + ], + "3min_transcript": "Let's try to compute 6 times 37. And I'll show you one way of doing this. And then in future videos, we can look at other ways of doing this and think about why this is actually working. So what I like to do-- and this is often called the standard method-- is take the larger of the two numbers. It doesn't matter if you're doing 6 times 37 or 37 times 6. They equal each other. 6 times 37 is the same thing as 37 times 6. So what I like to do is I take the larger of the two numbers, and I write it on top. So I'll write 37. And then the smaller of the two numbers, which is 6, I'll write it on the bottom. And I'll align it by the correct place. This only has one digit. It's in the ones place, obviously. So I can write the 6 right over here. And I'll write the multiplication symbol like that. And this is just another way of expressing 37 times 6, which is the same thing as 6 times 37. Now, what we do is we go with this, the first place in this lower number. And there's only one place here. It's only the number 6 right over here. And we're going to multiply that times each of the digits So first, we will start with 6 times 7. So we're going to first multiply 6 times 7. Well, you remember from your multiplication tables, 6 times 7 is equal to 42. Now, we don't just write 42 here. At least, not in the standard method we wouldn't write 42 here. We'd write the 2 in 42 in the ones place. So we'd write that right over there. And then we'd carry the 4 in 42 up to the tens place. Now we need to think about what 6 times 3 is. Well, once again, we know 6 times 3. 6 times 3 is equal to 18. But we can't just write an 18 down here. We still have this 4 to deal with. So 6 times 3 is 18, but we've got to then add the 4. So 6 times 3 is 18, plus 4 is 22. So it's 6 times 3, and then we're And that's how we get our answer-- so 222." + }, + { + "Q": "\n9:20, couldn't he also have used HL Theorem to prove DBE and FBE are congruent?", + "A": "Yes, that is what Sal is calling the RSH theorem at around 9:00. Either strategy is effective for showing that the two triangles are congruent.", + "video_name": "yj4oS-27Q3k", + "timestamps": [ + 560 + ], + "3min_transcript": "between the point and those 2 respective sides. We've just proven the first case, if a point lies on an angle bisector, it is equidistant from the 2 sides of the angle, now let's go the other way around, let's say that I have, so let me draw another angle here, so let me draw another angle over here and let's call this A, B, and C and let's pick some arbitrary point E, let's point some arbitrary point E right over here and let's say we start off with the assumption that E is equidistant to BC and BA and what we want to do is prove that E must be on the angle bisector, so here, if you're on the angle bisector you're equidistant, over here we're going to show if you're equidistant you're on the angle bisector, so if it's equidistant to BC and BA, then this perpendicular perpendicular right over there, and let me give these points' labels, so let call this point D, and let's call this point right over F, and let's just draw segment BE here, let's just draw segment BE right over here, so once again, we have 2 right triangles, we already know that 2 of the legs are congruent to each other, they both share the hypotenuse this hypotenuse is equal to its self, we know from the pythagorean theorm if you know 2 sides of a triangle, it determines the third side, so and we know 2 sides of both of these, so the third sides must be the same, so this side must be equal to that side so you could invoke SSS, Side Side Side, to show these 2 triangles are congruent or you actually didn't need to -have to good there you could have used a special case, the RSH case, where if you have a right triangle, so if you have the right triangle then you're also okay, you could use RSH to prove congruency as well, and so either way we know that triangle EBD, triangle EBD, is congruent to triangle EBF congruent to triangle EBF, we used Side Side Side here but you could have used RSH, let me write that RSH, which is -we know that Angle Side Side can be used for any general triangle but it can be, -RSH is essentially Angle Side Side for right triangles, if you have 2 sides of a right triangle and- in common, --if 2 sides of a right triangle are congruent then the 2 triangles are definitely congruent but once you know that two triangles are congruent, then they're corresponding angles have to be congruent, and angle EBD, angle EBD" + }, + { + "Q": "At 7:55, Sal begins the proof of the reverse case, where: if a point is equidistant from both sides of an angle, then that point is on the angle bisector. So that is clear, but when I went on to the videos about medians, I thought, if the median bisects the side opposite the vertex, then that point is also equidistant from both sides of the angle from that vertex, so that makes it an angle bisector, right? Well, apparently not, so I'm confused. What is the relationship of angle bisectors and medians?\n", + "A": "As far as I know a median is set to see if the sides are equal to each other, where an angle bisector splits a angle in half making said angle into 2 equal angles. If I am wrong sorry its been awhile seance i went over medians.", + "video_name": "yj4oS-27Q3k", + "timestamps": [ + 475 + ], + "3min_transcript": "EBG and we can use AAS, by Angle Angle Side congruency or you could say hey if two angles, -if corresponding angles are the same then the third angle is also going to be the same so this angle right over here could also be the same and you could use ASA, but either way these two things are going to be congruent but if these two things are congruent, then the corresponding sides are going to be congruent so then, then length of EF, segment EF, is going to be congruent to segment EF is going to be congruent to segment EG which is the same thing as the length of EF is equal to the length of EG, these are really, these are really equivalent statements right over there, so the length of EF, the length of EF between the point and those 2 respective sides. We've just proven the first case, if a point lies on an angle bisector, it is equidistant from the 2 sides of the angle, now let's go the other way around, let's say that I have, so let me draw another angle here, so let me draw another angle over here and let's call this A, B, and C and let's pick some arbitrary point E, let's point some arbitrary point E right over here and let's say we start off with the assumption that E is equidistant to BC and BA and what we want to do is prove that E must be on the angle bisector, so here, if you're on the angle bisector you're equidistant, over here we're going to show if you're equidistant you're on the angle bisector, so if it's equidistant to BC and BA, then this perpendicular perpendicular right over there, and let me give these points' labels, so let call this point D, and let's call this point right over F, and let's just draw segment BE here, let's just draw segment BE right over here, so once again, we have 2 right triangles, we already know that 2 of the legs are congruent to each other, they both share the hypotenuse this hypotenuse is equal to its self, we know from the pythagorean theorm if you know 2 sides of a triangle, it determines the third side, so and we know 2 sides of both of these, so the third sides must be the same, so this side must be equal to that side so you could invoke SSS, Side Side Side, to show these 2 triangles are congruent or you actually didn't need to -have to good there you could have used a special case, the RSH case, where if you have a right triangle, so if you have the right triangle" + }, + { + "Q": "At 7:07, Vi says that the cardioid is like an \"Anti-parabola\", and I don't quite understand what that mean. Please explain.\n", + "A": "If you turn positive to negative, infinity to zero while graphing a parabola you get a cardioid", + "video_name": "v-pyuaThp-c", + "timestamps": [ + 427 + ], + "3min_transcript": "both are like extreme ellipses, because a circle is like taking one focus of an ellipse and putting the other focus zero distance away. And a parabola is like an ellipse where one focus is infinity distance away. Which is why everyone lies to you and says throwing balls or shooting birds is all about parabolas when really it's about ellipses because the earth is a sphere and gravity doesn't actually go straight down. And the other focus of the elliptical orbit of your thrown object of choice may be very far away, but very far away is a great bit closer than infinity. So let's not fool ourselves. You can't look at everything that seems kind of parabolic and call it a parabola. Sure, if you connect two dots with a hanging string or chain, it looks parabolic, and so do structurally strong arches, but they're actually catenary arches and maybe you can't tell by sight, but if you're an architect you'd better know the difference. Though caternarys are quite related to parabolas, you get them by rolling around a parabola and tracing the focus which makes them a cousin of the ellipse and even a hyperbola is like an ellipse that got turned inside out or whose focus went through infinity and came out the other side or something. And of course parabolas and hyperbolas and ellipses they all come from a line that got spun around. And a line is just what happens when a couple dots get connected. Or maybe what happens when your circle is so big that, like the extreme ellipse that becomes a parabola, the extreme circle is broken at infinity and becomes a line before getting larger than infinitely big which brings it back to the other side. This linear circle, the infinite in-between. Or maybe a line is what happens when you roll around a circle and trace the focus. Or rather the 2 foci, which are zero distance apart, in ellipse terms. Which makes you wonder what you get when you roll around an ellipse and trace the foci. In fact, there's lots of great shapes you can get by rolling around shapes on other shapes, like if you roll a circle around a circle and trace the focus, you get just another circle. But, if you trace a point on the edge, you get our awesome friend the cardioid again. So now it's related to circles in three ways , which means it's a close cousin of the ellipse and a second cousin to the infinite ellipse or, parabola. Except, not just that, if you take a parabola and invert it around the unit circle, reversing inside and outside, one half becomes two, one hundred becomes a hundredth, you get once again, the cardioid. The cardioid is the anti-parabola which is good because parabolas make you sad but you heart cardioids. And of course, any time you want to connect two dots on a piece of paper, instead of drawing the line you could fold the line. Here's the thing about connecting dots. You can have all the steps laid out for you, taking whatever next step is easiest and closest and be sure of what you're getting the whole time. This way is safe and comfortable. Or, you can try new ways of connecting dots and not know what you're going to get. Maybe it will be something great, maybe it will fail. And when it fails it will be your fault. You can't blame anyone else, not mathematics or the system or the check-boxes. But if I am to have faults I would rather they be my own." + }, + { + "Q": "at 3:00 whats a prolabala?\n", + "A": "Did you mean parabola? A parabola is basically the opposite of a hyperbola. If you took two identical and perfectly symmetrical cones and lined then up perfectly, one on top of the other, meaning one right-side-up and the one on top upside-down, so it should look some what related to the shape of an hour glass. In a hyperbola, no matter how far you draw out the lines, the ends will never meet. In a parabola it is quite the opposite, or a least that s what I think.", + "video_name": "v-pyuaThp-c", + "timestamps": [ + 180 + ], + "3min_transcript": "and how many dots you skip. But then there are other shapes. Circles are good friends with sine waves. And sine waves are good friends with square waves. And let's admit it, that's pretty cool looking. In fact, just two simple straight lines of dots connecting the dots from one line to the other in order somehow gives you this awesome woven curve shape. Another student is asking the teacher when he's ever going to need to know how to graph a parabola \u2013 even as he hides his multi-million dollar enterprise of a parabola graphing game under his desk. If your teacher thought about it, he would probably think shooting birds at things is a great reason to learn about parabolas because he's come to understand that education is about money and prestige and not about becoming a better human able to do great things. You yourself haven't done anything really great yet but you figure the path to your future greatness lies more in inventing awesome new connect-the-dot arrangements than in graphing parabolas or shooting birds at things. And that's when you begin to worry. What if this cool liney curvey thing you drew approximates a parabola? As if your teacher doesn't realize everyone has their phones under their desks, his whole word runs on plausible deniability, so he shouts state-mandated, pass the test, teach-to-the-middle nonsense at students who are not at all fooled by his false enthusiasm or false mathematics and he pretends he's teaching algebra and the students pretend to be taught algebra and everyone else involved in the system is too invested to do anything but pretend to believe them both. You think maybe it's a hyperbola, which is similar to the parabola in that they are both conic sections. A hyperbola is a nice vertical slice of cone, the cone itself being just like a line swirled around in a circle, which is why the cone is like two cones radiating both ways; the lovely hyperbola insecting both parts. Two perfect curves, looking disconnected when seen alone but sharing their common conic heritage. While the boring old parabola is a slice taken at an angle completely meant to miss the top part of the cone and to miss wrapping around the bottom like an ellipse would. And it's such a special, specific case of conic section that all parabolas are exactly the same, just bigger or smaller or moved around. Your teacher could just as well hand you parabolas already drawn and have you draw coordinate grids on parabolas And it's stupid, and you hate it, and you don't wanna learn to graph them, even if it means not making a billion dollars from a game about shooting birds at things. Meanwhile anyone who actually learns how to think mathematically can then learn to graph a parabola or anything else they need in like five minutes. But teaching how to think is an individualized process that gives power and responsibility to individuals while teaching what to think can be done with one-size-fits-all bullet points and check-boxes and our culture of excuses demands that we do the latter, keeping ourselves placated in the comforting structure of tautology and clear expectations. Algebra has become a check-box subject and mathematics weeps alone in the top of the ivory tower prison to which she has been condemned. But you're not interested in check-boxes; you're interested in dots, and lines that connect them. Or maybe you could connect them with semicircles, to give visual structure to lines that would otherwise overlap. Or you could say one dot is the center of a circle and another defines a radius and draw the entire circle and do things that way. You could make rules about how every dot is the center of a circle with its neighbor being the radius," + }, + { + "Q": "how does she do the plaid thing at 4:27\n", + "A": "The circles overlapped over each other, and created a pseudo-grid of shapes. She just alternated colors between them.", + "video_name": "v-pyuaThp-c", + "timestamps": [ + 267 + ], + "3min_transcript": "his whole word runs on plausible deniability, so he shouts state-mandated, pass the test, teach-to-the-middle nonsense at students who are not at all fooled by his false enthusiasm or false mathematics and he pretends he's teaching algebra and the students pretend to be taught algebra and everyone else involved in the system is too invested to do anything but pretend to believe them both. You think maybe it's a hyperbola, which is similar to the parabola in that they are both conic sections. A hyperbola is a nice vertical slice of cone, the cone itself being just like a line swirled around in a circle, which is why the cone is like two cones radiating both ways; the lovely hyperbola insecting both parts. Two perfect curves, looking disconnected when seen alone but sharing their common conic heritage. While the boring old parabola is a slice taken at an angle completely meant to miss the top part of the cone and to miss wrapping around the bottom like an ellipse would. And it's such a special, specific case of conic section that all parabolas are exactly the same, just bigger or smaller or moved around. Your teacher could just as well hand you parabolas already drawn and have you draw coordinate grids on parabolas And it's stupid, and you hate it, and you don't wanna learn to graph them, even if it means not making a billion dollars from a game about shooting birds at things. Meanwhile anyone who actually learns how to think mathematically can then learn to graph a parabola or anything else they need in like five minutes. But teaching how to think is an individualized process that gives power and responsibility to individuals while teaching what to think can be done with one-size-fits-all bullet points and check-boxes and our culture of excuses demands that we do the latter, keeping ourselves placated in the comforting structure of tautology and clear expectations. Algebra has become a check-box subject and mathematics weeps alone in the top of the ivory tower prison to which she has been condemned. But you're not interested in check-boxes; you're interested in dots, and lines that connect them. Or maybe you could connect them with semicircles, to give visual structure to lines that would otherwise overlap. Or you could say one dot is the center of a circle and another defines a radius and draw the entire circle and do things that way. You could make rules about how every dot is the center of a circle with its neighbor being the radius, and all of the others define radii. But then you just get concentric circles, which I suppose should have been obvious. But what if you did it the other way around and said one dot always stays on the circle and all the other dots are centers, like this. Looks more promising. So you try putting all the dots in a circle and using them as circle centers and choose just one dot for the circles to go through and you get this awesome shape that looks kind of like a heart. So let's call it, oh I don't know, a cardioid. Which happens to be the same curve that you get when parallel lines like rays of light reflect off a circle the same heart of sunshine in a cup. Or maybe instead of circle centers you could have points all on the curve of a circle, which means you need three points to define a circle, maybe just a point and its two closest neighbors to start with. And of course, any collection of circles is two-colorable, which means you can contrast light and dark colors for a classy color scheme. Or maybe you could throw down some random points to make all possible circles. Only that would be a lot of circles, so you choose just ones you like. And then, against your will, you begin to wonder how many points it takes to define the boring old parabola." + }, + { + "Q": "3:37 - 3: 40 why would you divide by 10 when the scientific notion is 10^-2??\n", + "A": "Actually, Sal was multiplying by 10, so 10^-3 * 10 is 10^-2.", + "video_name": "ios3QL9t9LQ", + "timestamps": [ + 217 + ], + "3min_transcript": "So now we have two things. We have 41 X 10 to the -3 - 2.6 x 10 to the -3. Well this is going to be the same thing as 41 - 2.6. - 2.6, let me do it in that same color. That was purple. - 2.6, 10 to the -3. 10, whoops, 10 to the -3. There's 10 to the -3 there, 10 to the -3 there. One way to think about it, I have just factored out a 10 to the -3. Now what's 41 - 2.6? Well 41 - 2 is 39, and then -.6 is going to be 38.4. times 10 to the -3 power. x 10 to the -3 power. Now, this is what the product, or this is what the difference of these two numbers is but this is no longer in scientific notation. In order to be scientific notation this number right over here has to be between 1, has to be greater than or equal to 1 and less than 10. So what we could do is we could divide this number right over here by 10. We can divide this by 10 and then we can multiply, then we can multi...so we could do this. We could, kind of the opposite of what we did up here. We could divide this by 10 and then multiply by 10. So if you divide by 10 and multiply by 10 you're not changing the value. So 38.4 divided by 10 is 3.84 and then all of this business, 10 to the -3, 10 to the -3 x 10 is 10 to the -2 power." + }, + { + "Q": "\nWhen you move the decimal point in 4.1 to the right, does it increase or decrease the exponent of 10? How about moving it to the left? At 1:41, Sal is doing it a way I wasn't taught so I am getting confused.", + "A": "When you move the decimal to the right in 4.1, it decreases the exponent by I or how many spaces you move the decimal to the right. When you move it to the left, the exponent of 10 will increase. This is one way you can think about it, when you are moving the decimal point to the right, you are putting a SMALLER number into scientific notation. When you are moving the decimal point to the left , you are putting a LARGER number into Scientific notation.", + "video_name": "ios3QL9t9LQ", + "timestamps": [ + 101 + ], + "3min_transcript": "- [Voiceover] What I want to do in this video is get a little bit of practice subtracting in scientific notation. So let's say that I have 4.1 x 10 to the -2 power. 4.1 x 10 to the -2 power and from that I want to subtract, I want to subtract 2.6, 2.6 x 10 to the -3 power. Like always, I encourage you to pause this video and see if you can solve this on your own and then we could work through it together. All right, I'm assuming you've had a go it. So the easiest thing that I can think of doing is try to convert one of these numbers so that it has the same, it's being multiplied by the same power of ten as the other one. What I could think about doing, well can we express 4.1 times 10 to the -2? Can we express it as something times 10 to the -3? So we have 4.1 times 10 to the -2. we would divide by 10, but we can't just divide by 10. That would literally change the value of the number. In order to not change it, we want to multiply by 10 as well. So we're multiplying by 10 and dividing by 10. I could have written it like this. I could have written 10/10 times, let me write this a little bit neater. I could have written 10/10 x this and then you take 10 x 4.1, you get 41, and then 10 to the -2 divided by 10 is going to be 10 to the -3. So this right over here, this is equal to 10 x 4.1 is 41 times 10 to the -3. And that makes sense. 41 thousandths is the same thing as 4.1 hundredths and all we did is we multiplied this times 10 and we divided this times 10. So let's rewrite this. We can rewrite it now as 41 X 10 to the -3 So now we have two things. We have 41 X 10 to the -3 - 2.6 x 10 to the -3. Well this is going to be the same thing as 41 - 2.6. - 2.6, let me do it in that same color. That was purple. - 2.6, 10 to the -3. 10, whoops, 10 to the -3. There's 10 to the -3 there, 10 to the -3 there. One way to think about it, I have just factored out a 10 to the -3. Now what's 41 - 2.6? Well 41 - 2 is 39, and then -.6 is going to be 38.4." + }, + { + "Q": "\nat 3:50, why does he divide by 10?", + "A": "You Cannot have a scientific notation over 10 so he divides 38.4 by 10 to make it valid for a scientific notation, giving you the answer of 3.84 x 10^-2", + "video_name": "ios3QL9t9LQ", + "timestamps": [ + 230 + ], + "3min_transcript": "So now we have two things. We have 41 X 10 to the -3 - 2.6 x 10 to the -3. Well this is going to be the same thing as 41 - 2.6. - 2.6, let me do it in that same color. That was purple. - 2.6, 10 to the -3. 10, whoops, 10 to the -3. There's 10 to the -3 there, 10 to the -3 there. One way to think about it, I have just factored out a 10 to the -3. Now what's 41 - 2.6? Well 41 - 2 is 39, and then -.6 is going to be 38.4. times 10 to the -3 power. x 10 to the -3 power. Now, this is what the product, or this is what the difference of these two numbers is but this is no longer in scientific notation. In order to be scientific notation this number right over here has to be between 1, has to be greater than or equal to 1 and less than 10. So what we could do is we could divide this number right over here by 10. We can divide this by 10 and then we can multiply, then we can multi...so we could do this. We could, kind of the opposite of what we did up here. We could divide this by 10 and then multiply by 10. So if you divide by 10 and multiply by 10 you're not changing the value. So 38.4 divided by 10 is 3.84 and then all of this business, 10 to the -3, 10 to the -3 x 10 is 10 to the -2 power." + }, + { + "Q": "At 3:13, did Sal just invert and multiply? Also can I cross multiply and then transfer the 4 to divide the other side by 4?\n", + "A": "Yeah it looks to me like he just multiplied the reciprocals of the two numbers. I don t think you could just transfer the 4 to the other side, I think you d have to divide BOTH sides by four which would get X alone, and then you could divide the other side by 4. Hope this helps!", + "video_name": "lEGS5ECgFxE", + "timestamps": [ + 193 + ], + "3min_transcript": "So let's say C is-- I'm just trying to eyeball it right now-- let's put C right over here. And we don't know C's coordinates. Well the way we could think about it is to break it up into horizontal change in coordinate and vertical change in coordinate, and apply the same ratio. So for example, what is the horizontal change in coordinates going from A to B? Well let's draw that. Going from A to B, so this is A's x-coordinate, it's at negative 6. B's x-coordinate is at negative 2. So it's this change right over here. This is the horizontal change that we care about. Now what is that? Well if you start at negative 6 and you go to negative 2, you have increased by 4. Another way of thinking about it is negative 2 minus negative 6 Now the ratio between this change and the change of the x-coordinate between A and C is going to be 2 to 5. So let's call that change, this entire change, let's call this x. So we could say that the ratio between 4 and x is equal to-- and notice this is the horizontal change from A to B just if you look on the horizontal axis-- so the ratio of that, which is 4, to the horizontal change between A and C, well that's going to have to be the same ratio. So it's going to be 2/5. Now to solve for x, a fun thing might be to just take the reciprocal of both sides. So x/4 is equal to 5/2. We can multiply both sides times 4 and we are left with x is equal to 5 times 4 divided by 2, So the change in x from A to C is going to be 10. So what does that tell us about C's x-coordinate? So we could start with A's x-coordinate, which is negative 6, add 10 to it, negative 6 plus 10 is 4. So we figured out the x-coordinate, now we just have to do the same thing for the y. So what is the change in y going from A to B? Well here, we go from 9 to 3, we've gone down 6. Another way you could say it is, 3 minus 9 is negative 6. To find the change, you could think, I'm just taking the end point and subtracting from that the starting point. Negative 2 minus negative 6 was positive 4. 3 minus 9 is negative 6. Or you could just look at it." + }, + { + "Q": "\nWhat about when im trying to solve with fractions or is the form 2:5 the same as 2/5", + "A": "No, it isn t. It should be 2/7 because you re dividing into 7 sections. If you need more explanation, just ask me.", + "video_name": "lEGS5ECgFxE", + "timestamps": [ + 125 + ], + "3min_transcript": "A, B and C are collinear, and B is between A and C. The ratio of AB to AC is 2 to 5. If is A is at negative 6 comma 9, and B is at negative 2 comma 3, what are the coordinates of point C? And I encourage you to now pause this video and try this on your own. So let's try to visualize this. So A is at negative 6 comma 9. B, let's see, it's less negative in the horizontal direction, it's lower in the vertical direction. So we could put B right over here. B is at the point negative 2 comma 3. And that C, it's going to be collinear, so we're going to go along the same line, let me draw that line right now. So it's going to be collinear and they tell us that the ratio of AB to AC is 2 to 5. So let's say C is-- I'm just trying to eyeball it right now-- let's put C right over here. And we don't know C's coordinates. Well the way we could think about it is to break it up into horizontal change in coordinate and vertical change in coordinate, and apply the same ratio. So for example, what is the horizontal change in coordinates going from A to B? Well let's draw that. Going from A to B, so this is A's x-coordinate, it's at negative 6. B's x-coordinate is at negative 2. So it's this change right over here. This is the horizontal change that we care about. Now what is that? Well if you start at negative 6 and you go to negative 2, you have increased by 4. Another way of thinking about it is negative 2 minus negative 6 Now the ratio between this change and the change of the x-coordinate between A and C is going to be 2 to 5. So let's call that change, this entire change, let's call this x. So we could say that the ratio between 4 and x is equal to-- and notice this is the horizontal change from A to B just if you look on the horizontal axis-- so the ratio of that, which is 4, to the horizontal change between A and C, well that's going to have to be the same ratio. So it's going to be 2/5. Now to solve for x, a fun thing might be to just take the reciprocal of both sides. So x/4 is equal to 5/2. We can multiply both sides times 4 and we are left with x is equal to 5 times 4 divided by 2," + }, + { + "Q": "So in one of my questions it says the ratio between AC and BC is 2:3 and to find point B. Then when I click on hint it says the ratio between AC and BC is 2:3 therefore the ratio between AB and AC is 2:5, how does that work?\n", + "A": "AC to BC is 2 : 3 and since they are the only two parts of the line segment AC, the segment has a total of 2 + 3 = 5 parts. Therefore, AB to the whole thing (AC) is 2 : 5.", + "video_name": "lEGS5ECgFxE", + "timestamps": [ + 123, + 123, + 125 + ], + "3min_transcript": "A, B and C are collinear, and B is between A and C. The ratio of AB to AC is 2 to 5. If is A is at negative 6 comma 9, and B is at negative 2 comma 3, what are the coordinates of point C? And I encourage you to now pause this video and try this on your own. So let's try to visualize this. So A is at negative 6 comma 9. B, let's see, it's less negative in the horizontal direction, it's lower in the vertical direction. So we could put B right over here. B is at the point negative 2 comma 3. And that C, it's going to be collinear, so we're going to go along the same line, let me draw that line right now. So it's going to be collinear and they tell us that the ratio of AB to AC is 2 to 5. So let's say C is-- I'm just trying to eyeball it right now-- let's put C right over here. And we don't know C's coordinates. Well the way we could think about it is to break it up into horizontal change in coordinate and vertical change in coordinate, and apply the same ratio. So for example, what is the horizontal change in coordinates going from A to B? Well let's draw that. Going from A to B, so this is A's x-coordinate, it's at negative 6. B's x-coordinate is at negative 2. So it's this change right over here. This is the horizontal change that we care about. Now what is that? Well if you start at negative 6 and you go to negative 2, you have increased by 4. Another way of thinking about it is negative 2 minus negative 6 Now the ratio between this change and the change of the x-coordinate between A and C is going to be 2 to 5. So let's call that change, this entire change, let's call this x. So we could say that the ratio between 4 and x is equal to-- and notice this is the horizontal change from A to B just if you look on the horizontal axis-- so the ratio of that, which is 4, to the horizontal change between A and C, well that's going to have to be the same ratio. So it's going to be 2/5. Now to solve for x, a fun thing might be to just take the reciprocal of both sides. So x/4 is equal to 5/2. We can multiply both sides times 4 and we are left with x is equal to 5 times 4 divided by 2," + }, + { + "Q": "At 3:29,How did Sal used the chain rule and got the derivative to be (e^((ln a)x) * ln a).\nShouldn't we use the power rule? I mean, if Sal is using the chain rule, (d [f(g(x))]) / dx =f \u00e2\u0080\u00b2(g(x)) g \u00e2\u0080\u00b2(x), then what is the f(x) and g(x)?\nThank you for you reply\n", + "A": "He let \u00f0\u009d\u0091\u0094(\u00f0\u009d\u0091\u00a5) = \u00f0\u009d\u0091\u00a5ln \u00f0\u009d\u0091\u008e and \u00f0\u009d\u0091\u0093(\u00f0\u009d\u0091\u00a5) = \u00f0\u009d\u0091\u0092\u00cb\u00a3. Power rule only works when the power is a constant, not a variable which we are differentiating WRT.", + "video_name": "gHzLHknEk1M", + "timestamps": [ + 209 + ], + "3min_transcript": "you could view a as being equal to e. Let me write it this way. Well all right, a as being equal to e to the natural log of a. Now if this isn't obvious to you, I really want you to think about it. What is the natural log of a? The natural log of a is the power you need to raise e to, to get to a. So if you actually raise e to that power, if you raise e to the power you need to raise e too to get to a. Well then you're just going to get to a. So really think about this. Don't just accept this as a leap of faith. It should make sense to you. And it just comes out of really what a logarithm is. And so we can replace a with this whole expression here. If a is the same thing as e to the natural log of a, well then this is going to be, with respect to x of e to the natural log, I keep writing la (laughs), to the natural log of a and then we're going to raise that to the xth power. We're going to raise that to the x power. And now this, just using our exponent properties, this is going to be equal to the derivative with respect to x of, and I'll just keep color-coding it. If I raise something to an exponent and then raise that to an exponent, that's the same thing as raising our original base to the product of those exponents. That's just a basic exponent property. So that's going to be the same thing as e to the natural log of a, natural log of a times x power. Times x power. And now we can use the chain rule to evaluate this derivative. we will first take the derivative of the outside function. So e to the natural log of a times x with respect to the inside function, with respect to natural log of a times x. And so, this is going to be equal to e to the natural log of a times x. And then we take the derivative of that inside function with respect to x. Well natural log of a, it might not immediately jump out to you, but that's just going to be a number. So that's just going to be, so times the derivative. If it was the derivative of three x, it would just be three. If it's the derivative of natural log a times x, it's just going to be natural log of a. And so this is going to give us the natural log of a times e to the natural log of a. And I'm going to write it like this." + }, + { + "Q": "At 3:51, how would you write 8^ -1/3 as a radical? Would it be the negative third root (with the three over the root symbol being negative)?\n", + "A": "Any value raised to a negative exponent is the reciprocal of that value raised to the same positive exponent. For example: 8^(-1/3)=1/(8^(1/3))=1/2", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 231 + ], + "3min_transcript": "Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3. What would be the log base 8 of 1/2? What does this evaluate to? Let me clean this up so that we have some space to work with. So as always, we're saying, what power do I have to raise 8 to to get to 1/2? So let's think about that a little bit. We already know that 8 to the one-third power is equal to 2. If we want the reciprocal of 2 right over here, we have to just raise 8 to the negative one-third. So let me write that down. 8 to the negative one-third power is going to be equal to 1 over 8 to the one-third power. And we already know the cube root of 8, or 8 to the one-third power, is equal to 2. This is equal to 1/2. do I have to raise 8 to to get to 1/2-- is negative 1/3. Equal to negative 1/3. I hope you enjoyed that as much as I did." + }, + { + "Q": "\nWow, that was a lot of information to soke in. I don't really get the last example. At 3:36, Sal was giving another example that was very complicate. Can someone help me with it?", + "A": "I got lost when Sal talked about he reciprocal of 2 and then he quickly moved on to 1/3 being -3.", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 216 + ], + "3min_transcript": "Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3. What would be the log base 8 of 1/2? What does this evaluate to? Let me clean this up so that we have some space to work with. So as always, we're saying, what power do I have to raise 8 to to get to 1/2? So let's think about that a little bit. We already know that 8 to the one-third power is equal to 2. If we want the reciprocal of 2 right over here, we have to just raise 8 to the negative one-third. So let me write that down. 8 to the negative one-third power is going to be equal to 1 over 8 to the one-third power. And we already know the cube root of 8, or 8 to the one-third power, is equal to 2. This is equal to 1/2. do I have to raise 8 to to get to 1/2-- is negative 1/3. Equal to negative 1/3. I hope you enjoyed that as much as I did." + }, + { + "Q": "\nAt 1:37 how did you get 1/3 im not really understanding how I would get that?", + "A": "Look ! it was a good question but hear him again . Sal said you need to think in a different manner so, if you look at the question with log 2 with base 8 you have an anti-log or the exponential form as 2= 8^x so for cube Root we have 8 to the power 1/3 . For example 2 to the power 1/2 means an under root of 2 . so same way is for cube root i.e, power 1/3. hope you like the answer", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 97 + ], + "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3." + }, + { + "Q": "\nat 1:29 he says since 2x2x2=8 so there fore the answer is 1/3 i dont really get this can you help me? ( im not even learning this yet I'm in 7th grade algebra 1 honors, i just want to be prepared)", + "A": "Logx(2)=8 can also be told as 8^x=2. For 8 to be less than itself as 2, the exponent must be a fraction or negative as in powers, fractions and negative integers make the number smaller while whole numbers make the number bigger in this situation. This is not true for all situations. The reason x=1/3 is because it means 8^(1/3)=2. The cube root of 8 is 2 because 2x2x2=8. So that means log 1/3(2)=8, which is true. I m also in 7th grade algebra 1 honors.", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 89 + ], + "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3." + }, + { + "Q": "At 1:27, how did he get 1/3?\n", + "A": "The question is effectively what power do you need to put 8 to, to get 2. As 2^3=8, 8^(1/3) must equal 2.", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 87 + ], + "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3." + }, + { + "Q": "At around 5:39 he put 5x-5. But why did he do that? Why didn't he distribute all the way?\n", + "A": "Sal is multiplying the trinomial with a fraction that has the trinominal in the denominator. Thetrinomials cancel out. This is just like if you multiply: 7 * 5/7, the 7 s cancel out leaving just the 5 behind. Hope this helps.", + "video_name": "pLCmwHsDYqo", + "timestamps": [ + 339 + ], + "3min_transcript": "but we're saying it's not. It's now a lower degree than this down here. So we could say it's x plus 1 plus whatever this remainder is, divided by this thing over here. So our answer-- I'm going to write it one more time. It's x plus 1 plus 5x minus 5, over x squared minus x plus 1. And we can check that this works. If we take this thing over here, and we multiply it by this thing over here, we should get the x to the third plus 5x minus 4. So let's do that. Let's multiply this thing by x squared minus x plus 1. And to do that, let's just distribute this whole trinomial times each of these terms. When we do the first term, we have So that's going to be x times x squared, which is x to the third; x times negative x, which is negative x squared; x times 1, which is plus x. Then we can multiply this whole thing times 1. So it's going to be plus x squared minus x plus 1. I'm just multiplying all of these times 1. And then we can multiply this whole thing times this thing. Now, this is the same as the denominator here. So it'll cancel out. This will cancel with that. And we're just going to be left with the numerator over here, so plus 5x minus 5. And now we can try to simplify it. We only have one third-degree term, the x to the third. So we have x to the third here. Second-degree terms-- we have a negative x squared. And then we also have a positive x squared. So they cancel out with each other. First-degree terms-- let's see. We have a positive x and a negative x. Those cancel out with each other. So we're just going to have this 5x. So then we have plus 5x. And then we have the 0-th degree terms, or the constant terms. We have a positive 1 and a negative 5. You get negative 4. So you get x to the third plus 5x minus 4, which is exactly what we had over here." + }, + { + "Q": "\nat 5:36, Sal canceled the denominator and is only left with the numerator. Why he doesn't distribute the x^2-x+1 into 5x-5 as well? Thanks!", + "A": "The entire fraction is one term, so he is distributing x^2-x+1 into the entire fraction, not just the denominator. Let s say 5x-5 = a and x^2-x+1 = b. Distributing into the last term would leave you with (a/b) * b. Simplifying this would just leave you with a, or 5x-5. This question was asked last year, so my answer is a bit late - you may have already figured it out yourself - but maybe it ll help with someone else who has this question, since there are no other answers here.", + "video_name": "pLCmwHsDYqo", + "timestamps": [ + 336 + ], + "3min_transcript": "but we're saying it's not. It's now a lower degree than this down here. So we could say it's x plus 1 plus whatever this remainder is, divided by this thing over here. So our answer-- I'm going to write it one more time. It's x plus 1 plus 5x minus 5, over x squared minus x plus 1. And we can check that this works. If we take this thing over here, and we multiply it by this thing over here, we should get the x to the third plus 5x minus 4. So let's do that. Let's multiply this thing by x squared minus x plus 1. And to do that, let's just distribute this whole trinomial times each of these terms. When we do the first term, we have So that's going to be x times x squared, which is x to the third; x times negative x, which is negative x squared; x times 1, which is plus x. Then we can multiply this whole thing times 1. So it's going to be plus x squared minus x plus 1. I'm just multiplying all of these times 1. And then we can multiply this whole thing times this thing. Now, this is the same as the denominator here. So it'll cancel out. This will cancel with that. And we're just going to be left with the numerator over here, so plus 5x minus 5. And now we can try to simplify it. We only have one third-degree term, the x to the third. So we have x to the third here. Second-degree terms-- we have a negative x squared. And then we also have a positive x squared. So they cancel out with each other. First-degree terms-- let's see. We have a positive x and a negative x. Those cancel out with each other. So we're just going to have this 5x. So then we have plus 5x. And then we have the 0-th degree terms, or the constant terms. We have a positive 1 and a negative 5. You get negative 4. So you get x to the third plus 5x minus 4, which is exactly what we had over here." + }, + { + "Q": "@ 0:51 How He Knows That Slope is -4 by just seeing the Equation of Function?\n", + "A": "The equation of a line is y = mx + b. m is the slope of the line. In the video the equation is g(x) = -4x + 7, so -4 is the slope of the line.", + "video_name": "nGCW5teACC0", + "timestamps": [ + 51 + ], + "3min_transcript": "Let g of x equal negative 4x plus 7. What is the value of the limit as x approaches negative 1 of all of this? So before we think about this, let's just visualize the line. And then we can think about what they're asking here. So let me draw some axes here. So this is my vertical axis and this is my horizontal axis. And let's say this is my x-axis. We'll label that the x-axis. I'll graph g of x. g of x is going to have a positive-- I guess you would say y-intercept. or vertical axis intercept. It's going to have a slope of negative 4, so it's going to look something like this. Let me draw my best. So it's going to look something like that. And we already know the slope here is going to be negative 4. We get that right from this slope intercept form of the equation, slope is equal to negative 4. And they ask us, what is the limit as x approaches negative 1 of all of this kind of stuff? So let's plot the point negative 1. this point right over here. And this point right over here would be the point negative 1, g of negative 1. Let me label everything else. So I could call this my y-axis. I could call this graph. This is the graph of y is equal to g of x. So what they're doing right over here is they're finding the slope between an arbitrary point x, g of x, and this point right over here. So let's do that. So let's take another x. So let's say this is x. This would be the x, g of x. And this expression right over here, notice it is your change in the vertical axis. That would be your g of x. Let me make it this way. So this would be your change in the vertical axis. That would be g of x minus g of negative 1. And then that's over-- actually, let me write it this way all of that over your change in the horizontal axis. Well, your change in the horizontal axis is this distance, which is the same thing as this distance. Notice your change in vertical over change in horizontal, change in vertical over change in horizontal, reviewing the green point as the endpoint. So it's going to be x minus negative 1. And this is the exact same expression. These are the exact same expression. You can simplify them, minus negative 1, and this becomes plus. This could become a plus 1. But these are the exact same expression. So this is the expression, really, for the slope between negative 1 and g of negative 1 and an arbitrary x. Well, we already know that no matter what x you pick, the slope between x, g of , and this point right over here is" + }, + { + "Q": "\nAt 3:38, Sal says that the slope will be the same no matter what points are picked in the line. Does that mean the derivative of f or f'(x) will always equal to the slope of the secant line in the function? I just thought even if f(x) and f(x+h) are extremely close (derivative), they would have the same slope as the general slope of the function, therefore derivative equals to the slope.", + "A": "The function Sal is talking about here is linear, so its graph is a straight line. In this case we can t take a secant in the normal sense of the word. Every line that passes through two points on the graph of this function will be the same as the line that graphs the function. This is the only situation where the slope will be the same no matter what points are picked.", + "video_name": "nGCW5teACC0", + "timestamps": [ + 218 + ], + "3min_transcript": "all of that over your change in the horizontal axis. Well, your change in the horizontal axis is this distance, which is the same thing as this distance. Notice your change in vertical over change in horizontal, change in vertical over change in horizontal, reviewing the green point as the endpoint. So it's going to be x minus negative 1. And this is the exact same expression. These are the exact same expression. You can simplify them, minus negative 1, and this becomes plus. This could become a plus 1. But these are the exact same expression. So this is the expression, really, for the slope between negative 1 and g of negative 1 and an arbitrary x. Well, we already know that no matter what x you pick, the slope between x, g of , and this point right over here is It's going to be the slope of the line. It's going to be equal to negative 4. This thing is going to be equal to negative 4. It's going to be equal to negative 4. Doesn't matter how close x gets, and weather x comes from the right or whether x comes from the left. So this thing, taking the limit of this, this just gets you to negative 4. It's really just the slope of the line. So even if you were to take the limit as x approaches negative 1, as x gets closer and closer and closer to negative 1, well then, these points are just going to get closer and closer and closer. But every time you calculate the slope, it's just going to be the slope of the line, which Now, you could also do this algebraically. And let's try to do it algebraically. So let's actually just take the limit as x approaches negative 1 of g of x. Well, they already told us what g of x is. It is negative 4x plus 7, minus g of negative 1. Negative 1 times 4 is positive 4. Positive 4 plus 7 is 11. All of that over x plus 1, all of that over x plus 1. And that's really x minus negative 1, is you want to think of it that way. But I'll just write x plus 1 this way here. So this is going to be equal to the limit as x approaches negative 1 of, in our numerator-- let's see. 7 minus 11 is negative 4. We can factor out a negative 4. It's a negative 4 times x plus 1, all of that over x plus 1. And then since we're just trying to find the limit as x approaches negative 1, so we can cancel those out. And this is going to be non-zero for any x value other than negative 1. And so this is going to be equal to negative 4." + }, + { + "Q": "at 0:40 I dont get how did he draw that line...he says the line has a slope of -4 but i dont get how that works...pls help meunderstand this :/\n", + "A": "The slope of a line can be found from its coefficient before x in its equasion. For example, the line y = -4x + 7 has a slope of -4 because it is the coefficient before x. It also represents the tangent of the angle created by the line and the positive direction of the X axis. However, most curves don t behave like that.", + "video_name": "nGCW5teACC0", + "timestamps": [ + 40 + ], + "3min_transcript": "Let g of x equal negative 4x plus 7. What is the value of the limit as x approaches negative 1 of all of this? So before we think about this, let's just visualize the line. And then we can think about what they're asking here. So let me draw some axes here. So this is my vertical axis and this is my horizontal axis. And let's say this is my x-axis. We'll label that the x-axis. I'll graph g of x. g of x is going to have a positive-- I guess you would say y-intercept. or vertical axis intercept. It's going to have a slope of negative 4, so it's going to look something like this. Let me draw my best. So it's going to look something like that. And we already know the slope here is going to be negative 4. We get that right from this slope intercept form of the equation, slope is equal to negative 4. And they ask us, what is the limit as x approaches negative 1 of all of this kind of stuff? So let's plot the point negative 1. this point right over here. And this point right over here would be the point negative 1, g of negative 1. Let me label everything else. So I could call this my y-axis. I could call this graph. This is the graph of y is equal to g of x. So what they're doing right over here is they're finding the slope between an arbitrary point x, g of x, and this point right over here. So let's do that. So let's take another x. So let's say this is x. This would be the x, g of x. And this expression right over here, notice it is your change in the vertical axis. That would be your g of x. Let me make it this way. So this would be your change in the vertical axis. That would be g of x minus g of negative 1. And then that's over-- actually, let me write it this way all of that over your change in the horizontal axis. Well, your change in the horizontal axis is this distance, which is the same thing as this distance. Notice your change in vertical over change in horizontal, change in vertical over change in horizontal, reviewing the green point as the endpoint. So it's going to be x minus negative 1. And this is the exact same expression. These are the exact same expression. You can simplify them, minus negative 1, and this becomes plus. This could become a plus 1. But these are the exact same expression. So this is the expression, really, for the slope between negative 1 and g of negative 1 and an arbitrary x. Well, we already know that no matter what x you pick, the slope between x, g of , and this point right over here is" + }, + { + "Q": "Regarding, the y constraint at 2:30 - 2:44. I don't really get it why it is >= 1...\n", + "A": "If y is a number less than 1, it will make the value sqrt (x - 1) not able to be graphed in the Cartesian plane. (eg you cannot plot sqrt (-4) etc) There s no x-value that would then satisfy the condition sqrt (x - 1) = x + 2. So the constraint for y >= 1.", + "video_name": "aeyFb2eVH1c", + "timestamps": [ + 150, + 164 + ], + "3min_transcript": "We've got the function f of x is equal to x plus 2 squared plus 1, and we've constrained our domain that x has to be greater than or equal to negative 2. That's where we've defined our function. And we want to find its inverse. And I'll leave you to think about why we had to constrain it to x being a greater than or equal to negative 2. Wouldn't it have been possible to find the inverse if we had just left it as the full parabola? I'll leave you -- or maybe I'll make a future video about that. But let's just figure out the inverse here. So, like we've said in the first video, in the introduction to inverses, we're trying to find a mapping. Or, if we were to say that y -- if we were to say that y is equal to x plus 2 squared plus 1. This is the function you give me an x and it maps to y. We want to go the other way. We want to take, I'll give you a y and then map it to an x. So what we do is, we essentially just solve for x in terms of y. So let's do that one step at a time. So, the first thing to do, we could subtract 1 from both sides of this equation. And now to solve here, you might want to take the square root. And that actually will be the correct thing to do. But it's very important to think about whether you want to take the positive or the negative square root at this step. So we've constrained our domain to x is greater than or equal to negative 2. So this value right here, x plus 2, if x is always greater than or equal to negative 2, x plus 2 will always be greater than or equal to 0. So this expression right here, this right here is positive. This is positive. So we have a positive squared. So if we really want to get to the x plus 2 in the appropriate domain, we want to take the positive square root. And in the next video or the video after that, we'll solve an example where you want to take the negative square root. So we're going to take theundefined positive square root, or just the principal root, which is just the square root sign, of both sides. So you get the square root of y minus 1 is equal to x plus 2. the beginning we had a constraint on x. We had for x is greater than or equal to negative 2. But what constraint could we have on y? If you look at the graph right here, x is greater than equal to negative 2. But what's why? What is the range of y-values that we can get here? Well, if you just look at the graph, y will always be greater than or equal to 1. And that just comes from the fact that this term right here is always going to be greater than or equal to 0. So the minimum value that the function could take on is 1. So we could say for x is greater than or equal to negative 2, and we could add that y is always going to be greater than or equal to 1. y is always greater than or equal to 1. The function is always greater than or equal to that right there. To 1. And the reason why I want to write it at the stage is" + }, + { + "Q": "\nAt 04:28, why don't we take the positive and negative sqrt of ( y -1 ) ?", + "A": "As Sal explained earlier in the video, (x+2)^2 can only be positive, due to the restriction on the function x>=2. Therefore, the square-root of (y-1) can only be positive. The next steps after this are purely algebra, and have nothing to do with a positive or negative square root.", + "video_name": "aeyFb2eVH1c", + "timestamps": [ + 268 + ], + "3min_transcript": "the beginning we had a constraint on x. We had for x is greater than or equal to negative 2. But what constraint could we have on y? If you look at the graph right here, x is greater than equal to negative 2. But what's why? What is the range of y-values that we can get here? Well, if you just look at the graph, y will always be greater than or equal to 1. And that just comes from the fact that this term right here is always going to be greater than or equal to 0. So the minimum value that the function could take on is 1. So we could say for x is greater than or equal to negative 2, and we could add that y is always going to be greater than or equal to 1. y is always greater than or equal to 1. The function is always greater than or equal to that right there. To 1. And the reason why I want to write it at the stage is the the x's and y's. So let's just leave that there. So here we haven't explicitly solved for x and y. But we can write for y is greater than or equal to 1, this is going to be the domain for our inverse, so to speak. And so here we can keep it for y is greater This y constraint's going to matter more. Because over here, the domain is x. But for the inverse, the domain is going to be the y-value. And then, let's see. We have the square root of y minus 1 is equal to x plus 2. Now we can subtract 2 from both sides. We get the square root of y minus 1 minus 2, is equal to x for y is greater than or equal to 1. And so we've solved for x in terms of y. Or, we could say, let me just write it the other way. We could say, x is equal to, I'm just swapping this. x is equal to the square root of y minus one minus 2, for y is So you see, now, the way we've written it out. y is the input into the function, which is going to be the inverse of that function. x the output. x is now the range. So we could even rewrite this as f inverse of y. That's what x is, is equal to the square root of y minus 1 minus 2, for y is greater than or equal to 1. And this is the inverse function. We could say this is our answer. But many times, people want the answer in terms of x. And we know we could put anything in here. If we put an a here, we take f inverse of a. It'll become the square root of a minus 1 minus 2, 4. Well, assuming a is greater than or equal to 1. But we could put an x in here. So we can just rename the the y for x. So we could just do a renaming here. So we can just rename y for x. And then we would get -- let me scroll down a little bit. We would f inverse of x." + }, + { + "Q": "At 1:42, Sal said that if x was greater than or equal to -2 then the underlined expression would be positive, but zero isn't positive.\n", + "A": "So the inverse of y is x it s opposite.", + "video_name": "aeyFb2eVH1c", + "timestamps": [ + 102 + ], + "3min_transcript": "We've got the function f of x is equal to x plus 2 squared plus 1, and we've constrained our domain that x has to be greater than or equal to negative 2. That's where we've defined our function. And we want to find its inverse. And I'll leave you to think about why we had to constrain it to x being a greater than or equal to negative 2. Wouldn't it have been possible to find the inverse if we had just left it as the full parabola? I'll leave you -- or maybe I'll make a future video about that. But let's just figure out the inverse here. So, like we've said in the first video, in the introduction to inverses, we're trying to find a mapping. Or, if we were to say that y -- if we were to say that y is equal to x plus 2 squared plus 1. This is the function you give me an x and it maps to y. We want to go the other way. We want to take, I'll give you a y and then map it to an x. So what we do is, we essentially just solve for x in terms of y. So let's do that one step at a time. So, the first thing to do, we could subtract 1 from both sides of this equation. And now to solve here, you might want to take the square root. And that actually will be the correct thing to do. But it's very important to think about whether you want to take the positive or the negative square root at this step. So we've constrained our domain to x is greater than or equal to negative 2. So this value right here, x plus 2, if x is always greater than or equal to negative 2, x plus 2 will always be greater than or equal to 0. So this expression right here, this right here is positive. This is positive. So we have a positive squared. So if we really want to get to the x plus 2 in the appropriate domain, we want to take the positive square root. And in the next video or the video after that, we'll solve an example where you want to take the negative square root. So we're going to take theundefined positive square root, or just the principal root, which is just the square root sign, of both sides. So you get the square root of y minus 1 is equal to x plus 2. the beginning we had a constraint on x. We had for x is greater than or equal to negative 2. But what constraint could we have on y? If you look at the graph right here, x is greater than equal to negative 2. But what's why? What is the range of y-values that we can get here? Well, if you just look at the graph, y will always be greater than or equal to 1. And that just comes from the fact that this term right here is always going to be greater than or equal to 0. So the minimum value that the function could take on is 1. So we could say for x is greater than or equal to negative 2, and we could add that y is always going to be greater than or equal to 1. y is always greater than or equal to 1. The function is always greater than or equal to that right there. To 1. And the reason why I want to write it at the stage is" + }, + { + "Q": "\nat 7:12 what is the range of the inverse O_o ?", + "A": "The range of the inverse is the domain of the original function.", + "video_name": "aeyFb2eVH1c", + "timestamps": [ + 432 + ], + "3min_transcript": "You could rename it with anything really, is equal to the square root of x minus 1. Of x minus 1. Minus 2 for, we have to rename this to, for x being greater than or equal to 1. And so we now have our inverse function as a function of x. And if we were to graph it, let's try our best to graph it. Maybe the easiest thing to do is to draw some points here. So the smallest value x can take on is 1. If you put a 1 here, you get a 0 here. So the point 1, negative 2, is on our inverse graph. So 1, negative 2 is right there. And then if we go to 2, let's see, 2 minus 1 is 1. The principle root of that is 1. Minus 2. So it's negative 1, so the point, 2, negative 1 is right there. And let's think about it. If we did 5, I'm trying take perfect squares. 5 minus 1 is 4, minus 2. So the point 5, 2 is, let me make sure. 5 minus 1 is 4. Square root is 2. Minus 2 is 0. So the point 5, 0 is here. And so the inverse graph, it's only defined for x greater than or equal to negative 1. So the inverse graph is going to look something like this. It's going to look something like, I started off well, and it got messy. So it's going to look something like that. Just like that. And just like we saw, in the first, the introduction to function inverses, these are mirror images around the line y equals x. Let me graph y equals x. y equals x. y equals x is that line right there. Notice, they're mirror images around that line. Over here, we map the value 0 to 5. If x is 0, it gets mapped to 5. Here we go the other way. So that's why they're mirror images. We've essentially swapped the x and y." + }, + { + "Q": "ok,...at 1:04 does anyone understand moving the decimal?\n", + "A": "You multiply like you always do, & when you are done, take that answer and put in the decimal. Oh you want me to also tell you where the decimal goes? Sure, just count how many numbers where behind the decimal in the 2 numbers that you multiplied, and make sure that your answer has the same amount of numbers behind the new decimal. (Hint, when you get harder problems, your going to have to get comfortable putting extra zeros right behind the decimal, before your answer, sometimes!) -Cheers!", + "video_name": "JEHejQphIYc", + "timestamps": [ + 64 + ], + "3min_transcript": "We're asked to multiply 32.12, or 32 and 12 hundredths, times 0.5, or just 5 tenths. Now when you multiply decimals, you multiply them the exact same way you would multiply whole numbers, and then you count the number of spaces behind the decimal you have in your two numbers you're multiplying, and you're going to have that many spaces in your product. Let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You could almost ignore the decimal. Right now, you should write the decimal where they belong, but you can almost pretend that this is 3,212 times 5, and then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15, plus 1 is 16. And then we don't have any other places. If we were just doing this as 05, we wouldn't multiply 0 times this whole thing. We would just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal point in the two numbers we're multiplying. So we have one, two, three spaces, or three numbers, to the right of the decimals in the two numbers that we're So we need that many numbers to the right of the decimal in our answer. So we go one, two, three, put the decimal right over there. And this trailing zero right here we can ignore, because it's really not adding any information there. So we could just write this as 16.06. The last thing you want to do is just make sure that this makes sense. You have a number that's almost 32, and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1/2. So we're really multiplying 32.12 times 1/2. We're trying to figure out what one half of 32.12 is. And half of 32 is 16, and half of 0.12 0.06, so this makes complete sense." + }, + { + "Q": "\n@1:26, what do you do if there is a digit instead of a zero? Like say 32.12 multiplied by 35. I'm stuck on that part!", + "A": "You just add as many zero where there are no numbers for example 63.46 +02.59", + "video_name": "JEHejQphIYc", + "timestamps": [ + 86 + ], + "3min_transcript": "We're asked to multiply 32.12, or 32 and 12 hundredths, times 0.5, or just 5 tenths. Now when you multiply decimals, you multiply them the exact same way you would multiply whole numbers, and then you count the number of spaces behind the decimal you have in your two numbers you're multiplying, and you're going to have that many spaces in your product. Let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You could almost ignore the decimal. Right now, you should write the decimal where they belong, but you can almost pretend that this is 3,212 times 5, and then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15, plus 1 is 16. And then we don't have any other places. If we were just doing this as 05, we wouldn't multiply 0 times this whole thing. We would just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal point in the two numbers we're multiplying. So we have one, two, three spaces, or three numbers, to the right of the decimals in the two numbers that we're So we need that many numbers to the right of the decimal in our answer. So we go one, two, three, put the decimal right over there. And this trailing zero right here we can ignore, because it's really not adding any information there. So we could just write this as 16.06. The last thing you want to do is just make sure that this makes sense. You have a number that's almost 32, and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1/2. So we're really multiplying 32.12 times 1/2. We're trying to figure out what one half of 32.12 is. And half of 32 is 16, and half of 0.12 0.06, so this makes complete sense." + }, + { + "Q": "at 0:07 the problem has a dot in between 32.12 and 0.5 what does that dot mean?\n", + "A": "It is the same as the multiplication sign. Once you start using variables, it gets really confusing.", + "video_name": "JEHejQphIYc", + "timestamps": [ + 7 + ], + "3min_transcript": "We're asked to multiply 32.12, or 32 and 12 hundredths, times 0.5, or just 5 tenths. Now when you multiply decimals, you multiply them the exact same way you would multiply whole numbers, and then you count the number of spaces behind the decimal you have in your two numbers you're multiplying, and you're going to have that many spaces in your product. Let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You could almost ignore the decimal. Right now, you should write the decimal where they belong, but you can almost pretend that this is 3,212 times 5, and then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15, plus 1 is 16. And then we don't have any other places. If we were just doing this as 05, we wouldn't multiply 0 times this whole thing. We would just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal point in the two numbers we're multiplying. So we have one, two, three spaces, or three numbers, to the right of the decimals in the two numbers that we're So we need that many numbers to the right of the decimal in our answer. So we go one, two, three, put the decimal right over there. And this trailing zero right here we can ignore, because it's really not adding any information there. So we could just write this as 16.06. The last thing you want to do is just make sure that this makes sense. You have a number that's almost 32, and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1/2. So we're really multiplying 32.12 times 1/2. We're trying to figure out what one half of 32.12 is. And half of 32 is 16, and half of 0.12 0.06, so this makes complete sense." + }, + { + "Q": "\nAt 1:00 she says \"two separate... um, two separate...\". What does actually result from cutting it like so?", + "A": "its a mobius strip.", + "video_name": "AmN0YyaTD60", + "timestamps": [ + 60 + ], + "3min_transcript": "Hexaflexagons-- they're cool, hip, and hexa-fun to play with, right? Wrong. Hexaflexagons are not toys. With the increasing number of hexaflexagons finding their way into homes and schools, it's important to be aware of proper flexagation regulations when engaging in flexagon construction and use. Taking proper precautions can help avoid a flexa-catastrophe. Do not wear loose clothing when engaging in flexagation. If you have long hair, tie it back, so it doesn't get caught in a flexagation device. Ties are also a common source of incidents. Stay alert. Never flexagate while under the influence. When using a hexaflexagon, sudden unexpected sides may appear, and drugs like alcohol can slow reaction time. If you aren't sure what kind of flexagon you're dealing with, it's safer to temporarily disable the flexagon. Flexagons can be disarmed by using scissors to cut them apart. You can cut across the original seam where the paper strip was taped together, which may appear on the edge or through the face of the flexagon. In an emergency, however, flexagons can be cut apart right through a triangle, or on three edges if you want to retain symmetry, or into nine separate triangles if you really want to be safe. You can even cut them in half down the length of the paper strip like this, into two separate-- you can figure out what kind it is. If it has nine triangles, that's 18 triangle sides. So at six triangles per hexagon side, that's three sides of trihexaflexagon. Note that some flexagons might be made from a double strip of triangles that have been folded in half, so that marker doesn't bleed through. Don't let yourself be fooled by the extra triangles. Avoid danger during hexaflexagon construction. If you're not working from a printed pattern, you might start your flexagon by picking a point on the edge of a strip of paper, folding that 180 degree angle into thirds to create 360 degree angles, and then using the equilateral triangle that results as a guide to fold the rest of the strip of paper, zigzagging back and forth. Without proper attention and focus, this could easily lead to becoming unreasonably amused with the springy spring of happy triangles that results. Always keep your hexaflexagon in good working order. Pre-creasing all the triangles both ways before configuring them into hexaflexagonal formation will help your flexagon operate properly and avoid accidents. Keep a close watch on the chirality of your hexaflexagon. That is, whether it is right or left handed. clockwise down under the flaps, even if you flip it over or flex it. Well, in this hexaflexagon, it flows counter-clockwise. They're mirror images. The chirality is decided when you fold and tape your triangles into a twisty loop, and once taped, it is impossible to change from one to the other without cutting it apart, at least in three-dimensional euclidean space. A change in chirality could be a sign that your flexagon has been flipped through four-dimensional space and is possibly a highly dangerous multi-dimensional portal. With experience, a hexaflexagon master can construct a hexaflexagon in mere seconds. Some forgo tape and scissors entirely by folding a double strip that's too long and tucking the extra in. This is an advanced technique that should not be attempted without prior training. Beware topological changes. This family seems safe from this philosoraptor, because they live on separate planets with a cold, empty vacuum of space between them. But after a single flex, the unfortunate victims are now doomed, protected only by the inconsequential barrier of their domicile. Your stars might explode, your frowns may become smiles," + }, + { + "Q": "did anyone notice the mistake in 1:01 - 1:02\n", + "A": "I honestly couldn t because she was talking too fast for me to understand.", + "video_name": "AmN0YyaTD60", + "timestamps": [ + 61, + 62 + ], + "3min_transcript": "Hexaflexagons-- they're cool, hip, and hexa-fun to play with, right? Wrong. Hexaflexagons are not toys. With the increasing number of hexaflexagons finding their way into homes and schools, it's important to be aware of proper flexagation regulations when engaging in flexagon construction and use. Taking proper precautions can help avoid a flexa-catastrophe. Do not wear loose clothing when engaging in flexagation. If you have long hair, tie it back, so it doesn't get caught in a flexagation device. Ties are also a common source of incidents. Stay alert. Never flexagate while under the influence. When using a hexaflexagon, sudden unexpected sides may appear, and drugs like alcohol can slow reaction time. If you aren't sure what kind of flexagon you're dealing with, it's safer to temporarily disable the flexagon. Flexagons can be disarmed by using scissors to cut them apart. You can cut across the original seam where the paper strip was taped together, which may appear on the edge or through the face of the flexagon. In an emergency, however, flexagons can be cut apart right through a triangle, or on three edges if you want to retain symmetry, or into nine separate triangles if you really want to be safe. You can even cut them in half down the length of the paper strip like this, into two separate-- you can figure out what kind it is. If it has nine triangles, that's 18 triangle sides. So at six triangles per hexagon side, that's three sides of trihexaflexagon. Note that some flexagons might be made from a double strip of triangles that have been folded in half, so that marker doesn't bleed through. Don't let yourself be fooled by the extra triangles. Avoid danger during hexaflexagon construction. If you're not working from a printed pattern, you might start your flexagon by picking a point on the edge of a strip of paper, folding that 180 degree angle into thirds to create 360 degree angles, and then using the equilateral triangle that results as a guide to fold the rest of the strip of paper, zigzagging back and forth. Without proper attention and focus, this could easily lead to becoming unreasonably amused with the springy spring of happy triangles that results. Always keep your hexaflexagon in good working order. Pre-creasing all the triangles both ways before configuring them into hexaflexagonal formation will help your flexagon operate properly and avoid accidents. Keep a close watch on the chirality of your hexaflexagon. That is, whether it is right or left handed. clockwise down under the flaps, even if you flip it over or flex it. Well, in this hexaflexagon, it flows counter-clockwise. They're mirror images. The chirality is decided when you fold and tape your triangles into a twisty loop, and once taped, it is impossible to change from one to the other without cutting it apart, at least in three-dimensional euclidean space. A change in chirality could be a sign that your flexagon has been flipped through four-dimensional space and is possibly a highly dangerous multi-dimensional portal. With experience, a hexaflexagon master can construct a hexaflexagon in mere seconds. Some forgo tape and scissors entirely by folding a double strip that's too long and tucking the extra in. This is an advanced technique that should not be attempted without prior training. Beware topological changes. This family seems safe from this philosoraptor, because they live on separate planets with a cold, empty vacuum of space between them. But after a single flex, the unfortunate victims are now doomed, protected only by the inconsequential barrier of their domicile. Your stars might explode, your frowns may become smiles," + }, + { + "Q": "1:29 shows that vi hart is just awesome\n", + "A": "This should be in Tips and Thanks, not Questions.", + "video_name": "AmN0YyaTD60", + "timestamps": [ + 89 + ], + "3min_transcript": "Hexaflexagons-- they're cool, hip, and hexa-fun to play with, right? Wrong. Hexaflexagons are not toys. With the increasing number of hexaflexagons finding their way into homes and schools, it's important to be aware of proper flexagation regulations when engaging in flexagon construction and use. Taking proper precautions can help avoid a flexa-catastrophe. Do not wear loose clothing when engaging in flexagation. If you have long hair, tie it back, so it doesn't get caught in a flexagation device. Ties are also a common source of incidents. Stay alert. Never flexagate while under the influence. When using a hexaflexagon, sudden unexpected sides may appear, and drugs like alcohol can slow reaction time. If you aren't sure what kind of flexagon you're dealing with, it's safer to temporarily disable the flexagon. Flexagons can be disarmed by using scissors to cut them apart. You can cut across the original seam where the paper strip was taped together, which may appear on the edge or through the face of the flexagon. In an emergency, however, flexagons can be cut apart right through a triangle, or on three edges if you want to retain symmetry, or into nine separate triangles if you really want to be safe. You can even cut them in half down the length of the paper strip like this, into two separate-- you can figure out what kind it is. If it has nine triangles, that's 18 triangle sides. So at six triangles per hexagon side, that's three sides of trihexaflexagon. Note that some flexagons might be made from a double strip of triangles that have been folded in half, so that marker doesn't bleed through. Don't let yourself be fooled by the extra triangles. Avoid danger during hexaflexagon construction. If you're not working from a printed pattern, you might start your flexagon by picking a point on the edge of a strip of paper, folding that 180 degree angle into thirds to create 360 degree angles, and then using the equilateral triangle that results as a guide to fold the rest of the strip of paper, zigzagging back and forth. Without proper attention and focus, this could easily lead to becoming unreasonably amused with the springy spring of happy triangles that results. Always keep your hexaflexagon in good working order. Pre-creasing all the triangles both ways before configuring them into hexaflexagonal formation will help your flexagon operate properly and avoid accidents. Keep a close watch on the chirality of your hexaflexagon. That is, whether it is right or left handed. clockwise down under the flaps, even if you flip it over or flex it. Well, in this hexaflexagon, it flows counter-clockwise. They're mirror images. The chirality is decided when you fold and tape your triangles into a twisty loop, and once taped, it is impossible to change from one to the other without cutting it apart, at least in three-dimensional euclidean space. A change in chirality could be a sign that your flexagon has been flipped through four-dimensional space and is possibly a highly dangerous multi-dimensional portal. With experience, a hexaflexagon master can construct a hexaflexagon in mere seconds. Some forgo tape and scissors entirely by folding a double strip that's too long and tucking the extra in. This is an advanced technique that should not be attempted without prior training. Beware topological changes. This family seems safe from this philosoraptor, because they live on separate planets with a cold, empty vacuum of space between them. But after a single flex, the unfortunate victims are now doomed, protected only by the inconsequential barrier of their domicile. Your stars might explode, your frowns may become smiles," + }, + { + "Q": "I calculated the answer to the formula at 12:33, and I got 0.004998. can someone explain the intuition here? I also tried calculating what would be p(x=9), and I got ~0.13\n", + "A": "OK, so I used excel to graph all possible results from X=0 => X=20. I got a nice bell curve, and the sum inside is indeed 1...pretty cool. Indeed 13% (for 9, 10) seems to be the most likely outcome.", + "video_name": "Jkr4FSrNEVY", + "timestamps": [ + 753 + ], + "3min_transcript": "As this approaches infinity, this term's going So you have 1 to the minus k. 1 to any power is 1, so that term becomes 1. So we have another 1 there. So there you have it. We're done. The probability that our random variable, the number of cars that passes in an hour, is equal to a particular number. You know, it's equal to 7 cars pass in an hour. Is equal to the limit as n approaches infinity of n choose k times-- well, we said it was lambda over n to the k successes times 1 minus lambda over n to the n minus k failures. And we just showed that this is equal to lambda to the kth power over k factorial times e to the minus lambda. And that's pretty neat because when you just see it in kind of a vacuum-- if you have no context for it, you wouldn't binomial theorem. I mean, it's got an e in there. It's got a factorial, but a lot of things have factorials in life, so not clear that that would make it But this is just the limit as you take smaller and smaller and smaller intervals, and the probability of success in each interval becomes smaller. But as you take the limit you end up with e. And if you think about it it makes sense because one of our derivations of e actually came out of compound interest and we kind of did something similar there. We took smaller and smaller intervals of compounding and over each interval we compounded by a much smaller number. And when you took the limit you got e again. And that's actually where that whole formula up here came from to begin with. But anyway, just so that you know how to use this thing. So let's say that I were to go out, I'm the traffic engineer, and I figure out that on average, 9 cars passed per hour. And I want to know the probability that-- so this is my expected value. In a given hour, on average, 9 cars are passing. pass in a given hour, exactly 2 cars pass. That's going to be equal to 9 cars per hour to the 2'th power or squared, to the 2'th power. Divided by 2 factorial times e to the minus 9 power. So it's equal to 81 over 2 times e to the minus 9 power. And let's see, maybe I should just get the graphing Well, I'll let you do that exercise to figure out what that is, but I'll see you in the next video." + }, + { + "Q": "At 1:50 you said that 800 + 0 + 4 is 804. That is correct, but what is the reason to say 800 + 0 + 4 if the 0 doesn't have any value?\n", + "A": "He s trying to reinforce the concept that the numbers have different values based on their place, and how we keep track while we multiply. If it was 231 instead, it would work out to 800 + 30 + 4.", + "video_name": "4BtXvopHXI8", + "timestamps": [ + 110 + ], + "3min_transcript": "Let's multiply 4 times 2,012. Actually, let's make it a little bit simpler. Let's multiply 4 times 201 just to simplify things a little bit. So 4 times 201. So as we've seen in previous videos, I like to write the larger number on top. This is just one of many ways of tackling a calculation like this. I'll write the 201. And then I'll write the 4 right below it, and I'll write it right below the ones place. And so I have 201 times 4. Now, just like we did when we were multiplying a one digit times a two digit, we do essentially the same process. We first multiply 4 times the 1. Well, 4 times 1 we know is equal to 4. So we put a 4 right over there in the ones place. Then we can multiply our 4 times the digit that we have in the tens place. In this case, we have a 0 in the tens place. So 4 times 0, well, that's just 0. 4 times 0 is 0. We put the 0 in the tens place right over here. And then last, we have 4 times this 2 right over here. And we put the 8 right over here. And we get our answer-- 804. Now, why did this work? Well, remember, when we multiplied 4 times 1, that was literally just 4. And we've got that 4 right over here. When we multiply 4 times 0, that's 0 tens. So we've got 0 tens right over here. And when we multiplied 4 times 2, this was actually a 200. It's in the hundreds place. So 4 times 200 is 800. So what we're essentially doing by writing it in the right place is we're saying, 4 times 201, that's the same thing as 4 times 200, which is 800, plus 4 times 0 tens, which is 0 tens, plus 4 times 1, which is 4. So 800 plus 0 plus 4 is 804." + }, + { + "Q": "\nAt 8:13, he says 1/2^2 is just 1/2. but isn't that 1/4 since it's 1/2 x 1/2?? and he did the same thing for 1/3. shouldn't the answer have been 1/27?", + "A": "it s only squaring one in the numerator, not the quotient as a whole. so (1^2)/2 is just 1/2", + "video_name": "vhMl755vR5Q", + "timestamps": [ + 493 + ], + "3min_transcript": "to be equal to the value of this inner function, which is just x. And so this is just y is equal to x. And then we're going to multiply it times the depth, times the depth of each of these disks. And each of these disks are going to have a depth of dx. If you imagine a quarter that has an infinitely-thin depth right over here. So it's going to be dx. And so the volume our, kind of our truffle with a cone carved out, is going to be this integral minus this integral right over here. And we could evaluate it just like that. Or we could even say, OK we could factor out a pi out of both of them. There are actually, there's multiple ways But let's just evaluate it like this, and then I'll generalize it in the next video. So this is going to be equal to the definite integral from 0 to 1. You take the pi outside. Square root of x squared is going to be x dx minus the integral, we can factor the pi out. And we could say this is going to be equal to pi times the antiderivative of x, which is just x squared over 2 evaluated from 0 to 1, minus pi, times the antiderivative of x squared, which is x to the third over 3 evaluated from 0 to 1. This expression is equal to-- and I'm going to arbitrarily switch colors just because the green's getting monotonous-- pi times 1 squared over 2 minus 0 squared over 2. I could write squared. 1 squared over 2 minus 0 squared over 2, minus pi times 1 to the third over 3 minus 0 to the third over 3. me do it in that same blue color-- so this is this simplified. This is just 0 right over here. This is 1 squared over 2, which is just 1/2. So it's just pi over 2, 1/2 times pi minus-- well this is just 0, this is 1/3, minus pi over 3. And then to simplify this, it's just really subtracting fractions. So we can find a common denominator. Common denominator is 6. This is going to be 3 pi over 6. This is 3 pi over 6 minus 2 pi over 6. pi over 3 is 2 pi over 6, pi over 2 is 3 pi over 6. And we end up with, we end up with 3 of something minus 2 of something, you end up with 1 of something. We end up with 1 pi over 6. And we are done. We were able to find the volume of that wacky kind" + }, + { + "Q": "AT 0:56 on the video you multiply all numbers and how do you get the second number\n", + "A": "When Sal does 6x8x7, instead of doing 48x7, he does 40x7, which is 280, and 8x7, which is 56. Adding the two up gets 336. You get the same answer doing 6x8x7, or 48x7.", + "video_name": "I9efKVtLCf4", + "timestamps": [ + 56 + ], + "3min_transcript": "What is the volume of this box? Drag on the box to rotate it. So this is pretty neat. We can actually sit and rotate this box. And here it looks like everything's being measured in meters. So we want to measure our volume in terms of cubic meters. That's going to be our unit cube here. So when we want to think about how many cubic meters could fit in this box, we've already seen examples. You really just have to multiply the three different dimensions of this box. So if you wanted the number of cubic meters that could fit in here, it's going to be six meters times 8 meters times 7 meters which is going to give you something in cubic meters. So let's think about what that is. 6 times 8 is 48. Let me see if I can do this in my head. 48 times 7, that's 40 times 7, which is going to be 280 plus 8 times 7, which is 56, Let's check our answer. Let's do one more of these. So what's the volume of this box? We'll once again, we have its height at six feet. Now everything is being measured in feet. We have it's width being four feet. So we could multiple the height times the width of four feet. And then we can multiply that times its depth of two feet. So 6 times 4 is 24 times 2 is 48 feet. 48, and I should say cubic feet. We're saying how many cubic feet can fit in here? When we multiply the various dimensions measured in feet, we're counting almost how many of those cubic feet can fit into this box." + }, + { + "Q": "At 2:08, if the x-axis is inches, and later on it is shown that the area under the curve is the probability, what is the y-axis representative of, as far as units or meaning?\n", + "A": "Y is the rate of change of the area as you change the range of desired outcomes (x). E.g. if y is .5 then probability (area) is increasing at the rate of .5 per additional inch of rain in the interval of chosen outcomes.", + "video_name": "Fvi9A_tEmXQ", + "timestamps": [ + 128 + ], + "3min_transcript": "In the last video, I introduced you to the notion of-- well, really we started with the random variable. And then we moved on to the two types of random variables. You had discrete, that took on a finite number of values. And the these, I was going to say that they tend to be integers, but they don't always have to be integers. You have discrete, so finite meaning you can't have an infinite number of values for a discrete random variable. And then we have the continuous, which can take on an infinite number. And the example I gave for continuous is, let's say random variable x. And people do tend to use-- let me change it a little bit, just so you can see it can be something other than an x. Let's have the random variable capital Y. They do tend to be capital letters. Is equal to the exact amount of rain tomorrow. It's actually raining quite hard right now. We're short right now, so that's a positive. We've been having a drought, so that's a good thing. But the exact amount of rain tomorrow. And let's say I don't know what the actual probability distribution function for this is, but I'll draw one and then we'll interpret it. Just so you can kind of think about how you can think about continuous random variables. So let me draw a probability distribution, or they call it its probability density function. And we draw like this. And let's say that there is-- it looks something like this. Like that. All right, and then I don't know what this height is. So the x-axis here is the amount of rain. this is 3 inches, 4 inches. And then this is some height. Let's say it peaks out here at, I don't know, let's say this 0.5. So the way to think about it, if you were to look at this and I were to ask you, what is the probability that Y-- because that's our random variable-- that Y is exactly equal to 2 inches? That Y is exactly equal to two inches. What's the probability of that happening? Well, based on how we thought about the probability distribution functions for the discrete random variable, you'd say OK, let's see. 2 inches, that's the case we care about right now. Let me go up here. You'd say it looks like it's about 0.5." + }, + { + "Q": "\nAt 2:00, is .5 a bad example for the height? In a continuous random variable, it's very unlikely any observation is this common. If the y axis is not probability of a particular rainfall level, what is it?", + "A": "the rainfall amounts are on the x axis. The probability is for an interval (range of possible outcomes) and is the area under the curve for that interval. Y is the rate of change of probability (area) for an interval as it changes limits (as x changes).", + "video_name": "Fvi9A_tEmXQ", + "timestamps": [ + 120 + ], + "3min_transcript": "In the last video, I introduced you to the notion of-- well, really we started with the random variable. And then we moved on to the two types of random variables. You had discrete, that took on a finite number of values. And the these, I was going to say that they tend to be integers, but they don't always have to be integers. You have discrete, so finite meaning you can't have an infinite number of values for a discrete random variable. And then we have the continuous, which can take on an infinite number. And the example I gave for continuous is, let's say random variable x. And people do tend to use-- let me change it a little bit, just so you can see it can be something other than an x. Let's have the random variable capital Y. They do tend to be capital letters. Is equal to the exact amount of rain tomorrow. It's actually raining quite hard right now. We're short right now, so that's a positive. We've been having a drought, so that's a good thing. But the exact amount of rain tomorrow. And let's say I don't know what the actual probability distribution function for this is, but I'll draw one and then we'll interpret it. Just so you can kind of think about how you can think about continuous random variables. So let me draw a probability distribution, or they call it its probability density function. And we draw like this. And let's say that there is-- it looks something like this. Like that. All right, and then I don't know what this height is. So the x-axis here is the amount of rain. this is 3 inches, 4 inches. And then this is some height. Let's say it peaks out here at, I don't know, let's say this 0.5. So the way to think about it, if you were to look at this and I were to ask you, what is the probability that Y-- because that's our random variable-- that Y is exactly equal to 2 inches? That Y is exactly equal to two inches. What's the probability of that happening? Well, based on how we thought about the probability distribution functions for the discrete random variable, you'd say OK, let's see. 2 inches, that's the case we care about right now. Let me go up here. You'd say it looks like it's about 0.5." + }, + { + "Q": "\n8:50 This might be a stupid question, but is a probability distribution function related only to discrete random variables and a probability density function is for continuous random variables?", + "A": "From Wikipedia: The terms probability distribution function and probability function have also sometimes been used to denote the [continuous] probability density function. However, this use is not standard among probabilists and statisticians. I don t believe that a probability distribution of a discrete random variable is referred to as a function. Maybe it could be made into a step function, but I don t know if that is common.", + "video_name": "Fvi9A_tEmXQ", + "timestamps": [ + 530 + ], + "3min_transcript": "is pretty much 0. That you really have to say, OK what's the probably that we'll get close to 2? And then you can define an area. And if you said oh, what's the probability that we get someplace between 1 and 3 inches of rain, then of course the probability is much higher. The probability is much higher. It would be all of this kind of stuff. You could also say what's the probability we have less than 0.1 of rain? Then you would go here and if this was 0.1, you would calculate this area. And you could say what's the probability that we have more than 4 inches of rain tomorrow? Then you would start here and you'd calculate the area in the curve all the way to infinity, if the curve has area all And hopefully that's not an infinite number, right? Then your probability won't make any sense. But hopefully if you take this sum it comes to some number. And we'll say there's only a 10% chance that you have more than 4 inches tomorrow. And all of this should immediately lead to one light events that might occur can't be more than 100%. Right? All the events combined-- there's a probability of 1 that one of these events will occur. So essentially, the whole area under this curve has to be equal to 1. So if we took the integral of f of x from 0 to infinity, this thing, at least as I've drawn it, dx should be equal to 1. For those of you who've studied calculus. For those of you who haven't, an integral is just the area under a curve. And you can watch the calculus videos if you want to learn a little bit more about how to do them. And this also applies to the discrete probability distributions. Let me draw one. The sum of all of the probabilities have to be equal to 1. And that example with the dice-- or let's say, since it's faster to draw, the coin-- the two probabilities have So this is 1, 0, where x is equal to 1 if we're heads or 0 if we're tails. Each of these have to be 0.5. Or they don't have to be 0.5, but if one was 0.6, the other would have to be 0.4. They have to add to 1. If one of these was-- you can't have a 60% probability of getting a heads and then a 60% probability of getting a tails as well. Because then you would have essentially 120% probability of either of the outcomes happening, which makes no sense at all. So it's important to realize that a probability distribution function, in this case for a discrete random variable, they all have to add up to 1. So 0.5 plus 0.5. And in this case the area under the probability density function also has to be equal to 1. Anyway, I'm all the time for now. In the next video I'll introduce you to the idea of an expected value. See you soon." + }, + { + "Q": "At 4:27 it is said that the probability of something happening is \"actually zero\". Which makes sense when you are talking about area under a curve, but it doesn't make intuitive sense because doesn't that mean that the probability of ANYTHING happening in a continuous variable is zero? That doesn't make sense because things are still going to happen, even though we are saying the probability is zero. So if it rains in California, the exact amount of rain that came down had a zero probability of happening?\n", + "A": "This is quite an interesting question! While all impossible events definitely have probability zero of occurring, the reverse is not always true! An event that has probability zero of occurring could still be possible, though extremely unlikely. The probability that a continuous random variable equals any specific, exact value is always zero, but the probability that a continuous random variable lies in a specific interval of real numbers could be greater than zero. Have a blessed, wonderful day!", + "video_name": "Fvi9A_tEmXQ", + "timestamps": [ + 267 + ], + "3min_transcript": "this is 3 inches, 4 inches. And then this is some height. Let's say it peaks out here at, I don't know, let's say this 0.5. So the way to think about it, if you were to look at this and I were to ask you, what is the probability that Y-- because that's our random variable-- that Y is exactly equal to 2 inches? That Y is exactly equal to two inches. What's the probability of that happening? Well, based on how we thought about the probability distribution functions for the discrete random variable, you'd say OK, let's see. 2 inches, that's the case we care about right now. Let me go up here. You'd say it looks like it's about 0.5. And I would say no, it is not a 0.5 chance. And before we even think about how we would interpret it visually, let's just think about it logically. What is the probability that tomorrow we have exactly 2 inches of rain? Not 2.01 inches of rain, not 1.99 inches of rain. Not 1.99999 inches of rain, not 2.000001 inches of rain. Exactly 2 inches of rain. I mean, there's not a single extra atom, water molecule above the 2 inch mark. And not as single water molecule below the 2 inch mark. It's essentially 0, right? It might not be obvious to you, because you've probably heard, oh, we had 2 inches of rain last night. But think about it, exactly 2 inches, right? Normally if it's 2.01 people will say that's 2. But we're saying no, this does not count. We want exactly 2. 1.99 does not count. Normally our measurements, we don't even have tools that can tell us whether it is exactly 2 inches. No ruler you can even say is exactly 2 inches long. At some point, just the way we manufacture things, there's going to be an extra atom on it here or there. So the odds of actually anything being exactly a certain measurement to the exact infinite decimal point is actually 0. The way you would think about a continuous random variable, you could say what is the probability that Y is almost 2? So if we said that the absolute value of Y minus is 2 is less than some tolerance? Is less than 0.1. And if that doesn't make sense to you, this is essentially just saying what is the probability that Y is greater than 1.9 and less than 2.1?" + }, + { + "Q": "\nat 1:55, why does he do it like a(a+b) + b(a+b) instead of (a x a) + (ab) + (ab) + (b x b)?", + "A": "He is just showing a simpler way of doing it..it looks a little less confusing, :) Hope that helps!", + "video_name": "xjkbR7Gjgjs", + "timestamps": [ + 115 + ], + "3min_transcript": "We're asked to simply, or expand (7x + 10) ^ 2 Now the first thing I will show you is exactly what you should NOT do, well there's this huge temptation. A lot of people will look at this and say oh, that's (7x)^2 + 10^2. This is WRONG. And I'll write it in caps. This is WRONG! What your brain is doing is thinking if I had 7x times 10 and I squared that, this would be (7x)^2 times 10^2. We aren't multiplying here, we're adding 7x to 10. So you can't just square each of these terms. I just wanted to highlight, this is completely wrong, and to see why it's wrong, you have to remind yourself that (7x + 10)^2 is the exact same thing as (7x + 10)(7x + 10). That's what it means to square something. So this is what it is, so we're really just multiplying a binomial, or two binomials, they just happen to be the same one, and you could use F.O.I.L., you could use the distributive method, but this is actually a special case: when you're squaring a binomial, so let's just think about it as a special case first then we can apply whatever we learn to this. So we could've just done it straight here, but I want to learn the general case so you can apply it to any problem that you might see. If I have (a+b) squared We already realised that it's not a squared plus b squared That is a plus b times a plus b. and now we can use the distributive property We can distribute this a + b times this a So we get, we get a times a plus b and we can distribute the a plus b times this b plus b times a plus b, and we distribute this a And I'm just swapping the order so it's the same as this. plus b times b which is b squared. These are the same or these are like terms. So we can add them. One of something plus another of that something will give you two of that something. 2 ab. We have a squared plus 2 ab plus b squared. So the pattern here, the pattern here, if I have a plus b squared it's equal to a squared plus 2 times the product of these numbers plus b squared. So over here I have seven x plus ten squared So this is going to be equal to seven x squared seven x squared plus 2 times the product of seven x and 10. 2 times seven x times 10 plus 10 squared. So, the difference between the right answer and the wrong answer is that you have this middle term here that you might have" + }, + { + "Q": "\nAt 0:23 isn't pretty redundant to point this out this late in the course wouldn't it help to emphasize this point earlier?", + "A": "Well not everyone watches all the videos on order. If people are just watching this specific video because they re not so strong at this topic, you don t know what they already know so it might be a helpful reminder for them", + "video_name": "xjkbR7Gjgjs", + "timestamps": [ + 23 + ], + "3min_transcript": "We're asked to simply, or expand (7x + 10) ^ 2 Now the first thing I will show you is exactly what you should NOT do, well there's this huge temptation. A lot of people will look at this and say oh, that's (7x)^2 + 10^2. This is WRONG. And I'll write it in caps. This is WRONG! What your brain is doing is thinking if I had 7x times 10 and I squared that, this would be (7x)^2 times 10^2. We aren't multiplying here, we're adding 7x to 10. So you can't just square each of these terms. I just wanted to highlight, this is completely wrong, and to see why it's wrong, you have to remind yourself that (7x + 10)^2 is the exact same thing as (7x + 10)(7x + 10). That's what it means to square something. So this is what it is, so we're really just multiplying a binomial, or two binomials, they just happen to be the same one, and you could use F.O.I.L., you could use the distributive method, but this is actually a special case: when you're squaring a binomial, so let's just think about it as a special case first then we can apply whatever we learn to this. So we could've just done it straight here, but I want to learn the general case so you can apply it to any problem that you might see. If I have (a+b) squared We already realised that it's not a squared plus b squared That is a plus b times a plus b. and now we can use the distributive property We can distribute this a + b times this a So we get, we get a times a plus b and we can distribute the a plus b times this b plus b times a plus b, and we distribute this a And I'm just swapping the order so it's the same as this. plus b times b which is b squared. These are the same or these are like terms. So we can add them. One of something plus another of that something will give you two of that something. 2 ab. We have a squared plus 2 ab plus b squared. So the pattern here, the pattern here, if I have a plus b squared it's equal to a squared plus 2 times the product of these numbers plus b squared. So over here I have seven x plus ten squared So this is going to be equal to seven x squared seven x squared plus 2 times the product of seven x and 10. 2 times seven x times 10 plus 10 squared. So, the difference between the right answer and the wrong answer is that you have this middle term here that you might have" + }, + { + "Q": "At 3:15, what was wrong with leaving 7x^2 as is? When he simplified it, he still got 49x^2 so it didn't get rid of the x or the square, so i don't know why he did it. But it seems like it changes the answer, so could someone help me understand?\nThanks in advance.\n", + "A": "He was solving not simplifying. Distribute the ^2 in (7x)^2 and you have 7 squared, which is 49, and x squared, which is x^2.", + "video_name": "xjkbR7Gjgjs", + "timestamps": [ + 195 + ], + "3min_transcript": "So this is what it is, so we're really just multiplying a binomial, or two binomials, they just happen to be the same one, and you could use F.O.I.L., you could use the distributive method, but this is actually a special case: when you're squaring a binomial, so let's just think about it as a special case first then we can apply whatever we learn to this. So we could've just done it straight here, but I want to learn the general case so you can apply it to any problem that you might see. If I have (a+b) squared We already realised that it's not a squared plus b squared That is a plus b times a plus b. and now we can use the distributive property We can distribute this a + b times this a So we get, we get a times a plus b and we can distribute the a plus b times this b plus b times a plus b, and we distribute this a And I'm just swapping the order so it's the same as this. plus b times b which is b squared. These are the same or these are like terms. So we can add them. One of something plus another of that something will give you two of that something. 2 ab. We have a squared plus 2 ab plus b squared. So the pattern here, the pattern here, if I have a plus b squared it's equal to a squared plus 2 times the product of these numbers plus b squared. So over here I have seven x plus ten squared So this is going to be equal to seven x squared seven x squared plus 2 times the product of seven x and 10. 2 times seven x times 10 plus 10 squared. So, the difference between the right answer and the wrong answer is that you have this middle term here that you might have And this comes out when you are multiplying all the different combinations of the terms here. if we simplify this, if we simplify seven x squared That's seven squared times x squared. So seven squared is 49 times x squared When you multiply this part out 2 times 7 times 10 which is 140 and then we have our x. No other x there. And then plus 10 squared. So plus 100. And we are done." + }, + { + "Q": "\nFrom 6:30 to 6:56, Sal lists non linear equations Why are they nonlinear?", + "A": "A linear equation is only one where x is to the power of 1, generally some form of y=x Its graph is a straight line. In all the equations listed, x was being raised to a power that was either less than or greater than 1 - for example, x^2, or 1/x, which is the same as x^-1. If you look at the graphs of any of those equations, they re curves, not straight lines.", + "video_name": "AOxMJRtoR2A", + "timestamps": [ + 390, + 416 + ], + "3min_transcript": "And it's indeed-- that's indeed the case. Two times five is ten, minus three is seven. The point-- the point five comma seven is on, or it satisfies this linear equation. So if you take all of the xy pairs that satisfy it, you get a line. That is why it is called a linear equation. Now, this isn't the only way that we could write a linear equation. You could write a linear equation like-- let me do this in a new color. You could write a linear equation like this: Four x minus three y is equal to twelve. This also is a linear equation. And we can see that if we were to graph the xy pairs that satisfy this we would once again get a line. X and y. If x is equal to zero, then this goes away and you have negative three y is equal to twelve. Let's see, if negative three y equals twelve then y would be equal to negative four. You can verify that. Four times zero minus three times negative four well that's gonna be equal to positive twelve. And let's see, if y were to equal zero, if y were to equal zero then this is gonna be four times x is equal to twelve, well then x is equal to three. And so you have the point zero comma negative four, zero comma negative four on this line, and you have the point three comma zero on this line. Three comma zero. Did I do that right? So zero comma negative four and then three comma zero. These are going to be on this line. Three comma zero is also on this line. So this is, this line is going to look something like-- something like, I'll just try to hand draw it. Something like that. So once again, all of the xy-- all of the xy pairs that satisfy this, Now what are some examples, maybe you're saying \"Wait, wait, wait, isn't any equation a linear equation?\" And the simple answer is \"No, not any equation is a linear equation.\" I'll give you some examples of non-linear equations. So a non-- non-linear, whoops let me write a little bit neater than that. Non-linear equations. Well, those could include something like y is equal to x-squared. you will see that this is going to be a curve. it could be something like x times y is equal to twelve. This is also not going to be a line. Or it could be something like five over x plus y is equal to ten. This also is not going to be a line. So now, and at some point you could-- I encourage you to try to graph these things, they're actually quite interesting. But given that we've now seen examples of linear equations and non-linear equations," + }, + { + "Q": "\nat 0:36 he explains that negitive 3p minus p is equal to negative 4 p. how is that possible.", + "A": "To combine like terms, you add / subtract the coefficients (the numbers in front of the variables). The coefficient of -3p is -3 The coefficient of - p is -1 -3 - 1 = -4 If you don t get this part: -- use a number line: got to -3. To take away 1, you move 1 unit to the left. You will now be on -4. -- and, you need to review the lessons on adding and subtracting negative numbers. Thus, -3p - p = -4p Hope this helps.", + "video_name": "SgKBBUFaGb4", + "timestamps": [ + 36 + ], + "3min_transcript": "We're asked to solve for p. And we have the inequality here negative 3p minus 7 is less than p plus 9. So what we really want to do is isolate the p on one side of this inequality. And preferably the left-- that just makes it just a little easier to read. It doesn't have to be, but we just want to isolate the p. So a good step to that is to get rid of this p on the right-hand side. And the best way I can think of doing that is subtracting p from the right. But of course, if we want to make sure that this inequality is always going to be true, if we do anything to the right, we also have to do that to the left. So we also have to subtract p from the left. And so the left-hand side, negative 3p minus p-- that's negative 4p. And then we still have a minus 7 up here-- is going to be less than p minus p. Those cancel out. It is less than 9. Now the next thing I'm in the mood to do is get rid of this negative 7, or this minus 7 here, so that we can better isolate the p on the left-hand side. So the best way I can think of to get rid of a negative 7 is to add 7 to it. Then it will just cancel out to 0. So let's add 7 to both sides of this inequality. All we're left with is negative 4p. On the right-hand side, we have 9 plus 7 equals 16. And it's still less than. Now, the last step to isolate the p is to get rid of this negative 4 coefficient. And the easiest way I can think of to get rid of this negative 4 coefficient is to divide both sides by negative 4. So if we divide this side by negative 4, these guys are going to cancel out. We're just going to be left with p. We also have to do it to the right-hand side. Now, there's one thing that you really have to remember, since this is an inequality, not an equation. If you're dealing with an inequality and you multiply or divide both sides of an equation by a negative number, you have to swap the inequality. So in this case, the less than becomes greater than, since we're dividing by a negative number. And so negative 4 divided by negative 4-- those cancel out. We have p is greater than 16 divided by negative 4, which is negative 4. And then we can try out some values to help us feel good about the idea of it working. So let's say this is negative 5, negative 4, negative 3, negative 2, negative 1, 0. Let me write that a little bit neater. And then we can keep going to the right. And so our solution is p is not greater than or equal, so we have to exclude negative 4. p is greater than negative 4, so all the values above that. So negative 3.9999999 will work. Negative 4 will not work. And let's just try some values out to feel good that this is really the solution set. So first let's try out when p is equal to negative 3. This should work. The way I've drawn it, this is in our solution set. p equals negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative 3. The first negative 3 is this one," + }, + { + "Q": "\nat 1:32 why did sal attach the rope of the mast all the way at the top\nhow does he know where to put the rope on the mast?\nthanks for whoever answers my question in the future!!\nthanks future answers", + "A": "The question clearly states that the rope is attached to the top of the mast.", + "video_name": "JVrkLIcA2qw", + "timestamps": [ + 92 + ], + "3min_transcript": "The main mast of a fishing boat is supported by a sturdy rope that extends from the top of the mast to the deck. If the mast is 20 feet tall and the rope attaches to the deck 15 feet away from the base of the mast, how long is the rope? So let's draw ourselves a boat and make sure we understand what the deck and the mast and all of that is. So let me draw a boat. I'll start with yellow. So let's say that this is my boat. That is the deck of the boat. And the boat might look something like this. It's a sailing boat. This is the water down here. And then the mast is the thing that holds up the sail. So let me draw ourselves a mast. And they say the mast is 20 feet tall. So this distance right here is 20 feet. I can draw it as a pole so it's a little bit clearer. Even shade it in if we like. And then they say a rope attaches to the deck 15 feet away from the base of the mast. So this is the base of the mast. This is the deck right here. The rope attaches 15 feet away from the base of the mast. So if this is the base of the mast, we go 15 feet, might be about that distance right there. Let me mark that. And the rope attaches right here. From the top of the mast all the way that base. So the rope goes like that. And then they ask us, how long is the rope? So there's a few things you might realize. We're dealing with a triangle here. And it's not any triangle. We're assuming that the mast goes straight up and that the deck is straight left and right. So this is a right triangle. This is a 90 degree angle right here. we can always figure out the third side of a right triangle using the Pythagorean theorem. And all that tells us is it the sum of the squares of the shorter sides of the triangle are going to be equal to the square of the longer side. And that longer side is call the hypotenuse. And in all cases, the hypotenuse is the side opposite the 90 degree angle. It is always going to be the longest side of our right triangle. So we need to figure out the hypotenuse here. We know the lengths of the two shorter sides. So we can see that if we take 15 squared, that's one of the short sides, I'm squaring it. And then add that to the square of the other shorter side, to 20 feet squared." + }, + { + "Q": "i don't understand how this works does the dot such as in : 4'5 ( the ' means dot . look at 1:11 to 3:30 ) does that dot system work fo more than 1 digit numbers ??\n\nThanks\nYaz Lightning\n", + "A": "The dot is used for any number, large or small. It is mainly used in algebra, where x could be a variable, not an operator. Hope it helped!", + "video_name": "Yw3EoxC_GXU", + "timestamps": [ + 71, + 210 + ], + "3min_transcript": "Rewrite 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5 as a multiplication expression. And then they want us to write the expression three times using different ways to write multiplication. So let's do the first part. Let's write it as a multiplication expression. So how many times have we added 5 here? Well, we've got it at one, two, three, four, five, six, seven. So one way to think of it, if I just said what is here? How many 5's are there? You'd say, well, I added 5 to itself seven times, right? You could literally say this is 7 times 5. We could literally write, this is 7 times 5, or you could view it as 5 seven times. I'm not even writing it mathematically yet. I'm just saying, look, if I saw seven of something, you would literally say, if these were apples, you would say apples seven times, or you'd say seven times the apple, whatever it is. Now, in this case, we're actually adding the number to each other, and we could figure out what that is, and But the way we would write this mathematically, we would say this is 7 times 5. We could also write it like this. We could write it 7 dot 5. This and this mean the exact same thing. It means we're multiplying 7 times 5 or 5 times 7. You can actually switch the order, and you get the exact same value. You could actually write it 5 times 7. So you could interpret this as 7 five times or 5 seven times, however you like to do it, or 5 seven times. I don't want to confuse you. I just want to show you that these are all equivalent. This is also equivalent. 5 times 7. Same thing. You could write them in parentheses. You could write it like this. This all means the same thing. That's 7 times 5, and so is this. These all evaluate to the same thing: 5 times 7. So these are all equivalent, and since we've worked with it so much, let's just figure out the answer. So if we add up 5 to itself seven times, what do we get? Well, 5 plus 5 is 10. plus 5 is 35. So all of these evaluate to 35, just so you see that they're the same thing. These are all equivalent to 35. And just something to think about, this is also the exact same thing, depending on how you want to interpret this, as 7 five times. They didn't ask us to do it, but I thought I would point it out to you. 7 five times would look like this: 7 plus 7 plus 7 plus 7 plus 7, right? I have 7 five times. I added it to itself five different times. There's five 7's here added to each other. And when you add these up, you'll also get 35. And that's why 5 times 7 and 7 times 5 is the same thing." + }, + { + "Q": "At 1:14 that looks like a decimal point?\n", + "A": "Yes. The dot, point or period can be used to mean the decimal point, or it can be used instead of the multiplication symbol to mean multiplied by , just like x . You can usually tell from the context what the dot is being used for. As a decimal point, it is normally written at the bottom, where commas and periods go in writing regular sentences. As a multiplication symbol, it is normally written higher up, like the hyphen as in break-dance .", + "video_name": "Yw3EoxC_GXU", + "timestamps": [ + 74 + ], + "3min_transcript": "Rewrite 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5 as a multiplication expression. And then they want us to write the expression three times using different ways to write multiplication. So let's do the first part. Let's write it as a multiplication expression. So how many times have we added 5 here? Well, we've got it at one, two, three, four, five, six, seven. So one way to think of it, if I just said what is here? How many 5's are there? You'd say, well, I added 5 to itself seven times, right? You could literally say this is 7 times 5. We could literally write, this is 7 times 5, or you could view it as 5 seven times. I'm not even writing it mathematically yet. I'm just saying, look, if I saw seven of something, you would literally say, if these were apples, you would say apples seven times, or you'd say seven times the apple, whatever it is. Now, in this case, we're actually adding the number to each other, and we could figure out what that is, and But the way we would write this mathematically, we would say this is 7 times 5. We could also write it like this. We could write it 7 dot 5. This and this mean the exact same thing. It means we're multiplying 7 times 5 or 5 times 7. You can actually switch the order, and you get the exact same value. You could actually write it 5 times 7. So you could interpret this as 7 five times or 5 seven times, however you like to do it, or 5 seven times. I don't want to confuse you. I just want to show you that these are all equivalent. This is also equivalent. 5 times 7. Same thing. You could write them in parentheses. You could write it like this. This all means the same thing. That's 7 times 5, and so is this. These all evaluate to the same thing: 5 times 7. So these are all equivalent, and since we've worked with it so much, let's just figure out the answer. So if we add up 5 to itself seven times, what do we get? Well, 5 plus 5 is 10. plus 5 is 35. So all of these evaluate to 35, just so you see that they're the same thing. These are all equivalent to 35. And just something to think about, this is also the exact same thing, depending on how you want to interpret this, as 7 five times. They didn't ask us to do it, but I thought I would point it out to you. 7 five times would look like this: 7 plus 7 plus 7 plus 7 plus 7, right? I have 7 five times. I added it to itself five different times. There's five 7's here added to each other. And when you add these up, you'll also get 35. And that's why 5 times 7 and 7 times 5 is the same thing." + }, + { + "Q": "\nSal keeps on saying Thrill (cola) instead of Thrill (soda), right?\nI think its at 2:02 one of the times.", + "A": "well really that s not the point and it doesn t really matter, but yeah.", + "video_name": "gs-OPF3KEGU", + "timestamps": [ + 122 + ], + "3min_transcript": "Thrill Soda hired a marketing company to help them promote their brand against Yummy Cola. The company gathered the following data about consumers' preference of soda. So they have, year by year, percentage of respondents who preferred Yummy Cola, percentage of respondents who preferred Thrill Cola, and then these are people who had no preference. So in 2006, 80% liked Yummy, only 12% liked Thrill, and 8% didn't like either one or didn't have any preference. And so actually just from here you see that many, many more people liked Yummy Cola than Thrill Cola, actually every year over year. So, Thrill Cola definitely has something. They have an uphill battle. But then they said the advertising company created the following two graphs to promote Thrill Soda. And so let's see what's happening over here. And let's think about whether this is misleading or not. So if we look at this graph over here, in 2006, sure enough, 80% liked Yummy Cola. Then in 2007, 76%. Then it keeps going to then 77%, then 73%, then 73% to 68%. It actually represents the data that's given right over here. I'll do it in the same. It actually represents this data very faithfully. Then right over here, if we look at this chart, Percentage of People who Prefer Thrill Soda, so over here in 2006, 12% preferred Thrill Soda. 2007, 19%. 2008, 19%. Then we go up to 20%, 21%, and 25%. So the graphs are actually accurate. They're not lying. These are actually the data points of the percentage who prefer Thrill Soda. Now what's misleading is if someone were to just look at these two graphs without actually looking at the scales over here, they'll see two things. They'll say, oh, look, you see a declining trend. And that's what line graphs are good for, for seeing trends. They say, look, I see a declining trend in the percentage of people who prefer Yummy Cola. And I see this increasing trend in the percentage And that's true. You have a declining trend here. And you have an increasing trend here. But what's misleading here is the way that they've plotted the scales. These scales are not the same. So when you look at this, you say not only is there an increasing trend of people who prefer Thrill Soda, but the way they set up the scale, it looks like the trend is above. The human brain is tempted to compare these and to say, look, not only is this an upward trend, but it's above this trend right over here. Even in 2006, this data point looks higher than these data points right over here. But the reality is that it's only because the scale is distorted. Now this is the oldest trick in the book when plotting line graphs. It all depends on the scale. So this just looks good because they used this scale that went from 0 to 30 as opposed to 0 to 100. The better thing to do, or the more genuine thing to do, or the more honest thing to do, would have actually been to plot them on the same graph. Although if they did that, that wouldn't have painted a very good picture for Thrill Soda. So if we plotted on the same graph Thrill Soda," + }, + { + "Q": "\n@ 3:00 we play, what seems to me to be, a dirty trick and move the dx to before the e^x... term. How is that OK? If this is a valid algebraic tool then why not do that every time we have a product so that \u00e2\u0088\u00ab a \u00e2\u0080\u00a2 b \u00e2\u0080\u00a2 c \u00e2\u0080\u00a2 dx can be simplified to \u00e2\u0088\u00ab a \u00e2\u0080\u00a2 dx \u00e2\u0080\u00a2 b \u00e2\u0080\u00a2 c? This of course isn't right because we could just insert a 1 x in the front of any expression and play the same trick, e.g. \u00e2\u0088\u00ab a = \u00e2\u0088\u00ab 1 \u00e2\u0080\u00a2 a dx = \u00e2\u0088\u00ab 1 dx a = x \u00e2\u0080\u00a2 a + C which can't be right.", + "A": "A notationally better version of your equation is \u00e2\u0088\u00ab a \u00e2\u0080\u00a2 dx = \u00e2\u0088\u00ab a \u00e2\u0080\u00a2 1 \u00e2\u0080\u00a2 dx = a \u00e2\u0080\u00a2 \u00e2\u0088\u00ab 1 \u00e2\u0080\u00a2 dx = a \u00e2\u0080\u00a2 x + C , so your example equation is essentially correct, apart from forgetting the dx in the first integral.", + "video_name": "b76wePnIBdU", + "timestamps": [ + 180 + ], + "3min_transcript": "But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared, and this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is 3x squared, derivative of x squared is 2x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now, what is going to be the derivative of u with respect to x? du dx. Well, we've done this multiple times. It's going to be 3x squared plus 2x. And now we can write this in differential form. And du dx, this isn't really a fraction of the differential It really is a form of notation, but it is often useful to kind of pretend that it is a fraction, and you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here, how much does u change for a given change in x? You could multiply both sides times a dx. So both sides times a dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to 3x squared plus 2x dx. Now why is this over here? Why did I go to the trouble of doing that? Well we see we have a 3x squared plus 2x, and then it's being multiplied by a dx right over here. I could rewrite this as the integral of-- and let me do it in that color-- of 3x squared plus 2x times dx times e-- let me do that in that other color-- times e to the x to the third plus x squared. Now what's interesting about this? Well the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit, it's going to be equal to-- and what I'm going to do is I'm going to change the order. I'm going to put the du, this entire du, I'm gonna stick it on the other side here, so it looks like more of the standard form that we're used to seeing our indefinite integrals in." + }, + { + "Q": "\nAt time 1:44 why do you swap the sign if you aren't dividing by a negative number to solve the right side of the inequality? How come the sign switches from less than or equal to, to greater than or equal to? what is the logic behind that?", + "A": "He moved all of the constants from the right side of the inequality to the left. He left the variable on the right side if the variable is on the right side you swap the sign. I just isolate my variable and bring it to the left and all the constants on the right it s a lot less confusing! Hope that helped!!", + "video_name": "A3xPhzs-KBI", + "timestamps": [ + 104 + ], + "3min_transcript": "Let's do some compound inequality problems, and these are just inequality problems that have more than one set of constraints. You're going to see what I'm talking about in a second. So the first problem I have is negative 5 is less than or equal to x minus 4, which is also less than or equal to 13. So we have two sets of constraints on the set of x's that satisfy these equations. x minus 4 has to be greater than or equal to negative 5 and x minus 4 has to be less than or equal to 13. So we could rewrite this compound inequality as negative 5 has to be less than or equal to x minus 4, and x minus 4 needs to be less than or equal to 13. And then we could solve each of these separately, and then we have to remember this \"and\" there to think about the equation and this equation. So let's solve each of them individually. So this one over here, we can add 4 to both sides of the equation. The left-hand side, negative 5 plus 4, is negative 1. Negative 1 is less than or equal to x, right? These 4's just cancel out here and you're just left with an x on this right-hand side. So the left, this part right here, simplifies to x needs to be greater than or equal to negative 1 or negative 1 is less than or equal to x. So we can also write it like this. X needs to be greater than or equal to negative 1. I just swapped the sides. Now let's do this other condition here in green. Let's add 4 to both sides of this equation. And then the right-hand side, we get 13 plus 14, which is 17. So we get x is less than or equal to 17. So our two conditions, x has to be greater than or equal to negative 1 and less than or equal to 17. So we could write this again as a compound inequality if we want. We can say that the solution set, that x has to be less than or equal to 17 and greater than or equal to negative 1. It has to satisfy both of these conditions. So what would that look like on a number line? So let's put our number line right there. Let's say that this is 17. Maybe that's 18. You keep going down. Maybe this is 0. I'm obviously skipping a bunch of stuff in between. Then we would have a negative 1 right there, maybe a negative 2. So x is greater than or equal to negative 1, so we would" + }, + { + "Q": "\nwhy does he add 4 to every equation at 1:30?", + "A": "He does so to isolate x. By adding 4 to both sides, the x - 4 becomes x because (x - 4) + 4 = x.", + "video_name": "A3xPhzs-KBI", + "timestamps": [ + 90 + ], + "3min_transcript": "Let's do some compound inequality problems, and these are just inequality problems that have more than one set of constraints. You're going to see what I'm talking about in a second. So the first problem I have is negative 5 is less than or equal to x minus 4, which is also less than or equal to 13. So we have two sets of constraints on the set of x's that satisfy these equations. x minus 4 has to be greater than or equal to negative 5 and x minus 4 has to be less than or equal to 13. So we could rewrite this compound inequality as negative 5 has to be less than or equal to x minus 4, and x minus 4 needs to be less than or equal to 13. And then we could solve each of these separately, and then we have to remember this \"and\" there to think about the equation and this equation. So let's solve each of them individually. So this one over here, we can add 4 to both sides of the equation. The left-hand side, negative 5 plus 4, is negative 1. Negative 1 is less than or equal to x, right? These 4's just cancel out here and you're just left with an x on this right-hand side. So the left, this part right here, simplifies to x needs to be greater than or equal to negative 1 or negative 1 is less than or equal to x. So we can also write it like this. X needs to be greater than or equal to negative 1. I just swapped the sides. Now let's do this other condition here in green. Let's add 4 to both sides of this equation. And then the right-hand side, we get 13 plus 14, which is 17. So we get x is less than or equal to 17. So our two conditions, x has to be greater than or equal to negative 1 and less than or equal to 17. So we could write this again as a compound inequality if we want. We can say that the solution set, that x has to be less than or equal to 17 and greater than or equal to negative 1. It has to satisfy both of these conditions. So what would that look like on a number line? So let's put our number line right there. Let's say that this is 17. Maybe that's 18. You keep going down. Maybe this is 0. I'm obviously skipping a bunch of stuff in between. Then we would have a negative 1 right there, maybe a negative 2. So x is greater than or equal to negative 1, so we would" + }, + { + "Q": "\nWait....at like 3:10, in the vid, it says 54/10=5 remainder 4. So...since 4/10=.4, can we write it as 5.4? Are they the same thing?", + "A": "No, you can t. When someone says remainder 4 don t think it s .4. remainder 4 is going to be a whole number!", + "video_name": "STyoP3rCmb0", + "timestamps": [ + 190 + ], + "3min_transcript": "0.6 is 6 divided by 10. And this expression right over here, we could either do the 6 divided by 10 first, in which case we would get 0.6, and this would turn into this problem. Or we could do the 9 times 6 first. And so let's do 9 times 6, which we know how to calculate, and then divide by 10, which we also know how to do. That. all about just moving the decimal place. So we could write 9 times 6. 9 times 6, we already know, is 54. I'll do that in orange-- is going to be 54. So this right over here is 54. And now to get to this expression, we have to divide by 10. We have to divide by 10. And what happens when we divide something by 10? And we've seen this in previous videos, why this is the case. Each place represents 10 times as much as the place to its right, or each place represents 1/10 of the place So 54 divided by 10, this is going to be-- you could start with 54. And I'll put a 0 here after the decimal. And when you divide by 10, that's equivalent of shifting the decimal one to the left. This is going to be equal to 5.4. And that should make sense to you. 5 times 10 is 50. 0.4 times 10 is 4. So it makes sense that 54 divided by 10-- I shouldn't say equal. I'd write 54 divided by 10 is equal to 5.4. So this right over here is equal to 5.4, and that's what this is. This is equal to 5.4. Notice, 9 times 6 is 54. Now you might see a little pattern here. Between these two numbers, I had exactly one number to the right of the decimal. When I take its product, let's say I ignored the decimal. I just said 9 times 6, I would've gotten 54. But then I have to divide by 10 in order to take account of the decimal, take account of the fact this wasn't a 6. This was a 6/10. And so I have one number to the right of the decimal here. And I want to you to think about that whether that's a general principle. Can we just count the total numbers of digits to the right of the decimals and then our product is going to have the same number of digits to right of the decimal? I'll let you to think about that." + }, + { + "Q": "\nAt 7:57, why can we not simply (3 * X ^ 2) - (4 * X ^ -5) ? I know it's a stupid question, but I am really confused. Will the answer not be = -1X^7?", + "A": "You cannot subtract exponents like that. In order to do a subtraction, the exponent needs to match. Take a look below: 2^5 - 2^2 You re saying this would equal: 2^3 = 8 But if you evaluate it: 2^5 - 2^2 32 - 4 = 28 The exponents have to match in order for a subtraction or addition to work.", + "video_name": "mzOBlH32qdk", + "timestamps": [ + 477 + ], + "3min_transcript": "being the same thing as the derivative of f of x plus the derivative of g of x. So this is the same thing as f-- actually, let me use that derivative operator just to make it clear. It's the same thing as the derivative with respect to x of f of x plus the derivative with respect to x of g of x. So we'll put f of x right over here and put g of x right over there. And so with the other notation, we can say this is going to be the same thing. Derivative with respect to x of f of x, we can write as f prime of x. And the derivative with respect to x of g of x, we can write as g prime of x. Now, once again, this might look like kind of fancy notation to you. But when you see an example, it'll make it pretty clear. If I want to take the derivative with respect to x of let's say x to the third power tells us that the derivative of the sum is just the sum of the derivatives. So we can take the derivative of this term using the power rule. So it's going to be 3x squared. And to that, we can add the derivative of this thing right over here. So it's going to be plus-- that's a different shade of blue-- and over here is negative 4. So it's plus negative 4 times x to the negative 4 minus 1, or x to the negative 5 power. So we have-- and I could just simplify a little bit. This is going to be equal to 3x squared minus 4x to the negative 5. And so now we have all the tools we need in our toolkit to essentially take the derivative of any polynomial. So let's say that I have-- and I'll do it in white. Let's say that f of x is equal to 2x to the third power minus 7x squared plus 3x minus 100. What is f prime of x? What is the derivative of f with respect to x going to be? Well, we can use the properties that we just said. The derivative of this is just going to be 2 times the derivative of x to the third. Derivative of x to the third is going to be 3x squared, so it's just going to be 2 times 3x squared. What's the derivative of negative 7x squared going to be? Well, it's just going to be negative 7 times the derivative of x squared, which is 2x. What is the derivative of 3x going to be? Well, it's just going to be 3 times the derivative of x, or 3 times the derivative of x to the first." + }, + { + "Q": "\nSurely, at 0:30, the rule does also works if n = 0, since the derivative will equal 0.x^-1, which equals 0, and that is the derivative of a constant.", + "A": "It s pretty nitpicky, but 0 * x ^ (-1) is not always equal to 0. When x equals 0, it s undefined.", + "video_name": "mzOBlH32qdk", + "timestamps": [ + 30 + ], + "3min_transcript": "Now that we know the power rule, and we saw that in the last video, that the derivative with respect to x, of x to the n, is going to be equal to n times x to the n minus 1 for n not equal 0. I thought I would expose you to a few more rules or concepts or properties of derivatives that essentially will allow us to take the derivative of any polynomial. So this is powerful stuff going on. So the first thing I want to think about is, why this little special case for n not equaling 0? What happens if n equals 0? So let's just think of the situation. Let's try to take the derivative with respect to x of x to the 0 power. Well, what is x to the 0 power going to be? And we can assume that x for this case right over here is not equal to 0. 0 to the 0, weird things happen at that point. But if x does not equal 0, what is x to the 0 power going to be? Well, this is the same thing as the derivative with respect to x of 1. And so what is the derivative with respect to x of 1? And to answer that question, I'll just graph it. I'll just graph f of x equals 1 to make it a little bit clearer. So that's my y-axis. This is my x-axis. And let me graph y equals 1, or f of x equals 1. So that's 1 right over there. f of x equals 1 is just a horizontal line. So that right over there is the graph, y is equal to f of x, which is equal to 1. Now, remember the derivative, one way to conceptualize is just the slope of the tangent line at any point. So what is the slope of the tangent line at this point? And actually, what's the slope at every point? Well, this is a line, so the slope doesn't change. It has a constant slope. And it's a completely horizontal line. It has a slope of 0. So the slope at every point over here, slope is going to be equal to 0. is just going to be equal to 0. And that's actually going to be true for any constant. The derivative, if I had a function, let's say that f of x is equal to 3. Let's say that's y is equal to 3. What's the derivative of y with respect to x going to be equal to? And I'm intentionally showing you all the different ways of the notation for derivatives. So what's the derivative of y with respect to x? It can also be written as y prime. What's that going to be equal to? Well, it's the slope at any given point. And you see that no matter what x you're looking at, the slope here is going to be 0. So it's going to be 0. So it's not just x to the 0. If you take the derivative of any constant, you're going to get 0. Derivative with respect to x of any constant-- so let's say of a where this is just a constant, that's going to be equal to 0." + }, + { + "Q": "\nWhat is the reasoning behind the taking the constant out of the dy/dx at 4:08?", + "A": "It is one of the rules/properties of derivatives so we are allowed to do that. Now, why would you do that? It is to make the computation easier. It is not very noticeable when the constant is a nice small integer and you are taking the derivative of a power. But things can get really messy when you take the derivatives of a quotient with chain rules (just an example). So it is easier to take a constant out. You do not need to take out a constant if you can compute with it.", + "video_name": "mzOBlH32qdk", + "timestamps": [ + 248 + ], + "3min_transcript": "is just going to be equal to 0. And that's actually going to be true for any constant. The derivative, if I had a function, let's say that f of x is equal to 3. Let's say that's y is equal to 3. What's the derivative of y with respect to x going to be equal to? And I'm intentionally showing you all the different ways of the notation for derivatives. So what's the derivative of y with respect to x? It can also be written as y prime. What's that going to be equal to? Well, it's the slope at any given point. And you see that no matter what x you're looking at, the slope here is going to be 0. So it's going to be 0. So it's not just x to the 0. If you take the derivative of any constant, you're going to get 0. Derivative with respect to x of any constant-- so let's say of a where this is just a constant, that's going to be equal to 0. Now let's explore a few more properties. Let's say I want to take the derivative with respect to x of-- let's use the same A. Let's say I have some constant times some function. Well, derivatives work out quite well. You can actually take this little scalar multiplier, this little constant, and take it out of the derivative. This is going to be equal to A. I didn't want to do that magenta color. It's going to be equal to A times the derivative of f of x. Let me do that blue color. And the other way to denote the derivative of f of x is to just say that this is the same thing. This is equal to A times this thing right over here is the exact same thing as f prime of x. but I think if I gave you an example it might make some sense. So what about if I were to ask you the derivative with respect to x of 2 times x to the fifth power? Well, this property that I just articulated says, well, this is going to be the same thing as 2 times the derivative of x to the fifth, 2 times the derivative with respect to x of x to the fifth. Essentially, I could just take this scalar multiplier and put it in front of the derivative. So this right here, this is the derivative with respect to x of x to the fifth. And we know how to do that using the power rule. This is going to be equal to 2 times-- let me write that. I want to keep it consistent with the colors." + }, + { + "Q": "At 4:40, why Sal doesn't make a shortcut, and use the average of the past 4 exams as is? if we solve for - ( 82 + X ) divided by 2 = 88 , we could spare ourselves a step or two. is the number of past exams really necessary to the equation?\n", + "A": "now i noticed that Sal s way didn t work out. if we go the way i mentioned we get that X=92, and 92 + 84 = 176, then finally 176 / 2 = 88. so, if on past 4 exams you got an average of 84, to get an average of 88 - you ll have to get a score of 92 points on the next exam... isn t it?", + "video_name": "9VZsMY15xeU", + "timestamps": [ + 280 + ], + "3min_transcript": "8 plus 8 is 16. I just ran eight miles, so I'm a bit tired. And, 4/8, so that's 32. Plus 1 is 33. And now we divide this number by 4. 4 goes into 336. Goes into 33, 8 times. 8 times 4 is 32. 33 minus 32 is 1, 16. So the average is equal to 84. So depending on what school you go to that's either a B or a C. So, so far my average after the first four exams is an 84. Now let's make this a little bit more difficult. We know that the average after four exams, at four exams, is equal to 84. average an 88, to average an 88 in the class. So let's say that x is what I get on the next test. So now what we can say is, is that the first four exams, I could either list out the first four exams that I took. Or I already know what the average is. So I know the sum of the first four exams is going to 4 times 84. And now I want to add the, what I get on the 5th exam, x. And I'm going to divide that by all five exams. So in other words, this number is the average of my first five exams. We just figured out the average of the first four exams. We add what I got on the fifth exam, and then we divide it by 5, because now we're averaging five exams. And I said that I need to get in an 88 in the class. And now we solve for x. Let me make some space here. So, 5 times 88 is, let's see. 5 times 80 is 400, so it's 440. 440 equals 4 times 84, we just saw that, is 320 plus 16 is 336. 336 plus x is equal to 440. Well, it turns out if you subtract 336 from both sides, you get x is equal to 104. So unless you have a exam that has some bonus problems on it, it's probably impossible for you to get ah an 88 average in the class after just the next exam." + }, + { + "Q": "\n42:14=x:2 what does x equal", + "A": "X would equal 6.", + "video_name": "MaMk6-f3T9k", + "timestamps": [ + 2534 + ], + "3min_transcript": "" + }, + { + "Q": "So according, to this video,is 2/3 is equal to 2:3\n", + "A": "Yes you are understanding it very well", + "video_name": "MaMk6-f3T9k", + "timestamps": [ + 123 + ], + "3min_transcript": "We're told this table shows equivalent ratios to 24 to 40. Fill in the missing values. And they write the ratio 24 to 40 right over here. 24-- when the numerator is 24, the denominator is 40. So in that way, you could think of 24/40. But then they want us to write equivalent ratios where we have to fill in different blanks over here-- here in the denominator and here in the numerator. And there's a bunch of ways that we could actually tackle this. But maybe the easiest is to start with the ratio that they gave us, where they gave us both the numerator and the denominator, and then move from there. So for example, if we look at this one right over here, the numerator is 12. It is half of the 24. So the denominator is also going to be half of the denominator It's going to be half of 40. So we could stick a 20 right over there. And then we could go up here. If you compare the 3 to the 12, to go from 12 to 3, you have to divide by 4. So in the numerator, you're dividing by 4. So in the denominator, you also want to divide by 4. So 20 divided by 4 is 5. this numerator right over here. And we see from the denominator, we doubled the denominator. We went from 40 to 80. So we would double the numerator as well, and so you would get 48. And what we just did here is we wrote four equivalent ratios. The ratio 3 to 5 or 3/5 is the same thing as 12 to 20, is the same thing as 24 to 40, is the same thing as 48 to 80. Let's make sure we got the right answer. Let's do a couple more of these. The following table shows equivalent fractions to 27/75. So then they wrote all of the different equivalent fractions. This table shows ratios equivalent to 18/55. All right, so these are all equivalent to 27/75. These are all equivalent to 18/55, so all of these. Which fraction is greater, 27/75 or 18/55? What we want to do-- because you look at these two things. And you're like, well, I don't know. Their denominators are different. How do I compare them? And the best way that I can think of comparing them is look at a point where you're getting an equivalent fraction. And either the numerators are going to be the same, or the denominators are going to be the same. So let's see if there's any situation here. So you have this situation where we see 27/75 is 54/150. And over here, we see that 18/55 is 54-- and this 54 jumped out at me because it's the same numerator-- over 165. And that makes the comparison much easier. What is smaller? 54/150 or 54/165? Well, if you have the same numerator, having a larger denominator will make the number smaller. So 54/165 is smaller than 54/150, which tells us that 18/55 is smaller than 27/75." + }, + { + "Q": "\nWhat do you do to find out the rate of 14:10?", + "A": "You first find the greatest common factor between the 2 numbers and divide each number by the factor. To find the missing rates your use the unit rate to find the proportional number to the number given.", + "video_name": "MaMk6-f3T9k", + "timestamps": [ + 850 + ], + "3min_transcript": "" + }, + { + "Q": "At 3:23, how is 5 over x (100) + 5 = 5.05? How does the 5 become 0.5?\n", + "A": "I do not see where in the video 5 becoming 0.5, but dividing by powers of 10 moves decimal places, so 5/10 would be 0.5, 5/100 = 0.05, 5/1000 = 0.005, etc", + "video_name": "UvDcEvDC4vg", + "timestamps": [ + 203 + ], + "3min_transcript": "Say, x is zero, x is 50, x is 100. Well, when an x is zero, 100 - x is 100 - zero. So, it's 100. When x is 50, it's going be 100 - 50, so it's going to be 50. When x is 100, it's 100 - 100, so it's zero. So, it's pretty clear here that as x is increasing, as x is increasing, I'll just write incr. for increasing, we see that 100 - x is decreasing. We see that that is decreasing. Let's do this with a couple more expressions that have different forms. So, let's say that I have the expression. Let's say that I have the expression five over x plus five and x is decreasing, x is decreasing, but we also know that it is positive and even while it's decreasing, it's staying above zero. So, we're saying x is staying greater than zero. where x is decreasing from 10 to nine or a million to 100,000, but it's staying positive while it's decreasing. Let's think about that. We are going to be dividing by smaller and smaller positive values. So, as you have smaller and smaller positive values of the denominator, you're dividing by smaller and smaller positive values. So, if you're dividing by smaller positive values, you're going to know that this thing is going to get larger. This entire expression is going to get larger as you divide by smaller and smaller positive values. So, if that expression gets larger, then you're just adding five to it. The whole thing is going to increase. The whole thing is going to increase as, the whole thing is going to increase as x decreases while staying positive. And once again, we can make a little table to take a look at that. So, this is x, this is five over x plus five. Let's see, I'll go from x 100 to x is five, to x is one. So, this is clearly x is decreasing. X is decreasing. When x is 100, you're gonna have five divided by 100, which would be five hundredths plus five, so it would be 5.05. When x is five, you're gonna have five divided by five, which is one plus five, which is six. When x is one, you're gonna have five divided by one, which is five plus five. You just, it's 10. So notice, when x is staying positive, but decreasing, the whole expression, five over x plus five, this thing right over here is increasing. Let's do one more of these. So, let's say that we have the expression and we'll change up the variable here. 3y over 2y and I'm curious what happens" + }, + { + "Q": "\nAt 1:35, when we count do we always count starting from the right side if the rectangle?\nBecause if you see, when you count from the left side of the first blue rectangle, the total count is 9. But if i start counting from the right side of the firzt blue triangle it adds up to be 8.\nCan someone please explain this to me", + "A": "There are 8 rectangles, counting either from the left or from the right. I guess you re counting the sides rather than the rectangles. By doing so, you ll get 9 walls . Try counting the upper side of each rectangle...", + "video_name": "WeVWv_OEJsY", + "timestamps": [ + 95 + ], + "3min_transcript": "- [Voiceover] The graph of F is shown below. A total of 24 right hand rectangles are shown. So, what do I mean by right hand rectangles? So, there's clearly 24 rectangles. You can count them. And right hand rectangle means that for each of these rectangles the height of the rectangle is defined by the value of the function on the right hand side of the rectangle. So you can see this is the right hand side of this first rectangle and if you take the value of the function of that point that is the height of the rectangle. A left hand rectangle would define the height of the rectangle by the value of the function on the left hand side So, a left handed rectangle's height, the first rectangle's height would look like that. That's what they mean by right handed rectangle Fair enough, eight in blue. We see that. 16 in red. All right. All 24 of the rectangles have the same width. Which of the statements below is or are true? They give us three expressions in sigma notation and they say, like this first one is the sum of the areas of the blue rectangles. the red rectangles. This is the sum of the areas of all of the rectangles. So, I encourage you now to pause the video and try to determine on your own, which of the statements is or are true. So, I assume you've had a go at it. Let's just go through each of these and see whether they make sense. So this first one, the sum of the areas of the blue rectangles. Well, we know we have one, two, three, four, five, six, seven, eight blue rectangles, and we're summing from one to eight. So it seems like we're summing eight things right over here. This is one, two, three, four, five, six, seven, eight. So, this is looking good right over here. And then we take F of something times one half. So, we're not even looking at this yet. It looks like this would be the height of each of the rectangles. Remember, we're taking the value of the function on the right hand side for the height, and this would be the width. So does it make sense that the width Well, the total distance between X equals negative five and X equals seven, that distance is 12. Five plus seven, that's 12, and we're dividing it into 24 rectangles of equal width. So, if you divide 12 divided by 24 each of these are going to have a width of one half. So this is checking out that the one half. Now let's think about this part. Let's think about the F of negative five plus I over two. So, let's see. When I is equal to one, so we're going to take one half times F of negative five plus one over two. Right? I is one. So negative five plus one half is going to get us to this point right over here. F of that is going to be this distance, this height right over here. This is consistent with these being right handed rectangles." + }, + { + "Q": "\nNewbie here, how do you decide which side to initially subtract? 1:13 he deducts the left-hand variable using the right-hand equation's first \"9X - 7X\". What is a rule of thumb for deciding which side you are supposed to use? Thanks!", + "A": "It actually doesn t matter which side you subtract first - however, one way is easier then the other. When he subtracts 7x from 9x, he gets positive 2x. If he had subtracted 9x from 7x, he would ve gotten negative 2x, which is more confusing to work with. So, you can work out the equation and come to the same answer either way, but it is generally easier to work with positive numbers. Hope this helps!", + "video_name": "2CZrkdtgeNU", + "timestamps": [ + 73 + ], + "3min_transcript": "Let's say we have two intersecting lines. So that's one of the lines right over there. And then I have another line right over here. So those are my two intersecting lines. And let's say we know that the measure of this angle right over here is equal to 7x plus 182. And this is being given in degrees, so it's 7x plus 182 degrees. And we know that the measure of this angle right over here is 9x plus 194 degrees. So my question to you is, what is the measure of each of these angles? And I encourage you to pause the video and to think about it. Well, the thing that might jump out at you is that these two things are vertical angles. They're the opposite angles when we have these intersecting lines right over here. And vertical angles are equal to each other. So we know, because these are vertical angles, that 9x plus 194 degrees must be equal to 7x plus 182 degrees. So if we want all the x-terms on the left-hand side, we could subtract 7x from here. We've got to do it to both sides, of course, in order to maintain the equality. And then we could put all of our constant terms on the right-hand side. So we can subtract 194 from the left. We have to subtract 194 from the right in order to maintain the inequality. And on the left, what we're left with is just 2x. And on the right, what we're left with-- let's see. 182 minus 194. So if it was 194 minus 182, it would be positive 12. But now it's going to be negative 12. We're subtracting the larger from the smaller, so it's equal to negative 12. And then divide both sides by 2. And we get x is equal to negative 6. And now we can use that information to find out the measure of either one of these angles, which is the same as the other one. 182, so 7 times negative 6 is negative 42, plus 182 is going to be equal to 140 degrees. And you'll see the same thing over here. If we say 9 times negative 6, which is negative 54, plus 194, this also equals 140 degrees." + }, + { + "Q": "At 2:19 how in the world does it equal to 140? I get 248?\n", + "A": "On 9x + 194, you made the mistake of adding, 54 to 194, correct? You should have done 9(-6) + 194 = -54 + 194 = 140 Instead you did 9(6) + 194 = 54 + 194 = 248 hope this helps (sorry I m a bit late)", + "video_name": "2CZrkdtgeNU", + "timestamps": [ + 139 + ], + "3min_transcript": "Let's say we have two intersecting lines. So that's one of the lines right over there. And then I have another line right over here. So those are my two intersecting lines. And let's say we know that the measure of this angle right over here is equal to 7x plus 182. And this is being given in degrees, so it's 7x plus 182 degrees. And we know that the measure of this angle right over here is 9x plus 194 degrees. So my question to you is, what is the measure of each of these angles? And I encourage you to pause the video and to think about it. Well, the thing that might jump out at you is that these two things are vertical angles. They're the opposite angles when we have these intersecting lines right over here. And vertical angles are equal to each other. So we know, because these are vertical angles, that 9x plus 194 degrees must be equal to 7x plus 182 degrees. So if we want all the x-terms on the left-hand side, we could subtract 7x from here. We've got to do it to both sides, of course, in order to maintain the equality. And then we could put all of our constant terms on the right-hand side. So we can subtract 194 from the left. We have to subtract 194 from the right in order to maintain the inequality. And on the left, what we're left with is just 2x. And on the right, what we're left with-- let's see. 182 minus 194. So if it was 194 minus 182, it would be positive 12. But now it's going to be negative 12. We're subtracting the larger from the smaller, so it's equal to negative 12. And then divide both sides by 2. And we get x is equal to negative 6. And now we can use that information to find out the measure of either one of these angles, which is the same as the other one. 182, so 7 times negative 6 is negative 42, plus 182 is going to be equal to 140 degrees. And you'll see the same thing over here. If we say 9 times negative 6, which is negative 54, plus 194, this also equals 140 degrees." + }, + { + "Q": "at 10:44 if we were just asked to write the quadratic function would you just subtract the 61/20 from the right side to get it equal to 0\n", + "A": "If you wanted to set this up to use the quadratic equation, you would have just used the original problem given (10x\u00c2\u00b2 - 30x -8 = 0) and not bothered with all the complete the square stuff.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 644 + ], + "3min_transcript": "got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it. Crazy number. Now this, at least on the left hand side, is a perfect square. This is the same thing as x minus 3/2 squared. And it was by design. Negative 3/2 times negative 3/2 is positive 9/4. Negative 3/2 plus negative 3/2 is equal to negative 3. So this squared is equal to 61/20. We can take the square root of both sides and we get x minus 3/2 is equal to the positive or the negative square root of 61/20. And now, we can add 3/2 to both sides of this equation and you get x is equal to positive 3/2 plus or minus the And this is a crazy number and it's hopefully obvious you would not have been able to-- at least I would not have been able to-- get to this number just by factoring. And if you want their actual values, you can get your calculator out. And then let me clear all of this. And 3/2-- let's do the plus version first. So we want to do 3 divided by 2 plus the second square root. We want to pick that little yellow square root. So the square root of 61 divided by 20, which is 3.24. This crazy 3.2464, I'll just write 3.246. So this is approximately equal to 3.246, and that was just" + }, + { + "Q": "At 10:20 it seems as if Sal forgot to right 3/2x\nI may be worng\n", + "A": "There is no 3/2 x. To complete the square, divide the coefficient of x by 2 and square it. It is just the coefficient that is divided by 2, not the whole middle term. Hope this helps.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 620 + ], + "3min_transcript": "sides of this equation? The left-hand hand side of the equation just becomes x squared minus 3x, no 4/5 there. I'm going to leave a little bit of space. And that's going to be equal to 4/5. Now, just like the last problem, we want to turn this left-hand side into the perfect square of a binomial. How do we do that? Well, we say, well, what number times 2 is equal to negative 3? So some number times 2 is negative 3. Or we essentially just take negative 3 and divide it by 2, which is negative 3/2. And then we square negative 3/2. So in the example, we'll say a is negative 3/2. And if we square negative 3/2, what do we get? We get positive 9/4. I just took half of this coefficient, squared it, got positive 9/4. The whole purpose of doing that is to turn this left-hand side into a perfect square. got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it. Crazy number. Now this, at least on the left hand side, is a perfect square. This is the same thing as x minus 3/2 squared. And it was by design. Negative 3/2 times negative 3/2 is positive 9/4. Negative 3/2 plus negative 3/2 is equal to negative 3. So this squared is equal to 61/20. We can take the square root of both sides and we get x minus 3/2 is equal to the positive or the negative square root of 61/20. And now, we can add 3/2 to both sides of this equation and you get x is equal to positive 3/2 plus or minus the" + }, + { + "Q": "At 8:54, why would you half the coefficient?\n", + "A": "The easiest way to see why is to work backwards, expand (x+a)^2 to get (x+a)(x+a) using foil, you have x^2 + ax + ax + a^2 or x^2 + 2ax + a^2, so to complete the square, since b term is 2ax, we have to divide whatever coefficient is there by 2 so that 2ax/2 will leave us with ax. Hope this helps", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 534 + ], + "3min_transcript": "we can get rid of that. We can always turn this into a 1 or a negative 1 if we divide by the right number. But notice, by doing that we got this crazy 4/5 here. So this is super hard to do just using factoring. You'd have to say, what two numbers when I take the product is equal to negative 4/5? It's a fraction and when I take their sum, is equal to negative 3? This is a hard problem with factoring. This is hard using factoring. So, the best thing to do is to use completing the square. So let's think a little bit about how we can turn this into a perfect square. What I like to do-- and you'll see this done some ways and I'll show you both ways because you'll see teachers do it both ways-- I like to get the 4/5 on the other side. So let's add 4/5 to both sides of this equation. You don't have to do it this way, but I like to get the 4/5 out of the way. sides of this equation? The left-hand hand side of the equation just becomes x squared minus 3x, no 4/5 there. I'm going to leave a little bit of space. And that's going to be equal to 4/5. Now, just like the last problem, we want to turn this left-hand side into the perfect square of a binomial. How do we do that? Well, we say, well, what number times 2 is equal to negative 3? So some number times 2 is negative 3. Or we essentially just take negative 3 and divide it by 2, which is negative 3/2. And then we square negative 3/2. So in the example, we'll say a is negative 3/2. And if we square negative 3/2, what do we get? We get positive 9/4. I just took half of this coefficient, squared it, got positive 9/4. The whole purpose of doing that is to turn this left-hand side into a perfect square. got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it." + }, + { + "Q": "I'm confused about the part where he equates things at 01:59. If you have x2-4x+?=x2-2ax+a2\nshouldn't (-4x) correspond to (-2ax), meaning that a=2 (not -2)?\n\n(by \"x2\" I mean x squared)\n", + "A": "You re right. The first time I watched this video I didn t notice the problem because I knew what Sal intended to say and assumed he actually said what he intended -- but there s a mistake here. What he meant to do was set up (x+a)^2 (not (x-a)^2) on the right. That s the normal way to explain the process of completing the square. When he makes the substitution you mention, he does it as if he had written (x+a)^2 on the right, which is what he meant to do.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 119 + ], + "3min_transcript": "In this video, I'm going to show you a technique called completing the square. And what's neat about this is that this will work for any quadratic equation, and it's actually the basis for the And in the next video or the video after that I'll prove the quadratic formula using completing the square. But before we do that, we need to understand even what it's all about. And it really just builds off of what we did in the last video, where we solved quadratics using perfect squares. So let's say I have the quadratic equation x squared minus 4x is equal to 5. And I put this big space here for a reason. In the last video, we saw that these can be pretty straightforward to solve if the left-hand side is a perfect square. You see, completing the square is all about making the quadratic equation into a perfect square, engineering So how can we do that? Well, in order for this left-hand side to be a perfect square, there has to be some number here. There has to be some number here that if I have my number squared I get that number, and then if I have two times my number I get negative 4. Remember that, and I think it'll become clear with a few examples. I want x squared minus 4x plus something to be equal to x minus a squared. We don't know what a is just yet, but we know a couple of things. When I square things-- so this is going to be x squared minus 2a plus a squared. So if you look at this pattern right here, that has to be-- sorry, x squared minus 2ax-- this right here has to be 2ax. And this right here would have to be a squared. to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2, and if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2-- same idea there-- and then you square it. You square a, you get positive 4. So we add positive 4 here. Add a 4. Now, from the very first equation we ever did, you should know that you can never do something to just one side You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4" + }, + { + "Q": "I don't get it why does he do 5+4 instead of 9 ?\n\nat 3:00\n", + "A": "It is 9, but he wants to show that he added 4 to both sides in order to complete the square.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 180 + ], + "3min_transcript": "So how can we do that? Well, in order for this left-hand side to be a perfect square, there has to be some number here. There has to be some number here that if I have my number squared I get that number, and then if I have two times my number I get negative 4. Remember that, and I think it'll become clear with a few examples. I want x squared minus 4x plus something to be equal to x minus a squared. We don't know what a is just yet, but we know a couple of things. When I square things-- so this is going to be x squared minus 2a plus a squared. So if you look at this pattern right here, that has to be-- sorry, x squared minus 2ax-- this right here has to be 2ax. And this right here would have to be a squared. to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2, and if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2-- same idea there-- and then you square it. You square a, you get positive 4. So we add positive 4 here. Add a 4. Now, from the very first equation we ever did, you should know that you can never do something to just one side You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4 It's going to be equal to 5 plus 4. We added 4 on the left-hand side because we wanted this to be a perfect square. But if you add something to the left-hand side, you've got to add it to the right-hand side. And now, we've gotten ourselves to a problem that's just like the problems we did in the last video. What is this left-hand side? Let me rewrite the whole thing. We have x squared minus 4x plus 4 is equal to 9 now. All we did is add 4 to both sides of the equation. But we added 4 on purpose so that this left-hand side becomes a perfect square. What number when I multiply it by itself is equal to 4 and when I add it to itself I'm equal to negative 2? Well, we already answered that question. It's negative 2. So we get x minus 2 times x minus 2 is equal to 9. Or we could have skipped this step and written x minus 2 squared is equal to 9." + }, + { + "Q": "Around 6:40, Sal divides the quadratic equation by 5. This process makes the coefficient of x^2 equal to 1. My question is does the coefficient of x^2 need to be 1 to complete the square.\n", + "A": "no it doesn t have to. for example; 2x^2+18x+16 one can factor this by.. (x+8)(2x+2) but if you divide everthing by 2, you can make 2x^2+18x+16 to x^2+9x+8 then you can factor this to (x+8)(x+1) you see, this is the same as (x+8)(2x+1) but simpler. so to answer your question; it doesn t matter, but is s the matter of which one is simpler hope this helps :)", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 400 + ], + "3min_transcript": "negative 1. And in this case, this actually probably would have been a faster way to do the problem. But the neat thing about the completing the square is it will always work. It'll always work no matter what the coefficients are or no matter how crazy the problem is. And let me prove it to you. Let's do one that traditionally would have been a pretty painful problem if we just tried to do it by factoring, especially if we did it using grouping or something like that. Let's say we had 10x squared minus 30x minus 8 is equal to 0. Now, right from the get-go, you could say, hey look, we could maybe divide both sides by 2. That does simplify a little bit. Let's divide both sides by 2. So if you divide everything by 2, what do you get? But once again, now we have this crazy 5 in front of this coefficent and we would have to solve it by grouping which is a reasonably painful process. But we can now go straight to completing the square, and to do that I'm now going to divide by 5 to get a 1 leading coefficient here. And you're going to see why this is different than what we've traditionally done. So if I divide this whole thing by 5, I could have just divided by 10 from the get-go but I wanted to go to this the step first just to show you that this really didn't give us much. Let's divide everything by 5. So if you divide everything by 5, you get x squared minus 3x minus 4/5 is equal to 0. So, you might say, hey, why did we ever do that factoring we can get rid of that. We can always turn this into a 1 or a negative 1 if we divide by the right number. But notice, by doing that we got this crazy 4/5 here. So this is super hard to do just using factoring. You'd have to say, what two numbers when I take the product is equal to negative 4/5? It's a fraction and when I take their sum, is equal to negative 3? This is a hard problem with factoring. This is hard using factoring. So, the best thing to do is to use completing the square. So let's think a little bit about how we can turn this into a perfect square. What I like to do-- and you'll see this done some ways and I'll show you both ways because you'll see teachers do it both ways-- I like to get the 4/5 on the other side. So let's add 4/5 to both sides of this equation. You don't have to do it this way, but I like to get the 4/5 out of the way." + }, + { + "Q": "Where did Sal get the -2 two from at about 8:45?\n", + "A": "I assume you are referring to where he wrote: a = -3/2 The -3 comes from the equation. It is the coefficient of X Dividing by 2 is part of the process of completing the square. Note: Sal used 2, not -2. The minus sign is from the -3. I suggest you rewatch the video. You will see every time Sal completes the square that he takes the coefficient of X and divides it by 2.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 525 + ], + "3min_transcript": "we can get rid of that. We can always turn this into a 1 or a negative 1 if we divide by the right number. But notice, by doing that we got this crazy 4/5 here. So this is super hard to do just using factoring. You'd have to say, what two numbers when I take the product is equal to negative 4/5? It's a fraction and when I take their sum, is equal to negative 3? This is a hard problem with factoring. This is hard using factoring. So, the best thing to do is to use completing the square. So let's think a little bit about how we can turn this into a perfect square. What I like to do-- and you'll see this done some ways and I'll show you both ways because you'll see teachers do it both ways-- I like to get the 4/5 on the other side. So let's add 4/5 to both sides of this equation. You don't have to do it this way, but I like to get the 4/5 out of the way. sides of this equation? The left-hand hand side of the equation just becomes x squared minus 3x, no 4/5 there. I'm going to leave a little bit of space. And that's going to be equal to 4/5. Now, just like the last problem, we want to turn this left-hand side into the perfect square of a binomial. How do we do that? Well, we say, well, what number times 2 is equal to negative 3? So some number times 2 is negative 3. Or we essentially just take negative 3 and divide it by 2, which is negative 3/2. And then we square negative 3/2. So in the example, we'll say a is negative 3/2. And if we square negative 3/2, what do we get? We get positive 9/4. I just took half of this coefficient, squared it, got positive 9/4. The whole purpose of doing that is to turn this left-hand side into a perfect square. got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it." + }, + { + "Q": "\nAround 0:47 he mentions a \"plain vanilla one for real numbers.\" Does that mean there is a square root for negative numbers?", + "A": "There is a square root for negative numbers but it is not yet taught at this level of math.", + "video_name": "4h54s7BBPpA", + "timestamps": [ + 47 + ], + "3min_transcript": "Find the domain of f of x is equal to the principal square root of 2x minus 8. So the domain of a function is just the set of all of the possible valid inputs into the function, or all of the possible values for which the function is defined. And when we look at how the function is defined, right over here, as the square root, the principal square root of 2x minus 8, it's only going to be defined when it's taking the principal square root of a non-negative number. And so 2x minus 8, it's only going to be defined when 2x minus 8 is greater than or equal to 0. It can be 0, because then you just take the square root of 0 is 0. It can be positive. But if this was negative, then all of a sudden, this principle square root function, which we're assuming is just the plain vanilla one for real numbers, it would not be defined. So this function definition is only defined when 2x minus 8 is greater than or equal to 0. And then we could say if 2x minus 8 has to be greater than or equal to 0, what it's saying about what x has to be. So if we add 8 to both sides of this inequality, you get-- so let me just add 8 to both sides. These 8's cancel out. You get 2x is greater than or equal to 8. 0 plus 8 is 8. And then you divide both sides by 2. Since 2 is a positive number, you don't have to swap the inequality. So you divide both sides by 2. And you get x needs to be greater than or equal to 4. So the domain here is the set of all real numbers that are greater than or equal to 4. x has to be greater than or equal to 4. Or another way of saying it is this function is defined when x is greater than or equal to 4. And we're done." + }, + { + "Q": "At 0:34, why is the sign \"less than or equal to\"? Shouldn't it be just \"greater than\" because if x was 0, then it'd be the square root of -8?\n", + "A": "Sal IS using the greater than or equal to sign ( not the less than or equal to sign). But he isn t saying that x can equal zero. He s saying that the quantity ( 2x - 8 ) can equal zero. Then he has to solve that inequality to finally find out that x must be GREATER than or equal to 4.", + "video_name": "4h54s7BBPpA", + "timestamps": [ + 34 + ], + "3min_transcript": "Find the domain of f of x is equal to the principal square root of 2x minus 8. So the domain of a function is just the set of all of the possible valid inputs into the function, or all of the possible values for which the function is defined. And when we look at how the function is defined, right over here, as the square root, the principal square root of 2x minus 8, it's only going to be defined when it's taking the principal square root of a non-negative number. And so 2x minus 8, it's only going to be defined when 2x minus 8 is greater than or equal to 0. It can be 0, because then you just take the square root of 0 is 0. It can be positive. But if this was negative, then all of a sudden, this principle square root function, which we're assuming is just the plain vanilla one for real numbers, it would not be defined. So this function definition is only defined when 2x minus 8 is greater than or equal to 0. And then we could say if 2x minus 8 has to be greater than or equal to 0, what it's saying about what x has to be. So if we add 8 to both sides of this inequality, you get-- so let me just add 8 to both sides. These 8's cancel out. You get 2x is greater than or equal to 8. 0 plus 8 is 8. And then you divide both sides by 2. Since 2 is a positive number, you don't have to swap the inequality. So you divide both sides by 2. And you get x needs to be greater than or equal to 4. So the domain here is the set of all real numbers that are greater than or equal to 4. x has to be greater than or equal to 4. Or another way of saying it is this function is defined when x is greater than or equal to 4. And we're done." + }, + { + "Q": "\nAt 2:00, why did you subtract 64 from the other side, where in other videos you added 64? What you do to one side you do to the other?", + "A": "Sal doesn t have an equation, so there is no other side. He is working with an expression. To ensure he maintains the original value of the expression, he can only add a value that equal 0 (this is the identify property of addition). This is done by adding the 64 and then also subtracting the 64. 64 - 64 = 0. Thus, he hasn t the expression. He s just making it look different. Hope this helps.", + "video_name": "sh-MP-dVhD4", + "timestamps": [ + 120 + ], + "3min_transcript": "- [Voiceover] Let's see if we can take this quadratic expression here, X squared plus 16 X plus nine and write it in this form. You might be saying, hey Sal, why do I even need to worry about this? And one, it is just good algebraic practice to be able to manipulate things, but as we'll see in the future, what we're about to do is called completing the square. It's a really valuable technique for solving quadratics and it's actually the basis for the proof of the quadratic formula which you'll learn in the future. So it's actually a pretty interesting technique. So how do we write this in this form? Well one way to think about is if we expanded this X plus A squared, we know if we square X plus A it would be X squared plus two A X plus A squared, and then you still have that plus B, right over there. So one way to think about it is, let's take this expression, this X squared plus 16 X plus nine, and I'm just gonna write it with a few spaces in it. X squared plus 16 X And so, if we say alright, we have an X squared here. We have an X squared here. If we say that two A X is the same thing as that, then what's A going to be? So this is two A times X. Well, that means two A is 16 or that A is equal to 8. And so if I want to have an A squared over here, well if A is eight, I would add an eight squared which would be a 64. Well I can't just add numbers willy nilly to an expression without changing the value of an expression so if don't want to change the value of the expression, I still need to subtract 64. So notice, all that I have done now, is I just took our original expression and I added 64 and I subtracted 64, so I have not changed the value of that expression. But what was valuable about me doing that, is now this first part of the expression, it fits the pattern of a perfect square quadratic right over here. We have X squared plus two A X, where A is 8, plus A squared, 64. Once again, how did I get 64? I took half of the 16 and I squared it to get to the 64. And so the stuff that I just squared off, this is going to be X plus eight squared. X plus eight, squared. Once again I know that because A is eight, A is eight, so this is X plus eight squared, and then all of this business on the right hand side. What is nine minus 64? Well 64 minus nine is 55, so this is going to be negative 55. So minus 55, and we're done. We've written this expression in this form, and what's also called completing the square." + }, + { + "Q": "\nAt 3:09, Sal multiplied it by negative 1, but couldn't you just also subtract it with changing it in to a negative?", + "A": "but the can also mess you up right if you interpret it wrong right?", + "video_name": "-nlMXVrgtjw", + "timestamps": [ + 189 + ], + "3min_transcript": "How did that work? The reason why we were able to mindlessly drop this 3 straight down is because we assumed, to do this basic type of synthetic division, that we had just an x right over here. We didn't have a 3x. We didn't have a 4x. We didn't have an x squared. We just had an x. And if you divide an x into whatever this highest degree term is, your coefficient is going to be the exact same thing. It's just going to be one degree lower. So you go from a 3x to the third to 3x squared-- so the exact same coefficient. And now, in our little synthetic division right over here, that right over there is the x squared term. So you went from 3x to the third to 3x squared. We essentially divided it by x. And we could blindly do it because we knew, we assumed, that we were dealing with just a 1x to the first right over here. and see why we're essentially doing the exact same thing. Now let's take this 3x squared and multiply that times x plus 4. So 3x squared times x-- I'll do it in that white color-- it's going to be 3x to the third. And 3x squared times 4 is going to be 12x squared. And now we'll want to subtract this. So now we subtract. We subtract this out. These guys cancel out, and you have 4x squared minus 12x squared. And so that will give you negative 8x squared. So once again, you're probably seeing some parallels. You had the 4x squared over here. You have the 4x squared over there. We just wrote the coefficient, but that's what it represented. 4x squared, we wrote the 4 there. Then we essentially subtracted 12x squared. And the way we got that 12, we multiplied 3 times 4, Here we're multiplying 3 times negative 4. We're essentially multiplying 3 times 4 and then subtracting. That's why we put that negative there, so we don't have to keep remembering to subtract this row. So we could just keep adding them. But that's essentially what we did. We multiplied 3 times this 4. And now we subtract it. We get that negative 12x squared. And then we subtracted, and you got negative 8x squared. And you might say, oh, is this the same negative 8 as this right over here? Not quite yet, because over here, this negative 8 literally represents negative 8x. This is actually part of our simplification. When we divide this into that, we got 3x squared minus 8x plus 30. So over here in the algebraic long division, we then say, how many times does x plus 4 go into negative 8x squared? Well x goes into negative 8x squared negative 8x times." + }, + { + "Q": "Can this work?\n\n1. First apply a rotation until the projected vector aligns with the vector we want to project to.\n2. Second scale the rotated vector accordingly.\n\nWill the matrix of product of the transformations in 1) and 2) be the same with the one Sal defined at at 13:01?\n", + "A": "This would theoretically work. However, you also require an expression for the angle between every possible vector and the vector you want to project onto. By the time you figure that out and get a way to convert it back into a linear transformation, you may as well have done what Sal did.", + "video_name": "JK-8XNIoAkI", + "timestamps": [ + 781 + ], + "3min_transcript": "ready to take the projection of this guy. The definition of our projection is you dot this guy with our unit vector. So we dot it. We're taking the dot product of 0, 1. 0, 1 dot my unit vector dot u1, u2. I'm going to multiply that times my unit vector, times u1, u2. This seems very complicated, but it should simplify when we actually try to work out our transformation matrix. Let's do it. When I dot these two guys, what do I get. Let me write it here. My matrix A will become 1 times u1, plus 0 times u2. That's just u1. This whole thing just simplifies to u1, when I take the dot product of these two things. Times u1, u2. My second column, if I dot these two guys, I get 0 times u1 plus 1 times u2. So I'm going to get u2 times my unit vector, u1, u2. If I multiply that out, this will be equal to what. I can just write them as columns. u1 times u1 is u1 squared. u1 times u2 is u1, u2. u2 times u1 is u2 times u1. Then, u2 times u2 is u2 squared. You give me any unit vector and I will give you the transformation that gives you any projection of some other vector onto the line defined by that. That was a very long way of saying that. Let's go back to what I did before. Let's say we want to find any projection onto the line, onto the vector, I'll draw it here. We'll do the same example that we did in the last video. We said the vector v was equal to the vector 2, 1. That was my vector v. How can we find sum transformation for the projection onto the line defined by v? Onto this line right here. The line defined by v. What we can first do is convert v into a unit vector. We can convert v into a unit vector that goes in the same direction. Some unit vector u. We did that already up here. Where we essentially just divided [? bv ?] by it's length. Let's take v and divide by it's length. The unit vector is this, 1 over the square root of 5 times our vector v. It was 1 over the square root of 5 times our vector v, right there. You start with a unit vector there. You just create this matrix, and then we will have our" + }, + { + "Q": "\nAt 5:43, Sal says that dx/dy is the slope of the tangent line at any point. I didn't see where it was explained how/why this represents the slope prior to this point; it was just injected as a given into the explanation midstream.\n\nIs this explained elsewhere? How did we go from having d/dx represent the slope to having dx/dy represent it here?", + "A": "I haven t watched the video but... d/dx is a differential operator. d/dx[4x] means take the derivative of the expression 4x with respect to x. If instead we write d/dx[f(x)] we would write df/dx as the derivative (or f (x)). The derivative represents the slope of the tangent line. dx means a very small change in x (infinitesimally small). and dy means a very small change in y. If you write dy/dx and y is your dependent variable then you have \u00e2\u0088\u0086y/\u00e2\u0088\u0086x which is the formula for slope.", + "video_name": "mSVrqKZDRF4", + "timestamps": [ + 343 + ], + "3min_transcript": "It might be a little bit clearer if you kind of thought of it as the derivative with respect to x of y, as a function of x. This might be, or y is a function of x squared, which is essentially another way of writing what we have here. This might be a little bit clearer in terms of the chain rule. The derivative of y is a function of x squared with respect to y of x. So the derivative of something squared with respect to that something, times the derivative of that something, with respect to x. This is just the chain rule. I want to say it over and over again. This is just the chain rule. So let's do that. What do we get on the right hand side over here? And I'll write it over here as well. This would be equal to the derivative of y squared with respect to y, is just going to be 2 times y. And the derivative of y with respect to x? Well, we don't know what that is. So we're just going to leave that as times the derivative of y with respect to x. So let's just write this down over here. So we have is 2x plus the derivative of something squared, with respect to that something, is 2 times the something. In this case, the something is y, so 2 times y. And then times the derivative of y with respect to x. And this is all going to be equal to 0. Now that was interesting. Now we have an equation that has the derivative of a y with respect to x in it. And this is what we essentially want to solve for. This is the slope of the tangent line at any point. So all we have to do at this point is solve for the derivative of y with respect to x. Solve this equation. And actually just so we can do this whole thing on the same page so we can see where we started, let me copy and paste this up here. This is where we left off. And let's continue there. So let's say let's subtract 2x from both sides. So we're left with 2y times the derivative of y, with respect to x, is equal to-- we're subtracting 2x from both sides-- so it's equal to negative 2x. And if we really want to solve for the derivative of y with respect to x, we can just divide both sides by 2y. And we're left with the derivative of y with respect to x. Let's scroll down a little bit. The derivative of y with respect to x is equal to, well the 2s cancel out. We we're left with negative x over y. So this is interesting. We didn't have to us explicitly define y" + }, + { + "Q": "\nAt 2:24 why is 4 divided in to 36 and why does 4 become a 1 ?", + "A": "it is because 36/4 equals 9", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 144 + ], + "3min_transcript": "Multiply 1 and 3/4 times 7 and 1/5. Simplify your answer and write it as a mixed fraction. So the first thing we want to do is rewrite each of these mixed numbers as improper fractions. It's very difficult, or at least it's not easy for me, to directly multiply mixed numbers. One can do it, but it's much easier if you just make them improper fractions. So let's convert each of them. So 1 and 3/4 is equal to-- it's still going to be over 4, so you're still going to have the same denominator, but your numerator as an improper fraction is going to be 4 times 1 plus 3. And the reason why this makes sense is 1 is 4/4, or 1 is 4 times 1 fourths, right? 1 is the same thing as 4/4, and then you have three more fourths, so 4/4 plus 3/4 will give you 7/4. So that's the same thing as 1 and 3/4. Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5." + }, + { + "Q": "in Gulshan the ratio of private homes to apartment is 3:5.if all apartments are bricks made and 1/6 of the private homes are wooden, that is maximum portion of houses that may be brick...........??\n", + "A": "3+5=8. so apartments are 5/8 of all houses.and all of them are brick made. 3/8 are private homes and 1-1/6 of them may be brick made(1/6 are wooden). that means the maximum of private houses from brick are 3/8 * 5/6=15/48=5/16. adding the apartments we get 5/8+5/16=15/16", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 185 + ], + "3min_transcript": "Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5. 6 minus 5 is 1. Bring down the 3. 5 goes into 13 two times. And you could have immediately said 5 goes into 63 twelve times, but this way, at least to me, it's a little bit more obvious. And then 2 times 5 is 12, and then we have sorry! 2 times 5 is 10. That tells you not to switch gears in the middle of a math problem. 2 times 5 is 10, and then you subtract, and you have a remainder of 3. So 63/5 is the same thing as 12 wholes and 3 left over, or 3/5 left over. And if you wanted to go back from this to that, just think: 12 is the same thing as 60 fifths, or 60/5." + }, + { + "Q": "At 0:07, what does he mean by \"mixed fraction?\"\n", + "A": "no, this is a mixed fraction 5 1/2", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 7 + ], + "3min_transcript": "Multiply 1 and 3/4 times 7 and 1/5. Simplify your answer and write it as a mixed fraction. So the first thing we want to do is rewrite each of these mixed numbers as improper fractions. It's very difficult, or at least it's not easy for me, to directly multiply mixed numbers. One can do it, but it's much easier if you just make them improper fractions. So let's convert each of them. So 1 and 3/4 is equal to-- it's still going to be over 4, so you're still going to have the same denominator, but your numerator as an improper fraction is going to be 4 times 1 plus 3. And the reason why this makes sense is 1 is 4/4, or 1 is 4 times 1 fourths, right? 1 is the same thing as 4/4, and then you have three more fourths, so 4/4 plus 3/4 will give you 7/4. So that's the same thing as 1 and 3/4. Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5." + }, + { + "Q": "\nAt 0:08 he meant improper fraction not mixed fraction right?", + "A": "Hi Carolyn, He actually meant mixed fraction. You have to convert 2 mixed fractions to fractions, then multiply them, then convert the result (fraction) into mixed fraction again.", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 8 + ], + "3min_transcript": "Multiply 1 and 3/4 times 7 and 1/5. Simplify your answer and write it as a mixed fraction. So the first thing we want to do is rewrite each of these mixed numbers as improper fractions. It's very difficult, or at least it's not easy for me, to directly multiply mixed numbers. One can do it, but it's much easier if you just make them improper fractions. So let's convert each of them. So 1 and 3/4 is equal to-- it's still going to be over 4, so you're still going to have the same denominator, but your numerator as an improper fraction is going to be 4 times 1 plus 3. And the reason why this makes sense is 1 is 4/4, or 1 is 4 times 1 fourths, right? 1 is the same thing as 4/4, and then you have three more fourths, so 4/4 plus 3/4 will give you 7/4. So that's the same thing as 1 and 3/4. Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5." + }, + { + "Q": "At the end (+- 9:00) he says that one of the equations is impractical, because it would entail the calculation of all the pop, is that correct?\n", + "A": "Exactly. It is pretty hard to calculate the heights of all the 150 million people, after all.", + "video_name": "k5EbijWu-Ss", + "timestamps": [ + 540 + ], + "3min_transcript": "They could call this x subscript 4. They could call this x subscript 5. And so if you had n of these you would just keep going. x subscript 6, x subscript seven, all the way to x subscript n. And so to take the sum of all of these, they would denote it as let me write it right over here. So they will say that the sample mean is equal to the sum of all my x sub i's-- so the way you can conceptualize it, these i's will change. In this case, the i started at 1. The i's are going to start at 1 until the size of our actual sample. So all the way until n. In this case n was equal to 5. So this is literally saying this is equal to x sub 1 plus x sub 2 plus x sub 3, all Once again, in this case, we only had five. Now, are we done? Is this what the sample mean is? Well, no, we aren't done. We don't just add up all of the data points. We then have to divide by the number of data points there are. So this might look like very fancy notation, but it's really just saying, add up your data points and divide by the number of data points you have. And this capital Greek letter sigma literally means sum. Sum all of the x i's, from x sub 1 all the way to x sub n, and then divide by the number of data points you have. Now let's think about how we would denote the same thing but, instead of for the sample mean, doing it for the population mean. So the population mean, they will denote it with mu, we already talked about that. but this time it's going to be the sum of all of the elements in your population. So your x sub i's-- and you'll still start at i equals 1. But it usually gets denoted that, hey you're taking the whole population, so they'll often put a capital N right over here to somehow denote that this is a bigger number than maybe this smaller n. But once again, we are not done. We have to divide by the number of data points that we are actually summing. And so this, once again, is the same thing as x sub 1 plus x sub 2 plus x sub 3-- all the way to x sub capital N, all of that divided by capital N. And once again, in this situation, we found this practical. We found this impractical. We can debate whether we took enough data points on our sample mean right over here. But we're hoping that it's at least somehow indicative of our population mean." + }, + { + "Q": "At 4:54 how it is 12. The equation is supposed to be 5-(- -7). The answer is supposed to be -2.\n", + "A": "no. It s supposed to be 5-(-7), in which the minus symbol and the negative symbol cancel out each other, making the equation then 5+7, equalling 12.", + "video_name": "XkRD9lv_y44", + "timestamps": [ + 294 + ], + "3min_transcript": "So, actually first of all, I can figure out what three plus six is. Three plus six is equal to nine. And then I have this minus four here. So I could say nine minus four, actually I want to be careful not to skip any steps. So three plus six, three plus six, in a color that you can see, so three plus six is nine. So that's nine minus H minus four, minus four. Now I could change the order in which I do this addition or subtraction, so this is going to be the equivalent of nine minus four. Nine minus four minus H, minus H. And I just did that so I can simplify and figure out what nine minus four is. Nine minus four, of course, let me do this in blue, navy blue. Nine minus four is five. before I even did the substitution. And now I can substitute H with negative seven. So this is going to be equal to, this is going to be equal to, when I do the substitution, I'll write it up here, it's going to be five minus, I'll do the minus in that magenta color, minus and now where I see an H, I'm gonna replace it with negative seven. Five minus negative seven. You want to be very careful there, you might be tempted to say, oh I have a negative here, negative here, let me just replace H with a seven. Remember, H is negative seven, so you're subtracting H. You're gonna subtract negative seven. So this is five minus negative seven, which is the same thing, which is the same thing as five plus seven. Five plus seven, which we all know is equal to 12. And we're done. Let's do a few more of these. You can't really get enough practice here, this is some important foundational skills (laughing) Alright. So consider, I don't make you too stressed about it. Consider the following number line. Alright, so we've got a number line here and let's see, they didn't mark off all the numbers here. This is negative four, E is at this point, this is then we go to two. So it looks like we're counting by twos here, that this is negative two, this is zero, yep. Negative four, if you increase by two, negative four, negative two, zero, two, four, this would be a six. This would be a negative six. They intentionally left those numbers off, so we had to figure out that hey, look, between negative four and two, to go from negative four to two, you have to increase by six and we only have one, two, three hashmarks. So each of those hashmarks must be increasing by two. Well anyway, now we know, now we know what all the points in the number line are. Evaluate E minus F. Well we know, we know that E is equal to negative two. And we know, we know that F" + }, + { + "Q": "at around 1:17 I don't really understand how at least classic principal root is a defined for a negative number. Can someone please explain it to me?\n", + "A": "Are you trying to find the square root of a negative number?", + "video_name": "n17q8CBiMtQ", + "timestamps": [ + 77 + ], + "3min_transcript": "Let's do some more examples finding do mains of functions. So let's say we have a function g of x. So this is our function definition here tells us, look, if we have an input x, the output g of x is going to be equal to 1 over the square root of 6 minus -- we write this little bit neater, 1 over the square root of 6 minus the absolute value of x So like always, pause this video and see if you can figure out what what is the domain of this function. Based on this function definition what is the domain of g? What is the set of all inputs for which this function is defined? Alright. So, to think about all of the inputs that would allow this function to be defined, it may be easier to state when is this function not get defined. Well if we divide by 0 then we're not going to be defined. Or if we have a negative under the radical. So if you think about it -- if if what we have under the radical 0, you can take the, you can take the principal root of 0. It's going to be 0. But then you are going to divide by 0. That's going to be undefined. And if what you have another radical is negative, the principal root isn't defined for native number, at least the classic principal root is a defined for a negative number. So if 6 minus the absolute value of x is zero or negative, this thing is not going to be defined. Or another way to think about it is it's going to be defined -- so g is is defined -- is defined if -- g is defined if 6 minus, maybe I can write if and only if. Sometimes people write if and only if with two f's right there, iff. g is defined if and only if -- this is kind of mathy way of saying if and only if -- 6 minus the absolute value of x is greater than 0 we're going take the square root of 0 is 0. Then you divide by 0. That's undefined. And if it's less than zero, then you're taking, you're trying to find the principal root of a negative number, that's not defined. So let's see, it's g is defined if and only if this is true. And let's see. We could add the absolute value of x to both sides. We could add the absolute value of x to both sides, then that would give us 6 is greater than the absolute value of x, or that the absolute value of x is less than 6. Or we could say that, you know, let me write that way. The absolute value of x is less than 6. Another way of saying that is x would have to be less than 6 and greater than negative 6. or x is between negative 6 and 6. These two things -- these two things are equivalent. If the magnitude of x is less than 6 then x is greater than negative 6 and less than positive 6." + }, + { + "Q": "\nHow did he get that a negative wouldn't work, at 0:56 & 1:15? If it was the absolute value of x, or whatever the number is, wouldn't the absolute value have canceled out the negative ( - ) and turned it to positive, so that it'd work? Also, How did he get -6 < x at about 3:00?? He had just said earlier that it COULDN'T be a negative number, but -6 < x < 6 implies that it could be anything from -5 to 5.", + "A": "At 0:56 and 1:15 he s not saying that a negative value won t work for x, since you are correct -- a negative value would not affect anything since we are taking the absolute value of it anyway. Rather, he s saying that we can t have the expression in the radical -- 6-x -- be equal to a negative, since we would then be taking the square root of a negative of number. This translates to |x|<6 (or -6 +1 BUT If NOT => -1. \u00e2\u0080\u00a2Secondly, any ODD exponent is either -i or +i. i^99 IS i^98 * i => 98 IS NOT divisible by 4 so => -i", + "video_name": "QiwfF83NWNA", + "timestamps": [ + 296 + ], + "3min_transcript": "So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over. i to the first power. This is a multiple of 4-- this right here is a multiple of 4-- and I know that because any 1,000 is multiple of 4, any 100 is a multiple of 4, and then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the first power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1, and we're just going to be left with i to the first power, or just i. Let's do another one. i to the 90-- let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99? So this is the same thing as i to the 96th power times i to the third power, right? If you multiply these, same base, add the exponent, you would get i to the 99th power. i to the 96th power, since this is a multiple of 4, this is i to the fourth, and then that to the 16th power. So that's just 1 to the 16th, so this is just 1. And then you're just left with i to the third power. And you could either remember that i to the third power is equal to-- you can just remember that it's equal to negative i. Or if you forget that, you could just say, look, this is the same thing as i squared times i. This is equal to i squared times i. i squared, by definition, is equal to negative 1. So you have negative 1 times i is equal to negative i. Let me do one more just for the fun of it. Let's take i to the 38th power." + }, + { + "Q": "\nAt 4:10 sal wrote i^i ?", + "A": "He wrote i^i but said i^1, so he meant i^1.", + "video_name": "QiwfF83NWNA", + "timestamps": [ + 250 + ], + "3min_transcript": "as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over. i to the first power. This is a multiple of 4-- this right here is a multiple of 4-- and I know that because any 1,000 is multiple of 4, any 100 is a multiple of 4, and then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the first power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1, and we're just going to be left with i to the first power, or just i. Let's do another one. i to the 90-- let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99?" + }, + { + "Q": "At 4:09, Sal wrote i^i. Just out of curiosity, what is i^i?\n", + "A": "Euler s formula says that e^(ix)=cos(x)+isin(x) Let x=\u00cf\u0080/2 and we get e^(i\u00cf\u0080/2)=cos(\u00cf\u0080/2)+isin(\u00cf\u0080/2) e^(i\u00cf\u0080/2)=0+i e^(i\u00cf\u0080/2)=i Now raise both sides to the power of i and we get [e^(i\u00cf\u0080/2)]^i=i^i e^(i\u00c2\u00b2\u00cf\u0080/2)=i^i e^(-\u00cf\u0080/2)=i^i", + "video_name": "QiwfF83NWNA", + "timestamps": [ + 249 + ], + "3min_transcript": "as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over. i to the first power. This is a multiple of 4-- this right here is a multiple of 4-- and I know that because any 1,000 is multiple of 4, any 100 is a multiple of 4, and then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the first power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1, and we're just going to be left with i to the first power, or just i. Let's do another one. i to the 90-- let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99?" + }, + { + "Q": "\nAt 1:35-\n5 is also a multiple of 100, so why would i^100 =1 and not i?", + "A": "i^100 can be written as (i^4)^25 which is equal to 1^25 = 1. For what you re asking - even though 5 is a factor of 100 (not multiple) - we could take it like this: i^100 = (i^5)^20 which can written again as (i^5)^4 = i^4 = 1.", + "video_name": "QiwfF83NWNA", + "timestamps": [ + 95 + ], + "3min_transcript": "Now that we've seen that as we take i to higher and higher powers, it cycles between 1, i, negative 1, negative i, then back to 1, i, negative 1, and negative i. I want to see if we can tackle some, I guess you could call them, trickier problems. And you might see these surface. And they're also kind of fun to do to realize that you can use the fact that the powers of i cycle through these values. You can use this to really, on a back of an envelope, take arbitrarily high powers of i. So let's try, just for fun, let's see what i to the 100th power is. And the realization here is that 100 is a multiple of 4. So you could say that this is the same thing as i to the 4 times 25th power. And this is the same thing, just from our exponent properties, as i to the fourth power raised to the 25th power. If you have something raised to an exponent, and then that is raised to an exponent, that's the same thing as multiplying the two exponents. that's pretty straightforward. i to the fourth is just 1. i to the fourth is 1, so this is 1. So this is equal to 1 to the 25th power, which is just equal to 1. So once again, we use this kind of cycling ability of i when you take its powers to figure out a very high exponent of i. Now let's say we try something a little bit stranger. Let's try i to the 501st power. Now in this situation, 501, it's not a multiple of 4. So you can't just do that that simply. But what you could do, is you could write this as a product of two numbers, one that is i to a multiple of fourth power. And then one that isn't. And so you could rewrite this. 500 is a multiple of 4. So you could write this as i to the 500th power times i Right? You have the same base. When you multiply, you can add exponents. So this would be i to the 501st power. as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k" + }, + { + "Q": "\nAt 3:02, Sal mentions that K \u00e2\u0089\u00a5 0. I understand that for now K can't be a negative number, but can it be a positive decimal? For example, is it possible to find and answer to i to the 7.89th power? (It is between the 7th and 8th power, so it couldn't follow any of the principles making the value either 1, i, -1, or -i.)", + "A": "You can surely raise any number to a decimal.. I ll tell you what that means... You can convert any decimal number into a fractional number very easily and when you do that understanding the power of decimals becomes meaningful. For eg. If u r saying 2^1.5, you can write it as 2^3/2 which means it is square root of 2^3.. Right? Similarly you can convert any decimal power into a fractional number and understand its meaning...", + "video_name": "QiwfF83NWNA", + "timestamps": [ + 182 + ], + "3min_transcript": "that's pretty straightforward. i to the fourth is just 1. i to the fourth is 1, so this is 1. So this is equal to 1 to the 25th power, which is just equal to 1. So once again, we use this kind of cycling ability of i when you take its powers to figure out a very high exponent of i. Now let's say we try something a little bit stranger. Let's try i to the 501st power. Now in this situation, 501, it's not a multiple of 4. So you can't just do that that simply. But what you could do, is you could write this as a product of two numbers, one that is i to a multiple of fourth power. And then one that isn't. And so you could rewrite this. 500 is a multiple of 4. So you could write this as i to the 500th power times i Right? You have the same base. When you multiply, you can add exponents. So this would be i to the 501st power. as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over." + }, + { + "Q": "At 9:27 , how did you come up with how many songs and game he would buy?\n", + "A": "Since the shaded area (the darker blueish color) is the solution to both equations (the first being the one about how much money he has and the second being how many songs and games he can buy) any coordinates in that region would satisfy the problem. So 4 games and 14 songs would work, as Sal mentioned. Obviously, the two values need to be integers because you can never buy a half game or song. Hope that helps. : )", + "video_name": "BUmLw5m6F9s", + "timestamps": [ + 567 + ], + "3min_transcript": "Just a little over 28. So 28.08. So that is, g is 0, s is 28. So that is 2, 4, 24, 6, 8. A little over 28. So it's right over there. So this line, 0.89s plus 1.99g is equal to 25 is going to go from this coordinate, which is 0, 28. So that point right there. All the way down to the point 12.56,0. So let me see if I can draw that. It's going to go-- I'll draw up one more attempt. Maybe if I start from the bottom it'll be easier. That was a better attempt. Let me bold that in a little bit, so you can make sure you can see it. So that line represents this right over here. that imply? So if we think about it, when g is equal to 0, 0.89s is less than 25. So when g is equal to 0, if we really wanted the less than there, we could think of it this way. It's less than instead of just doing less than or equal to. So s is less than 28.08. So it'll be the region below. When s is 0, g-- so if we think s is 0, if we use this original equation, 1.99g will be less than or equal to. I use this just to plot the graph, but if we actually care about the actual inequality, we get 1.99g is less than 25. g would be less than or equal to 12.56. So when s is equal to 0, g is less than 12.56. So the area that satisfies this second constraint is everything below this graph. So it's going to be the overlap of the regions that satisfy one of the two. So the overlap is going to be this region right here. Below the orange graph and above the blue graph, including both of them. So if you pick any combination-- so if he buys 4 games and 14 songs, that would work. Or if he bought 2 games and 16 songs, that would work. So you can kind of get the idea. Anything in that region-- and he can only buy integer values-- would satisfy" + }, + { + "Q": "\nAt 9:33, Sal says that if you buy 4 games and 14 songs that would work. Sure, on the equation it makes sense but on the graph it looks like it's an impossible point. Or is it just the line....I don't know.", + "A": "no it seems like it could work on the graph too. its just that there is not 14 mark on his graph. but we know 14 is halfway between 12 and 16 so he put his point there.... Hope This Helps =)", + "video_name": "BUmLw5m6F9s", + "timestamps": [ + 573 + ], + "3min_transcript": "Just a little over 28. So 28.08. So that is, g is 0, s is 28. So that is 2, 4, 24, 6, 8. A little over 28. So it's right over there. So this line, 0.89s plus 1.99g is equal to 25 is going to go from this coordinate, which is 0, 28. So that point right there. All the way down to the point 12.56,0. So let me see if I can draw that. It's going to go-- I'll draw up one more attempt. Maybe if I start from the bottom it'll be easier. That was a better attempt. Let me bold that in a little bit, so you can make sure you can see it. So that line represents this right over here. that imply? So if we think about it, when g is equal to 0, 0.89s is less than 25. So when g is equal to 0, if we really wanted the less than there, we could think of it this way. It's less than instead of just doing less than or equal to. So s is less than 28.08. So it'll be the region below. When s is 0, g-- so if we think s is 0, if we use this original equation, 1.99g will be less than or equal to. I use this just to plot the graph, but if we actually care about the actual inequality, we get 1.99g is less than 25. g would be less than or equal to 12.56. So when s is equal to 0, g is less than 12.56. So the area that satisfies this second constraint is everything below this graph. So it's going to be the overlap of the regions that satisfy one of the two. So the overlap is going to be this region right here. Below the orange graph and above the blue graph, including both of them. So if you pick any combination-- so if he buys 4 games and 14 songs, that would work. Or if he bought 2 games and 16 songs, that would work. So you can kind of get the idea. Anything in that region-- and he can only buy integer values-- would satisfy" + }, + { + "Q": "Hi,\nI am with you until 2:14 in the video,\nWhere pi \"f\" 1/u du,\nThen that is = to pi ln u (but what happens to the du?)...+C\nAs i understand it there is two term after the \"f\" (1/u) x (du),\nRegards\nPs: thank you very much for all the math help :)\n", + "A": "When you integrate with respect to u, the du disappears, just as dx disappears when you integrate with respect to x. For example, the integral of 2x dx is x\u00c2\u00b2 (not x\u00c2\u00b2 dx). There are different ways to think about why this happens. One is to consider dx or du as part of the integral symbol. It s probably more helpful, especially when doing u-substitution, to understand dx or du as a distinct element, but one that gets incorporated into the answer when you integrate.", + "video_name": "OLO64d4Y1qI", + "timestamps": [ + 134 + ], + "3min_transcript": "We are faced with a fairly daunting-looking indefinite integral of pi over x natural log of x dx. Now, what can we do to address this? Is u-substitution a possibility here? Well for u-substitution, we want to look for an expression and its derivative. Well, what happens if we set u equal to the natural log of x? Now what would du be equal to in that scenario? du is going to be the derivative of the natural log of x with respect to x, which is just 1/x dx. This is an equivalent statement to saying that du dx is equal to 1/x. So do we see a 1/x dx anywhere in this original expression? Well, it's kind of hiding. It's not so obvious, but this x in the denominator is essentially a 1/x. And then that's being multiplied by a dx. Let me rewrite this original expression to make a little bit more sense. So the first thing I'm going to do is I'm going to take the pi. since I've already used-- let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral. And so this becomes the integral of-- and let me write the 1 over natural log of x first. 1 over the natural log of x times 1/x dx. Now it becomes a little bit clearer. These are completely equivalent statements. But this makes it clear that, yes, u-substitution will work over here. If we set our u equal to natural log of x, then our du is 1/x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of 1/u. Natural log of x is u-- we set that equal to natural log of x-- times du. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log of the absolute value of u so that we can handle even negative values of u. The natural log of the absolute value of u plus c, just in case we had a constant factor out here. And we're almost done. We just have to unsubstitute for the u. u is equal to natural log of x. So we end up with this kind of neat-looking expression. The anti of this entire indefinite integral we have simplified. We have evaluated it, and it is now equal to pi times the natural log of the absolute value of u. But u is just the natural log of x. And then we have this plus c right over here." + }, + { + "Q": "\nI get all that happened, but wouldn't the answer at 2:48 be \u00cf\u0080(ln |lnx| + C) = \u00cf\u0080(ln |lnx|) + \u00cf\u0080C instead of \u00cf\u0080(ln|lnx|) + C?", + "A": "It could be, but the two C s would have different values. Imagine if the pi had been left inside the integral--that s the way Sal is handling it. Typically we choose the constant to be the thing added on after computing the integral as a whole rather than at an intermediate step.", + "video_name": "OLO64d4Y1qI", + "timestamps": [ + 168 + ], + "3min_transcript": "since I've already used-- let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral. And so this becomes the integral of-- and let me write the 1 over natural log of x first. 1 over the natural log of x times 1/x dx. Now it becomes a little bit clearer. These are completely equivalent statements. But this makes it clear that, yes, u-substitution will work over here. If we set our u equal to natural log of x, then our du is 1/x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of 1/u. Natural log of x is u-- we set that equal to natural log of x-- times du. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log of the absolute value of u so that we can handle even negative values of u. The natural log of the absolute value of u plus c, just in case we had a constant factor out here. And we're almost done. We just have to unsubstitute for the u. u is equal to natural log of x. So we end up with this kind of neat-looking expression. The anti of this entire indefinite integral we have simplified. We have evaluated it, and it is now equal to pi times the natural log of the absolute value of u. But u is just the natural log of x. And then we have this plus c right over here. this original expression was only defined for positive values of x because you had to take the natural log here, and it wasn't an absolute value. So we can leave this as just a natural log of x, but this also works for the situations now because we're doing the absolute value of that where the natural log of x might have been a negative number. For example, if it was a natural log of 0.5 or, who knows, whatever it might be. But then we are all done. We have simplified what seemed like a kind of daunting expression." + }, + { + "Q": "\nAt 5:33, Sal said that the area of the parallelogram is 5x6, whereas before, following the previously spoken format it would be 2x5x6. Is this right?", + "A": "Yes! Correct! 2x5x6 is pretty much the same thing as what sal previously said.", + "video_name": "tFhBAeZVTMw", + "timestamps": [ + 333 + ], + "3min_transcript": "We could call that the height right over there. So if you want the total area of parallelogram ABCD, it is equal to two times one half times base times height. Well, two times one half is just one. And so you're just left with base times height. So we can call this b. So it's just b times this height over here, base times height. So that's a neat result. And you might have guessed that this would be the case. But if you want to find the area of any parallelogram, and if you can figure out the height, it is literally you just take one of the bases, because both bases are going to be the-- opposite sides are equal, so it could have been either that side or that side, times the height. So that's one way you could have found the area. Or you could have multiplied. The other way to think about it is you could have multiplied. So if I were to turn this parallelogram over, it would look something like this. It would look something like this. So if I were to rotate it like that, let me draw the points-- this would be point A. Let me make sure I'm doing this right. This would be point A. This would be point D. This would be point C. And then this would be point B. You could also do it this way. You could say it's 1-- sorry, not 1/2. That would be for a triangle. The area of this would be base times height. So you could say it's h times DC. So you could say this is going to be equal to h times the length DC. That's one way to do it. That's this base times this height. Or you could say it's equal to AD times, I'll call this altitude right here, I'll call this height 2. Times h2. Maybe I'll call this h1. h1, h2. So you could take this base times this height, or you could take this base times This is h2. Either way. So if someone were to give you a parallelogram, just to make things clear, obviously, you'd have to be have some way to be able to figure out the height. So if someone were to give you a parallelogram like this, they would tell you this is a parallelogram. If they were to tell you that this length right over here is 5, and if they were to tell you that this distance is 6, then the area of this parallelogram would literally be 5 times 6. I drew the altitude outside of the parallelogram. I could have drawn it right over here as well. That would also be 6. So the area of this parallelogram would be 30." + }, + { + "Q": "\nat 1:22 doesn't that sign mean delta as well?", + "A": "You re right, it s capital delta (at least it looks like it). Here it just means triangle of course.", + "video_name": "tFhBAeZVTMw", + "timestamps": [ + 82 + ], + "3min_transcript": "We know that quadrilateral ABCD over here is a parallelogram. And what I want to discuss in this video is a general way of finding the area of a parallelogram. In the last video, we talked about a particular way of finding the area of a rhombus. You could take half the product of its diagonals. And a rhombus is a parallelogram. But you can't just generally take half the product of the diagonals of any parallelogram. It has to be a rhombus. And now we're just going to talk about parallelograms. So what do we know about parallelograms? Well, we know the opposite sides are parallel. So that side is parallel to that side, and this side is parallel to this side. And we also know that opposite sides are congruent. So this length is equal to this length, and this length is equal to this length over here. Now, if we draw a diagonal-- I'll draw a diagonal AC-- we can split our parallelogram into two triangles. And we've proven this multiple times. These two triangles are congruent. But we can do it in a pretty straightforward way. Obviously, AD is equal to BC. We have DC is equal to AB. And then both of these triangles share this third side right over here. They both share AC. So we can say triangle-- let me write this in yellow-- we could say triangle ADC is congruent to triangle-- let me get this right. So it's going to be congruent to triangle-- so I said A, D, C. So I went along this double magenta slash first, then the pink, and then I went D, and then I went the last one. So I'm going to say CBA. Because I went the double magenta, then pink, then the last one. So CBA, triangle CBA. And this is by side, side, side congruency. All three sides, they have three corresponding sides that are congruent to each other. So the triangles are congruent to each other. And what that tells us is that the areas of these two So if I want to find the area of ABCD, the whole parallelogram, it's going to be equal to the area of triangle-- let me just write it here-- it's equal to the area of ADC plus the area of CBA. But the area of CBA is just the same thing as the area of ADC because they are congruent, by side, side, side. So this is just going to be two times the area of triangle ADC. Which is convenient for us, because we know how to find the areas of triangles. Area of triangles is literally just 1/2 times base times height. So it's 1/2 times base times height of this triangle. And we are given the base of ADC. It is this length right over here. It is DC. You could view it as the base of the entire parallelogram. And if we wanted to figure out the height, we could draw an altitude down like this." + }, + { + "Q": "At 0:54, Khan says to factor out the 2. Is that necessary?\n", + "A": "We factor out 2 to bring expressions in the brackets to similar form. Thus, you can easily see that parts inside the brackets are similar on both sides of the equation. In this form it is quite easy to identify and make a substitution. You don t have to do this step if you can identify similarities right away, remember (4x-6)=2(2x-3).", + "video_name": "ZIqW_sXymrM", + "timestamps": [ + 54 + ], + "3min_transcript": "- [Voiceover] So let's try to find the solutions to this equation right over here. We have the quantity two X minus three squared, and that is equal to four X minus six, and I encourage you to pause the video and give it a shot. And I'll give you a little bit of a hint, you could do this in the traditional way of expanding this out, and then turning it into kind of a classic quadratic form, but there might be a faster or a simpler way to do this if you really pay attention to the structure of both sides of this equation. Well let's look at this, we have two X minus three squared on the left-hand side, on the right-hand side we have four X minus six. Well four X minus six, that's just two times two X minus three, let me be clear there, so this is the same thing as two X minus three squared is equal to, four X minus six, if I factor out a two, that's two times two X minus three. And so this is really interesting, we have something squared is equal to two times that something. let me be very clear here, so the stuff in blue squared is equal to two times the stuff in blue. So if we can solve for what the stuff in blue could be equal to, then we could solve for X, and I'll show you that right now. So let's say, let's just replace two X minus three, we'll do a little bit of a substitution, let's replace that with P. So let's say that P is equal to two X minus three. Well then this equation simplifies quite nicely, the left-hand side becomes P squared, P squared is equal to two times P, 'cause once again two X minus three is P, two times P. And now we just have to solve for P. And I'll switch to just one color now. So we can write this as, if we subtract two P from both sides, we can get P squared minus two P and we can factor out a P, so we get P times P minus two is equal to zero. And we've seen this shown multiple times, if I have the product of two things and they equal to zero, at least one of them needs to be equal to zero, so either P is equal to zero, or P minus two is equal to zero. Well if P minus two is equal to zero, then that means P is equal to two. So either P equals zero, or P equals two. Well we're not quite done yet, because we wanted to solve for X, and not for P. But luckily we know that two X minus three is equal to P. So now we could say either two X minus three is going to be equal to this P value, is going to be equal to zero, or two X minus three is going to be equal to this P value, is going to be equal to two. And so this is pretty straightforward to solve," + }, + { + "Q": "\nAt 3:48 of the video why does he want to multiply and take 1/3 ?", + "A": "He s trying to get the x by itself on one side of the equation and as long as you do the same thing to both sides then everything equals out. So to get from 3x to x you divide by 3 which is the same as multiplying it by 1/3. Since want to do the same thing to both sides you multiply 12 by 1/3 (which is the same as dividing it by 3). 12 * 1/3 = 12/3 = 4. So x=4.", + "video_name": "_y_Q3_B2Vh8", + "timestamps": [ + 228 + ], + "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced." + }, + { + "Q": "\nAt 4:11, why does he multiply by 1/3? Wouldn't it be easier to divide by 3?", + "A": "Multiplying with 1/3 is the same with dividing by 3.", + "video_name": "_y_Q3_B2Vh8", + "timestamps": [ + 251 + ], + "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced." + }, + { + "Q": "I just noticed at 4:26, there is 6 yellow blocks on the right when there's supposed to be 4 blocks on the right.\n", + "A": "The video has amended this. If you watch it again, you should notice the box in the lower right corner.", + "video_name": "_y_Q3_B2Vh8", + "timestamps": [ + 266 + ], + "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced." + }, + { + "Q": "at 4:23 There were 6 boxes on the right side. WAs that on purpose or what.\n", + "A": "that was an accident", + "video_name": "_y_Q3_B2Vh8", + "timestamps": [ + 263 + ], + "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced." + }, + { + "Q": "\n@4:26 he left 6 on the right did anyone else notice that?", + "A": "Yeah. I noticed it.", + "video_name": "_y_Q3_B2Vh8", + "timestamps": [ + 266 + ], + "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced." + }, + { + "Q": "Any one else notice Sal repeats himself a lot? 2:40 to 2:53\n", + "A": "Repetition is the mother of learning :)", + "video_name": "_y_Q3_B2Vh8", + "timestamps": [ + 160, + 173 + ], + "3min_transcript": "We can simply count them: [counts to 14] Fourteen blocks, each has a mass of 1 kg, so the total mass will be 14 kg. And we see that the scale is balanced, not tilting down or upwards. So this mass over here must be equal to this total mass. The scale is balanced, so we can write an 'equal'-sign. (let me do that in a white coulour, I do not like that brown) Now, what I want you to think about, and you can think about it either through the symbols or through the scales, is: how would you go about -- let us think about a few things: how would you first go about at least getting rid of these little 1kg blocks? I will give you a second to think about that... Well, the simplest thing is: but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these." + }, + { + "Q": "\nFor 1:40 of the video, how do you know if all three are 30 degrees? How do you find those angles if they were different angles? If they weren't the 30, 60, 90, then how would you find the angles? I get the 90 degrees part, but what about the rest?", + "A": "The reason that Sal knows those angles all equal 30\u00c2\u00b0 is because the problem states that the 90\u00c2\u00b0 angle is trisected or split into 3 equal parts. So 90\u00c2\u00b0/3=30\u00c2\u00b0.", + "video_name": "dgHksfBFbjk", + "timestamps": [ + 100 + ], + "3min_transcript": "So we have this rectangle right over here, and we're told that the length of AB is equal to 1. So that's labeled right over there. AB is equal to 1. And then they tell us that BE and BD trisect angle ABC. So BE and BD trisect angle ABC. So trisect means dividing it into 3 equal angles. So that means that this angle is equal to this angle is equal to that angle. And what they want us to figure out is, what is the perimeter of triangle BED? So it's kind of this middle triangle in the rectangle right over here. So at first this seems like a pretty hard problem, because you're like well, what is the width of this rectangle. How can I even start on this? They've only given us one side here. But they've actually given us a lot of information, given that we do know this is a rectangle. We have four sides, and that we have four angles. The sides are all parallel to each other and that the angles are all 90 degrees. Which is more than enough information And so one thing we do know is that opposite sides of a rectangle are the same length. So if this side is 1, then this side right over there is also 1. The other thing we know is that this angle is trisected. Now we know what the measure of this angle is. It was a right angle, it was a 90 degree angle. So if it's divided into three equal parts, that tells us that this angle right over here is 30 degrees, this angle right over here is 30 degrees, and then this angle right over here is 30 degrees. And then we see that we're dealing with a couple of 30-60-90 triangles. This one is 30, 90, so this other side right over here needs to be 60 degrees. This triangle right over here, you have 30, you have 90, so this one has to be 60 degrees. They have to add up to 180, 30-60-90 triangle. And you can also figure out the measures of this triangle, although it's not going to be a right triangle. But knowing what we know about 30-60-90 triangles, if we just have one side of them, So for example, here we have the shortest side. We have the side opposite of the 30 degree side. Now, if the 30 degree side is 1, then the 60 degree side is going to be square root of 3 times that. So this length right over here is going to be square root of 3. And that's pretty useful because we now just figured out the length of the entire base of this rectangle right over there. And we just used our knowledge of 30-60-90 triangles. If that was a little bit mysterious, how I came up with that, I encourage you to watch that video. We know that 30-60-90 triangles, their sides are in the ratio of 1 to square root of 3 to 2. So this is 1, this is a 30 degree side, this is going to be square root of 3 times that. And the hypotenuse right over here is going to be 2 times that. So this length right over here is going to be 2 times this side right over here. So 2 times 1 is just 2. So that's pretty interesting." + }, + { + "Q": "I understand this topic until the last part. I thought that the opposite of the 60 degree angle was supposed to be sqrt(3) and that the side opposite the 90 degree angle is 2 times the number. I don't understand how he got 2/sqrt(3) and 1/sqrt(3). I become confused around 3:16.\n", + "A": "Yes, you multiply the short side by sqrt(3) to find the side opposite the 60\u00c2\u00b0 angle. This problem is different because they gave the you side opposite the 60\u00c2\u00b0 angle and you need to find the short side. So essentially you are saying that the short side times sqrt(3) = 1; therefore to find the length of the short side you have to divide by sqrt(3) to find the short side. This gives you the 1/sqrt(3) for the short side and you multiply this by 2/sqrt(3).", + "video_name": "dgHksfBFbjk", + "timestamps": [ + 196 + ], + "3min_transcript": "And so one thing we do know is that opposite sides of a rectangle are the same length. So if this side is 1, then this side right over there is also 1. The other thing we know is that this angle is trisected. Now we know what the measure of this angle is. It was a right angle, it was a 90 degree angle. So if it's divided into three equal parts, that tells us that this angle right over here is 30 degrees, this angle right over here is 30 degrees, and then this angle right over here is 30 degrees. And then we see that we're dealing with a couple of 30-60-90 triangles. This one is 30, 90, so this other side right over here needs to be 60 degrees. This triangle right over here, you have 30, you have 90, so this one has to be 60 degrees. They have to add up to 180, 30-60-90 triangle. And you can also figure out the measures of this triangle, although it's not going to be a right triangle. But knowing what we know about 30-60-90 triangles, if we just have one side of them, So for example, here we have the shortest side. We have the side opposite of the 30 degree side. Now, if the 30 degree side is 1, then the 60 degree side is going to be square root of 3 times that. So this length right over here is going to be square root of 3. And that's pretty useful because we now just figured out the length of the entire base of this rectangle right over there. And we just used our knowledge of 30-60-90 triangles. If that was a little bit mysterious, how I came up with that, I encourage you to watch that video. We know that 30-60-90 triangles, their sides are in the ratio of 1 to square root of 3 to 2. So this is 1, this is a 30 degree side, this is going to be square root of 3 times that. And the hypotenuse right over here is going to be 2 times that. So this length right over here is going to be 2 times this side right over here. So 2 times 1 is just 2. So that's pretty interesting. with this side right over here. Here the 1 is not the side opposite the 30 degree side. Here the 1 is the side opposite the 60 degree side. So once again, if we multiply this side times square root of 3, we should get this side right over This is the 60, remember this 1, this is the 60 degree side. So this has to be 1 square root of 3 of this side. Let me write this down, 1 over the square root of 3. And the whole reason, the way I was able to get this is, well, whatever this side, if I multiply it by the square root of 3, I should get this side right over here. I should get the 60 degree side, the side opposite the 60 degree angle. Or if I take the 60 degree side, if I divide it by the square root of 3 I should get the shortest side, the 30 degree side. So if I start with the 60 degree side, divide by the square root of 3, I get that right over there. And then the hypotenuse is always" + }, + { + "Q": "How did Sal get 3sqrt(3) from just sqrt(3) at 6:05. Can someone please explain and link the appropriate khan lesson for that.\n", + "A": "Sal actually got 3sqrt(3)/3 because he multiplied both the numerator and denominator by 3.", + "video_name": "dgHksfBFbjk", + "timestamps": [ + 365 + ], + "3min_transcript": "So this is the side opposite the 30 degree angle. The hypotenuse is always twice that. So this is the side opposite the 30 degree angle. The hypotenuse is going to be twice that. It is going to be 2 over the square root of 3. So we're doing pretty good. We have to figure out the perimeter of this inner triangle right over here. We already figured out one length is 2. We figured out another length is 2 square roots of 3. And then all we have to really figure out is, what ED is. And we can do that because we know that AD is going to be the same thing as BC. We know that this entire length, because we're dealing with a rectangle, is the square root of 3. If that entire length is square root of 3, if this AE is 1 over the square root of 3, then this length right over here, ED is going to be square root of 3 minus 1 over the square root of 3. That length minus that length right over there. And how to find the perimeter is pretty straight forward. We just have to add these things up and simplify it. this, perimeter of triangle BED is equal to-- This is short for perimeter. I just didn't feel like writing the whole word.-- is equal to 2 over the square root of 3 plus square root of 3 minus 1 over the square root of 3 plus 2. And now this just boils down to simplifying radicals. You could take a calculator out and get some type of decimal approximation for it. Let's see, if we have 2 square root of 3 minus 1 square root of 3, that will leave us with 1 over the square root of 3. 2 over the square of 3 minus 1 over the square root of 3 is 1 over the square root of 3. And then you have the square root of 3 plus 2. And let's see, I can rationalize this. If I multiply the numerator and the denominator by the square root of 3, this gives me the square root of 3 over 3 plus the square root of 3, square roots of 3 over 3. I just multiplied this times 3 over 3 plus 2. And so this gives us-- this is the drum roll part now-- so one square root of 3 plus 3 square roots of 3, and all of that over 3, gives us 4 square roots of 3 over 3 plus 2. Or you could put the 2 first. Some people like to write the non-irrational part before the irrational part. But we're done. We figured out the perimeter. We figured out the perimeter of this inner triangle BED, right there." + }, + { + "Q": "at 3:59, why do we have to subtract 6100 from 400 & 6100?\n", + "A": "Yes, we are isolating the cosine of theta (at 3:59) . You are right.", + "video_name": "Ei54NnQ0FKs", + "timestamps": [ + 239 + ], + "3min_transcript": "one of these to be A or B. We could say that this A is 50 meters and B is 60 meters. And now we could just apply the law of cosines. So the law of cosines tells us that 20-squared is equal to A-squared, so that's 50 squared, plus B-squared, plus 60 squared, minus two times A B. So minus two times 50, times 60, times 60, times the cosine of theta. This works out well for us because they've given us everything. There's really only one unknown. There's theta here. So let's see if we can solve for theta. So 20 squared, that is 400. 50 squared is 2,500. And then 50 times... Let's see, two times 50 is 100, times 60, this is all equal to 6,000. So let's see, if we simplify this a little bit we're going to get 400 is equal to 2,500 plus 3,600. Let's see, that'd be 6,100. That's equal to 6,000... Let me do this in a new color. So when I add these two, I get 6,100. Did I do that right? So it's 2,000 plus 3,000, plus 5,000. 500 plus 600 is 1,100. So I get 6,100 minus 6,000, times the cosine of theta. And let's see, now we can subtract 6,100 from both sides. So I'm just gonna subtract 6,100 from both sides So let's do that. So this is going to be negative 5,700. Is that right? 5,700 plus... Yes, that is right. Right, because if this was the other way around, if this was 6,100 minus 400 it would be positive 5,700. Alright. And then these two of course cancel out. And this is going to be equal to negative 6,000 times the cosine of theta. Now we can divide both sides by negative 6,000. And we get... I'm just gonna swap the sides. We get cosine of theta is equal to... Let's see we could divide the numerator and the denominator by essentially negative 100." + }, + { + "Q": "\nI have a question about when you are doing the equation and you are crossing out and everything when you was getting on the part when you add on one side and then you subtracted on the other why did you do that? I have always been taught that what you do to one side that you do to the other can you please explain that to me? I do believe it is on minute 5:04 Thank you for the help in advance", + "A": "I don t really understand what you are asking, but I think I have an idea. When the equation was -5700 = -6000 cos (Theta) It also meant -5700 = -6000 times cos (Theta). He divided -6000 from the right, so he did the same to the left. It equaled -5700/-6000 on the left, which is the same as 5700/6000. I think you were just confused by the way he drew the positive signs. I hope I helped! Good luck!", + "video_name": "Ei54NnQ0FKs", + "timestamps": [ + 304 + ], + "3min_transcript": "And then 50 times... Let's see, two times 50 is 100, times 60, this is all equal to 6,000. So let's see, if we simplify this a little bit we're going to get 400 is equal to 2,500 plus 3,600. Let's see, that'd be 6,100. That's equal to 6,000... Let me do this in a new color. So when I add these two, I get 6,100. Did I do that right? So it's 2,000 plus 3,000, plus 5,000. 500 plus 600 is 1,100. So I get 6,100 minus 6,000, times the cosine of theta. And let's see, now we can subtract 6,100 from both sides. So I'm just gonna subtract 6,100 from both sides So let's do that. So this is going to be negative 5,700. Is that right? 5,700 plus... Yes, that is right. Right, because if this was the other way around, if this was 6,100 minus 400 it would be positive 5,700. Alright. And then these two of course cancel out. And this is going to be equal to negative 6,000 times the cosine of theta. Now we can divide both sides by negative 6,000. And we get... I'm just gonna swap the sides. We get cosine of theta is equal to... Let's see we could divide the numerator and the denominator by essentially negative 100. So cosine of theta is equal to 57 over 60. And actually that can be simplified even more. Three goes into 57, is that 19 times? Yep, so this is actually... This could be simplified. This is equal to 19 over 20. We actually didn't have to do that simplification step because we're about to use our calculators, but that makes the math a little more tractable. Right, 3 goes into 57, yeah, 19 times. And so now we can take the inverse cosine of both sides. So we could get theta is equal to the inverse cosine, or the arc cosine, of 19 over 20. So let's get our calculator out and see if we get something that makes sense. So we wanna do the inverse cosine of 19 over 20." + }, + { + "Q": "\nAt 2:14, Sal says,\"The best way to get 12x on the left is to subtract.\" Was he implying that there's more than one way to balance the equation? If so, how?", + "A": "It was just a figure of speech. The only way to move that 12x from one side to the other is to subtract it. However, you could add 9x to both sides instead. Your x s will end up on the right hand side instead of the left but that s OK. x = -1 means the exact same thing as -1 = x", + "video_name": "YZBStgZGyDY", + "timestamps": [ + 134 + ], + "3min_transcript": "We have the equation negative 9 minus this whole expression, 9x minus 6-- this whole thing is being subtracted from negative 9-- is equal to 3 times this whole expression, 4x plus 6. Now, a good place to start is to just get rid of these parentheses. And the best way to get rid of these parentheses is to kind of multiply them out. This has a negative 1-- you just see a minus here, but it's just really the same thing as having a negative 1-- times this quantity. And here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left-hand side of our equation, we have our negative 9. And then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to-- let's distribute the 3-- 3 times 4x is 12x. Now what we want to do, let's combine our constant terms, if we can. We have a negative 9 and a 6 here, on this side, we've combined all of our like terms. We can't combine a 12x and an 18, so let's combine this. So let's combine the negative 9 and the 6, our two constant terms on the left-hand side of the equation. So we're going to have this negative 9x. So we're going to have negative 9x plus-- let's see, we have a negative 9 and then plus 6-- so negative 9 plus 6 is negative 3. So we're going to have a negative 9x, and then we have a negative 3, so minus 3 right here. That's the negative 9 plus the 6, and that is equal to 12x plus 18. Now, we want to group all the x terms on one side of the equation, and all of the constant terms-- the negative 3 and the positive 18 on the other side-- I like to always You don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right. And the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now, on the left-hand side, I have negative 9x minus 12x. So negative 9 minus 12, that's negative 21. Negative 21x minus 3 is equal to-- 12x minus 12x, well, that's just nothing. So I could write a 0 here, but I don't That was the whole point of subtracting the 12x from the left-hand side. And that is going to be equal to-- so on the right-hand side, we just are left with an 18. We are just left with that 18 here. These guys canceled out. Now, let's get rid of this negative 3 from the left-hand side. So on the left-hand side, we only have x terms, and on the right-hand side, we only have constant terms. So the best" + }, + { + "Q": "At 2:12, Sal says that -12x is the best way to do it. I thought that was the only way- have I been missing something?\n", + "A": "You could also add 9x to both sides.", + "video_name": "YZBStgZGyDY", + "timestamps": [ + 132 + ], + "3min_transcript": "We have the equation negative 9 minus this whole expression, 9x minus 6-- this whole thing is being subtracted from negative 9-- is equal to 3 times this whole expression, 4x plus 6. Now, a good place to start is to just get rid of these parentheses. And the best way to get rid of these parentheses is to kind of multiply them out. This has a negative 1-- you just see a minus here, but it's just really the same thing as having a negative 1-- times this quantity. And here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left-hand side of our equation, we have our negative 9. And then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to-- let's distribute the 3-- 3 times 4x is 12x. Now what we want to do, let's combine our constant terms, if we can. We have a negative 9 and a 6 here, on this side, we've combined all of our like terms. We can't combine a 12x and an 18, so let's combine this. So let's combine the negative 9 and the 6, our two constant terms on the left-hand side of the equation. So we're going to have this negative 9x. So we're going to have negative 9x plus-- let's see, we have a negative 9 and then plus 6-- so negative 9 plus 6 is negative 3. So we're going to have a negative 9x, and then we have a negative 3, so minus 3 right here. That's the negative 9 plus the 6, and that is equal to 12x plus 18. Now, we want to group all the x terms on one side of the equation, and all of the constant terms-- the negative 3 and the positive 18 on the other side-- I like to always You don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right. And the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now, on the left-hand side, I have negative 9x minus 12x. So negative 9 minus 12, that's negative 21. Negative 21x minus 3 is equal to-- 12x minus 12x, well, that's just nothing. So I could write a 0 here, but I don't That was the whole point of subtracting the 12x from the left-hand side. And that is going to be equal to-- so on the right-hand side, we just are left with an 18. We are just left with that 18 here. These guys canceled out. Now, let's get rid of this negative 3 from the left-hand side. So on the left-hand side, we only have x terms, and on the right-hand side, we only have constant terms. So the best" + }, + { + "Q": "At 6:21 how come pi as a coefficient is allowed? Isn't pi an irrational number?\n", + "A": "Coefficients can be any real number including Pi.", + "video_name": "Vm7H0VTlIco", + "timestamps": [ + 381 + ], + "3min_transcript": "So, this right over here is a coefficient. It can be, if we're dealing... Well, I don't wanna get too technical. Positive, negative number. Could be any real number. We have our variable. And then the exponent, here, has to be nonnegative. Nonnegative integer. So here, the reason why what I wrote in red is not a polynomial is because here I have an exponent that is a negative integer. Let's give some other examples of things that are not polynomials. So, if I were to change the second one to, instead of nine a squared, if I wrote it as nine a to the one half power minus five, this is not a polynomial because this exponent right over here, it is no longer an integer; it's one half. the square root of a minus five. This also would not be a polynomial. Or, if I were to write nine a to the a power minus five, also not a polynomial because here the exponent is a variable; it's not a nonnegative integer. All of these are examples of polynomials. There's a few more pieces of terminology that are valuable to know. Polynomial is a general term for one of these expression that has multiple terms, a finite number, so not an infinite number, and each of the terms has this form. But there's more specific terms for when you have only one term or two terms or three terms. When you have one term, it's called a monomial. This is a monomial. which we could write as six x to the zero. Another example of a monomial might be 10z to the 15th power. That's also a monomial. Your coefficient could be pi. Pi. Whoops. Could be pi. So we could write pi times b to the fifth power. Any of these would be monomials. So what's a binomial? Binomial's where you have two terms. Monomial, mono for one, one term. Binomial is you have two terms. This right over here is a binomial. Binomial. You have two terms. All these are polynomials but these are subclassifications. It's a binomial; you have one, two terms. Another example of a binomial would be three y to the third plus five y. Once again, you have two terms that have this form right over here." + }, + { + "Q": "\nAt 3:45, Sal says that he could change it to x to the power of zero. Is this incorrect ? Anything to the power of zero is one, and 9x is not always equal to nine.", + "A": "9x^0 does not be come 9x. 9x means 9x^1 Having x to the 1st power is not the same as x^0 power. x^0 = 1, not x 9x^0 = 9(1) = 9 Hope this helps.", + "video_name": "Vm7H0VTlIco", + "timestamps": [ + 225 + ], + "3min_transcript": "for what makes something a polynomial. You have to have nonnegative powers of your variable in each of the terms. I just used that word, terms, so lemme explain it, 'cause it'll help me explain what a polynomial is. A polynomial is something that is made up of a sum of terms. And so, for example, in this first polynomial, the first term is 10x to the seventh; the second term is negative nine x squared; the next term is 15x to the third; and then the last term, maybe you could say the fourth term, is nine. You can see something. Let me underline these. These are all terms. This is a four-term polynomial right over here. You could say: \"Hey, wait, this thing you wrote in red, \"this also has four terms.\" We have to put a few more rules for it to officially be a polynomial, especially a polynomial in one variable. Each of those terms are going to be made up of a coefficient. This is the thing that multiplies So in this first term the coefficient is 10. Lemme write this word down, coefficient. It's another fancy word, but it's just a thing that's multiplied, in this case, times the variable, which is x to seventh power. The first coefficient is 10. The next coefficient. Actually, lemme be careful here, because the second coefficient here is negative nine. We are looking at coefficients. The third coefficient here is 15. You can view this fourth term, or this fourth number, as the coefficient because this could be rewritten as, instead of just writing as nine, you could write it as nine x to the zero power. And then it looks a little bit clearer, like a coefficient. So, in general, a polynomial is the sum of a finite number of terms where each term has a coefficient, which I could represent with the letter A, being multiplied by a variable So, this right over here is a coefficient. It can be, if we're dealing... Well, I don't wanna get too technical. Positive, negative number. Could be any real number. We have our variable. And then the exponent, here, has to be nonnegative. Nonnegative integer. So here, the reason why what I wrote in red is not a polynomial is because here I have an exponent that is a negative integer. Let's give some other examples of things that are not polynomials. So, if I were to change the second one to, instead of nine a squared, if I wrote it as nine a to the one half power minus five, this is not a polynomial because this exponent right over here, it is no longer an integer; it's one half." + }, + { + "Q": "\nAt 8:15 Sal mentions that the highest degree will be the degree of the polynomial. What if there was an equation where there were two terms and they both had the same degree? What will the degree of the polynomial be?", + "A": "The degree of the entire polynomial would just be that degree", + "video_name": "Vm7H0VTlIco", + "timestamps": [ + 495 + ], + "3min_transcript": "which we could write as six x to the zero. Another example of a monomial might be 10z to the 15th power. That's also a monomial. Your coefficient could be pi. Pi. Whoops. Could be pi. So we could write pi times b to the fifth power. Any of these would be monomials. So what's a binomial? Binomial's where you have two terms. Monomial, mono for one, one term. Binomial is you have two terms. This right over here is a binomial. Binomial. You have two terms. All these are polynomials but these are subclassifications. It's a binomial; you have one, two terms. Another example of a binomial would be three y to the third plus five y. Once again, you have two terms that have this form right over here. Trinomial's when you have three terms. Trinomial. This right over here is an example. This is the first term; this is the second term; and this is the third term. Now, the next word that you will hear often in the context with polynomials is the notion of the degree of a polynomial. You might hear people say: \"What is the degree of a polynomial?\", or \"What is the degree of a given term of a polynomial?\" Let's start with the degree of a given term. Let's go to this polynomial here. We have this first term, 10x to the seventh. The degree is the power that we're raising the variable to. So this is a seventh-degree term. The second term is a second-degree term. The third term is a third-degree term. And you could view this constant term, which is really just nine, you could view that as, sometimes people say the constant term. If people are talking about the degree of the entire polynomial, they're gonna say: \"What is the degree of the highest term? \"What is the term with the highest degree?\" That degree will be the degree of the entire polynomial. So, this first polynomial, this is a seventh-degree polynomial. This one right over here is a second-degree polynomial because it has a second-degree term and that's the highest-degree term. This right over here is a third-degree. You could even say third-degree binomial because its highest-degree term has degree three. If this said five y to the seventh instead of five y, then it would be a seventh-degree binomial. This right over here is a 15th-degree monomial. This is a second-degree trinomial. A few more things I will introduce you to is the idea of a leading term and a leading coefficient." + }, + { + "Q": "At 1:40 why did he multiply 4 * 0 ?\n", + "A": "In the original equation it was 4*Y and since he was solving for the X intercept, Y would be equal to 0 and so 4*0", + "video_name": "xGmef7lFc5w", + "timestamps": [ + 100 + ], + "3min_transcript": "" + }, + { + "Q": "at 1:40 what is the somthing ?\n", + "A": "The something is x , what you re solving for.", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 100 + ], + "3min_transcript": "Let's do a few more examples of solving equations. And I think you're going to see that these equations require a few more steps than the ones we did in the last video. But the fun thing about these is that there's more than one way to do it. But as long as you do legitimate steps, as long as anything you do to the left-hand side, you also do to the right-hand side, you should move in the correct direction, or you shouldn't get the wrong answer. So let's do a couple of these. So the first one says-- I'll rewrite it-- 1.3 times x minus 0.7 times x is equal to 12. Well, here the first thing that my instinct is to do, is to merge these two terms. Because I have 1.3 of something minus 0.7 of that same something. This is the same variable. If I have 1.3 apples minus 0.7 apples, well, why don't I subtract 0.7 from 1.3? And I will get 1.3 minus 0.7 x's, or apples, or whatever So is equal to 12. You could imagine that I did the reverse distributive property out here. I factored out an x. But the way my head thinks about it is, I have 1.3 of something minus 0.7 of something, that's going to be equal to 1.3 minus 0.7 of those somethings, that x. And of course 1.3 minus 0.7 is 0.6 times x of my somethings is equal to 12. And now, this looks just like one of the problems we did in the last video. We have a coefficient times x is equal to some other number. Well, let's divide both sides of this equation by that coefficient. Divide both sides by 0.6. So the left-hand side will just become an x. X is equal to-- and what is 12 divided by 0.6? 0.6 goes into 12-- let's add some decimal points here-- 6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12." + }, + { + "Q": "\nAt 12:34 Sal said ten minus -2.35 but he had a plus sign next to -2.35? Im confused, can someone help me?", + "A": "adding a negative number and subtracting a positive number are the same thing :)", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 754 + ], + "3min_transcript": "So the total cost of a cab ride is going to be equal to just the initial hire charge, which is $2.35, plus the $0.75 per mile, times the number of miles. We're letting m is equal to the miles she travels. Miles traveled. So this is the equation. We know that she only has $10 to get home. So her cost has to be $10. So we have to say, the cost has to be $10. So 10 is equal to 2.35 plus 0.75m. So how do we solve for m, or the number of miles Jade can travel? Well, we can get rid of the 2.35 on this right-hand side by subtracting that amount from both So let's do that. So let's subtract minus 2.35 from both sides. These will cancel out. That was the point. The left-hand side-- what is 10 minus 2.35? Now, these will cancel out. Now what is 10 minus 2.35? 10 minus 2 is 8. 10 minus 2.3 is 7.7. So it's going to be 7.65. If you want to believe me, let's do it. 10 minus 2.35. 7.65. And that is going to be equal to 0.75m. Let me write that in that same color. It's nice to see where different things came from. I have, like, five shades of this purple here. so this is that, that is that, and then these two guys canceled out. Now to solve for m, I can just divide both sides by 0.75. So if I divide that side by 0.75, I have to do it to the left-hand side as well. 0.75. That cancels out, so on the right-hand side, I'm left with just an m. And on the left-hand side-- I'll have to get my calculator out for this one-- I have 7.65 divided by 0.75, which is equal to 10.2. m is 10.2, so Jade can travel 10.2 miles." + }, + { + "Q": "\nin 8:18 he say 7 can be written 35/5 , how is that?", + "A": "35 divided by five equals seven, he was simply just showing another way to write whole numbers. But by using fractions! Like 25/5 equals 5, so that s a way of representing five. In my mind, it looks like you know pretty cool things then you put this stuff on a test!", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 498 + ], + "3min_transcript": "inverse of this coefficient. So I multiply 8 over 5 times 5/8s. If I do it to the left-hand side, I have to do it to the right-hand side. 8 over 5. I multiplied by 8 over 5 so that those cancel out and those cancel out. And you are left with s is equal to-- right, this is just a 1-- is equal to-- well, the 5's we can divide. Divide the numerator and the denominator by 5. Divide the numerator by 2 and the denominator by 2. You're left with-- sorry, divide the denominator by 2, you get 6 divided by 2 is 3. You're left with 4/3. s is equal to 4/3. Let's do one more of these. So here I have 5 times q minus 7 over 12 is equal to 2/3. So let me write this. And I could rewrite this as just 5 over 12 times q minus 7 is equal to 2/3. that I can do it two different ways. But as long as I do legitimate operations, I should get the same answer. So the first way I'm going to do it, is I'm going to multiply both sides of this equation by the inverse of 5/12. So I'm going to multiply both sides by 12 over 5. Because I wanted to get rid of this 5/12 on It makes everything look a little bit messy. And I multiply it by 12 over 5, because these are going to The 5 and the 5 cancel out, the 12 and the 12 cancel out. So the left-hand side of my equation becomes q minus 7 is equal to the right-hand side, 2/3 times 5/12. If you divide the 12 by 3, you get a 4. You divide the 3 by 3, you get a 1. So 2 times 4 is 8 over 5. And now we can add 7 to both sides of this equation. So let's add-- I want to do that in a different color-- add 7 to both sides of this equation. That was the whole point of adding the 7. And you are left with q is equal to 8/5 plus 7. Or we could write 8/5 plus 7 can be written as 35/5. And so this is going to be equal to 8-- well, the denominator is 5. 8 plus 35 is 43. So my answer, going this way, is q is equal to 43/5. And I said I would do it two ways. Let's do it another way. So let me write the same problem down. So I have 5/12-- actually, let me just do it a completely different way. Let me write it the way they wrote it. 5 times q minus 7, over 12 is equal to 2/3. Let me just get rid of the 12 first. Let me multiply both" + }, + { + "Q": "at 3:34 how did he get the negative 2\n", + "A": "here is the full equation: 5x -1(3x + 2) = so then you distribute the -1 and you get: 5x - 3x - 2 = hope this helps!", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 214 + ], + "3min_transcript": "6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12. We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that?" + }, + { + "Q": "At 5:34 why did you or how did you rewrite it to 8/8?\n", + "A": "So that he could subtract his number that was over 8. You probably know that 1=8/8, as well as 4/4 etc... So by making the denominators the same, therefore being able to subtract a number from 1. Hope this helps! :)", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 334 + ], + "3min_transcript": "And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something. inverse of this coefficient. So I multiply 8 over 5 times 5/8s. If I do it to the left-hand side, I have to do it to the right-hand side. 8 over 5. I multiplied by 8 over 5 so that those cancel out and those cancel out. And you are left with s is equal to-- right, this is just a 1-- is equal to-- well, the 5's we can divide. Divide the numerator and the denominator by 5. Divide the numerator by 2 and the denominator by 2. You're left with-- sorry, divide the denominator by 2, you get 6 divided by 2 is 3. You're left with 4/3. s is equal to 4/3. Let's do one more of these. So here I have 5 times q minus 7 over 12 is equal to 2/3. So let me write this. And I could rewrite this as just 5 over 12 times q minus 7 is equal to 2/3." + }, + { + "Q": "@ 4:43 how did he end up with 3/2?\n", + "A": "The problem was 5x-(3x+2)=1. He first, to get rid of the parentheses, reversed the sign inside (because the sign outside was negative). He then subtracted 3x from 5x, and got 2x. So far it is 2x-2=1. You can then add 2 to both sides, to get 2x=3, and divided both sides by 2, to get x=3/2. You can look at the previous videos if you re still confused. Hope it helped!", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 283 + ], + "3min_transcript": "We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something." + }, + { + "Q": "\nAt 3:35, where does he get the -2 from? Using the Distributive property doesn't change the sign, right? Why would re-arranging the numbers turn a +2 into a -2?", + "A": "Because, he is having to multiply EVERYTHING in the parenthesis by a negative. There is no number written, but in between the negative and parenthesis symbol is an imaginary number 1. So it is a negative 1.", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 215 + ], + "3min_transcript": "6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12. We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that?" + }, + { + "Q": "\nat 5:28 why does he change it to 1S to S(1-3/8)", + "A": "He is using the distributive property to combine like terms. He isn t changing 1S to S(1-3/8) he s changing 1S-3/8S to S(1-3/8).", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 328 + ], + "3min_transcript": "We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something." + }, + { + "Q": "\nAt 3:35, how did you change (3x+2) into -3x-2?", + "A": "He distributed -1 to the expression. There was a - before the 3x+2. That s actually a hidden -1. -1(3x+2)=-3x-2. Hope this helps!", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 215 + ], + "3min_transcript": "6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12. We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that?" + }, + { + "Q": "And why 8/8 on 5:43?\n", + "A": "Because he was subtracting 3/8 from one so 1 - 3/8 = 1/1 - 3/8, since he needs a common denominator he multiplied the denominator by 8 and whatever you do to the denominator you must do to the nummerator thus he got 8/8 - 3/8", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 343 + ], + "3min_transcript": "And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something. inverse of this coefficient. So I multiply 8 over 5 times 5/8s. If I do it to the left-hand side, I have to do it to the right-hand side. 8 over 5. I multiplied by 8 over 5 so that those cancel out and those cancel out. And you are left with s is equal to-- right, this is just a 1-- is equal to-- well, the 5's we can divide. Divide the numerator and the denominator by 5. Divide the numerator by 2 and the denominator by 2. You're left with-- sorry, divide the denominator by 2, you get 6 divided by 2 is 3. You're left with 4/3. s is equal to 4/3. Let's do one more of these. So here I have 5 times q minus 7 over 12 is equal to 2/3. So let me write this. And I could rewrite this as just 5 over 12 times q minus 7 is equal to 2/3." + }, + { + "Q": "\nAt around 4:50, you say that x^l * x^m = x^(l+m)\nI thought it would be 2x^(l+m)\n\nWHat am I missing?", + "A": "Where would you get the 2 from? You are multiplying, not adding. Try this with some actual number, maybe that might make the concept more clear: 4\u00c2\u00b2 * 4\u00c2\u00b3 = 4*4 * 4*4*4 = 4\u00e2\u0081\u00b5 = 1024", + "video_name": "FP2arCfAfBY", + "timestamps": [ + 290 + ], + "3min_transcript": "It's pretty neat. x to the n is equal to a times b. x to the n is equal to a times b. And that's just like saying that log base x is equal to a times b. So what can we do with all of this? Well, let's start with with this right here. x to the n is equal to a times b. So, how could we rewrite this? Well, a is this. And b is this, right? So let's rewrite that. So we know that x to the n is equal to a. a is this. x to the l. x to the l. And what's b? Times b. Not doing anything fancy right now. But what's x to the l times x to the m? Well, we know from the exponents, when you multiply two expressions that have the same base and different exponents, you just add the exponents. So this is equal to, let me take a neutral color. I don't know if I said that verbally correct, but you get the point. When you have the same base and you're multiplying, you can just add the exponents. That equals x to the, I want to keep switching colors, because I think that's useful. l, l plus m. That's kind of onerous to keep switching colors, but. You get what I'm saying. So, x to the n is equal to x to the l plus m. Let me put the x here. Oh, I wanted that to be green. x to the l plus n. So what do we know? We know x to the n is equal to x to the l plus m. Right? Well, we have the same base. So we know that n is equal to l l plus m. What does that do for us? I've kind of just been playing around with logarithms. Am I getting anywhere? I think you'll see that I am. Well, what's another way of writing n? So we said, x to the n is equal to a times b -- oh, I actually skipped a step here. So that means -- so, going back here, x to the n That means that log base x of a times b is equal to n. You knew that. I hope you don't realize I'm not backtracking or anything. I just forgot to write that down when I first did it. But, anyway. So, what's n? What's another way of writing n? Well, another way of writing n is right here. Log base x of a times b." + }, + { + "Q": "at 7:08 Sal does not multiply (a+b) squared and then on the right sides multiplies ab x 2. Why?\n", + "A": "You can t multiply two different variables.", + "video_name": "EINpkcphsPQ", + "timestamps": [ + 428 + ], + "3min_transcript": "The average of the bottom and the top gives you the area of the trapezoid. Now, how could we also figure out the area with its component parts? Regardless of how we calculate the area, as long as we do correct things, we should come up with the same result. So how else can we come up with this area? Well, we could say it's the area of the two right triangles. The area of each of them is one half a times b, but there's two of them. Let me do that b in that same blue color. But there's two of these right triangle. So let's multiply by two. So two times one half ab. That takes into consideration this bottom right triangle and this top one. And what's the area of this large one that I will color in in green? Well, that's pretty straightforward. It's just one half c times c. So plus one half c times c, which is one half c squared. Now, let's simplify this thing and see what we come up with, and you might guess where all of this is going. So let's see what we get. So we can rearrange this. Let me rearrange this. So one half times a plus b squared is going to be equal to 2 times one half. Well, that's just going to be one. So it's going to be equal to a times b, plus one half c squared. Well, I don't like these one halfs laying around, so let's multiply both sides of this equation by 2. I'm just going to multiply both sides of this equation by 2. So let me write that. And on the right-hand side, I am left with 2ab. Trying to keep the color coding right. And then, 2 times one half c squared, that's just going to be c squared plus c squared. Well, what happens if you multiply out a plus b times a plus b? What is a plus b squared? Well, it's going to be a squared plus 2ab plus 2ab plus b squared. And then, our right-hand side it's going to be equal to all of this business. And changing all the colors is difficult for me, so let me copy and let me paste it. So it's still going to be equal to the right-hand side." + }, + { + "Q": "at around 7:00 how and where did he get the 2's with?\n", + "A": "Those are power of twos, the formula for pythagorean theorem is a^2+B^2= C^2", + "video_name": "EINpkcphsPQ", + "timestamps": [ + 420 + ], + "3min_transcript": "The average of the bottom and the top gives you the area of the trapezoid. Now, how could we also figure out the area with its component parts? Regardless of how we calculate the area, as long as we do correct things, we should come up with the same result. So how else can we come up with this area? Well, we could say it's the area of the two right triangles. The area of each of them is one half a times b, but there's two of them. Let me do that b in that same blue color. But there's two of these right triangle. So let's multiply by two. So two times one half ab. That takes into consideration this bottom right triangle and this top one. And what's the area of this large one that I will color in in green? Well, that's pretty straightforward. It's just one half c times c. So plus one half c times c, which is one half c squared. Now, let's simplify this thing and see what we come up with, and you might guess where all of this is going. So let's see what we get. So we can rearrange this. Let me rearrange this. So one half times a plus b squared is going to be equal to 2 times one half. Well, that's just going to be one. So it's going to be equal to a times b, plus one half c squared. Well, I don't like these one halfs laying around, so let's multiply both sides of this equation by 2. I'm just going to multiply both sides of this equation by 2. So let me write that. And on the right-hand side, I am left with 2ab. Trying to keep the color coding right. And then, 2 times one half c squared, that's just going to be c squared plus c squared. Well, what happens if you multiply out a plus b times a plus b? What is a plus b squared? Well, it's going to be a squared plus 2ab plus 2ab plus b squared. And then, our right-hand side it's going to be equal to all of this business. And changing all the colors is difficult for me, so let me copy and let me paste it. So it's still going to be equal to the right-hand side." + }, + { + "Q": "\nAt 7:44 in the video when you were solving the equation you managed to get a \"2ab\" after solving (a+b)^2. Why is that? Please explain.", + "A": "Try rewriting (a+b)^2 as (a+b)(a+b) instead. Then, use FOIL and see what happens. First: a times a equals a^2. Outer: a times b equals ab Inner: b times a equals ab Last: b times b equals b^2 Add them all up, and you get a^2 + 2ab + b^2. That s where the 2ab came from.", + "video_name": "EINpkcphsPQ", + "timestamps": [ + 464 + ], + "3min_transcript": "Well, that's pretty straightforward. It's just one half c times c. So plus one half c times c, which is one half c squared. Now, let's simplify this thing and see what we come up with, and you might guess where all of this is going. So let's see what we get. So we can rearrange this. Let me rearrange this. So one half times a plus b squared is going to be equal to 2 times one half. Well, that's just going to be one. So it's going to be equal to a times b, plus one half c squared. Well, I don't like these one halfs laying around, so let's multiply both sides of this equation by 2. I'm just going to multiply both sides of this equation by 2. So let me write that. And on the right-hand side, I am left with 2ab. Trying to keep the color coding right. And then, 2 times one half c squared, that's just going to be c squared plus c squared. Well, what happens if you multiply out a plus b times a plus b? What is a plus b squared? Well, it's going to be a squared plus 2ab plus 2ab plus b squared. And then, our right-hand side it's going to be equal to all of this business. And changing all the colors is difficult for me, so let me copy and let me paste it. So it's still going to be equal to the right-hand side. How can we simplify this? Is there anything that we can subtract from both sides? Well, sure there is. You have a 2ab on the left-hand side. You have a 2ab on the right-hand side. Let's subtract 2ab from both sides. If you subtract 2ab from both sides, what are you left with? You are left with the Pythagorean theorem. So you're left with a squared plus b squared is equal to c squared. Very, very exciting. And for that, we have to thank the 20th president of the United States, James Garfield. This is really exciting. The Pythagorean theorem, it was around for thousands of years before James Garfield, and he was able to contribute just kind of fiddling around while he was a member of the US House of Representatives." + }, + { + "Q": "at 2:16 what is that symbol he drew and what does it do?\n", + "A": "It s a greek symbol called theta", + "video_name": "EINpkcphsPQ", + "timestamps": [ + 136 + ], + "3min_transcript": "What we're going to do in this video is study a proof of the Pythagorean theorem that was first discovered, or as far as we know first discovered, by James Garfield in 1876, and what's exciting about this is he was not a professional mathematician. You might know James Garfield as the 20th president He was elected president. He was elected in 1880, and then he became president in 1881. And he did this proof while he was a sitting member of the United States House of Representatives. And what's exciting about that is that it shows that Abraham Lincoln was not the only US politician or not the only US President who was really into geometry. And what Garfield realized is, if you construct a right triangle-- so I'm going to do my best attempt to construct one. So let me construct one right here. So let's say this side right over here is length b. Let's say this side is length a, and let's say has length c. So I've just constructed enough a right triangle, and let me make it clear. It is a right triangle. He essentially flipped and rotated this right triangle to construct another one that is congruent to the first one. So let me construct that. So we're going to have length b, and it's collinear with length It's along the same line, I should say. They don't overlap with each other. So this is side of length b, and then you have a side of length-- let me draw a it so this will be a little bit taller-- side of length b. And then, you have your side of length a at a right angle. Your side of length a comes in at a right angle. And then, you have your side of length c. So the first thing we need to think about is what's the angle between these two sides? What's that mystery angle going to be? Well, it looks like something, but let's see if we can prove to ourselves that it really is what we think it looks like. If we look at this original triangle and we call this angle \"theta,\" what's this angle over here, the angle that's between sides of length a and length c? What's the measure of this angle going to be? Well, theta plus this angle have to add up to 90. Because you add those two together, they add up to 90. And then, you have another 90. You're going to get 180 degrees for the interior angles of this triangle. So these two have to add up to 90. This angle is going to be 90 minus theta. Well, if this triangle appears congruent-- and we've constructed it so it is congruent-- the corresponding angle to this one is this angle right over here. So this is also going to be theta, and this right over here is going to be 90 minus theta. So given that this is theta, this is 90 minus theta, what is our angle going to be? Well, they all collectively go 180 degrees." + }, + { + "Q": "what do you mean at 2:19 ?\n", + "A": "She protested with a sign. That s all", + "video_name": "sxnX5_LbBDU", + "timestamps": [ + 139 + ], + "3min_transcript": "[SINGING] On the first day of Christmas, my true love gave to me, the multiplicative identity. I always hated the song the \"12 days of Christmas\" when I was younger. Not that the tune or the words are any worse than any other Christmas song, but it's just so long and repetitive. Singing it sucks too, because like math class, it seems no matter how hard you try to pay attention, you lose focus out of the sheer tedium of it all and forget where you are. Unless you keep vigorous records by drawing complicated graphs. [SINGING] On the second day of Christmas my true love gave to me, the only even prime, and the absolute value of e to the i pi. See? It doesn't have to be repetitive, but the \"12 days of Christmas\" is more fun to think about, I do like these kinds of visualizations. And you can think of other ways to visualize the process of singing the \"12 Days.\" Which might come in useful next time you're at a family Christmas party, and someone insists on singing through the whole thing. [SINGING] On the third day of Christmas my true love gave to me, the number of spatial dimensions, at least macroscopically-- don't yell at me string theorists-- the number of points that define a line, and the limit of sine x over x as x goes to zero. Unlike a normal, reasonable songs, adding 12 more verses wouldn't just make it twice as long, because the stupid thing just grows and grows. Even 99 bottles of beer on the wall has the promise of getting to zero, unless you believe in anti-beer. But \"12 Days\" is just disheartening, because the closer you think you're getting to the end, the longer the verses get. Dragging it out to the bitter end. [SINGING] On the fourth day of Christmas my true love gave to me, the smallest possible number of points that define a plane, the divisor of even numbers, and any other number to the power of 0. If I had a time machine, and was not bitterly anti-time travel-- and yes I've actually protested with a sign-- one of the things on my list would be to go back to the year 0, pick your era, and be like, hey three kings of orient, hurry it up, will ya? Also what's myrrh? Because it sounds like the noise a camel makes. Really though, five days of Christmas would be more reasonable. Even eight I could live with. [SINGING] On the fifth day of Christmas, my true love gave to me, five golden ratio producing pentagons, the number of sides on a square, the number of sides on that rigid, functional, and beautiful creature called the triangle, and I guess a two, and the number" + }, + { + "Q": "at 3:06, what is a mobius strip?\n", + "A": "A two dimensional object that can exist in three dimensional space.", + "video_name": "sxnX5_LbBDU", + "timestamps": [ + 186 + ], + "3min_transcript": "I do like these kinds of visualizations. And you can think of other ways to visualize the process of singing the \"12 Days.\" Which might come in useful next time you're at a family Christmas party, and someone insists on singing through the whole thing. [SINGING] On the third day of Christmas my true love gave to me, the number of spatial dimensions, at least macroscopically-- don't yell at me string theorists-- the number of points that define a line, and the limit of sine x over x as x goes to zero. Unlike a normal, reasonable songs, adding 12 more verses wouldn't just make it twice as long, because the stupid thing just grows and grows. Even 99 bottles of beer on the wall has the promise of getting to zero, unless you believe in anti-beer. But \"12 Days\" is just disheartening, because the closer you think you're getting to the end, the longer the verses get. Dragging it out to the bitter end. [SINGING] On the fourth day of Christmas my true love gave to me, the smallest possible number of points that define a plane, the divisor of even numbers, and any other number to the power of 0. If I had a time machine, and was not bitterly anti-time travel-- and yes I've actually protested with a sign-- one of the things on my list would be to go back to the year 0, pick your era, and be like, hey three kings of orient, hurry it up, will ya? Also what's myrrh? Because it sounds like the noise a camel makes. Really though, five days of Christmas would be more reasonable. Even eight I could live with. [SINGING] On the fifth day of Christmas, my true love gave to me, five golden ratio producing pentagons, the number of sides on a square, the number of sides on that rigid, functional, and beautiful creature called the triangle, and I guess a two, and the number Graphing the numbers like this may seem trivial, but I think it's nice to be reminded that these numbers aren't just squiggles on a page, symbols to be manipulated, but actually represent something. And I think it's nice that it results in such a lovely triangle. You know how I feel about triangles. [SINGING] On the sixth day of Christmas, my true love gave to me, my name in Roman numerals, number of feet in iambic pentameter, my name in Roman numerals backwards, the first Mersenne prime, the number of syllables in a foot of iambic pentameter, and sine x squared plus cosine x squared. Right. So if you want to know how many times you're going to have to sing a line about whatever stuff your true love got you, it's like counting up all the things in one of these triangles. That has however many layers, in this case 12." + }, + { + "Q": "At about 5:45, what's a frieze pattern?\n", + "A": "They are patterns that have all the kinds of symmetry, I think. Go to Vi s video about freizepatterns called : Math Improv:Fruit By The Foot.", + "video_name": "sxnX5_LbBDU", + "timestamps": [ + 345 + ], + "3min_transcript": "number, for obvious reasons. You can also shift the things around to get an equilateral triangle, which is how triangular numbers are usually explained. So the first triangular number is 1, the second is 3, the third is 6, the fourth is 10. [SINGING] On the seventh day of Christmas, my true love gave to me, the most common lucky number, the first perfect number, the only prime ending in five. The number of colors sufficient for coloring in a map, the only prime triangular number, the highest number that is its own factorial, and 1/2 plus 1/4 plus 1/8 plus 1/16 and so on. Almost. Here's a famous story that I've heard told a few different ways, and the actual facts are fuzzy, but basically, here's the gist of it. Carl Gauss was bored in his math class. but math classes sucked 300 years ago, just as much as they suck today. And Gauss would get himself in trouble when he was bored. Maybe he also liked to escape via the window. So his teacher got fed up and was like, Gauss-- I mean, he probably called him Carl, but that's not the point-- Guass, if you're so freaking bored with my class, how about you go sit in the corner and add up all the numbers between 1 and 100. That ought to keep you busy for a while. So Gauss goes to the corner, but he's just sitting there. And the teacher gets all mad and he's like, hey Gauss. I guess that means you've already added up all those numbers, right? And Gauss is like, sure, it's 5,050. [SINGING] On the eighth day of Christmas, my true love gave to me, the only perfect cube that's a Fibonacci number besides one, the number of Frieze patterns, the number of sides on a cube, the number of platonic solids, the first composite number, the number of regular polytropes in all dimensions greater than four, the Euler characteristic of polyhedra homeomorphic to a sphere, it's undefined. His teacher of course, didn't believe him. I think the teacher spent the next 10 minutes adding up the numbers by hand in an effort to catch Gauss in a lie, and when he saw that Gauss was right, he probably gave him detention any way. Or more likely whacked him with a ruler a few times for having the gall to be smarter than him. Or it could be that the story's mostly made up. Nevertheless, here's how he did it. Instead of adding up the numbers individually, like his teacher did, which would have been super boring, he realized this fact. The numbers 1 through 100 come in pairs that add up to 101. 1 plus 100. 2 plus 99. 3 plus 98. 4 plus 97. And so on. There's 50 such pairs of numbers, and 50 times 101 is really easy to do in your head, because it's 50 times 100 plus 50, or 5050. [SINGING] On the ninth day of Christmas my true love gave to me, an upside down six, infinity sideways, flipped over L, an upside" + }, + { + "Q": "(0:35) - specifically the word placeholder gave me this question\nIf you were to \"say\" 1 is \"y\", and anything times 1 is the same, would 1 actually represent the \"laws of physics\"? Like, the laws of physics is based on \"common perception/observation\", which should always equal the same thing(s)... Like when you multiply by one... Is this an accurate/logical way to look at it?\n", + "A": "And does 567x1 is. 567", + "video_name": "6nZp2QGeQ9k", + "timestamps": [ + 35 + ], + "3min_transcript": "We're asked to multiply 65 times 1. So literally, we just need to multiply 65-- we could write it is a times sign like that or we could write it as a dot like that-- but this means 65 times 1. And there's two ways to interpret this. You could view this as the number 65 one time or you could view this as the number 1 sixty-five times, all added up. But either way, if you have one 65, this is literally just going to be 65. Anything times 1 is going to be that anything, whatever this is. Whatever this is times 1 is going to be that same thing again. If I have just some kind of placeholder here times 1, and I could even write it as the times symbol times 1, that's going to be that same placeholder. If I have 5 times 1, I'm going to get 5, because literally, all this is saying is 5 one time. If I put-- I don't know-- 157 times 1, that'll be 157. I think you get the idea." + }, + { + "Q": "\nwhat is the in don't know what to call it but at 6:44 when u ad the numbers together what if its 5/4 how do u divide", + "A": "Well when you divide a fraction you must flip the denominator (Bottom number of the fraction) , but when you flip it has to be the SECOND fraction. Ex 4/5 \u00c3\u00b7 6/7 to 4/5 \u00c3\u00b7 7/6 A lot of people forget to add the \u00c3\u0097 sign when you are dividing. If this sounds confusing then let me show you Ex 4/5 \u00c3\u00b7 6/7 to 4/5 \u00c3\u0097 7/6", + "video_name": "GdIkEngwGNU", + "timestamps": [ + 404 + ], + "3min_transcript": "So we say, two minus the mean. Two minus the mean, and we take the absolute value. So that's its absolute deviation. Then we have another two, so we find that absolute deviation from three. Remember, if we're just taking two minus three, taking the absolute value, that's just saying its absolute deviation. How far is it from three? It's fairly easy to calculate in this case. Then we have a four and another four. Let me write that. Then we have the absolute deviation of four from three, from the mean. Then plus, we have another four. We have this other four right up here. Four minus three. because once again, it's absolute deviation. And then we divide it, and then we divide it by the number of data points we have. So what is this going to be? Two minus three is negative one, but we take the absolute value. It's just going to be one. Two minus three is negative one. It's just gonna be one. And you see that here visually. This point is just one away. It's just one away from three. This point is just one away from three. Four minus three is one. Absolute value of that is one. This point is just one away from three. Four minus three, absolute value. That's another one. every data point was exactly one away from the mean. And we took the absolute value so that we don't have negative ones here. We just care how far it is in absolute terms. So you have four data points. Each of their absolute deviations is four away. So the mean of the absolute deviations are one plus one plus one plus one, which is four, over four. So it's equal to one. One way to think about it is saying, on average, the mean of the distances of these points away from the actual mean is one. And that makes sense because all of these are exactly one away from the mean. Now, let's see how, what results we get for this data set right over here. Let me actually get some space over here. At any point, if you get inspired, I encourage you to calculate the Mean Absolute Deviation on your own. So let's calculate it. The Mean Absolute Deviation here, I'll write MAD, is going to be equal to ... Well, let's figure out the absolute deviation of each of these points from the mean. It's the absolute value of one minus three, that's this first one, plus the absolute deviation, so one minus three, that's the second one, then plus the absolute value of six minus three, that's the six, then we have the four, plus the absolute value of four minus three. Then we have four points. So one minus three is negative two. Absolute value is two. And we see that here. This is two away from three. We just care about absolute deviation. We don't care if it's to the left or to the right. Then we have another one minus three is negative two. It's absolute value, so this is two. That's this. This is two away from the mean. Then we have six minus three. Absolute value of that is going to be three." + }, + { + "Q": "\nAt 1:19 when he wrote P-2w/2 = l, why can't sal divide 2w/2 to just get w?", + "A": "The way he arranged the equation (which is hard to represent here because you can t represent a fraction with a horizontal bar) is L= (P-2w)/2....so both the P and the 2w need to be divided by 2. Think of it with just numbers... - let P=6 and w = 1 so the way you wrote it: P-2w/2 = L says that 6- 2(1)/2 = 5 but the way he represented it L= (P-2w)/2 : again let P=6 and w = 1: L = (6-2)/2 = 2 not the same expression without the ( )", + "video_name": "fnuIT7EhAvs", + "timestamps": [ + 79 + ], + "3min_transcript": "Solve P equals 2l plus 2w for l. So this right here, this is just the formula for the perimeter of a rectangle. Perimeter is equal to 2 times the length plus 2 times the width. But they just want us to solve this equation right here, solve for l. So let's do that. So we have P is equal to 2 times l plus 2 times w. So we need to solve for l. So let's isolate all of the l terms on one side. And the best way, we could just do that by leaving it here on the right and then getting rid of this 2w. And the best way to get rid of this 2w is to subtract it from the right. But if you're going to subtract it from the right, you also have to subtract it from the left if you want this equality to hold. So you have to also subtract it from the left. And so the left-hand side becomes P minus 2w. And the right-hand side, you get-- this 2w minus 2w cancels out. You just have a 2l. And then if you want to solve for l, you just have to divide both sides of this equation by 2. And we have isolated our l. We get l is equal to P minus 2w over 2. Or if we wanted to write it the other way, you could write l is equal to P minus 2w over 2. And we are done." + }, + { + "Q": "At1:09 when we divide both sides by 2. It cancells out the 2 on the right so we are left with (l). Why doesn't it cancel out on the left and leave us with (p-w)??\n", + "A": "You are dividing both sides by 2. Another way to write the final answer would be: l = P/2 - w", + "video_name": "fnuIT7EhAvs", + "timestamps": [ + 69 + ], + "3min_transcript": "Solve P equals 2l plus 2w for l. So this right here, this is just the formula for the perimeter of a rectangle. Perimeter is equal to 2 times the length plus 2 times the width. But they just want us to solve this equation right here, solve for l. So let's do that. So we have P is equal to 2 times l plus 2 times w. So we need to solve for l. So let's isolate all of the l terms on one side. And the best way, we could just do that by leaving it here on the right and then getting rid of this 2w. And the best way to get rid of this 2w is to subtract it from the right. But if you're going to subtract it from the right, you also have to subtract it from the left if you want this equality to hold. So you have to also subtract it from the left. And so the left-hand side becomes P minus 2w. And the right-hand side, you get-- this 2w minus 2w cancels out. You just have a 2l. And then if you want to solve for l, you just have to divide both sides of this equation by 2. And we have isolated our l. We get l is equal to P minus 2w over 2. Or if we wanted to write it the other way, you could write l is equal to P minus 2w over 2. And we are done." + }, + { + "Q": "\nat 3:40 Sal writes yos does that mean yards?", + "A": "Yes at 3:40 he wrote yos but it actually meant yards.", + "video_name": "ZS1OZj_oWao", + "timestamps": [ + 220 + ], + "3min_transcript": "So we took the remaining distance and we divided it into fourths. And they say, then she skipped for 1/4 of the remaining distance. So this is the remaining distance after running. She skipped for 1/4 of it. And then she walked the rest of the way home. So she walked to the rest of the way home, this is the distance that she walked. And this distance right over here is 120 yards. What is the distance between school and Judy's home? So there's a bunch of ways that you could think about this. When we talked about the remaining distance after running, so this was the remaining distance after she ran, after this orange part right over here. You see that this blue section is 3/4 of that remaining distance. 120 yards is 3/4 of this remaining distance. So this is equal to 3/4 of the remaining distance. Well since this is 3/4, 1/4 is going to be 1/3 of this. 1/3 of 120 is 40. So this distance right over here is 40. This is 40. This is 40. Notice 40 plus 40 plus 40 is 120. And this right over here is also going to be 40. These are all in terms of yards. So the remaining distance is going to be 40 times 4 or 160 So this is 160 yards. Now, what is the total distance? Well this 160 yards, the remaining distance after she ran, that's half of the total distance. So this is 160 yards, then this is going to be 160 yards as well. Gives us a total distance from school to home of 320 yards." + }, + { + "Q": "\nAt 2:07 Sal substitutes (4x^3 dx) for du. But why does he not have to write another dx at the end of the Integral?", + "A": "Because of du. When you use u-substitution, you basically are writing the integral in terms of u instead of (in this case) x. du and dx ultimately serve the same function as well as representing an infinitesimal change in x or u and also being the dummy variable that integrates to +C .", + "video_name": "Zp5z0wa0kgo", + "timestamps": [ + 127 + ], + "3min_transcript": "So we want to take the indefinite integral of 4x^3 over x^4 plus 7 dx. So how can we tackle this? It seems like a hairy integral. Now the key inside here is to realize you have this expression x^4 + 7 and you also have its derivative up here. The derivative of x^4 plus 7 is equal to 4x^3. Derivative of x^4 is 4x^3; derivative of 7 is just 0. So that's a big clue that u-substitution might be the tool of choice here. U-sub -- I'll just write u- -- I'll write the whole thing. U-Substitution could be the tool of choice. So given that, what would you want to set your u equal to? And I'll let you think about that 'cause it can figure out this part and the rest will just boil down to a fairly straightforward integral. Well, you want to set u be equal to the expression that you have its derivative laying around. So we could set u equal to x^4 plus 7. Now, what is du going to be equal to? so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du?" + }, + { + "Q": "\nIn 8:35 - isn't it supposed to be -(34/3)? So that A will equal -(5/17)?", + "A": "Actually I think it should be -5/17, he dropped the minus sign when he subtracted 9 from -25. It should be -34/3 which reduces to -5/17.......right??? :/", + "video_name": "hbJ2o9EUmJ0", + "timestamps": [ + 515 + ], + "3min_transcript": "" + }, + { + "Q": "If you had a ratio of 3:5, would you plot it a (3,5), (6,10), etc...\n", + "A": "Yes, because the ratio of x:y is 3/5.", + "video_name": "dmcVzFbXMCU", + "timestamps": [ + 185 + ], + "3min_transcript": "In a recent gum-chewing contest, Violet tried to blow the biggest blueberry-flavored bubble ever. Every 3 seconds, she added 1 piece of gum to her huge bubble. On the graph below, the x-axis represents time in seconds. So this right over here is our x-axis. And the y-axis represents pieces of gum in Violet's bubble. So this is the y-axis right over here. Plot points to show how many pieces of gum were in Violet's bubble after 3, 6, and 9 seconds. So every 3 seconds, she adds one piece of gum. So let me throw all of my-- so what we care is about 3 seconds, 6 seconds, and 9 seconds. So every 3 seconds, we know that she adds one piece of gum. This is 0 right over here. We're not going to worry about negative time. So after 3 seconds, she would have added one piece of gum to her bubble. Now, if 3 more seconds pass by, she would have added 1 more, And then if 3 more seconds pass by, now we're at 9 seconds, she would have added 1 more piece of gum, and now we're at 3 total pieces of gum. And notice, the ratio between the pieces of gum and the time that has passed has stayed constant, because she's doing it at a constant rate. 1 to 3 is the same ratio as 2 to 6, which is the same ratio as 3 to 9. So, let's check our answer. Very good." + }, + { + "Q": "\nat 1:20 what does numerical expression mean?", + "A": "A numerical expression is a mathematical phrase that involves only numbers and one or more operational symbols. The expression represents a particular number. For instance the numerical expression 10 + 6 \u00e2\u0080\u0093 8 simplifies to the number 8.", + "video_name": "arY-EUZDNfk", + "timestamps": [ + 80 + ], + "3min_transcript": "Alan found 4 marbles to add to his 5 marbles currently in his pocket. He then had a competition with his friends and tripled his marbles. Write a numerical expression to model the situation without performing any operations. So let's think about what's going on. So he already had 5 marbles in his pocket. And then he found 4 more marbles to add to that. So we can add the 4 marbles to the 5 marbles. So 4 marbles plus the 5 marbles. So that's what happens after the first sentence. He found 4 marbles to add to his 5 marbles currently in his pocket. He then had a competition with his friends and tripled his marbles. So this is how many marbles he had before tripling. And now he's tripling his marbles. So we want to multiply 3 times the total number of marbles he has now-- times 4 plus 5. So this right over here is the numerical expression any operations. We, of course, could then actually calculate this. He has 9 marbles before tripling. And then you multiply it by 3 and he has 27. But this is what they're asking for. They want us to write this expression." + }, + { + "Q": "For the last example at 5:30 can I also define like that? Is it true?\n{x \u00e2\u0088\u0088 \u00e2\u0084\u009d | x = \u00cf\u0080 or x=3}\n", + "A": "You could, but it s a bit more complicated than it needs to be since the set has only 2 values.", + "video_name": "-DTMakGDZAw", + "timestamps": [ + 330 + ], + "3min_transcript": "most of the real numbers except it cannot be 0 because we don't know -- this definition is undefined when you put the input as 0 So x is a member of the real numbers, and we write real numbers -- we write it with this kind of double stroke right over here. That's the set of all real numbers such that -- we have to put the exception. 0 is not a -- x equals to 0 is not a member of that domain -- such that x does not -- does not equal 0. Now let's make this a little bit more concrete by do some more examples So more examples we do, hopefully the clearer this will become. So let's say we have another function. Just be clear, we don't always have to use f's and x's. We could say, let's say we have g of y is equal to the square root of y minus 6. So what is the domain here? What is the set of all inputs over which this function g is defined? So here we are in putting a y it to function g g of y. Well it's going to be defined as long as whatever we have under the radical right over here is non-negative. If this becomes a negative, our traditional principal root operator here is not defined. We need something that -- if this was a negative number, how would you take the principal root of a negative number? We just think this is kind of the the traditional principal root operator. So y minus 6, y minus 6 needs to be greater than or equal to 0, in order for, in order for g to be defined for that input y. Or you could say add six to both sides. y is to be greater than or equal 6. Or you could say g is defined for any inputs y that are greater than or equal to 6. So you could say the domain here, we could say the domain here is the set of all y's that are members of the real numbers such that y, such that they're also greater than or equal, such that they're also You're all used to a function that is defined this way. You could even see functions that are divided fairly exotic ways. You could see a function -- let me say h of x -- h of x could be defined as -- it literally could be defined as, well h of x is gonna be 1 if x is equal to pi and it's equal to 0 if, if, x is equal to 3. Now what's the domain here? And I encourage you to pause the video and think about it. Well, this function is actually only defined for two input. If you, we know h of -- we know h of pi -- if you input pi into it we know you're gonna output 1, and we know that if you input 3 into it h of 3, when x equals 3, you're going to -- you're going to -- put some commas here. You're gonna get 0. But if you input anything else, what's h of 4 going to be? Well, it hasn't defined. It's undefined. What's h of negative 1 going to be? It hasn't defined. So the domain, the domain here," + }, + { + "Q": "Hi, Can somebody please explain in \"domain and range of a function\" video (at 5:02 minutes). How can y-6 >= 0 because greater than i understood but how can it equals to 0 as if y= 6 than the equation goes like 6-6 which is equals to 0 and 0 is not defined output.\n\nPlease answer.\n", + "A": "sqrt(0) is defined. It = 0. So, there is no issue have a zero inside the square root. You don t want a negative inside the square root after you have simplified because it is not a real number.", + "video_name": "-DTMakGDZAw", + "timestamps": [ + 302 + ], + "3min_transcript": "most of the real numbers except it cannot be 0 because we don't know -- this definition is undefined when you put the input as 0 So x is a member of the real numbers, and we write real numbers -- we write it with this kind of double stroke right over here. That's the set of all real numbers such that -- we have to put the exception. 0 is not a -- x equals to 0 is not a member of that domain -- such that x does not -- does not equal 0. Now let's make this a little bit more concrete by do some more examples So more examples we do, hopefully the clearer this will become. So let's say we have another function. Just be clear, we don't always have to use f's and x's. We could say, let's say we have g of y is equal to the square root of y minus 6. So what is the domain here? What is the set of all inputs over which this function g is defined? So here we are in putting a y it to function g g of y. Well it's going to be defined as long as whatever we have under the radical right over here is non-negative. If this becomes a negative, our traditional principal root operator here is not defined. We need something that -- if this was a negative number, how would you take the principal root of a negative number? We just think this is kind of the the traditional principal root operator. So y minus 6, y minus 6 needs to be greater than or equal to 0, in order for, in order for g to be defined for that input y. Or you could say add six to both sides. y is to be greater than or equal 6. Or you could say g is defined for any inputs y that are greater than or equal to 6. So you could say the domain here, we could say the domain here is the set of all y's that are members of the real numbers such that y, such that they're also greater than or equal, such that they're also You're all used to a function that is defined this way. You could even see functions that are divided fairly exotic ways. You could see a function -- let me say h of x -- h of x could be defined as -- it literally could be defined as, well h of x is gonna be 1 if x is equal to pi and it's equal to 0 if, if, x is equal to 3. Now what's the domain here? And I encourage you to pause the video and think about it. Well, this function is actually only defined for two input. If you, we know h of -- we know h of pi -- if you input pi into it we know you're gonna output 1, and we know that if you input 3 into it h of 3, when x equals 3, you're going to -- you're going to -- put some commas here. You're gonna get 0. But if you input anything else, what's h of 4 going to be? Well, it hasn't defined. It's undefined. What's h of negative 1 going to be? It hasn't defined. So the domain, the domain here," + }, + { + "Q": "\nAt 4:58 can't ti also be 4.5 instead of 4 1/2?", + "A": "Since 4.5 = 4 1/2, yes you can write it the way you prefer.", + "video_name": "hq1bUM2tyg0", + "timestamps": [ + 298 + ], + "3min_transcript": "" + }, + { + "Q": "\nCould you include a fraction in the answer in 4:24? Could you also get a rational number?", + "A": "Fractions are rational numbers. Yes, there s no reason why you couldn t use fractions here.", + "video_name": "hq1bUM2tyg0", + "timestamps": [ + 264 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 1:59 Sal took the triangle to the other side of the parallelogram to make a rectangle, can't you just multiply base times height to get the area?", + "A": "Yes you can. Sal most likely did that move so that you could visually see why multiplying base times height works.", + "video_name": "hm17lVaor0Q", + "timestamps": [ + 119 + ], + "3min_transcript": "- If we have a rectangle with base length b and height length h, we know how to figure out its area. Its area is just going to be the base, is going to be the base times the height. The base times the height. This is just a review of the area of a rectangle. Just multiply the base times the height. Now let's look at a parallelogram. And in this parallelogram, our base still has length b. And we still have a height h. So when we talk about the height, we're not talking about the length of these sides that at least the way I've drawn them, move diagonally. We're talking about if you go from this side up here, and you were to go straight down. If you were to go at a 90 degree angle. If you were to go perpendicularly straight down, you get to this side, that's going to be, that's going to be our height. So in a situation like this when you have a parallelogram, you know its base and its height, what do we think its area is going to be? So at first it might seem well this isn't as obvious But we can do a little visualization that I think will help. So what I'm going to do is I'm going to take a chunk of area from the left-hand side, actually this triangle on the left-hand side that helps make up the parallelogram, and then move it to the right, and then we will see something somewhat amazing. So I'm going to take this, I'm going to take this little chunk right there, Actually let me do it a little bit better. So I'm going to take that chunk right there. And let me cut, and paste it. So it's still the same parallelogram, but I'm just going to move this section of area. Remember we're just thinking about how much space is inside of the parallelogram and I'm going to take this area right over here and I'm going to move it to the right-hand side. And what just happened? What just happened? Let me see if I can move it a little bit better. What just happened when I did that? Well notice it now looks just like my previous rectangle. by taking some of the area from the left and moving it to the right, I have reconstructed this rectangle so they actually have the same area. The area of this parallelogram, or well it used to be this parallelogram, before I moved that triangle from the left to the right, is also going to be the base times the height. So the area here is also the area here, is also base times height. I just took this chunk of area that was over there, and I moved it to the right. So the area of a parallelogram, let me make this looking more like a parallelogram again. The area of a parallelogram is just going to be, if you have the base and the height, it's just going to be the base times the height. So the area for both of these, the area for both of these, are just base times height." + }, + { + "Q": "\nWhat is the name of the notation \"S-like\" figure Sal describes at 2:36?", + "A": "The integral symbol or the integral sign. (I know, it seems almost too obvious.)", + "video_name": "MMv-027KEqU", + "timestamps": [ + 156 + ], + "3min_transcript": "Now, let's go the other way around. Let's think about the antiderivative. And one way to think about it is we're doing the opposite of the derivative operator. The derivative operator, you get an expression and you find it's derivative. Now, what we want to do, is given some expression, we want to find what it could be the derivative of. So if someone were to tell-- or give you 2x-- if someone were to say 2x-- let me write this. So if someone were to ask you what is 2x the derivative of? They're essentially asking you for the antiderivative. And so you could say, well, 2x is the derivative of x squared. You could also say that 2x is the derivative of x squared plus pi, I think you get the general idea. So if you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant. So this is what you would consider the antiderivative of 2x. Now, that's all nice, but this is kind of clumsy to have to write a sentence like this, so let's come up with some notation for the antiderivative. And the convention here is to use kind of a strange looking notation, is to use a big elongated s looking thing like that, and a dx around the function that you're trying to take the antiderivative of. So in this case, it would look something like this. This is just saying this is equal to the antiderivative of 2x, and the antiderivative of 2x, we have already seen, is x squared plus c. use this type of crazy notation. It'll become more obvious when we study the definite integral and areas under curves and taking sums of rectangles in order to approximate the area of the curve. Here, it really should just be viewed as a notation for antiderivative. And this notation right over here, this whole expression, is called the indefinite integral of 2x, which is another way of just saying the antiderivative of 2x." + }, + { + "Q": "\nAt 2:13, how did he convert 3/10 into 30/100?", + "A": "He multiplied them both by 10. don t worry, you l get used to it. its a common technique.", + "video_name": "YZD5ifHZILE", + "timestamps": [ + 133 + ], + "3min_transcript": "Let's see if we can add 3/10 to 7/100. And I encourage you to try adding these two fractions on your own first before I work through it, and I'll give you one hint. Right now, it's very hard to add these fractions. You're adding 3/10 to 7/100. You're adding two different fractions with two different denominators. So what I would encourage you to do is try to rewrite 3/10 in a way that it has 100 as a denominator, or so it's expressed in terms of hundredths, and then see if you can add them. Well, I'm assuming you've given a go at it. Let's see how we can rewrite 3/10, and I'll try to visualize it. So what I've done here, so this you could consider a whole, and this is a whole as well. And this whole is divided into tenths-- 1,2, 3, 4, 5, 6, 7, 8, 9, 10. So what would 3/10 look like? Well, it would be one, two, and three of the tenths. and you divided them into 10 more sections? So you're essentially taking each of those tenths and you're dividing them into 10 sections. Well, you have 10 sections and then each of them will have 10 subsections in it. So you're going to have hundredths. Then the sections are going to describe hundredths. So 10-- let me write it this way. 10 times 10. You are then going to have hundredths. And 3/10 would be equivalent to how many of these hundredths? Well, each of these tenths will now become 10. So you're going to have 10, 20, and-- let me color them in better. So this part right over here. This-- let me get my tool right-- this part right over here, this is going to be 3 times 10, which is going to be equal to 30. 10 times 10 is equal to 100. So this is how we're going to change the denominator. Instead of thinking in terms of tenths, we're going to think in terms of hundredths. And now our numerator, 3/10, is equivalent to 30 hundredths so we can rewrite this fraction. We essentially multiplied the numerator by 10, and we multiplied the denominator by 10, which didn't change the value of the fraction. It still represents the same-- it still represents 3/10 of the whole. So when you do that, you end up-- we can rewrite this thing as 30 over 100." + }, + { + "Q": "\nAt 5:24, how would you solve the C part? Do you spread it out like you did? Or is there another way?\nCan anyone help me quick? Sal... you there?", + "A": "When you say the C part, do you mean the formula for a combination? Sal actually explains that in the following video. But perhaps you meant something else?", + "video_name": "iKy-d5_erhI", + "timestamps": [ + 324 + ], + "3min_transcript": "I'm going to be a little bit more systematic. F, uh lemme do it. B, B, F, C, B, C, F. And obviously I could keep doing. I can do 120 of these. I'll do two more. You could have C, F, B. And then you could have C, B, F. So in the permutation world, these are, these are literally 12 of the 120 permutations. But what if we, what if all we cared about is the three people we're choosing to sit down, but we don't care in what order that they're sitting, or in which chair they're sitting. So in that world, these would all be one. This is all the same set of three people if we don't care which chair they're sitting in. This would also be the same set of three people. And so this question. If I have six people sitting in three chairs, how many ways can I choose three people out of the six And I encourage you to pause the video, and try to think of what that number would actually be. Well a big clue was when we essentially wrote all of the permutations where we've picked a group of three people. We see that there's six ways of arranging the three people. When you pick a certain group of three people, that turned into six permutations. And so if all you want to do is care about well how many different ways are there to choose three from the six? You would take your whole permutations. You would take your number of permutations. You would take your number of permutations. And then you would divide it by the number of ways to arrange three people. Number of ways to arrange, arrange three people. And we see that you can arrange three people, or even three letters. You can arrange it in six different ways. So this would be equal to 120 divided by six, So there are 120 permutations here. If you said how many different arrangements are there of taking six people and putting them into three chairs? That's 120. But now we're asking another thing. We're saying if we start with 120 people, and we want to choose. Sorry if we're starting with six people and we want to figure out how many ways, how many combinations, how many ways are there for us to choose three of them? Then we end up with 20 combinations. Combinations of people. This right over here, once again, this right over here is just one combination. It's the combination, A, B, C. I don't care what order they sit in. I have chosen them. I have chosen these three of the six. This is a combination of people. I don't care about the order. This right over here is another combination. It is F, C, and B. Once again I don't care about the order. I just care that I've chosen these three people." + }, + { + "Q": "can anyone explain what sal tried to do after 4:00 ??\n\u00e2\u0096\u00baQ2 whats the difference between permutations and combinations.. ?\n", + "A": "He was showing how to find the number of ways to choose a subset of a set of things. In this case how many ways you can choose 3 people from a pool of 6 people (where order doesn t matter). In permutations order matters, in combinations it does not.", + "video_name": "iKy-d5_erhI", + "timestamps": [ + 240 + ], + "3min_transcript": "Permutations. Now lemme, permutations. Now it's worth thinking about what permutations are counting. Now remember we care, when we're talking about permutations, we care about who's sitting So for example, for example this is one permutation. And this would be counted as another permutation. And this would be counted as another permutation. This would be counted as another permutation. So notice these are all the same three people, but we're putting them in different chairs. And this counted that. That's counted in this 120. I could keep going. We could have that, or we could have that. So our thinking in the permutation world. We would count all of these. Or we would count this as six different permutations. These are going towards this 120. And of course we have other permutations where we would involve other people. Where we have, it could be F, B, C, F, C, B, F, A, C, F, F. I'm going to be a little bit more systematic. F, uh lemme do it. B, B, F, C, B, C, F. And obviously I could keep doing. I can do 120 of these. I'll do two more. You could have C, F, B. And then you could have C, B, F. So in the permutation world, these are, these are literally 12 of the 120 permutations. But what if we, what if all we cared about is the three people we're choosing to sit down, but we don't care in what order that they're sitting, or in which chair they're sitting. So in that world, these would all be one. This is all the same set of three people if we don't care which chair they're sitting in. This would also be the same set of three people. And so this question. If I have six people sitting in three chairs, how many ways can I choose three people out of the six And I encourage you to pause the video, and try to think of what that number would actually be. Well a big clue was when we essentially wrote all of the permutations where we've picked a group of three people. We see that there's six ways of arranging the three people. When you pick a certain group of three people, that turned into six permutations. And so if all you want to do is care about well how many different ways are there to choose three from the six? You would take your whole permutations. You would take your number of permutations. You would take your number of permutations. And then you would divide it by the number of ways to arrange three people. Number of ways to arrange, arrange three people. And we see that you can arrange three people, or even three letters. You can arrange it in six different ways. So this would be equal to 120 divided by six," + }, + { + "Q": "\nI have a word problem it says \"the angle of the roof on Wendy's dollhouse is 56 degrees. She built a scale model of the dollhouse with a scale ratio of 1:4. What is the angle of the roof of the model she built?\" I'm curious on how I would work this out. Do I do the same as on this video?", + "A": "Well If You think it is correct then you do it but I think you should do the video", + "video_name": "tOd2T72eJME", + "timestamps": [ + 64 + ], + "3min_transcript": "The scale on a map is 7 centimeters for every 10 kilometers, or 7 centimeters for 10 kilometers. If the distance between two cities is 60 kilometers-- so that's the actual distance-- how far apart in centimeters are the two cities on the map? Well, they give us the scale. For every 10 kilometers in the real world, the map is going to show 7 centimeters. Or another way to think about it is if you see 7 centimeters on the map, that represents 10 kilometers in the real world. Now, they're saying that the distance between two cities is 60 kilometers. So it's essentially 6 times 10 kilometers, so times 6. So if you have to 6 times 10 kilometers on the map, you're going to have 6 times 7 centimeters, so times 6. And 6 times 7 centimeters gets you to 42 centimeters. Well, 42 centimeters." + }, + { + "Q": "At 1:00 Sal talks of \"g of negative 6.999.\" Does he mean to say \"negative?\" Isn't the 6.999 positive?\n", + "A": "He said g of negative 6.999 but it should be g of 6.999 . I m guessing what he was trying to say that g of 6.999 is still negative but got mixed up. Report it so he can add an annotation for correction.", + "video_name": "hWJLd6bRthI", + "timestamps": [ + 60 + ], + "3min_transcript": "- [Voiceover] So, here we have the graph y is equal to g of x. We have a little point discontinuity right over here at x is equal to seven, and what we want to do is figure out what is the limit of g of x as x approaches seven. So, since you say, well, what is the function approaching as the inputs in the function are approaching seven? So, let's see, so if we input as the input to the function approaches seven from values less than seven, so if x is three, g of three is here, g of three is right there, g of four is right there, g of five is right there, g of six looks like it's a little bit more than, or a little bit less than negative one, g of 6.5 looks like it's around negative half, g of negative 6.9 is right over there, looks like it's a little bit less than zero, g of negative 6.999 looks like it's a little bit, closer to zero so it looks like we're getting closer as x gets closer and closer but not quite at seven, it looks like the value of our function is approaching zero. Let's see if that's also true from values for x values greater than seven, so g of nine is up here, looks like it's around six, g of eight looks like it's a little bit more than two, g of 7.5 looks like it's a little bit more than one, g of 7.1 looks like it's a little bit more than zero, the g of 7.1 looks like it's a little bit more than zero, g of 7.01 is even closer to zero, g of 7.0000001 looks like it'll be even closer to zero so once again it looks like we are approaching zero as x as approaches seven, in this case as we approach from larger values of seven. And this is interesting because the limit as x approaches seven of g of x is different than the function's actual value, g of seven, When we actually input seven into the function, we can see the graph tells us that the value of the function is equal to three. So we actually have this point discontinuity, or sometimes called a removable discontinuity, right over here, and this is, I'm not gonna do a lot of depth here, but this is starting to touch on how we, one of the ways that we can actually test for continuity is if the limit as we approach a value is not the same as the actual value of the function of that point, well, then we're probably talking about, or actually we are talking about, a discontinuity." + }, + { + "Q": "\nAt 3:40 ish wouldn't the probability be a dependent event since there is a set amount of fish in the pond/lake?", + "A": "Fish after catching are being released back into the pond that is why the probability is independent of what was caught.", + "video_name": "86nb02Bx_5w", + "timestamps": [ + 220 + ], + "3min_transcript": "Or I should, well you're going to pay that. Since you're paying it we'll put it as negative 100 because we're saying that this is your expected profit, so you're going to lose money there. That's going to be one minus this probability, the probability that Jeremy catches three sunfish. In that situation he'll pay you 20 dollars. You get 20 dollars there. The important thing is to figure out the probability that Jeremy catches three sunfish. Well the sunfish are 10 out of the 20 fish so any given time he's trying to catch fish there's a 10 in 20 chance, or you could say one half probability that it's going to be a sunfish. The probability that you get three sunfish in a row is going to be one half, times one half, times one half. of the 20 fish. If he wasn't putting the fish back in then the second sunfish you would have a nine out of 20 chance of the second one being a sunfish. In this case they keep replacing the fish every time they catch it. There is a one eighth chance that Jeremy catches three sunfish, so this right over here is one eighth. And one minus one eighth, this is seven eighths. You have a one eighth chance of paying 100 dollars and a seven eights chance of getting twenty dollars so this gets us to ... Your expected profit here, there's a one eighths chance, one eighth probability, that you lose 100 dollars here, so times negative 100. But then there is a seven eighths chance that you get -- I think the order of operations of the calculator would have taken care of it but I'll just do it so that it looks the same. Seven eighths, there's a seven eighths chance that you get 20 dollars. Your expected payoff here is positive five dollars. Your expected payoff here is equal to five dollars. This is your expected value from bet one. Now let's think about bet two. If you catch at least two sunfish of the next three fish you catch he will pay you 50, otherwise you will pay him 25. Let's think about the probability of catching at least two sunfish of the next three fish that you catch. There's a bunch of ways to think about this but since there's only three times that you're trying to catch the fish and there's only one of two outcomes you could actually" + }, + { + "Q": "At 1:12, Sal defines the random variable X as \"what your profit is from bet 1.\" He then goes on to define a function E(X) that gives the expected profit from taking the first bet. Shouldn't X be defined as the outcome of Jeremy's next three fish?\nAgain at 6:55, Sal similarly defines Y as the expected profit from the second bet, which is what E(Y) is. Shouldn't Y be defined as the outcome of my next three fish?\n", + "A": "He makes a small mistake at 6:55 when he says Y is the expected profit. Y should be defined the same as X as just the profit of the bet. E(Y) is the expected profit.", + "video_name": "86nb02Bx_5w", + "timestamps": [ + 72, + 415 + ], + "3min_transcript": "You and your friend Jeremy are fishing in a pond that contains ten trout and ten sunfish. Each time one of you catches a fish you release it back into the water. Jeremy offers you the choice of two different bets. Bet number one. We don't encourage betting but I guess Jeremy wants to bet. If the next three fish he catches are all sunfish you will pay him 100 dollars, otherwise he will pay you 20 dollars. Bet two, if you catch at least two sunfish of the next three fish that you catch he will pay you 50 dollars, otherwise you will pay him 25 dollars. What is the expected value from bet one? Round your answer to the nearest cent. I encourage you to pause this video and try to think about it on your own. Let's see. The expected value of bet one. The expected value of bet one where we'll say bet one is -- just to be a little bit better about this. Let's say x is equal what you pay, or I guess you could say , because you might get something, what your profit is from bet one. It's a random variable. The expected value of x is going to be equal to, let's see. What's the probability, it's going to be negative 100 dollars times the probability that he catches three fish. The probably that Jeremy catches three sunfish, the next three fish he catches are Or I should, well you're going to pay that. Since you're paying it we'll put it as negative 100 because we're saying that this is your expected profit, so you're going to lose money there. That's going to be one minus this probability, the probability that Jeremy catches three sunfish. In that situation he'll pay you 20 dollars. You get 20 dollars there. The important thing is to figure out the probability that Jeremy catches three sunfish. Well the sunfish are 10 out of the 20 fish so any given time he's trying to catch fish there's a 10 in 20 chance, or you could say one half probability that it's going to be a sunfish. The probability that you get three sunfish in a row is going to be one half, times one half, times one half." + }, + { + "Q": "\nAt about 0:50, give or take, in the video, Sal figures out the square root so easily. Now, I know this is an easy example, but it there a formula for finding the square root?", + "A": "a square root of a number a is a number y such that y^2 = a, in other words, a number y whose square (the result of multiplying the number by itself, or y \u00c3\u0097 y) is a. For example, 4 and \u00e2\u0088\u00924 are square roots of 16 because 4^2 = (\u00e2\u0088\u00924)^2 = 16. Wikipedia", + "video_name": "ROIfbUQrSY4", + "timestamps": [ + 50 + ], + "3min_transcript": "We're asked to find the square root of 100. Let me write this down bigger. So the square root is this big check-looking thing. The square root of 100. When you see it like this, this means the positive square root. If you're familiar with negative numbers, you know that there's also a negative square root, but when you just see this symbol, that means the positive square root. So let's think about what this is saying. This is asking us find the number, the positive number, that when I multiply that number by itself, I get 100. So what number when I multiply it by itself do I get 100? Well, let's see, if I multiply 9 by itself, that's only going to be 81. If I multiply 10 by itself, that is 100. So this is equal to-- and let me write it this way. Normally, you could skip this step. But you could write this as the square root of-- and instead of 100, 100 is the same thing as 10 times 10. And then you know, the square root of something times itself, that's just going to be that something. This is just equal to 10. Or another way you could write, I guess, this same truth is that 10 squared, which is equal to 10 times 10, is equal to 100." + }, + { + "Q": "\nAt 0:45 Sal explains that the square root of 100 is 10 times 10. This is pretty simple to do figure out because if you know your times tables, you will have this memorized. But what if I'm trying to find the square root of a larger number? Any tried and true methods OTHER than a calculator? Thanks!", + "A": "Use factoring. Break the number down into prime factors. For example: 2025 = 3*3*3*3*5*5 Each pair of matching factors is a perfect square. Group the pairs. 2025 = (3*3)*(3*3)*(5*5) OR, you can think of this as 2025 = 9*9*25 This number contains 3 perfect squares. You can take the square root of each perfect square, then multiply the results. sqrt(2025) = sqrt(9) * sqrt(9) * sqrt(25) = 3*3*5 = 45 Hope this helps.", + "video_name": "ROIfbUQrSY4", + "timestamps": [ + 45 + ], + "3min_transcript": "We're asked to find the square root of 100. Let me write this down bigger. So the square root is this big check-looking thing. The square root of 100. When you see it like this, this means the positive square root. If you're familiar with negative numbers, you know that there's also a negative square root, but when you just see this symbol, that means the positive square root. So let's think about what this is saying. This is asking us find the number, the positive number, that when I multiply that number by itself, I get 100. So what number when I multiply it by itself do I get 100? Well, let's see, if I multiply 9 by itself, that's only going to be 81. If I multiply 10 by itself, that is 100. So this is equal to-- and let me write it this way. Normally, you could skip this step. But you could write this as the square root of-- and instead of 100, 100 is the same thing as 10 times 10. And then you know, the square root of something times itself, that's just going to be that something. This is just equal to 10. Or another way you could write, I guess, this same truth is that 10 squared, which is equal to 10 times 10, is equal to 100." + }, + { + "Q": "At 0:46 why does Sal say the answer would be a positive number? B could be -3 and A could be -2 and the answer would we -1 which is NOT a positive value.\n", + "A": "We always measure distances in positive numbers. We have no rulers with negative numbers. We don t say we just walked -3 miles. We don t measure the objects and come up with negative inches or negative centimeters. The absolute value of any number represents its distance from zero. We use it to help ensure we create distances in positive values. If A = -2 and B = -3, the distance between them is 1 unit (a positive number. You can find this by doing: | -3 -(-2)| = |-1| = 1 or |-2 -(-3)| = |1| = 1 Hope this helps.", + "video_name": "t4xOkpP8FgE", + "timestamps": [ + 46 + ], + "3min_transcript": "- [Voiceover] Let's say that I have two numbers on a number line. So let me draw a little quick number line right over here. The two numbers on my number line that I care about, the number a and the number represented by b here. The way I've drawn it, b is to the right of a on our number line, and by our convention, b is going to be greater than a. So if I were to figure out the distance between a and b, what is this distance going from a, I want to draw a straight line here, this distance going from a to b, so this distance right over there, how would I figure it out? Well I could just take the larger of these two numbers, which is going to be b, and then subtract out the smaller. So I subtract out a, and I'll be left with this distance. This will give me a positive value. When I want a distance, I just think in terms of a positive value. How far apart are these two things? But I was only able to know to do b minus a because I knew that b was greater than a. This was going to give me a positive value. What if I knew that a was greater than b? So let me draw that again. Let me draw another number line right over here. In this world, in this world, I'm going to make a greater than b. This is b, that is a, and if I wanted to calculate the distance between b and a here, well now I would take the larger of the two, a, remember I want the positive distance here, and then I would subtract out the smaller. I would do a minus b. Well so here I did b minus a, here I did a minus b, but what if I didn't know which one was greater? If I didn't know whether b or a was greater, what could I do? Well what you could do is just take either a minus b or b minus a and take the absolute value. If you do that, it doesn't matter if you take b minus a or a minus b. It turns out that regardless of whether a is greater than b, or b is greater than a, or they're equivalent, that the absolute value of b minus a, and this is equivalent, either of these expressions is the distance between these numbers. I encourage you to play around with the negatives to see if you can factor out some negatives and think about the absolute value. It will actually make a lot of sense why this is true. In another video, I might do a little bit more of a rigorous justification for it. for this video is to see that this is actually true. So let's say we're in a world, let's get a number line out, and let's look at some examples. So let's say that we want to figure out the distance between, between, let's say negative two, the distance between negative two and positive three. So we can look at the number line and figure out what that distance is. To go from negative two to positive three, or the distance between them, we see is one, two, three, four, five." + }, + { + "Q": "\nAt 6:05 - 6:10, I believe you meant to say that the subject terms are equivalent to the Identity Matrix time vector 'a' as opposed to the Identity Matrix times 'a1'.", + "A": "yup, i m pretty sure he meant matrix a", + "video_name": "PErhLkQcpZ8", + "timestamps": [ + 365, + 370 + ], + "3min_transcript": "of these-- so let's say e1, e2, all the way to en-- this is called the standard basis for Rn. So why is it called that? Well, the word basis is there, so two things must be true. These things must span Rn and they must be linearly It's pretty obvious from inspection they're linearly If this guy has a 1 here and no one else has a 1 there, there's no way you can construct that 1 with some combination of the rest of the guys. And you can make that same argument for each of the ones So it's clearly linearly independent. And then to see that you can span, that you can construct any vector with a linear combination of these guys, you just really have to-- you know, whatever vector you want to construct, if you want to construct x1-- let me put it this way. Let me pick a different one. Let's say you want to construct the vector a1, a2, a3 all the way down to an. So this is some member of Rn, you want to construct this vector. Well, the linear combination that would get you this is literally a1 times e1 plus a2 times e2 plus all the way to an times en. This scalar times this first column vector will essentially just get you-- what will this look like? This will look like a1 and then you'd have a bunch of zeroes. You'd have n minus 1 zeroes plus 0 and you'd have an a2 and then you'd have a bunch of zeroes. And then you'd keep doing that, and then you would have a bunch of zeroes, and then you would have an an. Obviously, by our definition of vector addition, you add all these things up, you get this guy right here. And it's kind of obvious, because this right here is the I just wanted to expose you to that idea. Now, let's apply what we already know about linear transformations to what we've just learned about this identity matrix. I just told you that I can represent any vector like this. Let me rewrite it in maybe terms of x. I can write any vector x as a linear combination of the standard basis, which are really just the columns of the identity matrix. I can write that as x1 times e1 plus x2 times e2, all the way to xn times en. And remember, each of these column vectors right here, like for e1, is just 1 in the first entry and then all the rest are zeroes. e2 is a 1 in the second entry and everything else is 0. e5 is a 1 in the fifth entry and everything else is 0." + }, + { + "Q": "At 0:45, why does Sal draw the Vertical Line?\n", + "A": "It s kind of working as the = signs of the equations, so it separates the variable constants from the number on the other side of the equation. I think it s just so you remember which are which.", + "video_name": "lP1DGtZ8Wys", + "timestamps": [ + 45 + ], + "3min_transcript": "I figure it never hurts getting as much practice as possible solving systems of linear equations, so let's What I'm going to do is I'm going to solve it using an augmented matrix, and I'm going to put it in reduced row echelon form. So what's the augmented matrix for this system of equations? Three unknowns with three equations. I just have to do the coefficents. So the coefficients of x terms are just 1, 1, 1. Coefficients of the y terms are 1, 2, and 3. Coefficients of the z terms are 1, 3, and 4. And let me show that it's augmented. And then they equal 3, 0, and minus 2. Now, I want to get this augmented matrix into reduced row echelon form. So the first thing, I have a leading 1 here that's a pivot entry. Let me make everything else in that column equal to a 0. So I'm not going to change my first row. and then I have a 3. Now, to zero this out, let me just replace the second row with the first row minus the second row. So 1 minus 1 is 0. 1 minus 2 is-- actually, a better thing to do, because I eventually want this to be 1 anyway, let me replace this row with this row, with the second row minus the first row instead of the first row minus the second row. I can do it either way. So the second row minus the first row. So 1 minus 1 is 0. 2 minus 1 is 1. 3 minus 1 is 2. And then 0 minus 3 is minus 3. Now I want to also zero this out. So let me replace this guy with this equation minus that equation. So 1 minus 1 is 0. 4 minus 1 is 3. Minus 2 minus 3 is minus 5. Fair enough. So I got my pivot entry here. I have another pivot entry here. It's to the right of this one, which is what I want for reduced row echelon form. Now, I need to target this entry and that entry. I need to zero them out. So let's do it. So I'm going to keep my second row the same. My second row is 0, 1, 2, and then I have a minus 3, the augmented part of it. And to zero this guy out, what I can do is I can replace the first row with the first row minus the second row. So I get 1 minus 0 is 1. 1 minus 1-- there's a bird outside. Let me close my window. So where was I? I'm replacing the first row with the first row minus the" + }, + { + "Q": "5:05 Which ordered pair I know that is on the y or x axis?\n", + "A": "For points on a graph, you define them by (distance on the x-axis, distance on the y-axis). So even if you didn t know which point it was, you would figure out how far it was on the different axis, put them in x, y order and in parentheses, and you ve got the point. Hopefully that helps!", + "video_name": "WkspBxrzuZo", + "timestamps": [ + 305 + ], + "3min_transcript": "over change in x, which is-- our change in y is 14. And our change in x is negative 7. And then if we want to simplify this, 14 divided by negative 7 is negative 2. Now, what I want to show you is, is that we could have done it the other way around. We could have made this the starting point and this the endpoint. And what we would have gotten is the negative values of each of these, but then they would've canceled out and we would still get negative 2. Let's try it out. So let's say that our start point was negative 3 comma 16. And let's say that our endpoint is the 4 comma 2. 4 comma 2. So in this situation, what is our change in x? Our change in x. If I start at negative 3 and I go to 4, that means I went up 7. Or if you want to just calculate that, you would do 4 minus negative 3. 4 minus negative 3. And what is our change in y? Our change in y over here, or we could say our rise. If we start at 16 and we end at 2, that means we went down 14. Or you could just say 2 minus 16 is negative 14. We went down by 14. This was our run. So if you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7. So notice, these are just the negatives of these values from when we swapped them. So once again, this is equal to negative 2. And let's just visualize this. Let me do a quick graph here just to show you what a downward slope would look like. So let me draw our two points. So this is my x-axis. That is my y-axis. So this point over here, 4 comma 2. So let me graph it. So let me save some space here. So we have 1, 2, 3, 4. It's 4 comma-- 1, 2. So 4 comma 2 is right over here. 4 comma 2. Then we have the point negative 3 comma 16. So let me draw that over here. So we have negative 1, 2, 3. And we have to go up 16. So this is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. So it goes right over here. So this is negative 3 comma 16. Negative 3 comma 16. So the line that goes between them is going to look something like this. Try my best to draw a relatively straight line. That line will keep going. So the line will keep going. So that's my best attempt. And now notice, it's downward sloping. As you increase an x-value, the line goes down. It's going from the top left to the bottom right. As x gets bigger, y gets smaller." + }, + { + "Q": "\nwhat does the triangle mean at 2:10.", + "A": "The triangle is denotes change. So, slope equals the change in the y values divided by the change in the x values.", + "video_name": "WkspBxrzuZo", + "timestamps": [ + 130 + ], + "3min_transcript": "Find the slope of the line that goes through the ordered pairs 4 comma 2 and negative 3 comma 16. So just as a reminder, slope is defined as rise over run. Or, you could view that rise is just change in y and run is just change in x. The triangles here, that's the delta symbol. It literally means \"change in.\" Or another way, and you might see this formula, and it tends to be really complicated. But just remember it's just these two things over here. Sometimes, slope will be specified with the variable m. And they'll say that m is the same thing-- and this is really the same thing as change in y. They'll write y2 minus y1 over x2 minus x1. And this notation tends to be kind of complicated, but all this means is, is you take the y-value of your endpoint and subtract from it the y-value of your starting point. That will essentially give you your change in y. And it says take the x-value of your endpoint And that'll give you change in x. So whatever of these work for you, let's actually figure out the slope of the line that goes through these two points. So we're starting at-- and actually, we could do it both ways. We could start at this point and go to that point and calculate the slope or we could start at this point and go to that point and calculate the slope. So let's do it both ways. So let's say that our starting point is the point 4 comma 2. And let's say that our endpoint is negative 3 comma 16. So what is the change in x over here? What is the change in x in this scenario? So we're going from 4 to negative 3. If something goes from 4 to negative 3, what was it's change? You have to go down 4 to get to 0, and then you have to go down another 3 to get to negative 3. So our change in x here is negative 7. Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. Let's do the same thing for the change in y. And notice, I implicitly use this formula over here. Our change in x was this value, our endpoint, our end x-value minus our starting x-value. Let's do the same thing for our change in y. Our change in y. If we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y-value and subtract from that your starting y-value and you get 14. So what is the slope over here? Well, the slope is just change in y over change in x." + }, + { + "Q": "At the 4:24 mark in the video, shouldn't you say \"this is our rise\" instead of \"this is our run\"?\n", + "A": "Yes. If you look at it again, you ll see a box pop up in the lower right corner with the correction.", + "video_name": "WkspBxrzuZo", + "timestamps": [ + 264 + ], + "3min_transcript": "Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. Let's do the same thing for the change in y. And notice, I implicitly use this formula over here. Our change in x was this value, our endpoint, our end x-value minus our starting x-value. Let's do the same thing for our change in y. Our change in y. If we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y-value and subtract from that your starting y-value and you get 14. So what is the slope over here? Well, the slope is just change in y over change in x. over change in x, which is-- our change in y is 14. And our change in x is negative 7. And then if we want to simplify this, 14 divided by negative 7 is negative 2. Now, what I want to show you is, is that we could have done it the other way around. We could have made this the starting point and this the endpoint. And what we would have gotten is the negative values of each of these, but then they would've canceled out and we would still get negative 2. Let's try it out. So let's say that our start point was negative 3 comma 16. And let's say that our endpoint is the 4 comma 2. 4 comma 2. So in this situation, what is our change in x? Our change in x. If I start at negative 3 and I go to 4, that means I went up 7. Or if you want to just calculate that, you would do 4 minus negative 3. 4 minus negative 3. And what is our change in y? Our change in y over here, or we could say our rise. If we start at 16 and we end at 2, that means we went down 14. Or you could just say 2 minus 16 is negative 14. We went down by 14. This was our run. So if you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7. So notice, these are just the negatives of these values from when we swapped them. So once again, this is equal to negative 2. And let's just visualize this. Let me do a quick graph here just to show you what a downward slope would look like. So let me draw our two points. So this is my x-axis. That is my y-axis. So this point over here, 4 comma 2. So let me graph it." + }, + { + "Q": "\nAt 6:22, Sal writes r(theta). Does he mean f(theta) as r = f(theta) or am I getting mixed up?", + "A": "Yup he just used both r (theta) and f (theta) as representations of the polar function.", + "video_name": "qVn_Lfec-Ac", + "timestamps": [ + 382 + ], + "3min_transcript": "So each of these things that I've drawn, let's focus on just one of these wedges. I will highlight it in orange. So instead of the angle being theta let's just assume it's a really, really, really small angle. We'll use a differential although this is a bit of loosey-goosey mathematics but the important here is to give you the conceptual understanding. I could call it a delta theta and then eventually take the limit as our delta theta approaches zero. But just for conceptual purposes when we have a infinitely small or super small change in theta, so let's call that d theta, and the radius here or I guess we could say this length right over here. You could view it as the radius of at least the arc right at that point. It's going to be r as a function of the thetas that to call that our r right over there. And so what is going to be the area of this little sector? Well the area of this little sector is instead of my angle being theta I'm calling my angle d theta, this little differential. So instead of one half r squared it's going to be, let me do that in a color you can see. This area is going to be one half r squared d theta. Notice here the angle was theta, here the angle was d theta, super, super small angle. Now if I wanted to take the sum of all of these from theta is equal to alpha to theta is equal to beta and literally there is an infinite number of these. This is an infinitely small angle. Well then for the entire area right over here I could just integrate all of these. So that's going to be the integral from alpha to of course, is a function of theta. So you could even write it this way, you could write it as the integral from alpha to beta of one half r of theta squared d theta. Just to remind ourselves or assuming r is a function of theta in this case." + }, + { + "Q": "At 0:00, we learn?\n", + "A": "Yes, we learn!", + "video_name": "daCT_24RnIY", + "timestamps": [ + 0 + ], + "3min_transcript": "I have this figure here. You could call it a rectangular prism. And I want to measure its volume. And I'm defining my unit cube as being a 1 centimeter by 1 centimeter by 1 centimeter cube. It has 1 centimeter width, 1 centimeter depth, 1 centimeter And I will call this, this is equal to 1 cubic centimeter. So I want to measure this volume in terms of cubic centimeters. We've already seen that we can do that by saying, hey, how many of these cubic centimeters can fit into this figure without them overlapping in any way? So if we had this in our hands, we could kind of try to go around it and try to count it, but it's hard to see here because there's some cubes that we can't see behind the ones that we are seeing. So I'm going to try different tactics at it. So first, let's just think about what we can observe. So we see that this one, if we measure its different dimensions, its width, it's 2 of the unit length wide. So it's 2 centimeters wide. It's 4 of our unit length-- we're defining our unit length as a centimeter-- it's 4 of our unit length high. And it is 3 of our unit length deep. So this dimension right over here is 3 centimeters. So I want to explore if we can somehow use these numbers to figure out how many of these cubic centimeters would fit into this figure. And the first way I'm going to think about it is by looking at slices. So I'm going to take this slice right over here of our original figure. And let's think about how using these numbers, we can figure out how many unit cubes were in that slice. Well this is 2 centimeters wide and it is 4 centimeters high. And you might be saying, hey Sal, I could just count these things. I could get 8 squares here. But what if there was a ton there? It would be a lot harder. And you might realize well I could just the area of this surface right over here. And it's only 1 deep so that also will give me the number of cubes. Let's find the area here. Well that's going to be 2 centimeters times 4 centimeters. That gives us the area of this. And then if we want to find out the number of cubes, well that's also going to be equivalent to the number of cubes. So we have 8 square centimeters is this area, and the number of cubes is 8. And if we want the number of cubes in the whole thing, we just have to multiply by the number of slices. And we see that we need one, two, three slices. This is 3 centimeters deep. So we're going to multiply that times 3. So we took the area of one surface. We took the area of this surface right over here. And then we multiply by the depth, that essentially gives us the number of cubes because the area of this surface gives us the number of cubes in an slice that is 1 cube deep." + }, + { + "Q": "I already know what wide is, but what does it mean by 'deep'? Sal was saying it at 5:05.\n", + "A": "anything with 3 dimensions. A rectangle has length and width , a box has depth as well. Think of a swimming pool , depth would be the bottom to the top although normally a pool isn t filled to the brim. In khan you will see swimming pool questions and you have to watch out for the depth of the water as well as the depth of the container", + "video_name": "daCT_24RnIY", + "timestamps": [ + 305 + ], + "3min_transcript": "So we would have to have this is 1 slice. We would have to have another slice, another slice and then another slice in order to construct the original figure. So 2 centimeters times 4 centimeters times 3 centimeters would give us our volume. Let's see if that works out. 2 times 4 is 8 times 3 is 24. Let me do that in that pink color. 24 centimeters cubed, or I could say cubic centimeters. So that's one way to measure the volume. Now there's multiple surfaces here. I happened to pick this surface, but I could have picked another one. I could have picked this surface right over here and done the exact same thing. So let's pick this surface and do the exact same thing. This surface is 3 centimeters by 4 centimeters. Let me do that in that blue color. So its area is going to be 12 square centimeters is the area of this surface. And 12 is also the number of cubes that we have in that slice. And so how many slices do we need like this in order to construct the original figure? Well we need, it's 2 centimeters deep. This is only 1 centimeter deep so we need two of them to construct the original figure. So we can essentially find the area of that first surface which was 3 times 4, and then multiply that times the width, times how many of those slices you need, so times 2. And once again, this is going to be 3 times 4 is 12 times 2 is 24. I didn't write the units this first time. But that's going to give us the count of how many cubic centimeters we have, how many unit cubes we can fit. So once again, this is 24 cubic centimeters. And you could imagine, you could do the same thing, but with the top surface. The top surface is 3 centimeters deep. And 2 centimeters wide. So you could view its area or its area is going to be 3 centimeters times 2 centimeters. So that area is-- let me do it in the same colors-- 3 centimeters times 2 centimeters which is 6 square centimeters. And that also tells you that there's going to be 6 cubes in this one cube deep slice. But how many of these slices do you need? Well you have this whole thing is 4 centimeters tall, and this thing is only 1 centimeter so you're going to need four of them. So that's 2, 3, try to draw it as neatly as I can, and 4. You're going to need 4 of these." + }, + { + "Q": "Why does Sal include a number line at 2:30 to represent is process to tackle this problem when he could just do count to 73 from 68 to show that we added 5 to get there?\n", + "A": "Because Sal just wanna show how we get 73, by breaking up 5 into 3+2, you can add 2 to 68 first,and then add 3 to 70 to get 73. This is an easy method of solving the problem like that, technically. If we just count from 68 to 73, how could we do if we need to add 68 to 29? continue to count?", + "video_name": "DzJvR56Suss", + "timestamps": [ + 150 + ], + "3min_transcript": "- [Voiceover] What I wanna show you now is a way to add numbers, or a way that I sometimes add numbers, when I'm doing it in my head. So let's say I wanna add five to 68. And we've already seen other ways of doing it, so what I'm showing in this video isn't the only way to do it but it might be helpful when you're doing it in your head. Well when I look at a number like 68 I think well gee, If I added just two to that I would get to 70. So why don't I break up five into three and two, add the two to 68 and then I have to add the three. So what's that going to get me? Well if I break up the five into three plus two, and the whole reason why I broke it up into three and two is so I have this two here to add to 68. So five is three plus two, and to that I'm going to add 68, 68. And once again the reason why I took this two out is because I said hey, what do I have to add to 68 to get to 70? So now I could just rewrite this as, But now I can add the two to the 68. So two plus 68 is going to be Two plus 68 is 70. 70 and I still have this three here, so I have three plus 70 which is equal to 73. 73. Now it might seem like this really long way of doing it, but this is just a way to think about it. In your head you'd say okay five plus 68, let's see if I add two to 68 I'll get to 70 and so let's see, five is three plus two, I add the two to 68, I get to 70, and I have three left over. So it's going to be 73. Another way you could think about it is on a number line. So let's draw ourselves a number line here. Let's draw a number line and let me draw some And let's say that this right over here, this right over here is 68, and so this is going to be 70, this is going to be 70, and we're gonna add five. We're gonna add five. So if you add five you say well let me add two first. So if I add two first I get to 70. So plus two, that's what I did right over here to get to 70, and then if I wanna add five then I have to add three more. Then I have to add three more. And so it's going to be 70 plus three. Which of course is 73. Hopefully you found that interesting." + }, + { + "Q": "\nAt 1:28 into the video you wrote a symbol which had a = sign and something that looks like a vertical line and a greater than symbol. Would you please explane what it is? Thanks", + "A": "I think Sal is just drawing an arrow to show that he is changing the expression into a simplified form.", + "video_name": "Q1vMNyIP4Us", + "timestamps": [ + 88 + ], + "3min_transcript": "What I want to do in this video is write the algebraic expressions that represent the same thing that these statements are saying. So this first statement, they say the sum of negative 7 and the quantity 8 times x. So the sum-- so we're going to have an addition here-- of negative 7 and the quantity 8 times x. So the quantity 8 times x, well, that's just 8x. So I can just write 8x over there. So it is negative 7 plus 8x. Or you could view this as the sum of negative 7 and the quantity 8 times x. Let's do the next one. Take the quantity negative 3 times x and then add 1. So the quantity negative 3 times x, we can write that as negative 3x. And then we need to add 1 to that. So that's going to be plus 1. Negative 6 plus-- so we can write negative 6 plus something-- the product of negative 1 and x. So the product of negative 1 and x, that's just going to be negative 1x, which is the same thing as negative x. So we can write this as negative 6 plus negative x. Or we can just write this-- this is the exact same thing as negative 6 minus x. And we are done." + }, + { + "Q": "\nWhat the heck is a reciprocal? 4:00", + "A": "Example : 1/2 and 2/1= 2/2=1", + "video_name": "f3ySpxX9oeM", + "timestamps": [ + 240 + ], + "3min_transcript": "We could write times 3 like that. Or, if we want to write 3 as a fraction, we know that 3 is the same thing as 3/1. And we already know how to multiply fractions. Multiply the numerators. 8 times 3. So you have 8-- let me do that that same color. You have 8 times 3 in the numerator now, 8 times 3. And then you have 3 times 1 in the denominator. Which would give you 24/3, which is the same thing as 24 divided by 3, which once again is equal to 8. Now let's see if this still makes sense. Instead of dividing by 1/3, if we were to divide by 2/3. So let's think about what 8/3 divided by 2/3 is. if we wanted to break up this section from 0 to 8/3 into sections of 2/3, or jumps of 2/3, how many sections, or how many jumps, would I have to make? Well, think about it. 1 jump-- we'll do this in a different color. We could make 1 jump. No, that's the same color as my 8/3. We could do 1 jump. My computer is doing something strange. We could do 1 jump, 2 jumps, 3 jumps, and 4 jumps. So we see 8/3 divided by 2/3 is equal to 4. Now, does this make sense in this world right over here? Well, if we take 8/3 and we do the same thing, saying hey, look, dividing by a fraction is the same thing as multiplying by a reciprocal. Well, let's multiply by 3/2. So we swap the numerator and the denominator. So we multiply it times 3/2. And then what do we get? In the numerator, once again, we get 8 times 3, which is 24. And in the denominator, we get 3 times 2, which is 6. So now we get 24 divided by 6 is equal to 4. Now, does it make sense that we got half the answer? If you think about the difference between what we did here and what we did here, these are almost the same, except here we really just didn't divide. Or you could say you divided by 1, while here you divided by 2. Well, does that make sense? Well, sure. Because here you jumped twice as far. So you had to take half the number of steps. And so in the first example, you saw why it makes sense to multiply by 3. When you divide by a fraction, for every whole, you're making 3 jumps. So that's why when you divide by this fraction," + }, + { + "Q": "\nat 0:28,why is 5+5+8,5+8+5,and 8+5+5 exactly the same?", + "A": "how do you have 100,00 and you joined 2 m ago??", + "video_name": "HwSszh3L358", + "timestamps": [ + 28 + ], + "3min_transcript": "Use the commutative law of addition-- let me underline that-- the commutative law of addition to write the expression 5 plus 8 plus 5 in a different way and then find the sum. Now, this commutative law of addition sounds like a very fancy thing, but all it means is if you're just adding a bunch of numbers, it doesn't matter what order you add the numbers in. So we could add it as 5 plus 8 plus 5. We could order it as 5 plus 5 plus 8. We could order it 8 plus 5 plus 5. These are all going to add up to the same things, and it makes sense. If I have 5 of something and then I add 8 more and then I add 5 more, I'm going to get the same thing as if I had took 5 of something, then added the 5, then added the 8. You could try all of these out. You'll get the same thing. Now, they say in a different way, and then find the sum. The easiest one to find the sum of-- actually, let's do But the easiest one, just because a lot of people immediately know that 5 plus 5 is 10, is to maybe start with So if you have 5 plus 5, that's 10, plus 8 is equal to 18. Now, let's verify that these two are the same exact thing. Up here, 5 plus 8 is 13. 13 plus 5 is also 18. That is also 18. If we go down here, 8 plus 5 is 13. 13 plus 5 is also equal to 18. So no matter how you do it and no matter what order you do it in-- and that's the commutative law of addition. It sounds very fancy, but it just means that order doesn't matter if you're adding a bunch of things." + }, + { + "Q": "At 2:30 he says that the shape can be a rhombus. I thought rhombuses could not have any right angles!?\n", + "A": "Definition of rhombus is all sides are equal, and opposite sides are parallel. So a square is a particular type of rhombus.", + "video_name": "wPZIa3SjPF0", + "timestamps": [ + 150 + ], + "3min_transcript": "What is the type of this quadrilateral? Be as specific as possible with the given data. So it clearly is a quadrilateral. We have four sides here. And we see that we have two pairs of parallel sides. Or we could also say there are two pairs of congruent sides here as well. This side is parallel and congruent to this side. This side is parallel and congruent to that side. So we're dealing with a parallelogram. Let's do more of these. So here it looks like a same type of scenario we just saw in the last one. We have two pairs of parallel and congruent sides, but all the sides aren't equal to each other. If they're all equal to each other, we'd be dealing with a rhombus. But here, they're not all equal to each other. This side is congruent to the side opposite. This side is congruent to the side opposite. That's another parallelogram. Now this is interesting. We have two pairs of sides that are parallel to each other, but now all the sides have an equal length. So this would be a parallelogram. And it is a parallelogram, but they're So saying it's a rhombus would be more specific than saying it's a parallelogram. This does satisfy the constraints for being a parallelogram, but saying it's a rhombus tells us even more. Not every parallelogram is a rhombus, but every rhombus is a parallelogram. Here, they have the sides are parallel to the side opposite and all of the sides are equal. Let's do a few more of these. What is the type of this quadrilateral? Be as specific as possible with the given data . So we have two pairs of sides that are parallel, or I should say one pair. We have a pair of sides that are parallel. And then we have another pair of sides that are not. So this is a trapezoid. But then they have two choices here. They have trapezoid and isosceles trapezoid. Now an isosceles trapezoid is a trapezoid where the two non-parallel sides have the same length, just like an isosceles triangle, you have Well we could see these two non-parallel sides do not have the same length. So this is not an isosceles trapezoid. If they did have the same length, then we would pick that because that would be more specific than just trapezoid. But this case right over here, this is just a trapezoid. Let's do one more of these. What is the type of this quadrilateral? Well we could say it's a parallelogram because all of the sides are parallel. But if we wanted to be more specific, you could also see that all the sides are the same. So you could say it's a rhombus, but you could get even more specific than that. You notice that all the sides are intersecting at right angles. So this is-- if we wanted to be as specific as possible-- this is a square. Let me check the answer. Got it right." + }, + { + "Q": "\nAt 1:36, Sal says the lines are parallel. How do you know for sure that they are parallel?", + "A": "Great point. We do not know for sure. However, we can take it as a given from the problem.", + "video_name": "wPZIa3SjPF0", + "timestamps": [ + 96 + ], + "3min_transcript": "What is the type of this quadrilateral? Be as specific as possible with the given data. So it clearly is a quadrilateral. We have four sides here. And we see that we have two pairs of parallel sides. Or we could also say there are two pairs of congruent sides here as well. This side is parallel and congruent to this side. This side is parallel and congruent to that side. So we're dealing with a parallelogram. Let's do more of these. So here it looks like a same type of scenario we just saw in the last one. We have two pairs of parallel and congruent sides, but all the sides aren't equal to each other. If they're all equal to each other, we'd be dealing with a rhombus. But here, they're not all equal to each other. This side is congruent to the side opposite. This side is congruent to the side opposite. That's another parallelogram. Now this is interesting. We have two pairs of sides that are parallel to each other, but now all the sides have an equal length. So this would be a parallelogram. And it is a parallelogram, but they're So saying it's a rhombus would be more specific than saying it's a parallelogram. This does satisfy the constraints for being a parallelogram, but saying it's a rhombus tells us even more. Not every parallelogram is a rhombus, but every rhombus is a parallelogram. Here, they have the sides are parallel to the side opposite and all of the sides are equal. Let's do a few more of these. What is the type of this quadrilateral? Be as specific as possible with the given data . So we have two pairs of sides that are parallel, or I should say one pair. We have a pair of sides that are parallel. And then we have another pair of sides that are not. So this is a trapezoid. But then they have two choices here. They have trapezoid and isosceles trapezoid. Now an isosceles trapezoid is a trapezoid where the two non-parallel sides have the same length, just like an isosceles triangle, you have Well we could see these two non-parallel sides do not have the same length. So this is not an isosceles trapezoid. If they did have the same length, then we would pick that because that would be more specific than just trapezoid. But this case right over here, this is just a trapezoid. Let's do one more of these. What is the type of this quadrilateral? Well we could say it's a parallelogram because all of the sides are parallel. But if we wanted to be more specific, you could also see that all the sides are the same. So you could say it's a rhombus, but you could get even more specific than that. You notice that all the sides are intersecting at right angles. So this is-- if we wanted to be as specific as possible-- this is a square. Let me check the answer. Got it right." + }, + { + "Q": "At 1:38 she said that they found the highest prime number so far? How have they not found more?\n", + "A": "Because finding primes is hard. I could explain exactly why it is hard, but I will instead ask you a question: is 1000000007 (one billion seven) prime? If you try to find it out without a computer, you will soon see the difficulties behind finding primes.", + "video_name": "Yhlv5Aeuo_k", + "timestamps": [ + 98 + ], + "3min_transcript": "Pretend you're me and you're in math class. Actually... nevermind, I'm sick so I'm staying home today so pretend you are Stanislaw Ulam instead. What I am about to tell you is a true story. So you are Stan Ulam and you're at a meeting but there's this really boring presentation so of course you're doodling and, because you're Ulam and not me, you really like numbers... I mean super like them. So much that what you're doodling is numbers, just counting starting with one and spiralling them around. I'm not too fluent in mathematical notation so so i find things like numbers to be distracting, but you're a number theorist and if you love numbers who am I to judge? Thing is, because you know numbers so intimately, you can see beyond the confusing, squiggly lines you're drawing right into the heart of numbers. And, because you're a number theorist, and everyone knows that number theorists are enamoured with prime numbers( which is probably why they named them \"prime numbers\"), the primes you've doodled suddenly jump out at you like the exotic indivisible beasts they are... So you start drawing a heart around each prime. Well... it was actually boxes but in my version of the story it's hearts because you're not afraid to express your true feelings about prime numbers. You can probably do this instantly but it's going to take me a little longer... I'm all like - \"Hmmm what about 29...? pretty sure it's prime.\" But as a number theorist, you'll be shocked to know it takes me a moment to figure these out. But, even though you have your primes memorised up to at least 1000 that doesn't change that primes, in general, are difficult to find. I mean if I ask you to find the highest even number, you'd say, \"that's silly, just give me the number you think is the highest and i'll just add 2.... BAM!!\" But guess what the highest prime number we know is? 2 to the power of 43,112,609 - 1. Just to give you an idea about how big a deal primes are, the guy that found this one won a $100,000 prize for it! We even sent our largest known prime number into space because scientists think aliens will recognise it as something important and not just some arbitrary number. So they will be able to figure out our alien space message... So if you ever think you don't care about prime numbers because they're 'not useful', remember that we use prime numbers to talk to aliens, I'm not even making this up! Anyway, the point is you started doodling because you were bored but ended up discovering some neat patterns. See how the primes tend to line up on the diagonals? Why do they do that?... also this sort of skeletal structure reminds me of bones so lets call these diagonal runs of primes: Prime Ribs! But how do you predict when a Prime Rib will end? I mean, maybe this next number is prime... (but my head is too fuzzy for now this right now so you tell me.) Anyway...Congratulations, You've discovered the Ulam Spiral! So that's a little mathematical doodling history for you. Yyou can stop being Ulam now... or you can continue. Maybe you like being Ulam. (thats fine) However you could also be Blaise Pascal. Here's another number game you can do using Pascal's triangle.(I don't know why I'm so into numbers today but I have a cold so if you'll just indulge my sick predelections maybe I'll manage to infect you with my enthusiasm :D Pascal's Triangle is the one where you get the next row in the triangle by adding two adjacent numbers. Constructing Pascal's Triangle is, in itself a sort of number game because it's not just" + }, + { + "Q": "\nat 2:31 what is a prespictipal", + "A": "Simply put, a reciprocal of a fraction is when you flip the fraction upside down. For example, the reciprocal of 5/7 is 7/5. The reciprocal of 3/4 is 4/3.", + "video_name": "d8vvVjfTbYY", + "timestamps": [ + 151 + ], + "3min_transcript": "Odie the octopus swam home. He swam at a constant speed. He plotted how far he'd gone every 1/7 of a second on the graph below. Not all octopi can create graphs, but Odie was no ordinary octopus. And actually, octopi are some of the smartest creatures on this planet. What was Odie's speed-- after probably dolphins and great apes and probably most primates, but they're still fairly smart, smarter than most fish. What was Odie's speed in meters per second? Note we use m/s to show the units of meters per second. All right. So to figure out speed in meters per second, we essentially need to divide distance in meters divided by time in seconds. So the easiest thing-- let me find a point that I can clearly read how far Odie has traveled. So it looks like this point right over here. It's clearly that the distance in meters is 4/7. So the distance in meters is 4/7, so I'll do that in the numerator. the distance in meters. Gee, I don't like that con-- well, I'll You know, the contrast with the white background is a little hard on the eye. And It took him 5/7 of a second to get there. So 4/7 meters-- actually, let me make sure I put the units here. 4/7 meters in 5/7 seconds. And so literally, we just need to divide 4/7 by 5/7, and we're going to get it in meters per second. So this is going to be the same thing as 4/7 over 5/7 meters per second. So let me write that down. So 4/7 times 7/5. Dividing by a fraction is the same thing as multiplying by its reciprocal. So times 7/5, and we get this-- well, this 7 and this 7 cancel out, and you get 4 over 5. You could have also done that by multiplying the numerator and the denominator here by 7. If you multiply it by 7, multiply by 7, you would get 4/5. But either way, his speed in meters per second is 4/5 meters per second." + }, + { + "Q": "At 8:10, you integrate (5/2)(1/x+1) to 5/2*ln(x+1) explaining that the derivative of the denominator (x+1) is equal to the numerator 1 . Could you please explain this further as I am still very confused?\n", + "A": "F(x) = \u00e2\u0088\u00ab(5/2\u00e2\u0080\u00a21/(x + 1))dx F(x) = 5/2\u00e2\u0080\u00a2\u00e2\u0088\u00ab(1/(x + 1))dx u = x + 1 (d/dx(x + 1))dx = du (d/dx(x) + d/dx(1))dx = du (d/dx(x) + 0)dx = du (d/dx(x))dx = du (dx/dx)dx = du (1)dx = du dx = (1)du dx = du F(u) = 5/2\u00e2\u0080\u00a2\u00e2\u0088\u00ab(1/u)du F(u) = 5/2\u00e2\u0080\u00a2ln(u) + C u = x + 1 F(x) = 5/2\u00e2\u0080\u00a2ln(x + 1) + C", + "video_name": "7IkufOBIw5g", + "timestamps": [ + 490 + ], + "3min_transcript": "I'll do it up here since I have a little bit of real estate. A plus B is going to be equal to one, and B minus A or I could write that as negative A plus B is equal to negative four. We could add the left hand sides and add the right hand sides and then the A's would disappear. We would get two B is equal to negative three or B is equal to negative three halves. We know that A is equal to one minus B, which would be equal to one plus three halves, since B is negative three halves, which is equal to five halves. A is equal to five halves. B is equal to negative three halves. And just like that we can rewrite this whole integral in a way that is a little bit easier to take the anti or this whole expression so it's easier to integrate. So it's going to be the integral of A is five halves and so I could just write that as, let me write it this way, five halves times one over X plus one. I wrote it that way because it's very straight forward to take the antiderivative of this. Then plus B over X minus one. Which is going to be negative three halves. So I'll just write it as minus three halves times one over X minus one. That was this right over here, DX. Notice all I did is I took this expression right over here and I did a little bit of partial fraction expansion into these two, I guess you could say, expressions or terms right over there. It's fairly straight forward to integrate this. Antiderivative of one, it's just going to be X. is going to be plus five halves, the natural log of the absolute value of X plus one. We're able to do that because the derivative of X plus one is just one, so the derivative is there so that we can take the antiderivative with respect to X plus one. You could also do u-substitution like we've done in previous examples, U is equal to X plus one. And over here, this is going to be minus three halves times the natural log of the absolute value of X minus one, by the same exact logic with how we were able to take the antiderivative there. And of course we cannot forget our constant. And there we have it. We've been able to integrate, we were able to evaluate this expression." + }, + { + "Q": "\nAt 1:42, Sal says \"the fewest number of candy bars we can buy is zero candy bars\". Why is this so? The definition of a purchase is receiving goods in exchange for money. Surely no money and no candy bars equals no purchase. Do all functions just have to accept zero as an input?", + "A": "Not all functions have the same words like purchase so this makes it easier to understand so in a less mathy way of saying it is you can buy No candy bars for $0.00( my keyboard does not have the cent symbol). So manly all functions have different meaning of the words so they can all accept 0 to make it easier to understand p(0) completely means 0 candy bars so $0.00 was spent for it. Hope this helps..........Sorry if you already had it. It been a month xD.", + "video_name": "AiW7syKXfJM", + "timestamps": [ + 102 + ], + "3min_transcript": "Thomas has 400 candy bars in his shop and each cost 50 cents. Let p of b denote the price, p, measured in dollars of a purchase of b candy bars. Alright I input b, the number candy bars I wanna buy, and p(b) will tell me what's the purchase price is really just taking the number of candy bars multiplied by 50 cents, but we won't have to worry about that just yet. Which number type is more appropriate for the domain of the function? So just to remind ourselves, what is the domain of a function? A domain is a set of all inputs over which the function is defined. So it is the set of all b's. It is the set of all inputs over which p of b will produce a defined response So let's think about it. Is it integers or real numbers? So I could buy -- b could be 0 candy bars, 1 candy bars, 2 candy bars, all up to 400 candy bars. Could I -- Could I have a fractional can-- Could be b 0.372 of a candy bar? Well, this is a normal candy shop. chunk. You're not going to be able to buy 0.372 of a candy bar. You can either buy a one more or none more, so you buy your 1, 2, 3 all the way up to 400. So I would say integers -- that the domain of this function is going to be is going to be a subset of integers. It's not -- you not, you can't have a real, all real number, but integers are obviously a subset real numbers. But you can't say, hey, I'm gonna buy pi candy bars, or I'm gonna buy the square root of two candy bars. You're gonna buy integer number candy bars. Now they say, define the interval of the domain. So the fewest candy bars I could are 0 candy bars, and I have to decide whether I put a bracket or I put a parenthesis. I can actually buy 0 candy bars so I'm gonna put a bracket. If I put a parentheses, that means I could have values above zero but not including 0, but I want to include 0 so I'm gonna put the bracket there. So the least I could buy can buy. The most I could buy are 400 candy bars, and I can buy 400. So I would put brackets there as well. So the interval of the domain, I would want to select integers. So b is a member of integers such that b is also a member of this interval. It could be as low as 0 including 0, and as high as 400 including 400. Got it right." + }, + { + "Q": "At 3:26 why did Sal put the x's on the right and not the left?\n", + "A": "It doesn t matter which side you move the x s to. Either side will work. Sal likely moved the x s to the right side because it keeps the value positive (which some people prefer).", + "video_name": "EHR-YDwrrhM", + "timestamps": [ + 206 + ], + "3min_transcript": "x is negative five, five times negative five, plus two y. See, negative three times negative five is positive 15, plus seven y, is equal to negative 25 plus two y. And now, to solve for y, let's see, I could subtract two y from both sides, so that I get rid of the two y here on the right. So let me subtract two y, subtract two y from both sides. And then if I want all my constants on the right hand side, I can subtract 15 from both sides. So let me subtract 15 from both sides. And I'm going to be left with 15 minus 15, that's zero, that's the whole point of subtracting 15 from both sides, so I get rid of this 15 here. Seven y minus two y. Seven of something minus two of that same something is gonna be five of that something. It's gonna be equal to five y, is equal to negative 25 minus 15. And then two y minus two y, well, that's just gonna be zero. That was the whole point of subtracting two y from both sides. So you have five times y is equal to negative 40. Or, if we divide both sides by five, we divide both sides by five, we would get y is equal to negative eight. So when x is equal to negative five, y is equal to negative eight. Y is equal to negative eight. And actually we can fill that in. So this y is going to be equal to negative eight. And now we gotta figure this out. What does x equal when y is positive eight? Well, we can go back to our scratchpad here. And I'll take the same equation, but let's make y equal to positive eight. So you have negative three x plus seven, now y is going to be eight, y is eight, seven times eight is equal to five times x, y is eight, two times eight. So we get negative three x plus 56, that's 56, is equal to five x plus 16. Now, if we wanna get all of our constants on one side, and of all of our x terms on the other side, well, what could we do? Let's see, we could add three x to both sides. That would get rid of all the xs on this side, and put 'em all on this side. So we're gonna add three x to both sides. And, let's see, if we want to get all the constants on the left hand side, we'd wanna get rid of the 16, so we could subtract 16 from the right hand side, if we do it from the right, we're gonna have to do it from the left as well. And we're gonna be left with, these cancel out, 56 minus 16 is positive 40. And then, let's see, 16 minus 16 is zero. Five x plus three x is equal to eight x. We get eight x is equal to 40." + }, + { + "Q": "\nAt 6:43, when you did 8/5 times -5 how did you end up with -8?", + "A": "8/5 (-5) = 8/5 * (-5/1 ) = 8 (-5) / (5 * 1) = -40/5 = -8 hope this helps.", + "video_name": "EHR-YDwrrhM", + "timestamps": [ + 403 + ], + "3min_transcript": "We can check our answer, if we like. We got it right. Now, I said there was two ways to tackle it, I kind of just did it, I guess you could say, the naive way. I just substituted negative five directly into this and solved for y. And then I substituted y equals positive eight directly into this, and then solved for x. Another way that I could have done it, that actually probably would have been, or, it would for sure, would have been the easier way to do it, is ahead of time to try to simplify this expression. So what I could have done, right from the get-go, is said, \"Hey, let's put all my xs on one side, \"and all my ys on the other side.\" So this is negative three x plus seven y is equal to five x plus two y. Now let's say I wanna get all my ys on the left and all my xs on the right. So I don't want this negative three x on the left, so I'd wanna add three x. Adding three x would cancel this out, but I can't just do it on the left hand side, I have to do it on the right hand side as well. And then, if I wanna get rid of this two y on the right, I could subtract two y from the right, but, of course, I'd also wanna do it from the left. So negative three x plus three x is zero, seven y minus two y is five y. And then I have five x plus three x is eight x. Two y minus two y is zero. And then if I wanted to, I could solve for y, I could divide both sides by five and I'd get y is equal to 8/5 x. So, this right over here represents the same exact equation as this over here, it's just written in a different way. All of the xy pairs that satisfy this, would satisfy this, and vice versa. And this is much easier. Because if x is now negative five, if x is negative five, y would be 8/5 times negative five, well, that's going to be negative eight. And when y is equal to eight, well, you actually could even do this up here, you could say five times eight is equal to eight x, and then you could see, well five times eight the same thing as eight times five, so x would be equal to five. So I think this would actually have been You see it all, I was able to do the entire problem instead of having to do all of this, slightly, slightly hairier, algebra." + }, + { + "Q": "Hello! Okay, so at 1:03 in the video, how did he know that y>= -2? I get how he got it from the graph, but how would one find that algebraically? Please respond quickly, thank you so much!\n", + "A": "Algebraically, the equation was y = ( x - 1 )^2 - 2. Whatever value of x is substituted, the quantity ( x - 1 )^2 will be positive or zero; it will never be negative. Therefore, we ll always be adding a positive number (or zero) to -2, so the total value will be getting larger than (or staying equal to) -2.", + "video_name": "Bq9cq9FZuNM", + "timestamps": [ + 63 + ], + "3min_transcript": "We have the function f of x is equal to x minus 1 squared minus 2. And they've constrained the domain to x being less than or equal to 1. So we have the left half of a parabola right here. They've constrained so that it's not a full U parabola. And I'll let you think about why that would make finding the inverse difficult. But let's try to find the inverse here. And a good place to start -- let's just set y being equal to f of x. You could say y is equal to f of x or we could just write that y is equal to x minus 1 squared minus 2. We know it's 4. x is less than or equal to 1. But right now we have y solved for in terms of x. Or we've solved for y, to find the inverse we're going to want to solve for x in terms of y. And we're going to constrain y similarly. We could look at the graph and we could say, well, in this graph right here, this is defined for y being greater than or equal to negative 2. than or equal to negative 2. Because this is going to be -- right now this is our range. But when we swap the x's and y's, this is going to be our domain. So let's just keep that in parentheses right there. So let's solve for x. So that's all you have to do, to find the inverse. Solve for x and make sure you keep track of the domains and the ranges. So, let's see. We could add 2 to both sides of this equation. We get y plus 2 is equal to x minus 1 squared. Minus 2, plus 2, so those cancel out. And then, I'm just going to switch to the y constraint. Because now it's not clear what we're -- whether x is the domain or the range. But we know by the end of this problem, y is going to be the domain. So let's just swap this here. So, 4 for y is greater than or equal to negative 2. And we could also say, in parentheses, x is less than 1. This is -- we haven't solved it explicitly for either one, so we'll keep both of them around right now. Now, to solve for x, you might be tempted to just take the And you wouldn't be completely wrong. But we have to be very, very, very careful here. And this might not be something that you've ever seen before. So this is an interesting point here. We want the right side to just be x minus 1. That's our goal here, in taking the square root of both sides. We want to just have an x minus 1 over there. Now, is x minus 1 a positive or a negative number? Well, we've constrained our x's to being less than 1. So we're dealing only in a situation where x is less than or equal to 1. So if x is less than or equal to 1, this is negative. This is negative. So we want to take the negative square root. Let me just be very clear here. If I take negative 3 -- I take negative 3 and I were to square it, that is equal to 9. Now, if we take the square root of 9 -- if we take the square" + }, + { + "Q": "At 6:53 in, shouldn't the point be (3,-1), not (2,-1)?\n", + "A": "The point must be (3,1), not (-1,2).", + "video_name": "Bq9cq9FZuNM", + "timestamps": [ + 413 + ], + "3min_transcript": "You get negative square root of y plus 2 plus 1 is equal to x for y is greater than or equal to negative 2. Or, if we want to rewrite it, we could say that x is equal to the negative square root of y plus 2 plus 1 for y is greater than or equal to negative 2. Or if we want to write it in terms, as an inverse function of y, we could say -- so we could say that f inverse of y is equal to this, or f inverse of y is equal to the negative square root of y plus 2 plus 1, for y is greater than or equal to negative 2. And now, if we wanted this in terms of x. If we just want to rename y as x we just replace the y's with x's. So we could write f inverse of x -- I'm just Is equal to the negative square root of x plus 2 plus 1 for, I'm just renaming the y, for x is greater than or equal to negative 2. And if we were to graph this, let's see. If we started at x is equal to negative 2, this is 0. So the point negative 2, 1 is going to be on our graph. So negative 2, 1 is going to be on our graph. Let's see, if we go to negative 1, negative 1 this will become a negative 1. Negative 1 is 0 on our graph. Negative 1, 0 is on our graph. And then, let's see. If we were to do, if we were to put x is equal to 2 here. So x is equal to 2 is 4. 4 square root, principle root is 2. It becomes a negative 2 So it becomes 2, negative 1. So that's on our graph right there. So the graph is going to look something like this, of f inverse. It's going to look something like that right there. f of x along the line y is equal to x. Along the line y is equal to x. Because we've essentially just swapped the x and the y. This is about as hard of an inverse problem that I expect you to see. Especially in a precalculus class because it really is tricky to realize that you have to take the negative square root here. Because the way our domain was constrained, this value right here is going to be negative. So to solve for it, you want to have the negative square root." + }, + { + "Q": "At 5:05 Sal mentions interval. Can someone explain to me what this 'interval' means? I keep on hearing this term more and more often.\n", + "A": "In this context, interval just refers to the section of points (along the x-axis) between x1 and x2. If it were an interval of time, for example, it could be the segment of time between two events. Hope this helps!", + "video_name": "8r8Vp_1iB4k", + "timestamps": [ + 305 + ], + "3min_transcript": "You could have done that. Then you would have just gotten the negative of each of these values in the numerator and denominator, but they would have canceled out. The important thing is that you're consistent. If you're subtracting you're starting value from your ending value in the numerator, you have to subtract your starting value from your ending value in the denominator as well. So this right here you probably remember from algebra class. The definition of slope is the rate of change of y with respect to x. Or it's the rate of change of our vertical axis, I should say, with respect to our horizontal axis. Or change in y, or change in our vertical axis over change in a horizontal axis. Now I'm going to introduce a little bit of a conundrum. So let me draw another axis right over here. Scroll over a little bit just so we have some space to work with. So that was for a line. And a line, by definition, has a constant slope. If you calculate this between any two points on the line, it's going to be constant for that line. But what happens when we start dealing with curves? So let's imagine a curve that looks something like this. So what is the rate of change of y with respect to x of this curve? Well, let's look at it at different points. And we could at least try to approximate what it might be in any moment. So let's say that this is one point on a curve. Let's call that x1, and then this is y1. And let's say that this is another point on a curve right over here, x2. And let's call this y2. So this is a point x1, y1, this is a point x2, y2. So we don't have the tools yet. And this is what's exciting about calculus, we will soon have the tools to figure out, what is the rate of change of y with respect to x at exactly this point? But we don't have that tool yet. But using just the tools from algebra, what is the average rate of change over the interval from x1 to x2? Well, what's the average rate of change? Well, that's just how much did my y change-- so that's my change in y-- for this change in x. And so we would calculate it the same way. y2 minus y1 over x2 minus x1. So our change in y over this interval is equal to y2 minus y1, and our change in x is going to be equal to x2 minus x1. So just like that we were able to figure out the rate of change between these two points. Or another way of thinking about it is, this is the average rate of change for the curve between x equals x1, and x is equal to x2. This is the average rate of change of y with respect to x over this interval." + }, + { + "Q": "\nWhy is he multiplying the -(3x - 4) @ 1:10? Can he not also just add that?", + "A": "The minus in front of the (3x - 4) says to subtract the entire binomial. The minus sign must be distributed to accomplish that subtraction. If you just drop the parentheses and add, then the -4 is being added, not subtracted (you have +(-4) when you need -(-4)). Hope this helps.", + "video_name": "DMyhUb1pZT0", + "timestamps": [ + 70 + ], + "3min_transcript": "We're asked to simplify this huge, long expression here. x to the third plus 3x minus 6-- that's in parentheses-- plus negative 2x squared plus x minus 2. And then minus the quantity 3x minus 4. So a good place to start, we'll just rewrite this and see if we can eliminate the parentheses in this step. So let's just start at the beginning. We have the x to the third right over there. So x to the third and then plus 3x-- I'll do that in pink-- plus 3x. And then we have a minus 6. And we don't have to put the parentheses around there, those don't really change anything. And we don't have to even write these-- do anything with these parentheses. We can eliminate them. Just because there's a positive sign out here we don't have to distribute anything. Distributing a positive sign doesn't do anything to these numbers. So then plus, we have a negative 2x squared. So this term right here is negative 2x squared, or minus x squared. And then we have a plus x. We have a plus x. Then we have a negative sign times this whole expression. So we're going to have to distribute the negative sign. So it's a positive 3x, but it's being multiplied by negative 1. So it's really a negative 3x. So minus 3x, then you have a negative-- you can imagine this is a negative 1 implicitly out here-- negative 1 times negative 4. That's a positive 4. So plus 4. Now, we could combine terms of similar degree, of the same degree. Now, first we have an x to the third term and I think it's the only third degree term here, because we have x being raised to the third power. So let me just rewrite it here. We have x to the third. And now let's look at our x squared terms. Looks like we only have one. We only have this term right here. So we have minus 2x squared. And then what about our x terms? We have a 3x plus an x minus a 3x again. So that 3x minus the 3x would cancel out, and you're just So plus x. And then finally our constant terms. Negative 6 minus 2 plus 4. Negative 6 minus 2 gets us to negative 8. Plus 4 is negative 4. We have simplified the expression. Now we just have a four term polynomial." + }, + { + "Q": "\nAt 3:08, Vertically, I don't see 2 blocks.\nThanks\n---- Moon trainer", + "A": "I code on Scratch too! My username is Song_of_the_Cats", + "video_name": "gkifo46--JA", + "timestamps": [ + 188 + ], + "3min_transcript": "And so instead, let me see if I can get away with making a triangle just like that. Now, I just did that. So let me try to raise this up to see if I can make another triangle where it would be easy to figure out its dimensions. So once again, I don't want to go all the way up here because now I'm not at a whole unit. Instead, let me take a right and go right over here. And notice, both of these are very easy to figure out its dimensions. This is 1, 2, 3, 4, 5 units long and 1 unit high. This one right over here is 1, 2, 3, 4 units long and 1, 2 units wide. So let's see if we can cover the entire quadrilateral, if we can break it up, I should say, into a bunch of figures like this. And then I could drop this down, and then we're done. All of these are pretty straightforward to figure out what their dimensions are. This is 5 by 1. This is 4 by 2. This is 1, 2, 3, 4, 5, 6 by 1, 2. And this is 1 by 1, 2, 3, 4, 5. So what is the area of this figure? And of course, we have this center rectangle right over here. Well, a triangle that is 5 units long and 1 unit high, its area is going to be 1/2 times 1 times 5. Or I could write it 1/2 times 1 times 5, depending on what multiplication symbol you are more comfortable with. Well that's just going to be 1/2 times 5, which is going to be equal to 2.5. So that's 2.5 right over there. This one is going to be 1/2 times 4 times 2. This one is going to be 1/2 times 2 times 1, 2, 3, 4, 5, 6. Well, 1/2 half times 2 is 1 times 6 is just 6. And then this one's going to be 1/2 times 1 times 1, 2, 3, 4, 5. So once again, the area of this one is going to be 2.5. And then finally, this is a 3 by 4 rectangle. And you could even count the unit squares in here. But it has 12 of those unit square, so it has an area of 12. So if we want to find the total area, we just add all of these together. So 2.5 plus 2.5 is 5, plus 4 is 9, plus 6 is 15, plus 12 is 27. So it has a total area of 27." + }, + { + "Q": "\n@ 4:50 - I would have thought that since we are solving for where the greater change occurs, g'(4) < g'(6) would be correct, as the change at g'(6) is far more 'steep'. Its obviously that -1 > -3, however I thought that we're really solving for here is the rate of change.", + "A": "It is descending more at 6 but the derivative gives how much something increases as we move along the graph. So if it decreases then it will be negative.", + "video_name": "S-dcMvJlMJs", + "timestamps": [ + 290 + ], + "3min_transcript": "the slope of the tangent line right at that bottom point would have a slope of zero. So I feel really good about that response. Let's do one more of these. So alright, so they're telling us to compare the derivative of G at four to the derivative of G at six and which of these is greater and like always, pause the video and see if you can figure this out. Well this is just an exercise let's see if we were to if we were to make a line that indicates the slope there you can do this as a tangent line let me try to do that. So now that wouldn't, that doesn't do a good job so right over here at that looks like a I think I can do a better job than that no that's too shallow to see not shallow's not the word, that's too flat. So let me try to really okay, that looks pretty good. seems to be indicative of the rate of change of Y with respect to X or the slope of that curve or that line you can view it as a tangent line so we could think about what its slope is going to be and then if we go further down over here this one is, it looks like it is steeper but in the negative direction so it looks like it is steeper for sure but it's in the negative direction. As we increase, think of it this way as we increase X one here it looks like we are decreasing Y by about one. So it looks like G prime of four G prime of four, the derivative when X is equal to four is approximately, I'm estimating it negative one while the derivative here when we increase X if we increase X by if we increase X by one so G prime of six looks like it's closer to negative three. So which one of these is larger? Well, this one is less negative so it's going to be greater than the other one and you could have done this intuitively if you just look at the curve this is some type of a sinusoid here you have right over here the curve is flat you have right at that moment you have no change in Y with respect to X then it starts to decrease then it decreases at an even faster rate then it decreases at a faster rate then it starts, it's still decreasing but it's decreasing at slower and slower rates decreasing at slower rates and right at that moment you have your slope of your tangent line is zero then it starts to increase, increase, so on and so forth" + }, + { + "Q": "\nAt 0:33, what does translate mean?", + "A": "Moving the entire figure, without changing its size or orientation.", + "video_name": "6p1lweGactg", + "timestamps": [ + 33 + ], + "3min_transcript": "These two quadrilaterals, EFGH and ABCD, these are similar. If through some combination of translations, rotations, reflections, and dilations, I can make them sit on top of each other. And similarity allows us to use dilations, which essentially means scale up one of the figures up or down. If we're talking about congruence, then we wouldn't be scaling the figures up or down. We would only be translating, rotating, and reflecting. But let's see if we can do it. So let's first translate it. Let's first translate this figure. So let's make that point on top of that point. And now let's rotate it. I'm going to rotate it around point E right over here. So I get right over there. So I got two of the sides to kind of match up. But now let me dilate it down. Let me scale it down. And so this is-- let me put it on what I want to not change-- and then let me see if I can scale this down. And I was able to. So through just purely translations, rotations-- rotations, and dilations, I was able to make these sit on top of each other. So these two figures are similar. So yes, the rectangles are similar. Let's do one more of these. So these two triangles-- and just eyeballing it, this one looks kind of taller than this one does right over here. So they don't feel similar, but let me at least try. So let me translate, maybe make this point C might correspond to point F, although it already looks pretty clear that it won't. But now let me dilate it. So I'll keep that point where it is. And let me try to scale this up. And then we see pretty clearly we made segment CB sit on top of, after scaling it up, sit on top of segment FE. But then everything else is not matching up. Point D is in a very different place than where point A is once I try to scale things So these two triangles are clearly, clearly not similar." + }, + { + "Q": "At 5:36, the outer radius is defines as 2-y^2.\nCan anybody explain why ones takes 2 - the function. My first guess was the function + 2 so that i would reach the line where x =2. This is wrong, but i can\u00c2\u00b4t say i really understand why.\n\nAnd, taking 2-y^2 would that be the gap between the undefined point on the x-axes and x=2? Where is this on our shape?\nThanks.\n", + "A": "At this stage in the process we re trying to find the radius of the outer disk. The radius is the distance from the center of the disk to the perimeter. At the beginning of the video we re told that the figure is created by rotation around the line x=2, so we know the center of the disk is at x=2. The outer edge of the larger disk is at the point x=y^2, so the distance between the two points is 2-y^2.", + "video_name": "WAPZihVUmzE", + "timestamps": [ + 336 + ], + "3min_transcript": "going to be pi times outer radius squared minus pi-- let me write it this way-- outer-- well, I'll just write it-- outer radius-- this is going to be as a function of y-- outer radius squared minus pi times inner radius squared. And we want this all to be a function of y. So the outer radius as a function of y is going to be what? Well, it might be easier to visualize-- actually, I'll try it both places. So it's this entire distance right over here, essentially, the distance between the vertical line, the horizontal distance between our vertical line, and our outer function. It's going to be 2 minus whatever the x value is right over here. So the x value right over here is going to be y squared. Remember, we want this is a function of y. So our outer radius, this whole distance, is going to be 2 minus the x value here as a function of y. That x value is y squared. So the outer radius is 2 minus x-- sorry. 2 minus y squared. We want it as a function of y. And then our inner radius is going to be equal to what? Well, that's going to be the difference, the horizontal distance, between this vertical line and our inner function, our inner boundary. So it's going to be the horizontal distance between two and whatever x value this is. But this x value as a function of y is just square root of y. So it's going to be 2 minus square root of y. And so now, we can come up with an expression for area. I'll leave the pi there. So it's going to be pi-- right over here-- it's going to be pi times outer radius squared. Well, the outer radius is 2 minus y squared-- and let me just-- well, I'll just write it. 2 minus y squared-- and we're going to square that-- squared, minus pi times the inner radius squared. Well, we already figured that out. The inner radius is 2 minus square root of y. And we're going to square that one, too. So this gives us the area of one of our rings as a function of y, the top of the ring, where I shaded in orange. And now, if we want the volume of one of those rings, we have to multiply it by its depth or its height the way we've drawn it right over here. And its height-- we've done this multiple times already right over here-- is an infinitesimal change in y. So we're going to multiply all that business times dy. This is the volume of one of our rings." + }, + { + "Q": "\nAt 5:29, I was taught to do it in the reverse order like (y^2-2), is that wrong?", + "A": "In that case the radius would be negative, but it is just a distance, it shouldn t be negative, but it has the same magnitude so when you square it, I guess it works out just the same. Anyway how Sal does it is the most intuitive and rigorously correct manner.", + "video_name": "WAPZihVUmzE", + "timestamps": [ + 329 + ], + "3min_transcript": "x is equal to y squared. That's our top function the way we've drawn a kind of outer shell for our figure. And then y is equal to x squared. If you take the principal root of both sides of that and it all works out because we're operating in the first quadrant That's the part of it that we care about. So you're going to get x is equal to the square root of y. That is our yellow function right over there. Now, how do we figure out the area of the surface of one of these rings or one of these washers? Well, the area-- let me do this in orange because I drew that ring in orange. So the area of the surface right over here in orange as a function of y is going to be equal to the area of the circle if I just consider the outer radius, and then I subtract out the area of the circle constructed by the inner radius, just kind of subtract it out. going to be pi times outer radius squared minus pi-- let me write it this way-- outer-- well, I'll just write it-- outer radius-- this is going to be as a function of y-- outer radius squared minus pi times inner radius squared. And we want this all to be a function of y. So the outer radius as a function of y is going to be what? Well, it might be easier to visualize-- actually, I'll try it both places. So it's this entire distance right over here, essentially, the distance between the vertical line, the horizontal distance between our vertical line, and our outer function. It's going to be 2 minus whatever the x value is right over here. So the x value right over here is going to be y squared. Remember, we want this is a function of y. So our outer radius, this whole distance, is going to be 2 minus the x value here as a function of y. That x value is y squared. So the outer radius is 2 minus x-- sorry. 2 minus y squared. We want it as a function of y. And then our inner radius is going to be equal to what? Well, that's going to be the difference, the horizontal distance, between this vertical line and our inner function, our inner boundary. So it's going to be the horizontal distance between two and whatever x value this is. But this x value as a function of y is just square root of y. So it's going to be 2 minus square root of y. And so now, we can come up with an expression for area." + }, + { + "Q": "At 5:15,what is associative property?\n", + "A": "The associative property of multiplication just means that when you multiply the numbers, it doesn t matter what order you do it in. (4 x 5) x 6 is (20) x 6, which is 120. 4 x (5 x 6) is 4 x 30, which is also 120.", + "video_name": "VXrn5HOQmHQ", + "timestamps": [ + 315 + ], + "3min_transcript": "So you could do that as 2 times 3. And we have one 3 and another 3. So in this purple zone, this is another 2 times 3. We have another 2 times 3. I wrote 2 times 2. 2 times 3. We have another 2 times 3. And then finally, we have a fourth 2 times 3. So how many 2 times 3's do we have here? Well, we have one, two, three, four 2 times 3's. So this whole thing could be written as 4 times 2 times 3. Now, what's this going to be equal to? Well, it needs to be equal to 24. And we can verify 2 times 3 is 6 times 4 is, indeed, 24. So the whole idea of what I'm trying to show here is that the order in which you multiply does not matter. Let me pick a different example, a completely new example. So let's say that I have 4 times 5 times 6. You can do this multiplication in multiple ways. You could do 4 times 5 first. Or you could do 4 times 5 times 6 first. And you can verify that. I encourage you to pause the video and verify that these two things are equivalent. And this is actually called the associative property. It doesn't matter how you associate these things, which of these that you do first. Also, order does not matter. And we've seen that multiple times. Whether you do this or you do 5 times 4 times 6-- notice I swapped the 5 and 4-- this doesn't matter. Or whether you do this or 6 times 5 times 4, Here I swapped the 6 and the 5 times 4. And I encourage you to pause the video. So when we're talking about which one we do first, whether we do the 4 times 5 first or the 5 times 6, that's called the associative property. It's kind of fancy word for a reasonably simple thing. And when we're saying that order doesn't matter, when it doesn't matter whether we do 4 times 5 or 5 times 4, that's called the commutative property. And once again, fancy word for a very simple thing. It's just saying it doesn't matter what order I do it in." + }, + { + "Q": "At 2:00, what are they trying to say?\n", + "A": "in the whole column there are 8 balls but it is divided into 2 groups of four. So 2 times four = 8", + "video_name": "VXrn5HOQmHQ", + "timestamps": [ + 120 + ], + "3min_transcript": "So if you look at each of these 4 by 6 grids, it's pretty clear that there's 24 of these green circle things in each of them. But what I want to show you is that you can get 24 as the product of three numbers in multiple different ways. And it actually doesn't matter which products you take first or what order you actually do them in. So let's think about this first. So the way that I've colored it in, I have these three groups of 4. If you look at the blue highlighting, this is one group of 4, two groups of 4, three groups of 4. Actually, let me make it a little bit clearer. One group of 4, two groups of 4, and three groups of 4. So these three columns you could view as 3 times 4. Now, we have another 3 times 4 right over here. This is also 3 times 4. We have one group of 4, two groups of 4, and three groups of 4. So you could view these combined as 2 times 3 times 4. We have one 3 times 4. And then we have another 3 times 4. some more space-- as 2 times-- let me do that in blue-- 2 times 3 times 4. That's the total number of balls here. And you could see it based on how it was colored. And of course, if you did 3 times 4 first, you get 12. And then you multiply that times 2, you get 24, which is the total number of these green circle things. And I encourage you now to look at these other two. Pause the video and think about what these would be the product of, first looking at the blue grouping, then looking at the purple grouping in the same way that we did right over here, and verify that the product still equals 24. Well, I assume that you've paused the video. So you see here in this first, I guess you could call it a zone, we have two groups of 4. So this is 2 times 4 right over here. We have one group of 4, another group of 4. We have one group of 4, another group of 4. So this is also 2 times 4 if we look in this purple zone. One group of 4, another group of 4. So this is also 2 times 4. So we have three 2 times 4's. So if we look at each of these, or all together, this is 3 times 2 times 4, so 3 times 2 times 4. Notice I did a different order. And here I did 3 times 4 first. Here I'm doing 2 times 4 first. But just like before, 2 times 4 is 8. 8 times 3 is still equal to 24, as it needs to, because we have exactly 24 of these green circle things. Once again, pause the video and try to do the same here. Look at the groupings in blue, then look at the groupings in purple, and try to express these 24 as some kind of product of 2, 3, and 4. Well, you see first we have these groupings of 3. So we have one grouping of 3 in this purple zone," + }, + { + "Q": "At 1:38, I got kinda lost.\n", + "A": "Andrew i can t find where you got stuck on show me pls.", + "video_name": "VXrn5HOQmHQ", + "timestamps": [ + 98 + ], + "3min_transcript": "So if you look at each of these 4 by 6 grids, it's pretty clear that there's 24 of these green circle things in each of them. But what I want to show you is that you can get 24 as the product of three numbers in multiple different ways. And it actually doesn't matter which products you take first or what order you actually do them in. So let's think about this first. So the way that I've colored it in, I have these three groups of 4. If you look at the blue highlighting, this is one group of 4, two groups of 4, three groups of 4. Actually, let me make it a little bit clearer. One group of 4, two groups of 4, and three groups of 4. So these three columns you could view as 3 times 4. Now, we have another 3 times 4 right over here. This is also 3 times 4. We have one group of 4, two groups of 4, and three groups of 4. So you could view these combined as 2 times 3 times 4. We have one 3 times 4. And then we have another 3 times 4. some more space-- as 2 times-- let me do that in blue-- 2 times 3 times 4. That's the total number of balls here. And you could see it based on how it was colored. And of course, if you did 3 times 4 first, you get 12. And then you multiply that times 2, you get 24, which is the total number of these green circle things. And I encourage you now to look at these other two. Pause the video and think about what these would be the product of, first looking at the blue grouping, then looking at the purple grouping in the same way that we did right over here, and verify that the product still equals 24. Well, I assume that you've paused the video. So you see here in this first, I guess you could call it a zone, we have two groups of 4. So this is 2 times 4 right over here. We have one group of 4, another group of 4. We have one group of 4, another group of 4. So this is also 2 times 4 if we look in this purple zone. One group of 4, another group of 4. So this is also 2 times 4. So we have three 2 times 4's. So if we look at each of these, or all together, this is 3 times 2 times 4, so 3 times 2 times 4. Notice I did a different order. And here I did 3 times 4 first. Here I'm doing 2 times 4 first. But just like before, 2 times 4 is 8. 8 times 3 is still equal to 24, as it needs to, because we have exactly 24 of these green circle things. Once again, pause the video and try to do the same here. Look at the groupings in blue, then look at the groupings in purple, and try to express these 24 as some kind of product of 2, 3, and 4. Well, you see first we have these groupings of 3. So we have one grouping of 3 in this purple zone," + }, + { + "Q": "\n10:51pm solve for w in the formula -2.4=w/8+10.4", + "A": "-2.4 = w/8 + 10.4 -2.4 - 10.4 = w/8 + 10.4 - 10.4 -12.8 = w/8 8*(-12.8) = w/8 * 8 -102.4 = w", + "video_name": "Aig1hkq3OsU", + "timestamps": [ + 651 + ], + "3min_transcript": "" + }, + { + "Q": "\n@ 3:54, Sal says there are other ways to write w = (P-2l)/2.\nWhy can't you cancel the 2's and have w = P - l?", + "A": "(P-2l)/2 = P/2 - (2l)/2 = P/2 - l", + "video_name": "Aig1hkq3OsU", + "timestamps": [ + 234 + ], + "3min_transcript": "So you subtract 2l over here. Minus 2l. You're also going to have to do that on the left-hand side. So you're going to have minus 2l. We're doing it on both sides of the equation. And remember, an equation says P is equal to that, so if you do anything to that, you have to do it to P. So if you subtract 2l from this, you're going to have to subtract 2l from P in order for the equality to keep being true. So the left-hand side is going to be P minus 2l, and then that is going to be equal to-- well, 2l minus 2l, the whole reason why we subtracted 2l is because these are going to cancel out. So these cancel out, and you're just left with a 2w here. You're just left with a 2w. We're almost there. We've almost solved for w. To finish it up, we just have to divide both sides of this equation by 2. equation by 2 is to get rid of this 2 coefficient, this 2 that's multiplying w. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything, so this is going to be a w. And then we have our left-hand side. So we're done. If we flip these two sides, we have our w will be equal to this thing over here-- equals P minus 2l, all of that over 2. Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this, so let me square this off, because this is completely the correct answer. you might have been able to get this answer, other expressions for this answer. You might have also, you know, another completely legitimate way to do this problem-- let me write it this way-- so our original problem is P is equal to 2l plus 2w-- is on this right-hand side, what if we factor out a 2? So let me make this clear. You have a 2 here, and you have a 2 here. So you could imagine undistributing the 2. So we would get P is equal to 2 times l plus w. This is an equally legitimate way to do this problem. Now, we can divide both sides of this equation by 2, so that we get rid of this 2 on the right-hand side. So if you divide both sides of this equation by 2, these 2's are going to cancel out-- 2 times anything divided by 2 is just going to be the anything-- is" + }, + { + "Q": "at 3:36, sal has his formula for W written as p-2l / 2 = W\ncouldn't you just do the division and make the formula into 1/2P - L = W?\n", + "A": "Yes, you can simplify it that way. Sal just decided to leave it as (P-2l)/2.", + "video_name": "Aig1hkq3OsU", + "timestamps": [ + 216 + ], + "3min_transcript": "So you subtract 2l over here. Minus 2l. You're also going to have to do that on the left-hand side. So you're going to have minus 2l. We're doing it on both sides of the equation. And remember, an equation says P is equal to that, so if you do anything to that, you have to do it to P. So if you subtract 2l from this, you're going to have to subtract 2l from P in order for the equality to keep being true. So the left-hand side is going to be P minus 2l, and then that is going to be equal to-- well, 2l minus 2l, the whole reason why we subtracted 2l is because these are going to cancel out. So these cancel out, and you're just left with a 2w here. You're just left with a 2w. We're almost there. We've almost solved for w. To finish it up, we just have to divide both sides of this equation by 2. equation by 2 is to get rid of this 2 coefficient, this 2 that's multiplying w. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything, so this is going to be a w. And then we have our left-hand side. So we're done. If we flip these two sides, we have our w will be equal to this thing over here-- equals P minus 2l, all of that over 2. Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this, so let me square this off, because this is completely the correct answer. you might have been able to get this answer, other expressions for this answer. You might have also, you know, another completely legitimate way to do this problem-- let me write it this way-- so our original problem is P is equal to 2l plus 2w-- is on this right-hand side, what if we factor out a 2? So let me make this clear. You have a 2 here, and you have a 2 here. So you could imagine undistributing the 2. So we would get P is equal to 2 times l plus w. This is an equally legitimate way to do this problem. Now, we can divide both sides of this equation by 2, so that we get rid of this 2 on the right-hand side. So if you divide both sides of this equation by 2, these 2's are going to cancel out-- 2 times anything divided by 2 is just going to be the anything-- is" + }, + { + "Q": "At 4:25, what does 'factoring out a two' mean?\n", + "A": "He explains it as reversing the distributive property. 2(L + W) = 2*L + 2*W = 2L + 2W He is reversing this and since 2 is a factor of both terms, it is called factoring out a 2.", + "video_name": "Aig1hkq3OsU", + "timestamps": [ + 265 + ], + "3min_transcript": "So you subtract 2l over here. Minus 2l. You're also going to have to do that on the left-hand side. So you're going to have minus 2l. We're doing it on both sides of the equation. And remember, an equation says P is equal to that, so if you do anything to that, you have to do it to P. So if you subtract 2l from this, you're going to have to subtract 2l from P in order for the equality to keep being true. So the left-hand side is going to be P minus 2l, and then that is going to be equal to-- well, 2l minus 2l, the whole reason why we subtracted 2l is because these are going to cancel out. So these cancel out, and you're just left with a 2w here. You're just left with a 2w. We're almost there. We've almost solved for w. To finish it up, we just have to divide both sides of this equation by 2. equation by 2 is to get rid of this 2 coefficient, this 2 that's multiplying w. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything, so this is going to be a w. And then we have our left-hand side. So we're done. If we flip these two sides, we have our w will be equal to this thing over here-- equals P minus 2l, all of that over 2. Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this, so let me square this off, because this is completely the correct answer. you might have been able to get this answer, other expressions for this answer. You might have also, you know, another completely legitimate way to do this problem-- let me write it this way-- so our original problem is P is equal to 2l plus 2w-- is on this right-hand side, what if we factor out a 2? So let me make this clear. You have a 2 here, and you have a 2 here. So you could imagine undistributing the 2. So we would get P is equal to 2 times l plus w. This is an equally legitimate way to do this problem. Now, we can divide both sides of this equation by 2, so that we get rid of this 2 on the right-hand side. So if you divide both sides of this equation by 2, these 2's are going to cancel out-- 2 times anything divided by 2 is just going to be the anything-- is" + }, + { + "Q": "How come you do not divide by two on the other side? Like I 3:15\n", + "A": "It s a matter of preference. Personally, I prefer Sal s way: (p-2l)/2 However, you seem to prefer this way: (p/2)-l The question is whether you want there to be one big fraction or one little fraction.", + "video_name": "Aig1hkq3OsU", + "timestamps": [ + 195 + ], + "3min_transcript": "that's going to give you, if you add them together, that's going to give you 2 l's. So the perimeter is going to be 2 l's plus 2 w's. They just wrote it in a different order than the way I wrote it. But the same thing, so hopefully that makes sense. Now, their question is, rewrite the formula so that it solves for width. So the formula, the way it's written now, it says P is equal to something. They want us to write it so it's, this w, right here, they want it to be w is equal to a bunch of stuff with l's and P's in it, and maybe some numbers there. So let's think about how we can do this. So they tell us that P is equal to 2 times l, plus 2 times w. We want to solve for w. Well, a good starting point might be to get rid of the l on this side of the equation. And to get rid of it on that side of the equation, we could subtract the 2l from both sides of the equation. So you subtract 2l over here. Minus 2l. You're also going to have to do that on the left-hand side. So you're going to have minus 2l. We're doing it on both sides of the equation. And remember, an equation says P is equal to that, so if you do anything to that, you have to do it to P. So if you subtract 2l from this, you're going to have to subtract 2l from P in order for the equality to keep being true. So the left-hand side is going to be P minus 2l, and then that is going to be equal to-- well, 2l minus 2l, the whole reason why we subtracted 2l is because these are going to cancel out. So these cancel out, and you're just left with a 2w here. You're just left with a 2w. We're almost there. We've almost solved for w. To finish it up, we just have to divide both sides of this equation by 2. equation by 2 is to get rid of this 2 coefficient, this 2 that's multiplying w. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything, so this is going to be a w. And then we have our left-hand side. So we're done. If we flip these two sides, we have our w will be equal to this thing over here-- equals P minus 2l, all of that over 2. Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this, so let me square this off, because this is completely the correct answer." + }, + { + "Q": "At 4:28, instead of 9 minus -49, Sal writes 9 minus 49. Why is this? Doesn't the -7 multiply the 7, making it -49?\n", + "A": "I think you are using the minus sign twice and trying to get -9 -(-49)? It s not clear from your words. The problem has -9 -7 (7). There is one minus sign with the 7 s. You either use it as part of the multiplication and create -9 -49. Or, you multiply 7(7) = +49, then apply the minus and get -9 - (+49) = -9 -49. Hope this helps.", + "video_name": "TIwGXn4NalM", + "timestamps": [ + 268 + ], + "3min_transcript": "What is this equal to? And I'll give you a few seconds to give it a little bit of thought. Well, you could imagine. They've given three variables with one equation up here, another two variables with another equation. We have five unknowns with two equations. There's no way we're going to be able to individually figure out what a, or b, or c, or x, or y is. But maybe we can use what we saw in the last example to solve this. And so what we might want do is rearrange it so that we have the x's and the y's kind of grouped together, and the a's, b'c, and c's kind of grouped together. So let's try that out. So let's focus first on the a's, b's, and c's. So we have negative 9a, negative 9b-- I'll go in order, alphabetical order-- negative 9b, and we have negative 9c, and I think you see what's emerging. And then let's work on the x's and the y's. And then we have negative 7x, minus 7x, and then we have a minus 7y. So all I have done is rearranged this expression here. of what's going on here. These first three terms, I can factor out a negative 9, and I get negative 9 times a plus b plus c. And these second two terms, I can factor out a 7, minus 7 times x plus y. And just to verify, if you wanted to go the other way, multiply this negative 7 times x plus y, you'll get this. Multiply negative 9 times a plus b plus c, distribute it, you'll get this right over here. And so this makes it a little bit clearer. What is a plus b plus c equal to? Well, they tell us right over here. a plus b plus c is equal to negative 1. This whole expression is negative 1, at least in the parentheses. And what is x plus y equal to? Let me do this in a new color. What is x plus y equal to? Well, they tell us right over here. x plus y is equal to 7. 9 times negative 1 minus 7 times 7. And so this is equal to-- we're in the home stretch-- negative 9 times negative 1. Well, that's positive 9. And then negative 7 times 7 is negative 49. So it's 9 minus 49, which is equal to negative 40. And we are done." + }, + { + "Q": "at 4:38 you say you can factor out a -9 but why is it a negative? If the original equation is a+b+c = -1 then you would need to factor in a positive 9, right? I'm so confused\n", + "A": "can i just say probably try to rewatch the video and pay attentiion close to what he is saying and watch his steps and if that still doesnt answer your question try to research it", + "video_name": "TIwGXn4NalM", + "timestamps": [ + 278 + ], + "3min_transcript": "What is this equal to? And I'll give you a few seconds to give it a little bit of thought. Well, you could imagine. They've given three variables with one equation up here, another two variables with another equation. We have five unknowns with two equations. There's no way we're going to be able to individually figure out what a, or b, or c, or x, or y is. But maybe we can use what we saw in the last example to solve this. And so what we might want do is rearrange it so that we have the x's and the y's kind of grouped together, and the a's, b'c, and c's kind of grouped together. So let's try that out. So let's focus first on the a's, b's, and c's. So we have negative 9a, negative 9b-- I'll go in order, alphabetical order-- negative 9b, and we have negative 9c, and I think you see what's emerging. And then let's work on the x's and the y's. And then we have negative 7x, minus 7x, and then we have a minus 7y. So all I have done is rearranged this expression here. of what's going on here. These first three terms, I can factor out a negative 9, and I get negative 9 times a plus b plus c. And these second two terms, I can factor out a 7, minus 7 times x plus y. And just to verify, if you wanted to go the other way, multiply this negative 7 times x plus y, you'll get this. Multiply negative 9 times a plus b plus c, distribute it, you'll get this right over here. And so this makes it a little bit clearer. What is a plus b plus c equal to? Well, they tell us right over here. a plus b plus c is equal to negative 1. This whole expression is negative 1, at least in the parentheses. And what is x plus y equal to? Let me do this in a new color. What is x plus y equal to? Well, they tell us right over here. x plus y is equal to 7. 9 times negative 1 minus 7 times 7. And so this is equal to-- we're in the home stretch-- negative 9 times negative 1. Well, that's positive 9. And then negative 7 times 7 is negative 49. So it's 9 minus 49, which is equal to negative 40. And we are done." + }, + { + "Q": "\nI am still confused about the speed and velocity thing at 7:12. What is that about?", + "A": "Speed indicates how fast something is moving. Velocity indicates how fast something is moving in a certain direction. Hence, speed is a positive number while velocity is positive or negative depending on the direction.", + "video_name": "ppBJWf_Wdmc", + "timestamps": [ + 432 + ], + "3min_transcript": "then that means the speed is increasing. The magnitude is increasing in the leftward direction. So what we really need to do, beyond just evaluating the acceleration at time 5.5, we also have to evaluate the velocity to see if it's going in the left or the rightward direction. So let's evaluate the velocity. The velocity at 5.5-- and we'll just get our calculator out again. Velocity at 5.5-- this is our velocity function-- is going to be equal to 2 times the sine of-- let me write it this way. Just because I want to make sure I get my parentheses right. 2 times the sine-- let me write it this way-- 2 sine of e to the 5.5, that's our time, divided by 4. So I did that part right over here. And I'm going to close the 2 as well. And then plus 1. So our velocity at time 5.5 is negative. So negative 0.45. Negative 0.45 roughly. So this is negative. The velocity is negative. So we have this scenario where the velocity is negative, which means we're going in the left direction. And the fact that the velocity is also decreasing means that over time-- at least at this point in time-- as we go forward in time, it'll become even more negative. And it'll become even more negative if we wait a little bit longer. So that means that the magnitude of the velocity is increasing. It's just going in the leftward direction. So if the magnitude of the velocity is increasing although it's going in the leftward direction, that means that the speed is increasing. So the velocity-- so this is one of those interesting scenarios-- the velocity is decreasing. But the speed, which is what they're asking us in the question, speed is increasing. of this explanation I gave you, you would say, hey, look. What's acceleration? Is it positive or negative? You would evaluate it. You'd say, hey, it's negative. So you know velocity is decreasing. And then you would say, hey, what is velocity? Is it positive or negative? You evaluate it. You say it's negative. So you have a negative value that is decreasing. So it's becoming more negative. So that means its magnitude is increasing, or speed is increasing." + }, + { + "Q": "At around 1:20: What is the magnitude?\n", + "A": "Remember, velocity is a vector quantity, so it has both direction and magnitude. In the example at 1:20, the particle said to be moving in the negative direction and with a magnitude of 5.", + "video_name": "ppBJWf_Wdmc", + "timestamps": [ + 80 + ], + "3min_transcript": "Problem 1. For 0 is less than or equal to t is less than or equal to 6, a particle is moving along the x-axis. The particle's position x of t is not explicitly given. The velocity of the particle is given by v of t is equal to all of this business right over here. The acceleration of the particle is given by a of t is equal to all of this business over here. They actually didn't have to give us that because the acceleration is just the derivative of the velocity. And they also gave us-- they don't tell us the position function, but they tell us where we start off, x of 0 is equal to 2. Fair enough. Now let's do part A. Is the speed of the particle increasing or decreasing at time t equals 5.5? Give a reason for your answer. It looks like they did something a little sneaky here because they gave us a velocity function and then they ask about a speed. And you might say, wait, aren't those the same thing? And I would say, no, they aren't quite the same thing. Velocity is a magnitude and a direction. It is a vector quantity. Speed is just a magnitude. It is a scalar quantity. And to see the difference, you could have a velocity-- because they don't give us the units-- but you could have a velocity of negative 5 meters per second. And maybe if we're talking about on the x-axis, this would mean we're moving leftward at 5 meters per second on the x-axis. So the magnitude is 5 meters per second. This is the magnitude. And the direction is specified by the negative number. And that is the direction. Your velocity could be negative 5 meters per second, but your speed would just be 5 meters per second. So your speed is 5 meters per second whether you're going to the left or to the right. Your velocity, you actually care whether you're going to the left or the right. So let's just keep that in mind while we try to solve this problem. So the best way to figure out whether our rate of change is increasing or decreasing is to look at the acceleration. Because acceleration is really just the rate of change of velocity. And then, we can think a little bit about this velocity versus speed question. Get the calculator out. We can use calculators for this part of the AP exam. And I assume they intend us to because this isn't something that's easy to calculate by hand. So the acceleration at time 5.5, we just have to say t is 5.5 and evaluate this function. So 1/2-- I'll just write 0.5-- times e to the t over 4. Well t is 5.5. 5.5 divided by 4. And then, times cosine of 5.5 divided by 4 gives us 0.38. Did I do that right? We have 0.5 times e to the 5.5 divided by 4," + }, + { + "Q": "At 1:37 sal talks about a magenta curve, what does he mean by that\n", + "A": "There is a box that pops up and tells you Sal is talking about the magenta line, not curve.", + "video_name": "xR9r38mZjK4", + "timestamps": [ + 97 + ], + "3min_transcript": "what i want to do with this video is think about the relationship between variables and then think about what the graph of that relationship should look like. So let's say these two axis, the horizontal axis over here I plot the price of a product and lets say this vertical axis over here i plot the demand for the product, and i'm only plotting the first quadrant here because i'm assuming that the price can only be positive and i'm assuming that the demand can only be positive, that people aren't going to pay someone to take the product away from them. So let's think about what would happen for the price and demand for most normal products. So, if the price is low, you would expect that a lot of poeple are willing to buy that thing they're like \"Oh it's a good price, i would like to buy it.\" So, if the price is low, then the demand would be high so maybe it would be somplace over here, all the way that you would have really high demand if the price was zero. so if the price was low the demand would be high. Now what happens is the price -- so right here the price is low, demand is high, if the price were to price went up a little bit demand went down a little bit. if the price went up a little bit more then maybe demand goes down a little bit more. as the price went up a bunch then demand would go down a bunch and so the line that represents how the demand relates to price might look something like this, and i'm just going to assume it's it a line. It might not be a line, it might be a curve. It might look something like that. Or it might look something like that. But in general is someone were to ask you, if you saw this magenta curve that as price increases what happens to demand. You just say \"Well look price increases, as price increases what happens to demand?\" Well demand is decreasing. Now let's think about a different scenario. Let's talk about the demand for real estate. For actual property, and lets say that on this axis demand for land. So when the population is very low, you can imagine, if the population is zero there is no one there that would want to buy land. So if the population is very low the demand is going to be very low. And as population increases, demand should increase. If the population increases, more people are going to want to buy land. And if the population goes up a bunch then a lot of people are going to want to buy land. So you'll see a line that looks something like this. And once again I drew a line, it doesn't have to be a line it could be a curve of some kind. It could be a curve that looks something like that, or a curve that looks something like that. We don't know but the general idea is that if someone showed you a graph that looked like this. And as population increases what happens to demand. We'll you'd say \"Look, this is population increasing, what happens to demand?\" Demand is going up. Where as price increased the demand went down. Here as population is increasing" + }, + { + "Q": "At 5:19, why does Sal divide 350 by 2?\n", + "A": "(1994/2) * 350 was changed to 1994 * (350/2) for easier multiplication", + "video_name": "0-wa7voc0uM", + "timestamps": [ + 319 + ], + "3min_transcript": "we have 54, 6 goes into 54 nine times, 9 times 6 is 54 and we are done. So to go from negative 50 to 2,044, I have to add 6 to 349 times, so I add it once, I add it twice, and then this right over here, this is the 349th time that I'm adding 6, so how many terms do I have? Now, you might be tempted to say 349 terms, but really, you have 349 plus 1 terms. You have the 349 for every time you added 6, so this is the first time you added 6, second time you added 6, all the way to the 349th time you added 6, so let me make it clear, this, this is, oh, actually, this is the 349th time I added 6 to get to this, but we haven't counted the first term just yet, so we're going to have, so we have 300, we have the first term and then we add 6 349 times, so we have 350 terms in this sum, N is equal to 350. And so we can say the sum of the first 300... I'll do this in green, the sum of the first 350 terms is going to be equal to the average of the first and last term, so negative 50 plus 2,044 over 2, over 2, times 350, so let's see, negative 50 plus 2,044, that's going to be what? 2,094, 2,094 divided by 2 times, times 350, so let's see, if I just take, so this is going to be, 'cause right, this is 294 times, 350 divided, oh, sorry, not 294, what am I? My brain is not working. 2,000, actually, this is going to be 1994. My brain really wasn't working a little while ago, so this is going to be 1994 divided by 2 times 350 and so let's see, 350 divided by 2 is 175, so this is going to be 1,900-- 1,994 times 175. Which is equal to, and I'll use a calculator for this one, so, let me get the calculator out, so, I have 1,994 times 175 gives us 348,950. 348,950. And we could express this in sigma notation now, now that we know what the n is. We found our answer, this is what we were looking for, but just in case you're curious," + }, + { + "Q": "\nAt 4:00, why doesn't he multiplicate mu with i squared?", + "A": "i isn t squared in this instance, it remains an imaginary number. Additionally, when the general solution is written, mu plays an important role, so keeping it separate allows you to plug it in without doing extra work.", + "video_name": "6xEO4BeawzA", + "timestamps": [ + 240 + ], + "3min_transcript": "Well, if this expression right here-- if this B squared minus 4AC-- if that's a negative number, then I'm going to have to take the square root of a negative number. So it will actually be an imaginary number, and so this whole term will actually become complex. We'll have a real part and an imaginary part. And actually, the two roots are going to be conjugates of each other, right? We could rewrite this in the real and imaginary parts. We could rewrite this as the roots are going to be equal to minus B over 2A, plus or minus the square root of B squared minus 4AC over 2A. And if B squared minus 4AC is less than 0, this is going to be an imaginary number. So in that case, let's just think about what the roots look like generally and then we'll actually do some problems. So let me go back up here. numbers like that. The roots, we can write them as two complex numbers that are conjugates of each other. And I think light blue is a suitable color for that. So in that situation, let me write this, the complex roots-- this is a complex roots scenario-- then the roots of the characteristic equation are going to be, I don't know, some number-- Let's call it lambda. Let's call it mu, I think that's the convention that people use-- actually let me see what they tend to use, it really doesn't matter-- let's say it's lambda. So this number, some constant called lambda, and then plus or minus some imaginary number. And so it's going to be some constant mu. That's just some constant, I'm not trying to be fancy, but this is I think the convention used in most differential So it's mu times i. So these are the two roots, and these are true roots, right? Because we have lambda plus mu i, and lambda minus mu i. So these would be the two roots, if B squared minus 4AC is less than 0. So let's see what happens when we take these two roots and we put them into our general solution. So just like we've learned before, the general solution is going to be-- I'll stay in the light blue-- the general solution is going to be y is equal to c1 times e to the first root-- let's make that the plus version-- so lambda plus mu i. All of that times x, plus c2 times e to the second root. So that's going to be lambda minus mu i times x." + }, + { + "Q": "I got confused at 1:03, can u help?\n", + "A": "So first, you multiply x and y which is 3 * 2 = 6. Then you subtract y which is 2. So, 6 - 2 = 4. Then you add 3 * x which is 3 * 3 = 9. So, the answer is 13 as stated at 1:59.", + "video_name": "S_OX3ByvBSc", + "timestamps": [ + 63 + ], + "3min_transcript": "Now, let's think about expressions with more than one variable. So say I had the expression a plus-- I'll do a really simple one, a plus b. And I want to evaluate this expression when a is equal to 7 and b is equal to 2. And I encourage you to pause this and try this on your own. Well, wherever we see the a, we would just replace it with the 7. And wherever we see the b, we'd replace it with the 2. So when a equals 7 and b equals 2, this expression will be 7 plus 2, which, of course, is equal to 9. So this expression would be equal to 9 in this circumstance. Let's do a slightly more complicated one. Let's say we have the expression x times y minus y plus x. Actually, let's make it plus 3x. Or another way of saying it plus 3 times x. And once again, I encourage you to pause this video and try this on your own. Well, everywhere we see an x, let's replace it with a 3. Every place we see a y, let's replace it with a 2. So this is going to be equal to 3 times y. And y is 2 in this case. 3 times 2 minus 2 plus this 3 times x. But x is also now equal to 3. So what is this going to be equal to? Well, this is going to be equal to 3 times 2 is 6. This 3 times 3 is 9. So it simplifies to 6 minus 2, which is 4, plus 9, which is equal to 13. So in this case, it is equal to 13." + }, + { + "Q": "\nAt 3:25 the equation x+180-x+z=180... Wouldn't you put 180-x in parentheses so it would look like x+(180-x)+z=180? Wouldn't that make more sense?", + "A": "Yes, using the parenthesis makes it clearer. You will still get the same answer either way.", + "video_name": "9_3OxtdqmqE", + "timestamps": [ + 205 + ], + "3min_transcript": "So I'm going to assume that x is equal to y and l is not parallel to m. So let's think about what type of a reality that would create. So if l and m are not parallel, and they're different lines, then they're going to intersect at some point. So let me draw l like this. This is line l. Let me draw m like this. They're going to intersect. By definition, if two lines are not parallel, they're going to intersect each other. And that is going to be m. And then this thing that was a transversal, I'll just draw it over here. So I'll just draw it over here. And then this is x. This is y. And we're assuming that y is equal to x. So we could also call the measure of this angle x. So given all of this reality, and we're assuming in either case that this is some distance, that this line is not of 0 length. Or this line segment between points A and B. I guess we could say that AB, the length of that line segment is greater than 0. I think that's a fair assumption in either case. AB is going to be greater than 0. So when we assume that these two things are not parallel, we form ourselves a nice little triangle here, where AB is one of the sides, and the other two sides are-- I guess we could label this point of intersection C. The other two sides are line segment BC and line segment AC. And we know a lot about finding the angles of triangles. So let's just see what happens when we just apply what we already know. Well first of all, if this angle up here is x, we know that it is supplementary to this angle right over here. So this angle over here is going to have measure 180 minus x. And then we know that this angle, this angle z-- we know that the sum of those interior angles of a triangle are going to be equal to 180 degrees. So we know that x plus 180 minus x plus 180 minus x plus z is going to be equal to 180 degrees. Now these x's cancel out. We can subtract 180 degrees from both sides. And we are left with z is equal to 0. So if we assume that x is equal to y but that l is not parallel to m, we get this weird situation where we formed this triangle, and the angle at the intersection of those two lines that are definitely not parallel all of a sudden becomes 0 degrees. But that's completely nonsensical. If this was 0 degrees, that means that this triangle wouldn't open up at all, which means that the length of AB" + }, + { + "Q": "At around 2:55: Why do you subtract 3 sin(5x-3y) from both sides?\n", + "A": "You want to get all values of dy/dx isolated on one side.", + "video_name": "-EG10aI0rt0", + "timestamps": [ + 175 + ], + "3min_transcript": "Negative 3 times the derivative of y with respect to x. And now we just need to solve for dy/dx. And as you can see, with some of these implicit differentiation problems, this is the hard part. And actually, let me make that dy/dx the same color. So that we can keep track of it easier. So this is going to be dy/dx. And then I can close the parentheses. So how can we do it? It's just going to be a little bit of algebra to work through. Well, we can distribute the sine of 5x minus 3y. So let me rewrite everything. We get dy-- whoops, I'm going to do that in the yellow color-- we get dy/dx is equal to-- you distribute the negative sine of 5x minus 3y. You get-- so let me make sure we know what we're doing. It's going to be, we're going to distribute that, and we're going to distribute that. So you're going to have 5 times all of this. So you're going to have negative this 5 times And then you're going to have the negative times a negative, those are going to, you're going to end up with a positive. And so you're going to end up with plus 3 times the sine of 5x minus 3y dy/dx. Now what we can do is subtract 3 sine of 5x minus 3y from both sides. So just to be clear, this is essentially a 1 dy/dx. So if we subtract this from both sides, we are left with-- So on the left-hand side, we're going to have a 1 dy/dx, and we're going to subtract from that 3 sine of 5x minus 3y dy/dx's. So you're going to have 1 minus 3-- I'll keep the color for the 3 for fun-- is going to be equal to, well, we subtracted this from both sides. So on the right-hand side, this is going to go away. So we're just going to be left with a negative 5 sine of 5x minus 3y. And we're in the home stretch now. To solve for dy/dx, we just have to divide both sides of the equation by this. And we are left with dy/dx is equal to this thing, negative 5 times the sine of 5x minus 3y. All of that over 1 minus 3 sine of 5x minus 3y." + }, + { + "Q": "\nat 5:49 why does he choose -2 to multiply both a and b by? Is it only to be rid of the 2b?", + "A": "simultaneous equations. You multiply one of the equations by a number, so that when you add the two equations, one of the variables is cancelled out. You can then solve for one variable, and use it s answer for solving for the other variable. You have to do the same thing on both sides of the equals sign, otherwise it wouldn t be an equation!", + "video_name": "EdQ7Q9VoF44", + "timestamps": [ + 349 + ], + "3min_transcript": "Nothing fancier than there. That was Algebra two. Actually, I think I should do an actual video on that as well. But that's going to equal this thing. 2s plus 13, all of that over s plus 2 times s plus 3. Notice in all differential equations, the hairiest part's always the algebra. So now what we do is we match up. We say, well, let's add the s terms here. And we could say that the numerators have to equal each other, because the denominators are equal. So we have A plus Bs plus 3A plus 2B is equal to 2s plus B. So the coefficient on s, on the right-hand side, is 2. The coefficient on the left-hand side is A plus B, so we know that A plus B is equal to 2. be equal to-- oh, this is a 13. This is a 13. That's a 13. It looks just like a B, right? That was 2s plus 13. Anyway, so on the right-hand side I get, it was 3A plus 2B is equal to 13. Now we have two equations with two unknowns, and what do we get? I know this is very tiresome, but it'll be satisfying in the end. Because you'll actually solve something with the Laplace Transform. So let's multiply the top equation by 2, or let's just say minus 2. So we get minus 2A minus 2B equals minus 4. And then we get-- add the two equations-- you get A is equal to-- these cancel out-- A is equal to 9. Great. If A is equal to 9, what is B equal to? B is equal to 9 plus what is equal to 2? And we have done some serious simplification. Because now we can rewrite this whole expression as the Laplace Transform of y is equal to A over s plus 2, is equal to 9 over s plus 2, minus 7 over s plus 3. Or another way of writing it, we could write it as equal to 9 times 1 over s plus 2, minus 7 times 1 over s plus 3. Why did I take the trouble to do this? Well hopefully, you'll recognize this was actually the second Laplace Transform we figured out. What was that? I'll write it down here just so you remember it." + }, + { + "Q": "\nat 4:00 how do i make it a fraction", + "A": "have an equal shape to make an equal fration", + "video_name": "jgWqSjgMAtw", + "timestamps": [ + 240 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:30 that golden thingy looks a lot like a hexaflexagon. Does anyone else see that?\n", + "A": "Yep. I bet you could make an interesting snowflake pattern with a hexaflexagon.", + "video_name": "toKu2-qzJeM", + "timestamps": [ + 150 + ], + "3min_transcript": "So in my last video I joked about folding and cutting spheres instead of paper. But then I thought, why not? I mean, finite symmetry groups on the Euclidean plane are fun and all, but there's really only two types. Some amount of mirror lines around a point, and some amount of rotations around a point. Spherical patterns are much more fun. And I happen to be a huge fan of some of these symmetry groups, maybe just a little bit. Although snowflakes are actually three dimensional, this snowflake doesn't just have lines of mirror symmetry, but planes of mirror symmetry. And there's one more mirror plane. The one going flat through the snowflake, because one side of the paper mirrors the other. And you can imagine that snowflake suspended in a sphere, so that we can draw the mirror lines more easily. Now this sphere has the same symmetry as this 3D paper snowflake. If you're studying group theory, you could label this with group theory stuff, but whatever. I'm going to fold this sphere on these lines, and then cut it, and it will give me something with the same symmetry as a paper snowflake. Except on a sphere, and it's a mess, so let's glue it to another sphere. And now it's perfect and beautiful in every way. But the point is it's equivalent to the snowflake OK, so that's a regular, old 6-fold snowflake, but I've seen pictures of 12-fold snowflakes. How do they work? Sometimes stuff goes a little oddly at the very beginning of snowflake formation and two snowflakes sprout. Basically on top of each other, but turned 30 degrees. If you think of them as one flat thing, it has 12-fold symmetry, but in 3D it's not really true. The layers make it so there's not a plane of symmetry here. See the branch on the left is on top, while in the mirror image, the branch on the right is on top. So is it just the same symmetry as a normal 6-fold snowflake? What about that seventh plane of symmetry? But no, through this plane one side doesn't mirror the other. There's no extra plane of symmetry. But there's something cooler. Rotational symmetry. If you rotate this around this line, you get the same thing. The branch on the left is still on top. If you imagine it floating in a sphere you can draw the mirror lines, and then 12 points of rotational symmetry. So I can fold, then slit it so it can swirl around the rotation point. And cut out a sphereflake with the same symmetry as this. And you can fold spheres other ways to get other patterns. OK what about fancier stuff like this? Well, all I need to do is figure out the symmetry to fold it. So, say we have a cube. What are the planes of symmetry? It's symmetric around this way, and this way, and this way. Anything else? How about diagonally across this way? But in the end, we have all the fold lines. And now we just need to fold a sphere along those lines to get just one little triangle thing. And once we do, we can unfold it to get something with the same symmetry as a cube. And of course, you have to do something with tetrahedral symmetry as long as you're there. And of course, you really want to do icosahedral, but the plastic is thick and imperfect, and a complete mess, so who knows what's going on. But at least you could try some other ones with rotational symmetry. And other stuff and make a mess. And soon you're going to want to fold and cut the very fabric of space itself to get awesome, infinite 3D symmetry groups, such as the one water molecules follow when they pack in together into solid ice crystals. And before you know it, you'll be playing with multidimensional, quasi crystallography, early algebra's, or something. So you should probably just stop now." + }, + { + "Q": "At 1:38, how ca Vi say she is lazy!?\n", + "A": "She said she was lazy, but it was a joke, because right after that she said they are the binary expansion of pi. That shows she is not lazy.", + "video_name": "Gx5D09s5X6U", + "timestamps": [ + 98 + ], + "3min_transcript": "Snakes. Lots and lots of snakes. These snakes are just writhing with potential, similar parts linked together. They move in a specific and limited way. Part of the potential of things is how they break. These snakes break fantastically into these snake modules. You can put them back together too, allowing the existence of the super snake. Super snakes are obviously desirable for many reasons, besides being inherently awesome. You can wear them, and put them on things, and drop them, which I find amusing for some reason. You can arrange them into a space-filling, fractal curve, if that's what you like, which I do. You can even jump snake. But let's not forget. You can make mini-snakes too, which enables the snake stash, and starting from mini-snake gives you room to grow. Snake. Snake. Snake. Snake. They like to bend into this angle. Who cares what it actually is, besides about 90 degrees? But it begs the question, how many ways can I fold this snake if it has 10 segments? I can notate the way it slithers back and forth from tail to head, left, right, right, left, left, right, left. This is a valid slither. This is an invalid one, since anyone who's played snake knows that a snake isn't allowed to run into itself. Given a slither, how can I tell whether it's-- of course, snake lets you go straight too. So you can do another version of this that allows going straight and notate it like this, and wonder whether this snake is a loser snake or not. Snake. Snake. Snake. Snake. Snake. Don't forget to try putting the snake modules together in ways they were never meant to go. You can mix colors. So in theory, I could be hiding a secret message in the color pattern of this snake. But I'm not, because I'm lazy. So it's just the digits of the binary expansion of pi. Even better, you can attach more than one segment at a point. I can have a two-headed serpent. I can play the game where I cut off the heads of a Hydra, adding two more in the old head's place and see how far that gets me. I can put the snake modules on my fingertips and have snaky fingers. That's cool too. I can even do super snaky fingers. Snake. Snake. Snake. Snake. Snake. Snake. Snake. Snake. Snake." + }, + { + "Q": "How do those snake modules at 0:45 add to the middle of the chain? It doesn't seems like she attaching them on. It is so cool!\n", + "A": "The part where she adds them is cut from the video.", + "video_name": "Gx5D09s5X6U", + "timestamps": [ + 45 + ], + "3min_transcript": "Snakes. Lots and lots of snakes. These snakes are just writhing with potential, similar parts linked together. They move in a specific and limited way. Part of the potential of things is how they break. These snakes break fantastically into these snake modules. You can put them back together too, allowing the existence of the super snake. Super snakes are obviously desirable for many reasons, besides being inherently awesome. You can wear them, and put them on things, and drop them, which I find amusing for some reason. You can arrange them into a space-filling, fractal curve, if that's what you like, which I do. You can even jump snake. But let's not forget. You can make mini-snakes too, which enables the snake stash, and starting from mini-snake gives you room to grow. Snake. Snake. Snake. Snake. They like to bend into this angle. Who cares what it actually is, besides about 90 degrees? But it begs the question, how many ways can I fold this snake if it has 10 segments? I can notate the way it slithers back and forth from tail to head, left, right, right, left, left, right, left. This is a valid slither. This is an invalid one, since anyone who's played snake knows that a snake isn't allowed to run into itself. Given a slither, how can I tell whether it's-- of course, snake lets you go straight too. So you can do another version of this that allows going straight and notate it like this, and wonder whether this snake is a loser snake or not. Snake. Snake. Snake. Snake. Snake. Don't forget to try putting the snake modules together in ways they were never meant to go. You can mix colors. So in theory, I could be hiding a secret message in the color pattern of this snake. But I'm not, because I'm lazy. So it's just the digits of the binary expansion of pi. Even better, you can attach more than one segment at a point. I can have a two-headed serpent. I can play the game where I cut off the heads of a Hydra, adding two more in the old head's place and see how far that gets me. I can put the snake modules on my fingertips and have snaky fingers. That's cool too. I can even do super snaky fingers. Snake. Snake. Snake. Snake. Snake. Snake. Snake. Snake. Snake." + }, + { + "Q": "At 1:18 you multiply the fraction by the denominator, I understand that this in the inverse of x/4 but I don't understand how it isolates the x. It feels like the two operations would cancel eachother out eliminating the x instead of isolating it. ?\n", + "A": "The inverse of x/4 is actually 4/x. Multiplying these together does give 1, eliminating the x, as you suggested. However, Sal is multiplying x/4 by 4/1, which gives 4x/4, allowing us to cancel the 4 leaving x by itself. On the other side -18 x 4 gives -72.", + "video_name": "p5e5mf_G3FI", + "timestamps": [ + 78 + ], + "3min_transcript": "We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then that is equal to x over 4. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. unnecessary. And so we have negative 18 is equal to x over 4. And our whole goal here is to isolate the x, to solve for the x. And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4, times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4, you're just going to be left with an x. And on the other side, 4 times negative 18. Let's see, that's 40. Well, let's just write it out. So 18 times 4. If we were to multiply 18 times 4, 4 times 8 is 32. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to-- instead of writing x, I'm going to write negative 72-- is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true. This right-hand side, when x is equal to negative 72, does indeed equal negative 16." + }, + { + "Q": "\n0:28 i don't know how to solve the X over 4 fraction.", + "A": "Using the inverse property, you can multiply any number by its inverse to get 1. Therefore, x/4 *4 = x", + "video_name": "p5e5mf_G3FI", + "timestamps": [ + 28 + ], + "3min_transcript": "We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then that is equal to x over 4. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. unnecessary. And so we have negative 18 is equal to x over 4. And our whole goal here is to isolate the x, to solve for the x. And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4, times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4, you're just going to be left with an x. And on the other side, 4 times negative 18. Let's see, that's 40. Well, let's just write it out. So 18 times 4. If we were to multiply 18 times 4, 4 times 8 is 32. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to-- instead of writing x, I'm going to write negative 72-- is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true. This right-hand side, when x is equal to negative 72, does indeed equal negative 16." + }, + { + "Q": "At 1:04 how did he get -18 when subtracted -16-2? Sorry I know this is a stupid question.\n", + "A": "When you subtract, you count down. Because it s in the negatives, you count down twice. -16 to -17 and then to -18.", + "video_name": "p5e5mf_G3FI", + "timestamps": [ + 64 + ], + "3min_transcript": "We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then that is equal to x over 4. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. unnecessary. And so we have negative 18 is equal to x over 4. And our whole goal here is to isolate the x, to solve for the x. And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4, times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4, you're just going to be left with an x. And on the other side, 4 times negative 18. Let's see, that's 40. Well, let's just write it out. So 18 times 4. If we were to multiply 18 times 4, 4 times 8 is 32. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to-- instead of writing x, I'm going to write negative 72-- is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true. This right-hand side, when x is equal to negative 72, does indeed equal negative 16." + }, + { + "Q": "\nIn the video; Patterns in raising 1 and -1 to different powers, I believe there is a mistake. Starting at 5:41 we are shown that (-1) to the 1,000,000 power = 1. At 5:43 we are told that \" \"1,000,000 is an even number so the answer is positive 1\". But, to the left of the problem we see that (-1) to the 1 power = -1, my point being; 1,000,000 is an odd number, is it not?. (-1) x 1= -1 then add the six zeroes. So, (-1) to the 1,000,000 power = -1. correct?", + "A": "Can 1,000,000 be evenly divided by 2? If so, then 1,000,000 is even. 1,000,000 /2 = 500,000 therefore 1,000,000 is even.", + "video_name": "jYOfMszfzAQ", + "timestamps": [ + 341, + 343 + ], + "3min_transcript": "times negative 1. And this is, of course, going to be equal to negative 1. Now let's take negative 1, and let's take it to the second power. We often say that we are squaring it when we take something to the second power. So negative 1 to the second power-- well, we could start with a 1. We could start with a 1, and then multiply it by negative 1 two times-- multiply it by negative 1 twice. And what's this going to be equal to? And once again, by our old definition, you could also just say, hey, ignoring this one, because that's not going to change the value, we took two negative 1's and we're multiplying them. Well, negative 1 times negative 1 is 1. And I think you see a pattern forming. Let's take negative 1 to the third power. What's this going to be equal to? Well, by this definition, you start with a 1, and then you multiply it by negative 1 three times, Or you could just think of it as you're taking three negative 1's and you're multiplying it, because this 1 doesn't change the value. And this is going to be equal to negative 1 times negative 1 is positive 1, times negative 1 is negative 1. So you see the pattern. Negative 1 to the 0 power is 1. Negative 1 to the first power is negative 1. Then you multiply it by negative 1, you're going to get positive 1. Then you multiply it by negative 1 again to get negative 1. And the pattern you might be seeing is if you take negative 1 to an odd power you're going to get negative 1. And if you take it to an even power, you're going to get 1 because a negative times a negative is going to be the positive. And you're going to have an even number of negatives, so that you're always going to have negative times negatives. So this right over here, this is even. Even is going to be positive 1. And then you could see that if you went to negative 1 to the fourth power. Well, you could start with a 1 and then multiply it by negative 1 four times, so a negative 1 times negative 1, times negative 1, times negative 1, which is just going to be equal to positive 1. So if someone were to ask you-- we already established that if someone were to take 1 to the, I don't know, 1 millionth power, this is just going to be equal to 1. If someone told you let's take negative 1 and raise it to the 1 millionth power, well, 1 million is an even number, so this is still going to be equal to positive 1. But if you took negative 1 to the 999,999th power, this is an odd number. So this is going to be equal to negative 1." + }, + { + "Q": "At 3:46, Sal says that -1 squared equals one. So every answer above squaring is positive? I don't get how it changes when its squared.\nInstead, it changes to a positive number. How can that be possible when we are squaring a negative number?\n", + "A": "Multiplying two negative numbers always results in a positive product- think of the two negatives as canceling each other out. So, it s the same if you multiply a negative by itself.", + "video_name": "jYOfMszfzAQ", + "timestamps": [ + 226 + ], + "3min_transcript": "And that's why anything that's not 0 to the 1 power is going to be equal to 1. Now let's try some other interesting scenarios. Let's start try some negative numbers. So let's take negative 1. And let's first raise it to the 0 power. So once again, this is just going, based on this definition, this is starting with a 1 and then multiplying it by this number 0 times. Well, that means we're just not going to multiply it by this number. So you're just going to get a 1. Let's try negative 1. Let's try negative 1 to the first power. Well, anything to the first power, you could view this-- and I like going with this definition as opposed to this one right over here. If we were to make them consistent, if you were to make this definition consistent with this, you would say hey, let's start with a 1, and then multiply it by 1 eight times. And you're still going to get a 1 right over here. But let's do this with negative 1. So we're going to start with a 1, times negative 1. And this is, of course, going to be equal to negative 1. Now let's take negative 1, and let's take it to the second power. We often say that we are squaring it when we take something to the second power. So negative 1 to the second power-- well, we could start with a 1. We could start with a 1, and then multiply it by negative 1 two times-- multiply it by negative 1 twice. And what's this going to be equal to? And once again, by our old definition, you could also just say, hey, ignoring this one, because that's not going to change the value, we took two negative 1's and we're multiplying them. Well, negative 1 times negative 1 is 1. And I think you see a pattern forming. Let's take negative 1 to the third power. What's this going to be equal to? Well, by this definition, you start with a 1, and then you multiply it by negative 1 three times, Or you could just think of it as you're taking three negative 1's and you're multiplying it, because this 1 doesn't change the value. And this is going to be equal to negative 1 times negative 1 is positive 1, times negative 1 is negative 1. So you see the pattern. Negative 1 to the 0 power is 1. Negative 1 to the first power is negative 1. Then you multiply it by negative 1, you're going to get positive 1. Then you multiply it by negative 1 again to get negative 1. And the pattern you might be seeing is if you take negative 1 to an odd power you're going to get negative 1. And if you take it to an even power, you're going to get 1 because a negative times a negative is going to be the positive. And you're going to have an even number of negatives, so that you're always going to have negative times negatives. So this right over here, this is even. Even is going to be positive 1. And then you could see that if you went to negative 1 to the fourth power." + }, + { + "Q": "\nAt 2:58, wouldn't (-1)^0 equal -1?. Its 1*-1 which equals -1 right?", + "A": "if (x < 0) {Yes your right! }", + "video_name": "jYOfMszfzAQ", + "timestamps": [ + 178 + ], + "3min_transcript": "except for 0-- that's where we're going to-- it's actually up for debate. But anything to the 0 power is going to be equal to 1. And just as a little bit of intuition here, you could literally view this as our other definition of exponentiation, which is you start with a 1, and this number says how many times you're going to multiply that 1 times this number. So 1 times 1 zero times is just going to be 1. And that was a little bit clearer when we did it like this, where we said 2 to the, let's say, fourth power is equal to-- this was the other definition of exponentiation we had, which is you start with a 1, and then you multiply it by 2 four times, so times 2, times 2, times 2, times 2, which is equal to-- let's see, this is equal to 16. So here if you start with a 1 and then you multiply it by 1 zero times, you're And that's why anything that's not 0 to the 1 power is going to be equal to 1. Now let's try some other interesting scenarios. Let's start try some negative numbers. So let's take negative 1. And let's first raise it to the 0 power. So once again, this is just going, based on this definition, this is starting with a 1 and then multiplying it by this number 0 times. Well, that means we're just not going to multiply it by this number. So you're just going to get a 1. Let's try negative 1. Let's try negative 1 to the first power. Well, anything to the first power, you could view this-- and I like going with this definition as opposed to this one right over here. If we were to make them consistent, if you were to make this definition consistent with this, you would say hey, let's start with a 1, and then multiply it by 1 eight times. And you're still going to get a 1 right over here. But let's do this with negative 1. So we're going to start with a 1, times negative 1. And this is, of course, going to be equal to negative 1. Now let's take negative 1, and let's take it to the second power. We often say that we are squaring it when we take something to the second power. So negative 1 to the second power-- well, we could start with a 1. We could start with a 1, and then multiply it by negative 1 two times-- multiply it by negative 1 twice. And what's this going to be equal to? And once again, by our old definition, you could also just say, hey, ignoring this one, because that's not going to change the value, we took two negative 1's and we're multiplying them. Well, negative 1 times negative 1 is 1. And I think you see a pattern forming. Let's take negative 1 to the third power. What's this going to be equal to? Well, by this definition, you start with a 1, and then you multiply it by negative 1 three times," + }, + { + "Q": "\nAt 0:26, how do you know which dot is x and which one is y? Would they be labeled, usually?", + "A": "A dot (or point) isn t x or y. A point is just a location. Using the Cartesian coordinate system (the x,y-grid), we assign the point a x-coordinate and a y-coordinate, which describes the location of the point. Hope this helped.", + "video_name": "6_9xNMtwnfs", + "timestamps": [ + 26 + ], + "3min_transcript": "Graph a line that has a slope that is negative and greater than the slope of the blue line. So let's think for a second about what slope means. So if you use the word slope in your everyday life, you're really talking about how inclined something is, like a ski slope. So for example, this orange line isn't inclined at all. It's flat, so this one actually has a slope of 0. Another way of thinking about it is as x increases, what is happening to y? And you see here that y isn't changing at all, so this orange line right now has a slope of 0. If the orange line looked like this, it now has a positive slope. Notice, when x is negative, your y value-- or say when x is negative 5, your y value is here, and then when x is positive 5, your y value has increased. As x increases, y is increasing, so this has a positive slope. This has an even more positive slope, an even more positive This has an even more positive slope. As x is increasing, the y value is increasing really fast. as we move towards the right, so it's a very positive slope. This is less positive, less positive. This is a 0 slope, and then this is a negative slope. Notice, as x is increasing, the line is going down. Your y value is decreasing. When x is negative 5, your y is 7. Well, when x is 5, your y is 5. So x is increased, but y has gone down, so this is a negative slope. So they say graph a line that has a slope that is negative and greater than the slope of the blue line. So the blue line also has a negative slope. As x is increasing, your blue line is going down. Over here, when x has negative values, your y value is quite high. And here, when x has positive values, your y value has gone all the way down. So this has a negative slope, but we want to have a slope greater than this one. So we still want to have a-- they so my orange line currently has a negative slope-- and greater than the slope of the blue line. So it has a negative slope, but it is less negative than this blue line right over here. If I wanted to be more negative than the blue line, I'd have to do something like that. But I want to be negative but less negative than the blue line, so that would be like that. If we wanted a 0 slope, once again, something like this. If we wanted a positive slope, something like this. So once again, negative slope, less negative than the blue line. Check our answer." + }, + { + "Q": "At 6:45 , can't it be like we are in a building , & are trying to get to the exit of the building.\n", + "A": "Yes, its basically like your in a building, but not quite. It does however look like you are trying to exit the building.", + "video_name": "wRxzDOloS3o", + "timestamps": [ + 405 + ], + "3min_transcript": "That's the only way to get there. Or I could go two to the right there. And that's the only way to get there. And now if you watched the two dimensional path counting brain teaser, you know that there's two ways to get here. And the logic is, well you could draw it out. You could go like that. One, two. And that's the same thing as saying, one, two. Though it's easier to visualize here. But the general logic was, well, to know how many ways to get to any square, think about the squares that lead to it, and how many ways can I get to those two squares? And then sum them up. And by the same logic-- so there's two ways to get here. That's that cell. Three ways to get here, right? Two plus one is three. One plus two is three. And three plus three is six. So there were six ways to get to this cube right there, from that one. So this isn't too different from the two dimensional problem so far. But now it gets interesting. done it in the color of that layer. How many ways are there to get to this cell right here? This cell is that one right there. Well, I start here. And I can just go straight down. There's only one way to be there. But I go straight down. So there's only one way to get there. Actually let's extend. There's only one way to get here, if I'm going straight down. And so there's only one way to get to this cell too. I'd have to go straight down again. So there's only one way to get there. Hopefully you understand the way we're visualizing it. This is the bottom row. And there's only one way. You go from here, straight down to there, And that's the only way to get there. Fair enough. Now this is where it gets interesting. How many ways are there to get to this cell? Well in our old example, there was only one way in two dimensions to go from this cell. But now we can go from this cell, and we could come from above. And where's above? Above is right there. So now we add this cell to that cell. How many ways to get to this? This is kind of in the back middle of this cube. How many ways to get there? Well, there's two ways to come from this direction. And I can also come from above right there. So two plus one is three. How many ways to get here? Well, one from behind. And then one from above. So that's two. And you see a little bit of symmetry. And how many ways to come here. Well there's two here. From going straight forward. Two ways to go that way. And then one way to come from above. This is two, and we're on this cell. So if we wanted to know how many ways to get to this cell? There's two ways to go from there. And then one way from above. So that's three. And now, right here, how many ways to get to this cell? There's three ways. I could come from here, from here, or from above." + }, + { + "Q": "in 1:49 it said plus or negative square root of 8, I see the point because you can get it either negative or positive square root of 8 but why don't you do the same thing in the left side? like + or - 4x+1\n", + "A": "You don t need to because you would still end up with the same 2 equations. Sal shows 4x+1 = Sqrt(8) or 4x+1 = -Sqrt(8) If we did the same for the left side we would also have -(4x+1) = Sqrt(8) and -(4x+1) = -Sqrt(8), but you can see if you just multiply these though by -1 you end up with the same 2 equations we already had.", + "video_name": "55G8037gsKY", + "timestamps": [ + 109 + ], + "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously." + }, + { + "Q": "where does the +1 come in at5:00\n", + "A": "Colby, The original problem was (4x+1)\u00c2\u00b2 - 8 = 0 Then at about 4:50 he said he was going to put his answer ((-1+2\u00e2\u0088\u009a2)/4) in for the x in the original equation to prove the answer actually works. And the +1 he wrote at 5:00 is the +1 in the original equation (4[x}+1)\u00c2\u00b2 - 8 = 0 When he put the answer in for x he had (4[(-1+2\u00e2\u0088\u009a2)/4}+1)\u00c2\u00b2 - 8 = 0 I hope that helps.", + "video_name": "55G8037gsKY", + "timestamps": [ + 300 + ], + "3min_transcript": "equation, what do I have? On the left-hand side, I'm just left with 4x. On the right-hand side, I have-- you can't really mathematically, I mean, you could do them if you had a calculator, but I'll just leave it as negative 1 plus or minus the square root, or 2 times the square root of 2. That's what 4x is equal to. If we did it here, as two separate equations, same idea. Subtract 1 from both sides of this equation, you get 4x is equal to negative 1 plus 2, times the square root of 2. This equation, subtract 1 from both sides. 4x is equal to negative 1 minus 2, times the square root of 2. This statement right here is completely equivalent to these two statements. Now, last step, we just have to divide both sides by 4, so you divide both sides by 4, and you get x is equal to root of 2, over 4. Now, this statement is completely equivalent to dividing each of these by 4, and you get x is equal to negative 1 plus 2, times the square root 2, over 4. This is one solution. And then the other solution is x is equal to negative 1 minus 2 roots of 2, all of that over 4. That statement and these two statements are equivalent. And if you want, I encourage you to-- let's substitute one of these back in, just so you feel confident that something as bizarro as one of these expressions can be a solution to a nice, vanilla-looking equation like this. So let's substitute it back in. 4 times x, or 4 times negative 1, plus 2 root 2, over 4, plus 1 squared, minus 8 is equal to 0. plus 2 roots 2, plus 1, squared, minus 8 is equal to 0. This negative 1 and this positive 1 cancel out, so you're left with 2 roots of 2 squared, minus 8 is equal to 0. And then what are you going to have here? So when you square this, you get 4 times 2, minus 8 is equal to 0, which is true. 8 minus 8 is equal to 0. And if you try this one out, you're going to get the exact same answer. Let's do another one like this. And remember, these are special forms where we have squares of binomials in our expression. And we're going to see that the entire quadratic formula is actually derived from a notion like this, because you can actually turn any, you can turn any, quadratic equation into a perfect square equalling something else." + }, + { + "Q": "At 8:45, what is the step-by-step procedure to solving (x-8)(x-2)=0 when solving for x? How did Sal immediately know that the answer would be x=8 or x=2? The only way I see him working it out is if he divided the expression up into two parts like so; (x-8)=0 and (x-2)=0 and then solving for x. If so, why is this allowed? It doesn't make intuitive sense.\n", + "A": "Basically the expression (x+8)(x+2)=0 means that one of those two factor [you have to look at (x+8) as one factor and at (x+2) as another factor] must be equal to zero to satisfy the equation. Either (x+8) must be zero or (x+2) must be zero, or both. Therefore if x+8=0 then x = -8 and this is one valid solution, similarly if x+2=0 then x= -2 and this is the second valid solution. When you tackle problems like this remember that, for a product to be equal to zero there must be a factor which is 0. Hope it helped", + "video_name": "55G8037gsKY", + "timestamps": [ + 525 + ], + "3min_transcript": "on factoring so far. We can only do this when this is a perfect square. If you got, like, x minus 3, times x plus 4, and that would be equal to 9, that would be a dead end. You wouldn't be able to really do anything constructive with that. Only because this is a perfect square, can we now say x minus 5 squared is equal to 9, and now we can take the square root of both sides. So we could say that x minus 5 is equal to plus or minus 3. Add 5 to both sides of this equation, you get x is equal to 5 plus or minus 3, or x is equal to-- what's 5 plus 3? Well, x could be 8 or x could be equal to 5 minus 3, or x is equal to 2. Now, we could have done this equation, this quadratic equation, the traditional way, the way you were tempted to do it. What happens if you subtract 9 from both You'll get x squared minus 10x. And what's 25 minus 9? 25 minus 9 is 16, and that would be equal to 0. And here, this would be just a traditional factoring problem, the type that we've seen in the last few videos. What two numbers, when you take their product, you get positive 16, and when you sum them you get negative 10? And maybe negative 8 and negative 2 jump into your brain. So we get x minus 8, times x minus 2 is equal to 0. And so x could be equal to 8 or x could be equal to 2. That's the fun thing about algebra: you can do things in two completely different ways, but as long as you do them in algebraically-valid ways, you're not going to get different answers. And on some level, if you recognize this, this is easier because you didn't have to do that little game in your head, in terms of, oh, what two numbers, when you multiply Here, you just said, OK, this is x minus 5-- oh, I guess you did have to do it. You had to say, oh, 5 times 5 is 25, and negative 10 is negative 5 plus negative 5. So I take that back, you still have to do that little game in your head. So let's do another one. Let's do one more of these, just to really get ourselves nice and warmed up here. So, let's say we have x squared plus 18x, plus 81 is equal to 1. So once again, we can do it in two ways. We could subtract 1 from both sides, or we could recognize that this is x plus 9, times x plus 9. This right here, 9 times 9 is 81, 9 plus 9 is 18. So we can write our equation as x plus 9" + }, + { + "Q": "at about 10:53 why is (x+8)(x+10) equivalent to x= -8 and x= -10? if you add -8 and -10 together u get -18 not +18, shouldn't it be x= +8 and x= +10?\n", + "A": "Because if you are trying to make them both equal 0 you have to do this: (x+8) if you replace x with 8 you get 16, so you replace x with -8 and you get zero. That is why you get -8 and -10", + "video_name": "55G8037gsKY", + "timestamps": [ + 653 + ], + "3min_transcript": "You'll get x squared minus 10x. And what's 25 minus 9? 25 minus 9 is 16, and that would be equal to 0. And here, this would be just a traditional factoring problem, the type that we've seen in the last few videos. What two numbers, when you take their product, you get positive 16, and when you sum them you get negative 10? And maybe negative 8 and negative 2 jump into your brain. So we get x minus 8, times x minus 2 is equal to 0. And so x could be equal to 8 or x could be equal to 2. That's the fun thing about algebra: you can do things in two completely different ways, but as long as you do them in algebraically-valid ways, you're not going to get different answers. And on some level, if you recognize this, this is easier because you didn't have to do that little game in your head, in terms of, oh, what two numbers, when you multiply Here, you just said, OK, this is x minus 5-- oh, I guess you did have to do it. You had to say, oh, 5 times 5 is 25, and negative 10 is negative 5 plus negative 5. So I take that back, you still have to do that little game in your head. So let's do another one. Let's do one more of these, just to really get ourselves nice and warmed up here. So, let's say we have x squared plus 18x, plus 81 is equal to 1. So once again, we can do it in two ways. We could subtract 1 from both sides, or we could recognize that this is x plus 9, times x plus 9. This right here, 9 times 9 is 81, 9 plus 9 is 18. So we can write our equation as x plus 9 Take the square root of both sides, you get x plus 9 is equal to plus or minus the square root of 1, which is just 1. So x is equal to-- subtract 9 from both sides-- negative 9 plus or minus 1. And that means that x could be equal to-- negative 9 plus 1 is negative 8, or x could be equal to-- negative 9 minus 1, which is negative 10. And once again, you could have done this the traditional way. You could have subtracted 1 from both sides and you would have gotten x squared plus 18x, plus 80 is equal to 0. And you'd say, hey, gee, 8 times 10 is 80, 8 plus 10 is 18, so you get x plus 8, times x plus 10 is equal to 0. And then you'd get x could be equal to negative 8, or x could be equal to negative 10. That was good warm up. Now, I think we're ready to tackle completing the square." + }, + { + "Q": "I do not understand at 1:43 why Sal said it could be both the positive and negative square roots of 8?\n", + "A": "When you square any real number, either negative or positive, you always get a positive number. So the sqrt of 4 can either be -2 or 2. Therefore the square root of any number can be either negative or positive. You do not know for a fact that it is the principle square root (i.e that the answer is just 2 squared), it could easily be -2!", + "video_name": "55G8037gsKY", + "timestamps": [ + 103 + ], + "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously." + }, + { + "Q": "At 1:25, he square roots both sides. If you had another equation like (4x + 1)^2 = 0,\ncan you just square root both sides into 4x + 1 = 0?\n", + "A": "Yes you can. Another way to solve it which produces same result as square rooting both sides: (4x+1)^2 =0 is the same as (4x+1)(4x+1) = 0 One of the factors has to be 0 for the expression to equal 0 as anything multiplied by 0 is 0. So we end up with 4x+1=0, you can then use simple algebra to solve for x.", + "video_name": "55G8037gsKY", + "timestamps": [ + 85 + ], + "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously." + }, + { + "Q": "\nAt 1:50, isn't the square root positive?", + "A": "Square roots can be positive or negative, so he marks it with the correct symbol", + "video_name": "55G8037gsKY", + "timestamps": [ + 110 + ], + "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously." + }, + { + "Q": "At 3:19 when you subtracted 1 from both sides, how did you get -1 on the right side?\n", + "A": "Since he can t really subtract it, he just keeps it there for when he does find out what x is equal to.", + "video_name": "55G8037gsKY", + "timestamps": [ + 199 + ], + "3min_transcript": "squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously. equation, what do I have? On the left-hand side, I'm just left with 4x. On the right-hand side, I have-- you can't really mathematically, I mean, you could do them if you had a calculator, but I'll just leave it as negative 1 plus or minus the square root, or 2 times the square root of 2. That's what 4x is equal to. If we did it here, as two separate equations, same idea. Subtract 1 from both sides of this equation, you get 4x is equal to negative 1 plus 2, times the square root of 2. This equation, subtract 1 from both sides. 4x is equal to negative 1 minus 2, times the square root of 2. This statement right here is completely equivalent to these two statements. Now, last step, we just have to divide both sides by 4, so you divide both sides by 4, and you get x is equal to" + }, + { + "Q": "\nat around 3:37, sal says 56cm squared but puts the exponent 3 [A.K.A. cubed] when he should have put 2 [A.K.A. squared.]", + "A": "I think it was an accident. I think they meant to put a 2.", + "video_name": "b8q6i_XPyhk", + "timestamps": [ + 217 + ], + "3min_transcript": "So I have this little red section here. I have this red section here. And actually, I'm going to try to color the actual lines here so that we can keep track of those as well. So I'll make this line green and I'll make this line purple. So imagine taking this little triangle right over here-- and actually, let me do this one too in blue. So this one over here is blue. You get the picture. Let me try to color it in at least reasonably. So I'll color it in. And then I could make this segment right over here, I'm going to make orange. So let's start focusing on this red triangle here. Imagine flipping it over and then moving it down here. So what would it look like? Well then the green side is going to now be over here. And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not So it's 8 centimeters times 1/2 or 4 centimeters high. And we know how to find the area of this. This is 4 centimeters times 14 centimeters. So the area is equal to 4 centimeters times 14 centimeters which is equal to-- let's see, that's 40 plus 16-- 56 square centimeters. So if you're taking the area of a kite, you're really just taking 1/2 the width times the height, or 1/2 the width times the height, any way you want to think about it." + }, + { + "Q": "At 0:28, Sal said that kites are symmetrical. Does this mean that all rhombuses are also kites?\n", + "A": "Yes. A kite can become a rhombus In the special case where all 4 sides are the same length, the kite satisfies the definition of a rhombus.", + "video_name": "b8q6i_XPyhk", + "timestamps": [ + 28 + ], + "3min_transcript": "What is the area of this figure? And this figure right over here is sometimes called a kite for obvious reasons. If you tied some string here, you might want to fly it at the beach. And another way to think about what a kite is, it's a quadrilateral that is symmetric around a diagonal. So this right over here is the diagonal of this quadrilateral. And it's symmetric around it. This top part and this bottom part are mirror images. And to think about how we might find the area of it given that we've been given essentially the width of this kite, and we've also been given the height of this kite, or if you view this as a sideways kite, you could view this is the height and that the eight centimeters as the width. Given that we've got those dimensions, how can we actually figure out its area? So to do that, let me actually copy and paste half of the kite. So this is the bottom half of the kite. And then let's take the top half of the kite So I have this little red section here. I have this red section here. And actually, I'm going to try to color the actual lines here so that we can keep track of those as well. So I'll make this line green and I'll make this line purple. So imagine taking this little triangle right over here-- and actually, let me do this one too in blue. So this one over here is blue. You get the picture. Let me try to color it in at least reasonably. So I'll color it in. And then I could make this segment right over here, I'm going to make orange. So let's start focusing on this red triangle here. Imagine flipping it over and then moving it down here. So what would it look like? Well then the green side is going to now be over here. And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not" + }, + { + "Q": "At roughly 4:40, you cubed the centimeters instead of squaring them\n", + "A": "The maximum length of the video tape is 3:48. How come you said at 4:40?", + "video_name": "b8q6i_XPyhk", + "timestamps": [ + 280 + ], + "3min_transcript": "And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not So it's 8 centimeters times 1/2 or 4 centimeters high. And we know how to find the area of this. This is 4 centimeters times 14 centimeters. So the area is equal to 4 centimeters times 14 centimeters which is equal to-- let's see, that's 40 plus 16-- 56 square centimeters. So if you're taking the area of a kite, you're really just taking 1/2 the width times the height, or 1/2 the width times the height, any way you want to think about it." + }, + { + "Q": "\nAt 0:21, is calling the length a diagonal proper terminology?", + "A": "If Sal says that, then it certainly seems right.", + "video_name": "b8q6i_XPyhk", + "timestamps": [ + 21 + ], + "3min_transcript": "What is the area of this figure? And this figure right over here is sometimes called a kite for obvious reasons. If you tied some string here, you might want to fly it at the beach. And another way to think about what a kite is, it's a quadrilateral that is symmetric around a diagonal. So this right over here is the diagonal of this quadrilateral. And it's symmetric around it. This top part and this bottom part are mirror images. And to think about how we might find the area of it given that we've been given essentially the width of this kite, and we've also been given the height of this kite, or if you view this as a sideways kite, you could view this is the height and that the eight centimeters as the width. Given that we've got those dimensions, how can we actually figure out its area? So to do that, let me actually copy and paste half of the kite. So this is the bottom half of the kite. And then let's take the top half of the kite So I have this little red section here. I have this red section here. And actually, I'm going to try to color the actual lines here so that we can keep track of those as well. So I'll make this line green and I'll make this line purple. So imagine taking this little triangle right over here-- and actually, let me do this one too in blue. So this one over here is blue. You get the picture. Let me try to color it in at least reasonably. So I'll color it in. And then I could make this segment right over here, I'm going to make orange. So let's start focusing on this red triangle here. Imagine flipping it over and then moving it down here. So what would it look like? Well then the green side is going to now be over here. And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not" + }, + { + "Q": "at 1:00 i don't understand how u would find the height if you copy only half of the kite\n", + "A": "Because if you find halve of the height then you can multiply it by 2 and get the height of the full kite :3", + "video_name": "b8q6i_XPyhk", + "timestamps": [ + 60 + ], + "3min_transcript": "What is the area of this figure? And this figure right over here is sometimes called a kite for obvious reasons. If you tied some string here, you might want to fly it at the beach. And another way to think about what a kite is, it's a quadrilateral that is symmetric around a diagonal. So this right over here is the diagonal of this quadrilateral. And it's symmetric around it. This top part and this bottom part are mirror images. And to think about how we might find the area of it given that we've been given essentially the width of this kite, and we've also been given the height of this kite, or if you view this as a sideways kite, you could view this is the height and that the eight centimeters as the width. Given that we've got those dimensions, how can we actually figure out its area? So to do that, let me actually copy and paste half of the kite. So this is the bottom half of the kite. And then let's take the top half of the kite So I have this little red section here. I have this red section here. And actually, I'm going to try to color the actual lines here so that we can keep track of those as well. So I'll make this line green and I'll make this line purple. So imagine taking this little triangle right over here-- and actually, let me do this one too in blue. So this one over here is blue. You get the picture. Let me try to color it in at least reasonably. So I'll color it in. And then I could make this segment right over here, I'm going to make orange. So let's start focusing on this red triangle here. Imagine flipping it over and then moving it down here. So what would it look like? Well then the green side is going to now be over here. And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not" + }, + { + "Q": "\nAt 1:00 does he mean the 0 vector in R2?", + "A": "Yes, that was Sal s mistake.", + "video_name": "qBfc57x_RSg", + "timestamps": [ + 60 + ], + "3min_transcript": "I've got this matrix, A, here, it's a 2 by 3 matrix. And just as a bit of review, let's figure out its nullspace and its columnspace. So the nullspace of A is the set of all vectors x that are member of-- let's see we have 3 columns here-- so a member of R3, such that A times the vector are going to be equal to the 0 vector. So we can just set this up. Let me just-- we just need to figure out all of the x's that satisfy this in R3. So we take our matrix A 2, minus 1, minus 3, minus 4, 2, 6. Multiply them times some arbitrary vector in R3 here. So you get x1, x2, x3. And you set them equal to the 0 vector. It's going to be the 0 vector in R2. Because we have 2 rows here. You multiply a 2 by 3 matrix times a vector in R3, you're going to get a 2 by 1 vector or 2 by 1 matrix. So you're going to get the 0 vector in R3. equations-- you get 2 x1 minus x2 minus 3 x3 is equal to 0 and so on and so forth. We can just set up an augmented matrix. So we can just set up this augmented matrix right here. 2 minus 1 minus 3 minus 4, 2, 6. And then augment it with what we're trying to set it equal to to solve the system. And you know we're going to perform a bunch of row operations here to put this in reduced row echelon form. And they're not going to change the right-hand side of this augmented matrix. And that's essentially the argument as to why the nullspace of the reduced row echelon form of A is the same thing as the nullspace of A. But anyway, that's just a bit of review. So let's perform some row operations to solve this a little bit better. So, the first thing I might want to do is divide the first row by 2. So if I divide the first row by 2 I get a 1 minus 1/2 minus And let's just divide this row right here by-- I don't know just to simplify things-- let's divide it by 4. So I'm doing two row operations in one step. And you can do that. I could have done it in two separate steps. So if we divide it by 4, this becomes minus 1, 1/2 and then you get 3/2 and then you get 0. And now, let's keep my first row the same. I'm going to keep my first row the same. It's 1 minus 1/2 minus 3/2 and of course the 0 is the right-hand side. Now let's replace my second row with my second row plus my first row. So these are just linear operations on these guys. So negative 1 plus 1 is 0. 1/2 plus minus 1/2 is 0. 3/2 plus minus 3/2 is 0. And of course, 0 plus 0 is 0." + }, + { + "Q": "At 0:55, he says the ratio for a 30-60-90 triangle is x/2: x*square root of 3/2: x. however, my math book says the ratio is x*square root of 3/2: x: 2x....Am i misunderstanding something?\n", + "A": "The video: x/2 : (x\u00e2\u0088\u009a3)/2 : x The book: x : x\u00e2\u0088\u009a3 : 2x Both are equivalent statements. Multiply the terms of the video s ratio by 2 and you will get the book s ratio. Multiplying by all terms by a common factor will not change the value of a ratio.", + "video_name": "SFL4stapeUs", + "timestamps": [ + 55 + ], + "3min_transcript": "What I want to do in this video is discuss a special class of triangles called 30-60-90 triangles. And I think you know why they're called this. The measures of its angles are 30 degrees, 60 degrees, and 90 degrees. And what we're going to prove in this video, and this tends to be a very useful result, at least for a lot of what you see in a geometry class and then later on in trigonometry class, is the ratios between the sides of a 30-60-90 triangle. Remember, the hypotenuse is opposite the 90-degree side. If the hypotenuse has length x, what we're going to prove is that the shortest side, which is opposite the 30-degree side, has length x/2, and that the 60 degree side, or the side that's opposite the 60-degree angle, I should say, is going to be square root of 3 times the shortest side. So square root of 3 times x/2, that's going to be its length. So that's where we're going to prove in this video. And then in other videos, we're just going to apply this. We're going to show that this is actually a pretty useful result. Now, let's start with a triangle that we're very familiar with. So drawing the triangles is always the hard part. This is my best shot at a equilateral triangle. So let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now, what I'm going to do is I'm going So I'm going to drop an altitude right down, and by definition, when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that not only is this an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it, if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. This angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other," + }, + { + "Q": "At 6:20, Sal mentions taking the \"principal root\" of something. What is a principal root, and how is it different than square root? Thanks!\n", + "A": "The square root of a number is both positive and negative: \u00e2\u0088\u009a9 = \u00c2\u00b13 Because both 3 and -3 squared give you 9. The principal root is just the positive answer. So the principal square root of 9 is 3.", + "video_name": "SFL4stapeUs", + "timestamps": [ + 380 + ], + "3min_transcript": "and they add up to x-- remember, this was an equilateral triangle of length x-- we know that this side right over here, is going to be x/2. We know this is going to be x/2. Not only do we know that, but we also knew when we dropped this altitude, we showed that this angle has to be congruent to that angle, and their measures have to add up to 60. So if two things are the same and they add up to 60, this is going to be 30 degrees, and this is going to be 30 degrees. So we've already shown one of the interesting parts of a 30-60-90 triangle, that if the hypotenuse-- notice, and I guess I didn't point this out. By dropping this altitude, I've essentially split this equilateral triangle into two 30-60-90 triangles. And so we've already shown that if the side opposite the 90-degree side is x, that the side opposite the 30-degree side is going to be x/2. That's what we showed right over here. Now we just have to come up with the third side, the side that is opposite the 60-degree side. This is BD. And we can just use the Pythagorean theorem right here. BD squared plus this length right over here squared plus x/2 squared is going to be equal to the hypotenuse squared. So we get BD squared plus x/2 squared-- this is just straight out of the Pythagorean theorem.-- plus x/2 squared is going to equal this hypotenuse squared. It's going to equal x squared. And just to be clear, I'm looking at this triangle right I'm looking at this triangle right over here on the right, and I'm just applying the Pythagorean theorem. This side squared plus this side squared is going to equal the hypotenuse squared. And let's solve now for BD. You get BD squared plus x squared over 4. x squared over 4 is equal to x squared. You could view this as 4x squared over 4. That's the same thing, obviously, as x squared. or x squared over 4 from both sides, you get BD squared is equal to-- 4x squared over 4 minus x squared over 4 is going to be 3x squared over 4. So it's just going to be 3x squared over 4. Take the principal root of both sides. You get BD is equal to the square root of 3 times x. The principal root of 3 is square root of 3. the principal root of x squared is just x, over the principal root of 4 which is 2. And BD is the side opposite the 60-degree side. So we're done. If this hypotenuse is x, the side opposite the 30-degree side is going to be x/2, and the side opposite the 60-degree side is going to be square root of 3 over 2 times x, or the square root of 3x over 2, depending on how you want to view it." + }, + { + "Q": "\nAt 5:59, how did you get (4x^2)/(4)?", + "A": "He changed x^2 to (4x^2)/(4) for calculation purposes, he will subtract (x^2)/(4) later, and changing x^2 will make the subtraction easier to visualize. (4x^2)/(4) is the same thing as x^2 because the 4 in the numerator and 4 in the denominator cancel out (4 / 4 = 1)", + "video_name": "SFL4stapeUs", + "timestamps": [ + 359 + ], + "3min_transcript": "and they add up to x-- remember, this was an equilateral triangle of length x-- we know that this side right over here, is going to be x/2. We know this is going to be x/2. Not only do we know that, but we also knew when we dropped this altitude, we showed that this angle has to be congruent to that angle, and their measures have to add up to 60. So if two things are the same and they add up to 60, this is going to be 30 degrees, and this is going to be 30 degrees. So we've already shown one of the interesting parts of a 30-60-90 triangle, that if the hypotenuse-- notice, and I guess I didn't point this out. By dropping this altitude, I've essentially split this equilateral triangle into two 30-60-90 triangles. And so we've already shown that if the side opposite the 90-degree side is x, that the side opposite the 30-degree side is going to be x/2. That's what we showed right over here. Now we just have to come up with the third side, the side that is opposite the 60-degree side. This is BD. And we can just use the Pythagorean theorem right here. BD squared plus this length right over here squared plus x/2 squared is going to be equal to the hypotenuse squared. So we get BD squared plus x/2 squared-- this is just straight out of the Pythagorean theorem.-- plus x/2 squared is going to equal this hypotenuse squared. It's going to equal x squared. And just to be clear, I'm looking at this triangle right I'm looking at this triangle right over here on the right, and I'm just applying the Pythagorean theorem. This side squared plus this side squared is going to equal the hypotenuse squared. And let's solve now for BD. You get BD squared plus x squared over 4. x squared over 4 is equal to x squared. You could view this as 4x squared over 4. That's the same thing, obviously, as x squared. or x squared over 4 from both sides, you get BD squared is equal to-- 4x squared over 4 minus x squared over 4 is going to be 3x squared over 4. So it's just going to be 3x squared over 4. Take the principal root of both sides. You get BD is equal to the square root of 3 times x. The principal root of 3 is square root of 3. the principal root of x squared is just x, over the principal root of 4 which is 2. And BD is the side opposite the 60-degree side. So we're done. If this hypotenuse is x, the side opposite the 30-degree side is going to be x/2, and the side opposite the 60-degree side is going to be square root of 3 over 2 times x, or the square root of 3x over 2, depending on how you want to view it." + }, + { + "Q": "At 3:21, what does the side-angle side, etc. congruence mean?\n", + "A": "That between the two triangles, 2 sides are congruent, and the angle between them is also congruent. It is referred to as SAS There is AAS, SSS, ASA, and HL These will come up in geometry typically in your first or second year of high school.", + "video_name": "SFL4stapeUs", + "timestamps": [ + 201 + ], + "3min_transcript": "So drawing the triangles is always the hard part. This is my best shot at a equilateral triangle. So let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now, what I'm going to do is I'm going So I'm going to drop an altitude right down, and by definition, when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that not only is this an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it, if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. This angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other, So this angle right over here needs to be congruent to that angle right over there. So these two are congruent. And so you can use actually a variety of our congruence postulates. We could say, side-angle-side congruence. We could use angle-side-angle, any of those to show that triangle ABD is congruent to triangle CBD. And what that does for us, and we could use, as I said, we could use angle-side-angle or side-angle-side, whatever we like to use for this. What that does for us is it tells us that the corresponding sides of these triangles are going to be equal. In particular, AD is going to be equal to CD. These are corresponding sides. So these are going to be equal to each other." + }, + { + "Q": "\nAt 3:14, what does \"postulates\" mean when Sal says \"...of our congruence postulates.\"", + "A": "A postulate (or axiom) is just a simple statement that is accepted as true. It is similar to theorems, but theorems must be proven, while postulates do not.", + "video_name": "SFL4stapeUs", + "timestamps": [ + 194 + ], + "3min_transcript": "So drawing the triangles is always the hard part. This is my best shot at a equilateral triangle. So let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now, what I'm going to do is I'm going So I'm going to drop an altitude right down, and by definition, when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that not only is this an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it, if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. This angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other, So this angle right over here needs to be congruent to that angle right over there. So these two are congruent. And so you can use actually a variety of our congruence postulates. We could say, side-angle-side congruence. We could use angle-side-angle, any of those to show that triangle ABD is congruent to triangle CBD. And what that does for us, and we could use, as I said, we could use angle-side-angle or side-angle-side, whatever we like to use for this. What that does for us is it tells us that the corresponding sides of these triangles are going to be equal. In particular, AD is going to be equal to CD. These are corresponding sides. So these are going to be equal to each other." + }, + { + "Q": "At 1:30 does he mean equal in length\n", + "A": "Yes. equilateral triangles have exactly the same measures of sides and the same angles.", + "video_name": "SFL4stapeUs", + "timestamps": [ + 90 + ], + "3min_transcript": "What I want to do in this video is discuss a special class of triangles called 30-60-90 triangles. And I think you know why they're called this. The measures of its angles are 30 degrees, 60 degrees, and 90 degrees. And what we're going to prove in this video, and this tends to be a very useful result, at least for a lot of what you see in a geometry class and then later on in trigonometry class, is the ratios between the sides of a 30-60-90 triangle. Remember, the hypotenuse is opposite the 90-degree side. If the hypotenuse has length x, what we're going to prove is that the shortest side, which is opposite the 30-degree side, has length x/2, and that the 60 degree side, or the side that's opposite the 60-degree angle, I should say, is going to be square root of 3 times the shortest side. So square root of 3 times x/2, that's going to be its length. So that's where we're going to prove in this video. And then in other videos, we're just going to apply this. We're going to show that this is actually a pretty useful result. Now, let's start with a triangle that we're very familiar with. So drawing the triangles is always the hard part. This is my best shot at a equilateral triangle. So let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now, what I'm going to do is I'm going So I'm going to drop an altitude right down, and by definition, when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that not only is this an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it, if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. This angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other," + }, + { + "Q": "\nI don't understand how the one comes in at about 4:51 . How does Sal get 1.08?", + "A": "The chain is increasing at a rate of eight percent so, you have .08. You also have the original 100 percent of 200, that is where he gets the 1.", + "video_name": "m5Tf6vgoJtQ", + "timestamps": [ + 291 + ], + "3min_transcript": "So we have 100 times 0.965 to the sixth power, which is equal to 80.75. This is all in percentages. So it's 80.75% of our original substance. Let's do another one of these. So we have, Nadia owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is-- oh actually, there's a typo here, it should be 8%-- the rate of increase is 8% annually, how many stores does the restaurant operate in 2007? So let's say years after 1999. And let's talk about how many stores Nadia is operating, her fast food chain. So 1999 itself is 0 years after 1999. And she is operating 200 stores. Then in 2000, which is 1 year after 1999, how many is she going to be operating? Well, she grows at the rate of 8% annually. So she'll be operating all the stores that she had before plus 8% of the store she had before. So 1.08 times the number of stores she had before. And you're going to see, the common ratio here is 1.08. If you're growing by 8%, that's equivalent to Let me make that clear. 200 plus 0.08, times 200. Well, this is just 1 times 200 plus 0.08, times 200. That's 1.08 times 200. Then in 2001, what's going on? This is now 2 years after 1999, and you're going to grow 8% from this number. You're going to multiply 1.08 times that number, times 1.08 times 200. I think you get the general gist. If, after n years after 1999, it's going to be 1.08-- let me write it this way. It's going to be 200 times 1.08 to the nth power. After 2 years, 1.08 squared. 1 year, 1.08 to the first power. 0 years, this is the same thing as a 1 times 200, which" + }, + { + "Q": "\nAt 5:00, how did Sal get 1.08?", + "A": "If in 1999 Nadia had 200 stores and it increases by 8% per year, all you have to do is turn 8% into a decimal (by moving the decimal place 2 places to the left) you get 0.08. So for one year past 1999, you have 200 + 200(0.08), or a way to simplify this because they have the same base you simply multiply 200 by 1.08.", + "video_name": "m5Tf6vgoJtQ", + "timestamps": [ + 300 + ], + "3min_transcript": "So we have 100 times 0.965 to the sixth power, which is equal to 80.75. This is all in percentages. So it's 80.75% of our original substance. Let's do another one of these. So we have, Nadia owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is-- oh actually, there's a typo here, it should be 8%-- the rate of increase is 8% annually, how many stores does the restaurant operate in 2007? So let's say years after 1999. And let's talk about how many stores Nadia is operating, her fast food chain. So 1999 itself is 0 years after 1999. And she is operating 200 stores. Then in 2000, which is 1 year after 1999, how many is she going to be operating? Well, she grows at the rate of 8% annually. So she'll be operating all the stores that she had before plus 8% of the store she had before. So 1.08 times the number of stores she had before. And you're going to see, the common ratio here is 1.08. If you're growing by 8%, that's equivalent to Let me make that clear. 200 plus 0.08, times 200. Well, this is just 1 times 200 plus 0.08, times 200. That's 1.08 times 200. Then in 2001, what's going on? This is now 2 years after 1999, and you're going to grow 8% from this number. You're going to multiply 1.08 times that number, times 1.08 times 200. I think you get the general gist. If, after n years after 1999, it's going to be 1.08-- let me write it this way. It's going to be 200 times 1.08 to the nth power. After 2 years, 1.08 squared. 1 year, 1.08 to the first power. 0 years, this is the same thing as a 1 times 200, which" + }, + { + "Q": "\nat 5:37, what does gist mean?", + "A": "Here s the etymology way to understand it: From French gist en = lie in consist in ; from Latin jacet = to lie, rest (as when boards lie on a wooden beam) Now means the real point , important part . (See the Online Etymology Dictionary and Oxford English Dictionary .)", + "video_name": "m5Tf6vgoJtQ", + "timestamps": [ + 337 + ], + "3min_transcript": "So let's say years after 1999. And let's talk about how many stores Nadia is operating, her fast food chain. So 1999 itself is 0 years after 1999. And she is operating 200 stores. Then in 2000, which is 1 year after 1999, how many is she going to be operating? Well, she grows at the rate of 8% annually. So she'll be operating all the stores that she had before plus 8% of the store she had before. So 1.08 times the number of stores she had before. And you're going to see, the common ratio here is 1.08. If you're growing by 8%, that's equivalent to Let me make that clear. 200 plus 0.08, times 200. Well, this is just 1 times 200 plus 0.08, times 200. That's 1.08 times 200. Then in 2001, what's going on? This is now 2 years after 1999, and you're going to grow 8% from this number. You're going to multiply 1.08 times that number, times 1.08 times 200. I think you get the general gist. If, after n years after 1999, it's going to be 1.08-- let me write it this way. It's going to be 200 times 1.08 to the nth power. After 2 years, 1.08 squared. 1 year, 1.08 to the first power. 0 years, this is the same thing as a 1 times 200, which So they're asking us, how many stores does the restaurant operate in 2007? Well, 2007 is 8 years after 1999. So here n is equal to 8. So let us substitute n is equal to 8. The answer to our question will be 200 times 1.08 to the eighth power. Let's get our calculator out and calculate it. So we want to figure out 200 times 1.08 to the eighth power. She's going to be operating 370 restaurants, and she'll be in the process of opening a few more. So if we round it down, she's going to be operating 370 So 8% growth might not look like something that's so fast" + }, + { + "Q": "\n0:10 isn't it 2 units wide?", + "A": "Yes. This mistake has been noticed and now a pop up on the screen informs you that it is really 2 units 0.10", + "video_name": "1UQ5IbihJNI", + "timestamps": [ + 10 + ], + "3min_transcript": "- [Voiceover] So we have an interesting shape right over here. The base is a rectangular prism. And the dimensions of that rectangular prism, it's three units I guess we could say tall, four units wide, and then four units long. And then on top of that, on top of that, we have what you could call a right pyramid, where the height of this right pyramid, so if you start at the center of its base right over here, and you go to the top, this height right over here is one unit. And this hasn't been drawn completely to scale, and kind of the perspective skews a little bit. But our goal here, our goal here, is to figure out what is the length? What is the length of one of these edges right over here? So either that one, or this one right over here. What is that length? And we will call that, we will call that x. And so I encourage you to pause this video and try to think about it on your own. Remember, this is a right, this is a right pyramid. that's one unit long, it is perpendicular, it is perpendicular to this entire plane. It's perpendicular to the top of the rectangular prism. So with that in mind, I encourage you to pause the video and see if you can figure it out. And I will give you a hint. You will have to use the Pythagorean theorum, maybe more than once. Alright, so I am assuming you've at least given it a shot. So let's work through it together. So the key here is to realize, well okay this point, this base right over here, this point right over here it's half way in this direction and half way in this direction. So we can figure out, well this entire length right over here is length four. So half way, this is going to be, That's going to be two, and that's going to be two. Just like that. And then the other thing we can figure out, we can figure out what this length is going to be. 'Cause once again it's half way in that direction. So if this whole thing is two, and we see it right over here. This is a rectangular prism, so this length is going to be the same thing as this length. So if this whole thing is two, then each of these, this is going to be one, and this is going to be one right over there. Well how does that help us? Well using that information, we should be able to figure out this length. Actually, I'll keep it in this color 'cause this color's easy to see. We should be able to figure out this length. Well why is this length interesting? Well if we know that length, that length forms a right triangle. That length and the one are the two non-hypotenuse sides of a right triangle. And then the x would be the hypotenuse. So we could just apply the Pythagorean theorem." + }, + { + "Q": "\n0:01 ... that is ... very easy sir.", + "A": "for some,yes but others, not as much.", + "video_name": "4IWfJ7-CYfE", + "timestamps": [ + 1 + ], + "3min_transcript": "Let's see if we can calculate 2.91 times 3.2. And I encourage you to pause this video and try it out on your own. So the way I'm going to think about it is 2.91 is the same thing as 291 divided by 10. Or not divided by 10, divided by 100. And we know that if you divide something by 100, you are going to move the decimal place two places to the left-- one, two. And you would end up at 2.91. It also make sense, if I take 2, and I multiply it by 100, I'd get 200. Or if I take 200 and divided by 100, I would get 2. So it makes sense that 2.91 is the same thing as 291 divided by 100. Similarlarly-- I can never say that word-- 3.2 can be rewritten. It's the same thing as 32 divided by 10. Well, I could rewrite 2.91 times 3.2 as being the same thing as. Instead of 2.91, I can write 291 divided by 100. And then times-- instead of writing 3.2, I could write 32 divided by 10. And this can be rewritten as-- this is going to be equal to 291 times 32 divided by 100. I'm just reordering this-- divided by 100, divided by 10. This is equal to 291 times 32. If I divide by 100 and then I divide by 10 again, I'm essentially dividing by 1,000. So this part right over here, I could rewrite as dividing by 1,000. Now, why is this interesting? Well, I already know how to multiply 291 times 32. And then we know how to move the decimal so that when we divide by 1,000. So let's calculate 291 times 32. Let me write it right over here. 291 times 32. Notice, I've just essentially rewritten this without the decimals. So this right over here-- but of course, these are different quantities than this one is right over here. To go from this product to this product, I have to divide by 1,000. But let's just think about this. We already know how to compute this type of thing." + }, + { + "Q": "\nAt 02:24 why not 291/100*32/10 be equal to 291*32/(100/10)=291*32/(10)", + "A": "No, actually when you have 2 fractions being multiplied together (in this example 291/100 and 32/10) you multiply the numerators, then multiply the denominators. So in this case it would end up looking like this: (291/100)*(32/10)= (291*32)/(100*10= (291*32)*(1000)", + "video_name": "4IWfJ7-CYfE", + "timestamps": [ + 144 + ], + "3min_transcript": "Let's see if we can calculate 2.91 times 3.2. And I encourage you to pause this video and try it out on your own. So the way I'm going to think about it is 2.91 is the same thing as 291 divided by 10. Or not divided by 10, divided by 100. And we know that if you divide something by 100, you are going to move the decimal place two places to the left-- one, two. And you would end up at 2.91. It also make sense, if I take 2, and I multiply it by 100, I'd get 200. Or if I take 200 and divided by 100, I would get 2. So it makes sense that 2.91 is the same thing as 291 divided by 100. Similarlarly-- I can never say that word-- 3.2 can be rewritten. It's the same thing as 32 divided by 10. Well, I could rewrite 2.91 times 3.2 as being the same thing as. Instead of 2.91, I can write 291 divided by 100. And then times-- instead of writing 3.2, I could write 32 divided by 10. And this can be rewritten as-- this is going to be equal to 291 times 32 divided by 100. I'm just reordering this-- divided by 100, divided by 10. This is equal to 291 times 32. If I divide by 100 and then I divide by 10 again, I'm essentially dividing by 1,000. So this part right over here, I could rewrite as dividing by 1,000. Now, why is this interesting? Well, I already know how to multiply 291 times 32. And then we know how to move the decimal so that when we divide by 1,000. So let's calculate 291 times 32. Let me write it right over here. 291 times 32. Notice, I've just essentially rewritten this without the decimals. So this right over here-- but of course, these are different quantities than this one is right over here. To go from this product to this product, I have to divide by 1,000. But let's just think about this. We already know how to compute this type of thing." + }, + { + "Q": "\nAt 4:37, Sal said that (-48x\u00e2\u0081\u00b4-42x\u00c2\u00b3-15x\u00c2\u00b2-5x)/(8x+7)(3x+1) is the final answer. But isn't possible to factor it by grouping to simplify the answer to (-6x\u00c2\u00b3-5x) with this as the solution:\n(-48x\u00e2\u0081\u00b4-42x\u00c2\u00b3-15x\u00c2\u00b2-5x)/(8x+7)(3x+1)\n-6x\u00c2\u00b3(8x+7)-5x(3x+1)/(8x+7)(3x+1)\n=(-6x\u00c2\u00b3-5x)", + "A": "Factors are items that are being multiplied. When you get to: -6x\u00c2\u00b3(8x+7)-5x(3x+1) you have terms (2 items being added or subtracted). Each term is made up of factors. But, you haven t fully factored the polynomial. Since you don t have factors, you can t reduce the fraction. Hope this helps.", + "video_name": "evmDZkDvlNw", + "timestamps": [ + 277 + ], + "3min_transcript": "in a new color, do this in green. Our common denominator, we already established, is the product of our two denominators so it is going to be 8x plus seven times 3x plus one. Now if we multiply the denominator here was 3x plus one, we're multiplying it by 8x plus seven. So that mean's we have to multiply the numerator by 8x plus seven as well. 8x plus seven times negative 6x to the third power. Notice, 8x plus seven divided by 8x plus seven is one. If you were to do that, you would get back to your original expression right over here, the negative 6x to the third over 3x plus one. And now, we're ready to add. This is all going to be equal to, I'll write the denominator in white. So we have our common denominator, 8x plus seven times 3x plus one. to distribute the negative 5x, so negative 5x times positive 3x is negative 15x squared. And then, negative 5x times one is minus 5x. And then, in the green, I would have, let's see, I'll distribute the negative 6x to the third power, so negative 6x to the third times positive 8x is going to be, negative 48x to the fourth power. And then negative 6x to the third times positive seven is going to be negative 42, negative 42x to the third. And I think I'm done because there's no more, there's, you know, I only have one fourth-degree term, one third-degree term, one second-degree term, one first-degree term, and that's it. There's no more simplification here. Some of you might want to just write it as negative 48x to the fourth minus 42x to the third minus 15x squared minus 5x. All of that over 8x plus seven times 3x plus one. But either way, we are all done. I mean, it looks like up here, yeah, there's no, nothing to factor out. These two are divisible by five. These are divisible by six but even if I were to factor that out, nothing over here, down here, no five or six to factor out. Yeah, so it looks like we are all done." + }, + { + "Q": "\n@6:08 Probability of A and B are independent of each other.\n\nA --> Select a Blue garment\nB --> Select a Shirt\n\nThere is a Blue Shirt in the Sample Space. So if A happens to select the Blue shirt. Then for sure the chance of B selecting a Shirt is impacted since there is one less shirt to select from.\n\nIn this case, how is it that A and B are independent events ?", + "A": "Tomas chooses a garment at random. So if he happens to choose the blue shirt, then A and B are true. A and B are groups of possibilities, not selections. To be clearer, Tomas doesn t select a blue garment, and then select a shirt, he simply picks one at random. So A and B can be independent events. Tell me if you still have questions, or if this wasn t clear enough, please. :)", + "video_name": "R-NeYKSEqns", + "timestamps": [ + 368 + ], + "3min_transcript": "let's see if we can answer these questions. The probability of A given B equals the probability of A. And that does work out. Probability of A given B is 1/2. And that's the same thing as the probability of A. The probability that Tomas likes a blue garment given that he has chosen a shirt is equal to the probability that Tomas likes a blue garment. Yep, that's exactly. So I guess the words are just rephrasing what they wrote here in a more mathy notation. So this is absolutely true. The probability of B given A is equal to the probability of B. Yep, the probability of B given A is 1/3. The probability of B is 1/3. The probability that Tomas selects a shirt given that he has chosen a blue garment is equal to the probability that Tomas selects a shirt. Yep, that's right. Events A and B are independent events. Independent events. So two events are independent These are independent if the probability of A given B is equal to the probability of A. Then we can say A and B are independent. Because the probability of A, then if this is true then this means the probability of A given B isn't dependent on whether B happened or not. It's the same thing as the probability of A. This would lead to these events being indpendent. So if you had the probability of B given A is equal to the probability of B. Same argument. That would mean they are independent. Or, if we said that the probability of A and B is equal to the probability of A times the probability of B then this also means they are independent. We know that this one is true. The probability of A and B is 1/6. The probability of A times the probability of B is 1/2 times 1/3 which is 1/6. So all of these are clearly true. So we can say that A and B are independent. of whether B has happened or not. The probability of B happening is independent of whether A has happened or not. The outcome of events A and B are dependent on each other. No. That's the opposite of saying they are independent. So we can cross that out. Probably of A and B is equal to the probability of A times the probability of B We already said that to be true. 1/6 is 1/2 times 1/3. The probability that Tom selects a blue garment that is a shirt is equal to the probability that tom selects a blue garment multiplied by the probability that he selects a shirt. Yep. That is absolutely right. So actually this is, a lot of these statements are true. The only one that is not is that the outcome of events A and B are dependent on each other." + }, + { + "Q": "At 0:21. Why are radian angles much smaller than degree angles?\n", + "A": "Actually, 1 radian is much larger than 1 degree. You can equate the two as: \u00cf\u0080 rad = 180\u00cb\u009a", + "video_name": "C3HFAyigqoY", + "timestamps": [ + 21 + ], + "3min_transcript": "Voiecover:One angle whose tangent is half is 0.46 radians. So we're saying that the tangent right over here is... So the tangent... So we're gonna write this down. So we're saying that the tangent of 0.46 radians is equal to half. And another way of thinking about the tangent of an angle is that's the slope of that angle's terminal ray. So it's the slope of this ray right over here. Yeah that makes sense that that slope is about half. Now what other angles have a tangent of 1 half? So let's look at these choices. So this is our original angle, 0.46 radians, plus pi over 2. If you think in degrees, pi is 180. pi over 2 is 90 degrees. So this one... Actually let me do in a color you're more likely to see. This one is gonna look like this. Where this is an angle of pi over 2. And just eyeballing it, you immediately see that the slope of this ray right over here. In fact they look like they are. They are perpendicular because they have an angle of pi over 2 between them. But they're definitely not going to have the same tangent. They don't have the same slope. Let's think about pi minus 0.46. So that's essentially pi is going along the positive x axis. You go all the way around. Or half way around to your pi radians. But then we're gonna subtract 0.46. So it's gonna look something like this. It's gonna look something like that where this is 0.46 that we have subtracted. Another way to think about it, if we take our original terminal ray and we flip it over the y axis, we get to this terminal ray right over here. And you could immediately see that the slope of the terminal ray is not the same as the slope of this one, of our first one, of our original, in fact they look like the negatives of each other. So we can rule that one out as well. So that's going to take us... If you add pi to this you're essentially going half way around the unit circle and you're getting to a point that is... Or you're forming a ray that is collinear with the original ray. So that's that angle right over here. So pi plus 0.46 is this entire angle right over there. And when you just look at this ray, you see its collinear is going to have the exact same slope as the terminal ray for the 0.46 radion. So just that tells you that the tangent is going to be the same. So I could check that there. And in previous videos when we explore the symmetries of the tangent function, we in fact saw that. That if you took an angle and you add pi to it, you're going to have the same tangent. And if you wanna dig a little bit deeper, I encourage you to look at that video on the symmetries of unit circle symmetries for the tangent function. So let's look at these other choices." + }, + { + "Q": "\nWhy does Sal move the graph 2 units to the right when it says -2 at 1:05?\n\nShouldn't he move it to the left by 2 units?", + "A": "Think about it on a coordinate plane: if I shift 0 two to the right, I get 2. Because you have to think about it in terms of 0, not 2, you subtract 2 instead of adding 2.", + "video_name": "5DLkB-g8Rr8", + "timestamps": [ + 65 + ], + "3min_transcript": "- [Voiceover] We\u2019re told, \"the graph of the function f of x \"is equal to x-squared.\" We see it right over here in grey. It\u2019s shown in the grid below. \"Graph the function g of x is equal to \"x minus two-squared, minus four \"in the interactive graph.\" This is from the shifting functions exercise on Khan Academy, and we can see we can change the graph of g of x. But let\u2019s see, we want to graph it properly, so let\u2019s see how they relate. Well, let\u2019s think about a few things. Let\u2019s first just make g of x completely overlap. Now they're completely overlapping. And let\u2019s see how they\u2019re different. Well, g of x, if you look at what's going on here, instead of having an x-squared, we have an x minus two-squared. So, one way to think about it is, when x is zero, you have zero-squared is equal to zero. But how do you get zero here? Well, x has got to be equal to two. Two minus two-squared is zero-squared, if we don\u2019t look at the negative four just yet. And so, we would want to shift this graph This is essentially how much do we shift to the right. It\u2019s sometimes a little bit counterintuitive that we have a negative there, because you might say, well, negative, that makes me think that I want to shift to the left. But you have to remind yourself is like, well okay, for the original graph, when it was just x-squared, to get the zero-squared, I just have to put x equals zero. Now to get a zero-squared, I have to put in a two. So this is actually shifting the graph to the right. And so, what do we do with this negative four? Well, this is a little bit more intuitive, or at least for me when I first learned it. This literally will just shift the graph down. Whatever your value is of x minus two-squared, it's gonna shift it down by four. So what we wanna do is just shift both of these points down by four. So this is gonna go from nine, and this is gonna go from the coordinate five comma nine, to five comma, if we go down four, five comma five. to two comma negative four. Two comma negative four. Did I do that right? I think that\u2019s right. What, essentially, what we have going on is, g of x is f of x shifted two to the right and four down. Two to the right and four down. And notice, if you look at the vertex here, we shifted two to the right and four down. And I shifted this one also, this one also I shifted two to the right, and four down. And, there you have it. We have graphed g of x, which is a shifted version of f of x." + }, + { + "Q": "\n5:31 what is 80 divided by 2/5", + "A": "80/2/5=80*5/2=40*5=200", + "video_name": "xoXYirs2Mzw", + "timestamps": [ + 331 + ], + "3min_transcript": "" + }, + { + "Q": "\nIn the video at 3:09 , the subtitles say c (comedy) instead of (c, d). Someone should fix that.", + "A": "The subtitle is auto-generated by youtube I believe. Sal said, c comma d which I guess it picked up as comedy", + "video_name": "S-agS4YaxxU", + "timestamps": [ + 189 + ], + "3min_transcript": "with a hypotenuse of the isosceles right triangle sits along the base. Sits along the base. So, it's isosceles. So that's equal to that. It's a right triangle, and then this distance this distance between that point and this point is the same as the distance between F of X and G of X. F of X and G of X for this X value right over there. Now obviously that changes as we change our X value. To help us visualize this shape here, I've kind of drawn a picture of our coordinate plane. If we've viewed as an angle, if we're kind of above it, you can kind of start to see how this figure would look. Once again, I've drawn the base. I've drawn the base of it. I've drawn the base of it right over there. Maybe I should to make it clear. Shade it in kind of parallel to these cross sections. So, I've drawn the base right over there. There's two other sides. There the side that's on ... I've drawn it here. I guess you could view it on its top side or the left side right over there. Over on this picture that would be this. When we're looking at it from above. Then you have this other side. I guess on this view this one you could call kind of the right side. Over here this is kind of the ... when you viewed over here this is the bottom side. The whole reason why I set this up, and we tempting to visualize this figure. I want to see if you can come up with a definite integral that describes the volume of this figure. That kind of almost looks like a football if you cut it in half or a rugby ball. It's skewed a little bit as well. What's an expression a definite integral that expresses the volume of this. that it intersects at the points. These functions intersect at the point zero zero and (c,d). Can you come up with some expression a definite in row of terms of zeros, and Cs, and Ds, and Fs, and Gs that describe the volume of this figure? Assuming you've paused the video, and have had a go at it, let's think about it. If we want to find the volume, one way to think about it is we could take the volume of, we could approximate the volume as the volume of these individual triangles. That would be the area of each of these triangles times some very small depth. Some very small depth. I'll just shade it in to show the depth. Some very small depth which we could call DX. Once again we could find the volume of each of these by finding the area, the cross sectional area there, and then multiplying that times a little DX." + }, + { + "Q": "At about the 5:40 sec mark, Khan talks about the product of dividing:\n\n9.2 * 10^5\n11.5 * 10^2\n\nI worked the equation before Khan. Instead of dividing 9.2 by 11.5 * 10^2, I said divide by\n1.15 * 10^3 (.) We came up with the same answer; 8 * 10^2.\n\nMy doing scientific math problems that way (changing any multiplication involved in the numerator or denominator to correct scientific notation, will that cause me any problems down the road (?))\n", + "A": "I think it is good to know how to do a math question two different ways, so if you learn to do it both ways and remember how to do it, then I think you are set!", + "video_name": "EbmgLiSVACU", + "timestamps": [ + 340 + ], + "3min_transcript": "And now I can divide these two things. So this is going to be equal to-- we'll have to think about what 9.2 over 11.5 is. But actually let me just do that right now, get a little practice dividing decimals. Let me get some real estate here. Let me do that in the same color. 9.2 divided by 11.5-- well if we multiply both of these times 10, that's the exact same thing as 92 divided by 115. We're essentially moved the decimal to the right for both of them. And let me add some zeros here because I suspect that I'm going to get a decimal here. So let's think what this is going to be. Let's think about this. Well 115 doesn't go into 9. It doesn't go into 92. It does go into 920. Let's see if that works out. So I have my decimal here. That's a 0. 8 times-- 8 times 5 is 40. 8 times 11 is 88. And then 88 plus 4 is 92. Oh, it went in exactly, very good. So 920, we have no remainder. So 9.2 divided by 11.5 simplified to 0.80. And then 10 to the fifth divided by 10 to the second, we have the same base, and we're dividing. So we can subtract the exponents. That's going to be 10 to the 5 minus 2. So this right over here is going to be 10 to the third power-- so times 10 to the third power. Now, are we done? Well in order to be done, this number right over here needs to be greater than or equal to 1 and less than 10. It is clearly not greater than or equal to 1. that is greater than or equal to 1 and less than 10 and some power of 10? Well this 8 right over here, this is in the tenths place. It's 8/10, 8 times 1/10. So this is going to be the same thing as 8 times 10 to the negative 1 power. And then we have this 10 to the third here-- so times 10 to the third power. We'll do that in that other color. And now we have the same base. Just add the exponents. So this is going to be equal to 8 times 10 to the 3 minus 1-- so 8 times 10 squared. And we're done. We've simplified our original expression." + }, + { + "Q": "At 3:45, Sal talks about the \"orientation\" and \"magnitude\" of a vector. So, is \"orientation\" the direction of a vector? And is \"magnitude\" the length of a vector?\n", + "A": "Yes, that is correct. Sometimes you will hear angle instead of orientation or direction, but it is all the same.", + "video_name": "gsNgdVdAT1o", + "timestamps": [ + 225 + ], + "3min_transcript": "vector b, or 10 times victor a minus 6 times vector b-- some combination of vector a and b, where I can get vector c. And vector c is the vector 7, 6. So let me see if I can visually draw this problem. So let me draw the coordinate axes. Let's see this one. 3, negative 6. That'll be in quadrant-- these are both in the first quadrant. So I just want to figure out how much of the axes I need to draw. So let's see-- Let me do a different color. That's my y-axis. I'm not drawing the second or third quadrants, because I don't think our vectors show up there. And then this is the x-axis. So first I'll do vector a. That's 3, negative 6. 1, 2, 3, and then negative 6. 1, 2, 3, 4, 5, 6. So it's there. So if I wanted to draw it as a vector, usually start at the origin. And it doesn't have to start at the origin like that. I'm just choosing to. You can move around a vector. It just has to have the same orientation and the same magnitude. So that is vector a for the green. Now let me do in magenta, I'll do vector b. That is 2, 6. 1, 2, 3, 4, 5, 6. And that's vector b. So it'll look like this. That's vector b. And then let me write down vector a down there. That's vector a. And I want to take some combination of vectors a and b. And add them up and get vector c. So what does vector c look like? It's 7, 6. Let me do that in purple. So 1, 2, 3, 4, 5, 6, 7. Comma 6. So 7, 6 is right over there. That's vector c. Vector c looks like that." + }, + { + "Q": "\nWhere does the 9/3 come from at 2:32 ?", + "A": "Since you need common denominators in order to subtract two fractions, what he did is that after simplifying 12/4 to 3, he changed it into a fraction that equals 3, 9/3=3, in order to subtract the two fractions.", + "video_name": "BOIA9wsM4ok", + "timestamps": [ + 152 + ], + "3min_transcript": "the negative 10/3 and the positive 10/3, those cancel out to get a zero, and I'm just left with j/4. It's equal to j/4. Now you might recognize 9/3, that's the same thing as nine divided by three. So this is just going to be three. So that simplifies a little bit. Three, let me just rewrite it so you don't get confused. Three is equal to j/4. Now, to solve for j, I could just multiply both sides by four. 'Cause if I divide something by four and then multiply by four, I'm just going to be left with that something. If I start with j and I divide by four, and then I multiply, and then I multiply by four, so I'm just going to multiply by four, then I'm just going to be left with j on the right-hand side. But I can't just multiply the right-hand side by four. I have to do it with the left-hand side So I multiply the left-hand side by four as well. And what I will be left with, four times three is twelve. And then j divided by four times four, So we get j is equal to 12. And the neat thing about equations is you can verify that you indeed got the right answer. You can substitute 12 for j here, and verify that negative 1/3 is equal to 12/4 - 10/3. Does this actually work out? Well 12/4 is the same thing as three, and if I wanted to write that as thirds, this is the same thing as 9/3. And 9/3 - 10/3 is indeed equal to negative 1/3. So we feel very good about that. Let's do another example. So I have n/5 + 0.6 = 2. So let's isolate this term that involves n on the left-hand side. So let's get rid of this 0.6. So let's subtract 0.6 from the left-hand side. But I can't just do it from the left. I have to do it from both sides if I want the equality to hold true. So, subtract 0.6. I'm just going to be left with n/5, and on the right-hand side, 2 - 0.6, that;'s going to be 1.4. And if you don't want to do this in your head, you could work this out It's going to be 2.0 - 0.6. You could say, \"Oh, this is 20/10-6/10\" which is going to be 14/10, which is that there. Or if you want to do it a little bit kind of the traditional method, six from zero, let me re-group.\" That's going to be a 10. I'm going to take from the ones place. If I take a one from the one's place, and that's going to be equal to 10/10. 10/10 - 6/10 is 4/10. And then, bring down one one minus zero ones is just one. So it's 1.4. And now, to solve for n. Well on the left have n being divided by five. If I just want n here, I can just multiply by five. So, if I multiply by five," + }, + { + "Q": "Why cant we directly substitute and find at 2:28 where 3y+3=y+7\n3y-y=7-3and y=2\n", + "A": "Add 18r to -2r and add 12s to 6s", + "video_name": "vkhYFml0w6c", + "timestamps": [ + 148 + ], + "3min_transcript": "Alright, now we have a very interesting situation. On both sides of the scale, we have our mystery mass and now I'm calling the mystery mass having a mass of Y. Just to show you that it doesn't always have to be X. It can be any symbol as long as you can keep track of that symbol. But all these have the same mass. That's why I wrote Y on all of them. And we also have a little 1 kilogram boxes on both sides of the scale. So the first thing I wanna do, we're gonna do step by step and try to figure out what this mystery mass is. But the first thing I wanna do is, is, is, have you think about, whether you can represent this algebraically? Whether with, with a little bit of mathematic symbolry, you can represent what's going on this scale. Over here, I have three Ys and three of these boxes and their total mass is equal to this one Y. And I think I have about let's see, I have 7 boxes right over here. So I'll give you a few seconds to do that. So let's think about the total mass over here. We have 3 boxes and a mass Y. So they're going to have a mass of 3Y, and then you have 3 boxes with a mass of 1 kilogram. So that's going to be my 1Y right over there. I could've written 1Y but that's, I don't need to. A Y is the same thing as 1Y. So I have the Y kilograms right there. And I have seven of these, right? 1,2,3,4,5,6,7. Yup, seven of these. So I have Y plus 7 kilograms on the right-hand side. And once again, it's balanced. The scale is balanced. This mass, total mass is equal to this total mass. So we can write an equal sign, right over there. So that's a good starting point. We were able to represent this situation to this real-life simple situation. You know, back in the day, when people actually had to figure out the mass of things if you were to go the jewelry store, whatever. They actually did had problems like this. We were able to represent it mathematically. Now the next thing to do is, what are some reasonable next steps? How can we start to simplify this a little bit? Well, the neat thing about algebra is there's actually multiple paths that you could go down. You might say, well why won't we remove 3 of these what, of these yellow blocks from both sides? That would be completely legitimate. You might say, well, why won't we remove 1 of these Ys from both sides? That also would be legitimate. And we could do it in either order. So let's just pick one of them. Let's say that we've first want to remove, let's say that we first want to remove the, a Y from either sides. Just so that we feel a little bit more comfortable with all of our Ys sitting on the 1 side. And so the best way, if we don't want all of our Ys to sit on the 1 side, we can remove, we can remove a Y from each side. Remember, if you removed a Y from only 1 side, that would unbalance the scale. The scale was already balanced whatever you have to do to the one side after you do the other. So I'm gonna remove a Y, I'm gonna remove Y mass from both sides. Now what will that look like algebraically?" + }, + { + "Q": "at 0:43 where did you get 0 from\n", + "A": "You get the zero when you do not move either way on the x axis.", + "video_name": "5a6zpfl50go", + "timestamps": [ + 43 + ], + "3min_transcript": "Let's say I have the equation y is equal to x plus 3. And I want to graph all of the sets, all of the coordinates x comma y that satisfy this equation right there. And we've done this many times before. So we draw our axis, our axes. That's my y-axis. This is my x-axis. And this is already in mx plus b form, or slope-intercept form. The y-intercept here is y is equal to 3, and the slope here is 1. So this line is going to look like this. We intersect at 0 comma 3-- 1, 2, 3. At 0 comma 3. And we have a slope of 1, so every 1 we go to the right, we go up 1. So the line will look something like that. It's a good enough approximation. So the line will look like this. And remember, when I'm drawing a line, every point on this Or it represents a pair of x and y that satisfy this equation. So maybe when you take x is equal to 5, you go to the line, and you're going to see, gee, when x is equal to 5 on that line, y is equal to 8 is a solution. And it's going to sit on the line. So this represents the solution set to this equation, all of the coordinates that satisfy y is equal to x plus 3. Now let's say we have another equation. Let's say we have an equation y is equal to negative x plus 3. And we want to graph all of the x and y pairs that satisfy this equation. Well, we can do the same thing. This has a y-intercept also at 3, right there. But its slope is negative 1. So it's going to look something like this. to move down 1. Or if you move to the right a bunch, you're going to move down that same bunch. So that's what this equation will look like. Every point on this line represents a x and y pair that will satisfy this equation. Now, what if I were to ask you, is there an x and y pair that satisfies both of these equations? Is there a point or coordinate that satisfies both equations? Well, think about it. Everything that satisfies this first equation is on this green line right here, and everything that satisfies this purple equation is on the purple line right there. So what satisfies both? Well, if there's a point that's on both lines, or essentially, a point of intersection of the lines. So in this situation, this point is on both lines. And that's actually the y-intercept. So the point 0, 3 is on both of these lines. So that coordinate pair, or that x, y pair, must satisfy" + }, + { + "Q": "At 0:58, Sal draws a line segment instead of a line. Is this supposed to be the case?\n", + "A": "Should be a line, he just didn t draw the arrows.", + "video_name": "5a6zpfl50go", + "timestamps": [ + 58 + ], + "3min_transcript": "Let's say I have the equation y is equal to x plus 3. And I want to graph all of the sets, all of the coordinates x comma y that satisfy this equation right there. And we've done this many times before. So we draw our axis, our axes. That's my y-axis. This is my x-axis. And this is already in mx plus b form, or slope-intercept form. The y-intercept here is y is equal to 3, and the slope here is 1. So this line is going to look like this. We intersect at 0 comma 3-- 1, 2, 3. At 0 comma 3. And we have a slope of 1, so every 1 we go to the right, we go up 1. So the line will look something like that. It's a good enough approximation. So the line will look like this. And remember, when I'm drawing a line, every point on this Or it represents a pair of x and y that satisfy this equation. So maybe when you take x is equal to 5, you go to the line, and you're going to see, gee, when x is equal to 5 on that line, y is equal to 8 is a solution. And it's going to sit on the line. So this represents the solution set to this equation, all of the coordinates that satisfy y is equal to x plus 3. Now let's say we have another equation. Let's say we have an equation y is equal to negative x plus 3. And we want to graph all of the x and y pairs that satisfy this equation. Well, we can do the same thing. This has a y-intercept also at 3, right there. But its slope is negative 1. So it's going to look something like this. to move down 1. Or if you move to the right a bunch, you're going to move down that same bunch. So that's what this equation will look like. Every point on this line represents a x and y pair that will satisfy this equation. Now, what if I were to ask you, is there an x and y pair that satisfies both of these equations? Is there a point or coordinate that satisfies both equations? Well, think about it. Everything that satisfies this first equation is on this green line right here, and everything that satisfies this purple equation is on the purple line right there. So what satisfies both? Well, if there's a point that's on both lines, or essentially, a point of intersection of the lines. So in this situation, this point is on both lines. And that's actually the y-intercept. So the point 0, 3 is on both of these lines. So that coordinate pair, or that x, y pair, must satisfy" + }, + { + "Q": "\nDo you get the 1 in the rise over run from 3/1? 5:37 in the video", + "A": "3x is the same as saying go over 1 and up 3 which is like saying 3/1, so yes, you are corect", + "video_name": "5a6zpfl50go", + "timestamps": [ + 337 + ], + "3min_transcript": "graph both lines, both equations, and then look at their intersection. And that will be the solution to both of these equations. In the next few videos, we're going to see other ways to solve it, that are maybe more mathematical and less graphical. But I really want you to understand the graphical nature of solving systems of equations. Let's do another one. Let's say we have y is equal to 3x minus 6. That's one of our equations. And let's say the other equation is y is equal to negative x plus 6. And just like the last video, let's graph both of these. I'll try to do it as precisely as I can. There you go. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And then 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I should have just copied and pasted some graph paper here, but I think this'll do the job. So let's graph this purple equation here. Y-intercept is negative 6, so we have-- let me do another 1, 2, 3, 4, 5, 6. So that's y is equal to negative 6. And then the slope is 3. So every time you move 1, you go up 3. You moved to the right 1, your run is 1, your rise is 1, 2, 3. That's 3, right? 1, 2, 3. So the equation, the line will look like this. And it looks like I intersect at the point 2 comma 0, which is right. 3 times 2 is 6, minus 6 is 0. So our line will look something like that right there. What about this line? Our y-intercept is plus 6. 1, 2, 3, 4, 5, 6. And our slope is negative 1. So every time we go 1 to the right, we go down 1. And so this will intersect at-- well, when y is equal to 0, x is equal to 6. 1, 2, 3, 4, 5, 6. So right over there. So this line will look like that. The graph, I want to get it as exact as possible. And so we're going to ask ourselves the same question. What is an x, y pair that satisfies both of these equations? Well, you look at it here, it's going to be this point. This point lies on both lines. And let's see if we can figure out what that point is." + }, + { + "Q": "Are rates and ratios the same as fractions?\nFor example, 35:1 can also be written as 35/1, which is a fraction.\n", + "A": "Fractions explain how much a number is out of something, whereas ratios compare numbers", + "video_name": "qGTYSAeLTOE", + "timestamps": [ + 2101 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:09, why is decameter spelled wrong? Or am I wrong?\n", + "A": "Decimeter and Decameter are actually 2 different things. Decameters are bigger than Meters but smaller than Kilometers and are used less often in math.and science and such.", + "video_name": "I3kQJvR7ZIg", + "timestamps": [ + 249 + ], + "3min_transcript": "This would be even the distance from one city to another city. Even the radius of the planet is often measured in kilometers. And to once again get a sense of that, this right over here is a map of New York City. And right over here they give us the scale. This distance right over here is 5 kilometers. So 5 kilometers just to give us a sense of things-- and they wrote km for short so 5 km. This lowercase k, lowercase m is just shorthand for kilometers, this would be equal to-- well if 1 kilometer is 1,000 meters, 5 kilometers is going to be 5 times as many. It's going to be 5,000 meters is this distance right over here. If you were to try to imagine 1 kilometer, it would be 1/5 of that. So 1 kilometer on this map might look something like that. allow us to describe distances less than a kilometer but more But they're not as typically used. But I'll list them here, just so that you see that they exist. You have the \"hect-O-meter,\" or \"heck-TOM-eter\"-- it's so infrequently used that I really haven't heard a lot of people say it-- which is equal to 100 meters. And you have the \"DECK-a-meter\" or \"deck-AM-eter,\" and I actually think it's \"DECK-a-meter.\" Dekameter, which is 10 meters. And I'm going to write these in orange because they're not that frequently used. For example, it's not typical to hear someone say that the Empire State Building is 4.43 hectometers or for them to say that this is 44.3 dekameters. They would typically say that this is 443 meters. Now you're probably saying, OK, well, this is fine. Using these prefixes on the meter, are larger than a meter, that are multiples of tens of the meter or a multiple of 100 or a multiple of 1,000 of the meter. But what if I want to go smaller? Well, the metric system has units for that. And if we go just 1/10 of a meter, this is not used as typically, but I'll write it here-- the decimeter. The decimeter is 1/10 of a meter. Or another way of saying it is 1 meter is equal to 10 decimeters, lowercase dm for short. Once again, this is not so typically used. But if we go one scale even below that, we get to the centimeter, which is a heavily used unit. And here, the prefix centi means 1/100. So this is 1/100 of a meter. Another way of thinking about it-- 1 meter is equal to 100 centimeters." + }, + { + "Q": "At 4:40, how did someone found out all this?\n", + "A": "As Sal says, exponential functions and logarithmic functions are inverses so they appear as reflections on the graph. Basically, the x values and y values are swapped.", + "video_name": "K_PiPfYxtao", + "timestamps": [ + 280 + ], + "3min_transcript": "swap these two columns. x and y, so let me just do 1/4, 1/2, one, two, four, and eight. Here now we're saying if x is 1/4, what power do we have to raise two to, to get to 1/4. We have to raise it to the negative two power. Two to the negative one power is equal to 1/2. Two to the zero power is equal to one. Two to the first power is equal to two. Two to the second power is equal to four. Two to the third power is equal to eight. Notice all we did, as we essentially swapped these two columns, so let's graph this. When x is equal to 1/4, y is equal to negative two. When x is one, y is zero. When x is two, y is one. When x is four, y is two. When x is eight, y is three. It's going to look like this. Notice, I think you might already be seeing a pattern right over here. These two graphs are essentially the reflections of each other. What would you have to reflect about to get these two? Well you'd have to reflect about y is equal to x. If you swap the x's and the y's, another way to think about, if you swap the axis you would get the other graph. It's essentially what we're doing. Notice it's symmetric about that line and that's because these are essentially the inverse functions of each other. One way to think about it is we swapped Just as this, as x becomes more and more and more and more negative you see y approaching zero. Here you see is y is becoming more and more negative as x is approaching zero, or you could say as x approaches zero y becomes more and more and more negative. The whole point of this is just to give you an appreciation for the relationship between an exponential function and a logarithmic function. They're essentially inverses of each other. You see that in the graphs, they're reflections of each other about the line y is equal to x." + }, + { + "Q": "at 8:10 and 9:21 why is he multiplying by 1/2 ?\n", + "A": "He is finding the area of the triangles. The formula for finding the area of a triangle is 1/2 times base times height.", + "video_name": "EqNzr56h1Ic", + "timestamps": [ + 490, + 561 + ], + "3min_transcript": "because we can see that triangle ABH is actually similar to triangle ACG. They both have this angle here, and then they both have a right angle. ABH has a right angle there. ACG has a right angle right over there. So you have two angles. Two corresponding angles are equal to each other. You're now dealing with similar triangles. So we know that triangle ABH-- I'll just write it as AHB, since I already wrote it this way. AHB is similar to triangle AGC. You want to make sure you get the vertices in the right order. A is the orange angle. G is the right angle, and then C is the unlabeled angle. This is similar to triangle AGC. And what that does for us is now we can use the ratios to figure out what HG is equal to. So what could we say over here? BH over its corresponding side of the larger triangle-- so we say 8 over 24 is equal to 6 over not HG, but over a AG. 6 over AG, and I think you can see where this is going. You have 1/3. 1/3 is equal to 6 over AG, or we can cross-multiply here, and we can get AG is equal to 18. So this entire length right over here is 18. If AG is 18 and AH is 6, then HG is 12. This is what you might have guessed if you were just trying to guess the answer right over here. But now we have proven to ourselves that this base has length of-- well, we have 18 here, and then we have another 18 here. So it has a length of 36. So the entire base here is 36. So that is 36. of the entire isosceles triangle. So the area of ACE is going to be equal to 1/2 times the base, which is 36, times 24. And so this is going to be the same thing as 1/2 times 36 is 18. 18 times 24. I'll just do that over here on the top. So 18 times 24. 8 times 4 is 32. 1 times 4 is 4, plus 3 is 7. Then we put as 0 here, because we're now dealing not with 2, but 20. You have 2 times 8 is 16. 2 times 1 is 2, plus 1. So it's 360, and then you have a 2, 7 plus 6 is 13. 1 plus 3 is 4. So the area of ACE is equal to 432. But we're not done yet." + }, + { + "Q": "\nWhy does Log2(2t) become zero at 3:01?So confused", + "A": "Have a look in Top Questions for Theresa Johnson s answer (currently 14 votes) to dugee23 s question. She explains how Sal simplified log_2(2^t) and got t. (Note that there is no zero at 3:01.)", + "video_name": "7Ig6kVZaWoU", + "timestamps": [ + 181 + ], + "3min_transcript": "the variable that we're trying to solve for, we're trying to find what t value will make this equal that right over there. A good first step would maybe try to get this five out of the left hand side, so let's divide the left by five. If we want to keep this being in equality we have to do the same thing to both sides. We get two to the t power is equal to 1,111 over five. How do we solve for t here? What function is essentially the inverse of the exponential function? Well it would be the logarithm. If we say that a to the b power is equal to c then that means that log base a of c is equal to b. a to the b power is equal to c. Log base a of c says what power do I need Well I need to raise a to the b power, to get to c. a to the b power is equal to c. These two are actually equivalent statements. Let's take log base two of both sides of this equation. On the left hand side you have log base two of two to the t power. On the right hand side, you have log base two of 1,111 over five. Why is this useful right over here? This is what power do we have to raise two to, to get two to t power? Well to get two to the t power, we have to raise two to the t power. This thing right over here just simplifies to t. That just simplifies to t. On the right hand side, we have log base two, we have all of these business right over here. t is equal to log base two of 1,111 over five. This is an expression that gives us our t value but then the next question is well how do we figure out what this is? If you take out your calculator, you'll quickly notice that there is no log base two button, so how do we actually compute it? Here we just have to apply a very useful property of exponents. If we have log base two of well really anything. Let me write it this way, if we have log base a of c, we can compute this as log base anything of c over log base that same anything of a. This anything has to be the same thing. Our calculator is useful because it has a log, you just press log, it's log base 10." + }, + { + "Q": "\nAt around 4:04, why does Sal say \"pi over 2\"? It's supposed to be pi over 4, right?", + "A": "yes", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 244 + ], + "3min_transcript": "arcsine of the square root of 2 over 2 is. What is the arcsine? You're like I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize, when they have this word arc in front of it-- This is also sometimes referred to as the inverse sine. This could have just as easily been written as: what is the inverse sine of the square root of 2 over 2? All this is asking is what angle would I have to take the sine of in order to get the value square root of 2 over 2. This is also asking what angle would I have to take the sine of in order to get square root of 2 over 2. I could rewrite either of these statements as saying square-- Let me do it. of what is equal to the square root of 2 over 2. And this, I think, is a much easier question for you to answer. Sine of what is square root of 2 over 2? Well I just figured out that the sine of pi over 4 is square root of 2 over 2. So, in this case, I know that the sine of pi over 4 is equal to square root of 2 over 2. So my question mark is equal to pi over 4. Or, I could have rewritten this as, the arcsine-- sorry --arcsine of the square root of 2 over 2 is equal to pi over 4. Now you might say so, just as review, I'm giving you a value and I'm saying give me an angle that gives me, when I take the sine of that angle that gives me that value. Let me go over here. You're like, look pi over 2 worked. 45 degrees worked. But I could just keep adding 360 degrees or I could keep just adding 2 pi. And all of those would work because those would all get me to that same point of the unit circle, right? And you'd be correct. And so all of those values, you would think, would be valid answers for this, right? Because if you take the sine of any of those angles-- You could just keep adding 360 degrees. If you take the sine of any of them, you would get square root of 2 over 2. And that's a problem. You can't have a function where if I take the function-- I can't have a function, f of x, where it maps to multiple values, right? Where it maps to pi over 4, or it maps to pi over 4 plus 2 pi or pi over 4 plus 4 pi. So in order for this to be a valid function-- In order for the inverse sine function to be valid, I have to restrict its range. And the way that-- We'll just restrict its range to" + }, + { + "Q": "Can someone explain to me why at 1:33 Sal says that the unit circle is sqrt2/2!!\n\nI don't understand, looking at the circle, knowing that 180 degrees equals pi, how what looks ro ME as pi/4, is actually the sqrt2/2!\n\nHELP!\n", + "A": "@1:33, you re right that the ANGLE is pi / 4 (or 45 degrees), but Sal was solving the equation for the length of the non-hypotenuse SIDES of the right triangle, each of which turned out to be sqrt2 / 2. @7:32, Sal indicates - sqrt3 / 2 on the Y-axis because the problem given is arcsin(-sqrt3/2). In other words we re looking for the angle whose sin (its measure on the Y-axis) is - sqrt3 / 2.", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 93 + ], + "3min_transcript": "If I were to walk up to you on the street and say you, please tell me what-- so I didn't want to write that thick --please tell me what sine of pi over 4 is. And, obviously, we're assuming we're dealing in radians. You either have that memorized or you would draw the unit circle right there. That's not the best looking unit circle, but you get the idea. You'd go to pi over 4 radians, which is the same thing as 45 degrees. You would draw that unit radius out. And the sine is defined as a y-coordinate on the unit circle. So you would just want to know this value right here. And you would immediately say OK. This is a 45 degrees. Let me draw the triangle a little bit larger. The triangle looks like this. This is 45. That's 45. This is 90. And you can solve a 45 45 90 triangle. The hypotenuse is 1. This is x. They're going to be the same values. Their base angles are the same. So you say, look. x squared plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1/2. x is equal to the square root of 1/2, which is one over the square root of 2. I can put that in rational form by multiplying that by the square root of 2 over 2. And I get x is equal to the square root of 2 over 2. So the height here is square root of 2 over 2. And if you wanted to know this distance too, it would also be the same thing. But we just cared about the height. Because the sine value, the sine of this, is just this height right here. The y-coordinate. And we got that as the square root of 2 over 2. This is all review. We learned this in the unit circle video. But what if someone else-- Let's say on another day, I arcsine of the square root of 2 over 2 is. What is the arcsine? You're like I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize, when they have this word arc in front of it-- This is also sometimes referred to as the inverse sine. This could have just as easily been written as: what is the inverse sine of the square root of 2 over 2? All this is asking is what angle would I have to take the sine of in order to get the value square root of 2 over 2. This is also asking what angle would I have to take the sine of in order to get square root of 2 over 2. I could rewrite either of these statements as saying square-- Let me do it." + }, + { + "Q": "At 6:10 the domain arcsin is restricted to the 1st and 4th quadrant. So far I follow. It starts at Pi/2 - ok I follow. the other range is - Pi/2 ? I do not understand. On the unit circle diagram I pulled up the point (0,-1) is 3pi/2. and I do not see - 2/pi any where on the unit circle. What gives?\n", + "A": "3pi/2 is the same place as -pi/2.", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 370 + ], + "3min_transcript": "Let me go over here. You're like, look pi over 2 worked. 45 degrees worked. But I could just keep adding 360 degrees or I could keep just adding 2 pi. And all of those would work because those would all get me to that same point of the unit circle, right? And you'd be correct. And so all of those values, you would think, would be valid answers for this, right? Because if you take the sine of any of those angles-- You could just keep adding 360 degrees. If you take the sine of any of them, you would get square root of 2 over 2. And that's a problem. You can't have a function where if I take the function-- I can't have a function, f of x, where it maps to multiple values, right? Where it maps to pi over 4, or it maps to pi over 4 plus 2 pi or pi over 4 plus 4 pi. So in order for this to be a valid function-- In order for the inverse sine function to be valid, I have to restrict its range. And the way that-- We'll just restrict its range to So let's restrict its range. Actually, just as a side note, what's its domain restricted to? So if I'm taking the arcsine of something. So if I'm taking the arcsine of x, and I'm saying that that is equal to theta, what's the domain restricted to? What are the valid values of x? x could be equal to what? Well if I take the sine of any angle, I can only get values between 1 and negative 1, right? So x is going to be greater than or equal to negative 1 and then less than or equal to 1. That's the domain. Now, in order to make this a valid function, I have to restrict the range. The possible values. I have to restrict the range. Now for arcsine, the convention is to restrict it to the first and fourth quadrants. To restrict the possible angles to this area right here along the unit circle. So theta is restricted to being less than or equal to pi over So given that, we now understand what arcsine is. Let's do another problem. Clear out some space here. Let me do another arcsine. So let's say I were to ask you what the arcsine of minus the square root of 3 over 2 is. Now you might have that memorized. And say, I immediately know that sine of x, or sine of theta is square root of 3 over 2. And you'd be done. But I don't have that memorized. So let me just draw my unit circle. And when I'm dealing with arcsine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I?" + }, + { + "Q": "At 1:13, how did Sal know that radius of the circle is 1? Like where did he get the 1 for the hyponthensus\n", + "A": "He refer to the unit circle radius 1", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 73 + ], + "3min_transcript": "If I were to walk up to you on the street and say you, please tell me what-- so I didn't want to write that thick --please tell me what sine of pi over 4 is. And, obviously, we're assuming we're dealing in radians. You either have that memorized or you would draw the unit circle right there. That's not the best looking unit circle, but you get the idea. You'd go to pi over 4 radians, which is the same thing as 45 degrees. You would draw that unit radius out. And the sine is defined as a y-coordinate on the unit circle. So you would just want to know this value right here. And you would immediately say OK. This is a 45 degrees. Let me draw the triangle a little bit larger. The triangle looks like this. This is 45. That's 45. This is 90. And you can solve a 45 45 90 triangle. The hypotenuse is 1. This is x. They're going to be the same values. Their base angles are the same. So you say, look. x squared plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1/2. x is equal to the square root of 1/2, which is one over the square root of 2. I can put that in rational form by multiplying that by the square root of 2 over 2. And I get x is equal to the square root of 2 over 2. So the height here is square root of 2 over 2. And if you wanted to know this distance too, it would also be the same thing. But we just cared about the height. Because the sine value, the sine of this, is just this height right here. The y-coordinate. And we got that as the square root of 2 over 2. This is all review. We learned this in the unit circle video. But what if someone else-- Let's say on another day, I arcsine of the square root of 2 over 2 is. What is the arcsine? You're like I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize, when they have this word arc in front of it-- This is also sometimes referred to as the inverse sine. This could have just as easily been written as: what is the inverse sine of the square root of 2 over 2? All this is asking is what angle would I have to take the sine of in order to get the value square root of 2 over 2. This is also asking what angle would I have to take the sine of in order to get square root of 2 over 2. I could rewrite either of these statements as saying square-- Let me do it." + }, + { + "Q": "\n@ 3:22 Couldn't he factor out (s-a) (s-b) and (s-c) to just s(-a*-b*-c)?\nI'm not sure. I'm rusty with my factoring...", + "A": "It would seem that way, but actually, if we factor it out that way, we get -s*a*b*c which is much different from (s-a) (s-b) and (s-c)", + "video_name": "-YI6UC4qVEY", + "timestamps": [ + 202 + ], + "3min_transcript": "And to do that we're going to apply something called Heron's Formula. And I'm not going to prove it in this video. I'm going to prove it in a future video. And really to prove it you already probably have the tools necessary. It's really just the Pythagorean theorem and a lot of hairy algebra. But I'm just going to show you the formula now and how to apply it, and then you'll hopefully appreciate that it's pretty simple and pretty easy to remember. And it can be a nice trick to impress people with. So Heron's Formula says first figure out this third variable S, which is essentially the perimeter of this triangle divided by 2. a plus b plus c, divided by 2. Then once you figure out S, the area of your triangle-- of this triangle right there-- is going to be equal to the square root of S-- this variable S right here that you just calculated-- times S minus a, times S minus b, times S minus c. This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18. the square root of S-- 18-- times S minus a-- S minus 9. 18 minus 9, times 18 minus 11, times 18 minus 16. And then this is equal to the square root of 18 times 9 times 7 times 2. Which is equal to-- let's see, 2 times 18 is 36. So I'll just rearrange it a bit. This is equal to the square root of 36 times 9 times 7, which is equal to the square root of 36 times the square" + }, + { + "Q": "at 4:22 Sal gets to the square root of \"7\" and stops and says, \"you don't deal with the negative square roots . . . so this is just the square root of 7.\" The square root of 7 is not negative, it's 2.645... So, why did he stop and not figure out the square root of 7?\n", + "A": "He said negative not about the square root of 7 but about all of those numbers because square root of a positive number can be either positive or negative eg. square root of 4 can be either 2 or -2 ( -2 *-2=4 ). He did not want it to be negative because you are dealing with area, which is not negative. Also by the way there is no need to calculate root of 7 unless you need to find cost of anything.", + "video_name": "-YI6UC4qVEY", + "timestamps": [ + 262 + ], + "3min_transcript": "This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18. the square root of S-- 18-- times S minus a-- S minus 9. 18 minus 9, times 18 minus 11, times 18 minus 16. And then this is equal to the square root of 18 times 9 times 7 times 2. Which is equal to-- let's see, 2 times 18 is 36. So I'll just rearrange it a bit. This is equal to the square root of 36 times 9 times 7, which is equal to the square root of 36 times the square The square root of 36 is just 6. This is just 3. And we don't deal with the negative square roots, because you can't have negative side lengths. And so this is going to be equal to 18 times the square root of 7. So just like that, you saw it, it only took a couple of minutes to apply Heron's Formula, or even less than that, to figure out that the area of this triangle right here is equal to 18 square root of seven. Anyway, hopefully you found that pretty neat." + }, + { + "Q": "For the triangle at 3:17, with lengths 9, 11 and 16, can you divide that in half and use pythagoras, so one side becomes 8,11 and x or 8,9 and x? and find the value of x, and that is your height?\n", + "A": "When we look at the triangle at 3;17, you cannot divide it in half and use the Pythagorean Theorem, because we don t know if it is a right triangle. So with a question were they don t give you the 90\u00c2\u00b0 right angle symbol, you can never assume anything in math. Hope this helps.", + "video_name": "-YI6UC4qVEY", + "timestamps": [ + 197 + ], + "3min_transcript": "And to do that we're going to apply something called Heron's Formula. And I'm not going to prove it in this video. I'm going to prove it in a future video. And really to prove it you already probably have the tools necessary. It's really just the Pythagorean theorem and a lot of hairy algebra. But I'm just going to show you the formula now and how to apply it, and then you'll hopefully appreciate that it's pretty simple and pretty easy to remember. And it can be a nice trick to impress people with. So Heron's Formula says first figure out this third variable S, which is essentially the perimeter of this triangle divided by 2. a plus b plus c, divided by 2. Then once you figure out S, the area of your triangle-- of this triangle right there-- is going to be equal to the square root of S-- this variable S right here that you just calculated-- times S minus a, times S minus b, times S minus c. This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18. the square root of S-- 18-- times S minus a-- S minus 9. 18 minus 9, times 18 minus 11, times 18 minus 16. And then this is equal to the square root of 18 times 9 times 7 times 2. Which is equal to-- let's see, 2 times 18 is 36. So I'll just rearrange it a bit. This is equal to the square root of 36 times 9 times 7, which is equal to the square root of 36 times the square" + }, + { + "Q": "hey isn't that any number to the 0th power is always 1 including negative number at 1:42\n", + "A": "Yes, but that is not what was in the problem \u00e2\u0088\u0092 b\u00e2\u0081\u00b0 means the same as \u00e2\u0088\u0092 (b\u00e2\u0081\u00b0) = \u00e2\u0088\u0092 (1) = \u00e2\u0088\u00921 However (\u00e2\u0088\u0092b)\u00e2\u0081\u00b0 would be equal to 1", + "video_name": "gR8-vRg6Yp0", + "timestamps": [ + 102 + ], + "3min_transcript": "Let's say that the position of some particle as a function of time is given by this expression right over here. Negative d to the negative t power plus c to the fourth over c squared plus 1, where c and d are constants and both of them are greater than one. So what I want to do over the course of this video is see what can we infer based on this expression, this function definition, that we have here. And the first thing that I want you to think about is, what is the initial position? If I were to express the initial position, in terms of c's and d's, and try to simplify it. So I encourage you to pause the video and try to find an expression for the initial position. Well, the initial position is the position we're at when our time is equal to 0. So we essentially just want to find p of 0. And p of 0 is going to be equal to negative d to the negative 0. Negative 0 plus c to the fourth over c squared plus 1. Well, d to the negative 0, that's the same thing as d to the 0, and since we know that d is non-zero, we know this is defined. Anything non-zero to the 0-th power is going to be 1. And the zero is actually under debate, what 0 to the 0-th power is. But we can safely say that this right over here is going to be equal to 1. And so the numerator here simplifies 2. This is equal to-- and I'll switch the order. c to the fourth minus 1 over c squared plus 1. And now this might jump out at you as a difference of squares. We could write this as c squared, squared, minus 1 squared over c squared plus 1. And that's the same thing as c squared plus 1 times c squared minus 1. And we have a c squared plus 1 in the numerator and that denominator so we can simplify. And so our initial position is going to be c squared minus 1. So that actually simplified out quite nicely. Now the next question I'm going to ask you is-- OK, we know that the initial position that time equals zero, the particle is going to be at c squared minus 1. But what happens after that? Does the position keep increasing? Does the position keep decreasing? Or does the position maybe increase and then decrease, or decrease and then increase as in keep swapping around? So I encourage you to pause the video now and think about what happens to the position. Does it keep increasing? Does it keep decreasing? Or does it do something else? Well, let's answer that question of what's happening to the position after our initial position. We really just have to focus on this term right over here. This d to the negative t." + }, + { + "Q": "\nAt 3:42 when it says the radical sign means the principle square root; what's the notation for either square root, or for the negative square root, in that case?\n\nAlso when talking about the third root, you wouldn't have to distinguish between principle and negative, right? Do you have to distinguish for all even roots, but not for odd roots?", + "A": "The notation for both square roots is \u00c2\u00b1\u00e2\u0088\u009ax. If you want only the negative square root, then you would say -\u00e2\u0088\u009ax. As for odd roots, you still have to distinguish between principal and non principal roots. The only difference is that a non principal odd root is not simply the negative of the principal root. For instance, the cube root of -8 is -2, but it can also be 1\u00c2\u00b1\u00e2\u0088\u009a3 i.", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 222 + ], + "3min_transcript": "And what you should then point out is that this was not the incorrect step. That it is true negative 1 is not equal to 1, but the faulty line of reasoning here was in using this property when both a and b are negative. If both a and b are negative, this will never be true. So a and b both cannot be negative. In fact normally when this property is given-- sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time-- but they'll usually give a little bit of a constraint They'll usually say for a and b greater than or equal to 0. So that's where they list this property. This is true for a and b greater than or equal 0. And in particular, it's false if both a and b are both, if they are both negative. saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. When we take traditional principal square roots. So when you take the principle square root of 4. We know that this is positive 2, that 4 actually has two square roots. There's negative 2 also is a square root of 4. If you have negative 2 times negative 2 it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non imaginary, non complex numbers, you could really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers-- and complex numbers and all the rest-- you have to expand the definition of what this radical means. So when you are taking the square root of really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function, or this is now defined for complex inputs or the domain, it can also generate imaginary or complex outputs, or I guess you could call that the range. And if you assume that, then really straight from this you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only-- and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true-- So we could apply this when x is greater than or equal to 0." + }, + { + "Q": "\nwait.. at 1:11, Sal says (principal square root = psr) psr of (a*b) = psr of a *psr of b, when in the last video he states that it does not work when the numbers are both negative, and in this case they are. so therefore, this is wrong, right?", + "A": "Square roots generally have two roots: the principal root and the negative root. For example: \u00e2\u0088\u009a9: psr = 3; negative root = -3. The default is to use the psr. If the negative root is wanted, then a minus sign is placed in front of the square root symbol. -\u00e2\u0088\u009a9 = -3", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 71 + ], + "3min_transcript": "In your mathematical careers you might encounter people who say it is wrong to say that i is equal to the principal square root of negative 1. And if you ask them why is this wrong, they'll show up with this kind of line of logic that actually seems pretty reasonable. They will tell you that, OK, well let's just start with negative 1. We know from definition that negative 1 is equal to i times i. Everything seems pretty straightforward right now. And then they'll say, well look, if you take this, if you assume this part right here, then we can replace each of these i's with the square root And they'd be right. So then this would be the same thing as the square root of negative 1 times the square root of negative 1. And then they would tell you that, hey, look just from straight up properties of the principal square root function, they'll tell you that the square root of a times b is the same thing as the principal square root of a times the principal square root of b. And so if you have the principal square root of a times the same thing as the square root of a times b. So based on this property of the radical of the principal root, they'll say that this over here is the same thing as the square root of negative 1 times negative 1. If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots. I'm doing this in the other order here. Here I have the principal root of the products, over here I have this on the right. And then from that we all know that negative 1 times negative 1 is 1. So this should be equal to the principal square root of 1. And then the principal square root of 1-- Remember, this radical means principal square root, positive square root, that is just going to be positive 1. And they'll say, this is wrong. Clearly, negative 1 and positive 1 are not the same thing. And they'll argue therefore, you can't make this substitution And what you should then point out is that this was not the incorrect step. That it is true negative 1 is not equal to 1, but the faulty line of reasoning here was in using this property when both a and b are negative. If both a and b are negative, this will never be true. So a and b both cannot be negative. In fact normally when this property is given-- sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time-- but they'll usually give a little bit of a constraint They'll usually say for a and b greater than or equal to 0. So that's where they list this property. This is true for a and b greater than or equal 0. And in particular, it's false if both a and b are both, if they are both negative." + }, + { + "Q": "at 3:50 sal says that each number has a positive and negative square root. And i was thinking, if that was the case, couldn't the square root of -1= -i, which is i to the 3rd power. If this was true then i to the 1st power would eqaul i to the 3rd power. in essence -i=i. I'm a little cunfused so could someone please explain this to me.\n", + "A": "Sal says that 4 has 2 square roots. Positive numbers have two square roots. Negative numbers do not.", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 230 + ], + "3min_transcript": "And what you should then point out is that this was not the incorrect step. That it is true negative 1 is not equal to 1, but the faulty line of reasoning here was in using this property when both a and b are negative. If both a and b are negative, this will never be true. So a and b both cannot be negative. In fact normally when this property is given-- sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time-- but they'll usually give a little bit of a constraint They'll usually say for a and b greater than or equal to 0. So that's where they list this property. This is true for a and b greater than or equal 0. And in particular, it's false if both a and b are both, if they are both negative. saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. When we take traditional principal square roots. So when you take the principle square root of 4. We know that this is positive 2, that 4 actually has two square roots. There's negative 2 also is a square root of 4. If you have negative 2 times negative 2 it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non imaginary, non complex numbers, you could really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers-- and complex numbers and all the rest-- you have to expand the definition of what this radical means. So when you are taking the square root of really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function, or this is now defined for complex inputs or the domain, it can also generate imaginary or complex outputs, or I guess you could call that the range. And if you assume that, then really straight from this you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only-- and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true-- So we could apply this when x is greater than or equal to 0." + }, + { + "Q": "\nAt 3:23, does that mean that every number actually has two square roots? Wouldn't that make algebra nearly impossible?", + "A": "Not really, it just means you have two possible solutions if you have a function with a square root. example: x^2=9 -> x=sqrt(9) -> x=3 or x=-3. After all 3*3 = 9 and -3*-3 = 9.", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 203 + ], + "3min_transcript": "the same thing as the square root of a times b. So based on this property of the radical of the principal root, they'll say that this over here is the same thing as the square root of negative 1 times negative 1. If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots. I'm doing this in the other order here. Here I have the principal root of the products, over here I have this on the right. And then from that we all know that negative 1 times negative 1 is 1. So this should be equal to the principal square root of 1. And then the principal square root of 1-- Remember, this radical means principal square root, positive square root, that is just going to be positive 1. And they'll say, this is wrong. Clearly, negative 1 and positive 1 are not the same thing. And they'll argue therefore, you can't make this substitution And what you should then point out is that this was not the incorrect step. That it is true negative 1 is not equal to 1, but the faulty line of reasoning here was in using this property when both a and b are negative. If both a and b are negative, this will never be true. So a and b both cannot be negative. In fact normally when this property is given-- sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time-- but they'll usually give a little bit of a constraint They'll usually say for a and b greater than or equal to 0. So that's where they list this property. This is true for a and b greater than or equal 0. And in particular, it's false if both a and b are both, if they are both negative. saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. When we take traditional principal square roots. So when you take the principle square root of 4. We know that this is positive 2, that 4 actually has two square roots. There's negative 2 also is a square root of 4. If you have negative 2 times negative 2 it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non imaginary, non complex numbers, you could really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers--" + }, + { + "Q": "How did you get the porabla at 3:02\nI'm still confused..\n", + "A": "I think I understand your question. The line describing the parabola is curved, not straight, but with only a few points drawn in, it s hard to see why. If you went through the equation and filled in x with all the points in between the points that he graphed, you would see a curve shape appear. When you only have two points it s easy to see why it looks like it should be straight. Sal just skipped over doing 2.5, and then 2.25, and so on. The more points you graph, the more it looks l", + "video_name": "hjigR_rHKDI", + "timestamps": [ + 182 + ], + "3min_transcript": "So that right there is the highest point of our parabola. Then, if we want, we can a graph a couple of other points, just to see what happens. So let's see what happens when x is equal to-- let me just draw a little table here-- 2, what is y? It's negative x squared plus 6. So when x is 2, what is y? You have 2 squared, which is 4, but you have negative 2 squared, so it's negative 4 plus 6-- it is equal to 2. It's the same thing when x is negative 2. You put negative 2 there, you square it, then you have positive 4, but you have a negative there, so it's negative 4 plus 6 is 2. You have both of those points there, so 2 comma 2, and then you have a negative 2 comma 2. If I were to graph it, Let's try it with 3, as well-- if we It then becomes a negative 9 plus 3, it becomes negative 3, and negative 3 will also become a negative 3. Negative 3 squared is positive 9, you have a negative out front, it becomes negative 9 plus 6, which is negative 3. You have negative 3, negative 3, and then you have 3, negative 3. So those are all good points. Now we can graph our parabola. Our parabola will look something-- I was doing well until that second part --like that, and let me just do the second part. That second part is hard to draw-- let me do it from here. It looks something like that. We connect to this dot right here, and then let me connect this. So that it looks something like that. That's what our parabola looks like, and obviously it keeps going down in that direction. Let's graph this second one over here: y is equal to negative 2x minus 2. This is just going to be a line. It's a linear equation, and the highest degree here is 1. Our y-intercept is negative 2, so 0, 1, 2. Our y-intercept is negative 2. Our slope is negative 2. If we move 1 in the x direction, we're going to go 2 in the y-direction, and if we move 2 in the x direction, we're going to move down 4 in the y direction. If we move back 2, we're going to move up 2 in the y direction, and it looks like we found one of our points of intersection. Let's just draw that line, so that line will look something like-- It's hard for my hand to draw that, but let me try as best as I can. This is the hardest part. It will look something like that right there." + }, + { + "Q": "\nWhy is the R^n space not straight? I thought it was made up of vectors. At 1:47?", + "A": "It is straight . The wiggly circle that Sal drew should be viewed very abstractly. Within that wiggly loop is all the vectors in R^n.", + "video_name": "pMFv6liWK4M", + "timestamps": [ + 107 + ], + "3min_transcript": "We now have the tools, I think, to understand the idea of a linear subspace of Rn. Let me write that down. I'll just always call it a subspace of Rn. Everything we're doing is linear. Subspace of Rn. I'm going to make a definition here. I'm going to say that a set of vectors V. So V is some subset of vectors, some subset of Rn. So we already said Rn, when we think about it, it's really just really an infinitely large set of vectors, where each of those vectors have n components. I'm going to not formally define it, but this is just a set of vectors. I mean sometimes we visualize it as multi-dimensional space and all that, but if we wanted to be just as abstract about it as possible, it's just all the set. It's the set of all of the -- you know we could call x1, x2, xi's are a member of the real numbers for all of the i's. That was our definition of Rn. It's just a huge set of vectors. An infinitely large set of vectors. V, I'm calling that, I'm going to call that a subset of Rn, and which means it's just some -- you know, it could be all of these vectors, and I'll talk about that in a second. Or it could be some subset of these vectors. Maybe it's all of them but one particular vector. In order for this V to be a subspace-- so I'm already saying it's a subset of Rn. Maybe this'll help you. If I draw all of Rn here as this big blob. So these are all of the vectors that are in Rn. V is some subset of it. It could be all of Rn. I'll show that a second. But let's just say that this is V. V is a subset of vectors. definition, if V is a subspace, or linear subspace of Rn, this means, this is my definition, this means three things. This means that V contains the 0 vector. I'll do it really, that's the 0 vector. This is equal to 0 all the way and you have n 0's. So V contains the 0 vector, and this is a big V right there. If we have some vector x in V. So let me write this, if my vector x is in V, if x is one of these vectors that's included in my V, then when I multiply x times any member of the reals." + }, + { + "Q": "\nsall, in 13:40 - 13:56, I think I don't get it. -a is a subset of R^2? since R^2 is all the cartesian plain . What we are saying here is -a is not a subset of our restriction of R^2 that we made and not R^2. Is this right? So we are saying that -a is not a subset of R^2 when x_1>0 right?", + "A": "I think I need to rethink what a subspace is, Thank you.", + "video_name": "pMFv6liWK4M", + "timestamps": [ + 820, + 836 + ], + "3min_transcript": "to each other, this thing is also going to be greater than 0. And we don't care what these, these can be anything, I didn't put any constraints on the second component of my vector. So it does seem like it is closed under addition. Now what about scalar multiplication? Let's take a particular case here. Let's take my a, b again. I have my vector a, b. Now I can pick any real scalar. So any real scalar. What if I just multiply it by minus 1? So minus 1. So if I multiply it by minus 1, I get minus a, minus b. If I were to draw it visually, if this is-- let's say a, b was the vector 2, 4. So it's like this. When I multiply it by minus 1, what do I get? I get this vector. Which you can be visually clearly see falls out of, if we view these as kind of position vectors, it falls out of our subspace. Or if you just view it not even visually, if you just do it mathematically, clearly if this is positive then this is going to-- and let's say if we assume this is positive, and definitely not 0. So it's definitely a positive number. So this is definitely going to be a negative number. So when we multiply it by negative 1, for really any element of this that doesn't have a 0 there, you're going to end up with something that falls out of it, right? This is not a member of this set, because to be a member of the set, your first component had to be greater than 0. This first component is less than 0. So this subset that I drew out here, the subset of R2, is not It's not closed under multiplication or scalar multiplication. Now I'll ask you one interesting question. What if I ask you just the span of some set of vectors? Let's say I want to know the span of, I don't know, let's sat I have vector v1, v2, and v3. I'm not even going to tell you how many elements each of these vectors have. Is this a valid subspace of Rn? Where n is the number of elements that each of these have. Let's pick one of the elements." + }, + { + "Q": "At 4:00.. why do you use 2pi for the period?\n", + "A": "2pi is once around a circle. Going around a circle makes a sine wave or a cosine wave.", + "video_name": "SBqnRja4CW4", + "timestamps": [ + 240 + ], + "3min_transcript": "" + }, + { + "Q": "At 5:46, how does Sal determine what the slope is? How does he know it's -(1/2)? Isn't -(1/2) the y-intercept?\n", + "A": "Both the slope and the y-intercept are - \u00c2\u00bd When a function of a line is in the form of mx +b, the slope is m. Since -x/2 = -\u00c2\u00bd(x) the slope is -\u00c2\u00bd", + "video_name": "wSiamij_i_k", + "timestamps": [ + 346 + ], + "3min_transcript": "like it's supposed to do. Let's do one more of these. So here I have g of x is equal to negative 2x minus 1. So just like the last problem, I like to set y equal to this. So we say y is equal to g of x, which is equal to negative 2x minus 1. Now we just solve for x. y plus 1 is equal to negative 2x. Just added 1 to both sides. Now we can divide both sides of this equation by negative 2, and so you get negative y over 2 minus 1/2 is equal to x, or we could write x is equal to negative y over 2 minus 1/2, or we could write f inverse as a function of y is equal to negative y over 2 minus 1/2, or we can just rename y as x. That shouldn't be an f. The original function was g , so let me be clear. That is g inverse of y is equal to negative y over 2 minus 1/2 because we started with a g of x, not an f of x. Make sure we get our notation right. Or we could just rename the y and say g inverse of x is equal to negative x over 2 minus 1/2. Now, let's graph it. Its y-intercept is negative 1/2. It's right over there. And it has a slope of negative 1/2. Let's see, if we start at negative 1/2, if we move over to 1 in the positive direction, it will go down half. If we move over 1 again, it will go down half again. If we move back-- so it'll go like that. look something like that. It'll just keep going, so it'll look something like that, and it'll keep going in both directions. And now let's see if this really is a reflection over y equals x. y equals x looks like that, and you can see they are a reflection. If you reflect this guy, if you reflect this blue line, it becomes this orange line. But the general idea, you literally just-- a function is originally expressed, is solved for y in terms of x. You just do some algebra. Solve for x in terms of y, and that's essentially your inverse function as a function of y, but then you can rename it as a function of x." + }, + { + "Q": "\nhow do we come to know wheter a curve is 1:1 function and does its has an inverse?", + "A": "The horizontal line test is the easiest way. If you can draw a perfectly horizontal line such that it touches two points on the function, then it is not 1 to 1.", + "video_name": "wSiamij_i_k", + "timestamps": [ + 61 + ], + "3min_transcript": "So we have f of x is equal to negative x plus 4, and f of x is graphed right here on our coordinate plane. Let's try to figure out what the inverse of f is. And to figure out the inverse, what I like to do is I set y, I set the variable y, equal to f of x, or we could write that y is equal to negative x plus 4. Right now, we've solved for y in terms of x. To solve for the inverse, we do the opposite. We solve for x in terms of y. So let's subtract 4 from both sides. You get y minus 4 is equal to negative x. And then to solve for x, we can multiply both sides of this equation times negative 1. And so you get negative y plus 4 is equal to x. Or just because we're always used to writing the dependent variable on the left-hand side, we could rewrite this as x is equal to negative y plus 4. Or another way to write it is we could say that f inverse of y is equal to negative y plus 4. it as a function of y, but we can just rename the y as x so it's a function of x. So let's do that. So if we just rename this y as x, we get f inverse of x is equal to the negative x plus 4. These two functions are identical. Here, we just used y as the independent variable, or as the input variable. Here we just use x, but they are identical functions. Now, just out of interest, let's graph the inverse function and see how it might relate to this one right over here. So if you look at it, it actually looks fairly identical. It's a negative x plus 4. It's the exact same function. So let's see, if we have-- the y-intercept is 4, it's going to be the exact same thing. The function is its own inverse. So if we were to graph it, we would put it right on top of this. In the first inverse function video, I talked about how a function and their inverse-- they are the reflection over the line y equals x. So where's the line y equals x here? Well, line y equals x looks like this. And negative x plus 4 is actually perpendicular to y is equal to x, so when you reflect it, you're just kind of flipping it over, but it's going to be the same line. It is its own reflection. Now, let's make sure that that actually makes sense. When we're dealing with the standard function right there, if you input a 2, it gets mapped to a 2. If you input a 4, it gets mapped to 0. What happens if you go the other way? If you input a 2, well, 2 gets mapped to 2 either way, so that makes sense. For the regular function, 4 gets mapped to 0. For the inverse function, 0 gets mapped to 4." + }, + { + "Q": "In 5:00 onwards, why doesn't the -2 turn the whole function into -y/2 +1/2? Shouldn't it invert for both numbers?\n", + "A": "It doesn t invert any of the numbers actually. He had y+1 = -2x, so he divided both sides by -2. Think of it as having to divide both terms on the left side (y and 1) individually by -2. So y divided by -2 is -y/2 and 1 divided by -2 is -1/2. SO you end up with -y/2 - 1/2 = x", + "video_name": "wSiamij_i_k", + "timestamps": [ + 300 + ], + "3min_transcript": "Let's think about it another way. For the regular function-- let me write it explicitly down. This might be obvious to you, but just in case it's not, it might be helpful. Let's pick f of 5. f of 5 is equal to negative 1. Or we could say, the function f maps us from 5 to negative 1. Now, what does f inverse do? What's f inverse of negative 1? f inverse of negative 1 is 5. Or we could say that f maps us from negative 1 to 5. So once again, if you think about kind of the sets, they're our domains and our ranges. So let's say that this is the domain of f, this is the range of f. f will take us from to negative 1. That's what the function f does. And we see that f inverse takes us back from negative 1 to 5. like it's supposed to do. Let's do one more of these. So here I have g of x is equal to negative 2x minus 1. So just like the last problem, I like to set y equal to this. So we say y is equal to g of x, which is equal to negative 2x minus 1. Now we just solve for x. y plus 1 is equal to negative 2x. Just added 1 to both sides. Now we can divide both sides of this equation by negative 2, and so you get negative y over 2 minus 1/2 is equal to x, or we could write x is equal to negative y over 2 minus 1/2, or we could write f inverse as a function of y is equal to negative y over 2 minus 1/2, or we can just rename y as x. That shouldn't be an f. The original function was g , so let me be clear. That is g inverse of y is equal to negative y over 2 minus 1/2 because we started with a g of x, not an f of x. Make sure we get our notation right. Or we could just rename the y and say g inverse of x is equal to negative x over 2 minus 1/2. Now, let's graph it. Its y-intercept is negative 1/2. It's right over there. And it has a slope of negative 1/2. Let's see, if we start at negative 1/2, if we move over to 1 in the positive direction, it will go down half. If we move over 1 again, it will go down half again. If we move back-- so it'll go like that." + }, + { + "Q": "\nat 5:50 he says the slope is -1/2. he moves over in the positive direction on the x axis which i don't quite get and also slope i thought was change in Y over change in X so when he moves down by 1/2 i am at a loss. over by 1 down 1/2 doesn't make sense to me. anybody got another way of explaining by chance?", + "A": "nevermind, i think i got it. i get the same inverse line as Sal if i go down -1 on Y then over 2 on X.", + "video_name": "wSiamij_i_k", + "timestamps": [ + 350 + ], + "3min_transcript": "like it's supposed to do. Let's do one more of these. So here I have g of x is equal to negative 2x minus 1. So just like the last problem, I like to set y equal to this. So we say y is equal to g of x, which is equal to negative 2x minus 1. Now we just solve for x. y plus 1 is equal to negative 2x. Just added 1 to both sides. Now we can divide both sides of this equation by negative 2, and so you get negative y over 2 minus 1/2 is equal to x, or we could write x is equal to negative y over 2 minus 1/2, or we could write f inverse as a function of y is equal to negative y over 2 minus 1/2, or we can just rename y as x. That shouldn't be an f. The original function was g , so let me be clear. That is g inverse of y is equal to negative y over 2 minus 1/2 because we started with a g of x, not an f of x. Make sure we get our notation right. Or we could just rename the y and say g inverse of x is equal to negative x over 2 minus 1/2. Now, let's graph it. Its y-intercept is negative 1/2. It's right over there. And it has a slope of negative 1/2. Let's see, if we start at negative 1/2, if we move over to 1 in the positive direction, it will go down half. If we move over 1 again, it will go down half again. If we move back-- so it'll go like that. look something like that. It'll just keep going, so it'll look something like that, and it'll keep going in both directions. And now let's see if this really is a reflection over y equals x. y equals x looks like that, and you can see they are a reflection. If you reflect this guy, if you reflect this blue line, it becomes this orange line. But the general idea, you literally just-- a function is originally expressed, is solved for y in terms of x. You just do some algebra. Solve for x in terms of y, and that's essentially your inverse function as a function of y, but then you can rename it as a function of x." + }, + { + "Q": "At 8:50, how can (x+2) be considered positive when -1 still falls in the range of x>-2?\n", + "A": "SG, When x=-1 x+2 = -1+2 = +1 so even though x is not positive, (x+2) is positive. I hope that is of some help.", + "video_name": "ZjeMdXV0QMg", + "timestamps": [ + 530 + ], + "3min_transcript": "I'll just keep it the same way and maybe in the next video I'll do the case where it's greater than or equal just because I really don't want to-- maybe I want to incrementally step up the level of difficulty. We're just saying x minus 1 over x plus 2 is just straight up greater than 0. Now one thing you might say is well, if I have a rational expression like this, maybe I multiply both sides of this equation by x plus 2. So I get rid of it in the denominator and I can multiply it times 0 and get it out of the way. But the problem is when you multiply both sides of an inequality by a number-- if you're multiplying by a positive you can keep the inequality the same. But if you're multiplying by a negative you have to switch the inequality, and we don't know whether x plus 2 is positive or negative. So let's do both situations. Let's do one situation where x plus 2-- let me write it this way. x plus 2 is greater than 0. And then another situation where-- let me do that in a different color. Where x plus 2 is less than 0. Actually, in those situations can x plus 2 equal 0? If x plus 2 were to be equal to 0 than this whole expression would be undefined. And so that definitely won't be a situation that we want to deal with it. It would an undefined situation. So these are our two situations when we're multiplying both sides. So if x plus 2 is greater than 0 that means that x is greater than minus 2. We can just subtract 2 from both sides of this equation. So if x is greater than minus 2, then x plus 2 is greater than 0. And then we can multiply both sides of this equation times x plus 2. So you have x minus 1 over x plus 2 greater than 0. I'm going to multiply both sides by x plus 2, which I'm assuming is positive because x is greater than minus 2. Multiply both sides by x plus 2. These cancel out. 0 times x plus 2 is it just 0. just simplified to 0. Solve for x, add 1 to both sides, you get x is greater than 1. So we saw that if x plus 2 is greater than 0, or we could say, if x is greater than minus 2, then x also has to be greater than 1. Or you could say if x is great-- well, you could go both ways in that. But we say, look, both of these things have to be true. If for x to satisfy both of these, it just has to be greater than 1. Because if it's greater than 1 it's definitely going to satisfy this constraint over here. So for this branch we come up with the solution x is greater than 1. So this is one situation where x plus 2 is greater than 0. The other situation is x plus 2 being less than 0. If x plus 2 is less than 0 that's equivalent to saying that x is less than minus 2. You just subtract 2 from both sides. Now, if x plus 2 is less than 0, what we'll have to do when we multiply both sides-- let's do that." + }, + { + "Q": "At around 00:58: Why is a>0 and b>0 the consequence of a/b>0?\n", + "A": "A/B can only be greater than 0 (in other words, positive) only if A and B are both positive (A>0 and A>0) or A and B are both negative (A<0 and B<0). Any other combination will result in a negative fraction (A/B<0 instead of A/B>0).", + "video_name": "ZjeMdXV0QMg", + "timestamps": [ + 58 + ], + "3min_transcript": "In this video I want to do a couple of inequality problems that are deceptively tricky. And you might be saying, hey, aren't all inequality problems deceptively tricky? And on some level you're probably right. But let's start with the first problem. We have x minus 1 over x plus 2 is greater than 0. And I'm actually going to show you two ways to do this. The first way is, I think, on some level, the simpler way. But I'll show you both methods and whatever works for you, well, it works for you. So the first way you can think about this, if I have just any number divided by any other number and I say that they're going to be greater than 0. Well, we just have to remember our properties of multiplying and dividing negative numbers. In what situation is this fraction going to be greater than 0? Well, this is going to be greater than 0 only if both a and-- so we could write both a is greater than 0 and b is greater than 0. So this is one circumstance where this'll definitely be true. definitely be a positive. It'll definitely be greater than 0. Or we could have the situation where we have a negative divided by a negative. If we have the same sign divided by the same sign we're also going to be positive. So or-- a is less than 0 and b is less than 0. So whenever you have any type of rational expression like this being greater than 0, there's two situations in which it will be true. The numerator and the denominator are both greater than 0, or they're both less than 0. So let's remember that and actually do this problem. So there's two situations to solve this problem. The first is where both of them are greater than 0. If that and that are both greater than 0, we're cool. So we could say our first solution-- maybe I'll draw a little tree like that-- is x minus 1 greater than 0 and x plus 2 greater than 0. That's equivalent to this. The top and the bottom-- if they're both greater than 0 something greater than 0. The other option-- we just saw that-- is if both of them were less than 0. So the other option is x minus 1 less than 0 and x plus 2 less than 0. If both of these are less than 0 then you have a negative divided by a negative, which will be positive. Which will be greater than 0. So let's actually solve in both of these circumstances. So x minus 1 greater than 0. If we add 1 to both sides of that we get x is greater than 1. And if we do x plus 2 greater than 0, if we subtract 2 from both sides of that equation-- remember I'm doing this right now-- we get x is greater than minus 2. So for both of these to hold true-- so in this little brown or red color, whatever you want to think of it-- in order for both of these to go hold true, x has to be greater than 1 and" + }, + { + "Q": "\nAt around 3:40 confused me a lot. He says you can re-write the equation as \"|x - -2|=6\", and then find that the answers are 6 away from -2. While that does appear true, why does it work for that equation, and not the original equation \"|x+2|=6\"? How did Sal know to re-write the \"+2\" as \"- -2\"?", + "A": "The minuses cancel out. Thus, subtracting negative 2 is eqall to adding 2.", + "video_name": "u6zDpUL5RkU", + "timestamps": [ + 220 + ], + "3min_transcript": "Now, to solve this one, add 5 to both sides of this equation. You get x is equal to 15. To solve this one, add 5 to both sides of this equation. x is equal to negative 5. So our solution, there's two x's that satisfy this equation. x could be 15. 15 minus 5 is 10, take the absolute value, you're going to get 10, or x could be negative 5. Negative 5 minus 5 is negative 10. Take the absolute value, you get 10. And notice, both of these numbers are exactly 10 away from the number 5. Let's do another one of these. Let's do another one. Let's say we have the absolute value of x plus 2 is equal to 6. So what does that tell us? That tells us that either x plus 2, that the thing inside Or the thing inside of the absolute value sign, the x plus 2, could also be negative 6. If this whole thing evaluated to negative 6, you take the absolute value, you'd get 6. So, or x plus 2 could equal negative 6. And then if you subtract 2 from both sides of this equation, you get x could be equal to 4. If you subtract 2 from both sides of this equation, you get x could be equal to negative 8. So these are the two solutions to the equation. And just to kind of have it gel in your mind, that absolute value, you can kind of view it as a distance, you could rewrite this problem as the absolute value of x minus negative 2 is equal to 6. And so this is asking me, what are the x's that are exactly 6 away from negative 2? Remember, up here we said, what are the x's that are exactly 10 away from positive 5? these are both 10 away from positive 5. This is asking, what is exactly 6 away from negative 2? And it's going to be 4, or negative 8. You could try those numbers out for yourself. Let's do another one of these. Let's do another one, and we'll do it in purple. Let's say we have the absolute value of 4x-- I'm going to change this problem up a little bit. 4x minus 1. The absolute value of 4x minus 1, is equal to-- actually, I'll just keep it-- is equal to 19. So, just like the last few problems, 4x minus 1 could be equal to 19. Or 4x minus 1 might evaluate to negative 19. Because then when you take the absolute value, you're going to get 19 again. Or 4x minus 1 could be equal to negative 19. Then you just solve these two equations. Add 1 to both sides of this equation-- we could do them" + }, + { + "Q": "\nat 1:40 how did he remove the absolute value portion? Did he just make it into an equation? Can someone please clarify.", + "A": "Sal is solving the absolute value equation of |x-5|=10. To solve this, you have two possible choices (since we have the absolute value there). Either x-5 could be 10 or x-5 could be -10. (because | -10 | = 10. So he is solving the absolute value equation by taking what s inside of it and setting it equal to 10 and -10. x-5=10 and therefore x = 15. x-5=-10 and therefore x= -5. So our two answers are 15 and -5.", + "video_name": "u6zDpUL5RkU", + "timestamps": [ + 100 + ], + "3min_transcript": "Let's do some equations that deal with absolute values. And just as a bit of a review, when you take the absolute value of a number. Let's say I take the absolute value of negative 1. What you're really doing is you're saying, how far is that number from 0? And in the case of negative 1, if we draw a number line right there-- that's a very badly drawn number line. If we draw a number line right there, that's 0. You have a negative 1 right there. Well, it's 1 away from 0. So the absolute value of negative 1 is 1. And the absolute value of 1 is also 1 away from 0. It's also equal to 1. So on some level, absolute value is the distance from 0. But another, I guess simpler way to think of it, it always results in the positive version of the number. The absolute value of negative 7,346 is equal to 7,346. So with that in mind, let's try to solve some equations So let's say I have the equation the absolute value of x minus 5 is equal to 10. And one way you can interpret this, and I want you to think about this, this is actually saying that the distance between x and 5 is equal to 10. So how many numbers that are exactly 10 away from 5? And you can already think of the solution to this equation, but I'll show you how to solve it systematically. Now this is going to be true in two situations. Either x minus 5 is equal to positive 10. If this evaluates out to positive 10, then when you take the absolute value of it, you're going to get positive 10. Or x minus 5 might evaluate to negative 10. If x minus 5 evaluated to negative 10, when you take the absolute value of it, you would get 10 again. So x minus 5 could also be equal to negative 10. Now, to solve this one, add 5 to both sides of this equation. You get x is equal to 15. To solve this one, add 5 to both sides of this equation. x is equal to negative 5. So our solution, there's two x's that satisfy this equation. x could be 15. 15 minus 5 is 10, take the absolute value, you're going to get 10, or x could be negative 5. Negative 5 minus 5 is negative 10. Take the absolute value, you get 10. And notice, both of these numbers are exactly 10 away from the number 5. Let's do another one of these. Let's do another one. Let's say we have the absolute value of x plus 2 is equal to 6. So what does that tell us? That tells us that either x plus 2, that the thing inside" + }, + { + "Q": "6:27 Why isn't it x+3=y or x+3=-y?\n", + "A": "think of what a basic absolute value y = | x | looks like, it is a V shape with the vertex at the origin. Thus for any value of y except 0 (y values are horizontal lines), the value of x would be +y or -y. If y = 1, we would have two points (1,1) and (-1,1), We expand this property into more complicated square roots, so y = x+3 or y = - (x+3)", + "video_name": "u6zDpUL5RkU", + "timestamps": [ + 387 + ], + "3min_transcript": "these are both 10 away from positive 5. This is asking, what is exactly 6 away from negative 2? And it's going to be 4, or negative 8. You could try those numbers out for yourself. Let's do another one of these. Let's do another one, and we'll do it in purple. Let's say we have the absolute value of 4x-- I'm going to change this problem up a little bit. 4x minus 1. The absolute value of 4x minus 1, is equal to-- actually, I'll just keep it-- is equal to 19. So, just like the last few problems, 4x minus 1 could be equal to 19. Or 4x minus 1 might evaluate to negative 19. Because then when you take the absolute value, you're going to get 19 again. Or 4x minus 1 could be equal to negative 19. Then you just solve these two equations. Add 1 to both sides of this equation-- we could do them Add 1 to both sides of this equation, you get 4x is equal to negative 18. Divide both sides of this by 4, you get x is equal to 5. Divide both sides of this by 4, you get x is equal to negative 18/4, which is equal to negative 9/2. So both of these x values satisfy the equation. Try it out. Negative 9/2 times 4. This will become a negative 18. Negative 18 minus 1 is negative 19. Take the absolute value, you get 19. You put a 5 here, 4 times 5 is 20. Minus 1 is positive 19. So you take the absolute value. Once again, you'll get a 19. Let's try to graph one of these, just for fun. So let's say I have y is equal to the absolute So this is a function, or a graph, with an absolute value in it. So let's think about two scenarios. There's one scenario where the thing inside of the absolute value is positive. So you have the scenario where x plus 3-- I'll write it over here-- x plus 3 is greater than 0. And then you have the scenario where x plus 3 is less than 0. When x plus 3 is greater than 0, this graph, or this line-- or I guess we can't call it a line-- this function, is the same thing as y is equal to x plus 3. If this thing over here is greater than 0, then the absolute value sign is irrelevant. So then this thing is the same thing as y is equal to x plus 3. But when is x plus 3 greater than 0? Well, if you subtract 3 from both sides, you get x is greater than negative 3. So when x is greater than negative 3, this graph is" + }, + { + "Q": "\nyou lost me at around 1:35 to 2:00. I don't get how 1to the 0 power could equal to 1. isn't the power of powers multiplication? shouldn't 1x0=0?", + "A": "Think of it this way: x\u00c2\u00b3/x\u00c2\u00b2 = x\u00c2\u00b3\u00e2\u0081\u00bb\u00c2\u00b2 = x\u00c2\u00b9 = x Similarly: x\u00c2\u00b3/x\u00c2\u00b3 = x\u00c2\u00b3\u00e2\u0081\u00bb\u00c2\u00b3 = x\u00e2\u0081\u00b0 but, of course, x\u00c2\u00b3/x\u00c2\u00b3 = 1, Thus, x\u00e2\u0081\u00b0=1 Of course, this doesn t work if x=0 , 0\u00e2\u0081\u00b0 is undefined.", + "video_name": "NEaLgGi4Vh4", + "timestamps": [ + 95, + 120 + ], + "3min_transcript": "Evaluate the expression 4n to the first power minus 2n to the zeroth power for n equals 1 and n equals 5. So, let's do n equals 1 first. So, every place we see an n, let's put a 1 in there. So this is equal to 4 times 1 to the first power minus 2 times 1 to the zeroth power. Every place we saw an n, we have an n there. We have an n to the first, one to the first. So remember, we want to do our order of operations. And I'm going to write it this way just to make it clear that we do parentheses then exponentiation. Then multiplication and division are at the same level, they get the same priority. And then addition and subtraction get the same priority. So, we want to do our exponents before we do anything else. There are no parentheses here. So, we first want to evaluate one to the first power. And anything to the first power is just that anything. If I just have any number to the first power, that is just going to be equal to that number. So, 1 to the first power is just equal to 1. Let me write it. 4 times 1 minus 2 times 1 to the zeroth power. Now, any number other than 0 to the zeroth power is going to be 1. So, if I tell you x to the zeroth power is equal to what? And I tell you that x is not equal to 0. What is x to the zeroth power? You should immediately know that it is equal to 1. I exclude 0 because 0 to the zeroth power is usually undefined. Or for our purposes, you can accept it to be undefined. And maybe in the future, I'll do a video on explaining why it's undefined. There's some good arguments for why this should be 0. Because 0 to pretty much any other power is a 0. But then there's a really good argument for why it should be 1. Because any other number to the zeroth power is 1. So that's why, for the most part, we like to keep this undefined. So with that said, 1 to the zeroth power Anything other than 0 to the zeroth power is 1. So, we have 4 times 1, which is 4, minus 2 times 1, which is 2. So, 4 minus 2 is just equal to 2. Now, let's do the same thing with n equals 5. So it becomes 4 times 5 to the first power minus 2 times 5 to the zeroth power. Same thing. 5 to the first power is just going to be equal to 5. So, I want to do that in a different color. 5 to the first power is just going to be equal to 5. 5 to the zeroth power is just going to be equal to 1. And so, this expression becomes 4 times 5 minus 2 times 1. Or 4 times 5 is 20, minus 2 times 1, that is 2. And this is equal to 18. And we're done." + }, + { + "Q": "As well as my previous question, at 1:19 sal said that any number to the zeroth power is going to be 1. Does that mean that 8 to the zeroth power will be 1, or that 8 to the zeroth power will be 8?\n", + "A": "It means that 8^0 = 1, even 100^0 = 1, any number to the zero-th power is equal to 1.", + "video_name": "NEaLgGi4Vh4", + "timestamps": [ + 79 + ], + "3min_transcript": "Evaluate the expression 4n to the first power minus 2n to the zeroth power for n equals 1 and n equals 5. So, let's do n equals 1 first. So, every place we see an n, let's put a 1 in there. So this is equal to 4 times 1 to the first power minus 2 times 1 to the zeroth power. Every place we saw an n, we have an n there. We have an n to the first, one to the first. So remember, we want to do our order of operations. And I'm going to write it this way just to make it clear that we do parentheses then exponentiation. Then multiplication and division are at the same level, they get the same priority. And then addition and subtraction get the same priority. So, we want to do our exponents before we do anything else. There are no parentheses here. So, we first want to evaluate one to the first power. And anything to the first power is just that anything. If I just have any number to the first power, that is just going to be equal to that number. So, 1 to the first power is just equal to 1. Let me write it. 4 times 1 minus 2 times 1 to the zeroth power. Now, any number other than 0 to the zeroth power is going to be 1. So, if I tell you x to the zeroth power is equal to what? And I tell you that x is not equal to 0. What is x to the zeroth power? You should immediately know that it is equal to 1. I exclude 0 because 0 to the zeroth power is usually undefined. Or for our purposes, you can accept it to be undefined. And maybe in the future, I'll do a video on explaining why it's undefined. There's some good arguments for why this should be 0. Because 0 to pretty much any other power is a 0. But then there's a really good argument for why it should be 1. Because any other number to the zeroth power is 1. So that's why, for the most part, we like to keep this undefined. So with that said, 1 to the zeroth power Anything other than 0 to the zeroth power is 1. So, we have 4 times 1, which is 4, minus 2 times 1, which is 2. So, 4 minus 2 is just equal to 2. Now, let's do the same thing with n equals 5. So it becomes 4 times 5 to the first power minus 2 times 5 to the zeroth power. Same thing. 5 to the first power is just going to be equal to 5. So, I want to do that in a different color. 5 to the first power is just going to be equal to 5. 5 to the zeroth power is just going to be equal to 1. And so, this expression becomes 4 times 5 minus 2 times 1. Or 4 times 5 is 20, minus 2 times 1, that is 2. And this is equal to 18. And we're done." + }, + { + "Q": "At 0:00 who is the ancient philosopher?\n", + "A": "The video starts out at 0:00 with So you, as the ancient philosopher . So is he you?", + "video_name": "pzQY-9Nmtws", + "timestamps": [ + 0 + ], + "3min_transcript": "So you, as the ancient philosopher in mathematics have concluded in order for the multiplication of positive and negative numbers to be consistent with everything you've been constructing so far with all the other properties of multiplication that you know so far that you need a negative number times a positive number or a positive times a negative to give you a negative number and a negative times a negative to give you a positive number and so you accept it's all consistent so far.. this deal does not make complete concrete sense to you, you want to have a slightly deeper institution than just having to accept its consistent with the distributive property and whatever else and so you try another thought experiment, you say \"well what is just a basic multiplication way of doing it?\" So if I say, two times three, one way to to conceptualize is basic multiplication is really repeating addition, so you could view this as two threes so let me write three plus three or you could view this as three twos, and so this is the same thing as two plus two plus two and there are three of them, and either way you can conceptualize as you get the same exact answer. This is going to be equal to six, fair enough! Now, you knew this before you even tried to tackle negative numbers. Now let's try to make one of these negatives and see what happens. Let's do two times negative three, I want to make the negative into a different color. Two times negative three. Well, one way you could view this is the same analogy here, it's negative three twice so it would be negative.. I'll try to color code it negative three and then another negative or, and this is the interesting thing, instead of over here there's a two times positive three you added two, three times. But since here is two times negative three you could also imagine you are going to subtract two, three times So instead of up here, I could written two plus two plus two because this is a positive two right over here, but since we're doing this over negative three we could imagine subtracting two, three times, so this would be subtracting two (repeated) subtract another two right over here, subtract another two and then you subtract another two notice you did it, once again, you did it three times, so this is a negative three, so essentially you are subtracting two, three times. And either way, you can conceptualize" + }, + { + "Q": "\nAt 5:07 Why does he simplify (1/x)(x-1) to (x-1)/x. Could he also have multiplied (1/x) by both halves? so (1/x)(x-1) = (x/x)-(1/x)= -(1/x). I am wondering if these are two different simplifications or if my answer is wrong.", + "A": "You ve made an algebra error where you say (x/x)-(1/x) = -(1/x). The expression (x/x) is equal to 1, so after the equals sign you should have 1 - (1/x), which evaluates to 0 when x = 1.", + "video_name": "MeVFZjT-ABM", + "timestamps": [ + 307 + ], + "3min_transcript": "over the derivative of that, that that limit exists. So let's try to do it. So this is going to be equal to, if the limit exists, this is going to be equal to the limit as x approaches 1. And let's take the derivative in magenta, I'll take the derivative of this numerator right over here. And for this first term, just do the product rule. Derivative of x is one, and then so 1 times the natural log of x, the derivative of the first term times the second term. And then we're going to have plus the derivative of the second term plus 1 over x times the first term. It's just the product rule. So 1 over x times x, we're going to see, that's just 1, and then we have minus the derivative of x minus 1. Well, the derivative of x minus 1 is just 1, so it's just going to be minus 1. So let's take the derivative of that, over here. So the derivative of the first term, of x minus 1, is just 1. Multiply that times the second term, you get natural log of x. And then plus the derivative of the second term, derivative of natural log of x is one over x, times x minus 1. I think we can simplify this a little bit. This 1 over x times x, that's a 1. We're going to subtract one from it. So these cancel out, right there. And so this whole expression can be rewritten as the limit as approaches 1, the numerator is just natural log of x, do that in magenta, and the denominator is the natural log of x plus x minus 1 over x So if we take x approaches one of natural log of x, that will give us a, well, natural log of 1 is 0. And over here, we get natural log of 1, which is 0. And then plus 1 minus 1 over plus 1 minus 1 over 1, well, that's just going to be another 0. 1 minus 1 is zero. So you're going to have 0 plus 0. So you're going to get a 0 over 0 again. 0 over 0. So once again, let's apply l'Hopital's rule again. Let's take the derivative of that, put it over the derivative of that. So this, if we're ever going to get to a limit, is going to be equal to the limit as x approaches 1 of the derivative of the numerator, 1 over x, right, the derivative of ln of x is 1/x, over the derivative of the denominator. And what's that?" + }, + { + "Q": "4:30 Why doesn't the lnx in the numerator cancel with the lnx in the denominator?\n", + "A": "In the denominator, you are adding (+) two things and thus you can not cancel it out with the numerator", + "video_name": "MeVFZjT-ABM", + "timestamps": [ + 270 + ], + "3min_transcript": "over x minus 1, because the natural log of x's cancel out. Let me get rid of that. And then this right here is the same thing as 1 over natural log of x, because the x minus 1's cancel out. So hopefully you realize, all I did is I added these two expressions. So given that, let's see what happens if I take the limit as x approaches 1 of this thing. Because these are the same thing. Do we get anything more interesting? So what do we have here? We have one times the natural log of 1. The natural log of 1 is 0, so we have 0 here, so that is a 0. Minus 1 minus 0, so that's going to be another 0, minus 0. So we get a 0 in the numerator. And in the denominator we get a 1 minus 1, which is 0, times the natural log of 1, which is 0, so 0 times 0, that is 0. We have indeterminate form that we need for l'Hopital's rule, over the derivative of that, that that limit exists. So let's try to do it. So this is going to be equal to, if the limit exists, this is going to be equal to the limit as x approaches 1. And let's take the derivative in magenta, I'll take the derivative of this numerator right over here. And for this first term, just do the product rule. Derivative of x is one, and then so 1 times the natural log of x, the derivative of the first term times the second term. And then we're going to have plus the derivative of the second term plus 1 over x times the first term. It's just the product rule. So 1 over x times x, we're going to see, that's just 1, and then we have minus the derivative of x minus 1. Well, the derivative of x minus 1 is just 1, so it's just going to be minus 1. So let's take the derivative of that, over here. So the derivative of the first term, of x minus 1, is just 1. Multiply that times the second term, you get natural log of x. And then plus the derivative of the second term, derivative of natural log of x is one over x, times x minus 1. I think we can simplify this a little bit. This 1 over x times x, that's a 1. We're going to subtract one from it. So these cancel out, right there. And so this whole expression can be rewritten as the limit as approaches 1, the numerator is just natural log of x, do that in magenta, and the denominator is the natural log of x plus x minus 1 over x" + }, + { + "Q": "\nSo, wait at 5:05 he got rid of a 1? How can he just remove a 1 that he was adding?", + "A": "It was +1 - 1 so he just canceled them out", + "video_name": "MeVFZjT-ABM", + "timestamps": [ + 305 + ], + "3min_transcript": "over the derivative of that, that that limit exists. So let's try to do it. So this is going to be equal to, if the limit exists, this is going to be equal to the limit as x approaches 1. And let's take the derivative in magenta, I'll take the derivative of this numerator right over here. And for this first term, just do the product rule. Derivative of x is one, and then so 1 times the natural log of x, the derivative of the first term times the second term. And then we're going to have plus the derivative of the second term plus 1 over x times the first term. It's just the product rule. So 1 over x times x, we're going to see, that's just 1, and then we have minus the derivative of x minus 1. Well, the derivative of x minus 1 is just 1, so it's just going to be minus 1. So let's take the derivative of that, over here. So the derivative of the first term, of x minus 1, is just 1. Multiply that times the second term, you get natural log of x. And then plus the derivative of the second term, derivative of natural log of x is one over x, times x minus 1. I think we can simplify this a little bit. This 1 over x times x, that's a 1. We're going to subtract one from it. So these cancel out, right there. And so this whole expression can be rewritten as the limit as approaches 1, the numerator is just natural log of x, do that in magenta, and the denominator is the natural log of x plus x minus 1 over x So if we take x approaches one of natural log of x, that will give us a, well, natural log of 1 is 0. And over here, we get natural log of 1, which is 0. And then plus 1 minus 1 over plus 1 minus 1 over 1, well, that's just going to be another 0. 1 minus 1 is zero. So you're going to have 0 plus 0. So you're going to get a 0 over 0 again. 0 over 0. So once again, let's apply l'Hopital's rule again. Let's take the derivative of that, put it over the derivative of that. So this, if we're ever going to get to a limit, is going to be equal to the limit as x approaches 1 of the derivative of the numerator, 1 over x, right, the derivative of ln of x is 1/x, over the derivative of the denominator. And what's that?" + }, + { + "Q": "\nAt 3:49, how is the derivative of x-1 = 1. I thought the derivative of x is 1 so, that 1 minus the 1 should equal 0.", + "A": "You re really taking the derivative of x (which is 1) and the derivative of -1 (which is 0). (1) - (0) = 1. The derivative of -1 is 0 because -1 is a constant. The derivative of any constant is 0. If you imagine a graph with y=-1 (or any constant) the line is horizontal. The tangent line of any point on the graph will also be horizontal, and the slope of a horizontal line is zero.", + "video_name": "MeVFZjT-ABM", + "timestamps": [ + 229 + ], + "3min_transcript": "over x minus 1, because the natural log of x's cancel out. Let me get rid of that. And then this right here is the same thing as 1 over natural log of x, because the x minus 1's cancel out. So hopefully you realize, all I did is I added these two expressions. So given that, let's see what happens if I take the limit as x approaches 1 of this thing. Because these are the same thing. Do we get anything more interesting? So what do we have here? We have one times the natural log of 1. The natural log of 1 is 0, so we have 0 here, so that is a 0. Minus 1 minus 0, so that's going to be another 0, minus 0. So we get a 0 in the numerator. And in the denominator we get a 1 minus 1, which is 0, times the natural log of 1, which is 0, so 0 times 0, that is 0. We have indeterminate form that we need for l'Hopital's rule, over the derivative of that, that that limit exists. So let's try to do it. So this is going to be equal to, if the limit exists, this is going to be equal to the limit as x approaches 1. And let's take the derivative in magenta, I'll take the derivative of this numerator right over here. And for this first term, just do the product rule. Derivative of x is one, and then so 1 times the natural log of x, the derivative of the first term times the second term. And then we're going to have plus the derivative of the second term plus 1 over x times the first term. It's just the product rule. So 1 over x times x, we're going to see, that's just 1, and then we have minus the derivative of x minus 1. Well, the derivative of x minus 1 is just 1, so it's just going to be minus 1. So let's take the derivative of that, over here. So the derivative of the first term, of x minus 1, is just 1. Multiply that times the second term, you get natural log of x. And then plus the derivative of the second term, derivative of natural log of x is one over x, times x minus 1. I think we can simplify this a little bit. This 1 over x times x, that's a 1. We're going to subtract one from it. So these cancel out, right there. And so this whole expression can be rewritten as the limit as approaches 1, the numerator is just natural log of x, do that in magenta, and the denominator is the natural log of x plus x minus 1 over x" + }, + { + "Q": "\nMy memory of math is a bit fuzzy so this might be a stupid question. At 9:00, why did Sal randomly divide the equation 2C1 + 3C1 = 0 by one half?", + "A": "He didn t divide by 1/2, he multiplied by 1/2. He did that to get the same coefficient on the C1 terms in both equations. After that, he can subtract one equation from the other to get rid of C1 altogether. Another possibility would have been to multiply the second equation by 2.", + "video_name": "Alhcv5d_XOs", + "timestamps": [ + 540 + ], + "3min_transcript": "In order for them to be linearly dependent, that means that if some constant times 2,1 plus some other constant times this second vector, 3,2 where this should be equal to 0. Where these both aren't necessarily 0. Before I go up for this problem, let's remember what we're going to find out. If either of these are non-zero, if c1 or c2 are non-zero, then this implies that we are dealing with a dependent, linearly dependent set. If c1 and c2 are both 0, if the only way to satisfy this equation -- I mean you can always satisfy it by sitting guys 0, then we're dealing with a linearly independent set. Let's try to do some math. And this'll just take us back to our Algebra 1 days. In order for this to be true, that means 2 times c1 plus 3 times c2 is equal to -- when I say this is equal to 0, it's really the 0 vector. I can rewrite this as 0,0. So 2 times c1 plus 3 times c2 would be equal to that 0 there. And then we'd have 1 times c1 plus 2 times c2 is equal to that 0. And now this is just a system, two equations, two unknowns. A couple of things we could do. Let's just multiply this top equation by 1/2. equal to 0. And then if we subtract the green equation from the red equation this becomes 0. 2 minus 1 and 1/2-- 3/2 is 1 and 1/2 --of this is just 1/2 c2 is equal to 0. And this is easy to solve. c2 is equal to 0. So what's c1? Well, just substitute this back in. c2 is equal to 0. So this is equal to 0. So c1 plus 0 is equal to 0. So c1 is also equal to 0. We could have substituted it back into that top equation as well. So the only solution to this equation involves both c1 and c2 being equal to 0. So they both have to be 0. So this is a linearly independent set of vectors." + }, + { + "Q": "At 3:00 why did you not start writing from a1v1 but rather a2v2?\n", + "A": "It is his way of showing that v1 can be made from a linear combination of the other vectors. If he included v1 in there, then that would be saying that v1 is a linear combination of itself and the other vectors, which is true for all vectors, all the time.", + "video_name": "Alhcv5d_XOs", + "timestamps": [ + 180 + ], + "3min_transcript": "this into the 0 vector. Sometimes it's just written as a bold 0, and sometimes you could just write it -- I mean we don't know the dimensionality of this vector. It would be a bunch of 0's. We don't know how many actual elements are in each of these vectors, but you get the idea. My set of vectors is linearly dependent-- remember I'm saying dependent, not independent --is linearly dependant, if and only if I can satisfy this equation for some ci's where not all of them are equal to 0. This is key, not all are 0. Or you could say it the other way. You could say at least one is non-zero. previous video where I said look, a set is linearly dependent if one of the of vectors can be represented by the combination of the other vectors? Let me write that down. In the last few I said, look, one vector can be -- Let me write it this way. One vector being represented by the some of the other vectors, I can just write it like this. I can write it a little bit more math-y. In the last video, I said that linear dependence means that-- let me just pick an arbitrary vector, v1. Let's say that v1, you know this is arbitrary, v1 one could be represented by some combination of the other vectors. Let me call them a1 times v -- let me be careful -- a2 times This is what we said in the previous video. If this is linear dependence, any one of these guys can be represented as some combination of the other ones. So how does this imply that? In order show this if and only if, I have to show that this implies that and I have to show that that implies this. So this is almost a trivially easy proof. Because if I subtract v1 from both sides of this equation I get 0 is equal to minus 1 v1 plus a2 v2 plus a3 v3 all the way to an vn. And clearly I've just said, well, this is linearly dependent. That means that I can represent this vector as a sum of the other vectors, which means that minus 1 times v1" + }, + { + "Q": "\nI have a question regarding the giveaway problem (around 12:00). Sal says that if there are three two-dimensional vectors in a set, the set will definitely be linearly dependent.\nWhat if:\nv1 = [1,0]\nv2=[2,0]\nv3=[7,8]\nin this case, v1 and v2 are linearly dependent since 2*v1 = v2\nBut there is no way v3 can be represented by the linear combination of v1 and v2, how could this set of these three vectors be linearly dependent?", + "A": "That is an example of a set of three linearly dependent vectors, for the reason you describe. If you understood Sal to mean that a linearly dependent set means that ANY of the vectors is a linear combination of the others, that is incorrect. It is just necessary that at least one of the vectors is a linear combination of the others.", + "video_name": "Alhcv5d_XOs", + "timestamps": [ + 720 + ], + "3min_transcript": "of the other one. You can't represent one as a combination of the other. And since we have two vectors here, and they're linearly independent, we can actually know that this will span r2. The span of my r vectors is equal to r2. If one of these vectors was just some multiple of the other, than the span would have been some line within r2, not all of. But now I can represent any vector in r2 as some combination of those. Let's do another example. Let me scroll to the right, because sometimes this thing, when I go too far down, I haven't figured out why, when I go too far down it starts messing up. So my next example is the set of vectors. So I have the vector 2,1. I have the vector 3,2. And I want to know are these linearly dependent or linearly independent. So I go to through the same drill. I use that little theorem that I proved at the beginning of this video. In order for them to be linearly dependent there must be some set of weights that I can multiply these guys. So c1 times this vector plus c2 times this vector plus c3 times that vector, that will equal the 0 vector. And if one of these is non-zero then we're dealing with a linearly dependent set of vectors. And if all of them are 0, then it's independent. Let's just do our linear algebra. So this means that 2 times c1 plus 3 times c2 plus c3 is And then if we do the bottom rows-- Remember when you multiply a scalar times a vector you multiply it by each of these terms. So c1 times 1. 1c1 plus 2c2 plus 2c3 is equal to 0. There's a couple of giveaways on this problem. If you have three two-dimensional vectors, one of them is going to be redundant. Because, in the very best case, even if you assume that that vector and that vector are linearly independent, then these would span r2. Which means that any point, any vector, in your two-dimensional space can be represented by some combination of those two. In which case, this is going to be one of them because this is just a vector in two-dimensional space. So it would be linearly dependent. And then, if you say, well, these aren't linearly" + }, + { + "Q": "At 0:39, how is this about subtracting formula for a cosine? I am interested in knowing how to solve for cos(u+v)=cos u cos v + sin u sin v\n", + "A": "cos(u + v) is actually cos(u)cos(v) - sin(u)sin(v).", + "video_name": "D_smr0GBPvA", + "timestamps": [ + 39 + ], + "3min_transcript": "We have triangle ABC here, which looks like a right triangle. And we know it's a right triangle because 3 squared plus 4 squared is equal to 5 squared. And they want us to figure out what cosine of 2 times angle So that's this angle-- ABC. Well, we can't immediately evaluate that, but we do know what the cosine of angle ABC is. We know that the cosine of angle ABC-- well, cosine is just adjacent over hypotenuse. It's going to be equal to 3/5. And similarly, we know what the sine of angle ABC is. That's opposite over hypotenuse. That is 4/5. So if we could break this down into just cosines of ABC and sines of ABC, then we'll be able to evaluate it. And lucky for us, we have a trig identity at our disposal that does exactly that. We know that the cosine of 2 times an angle is equal to cosine of that angle squared And we've proved this in other videos, but this becomes very helpful for us here. Because now we know that the cosine-- Let me do this in a different color. Now, we know that the cosine of angle ABC is going to be equal to-- oh, sorry. It's the cosine of 2 times the angle ABC. That's what we care about. 2 times the angle ABC is going to be equal to the cosine of angle ABC squared minus sine of the angle ABC squared. And we know what these things are. This thing right over here is just going to be equal to 3/5 squared. Cosine of angle a ABC is 3/5. So we're going to square it. And this right over here is just 4/5 squared. And so this simplifies to 9/25 minus 16/25, which is equal to 7/25. Sorry. It's negative. Got to be careful there. 16 is larger than 9. Negative 7/25. Now, one thing that might jump at you is, why did I get a negative value here when I doubled the angle here? Because the cosine was clearly a positive number. And there you just have to think of the unit circle-- which we already know the unit circle definition of trig functions is an extension of the Sohcahtoa definition. Y-axis. Let me draw a unit circle here. My best attempt. So that's our unit circle. So this angle right over here looks like something like this." + }, + { + "Q": "At 9:30, was Sal supposed to write a \"t\" after the \"cos^2\"?\n", + "A": "yes he was trying to write cos^2t", + "video_name": "AFF8FXxt5os", + "timestamps": [ + 570 + ], + "3min_transcript": "but let me rewrite our vector field in terms of in terms of t, so to speak. So what's our field going to be doing at any point t? We don't have to worry about every point. We don't have to worry, for example, that over here the vector field is going to be doing something like that because that's not on our path. That force never had an impact on the particle. We only care about what happens along our path. So we can find a function that we can essentially substitute y and x for, their relative functions with respect to t, and then we'll have the force from the field at any point or any time t. So let's do that. So this guy right here, if I were to write it as a function of t, this is going to be equal to y of t, right? y is a function of t, so it's sine of t, right? that's that. Sine of t times i plus-- or actually minus x, or x of So minus cosine of t times j. And now all of it seems a little bit more straightforward. If we want to find this line integral, this line integral is going to be the same thing as the integral-- let me pick a nice, soothing color. Maybe this is a nice one. The integral from t is equal to 0 to t is equal to 2pi of f dot dr. Now, when you take the dot product, you just multiply the corresponding components, and add it up. So we take the product of the minus sign and the sine of t-- or the sine of t with the minus sine of t dt, I get-- you're going to get minus sine squared t dt, and then you're going to add that to-- so you're going to have that plus. Let me write that dt a little bit. dt, and then you're going to have that plus these two guys multiplied by each other. So that's-- well, there's a minus to sign here so plus. Let me just change this to a minus. Minus cosine squared dt. And if we factor out a minus sign and a dt, what is this going to be equal to? This is going to be equal to the integral from 0 to 2pi of, we could say, sine squared plus-- I want to put the t -- sine squared of t plus cosine squared of t. And actually, let me take the minus sign out to the front. So if we just factor the minus sign, and put a minus there, make this a plus. So the minus sign out there, and then we factor dt out. I did a couple of steps in there, but I think you got it. Now this is just algebra at this point. Factoring out a minus sign, so this becomes positive. And then you have a dt and a dt. Factor that out, and you get this." + }, + { + "Q": "\nat 08:25, why the equation of normal line is: y-x0^2=-(1/2x0)*(x-x0)?", + "A": "i don t get something here if the y =2x and we want to get a perpendicular line to that one shouldn t it be (-1/2)*x ? as i know the to perpendicular lines have some equations like: m1*m2=-1 so 2*m2=-1 m2 would be -1/2 and then the new line perpendicular to the first one would be just (-1/2)*x please tell me where i am wrong", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 505 + ], + "3min_transcript": "This is the slope of the tangent at any point x. So if I want to know the slope of the tangent at x0, at some particular x, I would just say, well, let me just say, slope, it would be 2 x0. Or let me just say, f of x0 is equal to 2 x0. This is the slope at any particular x0 of the tangent line. Now, the normal line slope is perpendicular to this. So the perpendicular line, and I won't review it here, but the perpendicular line has a negative inverse slope. So the slope of normal line at x0 will be the negative inverse of this, because this is the slope of the tangent line x0. So it'll be equal to minus 1 over 2 x0. Now, what is the equation of the normal line at x0 let's say that this is my x0 in question. What is the equation of the normal line there? Well, we can just use the point-slope form of our equation. So this point right here will be on the normal line. And that's the point x0 squared. Because this the graph of y equals x0, x squared. So this normal line will also have this point. So we could say that the equation of the normal line, let me write it down, would be equal to, this is just a point-slope definition of a line. You say, y minus the y-point, which is just x0 squared, that's that right there, is equal to the slope of the normal line minus 1 over 2 x0 times x minus the Minus x, minus x0. This is the equation of the normal line. So let's see. And what we care about is when x0 is greater than 0, right? We care about the normal line when we're in the first quadrant, we're in all of these values right there. So that's my equation of the normal line. And let's solve it explicitly in terms of x. So y is a function of x. Well, if I add x0 squared to both sides, I get y is equal to, actually, let me multiply this guy out. I get minus 1/2 x0 times x, and then I have plus, plus, because I have a minus times a minus, plus 1/2. The x0 and the over the x0, they cancel out. And then I have to add this x0 to both sides." + }, + { + "Q": "At 2:10 and 4:30, when the questions says the parabola gets smaller and after that starts to increase, isn't it referring to the y values?\n", + "A": "The problem refers to the x coordinates but have the poor use of words and Sal explains what it means by getting smaller and increase . Although, there is a relation as x is getting smaller so is y.", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 130, + 270 + ], + "3min_transcript": "I just got sent this problem, and it's a pretty meaty problem. A lot harder than what you'd normally find in most textbooks. So I thought it would help us all to work it out. And it's one of those problems that when you first read it, your eyes kind of glaze over, but when you understand what they're talking about, it's reasonably interesting. So they say, the curve in the figure above is the parabola y is equal to x squared. So this curve right there is y is equal to x squared. Let us define a normal line as a line whose first quadrant intersection with the parabola is perpendicular to the parabola. So this is the first quadrant, right here. And they're saying that a normal line is something, when the first quadrant intersection with the parabola is normal to the parabola. So if I were to draw a tangent line right there, this line is normal to that tangent line. That's all that's saying. So this is a normal line, right there. Normal line. Fair enough. 5 normal lines are shown in the figure. 1, 2, 3, 4, 5. Good enough. And these all look perpendicular, or normal to so that makes sense. For a while, the x-coordinate of the second quadrant intersection of the normal line of the parabola gets smaller, as the x-coordinate of the first quadrant intersection gets smaller. So let's see what happens as the x-quadrant of the first intersection gets smaller. So this is where I left off in that dense text. So if I start at this point right here, my x-coordinate right there would look something like this. Let me go down. my x-coordinate is right around there. And then as I move to a smaller x-coordinate to, say, this one right here, what happened to the normal line? Or even more important, what happened to the intersection of the normal line in the second quadrant? This is the second quadrant, right here. So when I had a larger x-value here, my normal line intersected here, in the second quadrant. Then when I brought my x-value in, when I lowered my x-value, my x-value here, because this is the next point, right here, their wording is bad. They're saying that the second quadrant intersection gets smaller. But actually, it's not really getting smaller. It's getting less negative. I guess smaller could be just absolute value or magnitude, but it's just getting less negative. It's moving there, but it's actually becoming a larger number, right? It's becoming less negative, but a larger number. But if we think in absolute value, I guess it's getting smaller, right? As we went from that point to that point, as we moved the x in for the intersection of the first quadrant, the second quadrant intersection also moved in a bit, from that line to that line. Fair enough. But eventually, a normal line second quadrant intersection gets as small as it can get. So if we keep lowering our x-value in the first quadrant, so we keep on pulling in the first quadrant, as we get to this point. And then this point intersects the second quadrant, right there. And then, if you go even smaller x-values in the first" + }, + { + "Q": "\nAt around 11:23: Why do you use the quadratic formula?", + "A": "Sal wants to find the x coordinates of the intersections, and these are the solution of the equation he built by considering the y values of the normaline and the parabola are the same (which happens only at the intersections). But I think he could have factorized the polynomial, because we already know one solution is x0. x\u00c2\u00b2 + (1/(2.x0)).x - ((1/2)+x0\u00c2\u00b2) = (x-x0)(x+(1/(2.x0) + x0) Thus the other solution is -(1/(2.x0) + x0)", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 683 + ], + "3min_transcript": "That's this right there. And then I have to add this to both sides of the equation, so then I have plus x0 squared. So this is the equation of the normal line, in mx plus b form. This is its slope, this is the m, and then this is its y-intercept right here. That's kind of the b. Now, what do we care about? We care about where this thing intersects. We care about where it intersects the parabola. And the parabola, that's pretty straightforward, that's just y is equal to x squared. So to figure out where they intersect, we just have to set the 2 y's to be equal to each other. So they intersect, the x-values where they intersect, x squared, this y would have to be equal to that y. Or we could just substitute this in for that y. So you get x squared is equal to minus 1 over 2 x0 times x, Fair enough. And let's put this in a quadratic equation, or try to solve this, so we can apply the quadratic equation. So let's put all of this stuff on the lefthand side. So you get x squared plus 1 over 2 x0 times x minus all of this, 1/2 plus x0 squared is equal to 0. All I did is, I took all of this stuff and I put it on the lefthand side of the equation. Now, this is just a standard quadratic equation, so we can figure out now where the x-values that satisfy this quadratic equation will tell us where our normal line and our parabola intersect. So let's just apply the quadratic equation here. So the potential x-values, where they intersect, x is equal to minus b, I'm just applying the So minus b is minus 1 over 2 x0, plus or minus the square root of b squared. So that's that squared. So it's one over four x0 squared minus 4ac. So minus 4 times 1 times this minus thing. So I'm going to have a minus times a minus is a plus, so it's just 4 times this, because there was one there. So plus 4 times this, right here. 4 times this is just 2 plus 4 x0 squared. All I did is, this is 4ac right here. Well, minus 4ac. The minus and the minus canceled out, so There's a 1. So 4 times c is just 2 plus 4x squared. I just multiply this by two, and of course all of this" + }, + { + "Q": "At 12:36, Sal says \"let's divide everything by 1/2\" shouldn't it be 2, not 1/2?\n", + "A": "You are right, he has a pronunciation error, he should have said either divide everything by 2 or multiply everything by 1/2 . Luckily he did wrote everything correctly.", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 756 + ], + "3min_transcript": "So minus b is minus 1 over 2 x0, plus or minus the square root of b squared. So that's that squared. So it's one over four x0 squared minus 4ac. So minus 4 times 1 times this minus thing. So I'm going to have a minus times a minus is a plus, so it's just 4 times this, because there was one there. So plus 4 times this, right here. 4 times this is just 2 plus 4 x0 squared. All I did is, this is 4ac right here. Well, minus 4ac. The minus and the minus canceled out, so There's a 1. So 4 times c is just 2 plus 4x squared. I just multiply this by two, and of course all of this So let's see if I can simplify this. We're just figuring out where the normal line and the parabola intersect. Now, what do we get here. This looks like a little hairy beast here. Let me see if I can simplify this a little bit. So let us factor out-- let me rewrite this. I can just divide everything by 1/2, so this is minus 1 over 4 x0, I just divided this by 2, plus or minus 1/2, that's just this 1/2 right there, times the square root, let me see what I can simplify out of here. So if I factor out a 4 over x0 squared, then what does This term right here will become an x to the fourth, x0 to the fourth, plus, now, what does this term become? This term becomes a 1/2 x0 squared. And just to verify this, multiply 4 times 1/2, you get 2, and then the x0 squares cancel out. So write this term times that, will equal 2, and then you have plus-- now we factored a four out of this and the x0 squared, so plus 1/16. Let me scroll over a little bit. And you can verify that this works out. If you were to multiply this out, you should get this business right here. I see the home stretch here, because this should actually factor out quite neatly. So what does this equal?" + }, + { + "Q": "\nAt 7:02 I thought the line perpendicular to y=2x is y=(-1/2)x not y=-1/(2x). I might not be understanding something, but -1/(2x) just doesn't make sense to me.", + "A": "Indeed. I agree entirely. I was under the impression that only the slope was affected, and thus 2x would become -x/2...", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 422 + ], + "3min_transcript": "is below it. Right, fair enough. For example, this guy right here, this is when we had a large x-value. He intersects with the second quadrant there. Then if you lower and lower the x-value, if you lower it enough, you pass the extreme normal line, and then you get to this point, and then this point, he intersects, or actually, you go to this point. So if you pull in your x-value enough, you once again intersect at that same point in the second quadrant. So hopefully I'm making some sense to you, as I try to make some sense of this problem. Now what do they want to know? And I think I only have time for the first part of this. Maybe I'll do the second part in the another video. Find the equation of the extreme normal line. Well, that seems very daunting at first, but I think our toolkit of derivatives, and what we know about equations of a line, should be able to get us there. So what's the slope of the tangent line at any point on this curve? Well, we just take the derivative of y equals This is the slope of the tangent at any point x. So if I want to know the slope of the tangent at x0, at some particular x, I would just say, well, let me just say, slope, it would be 2 x0. Or let me just say, f of x0 is equal to 2 x0. This is the slope at any particular x0 of the tangent line. Now, the normal line slope is perpendicular to this. So the perpendicular line, and I won't review it here, but the perpendicular line has a negative inverse slope. So the slope of normal line at x0 will be the negative inverse of this, because this is the slope of the tangent line x0. So it'll be equal to minus 1 over 2 x0. Now, what is the equation of the normal line at x0 let's say that this is my x0 in question. What is the equation of the normal line there? Well, we can just use the point-slope form of our equation. So this point right here will be on the normal line. And that's the point x0 squared. Because this the graph of y equals x0, x squared. So this normal line will also have this point. So we could say that the equation of the normal line, let me write it down, would be equal to, this is just a point-slope definition of a line. You say, y minus the y-point, which is just x0 squared, that's that right there, is equal to the slope of the normal line minus 1 over 2 x0 times x minus the" + }, + { + "Q": "I find it easier to think of conversion problems in terms of proportions.\nfor example: At 4:36 It make more sense to think of it as 45 degree/ x radii is in proportion to \u00cf\u0080 radii / 180 degrees. So x(180)= 45(\u00cf\u0080), then x=45/180\u00cf\u0080, but that gives 1/4\u00cf\u0080 which is not the answer, how can you solve these with proportions and how would you solve that example from the video using a proportion?\n", + "A": "x/\u00cf\u0080 = 45/180 Is the form I normally use. x is to \u00cf\u0080 as 45 is to 180. 180x = 45\u00cf\u0080 This is the same equation you used, and it works. x = 45\u00cf\u0080/180 x = \u00cf\u0080/4 This is the right answer. 45\u00c2\u00b0 = \u00cf\u0080/4 radians", + "video_name": "z8vj8tUCkxY", + "timestamps": [ + 276 + ], + "3min_transcript": "pi radians for every 180 degrees or pi/180 radians/degree. This is going to get us to...we're going to get 30 times pi/180 30 times pi/180 which will simplify to 30/180 is 1/6 so this is equal to pi/6 let me write the units out this is 30 radians which is equal to pi/6 radians. Now lets go the other way lets think about if we have pi/3 radians and I wanna convert that to degrees. So what am I going to get if I convert that to degrees? Well here we're gonna want to figure out how many degrees are there per radian? think about the pi and the 180 for every 180 degrees you have pi radians. 180 degrees/pi radians these are essentially the equivalent thing essentially you're just multiplying this quantity by 1 but you're changing the units the radians cancel out and then the pi's cancel out and you're left with 180/3 degrees 180/3 is 60 and we can either write out the word degrees or you can write degrees just like that. Now lets think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out with that notation as well. How many radians will this be equal to? Well once again we're gonna want to think about how many radians do we have per degree? So we're going to multiply this times for every 180 degrees or we can even write it this way pi radians for every 180 degrees. And here this might be a little less intuitive the degrees cancel out and that's why I usually like to write out the word and you're left with 45 pi/180 radians. Actually let me write this with the words written out for me that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with when you multiply 45 times pi over 180 the degrees have canceled out and you're just left with" + }, + { + "Q": "\nto be included in N, doesn't \u00c2\u00b5(x) also have to be multiplied by y' ?\nAlthough this isn't in this case- at 5:25 Sal is still treating \u00c2\u00b5(x) as part of N. how can he do that?", + "A": "It is actually part of x. \u00c2\u00b5(x) is the factor that needs to be multiplied to M and N to make it integratable not y . If you reason like that, there is no y in M then but you can get dx if you multiply through by dx becayse y = dy/dx. Please vote if that helped.", + "video_name": "j511hg7Hlbg", + "timestamps": [ + 325 + ], + "3min_transcript": "So I get mu of x times 3xy plus y squared plus mu of x times x squared plus xy times y prime. And then, what's 0 times any function? Well, it's just going to be 0, right? 0 times mu of x is just going to be 0. But I did multiply the right hand side times mu of x. And remember what we're doing. This mu of x is-- when we multiply it, the goal is, after multiplying both sides of the equation by it, we should have an exact equation. So now, if we consider this whole thing our new M, the partial derivative of this with respect to y should be equal to the partial derivative of this with respect to x. So what's the partial derivative of this with respect to y? here, mu of x, which is only a function of x, it's not a function of y, it's just a constant term, right? We take a partial with respect to y. x is just a constant, or a function of x can be viewed just as a constant. So the partial of this with respect to y is going to be equal to mu of x, you could just say, times 3x plus 2y. That's the partial of this with respect to y. And what's the partial of this with respect to x? Well, here, we'll use the product rule. So we'll take the derivative of the first expression with respect to x. mu of x is no longer a constant anymore, since we're taking the partial with respect to x. So the derivative of mu of x with respect to x. Well, that's just mu prime of x, mu prime, not U. mu prime of x. mu is the Greek letter. It's for the muh sound, but it looks a lot like a U. plus xy, plus just the first expression. This is just the product rule, mu of x. Times the derivative of the second expression with respect to x. So times-- ran out of space on that line-- 2x plus y. And now for this new equation, where I multiplied both sides by mu. In order for this to be exact, these two things have to be equal to each other. So let's just remember the big picture. We're kind of saying, this is going to be exact. And now, we're going to try to solve for mu. So let's see if we can do that. So let's see, on this side, we have mu of x times 3x plus 2y. And let's subtract this expression from both sides. So it's minus mu of x times 2x plus y." + }, + { + "Q": "\nAt 5:26 ,shouldn't you take the derivative of (2x+y)? Since he is using the chain rule?", + "A": "At 5:13 Sal mentions that he is using the product rule which states that d/dx[f(x)*g(x)] = d/dx(f(x))*g(x) + f(x)*d/dx(g(x)) in this case f(x)=mu(x) and g(x)=x^2+xy All of that to say that (2x+y) is the derivative of g(x). Notice: To find this derivative with respect to x simply split the function at the + sign and take the derivative of x^2 and xy seperatly to get (2x+y)", + "video_name": "j511hg7Hlbg", + "timestamps": [ + 326 + ], + "3min_transcript": "So I get mu of x times 3xy plus y squared plus mu of x times x squared plus xy times y prime. And then, what's 0 times any function? Well, it's just going to be 0, right? 0 times mu of x is just going to be 0. But I did multiply the right hand side times mu of x. And remember what we're doing. This mu of x is-- when we multiply it, the goal is, after multiplying both sides of the equation by it, we should have an exact equation. So now, if we consider this whole thing our new M, the partial derivative of this with respect to y should be equal to the partial derivative of this with respect to x. So what's the partial derivative of this with respect to y? here, mu of x, which is only a function of x, it's not a function of y, it's just a constant term, right? We take a partial with respect to y. x is just a constant, or a function of x can be viewed just as a constant. So the partial of this with respect to y is going to be equal to mu of x, you could just say, times 3x plus 2y. That's the partial of this with respect to y. And what's the partial of this with respect to x? Well, here, we'll use the product rule. So we'll take the derivative of the first expression with respect to x. mu of x is no longer a constant anymore, since we're taking the partial with respect to x. So the derivative of mu of x with respect to x. Well, that's just mu prime of x, mu prime, not U. mu prime of x. mu is the Greek letter. It's for the muh sound, but it looks a lot like a U. plus xy, plus just the first expression. This is just the product rule, mu of x. Times the derivative of the second expression with respect to x. So times-- ran out of space on that line-- 2x plus y. And now for this new equation, where I multiplied both sides by mu. In order for this to be exact, these two things have to be equal to each other. So let's just remember the big picture. We're kind of saying, this is going to be exact. And now, we're going to try to solve for mu. So let's see if we can do that. So let's see, on this side, we have mu of x times 3x plus 2y. And let's subtract this expression from both sides. So it's minus mu of x times 2x plus y." + }, + { + "Q": "At 1:31, how does- 3x -(+2x) equal -3x+2x and result in a sum of x?\n", + "A": "The above equation was not solved right. It should be: - 3x - (+2x) = - 3x - 2x = - 5x", + "video_name": "8Wxw9bpKEGQ", + "timestamps": [ + 91 + ], + "3min_transcript": "Divide x squared minus 3x plus 2 divided by x minus 2. So we're going to divide this into that. And we can do this really the same way that you first learned long division. So we have x minus 2 being divided into x squared minus 3x plus 2. Another way we could have written the same exact expression is x squared minus 3x plus 2, all of that over x minus 2. That, that, and that are all equivalent expressions. Now, to do this type of long division-- we can call it algebraic long division-- you want to look at the highest degree term on the x minus 2 and the highest degree term on the x squared minus 3x plus 2. And here's the x, and here's the x squared. x goes into x squared how many times? Or x squared divided by x is what? Well, that's just equal to x. So x goes into x squared x times. And I'm going to write it in this column right here above all of the x terms. And then we want to multiply x times x minus 2. That gives us-- x times x is x squared. And just like you first learned in long division, you want to subtract this from that. But that's completely the same as adding the opposite, or multiplying each of these terms by negative 1 and then adding. So let's multiply that times negative 1. And negative 2x times negative 1 is positive 2x. And now let's add. x squared minus x squared-- those cancel out. Negative 3x plus 2x-- that is negative x. And then we can bring down this 2 over here. So it's negative x plus 2 left over, when we only go x times. So then we say, can x minus 2 go into negative x plus 2? Well, x goes into negative x negative one times. You can look at it right here. Negative x divided by x is negative 1. These guys cancel out. Those guys cancel out. So negative 1 times x minus 2-- you have negative 1 times x, which is negative x. Negative 1 times negative 2 is positive 2. just like you do in long division. But that's the same thing as adding the opposite, or multiplying each of these terms by negative 1 So negative x times negative 1 is positive x. Positive 2 times negative 1 is negative 2. These guys cancel out, add up to 0. These guys add up to 0. We have no remainder. So we got this as being equal to x minus 1. And we can verify it. If we multiply x minus 1 times x minus 2, we should get this. So let's actually do that. So let's multiply x minus 1 times x minus 2. So let's multiply negative 2 times negative 1. That gives us positive 2. Negative 2 times x-- that's negative 2x. Let's multiply x times negative 1. That is negative x." + }, + { + "Q": "\nAt 0:52, why did you write x over the 3x? I was taught to write it over the x squared. Sorry, just a little confused.", + "A": "you want to align all like terms wherever possible to keep it organized. if you write x_ over the _x squared , it s more difficult to keep track of your different terms", + "video_name": "8Wxw9bpKEGQ", + "timestamps": [ + 52 + ], + "3min_transcript": "Divide x squared minus 3x plus 2 divided by x minus 2. So we're going to divide this into that. And we can do this really the same way that you first learned long division. So we have x minus 2 being divided into x squared minus 3x plus 2. Another way we could have written the same exact expression is x squared minus 3x plus 2, all of that over x minus 2. That, that, and that are all equivalent expressions. Now, to do this type of long division-- we can call it algebraic long division-- you want to look at the highest degree term on the x minus 2 and the highest degree term on the x squared minus 3x plus 2. And here's the x, and here's the x squared. x goes into x squared how many times? Or x squared divided by x is what? Well, that's just equal to x. So x goes into x squared x times. And I'm going to write it in this column right here above all of the x terms. And then we want to multiply x times x minus 2. That gives us-- x times x is x squared. And just like you first learned in long division, you want to subtract this from that. But that's completely the same as adding the opposite, or multiplying each of these terms by negative 1 and then adding. So let's multiply that times negative 1. And negative 2x times negative 1 is positive 2x. And now let's add. x squared minus x squared-- those cancel out. Negative 3x plus 2x-- that is negative x. And then we can bring down this 2 over here. So it's negative x plus 2 left over, when we only go x times. So then we say, can x minus 2 go into negative x plus 2? Well, x goes into negative x negative one times. You can look at it right here. Negative x divided by x is negative 1. These guys cancel out. Those guys cancel out. So negative 1 times x minus 2-- you have negative 1 times x, which is negative x. Negative 1 times negative 2 is positive 2. just like you do in long division. But that's the same thing as adding the opposite, or multiplying each of these terms by negative 1 So negative x times negative 1 is positive x. Positive 2 times negative 1 is negative 2. These guys cancel out, add up to 0. These guys add up to 0. We have no remainder. So we got this as being equal to x minus 1. And we can verify it. If we multiply x minus 1 times x minus 2, we should get this. So let's actually do that. So let's multiply x minus 1 times x minus 2. So let's multiply negative 2 times negative 1. That gives us positive 2. Negative 2 times x-- that's negative 2x. Let's multiply x times negative 1. That is negative x." + }, + { + "Q": "Can a snail live underwater?@12:30pm\n", + "A": "Snails CAN stay underwater for a while but CAN T live underwater.", + "video_name": "gBxeju8dMho", + "timestamps": [ + 750 + ], + "3min_transcript": "" + }, + { + "Q": "at 1:40, hilarious! his pineapple is BLASPHEMY! rotfl\n", + "A": "The pineapple is blasphemy", + "video_name": "gBxeju8dMho", + "timestamps": [ + 100 + ], + "3min_transcript": "Dear Nickelodeon, I've gotten over how SpongeBob's pants are not actually square. I can ignore most of the time that Gary's shell is not a logarithmic spiral. But what I cannot forgive is that SpongeBob's pineapple house is a mathematical impossibility. There's three easy ways to find spirals on a pineapple. There's the ones that wind up it going right, the ones that spiral up to the left, and the ones that go almost straight up-- keyword almost. If you count the number of spirals going left and the number of spirals going right, they'll be adjacent Fibonacci numbers-- 3 and 5, or 5 and 8, 8 and 13, or 13 and 21. You claim that SpongeBob Squarepants lives in a pineapple under the sea, but does he really? A true pineapple would have Fibonacci spiral, so let's take a look. Because these images of his house don't let us pick it up and turn it around to count the number of spirals going around it, it might be hard to figure out whether it's mathematically a pineapple or not. But there's a huge clue in the third spiral, the one going upwards. In this pineapple there's 8 to the right, 13 to the left. You can add those numbers together to get how many spirals are in the set spiraling In this case, 21. The three sets of spirals in any pineapple are pretty much always adjacent Fibonacci numbers. The rare mutant cases might show Lucas numbers or something, but it will always be three adjacent numbers in a series. What you'll never have is the same number of spirals both ways. Pineapples, unlike people, don't have bilateral symmetry. You'll never have that third spiral be not a spiral, but just a straight line going up a pineapple. Yet, when we look at SpongeBob's supposed pineapple under the sea, it clearly has lines of pineapple things going straight up. It clearly has bilateral symmetry. It clearly is not actually a pineapple at all, because no pineapple could possibly grow that way. Nickelodeon, you need to take a long, hard look in the mirror and think about the way you're misrepresenting the universe to your viewers. This kind of mathematical oversight is simply irresponsible. Sincerely, Vi Hart." + }, + { + "Q": "why is there a snail wandering around between 0:00 and 1:00\n", + "A": "because she likes it and it is cool, and looks like spongbobs pet garey", + "video_name": "gBxeju8dMho", + "timestamps": [ + 0, + 60 + ], + "3min_transcript": "Dear Nickelodeon, I've gotten over how SpongeBob's pants are not actually square. I can ignore most of the time that Gary's shell is not a logarithmic spiral. But what I cannot forgive is that SpongeBob's pineapple house is a mathematical impossibility. There's three easy ways to find spirals on a pineapple. There's the ones that wind up it going right, the ones that spiral up to the left, and the ones that go almost straight up-- keyword almost. If you count the number of spirals going left and the number of spirals going right, they'll be adjacent Fibonacci numbers-- 3 and 5, or 5 and 8, 8 and 13, or 13 and 21. You claim that SpongeBob Squarepants lives in a pineapple under the sea, but does he really? A true pineapple would have Fibonacci spiral, so let's take a look. Because these images of his house don't let us pick it up and turn it around to count the number of spirals going around it, it might be hard to figure out whether it's mathematically a pineapple or not. But there's a huge clue in the third spiral, the one going upwards. In this pineapple there's 8 to the right, 13 to the left. You can add those numbers together to get how many spirals are in the set spiraling In this case, 21. The three sets of spirals in any pineapple are pretty much always adjacent Fibonacci numbers. The rare mutant cases might show Lucas numbers or something, but it will always be three adjacent numbers in a series. What you'll never have is the same number of spirals both ways. Pineapples, unlike people, don't have bilateral symmetry. You'll never have that third spiral be not a spiral, but just a straight line going up a pineapple. Yet, when we look at SpongeBob's supposed pineapple under the sea, it clearly has lines of pineapple things going straight up. It clearly has bilateral symmetry. It clearly is not actually a pineapple at all, because no pineapple could possibly grow that way. Nickelodeon, you need to take a long, hard look in the mirror and think about the way you're misrepresenting the universe to your viewers. This kind of mathematical oversight is simply irresponsible. Sincerely, Vi Hart." + }, + { + "Q": "at 0:42, why is the first dot on x axis on zero? It said 1 question = 5 points so shouldn't it be 1 on the x axis and 5 on the y axis?\n", + "A": "no because if you have those points (1,0) and (0,5) the slope would be negative (5-0)/(0-1), and you would quickly go into the negatives for y as x increases. If you get 0 questions right, why would get a 1 point credit? If you get 0 questions correct, you should get 0 points (0,0), if you get 1 question right, 5 points (1,5), 2 questions is 10 points (2,10) etc.", + "video_name": "0eWm-LY23W0", + "timestamps": [ + 42 + ], + "3min_transcript": "On your math quiz, you earn 5 points for each question that you answer correctly. In the table below, x represents the number of questions that you answer correctly, and y represents the total number of points that you score on your quiz. Fair enough? The relationship between these two variables can be expressed by the following equation-- y is equal to 5x. Graph the equation below. So you could look at a couple of your points. You could say, well, look, if I got 0 questions right I'm going to have 0 points. So you could literally graph that point-- if I got 0 questions right, I get 0 points. And then if I get one question right, and the table tells us that, or we could logically think about it, every question I would get right I'm going to get 5 more points. So if I get one question right, I'm going to get 5 more points, and we saw that in our table as well. We saw that right over here-- one question, 5 points. We could also plot it two questions, 10 points, But you only have to do two points to define a line. So it looks like we are actually done here." + }, + { + "Q": "Why is 1 over 25/64 equal to 64/25? (at 6:23)? Thank you.\n", + "A": "Ah, that s 1/(25/64) or 1 divided by 25/64 which is the same as 1 * 64/25 or 64/25.", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 383 + ], + "3min_transcript": "" + }, + { + "Q": "I still don't get the point of the 1 at 0:14. Can someone please help me?\n", + "A": "its just showing that the one doesn t make a difference", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 14 + ], + "3min_transcript": "" + }, + { + "Q": "i like how at 0:51 it gives that little point out XD\n", + "A": "Me too! it s pretty cool.", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 51 + ], + "3min_transcript": "" + }, + { + "Q": "\nWhy is at 1:30 the negitave turns into the fraction?", + "A": "The method Sal is using may be confusing, so you may want to solve a negative exponent problem this way: 4 to the negative 3rd power=1*4*4*4(find the reciprocal)> 1/64 Why? Well, we know that 4 to the positive 3rd power is 64, or 64/1, and a negative is the opposite of a positive. Since a negative is the opposite of a positive, we will have to find the opposite, or reciprocal of 64 or 64/1, which is 1/64.", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 90 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:08 Why does he put the 1. (2^4 = 1 x 2 x 2 x 2 x 2) It does'nt really do anything and it confuses me because I did that on a test and i lost a point so I don't know if it supposed to be like that or not\n", + "A": "he puts the 1 to show you 2^0 still leaves a 1 2^0 = 1 2^1 = 1 * 2 2^2 = 1 * 2 * 2", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 8 + ], + "3min_transcript": "" + }, + { + "Q": "\nSal said \"negative one\" but meant \"one.\"\n0:54", + "A": "Yes.. The black box pops up and it tells you..", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 54 + ], + "3min_transcript": "" + }, + { + "Q": "So, at 6:24 Sal says that 5/8^ -2 is 64/25. Isn't that equal to 2 14/25?\n", + "A": "yes, but no use in here by simplifying it", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 384 + ], + "3min_transcript": "" + }, + { + "Q": "5:25 So, all I have to do to use a negative exponent on a fraction is swap the numerator and denominator on the fraction, and then apply a positive exponent to the faction?\n", + "A": "Yep, that s right!", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 325 + ], + "3min_transcript": "" + }, + { + "Q": "\nSo the numerator of the fraction will always be 1? At 4:05 I noticed that.", + "A": "Correct, the numerator of the fraction will be one, when a number is raised to a negative exponent.", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 245 + ], + "3min_transcript": "" + }, + { + "Q": "Isn't scalar only in matrices?? ( 0:40 )\n", + "A": "A scalar is just a number. When you multiply a scalar by a vector or matrix, you will just get a scaled up version of the vector or matrix.", + "video_name": "br7tS1t2SFE", + "timestamps": [ + 40 + ], + "3min_transcript": "A vector is something that has both magnitude and direction. Magnitude and direction. So let's think of an example of what wouldn't and what would be a vector. So if someone tells you that something is moving at 5 miles per hour, this information by itself is not a vector quantity. It's only specifying a magnitude. We don't know what direction this thing is moving 5 miles per hour in. So this right over here, which is often referred to as a speed, is not a vector quantity just by itself. This is considered to be a scalar quantity. If we want it to be a vector, we would also have to specify the direction. So for example, someone might say it's moving 5 miles per hour east. So let's say it's moving 5 miles per hour due east. So now this combined 5 miles per are due east, this is a vector quantity. And now we wouldn't call it speed anymore. So velocity is a vector. We're specifying the magnitude, 5 miles per hour, and the direction east. But how can we actually visualize this? So let's say we're operating in two dimensions. And what's neat about linear algebra is obviously a lot of what applies in two dimensions will extend to three. And then even four, five, six, as made dimensions as we want. Our brains have trouble visualizing beyond three. But what's neat is we can mathematically deal with beyond three using linear algebra. And we'll see that in future videos. But let's just go back to our straight traditional two-dimensional vector right over here. So one way we could represent it, as an arrow that is 5 units long. We'll assume that each of our units here is miles per hour. And that's pointed to the right, where we'll say the right is east. So for example, I could start an arrow right over here. And I could make its length 5. The length of the arrow specifies the magnitude. And then the direction that the arrow is pointed in specifies it's direction. So this right over here could represent visually this vector. If we say that the horizontal axis is say east, or the positive horizontal direction is moving in the east, this would be west, that would be north, and then that would be south. Now, what's interesting about vectors is that we only care about the magnitude in the direction. We don't necessarily not care where we start, where we place it when we think about it visually like this. So for example, this would be the exact same vector, or be equivalent vector to this. This vector has the same length. So it has the same magnitude. It has a length of 5. And its direction is also due east. So these two vectors are equivalent. Now one thing that you might say is, well, that's fair enough. But how do we represent it with a little bit more mathematical notation? So we don't have to draw it every time. And we could start performing operations on it." + }, + { + "Q": "\nAt 5:21, he says that you can use the pythagorean theorem to calculate the length of the vector, but wouldn't the length be sqr21, since 3^2+4^2=21?", + "A": "3^2 + 4^2 = 9 + 16 = 25. Where are you getting 21 from?", + "video_name": "br7tS1t2SFE", + "timestamps": [ + 321 + ], + "3min_transcript": "want a variable to represent a vector, is usually a lowercase letter. If you're publishing a book, you can bold it. But when you're doing it in your notebook, you would typically put a little arrow on top of it. And there are several ways that you could do it. You could literally say, hey 5 miles per hour east. But that doesn't feel like you can really operate on that easily. The typical way is to specify, if you're in two dimensions, to specify two numbers that tell you how much is this vector moving in each of these dimensions? So for example, this one only moves in the horizontal dimension. And so we'll put our horizontal dimension first. So you might call this vector 5, 0. It's moving 5, positive 5 in the horizontal direction. And it's not moving at all in the vertical direction. And the notation might change. You might also see notation, and actually in the linear algebra context, it's more typical to write it as a column vector like this-- 5, 0. represents how much we're moving in the horizontal direction. And the second coordinate represents how much are we moving in the vertical direction. Now, this one isn't that interesting. You could have other vectors. You could have a vector that looks like this. Let's say it's moving 3 in the horizontal direction. And positive 4. So 1, 2, 3, 4 in the vertical direction. So it might look something like this. So this could be another vector right over here. Maybe we call this vector, vector a. And once again, I want to specify that is a vector. And you see here that if you were to break it down, in the horizontal direction, it's shifting three in the horizontal direction, and it's shifting positive four in the vertical direction. about how much we're moving up and how much we're moving to the right when we start at the end of the arrow and go to the front of it. So this vector might be specified as 3, 4. 3, 4. And you could use the Pythagorean theorem to figure out the actual length of this vector. And you'll see because this is a 3, 4, 5 triangle, that this actually has a magnitude of 5. And as we study more and more linear algebra, we're going to start extending these to multiple dimensions. Obviously we can visualize up to three dimensions. In four dimensions it becomes more abstract. And that's why this type of a notation is useful. Because it's very hard to draw a 4, 5, or 20 dimensional arrow like this." + }, + { + "Q": "\n0:36 What does \"Scalar\" mean?", + "A": "A scalar is just a real number. This is as opposed to a vector (which is an ordered set of real numbers).", + "video_name": "br7tS1t2SFE", + "timestamps": [ + 36 + ], + "3min_transcript": "A vector is something that has both magnitude and direction. Magnitude and direction. So let's think of an example of what wouldn't and what would be a vector. So if someone tells you that something is moving at 5 miles per hour, this information by itself is not a vector quantity. It's only specifying a magnitude. We don't know what direction this thing is moving 5 miles per hour in. So this right over here, which is often referred to as a speed, is not a vector quantity just by itself. This is considered to be a scalar quantity. If we want it to be a vector, we would also have to specify the direction. So for example, someone might say it's moving 5 miles per hour east. So let's say it's moving 5 miles per hour due east. So now this combined 5 miles per are due east, this is a vector quantity. And now we wouldn't call it speed anymore. So velocity is a vector. We're specifying the magnitude, 5 miles per hour, and the direction east. But how can we actually visualize this? So let's say we're operating in two dimensions. And what's neat about linear algebra is obviously a lot of what applies in two dimensions will extend to three. And then even four, five, six, as made dimensions as we want. Our brains have trouble visualizing beyond three. But what's neat is we can mathematically deal with beyond three using linear algebra. And we'll see that in future videos. But let's just go back to our straight traditional two-dimensional vector right over here. So one way we could represent it, as an arrow that is 5 units long. We'll assume that each of our units here is miles per hour. And that's pointed to the right, where we'll say the right is east. So for example, I could start an arrow right over here. And I could make its length 5. The length of the arrow specifies the magnitude. And then the direction that the arrow is pointed in specifies it's direction. So this right over here could represent visually this vector. If we say that the horizontal axis is say east, or the positive horizontal direction is moving in the east, this would be west, that would be north, and then that would be south. Now, what's interesting about vectors is that we only care about the magnitude in the direction. We don't necessarily not care where we start, where we place it when we think about it visually like this. So for example, this would be the exact same vector, or be equivalent vector to this. This vector has the same length. So it has the same magnitude. It has a length of 5. And its direction is also due east. So these two vectors are equivalent. Now one thing that you might say is, well, that's fair enough. But how do we represent it with a little bit more mathematical notation? So we don't have to draw it every time. And we could start performing operations on it." + }, + { + "Q": "\n3:53 Why do you have to substitute the variable into an original equation?", + "A": "So that you can get y.", + "video_name": "vA-55wZtLeE", + "timestamps": [ + 233 + ], + "3min_transcript": "You would get Ax plus By, plus D is equal to C plus D. And we've seen that multiple, multiple times. Anything you do to one side of the equation, you have to do to the other side. But you're saying, hey, Sal, wait, on the left-hand side, you're adding 5x minus 4y to the equation. On the right-hand side, you're adding 25.5 to the equation. Aren't you adding two different things to both sides of the equation? And my answer would be no. We know that 5x minus 4y is 25.5. This quantity and this quantity are the same. They're both 25.5. This second equation is telling me that explicitly. So I can add this to the left-hand side. I'm essentially adding 25.5 to it. And I could add 25.5 to the right-hand side. So let's do that. If we were to add the left-hand side, 3x plus 5x is 8x. And this was the whole point. When I looked at these two equations, I said, oh, I have a 4y, I have a negative 4y. If you just add these two together, they are going to cancel out. They're going to be plus 0y. Or that whole term is just going to go away. And that's going to be equal to 2.5 plus 25.5 is 28. So you divide both sides. So you get 8x is equal to 28. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. That's equal to 7 over 2. That's our x value. Now we want to solve for our y value. And we could substitute this back into either of these two equations. Let's use the top one. You could do it with the bottom one as well. So we know that 3 times x, 3 times 7 over 2-- I'm just equation-- 3 times 7 over 2, plus 4y is equal to 2.5. Let me just write that as 5/2. We're going to stay in the fraction world. So this is going to be 21 over 2 plus 4y is equal to 5/2. Subtract 21 over 2 from both sides. So minus 21 over 2, minus 21 over 2. The left-hand side-- you're just left with a 4y, because these two guys cancel out-- is equal to-- this is 5 minus 21 over 2. That's negative 16 over 2. So that's negative 16 over 2, which is the same thing-- well, I'll write it out as negative 16 over 2. Or we could write that-- let's continue up here-- 4y-- I'm" + }, + { + "Q": "\nAt the time 0:19 9 1/2 ends up being reduced to 3 like you had said. But, what happens to 1/2. Since you are using 3 to divide/reduce you have to use 3 to reduce 1/2 which you can't do. So how does that work. Since 3 is bigger then 1/2 you cant do that because the difference is ginormous like ginormous like absolutely HUGE! Well not that big just 2 1/2 but i mean still please explain.......", + "A": "9 to the exponent 1/2 means principal square root of 9, which is 3 because 3\u00c2\u00b2=9. He didn t divide 9 by 3. Perhaps you need to watch review on exponents from previous videos then come back to this one.", + "video_name": "tn53EdOr6Rw", + "timestamps": [ + 19 + ], + "3min_transcript": "Let's do some slightly more complicated fractional exponent examples. So we already know that if I were to take 9 to the 1/2 power, this is going to be equal to 3, and we know that because 3 times 3 is equal to 9. This is equivalent to saying, what is the principal root of 9? Well, that is equal to 3. But what would happen if I took 9 to the negative 1/2 power? Now we have a negative fractional exponent, and the key to this is to just not get too worried or intimidated by this, but just think about it step by step. Just ignore for the second that this is a fraction, and just look at this negative first. Just breathe slowly, and realize, OK, I got a negative exponent. That means that this is just going to be 1 over 9 to the 1/2. That's what that negative is a cue for. This is 1 over 9 to the 1/2, and we know that 9 to the 1/2 is equal to 3. So this is just going to be equal to 1/3. What would this evaluate to? And I encourage you to pause the video after trying it, or pause the video to try it. Negative 27 to the negative 1/3 power. So I encourage you to pause the video and think about what this would evaluate to. So remember, just take a deep breath. You can always get rid of this negative in the exponent by taking the reciprocal and raising it to the positive. So this is going to be equal to 1 over negative 27 to the positive 1/3 power. And I know what you're saying. Hey, I still can't breathe easily. I have this negative number to this fractional exponent. But this is just saying what number, if I were to multiply it three times-- so if I have that number, so whatever the number this is, if I were to multiply it, if I took three of them and I multiply them together, if I multiplied 1 by that number three times, what number would I Well, we already know that 3 to the third, which is equal to 3 times 3 times 3, is equal to positive 27. So that's a pretty good clue. What would negative 3 to the third power be? Well, that's negative 3 times negative 3 times negative 3, which is negative 3 times negative 3 is positive 9. Times negative 3 is negative 27. So we've just found this number, this question mark. Negative 3 times negative 3 times negative 3 is equal to negative 27. So negative 27 to the 1/3-- this part right over here-- is equal to negative 3. So this is going to be equal to 1 over negative 3, which is the same thing as negative 1/3." + }, + { + "Q": "\nAt the minute 1:50 you came out with 1/(-27)^1/3 and you started looking for a number which could be multiplied by itself in order to get -27. So my question is why did not you put the -27 into the radical sign? ( I know that it will lead to imaginary number) But is there any rule to apply when dealing with negative bases?", + "A": "If you have a negative under a 2nd radical, then the answer to the problem is no solution. However, this problem does not need the 2nd radical (square root), but instead the cube root. The cube root can be positive or negative. If your calculator has this button, you can do that, but many simple calculators don t so that s why he was explaining how to find it without a calculator.", + "video_name": "tn53EdOr6Rw", + "timestamps": [ + 110 + ], + "3min_transcript": "Let's do some slightly more complicated fractional exponent examples. So we already know that if I were to take 9 to the 1/2 power, this is going to be equal to 3, and we know that because 3 times 3 is equal to 9. This is equivalent to saying, what is the principal root of 9? Well, that is equal to 3. But what would happen if I took 9 to the negative 1/2 power? Now we have a negative fractional exponent, and the key to this is to just not get too worried or intimidated by this, but just think about it step by step. Just ignore for the second that this is a fraction, and just look at this negative first. Just breathe slowly, and realize, OK, I got a negative exponent. That means that this is just going to be 1 over 9 to the 1/2. That's what that negative is a cue for. This is 1 over 9 to the 1/2, and we know that 9 to the 1/2 is equal to 3. So this is just going to be equal to 1/3. What would this evaluate to? And I encourage you to pause the video after trying it, or pause the video to try it. Negative 27 to the negative 1/3 power. So I encourage you to pause the video and think about what this would evaluate to. So remember, just take a deep breath. You can always get rid of this negative in the exponent by taking the reciprocal and raising it to the positive. So this is going to be equal to 1 over negative 27 to the positive 1/3 power. And I know what you're saying. Hey, I still can't breathe easily. I have this negative number to this fractional exponent. But this is just saying what number, if I were to multiply it three times-- so if I have that number, so whatever the number this is, if I were to multiply it, if I took three of them and I multiply them together, if I multiplied 1 by that number three times, what number would I Well, we already know that 3 to the third, which is equal to 3 times 3 times 3, is equal to positive 27. So that's a pretty good clue. What would negative 3 to the third power be? Well, that's negative 3 times negative 3 times negative 3, which is negative 3 times negative 3 is positive 9. Times negative 3 is negative 27. So we've just found this number, this question mark. Negative 3 times negative 3 times negative 3 is equal to negative 27. So negative 27 to the 1/3-- this part right over here-- is equal to negative 3. So this is going to be equal to 1 over negative 3, which is the same thing as negative 1/3." + }, + { + "Q": "Around 5:10-\nIsn't the probability to score 4 out of 10 shots the number of ways you can arrange those 4 scores (namely 10 choose 4) and then multipy by the score probability - 40% to the power equal to the total number of shots. I don't understand why we have to multiply by the probability of a miss too? (He multiplies by (1-P)^n-k I don't understand why this is necessary in order to get the probability of k scores)\n", + "A": "It is necessary to take in to consideration the effect of the missed shots or else you are just finding the probability of making four shots in a row if you hand a probability of making a single shot is 40%. If you don t consider the missed shots its as if you only took four shots all together and not ten. Hope this helps", + "video_name": "SqcxYnNlI3Y", + "timestamps": [ + 310 + ], + "3min_transcript": "So I multiply probability times the number of baskets or the number of shots I'm taking, which should be equal to 4. So I know I said-- and you really shouldn't necessarily strictly view expected value as the number of shots you should expect to make because sometimes probability distributions can be kind of weird. But in the binomial distribution you can kind of view it that way. That this is the number of shots you would expect to make. Or you can kind of view it as the most likely outcome. That if you have a 40% shot percentage, and you takes 10 shots, the most likely outcome is that you'll make 4 shots. You still might make 6 shots or 3 shots, but this is going to be the most likely outcome. and in my head, the way I think about, the way it makes intuitive sense is that every time you shoot you have a 40% chance of making the shot. So you could say that you always make 40% of a shot. And if you take 10 shots you're going to make 4 whole shots. So that's one way to think about it and why this might make a little intuitive sense. true for any a random variable that's described by a binomial distribution. So in a binomial distribution what is the probability-- so if I say, what is the probability that X is equal to k? And I know it just gets a little complicated sometimes. But I'm just saying, what's the probability say in this basketball analogy. Would be you know, what's the probability that I make-- k could be 3 shots or something like that. So that's what we're talking about. And that we learned was, if we're taking n shots we're going to choose k of them. And we did that several times in the last couple of videos. And then we multiply that times the probability of any one of those particular occurrences. So if I'm making k shots, it'll be the probability of me making any one shot, Which is p to the kth power. p times itself k times. That's the probability of me making k shots. And then the rest of the shots I have to miss. And then how many shots? If I've made k shots, the rest of the shots I have to miss. So I'm going to miss n minus k shots. So in any binomial distribution this is a probability that you get k successes. Now we know that the expected value, the way you calculate an expected value of a random variable is you just take the probability weighted sum. I don't want to confuse you too much and if you main take away from this video is just this, that's good enough. You should feel good. Now it'll get a little technical, but it'll hopefully make you a little bit more comfortable with sigma and sum notation as well. It'll make you a little bit more comfortable with binomial coefficients and things like that. But just going back, the expected value is a a probability weighted sum of each of these. So what you want to do is you want to take the probability that X is equal to k, times k, and then add that up for each of the possible k's. So how would I write that? So the expected value of X, the expected value of our random" + }, + { + "Q": "At 0:20 where does he get -1?\n", + "A": "the minus before the parenthesis is the same as times -1", + "video_name": "5ZdxnFspyP8", + "timestamps": [ + 20 + ], + "3min_transcript": "Simplify 16x plus 14 minus the entire expression 3x squared plus x minus 9. So when you subtract an entire expression, this is the exact same thing as having 16x plus 14. And then you're adding the opposite of this whole thing. Or you're adding negative 1 times 3x squared plus x minus 9. Or another way to think about it is you can distribute this negative sign along all of those terms. That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part-- I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x because that's positive 1x. Negative 1 times negative 9-- remember, That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second-degree term. We only have one of those. So let me write it over here-- negative 3x squared. And then what do we have in terms of first-degree terms, of just an x, x to the first power? Well, we have a 16x. And then from that, we're going to subtract an x, subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract 1 of them away, you're going to have 15 of that something. And then finally, you have 14. You could view that as 14 times x to the 0 or just 14. 14 plus 9-- they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23." + }, + { + "Q": "At 1:03, if you multiply 3x^2+x-9 by -1, don't you also have to multiply the 16x+14 by -1?\n", + "A": "Don t you have to do the same thing to both sides of the equation?", + "video_name": "5ZdxnFspyP8", + "timestamps": [ + 63 + ], + "3min_transcript": "Simplify 16x plus 14 minus the entire expression 3x squared plus x minus 9. So when you subtract an entire expression, this is the exact same thing as having 16x plus 14. And then you're adding the opposite of this whole thing. Or you're adding negative 1 times 3x squared plus x minus 9. Or another way to think about it is you can distribute this negative sign along all of those terms. That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part-- I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x because that's positive 1x. Negative 1 times negative 9-- remember, That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second-degree term. We only have one of those. So let me write it over here-- negative 3x squared. And then what do we have in terms of first-degree terms, of just an x, x to the first power? Well, we have a 16x. And then from that, we're going to subtract an x, subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract 1 of them away, you're going to have 15 of that something. And then finally, you have 14. You could view that as 14 times x to the 0 or just 14. 14 plus 9-- they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23." + }, + { + "Q": "\nAt 2:06 I did not receive my full points but i watched the whole video and only got 100 points also at 0:25 seconds sal said their was the negative one how did he get that", + "A": "So when you have the problem (4x^2+6x-5)-(2x^2-4x-7) to remove the parenthesis you have to multiply everything after the - sign in the middle by -1 because that is really what the - sign in the middle represents.", + "video_name": "5ZdxnFspyP8", + "timestamps": [ + 126, + 25 + ], + "3min_transcript": "Simplify 16x plus 14 minus the entire expression 3x squared plus x minus 9. So when you subtract an entire expression, this is the exact same thing as having 16x plus 14. And then you're adding the opposite of this whole thing. Or you're adding negative 1 times 3x squared plus x minus 9. Or another way to think about it is you can distribute this negative sign along all of those terms. That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part-- I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x because that's positive 1x. Negative 1 times negative 9-- remember, That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second-degree term. We only have one of those. So let me write it over here-- negative 3x squared. And then what do we have in terms of first-degree terms, of just an x, x to the first power? Well, we have a 16x. And then from that, we're going to subtract an x, subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract 1 of them away, you're going to have 15 of that something. And then finally, you have 14. You could view that as 14 times x to the 0 or just 14. 14 plus 9-- they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23." + }, + { + "Q": "1:12 why couldn't a2 be a3:a1?\n", + "A": "Because a^3*a^1 = a^4 the exponents add up.", + "video_name": "c-wtvEdEoVs", + "timestamps": [ + 72 + ], + "3min_transcript": "We're asked to simplify the cube root of 27a squared times b to the fifth times c to the third power. And the goal, whenever you try to just simplify a cube root like this, is we want to look at the parts of this expression over here that are perfect cubes, that are something raised to the third power. Then we can take just the cube root of those, essentially taking them out of the radical sign, and then leaving everything else that is not a perfect cube inside of it. So let's see what we can do. So first of all, 27-- you may or may not already recognize this as a perfect cube. If you don't already recognize it, you can actually do a prime factorization and see it's a perfect cube. 27 is 3 times 9, and 9 is 3 times 3. So 27-- its prime factorization is 3 times 3 times 3. So it's the exact same thing as 3 to the third power. So let's rewrite this whole expression down here. But let's write it in terms of things that are perfect cubes and things that aren't. So 27 can be just rewritten as 3 to the third power. a to the third would have been. So we're just going to write this-- let me write it over here. We can switch the order here because we just have a bunch of things being multiplied by each other. So I'll write the a squared over here. b to the fifth is not a perfect cube by itself, but it can be expressed as the product of a perfect cube and another thing. b to the fifth is the exact same thing as b to the third power times b to the second power. If you want to see that explicitly, b to the fifth is b times b times b times b times b. So the first three are clearly b to the third power. And then you have b to the second power after it. So we can rewrite b to the fifth as the product of a perfect cube. So I'll write b to the third-- let me do that in that same purple color. So we have b to the third power over here. And then it's b to the third times b squared. So I'll write the b squared over here. And then finally, we have-- I'll do in blue-- c to the third power. Clearly, this is a perfect cube. It is c cubed. It is c to the third power. So I'll put it over here. So this is c to the third power. And of course, we still have that overarching radical sign. So we're still trying to take the cube root of all of this. And we know from our exponent properties, or we could say from our radical properties, that this is the exact same thing. That taking the cube root of all of these things is the same as taking the cube root of these individual factors and then multiplying them. So this is the same thing as the cube root-- and I could separate them out individually. Or I could say the cube root of 3 to the third b to the third c to the third. Actually, let's do it both ways. So I'll take them out separately. So this is the same thing as the cube root of 3 to the third times the cube root-- I'll write them all in. Let me color-code it so we don't get confused-- times the cube" + }, + { + "Q": "\nAt 0:30, what if the number isn't a perfect cube?", + "A": "Both methods to do this are likely not in your skill set yet, as they are usually taught in Trig or Calc classes. However, there is the possibility of simplifying the radical down if the number under the radical is a multiple of a perfect cube, such as 16. 16 is 2*2*2*2. As such, you could pull the group of three 2 s out, and leave the last 2 behind under the radical. This would then be 2*(cubic-root 2)", + "video_name": "c-wtvEdEoVs", + "timestamps": [ + 30 + ], + "3min_transcript": "We're asked to simplify the cube root of 27a squared times b to the fifth times c to the third power. And the goal, whenever you try to just simplify a cube root like this, is we want to look at the parts of this expression over here that are perfect cubes, that are something raised to the third power. Then we can take just the cube root of those, essentially taking them out of the radical sign, and then leaving everything else that is not a perfect cube inside of it. So let's see what we can do. So first of all, 27-- you may or may not already recognize this as a perfect cube. If you don't already recognize it, you can actually do a prime factorization and see it's a perfect cube. 27 is 3 times 9, and 9 is 3 times 3. So 27-- its prime factorization is 3 times 3 times 3. So it's the exact same thing as 3 to the third power. So let's rewrite this whole expression down here. But let's write it in terms of things that are perfect cubes and things that aren't. So 27 can be just rewritten as 3 to the third power. a to the third would have been. So we're just going to write this-- let me write it over here. We can switch the order here because we just have a bunch of things being multiplied by each other. So I'll write the a squared over here. b to the fifth is not a perfect cube by itself, but it can be expressed as the product of a perfect cube and another thing. b to the fifth is the exact same thing as b to the third power times b to the second power. If you want to see that explicitly, b to the fifth is b times b times b times b times b. So the first three are clearly b to the third power. And then you have b to the second power after it. So we can rewrite b to the fifth as the product of a perfect cube. So I'll write b to the third-- let me do that in that same purple color. So we have b to the third power over here. And then it's b to the third times b squared. So I'll write the b squared over here. And then finally, we have-- I'll do in blue-- c to the third power. Clearly, this is a perfect cube. It is c cubed. It is c to the third power. So I'll put it over here. So this is c to the third power. And of course, we still have that overarching radical sign. So we're still trying to take the cube root of all of this. And we know from our exponent properties, or we could say from our radical properties, that this is the exact same thing. That taking the cube root of all of these things is the same as taking the cube root of these individual factors and then multiplying them. So this is the same thing as the cube root-- and I could separate them out individually. Or I could say the cube root of 3 to the third b to the third c to the third. Actually, let's do it both ways. So I'll take them out separately. So this is the same thing as the cube root of 3 to the third times the cube root-- I'll write them all in. Let me color-code it so we don't get confused-- times the cube" + }, + { + "Q": "At 04:08 what did you mean about, \"3 is too big\" how does it work?\n", + "A": "3 is too big or too large of a number cuz 3x7=21 but if your trying to find out what equels or what is close to equeling 17...........2 is the right answer cuz 2x7=14 so which is closest.....21 or 14?", + "video_name": "NcADzGz3bSI", + "timestamps": [ + 248 + ], + "3min_transcript": "And then you bring down this 9. And you saw in the last video exactly what this means. When you wrote this 5 up here-- notice we wrote So this is really a 500. But in this video I'm just going to focus more on the process, and you can think more about what it actually means in terms of where I'm writing the numbers. But I think the process is going to be crystal clear hopefully, by the end of this video. So we brought down the 9. 4 goes into 29 how many times? It goes into at least six times. What's 4 times 7? 4 times 7 is 28. So it goes into it at least seven times. What's 4 times 8? 4 times 8 is 32, so it can't go into it eight times so it's going to go into it seven. 4 goes into 29 nine seven times. 7 times 4 is 28. 29 minus 28 to get our remainder for this step in the problem is 1. And now we're going to bring down this 2. We're going to bring it down and you get a 12. That's easy. 4 times 3 is 12. 4 goes into 12 three times. 3 times 4 is 12. 12 minus 12 is 0. We have no remainder. So 4 goes into 2,292 exactly 573 times. So this 2,292 divided by 4 we can say is equal to 573. Or we could say that this thing right here is equal to 573. Let's do a couple of more. Let's do a few more problems. So I'll do that red color. Let's say we had 7 going into 6,475. Maybe it's called long division because you write it nice and long up here and you have this line. I don't know. There's multiple reasons why it could be called long division. So you say 7 goes into 6 zero times. So we need to keep moving forward. 7 goes into 64 how many times? Let's see. 7 times 7 is? Well, that's way too small. Let me think about it a little bit. Well 7 times 9 is 63. That's pretty close. And then 6 times 10 is going to be too big. 7 times 10 is 70. So that's too big. So 7 goes into 64 nine times. 9 times 7 is 63. 64 minus 63 to get our remainder of this stage 1. Bring down the 7. 7 goes into 17 how many times? Well, 7 times 2 is 14. And then 7 times 3 is 21. So 3 is too big. So 7 goes into 17 two times. 2 times 7 is 14. 17 minus 14 is 3. And now we bring down the 5." + }, + { + "Q": "\nIn this video he explains how to find if a number is divisible by 3 at 6:06 but than because it is changes it at 6:36, the first number when you used his trick added up to 27, the second number it added up to 26, at the end 9:52 the remainder was 2, if you divided 26 by 3 it would be 8 r 2 (same amount remaining as with Sal's problem) so is the method to find if a number is divisible by 3 also a way to find what the remainder would be if it isn't?", + "A": "Pretty much. Good pick up.", + "video_name": "NcADzGz3bSI", + "timestamps": [ + 366, + 396, + 592 + ], + "3min_transcript": "7 goes into 64 how many times? Let's see. 7 times 7 is? Well, that's way too small. Let me think about it a little bit. Well 7 times 9 is 63. That's pretty close. And then 6 times 10 is going to be too big. 7 times 10 is 70. So that's too big. So 7 goes into 64 nine times. 9 times 7 is 63. 64 minus 63 to get our remainder of this stage 1. Bring down the 7. 7 goes into 17 how many times? Well, 7 times 2 is 14. And then 7 times 3 is 21. So 3 is too big. So 7 goes into 17 two times. 2 times 7 is 14. 17 minus 14 is 3. And now we bring down the 5. That's in our 7 multiplication tables, five times. 5 times 7 is 35. And there you go. So the remainder is zero. So all the examples I did so far had no remainders. Let's do one that maybe might have a remainder. And to ensure it has a remainder I'll just make up the problem. It's much easier to make problems that have remainders than the ones that don't have remainders. So let's say I want to divide 3 into-- I'm going to divide it into, let's say 1,735,092. This will be a nice, beastly problem. So if we can do this we can handle everything. So it's 1,735,092. That's what we're dividing 3 into. And actually, I'm not sure if this will have a remainder. In the future video I'll show you how to figure out whether Actually, we can do it right now. We can just add up all these digits. 1 plus 7 is 8. 8 plus 3 is 11. 11 5 five is 16. 16 plus 9 is 25. 25 plus 2 is 27. So actually, this number is divisible by 3. So if you add up all of the digits, you get 27. And then you can add up those digits-- 2 plus 7 is 9. So that is divisible by 9. That's a trick that only works for 3. So this number actually is divisible by 3. So let me change it a little bit, so it's not divisible by 3. Let me make this into a 1. Now this number will not be divisible by 3. I definitely want a number where I'll end up with a remainder. Just so you see what it looks like. So let's do this one. 3 goes into 1 zero times. You could write a 0 here and multiply that out, but that" + }, + { + "Q": "\nAt 0:32 couldn't he just multiply 21*9=189 and then multiply the same for the other side\n28*6=176 and 189 > 176", + "A": "Not more complicated then what he did", + "video_name": "2dbasvm3iG0", + "timestamps": [ + 32 + ], + "3min_transcript": "Use less than, greater than, or equal to compare the two fractions 21/28, or 21 over 28, and 6/9, or 6 over 9. So there's a bunch of ways to do this. The easiest way is if they had the same denominator, you could just compare the numerators. Unlucky for us, we do not have the same denominator. So what we could do is we can find a common denominator for both of them and convert both of these fractions to have the same denominator and then compare the numerators. Or even more simply, we could simplify them first and then So let me do that last one, because I have a feeling that'll be the fastest way to do it. So 21/28-- you can see that they are both divisible by 7. So let's divide both the numerator and the denominator by 7. So we could divide 21 by 7. And we can divide-- so let me make the numerator-- and we can divide the denominator by 7. We're doing the same thing to the numerator and the denominator, so we're not going to change the value of the fraction. So 21 divided by 7 is 3, and 28 divided by 7 is 4. 3/4 is the simplified version of it. Let's do the same thing for 6/9. 6 and 9 are both divisible by 3. So let's divide them both by 3 so we can simplify this fraction. So let's divide both of them by 3. 6 divided by 3 is 2, and 9 divided by 3 is 3. So 21/28 is 3/4. They're the exact same fraction, just written a different way. This is the more simplified version. And 6/9 is the exact same fraction as 2/3. So we really can compare 3/4 and 2/3. So this is really comparing 3/4 and 2/3. And the real benefit of doing this is now this is much easier to find a common denominator for than 28 and 9. Then we would have to multiply big numbers. Here we could do fairly small numbers. The common denominator of 3/4 and 2/3 And 4 and 3 don't share any prime factors with each other. So their least common multiple is really just going to be the product of the two. So we can write 3/4 as something over 12. And we can write 2/3 as something over 12. And I got the 12 by multiplying 3 times 4. They have no common factors. Another way you could think about it is 4, if you do a prime factorization, is 2 times 2. And 3-- it's already a prime number, so you can't prime factorize it any more. So what you want to do is think of a number that has all of the prime factors of 4 and 3. So it needs one 2, another 2, and a 3. Well, 2 times 2 times 3 is 12. And either way you think about it, that's how you would get the least common multiple or the common denominator for 4 and 3. Well, to get from 4 to 12, you've got to multiply by 3." + }, + { + "Q": "In the video you provided \"converting decimals to fractions (ex 1) \" 1:13 I have a question, if there was a number greater than zero in front of the decimal (example 4.0727) would that change the place values to something higher than ten-thousand.\n", + "A": "The number 4.0727 is a mixed number. Any digits to the left of the decimal point are the whole number (the 4). Any digits to the right of the decimal point are the fraction. 4.0727 = 4 727/10,000 The whole number does not change the place value used to create the fraction.", + "video_name": "EGr3KC55sfU", + "timestamps": [ + 73 + ], + "3min_transcript": "Let's see if we can write 0.0727 as a fraction. Now, let's just think about what places these are in. This is in the tenths place. This is in the hundredths place. This 2 is in the thousandths place. And this 7 right here, this last 7, is in the ten-thousandths place. So there's a couple of ways we can do this. The way I like to think of this, this term over here is in the ten-thousandths place. We can view this whole thing right over here as 727 ten-thousandths because this is the smallest place right over here. So let's just rewrite it. This is equal to 727 over 10,000. And we've already written it as a fraction. And I think that's about as simplified as we can get. This number up here is not divisible by 2. It's not divisible by 5. which means it wouldn't be divisible by 6 or 9. It doesn't even seem to be divisible by 7. It might be a prime number. But I think we are done." + }, + { + "Q": "\nAt 0:00 why can't you just do this:\n31+50+64+x=180\nx=35", + "A": "Because connecting these two triangles would mean making a bigger triangle and the angle measurements would have to add up to 180\u00c2\u00b0, this would be a way to solve this problem as well. Doing this would allow you to find the missing measurement of other part of the third angle.", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 0 + ], + "3min_transcript": "What I want to do now is just a series of problems that really make sure that we know what we're doing with parallel lines and triangles and all the rest. And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here-- so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it just knowing what you know about the sums of the measures of the angles inside of a triangle, and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself because I'm about to give you the solution. So the first thing you might say-- and this is a general way to think about a lot of these problems where they give you some angles and you have to figure out some other angles based on the sum of angles and a triangle equaling 180, or this one doesn't have parallel lines on it. and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114." + }, + { + "Q": "\nAt 4:20 why is a+b=WHAT ARE THoSE", + "A": "a and b are variables; they re merely substituting in for a number that we don t know. You technically can use any letter. That s what Sal was using them for -- a demonstration where we didn t know the numbers, just that y = a + b (in the problem he solves at 4:20.)", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 260 + ], + "3min_transcript": "So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114. Notice, this 114 was the exact same sum of these 2 angles over here. And that's actually a general idea, and I'll do it on the side here just to prove it to you. If I have, let's say that these 2 angles-- let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle. So in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You could subtract 180 from both sides. You could add a plus b to both sides. Running out of space on the right hand side. And then you're left with-- these cancel out. On the left hand side, you're left with y. On the right hand side is equal to a plus b. So this is just a general property. You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees, and then you have a supplementary angles Or you could just say, look, if I have the exterior angles right over here, it's equal to the sum of the remote interior angles. That's just a little terminology you could see there. So y is equal to a plus b. 114 degrees, we've already shown to ourselves, is equal to 64 plus 50 degrees. But anyway, regardless of how we do it, if we just reason it out step by step or if we just knew this property from the get go, if we know that y is equal to 114 degrees-- and I like to reason it out every time just to make sure I'm not jumping to conclusions. So if y is 114 degrees, now we know this angle." + }, + { + "Q": "how can we findout the angles if they r in ratio\nAngles of a triangle are in the ratio 3:5:4 ,the samallest angle of the triangle is ?\n", + "A": "The ratio 3 : 5 : 4 means that if the first angle is 3\u00f0\u009d\u0091\u00a5, then the second angle is 5\u00f0\u009d\u0091\u00a5, and the third angle is 4\u00f0\u009d\u0091\u00a5. The sum of these three angles is 3\u00f0\u009d\u0091\u00a5 + 5\u00f0\u009d\u0091\u00a5 + 4\u00f0\u009d\u0091\u00a5 = 180\u00c2\u00b0 \u00e2\u0087\u0094 \u00f0\u009d\u0091\u00a5 = 180\u00c2\u00b0 \u00e2\u0088\u0095 12 = 15\u00c2\u00b0 So, the first angle is 3 \u00e2\u0088\u0099 15\u00c2\u00b0 = 45\u00c2\u00b0, the second angle is 5 \u00e2\u0088\u0099 15\u00c2\u00b0 = 75\u00c2\u00b0, and the third angle is 4 \u00e2\u0088\u0099 15\u00c2\u00b0 = 60\u00c2\u00b0", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 185 + ], + "3min_transcript": "and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114. Notice, this 114 was the exact same sum of these 2 angles over here. And that's actually a general idea, and I'll do it on the side here just to prove it to you. If I have, let's say that these 2 angles-- let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle. So in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You could subtract 180 from both sides. You could add a plus b to both sides." + }, + { + "Q": "\nAt 3:47, do angles a + b equal to y?", + "A": "Yes, actually, that s exactly right :)", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 227 + ], + "3min_transcript": "So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114. Notice, this 114 was the exact same sum of these 2 angles over here. And that's actually a general idea, and I'll do it on the side here just to prove it to you. If I have, let's say that these 2 angles-- let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle. So in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You could subtract 180 from both sides. You could add a plus b to both sides. Running out of space on the right hand side. And then you're left with-- these cancel out. On the left hand side, you're left with y. On the right hand side is equal to a plus b. So this is just a general property. You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees, and then you have a supplementary angles Or you could just say, look, if I have the exterior angles right over here, it's equal to the sum of the remote interior angles. That's just a little terminology you could see there. So y is equal to a plus b. 114 degrees, we've already shown to ourselves, is equal to 64 plus 50 degrees. But anyway, regardless of how we do it, if we just reason it out step by step or if we just knew this property from the get go, if we know that y is equal to 114 degrees-- and I like to reason it out every time just to make sure I'm not jumping to conclusions. So if y is 114 degrees, now we know this angle." + }, + { + "Q": "\nat 1:51, why did he minus 114 from 114? it would equal to zero anyways!", + "A": "He does this to show that he is subtracting the same amount from both sides, and therefore preserving the equality of the equation. Although it isn t necessary to show, it helps clarify how he is solving the equation.", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 111 + ], + "3min_transcript": "What I want to do now is just a series of problems that really make sure that we know what we're doing with parallel lines and triangles and all the rest. And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here-- so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it just knowing what you know about the sums of the measures of the angles inside of a triangle, and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself because I'm about to give you the solution. So the first thing you might say-- and this is a general way to think about a lot of these problems where they give you some angles and you have to figure out some other angles based on the sum of angles and a triangle equaling 180, or this one doesn't have parallel lines on it. and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114." + }, + { + "Q": "\n<---- push the up button if i am right, press down if ur a hatr.... My answer is 35.......\nPLZ RESPOND! I paused @ 1:50", + "A": "correct", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 110 + ], + "3min_transcript": "What I want to do now is just a series of problems that really make sure that we know what we're doing with parallel lines and triangles and all the rest. And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here-- so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it just knowing what you know about the sums of the measures of the angles inside of a triangle, and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself because I'm about to give you the solution. So the first thing you might say-- and this is a general way to think about a lot of these problems where they give you some angles and you have to figure out some other angles based on the sum of angles and a triangle equaling 180, or this one doesn't have parallel lines on it. and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114." + }, + { + "Q": "at 1:50 or so in the video, he says \"one hundred thirty-one thousand, six hundred and seventy-two\"... but isn't 2 to the 17th power 131,072?\n", + "A": "Yes, that was a mistake in the video.", + "video_name": "UCCNoXqCGZQ", + "timestamps": [ + 110 + ], + "3min_transcript": "Voiceover:Hey Britt. Voiceover:How are you? Voiceover:Good, looks like we have a game going on here. Yeah, kind of a challenge question for you. What I did is, I put 1 grain of rice in the first square. Voiceover:That's right. Voiceover:There's 64 squares on the board. Voiceover:Yup. Voiceover:And in each consecutive square I doubled the amount of rice. Voiceover:How much rice do you think would be on this square? Voiceover:On that square? Let me think about it a little bit. Actually, I'm going to take some ... Here you have 1 and we multiply that times 2, so this is going to be 2 times 2. No, no 2 times 1, what am I doing? Now this is 2 times that one so this is 2 times 2. Now this is 2 times that. So this is ... Okay, we're starting to take a lot of 2's here and multiplying them together. So this is 2 times 2 ... I'm trying to write sideways. Times 2. This one is going to be 5, 2's multiplied together. This is going to be 6, 2's multiplied together. 8, 2's multiplied together. 9, 2's. 10, 11, 12, 13. So all of this stuff multiplied together. 8,192 grains of rice is what we should see right over here. Voiceover:And you know, I had fun last night and I was up late, but there you go. Voiceover:Did you really count out 8,192 grains of rice? Voiceover:More or less. Voiceover:Okay. Let's just say you did. Voiceover:What if we just went, you know, 4 steps ahead. How much rice would be here? Voiceover:4 steps ahead, so we're going to multiple by 2, then multiple by 2 again, then multiply by 2 again, the multiply by 2 again. So it's this number times ... Let's see, 2 times 2 is 4. Times 2 is 8, times 2 is 16. So it's going to get us like 120, like 130,000 or around there. Voiceover:131,672. Voiceover:You had a lot of time last night. We're not even halfway across the board yet. Voiceover:We're not. Voiceover:This is a lot of ... You could throw a party. Voiceover:What about the last square? This is 63 steps. Voiceover:We're going to take 2 times 2 and we're going to do 63 of those. So this is going to be a huge number. And actually, it would be neat if there was a notation for that. Voiceover:I didn't count this one out but it is the size of Mount Everest, the pile of rice. And it would feed 485 trillion people. I mean, you know, this was a little bit of a pain for me to write all of these 2's. Voiceover:So was this. Voiceover:If I were the mathematical community I would want some type of notation. Voiceover:You kind of got on it here. I like this dot, dot, dot and the 63. This I understand this. Voiceover:Yeah, you could understand this but this is still a little bit ... This is a little bit too much. What if, instead, we just wrote ... Voiceover:Mathematicians love being efficient, right? They're lazy. Voiceover:Yeah, they have things to do. They have to go home and count grains of rice. (laughter) Voiceover:Yeah. So that is, take 63, 2's and multiply them" + }, + { + "Q": "my bad sorry at 0:13 he says it has 64 squares. But then at 2:07 he says 63 but there is 64 squares.\n", + "A": "He says there are 63 steps. This is measured from the first square. Therefore, the first square had 0 steps, the second square had 1 step, and so on up to the sixty-fourth square, which had 63 steps to get to it.", + "video_name": "UCCNoXqCGZQ", + "timestamps": [ + 13, + 127 + ], + "3min_transcript": "Voiceover:Hey Britt. Voiceover:How are you? Voiceover:Good, looks like we have a game going on here. Yeah, kind of a challenge question for you. What I did is, I put 1 grain of rice in the first square. Voiceover:That's right. Voiceover:There's 64 squares on the board. Voiceover:Yup. Voiceover:And in each consecutive square I doubled the amount of rice. Voiceover:How much rice do you think would be on this square? Voiceover:On that square? Let me think about it a little bit. Actually, I'm going to take some ... Here you have 1 and we multiply that times 2, so this is going to be 2 times 2. No, no 2 times 1, what am I doing? Now this is 2 times that one so this is 2 times 2. Now this is 2 times that. So this is ... Okay, we're starting to take a lot of 2's here and multiplying them together. So this is 2 times 2 ... I'm trying to write sideways. Times 2. This one is going to be 5, 2's multiplied together. This is going to be 6, 2's multiplied together. 8, 2's multiplied together. 9, 2's. 10, 11, 12, 13. So all of this stuff multiplied together. 8,192 grains of rice is what we should see right over here. Voiceover:And you know, I had fun last night and I was up late, but there you go. Voiceover:Did you really count out 8,192 grains of rice? Voiceover:More or less. Voiceover:Okay. Let's just say you did. Voiceover:What if we just went, you know, 4 steps ahead. How much rice would be here? Voiceover:4 steps ahead, so we're going to multiple by 2, then multiple by 2 again, then multiply by 2 again, the multiply by 2 again. So it's this number times ... Let's see, 2 times 2 is 4. Times 2 is 8, times 2 is 16. So it's going to get us like 120, like 130,000 or around there. Voiceover:131,672. Voiceover:You had a lot of time last night. We're not even halfway across the board yet. Voiceover:We're not. Voiceover:This is a lot of ... You could throw a party. Voiceover:What about the last square? This is 63 steps. Voiceover:We're going to take 2 times 2 and we're going to do 63 of those. So this is going to be a huge number. And actually, it would be neat if there was a notation for that. Voiceover:I didn't count this one out but it is the size of Mount Everest, the pile of rice. And it would feed 485 trillion people. I mean, you know, this was a little bit of a pain for me to write all of these 2's. Voiceover:So was this. Voiceover:If I were the mathematical community I would want some type of notation. Voiceover:You kind of got on it here. I like this dot, dot, dot and the 63. This I understand this. Voiceover:Yeah, you could understand this but this is still a little bit ... This is a little bit too much. What if, instead, we just wrote ... Voiceover:Mathematicians love being efficient, right? They're lazy. Voiceover:Yeah, they have things to do. They have to go home and count grains of rice. (laughter) Voiceover:Yeah. So that is, take 63, 2's and multiply them" + }, + { + "Q": "\nAt 0:19, Sal says two lines are parallel if they are on the same plane and will never intersect. If two lines are on different planes and never intersect, what is that called?", + "A": "They are called skew!", + "video_name": "aq_XL6FrmGs", + "timestamps": [ + 19 + ], + "3min_transcript": "Identify all sets of parallel and perpendicular lines in the image below. So let's start with the parallel lines. And just as a reminder, two lines are parallel if they're in the same plane, and all of these lines are clearly in the same plane. They're in the plane of the screen you're viewing right now. But they are two lines that are in the same plane that never intersect. And one way to verify, because you can sometimes-- it looks like two lines won't intersect, but you can't just always assume based on how it looks. You really have to have some information given in the diagram or the problem that tells you that they are definitely parallel, that they're definitely never going to intersect. And one of those pieces of information which they give right over here is that they show that line ST and line UV, they both intersect line CD at the exact same angle, at this angle right here. And in particular, it's at a right angle. And if you have two lines that intersect a third line at the same angle-- so these are actually called corresponding angles and they're the same-- then these two lines are parallel. So line ST is parallel to line UV. And we can write it like this. Line ST, we put the arrows on each end of that top bar to say that this is a line, not just a line segment. Line ST is parallel to line UV. And I think that's the only set of parallel lines in this diagram. Yep. Now let's think about perpendicular lines. Perpendicular lines are lines that intersect at a 90-degree angle. So, for example, line ST is perpendicular to line CD. So line ST is perpendicular to line CD. And we know that they intersect at a right angle or at a 90-degree angle because they gave us this little box here which literally means that the measure of this angle is 90 degrees. Let me make sure I specified these as lines. Line UV is perpendicular to CD. So I did UV, ST, they're perpendicular to CD. And then after that, the only other information where they definitely tell us that two lines are intersecting at right angles are line AB and WX. So AB is definitely perpendicular to WX, line WX. And I think we are done. And one thing to think about, AB and CD, well, they don't even intersect in this diagram. So you can't make any comment about perpendicular, but they're definitely not parallel. You could even imagine that it looks like they're about to intersect. And they give us no information that they intersect the same lines at the same angle. So if somehow they told us that this is a right angle, even" + }, + { + "Q": "So whenever (3:01) two lines intersect at a right angle, it's perpendicular? Is there a case where a line is perpendicular to another line without intersecting at a right angle?\n", + "A": "No, you can t have a line that is perpendicular to another line without them intersecting at a right angle, because that is the very definition of perpendicular. The fact that two lines intersect at right angles is what makes them perpendicular. (1:37) And don t forget, right angle just means a 90 degree angle. Also a box indicated at the intersection is what denotes a right angle.", + "video_name": "aq_XL6FrmGs", + "timestamps": [ + 181 + ], + "3min_transcript": "then these two lines are parallel. So line ST is parallel to line UV. And we can write it like this. Line ST, we put the arrows on each end of that top bar to say that this is a line, not just a line segment. Line ST is parallel to line UV. And I think that's the only set of parallel lines in this diagram. Yep. Now let's think about perpendicular lines. Perpendicular lines are lines that intersect at a 90-degree angle. So, for example, line ST is perpendicular to line CD. So line ST is perpendicular to line CD. And we know that they intersect at a right angle or at a 90-degree angle because they gave us this little box here which literally means that the measure of this angle is 90 degrees. Let me make sure I specified these as lines. Line UV is perpendicular to CD. So I did UV, ST, they're perpendicular to CD. And then after that, the only other information where they definitely tell us that two lines are intersecting at right angles are line AB and WX. So AB is definitely perpendicular to WX, line WX. And I think we are done. And one thing to think about, AB and CD, well, they don't even intersect in this diagram. So you can't make any comment about perpendicular, but they're definitely not parallel. You could even imagine that it looks like they're about to intersect. And they give us no information that they intersect the same lines at the same angle. So if somehow they told us that this is a right angle, even then we would have to suspend our judgment based on how it actually looks and say, oh, I guess maybe those things are perpendicular, or maybe these two things are parallel. But they didn't tell us that. And that would actually be bizarre because it looks so not parallel. And actually then this would end up being parallel to other things as well if that was done. It's a good thing that wasn't because it would look very strange. But based on the information they gave us, these are the parallel and the perpendicular lines." + }, + { + "Q": "\nAt 3:34, Is the slope of the tangent line constant at any point along the curve?", + "A": "The slope of the tangent line is constant on the tangent line, but only for that tangent line. Each point on the curve has a different tangent line, meaning each point on the curve will have a different slope.", + "video_name": "fqQ6sslzyhY", + "timestamps": [ + 214 + ], + "3min_transcript": "and so we don't know anything other than f but we can imagine what f looks like. Our function f could, so our function f, it could look something like this. It just has to be tangent so that line has to be tangent to our function right at that point. So our function f could look something like that. So when they say, find f prime of two, they're really saying, what is the slope of the tangent line when x is equal to two? So when x is equal to two, well the slope of the tangent line is the slope of this line. They gave us, they gave us the two points that sit on the tangent line. So we just have to figure out its slope because that is going to be the rate of change of that function right over there, It's going to be the slope of the tangent line because this is the tangent line. So let's do that. So as we know, slope is change in y over change in x. So if we change our, our change in x, change in x, we go from x equals two to x equals 7 so our change in x is equal to five. And our change in y, our change in y, we go from y equals three to y equals six. So our change in y is equal to three. So our change in y over change in x is going to be three over five which is the slope of this line, which is the derivative of the function at two because this is the tangent line at x equals two. Let's do another one of these. For a function g, we are given that g of negative one equals three and g prime of negative one is equal to negative two. What is the equation of the tangent line to the graph of g Alright, so once again I think it will be helpful to graph this. So we have our y axis, we have our x axis and let's see. We say for function g we are given that g of negative one is equal to three. So the point negative one comma three is on our function. So this is negative one and then we have, one, two, and three. So that's that right over there. That is the point. That is the point negative one comma three, it's going to be on our function. And we also know that g prime of negative one is equal to negative two. So the slope of the tangent line right at that point on our function is going to be negative two. That's what that tells us. The slope of the tangent line, when x is equal to negative one is equal to negative two. So I could use that information to actually draw the tangent line." + }, + { + "Q": "Did Sal mean to say -4/2=-2 instead of 4/2=2 at 4:07?\n", + "A": "No, he s actually right. I ll explain why. At 4:07 of the video, all he does is from y-4=2x, he brings the two from that side over to this side, and divides y-4 by 2. So that way it becomes: y/2 - 4/2 = x. and -4/2 =-2. Hope that helped.", + "video_name": "W84lObmOp8M", + "timestamps": [ + 247 + ], + "3min_transcript": "This right here, that is equal to f of 2. Same idea. You start with 3, 3 is being mapped by the function to 10. It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2, or is there a way to go back from the 10 to the 3? Or is there some other function? Is there some other function, we can call that the inverse of f, that'll take us back? Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as notation, and it'll take us back from 10 to 3. Is there a way to do that? Will that same inverse of f, will it take us back from-- if we apply 8 to it-- will that take us back to 2? What you'll find is it's actually very easy to solve for this inverse of f, and I think once we solve for it, it'll make it clear what I'm talking about. That the function takes you from 2 to 8, the inverse will take us back from 8 to 2. So to think about that, let's just define-- let's just say y is equal to f of x. So y is equal to f of x, is equal to 2x plus 4. So I can write just y is equal to 2x plus 4, and this once again, this is our function. You give me an x, it'll give me a y. But we want to go the other way around. We want to give you a y and get an x. So all we have to do is solve for x in terms of y. So let's do that. If we subtract 4 from both sides of this equation-- let me switch colors-- if we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x, and then if we minus 2-- 4 divided by 2 is 2-- is equal to x. Or if we just want to write it that way, we can just swap the sides, we get x is equal to 1/2y-- same thing as y over 2-- minus 2. So what we have here is a function of y that gives us an x, which is exactly what we wanted. We want a function of these values that map back to an x. So we can call this-- we could say that this is equal to-- I'll do it in the same color-- this is equal to f inverse as a function of y. Or let me just write it a little bit cleaner. We could say f inverse as a function of y-- so we can have 10 or 8-- so now the range is now the domain for f inverse. f inverse as a function of y is equal to 1/2y minus 2. So all we did is we started with our original function, y is equal to 2x plus 4, we solved for-- over here, we've" + }, + { + "Q": "\nAt 0:58, Sal says you could input any real number into the function. Couldn't you input an imaginary number as well?", + "A": "Yes, it could be a nonreal complex number. However, it is not typical to discuss complex domains and ranges at this level of study. Some teachers might cover that topic, but it is not standard. So, for now, assume that ranges and domains are specifically referring to real numbers unless instructed otherwise.", + "video_name": "W84lObmOp8M", + "timestamps": [ + 58 + ], + "3min_transcript": "Let's think about what functions really do, and then we'll think about the idea of an inverse of a function. So let's start with a pretty straightforward function. Let's say f of x is equal to 2x plus 4. And so if I take f of 2, f of 2 is going to be equal to 2 times 2 plus 4, which is 4 plus 4, which is 8. I could take f of 3, which is 2 times 3 plus 4, which is equal to 10. 6 plus 4. So let's think about it in a little bit more of an abstract sense. So there's a set of things that I can input into this function. You might already be familiar with that notion. It's the domain. The set of all of the things that I can input into that function, that is the domain. And in that domain, 2 is sitting there, you have 3 over there, pretty much you could input any real number into this function. So this is going to be all real, but we're making it a Now, when you apply the function, let's think about it means to take f of 2. We're inputting a number, 2, and then the function is outputting the number 8. It is mapping us from 2 to 8. So let's make another set here of all of the possible values that my function can take on. And we can call that the range. There are more formal ways to talk about this, and there's a much more rigorous discussion of this later on, especially in the linear algebra playlist, but this is all the different values I can take on. So if I take the number 2 from our domain, I input it into the function, we're getting mapped to the number 8. So let's let me draw that out. So we're going from 2 to the number 8 right there. And it's being done by the function. The function is doing that mapping. This right here, that is equal to f of 2. Same idea. You start with 3, 3 is being mapped by the function to 10. It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2, or is there a way to go back from the 10 to the 3? Or is there some other function? Is there some other function, we can call that the inverse of f, that'll take us back? Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as notation, and it'll take us back from 10 to 3. Is there a way to do that? Will that same inverse of f, will it take us back from-- if we apply 8 to it-- will that take us back to 2?" + }, + { + "Q": "@2:45 sal say that inverse of F so does he mean that the inverse is a different function\n", + "A": "Edgar Peixoto thnx a lot :)", + "video_name": "W84lObmOp8M", + "timestamps": [ + 165 + ], + "3min_transcript": "Now, when you apply the function, let's think about it means to take f of 2. We're inputting a number, 2, and then the function is outputting the number 8. It is mapping us from 2 to 8. So let's make another set here of all of the possible values that my function can take on. And we can call that the range. There are more formal ways to talk about this, and there's a much more rigorous discussion of this later on, especially in the linear algebra playlist, but this is all the different values I can take on. So if I take the number 2 from our domain, I input it into the function, we're getting mapped to the number 8. So let's let me draw that out. So we're going from 2 to the number 8 right there. And it's being done by the function. The function is doing that mapping. This right here, that is equal to f of 2. Same idea. You start with 3, 3 is being mapped by the function to 10. It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2, or is there a way to go back from the 10 to the 3? Or is there some other function? Is there some other function, we can call that the inverse of f, that'll take us back? Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as notation, and it'll take us back from 10 to 3. Is there a way to do that? Will that same inverse of f, will it take us back from-- if we apply 8 to it-- will that take us back to 2? What you'll find is it's actually very easy to solve for this inverse of f, and I think once we solve for it, it'll make it clear what I'm talking about. That the function takes you from 2 to 8, the inverse will take us back from 8 to 2. So to think about that, let's just define-- let's just say y is equal to f of x. So y is equal to f of x, is equal to 2x plus 4. So I can write just y is equal to 2x plus 4, and this once again, this is our function. You give me an x, it'll give me a y. But we want to go the other way around. We want to give you a y and get an x. So all we have to do is solve for x in terms of y. So let's do that. If we subtract 4 from both sides of this equation-- let me switch colors-- if we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x, and then if we" + }, + { + "Q": "@ 0:05 Sal says that 12 is arbitrary. What does that mean?\n", + "A": "Arbitrary means random, and how he uses it, he means he s picking a random number from his mind, there s really no specific reason he used the number 12. What he was basically saying was: let me just pick a random number, the number 12. instead of, let me just pick an arbitrary number, the number 12. Why does he say arbitrary instead of random ? Well, he s more sophisticated than you and me, and he has a bigger vocabulary than us, so he s probably used to saying arbitrary instead of random . Got it?", + "video_name": "I6TBBzIvgB8", + "timestamps": [ + 5 + ], + "3min_transcript": "- In earlier mathematics that you may have done, you probably got familiar with the idea of a factor. So for example, let me just pick an arbitrary number, the number 12. We could say that the number 12 is the product of say two and six; two times six is equal to 12. So because if you take the product of two and six, you get 12, we could say that two is a factor of 12, we could also say that six is a factor of 12. You take the product of these things and you get 12! You could even say that this is 12 in factored form. People don't really talk that way but you could think of it that way. We broke 12 into the things that we could use to multiply. And you probably remember from earlier mathematics the notion of prime factorization, where you break it up into all of the prime factors. So in that case you could break the six into a two and a three, and you have two times two times three is equal to 12. And you'd say, \"Well, this would be 12 \"in prime factored form or the prime factorization of 12,\" so these are the prime factors. is things that you can multiply together to get your original thing. Or if you're talking about factored form, you're essentially taking the number and you're breaking it up into the things that when you multiply them together, you get your original number. What we're going to do now is extend this idea into the algebraic domain. So if we start with an expression, let's say the expression is two plus four X, can we break this up into the product of two either numbers or two expressions or the product of a number and an expression? Well, one thing that might jump out at you is we can write this as two times one plus two X. And you can verify if you like that this does indeed equal two plus four X. We're just going to distribute the two. is equal to four X, so plus four X. So in our algebra brains, this will often be reviewed as or referred to as this expression factored or in a factored form. Sometimes people would say that we have factored out the two. You could just as easily say that you have factored out a one plus two X. You have broken this thing up into two of its factors. So let's do a couple of examples of this and then we'll think about, you know, I just told you that we could write it this way but how do you actually figure that out? So let's do another one. Let's say that you had, I don't know, let's say you had, six, let me just in a different color, let's say you had six X six X plus three," + }, + { + "Q": "\"You'll appreciate this even more when you get into calculus.\"-Sal, 2:53\nWas he referring to how the second derivative of x^2 is 2?\n", + "A": "I believe what he was getting at is that the second derivative of a quadratic equation is always a constant, and because his second step in looking at the change between adjacent numbers did not produce a constant change, the formula can t be quadratic, which is why we go on to determine that it must be a cubic function.", + "video_name": "i7iKLZQ-vCk", + "timestamps": [ + 173 + ], + "3min_transcript": "And now, for each of these, let's see what the output of our function should be. The output of the function should be this thing. It should be the sum from i equals 0 to n of i squared. So when n is 0-- well, that's just going to be 0 squared. We'd just stop right over there. So that's just 0. When n is 1, it's 0 squared plus 1 squared. So that is 1. When n is 2, it's 0 squared plus 1 squared plus 2 squared. So that's 1 plus 4, which is 5. When n is 3, now we go all the way to 3. So it's going to be 1 plus 4, which is 5, plus 9. So 5 plus 9 is 14. And then when n is 4, we're going to add the 16, 4 squared, So this gets us to 30. And of course, we could keep going on and on and on. So let's study this a little bit to think about what type of a function that, for each of these inputs, might give us this type of an output. So the difference here is 1. The difference here is 4. And this is obvious. We added 2 squared here. We added 3 squared, or 9, here. We added 4 squared, or 16, here. And the reason why I'm doing this is if this was a linear function, then the difference between successive terms would be the same. Now, if this is a quadratic function, then the differences between the differences would be the same. Let's see if that's the case. So the difference here is 1. The difference here is 4. So the difference between those is 3. The difference here is 5. The difference here is 7. So even the difference of the differences is increasing. But if this is a cubic function, then the differences of the difference of the difference should be constant. So let's see if that's the case. And you'll appreciate this even more when you start learning calculus. The difference between 3 and 5 is 2. The difference between 5 and 7 is 2. So the fact that the difference of the difference of the difference is fixed tells us that we should be able to express this as some type of a cubic function. So this we could write as this should be equal to some function in terms of n. And we could write it as An to the third plus Bn squared plus C times n plus D. And now we can just use what the inputs are and the outputs are of these to solve for A, B, C, and D. And I encourage you to do that. Well, let's first think about when n is equal to 0. When n is equal to 0, this function evaluates to D. So this function evaluates to D. But that function needs to evaluate to 0, so D needs to be 0. So I'm just trying to fix these letters here to get the right outputs." + }, + { + "Q": "\nAt 3:19 sal says the sum of 1^2 + 2^2 + ... is equal to An^3 + Bn^2 + Cn + D.\nI understand why he used An^3 + Bn^2 + Cn + D but any chance there's a video which explain that in further details?", + "A": "A^3 Bn^2 +Cn^ + D is just the general form of a cubic equation. if he had determined that the series was quadratic he would have used An^2 + Bn + C .. does that help?", + "video_name": "i7iKLZQ-vCk", + "timestamps": [ + 199 + ], + "3min_transcript": "And now, for each of these, let's see what the output of our function should be. The output of the function should be this thing. It should be the sum from i equals 0 to n of i squared. So when n is 0-- well, that's just going to be 0 squared. We'd just stop right over there. So that's just 0. When n is 1, it's 0 squared plus 1 squared. So that is 1. When n is 2, it's 0 squared plus 1 squared plus 2 squared. So that's 1 plus 4, which is 5. When n is 3, now we go all the way to 3. So it's going to be 1 plus 4, which is 5, plus 9. So 5 plus 9 is 14. And then when n is 4, we're going to add the 16, 4 squared, So this gets us to 30. And of course, we could keep going on and on and on. So let's study this a little bit to think about what type of a function that, for each of these inputs, might give us this type of an output. So the difference here is 1. The difference here is 4. And this is obvious. We added 2 squared here. We added 3 squared, or 9, here. We added 4 squared, or 16, here. And the reason why I'm doing this is if this was a linear function, then the difference between successive terms would be the same. Now, if this is a quadratic function, then the differences between the differences would be the same. Let's see if that's the case. So the difference here is 1. The difference here is 4. So the difference between those is 3. The difference here is 5. The difference here is 7. So even the difference of the differences is increasing. But if this is a cubic function, then the differences of the difference of the difference should be constant. So let's see if that's the case. And you'll appreciate this even more when you start learning calculus. The difference between 3 and 5 is 2. The difference between 5 and 7 is 2. So the fact that the difference of the difference of the difference is fixed tells us that we should be able to express this as some type of a cubic function. So this we could write as this should be equal to some function in terms of n. And we could write it as An to the third plus Bn squared plus C times n plus D. And now we can just use what the inputs are and the outputs are of these to solve for A, B, C, and D. And I encourage you to do that. Well, let's first think about when n is equal to 0. When n is equal to 0, this function evaluates to D. So this function evaluates to D. But that function needs to evaluate to 0, so D needs to be 0. So I'm just trying to fix these letters here to get the right outputs." + }, + { + "Q": "at 2:18, if your going to do a graph is it easier to do a bar graph?\n", + "A": "It can be, but there are other ways to do it too", + "video_name": "0ZKtsUkrgFQ", + "timestamps": [ + 138 + ], + "3min_transcript": "- [Voiceover] What I want to do in this video is think about all of all the different ways that we can represent data. So right over here, we have a list of, and I'm just using this as one form of data, a list of students' scores on, say, the last test, so Amy got 90 percent right, Bill got 95 percent right, Cam got 100 percent right, Efra also got 100 percent right, and Farah got 80 percent right. This is one way to show data. Remember, data is just recorded information, and it could be numeric like this, it could be quantitative, so you're recording actual numbers, or it could even be things you could record data on how do they like the test, and they could have scored it based on, I really liked it, I kind of liked it, I didn't like it, or they might have rated it on a scale of zero to five, which would have been numbers, but it's numbers that are measuring peoples' opinions, as opposed to, here, we have numbers that are measuring their actual scores. So there's all different types of data, and I don't want to get into all of that, but let's just start thinking about different ways to represent this data. So this is one way, you could view this as a table where you have the name, So you have your name column, and then you have your score column. And I can construct it as a table, so clearly, it looks like a table. Like that, that's one way, one very common way of representing data, just like that. That's actually how most traditional databases record data, in tables like this. But you could also do it in other ways. So you could record it as a... Often times called a bar graph, or sometimes, a histogram, so you could put score on the vertical axis here, and then you could have your names over here. And let's see the scores, let's see, maybe we'll make this a 50. Actually, let me just mark them off. So this is 10, 20, 30, that's too big. 60, 70, 80, 90, so that's... And then 100, so that's 100. One, two, three, four, five, that would be 50 right over there, and then you can go person by person. So Amy got a 90 on the exam, so the bar will go up to 90. So that is Amy, and then you have Bill, got a 95, so it's going to be between 90 and 100, so it's going to be right over there. Bill got a 95, and so it'll look like this. Bill, so that is Bill. And then you have Cam, who got 100, on the exam. So, make sure, you see I'm hand drawing it, so it's not as precise as if I were to do it on a computer. So this right over there, that is Cam's score." + }, + { + "Q": "Hmm... So at 5:08...why doesn't he take the product of 3 *3 ? Both numbers have 3 in common.. Because at 6:15, Sal says that you take the product of both numbers. In that case at 6:15, he multiplies the 3 and the 5... So why doesn't that apply to the situation at 5:08 with the numbers sharing the 3 in common?\n", + "A": "The numbers 21 and 30 have one factor in common: 3 If you tried to use 3*3, you would be saying that 9 is a common factor of 21 and 30. Or, that you can divide both numbers evenly by 9. You can t. The numbers 105 and 30 share 2 common factors: 3 and 5. You can evenly divide both numbers by 3 and 5. Thus, you can also evenly divide both of them by 3*5 = 15. So, the greatest common factor = 15. Do you see the difference? Comment back if you have more questions.", + "video_name": "bLTfBvkrfsM", + "timestamps": [ + 308, + 375, + 375, + 308 + ], + "3min_transcript": "Well, it's 1 and 21, and 3, and 7. I think I've got all of them. And 30 can be written as 1 and 30, 2 and 15, and 3-- actually, I'm going to run out of them. Let me write it this way so I get a little more space. So 1 and 30. 2 and 15. 3 and 10. And 5 and 6. So here are all of the factors of 30. And now what are the common factors? Well, 1 is a common factor. 3 is also a common factor. But what is the greatest common factor or the greatest common divisor? Well, it is going to be 3. Now, I keep talking about another technique. Let me show you the other technique, and that involves the prime factorization. So if you say the prime factorization of 21-- well, let's see, it's divisible by 3. It is 3 times 7. And the prime factorization of 30 is equal to 3 times 10, and 10 is 2 times 5. So what are the most factors that we can take from both 21 and 30 to make the largest possible numbers? So when you look at the prime factorization, the only thing that's common right over here is a 3. And so we would say that the greatest common factor or the greatest common divisor of 21 and 30 is 3. If you saw nothing in common right over here, then you say the greatest common divisor is one. Let me give you another interesting example, just so So let's say these two numbers were not 21 and 30, but let's say we care about the greatest common divisor not of 21, but let's say of 105 and 30. So if we did the prime factorization method, it might become a little clearer now. Actually figuring out, hey, what are all the factors of 105 might be a little bit of a pain, but if you do a prime factorization, you'd say, well, let's see, 105-- it's divisible by 5, definitely. So it's 5 times 21, and 21 is 3 times 7. So the prime factorization of 105 is equal to-- if I write them in increasing order-- 3 times 5 times 7. The prime factorization of 30, we already figured out is 30 is equal to 2 times 3 times 5. So what's the most number of factors" + }, + { + "Q": "at 4:59, when you say the greatest common divisor is one, that's wrong, it is actually three.\n", + "A": "no, if you go back a couple of seconds it actually says If you see that there s nothing in common here, the GCD or GCF would be 1", + "video_name": "bLTfBvkrfsM", + "timestamps": [ + 299 + ], + "3min_transcript": "Well, it's 1 and 21, and 3, and 7. I think I've got all of them. And 30 can be written as 1 and 30, 2 and 15, and 3-- actually, I'm going to run out of them. Let me write it this way so I get a little more space. So 1 and 30. 2 and 15. 3 and 10. And 5 and 6. So here are all of the factors of 30. And now what are the common factors? Well, 1 is a common factor. 3 is also a common factor. But what is the greatest common factor or the greatest common divisor? Well, it is going to be 3. Now, I keep talking about another technique. Let me show you the other technique, and that involves the prime factorization. So if you say the prime factorization of 21-- well, let's see, it's divisible by 3. It is 3 times 7. And the prime factorization of 30 is equal to 3 times 10, and 10 is 2 times 5. So what are the most factors that we can take from both 21 and 30 to make the largest possible numbers? So when you look at the prime factorization, the only thing that's common right over here is a 3. And so we would say that the greatest common factor or the greatest common divisor of 21 and 30 is 3. If you saw nothing in common right over here, then you say the greatest common divisor is one. Let me give you another interesting example, just so So let's say these two numbers were not 21 and 30, but let's say we care about the greatest common divisor not of 21, but let's say of 105 and 30. So if we did the prime factorization method, it might become a little clearer now. Actually figuring out, hey, what are all the factors of 105 might be a little bit of a pain, but if you do a prime factorization, you'd say, well, let's see, 105-- it's divisible by 5, definitely. So it's 5 times 21, and 21 is 3 times 7. So the prime factorization of 105 is equal to-- if I write them in increasing order-- 3 times 5 times 7. The prime factorization of 30, we already figured out is 30 is equal to 2 times 3 times 5. So what's the most number of factors" + }, + { + "Q": "\nAt 4:25 Sal says the prime factorisation of 30 is 3*10\n\nBut shouldn't it be 2*15? Aren't you supposed to start with the lowest prime that goes into the number in question, which would make it 2 and not 3?", + "A": "It doesn t matter which prime you start dividing with. The resulting prime factors will always be the same. Method 1: 30 = 3 * 10 -> 2 * 5 prime factors = 3, 2, 5 Method 2: 30 = 2 * 15 -> 3 * 5 prime factors = 2, 3, 5", + "video_name": "bLTfBvkrfsM", + "timestamps": [ + 265 + ], + "3min_transcript": "So what is the greatest common factor? Well, there's only one common factor here, 1. 1 is the only common factor. So the greatest common factor of 10 and 7, or the greatest common divisor, is going to be equal to 1. So let's write that down. 1. Let's do one more. What is the greatest common divisor of 21 and 30? And this is just another way of saying that. So 21 and 30 are the two numbers that we care about. So we want to figure out the greatest common divisor, and I could have written greatest common factor, of 21 and 30. So once again, there's two ways of doing this. And so there's the way I did the last time where I literally list all the factors. Let me do it that way really fast. Well, it's 1 and 21, and 3, and 7. I think I've got all of them. And 30 can be written as 1 and 30, 2 and 15, and 3-- actually, I'm going to run out of them. Let me write it this way so I get a little more space. So 1 and 30. 2 and 15. 3 and 10. And 5 and 6. So here are all of the factors of 30. And now what are the common factors? Well, 1 is a common factor. 3 is also a common factor. But what is the greatest common factor or the greatest common divisor? Well, it is going to be 3. Now, I keep talking about another technique. Let me show you the other technique, and that involves the prime factorization. So if you say the prime factorization of 21-- well, let's see, it's divisible by 3. It is 3 times 7. And the prime factorization of 30 is equal to 3 times 10, and 10 is 2 times 5. So what are the most factors that we can take from both 21 and 30 to make the largest possible numbers? So when you look at the prime factorization, the only thing that's common right over here is a 3. And so we would say that the greatest common factor or the greatest common divisor of 21 and 30 is 3. If you saw nothing in common right over here, then you say the greatest common divisor is one. Let me give you another interesting example, just so" + }, + { + "Q": "In 4:28, I don't understand how in the equation b=2(m - 1/2) - 1/2, the 1/2 in brackets becomes -1..... that is the equation becomes b=2m -1 - 1/2\n\nAnd also in 4:44, how does 2m -1 - 1/2 become 2m - 3/2?\n", + "A": "First, everything within the brackets is multiplied by 2, hence b = 2m, -2/2 (2/2 = 1). Second, the same principle applies, 2/2 is the same thing as 1, so -1 (or - 2/2) plus - 1/2 equals -3/2", + "video_name": "cNlwi6lUCEM", + "timestamps": [ + 268, + 284 + ], + "3min_transcript": "So that's what that sentence in orange is telling us. The top shelf needs to be 1/2 a foot shorter than the length of the middle shelf. Now, what does the next statement tell us? And the bottom shelf to be-- so the bottom shelf needs to be equal to 1/2 a foot shorter than-- so it's 1/2 a foot shorter than twice the length of the top shelf. So it's 1/2 a foot shorter than twice the length of the top shelf. These are the two statements interpreted in equal equation form. The top shelf's length has to be equal to the middle shelf's length minus 1/2. It's 1/2 foot shorter than the middle shelf. And the bottom shelf needs to be 1/2 a foot shorter than twice the length of the top shelf. And so how do we solve this? Well, you can't just solve it just with these two constraints, but they gave us more information. entire 12 feet of wood? So the length of all of the shelves have to add up to 12 feet. She's using all of it. So t plus m, plus b needs to be equal to 12 feet. That's the length of each of them. She's using all 12 feet of the wood. So the lengths have to add to 12. So what can we do here? Well, we can get everything here in terms of one variable, maybe we'll do it in terms of m, and then substitute. So we already have t in terms of m. We could, everywhere we see a t, we could substitute with m minus 1/2. But here we have b in terms of t. So how can we put this in terms of m? Well, we know that t is equal to m minus 1/2. So let's take, everywhere we see a t, let's substitute it with this thing right here. That is what t is equal to. So we can rewrite this blue equation as, the length of the but we know that t is equal to m minus 1/2. And if we wanted to simplify that a little bit, this would be that the bottom shelf is equal to-- let's distribute the 2-- 2 times m is 2m. 2 times negative 1/2 is negative 1. And then minus another 1/2. Or, we could rewrite this as b is equal to 2 times the middle shelf minus 3/2. 1/2 is 2/2 minus another 1/2 is negative 3/2, just like that. So now we have everything in terms of m, and we can substitute back here. So the top shelf-- instead of putting a t there, we could put m minus 1/2. So we put m minus 1/2, plus the length of the middle" + }, + { + "Q": "Why is it that 6- (minus) 1/2 (one half) = 5 1/2 (five and a half) at the time of the video 7:37 ???\n", + "A": "The reason why 6- (minus) 1/2 (one half)= 5 1/2 is because of this: You have 6 units for example this means you have six whole units. Now you want to take away 1/2 of a unit away. What is 1/2? 1/2 is half of a whole unit. So if you have six whole units and then you take away one half of a unit what will you have? 5 1/2. A whole unit is 1, half of that unit is 1/2. I hope this helped.", + "video_name": "cNlwi6lUCEM", + "timestamps": [ + 457 + ], + "3min_transcript": "Well, we already put that in terms of m. That's what we just did. This is the length of the bottom shelf in terms of m. So instead of writing b there, we could write 2m minus 3/2. Plus 2m minus 3/2, and that is equal to 12. All we did is substitute for t. We wrote t in terms of m, and we wrote b in terms of m. Now let's combine the m terms and the constant terms. So if we have, we have one m here, we have another m there, and then we have a 2m there. They're all positive. So 1 plus 1, plus 2 is 4m. So we have 4m. And then what do our constant terms tell us? We have a negative 1/2, and then we have a negative 3/2. So negative 1/2 minus 3/2, that is negative 4/2 or negative 2. So we have 4m minus 2. Now, we want to isolate just the m variable on one side of So let's add 2 to both sides to get rid of this 2 on the left-hand side. So if we add 2 to both sides of this equation, the left-hand side, we're just left with 4m-- these guys cancel out-- is equal to 14. Now, divide both sides by 4, we get m is equal to 14 over 4, or we could call that 7/2 feet, because we're doing everything in feet. So we solved for m, but now we still have to solve for t and b. Let's solve for t. t is equal to m minus 1/2. So it's equal to-- our m is 7/2 minus 1/2, which is equal to 6/2, or 3 feet. know it's feet there. So that's the top shelf is 3 feet. The middle shelf is 7/2 feet, which is the same thing as 3 and 1/2 feet. And then the bottom shelf is 2 times the top shelf, minus 1/2. So what's that going to be equal to? That's going to be equal to 2 times 3 feet-- that's what the length of the top shelf is-- minus 1/2, which is equal to 6 minus 1/2, or 5 and 1/2 feet. And you can verify that these definitely do add up to 12. 5 and 1/2 plus 3 and 1/2 is 9, plus 3 is 12 feet, and it meets all of the other constraints. The top shelf is 1/2 a foot shorter than the middle shelf, and the bottom shelf is 1/2 a foot shorter than 2 times the top shelf. And we are done. We know the lengths of the shelves that Devon needs to make." + }, + { + "Q": "Where did he get 11 from at 6:42 - when he said 11*300?\n", + "A": "Sal was doing a rough estimate of 3520 divided by 300. 10 *300 = 3000. 11 * 300 = 3300, and 12*300 = 3600.", + "video_name": "AGFO-ROxH_I", + "timestamps": [ + 402 + ], + "3min_transcript": "Now we want to figure out how many laps there are. We want this in terms of laps, not in terms of yards. So we want the yards to cancel out. And we want laps in the numerator, right? Because when you multiply, the yards will cancel out, and we'll just be left with laps. Now, how many laps are there per yard or yards per lap? Well, they say the distance around the field is 300 yards. So we have 300 yards for every 1 lap. So now, multiply this right here. The yards will cancel out, and we will get 3,520. Let me do that in a different color. We will get 3,520, that right there, times 1/300. When you multiply it times 1, it just becomes 3,520 divided by 300. And in terms of the units, the yards canceled out. We're just left with the laps. So 3,520 divided by 300. Well, we can eyeball this right here. What is 11 times 300? Let's just approximate this right here. So if we did 11 times 300, what is that going to be equal to? Well, 11 times 3 is 33, and then we have two zeroes here. So this will be 3,300. So it's a little bit smaller than that. If we have 12 times 300, what is that going to be? 12 times 3 is 36, and then we have these two zeroes, so it's equal to 3,600. So this is going to be 11 point something. It's larger than 11, right? 3,520 is larger than 3,300. So when you divide by 300 you're going to get something larger than 11. But this number right here is smaller than 3,600 so when you divide it by 300, you're going to get something a little bit smaller than 12. than 12 laps. So 2 miles is a little bit lower than 12 laps. But let's make sure we're answering their question. How many complete laps would he need to do to run at least 2 miles? So they're telling us that, look, this might be, 11 point something, something, something laps. That would be the exact number of laps to run 2 miles. But they say how many complete laps does he have to run? 11 complete laps would not be enough. He would have to run 12. So our answer here is 12 complete laps. That complete tells us that they want a whole number of laps. We can't just divide this. If we divide this, we're going to get some 11 point something, something. You can do with the calculator or do it by hand if you're interested. But we have to do at least 12 because that's the smallest whole number of laps that will get us to at least this distance right here, or this number of laps, or the" + }, + { + "Q": "\nSo how would I write an equation for the example at 2:21 ?", + "A": "Since the sequence is not being summed, then it can t be equated to anything so you can t write an equation (in the strictest sense of the term) for the example. If you are merely trying to notate the sequence then: 5\u00e2\u0080\u00a2(1/7)^n for n = 0 to \u00e2\u0088\u009e", + "video_name": "W2NnNKtquaE", + "timestamps": [ + 141 + ], + "3min_transcript": "- [Instructor] I'm going to construct a sequence. We're going to start with some number. Let's say I start with the number, a. And then each successive term of the sequence, I'm going to multiply the, to get each successive term of the sequence, I'm going to muliply the previous term by some fixed non-zero number, and I'm going to call that r. So the next term is going to be a, is going to be a times r. And then the term after that, I'm going to multipy this thing times r. So it's going to be a, if you multiply ar times r, that's going to be ar-squared. And then if you were going to multiply this term times r, you would get a times r to the third power. And you could keep going on and on and on and on. And this type of sequence or this type of progression is called a geometric. Geometric Sequence or Progression. Geometric Sequence. You start with some first value. Let me circle that in a different color, So you start with some first value, and then to get each successive term, you multiply by this fixed number. In this case, this fixed number is r. And so we call r, our common ratio. Our common ratio. Why is it called a common ratio? Well take any two successive terms. Take this term and this term, and divide this term by this term right over here, ar to the third, divided by ar-squared. So if you find the ratio between these two things. Let me rewrite this in the same colors. So if you took ar to the third power, and were to divide it by the term before it. So if you were to divide it by ar-squared, what are you going to be left with? Well a divided by a is one, r to the third divided by r-squared is just going to be r. And this is true if you divide any term by the term before it. Or if you find the ratio between any term And so that's why it's called a common ratio. And so let's look at some examples of geometric sequences. So if I start with the number, if I start with the number five, so my a is five and then each time I'm going to multiply, I'm going to multiply by, I don't know. Let's say I multiply by 1/7. So then the next term is going to be five over seven. What's the next term going to be? Well, I'm going to multiply this thing times 1/7. So that's going to be 5/7 times 1/7 is 5/49. So it's going to be five over seven-squared or 49. If I were to multiply this times 1/7, what am I going to get? I'll just change the notation. I'll get five times, I don't actually know in my head what seven to the third power is, I guess I could calculate it, 280 plus 63," + }, + { + "Q": "In the video at 3:57, Sal made a red number 7 and put a line thru it. But in the video at 4:30, Sal made a green number 7 and didn't put a line thru it. Is the red number 7 a cursive 7?\n", + "A": "Numbers cannot be written in cursive. Some people put a line through their 7s. Neither is right or wrong, and they both mean the same thing", + "video_name": "wx2gI8iwMCA", + "timestamps": [ + 237, + 270 + ], + "3min_transcript": "And maybe when you're first trying you might have even called it something called 37. You would just call it this, this number. This number of days since my birthday. What if there was an easier way to group the numbers? You know I have 10 fingers on my hands. What if I were to group them into 10s? And then I would say just how many groups of 10 I have and then how many ones do I have left over. Maybe that would be an easier way to represent, to represent this quantity here. And so let's do that. So one, two, three, four, five, six, seven, eight, nine, 10. So that's a group of 10 right over there. And then you have one, two, three, four, five, six, seven, eight, nine, 10. So this is another group of 10 right over here. And then let's see. We have one, two, three, four, five, six, seven, eight, nine, 10. And then finally you have one, two, three, four, five, six, seven. So you don't get a whole group of 10 so you don't circle em. So just by doing this very simple thing now all of a sudden it's much easier to realize how many days have passed. You don't have to count everything. You just have to say okay. One group of 10. Two groups of 10. Three groups of 10. Or you can say one, two, three 10s. And so that's essentially 30. And then I have another one, two, three, four, five, six, seven. And so you say oh I have 30 and then seven if you knew to use those words which we now use. Now this is essentially what our number system does using the 10 digits we know of. The 10 digits we know of are zero, one, two, three, four, five, six, seven, eight, nine. Now what our number system allows us to do we can essentially represent any number we want in a very quick way, a very easy way for our brains to understand it. So here if we want to represent three 10s, we would have put a three in what we would call the 10s place. We would put a three in the 10s place. And then we would put the ones, one, two, three, four, five, six, seven. We'd put the seven in the ones place. And so how do you know which place is which? Well the first place starting from the right, the first place is the ones place. And then you go one space to the left of it, you get to the 10s place. And as we'll see you go one more space you go to the 100s space. But we'll cover that in a future video. So this essentially tells us the exact same thing. This tells us the exact same thing as this does right over here. This tells us three 10s. One, two, three. Three 10s Three groups of 10." + }, + { + "Q": "\nHow do you get badges at 1:10 for asking a question.", + "A": "The more questions you ask, the more you ll end up learning, so Khan Academy wants to encourage you to ask questions when you feel stuck.", + "video_name": "wx2gI8iwMCA", + "timestamps": [ + 70 + ], + "3min_transcript": "Voiceover:Let's say that you wanna count the days since your last birthday because you just wanna know how long its been. And so one day after your birthday you put a mark on a wall. Then the next day you put another mark on the wall. The day after that you put another mark on the wall. So out that day you say well how many days has it been? Well you can say look there's been one, two, three days. So one way to think about it is this set of symbols right over here represents the number three. But then you keep going. The fourth day, you put another mark. Fifth day, you put another mark. And then you keep going like that day after day each day you add another mark. And this is actually the earliest way, the most basic way of representing numbers. The number is represented by the number of marks. So after bunch of days you get here and you're like on well how many days has it been? Well you just recount everything. You say one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17 days. it took me little bit of time to realize that this is 17 but it seems to be working so you just keep going. Day after day after day after day you just keep marking off the days on your wall just to sense you're counting the days since your last birthday. But at some point you realize every time you wanna know how may days its been to count it is a little bit painful. And not only that, is this is taking up a lot of space on your wall. You wish that there was an easier way to represent whatever number this is. So first of all let's just think about what number this actually is. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37. So you wish that there was a better way And maybe when you're first trying you might have even called it something called 37. You would just call it this, this number. This number of days since my birthday. What if there was an easier way to group the numbers? You know I have 10 fingers on my hands. What if I were to group them into 10s? And then I would say just how many groups of 10 I have and then how many ones do I have left over. Maybe that would be an easier way to represent, to represent this quantity here. And so let's do that. So one, two, three, four, five, six, seven, eight, nine, 10. So that's a group of 10 right over there. And then you have one, two, three, four, five, six, seven, eight, nine, 10. So this is another group of 10 right over here. And then let's see. We have one, two, three, four, five, six, seven, eight, nine, 10." + }, + { + "Q": "\nat 0:56 couldnt sal just multiply 2 and 3 and get 6 instead of breaking it down", + "A": "he could have, but I think his objective is not for you to memorize the procedure, but to gain intuition on how he came to the solution.", + "video_name": "Of8ezQj1hRk", + "timestamps": [ + 56 + ], + "3min_transcript": "Simplify x to the third, and then that raised to the fourth power times x squared, and then that raised to the fifth power. Now, here we're going to use the power property of exponents, sometimes called the \"power rule.\" And that just tells us if I have x to the a, and then I raise that to b-th power, this is the same thing as x to the a times b power. And to see why that works, let's try that with a-- well, I won't do it with these right here. I'll do it with a simpler example. Let's say we are taking x squared and then raising that to the third power. Well, in that situation, that literally means multiplying x squared by itself three times. So that literally means, x squared times x squared times x squared. I'm taking x squared and I'm raising it to the third power. Now, what does this mean over here? Well, there's a couple of ways to do it. You could just say, look, I have the same base, I'm taking the product, I can just add the exponents. So this is going to be equal to-- let me do it plus 2 plus 2 power, or essentially x to the 3 times 2. This right here is just 3 times 2. So we get x to the sixth power. And if you want, you say, hey, Sal, I don't understand why you can add those exponents. You just have to remember that x squared-- this thing right over here, we could rewrite as x times x times x times x-- in parentheses, I'm putting each of these x squareds-- times x times x. And this is just x times itself 6 times x times x times x times x times x times x. This is just x to the sixth power. So that's why we can add the exponents like that. So let's just use that power property on this expression right over here. We start off with we have x to the third raised to the fourth power. So that's just going to be x to the 3 times 4 power, or x Then, we're multiplying that by x squared raised to the fifth power. Well, that's just going to be x to the 2 times 5 power, or x to the 10th power. And now we have the same base, and we're taking the product, we can just add the exponents. This is going to be equal to-- this whole expression is going to be equal to x to the 12 plus 10th power, or x to the 22nd power. And we are done." + }, + { + "Q": "1:55, why add the two amounts when bees prefer a new solution\n", + "A": "When he added the 2 solutions, that was the new solution.", + "video_name": "JVlfQEhzLMM", + "timestamps": [ + 115 + ], + "3min_transcript": "Make a table and solve. A biologist is researching the impact of three different water-based sugar drinks on bees ability to make honey. He takes 2 liters of Drink A, which contains 40% sugar. So let me write this down. Let me make our table and then we can solve it. So let's take amount of drink. And then we'll say percent sugar. And then we can say sugar quantity, so the actual physical quantity of sugar. Maybe I should say sugar amount, or amount of sugar. Now this first drink, Drink A, it says he takes 2 liters of Drink A, which contains 40% sugar. The first column will be which drink we're talking about, so It's 40% sugar. So if we want the actual amount of sugar in liters, we just multiply 2 liters times 40%, or times 0.4. Let me write times with a dot so you don't think it's an x. 2 times 0.4, which is equal to 0.8 liters of sugar. So you have 0.8 liters of sugar. 1.2 liters of I guess the other stuff in there is water. But it's 0.8 of the 2 liters is sugar, which is 40%. Now,, he adds 1.2 liters of Drink B. He finds that bees prefer this new solution, Drink C. So when you add these two together, you end And we end up with how much of Drink C? 2 plus 1.2 is 3.2 liters of Drink C, which has 25% sugar content. So this is 25% sugar, which also says we know the amount of sugar in it. Because if we have 3.2 liters of it and it's 25% sugar, or it's 1/4 sugar, that means that we have 0.8 liters of sugar here. So this is 0.8 liters of sugar. Well, that I already wrote in the column name. That's the amount of sugar. It's 25% sugar. We have 3.2 liters of it. Now, they want to know what is the percentage of sugar in Drink B? So let's just call that x. So that's right over here. Now, if it's x percent sugar here, or this is the decimal" + }, + { + "Q": "why at 1:40 did you say you were not going to put x because we might think it was a variable, when we could still mistake the dot for a decimal point? you could have put an asterisk...\n", + "A": "because you put the dot in the middle like this 9 * 34", + "video_name": "JVlfQEhzLMM", + "timestamps": [ + 100 + ], + "3min_transcript": "Make a table and solve. A biologist is researching the impact of three different water-based sugar drinks on bees ability to make honey. He takes 2 liters of Drink A, which contains 40% sugar. So let me write this down. Let me make our table and then we can solve it. So let's take amount of drink. And then we'll say percent sugar. And then we can say sugar quantity, so the actual physical quantity of sugar. Maybe I should say sugar amount, or amount of sugar. Now this first drink, Drink A, it says he takes 2 liters of Drink A, which contains 40% sugar. The first column will be which drink we're talking about, so It's 40% sugar. So if we want the actual amount of sugar in liters, we just multiply 2 liters times 40%, or times 0.4. Let me write times with a dot so you don't think it's an x. 2 times 0.4, which is equal to 0.8 liters of sugar. So you have 0.8 liters of sugar. 1.2 liters of I guess the other stuff in there is water. But it's 0.8 of the 2 liters is sugar, which is 40%. Now,, he adds 1.2 liters of Drink B. He finds that bees prefer this new solution, Drink C. So when you add these two together, you end And we end up with how much of Drink C? 2 plus 1.2 is 3.2 liters of Drink C, which has 25% sugar content. So this is 25% sugar, which also says we know the amount of sugar in it. Because if we have 3.2 liters of it and it's 25% sugar, or it's 1/4 sugar, that means that we have 0.8 liters of sugar here. So this is 0.8 liters of sugar. Well, that I already wrote in the column name. That's the amount of sugar. It's 25% sugar. We have 3.2 liters of it. Now, they want to know what is the percentage of sugar in Drink B? So let's just call that x. So that's right over here. Now, if it's x percent sugar here, or this is the decimal" + }, + { + "Q": "at 4:35 why did you multiply by 100 but then not put the decimal back in its original place (divide by 100)? Isn't the answer then incorrect?\n", + "A": "If you think about it 4 / 2 = 400 / 200 so there is no reason to adjust the final answer. Since he multiplied both parts of the expression (the divisor and the dividend) they end up cancelling out. This is the same as multiplying the top and bottom of a fraction by the same number when trying to get matching denominators.", + "video_name": "DAikW24_O0A", + "timestamps": [ + 275 + ], + "3min_transcript": "can do all that, or you can just say, OK, that's not going to give us anything. So then how many times does 15 go into 78? So let's think about it. 15 goes into 60 four times. 15 times 5 is 75. That looks about right, so we say five times. 5 times 15. 5 times 5 is 25. Put the 2 up there. 5 times 1 is 5, plus 2 is 7. 75, you subtract. 78 minus 75 five is 3. Bring down a zero. 15 goes into 30 exactly two times. 2 times 15 is 30. Subtract. No remainder. Bring down the next zero. We're still to the left of the decimal point. The decimal point is right over here. If we write it up here, which we should, it's right over there, so we have one more place to go. So we bring down this next zero. 15 goes into 0 zero times. Subtract. No remainder. So 78 divided by 0.15 is exactly 520. So x is equal to 520. So 78 is 15% of 520. And if we want to use some of the terminology that you might see in a math class, the 15% is obviously the percent. 520, or what number before we figured out it was 520, that's what we're taking the percentage of. This is sometimes referred to as the base. And then when you take some percentage of the base, you get what's sometimes referred to as the amount. So in this circumstance, 78 would be the amount. You could view it as the amount is the percentage of the base, but we were able to figure that out. It's nice to know those, if that's the terminology you use just answer this question. And it makes sense, because 15% is a very small percentage. If 78 is a small percentage of some number, that means that number has to be pretty big, and our answer gels with that. This looks about right. 78 is exactly 15% of 520." + }, + { + "Q": "At around 5:21, is it ok to write \"obvious from diagram\" in formal proofs?\n", + "A": "At 5:21, I would say the reason should be the reflexive property since I was taught that was used in this situation.", + "video_name": "fSu1LKnhM5Q", + "timestamps": [ + 321 + ], + "3min_transcript": "Their corresponding sides have the same length, and so we know that they're congruent. So we know that triangle CDA is congruent to triangle CBA. And we know that by the side side side postulate and the statements given up here. Actually, let me number our statements just so we can refer back to this 1, 2, 3, and 4. And so side side side postulate and 1, 2, and 3-- statements 1, 2, and 3. So statements 1, 2, and 3 and the side side postulate let us know that these two triangles are congruent. And then if these are congruent, then we know, for example, we know that all of their corresponding angles are equivalent. So for example, this angle is going to be equal to that angle. So let's make that statement right over there. going to be statement 5-- we know that angle DCE, that's this angle right over here, is going to have the same measure, we could even say they're congruent. I'll say the measure of angle DCE is going to be equal to the measure of angle BCE. And this comes straight out of statement 4, congruency, I could put it in parentheses. Congruency of those triangles. This implies straight, because they're both part of this larger triangle, they are the corresponding angles, so they're going to have the exact same measure. Now it seems like we could do something pretty interesting with these two smaller triangles at the top left and the top right of this, looks like, a kite like figure. Because we have a side, two corresponding sides are congruent, two corresponding angles are congruent, and they have a side in common. They have this side in common right over here. So let's first just establish that they So I'll just write statement 6. We have CE, the measure or the length of that line, is equal to itself. Once again, this is just obvious. It's the same. Obvious from diagram it's the same line. Obvious from diagram. But now we can use that information. So we don't have three sides, we haven't proven to ourselves that this side is the same as this side, that DE has the same length as EB. But we do have a side, an angle between the sides, and then another side. And so this looks pretty interesting for our side angle side postulate. And so we can say, by the side angle side postulate, we can say that triangle DCE is congruent to triangle BCE." + }, + { + "Q": "at 2:59 how do you get absolute value\n", + "A": "by figuring out how far the number is from 0.", + "video_name": "Oo2vGhVkvDo", + "timestamps": [ + 179 + ], + "3min_transcript": "temperatures was negative 128 degrees Fahrenheit. So let's say that's right over here. This is negative 128 degrees Fahrenheit. And one of the warmest temperatures ever recorded was 134 degrees. This is a positive 134. So it's about that far and a little bit further. So it's a positive 134 degrees Fahrenheit. So when they're asking us how many degrees difference are there between the coldest and the warmest, they're essentially saying, well, what is this distance between the coldest and the warmest right over here? What is this distance? And there's a couple of ways you could think about it. You could say, hey, if I started at the coldest temperature and I wanted to go all the way up to the warmest, how much would I have to add? Or you could say, well, what's the difference between the coldest and the warmest? So you could take the larger number. And from that, you could subtract the smaller number, which is negative 128. So this essentially saying what's the difference between these two numbers? It's going to be positive, because we're subtracting the smaller one from the larger one. This is going to give you the exact same thing as this. Now, there's several ways to think about it. One is we know that if you subtract a negative number, that's the same thing as adding the positive of that number, or adding the absolute value. So this is the same thing. This is going to be equal to 134 plus positive 128 degrees. And what's the intuition behind that? Why does this happen? Well, look at this right over here. We're trying to figure out this distance. This distance is 134 minus negative 128. to be the absolute value of 134. It's going to be this distance right over here, which is just 134-- which is just that right over there-- plus this distance right over here. Now, what is this distance? Well, it's the absolute value of negative 128. It's just 128. So it's going to be that distance, 134, plus 128. And that's why it made sense. This way, you're thinking of what's the difference between a larger number and a smaller number. But since it's a smaller number and you're subtracting a negative, it's the same thing as adding a positive. And hopefully this gives you a little bit of that intuition. But needless to say, we can now figure out what's going to be. And this is going to be equal to-- let me figure this out separately over here. So if I were to add 134 plus 128, I get 4 plus 8" + }, + { + "Q": "\nAt 3:25, So, in general, in a equation you must have 2 positive and 1 negative, right?", + "A": "No. An equation has two mathematical expressions that are equal. Those expressions can have any numbers or variables as long as they are equal.", + "video_name": "Oo2vGhVkvDo", + "timestamps": [ + 205 + ], + "3min_transcript": "temperatures was negative 128 degrees Fahrenheit. So let's say that's right over here. This is negative 128 degrees Fahrenheit. And one of the warmest temperatures ever recorded was 134 degrees. This is a positive 134. So it's about that far and a little bit further. So it's a positive 134 degrees Fahrenheit. So when they're asking us how many degrees difference are there between the coldest and the warmest, they're essentially saying, well, what is this distance between the coldest and the warmest right over here? What is this distance? And there's a couple of ways you could think about it. You could say, hey, if I started at the coldest temperature and I wanted to go all the way up to the warmest, how much would I have to add? Or you could say, well, what's the difference between the coldest and the warmest? So you could take the larger number. And from that, you could subtract the smaller number, which is negative 128. So this essentially saying what's the difference between these two numbers? It's going to be positive, because we're subtracting the smaller one from the larger one. This is going to give you the exact same thing as this. Now, there's several ways to think about it. One is we know that if you subtract a negative number, that's the same thing as adding the positive of that number, or adding the absolute value. So this is the same thing. This is going to be equal to 134 plus positive 128 degrees. And what's the intuition behind that? Why does this happen? Well, look at this right over here. We're trying to figure out this distance. This distance is 134 minus negative 128. to be the absolute value of 134. It's going to be this distance right over here, which is just 134-- which is just that right over there-- plus this distance right over here. Now, what is this distance? Well, it's the absolute value of negative 128. It's just 128. So it's going to be that distance, 134, plus 128. And that's why it made sense. This way, you're thinking of what's the difference between a larger number and a smaller number. But since it's a smaller number and you're subtracting a negative, it's the same thing as adding a positive. And hopefully this gives you a little bit of that intuition. But needless to say, we can now figure out what's going to be. And this is going to be equal to-- let me figure this out separately over here. So if I were to add 134 plus 128, I get 4 plus 8" + }, + { + "Q": "\n@ 2:00 when he starts making the problem, would it matter if you put the smaller number first rather than the big number?", + "A": "For this particular problem no it wouldn t. You would get the same answer either way.", + "video_name": "Oo2vGhVkvDo", + "timestamps": [ + 120 + ], + "3min_transcript": "One of the coldest temperatures ever recorded outside was negative 128 degrees Fahrenheit in Antarctica. One of the warmest temperatures ever recorded outside was 134 degrees Fahrenheit in Death Valley, California. How many degrees difference are there between the coldest and warmest recorded outside temperatures? So let's think about this a little bit. Now, what I'll do is I'll plot them on a number line. But I'm going to plot it on a vertical number line that has a resemblance to a thermometer, since we're talking about temperature. So I'm going to make my number line vertical right over here. So there's my little vertical number line. And this right over here is 0 degrees Fahrenheit, which really is of no significance. If it was Celsius, we'd be talking about the freezing point. But for Fahrenheit, that happens at 32 degrees. But let's say this is 0 degrees Fahrenheit. And let's plot these two points. temperatures was negative 128 degrees Fahrenheit. So let's say that's right over here. This is negative 128 degrees Fahrenheit. And one of the warmest temperatures ever recorded was 134 degrees. This is a positive 134. So it's about that far and a little bit further. So it's a positive 134 degrees Fahrenheit. So when they're asking us how many degrees difference are there between the coldest and the warmest, they're essentially saying, well, what is this distance between the coldest and the warmest right over here? What is this distance? And there's a couple of ways you could think about it. You could say, hey, if I started at the coldest temperature and I wanted to go all the way up to the warmest, how much would I have to add? Or you could say, well, what's the difference between the coldest and the warmest? So you could take the larger number. And from that, you could subtract the smaller number, which is negative 128. So this essentially saying what's the difference between these two numbers? It's going to be positive, because we're subtracting the smaller one from the larger one. This is going to give you the exact same thing as this. Now, there's several ways to think about it. One is we know that if you subtract a negative number, that's the same thing as adding the positive of that number, or adding the absolute value. So this is the same thing. This is going to be equal to 134 plus positive 128 degrees. And what's the intuition behind that? Why does this happen? Well, look at this right over here. We're trying to figure out this distance. This distance is 134 minus negative 128." + }, + { + "Q": "\nAt about 0:20, Sal says \"We already know that if all three angle, all three of the corresponding angles are congruent to the corresponding angles of ABC, then we're dealing with congruent triangles\". But actually, we're dealing with similar triangles and NOT congruent triangles, right? Because AAA is not a postulate of congruent triangles.", + "A": "If the angles are congruent then the shapes would be congruent.", + "video_name": "7bO0TmJ6Ba4", + "timestamps": [ + 20 + ], + "3min_transcript": "Let's say we have triangle ABC. It looks something like this. I want to think about the minimum amount of information. I want to come up with a couple of postulates that we can use to determine whether another triangle is similar to triangle ABC. So we already know that if all three of the corresponding angles are congruent to the corresponding angles on ABC, then we know that we're dealing with congruent triangles. So for example, if this is 30 degrees, this angle is 90 degrees, and this angle right over here is 60 degrees. And we have another triangle that looks like this, it's clearly a smaller triangle, but it's corresponding angles. So this is 30 degrees. This is 90 degrees, and this is 60 degrees, we know that XYZ in this case, is going to be similar to ABC. So we would know from this because corresponding angles are congruent, we would know that triangle ABC And you've got to get the order right to make sure that you have the right corresponding angles. Y corresponds to the 90-degree angle. X corresponds to the 30-degree angle. A corresponds to the 30-degree angle. So A and X are the first two things. B and Y, which are the 90 degrees, are the second two, and then Z is the last one. So that's what we know already, if you have three angles. But do you need three angles? If we only knew two of the angles, would that be enough? Well, sure because if you know two angles for a triangle, you know the third. So for example, if I have another triangle that looks like this-- let me draw it like this-- and if I told you that only two of the corresponding angles are congruent. So maybe this angle right here is congruent to this angle, and that angle right there is congruent to that angle. Is that enough to say that these two triangles are similar? Because in a triangle, if you know two of the angles, If you know that this is 30 and you know that that is 90, then you know that this angle has to be 60 degrees. Whatever these two angles are, subtract them from 180, and that's going to be this angle. So in general, in order to show similarity, you don't have to show three corresponding angles are congruent, you really just have to show two. So this will be the first of our similarity postulates. We call it angle-angle. If you could show that two corresponding angles are congruent, then we're dealing with similar triangles. So for example, just to put some numbers here, if this was 30 degrees, and we know that on this triangle, this is 90 degrees right over here, we know that this triangle right over here is similar to that one there. And you can really just go to the third angle in this pretty straightforward way. You say this third angle is 60 degrees, so all three angles are the same. That's one of our constraints for similarity." + }, + { + "Q": "SOMEONE HELP ME IM CONFUSED D:\nAt the equation on 8:27, why or how did the 1/2 get in front of the log?\nAfter that I got lost...\n", + "A": "The half is the exponent of the log result. Now what I am trying to figure out myself is how did the exponent convert to a coefficient!", + "video_name": "TMmxKZaCqe0", + "timestamps": [ + 507 + ], + "3min_transcript": "We could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just type in 357 in your calculator and press the log button and you're going to get bada bada bam. Then you can clear it, or if you know how to use the parentheses on your calculator, you could do that. But then you type 17 on your calculator, press the log button, go to bada bada bam. And then you just divide them, and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth. This one, to me it's the most useful, but it doesn't completely-- it does fall out of, obviously, the exponent properties. But it's hard for me to describe the intuition simply, so you probably want to watch the proof on it, if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. Just so you know logarithms are useful. Let's do some examples. Let's just let's just rewrite a bunch of things in simpler forms. So if I wanted to rewrite the log base 2 of the square root of-- let me think of something. Of 32 divided by the cube-- no, I'll just take the square root. Divided by the square root of 8. How can I rewrite this so it's reasonably not messy? Well let's think about this. This is the same thing, this is equal to-- I don't know I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right?" + }, + { + "Q": "At 8:20 why does the base 2 disappear when changing the exponent to the coefficient? He says this is the 3rd rule we learnt but I don't remember seeing this?\n", + "A": "At 9:00, he realizes he forgot it and writes it in. As for the third rule, I don t know what you mean.", + "video_name": "TMmxKZaCqe0", + "timestamps": [ + 500 + ], + "3min_transcript": "We could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just type in 357 in your calculator and press the log button and you're going to get bada bada bam. Then you can clear it, or if you know how to use the parentheses on your calculator, you could do that. But then you type 17 on your calculator, press the log button, go to bada bada bam. And then you just divide them, and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth. This one, to me it's the most useful, but it doesn't completely-- it does fall out of, obviously, the exponent properties. But it's hard for me to describe the intuition simply, so you probably want to watch the proof on it, if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. Just so you know logarithms are useful. Let's do some examples. Let's just let's just rewrite a bunch of things in simpler forms. So if I wanted to rewrite the log base 2 of the square root of-- let me think of something. Of 32 divided by the cube-- no, I'll just take the square root. Divided by the square root of 8. How can I rewrite this so it's reasonably not messy? Well let's think about this. This is the same thing, this is equal to-- I don't know I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right?" + }, + { + "Q": "\nI lost him where he appears to have reduced 1/2log2,32 to 5/2, at 9:51. Help.", + "A": "Dan, log\u00e2\u0082\u0082(32) = 5 because 2*2*2*2*2 = 32 = 2\u00e2\u0081\u00b5 so 1/2 * log\u00e2\u0082\u0082(32) = 1/2 * 5 = 5/2 Also, log\u00e2\u0082\u0082(8) = 3 because 2*2*2 =8 = 2\u00c2\u00b3 so 1/4 * log\u00e2\u0082\u0082(8) = 1/4 * 3 = 3/4 I hope that helps make it click for you.", + "video_name": "TMmxKZaCqe0", + "timestamps": [ + 591 + ], + "3min_transcript": "I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right? Minus the logarithm base 2 of the square root of 8. Right? Let's see. Well here once again we have a square root here, so we could say this is equal to 1/2 times log base 2 of 32. Minus this 8 to the 1/2, which is the same thing is 1/2 log base 2 of 8. We learned that property in the beginning of this presentation. And then if we want, we can distribute this original 1/2. This equals 1/2 log base 2 of 32 minus 1/4-- because we have to distribute that 1/2-- minus 1/4 log base 2 of 8. This is 5/2 minus, this is 3. 3 times 1/4 minus 3/4. Or 10/4 minus 3/4 is equal to 7/4. I probably made some arithmetic errors, but you get the point." + }, + { + "Q": "At 4:35, Sal explains a way that he remembers concave and convex... Is there any other way?\n", + "A": "In Spanish con means with so concave can mean with cave. Another trick to determining concavity is to imagine an elastic band being stretched around the figure, does it touch all sides or does it have to span a gap. If it has to span a gap then the figure must be concave (having a cavity or cave).", + "video_name": "W9B3VYdC5T8", + "timestamps": [ + 275 + ], + "3min_transcript": "around the circle. And some of this angle, A+B+C+D+E is just going to be 360 degree And this is work for any convex polygon, and when I say convex polygon I mean it is not that dented words Just to be clear what I'm talking about, it would work for any convex polygon that is kind of I don't want to say regular, regular means it has the same size and angle, but it is not dented, this is a convex polygon. This right here is a concave polygon Let me draw this, right this way, so this would be a concave polygon Let me draw as it having the same number of side, So i just going to dent this two sides and then that's the same side over there, Let me do that and then like that. This has 1,2,3,4,5,6, sides and this has 1,2,3,4,5,6 sides. This is concave, sorry this is a convex polygon, this is concave polygon, All you have to remember is kind of cave in words And so, what we just did is applied to any exterior angle of any convex polygon. I Am a bit confused. This applied to any convex polygon and once again if you take this angle and added to this angle and added to this angle, this angle, that angle and that angle and I'm not applying that all that they are different angles, i could say that one green, and that one some other colour they can all be different but if you shift the angle like this you can see that they just go round the circle. So, once again, I'll just add up to 360 degrees" + }, + { + "Q": "\nAt 3:10 Sal said something like clockwise and counterclockwise..\nWhat does he mean by clockwise and counterclockwise ?", + "A": "Clockwise on the clock rotation goes as top-right-down-left. Counterclockwise on the clock rotation goes as top-left-down-right. Most clocks go Clockwise.", + "video_name": "W9B3VYdC5T8", + "timestamps": [ + 190 + ], + "3min_transcript": "right over here. So this going to be a convex angle right over here it's going to have a measure of A, now let me draw angle B, angle B, and i going to draw adjacent to angle A, and what you could do is just to think about it maybe if we draw a line over here, if we draw a line over here that is parallel to this line then the measure over here would also be B,because this is obviously a straight line, it would be like transversal, this of course a responding angles, so if u want to draw adjacent angle, the adjacent to A, do it like that, or whatever angle this is the measure of B and now it is adjacent to A, now let's draw the same thing to C We can draw a parallel line to that right over here. And this angle would also be C and if we want it to be adjacent to that, we could draw it there, so that angle is C we do it in different color, you could do D, right over here or you could shift it over here it'll look like that, or shift over here, it'll look like that If we just kept thinking of parallel, if all of this line were parallel to each other So, let's just draw D like this, so this line is going to parallel to that line Finally, you have E, and again u can draw a line that is parallel to this right over here and this right over here would be angle E or you could draw right over here, right over here And when you see it drawn this way, it's clear that when you add up, the measure, this angle A,B,C,D and E going all the way around the circle, either way around the circle. And some of this angle, A+B+C+D+E is just going to be 360 degree And this is work for any convex polygon, and when I say convex polygon I mean it is not that dented words Just to be clear what I'm talking about, it would work for any convex polygon that is kind of I don't want to say regular, regular means it has the same size and angle, but it is not dented, this is a convex polygon. This right here is a concave polygon Let me draw this, right this way, so this would be a concave polygon Let me draw as it having the same number of side, So i just going to dent this two sides" + }, + { + "Q": "At 1:01, you are assuming that the bases of the exponents are the same. How would you solve the expression, if the bases were different numbers? Would you still apply the quotient rule, and add the bases, or what?\n", + "A": "Sadly, you can t solve it the same way if the bases are different. If you really wanted to solve it, then you d have to actually calculate the exponents and then multiply/divide them. An explanation for this is that if you write out the exponents (3^5 * 4^3 = 3*3*3*3*3*4*4*4), you can t tie them together into one exponent, right? So it s not possible to just add the exponents when the bases are different, because a different number is being multiplied.", + "video_name": "tvj42WdKlH4", + "timestamps": [ + 61 + ], + "3min_transcript": "Simplify 3a to the fifth over 9a squared times a to the fourth over a to the third. So before we even worry about the a's, we can actually simplify the 3 and the 9. They're both divisible by 3. So let's divide the numerator and the denominator here by 3. So if we divide the numerator by 3, the 3 becomes a 1. If we divide the denominator by 3, the 9 becomes a 3. So this reduces to, or simplifies to 1a to the fifth times a to the fourth over-- or maybe I should say, a to the fifth over 3a squared times a to the fourth over a to the third. Now this, if we just multiply the two expressions, this would be equal to 1a to the fifth times a to the fourth in the numerator, and we don't have to worry about the one, it doesn't change the value. So it's a to the fifth times a to the fourth in the numerator. And then we have 3a-- let me write the 3 like this-- and then we have 3 times a squared times a to the third in the denominator. can simplify this from here. One sometimes is called the quotient rule. And that's just the idea that if you have a to the x over a to the y, that this is going to be equal to a to the x minus y. And just to understand why that works, let's think about a to the fifth over a squared. So a to the fifth is literally a times a times a times a times a. That right there is a to the fifth. And we have that over a squared. And I'm just thinking about the a squared right over here, which is literally just a times a. That is a squared. Now, clearly, both the numerator and denominator are both divisible by a times a. We can divide them both by a times a. So we can get rid of-- if we divide the numerator And if we divide the denominator by a times a, we just get a 1. So what are we just left with? We are left with just a times a times a over 1, which is just a times a times a. But what is this? This is a to the third power, or a to the 5 minus 2 power. We had 5, we were able to cancel out 2, that gave us 3. So we could do the same thing over here. We can apply the quotient rule. And I'll do two ways of actually doing this. So let's apply the quotient rule with the a to the fifth and the a squared. So let me do it this way. So let's apply with these two guys, and then let's apply it with these two guys. And of course, we have the 1/3 out front. So this can be reduced to 1/3 times-- if we apply the quotient rule with a" + }, + { + "Q": "\nAt 4:55 when he was solving a^9-3a^5 .... He got a a4 but it became a^4/3 ??? I kept thinking it would just be 3a^4. How did the a^4 turn into the numerator? and why did 3 stay as the denominator?", + "A": "Divided by", + "video_name": "tvj42WdKlH4", + "timestamps": [ + 295 + ], + "3min_transcript": "did it over here-- that becomes a to the third power. And if we apply it over here with the a to the fourth over a to the third, that'll give us a-- let me do it that same blue color. That'll give us a-- that's not the same blue color. There we go. This will give us a to the 4 minus 3 power, or a to the first power. And of course, we can simplify this as a to the third times a-- well, actually, let me just do it over here. Before I even rewrite it, we know that a to the third times a to the first is going to be a to the 3 plus 1 power. We have the same base, we can add the exponents. We're multiplying a times itself three times and then one more time. So that'll be a to the fourth power. So this right over here becomes a to the fourth power. a to the 3 plus 1 power. And then we have to multiply that by 1/3. So our answer could be 1/3 a to the fourth, or we could equally write it a to the fourth over 3. would have been to apply the product, or to add the exponents in the numerator, and then add the exponents in the denominator. So let's do it that way first. If we add the exponents in the numerator first, we don't apply the quotient rule first. We apply it second. We get in the numerator, a to the fifth times a to the fourth would be a to the ninth power. 5 plus 4. And then in the denominator we have a squared times a to the third. Add the exponents, because we're taking the product with the same base. So it'll be a to the fifth power. And of course, we still have this 3 down here. We have a 1/3, or we could just write a 3 over here. Now, we could apply the quotient property of exponents. We could say, look, we have a to the ninth over a to the fifth. a to the ninth over a to the fifth is equal to a to the 9 minus 5 power, or it's equal to a to the fourth power. And of course, we still have the divided by 3." + }, + { + "Q": "\nHow does it work to take away a ten from 10? At 1:27 he takes away one ten from 10, but then has 90.", + "A": "He is not taking a ten away from 10, he is taking a ten away from 10 tens, or 100. 100-10=90", + "video_name": "3lHBgFvr3yE", + "timestamps": [ + 87 + ], + "3min_transcript": "Let's try to subtract 164 from 301, and I encourage you to pause this video and try it on your own first. So let's go place by place, and we can realize where we have to do some borrowing or regrouping. So in the ones place, we have an issue. 4 is larger than 1. How do we subtract a larger number from a smaller number? We also an issue in the tens place. 6 is larger than 0. How do we subtract 6 from 0? So the answer that might be jumping into your head is oh, we've got to do some borrowing or some regrouping. But then you might be facing another problem. You'd say, OK, well, let's try to borrow from the tens place here. So we have a 1. If we could borrow 10 from the tens place, it could be 11. But there's nothing here in the tens place. There's nothing to borrow, so what do we do? So the way I would tackle it is first borrow for the tens place. So we have nothing here, so let's regroup 100 from the hundreds place. So that's equivalent to borrowing a 1 from the hundreds place. So that's now a 2. And now in the tens place, instead of a 0, Now, let's make sure that this still makes sense. This is 200 plus 10 tens. 10 tens is 100, plus 1. 200 plus 100 plus 1 is still 301. So this still makes sense. Now, the reason why this is valuable is now we have something to regroup from the tens place. If we take one of these tens, so now we're left with 9 tens, and we give it to the ones place, so you give 10 plus 1, you're going to be left with 11. And we can verify that we still haven't changed the value of the actual number. 200 plus 90 is 290, plus 11 is still 301. And what was neat about this is now up here, all of these numbers are larger than the corresponding number in the same place. So we're ready to subtract. 11 minus 4 is-- let's see, 10 minus 3 is 7, so 11 minus 4 9 minus 6 is 3, and 2 minus 1 is 1. So we are left with 137." + }, + { + "Q": "\nI've recently been learning about this at school. I came on here to study and the beginning of the video seemed right to me, but when it got to around 1:40 , I got a bit confused. I was taught to take the total of everything on the first table, which would be 109, and divide 28/109, then 35/109, then 97/109, then 104/109. I'm very confused on how this creates the same answer, though. I'm thoroughly confused, because my teacher does this concept way differently than this video.", + "A": "You re using the same total for both your columns.", + "video_name": "_ETPMszULXc", + "timestamps": [ + 100 + ], + "3min_transcript": "- [Voiceover] The two-way frequency table below shows data on type of vehicle driven. So, this is type of vehicle driven, and whether there was an accident the last year. So, whether there was an accident in the last year, for customers of All American Auto Insurance. Complete the following two-way table of column relative frequencies. So that's what they're talking here, this is a two-way table of column relative frequencies. If necessary, round your answers to the nearest hundredth. So, let's see what they're saying. They're saying, let's see... Of the accidents within the last year, 28 were the people were driving an SUV, a Sport Utility Vehicle and 35 were in a Sports car. Of the No accidents in the last year, 97 were in SUV and a 104 were in sports cars. Another way you could think of it, of the Sport utility vehicles that were driven and the total, let's see it's 28 plus 97 which is going to be 125. Of that 125, 28 had an accident within the last year Similarly, you could say of the 139 Sports cars 35 had an accident in the last year, 104 did not have an accident in the last year. So what they want us to do is put those relative frequencies in here. So the way we could think about it. One right over here, this represents all the Sport utility vehicles. So one way you could think about, that represents the whole universe of the Sport Utility Vehicles, at least the universe that this table shows. So, that's really representative of the 28 plus the 97. And so, in each of these we want to put the relative frequencies. So this right over here is going to be 28 divided by the total. So over here is 28, but we want this number to be a fraction of the total. Well the fraction of the total is gonna be, 28 over 97 plus 28. Which of course is going to be 125. This one right over here is going to be 97 over 125. And of course, when you add this one and this one, it should add up to one. Likewise, this one's going to be 35 over 139. 35 plus 104. So, 139. And this is going to be 104 over 104 plus 35. Which is 139. So, let me just calculate each of them using this calculator. Let me scroll down a little bit. And so, if I do 28 divided by 125, I get 0.224. They said round your answers to the nearest hundredth. So this is 0.22. No accident within the last year, 97 divided by 125. So 97 divided by 125 is equal to, see here if I rounded" + }, + { + "Q": "\nAt 14:28, it mentions using function notation for the answer. Would it be counted wrong of you put \"y\"?", + "A": "If you write y without saying what y is, then technically it s not right. At 10:57, Sal defines y=f(x), so after that, it s cool to write y wherever you d write f(x).", + "video_name": "5fkh01mClLU", + "timestamps": [ + 868 + ], + "3min_transcript": "So this is going to be equal to 2 minus negative 3. That's the same thing as 2 plus 3. So that is 5. Negative 1 minus 1.5 is negative 2.5. 5 divided by 2.5 is equal to 2. So the slope of this line is negative 2. Actually I'll take a little aside to show you it doesn't matter what order I do this in. If I use this coordinate first, then I have to use that coordinate first. Let's do it the other way. If I did it as negative 3 minus 2 over 1.5 minus negative 1, this should be minus the 2 over 1.5 minus the negative 1. This is equal to what? Negative 3 minus 2 is negative 5 over 1.5 minus negative 1. That's 1.5 plus 1. That's over 2.5. So once again, this is equal the negative 2. So I just wanted to show you, it doesn't matter which one you pick as the starting or the endpoint, as long as If this is the starting y, this is the starting x. If this is the finishing y, this has to be the finishing x. But anyway, we know that the slope is negative 2. So we know the equation is y is equal to negative 2x plus some y-intercept. Let's use one of these coordinates. I'll use this one since it doesn't have a decimal in it. So we know that y is equal to 2. So y is equal to 2 when x is equal to negative 1. Of course you have your plus b. So 2 is equal to negative 2 times negative 1 is 2 plus b. 2, minus 2, you're subtracting it from both sides of this equation, you're going to get 0 on the left-hand side is equal to b. So b is 0. So the equation of our line is just y is equal to negative 2x. Actually if you wanted to write it in function notation, it would be that f of x is equal to negative 2x. I kind of just assumed that y is equal to f of x. But this is really the equation. They never mentioned y's here. So you could just write f of x is equal to 2x right here. Each of these coordinates are the coordinates of x and f of x. So you could even view the definition of slope as change in f of x over change in x. These are all equivalent ways of viewing the same thing." + }, + { + "Q": "\nI don't get it. At 3:39 is confusing. Why can't we use the slope equation to get the slope and then the slope would be 2 not -2. Am I missing something? He said if it's switch then it would be be negative?", + "A": "OMG that is what I meant! Thank you! I forgot that the number would be negative 6. I can t believe I didn t see it!", + "video_name": "5fkh01mClLU", + "timestamps": [ + 219 + ], + "3min_transcript": "We could add a 4/5 to that side as well. The whole reason I did that is so that cancels out with that. You get b is equal to 4/5. So we now have the equation of the line. y is equal to negative 1 times x, which we write as negative x, plus b, which is 4/5, just like that. Now we have this one. The line contains the point 2 comma 6 and 5 comma 0. So they haven't given us the slope or the y-intercept explicitly. But we could figure out both of them from these So the first thing we can do is figure out the slope. So we know that the slope m is equal to change in y over change in x, which is equal to-- What is the change in y? Let's start with this one right here. So we do 6 minus 0. So that's a 6-- I want to make it color-coded-- minus 0. So 6 minus 0, that's our change in y. Our change in x is 2 minus 2 minus 5. The reason why I color-coded it is I wanted to show you when I used this y term first, I used the 6 up here, that I have to use this x term first as well. So I wanted to show you, this is the coordinate 2 comma 6. This is the coordinate 5 comma 0. I couldn't have swapped the 2 and the 5 then. Then I would have gotten the negative of the answer. But what do we get here? This is equal to 6 minus 0 is 6. 2 minus 5 is negative 3. So this becomes negative 6 over 3, which is the same thing as negative 2. So, so far we know that the line must be, y is equal to the slope-- I'll do that in orange-- negative 2 times x plus our y-intercept. Now we can do exactly what we did in the last problem. We can use one of these points to solve for b. We can use either one. Both of these are on the line, so both of these must satisfy this equation. I'll use the 5 comma 0 because it's always nice when you have a 0 there. The math is a little bit easier. So let's put the 5 comma 0 there. So y is equal to 0 when x is equal to 5. So y is equal to 0 when you have negative 2 times 5, when x is equal to 5 plus b. So you get 0 is equal to -10 plus b. If you add 10 to both sides of this equation, let's add 10 to both sides, these two cancel out. You get b is equal to 10 plus 0 or 10." + }, + { + "Q": "At 8:05 how do you get 5/2. Please explain.\n", + "A": "5/6(3) is the same as diving 6 by 3. 6/3=2 5/2", + "video_name": "5fkh01mClLU", + "timestamps": [ + 485 + ], + "3min_transcript": "So let's say it's 0 minus 5 just like that. So I'm using this coordinate first. I'm kind of viewing it as the endpoint. Remember when I first learned this, I would always be tempted to do the x in the numerator. No, you use the y's in the numerator. So that's the second of the coordinates. That is going to be over negative 3 minus 3. This is the coordinate negative 3, 0. This is the coordinate 3, 5. We're subtracting that. So what are we going to get? This is going to be equal to-- I'll do it in a neutral color-- this is going to be equal to the numerator is negative 5 over negative 3 minus 3 is negative 6. You get 5/6. So we know that the equation is going to be of the form y is equal to 5/6 x plus b. Now we can substitute one of these coordinates in for b. So let's do. I always like to use the one that has the 0 in it. So y is a zero when x is negative 3 plus b. So all I did is I substituted negative 3 for x, 0 for y. I know I can do that because this is on the line. This must satisfy the equation of the line. Let's solve for b. So we get zero is equal to, well if we divide negative 3 by 3, that becomes a 1. If you divide 6 by 3, that becomes a 2. So it becomes negative 5/2 plus b. plus 5/2, plus 5/2. I like to change my notation just so you get familiar with both. So the equation becomes 5/2 is equal to-- that's a 0-- is equal to b. b is 5/2. So the equation of our line is y is equal to 5/6 x plus b, which we just figured out is 5/2, plus 5/2. We are done. Let's do another one. We have a graph here. Let's figure out the equation of this graph. This is actually, on some level, a little bit easier. What's the slope? Slope is change in y over change it x. So let's see what happens. When we move in x, when our change in x is 1, so that is our change in x. So change in x is 1. I'm just deciding to change my x by 1, increment by 1." + }, + { + "Q": "\nAt 4:37 Sal says that x goes into 4 zero times. But x is an unknown - how does he know this?", + "A": "because it is a un known so he dosen t know but if it is x and x^2 we know that x goes into x x times.", + "video_name": "FXgV9ySNusc", + "timestamps": [ + 277 + ], + "3min_transcript": "So x times x plus 1 is what? x times x is x squared. x times 1 is x, so it's x squared plus x. And just like we did over here, we now subtract. And what do we get? x squared plus 3x plus 6 minus x squared-- let me be very careful-- this is minus x squared plus x. I want to make sure that negative sign only-- it applies to this whole thing. So x squared minus x squared, those cancel out. 3x, this is going to be a minus x. Let me put that sign there. So this is minus x squared minus x, just to be clear. We're subtracting the whole thing. 3x minus x is 2x. And then you bring down the 6, or 6 minus 0 is nothing. So 2x plus 6. Now, you look at the highest degree term, an x and a 2x. How many times does x go into 2x? 2 times x is 2x. 2 times 1 is 2. So we get 2 times x plus 1 is 2x plus 2. But we're going to want to subtract this from this up here, so we're going to subtract it. Instead of writing 2x plus 2, we could just write negative 2x minus 2 and then add them. These guys cancel out. 6 minus 2 is 4. And how many times does x go into 4? We could just say that's zero times, or we could say that 4 is the remainder. So if we wanted to rewrite x squared plus 3x plus 6 over x plus 1-- notice, this is the same thing as x squared plus 3x plus 6 divided by x plus 1, this thing divided by this, we it is equal to x plus 2 plus the remainder divided by x plus 1 plus 4 over x plus 1. This right here and this right here are equivalent. And if you wanted to check that, if you wanted to go from this back to that, what you could do is multiply this by x plus 1 over x plus 1 and it add the two. So this is the same thing as x plus 2. And I'm just going to multiply that times x plus 1 over x plus 1. That's just multiplying it by 1. And then to that, add 4 over x plus 1. I did that so I have the same common denominator. And when you perform this addition right here, when you multiply these two binomials and then add the 4 up here," + }, + { + "Q": "at 3:26, sal puts brackets and says that the equation is in -, how did he know and how did he come to this??\n", + "A": "The subtraction step of long division requires that you subtract the entire binomial: x^2 + x Subtraction of any polynomial requires that you distribute the minus sign across the polynomial being subtracted. To write the problem another way, you are doing: x^2+3x+6 - (x^2+x) = x^2+3x+6 - x^2 -x = 2x + 6 If you don t distribute the minus, you only subtract the x^2 term. The rest gets added. This would be like saying 57 - 32 = 29 (subtract the 3, but add the 2). Hope this helps.", + "video_name": "FXgV9ySNusc", + "timestamps": [ + 206 + ], + "3min_transcript": "for more complicated problems. So you could have also written this as 2 goes into 2x plus 4 how many times? And you would perform this the same way you would do traditional long division. You'd say 2-- you always start with the highest degree term. 2 goes into the highest degree term. You would ignore the 4. 2 goes into 2x how many times? Well, it goes into 2x x times and you put the x in the x place. x times 2 is 2x. And just like traditional long division, you now subtract. So 2x plus 4 minus 2x is what? It's 4, right? And then 2 goes into 4 how many times? It goes into it two times, a positive two times. Put that in the constants place. 2 times 2 is 4. You subtract, remainder 0. So this might seem overkill for what was probably a do it in a few steps. We're now going to see that this is a very generalizable process. You can do this really for any degree polynomial dividing into any other degree polynomial. Let me show you what I'm talking about. So let's say we wanted to divide x plus 1 into x squared plus 3x plus 6. So what do we do here? So you look at the highest degree term here, which is an x, and you look at the highest degree term here, which is an x squared. So you can ignore everything else. And that really simplifies the process. You say x goes into x squared how many times? Well, x squared divided by x is just x, right? x goes into x squared x times. You put it in the x place. This is the x place right here or the x to So x times x plus 1 is what? x times x is x squared. x times 1 is x, so it's x squared plus x. And just like we did over here, we now subtract. And what do we get? x squared plus 3x plus 6 minus x squared-- let me be very careful-- this is minus x squared plus x. I want to make sure that negative sign only-- it applies to this whole thing. So x squared minus x squared, those cancel out. 3x, this is going to be a minus x. Let me put that sign there. So this is minus x squared minus x, just to be clear. We're subtracting the whole thing. 3x minus x is 2x. And then you bring down the 6, or 6 minus 0 is nothing. So 2x plus 6. Now, you look at the highest degree term, an x and a 2x. How many times does x go into 2x?" + }, + { + "Q": "At 8:13 Sal removes x+4 as a factor from numerator and denominator. Why isn't a discontinuity marked at x=-4?\n", + "A": "When doing this polynomial division, it is automatically assumed that the denominator is not equal to zero, otherwise you cant divide. Yes you are correct that the expression is undefined at x=-4, but we assume that x is not equal to negative 4 so there is no need to write it.", + "video_name": "FXgV9ySNusc", + "timestamps": [ + 493 + ], + "3min_transcript": "Let's do another one of these. They're kind of fun. So let's say that we have-- we want to simplify x squared plus 5x plus 4 over x plus 4. So once again, we can do our algebraic long division. We can divide x plus 4 into x squared plus 5x plus 4. And once again, same exact process. Look at the highest degree terms in both of them. x goes into x squared how many times? It goes into it x times. Put it in the x place. This is our x place right here. X times x is x squared. x times 4 is 4x. So let me just put a negative sign there. And then these cancel out. 5x minus 4x is x. 4 minus 0 is plus 4. x plus 4, and then you could even see this coming. You could say x plus 4 goes into x plus 4 obviously one time, or if you were not looking at the constant terms, you would completely just say, well, x goes into x how many times? Well, one time. Plus 1. 1 times x is x. 1 times 4 is 4. We're going to subtract them from up here, so it cancels out, so we have no remainder. So this right here simplifies to-- this is equal to x plus 1. And there's other ways you could have done this. We could have tried to factor this numerator. x squared plus 5x plus 4 over x plus 4. This is the same thing as what? We could have factored this numerator as x plus 4 4 times 1 is 4. 4 plus 1 is 5, all of that over x plus 4. That cancels out and you're left just with x plus 1. Either way would have worked, but the algebraic long division will always work, even if you can't cancel out factors like that, even if you did have a remainder. In this situation, you didn't. So this was equal to x plus 1. Let's do another one of these just to make sure that you really-- because this is actually a very, very useful skill to have in your toolkit. So let's say we have x squared-- let me Let's say we had 2x squared-- I could really make these numbers up on the fly. 2x squared minus 20x plus 12 divided by-- actually, let's" + }, + { + "Q": "I do not really understand what Sal means by \"factoring the numerator\" at 7:51. Could someone please explain to me what he is doing?\n", + "A": "Sal didn t show all the steps. Option 1: (2x+4)/2 can be changed into 2x/2 + 4/2 -- Reduce each fraction and you get x+2 Option 2: Factor the numerator (factoring out GCF=2) (2x+4)/2 becomes 2(x+2)/2 Cancel the common factor of 2, and you get: (x+2)/1 or x+2 Hope this helps.", + "video_name": "FXgV9ySNusc", + "timestamps": [ + 471 + ], + "3min_transcript": "Let's do another one of these. They're kind of fun. So let's say that we have-- we want to simplify x squared plus 5x plus 4 over x plus 4. So once again, we can do our algebraic long division. We can divide x plus 4 into x squared plus 5x plus 4. And once again, same exact process. Look at the highest degree terms in both of them. x goes into x squared how many times? It goes into it x times. Put it in the x place. This is our x place right here. X times x is x squared. x times 4 is 4x. So let me just put a negative sign there. And then these cancel out. 5x minus 4x is x. 4 minus 0 is plus 4. x plus 4, and then you could even see this coming. You could say x plus 4 goes into x plus 4 obviously one time, or if you were not looking at the constant terms, you would completely just say, well, x goes into x how many times? Well, one time. Plus 1. 1 times x is x. 1 times 4 is 4. We're going to subtract them from up here, so it cancels out, so we have no remainder. So this right here simplifies to-- this is equal to x plus 1. And there's other ways you could have done this. We could have tried to factor this numerator. x squared plus 5x plus 4 over x plus 4. This is the same thing as what? We could have factored this numerator as x plus 4 4 times 1 is 4. 4 plus 1 is 5, all of that over x plus 4. That cancels out and you're left just with x plus 1. Either way would have worked, but the algebraic long division will always work, even if you can't cancel out factors like that, even if you did have a remainder. In this situation, you didn't. So this was equal to x plus 1. Let's do another one of these just to make sure that you really-- because this is actually a very, very useful skill to have in your toolkit. So let's say we have x squared-- let me Let's say we had 2x squared-- I could really make these numbers up on the fly. 2x squared minus 20x plus 12 divided by-- actually, let's" + }, + { + "Q": "\nat \"4:10\" Sal mentions anti derivitive. what is that?", + "A": "It s another name for the integral (without any bounds).", + "video_name": "OiNh2DswFt4", + "timestamps": [ + 250 + ], + "3min_transcript": "of numbers. A transform will take you from one set of functions to another set of functions. So let me just define this. The Laplace Transform for our purposes is defined as the improper integral. I know I haven't actually done improper integrals just yet, but I'll explain them in a few seconds. The improper integral from 0 to infinity of e to the minus st times f of t-- so whatever's between the Laplace Transform brackets-- dt. Now that might seem very daunting to you and very confusing, but I'll now do a couple of examples. So what is the Laplace Transform? Well let's say that f of t is equal to 1. So what is the Laplace Transform of 1? of time-- well actually, let me just substitute exactly the way I wrote it here. So that's the improper integral from 0 to infinity of e to the minus st times 1 here. I don't have to rewrite it here, but there's a times 1dt. And I know that infinity is probably bugging you right now, but we'll deal with that shortly. Actually, let's deal with that right now. This is the same thing as the limit. And let's say as A approaches infinity of the integral from 0 to Ae to the minus st. dt. Just so you feel a little bit more comfortable with it, you might have guessed that this is the same thing. could take the limit as something approaches infinity. So anyway, let's take the anti-derivative and evaluate this improper definite integral, or this improper integral. So what's anti-derivative of e to the minus st with respect to dt? Well it's equal to minus 1/s e to the minus st, right? If you don't believe me, take the derivative of this. You'd take minus s times that. That would all cancel out, and you'd just be left with e to the minus st. Fair enough. Let me delete this here, this equal sign. Because I could actually use some of that real estate. We are going to take the limit as A approaches infinity. You don't always have to do this, but this is the first time we're dealing with improper intergrals. So I figured I might as well remind you that we're taking a limit. Now we took the anti-derivative." + }, + { + "Q": "Why can you assume that s > 0 at 6:36?\n", + "A": "S is representing a frequency as in a per second or 1/second which is why you get 1/s for the Laplace of 1. From a physics perspective there is no such thing as a negative frequency because in reality it is the same thing as the frequency, saying something happens -3 times a second does not make a lot of sense. You can also tell that s cannot equal zero from the 1/s.", + "video_name": "OiNh2DswFt4", + "timestamps": [ + 396 + ], + "3min_transcript": "valued at 0. And then take the limit of whatever that ends up being as A approaches infinity. So this is equal to the limit as A approaches infinity. If we substitute A in here first, we get minus 1/s. Remember we're, dealing with t. We took the integral with respect to t. e to the minus sA, right? That's what happens when I put A in here. Now what happens when I put t equals 0 in here? So when t equals 0, it becomes e to the minus s times 0. This whole thing becomes 1. And I'm just left with minus 1/s. Fair enough. And then let me scroll down a little bit. to, but that's OK. So this is going to be the limit as A approaches infinity of minus 1/s e to the minus sA minus 1/s. So plus 1/s. So what's the limit as A approaches infinity? Well what's this term going to do? As A approaches infinity, if we assume that s is greater than 0-- and we'll make that assumption for now. Actually, let me write that down explicitly. Let's assume that s is greater than 0. So if we assume that s is greater than 0, then as A approaches infinity, what's going to happen? Well this term is going to go to 0, right? e to the minus-- a googol is a very, very small number. And an e to the minus googol is an even smaller number. So then this e to the minus infinity approaches 0, so this This term isn't affected because it has no A in it, so we're just left with 1/s. So there you go. This is a significant to moment in your life. You have just been exposed to your first Laplace Transform. I'll show you in a few videos, there are whole tables of Laplace Transforms, and eventually we'll prove all of them. But for now, we'll just work through some of the more basic ones. But this can be our first entry in our Laplace Transform table. The Laplace Transform of f of t is equal to 1 is equal to 1/s. Notice we went from a function of t-- although obviously this one wasn't really dependent on t-- to a function of s. I have about 3 minutes left, but I don't think that's enough time to do another Laplace Transform. So I will save that for the next video. See you soon." + }, + { + "Q": "at 4:02 it really looks hard.\nI don't understand it.\n", + "A": "He is just setting up a histogram. It looks sort of like a bar graph.", + "video_name": "4eLJGG2Ad30", + "timestamps": [ + 242 + ], + "3min_transcript": "So I want to look at the frequency of each of these numbers. So I have one, two, three, four 0's. I have one, two, three, four, five, six, seven 1's. I have one, two, three, four, five 2's. I have one, two 3's. I have one 4, and one 6. So we could write it this way. We could write the number, and then we can have the frequency. So I have the numbers 0, 1, 2, 3, 4-- we could even throw 5 in there, although 5 has a frequency of 0. And we have a 6. So the 0 showed up four times in this data set. 1 showed up seven times in this data set. showed up one time, 5 didn't show up, and 6 showed up one time. All I did is I counted this data set, and I did this first. But you could say how many times do I see a 0? I see it one, two, three, four times. How many times do I see a 1? One, two, three, four, five, six, seven times. That's what we mean by frequency. Now a histogram is really just a plot, kind of a bar graph, plotting the frequency of each of these numbers. It's going to look a lot like this original thing that I drew. So let me draw some axes here. So the different buckets here are the numbers. And that worked out because we're dealing with very clean integers that tend to repeat. If you're dealing with things that the exact number doesn't repeat, oftentimes people will put the numbers into buckets or ranges. But here they fit into nice little buckets. You have the numbers 0, 1, 2, 3, 4, 5, and 6. And then on the vertical axis we're going to plot the frequency. So one, two, three, four, five, six, seven. So that's 7, 6, 5, 4, 3, 2, 1. So 0 shows up four times. So we'll draw a little bar graph here. 0 shows up four times. Draw it just like that. 0 shows up four times. That is that information right there. 1 shows up seven times. So I'll do a little bar graph. 1 shows up seven times. Just like that. I want to make it a little bit straighter than that-- 1 shows up seven times. 2-- I'll do it in a different color-- 2 shows up five times." + }, + { + "Q": "At 1:35, does Sal says that he groups them into two groups?\n", + "A": "Yes he does -Andrew", + "video_name": "2nZsIeaiJUo", + "timestamps": [ + 95 + ], + "3min_transcript": "You and your friends are making name cards to assign seats for dinner. You each are given an identical strip of blank paper and begin coloring parts of it with crayons. You start with blue, or you start with blue. Your friend Micah starts off with green, Joelle starts with the red, Holden starts with pink, And Kate starts with purple. After one color is used, whose strip has the same part of its area colored as yours? So let's look at my strip right over here. It has how many equal sections? 1, 2, 3, 4, 5, 6 equal sections. And how many of them have been colored blue? Well, let's see. 1, 2, 3, 4 of my equal sections have been colored blue. So I need to figure out which of these other folks have colored exactly 4/6 of their strip blue. Or another way of thinking about it, I want to say, whoever has an equivalent fraction colored, not blue, colored whatever their color might be. So he's got 3 equal sections. He's got 3 equal sections, and he's colored, he's colored 2 of them, 2 of them green. Now is 4/6 and 2/3 the same thing? Well, he has 3 equal sections. I had 6. If I were to group each of these into groups of 2, so if I were to divide by 2, let me do that. If I were divide by 2, so I'm going to group this into groups of 2. Then I would have 1, 2-- so I have 1, 2, 3 equal sections. And then if I were to group the blue strips into groups of 2, then I would have 1, 2 of them that have actually been colored blue. So if I divide both the numerator and the denominator by 2, I get 2/3, which is the exact same fraction as Micah. Another way to think about it, is if you divide each of Micah's sections into 2. Well, then how many total sections does he have? Well, he had 3, now each of those 3 have been divided into 2. So we can multiply by 2 to get 6. So this is multiplying by 2. And each of the 2 in green, there now each of those sections have now been divided into 2. So you can multiply by 2 as well, and you get 4. So 2/3 and 4/6 are the same thing. So Micah has colored in the same amount that I have. Now let's see whether any of these other fractions are the same. So Joelle has colored in 4, has colored in 4 out of 1, 2, 3, 4, 5, 6, 7. 4 out of 7. So it's not clear to me that we can somehow multiply or divide" + }, + { + "Q": "At 6:30, Sal talks about the transversal line. What's that?\n", + "A": "A line that cuts across two or more (usually parallel) lines", + "video_name": "R0EQg9vgbQw", + "timestamps": [ + 390 + ], + "3min_transcript": "That is the opposite side of our angle x so we know sine of x is equal to DC over one, or DC over one is just DC. This length right over here is sine of x. Segment AC, same exact logic. Cosine of x is the length of AC over one which is just the length of AC. This length right over here, segment AC its length is cosine of x. That's kind of interesting. Now let's see what we can figure out about this triangle, triangle ACB right over here. How could we figure out CB? Well we know that sine of y, let me write this here. Sine of y is equal to what? It's equal to the length of segment CB over the hypotenuse. The hypotenuse here is the cosine of x, At any point if you get excited pause the video and try to finish the proof on your own. The length of segment CB if we just multiply both sides by cosine of x, the length of segment CB is equal to cosine of x times sine of y. Which is neat because we just showed that this thing right over here is equal to this thing right over here. To complete our proof we just need to prove that this thing is equal to this thing right over there. If that's equal to that and that's equal to that well we already know that the sum of these is equal to the length of DF which is sine of x plus y. Let's see if we can figure out, if we can express DE somehow. What angle would be useful? If somehow we could figure If we could figure out this angle then DE we could express in terms of this angle and sine of x. Let's see if we can figure out that angle. We know this is angle y over here and we also know that this is a right angle. EC is parallel to AB so you could view AC as a transversal. If this is angle y right over here then we know this is also angle y. These are once again, notice. If AC is a transversal here and EC and AB are parallel then if this is y then that is y. If that's y then this is 90 minus y. If this is 90 degrees and this is 90 minus y then these two angles combined add up to 180 minus y," + }, + { + "Q": "\nDid anyone see Vi Hart draw the dolphin at 1:23 above the 42? Anyone get the reference?", + "A": "Vi Hart also makes a reference to Douglas Adams and his book the Hitchhiker s Guide to the Galaxy in another one of her videos at her website, Silly Band Fight .", + "video_name": "a-e8fzqv3CE", + "timestamps": [ + 83 + ], + "3min_transcript": "While I'm working on some more ambitious projects, I wanted to quickly comment on a couple of mathy things that have been floating around the internet, just so you know I'm still alive. So there's this video that's been floating around about how to multiply visually like this. Pick two numbers, let's say, 12 times 3. And then you draw these lines. 12, 31. Then you start counting the intersections-- 1, 2, 3 on the left; 1, 2, 3, 4, 5, 6, 7 in the middle; 1, 2 on the right, put them together, 3, 7, 2. There's your answer. Magic, right? But one of the delightful things about mathematics is that there's often more than one way to solve a problem. And sometimes these methods look entirely different, but because they do the same thing, they must be connected somehow. And in this case, they're not so different at all. Let me demonstrate this visual method again. This time, let's do 97 times 86. So we draw our nine lines and seven lines time eight lines and six lines. Now, all we have to do is count the intersections-- 1,2, 3, 4, 5, 6, 7, 8, 9, 10. This is boring. How about instead of counting all the dots, we just figure out how many intersections there are. Let's see, there's seven going one way and six Forget everything I ever said about learning a certain amount of memorization in mathematics being useful, at least at an elementary school level. Because apparently, I've been faking my way through being a mathematician without having memorized 6 times 7. And now I'm going to have to figure out 5 times 7, which is half of 10 times 7, which is 70, so that's 35, and then add the sixth 7 to get 42. Wow, I really should have known that one. OK, but the point is that this method breaks down the two-digit multiplication problem into four one-digit multiplication problems. And if you do have your multiplication table memorized, you can easily figure out the answers. And just like these three numbers became the ones, tens, and hundreds place of the answer, these do, too-- ones, tens, hundreds-- and you add them up and voila! Which is exactly the same kind of breaking down into single-digit multiplication and adding that you do during the old boring method. The whole point is just to multiply every pair of digits, make sure you've got the proper number of zeroes on the end, and add them all up. But of course, seeing that what you're actually doing is not something your teachers want you to realize, or else you might remember the every combination concept when you get to multiplying binomials, and it might make it too easy. In the end, all of these methods of multiplication distract from what multiplication really is, which for 12 times 31 is this. All the rest is just breaking it down into well-organized chunks, saying, well, 10 times 30 is this, 10 times 1 this, 30 times 2 is that, and 2 times 1 is that. Add them all up, and you get the total area. Don't let notation get in the way of your understanding. Speaking of notation, this infuriating bit of nonsense has been circulating around recently. And that there has been so much discussion of it is sign that we've been trained to care about notation way too much. Do you multiply here first or divide here first? The answer is that this is a badly formed sentence. It's like saying, I would like some juice or water with ice. Do you mean you'd like either juice with no ice or water with ice? Or do you mean that you'd like either juice with ice or water with ice? You can make claims about conventions and what's right and wrong, but really the burden" + }, + { + "Q": "At 3:28, If you watch what Sal does/says, can you conclude that you can do this:\nthe square root of 60 times the square root of 6 = the square root of 60 times 6??\n", + "A": "Yes, that is correct.", + "video_name": "74iuGIaBgRc", + "timestamps": [ + 208 + ], + "3min_transcript": "If you wanted to do a full factorization of 10, a full prime factorization, it would be two times five. So there's no perfect squares in 10. And so we can work it out from here. This is the same thing as the negative of the square root of four times the square root of 10, plus the square root of nine, times the square root of 10. And when I say square root, I'm really saying principal root, the positive square root. So it's the negative of the positive square root of four, so that is, so let me do this is in another color. so it can be clear. So, this right here is two. This right here is three. So it's going to be equal to negative two square roots of 10 plus three square roots of 10. So if I have negative two of something and I add three of that same something to it, that's going to be what? Well that's going to be one square root of 10. Actually, let me slow it down a little bit. I could rewrite it this way. I could write it as three square roots of 10 minus two square roots of 10. That might jump out at you a little bit clear. If I have three of something and I were to take away two of that something, and that case it's squares of 10s, well, I'm going to be left with just one of that something. I'm just gonna be left with one square root of 10. Which we could just write as the square root of 10. Another way to think about it is, we could factor out a square root of 10 here. So you undistribute it, do the distributive property in reverse. That would be the square root of 10 times three minus two, which is of course, this is just one. So you're just left with the square root of 10. So all of this simplifies to square root of 10. Let's do a few more of these. So this says, simplify the expression by removing all factors that are perfect squares from inside So essentially the same idea. All right, let's see what we can do. So, this is interesting. We have a square root of 1/2. So can I, well actually, what could be interesting is since if I have a square root of something times the square root of something else. So the square root of 180 times the square root of 1/2, this is the same thing as the square root of 180 times 1/2. And this just comes straight out of our exponent properties. It might look a little bit more familiar if I wrote it as 180 to the 1/2 power, times 1/2 to the 1/2 power, is going to be equal to 180 times 1/2 to the 1/2 power, taking the square root, the principal root is the same thing as raising something to the 1/2 power. And so this is the square root of 80 times 1/2 which is going to be the square root of 90," + }, + { + "Q": "\nAt 5:35, Sal says you can put the answer in either way. On the practice, can you put either option? Thanks!", + "A": "It would be better to use the second way, because it is more simple. (8/3) sqrt of (2/3)", + "video_name": "74iuGIaBgRc", + "timestamps": [ + 335 + ], + "3min_transcript": "times 10, and we just simplified square root of 90 in the last problem, that's equal to the square root of nine times the square root of, principle root of 10, which is equal to three times the square root of 10. Three times the square root of 10. All right, let's keep going. So I have one more of these examples, and like always, pause the video and see if you can work through these on your own before I work it out with you. Simplify the expression by removing all factors that are perfect square, okay, these are just same directions that we've seen the last few times. And so let's see. If I wanted to do, if I wanted to simplify this, this is equal to the square root of, well, 64 times two is 128, and 64 is a perfect square, so I'm gonna write it as 64 times two, over 27 is nine times three. Nine is a perfect square. So this is going to be the same thing. We could say this is the same thing as the square root of 64 times two, over the square root of nine times three, which is the same thing as the square root of 64 times the square root of two, over square root of nine times the square root of three, which is equal to, this is eight, this is three, so it would be eight times the square root of two, over three times the square root of three. That's one way to say it. Or we could even view the square root of two over the square root of three as a square root of 2/3. So we could say this is eight over three times the square root of 2/3. So these are all possible ways of trying to tackle this. So we could just write it, let's see. Have we removed all factors that are perfect squares? Yes, from inside the radicals and we've combined terms. We weren't doing any adding or subtracting here, from inside the radicals and I think we've done that. So we could say this is going to be 8/3 times the square root of 2/3. And there's other ways that you could express this that would be equivalent but hopefully this makes some sense." + }, + { + "Q": "at 15:03 I was not sure if the answer to that was correct\n", + "A": "it is correct because 9+9+9+9+9+9+9+9=72 or 9x8=72", + "video_name": "xO_1bYgoQvA", + "timestamps": [ + 903 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt around 3:50 to 4:38, Sal constructs matrix A\nHowever he seems to write \"3Rot\"... I'm not too sure what the 3 part of the Rot is. In the previous video, he doesn't write \"2Rot\"... Is the 3 just notation to represent R3? Since intuitively it feels like a multiplication by 3", + "A": "@ around 2:40 he says (roughly): ...let me call it 3Rot(theta) to denote that it s in R^3 .", + "video_name": "gkyuLPzfDV0", + "timestamps": [ + 230, + 278 + ], + "3min_transcript": "z-component, it looks like that. Then when you rotate it, its z and its y-components will change, but its x-component will stay the same. So then it might look something like this. Let me see if I can give it justice. So then the vector when I rotate it around might look something like that. Anyway, I don't know if I'm giving it proper justice but this was rotated around the x-axis. I think you understand what that means. But just based on the last video, we want to build a transformation. Let me call this rotation 3 theta. Or let me call it 3 rotation theta now that we're dealing in R3. And what we want to do is we want to find some matrix, so I can write my 3 rotation sub theta transformation of x as being some matrix A times the vector x. Since this is a transformation from R3 to R3 this is of course going to be a 3 by 3 matrix. out, you just have to apply the transformation essentially to the identity matrix. So what we do is we start off with the identity matrix in R3, which is just going to be a 3 by 3. It's going to have 1, 1, 1, 0, 0, 0, 0, 0, 0. Each of these columns are the basis vectors for R3. That's e1, e2, e3-- I'm writing it probably too small for you to see-- but each of these are the basis vectors for R3. And what we need to do is just apply the transformation to each of these basis vectors in R3. So our matrix A will look like this. Our matrix A is going to be a 3 by 3 matrix. Where the first column is going to be our transformation, 3 rotation sub theta, applied to that column And then I'm going to apply it to this middle column vector right here. You get the idea, I don't want to write that whole thing again. I'm going to apply 3 rotation sub theta to 0, 1, 0. And then I'm going to apply it-- I'll do it here-- 3 rotation sub theta. I'm going to apply it to this last column vector, so 0, 0, 1. We've seen this multiple times. So let's apply it. Let's rotate each of these basis vectors for R3. Let's rotate them around the x-axis. So the first guy, if I were to draw an R3, what would he look like? He only has directionality in the x direction right? If we call this the x-dimension, if the first entry corresponds to our x-dimension, the second entry corresponds to our y-dimension. And the third entry corresponds to our z-dimension." + }, + { + "Q": "\nAt 1:37, how did he define f(x) and g(x)?", + "A": "Sal simply splitted the root frunction away from the remaining rest. f(x) is the root function, g(x) is the remaining rest.", + "video_name": "IiBC4ngwH6E", + "timestamps": [ + 97 + ], + "3min_transcript": "What I want to do in this video is start with the abstract-- actually, let me call it formula for the chain rule, and then learn to apply it in the concrete setting. So let's start off with some function, some expression that could be expressed as the composition of two functions. So it can be expressed as f of g of x. So it's a function that can be expressed as a composition or expression that can be expressed as a composition of two functions. Let me get that same color. I want the colors to be accurate. And my goal is to take the derivative of this business, the derivative with respect to x. And what the chain rule tells us is that this is going to be equal to the derivative of the outer function with respect to the inner function. And we can write that as f prime of not x, but f prime of g of x, of the inner function. f prime of g of x times the derivative of the inner function with respect to x. How do you actually apply it? Well, let's try it with a real example. Let's say we were trying to take the derivative of the square root of 3x squared minus x. So how could we define an f and a g so this really is the composition of f of x and g of x? Well, we could define f of x. If we defined f of x as being equal to the square root of x, and if we defined g of x as being equal to 3x squared minus x, then what is f of g of x? Well, f of g of x is going to be equal to-- I'm going to try to keep all the colors accurate, f of g of x is equal to-- where everywhere you see the x, you replace with the g of x-- the principal root of g of x, which is equal to the principal root of-- we defined g of x right over here-- 3x squared minus x. So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule. What is f prime of g of x going to be equal to, the derivative of f with respect to g? Well, what's f prime of x? f prime of x is equal to-- this is the same thing as x to the 1/2 power, so we can just apply the power rule. So it's going to be 1/2 times x to the-- and then we just take 1 away from the exponent, 1/2 minus 1" + }, + { + "Q": "At 5:10 would you have to simplify any farther or is that an adequte answer?\n", + "A": "The answer at 5:10 is in its most simplified form. (3x^2 - x)^(-1/2) can t be multiplied with (6x - 1).", + "video_name": "IiBC4ngwH6E", + "timestamps": [ + 310 + ], + "3min_transcript": "And so what is f prime of g of x? Well, wherever in the derivative we saw an x, we can replace it with a g of x. So it's going to be 1/2 times-- instead of an x to the negative 1/2, we can write a g of x to the 1/2. And this is just going to be equal to-- let me write it right over here. It's going to be equal to 1/2 times all of this business to the negative 1/2 power. So 3x squared minus x, which is exactly what we need to solve right over here. f prime of g of x is equal to this. So this part right over here I will-- let me square it off in green. What we're trying to solve right over here, f prime of g of x, we've just figured out So the derivative of f of the outer function with respect to the inner function. So let me write it. It is equal to 1/2 times g of x to the negative 1/2, times 3x squared minus x. This is exactly this based on how we've defined f of x and how we've defined g of x. Conceptually, if you're just looking at this, the derivative of the outer thing, you're taking something to the 1/2 power. So the derivative of that whole thing with respect to your something is going to be 1/2 times that something to the negative 1/2 power. That's essentially what we're saying. But now we have to take the derivative of our something with respect to x. And that's more straightforward. g prime of x-- we just use the power rule for each of these terms-- is equal to 6x to the first, or just 6x minus 1. So this part right over here is just going to be 6x minus 1. is this right over here and we're multiplying. And we're done. We have just applied the power rule. So just to review, it's the derivative of the outer function with respect to the inner. So instead of having 1/2x to the negative 1/2, it's 1/2 g of x to the negative 1/2, times the derivative of the inner function with respect to x, times the derivative of g with respect to x, which is right over there." + }, + { + "Q": "At 3:00 I don't understand why f'(x) = 1/2x^-1/2. How and why did we get there from f(x) = x^1/2? What's the rule for moving from f(x) to f'(x)?\n", + "A": "This is a function of the power rule: d/dx( x^n ) \u00e2\u0086\u0092 n\u00e2\u0080\u00a2x^(n - 1) Let: f(x) = sqrt(x) f (x) = d/dx(sqrt(x)) f (x) = d/dx(x^(1/2)) Apply power rule d/dx( x^n ) \u00e2\u0086\u0092 n\u00e2\u0080\u00a2x^(n - 1) where n = 1/2: f (x) = 1/2\u00e2\u0080\u00a2x^(1/2 - 1)\u00e2\u0080\u00a2d/dx(x) f (x) = 1/2\u00e2\u0080\u00a2x^(1/2 - 1)\u00e2\u0080\u00a2dx/dx f (x) = 1/2\u00e2\u0080\u00a2x^(1/2 - 1) f (x) = 1/2\u00e2\u0080\u00a2x^(1/2 - 2/2) f (x) = 1/2\u00e2\u0080\u00a2x^(-1/2) f (x) = 1/(2\u00e2\u0080\u00a2x^(1/2)) f (x) = [1/(2\u00e2\u0080\u00a2sqrt(x))]", + "video_name": "IiBC4ngwH6E", + "timestamps": [ + 180 + ], + "3min_transcript": "How do you actually apply it? Well, let's try it with a real example. Let's say we were trying to take the derivative of the square root of 3x squared minus x. So how could we define an f and a g so this really is the composition of f of x and g of x? Well, we could define f of x. If we defined f of x as being equal to the square root of x, and if we defined g of x as being equal to 3x squared minus x, then what is f of g of x? Well, f of g of x is going to be equal to-- I'm going to try to keep all the colors accurate, f of g of x is equal to-- where everywhere you see the x, you replace with the g of x-- the principal root of g of x, which is equal to the principal root of-- we defined g of x right over here-- 3x squared minus x. So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule. What is f prime of g of x going to be equal to, the derivative of f with respect to g? Well, what's f prime of x? f prime of x is equal to-- this is the same thing as x to the 1/2 power, so we can just apply the power rule. So it's going to be 1/2 times x to the-- and then we just take 1 away from the exponent, 1/2 minus 1 And so what is f prime of g of x? Well, wherever in the derivative we saw an x, we can replace it with a g of x. So it's going to be 1/2 times-- instead of an x to the negative 1/2, we can write a g of x to the 1/2. And this is just going to be equal to-- let me write it right over here. It's going to be equal to 1/2 times all of this business to the negative 1/2 power. So 3x squared minus x, which is exactly what we need to solve right over here. f prime of g of x is equal to this. So this part right over here I will-- let me square it off in green. What we're trying to solve right over here, f prime of g of x, we've just figured out" + }, + { + "Q": "\n@3:03 what is does the negative one-half power mean", + "A": "x^(-0.5) = 1/sqrt(x)", + "video_name": "IiBC4ngwH6E", + "timestamps": [ + 183 + ], + "3min_transcript": "How do you actually apply it? Well, let's try it with a real example. Let's say we were trying to take the derivative of the square root of 3x squared minus x. So how could we define an f and a g so this really is the composition of f of x and g of x? Well, we could define f of x. If we defined f of x as being equal to the square root of x, and if we defined g of x as being equal to 3x squared minus x, then what is f of g of x? Well, f of g of x is going to be equal to-- I'm going to try to keep all the colors accurate, f of g of x is equal to-- where everywhere you see the x, you replace with the g of x-- the principal root of g of x, which is equal to the principal root of-- we defined g of x right over here-- 3x squared minus x. So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule. What is f prime of g of x going to be equal to, the derivative of f with respect to g? Well, what's f prime of x? f prime of x is equal to-- this is the same thing as x to the 1/2 power, so we can just apply the power rule. So it's going to be 1/2 times x to the-- and then we just take 1 away from the exponent, 1/2 minus 1 And so what is f prime of g of x? Well, wherever in the derivative we saw an x, we can replace it with a g of x. So it's going to be 1/2 times-- instead of an x to the negative 1/2, we can write a g of x to the 1/2. And this is just going to be equal to-- let me write it right over here. It's going to be equal to 1/2 times all of this business to the negative 1/2 power. So 3x squared minus x, which is exactly what we need to solve right over here. f prime of g of x is equal to this. So this part right over here I will-- let me square it off in green. What we're trying to solve right over here, f prime of g of x, we've just figured out" + }, + { + "Q": "In 0:51, did Sal say tenth hundredths or ten hundredths?\n", + "A": "He said ten hundreds.", + "video_name": "o31cLUkS23E", + "timestamps": [ + 51 + ], + "3min_transcript": "Let's try to subtract 9.57 minus 8.09. So try to pause this video and figure this out first before we work through it together. Well, let's just rewrite it. Let's rewrite it. And when I rewrite it, I like to line up the decimals. This one it's a little intuitive. We have 8.09. Just like that, and now we're ready to subtract. And we want to subtract 9 hundredths from 7 hundredths. Well, we don't have enough hundredths up here so let's move over here. Let's see if we can do some regrouping so we always have a higher number on top. So over here we want to subtract 0 tenths from 5 tenths so we have enough tenths over here. So let's regroup. So instead of 5 tenths, I'm going to have 4 tenths, and then I'm going to give that other tenth, which is the same thing as 10 hundredths over here, so this becomes 17 hundredths. 17 minus 9 is 8. 4 minus 0 is 4. So this is going to be 1.48." + }, + { + "Q": "\nat 1:54 sal says one people in line", + "A": "He said it twice, probably on purpose.", + "video_name": "BIpsQIJUCC8", + "timestamps": [ + 114 + ], + "3min_transcript": "- [Voiceover] Let's say that you love frozen yogurt. So every day after school you decide to go to the frozen yogurt store at exactly four o'clock, four o'clock PM. Now, because you like frozen yogurt so much, you are not a big fan of having to wait in line when you get there, you're impatient, you want your frozen yogurt immediately. And so you decide to conduct a study. You want to figure out the probability of there being lines of different sizes when you go to the frozen yogurt store after school, exactly at four o'clock PM. So in your study, the next 50 times you observe, you go to the frozen yogurt store at four PM, you make a series of observations. You observe the size of the line. So, let me make two columns here, line size is the left column, and on the right column, let's say this is the number of times observed. So, times observed, observed. O-B-S-E-R-V-E-D, all right, times observed. All right, so let's first think about it. Okay, so you go and you say, hey look, I see no people in line, exactly, or you see no people in line, exactly 24 times. You see one person in line exactly 18 times, and you see two people in line exactly eight times. And, in your 50 visits, you don't see more, you never see more than two people in line. I guess this is a very efficient cashier at this frozen yogurt store. So based on this, based on what you have observed, what would be your estimate of the probabilities of finding no people in line, one people in line, or two people in line, at four PM on the days after school that you visit the frozen yogurt store? So what's the probability of there being no line, a one person line, or a two person line when you visit at four PM on a school day? Well, all you can do is estimate the true probability, the true theoretical probability. We don't know what that is, but you've done 50 observations here right. I know that this adds up to 50, 18 plus eight is 26, 26 plus 24 is 50, so you've done 50 observations here and so you can figure out, well what are the relative frequencies of having zero people? What is the relative frequency of one person, or the relative frequency of two people in line? And then we can use that as the estimates for the probability. So let's do that. So, probability estimate. I'll do it in the next column. So probability, probability estimate, and once again we can do that by looking at the relative frequency. The relative frequency of zero, well we" + }, + { + "Q": "\ncan we factor something out from a term which is not even present in every subterm...just a little bit confusing because i've heard we can't factor out something which is not present in each term.. sal just did that at 8:35", + "A": "You are right in that it is nonsensical to factor something out that is not present in every term undergoing the factoring. Sal didn t do that. At 8:35, each of the 3 terms had involved c_x (c sub x).", + "video_name": "b7JTVLc_aMk", + "timestamps": [ + 515 + ], + "3min_transcript": "Remember, this is just the x component of our triple product. Just the x component. But to do this, let me factor out. I'm going to factor out a bx. So let me do this, let me get the bx. So if I were to factor it out-- I'm going to factor it out of this term that has a bx. I'm going to factor it out of this term. And then I'm going to factor it out of this term. So if I take the bx out, I'm going to have an aycy. Actually, let me write it a little bit differently. Let me factor it out of this one first. So then it's going to have an axcx. a sub x, c sub x. So I used this one up. And then I'll do this one now. Plus, if I factor the bx out, I get ay cy. I've used that one now. And now I have this one. I'm going to factor the bx out. So that's all of those. So I've factored that out. And now, from these right over here, let me factor out a negative cx. And so, if I do that-- let me go to this term right over here-- I'm going to have an axbx when I factor it out. So an axbx, cross that out. And then, over here, I'm going to have an ayby. Remember, I'm factoring out a negative cx, so I'm going to have a plus ay, sub by. And then, finally, I'm going to have a plus az, az bz. And what is this? Well, this right here, in green, this is the exact same thing as the dot products of a and c. This is the dot product of the vectors a and c. So that's the dot of a and c times the x component of b minus-- I'll do this in the same-- minus-- once again, this is the dot product of a and b now, minus a dot b times the x component of c. And we can't forget, all of this was multiplied by the unit vector i. We're looking at the x component, or the i component of that whole triple product. So that's going to be all of this. All of this is times the unit vector i. Now, if we do this exact same thing-- and I'm not going to do it, because it's computationally intensive. But I think it won't be a huge leap of faith for you. This is for the x component." + }, + { + "Q": "\nQuestion. At 7:03, Sal said \"potential inflection point\" . Just curious, but how could it possibly not be an inflection point if x=0 is a critical number and the second derivative of 0 equals 0.", + "A": "Sometimes critical points are not inflection points. For example take the graph of y=x^3. At zero, we find that the first derivative is zero [y (0)=3(0)^2], indicating a critical point. However if you look at points slightly to the right of zero and slightly to the left of zero, you ll see that the rate of change (first derivative) is positive on both sides. The graph is not changing concavity, even though zero is an inflection point", + "video_name": "hIgnece9ins", + "timestamps": [ + 423 + ], + "3min_transcript": "" + }, + { + "Q": "At 12:50, do you have to be so close to 0? If it was 1 (which is still greater than zero) 3x-2 would be positive. So that value is not negative for all x>0.\n", + "A": "You don t. It just has to be between critical numbers. Let s say you had critical numbers of 1, 5, and 7. You could test 0, 2, 6, and 8 to cover all of your bases. Your test numbers don t have to be so close to your critical numbers.", + "video_name": "hIgnece9ins", + "timestamps": [ + 770 + ], + "3min_transcript": "" + }, + { + "Q": "@ 18:56 for x=0 , if x>0 shouldnt curve be upwards,x<0 be downwards?\n", + "A": "No. The best way to verify this is to graph the original equation. Perhaps a little more detail would help clarify. When x<0 f (x)>0 therefore the original function is concave up. BUT, when 02/3, f (x)>0 again and concave up. Since f has a positive leading coefficient the graph as concave up toward positive and negative infinity.", + "video_name": "hIgnece9ins", + "timestamps": [ + 1136 + ], + "3min_transcript": "" + }, + { + "Q": "5:45\u00e3\u0080\u0082\u00e3\u0080\u0082\u00e3\u0080\u0082why derivative zai equals 0...he only proved h(y)=0\n", + "A": "If you remember, the original function was in the form M(x,y)+N(x,y)y = 0. He also assumed u(x)M(x,y) = psi_x and u(x)N(x,y) = psi_y. Using the chain rule for partial derivatives that he showed us that psi = u(x)M(x,y)+u(x)N(x,y)y *. Since *M(x,y)+N(x,y)y = 0 is the same as u(x)M(x,y)+u(x)N(x,y)y = 0, you can say psi = 0.", + "video_name": "0NyeDUhKwBE", + "timestamps": [ + 345 + ], + "3min_transcript": "That's the partial of a function purely of y with respect to y. And then that has to equal our new N, or the new expression we got after multiplying by the integrating factor. So that's going to be equal to this right here. This is, hopefully, making sense to you at this point. And that should be equal to x to the third plus x squared y. And interesting enough, both of these terms are on this side. So let's subtract both of those terms from both sides. So x to the third, x to the third, x squared y, x squared y. And we're left with h prime of y is equal to 0. Or you could say that h of y is equal to some constant. There's just some constant left over. So for our purposes, we can just say that psi is equal to this. Because this is just a constant, we're going to take the antiderivative anyway, and get a constant on the right hand side. And in the previous videos, the constants all merged together. So we'll just assume that that is our psi. And we know that this differential equation, up here, can be rewritten as, the derivative of psi with respect to x, and that just falls out of the partial derivative chain rule. The derivative of psi with respect to x is equal to 0. If you took the derivative of psi with respect to x, it should be equal to this whole thing, just using the partial derivative chain rule. And we know what psi is. So we can write-- or actually we don't even have to. We could use this fact to say, well, if we integrate both sides, that a solution of this differential equation is that I just took the antiderivative of both sides. So, a solution to the differential equation is psi is equal to c. So psi is equal to x to the third y plus 1/2 x squared y squared. And we could have said plus c here, but we know the solution is that psi is equal to c, so we'll just write that there. I could have written a plus c here, but then you have a plus c here. You have another constant there. And you can just subtract them from both sides. And they just merge into another arbitrary constant. But anyway, there we have it. We had a differential equation that, at least superficially, looked exact. It looked exact, but then, when we tested the exactness of it, it was not exact. But we multiplied it by an integrating factor. And in the previous video, we figured out that a possible integrating factor is that we could just multiply both sides by x. And when we did that, we tested it. And true enough, it was exact." + }, + { + "Q": "\nAT 4:30 what happened to the y' how come it did not come down into the new equations, specifically to the right of the equals", + "A": "Because in the method to solving Exact Equations, you consider what is multiplying your y as your function N(x,y) and the rest as your function M(x,y). The original y is used just to separate those 2 parts, it no longer has any involvement in the rest of the solution.", + "video_name": "0NyeDUhKwBE", + "timestamps": [ + 270 + ], + "3min_transcript": "So we get 3x squared plus 2xy. And there we have it. The partial of this with respect to y is equal to the partial of this with respect to N. So we now have an exact equation whose solution should be the same as this. All we did is we multiplied both sides of this equation by x. So it really shouldn't change the solution of that equation, or that differential equation. So it's exact. Let's solve it. So how do we do that? Well, what we say is, since we've shown this exact, we know that there's some function psi where the partial derivative of psi with respect to x is equal to this expression right here. So it's equal to 3x squared y plus xy squared. Let's take the antiderivative of both sides with respect to x, and we'll get psi is equal to what? 1/2 x squared y squared. And of course, this psi is a function of x and y, so when you take the partial with respect to x, when you go that way, you might have lost some function that's only a So instead of a plus c here, it could've been a whole function of y that we lost. So we'll add that back when we take the antiderivative. But we're not completely done yet, because we have to somehow figure out what this function of y is. And the way we figure that out is we use the information that the partial of this with respect to y should be equal to this. So let's set that up. So what's the partial of this expression with respect to y? So I could write, the partial of psi with respect to y is equal to x to the third plus 2 times 1/2, so it's just x That's the partial of a function purely of y with respect to y. And then that has to equal our new N, or the new expression we got after multiplying by the integrating factor. So that's going to be equal to this right here. This is, hopefully, making sense to you at this point. And that should be equal to x to the third plus x squared y. And interesting enough, both of these terms are on this side. So let's subtract both of those terms from both sides. So x to the third, x to the third, x squared y, x squared y. And we're left with h prime of y is equal to 0. Or you could say that h of y is equal to some constant." + }, + { + "Q": "\nHow did you get the negative sixteen to become a positive at 0:51? How is it the same?", + "A": "remember if the problem say is 7- (-9) it makes no sense to solve it. So it s easier if you remember that if there are two minus signs next to each other then change them into a plus sign so there won t have to be any of the minus negative confusion. :) Example: 2+7- (-8)+13 9+15+13 24+13 37", + "video_name": "03yq7XsErqo", + "timestamps": [ + 51 + ], + "3min_transcript": "Evaluate 3x squared minus 8x plus 7 when x is equal to negative 2. So to evaluate this expression when x is equal to negative 2, everywhere that we see an x, we just have to substitute it with a negative 2. So it would be 3 times, instead of x squared, it would be 3 times negative 2 squared minus 8 times-- instead of 8 times x, 8 times negative 2 plus 7. And so what does this give us? So we have 3 times negative 2 squared. Well, that's just positive 4. A negative times a negative is a positive. Minus 8 times negative 2 is negative 16 plus 7. So 3 times 4 is 12. From that, you're subtracting negative 16. Subtracting negative 16 is the same thing as adding positive 16. So it's 12 plus 16 plus 7. And so this is equal to 12 plus 16 is 28." + }, + { + "Q": "Why is it 4 becquerel at 3:48 ? Is it because of c=8 and you need to half it because cesium are reduced to half after 30 days?\n", + "A": "Yes, you are right.", + "video_name": "polop-89aIA", + "timestamps": [ + 228 + ], + "3min_transcript": "Fair enough? So let's just be clear, this is days since release. In addition, assume that we know that the initial amount of caesium-137 released in the soil is 8 becquerels. Solve for the unknown constants c and r. So the initial in the soil. That's when t is equal to 0. When no days have passed. So we could say that the amount at times 0-- well, that's going to be equal to c times r to the 0 power, which is just going to be equal to c times 1, which is equal to c. And they tell us what A of 0 is. They say A of 0 is 8 becquerels. So this is going to be equal to 8. So our constant here, the c is just going to be equal to 8. What is the value of the constant? We could just write 8 right over there. So the value of the constant c is 8. What is the value of the constant r? So we're starting with 8. So A of 0 is 8. How much are we going to have after 30 days? And the reason why I'm picking 30 days is-- that is the half-life of caesium-137. So A of 30, remember our t is in-- let me just switch colors just for fun. Remember, t is in days. So A of 30. So after 30 days, I am going to-- if I want to use this formula right over here, if I wanted to use the description of this exponential function, we already know that c is 8. It's going to be 8 times r to the 30th power, which is going to be equal to what? Well, if we started with 8, 30 days later, we're going to have half as much. We're going to have 4 becquerels. And now, we can use this to solve for r. So you have 8 times r to the 30th power is equal to 4. Divide both sides by 8. which is the same thing as 1/2. And then, we can raise both sides to the 1/30th power. r to the 30th-- but then, you could think of the 30th root of that or raising that to the 1/30th power-- that's just going to give us r is equal to one half to the 1/30th power. And that is something that's very hard to compute in your head. So I suggest you use a calculator for that. And they hint because we're going to round to the nearest thousandth. So let's get a calculator right out. And so we're talking about one half to the 1/30 power. So we get 0.9771599-- it keeps going. But they tell us to round to the nearest thousandth, so 0.977" + }, + { + "Q": "at 2:20 how the heck does (4x^4-x)/(x^4) simplify to 4 - (1/x^3) ?? I'm missing something, that for sure.\n", + "A": "Recall that in general, (a - b)/c is equivalent to a/c - b/c. So (4x^4-x)/(x^4) = (4x^4)/(x^4) - x/(x^4) = 4 - 1/(x^3). If you are rusty with algebra, you will need to review algebra in order to have a realistic chance of performing well in calculus. Have a blessed, wonderful day!", + "video_name": "uPksX_O9ARo", + "timestamps": [ + 140 + ], + "3min_transcript": "- [Voiceover] Let's see if we can find the limit as x approaches negative infinity of the square root of 4x to the fourth minus x over 2x squared plus three. And like always, pause this video and see if you can figure it out. Well, whenever we're trying to find limits at either positive or negative infinity of rational expressions like this, it's useful to look at what is the highest degree term in the numerator or in the denominator, or, actually in the numerator and the denominator, and then divide the numerator and the denominator by that highest degree, by x to that degree. Because if we do that, then we're going to end up with some constants and some other things that will go to zero as we approach positive or negative infinity, and we should be able to find this limit. So what I'm talking about, let's divide the numerator by one over x squared and let's divide the denominator by one over x squared. Now, you might be saying, \"Wait, wait, \"I see an x to the fourth here. But remember, it's under the radical here. So if you wanna look at it at a very high level, you're gonna take the square root of this entire expression, so you can really view this as a second degree term. So the highest degree is really second degree, so let's divide the numerator and the denominator by x squared. And if we do that, dividing, so this is going to be the same thing as, so this is going to be the limit, the limit as x approaches negative infinity of, so let me just do a little bit of a side here. So if I have, if I have one over x squared, all right, let me write it. Let me just, one over x squared times the square root of 4x to the fourth minus x, like we have in the numerator here. This is equal to, this is the same thing as one over the square root of x to the fourth times the square root of 4x to the fourth minus x. And so this is equal to the square root of 4x to the fourth minus x and all I did is I brought the radical in here. You could view this as the square root of all this divided by the square root of this, which is equal to, just using our exponent rules, the square root of 4x to the fourth minus x over x to the fourth. And then this is the same thing as four minus, x over x to the fourth is one over x to the third. So this numerator is going to be, the numerator's going to be the square root of four minus one, x to the third power. And then the denominator is going to be equal to, well, you divide 2x squared by x squared. You're just going to be left with two. And then three divided by x squared is gonna be three over x squared. Now, let's think about the limit as we approach negative infinity. As we approach negative infinity, this is going to approach zero." + }, + { + "Q": "\nat 3:00 how can he go from x^2 right to x?", + "A": "It should be |x| < 1/2.", + "video_name": "6ynr9N-NQ8E", + "timestamps": [ + 180 + ], + "3min_transcript": "We can rewrite f of x as being equal to the sum from n equals 0 to infinity of-- let's see, our first term here is 2-- 2 times negative 4x squared to the n-th power. This is a geometric series where our common ratio is negative 4x squared to the n-th power. Now, when will this thing right over here converge? Well, we know that a geometric series will converge if the absolute value of its common ratio is less than 1. So let me write this down. So converge if the absolute value of the common ratio, negative 4x squared, is less than 1. Well, this-- the way it's written right now-- So the absolute value of this is just going to be 4x squared. Right? x squared is going to be non-negative, so 4x squared is going to be non-negative. Negative 4x squared is going to be non-positive. So if you're taking the absolute value of a non-positive thing, that's going to be the same thing as the absolute value of the negative of it. So this just has to be less than 1. And the absolute value of something that is strictly non-negative like this, well, that's just going to be 4x squared-- these two statements are equivalent-- and that has to be less than 1. Can divide both sides by 4, you get x squared is less than 1/4. And so we can say that the absolute value of x has to be less than 1/4, or we could say that negative 1/4 So expressed this way, we're giving the interval of convergence. This thing will converge as long as x is in this interval. Expressed this way, we're really saying the radius of convergence. This will converge as long as x is less than our radius of convergence, as long as the absolute value of x is less than our radius of convergence, as long as x stays less than 1/4 away from 0. To make it a little bit clearer, you could rewrite this as the distance between x and 0, as long as this- this, you could view this as the distance between x and 0-- as long as this stays less than 1/4, this thing is going to converge. So this is the interval of convergence, this, you could view, 1/4, you could view as the radius of convergence. Now with that out of the way, we've thought about where this thing converges, let's think about what it converges to. Well, we've done this multiple times." + }, + { + "Q": "\nAt 3:06, can we write the answer as - (m+4)/5\nSince both values have the same denominator, and we can take -1 as a common for the whole numerator, making both m and 4 positive.", + "A": "Yes, the answer could be written that way. It is equivalent to Sal s version.", + "video_name": "rtNuo7R3scY", + "timestamps": [ + 186 + ], + "3min_transcript": "as the negative of 8.55 minus 4.35, and 8.55 minus 4.35, let's see, eight minus four is going to be the negative, eight minus four is four, 55 hundredths minus 35 hundredths is 20 hundredths. So I could write 4.20, which is really just the same thing as 4.2. So all of this can be replaced with a -4.2. So my entire expression has simplified to -5.55, and instead of saying plus -4.2c, I can just write it as minus 4.2c, We can't simplify this anymore. We can't add this term that doesn't involve the variable to this term that does involve the variable. So this is about as simple as we're gonna get. So let's do another example. So here I have some more hairy numbers involved. And so, let's see, I have 2/5m minus 4/5 minus 3/5m. So how can I simplify? Well I could add all the m terms together. So let me just change the order. I could rewrite this as 2/5m minus 3/5m minus 4/5. All I did was I changed the order. We can see that I have these two m terms. I can add those two together. So this is going to be 2/5 minus 3/5 times m, and then I have the -4/5 still on the right hand side. Now what's 2/5 minus 3/5? Well that's gonna be -1/5. That's gonna be -1/5. So I have -1/5m minus 4/5. Minus 4/5. I can't simplify it anymore. I can't add this term that involves m somehow So we are done here. Let's do one more. Let's do one more example. So here, and this is interesting, I have a parentheses and all the rest. Like always, pause the video. See if you can simplify this. All right, let's work through it together. Now the first thing that I want to do is let's distribute this two so that we just have three terms that are just being added and subtracted. So if we distribute this two, we're gonna get two times 1/5m is 2/5m. Let me make sure you see that m. M is right here. Two times -2/5 is -4/5, and then I have plus 3/5. Now how can we simplify this more? Well I have these two terms here that don't involve the variable. Those are just numbers. I can add them to each other. So I have -4/5 plus 3/5. So what's negative four plus three? That's going to be negative one. So this is going to be -1/5," + }, + { + "Q": "At 1:03 is the expression like this: -5.55 - 8.55c + 4.35c or is it like this: -5.55 + (-8.55c + 4.35c)?\nWhich expression is the right one, or are they the same?\n", + "A": "These expressions are equal to each other due to the associative property of addition and the fact that subtraction is the same as adding a negative. This can be seen more clearly by replacing the terms with easier numbers to visualize. For example, 10 - 5 + 2 = 10 + (-5 + 2). You can play around with the groups and order in addition and even subtraction, as long as you remember that subtraction is adding a negative. Be sure to keep your positive and negative signs with their numbers.", + "video_name": "rtNuo7R3scY", + "timestamps": [ + 63 + ], + "3min_transcript": "- [Voiceover] What I wanna do in this video is get some practice simplifying expressions and have some hairier numbers involved. These numbers are kind of hairy. Like always, try to pause this video and see if you can simplify this expression before I take a stab at it. All right, I'm assuming you have attempted it. Now let's look at it. We have -5.55 minus 8.55c plus 4.35c. So the first thing I wanna do is can I combine these c terms, and I definitely can. We can add -8.55c to 4.35c first, and then that would be, let's see, that would be -8.55 plus 4.35, I'm just adding the coefficients, times c, and of course, we still have that -5.55 out front. -5.55. I'll just put a plus there. Now how do we calculate -8.55 plus 4.35? Well there's a couple of ways to think about it or visualize it. as the negative of 8.55 minus 4.35, and 8.55 minus 4.35, let's see, eight minus four is going to be the negative, eight minus four is four, 55 hundredths minus 35 hundredths is 20 hundredths. So I could write 4.20, which is really just the same thing as 4.2. So all of this can be replaced with a -4.2. So my entire expression has simplified to -5.55, and instead of saying plus -4.2c, I can just write it as minus 4.2c, We can't simplify this anymore. We can't add this term that doesn't involve the variable to this term that does involve the variable. So this is about as simple as we're gonna get. So let's do another example. So here I have some more hairy numbers involved. And so, let's see, I have 2/5m minus 4/5 minus 3/5m. So how can I simplify? Well I could add all the m terms together. So let me just change the order. I could rewrite this as 2/5m minus 3/5m minus 4/5. All I did was I changed the order. We can see that I have these two m terms. I can add those two together. So this is going to be 2/5 minus 3/5 times m, and then I have the -4/5 still on the right hand side. Now what's 2/5 minus 3/5? Well that's gonna be -1/5. That's gonna be -1/5. So I have -1/5m minus 4/5. Minus 4/5. I can't simplify it anymore. I can't add this term that involves m somehow" + }, + { + "Q": "\nAt 2:17 in this vid of Extraneous solutions of Rad Eq, Sal goes to Quad Formula. However, at this point aren't we able to factor 4x^2 - 3x -7 = 0 to (4x-7)(x+1) to get one of his answers which is a solution, 7/4, and, -1, which is an extraneous solution.", + "A": "Yes... you can use factoring. It is likely simpler than the quadratic formula.", + "video_name": "m4eiYHL3PP8", + "timestamps": [ + 137 + ], + "3min_transcript": "- [Voiceover] So let's say we have the radical equation two-x minus one is equal to the square root of eight minus x. So we already have the radical isolated on one side So, we might say, \"Well, let's just get rid of the radical. \"Let's square both sides of this equation.\" So we might say that this is the same thing as two-x minus one squared is equal to the square root of eight minus x, eight minus x squared, and then we would get, let's see, two-x minus one squared is four x-squared minus four-x plus one is equal to eight minus x. Now we have to be very, very, very careful here. We might feel, \"Okay we did legitimate operations. \"We did the same thing to both sides. \"That these are equivalent equations.\" But, they aren't quite equivalent. Because when you're squaring something, one way to think about it, is when you're squaring it, you're losing information. So, for example, this would be true even if the original equation were two-x... Even if the original equation were two-x minus one is equal to the negative of the square root of eight minus x. Because if you squared both sides of this, you would also get, you would also get that right over there, because a negative squared would be equal to a positive. So, when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here, and not the solution to this up here. If it's a solution to this right-hand side, and not the yellow one, then we would call that an extraneous solution. So, let's see if we can solve this. So let's write this as kind of a standard quadratic. Let's subtract eight from both sides. So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides. So, plus x, plus x, and we are going to get, minus three-x, minus seven, minus seven, is equal to, is equal to zero. And let's see, we would want to factor this right over here, and, let's see, maybe I could do this by, if I do it by... Well, I'll just use the quadratic formula, here. So the solutions are going to be... X is going to be equal to negative b, so three, plus or minus the square root of b-squared, so negative three squared is nine, minus four times a, which is four, times c, which is negative seven. So I could just say times... Well, I'll just write a seven here and then that negative is gonna make this a positive. All of that over two-a. So two times four is eight. So, this is gonna be three" + }, + { + "Q": "At 0:42, how does he get 4x^2-4x+1=8-x?\nMore specifically, how does he get the second negative four?\n", + "A": "=(2x-1)^2 =(2x-1)(2x-1) =4x^2-2x-2x+1 =4x^2-4x+1 By using the distributive properties on this perfect square we are able to come up with two 2x and then combine like terms to get -4x.", + "video_name": "m4eiYHL3PP8", + "timestamps": [ + 42 + ], + "3min_transcript": "- [Voiceover] So let's say we have the radical equation two-x minus one is equal to the square root of eight minus x. So we already have the radical isolated on one side So, we might say, \"Well, let's just get rid of the radical. \"Let's square both sides of this equation.\" So we might say that this is the same thing as two-x minus one squared is equal to the square root of eight minus x, eight minus x squared, and then we would get, let's see, two-x minus one squared is four x-squared minus four-x plus one is equal to eight minus x. Now we have to be very, very, very careful here. We might feel, \"Okay we did legitimate operations. \"We did the same thing to both sides. \"That these are equivalent equations.\" But, they aren't quite equivalent. Because when you're squaring something, one way to think about it, is when you're squaring it, you're losing information. So, for example, this would be true even if the original equation were two-x... Even if the original equation were two-x minus one is equal to the negative of the square root of eight minus x. Because if you squared both sides of this, you would also get, you would also get that right over there, because a negative squared would be equal to a positive. So, when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here, and not the solution to this up here. If it's a solution to this right-hand side, and not the yellow one, then we would call that an extraneous solution. So, let's see if we can solve this. So let's write this as kind of a standard quadratic. Let's subtract eight from both sides. So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides. So, plus x, plus x, and we are going to get, minus three-x, minus seven, minus seven, is equal to, is equal to zero. And let's see, we would want to factor this right over here, and, let's see, maybe I could do this by, if I do it by... Well, I'll just use the quadratic formula, here. So the solutions are going to be... X is going to be equal to negative b, so three, plus or minus the square root of b-squared, so negative three squared is nine, minus four times a, which is four, times c, which is negative seven. So I could just say times... Well, I'll just write a seven here and then that negative is gonna make this a positive. All of that over two-a. So two times four is eight. So, this is gonna be three" + }, + { + "Q": "\nAt 0:21, Sal said that when we've got a limit as x approaches infinity of a compilcated function, we can simplify the function to the most important terms, e.g.:\nlim x->\u00e2\u0088\u009e (5x\u00c2\u00b3+3x\u00c2\u00b2-8)/(x\u00c2\u00b3-60x\u00c2\u00b2+18)\u00e2\u0089\u00885x\u00c2\u00b3 /x\u00c2\u00b3 =5.\nMy question is, what's about x as an exponent? What are the most important terms in a function, where some numbers are in the x-th power, for example:\nlim x->\u00e2\u0088\u009e (5x\u00c2\u00b2 +18pi)/(e^x+5x)?", + "A": "As x goes to infinity, an exponential with a base greater than 1 will eventually outgrow any term with a constant exponent, so the limit in your example is 0 (and it would be 0 even if the first term in the numerator changed to 1000x^1000 and the first term in the denominator changed to 1.001^x). At infinity, exponentials win (unless they re up against something even more powerful, like factorials).", + "video_name": "KcqO1fX9b_I", + "timestamps": [ + 21 + ], + "3min_transcript": "" + }, + { + "Q": "\nwhy at 2:26 did sal say 3x cubed/6x to the fourth is equal to 1/2x?", + "A": "At 2:03. This is plain algebra; the 3 in the numerator and the 6 in the denominator simplify to 1/2. The x^3 in the numerator and the x^4 in the denominator simplifies to 1/x. Perhaps think of the original fraction as 3(x)(x)(x) / 6(x)(x)(x)(x). Can you see three of the x s cancel out leaving 3/6x which is 1/2x ?", + "video_name": "KcqO1fX9b_I", + "timestamps": [ + 146 + ], + "3min_transcript": "" + }, + { + "Q": "When we take limits at infinity we basically look if there is a horizontal asymptote. At 03:57 Sal talks about the case where the limit at infinity is equal to infinity. Wouldn't that look like a vertical rather than a horizontal asymptote?\n", + "A": "If the limit at infinity is infinity, there won t really be an asymptote, it will just look like the line s constantly moving up and to the right.", + "video_name": "KcqO1fX9b_I", + "timestamps": [ + 237 + ], + "3min_transcript": "" + }, + { + "Q": "\nWhy can he take out the constants at 6:35? I'm guessing it won't change the answer, but I don't get how", + "A": "the constant rule states that we can take the constant out because a derivative of a constant is 0", + "video_name": "Xe6YlrCgkIo", + "timestamps": [ + 395 + ], + "3min_transcript": "let me move this over. Let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well the derivative with respect to time of our volume, we could just rewrite that as dV dt, this thing right over here. This is dV dt, and this is going to be equal to-- well we could take the constants out of this-- this is going to be equal to pi over 12 times the derivative with respect to t of h, of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time. As time goes on, the height will change. Because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power. Just to make it clear that this is a function of t. h of t to the third power. Now what is the derivative with respect to t, of h of t to the third power. Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule. The chain rule tells us-- let me rewrite everything else. dV with respect to t, is going to be equal to pi over 12, times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something. So that's going to be-- let me write this in a different color, maybe in orange-- so that's going to be 3 times our something squared, Times dh-- I've already used that pink-- times dh dt. Let's just be very clear. This orange term right over here-- and I'm just using the chain rule-- this is the derivative of h of t to the third power with respect to h of t. And then we're going to multiply that times the derivative of h of t with respect to t. And then that gives us the derivative of this entire thing, h of t to the third power, with respect This will give us the derivative of h of t to the third power with respect to d with respect" + }, + { + "Q": "at 8:27 Saul says \"pi over 12\" but writes 'pi over 2' and carries that mistake through\n", + "A": "yes, yes it is. but he doesn t carry the mistake through. as he fixes it at 10:28", + "video_name": "Xe6YlrCgkIo", + "timestamps": [ + 507 + ], + "3min_transcript": "As time goes on, the height will change. Because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power. Just to make it clear that this is a function of t. h of t to the third power. Now what is the derivative with respect to t, of h of t to the third power. Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule. The chain rule tells us-- let me rewrite everything else. dV with respect to t, is going to be equal to pi over 12, times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something. So that's going to be-- let me write this in a different color, maybe in orange-- so that's going to be 3 times our something squared, Times dh-- I've already used that pink-- times dh dt. Let's just be very clear. This orange term right over here-- and I'm just using the chain rule-- this is the derivative of h of t to the third power with respect to h of t. And then we're going to multiply that times the derivative of h of t with respect to t. And then that gives us the derivative of this entire thing, h of t to the third power, with respect This will give us the derivative of h of t to the third power with respect to d with respect to do when we apply this operator. How fast is this changing? How is this changing with respect to time? So we can just rewrite this, just so gets a little bit cleaner. Let me rewrite everything I've done. So we've got dV, the rate at which our volume is changing with respect to time. The rate at which our volume is changing with respect to time is equal to pi over 12 times 3 h of t squared, or I could just write that as 3h squared, times the rate at which the height is changing with respect to time, times dh dt. And you might be a little confused. You might have been tempted to take the derivative over here with respect to h. But remember, we're thinking about how things are changing with respect to time. So we're assuming-- we did express volume as a function of height-- but we're" + }, + { + "Q": "\nI don't understand how at 6:30 he separates the function to take the derivative and how you know to do so.", + "A": "\u00cf\u0080/12 is a constant, so by definition d/dx [3x^2] is equal to 3*d/dx [x^2]. You might know that the derivative of 3x^2 is equal to 3*2x^2-1, 6x^1 or 6x. Likewise, 3 derivative of x^2 is 3*2x or 6x. Same result. So, d/dt [\u00cf\u0080/12*h^3] is the same of \u00cf\u0080/12*d/dx [h^3]. \u00cf\u0080 is not variable.", + "video_name": "Xe6YlrCgkIo", + "timestamps": [ + 390 + ], + "3min_transcript": "is going to be pi r squared, pi times the radius squared. h over 2 squared. That's the area of the surface of the water. And of course we still have the 1/3 out here. And we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1/3 times pi h squared over 4 times another h, which is equal to-- we have pi, h to the third power over 12. So that is our volume. Now what we want to do is relate the volume, how fast the volume is changing with respect to time, and how fast the height is changing with respect to time. So we care with respect to time, since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time. let me move this over. Let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well the derivative with respect to time of our volume, we could just rewrite that as dV dt, this thing right over here. This is dV dt, and this is going to be equal to-- well we could take the constants out of this-- this is going to be equal to pi over 12 times the derivative with respect to t of h, of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time. As time goes on, the height will change. Because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power. Just to make it clear that this is a function of t. h of t to the third power. Now what is the derivative with respect to t, of h of t to the third power. Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule. The chain rule tells us-- let me rewrite everything else. dV with respect to t, is going to be equal to pi over 12, times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something. So that's going to be-- let me write this in a different color, maybe in orange-- so that's going to be 3 times our something squared," + }, + { + "Q": "At 11:29 is the rate of change (Dh/dt) cm/s or cm^3/s\n", + "A": "cm/s. The unit of the rate depends on the units of the letters on top and bottom. h(cm)/ t (sec) => cm/s", + "video_name": "Xe6YlrCgkIo", + "timestamps": [ + 689 + ], + "3min_transcript": "So we're taking the derivative of everything with respect to time. So that's why the chain rule came into play when we were taking the derivative of h, or the derivative of h of t, because we're assuming that h is a function of time. Now what does this thing right over here get us? Well we're telling us at the exact moment that we set up this problem, we know what dV dt is, we know that it's 1 centimeter cubed per second. We know that this right over here is 1 centimeter cubed per second. We know what our height is right at this moment. We were told it is 2 centimeters. So the only unknown we have over here is the rate at which our height is changing with respect to time. Which is exactly what we needed to figure out in the first place. So we just have to solve for that. So we get 1 cubic centimeter-- let me make it clear-- we get 1 cubic centimeter per second-- equal to pi over 2. And I'll write this in a neutral color. Actually, let me write in the same color. Is equal to pi over 2, times 3, times h squared. h is 2 so you're going to get 4 squared centimeters if we kept the units. So 3 times 4. All right let me be careful, that wasn't pi over 2, that was pi over 12. This is a pi over 12 right over here. So you get pi over 12, times 3 times 2 squared, times dh dt. All of this is equal to 1. So now I'll switch to a neutral color. We get 1 is equal to, well 3 times 4 is 12, cancels out with that 12. We get one is equal to pi times dh dt. To solve for dh dt divide both sides by pi. And we get our drum roll now. The rate at which our height is changing with respect to time as we're putting 1 cubic centimeter of water per second the rate at which this height is changing with respect to time is 1 over pi. And I haven't done the dimensional analysis but this is going to be in centimeters per second. You can work through the dimensional analysis if you like by putting in the dimensions right over here. But there you have it. That's how fast our height is going to be changing at exactly that moment." + }, + { + "Q": "\nAt 3:15, if Sal had known that the log10 5 = sqrt(0.5), he would've saved time, but it is tough to calculate the square root of 0.5, because square root of 0.5 is sqrt2 / 2, and the square root of 2 is hard to calculate. Am I right?", + "A": "sqrt(.5)= 0.7071... log 5 = .698... so log 5 =\\= sqrt(.5)", + "video_name": "OkFdDqW9xxM", + "timestamps": [ + 195 + ], + "3min_transcript": "is that we can change our base. Here are our bases, a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying-- this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x, and e is obviously the number 2.71, keeps going on and on and on forever. Now let's apply it to this problem. We need to figure out the logarithm-- and I'll use colors-- base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100-- what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2. So it simplifies to 2 over log base 10 of 5. And we can now use our calculator, because the log function on a calculator is log base 10. So let's get our calculator out. We want to clear this. 2 divided by-- When someone just writes log, they mean base 10. If they press LN, that means base e. So log without any other information is log base 10. So this is log base 10 of 5 is equal to 2 point-- So this is approximately equal to 2.861. And we can verify it because in theory, if I raise 5 to this power, I should get 100. And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power. And this number is closer to 3 than it is to 2. Well, let's verify it. So if I take 5 to that power, and then let me type in-- let me just type in what we did to the nearest thousandth-- 5 to the 2.861. So I'm not putting in all of the digits. What do I get? I get 99.94. If I put all of these digits in, it should get pretty close to 100. So that's what makes you feel good. That this is the power that I have to raise 5 to to get to 100." + }, + { + "Q": "\nWhat are combinatorics, mentioned at 1:51?", + "A": "Combinatorics, often represented by n!, is a field in mathematics dealing with how many ways there are to arrange a certain amount of objects. For example, if I had 4 objects, I could use combinatorics to figure out how many ways I could arrange them. Don t worry about them for now, you will learn about them in later grades.", + "video_name": "RdehfQJ8i_0", + "timestamps": [ + 111 + ], + "3min_transcript": "- [Voiceover] There's a lot of times, there's a lot of situations in which we're studying something pretty straightforward and we can find an exact theoretical probability. So what am I talking about? Just let me write that down. Theoretical probabiity. Well, maybe the simplest example, or one of the simplest examples is if you're flipping a coin. And let's say in theory you're flipping a completely fair coin and you're flipping it in a way that is completely fair. Well, there you know you have two outcomes. Either heads will be on top or tails will be on top. So theoretically you say, \"well, look, \"if I want to figure out the probability \"of getting a heads, in theory I have two \"equally likely possibilities, and heads \"is one of those two equally likely possibilities.\" So you have a 1/2 probability. Once again, if in theory the coin is definitely fair, it's a fair coin and it's flipped in a very fair way, then this is true. You have a 1/2 probability. A fair six-sided die is going to have six possible outcomes: one, two, three, four, five and six. And if you said \"what is the probability of getting \"a result that is greater than or equal to three?\" Well, we have six equally likely possibilities. You see them there. In theory, if they're all equally likely, four of these possibilities meet our constraint of being greater than or equal to three. We have four out of the six of these possibilities meet our constraints. So we have a 2/3, 4/6 is the same thing as 2/3, probability of it happening. Now these are for simple things, like die or flipping a coin. And if you have fancy computers or spreadsheets you can even say \"hey, \"I'm gonna flip a coin a bunch of times \"and do all the combinatorics\" and all that. But there are things that are even beyond what a computer can find the exact theoretical probability for. Let's say you are playing a game, say football, American football, of scoring a certain number of points. Well that isn't very simple because that's going to involve what human beings are doing. Minds are very unpredictable, how people will respond to things. The weather might get involved. Someone might fall sick. The ball might be wet, or just how the ball might interact with some player's jersey. Who knows what might actually result in the score being one point this way, or seven points this way, or seven points that way. So for situations like that, it makes more sense to think more in terms of experimental probability. In experimental probability, we're really just trying to get an estimate of something happening, based on data and experience that we've had in the past. For example, let's say you had data from your football team and it's many games into the season." + }, + { + "Q": "At 1:51 you said \"do all the COMBINATORICS\". But you never said what it is a combinatoric?\n", + "A": "Combinatoric is just a fancy word for math that has to do with different combinations of something.", + "video_name": "RdehfQJ8i_0", + "timestamps": [ + 111 + ], + "3min_transcript": "- [Voiceover] There's a lot of times, there's a lot of situations in which we're studying something pretty straightforward and we can find an exact theoretical probability. So what am I talking about? Just let me write that down. Theoretical probabiity. Well, maybe the simplest example, or one of the simplest examples is if you're flipping a coin. And let's say in theory you're flipping a completely fair coin and you're flipping it in a way that is completely fair. Well, there you know you have two outcomes. Either heads will be on top or tails will be on top. So theoretically you say, \"well, look, \"if I want to figure out the probability \"of getting a heads, in theory I have two \"equally likely possibilities, and heads \"is one of those two equally likely possibilities.\" So you have a 1/2 probability. Once again, if in theory the coin is definitely fair, it's a fair coin and it's flipped in a very fair way, then this is true. You have a 1/2 probability. A fair six-sided die is going to have six possible outcomes: one, two, three, four, five and six. And if you said \"what is the probability of getting \"a result that is greater than or equal to three?\" Well, we have six equally likely possibilities. You see them there. In theory, if they're all equally likely, four of these possibilities meet our constraint of being greater than or equal to three. We have four out of the six of these possibilities meet our constraints. So we have a 2/3, 4/6 is the same thing as 2/3, probability of it happening. Now these are for simple things, like die or flipping a coin. And if you have fancy computers or spreadsheets you can even say \"hey, \"I'm gonna flip a coin a bunch of times \"and do all the combinatorics\" and all that. But there are things that are even beyond what a computer can find the exact theoretical probability for. Let's say you are playing a game, say football, American football, of scoring a certain number of points. Well that isn't very simple because that's going to involve what human beings are doing. Minds are very unpredictable, how people will respond to things. The weather might get involved. Someone might fall sick. The ball might be wet, or just how the ball might interact with some player's jersey. Who knows what might actually result in the score being one point this way, or seven points this way, or seven points that way. So for situations like that, it makes more sense to think more in terms of experimental probability. In experimental probability, we're really just trying to get an estimate of something happening, based on data and experience that we've had in the past. For example, let's say you had data from your football team and it's many games into the season." + }, + { + "Q": "So at 1:45 you had to simplify to get 2/3.\n", + "A": "Yeah, you had to", + "video_name": "RdehfQJ8i_0", + "timestamps": [ + 105 + ], + "3min_transcript": "- [Voiceover] There's a lot of times, there's a lot of situations in which we're studying something pretty straightforward and we can find an exact theoretical probability. So what am I talking about? Just let me write that down. Theoretical probabiity. Well, maybe the simplest example, or one of the simplest examples is if you're flipping a coin. And let's say in theory you're flipping a completely fair coin and you're flipping it in a way that is completely fair. Well, there you know you have two outcomes. Either heads will be on top or tails will be on top. So theoretically you say, \"well, look, \"if I want to figure out the probability \"of getting a heads, in theory I have two \"equally likely possibilities, and heads \"is one of those two equally likely possibilities.\" So you have a 1/2 probability. Once again, if in theory the coin is definitely fair, it's a fair coin and it's flipped in a very fair way, then this is true. You have a 1/2 probability. A fair six-sided die is going to have six possible outcomes: one, two, three, four, five and six. And if you said \"what is the probability of getting \"a result that is greater than or equal to three?\" Well, we have six equally likely possibilities. You see them there. In theory, if they're all equally likely, four of these possibilities meet our constraint of being greater than or equal to three. We have four out of the six of these possibilities meet our constraints. So we have a 2/3, 4/6 is the same thing as 2/3, probability of it happening. Now these are for simple things, like die or flipping a coin. And if you have fancy computers or spreadsheets you can even say \"hey, \"I'm gonna flip a coin a bunch of times \"and do all the combinatorics\" and all that. But there are things that are even beyond what a computer can find the exact theoretical probability for. Let's say you are playing a game, say football, American football, of scoring a certain number of points. Well that isn't very simple because that's going to involve what human beings are doing. Minds are very unpredictable, how people will respond to things. The weather might get involved. Someone might fall sick. The ball might be wet, or just how the ball might interact with some player's jersey. Who knows what might actually result in the score being one point this way, or seven points this way, or seven points that way. So for situations like that, it makes more sense to think more in terms of experimental probability. In experimental probability, we're really just trying to get an estimate of something happening, based on data and experience that we've had in the past. For example, let's say you had data from your football team and it's many games into the season." + }, + { + "Q": "What is that symbol he draws at 1:13?\n", + "A": "greater than or equal to", + "video_name": "RdehfQJ8i_0", + "timestamps": [ + 73 + ], + "3min_transcript": "- [Voiceover] There's a lot of times, there's a lot of situations in which we're studying something pretty straightforward and we can find an exact theoretical probability. So what am I talking about? Just let me write that down. Theoretical probabiity. Well, maybe the simplest example, or one of the simplest examples is if you're flipping a coin. And let's say in theory you're flipping a completely fair coin and you're flipping it in a way that is completely fair. Well, there you know you have two outcomes. Either heads will be on top or tails will be on top. So theoretically you say, \"well, look, \"if I want to figure out the probability \"of getting a heads, in theory I have two \"equally likely possibilities, and heads \"is one of those two equally likely possibilities.\" So you have a 1/2 probability. Once again, if in theory the coin is definitely fair, it's a fair coin and it's flipped in a very fair way, then this is true. You have a 1/2 probability. A fair six-sided die is going to have six possible outcomes: one, two, three, four, five and six. And if you said \"what is the probability of getting \"a result that is greater than or equal to three?\" Well, we have six equally likely possibilities. You see them there. In theory, if they're all equally likely, four of these possibilities meet our constraint of being greater than or equal to three. We have four out of the six of these possibilities meet our constraints. So we have a 2/3, 4/6 is the same thing as 2/3, probability of it happening. Now these are for simple things, like die or flipping a coin. And if you have fancy computers or spreadsheets you can even say \"hey, \"I'm gonna flip a coin a bunch of times \"and do all the combinatorics\" and all that. But there are things that are even beyond what a computer can find the exact theoretical probability for. Let's say you are playing a game, say football, American football, of scoring a certain number of points. Well that isn't very simple because that's going to involve what human beings are doing. Minds are very unpredictable, how people will respond to things. The weather might get involved. Someone might fall sick. The ball might be wet, or just how the ball might interact with some player's jersey. Who knows what might actually result in the score being one point this way, or seven points this way, or seven points that way. So for situations like that, it makes more sense to think more in terms of experimental probability. In experimental probability, we're really just trying to get an estimate of something happening, based on data and experience that we've had in the past. For example, let's say you had data from your football team and it's many games into the season." + }, + { + "Q": "7:15 - 7;23 you can't simplify 19/20?\n", + "A": "no. there is no number that can go into both 19 and 20.", + "video_name": "bcCLKACsYJ0", + "timestamps": [ + 435 + ], + "3min_transcript": "it's gonna be six plus 11 over 12, which is equal to, six plus 11 is 17/12. If we wanted to write it as a mixed number, that is what, 12 goes into 17 one time with a remainder of five, so 1 5/12. Let's do one more of these. This is strangely fun. Alright. Let's say that we wanted to add, We're gonna add 3/4 to, we're gonna add 3/4 to 1/5. To one over five. What is this going to be? And once again, pause the video and see if you could work it out. Well we have different denominators here, and we wanna find, we wanna rewrite these so they have the same denominators, so we have to find a common multiple, ideally the least common multiple. So what's the least common multiple of four and five? Well let's start with the larger number, them until we get one that's divisible by four. So five is not divisible by four. 10 is not divisible by four, or perfectly divisible by four is what we care about. 15 is not perfectly divisible by four. 20 is divisible by four, in fact, that is five times four. That is 20. So what we could do is, we could write both of these fractions as having 20 in the denominator, or 20 as the denominator. So we could write 3/4 is something over 20. So to go from four to 20 in the denominator, we multiplied by five. So we also do that to the numerator. We multiply by three times five to get 15. All I did to go from four to 20, multiplied by five. So I have to do the same thing to the numerator, three times five is 15. 3/4 is the same thing as 15/20, and over here. 1/5. What is that over 20? Well to go from five to 20, you have to multiply by four. So we have to do the same thing to the numerator. I have to multiply this numerator times four to get 4/20. it's now written as 15/20 plus 4/20. And what is that going to be? Well that's going to be 15 plus four is 19/20. 19/20, and we're done." + }, + { + "Q": "At 7:25pm, How to rewrite a mixed fraction number into a decimal?\n", + "A": "You turn the mixed number into an improper fraction and the turn it into a decimal. Hope this helps!", + "video_name": "Gn2pdkvdbGQ", + "timestamps": [ + 445 + ], + "3min_transcript": "93 goes into 170? Goes into it one time. 1 times 93 is 93. 170 minus 93 is 77. Bring down the 0. 93 goes into 770? Let's see. It will go into it, I think, roughly eight times. 8 times 3 is 24. 8 times 9 is 72. Plus 2 is 74. And then we subtract. 10 and 6. It's equal to 26. Then we bring down another 0. 93 goes into 26-- about two times. 2 times 3 is 6. 18. This is 74. We could keep figuring out the decimal points. You could do this indefinitely. But if you wanted to at least get an approximation, you would say 17 goes into 93 0.-- or 17/93 is equal to 0.182 and then the decimals will keep going. And you can keep doing it if you want. If you actually saw this on exam they'd probably tell you to stop at some point. You know, round it to the nearest hundredths or thousandths place. And just so you know, let's try to convert it the other way, from decimals to fractions. Actually, this is, I think, you'll find a much easier thing to do. If I were to ask you what 0.035 is as a fraction? Well, all you do is you say, well, 0.035, we could write it this way-- we could write that's the same thing as 03-- That's the same thing as 35/1,000. And you're probably saying, Sal, how did you know it's 35/1000? Well because we went to 3-- this is the 10's place. Tenths not 10's. This is hundreths. This is the thousandths place. So we went to 3 decimals of significance. So this is 35 thousandths. If the decimal was let's say, if it was 0.030. There's a couple of ways we could say this. Well, we could say, oh well we got to 3-- we went to the thousandths Place. So this is the same thing as 30/1,000. We could have also said, well, 0.030 is the same thing as 0.03 because this 0 really doesn't add any value." + }, + { + "Q": "at 3:00 minutes i got stuck\n", + "A": "ok. you can complain it. that is not a problem. or you have no internet. that is why you are stuck.", + "video_name": "Gn2pdkvdbGQ", + "timestamps": [ + 180 + ], + "3min_transcript": "Let's do a slightly harder one. Let's figure out 1/3. Well, once again, we take the denominator, 3, and we divide it into the numerator. And I'm just going to add a bunch of trailing 0's here. 3 goes into-- well, 3 doesn't go into 1. 3 goes into 10 three times. 3 times 3 is 9. Let's subtract, get a 1, bring down the 0. 3 goes into 10 three times. Actually, this decimal point is right here. 3 times 3 is 9. Do you see a pattern here? We keep getting the same thing. As you see it's actually 0.3333. It goes on forever. And a way to actually represent this, obviously you can't write an infinite number of 3's. Is you could just write 0.-- well, you could write 0.33 repeating, which means that the 0.33 will go on forever. Although I tend to see this more often. Maybe I'm just mistaken. But in general, this line on top of the decimal means that this number pattern repeats indefinitely. So 1/3 is equal to 0.33333 and it goes on forever. Another way of writing that is 0.33 repeating. Let's do a couple of, maybe a little bit harder, but they all follow the same pattern. Let me pick some weird numbers. Let me actually do an improper fraction. Let me say 17/9. So here, it's interesting. The numerator is bigger than the denominator. So actually we're going to get a number larger than 1. But let's work it out. So we take 9 and we divide it into 17. And let's add some trailing 0's for the decimal point here. 1 times 9 is 9. 17 minus 9 is 8. Bring down a 0. 9 goes into 80-- well, we know that 9 times 9 is 81, so it has to go into it only eight times because it can't go into it nine times. 8 times 9 is 72. 80 minus 72 is 8. Bring down another 0. I think we see a pattern forming again. 9 goes into 80 eight times. 8 times 9 is 72. And clearly, I could keep doing this forever and we'd keep getting 8's. So we see 17 divided by 9 is equal to 1.88 where the 0.88 actually repeats forever. Or, if we actually wanted to round this we could say that that is also equal to 1.-- depending where we wanted" + }, + { + "Q": "My calc textbook just told me whenever I see this format of integral that I should just apply a formula. It gave the exact formula Sal achieved at 17:35, and just sub what ever x^2 value I get. IE when it is a a^2-x^2, just sub the x^2 into the formula which is exactly the same as what Sal got at 17:35. Given that this video took 20 mins and the formula takes 2 mins, I don't see any reason why I should disagree.\n", + "A": "Exactly, I have formula for all the problems worked through this chapter. The videos make a nice proof of showing why the formula works but it seems I d never have to do it as Sal does.", + "video_name": "sw2p2tUIFpc", + "timestamps": [ + 1055, + 1055 + ], + "3min_transcript": "this times 2 over 2. You might say, Sal, why are you doing that? Because I can rewrite this, let me write my whole thing here. So I have 2 arc sine of x minus 3 over 2, and then I have, I could take this denominator 2 right here. So I say, plus x minus 3 over 2. That 2 is that 2 right there. And then I could write this 2 right here as a square root of 4. Times the square root of 4, times the square root of all of this. 1 minus x minus 3 squared over 4. I think you see where I'm going. I'm kind of reversing everything that I did at the beginning of this problem. And maybe I'm getting a little fixated on making this as simple as possible, but I'm so close, so let me finish. So I get 2 times the arc sine of x minus 3 over 2, which I'm tired of writing, plus x minus 3 over 2, and if we bring this is equal to the square root of 4 times these things. So it's 4 minus x minus 3 squared, all of that that plus c. And we're at the home stretch. This is equal to 2 times the arc sine of x minus 3 over 2 plus x minus 3 over 2 times the radical of 4 minus, let's expand this, x squared minus 6x plus 9. And then this expression right here simplifies to minus minus, it's 6x minus x squared, and then you have a 4 minus 9 minus 5. Which is our original antiderivative. So finally, we're at the very end. plus x minus 3 over 2 times, time the radical of 6x minus x squared minus 5. That right there is the antiderivative of what this thing that we had at the very top of my little chalkboard, which is right there. So that is equal to the antiderivative of the square root of 6x minus x squared minus 5 dx. And I can imagine that you're probably as tired as I am. My hand actually hurts. But hopefully you find that to be vaguely satisfying. Sometimes I get complaints that I only do easy problems. Well, this was quite a hairy and not-so-easy problem." + }, + { + "Q": "i need help on 3:19\n", + "A": "what do you not get?", + "video_name": "x6xtezhuCZ4", + "timestamps": [ + 199 + ], + "3min_transcript": "So I'm going to draw 2/3, and I'm going to take 4/5 of it. So 2/3, and I'm going to make it pretty big. Just like this. So this is 1/3. And then this would be 2/3. Which I could do a little bit better job making those equal, or at least closer to looking equal. So there you go. I have thirds. Let me do it one more time. So here I have drawn thirds. 2/3 represents 2 of them. It represents 2 of them. One way to think about this is 2/3 times 4/5 is 4/5 of this 2/3. So how do we divide this 2/3 into fifths? Well, what if we divided each of these sections into 5. So let's do that. So let's divide each into 5. 1, 2, 3, 4, 5. And I could even divide this into 5 if I want. 1, 2, 3, 4, 5. And we want to take 4/5 of this section here. So how many fifths do we have here? We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And we've got to be careful. These really aren't fifths. These are actually 15ths, because the whole is this thing over here. So I should really say how many 15ths do we have? And that's where we get this number from. But you see if 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. Where did that come from? I had 3, I had thirds. And then I took each of those thirds, and I split them into fifths. So then I have five times as many sections. 3 times 5 is 15. But now we want 4/5 of this right over here. This is 10/15 right over here. Notice it's the same thing as 2/3. Now if we want to take 4/5 of that, So we're going to take 8 of them. So 1, 2, 3, 4, 5, 6, 7, 8. We took 8 of the 15, so that is 8/15. You could have thought about it the other way around. You could have started with fifths. So let me draw it that way. So let me draw a whole. So this is a whole. Let me cut it into five equal pieces, or as close as I can draw five equal pieces. 1, 2, 3, 4, 5. 4/5, we're going to shade in 4 of them. 4 of the 5 equal pieces. 3, 4. And now we want to take 2/3 of that. Well, how can we do that?" + }, + { + "Q": "I can take the derivative of V to find dv:\n[1/\u00cf\u0080 sin \u00cf\u0080x] => (cos \u00cf\u0080x)(\u00cf\u0080) / \u00cf\u0080\nbut (at 11:25) how did Sal find 1/\u00cf\u0080 sin \u00cf\u0080x from cos \u00cf\u0080x to solve for the anti-derivative?\n", + "A": "Yes, I understand now... Thank you for your answer! (cos \u00cf\u0080x) = (cos \u00cf\u0080x)(1) = (cos \u00cf\u0080x)(\u00cf\u0080 / \u00cf\u0080); D[sin \u00cf\u0080x] = \u00cf\u0080 cos \u00cf\u0080x; divide both sides by \u00cf\u0080 which yields: D[(sin \u00cf\u0080x) / \u00cf\u0080] = D[(1/\u00cf\u0080)(sin \u00cf\u0080x)] = (\u00cf\u0080 cos \u00cf\u0080x) / \u00cf\u0080 = cos \u00cf\u0080x", + "video_name": "CZdziIlYIfI", + "timestamps": [ + 685 + ], + "3min_transcript": "Well, we just figured out, from 0 to 1, f of x is just 1 minus x. f of x is just 1 minus x from 0 to 1 times cosine of pi x, cosine of pi x dx. And now we just have to evaluate this integral right over here. So let's do that. So 1 minus x times cosine of pi x is the same thing as cosine of pi x minus x cosine of pi x. Now, this right here, well, let's just focus on taking the antiderivative. This is pretty easy. But let's try to do this one, because it seems a little bit more complicated. So let's take the antiderivative of x cosine of pi x dx. And what should jump in your mind is, well, this isn't that simple. But if I were able to take the derivative of x, that would simplify. It's very easy to take the antiderivative of cosine of pi x without making it more complicated. And remember, integration by parts tells us that the integral-- I'll write it up here-- the integral of udv is equal to uv minus the integral of vdu. And we'll apply that here. But I've done many, many videos where I prove this and show examples of exactly what that means. But let's apply it right over here. And in general, we're going to take the derivative of whatever the u thing is. So we want u to be something that's simpler when I take the derivative. And then we're going to take the antiderivative of dv. So we want something that does not become more complicated when I take the antiderivative. So the thing that becomes simpler when I take this derivative is x. So if I set u is equal to x, then clearly du is equal to just dx. Or you say du dx is equal to 1. So du is equal to dx. And then dv is going to be the rest of this. dv is equal to cosine pi x dx. And so v would just be the antiderivative of this with respect to x. v is going to be equal to 1 over pi sine of pi x. If I took the derivative here, derivative of the inside, you get a pi, times 1 over pi, cancels out. Derivative of sine of pi x becomes cosine of pi x. So that's our u, that's our v. This is going to be equal to u times v. So it's equal to x, this x times this. So x over pi sine of pi x minus the integral of v, which is 1 over pi sine of pi x du." + }, + { + "Q": "I'm thoroughly confused. At 10:06, Sal gives a formula for integration by parts which he says he's gone through many times, but I just watched the integration by parts videos and didn't find it written it in this form? And I really thought I was getting the hang of these du, dv notations :(\n\nCan someone please explain how \u00e2\u0088\u00abf(x)g'(x)dx=\u00e2\u0088\u00abudv? I mean, I get it's some sort of substitution, but can't understand how f, g, and dx can be reduced to u and dv.\n\nAny help is greatly appreciated! :)\n", + "A": "Nvm, I was able to derive it: (uv) =uv +u v d(uv)/dx=u*dv/dx+du/dx*v u*dv/dx=d(uv)/dx-du/dx*v \u00e2\u0088\u00ab(u*dv/dx)dx=\u00e2\u0088\u00abd(uv)/dx*dx-\u00e2\u0088\u00ab(du/dx*v)dx \u00e2\u0088\u00abu*dv=\u00e2\u0088\u00abd(uv)-\u00e2\u0088\u00abdu*v \u00e2\u0088\u00abudv=uv-\u00e2\u0088\u00abvdu", + "video_name": "CZdziIlYIfI", + "timestamps": [ + 606 + ], + "3min_transcript": "So times 20 times the integral from 0 to 1 of f of x cosine of pi x dx. I forgot to write the dx over there. I want to make sure you understand, because this is really the hard part of the problem, just realizing that the integral over this interval is just 1/20 of the whole thing, because over every interval, from 0 to 1, the integral is going to evaluate to the same thing as going from 1 to 2, which will be the same thing as going from 2 to 3, or going from negative 2 to negative 1. So instead of doing the whole interval from negative 10 to 10, we're just doing 20 times the interval from 0 to 1. From negative 10 to 10, there's a difference of 20 here. So we're multiplying by 20. And this simplifies it a good bit. First of all, this part over here simplifies to 20 divided by 10 is 2. So it's 2 pi squared. So it becomes 2 pi squared-- that's just this part over here-- times the integral from 0 to 1. Well, we just figured out, from 0 to 1, f of x is just 1 minus x. f of x is just 1 minus x from 0 to 1 times cosine of pi x, cosine of pi x dx. And now we just have to evaluate this integral right over here. So let's do that. So 1 minus x times cosine of pi x is the same thing as cosine of pi x minus x cosine of pi x. Now, this right here, well, let's just focus on taking the antiderivative. This is pretty easy. But let's try to do this one, because it seems a little bit more complicated. So let's take the antiderivative of x cosine of pi x dx. And what should jump in your mind is, well, this isn't that simple. But if I were able to take the derivative of x, that would simplify. It's very easy to take the antiderivative of cosine of pi x without making it more complicated. And remember, integration by parts tells us that the integral-- I'll write it up here-- the integral of udv is equal to uv minus the integral of vdu. And we'll apply that here. But I've done many, many videos where I prove this and show examples of exactly what that means. But let's apply it right over here. And in general, we're going to take the derivative of whatever the u thing is. So we want u to be something that's simpler when I take the derivative. And then we're going to take the antiderivative of dv. So we want something that does not become more complicated when I take the antiderivative. So the thing that becomes simpler when I take this derivative is x. So if I set u is equal to x, then clearly du is equal to just dx. Or you say du dx is equal to 1. So du is equal to dx. And then dv is going to be the rest of this." + }, + { + "Q": "1:10\nso there are a infinate number of raidus?\n", + "A": "yes, a circle has infinite radii, as you can keep on changing the angle by a small amount", + "video_name": "04N79tItPEA", + "timestamps": [ + 70 + ], + "3min_transcript": "Draw a circle and label the radius, diameter, center and the circumference. Let me draw a circle, it won't be that well drawn of a circle but I think you get the idea so that is my circle. I'm going to label the center..over here. So I'll do the center I'll call it 'c'. So that is my center....and I'll draw an arrow there that is the center..of the circle and actually the circle itself is the set of all points that are a fixed distance away from that center and that fixed distance away, they're all from that center..that is the radius. So let me draw..the radius. So this distance right over here is the radius That is the radius and..that's going to be the same as this distance! ..which is the same as that distance! I can draw multiple radii. All of these are radii The distance between the center and any point on the circle now a diameter just goes straight across the circle and it's essentially two radii put toghether. So for example this would be a diameter.. that would be a diameter, you have one radii and another radii all on one line going from one side of the circle to another going through the center. So that is a diameter! ^_^........ that is a diameter. and I could have drawn it other ways, I could have drawn it like this that would be another diameter it would have the exact same length and finally, we have to think about the circumference, and the circumference is really just how far you have to go to go around the circle or if you put a string on that circle, how long would the string have to be? so what I'm tracing out in blue right now the length of what im tracing out is the circumference so....right over here..that is the circumference cir-cum-ference. Cir-cum-ference. And we're done!" + }, + { + "Q": "at 1:34 what dont undertsand bro\n", + "A": "The circumference of a circle is basically the distance around a circle. For example, if you had a park or other outdoor area that was shaped in a perfect circle, and you walked all the way around the edge of it, you would have walked along the circumference of the circle. Basically, you can think of the circumference as the perimeter of a circle.", + "video_name": "04N79tItPEA", + "timestamps": [ + 94 + ], + "3min_transcript": "Draw a circle and label the radius, diameter, center and the circumference. Let me draw a circle, it won't be that well drawn of a circle but I think you get the idea so that is my circle. I'm going to label the center..over here. So I'll do the center I'll call it 'c'. So that is my center....and I'll draw an arrow there that is the center..of the circle and actually the circle itself is the set of all points that are a fixed distance away from that center and that fixed distance away, they're all from that center..that is the radius. So let me draw..the radius. So this distance right over here is the radius That is the radius and..that's going to be the same as this distance! ..which is the same as that distance! I can draw multiple radii. All of these are radii The distance between the center and any point on the circle now a diameter just goes straight across the circle and it's essentially two radii put toghether. So for example this would be a diameter.. that would be a diameter, you have one radii and another radii all on one line going from one side of the circle to another going through the center. So that is a diameter! ^_^........ that is a diameter. and I could have drawn it other ways, I could have drawn it like this that would be another diameter it would have the exact same length and finally, we have to think about the circumference, and the circumference is really just how far you have to go to go around the circle or if you put a string on that circle, how long would the string have to be? so what I'm tracing out in blue right now the length of what im tracing out is the circumference so....right over here..that is the circumference cir-cum-ference. Cir-cum-ference. And we're done!" + }, + { + "Q": "\nFrom 0:44 to 0:46, Sal says \"then we add 3/4 to -10/6.\" But why add? Should the distributive property apply and we multiply? Vote up if this is a question you have too, please.", + "A": "The distributive property only applies if you re multiplying. Here, you re just adding the three fractions together. It looks confusing because Sal separated two of the fractions and added them together and then he added the third one to those two, but what you re doing to the fractions didn t change -- it s still just three fractions being added.", + "video_name": "9tmtDBpqq9s", + "timestamps": [ + 44, + 46 + ], + "3min_transcript": "We have negative 3/4 minus 7/6 minus 3/6. And there's many ways to do this. But it immediately jumps out at me that these last two numbers have a 6 in the denominator. So I'm going to worry about these first. I'm going to view this as negative 7/6 minus 3/6. So if we have negative 7/6 minus 3/6, that's going to be the same thing as negative 7 minus 3 over 6. And of course, we have this negative 3/4 out front that we're going to add to whatever we get over here. So this is these two terms that I'm adding together. Negative 7 minus 3 is negative 10. So it's negative 10 over 6. And then I'm going to have to add that to negative 3/4. And now I have to worry about finding a common denominator. Let me write that so they have a similar size. What is the smallest number that is a multiple of both 4 and 6? Well, it might jump out at you that it's 12. You can literally just go through the multiples of 4. Or you could look at the prime factorization of both of these numbers. And what's the smallest number that has all of the prime factors of both of these? So you need two 2s, and you need a 2 and a 3. So if you have two 2s and a 3, that's 4 times 3 is 12. So let's rewrite this as something over 12 plus something over 12. Well, to get your denominator from 4 to 12, you have to multiply by 3. So let's multiply our numerator by 3 as well. So if we multiply negative 3 times 3, you're going to have negative 9. And to get your denominator from 6 to 12, you have to multiply by 2. So let's multiply our numerator by 2 as well so that we don't change the value of the fraction. So that's going to be negative 20. Our common denominator is 12. And so this is going to be negative 9 plus negative 20, or we could even write that as minus 20, over 12, which is equal to-- and we deserve a drum roll now. This is negative 29 over 12. And 29 is a prime number, so it's not going to share any common factors other than 1 with 12. So we also have this in the most simplified form." + }, + { + "Q": "\nat 2:27 it said negative 9/12 and 20/12 but it forgot to write the plus sign.why is thst", + "A": "Because they are equivalent terms. The rule for subtraction is that you invert/change the sign of the second number and then add. So (-9/12) + (-20/12) is the same as (-9/12) - (20/12). He didn t need to add the plus sign at the end because it didn t change the value of the statement. Hope it helps.", + "video_name": "9tmtDBpqq9s", + "timestamps": [ + 147 + ], + "3min_transcript": "We have negative 3/4 minus 7/6 minus 3/6. And there's many ways to do this. But it immediately jumps out at me that these last two numbers have a 6 in the denominator. So I'm going to worry about these first. I'm going to view this as negative 7/6 minus 3/6. So if we have negative 7/6 minus 3/6, that's going to be the same thing as negative 7 minus 3 over 6. And of course, we have this negative 3/4 out front that we're going to add to whatever we get over here. So this is these two terms that I'm adding together. Negative 7 minus 3 is negative 10. So it's negative 10 over 6. And then I'm going to have to add that to negative 3/4. And now I have to worry about finding a common denominator. Let me write that so they have a similar size. What is the smallest number that is a multiple of both 4 and 6? Well, it might jump out at you that it's 12. You can literally just go through the multiples of 4. Or you could look at the prime factorization of both of these numbers. And what's the smallest number that has all of the prime factors of both of these? So you need two 2s, and you need a 2 and a 3. So if you have two 2s and a 3, that's 4 times 3 is 12. So let's rewrite this as something over 12 plus something over 12. Well, to get your denominator from 4 to 12, you have to multiply by 3. So let's multiply our numerator by 3 as well. So if we multiply negative 3 times 3, you're going to have negative 9. And to get your denominator from 6 to 12, you have to multiply by 2. So let's multiply our numerator by 2 as well so that we don't change the value of the fraction. So that's going to be negative 20. Our common denominator is 12. And so this is going to be negative 9 plus negative 20, or we could even write that as minus 20, over 12, which is equal to-- and we deserve a drum roll now. This is negative 29 over 12. And 29 is a prime number, so it's not going to share any common factors other than 1 with 12. So we also have this in the most simplified form." + }, + { + "Q": "\nln(sinx)/lnx=sinx/x right?so can i omit the step 6:30-11:00?and concluded that limit of ln(sinx)/lnx=1?\nthanks in anvance", + "A": "No, ln (sin x) / ln x DOES NOT equal sin x / x For example, when x = e: ln (sin e) / ln e = ln (sin e) \u00e2\u0089\u0088 -0.8897 but sin (e) / e \u00e2\u0089\u0088 0.15112", + "video_name": "CDf_aE5yg3A", + "timestamps": [ + 390, + 660 + ], + "3min_transcript": "what's the limit as X approaches zero from the positive direction of Y and once again we don't know. Maybe it's zero to the zero but we don't know what zero to zero actually is. What we could do, what we could do and this is a trick that you see a lot and anytime you get kind of weird things with exponents and whether you're doing limits or derivatives, as you'll see it's often times useful to take the natural log of both sides. Well what happens if you take the natural log of both sides here? On the left-hand side you're going to have the natural log, the natural log, and whenever I think of natural log and E the way I always think about them, the color green for some bizarre reason but we'll say the natural log of Y is equal to. If you take the natural log of this thing, actually let me just, I don't want to skip steps here because this is interesting. This is going to be the natural log of all of this business of sine of X, let me write this way. Sine of X, sine, I want to do this in that orange color. The natural log of over the natural log of X. Well we know from our exponent prior our logarithm properties, the logarithm of something to a power, that's the same thing as the power. One over natural log of X times the logarithm, this case the natural logarithm of whatever taking the sine of X here. Sine of X or we could say the natural log of Y. Want to keep, stay color consistent for at least one more step. The natural log of Y is equal to, if we just rewrite this this is going to be the natural log of sine of X, the natural log of sine of X over the natural log of X. Well this is all interesting but why do we care about this? Well instead of thinking about what is the limit? What is the limit as X approaches zero from the positive direction of Y? Let's think about what the natural log of Y is approaching as we approach, as X approaches zero from the positive direction. Let's figure out what the limit of this expression right over here is as X approaches zero from the positive direction. What is a natural log of Y? What is this whole thing? Not Y, what is the natural log of Y approaching? Let's think about that scenario. Let me write, do this in a new color. We want to figure out what is the limit as X approaches zero from the positive direction of this business and I'll just write it in one color. The natural log of sine of X over the natural log of X. I don't know, I wrote one time in print, one time in cursive. I'll just be consistent right over there. Now why is this interesting? Let's see in the numerator here, this thing's going to approach zero, natural log of zero you're going to approach This thing right over here natural log of," + }, + { + "Q": "At around 10:30 in the video, would it be valid to, instead of splitting up the single limit into two and taking the first derivative of one, just take the second derivative of x*cos(x) / sin(x) to get\ncos(x) + (-x*sin(x)) / cos (x) ? When evaluating it, you still get 1 + 0 / 1, but is that just a coincidence?\n", + "A": "If I followed what you meant correctly, yes, that is a valid way of doing it. You can continue using multiple iterations of l Hopital s provided the derivatives exist AND you continue to have 0/0 or \u00e2\u0088\u009e/\u00e2\u0088\u009e forms. If you have those forms, you do not necessarily have to split up the limit -- though doing so is often easier. Remember that l Hopital s is NOT valid or true if you don t have 0/0 or \u00e2\u0088\u009e/\u00e2\u0088\u009e . So, you must be careful about that.", + "video_name": "CDf_aE5yg3A", + "timestamps": [ + 630 + ], + "3min_transcript": "one over X. This is all going to be equal to, this is equal to the limit as X approaches zero from the positive direction of I could write this as cosine of X, cosine of X. Let's see, if I'm dividing by X. I'm dividing by X, I am going to get, this is going to be X over sine of X, X over sine of X. When I apply and when I try to take the limit here I'm going to get a zero, once again we got a zero over zero. This doesn't feel too satisfying but once again this is where our limit properties might be useful. As you can tell, this is not the most trivial of problems but this is going to be the same thing and this will take a little bit of pattern recognition. because we know that the limit of the product of two functions is equal to the product of their limits, this is the same thing as the limit, the limit as X approaches zero from the positive direction of, if we take this part, let me do this in a different color. If we take this part, that's not a different color. If we take this part right over here, so that's going to be X over sine of X and then times the limit. Let me put parenthesis here times the limit, the limit as X approaches zero from the positive direction of cosine of X, of cosine of X. Now this thing right over here is pretty straightforward. You can just evaluate it at zero, you get one. This thing right over here is equal to one This might ring a bell. You might have seen the limit as X approaches zero, of sine of X over X. This is just the reciprocal of that. This is X over sine of X but when you just superficially try to evaluate it. You get zero over zero so you can then apply L'Hopital's rule to this thing. Once again this is quite an interesting scenario we find ourselves in. This is going to be the same thing as the limit as X approaches zero from the positive direction. Derivative at the top is one, derivative at the bottom, is cosine of X. Well, this is just going to be one over cosine of zero is one, so this is just going to be equal to one. We got to apply L'Hopital's rule again and realize that this limit is going to be equal to one. One times one is one so this thing right over here is equal to one. This thing right over here, this thing right over here is going to be, this thing right over here is going to approach one" + }, + { + "Q": "well actually, at 4:40, there is evidence. both of the triangles that were produced by the parallelogram are the exact same size.\n", + "A": "While the triangles from the parallelogram were congruent, the sides in question were not corresponding sides. DC is congruent to AB Thus, DC/BC equals the cosine, not the sine of \u00e2\u0088\u00a0CBA Thus, none of the options was correct.", + "video_name": "TugWqiUjOU4", + "timestamps": [ + 280 + ], + "3min_transcript": "might be the same, but it doesn't tell us what this number right over here, doesn't tell us that this side is somehow congruent to DC. So we can't go with this one. Now let's think about the sine of CBA. So the sine-- let me do this in a different color. So the sine of angle CBA. So that's this angle right over here, CBA. Well, sine is opposite over hypotenuse. I guess let me make it clear which I'll do this in yellow. We're now looking at this triangle right over here. The opposite side is AC. That's what the angle opens up into. So it's going to be equal to AC. And what is the hypotenuse? What is the hypotenuse here? Well, the hypotenuse-- so let me see, It's the side opposite the 90 degree side. So this, it's BC. Sine is opposite over hypotenuse, so over BC. Is that what they wrote over here? No. They have DC over BC. Now what is DC equal to? Well, DC is this. And DC is not-- there's no evidence on this drawing right over here that DC is somehow equivalent to AC. So given this information right over here, we can't make this statement, either. So neither of these are true. So let's make sure we got this right. We can go back to our actual exercise, and we get-- oh, that's not the actual exercise. Let me minimize that. And we got it right." + }, + { + "Q": "\nAt 3:13, Sal says that tan(\u00e2\u0088\u00a0ABC) is unequal to AC/EF, right? But what if both the sides are in the same figure, like \u00e2\u0088\u00a0EFG? Does that make tan(\u00e2\u0088\u00a0ABC) = EG/EF?", + "A": "tan(\u00e2\u0088\u00a0ABC) is equal to EG/EF , because triangles ABC and EFG are similar. However, the solution that Sal ruled out said that tan(\u00e2\u0088\u00a0ABC) = AC/EF, which is untrue because there is no way to tell if EG is the same length as AC, because even though the triangles are similar, the size of the triangles can still differ dramatically.", + "video_name": "TugWqiUjOU4", + "timestamps": [ + 193 + ], + "3min_transcript": "we're dealing with this right triangle right over here. That's the only right triangle that angle ADC is part of. And so what side is opposite angle ADC? Well, it's side CA, or I guess I say AC, side AC. So that is opposite. And what side is adjacent? Well, this side, CD. CD, or I guess I could call it DC, whatever I want to call it. DC, or CD, is adjacent. Now how did I know that this side is adjacent and not side DA? Because DA is the hypotenuse. They both, together, make up the two sides of this angle. But the adjacent side is one of the sides of the angle that is not the hypotenuse. AD or DA in the sohcahtoa context we would consider to be the hypotenuse. For this angle, this is opposite, this is adjacent, this is hypotenuse. Tangent of this angle is opposite over adjacent-- AC Now is that what they wrote here? No. They wrote AC over EF. Well, where's EF? EF is nowhere to be seen either in this triangle, or even in this figure. EF is this thing right over here. EF is this business right over here. That's EF. It's in a completely different triangle in a completely different figure. We don't even know what scale this is drawn at. There's no way the tangent of this angle is related to this somewhat arbitrary number that's over here. They haven't labelled it. This thing might be a million miles long for all we know. This thing really could be any number. So this isn't the case. We would have to relate it to something within this triangle, or something that's the same length. So if somehow we could prove that EF is the same length as DC, then we could go with that. But there's no way. This is a completely different figure, a completely different diagram. This is a similar triangle to this, but we don't know anything about the lengths. A similar triangle just lets us know that the angles are all might be the same, but it doesn't tell us what this number right over here, doesn't tell us that this side is somehow congruent to DC. So we can't go with this one. Now let's think about the sine of CBA. So the sine-- let me do this in a different color. So the sine of angle CBA. So that's this angle right over here, CBA. Well, sine is opposite over hypotenuse. I guess let me make it clear which I'll do this in yellow. We're now looking at this triangle right over here. The opposite side is AC. That's what the angle opens up into. So it's going to be equal to AC. And what is the hypotenuse? What is the hypotenuse here? Well, the hypotenuse-- so let me see," + }, + { + "Q": "at 4:13 why did he put a plus sign and not the minus sign?\n", + "A": "Sal is doing: -2x^2 + 3x^2 -2 + 3 = +1, not -1. If you aren t sure, use the number line. Go left to -2 on the number line. Then, to do +3, move 3 units to the right. You end up on +1. That is why Sal has +x^2 . Hope this helps.", + "video_name": "FNnmseBlvaY", + "timestamps": [ + 253 + ], + "3min_transcript": "Do I have any other x squared? Yes, I do. I have this 3x squared right over there. So plus 3x squared. And then let's see, I have an x term right over here, and that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term-- I'll circle that in orange-- so plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have two of something and I subtract 3 of that, what am I left with? Well, I'm left with negative 1 of that something. So I could write negative 1y, or I could just write negative y. And another way you could think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. It's 2. Where obviously both are dealing-- they're both y terms, not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1, or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms. If I have 4 of this, 4 xy's and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say add the coefficients, 4 plus negative 4, gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. And so I'm left with no xy's. And then I have right over here-- I could have written 0xy, but that seems unnecessary-- then right over here I have my x squared terms. Negative 2 plus 3 is 1. Or another way of saying it, if I have 3x squared squared, so I'm left with the 1x squared. So this right over here simplifies to 1x squared. Or I could literally just write x squared. 1x squared is the same thing as x squared. So plus x squared, and then these there's nothing really left to simplify. So plus 2x plus y squared. And obviously you might have gotten an answer in some other order, but the order in which I write these terms don't matter. It just matters that you were able to simplify it to these four terms." + }, + { + "Q": "At 2:56 why does he put -y instead of -1y?\n", + "A": "-y and -1y are the same thing because -1 times y is -y", + "video_name": "FNnmseBlvaY", + "timestamps": [ + 176 + ], + "3min_transcript": "A y is different than a y squared, is different than an xy. Now with that said, let's see if there is anything that we can simplify. So first, let's think about the y terms. So you have a negative 3y there. Do we have any more y term? We have this 2y right over there. So I'll just write it out-- I'll just reorder it. So we have negative 3y plus 2y. Now, let's think about-- and I'm just going in an arbitrary order, but since our next term is an xy term-- let's think about all of the xy terms. So we have plus 4xy right over here. So let me just write it down-- I'm just reordering the whole expression-- plus 4xy. And then I have minus 4xy right over here. Then let's go to the x squared terms. I have negative 2 times x squared, or minus 2x squared. So let's look at this. Do I have any other x squared? Yes, I do. I have this 3x squared right over there. So plus 3x squared. And then let's see, I have an x term right over here, and that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term-- I'll circle that in orange-- so plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have two of something and I subtract 3 of that, what am I left with? Well, I'm left with negative 1 of that something. So I could write negative 1y, or I could just write negative y. And another way you could think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. It's 2. Where obviously both are dealing-- they're both y terms, not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1, or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms. If I have 4 of this, 4 xy's and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say add the coefficients, 4 plus negative 4, gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. And so I'm left with no xy's. And then I have right over here-- I could have written 0xy, but that seems unnecessary-- then right over here I have my x squared terms. Negative 2 plus 3 is 1. Or another way of saying it, if I have 3x squared" + }, + { + "Q": "At around 4:30, it says that by using *(1+1/n)^n* you get e. I understand that, and why, but why do you need the 1 in the beginning? Could you get e just by doing *(1/n)^n* ?\n", + "A": "No, you could not get e without the 1 in the beginning. Since, as n goes to infinity, 1/n approaches 0 and n grows to infinity, (1/n)^n would approach 0^infinity which is definitely 0. Have a blessed, wonderful day!", + "video_name": "oQhp3ndj28Y", + "timestamps": [ + 270 + ], + "3min_transcript": "Now, let's go even higher. Let's take it ... let's do 1+1/ and actually I can now use scientific notation. Let's just say (1+1/1'10^7)^1'10^7, so what do we get here? So, now we went 2.718281692. Let's go even larger. Let's get our last entry here. Let's go, instead of the 7th power, let's go to the eighth power, so now we're (1+1/100,000,000^100,000,000) I don't even know if this calculator can handle this and we get 2.71828181487 and you see that we are quickly approaching, to a very large power, to the number e. The number e in our calculator. You see we've already gotten 1, 2, 3, 4, 5, 6, 7 digits to the right of the decimal point by taking it to the 100 millionth power. So, we are approaching this number. We are approaching, so one way to talk about it is we could say the limit, as n approaches infinity. As n becomes larger and larger, it's not becoming unbounded. It's not going to infinity. It seems to be approaching this number and we will call this number, we will call this magical and mystical number e. We'll call this number e and we see from our calculator that this number and these are kind of, these are almost as famous digits as the digits for Pi, we are getting 2.7182818 and it just Never, never repeating, so it's an infinite string of digits, never, never repeating. Just like Pi. Pi, you remember, is the ratio of the circumference to the diameter of the circle. e is another one of these crazy numbers that shows up in the universe. And in other videos on Khan Academy we go into depth, why this is so magical and mystical. Already this is kind of cool. That I can take an infinite ... If I just add 1 over a number to 1 and take it to that number and I make that number larger and larger and larger, it's approaching this number, but what's even crazier about it is we'll see that this number, which you can view, one way of it, it is coming out of this compound interest. That number, Pi, the imaginary unit which is defined as that imaginary unit squared is a negative 1, that they all fit together in this magical and mystical way and we'll see that again in future videos. But just for the sake of e, what you could imagine what's happening here" + }, + { + "Q": "At 11:40, I'm a little confused as to how Sal simplified the equation on the final step. I get that 1/9 becomes 4/9; but what happened to the exponents that were attached to the integers? It doesn't make sense with the exponent rules that are taught here. What am I missing?\n", + "A": "Because 4 and 1/9 both have the same exponent (1st power) they can be multiplied together.", + "video_name": "64bH_27Ehoc", + "timestamps": [ + 700 + ], + "3min_transcript": "And then let me rewrite everything. Because I don't want to do too much on this one step right over here. So this part right over here gives us square root of 3s squared over 16. And then that's going to be times-- I'll open and close the parentheses. So then we have 4 plus. Then in blue, I'll write 3 times 4 to the first power. And then I can rewrite this as times 1/3. We could view this is 1/3 squared. We could view this as 1 over 3 to the first power, squared, or we could view this as 1 over 3 squared to the first power. And I'm going to write it that way. So times 1/9 to the first power. And then plus 3 times 4 squared. And then this we can write as, times 1/9 to the second power. And then this one we could write, plus 3 times 4 to the third times-- But we could also write this based on what we saw over here. We could write this as 1 over 3 squared to the third power, based on this right over here. Let me make this clear. 1 over 3 to the third, to the second power. This is the same thing as 1 over 3 squared to the third power. That's what we showed right over here. So this is the equivalent to 1/9 to the third power. And now we start to see the pattern is starting to clean up a little bit. And let me just do one more step. And then we'll finish this in the next video. So this is equal to square root of 3s squared over 16 times 4 plus 3 times, this is 4/9, plus the next term is three times 4/9 squared. And then we have plus 3 times 4 over 9 to the third power. and on, and on, and on taking the 3 times 4/9 to the successively larger, and larger powers. So this is what we have to find the sum of to find our area. And we're going to do that in the next video. And we're going to use some of the tools we've used to find the sums of infinite geometric series. But we're going to re-do it in this video just so that you don't have to remember that formula or that proof." + }, + { + "Q": "\nAt 2:17 in the video, it is shown how the vector is pointed down in the eastwards direction; then if we are given simple numbers in the form of a column vector, how would we know if the arrow should point westwards or otherwise?", + "A": "If the first number in the column is positive, then it will point partly to the right. If the first number in the column is negative, then it will point partly to the left. If the second number in the column is positive, then it will point partly upward. If the second number in the column is negative, then it will point partly downward.", + "video_name": "8QihetGj3pg", + "timestamps": [ + 137 + ], + "3min_transcript": "So I have two 2-dimensional vectors right over here, vector a and vector b. And what I want to think about is how can we define or what would be a reasonable way to define the sum of vector a plus vector b? Well, one thing that might jump at your mind is, look, well, each of these are two dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just a sum of the first two components of these two vectors. So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two 2-dimensional vectors. You add them together, you get another two 2-dimensional vectors. If you think about it in terms of real coordinates bases, both of these are members of R2-- I'll write this down here just so we get used to the notation. which is just another way of saying that these are both two tuples. They are both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it, but how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector a, we could visualize, this tells us how far this vector moves in each of these directions-- horizontal direction and vertical direction. So if we put the, I guess you could say the tail of the vector at the origin-- remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction. 1, 2, 3, 4, 5, 6. And then negative 2 in the vertical. So negative 2. So vector a could look like this. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector. And the direction is the direction that it is pointed in. And also just to emphasize, I could have drawn vector a like that or I could have put it over here. These are all equivalent vectors. These are all equal to vector a. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector b. Vector b in the horizontal direction goes negative 4-- 1, 2, 3, 4, and in the vertical direction goes 4-- 1, 2, 3, 4. So its tail if we start at the origin, if its tail is at the origin, its head would be at negative 4, 4. So let me draw that just like that." + }, + { + "Q": "\nAt about 1:30 Sal writes a bunch of stuff I don't understand. What is the meaning?I was also wondering why are there braces around the vectors x and y.", + "A": "These are 2 vectors in column (vertical) form, and their sum.", + "video_name": "8QihetGj3pg", + "timestamps": [ + 90 + ], + "3min_transcript": "So I have two 2-dimensional vectors right over here, vector a and vector b. And what I want to think about is how can we define or what would be a reasonable way to define the sum of vector a plus vector b? Well, one thing that might jump at your mind is, look, well, each of these are two dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just a sum of the first two components of these two vectors. So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two 2-dimensional vectors. You add them together, you get another two 2-dimensional vectors. If you think about it in terms of real coordinates bases, both of these are members of R2-- I'll write this down here just so we get used to the notation. which is just another way of saying that these are both two tuples. They are both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it, but how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector a, we could visualize, this tells us how far this vector moves in each of these directions-- horizontal direction and vertical direction. So if we put the, I guess you could say the tail of the vector at the origin-- remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction. 1, 2, 3, 4, 5, 6. And then negative 2 in the vertical. So negative 2. So vector a could look like this. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector. And the direction is the direction that it is pointed in. And also just to emphasize, I could have drawn vector a like that or I could have put it over here. These are all equivalent vectors. These are all equal to vector a. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector b. Vector b in the horizontal direction goes negative 4-- 1, 2, 3, 4, and in the vertical direction goes 4-- 1, 2, 3, 4. So its tail if we start at the origin, if its tail is at the origin, its head would be at negative 4, 4. So let me draw that just like that." + }, + { + "Q": "When Sal is drawing vectors through the video, something came to my mind.\nHow do we know exactly what direction the vectors are going in? Most vectors seem to be drawn going outward from the origin, but couldn't they go inward? Like at 2:09, when he's drawing the purple vector. Couldn't it also be pointing in the opposite direction, but still have the same magnitude?\n\nSorry if the answer is extremely obvious and I just don't see it...\n", + "A": "The numbers in the video represent the vectors he draws. So when he draws a vector from the origin, it will point to the point (a,b). At 2:09, he s drawing the vector [6,-2], so it points from the origin to the right and down because it s positive (right) 6 and negative (down) 2.", + "video_name": "8QihetGj3pg", + "timestamps": [ + 129 + ], + "3min_transcript": "So I have two 2-dimensional vectors right over here, vector a and vector b. And what I want to think about is how can we define or what would be a reasonable way to define the sum of vector a plus vector b? Well, one thing that might jump at your mind is, look, well, each of these are two dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just a sum of the first two components of these two vectors. So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two 2-dimensional vectors. You add them together, you get another two 2-dimensional vectors. If you think about it in terms of real coordinates bases, both of these are members of R2-- I'll write this down here just so we get used to the notation. which is just another way of saying that these are both two tuples. They are both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it, but how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector a, we could visualize, this tells us how far this vector moves in each of these directions-- horizontal direction and vertical direction. So if we put the, I guess you could say the tail of the vector at the origin-- remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction. 1, 2, 3, 4, 5, 6. And then negative 2 in the vertical. So negative 2. So vector a could look like this. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector. And the direction is the direction that it is pointed in. And also just to emphasize, I could have drawn vector a like that or I could have put it over here. These are all equivalent vectors. These are all equal to vector a. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector b. Vector b in the horizontal direction goes negative 4-- 1, 2, 3, 4, and in the vertical direction goes 4-- 1, 2, 3, 4. So its tail if we start at the origin, if its tail is at the origin, its head would be at negative 4, 4. So let me draw that just like that." + }, + { + "Q": "So at 3:36 and a little before that, when Sal says that f(-2) is EQUAL TO g(-1) is that in relation to the translation? Like point \"a\" on g(x) has different coordinates than f(x) but they are still equal because they're both have point \"a\" just one is shifted so that's why it appears different?\n", + "A": "Yes, you are exactly right. This is a big concept in function translation; basically you are trying to find where one function is equal to another, and in this case he means that the function f at -2 is equivalent to function g at -1. Even more specifically, the y-value of x = -2 in f(x) is the same y-value for x= -1 in g(x) (and you can see him drawing horizontal lines across the graphs to illustrate this concept).", + "video_name": "ENFNyNPYfZU", + "timestamps": [ + 216 + ], + "3min_transcript": "I'll label it. f of x. And here is g of x. So let's think about it a little bit. Let's pick an arbitrary point here. Let's say we have in red here, this point right over there is the value of f of negative 3. This is negative 3. This is the point negative 3, f of 3. Now g hits that same value when x is equal to negative 1. So let's think about this. g of negative 1 is equal to f of negative 3. And we could do that with a bunch of points. We could see that g of 0, which is right there-- let me do it equivalent to f of negative 2. So let me write that down. g of 0 is equal to f of negative 2. We could keep doing that. We could say g of 1, which is right over here. This is 1. g of 1 is equal to f of negative 1. g of 1 is equal to f of negative 1. So I think you see the pattern here. g of whatever is equal to the function evaluated at 2 less than whatever is here. So we could say that g of x is equal to f of-- well So f of x minus 2. So this is the relationship. g of x is equal to f of x minus 2. And it's important to realize here. When I get f of x minus 2 here-- and remember the function is being evaluated, this is the input. x minus 2 is the input. When I subtract the 2, this is shifting the function to the right, which is a little bit counter-intuitive unless you go through this exercise right over here. So g of x is equal to f of x minus 2. If it was f of x plus 2 we would have actually shifted f to the left. Now let's think about this one. This one seems kind of wacky. So first of all, g of x, it almost looks like a mirror image but it looks like it's been flattened out. So let's think of it this way. Let's take the mirror image of what g of x is. So I'm going to try my best to take the mirror image of it. So let's see... It gets to about 2 there, then it gets pretty close to 1 right over there." + }, + { + "Q": "\nFrom the exponents of i it looks as if i is a negative number,but a negative number times a negative number equals a positive number.But then no number's square is a negative number.But now Sal says -i at 2:56 and that doesn't support that i is negative. Unless he's actually saying (mathematically )(he doesn't know)that -i is positive i.", + "A": "i is not a negative number, but i^2 is a negative number.....i^2=-1 and i=sqrt. of -1, So -i = -\u00e2\u0088\u009a-1 and hence -i^2=+1", + "video_name": "ysVcAYo7UPI", + "timestamps": [ + 176 + ], + "3min_transcript": "some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of \"i\" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of \"i\" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with \"i\" to the zeroth power. And so you might say, anything to the zeroth power is one, so \"i\" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including \"i\" is one. well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power, Well once again this is going to be \"i\" times \"i\" to the third power. So that's \"i\" times \"i\" to the third power. \"i\" times \"i\" to the third power Well what was \"i\" to the third power? \"i\" to the third power was negative \"i\" This over here is negative \"i\". And so \"i\" times \"i\" would get negative one, but you have a negative out here, so its \"i\" times \"i\" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is \"i\" times negative \"i\", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times \"i\" times \"i\". \"i\" times \"i\", by definition, is negative one. Negative one times negative one is equal to positive one. So \"i\" to the fourth is the same thing as \"i\" to the zeroth power." + }, + { + "Q": "\nThis is simple question: @3:40 how does i x i^3, or i x -i, = (-1)(i)(i) or (-1)(-1)?", + "A": "Do you mean why does i \u00e2\u00a8\u0089 i\u00c2\u00b3 = (-1)(i)(i) = 1? We will use the exponent addition formula for this: i \u00e2\u00a8\u0089 i\u00c2\u00b3 = i\u00e2\u0081\u00b4 = 1. Or we can use this way: i \u00e2\u00a8\u0089 i\u00c2\u00b3 = (-1)(i)(i) = (-1)(-1) = 1 Always remember powers of i: i\u00c2\u00b9 = i, i\u00c2\u00b2 = -1, i\u00c2\u00b3 = -i, i\u00e2\u0081\u00b4 = 1, i\u00e2\u0081\u00b5 = i Hope this helps! \u00e2\u0080\u0094CT-2/002-24", + "video_name": "ysVcAYo7UPI", + "timestamps": [ + 220 + ], + "3min_transcript": "well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power, Well once again this is going to be \"i\" times \"i\" to the third power. So that's \"i\" times \"i\" to the third power. \"i\" times \"i\" to the third power Well what was \"i\" to the third power? \"i\" to the third power was negative \"i\" This over here is negative \"i\". And so \"i\" times \"i\" would get negative one, but you have a negative out here, so its \"i\" times \"i\" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is \"i\" times negative \"i\", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times \"i\" times \"i\". \"i\" times \"i\", by definition, is negative one. Negative one times negative one is equal to positive one. So \"i\" to the fourth is the same thing as \"i\" to the zeroth power. \"i\" to the fifth power. Well that's just going to be \"i\" to to the fourth times \"i\". And we know what \"i\" to the fourth is. It is one. So its one times \"i\", or it is one times \"i\", or it is just \"i\" again. So once again it is exactly the same thing as \"i\" to the first power. Lets try again just to see the pattern keep going. Lets try \"i\" to the seventh power. Sorry, \"i\" to the sixth power. Well that's \"i\" times \"i\" to the fifth power, that's \"i\" times \"i\" to the fifth, \"i\" to the fifth we already established as just \"i\", so its \"i\" times \"i\", it is equal to, by definition,\"i\" times \"i\" is negative one. And then lets finish off, well we could keep going on this way We can keep putting high and higher powers of \"i\" here. An we'll see that it keeps cycling back. In the next video I'll teach you how taking an arbitrarily high power of \"i\", how you can figure out what that's going to be. But lets just verify that this cycle keeps going. \"i\" to the seventh power is equal to \"i\" times \"i\" to the sixth power." + }, + { + "Q": "At 0:22 Sal says E. What is that number and can i have a link to it?\n", + "A": "e is a constant that comes up in math and science all the time. It is an irrational and transcendental number the first few digits of which are 2.71828... e is officially defined as: lim h\u00e2\u0086\u00920 (1+h)^(1/h) This same definition can also be expressed as: lim h\u00e2\u0086\u0092 \u00e2\u0088\u009e (1+1/h)^(h)", + "video_name": "ysVcAYo7UPI", + "timestamps": [ + 22 + ], + "3min_transcript": "In this video, I want to introduce you to the number i, which is sometimes called the imaginary, imaginary unit What you're gonna see here, and it might be a little bit difficult, to fully appreciate, is that its a more bizzare number than some of the other wacky numbers we learn in mathematics, like pi, or e. And its more bizzare because it doesnt have a tangible value in the sense that we normally, or are used to defining numbers. \"i\" is defined as the number whose square is equal to negative 1. This is the definition of \"i\", and it leads to all sorts of interesting things. Now some places you will see \"i\" defined this way; \"i\" as being equal to the principle square root of negative one. I want to just point out to you that this is not wrong, it might make sense to you, you know something squared is negative one, then maybe its the principle square root of negative one. And so these seem to be almost the same statement, some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of \"i\" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of \"i\" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with \"i\" to the zeroth power. And so you might say, anything to the zeroth power is one, so \"i\" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including \"i\" is one. well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power," + }, + { + "Q": "\non 2:56 why did Sal keep it as -1*i instead of make it -i?{OK i know he corrected it at 3:27}", + "A": "he did that to clarify the the minus is a separate number from i.", + "video_name": "ysVcAYo7UPI", + "timestamps": [ + 176, + 207 + ], + "3min_transcript": "some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of \"i\" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of \"i\" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with \"i\" to the zeroth power. And so you might say, anything to the zeroth power is one, so \"i\" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including \"i\" is one. well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power, Well once again this is going to be \"i\" times \"i\" to the third power. So that's \"i\" times \"i\" to the third power. \"i\" times \"i\" to the third power Well what was \"i\" to the third power? \"i\" to the third power was negative \"i\" This over here is negative \"i\". And so \"i\" times \"i\" would get negative one, but you have a negative out here, so its \"i\" times \"i\" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is \"i\" times negative \"i\", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times \"i\" times \"i\". \"i\" times \"i\", by definition, is negative one. Negative one times negative one is equal to positive one. So \"i\" to the fourth is the same thing as \"i\" to the zeroth power." + }, + { + "Q": "at 5:20 Sal said tat i=square root of -1, then how is i=1!\n", + "A": "i = -1 it doesn t equal 1. I don t know if that answered your question...", + "video_name": "ysVcAYo7UPI", + "timestamps": [ + 320 + ], + "3min_transcript": "Well once again this is going to be \"i\" times \"i\" to the third power. So that's \"i\" times \"i\" to the third power. \"i\" times \"i\" to the third power Well what was \"i\" to the third power? \"i\" to the third power was negative \"i\" This over here is negative \"i\". And so \"i\" times \"i\" would get negative one, but you have a negative out here, so its \"i\" times \"i\" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is \"i\" times negative \"i\", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times \"i\" times \"i\". \"i\" times \"i\", by definition, is negative one. Negative one times negative one is equal to positive one. So \"i\" to the fourth is the same thing as \"i\" to the zeroth power. \"i\" to the fifth power. Well that's just going to be \"i\" to to the fourth times \"i\". And we know what \"i\" to the fourth is. It is one. So its one times \"i\", or it is one times \"i\", or it is just \"i\" again. So once again it is exactly the same thing as \"i\" to the first power. Lets try again just to see the pattern keep going. Lets try \"i\" to the seventh power. Sorry, \"i\" to the sixth power. Well that's \"i\" times \"i\" to the fifth power, that's \"i\" times \"i\" to the fifth, \"i\" to the fifth we already established as just \"i\", so its \"i\" times \"i\", it is equal to, by definition,\"i\" times \"i\" is negative one. And then lets finish off, well we could keep going on this way We can keep putting high and higher powers of \"i\" here. An we'll see that it keeps cycling back. In the next video I'll teach you how taking an arbitrarily high power of \"i\", how you can figure out what that's going to be. But lets just verify that this cycle keeps going. \"i\" to the seventh power is equal to \"i\" times \"i\" to the sixth power. And if you take \"i\" to the eighth, once again it'll be one, \"i\" to the ninth will be \"i\" again, and so on and so forth." + }, + { + "Q": "\nIs there any way to check the problem? Like at 2:32, how is he sure that the answer is 1436.8?", + "A": "He is completely sure since that is the answer the calculator gave him. You can verify by doing the problem again, or if you have a graphing calculator, you can start from the volume and use algebraic manipulation to get back to the radius.", + "video_name": "IelS2vg7JO8", + "timestamps": [ + 152 + ], + "3min_transcript": "Find the volume of a sphere with a diameter of 14 centimeters. So if I have a sphere-- so this isn't just a circle, this is a sphere. You could view it as a globe of some kind. So I'm going to shade it a little bit so you can tell that it's three-dimensional. They're giving us the diameter. So if we go from one side of the sphere straight through the center of it. So we're imagining that we can see through the sphere. And we go straight through the centimeter, that distance right over there is 14 centimeters. Now, to find the volume of a sphere-- and we've proved this, or you will see a proof for this later when you learn calculus. But the formula for the volume of a sphere is volume is equal to 4/3 pi r cubed, where r is the radius of the sphere. So they've given us the diameter. And just like for circles, the radius of the sphere is half of the diameter. So in this example, our radius is going to be 7 centimeters. And in fact, the sphere itself is the set the radius away from the center. But with that out of the way, let's just apply this radius being 7 centimeters to this formula right over here. So we're going to have a volume is equal to 4/3 pi times 7 centimeters to the third power. So I'll do that in that pink color. So times 7 centimeters to the third power. And since it already involves pi, and you could approximate pi with 3.14. Some people even approximate it with 22/7. But we'll actually just get the calculator out to get the exact value for this volume. So this is going to be-- so my volume is going to be 4 divided by 3. And then I don't want to just put a pi there, because that might interpret it as 4 divided by 3 pi. So 4 divided by 3 times pi, times 7 to the third power. before it does the multiplication, so this should work out. And the units are going to be in centimeters cubed or cubic centimeters. So we get 1,436. They don't tell us what to round it to. So I'll just round it to the nearest 10th-- 1,436.8. So this is equal to 1,436.8 centimeters cubed. And we're done." + }, + { + "Q": "1:21 can you explain why the sphere is radius cubed instead of raidus squared\n", + "A": "I think that it is because a sphere is 3 dimensional and a circle is 2 dimensional.", + "video_name": "IelS2vg7JO8", + "timestamps": [ + 81 + ], + "3min_transcript": "Find the volume of a sphere with a diameter of 14 centimeters. So if I have a sphere-- so this isn't just a circle, this is a sphere. You could view it as a globe of some kind. So I'm going to shade it a little bit so you can tell that it's three-dimensional. They're giving us the diameter. So if we go from one side of the sphere straight through the center of it. So we're imagining that we can see through the sphere. And we go straight through the centimeter, that distance right over there is 14 centimeters. Now, to find the volume of a sphere-- and we've proved this, or you will see a proof for this later when you learn calculus. But the formula for the volume of a sphere is volume is equal to 4/3 pi r cubed, where r is the radius of the sphere. So they've given us the diameter. And just like for circles, the radius of the sphere is half of the diameter. So in this example, our radius is going to be 7 centimeters. And in fact, the sphere itself is the set the radius away from the center. But with that out of the way, let's just apply this radius being 7 centimeters to this formula right over here. So we're going to have a volume is equal to 4/3 pi times 7 centimeters to the third power. So I'll do that in that pink color. So times 7 centimeters to the third power. And since it already involves pi, and you could approximate pi with 3.14. Some people even approximate it with 22/7. But we'll actually just get the calculator out to get the exact value for this volume. So this is going to be-- so my volume is going to be 4 divided by 3. And then I don't want to just put a pi there, because that might interpret it as 4 divided by 3 pi. So 4 divided by 3 times pi, times 7 to the third power. before it does the multiplication, so this should work out. And the units are going to be in centimeters cubed or cubic centimeters. So we get 1,436. They don't tell us what to round it to. So I'll just round it to the nearest 10th-- 1,436.8. So this is equal to 1,436.8 centimeters cubed. And we're done." + }, + { + "Q": "At 3:13, why Sal said \"clog\"?\n", + "A": "It was an accident. He means c Log but pronounced it incorrectly.", + "video_name": "yEAxG_D1HDw", + "timestamps": [ + 193 + ], + "3min_transcript": "Fair enough. I think you realize I have not done anything profound just yet. But let's go back. We said that this is the same thing as this. So let's experiment with something. Let's raise this side to the power of C. So I'm going to raise this side to the power of C. That's a kind of caret. And when you type exponents that's what you would use, a caret. So I'm going to raise it to the power of C. So then, this side is x to the B to the C power, is equal to A to the C. All I did is I raised both sides of this equation to the Cth power. And what do we know about when you raise something to an exponent and you raise that whole thing to another exponent, what happens to the exponents? Well, that's just exponent rule and you just multiply those two exponents. What can we do now? Well, I don't know. Let's take the logarithm of both sides. Or let's just write this-- let's not take the Let's write this as a logarithm expression. We know that x to the BC is equal to A to the C. Well, that's the exact same thing as saying that the logarithm base x of A to the C is equal to BC. Correct? Because all I did is I rewrote this as a logarithm expression. And I think now you realized that something interesting has happened. That BC, well, of course, it's the same thing as this BC. So this expression must be equal to this expression. That if I have some kind of a coefficient in front of the logarithm where I'm multiplying the logarithm, so if I have C log-- Clog base x of A, but that's C times the logarithm base x of A. That equals the log base x of A to the C. So you could take this coefficient and instead make it an exponent on the term inside the logarithm. That is another logarithm property. So let's review what we know so far about logarithms. We know that if I write-- let me say-- well, let me just with the letters I've been using. C times logarithm base x of A is equal to logarithm" + }, + { + "Q": "\nStarting from 6:28, I don't understand what Mr. Khan means when he says that the \"Distance between x and c is less than delta,\" and that the \"distance between f(x) and L is going to be less than epsilon.\" Can someone explain please? Thank you!", + "A": "You feed numbers called x into a function to get numbers called f(x) out. What he is saying is that as the number going in, i.e. x, gets closer and closer to c .. then the output f(x) is getting closer and closer to L. Now delta is just how far away from c a particular input value x is, and epsilon is how far away from L a particular output value f(x) is. The big idea about limits then, is that even if f(x) is not defined when x = c, you can see where it is going as x gets (arbitrarily) close to c.", + "video_name": "w70af5Ou70M", + "timestamps": [ + 388 + ], + "3min_transcript": "defined at C, so we think of ones that maybe aren't C, but are getting very close. If you find any x in that range, f of those x's are going to be as close as you want to your limit. They're going to be within the range L plus epsilon or L minus epsilon. So what's another way of saying this? Another way of saying this is you give me an epsilon, then I will find you a delta. So let me write this in a little bit more math notation. So I'll write the same exact statements with a little bit more math here. But it's the exact same thing. Let me write it this way. Given an epsilon greater than 0-- so that's kind of the first part of the game-- we can find a delta greater than 0, such So what's another way of saying that x is within delta of C? Well, one way you could say, well, what's the distance between x and C is going to be less than delta. This statement is true for any x that's within delta of C. The difference between the two is going to be less than delta. So that if you pick an x that is in this range between C minus delta and C plus delta, and these are the x's that satisfy that right over here, then-- and I'll do this in a new color-- then the distance between your f of x and your limit-- and this is just the distance between the f of x and the limit, it's going to be less than epsilon. So all this is saying is, if the limit truly does exist, it could be super, super small one, we can find a delta. So we can define a range around C so that if we take any x value that is within delta of C, that's all this statement is saying that the distance between x and C is less than delta. So it's within delta of C. So that's these points right over here. That f of those x's, the function evaluated at those x's is going to be within the range that you are specifying. It's going to be within epsilon of our limit. The f of x, the difference between f of x, and your limit will be less than epsilon. Your f of x is going to sit some place over there. So that's all the epsilon-delta definition is telling us. In the next video, we will prove that a limit exists by using this definition of limits." + }, + { + "Q": "\n7:25 side side side? who came up with that name", + "A": "The name refers to the method of finding congruence: it is quite logical. The SSS (or side, side, side) Congruence Postulate states that if all 3 sides of a triangle are congruent to that of another triangle, the two triangles are congruent.", + "video_name": "CJrVOf_3dN0", + "timestamps": [ + 445 + ], + "3min_transcript": "have the same measure, they're congruent We also know that these two corresponding angles I'll use a double arch to specify that this has the same measure as that So, we also know the measure of angle ABC is equal to the measure of angle XYZ And then finally we know that this angle, if we know that these two characters are congruent, then this angle is gonna have the same measure as this angle as a corresponding angle So, we know that the measure of angle ACB is gonna be equal to the measure of angle XZY Now what we're gonna concern ourselves a lot with is how do we prove congruence? 'Cause it's cool, 'cause if you can prove congruence of 2 triangles then all of the sudden you can make all of these assumptions we're gonna assume it for the sake of introductory geometry course This is an axiom or a postulate or just something you assume So, an axiom, very fancy word Postulate, also a very fancy word It really just means things we are gonna assume are true An axiom is sometimes, there's a little bit of distinction sometimes where someone would say \"an axiom is something that is self-evident\" or it seems like a universal truth that is definitely true and we just take it for granted You can't prove an axiom A postulate kinda has that same role but sometimes let's just assume this is true and see if we assume that it's true what can we derive from it, what we can prove if we assume its true But for the sake of introductory geometry class and really most in mathematics today, these two words are use interchangeably An axiom or a postulate, just very fancy words that things we take as a given Things that we'll just assume, we won't prove them, and then we're just gonna build up from there And one of the core ones that we'll see in geometry is the axiom or the postulate That if all of the sides are congruent, if the length of all the sides of the triangle are congruent, then we are dealing with congruent triangles So, sometimes called side, side, side postulate or axiom We're not gonna prove it here, we're just gonna take it as a given So this literally stands for side, side, side And what it tells is, if we have two triangles and So I say that's another triangle right over there And we know that corresponding sides are equal So, we know that this side right over here is equal into, like, that side right over there" + }, + { + "Q": "\nAt the beginning of the video this means that the definition of something being congruent would simply be that it has the same size and shape?\nAnd if if at 3:19, this means that if one side is congruent to another side of another triangle, then all 3 sides of both triangles are both congruent to each other?", + "A": "Congruent means to be the same size and shape. However, if one side of a triangle is congruent to another side of a triangle, it would not mean all 3 sides are congruent. You would need another side or angle to be congruent to the other triangle to be sure of that.", + "video_name": "CJrVOf_3dN0", + "timestamps": [ + 199 + ], + "3min_transcript": "And if you say that a triangle is congruent, let me label this So, let's call this triangle ABC Now let's call this D, let me call it XYZ XY and Z So, if we were to say, if we make the claim that both of these triangles are congruent So, if we say triangle ABC is congruent And the way you specify it, it almost look like an equal sign But it's equal sign with a curly thing on top Let me write it a little bit either So, we would write it like this If we know that triangle ABC is congruent to triangle XYZ That means their corresponding sides have the same length And their corresponding angles have the same measure So, if we make this assumption or someone tells us that this is true then we know, for example, that AB is going to equal to XY And we could do this like this, and I'm assuming this are the corresponding sides And you can see that actually we've defined these triangles A corresponds to X, B corresponds to Y and C corresponds to Z right over there So, side AB is gonna have the same length as XY Then you can sometimes if you don't have the colors you can denote it just like that These two length are- or this two lines segments have the same length And you can actually say this, you don't always see this written this way You could also make the statement that line segment AB is congruent to line segment XY But congruence of line segments really just means that their lengths are equivalent So, these two things mean the same thing that just means the measure of one line segment is equal to the measure of the other line segment And so we can go thru all the corresponding sides If these two characters are congruent, we also know that BC, we also know that the length BC is gonna be the length of YZ Assuming those are the corresponding sides And we can put these double hash marks right over here to show that these lengths are the same And when we go the third side, we also know that these are going to be has same length or the line segments are going to be congruent So, we also know that the length of AC is going to be equal to the length of XZ Not only do we know that all of the sides, the corresponding sides are gonna have the same length If someone tells that a triangle is congruent We also know that all the corresponding angles are going to have the same measure So, for example: we also know that this angle's measure" + }, + { + "Q": "\nAt 1:54, why does Sal put 3-1 over 4 instead of just doing the answer of 2 over four?", + "A": "because he s showing you the steps", + "video_name": "0njioQqIxKY", + "timestamps": [ + 114 + ], + "3min_transcript": "Pedro is supposed to practice piano for 3/4 of an hour every day. Today, he has practiced for 1/4 of an hour. What fraction of an hour does he need to practice? So let's visualize 3/4. So he needs to practice 3/4 of an hour. So if this represents an entire hour, so let's divide it into fourths. So now, divide it into halves. Now, let's make it into four equal sections. Now, it's into fourths. So he needs to practice 3/4 of an hour. That was my attempt at drawing an arrow. So let me shade it in. So he needs to practice 3/4 of an hour. This gets us to 3/4 right over here. So he needs to practice 3 out of the fourths of an hour. So this is 3/4. Now, it says he has already practiced for 1/4 of an hour. So how much more does he need to practice? Well, he needs to practice this much more. And you might already see the answer visually. But let's think about how to represent it as a fraction expression. So let me write it like this. So this is how much he needs to go. How much does he need to practice? So he needs to do a total of 3/4. He's already done 1/4. So if you subtract 1/4 from the 3/4, you're going to get this amount right over here. You're going to get the amount that he needs to practice. Now, this already has the same denominator. So this is going to be equal to 3 minus 1 over 4, which is equal to 2/4. And you see that right over there. He still needs to practice 1 and 2 out of the fourths, 2/4. There's a couple of ways to think about it. You could say, hey, look, this is half of the length of this entire thing. It has a little bit on this end and that end. Or if you make 4 equal blocks, and if you were to shade in 2 of them, you see that you have shaded in 1/2 of the blocks. This is the exact same amount as if I just divide it into 2 sections and I shaded in only 1 of them. 2/4 is the same thing as 1/2. And if you wanted to work it out mathematically, you just have to do the same thing to the numerator and the denominator. So let's divide the numerator and the denominator by 2, because they are both divisible by 2. That's actually their greatest common factor. So 2 divided by 2 is 1. 4 divided by 2 is 2. So what fraction of an hour does he need to practice? He's got to practice 1/2 an hour." + }, + { + "Q": "at 6:43 coordinates of vertices are 1,23/4. bt we knw that vertex is the point where the conic section and the axis itersect. how is it possible then?\n", + "A": "It is not necessary for the focus to always intersect the axes. The focus can be any point in 2D space. It can also obviously be at the origin.", + "video_name": "w56Vuf9tHfA", + "timestamps": [ + 403 + ], + "3min_transcript": "We can label 'em. One, two, three, four five, six and seven and so our vertex is right over here. One comma 23 over four, so that's five and three-fourths. So it's gonna be right around right around there and as we said, since we have a negative value in front of this x minus one squared term, I guess we could call it, this is going to be a downward opening parabola. This is going to be a maximum point. So our actual parabola is going to look is going to look something it's gonna look something like this. It's gonna look something like this and we could, obviously, I'm hand drawing it, so it's not going to be exactly perfect, but hopefully you get the general idea of what the parabola is going look like and actually, let me just do part of it, about the parabola just yet. I'm just gonna draw it like that. So we don't know just yet where the directrix and focus is, but we do know a few things. The focus is going to sit on the same, I guess you could say, the same x value as the vertex. So if we draw, this is x equals one, if x equals one, we know from our experience with focuses, foci, (laughs) I guess, that they're going to sit on the same axis as the vertex. So the focus might be right over here and then the directrix is going to be equidistant on the other side, equidistant on the other side. So the directrix might be something like this. Might be right over here. And once again, I haven't figured it out yet, but what we know is that because this point, the vertex, sits on the parabola, by definition has to be equidistant So. This distance has to be the same as this distance right over here and what's another way of thinking about this entire distance? Remember, this coordinate right over here is a, b and this is the line y is equal to k. This is y equals k. So what's this distance in yellow? What's this difference in y going to be? Well, you could call that, in this case, the directrix is above the focus, so you could say that this would be k minus b or you could say it's the absolute value of b minus k. This would actually always work. It'll always give you kind of the positive distance. So if we knew what the absolute value of b minus k is, if we knew this distance, then just split it in half with the directrix is gonna be that distance, half the distance above and then the focus is gonna be half the distance below." + }, + { + "Q": "At 17:30 isn't the 99% confidence interval 0.568 +/- 0.08 actually giving 56% to 57.6% ?\n", + "A": "Because he added the 0 before the . it did seem that way. It got me too :P", + "video_name": "SeQeYVJZ2gE", + "timestamps": [ + 1050 + ], + "3min_transcript": "So it is 2.58 times our best estimate of the standard deviation of the sampling distribution, so times 0.031 is equal to 0.0-- well let's just round this up because it's so close to 0.08-- is within 0.08 of the population proportion. Or you could say that you're confident that the population proportion is within 0.08 of your sample mean. That's the exact same statement. So if we want our confidence interval, our actual number that we got for there, our actual sample mean we got was 0.568. I can delete this right here. Let me clear it. I can replace this, because we actually did take a sample. So I can replace this with 0.568. So we could be confident that there's a 99% chance that 0.568 is within 0.08 of the population proportion, which is the same thing as the population mean, which is the same thing as the mean of the sampling distribution of the sample mean, so forth and so on. And just to make it clear we can actually swap these two. It wouldn't change the meaning. If this is within 0.08 of that, then that is within 0.08 of this. So let me switch this up a little bit. So we could put a p is within of-- let me switch this up-- of 0.568. And now linguistically it sounds a little bit more like a confidence interval. We are confident that there's a 99% chance that p is within So what would be our confidence interval? It will be 0.568 plus or minus 0.08. And what would that be? If you add 0.08 to this right over here, at the upper end you're going to have 0.648. And at the lower end of our range, so this is the upper end, the lower end. If we subtract 8 from this we get 0.488. So we are 99% confident that the true population proportion is between these two numbers. Or another way, that the true percentage of teachers who think those computers are good ideas is between-- we're 99% confident-- we're confident that there's a 99% chance that the true percentage of teachers that like the computers is between 48.8% and 64.8%. Now we answered the first part of the question. The second part, how could the survey be changed to narrow" + }, + { + "Q": "at around 5:00 you said x 10^4 . where did 4 come from? i thought it was 10^5.\n", + "A": "The term 0.3979 * 10^5 is not in scientific notation form. He is converting it to that. Scientific notation is a number with a single non-zero digit in front of the decimal times 10 raised to a power. So he is multiplying the 0.3979 by 10 which removes one of the 10s from the 10^5 turning it into 10^4", + "video_name": "XJBwJjP2_hM", + "timestamps": [ + 300 + ], + "3min_transcript": "" + }, + { + "Q": "1:54 Could we just have calculated both volumes and subtracted the answers together to find the volume of the gold ring\n", + "A": "Well,if you mean taking the volume of the rectangular glass with the ring and subtract the volume of the rectangular glass without the ring, well you are right. Bang on!", + "video_name": "ViFLPsLTO1k", + "timestamps": [ + 114 + ], + "3min_transcript": "Jamie wants to know the volume of his gold ring in cubic inches. He gets a rectangular glass with base 3 inches by 2 inches. So you see that here, the base is 3 inches by 2 inches. And he fills the glass 4 inches high with water. So you see that over here, 4 inches high with water. Jamie drops his gold ring in the glass and measures the new height of the water to be 4.25 inches. So this is after the gold ring is dropped. What is the volume of Jamie's ring in cubic inches? Well when you start with this water right over here and you add his ring, whatever that volume is of his ring is going to displace an equal volume of water and push it up. And so the incremental volume that you now have is essentially going to be the volume of his ring. Well what is the incremental volume here? Well it's going to be the volume. If you think about going from this before volume to the after volume, the difference is the base stays the same. It's 3 inches by 2 inches, the difference is-- to make it a little bit neater-- the base is the same. The height now is 4.25 inches after dropping in the ring So the water went up by 0.25 inches. Let me write that, 0.25 inches is what the water went up by. So we could just think about, what is this incremental volume going to be? So this incremental volume right over here, that I'm shading in with purple. Well to figure that out we just have to measure. We just have to multiply the length times the width times the height times 0.25. So it's just going to be 3 times 2 times 0.25. 3 times 2 is 6, times 0.25, and you could do that either on paper or you might be able do that in your head. 4 times 0.25 is going to be 1, and you have 2 more So this is going to be 1.50. And we multiply it inches times inches times inches. So this is going to be in terms of cubic inches. 1.5 cubic inches is the volume of Jamie's ring, which is actually a pretty sizable volume for a gold ring. Maybe he has a very big finger or he just likes to spend, or I guess is his, whoever bought him the ring likes to spend a lot on gold." + }, + { + "Q": "\nAt 2:43 he got the sum, but would he have to simplify it? or would that be the absolute answer?", + "A": "OMGGGOMGOMG i knew how do to normal division but when he got the reciprocal out of no where that was confusing!", + "video_name": "Mcm0Q3wGhMo", + "timestamps": [ + 163 + ], + "3min_transcript": "My wife and I have recently purchased an assortment pack of soap, and we both want to experience all 3 bars. And we are not willing to share soap with each other. So we have a little bit of a conundrum. How do we share these 3 bars of soap so that we each get to experience all of the smells? So this might be a nice rose smell. This might be some type of ivory smell. I don't even know if that's a legitimate smell. This might be some type of sandalwood, which is always very nice. And my wife has an idea. She says, look I'm going to take these 3 bars, so we're starting with 3 bars of soap. And I'm going to divide it into 2 equal groups. And I said, how are you going to do that? And she says, well, I'm just going to take out some type of carving saw or carving knife, or who knows what it is, and I'm just going to cut it right down the middle, so right down this. This is my wife cutting the soap right over here with her saw. So she's cutting the soap right over here. So she has divided the soap into 2 equal groups. So the interesting question here is, how many bars of soap do we each have now? And I encourage you to pause the video and think about that for a second. How many bars of soap do we each have now? Well, let's just visualize my share. Let's say I take this bottom half right over here. So this is Sal's share of the soap. Let me write this out. So this is Sal's share. My wife took this top half. Well, what do I have? Well, I have 1/2, I have another 1/2, and I have another 1/2 of a bar. So I have 3/2 bars of soap. Or you could say that I have 3 times 1/2 bars of soap. Notice, something very interesting happened here. 3 divided by 2 is equal to 3 times 1/2. And we could make it even more interesting, So 3 divided by 2/1 is equal to 3 times 1/2. Notice we went from a division to a multiplication, and we took the reciprocal. And that makes complete sense. We have 3/2 bars now. But what's 3 times 1/2? Well, 3 times 1/2 is equal to 3/2. So just doing this little simple, smelly soap example, we've got a very interesting result. 3 divided by 2 is the same thing as 3 times 1/2, which is the same thing as 3/2." + }, + { + "Q": "\nWhat would happen if you did the mobius strip - flake that she showed at 3:56? Would you get the pattern, then the pattern reflected, then the pattern again, and then the pattern reflected and so on?", + "A": "Yes, you would. That is called a Frieze pattern.", + "video_name": "8EmhGOQ-DNQ", + "timestamps": [ + 236 + ], + "3min_transcript": "Stars count as holiday spirit, right? And you can do seven-fold symmetry in a similar way but you're probably going to need your emergency protractor. But you could do nine-fold without a protractor because you can do thirds, and then thirds again. And if you can do fifths without a protractor, you can do tenths too, because it's a fifth times a half. Look, I said happy holidays but I never said which one. Valentine's Day is totally a winter holiday. 11 is prime, though, so time for the protractor again. Look, prime factorization. So now theoretically you can get all sorts of end-fold symmetry, but what about rotational symmetry? There's no mirror lines, which means no folding, so does it even make sense as a question? Cutting a snowflake design efficiently is all about putting the same cut lines on top of each other so you only have to cut them once. So how do you take a rotationally symmetric design like this and put all the layers on top of each other without overlapping anything else? Maybe it's not surprising to see that to get stuff with rotational symmetry to line up, you rotate it. If you make a cut to the center then you can rotate all the way and roll the symmetry up into one unique thing. It's hard to draw accurate rotational symmetry by hand. So to cut out a paper swirl flake, start with a cut, then curl your paper into a cone. You can swirl around once or twice or more. But the important thing is to make sure the cut lines up with itself, because as far as symmetry is concerned, that cut doesn't exist. I like to tape it in place so it doesn't unroll, then cut stuff out. I find that spiraly things work well. Folding the paper is a good way to start a cut, but remember that folding creates symmetry. So I like to use it just to get the scissors in there and then do something asymmetric. Voila, snowflake. For a starflake swirlflake you'll have to curl your paper around five times, or four times. It's funny because I think of this as going around once but really it's going around twice, and a flat sheet of paper goes around once. Anyway, yeah, do that. And then give it a nice spiraly arm or two. You can make a nice fancy starflake swirlflake snowflake, awesome flake. Of course, from snowflakes it's only one small step to folding and cutting freeze patterns, and then wallpaper patterns and, get if you start by folding stuff into a [INAUDIBLE] strip? And then maybe you'll want to start folding and cutting spheres and everything will be a mess," + }, + { + "Q": "At 5:23 , why does 2y' equal -6e^-3x and not positive 6e^-3x. since you multiply the function 2 times itself and negative times negative becomes positive?\n", + "A": "You actually aren t multiplying the function by itself. The notation is kind of confusing, but 2 is just the coefficient for the first derivative. So 2y = 2 (-3e^-3x) is just -6e^-3x. Because the coefficient (2) is positive, the sign stays the same. I hope this helps.", + "video_name": "6o7b9yyhH7k", + "timestamps": [ + 323 + ], + "3min_transcript": "Now let's make that a little more tangible. What would a solution to something like any of these three, which really represent the same thing, what would a solution actually look like? Actually let me move this over a little bit. Move this over a little bit. So we can take a look at what some of these solutions could look like. Let me erase this a little. This little stuff that I have right over here. So I'm just gonna give you examples of solutions here. We'll verify that these indeed are solutions for I guess this is really just one differential equation represented in different ways. But you'll hopefully appreciate what a solution to a differential equation looks like. And that there is often more than one solution. There's a whole class of functions that could be a solution. So one solution to this differential equation, and I'll just write it as our first one. So one solution, I'll call it y one. And I could even write it as y one of x to make it explicit that it is a function of x. to the negative three x. And I encourage you to pause this video right now and find the first derivative of y one, and the second derivative of y one, and verify that it does indeed satisfy this differential equation. So I'm assuming you've had a go at it. So let's work through this together. So that's y one. So the first derivative of y one, so we just have to do the chain rule here, the derivative of negative three x with respect to x is just negative three. And the derivative of e to the negative three x with respect to negative three x is just e to the negative three x. And if we take the second derivative of y one, this is equal to the same exact idea, the derivative of this is three times negative three is going to be nine e to the negative three x. And now we could just substitute these values into the differential equation, or these expressions into the differential equation to verify that this So let's verify that. So we first have the second derivative of y. So that's that term right over there. So we have nine e to the negative three x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative three x. Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative three x. Three e to the negative three x." + }, + { + "Q": "\nso\ni knew that (\u00e2\u0088\u009a2)^2 =2 and (-\u00e2\u0088\u009a2)^2=2 why did he chose to (\u00e2\u0088\u009a2)^2 =2\nat 0:48", + "A": "\u00e2\u0088\u009a2 is less complicated than -\u00e2\u0088\u009a2 Why complicate things!", + "video_name": "C3QPTCwpIZo", + "timestamps": [ + 48 + ], + "3min_transcript": "Voiceover:Let's see if we can factor 36a to the eighth, plus 2b to the sixth power and I encourage you to pause the video and try it out on your own. So let's see if we can express this, or re-express this as the difference of squares using imaginary numbers. So we can rewrite 36 as, six squared, and a to the 8th is the same thing as, a to the fourth squared, and so let me actually just rewrite it this way. We can rewrite it as 6a to the fourth squared. That's this first term right over here. And then the second term we can write it as, actually just let me write it this way first. Let me just write it as a square. Plus, the square root of 2, b to the third power, squared. Now we wanted to write it as a difference of squares. So instead of writing it this way, let's get rid of this plus, and let's - so let me clear that out, and I could write it as and negative one, we know, is the same thing as i squared, so we can rewrite this whole thing as 6a to the fourth, squared. And then we have this minus right over here, minus. And so this is i squared. Negative one is i squared. So we can rewrite this in this pink color as i times the square root of two, times b to the third, all of that squared. Notice i squared is negative one. Square root of two squared, is two. B to the third, squared, is b to the sixth power. If I raise something to an exponent, and then raise it to another exponent, I would multiply the two exponents. And so now I've expressed it as a difference of squares, so we're ready to factor. This is going to be equal to 6a to the fourth, minus i times the square root of two, and let me get myself space here, times 6a to the fourth, plus all of this business, i times the square root of 2, times b to the third power, and we are done." + }, + { + "Q": "\nAt about 5:47, Sal starts talking about how you can't ever add enough 9s onto 4 to get the exact x value of the maximum. But, if you keep putting on more and more 9s to infinity, you get 4.999... on forever. If I'm not mistaken, 0.999... = 1, so wouldn't the maximum be still at x = (4+0.999...) = (4+1) = 5?", + "A": "There is no such number as infinity, so you cannot actually add an infinite number of 9s. So, this is a limit, not an actual finite sum.", + "video_name": "bZYTDst1MOo", + "timestamps": [ + 347 + ], + "3min_transcript": "Here our maximum point happens right when we hit b. And our minimum point happens at a. For a flat function we could put any point as a maximum or the minimum point. And we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous, and why this needs to be a closed interval. So first let's think about why does f need to be continuous? Well I can easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult or you can't really pick out an absolute minimum or an absolute maximum value over that interval. So let's say that this right over here is my interval. Let's say that's a, that's b. Let's say our function did something like this. Let's say our function did something right where you would have expected to have a maximum value let's say the function is not defined. And right where you would have expected to have a minimum value, the function is not defined. And so right over here you could say, well look, the function is clearly approaching, as x approaches this value right over here, the function is clearly approaching this limit. But that limit can't be the maxima because the function never gets to that. So you could say, well let's a little closer here. Maybe this number right over here is 5. So you could say, maybe the maximum is 4.9. Then you could get your x even closer to this value and make your y be 4.99, or 4.999. You could keep adding another 9. So there is no maximum value. Let me draw it a little bit so it looks more like a minimum. There is-- you can get closer and closer to it, but there's no minimum. Let's say that this value right over here is 1. So you could get to 1.1, or 1.01, or 1.0001. And so you could keep drawing some 0s between the two 1s but there's no absolute minimum value there. Now let's think about why it being a closed interval matters. Why you have to include your endpoints as kind of candidates for your maximum and minimum values over the interval. Well let's imagine that it was an open interval. Let's imagine open interval. And sometimes, if we want to be particular, we could make this is the closed interval right of here And if we wanted to do an open interval right over here, that's a and that's b. And let's just pick very simple function," + }, + { + "Q": "at 4:10-4:12 minimum happens at A? I'm so confused, I thought minimum was F(c). What are the white lines for?\n", + "A": "The white lines are a completely different function superimposed on the illustration, it is not related at all the original function other than they both share the same domain interval.", + "video_name": "bZYTDst1MOo", + "timestamps": [ + 250, + 252 + ], + "3min_transcript": "like this over the interval. And I'm just drawing something somewhat arbitrary right over here. So I've drawn a continuous function. I really didn't have to pick up my pen as I drew this right over here. And so you can see at least the way this continuous function that I've drawn, it's clear that there's an absolute maximum and absolute minimum point over this interval. The absolute minimum point, well it seems like we hit it right over here, when x is, let's say this is x is c. And this is f of c right over there. And it looks like we had our absolute maximum point over the interval right over there when x is, let's say this is x is equal to d. And this right over here is f of d. So another way to say this statement right over here if f is continuous over the interval, we could say there exists a c and d that are in the interval. are in the interval such that-- and I'm just using the logical notation here. Such that f c is less than or equal to f of x, which is less than or equal to f of d for all x in the interval. Just like that. So in this case you're saying, look, we hit our minimum value when x is equal to c. That's that right over here. Our maximum value when x is equal to d. And for all the other Xs in the interval we are between those two values. Now one thing, we could draw other continuous functions. And once again I'm not doing a proof of the extreme value theorem. But just to make you familiar with it and why it's stated the way it is. And you could draw a bunch of functions here Here our maximum point happens right when we hit b. And our minimum point happens at a. For a flat function we could put any point as a maximum or the minimum point. And we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous, and why this needs to be a closed interval. So first let's think about why does f need to be continuous? Well I can easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult or you can't really pick out an absolute minimum or an absolute maximum value over that interval." + }, + { + "Q": "3:42 is so confusing\n", + "A": "Not really. All that means is this: (6x6x6) TIMES (6x6x6x6x6x6). You have to do the parentheses first and then multiply.", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 222 + ], + "3min_transcript": "multiplication in. So this is going to be the same thing as 3 times 3 times 3 times x times x times x. And just based on what we reviewed just here, that part right there, 3 times 3, three times, that's 3 to the third power. And this right here, x times itself three times. that's x to the third power. So this whole thing can be rewritten as 3 to the third times x to the third. Or if you know what 3 to the third is, this is 9 times 3, which is 27. This is 27 x to the third power. Now you might have said, hey, wasn't 3x times 3x times 3x. Wasn't that 3x to the third power? Right? You're multiplying 3x times itself three times. And I would say, yes it is. So this, right here, you could interpret that as 3x to the And just like that, we stumbled on one of our exponent properties. Notice this. When I have something times something, and the whole thing is to the third power, that equals each of those things to the third power times each other. So 3x to the third is the same thing is 3 to the third times x to the third, which is 27 to the third power. Let's do a couple more examples. What if I were to ask you what 6 to the third times 6 to the sixth power is? And this is going to be a really huge number, but I want to write it as a power of 6. Let me write the 6 to the sixth in a different color. 6 to the third times 6 to the sixth power, what is this going to be equal to? Well, 6 to the third, we know that's 6 times itself three times. So it's 6 times 6 times 6. green, so I'll do it in green. Maybe I'll make both of them in orange. That is going to be times 6 to the sixth power. Well, what's 6 to the sixth power? That's 6 times itself six times. So, it's 6 times 6 times 6 times 6 times 6. Then you get one more, times 6. So what is this whole number going to be? Well, this whole thing-- we're multiplying 6 times itself-- how many times? One, two, three, four, five, six, seven, eight, nine times, right? Three times here and then another six times here. So we're multiplying 6 times itself nine times. 3 plus 6. So this is equal to 6 to the 3 plus 6 power or 6 to the ninth power. And just like that, we/ve stumbled on another exponent property." + }, + { + "Q": "At 3:22 which is the simplest form? (3x)^3 , 3^3 *X^3, or 27x^3\n", + "A": "27x^3 would be simplified.", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 202 + ], + "3min_transcript": "5 times 5 times 5 times 5 times 5 times 5 times 5. One, two, three, four, five, six, seven. This is going to be a really, really, really, really, large number and I'm not going to calculate it right now. If you want to do it by hand, feel free to do so. Or use a calculator, but this is a really, really, really, large number. So one thing that you might appreciate very quickly is that exponents increase very rapidly. 5 to the 17th would be even a way, way more massive number. But anyway, that's a review of exponents. Let's get a little bit steeped in algebra, using exponents. So what would 3x-- let me do this in a different color-- what would 3x times 3x times 3x be? Well, one thing you need to remember about multiplication multiplication in. So this is going to be the same thing as 3 times 3 times 3 times x times x times x. And just based on what we reviewed just here, that part right there, 3 times 3, three times, that's 3 to the third power. And this right here, x times itself three times. that's x to the third power. So this whole thing can be rewritten as 3 to the third times x to the third. Or if you know what 3 to the third is, this is 9 times 3, which is 27. This is 27 x to the third power. Now you might have said, hey, wasn't 3x times 3x times 3x. Wasn't that 3x to the third power? Right? You're multiplying 3x times itself three times. And I would say, yes it is. So this, right here, you could interpret that as 3x to the And just like that, we stumbled on one of our exponent properties. Notice this. When I have something times something, and the whole thing is to the third power, that equals each of those things to the third power times each other. So 3x to the third is the same thing is 3 to the third times x to the third, which is 27 to the third power. Let's do a couple more examples. What if I were to ask you what 6 to the third times 6 to the sixth power is? And this is going to be a really huge number, but I want to write it as a power of 6. Let me write the 6 to the sixth in a different color. 6 to the third times 6 to the sixth power, what is this going to be equal to? Well, 6 to the third, we know that's 6 times itself three times. So it's 6 times 6 times 6." + }, + { + "Q": "10:18 what happend to the -1? (-1)^2 = just 1? how?\n", + "A": "A negative times a negative...", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 618 + ], + "3min_transcript": "So just to review the properties we've learned so far in this video, besides just a review of what an exponent is, if I have x to the a power times x to the b power, this is going to be equal to x to the a plus b power. We saw that right here. x squared times x to the fourth is equal to x to the sixth, 2 plus 4. We also saw that if I have x times y to the a power, this is the same thing is x to the a power times y to the a power. We saw that early on in this video. We saw that over here. 3x to the third is the same thing as 3 to the third times x to the third. That's what this is saying right here. 3x to the third is the same thing is 3 to the third times x to the third. is, if you have x to the a and then you raise that to the bth power, that's equal to x to the a times b. And we saw that right there. a to the third and then raise that to the fourth power is the same thing is a to the 3 times 4 or a to the 12th power. So let's use these properties to do a handful of more complex problems. Let's say we have 2xy squared times negative x squared y squared times three x squared y squared. And we wanted to simplify this. squared times y squared. So if we take this whole thing to the squared power, this is like raising each of these to the second power. So this part right here could be simplified as negative 1 squared times x squared squared, times y squared. And then if we were to simplify that, negative 1 squared is just 1, x squared squared-- remember you can just multiply the exponents-- so that's going to be x to the fourth y squared. That's what this middle part simplifies to. And let's see if we can merge it with the other parts. The other parts, just to remember, were 2 xy squared, and then 3x squared y squared. Well now we're just going ahead and just straight up multiplying everything. And we learned in multiplication that it doesn't matter which order you multiply things in. So I can just rearrange. We're just going and multiplying 2 times x times y" + }, + { + "Q": "\nShouldn't you put a paratheses for the x's on 7:04 of the video.", + "A": "You could, but it s not necessary. Remember our rule for associativity says if we are just multiplying it doesn t matter what order we do it in. This also means that the parentheses aren t necessary.", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 424 + ], + "3min_transcript": "number 6 is the base. We're taking the base to the exponent of 3. When you have the same base, and you're multiplying two exponents with the same base, you can add the exponents. Let me do several more examples of this. Let's do it in magenta. Let's say I had 2 squared times 2 to the fourth times 2 to the sixth. Well, I have the same base in all of these, so I can add the exponents. This is going to be equal to 2 to the 2 plus 4 plus 6, which is equal to 2 to the 12th power. And hopefully that makes sense, because this is going to be 2 times itself two times, 2 times itself four times, 2 times itself six times. When you multiply them all out, it's going to be 2 times itself, 12 times or 2 to the 12th power. Let's do it in a little bit more abstract way, using some What is x to the squared or x squared times x to the fourth? Well, we could use the property we just learned. We have the exact same base, x. So it's going to be x to the 2 plus 4 power. It's going to be x to the sixth power. And if you don't believe me, what is x squared? x squared is equal to x times x. And if you were going to multiply that times x to the fourth, you're multiplying it by x times itself four times. x times x times x times x. So how many times are you now multiplying x by itself? Well, one, two, three, four, five, six times. x to the sixth power. Let's do another one of these. The more examples you see, I figure, the better. mix and match it. Let's say I have a to the third to the fourth power. So I'll tell you the property here, and I'll show you why it makes sense. When you add something to an exponent, and then you raise that to an exponent, you can multiply the exponents. So this is going to be a to the 3 times 4 power or a to the 12th power. And why does that make sense? Well this right here is a to the third times itself four times. So this is equal to a to the third times a to the third times a to the third times a to the third. Well, we have the same base, so we can add the exponents. So there's going to be a to the 3 times 4, right? This is equal to a to the 3 plus 33 plus three plus 3 power, which is the same thing is a the 3 times 4 power or a" + }, + { + "Q": "@10:55, if it doesn't matter the order when you multiple, then what is FOIL?\n\nhe didn't need it for this video, i'm confused on when to use it.\n", + "A": "FOIL is just the standard method of multiplying two binomials (e.g. (a+b)(a-b) ) which you use to make sure you don t miss one of the multiplications necessary (because you need aa, ab, ba and bb) - technically you don t have to do it in that order (because as Sal says the order doesn t matter) but it just helps you make sure you get each of the components.", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 655 + ], + "3min_transcript": "is, if you have x to the a and then you raise that to the bth power, that's equal to x to the a times b. And we saw that right there. a to the third and then raise that to the fourth power is the same thing is a to the 3 times 4 or a to the 12th power. So let's use these properties to do a handful of more complex problems. Let's say we have 2xy squared times negative x squared y squared times three x squared y squared. And we wanted to simplify this. squared times y squared. So if we take this whole thing to the squared power, this is like raising each of these to the second power. So this part right here could be simplified as negative 1 squared times x squared squared, times y squared. And then if we were to simplify that, negative 1 squared is just 1, x squared squared-- remember you can just multiply the exponents-- so that's going to be x to the fourth y squared. That's what this middle part simplifies to. And let's see if we can merge it with the other parts. The other parts, just to remember, were 2 xy squared, and then 3x squared y squared. Well now we're just going ahead and just straight up multiplying everything. And we learned in multiplication that it doesn't matter which order you multiply things in. So I can just rearrange. We're just going and multiplying 2 times x times y x squared times y squared. So I can rearrange this, and I will rearrange it so that it's in a way that's easy to simplify. So I can multiply 2 times 3, and then I can worry about the x terms. Let me do it in this color. Then I have times x times x to the fourth times x squared. And then I have to worry about the y terms, times y squared times another y squared times another y squared. And now what are these equal to? Well, 2 times 3. You knew how to do that. That's equal to 6. And what is x times x to the fourth times x squared. Well, one thing to remember is x is the same thing as x to the first power. Anything to the first power is just that number. So you know, 2 to the first power is just 2." + }, + { + "Q": "\nAt 3:40, if Sal did 6^6*6^6 would that equal (6^6)^6", + "A": "6^6*6^6 would equal (6^6)^2 (6^6) gets multiplied two times with it self. (6^6)^6 would equal 6^6 * 6^6 * 6^6 * 6^6 * 6^6 * 6^6", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 220 + ], + "3min_transcript": "multiplication in. So this is going to be the same thing as 3 times 3 times 3 times x times x times x. And just based on what we reviewed just here, that part right there, 3 times 3, three times, that's 3 to the third power. And this right here, x times itself three times. that's x to the third power. So this whole thing can be rewritten as 3 to the third times x to the third. Or if you know what 3 to the third is, this is 9 times 3, which is 27. This is 27 x to the third power. Now you might have said, hey, wasn't 3x times 3x times 3x. Wasn't that 3x to the third power? Right? You're multiplying 3x times itself three times. And I would say, yes it is. So this, right here, you could interpret that as 3x to the And just like that, we stumbled on one of our exponent properties. Notice this. When I have something times something, and the whole thing is to the third power, that equals each of those things to the third power times each other. So 3x to the third is the same thing is 3 to the third times x to the third, which is 27 to the third power. Let's do a couple more examples. What if I were to ask you what 6 to the third times 6 to the sixth power is? And this is going to be a really huge number, but I want to write it as a power of 6. Let me write the 6 to the sixth in a different color. 6 to the third times 6 to the sixth power, what is this going to be equal to? Well, 6 to the third, we know that's 6 times itself three times. So it's 6 times 6 times 6. green, so I'll do it in green. Maybe I'll make both of them in orange. That is going to be times 6 to the sixth power. Well, what's 6 to the sixth power? That's 6 times itself six times. So, it's 6 times 6 times 6 times 6 times 6. Then you get one more, times 6. So what is this whole number going to be? Well, this whole thing-- we're multiplying 6 times itself-- how many times? One, two, three, four, five, six, seven, eight, nine times, right? Three times here and then another six times here. So we're multiplying 6 times itself nine times. 3 plus 6. So this is equal to 6 to the 3 plus 6 power or 6 to the ninth power. And just like that, we/ve stumbled on another exponent property." + }, + { + "Q": "At 9:02 when Sal tells us that the value of (Av) in c(Av) is zero, does that come from 2:57 when he was writing out the set notation for subspace N?\n", + "A": "Yes. You are assuming that the v is in your subspace and you know that Av=0 for all v in your subspace.", + "video_name": "jCwRV1QL_Xs", + "timestamps": [ + 542, + 177 + ], + "3min_transcript": "means they both satisfy this equation, then v1 plus v2 is definitely still a member of n. Because when I multiply a times that, I get the 0 vector again. So let me write that result, as well. So we now know that v1 plus v2 is also a member of n. And the last thing we have to show is that it's closed under multiplication. Let's say that v1 is a member of our space that I defined here, where they satisfy this equation. What about c times v1? Is that a member of n? Well let's think about it. What's our matrix a times the vector-- right? I'm just multiplying this times the scale. I'm just going to get another vector. Lowercase v, so it's a vector. What's this equal to? Well, once again, I haven't prove it to you yet, but it's actually a very straightforward thing to do, to show that when you're dealing with scalars, if you have a scalar here, it doesn't matter if you multiply the scalar times the vector before multiplying it times the matrix or multiplying the matrix times the vector, and then doing the scalar. So it's fairly straightforward to prove that this is equal to c times our matrix a-- I'll make that nice and bold, times our vector v. That these two things are equivalent. Maybe I should just churn out the video that does this, but I'll leave it to you. You, literally, just go through the mechanics component by component. And you show this. But clearly, if there's is true, we already know that v1 is a member of our set, which means that a times v1 is equal to the 0 vector. vector, which is still the 0 vector. So c[v,1] is definitely a member of n. So it's closed under multiplication. And I kind of assumed this right here. But maybe I'll prove that in a different video. But I want to do all this to show that this set n is a valid subspace. This is a valid subspace. It contains a 0 vector. It's close under addition. It's close under multiplication. And we actually have a special name for this. We call this right here, we call n, the null space of a. Or we could write n is equal to-- maybe I shouldn't have written an n. Let me write orange in there. Our orange n is equal to-- the notation is just the null space of a. Or we could write the null space is equal to the orange" + }, + { + "Q": "At 1:57, a homogeneous equation is mentioned. What exactly is a homogeneous equation and why is A times vector X =0 a homogeneous equation?\n", + "A": "Being more precise, A x = 0 is a homogeneous system of equations (a set of homogeneous equations). A linear equation is homogeneous when the constant part is zero. For example, ax + by = c is homogeneous only if c is zero. Ax=b is a way to write a system of equations using matrices.", + "video_name": "jCwRV1QL_Xs", + "timestamps": [ + 117 + ], + "3min_transcript": "Let's review our notions of subspaces again. And then let's see if we can define some interesting subspaces dealing with matrices and vectors. So a subspace-- let's say that I have some subspace-- oh, let me just call it some subspace s. This is a subspace if the following are true-- and this is all a review-- that the 0 vector-- I'll just do it like that-- the 0 vector, is a member of s. So it contains the 0 vector. Then if v1 and v2 are both members of my subspace, then v1 plus v2 is also a member of my subspace. So that's just saying that the subspaces are closed under addition. You can add any of their two members and you'll get another member of the subspace. And then the last requirement, if you remember, is that subspaces are closed under multiplication. So that if c is real number, and it's just a scalar. And if I multiply, and v1 is a member of my subspace, then if subspace, v1, I'm going to get another member of the subspace. So it's closed under multiplication. These were all of what a subspace is. This is our definition of a subspace. If you call something a subspace, these need to be true. Now let's see if we can do something interesting with what we understand about matrix vector multiplication. Let's say I have the matrix a-- I'll make it nice and bold-- and it's an m by n matrix. And I'm interested in the following situation; I want to set up the homogeneous equation. And we'll talk about why it's homogeneous. Well, I'll tell you in a second. So let's say we set up the equation. My matrix a times vector x is equal to the 0 vector. This is a homogeneous equation, And I want to ask the question-- I talked about subspaces. If I take all of the x's-- if I take the world, the universe, the set of all of the x's that satisfy this equation, do I have a valid subspace? Let's think about this. I want to take all of the x's that are a member of Rn. Remember, if our matrix a has n columns, then I've only defined this matrix vector multiplication. If x is a member of r, and if x has to have exactly n components, only then is it defined. So let me define a set of all the vectors that are a member of Rn where they satisfy the equation a times my vector x is equal to the 0 vector. So my question is, is this a subspace? Is this a valid subspace?" + }, + { + "Q": "\nAt 3:34, Sal uses the point (1,2) to find r, but can you use any of the other points? How would you find r by using the next point, (2, 4/3)?", + "A": "You could use any one of the given points. For (2, 4/3) you would substitute the y-value (so 4/3) into g(x) and the x-value ( so 2) into the x to end up with 4/3=3r^2. You then solve that equation to end up with the r. Just be careful because if the x-value is even, you could have two r s so it is more accurate to use a point with an odd x-value when it is available to eliminate any confusion.", + "video_name": "Qst1UVtq8pE", + "timestamps": [ + 214 + ], + "3min_transcript": "And you see that. When x increases by 1, our function increases by 2. So now we know the equation for f of x. f of x is going to be equal to 2 times 2x plus b, or 5. So we figured out what f of x is. Now we need to figure out what g of x is. So g of x is an exponential function. And there's really two things that we need to figure out. We need to figure out what a is, and we need to figure out what r is. And let me just rewrite that. So we know that g of x-- maybe I'll do it down here. g of x is equal to a times r to the x power. And if we know what g of 0 is, that's a pretty useful thing. Because r to the 0th power, regardless of what r is-- or I guess we could assume that r is not equal to 0. People can debate what 0 to the 0 power is. But if r is any non-zero number, we And so that essentially gives us a. So let's just write that down. g of 0 is a times r to the 0 power, which is just going to be equal to a times 1 or a. And they tell us what g of 0 is. g of 0 is equal to 3. So we know that a is equal to 3. So so far, we know that our g of x can be written as 3 times r to the x power. So now we can just use any one of the other values they gave us to solve for r. For example, they tell us that g of 1 is equal to 2. So let's write that down. g of 1, which would be 3 times r to the first power, or just 3-- let me just write it. It could be 3 times r to the first power, or we could just write that as 3 times r. They tell us that g of 1 is equal to 2. Or we get that r is equal to 2/3. Divide both sides of this equation by 3. So r is 2/3. And we're done. g of x is equal to 3 times 2/3. Actually, let me just write it this way. 3 times 2/3 to the x power. You could write it that way if you want, any which way. So 3 times 2/3 to the x power, and f of x is 2x plus 5. So let's actually just type that in. So f of x is 2x plus 5. And we can verify that that's the expression that we want. And g of x is 3 times 2 over 3 to the x power." + }, + { + "Q": "At 3:23 minutes, Sal says that the value of |x+10| is going to be greater than or equal to zero. But how can it ever be equal to zero? Shouldn't it just be greater than?\n", + "A": "when x = -10, then |x+10| = 0", + "video_name": "15s6B7K9paA", + "timestamps": [ + 203 + ], + "3min_transcript": "If these two things are equal, and if I want to keep them equal, if I subtract 6 from the right-hand side, I've got to subtract-- or if I subtract 6 times the absolute value of x plus 10 from the right-hand side, I have to subtract the same thing from the left-hand side. So we're going to have minus 6 times the absolute value of x plus 10. And likewise, I want to get all my constant terms, I want to get this 4 out of the left-hand side. So let me subtract 4 from the left, and then I have to also do it on the right, otherwise my equality wouldn't hold. And now let's see what we end up with. So on the left-hand side, the 4 minus 4, that's 0. You have 4 of something minus 6 of something, that means you're going to end up with negative 2 of that something. Negative 2 of the absolute value of x plus 10. Remember, this might seem a little confusing, but remember, if you had 4 apples and you subtract 6 apples, you now have negative 2 apples, Same way, you have 4 of this expression, you take away 6 of this expression, you now have negative 2 of this expression. Let me write it a little bit neater. So it's negative 2 times the absolute value of x plus 10 is equal to, well the whole point of this, of the 6 times the absolute value of x plus 10 minus 6 times the absolute value of x plus 10 is to make those cancel out, and then you have 10 minus 4, which is equal to 6. Now, we want to solve for the absolute value of x plus 10. So let's get rid of this negative 2, and we can do that by dividing both sides by negative 2. You might realize, everything we've done so far is just treating this red expression as almost just like a variable, and we're going to solve for that red expression and then take it from there. So negative 2 divided by negative 2 is 1. 6 divided by negative 2 is negative 3. So we get the absolute value of x plus 10 is equal to negative 3. You might say maybe this could be the positive version or the negative, but remember, absolute value is always non-negative. If you took the absolute value of 0, you would get 0. But the absolute value of anything else is going to be positive. So this thing right over here is definitely going to be greater than or equal to 0. Doesn't matter what x you put in there, when you take its absolute value, you're going to get a value that's greater than or equal to 0. So there's no x that you could find that's somehow-- you put it there, you add 10, you take the absolute value of it, you're actually getting a negative value. So this right over here has absolutely no solution. And I'll put some exclamation marks there for emphasis." + }, + { + "Q": "Please can anyone explain how Sal gets from Y/X = -3 to 1/X = -3 x 1/y @ 4:42 and/ or how he gets from X=2/y to 2 x 1/y @ 8:30? Thanks.\n", + "A": "It s okay, I get it now", + "video_name": "92U67CUy9Gc", + "timestamps": [ + 282, + 510 + ], + "3min_transcript": "2 is negative 6. So notice, we multiplied. So if we scaled-- let me do that in that same green color. If we scale up x by 2-- it's a different green color, but it serves the purpose-- we're also scaling up y by 2. To go from 1 to 2, you multiply it by 2. To go from negative 3 to negative 6, you're also multiplying by 2. So we grew by the same scaling factor. And if you wanted to go the other way-- let's try, I don't know, let's go to x is 1/3. If x is 1/3, then y is going to be-- negative 3 times 1/3 is negative 1. So notice, to go from 1 to 1/3, we divide by 3. To go from negative 3 to negative 1, we also divide by 3. We also scale down by a factor of 3. So whatever direction you scale x in, That's what it means to vary directly. Now, it's not always so clear. Sometimes it will be obfuscated. So let's take this example right over here. y is equal to negative 3x. And I'm saving this real estate for inverse variation in a second. You could write it like this, or you could algebraically manipulate it. You could maybe divide both sides of this equation by x, and then you would get y/x is equal to negative 3. Or maybe you divide both sides by x, and then you divide both sides by y. So from this, so if you divide both sides by y now, you could get 1/x is equal to negative 3 times 1/y. These three statements, these three equations, are all saying the same thing. So sometimes the direct variation isn't quite in your face. But if you do this, what I did right here with any of these, you will get the exact same result. to this form over here. And there's other ways we could do it. We could divide both sides of this equation by negative 3. And then you would get negative 1/3 y is equal to x. And now, this is kind of an interesting case here because here, this is x varies directly with y. Or we could say x is equal to some k times y. And in general, that's true. If y varies directly with x, then we can also say that x varies directly with y. It's not going to be the same constant. It's going to be essentially the inverse of that constant, but they're still directly varying. Now with that said, so much said, about direct variation, let's explore inverse variation a little bit. Inverse variation-- the general form, if we use the same variables. And it always doesn't have to be y and x. It could be an a and a b. It could be a m and an n. If I said m varies directly with n, we would say m is equal to some constant times n." + }, + { + "Q": "\nat 6:53, why is it called \"inverse?\" is it because the greater \"x\" is, the smaller \"y\" gets?", + "A": "Yes. One variable increases by the same factor that the other decreases. In his example, X multiplies by a factor of 2 while Y divides by a factor of two. If you re graphing the functions, The inverse variation indicates the variables separate from a central point.", + "video_name": "92U67CUy9Gc", + "timestamps": [ + 413 + ], + "3min_transcript": "to this form over here. And there's other ways we could do it. We could divide both sides of this equation by negative 3. And then you would get negative 1/3 y is equal to x. And now, this is kind of an interesting case here because here, this is x varies directly with y. Or we could say x is equal to some k times y. And in general, that's true. If y varies directly with x, then we can also say that x varies directly with y. It's not going to be the same constant. It's going to be essentially the inverse of that constant, but they're still directly varying. Now with that said, so much said, about direct variation, let's explore inverse variation a little bit. Inverse variation-- the general form, if we use the same variables. And it always doesn't have to be y and x. It could be an a and a b. It could be a m and an n. If I said m varies directly with n, we would say m is equal to some constant times n. So if I did it with y's and x's, this would be y is equal to some constant times 1/x. So instead of being some constant times x, it's some constant times 1/x. So let me draw you a bunch of examples. It could be y is equal to 1/x. It could be y is equal to 2 times 1/x, which is clearly the same thing as 2/x. It could be y is equal to 1/3 times 1/x, which is the same thing as 1 over 3x. it could be y is equal to negative 2 over x. And let's explore this, the inverse variation, the same way that we explored the direct variation. So let's pick-- I don't know/ let's pick y is equal to 2/x. And let me do that same table over here. So I have my table. I have my x values and my y values. If x is 1, then y is 2. So if you multiply x by 2, if you scale it up by a factor of 2, what happens to y? y gets scaled down by a factor of 2. You're dividing by 2 now. Notice the difference. Here, however we scaled x, we scaled up y by the same amount. Now, if we scale up x by a factor, when we have inverse variation, we're scaling down y by that same. So that's where the inverse is coming from. And we could go the other way. If we made x is equal to 1/2. So if we were to scale down x, we're going to see that it's going to scale up y. Because 2 divided by 1/2 is 4. So here we are scaling up y. So they're going to do the opposite things. They vary inversely. And you could try it with the negative version of it, as well. So here we're multiplying by 2. And once again, it's not always neatly written for you like this." + }, + { + "Q": "At 4:40, how did he divide \"y/x\" by \"y\"? I know that multiplying by the inverse is how you divide a fraction; but I don't understand how \"1/y\" times \"y/x\" would equal \"1/x\" unless you cross multiply; but that wouldn't work because \"1/y\" is already inverted, right?\n", + "A": "y/x / y = y/x * 1/y = 1y/xy [Simplify by dividing by y] = 1/x I hope this helps!", + "video_name": "92U67CUy9Gc", + "timestamps": [ + 280 + ], + "3min_transcript": "2 is negative 6. So notice, we multiplied. So if we scaled-- let me do that in that same green color. If we scale up x by 2-- it's a different green color, but it serves the purpose-- we're also scaling up y by 2. To go from 1 to 2, you multiply it by 2. To go from negative 3 to negative 6, you're also multiplying by 2. So we grew by the same scaling factor. And if you wanted to go the other way-- let's try, I don't know, let's go to x is 1/3. If x is 1/3, then y is going to be-- negative 3 times 1/3 is negative 1. So notice, to go from 1 to 1/3, we divide by 3. To go from negative 3 to negative 1, we also divide by 3. We also scale down by a factor of 3. So whatever direction you scale x in, That's what it means to vary directly. Now, it's not always so clear. Sometimes it will be obfuscated. So let's take this example right over here. y is equal to negative 3x. And I'm saving this real estate for inverse variation in a second. You could write it like this, or you could algebraically manipulate it. You could maybe divide both sides of this equation by x, and then you would get y/x is equal to negative 3. Or maybe you divide both sides by x, and then you divide both sides by y. So from this, so if you divide both sides by y now, you could get 1/x is equal to negative 3 times 1/y. These three statements, these three equations, are all saying the same thing. So sometimes the direct variation isn't quite in your face. But if you do this, what I did right here with any of these, you will get the exact same result. to this form over here. And there's other ways we could do it. We could divide both sides of this equation by negative 3. And then you would get negative 1/3 y is equal to x. And now, this is kind of an interesting case here because here, this is x varies directly with y. Or we could say x is equal to some k times y. And in general, that's true. If y varies directly with x, then we can also say that x varies directly with y. It's not going to be the same constant. It's going to be essentially the inverse of that constant, but they're still directly varying. Now with that said, so much said, about direct variation, let's explore inverse variation a little bit. Inverse variation-- the general form, if we use the same variables. And it always doesn't have to be y and x. It could be an a and a b. It could be a m and an n. If I said m varies directly with n, we would say m is equal to some constant times n." + }, + { + "Q": "\ncan someone tell me the mathematical reason as for why Sir Khan multiplied 108 by 139 at 5:09 ?", + "A": "we call also do Cross-Multiplication to get the value of x . which means , multiplying the numerator of LHS with the denominator of RHS and multiplying the numerator of RHS with the denominator of LHS..... provided that there is an equal-to sign between the fractions .", + "video_name": "Z5EnuVJawmY", + "timestamps": [ + 309 + ], + "3min_transcript": "And this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? Well, we just write SOHCAHTOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse. So we get the cosine of theta is going to be equal to the height that we care about. That's the adjacent side of this right triangle over the length of the hypotenuse, OVER 108. Well, that doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here. Theta is also sitting up here. So maybe if we can figure out what cosine of theta is based up here, then we can solve for h. So if we look at this data, what is the cosine of theta? And now we're looking at a different right triangle. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here. We already know that's 139 meters. So it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here. It's 72 plus 108. Oh, we already have it labeled here. It's 180. We can assume that this is an isosceles-- that this pyramid is an isosceles triangle. So 180 on that side and 180 on that side. So the cosine is adjacent-- 139-- over the hypotenuse, which is 180, over 180. And these data are the same data. We just showed that. So now we have cosine of theta is h/108. Cosine of theta is 139/180. Or we could say that h/108, which is cosine of theta, Both of these things are equal to cosine of theta. Now to solve for h, we just multiply both sides by 108. So h is equal to 139 times 108/180. So let's get our calculator out and calculate that. So that is going to be 139 times 108, divided by 180, gets us to 83.4 meters. So h is equal to 83.4 meters. The height of the water is 83.4." + }, + { + "Q": "\nat around 5:17 why do we multiply by 108 on all sides? Why wouldn't we multiply by 139, 180, or h?", + "A": "To make h not a fraction, or to remove the 108 from underneath h. Anything else wouldn t have done very well. That is what you need to do to solve for h.", + "video_name": "Z5EnuVJawmY", + "timestamps": [ + 317 + ], + "3min_transcript": "And this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? Well, we just write SOHCAHTOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse. So we get the cosine of theta is going to be equal to the height that we care about. That's the adjacent side of this right triangle over the length of the hypotenuse, OVER 108. Well, that doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here. Theta is also sitting up here. So maybe if we can figure out what cosine of theta is based up here, then we can solve for h. So if we look at this data, what is the cosine of theta? And now we're looking at a different right triangle. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here. We already know that's 139 meters. So it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here. It's 72 plus 108. Oh, we already have it labeled here. It's 180. We can assume that this is an isosceles-- that this pyramid is an isosceles triangle. So 180 on that side and 180 on that side. So the cosine is adjacent-- 139-- over the hypotenuse, which is 180, over 180. And these data are the same data. We just showed that. So now we have cosine of theta is h/108. Cosine of theta is 139/180. Or we could say that h/108, which is cosine of theta, Both of these things are equal to cosine of theta. Now to solve for h, we just multiply both sides by 108. So h is equal to 139 times 108/180. So let's get our calculator out and calculate that. So that is going to be 139 times 108, divided by 180, gets us to 83.4 meters. So h is equal to 83.4 meters. The height of the water is 83.4." + }, + { + "Q": "at 3:44, how do we know that the second derivative is less than zero? and how do we even know if this function can have a second derivative? what is needed to, what defines a second derivative? thank you!\n", + "A": "The second derivative is the derivative of the first derivative. e.g. f(x) = x^3 - x^2 f (x) = 3x^2 - 2x f (x) = 6x - 2 So, to know the value of the second derivative at a point (x=c, y=f(c)) you: 1) determine the first and then second derivatives 2) solve for f (c) e.g. for the equation I gave above f (x) = 0 at x = 0, so this is a critical point. f (0) = 6\u00e2\u0080\u00a20 - 2 = -2 Therefore, f(x) is concave downward at x=0 and this critical point is a local maximum. Can you do the same for the other critical point?", + "video_name": "-cW5hCsc9Yc", + "timestamps": [ + 224 + ], + "3min_transcript": "So our first derivative should still be equal to zero 'cause our slope of a tangent line right over there is still zero. So F prime of C is equal to zero. But in this second situation, we are concave upwards. The slope is constantly increasing. We have an upward opening bowl and so here we have a relative minimum value or we could say our second derivative is greater than zero. Visually we see it's a relative minimum value and we can tell just looking at our derivatives, at least the way I've drawn it, first derivative is equal to zero and we are concave upwards. Second derivative is greater than zero. And so this intuition that we hopefully just built up is what the second derivative test tells us. So it says hey look, if we're dealing with some function F, let's say it's a twice differentiable function. So that means that over some interval. and second derivatives are defined and so let's say there's some point, X equals C, where its first derivative is equal to zero, so the slope of the tangent line is equal to zero, and the derivative exists in a neighborhood around C and most of the functions we deal with, if it's differentiable at C, it tends to be differentiable in the neighborhood around C and then we also assume that second derivative exists is twice differentiable. Well then we might be dealing with a maximum point, we might be dealing with a minimum point, or we might not know what we're dealing with and it might be neither a minimum or a maximum point. But using the second derivative test, if we take the second derivative and if we see that the second derivative is indeed less than zero, then we have a relative maximum point. Then so this is a situation that we started with right up there. If our second derivative is greater than zero, then we are in this situation right here, we're concave upwards. We have a relative minimum point and if our second derivative is zero, it's inconclusive. We don't know what is actually going on at that point. We can't make any strong statement. So with that out of the way, let's just do a quick example just to see if this has gelled. Let's say that I have some twice differentiable function H and let's say that I tell you that H of eight is equal to five, I tell you that H prime of eight is equal to zero, and I tell you that the second derivative at X equals eight is equal to negative four. So given this, can you tell me whether the point eight comma five, so the point eight comma five, is it a relative minimum, relative minimum, maximum point or not enough info? Not enough info or inconclusive?" + }, + { + "Q": "\nat 0:55, wouldn't the notation for the arc be ABC then? how would you specify the major arc if there isn't a point on the circle to differentiate between the two?", + "A": "Hi angela braukman, The notation of the arc would still be two letters because an arc is not a major arc till it is either 180\u00cb\u0099 or more. And to specify it: I would recommend marking your own point if there is no third point. Hope that helps! - Sam", + "video_name": "GOA9XWEo7QI", + "timestamps": [ + 55 + ], + "3min_transcript": "- So I have some example questions here from Khan Academy on arc measure. And like always, I encourage you to pause the video after you see each of these questions, and try to solve them before I do. So this first question says what is the arc measure, in degrees, of arc AC on circle P below. So this is point A, that is point C, and when they're talking about arc AC, since they only have two letters here, we can assume that it's going to be the minor arc. When we talk about the minor arc. There's two potential arcs that connect point A and point C. There's the one here on the left, and then there's the one, there is the one on the right. And since C isn't exactly straight down from A, it's a little bit to the right, the shorter arc, the arc with the smaller length, or the minor arc is going to be this one that I'm depicting here right on the right. So what is this arc measure going to be? Well, the measure of this arc is going to be exactly the same thing as, in degrees, as the measure of the central angle that intercepts the arc. I'll do it in this blue color, that central angle is angle C, P, A. Angle C, P, A, and the measure of that central angle is going to be 70 degrees plus 104 degrees. It's going to be this whole thing right over there. So it's going to be 174 degrees. One hundred and seventy four degrees, that's the arc measure, in degrees, of arc AC. Let's keep doing these. So let me do another one. So, this next one asks us, in the figure below, in the figure below, segment AD-- so this is point A, this is point D, so segment AD is this one right over here. Let me see if I can draw that. That's AD right over there, AD and CE are diameters of the circle. So let me draw CE, we're going to connect point C and E. These are diameters. So, let me, so they go straight. Whoops, I'm using the wrong tool, let me... So those are, somehow I should, alright. So, those are di-- whoops, how did that happen? So let me, somehow my pen got really big, alright. That'll be almost there, ok. So CE, there you go. So those are both diameters of the circle P. What is the arc measure of AB, of arc AB in degrees? So arc AB, once again there's two potential arcs that connect point A and B. There's the minor arc, and since this only has two letters we'll assume it's the minor arc. It's going to be this one over here. There's a major arc, but to not the major arc they would've said something like A, E, B or A, D, B" + }, + { + "Q": "At 1:30, why does Sal give the measure of the minor arc? I thought the question was asking for the major arc on the other side.\n", + "A": "AC by notation is a minor arc, it just happens to have another letter between them, A major arc is designated by three letters always to distinguish it from a minor arc, so you would need an additional point on the left side (D) to be talking about major arc ADC.", + "video_name": "GOA9XWEo7QI", + "timestamps": [ + 90 + ], + "3min_transcript": "- So I have some example questions here from Khan Academy on arc measure. And like always, I encourage you to pause the video after you see each of these questions, and try to solve them before I do. So this first question says what is the arc measure, in degrees, of arc AC on circle P below. So this is point A, that is point C, and when they're talking about arc AC, since they only have two letters here, we can assume that it's going to be the minor arc. When we talk about the minor arc. There's two potential arcs that connect point A and point C. There's the one here on the left, and then there's the one, there is the one on the right. And since C isn't exactly straight down from A, it's a little bit to the right, the shorter arc, the arc with the smaller length, or the minor arc is going to be this one that I'm depicting here right on the right. So what is this arc measure going to be? Well, the measure of this arc is going to be exactly the same thing as, in degrees, as the measure of the central angle that intercepts the arc. I'll do it in this blue color, that central angle is angle C, P, A. Angle C, P, A, and the measure of that central angle is going to be 70 degrees plus 104 degrees. It's going to be this whole thing right over there. So it's going to be 174 degrees. One hundred and seventy four degrees, that's the arc measure, in degrees, of arc AC. Let's keep doing these. So let me do another one. So, this next one asks us, in the figure below, in the figure below, segment AD-- so this is point A, this is point D, so segment AD is this one right over here. Let me see if I can draw that. That's AD right over there, AD and CE are diameters of the circle. So let me draw CE, we're going to connect point C and E. These are diameters. So, let me, so they go straight. Whoops, I'm using the wrong tool, let me... So those are, somehow I should, alright. So, those are di-- whoops, how did that happen? So let me, somehow my pen got really big, alright. That'll be almost there, ok. So CE, there you go. So those are both diameters of the circle P. What is the arc measure of AB, of arc AB in degrees? So arc AB, once again there's two potential arcs that connect point A and B. There's the minor arc, and since this only has two letters we'll assume it's the minor arc. It's going to be this one over here. There's a major arc, but to not the major arc they would've said something like A, E, B or A, D, B" + }, + { + "Q": "At 1:08, Sal says..\"The actual coordinate in R2 on the Cartesian coordinate..\" what does R2 mean?\n", + "A": "It means the x-y coordinate plane. R for the real numbers, and 2 for the number of dimensions (loosely speaking)", + "video_name": "7fYDCUIvZnM", + "timestamps": [ + 68 + ], + "3min_transcript": "- [Voiceover] Let's say I have some curve C, and it's described, it can be parameterized, I can't say that word, as, let's say x is equal to x of t, y is equal to some function y of t, and let's say that this is valid for t is between a and b. So, t is greater than or equal to a and then less than or equal to b. So, if I were to just draw this on, let me see, I could draw it like this. I'm staying very abstract right now. This is not a very specific example. This is the x axis, this is the y axis. My curve, let's say this is when t is equal to a and then the curve might do something like this. I don't know what it does, let's say it's over there. This is t is equal to b. This actual point right here will be x of b. That would be the x coordinate, And this is, of course, when t is equal to a. The actual coordinate in R2 on the Cartesian coordinates will be x of a, which is this right here, and then y of a, which is that right there. And we've seen that before, that's just a standard way of describing a parametric equation or curve using two parametric equations. What I want to do now is describe this same exact curve using a vector valued function. So, if I define a vector valued function, and if you don't remember what those are we'll have a little bit of review here. Let me say I have a vector valued function r, and I'll put a little vector arrow on top of it. In a lot of textbooks, they'll just bold it and they'll leave scalar valued functions unbolded, but it's hard to draw bold so I'll put a little vector on top. And let's say that r is a function of t. And these are going to be position vectors. Position vectors. And I'm specifying that because in general, when someone talks about a vector, this vector and this vector are considered equivalent as long as they have the same magnitude and direction. No one really cares about what their start and end points are as long as their direction is the same and their length is the same. But, when you talk about position vectors, you're saying, \"No, these vectors are all going to start at zero, at the origin.\" And when you state the position vector you're implicitly saying, \"This is specifying a unique position.\" In this case, it's going to be in two dimensional space, but it could be in three dimensional space, or really four, five, whatever, n dimensional space. So, when you state the position vector, you're literally saying, \"Okay, this vector literally specifies that point in space.\" So, let's see if we can describe this curve as a vector, a position vector valued function. So, we can say r of t, let me switch back to that pink color," + }, + { + "Q": "at 0:57 Sal introduces a new word to me, degenerate, but doesn't explain it thoroughly. Can someone please help me?\n", + "A": "Imagine a line segment with a point on a side so it s barely a triangle. That s a degenerate triangle in a nutshell", + "video_name": "KlKYvbigBqs", + "timestamps": [ + 57 + ], + "3min_transcript": "Let's draw ourselves a triangle. Let's say this side has length 6. Let's say this side right over here has length 10. And let's say that this side right over here has length x. And what I'm going to think about is how large or how small that value x can be. How large or small can this side be? So the first question is how small can it get? Well, if we want to make this small, we would just literally have to look at this angle right over here. So let me take a look at this angle and make it smaller. So let's try to make that angle as small as possible. So we have our 10 side. Actually let me do it down here. So you have your 10 side, the side of length 10, and I'm going to make this angle really, really, really small, approaching 0. If that angle becomes 0, we end up with a degenerate triangle. It essentially becomes one dimension. But as we approach 0, this side starts to coincide or get closer and closer to the 10 side. And you could imagine the case where it actually coincides with it and you actually get the degenerate. So if want this point right over here to get as close as possible to that point over there, essentially minimizing your distance x, the closest way is if you make the angle the way equal to 0, all the way. So let's actually-- let me draw a progression. So now the angle is getting smaller. This is length 6. x is getting smaller. Then we keep making that angle smaller and smaller and smaller all the way until we get a degenerate triangle. So let me draw that pink side. So you have the side of length 10. Now the angle is essentially 0, this angle that we care about. So this side is length 6. And so what is the distance between this point and this point? And that distance is length x. We know that 6 plus x is going to be equal to 10. So in this degenerate case, x is going to be equal to 4. So if you want this to be a real triangle, at x equals 4 you've got these points as close as possible. It's degenerated into a line, into a line segment. If you want this to be a triangle, x has to be greater than 4. Now let's think about it the other way. How large can x be? Well to think about larger and larger x's, we need to make this angle bigger. So let's try to do that. So let's draw my 10 side again. So this is my 10 side. I'm going to make that angle bigger and bigger. So now let me take my 6 side and put it like that. And so now our angle is getting bigger and bigger and bigger. It's approaching 180 degrees. At 180 degrees, our triangle once again" + }, + { + "Q": "What does timesish mean? She says it at 5:06.\n", + "A": "She means taking steps on the number line by multiplying, not adding.", + "video_name": "N-7tcTIrers", + "timestamps": [ + 306 + ], + "3min_transcript": "From where you're looking, 5 is plus 1. 3 is also just plus 1 away, but against the flow of time. So it's an anti-1. Minus 4 is twice as far away back in time as it would be if you were looking from a 0 perspective. To get to negative 8, sometimes to make the harder things simple, first, you have to make the simple things harder. Addition is a process of plus 1 plus 1 plus 1. Multiplication is a process of plus n plus n plus n. For multiples of 2, instead of counting plus 1 plus 1 plus 1, you're counting by plus 1 plus 1, plus 1 plus 1, or plus 2, for short. Plus 2 plus 2 plus 2 plus 2. Counting in a plus 2 kind of way. Counting in a plus 5 kind of way. Counting in a plus 10 kind of way. What is 5 times 7? All that means is count in a plus 5 sort of way seven times, which happens to be the same as counting in a plus 7 sort of way 5 times. 1, 2, 3, 4, 5. This is why 8 smells like 2 and 4. 2 sort of way. And 8 is 2 if you're counting in a plus 4 sort of way. Division is fancy counting. A divided by B asks the question, what is A, when counting in a plus B sort of way? 27 divided by 9 means, what is 27, in a plus 9 sort of way? Plus 9 plus 9 plus 9, it's 3. What is seven in a plus 2 sort of way? Just count, 1, 2, 3, and 1/2. Division means counting by plus denominators until you get to the numerator and seeing how long it takes. 7 divided by 1/3 means counting by plus 1/3's until you get to seven. Plus 1/3 plus 1/3 plus 1/3 plus 1/3 plus 1/3, all the way to 21. Exponentiation-- powers-- also fancy counting. This time, instead of counting by plus whatever, you count by times whatever. Powers of 2 are just counting in a times 2 sort of way. The size of the times-ish step is the bottom number. And the number of times-ish steps is a little exponent. Powers of 10 count in a times 10 sort of way. To take 10 to the sixth, you just take six steps of size times 10, 10 times 10 times 10 times 10 times 10 times 10. That's why 8 has a whiff of 3. 8 is 3 when counting in a times 2-ish way. Then, there's roots. To take the fourth root of 81, you're just saying, hey, if 81 is 4 in a timesy sort of way, then how are we counting? It's like going a quarter of the way there, but a timesy quarter, not an addish quarter, times 3 times 3 times 3 times 3. As far as times-ish counting is concerned, 3 is a quarter of 81, which is why 81 to the 1/4 is the same as the fourth root of 81. And personally, I think it's kind of weird that we keep around this root notation, when fractional powers are so much more descriptive. 81 to this 3/4 means you're going 3/4 of the way" + }, + { + "Q": "Maybe he said it and I missed it, but how do we know that the zero vector is the only solution to Rx=0, if he only used the particular solution of c1,c2,0,c4,0 around 3:15?\n", + "A": "It s not, but that s beside the point. We aren t trying to solve c1r1 + c2r2 + c3r3 + c4r4 + c5r5 = 0, only c1r1 + c2r2 + c4r4 = 0. We already know that the other columns are extras .", + "video_name": "BfVjTOjvI30", + "timestamps": [ + 195 + ], + "3min_transcript": "combination of my basis vectors? So the first thing that wasn't too much of a stretch of the imagination in the last video, was the idea that these pivot vectors are linearly independent. So r1, r2, and r4. And everything I'm doing, I'm kind of applying to the special case just so that it's easier to understand. But it should be generalizable. In fact, it definitely is generalizable. That all of the pivot columns in reduced row echelon form are linearly independent. And that's because the very nature of reduced row echelon form, is that you are the only pivot column that has a 1 in that respective row. So the only way to construct it is with that vector. You can't construct it with the other pivot columns because they're all going to have 0 in that row. And when I say it's linearly independent, I'm just saying the set of pivot columns. So let me say this in general. The set of pivot columns for any reduced row echelon form And it's just a very straightforward argument. Because each column is going to have a 1 in a very unique place. All of the other pivot columns are going to have a 0 in that same place. And so you can't take any linear combinations to get to that 1 because 0 times anything, minus or plus 0 times anything, can never be equal to 1. So I think you can accept that. Now, that means that the solution to c1 times r1, plus c2 times r2, plus, let me say, c4 times r4. The solution to this equation, because these guys are linearly independent, we know that this only has one solution, and that's c1, c2, and c4 is equal to 0. So another way we could say it is, if we write r times some vector x-- well I'll just write it times this particular x-- where I write it as c1, c2, 0, c4, and 0 is equal to 0. So this will be some special member of your null space. It's a particular solution to the equation. This is equal to one, two, three, four 0's because we have four rows here. Now, if we just expand this out. If we just multiply 1 times c1, plus 0 times c2, minus 1 times 0, plus 4 times 0, you'll get-- or actually a better way to explain it-- this multiplication right here can be written as-- and we've seen this multiple times-- c1" + }, + { + "Q": "\nAt 6:58, isn't the formula for finding the area of a circle, pi*diameter?", + "A": "NO. The circumference is pi x the diameter. Area is pi x r^2.", + "video_name": "mLE-SlOZToc", + "timestamps": [ + 418 + ], + "3min_transcript": "so let's say the circle looks-- I can draw a neater circle than that. So let's say the circle looks something like that. And its circumference-- we have to be careful here, they're giving us interesting-- the circumference is 36 pi. Then they tell us that contained in that circle is a smaller circle with area 16 pi. So inside the bigger circle, we have a smaller circle that has an area of 16 pi. A point is selected at random from inside the larger circle, so we're going to randomly select some point in this larger circle. What is the probability that the point also lies in this smaller circle? So here's a little bit interesting, because you actually have an infinite number of points in both of these circles, because it's not kind of a separate balls or marbles, There's actually an infinite number of points you could pick here. And so, when we talk about the probability that the point also lies in the smaller circle, we're really thinking about the percentage of the points in the larger circle that are also in the smaller circle. Or another way to think about it is the probability that if we pick a point from this larger circle, the probably that it's also in the smaller circle is really just going to be the percentage of the larger circle that is the smaller circle. I know that might sound confusing, but we're really just have to figure out the areas for both of them, and it's really just going to be the ratios so let's think about that. So there's a temptation to just use this 36 pi up here, but we have to remember, this was the circumference, and we need to figure out the area of both of these circles. And so for area, we need to know the radius, because area is pi r squared. So we can figure out the radius from the circumference by saying, well, circumference is equal to 2 times pi times the radius of the circle. is equal to 2 times pi times the radius, we can divide both sides by 2 pi, and on the left hand side, 36 divided by 2 is 18 the pi's cancel out, we get our radius as being equal to 18 for this larger circle. So if we want to know its area, its area is going to be pi r squared, which is equal to pi times 18 squared. And let's figure out what 18 squared is. 18 times 18, 8 times 8 is 64, eight times 1 is 8 plus 6 is 14, and then we put that 0 there because we're now in the tens place, 1 times 8 is 8, 1 times 1 is 1. And really, this is a 10 times the 10, and that's why it gives us 100. Anyway, 4 plus 0 is a 4, 4 plus 8 is a 12," + }, + { + "Q": "\nHello! I have a question. In Finding probability example 2 at 9:33 you talk about probability. Why you don't calculated area of the smallest circumference?\nSincerely,Alex.", + "A": "The area of the smaller circle is already calculated as far as we need it to be in order to construct the ratio of areas. It was given as A=16\u00cf\u0080. The larger circles area was not given, so it needed to be calculated from the given circumference.", + "video_name": "mLE-SlOZToc", + "timestamps": [ + 573 + ], + "3min_transcript": "is equal to 2 times pi times the radius, we can divide both sides by 2 pi, and on the left hand side, 36 divided by 2 is 18 the pi's cancel out, we get our radius as being equal to 18 for this larger circle. So if we want to know its area, its area is going to be pi r squared, which is equal to pi times 18 squared. And let's figure out what 18 squared is. 18 times 18, 8 times 8 is 64, eight times 1 is 8 plus 6 is 14, and then we put that 0 there because we're now in the tens place, 1 times 8 is 8, 1 times 1 is 1. And really, this is a 10 times the 10, and that's why it gives us 100. Anyway, 4 plus 0 is a 4, 4 plus 8 is a 12, So the area here is equal to pi times 324, or we could say 324 pi. So the area of the entire larger circle, the part that I shaded in yellow, including what's kind of under this orange circle, if you want to view it that way, this area right over here is equal to 324 pi. So the probability that a point that we select from this larger circle is also in the smaller circle is really just a percentage of the larger circle that is the smaller circle. So our probability-- I'll just write it like this-- the probability that the point also lies in the smaller circle-- so all of that stuff The probability of that is going to be equal to the percentage of this larger circle that is this smaller one, and that's going to be-- that is the smaller circle's area. So it's going to be 16 pi over 324 pi. And the pi's cancel out, and it looks like both of them are divisible by 4. If we divide the numerator by 4, we get 4, if we divide the denominator by 4, what do we get? 4 goes into 320 80 times, it goes into 4 once, so we get an 81. So a probability-- I didn't even draw this to scale, this area is actually much smaller when you do it to scale-- the probability that if you were to randomly select a point from the larger circle, that it also lies in the smaller one is the ratio of their areas, the ratio of the smaller circle to the larger one. And that is 4/81, I guess is the best way to say it." + }, + { + "Q": "\n6:05\nWhy is there an infinite number of points contained in the circle ?", + "A": "There is an infinite number of points because the distance between two points can always be cut in half, whether it is an arc or a line. If we were standing 64 feet apart and every minute we cut the distance in half, within 6 minutes we will be within one foot, then .5ft, .25.ft, .125ft, .0625ft, .03125ft... in theory, and that is what we are talking about, theory, we would never touch. As it should be...", + "video_name": "mLE-SlOZToc", + "timestamps": [ + 365 + ], + "3min_transcript": "55 is 11 times 5. Not 56, not 28. This is clearly 5 times 10, this is 8 times 5, this is the same number again, also 8 times 5. So all of these are multiples of 5. 45, that's 9 times 5. 3 is not a multiple of 5. 25, clearly 5 times 5. So I've circled all the multiples of 5. So of all the possibilities, the ones that meet our constraint of being a multiple of 5, there are 1, 2, 3, 4, 5, 6, 7 possibilities. So 7 meet our constraint. So in this example, the probability of a selecting a number that is a multiple of 5 is 7/12. Let's do another one. the circumference of a circle is 36 pi. Let's draw this circle. so let's say the circle looks-- I can draw a neater circle than that. So let's say the circle looks something like that. And its circumference-- we have to be careful here, they're giving us interesting-- the circumference is 36 pi. Then they tell us that contained in that circle is a smaller circle with area 16 pi. So inside the bigger circle, we have a smaller circle that has an area of 16 pi. A point is selected at random from inside the larger circle, so we're going to randomly select some point in this larger circle. What is the probability that the point also lies in this smaller circle? So here's a little bit interesting, because you actually have an infinite number of points in both of these circles, because it's not kind of a separate balls or marbles, There's actually an infinite number of points you could pick here. And so, when we talk about the probability that the point also lies in the smaller circle, we're really thinking about the percentage of the points in the larger circle that are also in the smaller circle. Or another way to think about it is the probability that if we pick a point from this larger circle, the probably that it's also in the smaller circle is really just going to be the percentage of the larger circle that is the smaller circle. I know that might sound confusing, but we're really just have to figure out the areas for both of them, and it's really just going to be the ratios so let's think about that. So there's a temptation to just use this 36 pi up here, but we have to remember, this was the circumference, and we need to figure out the area of both of these circles. And so for area, we need to know the radius, because area is pi r squared. So we can figure out the radius from the circumference by saying, well, circumference is equal to 2 times pi times the radius of the circle." + }, + { + "Q": "\nAt 0:31-- Why does it ask -2+(-7) when you could do -2-(7)?", + "A": "It is working with negative numbers. This question is showing that adding a negative is the same as subtracting a positive. Either way of writing it is perfectly acceptable.", + "video_name": "3CKpidALDEg", + "timestamps": [ + 31 + ], + "3min_transcript": "- [Voiceover] So we've already spent some time introducing ourselves to the idea of adding or subtracting positive and negative numbers. What I now what do in this video is do a bunch of examples using the adding negative numbers on the number line exercise on Khan Academy, so that we can think about different ways to model, or think about, or visualize adding and subtracting positive and negative numbers. So this question asks us, \"Which number line model represents \"the expression negative two \"plus negative seven?\" And a number line model seems like this really fancy thing, but they're just saying, which of these diagrams, which of these drawings are a way to think about what negative two plus negative seven is. So let's think about it. I want to start at negative two and then to that I want to add negative seven. So let's see what's going on here. This number line model I guess we could call it, it's saying this is positive two right over here. Then it subtracts seven. This is the model that would be for positive two plus negative seven, or two, or positive two minus seven. But that's not what we have over here, we have negative two minus seven. This one right over here, let's see, this first... I guess you could say this thing, this arrow, it's going one, two, three, four, five to the left of zero. So starting at negative five, and then it goes another one, two, three, four to the left of that. So you could view this as negative five minus four. Or you could view this as negative five plus negative four, but that's not what we have up here. And then this last one, hopefully this is the answer, otherwise there would be a mistake in the exercise. Let's see, we have this first arrow that takes us two steps to the left of zero. So, one, two. So this is negative two, and then we take another seven steps So one, two, three, four, five, six, seven. So you can think of this arrow as representing negative two, and then we're going to that, so we're starting from this tip of this arrow. We're going to add a negative seven which makes us move seven to the left again to get to negative nine. So that's definitely the model that represents that expression. Let's keep going. All right. Which number line model represents the expression six plus negative two. So here, they've taken our number lines but they've made them go up and down, they've made them vertical. So we wanna really think about it. I want to start at six and I want to add negative two. So on a vertical number line like this, it seems like we're increasing as we go up. I want to start at six above zero. So, I want to start at six above zero, but then I want to add negative two, which should take me two steps back down. This one takes me two steps even higher, so this one right over here is positive six plus two, or positive six plus positive two." + }, + { + "Q": "\nAt 1:01, how do you round down?", + "A": "you round down if the number is 4 or less", + "video_name": "_MIn3zFkEcc", + "timestamps": [ + 61 + ], + "3min_transcript": "0.710 Round 9.564, or nine and five hundred sixty-four Thousandths, to the nearest tenth. So lie me write it a bit larger, 9.564 And we need to round to the nearest tenth. So what's the tenth place? The tenths place is right here This represents 5 tenths. This is the ones place, this is the tenths place, this is The hundredths place, and this is the Thousandths place right here So we need to round to the nearest tenth. So if we round up, this will be 9.6 If we round down, this will be 9.5 And just like regular rounding, when we're not Dealing with decimals, you move to one spot, or you look At one place to the right or one place lower, I guess, and You say is that 5 larger If it is, you round if it isn't,you round down So this 9.564 becomes 9.6, or we can call this Nine and sixth tenths. And then we're done!" + }, + { + "Q": "\nAt 6:20 ish, how does Sal say that that function meets all three requirements? if c = 0, the lim as x approaches c of either of the functions neither gives us positive or negitive infinity.", + "A": "Earlier in the video Sal sets up two different ways to meet the requirements for application of l Hospital s rule. One involves the indeterminate form 0/0, and the other involves an indeterminate form with plus or minus infinity in both the numerator and denominator. He s working with an example that meets the first set of conditions, so it doesn t have to meet the other set of conditions.", + "video_name": "PdSzruR5OeE", + "timestamps": [ + 380 + ], + "3min_transcript": "Edit, copy, and then let me paste it. So in either of these two situations just to kind of make sure you understand what you're looking at, this is the situation where if you just tried to evaluate this limit right here you're going to get f of c, which is 0. Or the limit as x approaches c of f of x over the limit as x approaches c of g of x. That's going to give you 0/0. And so you say, hey, I don't know what that limit is? But this says, well, look. If this limit exists, I could take the derivative of each of these functions and then try to evaluate that limit. And if I get a number, if that exists, then they're going to be the same limit. This is a situation where when we take the limit we get infinity over infinity, or negative infinity or positive infinity over positive or negative infinity. So these are the two indeterminate forms. And to make it all clear let me just show you an example because I think this will make things a lot more clear. do this in a new color. Let me do it in this purplish color. Let's say we wanted to find the limit as x approaches 0 of sine of x over x. Now if we just view this, if we just try to evaluate it at 0 or take the limit as we approach 0 in each of these functions, we're going to get something that looks like 0/0. Sine of 0 is 0. Or the limit as x approaches 0 of sine of x is 0. And obviously, as x approaches 0 of x, that's also going to be 0. So this is our indeterminate form. And if you want to think about it, this is our f of x, that f of x right there is the sine of x. And our g of x, this g of x right there for this first case, is the x. g of x is equal to x and f of x is equal to sine of x. first two constraints. The limit as x, and in this case, c is 0. The limit as x approaches 0 of sine of sine of x is 0, and the limit as x approaches 0 of x is also equal to 0. So we get our indeterminate form. So let's see, at least, whether this limit even exists. If we take the derivative of f of x and we put that over the derivative of g of x, and take the limit as x approaches 0 in this case, that's our c. Let's see if this limit exists. So I'll do that in the blue. So let me write the derivatives of the two functions. So f prime of x. If f of x is sine of x, what's f prime of x? Well, it's just cosine of x. You've learned that many times. And if g of x is x, what is g prime of x?" + }, + { + "Q": "Shouldn't it be lim x->c of f(c)? not f(x)? 2:05\n", + "A": "No. The limit is of the function f(x), not the limit of f(c) because that is already evaluated at a value.", + "video_name": "PdSzruR5OeE", + "timestamps": [ + 125 + ], + "3min_transcript": "Most of what we do early on when we first learn about calculus is to use limits. We use limits to figure out derivatives of functions. In fact, the definition of a derivative uses the notion of a limit. It's a slope around the point as we take the limit of points closer and closer to the point in question. And you've seen that many, many, many times over. In this video I guess we're going to do it in the opposite direction. We're going to use derivatives to figure out limits. And in particular, limits that end up in indeterminate form. And when I say by indeterminate form I mean that when we just take the limit as it is, we end up with something like 0/0, or infinity over infinity, or negative infinity over infinity, or maybe negative infinity over negative infinity, or positive infinity over negative infinity. All of these are indeterminate, undefined forms. And in this video I'm just going to show you what l'Hoptial's rule says and how to apply it because it's fairly straightforward, and it's actually a very useful tool sometimes if you're in some type of a math competition and they ask you to find a difficult limit that when you just plug the numbers in you get something like this. L'Hopital's rule is normally what they are testing you for. And in a future video I might prove it, but that gets a The application is actually reasonably straightforward. So what l'Hopital's rule tells us that if we have-- and I'll do it in abstract form first, but I think when I show you the example it will all be made clear. That if the limit as x roaches c of f of x is equal to 0, and this is another and-- and the limit as x approaches c of f prime of x over g prime of x exists and it equals L. then-- so all of these conditions have to be met. This is the indeterminate form of 0/0, so this is the first case. Then we can say that the limit as x approaches c of f of x over g of x is also going to be equal to L." + }, + { + "Q": "Why at 6:35 does Sal considers that 1/cos(theta) =cos(theta)?\n", + "A": "He doesn t \u00e2\u0080\u0093 if you go back and listed from 6:30, he is taking the reciprocal of each element in the inequality \u00e2\u0080\u0093 this is also why he switches the direction of the inequalities.", + "video_name": "5xitzTutKqM", + "timestamps": [ + 395 + ], + "3min_transcript": "of tangent of theta, and let's see. Actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's gonna be the same thing as the absolute value of tangent of theta. is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta and then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one and on the right-hand side, I get a one over the absolute value of cosine theta. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one but now, since I'm taking the reciprocal of this here, it's gonna be greater than or equal to the absolute value of the sine of theta over the absolute value of theta, and that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there, so that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. well, sine of theta is gonna have the same sign. It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value, so I can erase those. If we're in the first or fourth quadrant, our X value is not negative, and so cosine of theta, which is the x-coordinate on our unit circle, is not going to be negative, and so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say for negative pi over two is less than theta is less than pi over two, but over this interval, this is true for any theta over which these functions are defined." + }, + { + "Q": "\nWhen Theta was divided by sin theta, shouldn't it be theta/sin theta, why it's sin theta/ theta in the video at 6:25 ?", + "A": "He took the reciprocal of all 3 terms, which made \u00ce\u00b8/sin\u00ce\u00b8 into sin\u00ce\u00b8/\u00ce\u00b8.", + "video_name": "5xitzTutKqM", + "timestamps": [ + 385 + ], + "3min_transcript": "which is the absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two. Now, how would you compare the areas of this pink or this salmon-colored triangle which sits inside of this wedge and how do you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge and the area of the wedge is less than or equal to the area of the big, blue triangle. The wedge includes the salmon triangle plus this area right over here, and then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true a little bit of algebraic manipulation. Let me multiply everything by two so I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta of tangent of theta, and let's see. Actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's gonna be the same thing as the absolute value of tangent of theta. is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta and then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one and on the right-hand side, I get a one over the absolute value of cosine theta. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one but now, since I'm taking the reciprocal of this here, it's gonna be greater than or equal to the absolute value of the sine of theta over the absolute value of theta, and that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there, so that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well." + }, + { + "Q": "At 3:20, why does Sal take only the positive square root of four, and not both negative and positive two? (I get that eventually it will not matter since you will add or subtract it from 6)\n", + "A": "Because he is looking at sqrt(4), which means the principal square root of 4 (the positive value you square to get 4). So that is positive 2 only. You ll notice, however, that in the expression he derives using the quadratic formula, there is a +/- symbol in front of the principal square root sign, so you do get both roots coming into play there. Does that help?", + "video_name": "dnjK4DPqh0k", + "timestamps": [ + 200 + ], + "3min_transcript": "So let's do that in this scenario. And the quadratic formula tells us that if we have something in standard form like this, that the roots of it are going to be negative b plus or minus-- so that gives us two roots right over there-- plus or minus square root of b squared minus 4ac over 2a. So let's apply that to this situation. Negative b-- this right here is b. So negative b is negative negative 6. So that's going to be positive 6, plus or minus the square root of b squared. Negative 6 squared is 36, minus 4 times a-- which is 2-- times 2 times c, which is 5. Times 5. All of that over 2 times a. a is 2. So 2 times 2 is 4. So this is going to be equal to 6 plus or minus 36 minus-- so this is 4 times 2 times 5. This is 40 over here. So 36 minus 40. And you already might be wondering what's going to happen here. All of that over 4. Or this is equal to 6 plus or minus the square root of negative 4. 36 minus 40 is negative 4 over 4. And you might say, hey, wait Sal. Negative 4, if I take a square root, I'm going to get an imaginary number. And you would be right. The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. So we're essentially going to get two complex numbers when we take the positive and negative version of this root. So let's do that. So the square root of negative 4, that is the same thing as 2i. And we know that's the same thing as 2i, or if you want to think of it this way. Square root of negative 4 is the same thing I could even do it one step-- that's the same thing as negative 1 times 4 under the radical, which is the same thing as the square root of negative 1 times the square root of 4. And the principal square root of negative 1 is i times the principal square root of 4 is 2. So this is 2i, or i times 2. So this right over here is going to be 2i. So we are left with x is equal to 6 plus or minus 2i over 4. And if we were to simplify it, we could divide the numerator and the denominator by 2. And so that would be the same thing as 3 plus or minus i over 2. Or if you want to write them as two distinct complex numbers, you could write this as 3 plus i over 2, or 3/2 plus 1/2i. That's if I take the positive version of the i there. Or we could view this as 3/2 minus 1/2i." + }, + { + "Q": "\nWhy did Sal divide (8-6i)/(4) by 2 at 9:17?", + "A": "Because the numerator can easily be divided by 2, and what s left of the denominator will cancel out when multiplied by the 2 outside the parentheses. 2((8 - 6i) / 4) = 2((4 - 3i) / 2) = 2(4 - 3i) / 2 = 4 - 3i You could, of course, do it the other way around: 2((8 - 6i) / 4) = 2(8 - 6i) / 4 = (8 - 6i) / 2 = 4 - 3i", + "video_name": "dnjK4DPqh0k", + "timestamps": [ + 557 + ], + "3min_transcript": "So once again, just looking at the original equation, 2x squared plus 5 is equal to 6x. Let me write it down over here. Let me rewrite the original equation. We have 2x squared plus 5 is equal to 6x. And now we're going to try this root, verify that it works. So we have 2 times 3 minus i over 2 squared plus 5 needs to be equal to 6 times this business. 6 times 3 minus i over 2. Once again, a little hairy. But as long as we do everything, we put our head down and focus on it, we should be able to get the right result. So 3 minus i squared. 3 minus i times 3 minus i, which is-- and you could get practice taking squares of two termed expressions, or complex numbers in this case actually-- it's going to be 9, that's 3 squared, and then 3 times negative i And then you're going to have two of those. So negative 6i. So negative i squared is also negative 1. That's negative 1 times negative 1 times i times i. So that's also negative 1. Negative i squared is also equal to negative 1. Negative i is also another square root. Not the principal square root, but one of the square roots of negative 1. So now we're going to have a plus 1, because-- oh, sorry, we're going to have a minus 1. Because this is negative i squared, which is negative 1. And all of that over 4. All of that over-- that's 2 squared is 4. Times 2 over here, plus 5, needs to be equal to-- well, before I even multiply it out, we could divide the numerator and the denominator by 2. So 6 divided by 2 is 3. 2 divided by 2 is 1. So 3 times 3 is 9. 3 times negative i is negative 3i. is going to be-- I'll do this in blue. 9 minus 1 is going to be 8. We have 8 minus 6i. And then if we divide 8 minus 6i by 2 and 4 by 2, in the numerator, we're going to get 4 minus 3i. And in the denominator over here, we're going to get a 2. We divided the numerator and the denominator by 2. Then we have a 2 out here. And we have a 2 in the denominator. Those two characters will cancel out. And so this expression right over here cancels or simplifies to 4 minus 3i. Then we have a plus 5 needs to be equal to 9 minus 3i. I We have a negative 3i on the left, a negative 3i on the right. We have a 4 plus 5. We could evaluate it. This left hand side is 9 minus 3i, which is the exact same complex number as we have" + }, + { + "Q": "At 1:43, Sal takes -6 squared to be positive 36 but if you press -6^2 on a calculator, you get -36. I remember working out a question wrong on Khan Academy. I had at some point in the problem work out g= -1^2 +4. I got 5 as an answer to be wrong. Khan Academy took -1^2 to be -1 and the correct answer was 3. Can somebody please explain?\n", + "A": "You wrote it like -6\u00c2\u00b2, but you need to write it like (-6)\u00c2\u00b2", + "video_name": "dnjK4DPqh0k", + "timestamps": [ + 103 + ], + "3min_transcript": "We're asked to solve 2x squared plus 5 is equal to 6x. And so we have a quadratic equation here. But just to put it into a form that we're more familiar with, let's try to put it into standard form. And standard form, of course, is the form ax squared plus bx plus c is equal to 0. And to do that, we essentially have to take the 6x and get rid of it from the right hand side. So we just have a 0 on the right hand side. And to do that, let's just subtract 6x from both sides of this equation. And so our left hand side becomes 2x squared minus 6x plus 5 is equal to-- and then on our right hand side, these two characters cancel out, and we just are left with 0. And there's many ways to solve this. We could try to factor it. And if I was trying to factor it, I would divide both sides by 2. If I divide both sides by 2, I would get integer coefficients on the x squared in the x term, but I would get 5/2 for the constant. So it's not one of these easy things to factor. We could complete the square, or we could apply the quadratic formula, which is really So let's do that in this scenario. And the quadratic formula tells us that if we have something in standard form like this, that the roots of it are going to be negative b plus or minus-- so that gives us two roots right over there-- plus or minus square root of b squared minus 4ac over 2a. So let's apply that to this situation. Negative b-- this right here is b. So negative b is negative negative 6. So that's going to be positive 6, plus or minus the square root of b squared. Negative 6 squared is 36, minus 4 times a-- which is 2-- times 2 times c, which is 5. Times 5. All of that over 2 times a. a is 2. So 2 times 2 is 4. So this is going to be equal to 6 plus or minus 36 minus-- so this is 4 times 2 times 5. This is 40 over here. So 36 minus 40. And you already might be wondering what's going to happen here. All of that over 4. Or this is equal to 6 plus or minus the square root of negative 4. 36 minus 40 is negative 4 over 4. And you might say, hey, wait Sal. Negative 4, if I take a square root, I'm going to get an imaginary number. And you would be right. The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. So we're essentially going to get two complex numbers when we take the positive and negative version of this root. So let's do that. So the square root of negative 4, that is the same thing as 2i. And we know that's the same thing as 2i, or if you want to think of it this way. Square root of negative 4 is the same thing" + }, + { + "Q": "\nAt 5:45 how 9 + 6i is resulted from 2(3+i/2)^2 ? I did the distributive property and came out with (36 + i^4) / 2\nWhy did Sal multiply \"3\" and \"i\" while there is a plus sign between them at 5:35?", + "A": "You have some errors... To square a binomial, use FOIL. It looks like you only squared the end values. and I have no idea how i became i^4 in your version as i*i = i^2 = -1. It also looks like you never squared the denominator. [(3+i)/2]^2 = (3+i)(3+i)/[2*2] = [9 + 3i + 3i + i^2]/4 = [9 + 6i -1]/4 = [8 + 6i]/4 Multiply it by 2 and it becomes: [8 + 6i]/2 The fraction can then be reduced: 8/2 + 6i/2 = 4 + 3i Hope this helps.", + "video_name": "dnjK4DPqh0k", + "timestamps": [ + 345, + 335 + ], + "3min_transcript": "Those are the two roots. Now what I want to do is a verify that these work. Verify these two roots. So this one I can rewrite as 3 plus i over 2. These are equivalent. All I did-- you can see that this is just dividing both of these by 2. Or if you were to essentially factor out the 1/2, you could go either way on this expression. And this one over here is going to be 3 minus i over 2. Or you could go directly from this. This is 3 plus or minus i over 2. So 3 plus i over 2. Or 3 minus i over 2. This and this or this and this, or this. These are all equal representations of both of the roots. But let's see if they work. So I'm first going to try this character right over here. It's going to get a little bit hairy, because we're going to have to square it and all the rest. But let's see if we can do it. 2 times this quantity squared. So 2 times 3 plus i over 2 squared plus 5. And we want to verify that that's the same thing as 6 times this quantity, as 6 times 3 plus i over 2. So what is 3 plus i squared? So this is 2 times-- let me just square this. So 3 plus i, that's going to be 3 squared, which is 9, plus 2 times the product of three and i. So 3 times i is 3i, times 2 is 6i. So plus 6i. And if that doesn't make sense to you, I encourage you to kind of multiply it out either with the distributive property or FOIL it out, and you'll get the middle term. You'll get 3i twice. When you add them, you get 6i. I And then plus i squared, and i squared is negative 1. Minus 1. All of that over 4, plus 5, is equal to-- well, you get a 3 here and you get a 1 here. And 3 distributed on 3 plus i is equal to 9 plus 3i. And what we have over here, we can simplify it just to save some screen real estate. 9 minus 1 is 8. So if I get rid of this, this is just 8 plus 6i. We can divide the numerator and the denominator right here by 2. So the numerator would become 4 plus 3i, if we divided it by 2, and the denominator here is just going to be 2. This 2 and this 2 are going to cancel out. So on the left hand side, we're left with 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works." + }, + { + "Q": "At 8:24 how does Sal get (-i)^2=-1?\n", + "A": "(-i)^2 = i^2, just like (-2)^2 = 2^2, because when you multiply negative by negative you get positive. i^2 = -1, because that s the definition of i.", + "video_name": "dnjK4DPqh0k", + "timestamps": [ + 504 + ], + "3min_transcript": "you get a 3 here and you get a 1 here. And 3 distributed on 3 plus i is equal to 9 plus 3i. And what we have over here, we can simplify it just to save some screen real estate. 9 minus 1 is 8. So if I get rid of this, this is just 8 plus 6i. We can divide the numerator and the denominator right here by 2. So the numerator would become 4 plus 3i, if we divided it by 2, and the denominator here is just going to be 2. This 2 and this 2 are going to cancel out. So on the left hand side, we're left with 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works. So once again, just looking at the original equation, 2x squared plus 5 is equal to 6x. Let me write it down over here. Let me rewrite the original equation. We have 2x squared plus 5 is equal to 6x. And now we're going to try this root, verify that it works. So we have 2 times 3 minus i over 2 squared plus 5 needs to be equal to 6 times this business. 6 times 3 minus i over 2. Once again, a little hairy. But as long as we do everything, we put our head down and focus on it, we should be able to get the right result. So 3 minus i squared. 3 minus i times 3 minus i, which is-- and you could get practice taking squares of two termed expressions, or complex numbers in this case actually-- it's going to be 9, that's 3 squared, and then 3 times negative i And then you're going to have two of those. So negative 6i. So negative i squared is also negative 1. That's negative 1 times negative 1 times i times i. So that's also negative 1. Negative i squared is also equal to negative 1. Negative i is also another square root. Not the principal square root, but one of the square roots of negative 1. So now we're going to have a plus 1, because-- oh, sorry, we're going to have a minus 1. Because this is negative i squared, which is negative 1. And all of that over 4. All of that over-- that's 2 squared is 4. Times 2 over here, plus 5, needs to be equal to-- well, before I even multiply it out, we could divide the numerator and the denominator by 2. So 6 divided by 2 is 3. 2 divided by 2 is 1. So 3 times 3 is 9. 3 times negative i is negative 3i." + }, + { + "Q": "I don't get why it is 1/a (3:11)? Why not -a?\n", + "A": "hmm.. let s suppose a was 2 2^4=16 2^3=8 2^2=4 2^1=2 2^0=1 at each step, we re dividing by 2 (that is, a) therefore, 2^-1 should be 1 divided by 2 2^-1= 1/2 replace 2 by a and you get a^-1= 1/a (if you have a question comment below)", + "video_name": "Tqpcku0hrPU", + "timestamps": [ + 191 + ], + "3min_transcript": "We multiplied by a, right? a squared is just a times a. And then to get to a cubed, what did we do? We multiplied by a again. And then to get to a to the fourth, what did we do? We multiplied by a again. Or the other way, you could imagine, is when you decrease the exponent, what are we doing? We are multiplying by 1/a, or dividing by a. And similarly, you decrease again, you're dividing by a. And to go from a squared to a to the first, you're dividing by a. So let's use this progression to figure out what a to the 0 is. So this is the first hard one. So a to the 0. So you're the inventor, the founding mother of mathematics, and you need to define what a to the 0 is. And, you know, maybe it's 17, maybe it's pi. I don't know. It's up to you to decide what a to the 0 is. But wouldn't it be nice if a to the 0 retained this pattern? dividing by a, right? So if you're going from a to the first to a to the zero, wouldn't it be nice if we just divided by a? So let's do that. So if we go from a to the first, which is just a, and divide by a, right, so we're just going to go-- we're just going to divide it by a, what is a divided by a? Well, it's just 1. So that's where the definition-- or that's one of the intuitions behind why something to the 0-th power is equal to 1. Because when you take that number and you divide it by itself one more time, you just get 1. So that's pretty reasonable, but now let's go into So what should a to the negative 1 equal? Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent we're dividing by a. So let's divide by a again, so 1/a. a to the 0 is one, so what's 1 divided by a? It's 1/a. Now, let's do it one more time, and then I think you're going to get the pattern. Well, I think you probably already got the pattern. What's a to the minus 2? Well, we want-- you know, it'd be silly now to Every time we decrease the exponent, we're dividing by a, so to go from a to the minus 1 to a to the minus 2, let's just divide by a again. And what do we get? If you take 1/2 and divide by a, you get 1 over a squared. And you could just keep doing this pattern all the way to the left, and you would get a to the minus b is equal to 1 over a to the b. Hopefully, that gave you a little intuition as to why-- well, first of all, you know, the big mystery is, you know, something to the 0-th power, why does that equal 1? First, keep in mind that that's just a definition. Someone decided it should be equal to 1, but they had a good reason. And their good reason was they wanted to keep" + }, + { + "Q": "At 3:33, Sal says \"when you take 1/a and divide by a you get 1/a^2\". Why is this?\n", + "A": "Walk it thru using the rules to divide fractions. 1/a divided by a/1 = 1/a * 1/a = 1/a^2 Hope this helps.", + "video_name": "Tqpcku0hrPU", + "timestamps": [ + 213 + ], + "3min_transcript": "dividing by a, right? So if you're going from a to the first to a to the zero, wouldn't it be nice if we just divided by a? So let's do that. So if we go from a to the first, which is just a, and divide by a, right, so we're just going to go-- we're just going to divide it by a, what is a divided by a? Well, it's just 1. So that's where the definition-- or that's one of the intuitions behind why something to the 0-th power is equal to 1. Because when you take that number and you divide it by itself one more time, you just get 1. So that's pretty reasonable, but now let's go into So what should a to the negative 1 equal? Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent we're dividing by a. So let's divide by a again, so 1/a. a to the 0 is one, so what's 1 divided by a? It's 1/a. Now, let's do it one more time, and then I think you're going to get the pattern. Well, I think you probably already got the pattern. What's a to the minus 2? Well, we want-- you know, it'd be silly now to Every time we decrease the exponent, we're dividing by a, so to go from a to the minus 1 to a to the minus 2, let's just divide by a again. And what do we get? If you take 1/2 and divide by a, you get 1 over a squared. And you could just keep doing this pattern all the way to the left, and you would get a to the minus b is equal to 1 over a to the b. Hopefully, that gave you a little intuition as to why-- well, first of all, you know, the big mystery is, you know, something to the 0-th power, why does that equal 1? First, keep in mind that that's just a definition. Someone decided it should be equal to 1, but they had a good reason. And their good reason was they wanted to keep And that's the same reason why they defined negative exponents in this way. And what's extra cool about it is not only does it retain this pattern of when you decrease exponents, you're dividing by a, or when you're increasing exponents, you're multiplying by a, but as you'll see in the exponent rules videos, all of the exponent rules hold. All of the exponent rules are consistent with this definition of something to the 0-th power and this definition of something to the negative power. Hopefully, that didn't confuse you and gave you a little bit of intuition and demystified something that, frankly, is quite mystifying the first time you learn it." + }, + { + "Q": "at 4:30pm, how do you write a percentage as a fraction or mix number in simplest form? for ex: 146.8 into a fraction or mixed number in simplest form\n", + "A": "Since the 8 is in the tenths place you create a fraction with 1468 in the numerator and 10 (for the tenths place) in the denominator. This gives you the fraction 1468/10. Now you reduce the fraction by dividing numerator and denominator by 2. This gives you 734/5 as an improper fraction. If you want a mixed number you leave the 146 as is and create a fraction using the 8. This gives you 146 8/10 which reduces to 146 4/5.", + "video_name": "-gB1y-PMWfs", + "timestamps": [ + 270 + ], + "3min_transcript": "It's 0.18. You could view this as 1 tenth and 8 hundredths, which is the same thing, or 10 hundredths and 8 hundredths, which is 18 hundredths. So this is written in decimal form. And if we write it as a simplified fraction, we need to see if there is a common factor for 18 and 100. And they're both even numbers, so we know they're both divisible by 2, so let's divide both the numerator and the denominator by 2. So we have 18 divided by 2 over 100 divided by 2. And we're going to get 18 divided by 2 is 9. 100 divided by 2 is 50. And I don't think these guys share any common factors. 50 is not divisible by 3. 9 is only divisible by 3 and 1 and 9. So this is the fraction in simplest form. So we have 18% is the same thing as 0.18, which is the Now, I went through a lot of pain here to show you that this really just comes from the word, from percent, from per 100. But if you ever were to see this in a problem, the fast way to do this is to immediately say, OK, if I have 18%, you should immediately say, anything in front of the percent-- that's that anything, whatever this anything is-- it should be equal to that anything. In this case it's 18/100. And another way to think about it, you could view this as 18.0%. I just added a trailing zero there, just so that you see the decimal, really. But if you want to express this as a decimal without the percent, you just move the decimal to the left two spaces. this becomes 0.18. Or you could immediately say that 18% as a fraction is 18/100. When you put it in simplified form, it's 9/50. But you should also see that 18/100, and we have seen this, is the exact same thing as 18 hundredths, or 0.18. Hopefully, this made some connections for you and didn't confuse you." + }, + { + "Q": "At 0:57 why do you devide 36 and 100 by 4?\n", + "A": "Dividing 36 and 100 by 4 is just simplifying it to its simplest form, which is 9/20.", + "video_name": "OS1g4PDdNdM", + "timestamps": [ + 57 + ], + "3min_transcript": "Let's see if we can write 0.36 as a fraction. There are several ways of doing it. The way I like to do it is to say, well, 0.36, this is the same thing as 36 hundredths. Or one way to think about it is this is in the hundredths place. This is in the tens place, or you could view this as 30 hundredths. You could view this as 3 tens, or 30 hundredths. So we could say that this is the same thing as 36 hundredths, or this is equal to 36/100. And we've already expressed it as a fraction, but now we could actually simplify it because both 36 and 100 have some common factors. They're both divisible by, well, looks like they're both divisible by 4. So if we could divide the numerator by 4 and the denominator by 4, we're doing the same thing to both. So we're not changing the value of the fraction. 36 divided by 4 is 9, and then 100 divided by 4 is 25. seem to share any other common factors. And so we've written it in simplified form, and we're done." + }, + { + "Q": "at 5:31 how is the moon larger enough to block the sun? Isn't it WAY bigger?\n", + "A": "The sun is bigger! But the Moon is closer to us, so when the Moon is in line with the Earth and Sun, the Sun appears to be blocked out because the Moon is right in it s way. If something is super far away, but you can still see it, it appears really small. But as it moves closer to you, it gets bigger. So because the Moon is closer to us than the Sun, it appears as if the Moon is just as big as the Sun.", + "video_name": "s8_14yxp1lQ", + "timestamps": [ + 331 + ], + "3min_transcript": "35 is five times seven and 21 is three times seven. So, you're multiplying by seven up here and here, you have a seven in the denominator, you're dividing by seven, so they're going to cancel out. So, this is going to simplify to five times 66 over 3, and then we could simplify it even more, because 66 is the same thing as three times 22. Three times 22 and so, you have a three in the numerator, you're multiplying by three and three in the denominator, dividing by three. Three divided by three is one, so you're left with five times 22, which is 110. So, it would take her m minutes to eat 35 hot dogs at the same pace. Now, when some of you might have tackled it, you might have had a different equation set up here. Instead of thinking of hot dogs per minute, you might have thought about minutes per hot dog. of minutes per hot dog, you might have said, ok look, it took Mika 66 minutes to eat 20, to eat 21 hot dogs,and it's gonna take her m minutes to eat 35 hot dogs and if it's the same pace, then these two rates are going to be equal. They have to be the same pace. And so, then you can solve for m and actually, this one's easier to solve for m, you just multiply both sides by 35. Multiply both sides by 35 and you're left with, on the right hand side you're left with just an m, and on the left hand side, same, same idea. You're taking 35, you have 35 times 66/21, which we already figured out is 110. So, 110 is equal, is equal to m. So, once again, multiple ways to tackle it, but it's important that we got the same answer." + }, + { + "Q": "\nAt 0:09, what does Sal mean by \" one term\"?", + "A": "a term is anything in a equation for example: x is a term x + 3 are 2 terms remember, terms are only separated by addition and subtraction signs", + "video_name": "p_61XhXdlxI", + "timestamps": [ + 9 + ], + "3min_transcript": "Express the area of a rectangle with length 4xy and width 2y as a monomial. Monomial just means just one term. So let's think about a rectangle. So let me draw a rectangle here. And they're telling us that the length is 4xy and they're telling us that the width is 2y. And just as a bit of a refresher, we know the area of a rectangle is just the width times the length, or the height times the width, or however you want to view it-- just the product of its two dimensions. So the area of this rectangle, the area is going to be equal to this length, 4xy times the width times 2y. We can simplify this. We have a 4 times a 2. you can switch the order however you like, as long as it's all multiplication. So 4 times 2, that gives us 8. Then we have this x sitting here. That is the only x we have there, so it's 8 times x. And then we have a y here. We could view that as y to the first. And then we have another y there. We could view that as y to the first. So y times y. You could view that as y squared. Or you could say look, y to the first times y to the first-- same base, add the two exponents-- 1 plus 1 is 2. So it's 8xy squared. This y squared covers both that y and that y over there. And we're done. We've expressed the area of this rectangle as a monomial." + }, + { + "Q": "At 6:30\n\nCould you also find the vertex by using the -b/2a that Sal used in the previous video?\n", + "A": "Yes, but I m pretty sure that you use -b/2a only to find the x-value of the vertex. Well, I guess you do end up finding the vertex, because you could just then plug the x-value into the quadratic to find the y-value of the vertex.", + "video_name": "FksgVpM_iXs", + "timestamps": [ + 390 + ], + "3min_transcript": "The discriminant is this part-- b squared minus 4ac. We see the discriminant is negative, there's no solution, which means that these two guys-- these two equations-- never intersect. There is no solution to the system. There are no x values that when you put into both of these equations give you the exact same y value. Let's think a little bit about why that happened. This one is already in kind of our y-intercept form. It's an upward opening parabola, so it looks something like this. I'll do my best to draw it-- just a quick and dirty version of it. Let me draw my axes in a neutral color. Let's say that this right here is my y-axis, that right there is my x-axis. x and y. This vertex-- it's in the vertex form-- occurs when x is equal to 4 and y is equal to 3. So x is equal to 4 and y is equal to 3. It's an upward opening parabola. So this will look something like this. I don't know the exact thing, but that's close enough. Now, what will this thing look like? It's a downward opening parabola and we can actually put this in vertex form. Let me put the second equation in vertex form, just so we have it. So we have a good sense. So, y is equal to-- we could factor in a negative 1-- negative x squared minus 2x plus 2. Actually, let me put the plus 2 further out-- plus 2, all the way up out there. Then we could say, half of negative 2 is negative 1. You square it, so you have a plus 1 and then a minus 1 there. This part right over here, we can rewrite as x minus 1 squared, so it becomes negative x minus 1 squared. I don't want to skip steps. Negative x minus 1 squared minus 1 plus 2. So that's plus 1 out here. Or if we want to distribute the negative, we get y is equal to negative x minus 1 squared minus 1. Here the vertex occurs at x is equal to 1, y is equal to negative 1. The vertex is there, and this is a downward opening parabola. We have a negative coefficient out here on the second degree term, so it's going to look something like this. So as you see, they don't intersect. This vertex is above it and it opens upward. This is its minimum point. And it's above this guy's maximum point. So they will never intersect, so there is no solution to this system of equations." + }, + { + "Q": "at2:15-2:45 i am confused\n", + "A": "It s easy, just multiply the numerator by the numerator and the denominator by the denominator, but first convert the mixed number to an improper fraction if neccesary", + "video_name": "XaJQse2u5TQ", + "timestamps": [ + 135, + 165 + ], + "3min_transcript": "We've already seen that the fraction 2/5, or fractions like the fraction 2/5, can be literally represented as 2 times 1/5, which is the same thing, which is equal to literally having two 1/5s. So 1/5 plus 1/5. And if we wanted to visualize it, let me make a hole here and divide it into five equal sections. And so this represents two of those fifths. This is the first of the fifths, and then this is the second of the fifths, Literally 2/5, 2/5, 2/5. Now let's think about something a little bit more interesting. What would 3 times 2/5 represent? 3 times 2/5. And I encourage you to pause this video and, based on what we just did here, think about what you think this would be equivalent to. as-- so let me just rewrite this as instead of 3 times 2/5 written like this, let me write 2/5 like that-- so this is the same thing as 3 times 2 times 1/5. And multiplication, we can multiply the 2 times the 1/5 first and then multiply by the 3, or we can multiply the 3 times the 2 first and then multiply by the 1/5. So you could view this literally as being equal to 3 times 2 is, of course, 6, so this is the same thing as 6 times 1/5. And if we were to try to visualize that again, so that's a whole. That's another whole. Each of those wholes have been divided into five equal sections. And so we're going to color in six of them. fifth 1/5-- and that gets us to a whole-- and then we have 6/5 just like that. So literally 3 times 2/5 can be viewed as 6/5. And of course, 6 times 1/5, or 6/5, can be written as-- so this is equal to, literally-- let me do the same color-- 6/5, 6 over 5. Now you might have said, well, what if we, instead of viewing 2/5 as this, as we just did in this example, we view 2/5 as 1/5 plus 1/5, what would happen then? Well, let's try it out. So 3 times 2/5-- I'll rewrite it-- 3 times 2/5, 2 over 5, is the same thing as 3 times 1/5 plus 1/5." + }, + { + "Q": "Why, at 5:30, did Sal use the word \"Dimension?\" It makes it sound like the rectangle's some sort of galactic unknown universe where extra-terrestial life lurks unseen! Shouln't it be \"Area?\"\n", + "A": "Dimension is the width, length, or height of an object. At 5:30 he was referring to the width of the square, saying that it s eight units", + "video_name": "FKJjqEdfB9Y", + "timestamps": [ + 330 + ], + "3min_transcript": "I can take each of them and divide by two. So what is, what is this, what is this dimension going to be, right over here? Actually let me do that in a different color. What is this dimension right over here going to be? Well two times that is going to be a hundred, so this is going to be 100. And how did we get that? Well we got that by 200 divided by two. 200 divided by two is 100. What's 60 divided by two? Well, 60 divided by two is going to be 30. So this part of the field is going to be 30 in that direction and two in this direction. And once again, not drawn to scale. And then finally, what's this section going to be? It's going to be, it's going to be, eight divided by two, which is four. Notice, hundred times two is 200, 30 times two is 60, four times two is eight. is going to be 100, plus 30, plus four, or 134. Now we've already seen other ways of coming up with this. You say, \"Look, two hundreds divided by two is 100, \"six tens divided by two is three tens, \"eight ones divided by two is four ones.\" And that's exactly what we just did over here, but we visualized it using this kind of a rectangle, breaking it up into chunks that are, maybe easier to imagine dividing by two. We broke it up into two hundreds, two hundreds right over here. We broke it up into six tens, or 60, or 60 right over here. And we broke it up into eight ones, eight ones. We broke up the area, and then we took each of those areas and we divided it by two to find that part of the length. And when you add them all together, you get the entire, you get the entire length. where you take each of the place values and you break up your field or your rectangle like that, but you could do it other ways. You don't always have to break it up that much. For example, let's say, let's say that this... let's say that this area is 856 square units. Square units... Now let's say that this dimension right over here is eight, is eight units. Is eight units. So how could we break this up so it's easier to think about what the other dimension would be, what the length is going to be? is going to be 856 divided by eight. 856 divided by eight is this length right over here. Well you could break it up in to eight hundreds, five tens, and six ones, but you might notice, \"Well five tens, it's not so easy to divide that by eight.\"" + }, + { + "Q": "where does 0 and 1 come from in 2:01? Is this a formula or rule? Why do we use 0 and 1 in this example and can other numbers be substituted?\n", + "A": "Sal just plucked them out the air. Any 2 numbers would do, which could be substituted in to the formula to workout what y should equal. I m guessing he chose small numbers so the graph he had to draw wouldn t be to big, but any 2 numbers would do 5, 7, 100 or 1000000 as long as the value of x can be used to calculate the corresponding value of y", + "video_name": "SSNA9gaAOVc", + "timestamps": [ + 121 + ], + "3min_transcript": "Is 3 comma negative 4 a solution to the equation 5x plus 2y is equal to 7? So there's two ways to think about it. One, you could just substitute this x and y value into this equation to see if it satisfies-- and then we'll do that way first-- and the other way is if you had a graph of this equation, you could see if this point sits on that graph, which would also mean that it is a solution to this equation. So let's do it the first way. So we have 5x plus 2y is equal to 7, so let's substitute. Instead of x, let us put in 3 for x. So 5 times 3 plus 2 times y-- so y is negative 4-- plus 2 times negative 4 needs to be equal to 7. I'll put a question mark here, because we're not sure yet if it does. So 5 times 3 is 15, and then 2 times negative 4 is negative 8. and this needs to be equal to 7. And of course, 15 minus 8 does equal 7, so this all works out. This is a solution, so we've answered the question. But I also want to show you, this way we just did it by substitution. If we had the graph of this equation, we could also do it graphically. So let's give ourselves the graph of this equation, and I'll do that by setting up a table. There's multiple ways to graph this. You could put it in a slope-intercept form and all of the rest, but I'll just set up a table of x and y values. And I'll graph it, and then given the graph, I want to see if this actually sits on it. And obviously it will, because we've already shown that this works. In fact, we could try the point 3, negative 4, and that actually is on the graph. We could do it on our table, but I won't do that just yet. I'm just going to do this to give ourselves a graph. So let's try it when x is equal to 0. We have 5 times 0 plus 2 times y is equal to 7. to be-- so you're going to have 0 plus 2y is equal to 7. y is going to be equal to 3.5. When x is equal to 1, you have 5 plus 2y is equal to 7. If you subtract 5 from both sides, you get 2y is equal to 2. You get y is equal to 1. So when x is 1, y is 1, and when x is-- well let's try-- well that's actually enough for us to graph. We could keep doing more points. We could even put the point 3, negative 4 there, but let's just try to graph it in this very rough sense right here. So let me draw my x-axis, and then this right over here is my y-axis." + }, + { + "Q": "\nAt 1:44, why cannot the answer be -1 to both equations. It should be simple right?", + "A": "Hey Asish, Sal is trying to demonstrated the fact that one of them must be equal to zero. -1 is not even somewhat related to making the equations zero. I will demonstrate in the equation he is using. (2x - 1) = 0 substitute -` (2 * -1 - 1) = 0 (-2 - 1) = 0 -3 = 0 This is not true, so -1 does not work in the first part. (x + 4) = 0 substitute -1 (-1 + 4) = 0 (3) = 0 3 = 0 That is not possible, so -1 does not work in this part too. Hope that helps! - JK #YouCanLearnAnything", + "video_name": "-lWVpoPaPBc", + "timestamps": [ + 104 + ], + "3min_transcript": "- [Instructor] Let's say that we've got the equation two X minus one times X plus four is equal to zero. Pause this video and see if you can figure out the X values that would satisfy this equation, essentially our solutions to this equation. Alright, now let's work through this together. So at first, you might be tempted to multiply these things out, or there's multiple ways that you might have tried to approach it, but the key realization here is that you have two things being multiplied, and it's being equal to zero. So you have the first thing being multiplied is two X minus one. This is expression is being multiplied by X plus four, and to get it to be equal to zero, one or both of these expressions needs to be equal to zero. Let me really reinforce that idea. If I had two variables, let's say A and B, and I told you A times B Well, can you get the product of two numbers to equal zero without at least one of them being equal to zero? And the simple answer is no. If A is seven, the only way that you would get zero is if B is zero, or if B was five, the only way to get zero is if A is zero. So you see from this example, either, let me write this down, either A or B or both, 'cause zero times zero is zero, or both must be zero. The only way that you get the product of two quantities, and you get zero, is if one or both of them is equal to zero. I really wanna reinforce this idea. I'm gonna put a red box around it so that it really gets stuck in your brain, and I want you to think about why that is. Try to come up with two numbers. Try to multiply them so that you get zero, is going to need to be zero. So we're gonna use this idea right over here. Now this might look a little bit different, but you could view two X minus one as our A, and you could view X plus four as our B. So either two X minus one needs to be equal to zero, or X plus four needs to be equal to zero, or both of them needs to be equal to zero. So I could write that as two X minus one needs to be equal to zero, or X plus four, or X, let me do that orange. Actually, let me do the two X minus one in that yellow color. So either two X minus one is equal to zero, or X plus four is equal to zero. X plus four is equal to zero," + }, + { + "Q": "At 10:39, you take the square root of both sides - yet keep the less than/equal sign unchanged.\nCouldn't it be that there might be a case which \"less than\" should turn into \"greater than\" ? In case ||x+y|| <0 ?\n", + "A": "|x+y| is never < 0.", + "video_name": "PsNidCBr5II", + "timestamps": [ + 639 + ], + "3min_transcript": "I said that this thing that I wrote over here, this is the same as that. So this thing up here, which is the same as that, which is less than that also. So we can write that the magnitude of x plus y squared and not the magnitude, the length of the vector x plus y squared is less than this whole thing that I wrote out here. Or less than or equal to. Now, what is this thing? Remember, I mean this might look all fancy with my little double lines around everything. But these are just numbers. This length of x squared, this is just a number. Each of these are numbers and I can just say hey, look, this looks like a perfect square to me. This term on the right-hand side is the exact same thing as the length of x plus the length of y. If you just squared this out you'll get x squared plus 2 times the length of x times the length of y plus y squared. So our length of the vector x plus y squared is less than or equal to this quantity over here. And if we just take the square root of both sides of this, you get the length of our vector x plus y is less than or equal to the length of the vector x by itself plus the length of the vector y. And we call this the triangle inequality, which you might have remembered from geometry. Now why is it called the triangle inequality? Well you could imagine each of these to be separate side of a triangle. We can draw this in R2. Let me turn my graph paper on. Let me see where the graphs show up. If I turn my graph paper on right there, maybe I'll draw it here. So let's draw my vector x. So let's say that my vector x look something like this. Let's say that's my vector x. It's a vector 2, 4. So that's my vector x. And let's say my vector y-- well I'm just going to do it head to tail because I'm going to add the two. So my vector y-- I'm going to do it in nonstandard position. Let's say it's look something like-- let's say my vector y looks something like this. Draw it properly. That's my vector y. What does x plus y look like? And remember, I can't necessarily draw any two vectors on a two-dimensional space like this." + }, + { + "Q": "\nAt 1:34, Sal multiplied both numerator and denominator by 10, however he did not multiplied the left hand side of the equation by 10 too! And so, how is the equality still maintained?", + "A": "When he multiplies the top and bottom by 10, it is equivalent to multiplying by 1, so you don t need to multiply the left hand side to maintain equality. Take a really simple example: 5 = 10/2 I can multiply the RHS by 10/10 to give: 5= 100/20 (which still holds) But I cannot multiply the LHS by just 10, ie: 50 = 100/20 which is the wrong answer But I can do this: 50/10 = 100/20 The key here is as long as you re multiplying by what is essentially 1, you don t need to multiply the other side by anything.", + "video_name": "a3acutLstF8", + "timestamps": [ + 94 + ], + "3min_transcript": "Let's get some practice solving some equations, and we're gonna set up some equations that are a little bit hairier than normal, they're gonna have some decimals and fractions in them. So let's say I had the equation 1.2 times c is equal to 0.6. So what do I have to multiply times 1.2 to get 0.6? And it might not jump out immediately in your brain but lucky for us we can think about this a little bit methodically. So one thing I like to do is say okay, I have the c on the left hand side, and I'm just multiplying it by 1.2, it would be great if this just said c. If this just said c instead of 1.2c. So what can I do there? Well I could just divide by 1.2 but as we've seen multiple times, you can't just do that to the left hand side, that would change, you no longer could say that this is equal to that if you only operate on one side. So you have to divide by 1.2 on both sides. So on your left hand side, 1.2c divided by 1.2, well that's just going to be c. You're just going to be left with c, Now what is that equal to? There's a bunch of ways you could approach it. The way I like to do it is, well let's just, let's just get rid of the decimals. Let's just multiply the numerator and denominator by a large enough number so that the decimals go away. So what happens if we multiply the numerator and the denominator by... Let's see if we multiply them by 10, you're gonna have a 6 in the numerator and 12 in the denominator, actually let's do that. Let's multiply the numerator and denominator by 10. So once again, this is the same thing as multiplying by 10 over 10, it's not changing the value of the fraction. So 0.6 times 10 is 6, and 1.2 times 10 is 12. So it's equal to six twelfths, and if we want we can write that in a little bit of a simpler way. We could rewrite that as, divide the numerator and denominator by 6, you get 1 over 2, And if you look back at the original equation, 1.2 times one half, you could view this as twelve tenths. Twelve tenths times one half is going to be equal to six tenths, so we can feel pretty good that c is equal to one half. Let's do another one. Let's say that we have 1 over 4 is equal to y over 12. So how do we solve for y here? So we have a y on the right hand side, and it's being divided by 12. Well the best way I can think of of getting rid of this 12 and just having a y on the right hand side is multiplying both sides by 12. We do that in yellow. So if I multiply the right hand side by 12, I have to multiply the left hand side by 12. And once again, why did I pick 12? Well I wanted to multiply by some number, that when I multiply it by y over 12" + }, + { + "Q": "At 2:21 my answer was 0.5 instead of 1/2 because I chose to divide the decimals. Since that's equivalent would it still be correct to write it in decimal form?\n", + "A": "I think that it would be fine if you got 0.5 instead of 1/2 because they are equivalent.", + "video_name": "a3acutLstF8", + "timestamps": [ + 141 + ], + "3min_transcript": "Let's get some practice solving some equations, and we're gonna set up some equations that are a little bit hairier than normal, they're gonna have some decimals and fractions in them. So let's say I had the equation 1.2 times c is equal to 0.6. So what do I have to multiply times 1.2 to get 0.6? And it might not jump out immediately in your brain but lucky for us we can think about this a little bit methodically. So one thing I like to do is say okay, I have the c on the left hand side, and I'm just multiplying it by 1.2, it would be great if this just said c. If this just said c instead of 1.2c. So what can I do there? Well I could just divide by 1.2 but as we've seen multiple times, you can't just do that to the left hand side, that would change, you no longer could say that this is equal to that if you only operate on one side. So you have to divide by 1.2 on both sides. So on your left hand side, 1.2c divided by 1.2, well that's just going to be c. You're just going to be left with c, Now what is that equal to? There's a bunch of ways you could approach it. The way I like to do it is, well let's just, let's just get rid of the decimals. Let's just multiply the numerator and denominator by a large enough number so that the decimals go away. So what happens if we multiply the numerator and the denominator by... Let's see if we multiply them by 10, you're gonna have a 6 in the numerator and 12 in the denominator, actually let's do that. Let's multiply the numerator and denominator by 10. So once again, this is the same thing as multiplying by 10 over 10, it's not changing the value of the fraction. So 0.6 times 10 is 6, and 1.2 times 10 is 12. So it's equal to six twelfths, and if we want we can write that in a little bit of a simpler way. We could rewrite that as, divide the numerator and denominator by 6, you get 1 over 2, And if you look back at the original equation, 1.2 times one half, you could view this as twelve tenths. Twelve tenths times one half is going to be equal to six tenths, so we can feel pretty good that c is equal to one half. Let's do another one. Let's say that we have 1 over 4 is equal to y over 12. So how do we solve for y here? So we have a y on the right hand side, and it's being divided by 12. Well the best way I can think of of getting rid of this 12 and just having a y on the right hand side is multiplying both sides by 12. We do that in yellow. So if I multiply the right hand side by 12, I have to multiply the left hand side by 12. And once again, why did I pick 12? Well I wanted to multiply by some number, that when I multiply it by y over 12" + }, + { + "Q": "2:07 How can you cross both of those lines when they are not equal to each other and call them parallel?\n", + "A": "I am not sure what you mean by not equal to each other. Lines, while he does not show it with arrows on the end, go forever.", + "video_name": "V0xounKGEXs", + "timestamps": [ + 127 + ], + "3min_transcript": "Let's think a little bit about two terms that you'll see throughout your geometry, and really, mathematical career. One is the idea of things being perpendicular. And usually, people are talking about perpendicular. Actually I'm misspelling it-- perpendicular lines, and the idea of parallel lines. So perpendicular lines are two lines that intersect at a right angle. So what am I talking about? So let's say that this is one line right over here and that this is another line right over here. We would say these two lines are perpendicular if they intersect at a right angle. So they clearly intersect. In order for them to intersect at a right angle, the angle formed between these two lines needs to be 90 degrees. And if any one of these angles is 90 degrees, the rest of them are going to be 90 degrees. And if that's 90 degrees, then that's going to be 90 degrees, that's going to be 90 degrees, and that's going to be 90 degrees. So if any of them are 90 degrees, the rest of them are 90 degrees, and we have perpendicular lines. If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines. So this line right over here and this line right over here, the way I've drawn them, are parallel lines. They aren't intersecting. They're both kind of going in the same direction, but they're kind of shifted versions of each other. They will never intersect with each other. So these two are parallel. If we have two lines that, let's say, they intersect, but they don't intersect at a right angle, so let's say we have that line and we have this line right over here, and they're clearly not intersecting at a right angle, then we call these neither perpendicular These lines just intersect." + }, + { + "Q": "\nat 1:00 doesn't the numbers in A add by itself too?", + "A": "In the example, the pattern of the numbers in A is a doubling one, that is (1, 2, 4, 8, 16, 32). This means that the increase is one of multiplication (each interval is 2 times larger than the previous one) rather than one of simple addition.", + "video_name": "Muba9-W2FOQ", + "timestamps": [ + 60 + ], + "3min_transcript": "Below are ordered pairs that represent the first six terms of two given patterns. The first value in each pair is a term from pattern A. And the second value is a term from pattern B. In the answer box, there are different statements about the two patterns. Choose all correct statements. So let's think about what's going on here. They said the first term is pattern A. So the first term in each of these coordinates is pattern A, or in each pair is pattern A. So pattern A goes from 1, to 2, to 4, to 8, to 16, to 32. So it looks like pattern A, to go from the first term to the second term, we multiplied by 2. And then to go from the second to the third term, we also multiplied by 2. And we just keep multiplying by 2. And we just keep doing that. 8 times 2 is 16. 16 times 2 is 32. Now let's think about what's going on with pattern B. And it's just always 3. So there's a couple of ways you can think about it. You could just say, pattern B's always 3. You could say pattern B starts at 3, and we're just adding 0 every time. Or you could say that pattern B starts at 3, and we are multiplying by 1 every time. Either of those would give you just 3 showing up over and over again. So now that we've looked at these pairs, we show the corresponding terms for pattern A and pattern B, let's look at the choices here and see which of these apply. In pattern A, you can get from any term to the next by multiplying by a constant number. Well, that looks right. We go from the first term to the second term by multiplying by 2. Then we multiply by 2 again to get to the third term. Then we keep multiplying by 2. So that constant number that we're multiplying by to get to the next term is 2. So this looks right. The next pair should be 52 comma 3. If we keep doubling for pattern A-- so this is going to be times 2. 32 times 2 is 64. And then if we'd say that this is 1 times the previous term, we're just going to get a 3 again. So it should be 64 comma 3 should be the next one. They say the next pair should be 52 comma 3. So that's not right. If we graph the pairs, the points will be on the same line. So let's think about that a little bit. Let's think about that. So this is my vertical axis. This is my horizontal axis. On the horizontal axis, I will graph pattern A. And on my vertical axis, I will graph pattern B. And let's see. Pattern A goes all the way up to 32. So I'm going to try my best here. So let's say that this is 32. Then half of that is going to be 16. Half of that is going to be 8." + }, + { + "Q": "At 1:17, Sal said something about a \"negative squared is just going to be a 1\". I can't wrap my head around how this actually happened in that particular equation (y= -sqrt(x-3)). Anyone who can clarify this one?\n", + "A": "What he s talking about is the assumed coefficient in front of the radical. Writing -sqrt(x-3) is the same thing as writing -1 * sqrt(x-3). So if you square -1, you get a positive 1.", + "video_name": "QWLcNxQ3KvQ", + "timestamps": [ + 77 + ], + "3min_transcript": "In the relation x is equal to y squared plus 3, can y be represented as a mathematical function of x? So the way they've written it, x is being represented as a mathematical function of y. We could even say that x as a function of y is equal to y squared plus 3. Now, let's see if we can do it the other way around, if we can represent y as a function of x. So one way you could think about it is you could essentially try to solve for y here. So let's do that. So I have x is equal to y squared plus 3. Subtract 3 from both sides, you get x minus 3 is equal to y squared. Now, the next step is going to be tricky, x minus 3 is equal to y squared. So y could be equal to-- and I'm just going to swap the sides. y could be equal to-- if we take the square root of both sides, it could be the positive square root of x minus 3, or it could be the negative square root. If you don't believe me, square both sides of this. You'll get y squared is equal to x minus 3. Square both sides of this, you're going to get y squared is equal to-- well, the negative squared is just going to be a positive 1. And you're going to get y squared is equal to x minus 3. So this is a situation here where for a given x, you could actually have 2 y-values. Let me show you. Let me attempt to sketch this graph. So let's say this is our y-axis. I guess I could call it this relation. This is our x-axis. And this right over here, y is a positive square root of x minus 3. That's going to look like this. So if this is x is equal to 3, it's going to look like this. That's y is equal to the positive square root of x minus 3. And this over here, y is equal to the negative square root of x minus 3, is going to look something like this. because it's going to essentially be the mirror image if you flip over the x-axis. So it's going to look something like this-- y is equal to the negative square root of x minus 3. And this right over here, this relationship cannot be-- this right over here is not a function of x. In order to be a function of x, for a given x it has to map to exactly one value for the function. But here you see it's mapping to two values of the function. So, for example, let's say we take x is equal to 4. So x equals 4 could get us to y is equal to 1. 4 minus 3 is 1. Take the positive square root, it could be 1. Or you could have x equals 4, and y is equal to negative 1. So you can't have this situation. If you were making a table x and y as a function of x," + }, + { + "Q": "\nAt 5:44, where did the three came from?", + "A": "There was three chairs", + "video_name": "DROZVHObeko", + "timestamps": [ + 344 + ], + "3min_transcript": "five different people in five different chairs, and we cared which seat they sit in, we had this five factorial. Factorial is kind of neat little operation there. How can I relate factorial to what we did just now? It looks like we kind of did factorial, but then we stopped. We stopped at, we didn't go times two times one. So one way to think about what we just did, is we just did five times four times three times two times one, but of course we actually didn't do the two times one, so you could take that and you could divide by two times one. If you did that, this two times one would cancel with that two times one and you'd be left with five times four times three. The whole reason I'm writing this way is that now I can write it in terms of factorial. I could write this as five factorial, five factorial, over two factorial, over two factorial. But then you might have the question I have three seats. Where did this two come from? Well, think about it. I multiplied five times four times three, I kept going until I had that many seats, and then I didn't do the remainder. So the things that I left out, the things that I left out, that was essentially the number of people minus the number of chairs, so I was trying to put five things in three places. Five minus three, that gave me two left over. So I could write it like this. I could write it as five ... Let me use the same colors. I could write it as five factorial over, over five minus three, which of course is two, five minus three factorial. Another way of thinking about it, if we wanted to generalize, is if you're trying to figure out the number of permutations and there's a bunch of notations for writing this, if you're trying to figure out the number of permutations or the number of permutations you could put n people in r seats, and there's other notations as well, well, this is just going to be n factorial over n minus r factorial. Here n was five, r was three. Five minus three is two. Now, you'll see this in a probability or a statistics class, and people might memorize this thing. It seems like this kind of daunting thing. I'll just tell you right now, the whole reason why I just showed this to you is so that you could connect it with what you might see in your textbook, or what you might see in a class, or when you see this type of formula, you see that it's not some type of voodoo magic. But I will tell that for me, personally, I never use this formula. I always reason it through, because if you just memorize the formula, you're always going, wait, does this formula apply there? What's n? What's r?" + }, + { + "Q": "\nAt 4:51, when Sal writes 5!/2!, couldn't he just write (5/2)! ?", + "A": "No, Sal could not do this. Factorial does not distribute over division. Example: 4!/2! = (1*2*3*4)/(1*2) = 12, but (4/2)! = 2! = 2*1 = 2. In this particular situation, 5!/2! = (1*2*3*4*5)/(1*2) = 60, but (5/2)! = 2.5! which turns out be approximately 3.32 from using a calculator (the definition of fractional factorials is much more complex than the definition of integer factorials). Have a blessed, wonderful day!", + "video_name": "DROZVHObeko", + "timestamps": [ + 291 + ], + "3min_transcript": "so four people could sit in seat two. So we have five times four scenarios where we've seated seats one and seat two. For each of those 20 scenarios, how many people could sit in seat three? Well, we haven't sat, we haven't seaten or sat three of the people yet, so for each of these 20, we could put three different people in seat three, so that gives us five times four times three scenarios. So this is equal to five times four times three scenarios, which is equal to, this is equal to 60. So there's 60 permutations of sitting five people in three chairs. Now this, and my brain, whenever I start to think in terms of permutations, I actually think in these ways. I just literally draw it out because I don't like formulas. I like to actually conceptualize and visualize what I'm doing. five different people in five different chairs, and we cared which seat they sit in, we had this five factorial. Factorial is kind of neat little operation there. How can I relate factorial to what we did just now? It looks like we kind of did factorial, but then we stopped. We stopped at, we didn't go times two times one. So one way to think about what we just did, is we just did five times four times three times two times one, but of course we actually didn't do the two times one, so you could take that and you could divide by two times one. If you did that, this two times one would cancel with that two times one and you'd be left with five times four times three. The whole reason I'm writing this way is that now I can write it in terms of factorial. I could write this as five factorial, five factorial, over two factorial, over two factorial. But then you might have the question I have three seats. Where did this two come from? Well, think about it. I multiplied five times four times three, I kept going until I had that many seats, and then I didn't do the remainder. So the things that I left out, the things that I left out, that was essentially the number of people minus the number of chairs, so I was trying to put five things in three places. Five minus three, that gave me two left over. So I could write it like this. I could write it as five ... Let me use the same colors. I could write it as five factorial over, over five minus three, which of course is two, five minus three factorial. Another way of thinking about it, if we wanted to generalize, is if you're trying to figure out the number of permutations and there's a bunch of notations for writing this, if you're trying to figure out the number of permutations" + }, + { + "Q": "At 2:10 , what does Sal mean by \"0 degree terms?\"\n", + "A": "Basically, the degree refers to the power of the term s variable. So 2x is a 1-degree term and 3x^2 is a second degree term. 0-degree terms simply are constant terms, like the 5 in 4x + 5. They are 0-degree because you can think of the 5 as 5x^0, which is just 5*1.", + "video_name": "n34dqyVCXs4", + "timestamps": [ + 130 + ], + "3min_transcript": "f of x is equal to 9 minus x. Actually, it should be 9 minus x squared. And g of x is equal to 5x squared plus 2x plus 1. And then they say find f plus g of x. And this looks a little bit bizarre. What kind of notation is this? And that's really the core of this problem is just to realize that when someone writes f plus g in parentheses like this of x this is just notation for-- so f plus g of x, I just rewrote it. This is the same thing. This is just a kind of shorthand notation for f of x plus g of x. You could view f plus g as a new function that's created by adding the other two functions. But when you view it like this-- so this is really what we have to find. Then, you just have to add these two functions. So f of x, they've given the definition right over there, is 9 minus x squared. And g of x, they've given the definition right over here, So when you add f of x to g of x, this is going to be equal to-- and I'm just rewriting a lot of things just to make it clear. The f of x part is 9 minus x squared. And then you have the plus. I'll do that in yellow. Plus the g of x part, which is 5x squared plus 2x plus 1. And now we can just simplify this a little bit. We can say-- let me just get rid of the parentheses. I'll just rewrite it. So this is equal to 9 minus x squared plus-- since this is a positive, we don't have to worry about the parentheses. So plus 5x squared plus 2x plus 1. And then we have two x squared terms or second degree terms. We have 5x squared. And then we have negative x squared. 5x squared minus x squared is 4x squared. So you get 4x squared when you combine these two terms. And then you have two constant terms, or 0 degree terms. So you have the 9 and you have the 1. And 9 plus 1 gets us to 10. So we're done. f plus g of x is equal to 4x squared plus 2x plus 10. Notice this is a new function that's created by summing the function definitions of f and g." + }, + { + "Q": "At 1:10, how did you determine the number range for the buckets?\n", + "A": "Looking that the time you referenced, do you mean the ranges of ages that Sal put in the t-chart? If so, the reason why is that is it is really easy to group everything in tens. There is no right way to determine the number range. It is just a matter of preference.", + "video_name": "gSEYtAjuZ-Y", + "timestamps": [ + 70 + ], + "3min_transcript": "- [Voiceover] So let's say you were to go to a restaurant and just out of curiosity you want to see what the makeup of the ages at the restaurant are. So you go around the restaurant and you write down everyone's age. And so these are the ages of everyone in the restaurant at that moment. And so you're interested in somehow presenting this, somehow visualizing the distribution of the ages, because you want just say, well, are there more young people? Are there more teenagers? Are there more middle-aged people? Are there more seniors here? And so when you just look at these numbers it really doesn't give you a good sense of it. It's just a bunch of numbers. And so how could you do that? Well one way to think about it, is to put these ages into different buckets, and then to think about how many people are there in each of those buckets? Or sometimes someone might say how many in each of those bins? So let's do that. So let's do buckets or categories. So, I like, sometimes it's called a bin. So the bucket, I like to think of it more of as a bucket, the bucket and then the number in the bucket. The number in the bucket. It's the, oops. It's the number (laughing), it's the number in the bucket. Alright. So let's just make buckets. Let's make them 10 year ranges. So let's say the first one is ages zero to nine. So how many people... Why don't we just define all of the buckets here? So the next one is ages 10 to 19, then 20 to 29, then 30 to 39, and 40 to 49, 50 to 59, let me make sure you can read that properly, then you have 60 to 69. And I think that covers everyone. I don't see anyone 70 years old or older here. So then how many people fall into the zero to nine-year-old bucket? Well it's gonna be one, two, three, four, five, six people fall into that bucket. How many people fall into the... How many people fall into the 10 to 19-year-old bucket? One, two, three. Three people. And I think you see where this is going. What about 20 to 29? So that's one, two, three, four, five people. Five people fall into that bucket. Alright, what about 30 to 39? We have one, and that's it. Only one person in that 30 to 39 bin or bucket or category. Alright, what about 40 to 49? We have one, two people. Two people are in that bucket. And then 50 to 59. Let's see, you have one, two people. Two people. And then finally, finally, ages 60-69. Let me do that in a different color. 60 to 69. There is one person, right over there." + }, + { + "Q": "\nIsn't it enough to just stop at 4:35, the part where you claim a is a multiple of an integer k times p. Since a/b is said to be 2 integers with no factors in common in the beginning of the video wouldn't a=kp be impossible, since p is prime then this claims p=a/k, and a prime cannot be defined as a integer divided by an integer?", + "A": "Any prime p can be represented as p/1, both of which are integers.", + "video_name": "W-Nio466Ek4", + "timestamps": [ + 275 + ], + "3min_transcript": "I'm just saying that a is some integer right over here. So that's the prime factorization of a. What is the prime factorization of a squared going to be? Well, a squared is just a times a. Its prime factorization is going to be f1 times f2, all the way to fn. And then that times f1 times f2 times, all the way to fn. Or I could rearrange them if I want. f1 times f1 times f2 times f2, all the way to fn times fn. Now, we know that a squared is a multiple of p. p is a prime number, so p must be one of these numbers in the prime factorization. p could be f2, or p could be f1, but p needs to be one of these numbers in the prime factorization. So p needs to be one of these factors. pick one of these arbitrarily. Let's say that p is f2. If p is f2, then that means that p is also a factor of a. So this allows us to deduce that a is a multiple of p. Or another way of saying that is that we can represent a as being some integer times p. Now, why is that interesting? And actually, let me box this off, because we're going to reuse this part later. But how can we use this? Well, just like we did in the proof of the square root of 2 being irrational, let's now substitute this back into this equation right over here. So we get b squared times p. We have b squared times p is equal to a squared. that as some integer k times p. So we can rewrite that as some integer k times p. And so, let's see, if we were to multiply this out. So we get b squared times p-- and you probably see where this is going-- is equal to k squared times p squared. We can divide both sides by p, and we get b squared is equal to p times k squared. Or k squared times p. Well, the same argument that we used, if a squared is equal to b squared times p, that let us know that a squared is a multiple of p. So now we have it the other way around." + }, + { + "Q": "i didn't understand how b\u00c2\u00b2=k\u00c2\u00b2p in 5:45\n", + "A": "If you understand why a = kp, then it s just simple algebra from the video, starting around 5:14. If you don t know why a = kp, ask about that.", + "video_name": "W-Nio466Ek4", + "timestamps": [ + 345 + ], + "3min_transcript": "pick one of these arbitrarily. Let's say that p is f2. If p is f2, then that means that p is also a factor of a. So this allows us to deduce that a is a multiple of p. Or another way of saying that is that we can represent a as being some integer times p. Now, why is that interesting? And actually, let me box this off, because we're going to reuse this part later. But how can we use this? Well, just like we did in the proof of the square root of 2 being irrational, let's now substitute this back into this equation right over here. So we get b squared times p. We have b squared times p is equal to a squared. that as some integer k times p. So we can rewrite that as some integer k times p. And so, let's see, if we were to multiply this out. So we get b squared times p-- and you probably see where this is going-- is equal to k squared times p squared. We can divide both sides by p, and we get b squared is equal to p times k squared. Or k squared times p. Well, the same argument that we used, if a squared is equal to b squared times p, that let us know that a squared is a multiple of p. So now we have it the other way around. which is still going to be an integer, times p. So b squared must be a multiple of p. So this lets us know that b squared is a multiple of p. And by the logic that we applied right over here, that lets us know that b is a multiple of p. And that's our contradiction, or this establishes our contradiction that we assumed at the beginning. We assumed that a and b are co-prime, that they share no factors in common other than 1. We assumed that this cannot be reduced. But we've just established, just from this, we have deduced that is a multiple of p and b is a multiple of p. Which means that this fraction can be reduced. We can divide the numerator and the denominator by p. So that is our contradiction. We started assuming it cannot be reduced," + }, + { + "Q": "Just noticed that in this vid about 9:13, when simplifying and rearranging, the \"b\" term was dropped. Just so everyone watching does not get too lost.\nThanks for the vids tho. Actually beginning to understand why I did this the first time.\n", + "A": "I saw that too, good that you pointed it out.", + "video_name": "9kW6zFK5E5c", + "timestamps": [ + 553 + ], + "3min_transcript": "What I want to do is I want to multiply this bottom equation times 3 and add it to this middle equation to eliminate this term right here. So if I multiply this bottom equation times 3-- let me just do-- well, actually, I don't want to make things messier, so this becomes a minus 3 plus a 3, so those cancel out. This becomes a 12 minus a 1. So this becomes 12c3 minus c3, which is 11c3. And then this becomes a-- oh, sorry, I was already done. When I do 3 times this plus that, those canceled out. And then when I multiplied 3 times this, I get 12c3 minus a c3, so that's 11c3. And I multiplied this times 3 plus this, so I get 3c minus this, plus b plus a. So what can I rewrite this by? Actually, I want to make something very clear. This c is different than these c1's, c2's and c3's that I had up here. I think you realize that. But I just realized that I used the letters c twice, and I just didn't want any confusion here. So this c that doesn't have any subscript is a different constant then all of these things over here. Let's see if we can simplify this. We have an a and a minus 6a, so let's just add them. So let's get rid of that a and this becomes minus 5a. If we divide both sides of this equation by 11, what do we get? We get c3 is equal to 1/11 times 3c minus 5a. tell you what c3 is. What is c2? c2 is equal to-- let me simplify this equation right here. Let me do it right there. So if I just add c3 to both sides of the equation, I get 3c2 is equal to b plus a plus c3. And if I divide both sides of this by 3, I get c2 is equal to 1/3 times b plus a plus c3. I'll just leave it like that for now. Then what is c1 equal to? I could just rewrite this top equation as if I subtract 2c2 and add c3 to both sides, I get c1 is equal to a" + }, + { + "Q": "\naround 1:54, could you be more specific when you say NO X Value is equal to -14", + "A": "The question was 24x + 80 = 24x - 14 What he is saying is that it doesn t matter what value you give to 24x, there is no value that you could give to the x and then multiply that number by 24 so that 24x would then make 80 = -14", + "video_name": "zKotuhQWIRg", + "timestamps": [ + 114 + ], + "3min_transcript": "Solve for x. We have 8 times the quantity 3x plus 10 is equal to 28x minus 14 minus 4x. So like every equation we've done so far, we just want to isolate all of the x's on one side of this equation. But before we do that, we can actually simplify each of these sides. On the left-hand side, we can multiply the quantity 3x plus 10 times 8. So we're essentially just distributing the 8, the distributive property right here. So this is the same thing as 8 times 3x, which is 24x, plus 8 times 10, which is 80, is equal to-- and over here, we have 28x minus 14 minus 4x. So we can combine the 28x and the minus 4x. If we have 28x minus 4x, that is 24x And then you have the minus 14 right over here. Now, the next thing we could-- and it's already looking a little bit suspicious, but just to confirm that it's as suspicious as it looks, let's try to subtract 24x from both sides of this equation. And if we do that, we see that we actually remove the x's and we have a 24x there. You might say, hey, let's put all the x's on the left-hand side. So let's get rid of this 24x. So you subtract 24x right over there, but you have to do it to the left-hand side as well. On the left-hand side, these guys cancel out, and you're left with just 80-- these guys cancel out as well-- is equal to a negative 14. Now, this looks very bizarre. It's making a statement that 80 is equal to negative 14, which we know is not true. This does not happen. 80 is never equal to negative 14. They're just inherently inequal. So this equation right here actually has no solution. This has no solution. There is a no x-value that will make 80 equal to negative 14." + }, + { + "Q": "\nIs there a proper name for the operation Sal calls a \"slash\" at 4:19?", + "A": "Not like. That is a backslash. Pretty simple character. There can t really be any misinterpreting it, though there could always be additional names for it like anything.", + "video_name": "2B4EBvVvf9w", + "timestamps": [ + 259 + ], + "3min_transcript": "relative complement of set B in A. And we're going to talk a lot more about complements in the future. But the complement is the things that are not in B. And so this is saying, look, what are all of the things that are not in B-- so you could say what are all of the things not in B but are in A? So once again, if you said all of the things that aren't in B, then you're thinking about all of the numbers in the whole universe that aren't 17, 19, or 6. And actually, you could even think broader. You're not even just thinking about numbers. It could even be the color orange is not in set B, so that would be in the absolute complement of B. I don't see a zebra here in set B, so that would be its complement. But we're saying, what are the things that are not in B but are in A? And that would be the numbers 5, 3, and 12. Now, when we visualized this as B subtracted from A, OK. I could imagine you took the 17 out. You took the 19 out. But what about taking the 6 out? Shouldn't you have taken a 6 out? Or in traditional subtraction, maybe we would end up with a negative number or something. And when you subtract a set, if the set you're subtracting from does not have that element, then taking that element out of it doesn't change it. If I start with set A, and if I take all the 6s out of set A, it doesn't change it. There was no 6 to begin with. I could take all the zebras out of set A; it will not change it. Now, another way to denote the relative complement of set B in A or B subtracted from A, is the notation that I'm about to write. We could have written it this way. A and then we would have had this little figure like this. That looks eerily like a division sign, but this also means the difference between set A and B where we're talking about-- when we write it this way, we're talking about all the things in set A that are not in Or the relative complement of B in A. Now, with that out of the way, let's think about things the other way around. What would B slash-- I'll just call it a slash right over here. What would B minus A be? So what would be B minus A? Which we could also write it as B minus A. What would this be equal to? Well, just going back, we could view this as all of the things in B with all of the things in A taken out of it. Or all of the things-- the complement of A that happens to be in B. So let's think of it as the set B with all of the things in A taken out of it. So if we start with set B, we have a 17." + }, + { + "Q": "At 6:05 the symbol for a null set is written as a 0 with a slash going through it. The slash is with the top to the right ( / ). Can it also be written with the top on the left ( \\ ) ?\n", + "A": "It s better to donate null set as { } or {0}. However, both slash thing is also correct.", + "video_name": "2B4EBvVvf9w", + "timestamps": [ + 365 + ], + "3min_transcript": "Or the relative complement of B in A. Now, with that out of the way, let's think about things the other way around. What would B slash-- I'll just call it a slash right over here. What would B minus A be? So what would be B minus A? Which we could also write it as B minus A. What would this be equal to? Well, just going back, we could view this as all of the things in B with all of the things in A taken out of it. Or all of the things-- the complement of A that happens to be in B. So let's think of it as the set B with all of the things in A taken out of it. So if we start with set B, we have a 17. Then we have a 19. But there's a 19 in set A, so we have to take the 19 out. Then we have a 6. Oh, well, we don't have to take a 6 out of B because the 6 is not in set A. So we're left with just the 6. So this would be just the set with a single element in it, set 6. Now let me ask another question. What would the relative complement of A in A be? Well, this is the same thing as A minus A. And this is literally saying, let's take set A and then take all of the things that are in set A out of it. Well, I start with the 5. Oh, but there's already a 5. There's a 5 in set A. So I have to take the 5 out. Well, there's a 3, but there's a 3 in set A, so I have to take a 3 out. So I'm going to take all of these things out. And so I'm just going to be left with the empty set, often called the null set. will look like this, the null set, the empty set. There's a set that has absolutely no objects in it." + }, + { + "Q": "At 3:35, he uses (f(c - h), f(c + h)) to symbolize getting near f(c). However, wouldn't it be more accurate to write:\nlimit as h approaches c of f(h) < f(c) For maximum\nlimit as h approaches c of f(h) > f(c) For minimum\n", + "A": "I agree with the strictly less than/greater than, otherwise f(c) could be equal to surrounding f(x) values (as implied by the or equal to notation on the inequalities)", + "video_name": "Hoyv3-BMAGc", + "timestamps": [ + 215 + ], + "3min_transcript": "there's definitely points that are lower. And we hit an absolute minimum for the interval at x is equal to b. But this is a relative minimum or a local minimum because it's lower than the-- if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values. And you're at a minimum if you're at a smaller value than any of the surrounding areas. But how could we write that mathematically? So here I'll just give you the definition that really is just a more formal way of saying what we just said. So we say that f of c is a relative max, relative maximum value, if f of c x that-- we could say in a casual way, for all x near c. So we could write it like that. But that's not too rigorous because what does it mean to be near c? And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0. So does that make sense? Well, let's look at it. So let's construct an open interval. So it looks like for all of the x values in-- and you just have to find one open interval. There might be many open intervals where this is true. But if we construct an open interval that is c plus h. That value right over here c minus h. And you see that over that interval, the function at c, f of c is definitely greater than or equal to the value of the function over any other part of that open interval. And so you could imagine-- I encourage you to pause the video, and you could write out what the more formal definition of a relative minimum point would be. Well, we would just write-- let's take d as our relative minimum. We can say that f of d is a relative minimum point if f of d is less than or equal to f of x for all x in an interval, in an open interval, between d minus h and d plus h for h is greater than 0. So you can find an interval here. So let's say this is d plus h. This is d minus h. The function over that interval, f of d is always" + }, + { + "Q": "\nAt 2:46 the lines are not parallel because they have different slope.\nJust find the slope for both lines.", + "A": "No, they have the same slope - it is 4 for both.", + "video_name": "BNHLzEv6Mjg", + "timestamps": [ + 166 + ], + "3min_transcript": "Or I guess another way to think about it, it represents an x and y value that satisfy both of these constraints. So one system that has one solution is the system that has y is equal to 0.1x plus 1, and then this blue line right here, which is y is equal to 4x plus 10. Now, they only want us to identify one system of two lines that has a single solution. We've already done that. But just so you see it, there's actually another system here. So this is one system right here, or another system would be the green line and this red line. This point of intersection right here, once again, that represents an x and y value that satisfies both the equation y is equal to 0.1x plus 1, and this point right here satisfies the equation y is equal to 4x minus 6. there's one point of intersection of these two equations or these two lines, and this system also has one solution because it has one point of intersection. Now, the second part of the problem, they say identify one system of two lines that does not have a single solution or does not have a solution, so no solution. So in order for there to be no solution, that means that the two constraints don't overlap, that there's no point that is common to both equations or there's no pair of x, y values that's common to both equations. And that's the case of the two parallel lines here, this blue line and this green line. Because they never intersect, there's no coordinate on the coordinate plane that satisfies both equations. So there's no x and y that satisfy both. So the second part of the question, a system that has no solution is y is equal to 4x plus 10, and then the other And notice, they have the exact same slope, and they're two different lines, they have different intercepts, so they never, ever intersect, and that's why they have no solutions." + }, + { + "Q": "At 0:48, Sal divides it by 8 but why does he multiply it by 8 also?\n", + "A": "He is just factoring out the 8; it is still the same answer. -> 8+4+2 is the same as 2(4+2+1).", + "video_name": "GMoqg_s4Dl4", + "timestamps": [ + 48 + ], + "3min_transcript": "Factor 8k squared minus 24k minus 144. Now the first thing we can do here, just eyeballing each of these terms, if we want to simplify it a good bit is all of these terms are divisible by 8. Clearly, 8k squared is divisible by 8, 24 is divisible by 8, and 144-- it might not be as obvious is divisible by 8-- but it looks like it is. 8 goes into one and 144, 8 goes into 14 one time. 1 times 8 is 8. Subtract, you get a 6. 14 minus 8 is 6. Bring down the 4. 8 goes into 64 eight times. So it goes into 144 18 times. So let's just factor out an 8 of this. And then that will simplify our expression. It will actually give us a leading 1 coefficient. So this will become 8 times k squared minus 24 divided by 8 is 3k minus 18. And remember if anything has the form x squared plus bx plus c, where you have a leading 1 coefficient-- this is implicitly a one-- we have that here in this expression in parentheses. Then we literally just need to-- and we can do this multiple ways-- but we need to find two numbers whose sum is equal to the coefficient on x. So two numbers whose sum is equal to negative 3 and whose product is equal to the constant term. And whose product is equal to negative 18. So let's just think about the factors of negative 18 here. Let's see if we can do something interesting. So it could be 1. And since it's negative, one of the numbers has to be positive, one has to be negative 1 and 18 is if it was positive. And then one of these could be positive and then one of these could be negative. But no matter what if this is negative If you switch them, then they add up to negative 17. So those won't work. So either we could write it this way, positive or negative 1, and then negative or positive 18 to show that they have to be different signs. So those don't work. Then you have positive or negative 3. And then negative or positive 6, just to know that they are different signs. So if you have positive 3 and negative 6, they add up to negative 3 which is what we need them to add up to. And clearly, positive 3 and negative 6, their product is negative 18. So it works. So we're going to go with positive 3 and negative 6 as our two numbers. Now, for this example-- just for the sake of this example-- We'll do this by grouping. So what we can do is we can separate this middle term right here as the sum of 3k negative 6k. So I could write the negative 3k as plus 3k minus 6k. And then let me write the rest of it." + }, + { + "Q": "\nAre the charts mentioned at the 9:00 marker something that is given or would I have to fill it in myself while solving the problem. If so, how would i go about doing that?", + "A": "You do not memorize the values. If you are doing homework/classwork, the tables are provided at the end of the book. On a test, the table should be attached to the test or the teacher should allow you to look up the critical value for a specified level of significance and degrees of freedom using the text. I hope this helps.", + "video_name": "2QeDRsxSF9M", + "timestamps": [ + 540 + ], + "3min_transcript": "Plus 45 minus 40 is 5 squared is 25. So plus 25 over 40. Plus the difference here is 3 squared is 9, so it's 9 over 60. Plus we have a difference of 10 squared is plus 100 over 30. And this is equal to-- and I'll just get the calculator out for this-- this is equal to, we have 100 divided by 20 plus 36 divided by 20 plus 16 divided by 30 plus 25 divided by 40 plus 9 divided by 60 plus 100 divided by 30 gives us 11.44. So this right here is going to be 11.44. This is my chi-square statistic, or we could call it a big capital X squared. Sometimes you'll have it written as a chi-square, but this statistic is going to have approximately a chi-square distribution. Anyway, with that said, let's figure out, if we assume that it has roughly a chi-square distribution, what is the probability of getting a result this extreme or at least this extreme, I guess is another way of thinking about it. Or another way of saying, is this a more extreme result than the critical chi-square value that there's a 5% chance of getting a result that extreme? So let's do it that way. Let's figure out the critical chi-square value. And if this is more extreme than that, then we will reject our null hypothesis. So let's figure out our critical chi-square values. So we have an alpha of 5%. And actually the other thing we have to figure out is the degrees of freedom. five, six sums, so you might be tempted to say the degrees of freedom are six. But one thing to realize is that if you had all of this information over here, you could actually figure out this last piece of information, so you actually have five degrees of freedom. When you have just kind of n data points like this, and you're measuring kind of the observed versus expected, your degrees of freedom are going to be n minus 1, because you could figure out that nth data point just based on everything else that you have, all of the other information. So our degrees of freedom here are going to be 5. It's n minus 1. So our significance level is 5%. And our degrees of freedom is also going to be equal to 5. So let's look at our chi-square distribution. We have a degree of freedom of 5. We have a significance level of 5%. And so the critical chi-square value is 11.07." + }, + { + "Q": "\nAt 5:31, Sal decides to use the line as a transversal. If the point of this video is to prove that a line has a constant slope, wouldn't it be bad science to use an assumption of a line's constant slope to prove it has a constant slope?", + "A": "Using the line as a transversal is making no assumptions about the slope. All he is doing here is using the fact that if you have a line that crosses two parallel lines (in other words, if you have a transversal) that the corresponding angles are congruent. This is something we know from geometry. From here he is able to prove that the triangles are similar and thus the equation for the slope is the same for both segments.", + "video_name": "24WMbh1BBKc", + "timestamps": [ + 331 + ], + "3min_transcript": "if you can establish that 2 triangles are similar, then the ratio between corresponding sides is going to be the same. So if these 2 are similar, then the ratio of this side to this side is going to be the same as the ratio of-- let me do that pink color-- this side to this side. And so you can see why that will be useful in proving that the slope is constant here, because all we have to do is look. If these 2 triangles are similar, then the ratio between corresponding sides is always going to be the same. We've picked 2 arbitrary sets of points. Then this would be true, really, for any 2 arbitrary set of points across the line. It would be true for the entire line. So let's try to prove similarity. So the first thing we know is that both of these are right triangles. These green lines are perfectly horizontal. These purple lines are perfectly vertical because the green lines literally go in the horizontal direction. The purple lines go in the vertical direction. So we know that these are both right angles. So we have 1 corresponding angle that is congruent. Now we have to show that the other ones are. And we can show that the other ones are using our knowledge of parallel lines and transversals. Let's look at these 2 green lines. So I'll continue them. These are line segments, but if we view them as lines and we just continue them, on and on and on. So let me do that, just like here. So this line is clearly parallel to that 1. They essentially are perfectly horizontal. And now you can view our orange line as a transversal. And if you view it as a transversal, then you know that this angle corresponds to this angle. And we know from transversals of parallel lines that corresponding angles are congruent. So this angle is going to be congruent to that angle right over there. Now, we make a very similar argument for this angle, but now we use the 2 vertical lines. We know that this segment, we could continue it as a line. So we could continue it, if we wanted, as a line, And we could continue this one as a vertical line. We know that these are both vertical. They're just measuring-- they're exactly in the y direction, the vertical direction. So this line is parallel to this line right over here. Once again, our orange line is a transversal of it. And this angle corresponds to this angle right over here. And there we have it. Corresponding angles of the transversal of 2 parallel lines are congruent. We learned that in geometry class. And there you have it. All of the corresponding-- this angle is congruent to this angle. This angle is congruent to that angle. And then both of these are 90 degrees. So both of these are similar triangles. Just let me write that down so we know that these are both similar triangles. And now we can use the common ratio of both sides. So for example, if we called this side length a." + }, + { + "Q": "at 4:30 why does Sal move 4,-1 clock wise for the rotation example? Is negative clockwise and positive counter?\n\nThanks\n", + "A": "Just to answer your question, when you rotate it counter-clockwise it is positive. when it is rotated clockwise it is negative. I hope this answers your question :)", + "video_name": "_eAWDuLYVfg", + "timestamps": [ + 270 + ], + "3min_transcript": "So that one. And once again, I'm just eyeballing it. So a line that has slope of positive 3/2. So this one looks right in between the two. Or actually it could be someplace in between. But either way, we just have to think about it qualitatively. If you had a line that looked something like that, and if you were to reflect over this line, then this point would map to this point, which is what we want. And this purple point, negative 5 comma 5, would map to that point. It would be reflected over. So it's pretty clear that this could be a reflection. Now rotation actually makes even more sense, or at least in my brain makes a little more sense. If you were to rotate around to this point right over here, this point would map to that point, and that point would map to that point. So a rotation also seems like a possibility for transformation C. Now let's think about transformation Actually maybe I'll put that in magenta, as well. To 7, negative 3, just like that. And we want to go from negative 5, 5 to negative 2, 3. So I could definitely imagine a translation right over here. This point went 3 to the right and 2 down. This point went 3 to the right and 2 down. So a translation definitely makes sense. Now let's think about a reflection. So it would be tempting to-- let's I could reflect around that, but that won't help this one over here. And to get from that point to that point, I could reflect around that, but once again, that's not going to help that point over there. So a reflection really doesn't seem to do the trick. And what about a rotation? Well to go from this point to this point, we could rotate around this point. We could go there, but that won't help this point right While this is rotating there, this point is going to rotate around like that and it's going to end up someplace out here. So that's not going to help. So it looks like this one can only be a translation." + }, + { + "Q": "\nIn 1:15, why can't it be a translation? I didn't understand.", + "A": "Because one point would have to move 6 to the right and 4 down. While the other would move 12 to the right and 8 down. In a translation, all points should be moved the same distance.", + "video_name": "_eAWDuLYVfg", + "timestamps": [ + 75 + ], + "3min_transcript": "Transformation C maps negative 2, 3 to 4, negative 1. So let me do negative 2 comma 3, and it maps that to 4, negative 1. And point negative 5 comma 5, it maps that to 7, negative 3. And so let's think about this a little bit. How could we get from this point to this point, and that point to that point? Now it's tempting to view this that maybe a translation is possible. Because if you imagined a line like that, you could say, hey, let's just shift this whole thing down and then to the right. These two things happen to have the same slope. They both have a slope of negative 2/3, and that point would map to that point. But that's not what we want. We don't want negative 2, 3 to map to 7, negative 3. We want negative 2, 3 to map to 4, negative 1. So you could get this line over this line, but we won't map the points that we want to map. So this can't be, at least I can't think of a way, that this could actually be a translation. Now let's think about whether our transformation could be a reflection. Well, if we imagine a line that has-- let's see, these both have a slope of negative 3. These both have a slope of negative 2/3. So if you imagined a line that had a slope of positive 3/2 that was equidistant from both-- and I don't know if this is. Let's see, is this equidistant? Is this equidistant from both of them? It's either going to be that line or this line right over-- So that one. And once again, I'm just eyeballing it. So a line that has slope of positive 3/2. So this one looks right in between the two. Or actually it could be someplace in between. But either way, we just have to think about it qualitatively. If you had a line that looked something like that, and if you were to reflect over this line, then this point would map to this point, which is what we want. And this purple point, negative 5 comma 5, would map to that point. It would be reflected over. So it's pretty clear that this could be a reflection. Now rotation actually makes even more sense, or at least in my brain makes a little more sense. If you were to rotate around to this point right over here, this point would map to that point, and that point would map to that point. So a rotation also seems like a possibility for transformation C. Now let's think about transformation" + }, + { + "Q": "\nI got lost at around 6:00\nCan someone please help me?", + "A": "Do you have a more specific question? I can help if I know how : )", + "video_name": "dvoHB9djouc", + "timestamps": [ + 360 + ], + "3min_transcript": "And to that, I'm going to add the squared distance between 5 and my mean. And since I'm squaring, it doesn't matter if I do 5 minus 6, or 6 minus 5. When I square it, I'm going to get a positive result regardless. And then, to that I'm going to add the squared distance between 7 and my mean. So 7 minus 6 squared. All of this, this is my population mean that I'm finding the difference between. And then, finally, the squared difference between 14 and my mean. And then, I'm going to find, essentially, the mean of these squared distances. So I have five squared distances right over here. So let me divide by 5. So what will I get when I make this calculation, right over here? This is going to be equal to 1 minus 6 is negative 5, negative 5 squared is 25. 3 minus 6 is negative 3, now if I square that, I get 9. 5 minus 6 is negative 1, if I square it, I get positive 1. 7 minus 6 is 1, if I square it, I get positive 1. And 14 minus 6 is 8, if I square it, I get 64. And then, I'm going to divide all of that by 5. And I don't need to use a calculator, but I tend to make a lot of careless mistakes when I do things while making a video. So I get 25 plus 9 plus 1 plus 1 plus 64 divided by 5. So I get 20. So the average squared distance, or the mean squared distance, from our population mean is equal to 20. Remember, it's the squared distance away from my population mean. So I squared each of these things. I liked it, because it made it positive. And we'll see later it has other nice properties about it. Now the last thing is, how can we represent this mathematically? We already saw that we know how to represent a population mean, and a sample mean, mathematically like this, and hopefully, we don't find it that daunting anymore. But how would we do the exact same thing? How would we denote what we did, right over here? Well, let's just think it through. We're just saying that the population variance, we're taking the sum of each-- so we're going to take each item, we'll start with the first item. And we're going to go to the n-th item in our population. We're talking about a population here. And we're going to take-- we're not" + }, + { + "Q": "\nAround 6:10 he's talking about the 20 being the squared distance away from the population mean - is there a time when you would take the square root?", + "A": "Yep. The square root of the variance is called the standard deviation, which will be another crucial concept that you ll get to pretty soon.", + "video_name": "dvoHB9djouc", + "timestamps": [ + 370 + ], + "3min_transcript": "And to that, I'm going to add the squared distance between 5 and my mean. And since I'm squaring, it doesn't matter if I do 5 minus 6, or 6 minus 5. When I square it, I'm going to get a positive result regardless. And then, to that I'm going to add the squared distance between 7 and my mean. So 7 minus 6 squared. All of this, this is my population mean that I'm finding the difference between. And then, finally, the squared difference between 14 and my mean. And then, I'm going to find, essentially, the mean of these squared distances. So I have five squared distances right over here. So let me divide by 5. So what will I get when I make this calculation, right over here? This is going to be equal to 1 minus 6 is negative 5, negative 5 squared is 25. 3 minus 6 is negative 3, now if I square that, I get 9. 5 minus 6 is negative 1, if I square it, I get positive 1. 7 minus 6 is 1, if I square it, I get positive 1. And 14 minus 6 is 8, if I square it, I get 64. And then, I'm going to divide all of that by 5. And I don't need to use a calculator, but I tend to make a lot of careless mistakes when I do things while making a video. So I get 25 plus 9 plus 1 plus 1 plus 64 divided by 5. So I get 20. So the average squared distance, or the mean squared distance, from our population mean is equal to 20. Remember, it's the squared distance away from my population mean. So I squared each of these things. I liked it, because it made it positive. And we'll see later it has other nice properties about it. Now the last thing is, how can we represent this mathematically? We already saw that we know how to represent a population mean, and a sample mean, mathematically like this, and hopefully, we don't find it that daunting anymore. But how would we do the exact same thing? How would we denote what we did, right over here? Well, let's just think it through. We're just saying that the population variance, we're taking the sum of each-- so we're going to take each item, we'll start with the first item. And we're going to go to the n-th item in our population. We're talking about a population here. And we're going to take-- we're not" + }, + { + "Q": "At 3:20, how did Sal go from cos(2*theta) to cos^2(theta) - sin^2(theta)? I didn't get that part, can someone please help me? Thanks!\n", + "A": "That s a double angle formula. It s a known trig identity.", + "video_name": "lXShNH1G6Pk", + "timestamps": [ + 200 + ], + "3min_transcript": "this expression evaluated at negative pi over 4, so 1 plus square root of 2 times sine of negative pi over 4, over cosine of 2 times negative pi over 4. Now, negative pi over 4, sine of negative pi over 4 is going to be negative square root of 2 over 2, so this is negative square root of 2 over 2, we're assuming this is in radians, if we're thinking in degrees, this would be a negative 45-degree angle, so this is one of the, one of the trig values that it's good to know and so if you have, if you have 1, so let's see, actually, let me just rewrite it, so this is going to be equal to 1 plus square root of 2 times that is going to be negative 2 over 2, so this is going to be minus 1, that's the numerator over here. All of this stuff simplifies to negative 1 this is going to be cosine of negative pi over 2, right? This is negative pi over 2, cosine of negative pi over 2, if you thought in degrees, that's going to be negative 90 degrees. Well, cosine of that is just going to be zero, so what we end up with is equal to zero over zero, and as we've talked about before, if we had something non-zero divided by zero, we'd say, okay, that's undefined. We might as well give up, but we have this indeterminate form, it does not mean the limit does not exist. It's usually a clue that we should use some tools in our toolkit, one of which is to do some manipulation here to get an expression that maybe is defined at theta is equal to, or does not, is not an indeterminate form, that theta is equal to pi over 4 and we'll see other tools in our toolkit in the future. So let me algebraically manipulate this a little bit. So if I have 1 plus the square root of 2, sine theta, over cosine 2 theta, the things that might be useful here are our trig identities and in particular, cosine of 2 theta seems interesting. Let me write some trig identities involving cosine of 2 theta. I'll write it over here. So we know that cosine of 2 theta is equal to cosine squared of theta minus sine squared of theta which is equal to 1 minus 2 sine squared of theta which is equal to 2 cosine squared theta minus 1, and you can go from this one to this one to this one just using the Pythagorean identity. We proved that in earlier videos in trigonometry on Khan Academy. Now, do any of these look useful? Well, all of these three are going to be differences of squares, so we can factor them in interesting ways," + }, + { + "Q": "At 10:05, why is f(x) continuous at a when a = -\u00cf\u0080/4? Isn't f(-\u00cf\u0080/4) undefined?\n", + "A": "Sal is just making a general statement about limit equality between functions. E.g if f(x)=a for all x, and g(x)=a for all x!=10 and g(10)=1234 then g(x) -> 4 when x -> 10. The functions have exactly the same limits, regardless of x-value.", + "video_name": "lXShNH1G6Pk", + "timestamps": [ + 605 + ], + "3min_transcript": "1 minus square root of 2 times the negative square root of 2 over 2, so negative, negative, you get a positive, square root of 2 times square root of 2 is 2, over 2 is going to be 1. So this is going to be equal to 1/2. And so, I want to be very clear. This expression is not the same thing as this expression. They are the same thing at all values of theta, especially if we're dealing in this open interval except at theta equals negative pi over 4. This one is not defined and this one is defined, but as we've seen multiple times before, if we find a function that is equal to our original or an expression, is equal to our original expression, and all values of theta except, except where the original one was not defined at a certain point, but this new one is defined and is continuous there, well then these two limits are going to be equal, then this limit is going to be 1/2, and I've said this in previous videos. It might be very tempting to say, well, I'm just going to algebraically simplify this in some way to get this, and I'm not going to worry about too much about these constraints, and then I'm just going to substitute negative pi over 4, and you will get this answer which is the correct answer but it's really important to recognize that this expression and this expression are not the same thing and what allows you to do this is, is the truth that if you have two functions, if you have f and g, two functions equal, let me write it this way, equal, equal for all x, except for all, wait, let me just write this this way, for all x except for a, then the limit, then, and let me write it this way, equal for all except, for all x except a continuous at a, then, then the limit of f of x as x approaches a is going to be equal to the limit of g of x as x approaches a, and I said this in multiple videos and that's what we are doing right here, but just so you can make sure you got it right, the answer here is 1/2." + }, + { + "Q": "\nAt 1:21 what does it mean by past the . that the numbers are smaller than the 1's?", + "A": "The numbers to the right of the decimal point represent fractions, anything less than 1. For instance, 0.1 is one tenth and is the same as 1/10.", + "video_name": "BItpeFXC4vA", + "timestamps": [ + 81 + ], + "3min_transcript": "So I have a number written here. It's a 2, a 3, and a 5. And we already have some experience with numbers like this. We can think about 'what does it represent'. And to think about that we just have to look at the actual place values. So this right-most place right over here. This is the ones place. So this 5 represents five ones, or I guess you could say that's just going to be 5. This 3, this is in the tens place. This is the tens place, so we have three tens. So that's just going to be 30. And the 2 is in the hundreds place. So putting a 2 there means that we have two hundreds. So this number we can view as two hundred, thirty, five. Or you could view it as two hundred plus thirty plus five. Now what I want to do in this video is think about place values to the right of the ones place. And you might say 'wait, wait, I always thought that the ones place was the place furthest to the right.' Well everything that we've done so far, it has been. But to show that you can go even further to the right We call that a 'decimal point'. And that dot means that anything to the right of this is going to be place values that are smaller, I guess you could say, than the ones place. So right to the left you have the ones place and the tens place and the hundreds place, and if you were to keep going you'd go to the thousands place and the ten thousands place. But then if you go to the right of the decimal point now you're going to divide by 10. So what am I talking about? Well, right to the right of the decimal point you are going to have-- find a new color-- this is going to be the tenths place. Well what does that mean? Well whatever number I write here that tells us how many tenths we're dealing with. So if I were to write the number 4 right over here, now my number is 2 hundreds plus 3 tens plus 5 ones plus 4 tenths. Or you could write this as 4 tenths. Not tens, 4 tenths. Or 4 tenths is the same thing as this right over here. So this is a super important idea in mathematics. I can now use our place values to represent fractions. So this right over here, this 'point 4', this is 4/10. So another way to write this number-- I could write it this way, I could write it as two hundred, thirty-- let me do the thirty in blue-- two hundred and thirty five and four tenths. So I could write it like this, as a mixed number. So this up here would be a decimal representation: 235.4 And this right over here would be a mixed number representation: 235 and 4/10 but they all represent 200 plus 30 plus 5 plus 4/10." + }, + { + "Q": "at 2:51 it says multiply both by 2/3 cant you just divide both by 3 and then multiply them by 2?\n", + "A": "Yes you can but that s the same thing as multiplying both sides by 2/3. Because 2*1/3 = 2 divided 3 = 2/3", + "video_name": "wo7DSaPP8hQ", + "timestamps": [ + 171 + ], + "3min_transcript": "So this first sentence, they say-- let me do this in a different color-- they say for the past few years, Old Maple Farms has grown about 1,000 more apples than their chief rival in the region, River Orchards. So we could say, hey, Maple is approximately Old River, or M is approximately River plus 1,000. Or since we don't know the exact amount-- it says it's about 1,000 more, so we don't know it's exactly 1,000 more-- we can just say that in a normal year, Old Maple Farms, which we denote by M, has a larger amount of apples than River Orchard. So in a normal year, M is greater than R, right? It has about 1,000 more apples than Old Maple Farms. Now, they say due to cold weather this year-- so let's talk about this year now-- the harvests at both farms were So this isn't a normal year. Let's talk about what's going to happen this year. In this year, each of these characters are going to be down by 1/3. Now if I go down by 1/3, that's the same thing as being 2/3 of what I was before. Let me do an example. If I'm at x, and I take away 1/3x, I'm left with 2/3x. So going down by 1/3 is the same thing as multiplying the quantity by 2/3. So if we multiply each of these quantities by 2/3, we can still hold this inequality, because we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number. If we were multiplying by a negative number, we would have to swap the inequality. So we can multiply both sides of this by 2/3. So 2/3 of M is still going to be greater than 2/3 of R. And you could even draw that in a number line if you like. This all might be a little intuitive for you, and if it is, I apologize, but if it's not, it never hurts. So that's 0 on our number line. So in a normal year, M is has 1,000 more than R. So in a normal year, M might be over here and maybe R is over here. I don't know, let's say R is over there. Now, if we take 2/3 of M, that's going to stick us some place around, oh, I don't know, 2/3 is right about there. So this is M-- let me write this-- this is 2/3 M. And what's 2/3 of R going to be? Well, if you take 2/3 of this, you get to right about there, that is 2/3R. So you can see, 2/3R is still less than 2/3M, or 2/3M is greater than 2/3R. Now, they say both farms made up for some of the shortfall" + }, + { + "Q": "\nI don't know if this is a mistake, but at 3:23 in the video, Sal says Maple Farms lost 2/3 of their farm, not 1/3.Is this a mistake, or did i miss something?", + "A": "At 2.22, Sal explains why it is 2/3 : 1/3 less than an amount x is x-1/3x which equals 2/3x. Hope this helps! :)", + "video_name": "wo7DSaPP8hQ", + "timestamps": [ + 203 + ], + "3min_transcript": "So this first sentence, they say-- let me do this in a different color-- they say for the past few years, Old Maple Farms has grown about 1,000 more apples than their chief rival in the region, River Orchards. So we could say, hey, Maple is approximately Old River, or M is approximately River plus 1,000. Or since we don't know the exact amount-- it says it's about 1,000 more, so we don't know it's exactly 1,000 more-- we can just say that in a normal year, Old Maple Farms, which we denote by M, has a larger amount of apples than River Orchard. So in a normal year, M is greater than R, right? It has about 1,000 more apples than Old Maple Farms. Now, they say due to cold weather this year-- so let's talk about this year now-- the harvests at both farms were So this isn't a normal year. Let's talk about what's going to happen this year. In this year, each of these characters are going to be down by 1/3. Now if I go down by 1/3, that's the same thing as being 2/3 of what I was before. Let me do an example. If I'm at x, and I take away 1/3x, I'm left with 2/3x. So going down by 1/3 is the same thing as multiplying the quantity by 2/3. So if we multiply each of these quantities by 2/3, we can still hold this inequality, because we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number. If we were multiplying by a negative number, we would have to swap the inequality. So we can multiply both sides of this by 2/3. So 2/3 of M is still going to be greater than 2/3 of R. And you could even draw that in a number line if you like. This all might be a little intuitive for you, and if it is, I apologize, but if it's not, it never hurts. So that's 0 on our number line. So in a normal year, M is has 1,000 more than R. So in a normal year, M might be over here and maybe R is over here. I don't know, let's say R is over there. Now, if we take 2/3 of M, that's going to stick us some place around, oh, I don't know, 2/3 is right about there. So this is M-- let me write this-- this is 2/3 M. And what's 2/3 of R going to be? Well, if you take 2/3 of this, you get to right about there, that is 2/3R. So you can see, 2/3R is still less than 2/3M, or 2/3M is greater than 2/3R. Now, they say both farms made up for some of the shortfall" + }, + { + "Q": "At 1:07 to 2:59, couldn't he find the area and then multiply it by 3/4 and you get an answer?\n", + "A": "Sure, if what you needed was the area. But the exercise is asking for the arc length which is part of the circumference and not the area. Just remember that arc length, and circumference for that matter, is a measurement of length as the name implies while area is not.", + "video_name": "YjWCDdNlXxc", + "timestamps": [ + 67, + 179 + ], + "3min_transcript": "- [Instructor] Find the area of the semicircle. So pause this video and see if you can figure it out. So let's see. We know that the area of a circle is equal to pi times our radius squared. So, if we think about the entire circle, what is the area going to be? Well, they tell us what our radius is. Our radius is equal to two, so the area, if we're talking about the whole circle, it would be equal to pi times two squared. Pi times two squared. Two squared is of course two times two, which is equal to four, so our area is going to be equal to four times pi. Now, I wouldn't put four pi here, because that would be the entire circle. They want the area of just the semicircle, of just this region right over here. Well, the semicircle is half of the circle, so if I want the area of the semicircle, this is gonna be half this. So instead of four pi, it is going to be two pi square units. Let's do another example. So here, instead of area, we're asked to find the arc length of the partial circle, and that's we have here in this bluish color right over here, find this arc length. And you can see this is going three fourths of the way around the circle, so this arc length is going to be three fourths of the circumference. So what is the circumference? Well, we know the circumference is equal to two pi times the radius. They tell us what the radius is. It's equal to four, so our circumference is equal to two pi times four. Let's see, we can just change the order in which we multiply so it's two times four times pi. This is going to be equal to eight pi. This is going to be equal to eight pi. Now, that is the circumference of the entire circle. it's going to be three fourths times the circumference of the entire circle. So three over four times eight pi. What is that going to be? Well, what's three fourths times eight? Well, three times eight is 24 divided by four is six. So this is going to be equal to six pi. Another way to think about it, one fourth of eight is two, so three fourths is going to be six. Or another way to think about it is, one fourth of eight pi is two pi, and so three of those is going to be equal to six pi. So the arc length of the partial circle is six pi, and once again we knew that because it was three fourths of the way around. The way that I knew it was three fourths is that this is a 90 degree angle. This is 90 degrees, which is one fourth of the way around a circle, so the arc length that we care about is the three fourths of our circumference." + }, + { + "Q": "At 2:03 to 2:13, I wish he would have wrote out the 10^5-10^7 to show what he did in his head.\n", + "A": "That would certainly be helpful!", + "video_name": "497oIjqRPco", + "timestamps": [ + 123, + 133 + ], + "3min_transcript": "We have 7 times 10 to the fifth over 2 times 10 to the negative 2 times 2.5 times 10 to the ninth. So let's try to simplify this a little bit. And I'll start off by trying to simplify this denominator here. So the numerator's just 7 times 10 to the fifth. And the denominator, I just have a bunch of numbers that are being multiplied times each other. So I can do it in any order. So let me swap the order. So I'm going to do over 2 times 2.5 times 10 to the negative 2 times 10 to the ninth. And this is going to be equal to-- so the numerator I haven't changed yet-- 7 times 10 to the fifth over-- and here in the denominator, 2 times-- let me do this in a new color now. 2 times 2.5 is 5. And then 10 to the negative 2 times 10 to the ninth, when you multiply two numbers that are being raised to exponents and have the exact same base-- we can add the exponents. So this is going to be 10 to the 9 minus 2, or 10 to the seventh. So times 10 to the seventh. And now we can view this as being equal to 7 over 5 times 10 to the fifth over 10 to the seventh. Let me do that in that orange color to keep track of the colors. 10 to the seventh. Now, what is 7 divided by 5? 7 divided by 5 is equal to-- let's see, it's 1 and 2/5, or 1.4. So I'll just write it as 1.4. And then 10 to the fifth divided by 10 to the seventh. So that's going to be the same thing as-- and there's two ways to view this. You could view this as 10 to the fifth times 10 You get 10 to the negative 2. Or you say, hey, look, I'm dividing this by this. We have the same base. We can subtract exponents. So it's going to be 10 to the 5 minus 7, which is 10 to the negative 2. So this part right over here is going to simplify to times 10 to the negative 2. Now, are we done? Have we written what we have here in scientific notation? It looks like we have. This value right over here is greater than or equal to 1, but it is less than or equal to 9. It's a digit between 1 and 9, including 1 and 9. And it's being multiplied by 10 to some power. So it looks like we're done. This simplified to 1.4 times 10 to the negative 2." + }, + { + "Q": "\nAt 1:40 Sal says that (7 * 10^5) / (5 * 10^7) = (7 / 5) * (10^5 / 10^7). Why can we detach 7 / 5 like this? Shouldn't we get a different answer? I'm sure Sal says the right thing, I just don't get why it works.", + "A": "How many sides does your polygon have?", + "video_name": "497oIjqRPco", + "timestamps": [ + 100 + ], + "3min_transcript": "We have 7 times 10 to the fifth over 2 times 10 to the negative 2 times 2.5 times 10 to the ninth. So let's try to simplify this a little bit. And I'll start off by trying to simplify this denominator here. So the numerator's just 7 times 10 to the fifth. And the denominator, I just have a bunch of numbers that are being multiplied times each other. So I can do it in any order. So let me swap the order. So I'm going to do over 2 times 2.5 times 10 to the negative 2 times 10 to the ninth. And this is going to be equal to-- so the numerator I haven't changed yet-- 7 times 10 to the fifth over-- and here in the denominator, 2 times-- let me do this in a new color now. 2 times 2.5 is 5. And then 10 to the negative 2 times 10 to the ninth, when you multiply two numbers that are being raised to exponents and have the exact same base-- we can add the exponents. So this is going to be 10 to the 9 minus 2, or 10 to the seventh. So times 10 to the seventh. And now we can view this as being equal to 7 over 5 times 10 to the fifth over 10 to the seventh. Let me do that in that orange color to keep track of the colors. 10 to the seventh. Now, what is 7 divided by 5? 7 divided by 5 is equal to-- let's see, it's 1 and 2/5, or 1.4. So I'll just write it as 1.4. And then 10 to the fifth divided by 10 to the seventh. So that's going to be the same thing as-- and there's two ways to view this. You could view this as 10 to the fifth times 10 You get 10 to the negative 2. Or you say, hey, look, I'm dividing this by this. We have the same base. We can subtract exponents. So it's going to be 10 to the 5 minus 7, which is 10 to the negative 2. So this part right over here is going to simplify to times 10 to the negative 2. Now, are we done? Have we written what we have here in scientific notation? It looks like we have. This value right over here is greater than or equal to 1, but it is less than or equal to 9. It's a digit between 1 and 9, including 1 and 9. And it's being multiplied by 10 to some power. So it looks like we're done. This simplified to 1.4 times 10 to the negative 2." + }, + { + "Q": "\nAround 2:25 he mentions \"f of x\", this mightt sound like a stupid question but what does this mean? Function?", + "A": "yup yup yups!", + "video_name": "W0VWO4asgmk", + "timestamps": [ + 145 + ], + "3min_transcript": "Welcome to the presentation on limits. Let's get started with some-- well, first an explanation before I do any problems. So let's say I had-- let me make sure I have the right color and my pen works. OK, let's say I had the limit, and I'll explain what a limit is in a second. But the way you write it is you say the limit-- oh, my color is on the wrong-- OK, let me use the pen and yellow. OK, the limit as x approaches 2 of x squared. Now, all this is saying is what value does the expression x squared approach as x approaches 2? Well, this is pretty easy. If we look at-- let me at least draw a graph. I'll stay in this yellow color. So let me draw. x squared looks something like-- let me use x square looks something like this, right? And when x is equal to 2, y, or the expression-- because we don't say what this is equal to. It's just the expression-- x squared is equal to 4, right? So a limit is saying, as x approaches 2, as x approaches 2 from both sides, from numbers left than 2 and from numbers right than 2, what does the expression approach? And you might, I think, already see where this is going and be wondering why we're even going to the trouble of learning this new concept because it seems pretty obvious, but as x-- as we get to x closer and closer to 2 from this direction, and as we get to x closer and closer to 2 to this direction, what does this expression equal? Well, it essentially equals 4, right? The way I think about it is as you move on the curve closer and closer to the expression's value, what does the expression equal? In this case, it equals 4. You're probably saying, Sal, this seems like a useless concept because I could have just stuck 2 in there, and I know that if this is-- say this is f of x, that if f of x is equal to x squared, that f of 2 is equal to 4, and that would have been a no-brainer. Well, let me maybe give you one wrinkle on that, and hopefully now you'll start to see what the use of a limit is. Let me to define-- let me say f of x is equal to x squared when, if x does not equal 2, and let's say it equals 3 when x equals 2." + }, + { + "Q": "So around the time 4:43 in the video, he is talking about simplifying your answer. The answer I got that I think I need to simplify is -3x+9y=-6. My question is would it be possible to divide the positive 9 by a negative 3 or the only way you would be able to divide that positive is by a positive vice versa.\n", + "A": "Yes it would be possible. I would divide it by -3. So you would have -3x/-3=x and 9y/-3=-3y and -6/-3=2 . The simplified version would be x+3y+2 :)", + "video_name": "XOIhNVeLfWs", + "timestamps": [ + 283 + ], + "3min_transcript": "And then 21 divided by seven is three times four is 12. So just like that I was able to get rid of the fractions. And now I wanna get all the X's and Y's on one side. So I wanna get this 14x onto the left side. So let's see if I can do that. So I'm gonna do that by. To get rid of this I would want to subtract 14x. I can't just do it on the right hand side I have to do it on the left hand side as well. So I wanna subtract 14x, and then what am I left with? Let me give myself a little bit more space. So on the left hand side I have negative 14x plus 21y. Plus 21y is equal to. Let's see and I subtracted 14x to get rid of this. And then I have this is equal to 12. Now let's see, am I done? Do these share any, do 14, 21, and 12 Let's see, 14 is divisible by two and seven. 21 is divisible by three and seven. 12 is divisible by two, six, three, four. But all of these aren't divisible by the same number. 14, let's see. 14 is divisible two, so is 12, but 21 isn't. 14 is divisible by seven, so is 21, but 12 isn't. And 21 and 12 are divisible by three but 14 isn't. So I think this is about as simplified as I could get. If there was a common factor for all three of these numbers then I would divide all of them the way I did in that previous example. But that's not the case right over here. So it's negative 14x plus 21y is equal to 12. So let me see if I can remember that and type that in. So it is, negative 14x plus 21y is equal to 12. It worked out." + }, + { + "Q": "At 2:44 , SAl said the slope is 7. How is it 7 ? is he assuming ?\n", + "A": "In the video there s a graphic clip with given information: Let f be a differentiable function for all x. If f(-2) = 3 and f (x) \u00e2\u0089\u00a4 7 for all x, then what is the largest possible value of f(10). f (x) is the slope of f(x).", + "video_name": "EXLVMGSDQbI", + "timestamps": [ + 164 + ], + "3min_transcript": "So really, the way to get the largest possible value of f-- we don't have to necessarily invoke the mean value theorem, although the mean value theorem will help us know for sure-- is to say well, look, the largest possible value of f of 10 is essentially if we max this thing out. If we assume that the instantaneous rate of change just stays at the ceiling right at 7. So if we assumed that our function, the fastest growing function here would be a line that has a slope exactly equal to 7. So the slope of 7 would look-- and obviously, I'm not drawing this to scale. Visually, this looks more like a slope of 1, but we'll just assume this is a slope of 7 because it's not at the same-- the x and y are not at the same scale. So slope is equal to 7. where do we get to when x is equal to 10? When x is equal to 10, which is right over here, well what's our change in x? So what's our change in x? Let's just think about it this way. Our change in y over change in x is going to be what? Well our change in y is going to be f of 10 minus f of 2. f of 2 is 3, so minus 3, over our change in x. Our change in x is 10 minus negative 2. 10 minus negative 2 is going to be equal to 7. This is the way to max out what our value of f of 10 might be. If at any point the slope were anything less than that, because remember, the instantaneous rate of change can never be more than that. So if we start off even a little bit lower, than the best we can do is get to that. That would get us too steep. So it has to be like that. And then we would get to a lower f of 10. Every time you have a slightly lower rate of change, then it kind of limits what happens to you. So remember, our slope can never be more than 7. So this part should be parallel. So this should be parallel to that right over there. This should be parallel. But we can never have a higher slope than that. So the way to max it out is to actually have a slope of 7. And so what is f of 10 going to be? So let's see, 10 minus negative 2, that is 12. Multiply both sides by 12, you get 84. So f of 10 minus 3 is going to be equal to 84. Or f of 10 is going to be equal to 87. So if you have a slope of 7, the whole way, you travel 12." + }, + { + "Q": "at 3:30 sal multiplies vector a with a 'scalar quantity'-1.....but how come -1 is scalar....1 represents magnitude and the negative sign represents direction so it has to be a vector. and if it is a vector how do we plot it on a graph\n", + "A": "That s creative! But not exactly how that works. You could think of the number 2 as having a + in front of it, that we just don t write. The negative symbol is a part of the number. It says we are going negative one in some direction, it does not say whether we are going negative 1 up, or left or right or down or diagonally, just that we are going negative 1. Does that make sense?", + "video_name": "ZN7YaSbY3-w", + "timestamps": [ + 210 + ], + "3min_transcript": "But how do we define multiplying 3 times this vector? Well one reasonable thing that might jump out at you is, why don't we just multiply the 3 times each of these components? So this could be equal to.. we have 2 and 1.. And we're going to multiply each of these with 3. So 3 times 2 and 3 times 1. And then the resulting vector is still going to be a 2-dimensional vector. And it's going to be the 2-dimensional vector (6,3). Now I encourage you to get some graph paper out and to actually plot this vector, and think about how it relates to this vector right over here. So let me do that.. So the vector (6,3), if we started at the origin.. We would move 6 in the horizontal direction.. 1, 2, 3, 4, 5, 6.. And 3 in the vertical.. 1, 2, 3.. So it gets us right over there, so it would look like this. Well notice, one way to think about it is what has changed, and what has not changed about this vector? Well what's not changed is still pointing in the same direction. So this right over here has the same direction. Multiplying by the scalar, at least the way we defined it.. did not change the direction that my vector is going in. Or at least in this case it didn't.. But it did change its magnitude. Its magnitude is now 3 times longer, which makes sense! Because we multiplied it by 3. One way to think about it is we scaled it up by 3. The scalar scaled up the vector. That might make sense. Or it might make an intuition of where that word scalar came from. The scalar, when you multiply it, it scales up a vector. It Increased its magnitude by 3 without changing its direction. Well let's do something interesting.. Let's multiply our vector a by a negative number. So let's just multiply -1 times a. Well using the convention that we just came up with.. We would multiply each of the components by -1. So 2 times -1 is -2, and 1 times -1 is -1. So now -1 times a is going to be (-2,-1) So if we started at the origin, we would move in the horizontal direction -2, and in the vertical -1 So now what happened to the vector? When I did that? Well now it flipped its direction! Multiplying it by this -1, it flipped it's direction. Its magnitude actually has not changed, but its direction is now in the exact opposite direction. Which makes sense, that multiplying by a negative number would do that." + }, + { + "Q": "\nAt 4:03 Sal said the magnitude didn't change. But he added, essentially, 1 more of the original vector in the negative direction. Am Imissing something?", + "A": "The magnitude remained the same only for the -1 scalar.", + "video_name": "ZN7YaSbY3-w", + "timestamps": [ + 243 + ], + "3min_transcript": "Well notice, one way to think about it is what has changed, and what has not changed about this vector? Well what's not changed is still pointing in the same direction. So this right over here has the same direction. Multiplying by the scalar, at least the way we defined it.. did not change the direction that my vector is going in. Or at least in this case it didn't.. But it did change its magnitude. Its magnitude is now 3 times longer, which makes sense! Because we multiplied it by 3. One way to think about it is we scaled it up by 3. The scalar scaled up the vector. That might make sense. Or it might make an intuition of where that word scalar came from. The scalar, when you multiply it, it scales up a vector. It Increased its magnitude by 3 without changing its direction. Well let's do something interesting.. Let's multiply our vector a by a negative number. So let's just multiply -1 times a. Well using the convention that we just came up with.. We would multiply each of the components by -1. So 2 times -1 is -2, and 1 times -1 is -1. So now -1 times a is going to be (-2,-1) So if we started at the origin, we would move in the horizontal direction -2, and in the vertical -1 So now what happened to the vector? When I did that? Well now it flipped its direction! Multiplying it by this -1, it flipped it's direction. Its magnitude actually has not changed, but its direction is now in the exact opposite direction. Which makes sense, that multiplying by a negative number would do that. If you took 5 times -1, well now you're going in the other direction you're at -5, you're 5 to the left of zero. So it makes sense that this would flip its direction. So you could imagine, if you were to take something like -2 times your vector a, -2 times your vector a.. And I encourage you to pause this video and try this on your own.. What would this give? And what would be the resulting visualization of the vector? Well let's see, this would be equal to -2 times 2 is -4, -2 times 1 is -2, so this vector.. if you were to start at the origin! remember you don't have to start at the origin.. but if you were.. it would go 0, 1, 2, 3, 4.. 1, 2.. It looks just like this.. And so just to remind ourselves.. our original vector a looked like this.." + }, + { + "Q": "\n10:50 what are pictograms?", + "A": "Pictograms are the pictures/symbols themselves. A pictograph is the entire graph.", + "video_name": "qrVvpYt3Vl0", + "timestamps": [ + 650 + ], + "3min_transcript": "" + }, + { + "Q": "\nWhy at 1:58 is 8x-7x= to x ?", + "A": "8-7=1. 1x is the same thing as x. It is easier to say x than 1x, but you could actually say 1x, and it wouldn t be wrong.", + "video_name": "E0TNh9uWesw", + "timestamps": [ + 118 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:57 in the video, he says that 8x - 7x = x, which I think I understand. But what if my number was larger than just x? What would I do from there, divide both sides?\n\nThanks\n", + "A": "Yes, you would. It s nice that you can know (or take a good guess at) what to do.", + "video_name": "E0TNh9uWesw", + "timestamps": [ + 117 + ], + "3min_transcript": "" + }, + { + "Q": "\nAround 2:00, 8x - 7x = x? I assume the missing 1 is just a place holder?", + "A": "Yeah, if there is no coefficient in front of a variable, that means that there is a coefficient of 1: 1x = x", + "video_name": "E0TNh9uWesw", + "timestamps": [ + 120 + ], + "3min_transcript": "" + }, + { + "Q": "On that last one (5:31) shouldn't we have multiplied 2/1 by three to get the denominators to be the same before we multiplied it... to get our final answer?\n", + "A": "if you add 2/1 to -(1/3), then yes. But this is a multiplication problem. So just multiply the numerators and multiply the denominators respectively.", + "video_name": "a_Wi-6SRBTc", + "timestamps": [ + 331 + ], + "3min_transcript": "" + }, + { + "Q": "\nx < -1 (-infinity,-1) since the -1 is in the parenthesis, shouldn't it be less than or equal to? @1:29", + "A": "The parentheses mean only less than . To have x <= -1, the interval would be written as (-infinity, -1] The square bracket indicates that x can also be equal to -1.", + "video_name": "xOxvyeSl0uA", + "timestamps": [ + 89 + ], + "3min_transcript": "Let's do a few more problems that bring together the concepts that we learned in the last two videos. So let's say we have the inequality 4x plus 3 is less than negative 1. So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a zero. No reason to change the inequality just yet. We're just adding and subtracting from both sides, in this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to-- let's see, we can divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1, so we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. So let's get all our x's on the left-hand side, and the best way to do that is subtract 8x from both sides. So you subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantities on both sides. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by" + }, + { + "Q": "\nat 5:40, why do you subtract -8x instead of adding 12x to both sides or adding 5 or 7 to both first. Is there a reason you pick 8x to subtract first?", + "A": "That is what I would have advised my students to do so that the coefficient of the variable stays positive. There is no reason for subtracting 8 except to possibly remind us of the need to flip the inequality sign when dividing by a negative. You should have gotten the same answer.", + "video_name": "xOxvyeSl0uA", + "timestamps": [ + 340 + ], + "3min_transcript": "Now, this might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. So let's just simplify this. You get 8x minus-- let's distribute this negative 5. So let me say 8x, and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5-- when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5, and then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. And now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to-- we that's negative 7, and then we have this plus 8x left over. Now, I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides, or adding and subtracting from them. The right-hand side becomes-- this thing cancels out, 8x minus 8x, that's 0. So you're just left with a negative 7. And now I want to get rid of this negative 5. So let's add 5 to both sides of this equation. These 5's cancel out. No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2. Now, we're at an interesting point. We have negative 20x is greater than or equal to negative 2. If this was an equation, or really any type of an inequality, we want to divide both sides by negative 20. But we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality. So let's remember that. So if we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides divided by negative 20, we have to swap the inequality. The greater than or equal to has to become a less than or equal sign. And, of course, these cancel out, and you get x is less" + }, + { + "Q": "\nwhy do you flip the sign at 2:35???", + "A": "oh ok thank you for explaining this to me", + "video_name": "xOxvyeSl0uA", + "timestamps": [ + 155 + ], + "3min_transcript": "So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1, so we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. So let's get all our x's on the left-hand side, and the best way to do that is subtract 8x from both sides. So you subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantities on both sides. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by to less than. And the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9, maybe this would be negative 8, maybe this would be negative 10. You would start at negative 9, not included, because we don't have an equal sign here, and you go everything less than that, all the way down, as we see, to negative infinity. Let's do a nice, hairy problem. So let's say we have 8x minus 5 times 4x plus 1 is greater" + }, + { + "Q": "\nAt 4:00, while setting up the problem, verbally \"8x minus 5\", then at 4:24 while simplifying the inequality, verbally says \"distribute the negative 5\".\nWhile trying to respect the order of operations, this interchangeable \"subtraction\" and \"negative\" concept is very confusing. Any help would be appreciated, thank you in advance!", + "A": "Do you already know that subtraction is adding the opposite? For example, having $10 and then spending $4 could be written 10 - 4 (ten take away four) or it could be written 10 + -4 (having ten and then losing 4). Think of that for this distribution problem: 8x - 5(4x + 1) could be written 8x + -5(4x + 1). Now can you see how it is a negative 5 that needs to be distributed?", + "video_name": "xOxvyeSl0uA", + "timestamps": [ + 240, + 264 + ], + "3min_transcript": "So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by to less than. And the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9, maybe this would be negative 8, maybe this would be negative 10. You would start at negative 9, not included, because we don't have an equal sign here, and you go everything less than that, all the way down, as we see, to negative infinity. Let's do a nice, hairy problem. So let's say we have 8x minus 5 times 4x plus 1 is greater Now, this might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. So let's just simplify this. You get 8x minus-- let's distribute this negative 5. So let me say 8x, and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5-- when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5, and then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. And now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to-- we" + }, + { + "Q": "At 1:13, I thought that multiplying or dividing in an inequality you were supposed to flip the sign?\n", + "A": "For an inequality, you only flip the sign if you re multiplying or dividing by a negative number. At that point in the video, he s dividing by positive 4, so there s no need to flip the inequality sign.", + "video_name": "xOxvyeSl0uA", + "timestamps": [ + 73 + ], + "3min_transcript": "Let's do a few more problems that bring together the concepts that we learned in the last two videos. So let's say we have the inequality 4x plus 3 is less than negative 1. So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a zero. No reason to change the inequality just yet. We're just adding and subtracting from both sides, in this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to-- let's see, we can divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1, so we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. So let's get all our x's on the left-hand side, and the best way to do that is subtract 8x from both sides. So you subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantities on both sides. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by" + }, + { + "Q": "At 4:34 Sal calculates the probability of small to be (1/26-1/2600). Why isn't this probability equal to (9/10*9/10*1/26)?\nSince, Probability of losing number = 9/10 and Probability of winning alphabet = 1/26.\n", + "A": "I did the problem like you say. Once you buy a ticket, the expected values are as follows: Expected Value of $5 payout = probability*value = 1 * (-$5) EV(grand prize) = P(x)*x = (1/10*1/10*1/26) * (10405) = 4.0019 EV(small prize) = (1/10*9/10*1/26) * 100 + (9/10*1/10*1/26) * 100 + (9/10*9/10*1/26) * 100 [ there are 2 ways to get 1 number, 1 way to get no numbers] = 0.34615 + 0.34615 + 3.1154 Total is 7.81 - 5 = 2.81", + "video_name": "6vlBOHckmzU", + "timestamps": [ + 274 + ], + "3min_transcript": "What's the probability of the grand prize? I'll do that over here, probability of grand prize. Well the probability that he gets the first letter right is one in 10, there's 10 digits there. Probability he gets the second letter right is one in 10, these are all independent and probability he gets the letter right, there's 26 equally likely letters that might be in the actual one so he has a one in 26 chance of that one as well. The probability of the grand prize is one in 2600. This is one in 2600. Now what's the probability of getting the small price? Well let's see, he has a one in 26 chance. The small prize is getting the letter right but not getting both of the numbers right. but we're not done here just with the one in 26 because this one in 26, this includes all the scenarios where he gets the letter right, including the scenarios where he wins the grand prize, where he gets the letter and he gets the two numbers right. We need to do is we need to subtract out the situation, the probability of getting the two numbers, getting the letter and the two numbers right and we already know what that is, it's one in 2600. It's one and 26 minus one and 2600. The reason why I have to subtract out at this 2600 is he has one in 26 chance of getting this letter right. That includes the scenario where he gets everything right but the small prize is only where you get the letter and one or none of these. If you get both of these then you're at the grand prize case. You essentially have to subtract out the probability that you won the grand prize, to figure out the probability of the small prize. Now what's the probability of essentially losing? The probability of neither. Well it's just kind of that's everything else. It would be one minus these probabilities right over here. It would be one minus the probability of the small prize. The probability of the small minus the probability of the grand, these are the possible outcomes so they have to add up to one or a 100%. This is one less probability small minus probability of large or I'll say grand prize. Let's fill this in. The probability of the small one, this right over ... I'm using that red too much. This right over here is one in 26 minus one in 2600" + }, + { + "Q": "at 1-1:20, how did you choose the bounds? The upper limit can be anything right? I was confused because 3 is the value you get when you plug X= -2 into the y=x^2-1 formula. I thought it had extra significance. As for the lower bound, I understand y= -1 is the minimum of the parabola. Is that the reason for it being the bound? Thank you.\n", + "A": "Since the only part of the function that is being rotated is the one for positive x values, it is a mere coincidence that 3 is the intersection of the whole function and x = -2. The part of the function for negative x values was never considered and the number 3 was chosen at random.", + "video_name": "jxf7XqvZWWg", + "timestamps": [ + 80 + ], + "3min_transcript": "Let's do another example, and this time we're going to rotate our function around a vertical line that And if we do that-- so we're going to rotate y is equal to x squared minus 1-- or at least this part of it-- we're going to rotate it around the vertical line x is equal to negative 2. And if we do that, we get this gumball shape that looks something like this. So what I want to do is I want to find the volume of this using the disc method. So what I want to do is construct some discs. So that's one of the discs right over here. It's going to have some depth, and that depth is going to be dy right over there. And it's going to have some area on top of it that is a function of any given y that I have. So the volume of a given disc is going to be the area as a function of y times the depth of the disc times dy. And then we just have to integrate it over the interval that we care about, and we're doing it all in terms of y. And in this case, we're going to integrate going to hit-- this y-intercept right over here is y is equal to negative 1. And let's go all the way to y is equal to, let's say y is equal to 3 right over here. So from y equal negative 1 to y equals 3. And that's going to give us the volume of our upside-down gumdrop-type-looking thing. So the key here, so that we can start evaluating the double integral, is to just figure out what the area of each of these discs are as a function of y. And we know that area as a function of y is just going to be pi times radius as a function of y squared. So the real key is, what is the radius as a function of y for any one of these y's? So what is the radius as a function of y? So let's think about that a little bit. What is this curve? Well, let's write it as a function of y. so you'll get x squared is equal to y plus 1. I just added 1 to both sides and then swapped sides. And you get x is equal to the principal root of the square root of y plus 1. So this we can write as x-- or we can even write it as f of y if we want-- f of y is equal to the square root of y plus 1. Or we could say x is equal to a function of y, which is the square root of y plus 1. So what's the distance here at any point? Well, this distance-- let me make it very clear. So it's going to be our total distance in the horizontal direction. So this first part as we're-- and I'm going to do it in a different color so we can see. So this part right over here is just going to be the value of the function. It's going to give you an x value. But then you have to add another 2 to go all the way over here. So your entire radius as a function of y" + }, + { + "Q": "At 7:57, Sal mentions that dA is used as a shorthand. Isn't it also used to generalize dxdy so that the order of integration (and the coordinate system) is not specified? I feel like this would make it more applicable in proofs and theorems and such.\n", + "A": "dA is often used to indicate integration over an area without specifying how the integration will be performed. dA can even indicate integration over a curved surface in 3-D.", + "video_name": "twT-WZChfZ8", + "timestamps": [ + 477 + ], + "3min_transcript": "So we could take, we're summing in the y-direction first. We would get a sheet that's parallel to the y-axis, now. So the top of the sheet would look something like that. So if we're coming the dy's first, we would take the sum, we would take the integral with respect to y, and it would be, the lower bound would be y is equal to c, and the upper bound is y is equal to d. And then we would have that sheet with a little depth, the depth is dx, and then we could take the sum of all of those, sorry, my throat is dry. I just had a bunch of almonds to get power to be able to record these videos. But once I have one of these sheets, and if I want to sum up all of the x's, then I could take the infinite sum of infinitely small columns, or in this view, sheets, infinitely the upper bound is x is equal to b. And once again, I would have the volume of the figure. And all I showed you here is that there's two ways of doing the order of integration. Now, another way of saying this, if this little original square was da, and this is a shorthand that you'll see all the time, especially in physics textbooks, is that we are integrating along the domain, right? Because the x-y plane here is our domain. So we're going to do a double integral, a two-dimensional integral, we're saying that the domain here is two-dimensional, and we're going to take that over f of x and y times da. And the reason why I want to show you this, is you see this in physics books all the time. I don't think it's a great thing to do. Because it is a shorthand, and maybe it looks simpler, but for me, whenever I see something that I don't know how to compute or that's not obvious for me to know how to compute, So I wanted to just show you that what you see in this physics book, when someone writes this, it's the exact same thing as this or this. The da could either be dx times dy, or it could either be dy times dx, and when they do this double integral over domain, that's the same thing is just adding up all of these squares. Where we do it here, we're very ordered about it, right? We go in the x-direction, and then we add all of those up in the y-direction, and we get the entire volume. Or we could go the other way around. When we say that we're just taking the double integral, first of all, that tells us we're doing it in two dimensions, over a domain, that leaves it a little bit ambiguous in terms of how we're going to sum up all of the da's. And they do it intentionally in physics books, because you don't have to do it using Cartesian coordinates, using x's and y's. You can do it in polar coordinates, you could do it a ton of different ways. But I just wanted to show you, this is another way to having an intuition of the volume under a surface. And these are the exact same things as this type of" + }, + { + "Q": "\nAt 8:52 Sal says we can do this in polar coordinates. Can someone explain to me how this would work?", + "A": "3D polar could be either with an extra angle coordinate or another radius coordinate , spherical or cylindrical respectively. You might try 2D polar first, I would go back and watch integration videos in 2D and do a parallel situation in polar. But basically, the idea is to split the area into sectors of a circle where dTHETA becomes infinitely small. The area of a sector is 1/2*pi*r^2. Be mindful of your endpoints - integrating from pi/4 to 3pi/2 is much different than -pi/4 to pi/4. Hope this helps.", + "video_name": "twT-WZChfZ8", + "timestamps": [ + 532 + ], + "3min_transcript": "the upper bound is x is equal to b. And once again, I would have the volume of the figure. And all I showed you here is that there's two ways of doing the order of integration. Now, another way of saying this, if this little original square was da, and this is a shorthand that you'll see all the time, especially in physics textbooks, is that we are integrating along the domain, right? Because the x-y plane here is our domain. So we're going to do a double integral, a two-dimensional integral, we're saying that the domain here is two-dimensional, and we're going to take that over f of x and y times da. And the reason why I want to show you this, is you see this in physics books all the time. I don't think it's a great thing to do. Because it is a shorthand, and maybe it looks simpler, but for me, whenever I see something that I don't know how to compute or that's not obvious for me to know how to compute, So I wanted to just show you that what you see in this physics book, when someone writes this, it's the exact same thing as this or this. The da could either be dx times dy, or it could either be dy times dx, and when they do this double integral over domain, that's the same thing is just adding up all of these squares. Where we do it here, we're very ordered about it, right? We go in the x-direction, and then we add all of those up in the y-direction, and we get the entire volume. Or we could go the other way around. When we say that we're just taking the double integral, first of all, that tells us we're doing it in two dimensions, over a domain, that leaves it a little bit ambiguous in terms of how we're going to sum up all of the da's. And they do it intentionally in physics books, because you don't have to do it using Cartesian coordinates, using x's and y's. You can do it in polar coordinates, you could do it a ton of different ways. But I just wanted to show you, this is another way to having an intuition of the volume under a surface. And these are the exact same things as this type of Sometimes they won't write a domain, sometimes they'd write over a surface. And we'll later do those integrals. Here the surface is easy, it's a flat plane, but sometimes it'll end up being a curve or something like that. But anyway, I'm almost out of time. I will see you in the next video." + }, + { + "Q": "at 0:10, Why did you call the space \"blank\"?\n", + "A": "Blank represents any unknown or missing value. It is commonly used in language and math lessons. The goal is to find the/an answer that fits in the blank and completes the sentence or problem.", + "video_name": "OPpmp-kAuE4", + "timestamps": [ + 10 + ], + "3min_transcript": "- [Voiceover] So, we have blank plus one-16th is equal to three halves. So, I encourage you to pause the video and just figure out what blank is. What plus one-16th is equal to three-halves? All right. Now, let's work through this together. One thing that might make it a little bit simpler for our brains is if we were to express one-16th and three-halves with a common denominator. When we think about a common denominator, we will look at a common multiple of the denominators here. And lucky for us, 16 is already divisible by 2. It's divisible by 16 as well, so it is the common multiple, or it is the least common multiple of 16 and 2. There are other common multiples, but the smallest one is going to be 16 times 1, which is also divisible by 2. So, let's try both of these, let's try both of these fractions. Let's rewrite this equation where both of these fractions have 16 as their denominators. This one obviously already has it. So, let's write that. So, we're going to have, we're going to have blank plus one-16th, is equal to, is equal to, let's see, let's write three-halves as something over 16. Well, to get our denominator from 2 to 16, we have to multiply by 8. So, we have to multiply the numerator by 8 as well. So, 3 times 8 is going to be 24. Now, at this point, you might be able to do it in your head. Blank plus one-16th is equal to 24-16ths. We could say, okay, this could, this could... We could do this as a certain number of 16ths. So, how many 16ths plus one-16th is going to be 24-16ths? Well, 23-16ths. That's 23-16ths, and I add one more 16th, I'm going to have 24-16ths. Another way that you could have thought about it, Actually, you could have even thought about it from the first step, is you could say, look, if blank plus one-16th is equal to three-halves, then you could say that blank, three-halves, three-halves minus one-16th, minus one-16th. This is another way that you could have tackled it. If blank plus one-16th is three-halves, then blank is going to be equal to three-halves minus one-16th. And we know that this is going to be equal to... Three-halves, we already know, it's the same thing as 24-16ths. So, 24 over 16 minus one-16th, minus one-16th, which we figured out is 23/16ths, which is equal to... I'll just rewrite it again, 23 over 16. Let's do another example. So, here, this is a little bit different. I have blank minus three-fourths is equal to two-thirds. And there's a couple of ways to think about it. If blank minus three-fourths is equal to two-thirds," + }, + { + "Q": "shouldnt the x+5 over 100 @ 16:12 be a 10 because of ([x--5]/10)^2 ?!\n", + "A": "Yes, in the exponential term of the yellow integral either the denominator should be 10 or the square should be limited to the numerator", + "video_name": "hgtMWR3TFnY", + "timestamps": [ + 972 + ], + "3min_transcript": "too much-- because the area under the curve just under 0-- there's no area. It's just a line. You have to say between a range. So you have to say the probability between, let's say, minus-- and actually, I can type it in here on our-- I can say, the probability between, let's say, minus 0.005 and plus 0.05 is-- well, it rounded, so it says there, close to 0. Let me do it-- between minus 1 and between 1, all right? It calculated at 7%, and I'll show you how I calculated this in a second. So let me get the Screen Draw tool. This between minus 1 and 1-- and I'll show you the behind the scenes what Excel is doing-- we're going from minus 1, which is roughly right here, to 1. And we're calculating the area under the curve, all right? Or, for those of you who know calculus, we're calculating the integral from minus 1 to 1 of this function, where the standard deviation is right here, is 10, and the mean is minus 5. And actually, let me put that in. So we're calculating for this example, the way it's drawn right here, the normal distribution Let's see. Our standard deviation is 10 times the square root of 2 pi, times e to the minus 1/2, times x minus our mean. Our mean is negative right now, right? Our mean is minus 5. So it's x plus 5 over the standard deviation squared, which is the variance. So that's 100 squared dx. This is what this number is right here. This 7%, or actually 0.07, is the area right under there. this isn't an easy integral to evaluate analytically, even for those of us who know our calculus. So this tends to be done numerically. And kind of an easy way to do this-- well, not an easy way-- but a function has been defined, called the cumulative distribution function, that is a useful tool for figuring out this area. So what the cumulative distribution function is, is essentially-- let me call it the cumulative distribution function-- it's a function of x. It gives us the area under this curve. So let's say that this is x right here. That's our x. It tells you the area under the curve up to x. Or so another way to think about it-- it tells you, what is the probability that you land at some value less than your x value? So it's the area from minus infinity to x of our probability density function, dx. When you actually use the Excel normal distribution function," + }, + { + "Q": "why is the middle one horizontal at 1:35 in the video?\n", + "A": "As Sal explain from 1:36 onward, the reason it was initially horizontal, is to remind us, how we have to put the numbers vertically, with the ones aligned, and the tens aligned. It s just a reminder that an addition might be presented in a test in this (horizontal) notation, and that we have to arrange them in the specific way to carry out the addition procedure.", + "video_name": "9hM32lsQ4aI", + "timestamps": [ + 95 + ], + "3min_transcript": "We have three different addition problems right over here. And what I want you to do so you get the hang of things is to pause the video and try them on your own. But as you do them, I want you to really keep in mind and think about what the carrying actually means. So I assume you've tried on your own. Now I'll work through them with you. So we have 9 + 6. 9 ones + 6 ones. Well 9 + 6 is 15. Well, we could write the 5 in the ones place, and then we carry the 1. But what did we just do? What does this 1 represent? Well, we put it in the tens place. 1 ten represents 10. So all we said is that 9 + 6 is equal to 10 + 5, is equal to 1 ten + 5, which is equal to 15. Now, in the tens place, we have 1 + 0 + 9, which is 10. And so we can write 0 and carry the 1. 1 + 0 + 9 is 10. Now what does that really mean? Well this is 1 ten + 0 tens + 9 tens, which is 10 tens. Or another way of thinking about 100, it's 1 hundred, and 0 tens. So that's all that carrying represents. So now we have 1 + 7 + 9. That is going to be 17. Now, we have to remind ourselves, this is in the hundreds place. So this is actually 1 hundred, plus 7 hundreds, plus 9 hundreds, or 17 hundreds. Or 1 thousand, 7 hundreds. And of course we have this 5 here. And we are done. Now let's try to tackle this one. Now the reason why I wrote it that way is to make sure that we remind ourselves to align the proper places under the appropriate places. So this problem we could rewrite as 373 + 88 (stacked vertically). We want to write the ones under the ones place, and the tens under the tens place, so that we're adding the appropriate place values. The 1 in the ones place, and the 1 in the tens place. 10 + 1 is 11. 1 + 7 is 8. 8 + 8 is 16. But this is actually 16 tens. So let me just write down the 16. What is 16 tens? Well, that's 160. So this 6 is a sixty. And then I have 100. 1 + 3 is 4. But these are actually hundreds, so it's 4 hundreds. So we get 461. Now finally, 9 + 3 is 12. 2 ones, and 1 ten. 12 is the same thing as 10 + 2. Now in the tens place: 1 + 4 + 9 is 14. So we write the 4, and carry the 1. But remind ourselves, this is actually 10 + 40 + 90, which is 140. Which is the same thing as 40 + 100. And then 1 + 1 + 2 is 4. But this is the hundreds place. So this is actually 1 hundred, plus 1 hundred, plus 2 hundreds or 4 hundreds." + }, + { + "Q": "At 2:00 what if it remained as dt? Could we still do it with respect to another variable, like in differentiation?\n", + "A": "Yes! You can still do the problem but treating all the other variables as constants. So you re answer would be: (e^a + 1/a)t + c Everything within the parenthesis can be thought of as a single constant. So the problem would be the same as say integral(3) = 3t + c Hope that helps!", + "video_name": "hXg-6YgAARk", + "timestamps": [ + 120 + ], + "3min_transcript": "I thought I would do a few more examples of taking antiderivatives, just so we feel comfortable taking antiderivatives of all of the basic functions that we know how to take the derivatives of. And on top of that, I just want to make it clear that it doesn't always have to be functions of x. Here we have a function of t, and we're taking the antiderivative with respect to t. And so you would not write a dx here. That is not the notation. You'll see why when we focus on definite integrals. So what's the antiderivative of this business right over here? Well, it's going to be the same thing as the antiderivative of sine of t, or the indefinite integral of sine of t, plus the indefinite integral, or the antiderivative, of cosine of t. So let's think about what these antiderivatives are. And we already know a little bit about taking the derivatives of trig functions. We know that the derivative with respect to t of cosine of t is equal to negative sine of t. would just have to take the derivative of negative cosine t. If we take the derivative of negative cosine t, then we get positive sine of t. The derivative with respect to t of cosine t is negative sine of t. We have the negative out front. It becomes positive sine of t. So the antiderivative of sine of t is negative cosine of t. So this is going to be equal to negative cosine of t. And then what's the antiderivative of cosine of t? Well, we already know that the derivative with respect to t of sine of t is equal to cosine of t. So cosine of t's antiderivative is just sine of t-- so plus sine of t. We've found the antiderivative. Now we don't have a t. We're taking the indefinite integral with respect to-- actually, this is a mistake. This should be with respect to a. If we were taking this with respect to t, then we would treat all of these things as just constants. But I don't want to confuse you right now. Let me make it clear. This is going to be da. That's what we are integrating or taking the antiderivative with respect to. So what is this going to be equal to? Well once again, we can rewrite it as the sum of integrals. This is the indefinite integral of e to the a da, so this one right over here-- a d I'll do it in green-- plus the indefinite integral, or the antiderivative, of 1/a da. Now, what is the antiderivative of e to the a? Well, we already know a little bit about exponentials. The derivative with respect to x of e to the x is equal to e to the x. That's one of the reasons why e in the exponential function in general is so amazing. And if we just replaced a with x or x with a, you get the derivative with respect to a of e to the a" + }, + { + "Q": "\nAt 3:45 couldn't the indefinite integral of (1/a) also be ln(a)+c or does the (a) part have to be its absolute value like Sal put it, ln(IaI)+c?\nThanks in advance", + "A": "Sometimes people don t bother with the absolute value for this integral, but in that case you re stating the integral only for positive values of x, because ln(x) is undefined for x<=0. When you include the absolute value, you state the integral for all x in the domain of 1/x (that is, all x except 0).", + "video_name": "hXg-6YgAARk", + "timestamps": [ + 225 + ], + "3min_transcript": "would just have to take the derivative of negative cosine t. If we take the derivative of negative cosine t, then we get positive sine of t. The derivative with respect to t of cosine t is negative sine of t. We have the negative out front. It becomes positive sine of t. So the antiderivative of sine of t is negative cosine of t. So this is going to be equal to negative cosine of t. And then what's the antiderivative of cosine of t? Well, we already know that the derivative with respect to t of sine of t is equal to cosine of t. So cosine of t's antiderivative is just sine of t-- so plus sine of t. We've found the antiderivative. Now we don't have a t. We're taking the indefinite integral with respect to-- actually, this is a mistake. This should be with respect to a. If we were taking this with respect to t, then we would treat all of these things as just constants. But I don't want to confuse you right now. Let me make it clear. This is going to be da. That's what we are integrating or taking the antiderivative with respect to. So what is this going to be equal to? Well once again, we can rewrite it as the sum of integrals. This is the indefinite integral of e to the a da, so this one right over here-- a d I'll do it in green-- plus the indefinite integral, or the antiderivative, of 1/a da. Now, what is the antiderivative of e to the a? Well, we already know a little bit about exponentials. The derivative with respect to x of e to the x is equal to e to the x. That's one of the reasons why e in the exponential function in general is so amazing. And if we just replaced a with x or x with a, you get the derivative with respect to a of e to the a So the antiderivative here, the derivative of e to the a, the antiderivative is going to be e to the a. And maybe you can shift it by some type of a constant. Oh, and let me not forget, I have to put my constant right over here. I could have a constant factor. So let me-- always important. Remember the constant. So you have a constant factor right over here. Never forget that. I almost did. So once again, over here, what's the antiderivative of e to the a? It is e to the a. What's the antiderivative of 1/a? Well, we've seen that in the last video. It is going to be the natural log of the absolute value of a. And then we want to have the most general antiderivative, so there could be a constant factor out here as well. And we are done. We found the antiderivative of both of these expressions." + }, + { + "Q": "At 5:57 Vi mentions an equilateral right triangle. How can an equilateral triangle also be a right triangle? Doesn't each angle have to be 60 degrees?\n", + "A": "Whenever Vi says equilateral right triangle, she means isosceles right triangle. She actually makes that mistake a lot.", + "video_name": "Oc8sWN_jNF4", + "timestamps": [ + 357 + ], + "3min_transcript": "And you know how line segments behave. And you convince yourself all 2D things scaled up by two get four times as much stuff, because 2D things can be thought of as being mad of squares, and you know how squares behave. But then there is this which has no straight lines in it. And there's no square areas in it, either. More than three to the one, less than three to the two. It behaves as if it's between one and two dimensions. You think back to Sierpinski's triangle. Maybe it can be thought of as being made out of straight line segments, though there's an infinite amount of them and they get infinitely small. When you make it twice as tall, if you just make all the lines of this drawing twice as long, you're missing detail again. But the tiny lines too small to draw are also twice as long and now visible. And so on, all the way down to the infinitely small line segments. You wonder if your similar line thing works on lines that don't actually have length. Wait. Lines that don't have length? Is that a thing? First, though, you figure out that when you make it twice as Not two, like a 1D triangle outline. Not four, like a solid 2D triangle. But somewhere in between. And the in between-ness seems to be true, no matter which way you make it-- out of lines, or by subtracting 2D triangles, or with squiggles. They all end up the same. An object in fractional dimension. No longer 1D because of infinity infinitely small lines. Or no longer 2D because of subtracting out all the area. Or being an infinitely squiggled up line that's too infinate and squiggled to be a line anymore, but doesn't snuggle into itself enough to have any 2D area, either. Though in the dragon curve, it does seem to snuggle up into itself. Hm. If you pretend this is the complete dragon curve and iterate this way, there's twice as much stuff. That's what you'd expect from a 1D line if it were scaling it by two. But let's see, this is scaling up by, well, not quite two. Let's see. I suppose if you did it perfectly, it's supposed to be an equilateral right triangle. So square root 2. If it were two dimensional, you'd square root 2 squared as much stuff. And square root 2 squared is, of course, two, which is the amount of stuff you got. Odd how dragons turn out to be exactly two dimensional. But in the end, you get a fill-up thing with a fractal edge. And that's a lot like a filled in dragon dungeon. 2D area of pink steel on the inside, infinite fractal patina on the outside. Except the dragon curve still gets weirdness points for getting its 2D-ness from an infinitely squiggled up line rather than triangles that are 2D to begin with. And now you're plagued by another thought. What about an infinitely long line, a true line, rather than the line segments you've been dealing with? In a way, an infinitely long line doesn't have a length. Not a defined one. Like the infinitely short line, there's no real number capable of describing it. And if there's no number for it, how can you multiply that number by 2? If, when you make a line twice as long, you don't get a line with exactly twice the length, is a line really one dimensional? And if the Koch curve is made out" + }, + { + "Q": "\nAt 2:50what would be an example of what he's explaining?", + "A": "Sal is saying that given two functions, if each is substituted into the other, different answers will (usually) result. Say we had f(x) = x - 3 and g(x) = - 4x + 7. Substituting f(x) into g will give us - 4 ( x - 3 ) +7 which simplifies to - 4x +19. But substituting g(x) into f will give us ( - 4x + 7 ) - 3 which simplifies to - 4x +4. Hope this helps!", + "video_name": "_b-2rZpX5z4", + "timestamps": [ + 170 + ], + "3min_transcript": "Voiceover:When we first got introduced to function composition, we looked at actually evaluating functions at a point, or compositions of functions at a point. What I wanna do in this video is come up with expressions that define a function composition. So, for example, I wanna figure out, what is, f of, g of x? f of, g of x. And I encourage you to pause the video, and try to think about it on your own. Well, g of x in this case, is the input to f of x. So, wherever we see the x in this definition, that's the input. So we're going to replace the input with g of x. We're going to replace the x with g of x. So, f of g of x is going to be equal to the square root of- Well instead of an x, we would write a g of x. g of x, g of x squared. g of x squared, minus one. Now what is g of x equal to? So this is going to be equal to the square root of, g of x, is x over 1 plus x. We're going to square that. We're going to square that, minus 1. So f of g of x, is also a function of x. So f of g of x is a square root of, and we could write this as x squared over 1 plus x squared, but we could just leave it like this. It's equal to the square root of this whole thing, x over 1 plus x, squared, minus one. Now let's go the other way round. What is g of f of x? What is g of f of x? And once again, I encourage you to pause the video, and try to think about it on your own. Well, f of x is now the input into g of x. So everywhere we see the x here, we'll replace it with f of x. So this is going to be equal to, this is going to be equal to, f of x, over- so you can appreciate it better. f of x over, one plus f of x. One plus f of x. And what's that equal to? Well, f of x is equal to the square root, of x squared minus one. x squared minus one. So it's gonna be that over 1, plus the square root. One plus the square root of x squared minus one. So this is a composition f of g of x, you get this thing. This is g of f of x, where you get this thing. And to be clear, these are very different expressions. So typically, you want the composition one way. This isn't gonna be the same as the composition the other way, unless the functions are designed in a fairly special way." + }, + { + "Q": "\nAt 2:35, can you cancel out the square root of x - 1 in the numerator and the denominator? What would the answer be if you could?", + "A": "First, it is sqrt(x^2-1), not x-1 in both numerator & denominator. Second, you can t cancel them out because the sqrt(x^2-1) in the denominator is being added to 1. You can only cancel items being multiplied (factors), not terms (items being added/subtracted).", + "video_name": "_b-2rZpX5z4", + "timestamps": [ + 155 + ], + "3min_transcript": "Voiceover:When we first got introduced to function composition, we looked at actually evaluating functions at a point, or compositions of functions at a point. What I wanna do in this video is come up with expressions that define a function composition. So, for example, I wanna figure out, what is, f of, g of x? f of, g of x. And I encourage you to pause the video, and try to think about it on your own. Well, g of x in this case, is the input to f of x. So, wherever we see the x in this definition, that's the input. So we're going to replace the input with g of x. We're going to replace the x with g of x. So, f of g of x is going to be equal to the square root of- Well instead of an x, we would write a g of x. g of x, g of x squared. g of x squared, minus one. Now what is g of x equal to? So this is going to be equal to the square root of, g of x, is x over 1 plus x. We're going to square that. We're going to square that, minus 1. So f of g of x, is also a function of x. So f of g of x is a square root of, and we could write this as x squared over 1 plus x squared, but we could just leave it like this. It's equal to the square root of this whole thing, x over 1 plus x, squared, minus one. Now let's go the other way round. What is g of f of x? What is g of f of x? And once again, I encourage you to pause the video, and try to think about it on your own. Well, f of x is now the input into g of x. So everywhere we see the x here, we'll replace it with f of x. So this is going to be equal to, this is going to be equal to, f of x, over- so you can appreciate it better. f of x over, one plus f of x. One plus f of x. And what's that equal to? Well, f of x is equal to the square root, of x squared minus one. x squared minus one. So it's gonna be that over 1, plus the square root. One plus the square root of x squared minus one. So this is a composition f of g of x, you get this thing. This is g of f of x, where you get this thing. And to be clear, these are very different expressions. So typically, you want the composition one way. This isn't gonna be the same as the composition the other way, unless the functions are designed in a fairly special way." + }, + { + "Q": "\nat 11:20, I dont understand, like why does the y axis have to have the price, and the x axis is months why cant the y axis be months, and x axis be price ?", + "A": "X is usually used for representing an independent variable. Y is usually for dependent variables. Since time is independent (time goes on no matter what) the months are put on the X axis. I don t expect text books to do it the other way around.", + "video_name": "36v2EXZRzUE", + "timestamps": [ + 680 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 5:19, Why didn't he use positive 4 instead of \"I minus x\"? which is from what I can understand here is, 1 minus 4? And how did he get the answer 4 instead?", + "A": "Using the lower value of the domain, -3. 1-(-3) = 4", + "video_name": "rpI-X9Gn5a4", + "timestamps": [ + 319 + ], + "3min_transcript": "So when does this thing hit its low point? o this thing hits, hits its low point when x is as small as possible. An x is going to be as small as possible when x is approaching negative 6. So if x were equal to negative 6, it can't equal negative 6 herer but if x is equal to negative 6, then this thing over here would be equal to negative 6 plus 7, would be, would be 1. So if x is greater than negative 6, g of x is going to be greater than 1, or another way to think about it is if negative 6 is less than x, then 1 is going to be less than g of x. And the reason I said that is if I put negative 6 into this, negative 6 plus 7 is equal to 1. Now this gonna hit a high end when it as large as possible. The largest value in this interval that we can take on is x being equal to negative 3. So when x is equal to negative 3, negative 3 plus 7 is equal to equal to, so we can actually take on x equals negative 3 in which case g of x actually will take on positive 4. So, let' do that for each of these. Now here we have 1 minus x, so this is going to take on its smallest value when x is as large as possible. So the largest value x can approach for, it can't quite take on for, but it's going to approach for. So if x, let's see, if we said x was 4, although that's not this clause here, 1 minus x, 1 minus 4 is negative 3. So as long as x is less than 4, then negative 3 is going to be less than g of x. I wanna make sure that makes sense to you because it can be little bit confusing because this takes on its minimum value when x is approaching, or it's approaching its minimum value when x is approaching its, when x is approaching its maximum because So if you take the upper end, even though this doesn't actually include 4, but as we approach for, we could say, OK, 1 negative 4 is negative 3 so that's, so g of x is always going to be greater than that, as well it's going to be going to be a less than. Well what happens as we approach x being equal to negative 3? So, 1 minus negative 3 is going to be positive 4. So this is going to be positive 4 right over here. And these are both less than, not less than or equal to because these are both less than right over here. And now let's think about this right over here. So 2x minus 11 is gonna hit its maximum value when x is as large as possible. So its maximum value's going to be hit when x is equal to 6 So 2 times 6 is 12, minus 11. Well that's going to be 1. So its maximum value's going to be 1. It's actually going to be able to hit because x can be equal to 6. Its minimum value is going to be when x is equal to 4, and actually can be equal to" + }, + { + "Q": "When Salman does range past 3:00, why are x values inputted?\n", + "A": "Since this is a piecewise function, you have to figure out the range for each piece . And, each piece is used depending upon the value of X. So, if you are looking for the range of the 1st piece, you look at the X-values that would use that piece and what they create for output values. Hope this helps.", + "video_name": "rpI-X9Gn5a4", + "timestamps": [ + 180 + ], + "3min_transcript": "is less than x and I'm leaving -- so let's write it here. All real numbers -- actually let me write this way x, I could write it more math-y. I could say x is a member of the real numbers such that, such that negative 6 is less than x. Negative 6 is less than x and I also think about the upper bound. So as x goes, I just wanna make sure that we fill in all the gaps between x being a greater than negative 6 and x is less than or equal to 6. So let's see. As we go up to and including negative 3, we're in this clause. As soon as we cross negative 3, we fall into this clause up to 4, but as soon as we get 4, we're in this clause up to and including 6. So x at the high end is said to be less than or equal to 6, less then or equal to 6. Now another way to say this and kind of less math-y notation is x, x real number, any the real number such that, such that negative 6 is less than x is less than or equal to 6. These two statements are equivalent. So now let's think about the range of this function. Let's think about the range, and the range is, this is the set of all inputs , oh sorry, this is the set of all outputs that this function can take on, or all the values that this function can take on. And to do that, let's just think about as x goes, but x varies or x can be any values in this interval. What are the different values that g of x could take on? Let's think about that. g of x is going to be between what and what? g of x is going to be between what and what? g of x is going to be between what and what? And it might actually, this might be some equal signs there but I'm gonna worry about So when does this thing hit its low point? o this thing hits, hits its low point when x is as small as possible. An x is going to be as small as possible when x is approaching negative 6. So if x were equal to negative 6, it can't equal negative 6 herer but if x is equal to negative 6, then this thing over here would be equal to negative 6 plus 7, would be, would be 1. So if x is greater than negative 6, g of x is going to be greater than 1, or another way to think about it is if negative 6 is less than x, then 1 is going to be less than g of x. And the reason I said that is if I put negative 6 into this, negative 6 plus 7 is equal to 1. Now this gonna hit a high end when it as large as possible. The largest value in this interval that we can take on is x being equal to negative 3. So when x is equal to negative 3, negative 3 plus 7 is equal to" + }, + { + "Q": "\nat 1:55, you crossed out the 9 and made it a 3, and you made the 3 into a 1. when you did the simplified equation, instead of writing 3/100+1/1*, you wrote 3/100+1/3* shown at 2:22. when i did the problem 3/100+1/1, the answer was 103/100. when you did the problem you answered 109/300.\ndid you make a mistake? or did i miss something in a previous video?", + "A": "Why do you have the extra * at the end of two numbers? Sal simplified 1/3 + 9/100 * 1/3 into 1/3 + 3/100 * 1/1. Because *1/1 is the same as *1 and doesn t change anything, he wrote it as 1/3 + 3/100 instead, leaving out the *1/1.", + "video_name": "MZpULgKhaEU", + "timestamps": [ + 115, + 142 + ], + "3min_transcript": "Most liquids, when cooled, will simply shrink. Water, on the other hand, actually expands when it is frozen. Its volume will increase by about 9%. Suppose you have 1/3 of a gallon of water that gets frozen. What is the volume of the ice that you now have? So you're starting with 1/3 of a gallon of water. They tell us that when it gets frozen, when it turns into ice, its volume is going to expand by 9%. So the new volume is going to be your existing volume. So this is the original volume, 1/3 of a gallon, and it's going to expand by 9%. So your frozen volume is going to be your original volume plus 9% of your original volume. So you could say it's 9% times 1/3. So this right over here is going to be the expanded volume. We could turn things to decimals or whatever else, but they tell us to express your answer as a fraction. So let's make sure that everything here is a fraction, and then we'll just try to simplify. So the one thing that's sitting here that is not a fraction is our 9%. Well, what does 9% actually represent? Well, 9% literally means 9 per 100. So we could rewrite this as-- so this is going to be equal to 1/3 plus, instead of writing 9%, I'll write that as 9 per 100, and then once again times 1/3. And we can simplify this expression right over here. We have a 9 in the numerator, a 3 in the denominator. If we divide both of them by 3, we get a 3 and a 1. And so we're left with 1/3 plus 300 times 1/1. Well, that's just going to be 3/100. write this in orange still, or maybe I'll do it in a new color-- plus 3/100. And now we have to add something, or two numbers that have different denominators. So let's find a common denominator. So this is going to be equal to, well, the least common multiple of 3 and 100. And they share no common factor, so it's really just going to be the product of 3 and 100-- the least common multiple is So it's going to be something over 300 plus something over 300. Now to go from 3 to 300, in the denominator you multiply by 100, so you have to multiply the numerator by 100 as well. So 1/3 is the same thing as 100/300. And to go from 100 to 300, we have to multiply by 3 in the denominator, so we have to multiply by 3 in the numerator as well. So 3/100 is the same thing as 9/300." + }, + { + "Q": "\nhow can the temperature be same on a particular day every year. At 2:37 you can observe Sal saying in future January 7 also you will have the highest average temperature as 29C .", + "A": "It isn t the same on a particular day every year, but if you averaged all of the temperatures that it had previously been on that day of the year, it would be the temperature that Sal used. Therefore, the temperature could really have been 30C or 26C, but it is most likely to be 29C.", + "video_name": "mVlCXkht6hg", + "timestamps": [ + 157 + ], + "3min_transcript": "pretty apparent why they are suggesting that we use a trigonometric function to model this. Because our seasonal variations they're cyclical. They go up and down. Actually, if you look at the average temperature for any city over the course of the year, it really does look like a trigonometric function. This axis right over here. This is the passage of the days. Let's do d for days and that's going to be in days after January seventh. So this right over here would be January seventh. And the vertical axis, this is the horizontal axis. The vertical axis is going to be in terms of Celsius degrees. The high is 29 and I could write 29 degrees Celsius. The highest average day. If this is zero then 14, which is the lowest average day. 14 degrees Celsius. The highest average day, which they already told us, is January seventh we get to 29 degrees Celsius. And then the coldest day of the year, on average, you get to an average high of 14 degrees Celsius. So it looks like this. We're talking about average highs on a given day and the reason why a trigonometric function is a good idea is because it's cyclical. If this is January seventh, if you go 365 days in the future, you're back at January seventh. If the average high temperature is 29 degrees Celsius on that day, the average high temperature is going to be 29 degrees Celsius on that day. Now, we're using a trigonometric function so we're going to hit our low point exactly halfway in between. So we're going to hit our low point exactly halfway in between. Something right like that. So our function is going to look like this. I'm going to draw the low point right over there and this is a high point. That's a high point right over there. That looks pretty good. Then, I have the high point right over here and then, I just need to connect them and there you go. I've drawn one period of our trigonometric function and our period is 365 days. If we go through 365 days later we're at the same point in the cycle, we are at the same point in the year. We're at the same point in the year. So, what I want you to do right now is, given what I've just drawn, try to model this right. So, this right over here, let's write this as T as a function of d. Try to figure out an expression for T as a function of d and remember it's going to be some trigonometric function. So, I'm assuming you've given a go at it and you might say, \"Well this looks like a cosine curve, maybe it could be \"a sine curve, which one should I use?\" You could actually use either one, but I always like to go with the simpler one." + }, + { + "Q": "At 2:48, why did he take 25^2 and not -25^2 like he did with the -b part of the abc formula?\n", + "A": "He made a mistake saying it, it is supposed to be (-25)\u00c2\u00b2 but because when you square a number or its negative, it will be the same thing, so he got away with it. Good job spotting it. You should report so they can add an annotation to youtube that he meant (-25)\u00c2\u00b2 to clear the confusion for others.", + "video_name": "711pdW8TbbY", + "timestamps": [ + 168 + ], + "3min_transcript": "So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8. this is over here. So let's say so we have 625 minus-- let's see., this is going to be 16 times 36. 16 times 36 is equal to 49. It's a nice perfect square. We know what the square root of 49 is. It's 7. So let me go back to the problem. So this in here simplified to 49. So x is equal to 25 plus or minus the square root of 49, which is 7, all of that over 8. So our two solutions here, if we add 7, we get x is equal to 25 plus 7 is 32, 32/8, which is equal to 4. And then our other solution, let me do that in a different color." + }, + { + "Q": "1:27 is really confusing. Need some clarification\n", + "A": "He expanded (2x-6)^2, making it now 4x^2-24x+36. When you have the square of a binomial, you take square of the first monomial=4x^2; multiply (2)(the first)(the second): (2)(2x)(-6): -24x. Then you square the second=36. 4x^-24x+36. If you didn t get that you can multiply (2x-6)(2x-6). You will get the same.", + "video_name": "711pdW8TbbY", + "timestamps": [ + 87 + ], + "3min_transcript": "In this video, we're going to start having some experience solving radical equations or equations that involve square roots or maybe even higher-power roots, but we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's say I have the equation the square root of x is equal to 2 times x minus 6. Now, one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that, I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8." + }, + { + "Q": "at 2:17 after he subtracted x from both sides why did he right -25x when in the equation above it he had written -24x, and where did he subtract the x on the right hand side from\n", + "A": "becuz hes subtracting a -24 by a -x and the x on the right hand side hes combining with the -24. so a -24 minus -x=-25", + "video_name": "711pdW8TbbY", + "timestamps": [ + 137 + ], + "3min_transcript": "In this video, we're going to start having some experience solving radical equations or equations that involve square roots or maybe even higher-power roots, but we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's say I have the equation the square root of x is equal to 2 times x minus 6. Now, one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that, I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8." + }, + { + "Q": "\nAt 0:55, isn't PEMDAS the same as Please Excuse My Dear Aunt Sally?", + "A": "Yes. It is the same thing. :) some people find it easier to remember PEMDAS, others say GEMDAS (where G stands for Grouping symbols), and yet others memorize the mnemonic Please Excuse My Dear Aunt Sally. Another version of it is Please Excuse My Dear Aunt Sally, she limps from left to right. This helps the student remember which direction do do these symbols in. :) Hope this helps! Sylvia.", + "video_name": "0uCslW40VHQ", + "timestamps": [ + 55 + ], + "3min_transcript": "Evaluate the expression 5y to the fourth minus y squared when y is equal to 3. So every place we see a y here, we could just replace it with a 3 to evaluate it. So it becomes 5 times 3 to the fourth power minus 3 squared. All I did is every time we saw a y here, I put a 3 there. Every time we saw a y, I put a 3. So what does this evaluate to? And we have to remember our order of operations. Remember, parentheses comes first. Sometimes it's referred to as PEMDAS. Let me write that down. PEMDAS, PEMDAS. P is for parentheses. E is for exponents. M and D are for Multiplication and Division. They're really at the same level of priority. And then addition and subtraction If you really want to do it properly, it should be P-E, and then multiplication and division are really at the same level. And addition and subtraction are at the same level. But what this tells us is that we do parentheses first. But then after that, exponentiation takes priority over everything else here. before we multiply anything or before we subtract anything. So the one exponent we'd have to evaluate is 3 squared. So let's remember. 3 to the first is just 3. It's just 3 times itself once. 3 squared is equal to 3 times 3, 3 multiplied by itself twice. That's equal to 9. 3 to the third power is equal to 3 times 3 times 3. Or you could view it as 3 squared times 3. So it'll be 9. 3 times 3 is 9. 9 times 3 is equal to 27. 3 to the fourth is equal to 3 times 3 times 3 times 3. So 3 times 3 is 9. 3 times 3 is 9. So it's going to be the same thing as 9 times 9. So this is going to be equal to 81. So we now know what 3 to the fourth is. We know what 3 squared is. Let's just put it in the expression. So this is going to be equal to 5 times 3 to the fourth. 3 to the fourth is 81. And we have 3 squared right over here. It is equal to 9. 5 times 81 minus 9. Let's figure out what 5 times 81 is. So 81 times 5. 1 times 5 is 5. 8 times 5 is 40. So this right over here is 405. So it becomes 405 minus 9. So that is going to be equal to-- if we were subtracting 10, it would be 395. But we're subtracting one less than that. So it's 396. And we're done." + }, + { + "Q": "\nIn the video on calculating the square footage of a house, at 2:50 how did you get the width of the green rectangle to be 25ft? Shouldn't it be 26ft, the same as the width of the blue one?", + "A": "It isn t drawn to scale.", + "video_name": "xo4VpX2IIMk", + "timestamps": [ + 170 + ], + "3min_transcript": "It has a width of 20 feet, and it has a length of 20 feet. So that would be a rectangle right over there. We should be able to figure out its area. Then I could set up another rectangle that has a width of 26 feet and has a length. That's its length right over there. And we could think about in a second what that length actually is. Actually, let's think about that. How would we figure out what this length actually is? Well, this length plus 5 feet is going to be the same thing as this length over here. It's the same as the opposite wall of this rectangle. So this length plus 5 feet is 20 feet. Well, this must be 15 feet. So this blue rectangle is 15 feet long and 26 feet wide. Now let's add another rectangle. We could have one that's 18 feet long and then goes the entire length of the house. Goes the entire length of the house like that. How do we figure out the width of this rectangle? Well, we know that this is 8 feet. We know that this is 20 feet, and we know that this is 26 feet. So the entire width is going to be 26 feet plus 20 feet. So 26 plus 20 gets us to 46. Plus 8 gets us to 54 feet. So this is 54 feet right over here. Did I do that right? 8 plus 36 would be 34, plus 20 is 54 feet. And then finally we have one last rectangle to deal with, this rectangle right over here, which is 15 feet long and 25 feet wide. And so now we can calculate the areas of the different rectangles. So the total area is going to be the 20 feet by the 20 feet. So let's just multiply them. So it's going to be 20 times 20. Plus 15 times 26. That's this area right over there. Plus 18 times 54, which is this area right over there. And then finally, plus 15 times 25, which is this area right over here. So we just have to now evaluate these things. So what is 20 times 20? Well, this is going to be 400. What's 15 times 26? Well, let's multiply it out. 26 times 15. 6 times 5 is 30. 2 times 5 is 10, plus 3 is 13." + }, + { + "Q": "At 1:00, couldn't there be another line that bisects angle BCA starting from point C and ending at a point that is between A and B?\n", + "A": "Yes. There could be another line.", + "video_name": "21vbBiCVijE", + "timestamps": [ + 60 + ], + "3min_transcript": "I have triangle ABC here. And in the last video, we started to explore some of the properties of points that are on angle bisectors. And now, what I want to do in this video is just see what happens when we apply some of those ideas to triangles or the angles in triangles. So let's bisect this angle right over here-- angle BAC. And let me draw an angle bisector. So the angle bisector might look something-- I want to make sure I get that angle right in two. Pretty close. So that looks pretty close. So that's the angle bisector. Let me call this point right over here-- I don't know-- I could call this point D. And then, let me draw another angle bisector, the one that bisects angle ABC. So let me just draw this one. It might look something like that right over there. And I could maybe call this point E. So AD bisects angle BAC, and BE bisects angle ABC. So the fact that this green line-- AD bisects this angle right over here-- be equal to that angle right over there. They must have the same measures. And the fact that this bisects this angle-- angle ABC-- tells us that the measure of this angle-- angle ABE-- must be equal to the measure of angle EBC. Now, we see clearly that they have intersected at a point inside of the triangle right over there. So let's call that point I just for fun. I'm skipping a few letters, but it's a useful letter based on what we are going to call this in very short order. And there's some interesting things that we know about I. I sits on both of these angle bisectors. And we saw in the previous video that any point that sits on an angle bisector is equidistant from the two sides of that angle. So for example, I sits on AD. So it's going to be equidistant from the two sides of angle BAC. So this is one side right over here. And then this is the other side right over there. So because I sits on AD, we know that these two distances are going to be the same, assuming that this is the shortest distance between I and the sides. And then, we've also shown in that previous video that, when we talk about the distance between a point and a line, we're talking about the shortest distance, which is the distance you get if you drop a perpendicular. So that's why I drew the perpendiculars right And let's label these. This could be point F. This could be point G right over here. So because I sits on AD, sits on this angle bisector, we know that IF is going to be equal to IG. Fair enough. Now, I also sits on this angle bisector. It also sits on BE, which says that it must be equidistant. The distance to AB must be the same as I's distance to BC. I's distance to AB we already just said is this right over here." + }, + { + "Q": "ok, be sure to watch all the way to the end ! sal makes an important correction after minute 5:26. for a minute there i thought i still didn't get how to simplify fractions! whew... :))\n", + "A": "This is a duplicate - see the question above", + "video_name": "CLrImGKeuEI", + "timestamps": [ + 326 + ], + "3min_transcript": "All of that over negative 2. And let's see if we can simplify the radical expression here, the square root of 60. Let's see, 60 is equal to 2 times 30. 30 is equal to 2 times 15. And then 15 is 3 times 5. So we do have a perfect square here. We do have a 2 times 2 in there. It is 2 times 2 times 15, or 4 times 15. So we could write, the square root of 60 is equal to the square root of 4 times the square root of 15, right? The square root of 4 times the square root of 15, that's what 60 is. 4 times 15. And so this is equal to-- square root of 4 is 2 times the square of 15. So we can rewrite this expression, right here, as being equal to negative 8 plus or minus 2 times the square Now both of these terms right here are divisible by either 2 or negative 2. So let's divide it. So we have negative 8 divided by negative 2, which is positive 4. So let me write it over here. Negative 8 divided by negative 2 is positive 4. And then you have this weird thing. Plus or minus 2 divided by negative 2. And really what we have here is 2 expressions. But if we're plus 2 and we divide by negative 2, it will be negative 1. And if we take negative 2 and divide by negative 2, we're going to have positive 1. So instead of plus or minus, you could imagine it is going to be minus or plus. But it's really the same thing. It's really now minus or plus. If it was plus, it's now going to be a minus. If it was a minus, it's now going to be a plus. Minus or plus 2 times the square root of 15. Or another way to view it is that the two solutions here These are both values of x that'll satisfy this equation. And if this confuses you, what I did, turning a plus or minus into minus plus. Let me just take a little bit of an aside there. I could write this expression up here as two expressions. That's what the plus or minus really is. There's a negative 8 plus 2 roots of 15 over negative 2. And then there's a negative 8 minus 2 roots of 15 over negative 2. This one simplifies to-- negative 8 divided by negative 2 is 4. 2 divided by negative 2 is negative 1. 2 times a 4 minus the square root of 15. And then over here you have negative 8 divided by negative 2, which is 4. And then negative 2 divided by negative 2, which is plus the square of 15." + }, + { + "Q": "\nAt 2:06 how did he get the four in the equation? (8^2 -4 (-1)(-1)...", + "A": "It is part of the quadratic formula: x = [-B +/= sqrt(B^2 - 4AC)] / 2A", + "video_name": "CLrImGKeuEI", + "timestamps": [ + 126 + ], + "3min_transcript": "Use the quadratic formula to solve the equation, negative x squared plus 8x is equal to 1. Now, in order to really use the quadratic equation, or to figure out what our a's, b's and c's are, we have to have our equation in the form, ax squared plus bx plus c is equal to 0. And then, if we know our a's, b's, and c's, we will say that the solutions to this equation are x is equal to negative b plus or minus the square root of b squared minus 4ac-- all of that over 2a. So the first thing we have to do for this equation right here is to put it in this form. And on one side of this equation, we have a negative x squared plus 8x, so that looks like the first two terms. But our constant is on the other side. So let's get the constant on the left hand side and get a 0 here on the right hand side. So let's subtract 1 from both sides of this equation. The left hand side of the equation will become negative x squared plus 8x minus 1. Now we have it in that form. We have ax squared a is negative 1. So let me write this down. a is equal to negative 1. a is equal to negative 1. It's implicit there, you could put a 1 here if you like. A negative 1. Negative x squared is the same thing as negative 1x squared. b is equal to 8. So b is equal to 8, that's the 8 right there. And c is equal to negative 1. That's the negative 1 right there. So now we can just apply the quadratic formula. The solutions to this equation are x is equal to negative b. Plus or minus the square root of b squared, of 8 squared, the green is the part of the formula. The colored parts are the things that we're substituting into the formula. Minus 4 times a, which is negative 1, times negative 1, times c, which is also negative 1. And then all of that-- let me extend the square root sign a little bit further --all of that is going to be over 2 times a. In this case a is negative 1. So let's simplify this. So this becomes negative 8, this is negative 8, plus or minus the square root of 8 squared is 64. And then you have a negative 1 times a negative 1, these just cancel out just to be a 1. So it's 64 minus is 4. That's just that 4 over there. All of that over negative 2. So this is equal to negative 8 plus or minus the" + }, + { + "Q": "\nAt 0:35 what would be the way to solve for x?", + "A": "3x + 5 = 17 To solve for x, we need to be on one side of the equation by itself. To do this we could first subtract 5 from both sides of the equation so that we would get 3x = 12. Then we can divide both sides of the equation by 3 so that the equation then becomes x = 4 Did that help?", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 35 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 0:34 Arnt you supposed to do minus 3 on both sides? 3x=17 im new to this thing", + "A": "For 3x=17, we need to divide by 3 because x is being multiplied by 3 and the opposite of multiplication is division, not subtraction. Does that make sense?", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 34 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 8:21 can you make the fraction -8/7 a decimal?", + "A": "Yes! It is also equal to -(8/7). So you can make it a negative mixed number -(1 and 1/7). You ignore the 1 for a moment, and you do 1 divided by 7, which is 0.14285714285, and 0.14285714285 plus the ignored 1 is equal to 1.14285714285.", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 501 + ], + "3min_transcript": "" + }, + { + "Q": "xD is that all you simplify it to!? seems like you alreddy got the awnser at 8:21\n", + "A": "yes you want to make sure u simplify everything if u want it to be correct", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 501 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:58 What if the outcomes werent all equally likely? How would I solve it?\n", + "A": "As long as the other holly wands have a probability not all zero, the answer would be the same. This is because the probability of a set of events happening must be greater than the probability of a proper subset of them happening.", + "video_name": "0uHhk7P9SNo", + "timestamps": [ + 118 + ], + "3min_transcript": "- So, this right over here is a screenshot of the Describing Subsets of Sample Spaces exercise on Khan Academy, and I thought I would do a couple of examples, just because it's good practice just thinking about how do we describe sets and subsets. So it reads, Harry Potter is at Ollivanders Wand Shop. As we all know, the wand must choose the wizard, so Harry cannot make the choice himself. He interprets the wand selection as a random process so he can compare the probabilities of different outcomes. The wood types available are holly, elm, maple, and wenge. The core materials on offer are phoenix feather, unicorn hair, dragon scale, raven feather and thestral tail. All right! Based on the sample space of possible outcomes listed below, what is more likely? And so, we see here, we have four different types of woods for the wand, and then each of those could be combined with five different types of core, Phoenix Feather, Unicorn Hair, Dragon Scale, And so, that gives us four different woods and each of those can be combined with five different cores. 20 possible outcomes. And they don't say it here but they way they're talking I guess we can, I'm going to go with the assumption that they're equally likely outcomes, although it would have been nice if they said that, \"These are all equally likely\" but these are the 20 outcomes. And so, which of these are more likely? The wand that selects Harry will be made of holly or unicorn hair. So, how many of those outcomes involve this? So if, Holly are these five outcomes and if you said, \"Holly or Unicorn Hair\" it's going to be these five outcomes plus, well this one involves Unicorn Hair but we've already included this one, but the other ones that's not included for the Holly, that involve Unicorn Hair, are the Elm Unicorn, the Maple Unicorn and the Wenge Unicorn. So it's these five, plus these three, right over here. So eight of these 20 outcomes. And if these are all equally likely outcomes, that means there is an 8/20 probability of a wand So this is 8/20 or, that's the same thing as 4/10 or 40% chance. Now, the wand that selects Harry will be made of Holly and Unicorn Hair. Well, Holly and Unicorn Hair, that's only one out of the 20 outcomes. So this, or course, is going to be a higher probability. It actually includes this outcome and then seven other outcomes. So, the first choice includes the outcome for the second choice plus seven other outcomes. So this is definitely going to be a higher probability. Let's do a couple more of these or at least one more of these. You and a friend are playing \"Fire-Water-Sponge\". I've never played that game. In this game, each of the two players chooses fire, water or sponge. Both players reveal their choice at the same time and the winner is determined based on the choices. I guess this is like Rock, Paper, Scissors. Fire beats sponge by burning it." + }, + { + "Q": "\nAt 1:34, why did he switched around y-x=5 to y=5+x?", + "A": "if you add x to both sides of the equal sign, then you can eliminate the negative x on the left side, and add a positive x on the right side (-x + x = 0, and 0 + x = x)", + "video_name": "0BgUKHTW37E", + "timestamps": [ + 94 + ], + "3min_transcript": "We're asked to solve and graph this system of equations here. And just as a bit of a review, solving a system of equations really just means figuring out the x and y value that will satisfy both of these equations. And one way to do it is to use one of the equations to solve for either the x or the y, and then substitute for that value in the other one. That makes sure that you're making use of both constraints. So let's start with this bottom equation right here. So we have y minus x is equal to 5. It's pretty straightforward to solve for y here. We just have to add x to both sides of this equation. So add x. And so the left-hand side, these x's cancel out, the negative x and the positive x, and we're left with y is equal to 5 plus x. Now, the whole point of me doing that, is now any time we see a y in the other equation, we can replace it with a 5 plus x. So the other equation was-- let me do it in this orange This second equation told us, if we just rearranged it, that y is equal to 5 plus x, so we can replace y in the second equation with 5 plus x. That makes sure we're making use of both constraints. So let's do that. We're going to replace y with 5 plus x. So this 9x plus 3y equals 15 becomes 9x plus 3 times y. The second equation says y is 5 plus x. So we're going to put 5 plus x there instead of a y. 3 times 5 plus x is equal to 15. And now we can just solve for x. We get 9x plus 3 times 5 is 15 plus 3 times x is 3x is equal to 15. So we can add the 9x and the 3x, so we get 12x, plus 15, is Now we can subtract 15 from both sides, just so you get only x terms on the left-hand side. These guys cancel each other out, and you're left with 12x is equal to 0. Now you divide both sides by 12, and you get x is equal to 0/12, or x is equal to 0. So let me scroll down a little bit. So x is equal to 0. Now if x equals 0, what is y? Well, we could substitute into either one of these equations up here. If we substitute x equals 0 in this first equation, you get 9 times 0 plus 3y is equal to 15. Or that's just a 0, so you get 3y is equal to 15. Divide both sides by 3, you get y is equal to 15/3 or 5. y is equal to 5." + }, + { + "Q": "\nat 1:30 of the video why do we need to subtract 5 or add -5 to the problem 3 times.", + "A": "Because 3x has an x in it, and x is a variable meaning it can change. so if: x=2 3x = 6 there are too many possibilites, so we would call that another term But -5 is just -5! So it would not change. That is why we do that. Hope that helped!", + "video_name": "cvB8b4AACyE", + "timestamps": [ + 90 + ], + "3min_transcript": "We're asked to solve for x. And we have this compound inequality here, negative 16 is less than or equal to 3x plus 5, which is less than or equal to 20. And really, there's two ways to approach it, which are really the same way. And I'll do both of them. And I'll actually do both of them simultaneously. So one is to just solve this compound inequality And I'll just rewrite it. Negative 16 is less than or equal to 3x plus 5, which is less than or equal to 20. And the other way is to think of it as two separate inequalities, but both of them need to be true. So you could also view it as negative 16 has to be less than or equal to 3x plus 5. And 3x plus 5 needs to be less than or equal to 20. This statement and this statement are equivalent. This one may seem a little bit more familiar because we can independently solve each of these inequalities and just remember the \"and.\" This one might seem a little less traditional because now we have three sides to the statement. We have three parts of this compound inequality. But what we can see is that we're actually In any situation, we really just want to isolate the x on one side of the inequality, or in this case, one part of the compound inequality. Well, the best way to isolate this x right here is to first get rid of this positive 5 that's sitting in the middle. So let's subtract 5 from every part of this compound inequality. So I'm going to subtract 5 there, subtract 5 there, and subtract 5 over there. And so we get negative 16 minus 5 is negative 21, is less than or equal to 3x plus 5 minus 5 is 3x, which is less than or equal to 20 minus 5, which is 15. And we could essentially do the same thing here. If we want to isolate the 3x, we can subtract 5 from both sides. We get negative 21. Negative 21 is less than or equal to 3x. And we get, subtracting 5 from both sides. And notice, we're just subtracting 5 from every part of this compound inequality. So this statement and this statement, once again, are the exact same thing. Now, going back here, if we want to isolate the x, we can divide by 3. And we have to do it to every part of the inequality. And since 3 is positive, we don't have to change the sign. So let's divide every part of this compound inequality by 3. You divide every part by 3. This is equivalent to dividing every part of each of these inequalities by 3. And then we get negative 21 divided by 3 is negative 7, is less than or equal to x, which is less than or equal to 15 divided by 3 is 5. You get negative 7 is less than or equal to x, and x is less than or equal to 15/3, which is 5. This statement and this statement are completely equal. And we've solved for x. We've given you the solution set. And if we want to graph it on a number line," + }, + { + "Q": "\nAt 7:41 Sal says that an ellipse is the locus of all points. What does Sal mean when he says locus?", + "A": "No, he does not mean focus. A locus can be defined as a set of points satisfying a given condition. Sal defines a locus as a set of points which are such that the sum of the distance between the point and two fixed points (f1 and f2) is constant.", + "video_name": "QR2vxfwiHAU", + "timestamps": [ + 461 + ], + "3min_transcript": "that focus, is equal to g plus h, or this big green part, which is the same thing as the major diameter of this ellipse, which is the same thing as 2a. Hopefully that that is good enough for you. Now, the next thing, now that we've realized that, is how do we figure out where these foci stand. Or, if we have this equation, how can we figure out what these two points are? Let's figure that out. So, the first thing we realize, all of a sudden is that no matter where we go, it was easy to do it with these points. But even if we take this point right here and we say, OK, what's this distance, and then sum it to that distance, that should also be equal to 2a. And we could use that information to actually figure out where the foci lie. So, let's say I have -- let me draw another one. And then we want to draw the axes. For clarity. Let me write down the equation again. Just so we don't lose it. x squared over a squared plus y squared over b squared is equal to 1. Let's take this point right here. These extreme points are always useful when you're trying to prove something. Or they can be, I don't want to say always. Now, we said that we have these two foci that are symmetric around the center of the ellipse. This is f1, this is f2. And we've already said that an ellipse is the locus of all points, or the set of all points, that if you take each of these points' distance from each of the focuses, and add them up, you get a constant number. And we've figured out that that constant number is 2a. So we've figured out that if you take this distance right here and add it to this distance right here, So we could say that if we call this d, d1, this is d2. We know that d1 plus d2 is equal to 2a. And an interesting thing here is that this is all symmetric, right? This length is going to be the same, d1 is is going to be the same, as d2, because everything we're doing is symmetric. These two focal lengths are symmetric. This distance is the same distance as this distance right there. So, d1 and d2 have to be the same. There's no way that you could -- this is the exact center point the ellipse. The ellipse is symmetric around the y-axis. So if d1 is equal to d2, and that equals 2a, then we know that this has to be equal to a. And this has to be equal to a. I think we're making progress. And the other thing to think about, and we already did that in the previous drawing of the ellipse is, what" + }, + { + "Q": "\nhow can it be reduced to a scalar multiple of the first vector?\n2:18", + "A": "the first vector is 2 3 and second vector 4 6 let C1 be the scalar multiple of first vector and C2 be the scalar multiple of second vector but, second vector = [4 6] = 2 [2 3] = 2 first vector therefore, Linear combination = (C1 + 2C2) [2 3] by substituting *(C1 + 2C2)* as C3, C3 becomes a new scalar multiple for the first vector", + "video_name": "CrV1xCWdY-g", + "timestamps": [ + 138 + ], + "3min_transcript": "Let's say I had the set of vectors-- I don't want to do it that thick. Let's say one of the vectors is the vector 2, 3, and then the other vector is the vector 4, 6. And I just want to answer the question: what is the span of And let's assume that these are position vectors. What are all of the vectors that these two vectors can represent? Well, if you just look at it, and remember, the span is just all of the vectors that can be represented by linear combinations of these. So it's the set of all the vectors that if I have some constant times 2 times that vector plus some other constant times this vector, it's all the possibilities that I can represent when I just put a bunch of different real numbers for c1 and c2. Now, the first thing you might realize is that, look, this vector 2, this is just the same thing as So I could just rewrite it like this. I could just rewrite it as c1 times the vector 2, 3 plus c2 times the vector-- and here, instead of writing the vector 4, 6, I'm going to write 2 times the vector 2, 3, because this vector is just a multiple of that vector. So I could write c2 times 2 times 2, 3. I think you see that this is equivalent to the 4, 6. 2 times 2 is 4. 2 times 3 is 6. Well, then we can simplify this a little bit. We can rewrite this as just c1 plus 2c2, all of that, times 2, 3, times our vector 2, 3. And this is just some arbitrary constant. It's some arbitrary constant plus 2 times some other arbitrary constant. So we can just call this c3 times my vector 2, 3. So in this situation, even though we started with two vectors is equal to all of the vectors that can be constructed with some linear combination of these, any linear combination of these, if I just use this substitution right here, can be reduced to just a scalar multiple of my first vector. And I could have gone the other way around. I could have substituted this vector as being 1/2 times this, and just made any combination of scalar multiple of the second vector. But the fact is, that instead of talking about linear combinations of two vectors, I can reduce this to just a scalar combination of one vector. And we've seen in R2 a scalar combination of one vector, especially if they're position vectors. For example, this vector 2, 3. It's 2, 3. It looks like this. All the scalar combinations of that vector are just going to lie along this line. So 2, 3, it's going to be right there. They're all just going to lie along that line right there, so along this line going in both directions forever." + }, + { + "Q": "at 5:23, whats place values?\n", + "A": "Place values are which place a specific digit is in a number. It goes ones, tens, hundreds, thousands, hundred thousands, and so on. So, for a number 7,932, the 2 is in the ones place, the 3 is in the tens place, the 9 is in the hundreds place, and the 7 is in the thousands place.", + "video_name": "UzXoxglkr98", + "timestamps": [ + 323 + ], + "3min_transcript": "is eight 10s, or 80. One other way we could've thought about this is 240, as we've already said, is 24 10s. If we divide 24 10s by three, we end up with eight 10s, and eight 10s is equal to 80. If we have eight 10s, that equals 80. So this is one other way that we could've thought about it, both ways using the zero or our knowledge of 10s to break this division problem down so we didn't have to deal with a large three-digit number but could deal with simpler, smaller numbers. Let's try another one, this time let's do thousands, let's make this one trickier. What about something like 42 hundred, or 4,200, divided by seven. So here, again we can break this number down. 42 hundred, this number 42 hundred, or 4,200, can be written as 42 times 100, because our pattern tells us when we have a number, like a whole number like 42, and we multiply by 100, we keep our whole number of 42 and we add two zeroes now. So 42 times 100, and then we still need to have our divided by seven at the end. Reverse these numbers so that 42 and seven can be next to each other, 42 divided by seven, because that's a division problem, a division factor we might already know. 42 divided into groups of seven is six, 100 times six is 600. So our solution, going back up here to 4,200 divided by seven, is 600. Or we could've thought about it again, still thinking about place value but using words here instead of digits, 4,200 is 42 hundreds, 42, I can write that out, 42 hundreds, and if you divide 42 hundreds into groups of seven or into seven groups, each group will have six hundreds, or 600 in it. So either way, 42 hundred, 4,200 divided by seven is 600. So here again, we were able to solve a tricky problem," + }, + { + "Q": "\nAt 1:15, why does Sal say \"this is the probability of 5 coin flips, NOT of the outcomes of X\"? I thought X was defined as the outcome of 5 coin flips.\nPlease answer ASAP and thanks in advance.", + "A": "He isn t finding the probability of X at that moment. He says that he needs to FIRST find out what the probability is of 5 coin flips is so that LATER he can find the probability of X. X is defined as the # of heads out of five flips.", + "video_name": "WWv0RUxDfbs", + "timestamps": [ + 75 + ], + "3min_transcript": "- [Voiceover] Let's define a random variable x as being equal to the number of heads, I'll just write capital H for short, the number of heads from flipping coin, from flipping a fair coin, we're gonna assume it's a fair coin, from flipping coin five times. Five times. Like all random variables this is taking particular outcomes and converting them into numbers. And this random variable, it could take on the value x equals zero, one, two, three, four or five. And I what want to do is figure out what's the probability that this random variable takes on zero, can be one, can be two, can be three, can be four, can be five. To do that, first let's think about how many possible outcomes are there from flipping a fair coin five times. Let's think about this. Let's write possible outcomes. Possible outcomes from five flips. These aren't the possible outcomes for the random variable, this is literally the number of possible outcomes from flipping a coin five times. For example, one possible outcome could be tails, heads, tails, heads, tails. Another possible outcome could be heads, heads, heads, tails, tails. That is one of the equally likely outcomes, that's another one of the equally likely outcomes. How many of these are there? For each flip you have two possibilities. Let's write this down. Let me... The first flip, the first flip there's two possibilities, times two for the second flip, times two for the third flip. Actually maybe we'll not use the time notation, you might get confused with the random variable. Two possibilities for the first flip, two possibilities for the second flip, two possibilities for the third flip, two possibilities for the fourth flip, and then two possibilities for the fifth flip, or two to the fifth equally likely possibilities which is, of course, equal to 32. This is going to be helpful because for each of the values that the random variable can take on, we just have to think about how many of these equally likely possibilities would result in the random variable taking on that value. Let's just delve into it to see what we're actually talking about. I'll do it in this light, let me do it in... I'll start in blue. Let's think about the probability that our random variable x is equal to one. Well actually, let me start with zero. The probability that our random variable x is equal to zero. That would mean that you got no heads out of the five flips. Well there's only one way, one out of the 32 equally likely possibilities, that you get no heads. That's the one where you just get five tails. So this is just going to be, this is going to be equal to" + }, + { + "Q": "why At 2:27, Sal puts +10 below the 15 when the 15 is positive? shouldn't have been negative the 10?\n", + "A": "Sal wanted to eliminate the - 10 from the left-hand side of the equation so that only the variable term would remain there. That s why he added + 10 to both sides of the equation.", + "video_name": "Z7C69xP08d8", + "timestamps": [ + 147 + ], + "3min_transcript": "So I have the equation 7 minus 10/x is equal to 2 plus 15/x. And so this isn't the type of equation that you might think that you're used to solving. But I'll give you a few moments to see if you can solve it on your own. Well, what we'll see is we can do a quick multiplication of both sides to actually simplify this to a form that we are more used to looking at. So what's probably bothering you, because it's bothering me, is these x's that we have in the denominators right over here. We're like, well, how do we deal with that? Well, whenever we see an x in the denominator, the temptation is to multiply it by x. But we can't just multiply one of the terms by x. We have to multiply the entire side by x. So we could multiply this entire side by x. But we can't just multiply the left-hand side by x. We'd also want to multiply the right-hand side by x. And so what will that give us? Well, we distribute the x. We get x times 7 is 7x. And then x times negative 10/x, well, So you get negative 10 right over there. So the left-hand side simplifies to 7x minus 10. And then your right-hand side, once again, distribute the x. x times 2 is 2x. x times 15/x, well, x times something over x is just going to be the something. x times 15/x is just going to be 15-- plus 15. So now we've simplified this to a linear equation. We have the variable on both sides. So we just have to do some of the techniques that we already know. So the first thing that I like to do is maybe get all my x's on the left-hand side. So I want to get rid of this 2x right over here. So I subtract 2x from the right-hand side. Now, and I always remind you, I can't do that just to the right-hand side. If I did it just to the right-hand side, it wouldn't be an equality anymore. You have to do that to the left-hand side as well. And so we are left with-- let me get that pink color again. well, you're going to have 5 of that something, minus 10. These two x's negate each other. And you're left with equals 15. Now we can get rid of this negative 10 by adding 10 to both sides. You know, I like that green color when I do stuff to both sides. So I can add 10 to both sides. And I'm left with 5x-- these negate each other-- is equal to 25. And this is the home stretch. You see where this is going. We can divide both sides by 5. And we are left with x is equal to 5. Now let's verify that this actually worked. So let's go back to the original equation. We have 7 minus 10/5. This needs to be equal to-- I'm just taking our 5 and substituting it back here. This needs to be equal to 2 plus 15/5." + }, + { + "Q": "At 4:30 he just \"sees\", that \u00ce\u00bb\u00c2\u00b2-4\u00ce\u00bb-5 is equal to (\u00ce\u00bb-5)(\u00ce\u00bb+1). How does one just see something like this? Did I miss something?\n", + "A": "-5 and 1 are two factors of -5 that sum to -4", + "video_name": "pZ6mMVEE89g", + "timestamps": [ + 270 + ], + "3min_transcript": "The second term is 0 minus 2, so it's just minus 2. The third term is 0 minus 4, so it's just minus 4. And then the fourth term is lambda minus 3, just like that. So kind of a shortcut to see what happened. The terms along the diagonal, well everything became a negative, right? We negated everything. And then the terms around the diagonal, we've got a lambda out front. That was essentially the byproduct of this expression right there. So what's the determinant of this 2 by 2 matrix? Well the determinant of this is just this times that, minus this times that. So it's lambda minus 1, times lambda minus 3, minus these two guys multiplied by each other. So minus 2 times minus 4 is plus eight, minus 8. matrix right here, which simplified to that matrix. And this has got to be equal to 0. And the whole reason why that's got to be equal to 0 is because we saw earlier, this matrix has a non-trivial null space. And because it has a non-trivial null space, it can't be invertible and its determinant has to be equal to 0. So now we have an interesting polynomial We can multiply it out. We get what? We get lambda squared, right, minus 3 lambda, minus lambda, plus 3, minus 8, is equal to 0. Or lambda squared, minus 4 lambda, minus 5, is equal to 0. expression right here is known as the characteristic polynomial. Just a little terminology, polynomial. But if we want to find the eigenvalues for A, we just have to solve this right here. This is just a basic quadratic problem. And this is actually factorable. Let's see, two numbers and you take the product is minus 5, when you add them you get minus 4. It's minus 5 and plus 1, so you get lambda minus 5, times lambda plus 1, is equal to 0, right? Minus 5 times 1 is minus 5, and then minus 5 lambda plus 1 lambda is equal to minus 4 lambda. So the two solutions of our characteristic equation being set to 0, our characteristic polynomial, are lambda is equal to 5 or lambda is equal to minus 1." + }, + { + "Q": "\nWhat does the symbol that Sal draws at 2:34 mean?", + "A": "If you are referring to the \u00e2\u0088\u00a9 symbol, that means intersection. For instance, let A and B be to possible events. If we want to talk about the probability of event A taking place, we would write P(A), and similarly P(B) for B. What if we want to talk about the probability of both A and B happening at the same time? We would write that as P(A \u00e2\u0088\u00a9 B) - the probability of A and B. C = A \u00e2\u0088\u00a9 B is a new set containing all the elements common to both sets A and B - the intersection of the two.", + "video_name": "VjLEoo3hIoM", + "timestamps": [ + 154 + ], + "3min_transcript": "You say, well, this sounds like an interesting game. How much does it cost to play? And the guy tells you, it is $0.35 to play, so obviously a fairly low stakes casino. So my question to you is, would you want to play this game? And don't put the fun factor into it. Just economically, does it make sense for you to actually play this game? Well, let's think through the probabilities a little bit. So, first of all, what's the probability that the first marble you pick is green? Actually, let me just write, first green, probability first green. Well, the total possible outcomes-- there's 5 marbles here, all equally likely. So there's 5 possible outcomes. So there's a 3/5 probability that the first is green. So you have a 3/5 chance, 3/5 probability I should say, that after that first pick you're kind of still in the game. Now, what we really care about is your probability of winning the game. You want the first to be green, and the second green. Well, let's think about this a little bit. What is the probability that the first is green-- first, I'll just write g for green-- and the second is green? Now, you might be tempted to say, oh, well, maybe the second being green is the same probability. It's 3/5. I can just multiply 3/5 times 3/5, and I'll get 9/25, seems like a pretty straightforward thing. But the realization here is what you do with that first green marble. You don't take that first green marble out, look at it, and put it back in the bag. So when you take that second pick, depends on what you got on the first pick. Remember, we take the marble out. If it's a green marble, whatever marble it is, at whatever after the first pick, we leave it on the table. We are not replacing it. So there's not any replacement here. So these are not independent events. Let me make this clear, not independent. Or in particular, the second pick is dependent on the first. Dependent on the first pick. If the first pick is green, then you don't have 3 green marbles in a bag of 5. If the first pick is green, you now have 2 green marbles in a bag of 4. So the way that we would refer to this, the probability of both of these happening, yes, it's definitely equal to the probability of the first green times-- now, this is kind of the new idea--" + }, + { + "Q": "at 6:30 , why did Sal write the \"0.30*$1=0.30\"??\n", + "A": "Because he had just worked out that you had a 30% chance of winning the game which when multiplied by what your prize will give you your average gain (30 cents as opposed to playing 35 cents to play the game).", + "video_name": "VjLEoo3hIoM", + "timestamps": [ + 390 + ], + "3min_transcript": "right over here, just a straight up vertical line, just means given-- given that the first was green. Now, what is the probability that the second marble is green given that the first marble was green? Well, we drew the scenario right over here. If the first marble is green, there are 4 possible outcomes, not 5 anymore, and 2 of them satisfy your criteria. So 2 of them satisfy your criteria. So the probability of the first marble green being green and the second marble being green, is going to be the probability that your first is green, so it's going to be 3/5, times the probability that the second is green, given that the first was green. Now you have 1 less marble in the bag, and we're assuming that the first pick was green, so you'll only have 2 green marbles left. Let's see, 3/5 times 2/4. Well, 2/4 is the same thing as 1/2. This is going to be equal to 3/5 times 1/2, which is equal to 3/10. Or we could write that as 0.30, or we could say there's a 30% chance of picking 2 green marbles, when we are not replacing. So given that, let me ask you the question again. Would you want to play this game? Well, if you played this game many, many, many, many, many times, on average you have a 30% chance of winning $1. And we haven't covered this yet, but so your expected value is really going to be 30% times $1-- this gives you a little bit of a preview-- which is going to be $0.30 30% chance of winning $1. many, many times, that playing the game is going to give you $0.30. Now, would you want to give someone $0.35 to get, on average, $0.30? No, you would not want to play this game. Now, one thing I will let you think about is, would you want to play this game if you could replace the green marble, the first pick. After the first pick, if you could replace the green marble, would you want to play the game in that scenario?" + }, + { + "Q": "What does the upside down U symbol at 2:32-2:35 mean?\n", + "A": "In this example, Sal is asking what is the probability of both the first AND second being green . The upside down U symbol in this case stands for the AND. The symbol typically stands for intersection and is used in set theory to refer to common numbers or letters in sets. For example, the intersection of {1, 3, 5, 7} and {4, 5, 6, 7} is {5, 7} because those are the numbers that you can find in both sets.", + "video_name": "VjLEoo3hIoM", + "timestamps": [ + 152, + 155 + ], + "3min_transcript": "You say, well, this sounds like an interesting game. How much does it cost to play? And the guy tells you, it is $0.35 to play, so obviously a fairly low stakes casino. So my question to you is, would you want to play this game? And don't put the fun factor into it. Just economically, does it make sense for you to actually play this game? Well, let's think through the probabilities a little bit. So, first of all, what's the probability that the first marble you pick is green? Actually, let me just write, first green, probability first green. Well, the total possible outcomes-- there's 5 marbles here, all equally likely. So there's 5 possible outcomes. So there's a 3/5 probability that the first is green. So you have a 3/5 chance, 3/5 probability I should say, that after that first pick you're kind of still in the game. Now, what we really care about is your probability of winning the game. You want the first to be green, and the second green. Well, let's think about this a little bit. What is the probability that the first is green-- first, I'll just write g for green-- and the second is green? Now, you might be tempted to say, oh, well, maybe the second being green is the same probability. It's 3/5. I can just multiply 3/5 times 3/5, and I'll get 9/25, seems like a pretty straightforward thing. But the realization here is what you do with that first green marble. You don't take that first green marble out, look at it, and put it back in the bag. So when you take that second pick, depends on what you got on the first pick. Remember, we take the marble out. If it's a green marble, whatever marble it is, at whatever after the first pick, we leave it on the table. We are not replacing it. So there's not any replacement here. So these are not independent events. Let me make this clear, not independent. Or in particular, the second pick is dependent on the first. Dependent on the first pick. If the first pick is green, then you don't have 3 green marbles in a bag of 5. If the first pick is green, you now have 2 green marbles in a bag of 4. So the way that we would refer to this, the probability of both of these happening, yes, it's definitely equal to the probability of the first green times-- now, this is kind of the new idea--" + }, + { + "Q": "\nAt 0:40, does it matter what kind of brackets to draw? Or can any be used?", + "A": "Set notation traditionally uses the curly brackets { }", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 40 + ], + "3min_transcript": "What I want to do in this video is familiarize ourselves with the notion of a sequence. And all a sequence is is an ordered list of numbers. So for example, I could have a finite sequence-- that means I don't have an infinite number of numbers in it-- where, let's say, I start at 1 and I keep adding 3. So 1 plus 3 is 4. 4 plus 3 is 7. 7 plus 3 is 10. And let's say I only have these four terms right over here. So this one we would call a finite sequence. I could also have an infinite sequence. So an example of an infinite sequence-- let's say we start at 3, and we keep adding 4. So we go to 3, to 7, to 11, 15. And you don't always have to add the same thing. We'll explore fancier sequences. The sequences where you keep adding the same amount, we call these arithmetic sequences, which we will also explore in more detail. But to show that this is infinite, to show that we keep this pattern going on and on and on, I'll put three dots. This just means we're going to keep going on and on and on. Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1" + }, + { + "Q": "Why did he add the K-1 thing at about 3:11\n", + "A": "Around 3:11 Sal put in the k - 1 in to show the term before K. hope it helps", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 191 + ], + "3min_transcript": "Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1 should write 3 times k minus 1-- same thing. And you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with kind of this function notation. I want to make it clear-- I have essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about. a of k is equal to 1 plus 3 times k minus 1." + }, + { + "Q": "at 4:28 Sal writes {a sub k} k = 1, is it correct for him to write that k = 1 because if you look at the sequence then you can see that (a sub 1) is 3?\n", + "A": "Be careful here. k is the number of the term (first term, second term, third term, fourth term), not the value of the term ( 1, 4, 7, and 10). In Sal s first example, k went from 1 to 4 because there are 4 terms in the sequence., but the values of those terms were 1 , 4, 7, and 10. The number of the first term ( 1 ) just happened to be the same as the value of the first term ( 1 ).", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 268 + ], + "3min_transcript": "I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1 should write 3 times k minus 1-- same thing. And you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with kind of this function notation. I want to make it clear-- I have essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about. a of k is equal to 1 plus 3 times k minus 1. an allowable input, the domain, is restricted to positive integers. Now, how would I denote this business right over here? Well, I could say that this is equal to-- and people tend to use a. But I could use the notation b sub k or anything else. But I'll do a again-- a sub k. And here, we're going from our first term-- so this is a sub 1, this is a sub 2-- all the way to infinity. Or we could define it-- if we wanted to define it explicitly as a function-- we could write this sequence as a sub k, where k starts at the first term and goes to infinity, with a sub k is equaling-- so we're starting at 3. And we are adding 4 one less time. For the second term, we added 4 once. For the third term, we add 4 twice." + }, + { + "Q": "\nAt 3:09, why does he put \"k-1\" into that function?", + "A": "Starting on the first term, we need to move (\u00f0\u009d\u0091\u0098 \u00e2\u0088\u0092 1) spaces in the sequence in order to get to the \u00f0\u009d\u0091\u0098:th term. (Moving 1 space gets us to the 2nd term, moving 2 spaces gets us to the 3rd term, and so on...) And each time we move a space, we add 3, so by the time we get to the \u00f0\u009d\u0091\u0098:th term we have added 3(\u00f0\u009d\u0091\u0098 \u00e2\u0088\u0092 1).", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 189 + ], + "3min_transcript": "Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1 should write 3 times k minus 1-- same thing. And you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with kind of this function notation. I want to make it clear-- I have essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about. a of k is equal to 1 plus 3 times k minus 1." + }, + { + "Q": "at time 3:42, why the radius of the surface is radical 2?\n", + "A": "The radius isn t set at the square root of 2: it is the square root of y. The radius is variable as the function increases in either direction, and is equal to the distance between the y axis and f(x). The radius is only the square root of 2 when f(x) is 2.", + "video_name": "43AS7bPUORc", + "timestamps": [ + 222 + ], + "3min_transcript": "And then the x-axis is going like this. So I just tilted this over. I tilted it over a little bit to be able to view it at a different angle. This top right over here is this top right over there. So that gives you an idea of what it looks like. But we still haven't thought about how do we actually find the volume of this thing? Well, what we can do, instead of creating discs where the depth is in little dx's, what if we created discs where the depth is in dy? So let's think about that a little bit. So let's create-- let's think about constructing a disc at a certain y-value. So let's think about a certain y-value, and we're going to construct a disc right over there that has the same radius of the shape at that point. So that's our disc. That's our disc right over here. of saying it has a depth of dx, let's say it has a depth of dy. So this depth right over here is dy. So what is the volume of this disc in terms of y? And as you could imagine, we're going to do this definite integral, and it is a definite integral, with respect to y. So what's the volume of this thing? Well, like we did in the last video, we have to figure out the area of the top of each of these discs. Or I guess you could say the face of this coin. Well, to find that area it's pi r squared. If we can figure out this radius right over here, we know the area. So what's that radius? So to think about that radius in terms of y, we just have to solve this equation explicitly in terms of y. So instead of saying it's y is equal to x squared, we can take the principal root of both sides, and we could say that the square root of y is equal to x. And this right over here is only defined for non-negative y's, over here. So we can also call of this function right over here x is equal to the square root of y. And we're essentially looking at this side of it. We're not looking at this stuff right over here. So we're only looking at this side right over here. We've now expressed this graph, this curve, as x as a function of y. So if we do it that way, what's our radius right over here? Well, our radius right over here is going to be f of y. It's going to be the square root of y. It's going to be the square root of y is our radius. So it's going to be a function of y. I don't want to confuse you if you thought this was f of x, and actually this is f of y. No, it would be a function of y. We could call it g of y. It's going to be the square root of y. So area is equal to pi r squared, which means that the area of this thing is going to be pi times our radius, radius squared." + }, + { + "Q": "did he graph it in a wrong way at 6:03\n", + "A": "He did not. Since 1/2x was there, he started the grAph at (0, -6) , before going down by 1/2 from there.", + "video_name": "unSBFwK881s", + "timestamps": [ + 363 + ], + "3min_transcript": "So if you were to do this for all the possible x's, you would not only get all the points on this line which we've drawn, you would get all the points below the line. So now we have graphed this inequality. It's essentially this line, 4x plus 3, with all of the area below it shaded. Now, if this was just a less than, not less than or equal sign, we would not include the actual line. And the convention to do that is to actually make the line a dashed line. This is the situation if we were dealing with just less than 4x plus 3. Because in that situation, this wouldn't apply, and we would just have that. So the line itself wouldn't have satisfied it, just the area below it. Let's do one like that. So let's say we have y is greater than negative x So a good way to start-- the way I like to start these problems-- is to just graph this equation right here. So let me just graph-- just for fun-- let me graph y is equal to-- this is the same thing as negative 1/2 minus 6. So if we were to graph it, that is my vertical axis, that is my horizontal axis. And our y-intercept is negative 6. So 1, 2, 3, 4, 5, 6. So that's my y-intercept. And my slope is negative 1/2. Oh, that should be an x there, negative 1/2 x minus 6. So my slope is negative 1/2, which means when I go 2 to the right, I go down 1. So if I go 2 to the right, I'm going to go down 1. going to go up 1. So negative 2, up 1. So my line is going to look like this. My line is going to look like that. That's my best attempt at drawing the line. So that's the line of y is equal to negative 1/2 x minus 6. Now, our inequality is not greater than or equal, it's just greater than negative x over 2 minus 6, or greater than negative 1/2 x minus 6. So using the same logic as before, for any x-- so if you take any x, let's say that's our particular x we want to pick-- if you evaluate negative x over 2 minus 6, you're going to get that point right there. You're going to get the point on the line. But the y's that satisfy this inequality are the y's greater than that. So it's going to be not that point-- in fact, you draw an open circle there-- because you can't include the point of negative 1/2 x minus 6. But it's going to be all the y's greater than that." + }, + { + "Q": "\ni'm confused,at1:25 do you graph the whole equation", + "A": "You do graph the line just as you would if you were given the equality: y = 4x + 3 Once you have graphed the line, you look at the inequality to see if you should shade the area above the line or below it. For y > 4x + 3, you would shade above, and make the line a dotted line For y >= 4x+3, you would shade above, and make the line a solid line For y < 4x + 3, you would shade below, and make the line a dotted line For y <= 4x+3, you would shade below, and make the line a solid line", + "video_name": "unSBFwK881s", + "timestamps": [ + 85 + ], + "3min_transcript": "Let's graph ourselves some inequalities. So let's say I had the inequality y is less than or equal to 4x plus 3. On our xy coordinate plane, we want to show all the x and y points that satisfy this condition right here. So a good starting point might be to break up this less than or equal to, because we know how to graph y is equal to 4x plus 3. So this thing is the same thing as y could be less than 4x plus 3, or y could be equal to 4x plus 3. That's what less than or equal means. It could be less than or equal. And the reason why I did that on this first example problem is because we know how to graph that. So let's graph that. Try to draw a little bit neater than that. So that is-- no, that's not good. So that is my vertical axis, my y-axis. And then we know the y-intercept, the y-intercept is 3. So the point 0, 3-- 1, 2, 3-- is on the line. And we know we have a slope of 4. Which means if we go 1 in the x-direction, we're going to go up 4 in the y. So 1, 2, 3, 4. So it's going to be right here. And that's enough to draw a line. We could even go back in the x-direction. If we go 1 back in the x-direction, we're going to go down 4. 1, 2, 3, 4. So that's also going to be a point on the line. So my best attempt at drawing this line is going to look something like-- this is the hardest part. It's going to look something like that. That is a line. It should be straight. I think you get the idea. That right there is the graph of y is equal to 4x plus 3. So all of these points satisfy this inequality, but we have more. This is just these points over here. What about all these where y ix less than 4x plus 3? So let's think about what this means. Let's pick up some values for x. When x is equal to 0, what does this say? When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3. When x is equal to negative 1, what is this telling us? 4 times negative 1 is negative 4, plus 3 is negative 1. y would be less than negative 1. When x is equal to 1, what is this telling us? 4 times 1 is 4, plus 3 is 7. So y is going to be less than 7. So let's at least try to plot these. So when x is equal to-- let's plot this one first. When x is" + }, + { + "Q": "@ around 4:15 Sal talks about:\n\n\"...the set of all of the positions or all of the position vectors that specify the triangle that is essentially formed by connecting these dots.\"\n\nI'm thinking that yes, the \"...set of all of the positions...\"; but not \"...all of the position vectors...\" which seem to span the interior of the (transformed or untransformed) triangle (if their tails are at the origin). Anyone disagree?\n", + "A": "When it comes to vectors in standard position (starting at the origin) (i.e. position vectors), we only care about the position of the head. The line which composes the body of the vector is immaterial, merely a graphical representation which is not preserved in the numerical representation.", + "video_name": "qkfODKmZ-x4", + "timestamps": [ + 255 + ], + "3min_transcript": "very easy to operate any transformation on each of these basis vectors that only have a 1 in its corresponding dimension, or with respect to the corresponding variable, and everything else is 0. So all of this is review. Let's actually use this information to construct some interesting transformations. So let's start with some set in our Rn. And actually everything I'm going to do is going to be in R2, but you can extend a lot of this into just general dimensions. But we're dealing with R2 right here. Obviously, it's only 2 dimensions right here. Let's say we have a triangle formed by the points, let's say the first point is right here. Let's say it's the point 3, 2. And then you have the point, let's say that your next point in your triangle, is the point, let's just make it the I shouldn't have written that as a fraction. I don't know why I did that. 3, 2. Then you have the point minus 3, 2. And that's this point right here. Minus 3, 2. And then let's say, just for fun, let's say you have the point, or the vector-- the position vector, right? 3, minus 2. Which is right here. Now each of these are position vectors, and I can draw them. I could draw this 3, 2 as in the standard position by drawing an arrow like that. I could do the minus 3, 2 in its standard position like that. And 3, minus 2 I could draw like that. But more than the actual position vectors, I'm more concerned with the positions that they specify. And we know that if we take the set of all of the the triangle that is essentially formed by connecting these dots. The transformation of this set-- so we're going to apply some transformation of that-- is essentially, you can take the transformation of each of these endpoints and then you connect the dots in the same order. And we saw that several videos ago. But let's actually design a transformation here. So let's say we want to-- let's just write down and words what we want to do with whatever we start in our domain. Let's say we want to reflect around the x-axis. Reflect around-- well actually let's reflect around the y-axis. We essentially want to flip it over. We want to flip it over that way. So I'm kind of envisioning something that'll look something like that when we flip it over." + }, + { + "Q": "\n3:25 Isn't this essentially another logarithmic property?", + "A": "Yes. \u00f0\u009d\u0091\u008f^(log[\u00f0\u009d\u0091\u008f] (4)) = 4. 4^(log[4] (\u00f0\u009d\u0091\u008f)) = \u00f0\u009d\u0091\u008f \u00e2\u0087\u0094 4 = \u00f0\u009d\u0091\u008f^(1/(log[4] (\u00f0\u009d\u0091\u008f))). Thereby log[\u00f0\u009d\u0091\u008f] (4) = 1/(log[4] (\u00f0\u009d\u0091\u008f))", + "video_name": "qtsMgdZ98Yg", + "timestamps": [ + 205 + ], + "3min_transcript": "write what the base is. Well this is going to be true regardless of which base you choose as long as you pick the same base. This could be base nine, base nine in either case. Now typically, people choose base 10. So 10 is the most typical one to use and that's because most peoples calculators or they might be logarithmic tables for base 10. So here you're saying the exponent that I have to raise A to to get to B is equal to the exponent I have to raise 10 to to get to B, divided by the exponent I have to raise 10 to to get to A. This is a really really useful thing to know if you are dealing with logarithms. And we prove it in another video. But now we'll see if we can apply it. So now going back to this yellow expression, this once again, is the same thing as one divided by this right over here. So let me write it that way actually. This is one divided by log base B of four. to re-write it. So this is going to be equal to, this is going to be equal to one, divided by, instead of writing it log base B of four, we could write it as log of four, and if I just, if I don't write the base there we can assume that it's base 10, log of four over log of B. Now if I divide by some fraction, or some rational expression, it's the same thing as multiplying by the reciprocal. So this is going to be equal one times the reciprocal of this. Log of B over log of four, which of course is just going to be log of B over log of four, I just multiplied it by one, and so we can go in the other direction now, using this little tool we established at the beginning of the video. This is the same thing as log base four of B. So we have a pretty neat result that actually came out here, we didn't prove it for any values, although we have a pretty general B here. If I take the, If I take the reciprocal of a logarithmic expression, I essentially have swapped the bases. This is log base B, what exponent do I have to raise B to to get to four? And then here I have what exponent do I have to raise four to to get to B? Now it might seem a little bit magical until you actually put some tangible numbers here. Then it starts to make sense, especially relative to fractional exponents. For example, we know that four to the third power is equal to 64. So if I had log base four of 64, that's going to be equal to three. And if I were to say log base 64 of four," + }, + { + "Q": "At 0:58, Sal mentions \"mu\". Is that the symbol for the arithmetic mean?\n", + "A": "In the context of statistics, the letter mu (\u00ce\u00bc) often indeed refers to the arithmetic mean. In mathematics in general, it is used for many things in mathematics - sometimes even just as a variable (much like n for a natural number). The context should make it clear which \u00ce\u00bc is meant.", + "video_name": "PWiWkqHmum0", + "timestamps": [ + 58 + ], + "3min_transcript": "Let's say that you're curious about studying the dimensions of the cars that happen to sit in the parking lot. And so you measure their lengths. Let's just make the computation simple. Let's say that there are five cars in the parking lot. The entire size of the population that we care about is 5. And you go and measure their lengths-- one car is 4 meters long, another car is 4.2 meters long, another car is 5 meters long, the fourth car is 4.3 meters long, and then, let's say the fifth car is 5.5 meters long. So let's come up with some parameters for this population. So the first one that you might want to figure out is a measure of central tendency. And probably the most popular one is the arithmetic mean. So let's calculate that first. So we're going to do that for the population. So we're going to use mu. Well, we just have to add all of these data points up and divide by 5. And I'll just get the calculator out just so it's a little bit quicker. This is going to be for 4 plus 4.2 plus 5 plus 4.3 plus 5.5. And then, I'm going to take that sum and then divide by 5. And I get an arithmetic mean for my population of 4.6. So that's fine. And if we want to put some units there, it's 4.6 meters. Now, that's the central tendency or measure of central tendency. We also might be curious about how dispersed is the data, especially from that central tendency. So what would we use? the population variance. And the population variance is one of many ways of measuring dispersion. It has some very neat properties the way we've defined it as the mean of the squared distances from the mean. It tends to be a useful way of doing it. So let's just a bit. Let's actually calculate the population variance for this population right over here. Well, all we need to do is find the distance from each of these points to our mean right over here. And then, square them. And then, take the mean of those two squared distances. So let's do that. So it's going to be 4 minus 4.6 squared plus 4.2 minus 4.6 squared plus 5 minus 4.6 squared" + }, + { + "Q": "How is \u00cf\u0086 - 1 = 1/\u00cf\u0086 ? How is \u00cf\u0086 ^2-\u00cf\u0086 -1=0?? (At 1:38) and (4:17)\n", + "A": "\u00cf\u0086 = 1 + (1/ \u00cf\u0086) He subtracts 1 from both sides. \u00cf\u0086 - 1 = 1 + (1/ \u00cf\u0086) - 1. On the right side, 1 - 1 is just 0. You are left with \u00cf\u0086 - 1 = 1/ \u00cf\u0086 \u00cf\u0086 = 1 + (1/ \u00cf\u0086) He multiplies both sides by \u00cf\u0086, like this: (\u00cf\u0086)\u00cf\u0086 = (1 + 1/\u00cf\u0086) (\u00cf\u0086). You get \u00cf\u0086^2 = \u00cf\u0086 + 1 To make a quadratic equation, subtract \u00cf\u0086 and 1 from the right side. You get \u00cf\u0086^2 - \u00cf\u0086 - 1 = 0", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 98, + 257 + ], + "3min_transcript": "What I want to explore in this video is, given some length of string or a line or some line segment right here, b, can I set up an a, so that the ratio of a to b is equal to the ratio of the sum of these two to the longer side? So it's equal to the ratio of a plus b to a. So I want to sit and think about this a little bit. I want to see is can I construct some a that's on this ratio, this perfect ratio that I'm somehow referring to right here, so that the ratio of the longer side to the shorter side is equal to the ratio of the whole thing to the longer side. And let's just assume that we can find a ratio like that. And we'll call it phi. We'll use the Greek letter phi for that ratio over there. So let's see what we can learn about this special ratio phi. Well if phi is equal to a over b, which is equal to a plus b over a, we know that a plus b over a is the same thing as a over a plus b over a. And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let" + }, + { + "Q": "At 11:30 Sal says b/(a-b) = 1/((a-b)/b)\nHe says that he took the reciprocal, I guess I don't understand the mechanics behind taking reciprocals of fractures? Can someone explain?\n", + "A": "Reciprocals are basically when you divided 1 by a certain fraction. What this does is it swaps both the top and the bottom parts of the fraction. So, the reciprocal of, to use this example, b/(a-b) is 1/b/(a-b), or (a-b)/b. However, you might note that the two are not equal. Therefore, in order to use it in an equation, you need to have 1/(a-b)/b, which is equal to the original fraction.", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 690 + ], + "3min_transcript": "It just keeps on showing up in a ton of different ways when you look at a pentagram like this. If you look at something like a pentagon, a regular pentagon where all the angles are the same and all the sides are the same, a regular pentagon. If you take any of the diagonals of a regular pentagon, so right over here, if you take this diagonal right over here, the ratio of this green side to--and when I'm talking about the diagonals, ones that actually aren't one of the edges-- the ratio of any of the diagonals to any of the sides is once again, this golden ratio. So it keeps showing up on and on and on. And we can do interesting things with the golden ratio. Let's say that we had a rectangle, where the ratio of the width to the height is the golden ratio. So let's try that out. So let's say that this is its height. This is its width. And that the ratio-- So let's call this a. Let's call this b. That 1.61 so on and so forth. Let me scroll down a little bit. So that is going to be equal to phi. So that's something interesting to do. Maybe that's a nice looking rectangle of some sort. But let me put out a square here. So let me separate this into a b by b square. So this is a b by b square right over here. And then-- actually let me do it a little bit, let me draw it a little bit differently, this rectangle actually isn't exactly the way I would want to draw it-- so the ratio might look a little bit like this. So the ratio of the width to the length, or the width to the height, is going to be the golden ratio. So a over b is going to be that golden ratio. And let me separate out a little b by b square over here. So this has width b as well. And so this distance right over here is going to be a minus b. Actually, I should say, we have a b by b square, right over here. This is b by b. And then we're left with a b by a minus b rectangle. Now wouldn't it be cool if this was also the golden ratio? And so let's try it out. Let's find the ratio of b to a minus b. So the ratio of b to a minus b. Well, that's going to be equal to 1 over the ratio of a minus b to b. I just took the reciprocal of this right over here. And this is just going to be equal to 1 over a over b. Let me write this, a over b minus 1. I just rewrote this right there. And that's just going to be equal to 1 over phi. The ratio of a to b, we said, by definition was phi minus 1. But what is phi minus 1? Well phi minus 1 is 1 over phi. It's this cool number. So it's equal to 1 over 1 over 1 over phi, which is once again," + }, + { + "Q": "\n@ 4:57, how come a is = to it's coefficient, and b = to it's coefficient?", + "A": "He is using the quadratic formula to solve this quadratic equation (if you changed it to x s to make it more clear: x^2 - x - 1 = 0 The letters in the video are phi s, but it really makes no difference.", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 297 + ], + "3min_transcript": "equal to phi plus 1. phi squared is equal to phi plus 1. And then, actually, I'm going to take a little bit of a side But even this is interesting, because then if we take the square root of both sides of this, you get-- let me scroll down a little bit-- you get phi is equal to the square root of-- and I'll just switch the order here-- the square root of 1 plus phi. So once again, we can set up another recursive definition. phi is equal to the square root of 1 plus phi. And I could write phi there, but hey, phi is equal to the square root of 1 plus and I could write phi there, but hey, phi is just equal to the square root of 1 plus, the square root of 1 plus. And we could just keep going on and on like this forever. So even this is neat. The same number that can be expressed this way, the same number where if I just subtract 1 from it, It can also be expressed in these kind of recursive square roots underneath each other. So this is already starting to get very, very, very intriguing, but let's get back to business. Let's actually solve for this magic number, this magic ratio that we started thinking about. And really from a very simple idea, that the ratio of the longer side to the shorter side is equal to the ratio of the sum of the two to the longer side. So let's just solve this as a traditional quadratic. Let's get everything on the left-hand side. So we're going to subtract phi plus 1 from both sides. And we get phi squared minus phi minus 1 is equal to 0. And we can solve for phi now using the quadratic formula, which we've proven in other videos. You can prove using completing the square. But the quadratic formula you say, negative b. Negative b is the coefficient on this term right here. So let me just write it down, a is equal to 1, that's the coefficient on this term. b is equal to negative 1, that's the coefficient on this term. or it's really the constant term right over there. So the solutions to this, phi-- and we're actually only going to care about the positive solution because we're thinking about a positive-- when we go to our original problem here, we're assuming that these are both positive distances, so we care about a positive value right over here. We get phi is equal to-- do it in orange-- negative b. Well negative negative 1 is 1 plus or minus the square root of b squared. b squared is going to be 1 minus 4ac. a is 1, c is negative 1. So negative 4 times negative 1 is positive 4. So 1 plus 4, all of that over 2a. So a is 1, so all of that over 2. So phi is equal to 1. And once again, we only care about the positive solution This is going to be the square root of 5. If you have 1 minus the square root of 5, you're going to get a negative in the numerator. So we only care about the positive solution. 1 plus the square root of 5 over 2." + }, + { + "Q": "\nHow does Phi=1+1/Phi = Phi^2=Phi+1\nDoes 1/(x^2)=x\n(3:14)", + "A": "If 1 + 1/phi = phi then you can multiply both sides by phi: phi(1 + 1/phi) = phi x phi (now use the distributive property on the left side: phi + 1 = phi^2 I hope my answer is clear enough", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 194 + ], + "3min_transcript": "And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let equal to phi plus 1. phi squared is equal to phi plus 1. And then, actually, I'm going to take a little bit of a side But even this is interesting, because then if we take the square root of both sides of this, you get-- let me scroll down a little bit-- you get phi is equal to the square root of-- and I'll just switch the order here-- the square root of 1 plus phi. So once again, we can set up another recursive definition. phi is equal to the square root of 1 plus phi. And I could write phi there, but hey, phi is equal to the square root of 1 plus and I could write phi there, but hey, phi is just equal to the square root of 1 plus, the square root of 1 plus. And we could just keep going on and on like this forever. So even this is neat. The same number that can be expressed this way, the same number where if I just subtract 1 from it," + }, + { + "Q": "Hey Guys, and Khan, I just have one question at aaround 3:10 3:09, just say that you times both side by \u00c3\u0098 (\u00c3\u0098=phi) , but I don't understand how u get \u00c3\u0098^2= \u00c3\u0098+1, because the origional equation is \u00c3\u0098= 1+1/\u00c3\u0098, and then when you times both side by \u00c3\u0098, \u00c3\u0098^2= 1+ 1/\u00c3\u0098*\u00c3\u0098 = \u00c3\u0098^2 = 1+1, because 1/\u00c3\u0098*\u00c3\u0098 = 1/1 ... or have I got a brain malfunction ??...\n", + "A": "I believe you re missing one step here. You are forgetting to fully distribute phi on the right side of the equation, (1+ 1/phi), for I don t believe you are multiplying the constant 1 by phi. If you multiply the whole equation by phi , then you ll first have (phi = 1 + 1/phi) * phi; then after distribution (phi * phi) = (1 * phi) + (phi * 1/phi), which of course leads to the equation phi^2 = phi +1. I this still isn t making sense I would suggest watching some of Sal s Algebra videos. Cheers!", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 190, + 189 + ], + "3min_transcript": "And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let equal to phi plus 1. phi squared is equal to phi plus 1. And then, actually, I'm going to take a little bit of a side But even this is interesting, because then if we take the square root of both sides of this, you get-- let me scroll down a little bit-- you get phi is equal to the square root of-- and I'll just switch the order here-- the square root of 1 plus phi. So once again, we can set up another recursive definition. phi is equal to the square root of 1 plus phi. And I could write phi there, but hey, phi is equal to the square root of 1 plus and I could write phi there, but hey, phi is just equal to the square root of 1 plus, the square root of 1 plus. And we could just keep going on and on like this forever. So even this is neat. The same number that can be expressed this way, the same number where if I just subtract 1 from it," + }, + { + "Q": "At 1:33 is that the only symbol for phi?\n", + "A": "yes like this:\u00cf\u0086", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 93 + ], + "3min_transcript": "What I want to explore in this video is, given some length of string or a line or some line segment right here, b, can I set up an a, so that the ratio of a to b is equal to the ratio of the sum of these two to the longer side? So it's equal to the ratio of a plus b to a. So I want to sit and think about this a little bit. I want to see is can I construct some a that's on this ratio, this perfect ratio that I'm somehow referring to right here, so that the ratio of the longer side to the shorter side is equal to the ratio of the whole thing to the longer side. And let's just assume that we can find a ratio like that. And we'll call it phi. We'll use the Greek letter phi for that ratio over there. So let's see what we can learn about this special ratio phi. Well if phi is equal to a over b, which is equal to a plus b over a, we know that a plus b over a is the same thing as a over a plus b over a. And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let" + }, + { + "Q": "\nAt about 11:35 Sal determines that the reciprocal of b/a-b= 1/a-b/b. Why isn't it just a-b/b?", + "A": "b/(a-b)=1/((a-b)/b) This is equivalent to b/(a-b). The second expression is the reciprocal of the reciprocal, not just the reciprocal, of b/(a-b), so it is equivalent to it. I hope this helps!", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 695 + ], + "3min_transcript": "That 1.61 so on and so forth. Let me scroll down a little bit. So that is going to be equal to phi. So that's something interesting to do. Maybe that's a nice looking rectangle of some sort. But let me put out a square here. So let me separate this into a b by b square. So this is a b by b square right over here. And then-- actually let me do it a little bit, let me draw it a little bit differently, this rectangle actually isn't exactly the way I would want to draw it-- so the ratio might look a little bit like this. So the ratio of the width to the length, or the width to the height, is going to be the golden ratio. So a over b is going to be that golden ratio. And let me separate out a little b by b square over here. So this has width b as well. And so this distance right over here is going to be a minus b. Actually, I should say, we have a b by b square, right over here. This is b by b. And then we're left with a b by a minus b rectangle. Now wouldn't it be cool if this was also the golden ratio? And so let's try it out. Let's find the ratio of b to a minus b. So the ratio of b to a minus b. Well, that's going to be equal to 1 over the ratio of a minus b to b. I just took the reciprocal of this right over here. And this is just going to be equal to 1 over a over b. Let me write this, a over b minus 1. I just rewrote this right there. And that's just going to be equal to 1 over phi. The ratio of a to b, we said, by definition was phi minus 1. But what is phi minus 1? Well phi minus 1 is 1 over phi. It's this cool number. So it's equal to 1 over 1 over 1 over phi, which is once again, So once again, the ratio of this smaller rectangle, of its height to its width, is once again this golden ratio, this number that keeps showing up. And then we could do the same thing again. We could separate this into an a minus b by a minus b square. Just like that. And then we'll have another golden rectangle, sometimes it's called, right over there. And then we could separate that into a square and another golden rectangle. Then we could separate that into a square and then another golden rectangle. Then another golden rectangle. Actually let me do it like this. This would be better. So let me separate. Let me do the square up here. So this is an a minus b by a minus b square and then we have another golden rectangle right over here. I could put a square right in there. Then we'll have another golden rectangle. Then we could put another square right over there, you have another golden rectangle. I think you see where this is going." + }, + { + "Q": "at 1:41, when would we use the golden ratio?\n", + "A": "Music, art, and nature according to Sal. Check out the pics on the end!", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 101 + ], + "3min_transcript": "What I want to explore in this video is, given some length of string or a line or some line segment right here, b, can I set up an a, so that the ratio of a to b is equal to the ratio of the sum of these two to the longer side? So it's equal to the ratio of a plus b to a. So I want to sit and think about this a little bit. I want to see is can I construct some a that's on this ratio, this perfect ratio that I'm somehow referring to right here, so that the ratio of the longer side to the shorter side is equal to the ratio of the whole thing to the longer side. And let's just assume that we can find a ratio like that. And we'll call it phi. We'll use the Greek letter phi for that ratio over there. So let's see what we can learn about this special ratio phi. Well if phi is equal to a over b, which is equal to a plus b over a, we know that a plus b over a is the same thing as a over a plus b over a. And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let" + }, + { + "Q": "At 1:50, sal said 8-24-3= negative 19. 8-24 is 16, not 22.\n", + "A": "just listen to the guy", + "video_name": "-rxUip6Ulnw", + "timestamps": [ + 110 + ], + "3min_transcript": "What I want to do is think about whether this expression right over here would evaluate the same way whether or not we had parentheses. So to think about that, let's first think about how it would evaluate if we add the parentheses. So if we add the parentheses, we want to do what's ever in the parentheses first. And so here we have 8 minus 3, which is equal to 5. So this simplifies to 5 times 5 times 8 minus 3. And now we want to do the multiplication before we do subtraction. This goes back to order of operations. You do your multiplication and division first. Well, you do your parentheses first. Then if you have multiplication, division, addition, and subtraction all in a row, you want to do your multiplication and your division first. So here we're going to multiply 5 times 8 to get 40, and then we're going to subtract 3 to get 37. Now, let's think about what this would evaluate to if we did not have the parentheses. So it would be 8 minus 3 times 8 minus 3. about the order of operations. The convention is to do your multiplication first. So you're actually going to multiply the 3 times the 8 before you subtract it from this 8 and then before you subtract this 3. So we took away the parentheses, but the order of operations say, hey, do this multiplication first. We could even put a parentheses here to emphasize that. So this will become 8 minus 8 minus 24. Let me write it this way. 8 minus 24 minus 3. 8 minus 24 minus 3. Now, 8 minus 24 is negative 16. You subtract another 3, you're going to get to negative 19. So clearly, you get very, very different values" + }, + { + "Q": "\nat 0:18 why do we put a sqwiggle under (8-3)", + "A": "That means we will solve that part first. (8-3) Also, you have to solve that part first, because it is in parentheses.", + "video_name": "-rxUip6Ulnw", + "timestamps": [ + 18 + ], + "3min_transcript": "What I want to do is think about whether this expression right over here would evaluate the same way whether or not we had parentheses. So to think about that, let's first think about how it would evaluate if we add the parentheses. So if we add the parentheses, we want to do what's ever in the parentheses first. And so here we have 8 minus 3, which is equal to 5. So this simplifies to 5 times 5 times 8 minus 3. And now we want to do the multiplication before we do subtraction. This goes back to order of operations. You do your multiplication and division first. Well, you do your parentheses first. Then if you have multiplication, division, addition, and subtraction all in a row, you want to do your multiplication and your division first. So here we're going to multiply 5 times 8 to get 40, and then we're going to subtract 3 to get 37. Now, let's think about what this would evaluate to if we did not have the parentheses. So it would be 8 minus 3 times 8 minus 3. about the order of operations. The convention is to do your multiplication first. So you're actually going to multiply the 3 times the 8 before you subtract it from this 8 and then before you subtract this 3. So we took away the parentheses, but the order of operations say, hey, do this multiplication first. We could even put a parentheses here to emphasize that. So this will become 8 minus 8 minus 24. Let me write it this way. 8 minus 24 minus 3. 8 minus 24 minus 3. Now, 8 minus 24 is negative 16. You subtract another 3, you're going to get to negative 19. So clearly, you get very, very different values" + }, + { + "Q": "At 4:30 Sal says that all bi's should be members of R3. Shouldn't be all bi's members of R4 as B has 4 column components ? Or my understanding is wrong ?\n", + "A": "How many components do B s column vectors have? What vector space are they in? How many components do B s row vectors have? What vector space are they in? (3, R3. 4, R4.) (Rn is defined as the set of all the vectors with n real number components.) This matrix B itself is 4x3 , and it s not an Rn or an Rm vector - although parts of it (e.g., its rows and columns) are. An mx1 matrix is an Rm column vector. A 1xn matrix is an Rn row vector.", + "video_name": "x1z0hOyjapU", + "timestamps": [ + 270 + ], + "3min_transcript": "And I showed you that in the last video. With that said, let's actually compute some matrix-matrix products just so you get the hang of it. So let's say that I have the matrix A. Let's say that A is equal to the matrix 1, minus 1, 2, 0, minus 2, and 1. I keep the numbers low to keep our arithmetic fairly straightforward. And let's say that I have the matrix B, and let's say that it is equal to 1, 0, 1, 1, 2, 0, 1, minus 1, and then 3, 1, 0, 2. So A is a 2 by 3 matrix, 2 rows, 3 columns. So by our definition, what is the product AB going to be equal to? Well, we know it's well-defined because the number of columns here is equal to the number of rows, so we can actually take these matrix vector products-- you'll see that in a second-- so AB is equal to the matrix A times the column vector, 1, 2, 3. That's going to be the first column in our product matrix. And the second one is going to be the matrix A times the column 0, 0, 1. The third column is going to be the matrix A times the column vector 1, 1, 0. And then the fourth column in our product vector is going to be the matrix A times the column vector 1, minus 1, 2. And this, when we write it like this, it should be clear be the number of rows in B, because the column vectors in B are going to have the same number of components as the number of rows in B, so all of the column vectors in B-- so if we call this B1, B2, B3, B4, all of my bi's-- let me write it this way-- all of my bi's where this i could be 1, 2, or 3, or 4, are all members of R3. So we only have matrix vector products well-defined when the number of columns in your matrix are equivalent to essentially the dimensionality of your vectors. That's why that number and that number has to be the same. Well, now we've reduced our matrix-matrix product problem to just four different matrix vector product problems, so we can just multiply these. This is nothing new to us, so let's do it." + }, + { + "Q": "\ni did not clearly understand how Sal undistributed s(s+5) -7(s+5)=0\nat 2:15 .. could someone help me out please ?", + "A": "so simple see that video once again seeit was s^2 +5s -7s -35 then he (s^2 +5s)_ _ _ _ _ _ (i) -(7s +35) _ _ _ _ _ _(ii) in (i) s is common so, s(s+5) in (ii) -7 is common so, -7(s+5) done this is the easiest way", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 135 + ], + "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers" + }, + { + "Q": "\nDo you come out with two answers when you do these kinds of equations? If S=S is a true statement, how could it be possible that S could equal both -5 or 7 as shown at 4:00? I'm so confused! HELP! :-O", + "A": "S can t equal both values at the same time. But, both values (-5 and 7) will make the equation be true. This means each value on its own is a valid solution to the equation. Hope this helps.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 240 + ], + "3min_transcript": "So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5. both sides of that equation, and you get s is equal to 7. So if s is equal to negative 5, or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is minus 35. That does equal zero. If you have 7, 49 minus 14 minus 35 does equal zero. So we've solved for s. Now, I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what is that equal to? x times x is x squared, x times b is bx." + }, + { + "Q": "\nAt 2:39 isn't -2*8 equals -16?", + "A": "-2*8 is -16 but he added the -2, making it -18", + "video_name": "jlID_mIJXi4", + "timestamps": [ + 159 + ], + "3min_transcript": "And what we will have will be h of what we inputted, h of g of 8. So let's just do it one step at a time. Let's figure out what g of 8 is. And I'm going to color code it, so we can keep track of things. g of 8 is equal to-- well, g of t, we have our definition here. So our input now, 8 is going to be our t, so our input is 8. So every place where we see a t in this function definition, we replace it with an 8. So it's going to be negative 2 times 8 minus 2 minus-- and this might be a little daunting, but let's just replace this t with an 8 and then see if we can make sense of it. h minus-- and let me do it in the right color-- minus 2 minus h of 8. we saw a t, we replaced it with the input 8. Now let's see if we can calculate this. This is going to be equal to negative 2 times 8 is negative 16. Minus 2 is negative 18. Let me do that the same. So this is going to be equals negative 18 minus-- what is h of 8 going to be equal to? So let's do that over here. So h of 8-- this thing, h of 8. Now we go to the definition of h. Don't worry about later we're going to input all this business into h again. Just let's worry about it one step at a time. We need to calculate h of 8. So h of 8 is just going to be-- well, every time we see an x, we replace it with an 8-- it's going to be 3 times 8, which is equal to 24. So this value right over here is 24. We are subtracting it, so we have minus 24. That's negative 42. So all of this business is going to be equal to-- did I do that right? Yeah-- negative 42. So we figured out what g of 8 is. It is negative 42. So this right over here is negative 42. And now we can input negative 42 into h. Let me do it right over here. h of negative 42-- remember, negative 42 is the same thing as g of 8. So this is h of g of 8 is the same thing as h of negative 42. Let me do that in the same color. This is going to be equal to 3 times negative 42, which is equal to-- this is negative 126. And we are done. So it seemed convoluted at first," + }, + { + "Q": "At 8:20 when x=1, why is that y is incremented by 'half' of dy/dx( 1.5 ) and not 1.5 itself?\n", + "A": "Because our \u00ce\u0094x = 0.5, so we only advance half a unit. dy/dx = 1.5 means that an increment of 1 in x would carry an increment of 1.5 in y, but since we are only incrementing x by 0.5, we only increment y by 0.75.", + "video_name": "q87L9R9v274", + "timestamps": [ + 500 + ], + "3min_transcript": "which is right over here. And so, for this next stretch, the next stretch is going to look like that. And as you can see just by doing this, we haven't been able to approximate what the particular solution looks like and you might say, \"Hey, so how do we know \"that's not so good of an approximation?\" And my reply to you is well, yeah I mean, depends on what your goals are. But I did this by hand. I didn't even do this using a computer. And because I wanted to do it by hand I took fairly large delta X steps. If I wanted a better approximation I could have lowered the delta X and let's do that. So let's take another scenario. So let's do another scenario where instead of delta X equal one, let's say delta X equals 1/2. So once again, X, Y and the derivative of Y with respect to X. So we know this first point. We're given this initial condition. When the X is zero, Y is one and so the slope of the tangent line is going to be one. But then if we're incrementing by 1/2 so then when X is, I'll just write it as 0.5. 0.5. What is our new Y going to be? Well we're gonna assume that our slope from this to this is this slope right over here. So our slope is one, so if we increase X by 0.5 we're gonna increase Y by 0.5 and we're going to get to 1.5. So, we can get 0.5, 1.5. We get to that point right over there. Actually you're having trouble seeing that. This stuff right over here is this point right over here and now our new slope is going to be 1.5. Which is going to look like. Which is going to look like actually not quite that steep. and it's starting to get a little bit messy but it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment X by another 0.5 where you get to one. So now if you increment by 0.5 and your slope is 1.5, your Y is going to increment by half of that by 0.75 and so, you're gonna get to 2.25. So now you get to one, 2.25 which is this point right over here. Once again, this is a better approximation. Remember, in the original one Y of one you know should be equal to E. Y of one in the actual solution should be equal to E. 2.7 on and on and on and on and on. Now in this one, Y of one got us to two. In this one Y of one got us to 2.25. Once again, closer to the actual reality, closer to E. Instead of stepping by 0.5, if we stepped by 0.1 we would get even closer." + }, + { + "Q": "\nAt 8:29, how did Sal get y=2.5 when x=1? I thought it would be 1+(1.5*.5)", + "A": "Sal got y=2.25 when x=1. 1.5 + (1.5*.5) =2.25", + "video_name": "q87L9R9v274", + "timestamps": [ + 509 + ], + "3min_transcript": "which is right over here. And so, for this next stretch, the next stretch is going to look like that. And as you can see just by doing this, we haven't been able to approximate what the particular solution looks like and you might say, \"Hey, so how do we know \"that's not so good of an approximation?\" And my reply to you is well, yeah I mean, depends on what your goals are. But I did this by hand. I didn't even do this using a computer. And because I wanted to do it by hand I took fairly large delta X steps. If I wanted a better approximation I could have lowered the delta X and let's do that. So let's take another scenario. So let's do another scenario where instead of delta X equal one, let's say delta X equals 1/2. So once again, X, Y and the derivative of Y with respect to X. So we know this first point. We're given this initial condition. When the X is zero, Y is one and so the slope of the tangent line is going to be one. But then if we're incrementing by 1/2 so then when X is, I'll just write it as 0.5. 0.5. What is our new Y going to be? Well we're gonna assume that our slope from this to this is this slope right over here. So our slope is one, so if we increase X by 0.5 we're gonna increase Y by 0.5 and we're going to get to 1.5. So, we can get 0.5, 1.5. We get to that point right over there. Actually you're having trouble seeing that. This stuff right over here is this point right over here and now our new slope is going to be 1.5. Which is going to look like. Which is going to look like actually not quite that steep. and it's starting to get a little bit messy but it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment X by another 0.5 where you get to one. So now if you increment by 0.5 and your slope is 1.5, your Y is going to increment by half of that by 0.75 and so, you're gonna get to 2.25. So now you get to one, 2.25 which is this point right over here. Once again, this is a better approximation. Remember, in the original one Y of one you know should be equal to E. Y of one in the actual solution should be equal to E. 2.7 on and on and on and on and on. Now in this one, Y of one got us to two. In this one Y of one got us to 2.25. Once again, closer to the actual reality, closer to E. Instead of stepping by 0.5, if we stepped by 0.1 we would get even closer." + }, + { + "Q": "\nat 3:35, how does he know that it's x squared and not anything else", + "A": "Okay, well he knows that he has to find the greatest common factor (GCF) among x^2, x^3, and x^4. So what terms can he evenly divide each of the above terms by, so that there is no remainder? First of all, there s x. Another option would be x-squared (x^2). x^2 divided by x^2 = 1 x^3 divided by x^2 = x x^4 divided by x^2 = x^2 But nothing larger than x^2 can fit evenly into x-squared. The only options are x and x^2. x^2 is the GREATEST, so that is what he will use to factor out of all the numbers.", + "video_name": "tYknkDjp-bQ", + "timestamps": [ + 215 + ], + "3min_transcript": "Three is divisible into all of them. And that's it 'cause we can't say a three and a two. A three and a two would be divisible into 12 and six, but there's no two that's divisible into 15. We can't say a three and a five 'cause five isn't divisible into 12 or six; so the greatest common factor is going to be three. Another way we could have done this is we could have said where are the non-prime factors of each of these numbers. 12 you could have said, OK I can get 12 by saying one times 12 or two times six or three times four. Six you could have said, let's see, that could be one times six or two times three. So those are the factors of six. And then 15 you could have said well one times 15 or three times five. And so you say the greatest common factor? that is common to all three of these factors. So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the greatest common monomial factor the coefficient is going to be three. And then we look at these powers of X. We have X to the fourth. I'm using a different color. We have X to the fourth, X to the third, and X squared. Well what's the largest power of X that's divisible into all of those? Well it's going to be X squared. X squared is divisible into X to the fourth and X to the third and of course X squared itself. So the greatest common monomial factor is 3x squared. This length right over here, this is 3x squared. So if this is 3x squared, we can then figure out what the width is. If we were to divide 12x to the fourth by 3x squared, what do we get? Well 12 divided by three is four is X squared. Notice 3x squared times 4x squared is 12x to the fourth. And then we move over to this purple section. If we take 6x to the third divided by 3x squared, six divided by three is two. And then X to the third divided by X squared is just going to be X. And then last but not least, we have 15 divided by three is going to be five. X squared divided by X squared is just one, so it's just gonna be five. So the width is going to be 4x squared, plus 2x, plus five. So once again, the length, we figure that out, is the greatest common monomial factor of these terms. It's 3x squared. And the width is 4x squared plus 2x, plus five. And one way to think about it is we just factored this expression over here. We could write that." + }, + { + "Q": "At 1:50 why bother multiplying by 1/2? You could have multiplied the right by 2; it wouldn't have been easier than adding the extra step?\n", + "A": "They are not a bother. Both ways are correct.", + "video_name": "STcsaKuW-24", + "timestamps": [ + 110 + ], + "3min_transcript": "The height of a triangle is four inches less than the length of the base. The area of the triangle is 30 inches squared. Find the height and base. Use the formula area equals one half base times height for the area of a triangle. OK. So let's think about it a little bit. We have the-- let me draw a triangle here. So this is our triangle. And let's say that the length of this bottom side, that's the base, let's call that b. And then this is the height. This is the height right over here. And then the area is equal to one half base times height. Now in this first sentence they tell us at the height of a triangle is four inch is less than the length of the base. So the height is equal to the base minus 4. That's what that first sentence tells us. The area of the triangle is 30 inches squared. So if we take one half the base times the height Or we could say that 30 inches squared is equal to one half times the base, times the height. Now instead of putting an h in for height, we know that the height is the same thing as 4 less than the base. So let's put that in there. 4 less than the base. And then let's see what we get here. We get-- let me do this in yellow. We get 30 is equal to one half times-- let's distribute the b-- times b, let me make it clear. So let's do it this way. Times b over 2, times b, minus 4. I just multiplied the one half times the b. Now let's distribute the b over 2. So 30 is equal to b squared over 2, be careful. b over 2 times b is just b squared over 2. And then b over 2, times negative 4 is negative 2b. here let's multiply both sides of this equation by 2. So let's multiply that side by 2. And let's multiply that side by 2. On the left hand side you get 60. On the right hand side 2 times b squared over 2 is just b squared. Negative 2b times 2 is negative 4b. And now we have a quadratic here. And the best way to solve a quadratic-- we have a second degree term right here-- is to get all of the terms on one side of the equation, having them equal 0. So let's subtract 60 from both sides of this equation. And we get 0 equal to b squared, minus 4b, minus 60. And so what we need to do here is just factor this thing right now, or factor it. And then, no-- if I have the product of some things, and that equals 0, that means that either one or both" + }, + { + "Q": "At 1:06, why wouldn't it be 30^2 instead of 30 since they directions tell you that area equals 30in^2?\n", + "A": "in^2 is a term for telling you the area of something. That is virtually all it is, just like you would use My room has 50 feet of flooring you could easily say My room s floor is 50 ft^2 and it would mean the same thing. You only have to worry about the in^2 thing when you get into physics and such. I hope that helps!", + "video_name": "STcsaKuW-24", + "timestamps": [ + 66 + ], + "3min_transcript": "The height of a triangle is four inches less than the length of the base. The area of the triangle is 30 inches squared. Find the height and base. Use the formula area equals one half base times height for the area of a triangle. OK. So let's think about it a little bit. We have the-- let me draw a triangle here. So this is our triangle. And let's say that the length of this bottom side, that's the base, let's call that b. And then this is the height. This is the height right over here. And then the area is equal to one half base times height. Now in this first sentence they tell us at the height of a triangle is four inch is less than the length of the base. So the height is equal to the base minus 4. That's what that first sentence tells us. The area of the triangle is 30 inches squared. So if we take one half the base times the height Or we could say that 30 inches squared is equal to one half times the base, times the height. Now instead of putting an h in for height, we know that the height is the same thing as 4 less than the base. So let's put that in there. 4 less than the base. And then let's see what we get here. We get-- let me do this in yellow. We get 30 is equal to one half times-- let's distribute the b-- times b, let me make it clear. So let's do it this way. Times b over 2, times b, minus 4. I just multiplied the one half times the b. Now let's distribute the b over 2. So 30 is equal to b squared over 2, be careful. b over 2 times b is just b squared over 2. And then b over 2, times negative 4 is negative 2b. here let's multiply both sides of this equation by 2. So let's multiply that side by 2. And let's multiply that side by 2. On the left hand side you get 60. On the right hand side 2 times b squared over 2 is just b squared. Negative 2b times 2 is negative 4b. And now we have a quadratic here. And the best way to solve a quadratic-- we have a second degree term right here-- is to get all of the terms on one side of the equation, having them equal 0. So let's subtract 60 from both sides of this equation. And we get 0 equal to b squared, minus 4b, minus 60. And so what we need to do here is just factor this thing right now, or factor it. And then, no-- if I have the product of some things, and that equals 0, that means that either one or both" + }, + { + "Q": "This question has been asked before, I think but it hasn't been answered yet so I'll just bring it up again. At 3:42 Sal mentions that he'll explain why he's picking the numbers he's picking. My question is that could you have made the height of the rectangle any fraction with the tau in the denominator as long as the limit to it went to infinity? Are we trying to force the area to 1 perhaps?\n", + "A": "Yes, we are trying to get the area to be 1, so that it will approach the dirac delta function as \u00cf\u0084 approaches 0.", + "video_name": "4qfdCwys2ew", + "timestamps": [ + 222 + ], + "3min_transcript": "This is part of the definition of the function. I'm going to tell you that if I were to take the integral of this function from minus infinity to infinity, so essentially over the entire real number line, if I take the integral of this function, I'm defining it to be equal to 1. I'm defining this. Now, you might say, Sal, you didn't prove it to me. No, I'm defining it. I'm telling you that this delta of x is a function such that its integral is 1. So it has this infinitely narrow base that goes infinitely high, and the area under this-- I'm telling you-- is of area 1. And you're like, hey, Sal, that's a crazy function. I want a little bit better understanding of how someone can construct a function like this. So let's see if we can satisfy that a little bit more. But then once that's satisfied, then we're going to start taking the Laplace transform of this, and then we'll start manipulating it and whatnot. Let's say that I constructed another function. Let's call it d sub tau And this is all just to satisfy this craving for maybe a better intuition for how this Dirac delta function can be constructed. And let's say my d sub tau of-- well, let me put it as a function of t because everything we're doing in the Laplace transform world, everything's been a function of t. So let's say that it equals 1 over 2 tau, and you'll see why I'm picking these numbers the way I am. 1 over 2 tau when t is less then tau and greater than minus tau. And let's say it's 0 everywhere else. So this type of equation, this is more reasonable. functions, and we can actually define it as a combination of unit step functions. So if I draw, that's my x-axis. And then if I put my y-axis right here. That's my y-axis. Sorry, this is a t-axis. I have to get out of that habit. This is the t-axis, and, I mean, we could call it the y-axis or the f of t-axis, or whatever we want to call it. That's the dependent variable. So what's going to happen here? It's going to be zero everywhere until we get to minus t, and then at minus t, we're going to jump up to some level. Just let me put that point here. So this is minus tau, and this is plus tau. So it's going to be zero everywhere, and then at minus tau, we jump to this level, and then we stay constant at that level until we get to plus tau." + }, + { + "Q": "\nwhat is an abaquabapattern 2:05", + "A": "It s abacaba or abacabadabacaba. You can learn about it in Vi s other video, Fractal Fractions. Hope this helps :)", + "video_name": "pjrI91J6jOw", + "timestamps": [ + 125 + ], + "3min_transcript": "You may have heard of turkducken, a turkey stuffed with a duck, stuffed with a chicken. Yeah, it's a cute idea, but mathematically uninspiring. Much more interesting would be some sort of fractal-fowl arrangement. Say the turkey were stuffed with two ducks, and say each duck were stuffed with two hens. You'd get a turduckduckenenenen, or turduckenenduckenen. I'm not sure yet. Then each hen could be stuffed with two quails, or whatever. And by the time you get down to sparrows, you have a whole flock of them in there. That sounds fun and practical. Let's do it! But being an extremely practical person who realizes that exponential quails will not fit inside of one turkey, I am using tiny unborn quails. Eight of them inside four hens, inside two ducks, inside one turkey. This structure is much more interesting because birds belong in trees, binary trees, that is. Obviously this is superior to the old linear bird-stuffing paradigm, and while you're deboning seven birds, you have lots of time to consider the question of binary bird stuffing nomenclature. How do you traverse this tree of syllables? you'll have to say the syllables in a linear order. You could go layer by layer, biggest birds first, like turduckenenenenailailailailailailailail. Or, you could go down through the layers, like turduckenailailenailailduckenailailenailail. The second one is certainly more complicated to say, but I like the way the structure of it suggests the structure of the whole. And say you've made two of these and put them in a goose. In the first scheme, the new bird names get inserted into the name to get gooturturduckduckduckenenenenenenenenailailailailailailailailailailailailailailailail, while for the second, you start with goose and then just repeat the old word twice to get gooturduckenailailenailailduckenailailenailailturduckenailailenailailduckenailailenailail. So it's nice that part of it stays the same. Of course, those aren't the only possible naming schemes. Maybe you go from left to right on the tree. So this would be, quailenquailduckquailenquailkeyquailenquailduckquailenquail OK, so in 1807, a guy roasted a bustard-- whatever that is-- stuffed with a turkey, stuffed with a goose, stuffed with a pheasant, chicken, duck, guinea fowl, teal, woodcock, partridge, plover, lapwing quail, thrush, lark, bunting, and warbler to get at buskeygooseantenduckneatealcockridgeerwingailusharktinbler. But say he had done this with the new exponential bird-stuffing paradigm. I mean, you'd need over 100,000 birds to do it, but the world has a lot of birds in it. And if you can consistently say four syllables a second, you can say the name of it in only like nine hours. It might seem like a lot, but it's really not compared to if you used, say just twice as many kinds of birds. Then you'd need over 8.5 billion individual birds. And if you started naming it as soon as you learned to talk and take breaks to sleep at night, you'd still probably die before you finish. So I can't say I recommend it, buy hey, maybe with advances in medicine, it will become a more feasible goal. And even there it is-- two quail eggs" + }, + { + "Q": "\nAt 3:54, it shows the egg with the whites still surrounding the yolk. How did she get it to stay together instead of all smushing around? Did she leave it inside of the shells? If so, then how do you eat it?", + "A": "They were hard-boiled and peeled.", + "video_name": "pjrI91J6jOw", + "timestamps": [ + 234 + ], + "3min_transcript": "OK, so in 1807, a guy roasted a bustard-- whatever that is-- stuffed with a turkey, stuffed with a goose, stuffed with a pheasant, chicken, duck, guinea fowl, teal, woodcock, partridge, plover, lapwing quail, thrush, lark, bunting, and warbler to get at buskeygooseantenduckneatealcockridgeerwingailusharktinbler. But say he had done this with the new exponential bird-stuffing paradigm. I mean, you'd need over 100,000 birds to do it, but the world has a lot of birds in it. And if you can consistently say four syllables a second, you can say the name of it in only like nine hours. It might seem like a lot, but it's really not compared to if you used, say just twice as many kinds of birds. Then you'd need over 8.5 billion individual birds. And if you started naming it as soon as you learned to talk and take breaks to sleep at night, you'd still probably die before you finish. So I can't say I recommend it, buy hey, maybe with advances in medicine, it will become a more feasible goal. And even there it is-- two quail eggs two ducks in this turkey-- if I can close it. Eventually I had to go for sewing up the turkey most of the way with the one duckenailailenailail, and then stuffing the other duckenailailenailail in. Now you can arrange it nicely with the original legs, and wings, and stuffing to make it look perfectly natural. There. Anyway, once you've got that, you're pretty much good to go as far as Thanksgiving is concerned. You should already have your gelatinous cranberry cylinder, bread spheres with butter prism, masked potatoes with organic hyperbolic plain, string bean vector field with Borromean onion rings on top, double helix cut ham, pi, tau, and so on. And now, finally, you've got your turduckenailailenailailduckenailailenailail, or quailenquailduckquailenquailkeyquailenquailduckquailenquail, or turduckduckenenenenailailailailailailailail whatever. Each slice gives a different cross section of this binary bird. Here, you can see a quail egg wrapped in the light meat of the hen, wrapped in the darker meat of the duck, wrapped in the light meat of the turkey. Another cross section shows two eggs. Now you can sit down, eat your mathematically-inspired food at least mathematics is always there for you, making sense, and truth, and beauty, and birds. Lots of birds." + }, + { + "Q": "At 3:07, when she sows the turkey to hold everything, I wondered if she was actually going to eat that, considering the fact that there is now string somewhere in there. Or will she just take the string out when she gets to it?\n", + "A": "She probably took the string out once it was cooked.", + "video_name": "pjrI91J6jOw", + "timestamps": [ + 187 + ], + "3min_transcript": "you'll have to say the syllables in a linear order. You could go layer by layer, biggest birds first, like turduckenenenenailailailailailailailail. Or, you could go down through the layers, like turduckenailailenailailduckenailailenailail. The second one is certainly more complicated to say, but I like the way the structure of it suggests the structure of the whole. And say you've made two of these and put them in a goose. In the first scheme, the new bird names get inserted into the name to get gooturturduckduckduckenenenenenenenenailailailailailailailailailailailailailailailail, while for the second, you start with goose and then just repeat the old word twice to get gooturduckenailailenailailduckenailailenailailturduckenailailenailailduckenailailenailail. So it's nice that part of it stays the same. Of course, those aren't the only possible naming schemes. Maybe you go from left to right on the tree. So this would be, quailenquailduckquailenquailkeyquailenquailduckquailenquail OK, so in 1807, a guy roasted a bustard-- whatever that is-- stuffed with a turkey, stuffed with a goose, stuffed with a pheasant, chicken, duck, guinea fowl, teal, woodcock, partridge, plover, lapwing quail, thrush, lark, bunting, and warbler to get at buskeygooseantenduckneatealcockridgeerwingailusharktinbler. But say he had done this with the new exponential bird-stuffing paradigm. I mean, you'd need over 100,000 birds to do it, but the world has a lot of birds in it. And if you can consistently say four syllables a second, you can say the name of it in only like nine hours. It might seem like a lot, but it's really not compared to if you used, say just twice as many kinds of birds. Then you'd need over 8.5 billion individual birds. And if you started naming it as soon as you learned to talk and take breaks to sleep at night, you'd still probably die before you finish. So I can't say I recommend it, buy hey, maybe with advances in medicine, it will become a more feasible goal. And even there it is-- two quail eggs two ducks in this turkey-- if I can close it. Eventually I had to go for sewing up the turkey most of the way with the one duckenailailenailail, and then stuffing the other duckenailailenailail in. Now you can arrange it nicely with the original legs, and wings, and stuffing to make it look perfectly natural. There. Anyway, once you've got that, you're pretty much good to go as far as Thanksgiving is concerned. You should already have your gelatinous cranberry cylinder, bread spheres with butter prism, masked potatoes with organic hyperbolic plain, string bean vector field with Borromean onion rings on top, double helix cut ham, pi, tau, and so on. And now, finally, you've got your turduckenailailenailailduckenailailenailail, or quailenquailduckquailenquailkeyquailenquailduckquailenquail, or turduckduckenenenenailailailailailailailail whatever. Each slice gives a different cross section of this binary bird. Here, you can see a quail egg wrapped in the light meat of the hen, wrapped in the darker meat of the duck, wrapped in the light meat of the turkey. Another cross section shows two eggs. Now you can sit down, eat your mathematically-inspired food" + }, + { + "Q": "\nI am extremely confused as to how he got s=1.04 at time 1:52... He says that s=sqrt(sum(x-xbar)^2/(n-1)), however, when I do this, I get s=2.8074. I've tried looking at other sites and they say the same thing about the formula that I wrote above. I've used the formula many times, and I still don't understand how he gets 1.04. Any help?", + "A": "I m getting 1.04 the same as Sal. In Excel it would be: =SQRT(((1.5-2.34)^2+(2.9-2.34)^2+(0.9-2.34)^2+(3.9-2.34)^2+(3.2-2.34)^2+(2.1-2.34)^2+(1.9-2.34)^2)/6) and this gives you 1.042209", + "video_name": "K4KDLWENXm0", + "timestamps": [ + 112 + ], + "3min_transcript": "7 patients blood pressures have been measured after having been given a new drug for 3 months. They had blood pressure increases of, and they give us seven data points right here-- who knows, that's in some blood pressure units. Construct a 95% confidence interval for the true expected blood pressure increase for all patients in a population. So there's some population distribution here. It's a reasonable assumption to think that it is normal. It's a biological process. So if you gave this drug to every person who has ever lived, that will result in some mean increase in blood pressure, or who knows, maybe it actually will decrease. And there's also going to be some standard deviation here. It is a normal distribution. And the reason why it's reasonable to assume that it's a normal distribution is because it's a biological process. It's going to be the sum of many thousands and millions of random events. And things that are sums of millions and thousands of random events tend to be normal distribution. And we don't know anything really about it outside of the sample that we have here. Now, what we can do is, and this tends to be a good thing to do, when you do have a sample just figure out everything that you can figure out about that sample from the get-go. So we have our seven data points. And you could add them up and divide by 7 and get your sample mean. So our sample mean here is 2.34. And then you can also calculate your sample standard deviation. Find the square distance from each of these points to your sample mean, add them up, divide by n minus 1, because it's a sample, then take the square root, and you get your sample standard deviation. I did this ahead of time just to save time. Sample standard deviation is 1.04. And when you don't know anything about the population distribution, the thing that we've been doing from the get-go is estimating that character with our sample So we've been estimating the true standard deviation of the population with our sample standard deviation. Now in this problem, this exact problem, we're going to run into a problem. We're estimating our standard deviation with an n of only 7. So this is probably going to be a not so good estimate because-- let me just write-- because n is small. In general, this is considered a bad estimate if n is less than 30. Above 30 you're dealing in the realm of pretty good estimates. So the whole focus of this video is when we think about the sampling distribution, which is what we're going to use to generate our interval, instead of assuming that the sampling distribution is normal like we did in many" + }, + { + "Q": "\n@3:32 why are we movie 1 step toward x and 7 towards y?? or 1 step back towards negative x and 7 steps towards positive y??", + "A": "There are 2 different ways to write the slope 7, you can just say 7, or you can say 7/1 because they are the same thing, so when you move 7 up you move 1 to the right, the reason you can move 1 to the left and 7 down is because the line will be in the same spot either way.", + "video_name": "MxiqyE2uMCo", + "timestamps": [ + 212 + ], + "3min_transcript": "So that tells us that when x is equal to negative 4, then y is equal to negative 11. So we can use this information in what we have or the part of our equation that we've been able to figure out so far. We know that when x is equal to negative 4, y is going to be equal to negative 11. So what b do we need to make that happen? So y is negative 11 when x is equal to negative 4. So negative 11 is equal to 7 times x-- and in this case x is negative 4-- plus b. And now we can just solve for b. A b that makes this equation, or that satisfies the constraint that when x is equal to negative 4, y is equal to negative 11. So let's see, we get negative 11 is equal to 7 times negative 4 can add a 28 to both sides of this equation. So let's add a 28. I'm just trying to isolate the b on the right-hand side. And so on the left-hand side, negative 11 plus 28, that is just positive 17. These guys cancel out on purpose. And I just have a b on the right-hand side. So I get b is equal to 17. Let me write it in green. That's not green. We get b is equal to 17. So we know m is 7, they told us that right at the beginning. And now we know b is 17. So the equation of our line is y is equal to 7x, that's our slope. 7 times x plus b, and b here is 17. And if we wanted to graph it, it would look something like this. I'll just do a real rough graph. and this is my y-axis. The y-intercept is 17. So that means that the point 0, 17 is on this line. So this point right over here is going to be 0, 17. And our slope is 7. So that means if we move to the right one, we move up t seven. So it's a high slope. So if we move to the right one, we move up seven. Or if we move back one, will move down seven. So we'll move down seven, so the line will look roughly like this. Obviously, haven't done it very exactly, but our line is going to look like. That's going to be a pretty steep upward-sloping line. It has a very high slope, slope of seven. If you move one in the x direction, you have to move up seven. And its y-intercept is at y is 17. When x is 0, y is 17." + }, + { + "Q": "\nAt 5:19, he says ' x - -3, is the same thing as x + 3, I don't understand... I always have problems trying to simplify the point-slope form...", + "A": "First you must understand that the opposite of the opposite of 3 is 3. Think of it this way: First you find the opposite of 3 (-3) and then find the opposite of that (3). I hope that answers your question.", + "video_name": "-6Fu2T_RSGM", + "timestamps": [ + 319 + ], + "3min_transcript": "there, or that 6 right there-- and we want to subtract from that our starting x value. Well, our starting x value is that right over there, that's that negative 3. And just to make sure we know what we're doing, this negative 3 is that negative 3, right there. I'm just saying, if we go from that point to that point, our y went down by 6, right? We went from 6 to 0. Our y went down by 6. So we get 0 minus 6 is negative 6. That makes sense. Y went down by 6. And, if we went from that point to that point, what happened to x? We went from negative 3 to 6, it should go up by 9. And if you calculate this, take your 6 minus negative 3, that's the same thing as 6 plus 3, that is 9. And what is negative 6/9? Well, if you simplify it, it is negative 2/3. So that is our slope, negative 2/3. So we're pretty much ready to use point slope form. We have a point, we could pick one of these points, I'll just go with the negative 3, 6. And we have our slope. So let's put it in point slope form. All we have to do is we say y minus-- now we could have taken either of these points, I'll take this one-- so y minus the y value over here, so y minus 6 is equal to our slope, which is negative 2/3 times x minus our Well, our x-coordinate, so x minus our x-coordinate is negative 3, x minus negative 3, and we're done. We can simplify it a little bit. x minus negative 3 is the same thing as x plus 3. This is our point slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope intercept form. Let's do that. So let's do slope intercept in orange. So we have slope intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2/3, so you get y minus 6 is equal to-- I'm just distributing the negative 2/3-- so negative 2/3 times x is negative 2/3 x. And then negative 2/3 times 3 is negative 2. And now to get it in slope intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side, so let's add 6 to both" + }, + { + "Q": "Do you have to do what Sal does at about 7:39? He simplified the fraction. Is this necessary?\n", + "A": "Doesn t seem necessary", + "video_name": "-6Fu2T_RSGM", + "timestamps": [ + 459 + ], + "3min_transcript": "x minus negative 3 is the same thing as x plus 3. This is our point slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope intercept form. Let's do that. So let's do slope intercept in orange. So we have slope intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2/3, so you get y minus 6 is equal to-- I'm just distributing the negative 2/3-- so negative 2/3 times x is negative 2/3 x. And then negative 2/3 times 3 is negative 2. And now to get it in slope intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side, so let's add 6 to both Left-hand side of the equation, we're just left with a y, these guys cancel out. You get a y is equal to negative 2/3 x. Negative 2 plus 6 is plus 4. So there you have it, that is our slope intercept form, mx plus b, that's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2/3 x to both sides of this equation. So I'll start it here. So we have y is equal to negative 2/3 x plus 4, that's slope intercept form. Let's added 2/3 x, so plus 2/3 x to both sides of this equation. I'm doing that so it I don't have this 2/3 x on the right-hand side, this negative 2/3 x. little bit, maybe more than I should have-- the left-hand side of this equation is what? It is 2/3 x, because 2 over 3x, plus this y, that's my left-hand side, is equal to-- these guys cancel out-- is equal to 4. So this, by itself, we are in standard form, this is the standard form of the equation. If we want it to look, make it look extra clean and have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get? 2/3 x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12. These are the same equations, I just multiplied every term by 3. If you do it to the left-hand side, you can do to the right-hand side-- or you have to do to the right-hand side-- and we are in standard form." + }, + { + "Q": "At 0:08 what does the line over the letter stand 4?\n", + "A": "That line indicates that you are talking about a segment defined by the points A and B. Those are the two ends of the line you are talking about. You may also see other similar forms of notation, like a small triangle next to the letters ABC describing triangle ABC, or something that looks like a small angle next to some letters indicating that the angle is defined by whatever is after it, like XYZ.", + "video_name": "bJF9R8_-0O0", + "timestamps": [ + 8 + ], + "3min_transcript": "Are line segment AB and line segment CD congruent? So let's look at these right over here. AB is this line segment right over here, and CD is this one right over here. AB has length of 1. It goes from 2 to 3. And CD has length of 1. It goes from 4 to 5. So they have the exact same length. These are just line segments with the exact same length. So yes, they are congruent. Yes. Let's do a couple more. Are AB and CD congruent? And you can just eyeball this. AB is much longer than CD. AB is of length 4. It goes from negative 5 to negative 1. CD is only of length 1. It goes from 1 to 2. So no, they are not congruent. They have different lengths. Let's do one more. Are AB and CD congruent? Let's see-- AB has length 2. It goes from negative 1 to 1. CD is of length 3. So they're not congruent. No." + }, + { + "Q": "At 1:12, why is the derivative of sqrt(3)x^2/36 = root 3x/18? I tried using chain rule and quotient rule on that term and I ended up getting 4 root 3x as my final answer. Was there an easier way to do that, or a correct way rather?\n", + "A": "No need for the chain rule or the quotient rule here. All you have is x^2 times a constant (which happens to be sqrt(3)/36). The derivative of x^2 is 2x, and you multiply that by the constant to get the result indicated.", + "video_name": "eS-_ZFzHjYA", + "timestamps": [ + 72 + ], + "3min_transcript": "Where we left off in the last video, we had come up with an expression as a function of x of our combined area based on where we make the cut. And now we just need to figure out where this hits a minimum value. And to do that, we just have to take the derivative of this business, figure out where our derivative is either undefined or 0, and then just make sure that that is a minimum value, and then we'll be all set. So let me rewrite this. So our combined area as a function of x, let me just rewrite this so it's a little bit easier to take the derivative. So this is going to be the square root of 3 times x squared over-- let's see, this is 4 times 9. This is x squared over 9. So this is going to be 4 times 9 is 36. And then over here in blue, this is going to be plus 100 minus x squared over 16. Now let's take the derivative of this. So A prime, the derivative of our combined area equal to-- well, the derivative of this with respect to x is just going to be square root of 3x over 18. The derivative of this with respect to x, well, it's the derivative of something squared over 16 with respect to that something. So that's going to be that something to the first power times 2/16, which is just over 8. And then times-- we're just doing the chain rule-- times the derivative of the something with respect to x. The derivative of 100 minus x with respect to x is just negative 1, so times negative 1. So we'll multiply negative 1 right over here. And so we can rewrite all of that as-- this is going to be equal to the square root of 3/18 x plus-- let's see, I could write this as positive x/8. So I could write this as 1/8 x, right? And then minus 100/8, which is negative 12.5-- minus 12.5. And we want to figure out an x that minimizes this area. So this derivative right over here is defined for any x. So we're not going to get our critical point by figuring out where the derivative is undefined. But we might get a critical point by setting this derivative equal to 0 to figure out what x-values make our derivative 0. When do we have a 0 slope for our original function? And then we just have to verify that this is going to be a minimum point if we can find an x that makes this thing equal to 0. So let's try to solve for x. So if we add 12.5 to both sides, we get 12.5 is equal to-- if you add the x terms, you get square root of 3/18 plus 1/8 x. To solve for x, divide both sides by this business. You get x is equal to 12.5 over square root of 3 over 18" + }, + { + "Q": "\nStarting at 1:05, why did you just divide by 2 to get the derivative?", + "A": "There s 2 things going on here. First, to get from sqrt(3)x/36 to sqrt(3)x/18, you multiply by 2, not divide. He does this due to the Power Rule for derivatives. Suppose that you have a function a*x^n, where a and n are both constants. The derivative of this function is (a*n)*x^(n-1). You multiply the constant by the exponent and then reduce the exponent by 1. As you see in the video, we go from having an x^2 to an x^1. There s a video on this website that describes this rule in more detail.", + "video_name": "eS-_ZFzHjYA", + "timestamps": [ + 65 + ], + "3min_transcript": "Where we left off in the last video, we had come up with an expression as a function of x of our combined area based on where we make the cut. And now we just need to figure out where this hits a minimum value. And to do that, we just have to take the derivative of this business, figure out where our derivative is either undefined or 0, and then just make sure that that is a minimum value, and then we'll be all set. So let me rewrite this. So our combined area as a function of x, let me just rewrite this so it's a little bit easier to take the derivative. So this is going to be the square root of 3 times x squared over-- let's see, this is 4 times 9. This is x squared over 9. So this is going to be 4 times 9 is 36. And then over here in blue, this is going to be plus 100 minus x squared over 16. Now let's take the derivative of this. So A prime, the derivative of our combined area equal to-- well, the derivative of this with respect to x is just going to be square root of 3x over 18. The derivative of this with respect to x, well, it's the derivative of something squared over 16 with respect to that something. So that's going to be that something to the first power times 2/16, which is just over 8. And then times-- we're just doing the chain rule-- times the derivative of the something with respect to x. The derivative of 100 minus x with respect to x is just negative 1, so times negative 1. So we'll multiply negative 1 right over here. And so we can rewrite all of that as-- this is going to be equal to the square root of 3/18 x plus-- let's see, I could write this as positive x/8. So I could write this as 1/8 x, right? And then minus 100/8, which is negative 12.5-- minus 12.5. And we want to figure out an x that minimizes this area. So this derivative right over here is defined for any x. So we're not going to get our critical point by figuring out where the derivative is undefined. But we might get a critical point by setting this derivative equal to 0 to figure out what x-values make our derivative 0. When do we have a 0 slope for our original function? And then we just have to verify that this is going to be a minimum point if we can find an x that makes this thing equal to 0. So let's try to solve for x. So if we add 12.5 to both sides, we get 12.5 is equal to-- if you add the x terms, you get square root of 3/18 plus 1/8 x. To solve for x, divide both sides by this business. You get x is equal to 12.5 over square root of 3 over 18" + }, + { + "Q": "At 3:42, the second missing number that you were explaining with the help of the number line.\nI am someone who gets very confused with Math, and the second 'number line' explanation, got me very confused. Rather I'd like to get this clarified.\nWhen trying to find a missing number i.e. ___ -(-2) = -7, as mentioned above. If minus of (minus some number) is nothing but a positive number, can we start with +2 on the number line and find out how much is needed to get to -7?\n", + "A": "Yes, -(-2) becomes +2 and you can start with +2 on the number line to get to -7. Hopefully, I didn t confuse you any further.", + "video_name": "KNGa11O2uLE", + "timestamps": [ + 222 + ], + "3min_transcript": "four is equal to, let me just write it without my boxes there, is equal to negative eight minus negative 12. Well what happens when you subtract a negative? Well that ends up being adding the positive. So this is gonna be the same thing as negative eight plus positive 12, or four is equal to negative 8 plus 12, which we know is true. You add eight you get to zero, negative eight plus eight is zero, then you add another four, you get to positive four. So that one was pretty interesting. Let's do another one, we can't get enough practice, this is too much fun. All right. So let's say we had... let's say we had something blank minus negative two is equal to negative seven. How do we think about this? Well the first thing my brain wants to do is to simplify subtracting a negative. that's the same thing as adding positive two. So this is an equivalent statement to say, something subtracting a negative two is the same thing as adding a positive two. Something plus two is equal to negative seven. And now we can get the number line out to think about it. So, get my number line out. So, my goal, my goal is I wanna get to negative seven. So there's negative five, negative six, negative seven right there. And I wanna add two to something to get to negative seven. So if I'm adding two that means I'm jumping two, I'm jumping two to the right, that's what adding two looks like. So what number do I need to start at that if I jump two to the right, if I add two I end up at negative seven? Well I would have to start two to the left of negative seven. If I start two to the left, one, two, if I started... Let me do this in a different color. If I start at this number right here I end up at negative seven. Well what number is this? This number that is two to the left of negative seven. Well it's negative nine. So this right over here is negative nine. Negative nine plus two, so if you start at negative nine and you move two to the right you end up at negative seven. Or, you could say, negative nine minus negative two, 'cause if you subtract negative two you're going to move to the right. If you subtract positive two, you move to the left. But then if you subtract negative two, you move two to the right, to negative seven. Another fascinating one. Let's do one more of these, I'm having maybe a little bit too much fun. All right. So let's say we have negative four is equal to blank minus nine. So let's get the number line out and always pause the video, see if you can try it on your own. But let's get my number line out" + }, + { + "Q": "what is si in 0:25 and deta in 2:04\n", + "A": "Psi and theta are Greek letters. Greek letters are often used to denote angles.", + "video_name": "MyzGVbCHh5M", + "timestamps": [ + 25, + 124 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 0:25, is there a special meaning si has? Oh, and theta too. (2:01.)", + "A": "psi and theta are just like using x in algebra they are just variables", + "video_name": "MyzGVbCHh5M", + "timestamps": [ + 25, + 121 + ], + "3min_transcript": "" + }, + { + "Q": "how does sal gets 1/2 at 5:50 - 5:56\n", + "A": "He divided both sides of the equation by 2. i.e. if 2 pens are worth 1 dollar, then 1 pen is worth 1/2 a dollar.", + "video_name": "MyzGVbCHh5M", + "timestamps": [ + 350, + 356 + ], + "3min_transcript": "" + }, + { + "Q": "At 3:56, what does base angle mean?\n", + "A": "Let s back up for a second. Sal has a triangle with an unknown side (a chord) and two radii. Because the two radii are congruent, then we know the triangle is isosceles. A base angle in an isosceles triangle is one of the two angles formed by the base (non congruent side) and one congruent side. A vertex angle is the angle formed by the two congruent sides of the triangle, in this case two radii.", + "video_name": "MyzGVbCHh5M", + "timestamps": [ + 236 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 1: 40 ish what's an alternate interior angle it means whats in alternate interior angle it said it at around 1:40 in the video", + "A": "An Alternate interior angle are angles that are on opposite sides of the transversal, and are between the lines cut by the ttransversal", + "video_name": "LhrGS4-Dd9I", + "timestamps": [ + 100 + ], + "3min_transcript": "What we're going to prove in this video is a couple of fairly straightforward parallelogram-related proofs. And this first one, we're going to say, hey, if we have this parallelogram ABCD, let's prove that the opposite sides have the same length. So prove that AB is equal to DC and that AD is equal to BC. So let me draw a diagonal here. And this diagonal, depending on how you view it, is intersecting two sets of parallel lines. So you could also consider it to be a transversal. Actually, let me draw it a little bit neater than that. I can do a better job. Nope. That's not any better. That is about as good as I can do. So if we view DB, this diagonal DB-- we can view it as a transversal for the parallel lines AB and DC. And if you view it that way, you can pick out that angle ABD is going to be congruent-- so angle ABD. That's that angle right there-- is going to be congruent to angle BDC, because they are alternate interior angles. So we know that angle ABD is going to be congruent to angle BDC. Now, you could also view this diagonal, DB-- you could view it as a transversal of these two parallel lines, of the other pair of parallel lines, AD and BC. And if you look at it that way, then you immediately see that angle DBC right over here is going to be congruent to angle ADB for the exact same reason. They are alternate interior angles of a transversal intersecting these two parallel lines. So I could write this. This is alternate interior angles are congruent when you have a transversal intersecting And we also see that both of these triangles, triangle ADB and triangle CDB, both share this side over here. It's obviously equal to itself. Now, why is this useful? Well, you might realize that we've just shown that both of these triangles, they have this pink angle. Then they have this side in common. And then they have the green angle. Pink angle, side in common, and then the green angle. So we've just shown by angle-side-angle that these two triangles are congruent. So let me write this down. We have shown that triangle-- I'll go from non-labeled to pink to green-- ADB is congruent to triangle-- non-labeled to pink to green-- CBD. And this comes out of angle-side-angle congruency." + }, + { + "Q": "\nWhy do we have to go down by one and a half ?\nduring 3:45", + "A": "To make a right triangle between point N and the line of reflection", + "video_name": "kj3ZfOQGKdE", + "timestamps": [ + 225 + ], + "3min_transcript": "I want to drop it to, I want to drop it to the line that I'm going to reflect on, and then I'm going to go the same distance onto the other side to find to find the corresponding point in the image. So how do I do that? Well if this line, if this purple line has a slope of negative one, a line that is perpendicular to it a line that is perpendicular to it so this thing that I'm drawing in purple right over here, its slope is going to be the negative reciprocal of this. So the reciprocal of negative one is still just negative one. One over negative one is still negative one. But we want the negative of that. So the slope here needs to be one. And luckily, that's how I drew it. The slope here needs to be equal to one which is however much I change in the X direction I change in the Y direction. We see that. To go from this point to this point right over here, we decrease Y by four and we decrease X by four. Now, if we want to stay on this line to find the reflection, we just do the same thing. so we'll go from negative two to negative six, and decrease Y by four, and we end up at this point right over here. So we end up at the point, this is X equals negative six, Y is equal to negative four. So this is, this point corresponds to this point right over there. Now, let's do the same thing. Let's do the same thing for point N. For point N, we already know, if we drop a perpendicular, and this is perpendicular, it's going to have a slope of one because this purple line has a slope of negative one. The negative reciprocal of negative one is positive one. And let's see, to go from this point to this point of intersection, we have to go down one and a half. We're going down one and a half, and we're going to the left one and a half. So we want to do that on the other side. We want to stay on this perpendicular line. So we want to go left one and a half, and down one and a half. And we get to this point right over here, which is the point So we are now equidistant. We're on this perpendicular line, still, but we're equidistant on the other side. So the image of IN is going to go through negative six, negative four, and three, negative eight. So let me draw that. Let me see if I can remember, negative six, negative four, and three, negative eight. So I have a bad memory. So negative six, negative four, and three, negative eight. And I was close when I estimated, but I wasn't exactly right. So that's looking pretty good. And actually, we can do the exact same thing with points T and point O. Let me do that. So point T, to point T, To get from point T to the line in the shortest distance, once again, we drop a perpendicular. This line is going to have a slope of one, because it's perpendicular to the line that has a slope of negative one. And so to get there, we have to decrease our X by" + }, + { + "Q": "At about 1:40 Sal shows that you can multiply the numerator by the reciprocal (8 x 36/8, or simplified which is 8 x 9/2) to get the denominator. But why? I mean how does he know he can do that? Where does this idea come from? I feel like there's something that I am missing here.\n", + "A": "they have top find out what to multiply the next ratio by so you flip the numerator and the denominator (pardon my spelling) which you can sliplify to 9/2 . WHich means you multiply thenext ratio by 9 over 2 that is how they know this.", + "video_name": "GO5ajwbFqVQ", + "timestamps": [ + 100 + ], + "3min_transcript": "We're asked to solve the proportion. We have 8 36ths is equal to 10 over what. Or the ratio of 8/36 is equal to the ratio of 10 to what. And there's a bunch of different ways to solve this. And I'll explore really all of them, or a good selection of them. So one way to think about it is, these two need to be equivalent ratios, or really, equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10? Well you could multiply 8 times 10/8. It will definitely give you 10. So we're multiplying by 10/8 over here. Or another way to write 10/8, 10/8 is the same thing as 5/4. So we're multiplying by 5/4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it. You have to multiply it times 5/4. And so we could say this n, this thing that we just solved for, Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We could divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9. Divide the denominator by 4 you get 1. You get 45. So that's one way to think about it. 8/36 is equal to 10/45. Another way to think about it is, what do we have to multiply 8 by to get its denominator. How much larger is the denominator 36 than 8? Well let's just divide 36/8. So 36/8 is the same thing as-- so we can simplify, dividing the numerator and the denominator by 4. That's the greatest common divisor. That's the same thing as 9/2. you get the denominator. So we're multiplying by 9/2 to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9/2 times 8, let me write this. 8 times 9/2 is equal to 36. That's how we go from the numerator to the denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9/2 again. So then we'll get 10 times 9/2 is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9/2. Divide the numerator and the denominator by 2, you get 5/1, which is 45. So 45 is equal to n. Once again, we got the same way, completely legitimate way, to solve it. Now sometimes when you see proportion like this," + }, + { + "Q": "\nCouldn't Sal have just converted 5/4 into 1.25 instead of going through all the motions at 0:60?", + "A": "I like the other ways because it gives me more than one POV and more ways to solve it.", + "video_name": "GO5ajwbFqVQ", + "timestamps": [ + 60 + ], + "3min_transcript": "We're asked to solve the proportion. We have 8 36ths is equal to 10 over what. Or the ratio of 8/36 is equal to the ratio of 10 to what. And there's a bunch of different ways to solve this. And I'll explore really all of them, or a good selection of them. So one way to think about it is, these two need to be equivalent ratios, or really, equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10? Well you could multiply 8 times 10/8. It will definitely give you 10. So we're multiplying by 10/8 over here. Or another way to write 10/8, 10/8 is the same thing as 5/4. So we're multiplying by 5/4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it. You have to multiply it times 5/4. And so we could say this n, this thing that we just solved for, Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We could divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9. Divide the denominator by 4 you get 1. You get 45. So that's one way to think about it. 8/36 is equal to 10/45. Another way to think about it is, what do we have to multiply 8 by to get its denominator. How much larger is the denominator 36 than 8? Well let's just divide 36/8. So 36/8 is the same thing as-- so we can simplify, dividing the numerator and the denominator by 4. That's the greatest common divisor. That's the same thing as 9/2. you get the denominator. So we're multiplying by 9/2 to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9/2 times 8, let me write this. 8 times 9/2 is equal to 36. That's how we go from the numerator to the denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9/2 again. So then we'll get 10 times 9/2 is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9/2. Divide the numerator and the denominator by 2, you get 5/1, which is 45. So 45 is equal to n. Once again, we got the same way, completely legitimate way, to solve it. Now sometimes when you see proportion like this," + }, + { + "Q": "\nI'm doing proportions right now, and I'm trying the first method that he used at 0:25 , but when I use the method on the problem, 9/8 = 11/r, and I try to divide 11 by 9 to see what times 9 is equal to 11, I'm getting a repeating decimal! What should I do? Thanks.", + "A": "Thanks this is super helpful!", + "video_name": "GO5ajwbFqVQ", + "timestamps": [ + 25 + ], + "3min_transcript": "We're asked to solve the proportion. We have 8 36ths is equal to 10 over what. Or the ratio of 8/36 is equal to the ratio of 10 to what. And there's a bunch of different ways to solve this. And I'll explore really all of them, or a good selection of them. So one way to think about it is, these two need to be equivalent ratios, or really, equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10? Well you could multiply 8 times 10/8. It will definitely give you 10. So we're multiplying by 10/8 over here. Or another way to write 10/8, 10/8 is the same thing as 5/4. So we're multiplying by 5/4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it. You have to multiply it times 5/4. And so we could say this n, this thing that we just solved for, Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We could divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9. Divide the denominator by 4 you get 1. You get 45. So that's one way to think about it. 8/36 is equal to 10/45. Another way to think about it is, what do we have to multiply 8 by to get its denominator. How much larger is the denominator 36 than 8? Well let's just divide 36/8. So 36/8 is the same thing as-- so we can simplify, dividing the numerator and the denominator by 4. That's the greatest common divisor. That's the same thing as 9/2. you get the denominator. So we're multiplying by 9/2 to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9/2 times 8, let me write this. 8 times 9/2 is equal to 36. That's how we go from the numerator to the denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9/2 again. So then we'll get 10 times 9/2 is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9/2. Divide the numerator and the denominator by 2, you get 5/1, which is 45. So 45 is equal to n. Once again, we got the same way, completely legitimate way, to solve it. Now sometimes when you see proportion like this," + }, + { + "Q": "At 4:55, I'm not sure where Sal gets the extra ten. Is it one of the two 10s left over from the beginning?\n", + "A": "He got an extra ten at that time because Sal rewrote 29.12 into 2.912. Since 29.12 equals 2.912 x 10.", + "video_name": "xxAFh-qHPPA", + "timestamps": [ + 295 + ], + "3min_transcript": "1 +1 = 2 and 2 + 7 = 9 and we have a 2 here. So 91 x 32 is 2912 But I didn't have 91 by 32 I had 9.1 x 3.2 So what I am going to do is count the number of digits behind the decimal point. I have 2 digits behind the point, so that's how many I need in the answer. I will stick the decimal right over there. So this part here comes out as 29.12 So you might say we are done as this looks like sientific notation As we have a number times by a power of ten. But remember this number has to be greater than or equal to one but this is not less than ten, so what we can do is write this number in scientific notation. and use the power of ten part to multiply this power of ten part. so 29.12 is the same thing as 2.912 x 10 notice what I had to do to go from there to there. I just moved the decimal to the left, so what can I do to this to get back I could multiply this (2.912) by 10 or move the decimal to the right. So I want to write this so 2.912 x 10 is the same as 29.12 so 2.912 is in scientific notation but I still have to multiply it by another part times another ten so to finnish up this problem I have Whats that, well that is ten Well that is going to be this part over here That's just ten squared. So it is 2.912 times ten to the second power. And we are done." + }, + { + "Q": "at 1:23, he said the oval had no sides, but doesnt it have 1?\n", + "A": "it has infinite tangents, but no sides, just like a circle.", + "video_name": "8xbIS2UkQxI", + "timestamps": [ + 83 + ], + "3min_transcript": "- [Voiceover] Oh look, it's my cousin Fal! Cousin Fal, what can I do for you? - [Voiceover] Hey Sal, what's up? - [Voiceover] Oh, well not much, I'm just sitting here thinking about what video I should do next. - [Voiceover] Well then maybe you can help me. I'm starting a shape collection. - [Voiceover] Okay, so how do you need help? - [Voiceover] Well look at these shapes over here! I have no way to classify them! - [Voiceover] Okay, well there's a bunch of ways that you could classify them. The first way, you could think about how many sides each of these shapes have, so that's one way to think about it, how many sides they have. - [Voiceover] How do I figure that out? - [Voiceover] Okay, well let's look at each of these and think about how many sides they have. So this first shape right over here this is one side. We see this kind of straight right over here, it's the straight line between these two corners, so that's one side. Two sides, three sides, and four sides. So this shape right over here has four sides. - [Voiceover] Okay, I think I get it. - [Voiceover] Okay,well let's see. This one, well same idea, this one also looks like it has one, two, three, and four. Four sides. - [Voiceover] What about this one? - [Voiceover] Okay, well this one, actually this is interesting. This one actually had no corners, and it has no sides, there's no straight edges over here, this thing is all curved. So since it has no corners, it also has no sides, or at least the way that I'm thinking about it. So I would say that this one over here has zero sides. Let me write that a little bit neater for you. - [Voiceover] Okay yeah, I like to write things really neat. - [Voiceover] Okay, zero sides. And it has zero corners. Actually let me write them for all of them. This one has four sides, and we could also count the corners. We have one, that's where the two sides meet. One, two, three, four corners. it also has four sides, we already counted that and it has one, two, three, four corners. So four sides and four corners actually describes both of these shapes, this one has zero sides and zero corners-- - [Voiceover] What about the green one? - [Voiceover] Okay, well let's just see. One side, two sides, three sides, and four sides, and it also has one, two, three, four corners. So this is also four sides and four corners is also true of that one there. - [Voiceover] Do all shapes have four sides and four corners? - [Voiceover] No no, not at all. I mean this one right over here had zero sides and zero corners, and actually this last shape right over here, doesn't seem to have four sides or four corners. If we count them, we have one side, two sides, and three sides. So this one has three sides, and if we count the corners we have one corner, two corners, and three," + }, + { + "Q": "\nAt the end of the video he messed up. It was around 3:34. It confused me", + "A": "he said so all the side are not the same length when ther are 2 small and 2 long sides", + "video_name": "8xbIS2UkQxI", + "timestamps": [ + 214 + ], + "3min_transcript": "it also has four sides, we already counted that and it has one, two, three, four corners. So four sides and four corners actually describes both of these shapes, this one has zero sides and zero corners-- - [Voiceover] What about the green one? - [Voiceover] Okay, well let's just see. One side, two sides, three sides, and four sides, and it also has one, two, three, four corners. So this is also four sides and four corners is also true of that one there. - [Voiceover] Do all shapes have four sides and four corners? - [Voiceover] No no, not at all. I mean this one right over here had zero sides and zero corners, and actually this last shape right over here, doesn't seem to have four sides or four corners. If we count them, we have one side, two sides, and three sides. So this one has three sides, and if we count the corners we have one corner, two corners, and three, - [Voiceover] Oh, thanks a bunch Uncle-- Cousin Sal. But I always get confused whether you're my cousin or my uncle. But I'll call you Cousin Sal. But I read in a shape magazine that there was other ways that I can classify. I can classify based on whether the sides have the same size. So which of these have all the sides are the same size? - [Voiceover] Okay, well calm down. Let's look at this. So if you look at this one, it has four sides. But we see, that this side right over here, this green side, is clearly shorter than this purple side. So all the four sides here are not the same length. But if we look at this shape right over here, this side at least looks like the same length as this side. And that looks like the same length as this side, and that looks like the same length as that side. So it looks over here like all four sides are the same length. And that's also true for this shape. This side looks the same length as that side, looks the same length as that side, looks the same length as that side. This side looks the same length as that side, This is the same length as that one. - [Voiceover] Oh, thank you so much Cousin Sal! I'm ready to go classify my shape collection. This is awesome! - [Voiceover] Well I'm glad that I could help!" + }, + { + "Q": "\nCan a circle or an oval be considered to have 1 side? At 1:32 Sal said, \"Or at least the way that I am thinking about it.\" It makes me wonder if there is a rule about this.", + "A": "no because it must have a starting and ending point to be a side but if we flatten the perimete of the circle (for example put a thread around a circle then measure the length ) to be a side then that s what we call a circumference", + "video_name": "8xbIS2UkQxI", + "timestamps": [ + 92 + ], + "3min_transcript": "- [Voiceover] Oh look, it's my cousin Fal! Cousin Fal, what can I do for you? - [Voiceover] Hey Sal, what's up? - [Voiceover] Oh, well not much, I'm just sitting here thinking about what video I should do next. - [Voiceover] Well then maybe you can help me. I'm starting a shape collection. - [Voiceover] Okay, so how do you need help? - [Voiceover] Well look at these shapes over here! I have no way to classify them! - [Voiceover] Okay, well there's a bunch of ways that you could classify them. The first way, you could think about how many sides each of these shapes have, so that's one way to think about it, how many sides they have. - [Voiceover] How do I figure that out? - [Voiceover] Okay, well let's look at each of these and think about how many sides they have. So this first shape right over here this is one side. We see this kind of straight right over here, it's the straight line between these two corners, so that's one side. Two sides, three sides, and four sides. So this shape right over here has four sides. - [Voiceover] Okay, I think I get it. - [Voiceover] Okay,well let's see. This one, well same idea, this one also looks like it has one, two, three, and four. Four sides. - [Voiceover] What about this one? - [Voiceover] Okay, well this one, actually this is interesting. This one actually had no corners, and it has no sides, there's no straight edges over here, this thing is all curved. So since it has no corners, it also has no sides, or at least the way that I'm thinking about it. So I would say that this one over here has zero sides. Let me write that a little bit neater for you. - [Voiceover] Okay yeah, I like to write things really neat. - [Voiceover] Okay, zero sides. And it has zero corners. Actually let me write them for all of them. This one has four sides, and we could also count the corners. We have one, that's where the two sides meet. One, two, three, four corners. it also has four sides, we already counted that and it has one, two, three, four corners. So four sides and four corners actually describes both of these shapes, this one has zero sides and zero corners-- - [Voiceover] What about the green one? - [Voiceover] Okay, well let's just see. One side, two sides, three sides, and four sides, and it also has one, two, three, four corners. So this is also four sides and four corners is also true of that one there. - [Voiceover] Do all shapes have four sides and four corners? - [Voiceover] No no, not at all. I mean this one right over here had zero sides and zero corners, and actually this last shape right over here, doesn't seem to have four sides or four corners. If we count them, we have one side, two sides, and three sides. So this one has three sides, and if we count the corners we have one corner, two corners, and three," + }, + { + "Q": "What does Sal mean when he says \"we can take the scalar out\" at 2:49? What does a scalar mean in this case? Thank you!\n", + "A": "coefficient on the h", + "video_name": "t4GfuftHH5c", + "timestamps": [ + 169 + ], + "3min_transcript": "we're approaching negative one, as we approach from the right, the value of the function seems to be approaching negative one. Now what about h of x? Well, h of x we have down here. As x approaches zero, as x approaches zero, the function is defined at x equals zero, it looks like it is equal to one, and you could, and the limit is also equal to one, we could see that as we approach it from the left, we are approaching one, as we approach from the right, we are approaching one. As we approach x equals zero from the left, we approach, the function approaches one. As we approach x equals zero from the right, the function itself is approaching one. And it makes sense that the function is defined there, and is defined at x equals zero, and the limit as x approaches zero is equal to the same as, is equal to the value of the function at that point, because this is a continuous function. So this is, this is one, and so negative one times one is going to be equal to, is equal to, negative one. Let's do a few more of these. So we have the limit of negative two times f of x plus three times h of x, as x approaches negative three. Well, once again, we can use our limit properties. We know that this is the same thing as the limit, as x approaches negative three of negative two f of x, plus the limit as x approaches negative three of three times h of x. And this is the same thing as the limit, or I should say this is equal to negative two times the limit as x approaches negative three, the limit of f of x, as x approaches negative three, plus, we can take this scaler out, three times the limit of h of x, as x approaches negative three. And so we just have to figure out what the limit and the limit of h of x as x approaches negative three. So I'll first do f of x, so f of x right over here, limit is x approaches negative three, when we approach negative three from the left-hand side, it seems the value of the function is approaching three, and as we approach it from the right-hand side, it seems like the value of the function is zero. So our left-handed and right-handed limits are approaching different things, so this limit actually does not exist. Does not exist. This limit actually here does exist. But since this limit doesn't exist, and we need to figure out this limit in order to figure out this entire limit, this whole thing does not exist. Does not, does not exist. The limits that it's made up of need to exist, this is a combined limit, so each of the pieces need to exist in order for the, in this case, the scaled up sum, to actually exist." + }, + { + "Q": "Am I the only one who was blown away at how quickly Sal calculated the height of the container in his head at 12:22 when he said, \"5 divided by 1.65 squared ... is going to be roughly a little under 2 meters tall\" ? (actual answer is h=1.84 meters). Once again, Sal is AMAZING !\n", + "A": "It was kind of easy to estimate that. 5/(1.65)^2 is approximately 5/(2)^2. 5/(2)^2=5/4=1.25 Since we made the denominator bigger, our answer is smaller than the actual, so it was probably more around (1.75, 2). This wasn t meant to insult or downgrade Sal s awesomeness, though. smile", + "video_name": "tSMuKcN-RKM", + "timestamps": [ + 742 + ], + "3min_transcript": "is exactly this right over here. So when x is equal to 1.65, this is going to be positive. This is going to be positive. So let me write this down. c prime prime of 1.65 is definitely greater than 0. So we're definitely concave upwards when x is 1.65. Concave upwards, which means that our graph is going to look something like this. And so where the derivative equal to 0, which is right over there, we are at a minimum point. We are minimizing our cost. And so if we go back to the question, the only thing that we have to do now-- We know the x value that minimizes our cost. We now have to find the cost of the material for the cheapest So we just have to figure out what our cost is. And we already know what our cost is as a function of x, so we just have to put 1.65 into this equation. Evaluate the function at 1.65. So let's do that. Our cost is going to be equal to 20 times 1.65. because I'm using an approximation of this original value. 1.65 squared plus 180. I could say divided by 1.65. That's the same thing as multiplying by 1.65 to the negative 1. So divided by 1.65, which is equal to 163. I'll just say $163.5. So it's approximately. So the cost-- let me do this in a new color. We deserve a drum roll now. The cost when x is 1.65 is approximately equal to $163.54. So $163.54, which is quite an expensive box. This is fairly expensive material here. Although it's a fairly large box. 1.65 meters in width, and it's going to be twice that in length. And then you could figure out what its height is going to be. Although it's not going to be too tall. 5 divided by 1.65 squared. I don't know, it'll be roughly a little under two meters tall. So it actually is quite a large box made out of quite expensive material. The minimum cost to make this box is going to be $163.54." + }, + { + "Q": "At 3:47 can someone explain why he set the radian measurement sin(theta+pi/2) as the y coordinate?\n", + "A": "He s comparing cos(theta) with sin(theta+pi/2). For example, what s sin(pi/6)? What s cos(pi/6+pi/2)?", + "video_name": "h-TPSylHrvE", + "timestamps": [ + 227 + ], + "3min_transcript": "the opposite side, and what's the hypotenuse? This is a unit circle, so it's going to be one. In this case, sine of theta is equal to the length of the opposite side. The length of the opposite side is equal to sine theta. And same exact logic. The cosine of theta is equal to adjacent over hypotenuse, is equal to adjacent over hypotenuse. And so that's... since the hypotenuse is equal to one, it's just the length of the adjacent side, so cosine of theta is the length of the adjacent side. So this is all a little bit of review, just showing how the unit circle definition is an extension of the Soh Cah Toa definition. But now let's do something interesting. This is the angle theta. Let's think about the angle theta plus pi over two. So the angle theta plus pi over two. So if I were to essentially add pi over two to this, I'm going to get a ray that is perpendicular to the first ray, pi over two. If we think in degrees, pi over two radians, I'm talking in radians. Pi over two radians is equivalent to 90 degrees. So we're essentially adding 90 degrees to it. So this angle right over here, that angle right over here is theta plus pi over two. Now, what I want to explore in this video, and I guess this is the interesting part of the video, is can we relate sin of theta plus pi over two to somehow sine of theta or cosine of theta? I encourage you to pause this video and try to think this through on your own before I work it out. Well let's think about what sine of theta plus pi over two is. We know from the unit circle definition, the sine of this angle, which is theta plus pi over two, is the Y coordinate. It's that, it's this value right over here. Or another way of thinking about it, it's the length of this line in magenta. This right over here is the sine of theta, So that right over there. Now how does that relate to what we have over here? Well when you look at it, it looks like we just took this triangle, and we just kind of... we rotated it. We rotated it counter clockwise by 90 degrees, which essentially what we did do. Because we took this terminal side, and we added 90 degrees to it, or pi over two radians. a little bit more rigorous about it, if this whole white angle here is theta plus pi over two, and the part that's in the first quadrant is pi over two, then this part right over here, that must be equal to theta. And if we think about it, if we try to relate the side this side that I've put in magenta relative to this angle theta using the Soh Cah Toa definition, here, relative to this angle theta in yellow, this is the adjacent side. So let's think about it a little bit. So if we were.. so what deals with the adjacent and the hypotenuse," + }, + { + "Q": "does anyone else sees a realistic calculator at 3:43 to 4:27?\n", + "A": "No I think that you are the only one.", + "video_name": "ZElOxG7_m3c", + "timestamps": [ + 223, + 267 + ], + "3min_transcript": "If they gave us another angle right over here, that's not the angle that we would use. We care about the angle that opens up into the side that we are going to solve for. So now let's solve for a, because we know what bc and theta actually are. So a squared is going to be equal to b squared... so it's going to be equal to 144, plus c squared which is 81, so plus 81, minus two times b times c. So, it's minus two, I'll just write it out. Minus two times 12 times nine, times the cosine of 87 degrees. And this is going to be equal to, let's see, this is 225 minus, let's see, 12 times nine is 108. Minus 216 times the cosine of 87 degrees. Now, let's get our calculator out in order to approximate this. And remember, this is a squared. Actually, before I get my calculator out, let's just solve for a. So a is just going to be the square root of this. So a is going to be equal to the square root... of all of this business, which I can just copy and paste. It's going to be equal to the square root of that. So let me copy and paste it. So a is going to be equal to the square root of that, which we can now use the calculator to figure out. Let me increase this radical a little bit, so that we make sure we're taking the square root of this whole thing. So let me get my calculator out. So I want to find that square root of 220. Actually, before I do that, let me just make sure I'm in degree mode, and I am in degree mode. Because we're evaluating a trig function in degrees here. So that's fine, so let me exit. So it's going to be 225 minus 216, of 87 degrees. Not 88 degrees, 87 degrees. And we deserve a drumroll now. This is going to be 14.61, or 14.618. If, say, we wanted to round to the nearest tenth, just to get an approximation, it would be approximately 14.6. So a is approximately equal to 14.6, whatever units we're using long." + }, + { + "Q": "\nstarting at 0:17 seconds within the video....\n\nThere are situations where two sides are known but there are no labels for which side is side a or b or c. Should I always assume the largest-looking side is side c?", + "A": "The side that says a = ? is the A side, the side that says b = 12 is the B side, and the side that says c = 9 is the C side. If you are trying to solve a side on a triangle that is not labeled for use of the law of cosines, then A is automatically the side you are trying to solve, and B and C are the other sides. It doesn t matter which side is B and which side is C, as they are interchangeable. I hope this helps!", + "video_name": "ZElOxG7_m3c", + "timestamps": [ + 17 + ], + "3min_transcript": "- [Voiceover] Let's say that I've got a triangle, and this side has length b, which is equal to 12, 12 units or whatever units of measurement we're using. Let's say that this side right over here, this side right over here, has length c, and that happens to be equal to nine. And that we want to figure out the length of this side, and this side has length a, so we need to figure out what a is going to be equal to. Now, we won't be able to figure this out unless we also know the angle here, because you could bring the blue side and the green side close together, and then a would be small, but if this angle was larger than a would be larger. So we need to know what this angle is as well. So let's say that we know that this angle, which we will call theta, is equal to 87 degrees. So how can we figure out a? Well, lucky for us, we have the Law of Cosines, which gives us a way for determining a third side if we know two of the sides and the angle between them. The Law of Cosines tells us that a squared is going to be equal b squared plus c squared. Now, if we were dealing with a pure right triangle, if this was 90 degrees, then a would be the hypotenuse, and we would be done, this would be the Pythagorean Theorem. But the Law of Cosines gives us an adjustment to the Pythagorean Theorem, so that we can do this for any arbitrary angle. So Law of Cosines tell us a squared is going to be b squared plus c squared, minus two times bc, times the cosine of theta. And this theta is the angle that opens up to the side that we care about. If they gave us another angle right over here, that's not the angle that we would use. We care about the angle that opens up into the side that we are going to solve for. So now let's solve for a, because we know what bc and theta actually are. So a squared is going to be equal to b squared... so it's going to be equal to 144, plus c squared which is 81, so plus 81, minus two times b times c. So, it's minus two, I'll just write it out. Minus two times 12 times nine, times the cosine of 87 degrees. And this is going to be equal to, let's see, this is 225 minus, let's see, 12 times nine is 108." + }, + { + "Q": "At 7:25, he says \"X sub i-1.\" What does \"sub\" mean?\n", + "A": "Sub just means that it is in the subscript, the smaller text following the bottom half of the X.", + "video_name": "ViqrHGae7FA", + "timestamps": [ + 445 + ], + "3min_transcript": "the conventions in order to actually calculate the area, or our approximation of the area. So our approximation, approximate area, is going to be equal to what? Well, it's going to be the area of the first rectangle-- so let me write this down. So it's going to be rectangle one-- so the area of rectangle one-- so rectangle one plus the area of rectangle two plus the area of rectangle three-- I think you get the point here-- plus all the way to the area of rectangle n. And so what are these going to be? Rectangle one is going to be its height, which is f of x0 or f of a. Either way. x0 and a are the same thing. So it's f of a times our delta x, times our width, our height times our width. So times delta-- actually I can write as f of x0, What is our height of rectangle two? It's f of x1 times delta x. What's our area of rectangle three? It's f of x2 times delta x. And then we go all the way to our area. We're taking all the sums, all the way to rectangle n. What's its area? It's f of x sub n minus 1. Actually, that's a different shade of orange. I'll use that same shade. It is f of x sub n minus 1 times delta x. We've written it in a very general way. But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating the areas or sums in general, I'm going to use the traditional sigma notation. So another way we could write this, as the sum, this is just based on the conventions that I set up. I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle. So the first rectangle, that's rectangle one. So it's going to be f of-- well, if we're in the i-th rectangle, then the left boundary is going to be x sub i minus 1 times delta x. And so here, right over here, is a general way of thinking about approximating the area under a curve using rectangles, where the height of the rectangles are defined by the left boundary. And this tells us it's the left boundary. And we see for each, if this is the i-th rectangle right over here, if this is rectangle i, then this right over here is x sub i minus 1, and this height right over here" + }, + { + "Q": "\nIn the Video at mark 2:36 I do not understand how it is true when f(x) when x=0 is equal to 0. wouldn't f(x) be equal to 0?", + "A": "No, he does not say that f(x) is equal to 0, when x=0. The question was, what is the limit as x--->0- (coming from the left of 0), which is 1. This is why he says it is true. lim x--->0- f(x) = 1 => True.", + "video_name": "_WOr9-_HbAM", + "timestamps": [ + 156 + ], + "3min_transcript": "This would be true if instead of saying from the positive direction, we said from the negative direction. From the negative direction, the value of the function really does look like it is approaching 0. For approaching 1 from the negative direction, when x is right over here, this is f of x. When x is right over here, this is f of x. When x is right over here, this is f of x. And we see that the value of f of x seems to get closer and closer to 0. So this would only be true if they were approaching from the negative direction. Next question. Limit of f of x, as x approaches 0 from the negative direction, is the same as limit of f of x as x approaches 0 from the positive direction. Is this statement true? Well, let's look. Our function, f of x, as we approach 0 from the negative direction-- I'm using a new color-- as we approach 0 from the negative direction, so right over here, this is our value of f of x. Then as we get closer, this is our value of f of x. As we get even closer, this is our value of f of x. like it is approaching positive 1. From the positive direction, when x is greater than 0, let's try it out. So if, say, x is 1/2, this is our f of x. If x is, let's say, 1/4, this is our f of x. If x is just barely larger than 0, this is our f of x. So it also seems to be approaching f of x is equal to 1. So this looks true. They both seem to be approaching the limit of 1. The limit here is 1. So this is absolutely true. Now let's look at this statement. The limit of f of x, as x approaches 0 from the negative direction, is equal to 1. Well, we've already thought about that. The limit of f of x, as x approaches 0 from the negative direction, we see that we're getting closer and closer to 1. As x gets closer and closer to 0, f of x gets closer and closer to 1. So this is also true. Well, it definitely exists. We've already established that it's equal to 1. So that's true. Now the limit of f of x as x approaches 1 exists, is that true? Well, we already saw that when we were approaching 1 from the positive direction, the limit seems to be approaching 1. We get when x is 1 and 1/2, f of x is 1. When x is a little bit more than 1, it's 1. So it seems like we're getting closer and closer to 1. So let me write that down. The limit of f of x, as x approaches 1 from the positive direction, is equal to 1. And now what's the limit of f of x as x approaches 1 from the negative direction? Well, here, this is our f of x. Here, this is our f of x. It seems like our f of x is getting closer and closer to 0, when we approach 1 from values less than 1. So over here it equals 0. So if the limit from the right-hand side is a different value than the limit from the left-hand side," + }, + { + "Q": "6:45 Sal's referring to HL (Hypotenuse-Leg) postulate right?\n", + "A": "He doesn t mention that at 6:45", + "video_name": "f8svAm237xM", + "timestamps": [ + 405 + ], + "3min_transcript": "So this side could pivot over here. We can kind of rotate it over there. But there's only one way, now, that this orange side can reach this green side. Now the only way is this way over here. And we were more constrained, or this case isn't ambiguous, because we used up our obtuse angle here. The A here is an obtuse one. And so then it constrains what the triangle can become. So I don't want to make you say, in general, SSA, you do not want to use it as a postulate. I just wanted to make it clear that there is the special case where if you know that the A in the SSA is obtuse, then it becomes a little bit less ambiguous. And then finally, there's a circumstance that this is an acute angle where it would be ambiguous. You have the obtuse angle, and then you have something in between, which is the right angle. So where you have the A in SSA is a right angle. So if you had it like this. If you have a right angle and you have some base of unknown length but you fix this length right saying it's congruent to some other triangle, and if you know that the next length is fixed-- and if you think about it, this next side is going to be the side opposite the right angle. It's going to have to be the hypotenuse of the right angle. Then you know that the only way you can construct this, and similar to the obtuse case, and if you know the length of this, the only way you could do it is to bring it down over here. So that actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. And here, they wrote the angle first. You could view this as angle-side-side. And they were able to do it because now they can write \"right angle,\" and so it doesn't form that embarrassing acronym. And this would also be a little bit common sense. Because if you know two sides of a right triangle-- and we haven't gone into depth on this in the geometry by Pythagorean theorem, you can always figure out the third side. So if you have this information about any triangle, you can always figure out the third side. And then you can use side-side-side. So I just wanted to show you this little special case. But in general, the important thing is that you can't just use SSA unless you have more information." + }, + { + "Q": "\nAt 0:45, Mr. Sal has gotten me confused.\nHe says \" twenty per hundred.\"\nTwenty WHAT?", + "A": "What he means when he said that: -cent means hundred . He was explaining what percent means per and cent = per and hundred. Get what I m saying? what s he s basically saying is twenty per-cent. Hope this cleared it up for you!", + "video_name": "Lvr2YsxG10o", + "timestamps": [ + 45 + ], + "3min_transcript": "We're asked to shade 20% of the square below. Before doing that, let's just even think about what percent means. Let me just rewrite it. 20% is equal to-- I'm just writing it out as a word-- 20 percent, which literally means 20 per cent. And if you're familiar with the word century, you might already know that cent comes from the Latin for the word hundred. This literally means you can take cent, and that literally means 100. So this is the same thing as 20 per 100. If you want to shade 20%, that means, if you break up the square into 100 pieces, we want to shade 20 of them. 20 per 100. So how many squares have they drawn here? four, five, six, seven, eight, nine, ten squares. If we go vertically, we have one, two, three, four, five, six, seven, eight, nine, ten. So this is a 10 by 10 square. So it has 100 squares here. Another way to say it is that this larger square-- I guess that's the square that they're talking about. This larger square is a broken up into 100 smaller squares, so it's already broken up into the 100. So if we want to shade 20% of that, we need to shade 20 of every 100 squares that it is broken into. So with this, we'll just literally shade in 20 squares. So let me just do one. So if I just do one square, just like that, I have just shaded 1 per 100 of the squares. 100 out of 100 would be the whole. I've shaded one of them. That one square by itself would be 1% If I were to shade another one, if I were to shade that and that, then those two combined, that's 2% of the It's literally 2 per 100, where 100 would be the entire square. If we wanted to do 20, we do one, two, three, four-- if we shade this entire row, that will be 10%, right? One, two, three, four, five, six, seven, eight, nine, ten. And we want to do 20, so that'll be one more row. So I can shade in this whole other row right here. And then I would have shaded in 20 of the 100 squares. Or another way of thinking about it, if you take this larger square, divide it into 100 equal pieces, I've shaded in 20 per 100, or 20%, of the entire larger square. Hopefully, that makes sense." + }, + { + "Q": "\nCouldn't you have substituted 180 = pi into -pi/3 at 2:39?", + "A": "Yes. And that is essentially what Sal did. He multiplied by 180 and divided by pi; thus, he cancelled out the pi, and put in a 180, so he basically just substituted it. But your way is a nice shortcut; so if you can do it that way in your head, feel free to.", + "video_name": "z0-1gBy1ykE", + "timestamps": [ + 159 + ], + "3min_transcript": "and on the righthand side here, 360 divided by two is 180. And we have still the units which are degrees. So we get pi radians are equal to 180 degrees. Which actually answers the first part of our question. We wanted to convert pi radians, well we just figured out! Pi radians are equal to 180 degrees. If you want to think about it, pi radians are halfway around the circle Halfway around the circle like that, and it is the same thing as 180 degrees. So now lets think about the second part. We want to convertnegative pi over three radians. --Switch to a new color-- so negative pi over three, so how do we convert that? Well, to figure this out we need to know how many degrees there are per radian. We need to multiply this by degrees -- I'm going to write the word out instead of the circle here -- It would be really hard to visualize that, degrees per radian So how many degrees are there per radian? well we know that for 180 degrees we have pi radians. Or you can say there are 180 over pi degrees per radian. This is going to work out: We have however manyradians we have times the number of degrees per radian. So of course the units are going to work out. The radians cancel out, the pi also cancels. And you are left with 180 divided by 3, leaving us with Negative 60, and we don't want to forget the units We could write them out, the only unit left is degrees. WE could write out the word degrees or just put that symbol there." + }, + { + "Q": "At 3:09, why is the answer -60? I thought degree measurements could not be negative, like length measurements.\n", + "A": "Angle measurement can be negative. Starting on the positive x-axis, it is negative when you draw it clockwise, and positive when you draw it counter-clockwise.", + "video_name": "z0-1gBy1ykE", + "timestamps": [ + 189 + ], + "3min_transcript": "and on the righthand side here, 360 divided by two is 180. And we have still the units which are degrees. So we get pi radians are equal to 180 degrees. Which actually answers the first part of our question. We wanted to convert pi radians, well we just figured out! Pi radians are equal to 180 degrees. If you want to think about it, pi radians are halfway around the circle Halfway around the circle like that, and it is the same thing as 180 degrees. So now lets think about the second part. We want to convertnegative pi over three radians. --Switch to a new color-- so negative pi over three, so how do we convert that? Well, to figure this out we need to know how many degrees there are per radian. We need to multiply this by degrees -- I'm going to write the word out instead of the circle here -- It would be really hard to visualize that, degrees per radian So how many degrees are there per radian? well we know that for 180 degrees we have pi radians. Or you can say there are 180 over pi degrees per radian. This is going to work out: We have however manyradians we have times the number of degrees per radian. So of course the units are going to work out. The radians cancel out, the pi also cancels. And you are left with 180 divided by 3, leaving us with Negative 60, and we don't want to forget the units We could write them out, the only unit left is degrees. WE could write out the word degrees or just put that symbol there." + }, + { + "Q": "\nFor the formula [Sn=a/(1-r)], the beginning value doesn't have to necessarily be k=1, for example at 5:02? Does that imply that any value for \"k\" can be used, such as starting at 2 or 1000?", + "A": "That formula only works for k=0 to infinity.", + "video_name": "2BgWWsypzLA", + "timestamps": [ + 302 + ], + "3min_transcript": "And that gives us that value right over there. Plus 0.4008 times 10 to the negative 4 to the second power. And we keep on going. And so in this form, it looks a little bit clearer, like a geometric series, an infinite geometric series. And if we wanted to write that out with sigma notation, we could write this as the sum from k equals 0 to infinity of, well, what's our first term going to be? It's going to be 0.4008 times our common ratio, which we could write out as either 10 to the negative fourth or 0.0001. I'll just write it as 10 to the negative fourth. 10 to the negative fourth to the k-th power, to the k-th power. this clearly can be represented as a geometric series-- is, well, what is the sum? You might say, well, that's just going to be 4008 repeating over and over. But I want to express it as a fraction. And so I want you to pause the video. Use what would you already know about finding the sum of an infinite geometric series to try to express this thing right over here as a fraction. So I'm assuming you've had a go at it. So let's think about it. We've already seen, we've already derived in previous videos, that the sum of an infinite geometric series-- let me do this in a neutral color. If I have a series like this, k equals 0 to infinity of ar to the k power, that this sum is going to be equal to a over 1 minus r. We've derived this actually in several other videos. So in this case, this is going to be-- well, our a here is 0.4008. And it's going to be that over 1 minus our common ratio, minus-- So what's this going to be? Well, this is going to be the same thing as 0.4008. If you take 1 minus 1 ten thousandths, or you could do this as 10,000 ten thousandths minus 1 ten thousandth, you're going to have 9,999 ten thousandths. Once again, you could view-- let me write this out just so this doesn't look confusing. 1 is the same thing as 10,000/10,000. And you're subtracting 1/10,000. And so you're going to get 9,999/10,000. And so this is going to be the same thing" + }, + { + "Q": "\nAt 6:05, Sal said that the equation of a circle is x^2 + y^2 = r^2, but I remember in the previous lessons that the standard form is (x \u00e2\u0080\u0093 a)^2 + (y \u00e2\u0080\u0093 b)^2 = r^2. Can someone please clarify the differences?", + "A": "a and b (a,b) are used to determine the center of the circle. In the equation x^2 + y^2 = r^2, we have a and b equal to zero, so the center of the circle would be (0,0). If the equation of the circle were to be (x-3)^2 + (y-4)^2 = r^2, the center of the circle would be (3,4). So basically, the 2 equations that Sal gave mean the same thing, but the a and b are used to define the center of the circle. Comment if you have any further questions.", + "video_name": "lvAYFUIEpFI", + "timestamps": [ + 365 + ], + "3min_transcript": "I think that makes more sense. And you can call this the minor radius. So let's just do an example. And I think when I've done an example with actual numbers, it'll make it all a little bit clearer. So let's say I were to show up at your door with the following: If I were to say x squared over 9 plus y squared over 25 is equal to 1. So what is your radius in the x-direction? This is your radius in the x-direction squared. So your radius in the x-direction if we just map it, we would say that a is equal to 3. Because this is a squared. And if we were just map it we'd say this is b squared than this tells us that b is equal to 5. is centered at the origin. Let me draw the ellipse first. So, first of all, we have our radius in the y-direction is larger than our radius in the x-direction. The ellipse is going to be taller and skinnier. It's going to look something like that. Draw some axes, so that could be your x-axis, your y-axis. This is your radius in the y-direction. So this distance right here is going to be 5, and so will this distance. And this is your radius in the x-direction. So this will be 3, and this will be 3. You have now plotted this ellipse. Nothing too fancy about it. And actually just to kind of hit the point home that the circle is a special case of an ellipse. We learned in the last video that the equation of a circle x squared plus y squared is equal to r squared. So if we were to divide both sides of this by r squared, we would get-- and this is just little algebraic manipulation-- x squared over r squared plus y squared over r squared is equal to 1. Now in this case, your a is r and so is your b. So your semi-minor axis is r and so is your semi-major axis of r. Or, in other words, this distance is the same as that distance, and so it will neither be short and fat nor tall and skinny. It'll be perfectly round. And so that's why the circle is a special case of an ellipse. So let me give you a slightly-- It'll look a lot more complicated, and this is something you might see on exam. But I just want to show you that this is just a shifting. Let's say we wanted to shift this ellipse." + }, + { + "Q": "at 9:15 Sal says that. if we add 2 to any function then we shift it down by two. but in reality that is not true.\nwe should subtract 2 in function to shift it down.\n", + "A": "I would usually make the same mistake. You see, what helped me was writing the equation like this, for the standard form: (x-(h))^2/a^2 + (y-(k))^2/b^2.... So if either h or k is negative, it would then be a double negative, which in turn makes it positive, like so: (x-(-h))/a^2 + (y-(-k))/b^2=1 or: (x+h)^2/a^2 + (y+k)^2/b^2. Really hope this helps you like it did me!!", + "video_name": "lvAYFUIEpFI", + "timestamps": [ + 555 + ], + "3min_transcript": "So instead of the origin being at x is equal to 0, the origin will now be at x is equal 5. So a way to think about that is what does this term have to be so that at 5 this term ends up being 0. Well I'll actually draw it for you, because I think that might be confusing. So if we shift that over the right by 5, the new equation of this ellipse will be x minus 5 squared over 9 plus y squared over 25 is equal to 1. So if I were to just draw this ellipse right now, it would look like this. I want to make it look fairly similar to the ellipse I had before. Just shifted it over by five. And the intuition we learned a little bit in the circle video where I said, oh well, you know, if you have x minus something that means that the new origin is now at positive 5. And you could memorize that. You could always say, oh, if I have a minus here, that the origin is at the negative of whatever this number is, so it would be a positive five. You know, if you had a positive it would be the opposite that. But the way to really think about it is now if you go to x is equal to 5, when x is equal to 5, this whole term, x minus 5, will behave just like this x term will here. When x is equal to 5 this term is 0, just like when x was 0 here. So when x is equal to 5, this term is 0, and then y squared over 25 is equal 1, so y has to be equal five. Just like over here when x is equal is 0, y squared over 25 had to be equal to 1, y is equal to either And I really want to give you that intuition. And then, let's say we wanted to shift this equation down by two. So our new ellipse looks something like this. A lot of times you learned this in conic sections. But this is true any function. When you shift things, you shift it this way. If you shift this graph to the right by five, you replace all of the x's with x minus 5. And if you were to shift it down by two, you would replace all the y's with y plus 2. So let me draw our new ellipse first, just to show you what I'm doing. So our new ellipse is going to look something like that. I'm shifting the yellow ellipse down by two." + }, + { + "Q": "So when Sal says at 1:49 that the area is 1/2 b\u00c2\u00b7h it is the answer of b\u00c2\u00b7h cut in half?\n", + "A": "You can think of it as half of a rectangle", + "video_name": "YTRimTJ5nX4", + "timestamps": [ + 109 + ], + "3min_transcript": "Let's say we've got a rectangle and we have two diagonals across the rectangle-- that's one of them, and then we have the other diagonal --and this rectangle has a height of h-- so that distance right there is h --and it has a width of w. What we're going to show in this video is that all of these four triangles have the same area. Now right when you look at it, it might be reasonably obvious that this bottom triangle will have the same area as the top triangle, as this top kind of upside down triangle. That these to have the same area, that might be reasonably obvious. they have the same dimension for their base, this width, and they have the same height because this distance right here is exactly half of the height of the rectangle. They have the same proportions. Now it's probably equally obvious that this triangle on the left has the same area as this triangle on the right. That's probably equally obvious. What is not obvious is that these orange triangles angles have the same area as these green, blue triangles. And that's what we're going to show right here. So all we have to do is really calculate the areas of the different triangles. So let's do the orange triangles first. and before doing that let's just remind ourselves what the area of a triangle is. Area of a triangle is equal to 1/2 times the base of the triangle times the height of the triangle. That's just basic geometry. Not with that said, let's figure out the area of the orange triangle. It's going to be 1/2 times the base. right here: it is w. So 1/2 times w. I want to do that in a different color; the color I wrote the w in. Now what's the height here? Well we already talked about it: it's exactly half way up the height of the rectangle. So times 1/2 times the height of the rectangle. So what's that going to be? You have 1/2 times 1/2 is 1/4 times width times height. So the area of that triangle is 1/4 width height. So is that one. Same exact argument; they have equal area. Now what's the area of these green or these green/blue triangles? Well once again-- we'll do this in a green color --area" + }, + { + "Q": "At 2:40, you begin to explain that you want to change dy / dx = 1 / cos y into something in terms of x by using the Pythagorean Identity and setting that (the Pythagorean Identity) equal to cos y . Could you just have inserted your original definition of y (y = sin^-1 x) instead? You would end up with:\n\ndy / dx = 1 / (cos (sin^-1 x)).\n\nDoes that work ? tried it with a few arbitrary numbers for x and was getting the same answer for both equations, but wasn't sure if it technically was correct.\n", + "A": "Yes. If you want to visualize, try to type cos(arcsin(x)) and sqrt(1-x^2) into Google or is cos(arcsin(x)) equal to sqrt(1-x^2) into WolframAlpha.", + "video_name": "yIQUhXa-n-M", + "timestamps": [ + 160 + ], + "3min_transcript": "So now we have things that we're a little bit more familiar with, and now we can do a little bit of implicit differentiation. We can take the derivative of both sides with respect to X. So, derivative of the left-hand side with respect to X and the derivative of the right-hand side with respect to X. But what's the derivative of the left-hand side with respect to X going to be? And here we just apply the chain rule. It's going to be the derivative of sine of Y with respect to Y. Which is going to be, which is going to be cosine of Y times the derivative of Y with respect to X. So times dy, dx, times dy,dx. And the right-hand side, what's the derivate of X with respect to X, well that's obviously just going to be equal to one. And so we could solve for dy,dx, divide both sides by cosine of Y. to X is equal to one over cosine of Y. Now this still isn't that satisfying cuz I have the derivative in terms of Y. So let's see if we can re-express it in terms of X. So, how could we do that? Well, we already know that X is equal to sine of Y. We already know that X is equal to sine of Y. So, if we could rewrite this bottom expression in terms, instead of cosine of Y. If we could use our trigonometric identities to rewrite it in terms of sine of Y, then we'll be in good shape because X is equal to sine of Y. Well, how can we do that? Well, we know from our trigonometric identities, we know that sine squared of Y plus cosine squared of Y is equal to one. Or, if we want to solve for cosine of Y, subtract sine squared of Y from both sides. We know that cosine squared of Y is equal to one minus sine squared of Y. is equal to the principal root of one minus sine squared of Y. So, we could rewrite this as being equal to one over, one over, instead of cosine of Y, we could rewrite it as one minus sine squared of Y. Now why is this useful? Well, sine of Y is just X. So this is the same if we just substitute back in, let me just write it that way so it's a little bit clear. I could write it as sine Y squared. We know that this thing right over here is X. So this is going to be equal to, and we deserve a little bit of a drumroll. One over the square root of one minus, instead of sin of Y, we know that X is equal to sin of Y. So, one minus X squared. And so, there you have it. The derivative with respect to X of the inverse sine of X is equal to one over" + }, + { + "Q": "I'm confused...\n\n1) In the last 2 videos \"Proof: product of rational & irrational is irrational\" and \"Proof: sum of rational and irrational is irrational\" state in the title that the sum or product of rational and irrational is irrational.\n\n2) BUT, @5:31 in this video Sal says, \"You don't know whether the product is going to be rational or irrational unless someone tells you the specific numbers\".\n\nSo which is correct, 1 or 2?\n", + "A": "There is no conflict. The last 2 videos were about sum/products of a rational and an irrational number. At 5:31, Sal is talking about the product of 2 irrational numbers. So, this is a whole different scenario. Hope this helps.", + "video_name": "16-GZWi66CI", + "timestamps": [ + 331 + ], + "3min_transcript": "and people don't tell you anything else, they don't tell you which specific irrational numbers they are, you don't know whether their sum is going to be rational or irrational. Now let's think about products. Similar exercise, let's say we have a times b is equal to c, ab is equal to c, a times b is equal to c. And once again, let's say someone tells you that both a and b are irrational. Pause this video and think about whether c must be rational, irrational, or whether we just don't know. Try to figure out some examples like we just did when we looked at sums. Alright, so let's think about, let's see if we can construct examples where c ends up being rational. Well one thing, as you can tell I like to use pi, pi might be my favorite irrational number. If a was one over pi well, what's their product going to be? Well, their product is going to be one over pi times pi, that's just going to be pi over pi, which is equal to one. Here we got a situation where the product of two irrationals became, or is, rational. But what if I were to multiply, and in general you could this with a lot of irrational numbers, one over square root of two times the square of two, that would be one. What if instead I had pi times pi? Pi times pi, that you could just write as pi squared, and pi squared is still going to be irrational. This is irrational, irrational. It isn't even always the case that if you multiply the same irrational number, that it's always going to be irrational. For example, if I have square root of two times, I think you see where this is going, times the square root of two, I'm taking the product of two irrational numbers. In fact, they're the same irrational number, but the square root of two times the square root of two, well, that's just going to be equal to two, which is clearly a rational number. So once again, when you're taking the product of two irrational numbers, you don't know whether the product is going to be rational or irrational unless someone tells you the specific numbers. Whether you're taking the product or the sum of irrational numbers, in order to know whether the resulting number is irrational or rational, you need to know something about what you're taking the sum or the product of." + }, + { + "Q": "\nI don' understand what happened at 6:57 - 7:42. I tried to follow by reading what you said, and I couldn't follow", + "A": "He said: -2+2*1=0, 4+2(-2)=0, and -10+2(5)=0. This came from the row operation: 2R2+R3 replace R3.", + "video_name": "L0CmbneYETs", + "timestamps": [ + 417, + 462 + ], + "3min_transcript": "I'm going to replace this row with that. 2 minus 2 times 1 is 0. That was the whole point. 4 minus 2 times 2 is 0. 0 minus 2 times 1 is minus 2. 6 minus 2 times 1 is 6 minus 2, which is 4. 4 minus 2 times 7, is 4 minus 14, which is minus 10. Now what can I do next. You can kind of see that this row, well talk more about what this row means. When all of a sudden it's all been zeroed out, there's nothing here. If I had non-zero term here, then I'd want to zero this guy out, although it's already zeroed out. I'm just going to move over to this row. The first thing I want to do is I want to make this leading coefficient here a 1. row by minus 1. If I multiply this entire row times minus 1. I don't even have to rewrite the matrix. This becomes plus 1, minus 2, plus 5. I think you can accept that. Now what can we do? Well, let's turn this right here into a 0. Let me rewrite my augmented matrix in the new form that I have. I'm going to keep the middle row the same this time. My middle row is 0, 0, 1, minus 2, and then it's augmented, and I get a 5 there. What I want to do is I want to eliminate this minus 2 here. Why don't I add this row to 2 times that row. Then I would have minus 2, plus 2, and that'll work out. What do I get. Well, these are just leading 0's. That's just 0. 4 plus 2 times minus 2, that is minus 4. That's 4 plus minus 4, that's 0 as well. Then you have minus 10 plus 2 times 5. Well, that's just minus 10 plus 10, which is 0. That one just got zeroed out. Normally, when I just did regular elimination, I was happy just having the situation where I had these Everything below it were 0's. I wasn't too concerned about what was above our 1's. What I want to do is, I want to make those into a 0 as well. I want to make this guy a 0 as well. What I can do is, I can replace this first row with that first row minus this second row." + }, + { + "Q": "\nAt 4:21, why is Khan allowed to subtract the second row from the first row and put it in the second row? I thought he could only do that vice versa: subtracting first row from the second row and making that the new second row.", + "A": "I agree with you and I do it the way you thought is correct, because it s less confusing for me; however: Both operations are equivalent, because if you multiply both sides of row 1 - row 2 by -1, you get row 2 - row 1 (try it), and these are just scalar multiples of each other. a-c = b-d <=> c-a = d-b", + "video_name": "L0CmbneYETs", + "timestamps": [ + 261 + ], + "3min_transcript": "coefficients on the x1 terms. We know that these are the coefficients on the x2 terms. And what this does, it really just saves us from having to write x1 and x2 every time. We can essentially do the same operations on this that we otherwise would have done on that. What we can do is, we can replace any equation with that equation times some scalar multiple, plus another equation. We can divide an equation, or multiply an equation by a scalar. We can subtract them from each other. We can swap them. Let's do that in an attempt to solve this equation. The first thing I want to do, just like I've done in the past, I want to get this equation into the form of, where if I can, I have a 1. My leading coefficient in any of my rows is a 1. And that every other entry in that column is a 0. In the past, I made sure that every other entry below it is a 0. That's what I was doing in some of the previous videos, when we tried to figure out of things were linearly Now I'm going to make sure that if there is a 1, if there is a leading 1 in any of my rows, that everything else in that column is a 0. That form I'm doing is called reduced row echelon form. Let me write that. Reduced row echelon form. If we call this augmented matrix, matrix A, then I want to get it into the reduced row echelon form of matrix A. And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters, instead of lowercase letters. We'll talk more about how matrices relate to vectors in the future. Let's just solve this system of equations. The first thing I want to do is, in an ideal world I would get all of these guys right here to be 0. Let me replace this guy with that guy, with the first entry Let me do that. The first row isn't going to change. It's going to be 1, 2, 1, 1. And then I get a 7 right there. That's my first row. Now the second row, I'm going to replace it with the first row minus the second row. So what do I get. 1 minus 1 is 0. 2 minus 2 is 0. 1 minus 2 is minus 1. And then 1 minus minus 1 is 2. That's 1 plus 1. And then 7 minus 12 is minus 5. Now I want to get rid of this row here. I don't want to get rid of it. I want to get rid of this 2 right here. I want to turn it into a 0. Let's replace this row with this row minus 2 times that row. What I'm going to do is, this row minus 2" + }, + { + "Q": "Where did Sal get x2 = 0 1 0 and x4= 0 0 1 at 14:30?\n", + "A": "I m still a little confused about this also. As far as I can tell the solution is correct with or without x2 and x4 components.", + "video_name": "L0CmbneYETs", + "timestamps": [ + 870 + ], + "3min_transcript": "I just subtracted these from both sides of the equation. This right here is essentially as far as we can go to the solution of this system of equations. I can pick, really, any values for my free variables. I can pick any values for my x2's and my x4's and I can solve for x3. What I want to do right now is write this in a slightly different form so we can visualize a little bit better. Of course, it's always hard to visualize things in four dimensions. So we can visualize things a little bit better, as to the set of this solution. Let's write it this way. If I were to write it in vector form, our solution is the vector x1, x3, x3, x4. Well it's equal to-- let me write it like this. It's equal to-- I'm just rewriting, I'm just essentially rewriting this solution set in vector form. So x1 is equal to 2-- let me write a little column there-- plus x2. Let me write it this way. Plus x2 times something plus x4 times something. x1 is equal to 2 minus 2 times x2, or plus x2 minus 2. I put a minus 2 there. I can say plus x4 times minus 3. I can put a minus 3 there. This right here, the first entries of these vectors literally represent that equation right there. x1 is equal to 2 plus x2 times minus 2 plus x4 times minus 3. x3 is equal to 5. Put that 5 right there. Plus x4 times 2. x2 doesn't apply to it. We can just put a 0. 0 times x2 plus 2 times x4. Now what does x2 equal? You could say, x2 is equal to 0 plus 1 times x2 plus 0 times x4. x2 is just equal to x2. It's a free variable. Similarly, what does x4 equal to? x4 is equal to 0 plus 0 times x2 plus 1 times x4. What does this do for us? Well, all of a sudden here, we've expressed our solution set as essentially the linear combination of the linear combination of three vectors. This is a vector." + }, + { + "Q": "At 2:58, Sal mentions \"previous videos where I was trying to figure out whether things were linearly independent or not.\" Where are these videos?\n", + "A": "Videos about linear independence within, I suspect, the linear algebra subjects.", + "video_name": "L0CmbneYETs", + "timestamps": [ + 178 + ], + "3min_transcript": "The coefficient there is 1. The coefficient there is 1. The coefficient there is 2. You have 2, 2, 4. 2, 2, 4. 1, 2, 0. 1, 2, there is no coefficient the x3 term here, because there is no x3 term there. We'll say the coefficient on the x3 term there is 0. And then we have 1, minus 1, and 6. Now if I just did this right there, that would be the coefficient matrix for this system of equations right there. What I want to do is I want to augment it, I want to augment it with what these equations need to be equal to. Let me augment it. What I am going to do is I'm going to just draw a little line here, and write the 7, the 12, and the 4. I think you can see that this is just another way of writing this. coefficients on the x1 terms. We know that these are the coefficients on the x2 terms. And what this does, it really just saves us from having to write x1 and x2 every time. We can essentially do the same operations on this that we otherwise would have done on that. What we can do is, we can replace any equation with that equation times some scalar multiple, plus another equation. We can divide an equation, or multiply an equation by a scalar. We can subtract them from each other. We can swap them. Let's do that in an attempt to solve this equation. The first thing I want to do, just like I've done in the past, I want to get this equation into the form of, where if I can, I have a 1. My leading coefficient in any of my rows is a 1. And that every other entry in that column is a 0. In the past, I made sure that every other entry below it is a 0. That's what I was doing in some of the previous videos, when we tried to figure out of things were linearly Now I'm going to make sure that if there is a 1, if there is a leading 1 in any of my rows, that everything else in that column is a 0. That form I'm doing is called reduced row echelon form. Let me write that. Reduced row echelon form. If we call this augmented matrix, matrix A, then I want to get it into the reduced row echelon form of matrix A. And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters, instead of lowercase letters. We'll talk more about how matrices relate to vectors in the future. Let's just solve this system of equations. The first thing I want to do is, in an ideal world I would get all of these guys right here to be 0. Let me replace this guy with that guy, with the first entry" + }, + { + "Q": "What determines the row you will use to minus with another row to replace a particular row? for example at 3:00 you replaced row 1 with (1st row - 2nd row)...why didn't you use the (3rd row - 2 times 1st row)? since it will still give you the same answer...?\n", + "A": "You can replace any row with: 1) its multiple 2) its difference from any other row or from any multiple of a row.", + "video_name": "L0CmbneYETs", + "timestamps": [ + 180 + ], + "3min_transcript": "The coefficient there is 1. The coefficient there is 1. The coefficient there is 2. You have 2, 2, 4. 2, 2, 4. 1, 2, 0. 1, 2, there is no coefficient the x3 term here, because there is no x3 term there. We'll say the coefficient on the x3 term there is 0. And then we have 1, minus 1, and 6. Now if I just did this right there, that would be the coefficient matrix for this system of equations right there. What I want to do is I want to augment it, I want to augment it with what these equations need to be equal to. Let me augment it. What I am going to do is I'm going to just draw a little line here, and write the 7, the 12, and the 4. I think you can see that this is just another way of writing this. coefficients on the x1 terms. We know that these are the coefficients on the x2 terms. And what this does, it really just saves us from having to write x1 and x2 every time. We can essentially do the same operations on this that we otherwise would have done on that. What we can do is, we can replace any equation with that equation times some scalar multiple, plus another equation. We can divide an equation, or multiply an equation by a scalar. We can subtract them from each other. We can swap them. Let's do that in an attempt to solve this equation. The first thing I want to do, just like I've done in the past, I want to get this equation into the form of, where if I can, I have a 1. My leading coefficient in any of my rows is a 1. And that every other entry in that column is a 0. In the past, I made sure that every other entry below it is a 0. That's what I was doing in some of the previous videos, when we tried to figure out of things were linearly Now I'm going to make sure that if there is a 1, if there is a leading 1 in any of my rows, that everything else in that column is a 0. That form I'm doing is called reduced row echelon form. Let me write that. Reduced row echelon form. If we call this augmented matrix, matrix A, then I want to get it into the reduced row echelon form of matrix A. And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters, instead of lowercase letters. We'll talk more about how matrices relate to vectors in the future. Let's just solve this system of equations. The first thing I want to do is, in an ideal world I would get all of these guys right here to be 0. Let me replace this guy with that guy, with the first entry" + }, + { + "Q": "\nWhy did Sal say 1 minus minus 1 at 4:37?", + "A": "It s just another way of saying 1 minus negative 1. Or 1 - (-1). He s subtracting row 2 from row 1. The 4th column in row 1 is 1, the 4th column in row to is -1. So 1 - (-1) = 2.", + "video_name": "L0CmbneYETs", + "timestamps": [ + 277 + ], + "3min_transcript": "Now I'm going to make sure that if there is a 1, if there is a leading 1 in any of my rows, that everything else in that column is a 0. That form I'm doing is called reduced row echelon form. Let me write that. Reduced row echelon form. If we call this augmented matrix, matrix A, then I want to get it into the reduced row echelon form of matrix A. And matrices, the convention is, just like vectors, you make them nice and bold, but use capital letters, instead of lowercase letters. We'll talk more about how matrices relate to vectors in the future. Let's just solve this system of equations. The first thing I want to do is, in an ideal world I would get all of these guys right here to be 0. Let me replace this guy with that guy, with the first entry Let me do that. The first row isn't going to change. It's going to be 1, 2, 1, 1. And then I get a 7 right there. That's my first row. Now the second row, I'm going to replace it with the first row minus the second row. So what do I get. 1 minus 1 is 0. 2 minus 2 is 0. 1 minus 2 is minus 1. And then 1 minus minus 1 is 2. That's 1 plus 1. And then 7 minus 12 is minus 5. Now I want to get rid of this row here. I don't want to get rid of it. I want to get rid of this 2 right here. I want to turn it into a 0. Let's replace this row with this row minus 2 times that row. What I'm going to do is, this row minus 2 I'm going to replace this row with that. 2 minus 2 times 1 is 0. That was the whole point. 4 minus 2 times 2 is 0. 0 minus 2 times 1 is minus 2. 6 minus 2 times 1 is 6 minus 2, which is 4. 4 minus 2 times 7, is 4 minus 14, which is minus 10. Now what can I do next. You can kind of see that this row, well talk more about what this row means. When all of a sudden it's all been zeroed out, there's nothing here. If I had non-zero term here, then I'd want to zero this guy out, although it's already zeroed out. I'm just going to move over to this row. The first thing I want to do is I want to make this leading coefficient here a 1." + }, + { + "Q": "At 11:16, can't you solve for the free variables if you switch the order of the variables and make them your pivot variables by changing how you change the matrix?\n", + "A": "If you use the matrix method in this video, you have to swap entire column vectors (but not the last, constant one) with each other. It seems to me that then the equations are the same, and the variables remain in their own columns, and the result should be equivalent, but as you suggest expressed as x2 and x4 in terms of x1 and x3. I put the column vectors in the order x2, x4, x1, x3 , and got: x2 = 19/4 - (x1)/2 - 3(x3)/4, x4 = -5/2 + (x3)/2. What do you get?", + "video_name": "L0CmbneYETs", + "timestamps": [ + 676 + ], + "3min_transcript": "If I have any zeroed out rows, and I do have a zeroed out row, it's right there. This is zeroed out row. Just the style, or just the convention, is that for reduced row echelon form, that has to be your last row. We have the leading entries are the only -- they're all 1. That's one case. You'd want to divide that equation by 5 if this was a 5. So your leading entries in each row are a 1. That the leading entry in each successive row is to the right of the leading entry of the row before it. This guy right here is to the right of that guy. This is just the style, the convention, of reduced row echelon form. If you have any zeroed out rows, it's in the last row. And finally, of course, and I think I've said this multiple times, this is the only non-zero entry in the row. What does this do for me? Now I can go back from this world, back to my linear equations. We remember that these were the coefficients on x1, these were the coefficients on x2. my constants out here. I can rewrite this system of equations using my reduced row echelon form as x1, x1 plus 2x2. There's no x3 there. So plus 3x4 is equal to 2. This equation, no x1, no x2, I have an x3. I have x3 minus 2x4 is equal to 5. I have no other equation here. This one got completely zeroed out. I was able to reduce this system of equations to this system of equations. The variables that you associate with your pivot entries, we call these pivot variables. x1 and x3 are pivot variables. The variables that aren't associated with the pivot x2 and x4 are free variables. Now let's solve for, essentially you can only solve for your pivot variables. The free variables we can set to any variable. I said that in the beginning of this equation. We have fewer equations than unknowns. This is going to be a not well constrained solution. You're not going to have just one point in R4 that solves You're going to have multiple points. Let's solve for our pivot variables, because that's all we can solve for. This equation tells us, right here, it tells us x3, let me do it in a good color, x3 is equal to 5 plus 2x4. Then we get x1 is equal to 2 minus x2, 2 minus 2x2." + }, + { + "Q": "\nAt 2:25, why is the exponent on 7 zero? Wouldn't it be one? I know that everything to the first power is just itself, but why does that x^0 have to go there instead of 7^1?", + "A": "He writes it to show you that there is an x with a degree on the 7, even though we usually don t write it.", + "video_name": "REiDXCN0lGU", + "timestamps": [ + 145 + ], + "3min_transcript": "In the following polynomial, identify the terms along with the coefficient and exponent of each term. So the terms are just the things being added up in this polynomial. So the terms here-- let me write the terms here. The first term is 3x squared. The second term it's being added to negative 8x. You might say, hey wait, isn't it minus 8x? And you could just view that as it's being added to negative 8x. So negative 8x is the second term. And then the third term here is 7. It's called a polynomial. Poly, it has many terms. Or you could view each term as a monomial, as a polynomial with only one term in it. So those are the terms. Now let's think about the coefficients of each of the terms. The coefficient is what's multiplying the power of x or what's multiplying in the x part of the term. So over here, the x part is x squared. That's being multiplied by 3. On the second term, we have negative 8 multiplying x. And we want to be clear, the coefficient isn't just 8. It's a negative 8. It's negative 8 that's multiplying x. So that's the coefficient right over here. And here you might say, hey wait, nothing is multiplying x here. I just have a 7. There is no x. 7 isn't being multiplied by x. But you can think of this as 7 being multiplied by x to the 0 because we know that x to the zeroth power is equal to 1. So we would even call this constant, the 7, this would be the coefficient on 7x to the 0. So you could view this as a coefficient. So this is also a coefficient. So let me make it clear, these three things are coefficients. Now the last part, they want us to identify So the exponent of this first term is 2. It's being raised to the second power. The exponent of the second term, remember, negative 8x, x is the same thing as x to the first power. So the exponent here is 1. And then on this last term, we already said, 7 is the same thing as 7x to the 0. So the exponent here on the constant term on 7 is 0. So these things right over here, those are our exponents. And we are done." + }, + { + "Q": "\nAround the 1:45 mark, it's mentioned that the 7 is a coefficient of the x^0. I've always understood that the lonely number at the end is a constant, not a coefficient. Which is it and why?", + "A": "x^0 = 1 and anything multiplied by 1 is itself. Therefore, it s perfectly logical to say that the constant is the coefficient of x^0 power. It helps in solving certain problems as you go further in your mathematical career. For now, you should just understand the logic behind it and why it makes sense.", + "video_name": "REiDXCN0lGU", + "timestamps": [ + 105 + ], + "3min_transcript": "In the following polynomial, identify the terms along with the coefficient and exponent of each term. So the terms are just the things being added up in this polynomial. So the terms here-- let me write the terms here. The first term is 3x squared. The second term it's being added to negative 8x. You might say, hey wait, isn't it minus 8x? And you could just view that as it's being added to negative 8x. So negative 8x is the second term. And then the third term here is 7. It's called a polynomial. Poly, it has many terms. Or you could view each term as a monomial, as a polynomial with only one term in it. So those are the terms. Now let's think about the coefficients of each of the terms. The coefficient is what's multiplying the power of x or what's multiplying in the x part of the term. So over here, the x part is x squared. That's being multiplied by 3. On the second term, we have negative 8 multiplying x. And we want to be clear, the coefficient isn't just 8. It's a negative 8. It's negative 8 that's multiplying x. So that's the coefficient right over here. And here you might say, hey wait, nothing is multiplying x here. I just have a 7. There is no x. 7 isn't being multiplied by x. But you can think of this as 7 being multiplied by x to the 0 because we know that x to the zeroth power is equal to 1. So we would even call this constant, the 7, this would be the coefficient on 7x to the 0. So you could view this as a coefficient. So this is also a coefficient. So let me make it clear, these three things are coefficients. Now the last part, they want us to identify So the exponent of this first term is 2. It's being raised to the second power. The exponent of the second term, remember, negative 8x, x is the same thing as x to the first power. So the exponent here is 1. And then on this last term, we already said, 7 is the same thing as 7x to the 0. So the exponent here on the constant term on 7 is 0. So these things right over here, those are our exponents. And we are done." + }, + { + "Q": "At 1:05, can we say that the exponent on x always counts as part of x, and has nothing to do with the coefficient?\n", + "A": "Yes, exactly.", + "video_name": "REiDXCN0lGU", + "timestamps": [ + 65 + ], + "3min_transcript": "In the following polynomial, identify the terms along with the coefficient and exponent of each term. So the terms are just the things being added up in this polynomial. So the terms here-- let me write the terms here. The first term is 3x squared. The second term it's being added to negative 8x. You might say, hey wait, isn't it minus 8x? And you could just view that as it's being added to negative 8x. So negative 8x is the second term. And then the third term here is 7. It's called a polynomial. Poly, it has many terms. Or you could view each term as a monomial, as a polynomial with only one term in it. So those are the terms. Now let's think about the coefficients of each of the terms. The coefficient is what's multiplying the power of x or what's multiplying in the x part of the term. So over here, the x part is x squared. That's being multiplied by 3. On the second term, we have negative 8 multiplying x. And we want to be clear, the coefficient isn't just 8. It's a negative 8. It's negative 8 that's multiplying x. So that's the coefficient right over here. And here you might say, hey wait, nothing is multiplying x here. I just have a 7. There is no x. 7 isn't being multiplied by x. But you can think of this as 7 being multiplied by x to the 0 because we know that x to the zeroth power is equal to 1. So we would even call this constant, the 7, this would be the coefficient on 7x to the 0. So you could view this as a coefficient. So this is also a coefficient. So let me make it clear, these three things are coefficients. Now the last part, they want us to identify So the exponent of this first term is 2. It's being raised to the second power. The exponent of the second term, remember, negative 8x, x is the same thing as x to the first power. So the exponent here is 1. And then on this last term, we already said, 7 is the same thing as 7x to the 0. So the exponent here on the constant term on 7 is 0. So these things right over here, those are our exponents. And we are done." + }, + { + "Q": "at 2:56 he adds an h(y). I don't understand his explanation\n", + "A": "If you were taking the indefinite anti-derivative of some function, you would add a +C on the end, to account for any constants that may have been lost. In this case, we took the indefinite anti-derivative of a partial function. So we add the h(y) on the end to account for any function of y that may have been lost. If you think about it backwards, if you took the partial derivative of Psi with respect to x, the function of y goes to zero.", + "video_name": "eu_GFuU7tLI", + "timestamps": [ + 176 + ], + "3min_transcript": "We just divide both sides of this equation by dx, right? And then we get 3x squared minus 2xy plus 2. We're dividing by dx, so that dx just becomes a 1. Plus 6y squared minus x squared plus 3. And then we're dividing by dx, so that becomes dy dx, is equal to-- what's 0 divided by dx? Well it's just 0. And there we have it. We have written this in the form that we need, in this form. And now we need to prove to ourselves that this is an exact equation. So let's do that. So what's the partial of M? This is the M function, right? This was a plus here. What's the partial of this with respect to y? This would be 0. This would be minus 2x, and then just a 2. So the partial of this with respect to y is minus 2x. This would be 0, this would be minus 2x. So there you have it. The partial of M with respect to y is equal to the partial of N with respect to x. My is equal to Nx. So we are dealing with an exact equation. So now we have to find psi. The partial of psi with respect to x is equal to M, which is equal to 3x squared minus 2xy plus 2. Take the anti-derivative with respect to x on both sides, and you get psi is equal to x to the third minus x squared y-- because y is just a constant-- plus 2x, plus some function of y. Because we know psi is a function of x and y. So when you take a derivative, when you take a partial with lost. So it's like the constant, when we first learned taking anti-derivatives. And now, to figure out psi, we just have to solve for h of y. And how do we do that? Well let's take the partial of psi with respect to y. That's going to be equal to this right here. So The partial of psi with respect to y, this is 0, this is minus x squared. So it's minus x squared-- this is o-- plus h prime of y, is going to be able to what? That's going to be equal to our n of x, y. It's going to be able to this. And then we can solve for this. So that's going to be equal to 6y squared minus x squared plus 3. You can add x squared to both sides to get rid of this and this. And then we're left with h prime of y is equal to 6y squared plus 3. Anti-derivative-- so h of y is equal to what is this-- 2y" + }, + { + "Q": "At 0:56, I'm confused about the form. Some places I see have the from M(x,y)dx + N(x,y)dy = 0, while others (I have the same textbook as Khan) have the form like he has it, M(x,y) +N(x,y)dy/dx. Anyone know which way is correct? I've been using the dx and dy at the end to figure out which function is M and which function is N too. How do you pick your Ms and Ns without it?\n", + "A": "M(x,y)dx + N(x,y)dy = 0 divide by dx, and you get: M(x,y) +N(x,y)dy/dx = 0.", + "video_name": "eu_GFuU7tLI", + "timestamps": [ + 56 + ], + "3min_transcript": "Welcome back. I'm just trying to show you as many examples as possible of solving exact differential equations. One, trying to figure out whether the equations are exact. And then if you know they're exact, how do you figure out the psi and figure out the solution of the differential equation? So the next one in my book is 3x squared minus 2xy plus 2 times dx, plus 6y squared minus x squared plus 3 times dy is equal to 0. So just the way it was written, this isn't superficially in that form that we want, right? What's the form that we want? We want some function of x and y plus another function of x and y, times y prime, or dy dx, is equal to 0. We're close. We just divide both sides of this equation by dx, right? And then we get 3x squared minus 2xy plus 2. We're dividing by dx, so that dx just becomes a 1. Plus 6y squared minus x squared plus 3. And then we're dividing by dx, so that becomes dy dx, is equal to-- what's 0 divided by dx? Well it's just 0. And there we have it. We have written this in the form that we need, in this form. And now we need to prove to ourselves that this is an exact equation. So let's do that. So what's the partial of M? This is the M function, right? This was a plus here. What's the partial of this with respect to y? This would be 0. This would be minus 2x, and then just a 2. So the partial of this with respect to y is minus 2x. This would be 0, this would be minus 2x. So there you have it. The partial of M with respect to y is equal to the partial of N with respect to x. My is equal to Nx. So we are dealing with an exact equation. So now we have to find psi. The partial of psi with respect to x is equal to M, which is equal to 3x squared minus 2xy plus 2. Take the anti-derivative with respect to x on both sides, and you get psi is equal to x to the third minus x squared y-- because y is just a constant-- plus 2x, plus some function of y. Because we know psi is a function of x and y. So when you take a derivative, when you take a partial with" + }, + { + "Q": "\nHow can you understand this at 6:38 with out more help?", + "A": "i am asuming you mean at 1:38 becasue the video is less than 2 minutes long. At that time he is placing the 4.1 on the number line and you would find 4 and then realise that youneed to go .1 positive (to the right) from that point.", + "video_name": "uC09taczvOo", + "timestamps": [ + 398 + ], + "3min_transcript": "" + }, + { + "Q": "\nWhy do we want x and y values that give zero in perfect squares? #3:35\nThanks for help.", + "A": "Technically, we could have the x and y values be anything, and that would probably give us a random point along the circle we graph. However, the zeroes of a graph are generally a good place to start as they tend to be easier to find than any other point.", + "video_name": "XyDMsotfJhE", + "timestamps": [ + 215 + ], + "3min_transcript": "We had an equality before, and just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2. If we square negative 2, it becomes a positive 4. We can't just do that on the left-hand side. We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared a, b, essentially, the point that makes both of these equal to 0. And that the radius is going to be r. So if we look over here, what is our a? We have to be careful here. Our a isn't 2. Our a is negative 2. x minus negative 2 is equal to 2. So the x-coordinate of our center is going to be negative 2, and the y-coordinate of our center is going to be 2. Remember, we care about the x value that makes this 0, and the y value that makes this 0. So the center is negative 2, 2. And this is the radius squared. So the radius is equal to 5. So let's go back to the exercise and actually plot this. So it's negative 2, 2. So our center is negative 2, 2. So that's right over there. X is negative 2, y is positive 2. So let's see, this would be 1, 2, 3, 4, 5. So you have to go a little bit wider than this. My pen is having trouble. There you go. 1, 2, 3, 4, 5. Let's check our answer. We got it right." + }, + { + "Q": "At 1:00, Sal says that it has to be plus y squared minus 4y, but isn't the y squared a minus?\n", + "A": "The original problem has a + y^2. It s only when Sal circles it that the upstroke on the + disappears so that it looks like it s a minus sign.", + "video_name": "XyDMsotfJhE", + "timestamps": [ + 60 + ], + "3min_transcript": "We're asked to graph the circle. And they give us this somewhat crazy looking equation. And then we could graph it right over here. And to graph a circle, you have to know where its center is, and you have to know what its radius is. So let me see if I can change that. And you have to know what its radius is. So what we need to do is put this in some form where we can pick out its center and its radius. Let me get my little scratch pad out and see if we can do that. So this is that same equation. And what I essentially want to do is I want to complete the square in terms of x, and complete the square in terms of y, to put it into a form that we can recognize. So first let's take all of the x terms. So you have x squared and 4x on the left-hand side. So I could rewrite this as x squared plus 4x. And I'm going to put some parentheses around here, because I'm going to complete the square. And then I have my y terms. I'll circle those in-- well, the red looks too much like the purple. I'll circle those in blue. y squared and negative 4y. So we have plus y squared minus 4y. And I'll just do that in a neutral color. So minus 17 is equal to 0. Now, what I want to do is make each of these purple expressions perfect squares. So how could I do that here? Well, this would be a perfect square if I took half of this 4 and I squared it. So if I made this plus 4, then this entire expression would be x plus 2 squared. And you can verify that if you like. If you need to review on completing the square, there's plenty of videos on Khan Academy on that. All we did is we took half of this coefficient and then squared it to get 4. Half of 4 is 2, square it to get 4. And that comes straight out of the idea if you take x plus 2 and square it, it's going to be x squared plus twice the product of 2 and x, plus 2 squared. We had an equality before, and just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2. If we square negative 2, it becomes a positive 4. We can't just do that on the left-hand side. We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared" + }, + { + "Q": "\n3:15, how did he know that it 6.5 equals to 3.5 and 3?", + "A": "Because, 3.5 plus 3 equals 6.5.", + "video_name": "loAA3TCNAvU", + "timestamps": [ + 195 + ], + "3min_transcript": "especially a triangle like this, a right triangle, is just going to be half of a rectangle like this. We just care about half of its area. So this area is going to be 1/2 times 2 times 3.5. 1/2 times 2 is equal to 1. 1 times 3.5 is 3.5 square units. So the area of that part is going to be 3.5 square units. Let's think about the area of this triangle right over here. Well, once again we have its height is 3.5. Its base is 7. So its area is going to be 1/2 times 7 times 3.5. 1/2 times 7 is 3.5 times 3.5. So this part is 3.5, and I'm going to multiply that times 3.5 again. Let's figure out what that product is equal to. 3.5 times 3.5. 5 times 5 is 25. Let's cross that out. Move one place over to the left. 3 times 5 is 15. 3 times 3 is 9, plus 1 is 10. So that gets us to 5 plus 0 is 5. 7 plus 5 is 12, carry the 1. 1 plus 1 is 2. And we have a 1. We have two digits to the right of the decimal, one, two. So we're going to have two digits to the right of the decimal in the answer. The area here is 12.25 square units. Now this region may be a little bit more difficult, because it's kind of us this weird trapezoid looking thing. But one thing that might pop out at you is that you can divide it very easily into a rectangle and a triangle. And we can actually figure out the dimensions that we need to figure out the areas of each of these. We know what the width of this rectangle you want to call it. It's going to be 2 units plus 7 units. So this is going to be 9. We know that this distance is 3.5. If this distance right over here is 3.5, then this distance down here has to add up with 3.5 to 6.5, so this must be 3. Now we can actually figure out the area. The area of this rectangle is just going to be its height times its length, or 9 times 3.5. 9 times 3.5. And one way you could do it-- we could even try to do this in our head-- this is going to be 9 times 3 plus 9 times 0.5. 9 times 3 is 27. 9 times 0.5, that's just half of nine, so it's going to be 4.5. 27 plus 4 will get us to 31, so that's going to be equal to 31.5." + }, + { + "Q": "\nAt 3:39,why sai said because of SAS so those triangles are similar?isn't SAS is for congruent?", + "A": "SAS applies both to similarity and congruence. In this case, Sal was using SAS to show similarity.", + "video_name": "Ly86lwq_2gc", + "timestamps": [ + 219 + ], + "3min_transcript": "Let's look at the shorter side on either side of this angle. So the shorter side is two, and let's look at the shorter side on either side of the angle for the larger triangle. Well, then the shorter side is on the right-hand side, and that's going to be XT. So what we want to compare is the ratio between-- let me write it this way. We want to see, is XY over XT equal to the ratio of the longer side? Or if we're looking relative to this angle, the longer of the two, not necessarily the longest of the triangle, although it looks like that as well. Is that equal to the ratio of XZ over the longer of the two sides-- when you're looking at this angle right here, on either side of that angle, for the larger triangle-- over XS? And it's a little confusing, because we've kind of flipped which side, but I'm just thinking about the shorter side on either side of this angle in between, and then the longer side on either side of this angle. and the larger triangle. These are the longer sides for the smaller triangle and the larger triangle. And we see XY. This is two. XT is 3 plus 1 is 4. XZ is 3, and XS is 6. So you have 2 over 4, which is 1/2, which is the same thing as 3/6. So the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle, for both triangles, the ratio is the same. So by SAS we know that the two triangles are congruent. But we have to be careful on how we state the triangles. We want to make sure we get the corresponding sides. And I'm running out of space here. Let me write it right above here. We can write that triangle XYZ is similar to triangle-- and we went to the shorter side first. So now we want to start at X and go to the shorter side on the large triangle. So you go to XTS. XYZ is similar to XTS. Now, let's look at this right over here. So in our larger triangle, we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles. Even though this looks like a right angle, we cannot assume it. And if we look at this smaller triangle right over here, it shares one side with the larger triangle, but that's not enough to do anything. And then this triangle over here also shares another side, but that also doesn't do anything. So we really can't make any statement here about any kind of similarity. So there's no similarity going on here. There are some shared angles. This guy-- they both share that angle, the larger triangle" + }, + { + "Q": "\nAt 5:47, why is it written as a square root, it makes no sense. is it representing a long line of decimals. Someone tell me if I'm having a stupid moment.", + "A": "The square root is used because it s more accurate and easier to write than a list of decimals.", + "video_name": "Ly86lwq_2gc", + "timestamps": [ + 347 + ], + "3min_transcript": "and we went to the shorter side first. So now we want to start at X and go to the shorter side on the large triangle. So you go to XTS. XYZ is similar to XTS. Now, let's look at this right over here. So in our larger triangle, we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles. Even though this looks like a right angle, we cannot assume it. And if we look at this smaller triangle right over here, it shares one side with the larger triangle, but that's not enough to do anything. And then this triangle over here also shares another side, but that also doesn't do anything. So we really can't make any statement here about any kind of similarity. So there's no similarity going on here. There are some shared angles. This guy-- they both share that angle, the larger triangle So there could be a statement of similarity we could make if we knew that this definitely was a right angle. Then we could make some interesting statements about similarity, but right now, we can't really do anything as is. Let's try this one out, this pair right over here. So these are the first ones that we have actually separated out the triangles. So they've given us the three sides of both triangles. So let's just figure out if the ratios between corresponding sides are a constant. So let's start with the short side. So the short side here is 3. The shortest side here is 9 square roots of 3. So we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here, is 3 square roots of 3 over the next longest side over here, which is 27. And then see if that's going to be equal to the ratio of the longest side. So the longest side here is 6, and then So this is going to give us-- let's see, this is 3. Let me do this in a neutral color. So this becomes 1 over 3 square roots of 3. This becomes 1 over root 3 over 9, which seems like a different number, but we want to be careful here. And then this right over here-- if you divide the numerator and denominator by 6, this becomes a 1 and this becomes 3 square roots of 3. So 1 over 3 root 3 needs to be equal to square root of 3 over 9, which needs to be equal to 1 over 3 square roots of 3. At first they don't look equal, but we can actually rationalize this denominator right over here. We can show that 1 over 3 square roots of 3, if you multiply it by square root of 3" + }, + { + "Q": "At 0:57, what is mx?\n", + "A": "mx is the slope of the line, or the steepness (is that a word??)", + "video_name": "YBYu5aZPLeg", + "timestamps": [ + 57 + ], + "3min_transcript": "Write an inequality that fits the graph shown below. So here they've graphed a line in red, and the inequality includes this line because it's in bold red. It's not a dashed line. It's going to be all of the area above it. So it's all the area y is going to be greater than or equal to this line. So first we just have to figure out the equation of this line. We can figure out its y-intercept just by looking at it. Its y-intercept is right there. Let me do that in a darker color. Its y-intercept is right there at y is equal to negative 2. That's the point 0, negative 2. So if you think about this line, if you think about its equation as being of the form y is equal to mx plus b in slope-intercept form, we figured out b is equal to negative 2. So that is negative 2 right there. And let's think about its slope. If we move 2 in the x-direction, if delta x is equal to 2, if our change in x is positive 2, what is our Our change in y is equal to negative 1. Slope, or this m, is equal to change in y over change in x, which is equal to, in this case, negative 1 over 2, or negative 1/2. And just to reinforce, you could have done this anywhere. You could have said, hey, what happens if I go back 4 in x? So if I went back 4, if delta x was negative 4, if delta x is equal to negative 4, then delta y is equal to positive 2. And once again, delta y over delta x would be positive 2 over negative 4, which is also negative 1/2. I just want to reinforce that it's not dependent on how far I move along in x or whether I go forward or backward. You're always going to get or you should always get, the same slope. So the equation of that line is y is equal to the slope, negative 1/2x, plus the y-intercept, minus 2. That's the equation of this line right there. Now, this inequality includes that line and everything above it for any x value. Let's say x is equal to 1. This line will tell us-- well, let's take this point so we get to an integer. Let's say that x is equal to 2. Let me get rid of that 1. When x is equal to 2, this value is going to give us negative 1/2 times 2, which is negative 1, minus 2, is going to give us negative 3. But this inequality isn't just y is equal to negative 3. y would be negative 3 or all of the values greater than I know that, because they shaded in this whole area up here. So the equation, or, as I should say, the inequality that fits the graph here below is-- and I'll do it in a bold color-- is y is greater than or equal to" + }, + { + "Q": "\nAt 2:50 why is it x- -3 as oppose to just x-3?", + "A": "Because in the standard formula of a circle, it should be like the following: (x - h)^2 + (y - k)^2 = r^2 And the problem gives you (x + 3)........................................ Mr. Khan just likes to rewrite it in the subtraction like in the standard formula: (x - -3) Let me write this way, it be clearer to you: (x - (-3)). Now, you can see -3 is the x component of the center of the circle in the problem. For the y , you don t need to rewrite, because it is already in the format of subtraction (y - 4).", + "video_name": "JvDpYlyKkNU", + "timestamps": [ + 170 + ], + "3min_transcript": "and y that describes all these points? Well, we know how to find the distance between two points on a coordinate plane. In fact, it comes straight out of the Pythagorean theorem. If we were to draw a vertical line right over here, that essentially is the change in the vertical axis between these two points, up here, we're at y, here we're k, so this distance is going to be y minus k. We can do the exact same thing on the horizontal axis. This x-coordinate is x while this x-coordinate is h. So this is going to be x minus h is this distance. And this is a right triangle, because by definition, we're saying, hey, we're measuring vertical distance here. We're measuring horizontal distance here, so these two things are perpendicular. And so from the Pythagorean theorem, be equal to our distance squared, and this is where the distance formula comes from. So we know that x minus h squared plus y minus k squared must be equal to r squared. This is the equation for the set-- this describes any x and y that satisfies this equation will sit on this circle. Now, with that out of the way, let's go answer their question. The equation of the circle is this thing. And this looks awfully close to what we just wrote, we just have to make sure that we don't get confused with the negatives. Remember, it has to be in the form x minus h, y minus k. So let's write it a little bit differently. Instead of x plus 3 squared, we can write that as x minus negative 3 squared. And then plus-- well this is already in the form-- plus y minus 4 squared is equal to, instead of 49, we can just call that 7 squared. that our h is negative 3-- I want to do that in the red color-- that our h is negative 3, and that our k is positive 4, and that our r is 7. So we could say h comma k is equal to negative 3 comma positive 4. Make sure to get-- you know you might say, hey, there's a negative 4 here, no. But look, it's minus k, minus 4. So k is 4. Likewise, it's minus h. You might say, hey, maybe h is a positive 3, but no you're subtracting the h. So you'd say minus negative 3, and similarly, the radius is 7." + }, + { + "Q": "@ 09:02 he multiplies by 15 but shouldn't it be 15/64 ?\nI think he made a typo\n", + "A": "The 1/64 comes from flipping a fair coin (1/2 chance of heads or tails each time) 6 times, because (1/2)^6 = 1/64. At 9:02, he s talking about the unfair coin (4/5 chance of heads & 1/5 chance of tails each time,) so instead of (1/2)^6, he uses (4/5)^4 * (1/5)^2. That number (~1.64%) represents the chance of a particular permutation e.g. HTHHTH, but since we don t care about the order of the flips, he multiplies by 15 because there are 15 ways to arrange the 4 heads within the 6 flips.", + "video_name": "xw6utjoyMi4", + "timestamps": [ + 542 + ], + "3min_transcript": "So the probability of getting four out of six heads, given a fair coin, is 15 out of 64. And if you look at it, based on our definition of b and a, this is the probability of b given a. Right? b is four out of six heads, given a fair coin. Fair enough. So let's figure out the probability of-- because there's two ways of getting four out of six heads. One, that we picked a fair coin, and then times 15 out of 64. And then there's the probability that we picked an unfair coin. So what's the probability of the unfair coin? Of getting four out of six heads, given the unfair coin? Well, once again, what's the probability of each of the combinations where we got four out of six? So in this situation, let's do the same one. Heads, tails, heads, tails, heads, heads. But in this situation, it's not a 50% chance of getting heads. It's 80%. So it would be .8 times .2 times .8 times .2 times .8 times .8. Now, essentially, we have-- you know, this multiplication, we can rearrange it, because it doesn't matter what order you multiply things in. So it's .8 to the fourth power times .2 squared. And it doesn't matter. You know, any of the unique combinations will each have Because we can just rearrange the order in which we multiply. And then how many of these combinations are there? If we are, once again, the God of probability, and out of six flips we're picking four-- we're choosing four that are going to end up heads? How many ways can I pick a group of four? Well, once again, that's times 6 choose 4. We figured out what that is. 6 choose 4 is 15. And this is the probability of four out of six heads, given an unfair coin. So what's the total probability of getting four out of six heads? Well, it's going to be the probability of getting the fair coin-- which is 1/3-- times the probability of getting four out of six heads, given the fair coin-- and that's this 15/64. Plus the probability of getting an unfair coin-- 2/3-- times the probability of getting four out of six heads, given the unfair coin-- and that's what we figured out here. Times 15 times .8 to the fourth, times .2 squared." + }, + { + "Q": "\nAt 5:22, he switches from \"-2cos2x\" to \"+4sin2x\". Where exactly did he get the \"4\" from?", + "A": "It came from the derivative of the inner function, 2x. d/dx -2 cos 2x = 2 * 2 sin 2x = 4 sin 2x", + "video_name": "BiVOC3WocXs", + "timestamps": [ + 322 + ], + "3min_transcript": "So minus 2 times cosine of 0, which is 2. All of that over 1 minus the cosine of 0, which is 1. So once again, we get 0/0. So does this mean that the limit doesn't exist? No, it still might exist, we might just want to do L'Hopital's rule again. Let me take the derivative of that and put it over the derivative of that. And then take the limit and maybe L'Hopital's rule will help us on the next [INAUDIBLE]. So let's see if it gets us anywhere. So this should be equal to the limit if L'Hopital's We're not 100% sure yet. This should be equal to the limit as x approaches 0 of the derivative of that thing over the derivative of that thing. So what's the derivative of 2 cosine of x? Well, derivative of cosine of x is negative sine of x. So it's negative 2 sine of x. So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times the 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times-- the negative right there's a plus. You have a positive sine, so it's the sine of 2x. That's the numerator when you take the derivative. And the denominator-- this is just an exercise in What's the derivative of the denominator? Derivative of 1 is 0. And derivative negative cosine of x is just-- well, that's just sine of x. So let's take this limit. So this is going to be equal to-- well, immediately if I take x is equal to 0 in the denominator, I know that Let's see what happens in the numerator. Negative 2 times sine of 0. That's going to be 0. And then plus 4 times sine of 2 times 0. Well, that's still sine of 0, so that's still going to be 0. So once again, we got indeterminate form again. Do we give up? Do we say that L'Hopital's rule didn't work? No, because this could have been our first limit problem. And if this is our first limit problem we say, hey, maybe we could use L'Hopital's rule here because we got an indeterminate form. Both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to-- if the limit exist, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then, plus the derivative of 4 sine of 2x. Well, it's 2 times 4, which is 8." + }, + { + "Q": "\n@4:48 why is the right side of the numerator 4sin2x?\nwhy I use chain rule i get -4sin(2cos)?\n\nf'(x) = g'(x).h'(g(x))\nso\ng(x)= 2cos and h(x)=2x\n\nthen\n-2sin*2(2cos) is the same as -4sin(2cos)\n\nIf some replies, could you workout the steps.", + "A": "Hello excuse me for butting- I see what you ve done but could you explain this in more mathematical terms? Ideally Sal could explain this. I know how to use the chain rule but it seems the usage is different in this case. Maybe I m missing something about trigonometry? Chain Rule: f(x)=h[g(x)] -> f (x)=g (x) * h [g(x)] I am quite confused.", + "video_name": "BiVOC3WocXs", + "timestamps": [ + 288 + ], + "3min_transcript": "So minus 2 times cosine of 0, which is 2. All of that over 1 minus the cosine of 0, which is 1. So once again, we get 0/0. So does this mean that the limit doesn't exist? No, it still might exist, we might just want to do L'Hopital's rule again. Let me take the derivative of that and put it over the derivative of that. And then take the limit and maybe L'Hopital's rule will help us on the next [INAUDIBLE]. So let's see if it gets us anywhere. So this should be equal to the limit if L'Hopital's We're not 100% sure yet. This should be equal to the limit as x approaches 0 of the derivative of that thing over the derivative of that thing. So what's the derivative of 2 cosine of x? Well, derivative of cosine of x is negative sine of x. So it's negative 2 sine of x. So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times the 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times-- the negative right there's a plus. You have a positive sine, so it's the sine of 2x. That's the numerator when you take the derivative. And the denominator-- this is just an exercise in What's the derivative of the denominator? Derivative of 1 is 0. And derivative negative cosine of x is just-- well, that's just sine of x. So let's take this limit. So this is going to be equal to-- well, immediately if I take x is equal to 0 in the denominator, I know that Let's see what happens in the numerator. Negative 2 times sine of 0. That's going to be 0. And then plus 4 times sine of 2 times 0. Well, that's still sine of 0, so that's still going to be 0. So once again, we got indeterminate form again. Do we give up? Do we say that L'Hopital's rule didn't work? No, because this could have been our first limit problem. And if this is our first limit problem we say, hey, maybe we could use L'Hopital's rule here because we got an indeterminate form. Both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to-- if the limit exist, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then, plus the derivative of 4 sine of 2x. Well, it's 2 times 4, which is 8." + }, + { + "Q": "\nat 5:37 are you allowed to cancel out the sin(x) to make it -2+4(2)/1 which is -2+8 which also equals 6 to find your answer?", + "A": "No, because they are not all sines of the same value. Two of them are sin(x) the other is sin(2x).", + "video_name": "BiVOC3WocXs", + "timestamps": [ + 337 + ], + "3min_transcript": "So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times the 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times-- the negative right there's a plus. You have a positive sine, so it's the sine of 2x. That's the numerator when you take the derivative. And the denominator-- this is just an exercise in What's the derivative of the denominator? Derivative of 1 is 0. And derivative negative cosine of x is just-- well, that's just sine of x. So let's take this limit. So this is going to be equal to-- well, immediately if I take x is equal to 0 in the denominator, I know that Let's see what happens in the numerator. Negative 2 times sine of 0. That's going to be 0. And then plus 4 times sine of 2 times 0. Well, that's still sine of 0, so that's still going to be 0. So once again, we got indeterminate form again. Do we give up? Do we say that L'Hopital's rule didn't work? No, because this could have been our first limit problem. And if this is our first limit problem we say, hey, maybe we could use L'Hopital's rule here because we got an indeterminate form. Both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to-- if the limit exist, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then, plus the derivative of 4 sine of 2x. Well, it's 2 times 4, which is 8. Derivative of sine of 2x is 2 cosine of 2x. And that first 2 gets multiplied by the 4 to get the 8. And then the derivative of the denominator, derivative of sine of x is just cosine of x. So let's evaluate this character. So it looks like we've made some headway or maybe L'Hopital's rule stop applying here because we take the limit as x approaches 0 of cosine of x. That is 1. So we're definitely not going to get that indeterminate form, that 0/0 on this iteration. Let's see what happens to the numerator. We get negative 2 times cosine of 0. Well that's just negative 2 because cosine of 0 is 1. Plus 8 times cosine of 2x. Well, if x is 0, so it's going to be cosine of 0, which is 1. So it's just going to be an 8. So negative 2 plus 8. Well this thing right here, negative 2 plus 8 is 6. 6 over 1. This whole thing is equal to 6." + }, + { + "Q": "Hi!\n\nAround 12:12, Sal starts transforming 0.5 to 5. Why can't we just leave the 0.5 as is, given that, once multiplied by 10 to the 17th power, we will get the same answer? I mean, isn't 0.5 times 10 to the 17th power just as correct as 5 times 10 to the 16th power?\n\nCheers!\n", + "A": "Scientific notation requires that the mantissa (the first bit!) is greater than or equal to 1, and less than 10. So if you put 0.5 x 10^17, that is not valid scientific notation because 0.5 is not greater than or equal to 1. You are correct that 0.5 x 10^17 is equal to 5 x 10^16, but one is valid scientific notation and the other is not.", + "video_name": "0Dd-y_apbRw", + "timestamps": [ + 732 + ], + "3min_transcript": "Well, this is equal to 3.2 over 6.4. We can just separate them out because it's associative. So, it's this times 10 to the 11th over 10 to the minus six, right? If you multiply these two things, you'll get that right there. So 3.2 over 6.4. This is just equal to 0.5, right? 32 is half of 64 or 3.2 is half of 6.4, so this is 0.5 right there. And what is this? This is 10 to the 11th over 10 to the minus 6. So when you have something in the denominator, you could write it this way. This is equivalent to 10 to the 11th over 10 to the minus 6. It's equal to 10 to the 11th times 10 to the minus 6 to the minus 1. Or this is equal to 10 to the 11th times 10 to the sixth. This is 1 over 10 to the minus 6. So 1 over something is just that something to the negative 1 power. And then I multiplied the exponents. You can think of it that way and so this would be equal to 10 to the 17th power. Or another way to think about it is if you have 1 -- you have the same bases, 10 in this case, and you're dividing them, you just take the 1 the numerator and you subtract the exponent in the denominator. So it's 11 minus minus 6, which is 11 plus 6, which is equal to 17. So this division problem ended up being equal to 0.5 times 10 to the 17th. Which is the correct answer, but if you wanted to be a stickler and put it into scientific notation, we want something maybe greater than 1 right here. So the way we can do that, let's multiply it by 10 on this side. And divide by 10 on this side or multiply by 1/10. by 10 and divide by 10. We're just doing it to different parts of the product. So this side is going to become 5 -- I'll do it in pink -- 10 times 0.5 is 5, times 10 to the 17th divided by 10. That's the same thing as 10 to the 17th times 10 to the minus 1, right? That's 10 to the minus 1. So it's equal to 10 to the 16th power. Which is the answer when you divide these two guys right there. So hopefully these examples have filled in all of the gaps or the uncertain scenarios dealing with scientific notation. If I haven't covered something, feel free to write a comment on this video or pop me an e-mail." + }, + { + "Q": "At 12:00 I thought you had to multiply or divide equaly on both sides of the equation. Here you multiply on one side and divide on the other.\n", + "A": "He is making two changes to the same side of the equation. The only way that is legal is when the changes cancel each other.", + "video_name": "0Dd-y_apbRw", + "timestamps": [ + 720 + ], + "3min_transcript": "Well, this is equal to 3.2 over 6.4. We can just separate them out because it's associative. So, it's this times 10 to the 11th over 10 to the minus six, right? If you multiply these two things, you'll get that right there. So 3.2 over 6.4. This is just equal to 0.5, right? 32 is half of 64 or 3.2 is half of 6.4, so this is 0.5 right there. And what is this? This is 10 to the 11th over 10 to the minus 6. So when you have something in the denominator, you could write it this way. This is equivalent to 10 to the 11th over 10 to the minus 6. It's equal to 10 to the 11th times 10 to the minus 6 to the minus 1. Or this is equal to 10 to the 11th times 10 to the sixth. This is 1 over 10 to the minus 6. So 1 over something is just that something to the negative 1 power. And then I multiplied the exponents. You can think of it that way and so this would be equal to 10 to the 17th power. Or another way to think about it is if you have 1 -- you have the same bases, 10 in this case, and you're dividing them, you just take the 1 the numerator and you subtract the exponent in the denominator. So it's 11 minus minus 6, which is 11 plus 6, which is equal to 17. So this division problem ended up being equal to 0.5 times 10 to the 17th. Which is the correct answer, but if you wanted to be a stickler and put it into scientific notation, we want something maybe greater than 1 right here. So the way we can do that, let's multiply it by 10 on this side. And divide by 10 on this side or multiply by 1/10. by 10 and divide by 10. We're just doing it to different parts of the product. So this side is going to become 5 -- I'll do it in pink -- 10 times 0.5 is 5, times 10 to the 17th divided by 10. That's the same thing as 10 to the 17th times 10 to the minus 1, right? That's 10 to the minus 1. So it's equal to 10 to the 16th power. Which is the answer when you divide these two guys right there. So hopefully these examples have filled in all of the gaps or the uncertain scenarios dealing with scientific notation. If I haven't covered something, feel free to write a comment on this video or pop me an e-mail." + }, + { + "Q": "At 2:27, at the top left says 10^-3. Since the number is getting larger due to the decimal moving to the right, why is the exponent a negative?\n", + "A": "0.00852 = 8.52* 10^(-3) the exponent -3 is negative because the number 8.52 is 1000 times larger than 0.00852 . The negative exponent -3 means you divide by 1000 and is equal to the multiplication with 1/1000 some examples: 8.52* 10^(-3) = 8.52 * 1/1000 = 8.52 / 1000 = 0.00852 hope this will help you", + "video_name": "0Dd-y_apbRw", + "timestamps": [ + 147 + ], + "3min_transcript": "It always helps me to see a lot of examples of something so I figured it wouldn't hurt to do more scientific notation examples. So I'm just going to write a bunch of numbers and then write them in scientific notation. And hopefully this'll cover almost every case you'll ever see and then at the end of this video, we'll actually do some computation with them to just make sure that we can do computation with scientific notation. Let me just write down a bunch of numbers. 0.00852. That's my first number. My second number is 7012000000000. I'm just arbitrarily stopping the zeroes. The next number is 0.0000000 I'll just draw a couple more. If I keep saying 0, you might find that annoying. 500 The next number -- right here, there's a The next number I'm going to do is the number 723. The next number I'll do -- I'm having a lot of 7's here. Let's do 0.6. And then let's just do one more just for, just to make sure we've covered all of our bases. Let's say we do 823 and then let's throw some -- an arbitrary number of 0's there. So this first one, right here, what we do if we want to write in scientific notation, we want to figure out the largest exponent of 10 that fits into it. So we go to its first non-zero term, which is that right there. We count how many positions to the right of the decimal point we have including that term. So we have one, two, three. So it's going to be equal to this. So it's going to be equal to 8 -- that's that guy right there -- 0.52. So everything after that first term is going to be behind the decimal. So 0.52 times 10 to the number of terms we have. 10 to the minus 3. Another way to think of it: this is a little bit more. This is like 8 1/2 thousands, right? Each of these is thousands. We have 8 1/2 of them. Let's do this one. Let's see how many 0's we have. We have 3, 6, 9, 12. So we want to do -- again, we start with our largest term that we have. Our largest non-zero term. In this case, it's going to be the term all the way to the left. That's our 7. So it's going to be 7.012. It's going to be equal to 7.012 times 10 to the what? Well it's going to be times 10 to the 1 with this many 0's. So how many things? We had a 1 here. Then we had 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 0's. I want to be very clear. You're not just counting the 0's. You're counting everything after this first" + }, + { + "Q": "At at 6:14, why would x^2\u00c2\u00b72 = 4x? Wouldn't it be 2x^2?\n", + "A": "x^(2) * 2 is equal to 2x^(2). However, in this case we are also taking \u00e2\u0088\u0082/\u00e2\u0088\u0082x(x^(2) * 2) which is equal to 4x. \u00e2\u0088\u0082/\u00e2\u0088\u0082x(x^(2) * 2) = \u00e2\u0088\u0082/\u00e2\u0088\u0082x(2x^(2)) = 4x", + "video_name": "AXqhWeUEtQU", + "timestamps": [ + 374 + ], + "3min_transcript": "Again, it depends on the function. And I'll show you how you can compute something like this in just a moment here. But, first there's kind of an annoying thing associated with partial derivatives, where we don't write them with D's in DX/DF. People came up with this new notation, mostly just to emphasize to the reader of your equation that it's a multi-variable function involved. And what you do, is you say, you write a D, but it's got kind of a curl at the top. It's this new symbol and people will often read it as partial. So, you might read like partial F, partial Y. If you're wondering, by the way, why we call these partial derivatives, it's sort of like, this doesn't tell the full story of how F changes 'cause it only cares about the X direction. Neither does this, this only cares about the Y direction. So, each one is only a small part of the story. So, let's actually evaluate something like this. I'm gonna go ahead and clear the board over here. I think the one-dimensional analogy is something we probably have already. So, if you're actually evaluating something like this, here, I'll write it up here again up here. Partial derivative of F, with respect to X, and we're doing it at one, two. It only cares about movement in the X direction, so it's treating Y as a constant. It doesn't even care about the fact that Y changes. As far as it's concerned, Y is always equal to two. So, we can just plug that in ahead of time. So, I'm gonna say partial, partial X, this is another way you might write it, put the expression in here. And I'll say X squared, but instead of writing Y, I'm just gonna plug in that constant ahead of time. 'Cause when you're only moving in the X direction, this is kind of how the multi-variable function sees the world. And I'll just keep a little note that we're evaluating this whole thing at X equals one. And here, this is actually just an ordinary derivative. This is an expression that's an X, you're asking how it changes as you shift around X and you know how to do this. This is just taking the derivative is gonna be 4x 'cause X squared goes to 2x. And then the derivative of a constant, sin of two is just a constant, is zero. And of course we're evaluating this at X equals one, so your overall answer is gonna be four. And as for practice, let's also do that with derivative with respect to Y. So, we look over here, I'm gonna write the same thing. You're taking the partial derivative of F with respect to Y. We're evaluating it at the same point one, two. This time it doesn't care about movement in the X direction. So, as far as it's concerned, that X just stays constant at one. So, we'd write one squared times Y, plus sin(Y). Sin(Y). And you're saying, oh, I'm keeping track of this at Y=2. So, it's kind of, you're evaluating at Y=2. When you take the derivative, this is just 1xY." + }, + { + "Q": "at 1:41, how do you divide 1/6 by side and dividing both side also with 6 can someone explain.\n", + "A": "He says multiply both sides by 1/6, or another way you could think about it is we re dividing both sides by 6.", + "video_name": "u5dPUHjagSI", + "timestamps": [ + 101 + ], + "3min_transcript": "We never know when we might have to do a little bit more party planning. So it doesn't hurt to have some practice. And that's what this exercise is doing for us, is generating problems so that we can try solving systems of equations with elimination. And so in this first problem, it says solve for x and y using elimination. And then this is what they have-- 6x minus 6y is equal to negative 24. Negative 5x minus 5y is equal to negative 60. So let me get my scratch pad out to solve this. Let me rewrite it. So they gave us 6x minus 6y is equal to negative 24. And negative 5x minus 5y is equal to negative 60. So what we have to think about, and we saw this in several of the other videos, is when we want to eliminate a variable, we want to manipulate these two equations. And if we were to add the corresponding sides, that variable might disappear. So if we just added a 6x to a negative 5x, If this was a negative 6x, that would work out. Or if this was a positive 5x, that would work out. But this isn't exactly right. So if I want to eliminate the x, I have to manipulate these equations so that these two characters might cancel out. And one thing that pops into my brain is it looks like all of this stuff up here is divisible by 6, and all of this stuff down here is divisible by 5. And if we were to divide all this stuff up here by 6, we'd be left with an x over here. And if we were to divide all this bottom stuff by 5, we'd be left with a negative x right over here. And then they just might cancel out. So let's try that out. Let's take this first equation. And we're going to multiply both sides by 1/6. Or another way you could think about it is we're dividing both sides by 6. And as long as we do the same thing to both sides, the equation holds. The equality holds. So if you multiply everything by 1/6, 6x times 1/6 is just going to be x. 6y times 1/6 is just y. So it's negative y. Or you could just view it as negative 24 divided by 6 is negative 4. So this equation, the blue one, we've simplified as x minus y is equal to negative 4. Let's do something similar with the second one. Here we could say we're going to multiply everything times 1/5. Or you could say that we're dividing everything by 5. If we do that, negative 5x divided by 5 is just negative x. Negative 5y divided by 5 is negative y. And then negative 60 divided by 5 is negative 12. And now, this looks pretty interesting. If we add the two left-hand sides-- and remember, we can keep the equality, because we're essentially adding the same thing to both sides. You can imagine we're starting with the blue equation. And on the left-hand side, we're adding negative x minus y. And on the right-hand side, we're adding negative 12. But the second equation tells us that those two things" + }, + { + "Q": "At 1:53, how can we figure out what number to multiply by the equation? Every equation isnt this simple, Khan should do one easy equation and one really hard equation to show that its possible.\n", + "A": "multiplying everything by 1/6 is the same as dividing it all by 6. The equation was 6x-6y=24. 6 goes into all the numbers equally therefore we are able to divide it by 6. The end product is x-y=4", + "video_name": "u5dPUHjagSI", + "timestamps": [ + 113 + ], + "3min_transcript": "We never know when we might have to do a little bit more party planning. So it doesn't hurt to have some practice. And that's what this exercise is doing for us, is generating problems so that we can try solving systems of equations with elimination. And so in this first problem, it says solve for x and y using elimination. And then this is what they have-- 6x minus 6y is equal to negative 24. Negative 5x minus 5y is equal to negative 60. So let me get my scratch pad out to solve this. Let me rewrite it. So they gave us 6x minus 6y is equal to negative 24. And negative 5x minus 5y is equal to negative 60. So what we have to think about, and we saw this in several of the other videos, is when we want to eliminate a variable, we want to manipulate these two equations. And if we were to add the corresponding sides, that variable might disappear. So if we just added a 6x to a negative 5x, If this was a negative 6x, that would work out. Or if this was a positive 5x, that would work out. But this isn't exactly right. So if I want to eliminate the x, I have to manipulate these equations so that these two characters might cancel out. And one thing that pops into my brain is it looks like all of this stuff up here is divisible by 6, and all of this stuff down here is divisible by 5. And if we were to divide all this stuff up here by 6, we'd be left with an x over here. And if we were to divide all this bottom stuff by 5, we'd be left with a negative x right over here. And then they just might cancel out. So let's try that out. Let's take this first equation. And we're going to multiply both sides by 1/6. Or another way you could think about it is we're dividing both sides by 6. And as long as we do the same thing to both sides, the equation holds. The equality holds. So if you multiply everything by 1/6, 6x times 1/6 is just going to be x. 6y times 1/6 is just y. So it's negative y. Or you could just view it as negative 24 divided by 6 is negative 4. So this equation, the blue one, we've simplified as x minus y is equal to negative 4. Let's do something similar with the second one. Here we could say we're going to multiply everything times 1/5. Or you could say that we're dividing everything by 5. If we do that, negative 5x divided by 5 is just negative x. Negative 5y divided by 5 is negative y. And then negative 60 divided by 5 is negative 12. And now, this looks pretty interesting. If we add the two left-hand sides-- and remember, we can keep the equality, because we're essentially adding the same thing to both sides. You can imagine we're starting with the blue equation. And on the left-hand side, we're adding negative x minus y. And on the right-hand side, we're adding negative 12. But the second equation tells us that those two things" + }, + { + "Q": "\nAt 0:08 why was a dot there insted of a X?", + "A": "Because some people use dots for multiplication instead of an X!", + "video_name": "0WUXQNjdRvM", + "timestamps": [ + 8 + ], + "3min_transcript": "Compute 23 times 44. And maybe the hardest part of this problem, or maybe the first hard part, is to recognize that that dot even means multiplication. This could have also been written as 23 times 44, or they could have written it as 23 in parentheses times 44, so you just put the two parentheses That also implies multiplication. So now that we know we're multiplying, let's actually do the problem. So we're going to multiply 23-- I'll write it bigger. We're going to multiply 23 by 44. I'll write the traditional multiplication sign there, just so that we know we're multiplying. When you write it vertically like this, you very seldom put a dot there. So let's do some multiplication. Let's start off multiplying this 4 in the ones place times 23. So you have 3 times 4 is 12. We can write 2 in the ones place, but then we want to carry the 1, or we want to regroup that So it's 12, so you put the 1 over here. And now you have 4 times 2 is 8 plus 1 is 9. So you can think about it as 4, this 4 right here, times 23 is 92. That's what we just solved for. Now, we want to figure out what this 4 times 23 is. Now what we do here is, when you just do it mechanically, when you just learn the process, you stick a 0 here. But the whole reason why you're putting a 0 here is because you're now dealing with a 4 in the tens place. If you had another-- I don't know, a 3 or a 4 or whatever digit, and you're dealing with the hundreds place, you'd put more zeroes here, because we're going to find out 4 times 23 is 92. We just figured that out. If we just multiplied this 4 times 23 again, we would get 92 again. But this 4 is actually a 40, so it actually should be 920, and that's why we're putting that 0. Now you're going to see it in a second. So we have-- so let me put this in a different color. 4 times 3 is 12. Let's put the 2 right here. It should be in the tens place because this is really a 40 times the 3. Just think about it, or you could just think of the process. It's the next space that's free. 4 times 3 is 12. Carry the 1. This blue 1 is from last time. You ignore it now. You don't want to make that mess it up. That's when we multiplied this 4. So now we have 4 times 2 is 8 plus 1 is 9. So what we figured out so far is 4 times 23 is 92, and this green 4 times 23 is 920, and that's because this green 4 actually represents 40. It's in the tens place. So when you multiply 44 times 23, it's going to be 4 times 23, which is 92, plus 40 times 23, which is 920. I just want to make sure we understand what we're doing here. And so we can take their sum now. Let's add them up." + }, + { + "Q": "At about 1:44 Sal said when we move to the tens place we put a zero under the quitiont. But can you also put an X?\n", + "A": "Yes. Both are correct and valid strategies. Some people use a 0 and some use an X, both are meant to help you not to mix up which column to put your current step s number in - otherwise known as a placeholder. I learned the 0 method, but if you re comfortable with X, or that s what your teacher/tutor/coach expects, then that s fine, too.", + "video_name": "0WUXQNjdRvM", + "timestamps": [ + 104 + ], + "3min_transcript": "Compute 23 times 44. And maybe the hardest part of this problem, or maybe the first hard part, is to recognize that that dot even means multiplication. This could have also been written as 23 times 44, or they could have written it as 23 in parentheses times 44, so you just put the two parentheses That also implies multiplication. So now that we know we're multiplying, let's actually do the problem. So we're going to multiply 23-- I'll write it bigger. We're going to multiply 23 by 44. I'll write the traditional multiplication sign there, just so that we know we're multiplying. When you write it vertically like this, you very seldom put a dot there. So let's do some multiplication. Let's start off multiplying this 4 in the ones place times 23. So you have 3 times 4 is 12. We can write 2 in the ones place, but then we want to carry the 1, or we want to regroup that So it's 12, so you put the 1 over here. And now you have 4 times 2 is 8 plus 1 is 9. So you can think about it as 4, this 4 right here, times 23 is 92. That's what we just solved for. Now, we want to figure out what this 4 times 23 is. Now what we do here is, when you just do it mechanically, when you just learn the process, you stick a 0 here. But the whole reason why you're putting a 0 here is because you're now dealing with a 4 in the tens place. If you had another-- I don't know, a 3 or a 4 or whatever digit, and you're dealing with the hundreds place, you'd put more zeroes here, because we're going to find out 4 times 23 is 92. We just figured that out. If we just multiplied this 4 times 23 again, we would get 92 again. But this 4 is actually a 40, so it actually should be 920, and that's why we're putting that 0. Now you're going to see it in a second. So we have-- so let me put this in a different color. 4 times 3 is 12. Let's put the 2 right here. It should be in the tens place because this is really a 40 times the 3. Just think about it, or you could just think of the process. It's the next space that's free. 4 times 3 is 12. Carry the 1. This blue 1 is from last time. You ignore it now. You don't want to make that mess it up. That's when we multiplied this 4. So now we have 4 times 2 is 8 plus 1 is 9. So what we figured out so far is 4 times 23 is 92, and this green 4 times 23 is 920, and that's because this green 4 actually represents 40. It's in the tens place. So when you multiply 44 times 23, it's going to be 4 times 23, which is 92, plus 40 times 23, which is 920. I just want to make sure we understand what we're doing here. And so we can take their sum now. Let's add them up." + }, + { + "Q": "At 2:13 how can you just split the triangle into two right triangles? Are you allowed to do that for a equilateral triangle too?\n", + "A": "You can do it for any triangle. To do it, pick a side of the triangle. Obviously, we can draw perpendiculars to that side wherever we like. And surely one of those perpendiculars will go through the opposite vertex. This divides the triangle into two right triangles.", + "video_name": "pGaDcOMdw48", + "timestamps": [ + 133 + ], + "3min_transcript": "In the last video, we had a word problem where we had-- we essentially had to figure out the sides of a triangle, but instead of, you know, just being able to do the Pythagorean theorem and because it was a right triangle, it was just kind of a normal triangle. It wasn't a right triangle. And we just kind of chugged through it using SOHCAHTOA and just our very simple trig functions, and we got the right answer. What I want to do now is to introduce you to something called the law of cosines, which we essentially proved in the last video, but I want to kind of prove it in a more-- you know, without the word problem getting in the way, and I want to show you, once you know the law of cosines, so you can then apply it to a problem, like we did in the past, and you'll do it faster. I have a bit of a mixed opinion about it because I'm not a big fan of memorizing things. You know, when you're 40 years old, you probably won't have the law of cosines still memorized, but if you have that ability to start with the trig functions and just move forward, then you'll always be set. And I'd be impressed if you're still doing trig at 40, but who knows? So let's go and let's see what this law of cosines is all about. So let's say that I know this angle theta. No, let's call this side b. I'm being a little arbitrary here. Actually, let me stay in the colors of the sides. Let's call that b and let's call this c, and let's call this side a. So if this is a right triangle, then we could have used the Pythagorean theorem somehow, but now we can't. So what do we do? So we know a-- well, let's assume that we know b, we know c, we know theta, and then we want to solve for a. But, in general, as long as you know three of these, you can solve for the fourth once you know the law of cosines. So how can we do it? Well, we're going to do it the exact same way we did that last problem. We can drop a line here to make-- oh, my I thought I was using the line tool. Edit, undo. So I can drop a line like that. So I have two right angles. And then once I have right triangles, then now I can start to use trig functions and the Pythagorean theorem, et cetera, et cetera. So, let's see, this is a right angle, this is a right angle. So what is this side here? Let me pick another color. I'm probably going to get too involved with all the colors, but it's for your improvement. So what is this side here? What is the length of that side, that purple side? Well, that purple side is just, you know, we use SOHCAHTOA. I was just going to write SOHCAHTOA up here. So this purple side is adjacent to theta, and then this blue or mauve side b is the hypotenuse of this right triangle." + }, + { + "Q": "\nAt 4:40, how does Sal know to solve for m (the magenta line) in terms of b and theta? For example, we can also say that tan theta = m/c, no?", + "A": "maybe because tan theta = m/d actually, and that would introduce another variable that he doesn t want. He wants to relate the existing variables to keep it simple. Also possibly because he knows sin and cos functions are complimentary and so hopes they will cancel or simplify easier down the track in the math..", + "video_name": "pGaDcOMdw48", + "timestamps": [ + 280 + ], + "3min_transcript": "because it'll take me forever if I keep switching colors. We know that cosine of theta-- let's call this side, let's call this kind of subside-- I don't know, let's call this d, side d. We know that cosine of theta is equal to d over b, right? And we know b. Or that d is equal to what? It equals b cosine theta. Now, let's call this side e right here. Well, what's e? Well, e is this whole c side-- c side, oh, that's interesting-- this whole c side minus this d side, right? So e is equal to c minus d. minus b cosine of theta. So that's e. We got e out of the way. Well, what's this magenta side going to be? Well, let's call this magenta-- let's call it m from magenta. Well, m is opposite to theta. Now, we know it. We've solved for c as well, but we know b, and b is simple. So what relationship gives us m over b, or involves the opposite and the hypotenuse? Well, that's sine: opposite over hypotenuse. So we know that m over b is equal to sine of theta. We know that-- let me go over here. m over b, right, because this is the hypotenuse, is equal to sine of theta, or that m is equal to b sine So we figured out m, we figured out e, and now we want to figure out a. And this should jump out at you. We have two sides of a right triangle. We want to figure out the hypotenuse. We can use the Pythagorean theorem. The Pythagorean theorem tells us a squared is equal to m squared plus e squared, right? Just the square of the other two sides. Well, what's m squared plus e squared? Let me switch to another color just to be arbitrary. a squared is equal to m squared. m is b sine of theta. So it's b sine of theta squared plus e squared. Well, e we figure out is this. So it's plus c minus b cosine theta squared." + }, + { + "Q": "\nI was able to follow his process up until 6:24 when he wrote Csquared-2cbcos(theta). Where did he get that part of the equation from? To me it looked like he pulled it out of nowhere.", + "A": "He simply FOILed the (c-bcos(theta))^2, i.e. multipled out the (c-bcos(theta))^2, which is just (c-bcos(theta))(c-bcos(theta)). That multiples out to c^2-cbcos(theta)-cbcos(theta)+(bcos(theta))^2, and there are two -cbccos(theta), so it simplifies to c^2-2bcos(theta)+(bcos(theta)^2.", + "video_name": "pGaDcOMdw48", + "timestamps": [ + 384 + ], + "3min_transcript": "minus b cosine of theta. So that's e. We got e out of the way. Well, what's this magenta side going to be? Well, let's call this magenta-- let's call it m from magenta. Well, m is opposite to theta. Now, we know it. We've solved for c as well, but we know b, and b is simple. So what relationship gives us m over b, or involves the opposite and the hypotenuse? Well, that's sine: opposite over hypotenuse. So we know that m over b is equal to sine of theta. We know that-- let me go over here. m over b, right, because this is the hypotenuse, is equal to sine of theta, or that m is equal to b sine So we figured out m, we figured out e, and now we want to figure out a. And this should jump out at you. We have two sides of a right triangle. We want to figure out the hypotenuse. We can use the Pythagorean theorem. The Pythagorean theorem tells us a squared is equal to m squared plus e squared, right? Just the square of the other two sides. Well, what's m squared plus e squared? Let me switch to another color just to be arbitrary. a squared is equal to m squared. m is b sine of theta. So it's b sine of theta squared plus e squared. Well, e we figure out is this. So it's plus c minus b cosine theta squared. So that equals b sine-- b squared sine squared of theta. theta squared, right? Plus, and we just foiled this out, although I don't like using foil. I just multiply it out. c squared minus 2cb cosine theta plus b squared cosine theta, right? I just expanded this out by multiplying it out. And now let's see if we can do anything interesting. Well, if we take this term and this term, we get-- those two terms are b squared sine squared of theta plus b squared cosine-- this should be squared there, right, because we squared it. b squared cosine squared of theta, and then we have plus c" + }, + { + "Q": "What happens to the -1 = 15 at 8:37? He glances his cursor towards the direction -1=15 and utters the number 16 but he loses me at that point.\n", + "A": "He adds 1 to both sides, resulting in 16 for the right hand side, and then subtracts 8 from both sides, which yields 8 for the right hand side. He could have subtracted 1 from 8 for the left hand side and then subtracted 7 from both sides. Same result.", + "video_name": "A52fEdPn9lg", + "timestamps": [ + 517 + ], + "3min_transcript": "And now for B, there's no obvious way to make the other two cancel out without making the B cancel out. But we've solved for everything else, so we can really just pick an arbitrary value for x that'll make things easy to compute. So if we just pick x is equal to 0, that's always something that can clear out a lot of the hairiness of an algebra problem. x is equal to 0, not 3. x is equal to 0. Let me do that in a different color. If x is equal to 0, then what do we have? We have A, but A is 2, right? 2 times 0 minus 2 squared. So that's minus 2 squared, so that's 2 times 4. Plus B-- that's what we're trying to solve for; we already solved for everything-- B times minus 1, right? 0 minus 1 is minus 1. Times minus 2, plus C times minus 1 is equal to-- well, Then we just solve for B. We get 8 plus 2B, right? Minus 1 times minus 2 is plus 2. Oh, and we shouldn't have written the C here, we know what C is. C is 1. So this is just a 1. So 1 times minus 1 is minus 1, is equal to 15. And let's see, we have 2B is equal to, lets see, 16 is equal to 8, B is equal to 4, dividing both sides by 2. So we're done. So the partial fraction decomposition of this right here is A, which we've solved for, which is 2. So it equals 2 over x minus 1 plus B, which is 4-- plus 4 And what we did in this with the repeated factor is true if we went to a higher degree term. So if we had blah blah blah, some polynomial there, and it was all over x minus-- I don't know, some number. x minus a to the 10th power. Well, to the-- well, yeah, sure. To the 10th power. If we wanted to decompose this partial fraction, or expand it, it would be A over x minus a-- it's a different a. I'm just showing this. Plus B times x minus a squared, plus, and you go blah blah blah. You'd have 10 terms, plus-- I don't know what the 10th letter of the alphabet is; maybe it's H or I or something. Maybe it's J. J over x minus a to the 10th." + }, + { + "Q": "\nWhy is it (x-2)^2? It should be 2(x-2)! If (x-2)^2=(x-2)(x-2). 2(x-2)=(x-2)+(x-2). (at 00:56):D", + "A": "(x-2)^2 = (x-2)*(x-2) = x^2-4x+4 what you wrote is: 2(x-2) = 2x-4 (x-2)+(x-2) = 2x-4 these are completely different polynomials from the one in the problem.", + "video_name": "A52fEdPn9lg", + "timestamps": [ + 56 + ], + "3min_transcript": "There's one more case of partial fraction expansion or decomposition problems that you might see, so I thought I would cover it. And that's the situation where you have a repeated factor in the denominator. So let's see, I've constructed a little problem here. It's 6x squared. Let me make sure my pen is right. 6x squared minus 19x plus 15. All of that over x minus 1 times-- and this is the thing that makes this one especially interesting-- x minus 2 squared. So you might say, gee, this is a little bit different. Because what do I do here? I have this first degree factor, but it shows up twice. It doesn't make sense, for example, it wouldn't make sense to do this. A over x minus 1 plus B over x minus 2 plus C over x minus 2. Because if you did this, the B and the C would just add You could just view them as one variable. You wouldn't have to have two separate variables here. So this wouldn't make sense as a partial fraction expansion of this. You might want to maybe square this and view it as a second degree term, and do it like you did in the previous example. But then you wouldn't have fully decomposed this problem. And so the answer here-- and I'll just show you how to do it, I'll maybe leave it to you a little bit to think about why it works-- is to decompose it almost like this, but instead of having C over x minus 2, you're going to have C over x minus 2 squared. And I'll try to give you a little intuition about why that happens. So the decomposition here is going to be A over x minus 1 plus B over x minus 2 plus C over x minus 2 squared. And the intuition here is that if you were to-- let's ignore this term right here. But if you were to add these two terms right there, you would get a rational function that would end up having And so it would be consistent with what we did in the second partial fraction video, where if you would have a second degree term on the bottom, and the numerator, when you add just these two parts-- let me clarify what I'm saying. And this is just for intuition. If you were to add just these two parts here, you would get something times x plus something, all of that over x minus 2 squared. And that's consistent with what we did in the previous video, where we said if we have a second degree term in the denominator-- which this really is if you were to expand this-- you should have a first degree term in the numerator. So I'll leave that there. That's a little bit of a nuance, and it's good to think about why this works. But with that said, let's just solve this problem. Let me erase some of this stuff that I've written." + }, + { + "Q": "\nconfused at 0:55 NEED HELP!", + "A": "If you re really having difficulty, please try to stay on topic (refrain from gratuitous woodchuck references in the future). Are you having trouble understanding why b -48 4\u00e2\u0080\u00a2y > -48 - 8 4\u00e2\u0080\u00a2y > -56 4\u00e2\u0080\u00a2y/4 > -56/4 y > -14", + "video_name": "d2cnQ5ahHgE", + "timestamps": [ + 143 + ], + "3min_transcript": "Solve for y. We have 3y plus 7 is less than 2y and 4y plus 8 is greater than negative 48. So we have to find all the y's that meet both of these constraints. So let's just solve for y in each of the constraints and just remember that this \"and\" is here. So we have 3y plus 7 is less than 2y. So let's isolate the y's on the left-hand side. So let's get rid of this 2y on the right-hand side, and we can do that by subtracting 2y from both sides. So we're going to subtract 2y from both sides. The left-hand side, we have 3y minus 2y, which is just y, plus 7 is less than 2y minus 2y. And there's nothing else there. That's just going to be 0. And then we can get rid of this 7 here by subtracting 7 from both sides. So let's subtract 7 from both sides. Left-hand side, y plus 7 minus 7. Those cancel out. We just have y is less than 0 minus 7, which is negative 7. That's this constraint right over here. Now let's work on this constraint. We have 4y plus 8 is greater than negative 48. So let's get rid of the 8 from the left-hand side. So we can subtract 8 from both sides. The left-hand side, we're just left with a 4y because these guys cancel out. 4y is greater than negative 48 minus 8. So we're going to go another 8 negative. So 48 plus 8 would be a 56, so this is going to be negative 56. And now to isolate the y, we can divide both sides by positive 4, and we don't have to swap the inequality since we're dividing by a positive number. So it's divide both sides by 4 over here. So we get y is greater than-- what is 56/4, or negative 56/4? So it's 14 times 4. So y is greater than negative 14. Is that right? 4 times 10 is 40, 4 times 4 is 16. Yep, 56. So y is greater than negative 14 and-- let's remember, we have this \"and\" here-- and y is less than negative 7. So we have to meet both of these constraints over here. So let's draw them on the number line. So I have my number line over here. And let's say negative 14 is over here. So you have negative 13, 12, 11, 10, 9, 8, 7-- that's negative 7-- and then negative 6, 5, 4, 3, 2, 1. This would be 0, and then you could keep going up more positive. And so we're looking for all of the y's that are less than negative 7." + }, + { + "Q": "\nAt 0:58 Sal talks about \"when x is 0, y is 3 - that's our y intercept\" and then talks about how the slope goes down from there. I've been following everything I can on geometry but I seemed to have missed exactly how these slopes work. Is there another unit I can look at that describes how the whole y = 3 - x thing works?", + "A": "I would suggest looking up equations of a line and slope-intercept form on the KA search bar.", + "video_name": "-nufZ41Kg5c", + "timestamps": [ + 58 + ], + "3min_transcript": "Two of the points that define a certain quadrilateral are 0 comma 9 and 3 comma 4. The quadrilateral is left unchanged by a reflection over the line y is equal to 3 minus x. Draw and classify the quadrilateral. Now, I encourage you to pause this video and try to draw and classify it on your own before I'm about to explain it. So let's at least plot the information they give us. So the point 0 comma 9, that's one of the vertices of the quadrilateral. So 0 comma 9. That's that point right over there. And another one of the vertices is 3 comma 4. That's that right over there. And then they tell us that the quadrilateral is left unchanged by reflection over the line y is equal to 3 minus x. So when x is 0, y is 3-- that's our y-intercept-- and it has a slope of negative 1. You could view this as 3 minus 1x. So the line looks like this. So every time we increase our x by 1, we decrease our y by 1. So the line looks something like this. y is equal to 3 minus x. Try to draw it relatively, pretty carefully. So that's what it looks like. y is equal to 3 minus x. So that's my best attempt at drawing it. y is equal to 3 minus x. So the quadrilateral is left unchanged by reflection over this. So that means if I were to reflect each of these vertices, I would, essentially, end up with one of the other vertices on it, and if those get reflected you're going to end up with one of these so the thing is not going to be different. So let's think about where these other two vertices of this quadrilateral need to be. So this point, let's just reflect it over this line, over y is equal to 3 minus x. So if we were to try to drop a perpendicular to this line-- notice, we have gone diagonally across one, need to go diagonally across three of them on the left-hand side. So one, two, three gets us right over there. This is the reflection of this point across that line. Now, let's do the same thing for this blue point. To drop a perpendicular to this line, we have to go diagonally across two of these squares. So let's go diagonally across two more of these squares just like that to get to that point right over there. And now we've defined our quadrilateral. Our quadrilateral looks like this. Both of these lines are perpendicular to that original line, so they're going to have the same slope. So that line is parallel to that line over there. And then we have this line and then we have this line. So what type of quadrilateral is this? Well, I have one pair of parallel sides," + }, + { + "Q": "Isn't the remainder 21 instead of 31 at 7:12?\n", + "A": "It is 31, 31 + 291 = 322", + "video_name": "omFelSZvaJc", + "timestamps": [ + 432 + ], + "3min_transcript": "So it seems that we can get pretty close if we do 291 times 6, so if you do a 1,746 and then add two zeros to it. This is going to be times 6 with two zeros, so this is times 600. Once again, you subtract. And let's say I'm only using the sixes and the threes, because I figured those out ahead of time, so I didn't have to do any extra math. So 2 minus 0 is 2. 5 minus 0 is 5. 9 minus 6 is 3. 0 minus 4-- well, there's a couple of ways you could think about doing this. You could borrow from here. That will become a 6. This becomes a 10. 10 minus 4 is 6. Now this one's lower, so it has to borrow as well. Make this into a 16. 16 minus 7-- and I have multiple videos on how to borrow, if I'm doing that part too fast. But the idea here is to show you a different way of long division. So 16 minus 7 is 9. So now we're at 96,352. is about as close as we can get. So let me put a 873 over here with two zeros. So that would literally be 291 times 3 with two zeros times 300. And so once again, we want to subtract here. 2 minus 0-- you get a 2, a 5, a 0. Make this a 16. Make this an 8. 16 minus 7 is 9. And then we have to get close to 9,052. Once again, that 873, those digits look pretty good, 873. We have to multiply 3 and then 10, so this is going to be times 30 right over here. We subtract again. 2 minus 0 is 2. 5 minus 3 is 2. And then you have 90 minus 87 is 3. I'm doing the subtraction a little fast, just so that we can get the general idea. Then we have to go into 322. And how can we get close to that? Well, actually, 291 is pretty darn close to that. 1 times 291 is 291. 2 minus 1 is 1. 32 minus 29 is 3. So you have a remainder of 31. 291 cannot go into 31 any more, so that's our remainder. But how many times did it actually go into this big, beastly number? This 9,873,952? Well there, we just have to add up all of these right over here. 30 plus 3,000-- we can even do it in our head-- 30 plus 3,000 is 33,000. 33,600, 33,900, 33,931, and we are done, assuming I haven't made some silly mistake. 291 goes into this thing 33,931 times with a remainder of 31." + }, + { + "Q": "\nAt 6:55: isn't that shape an octagon instead of a decagon? Isn't Sal supposed to connect the far right purple line with the far right yellow line? (And do the same with the left?)", + "A": "It s just, if you drew a square with the top first and the bottom next, you wouldn t say you have a 6 sided figure.", + "video_name": "qG3HnRccrQU", + "timestamps": [ + 415 + ], + "3min_transcript": "So we can assume that s is greater than 4 sides. Let's say I have an s-sided polygon, and I want to figure out how many non-overlapping triangles will perfectly cover that polygon. How many can I fit inside of it? And then I just have to multiply the number of triangles times 180 degrees to figure out what are the sum of the interior angles of that polygon. So let's figure out the number of triangles as a function of the number of sides. So once again, four of the sides are going to be used to make two triangles. So those two sides right over there. And then we have two sides right over there. I can draw one triangle over-- and I'm not even going to talk about what happens on the rest of the sides of the polygon. You could imagine putting a big black piece of construction paper. There might be other sides here. I'm not going to even worry about them right now. So out of these two sides I can draw one triangle, just like that. Out of these two sides, I can draw another triangle right over there. So four sides used for two triangles. I've already used four of the sides, but after that, if I have all sorts of craziness here. I could have all sorts of craziness here. Let me draw it a little bit neater than that. So I could have all sorts of craziness right over here. It looks like every other incremental side I can get another triangle out of it. So that's one triangle out of there, one triangle out of that side, one triangle out of that side, one triangle out of that side, and then one triangle out of this side. So for example, this figure that I've drawn is a very irregular-- one, two, three, four, five, six, seven, eight, nine, 10. Is that right? One, two, three, four, five, six, seven, eight, nine, 10. It is a decagon. And in this decagon, four of the sides were used for two triangles. So I got two triangles out of four of the sides. And out of the other six sides I was These are six. This is one, two, three, four, five. Actually, let me make sure I'm counting the number of sides right. So I have one, two, three, four, five, six, seven, eight, nine, 10. So let me make sure. Did I count-- am I just not seeing something? Oh, I see. I actually didn't-- I have to draw another line right over These are two different sides, and so I have to draw another line right over here. I can get another triangle out of that right over there. And so there you have it. I have these two triangles out of four sides. And out of the other six remaining sides I get a triangle each. So plus six triangles. I got a total of eight triangles. And so we can generally think about it. The first four, sides we're going to get two triangles. So let me write this down. So our number of triangles is going to be equal to 2." + }, + { + "Q": "at 4:46 the hexagon looks like cone! : )\n", + "A": "Hexagons can be in any form as long as it has the right amount of sides and is connected", + "video_name": "qG3HnRccrQU", + "timestamps": [ + 286 + ], + "3min_transcript": "of the polygon as a whole. And to see that, clearly, this interior angle is one of the angles of the polygon. This is as well. But when you take the sum of this one and this one, then you're going to get that whole interior angle of the polygon. And when you take the sum of that one and that one, you get that entire one. And then when you take the sum of that one plus that one plus that one, you get that entire interior angle. So if you take the sum of all of the interior angles of all of these triangles, you're actually just finding the sum of all of the interior angles of the polygon. So in this case, you have one, two, three triangles. So three times 180 degrees is equal to what? 300 plus 240 is equal to 540 degrees. Now let's generalize it. And to generalize it, let's realize that just to get our first two triangles, we have to use up four sides. We have to use up all the four sides in this quadrilateral. We had to use up four of the five sides-- One, two, and then three, four. So four sides give you two triangles. And it seems like, maybe, every incremental side you have after that, you can get another triangle out of it. Let's experiment with a hexagon. And I'm just going to try to see how many triangles I get out of it. So one, two, three, four, five, six sides. I get one triangle out of these two sides. One, two sides of the actual hexagon. I can get another triangle out of these two sides of the actual hexagon. And it looks like I can get another triangle out of each of the remaining sides. So one out of that one. And then one out of that one, right over there. So in general, it seems like-- let's say. So let's say that I have s sides. s-sided polygon. And I'll just assume-- we already So we can assume that s is greater than 4 sides. Let's say I have an s-sided polygon, and I want to figure out how many non-overlapping triangles will perfectly cover that polygon. How many can I fit inside of it? And then I just have to multiply the number of triangles times 180 degrees to figure out what are the sum of the interior angles of that polygon. So let's figure out the number of triangles as a function of the number of sides. So once again, four of the sides are going to be used to make two triangles. So those two sides right over there. And then we have two sides right over there. I can draw one triangle over-- and I'm not even going to talk about what happens on the rest of the sides of the polygon. You could imagine putting a big black piece of construction paper. There might be other sides here. I'm not going to even worry about them right now. So out of these two sides I can draw one triangle, just like that. Out of these two sides, I can draw another triangle right over there. So four sides used for two triangles." + }, + { + "Q": "where did the 4 come from in the s-4 @ 8:07?\n", + "A": "He states that the first 4 sides of the shapes will produce 2 triangles, then after the 4 sides, every side will add an additional triangle, So he adds two in the beginning (my two triangles) then the 5th side would add one (5-4), a 6th side would add two (6-4), etc. The s-4 just shows the first four sides creating the 2 triangles.", + "video_name": "qG3HnRccrQU", + "timestamps": [ + 487 + ], + "3min_transcript": "I've already used four of the sides, but after that, if I have all sorts of craziness here. I could have all sorts of craziness here. Let me draw it a little bit neater than that. So I could have all sorts of craziness right over here. It looks like every other incremental side I can get another triangle out of it. So that's one triangle out of there, one triangle out of that side, one triangle out of that side, one triangle out of that side, and then one triangle out of this side. So for example, this figure that I've drawn is a very irregular-- one, two, three, four, five, six, seven, eight, nine, 10. Is that right? One, two, three, four, five, six, seven, eight, nine, 10. It is a decagon. And in this decagon, four of the sides were used for two triangles. So I got two triangles out of four of the sides. And out of the other six sides I was These are six. This is one, two, three, four, five. Actually, let me make sure I'm counting the number of sides right. So I have one, two, three, four, five, six, seven, eight, nine, 10. So let me make sure. Did I count-- am I just not seeing something? Oh, I see. I actually didn't-- I have to draw another line right over These are two different sides, and so I have to draw another line right over here. I can get another triangle out of that right over there. And so there you have it. I have these two triangles out of four sides. And out of the other six remaining sides I get a triangle each. So plus six triangles. I got a total of eight triangles. And so we can generally think about it. The first four, sides we're going to get two triangles. So let me write this down. So our number of triangles is going to be equal to 2. So the remaining sides I get a triangle each. So the remaining sides are going to be s minus 4. So the number of triangles are going to be 2 plus s minus 4. 2 plus s minus 4 is just s minus 2. So if I have an s-sided polygon, I can get s minus 2 triangles that perfectly cover that polygon and that don't overlap with each other, which tells us that an s-sided polygon, if it has s minus 2 triangles, that the interior angles in it are going to be s minus 2 times 180 degrees. Which is a pretty cool result. So if someone told you that they had a 102-sided polygon-- so s is equal to 102 sides. You can say, OK, the number of interior angles are going to be 102 minus 2. So it's going to be 100 times 180 degrees, which is equal to 180 with two more zeroes behind it." + }, + { + "Q": "At about 5:20 Sal says \"Mu of X\", this makes it sound like Mu is some function of X, which may or may not end up being true, he should be saying \"Mu times X\". Sal makes this mistake twice when verbalizing the equation.\n", + "A": "You are right, it s a slight mistake when verbalizing the equations.", + "video_name": "jJyRrIZ595c", + "timestamps": [ + 320 + ], + "3min_transcript": "algebraic stamina, but as long you don't make careless mistakes you'll find it reasonably rewarding, because you'll see where things are coming from. So we get y-- the general solution is y is equal to e to the lambda x, times-- let's add up the two cosine mu x terms. So it's c1 plus c2 times cosine of mu x. And let's add the two sine of mu x terms. So plus i-- we could call that c1i-- that's that-- minus c2i times sine of mu x. And we're almost done simplifying. And the last thing we can simplify is-- well you know c1 and c2 are arbitrary constants. I don't know, let's call it-- I'll just call it c3, just to not confuse you by using c1 twice, I'll call this c3. And now this might be a little bit of a stretch for you, but if you think about it, it really makes sense. This is still just a constant, right? Especially if I say, you know what, I'm not restricting the constants to the reals. c could be an imaginary number. So if c is an imaginary number, or some type of complex number, we don't even know whether this is necessarily an imaginary number. So we're not going to make any assumptions about it. Let's just say that this is some other arbitrary constant. Call this c4, and we can worry about it when we're actually given the initial conditions. But what this gives us, if we make that simplification, we actually get a pretty straightforward, general solution to our differential equation, where the characteristic equation has complex roots. And that I'll do it in a new color. That is y is equal to e to the lambda x, times some It could be c a hundred whatever. Some constant times cosine of mu of x, plus some other constant-- and I called it c4, doesn't have to be c4, I just didn't want to confuse it with these-- plus some other constant times the sine of mu of x. So there's really two things I want you to realize. One is, we haven't done anything different. At the end of the day, we still just took the two roots and substituted it back into these equations for r1 and r2. The difference is, we just kept algebraically simplifying it so that we got rid of the i's. There was really nothing new here except for some algebra, and the use of Euler's formula. But when r1 and r2 involved complex numbers, we got to this simplification." + }, + { + "Q": "Why \"dv\" at 8:30?\n", + "A": "because you re summing very small bits of volume. (v for volume)", + "video_name": "XyiQ2dwJHXE", + "timestamps": [ + 510 + ], + "3min_transcript": "have positive divergence of our vector field within this region right over here. So we have positive divergence. So you can imagine that it's kind of-- the vector field within the region, it's a source of the vector field, or the vector field is diverging out. That's just the case I drew right over here. And the other thing we want to say about vector field S, it's oriented in a way that its normal vector is outward facing, so outward normal vector. So the normal, it's oriented so that the surface-- the normal vector is like that. The other option is that you have an inward-facing normal vector. But we're assuming it's an outward-facing N. Well, then we just extrapolate this to three dimensions. We essentially say the flux across the surface. So the flux across the surface, you would take your vector field, dot it with the normal vector at the surface, so multiply that times a little chunk of surface, and then sum it up along the whole surface, so sum it up. So it's going to be a surface integral. So this is flux across the surface. It's going to be equal to-- if we were to sum up the divergence, if we were to sum up across the whole volume, so now if we're summing up things on every little chunk of volume over here in three dimensions, we're going to have to take integrals along each dimension. So it's going to be a triple integral over the region of the divergence of F. So we're going to say, how much is F? What is the divergence at F at each point? And then multiply it times the volume of that little chunk to sense of how much is it totally diverging in that volume. That should be equal to the flux. It's completely analogous to what's here. Here we had a flux across the line. We had essentially a two-dimensional-- or I guess we could say it's a one-dimensional boundary, so flux across the curve. And here we have the flux across a surface. Here we were summing the divergence in the region. Here we're summing it in the volume. But it's the exact same logic. If you had a vector field like this that was fairly constant going through the surface, on one side you would have a negative flux. On the other side, you would have a positive flux, and they would roughly cancel out. And that makes sense, because there would be no diverging going on. If you had a converging vector field, where it's coming in, the flux would be negative, because it's going in the opposite direction of the normal vector. And so the divergence would be negative as well, because essentially the vector field would be converging. So hopefully this gives you an intuition of what the divergence theorem is actually saying something very, very, very, very-- almost common sense or intuitive." + }, + { + "Q": "At 6:46 Sal says -45/45 = 1. Is -45/45=-1?\n", + "A": "Listen more carefully to what he says. He says you divide 45/45 and you get 1 and you are left with -1 ... When you are doing operations, it is easier to think about the signs and numbers separate from each other and then combine them in the answer which is what he did", + "video_name": "O3jvUZ8wvZs", + "timestamps": [ + 406 + ], + "3min_transcript": "(typing) We care about how many radians there are per degree. We'll do that same green color. Per degree. How many radians are there per degree? Well, we already know, there's pi radians for every 180 degrees, or there are pi... Let me do that yellow color. There are pi over 180 radians per degree. And so, if we multiply, and this all works out because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians. So what do we get? This becomes, let me just rewrite it. 150 times pi. All of that over 180, And so, if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like, 30. So if you divide the numerator by 30, you get five. You divide the denominator by 30, you get six. So you get five pi over six radians, or 5/6 pi radians, depending how you wanna do it. Now let's do the same thing for negative 45 degrees. What do you get for negative 45 degrees if you were to convert that to radians? Same exact process. You have negative, and I'll do this one a little quicker. Negative 45 degrees. I'll write down the word. Times, times pi radians, pi radians for every 180 degrees. The degrees cancel out, and you're left So this is equal to negative 45 pi over 180, over 180 radians. How can we simplify this? Well it looks like they're both, at minimum, divisible by nine, nine times five is 45, this is nine times 20, so actually it's gonna be divisible by more than just, let's see... Actually, they're both divisible by 45. What am I doing? If you divide the numerator by 45, you get one. You divide the denominator by 45, 45 goes into 180 four times. You're left with negative pi over four radians. This is equal to negative pi over four radians. And we are done." + }, + { + "Q": "At 6:20, why does Salman call X and B vectors?\n", + "A": "ya why does it call it that.... is it just to confuse us all", + "video_name": "EC2mgUZyzoA", + "timestamps": [ + 380 + ], + "3min_transcript": "is equal to 7. All I did is I multiplied, I dealt with the first row, first column and said, when I take essentially the dot product of those, and if you don't know what a dot product is, don't worry. We'll explain it at other places. It's essentially what I just did here, the first entry here times the first entry, the second entry here times the second entry, and we add them together, that that must be equal to 7, but when you do that, you essentially construct this first equation. Now when you do it with the second row and this column, you construct the second equation. You get negative 2 times s, negative 2 times s plus 4 times t, 4 times t, is equal to negative 6. Hopefully you appreciate that this contains the same information as that. And there is other ways that I could have done it. For example, you could have, instead of writing it this way, this system is obviously the same thing, and actually, let me just copy and paste it, is the same thing as ... so copy and paste, is the same thing as this system, where I'm really just swapping. So once again, copy and paste. Obviously, I've written the second one first, and I've written the first one second, so this is obviously the same system. If I wanted to construct a matrix equation with this system, I would just swap all of the rows. The first row here would be negative 2, 4. I would swap the rows for the coefficients, but I would still keep the s and ts in the same order, and you could do that. Try to represent this right over here as a matrix equation. You would have the matrix here would be negative 2, 4, 2, negative 5, and this would be negative 6, 7. But now that we have set this up, how do we actually solve something like this? Why do we even do this? To think about this, let's actually think about it in terms of literally a matrix equation. This thing over here is the matrix A. Let's say that this right over here, this is the column vector x. I'll write it as a vector x right over here. You have the column vector x, and then this right over here, you could say that this is equal, and let's call this the column vector b. This is equal to the column vector b. We're essentially saying that A, the matrix A times the column vector x is equal to, is equal to the column vector b. Let me write that again right over here, just to emphasize it. The matrix A times the column vector x is going to be equal to, is equal to the column vector B. This is what they're talking about when they say a matrix equation. Actually, before we even think about computation and computer graphics and all of that, you will see a lot of things like this in physics," + }, + { + "Q": "I don't understand why Khan wrote at 05:34 wrote 3*(7/16) after 55/16, would someone please explain it?\n", + "A": "Sal just converted the improper fraction 55/16 into the mixed number 3 plus 7/16 . If that process is confusing to you, try looking up the videos and exercise about improper fractions and mixed numbers.", + "video_name": "pPnxPrhf6Ww", + "timestamps": [ + 334 + ], + "3min_transcript": "plus 7. And now, how many data points did we have? We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 data points. So we're going to divide this by 16. So what is this going to be? This is just 0. Let's see. This is, just right over, that's 0. This is 4. This is 15. This is 12. This is 10. So we have 1 plus 4 is 5 plus 15 is 20 plus 12 is 32 plus 10 is 42. 42 plus 6 is 48, 48. Am I doing ... 42 plus 6 is 48 plus 7, 48 plus 7 is 55. Did I do that right? 1 plus 4 is 5 plus 15 is 20, 42 plus 13 is 55. So this is equal to 55 over 16, which is the same thing as, let's see, that's the same thing as 3 and 3 that ... 3 times 16 is 48, so 3 and 7/16. So the mean for the seniors, 3 and 7/16, that's right around ... This is 3, that's 4, so 7/16, it's a little less than a half. It's right around there. So the mean number of fruits is defnitely greater for the freshmen. They have 4 ... Their mean number of fruit eaten per day is 4 versus 3 and 7/16. The mean is a good measure for the center of the distribution of. So when we think about whether it's freshmen or seniors, the mean is fairly sensitive to when you have outliers here. 19 pieces of fruit per day. That's an enormous amount of fruit. They must be only eating fruit. You can imagine if it was even a bigger number, if someone was eating 20 or 30 pieces of fruit, just that one data point will skew the entire mean upwards. That wouldn't be the effect on the mode because the mode is a middle number. Even if you change this one point all the way out here, it's not going to change what the middle number is. So the mean is more sensitive to these outliers, to these really, these points that are really, really high, really, really low. And because the seniors don't seem to have any outliers like that, I would say that the mean is a good measure for the center of distribution for the seniors, or a better measure for the center of distribution for the seniors. Let's fill both of those out. The mean number of fruit is greater for the freshmen, and the mean is a good measure for the center of distribution for the seniors." + }, + { + "Q": "\nDid you (Khan) mean to say the median is the middle number at 6:24? It's hard to tell what you meant, because the mode is the same as the median: 3. But it seems odd to call the mode the middle number.", + "A": "Median is the middle number, and the mode is the most commonly occurring number. (Occurs the most in a data set) The mode can be the same as the median if the middle number is also the most commonly occurring number. Does this clear things up?", + "video_name": "pPnxPrhf6Ww", + "timestamps": [ + 384 + ], + "3min_transcript": "plus 7. And now, how many data points did we have? We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 data points. So we're going to divide this by 16. So what is this going to be? This is just 0. Let's see. This is, just right over, that's 0. This is 4. This is 15. This is 12. This is 10. So we have 1 plus 4 is 5 plus 15 is 20 plus 12 is 32 plus 10 is 42. 42 plus 6 is 48, 48. Am I doing ... 42 plus 6 is 48 plus 7, 48 plus 7 is 55. Did I do that right? 1 plus 4 is 5 plus 15 is 20, 42 plus 13 is 55. So this is equal to 55 over 16, which is the same thing as, let's see, that's the same thing as 3 and 3 that ... 3 times 16 is 48, so 3 and 7/16. So the mean for the seniors, 3 and 7/16, that's right around ... This is 3, that's 4, so 7/16, it's a little less than a half. It's right around there. So the mean number of fruits is defnitely greater for the freshmen. They have 4 ... Their mean number of fruit eaten per day is 4 versus 3 and 7/16. The mean is a good measure for the center of the distribution of. So when we think about whether it's freshmen or seniors, the mean is fairly sensitive to when you have outliers here. 19 pieces of fruit per day. That's an enormous amount of fruit. They must be only eating fruit. You can imagine if it was even a bigger number, if someone was eating 20 or 30 pieces of fruit, just that one data point will skew the entire mean upwards. That wouldn't be the effect on the mode because the mode is a middle number. Even if you change this one point all the way out here, it's not going to change what the middle number is. So the mean is more sensitive to these outliers, to these really, these points that are really, really high, really, really low. And because the seniors don't seem to have any outliers like that, I would say that the mean is a good measure for the center of distribution for the seniors, or a better measure for the center of distribution for the seniors. Let's fill both of those out. The mean number of fruit is greater for the freshmen, and the mean is a good measure for the center of distribution for the seniors." + }, + { + "Q": "in 3:20-4:05 in the video, why is 6 and 7 not with a decimal because 2 had a . which is the amount behind it so why not 6 or 7? Is there a reason for it or is it just because the number has only one dot?\n", + "A": "There isn t any decimal points, those dots are multiplication as a different symbol.", + "video_name": "pPnxPrhf6Ww", + "timestamps": [ + 200, + 245 + ], + "3min_transcript": "And then we have two data points at 2, so you write plus 2 times 2. And then, let's see, we have a bunch of data. We have four data points at 3, so we could say we have four 3s. Let me circle that. So we have four 3s, plus 4 times 3. And then we have three 4s, so plus 3 times 4. And then we have a 5, so plus 5, and then we have a 6. Let me do this in a color that you can see. And then we have a 6 right over here, plus 6. How many total points did we have? We had 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, oh, actually, be careful. We had 15 points and I didn't put that one in there. So we have 15 points, and I can't forget this one over here, so plus ... my pen is acting a little funny right now, but we'll power through that, plus 19. So what is this going to be? This is just going to be 0. This is going to be 2. This is going to be 4. This is going to be 12. My pen is really acting up. It's almost like it's running out of digital ink or something. This is going to be another 12, and then we have 5, 6, and 19. 2 plus 4 is 6, plus 24 is 30, plus 11 is 41, plus 19 gets us to 60. 60 divided by 15 is 4, so the mean number of fruit per day for the freshmen is 4 pieces of fruit per day. This right over here, that right over there is our mean for the ... Now let's do the same calculation for the seniors. We have one data point where they didn't eat any fruit at all each day, not too healthy. Then you have one 1, so I'll just write that as, we could actually write that as 1 times 1, but I'll just write that as 1. Then we have two 2s, so plus 2 times 2. Then we have one, two , three, four, five 3s, five 3s, so plus 5 times 3. And then we have three 4s, so plus 3 times 4. And then we have two 5s, plus 2 times 5, and then we have a 6. We have a 6, plus 6, and we have a 7, someone eats 7 pieces of fruit each day," + }, + { + "Q": "\nWhy does he have to multiply from 1:04 to 1:36?", + "A": "at those points, multiple people ate the same number of fruit. so he was calculating each person s contribution to the arithmetic mean.", + "video_name": "pPnxPrhf6Ww", + "timestamps": [ + 64, + 96 + ], + "3min_transcript": "Voiceover:Kenny interviewed freshmen and seniors at his high school, asking them how many pieces of fruit they eat each day. The results are shown in the 2 plots below. The first statement that we have to complete is the mean number of fruits is greater for, and actually, let me go down the actual screen, is greater for, we have to pick between freshmen and seniors. Then they said the mean is a good measure for the center of distribution of, and we pick either freshmen or seniors. Let me go back to my scratch pad here, and let's think about this. Let's first think about the first part. Let's just calculate the mean for each of these distributions. I encourage you to pause the video and try to calculate it out on your own. Let's first think about the mean number of fruit for freshmen. Essentially, we're just going to take each of these data points, add them all together, and then divide by the number of data points that we have. We have one data point at 0. We have one data point at 0, so I'll write 0. And then we have two data points at 1, And then we have two data points at 2, so you write plus 2 times 2. And then, let's see, we have a bunch of data. We have four data points at 3, so we could say we have four 3s. Let me circle that. So we have four 3s, plus 4 times 3. And then we have three 4s, so plus 3 times 4. And then we have a 5, so plus 5, and then we have a 6. Let me do this in a color that you can see. And then we have a 6 right over here, plus 6. How many total points did we have? We had 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, oh, actually, be careful. We had 15 points and I didn't put that one in there. So we have 15 points, and I can't forget this one over here, so plus ... my pen is acting a little funny right now, but we'll power through that, plus 19. So what is this going to be? This is just going to be 0. This is going to be 2. This is going to be 4. This is going to be 12. My pen is really acting up. It's almost like it's running out of digital ink or something. This is going to be another 12, and then we have 5, 6, and 19. 2 plus 4 is 6, plus 24 is 30, plus 11 is 41, plus 19 gets us to 60. 60 divided by 15 is 4, so the mean number of fruit per day for the freshmen is 4 pieces of fruit per day. This right over here, that right over there is our mean for the ..." + }, + { + "Q": "At 1:01 how does he know whether to subtract or add the terms? Would you use the sign in front of or behind the term?\n", + "A": "Yes you use the sign in front of the term. In the example Sal uses 5x^2 + 2x^2 it results in 7x^2 however later in the question we get a +8x, -7x, +13x using the signs 8x-7x = x, x+13x = 14x hope that helps :)", + "video_name": "ahdKdxsTj8E", + "timestamps": [ + 61 + ], + "3min_transcript": "We're asked to simplify 5x squared plus 8x minus 3 plus 2x squared minus 7x plus 13x. So really, all we have to do is we have to combine like terms-- terms that have x raised to the same power. And the first thing we can do, we can actually get rid of these parentheses right here, because we have this whole expression, and then we're adding it to this whole The parentheses really don't change our order of operations here. So let me just rewrite it once without the parentheses. So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a minus then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x. Let's start with the x squared terms. So you have a 5x squared term here and you have a 2x squared term right there. So 5 of something plus 2 of that same something is going So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right there. We have a minus 7x. And then we have a plus 13x. So if you have 8 of something minus 7 of something, you're just going to have 1 of that something. And then if you add 14 of that something more, you're going to 15. So this is going to be plus 15x. 8x minus 7x-- oh, sorry. You're going to have 14x. 8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms. 8x minus 7x plus 13x. And then finally, you have a negative 3-- or minus 3, depending on how you want to view it. And that's the only constant term. You could say it's x times x to the 0. But it's a constant term. It's not be multiplied by x. And that's the only one there, so minus 3. And we've simplified it as far as we can go. We are done." + }, + { + "Q": "At 0:41 he says that he would have to distribute the negative; what is the process/operation for doing that?\n", + "A": "Multiply the polynomial by -1. This is what he is talking about, I think. :)", + "video_name": "ahdKdxsTj8E", + "timestamps": [ + 41 + ], + "3min_transcript": "We're asked to simplify 5x squared plus 8x minus 3 plus 2x squared minus 7x plus 13x. So really, all we have to do is we have to combine like terms-- terms that have x raised to the same power. And the first thing we can do, we can actually get rid of these parentheses right here, because we have this whole expression, and then we're adding it to this whole The parentheses really don't change our order of operations here. So let me just rewrite it once without the parentheses. So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a minus then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x. Let's start with the x squared terms. So you have a 5x squared term here and you have a 2x squared term right there. So 5 of something plus 2 of that same something is going So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right there. We have a minus 7x. And then we have a plus 13x. So if you have 8 of something minus 7 of something, you're just going to have 1 of that something. And then if you add 14 of that something more, you're going to 15. So this is going to be plus 15x. 8x minus 7x-- oh, sorry. You're going to have 14x. 8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms. 8x minus 7x plus 13x. And then finally, you have a negative 3-- or minus 3, depending on how you want to view it. And that's the only constant term. You could say it's x times x to the 0. But it's a constant term. It's not be multiplied by x. And that's the only one there, so minus 3. And we've simplified it as far as we can go. We are done." + }, + { + "Q": "\ndo yo know about the math error starting at about 1:11 in the video?", + "A": "There is no error at that point in the video. Sal briefly makes an error later on at about 1:28 when he writes 15x, but he quickly corrects it to 14x. Other than that, there are no errors in the video. What do you think is in error? Maybe I can help clarify.", + "video_name": "ahdKdxsTj8E", + "timestamps": [ + 71 + ], + "3min_transcript": "We're asked to simplify 5x squared plus 8x minus 3 plus 2x squared minus 7x plus 13x. So really, all we have to do is we have to combine like terms-- terms that have x raised to the same power. And the first thing we can do, we can actually get rid of these parentheses right here, because we have this whole expression, and then we're adding it to this whole The parentheses really don't change our order of operations here. So let me just rewrite it once without the parentheses. So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a minus then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x. Let's start with the x squared terms. So you have a 5x squared term here and you have a 2x squared term right there. So 5 of something plus 2 of that same something is going So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right there. We have a minus 7x. And then we have a plus 13x. So if you have 8 of something minus 7 of something, you're just going to have 1 of that something. And then if you add 14 of that something more, you're going to 15. So this is going to be plus 15x. 8x minus 7x-- oh, sorry. You're going to have 14x. 8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms. 8x minus 7x plus 13x. And then finally, you have a negative 3-- or minus 3, depending on how you want to view it. And that's the only constant term. You could say it's x times x to the 0. But it's a constant term. It's not be multiplied by x. And that's the only one there, so minus 3. And we've simplified it as far as we can go. We are done." + }, + { + "Q": "\nAt 6:42, why did Sal put a negative sign for cosine but not sine, aren't the ratios the same?", + "A": "In the second and third quadrant, cosine values are always negative. When sine is being taken in the third and fourth quadrant, the sine values will be negative. Does this help?", + "video_name": "tzQ7arA917E", + "timestamps": [ + 402 + ], + "3min_transcript": "That's a pretty interesting result. But what about their sines? Well, here, the sine of theta is this distance above the X-axis, and here, the sine of negative theta is the same distance below the X-axis, so they're going to be the negatives of each other. We could say that sine of negative theta, sine of negative theta is equal to, is equal to the negative sine of theta, equal to the negative sine of theta. It's the opposite. If you go the same amount above or below the X-axis, you're going to get the negative value for the sine. We could do the same thing over here. How does this one relate to that? These two are going to have the same sine values. The sine of this, the Y-coordinate, is the same as the sine of that. We see that this must be equal to that. Let's write that down. Now let's think about how do the cosines relate. The same argument, they're going to be the opposites of each other, where the X-coordinates are the same distance but on opposite sides of the origin. We get cosine of theta is equal to the negative of the cosine of ... let me do that in same color. Actually, let me make sure my colors are right. We get cosine of theta is equal to the negative of the cosine of pi minus theta. Now finally, let's think about how this one relates. Here, our cosine value, our X-coordinate is the negative, and our sine value is also the negative. We've flipped over both axes. Let's write that down. which is the same thing as pi plus theta, is equal to the negative of the sine of theta, and we see that this is sine of theta, this is sine of pi plus theta, or sine of theta plus pi, and we get the cosine of theta plus pi. Cosine of theta plus pi is going to be the negative of cosine of theta, is equal to the negative of cosine of theta. Even here, and you could see, You could try to relate this one to that one or that one to that one. You can get all sorts of interesting results. I encourage you to really try to think this through on your own and think about how all of these are related to each other based on essentially symmetries or reflections around the X or Y-axis." + }, + { + "Q": "At 2:43, you said, 'this angle right over here is theta'. I don't understand why, shouldn't it be negative theta? Thank you for the video anyway; it was really informative!\n", + "A": "It s going down, and it s below the x axis, but notice that it s still going counterclockwise, so it is positive. On the unit circle, the angle is not x or y in the xy plane. It s just stuck in there and goes around. It has its own relationships to x and y. It s not graphed in the xy plane like the circle is.", + "video_name": "tzQ7arA917E", + "timestamps": [ + 163 + ], + "3min_transcript": "so we would call this, by our convention, an angle of negative theta. Now let's flip our original green ray. Let's flip it over the positive Y-axis. If you flip it over the positive Y-axis, we're going to go from there all the way to right over there then we can draw ourselves a ray. My best attempt at that is right over there. What would be the measure of this angle right over here? What was the measure of that angle in radians? We know if we were to go all the way from the positive X-axis to the negative X-axis, that would be pi radians because that's halfway around the circle. This angle, since we know that that's theta, this is theta right over here, the angle that we want to figure out, this is going to be all the way around. It's going to be pi minus, Notice, pi minus theta plus theta, these two are supplementary, and they add up to pi radians or 180 degrees. Now let's flip this one over the negative X-axis. If we flip this one over the negative X-axis, you're going to get right over there, and so you're going to get an angle that looks like this, that looks like this. Now what is going to be the measure of this angle? If we go all the way around like that, what is the measure of that angle? To go this far is pi, and then you're going another theta. This angle right over here is theta, so you're going pi plus another theta. This whole angle right over here, this whole thing, this whole thing is pi plus theta radians. Pi plus theta, let me just write that down. This is pi plus theta. Now that we've figured out let's think about how the sines and cosines of these different angles relate to each other. We already know that this coordinate right over here, that is sine of theta, sorry, the X-coordinate is cosine of theta. The X-coordinate is cosine of theta, and the Y-coordinate is sine of theta. Or another way of thinking about it is this value on the X-axis is cosine of theta, and this value right over here on the Y-axis is sine of theta. Now let's think about this one down over here. By the same convention, this point, this is really the unit circle definition of our trig functions. This point, since our angle is negative theta now, this point would be cosine of negative theta, comma, sine of negative theta. And we can apply the same thing over here. This point right over here, the X-coordinate is cosine of pi minus theta." + }, + { + "Q": "at 4:35 how can the angle in yellow be equal to cos(pie minus theta),sin( pie minus theta)\nand how can it be equal to sin of theta and cos of theta at 6:09\n", + "A": "Be careful. Sal isn t saying that the angle equals cos or sine. Sal is labelling the x and y co-ordinates for points where the terminal ray of the angle in question intersects the unit circle. So, for the yellow angle of ( \u00cf\u0080 - theta ), its x co-ordinate is cos ( \u00cf\u0080 - theta) and its x co-ordinate is sin ( \u00cf\u0080 - theta. Likewise for the green angle of theta. Its x co-ordinate is cos (theta) and its y co-ordinate is sin (theta).", + "video_name": "tzQ7arA917E", + "timestamps": [ + 275, + 369 + ], + "3min_transcript": "let's think about how the sines and cosines of these different angles relate to each other. We already know that this coordinate right over here, that is sine of theta, sorry, the X-coordinate is cosine of theta. The X-coordinate is cosine of theta, and the Y-coordinate is sine of theta. Or another way of thinking about it is this value on the X-axis is cosine of theta, and this value right over here on the Y-axis is sine of theta. Now let's think about this one down over here. By the same convention, this point, this is really the unit circle definition of our trig functions. This point, since our angle is negative theta now, this point would be cosine of negative theta, comma, sine of negative theta. And we can apply the same thing over here. This point right over here, the X-coordinate is cosine of pi minus theta. from the positive X-axis. This is cosine of pi minus theta. And the Y-coordinate is the sine of pi minus theta. Then we could go all the way around to this point. I think you see where this is going. This is cosine of, I guess we could say theta plus pi or pi plus theta. Let's write pi plus data and sine of pi plus theta. Now how do these all relate to each other? Notice, over here, out here on the right-hand side, our X-coordinates are the exact same value. It's this value right over here. So we know that cosine of theta must be equal to the cosine of negative theta. That's pretty interesting. Let's write that down. Cosine of theta is equal to ... let me do it in this blue color, That's a pretty interesting result. But what about their sines? Well, here, the sine of theta is this distance above the X-axis, and here, the sine of negative theta is the same distance below the X-axis, so they're going to be the negatives of each other. We could say that sine of negative theta, sine of negative theta is equal to, is equal to the negative sine of theta, equal to the negative sine of theta. It's the opposite. If you go the same amount above or below the X-axis, you're going to get the negative value for the sine. We could do the same thing over here. How does this one relate to that? These two are going to have the same sine values. The sine of this, the Y-coordinate, is the same as the sine of that. We see that this must be equal to that. Let's write that down." + }, + { + "Q": "\nAt 0:33 , a constant , '9' ,has been used .Can we use a variable for the value of 'c'\nEx: what will be the value of 'c' when using the quadratic equation to solve the equation\nx^2 - x - 12 = 0", + "A": "What exactly do you mean by a variable for c? In your example, a=1, b= - 1, and c= - 12. To use the quadratic formula, all of a, b, and c have to be numbers (a and b are coefficients, and c is the constant). c can equal 0 such as x^2 + 2x, but then it is easier to factor x(x+2) so solutions are 0 and -2 than it is to use the quadratic formula. No, none of a, b or c can be a variable.", + "video_name": "iulx0z1lz8M", + "timestamps": [ + 33 + ], + "3min_transcript": "Use the quadratic formula to solve the equation, 0 is equal to negative 7q squared plus 2q plus 9. Now, the quadratic formula, it applies to any quadratic equation of the form-- we could put the 0 on the left hand side. 0 is equal to ax squared plus bx plus c. And we generally deal with x's, in this problem we're dealing with q's. But the quadratic formula says, look, if you have a quadratic equation of this form, that the solutions of this equation are going to be x is going to be equal to negative b plus or minus the square root of b squared minus 4ac-- all of that over 2a. And this is actually two solutions here, because there's one solution where you take the positive square root and there's another solution where you take the negative root. So it gives you both roots of this. So if we look at the quadratic equation that we need to solve We're dealing with q's, not x's, but this is the same general idea. It could be x's if you like. And if we look at it, negative 7 corresponds to a. That is our a. It's the coefficient on the second degree term. 2 corresponds to b. It is the coefficient on the first degree term. And then 9 corresponds to c. It's the constant. So, let's just apply the quadratic formula. The quadratic formula will tell us that the solutions-- the q's that satisfy this equation-- q will be equal to negative b. b is 2. Plus or minus the square root of b squared, of 2 squared, minus 4 times a times negative 7 times c, which is 9. All of that over 2 times a, which is once again negative 7. And then we just have to evaluate this. So this is going to be equal to negative 2 plus or minus the square root of-- let's see, 2 squared is 4-- and then if we just take this part right here, if we just take the negative 4 times negative 7 times 9, this negative and that negative is going to cancel out. So it's just going to become a positive number. And 4 times 7 times 9. 4 times 9 is 36. 36 times 7. Let's do it up here. 36 times 7. 7 times 6 is 42. 7 times 3, or 3 times 7 is 21." + }, + { + "Q": "\nAt 4:00, hwy did he put + instead of - ? Was there a certain reason? Also, thanks for the video! :-) Oh, nevermind! XD", + "A": "with the \u00c2\u00b1 sign, the logical way of doing it is to do the + first since it is on top, then do the - second, but it really does not matter. \u00c2\u00b1 reads as plus or minus and is two separate operations", + "video_name": "iulx0z1lz8M", + "timestamps": [ + 240 + ], + "3min_transcript": "All of that over 2 times a, which is once again negative 7. And then we just have to evaluate this. So this is going to be equal to negative 2 plus or minus the square root of-- let's see, 2 squared is 4-- and then if we just take this part right here, if we just take the negative 4 times negative 7 times 9, this negative and that negative is going to cancel out. So it's just going to become a positive number. And 4 times 7 times 9. 4 times 9 is 36. 36 times 7. Let's do it up here. 36 times 7. 7 times 6 is 42. 7 times 3, or 3 times 7 is 21. 252. So this becomes 4 plus 252. Remember, you have a negative 7 and you have a minus out front. Those cancel out, that's why we have a positive 252 for that part right there. And then our denominator, 2 times negative 7 is negative 14. Now what does this equal? Well, we have this is equal to negative 2 plus or minus the square root of-- what's 4 plus 252? It's just 256. All of that over negative 14. And what's 256? What's the square root of 256? It's 16. You can try it out for yourself. This is 16 times 16. So the square root of 256 is 16. So we can rewrite this whole thing as being equal to negative 2 plus 16 over negative 14. This is plus 16 over negative 14. Or minus 16 over negative 14. If you think of it as plus or minus, that plus is that plus right there. And if you have that minus, that minus is that minus right there. Now we just have to evaluate these two numbers. Negative 2 plus 16 is 14 divided by negative 14 is negative 1. So q could be equal to negative 1. Or negative 2 minus 16 is negative 18 divided by negative 14 is equal to 18 over 14. The negatives cancel out, which is equal to 9 over 7. So q could be equal to negative 1, or it could be equal to 9 over 7. And you could try these out, substitute these q's back into" + }, + { + "Q": "at 2:01 i dont get it\n", + "A": "Simple division. 42/3 is 14", + "video_name": "DqeMQHomwAU", + "timestamps": [ + 121 + ], + "3min_transcript": "Solve for x and check your solution. We have x divided by 3 is equal to 14. So to solve for x, to figure out what the variable x must be equal to, we really just have to isolate it on the left-hand side of this equation. It's already sitting there. We have x divided by 3 is equal to 14. We could also write this as 1/3 x is equal to 14. Obviously, x times 1/3 is going to be x/3. These are equivalent. So how can we just end up with an x on the left-hand side of either of these equations? These are really the same thing. Or another way, how can we just have a 1 in front of the x, a 1x, which is really just saying x over here? Well, I'm dividing it by 3 right now. So if I were to multiply both sides of this equation by 3, that would isolate the x. And the reason that would work is if I multiply this by 3 over here, I'm multiplying by 3 and dividing by 3. That's equivalent. That's equivalent to multiplying or dividing by 1. These guys cancel out. Remember, if you do it to the left-hand side, you also have to do it to the right-hand side. at the same time, because they're really the exact same equation. So what are we going to get over here on the left-hand side? 3 times anything divided by 3 is going to be that anything. We're just going to have an x left over on the left-hand side. And on the right-hand side, what's 14 times 3? 3 times 10 is 30, 3 times 4 is 12. So it's going to be 42. So we get x is equal to 42. And the same thing would happen here. 3 times 1/3 is just 1. So you get 1x is equal to 14 times 3, which is 42. Now let's just check our answer. Let's substitute 42 into our original equation. So we have 42 in place for x over 3 is equal to 14. So what's 42 divided by 3? And we could do a little bit of-- I guess we call it medium-long division. It's not really long division. 3 into 4. 1 times 3 is 3. You subtract. 4 minus 3 is 1. Bring down the 2. 3 goes into 12 four times. 3 goes into 42 14 times. So this right over here simplifies to 14. And it all checks out, so we're done." + }, + { + "Q": "From 0:56 to 1:40 you explain the whole process but, wouldn't it be easier if we just multiplied 14 times 3 and done?? Can someone respond to this because doing the whole process seems sorta like unnecessary in a way.\n", + "A": "The whole point of the video isn t just to intuit the entire process, but to break the process, and, as contrary as it seems, to analyse your intuition. That s why 0:50 to 1:40 exists.", + "video_name": "DqeMQHomwAU", + "timestamps": [ + 56, + 100 + ], + "3min_transcript": "Solve for x and check your solution. We have x divided by 3 is equal to 14. So to solve for x, to figure out what the variable x must be equal to, we really just have to isolate it on the left-hand side of this equation. It's already sitting there. We have x divided by 3 is equal to 14. We could also write this as 1/3 x is equal to 14. Obviously, x times 1/3 is going to be x/3. These are equivalent. So how can we just end up with an x on the left-hand side of either of these equations? These are really the same thing. Or another way, how can we just have a 1 in front of the x, a 1x, which is really just saying x over here? Well, I'm dividing it by 3 right now. So if I were to multiply both sides of this equation by 3, that would isolate the x. And the reason that would work is if I multiply this by 3 over here, I'm multiplying by 3 and dividing by 3. That's equivalent. That's equivalent to multiplying or dividing by 1. These guys cancel out. Remember, if you do it to the left-hand side, you also have to do it to the right-hand side. at the same time, because they're really the exact same equation. So what are we going to get over here on the left-hand side? 3 times anything divided by 3 is going to be that anything. We're just going to have an x left over on the left-hand side. And on the right-hand side, what's 14 times 3? 3 times 10 is 30, 3 times 4 is 12. So it's going to be 42. So we get x is equal to 42. And the same thing would happen here. 3 times 1/3 is just 1. So you get 1x is equal to 14 times 3, which is 42. Now let's just check our answer. Let's substitute 42 into our original equation. So we have 42 in place for x over 3 is equal to 14. So what's 42 divided by 3? And we could do a little bit of-- I guess we call it medium-long division. It's not really long division. 3 into 4. 1 times 3 is 3. You subtract. 4 minus 3 is 1. Bring down the 2. 3 goes into 12 four times. 3 goes into 42 14 times. So this right over here simplifies to 14. And it all checks out, so we're done." + }, + { + "Q": "\nWhen Sal used the variables A (1:27-1:49) and X (3:39-4:19). Sal says that the two variables are the same. What if (for example) we had 3exp-3 * 2exp-3. Then would we multiply the base and then add the exponents?", + "A": "The rules for exponents only work if you have a common base. In your expression: 3^(-3) * 2^(-3) you have different bases (the 3 and 2). You can only add exponents if you are multiplying with a common base. Since the exponents match, you can do: (3*2)^(-3) = 6^(-3) This only works because the exponents match. If the exponents were different, you would be required to follow PEMDAS rules and complete the exponents before doing any multiplication. Hope this helps.", + "video_name": "CZ5ne_mX5_I", + "timestamps": [ + 87, + 109, + 219, + 259 + ], + "3min_transcript": "- [Narrator] Let's get some practice with our exponent properties, especially when we have integer exponents. So, let's think about what four to the negative three times four to the fifth power is going to be equal to. And I encourage you to pause the video and think about it on your own. Well there's a couple of ways to do this. See look, I'm multiplying two things that have the same base, so this is going to be that base, four. And then I add the exponents. Four to the negative three plus five power which is equal to four to the second power. And that's just a straight forward exponent property, but you can also think about why does that actually make sense. Four to the negative 3 power, that is one over four to the third power, or you could view that as one over four times four times four. And then four to the fifth, that's five fours being multiplied together. So it's times four times four times four times four times four. And so notice, when you multiply this out, and three fours in the denominator. And so, three of these in the denominator with three of these in the numerator. And so you're going to be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have A to the negative fourth power times A to the, let's say, A squared. What is that going to be? Well once again, you have the same base, in this case it's A, and so since I'm multiplying them, you can just add the exponents. So it's going to be A to the negative four plus two power. Which is equal to A to the negative two power. And once again, it should make sense. This right over here, that is one over A times A times A times A times A times A, so that cancels with that, that cancels with that, and you're still left with one over A times A, which is the same thing as A to the negative two power. Now, let's do it with some quotients. So, what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So, this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent and so, this is going to be equal to 12 to the, subtracting a negative is the same thing as adding the positive," + }, + { + "Q": "\nWhy does the division problem at 2:40 become a multiplication problem when he's finished with it?", + "A": "12^(-5) can be converted to have a positive exponent if you use its reciprocal. 12^(-5) = 1/12^5 Divide: 12^(-7) / 1/12^5 = 12^(-7) * 12^5/1 Hope this helps.", + "video_name": "CZ5ne_mX5_I", + "timestamps": [ + 160 + ], + "3min_transcript": "and three fours in the denominator. And so, three of these in the denominator with three of these in the numerator. And so you're going to be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have A to the negative fourth power times A to the, let's say, A squared. What is that going to be? Well once again, you have the same base, in this case it's A, and so since I'm multiplying them, you can just add the exponents. So it's going to be A to the negative four plus two power. Which is equal to A to the negative two power. And once again, it should make sense. This right over here, that is one over A times A times A times A times A times A, so that cancels with that, that cancels with that, and you're still left with one over A times A, which is the same thing as A to the negative two power. Now, let's do it with some quotients. So, what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So, this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent and so, this is going to be equal to 12 to the, subtracting a negative is the same thing as adding the positive, And once again, we just have to think about, why does this actually make sense? Well, you could actually rewrite this. 12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power. If we take the reciprocal of this right over here, you would make exponent positive and then you would get exactly what we were doing in those previous examples with products. And so, let's just do one more with variables for good measure. Let's say I have X to the negative twentieth power divided by X to the fifth power. Well once again, we have the same base and we're taking a quotient. So, this is going to be X to the negative 20 minus five cause we have this one right over here in the denominator. So, this is going to be equal to X to the negative twenty-fifth power." + }, + { + "Q": "At 1:18 of the video why did Sal say that he figured out the sum of 35 and another number? He multiplied 7 by 5 to get part of the answer. Isn't the answer to a multiplication problem known as a product?\n", + "A": "Yes, however........ He multiplied 7 x 5 to get the product. Then 7 x 11 to get that product. Then he added the SUM of those products. 35+77", + "video_name": "xC-fQ0KEzsM", + "timestamps": [ + 78 + ], + "3min_transcript": "We're asked to rewrite the expression 7 times open parentheses 5 plus 11 close parentheses as the sum of 35 and another whole number. So really what they're asking us to do is just apply the distributive property. We have 7 times the quantity 5 plus 11. Now this is easy to calculate. You could just say 5 plus 11 is 16 and then 16 times 7 is what? That's 70 plus 42 which would be 112. But that's not what they're asking us to do. They're not saying just calculate this. They're saying express this as a sum of 35 and another whole number. So let's apply the distributive property and let's see if we can get a sum of 35 and another whole number. So 7 times 5 plus 11, that's the same thing as 7 times 5 plus 7 times 11. And you can see with this expression editor right over here, it tells you-- it puts it kind of the nice math formatting for what it looks like for the computer. So if we're distributing the 7 over the 5 and the 11, Well, 7 times 5 is 35. And 7 times 11 is 77. Now, have we done what they're asking us? They said rewrite this expression as the sum of 35 and another whole number. Well, we've done that. We've written it as a sum of 35 and another whole number and we were able to do it using the distributive property. So let's check to make sure that we got the right answer. Yes, we did. Let's do one more of these. Rewrite the expression 12 plus 75 in the following form-- a times 4 plus c where a and c represent whole numbers. Now this might look complex, but they're really asking us to factor out an a-- factoring out an a out of this expression right over here. Seeing how much we can factor out so that one of these two numbers becomes a 4. So let's think about how to do that. If we look at these two numbers, the greatest common divisor Both of them are divisible by 3. So you can write 12. 12 is the same thing as 3 times 4. And 75 is the same thing as 3 times 25. Now what we could do is we could essentially factor out the 3. So this is where you could say we're undistributing the 3. So that's the same thing. 3 times 4 plus 3 times 25, that's the same thing as 3 times 4 plus 25. And it looks like we've actually put it in the form that they want us to put it in, where 3 is a and 4 is right there. And then c is 25. So we've put it in the right form. Let's check our answer. We got it right." + }, + { + "Q": "For question 35 can't you use the side ratios for 30:60:90 triangles?\n", + "A": "yes, you can.", + "video_name": "BJSk1joCQsM", + "timestamps": [ + 1860 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:55, what does it mean when it says 8 liters per fish?\n", + "A": "It is saying how many liters there are for one fish. They are trying to find the unit rate.", + "video_name": "jOZ98FDyl2E", + "timestamps": [ + 55 + ], + "3min_transcript": "- [Voiceover] Let's get some practice comparing and computing rates. So they tell us the pet store has three fish tanks, each holding a different volume of water and a different number of fish. So Tank A has five fish, and it has 40 liters of water, Tank B, 12 fish, 100 liters of water, and Tank C, 23 fish, and it has 180 liters of water. Order the tanks by volume per fish from least to greatest. So let's think about what volume per fish, and we could think about this as volume divided by fish. Volume per fish. All right, so here for Tank A, it's going to be 40 liters for every five fish. 40 for every five fish, and let's see, 40 over five is eight, so you have eight liters per fish, is the rate at which they have to add water per fish for Tank A. Now, Tank B, you have 100 liters so what is this going to be? This is going to be, 12 goes into 100 eight times, so eight times 12 is 96, and then you have four left over. So this is going to be eight and 4/12, or eight and 1/3 liters per fish. And all I did is I converted this improper fraction, 100 over 12, to a mixed number, and I simplified it. 12 goes into 100 eight times with a remainder of four, so it's eight and 4/12, which is the same thing as eight and 1/3. And then finally in Tank C, I have 180 liters for 23 fish. So what is this going to be? 23 goes into 180. Let me try to calculate this. 23 goes into 180, it looks like it's going to be less than nine times. Is it eight times? Eight times, no, not eight times. Seven times three is 21, seven times two is 14, plus two is 16. When you subtract you get a remainder of 19. So this is going to be seven with a remainder of 19, or you could say this is seven and 19/23 liters per fish. So which one has, we're going to order the tanks by volume per fish, from least to greatest, so Tank B has the largest volume per fish, has eight and 1/3 liters per fish, so this is in first place. And then Tank A is in second place. Tank A is in second place. And then Tank C is in third place. Oh, actually we wanna go from least to greatest, so this is, let me write it this way. This is Tank C is the least, and Tank A is the greatest. So we really have to swap this order around. Now, I just copied and pasted this from the Khan Academy exercises. Let me actually bring the actual exercise up," + }, + { + "Q": "\nDid anyone notice that at 2:54 Sal accidentally put greatest on the middle number?", + "A": "Yeah, but the video explained for him", + "video_name": "jOZ98FDyl2E", + "timestamps": [ + 174 + ], + "3min_transcript": "so what is this going to be? This is going to be, 12 goes into 100 eight times, so eight times 12 is 96, and then you have four left over. So this is going to be eight and 4/12, or eight and 1/3 liters per fish. And all I did is I converted this improper fraction, 100 over 12, to a mixed number, and I simplified it. 12 goes into 100 eight times with a remainder of four, so it's eight and 4/12, which is the same thing as eight and 1/3. And then finally in Tank C, I have 180 liters for 23 fish. So what is this going to be? 23 goes into 180. Let me try to calculate this. 23 goes into 180, it looks like it's going to be less than nine times. Is it eight times? Eight times, no, not eight times. Seven times three is 21, seven times two is 14, plus two is 16. When you subtract you get a remainder of 19. So this is going to be seven with a remainder of 19, or you could say this is seven and 19/23 liters per fish. So which one has, we're going to order the tanks by volume per fish, from least to greatest, so Tank B has the largest volume per fish, has eight and 1/3 liters per fish, so this is in first place. And then Tank A is in second place. Tank A is in second place. And then Tank C is in third place. Oh, actually we wanna go from least to greatest, so this is, let me write it this way. This is Tank C is the least, and Tank A is the greatest. So we really have to swap this order around. Now, I just copied and pasted this from the Khan Academy exercises. Let me actually bring the actual exercise up, C was the least, and B was the most, and we check our answer and we got it right." + }, + { + "Q": "\nWhy (at 4:58) is (x-4)\u00c2\u00b2 \u00e2\u0089\u00a5 0?? Then why, when you multiply it by -3 is it \u00e2\u0089\u00a4 0? Am I missing something?", + "A": "Squares are always positive or equal to 0. Hence, (x-4)\u00c2\u00b2 \u00e2\u0089\u00a5 0. Yes, when you multiply by -3, it becomes \u00e2\u0089\u00a4 0 as +ve * -ve = -ve. But, there is also 0. Anything *0 = 0.", + "video_name": "IbI-l7mbKO4", + "timestamps": [ + 298 + ], + "3min_transcript": "And we just have to remind ourselves that if I have x plus a squared, that's going to be x squared plus 2ax plus a squared. So if I want to turn something that looks like this, 2ax, into a perfect square, I just have to take half of this coefficient and square it and add it right over here in order to make it look like that. So I'm going to do that right over here. So if I take half of negative 4, that's negative 2. If I square it, that is going to be positive 4. I have to be very careful here. I can't just willy nilly add a positive 4 here. I have equality here. If they were equal before adding the 4, then they're not going to be equal after adding the 4. So I have to do proper accounting here. I either have to add 4 to both sides or I should be careful. I have to add the same amount to both sides or subtract the same amount again. didn't just add 4 to the right hand side of the equation. Remember, the 4 is getting multiplied by 5. I have added 20 to the right hand side of the equation. So if I want to make this balance out, if I want the equality to still be true, I either have to now add 20 to y or I have to subtract 20 from the right hand side. So I'll do that. I'll subtract 20 from the right hand side. So I added 5 times 4. If you were to distribute this, you'll see that. I could have literally, up here, said hey, I'm adding 20 and I'm subtracting 20. This is the exact same thing that I did over here. If you distribute the 5, it becomes 5x squared minus 20x plus 20 plus 15 minus 20. Exactly what's up here. The whole point of this is that now I can write this in an interesting way. I could write this as y is equal to 5 times x minus 2 squared, and then 15 minus 20 is minus 5. So the whole point of this is now to be able to inspect this. Well, we know that this term right over here is always going to be non-negative. Or we could say it's always going to be greater than or equal to 0. This whole thing is going to hit a minimum value when this term is equal to 0 or when x equals 2. When x equals 2, we're going to hit a minimum value. And when x equals 2, what happens? Well, this whole term is 0 and y is equal to negative 5. The vertex is 2, negative 5." + }, + { + "Q": "\nI don't get how he got 16 around 2:10", + "A": "You have to think about how to achieve a perfect square. In this case, he looked at the coefficient of the second term (-8), divided it by 2 (which equals -4), and then squared it to get 16. You can also double check afterwards to make sure you have correctly calculated a perfect square. Is (x-4)*(x-4) equal to x2 -8x +16? It is, so you re good to move forward with the next steps of simplification.", + "video_name": "IbI-l7mbKO4", + "timestamps": [ + 130 + ], + "3min_transcript": "I have an equation right here. It's a second degree equation. It's a quadratic. And I know its graph is going to be a parabola. Just as a review, that means it looks something like this or it looks something like that. Because the coefficient on the x squared term here is positive, I know it's going to be an upward opening parabola. And I am curious about the vertex of this parabola. And if I have an upward opening parabola, the vertex is going to be the minimum point. If I had a downward opening parabola, then the vertex would be the maximum point. So I'm really trying to find the x value. I don't know actually where this does intersect the x-axis or if it does it all. But I want to find the x value where this function takes on a minimum value. Now, there's many ways to find a vertex. Probably the easiest, there's a formula for it. And we talk about where that comes from in multiple videos, where the vertex of a parabola or the x-coordinate of the vertex of the parabola. So the x-coordinate of the vertex And the negative b, you're just talking about the coefficient, or b is the coefficient on the first degree term, is on the coefficient on the x term. And a is the coefficient on the x squared term. So this is going to be equal to b is negative 20. So it's negative 20 over 2 times 5. Well, this is going to be equal to positive 20 over 10, which is equal to 2. And so to find the y value of the vertex, we just substitute back into the equation. The y value is going to be 5 times 2 squared minus 20 times 2 plus 15, which is equal to let's see. This is 5 times 4, which is 20, minus 40, which is negative 20, So just like that, we're able to figure out the coordinate. This coordinate right over here is the point 2, negative 5. Now it's not so satisfying just to plug and chug a formula like this. And we'll see where this comes from when you look at the quadratic formula. This is the first term. It's the x value that's halfway in between the roots. So that's one way to think about it. But another way to do it, and this probably will be of more lasting help for you in your life, because you might forget this formula. It's really just try to re-manipulate this equation so you can spot its minimum point. And we're going to do that by completing the square. So let me rewrite that. And what I'll do is out of these first two terms, I'll factor out a 5, because I want to complete a square here and I'm going to leave this 15 out to the right, because I'm going to have to manipulate that as well. So it is 5 times x squared minus 4x. And then I have this 15 out here." + }, + { + "Q": "\nAt 2:09, Sal completes the square by adding and subtracting 16. Could he have also done this by adding 27 to both sides and finding the perfect square?", + "A": "Your thinking is correct, but your method is not as good. Since we are looking to get in vertex form, moving the 27 to the other side serves no purpose since we would have to move it back later on. However (this is what you are getting at), leaving the -27 outside of the perfect square is acceptable and how a lot of math thinkers would do it. The answer would end up the same.", + "video_name": "IbI-l7mbKO4", + "timestamps": [ + 129 + ], + "3min_transcript": "I have an equation right here. It's a second degree equation. It's a quadratic. And I know its graph is going to be a parabola. Just as a review, that means it looks something like this or it looks something like that. Because the coefficient on the x squared term here is positive, I know it's going to be an upward opening parabola. And I am curious about the vertex of this parabola. And if I have an upward opening parabola, the vertex is going to be the minimum point. If I had a downward opening parabola, then the vertex would be the maximum point. So I'm really trying to find the x value. I don't know actually where this does intersect the x-axis or if it does it all. But I want to find the x value where this function takes on a minimum value. Now, there's many ways to find a vertex. Probably the easiest, there's a formula for it. And we talk about where that comes from in multiple videos, where the vertex of a parabola or the x-coordinate of the vertex of the parabola. So the x-coordinate of the vertex And the negative b, you're just talking about the coefficient, or b is the coefficient on the first degree term, is on the coefficient on the x term. And a is the coefficient on the x squared term. So this is going to be equal to b is negative 20. So it's negative 20 over 2 times 5. Well, this is going to be equal to positive 20 over 10, which is equal to 2. And so to find the y value of the vertex, we just substitute back into the equation. The y value is going to be 5 times 2 squared minus 20 times 2 plus 15, which is equal to let's see. This is 5 times 4, which is 20, minus 40, which is negative 20, So just like that, we're able to figure out the coordinate. This coordinate right over here is the point 2, negative 5. Now it's not so satisfying just to plug and chug a formula like this. And we'll see where this comes from when you look at the quadratic formula. This is the first term. It's the x value that's halfway in between the roots. So that's one way to think about it. But another way to do it, and this probably will be of more lasting help for you in your life, because you might forget this formula. It's really just try to re-manipulate this equation so you can spot its minimum point. And we're going to do that by completing the square. So let me rewrite that. And what I'll do is out of these first two terms, I'll factor out a 5, because I want to complete a square here and I'm going to leave this 15 out to the right, because I'm going to have to manipulate that as well. So it is 5 times x squared minus 4x. And then I have this 15 out here." + }, + { + "Q": "\nAt 1:50 Sal has broken down the expression 2 * 4/3 to 1/3+1/3+1/3+1/3+1/3+1/3+1/3+1/3. I am learning the process of thinking of it in this way for more difficult problems?\n\nI can do the mental math to know that 2 * 4/3 = 8/3 = 2 2/3. What is the importance of the process of breaking it down to 1/3+1/3+1/3+1/3+1/3+1/3+1/3+1/3 that I am not understanding?", + "A": "yah me too i never thought of it as this way", + "video_name": "ZlhrXO1-osA", + "timestamps": [ + 110 + ], + "3min_transcript": "So we have here, it says 2 times 4/3 is equal to 8 times blank. And what I encourage you to do is pause the video right now and try to think about what should go in this blank. So I'm assuming you've given your try. Now, let's think through this. So 2 times 4/3, we can literally view that as the same thing as-- if we rewrite the 4/3, this is the same thing as 2 times-- instead of writing 4/3 like this, I'm literally going to write it as four 1/3's. And I know it sounds like I just said the same thing over again. But I'm literally going to write 1/3 four times-- 1/3 plus 1/3 plus 1/3 plus 1/3. If you call each of these 1/3, you literally have four of them. This is four 1/3's. 2 times 4/3 is the same thing as 2 times, literally, four 1/3's. Now, what would this be? Well, this is going to be equal to-- let me just copy and paste So copy, and then let me paste it. So that's one group of those 1/3's, of those four 1/3's, or one group of one of these four 1/3's. And then, we'll have another one. And then, we'll have another one. And we're going to add them together. That's literally 2 times 4/3. So let's add these together. Now what do we have? Well, we have a bunch of 1/3's. And we need to count them up. We have one, two, three, four, five, six , seven, eight 1/3's. This is literally equal to-- and we could, just to make it clear what I've just done, we could ignore the parentheses and just add up all of these things together. So that might make it a little bit clearer. So let me do that just to make it clear that I literally take-- I've taken eight 1/3's and I'm adding them together, which is the exact same thing as 8/3. So let me clear that, and let me clear that, let me clear that. clearly, equal to 8 times 1/3. I have 8 1/3's there. So going back to the original question, what is this equal to? 2 times 4/3 is the same thing as 8 times 1/3. And we've already seen that 8 times 1/3, well, that's literally 8/3. So we could also write it like this-- 8 over 3. Let me do that 3 in that other color-- 8 over 3." + }, + { + "Q": "there is a extra decimal at 0:01 the question has a extra decimal\n", + "A": "no,actually that is just the period at the end of the sentence . :)", + "video_name": "Eq4mVCd-yyo", + "timestamps": [ + 1 + ], + "3min_transcript": "We need to calculate 9.005 minus 3.6, or we could view it as 9 and 5 thousandths minus 3 and 6 tenths. Whenever you do a subtracting decimals problem, the most important thing, and this is true when you're adding decimals as well, is you have to line up the decimals. So this is 9.005 minus 3.6. So we've lined up the decimals, and now we're ready to subtract. Now we can subtract. So we start up here. We have 5 minus nothing. You can imagine this 3.6, or this 3 and 6 tenths, we could add two zeroes right here, and it would be the same thing as 3 and 600 thousandths, which is the same thing as 6 tenths. And when you look at it that way, you'd say, OK, 5 minus 0 is nothing, and you just write a 5 right there. Or you could have said, if there's nothing there, it would have been 5 minus nothing is 5. Then you have 0 minus 0, which is just 0. And you can't subtract 6 from 0. So we need to get something into this space right here, and what we essentially are going to do is regroup. We're going to take one 1 from the 9, so let's do that. So let's take one 1 from the 9, so it becomes an 8. And we need to do something with that one 1. We're going to put it in the tenths place. Now remember, one whole is equal to 10 tenths. This is the tenths place. So then this will become 10. Sometimes it's taught that you're borrowing the 1, but you're really taking it, and you're actually taking 10 from the place to your left. So one whole is 10 tenths, we're in the tenths place. So you have 10 minus 6. Let me switch colors. 10 minus 6 is 4. You have your decimal right there, and then you have 8 minus 3 is 5. So 9.005 minus 3.6 is 5.405." + }, + { + "Q": "At 3:15 - How do you get y=0? Did you divide or multiply 5/2 ? Why?\n", + "A": "He got y=0 by substituting x=4 into our function rule. y=10-(5/2)x x=4. y=10-(5/2)(4) We multiply 4 by 5/2 because that s how our rule works. (5/2)(4)=10. y=10-10 10-10=0. y=0 I hope this explains what Sal did and why!", + "video_name": "86NwKBcOlow", + "timestamps": [ + 195 + ], + "3min_transcript": "is 10 minus 5x over 2 or minus 5/2 times x. And so now using this, let's just come up with a bunch of x values and see what the corresponding y values are, and then just plot them. So let me do this in a new color. So let me-- a slightly different shade of yellow. So we have x values, and then let's think about what the corresponding y value is going to be. So I'll start, well, I could start anywhere. I'll start at x is equal to 0, just because that tends to keep things pretty simple. If x is 0, then y is equal to 10 minus 5/2 times 0, which is equal to 5/2 times 0 is just a 0. So it's just 10 minus 0 or 10. So that gives us the coordinate, the point, 0 comma 10. When x is 0, y is 10. So x is 0. So it's going to be right here at the middle of the x-axis. And you go up 10 for the y-coordinate. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So that's the point 0 comma 10. Let's do another point. Let's say that x is 2. I'm going to pick multiples of 2 here just so that I get a nice clean answer here. So when x is 2, then y is equal to 10 minus 5/2 times 2, and the 2 in the denominator cancels out with this 2 in the numerator. So it simplifies to 10 minus 5, or just 5. So that tells us the point x equals 2, y is equal to 5, is on the line. So 2x is equal to 1, 2 right over here. And then y is equal to 5. You go up 5. 1, 2, 3, 4, 5, just like that. So that's the point 2, 5. And when you're drawing a line you actually just need two points. If you have a ruler or any kind of straight edge, we could just connect these two points. should satisfy this relationship right here. Just so we get practice, I'll do more points. So let me do, let's say when x is equal to 4, then y is equal to 10 minus 5/2 times 4. This is equal to 5/2 times 4. This is equal to 10, right? Because the 2, divide the denominator by 2 you get 1, divide the numerator by 2 you get 2, or 4 over 2 is the same thing as 2. So it becomes 2 times 5 is 10, 10 minus 10 is 0. So the point 4 comma 0 is on our line. So x is 1, 2, 3, 4, and then y is 0. So we don't move up at all, so we have 4 comma 0. And I could keep going. I could try other points. You could do them if you like, but this is plenty. Just two of these would have been enough to draw the line. So let me just draw it. So I'll do it in white." + }, + { + "Q": "Why did he change the 4 to a 2 3:40\n", + "A": "I don t entirely understand your question, but at 3:40, he was simplifying his multiplication", + "video_name": "86NwKBcOlow", + "timestamps": [ + 220 + ], + "3min_transcript": "So that's the point 0 comma 10. Let's do another point. Let's say that x is 2. I'm going to pick multiples of 2 here just so that I get a nice clean answer here. So when x is 2, then y is equal to 10 minus 5/2 times 2, and the 2 in the denominator cancels out with this 2 in the numerator. So it simplifies to 10 minus 5, or just 5. So that tells us the point x equals 2, y is equal to 5, is on the line. So 2x is equal to 1, 2 right over here. And then y is equal to 5. You go up 5. 1, 2, 3, 4, 5, just like that. So that's the point 2, 5. And when you're drawing a line you actually just need two points. If you have a ruler or any kind of straight edge, we could just connect these two points. should satisfy this relationship right here. Just so we get practice, I'll do more points. So let me do, let's say when x is equal to 4, then y is equal to 10 minus 5/2 times 4. This is equal to 5/2 times 4. This is equal to 10, right? Because the 2, divide the denominator by 2 you get 1, divide the numerator by 2 you get 2, or 4 over 2 is the same thing as 2. So it becomes 2 times 5 is 10, 10 minus 10 is 0. So the point 4 comma 0 is on our line. So x is 1, 2, 3, 4, and then y is 0. So we don't move up at all, so we have 4 comma 0. And I could keep going. I could try other points. You could do them if you like, but this is plenty. Just two of these would have been enough to draw the line. So let me just draw it. So I'll do it in white. And I could keep going in both directions. So there you have it. That is the graph of our linear equation. Let me make my line a little bit bolder, just in case you found that first line hard to read. So let me make it a little bit bolder. And I think you get the general idea." + }, + { + "Q": "\nlike at 3:16, those x in the table could be replace with numbers -2,-1, 0, 1, 2 in this kind of order can we solve this given equation? Our teacher said that we can only use this numbers to have a uniform answers in our problem.", + "A": "Linear equations have an infinite number of possible values for X. You can use any value of X to calculate Y and create a point on the line. Your teacher likely wants everyone finding the same points so there is consistency in the way the class is approaching the problem.", + "video_name": "86NwKBcOlow", + "timestamps": [ + 196 + ], + "3min_transcript": "is 10 minus 5x over 2 or minus 5/2 times x. And so now using this, let's just come up with a bunch of x values and see what the corresponding y values are, and then just plot them. So let me do this in a new color. So let me-- a slightly different shade of yellow. So we have x values, and then let's think about what the corresponding y value is going to be. So I'll start, well, I could start anywhere. I'll start at x is equal to 0, just because that tends to keep things pretty simple. If x is 0, then y is equal to 10 minus 5/2 times 0, which is equal to 5/2 times 0 is just a 0. So it's just 10 minus 0 or 10. So that gives us the coordinate, the point, 0 comma 10. When x is 0, y is 10. So x is 0. So it's going to be right here at the middle of the x-axis. And you go up 10 for the y-coordinate. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So that's the point 0 comma 10. Let's do another point. Let's say that x is 2. I'm going to pick multiples of 2 here just so that I get a nice clean answer here. So when x is 2, then y is equal to 10 minus 5/2 times 2, and the 2 in the denominator cancels out with this 2 in the numerator. So it simplifies to 10 minus 5, or just 5. So that tells us the point x equals 2, y is equal to 5, is on the line. So 2x is equal to 1, 2 right over here. And then y is equal to 5. You go up 5. 1, 2, 3, 4, 5, just like that. So that's the point 2, 5. And when you're drawing a line you actually just need two points. If you have a ruler or any kind of straight edge, we could just connect these two points. should satisfy this relationship right here. Just so we get practice, I'll do more points. So let me do, let's say when x is equal to 4, then y is equal to 10 minus 5/2 times 4. This is equal to 5/2 times 4. This is equal to 10, right? Because the 2, divide the denominator by 2 you get 1, divide the numerator by 2 you get 2, or 4 over 2 is the same thing as 2. So it becomes 2 times 5 is 10, 10 minus 10 is 0. So the point 4 comma 0 is on our line. So x is 1, 2, 3, 4, and then y is 0. So we don't move up at all, so we have 4 comma 0. And I could keep going. I could try other points. You could do them if you like, but this is plenty. Just two of these would have been enough to draw the line. So let me just draw it. So I'll do it in white." + }, + { + "Q": "\nAt 1:08 could Sal have simplified y = 10 - 5/2x even more?\n\ny = 5- 5x (cross reducing the numerator 10 and the denominator 2)", + "A": "Cross cancelling only works when the fractions are being multiplied. In this case, the problem is subtracting not multiplication. So, you can t do it.", + "video_name": "86NwKBcOlow", + "timestamps": [ + 68 + ], + "3min_transcript": "Create a graph of the linear equation 5x plus 2y is equal to 20. So the line is essentially the set of all coordinate, all x's and y's, that satisfy this relationship right over here. To make things simpler, what we're going to do is set up a table where we're going to put a bunch of x values in and then figure out the corresponding y value based on this relationship. But to make it a little bit simpler, I'm going to solve for y here. So it becomes easier to solve for y for any given x. So we have 5x plus 2y is equal to 20. If we want to solve for y, let's just get rid of the 5x on the left-hand side. So let's subtract 5x from both sides of this equation. The left-hand side, these guys cancel out, so we get 2y is equal to the right hand side, you have 20 minus 5x. And then you can divide both sides of this equation by 2. So you divide both sides by 2. The left-hand side, we just have a y, and then the right-hand side, we could leave it that way. That actually would be a pretty straightforward way is 10 minus 5x over 2 or minus 5/2 times x. And so now using this, let's just come up with a bunch of x values and see what the corresponding y values are, and then just plot them. So let me do this in a new color. So let me-- a slightly different shade of yellow. So we have x values, and then let's think about what the corresponding y value is going to be. So I'll start, well, I could start anywhere. I'll start at x is equal to 0, just because that tends to keep things pretty simple. If x is 0, then y is equal to 10 minus 5/2 times 0, which is equal to 5/2 times 0 is just a 0. So it's just 10 minus 0 or 10. So that gives us the coordinate, the point, 0 comma 10. When x is 0, y is 10. So x is 0. So it's going to be right here at the middle of the x-axis. And you go up 10 for the y-coordinate. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So that's the point 0 comma 10. Let's do another point. Let's say that x is 2. I'm going to pick multiples of 2 here just so that I get a nice clean answer here. So when x is 2, then y is equal to 10 minus 5/2 times 2, and the 2 in the denominator cancels out with this 2 in the numerator. So it simplifies to 10 minus 5, or just 5. So that tells us the point x equals 2, y is equal to 5, is on the line. So 2x is equal to 1, 2 right over here. And then y is equal to 5. You go up 5. 1, 2, 3, 4, 5, just like that. So that's the point 2, 5. And when you're drawing a line you actually just need two points. If you have a ruler or any kind of straight edge, we could just connect these two points." + }, + { + "Q": "\nat 1:17 he takes 8 away from 9 and calls it 10/10 what dos that mean?", + "A": "10/10 is equal to 1 because 10 10 s is", + "video_name": "lDXaju6JoQ0", + "timestamps": [ + 77 + ], + "3min_transcript": "Let's try to calculate 39.1 minus 0.794, and so pause the video and try this on your own. All right, I'm assuming you've given a go at it, so now let's work through it together. So I'm going to rewrite this. It's 39.1 minus-- I'm going to line up the decimals so that I have the right place values below the right place values-- minus-- this 0 is in the ones place, so I'll put it in the ones place-- 0.794. And now we're ready to subtract. Now, how do we subtract 4 from nothingness here, and 9 from nothingness here? Well, the same thing as nothing is a 0. And so now we can start to think about how to subtract. Well, we still have the problem. Well, we're trying to subtract 4 from 0, so we're trying to subtract 9 from 0. So what we could do is take this one tenth and try to regroup it into the hundredths place So let's think about this. If we make this-- actually that's not actually going to solve our problem. Well we could do it, but then we're going to have zero tenths, and we're still going to have a problem here. So actually let me go to the ones place. So let me get rid of a ones, so that's eight ones, which is going to be 10 tenths. So that's going to now-- we're going to have 11 tenths. The 10 tenths from here plus 1 is 11 tenths. Now let's take one of those tenths so that we have 10 tenths, and give it to the hundredths. So that's going to be 10 hundredths. And now let's take one of those hundredths-- so now we have nine hundredths-- and give it to the thousandths. So that's going to be 10 thousandths. Now we're ready to subtract. So 10-- let me do this in yellow-- 10 minus 4 is 6. 9 minus 9 is 0. 10 minus 7 is 3. We have our decimal point. 8 minus 0 is 8. And then we have 3 minus nothing is 3. So we're done, 38.306." + }, + { + "Q": "\nin 0:40... he said nothingness... what does nothingness mean??", + "A": "Basically nothingness is the equivalent of the word nothing . Hope this helps!", + "video_name": "lDXaju6JoQ0", + "timestamps": [ + 40 + ], + "3min_transcript": "Let's try to calculate 39.1 minus 0.794, and so pause the video and try this on your own. All right, I'm assuming you've given a go at it, so now let's work through it together. So I'm going to rewrite this. It's 39.1 minus-- I'm going to line up the decimals so that I have the right place values below the right place values-- minus-- this 0 is in the ones place, so I'll put it in the ones place-- 0.794. And now we're ready to subtract. Now, how do we subtract 4 from nothingness here, and 9 from nothingness here? Well, the same thing as nothing is a 0. And so now we can start to think about how to subtract. Well, we still have the problem. Well, we're trying to subtract 4 from 0, so we're trying to subtract 9 from 0. So what we could do is take this one tenth and try to regroup it into the hundredths place So let's think about this. If we make this-- actually that's not actually going to solve our problem. Well we could do it, but then we're going to have zero tenths, and we're still going to have a problem here. So actually let me go to the ones place. So let me get rid of a ones, so that's eight ones, which is going to be 10 tenths. So that's going to now-- we're going to have 11 tenths. The 10 tenths from here plus 1 is 11 tenths. Now let's take one of those tenths so that we have 10 tenths, and give it to the hundredths. So that's going to be 10 hundredths. And now let's take one of those hundredths-- so now we have nine hundredths-- and give it to the thousandths. So that's going to be 10 thousandths. Now we're ready to subtract. So 10-- let me do this in yellow-- 10 minus 4 is 6. 9 minus 9 is 0. 10 minus 7 is 3. We have our decimal point. 8 minus 0 is 8. And then we have 3 minus nothing is 3. So we're done, 38.306." + }, + { + "Q": "\nAt 1:02 it says meth why would she say that?", + "A": "She said math, not meth. --Blue Leaf", + "video_name": "WkmPDOq2WfA", + "timestamps": [ + 62 + ], + "3min_transcript": "So in my sphere flakes video, I joked about folding and cutting space time, but then I thought, hey, why not? So, how do you do that? Well, when we wanted to fold and cut only space, we chose a medium that takes place in space. That is, static paper cut outs, or sphere sculpture. But to fold and cut time, we need a medium that happens over time. I choose music. Music has two easily recognizable dimensions. One is time, and the other is pitch space. Not quite the same as space-space, but it's one dimensional, which makes things easier. But let's not be confused with the notation. There's a few things to notice about written music. Firstly, that it's not music. You can't listen to this. Or, well, you can, but it'll be like-- [PAPER RUSTLING] It's not music, it's music notation. And you can only interpret it into the beautiful music it represents. Kind of like how a book is squiggles on a page that your brain interprets into a meaningful story. And maybe you don't understand it at all, or understand just a literal surface meaning of the action. Or maybe you can read deep and critically into a story that's simple on the surface, and get more from it than even the author intended. Secondly, written music represents a two-dimensional space of pitch and time, but only represents it. Like, there's the suggestion that this is the time axis, but it's not. This is exactly the same as this. Even though you're changing the values on the x-axis. At least, in standard music notation. Some more modern composers do make use of spatial notation, just like some poets do intentionally stretch out words or play with formatting. But in standard notation, a stretched out word only means your text editor is terrible at justified margins, and has nothing to do with the word itself. Pitch also doesn't entirely depend on the notes placed on the y-axis. So I'm going to use something a little closer to reality. I've got this music box that plays a paper tape. As you put the strip through the box, it plays the punched holes. Hold on, it's too quiet. This is why music boxes usually come attached to wooden boxes. I don't have a wooden box, maybe this nice wooden bowl. Um. Music bowl. Anyway, here distance along the strip does translate directly to time, assuming constant crank speed. [MUSIC PLAYING] The box also has a set C major scale on one staff, so pitch space is represented pretty directly. You don't have to worry about sharps and flats, or the space between staves, or the infinite possibilities on a continuous logarithmic frequency scale. The box magically ignores all notational elements, except for where and when the holes are. Each hole, each note, is a point on this strip of space time. So now, let's fold and cut it. It's easy to fold time so that it goes both forwards and backwards simultaneously. Then we can punch in some notes, and unfold it, into a symmetric melody that goes first forwards, and then backwards. Or first backwards and then forwards. Point is, it's reversible. It sounds the same whether I play it like this-- [MUSIC PLAYING]" + }, + { + "Q": "At 1:23 how come 6 times 1/6 cancels?\n", + "A": "6 multiplied by 1/6 is the same thing as 6 divided by 6, which is equal to 1. You will learn how to do this in later classes. I hope this helps.", + "video_name": "CJyVct57-9s", + "timestamps": [ + 83 + ], + "3min_transcript": "Solve for x. And we have x minus 8 is equal to x/3 plus 1/6. Now the first thing I want to do here-- and there's multiple ways to do this problem-- but what I want to do is just to simplify the fraction. I'm going to multiply everything times the least common multiple of all of these guys' denominators. This is essentially x/1. This is 8/1, x/3, 1/6. The least common multiple of 1, 3, and 6 is 6. So if I multiply everything times 6, what that's going to do is going to clear out these fractions. So these weren't fractions to begin with, so we're just multiplying them by 6. So it becomes 6x minus 6 times negative 8, or 6 times 8 is 48. And we're subtracting it right over there. And then we have x/3 times 6. Let me just write it out here. So that's going to be 6 times x/3 plus 6 times 1/6. Or we get 6x minus 48 is equal to 6 times something divided That's the same thing as 6 divided by 3 times that something. That's just going to be equal to 2x plus 6 times 1/6 or 6 divided by 6 is just going to be 1. So that first step cleared out all of the fractions and now this is just a straightforward problem with all integer coefficients or integers on either side of the equation. And what we want to do is we want to isolate all of the x's on one side or the other. And we might as well isolate them all on the left hand side. So let's subtract 2x from both sides. We want to get rid of this 2x here. That's why I'm subtracting the 2x. So let's subtract 2x from both sides. And on the right hand side, I have 2x plus 1 minus 2x. Those cancel out. That was the whole point. So I'm left with just this 1 over here. On the left hand side I have 6x minus 2x. Well, that's just going to be 4x. If I have 6 of something minus 2 of that something, I have 4 of that something. Minus 48. And now I can-- let's see, I want to get rid of this 48 So let me add 48 to both sides of the equation. I'll do this in a new color. So let me add 48 to both sides of this equation. And on the left hand side 4x minus 48 plus 48, I'm left with just a 4x. And on the right hand side, 1 plus 48 is going to be 49. And now I've isolated the x but it's still multiplied by a 4. So to make that a 1 coefficient, let's multiply both sides by 1/4. Or you could also say, let's divide both sides by 4. Anything you do to one side you have to do to the other. And so you have-- what do we have over here? 4x/4 is just x. x is equal to 49 over 4. And that's about as far as we can simplify it because these don't have any common factors, 49 and 4. Let's check to see whether 49/4 is indeed the answer. So let's put it into the original equation. Remember, the original equation is" + }, + { + "Q": "At 2:51, he says that there's no way to simplify that. Why doesn't he use mixed numbers like 12 1/4?\n", + "A": "In algebra, it is easier to use improper fractions.", + "video_name": "CJyVct57-9s", + "timestamps": [ + 171 + ], + "3min_transcript": "That's the same thing as 6 divided by 3 times that something. That's just going to be equal to 2x plus 6 times 1/6 or 6 divided by 6 is just going to be 1. So that first step cleared out all of the fractions and now this is just a straightforward problem with all integer coefficients or integers on either side of the equation. And what we want to do is we want to isolate all of the x's on one side or the other. And we might as well isolate them all on the left hand side. So let's subtract 2x from both sides. We want to get rid of this 2x here. That's why I'm subtracting the 2x. So let's subtract 2x from both sides. And on the right hand side, I have 2x plus 1 minus 2x. Those cancel out. That was the whole point. So I'm left with just this 1 over here. On the left hand side I have 6x minus 2x. Well, that's just going to be 4x. If I have 6 of something minus 2 of that something, I have 4 of that something. Minus 48. And now I can-- let's see, I want to get rid of this 48 So let me add 48 to both sides of the equation. I'll do this in a new color. So let me add 48 to both sides of this equation. And on the left hand side 4x minus 48 plus 48, I'm left with just a 4x. And on the right hand side, 1 plus 48 is going to be 49. And now I've isolated the x but it's still multiplied by a 4. So to make that a 1 coefficient, let's multiply both sides by 1/4. Or you could also say, let's divide both sides by 4. Anything you do to one side you have to do to the other. And so you have-- what do we have over here? 4x/4 is just x. x is equal to 49 over 4. And that's about as far as we can simplify it because these don't have any common factors, 49 and 4. Let's check to see whether 49/4 is indeed the answer. So let's put it into the original equation. Remember, the original equation is But in theory, we should be able to put it into any of these steps and the x should satisfy. But let's do it in our original equation. So we have x minus 8. So we have 49/4 minus 8 should be equal to 49/4 over 3 plus 1/6. So let's see what we can do here. So we can multiply. Well, like we did before. We can multiply both sides of this equation by 6. That'll help simplify a lot of the fractions here. So if we multiply both sides of this equation by 6-- so we're going to multiply everything by 6-- what do we get on the left hand side? 6/4 is the same thing as 3/2, right? So this is going to be 3 times 49 over 2. 3 times 49 over 2 minus 48 will be equal to 6 divided by 3" + }, + { + "Q": "\nHes making no sense when he wrote down the exponent why did he bring down the numbers? can u just right down (2x2x2)x(2x2x2x2x2)? then solve? but do you HAVE to put parenthese?at 0:37", + "A": "You can do it that way but that was just a simplified example. It may be as complex as 2^67 x 2^89. Exponents make it easier. The parenthesis is needed if the number is negatice or if you need to find the opposite of the answer ex: -(x)^2 or if multiplying with another exponent", + "video_name": "kITJ6qH7jS0", + "timestamps": [ + 37 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:58, why is it 2^8 and not 2^15?\n", + "A": "Because you add 3+5 and not multiply 3x5.", + "video_name": "kITJ6qH7jS0", + "timestamps": [ + 118 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 7:16 why does this work", + "A": "Because it is the rule ofcancellation. x^200/x^50 if you wrote all these x s out and then cancelled them out, you d be able the cancel off 50 from each from before being left with no more x s on the bottom. So you d be left with x^150 as 50x s from the top would leave 150 x s.", + "video_name": "kITJ6qH7jS0", + "timestamps": [ + 436 + ], + "3min_transcript": "" + }, + { + "Q": "I don't understand at 8:08 how the 7^-5 can change into 7^5...?\nThanks!\n", + "A": "when you have a negative exponent in the denominator of the fraction you have to move the exponent and its base to the numerator and the exponent is now positive. If that doesn t help then try to watch a video on negative exponents. its complicated", + "video_name": "kITJ6qH7jS0", + "timestamps": [ + 488 + ], + "3min_transcript": "" + }, + { + "Q": "At 6:11 he says 2^9/2^10 is the same as 2^9*1/2^10 . Could someone explain how this works?\n", + "A": "Anything times one is going to be that number. 9*1 = 9. 54*1 = 54. That means that 2^9*1 = 2^9, so you might as well not have the *1 there.", + "video_name": "kITJ6qH7jS0", + "timestamps": [ + 371 + ], + "3min_transcript": "" + }, + { + "Q": "How in the world does he do that at 6:21 ?!!!?\n", + "A": "One of the earlier videos explains it, but i cant remember which one... 1/2^10 is the same as 2 to the 10th root, which can be written as 2^-10", + "video_name": "kITJ6qH7jS0", + "timestamps": [ + 381 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 1:22 he says \"the easiest way to split this whole into tenths is to take each of those fifths and turn them into 2 tenths.\" What's the other way to do it? (Even if it's not the easiest.)", + "A": "Not really, because the only way to turn fifths to tenths is to multiply 5 by 2. Hope this helps!", + "video_name": "XHLgY7Z3cb8", + "timestamps": [ + 82 + ], + "3min_transcript": "Let's see if we can write 3/5 as a decimal. And I encourage you to pause this video and think about if you can do it on your own. And I'll give you a hint here. Can we rewrite this fraction so, instead of it being in terms of fifths, it can be in terms of tenths? So I'm assuming you've given a go at it. Let's try to rewrite this as a fraction with 10 as the denominator. But let's just first visualize this. So we have fifths. So let's say that's 1/5. Actually, let me just copy and paste this. That is 2/5. That is 3/5, and that is 4/5. And that is 5/5, or this would be a whole now. So that is our whole. And we want to color in 3 of those 5, so we want to think about what 3/5 are. So let me get my magenta out. I can actually make this bigger even-- 2/5 and 3/5. There you go. Color that in. That is 3/5. Now, how could I write this in terms of tenths-- instead of 3/5, a certain number of tenths? Well, let's split this whole into tenths. And the easiest way to split this whole into tenths is to take each of those fifths and turn them into 2/10. So let's do that. So If we were to do this right over here, we now have twice as many sections. So another way of thinking about it, we are multiplying the number of sections by 2. We now have 10 sections. Each of these is a tenth. And the 3 of those sections are now going to be twice as many. What we have in magenta, we now have twice as many sections in magenta. Notice we just multiplied the numerator and the denominator by 2. But hopefully it makes conceptual sense. Every piece, when we're talking about fifths, we've now doubled so that instead of every 1/5 is now 2/10. You have a 1/10 now and a 1/10 now. And we could just keep writing 1/10 if we like. Each of these things right over here are a tenth. And then each of the 3 are now twice as many tenths. So the 3/5 is now 6/10. So let's write that down. So this is going to be equal to 6/10. Now why is this interesting? You can literally view this as 6/10-- let me write it this way-- 6 times 1/10." + }, + { + "Q": "At the end of the video: 5:25 Sal says \"converges to zero\"\n\nWhat makes this epsilon proof valid?\n\nWould not we be forced to count all the possibilities to prove this to be right??\nAnd as there are infinitely many cases for epsilon it seems to be impossible.\n\nWhat is the axiom behind this limit proof or\nin what is this epsilon limit proof based on??\nCommon sense??\n", + "A": "Yeah, but as we know for example that infinity +1 = infinity, one could ask then that if epsilon is infinitesimally small, would not then |an-L|= roughly epsilon and not strictly smaller than (0, there IS a positive M such that if n>M then |a (sub)n-L|right triangles>special right triangles.", + "video_name": "UKQ65tiIQ6o", + "timestamps": [ + 75 + ], + "3min_transcript": "Let's say that this triangle right over here is equilateral, which means all of its sides have the same length. And let's say that that length is s. What I want to do in this video is come up with a way of figuring out the area of this equilateral triangle, as a function of s. And to do that, I'm just going to split this equilateral in two. I'm just going to drop an altitude from this top vertex right over here. This is going to be perpendicular to the base. And it's also going to bisect this top angle. So this angle is going to be equal to that angle. And we showed all of this in the video where we proved the relationships between the sides of a 30-60-90 triangle. Well, in a regular equilateral triangle, all of the angles are 60 degrees. So this one right over here is going to be 60 degrees, let me do that in a different color. This one down here is going to be 60 degrees. This one down here is going to be 60 degrees. And then this one up here is 60 degrees, but we just split it in two. So this angle is going to be 30 degrees. And then the other thing that we know is that this altitude right over here also will bisect this side down here. So that this length is equal to that length. And we showed all of this a little bit more rigorously on that 30-60-90 triangle video. But what this tells us is well, if this entire length was s, because all three sides are going to be s, it's an equilateral triangle, then each of these, so this part right over here, is going to be s/2. And if this length is s/2, we can use what we know about 30-60-90 triangles to figure out this side right over here. So to figure out what the actual altitude is. And the reason why I care about the altitude is because the area of a triangle is 1/2 times the base times the height, or times the altitude. So this is s/2, the shortest side. The side opposite the 30 degree angle is s/2. Then the side opposite the 60 degree angle is going to be square root of 3 times that. And we know that because the ratio of the sides of a 30-60-90 triangle, if the side opposite the 30 degree side is 1, then the side opposite the 60 degree side is going to be square root of 3 times that. And the side opposite the 90 degree side, or the hypotenuse, is going to be 2 times that. So it's 1 to square root of 3 to 2. So this is the shortest side right over here. That's the side opposite the 30 degree side. The side opposite the 60 degree side is going to be square root of 3 times this. So square root of 3 s over 2. So now we just need to figure out what the area of this triangle is, using area of our triangle is equal to 1/2 times the base, times the height of the triangle. Well, what is the base of the triangle? Well, the entire base of the triangle right over here is s. So that is going to be s. And what is the height of the triangle?" + }, + { + "Q": "At 5:48 , how does Sal know that the intersecting point is exactly at pi/4? You cannot just assume that since the intersecting point is in between two points, that it is located in the middle. Can someone please help clarify this for me. Thank you.\n", + "A": "You can get an exact result: sin(x) = cos(x) sin^2(x) + cos^2(x) = 1 sin^2(x) = 1 - cos^2(x) sin(x) = sqrt(1 - cos^2(x)) sqrt(1 - cos^2(x)) = cos(x) sqrt(1 - cos^2(x))^2 = cos^2(x) 1 - cos^2(x) = cos^2(x) 1 = 2\u00e2\u0080\u00a2cos^2(x) 1/2 = cos^2(x) sqrt(1/2) = cos(x) sqrt(1)/sqrt(2) = cos(x) sqrt(2)/2 = cos(x) arccos(sqrt(2)/2) = x pi/4 = x", + "video_name": "fp9DZYmiSC4", + "timestamps": [ + 348 + ], + "3min_transcript": "The look of these curves should look somewhat familiar at this point. So this is the graph of y is equal to cosine of theta. Now let's do the same thing for sine theta. When theta is equal to 0, sine theta is 0. When theta is pi over 2, sine of theta is 1. When theta is equal to pi, sine of theta is 0. When theta is equal to 3 pi over 2, sine of theta is negative 1. When theta is equal to 2 pi, sine of theta is equal to 0. And so the graph of sine of theta is going to look something like this. My best attempt at drawing it. So just visually, we can think about the question. At how many points do the graphs of y equals sine of theta and y equals cosine of theta intersect for this range for theta? Well, you just look at this graph. You see there's two points of intersection. This point right over here and this point right over here. Just between 0 and 2 pi. These are cyclical graphs. If we kept going, they would keep intersecting with each other. But just over this 2 pi range for theta, you get two points of intersection. Now let's think about what they are, because they look to be pretty close between 0 and pi over 2. And right between pi and 3 pi over 2. So let's look at our unit circle if we can figure out what those values are. It looks like this is at pi over 4. So let's verify that. So let's think about what these values are at pi over 4. So pi over 4 is that angle, or that's the terminal side of it. So this is pi over 4. Pi over 4 is the exact same thing as a 45 degree angle. So we have to figure out what this point is what. What the coordinates are. So let's make this a right triangle. And so what do we know about this right triangle? And I'm going to draw it right over here, to make it a little clear. This is a typical type of right triangle. So it's good to get some familiarity with it. So let me draw my best attempt. Alright. So we know it's a right triangle. We know that this is 45 degrees. What is the length of the hypotenuse? Well this is a unit circle. It has radius 1. So the length of the hypotenuse here is 1. And what do we know about this angle right over here? Well, we know that it too must be 45 degrees, because all of these angles have to add up to 180. And since these two angles are the same, we know that these two sides are going to be the same. And then we could use the Pythagorean Theorem" + }, + { + "Q": "At 5:17 Sal found that the graphs of y=sin(theta) and y=cos(theta) intersect at two points in the given interval. So couldn\u00e2\u0080\u0099t the answer to the question asked (which is 2) be given then and there without doing anything else?\n", + "A": "yes, the answer could have been found easily but he is trying to explain why it is", + "video_name": "fp9DZYmiSC4", + "timestamps": [ + 317 + ], + "3min_transcript": "and think about where they might intersect. So first let's do cosine of theta. When theta is 0-- and let me mark this off. So this is going to be when y is equal to 1. And this is when y is equal to negative 1. So y equals cosine of theta. theta equals 0. Cosine of theta equals 1. So cosine of theta is equal to 1. When theta is equal to pi 2, cosine of theta is 0. When theta is equal to pi, cosine of theta is negative 1. When theta is equal to 3 pi over 2, cosine of theta is equal to 0. That's this right over here. And then finally when theta is 2 pi, cosine of theta is 1 again. And the curve will look something like this. My best attempt to draw it. Make it a nice smooth curve. The look of these curves should look somewhat familiar at this point. So this is the graph of y is equal to cosine of theta. Now let's do the same thing for sine theta. When theta is equal to 0, sine theta is 0. When theta is pi over 2, sine of theta is 1. When theta is equal to pi, sine of theta is 0. When theta is equal to 3 pi over 2, sine of theta is negative 1. When theta is equal to 2 pi, sine of theta is equal to 0. And so the graph of sine of theta is going to look something like this. My best attempt at drawing it. So just visually, we can think about the question. At how many points do the graphs of y equals sine of theta and y equals cosine of theta intersect for this range for theta? Well, you just look at this graph. You see there's two points of intersection. This point right over here and this point right over here. Just between 0 and 2 pi. These are cyclical graphs. If we kept going, they would keep intersecting with each other. But just over this 2 pi range for theta, you get two points of intersection. Now let's think about what they are, because they look to be pretty close between 0 and pi over 2. And right between pi and 3 pi over 2. So let's look at our unit circle if we can figure out what those values are. It looks like this is at pi over 4. So let's verify that. So let's think about what these values are at pi over 4. So pi over 4 is that angle, or that's the terminal side of it. So this is pi over 4. Pi over 4 is the exact same thing as a 45 degree angle." + }, + { + "Q": "\nSo basically in the whole row I can make any changes which would be equivalent in standard equation? (At 1:25 is mentioned the addition)", + "A": "Yes, it s like solving a system of linear equations using the elimination method. Same operations.", + "video_name": "obts_JDS6_Q", + "timestamps": [ + 85 + ], + "3min_transcript": "I will now show you my preferred way of finding an inverse of a 3 by 3 matrix. And I actually think it's a lot more fun. And you're less likely to make careless mistakes. But if I remember correctly from Algebra 2, they didn't teach it this way in Algebra 2. And that's why I taught the other way initially. But let's go through this. And in a future video, I will teach you why it works. Because that's always important. But in linear algebra, this is one of the few subjects where I think it's very important learn how to do the operations first. And then later, we'll learn the why. Because the how is very mechanical. And it really just involves some basic arithmetic for the most part. But the why tends to be quite deep. So I'll leave that to later videos. And you can often think about the depth of things when you have confidence that you at least understand the hows. So anyway, let's go back to our original matrix. And what was that original matrix that I did in the last video? It was 1, 0, 1, 0, 2, 1, 1, 1, 1. So this is what we're going to do. It's called Gauss-Jordan elimination, to find the inverse of the matrix. And the way you do it-- and it might seem a little bit like magic, it might seem a little bit like voodoo, but I think you'll see in future videos that it makes a lot of sense. What we do is we augment this matrix. What does augment mean? It means we just add something to it. So I draw a dividing line. Some people don't. So if I put a dividing line here. And what do I put on the other side of the dividing line? I put the identity matrix of the same size. This is 3 by 3, so I put a 3 by 3 identity matrix. So that's 1, 0, 0, 0, 1, 0, 0, 0, 1. All right, so what are we going to do? What I'm going to do is perform a series of elementary row operations. And I'm about to tell you what are valid elementary row But whatever I do to any of these rows here, I have to do to the corresponding rows here. And my goal is essentially to perform a bunch of operations on the left hand side. And of course, the same operations will be applied to the right hand side, so that I eventually end up with the identity matrix on the left hand side. And then when I have the identity matrix on the left hand side, what I have left on the right hand side will be the inverse of this original matrix. And when this becomes an identity matrix, that's actually called reduced row echelon form. And I'll talk more about that. There's a lot of names and labels in linear algebra. But they're really just fairly simple concepts. But anyway, let's get started and this should become a little clear. At least the process will become clear. Maybe not why it works. So first of all, I said I'm going to perform a bunch of operations here. What are legitimate operations? They're called elementary row operations. So there's a couple things I can do. I can replace any row with that row" + }, + { + "Q": "\nAt 0:08, why does Sal convert to an improper fraction, instead of just adding the 2/3 to the 8 1/3 to get 9. Wouldn't that be much easier? I just think it's a waste of time.", + "A": "becauseit s easier to multiply and divide with it, and he can convert at the end.", + "video_name": "y7QLay8wrW8", + "timestamps": [ + 8 + ], + "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12." + }, + { + "Q": "At 2:10\nCould you add +12y to both sides and subtract 2 you would get\n12y > -27\n= y > -27/12\n= y > -9/4\n", + "A": "That is correct. You could certainly do it that way.", + "video_name": "y7QLay8wrW8", + "timestamps": [ + 130 + ], + "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12." + }, + { + "Q": "\nAt 0:56, do you have to multiply by 3, or can you add 25/3 to both sides? Thanks!!", + "A": "If Sal had added 25/3 on both sides, he would still be left with a fraction (getting him no where, just moving the number) multiplying by 3 gets rid of the denominator of 3.", + "video_name": "y7QLay8wrW8", + "timestamps": [ + 56 + ], + "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12." + }, + { + "Q": "\nat 1:38 Sal says '3 times negative 25 over 3 is just negative 25'. I don't remember it ever being -25. Did Sal just say it wrong, or I'm I missing something?", + "A": "The equation starts out with -25/3. The minus in front of the fraction makes that fraction negative. -25/3 *3/1 = -25 Hope this helps.", + "video_name": "y7QLay8wrW8", + "timestamps": [ + 98 + ], + "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12." + }, + { + "Q": "can you explain it briefly? I couldn't understand from 2:31 to 4:04. pl. explain it again in a theoretical way.\n", + "A": "the third point is the right answer because the limit of f(x) when x approaches k from the negative direction = to the limit of f(x) when x approaches k from the positive direction.", + "video_name": "_bBAiZhfH_4", + "timestamps": [ + 151, + 244 + ], + "3min_transcript": "But then we jump right at x equals 8. And then we continue from 1 again. So this is our other candidate. So these are the three candidates where the function is not continuous. Now let's think about which of these points, which of these x values, does f of k exist. So if one of these is k, does f of k exist? Well f of negative 2 exists. f of 3 exists, right over here. That's f of 3. This is f of negative 2. And f of 8, all exist. So all of these potential k's meet this constraint-- f of k exists, and f is not continuous at k. So that's true for x equals 8, 3, or negative 2. Now let's look at this first constraint. The limit of f of x as x approaches k needs to exist. Well, if we tried to look at x equals 2, the limit of f as x approaches negative 2 here-- the limit from the left, the limit from values lower than negative 2, it looks like our function is approaching something a little higher. It looks like it's a little higher than 3. And the limit from the right, it looks like our function is approaching negative 3. So this one, the limit does not exist. You get a different limit from the left and from the right. Same thing for x equals positive 3. The limit from the left seems like it's approaching 4 and 1/2, while the limit from the right looks like it's approaching negative 4. So this is also not a candidate. So we only have one left. So for this one, the limit should exist. And we see the limit as f of x as x approaches 8 from the negative direction, it looks like f of x is approaching 1. And it looks like, as we approach 8 from the positive direction, the limit of f It's also equal to 1. So your left- and your right-sided limits approach the same value. So the limit of f of x as x approaches 8 is equal to 1. This limit exists. Now, the reason why the function isn't continuous there is that the limit of f of x as x approaches 8, which is equal to 1, it does not equal the value of f of 8. f of 8, we're seeing, is equal to 7. So that's why it meets the last constraint. The function is not continuous there. The function exists. It's defined, f of 8 is equal to 7. And the limit exists. But the limit of f of x as x approaches k is not the same thing, or is not the same as the value of the function evaluated at that point. And so x equals 8 meets all of our constraints. So we could say k is equal to 8." + }, + { + "Q": "\nTo make sure x=4 is a minimum value (at 4:22) can you also use the first derivative test and test around the point x=4?\nIf the derivative goes from decreasing to increasing at x=4 it would indicate a minimum at that point.", + "A": "Correct. I prefer your technique too.", + "video_name": "1TK8V_qmqrk", + "timestamps": [ + 262 + ], + "3min_transcript": "is where the derivative is either 0 or undefined, and see whether those critical points are possibly a minimum or a maximum point. They don't have to be, but those are the ones if we have a minimum or a maximum point, they're going to be one of the critical points. So let's take the derivative. So the derivative s prime-- let me do this in a different color-- s prime of x. I'll do it right over here, actually. The derivative s prime of x with respect to x is going to be equal to 2x times negative 2 times 2x plus 256 times negative 2. So that's minus 512x to the negative 3 power. Now, this is going to be undefined when x is equal to 0. But if x is equal to 0, then y is undefined. So this whole thing breaks down. So that isn't a useful critical point, x equals 0. Well, it's defined everywhere else. So let's think about where the derivative is equal to 0. So when does this thing equal 0? So when does 2x minus 512x to the negative 3 equal 0? Well, we can add 512x to the negative 3 to both sides. So you get 2x is equal to 512x to the negative third power. We can multiply both sides times x to the third power so all the x's go away on the right-hand side. So you get 2x to the fourth is equal to 512. We can divide both sides by 2, and you get x to the fourth power is equal to 256. And so what is the fourth root of 256? Well, we could take the square root of both sides just to help us here. So let's see. So it's going to be x squared is going to be equal to 256 So this is 16. This is going to be x squared is equal to 16 or x is equal to 4. Now that's our only critical point we have, so that's probably the x value that minimizes our sum of squares right over here. But let's make sure it's a minimum value. And to do that, we can just do our second derivative test. So let's figure out. Let's take the second derivative s prime prime of x and figure out if we are concave upwards or downwards when x is equal to 4. So s prime prime of x is going to be equal to 2. And then we're going to have negative 3 times negative 512. So I'll just write that as plus 3 times 512. That's going to be 1,536. Yeah, 3 times 500 is 1,500, 3 times 12 is 36, x to the negative 4 power. And this thing right over here is actually going to be positive for any x." + }, + { + "Q": "Sal in 7:34 86-30=56,and you said it is 50.That does not make any sense\n", + "A": "Yes, he did, and you are correct, but if you continue to watch, you will realize that he catches his mistake.", + "video_name": "ko-cYG3d6ec", + "timestamps": [ + 454 + ], + "3min_transcript": "And maybe we can do it, we can do it in a good color. Maybe we can do it in a couple of different ways. So one thing we could do is we could figure out what this angle is, so we could just subtract this green angle from 86 and we would get our answer. Well, this angle's easy, right, because we know two angles of this triangle, so we could figure that out. Let's just call this, I don't know, let's call this y. So y plus 122 plus 28 degrees is going to equal 180. So y plus 150 is equal to 180. So y is equal to 30 degrees, right? So this is equal to 30 degrees. So this is 30 degrees, and this big angle here is 86. So our goal, let's call that x, so x is going to just be equal out, minus 30. So x is going to be equal to 50 degrees. That was a pretty straightforward problem. Let's see if we could figure that out any other way. Well, we could say instead of doing it that way -- let's forget we just solved it that way. We could say this angle here is supplementary to this 122 degree angle, right, so it has to add up to 180. So this plus 122 is 180, so what does that make this? It makes this 58 degrees, right? This plus this is going to be 180. So we figured out this. If we could figure out this, then we could use this triangle. How do we figure out this angle? Well, we could look at this big triangle here, and we know Let's call this z. So we know that z plus this angle, plus 28, plus this big angle, plus 86 is equal to 180. So z plus, what is this, 106, 114 is equal to 180. So z is equal to, what is this, 66 degrees. I don't know if I'm doing any of my math correctly, but let's just hope. z equals 66. So z is 66, this angle is 58, and now we can use this triangle here to figure out what this angle is, our x. So x plus 66 plus 58 is equal to 180. I already think I might have made a mistake some" + }, + { + "Q": "\nAt 7:17, Sal said X=50' but probably meant X=56'.", + "A": "he fixed it around 7:20 there is a little note that pops up", + "video_name": "ko-cYG3d6ec", + "timestamps": [ + 437 + ], + "3min_transcript": "Let me draw some more. So this is going to be a pretty straightforward drawing. It's pretty much just two triangles next to each other. Like that and then let me draw another line that goes like that, and then we draw a line that goes like that, and I think I have done my drawing. There you go. I'm have done my drawing. So let's see. What do we know about this triangle and what do we need to figure out? I'm going to tell you that this angle here, this big angle here, is 86 degrees. We also know that this angle here is 28 degrees. And we also know that this angle here is 122 degrees. And our goal, our mission in this round is to figure And maybe we can do it, we can do it in a good color. Maybe we can do it in a couple of different ways. So one thing we could do is we could figure out what this angle is, so we could just subtract this green angle from 86 and we would get our answer. Well, this angle's easy, right, because we know two angles of this triangle, so we could figure that out. Let's just call this, I don't know, let's call this y. So y plus 122 plus 28 degrees is going to equal 180. So y plus 150 is equal to 180. So y is equal to 30 degrees, right? So this is equal to 30 degrees. So this is 30 degrees, and this big angle here is 86. So our goal, let's call that x, so x is going to just be equal out, minus 30. So x is going to be equal to 50 degrees. That was a pretty straightforward problem. Let's see if we could figure that out any other way. Well, we could say instead of doing it that way -- let's forget we just solved it that way. We could say this angle here is supplementary to this 122 degree angle, right, so it has to add up to 180. So this plus 122 is 180, so what does that make this? It makes this 58 degrees, right? This plus this is going to be 180. So we figured out this. If we could figure out this, then we could use this triangle. How do we figure out this angle? Well, we could look at this big triangle here, and we know" + }, + { + "Q": "\nAt 2:25 Sal mentions that factorial shows up in other topics of math. What might these other topics be?", + "A": "Some of the things where factorial comes up is really advanced math, but it does appear in what are known as Taylor Polynomials.", + "video_name": "HGoZfzz6dU0", + "timestamps": [ + 145 + ], + "3min_transcript": "If you've been paying particularly close attention to our use of the factorial operator and videos on permutations and combinations. You may or may not have noticed something that might be interesting. So let's just review factorial a little bit. So if I were to say n factorial, that of course is going to be n times n minus one times n minus two. And I would just keep going down, until I go to times one. So I'd keep decrementing n until I get to one and then I would multiply all of those things together. So for example: if I were to say three factorial: that's going to be three times two times one. If I were to say two factorial: that's going to be two times one. One factorial: well I just keep decrementing until I get to one but I don't even have to decrement here I'm already at one. So I just multiply one. So one logical thing is to say: \"Oh hey you know maybe zero factorial is zero.\" I'm just starting with itself, it's already below one, and maybe it is zero. Now what we will see is that this is not the case, that mathematics and mathematicians have decided. Remember this is what's interesting that the factorial operation is something that humans have invented. That they think it's just an interesting thing, it's a useful notation. So they can define what it does, and mathematicians have found it far more useful to define zero factorial as something else. To define zero factorial as -- and there's a little drum roll here... They believe zero factorial should be one, and I know, based on the reasoning, the conceptional reasoning of this, this doesn't make any sense. why this is a useful concept, especially in the world of permutations and combinations, which is frankly where factorial shows up the most. Most of the cases that I've ever seen factorial in anything has been in situations of combinations & permutations, and in a few other things but mainly permutations and combinations. So let's review a little bit, we've said if we have n things and we want to figure out the number of ways to permute them into k spaces it's going to be n factorial over n minus k factorial. Now we've also said that if we had n things that we wanna permute into n places, well this really should just be n factorial, the first place has -- Let's do this, so this is the first place, this is the second place," + }, + { + "Q": "Can, at 1:30, this be done if the coefficient in front is a negative one? If not, would we just use long division?\n", + "A": "It can be done just negative will become positive and same procedure applies.", + "video_name": "1byR9UEQJN0", + "timestamps": [ + 90 + ], + "3min_transcript": "In this expression, we're dividing this third degree polynomial by this first degree polynomial. And we could simplify this by using traditional algebraic long division. But what we're going to cover in this video is a slightly different technique, and we call it synthetic division. And synthetic division is going to seem like a little bit of voodoo in the context of this video. In the next few videos we're going to think about why it actually makes sense, why you actually get the same result as traditional algebraic long division. My personal tastes are not to like synthetic division because it is very, very, very algorithmic. I prefer to do traditional algebraic long division. But I think you'll see that this has some advantages. It can be faster. And it also uses a lot less space on your paper. So let's actually perform this synthetic division. Let's actually simplify this expression. Before we start, there's two important things to keep in mind. We're doing, kind of, the most basic form of synthetic division. And to do this most basic algorithm, this most basic process, you have to look for two The first is that it has to be a polynomial of degree 1. So you have just an x here. You don't have an x squared, an x to the third, an x to the fourth or anything like that. The other thing is, is that the coefficient here is a 1. There are ways to do it if the coefficient was different, but then our synthetic division, we'll have to add a little bit of bells and whistles to it. So in general, what I'm going to show you now will work if you have something of the form x plus or minus something else. So with that said, let's actually perform the synthetic division. So the first thing I'm going to do is write all the coefficients for this polynomial that's in the numerator. So let's write all of them. So we have a 3. We have a 4, that's a positive 4. We have a negative 2. And a negative 1. And you'll see different people draw different types of signs But this is the most traditional. And you want to leave some space right here for another row of numbers. So that's why I've gone all the way down here. And then we look at the denominator. And in particular, we're going to look whatever x plus or minus is, right over here. So we'll look at, right over here, we have a positive 4. Instead of writing a positive 4, we write the negative of that. So we write the negative, which would be negative 4. And now we are all set up, and we are ready to perform our synthetic division. And it's going to seem like voodoo. In future videos, we'll explain why this works. So first, this first coefficient, we literally just bring it straight down. And so you put the 3 there. Then you multiply what you have here times the negative 4. So you multiply it times the negative 4. 3 times negative 4 is negative 12. Then you add the 4 to the negative 12." + }, + { + "Q": "\nAround 2:24 in the video u mentioned turning the 4 in the denominator in -4. So what if the number in the denominator is already negative. would you keep it or change it positive.", + "A": "If you had x-4 in the denominator, you use +4. The X and the minus are ignored. Hope this helps.", + "video_name": "1byR9UEQJN0", + "timestamps": [ + 144 + ], + "3min_transcript": "In this expression, we're dividing this third degree polynomial by this first degree polynomial. And we could simplify this by using traditional algebraic long division. But what we're going to cover in this video is a slightly different technique, and we call it synthetic division. And synthetic division is going to seem like a little bit of voodoo in the context of this video. In the next few videos we're going to think about why it actually makes sense, why you actually get the same result as traditional algebraic long division. My personal tastes are not to like synthetic division because it is very, very, very algorithmic. I prefer to do traditional algebraic long division. But I think you'll see that this has some advantages. It can be faster. And it also uses a lot less space on your paper. So let's actually perform this synthetic division. Let's actually simplify this expression. Before we start, there's two important things to keep in mind. We're doing, kind of, the most basic form of synthetic division. And to do this most basic algorithm, this most basic process, you have to look for two The first is that it has to be a polynomial of degree 1. So you have just an x here. You don't have an x squared, an x to the third, an x to the fourth or anything like that. The other thing is, is that the coefficient here is a 1. There are ways to do it if the coefficient was different, but then our synthetic division, we'll have to add a little bit of bells and whistles to it. So in general, what I'm going to show you now will work if you have something of the form x plus or minus something else. So with that said, let's actually perform the synthetic division. So the first thing I'm going to do is write all the coefficients for this polynomial that's in the numerator. So let's write all of them. So we have a 3. We have a 4, that's a positive 4. We have a negative 2. And a negative 1. And you'll see different people draw different types of signs But this is the most traditional. And you want to leave some space right here for another row of numbers. So that's why I've gone all the way down here. And then we look at the denominator. And in particular, we're going to look whatever x plus or minus is, right over here. So we'll look at, right over here, we have a positive 4. Instead of writing a positive 4, we write the negative of that. So we write the negative, which would be negative 4. And now we are all set up, and we are ready to perform our synthetic division. And it's going to seem like voodoo. In future videos, we'll explain why this works. So first, this first coefficient, we literally just bring it straight down. And so you put the 3 there. Then you multiply what you have here times the negative 4. So you multiply it times the negative 4. 3 times negative 4 is negative 12. Then you add the 4 to the negative 12." + }, + { + "Q": "\nAt 1:48, why did Sal add 9 to each number and then cross out the 9's?", + "A": "This is because 9-9 equals 0. Crossing it out is another way to show that it is 0.", + "video_name": "qzsR7cChujg", + "timestamps": [ + 108 + ], + "3min_transcript": "We have the proportion x minus 9 over 12 is equal to 2/3. And we want to solve for the x that satisfies this proportion. Now, there's a bunch of ways that you could do it. A lot of people, as soon as they see a proportion like this, they want to cross multiply. They want to say, hey, 3 times x minus 9 is going to be equal to 2 times 12. And that's completely legitimate. You would get-- let me write that down. So 3 times x minus 9 is equal to 2 times 12. So it would be equal to 2 times 12. And then you can distribute the 3. You'd get 3x minus 27 is equal to 24. And then you could add 27 to both sides, and you would get-- let me actually do that. So let me add 27 to both sides. And we are left with 3x is equal to-- let's see, 51. And then x would be equal to 17. 17 minus 9 is 8. 8/12 is the same thing as 2/3. So this checks out. Another way you could do that, instead of just straight up doing the cross multiplication , you could say look, I want to get rid of this 12 in the denominator right over here. Let's multiply both sides by 12. So if you multiply both sides by 12, on your left-hand side, you are just left with x minus 9. And on your right-hand side, 2/3 times 12. Well, 2/3 of 12 is just 8. And you could do the actual multiplication. 2/3 times 12/1. 12 and 3, so 12 divided by 3 is 4. 3 divided by 3 is 1. So it becomes 2 times 4/1, which is just 8. And then you add 9 to both sides. So the fun of algebra is that as long as you do something that's logically consistent, you will get the right answer. There's no one way of doing it. So here you get x is equal to 17 again. and both sides by 3, and then that would be functionally equivalent to cross multiplying. Let's do one more. So here another proportion. And this time the x in the denominator. But just like before, if we want we can cross multiply. And just to see where cross multiplying comes from, it's not some voodoo, that you still are doing logical algebra, that you're doing the same thing to both sides of the equation, you just need to appreciate that we're just multiplying both sides by both denominators. So we have this 8 right over here on the left-hand side. If we want to get rid of this 8 on the left-hand side in the denominator, we can multiply the left-hand side But in order for the equality to hold true, I can't do something to just one side. I have to do it to both sides. Similarly, if I want to get this x plus 1 out of the denominator, I could multiply by x plus 1 right over here. But I have to do that on both sides if I want my equality to hold true." + }, + { + "Q": "at 5:24 why does vi wear the hilbert curve\n", + "A": "1. Because she SO rocks it. 2. I m not sure.", + "video_name": "ik2CZqsAw28", + "timestamps": [ + 324 + ], + "3min_transcript": "Three a-squig, a-squig, a-squiggle. Two, nine. Two a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle, two. Nine. Two a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. And 15 all the way over to here. And now, Yeah. I can talk that fast totally. OK, But let's not get too far from your original purpose, which was to nicely fill a page with this squiggle. The nicest page filling squiggles have kind of the same density of squiggle everywhere. You don't want to be clumped up here, but have left over space there, because monsters may start growing in the left over space. On graph paper, you can be kind of precise about it. Say you want a squiggle that goes through every box exactly once, and can be extended infinitely. So you try some of those, and decide that, since the point of them is to fill up all the space, you call them space filling curves. Yeah, that's actually a technical term, but be careful because your curve might actually be a snake, snake, snake, snake, snake, snake, snake, snake, snake, snake-- Also, to make it neater, you draw the lines on the lines, and shift the rules so that you go through each intersection on the graph paper exactly once. Which is the same thing, as far as space is concerned. Here's a space filling curve that a guy named Hilbert made up, because Hilbert was awesome, but he's dead now. Here's the first iteration. For the second one, we're going to build it piece-by-piece by connecting four copies of the first. So here's one. Put the second space away next to it, and connect those. Then turn the page to put the third sideways under the first, and connect those. And then the fourth will be the mirror image of that on the other side. Now you've got one nice curve. The third iteration will be made out of four copies of the second iteration. So first build another second iteration curve out of four copies of the first iteration-- one, two, three, four-- then put another next to it, then two sideways on the bottom. Connect them all up. There you go. The fourth iteration is made of four copies of the third iteration, the same way. If you learn to do the second iteration in one piece, it'll make this go faster. Then build two third iterations facing up next to each other, You can keep going until you run out of room, or you can make each new version the same size by making each line half the length. Or you can make it out of snakes. Or if you have friends, you can each make an iteration of the same size, and put them together. Or invent your own fractal curve so that you could be cool like Hilbert. Who was like, mathematics? I'm going to invent meta-mathematics like a boss." + }, + { + "Q": "\nIs there any design other than the one at 4:19 (The one that that person made?) Also, is Vi her real name, or is it Violet?", + "A": "There are uncountably infinitely many space-filling curves. Her name is Victoria. She prefers to be called Vi.", + "video_name": "ik2CZqsAw28", + "timestamps": [ + 259 + ], + "3min_transcript": "down a squiggle, up, wop, all the way over here. OK, but say you're me and you're in math class. This mean that you have graph paper. Opportunity for precision. You could draw that first curve like this. Squig-a, squig-a, squig-a, squig-a, squig-a, squig-a, squig-a, squig-a. The second iteration to fit squiggles going up and down will have a line three boxes across on top and bottom, if you want the squiggles as close on the grid as possible without touching. You might remind yourself by saying, three a-squig, a-squig, a-squiggle, three, a-squig, a-squig, a-squiggle. The next iteration has a woop, and you have to figure out how long that's going to be. Meanwhile, other lengths change to keep everything close. And, two a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. Two, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle, two, nine. We could write the pattern down like this. So what would the next pattern be? Five. Two a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. Two a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle, two. Nine. Two a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. And 15 all the way over to here. And now, Yeah. I can talk that fast totally. OK, But let's not get too far from your original purpose, which was to nicely fill a page with this squiggle. The nicest page filling squiggles have kind of the same density of squiggle everywhere. You don't want to be clumped up here, but have left over space there, because monsters may start growing in the left over space. On graph paper, you can be kind of precise about it. Say you want a squiggle that goes through every box exactly once, and can be extended infinitely. So you try some of those, and decide that, since the point of them is to fill up all the space, you call them space filling curves. Yeah, that's actually a technical term, but be careful because your curve might actually be a snake, snake, snake, snake, snake, snake, snake, snake, snake, snake-- Also, to make it neater, you draw the lines on the lines, and shift the rules so that you go through each intersection on the graph paper exactly once. Which is the same thing, as far as space is concerned. Here's a space filling curve that a guy named Hilbert made up, because Hilbert was awesome, but he's dead now. Here's the first iteration. For the second one, we're going to build it piece-by-piece by connecting four copies of the first. So here's one. Put the second space away next to it, and connect those. Then turn the page to put the third sideways under the first, and connect those. And then the fourth will be the mirror image of that on the other side. Now you've got one nice curve. The third iteration will be made out of four copies of the second iteration. So first build another second iteration curve out of four copies of the first iteration-- one, two, three, four-- then put another next to it, then two sideways on the bottom. Connect them all up. There you go. The fourth iteration is made of four copies of the third iteration, the same way. If you learn to do the second iteration in one piece, it'll make this go faster. Then build two third iterations facing up next to each other," + }, + { + "Q": "At 1:26, when he divides -5 times the cubed root of y by -5, why can the fives cancel and you are left with the cubed root of y when -5 times the cubed root of y divided by five is equal to -5/-5 times the cubed root of y/-5?\n", + "A": "[ 1/(-5) ] * [ (-5)*cbrt(y) ] = [ [ 1/(-5) ] * (-5) ] * cbrt(y) (associative law) = 1*cbrt(y).", + "video_name": "b6WtwQddAcY", + "timestamps": [ + 86 + ], + "3min_transcript": "We're asked to solve for y. So we're told that the negative of the cube root of y is equal to 4 times the cube root of y plus 5. So in all of these it's helpful to just be able to isolate the cube root, isolate the radical in the equation, and then solve from there. So let's see if we can isolate the radical. So the simplest thing to do, if we want all of the radical onto the left-hand side equation, we can subtract 4 times the cube root of y from both sides of this equation. So let's subtract 4. We want to subtract 4 times the cube root of y from both sides of this equation. And so your left-hand side, you already have negative 1 times the cube root of y, and you're going to subtract 4 more of the cube root of y. So you're going to have negative 5 times the cube root of y. That's your left-hand side. Now the right-hand side-- these two guys-- cancel out. So that cancels out and you're just left with a 5 there. You're just left with this 5 right over there. Now, we've almost isolated this cube root of y. We just have to divide both sides of the equation by negative 5. So you just divide both sides of this equation by negative 5. And these cancel out. That was the whole point. And we are left with the cube root of y is equal to-- 5 divided by negative 5 is negative 1. Now, the cube root of y is equal to negative 1. Well the easiest way to solve this is, let's take both sides of this equation to the third power. This statement right here is the exact same statement as saying y to the 1/3 is equal to negative 1. These are just two different ways of writing the same thing. This is equivalent to taking the 1/3 power. So if we take both sides of this equation to the third third power. And you can see here, y to the 1/3 to the third-- y to the 1/3 and then to the third, that's like saying y to the 1/3 times 3 power. Or y to the first power. That's the whole point of it. If you take the cube root of y to the third power, that's just going to be y. So the left-hand side becomes y. And then the right-hand side, what's negative 1 Negative 1 times negative 1 is 1. Times negative 1 again is negative 1. So we get y is equal to negative 1 as our solution. Now let's make sure that it actually works. Let's go back to our original equation. And I'll put negative 1 in for our y's. We had the negative of the cube root of-- this time, negative 1-- has to be equal to 4 times the cube root of negative 1 plus 5." + }, + { + "Q": "At 2:43, the third and fourth probability terms are identical. Sal says the fourth term correctly but writes it incorrectly. Am I correct?\n", + "A": "just saw that myself. it has to be a typo/writo(so). it should read ttth, not ttht.", + "video_name": "8TIben0bJpU", + "timestamps": [ + 163 + ], + "3min_transcript": "So we're going to have one flip, then another flip, then another flip, then another flip. And this first flip has two possibilities. It could be heads or tails. The second flip has two possibilities. It could be heads or tails. The third flip has two possibilities. It could be heads or tails. And the fourth flip has two possibilities. It could be heads or tails. So you have 2 times 2 times 2 times 2, which is equal to 16 possibilities. 16 possible outcomes when you flip a coin four times. And any one of the possible outcomes would be 1 of 16. So if I wanted to say, so if I were to just say the probability, and I'm just going to not talk about this one heads, if I just take a, just maybe this thing that has three heads right here. This exact sequence of events. This is the first flip, second flip, third flip, fourth flip. Getting exactly this, this is exactly one out of a possible of 16 events. Now with that out of the way, let's think possibilities, involve getting exactly one heads? Well, we could list them. You could get your heads. So this is equal to the probability of getting the heads in the first flip, plus the probability of getting the heads in the second flip, plus the probability of getting the heads in the third flip. Remember, exactly one heads. We're not saying at least one, exactly one heads. So the probability in the third flip, and then, or the possibility that you get heads in the fourth flip. Tails, heads, and tails. And we know already what the probability of each of these things are. There are 16 possible events, and each of these are one of those 16 possible events. So this is going to be 1 over 16, 1 over 16, 1 over 16, And so we're really saying the probability of getting exactly one heads is the same thing as the probability of getting heads in the first flip, or the probability of getting heads-- or I should say the probability of getting heads in the first flip, or heads in the second flip, or heads in the third flip, or heads in the fourth flip. And we can add the probabilities of these different things, because they are mutually exclusive. Any two of these things cannot happen at the same time. You have to pick one of these scenarios. And so we can add the probabilities. 1/16 plus 1/16 plus 1/16 plus 1/16. Did I say that four times? Well, assume that I did. And so you would get 4/16, which is equal to 1/4. Fair enough. Now let's ask a slightly more interesting question. Let's ask ourselves the probability of getting exactly two heads. And there's a couple of ways we can think about it." + }, + { + "Q": "I'm confused at 1:15, because I thought that (4x^2)/5 was equivalent to 4/5 and (x^2)/5. However, Sal sets it up as 4/5x^2 where x^2 is not being divided by 5. Any reasons why this is so?\n", + "A": "It s the difference between addition and multiplication. (\u00f0\u009d\u0091\u008e + \u00f0\u009d\u0091\u008f) \u00e2\u0088\u0095 \u00f0\u009d\u0091\u0090 = \u00f0\u009d\u0091\u008e \u00e2\u0088\u0095 \u00f0\u009d\u0091\u0090 + \u00f0\u009d\u0091\u008f \u00e2\u0088\u0095 \u00f0\u009d\u0091\u0090 (\u00f0\u009d\u0091\u008e \u00e2\u0088\u0099 \u00f0\u009d\u0091\u008f) \u00e2\u0088\u0095 \u00f0\u009d\u0091\u0090 = (\u00f0\u009d\u0091\u008e \u00e2\u0088\u0095 \u00f0\u009d\u0091\u0090)\u00f0\u009d\u0091\u008f", + "video_name": "6nALFmvvgds", + "timestamps": [ + 75 + ], + "3min_transcript": "- [Voiceover] So up here, we are multiplying two rational expressions. And here, we're dividing one rational expression by another one. Now what I encourage you do is pause this video and think about what these become when you multiply them. I don't know, maybe you simplify it a little bit, and I also want you to think about what constraints do you have to put on the x values in order for your resulting expression to be algebraically equivalent to your original expression. So let's work it out together just so you realize what I'm talking about. So this is going to be, in our numerator, we are going to get six x to the third power times two, and our denominator, we're going to have five times three x. And we can see both the numerator and the denominator are divisible by x, so let's divide the denominator by x. We get one there. Let's divide x to the third by x. We get x squared. And we can also see the both the numerator and denominator are divisible by three, so divide six by three, you get two. Divide three by three, you get one. And we are left with two x squared times two, over five times one times one over five. And we can also write that as 4/5 x squared. Now someone just presented you on the street with the expression 4/5 x squared and say, for what x is this defined? I could put any x here, x could be zero because zero squared is zero times 4/5 is just going to be zero, so it does seems to be defined for zero, and that is true. But if someone says, how would I have to constrain this in order for it to be algebraically equivalent to this first expression? Well then, you'd have to say, well, this first expression is not defined for all x. For example, if x were equal to zero, then you would be dividing by zero right over here, which would make this undefined. So you could explicitly call it out, x can not be equal to zero. And so if you want this to be algebraically equivalent, you would have to make that same condition, x cannot be equal to zero. defined this way, if you said, if you said f of x is equal to six x to the third over five times two over, times two over three x; and if someone said, well what is f of zero, you would say f of zero is undefined. Undefined. Why is that? Because you put x equals zero there, you're going to get two divided by zero and it's undefined. But if you said, okay, well, can I simplify this a little bit to get the exact same function? Well, we're saying you can say f of x is equal to 4/5 times x squared. But if you just left it at that, you would get f of zero is equal to zero. So now it would be defined at zero, but then this would make it a different function. These are two different functions the way they're written right over here. Instead, to make them, to make it clear that this is equivalent to that one, you would have to say x cannot be equal to zero." + }, + { + "Q": "\nAt 1:56, Sal writes that the limit of g(x) as x approaches 6 from the negative direction = \"not exist\". Is there an actual notation for it not existing? Maybe something like the zero with a slash through it would do it? A preemptive thanks to anyone who may have an answer to my question.", + "A": "it s kind of common to write DNE", + "video_name": "5f1-Rg3MmKs", + "timestamps": [ + 116 + ], + "3min_transcript": "- [Voiceover] Over here we have the graph of y is equal to g of x. What I wanna do is I wanna figure out the limit of g of x as x approaches positive six from values that are less than positive six or you could say from the left, from the, you could say the negative direction. So what is this going to be equal to? And if you have a sense of it, pause the video and give a go at it. Well, to think about this, let's just take different x-values that approach six from the left and look at what the values of the function are. So g of two, looks like it's a little bit more than one. G of three, it's a little bit more than that. G of four, looks like it's a little under two. G of five, it looks like it's around three. G of 5.5, looks like it's around five. G of, let's say 5.75, looks like it's like nine. And so, as x gets closer and closer to six becomes unbounded, it's just getting infinitely large. And so in some context, you might see someone write that, maybe this is equal to infinity. But infinity isn't, we're not talking about a specific number. If we're talking technically about limits the way that we've looked at it, what is, you'll sometimes see this in some classes. But in this context, especially on the exercises on Khan Academy, we'll say that this does not exist. Not exist. This thing right over here is unbounded. Now this is interesting because the left-handed limit here doesn't exist, but the right-handed limit does. If I were to say the limit of g of x as x approaches six from the right-hand side, well, let's see. We have g of eight is there, go of seven is there, g of 6.5, looks like it's a little less than negative three. G of 6.0000001 is very close to negative three. So it looks like this limit right over here, at least looking at it graphically, it looks like when we approach six from the right, looks like the function is approaching negative three. But from the left, it's just unbounded, so we'll say it doesn't exist." + }, + { + "Q": "\nAt 0:58, what is a p/2-gon?", + "A": "She said that wrong. She meant 2p-gon....I think. I think this because if you have two of the same Polygons (as in p) and overlap them. you have twice as many sides as the first polygons", + "video_name": "CfJzrmS9UfY", + "timestamps": [ + 58 + ], + "3min_transcript": "Let's say you're me, and you're in math class, and you're supposed to be learning about factoring. Trouble is, your teacher is too busy trying to convince you that factoring is a useful skill for the average person to know, with real-world applications ranging from passing your state exams all the way to getting a higher SAT score. And unfortunately, does not have the time to show you why factoring is actually interesting. It's perfectly reasonable for you to get bored in this situation. So like any reasonable person, you start doodling. Maybe it's because your teacher's soporific voice reminds you of a lullaby, but you're drawing stars. And because you're me, you quickly get bored of the usual five-pointed star and get to wondering, why five? So you start exploring. It seems obvious that a five-pointed star is the simplest one, the one that takes the least number of strokes to draw. Sure, you can make a start with four points, but that's not really a star the way you're defining stars. Then there's a six-pointed star, which is also pretty familiar, but totally different from the five-pointed star because it takes two separate lines to make. And then you're thinking about how, much like you can put two triangles together to make a six-pointed star, you can put two squares together to make an eight-pointed star. And any even-numbered star with p points can be made out of two p/2-gons. It is at this point that you realize maybe drawing stars was not the greatest idea. But wait, four would be an even number of points, but that would mean you could make it out of two 2-gons. Maybe you were taught polygons with only two sides But for the purposes of drawing stars, it works out rather well. Sure, the four-pointed star doesn't look too star-like. But then you realize you can make the six-pointed star out of three of these things, and you've got an asterisk, which is definitely a legitimate star. In fact, for any star where the number of points is divisible by 2, you can draw it asterisk style. But that's not quite what you're looking for. What you want is a doodle game, and here it is. Draw p points in a circle, evenly spaced. Pick a number Q. Starting at one point, go around the circle and connect to the point two places over. Repeat. If you get to the starting place before you've covered all the points, jump to a lonely point, and keep going. That's how you draw stars. And it's a successful game, in that previously you were considering running screaming from the room. Or the window was open, so that's an option, too. But now, you're not only entertained but beginning to become curious about the nature of this game. The interesting thing is that the more points you have, the more different ways there is to draw the star. two really good ways to draw them, but they're still simple. I would like to note here that I've never actually left a math class by the window, not that I can say the same for other subjects. Eight is interesting, too, because not only are there a couple nice ways to draw it, but one's a composite of two polygons, while another can be drawn without picking up the pencil. Then there's nine, which, in addition to a couple of other nice versions, you can make out of three triangles. And because you're me, and you're a nerd, and you like to amuse yourself, you decide to call this kind of star a square star because that's kind of a funny name. So you start drawing other square stars. Four 4-gons, two 2-gons, even the completely degenerate case of one 1-gon. Unfortunately, five pentagons is already difficult to discern. And beyond that, it's very hard to see and appreciate the structure of square stars. So you get bored and move on to 10 dots in a circle, which is interesting because this is the first number where you can make a star as a composite of smaller stars-- that is, two boring old five-pointed stars. Unless you count asterisk stars, in which case 8 was two 4s's or four 2's or two 2's and a 4. But 10 is interesting because you can make it as a composite in more than one way because it's divisible by 5, which itself can be made in two ways." + }, + { + "Q": "\nAt 4:24 how is -9 times 1 = -8. Doesn't he mean -9 + 1?", + "A": "He multiplied (10^1)*(10^-9). When you multiply numbers that are the same base to an exponent, you can add the exponents. Because of that, while he s multiplying 10^1 with 10^-9 to get 10^-8, he s adding 1 with -9 to get -8. I hope this helps!", + "video_name": "67jn5Zv-myg", + "timestamps": [ + 264 + ], + "3min_transcript": "and less likely to make a careless mistake-- times 9.2 times 3.01, which is equal to 40.1534. So this is equal to 40.1534. And of course, this is going to be multiplied times 10 to this thing. And so if we simplify this exponent, you get 40.1534 times 10 to the 8 minus 12 is negative 4, minus 5 is negative 9. 10 to the negative 9 power. Now you might be tempted to say that this is already in scientific notation because I have some number here times some power of 10. But this is not quite official scientific notation. And that's because in order for it to be in scientific notation, this number right over here has to be greater than or equal to 1 and less than 10. Essentially, for it to be in scientific notation, you want a non-zero digit right over here. And then you want your decimal and then the rest of everything else. So here-- and you want a non-zero single digit over here. Here we obviously have two digits. This is larger than 10-- or this is greater than or equal to 10. You want this thing to be less than 10 and greater than or equal to 1. So the best way to do that is to write this thing right over here in scientific notation. This is the same thing as 4.01534 times 10. And one way to think about it is to go from 40 to 4, we have to move this decimal over to the left. Moving a decimal over to the left to go from 40 to 4 you're dividing by 10. So you have to multiply by 10 so it all equals out. Divide by 10 and then multiply by 10. Or another way to write it, or another way to think about it, is 4.0 and all this stuff times 10 is going to be 40.1534. to the first power, that's the same thing as 10-- times this thing-- times 10 to the negative ninth power. And then once again, powers of 10, so it's 10 to the first times 10 to the negative 9 is going to be 10 to the negative eighth power. And we still have this 4.01534 times 10 to the negative 8. And now we have written it in scientific notation. Now, they wanted us to express it in both decimal and scientific notation. And when they're asking us to write it in decimal notation, they essentially want us to multiply this out, expand this out. And so the way to think about it-- write these digits out. So I have 4, 0, 1, 5, 3, 4. And if I'm just looking at this number, I start with the decimal right over here." + }, + { + "Q": "\nHi at 3:16 in the video he had the value (40.1534x10^-9) I was wondering why you cant simply move the decimal one to the left and add a -1 to the exponent. Is it because moving the decimal to the left and adding the negative exponent to the ten would change the true value of the numer by a tenth?", + "A": "He does move the decimal over and add a -1 to the exponent. First, he multiplied the numbers 1.45, 9.2 and 3.01. That s where he got the 40.1534. Then he multiplied the 10s. So then he had (40.1534x10^-9). Finally, he puts it into scientific notation in the way you described.", + "video_name": "67jn5Zv-myg", + "timestamps": [ + 196 + ], + "3min_transcript": "in that purple color-- times 10 to the eighth times 10 to the negative 12th power times 10 to the negative fifth power. And this is useful because now I have all of my powers of 10 right over here. I could put parentheses around that. And I have all my non-powers of 10 right over there. And so I can simplify it. If I have the same base 10 right over here, so I can add the exponents. This is going to be 10 to the 8 minus 12 minus 5 power. And then all of this on the left-hand side-- let me get a calculator out-- I have 1.45. and less likely to make a careless mistake-- times 9.2 times 3.01, which is equal to 40.1534. So this is equal to 40.1534. And of course, this is going to be multiplied times 10 to this thing. And so if we simplify this exponent, you get 40.1534 times 10 to the 8 minus 12 is negative 4, minus 5 is negative 9. 10 to the negative 9 power. Now you might be tempted to say that this is already in scientific notation because I have some number here times some power of 10. But this is not quite official scientific notation. And that's because in order for it to be in scientific notation, this number right over here has to be greater than or equal to 1 and less than 10. Essentially, for it to be in scientific notation, you want a non-zero digit right over here. And then you want your decimal and then the rest of everything else. So here-- and you want a non-zero single digit over here. Here we obviously have two digits. This is larger than 10-- or this is greater than or equal to 10. You want this thing to be less than 10 and greater than or equal to 1. So the best way to do that is to write this thing right over here in scientific notation. This is the same thing as 4.01534 times 10. And one way to think about it is to go from 40 to 4, we have to move this decimal over to the left. Moving a decimal over to the left to go from 40 to 4 you're dividing by 10. So you have to multiply by 10 so it all equals out. Divide by 10 and then multiply by 10. Or another way to write it, or another way to think about it, is 4.0 and all this stuff times 10 is going to be 40.1534." + }, + { + "Q": "\nAt 3:13, what does Sal mean when he says this:\n\"When arithmetic is a noun, we call it arithmetic. When arithmetic is an adjective like this, we call it arithmetic, arithmetic mean.\"\n?", + "A": "He is telling you how to pronounce the word, arithmetic, as in an arithmetic mean. Listen carefully to how he says the two words differently. The word has two different pronounciations, depending on their meaning. He puts the emphasis on a different syllable so that we know which way he is using the word.", + "video_name": "h8EYEJ32oQ8", + "timestamps": [ + 193 + ], + "3min_transcript": "in our garden. And let's say we have six plants. And the heights are 4 inches, 3 inches, 1 inch, 6 inches, and another one's 1 inch, and another one is 7 inches. And let's say someone just said-- in another room, not looking at your plants, just said, well, you know, how tall are your plants? And they only want to hear one number. They want to somehow have one number that represents all of these different heights of plants. How would you do that? Well, you'd say, well, how can I find something that-- maybe I want a typical number. Maybe I want some number that somehow represents the middle. Maybe I want the most frequent number. Maybe I want the number that somehow represents the center of all of these numbers. And if you said any of those things, you would actually have done the same things that the people who first came up with descriptive statistics said. They said, well, how can we do it? And we'll start by thinking of the idea of average. has a very particular meaning, as we'll see. When many people talk about average, they're talking about the arithmetic mean, which we'll see shortly. But in statistics, average means something more general. It really means give me a typical, or give me a middle number, or-- and these are or's. And really it's an attempt to find a measure of central tendency. So once again, you have a bunch of numbers. You're somehow trying to represent these with one number we'll call the average, that's somehow typical, or middle, or the center somehow of these numbers. And as we'll see, there's many types of averages. The first is the one that you're probably most familiar with. It's the one-- and people talk about hey, the average on this exam or the average height. And that's the arithmetic mean. I'll write in yellow, arithmetic mean. When arithmetic is a noun, we call it arithmetic. When it's an adjective like this, we call it arithmetic, arithmetic mean. And this is really just the sum of all the numbers divided by-- this is a human-constructed definition that we've found useful-- the sum of all these numbers divided by the number of numbers we have. So given that, what is the arithmetic mean of this data set? Well, let's just compute it. It's going to be 4 plus 3 plus 1 plus 6 plus 1 plus 7 over the number of data points we have. So we have six data points. So we're going to divide by 6. And we get 4 plus 3 is 7, plus 1 is 8, plus 6 is 14," + }, + { + "Q": "At 5:40, Sal says you have to repeat the one since it is repeated. Why? Why can't you just put one 1?\n", + "A": "In the example that Khan is doing, the numbers are 431617. There are two ones in this sequence so the first 1 is repeated. If I was trying to find the average of 58972, I wouldn t have to repeat any number because every number in that sequence was different. Hope this helps!", + "video_name": "h8EYEJ32oQ8", + "timestamps": [ + 340 + ], + "3min_transcript": "15 plus 7 is 22. Let me do that one more time. You have 7, 8, 14, 15, 22, all of that over 6. And we could write this as a mixed number. 6 goes into 22 three times with a remainder of 4. So it's 3 and 4/6, which is the same thing as 3 and 2/3. We could write this as a decimal with 3.6 repeating. So this is also 3.6 repeating. We could write it any one of those ways. But this is kind of a representative number. This is trying to get at a central tendency. Once again, these are human-constructed. No one ever-- it's not like someone just found some religious document that said, this is the way that the arithmetic mean must be defined. It's not as pure of a computation as, say, finding the circumference of the circle, which there really is-- that was kind of-- we studied the universe. And that just fell out of our study of the universe. It's a human-constructed definition Now there are other ways to measure the average or find a typical or middle value. The other very typical way is the median. And I will write median. I'm running out of colors. I will write median in pink. So there is the median. And the median is literally looking for the middle number. So if you were to order all the numbers in your set and find the middle one, then that is your median. So given that, what's the median of this set of numbers going to be? Let's try to figure it out. Let's try to order it. So we have 1. Then we have another 1. Then we have a 3. Then we have a 4, a 6, and a 7. So all I did is I reordered this. And so what's the middle number? Well, you look here. Since we have an even number of numbers, we have six numbers, there's not one middle number. You actually have two middle numbers here. You have two middle numbers right over here. And in this case, when you have two middle numbers, you actually go halfway between these two numbers. You're essentially taking the arithmetic mean of these two numbers to find the median. So the median is going to be halfway in-between 3 and 4, which is going to be 3.5. So the median in this case is 3.5. So if you have an even number of numbers, the median or the middle two, the-- essentially the arithmetic mean of the middle two, or halfway between the middle two. If you have an odd number of numbers, it's a little bit easier to compute. And just so that we see that, let me give you another data set. Let's say our data set-- and I'll order it for us-- let's say our data set was 0, 7, 50, I don't know, 10,000, and 1 million. Let's say that is our data set. Kind of a crazy data set. But in this situation, what is our median?" + }, + { + "Q": "\nat 7:55pm est...what is the shape, center, and mean in statistics?", + "A": "It is the number that is the answer when 2-3 or more are put together in an equation?", + "video_name": "h8EYEJ32oQ8", + "timestamps": [ + 475 + ], + "3min_transcript": "And in this case, when you have two middle numbers, you actually go halfway between these two numbers. You're essentially taking the arithmetic mean of these two numbers to find the median. So the median is going to be halfway in-between 3 and 4, which is going to be 3.5. So the median in this case is 3.5. So if you have an even number of numbers, the median or the middle two, the-- essentially the arithmetic mean of the middle two, or halfway between the middle two. If you have an odd number of numbers, it's a little bit easier to compute. And just so that we see that, let me give you another data set. Let's say our data set-- and I'll order it for us-- let's say our data set was 0, 7, 50, I don't know, 10,000, and 1 million. Let's say that is our data set. Kind of a crazy data set. But in this situation, what is our median? We have an odd number of numbers. So it's easier to pick out a middle. The middle is the number that is greater than two of the numbers and is less than two of the numbers. It's exactly in the middle. So in this case, our median is 50. Now, the third measure of central tendency, and this is the one that's probably used least often in life, is the mode. And people often forget about it. It sounds like something very complex. But what we'll see is it's actually a very straightforward idea. And in some ways, it is the most basic idea. So the mode is actually the most common number in a data set, if there is a most common number. If all of the numbers are represented equally, if there's no one single most common number, then you have no mode. But given that definition of the mode, what is the single most common number in our original data set, in this data set right over here? Well, we only have one 4. We only have one 3. We have one 6 and one 7. So the number that shows up the most number of times here is our 1. So the mode, the most typical number, the most common number here is a 1. So, you see, these are all different ways of trying to get at a typical, or middle, or central tendency. But they do it in very, very different ways. And as we study more and more statistics, we'll see that they're good for different things. This is used very frequently. The median is really good if you have some kind of crazy number out here that could have otherwise skewed the arithmetic mean. The mode could also be useful in situations like that, especially if you do have one number that's showing up a lot more frequently. Anyway, I'll leave you there. And we'll-- the next few videos, we will explore statistics even deeper." + }, + { + "Q": "At 2:11 I don't get how you move the x+1 over to the numerator. What happens to the denominator? Does the numerator become x+2 times x+1? Thanks.\n", + "A": "how do you know which one the denominator is?", + "video_name": "9IUEk9fn2Vs", + "timestamps": [ + 131 + ], + "3min_transcript": "Welcome to the presentation on level four linear equations. So, let's start doing some problems. Let's say I had the situation-- let me give me a couple of problems-- if I said 3 over x is equal to, let's just say 5. So, what we want to do -- this problem's a little unusual from everything we've ever seen. Because here, instead of having x in the numerator, we actually have x in the denominator. So, I personally don't like having x's in my denominators, so we want to get it outside of the denominator into a numerator or at least not in the denominator as So, one way to get a number out of the denominator is, if we were to multiply both sides of this equation by x, you see that on the left-hand side of the equation these two x's will cancel out. And in the right side, you'll just get 5 times x. So this equals -- the two x's cancel out. And you get 3 is equal to 5x. Now, we could also write that as 5x is equal to 3. We either just multiply both sides by 1/5, or you could just do that as dividing by 5. If you multiply both sides by 1/5. The left-hand side becomes x. And the right-hand side, 3 times 1/5, is equal to 3/5. So what did we do here? This is just like, this actually turned into a level two problem, or actually a level one problem, very quickly. All we had to do is multiply both sides of this equation by x. And we got the x's out of the denominator. Let's do another problem. Let's have -- let me say, x plus 2 over x plus 1 is equal to, let's say, 7. So, here, instead of having just an x in the denominator, we have a whole x plus 1 in the denominator. To get that x plus 1 out of the denominator, we multiply both sides of this equation times x plus 1 over 1 times this side. Since we did it on the left-hand side we also have to do it on the right-hand side, and this is just 7/1, times x plus 1 over 1. On the left-hand side, the x plus 1's cancel out. And you're just left with x plus 2. It's over 1, but we can just ignore the 1. And that equals 7 times x plus 1. And that's the same thing as x plus 2. And, remember, it's 7 times the whole thing, x plus 1. So we actually have to use the distributive property. And that equals 7x plus 7. So now it's turned into a, I think this is a level three linear equation. And now all we do is, we say well let's get all the x's on" + }, + { + "Q": "At 4:58 why is it Kt and not K^2/2?\n", + "A": "It s because K is a constant and we re integrating *with respect to t - the dt bit in \u00e2\u0088\u00ab K dt If we were integrating with respect to (a variable) K, then we d have \u00e2\u0088\u00ab K dK and you d be right, the integral would be K\u00c2\u00b2/2", + "video_name": "IYFkXWlgC_w", + "timestamps": [ + 298 + ], + "3min_transcript": "or differential equations in general, you can treat, you can treat these, this derivative in Leibniz notations like fractions, and you can treat these differentials like quantities because we will eventually integrate them. So let's do that. So, we want to put all the Ps and dPs on one side and all the, all the things that involve t or that I guess don't involve P on the other side. So, we could divide both sides by P. We could divide both sides by P, and so we'll have one over P, and you have one over P here and then those will cancel, and then you can multiply both sides times dt. We could multiply both sides times dt. Once again, treating the differential like a quantity which isn't, it really isn't a quantity. You really have to view this as a limit of change in P over change in time. The limit as we get smaller and smaller and smaller changes in time. we can do this, and when we do that we would be left with one over P dP is equal to, is equal to k dt, is equal to k dt. Now we can integrate, integrate both sides. Because this was a separable differential equation, we were able to completely separate the Ps and dPs from the things involving ts or, I guess, the things that aren't involving Ps, and then if we integrate this side, we would get the natural log, the natural log of the absolute value of our population, and we could say plus some constant if we want but we're going to get a constant on this side as well so we could just say that's going to be equal to, that's going to be equal to k, it's going to be equal to k times t, k times t plus some constant. I'll just call that C one, and once again I could've put a plus C two here, but I could've then subtracted the constant from both sides and I would just get the constant on the right hand side. Now, how can I solve for P? Well, the natural log of the absolute value of P is equal to this thing right over here. That means that's the same thing. That means that the absolute value of P, that means that the absolute value of P is equal to e to all this business, e to the, e to the, let me do in the same colors, kt, kt plus, plus C one, plus C one. Now this right over here is the same thing. Just using our exponent properties, this is the same thing as e to the kt, e to the k times t times e, times e to the C one," + }, + { + "Q": "0:08 you said they skipped 7 and 8 why did they skip them are their no tricks for 7 or 8\n", + "A": "sorry I did t read below", + "video_name": "2G_Jr_XpnY4", + "timestamps": [ + 8 + ], + "3min_transcript": "Determine whether 380 is divisible by 2, 3, 4, 5, 6, 9 or 10. They skipped 7 and 8 so we don't have to worry about those. So let's think about 2. So are we divisible by 2? Let me write the 2 here. Well, in order for something to be divisible by 2, it has to be an even number, and to be an even number, your ones digit-- so let me rewrite 380. To be even, your ones digit has to be even, so this has to be even. And for this to be even, it has to be 0, 2, 4, 6 or 8, and this is 0, so 380 is even, which means it is divisible by 2, so it works with 2. So 2 works out. Let's think about the situation for 3. Now, a quick way to think about 3-- so let me write just And if the sum that you get is divisible by 3, then you are divisible by 3. So let's try to do that. So 380, let's add the digits. 3 plus 8 plus 0 is equal to-- 3 plus 8 is 11 plus 0, so it's just 11. And if you have trouble figuring out whether this is divisible by 3, you could then just add these two numbers again, so you can actually add the 1 plus 1 again, and you would get a 2. Regardless of whether you look at the 1 or the 2, neither of these are divisible by 3. So not divisible by 3, and maybe in a future video, I'll explain why this works, and maybe you want to think about why this works. So these aren't divisible by 3, so 380 is not divisible. 380, not divisible by 3, so 3 does not work. Now, I'll think about the situation for 4, so we're thinking about 4 divisibility. So let me write it in orange. So we are wondering about 4. Now, something you may or may not already realize is that 100 is divisible by 4. It goes evenly. So this is 380. So the 300 is divisible by 4, so we just have to figure out whether the leftover, whether the 80, is divisible by 4. Another way to think about it is are the last two digits divisible by 4? And this comes from the fact that 100 is divisible by 4, so everything, the hundreds place or above, it's going to be divisible by 4. You just have to worry about the last part. So in this situation, is 80 divisible by 4?" + }, + { + "Q": "\nAt 3:19, why can we simply assume that cos(theta) is positive?", + "A": "Sal has explained this in one of the past videos, we have assumed x=asin(theta) and we have no problem with that, now based on this assumption sin(theta)=x/a now apply the arcsine function, theta= arcsine(x/a), the range of arcsine function is (-pi/2, pi/2) therefore based on our first assumption theta is between -pi/2 and pi/2, in this domain cos function has a positive value, thus cos(theta) is always positive, hope this helped", + "video_name": "nMrJ6nbOQhQ", + "timestamps": [ + 199 + ], + "3min_transcript": "So let's make the substitution that x is going to be equal to a sine theta, or three times sine of theta. And then we're going to also have to figure out what dx is equal to. So if you take the derivative, we will get dx. We could have dx d theta is equal to 3 cosine theta. Or if we wanted to write it in differential form, we could write that dx is equal to 3 times cosine theta d theta. This is just the derivative of this with respect to theta. And we're ready to substitute back. Our original expression now becomes-- I'll write it in that original green-- 3 sine theta to the third power, which is the same thing as 27 sine-- actually, let me color code it just so you know what parts I'm doing. So this part right over here, x to the third, is now going to become 27 sine to the third power of theta, And then all of this business, this is going to be the square root of 9 minus x squared, so minus 9 sine squared theta. And then dx-- let me do this in a new color-- dx right over here is going to be equal to-- that's not a new color. dx is going to be equal to all of this business, so times 3 cosine theta d theta. And now let's see if we can simplify this business a little bit. Let me do this over to the side. This thing right over here can be written as 9 times 1 minus sine squared theta, which is equal to the square root of 9 times cosine squared theta. as we did in the last video. And so this is going to be equal to 3 cosine theta in orange. So this right over here is 3 cosine theta. And so what does this simplify to? We have a 27 times 3 is 81 times 3 is going to be 243. So this is going to be 243-- I'll put it out front-- times the integral of-- let's see, we're going to have sine cubed theta. And then you're going to have cosine theta times cosine theta, or we could say cosine squared theta. That's this term right over here and this term right over there, and of course, d theta." + }, + { + "Q": "at 2:42 in the video, he is able to rule out cos and pick sin because he says one of them evaluates to 0. What does that mean, its very confusing.\n", + "A": "sin(kx) = 0 if x is 0. cos(kx) = 1 if x is 0. So where the function crosses the y-axis, the function will be at the midline for sine and the maximum for cosine. Since it s at the midline, we can eliminate cosine.", + "video_name": "yHo0CcDVHsk", + "timestamps": [ + 162 + ], + "3min_transcript": "Going from negative 2 to 1, it went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So immediately, we can say, well, look. This is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write \"cosine\" first. Cosine maybe some coefficient times x plus the midline. The midline-- we already figured out-- was minus 2 or negative 2. So it could take that form or it could take f of x is equal to 3 times-- it could be sine of x or sine of some coefficient times x. So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1. Whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0-- so if x is 0, k times 0 is going to be 0-- sine of 0 is 0. So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since, when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. And we just really need to figure out-- what could this constant actually be? at the period of this function. Let's see. If we went from this point-- where we intersect the midline-- and we have a positive slope, the next point that we do that is right over here. So our period is 8. So what coefficient could we have here to make the period of this thing be equal to 8? Well, let us just remind ourselves what the period of sine of x is. So the period of sine of x-- so I'll write \"period\" right over here-- is 2pi. You increase your angle by 2 pi radians or decrease it. you're back at the same point on the unit circle. So what would be the period of sine of kx? Well, now, your x, your input is increasing k times faster." + }, + { + "Q": "At 0:17 what does Sal mean by when he says the midline and the amplitude are not just the plain vanilla function?\n", + "A": "In the plain versions of the sine and cosine functions, the midline would be at y = 0, and the amplitude would be 1. Since this is not the case in the given example, these considerations will come into play.", + "video_name": "yHo0CcDVHsk", + "timestamps": [ + 17 + ], + "3min_transcript": "Write the equation of the function f of x graphed below. And so we have this clearly periodic function. So immediately you might say, well, this is either going to be a sine function or a cosine function. But its midline and its amplitude are not just the plain vanilla sine or cosine function. And we can see that right over here. The midline is halfway between the maximum point and the minimum point. The maximum point right over here, it hits a value of y equals 1. At the minimum points, it's a value of y is equal to negative 5. So halfway between those, the average of 1 and negative 5, 1 plus negative 5 is negative 4. Divided by 2 is negative 2. So this right over here is the midline. So this is y is equal to negative 2. So it's clearly shifted down. Actually, I'll talk in a second about what type of an expression it might be. But now, also, let's think about its amplitude. Its amplitude-- that's how far it might get away from the midline-- we see here. Going from negative 2 to 1, it went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So immediately, we can say, well, look. This is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write \"cosine\" first. Cosine maybe some coefficient times x plus the midline. The midline-- we already figured out-- was minus 2 or negative 2. So it could take that form or it could take f of x is equal to 3 times-- it could be sine of x or sine of some coefficient times x. So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1. Whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0-- so if x is 0, k times 0 is going to be 0-- sine of 0 is 0. So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since, when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. And we just really need to figure out-- what could this constant actually be?" + }, + { + "Q": "Sal begins explaining which function to choose @ about 2:00 .\nI'm still not sure why Cos(0) = 1 and Sin(0) = 0\nIs it because in the unit circle when an angle is 0 Cos = the x value, would still be one, but Sin = the y value which would be 0?\n", + "A": "That s exactly why, yes. For any angle you can refer to the unit circle definition of sine and cosine.", + "video_name": "yHo0CcDVHsk", + "timestamps": [ + 120 + ], + "3min_transcript": "Write the equation of the function f of x graphed below. And so we have this clearly periodic function. So immediately you might say, well, this is either going to be a sine function or a cosine function. But its midline and its amplitude are not just the plain vanilla sine or cosine function. And we can see that right over here. The midline is halfway between the maximum point and the minimum point. The maximum point right over here, it hits a value of y equals 1. At the minimum points, it's a value of y is equal to negative 5. So halfway between those, the average of 1 and negative 5, 1 plus negative 5 is negative 4. Divided by 2 is negative 2. So this right over here is the midline. So this is y is equal to negative 2. So it's clearly shifted down. Actually, I'll talk in a second about what type of an expression it might be. But now, also, let's think about its amplitude. Its amplitude-- that's how far it might get away from the midline-- we see here. Going from negative 2 to 1, it went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So immediately, we can say, well, look. This is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write \"cosine\" first. Cosine maybe some coefficient times x plus the midline. The midline-- we already figured out-- was minus 2 or negative 2. So it could take that form or it could take f of x is equal to 3 times-- it could be sine of x or sine of some coefficient times x. So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1. Whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0-- so if x is 0, k times 0 is going to be 0-- sine of 0 is 0. So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since, when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. And we just really need to figure out-- what could this constant actually be?" + }, + { + "Q": "\nat 3:01 how is it 625 i am getting confused", + "A": "It s 625 because it s 5^4 (he says 5 x 5 x 5 x 5) 5 x 5 (5 squared) =25 25 x 5 (5 cubed) =125 125 x 5 (5^4) = 625", + "video_name": "XZRQhkii0h0", + "timestamps": [ + 181 + ], + "3min_transcript": "You already know that we can view multiplication as repeated addition. So, if we had 2 times 3 (2 \u00d7 3), we could literally view this as 3 2's being added together. So it could be 2 + 2 + 2. Notice this is [COUNTING: 1, 2] 3 2's. And when you add those 2's together, you get 6. What we're going to introduce you to in this video is the idea of repeated multiplication \u2013 a new operation that really can be viewed as repeated multiplication. And that's the operation of taking an 'exponent.' And it sounds very fancy. But we'll see with a few examples that it's not too bad. So now, let's take the idea of 2 to the 3rd power (2^3) \u2013 which is how we would say this. (So let me write this down in the appropriate colors.) So 2 to the 3rd power. (2^3.) So you might be tempted to say, \"Hey, maybe this is 2 \u00d7 3, which would be 6.\" this is repeated multiplication. So if I have 2 to the 3rd power, (2^3), this literally means multiplying 3 2's together. So this would be equal to, not 2 + 2 + 2, but 2 \u00d7 ... (And I\u2019ll use a little dot to signify multiplication.) ... 2 \u00d7 2 \u00d7 2. Well, what's 2 \u00d7 2 \u00d7 2? Well that is equal to 8. (2 \u00d7 2 \u00d7 2 = 8.) So 2 to the 3rd power is equal to 8. (2^3 = 8.) Let's try a few more examples here. What is 3 to the 2nd power (3^2) going to be equal to? And I'll let you think about that for a second. I encourage you to pause the video. So let's think it through. This literally means multiplying 2 3's. So let's multiply 3 \u2013 (Let me do that in yellow.) So this is going to be equal to 9. Let\u2019s do a few more examples. What is, say, 5 to the \u2013 let's say \u2013 5 to the 4th power (5^4)? And what you'll see here is this number is going to get large very, very, very fast. So 5 to the 4th power (5^4) is going to be equal to multiplying 4 5's together. So 5^4 = 5 \u00d7 5 \u00d7 5 \u00d7 5. Notice, we have [COUNTING: 1, 2, 3] 4 5's. And we are multiplying them. This is not 5 \u00d7 4. This is not 20. This is 5 \u00d7 5 \u00d7 5 \u00d7 5. So what is this going to be? Well 5 \u00d7 5 is 25. (5 \u00d7 5 = 25.) 25 \u00d7 5 is 125. (25 \u00d7 5 = 125.) 125 \u00d7 5 is 625. (125 \u00d7 5 = 625.)" + }, + { + "Q": "\n2:03 , Is there a simple way to do exponents?", + "A": "Well, think of the exponent like this: the raised number is how many times you multiply the the other number. For example, 2 to the power of 3 is just 2x2x2.", + "video_name": "XZRQhkii0h0", + "timestamps": [ + 123 + ], + "3min_transcript": "You already know that we can view multiplication as repeated addition. So, if we had 2 times 3 (2 \u00d7 3), we could literally view this as 3 2's being added together. So it could be 2 + 2 + 2. Notice this is [COUNTING: 1, 2] 3 2's. And when you add those 2's together, you get 6. What we're going to introduce you to in this video is the idea of repeated multiplication \u2013 a new operation that really can be viewed as repeated multiplication. And that's the operation of taking an 'exponent.' And it sounds very fancy. But we'll see with a few examples that it's not too bad. So now, let's take the idea of 2 to the 3rd power (2^3) \u2013 which is how we would say this. (So let me write this down in the appropriate colors.) So 2 to the 3rd power. (2^3.) So you might be tempted to say, \"Hey, maybe this is 2 \u00d7 3, which would be 6.\" this is repeated multiplication. So if I have 2 to the 3rd power, (2^3), this literally means multiplying 3 2's together. So this would be equal to, not 2 + 2 + 2, but 2 \u00d7 ... (And I\u2019ll use a little dot to signify multiplication.) ... 2 \u00d7 2 \u00d7 2. Well, what's 2 \u00d7 2 \u00d7 2? Well that is equal to 8. (2 \u00d7 2 \u00d7 2 = 8.) So 2 to the 3rd power is equal to 8. (2^3 = 8.) Let's try a few more examples here. What is 3 to the 2nd power (3^2) going to be equal to? And I'll let you think about that for a second. I encourage you to pause the video. So let's think it through. This literally means multiplying 2 3's. So let's multiply 3 \u2013 (Let me do that in yellow.) So this is going to be equal to 9. Let\u2019s do a few more examples. What is, say, 5 to the \u2013 let's say \u2013 5 to the 4th power (5^4)? And what you'll see here is this number is going to get large very, very, very fast. So 5 to the 4th power (5^4) is going to be equal to multiplying 4 5's together. So 5^4 = 5 \u00d7 5 \u00d7 5 \u00d7 5. Notice, we have [COUNTING: 1, 2, 3] 4 5's. And we are multiplying them. This is not 5 \u00d7 4. This is not 20. This is 5 \u00d7 5 \u00d7 5 \u00d7 5. So what is this going to be? Well 5 \u00d7 5 is 25. (5 \u00d7 5 = 25.) 25 \u00d7 5 is 125. (25 \u00d7 5 = 125.) 125 \u00d7 5 is 625. (125 \u00d7 5 = 625.)" + }, + { + "Q": "At 2:46 we multipy 25 times 5 then 5 again?\n", + "A": "YES!! It is 5 x 5 = 25 25 x 5 = 125 125 x 5 = 625", + "video_name": "XZRQhkii0h0", + "timestamps": [ + 166 + ], + "3min_transcript": "You already know that we can view multiplication as repeated addition. So, if we had 2 times 3 (2 \u00d7 3), we could literally view this as 3 2's being added together. So it could be 2 + 2 + 2. Notice this is [COUNTING: 1, 2] 3 2's. And when you add those 2's together, you get 6. What we're going to introduce you to in this video is the idea of repeated multiplication \u2013 a new operation that really can be viewed as repeated multiplication. And that's the operation of taking an 'exponent.' And it sounds very fancy. But we'll see with a few examples that it's not too bad. So now, let's take the idea of 2 to the 3rd power (2^3) \u2013 which is how we would say this. (So let me write this down in the appropriate colors.) So 2 to the 3rd power. (2^3.) So you might be tempted to say, \"Hey, maybe this is 2 \u00d7 3, which would be 6.\" this is repeated multiplication. So if I have 2 to the 3rd power, (2^3), this literally means multiplying 3 2's together. So this would be equal to, not 2 + 2 + 2, but 2 \u00d7 ... (And I\u2019ll use a little dot to signify multiplication.) ... 2 \u00d7 2 \u00d7 2. Well, what's 2 \u00d7 2 \u00d7 2? Well that is equal to 8. (2 \u00d7 2 \u00d7 2 = 8.) So 2 to the 3rd power is equal to 8. (2^3 = 8.) Let's try a few more examples here. What is 3 to the 2nd power (3^2) going to be equal to? And I'll let you think about that for a second. I encourage you to pause the video. So let's think it through. This literally means multiplying 2 3's. So let's multiply 3 \u2013 (Let me do that in yellow.) So this is going to be equal to 9. Let\u2019s do a few more examples. What is, say, 5 to the \u2013 let's say \u2013 5 to the 4th power (5^4)? And what you'll see here is this number is going to get large very, very, very fast. So 5 to the 4th power (5^4) is going to be equal to multiplying 4 5's together. So 5^4 = 5 \u00d7 5 \u00d7 5 \u00d7 5. Notice, we have [COUNTING: 1, 2, 3] 4 5's. And we are multiplying them. This is not 5 \u00d7 4. This is not 20. This is 5 \u00d7 5 \u00d7 5 \u00d7 5. So what is this going to be? Well 5 \u00d7 5 is 25. (5 \u00d7 5 = 25.) 25 \u00d7 5 is 125. (25 \u00d7 5 = 125.) 125 \u00d7 5 is 625. (125 \u00d7 5 = 625.)" + }, + { + "Q": "at 7:12. when sal was copy pasting the diagram in the middle, what about the two right triangles sticking out. aren't those a part of the diagram too?\n", + "A": "Sal only used those triangles to show that the heights of the parallelograms are equal to \u00f0\u009d\u0091\u008e and \u00f0\u009d\u0091\u008f respectively, thereby showing that their respective areas are definitely equal to \u00f0\u009d\u0091\u008e\u00c2\u00b2 and \u00f0\u009d\u0091\u008f\u00c2\u00b2, which is of course imperative to the proof. After that the triangles can be discarded because they are no longer needed for the rest of the proof.", + "video_name": "rcBaqkGp7CA", + "timestamps": [ + 432 + ], + "3min_transcript": "That looks pretty good. And then the side of length a is going to be out here. So that's a. And then this right over here is b. And I wanted to do that b in blue. So let me do the b in blue. And then this right angle once we've rotated is just sitting right over here. Now, let's do the same exercise. Let's construct a parallelogram right over here. So this is height c. This is height c as well. So by the same logic we used over here, if this length is b, this length is b as well. These are parallel lines. We're going the same distance in the horizontal direction. We're rising the same in the vertical direction. We know that because they're parallel. So this is length b down here. This is length b up there. Now, what is the area of this parallelogram right over there? Well, once again to help us visualize it, we can draw it sitting flat. So this is that side. Then you have another side right over here. They both have length b. And you have the sides of length c. So that's c. That's c. What is its height? Well, you see it right over here. Its height is length b as well. It gives us right there. We know that this is 90 degrees. We did a 90-degree rotation. So this is how we constructed the thing. So given that, the area of a parallelogram is just the base times the height. The area of this parallelogram is b squared. So now, things are starting to get interesting. And what I'm going to do is I'm going to copy and paste the most interesting part of our diagram. Let me see how well I can select it. So let me select this part right over here. So let me copy. And then I'm going to scroll down. And then let me paste it. So this diagram that we've constructed right over here, it's pretty clear what the area of it is, the combined diagram. And let me delete a few parts of it. I want to do that in black so that it cleans it up. So let me clean this thing up, so we really get the part that we want to focus on. So cleaning that up and cleaning this up, cleaning this up right over there. And actually, let me delete this right down here as well," + }, + { + "Q": "\nIn around 0:02, Sal mentions \"hypotenuse\". What douse that mean?", + "A": "also the equation being A squared + B squared = C squared", + "video_name": "rcBaqkGp7CA", + "timestamps": [ + 2 + ], + "3min_transcript": "In case you haven't noticed, I've gotten somewhat obsessed with doing as many proofs of the Pythagorean theorem as I can do. So let's do one more. And like how all of these proofs start, let's construct ourselves a right triangle. So I'm going to construct it so that its hypotenuse sits on the bottom. So that's the hypotenuse of my right triangle. Try to draw it as big as possible, so that we have space to work with. So that's going to be my hypotenuse. And then let's say that this is the longer side that's not the hypotenuse. We can have two sides that are equal. But I'll just draw it so that it looks a little bit longer. Let's call that side length a. And then let's draw this side right over here. It has to be a right triangle. So maybe it goes right over there. That's side of length b. Let me extend the length a a little bit. So it definitely looks like a right triangle. And this is our 90-degree angle. So the first thing that I'm going to do it counterclockwise by 90 degrees. So if I rotate it counterclockwise by 90 degrees, I'm literally just going to rotate it like that and draw another completely congruent version of this one. So I'm going to rotate it by 90 degrees. And if I did that, the hypotenuse would then sit straight up. So I'm going to do my best attempt to draw it almost to scale as much as I can eyeball it. This side of length a will now look something like this. It'll actually be parallel to this over here. So let me see how well I can draw it. So this is the side of length a. And if we cared, this would be 90 degrees. The rotation between the corresponding sides are just going to be 90 degrees in every case. That's going to be 90 degrees. That's going to be 90 degrees. Now, let me draw side b. So it's going to look something like that And this and the right angle is now here. So all I did is I rotated this by 90 degrees counterclockwise. Now, what I want to do is construct a parallelogram. I'm going to construct a parallelogram by essentially-- and let me label. So this is height c right over here. Let me do that white color. This is height c. Now, what I want to do is go from this point and go up c as well. Now, so this is height c as well. And what is this length? What is the length over here from this point to this point going to be? Well, a little clue is this is a parallelogram. This line right over here is going to be parallel to this line." + }, + { + "Q": "Why x-c at an approximate 6:30???\n", + "A": "Do you remember how a parabola f(x) = x^2 can be translated? f(x) = x^2 is a parabola opening up, centered at 0 . A new function g(x) = (x - 2)^2 is just like f(x) = x^2, just moved over 2 units to the right on the x-axis. The same idea can be applied to this video. Instead of approximating the function at 0 , we approximate the function at a new point x = c. Hope that helps.", + "video_name": "1LxhXqD3_CE", + "timestamps": [ + 390 + ], + "3min_transcript": "" + }, + { + "Q": "\nSal Uses The F.O.I.L Method at 1:11 right? Just Making Sure!", + "A": "Yes! He is using the F.O.I.L Method!", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 71 + ], + "3min_transcript": "In this video I want to do a bunch of examples of factoring a second degree polynomial, which is often called a quadratic. Sometimes a quadratic polynomial, or just a quadratic itself, or quadratic expression, but all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, in all of the examples we'll do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x, plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a, and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9." + }, + { + "Q": "\ndoes it matter which variable is which number? for example, in 2:57, he says that a could equal 1, and b could equal 9. So, my question is, could he have said a=9 and b=1 and have the equation come out to be the same? (other than switching the values of a and b of course)", + "A": "Yes, he could have said it that way. The order of factors does not matter.", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 177 + ], + "3min_transcript": "Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9. times x plus 9. And if you multiply these two out, using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x, plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form, so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples. I think the more examples we do the more sense this'll make." + }, + { + "Q": "3:11- can the the number be 5 and 5 instead of 1 and 9\n", + "A": "5 and 5 would not work because the numbers have to fit two things. They have to add to be 10 and they have to multiply to be 9. 5+5 = 10 is okay but 5*5 = 25 which is not 9. Notice that 1+9 = 10 okay; and 1*9 = 9 okay as well. Good luck with your factoring.", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 191 + ], + "3min_transcript": "Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9. times x plus 9. And if you multiply these two out, using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x, plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form, so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples. I think the more examples we do the more sense this'll make." + }, + { + "Q": "At 15:10, why can't a * b be taken directly as (-1)(-72) ? Is that a wrong method?\n", + "A": "a and b also have to sum to -(-18) = 18. They don t, so they re wrong.", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 910 + ], + "3min_transcript": "we've been doing before. So this is the same thing as negative 1 times positive x squared, plus 5x, minus 24. Right? I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1. And you get that right there. Now, same game as before. I need two numbers, that when I take their product I get negative 24. So one will be positive, one will be negative. When I take their sum, it's going to be 5. So let's think about 24 is 1 and 24. Let's see, if this is negative 1 and 24, it'd be positive 23, if it was the other way around, it'd be negative 23. Doesn't work. What about 2 and 12? be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works! So if we pick negative 3 and 8, negative 3 and 8 work. Because negative 3 plus 8 is 5. Negative 3 times 8 is negative 24. So this is going to be equal to-- can't forget that negative 1 out front, and then we factor the inside. Negative 1 times x minus 3, times x plus 8. And if you really wanted to, you could multiply the negative 1 times this, you would get 3 minus x if you did. Or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared, minus 18x, plus 72. Now we just have to think of two numbers, that when I multiply them I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18. So they're the same sign, and their sum is a negative number, they both must be negative. And we could go through all of the factors of 72. But the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9, doesn't work. That turns into 17. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17." + }, + { + "Q": "is 2x the same thing as x squared? for example, around 1:30 when we use the foil method to do (x+a) (x+b) and we do x times x, does it become 2x or x squared?\n", + "A": "2x = x + x x^2 = x * x. So, you want x^2", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 90 + ], + "3min_transcript": "In this video I want to do a bunch of examples of factoring a second degree polynomial, which is often called a quadratic. Sometimes a quadratic polynomial, or just a quadratic itself, or quadratic expression, but all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, in all of the examples we'll do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x, plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a, and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9." + }, + { + "Q": "\nat 15:30 can't we do this without factoring out negative 1?", + "A": "Most likely, but factoring out the -1 is easy, and the tools we have for factoring are known to work on positive x-squared s. In the real world all behavior that is not specifically forbidden is allowed... in mathematics all that is not specifically allowed is forbidden. So I would first have to verify that the factoring procedure I was using was certain to work with negative leading coefficients.", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 930 + ], + "3min_transcript": "we've been doing before. So this is the same thing as negative 1 times positive x squared, plus 5x, minus 24. Right? I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1. And you get that right there. Now, same game as before. I need two numbers, that when I take their product I get negative 24. So one will be positive, one will be negative. When I take their sum, it's going to be 5. So let's think about 24 is 1 and 24. Let's see, if this is negative 1 and 24, it'd be positive 23, if it was the other way around, it'd be negative 23. Doesn't work. What about 2 and 12? be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works! So if we pick negative 3 and 8, negative 3 and 8 work. Because negative 3 plus 8 is 5. Negative 3 times 8 is negative 24. So this is going to be equal to-- can't forget that negative 1 out front, and then we factor the inside. Negative 1 times x minus 3, times x plus 8. And if you really wanted to, you could multiply the negative 1 times this, you would get 3 minus x if you did. Or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared, minus 18x, plus 72. Now we just have to think of two numbers, that when I multiply them I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18. So they're the same sign, and their sum is a negative number, they both must be negative. And we could go through all of the factors of 72. But the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9, doesn't work. That turns into 17. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17." + }, + { + "Q": "\nI have a question at 6:11 min in the video. I would like to know if the answers for x1 and x2 are negative or positive? I ask because I see that clearly they are -3 and -8 in the binomial answer (x-3)(x-8) but all the online Quadratic Equation Solvers show that the answers are positive numbers i.e. x1=3 and x2=8. thx", + "A": "They would be positive, because you want x-3 or x-8 to be equal to zero, so the right side is equal to zero. 3-3=0 and 8-8=0, so your answers would be positive.", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 371 + ], + "3min_transcript": "already did 10x, let's do a different number-- x squared plus 15x, plus 50. And we want to factor this. Well, same drill. We have an x squared term. We have a first degree term. This right here should be the sum of two numbers. And then this term, the constant term right here, should be the product of two numbers. So we need to think of two numbers that, when I multiply them I get 50, and when I add them, I get 15. And this is going to be a bit of an art that you're going to develop, but the more practice you do, you're going to see that it'll start to come naturally. So what could a and b be? Let's think about the factors of 50. It could be 1 times 50. 2 times 25. Let's see, 4 doesn't go into 50. It could be 5 times 10. Let's try out these numbers, and see if any of these add up to 15. So 1 plus 50 does not add up to 15. 2 plus 25 does not add up to 15. But 5 plus 10 does add up to 15. So this could be 5 plus 10, and this could be 5 times 10. So if we were to factor this, this would be equal to x plus 5, times x plus 10. And multiply it out. I encourage you to multiply this out, and see that this is indeed x squared plus 15x, plus 10. In fact, let's do it. x times x, x squared. x times 10, plus 10x. 5 times x, plus 5x. 5 times 10, plus 50. Notice, the 5 times 10 gave us the 50. The 5x plus the 10x is giving us the 15x in between. So it's x squared plus 15x, plus 50. signs in here. Let's say I had x squared minus 11x, plus 24. Now, it's the exact same principle. I need to think of two numbers, that when I add them, need to be equal to negative 11. a plus b need to be equal to negative 11. And a times b need to be equal to 24. Now, there's something for you to think about. When I multiply both of these numbers, I'm getting a positive number. I'm getting a 24. That means that both of these need to be positive, or both of these need to be negative. That's the only way I'm going to get a positive number here. Now, if when I add them, I get a negative number, if these were positive, there's no way I can add two positive numbers and get a negative number, so the fact that their sum is" + }, + { + "Q": "\nAt 0:45, where did you get the (x+a)(x+b) from?", + "A": "He was just showing you the general method, it wasn t directly part of the main problem.", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 45 + ], + "3min_transcript": "In this video I want to do a bunch of examples of factoring a second degree polynomial, which is often called a quadratic. Sometimes a quadratic polynomial, or just a quadratic itself, or quadratic expression, but all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, in all of the examples we'll do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x, plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a, and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9." + }, + { + "Q": "\nat 0:01 til 4:14 way do we need to emphasise this point in short way do i need this video !", + "A": "One real life situation that this is used for is exchanging money or making change. Because many people have not studied this idea, they get very nervous when exchanging money. They don t feel confident that Ten $20 bills is the same as Two $100 bills. Similarly they don t feel confident that 35 dimes is the same as 3 $1 bills and 10 nickles.", + "video_name": "3szFVS5p_7A", + "timestamps": [ + 1, + 254 + ], + "3min_transcript": "I want to think about all of the different ways we can represent value in the number 21.3. So one is to just look straight up at the place values. This 2 is in the tens place, so it literally represents 2 tens. So this is equal to 20, 2 times 10. This 1 is literally equal to 1. It's 1 one. And then this 3 is 3/10, so plus 3/10. But now I want to rearrange or regroup the value in these places. So, for example, I could take 1 from the ones place and give it to the tenths place. So let's see how that would work. So we're going to take 1 away from the ones place, and so it's going to become a 0. And we're going to give it to the tenths place. And what we're going to see is that that's going to make the tenths place into 13/10. Now, does that actually make sense that I took 1 from here and it essentially added 10 to the tenths place? So we still have 2 tens. So this is still going to be 2 tens. Now we have plus 0 ones. And we essentially wanted to write that 1 that we took away from the ones place in terms of tenths. So if we were to write this in terms of tenths, it would be 10/10 plus the 3/10 that were already there. And so this is going to be equal to 13/10. Let me write that down. So this is equal to 20. That's the color you can't see. This is equal to 20 plus 0 ones, so 2 tens plus 0 ones plus 13 tens. Let's do another example with this exact same number. So once again, 21.3. And I'll write it out again. This is equal to 20 plus 1. Plus 1 plus 3/10, plus 3 over 10. Now, I could take 1 from the tens place so that this becomes just 1. Now what do I do with that 10? Well, let's say with that 10 I give 9 of it to the ones place. So I give 9 of it to the ones place so that this becomes 10. And I still have 1 left over, and I give it to the tenths place, so that's going to become 13/10. So what did I just do? Well, I could rewrite this. Let me be clear what I did. This is the same thing as 1 plus 9. Actually, let me write it this way. 1 plus 9 plus 1. That's obviously the same thing-- 10" + }, + { + "Q": "At 8:05, Can anybody say please what means \"slope=6\"? What means that value?\n", + "A": "Slope is: (change in y)/(change in x) In linear notation: y = m\u00e2\u0080\u00a2x + b m is the slope derived from: (y(2) - y(1))/(x(2) - x(1)) b is the y intercept: b = y - m\u00e2\u0080\u00a2x Slope = 6 means that for every \u00c2\u00b11 in x, results in a \u00c2\u00b16 in y.", + "video_name": "IePCHjMeFkE", + "timestamps": [ + 485 + ], + "3min_transcript": "" + }, + { + "Q": "\nHow did he get 6 delta x as part of the y coordinate at 4:13?", + "A": "He foiled the (3 + delta X)^2. 3 * 3 + 3 * delta X + delta X * 3 + delta X * delta X = 9 + 3deltaX + 3deltaX + deltaX^2 = 9 + 6deltaX + deltaX^2", + "video_name": "IePCHjMeFkE", + "timestamps": [ + 253 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 6:33 how does it equal 6+deltax and not 6+deltax^2?", + "A": "Immediately before that point we ve determined that the slope is equal to (6\u00ce\u0094x + (\u00ce\u0094x)\u00c2\u00b2)/\u00ce\u0094x. Divide 6\u00ce\u0094x by \u00ce\u0094x to get 6 and divide (\u00ce\u0094x)\u00c2\u00b2 by \u00ce\u0094x to get \u00ce\u0094x.", + "video_name": "IePCHjMeFkE", + "timestamps": [ + 393 + ], + "3min_transcript": "" + }, + { + "Q": "1:58 You can solve the quadratic equation without using the quadratic formula like this:\n4r^2 - 8t + 3 = 0 |Divide by four:\nr^2 - (8/4)r + 3/4 = 0 |Multiply both numerator and denominator of 3/4 by four:\nr^2 - (8/4)r + 12/16 = 0 |Figure out two numbers a,b that satisfy a + b = -8 and ab = 12:\na = -2, b = -6 |Substitute a,b into (r + a/4) (r + b/4) = 0:\n(r - 2/4) (r - 6/4) = 0\n(r - 1/2) (r - 3/2) = 0\nr = 1/2 or r = 3/2\n", + "A": "It seems easier to just -> (2r - 1)(2r - 3)", + "video_name": "3uO_uPb9H8w", + "timestamps": [ + 118 + ], + "3min_transcript": "Let's solve another 2nd order linear homogeneous differential equation. And this one-- well, I won't give you the details before I So the differential equation is 4 times the 2nd derivative of y with respect to x, minus 8 times the 1st derivative, plus 3 times the function times y, is equal to 0. And we have our initial conditions y of 0 is equal to 2. And we have y prime of 0 is equal to 1/2. Now I could go into the whole thing y is equal to e to the rx is a solution, substitute it in, then factor out e to the rx, and have the characteristic equation. And if you want to see all of that over again, you might want to watch the previous video, just to see where that characteristic equation comes from. But in this video, I'm just going to show you, literally, how quickly you can do these type of problems mechanically. characteristic equation is going to be-- and I'll do this in a different color-- 4r squared minus 8r plus 3r is equal to 0. And watch the previous video if you don't know where this characteristic equation comes from. But if you want to do these problems really quick, you just substitute the 2nd derivatives with an r squared, the first derivatives with an r, and then the function with-- oh sorry, no. This is supposed to be a constant-- And then the coefficient on the original function is just a constant, right? I think you see what I did. 2nd derivative r squared. 1st derivative r. No derivative-- you could say that's r to the 0, or just 1. But this is our characteristic equation. And now we can just figure out its roots. This is not a trivial one for me to factor so, if it's not So we could say the solution of this is r is equal to negative b-- b is negative 8, so it's positive 8-- 8 plus or minus the square root of b squared. So that's 64, minus 4 times a which is 4, times c which is 3. All of that over 2a. 2 times 4 is 8. That equals 8 plus or minus square root of 64 minus-- what's 16 times 3-- minus 48. All of that over 8. What's 64 minus 48? Let's see, it's 16, right? 10 plus 48 is 58, then another-- so it's 16. So we have r is equal to 8 plus or minus the square root" + }, + { + "Q": "\nAt 6:04 is multiplying only top equation right? It seems intuitive, but doesn't that affect the solution?", + "A": "No, that one of the most common ways to save a system of equations. Since you are multiplying the whole equation (both sides of the equal sign), you are not modifying the equation at all.", + "video_name": "3uO_uPb9H8w", + "timestamps": [ + 364 + ], + "3min_transcript": "This differential equations problem was literally just a problem in using the quadratic equation. And once you figure out the r's you have your general solution. And now we just have to use our initial conditions. So to know the initial conditions, we need to know y of x, and we need to know y prime of x. Let's just do that right now. So what's y prime? y prime of our general solution is equal to 3/2 times c1 e to the 3/2 x, plus-- derivative of the inside-- 1/2 times c2 e to the 1/2 x. And now let's use our actual initial conditions. I don't want to lose them-- let me rewrite them down here so I can scroll down. So we know that y of 0 is equal to 2, and y prime of 0 is equal to 1/2. So let's use that information. So y of 0-- what happens when you substitute x is equal to 0 here.? You get c1 times e to the 0, essentially, so that's just 1, plus c2-- well that's just e to the 0 again, because x is 0-- is equal to-- so this is, when x is equal to 0, what is y? y is equal to 2. Y of 0 is equal to 2. And then let's use the second equation. So when we substitute x is equal to 0 in the derivative-- so when x is 0 we get 3/2 c1-- this goes to 1 again-- plus 1/2 c2-- this is 1 again, e to the 1/2 half times 0 is e to the 0, which is 1-- is equal to-- so when x is 0 for the derivative, y is equal to 1/2, or the derivative is 1/2 at And now we have two equations and two unknowns, and we could solve it a ton of ways. I think you know how to solve them. Let's multiply the top equation-- I don't know-- let's multiply it by 3/2, and what do we get? We get-- I'll do it in a different color-- we get 3/2 c1 plus 3/2 c2 is equal to-- what's 3/2 times 2? It's equal to 3. And now, let's subtract-- well, I don't want to confuse you, so let's just subtract the bottom from the top, so this cancels out. What's 1/2 minus 3/2? 1/2 minus 1 and 1/2. Well, that's just minus 1, right? So minus c2 is equal to-- what's 1/2 minus 3?" + }, + { + "Q": "\nWhat is the \"a\" he is talking about in 5:00?", + "A": "Sal is just using a as an arbitrary critical point. It can mean any critical point, and it s analogous to using letters such as x or \u00ce\u00b2 to represent any given quantity.", + "video_name": "lFQ4kMcODzU", + "timestamps": [ + 300 + ], + "3min_transcript": "The slope was undefined right at the point. But it did switch signs from positive to negative as we crossed that critical point. So these both meet our criteria for being a maximum point. So, so far our criteria seems pretty good. Now let's make sure that somehow this point right over here, which we identified in the last video as a critical point, let's make-- and I think we called this x0, this was x1, this was x2. So this is x3. Let's make sure that this doesn't somehow meet the criteria because we see visually that this is not a maximum point. So as we approach this, our slope is negative. And then as we cross it, our slope is still negative, we're still decreasing. So we haven't switched signs. So this does not meet our criteria, which is good. Now let's come up with the criteria for a minimum point. And I think you could see where this is likely to go. We can see that. It's a local minimum just by looking at it. And what's the slope of doing as we approach it? So the function is decreasing, the slope is negative as we approach it. f prime of x is less than 0 as we approach that point. And then right after we cross it-- this wouldn't be a minimum point if the function were to keep decreasing somehow. The function needs to increase now. So let me do that same green. So right after that, the function starts increasing again. f prime of x is greater than 0. So this seems like pretty good criteria for a minimum point. f prime of x switches signs from negative to positive as we cross a. If we have some critical point a, the function takes on a minimum value from negative to positive as we cross a, from negative to positive. Now once again, this point right over here, this critical point x sub 3 does not meet that criteria. We go from negative to 0 right at that point then go to negative again. So this is not a minimum or a maximum point." + }, + { + "Q": "\nIn 2:12 how did you change 5/5 into a whole?", + "A": "5/5 is equal to 1. (A whole) Imagine a pie cut into 5 pieces. If all five pieces are there, then the pie is whole.", + "video_name": "U44my48zgFE", + "timestamps": [ + 132 + ], + "3min_transcript": "I have a square here divided into one, two, three, four, five, six, seven, eight, nine equal sections. And we've already seen that if we were to shade in one of these sections, if we were to select one of these sections, let's say the middle one right over here, this is one out of the nine equal sections. So if someone said, what fraction of the whole does this purple square represent? Well, you would say, well, that represents 1/9 of the whole. This thing right over here represents 1/9. Now what would happen if we shaded in more than that? So let's say we shaded in this one and this one, let me shade it in a little bit better. And this one and this one right over here. Now what fraction of the whole have we shaded in? Well, each of these, we've already seen, each of these represent 1/9. So that's 1/9, that's 1/9. When I say 1/9, I could also say a ninth. So this is 1/9 or a ninth, so each of these But how many of these ninths do we have shaded in? Well we have one, two, three, four shaded in. So now we have a total of 4/9 shaded it. 4 of the 9 equal sections are shaded in. So 4/9 of the whole is shaded in. Now let's make things a little bit more interesting. Let's shade in. So here I have five equal sections. Let me write this down. I have five equal sections. And let me shade in five of them. So one, two, three, four, five. We already know that each of these sections, each of these situated in sections represent 1/5. So 1/5, another way of saying that is a fifth, is 1/5. But now how much do I have shaded in? Well I have five out of the five equal sections shaded, or I have 5/5 shaded in. And you might be saying, wait, wait, if I have 5/5 shaded in, I've got the whole thing shaded in. And you would be absolutely right. 5/5 is equal to the whole. Now what I want you to do is pause this video and write down on a piece of paper or at least think in your head, what fraction of each of these wholes is shaded in? So let's go to this first one. We have one, two, three, four, five, six equal sections. And we see that one, two, three, four are shaded in. So 4/6 of this figure is shaded in. Let's go over here. We have one, two, three, four, five equal sections. And one, two, three, four are shaded in. So here, 4/5 of this circle is shaded in. Now in this figure, I have two equal sections and both of them" + }, + { + "Q": "at (0:30), where did the 5t come from? I watched the last video and was it because the car is going 5 m/s?\n", + "A": "Yes. He is simply putting the base information from the previous video so that we can continue with the same problem.", + "video_name": "wToSIQJ2o_8", + "timestamps": [ + 30 + ], + "3min_transcript": "In the last video, we used a set of parametric equations to describe the position of a car as it fell off of a cliff, and the equations were x as a function of t was-- and I will write that-- x as a function of t was equal to 10, I believe. I did this a couple of hours ago, so I think 10 plus 5t. And y as a function of t was equal to 50 minus 5t squared. And the graph looks something like this. Let me redraw it. It never hurts. That was the y-axis. That's the x-axis. And we saw that at t equals 0, and we could try t equals 0 here, and we'll get the point 10 comma 50. So the point 10 comma 50 was right there. That was at t equal to 0, and then we plotted a few points in the last video. I think t equals 1 was there. t equals 2 was, like, there, and t equal 3 looked something it hurtled to the ground. But this parametric equation actually doesn't just describe this part of the curve. It describes a curve that goes in both directions forever. So it describes a curve that does something like this. If you actually plot t equals minus 1, what do you get here? You get minus 5. So 10 minus 5 is 5. If you put a minus 1 here, this becomes a plus 1. So minus 1, so you get 5, 45. So you get that point right there. And if you did minus 2, you're going to get a point that looks something like that. And minus 3, you're going to get a point something like that. So the whole curve described by this parametric equation will look something like this. Let me do it in a different color. It looks something like this. And the direction as x increases-- sorry, as This is t is equal to minus 3, minus 2, minus 1, 0, 1, 2, and so forth and so on. So, in the last example, our path was actually just a subset of the path described by this parametric equation. And I'm saying all of this because sometimes it's useful to just bound your parametric equation and say this is a path only for certain values of t. And in our example, it was when we leave this point, which is at t equals 0, all the way to when we hit the ground. So if we want to know those boundaries, well, we know the first boundary is t equals 0, so this in the last example, for when we actually talked about a car careening off of a cliff, it was between 0, so t is greater than or And then we have to figure out what t-value makes us hit the ground? Or when does y equal 0? Because when y equals 0-- y is our altitude, so when y equals 0, we've hit the ground." + }, + { + "Q": "\nI assume that at 4:34 he meant to say \"1.5\" rather than 1.25?", + "A": "Yeah, but I think he did it correctly, so...", + "video_name": "dEAk0BHBYCM", + "timestamps": [ + 274 + ], + "3min_transcript": "I'm giving up that area. I'm giving up that area there. But this is just an approximation, and maybe if I had many more rectangles, it would be a better approximation. So let's figure out what the areas of each of the rectangles So the area of this first rectangle is going to be the height, which is f of 1, times the base, which is delta x. The area of the second rectangle is going to be the height, which we already said was f of 1.5, times the base, times delta x. The height of the third rectangle is going to be the function evaluated at its left boundary, so f of 2-- so plus f of 2 times the base, times delta x. And then, finally, the area of the third rectangle, the height is the function evaluated at 2.5, so plus-- that's a different color than what I wanted to use. I wanted to use that orange color-- so This is going to be equal to our approximate area-- let me make it clear-- approximate area under the curve, just the sum of these rectangles. So let's evaluate this. So this is going to be equal to f of-- it's going to be equal to the function evaluated at 1. 1 squared plus 1 is just 2, so it's going to be 2 times 1/2. Plus the function evaluated at 1.25. 1.25 squared is 2.25. And then you add 1 to it, it becomes 3.25. So plus 3.25 times 1/2. And then we have the function evaluated at 2. Well, 2 squared plus 1 is 5, so it's 5 times 1/2. And then finally, you have the function evaluated at 2.5. 2.5 squared is 6.25 plus 1. And just to make the math simpler, we can factor out the 1/2. So this is going to be equal to-- write 1/2 in a neutral color-- 1/2 times 2 plus 3.25 plus 5 plus 7.25, which is equal to 1/2 times-- let's see if I can do this in my head. 2 plus 5 is easy. That's 7. 3 plus 7 is 10. And then we have 0.25 plus 0.25, so it's going to be 10.5 plus 7 is 17.5. So 1/2 times 17.5, which is equal to 8.75," + }, + { + "Q": "\nAt 4:58, shouldn't Sal put a comma instead of a addition sign?", + "A": "Good question! The complex plane isn t the same as the usual x, y plane because you plot complex numbers on it like 3 + 4i (3 in the x direction and 4 in the imaginary direction) while on the x, y plane you plot an ordered pair like (4,5) (4 in the x direction and 5 in the y). So no there shouldn t be a comma because it is one complex number rather than two real numbers. A complex number has a real part and an imaginary part so you plot them separately on the complex plane. I hope that answers your question :)", + "video_name": "Efoeqb6tC88", + "timestamps": [ + 298 + ], + "3min_transcript": "Well along the imaginary axis we're going from negative one to three so the distance there is four. So now we can apply the Pythagorean theorem. This is a right triangle, so the distance is going to be equal to the distance. Let's just say that this is x right over here. x squared is going to be equal to seven squared, this is just the Pythagorean theorem, plus four squared. Plus four squared or we can say that x is equal to the square root of 49 plus 16. I'll just write it out so I don't skip any steps. 49 plus 16, now what is that going to be equal to? That is 65 so x, that's right, 59 plus another 6 is 65. x is equal to the square root of 65. There's no factors that are perfect squares here, this is just 13 times five so we can just leave it like that. x is equal to the square root of 65 so the distance in the complex plane between these two complex numbers, square root of 65 which is I guess a little bit over eight. Now what about the complex number that is exactly halfway between these two? Well to figure that out, we just have to figure out what number has a real part that is halfway between these two real parts and what number has an imaginary part that's halfway between these two imaginary parts. So if we had some, let's say that some complex number, let's just call it a, is the midpoint, it's real part is going to be the mean of these two numbers. So it's going to be two plus negative five. Two plus negative five over two, over two, and it's imaginary part is going to be the mean of these two numbers so plus, plus three minus one. and this is equal to, let's see, two plus negative five is negative three so this is negative 3/2 plus this is three minus 1 is negative, is negative two over two is let's see three, make sure I'm doing this right. Three, something in the mean, three minus one is two divided by two is one, so three plus three. Negative 3/2 plus i is the midpoint between those two and if we plot it we can verify that actually makes sense. So real part negative 3/2, so that's negative one, negative one and a half so it'll be right over there and then plus i so it's going to be right over there." + }, + { + "Q": "\nAt 3:07 Sal says that A, B, and D are non-collinear. I get that part. But how are they not coplanar? Or am I seeing this wrong? If A and B exist on all plains, shouldn't they be exist on the same plane as D?", + "A": "They are coplanar because any three non-collinear points create a plane, but they are not coplanar on plane S in the diagram.", + "video_name": "J2Qz-7ZWDAE", + "timestamps": [ + 187 + ], + "3min_transcript": "So one point by itself does not seem to be sufficient to define a plane. Well, what about two points? Let's say I had a point, B, right over here. Well, notice the way I drew this, point A and B, they would define a line. For example, they would define this line right over here. So they would define, they could define, this line right over here. But both of these points and in fact, this entire line, exists on both of these planes that I just drew. And I could keep rotating these planes. I could have a plane that looks like this. I could have a plane that looks like this, that both of these points actually sit on. I'm essentially just rotating around this line that is defined by both of these points. So two points does not seem to be sufficient. Let's try three. So there's no way that I could put-- Well, let's be careful here. So I could put a third point right over here, point C. And C sits on that line, So it doesn't seem like just a random third point is sufficient to define, to pick out any one of these planes. But what if we make the constraint that the three points are not all on the same line. Obviously, two points will always define a line. But what if the three points are not collinear. So instead of picking C as a point, what if we pick-- Is there any way to pick a point, D, that is not on this line, that is on more than one of these planes? We'll, no. If I say, well, let's see, the point D-- Let's say point D is right over here. So it sits on this plane right over here, one of the first ones that I drew. So point D sits on that plane. Between point D, A, and B, there's only one plane that all three of those points sit on. So a plane is defined by three non-colinear points. So D, A, and B, you see, do not sit on the same line. A and B can sit on the same line. D and A can sit on the same line. But A, B, and D does not sit on-- They are non-colinear. So for example, right over here in this diagram, we have a plane. This plane is labeled, S. But another way that we can specify plane S is we could say, plane-- And we just have to find three non-collinear points on that plane. So we could call this plane AJB. We could call it plane JBW. We could call it plane-- and I could keep going-- plane WJA. But I could not specify this plane, uniquely, by saying plane ABW. And the reason why I can't do this is because ABW are all on the same line." + }, + { + "Q": "\n1:47 What does sufficient mean?", + "A": "Sufficient = enough to fulfill one s needs. It s not sufficient (/it s not enough) to say that three points define a plane, because we also need the points to be non-collinear.", + "video_name": "J2Qz-7ZWDAE", + "timestamps": [ + 107 + ], + "3min_transcript": "We've already been exposed to points and lines. Now let's think about planes. And you can view planes as really a flat surface that exists in three dimensions, that goes off in every direction. So for example, if I have a flat surface like this, and it's not curved, and it just keeps going on and on and on in every direction. Now the question is, how do you specify a plane? Well, you might say, well, let's see. Let's think about it a little bit. Could I specify a plane with a one point, right over here? Let's call that point, A. Would that, alone, be able to specify a plane? Well, there's an infinite number of planes that could go through that point. I could have a plane that goes like this, where that point, A, sits on that plane. I could have a plane like that. Or, I could have a plane like this. I could have a plane like this where point A sits on it, as well. So I could have a plane like that. So one point by itself does not seem to be sufficient to define a plane. Well, what about two points? Let's say I had a point, B, right over here. Well, notice the way I drew this, point A and B, they would define a line. For example, they would define this line right over here. So they would define, they could define, this line right over here. But both of these points and in fact, this entire line, exists on both of these planes that I just drew. And I could keep rotating these planes. I could have a plane that looks like this. I could have a plane that looks like this, that both of these points actually sit on. I'm essentially just rotating around this line that is defined by both of these points. So two points does not seem to be sufficient. Let's try three. So there's no way that I could put-- Well, let's be careful here. So I could put a third point right over here, point C. And C sits on that line, So it doesn't seem like just a random third point is sufficient to define, to pick out any one of these planes. But what if we make the constraint that the three points are not all on the same line. Obviously, two points will always define a line. But what if the three points are not collinear. So instead of picking C as a point, what if we pick-- Is there any way to pick a point, D, that is not on this line, that is on more than one of these planes? We'll, no. If I say, well, let's see, the point D-- Let's say point D is right over here. So it sits on this plane right over here, one of the first ones that I drew. So point D sits on that plane. Between point D, A, and B, there's only one plane that all three of those points sit on. So a plane is defined by three non-colinear points. So D, A, and B, you see, do not sit on the same line. A and B can sit on the same line. D and A can sit on the same line." + }, + { + "Q": "\nAt 0:18 sal said that the plane is something that is not curved goes off in every direction infinitely. He said that it is two dimensional and exists in three dimensions. How can it not have round edges if it goes off infinitely in every direction? Wouldn't it eventually become round, once it reaches infinity?", + "A": "Infinity does not have curves. The negative space the curves create are not everything and thus infinity is a square. Think of it like this: If you had a sandbox and wanted to cover all of it, and the sandbox went on forever, there would be no edge for curves to exist on.", + "video_name": "J2Qz-7ZWDAE", + "timestamps": [ + 18 + ], + "3min_transcript": "We've already been exposed to points and lines. Now let's think about planes. And you can view planes as really a flat surface that exists in three dimensions, that goes off in every direction. So for example, if I have a flat surface like this, and it's not curved, and it just keeps going on and on and on in every direction. Now the question is, how do you specify a plane? Well, you might say, well, let's see. Let's think about it a little bit. Could I specify a plane with a one point, right over here? Let's call that point, A. Would that, alone, be able to specify a plane? Well, there's an infinite number of planes that could go through that point. I could have a plane that goes like this, where that point, A, sits on that plane. I could have a plane like that. Or, I could have a plane like this. I could have a plane like this where point A sits on it, as well. So I could have a plane like that. So one point by itself does not seem to be sufficient to define a plane. Well, what about two points? Let's say I had a point, B, right over here. Well, notice the way I drew this, point A and B, they would define a line. For example, they would define this line right over here. So they would define, they could define, this line right over here. But both of these points and in fact, this entire line, exists on both of these planes that I just drew. And I could keep rotating these planes. I could have a plane that looks like this. I could have a plane that looks like this, that both of these points actually sit on. I'm essentially just rotating around this line that is defined by both of these points. So two points does not seem to be sufficient. Let's try three. So there's no way that I could put-- Well, let's be careful here. So I could put a third point right over here, point C. And C sits on that line, So it doesn't seem like just a random third point is sufficient to define, to pick out any one of these planes. But what if we make the constraint that the three points are not all on the same line. Obviously, two points will always define a line. But what if the three points are not collinear. So instead of picking C as a point, what if we pick-- Is there any way to pick a point, D, that is not on this line, that is on more than one of these planes? We'll, no. If I say, well, let's see, the point D-- Let's say point D is right over here. So it sits on this plane right over here, one of the first ones that I drew. So point D sits on that plane. Between point D, A, and B, there's only one plane that all three of those points sit on. So a plane is defined by three non-colinear points. So D, A, and B, you see, do not sit on the same line. A and B can sit on the same line. D and A can sit on the same line." + }, + { + "Q": "\nAt about 1:05 Sal said lumber line\nWhen he meant to say number line", + "A": "Yes, Sal ( the man speaking in the video ) was meant to say number line.", + "video_name": "LpLnmuAyNWg", + "timestamps": [ + 65 + ], + "3min_transcript": "Use a number line to compare 11.5 and 11.7. So let's draw a number line here. And I'm going to focus between 11 and 12, because that's where our two numbers are sitting. They're 11, and then something else, some number of 10ths. So this right here is 11. And this right here would be 12. And then let me draw the 10ths. So this would be smack dab in between. So that would be 11 and 5/10, or that would be 11.5. Well, I've already done the first part. I've figured out where 11.5 is. It's smack dab in between 11 and 12. It's 11 and 5/10. But let me find everything else. Let me mark everything else on this number line. So that's 1/10, 2/10, 3/10, 4/10, 5/10, 6/10, 7/10, 8/10, 9/10, and then 10/10, right on the 12. It's not completely drawn to scale. I'm hand-drawing it as good as I can. So where is 11.7 going to be? Well, this is 11.5, this is 11.6, this is 11.7. 11 and 7/10. This is 11.7. And the way we've drawn our number line, we are increasing as we go to the right. 11.7 is to the right of 11.5. It's clearly greater than 11.5. 11.7 is greater than 11.5. And really, seriously, you didn't have to draw a number line to figure that out. They're both 11 and something else. This is 11 and 5/10. This is 11 and 7/10. So clearly, this one is going to be greater. You both have 11, but this has 7/10, as opposed to 5/10." + }, + { + "Q": "At 0:17, the answer is 11! just kidding.... but isn't this basically a common sense video?\n", + "A": "Well, it s introducing you to the idea of adding variables to simple equations.", + "video_name": "P6_sK8hRWCA", + "timestamps": [ + 17 + ], + "3min_transcript": "Let's say that you started off with 3 apples. And then I were to give you another 7, another 7 apples. So my question to you-- and this might be very obvious-- is how many apples do you now have? And I'll give you a second to think about that. Well, this is fairly basic. You had 3 apples. Now, I'm going to give you 7 more. You now have 3 plus 7. You now have 10 apples. But let's say I want to do the same type of thinking, but I'm too lazy to write the word \"apples.\" Let's say instead of writing the word \"apples,\" I just use the letter a. And let's say this is, say, a different scenario. You start off with 4 apples. And to that, I add another 2 apples. How many apples do you now have? Instead of writing apples, I'm just going to write a's here. So how many of these a's do you now have? And once again, I'll give you a few seconds This also might be a little bit of common sense for you. If you had 4 of these apples or whatever these a's represented, if you had 4 of them and then you add 2 more of them, you're now going to have 6 of these apples. But once again, we started off assuming that a's represent apples. But they could have represented anything. If you have 4 of whatever a represents, and then you have another 2 of whatever a represents, you'll now have 6 of whatever a represents. Or if you just think of it if I have 4 a's, and then I add another 2 a's, I'm going to have 6 a's. You can literally think of 4 a's as a plus a plus a plus a. And if to that, I add another 2 a's-- so plus a plus a, that's 2 a's right over there-- how many a's do I now have? Well, that's 1, 2, 3, 4, 5, 6. I now have 6 a's. So thinking of it that way, let's get a little bit more abstract. Let's say that I have 5 x's, whatever x represents. So I have 5 of whatever that number is. And from that, I subtract 2 of whatever that number is. What would this evaluate to? How many of these x's would I now have? So it's essentially 5x minus 2x is going to be what times x? Once again, I'll give you a few seconds to think about it. Well, if I have 5 of something and I subtract 2 of those away, I'm going to have 3 of that something left. So this is going to be equal to 3x. 5x minus 2x is equal to 3x. And if you really think about what that means, five x's are just x plus x plus x plus x plus x. And then we're going to take away two of those x's. So take away one x, take away two x's. You are going to be left with three x's." + }, + { + "Q": "At 4:15, how did he make the expression -8 - 5a into the expression 8 + 5a, I don't understand how it is possible to do that, I thought that you can only make that kind of equation into -8 + (-5a)?\n", + "A": "He stated that he multiplies both sides of the equation by -1. so he changed the signs of all the terms on both sides of the equation which is valid", + "video_name": "adPgapI-h3g", + "timestamps": [ + 255 + ], + "3min_transcript": "We have a... Here we have a times the quantity, five minus x, is equal to b-x minus eight. So, once again, pause the video and see if you can solve for x. Well, the way I like to approach these is let's just expand everything all out. So, let me just distribute this a, and then I'm gonna collect all the x-terms on one side, and all of the non-x-terms on the other side, and essentially do what I just did in the last example. So, let's first distribute this a. So the left-hand side becomes five-a, I could say a times five or five-a, minus a-x, a-x, that is going to be equal to b-x minus eight. Now we can subtract b-x from both sides. So, we're gonna subtract b-x from the left-hand side, b-x from the right-hand side. And, once again, the whole reason I'm doing that, I want all the x-terms on the left, and all the non-x-terms on the right. And, actually, since I want all the non-x-terms on the right, So I'm kind of doing two steps at once, here, but hopefully it makes sense. I'm trying to get rid of the b-x here, and I'm trying to get rid of the five-a here. So, I subtract five-a there, and I'll subtract five-a there, and then let's see what this give us. So, the five-a's cancel out. And, on the left-hand side, I have negative a-x, negative a-x, minus b-x, minus, you know, in that same green color, minus b-x. And on the right-hand side, This is going to be equal to, the b-x's cancel out, and I have negative eight minus five-a. Negative eight minus, in that same magenta color, minus five-a. And let's see, I have all my x's on one side, all my non-x's on the other side. And here I can factor out an x, And, actually, one thing that might be nice. Let me just multiply both sides by negative one. If I multiple both sides by negative one, I get a-x plus b-x, plus b-x is equal to eight plus five-a. That just gets rid of all of those negative signs. And now I can factor out an x here. So let me factor out an x, and I get x times a plus b. A plus b is going to be equal to eight plus five-a. Eight plus five-a. And we're in the home stretch now. We can just divide both sides by a plus b. So we could divide both sides by a plus b. A plus b. And we're going to be left with, x is equal to" + }, + { + "Q": "at 2:30, Khan said something called Algebraic Manipulations. What does that mean?\n", + "A": "Algebraic Manipulation. The key to solving simple algebraic equations containing a single unknown (e.g. x + 6 = 10) is to realize that the equation is an equality.", + "video_name": "2REbsY4-S70", + "timestamps": [ + 150 + ], + "3min_transcript": "- [Voiceover] Anna wants to celebrate her birthday by eating pizza with her friends. For $42.50 total, they can buy p boxes of pizza. Each box of pizza costs $8.50. Select the equation that matches this situation. So before I even look at these, let's see if I can make sense of the sentence here. So for $42.50 total, and I'll just write 42.5, especially because in all these choices they didn't write 42.50, they just wrote 42.5 which is equivalent. So 42.50 that's the total amount they spent on pizza and if I wanted to figure out how many boxes of pizza they could buy, I could divide the total amount they spend, I could divide that by the price per box. That would give me the number of boxes. So this is the total, total dollars. This right over here is the dollar per box and then this would give me the number of boxes. Now other ways that I could think about it. I could say, well what's the total that they spend? So 42.50, but what's another way of thinking about the total they spend? Well you could have the amount they spend per box, times the number of boxes. So this is the total they spend and this another way of thinking about the total they spend, so these two things must be equal. So let's see, if I can see anything here that looks like this, well actually this first choice, this, is exactly, is exactly what I wrote over here. Let's see this choice right over here. P is equal to 8.5 x 42.5. Well we've already been able to write an equation that has explicitly, that has just a p on one side and so when you solve for just a p on one side, you get this thing over here, not this thing, so we could rule that out. Over here it looks kind of like this, except the p is on the wrong side. This has 8.5p is equal to 42.5, If we try to get the p on the other side here, you could divide both sides by p, but then you would get p divided by p is one. You would get 42.5 is equal to 8.5/p which is not true. We have 8.5 times p is equal to 42.5, so this is, this is not going to be the case. One thing to realize, no matter what you come up with, if you came up with this first, or if you came up with this first, you can go between these two with some algebraic manipulations. So for example, to go from this blue one to what I wrote in red up here, you just divide both sides by 8.5. So you divide by 8.5 on the left, you divide 8.5 on the right. Obviously to keep the equal sign you have to do the same thing to the left and right, but now you would have 42.5/8.5 is equal to, is equal to p. Which is exactly what we have over there. Let's do one more of these. Good practice." + }, + { + "Q": "\nSo the angle with a smaller area can have more degrees as long as they equal 180? (as noted at 1:54 in parallel lines 2)", + "A": "no, I think he just had it backwards, he switched the 60 and 120, I think he meant to do it the other way around", + "video_name": "0eDwckZOffc", + "timestamps": [ + 114 + ], + "3min_transcript": "Let's do a couple of examples dealing with angles between parallel lines and transversals. So let's say that these two lines are a parallel, so I can a label them as being parallel. That tells us that they will never intersect; that they're sitting in the same plane. And let's say I have a transversal right here, which is just a line that will intersect both of those parallel lines, and I were to tell you that this angle right there is 60 degrees and then I were to ask you what is this angle right over there? You might say, oh, that's very difficult; that's on a different line. But you just have to remember, and the one thing I always remember, is that corresponding angles are always equivalent. And so if you look at this angle up here on this top line where the transversal intersects the top line, what is the corresponding angle to where the transversal intersects this bottom line? that there's one, two, three, four angles. So this is on the bottom and kind of to the right a little bit. Or maybe you could kind of view it as the southeast angle if we're thinking in directions that way. And so the corresponding angle is right over here. And they're going to be equivalent. So this right here is 60 degrees. Now if this angle is 60 degrees, what is the question mark angle? Well the question mark angle-- let's call it x --the question mark angle plus the 60 degree angle, they go halfway around the circle. They are supplementary; They will add up to 180 degrees. So we could write x plus 60 degrees is equal to 180 degrees. And if you subtract 60 from both sides of this equation you get x is equal to 120 degrees. You could actually figure out every angle formed between the transversals and the parallel lines. If this is 120 degrees, then the angle opposite to it is also 120 degrees. If this angle is 60 degrees, then this one right here is also 60 degrees. If this is 60, then its opposite angle is 60 degrees. And then you could either say that, hey, this has to be supplementary to either this 60 degree or this 60 degree. Or you could say that this angle corresponds to this 120 degrees, so it is also 120, and make the same exact argument. This angle is the same as this angle, so it is also 120 degrees. Let's do another one. Let's say I have two lines. So that's one line. Let me do that in purple and let me do the other line in a different shade of purple. Let me darken that other one a little bit more. So you have that purple line and the other one" + }, + { + "Q": "at1:54 can you explain what in the world he's talking?\n", + "A": "He s saying that both of those angles are equal to 180 degrees. We already know one of them is 60, so to figure out what the other one is you just subtract 60 from 180. And that will leave you with 120. So that mystery angle is 120.", + "video_name": "0eDwckZOffc", + "timestamps": [ + 114 + ], + "3min_transcript": "Let's do a couple of examples dealing with angles between parallel lines and transversals. So let's say that these two lines are a parallel, so I can a label them as being parallel. That tells us that they will never intersect; that they're sitting in the same plane. And let's say I have a transversal right here, which is just a line that will intersect both of those parallel lines, and I were to tell you that this angle right there is 60 degrees and then I were to ask you what is this angle right over there? You might say, oh, that's very difficult; that's on a different line. But you just have to remember, and the one thing I always remember, is that corresponding angles are always equivalent. And so if you look at this angle up here on this top line where the transversal intersects the top line, what is the corresponding angle to where the transversal intersects this bottom line? that there's one, two, three, four angles. So this is on the bottom and kind of to the right a little bit. Or maybe you could kind of view it as the southeast angle if we're thinking in directions that way. And so the corresponding angle is right over here. And they're going to be equivalent. So this right here is 60 degrees. Now if this angle is 60 degrees, what is the question mark angle? Well the question mark angle-- let's call it x --the question mark angle plus the 60 degree angle, they go halfway around the circle. They are supplementary; They will add up to 180 degrees. So we could write x plus 60 degrees is equal to 180 degrees. And if you subtract 60 from both sides of this equation you get x is equal to 120 degrees. You could actually figure out every angle formed between the transversals and the parallel lines. If this is 120 degrees, then the angle opposite to it is also 120 degrees. If this angle is 60 degrees, then this one right here is also 60 degrees. If this is 60, then its opposite angle is 60 degrees. And then you could either say that, hey, this has to be supplementary to either this 60 degree or this 60 degree. Or you could say that this angle corresponds to this 120 degrees, so it is also 120, and make the same exact argument. This angle is the same as this angle, so it is also 120 degrees. Let's do another one. Let's say I have two lines. So that's one line. Let me do that in purple and let me do the other line in a different shade of purple. Let me darken that other one a little bit more. So you have that purple line and the other one" + }, + { + "Q": "\nHow did you get those angles for 1:26 ? I'm confused", + "A": "He made up those angles.", + "video_name": "0eDwckZOffc", + "timestamps": [ + 86 + ], + "3min_transcript": "Let's do a couple of examples dealing with angles between parallel lines and transversals. So let's say that these two lines are a parallel, so I can a label them as being parallel. That tells us that they will never intersect; that they're sitting in the same plane. And let's say I have a transversal right here, which is just a line that will intersect both of those parallel lines, and I were to tell you that this angle right there is 60 degrees and then I were to ask you what is this angle right over there? You might say, oh, that's very difficult; that's on a different line. But you just have to remember, and the one thing I always remember, is that corresponding angles are always equivalent. And so if you look at this angle up here on this top line where the transversal intersects the top line, what is the corresponding angle to where the transversal intersects this bottom line? that there's one, two, three, four angles. So this is on the bottom and kind of to the right a little bit. Or maybe you could kind of view it as the southeast angle if we're thinking in directions that way. And so the corresponding angle is right over here. And they're going to be equivalent. So this right here is 60 degrees. Now if this angle is 60 degrees, what is the question mark angle? Well the question mark angle-- let's call it x --the question mark angle plus the 60 degree angle, they go halfway around the circle. They are supplementary; They will add up to 180 degrees. So we could write x plus 60 degrees is equal to 180 degrees. And if you subtract 60 from both sides of this equation you get x is equal to 120 degrees. You could actually figure out every angle formed between the transversals and the parallel lines. If this is 120 degrees, then the angle opposite to it is also 120 degrees. If this angle is 60 degrees, then this one right here is also 60 degrees. If this is 60, then its opposite angle is 60 degrees. And then you could either say that, hey, this has to be supplementary to either this 60 degree or this 60 degree. Or you could say that this angle corresponds to this 120 degrees, so it is also 120, and make the same exact argument. This angle is the same as this angle, so it is also 120 degrees. Let's do another one. Let's say I have two lines. So that's one line. Let me do that in purple and let me do the other line in a different shade of purple. Let me darken that other one a little bit more. So you have that purple line and the other one" + }, + { + "Q": "at 2:30,sal said 3,6,12,24's pattern is*2,but itcould be+3,+6,+9,+12...\n", + "A": "I did changer the pattern", + "video_name": "l-6uEtTBH7g", + "timestamps": [ + 150 + ], + "3min_transcript": "- [Voiceover] What I want to in this video is get some practice figuring out patterns and numbers. In particular, patterns that take us from one number to a next number in a sequence. So over here, in this magenta color, I go from 4 to 25 to 46 to 67. So what's the pattern here? How did I get from 4 to 25 and can I get the same way from 25 to 46 and 46 to 67, and I could just keep going on and on and on? Well there's a couple of ways to think about it. When I see 4 and 25, let's see, 25 isn't an obvious multiple of 4. Another way to go from 4 to 25, I could add 21. Let's see, if I add 21, 4 plus 21 is 25. If I were to go from 25 to 46, well I could just add 21 again. It looks like to go from one number to the next I'm just adding. I wrote 12 by accident, 21. I'm just adding 21 over and over again. And if I were to keep going, if I add 21 I'm going to get to 89. If I add 21 to that I'm going to get 110, and I could keep going and going and going. I could just keep adding 21 over and over again. The pattern here is I'm adding 21. Now what about over here, in green? When I look at it at first, it's tempting to say, 3 plus 3 is 6. But then I'm not adding 3 anymore to get from 6 to 12, I'm adding 6. And then to get from 12 to 24, I'm not adding 6 anymore, I added 12. So every time I'm adding twice as much. But maybe an easier pattern might be, another way to go from 3 to 6, isn't to add 3, but to multiply it by 2. and if I multiply by 2 again, I go from 6 to 12. 6 times 2 is 12. If I multiply by 2 again, I'll go to 24. 2 times 12 is 24 and I could keep going on and on and on. 2 times 24 is 48, 96, I could go on and on and on. The pattern here, it's not adding a fixed amount, it's multiplying each number by a certain amount, by 2 in this case, to get the next number. So 3 times 2 is 6, 6 times 2 is 12, 12 times 2 is 24. Alright, now let's look at this last one. The first two terms here are the same, 3 and 6. The first two numbers here. I could say, maybe this is times 2, but then to go from 6 to 9, I'm not multiplying by 2. But maybe I am just adding 3 here. So 3 to 6, I just added 3. Then 6 to 9, I add 3 again, and then 9 to 12, I add 3 again." + }, + { + "Q": "3:16, what if you put x instead of nothing for the -8x+8x? Will it still be right?\n", + "A": "No, that would be incorrect as -8 + 8 does not = 1; it equals zero. If you make -8x + 8x = x, the x has a coefficient of 1. Hope this helps.", + "video_name": "vN0aL-_vIKM", + "timestamps": [ + 196 + ], + "3min_transcript": "Let's start with the highest degree So the highest degree or the highest exponent on an x here - is actually this x to the third here but it looks like the only one it's the only place where we're raising x to the third power so that can't be merged or added or subtracted to anything else so let's just write that down. so we have 2x to the third and let's look at the x squared terms. We have 3x squared over there and we have a minus or we can view as a negative x squared over there so if we want to simplify we can add these two terms - we can add- so let me just write it down we can add 3x squared to negative x squared so I'm just rearranging it really right now I'm putting the like terms next to each other so it'll be easy to simplify now let's just worry about the x to the first terms or just the x terms you have a negative 8x term right over here so let me write it over here so positive 8x and then finally let's look at the constant terms you can view those as times x to the zeroth power and the constant terms are - you have a positive 7 over here - so plus 7 and then you have a negative 3 over here you have a negative 3 so all I'll I've done is I've really just used the communative property of addition to just change the order - or addition and subtraction - to change the order in which I'm doing this I've just rearranged the things so the like terms are next to each other but now we can simplify so we have 2x to the third - nothing to simplify that with but then if we subtract - if we have - let me do that in the same blue color if we have 3x squared and from that we're taking away an x squared well, we're only gonna have 2x squared left so that's gonna be plus 2x squared and then we add 8 Xs to it or you can actually swap these around you could view it as you are subtracting 8 x from positive 8x we'll those are just going to cancel out so it's just going to be zero - I could just write plus zero here, but that'd just be redundant it wouldn't change the value and then finally I have a plus 7 minus 3 well that is just clearly 7 minus 3 is 4 so I have plus 4 And we're done! We've simplified it! 2x to the third plus 2 x squared plus 4" + }, + { + "Q": "\nAt around 2:30 Sal says that he used the commutative property of 'addition and subtraction' to change the order, but is there even a 'commutative property of subtraction'? (I understand that one use the commutative property of addition to add both positive and negative numbers in any order).", + "A": "That is what he meant. Treat subtraction as the addition of a negative and it becomes commutative. a - b = a + (-b) = -b + a", + "video_name": "vN0aL-_vIKM", + "timestamps": [ + 150 + ], + "3min_transcript": "simply 3x squared minus 8x plus 7 plus 2x to the third minus x squared plus eight x minus 3 so when we simplify this we're essentially going to add up like terms and just as a reminder we can only add or subtract like terms or simplify like terms and just a reminder and what I mean by that if I had an x squared to an x squared these are like terms they're both x terms raised to the same power the same degree so if I have one x squared and another x squared well then I have 2x squared - this is 2x squared If I have an x to the third - let's say I have 3x to thirds plus another 4x to the thirds Well that means I have 7x to the thirds - 7x to the thirds I can't take an x squared and add it to an x to the third I cannot simplify this in any way, so this you cannot simplify cannot simply - these are not like terms just because they both have Xs The Xs are not to - they are not to the same degree Let's start with the highest degree So the highest degree or the highest exponent on an x here - is actually this x to the third here but it looks like the only one it's the only place where we're raising x to the third power so that can't be merged or added or subtracted to anything else so let's just write that down. so we have 2x to the third and let's look at the x squared terms. We have 3x squared over there and we have a minus or we can view as a negative x squared over there so if we want to simplify we can add these two terms - we can add- so let me just write it down we can add 3x squared to negative x squared so I'm just rearranging it really right now I'm putting the like terms next to each other so it'll be easy to simplify now let's just worry about the x to the first terms or just the x terms you have a negative 8x term right over here so let me write it over here so positive 8x and then finally let's look at the constant terms you can view those as times x to the zeroth power and the constant terms are - you have a positive 7 over here - so plus 7 and then you have a negative 3 over here you have a negative 3 so all I'll I've done is I've really just used the communative property of addition to just change the order - or addition and subtraction - to change the order in which I'm doing this I've just rearranged the things so the like terms are next to each other but now we can simplify so we have 2x to the third - nothing to simplify that with but then if we subtract - if we have - let me do that in the same blue color if we have 3x squared and from that we're taking away an x squared well, we're only gonna have 2x squared left so that's gonna be plus 2x squared" + }, + { + "Q": "At 2:52, when Sal was subtracting the terms, does it mean the term without the numerical coefficient is equal to one?\n", + "A": "The term of -x^2 has a coefficient of -1. Whenever the coefficient is not shown, it s understood to be 1 or -1, depending on its sign.", + "video_name": "vN0aL-_vIKM", + "timestamps": [ + 172 + ], + "3min_transcript": "Let's start with the highest degree So the highest degree or the highest exponent on an x here - is actually this x to the third here but it looks like the only one it's the only place where we're raising x to the third power so that can't be merged or added or subtracted to anything else so let's just write that down. so we have 2x to the third and let's look at the x squared terms. We have 3x squared over there and we have a minus or we can view as a negative x squared over there so if we want to simplify we can add these two terms - we can add- so let me just write it down we can add 3x squared to negative x squared so I'm just rearranging it really right now I'm putting the like terms next to each other so it'll be easy to simplify now let's just worry about the x to the first terms or just the x terms you have a negative 8x term right over here so let me write it over here so positive 8x and then finally let's look at the constant terms you can view those as times x to the zeroth power and the constant terms are - you have a positive 7 over here - so plus 7 and then you have a negative 3 over here you have a negative 3 so all I'll I've done is I've really just used the communative property of addition to just change the order - or addition and subtraction - to change the order in which I'm doing this I've just rearranged the things so the like terms are next to each other but now we can simplify so we have 2x to the third - nothing to simplify that with but then if we subtract - if we have - let me do that in the same blue color if we have 3x squared and from that we're taking away an x squared well, we're only gonna have 2x squared left so that's gonna be plus 2x squared and then we add 8 Xs to it or you can actually swap these around you could view it as you are subtracting 8 x from positive 8x we'll those are just going to cancel out so it's just going to be zero - I could just write plus zero here, but that'd just be redundant it wouldn't change the value and then finally I have a plus 7 minus 3 well that is just clearly 7 minus 3 is 4 so I have plus 4 And we're done! We've simplified it! 2x to the third plus 2 x squared plus 4" + }, + { + "Q": "may someone explain to me what he is starting to do at 0:30 in the video. do I have to do that simplifying in all my problems.\nthanks in advanced.\n", + "A": "thanks again that helped", + "video_name": "vN0aL-_vIKM", + "timestamps": [ + 30 + ], + "3min_transcript": "simply 3x squared minus 8x plus 7 plus 2x to the third minus x squared plus eight x minus 3 so when we simplify this we're essentially going to add up like terms and just as a reminder we can only add or subtract like terms or simplify like terms and just a reminder and what I mean by that if I had an x squared to an x squared these are like terms they're both x terms raised to the same power the same degree so if I have one x squared and another x squared well then I have 2x squared - this is 2x squared If I have an x to the third - let's say I have 3x to thirds plus another 4x to the thirds Well that means I have 7x to the thirds - 7x to the thirds I can't take an x squared and add it to an x to the third I cannot simplify this in any way, so this you cannot simplify cannot simply - these are not like terms just because they both have Xs The Xs are not to - they are not to the same degree Let's start with the highest degree So the highest degree or the highest exponent on an x here - is actually this x to the third here but it looks like the only one it's the only place where we're raising x to the third power so that can't be merged or added or subtracted to anything else so let's just write that down. so we have 2x to the third and let's look at the x squared terms. We have 3x squared over there and we have a minus or we can view as a negative x squared over there so if we want to simplify we can add these two terms - we can add- so let me just write it down we can add 3x squared to negative x squared so I'm just rearranging it really right now I'm putting the like terms next to each other so it'll be easy to simplify now let's just worry about the x to the first terms or just the x terms you have a negative 8x term right over here so let me write it over here so positive 8x and then finally let's look at the constant terms you can view those as times x to the zeroth power and the constant terms are - you have a positive 7 over here - so plus 7 and then you have a negative 3 over here you have a negative 3 so all I'll I've done is I've really just used the communative property of addition to just change the order - or addition and subtraction - to change the order in which I'm doing this I've just rearranged the things so the like terms are next to each other but now we can simplify so we have 2x to the third - nothing to simplify that with but then if we subtract - if we have - let me do that in the same blue color if we have 3x squared and from that we're taking away an x squared well, we're only gonna have 2x squared left so that's gonna be plus 2x squared" + }, + { + "Q": "\nAt 1:47 how did you know the slope equaled 1?", + "A": "When Rise=Run (delta x = delta y), The slope is one", + "video_name": "EQoNfxToez0", + "timestamps": [ + 107 + ], + "3min_transcript": "- [Voiceover] Slope is defined as your change in the vertical direction, and I could use the Greek letter delta, this little triangle here is the Greek letter delta, it means change in. Change in the vertical direction divided by change in the horizontal direction. That is the standard definition of slope and it's a reasonable way for measuring how steep something is. So for example, if we're looking at the xy plane here, our change in the vertical direction is gonna be a change in the y variable divided by change in horizontal direction, is gonna be a change in the x variable. So let's see why that is a good definition for slope. Well I could draw something with a slope of one. A slope of one might look something like... so a slope of one, as x increases by one, y increases by one, so a slope of one... Notice, however much my change in x is, so for example here, my change in x is positive two, I'm gonna have the same change in y. My change in y is going to be plus two. So my change in y divided by change in x is two divided by two is one. So for this line I have slope is equal to one. But what would a slope of two look like? Well, a slope of two should be steeper and we can draw that. Let me start at a different point, so if I start over here a slope of two would look like... for every one that I increase in the x direction I'm gonna increase two in the y direction, so it's going to look like... that. If my change in x is equal to one, my change in y is two. So change in y over change in x is gonna be two over one, the slope here is two. And now, hopefully, you're appreciating why this definition of slope is a good one. The higher the slope, the steeper it is, the faster it increases, the faster we increase in the vertical direction as we increase in the horizontal direction. Now what would a negative slope be? So let's just think about what a line with a negative slope would mean. A negative slope would mean, well we could take an example. If we have our change in y over change in x was equal to a negative one. That means that if we have a change in x of one, then in order to get negative one here, that means that our change in y would have to be equal to negative one." + }, + { + "Q": "How did you get Negative 4 at 3:22? You are not starting from 0 you are starting from 1 to go down. wouldn't it be -5?\n", + "A": "Yes, the distance is -5, as Sal goes on to say. The confusion arises because in finding that distance, Sal subtracted the y-values, the first y-value being at -4, the second point lying at +1, So that subtraction gave him ( -4 ) - ( +1 ) = -5.", + "video_name": "iX5UgArMyiI", + "timestamps": [ + 202 + ], + "3min_transcript": "Something strange about my pen tool. It's making that very thin. Let me do it one more time. Okay, that's better. (laughs) The distance of that line right over there, that is going to be the radius. So how can we figure that out? Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean Theorem. To figure out the length of that line, so this is the radius, we could figure out a change in x. So, if we look at our change in x right over here. Our change in x as we go from the center to this point. So this is our change in x. And then we could say that this is our change in y. That right over there is our change in y. And so our change in x-squared plus our change in y-squared is going to be our radius squared. That comes straight out of the Pythagorean Theorem. This is a right triangle. is going to be equal to our change in x-squared plus our change in y-squared. Plus our change in y-squared. Now, what is our change in x-squared? Or, what is our change in x going to be? Our change in x is going to be equal to, well, when we go from the radius to this point over here, our x goes from negative one to six. So you can view it as our ending x minus our starting x. So negative one minus negative, sorry, six minus negative one is equal to seven. So, let me... So, we have our change in x, this right over here, is equal to seven. If we viewed this as the start point and this as the end point, it would be negative seven, but we really care about the absolute value in the change of x, and once you square it So our change in x right over here is going to be positive seven. And our change in y, well, we are starting at, we are starting at y is equal to one and we are going to y is equal to negative four. So it would be negative four minus one which is equal to negative five. And so our change in y is negative five. You can view this distance right over here as the absolute value of our change in y, which of course would be the absolute value of five. But once you square it, it doesn't matter. The negative sign goes away. And so, this is going to simplify to seven squared, change in x-squared, is 49. Change in y-squared, negative five squared, is 25. So we get r-squared, we get r-squared is equal to 49 plus 25. So what's 49 plus 25?" + }, + { + "Q": "at 7:42 I do not understand why it is our lower bound sqrt of y, from were we concluded that that's the funtion of our curved? thanks\n", + "A": "The curved shape that we are integrating at that moment of the video is bounded by x = 1 on the upper end and by the curve formed by y = x\u00c2\u00b2 at the lower end. As Sal says, we don t know what point we would chose, so we need a value of x . We need to put the bounds in terms of x so the next step is to find the inverse of y = x\u00c2\u00b2 . We can do that by taking the square root of both side of y = x\u00c2\u00b2 which gives \u00e2\u0088\u009ay = x", + "video_name": "hrIPO8mQqtw", + "timestamps": [ + 462 + ], + "3min_transcript": "And then you could imagine that we're looking at this thing from above. So the surface is up here some place and we're looking straight down on it, and so this is just this area. So let's say we wanted to take the integral with respect to x first. So we want to sum up, so if we want the volume above this column, first of all, is this area times dx, dy, right? So let's write the volume above that column. It's going to be the value of the function, the height at that point, which is xy squared times dx, dy. This expression gives us the volume above this area, or this column right here. And let's say we want the sum in the x direction first. So we want to sum that dx, sum one here, sum here, So we're going to sum in the x-direction. So my question to you is, what is our lower bound of integration? Well, we're kind of holding our y constant, right? And so if we go to the left, if we go lower and lower x's we kind of bump into the curve here. So the lower bound of integration is actually the curve. And what is this curve if we were to write x is a function of y? This curve is y is equal to x squared, or x is equal to the square root of y. So if we're integrating with respect to x for a fixed y right here-- we're integrating in the horizontal direction first --our lower bound is x is equal to the square root of y. That's interesting. I think it's the first time you've probably seen a variable bound integral. But it makes sense because for this row that we're adding up right here, the upper bound is easy. The upper bound is x is equal to 1. x is equal to the square root of y. Because you go back like, oh, I bump into the curve. Well the curve is x is equal to the square root of y because we don't know which y we picked. Fair enough. So once we've figured out the volume-- so that'll give us the volume above this rectangle right here --and then we want to add up the dy's. And remember, there's a whole volume above what I'm drawing right here. I'm just drawing this part in the xy plane. So what we've done just now, this expression, as it's written right now, figures out the volume above that rectangle. Now if we want to figure out the entire volume of the solid, we integrate along the y-axis. Or we add up all the dy's. This was a dy right here, not a dx. My dx's and dy's look too similar. So now what is the lower bound on the y-axis if I'm summing" + }, + { + "Q": "At 1:57, doesn't 5/2 = 2.5 instead of 2.4?\n", + "A": "Yes, well caught.", + "video_name": "LoKEPEPaNm4", + "timestamps": [ + 117 + ], + "3min_transcript": "We're asked to graph y is equal to 2.5 times x. So we really just have to think about two points that satisfy this equation here, and the most obvious one is what happens when x equals 0. When x equals 0, 2.5 times 0 is going to be 0. So when x is 0, y is going to be equal to 0. And then let's just pick another x that will give us a y that is a whole number. So if x increases by 1, y is going to increase by 2.5. It's going to go right over there, and I could graph it just like that. And we see just by what I just said that the unit rate of change of y with respect to x is 2.5. A unit increase in x, an increase of 1 and x, results in a 2.5 increase in y. You see that right over here. x goes from 0 to 1, and y goes from 0 to 2.5. But let's increase x by another 1, and then y is going to increase by 2.5 again to get to 5. Or you could say, hey, look, if x is equal to 2, 2.5 times 2 is equal to 5. but then they also tell us to select the statements that are true. So the first one is the equation does not represent a proportional relationship. Well, this is a proportional relationship. A proportional relationship is one where, first of all, if you have zero x's, you're going to have zero y's, where y is equal to some constant times x. And here, y is equal to 2.5 times x. So this is definitely a proportional relationship, so I'm not going to check that. The unit rate of the relationship is 2/5. So I'm assuming-- this is a little ambiguous the way they stated it. I'm assuming they're saying the unit rate of change of y with respect to x. And the unit rate of change of y with respect to x is, when x increases 1, y changed 2.5. So here they're saying when x changes by 1, y changes by 0.4, 2/5 is the same thing as 0.4. This should be 5/2. 5/2 would be 2.4. So this isn't right as well. The slope of the line is 2.5. Slope is change in y over change in x. When x changes 1, y changes 2.5. So change in y, 2.5, over change in x, 1. 2.5 over 1 is 2.5. And you could also see it looking at the form of this equation. y is equal to-- this is the slope times x. So that's right. A change of 5 units in x results in a change of 2 units in y. Well, let's test that idea. We know when x is 0, y is 0. So if x goes from 0 to 5, what's going to happen to y? Well, y is going to be 2.5 times 5. 2.5 times 5 is 12.5. So y would not just change 2. It actually would change 12.5. So this isn't right. A change of 2 units in x results in a change of 5 units in y. Well, we see that. A change in 2 units of x results in a change of 5 units in y. That's exactly what we graphed right over here." + }, + { + "Q": "\nFor the second triangle at 2:30, when he writes that the two base angles are 3x+5 and x+6, does that mean that 3x+5 is equal to x+6, therefore x= 0.5?", + "A": "When he calculates the second triangle at 3:30, he writes that the base angles are 3x+5 and x+16. This means the value of x = 11/2, or 5.5: 3x+5 = x+16 2x+5 = 16 2x = 11 x = 11/2", + "video_name": "ceDV0QBpcMA", + "timestamps": [ + 150 + ], + "3min_transcript": "The measure of two angles of an isosceles triangle are 3x plus 5 degrees, we'll say, and x plus 16 degrees. Find all possible values of x. So let's think about this. Let's draw ourselves an isosceles triangle or two. So it's an isosceles triangle, like that and like that. And actually, let me draw a couple of them just because we want to think about all of the different possibilities here. So we know, from what we know about isosceles triangles, that the base angles are going to be congruent. So that angle is going to be equal to that angle. That angle is going to be equal to that angle. And so what could the 3x plus 5 degrees and the x plus 16, what could they be measures of? Well, maybe this one right over here has a measure of 3x plus 5 degrees. And the vertex is the other one. So maybe this one up here is the x plus 16 degrees. The other possibility is that this is describing both base angles, in which case, So maybe this one is 3x plus 5, and maybe this one over here is x plus 16. And then the final possibility-- actually we haven't exhausted all of them-- is if we swap these two-- if this one is x plus 16, and that one is 3x plus 5. So let me draw ourselves another triangle. And obviously swapping these two aren't going to make a difference because they are equal to each other. And then we could make that one equal to 3x plus 5. But that's not going to change anything either because they're equal to each other. So the last situation is where this angle down here is x plus 16, and this angle up here is 3x plus 5. This is 3x plus 5. So let's just work through each of these. So in this situation, if this base angle is 3x plus 5, so is this base angle. And then we know that all three of these are going to have to add up to 180 degrees. So we get 3x plus 5 plus 3x plus 5 We have 3x. Let's just add up. You have 3x plus 3x, which gives you 6x, plus another x gives you 7x. And then you have 5 plus 5, which is 10, plus 16 is equal to 26. And that is going to be equal to 180. And then we have, let's see, 180 minus 26. If we subtract 26 from both sides, we get 180 minus 20 is 160, minus another 6 is 154. You have 7x is equal to 154. And let's see how many times-- if we divide both sides by 7," + }, + { + "Q": "\nAt 1:51, I still don't understand how you're able to express the integral in terms of \"n\" or \"m\"... could someone explain/refer me to a link? Otherwise, if I were to do practice problems where I convert the improper integral into a definite integral, I'll only have a slight idea of how to do so.", + "A": "A definite integral from a to b measures the area under a curve from x = a to x = b. If we have an indefinite integral, say from 0 to infinity, we are saying a = 0 and trying to figure out what the area is when b gets VERY LARGE. So in the context of the video, in the red integral he is setting the upper bound of the integral as a variable m, so we are measuring the area under the curve from 0 to m, and taking the limit as m goes to infinity to see what happens when the upper bound becomes infinitely large.", + "video_name": "9JX2s90_RNQ", + "timestamps": [ + 111 + ], + "3min_transcript": "Right here we have the graph of y is equal to 250 over 25 plus x squared. And what I'm curious about in this video is the total area under this curve and above the x-axis. So I'm talking about everything that I'm shading in white here, including what we can't see, as we keep moving to the right and we keep moving to the left. So I'm talking about from x at negative infinity all the way to x at infinity. So first, how would we actually denote this? Well, it would be an improper integral. We would denote this area as the indefinite integral from x is equal to negative infinity to x is equal to infinity of our function, 250 over 25 plus x squared, dx. Now, we've already seen improper integrals where one of our boundaries was infinity. But how do you do it when you have one boundary at positive infinity and one boundary at negative infinity? You can't take a limit to two different things. And so the way that we're going to tackle this into two different improper integrals, one improper integral that describes this area right over here in blue from negative infinity to 0. So we'll say that this is equal to the improper integral that goes from negative infinity to 0 of 250 over 25 plus x squared dx, plus the improper integral that goes from 0 to positive infinity. So plus the improper, or the definite, integral from 0 to positive infinity of 250 over 25 plus x squared dx. And now we can start to make sense of this. So what we have in blue can be rewritten. This is equal to the limit as n approaches negative infinity of the definite integral from n to 0 of 250 over 25 Plus-- and I'm running out of real estate here-- the limit as-- since I already used n, let me use m now-- the limit as m approaches positive infinity of the definite integral from 0 to m of 250 over 25 plus x squared dx. So now all we have to do is evaluate these definite integrals. And to do that, we just have to figure out an antiderivative of 250 over 25 plus x squared. So let's try to figure out what that is. I'll do it over here on the left. So we need to figure out the antiderivative of 250 over 25 plus x squared. And it might already jump out at you that trig substitution might be a good thing to do. You see this pattern of a squared plus x squared, where in this case, a would be 5." + }, + { + "Q": "At 03:28: What is the archtangent?\n", + "A": "It s the inverse of the tangent function. For example, tangent(Pi/4 + kPi) = 1. So arctangent(1)=Pi/4 + kPi. Where k is a whole number.", + "video_name": "9JX2s90_RNQ", + "timestamps": [ + 208 + ], + "3min_transcript": "into two different improper integrals, one improper integral that describes this area right over here in blue from negative infinity to 0. So we'll say that this is equal to the improper integral that goes from negative infinity to 0 of 250 over 25 plus x squared dx, plus the improper integral that goes from 0 to positive infinity. So plus the improper, or the definite, integral from 0 to positive infinity of 250 over 25 plus x squared dx. And now we can start to make sense of this. So what we have in blue can be rewritten. This is equal to the limit as n approaches negative infinity of the definite integral from n to 0 of 250 over 25 Plus-- and I'm running out of real estate here-- the limit as-- since I already used n, let me use m now-- the limit as m approaches positive infinity of the definite integral from 0 to m of 250 over 25 plus x squared dx. So now all we have to do is evaluate these definite integrals. And to do that, we just have to figure out an antiderivative of 250 over 25 plus x squared. So let's try to figure out what that is. I'll do it over here on the left. So we need to figure out the antiderivative of 250 over 25 plus x squared. And it might already jump out at you that trig substitution might be a good thing to do. You see this pattern of a squared plus x squared, where in this case, a would be 5. is equal to a tangent theta, 5 tangent theta. And since we're going to have to reverse substitute later on, we can also put in the constraint-- well, we'd say x/5 is equal to tangent of theta, which is completely consistent with this first statement. And so if we wanted to have theta expressed as a function of x, we can put the constraint that theta is equal to arctangent of x/5. So once again, this is completely consistent with this over here. x can be 5 tangent of theta, and theta can be equal to arctangent of x/5. So now let's do the substitution. Actually, before that, we also have to figure out what dx is equal to. So dx is equal to-- I'll do it right over here-- well, the derivative of this with respect to theta is 5 secant squared theta d theta. So now we're ready to substitute back in. So all of this business is going to be equal to 250 times dx." + }, + { + "Q": "Near 2:52 7/9 does not equal 2/9 + 3/9 x 2/9\n", + "A": "The second sign is an addition sign. You were mistaken. 2/9 + 3/9 + 2/9 = 7/9.", + "video_name": "_E9fG8BYcBo", + "timestamps": [ + 172 + ], + "3min_transcript": "2 over 9 plus 3 over 9, that's going to get us to 5 over 9 so we need 2 more, so that's going to be plus another 2 over 9. Plus another 2/9, so what would this look like? So let's just draw another grid here, so this is going to look like (so I'm going to put it right below it so we can look at what it looks like). So we have 2 ninths, this here is 2 ninths well we have nine equal sections. So 2 ninths is going to be, 1 and 2. And then 2 more ninths, that's 1 and 2. So notice, when I added 2 ninths to 3 ninths to 2 ninths, this equals 7 ninths. And we know that when we add a bunch of fractions like this that have the same denominator, we can just add the numerators. And this is why, this is 2/9 and 3/9 and 2/9 and they give me 7/9. Actually, let's do this again, this is a lot of fun. So let me draw my grid again. And then, let's see what we can do. So let me get my pen tool out and make sure my ink isn't too thick. And then we could add, that won't get us, let's see 1/9. I'm going to try to add 4 fractions here. So first I'm going to try to add 1/9 and see where that get's us. 1/9 is going to get us right over here. So that's 1/9, so let's say we add 2/9 to that. So that's 1 and 2, so that still doesn't get us there. So that's 3/9, 1 plus 2 is 3. So let's add 4 ninths." + }, + { + "Q": "at 8:03, why substitute t=b\u0002a?\n", + "A": "So that by 11:06 we would have a form similar to that we were looking for in the first place. Remember that the inequality was valid for any t, so Sal used one that he knew would lead to a simplification.", + "video_name": "r2PogGDl8_U", + "timestamps": [ + 483 + ], + "3min_transcript": "So these two terms result in that term right there. And then if you just rearrange these you have a minus 1 times a minus 1. They cancel out, so those will become plus and you're just left with plus x dot x. And I should do that in a different color as well. I'll do that in an orange color. So those terms end up with that term. Then of course, that term results in that term. And remember, all I did is I rewrote this thing and said, look. This has got to be greater than or equal to 0. So I could rewrite that here. This thing is still just the same thing. I've just rewritten it. So this is all going to be greater than or equal to 0. Now let's make a little bit of a substitution just to clean up our expression a little bit. And we'll later back substitute into this. Let's define this as a. So the whole thing minus 2x dot y. I'll leave the t there. And let's define this or let me just define this X dot x as c. So then, what does our expression become? It becomes a times t squared minus-- I want to be careful with the colors-- b times t plus c. And of course, we know that it's going to be greater than It's the same thing as this up here, greater than or equal to 0. I could write p of t here. Now this is greater than or equal to 0 for any t that I put in here. For any real t that I put in there. Let me evaluate our function at b over 2a. I just have to make sure I'm not dividing by 0 any place. So a was this vector dotted with itself. And we said this was a nonzero vector. So this is the square of its length. It's a nonzero vector, so some of these terms up here would end up becoming positively when you take its length. So this thing right here is nonzero. This is a nonzero vector. Then 2 times the dot product with itself is also going to be nonzero. So we can do this. We don't worry about dividing by 0, whatever else. But what will this be equal to? This'll be equal to-- and I'll just stick to the green. It takes too long to keep switching between colors. This is equal to a times this expression squared. So it's b squared over 4a squared. I just squared 2a to get the 4a squared. Minus b times this. So b times-- this is just regular multiplication. b times b over 2a." + }, + { + "Q": "\nFrom 0:40 onwards, I notice that whenever Sal has to find a derivative of a log with an arbitrary base, he always chooses to represent the log (via change of base formula) as a base-e log and not a base-10 log. Is that how you should always interpret a log with an arbitrary base when finding its derivative?", + "A": "It is not about interpreting (what ever you meant by that). The only reason, we transform to base-e is simply that we know its derivative. There is (as far as I know) no other way to do it.", + "video_name": "ssz6TElXEOM", + "timestamps": [ + 40 + ], + "3min_transcript": "We already know that the derivative with respect to x of the natural log of x is equal to 1/x. But what about the derivative, not of the natural log of x, but some logarithm with a different base? So maybe you could write log base b of x where b is an arbitrary base. How do we evaluate this right over here? And the trick is to write this using the change of base formula. So we could write it in terms of logarithms. We know that log-- I'm just going to restate the change of base formula. And I'm going to change from log base b to log base e, which is essentially the natural log. So the change of base formula, we prove it elsewhere on the site. Feel free to search for it on the Khan Academy. The change of base formula tells us that log base b of x is equal to the natural log, if we want to go to log base e. The natural log of x over the natural-- so it makes it clear what I'm doing. Log base e of x over log base e of b, which is the exact same thing as the natural log of x over the natural log of b. So all we have to do is rewrite this thing. This is equal to the derivative with respect to x of the natural log of x over the natural log of b. Or we could even write it as 1 over the natural log of b times the natural log of x. And now this becomes pretty straightforward. Because what we have right here, 1 over the natural log of b, this is just a constant that's multiplying the natural log of x. So we could take it out of the derivative. So this is the same thing as 1 over the natural log of b times the derivative with respect to x of the natural log of x. This thing right over here is just going to be equal to 1/x. So we end up with 1 over the natural log of b times 1/x. So we end up with 1 over the natural log of b times 1/x, or 1 over the natural log of b, which is just a number times x. So if someone asks you what is the derivative with respect to x of log base 5 of x, well, now you know. It's going to be 1 over the natural log of 5 times x, just like that." + }, + { + "Q": "At 2:55 , what is meant by derivative of something \" with respect to\" something else?\n", + "A": "It means that the variable on top of the derivative symbol is changing due to changes in variable on the bottom (the independent variable). For instance, a lot of calculus in physics uses dt as the bottom, meaning with respect to time. So dV/dt describes how volume changes in response to passing time.", + "video_name": "ssz6TElXEOM", + "timestamps": [ + 175 + ], + "3min_transcript": "We already know that the derivative with respect to x of the natural log of x is equal to 1/x. But what about the derivative, not of the natural log of x, but some logarithm with a different base? So maybe you could write log base b of x where b is an arbitrary base. How do we evaluate this right over here? And the trick is to write this using the change of base formula. So we could write it in terms of logarithms. We know that log-- I'm just going to restate the change of base formula. And I'm going to change from log base b to log base e, which is essentially the natural log. So the change of base formula, we prove it elsewhere on the site. Feel free to search for it on the Khan Academy. The change of base formula tells us that log base b of x is equal to the natural log, if we want to go to log base e. The natural log of x over the natural-- so it makes it clear what I'm doing. Log base e of x over log base e of b, which is the exact same thing as the natural log of x over the natural log of b. So all we have to do is rewrite this thing. This is equal to the derivative with respect to x of the natural log of x over the natural log of b. Or we could even write it as 1 over the natural log of b times the natural log of x. And now this becomes pretty straightforward. Because what we have right here, 1 over the natural log of b, this is just a constant that's multiplying the natural log of x. So we could take it out of the derivative. So this is the same thing as 1 over the natural log of b times the derivative with respect to x of the natural log of x. This thing right over here is just going to be equal to 1/x. So we end up with 1 over the natural log of b times 1/x. So we end up with 1 over the natural log of b times 1/x, or 1 over the natural log of b, which is just a number times x. So if someone asks you what is the derivative with respect to x of log base 5 of x, well, now you know. It's going to be 1 over the natural log of 5 times x, just like that." + }, + { + "Q": "\nAt 3:56, sal says 25 goes into zero 0 times. How does that work? That number was 60....", + "A": "You could think it this way... You have no candy, and 25 people want some from you. How are you going to give them candy? So 25 goes into 0, ZERO times!", + "video_name": "TvSKeTFsaj4", + "timestamps": [ + 236 + ], + "3min_transcript": "We can divide both sides of this equation by 0.25, or if you recognize that four quarters make a dollar, you could say, let's multiply both sides of this equation by 4. You could do either one. I'll do the first, because that's how we normally do algebra problems like this. So let's just multiply both by 0.25. That will just be an x. And then the right-hand side will be 150 divided by 0.25. And the reason why I wanted to is really it's just good practice dividing by a decimal. So let's do that. So we want to figure out what 150 divided by 0.25 is. And we've done this before. When you divide by a decimal, what you can do is you can make the number that you're dividing into the other number, you can turn this into a whole number by essentially shifting the decimal two to the right. But if you do that for the number in the denominator, you also have to do that to the numerator. So right now you can view this as 150.00. decimal two to the right. Then you'd also have to do that with 150, so then it becomes 15,000. Shift it two to the right. So our decimal place becomes like this. So 150 divided by 0.25 is the same thing as 15,000 divided by 25. And let's just work it out really fast. So 25 doesn't go into 1, doesn't go into 15, it goes into 150, what is that? Six times, right? If it goes into 100 four times, then it goes into 150 six times. 6 times 0.25 is-- or actually, this is now a 25. We've shifted the decimal. This decimal is sitting right over there. So 6 times 25 is 150. You subtract. You get no remainder. Bring down this 0 right here. 25 goes into 0 zero times. 0 times 25 is 0. Subtract. No remainder. 25 goes into 0 zero times. 0 times 25 is 0. Subtract. No remainder. So 150 divided by 0.25 is equal to 600. And you might have been able to do that in your head, because when we were at this point in our equation, 0.25x is equal to 150, you could have just multiplied both sides of this equation times 4. 4 times 0.25 is the same thing as 4 times 1/4, which is a whole. And 4 times 150 is 600. So you would have gotten it either way. And this makes total sense. If 150 is 25% of some number, that means 150 should be 1/4 of that number. It should be a lot smaller than that number, and it is. 150 is 1/4 of 600. Now let's answer their actual question. Identify the percent. Well, that looks like 25%, that's the percent. The amount and the base in this problem." + }, + { + "Q": "\n2:10 I still don't understand why it would be x--5, how did he get that from x+5?", + "A": "x - (-5) is the same as x + 5. The general formula for a circle with center (a, b) and radius r is (x-a)^2 + (y-b)^2 = r^2 for a center at (-5, 7) and a radius of 3, the equation is (x- -5)^2 + (y - 7)^2 = 3^2 or (x+5)^2 + (y - 7)^2 = 9", + "video_name": "thDrJvWNI8M", + "timestamps": [ + 130 + ], + "3min_transcript": "- [Voiceover] Whereas to graph the circle x plus five squared plus y minus 5 squared equals four. I know what you're thinking. What's all of this silliness on the right hand side? This is actually just the view we use when we're trying to debug things on Khan Academy. But we can still do the exercise. So it says drag the center point and perimeter of the circle to graph the equation. So the first thing we want to think about is well what's the center of this equation? Well the standard form of a circle is x minus the x coordinate of the center squared, plus y minus the y coordinate of the center squared is equal to the radius squared. So x minus the x coordinate of the center. So the x coordinate of the center must be negative five. Cause the way we can get a positive five here's by subtracting a negative five. So the x coordinate must be negative five and the y coordinate must be positive five. Cause y minus the y coordinate of the center. So y coordinate is positive five and then the radius squared is going to be equal to four. is equal to two. And the way it's drawn right now, we could drag this out like this, but this the way it's drawn, the radius is indeed equal to two. And so we're done. And I really want to hit the point home of what I just did. So let me get my little scratch pad out. Sorry for knocking the microphone just now. That equation was x plus five squared plus y minus five squared is equal to four squared. So I want to rewrite this as, this is x minus negative five, x minus negative five squared, plus y minus positive five, positive five squared is equal to, instead of writing it as four I'll write it as two squared. So this right over here tells us x equals negative five y equals five and the radius is going to be equal to two. And once again, this is no magic here. This is not just, I don't want you to just memorize this formula. I want you to appreciate that this formula comes straight out of the Pythagorean Theorem, straight out of the distance formula, which comes out of the Pythagorean Theorem. Remember, if you have some center, in this case is the point negative five comma five, so negative five comma five, and you want to find all of the x's and y's that are two away from it. So you want to find all the x's and y's that are two away from it. So that would be one of them, x comma y. This distance is two. And there's going to be a bunch of them. And when you plot all of them together, you're going to get a circle with radius two around that center. Plus think about how we got that actual formula. Well the distance between that coordinate, between any of these x's and y's, it could be an x and y here, it could be an x and y here," + }, + { + "Q": "At 0:26, why tan of a number is equal to sin of the number divided by the cosine of the number?\n", + "A": "(It s tangent of an a n g l e). Sketch an angle on the unit circle in the 1st quadrant, and include x, y, and r. (y/r)/(x/r) = sin/cos = y/x = tan. Do it for the other 3 quadrants.", + "video_name": "k_wJsio68D4", + "timestamps": [ + 26 + ], + "3min_transcript": "Voiceover:The previous video we explored how the cosine and sines of angles relate. We essentially take the terminal ray of the angle and we reflect it about the X or the Y axis, or both axes. What I want to do in this video is think a little bit about the tangent of these different angles. So just as a little bit of a reminder, we know that the tangent of theta is equal to the sine of an angle over the cosine of an angle, and by the Unit Circle Definition, it's essentially saying, \"What is the slope \"of the terminal ray right over here?\" We remind ourselves slope is rise over run. It is our change in the vertical axis over our change in the horizontal axis. If we're starting at the origin, what is our change in the vertical axis if we go from zero to sine theta? Well, our change in the vertical axis is sine theta. What is our change in the horizontal axis? It's cosine of theta. So this is change in Y over change in X for the terminal ray. over cosine of theta, or you could view it as the slope of this ray right over here. Lets think about what other angles are going to have the exact same tangent of theta? This ray is collinear with this ray right over here. In fact if you put them together you get a line. So the tangent of this angle right over here, this pink angle going all the way around, the tangent of pi plus theta, or the tangent of theta plus pi, obviously you could write theta plus pi instead of pi plus theta. This should be, just based on this slope argument, this should be equal to the tangent of theta. Lets see if this actually is the case. So these two things should be equal equal to the slope of the terminal ray. Of course the other side of the angle is going to be the positive X axis based on the conventions that we've set up. Lets think about what it is when the tangent of theta plus pi is in terms of sine and cosine. Let me write this down in the pink color. The tangent. That's not pink. The tangent of pi plus theta, that's going to be equal to, put the parentheses to avoid ambiguity, that's equal to the sine of pi plus theta, or theta plus pi, over the cosine of theta plus pi. And in the previous video we established that the sine of theta plus pi, that's the same thing as negative sine theta. So this is equal to negative sine theta." + }, + { + "Q": "Are there any other videos covering the area of equilateral triangles, like why the perpendicular bisector at 2:40 is equal to (\u00e2\u0088\u009a3*s)/2 ?\n", + "A": "Well you find the area of the circle than find the area of the square than subtract both of them together, in this case if the square is in the circle than the square would be smaller than the circle so you would subtract the area of the square from the area of the circle.", + "video_name": "QVxqgxVtKbs", + "timestamps": [ + 160 + ], + "3min_transcript": "the area of each of these equilateral triangles are. And so to do it, we remember that the area of a triangle is equal to 1/2 base times height. But how do we figure out the height of an equilateral triangle? So for example, if I have an equilateral triangle like this-- let me draw it big so I can dissect it little bit-- so I have an equilateral triangle like this. The length of each of the sides are s. And I always have to re-prove it for myself. Just because I always forget the formula. We remember that the angles are 60 degrees, 60 degrees, and 60 degrees. They're all equal. And what I like to do to find out the area of this, in order to figure out the height, is I drop an altitude. So I drop an altitude just like here, and it would split the side in two. I know it doesn't look like it perfectly because I didn't draw it to scale. But it would split it in two. It would form these right angles. split my equilateral triangle into two 30-60-90 triangles. And that's useful because I know the ratio of the sides of a 30-60-90 triangle. If this is s and I've just split this in two, this orange section right over here is going to be s/2. This is also going to be s/2 right over here. They obviously add up to s. And then we know from 30-60-90 triangles, that the side opposite the 60-degree side is square root of 3 times the shortest side. So this altitude right over here, is going to be square root of 3s/2. And now we can figure out a generalized formula for the area of an equilateral triangle. It's going to be equal to 1/2 times the base. Well the base is going to be s. So the base is s. And the height is square root of 3s over 2. see we have in the numerator we have the square root of 3s squared over four. And now we can apply this to figure out the areas of each of these triangles. So this is going to be equal to the area of the larger triangle, is going to be square root of 3/4 times 14 squared. And the area of the smaller triangle is going to be square root of 3/4 times 4 squared. And let's see, we could factor out a square root of 3/4. So this is going to be equal to square root of 3/4 times 14 squared minus 4 squared. Which of course we know is to be, is 16. But now let's actually evaluate this, to actually get a number here. And I could try to simplify it by hand. But instead let me actually just get my-- Actually," + }, + { + "Q": "\nAt 03:15, why is the denominator 4 instead of 2? I am still confused here.", + "A": "He is calculating the area of the entire equilateral triangle at this point. The formula for the area of an equilateral triangle is 1/2(base x height) He has calculated the height as being the square root of 3 divided by 2. Therefore, Area = 1/2(S x sqr(3)/2), so multiply out the denominators to get (S x sqr(3)) / 4", + "video_name": "QVxqgxVtKbs", + "timestamps": [ + 195 + ], + "3min_transcript": "the area of each of these equilateral triangles are. And so to do it, we remember that the area of a triangle is equal to 1/2 base times height. But how do we figure out the height of an equilateral triangle? So for example, if I have an equilateral triangle like this-- let me draw it big so I can dissect it little bit-- so I have an equilateral triangle like this. The length of each of the sides are s. And I always have to re-prove it for myself. Just because I always forget the formula. We remember that the angles are 60 degrees, 60 degrees, and 60 degrees. They're all equal. And what I like to do to find out the area of this, in order to figure out the height, is I drop an altitude. So I drop an altitude just like here, and it would split the side in two. I know it doesn't look like it perfectly because I didn't draw it to scale. But it would split it in two. It would form these right angles. split my equilateral triangle into two 30-60-90 triangles. And that's useful because I know the ratio of the sides of a 30-60-90 triangle. If this is s and I've just split this in two, this orange section right over here is going to be s/2. This is also going to be s/2 right over here. They obviously add up to s. And then we know from 30-60-90 triangles, that the side opposite the 60-degree side is square root of 3 times the shortest side. So this altitude right over here, is going to be square root of 3s/2. And now we can figure out a generalized formula for the area of an equilateral triangle. It's going to be equal to 1/2 times the base. Well the base is going to be s. So the base is s. And the height is square root of 3s over 2. see we have in the numerator we have the square root of 3s squared over four. And now we can apply this to figure out the areas of each of these triangles. So this is going to be equal to the area of the larger triangle, is going to be square root of 3/4 times 14 squared. And the area of the smaller triangle is going to be square root of 3/4 times 4 squared. And let's see, we could factor out a square root of 3/4. So this is going to be equal to square root of 3/4 times 14 squared minus 4 squared. Which of course we know is to be, is 16. But now let's actually evaluate this, to actually get a number here. And I could try to simplify it by hand. But instead let me actually just get my-- Actually," + }, + { + "Q": "\nAt 2:12, Sal mentions that the 1st quartile begins at 14 instead of assuming it would begin at 14 as originally mentioned in the video. Does that mean, that every quartile ends at the end of each one?", + "A": "In this problem box plot is given, Sal isn\u00e2\u0080\u0099t assuming 14. This data shows age of 100 trees. About 25 trees have the age between 8-14 years. Another 25 have the age between 14-21. You should not see gap between two quartiles(For e.g. Q1 and Q2) in box plots. Hope this helps.", + "video_name": "b2C9I8HuCe4", + "timestamps": [ + 132 + ], + "3min_transcript": "An ecologist surveys the age of about 100 trees in a local forest. He uses a box-and-whisker plot to map his data shown below. What is the range of tree ages that he surveyed? What is the median age of a tree in the forest? So first of all, let's make sure we understand what this box-and-whisker plot is even about. This is really a way of seeing the spread of all of the different data points, which are the age of the trees, and to also give other information like, what is the median? And where do most of the ages of the trees sit? So this whisker part, so you could see this black part is a whisker, this is the box, and then this is another whisker right over here. The whiskers tell us essentially the spread of all of the data. So it says the lowest to data point in this sample is an eight-year-old tree. I'm assuming that this axis down here is in the years. And it says at the highest-- the oldest tree right over here is 50 years. So if we want the range-- and when we think of range in a statistics point of view we're thinking of the highest data So it's going to be 50 minus 8. So we have a range of 42. So that's what the whiskers tell us. It tells us that everything falls between 8 and 50 years, including 8 years and 50 years. Now what the box does, the box starts at-- well, let me explain it to you this way. This line right over here, this is the median. And so half of the ages are going to be less than this median. We see right over here the median is 21. So this box-and-whiskers plot tells us that half of the ages of the trees are less than 21 and half are older than 21. And then these endpoints right over here, these are the medians for each of those sections. So this is the median for all the trees that are less than the real median or less than the main median. So this is in the middle of all of the ages of trees that are less than 21. This is the middle age for all the trees that are greater than 21 or older than 21. into four groups. This we would call the first quartile. So I'll call it Q1 for our first quartile. Maybe I'll do 1Q. This is the first quartile. Roughly a fourth of the tree, because the way you calculate it, sometimes a tree ends up in one point or another, about a fourth of the trees A fourth of the trees are between 14 and 21. A fourth are between 21 and it looks like 33. And then a fourth are in this quartile. So we call this the first quartile, the second quartile, the third quartile, and the fourth quartile. So to answer the question, we already did the range. There's a 42-year spread between the oldest and the youngest And then the median age of a tree in the forest is at 21. So even though you might have trees that are as old as 50, the median of the forest is actually closer to the lower end of our entire spectrum of all" + }, + { + "Q": "\nWhen he starts grouping @ 2:59 and he switches @ 3:33 and starts to factor out 3f does that mean we can do it both ways and still come out with the correct answer?", + "A": "Yes. You can do math in quite a few ways, but still end up with the same answers. :) Hope this helps! :)", + "video_name": "d-2Lcp0QKfI", + "timestamps": [ + 179, + 213 + ], + "3min_transcript": "times negative 11. So two numbers, so a times b, needs to be equal to 6 times negative 11, or negative 66. And a plus b needs to be equal to 19. So let's try a few numbers here. So let's see, 22, I'm just thinking of numbers that are roughly 19 apart, because they're going to be of different signs. So 22 and 3, I think will work. Right. If we take 22 times negative 3, that is negative 66, and 22 plus negative 3 is equal to 19. And the way I kind of got pretty close to this number is, well, you know, they're going to be of different signs, so the positive versions of them have to be about 19 apart, and that worked out. 22 and negative 3. So now we can rewrite this 19f right here as the sum of That's the same thing as 19f. I just kind of broke it apart. And, of course, we have the 6f squared and we have the minus 11 here. Now, you're probably saying, hey Sal, why did you put the 22 here and the negative 3 there? Why didn't you do it the other way around? Why didn't you put the 22 and then the negative 3 there? And my main motivation for doing it, I like to put the negative 3 on the same side with the 6 because they have the common factor of the 3. I like to put the 22 with the negative 11, they have the same common factor of 11. So that's why I decided to do it that way. So now let's do the grouping. And, of course, you can't forget this negative 2 that we have sitting out here the whole time. So let me put that negative 2 out there, but that'll just kind of hang out for awhile. But let's do some grouping. So let's group these first two. And then we're going to group this-- let me get a nice color here-- and then we're going to group this second two. Let me do it in this purple color. And then we can group that second two right there. So these first two, we could factor out a negative 3f, so it's negative 3f times-- 6f squared divided by negative 3f is negative 2f. And then negative 3f divided by negative 3f is just positive f. Actually, a better way to start, instead of factoring out a negative 3f, let's just factor out 3f, so we don't have a negative out here. We could do it either way. But if we just factor out a 3f, 6f squared divided by 3f is 2f. And then negative 3f divided by 3f is negative 1. So that's what that factors into. And then that second part, in that dark purple color, can factor out an 11. And if we factor that out, 22f divided by 11 is 2f, and negative 11 divided by 11 is negative 1." + }, + { + "Q": "\nAT 3:00 he said that 2-2= 1 but that is not true right ?", + "A": "It was a mistake", + "video_name": "uCBm8iDyg1s", + "timestamps": [ + 180 + ], + "3min_transcript": "So 8 minus 5 is 3 and 9 minus 6 is 3. And now we can bring down the next digit, this 1 here. And now this is where the art is going to come into play because we need to figure out how many times does 65 go into 331 without going over it. And you might just try to look at these numbers, try to approximate them a little bit. You might say, well, maybe 65, let me round this thing up. Maybe this is close to 70. And let's see, this is close to 300. So maybe we say, well, 70 would go into 300. So maybe we think about how many times does 70 go into 300? And we say without going over it, it doesn't go exactly into 300. Well you could say, well how many times does 7 go into 30? Well we know 7 goes into 30 four times. So maybe try a 4 right over here because then this will be 280, 4 times 70 is 280. You're still going to have a little bit left over, but what you have left over is going to be less than 70. It's going to be 20. So you say, well, if this is roughly 70 and if this is roughly 300, then maybe it's going to be the same thing. So let's try that out. Let's see if it goes four times. So 4 times 5 is 20, carry the 2. 4 times 6 is 24 plus 2 is 26. And now let's see how much we had left over. So when we subtract, we are left with-- I'll do this in a new color-- 1 minus 0 is 1. We have a 3 here and a 6 here so we're going to have to do a little regrouping. Let's take 100 from the hundreds place. It becomes 200. Give those 10 tens, that 100, to the tens place. So now we have 13 tens. 13 minus 6 is 7 and then 2 minus 2 is 1. Well no, our remainder, after we said it went in four times, we actually had 71 left over. 71, this right over here, is larger than 65. You don't want a situation where what you have left over is larger than what you're trying to divide into the number. You could have gone into it one more time because you had so much left over. So this 4 was actually too low. We should have probably approximated this as 60, and 60 goes into 300, if we were to estimate, we'd say, well that might be closer to five times. So this is where the art of this comes into play. So it was very reasonable to do what I just did, but it just turned out to not be the right way to think about it. I could just say, well the 4 wasn't enough. I had too much left over. Let me try 5 now. 5 times 5 is 25, carry the 2. 5 times 6 is 30, plus 2 is 32. We got much closer to 331 without going over. Now we can subtract." + }, + { + "Q": "At 7:30, in this case we are talking about a Normal Distribution, but what if the coin is an unfair coin? Would we get something completely different? or just something displaced to one of the two sides?\n\nIn other words, what kind of distribution we get when the possible outcomes are not equally likely?\n", + "A": "Even if the possible outcomes are not equally likely, it will still be a binomial distribution because there are only two possible outcomes from each flip. It doesn t matter what the actual probabilities are of each outcome as long as they sum up to 1.", + "video_name": "NF0lrkqXIkQ", + "timestamps": [ + 450 + ], + "3min_transcript": "You could have something that takes on discrete values, but in theory, it could take on an infinite number of discrete values. You could just keep counting higher and higher and higher. But this is discrete, in that it's these whole, these particular values. It can't take on any value in between, and it's also finite. It can only take on x equals zero, x equals one, x equals two, x equals three, x equals four, or x equals five, and you see when you plot its probability distribution, this discrete probability distribution, it starts at 1/32, it goes up, and then it comes back down, and it has this symmetry, and a distribution like this, this right over here, a discrete distribution like this, we call this a binomial distribution, and we'll talk in the future about why it's called a binomial distribution, but a big clue... Actually, I'll tell you why it's called a binomial distribution, is that these probabilities, you can get them using binomial coefficients, using combinatorics. In another video, we'll talk about, why we even call those things binomial coefficients. It's really based on taking powers of binomials in algebra, but this is a very, very, very, very important distribution. It's very important in statistics, because for a lot of discrete processes, one might assume that the underlying distribution is a binomial distribution, and when we get further into statistics, we'll talk why people do that. Now, if you were to have much more than five cases here, if, instead of saying that the number of heads from flipping a coin five times, you said, x is equal to the number of heads of flipping a coin five million times, then, you can imagine, you'd have much, much... The bars would get narrower and narrower relative to the whole hump, and what it would start to do, it would start to approach something that looks really, something that looks really like a bell curve. that you can see better, that I haven't used yet. So if you had more and more of these, if you had more and more of these possibilities, it's going to start approaching what looks like a bell curve, and you've probably heard the notion of a bell curve, and the bell curve is a normal distribution. So one way to think about it, is the normal distribution is a probability density function. It's a continuous case. So, the yellow one, that we're approaching a normal distribution, and a normal distribution, in kind of the classical sense, is going to keep going on and on, normal distribution, and it's related to the binomial. You know, a lot of times in statistics, people will assume a normal distribution, because you can say, okay, it's the product of kind of an almost an infinite number of random processes happening. Here, we're taking a coin, and we're flipping it five times, but if you imagine molecules interacting, or humans interacting, you're saying, oh, there's almost an infinite number of interactions," + }, + { + "Q": "\nAround 8:30, is a normal distribution by definition continuous? How does a very large discrete (binomial?) distribution (flip a coin 5 trillion times) differ from a normal distribution?", + "A": "It doesn t differ significantly beyond the fact that you have discrete results (which can be easily accounted for using a continuity correction). This is the premise of the Central Limit Theorem which states that the mean of many random variables independently drawn from the same distribution is distributed approximately normally, irrespective of the form of the original distribution (wikipedia.org)", + "video_name": "NF0lrkqXIkQ", + "timestamps": [ + 510 + ], + "3min_transcript": "why we even call those things binomial coefficients. It's really based on taking powers of binomials in algebra, but this is a very, very, very, very important distribution. It's very important in statistics, because for a lot of discrete processes, one might assume that the underlying distribution is a binomial distribution, and when we get further into statistics, we'll talk why people do that. Now, if you were to have much more than five cases here, if, instead of saying that the number of heads from flipping a coin five times, you said, x is equal to the number of heads of flipping a coin five million times, then, you can imagine, you'd have much, much... The bars would get narrower and narrower relative to the whole hump, and what it would start to do, it would start to approach something that looks really, something that looks really like a bell curve. that you can see better, that I haven't used yet. So if you had more and more of these, if you had more and more of these possibilities, it's going to start approaching what looks like a bell curve, and you've probably heard the notion of a bell curve, and the bell curve is a normal distribution. So one way to think about it, is the normal distribution is a probability density function. It's a continuous case. So, the yellow one, that we're approaching a normal distribution, and a normal distribution, in kind of the classical sense, is going to keep going on and on, normal distribution, and it's related to the binomial. You know, a lot of times in statistics, people will assume a normal distribution, because you can say, okay, it's the product of kind of an almost an infinite number of random processes happening. Here, we're taking a coin, and we're flipping it five times, but if you imagine molecules interacting, or humans interacting, you're saying, oh, there's almost an infinite number of interactions, which is very, very important in science and statistics. Binomial distribution is the discrete version of that, and one little point of notion, these are where the distributions are, this is where they come from, If you kind of think about as you get more and more trials, the binomial distribution is going to really approach the normal distribution, but it's really important to think about where these things come from, and we'll talk about it much more in a statistics, because it is reasonable to assume an underlying binomial distribution, or normal distribution, for a lot of different types of processes, but sometimes it's not, and even in things like economics, sometimes people assume a normal distribution when it's actually much more likely that the things on the ends are going to happen, which might lead to things like economic crises, or whatever else. But anyway, I don't want to get off-topic. The whole point here is just to appreciate, hey, we started with this random variable, the number of heads from flipping a coin five times, and we plotted it, and we were able to see, we were able to visualize this binomial distribution, and I'm kind of telling you," + }, + { + "Q": "At 0:37, what is the meaning of arbitrary point?\n", + "A": "An arbitrary point is a point of your choosing. You could pick any point on that line and Sal s proof would still work.", + "video_name": "KXZ6w91DioU", + "timestamps": [ + 37 + ], + "3min_transcript": "Let's start off with segment AB. So that's point A. This is point B right over here. And let's set up a perpendicular bisector of this segment. So it will be both perpendicular and it will split the segment in two. So thus we could call that line l. That's going to be a perpendicular bisector, so it's going to intersect at a 90-degree angle, and it bisects it. This length and this length are equal, and let's call this point right over here M, maybe M for midpoint. What I want to prove first in this video is that if we pick an arbitrary point on this line that is a perpendicular bisector of AB, then that arbitrary point will be an equal distant from A, or that distance from that point to A will be the same as that distance from that point to B. So let me pick an arbitrary point on this perpendicular bisector. So let's call that arbitrary point C. And so you can imagine we like to draw a triangle, and then another one from C to B. And essentially, if we can prove that CA is equal to CB, then we've proven what we want to prove, that C is an equal distance from A as it is from B. Well, there's a couple of interesting things we see here. We know that AM is equal to MB, and we also know that CM is equal to itself. Obviously, any segment is going to be equal to itself. And we know if this is a right angle, this is also a right angle. This line is a perpendicular bisector of AB. And so we have two right triangles. And actually, we don't even have to worry about that they're right triangles. If you look at triangle AMC, you have this side is congruent to the corresponding side on triangle BMC. Then you have an angle in between that corresponds to this angle over here, angle AMC corresponds to angle BMC, and they're both 90 degrees, so they're congruent. And then you have the side MC that's on both triangles, So we can just use SAS, side-angle-side congruency. So we can write that triangle AMC is congruent to triangle BMC by side-angle-side congruency. And so if they are congruent, then all of their corresponding sides are congruent and AC corresponds to BC. So these two things must be congruent. This length must be the same as this length right over there, and so we've proven what we want to prove. This arbitrary point C that sits on the perpendicular bisector of AB is equidistant from both A and B. And I could have known that if I drew my C over here or here, I would have made the exact same argument, so any C that sits on this line. So that's fair enough. So let me just write it." + }, + { + "Q": "\nAt 2:48 equidistant means equal to right?", + "A": "Equidistant means equal distance", + "video_name": "KXZ6w91DioU", + "timestamps": [ + 168 + ], + "3min_transcript": "and then another one from C to B. And essentially, if we can prove that CA is equal to CB, then we've proven what we want to prove, that C is an equal distance from A as it is from B. Well, there's a couple of interesting things we see here. We know that AM is equal to MB, and we also know that CM is equal to itself. Obviously, any segment is going to be equal to itself. And we know if this is a right angle, this is also a right angle. This line is a perpendicular bisector of AB. And so we have two right triangles. And actually, we don't even have to worry about that they're right triangles. If you look at triangle AMC, you have this side is congruent to the corresponding side on triangle BMC. Then you have an angle in between that corresponds to this angle over here, angle AMC corresponds to angle BMC, and they're both 90 degrees, so they're congruent. And then you have the side MC that's on both triangles, So we can just use SAS, side-angle-side congruency. So we can write that triangle AMC is congruent to triangle BMC by side-angle-side congruency. And so if they are congruent, then all of their corresponding sides are congruent and AC corresponds to BC. So these two things must be congruent. This length must be the same as this length right over there, and so we've proven what we want to prove. This arbitrary point C that sits on the perpendicular bisector of AB is equidistant from both A and B. And I could have known that if I drew my C over here or here, I would have made the exact same argument, so any C that sits on this line. So that's fair enough. So let me just write it. Now, let's go the other way around. Let's say that we find some point that is equidistant from A and B. Let's prove that it has to sit on the perpendicular bisector. So let's do this again. So I'll draw it like this. So this is my A. This is my B, and let's throw out some point. We'll call it C again. So let's say that C right over here, and maybe I'll draw a C right down here. So this is C, and we're going to start with the assumption that C is equidistant from A and B. So CA is going to be equal to CB. This is what we're going to start off with. This is going to be our assumption, and what we want to prove is that C sits on the perpendicular bisector of AB. So we've drawn a triangle here, and we've done this before." + }, + { + "Q": "\nAt 3:10, Sal says that that proof covers all the points C, but what if C is on segment AB?", + "A": "If C is on segment AB, then triangle CAM and triangle CBM would be degenerate triangles (triangles with the measure 0,0, 180) and will technically be congruent.", + "video_name": "KXZ6w91DioU", + "timestamps": [ + 190 + ], + "3min_transcript": "and then another one from C to B. And essentially, if we can prove that CA is equal to CB, then we've proven what we want to prove, that C is an equal distance from A as it is from B. Well, there's a couple of interesting things we see here. We know that AM is equal to MB, and we also know that CM is equal to itself. Obviously, any segment is going to be equal to itself. And we know if this is a right angle, this is also a right angle. This line is a perpendicular bisector of AB. And so we have two right triangles. And actually, we don't even have to worry about that they're right triangles. If you look at triangle AMC, you have this side is congruent to the corresponding side on triangle BMC. Then you have an angle in between that corresponds to this angle over here, angle AMC corresponds to angle BMC, and they're both 90 degrees, so they're congruent. And then you have the side MC that's on both triangles, So we can just use SAS, side-angle-side congruency. So we can write that triangle AMC is congruent to triangle BMC by side-angle-side congruency. And so if they are congruent, then all of their corresponding sides are congruent and AC corresponds to BC. So these two things must be congruent. This length must be the same as this length right over there, and so we've proven what we want to prove. This arbitrary point C that sits on the perpendicular bisector of AB is equidistant from both A and B. And I could have known that if I drew my C over here or here, I would have made the exact same argument, so any C that sits on this line. So that's fair enough. So let me just write it. Now, let's go the other way around. Let's say that we find some point that is equidistant from A and B. Let's prove that it has to sit on the perpendicular bisector. So let's do this again. So I'll draw it like this. So this is my A. This is my B, and let's throw out some point. We'll call it C again. So let's say that C right over here, and maybe I'll draw a C right down here. So this is C, and we're going to start with the assumption that C is equidistant from A and B. So CA is going to be equal to CB. This is what we're going to start off with. This is going to be our assumption, and what we want to prove is that C sits on the perpendicular bisector of AB. So we've drawn a triangle here, and we've done this before." + }, + { + "Q": "\nAt 0:52 why did he use the word (Concerned?)", + "A": "I guess because he was concerned about something", + "video_name": "ory05j2jgBM", + "timestamps": [ + 52 + ], + "3min_transcript": "- [Voiceover] What I wanna do in this video is compare the fractions 3/4 and 4/5, and I wanna do this visually. So what I'm gonna do is I'm gonna have two copies of the same whole, so let me just draw that, but I'm gonna divide the first one, so this is one whole right over here, this rectangle, when we draw the whole thing. So this is a whole, and right below that, we have the same whole. We have a rectangle of exactly the same size. Now you might notice that I've divided them into a different number of equal sections. In the top one, I've divided it into four equal sections because I am concerned with fourths so I've divided this top whole into fourths and I've divided this bottom whole, or this bottom bar or five equal sections. So let's think about what 3/4 represent. So that's gonna be one of the fourths, right over here, two of the fourths, and then three of the fourths. And what is 4/5 going to be? Well, 4/5 is going to be one fifth, two fifths, three fifths, and four fifths. So when you look at them visually, remember, we're taking fractions of the same whole. This is 3/4 of that rectangle, this is 4/5 of a same-sized rectangle. It wouldn't make any sense if you're doing it for different shapes or different sized rectangles. We just divided them into different sections and you see that if you have four of the fifths, than three of the fourths, and so 4/5 is greater than 3/4 or you could say 3/4 is less than 4/5, or any way you wanna think about it. The symbol you wanna use always opens to the larger number. 4/5 is larger than 3/4, so the large end of our symbol is facing the 4/5, so we would say 3/4 is less than 4/5." + }, + { + "Q": "at 0:17,what is a tape graph?!\n", + "A": "its to help you compare fractions.", + "video_name": "ory05j2jgBM", + "timestamps": [ + 17 + ], + "3min_transcript": "- [Voiceover] What I wanna do in this video is compare the fractions 3/4 and 4/5, and I wanna do this visually. So what I'm gonna do is I'm gonna have two copies of the same whole, so let me just draw that, but I'm gonna divide the first one, so this is one whole right over here, this rectangle, when we draw the whole thing. So this is a whole, and right below that, we have the same whole. We have a rectangle of exactly the same size. Now you might notice that I've divided them into a different number of equal sections. In the top one, I've divided it into four equal sections because I am concerned with fourths so I've divided this top whole into fourths and I've divided this bottom whole, or this bottom bar or five equal sections. So let's think about what 3/4 represent. So that's gonna be one of the fourths, right over here, two of the fourths, and then three of the fourths. And what is 4/5 going to be? Well, 4/5 is going to be one fifth, two fifths, three fifths, and four fifths. So when you look at them visually, remember, we're taking fractions of the same whole. This is 3/4 of that rectangle, this is 4/5 of a same-sized rectangle. It wouldn't make any sense if you're doing it for different shapes or different sized rectangles. We just divided them into different sections and you see that if you have four of the fifths, than three of the fourths, and so 4/5 is greater than 3/4 or you could say 3/4 is less than 4/5, or any way you wanna think about it. The symbol you wanna use always opens to the larger number. 4/5 is larger than 3/4, so the large end of our symbol is facing the 4/5, so we would say 3/4 is less than 4/5." + }, + { + "Q": "Re: 04:06. Is Sal saying that all (emphasis) 3 points cannot be on the same line to have a triangle, though it is true that there will always be 2 of the 3 points which are on the same line.\n", + "A": "Any two points, no matter how far apart or weirdly placed (even in 3D, if the two points were at any depth), will always form a line. But if you get three points on the same line, all it would form is a line, and not a triangle, which is a polygon and not a line. To better explain, if you had three points that fell on a same line, you couldn t make a triangle out of them, consequently preventing you from taking a circumcenter (because you don t have a triangle), and finally not letting you make a circle.", + "video_name": "4_xhiP6g2ow", + "timestamps": [ + 246 + ], + "3min_transcript": "and the length of OB, so OA is equal to OC is equal to OB, which is the c circumradius. And we've learned when we first talked about circles, if you give me a point, and if we find the locus of all points that are equidistant from that point, then that is a circle. And when I say a locus, all I mean is, the set of all points. If you give me any point right over here, so that's an arbitrary point, and you also specify a radius, and say what is the set of all the points on this two dimensional plane that are equidistant, that are that radius away from the center? It uniquely defines a circle. That's how we defined a circle right over here. And similarly, if you say, look, if you start with the center at O, and you say all of the points that are the circumradius away from O, And that circle will contain the points A, B, and C because those are the circumradius away from O. So they are included in that set. So the circle would look something like-- let me draw it. It would look something like this-- trying my best to draw it, just like that. Everything we've talked about, just now within the last few minutes, is all review. We know all of this. But I went over it just to kind of reinstate a pretty interesting idea, that if you give me three points that defines a unique triangle, and if you have a unique triangle-- And let me make it clear. This is three non-collinear points, so three points not on the same line. If you have three points that are not on the same line, that defines a unique triangle. and circumradius. I'll rewrite it, I don't want to get lazy and confuse you-- circumradius. And if you give me any point in space, any unique point, and a radius, the set of all points that are exactly that radius away from it, that defines a unique circle. So we went through all of this business of talking about the unique triangle, and the unique circumcenter, and the unique radius, to really just show you that if you give me any three points that eventually, really, just defines a unique circle. So just as you need three points to define a triangle, you also need three points to define a circle, two points won't do it. And one way to think about it is, if you give me two points," + }, + { + "Q": "At 1:30, why does using the fraction 9/13 work? I think I understand, but sometimes having other people's thoughts make my own clearer.\n", + "A": "The fraction 9/13 works because there are 13 integers between 36 and 49 (the closest square numbers). 45 is 9 integers greater than 36, so that gives you 9/13. In other words, you want to find out the relationship of the number you have with the close square numbers you can find nearby.", + "video_name": "EFVrAk61xjE", + "timestamps": [ + 90 + ], + "3min_transcript": "We are asked to approximate the principal root, or the positive square root of 45, to the hundredths place. And I'm assuming they don't want us to use a calculator. Because that would be too easy. So, let's see if we can approximate this just with our pen and paper right over here. So the square root of 45, or the principal root of 45. 45 is not a perfect square. It's definitely not a perfect square. Let's see, what are the perfect squares around it? We know that it is going to be less than-- the next perfect square above 45 is going to be 49 because that is 7 times 7-- so it's less than the square root of 49 and it's greater than the square root of 36. And so, the square root of 36, the principal root of 36 I should say, is 6. And the principal root of 49 is 7. So, this value right over here is going to be between 6 and 7. And it's nine away from 36. So, the different between 36 and 49 is 13. So, it's a total 13 gap between the 6 squared and 7 squared. And this is nine of the way through it. So, just as a kind of approximation maybe-- and it's not going to work out perfectly because we're squaring it, this isn't a linear relationship-- but it's going to be closer to 7 than it's going to be to 6. At least the 45 is 9/13 of the way. It looks like that's about 2/3 of the way. So, let's try 6.7 as a guess just based on 0.7 is about 2/3. It looks like about the same. Actually, we could calculate this right here if we want. So 9/13 as a decimal is going to be what? It's going to be 13 into 9. We're going to put some decimal places right over here. 13 doesn't go into 9 but 13 does go into 90. And it goes into 90-- let's see, does it go into it seven times-- it goes into it six times. So, 6 times 3 is 18. 6 times 1 is 6, plus 1 is 7. And then you subtract, you get 12. So, went into it almost exactly seven times. So, this value right here is almost a 0.7. And so if you say, how many times does 13 go into 120? It looks like it's like nine times? Yeah, it would go into it nine times. 9 times 3. Get rid of this. 9 times 3 is 27. 9 times 1 is 9, plus 2 is 11. You have a remainder of 3. It's about 0.69." + }, + { + "Q": "At 0:50 what does he mean when he says \"the principle root of 49 is 7?\" What is a principle root? Why not just say root?\n", + "A": "A principal (square) root is a unique, positive square root. 49 has two square roots: 7 & -7, but only has one principal root: 7.", + "video_name": "EFVrAk61xjE", + "timestamps": [ + 50 + ], + "3min_transcript": "We are asked to approximate the principal root, or the positive square root of 45, to the hundredths place. And I'm assuming they don't want us to use a calculator. Because that would be too easy. So, let's see if we can approximate this just with our pen and paper right over here. So the square root of 45, or the principal root of 45. 45 is not a perfect square. It's definitely not a perfect square. Let's see, what are the perfect squares around it? We know that it is going to be less than-- the next perfect square above 45 is going to be 49 because that is 7 times 7-- so it's less than the square root of 49 and it's greater than the square root of 36. And so, the square root of 36, the principal root of 36 I should say, is 6. And the principal root of 49 is 7. So, this value right over here is going to be between 6 and 7. And it's nine away from 36. So, the different between 36 and 49 is 13. So, it's a total 13 gap between the 6 squared and 7 squared. And this is nine of the way through it. So, just as a kind of approximation maybe-- and it's not going to work out perfectly because we're squaring it, this isn't a linear relationship-- but it's going to be closer to 7 than it's going to be to 6. At least the 45 is 9/13 of the way. It looks like that's about 2/3 of the way. So, let's try 6.7 as a guess just based on 0.7 is about 2/3. It looks like about the same. Actually, we could calculate this right here if we want. So 9/13 as a decimal is going to be what? It's going to be 13 into 9. We're going to put some decimal places right over here. 13 doesn't go into 9 but 13 does go into 90. And it goes into 90-- let's see, does it go into it seven times-- it goes into it six times. So, 6 times 3 is 18. 6 times 1 is 6, plus 1 is 7. And then you subtract, you get 12. So, went into it almost exactly seven times. So, this value right here is almost a 0.7. And so if you say, how many times does 13 go into 120? It looks like it's like nine times? Yeah, it would go into it nine times. 9 times 3. Get rid of this. 9 times 3 is 27. 9 times 1 is 9, plus 2 is 11. You have a remainder of 3. It's about 0.69." + }, + { + "Q": "\nWhat if you are doing a problem with an octagon, could it be used the same way? Question starting from 3:24", + "A": "Yes", + "video_name": "ZqzAOZ9pP9Q", + "timestamps": [ + 204 + ], + "3min_transcript": "So this right over here, this distance right over here, h, is equal to 3. And the length of the prism is equal to 4. So I'm assuming it's this dimension over here is equal to 4. So length is equal to 4. So in this situation, what you really just have to do is figure out the area of this triangle right over here. We could figure out the area of this triangle and then multiply it by how much you go deep, so multiply it by this length. So the volume is going to be the area of this triangle-- let me do it in pink-- the area of this triangle. We know that the area of a triangle is 1/2 times the base times the height. So this area right over here is going to be 1/2 times the base times the height. And then we're going to multiply it by our depth of this triangular prism. So we have a depth of 4. So then we're going to multiply that times the 4, And we get-- let's see, 1/2 times 4 is 2. So these guys cancel out. You'll just have a 2. And then 2 times 3 is 6. 6 times 7 is 42. And it would be in some type of cubic units. So if these were in-- I don't know-- centimeters, it would be centimeters cubed. But they're not making us focus on the units in this problem. Let's do another one. Shown is a cube. If each side is of equal length x equals 3, what is the total volume of the cube? So each side is equal length x, which happens to equal 3. So this side is 3. This side over here, x is equal to 3. Every side, x is equal to 3. So it's actually the same exercise as the triangular prism. It's actually a little bit easier when you're dealing with the cube, where you really just want to find the area of this surface right over here. Now, this is pretty straightforward. This is just a square, or it would Or essentially the same, it's just 3 times 3. So the volume is going to be the area of this surface, 3 times 3, times the depth. And so we go 3 deep, so times 3. And so we get 3 times 3 times 3, which is 27. Or you might recognize this from exponents. This is the same thing as 3 to the third power. And that's why sometimes, if you have something to the third power, they'll say you cubed it. Because, literally, to find the volume of a cube, you take the length of one side, and you multiply that number by itself three times, one for each dimension-- one for the length, the width, and-- or I guess the height, the length, and the depth, depending on how you want to define them. So it's literally just 3 times 3 times 3." + }, + { + "Q": "At 13:26, shouldn't it have been sqrt(b2-a2)? He added them, but I thought that was for finding the foci of an ellipse.\n", + "A": "For finding the foci of hyperbolas, the equation is like the Pythagorean theorum, sqrt(a2+b2) For an ellipse, its sqrt(a2-b2)", + "video_name": "S0Fd2Tg2v7M", + "timestamps": [ + 806 + ], + "3min_transcript": "the focal length is the same on either side of the center of the hyperbola depending on how you may view it, but I think that's not too much of a stretch of a statement for you to for you to accept. So if this distance is the same as this distance, then the magenta distance minus this blue distance is going to be equal to this green distance. And this green distance is what? That's 2a. We saw that at the beginning of this video. So this, once again, is also equal to 2a. Anyway, I'll leave you there right now. Actually, let's actually just do one problem, just because I like to make one concrete. Because I told you at the beginning that if you wanted to find the-- so if you have an ellipse-- so if you have-- this is an ellipse, x squared over a squared plus y squared over b squared is equal to 1, we learned that the-- that's over b squared-- this is an ellipse. square root of a squared minus b squared. Now for a hyperbola, you kind of see that there's a very close relation between the ellipse and the hyperbola, but it is kind of a fun thing to ponder about. And a hyperbola's equation looks like this. x squared over a squared minus y squared over b squared, or it could be y squared over b squared minus x squared over a square is equal to 1. It turns out, and I'll prove this to you in the next video, it's a little bit of a hairy math problem, that the focal length of a hyperbola is equal to the square root of the sum of these two numbers, is equal to the sum of a squared plus b squared. So if I were to give you-- so notice the difference. It's just a difference in sign. You're taking the difference of those two denominators, and now you're taking the sum of the two denominators. So if I were to give you the following hyperbola. x squared over 9 plus y squared over 16 is equal to 1. we could just figure out the focal length just by plugging into the formula. The focal length is equal to the square root of a squared plus b squared. This is squared, right? a is three. b is 4. So 9 plus 16 is 25, which is equal to 5. And so if we were to graph this-- that's my y-axis, that's my x-axis-- and the focal length is the distance to, in this case, to the left and the right of the origin. If it was kind of an up and down opening hyperbola, it would be above and below the origin, so this is a-- oh sorry, this should be a plus. We're doing with a hyperbola, that should be a minus. Don't want to confuse you. What I had written before, with a plus, that would have been an ellipse. A minus is the hyperbola. So the two asymptotes-- this is centered at the origin, it hasn't been shifted-- are going to be 16 over 9, so it's going" + }, + { + "Q": "\nI understand the difference between a leading coefficient and coefficient. In the expression at 4:40 wouldn't the leading coefficient be 1 since the x^2 is being multipled by 1.", + "A": "It appears you didn t quite understand it. Leading coefficient means the coefficient of the term with the highest exponent. It is usually the first term, but not necessarily always the case. As you see in the video, 7x\u00e2\u0081\u00b5 was not in the first order. If you were to rewrite them in order as Khan did, you d see it be in the first order. Just remember that the leading is the highest power, not the necessarily the one standing in first order.", + "video_name": "ZgFXL6SEUiI", + "timestamps": [ + 280 + ], + "3min_transcript": "properly is to understand the coefficients of a polynomial. So let me write a fifth degree polynomial here. And I'm going to write it in maybe a non-conventional form right here. I'm going to not do it in order. So let's just say it's x squared minus 5x plus 7x to the fifth minus 5. So, once again, this is a fifth degree polynomial. Why is that? Because the highest exponent on a variable here is the 5 So this tells us this is a fifth degree polynomial. And you might say, well why do we even care about that? And at least, in my mind, the reason why I care about the degree of a polynomial is because when the numbers get large, the highest degree term is what really dominates all of the other terms. It will grow the fastest, or go negative the fastest, depending on whether there's a But it's going to dominate everything else. It really gives you a sense for how quickly, or how fast the whole expression would grow or decrease in the case if it has a negative coefficient. Now I just used the word coefficient. What does that mean? Coefficient. And I've used it before, when we were just doing linear equations. And coefficients are just the constant terms that are multiplying the variable terms. So for example, the coefficient on this term right here is negative 5. You have to remember we have a minus 5, so we consider negative 5 to be the whole coefficient. The coefficient on this term is a 7. There's no coefficient here; it's just a constant term of negative 5. And then the coefficient on the x squared term is 1. The coefficient is 1. It's implicit. You're assuming it's 1 times x squared. idea of the standard form of a polynomial. Now none of this is going to help you solve a polynomial just yet, but when we talk about solving polynomials, I might use some of this terminology, or your teacher might use some of this terminology. So it's good to know what we're talking about. The standard form of a polynomial, essentially just list the terms in order of degree. So this is in a non-standard form. If I were to list this polynomial in standard form, I would put this term first. So I would write 7x to the fifth, then what's the next smallest degree? Well, they have this x squared term. I don't have an x to the fourth or an x to the third here. So that'll be plus 1-- well I don't have to write 1-- plus x squared. And then I have this term, minus 5x. And then I have this last term right here, minus 5." + }, + { + "Q": "at 1:42 Sal said x have to be raised to a non negative power or positive so that means 1/2 power or positive or principal square root should be acceptable as polynomial?\n", + "A": "The definition of a polynomial is a rational expression in which no variable occurs as a denominator. Perhaps Sal should have been more explicit stating that the powers a variable is raised to is limited to whole numbers which excludes negative numbers and fractional numbers. True that 4/2 is a rational and whole number but it is not allowed when writing a polynomial.", + "video_name": "ZgFXL6SEUiI", + "timestamps": [ + 102 + ], + "3min_transcript": "In this video I want to introduce you to the idea of a polynomial. It might sound like a really fancy word, but really all it is is an expression that has a bunch of variable or constant terms in them that are raised to non-zero exponents. So that also probably sounds complicated. So let me show you an example. If I were to give you x squared plus 1, this is a polynomial. This is, in fact, a binomial because it has two terms. The term polynomial is more general. It's essentially saying you have many terms. Poly tends to mean many. This is a binomial. If I were to say 4x to the third minus 2 squared plus 7. This is a trinomial. I have three terms here. Let me give you just a more concrete sense of what is and is not a polynomial. For example, if I were to have x to the negative 1/2 plus 1, That doesn't mean that you won't ever see it while you're doing algebra or mathematics. But we just wouldn't call this a polynomial because it has a negative and a fractional exponent in it. Or if I were to give you the expression y times the square root of y minus y squared. Once again, this is not a polynomial, because it has a square root in it, which is essentially raising something to the 1/2 power. So all of the exponents on our variables are going to have to be non-negatives. Once again, neither of these are polynomials. Now, when we're dealing with polynomials, we're going to have some terminology. And you may or may not already be familiar with it, so I'll expose it to you right now. The first terminology is the degree of the polynomial. in the polynomial. So for example, that polynomial right there is a third degree polynomial. Now why is that? No need to keep writing it. Why is that a third degree polynomial? Because the highest exponent that we have in there is the x to the third term. So that's where we get it's a third degree polynomial. This right here is a second degree polynomial. And this is the second degree term. Now a couple of other terminologies, or words, that we need to know regarding polynomials, are the constant versus the variable terms. And I think you already know, these are variable terms right here. This is a constant term. That right there is a constant term." + }, + { + "Q": "\nAt 6:16 on the video, what would you do if you were in a situation where 2 of the numbers had no exponents?", + "A": "Most likely that will not happen and if so just add the two numbers in each term", + "video_name": "ZgFXL6SEUiI", + "timestamps": [ + 376 + ], + "3min_transcript": "But it's going to dominate everything else. It really gives you a sense for how quickly, or how fast the whole expression would grow or decrease in the case if it has a negative coefficient. Now I just used the word coefficient. What does that mean? Coefficient. And I've used it before, when we were just doing linear equations. And coefficients are just the constant terms that are multiplying the variable terms. So for example, the coefficient on this term right here is negative 5. You have to remember we have a minus 5, so we consider negative 5 to be the whole coefficient. The coefficient on this term is a 7. There's no coefficient here; it's just a constant term of negative 5. And then the coefficient on the x squared term is 1. The coefficient is 1. It's implicit. You're assuming it's 1 times x squared. idea of the standard form of a polynomial. Now none of this is going to help you solve a polynomial just yet, but when we talk about solving polynomials, I might use some of this terminology, or your teacher might use some of this terminology. So it's good to know what we're talking about. The standard form of a polynomial, essentially just list the terms in order of degree. So this is in a non-standard form. If I were to list this polynomial in standard form, I would put this term first. So I would write 7x to the fifth, then what's the next smallest degree? Well, they have this x squared term. I don't have an x to the fourth or an x to the third here. So that'll be plus 1-- well I don't have to write 1-- plus x squared. And then I have this term, minus 5x. And then I have this last term right here, minus 5. it in descending order of degree. Now let's do a couple of operations with polynomials. And this is going to be a super useful toolkit later on in your algebraic, or really in your mathematical careers. So let's just simplify a bunch of polynomials. And we've kind of touched on this in previous videos. But I think this will give you a better sense, especially when we have these higher degree terms over here. So let's say I wanted to add negative 2x squared plus 4x minus 12. And I'm going to add that to 7x plus x squared. Now the important thing to remember when you simplify these polynomials is that you're going to add the terms of the same variable of like degree. I'll do another example in a second where I have multiple variables getting involved in the situation." + }, + { + "Q": "At 1:19 why did Sal say \"negative and FRACTIONAL exponent\", isn't it supposed to be RATIONAL exponent? Just need a clarification.\n\nThanks,\nAnkit Thumma\n", + "A": "No. A rational exponent is anything that does NOT have a square root in it. A fractional exponent is any fraction. When exponents are fractions, they have a square root portion. For example, x^(1/2)= \u00e2\u0088\u009ax You will most likely learn more about this later.", + "video_name": "ZgFXL6SEUiI", + "timestamps": [ + 79 + ], + "3min_transcript": "In this video I want to introduce you to the idea of a polynomial. It might sound like a really fancy word, but really all it is is an expression that has a bunch of variable or constant terms in them that are raised to non-zero exponents. So that also probably sounds complicated. So let me show you an example. If I were to give you x squared plus 1, this is a polynomial. This is, in fact, a binomial because it has two terms. The term polynomial is more general. It's essentially saying you have many terms. Poly tends to mean many. This is a binomial. If I were to say 4x to the third minus 2 squared plus 7. This is a trinomial. I have three terms here. Let me give you just a more concrete sense of what is and is not a polynomial. For example, if I were to have x to the negative 1/2 plus 1, That doesn't mean that you won't ever see it while you're doing algebra or mathematics. But we just wouldn't call this a polynomial because it has a negative and a fractional exponent in it. Or if I were to give you the expression y times the square root of y minus y squared. Once again, this is not a polynomial, because it has a square root in it, which is essentially raising something to the 1/2 power. So all of the exponents on our variables are going to have to be non-negatives. Once again, neither of these are polynomials. Now, when we're dealing with polynomials, we're going to have some terminology. And you may or may not already be familiar with it, so I'll expose it to you right now. The first terminology is the degree of the polynomial. in the polynomial. So for example, that polynomial right there is a third degree polynomial. Now why is that? No need to keep writing it. Why is that a third degree polynomial? Because the highest exponent that we have in there is the x to the third term. So that's where we get it's a third degree polynomial. This right here is a second degree polynomial. And this is the second degree term. Now a couple of other terminologies, or words, that we need to know regarding polynomials, are the constant versus the variable terms. And I think you already know, these are variable terms right here. This is a constant term. That right there is a constant term." + }, + { + "Q": "\nat 14:00 it doesnt seem very mathematically sound after working out this whole proof to just assume that (f^2-a^2) = b^2 is there no other way to connect the two without saying that looks like that other thing so lets assume this part of the equation equals that other part?", + "A": "How do we know f = a + b?", + "video_name": "HPRFmu7JsKU", + "timestamps": [ + 840 + ], + "3min_transcript": "Right, I divided both sides by this, so I just get a 1 on this right hand side. Let's see if I can simplify this. If I multiply the numerator and the denominator by a squared, right? As long as I multiply the numerator and the denominator by the same number I'm just multiplying my 1, so I'm not changing anything. So if I do that, the numerator becomes f-- if I multiply it, it becomes f squared minus a squared. I'm just multiplying that times a squared. And the denominator becomes a squared times f squared minus a squared. And all that times x squared. Minus y squared minus a squared mine is a squared is equal to 1. This cancels with this. And we get something to starting to look like the My energy is coming back! It seems like I see the light at the end of the tunnel. We get x squared over a squared minus y squared over f minus a squared is equal to 1. Now this looks a lot like our original equation of the hyperbola, which was x squared over a squared minus y squared over b squared is equal to 1. In fact this is equation of the hyperbola but instead set of writing b squared, since we wrote it, we essentially said, what is the locus of all points where the difference of the distances to those two foci is equal to 2a? And we just played with the algebra for while. It was pretty tiring, and I'm impressed if you've gotten this far into the video, and we got this equation, which should be the equation of the hyperbola, and it is the equation of the hyperbole. It is this equation. So this is the same thing is that. So f squared minus a square. Or the focal length squared minus a squared is You add a squared to both sides, and you get f squared is equal to b squared plus a squared or a squared plus b squared. Which tells us that the focal length is equal to the square root of this. Of a squared plus b squared. And that's what we set out to figure out in the beginning. So hopefully you're now satisfied that the focal length of a hyperbola is the sum of these two denominators. And it's also truth if it is an upward or vertical hyperbola. And if we're dealing with an ellipse, it's the difference of these two-- the square root of the difference of these two numbers. Anyway, I'll leave you there. That was an exhausting problem. I have to go get a glass of water now." + }, + { + "Q": "I don't understand how this proof can apply for an ellipse. I've tried the same proof, while naturally adding both d's instead of subtracting one from the other.\nUp until 9:07 the proof works fine, and in the ellipse's case fx and a^2 switch places. But once we raise the equation to the power of two, (fx-a^2)^2 is the exact same thing as (a^2-fx)^2 - and from that point on we can prove that the ellipse is - a hyperbola?!\n", + "A": "after you factor out a -1 from the denominator of the y term and simplify you get + y^2/(a^2-f^2). With an ellipse f^2 = a^2-b^2. or b^2=a^2-f^2. So y^2/b^2 = y^2/(a^2-f^2). It works. I did the entire proof only to end up with the same equation too. I was also confused about how the same equation could be both an ellipse and a hyperbola but it s the value of f that ultimately determines which one. If f > a, then it s a hyperbola.", + "video_name": "HPRFmu7JsKU", + "timestamps": [ + 547 + ], + "3min_transcript": "We have x squared on both sides of this, we subtract x squared from both sides of the equation. We have an f squared on both sides of the equation so let's cancel that out. And let's see, what can we do to simplify it. So we have a minus 2xf and a plus 2xf. Let's add 2xf to both sides this equation or bring this term over here. So if you add 2xf to both sides of this equation-- let's see, my phone is ringing, let me just turn it off --if you add 2xf to both sides of this equation, what do you get? You get 4xf-- remember I just brought this term over this left hand side --is equal to 4a squared plus 4a times a square root of x minus f squared plus y squared. It's easy to get lost in the algebra. Remember all we're doing, just to kind of remind you of what of the distances between these two points, and then see how it relates to the equation of the hyperbola itself. The a's and the b's. Let's take this 4a put it on this side, so you get 4xf minus 4a squared is equal to 4a times the square root of-- well let's just multiply this out 'cause we'll probably have to eventually --x squared minus 2xf plus f squared plus y squared. That's this just multiplied out. That's the y squared right there. We could divide both sides of this by 4. All I'm trying to do is just simplify this as much as possible, so then this becomes xf minus a squared is equal to a times the square root of this whole thing. x squared minus 2xf plus f squared plus why squared. equation right here. And then if you square both sides, this side becomes x squared f squared minus 2a squared xf plus a to the fourth. That's this side squared. And that's equal to, if you square the right hand side, a squared times the square of a square root is just that expression, x squared minus 2xf plus f squared plus y squared. This really is quite hairy. And let's see what we can do now. Let's divide both sides of this equation by a squared, and then you get x squared-- I'm really just trying to simplify this as much as possible --over a squared minus-- so the a" + }, + { + "Q": "\nAt 3:02 he devides both sides of the equation by nine, but doesn`t he only have to devide once on the left side of the equation in stead of deviding all numbers?", + "A": "Since the equation has an equals sign (=), the equation must be kept equal at all times. Multiply the left by 11, you have to multiply the right by 11. If not, then the equation will be false since both sides are no longer equal. However, if you set 1 side equal to 0, you can divide or multiply both sides and still have the 0 side set to 0 since 0 times/divided by anything is 0.", + "video_name": "vl9o9XEfXtw", + "timestamps": [ + 182 + ], + "3min_transcript": "That's, literally, the volume of the box. Now they also tell us that the volume of the box is 405 cubic units, is equal to 405. So now we just solve for x. So what do we get here? If we distribute this x into this x plus 4. Actually, if we distribute a 9x. Let me just rewrite it. This is the same thing as 9x times x plus 4 is equal to 405. 9x times x is equal to 9x squared. 9x times 4 is equal to 36x, is equal to 405. Now we want our quadratic expression to be equal to 0. So let's subtract 405 from both sides of this equation. So when you do that, your right-hand side equals 0, and Now, is there any common factor to these numbers right here? Well 405, 4 plus 0 plus 5 is 9, so that is divisible by 9. So all of these are divisible by 9. Let's just figure out what 405 divided by 9 is. So 9 goes into 405-- 9 goes into 40 4 times. 4 times 9 is 36. Subtract you get 45. 9 goes into 45 5 times. 5 times 9 is 45. Subtract, you get 0. So it goes 45 times. So if we factor out a 9 here, we get 9 times x squared-- actually even better, you don't even have to factor out of 9. If you think about it, you can divide both sides of this equation by 9. So if you can divide all of the terms by 9, it won't You're doing the same thing to both sides of equations, which we've learned long ago is a very valid thing to do. So here you get x squared-- if you just had this expression, here, and someone told you to factor it, then you'd have to factor out the 9. But because this is an equation, it equals 0, let's just divide everything by 9. It'll simplify things. So you get x squared plus 4x minus 45 is equal to 0. And now we can try to factor this right here. And this fits the pattern, where we don't have a leading 1 out here. So we don't even have to do it by grouping. You just have to think, what 2 numbers, when I take their product I get negative 45, and when I take their sum, I get positive 4. They are 4 apart. 1 has to be positive, 1 has to be negative. Their positive versions have to be 4 apart. Because when you take the sum, you are really taking their difference because 1 of them is negative. So let's think about it. When you have positive 9 and negative 5, I think that'll work." + }, + { + "Q": "at 9:27 what would the answer be if he had done 1000 instead of 100 what would it be?\n", + "A": "0.00865556 is the answer if sal had done it right. I used a calculator so that may be an abreviated version but that s what I got.", + "video_name": "ScvuRb6vsz4", + "timestamps": [ + 567 + ], + "3min_transcript": "This is really just a re-- we've just rewritten this dosage information in different units. And let's see how much solution we need per kilogram. So I want to cancel out the grams here and have a milliliters there. So to cancel out that grams, I'm going to have to have a gram in the denominator and a milliliter in the numerator. So in our solution, how many grams are there per milliliter? Well, they told us. There are 0.9 grams per milliliter. Or for every 1 milliliter, there are 0.9 grams. Notice, I just took the inverse of that. Because we want a milliliter in the numerator, grams in the denominator, so that these two cancel out. And let's do this multiplication now. So our grams cancel out. We have milliliters per kilogram. And then we multiply it out. So I'll just keep-- let me just write it like this. So there's going to be 11/1,000 times 0.9 milliliters of our solution per kilogram. So we've gotten this far. So this is per kilogram of patient body weight. And then finally, they tell us how many kilograms our patient weighs. So let's do that last multiplication, and then we can actually get our calculator out and do it all at once. So let's multiply this times-- we want to know how many milliliters per patient. We want the kilograms to cancel out. So we want kilograms per patient. Now we're talking about this particular patient. Not every patient is going to be the same number of kilograms. But if we do this, kilograms will cancel out. We'll have milliliters per patient-- milliliters of solution per patient-- which is exactly what we want. We want milliliters of solution per dose per patient. But everything we've assumed so far has been per dose. Well, there's 72.7 kilograms per patient. That's how much the patient weighs. So we just do this final multiplication and we'll be done. So our answer-- and as these two things are going to cancel out-- so our final answer is going to be 11 times 72.7 divided by 100 times-- actually, 100 times 0.9 is pretty easy to figure out. That's 900. Divided by 900 milliliters per patient. Or you can just say milliliters per dose. However you want to say it. Per dose per patient. Let's get our calculator out and do this. So we have 11 times 72.7 is equal to 799 divided by 900. Is equal to 0.88-- well, we could round up. 0.889. Hopefully the doctor won't mind. So that is equal to-- I'll write it in a nice, vibrant" + }, + { + "Q": "\nNot sure what to do with this. \"The 1st unit of PRBCs 250mL was started by the nurse at 7:30PM to be infused at 120 mL/hr in the first 15 minutes; then increased to 225mL/hr after the first 15 minutes. How long will the entire infusion take? (Provide answer in hours and minutes) \"", + "A": "you need to know the volume of the unit of blood. Typically it s written on the bag. In the equation they should have provided the volume in the unit (usually for an adult) between 280 and a bit over 300ml/unit of blood", + "video_name": "ScvuRb6vsz4", + "timestamps": [ + 450 + ], + "3min_transcript": "So let's do that. Pounds in the numerator, pounds in the denominator cancel out. And you multiply 5 times 2.2. This is equal to-- let's see. 5 times 2 is 10. 5 times 0.2 is 1. So this is equal to 11. And then in our numerator, we have milligrams. 11 milligrams per kilogram. So we just converted our dosage information to a pure metric system. It was actually a mix between the metric and the English system before. Now let's see what we can do. Well, let's see if we can get it in terms of how many milliliters we have to deliver per pound. So once again, we want this-- well, actually, let's go to grams first. Because we have milligrams here. We have grams up here. So let's see if we can convert this thing to grams. So just like we did before, we want a milligrams I'll do it in orange. We want a milligrams in the denominator and we want a gram in the numerator. Why did I say that? Because I want this and this to cancel. And I want a grams in the numerator. So how many grams are there per milligram? You can just think it through. There's 1 gram per 1,000 milligrams. Or 1,000 milligrams per gram. And you just multiply it out. So the milligrams cancels with the milligrams, and then we get-- this is equal to 11/1,000 grams per kilogram. So now we have everything in terms of grams, but we want it in terms of milliliters. The question is, how many milliliters of solution per dose? So let me go down here on this line right here. So we had this result. We have 11/1,000-- I won't do the division just yet-- grams This is really just a re-- we've just rewritten this dosage information in different units. And let's see how much solution we need per kilogram. So I want to cancel out the grams here and have a milliliters there. So to cancel out that grams, I'm going to have to have a gram in the denominator and a milliliter in the numerator. So in our solution, how many grams are there per milliliter? Well, they told us. There are 0.9 grams per milliliter. Or for every 1 milliliter, there are 0.9 grams. Notice, I just took the inverse of that. Because we want a milliliter in the numerator, grams in the denominator, so that these two cancel out. And let's do this multiplication now. So our grams cancel out. We have milliliters per kilogram. And then we multiply it out." + }, + { + "Q": "Do you mean 1000 times .9 @9:28?\n", + "A": "yes A do bleave that is true!", + "video_name": "ScvuRb6vsz4", + "timestamps": [ + 568 + ], + "3min_transcript": "This is really just a re-- we've just rewritten this dosage information in different units. And let's see how much solution we need per kilogram. So I want to cancel out the grams here and have a milliliters there. So to cancel out that grams, I'm going to have to have a gram in the denominator and a milliliter in the numerator. So in our solution, how many grams are there per milliliter? Well, they told us. There are 0.9 grams per milliliter. Or for every 1 milliliter, there are 0.9 grams. Notice, I just took the inverse of that. Because we want a milliliter in the numerator, grams in the denominator, so that these two cancel out. And let's do this multiplication now. So our grams cancel out. We have milliliters per kilogram. And then we multiply it out. So I'll just keep-- let me just write it like this. So there's going to be 11/1,000 times 0.9 milliliters of our solution per kilogram. So we've gotten this far. So this is per kilogram of patient body weight. And then finally, they tell us how many kilograms our patient weighs. So let's do that last multiplication, and then we can actually get our calculator out and do it all at once. So let's multiply this times-- we want to know how many milliliters per patient. We want the kilograms to cancel out. So we want kilograms per patient. Now we're talking about this particular patient. Not every patient is going to be the same number of kilograms. But if we do this, kilograms will cancel out. We'll have milliliters per patient-- milliliters of solution per patient-- which is exactly what we want. We want milliliters of solution per dose per patient. But everything we've assumed so far has been per dose. Well, there's 72.7 kilograms per patient. That's how much the patient weighs. So we just do this final multiplication and we'll be done. So our answer-- and as these two things are going to cancel out-- so our final answer is going to be 11 times 72.7 divided by 100 times-- actually, 100 times 0.9 is pretty easy to figure out. That's 900. Divided by 900 milliliters per patient. Or you can just say milliliters per dose. However you want to say it. Per dose per patient. Let's get our calculator out and do this. So we have 11 times 72.7 is equal to 799 divided by 900. Is equal to 0.88-- well, we could round up. 0.889. Hopefully the doctor won't mind. So that is equal to-- I'll write it in a nice, vibrant" + }, + { + "Q": "at the point 0:38, how are we expected to know which is concave and which is not?\n", + "A": "All curves are concave. The question is if they are upward or downward.", + "video_name": "UK2shgCXALo", + "timestamps": [ + 38 + ], + "3min_transcript": "If you were paying close attention in the last video, an interesting question might have popped up in your brain. We have talked about the intervals over which the function is concave downwards. And then we talked about the interval over which the function is concave upwards. But we see here that there's a point at which we transition from being concave downwards to concave upwards. Before that point, the slope was decreasing, and then the slope starts increasing. The slope was decreasing and then the slope started increasing. So that's one way to look at it. Right here in our function we go from being concave downwards to concave upwards. When you look at our derivative at that point, our derivative went from decreasing to increasing. And when you look at our second derivative at that point, it went from being negative to positive. So this must have some type of a special name you're probably And you'd be thinking correctly. This point at which we transition from being concave downwards to concave upwards, or the point at which our derivative has a extrema point, like this, we call it an inflection point. And the most typical way that people think about how could you test for an inflection point, it's a point, well, conceptually, it's where you go from being a downward opening u to an upward opening u. Or when you go from being concave downwards to concave upwards. But the easiest test is it's a point at which your second derivative switches signs. So in this case, we went from negative to positive. But we could have also switched from being positive to negative. So inflection point, your second derivative f prime prime of x switches signs. Goes from being positive to negative or negative to positive. Switches signs. So this is a case where we went from concave downwards to concave upwards. If we went from concave upwards to concave downwards, the slope was increasing. So the second derivative would to be positive. And then the slope is decreasing, so your second derivative would be negative. So here your second derivative is going from positive to negative. Here your second derivative is going from negative to positive. In either case, you are talking about an inflection point." + }, + { + "Q": "At 3:09 you described each of a function and an equation what would an example be of something that falls into both a function and an equation category?\n", + "A": "Here are some examples: y=f(x)=x\u00c2\u00b2 y=f(x)=x+1 y=f(x)=2 y=f(x)=cos(x) y=f(x)=sin(x) and so on. These equations are functions because they assign each real number x to ONE real number f(x).", + "video_name": "l3iXON1xEC4", + "timestamps": [ + 189 + ], + "3min_transcript": "So let me think of it that way. So I'm going to draw-- if this is the world of equations right over here, so this is equations. And then over here is the world of functions. That's the world of functions. I do think there is some overlap. We'll think it through where the overlap is, the world of functions. So an equation that is not a function that's sitting out here, a simple one would be something like x plus 3 is equal to 10. I'm not explicitly talking about inputs and outputs or relationship between variables. I'm just stating an equivalence. The expression x plus 3 is equal to 10. So this, I think, traditionally would just be an equation, would not be a function. Functions essentially talk about relationships between variables. You get one or more input variables, and we'll give you only one output variable. And you can define a function. And I'll do that in a second. You could define a function as an equation, but you can define a function a whole bunch of ways. You can visually define a function, maybe as a graph-- so something like this. And maybe I actually mark off the values. So that's 1, 2, 3. Those are the potential x values. And then on the vertical axis, I show what the value of my function is going to be, literally my function of x. And maybe that is 1, 2, 3. And maybe this function is defined for all non-negative values. So this is 0 of x. And so let me just draw-- so this right over here, at least for what I've drawn so far, defines that function. I didn't even have to use an equal sign. If x is 2, at least the way I drew it, y is equal to 3. You give me that input. I gave you the value of only one output. So that would be a legitimate function definition. Another function definition would be very similar to what you do in a computer program, And if day is equal to Monday, maybe you output cereal. So that's what we're going to eat that day. And otherwise, you output meatloaf. So this would also be a function. We only have one output. For any one day of the week, we can only tell you cereal or meatloaf. There's no days where you are eating both cereal and meatloaf, which sounds repulsive. And then if I were to think about something that could be an equation or a function, I guess the way I think about it is an equation is something that could be used to define a function. So for example, we could say that y is equal to 4x minus 10." + }, + { + "Q": "Is that suppose to be a '=' instead of a minus sign at 6:00 ?\n", + "A": "Yes, it s intended to be a =, not a -.", + "video_name": "4PCktDZJH8E", + "timestamps": [ + 360 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 4:59 he says \"you can always construct a faster-expanding function.\" Is that really true? What if you had a function that was only defined for the domain consisting of two x-values: 0, 1. And f(0) was defined to be negative infinity, and f(1) was defined to be infinity. How could you construct a faster-expanding function than that?", + "A": "Sal is specifically talking about real-valued functions. Since infinity is not defined in the context of real numbers, it would not make sense to define a function to have an infinite value. Infinites only make sense as a limiting value, not a functional value.", + "video_name": "6WMZ7J0wwMI", + "timestamps": [ + 299 + ], + "3min_transcript": "That's 10. That's 30. That'll be good for approximation. And then let's say that this is negative 5. This is positive 5 right here. And actually, let me stretch it out a little bit more. Let's say this is negative 1, negative 2, negative 3, negative 4. Then we have 1, 2, 3, and 4. So when x is equal to 0, we're equal to 1, right? When x is equal to 0, y is equal to 1, which is right around there. When x is equal to 1, y is equal to 3, which is right around there. When x is equal to 2, y is equal to 9, which is right around there. When x equal to 3, y is equal to 27, which is right around there. When x is equal to 4, y is equal to 81. If I did 5, we'd go to 243, which wouldn't even fit on my screen. When you go to negative 1, we get smaller and smaller. So at negative 1, we're at 1/9. Eventually, you're not even going to see this. It's going to get closer and closer to zero. As this approaches larger and larger negative numbers, or I guess I should say smaller negative numbers, so 3 to the negative thousand, 3 to the negative million, we're getting numbers closer and closer to zero without actually ever approaching zero. So as we go from negative infinity, x is equal to negative infinity, we're getting very close to zero, we're slowly getting our way ourselves away from zero, but then bam! Once we start getting into the positive numbers, we just explode. We just explode, and we keep exploding at an ever-increasing rate. So the idea here is just to show you that exponential functions are really, really dramatic. Well, you can always construct a faster expanding function. faster expanding, but out of the ones that we deal with in everyday life, this is one of the fastest. So given that, let's do some word problems that just give us an appreciation for exponential functions. So let's say that someone sends out a chain letter in week 1. In week 1, someone sent a chain letter to 10 people. And the chain letter says you have to now send this chain letter to 10 more new people, and if you don't, you're going to have bad luck, and your hair is going to fall out, and you'll marry a frog, or whatever else. So all of these people agree and they go and each send it to 10 people the next week. So in week 2, they go out and each send it to 10 more people. So each of those original 10 people are each sending out 10 more of the letters. So now 100 people have the letters, right?" + }, + { + "Q": "\nAt 4:28 Sal says \"were getting getting number closer and closer to 0, without actually ever approaching 0.\", yet the function does approach 0 when counting towards the negative direction of x.", + "A": "Poor choice of words on Sal s part. But, a box does pop up at 4:33 in the video and tells you that Sal meant without ever reaching 0 .", + "video_name": "6WMZ7J0wwMI", + "timestamps": [ + 268 + ], + "3min_transcript": "quickly we're exploding. Let me draw my axes here. So that's my x-axis and that is my y-axis. And let me just do it in increments of 5, because I really want to get the general shape of the graph here. So let me just draw as straight a line as I can. Let's say this is 5, 10, 15. Actually, I won't get to 81 that way. I want to get to 81. Well, that's good enough. Let me draw it a little bit differently than I've drawn it. So let me draw it down here because all of these values, you might notice, are positive values because I have a positive base. So let me draw it like this. Good enough. And then let's say I have 10, 20, 30, 40, 50, 60, 70, 80. That's 10. That's 30. That'll be good for approximation. And then let's say that this is negative 5. This is positive 5 right here. And actually, let me stretch it out a little bit more. Let's say this is negative 1, negative 2, negative 3, negative 4. Then we have 1, 2, 3, and 4. So when x is equal to 0, we're equal to 1, right? When x is equal to 0, y is equal to 1, which is right around there. When x is equal to 1, y is equal to 3, which is right around there. When x is equal to 2, y is equal to 9, which is right around there. When x equal to 3, y is equal to 27, which is right around there. When x is equal to 4, y is equal to 81. If I did 5, we'd go to 243, which wouldn't even fit on my screen. When you go to negative 1, we get smaller and smaller. So at negative 1, we're at 1/9. Eventually, you're not even going to see this. It's going to get closer and closer to zero. As this approaches larger and larger negative numbers, or I guess I should say smaller negative numbers, so 3 to the negative thousand, 3 to the negative million, we're getting numbers closer and closer to zero without actually ever approaching zero. So as we go from negative infinity, x is equal to negative infinity, we're getting very close to zero, we're slowly getting our way ourselves away from zero, but then bam! Once we start getting into the positive numbers, we just explode. We just explode, and we keep exploding at an ever-increasing rate. So the idea here is just to show you that exponential functions are really, really dramatic. Well, you can always construct a faster expanding function." + }, + { + "Q": "at 5:51: Why can f'(a) also be undefined?\n", + "A": "He is saying that it can either be 0 or undefined. Not 0 and also undefined. At any no endpoint max or min, the derivative has to be either 0 or undefined. This is part of the definition of a non endpoint max/min.", + "video_name": "lDY9JcFaRd4", + "timestamps": [ + 351 + ], + "3min_transcript": "We have a positive slope going into it, and then it immediately jumps to being a negative slope. So over here, f prime of x2 is not defined. Let me just write undefined. So we have an interesting-- and once again, I'm not rigorously proving it to you, I just want you to get the intuition here. We see that if we have some type of an extrema-- and we're not talking about when x is at an endpoint of an interval, just to be clear what I'm talking about when I'm talking about x as an endpoint of an interval. We're saying, let's say that the function is where you have an interval from there. So let's say a function starts right over there, and then This would be a maximum point, but it would be an end point. We're not talking about endpoints right now. We're talking about when we have points in between, or when our interval is infinite. So we're not talking about points like that, or points like this. We're talking about the points in between. it's going to be a minimum or maximum. And we see the intuition here. If you have-- so non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or a maximum point, at some point x is equal to a, and x isn't the endpoint of some interval, this tells you something interesting. Or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0. Or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0, derivative is 0, derivative is undefined. And we have a word for these points where the derivative is We called them critical points. So for the sake of this function, the critical points are, we could include x sub 0, we could include x sub 1. At x sub 0 and x sub 1, the derivative is 0. And x sub 2, where the function is undefined. Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. But can we say it the other way around? If we find a critical point, where the derivative is 0, or the derivative is undefined, is that going to be a maximum or minimum point? And to think about that, let's imagine this point right over here. So let's call this x sub 3." + }, + { + "Q": "\nAt 0:46, what does he mean by x0? Does he mean x to the 0th power?", + "A": "The subscript appended x_0 signifies a specific value of x. If we are trying to talk about specific values of x, we will just use the subscripts to differentiate them (x_0, x_1, x_2...). This is much easier and makes much more sense than assigning new variables/letters to these values of x.", + "video_name": "lDY9JcFaRd4", + "timestamps": [ + 46 + ], + "3min_transcript": "I've drawn a crazy looking function here in yellow. And what I want to think about is when this function takes on the maximum values and minimum values. And for the sake of this video, we can assume that the graph of this function just keeps getting lower and lower and lower as x becomes more and more negative, and lower and lower and lower as x goes beyond the interval that I've depicted right over here. So what is the maximum value that this function takes on? Well we can eyeball that. It looks like it's at that point right over there. So we would call this a global maximum. Function never takes on a value larger than this. So we could say that we have a global maximum at the point x0. Because f of of x0 is greater than, or equal to, f of x, for any other x in the domain. And that's pretty obvious, when you look at it like this. Now do we have a global minimum point, Well, no. This function can take an arbitrarily negative values. It approaches negative infinity as x approaches negative infinity. It approaches negative infinity as x approaches positive infinity. So we have-- let me write this down-- we have no global minimum. Now let me ask you a question. Do we have local minima or local maxima? When I say minima, it's just the plural of minimum. And maxima is just the plural of maximum. So do we have a local minima here, or local minimum here? Well, a local minimum, you could imagine means that that value of the function at that point is lower than the points around it. So right over here, it looks like we have a local minimum. And I'm not giving a very rigorous definition here. But one way to think about it is, we can say that we have a local minimum point at x1, is less than an f of x for any x in this region right over here. And it's pretty easy to eyeball, too. This is a low point for any of the values of f around it, right over there. Now do we have any other local minima? Well it doesn't look like we do. Now what about local maxima? Well this one right over here-- let me do it in purple, I don't want to get people confused, actually let me do it in this color-- this point right over here looks like a local maximum. Not lox, that would have to deal with salmon. Local maximum, right over there. So we could say at the point x1, or sorry, at the point x2, we have a local maximum point at x2. Because f of x2 is larger than f of x for any" + }, + { + "Q": "On 5:17, Sal said that there aren't any other solutions. Can't the values for a and b switch around? So what I am basically asking is that instead of 3 solutions, aren't there supposed to be 6 solutions because a and b will switch their values and be a repeat?\n", + "A": "He s assuming that such things are accounted for (i.e. order is irrelevant, because multiplication and addition are commutative).", + "video_name": "JPQ8cfOsYxo", + "timestamps": [ + 317 + ], + "3min_transcript": "So this is pretty interesting here. We can divide both sides by a plus b. We know that a plus b cannot be equal to 0 since all of these numbers have to be positive numbers. And the reason why I say that is if it was 0, dividing by 0 would give you an undefined answer. So if we divide both sides by a plus b, we get a times b is equal to 12. So all the constraints that they gave us boiled down to this right over here. The product of a and b is equal to 12. And there's only so many numbers, so many positive integers where you if you take the product, Let's try them out. So let me write some columns here. Let's say a, b, c. And then we care about their product, so I'll write that over here, so abc. c is the sum of those two, so c is going to be 13, 1 times 12 times 13. 12 times 12 is 144 plus another 12 is going to be 156. And just for fun, you can verify that this is going to be equal to 6 times their sum. Their sum is 26, 26 times 6 is 156. So this one definitely worked. It definitely worked for the constraints. And it should because we boiled down those constraints to a times b need to be equal to 12. So let's try another one. 2 times 6, their sum is 8. And then if I were to take the product of all of these, you get 2 times 6 is 12 times 8 is 96. And then we could try 3 and 4. 3 plus 4 is 7. Actually, I should have known the a times b is always 12, so you just have to multiply 12 times this last column. 12 times 7 is 84. And there aren't any others. You definitely can't go above 12 because then you would have to deal with non-integers. You would have to deal with fractions. You can't do the negative versions of these because they all have to be positive integers. So that's it. Those are all of the possible positive integers where if you take their products you get 12. We've essentially just factored 12. So they want us to find the sum of all possible values of N. Well, these are all the possible values of N. N was the product of those integers. So let's just take the sum. 6 plus 6 is 12 plus 4 is 16, 1 plus 5 is 6 plus 9 is 15 plus 8 is 23, 2 plus 1 is 3." + }, + { + "Q": "\nAt around 0:06, if Consistent solution #1 is independent, Consistent solution #2 is dependent, then what is an INconistent solution?", + "A": "He draws an inconsistent system at 1:20 - it s 2 parallel lines.", + "video_name": "WSpF5uvApLA", + "timestamps": [ + 6 + ], + "3min_transcript": "Is the system of linear equations below dependent or independent? And they give us two equations right here. And before I tackle this specific problem, let's just do a little bit a review of what dependent or independent means. And actually, I'll compare that to consistent and inconsistent. So just to start off with, if we're dealing with systems of linear equations in two dimensions, there's only three possibilities that the lines or the equations can have relative to each other. So let me draw the three possibilities. So let me draw three coordinate axes. So that's my first x-axis and y-axis. Let me draw another one. That is x and that is y. Let me draw one more, because there's only three possibilities in two dimensions. x and y if we're dealing with linear equations. So you can have the situation where the lines just intersect in one point. Let me do this. So you could have one line like that and they intersect at one point. You could have the situation where the two lines are parallel. So you could have a situation-- actually let me draw it over here-- where you have one line that goes like that and the other line has the same slope but it's shifted. It has a different y-intercept, so maybe it looks like this. And you have no points of intersection. And then you could have the situation where they're actually the same line, so that both lines have the same slope and the same y-intercept. So really they are the same line. They intersect on an infinite number of points. Every point on either of those lines is also a point on the other line. So just to give you a little bit of the terminology here, and we learned this in the last video, this type of system where they don't intersect, where you have no solutions, this is an inconsistent system. And by definition, or I guess just taking would be considered consistent. But then within consistent, there's obviously a difference. Here we only have one solution. These are two different lines that intersect in one place. And here they're essentially the same exact line. And so we differentiate between these two scenarios by calling this one over here independent and this one over here dependent. So independent-- both lines are doing their own thing. They're not dependent on each other. They're not the same line. They will intersect at one place. Dependent-- they're the exact same line. Any point that satisfies one line will satisfy the other. Any points that satisfies one equation will satisfy the other. So with that said, let's see if this system of linear equations right here is dependent or independent. So they're kind of having us assume that it's going to be consistent, that we're going to intersect in one place or going to intersect in an infinite number of places." + }, + { + "Q": "\nat 3:06, when 7 doesn't go into 5, why doesn't he write 0, and instead he immediately uses \"51\"?", + "A": "This is a stylistic preference. Placing a zero to the left of a number will not change its value, so you are free to place a zero there if you like. However, when reporting your final answer, I would not include superfluous digits, so you would have to change it again.", + "video_name": "cfr-yZxTH8Y", + "timestamps": [ + 186 + ], + "3min_transcript": "So if we move the decimal over to the right with the 0.7 to turn into a 7, we also need to move the decimal over to the right for 518. Now, you're probably saying, well, I don't see a decimal in 518. Well, there is one. You just didn't have to write it, because it's 518.00-- and we can add as many zeroes as we want. So if we move the decimal to the right, it becomes 5,180. So really what we're saying is 518 divided by 0.7 is the same thing as 5,180 divided by 7. Notice all we did by moving the decimal one place to the right, is we multiplied both of these numbers by 10, which is not going to change the actual value of the decimal. One other way of thinking about this, if you wanted to write this as a fraction, You multiply both the numerator and denominator by 10, you will get 5,180 over 7. So let's clean this up a little bit, just so we remember what we did. So we moved the decimal over to the right, one. So now this is just a 7. The decimal is there. In fact, we really don't have to write the decimal anymore. It's just a 7.0-- you could imagine 7.0, so we can just write this as a 7. And then the 518, the decimal is now out here. So this is 5,180. And let's increase the sign right over here. Now, this is just a straight-up long division problem. How many times does 7 go into 5? Well, it goes 0 times. 0 times 7-- actually, let's just cut to the chase. It does go into 51. 7 times 7 is 49. So it goes 7 times. 7 times 7 is 49. Subtract 51 minus 49 is 2. And now we can bring down this 8. 7 goes into 28 four times. 4 times 7 is 28. Subtract, you get a 0. Now we can bring down another 0. We want at least get to the decimal place. So we bring down another 0 right over here. When I say get to the decimal place, we could put the decimal place up here too, just to make sure we're keeping track of the right place values or that we can have the decimal in the right place. So notice, I'm very particular. When I'm doing 7 goes into 51, I put the 7 right above the 1 in the 51's place. When I'm saying 7 goes into 28, I'm putting the 4 right above the 8 in this one's place when we're doing the division." + }, + { + "Q": "At 2:57,why was it 1/2 of the whole triangle?\n", + "A": "The 1/2 is not based on the triangle, it is based on a triangle being exactly 1/2 of the rectangle that it is enclosed within. In the case that he is using, he is actually showing 1/2 of the parallelogram, not the triangle.", + "video_name": "rRTXKQpblEc", + "timestamps": [ + 177 + ], + "3min_transcript": "that the area of a parallelogram is base times height, because we're now going to use that to get the intuition for the area of a triangle. So let's look at some triangles here. So that is a triangle, and we're given the base and the height, and we're gonna try to think about what's the area of this triangle going to be, and you can imagine it's going to be dependent on base and height. Well, to think about that, let me copy and paste this triangle. So let me copy, and then let me paste it, and what I'm gonna do is, so now I have two of the triangles, so this is now going to be twice the area, and I'm gonna rotate it around, I'm gonna rotate it around like that, and then add it to the original area, and you see something very interesting is happening. I have now constructed a parallelogram. that has twice the area of our original triangle, 'cause I have two of our original triangles right over here, you saw me do it, I copied and pasted it, and then I flipped it over and I constructed the parallelogram. Now why is this interesting? Well, the area of the entire parallelogram, the area of the entire parallelogram is going to be the length of this base times this height. You also have height written with the \"h\" upside down over here. It's going to be base times height. That's going to be for the parallelogram, for the entire-- let me draw a parallelogram right over here. That's going to be the area of the entire parallelogram. So what would be the area of our original triangle? What would be the area of our original triangle? Well, we already saw that this area of the parallelogram, it's twice the area of our original triangle. So our original triangle is just going to have half the area. So this area right over here is going to be one half the area of the parallelogram. One half base times height. One half base times height. And you might say, \"OK, maybe it worked for this triangle, \"but I wanna see it work for more triangles.\" And so, to help you there, I've added another triangle right over here, you could do this as an obtuse triangle, this angle right over here is greater than 90 degrees, but I'm gonna do the same trick. We have the base, and then we have the height. if you start at this point right over here, and if you drop a ball, the length that the ball goes, if you had a string here, to kind of get to the ground level, you could view this as the ground level right over there, that that's going to be the height, it's not sitting in the triangle like we saw last time, but it's still the height of the triangle. If this was a building of some kind, you'd say, \"Well, this is the height.\" How far off the ground is it? Well, what's the area of this going to be? Well, you can imagine, it's going to be one half base times height. How do we feel good about that? Well, let's do the same magic here." + }, + { + "Q": "What was that thing that popped up at 2:08?\n\n(It seems to be something called ArtRage. . .)\n", + "A": "I agree because it said ART, and the person was drawing", + "video_name": "rRTXKQpblEc", + "timestamps": [ + 128 + ], + "3min_transcript": "- [Voiceover] We know that we can find the area of a rectangle by multiplying the base times the height. The area of a rectangle is equal to base times height. In another video, we saw that, if we're looking at the area of a parallelogram, and we also know the length of a base, and we know its height, then the area is still going to be base times height. Now, it's not as obvious when you look at the parallelogram, but in that video, we did a little manipulation of the area. We said, \"Hey, let's take this \"little section right over here.\" So we took that little section right over there, and then we move it over to the right-hand side, and just like that, you see that, as long as the base and the height is the same, as this rectangle here, I'm able to construct the same rectangle by moving that area over, and that's why the area of this parallelogram is base times height. I didn't add or take away area, I just shifted area from the left-hand side to the right-hand side to show you that the area of that parallelogram was the same as this area of the rectangle. It's still going to be base times height. that the area of a parallelogram is base times height, because we're now going to use that to get the intuition for the area of a triangle. So let's look at some triangles here. So that is a triangle, and we're given the base and the height, and we're gonna try to think about what's the area of this triangle going to be, and you can imagine it's going to be dependent on base and height. Well, to think about that, let me copy and paste this triangle. So let me copy, and then let me paste it, and what I'm gonna do is, so now I have two of the triangles, so this is now going to be twice the area, and I'm gonna rotate it around, I'm gonna rotate it around like that, and then add it to the original area, and you see something very interesting is happening. I have now constructed a parallelogram. that has twice the area of our original triangle, 'cause I have two of our original triangles right over here, you saw me do it, I copied and pasted it, and then I flipped it over and I constructed the parallelogram. Now why is this interesting? Well, the area of the entire parallelogram, the area of the entire parallelogram is going to be the length of this base times this height. You also have height written with the \"h\" upside down over here. It's going to be base times height. That's going to be for the parallelogram, for the entire-- let me draw a parallelogram right over here. That's going to be the area of the entire parallelogram. So what would be the area of our original triangle? What would be the area of our original triangle? Well, we already saw that this area of the parallelogram, it's twice the area of our original triangle. So our original triangle is just going to have half the area. So this area right over here is going to be one half the area of the parallelogram." + }, + { + "Q": "\nAt 2:18, why is there -B on the right side?", + "A": "It is the same thing as saying: A=50-B It also equals: A=50+-B", + "video_name": "3mimxluSVBo", + "timestamps": [ + 138 + ], + "3min_transcript": "Use graphing to solve the following problem. Abby and Ben did household chores last weekend. Together they earned $50, and Abby earned $10 more than Ben. How much did they each earn? So let's define some variables here. Let's let A equal Abby's earnings. And let's let B equal Ben's earnings. Then they tell us how these earnings relate. They first tell us that together they earned $50. So that statement can be converted mathematically into-- well, together, that means the sum of the two earnings. So A plus B needs to be equal to $50. Abby's plus Ben's earnings is $50. And then they tell us Abby earned $10 more than Ben. is equal to Ben's earnings plus 10. Abby earned $10 more than Ben. So we have a system of two equations and actually with two unknowns. And then they say, how much did each earn? So to do that, and they want us to solve this graphically. There's multiple ways to solve it, but we'll do what they ask us to do. Let me draw some axes over here. And I'll be in the first quadrant since we're dealing with earnings, so neither of their earnings can be negative. And let me just define the vertical axis as Abby's axis or the Abby's earnings axis. And let me define the horizontal axis as Ben's axis or Ben's earnings axis. And let me just graph each of these equations. And to do that, I'm going to take this first equation, and I'm going to put it in the equivalent of slope-intercept form. It might look a little unfamiliar to you, but it really is slope-intercept form. Let me rewrite it first. So we have A plus B is equal to 50. So let's subtract B from both sides. And then we get A is equal to negative B plus 50. So if you think about it this way, when B is equal to 0, A is going to be 50. So we know our A intercept, we could call it. We normally would call that a y-intercept, but now this is the A axis. So this right here, let me call this 10, 20, 30, 40, and 50. So if Ben made $0, then Abby would have to make $50 based on that first constraint. So we know that that's a point on the line right over there. And we also know that the slope is negative 1, that B is the independent variable, the way I've written it over here, and this coefficient is negative 1. Or another way to think about it is if A is 0, then B is going to be 50. If Abby made no money, then Ben would have to make $50." + }, + { + "Q": "\nAt 3:47 Sal said factor but didn't he mean simplify?", + "A": "Hi Knight of Stone / Ninjormon (Mormon ninja), Sal did say factor, but meant simplify. They have added the correction to the video in the box which shows up at 3:45. Hope that helps! - JK", + "video_name": "f-wz_ZzSDdg", + "timestamps": [ + 227 + ], + "3min_transcript": "In the numerator, we took care of our a plus 2's. That's the only one that's common, so in the numerator, we also have an a minus 2. Actually, we have an a plus 1-- let's write that there, too. We have an a plus 1 in the numerator. We also have an a plus 1 in denominator. In the numerator, we have an a minus 2, and in the denominator, we have an a minus 1. So all I did is I rearranged the numerator and the denominator, so if there was something that was of a similar-- if the same expression was in both, I just wrote them on top of each other, essentially. Now, before we simplify, this is a good time to think about the domain or think about the a values that aren't in the domain, the a values that would invalidate or make this expression undefined. Like we've seen before, the a values that would do that are the ones that would make the denominator equal 0. So the a values that would make that equal to 0 is a is equal to negative 2. You could say a plus 2 is equal to 0, or a is equal to negative 2. a plus 1 is equal to 0. Subtract 1 from both sides. a is equal to negative 1. Or a minus 1 is equal to 0. Add one to both sides, and you get a is equal to 1. For this expression right here, you have to add the constraint that a cannot equal negative 2, negative 1, or 1, that a can be any real number except for these. We're essentially stating our domain. We're stating the domain is all possible a's except for these things right here, so we'd have to add that little caveat right there. Now that we've done that, we can factor it. We have an a plus 2 over an a plus 2. We know that a is not going to be equal to negative 2, so that's always going to be defined. When you divide something by itself, that is going to just be 1. The same thing with the a plus 1 over the a plus 1. That's going to be 1. over a minus 1. So the simplified rational is a minus 2 over a minus 1 with the constraint that a cannot equal negative 2, negative 1, or 1. You're probably saying, Sal, what's wrong with it equaling, for example, negative 1 here? Negative 1 minus 1, it's only going to be a negative 2 here. It's going to be defined. But in order for this expression to really be the same as this expression up here, it has to have the same It has to have the same domain. It cannot be defined at negative 1 if this guy also is not defined at negative 1. And so these constraints essentially ensure that we're dealing with the same expression, not one that's just close." + }, + { + "Q": "at 2:10, to think of it as 4/8 is that the easiest way to think about 1/2?\n", + "A": "Yes, 4/8 is equal to 1/2, because 4 is half of 8 and 1 is half of 2. 1/2 is the lowest simplified form of 4/8.", + "video_name": "erZe85NrsK0", + "timestamps": [ + 130 + ], + "3min_transcript": "" + }, + { + "Q": "\nWhy did Sal take anti derivative at 8:25?", + "A": "Because derivative of arctan(2x) = f(x) at 2:46 So, antiderivative of f(x) = arctan(2x) at 8:25", + "video_name": "NgYrsqoKXpM", + "timestamps": [ + 505 + ], + "3min_transcript": "If f of x is equal to that then f of x is power series representation its just going to be taking this power series or at least the first four terms of it and replacing the x's with four x squares and then multiplying the whole thing times two, so let's do that. We can write that f of x, so we could write the f of x is going to be approximately equal to two times this thing, a value when x is equal to four x squared. It is one minus instead of an x, I'm going to write a four x squared plus x squared but instead of an x, I have a four x squared squared. This is plus four x squared squared. Well that's going to be 16 x to the fourth, so let me just write that, that's going to be plus 16 x to the fourth but now an x is four x squared so it's minus four x squared to the third power. Well that's going to be 64 x to the sixth. Let me write that, minus 64 x to the sixth power and then we could say that this is going to be so f of x is approximately equal to, distribute the two, two minus eight x squared plus 32 x to the fourth minus 128 x to the sixth. Just like that with a little bit of a substitution, I was able to reasonably a non hairy fashion find the first four nonzero terms of the power series of two over one plus four x squared which is the derivative, which is going to be the derivative of the power series of arctangent of two x. Let's write this down, so I'm going to write, I'm going to write down, arctangent of two x which is equal to an antiderivative of f of x, dx which is going to be equal to an antiderivative of this whole, all of this business. It's the antiderivative of two minus eight, minus eight x squared plus 32 x to the fourth minus 128 x to the sixth. Actually let me make this approximate because now of course we're doing approximation with the power series. Dx and what is this going to be equal to? We get approximately equal to well I'm going to have a constant there. Let me write the constant first because when we write our power series or Maclaurin Series, the first term is the constant term. It's our function evaluated to add zero. We're going to have a constant," + }, + { + "Q": "\nDear Sal, I think you made a mistake at 1:20 - 1:21 when you said that the third term is ar to the third power. Didn't you write ar^2/ar squared? I am slightly confused there. It might've been a careless mistake.", + "A": "yeah it was just a careless mistake", + "video_name": "dIGLhLMsy2U", + "timestamps": [ + 80, + 81 + ], + "3min_transcript": "Let's talk about geometric sequences, which is a class of sequences where we start at some number, then each successive number is the previous number multiplied by the same thing. So what am I talking about? Well let's multiply a times r. And then I'm going to get ar. Let's multiply it times, but to get the third term, let's multiply the second term times r. And then what am I going to have? I'm going to have-- it's a different shade of yellow-- I'm going to have ar squared. Multiply by r again, you're going to get ar to the third power, and you just keep on going like that. And this is, the way I've denoted this, this is an infinite geometric sequence. We just keep going on and on and on and on. And the different ways we can denote it, we can denote it explicitly. We could say that our sequence is a sub n starting with the first term going all the way to infinity, with a sub for any term-- is going to be a times r. And just to be clear, this right over here, a is the same thing as a times r to the zeroth power, r to the 0 is just 1. This second term is ar to the first power. The third term is ar to the third power. It looks like the nth term is going to be ar to the n minus 1 power. So ar to the n minus 1. And you could verify it. If you want the second term, you say a times r to the 2 minus 1, a times r to the first power. It works out. This is defining it explicitly. We could also define it recursively. We could say a sub n from n equals 1 to infinity, with a sub 1 being equal to a. That's the base case. a sub 1 is equal to a, ar to the 0 is just a. Or we could say for n equals 1, and then we could say a-- that because we're making it very clear that a sub 1 is equal to a-- and then we could say a sub n is equal to the previous term, a sub n minus 1, times r, for n is greater than or equal to 2. So this is saying, look, our first term is going to be a, that right over there is a, ar to the 0 is just a, and then each successive term is going to be the previous term times r, which is exactly what we did over there. So let's look at some geometric sequences. So I could have a geometric sequence like this. I could have a sub n, n is equal to 1 to infinity with, let's say, a sub n is equal to, let's say our first term is, I don't know, let's say it is equal to 20. And then r, the number that we're multiplying to get each successive term, let's say it's equal to 1/2. 1/2 to the n minus 1." + }, + { + "Q": "\nAt 3:47, what was that other sequence that wasn't geometric, and how do you solve it/put it into explicit form? For instance, my sequence is 1, 3, 6, 10, 15, and 21. How could I solve to find the 100th term?\nSide note just found out this is a triangular sequence. no idea wat that means...", + "A": "In general, there s no easy way to do this. But for a lot of cases, taking the sequence of differences or the sequence of quotients is a good way to approach this. In your case, the sequence of differences is 2,3,4,5,6,... From there we can say that a\u00e2\u0082\u0099 = a\u00e2\u0082\u0099\u00e2\u0082\u008b\u00e2\u0082\u0081 + n+1 and a\u00e2\u0082\u0081 = 1 so a\u00e2\u0082\u0099 = 1+2+3+4+...+n+(n+1) = n(n+1)/2.", + "video_name": "dIGLhLMsy2U", + "timestamps": [ + 227 + ], + "3min_transcript": "that because we're making it very clear that a sub 1 is equal to a-- and then we could say a sub n is equal to the previous term, a sub n minus 1, times r, for n is greater than or equal to 2. So this is saying, look, our first term is going to be a, that right over there is a, ar to the 0 is just a, and then each successive term is going to be the previous term times r, which is exactly what we did over there. So let's look at some geometric sequences. So I could have a geometric sequence like this. I could have a sub n, n is equal to 1 to infinity with, let's say, a sub n is equal to, let's say our first term is, I don't know, let's say it is equal to 20. And then r, the number that we're multiplying to get each successive term, let's say it's equal to 1/2. 1/2 to the n minus 1. Well let's think about it. The first term is 20. If you say, if n is 1, this is going to be 1/2 to the 0-th power. So it's going to be 1 times 20. So the first term is 20, and then each time we're multiplying by what? Well here each time we're multiplying by 1/2. So this could be 20 times 1/2 is 10, 10 times 1/2 is 5, 5 times 1/2 is 2.5-- actually let me just write that as a fraction, is 5/2, 5/2 times 1/2 is 5/4, and you can just keep going on and on and on. This is a geometric sequence. Now let me give you another sequence, and tell me if it is geometric. So let's say we start at 1, so then I'm going to go to 2, and then I'm going to go to 6, and then I'm going to go to-- let me see what I want to do-- I want to go to 24. And then I could go to 120, and I go on and on and on. Well let's think about what's going on. To go from 1 to 2, I multiplied by 2. To go from 2 to 6, I multiplied by 3. To go from 6 to 24, I multiplied by 4. So I'm always multiplying not by the same amount. You have to multiply by the same amount in order for it to be a geometric sequence. Here I'm multiplying it by a different amount. So this sequence that I just constructed has the form, I have my first term, and then my second term is going to be 2 times my first term, and then my third one is going to be 3 times my second term, so 3 times 2 times a. My fourth one is 4 times the third term, so 4 times 3 times 2 times a." + }, + { + "Q": "\ni dont get what you sal said at 5:36", + "A": "because he is smart", + "video_name": "t8m0NalQtEk", + "timestamps": [ + 336 + ], + "3min_transcript": "So you put the 10 there. There's nothing left. You put the 0 there. There's nothing left to put the 1, so you put the 10 there. So you essentially have solved the problem that 36-- let me do this is another color. That 36 times 3 is equal to 108. That's what we've solved so far, but we have this 20 sitting out here. We have this 20. We have to figure out what 20 times 360 is. Or sorry, what 20 times 36 is. This 2 is really a 20. And to make it all work out like that, what we do is we throw a 0 down here. We throw a 0 right there. In a second I'm going to explain why exactly we did that. So let's just do the same process as we did before with the 3. Now we do it with a 2, but we start filling up here and move to the left. So 2 times 6. That's 12. We put the 1 up here and we have to be very careful because we had this 1 from our previous problem, which doesn't apply anymore. So we could erase it or that 1 we could get rid of. If you have an eraser get rid of it, or you can just keep track in your head that the 1 you're about to write is a different 1. So what were we doing? We wrote 2 times 6 is 12. Put the 2 here. Put the 1 up here. And I got rid of the previous 1 because that would've just messed me up. Now I have 2 times 3. 2 times 3 is equal to 6. But then I have this plus 1 up here, so I have to add plus 1. So I get 7. So that is equal to 7. 2 times 3 plus 1 is equal to 7. So this 720 we just solved, that's literally-- let me write that down. That is 36 times 20. 36 times 20 is equal to 720. throw this 0 here. If we didn't throw that 0 here we would have just a 2-- we would just have a 72 here, instead of 720. And 72 is 36 times 2. But this isn't a 2. This is a 2 in the 10's place. This is a 20. So we have to multiply 36 times 20, and that's why we got 720 there. So 36 times 23. Let's write it this way. Let me get some space up here. So we could write 30-- well, actually, let me just finish the problem and then I'll explain to you why it worked. So now to finish it up we just add 108 to 720. So 8 plus 0 is 8. 0 plus 2 is 2. 1 plus 7 is 8. So 36 times 23 is 828. Now you're saying Sal, why did that work? why were we able to figure out separately 36 times 3 is equal to 108, and then 36 times 20 is equal to 720, and then" + }, + { + "Q": "At 1:39 Sal says that -10 and -4 will work for both 40 and -14. Would it matter if you used 8 and 5 too?\n", + "A": "It matters... The 2 numbers you pick must add to -14 AND multiply to 40. While 8 and 5 multiply to 40, they do not add to -14. 8+5 = +13. So, the 8 and 5 will not work. Hope this helps.", + "video_name": "1kfq0aR3ASs", + "timestamps": [ + 99 + ], + "3min_transcript": "To better understand how we can factor second degree expressions like this, I'm going to go through some examples. We'll factor this expression and we'll factor this expression. And hopefully it'll give you a background on how you could generally factor expressions like this. And to think about it, let's think about what happens if I were to multiply x plus something times x plus something else. Well, if I were to multiply this out, what do I get? Well, you're going to get x squared plus ax plus bx, which is the same thing as a plus bx plus a times b. So if you wanted to go from this form, which is what we have in these two examples, back to this, you really just have to think about well, what's our coefficient on our x term, and can I figure out two numbers that when I take their sum, are equal to that coefficient, and what's my constant term, and can I think of two numbers, those same two numbers, that when I take the product equal that constant term? If we look at our coefficient on x, can we think of an a plus ab that is equal to that number negative 14? And can we think of the same a and b that if we were to take its product, it would be equal to 40? So what's an a and a b that would work over here? Well, let's think about this a little bit. If I have 4 times 10 is 40, but 4 plus 10 is equal to positive 14. So that wouldn't quite work. What happens if we make them both negative? If we have negative 4 plus negative 10, well that's going to be equal to negative 14. And negative 4 times negative 10 is equal to 40. The fact that this number right over here is positive, this number right over here is positive, tells you that these are going to be the same sign. then we would have different signs. And so if you have 2 numbers that are going to be the same sign and they add up to a negative number, then that tells you that they're both going to be negative. So just going back to this, we know that a is going to be negative 4, b is equal to negative 10, and we are done factoring it. We can factor this expression as x plus negative 4 times x plus negative 10. Or another way to write that, that's x minus 4 times x minus 10. Now let's do the same thing over here. Can we think of an a plus b that's equal to the coefficient on the x term? Well, the coefficient on the x term here is essentially negative 1 times x. So we could say the coefficient is negative 1. And can we think of an a times b where it's going to be equal to negative 12?" + }, + { + "Q": "\nI Don't Get what he said at 2:03.", + "A": "He is writing the remainder part as a single fraction instead of two separate fractions.", + "video_name": "WqNc6My1aNU", + "timestamps": [ + 123 + ], + "3min_transcript": "The quotient of two polynomials-- a of x and b of x-- can be written in the form a of x over b of x is equal to q of x plus r of x over b of x-- where q of x and r of x are polynomials and the degree of r of x is less than the degree of b of x. Write the quotient 7x to the sixth plus x to the third plus 2x plus 1 over x squared in this form. Well, this one is pretty straightforward because we're dividing by x squared. So you could literally view this as 7x to the sixth divided by x squared plus x to the third divided by x squared plus 2x divided by x squared plus 1 divided by x squared. So we could just do this term by term. What's 7x to the sixth divided by x squared? Well, x to the sixth divided by x squared is x to the fourth. So it's going to be 7x to the fourth power. And then, same thing right over here. Plus x to the third divided by x squared. So plus x. And then, we're going to have 2x divided by x squared. But remember, we want to write it in a form of r of x over b of x-- where r of x has a lower degree than b of x. Well, 2x has a lower degree than x squared. Here this is degree 1. This is degree 2. So you could write it as plus 2x over x squared. Like that. And then, you could write plus 1 over x squared. So you could do this-- plus 1 over x squared. So you could write it like that. But that's not exactly the form that they want. They want us to write it q of x-- and you could view that as 7x to the fourth plus x. And then, they want plus r of x over b of x So plus some polynomial over x squared in this case. So instead of writing it as 2x over x squared plus 1 over x squared, we could just write it as 2x plus 1 over x squared. So let me just put some parentheses here so that it interprets my typing correctly. So notice, this part of the polynomial, these terms have an equal or higher degree than x squared. So I just divided those. 7x to the sixth divided by x squared is 7x to the fourth. x to the third divided by x squared is x. And then, once I got two terms that had a lower degree than x squared, I just left on there. I just said plus whatever 2x plus 1 divided by x squared is. And that's the form that they wanted us to write it in. We'll check our answer. And we got it right." + }, + { + "Q": "\nAt 2:11 You factor the 15x and not the 8 but at 4:45 you factor the 16 and 10x. Why was the 8 not a factor in the first part?", + "A": "In the first part we need two numbers such that their product is 16(8*2)(coefficient of x^2 and constant term) and their sum is 15(coefficient of x). The nos. are 16 and -1 Similarly, in the 2nd part. But, we factor out only 16 as coefficient of x^2 over there is just 1.", + "video_name": "u9v_bakOIcU", + "timestamps": [ + 131, + 285 + ], + "3min_transcript": "f of x is equal to 2x squared plus 15x minus 8. g of x is equal to x squared plus 10x plus 16. Find f/g of x. Or you could interpret this is as f divided by g of x. And so based on the way I just said it, you have a sense of what this means. f/g, or f divided by g, of x, by definition, this is just another way to write f of x divided by g of x. You could view this as a function, a function of x that's defined by dividing f of x by g of x, by creating a rational expression where f of x is in the numerator and g of x is in the denominator. And so this is going to be equal to f of x-- we have right up here-- is 2x squared 15x minus 8. And g of x-- I will do in blue-- is right over here, g of x. So this is all going to be over g And you could leave it this way, or you could actually try to simplify this a little bit. And the easiest way to simplify this would see if we could factor the numerator and the denominator expressions into maybe simpler expressions. And maybe some of them might be on-- maybe both the numerator and denominator is divisible by the same expression. So let's try to factor each of them. So first, let's try the numerator. And I'll actually do it up here. So let's do it. Actually, I'll do it down here. So if I'm looking at 2x squared plus 15x minus 8, we have a quadratic expression where the coefficient is not 1. And so one technique to factor this is to factor by grouping. You could also use the quadratic formula. And when you factor by grouping, you're going to split up this term, this 15x. And you're going to split up into two terms where the coefficients are, if I were to take the product of those coefficients, of the first and the last terms. And we proved that in other videos. So essentially, we want to think of two numbers that add up to 15, but whose product is equal to negative 16. And this is just the technique of factoring by grouping. It's really just an attempt to simplify this right over here. So what two numbers that, if I take their product, I get negative 16. But if I add them, I get 15? Well, if I take the product and get a negative number, that means they have to have a different sign. And so that means one of them is going to be positive, one of them is going to be negative, which means one of them is going to be larger than 15 and one of them is going to be smaller than 15. And the most obvious one there might be 16, positive 16, and negative 1. If I multiply these two things, I definitely get negative 16. If I add these two things, I definitely get 15. So what we can do is we can split this. We can rewrite this expression as 2x squared plus 2x squared" + }, + { + "Q": "\nAt 3:35, how did Sal get 2x(x+8) from 2x^2 + 16x?", + "A": "Sal is distributing the 2x and multiplying with the x and the 8. To get a better visualization, learn about the distributive property.", + "video_name": "u9v_bakOIcU", + "timestamps": [ + 215 + ], + "3min_transcript": "of the first and the last terms. And we proved that in other videos. So essentially, we want to think of two numbers that add up to 15, but whose product is equal to negative 16. And this is just the technique of factoring by grouping. It's really just an attempt to simplify this right over here. So what two numbers that, if I take their product, I get negative 16. But if I add them, I get 15? Well, if I take the product and get a negative number, that means they have to have a different sign. And so that means one of them is going to be positive, one of them is going to be negative, which means one of them is going to be larger than 15 and one of them is going to be smaller than 15. And the most obvious one there might be 16, positive 16, and negative 1. If I multiply these two things, I definitely get negative 16. If I add these two things, I definitely get 15. So what we can do is we can split this. We can rewrite this expression as 2x squared plus 2x squared All I did here is I took this middle term and, using this technique right over here, I split it into 16x minus x, which is clearly still just 15x. Now what's useful about this-- and this is why we call it factoring by grouping-- is we can see, are there any common factors in these first two terms right over here? Well, both 2x squared and 16x, they are both divisible by 2x. So you could factor out a 2x of these first two terms. This is the same thing as 2x times x plus x plus 8. 16 divided by 2 is 8, x divided by x is 1. So this is 2x times x plus 8. And then the second two terms right over here-- this is the whole basis of factoring by grouping-- we can factor out a negative 1. So this is equal to negative 1 times x plus 8. Both of them have an x plus 8 in them. So we can factor out an x plus 8. So if we factor out an x plus 8, we're left with 2x minus 1, put parentheses around it, times the factored out x plus 8. So we've simplified the numerator. The numerator can be rewritten. And you could have gotten here using the quadratic formula as well. The numerator is 2x minus 1 times x plus 8. And now see if you can factor the denominator. And this one's more straightforward. The coefficient here is 1. So we just have to think of two numbers that when I multiply them, I get 16. And when I add them, I get 10. And the obvious one is 8 and 2, positive 8 and positive 2. So we can write this as x plus 2 times x plus 8." + }, + { + "Q": "\nwhat is Sal doing, @ 2:28?", + "A": "He is trying to factor out f(x).", + "video_name": "u9v_bakOIcU", + "timestamps": [ + 148 + ], + "3min_transcript": "f of x is equal to 2x squared plus 15x minus 8. g of x is equal to x squared plus 10x plus 16. Find f/g of x. Or you could interpret this is as f divided by g of x. And so based on the way I just said it, you have a sense of what this means. f/g, or f divided by g, of x, by definition, this is just another way to write f of x divided by g of x. You could view this as a function, a function of x that's defined by dividing f of x by g of x, by creating a rational expression where f of x is in the numerator and g of x is in the denominator. And so this is going to be equal to f of x-- we have right up here-- is 2x squared 15x minus 8. And g of x-- I will do in blue-- is right over here, g of x. So this is all going to be over g And you could leave it this way, or you could actually try to simplify this a little bit. And the easiest way to simplify this would see if we could factor the numerator and the denominator expressions into maybe simpler expressions. And maybe some of them might be on-- maybe both the numerator and denominator is divisible by the same expression. So let's try to factor each of them. So first, let's try the numerator. And I'll actually do it up here. So let's do it. Actually, I'll do it down here. So if I'm looking at 2x squared plus 15x minus 8, we have a quadratic expression where the coefficient is not 1. And so one technique to factor this is to factor by grouping. You could also use the quadratic formula. And when you factor by grouping, you're going to split up this term, this 15x. And you're going to split up into two terms where the coefficients are, if I were to take the product of those coefficients, of the first and the last terms. And we proved that in other videos. So essentially, we want to think of two numbers that add up to 15, but whose product is equal to negative 16. And this is just the technique of factoring by grouping. It's really just an attempt to simplify this right over here. So what two numbers that, if I take their product, I get negative 16. But if I add them, I get 15? Well, if I take the product and get a negative number, that means they have to have a different sign. And so that means one of them is going to be positive, one of them is going to be negative, which means one of them is going to be larger than 15 and one of them is going to be smaller than 15. And the most obvious one there might be 16, positive 16, and negative 1. If I multiply these two things, I definitely get negative 16. If I add these two things, I definitely get 15. So what we can do is we can split this. We can rewrite this expression as 2x squared plus 2x squared" + }, + { + "Q": "Hey at 4:24, does the perpendicular sign have to be turned upside-down?\n", + "A": "no it is written correctly", + "video_name": "BTnAlNSgNsY", + "timestamps": [ + 264 + ], + "3min_transcript": "And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up. that's kind of related to right angles, when a right angle is formed, the two rays that form the right angle or the two lines that form that right angle or the two line segments that form that right angle are called perpendicular. So because we know that measure of angle DBC is 90 degrees or that angle DBC is a right angle, this tells us, we know that the line segment DB is perpendicular to line segment BC. Or we could even say ray BD is-- instead of using the word perpendicular, there's sometimes this symbol right here, which really just shows two perpendicular lines-- perpendicular to BC. And these come out of the fact that the angle formed between DB and BC, that is a 90-degree angle. Now, we have other words when our two angles add up to other things. So let's say, for example, I have one angle over here. Let me put some letters here so we can specify it. So let's say this is X, Y, and Z. And let's say that the measure of angle XYZ is equal to 60 degrees. And let's say that you have another angle that looks like this. And I'll call this, let's say, maybe MNO." + }, + { + "Q": "\nat 1:19 ,do adjacent angles have to add up to 90 degrees?", + "A": "No. They can, but they don t have to. Adjacent angles can sum up to any valid angle.", + "video_name": "BTnAlNSgNsY", + "timestamps": [ + 79 + ], + "3min_transcript": "Let's say I have an angle ABC, and it looks something like this. So its vertex is going to be at B. Maybe A sits right over here, and C sits right over there. And then also let's say that we have another angle called DBA. I want to have the vertex once again at B. So let's say it looks like this. So this right over here is our point D. That is our point D. And let's say that we know that the measure of angle DBA is equal to 40 degrees. So this angle right over here, its measure is equal to 40 degrees. And let's say that we know that the measure of angle ABC is equal to 50 degrees. So there's a bunch of interesting things happening here. The first interesting thing that you might realize is that both of these angles share a side. or rays-- but if you view them as rays, they both share this ray BA. And when you have two angles like this that share the same side, these are called adjacent angles. Because the word \"adjacent\" literally means next to. These are adjacent. They are adjacent angles. Now there's something else that you might notice that's interesting here. We know that the measure of angle DBA is 40 degrees and the measure of angle ABC is 50 degrees. And you might be able to guess what the measure of angle DBC is. If we drew a protractor over here-- I'm not going to draw it. It will make my drawing all messy. Well, maybe I'll draw it really fast. So if you had a protractor right over here, clearly this is opening up to 50 degrees. And this is going another 40 degrees. So if you wanted to say what the measure of angle DBC is, And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up." + }, + { + "Q": "\nWhy at 0:21 did Sal change the angle name from DAB to DBA? Does it matter because the subtitles don't change from DBA when he changes it in the first place?", + "A": "There is a difference: angle DAB is the angle made by lines DA and AB. The angle itself is at point (or vertex , see my answer to your other question) A. Angle DBA on the other hand is the angle made by lines DB and BA, and the vertex is at point B. Sal changed his mind and drew line DB (not line DA as he first intended), so DBA is the correct name for that angle. One side note: angle DBA is the same as angle ABD; the order doesn t matter as long as the point where the angle is remains in the middle.", + "video_name": "BTnAlNSgNsY", + "timestamps": [ + 21 + ], + "3min_transcript": "Let's say I have an angle ABC, and it looks something like this. So its vertex is going to be at B. Maybe A sits right over here, and C sits right over there. And then also let's say that we have another angle called DBA. I want to have the vertex once again at B. So let's say it looks like this. So this right over here is our point D. That is our point D. And let's say that we know that the measure of angle DBA is equal to 40 degrees. So this angle right over here, its measure is equal to 40 degrees. And let's say that we know that the measure of angle ABC is equal to 50 degrees. So there's a bunch of interesting things happening here. The first interesting thing that you might realize is that both of these angles share a side. or rays-- but if you view them as rays, they both share this ray BA. And when you have two angles like this that share the same side, these are called adjacent angles. Because the word \"adjacent\" literally means next to. These are adjacent. They are adjacent angles. Now there's something else that you might notice that's interesting here. We know that the measure of angle DBA is 40 degrees and the measure of angle ABC is 50 degrees. And you might be able to guess what the measure of angle DBC is. If we drew a protractor over here-- I'm not going to draw it. It will make my drawing all messy. Well, maybe I'll draw it really fast. So if you had a protractor right over here, clearly this is opening up to 50 degrees. And this is going another 40 degrees. So if you wanted to say what the measure of angle DBC is, And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up." + }, + { + "Q": "\nstarting at 0:34, there is an lowercase 'm' in front of all of the angles. What is it for, and why is it there?", + "A": "That means measure of . For example: m\u00e2\u0088\u00a0ABC = 50\u00cb\u009a That means the measure of angle ABC is 50 degrees.", + "video_name": "BTnAlNSgNsY", + "timestamps": [ + 34 + ], + "3min_transcript": "Let's say I have an angle ABC, and it looks something like this. So its vertex is going to be at B. Maybe A sits right over here, and C sits right over there. And then also let's say that we have another angle called DBA. I want to have the vertex once again at B. So let's say it looks like this. So this right over here is our point D. That is our point D. And let's say that we know that the measure of angle DBA is equal to 40 degrees. So this angle right over here, its measure is equal to 40 degrees. And let's say that we know that the measure of angle ABC is equal to 50 degrees. So there's a bunch of interesting things happening here. The first interesting thing that you might realize is that both of these angles share a side. or rays-- but if you view them as rays, they both share this ray BA. And when you have two angles like this that share the same side, these are called adjacent angles. Because the word \"adjacent\" literally means next to. These are adjacent. They are adjacent angles. Now there's something else that you might notice that's interesting here. We know that the measure of angle DBA is 40 degrees and the measure of angle ABC is 50 degrees. And you might be able to guess what the measure of angle DBC is. If we drew a protractor over here-- I'm not going to draw it. It will make my drawing all messy. Well, maybe I'll draw it really fast. So if you had a protractor right over here, clearly this is opening up to 50 degrees. And this is going another 40 degrees. So if you wanted to say what the measure of angle DBC is, And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up." + }, + { + "Q": "during 2:31 are complementary angles only made up of 2 different types angles?\n", + "A": "Yes. They only consist of two types of angles.", + "video_name": "BTnAlNSgNsY", + "timestamps": [ + 151 + ], + "3min_transcript": "or rays-- but if you view them as rays, they both share this ray BA. And when you have two angles like this that share the same side, these are called adjacent angles. Because the word \"adjacent\" literally means next to. These are adjacent. They are adjacent angles. Now there's something else that you might notice that's interesting here. We know that the measure of angle DBA is 40 degrees and the measure of angle ABC is 50 degrees. And you might be able to guess what the measure of angle DBC is. If we drew a protractor over here-- I'm not going to draw it. It will make my drawing all messy. Well, maybe I'll draw it really fast. So if you had a protractor right over here, clearly this is opening up to 50 degrees. And this is going another 40 degrees. So if you wanted to say what the measure of angle DBC is, And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up. that's kind of related to right angles, when a right angle is formed, the two rays that form the right angle or the two lines that form that right angle or the two line segments that form that right angle are called perpendicular. So because we know that measure of angle DBC is 90 degrees or that angle DBC is a right angle, this tells us, we know that the line segment DB is perpendicular to line segment BC. Or we could even say ray BD is-- instead of using the word perpendicular, there's sometimes this symbol right here, which really just shows two perpendicular lines-- perpendicular to BC." + }, + { + "Q": "\nAt 1:45, Sal begins to talk about Unit Vectors using the variables 'i' and 'j' with what he describes as 'little hats' on top. Could someone please explain the importance or relevance of these symbols? Can one only use letters i and j? Or does it matter?", + "A": "That notation means that those vectors are unit vectors , (the meaning is that does vectors have direction but their lengh is alwayes equal to 1 ). Any letter can be used to represent a vecor .", + "video_name": "9ylUcCOTH8Y", + "timestamps": [ + 105 + ], + "3min_transcript": "We've already seen that you can visually represent a vector as an arrow, where the length of the arrow is the magnitude of the vector and the direction of the arrow is the direction of the vector. And if we want to represent this mathematically, we could just think about, well, starting from the tail of the vector, how far away is the head of the vector in the horizontal direction? And how far away is it in the vertical direction? So for example, in the horizontal direction, you would have to go this distance. And then in the vertical direction, you would have to go this distance. Let me do that in a different color. You would have to go this distance right over here. And so let's just say that this distance is 2 and that this distance is 3. We could represent this vector-- and let's call this vector v. We could represent vector v as an ordered list or a 2-tuple of-- so we could say we and 3 in the vertical direction. So you could represent it like that. You could represent vector v like this, where it is 2 comma 3, like that. And what I now want to introduce you to-- and we could come up with other ways of representing this 2-tuple-- is another notation. And this really comes out of the idea of what it means to add and scale vectors. And to do that, we're going to define what we call unit vectors. And if we're in two dimensions, we define a unit vector for each of the dimensions we're operating in. If we're in three dimensions, we would define a unit vector for each of the three dimensions that we're operating in. And so let's do that. So let's define a unit vector i. And the way that we denote that is the unit vector we put this hat on top of it. So the unit vector i, if we wanted to write it in this notation right over here, we would say it only goes 1 unit in the horizontal direction, and it doesn't go at all in the vertical direction. So it would look something like this. That is the unit vector i. And then we can define another unit vector. And let's call that unit vector-- or it's typically called j, which would go only in the vertical direction and not in the horizontal direction. And not in the horizontal direction, and it goes 1 unit in the vertical direction. So this went 1 unit in the horizontal. And now j is going to go 1 unit in the vertical. So j-- just like that. Now any vector, any two dimensional vector, we can now represent as a sum of scaled up versions of i and j." + }, + { + "Q": "At 3:31, could x be zero? Or would the square root of zero be a non-real number?\n", + "A": "Yes... you can do sqrt(0), it just = 0 because 0^2 = 0.", + "video_name": "qFFhdLlX220", + "timestamps": [ + 211 + ], + "3min_transcript": "And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is, if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers-- then x has to be greater than or equal to 0. So maybe I could write it this way. The domain here is that x is any real number greater than or equal to 0. And the reason why I say that is, if you put a negative number in here and you cube it, And then at least in the real numbers, you won't get an actual value. You'll get a square root of a negative number here. So if you make this-- if you assume this right here, we're dealing with the real numbers. We're not dealing with any complex numbers. When you open it up to complex numbers, then you can expand the domain more broadly. But if you're dealing with real numbers, you can say that x is going to be greater than or equal to 0. And then the absolute value of x is just going to be x, because it's not going to be a negative number. And so if we're assuming that the domain of x is-- or if this expression is going to be evaluatable, or it's going to have a positive number, then this can be written as 30x times the square root of 5x. If you had the situation where we were dealing with complex numbers-- and if you don't know what a complex number is, or an imaginary number, But if you were dealing with those, then you would have to keep the absolute value of x there. Because then this would be defined for numbers that are less than 0." + }, + { + "Q": "\nAround 2:18 of the video Sal mentions the absolute value of | x | which he got from the x^2 under the radical. if he were to use the absolute value for the x^2 why would he not use the absolute value of the lone | x | under the radical?", + "A": "Because the lone x was not a perfect square, he could not simplify the radical. Only if you are taking the principle root and you are SIMPLIFYING the radical can you put in the absolute value. If he would have put the absolute value sign for the x under the radical it would ve become: sqrt(|x|) = +/- (x^1/2) Which still gives you the negative root which is extraneous for principle roots. Therefore putting the absolute value under the radical is never done when taking the principle root as it yields unwanted answers.", + "video_name": "qFFhdLlX220", + "timestamps": [ + 138 + ], + "3min_transcript": "What I want to do in this video is resimplify this expression, 3 times the principal root of 500 times x to the third, and take into consideration some of the comments that we got out on YouTube that actually give some interesting perspective on how you could simplify this. So just as a quick review of what we did in the last video, we said that this is the same thing as 3 times the principal root of 500. And I'm going to do it a little bit different than I did in the last video, just to make it interesting. This is 3 times the principal root of 500 times the principal root of x to the third. And 500-- we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5. Or even better, we could rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. Now, the one thing I'm going to do here-- actually, I won't talk about it just yet, of how we're going to do it differently than we did it in the last video. This radical right here can be rewritten as-- so this is going to be 3 times the square root, or the principal root, I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and then taking the product. And so then this over here is going to be times the square root of, or the principal root of, x squared times the principal root of x. And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30-- and I'm just going to switch the order here-- times the absolute value of x. And then you have the square root of 5, And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is, if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers-- then x has to be greater than or equal to 0. So maybe I could write it this way. The domain here is that x is any real number greater than or equal to 0. And the reason why I say that is, if you put a negative number in here and you cube it," + }, + { + "Q": "is there a mathematical pattern for 1:46 or is that just random?\n", + "A": "There is no pattern because it s the digits of tau, which like pi is transcendental (which is just a fancy term meaning that not only is it irrational, but you can t even have it as a solution for a polynomial).", + "video_name": "FtxmFlMLYRI", + "timestamps": [ + 106 + ], + "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211." + }, + { + "Q": "At 1:20, why does she start talking about the forth dimension?\n", + "A": "It s to show that the third dimension isn t the highest we can go.", + "video_name": "FtxmFlMLYRI", + "timestamps": [ + 80 + ], + "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211." + }, + { + "Q": "At around 1:30, you can see how (ab)^4 is the same as (a^4) * (b^4). Why is it that when you add different terms and square them, you need to use FOIL, but when multiplying you can simply do that? I mean, sure, it works on paper but a textbook explanation would be nice to think about. They just look so similar.\n", + "A": "Expanding (ab)^2, for example, yields (ab)(ab), which is (a^2)(b^2). Expanding (a+b)^2, however gives you (a+b)(a+b), which is aa+ab+ba+bb, or a^2+2ab+b^2", + "video_name": "SwqOrUWzDY8", + "timestamps": [ + 90 + ], + "3min_transcript": "And now I want to go over some of the other core exponent properties. But they really just fall out of what we already know about exponents. Let's say I have two numbers, a and b. And I'm going to raise it to-- I could do it in the abstract. I could raise it to the c power. But I'll do it a little bit more concrete. Let's raise it to the fourth power. What is that going to be equal to? Well that's going to be equal to-- I could write it like this. Copy and paste this, copy and paste. That's going to be equal to ab times ab times ab times ab times ab. But what is that equal to? Well when you just multiply a bunch of numbers like this it doesn't matter what order you're going to multiply it in. This right over here is going to be equivalent to a times a times a times a times-- We have four b's as well that we're multiplying together. And what is that equal to? Well this right over here is a to the fourth power. And this right over here is b to the fourth power. And so you see, if you take the product of two numbers and you raise them to some exponent, that's equivalent to taking each of the numbers to that exponent. And then taking their product. And here I just used the example with 4, but you could do this really with any arbitrary-- actually any exponent. This property holds. And you could satisfy yourself by trying different values, and using the same logic right over here. But this is a general property. That-- let me write it this way-- that if I have a to the b, to the c power, And we'll use this to throughout actually mathematics, when we try to simplify things or rewrite an expression in a different way. Now let me introduce you to another core idea here. And this is the idea of raising something to some power. And I'll just use example of 3. And then raising that to some power. What could this be simplified as? Well let's think about it. This is the same thing as a to the third-- let me copy and paste that-- as a to the third times a to the third. And what is a to the third times-- So this is equal to a to the third times a to the third. And that's going to be equal to a to the 3 plus 3 power." + }, + { + "Q": "At 10:59, why is Sal dividing 7:7 instead of multiplying (35*1)/(4*7) = (35/28) and then dividing the whole fraction by 7?\n", + "A": "its kinda late but hat 11:00 when he said you divide 7 by 7. 35/4 * 1/7 = 35/28. simplify it by dividing 7. sames as 5/4 so yeah.", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 659, + 427 + ], + "3min_transcript": "If you divided just straight up by 16, you would've gone straight to 5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these equations, or into one of the original equations. Let's substitute into the second of the original equations, where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? Is equal to 5. Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So the point of intersection of this right here is both x and y are going to be equal to 5/4. So if you looked at it as a graph, it'd be 5/4 comma 5/4. And let's verify that this satisfies the top equation. And if you take 5 times 5/4, plus 7 times 5/4, what do you get? It should be equal to 15. So this is equal to 25/4, plus-- what is this? This is plus 35/4. Which is equal to 60/4, which is indeed equal to 15. So it does definitely satisfy that top equation. And you could check out this bottom equation for yourself, but it should, because we actually used this bottom equation to figure out that x is equal to 5/4." + }, + { + "Q": "\nHow do you know which number to multiply it by at 2:20", + "A": "Because we can t see a clear variable to eliminate right away, we want to find the lowest common multiple of 5 and 7, in order to eliminate x. lcm(5, 7) = 5 x 7 = 35 Therefore, in order to get to 35x and -35x (one can be negative to cancel out when added together), we need to multiply the top equation by 7 and the bottom equation by -5. 7(5x+7) = 7(15) -5(7x-3y) = -5(5)", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 140 + ], + "3min_transcript": "Let's solve a few more systems of equations using elimination, but in these it won't be kind of a one-step elimination. We're going to have to massage the equations a little bit in order to prepare them for elimination. So let's say that we have an equation, 5x minus 10y is equal to 15. And we have another equation, 3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution, we could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, you know, with elimination, you were subtracting the left-hand side of one equation from another, or adding the two, and then adding the two right-hand sides. And I could do that, because it was essentially adding the same thing to both sides of the equation. But here, it's not obvious that that If we added these two left-hand sides, you would get 8x minus 12y. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand" + }, + { + "Q": "\nat about 3:20, he wrote 15, but it is supposed to be five. Just so you all know.", + "A": "Please review the video again. That did not happen. The video starts with the following system. . . . 5x - 10y = 15 3x - 2y = 3 Sal chooses to isolate the x term by multiplying the -2y in equation two by -5, so that the 10y produced, cancels with the -10y in equation one. Naturally, you need to keep the system balanced by multiplying both sides of equation two by -5 -5(3x - 2y) = (-5)3 and that gives -15x + 10y = -15 Everything is correct as he did it.", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 200 + ], + "3min_transcript": "And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand to both sides of the equation. Because this is equal to that. So let's do that. So 5x minus 15y-- we have this little negative sign there, we don't want to lose that-- that's negative 10x. The y's cancel out. Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to-- 15 minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative 10, and you get x is equal to 0. And now we can substitute back into either of these equations to figure out what y must be equal to. Let's substitute into the top equation. So we get 5 times 0, minus 10y, is equal to 15." + }, + { + "Q": "Sal makes a mistake at 3:21\n", + "A": "At 3:27: 5x + (-15x) = -10x - this is correct -10y + 10y = 0 - this is correct 15 + (-15) = 0 - this is also correct. No errors there.", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 201 + ], + "3min_transcript": "And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand to both sides of the equation. Because this is equal to that. So let's do that. So 5x minus 15y-- we have this little negative sign there, we don't want to lose that-- that's negative 10x. The y's cancel out. Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to-- 15 minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative 10, and you get x is equal to 0. And now we can substitute back into either of these equations to figure out what y must be equal to. Let's substitute into the top equation. So we get 5 times 0, minus 10y, is equal to 15." + }, + { + "Q": "\nAt 10:31 you said 7x = 20/4 + 15/4 but I thought we got rid of the 15/4 when we added it to both sides of the equation 7x - 15/4 = 5. It would look like this 7x-15/4+15/4 = 5+15/4. The 15/4's on the left side cancel out. 5+15/4 on the right side becomes 20/4. Why do you then add 15/4 to the 20/4?", + "A": "To add a whole number and a fraction, you have to get a common denominator, so 5 \u00e2\u0080\u00a2 4/4 = 20/4. What you are trying to do is (5+15)/4 which is incorrect order of operations. If switched to decimals, it would be 5 + 3.75 which is 8.75 or 35/4.", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 631 + ], + "3min_transcript": "If you divided just straight up by 16, you would've gone straight to 5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these equations, or into one of the original equations. Let's substitute into the second of the original equations, where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? Is equal to 5. Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So the point of intersection of this right here is both x and y are going to be equal to 5/4. So if you looked at it as a graph, it'd be 5/4 comma 5/4. And let's verify that this satisfies the top equation. And if you take 5 times 5/4, plus 7 times 5/4, what do you get? It should be equal to 15. So this is equal to 25/4, plus-- what is this? This is plus 35/4. Which is equal to 60/4, which is indeed equal to 15. So it does definitely satisfy that top equation. And you could check out this bottom equation for yourself, but it should, because we actually used this bottom equation to figure out that x is equal to 5/4." + }, + { + "Q": "\nWhat does Sal mean at 0:08 when he says \"Massage the Equation\"?", + "A": "Sal meant that to solve the systems of equations, you first must manipulate or change the equations. Once this is done, it will be much easier to solve the system.", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 8 + ], + "3min_transcript": "Let's solve a few more systems of equations using elimination, but in these it won't be kind of a one-step elimination. We're going to have to massage the equations a little bit in order to prepare them for elimination. So let's say that we have an equation, 5x minus 10y is equal to 15. And we have another equation, 3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution, we could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, you know, with elimination, you were subtracting the left-hand side of one equation from another, or adding the two, and then adding the two right-hand sides. And I could do that, because it was essentially adding the same thing to both sides of the equation. But here, it's not obvious that that If we added these two left-hand sides, you would get 8x minus 12y. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand" + }, + { + "Q": "at 2:50 sal talks about a \"zero matrix\" what does he mean by it?\n", + "A": "A zero matrix is a matrix all of whose entries are zero.", + "video_name": "OjF765iVuF8", + "timestamps": [ + 170 + ], + "3min_transcript": "Here you have Bernard, who says A times, C times B. We already know that that's going to be the equivalent to A, C, B which once again they've swapped the order between the B and the C, matrix multiplication is not commutative. You can't just swap order and expect to get the same product for any square matrices A, B, and C so we could rule that one out. A times, B, C, so we've already seen matrix multiplication is associative, so this is the same thing as A times B, times C which of course is the same thing as A, B, C. What Caren has right over here, that is right, that is equivalent for any square matrices A, B, and C that is equivalent to A, B, C. Now Ducheval, let's see, now this looks like a bit of a crazy expression but let's think it through a little bit. First of all matrix multiplication, the distributive property does hold. This first part right over here is equivalent to ... Let me write this down, this one's interesting. We have A times B, C plus A, minus A squared. You can actually distribute this A and I encourage you to prove it for yourself maybe using some two by two matrices for simplicity. This is going to be equal to, this part over here is going to be A, B, C plus A A, A times A which we could write as A squared and then we're going to subtract A squared. These two things are going to cancel out, they're going to end up being the zero matrix and if you take the zero, so these are going to be the zero matrix right over here. If you take the zero matrix and add it to A, B, C you're just going to end up with A, B, C. this one actually is equivalent. This one is right and this one is right. Here this is A times B, plus C so this is kind of not-y right over here. They're not even multiplying B and C, so this one's definitely not going to be true for all square matrices A, B, and C." + }, + { + "Q": "\nAt 1:00 why is it +4 and not *4? I don't understand above it was y^2/4. So wouldn't you multiply both sides by 4? And instead of getting (4/9)x^2 + 4 wouldn't it be (4/36)4x^2?\nI'm not doing to well in this chapter, so someone please explain this!", + "A": "Sal isn t showing the multiplication on each side. He is taking each term times 4. y^2/4 * 4 = y^2; x^2/9 * 4 = 4/9 x^2; 1 * 4 = 4. The +4 comes from taking the +1 times 4.", + "video_name": "hl58vTCqVIY", + "timestamps": [ + 60 + ], + "3min_transcript": "In the last hyperbola video I didn't get a chance to do some concrete examples. So I'll do that right now. So, let's say I had the hyperbola y squared over 4 minus x squared over, I don't know, let me think of a good number. Let's say, x squared over 9 is equal to 1. So the first thing to figure out about this hyperbola is, what are its asymptotes? And, once again, I always forget the formulas. And I just try to solve for y and see what happens when x approaches positive or negative infinity. So if you solve for y, you can add x squared over 9 to both sides. And you get y squared over 4 is equal to x squared over 9 plus 1. Now, I can multiply 4 times both sides. And you get y squared is equal to 4 over 9 times I distribute the 4, take the positive and negative square root both sides. y is equal to the plus or minus square root of 4 over 9x squared plus 4. And you can't really simplify this anymore. But we can think about, what does this approach as x approaches positive or negative infinity. So, as x approaches plus or minus infinity, what does this roughly equal? What does this approximate? What does the graph get a lot closer to? Well, then, y is approximately equal to just the square Because this becomes super huge and relative to this term, this starts to matter less and less and less. And that's why we get closer and closer to the asymptotes. Because when this number is, like, a trillion, or a google, then this number is almost insignificant. square root of this, and you'll just be a little bit above the graph. Because you have this extra plus-4 there. So as you approach positive or negative infinity, this equation is approximately equal to the plus or minus square root of 4 over 9x squared. And so, that is -- so y would be approximately equal We can take the square root of this. Plus or minus the square root of 4/9 is 2/3, right? Square root of 4 over square root of 9, times x. So, these are the asymptotes. There's two lines here. There's y is equal to 2/3 x. And then there's y is equal to minus 2/3 x. So let's draw those two lines. Let me draw my axes. Let's make that my y axis. Make that the x axis. Let me switch some colors, just to make things interesting. So let me draw the first one." + }, + { + "Q": "1:55 The +4 disappears just because it \"becomes irrelevant\" compared to the other term?\n", + "A": "The +4 is reverent to the equation, Its what makes the asymptotes that are the basis of the hyperbolas shape but when finding the asymptotes we want to find the line it get increasingly closer to. Since the +4 is what separates the function from the asymptotes if you remove the +4 you get the asymptotes", + "video_name": "hl58vTCqVIY", + "timestamps": [ + 115 + ], + "3min_transcript": "In the last hyperbola video I didn't get a chance to do some concrete examples. So I'll do that right now. So, let's say I had the hyperbola y squared over 4 minus x squared over, I don't know, let me think of a good number. Let's say, x squared over 9 is equal to 1. So the first thing to figure out about this hyperbola is, what are its asymptotes? And, once again, I always forget the formulas. And I just try to solve for y and see what happens when x approaches positive or negative infinity. So if you solve for y, you can add x squared over 9 to both sides. And you get y squared over 4 is equal to x squared over 9 plus 1. Now, I can multiply 4 times both sides. And you get y squared is equal to 4 over 9 times I distribute the 4, take the positive and negative square root both sides. y is equal to the plus or minus square root of 4 over 9x squared plus 4. And you can't really simplify this anymore. But we can think about, what does this approach as x approaches positive or negative infinity. So, as x approaches plus or minus infinity, what does this roughly equal? What does this approximate? What does the graph get a lot closer to? Well, then, y is approximately equal to just the square Because this becomes super huge and relative to this term, this starts to matter less and less and less. And that's why we get closer and closer to the asymptotes. Because when this number is, like, a trillion, or a google, then this number is almost insignificant. square root of this, and you'll just be a little bit above the graph. Because you have this extra plus-4 there. So as you approach positive or negative infinity, this equation is approximately equal to the plus or minus square root of 4 over 9x squared. And so, that is -- so y would be approximately equal We can take the square root of this. Plus or minus the square root of 4/9 is 2/3, right? Square root of 4 over square root of 9, times x. So, these are the asymptotes. There's two lines here. There's y is equal to 2/3 x. And then there's y is equal to minus 2/3 x. So let's draw those two lines. Let me draw my axes. Let's make that my y axis. Make that the x axis. Let me switch some colors, just to make things interesting. So let me draw the first one." + }, + { + "Q": "\n3:43 Inverse sin? Is there a video explaining raising trig functions to exponents?", + "A": "Inverse sine and sine to the power of -1 are different things. Sine Inverse: sin(x) = y arcsin(y) = x (the inverse function is commonly notated with arcsin) Sine to an exponent: sin^n(x) = [sin(x)]^n", + "video_name": "IJySBMtFlnQ", + "timestamps": [ + 223 + ], + "3min_transcript": "The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information So, let's get a calculator out and see if we can calculate it. Let me just verify, I am in degree mode. Very important. All right, now I'm going to take the inverse sine of 4/3 times sine of 40 degrees, and that gets me, and I deserve a little bit of a drum roll, 58, well if we round to the nearest, let's just maintain our precision here. So 58.99 degrees roughly. This is approximately equal to 58.99 degrees. So, if that is 58.99 degrees, what is this one? It's going to 180 minus this angle's measure minus that angle's measure. Let's calculate that. It's going to 180 degrees minus this angle, so minus 40," + }, + { + "Q": "\ni don't understand why did he use inverse sine to solve for it\nat around 4:15", + "A": "Inverse sin basically negates sin. Think of it like this: sin(90)=1 so arcsin(1)=90. It takes the output and gives the input. So in order to solve for sin(x)=(4/3)(sin(40), you have to take the inverse sin of both sides to get rid of the sin(x) and solve for x. Remember, (4/3)sin(40) is simply just a number.", + "video_name": "IJySBMtFlnQ", + "timestamps": [ + 255 + ], + "3min_transcript": "The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information So, let's get a calculator out and see if we can calculate it. Let me just verify, I am in degree mode. Very important. All right, now I'm going to take the inverse sine of 4/3 times sine of 40 degrees, and that gets me, and I deserve a little bit of a drum roll, 58, well if we round to the nearest, let's just maintain our precision here. So 58.99 degrees roughly. This is approximately equal to 58.99 degrees. So, if that is 58.99 degrees, what is this one? It's going to 180 minus this angle's measure minus that angle's measure. Let's calculate that. It's going to 180 degrees minus this angle, so minus 40," + }, + { + "Q": "\ncan someone please elaborate on why Sal took the inverse sine instead of just dividing both side of the equation by sine to isolate theta at 3:50? Thank you!", + "A": "sine is a function, not a variable", + "video_name": "IJySBMtFlnQ", + "timestamps": [ + 230 + ], + "3min_transcript": "The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information So, let's get a calculator out and see if we can calculate it. Let me just verify, I am in degree mode. Very important. All right, now I'm going to take the inverse sine of 4/3 times sine of 40 degrees, and that gets me, and I deserve a little bit of a drum roll, 58, well if we round to the nearest, let's just maintain our precision here. So 58.99 degrees roughly. This is approximately equal to 58.99 degrees. So, if that is 58.99 degrees, what is this one? It's going to 180 minus this angle's measure minus that angle's measure. Let's calculate that. It's going to 180 degrees minus this angle, so minus 40," + }, + { + "Q": "\nin 0:44 - 0:46 Sal says \"law of cosines \" twice (Though the second time the captions say \"sines\") does anyone know why?? I'm not sure if he just made a mistake or if I am hearing incorrectly.", + "A": "In this case, all Sal was trying to do was say that you could use EITHER the law of cosines or the law of sines for the problem shown above, but for the purpose of the video, he decided to use the law of sines. But, the second time he said the law of cosines was a mistake and it was corrected.", + "video_name": "IJySBMtFlnQ", + "timestamps": [ + 44, + 46 + ], + "3min_transcript": "Voiceover:Say you're out flying kites with a friend and right at this moment you're 40 meters away from your friend and you know that the length of the kite's string is 30 meters, and you measure the angle between the kite and the ground where you're standing and you see that it's a 40 degree angle. What you're curious about is whether you can use your powers of trigonometry to figure out the angle between the string and the ground. I encourage you to pause the video now and figure out if you can do that using just the information that you have. Whenever I see, I guess, a non right triangle where I'm trying to figure out some lengths of sides or some lengths of angles, I immediately think maybe the Law of Cosine might be useful or the Law of Sines might be useful. So, let's think about which one could be useful in this case. Law of Cosines, and I'll just rewrite them here. The Law of Cosine is c squared is equal to a squared plus b squared minus 2ab cosine of theta. 3 sides of a triangle. So a, b, c to an angle. So, for example, if I do 2 sides and the angle in between them, I can figure out the third side. Or if I know all 3 sides, then I can figure out this angle. But that's not the situation that we have over here. We're trying to figure out this question mark and we don't know 3 of the sides. We're trying to figure out an angle but we don't know 3 of the sides. The Law of Cosine just doesn't seem, at least in an obvious way, that it's going to help me. I could also try to find this angle. Once again, we don't know all 3 sides to be able to solve for the angle. So maybe Law of Sines could be useful. So the Law of Sines, the Law of Sines. Let's say that this is, the measure of this angle is a, the measure of this angle is lower case b, the measure of this angle is lower case c, length of this side is capital C, length of this side is capital A, The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta." + }, + { + "Q": "\nWhat is the troll he's talking about in 0:02?", + "A": "Watch the 2 previous videos Sal has made in that catagory. Hope that helps!", + "video_name": "VhH2nEDCd68", + "timestamps": [ + 2 + ], + "3min_transcript": "Just in case we encounter any more trolls who want us to figure out what types of money they have in their pockets, we have devised an exercise for you to practice with. And this is to solve systems of equations visually. So they say right over here, graph this system of equations And they give us two equations. This first one in blue, y is equal to 7/5x minus 5, and then this one in green, y is equal to 3/5x minus 1. So let's graph each of these, and we'll do it in the corresponding color. So first let's graph this first equation. So the first thing I see is its y-intercept is negative 5. Or another way to think about it, when x is equal to 0, y is going to be negative 5. So let's try this out. So when x is equal to 0, y is going to be equal to negative 5. So that makes sense. And then we see its slope is 7/5. This was conveniently placed in slope-intercept form for us. So it's rise over run. to move seven up. So if it moves 1, 2, 3, 4, 5 to the right, it's going to move 7 up. 1, 2, 3, 4, 5, 6, 7. So it'll get right over there. Another way you could have done it is you could have just tested out some values. You could have said, oh, when x is equal to 0, y is equal to negative 5. When x is equal to 5, 7/5 times 5 is 7 minus 5 is 2. So I think we've properly graphed this top one. Let's try this bottom one right over here. So we have when x is equal to 0, y is equal to negative 1. So when x is equal to 0, y is equal to negative 1. And the slope is 3/5. So if we move over 5 to the right, we will move up 3. So we will go right over there, and it looks like they intersect right at that point, right at the point x is equal to 5, y is equal to 2. And you could even verify by substituting those values into both equations, to show that it does satisfy both constraints. So let's check our answer. And it worked." + }, + { + "Q": "\n1:05 you were counting, why , I dont understand where you were counting?", + "A": "The slope of the equation was 7/5. Therefore, for every increase of 7 in y there s an increase of 5 in x. Sal counted the 7/5 slope and adjusted the line to follow it.", + "video_name": "VhH2nEDCd68", + "timestamps": [ + 65 + ], + "3min_transcript": "Just in case we encounter any more trolls who want us to figure out what types of money they have in their pockets, we have devised an exercise for you to practice with. And this is to solve systems of equations visually. So they say right over here, graph this system of equations And they give us two equations. This first one in blue, y is equal to 7/5x minus 5, and then this one in green, y is equal to 3/5x minus 1. So let's graph each of these, and we'll do it in the corresponding color. So first let's graph this first equation. So the first thing I see is its y-intercept is negative 5. Or another way to think about it, when x is equal to 0, y is going to be negative 5. So let's try this out. So when x is equal to 0, y is going to be equal to negative 5. So that makes sense. And then we see its slope is 7/5. This was conveniently placed in slope-intercept form for us. So it's rise over run. to move seven up. So if it moves 1, 2, 3, 4, 5 to the right, it's going to move 7 up. 1, 2, 3, 4, 5, 6, 7. So it'll get right over there. Another way you could have done it is you could have just tested out some values. You could have said, oh, when x is equal to 0, y is equal to negative 5. When x is equal to 5, 7/5 times 5 is 7 minus 5 is 2. So I think we've properly graphed this top one. Let's try this bottom one right over here. So we have when x is equal to 0, y is equal to negative 1. So when x is equal to 0, y is equal to negative 1. And the slope is 3/5. So if we move over 5 to the right, we will move up 3. So we will go right over there, and it looks like they intersect right at that point, right at the point x is equal to 5, y is equal to 2. And you could even verify by substituting those values into both equations, to show that it does satisfy both constraints. So let's check our answer. And it worked." + }, + { + "Q": "\nDoesn't \"constant\" begin with a c, so why do we use K? (0:17) I really do not get it.", + "A": "It does not matter what number you use. Unless you are dealing with a problem that has specific words like baseball, or cookies then you would probably want to use what ever letter it starts with. At least thats what I have always been told and I have an A in math.", + "video_name": "lkP-E2LUnjA", + "timestamps": [ + 17 + ], + "3min_transcript": "We're told in this question that on a string instrument, the length of a string-- so let's call that l-- the length of a string l varies inversely as the frequency, so varies inversely as the frequency. So l is going to be equal to some constant times the inverse of the frequency. And I'll use f for frequency-- the frequency of its vibrations. And then they tell us the vibrations are what give the string instruments their sound. That's nice. An 11-inch-- and it's actually the vibrations of the string affecting the air, and then the air compressions eventually get to our eardrum. And that's actually what gives us the perception of the sound. But we don't want to delve too much into the physics of it. An 11-inch inches string has a frequency. So 11-inch, so this is its length. So the 11-inch string has a frequency of 400 cycles per second. So this right here is the frequency. Find the constant of proportionality, and then find the frequency of a 10-inch string. So they say an 11-inch string. So 11-inch string is equal to some constant of proportionality times 1/400 cycles per second. So one over 400 cycles per second. I'll write second as sec. So to solve for the constant of proportionality, we need to multiply both sides by 400 cycles per second. Multiply the left hand side by 400 cycles per second, and the left hand side becomes 400 times 11. Well, 4 times 11 is 44. So 400 times 11 is 4,400. And then we have in our units, just for out of interest, So it's cycles times inches in the numerator of our units divided by seconds. And that is equal to our constant of proportionality. So we can say that the length is equal to 4,400 times-- or 4,400 cycles times inches per second-- I want to get the units right-- per second, times 1 over the frequency. So we solved for our constant of proportionality. And then we can use this to find the frequency of a 10-inch string. So now we're talking about a situation where our length is 10 inches, so 10 inches. So we get 10 inches are equal to 4,400--" + }, + { + "Q": "At 5:01, Sal uses three numbers for his example, do you NEED to have three numbers?\n", + "A": "Yes, in the scope of standard multivariable calculus, you can only use the cross product if you have vectors with three components.", + "video_name": "pJzmiywagfY", + "timestamps": [ + 301 + ], + "3min_transcript": "So it's a2 times b3 minus a3 times b2. That was hopefully pretty straightforward. Now, not to make your life any more complicated, when you do the second, when you do the middle row, when you do this one right here, so you cross that out. And you might want to do a1 times b3 minus a3 times b1. And that would be natural because that's what we did up there. But the middle row you do the opposite. You do a3 times b1 minus a1 times b3. Or you can kind of view it as the negative of what you would have done naturally. So you would have done a1 b3 minus a3 b1. Now we're going to do a3 b1 minus a1 b3. And then, for the bottom row, we cross that out again or ignore it. And we do a1 times b2, just like we do with the first row. Times a2 b1. Or minus a2 b1. This seems all hard to-- and it is hard to remember. That's why I kind of have to get that system in place like I just talked to you about. But this might seem pretty bizarre and hairy. So let me do a couple of examples with you, just so you get the hang of our definition of the dot product in R3. So let's say that I have the vector-- let's say I'm crossing the vector. I have the vector 1, minus 7, and 1. And I'm going to cross that with the vector 5, 2, 4. So this is going to be equal to a third vector. So for the first element in this vector, the first component, we just ignore the first components of these vectors and we say minus 7 times 4 minus 1 times 2. And these are just regular multiplication. I'm not taking the dot product. These are just regular numbers. Then for the middle term, we ignore the middle terms here and then we do the opposite. We do 1 times 5 minus 1 times 4. Remember, you might have been tempted to do 1 times 4 minus 1 times 5 because that's how we essentially did it in the But the middle term is the opposite. And then finally, the third term you ignore the third terms here and then you do it just like the first term. You start in the top left. 1 times 2 minus 7." + }, + { + "Q": "\nAt 0:33 how does she do that? Can anyone help me I really want to try it?!", + "A": "Yeah, I don t know but it is so cool. I am afraid you will have to explore it yourself.... I wish I knew.", + "video_name": "VIVIegSt81k", + "timestamps": [ + 33 + ], + "3min_transcript": "So say you just moved from England to the US and you've got your old school supplies from England and your new school supplies from the US and it's your first day of school and you get to class and find that your new American paper doesn't fit in your old English binder. The paper is too wide, and hangs out. So you cut off the extra and end up with all these strips of paper. And to keep yourself amused during your math class you start playing with them. And by you, I mean Arthur H. Stone in 1939. Anyway, there's lots of cool things you do with a strip of paper. You can fold it into Shapes and more shapes. Maybe spiral it around snugly like this. Maybe make it into a square. Maybe wrap it into a hexagon with a nice symmetric sort of cycle to the flappy parts. In fact, there's enough space here to keep wrapping the strip, and the your hexagon is pretty stable. and you're like. \"I don't know, hexagons aren't too exciting, but I guess it has symmetry or something.\" Maybe you could kinda fold it so the flappy parts are down and the unflappy parts are up. That's symmetric, and it collapses down into these three triangles, which collapse down into one triangle, and collapsible hexagons are, you suppose, cool enough to at least amuse you a little but during your class. you decide to try this three-way fold the other way, with flappy parts up, and are collapsing it down when suddenly the inside of your hexagon decides to open right up What, you close it back up and undo it. Everything seems the same as before, the center is not open-uppable. But when you fold it that way again, it, like, flips inside-out. Weird. This time, instead of going backwards, you try doing it again and again and again and again. And you want to make one that's a little less messy, so you try with another strip and tape it nicely into a twisty-foldy loop. You decide that it would be cool to colour the sides, so you get out a highlighter and make one yellow. Now you can flip from yellow side to white side. Yellow side, white side, yellow side, white side Hmm. White side? What? Where did the yellow side go? So you go back and this time you colour the white side green, and find that your piece of paper has three sides. Yellow, white and green. Now this thing is definitely cool. Therefore, you need to name it. And since it's shaped like a hexagon and you flex it and flex rhymes with hex, hexaflexagon it is. That night, you can't sleep because you keep thinking And the next day, as soon as you get to your math class you pull out your paper strips. You had made this sort of spirally folded paper that folds into again, the shape of a piece of paper, and you decide to take that and use it like a strip of paper to make a hexaflexagon. Which would totally work, but it feels sturdier with the extra paper. And you color the three sides and are like, orange, yellow, pink. And you're sort of trying to pay attention to class. Math, yeah. Orange, yellow, pink. Orange, yellow, white? Wait a second. Okay, so you colour that one green. And now it;s orange, yellow, green, Orange, yellow, green. Who knows where the pink side went? Oh, there it is. Now it's back to orange, yellow, pink. Orange, yellow, pink. Hmm. Blue. Yellow, pink, blue. Yellow, pink, blue. Yellow, pink, huh. With the old flexagon, you could only flex it one way, flappy way up. But now there's more flaps. So maybe you can fold it both ways. Yes, one goes from pink to blue, but the other, from pink to orange. And now, one way goes from orange to yellow, but the other way goes from orange to neon yellow." + }, + { + "Q": "\nAt 1:17, why didn't he take the (-) sign? Is there a reason?", + "A": "It s the definition: If (x - a) is a factor, f(a) = 0. Otherwise if (x + a) is a factor, you have to check wether f(-a) = 0. You have to revert the sign.", + "video_name": "JAdNNJynWM4", + "timestamps": [ + 77 + ], + "3min_transcript": "- [Voiceover] So we're asked, Is the expression x minus three, is this a factor of this fourth degree polynomial? And you could solve this by doing algebraic long division by taking all of this business and dividing it by x minus three and figuring out if you have a remainder. If you do end up with a remainder then this is not a factor of this. But if you don't have a remainder then that means that this divides fully into this right over here without a remainder which means it is a factor. So if the remainder is equal to zero, the remainder is equal to zero, if and only if, it's a factor. It is a factor. And we know a very fast way of calculating the remainder of when you take some polynomial and you divide it by a first degree expression like this. I guess you could say when you divide it by a first degree polynomial like this. The polynomial remainder theorem, the polynomial remainder theorem tells us that if we take some polynomial, p of x and we were to divide it by some x minus a our polynomial evaluated at our polynomial evaluated at a. So let's just see what's a in this case. Well in this case our a is positive three. So let's just evaluate our polynomial at x equals 3, if what we get is equal to zero that means our remainder is zero and that means that x minus three is a factor. If we get some other remainder that means well we have a non-zero remainder and this isn't a factor, so let's try it out. So, we're gonna have, so I'm just gonna do it all in magenta. It might be a little computationally intensive. So it's going to be two times three to the fourth power, three to the fourth, three to (mumbles), that's 81. 81. Minus 11 Yeah, this is gonna get a little computationally intensive but let's see if we can power through it. 11 times 27, I probably should have picked a simpler example, but let's just keep going. Plus four times three is 12. Minus 12 So lucky for us, at least those last two terms cancel out. And so this is going to be the rest from here is arithmetic. Two times 81 is 162. Now let's think about what 27 times 11 is. So let's see, 27 times 10 is going to be 270. 270 plus another 27 is minus 297. 297, did I do that, yeah, 270 So 27 times 10 is 270 plus 27, 297 Yep, that's right. And then we have, I'm prone to make careless errors here, see 90 plus 45 is 135. So plus 135. And let's see, if I were to take if I were to take 162 and 135, that's going to give me 297" + }, + { + "Q": "at 0:37, I am a bit confused on how that worked. How is X 2 and Y 5? How do we assign solution numbers to the variables? How do we know which ones to assign? Couldn't it be X 5 and Y 2? Checking the inequalities is really difficult for me.\n", + "A": "The question is asking to check for an x,y coordinate. I think the video assumes the watcher to understand the typical structure of a coordinate point: (x, y). So x = 2 and y = 5. Not the other way around. And after that, simply substitute the values into the inequalities.", + "video_name": "XzYNh2wpO0A", + "timestamps": [ + 37 + ], + "3min_transcript": "Is two comma five a solution of this system? And we have a system of inequalities right over here. We have Y is greater than or equal to 2x plus 1 and X is greater than 1. In order for two comma five to be a solution of this system, it just has to satisfy both inequalities. So, lets just try it out. So when X is equal to two and Y is equal to five, it has to satify both of these. So lets try it with the first one. So if we assume X is two and Y is five, we would get an inequality that says that five is greater than or equal to two times two plus one. X is two; Y is five. This gives us five is greater than or equal to two times two is four plus one is five. Y is greater than or equal to five. That's true! Five is equal to five. So that equal part of the greater than or equal saves us. That definitely satisfies the first inequality. Lets see the second one. X needs to be greater than one. So in two comma five, X is two. So it actually satisfies both of these inequalities. So two comma five is a solution for this system." + }, + { + "Q": "At 5:13, why doesn't he take the reciprocal of (2^-70) and make the exponent positive?\n\nHope that makes sense.\n", + "A": "I would agree that his final answer should have a positive exponent as that is the preferred form. He did leave his final answer as 2^(-98). This should have been changed to 1 / 2^98. However, while he is simplifying the expression, he can work with negative exponents.", + "video_name": "dC1ojsMi1yU", + "timestamps": [ + 313 + ], + "3min_transcript": "times this thing to the second power. Eight to the seventh to the second power, and then here, negative two times two is negative four, so that's A to the negative four times, eight to the seven times two is 14, eight to the 14th power. In other videos, we go into more depth about why this should hopefully make intuitive sense. Here you have eight to the seventh times eight to the seventh. Well, you would then add the two exponents, and you would get to eight to the 14th, so however many times you have eight to the seventh, you would just keep adding the exponents, or you would multiply by seven that many times. Hopefully that didn't sound too confusing, but the general idea is if you raise something to exponent and then another exponent, you can multiply those exponents. Let's do one more example where we are dealing with quotients, which that first example could have even been perceived as. So let's say we have divided by four squared, and we're gonna raise all of that to the seventh power. Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power, so if you have the difference of two things and you're raising it to some power, that's the same thing as a numerator raised to that power divided by the denominator raised to that power. Well, what's our numerator going to be? Well, we've done this drill before. It'd be two to the negative 10 times seventh power, so this would be equal to two to the negative 70th power, and then in the denominator, four to the second power, then that raised to the seventh power. Well, two times seven is 14, so that's going to be four to the 17th power. Now, we actually could think There's multiple ways that you could rewrite this, but one thing you could do is say, \"Hey, look, \"four is a power of two.\" So you could rewrite this as this is equal to two to the negative 70th power over, instead of writing four to the 17th power, why did I write the 17th power? It should be four to the 14th power. Let me correct that. Instead of writing four to the 14th power, I instead could write, so this is two, get the colors right. This is two to the negative 70th over, instead of writing four, I could write two squared to the 14th power. Four is the same thing as two squared, and so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second," + }, + { + "Q": "\nAt 6:43 Sal says that (3^-8*7^3) is equivalent to (7^3/3^8). Shouldn't it be (7^3/3^-8)?", + "A": "No. The negative exponent tells you to use the reciprocal to change it to a positive exponent. So: 3^(-8) = 1/3^8. Multiply that with 7^3, and you get: 7^3/3^8 Hope this helps.", + "video_name": "dC1ojsMi1yU", + "timestamps": [ + 403 + ], + "3min_transcript": "divided by four squared, and we're gonna raise all of that to the seventh power. Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power, so if you have the difference of two things and you're raising it to some power, that's the same thing as a numerator raised to that power divided by the denominator raised to that power. Well, what's our numerator going to be? Well, we've done this drill before. It'd be two to the negative 10 times seventh power, so this would be equal to two to the negative 70th power, and then in the denominator, four to the second power, then that raised to the seventh power. Well, two times seven is 14, so that's going to be four to the 17th power. Now, we actually could think There's multiple ways that you could rewrite this, but one thing you could do is say, \"Hey, look, \"four is a power of two.\" So you could rewrite this as this is equal to two to the negative 70th power over, instead of writing four to the 17th power, why did I write the 17th power? It should be four to the 14th power. Let me correct that. Instead of writing four to the 14th power, I instead could write, so this is two, get the colors right. This is two to the negative 70th over, instead of writing four, I could write two squared to the 14th power. Four is the same thing as two squared, and so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second, well, that's two to the 28th power, two to the 28th power. And so can I simplify this even more? Well, this is going to be equal to two to the, if I'm taking a quotient with the same base, I can subtract the exponent. So it's gonna be negative 70. It's going to be negative 70 minus 28th power, minus 28, and so this is going to simply two to the negative 98th power, and that's another way of viewing the same expression." + }, + { + "Q": "At 10:11, why does 1/2i become the square root of 1/4?\n", + "A": "because as we square 1/2, the numerator 1^2= 1 and denominator 2^2=4 and take the square root of 1/4, this is base on Pythagorean Theorem.", + "video_name": "FwuPXchH2rA", + "timestamps": [ + 611 + ], + "3min_transcript": "Now just to make this tangible, let's actually do this with an actual example. So let's say that I had, I don't know, let's say that I had to z1 is equal to square root of, let's say it's square root of 3/2 plus i. And so we want to figure out its magnitude, and we want to figure out its argument. So let's do that. So the magnitude of z1 is going to be equal to the square root of this squared. So this is going to be equal to 3/4 plus 1 squared-- or I should say plus 4/4. So this is going to be equal to square root of 7/4, which And now let's figure out its argument. So if I were to draw this on an Argand diagram, it would look like this. It's going to be in the first quadrant, so that's all I have to worry about. So let me draw it. Let me draw it like this. And so we have a situation. So it's going to be square root of 3, actually, let me change this up a little bit, just so the numbers get a little bit cleaner. Sorry about this. Let me make it a little bit, slightly cleaner. So just so that we have a slightly cleaner result, because we want to make our first example a simple one. So let's make this square root of 3/2 plus 1/2i.. So let's figure out the magnitude, the magnitude here is z1 is equal to the square root equal to 3/4 plus 1/2 squared is equal to 1/4,. This makes things a lot nicer. This is equal to the square root of 1, which is 1. And now let's think about it, let's draw it on an Argand diagram to visualize the argument. So this is my imaginary axis. This is my real axis. And so this complex number is square root of 3/2. The square root of 3 is like 1.7. So if we have like 1, it'll be like right over here, someplace right over here. This is square root of 3/2, the real part. The imaginary part is 1/2. So if this is 1, this is 1/2, the imaginary part is right over here, 1/2. And we actually also know its length, its length, or its magnitude is 1." + }, + { + "Q": "\nat 8:55 why is i squared +1 or 4/4? shouldn't it be minus 1?", + "A": "Sal is looking for a and b, which are the amplitude of the real and imaginary parts. b is the amplitude of the imaginary part by definition, so you remove the i before you square the coefficient. I m sure Sal could explain it better, but I hope that makes some sense to you.", + "video_name": "FwuPXchH2rA", + "timestamps": [ + 535 + ], + "3min_transcript": "And it's really one of the most profound results and all of mathematics, it still gives me chills. This is Euler's formula. Or this, by Euler's formula, is the same thing. And we show it by looking at the Taylor series representations of e to the x. And the Taylor series representations of cosine of x and sine of x. But this is, if we're dealing with radians, e to the i phi. So z is going to be equal to r times e to the i phi. So there's two ways to write a complex number. You could write it like this, where you have the real and imaginary part, that's maybe what we're used to. Or we can write it in exponential form, where you have the modulus, or the magnitude, being multiplied by a complex exponential. And we're going to see that this going to be super useful, Now just to make this tangible, let's actually do this with an actual example. So let's say that I had, I don't know, let's say that I had to z1 is equal to square root of, let's say it's square root of 3/2 plus i. And so we want to figure out its magnitude, and we want to figure out its argument. So let's do that. So the magnitude of z1 is going to be equal to the square root of this squared. So this is going to be equal to 3/4 plus 1 squared-- or I should say plus 4/4. So this is going to be equal to square root of 7/4, which And now let's figure out its argument. So if I were to draw this on an Argand diagram, it would look like this. It's going to be in the first quadrant, so that's all I have to worry about. So let me draw it. Let me draw it like this. And so we have a situation. So it's going to be square root of 3, actually, let me change this up a little bit, just so the numbers get a little bit cleaner. Sorry about this. Let me make it a little bit, slightly cleaner. So just so that we have a slightly cleaner result, because we want to make our first example a simple one. So let's make this square root of 3/2 plus 1/2i.. So let's figure out the magnitude, the magnitude here is z1 is equal to the square root" + }, + { + "Q": "at 0:56 you mentioned the principal root what does that mean\n", + "A": "This means ignoring the negative when you take a square root. That is, the number 4 has two square roots; 2 and -2. Principal means you only take the positive one. Cheers", + "video_name": "P1DJxuG7U9A", + "timestamps": [ + 56 + ], + "3min_transcript": "Let f be the function given by f of x is equal to the square root of x plus 4 minus 3 over x minus 5. If x does not equal 5, and it's equal to c if x equals 5. Then say, if f is continuous at x equals 5, what is the value of c? So if we know that f is continuous at x equals 5, that means that the limit as x approaches 5 of f of x is equal to f of 5. This is the definition of continuity. And they tell us that f of 5, when x equals 5, the value of the function is equal to c. So this must be equal to c. So what we really need to do is figure out what the limit of f of x as x approaches 5 actually is. Now, if we just try to substitute 5 into the expression right up here, in the numerator you have 5 plus 4 is 9. The square root of that is positive 3, the principal root is positive 3. 3 minus 3 is 0. So you get a 0 in the numerator. And then you get 5 minus 5 in the denominator, So you get this indeterminate form of 0/0. And in the future, we will see that we do have a tool that allows us, or gives us an option to attempt to find the limits when we get this indeterminate form. It's called L'Hopital's rule. But we can actually tackle this with a little bit of fancy algebra. And to do that, I'm going to try to get this radical out of the numerator. So let's rewrite it. So we have the square root of x plus 4 minus 3 over x minus 5. And any time you see a radical plus or minus something else, to get rid of the radical, what you can do is multiply by the radical-- or, if you have a radical minus 3, you multiply by the radical plus 3. So in this situation, you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3. We obviously have to multiply the numerator so that we actually don't change the value of the expression. If this right over here had a plus 3, then we would do a minus 3 here. This is a technique that we learn in algebra, or sometimes in pre-calculus class, to rationalize usually denominators, but to rationalize numerators or denominators. It's also a very similar technique that we use often times to get rid of complex numbers, usually in denominators. But if you multiply this out-- and I encourage you to do it-- you notice this has the pattern that you learned in algebra class. It's a difference of squares. Something minus something times something plus something. So the first term is going to be the first something squared. So square root of x plus 4 squared is x plus 4. And the second term is going to be the second something, or you're going to subtract the second something squared. So you're going to have minus 3 squared, so minus 9. And in the denominator, you're of course going to have x minus 5 times the square root of x" + }, + { + "Q": "At 1:11, shouldn't the unknown angle be 180-90-theta? Just for the sake of clarity.?\n", + "A": "that just makes it more confusing, we see the 90 degree angle. so the other two add to 90. so one is theta and one is 90-theta.", + "video_name": "QuZMXVJNLCo", + "timestamps": [ + 71 + ], + "3min_transcript": "We've got two right triangles here. And let's say we also know that they both have an angle whose measure is equal to theta. So angle A is congruent to angle D. What do we now know about these two triangles? Well for any triangle, if you know two of the angles, you're going to know the third angle, because the sum of the angles of a triangle add up to 180 degrees. So if you have two angles in common, that means you're going to have three angles in common. And if you have three angles in common, you are dealing with similar triangles. Let me make that a little bit clearer. So if this angle is theta, this is 90. They all have to add up to 180 degrees. That means that this angle plus this angle up here have to add up to 90. We've already used up 90 right over here, so angle A and angle B need to be complements. So this angle right over here needs to be 90 minus theta. Well we could use the same logic over here. We already use of 90 degrees over here. So we have a remaining 90 degrees So this angle is going to be 90 degrees minus theta. You have three corresponding angles being congruent. You are dealing with similar triangles. Now why is that interesting? Well we know from geometry that the ratio of corresponding sides of similar triangles are always going to be the same. So let's explore the corresponding sides here. Well, the side that jumps out-- when you're dealing with the right triangles-- the most is always the hypotenuse. So this right over here is the hypotenuse. This hypotenuse is going to correspond to this hypotenuse right over here. And then we could write that down. This is the hypotenuse of this triangle. This is the hypotenuse of that triangle. Now this side right over here, side BC, what side does that correspond to? Well if you look at this triangle, you can view it as the side that is opposite this angle theta. So it's opposite. So let's go opposite angle D. If you go opposite angle A, you get to BC. Opposite angle D, you get to EF. So it corresponds to this side right over here. And then finally, side AC is the one remaining one. We could view it as, well, there's two sides that make up this angle A. One of them is the hypotenuse. We could call this, maybe, the adjacent side to it. And so D corresponds to A, and so this would be the side that corresponds. Now the whole reason I did that is to leverage that, corresponding sides, the ratio between corresponding sides of similar triangles, is always going to be the same. So for example, the ratio between BC and the hypotenuse, BA-- so let me write that down-- BC/BA is going to be equal to EF/ED, the length of segment EF" + }, + { + "Q": "At 6:55, is he saying that sine, cosine, and tangent are all equal to each other?\n", + "A": "Actually, Sal is saying that the ratio of the lengths of two sides of one triangle is equal to the ratio of the corresponding sides of a similar triangle. He is certainly not implying that the sine, cosine and tangent are equivalent.", + "video_name": "QuZMXVJNLCo", + "timestamps": [ + 415 + ], + "3min_transcript": "And I keep stating from theta's point of view because that wouldn't be the case for this other angle, for angle B. From angle B's point of view, this is the adjacent side over the hypotenuse. And we'll think about that relationship later on. But let's just all think of it from theta's point of view right over here. So from theta's point of view, what is this? Well theta's right over here. Clearly AB and DE are still the hypotenuses-- hypoteni. I don't know how to say that in plural again. And what is AC, and what are DF? Well, these are adjacent to it. They're one of the two sides that make up this angle that is not the hypotenuse. So this we can view as the ratio, in either of these triangles, between the adjacent side-- so this is relative. Once again, this is opposite angle B, but we're only thinking about angle A right here, or the angle that measures theta, or angle D right over here-- relative to angle A, Relative to angle D, DF is adjacent. So this ratio right over here is the adjacent over the hypotenuse. And it's going to be the same for any right triangle that has an angle theta in it. And then finally, this over here, this is going to be the opposite side. Once again, this was the opposite side over here. This ratio for either right triangle is going to be the opposite side over the adjacent side. And I really want to stress the importance-- and we're going to do many, many more examples of this to make this very concrete-- but for any right triangle that has an angle theta, the ratio between its opposite side and its hypotenuse is going to be the same. That comes out of similar triangles. We've just explored that. The ratio between the adjacent side to that angle that is theta and the hypotenuse is going to be the same, for any of these triangles, as long as it has that angle theta in it. theta, between the opposite side and the adjacent side, between the blue side and the green side, is always going to be the same. These are similar triangles. So given that, mathematicians decided to give these things names. Relative to the angle theta, this ratio is always going to be the same, so the opposite over hypotenuse, they call this the sine of the angle theta. Let me do this in a new color-- by definition-- and we're going to extend this definition in the future-- this is sine of theta. This right over here, by definition, is the cosine of theta. And this right over here, by definition, is the tangent of theta. And a mnemonic that will help you remember this-- and these really are just definitions. People realized, wow, by similar triangles, for any angle theta, this ratio is always going to be the same." + }, + { + "Q": "\nAt 5:43,why is angle 1 a supplement of angle 2?", + "A": "The angle 1 and the angle that Sal put a magenta arc on are corresponding angles therefore they are equal. Since they are equal then you know that \u00e2\u0088\u00a01 + \u00e2\u0088\u00a02 = 180\u00c2\u00b0 since the magenta arc and \u00e2\u0088\u00a02 form a straight line.", + "video_name": "h0FFEBHBufo", + "timestamps": [ + 343 + ], + "3min_transcript": "Copy and paste this. OK. I try to avoid talking when I'm copying and pasting because I think it slows down my computer. In the accompanying diagram, parallel lines L and M are cut by transversal T. So it's a classic parallel line with a transversal problem. And they're parallel. That's why I did those arrows. Which statements about angle 1 and 2 must be true? I don't know if you've seen the Khan Academy videos of the angle game, but that's what we're going to play here, the angle game. So angle 1, if you want to look at its corresponding angle, its corresponding angle on the other parallel line, or with the transversal and the other parallel line, is right there. And they're going to be congruent. Those two angles are going to be congruent. So you could say this is equal to the measure of angle 1. I've picked up their terminology well, I think. And this is obviously angle 2. You see immediately that they have to be supplementary. Because when you add them together you get 180 degrees. Together they go all the way around and they kind of form a line. So you know that if this angle and this angle are supplementary. And this angle is congruent to angle 1, then angle 1 and angle 2 must be supplementary. So what do they say? Angle 1 is definitely not necessarily congruent to angle 2. It's congruent to this angle here. Angle 1 is a complement of angle 2. Complement means you add up to 90. No, we're talking about supplement. So it's not that. Angle 1 is a supplement of angle 2. There you go. And there's nothing that says that they're right angles, that's silly. All right, next problem. Let me copy and paste it. And paste it. Ready to go. What values-- let me pick a good color-- what values of A and B make the quadrilateral MNOP a parallelogram. For this to be a parallelogram, the opposite sides have to be equal. And I challenge you to experiment to draw a parallelogram where opposite sides are parallel where the opposite sides are also not equal. If you make two of the sides not equal, then the other two lines aren't going to be parallel anymore. And you can play with that if you like. But if opposite sides are going to be equal that means 4A plus B is equal to 21. Because they're opposite sides, so they should be equal to each other. Similarly, 3A minus 2B should be equal to 13 because they're opposite sides. So 3A minus 2B is equal to 13. And now we have two linear equations with two unknowns. So this is really an Algebra 1 problem in disguise. So let's see, they want us to solve for both." + }, + { + "Q": "\nAt 0:37, doesn't 5*3=3*3*3*3*3? I know it's the same thing, but I was wondering which one was the right way.", + "A": "5*3 is either 5+5+5 or 3+3+3+3+3. 5*3 = 3*5 because of the commutative property of multiplication. The commutative property of multiplication states that rearranging the order of the factors will not affect the product. Because of that, 5+5+5 = 3+3+3+3+3. They are both correct, but what you are asking is different. In your question, you multiplied 3 by itself 5 times. That would make it 3^5 and not 5*3. 5^3 = 5*5*5 = 125 is very different from 3^5 = 3*3*3*3*3 = 243.", + "video_name": "5qfOViJda_g", + "timestamps": [ + 37 + ], + "3min_transcript": "Find the value of 5 to the third power. Let me rewrite that. We have 5 to the third power. Now, it's important to remember, this does not mean 5 times 3. This means 5 times itself three times, so this is equal to 5 times 5 times 5. 5 times 3, just as a bit of a refresher so you realize the difference, 5 times 3-- let me write it over here. 5 times 3 is equal to 5 plus 5 plus 5. So when you multiply by 3, you're adding the number to itself three times. When you take it to the third power, you're multiplying the number by itself three times. So 5 times 3, you've seen that before, that's 15. But 5 to the third power, 5 times itself three times, is equal to-- well, 5 times 5 is 25, and then 25 times 5 is And we're done!" + }, + { + "Q": "\nIs the cochlea able to tune out any sound besides the frequency that is coming into the ear? 10:22 Confused me and made me wonder if the cochlea could do such a thing.", + "A": "Compare this to when you are watching a video, but an air conditioner is running in the background. You will subconsciously tune it out in order to focus on the video. The cochlea translates all frequencies into sounds, but the brain is where they are chosen to be ignored or not.", + "video_name": "i_0DXxNeaQ0", + "timestamps": [ + 622 + ], + "3min_transcript": "[PLAYS BACK PITCH] If we play the two at once, do you think we'll hear the two separate pitches? Or will our brain say, Hey, two pure frequencies an octave apart? The higher one must be an overtone of the lower one. So we're really hearing one note. [PLAYS BACK PITCH] Let's add the next overtime. 3 times 220 gives us 660. Here they are all at once. It sounds like a different instrument for the fundamental sine wave but the same pitch. Let's add 880 and now 1000. That sounds wrong. 880 plus 220 is 1100. There, that's better. We can keep going and now we have all these happy overtones. Zooming in to see the individual sine waves, I can highlight one little bump here and see how the first overtone perfectly fits two bumps. And the next has three, then four, and so on. By the way, knowing that the speed of sound is about 340 meters per second, and seeing that this wave takes about 0.0009 seconds to play, I can multiply those out to find that the distance between here and here is about 0.3 meters, or one foot. So C-sharp, 1100 is about a foot long. And each octave down is 1/2 the frequency or twice the length. That means the lowest C on a piano, which is five octaves lower than this C, has a sound wave 1 foot times 2 to the 5, or 32 feet long. OK, now I can play with the timbre of the sound by changing how loud the overtones are relative to each other. What your ears are doing right now is pretty complicated. All these sound waves get added up together into a single wave. And if I export this file, we can see what it looks like. Or I suppose you could graph it. Anyway, your speakers or headphones have this little diaphragm in them that pushes the air to make sound waves. To make this shape, it pushes forward fast here, then does this wiggly thing, and then another big push forwards. The speak, remember, is not pushing air from itself to your ears. It bumps against the air, which bumps against more air, and so on, until some air bumps into your ear drum, which moves in the same way that the diaphragm in the speaker did. And that pushes the little bones that push the cochlea, which pushes the fluid, which, depending is either going to push the basilar membrane in such a way that makes it vibrate a lot and push the little hairs, or it pushes with the wrong timing, just like someone bad at playgrounds. This sound wave will push in a way that makes the A220 part of your ear send off a signal, which is pretty easy to see. Some frequencies get pushed the wrong direction sometimes, but the pushes in the right direction more than make up for it. So now all these different frequencies that we added together and played are now separated out again. And in the meantime, many other signals are being sent out from other noise, like the sound of my voice and the sound of rain and traffic and noisy neighbors and air conditioner and so on. But then our brain is like, Yo, look at these! I found a pattern! And all these frequencies fit together into a series starting at this pitch. So I will think of them as one thing. And it is a different thing than these frequencies, which fit the patterns of Vi's voice. And oh boy, that's a car horn. Somehow this all works. And we're still pretty far from developing technology that can listen to lots of sound and separate it out into things anywhere near" + }, + { + "Q": "\nI think that Sal should't make up a word called \"furgle\" at 2:14~3:43\nbecause,then many people would find it very confusing weather or\nnot \"furgle is a real word.", + "A": "When you think about, someone, long ago, made up the words mile , inch , and pound . Sal is trying to explain an idea. He used a made up word so students wouldn t think the idea only worked with inches or centimeters or whatever real word he could have used instead.", + "video_name": "O1R4H3Ca82E", + "timestamps": [ + 134, + 223 + ], + "3min_transcript": "I have two identical rectangles here, and I want to measure how much space each of them take up on the plane of my screen, the screen that you are looking at right now. And I want to do it using two different units. It's clear, since they're two identical rectangles, that they take up the exact same amount of space. They will have the same area. But what we could see it that we can measure that area using different units. So first over here, let's say that this figure is 1 foot in width. And it is also 1 foot in height. So this right over here is equal to 1 square foot. It's clearly a square. It has the same width and the same height. And each of these dimensions are 1 foot, so we could call it 1 square foot. So let's see how many square feet we can get onto one of these rectangles, and essentially we'll be measuring its area And so we want to cover the entire space without overlapping and without going over the boundary. So that's 1, 2, 3, 4, 5, and 6. 6 in that first row, and then I have 7, 8, 9 10, 11, and 12. So that looks like this area, which is the same as this area If I were to measure it in square feet, the area is equal to, let me write this down, the area is equal to, we have-- let me write it down. We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 square feet. Now, I'm going to try to measure that same area in a different unit, and I'm going to just make up this unit. I'm going to call it a furgle. And a furgle in one dimension, so a furgle in one dimension is twice a foot. So that distance right over here, I'm going to call a furgle. That's one furgle. This is something that I made up just for the purpose of this video. Most people will not recognize what a furgle is. So its height is one furgle. Its width is a furgle. And so we could say that this is 1 square furgle. So let's see how many square furgles is this area, that same area that is 12 square feet. So let me copy and paste this. So let me copy, and then paste." + }, + { + "Q": "\nAt 2:51, when he was talking about 30 x 40, how did the 4 change to 40?", + "A": "It didn t. 30*4 is 120. If it was 40, his answer would ve been 1200.", + "video_name": "8bK-xfh8-rY", + "timestamps": [ + 171 + ], + "3min_transcript": "Number of times we get an elephant, out of the 210 times. So why don't you to have a go at it? All right, so you've, I'm assuming, like always, pause the video and then had a try. So one way to think about it, is, well, for one spin, what is the probability of getting an elephant? So let's do this, one spin. So for one spin, what is the probability of getting an elephant? Well let's see, we have already talked about, this is a fair spinner. There are seven equally likely possibilities. And then how many involve getting an elephant? Well, so we have one, two, three, four. Four out of the seven equally likely possibilities involve us getting an elephant. So one reasonable thing to do, and this is actually what I would do, is go, look, a 4/7 probability means I should expect that 4/7 of the time, and over, and over again, it's a reasonable expectation that, hey, 4/7 of the time, I will get an elephant. I've just calculated the theoretical probability here, based on this being a fair spinner. And that should inform, that if I were to do a bunch of experiments, that 4/7 of the time that I should see me getting the elephant. So it would be a reasonable prediction to say well, look, I'm going to spin this thing 210 times, and I would expect that 4/7 of those 210 times, I would get an elephant. And so, let's think about what this is. 210 times 4/7, 210 divided by seven is 30. 30 times 40 is 120. So 120 times. My prediction, or maybe your prediction was this as well, I think it's a reasonable prediction, is that if I spin it 210 times, that I'm going to get an elephant 120 times. what this is saying, and this is not saying. Is it possible that I get an elephant 121 times, or maybe 119 times? Sure, sure, it's completely reasonable that you might get something different than this. In fact, there's some probability that you get no elephant. If you consider getting an elephant lucky, that you just happen to keep landing on the monkey or one of the mice, and that's a very low probability that that would happen if you spun it 210 times, but it is possible. So it's important to realize that this is just a prediction. There's actually a possibility that you might get an elephant on all 210 spins. Once again, that's a low probability, but it is possible. So this isn't saying that you're definitely going to get the elephant 120 times. In fact, it's very reasonable, that you might get the elephant 123 times, or 128 times, or 110 times, or even 90 times." + }, + { + "Q": "\nCan the expression at 3:40 also be expressed as 5n+1 ?", + "A": "It was expressed as 5n + 1 when Sal simplified 6 + 5(n - 1) then he observed that 5n + 1 was equal to 1 + 5n. I prefer 5n + 1 because it can be used to visualize the pattern without resorting to a 0 term. 5n toothpicks can be used to almost construct n houses. Almost, because 1 more toothpick is needed for the right wall on the rightmost house.", + "video_name": "GvbrtnEYRpY", + "timestamps": [ + 220 + ], + "3min_transcript": "So let's say you are in the nth term of the sequence. If you are in the nth term, however more you are than 1, so if you're the nth term, you're going to be n minus 1 more than 1. If that is confusing, we'll do it with real numbers, so it gets a little bit more tangible. So the nth term, you are n minus 1 more or greater than-- I'll just say more-- than 1. For example, if n is 2, you are 1 more than 1. If n is 3, you are 3 minus 1, which is 2 more than 1. So however much more you are than 1, you multiply that by 5. We're 2 more than 1, so we add 10 to the number of toothpicks we have here. We're 3 more than 1 here, so we add 15 to the number of toothpicks we have there. toothpicks is equal to the number you're more than 1, < n minus 1 times 5. That's how much you're going to add above and beyond the amount of toothpicks in just the first sequence, so times 5. n minus 1 times 5 plus the number of toothpicks that you would just have in the first sequence or just this one house. And we already counted that: plus 6. So that's one way to think about it. And if this looks complicated, you just say, well, look, if I put a 4 here, I'm 3 more than 1, so 4 minus 1 is 3, so that'll be 3 times 5, which is 15. And then you add the number of toothpicks in 1 and then you get 6. Now another way, and many of you all might find this easier to think about, is even in 1, you could imagine a term here-- let me do it this way. You can imagine a 0th term. Let me just draw it here. Imagine a 0th term, and the 0th term would just be kind of a left wall of the house, or in this case, the left toothpick of a house. And then the first one, you're adding 5 toothpicks to that. In the second one, you're adding 5 toothpicks to that. And when you think about it this way, it actually becomes a little simpler to think in terms of n. Here, you could say, well, the nth term-- let me do this in a different color. You could say that the nth term is going to have-- or maybe we should say number of toothpicks in nth figure is going to be equal to 1. So in the 0th figure, which I just made up, you have at least 1 toothpick, and then whatever term in the sequence" + }, + { + "Q": "There is no longer a clarification for Sal's mistake at 2:20.\nHow do we fix this? This could really confuse a kid.\n", + "A": "Request a change, that might work", + "video_name": "muZmOiiukQE", + "timestamps": [ + 140 + ], + "3min_transcript": "- [Voiceover] Let's get some practice interpreting graphs of proportional relationships. This says the proportion relationship between the distance driven and the amount of time driving shown in the following graph, so we have the distance driven on the vertical axis, it's measured in kilometers, then we have the time driving and it's measured in hours along the horizontal axis. We can tell just visually that this, indeed, is a proportional relationship, how do we know that? Well, the point (0, 0) is on this graph, the graph goes through the origin. If we have zero time, then we have zero distance, and we can also see that it's a line, then it's a linear relationship. If you have a linear relationship that goes through the origin, you're dealing with a proportional relationship. You could also see that by taking out some points here. Let's see, I'm just eyeballing, so I'm gonna look at where does the graph kind of hit a very well-defined point. Actually, point A right over here, we see that when our time is five hours, our distance travelled or driven Then if we look at time, this point right over here, when our time is 2.5, we see that our distance driven is 200 km. And notice, the ratio between these variables at any one of these points is the same. 400 divided by five is 80, and 200 divided by 2.5 is also going to be 80. Or if you wanna go the other way around, to go from time to distance, we're always multiplying by 80. In fact, we can say that distance divided by time, our proportionality constant is going to be 80. Or if we wanted to include the units there, it might be a little more obvious than dealing with the rate, distance is in kilometers, time is in hours, 80 km per hour, We are going at this speed, 80 km per hour. This is also the proportionality constant. Anyway, with all of that out of the way, let's actually answer the questions. Which statements about the graph are true? Select all that apply. The vertical coordinate of point A represents the distance driven in four hours. The vertical coordinate. So point A is at the coordinate (5, 400). The vertical coordinate tells us how high to go up, how far to move in the vertical direction. That's gonna be the second coordinate right over here, so this is the vertical coordinate. This right over here tells us the distance we've driven in four hours, so yes, the vertical coordinate of point A represents the distance driven in four hours. We've driven 400 km, I like that one, I'll check that one. The distance driven in one hour is 80 km." + }, + { + "Q": "at 0:43, why did you use x instead of t for tiles\n", + "A": "variables represent values that you are solving for, often times the letter used to represent the unknown is arbitrary. he certainly could have used t, but most likely elected to avoid t because it is often used to represent time", + "video_name": "FZ2APP6-grU", + "timestamps": [ + 43 + ], + "3min_transcript": "A contractor is purchasing some stone tiles for a new patio. Each tile costs $3, and he wants to spend less than $1,000. And it's less than $1,000, not less than or equal to $1,000. The size of each tile is one square foot. Write an inequality that represents the number of tiles he can purchase with a $1,000 limit. And then figure out how large the stone patio can be. So let x be equal to the number of tiles purchased. And so the cost of purchasing x tiles, they're going to be $3 each, so it's going to be 3x. So 3x is going to be the total cost of purchasing the tiles. And he wants to spend less than $1,000. 3x is how much he spends if he buys x tiles. It has to be less than $1,000, we say it right there. sign right there. So if we want to solve for x, how many tiles can he buy? We can divide both sides of this inequality by 3. And because we're dividing or multiplying-- you could imagine we're multiplying by 1/3 or dividing by 3 -- because this is a positive number, we do not have to swap the inequality sign. So we are left with x is less than 1,000 over three, which is 333 and 1/3. So he has to buy less than 333 and 1/3 tiles, that's how many tiles, and each tile is one square foot. So if he can buy less than 333 and 1/3 tiles, then the patio also has to be less than 333 and 1/3 square feet. And we're done." + }, + { + "Q": "@1:00, Sal writes 3x<1000. Can we also write 3x-1000? Is that communicating the same thing.?- Thank You\n", + "A": "Adriene, If you have 3x<1000 you could subtract 1000 from both sides and then you would have 3x-1000<0", + "video_name": "FZ2APP6-grU", + "timestamps": [ + 60 + ], + "3min_transcript": "A contractor is purchasing some stone tiles for a new patio. Each tile costs $3, and he wants to spend less than $1,000. And it's less than $1,000, not less than or equal to $1,000. The size of each tile is one square foot. Write an inequality that represents the number of tiles he can purchase with a $1,000 limit. And then figure out how large the stone patio can be. So let x be equal to the number of tiles purchased. And so the cost of purchasing x tiles, they're going to be $3 each, so it's going to be 3x. So 3x is going to be the total cost of purchasing the tiles. And he wants to spend less than $1,000. 3x is how much he spends if he buys x tiles. It has to be less than $1,000, we say it right there. sign right there. So if we want to solve for x, how many tiles can he buy? We can divide both sides of this inequality by 3. And because we're dividing or multiplying-- you could imagine we're multiplying by 1/3 or dividing by 3 -- because this is a positive number, we do not have to swap the inequality sign. So we are left with x is less than 1,000 over three, which is 333 and 1/3. So he has to buy less than 333 and 1/3 tiles, that's how many tiles, and each tile is one square foot. So if he can buy less than 333 and 1/3 tiles, then the patio also has to be less than 333 and 1/3 square feet. And we're done." + }, + { + "Q": "\nat 1:44, you said ' if we subtract these 2 guy's', then said '90 plus 90 is 180'. it is, but weren't you going to subtract?", + "A": "I believe he said If we subtract... so these two guys.. not If we subtract these two guys . He was talking about subtracting theta from 180, not subtracting 90.", + "video_name": "iqeGTtyzQ1I", + "timestamps": [ + 104 + ], + "3min_transcript": "So in this diagram over here, I have this big triangle. And then I have all these other little triangles inside of this big triangle. And what I want to do is see if I can figure out the measure of this angle right here. And we'll call that measure theta. And they tell us a few other things. You might have seen this symbol before. That means that these are right angles or that they have a measure of 90 degrees. So that's a 90-degree angle, that is a 90-degree angle, and that is a 90-degree angle over there. And they also tell us that this angle over here is 32 degrees. So let's see what we can do. And maybe we can solve this in multiple different ways. That's what's really fun about these is there's multiple ways to solve these problems. So if this angle is theta, we have theta is adjacent to this green angle. And if you add them together, you're going to get this right angle. So this pink angle, theta, plus this green angle must be equal to 90 degrees. When you combine them, you get a right angle. So you could call this one-- its measure is And now we have three angles in the triangle, and we just have to solve for theta. Because we know this angle plus this angle plus this angle are going to be equal to 180 degrees. So you have 90 minus theta plus 90 degrees plus 32 degrees-- so I'm going to do that in a different color-- is going to be equal to 180 degrees. The sum of the measures of the angle inside of a triangle add up to 180 degrees. That's all we're doing over here. And so let's see if we can simplify this a little bit. So these two guys-- 90 plus 90's going to be 180, so you get 180 minus theta plus 32 is equal to 180 degrees. And then what else do we have? We have 180 on both sides. We can subtract that from both sides. So that cancels out. That goes to 0. You can add theta to both sides. And you get 32 degrees is equal to theta, or theta is equal to 32 degrees. So it's going to actually be the same measure as this angle right over here. That's one way to do the problem. There's other ways that we could have done the problem. Actually, there's a ton of ways we could have done this. We could have looked at this big triangle over here. And we could've said, look. If this is 90 degrees over here, this is 32 degrees over here, this angle up here is going to be 180 minus 90 degrees minus 32 degrees. Because they all have to add up to 180 degrees. And I just kind of skipped a step there. Actually, let me not skip a step. Let me call this x. If we call the measure of that angle x, we would have x plus 90. I'm looking at the biggest triangle in this diagram right here. x plus 90 plus 32 is going to be equal to 180 degrees." + }, + { + "Q": "At 2:20 he integrated g'(x)=1 to get g(x)=x, but shouldn't the integral of g'(x)=1 be g(x)=x+c?\n", + "A": "That is correct if that is where you were going to end your problem, but since there will be further integrals down the road, you can just add a +C at the end of the problem to encompass all the +C you would have had to put in.", + "video_name": "iw5eLJV0Sj4", + "timestamps": [ + 140 + ], + "3min_transcript": "The goal of this video is to try to figure out the antiderivative of the natural log of x. And it's not completely obvious how to approach this at first, even if I were to tell you to use integration by parts, you'll say, integration by parts, you're looking for the antiderivative of something that can be expressed as the product of two functions. It looks like I only have one function right over here, the natural log of x. But it might become a little bit more obvious if I were to rewrite this as the integral of the natural log of x times 1dx. Now, you do have the product of two functions. One is a function, a function of x. It's not actually dependent on x, it's always going to be 1, but you could have f of x is equal to 1. And now it might become a little bit more obvious to use integration by parts. Integration by parts tells us that if we have an integral that can be viewed as the product of one function, and the derivative of another function, and this is really just the reverse product rule, and we've shown that multiple times already. f of x times g of x minus the antiderivative of, instead of having f and g prime, you're going to have f prime and g. So f prime of x times g of x dx. And we've seen this multiple times. So when you figure out what should be f and what should be g, for f you want to figure out something that it's easy to take the derivative of and it simplifies things, possibly if you're taking the derivative of it. And for g prime of x, you want to find something where it's easy to take the antiderivative of it. So good candidate for f of x is natural log of x. If you were to take the derivative of it, it's 1 over x. Let me write this down. So let's say that f of x is equal to the natural log of x. Then f prime of x is equal to 1 over x. And let's set g prime of x is equal to 1. That means that g of x could be equal to x. And so let's go back right over here. So this is going to be equal to f of x times g of x. Well, f of x times g of x is x natural log of x. So g of x is x, and f of x is the natural log of x, I just like writing the x in front of the natural log of x to avoid ambiguity. So this is x natural log of x minus the antiderivative of f prime of x, which is 1 over x times g of x, which is x, which is xdx. Well, what's this going to be equal to? Well, what we have inside the integrand, this is just 1 over x times x, which is just equal to 1. So this simplifies quite nicely. This is going to end up equaling x natural log of x" + }, + { + "Q": "at 12:22 and earlier, could we have constructed the triangle to have the right angle at the tip of a? ..and would this have any other effect besides flipping the projection from a\u00e2\u0086\u0092b to b\u00e2\u0086\u0092a?\n", + "A": "a\u00c2\u00b7b and b\u00c2\u00b7a yield the same result, so switching the vectors is fine as long as you evaluate them consistently. For simplicity, I would call the shorter vector a so I m doing most of my work with smaller values. Sal was also unclear on what happens when the projection of a is shorter than the length of b. I doubt it would be an issue but am having trouble visualizing the area in that situation.", + "video_name": "tdwFdzVqito", + "timestamps": [ + 742 + ], + "3min_transcript": "Or if I put a light shining down from above and I'd say what's the shadow of a onto b? You'd get nothing. You'd get 0. This arrow has no width, even though I've drawn it to have It has no width. So you would have a 0 down here. The part of a that goes in the same direction as b. No part of this vector goes in the same direction as this vector. So you're going to have this 0 kind of adjacent side times b, so you're going to get something that's 0. So hopefully that makes a little sense. Now let's think about the cross product. The cross product tells us well, the length of a cross b, I painstakingly showed, you is equal to the length of a times the length of b times the sin of the angle between them. So let me do the same example. Let me draw my two vectors. That's my vector a and this is my vector b. And now sin-- SOH CAH TOA. Sin of theta-- SOH CAH TOA-- is equal to opposite over the hypotenuse. So if I were to draw a little right triangle here, so if I were to draw a perpendicular right there, this is theta. What is the sin of theta equal to in this context? The sin of theta is equal to what? It's equal to this side over here. Let me call that just the opposite. It's equal to the opposite side over the hypotenuse. So the hypotenuse is the length of this vector a right there. It's the length of this vector a. So the hypotenuse is the length over my vector a. So if I multiply both sides of this by my length of vector a, I get the length of vector a times the sin of theta is equal to the opposite side. So if we rearrange this a little bit, I can rewrite this as equal to-- I'm just going to swap them. This is equal to b, the length of vector b, times the length of vector a sin of theta. Well this thing is just the opposite side as I've defined it right here. So this right here is just the opposite side, this side right there. So when we're taking the cross product, we're essentially multiplying the length of vector b times the part of a that's going perpendicular to b. This opposite side is the part of a that's going perpendicular to b. So they're kind of opposite ideas. The dot product, you're multiplying the part of a that's going in the same direction as b with b. While when you're taking the cross product, you're multiplying the part of a that's going in the perpendicular direction to b with the length of b. It's a measure, especially when you take the length of this, it's a measure of how perpendicular these two guys are. And this is, it's a measure of how much do they move in the" + }, + { + "Q": "At about 5:10, wouldn't the \"- x\" (minus x) in \"-y * (x - x)\" be changed to \"-x\" (negative x)?\n", + "A": "No because the expression is -y * (x-x) which in words would read, negative y times x minus x. Since any number subtracted by itself is zero then your final product will equal zero.", + "video_name": "hmtJV49AWio", + "timestamps": [ + 310 + ], + "3min_transcript": "we multiply these things. So we could view this as negative S, that's that, let me write it a little bit neater. We could view it as negative S times S, times S times T. Times, oops, let me do that in a different color, times S times T. Times T. And do any of the choices look like that? Well almost. Instead of saying negative S times S, this says S times negative S. And because multiplication, once again, I'm not a big fan of using the word because it sounds complicated, but it's commutative. A times B is the same thing as B times A. So I can rewrite this as, I can rewrite this, I can swap these two and write this as S times, S times negative S times T, Times T. All I did is I swapped these two. This negative S and this S. I just swapped them and I got exactly what I have right over here. Now let's just make sure that this one does not apply and maybe the easiest way is to try to simplify this. And the best way I could think about that is by distributing this S. So if I distribute this S, what I'm going to get, this is going to be equal to S times T, which is ST. Or I could even, I can write it like this. I could write it S times T, like that. And then I have minus S times S, so minus S times S. I could write it that was or I could write minus S squared if I want to. That's the same thing as S times S. But this is very different. This is very different. Here I'm just taking the product of three variables here, I have two different terms, taking the product of two variables here and then the product I guess you could say So this is not, this is not the same thing. Which of the following expressions are equal to negative X times, and then in parentheses negative Y times X? And I forgot to mention it, but like always, pause the video, try to work them out by yourself before I do them. Alright, select all that apply. So let's just try to manipulate this a little bit. So once again, multiplication, it's associative. I could, so it's negative X times, times negative Y times X. So, the way it's written here, I could do these first, that's essentially what's written over here. Or it's associative. Instead, I could do these first. And the reason why I find this interesting is a negative times a negative is going to be a positive. So this is going to be the same thing, this thing over here is going to be the same thing as positive X times positive Y. Negative times a negative is a positive. So you're gonna get positive X times Y and then you're multiplying by an X again." + }, + { + "Q": "In 3:45 (or so) he says that the derivative of ln2 = ln2. How is this possible, if the derivative of lnx=1/x? shouldn't then the derivative of ln2 = 1/2?\n", + "A": "Gotcha, so you re saying there s a big difference in whether or not an x is involved with the ln. derivative of lnx=1/x therefore derivative of lncx (where c is a constant) = 1/cx. And then derivative of lnc=lnc, because its a number, without a variable. Is that right? (This is a response to Tysaki, not another answer)", + "video_name": "Mci8Cuik_Gw", + "timestamps": [ + 225 + ], + "3min_transcript": "And what's neat about this is now we've got this into a form of e to the something. So we can essentially use the chain rule to evaluate this. So this derivative is going to be equal to the derivative of e to the something with respect to that something. Well, the derivative e to the something with respect to that something is just e to that something. So it's going to be equal to e to the natural log of 2 times x. So let me make it clear what I just did here. This right over here is the derivative of e to the natural log of 2 times x with respect to the natural log of 2-- let me make it a little bit clearer-- with respect So we took the derivative of e to the something with respect to that something-- that's this right here, it's just e to that something. And then we're going to multiply that by, this is just an application of the chain rule, of the derivative of that something with respect to x. So the derivative of natural log of 2 times x with respect to x is just going to be natural log of 2. This is just going to be natural log of 2. The derivative of a times x is just going to be equal to a. This is just the coefficient on the x. And just to be clear, this is the derivative of natural log of 2 times x with respect to x. So we're essentially done. But we can simplify this even further. This thing right over here can be rewritten. And let me draw a line here just to make it clear that this equals sign is a continuation from what we did up there. But this e to the natural log of 2x, we can rewrite that, using this exact same exponent property, as e to the natural log of 2, and then And of course, we're multiplying it times the natural log of 2, so times the natural log of 2. Well, what is e to the natural log of 2? Well, we already figured that out. That is exactly equal to 2. This right over here is equal to 2. And so now we can simplify. This whole thing, the derivative of 2 to the x, is equal to-- and I'll switch the order a little bit-- it is the natural log of 2, that's this part right over here, times 2 to the x. Or we could write it as 2 to the x times the natural log of 2." + }, + { + "Q": "\nAt 3:12, when we get our final answer, why did we integrate just sinx and not 1? Shouldn't it be in the end \"xsinx + xcosx + c\" or x (sinx +cosx) + c?", + "A": "Yes yes! Thankyou :) I was terribly confused.", + "video_name": "bZ8YAHDTFJ8", + "timestamps": [ + 192 + ], + "3min_transcript": "If we assign g prime of x to be cosine of x, once again, if we take its antiderivative, that sine of x, it's not any more complicated. If we did it the other way around, if we set f of x to be cosine of x, then we're taking its derivative here. That's not that much more complicated. But if we set g prime of x equaling to x and then we had to take its antiderivative, we get x squared over 2, that is more complicated. So let me make it clear over here. We are assigning f of x to be equal to x. And that means that the derivative of f is going to be equal to 1. We are assigning-- I'll write it right here-- g prime of x to be equal to cosine of x, which means g of x is equal to sine of x, the antiderivative of cosine of x. Now let's see, given these assumptions, let's see if we can apply this formula. Let's see, the right-hand side says f of x times g of x. So f of x is x. g of x is sine of x. And then from that, we are going to subtract the antiderivative of f prime of x-- well, that's just 1-- times g of x, times sine of x dx. Now this was a huge simplification. Now I went from trying to solve the antiderivative of x cosine of x to now I just have to find the antiderivative of sine of x. And we know the antiderivative of sine of x dx is just equal to negative cosine of x. And of course, we can throw the plus c in now, now that we're pretty done with taking all of our antiderivatives. So all of this is going to be equal to x of this, which is just negative cosine of x. And then we could throw in a plus c right at the end of it. And doesn't matter if we subtract a c or add the c. We're saying this is some arbitrary constant which could even be negative. And so this is all going to be equal to-- we get our drum roll now-- it's going to be x times sine of x, subtract a negative, that becomes a positive, plus cosine of x plus c. And we are done. We were able to take the antiderivative of something that we didn't know how to take the antiderivative of before. That was pretty interesting." + }, + { + "Q": "At 1:48, why does g(x) = sinx. Why is it not g(x) = sinx + c?\n", + "A": "He is just noting what g (x) and g(x) are. The actual integration happens in the problem. When he does the integration the integral sign and dx disappear and that is your signal to add the +c to your answer before you put the little box around it. If the problem had more than one integral you would not put +c every time you did an integration. Only when the last integral disappears would you add a +c which represents all the arbitrary constants from all the integrations that happened.", + "video_name": "bZ8YAHDTFJ8", + "timestamps": [ + 108 + ], + "3min_transcript": "In the last video, I claimed that this formula would come handy for solving or for figuring out the antiderivative of a class of functions. Let's see if that really is the case. So let's say I want to take the antiderivative of x times cosine of x dx. Now if you look at this formula right over here, you want to assign part of this to f of x and some part of it to g prime of x. And the question is, well do I assign f of x to x and g prime of x to cosine of x or the other way around? Do I make f of x cosine of x and g prime of x, x? And that thing to realize is to look at the other part of the formula and realize that you're essentially going to have to solve this right over here. And here where we have the derivative of f of x times g of x. So what you want to do is assign f of x so that the derivative of f of x is actually simpler than f of x. And assign g prime of x that, if you were to take its antiderivative, it doesn't really become any more complicated. So in this case, if we assign f of x to be equal to x, f prime of x is definitely simpler, If we assign g prime of x to be cosine of x, once again, if we take its antiderivative, that sine of x, it's not any more complicated. If we did it the other way around, if we set f of x to be cosine of x, then we're taking its derivative here. That's not that much more complicated. But if we set g prime of x equaling to x and then we had to take its antiderivative, we get x squared over 2, that is more complicated. So let me make it clear over here. We are assigning f of x to be equal to x. And that means that the derivative of f is going to be equal to 1. We are assigning-- I'll write it right here-- g prime of x to be equal to cosine of x, which means g of x is equal to sine of x, the antiderivative of cosine of x. Now let's see, given these assumptions, let's see if we can apply this formula. Let's see, the right-hand side says f of x times g of x. So f of x is x. g of x is sine of x. And then from that, we are going to subtract the antiderivative of f prime of x-- well, that's just 1-- times g of x, times sine of x dx. Now this was a huge simplification. Now I went from trying to solve the antiderivative of x cosine of x to now I just have to find the antiderivative of sine of x. And we know the antiderivative of sine of x dx is just equal to negative cosine of x. And of course, we can throw the plus c in now, now that we're pretty done with taking all of our antiderivatives. So all of this is going to be equal to x" + }, + { + "Q": "\nat 2:33 I don't understand, why does he write expected value as m rather than E(X). Is it the same thing or not?", + "A": "he wrote mu, it just looked liked M", + "video_name": "ry81_iSHt6E", + "timestamps": [ + 153 + ], + "3min_transcript": "If this is 60% chance of success there has to be a 40% 70% chance of success, 30% chance of failure. Now with this definition of this-- and this is the most general definition of a Bernoulli Distribution. It's really exactly what we did in the last video, I now want to calculate the expected value, which is the same thing as the mean of this distribution, and I also want to calculate the variance, which is the same thing as the expected squared distance of a value from the mean. So let's do that. So what is the mean over here? What is going to be the mean? Well that's just the probability weighted sum of the values that this could take on. So there is a 1 minus p probability that we get So there's 1 minus p probability of getting 0, so times 0. And then there is a p probability of getting 1, plus p times 1. Well this is pretty easy to calculate. 0 times anything is 0. So that cancels out. And then p times 1 is just going to be p. So pretty straightforward. The mean, the expected value of this distribution, is p. And p might be here or something. So once again it's a value that you cannot actually take on in this distribution, which is interesting. But it is the expected value. Now what is going to be the variance? What is the variance of this distribution? Remember, that is the weighted sum of the squared distances from the mean. Now what's the probability that we get a 0? There's a 1 minus p probability that we get a 0. So that is the probability part. And what is the squared distance from 0 to our mean? Well the squared distance from 0 to our mean-- let me write it over here-- it's going to be 0, that's the value we're taking on-- let me do that in blue since I already wrote the 0-- 0 minus our mean-- let me do this in a new color-- minus our mean. That's too similar to that orange. Let me do the mean in white. 0 minus our mean, which is p plus the probability that we get a 1, which is just p-- this is the squared distance, It's the probability weighted sum of the squared distances from the mean. Now what's the distance-- now we've got a 1-- and what's the difference between 1 and the mean? It's 1 minus our mean, which is going to be p over here." + }, + { + "Q": "how did Sal get 3/3 when there is only one three present? @2:05 in the video.\n", + "A": "The average change is calculated using (y1-y1)/(x2-x1), or the slope, right? Here y is a function of f, that means we have (f(x2) - f(x1))/(x2 - x1) We are told that x1=2 and x2=5 We are told that f(x) = x\u00c2\u00b2 - 6x + 8. That means f(x2) = f(5) = 5\u00c2\u00b2 - (6)(5) + 8 = 25 - 30 + 8 = 3 And that f(x1) = f(2) = 2\u00c2\u00b2 - (6)(2) + 8 = 4 - 12 + 8 = 0. THEREFORE (y1-y1)/(x2-x1) = (f(x2) - f(x1))/(x2 - x1) = (3 - 0)/(5 - 2) = 3/3.", + "video_name": "S_YIUXy-WFM", + "timestamps": [ + 125 + ], + "3min_transcript": "Let's say I have some function f of x that is defined as being equal to x squared minus 6x plus 8 for all x. And what I want to do is show that for this function we can definitely find a c in an interval where the derivative at the point c is equal to the average rate of change over that interval. So let's give ourselves an interval right over here. Let's say we care about the interval between 2 and 5. And this function is definitely continuous over this closed interval, and it's also differentiable over it. And it just has to be differentiable over the open interval, but this is differentiable really for all x. And so let's show that we can find a c that's inside the open interval, that's a member of the open interval, that's in the open interval such that the derivative at c Or is equal to the slope of the secant line between the two endpoints of the interval. So it's equal to f of 5 minus f of 2 over 5 minus 2. And so I encourage you to pause the video now and try to find a c where this is actually true. Well to do that, let's just calculate what this has to be. Then let's just take the derivative and set them equal and we should be able to solve for our c. So let's see f of 5 minus f of 2, f of 5 is, let's see, f of 5 is equal to 25 minus 30 plus 8. So that's negative 5 plus 8 is equal to 3. f of 2 is equal to 2 squared minus 12. So it's 4 minus 12 plus 8. So this is equal to 3/3, which is equal to 1. f prime of c needs to be equal to 1. And so what is the derivative of this? Well let's see, f prime of x is equal to 2x minus 6. And so we need to figure out at what x value, especially it has to be in this open interval, at what x value is it equal to 1? So this needs to be equal to 1. So let's add 6 to both sides. You get 2x is equal to 7. x is equal to 7/2, which is the same thing as 3 and 1/2. So it's definitely in this interval right over here. So we've just found our c is equal to 7/2. And let's just graph this to really make sure that this makes sense. So this right over here is our y-axis." + }, + { + "Q": "At 3:01 shouldn't it be:\n2a^2-4a\n----------\n(a+2)(a+2)\n", + "A": "No, if you multiply it out using FOIL, you ll see (a+2)(a-2) = a^2 - 2a + 2a - 4. The a terms cancel out and you are left with a^2 - 4.", + "video_name": "IKsi-DQU2zo", + "timestamps": [ + 181 + ], + "3min_transcript": "is the least common multiple of this expression, and that expression, and it could be a good common denominator. Let's set that up. This will be the same thing as being equal to this first term right here, a minus 2 over a plus 2, but we want the denominator now to be a plus 2 times a plus 2-- we wanted it to be a plus 2 squared. So, let's multiply this numerator and denominator by a plus 2, so its denominator is the same thing as this. Let's multiply both the numerator and the denominator by a plus 2. We're going to assume that a is not equal to negative 2, that would have made this undefined, and it would have also made this undefined. Throughout this whole thing, we're going to assume that a cannot be equal to negative 2. The domain is all real numbers, a can be any real number except for negative 2. So, the first term is that-- extend the line a little bit-- denominator is already the common denominator. Minus a minus 3 over-- and we could write it either as a plus 2 times a plus 2, or as this thing over here. Let's write it in the factored form, because it'll make it easier to simplify later on: a plus 2 times a plus 2. And now, before we-- let's set this up like this-- now, before we add the numerators, it'll probably be a good idea to multiply this out right there, but let me write the denominator, we know what that is: it is a plus 2 times a plus 2. Now this numerator: if we have a minus 2 times a plus 2, we've seen that pattern before. We can multiply it out if you like, but we've seen it enough hopefully to recognize that this is going to be a squared minus 2 squared. This is going to be a squared minus 4. You can multiply it out, and the middle terms cancel out-- the negative 2 times a cancels out the a times 2, and you're just left with a squared minus 4-- that's that over there. be very careful here-- you're subtracting a minus 3, so you want to distribute the negative sign, or multiply both of these terms times negative 1. So you could put a minus a here, and then negative 3 is plus 3, so what does this simplify to? You have a squared minus a plus-- let's see, negative 4 plus 3 is negative 1, all of that over a plus 2 times a plus 2. We could write that as a plus 2 squared. Now, we might want to factor this numerator out more, to just make sure it doesn't contain a common factor with the denominator. The denominator is just 2a plus 2 is multiplied by themselves. And you can see from inspection a plus 2 will not" + }, + { + "Q": "At 1:17, isn't it supposed to be 150/1000 instead of 150/100? Or is it just for the decimals?\n", + "A": "Sal is starting with 150%. A percent by definition is a ratio that has a denominator of 100. So, it is 150/100. Hope this helps.", + "video_name": "xEDnwEOOf7Y", + "timestamps": [ + 77 + ], + "3min_transcript": "So we have 0.79 minus 4/3 minus 1/2 plus 150%. So we have four different numbers written in different formats. Here it's a decimal, here we have two fractions, and then here we have a percentage. So the easiest thing to do would be to write all of these in the same format. And for me, the easiest format to do this computation in would be to write them all as fractions. And the reason why I want to do that, in particular, is because 4/3, when you divide by 3, when you divide 1/3, 2/3, 4/3, you're going to have a repeating decimal. So to avoid that, I want to put all of these-- I want to rewrite all of these as fractions. So let's do them one at a time. So 0.79, this is the same thing as 79/100, so I'll just write it that way. So this is the same thing as 79 over 100. Then of course, we have minus 4/3. Then we have minus 1/2. And then finally, we have-- I don't Finally we have 150%. Well, 150%, percent literally means per cent, per hundred. So this is plus 150 per 100. So now we've written them all as fractions. And in order to do all the subtraction and addition, we have to find a common denominator. So what's the least common multiple of 100, 3, 2, and 100? Well, 100 is divisible by 2, so 100 is actually the least common multiple of 102. So we really have to just find the least common multiple between 100 and 300. And that's just going to be 300. There's no other common factors between 100 and 3. So let's write all of them with 300 as the common denominator. So let me do this in this reddish color. So 79 over 100 is the same thing. If I were to write it over 300, to go from 100 to 300 in the denominator, I'm multiplying by 3, so I have to multiply the numerator by 3 as well. Let's see, 80 times 3 would be 240. So it's going to be 3 less than that. So 240 minus 3 is 237. Now 4/3. Well, to get the denominator to be 300, we have to multiply the denominator by 100, so we have to multiply the numerator by 100 as well. 1/2, if our denominator is 300, we multiplied the denominator by 150 to go from 200 to 300, so we have to multiply the numerator by 150. And then finally, 150 over 100, well, we're multiplying the denominator by 3 to get to 300, to go from 100 to 300. So we have to do the same thing in the numerator. So 3 times 150 is 450." + }, + { + "Q": "\nAt 0:57, what is the difference between a zig and a zag?", + "A": "The direction.", + "video_name": "EdyociU35u8", + "timestamps": [ + 57 + ], + "3min_transcript": "So you're me and you're in math class, and you're supposed to be learning about logarithms, which you figure is probably something to do with avant-garde percussion performances. But every time you try to pay attention, you find that your teacher's explanations of logarithms inspires nothing but a bit of performance art involving drumming on wood. Anyway, your teacher gives you a dirty look. So you stop drumming and start doodling. You feel a need for motion, excitement, so you decide to do a flip book on the corner of your notebook. It's pretty easy because the paper is thin. You can trace the previous frame, modifying it just a little. And if you really want to get into the Zen zone of doodle bliss, you can make each new frame be modified according to some simple rule like, keep adding one new petal around a spiral, or add another dot in the next place in the spiral. Or how about, keep making the squiggle squigglier. But what does squigglier really mean? Like you could just increase the curvature of the squiggle, or you could make it squiggle more between the squiggles. You want to figure out an exact squiggle rule so that you can really get into the squiggle zone. So you discretize the squiggle into a zigzag line. On the next layer, you could just deepen the zag, Or maybe zig's too good to zig zag, and zag's too good to zag zig. So the next would be, zigzag zag zig, zigza-- wait, no. That was a zag, and those get turned into zags zigs. So it should be zigzag, zag zig zag zig zig zag. And next, zig zag zag zigzag, zig zig zigzag, zag zig-- wait, was this a zig? Maybe if we put it all onto some sort of reference diagram, that generates each new pattern based on-- No! This was supposed to be about mindless zigging in the squiggle zone. This is unacceptable. So maybe just pretend each time you get to a new stage, the old path is all just zigzag zigzag zigzag zigzag. And to keep it all neat and orderly, you decide to try to make all the lines the same length, always with right angles between them. Here we go. [SINGING] What if we tried starting with three lines? Zig zag zug. OK. So then each zig should get a zig zag zug, and each zag a zug zag zig? And then the zug gets, well, maybe it just goes back and forth, whether it goes on the inside or the outside. OK, wait now where did it start? And everything's running into itself, and getting bigger, and doesn't fit on the paper. OK, maybe if it were a little more open it wouldn't run into itself. Say like, half a hexagon, so you can keep all the angles perfect. And then trapezoids in and out. Yeah, this is totally going to work. Lookit. In fact, like this, you could even make it go inside first, and back and forth, and it wouldn't run into itself. Next, you could make it go even closer together here, and still not run into itself. Except that means you'd have to start on the outside this time. But that's OK, because next time you can go on the inside again. And now it's an easy pattern." + }, + { + "Q": "In 6:52, why does the paper's edge stay the same instead of shrinking just like the infinitely small paper is?\n", + "A": "because of how she folded it. if she folded it alternating sides the edge would shrink", + "video_name": "EdyociU35u8", + "timestamps": [ + 412 + ], + "3min_transcript": "OK, but there's probably some rule. Suddenly, a note lands on your desk from your friend Sam. Who writes, looks like you're concentrating pretty hard. Don't tell me you're actually doing math. As, if. You write back, no way. I'm just doodling this. And just to make extra clear it's not math, you turn it into a dragon, and name it the dragon curve. Yes. You don't want to crumple up your awesome dragon doodle, but you do have to throw it two rows over. So you neatly fold it into a note spear. Which just means you're folding it in half again and again, until it's easy to javelin across the room the moment the teachers back is turned-- Bam! Yes! Perfect landing. You watch as Sam unfolds it. And suddenly you feel like you see something familiar. Some sort of similarity between the paper and-- is that possible? You take your diagram, and fold it in half. And half again. And again. And wow, not only does it look like it's doing the same thing, probably, but it's also showing a new way to do it. you can just copy the old one, and add it 90 degrees from the other, which is totally traceable. Bam. Well, as long as you can keep track of what end to start from. And you don't even have to keep track of what order to draw the lines in. You just need to keep things roughly on a square grid so things line up. Until it gets too big for your paper and you have to dragon-ize it. It's funny, because one way it gets bigger and bigger. If you go on forever, it'll be infinitely big. But with the first way, it stays basically the same size. You just draw more details. Which means if you do it forever, the line will still get infinitely long, but the total size will stay the same. Will that even work? An infinitely long line all squiggled up into a finite area? And then with folding paper, the whole thing gets smaller and smaller until maybe it disappears entirely. Which you suppose makes sense because the edge of the paper stays the same length, no matter how you fold it. You can't make it longer and longer like copying it, where the length doubles each time. Maybe this grid really would fill up until that infinitely long line is squiggled up into an actual solid two dimensional triangle with no empty space left in it. Maybe that would make sense. Or be crazy. You know lines are infinitely thin, but if you had infinitely much of it, maybe the infinities would cancel out or something. Like the line gets closer and closer to itself until it actually touches, but doesn't overlap, but with no space between. Which would make no sense, unless you do it everywhere at once, so who can tell the difference. Yeah, that totally sounds legit. Although, in this one, there's holes that never get filled at any point, so you'd still have an infinite line, but it doesn't fill up space, it's just all holes forever, but it also never overlaps, So where is all that infinite line going? Anyway, class is over, so you pack up and save the question. After all, you've got math class again tomorrow." + }, + { + "Q": "\nAt 2:24 she mentions e. What is e?", + "A": "e is a mathematical constant that show up in many areas of mathematics.", + "video_name": "5iUh_CSjaSw", + "timestamps": [ + 144 + ], + "3min_transcript": "Voiceover:Hello and welcome to that one day of the year when, well, everyone else is building up how great Pi is. I'm here to tear it down, because you deserve the truth. Forget about the part where Pi isn't the correct circle concept. This Pi day, I'm not about how people worship Pi for being infinite for going on forever. First of all, Pi is not infinite. It is more three, but you know, less than four. There are cultures where three is the biggest number, so I don't want to be insensitive, but trust me on this four is not infinite and neither is Pi. I know it's not about it's magnitude, it's about all those digits, infinite digits going on forever, but first of all it doesn't go anywhere. It just is. There's no time element. If you had a number line, Pi would be exactly one point on that number line sitting perfect still right now. It's not going to start wondering off on an infinite journey that takes forever, or even on a finite journey that takes forever, or an infinite journey that takes finite time. Secondly, yeah, so it's got infinite digits. So what, one-third has infinite digits. There's exactly as exactly as many digits in one-third and in Pi as in 99.9999 repeating. Oh, and there's also as many digits as in numbers like, say, five, I know, big number. It's even more than four, so, it's piratically like double infinity. Which, it actuality kind of is, because in decimal innovation there's secretly infinite zeros in all of these places. Zero's going out to forever. Ooh! So mysterious, and then zero's going the other way too. Which is actually not any more zeros than if they only went one way. No. Pi is not especially infinite in any way, it's more like in-between-finite. There's an infinite number of rational numbers. for any two factions you can find another fraction that's between them again and again and again. There's never any fractions that are right next to each other on the number line. But, despite there's a infinite amount of rational numbers, Pi isn't one of them. and you can find an infinite number of rational numbers that are closer to Pi on either side. Pi is between all of them in one of the gaps. It isn't infinite. It's in-between-finite. So what, you think that's special, as if there's just one hole in the rational number line exactly where Pi is, and once you plug that in with a super special number, you're good to go? Maybe, a few more for E and [towel] and square root too. No! Super nope! The in-between-[finiteness] of Pi, its irrationality is an incredibility un-special property. Turns out, most real numbers are irrational. It's the nicely packaged rational numbers that are weird. In fact, if you threw a dart and picked a random number off the number line, the chance of getting a rational number is exactly zero. I'll get into kinds of infinities some other time, but [unintelligible] to say the number of rational numbers, like the number of digits in Pi, is the small and unimpressive countable infinity. While the number of irrational numbers is so much bigger than countable infinity," + }, + { + "Q": "\nin 2:18, can you make it until 'z' with using all the alphabets in a row?", + "A": "Sure you can! Neat stuff!", + "video_name": "a5z-OEIfw3s", + "timestamps": [ + 138 + ], + "3min_transcript": "OK, so fractal fractions. 5 equals 5. Bear with me now. Let's explode this 5 into fifths. 25 fifths to be precise. Now I want to split it into two parts. Say, 17/5 plus 8/5. Could have been 1 plus 24, or whatever, I don't care. OK, now I'm ready for the fun part. Since this 5 down here is just as much a 5 as any 5 is 5, such as this whole thing, which is equivalent to 5, let's go ahead and replace that 5 with this messier looking but still very fively 5. Oh, and now we've got more fives, so we can do it again. And again. And again. And then you can give someone this whole thing and be like, whoa, it's 5. Making things look more confusing than they actually are is a delicate art, but it speaks to the true heart of algebra, which is that you can shuffle numbers around all day and as long as you follow the rules, it all works out. OK, here's another one. Say you want to do something with 7. Maybe you could use 7/7, which is 1. So you need six more. Why not just add it? 7 equals 7/7 plus 6. 7 equals 7/7, plus 6/7, over 7 plus 6, plus 6. Instead of writing it all again, why not just extend this way? There we go. This equals 7. And you can actually take this all the way to infinity. The 7's kind of disappear. But then again, it didn't really matter what they were the first place, as long as they're the same. All these 7's could have been 3, or a billion, or pi to the i, and this would still equal 7. As long as this numerator equals this denominator this fraction equals 1. And whatever else you may think about algebra, at least it has the courtesy to make 1 plus 6 be 7 every time. The fractal structure of this first fraction was like a binary tree. Each layer with twice as many terms as the one above it, growing exponentially. And this one does too, but sideways. But awesomely enough, this is obviously an ABA CABA DABA CABA pattern. That's a fractal pattern that's actually found lots of places, but I'm not going to get into that right now. Point is, if you name this innermost layer and then try to read it from top to bottom, you get ABA CABA DABA CABA. And if your fraction was infinite, you'd get ABA CABA DABA CABA EABA CABA DABA CABA FABA CABA DABA CABA EABA CABA DABA CABA GABA CABA-- and so on. Anyway, a foolish algebra teacher would teach you that algebra is about solving equations. As if the goal of life were to get x on one side, and everything else on the other. As if every fiber of your being should cry out in protest when you see x on the left side, and yet more x on the right side. But you could replace that x with what it equals, and then you could do it again, and again, and each time your equation is still true. How's that for getting rid of the x on this side? And you can make equations even more confusing by remembering special numbers and identities. Write whatever you want, as long as you can sneak in a multiplied by 0, you don't even have to bother knowing what the rest is. Or, knowing that all you need is the top and bottom of the equation to be the same to get 1." + }, + { + "Q": "At 1:44, why do we use binary trees for situations like this in math? We can use some other, like just figuring out what kinds of parts and adding them.\n", + "A": "Binary trees are powerful tools for solving search space problems, but in the video at 1:44, Vi is pointing out the resemblence of the expressions to binary trees.", + "video_name": "a5z-OEIfw3s", + "timestamps": [ + 104 + ], + "3min_transcript": "OK, so fractal fractions. 5 equals 5. Bear with me now. Let's explode this 5 into fifths. 25 fifths to be precise. Now I want to split it into two parts. Say, 17/5 plus 8/5. Could have been 1 plus 24, or whatever, I don't care. OK, now I'm ready for the fun part. Since this 5 down here is just as much a 5 as any 5 is 5, such as this whole thing, which is equivalent to 5, let's go ahead and replace that 5 with this messier looking but still very fively 5. Oh, and now we've got more fives, so we can do it again. And again. And again. And then you can give someone this whole thing and be like, whoa, it's 5. Making things look more confusing than they actually are is a delicate art, but it speaks to the true heart of algebra, which is that you can shuffle numbers around all day and as long as you follow the rules, it all works out. OK, here's another one. Say you want to do something with 7. Maybe you could use 7/7, which is 1. So you need six more. Why not just add it? 7 equals 7/7 plus 6. 7 equals 7/7, plus 6/7, over 7 plus 6, plus 6. Instead of writing it all again, why not just extend this way? There we go. This equals 7. And you can actually take this all the way to infinity. The 7's kind of disappear. But then again, it didn't really matter what they were the first place, as long as they're the same. All these 7's could have been 3, or a billion, or pi to the i, and this would still equal 7. As long as this numerator equals this denominator this fraction equals 1. And whatever else you may think about algebra, at least it has the courtesy to make 1 plus 6 be 7 every time. The fractal structure of this first fraction was like a binary tree. Each layer with twice as many terms as the one above it, growing exponentially. And this one does too, but sideways. But awesomely enough, this is obviously an ABA CABA DABA CABA pattern. That's a fractal pattern that's actually found lots of places, but I'm not going to get into that right now. Point is, if you name this innermost layer and then try to read it from top to bottom, you get ABA CABA DABA CABA. And if your fraction was infinite, you'd get ABA CABA DABA CABA EABA CABA DABA CABA FABA CABA DABA CABA EABA CABA DABA CABA GABA CABA-- and so on. Anyway, a foolish algebra teacher would teach you that algebra is about solving equations. As if the goal of life were to get x on one side, and everything else on the other. As if every fiber of your being should cry out in protest when you see x on the left side, and yet more x on the right side. But you could replace that x with what it equals, and then you could do it again, and again, and each time your equation is still true. How's that for getting rid of the x on this side? And you can make equations even more confusing by remembering special numbers and identities. Write whatever you want, as long as you can sneak in a multiplied by 0, you don't even have to bother knowing what the rest is. Or, knowing that all you need is the top and bottom of the equation to be the same to get 1." + }, + { + "Q": "\nIn 2:29 Sal made delta= f(epsilon) but I think it should be epsilon=f(delta), shouldn't it ?", + "A": "In this kind of proof delta is a function of epsilon, even though epsilon refers to the y-axis. In his game, Sal has someone pick an epsilon first. Then he finds the delta to go with it. Thus, epsilon is the independent variable, not delta.", + "video_name": "0sCttufU-jQ", + "timestamps": [ + 149 + ], + "3min_transcript": "In the last video, we took our first look at the epsilon-delta definition of limits, which essentially says if you claim that the limit of f of x as x approaches C is equal to L, then that must mean by the definition that if you were given any positive epsilon that it essentially tells us how close we want f of x to be to L. We can always find a delta that's greater than 0, which is essentially telling us our distance from C such that if x is within delta of C, then f of x is going to be within epsilon of L. If we can find a delta for any epsilon, then we can say that this is indeed the limit of f of x as x approaches C. Now, I know what you're thinking. This seems all very abstract. I want to somehow use this thing. And what we will do in this video is use it and to rigorously prove that a limit actually exists. So, right over here, I've defined a function, f of x. It's equal to 2x everywhere except for x equals 5. but when x is equal to 5, it's just equal to x. So I could have really just written 5. It's equal to 5 when x is equal to 5. It's equal to itself. And so we've drawn the graph here. Everywhere else, it looks just like 2x. At x is equal to 5, it's not along the line 2x. Instead, the function is defined to be that point right over there. And if I were to ask you what is the limit of f of x as x approaches 5, you might think of it pretty intuitively. The closer I get to 5, the closer f of x seems to be getting to 10. And so, you might fairly intuitively make the claim that the limit of f of x as x approaches 5 really is equal to 10. It looks that way. the epsilon-delta definition to actually prove it. And the way that most of these proofs typically go is we define delta in the abstract. And then we essentially try to come up with a way that given any epsilon, we can always come up with a delta. Or another way is we're going to try to describe our delta as a function of epsilon, not to confuse you too much. But maybe I shouldn't use f again. But delta equals function of epsilon that is defined for any positive epsilon. So you give me an epsilon, I just put into our little formula or little function box. And I will always get you a delta. And if I can do that for any epsilon that'll always give you a delta, where this is true, that if x is within delta of C, then the corresponding f of x is going to be within epsilon of L. Then the limit definitely exists. So let's try to do that. So let's think about being within delta of our C." + }, + { + "Q": "At 6:46, why isn't delta divided by 2? He says that it is delta/2 but also multiplied by 2 so the 2s will cancel, but I don't see where the 2 that is multiplied is coming from.\n", + "A": "This is taken from the function. We want to show that |x-5|<\u00e2\u0088\u0082 in the form of |f(x)-L|<\u00ce\u00b5. That is to show that \u00e2\u0088\u0082 is a function of \u00ce\u00b5 .", + "video_name": "0sCttufU-jQ", + "timestamps": [ + 406 + ], + "3min_transcript": "this is just going to be the absolute value of 2x minus 10. And it's going to be less than on the right-hand side, you just end up with a 2 delta. Now, what do we have here on the left-hand side? Well, this is f of x as long as x does not equal 5. And this is our limit. So we can rewrite this as f of x minus L is less than 2 delta. And this is for x does not equal 5. This is f of x, this literally is our limit. Now this is interesting. This statement right over here is almost exactly what we want right over here, except the right sides are just different. In terms of epsilon, this has it in terms of delta. So, how can we define delta so that 2 delta is essentially going to be epsilon? Well, this is our chance. We're going to make 2 delta equal epsilon. Or if you divide both sides by 2, we're going to make delta equal to epsilon over 2. Let me switch colors just to ease the monotony. If you make delta equal epsilon over 2, then this statement right over here becomes the absolute value of f of x minus L is less than, instead of 2 delta, it'll be less than 2 times epsilon over 2. It's just going to be less than epsilon. So this is the key. If someone gives you any positive number epsilon for this function, as long as you make delta equal epsilon over 2, then any x within that range, that corresponding f of x is going And remember, it has to be true for any positive epsilon. But you could see how the game could go. If someone gives you the epsilon, let's say they want to be within 0.5 of our limit. So our limit is up here, so our epsilon is 0.5. So it would literally be the range I want to be between 10 plus epsilon would be 10.5. And then 10 minus epsilon would be 9.5. Well, we just came up with a formula. We just have to make delta equal to epsilon over 2, which is equal to 0.25. So that'll give us a range between 4.75 and 5.25. And as long as we pick an x between 4.75 and 5.25, but not x equals 5, the corresponding f of x will be between 9.5 and 10.5. And so, you give me any epsilon, I can just apply this formula right over here" + }, + { + "Q": "\nAt 4:17, when the calculator is shown in the video, is it correct to say that the symbol n! is like the one that can facilitate the factorial multiplication? Do all scientific calculators have it?", + "A": "Yes, that s correct. n! represents the factorial as long as n is non-negative integer. It is more than just like it. It IS it. I would assume that all scientific calculators have a factorial function, but I wouldn t rule out the possibility of the existence of a calculator deemed scientific to exclude it.", + "video_name": "SbpoyXTpC84", + "timestamps": [ + 257 + ], + "3min_transcript": "Or maybe another hand is I have the eight cards, 1, 2, 3, 4, 5, 6, 7, 8, and then I have the 9 of spades. If we were thinking of these as two different hands, because we have the exact same cards, but they're in different order, then what I just calculated would make a lot of sense, because we did it based on order. But they're telling us that the cards can be sorted however the player chooses, so order doesn't matter. So we're overcounting. We're counting all of the different ways that the same number of cards can be arranged. So in order to not overcount, we have to divide this by the ways in which nine cards can be rearranged. So we have to divide this by the way nine cards can be rearranged. So how many ways can nine cards be rearranged? If I have nine cards and I'm going to pick one of nine to be in the first slot, well, that means I have 9 ways to put something in the first slot. Then in the second slot, I have 8 ways of putting a card first, so I have 8 left. Then 7, then 6, then 5, then 4, then 3, then 2, then 1. That last slot, there's only going to be 1 card left to put in it. So this number right here, where you take 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1, or 9-- you start with 9 and then you multiply it by every number less than 9. Every, I guess we could say, natural number less than 9. This is called 9 factorial, and you express it as an exclamation mark. So if we want to think about all of the different ways that we can have all of the different combinations for hands, this is the number of hands if we cared about the order, but then we want to divide by the number of ways we can order things so that we don't overcount. And this will be an answer and this will be the correct answer. Now this is a super, super duper large number. Let's figure out how large of a number this is. times 35, times 34, times 33, times 32, times 31, times 30, times 29, times 28, divided by 9. Well, I can do it this way. I can put a parentheses-- divided by parentheses, 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1. Now, hopefully the calculator can handle this. And it gave us this number, 94,143,280. Let me put this on the side, so I can read it. So this number right here gives us 94,143,280." + }, + { + "Q": "Why is there a blank space on 2:12\n", + "A": "the pie was eaten 5 + 3 slices. and the pie initially have 9 slices. so, 8/9. (left 1 blank space). hope that help", + "video_name": "u2hLYcmI5y4", + "timestamps": [ + 132 + ], + "3min_transcript": "Brandon ate 5 slices of apple-- of pie. I'm just assuming it's apple pie. They didn't tell me that. Gabriela ate 3 slices. If there were originally 9 slices, what fraction of the pie was eaten? So let me see if I can draw this thing out. So let me draw the pie. I will draw the pie in a yellowish color. So let me try my best. So let's see how good I am at drawing a pie. So I'll just draw it from the top view as a circle. And there're 9 slices. I think it's a reasonable assumption to say that they're 9 equal slices. So we have 9 equal slices of pie. And I'll just make sure they're initially 9 equal slices. What fraction of the pie was eaten? So let's first divide this into 9 sections. So one way I could do that, I could divide it into 3 sections first, so it looks like a peace symbol. It actually looks more like the Mercedes emblem. So I'll draw it into 3 sections first. So let's see, I'll draw like that and like that. Keep in mind, I'm trying to make these as equal as possible. So bear with me if they don't look 100% equal, but I'm trying. I am trying my best. So 9 equal slices-- so that looks pretty respectable. So here's our pie that initially had 9 equal slices. Now, they tell us that Brandon ate 5 slices of pie. So Brandon eats-- he seems like a hungry young man-- so he eats 1, 2, 3, 4, 5. You could say that he ate 5/9 of the pie. But that's not it. That's not what they're asking. It's saying total, not just how much did Brandon eat, but how much was eaten total between Brandon and Gabriela. And they tell us Gabriela ate 3 slices. So she ate 1-- sounds like they didn't really So now let's answer their question. What fraction of the pie was eaten? Well, we know that there was a total of 9 equal slices of pie. What fraction was eaten? Well, as we see, 1, 2, 3, 4, 5, 6, 7, 8. 8/9 of the pie was eaten. So let's actually type that in. And then we will do it right over here. We'd say 8, and we use that little slash symbol on your computer like this. 8 over 9 or 8 divided by 9 or 8/9 of the pie was eaten. Let's do a couple more of these. Ishaan ate 2 slices of pizza. Omar ate 3 slices. If there were originally 8 slices, what fraction of the pie is remaining? So this is interesting. Actually, let me copy and paste this" + }, + { + "Q": "\nat 2:08, when he was rewriting the expressions, how come he did 6(2^2) instead of 6(2)^2?", + "A": "They mean the same thing. So, either form is acceptable.", + "video_name": "I9eLKDbc8og", + "timestamps": [ + 128 + ], + "3min_transcript": "The surface area of a cube is equal to the sum of the areas of its six sides. Let's just visualize that. I like to visualize things. So if that's the cube, we can see three sides. Three sides are facing us. But then if it was transparent, we see that there are actually six sides of a cube. So there's this one-- one, two, three in front-- and then one-- this is the bottom. This is in the back, and this is also in the back. So you have three sides of the cube. So I believe what they're saying. The surface area of a cube with side length x-- so if this is x, if this is x, if this is x-- is given by the expression 6x squared. That also makes sense. The area of each side is going to be x times x is x squared, and there's six of them. So it's going to be 6x squared. Jolene has two cube-shaped containers that she wants to paint. One cube has side length 2. So this is one cube right over here. I'll do my best to draw it. So this right over here has side length 2, The other cube has side length 1.5. So the other cube is going to be a little bit smaller. It has side length 1.5. So it's 1.5 by 1.5 by 1.5. What is the total surface area that she has to paint? Well, we know that the surface area of each cube is going to be 6x squared, where x is the dimensions of that cube. So the surface area of this cube right over here is going to be 6. And now-- let me do it in that color of that cube-- it's going to be 6 times x, where x is the dimension of the cube. And then the cube all has the same dimensions, so its length, width, and depth is all the same. So for this cube, the surface area is going to be 6 times 2 squared. And then the surface area of this cube is going to be 6 times 1.5 squared. it's going to be the sum of the two cubes. So we're just going to add these two things. And so if we were to compute this first one right over here, this is going to be 6 times 4. This is 24. And this one right over here, this is going to be a little bit hairier. Let's see. 15 times 15 is 225. So 1.5 times 1.5 is 2.25. So 1.5 squared is 2.25. And 2.25 times 6-- so let me just multiply that out. 2.25 times 6. Let's see. We're going to have 6 times 5 is 30. 6 times 2 is 12, plus 3 is 15. 6 times 2 is 12, plus 1 is 13. I have two numbers behind the decimal-- 13.5. So it's going to be 13.5." + }, + { + "Q": "At 1:27 am pacific time, what does it mean to evaluate an expression?\n", + "A": "make all the appropriate calculations to obtain the final result of the expression", + "video_name": "I9eLKDbc8og", + "timestamps": [ + 87 + ], + "3min_transcript": "The surface area of a cube is equal to the sum of the areas of its six sides. Let's just visualize that. I like to visualize things. So if that's the cube, we can see three sides. Three sides are facing us. But then if it was transparent, we see that there are actually six sides of a cube. So there's this one-- one, two, three in front-- and then one-- this is the bottom. This is in the back, and this is also in the back. So you have three sides of the cube. So I believe what they're saying. The surface area of a cube with side length x-- so if this is x, if this is x, if this is x-- is given by the expression 6x squared. That also makes sense. The area of each side is going to be x times x is x squared, and there's six of them. So it's going to be 6x squared. Jolene has two cube-shaped containers that she wants to paint. One cube has side length 2. So this is one cube right over here. I'll do my best to draw it. So this right over here has side length 2, The other cube has side length 1.5. So the other cube is going to be a little bit smaller. It has side length 1.5. So it's 1.5 by 1.5 by 1.5. What is the total surface area that she has to paint? Well, we know that the surface area of each cube is going to be 6x squared, where x is the dimensions of that cube. So the surface area of this cube right over here is going to be 6. And now-- let me do it in that color of that cube-- it's going to be 6 times x, where x is the dimension of the cube. And then the cube all has the same dimensions, so its length, width, and depth is all the same. So for this cube, the surface area is going to be 6 times 2 squared. And then the surface area of this cube is going to be 6 times 1.5 squared. it's going to be the sum of the two cubes. So we're just going to add these two things. And so if we were to compute this first one right over here, this is going to be 6 times 4. This is 24. And this one right over here, this is going to be a little bit hairier. Let's see. 15 times 15 is 225. So 1.5 times 1.5 is 2.25. So 1.5 squared is 2.25. And 2.25 times 6-- so let me just multiply that out. 2.25 times 6. Let's see. We're going to have 6 times 5 is 30. 6 times 2 is 12, plus 3 is 15. 6 times 2 is 12, plus 1 is 13. I have two numbers behind the decimal-- 13.5. So it's going to be 13.5." + }, + { + "Q": "\nat 7:57, why do the sun and moon travel across the sky?", + "A": "because it needs its orbit 24 hours and the moon goes down the sun comes up and when the sun goes down the moon comes up", + "video_name": "I9eLKDbc8og", + "timestamps": [ + 477 + ], + "3min_transcript": "" + }, + { + "Q": "\nat 0:30, Sal said the first one is irrational. but 8/2 is four and square of 4 is 2 which is rational.", + "A": "it was sqrt of 8 by 2 not 8/2", + "video_name": "d9pO2z2qvXU", + "timestamps": [ + 30 + ], + "3min_transcript": "Which of the following real numbers are irrational? Well, irrational just means it's not rational. It means that you cannot express it as the ratio of two integers. So let's see what we have here. So we have the square root of 8 over 2. If you take the square root of a number that is not a perfect square, it is going to be irrational. And then if you just take that irrational number and you multiply it, and you divide it by any other numbers, you're still going to get an irrational number. So square root of 8 is irrational. You divide that by 2, it is still irrational. So this is not rational. Or in other words, I'm saying it is irrational. Now, you have pi, 3.14159-- it just keeps going on and on and on forever without ever repeating. So this is irrational, probably the most famous of all of the irrational numbers. 5.0-- well, I can represent 5.0 as 5/1. So 5.0 is rational. It is not irrational. 0.325-- well, this is the same thing as 325/1000. So this is rational. Just as I could represent 5.0 as 5/1, both of these are rational. They are not irrational. Here I have 7.777777, and it just keeps going on and on and on forever. And the way we denote that, you could just say these dots that say that the 7's keep going. Or you could say 7.7. And this line shows that the 7 part, the second 7, just keeps repeating on forever. Now, if you have a repeating decimal-- in other videos, we'll actually convert them into fractions-- but a repeating decimal can be represented as a ratio of two integers. Just as 1/3 is equal to 0.333 on and on and on. Or I could say it like this. I could say 3 repeating. We can also do the same thing for that. I won't do it here, but this is rational. 8 and 1/2? Well, that's the same thing. 8 and 1/2 is the same thing as 17/2. So it's clearly rational. So the only two irrational numbers are the first two right over here." + }, + { + "Q": "At 0:18 Sal states that if you take the square root of a non-perfect square you will always end up with an irrational number. My question is this... 'What is a non-perfect square?'\n", + "A": "A perfect square is created when you multiply a number with itself. For example: 3 * 3 = 9. So, 9 is a perfect square. A number that does not have matching factors would be a non-perfect square. Hope this helps.", + "video_name": "d9pO2z2qvXU", + "timestamps": [ + 18 + ], + "3min_transcript": "Which of the following real numbers are irrational? Well, irrational just means it's not rational. It means that you cannot express it as the ratio of two integers. So let's see what we have here. So we have the square root of 8 over 2. If you take the square root of a number that is not a perfect square, it is going to be irrational. And then if you just take that irrational number and you multiply it, and you divide it by any other numbers, you're still going to get an irrational number. So square root of 8 is irrational. You divide that by 2, it is still irrational. So this is not rational. Or in other words, I'm saying it is irrational. Now, you have pi, 3.14159-- it just keeps going on and on and on forever without ever repeating. So this is irrational, probably the most famous of all of the irrational numbers. 5.0-- well, I can represent 5.0 as 5/1. So 5.0 is rational. It is not irrational. 0.325-- well, this is the same thing as 325/1000. So this is rational. Just as I could represent 5.0 as 5/1, both of these are rational. They are not irrational. Here I have 7.777777, and it just keeps going on and on and on forever. And the way we denote that, you could just say these dots that say that the 7's keep going. Or you could say 7.7. And this line shows that the 7 part, the second 7, just keeps repeating on forever. Now, if you have a repeating decimal-- in other videos, we'll actually convert them into fractions-- but a repeating decimal can be represented as a ratio of two integers. Just as 1/3 is equal to 0.333 on and on and on. Or I could say it like this. I could say 3 repeating. We can also do the same thing for that. I won't do it here, but this is rational. 8 and 1/2? Well, that's the same thing. 8 and 1/2 is the same thing as 17/2. So it's clearly rational. So the only two irrational numbers are the first two right over here." + }, + { + "Q": "\nAt 8:17, how is cos 1/3 x zero? Is it because the y-value is zero which makes the cos zero? And when you multiply -2.5 you get zero?\n\nThanks for the help!", + "A": "Yes, cos[(1/3)x] is 0 at that point because the y-value is 0. Also, yes, when you multiply that value, 0, by -2.5, you still get zero.", + "video_name": "uBVhtGL9y88", + "timestamps": [ + 497 + ], + "3min_transcript": "So given this change, we're now multiplying by negative 2.5, what is going to be-- well actually, let's think about a few things. What was the amplitude in the first two graphs right over here? Well there's two ways to think about it. You could say the amplitude is half the difference between the minimum and the maximum points. In either of these case, the minimum is negative 1, maximum is 1. The difference is 2, half of that is 1. Or you could just say it's the absolute value of the coefficient here, which is implicitly a 1. And the absolute value of 1 is, once again, 1. What's going to be the amplitude for this thing right over here? Well the amplitude is going to be the absolute value of what's multiplying the cosine function. the amplitude is going to be equal to the absolute value of negative 2.5, which is equal to 2.5. So given that, how is multiplying by negative 2.5 going to transform this graph right over here? Well let's think about it. If it was multiplying by just a positive 2.5, you would stretch it out. At each point it would go up by a factor of 2 and 1/2. But it's a negative 2.5, so at each point, you're going to stretch it out and then you're going to flip it over the x-axis. So let's do that. So when x was 0, you got 1 in this case. But now we're going to multiply that by negative 2.5, which means you're going to get to negative 2.5. So let me draw negative 2.5 right over there. So that's negative 2.5. That'd be negative-- let me make it clear. this would be positive 3. So that number right over there is negative 2.5. And let me draw a dotted line there. It could serve to be useful. Now when cosine of 1/3 x is 0, it doesn't matter what you multiply it by, you're still going to get 0 right over here. Now, when cosine of 1/3 x was negative 1, which was the case when x is equal to 3 pi, what's going to happen over here? Well cosine of 1/3 x, we see, is negative 1. Negative 1 times negative 2.5 is positive 2.5. So we're going to get to positive 2.5, which is right-- let me draw a dotted line over here. We're going to get to positive 2.5, which is right over there. And then when cosine of 1/3 x is equal to 0, doesn't matter what we multiply it by, we get to 0." + }, + { + "Q": "in 5:31 you said the M was negitive 800 why\n", + "A": "The negative 800 is paired with the W, not the M.", + "video_name": "VuJEidLhY1E", + "timestamps": [ + 331 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:43, wouldn't the logicians know if their forehead was painted? They would obviously feel that somebody painted/was painting their forehead.\n", + "A": "Yes, I agree. They would feel it, wouldn t they? Maybe the paint is special or something :)", + "video_name": "rBaCDC52NOY", + "timestamps": [ + 103 + ], + "3min_transcript": "There's a new reality television program, and it's called the Blue-- I should probably write it in blue-- but it's called the Blue Forehead Room. And what they do in this reality television program-- and you'll have to bear with me, because the show probably wouldn't be that interesting to watch-- but it's interesting to predict what happens. What they do is, they take a room. They'll call it the blue forehead room. And let's see, that's kind of a top view of the room. And let's say there's a door here. None of this is relevant to the actual problem. This is the door, right there. And what they do is they get 100 perfect logicians to sit in this room, in a circle. So they're all sitting in a circle in this room. Now, before the game even starts, before they even enter the room the first time, the logicians are told two things. forehead painted blue. At least one of you has your forehead-- And they all get their foreheads painted, so that obviously if you're the only guy who has your forehead painted. But you just don't know what color it is. So all of them have different color foreheads. Or, we don't know. But all they're told is, obviously I've painted your forehead. At least one of the people in the room that you will enter will have their forehead painted blue. And then they're also told that as soon as you deduce that your forehead is blue, you need to leave the room. that I set this up properly. They're all outside of the room. No-one's inside the room. And let's say they're blindfolded. And while they're blindfolded, they essentially have the thing painted onto their forehead. So they can't see the paintbrush or anything. So they really don't know what's on their forehead. And then after that, they all enter the room. And they all sit in a circle like this, so that they can all look at each other. And let's say when they enter, the lights are off. So the lights are off, and then the protocol is that the lights will be turned on, and then they can all look around at each other. There's no reflective surfaces. They can't look into each other's eyes and try to see No tricks like that. There are no mirrors in this room." + }, + { + "Q": "\n3:45 - 3:58 Can you think of a formula to help me with all this information you just gave me? I think it would really help me during school.", + "A": "What he is describing is the variance of a population, which he goes over in a video with the same title.", + "video_name": "iHXdzfF7UEs", + "timestamps": [ + 225, + 238 + ], + "3min_transcript": "Well, the sample mean, which we would denote by lowercase x with a bar over it, is just the sum of all of these divided by the number of data points we have. So let's see we have 1.5 plus 2.5 plus 4 plus 2 plus 1 plus 1. And all of that divided by 6, which gives-- let's see, the numerator 1.5 plus 2.5 is 4, plus 4 is 8, plus 2 is 10, plus 2 more is 12. So it's going to be 12 over 6, which is equal to 2 hours of television. So at least for your sample, you say, my sample mean is two hours of television. It's a statistic that is trying to estimate this parameter, this thing that's very hard to know. But it's our best shot. Maybe we get a better answer if we get more data points. But this is we have so far. well, I don't want to just estimate my population mean. I also want to estimate another parameter. I also am interested in estimating my population variance. So once again, since we can't survey every one in the population, this is pretty much impossible to know. But we're going to attempt to estimate of this parameter. We attempted to estimate the mean. Now we will also attempt to estimate this parameter, this variance parameter. So how would you do it? Well, reasonable logic would say, well, we maybe we'll do the same thing with a sample as we would have done with the population. When you're doing the population variance, you would take each data point in the population, find the distance between that and the normal population mean, take the square of that difference, and then add up all the squares of those differences, and then divide by the number of data points you have. So let's try that over here. So let's try to find-- take each of these data points, in a different color-- each of these data points, and find the difference between that data point and our sample mean-- not the population mean, we don't know what the population mean-- the sample mean. So that's that first data point plus the second data point-- so it's 4 minus 2 squared plus 1 minus 2 squared. And this is what you would have done if you were taking a population variance. If this was your entire population, this is how you would you find a population mean here, if this was your entire population. And you find the squared distances from each of those data points and then divide by the number of data points. So let's just think about this a little bit. 1 minus 2 squared. Then you have 2.5 minus 2-- 2 being the sample mean-- squared. Let me see, this green color. Plus 2 minus 2 squared." + }, + { + "Q": "\nat 9:32, Why didn't Sal write it as sqrt((a+b+c/2)*(a-b+c/2))", + "A": "because the second (a+b+c)/2 is a part of (a+b+c)/2 - 2b/2. for multiplying how you said he would have had to multiply (a+b+c)/2 with -2b/2 .", + "video_name": "nZu7IZLhJRI", + "timestamps": [ + 572 + ], + "3min_transcript": "This is the same thing as b minus c minus a. Right? And all of that over 4. Now, I can rewrite this whole expression. I don't want to run out of space. I can rewrite this whole expression as, well 4 is the product of 2 times 2. So our whole area expression has been, arguably, simplified to it equals the square root-- and this is really the home stretch-- of this right here, which I can just write as a plus b plus c over 2. That's that term right there. Times this term. Times that term. minus b, that's the same thing as a plus b plus c minus 2b. These two things are equivalent. You have an a, you have a c, and then b minus 2b is going to be equal to minus b. Right? b minus 2b, that's minus b. So this next term is going to be a plus b plus c minus 2b, over 2. Or instead of writing it like that, let me write this over 2 minus this over 2. And then our next term right here. Same exact logic. That's the same thing as a plus b plus c minus 2a, all of that over 2. If we add the minus 2a to the a we get minus a. So we get b plus c minus a. So all this over 2, or we can split the denominators just like that over 2. And then one last term. And you might already recognize the rule of Heron's formula popping up. I was thinking not the rule of Heron-- Heron's formula. That term right there is the exact same thing as a plus b plus c minus 2c. You take 2c away from the c you get a minus c, and then you still have the a and the b. And then all of that over 2. You could write that over 2 minus that over 2. And, of course, we're taking the square root of all of this stuff. Now, if we define an S to be equal to a plus b plus c over 2, then this equation simplifies a good bit. This right here is S. That right there is S. That right there is S." + }, + { + "Q": "I have noticed that when talking about Vectors Sal generally denotes them as a column matrix like he did at 3:00 when describing vector x?\nIS there a reason to do so or could your represent them as a row vector also?\n", + "A": "You could use a row vector without changing any meaning. It only becomes significant when mixing vectors with matrices, where a row and column are very different and in that context one or the other is used as appropriate.", + "video_name": "gAbadNuQEjI", + "timestamps": [ + 180 + ], + "3min_transcript": "Where you said, oh, a function is, you just give me some number and I'll give you another number. Or I'll do something to that number. While it can be much more general than that. It's an association between any member of one set and some other members of another set. Now, we know that vectors are members of sets. Right? In particular, if we say that some vector x is a member of some set -- let me just say it's a member of rn, because that's what we deal with -- all that means is that this is just a particular representation of an n-tuple. Remember what rn was. rn we defined way back at the beginning of the linear algebra playlist. We defined it as the set of all n-tuples -- x1, x2, xn, where your x1's, x2's, all the way to So your rn is most definitely a set. This could be rn. And obviously the use of the letter n is arbitrary. It could be rm, it could be rs. n is just kind of a placeholder for how many tuples we have. It could be r5. It could be 5 tuples. And when we say that a vector x is a member of rn, we're just saying that it's another way of writing one of these n-tuples. And all of our vectors so far are column vectors -- that's the only type that we've defined so far -- and we say it's this ordered list where each of the members are a member of r's. It's an ordered list of n -- it's an ordered list of n-components -- x1, x2, all the way to xn -- where each of those guys, or each of those x1's, x2's all the way to xn's, are a member of the real numbers. That's, by definition, what we mean when we say that x is a member of rn. here -- let's say that this set right here is rn and then let me just change, just to be general, let me create another set right there and call that set right there rm. Just a different number. It it could be the same as n, it could be different. This is m-tuples, that's n-tuples. We've defined that vectors can be members of rn. So you could have some vector here and then, if you associate with that vector in rn -- if you associate it with some vector in are rm -- if you associate it with, let's call that vector y, if you make this association, that too is a function. And that might have already been obvious to you and this would be a function that's mapping from rn to rm." + }, + { + "Q": "\nAt 1:16 Sal says that in order to have a vertical asymptote \"it cannot be defined there. \" What does he mean by this? What is he referring to as it? Why can \"it\" not be defined?", + "A": "It means that at that particular x value, the equation doesn t work so the y value is undefined. For example if the function is a fraction and at a particular x value the bottom of the fraction is equal to 0, then the output would be undefined since we can t divide by 0, so there would be an asymptote on the graph.", + "video_name": "onmNaDrxwmo", + "timestamps": [ + 76 + ], + "3min_transcript": "Voiceover:So, we've got three functions graphed here. The function f in magenta. We have the function g in this green color. And we have the function h in this dotted purple color. And then we have three potential expressions, or three expressions that could be potential definitions, for f, g, and h. And what I want to do in this video, is try to match them, try to match the function to a definition. I encourage you to pause this video and try to think about it on your own before I work through it. There's a couple of ways that we could approach this. One, we could think about what the graphs of each of these look like, and then think about which of these function's graphs look like that. Or, we could we could look at the function graph and think about the vertical and horizontal asymptotes, and think about which of these expressions would have a vertical or horizontal asymptotes at that point. So, I'm actually going to do it the second way. Let's look at the graphs. I tend to be a little bit more visual, or I like to look at graphs. So it looks like - f, I'll start with f. f looks like it has a vertical asymptote at x equals 5. Or, sorry, at x equals negative 5. at x is equal to negative five. Let's think about which of these would have a vertical asymptote at x equals negative 5. In order to have a vertical asymptote, it can't be defined there. So that'sthe first way I would think about it. And then even if isn't defined there, we need to make sure that it's an actual vertical asymptote, and not just a hole at that point, not just a point discontinuity. So, let's think about it. So, this first expression is actually defined at x equals negative 5. The only reason why it wouldn't be defined, if somehow we got a 0 in the denominator. But if you have negative 5 minus 5, that's negative 10. So this one's defined at that point, so that's not f. This one's also defined at x equals negative 5. The denominator does not become 0, so that's not f. This one, when x is equal to negative 5, this denominator does become 0. So this seems, I just used purely deductive reasoning, this looks like my best candidate for f of x. But let's confirm that it's consistent with the other things that we're seeing over here. If I look at the graph, it looks like there's a horizontal asymptote. Especially as x gets to larger and larger values, it looks like f of x is approaching 1. f of x is approaching 1. Now, is that the same case over here? As x gets larger and larger and larger, as x approaches infinity, well, then the negative 2 and the plus 5 don't matter. As x approaches infinity, this is going to approximate, for very large x's, it's going to be x over x. We look at the highest degree terms, and so that is going to approach 1, as x gets very very very large. Then the negative two, subtracting the 2 in the numerator and adding 5 in the denominator are going to matter less and less and less, because the x gets so big. So, this will approach 1. So, that seems consistent from that point of view. And let's see, is there anything else interesting? Well, when does this equal 0? Well, the numerator equals 0 when x is equal to 2. And we see that's indeed the case for f right over there. So, I'm feeling really good that this is our f of x." + }, + { + "Q": "\nAround 3:53, why does 10 to the power of both sides cancel out the base 10 and the log_10 exponent?", + "A": "Since that s how logarithms work; exponentiation and logarithms are inverse operations to each other. Since log(10) a is whatever 10 needs to be raised in to get a, 10^(log(10) a) = a.", + "video_name": "Kv2iHde7Xgw", + "timestamps": [ + 233 + ], + "3min_transcript": "log base 10 of 3 times x, of 3x. Then, based on this property right over here, this thing could be rewritten-- so this is going to be equal to-- this thing can be written as log base 10 of 4 to the second power, which is really just 16. So this is just going to be 16. And then we still have minus logarithm base 10 of 2. And now, using this last property, we know we have one logarithm minus another logarithm. This is going to be equal to log base 10 of 16 over 2, 16 divided by 2, which is the same thing as 8. So the right-hand side simplifies to log base 10 of 8. The left-hand side is log base 10 of 3x. And 10 to the same power is going to be equal to 8. So 3x must be equal to 8. 3x is equal to 8, and then we can divide both sides by 3. Divide both sides by 3, you get x is equal to 8 over 3. One way, this little step here, I said, look, 10 to the-- this is an exponent. If I raise 10 to this exponent, I get 3x, 10 to this exponent, I get 8. So 8 and 3x must be the same thing. One other way you could have thought about this is, let's take 10 to this power, on both sides. So you could say 10 to this power, and then 10 to this power over here. If I raise 10 to the power that I need to raise 10 to to get to 3x, well, I'm just going to get 3x. If I raise 10 to the power that I need to raise 10 to to get 8, So once again, you've got the 3x is equal to 8, and then you can simplify. You get x is equal to 8/3." + }, + { + "Q": "\nI'm getting confused at 3:42, when he offers an alternate way of solving the equation. Can someone theoretically explain to me how raising both sides of the equation to the log's base solves the problem?\nI understand the lesson up until that part.", + "A": "logarithms are just inverse functions of exponential functions so that the base and the exponents cancel and equal 1 .try this logany base (withthat number)=1 as well exponets leading coeffitient with raised with any logsame numbe =1 let say 10^x(power)=100 by logarithm rules it inverse it intern of x log(10_base)(100)=x so that x=2 log( 10^x(power))=log(100) this simplifies to x=log 100 or 2", + "video_name": "Kv2iHde7Xgw", + "timestamps": [ + 222 + ], + "3min_transcript": "log base 10 of 3 times x, of 3x. Then, based on this property right over here, this thing could be rewritten-- so this is going to be equal to-- this thing can be written as log base 10 of 4 to the second power, which is really just 16. So this is just going to be 16. And then we still have minus logarithm base 10 of 2. And now, using this last property, we know we have one logarithm minus another logarithm. This is going to be equal to log base 10 of 16 over 2, 16 divided by 2, which is the same thing as 8. So the right-hand side simplifies to log base 10 of 8. The left-hand side is log base 10 of 3x. And 10 to the same power is going to be equal to 8. So 3x must be equal to 8. 3x is equal to 8, and then we can divide both sides by 3. Divide both sides by 3, you get x is equal to 8 over 3. One way, this little step here, I said, look, 10 to the-- this is an exponent. If I raise 10 to this exponent, I get 3x, 10 to this exponent, I get 8. So 8 and 3x must be the same thing. One other way you could have thought about this is, let's take 10 to this power, on both sides. So you could say 10 to this power, and then 10 to this power over here. If I raise 10 to the power that I need to raise 10 to to get to 3x, well, I'm just going to get 3x. If I raise 10 to the power that I need to raise 10 to to get 8, So once again, you've got the 3x is equal to 8, and then you can simplify. You get x is equal to 8/3." + }, + { + "Q": "\nat 1:24 is 3-4 the same as 4-3", + "A": "3-4 is not the same as 4-3 because in subtracting, there is no Commutative Property. 3-4 would be -1, while 4-3 would be positive 1.", + "video_name": "AO9bHbUdg-M", + "timestamps": [ + 84 + ], + "3min_transcript": "Voiceover:Let's explore what it means to subtract numbers. So let's say that I want to figure out what 4, what 4 minus 3 is. 4 minus 3. So one way to think about this, is you start with 4 objects. And so let me just draw 4 objects. So there I have 1, 2, 3, and 4. And when I ... So this is the 4 objects right over here. And when I say minus 3, or if I'm going to subtract 3 from the 4, one way to think about it is, I'm going to take 3 of these 4 objects away. So let's do that. So I'm going to take away 1, I'm going to take away 2, and I'm going to take away 3. Notice, I took away 1, 2, 3 objects. Well, if I start with 4 and I take away 3, I subtract 3, 4 minus 3, I am left with, I am left with, 1 object, right over here. So 4 minus 3 is equal to, is equal to 1. Fascinating. Let's do another one of these. Let's figure out what 5, what 5, what 5 minus 2 is. And let's write it this way. Let's say we want to figure out something, some question mark, some question mark ... Actually, let me just clear this out. I'll just do it right over here, actually. So let's say we have some question mark. So there's some unknown number right over here. So I'll just put a question mark over here. And we say that unknown number is equal to, is equal to 5, 5 minus 2. Minus 2. So what is this going to be? Well let's visualize it. This means I have 5 things and I'm going to take away 2 of them and this is going to be what I have leftover after I start with 5 and take away 2. So I have 1, 2, 3, 4, 5 things. Now I am going to take away 2 of them. So I'm going to take away 1 and 2. So I took away 1, 2 objects. So how many do I have leftover? Well, I have leftover these, these ... Let me do this in a different color. I have leftover these purple things right over here. So how many is that? How many do I have leftover?" + }, + { + "Q": "At 12:46, how did he factor a -x out of +17x^2??\n", + "A": "Sal divided 17x^2 by -x. Another way to think about this is to say what does -x have to be multiplied by in order to get +17x^2.", + "video_name": "X7B_tH4O-_s", + "timestamps": [ + 766 + ], + "3min_transcript": "h times g times j. These are all the same things. Well, what is fh times gj? This is equal to fh times gj. Well, this is equal to the first coefficient times the constant term. So a plus b will be equal to the middle coefficient. And a times b will equal the first coefficient times the constant term. So that's why this whole factoring by grouping even works, or how we're able to figure out what a and b even are. Now, I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things. What I want to do is to teach you to factor things a little And this is a little bit of a add-on. I was going to make a whole video on this. But I think, on some level, it might be a little obvious for you. So let's say we had-- let me get a good one here. squared, minus 70x. Immediately, you say, gee, this isn't even a quadratic. I don't know how to solve something like this. It has an x to third power. And the first thing you should realize is that every term here is divisible by x. So let's factor out an x. Or even better, let's factor out a negative x. So if you factor out a negative x, this is equal to negative x times-- negative x to the third divided by negative x is x squared. 17x squared divided by negative x is negative 17x. Negative 70x divided by negative x is positive 70. The x's cancel out. And now, you have something that might look a little bit familiar. We have just a standard quadratic where the leading We just have to find two numbers whose product is 70, and that add up to negative 17. And the numbers that immediately jumped into my head are negative 10 and negative 7. You take their product, you get 70. You add them up, you get negative 17. So this part right here is going to be x minus 10, times x minus 7. And of course, you have that leading negative x. The general idea here is just see if there's anything you can factor out. And that'll get it into a form that you might recognize. Hopefully, you found this helpful. I want to reiterate what I showed you at the beginning of I think it's a really cool trick, so to speak, to be able to factor things that have a non-1 or non-negative 1 leading coefficient. But to some degree, you're going to find out easier ways to do this, especially with the quadratic formula, in not too long." + }, + { + "Q": "\nAt 9:36, you put fhx instead of fhx^2 when you multiplied fx and hx, why?", + "A": "Sal just made a mistake. It has been corrected already if you watch the video again.", + "video_name": "X7B_tH4O-_s", + "timestamps": [ + 576 + ], + "3min_transcript": "to have a plus here. But this second group, we just literally have a x plus 1. Or we could even write a 1 times an x plus 1. You could imagine I just factored out of 1 so to speak. Now, I have 6x times x plus 1, plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. And now, I'm going to actually explain why this little magical system actually works. Let me take an example. I'll do it in very general terms. Let's say I had ax plus b, times cx-- actually, I'm I think that'll confuse you, because I use a's and b's here. They won't be the same thing. So let me use completely different letters. Let's say I have fx plus g, times hx plus, I'll use j instead of i. You'll learn in the future why don't like using i as a variable. So what is this going to be equal to? Well, it's going to be fx times hx which is fhx. And then, fx times j. So plus fjx. And then, we're going to have g times hx. So plus ghx. And then g times j. Plus gj. Or, if we add these two middle terms, you have fh times x, Plus gj. Now, what did I do here? Well, remember, in all of these problems where you have a non-1 or non-negative 1 coefficient here, we look for two numbers that add up to this, whose product is equal to the product of that times that. Well, here we have two numbers that add up-- let's say that a is equal to fj. That is a. And b is equal to gh. So a plus b is going to be equal to that middle coefficient. And then what is a times b? a times b is going to be equal to fj times gh. We could just reorder these terms. We're just multiplying" + }, + { + "Q": "I am confused why at about 7:40, the 6x needed to come before the 1x. Why couldn't it have been the other way around??\n", + "A": "The 6x did not need to come first. You can absolutely write out the new polynomial as: 6x^2 + 1x + 6x + 1. If you finish the factoring process, you will get the same 2 binomial factors as Sal has in the video. Your intermediate steps will look a little different, but the end result will be the same. Hope this helps.", + "video_name": "X7B_tH4O-_s", + "timestamps": [ + 460 + ], + "3min_transcript": "Let's factor out the x plus 7. We get x plus 7, times 4x minus 3. Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping. And we factored it into two binomials. Let's do another example of that, because it's a little But once you get the hang of it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill. We want to find a times b that is equal to 1 times 6, which is equal to 6. And we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. What are the-- well, the obvious one is 1 and 6, right? 1 plus 6 is 7. So we have a is equal to 1. Or let me not even assign them. The numbers here are 1 and 6. Now, we want to split this into a 1x and a 6x. But we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squar ed here, plus-- and so I'm going to put the 6x first because 6 and 6 share a factor. And then, we're going to have plus 1x, right? 6x plus 1x equals 7x . That was the whole point. They had to add up to 7 . And then we have the final plus 1 there. Now, in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times-- 6x squar ed divided by 6x is just an x. 6x divided by 6x is just a 1. to have a plus here. But this second group, we just literally have a x plus 1. Or we could even write a 1 times an x plus 1. You could imagine I just factored out of 1 so to speak. Now, I have 6x times x plus 1, plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. And now, I'm going to actually explain why this little magical system actually works. Let me take an example. I'll do it in very general terms. Let's say I had ax plus b, times cx-- actually, I'm" + }, + { + "Q": "At, 1:55, if it is a*b, won't it be 4*25, instead of 4*-21?\n", + "A": "In this case a and b are not from the classical form a x\u00c2\u00b2 + b x + c, but rather two independent variables to use the method. You could call them p and q if thats easier for you. Or in short: For a x\u00c2\u00b2 + b x + c you search p and q such as p * q = a * c and p + q = b As Sal didn t use the a, b, c in the equation, he just chose them as variables.", + "video_name": "X7B_tH4O-_s", + "timestamps": [ + 115 + ], + "3min_transcript": "In this video, I want to focus on a few more techniques for factoring polynomials. And in particular, I want to focus on quadratics that don't have a 1 as the leading coefficient. For example, if I wanted to factor 4x squared plus 25x minus 21. Everything we've factored so far, or all of the quadratics we've factored so far, had either a 1 or negative 1 where this 4 is sitting. All of a sudden now, we have this 4 here. So what I'm going to teach you is a technique called, factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick. To some degree, it'll become obsolete once you learn the quadratic formula, because, frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique. And then at the end of this video, I'll actually show you why it works. So what we need to do here, is we need to think of two numbers, a and b, where a times b is equal 4 times So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a plus b, need to be equal to 25. Let me be very clear. This is the 25, so they need to be equal to 25. This is where the 4 is. So we go, 4 times negative 21. That's a negative 21. So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17. Doesn't work. Let's try some other combinations. 1 and 84, too far apart when you take their difference. Because that's essentially what you're going to do, if one is negative and one is positive. Too far apart. Let's see you could do 3-- I'm jumping the gun. 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40. 2 plus negative 42 is negative 40-- too far apart. 3 and-- Let's see, 3 goes into 84-- 3 goes into 8 2 times. 2 times 3 is 6. 8 minus 6 is 2. Bring down the 4. Goes exactly 8 times. So 3 and 28. This seems interesting. And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25." + }, + { + "Q": "\nSal mentions the Z-Score table at 12:56. How were the values on this table calculated - where did they come from?", + "A": "What could have been made more clear is that Sal is looking for the probability for X to be between 0 and up to 2+(2.02*0.099)=2.19998 liters of water. Since we know that it is a normal distribution we can write: X is N(2 , 0.099) and we look for P(0X and f from X-->Y shouldn't the composition of these functions be a mapping from Y-->Y?\n", + "A": "gof = g(f) maps from the domain of f (= X) to the codomain of f (= Y), and g maps from Y back to X (first you do f, then g - right to left ).", + "video_name": "-eAzhBZgq28", + "timestamps": [ + 734 + ], + "3min_transcript": "we're going to start dealing with these notions with transformations and matrices in the very near future. So it's good to be exposed to it in this more precise form. Now the first thing you might ask is let's say that I have a function f, and there does exist a function f inverse that satisfies these two requirements. So f is invertible. The obvious question, or maybe it's not an obvious question is, is f inverse unique? Actually probably the obvious question is how do you know when something's invertible. We're going to talk a lot about that in the very near future. But let's say we know that f is invertible. How do we know, or do we know whether f inverse is unique? To answer that question, let's assume it's not unique. that satisfy our two constraints that can act as inverse functions of f. Let's say that g is one of them. So let's say g is a mapping. Remember f is a mapping from X to Y. Let's say that g is a mapping from Y to X such that if I apply f to something and then apply g to it-- so this gets me from X to Y. Then when I do the composition with g, that gets me back into X. This is equivalent to the identity function. This was part of the definition of what it means to be an inverse. I'm assuming that g is an inverse of f. Now let's say that h is another inverse of f. By definition, by what I just called an inverse, h has to satisfy two requirements. It has to be a mapping from Y to X. Then if I take the composition of h with f, I have to get the identity matrix on the set X. Now that wasn't just part of the definition. It implies even more than that. If something is an inverse, it has to satisfy both of these. The composition of the inverse with the function has to become the identity matrix on x. Then the composition of the function with the inverse has to be the identity function on Y. Let's write that. So g is an inverse of f. It implies this. It also implies-- I'll do it in yellow-- that the" + }, + { + "Q": "At 6:33 what is a codomain?\n", + "A": "If the range of f (the set of all the y such that y = f(x)) is a subset of the real numbers, then the real numbers might be a codomain.", + "video_name": "-eAzhBZgq28", + "timestamps": [ + 393 + ], + "3min_transcript": "function, f inverse, that's a mapping from Y to X such that if I take the composition of f inverse with f, this is equal to the identity function over X. So let's think about what's happening. This is just part of it actually. Let me just complete the whole definition. This is true, this has to be true, and the composition of f with the inverse function has to be equal to the identity function over Y. So let's just think about what's this saying. There's some function-- I'll call it right now, this called the inverse of f-- and it's a mapping from Y to X. So f is a mapping from X to Y. We showed that. This is the mapping of f right there. It goes in that direction. We're saying there has be some other function, f inverse, that's a mapping from Y to X. So let's write it here. So f inverse is a mapping from Y to X. f inverse, if you give me some value in set Y, I go to set X. So this guy's domain is this guy's codomain, and this guy's codomain is this guy's domain. Fair enough. But let's see what it's saying. It's saying that the composition of f inverse with f, has to be equal to the identity matrix. So essentially it's saying if I apply f to some value in X-- right, if you think about what's this composition doing-- this guy's going from X to Y. This guy goes from Y to X. f is going from X to Y. Then f inverse is going from Y to X. So this composition is going to be a mapping from X to X, which the identity function needs to do. It needs to go from X to X. They're saying this equals the identity function. So that means when you apply f on some value in our domain, so you go here, and then you apply f inverse to that point over there, you go back to this original point. So another way of saying this is that f-- let me do it in another color-- the composition of f inverse with f of some member of the set X is equal to the identity" + }, + { + "Q": "At 2:40 he gives the number 0.714141414\nI fully understand the answer that he gives to solve this problem.\nBut it seems to me that the repeating section is 14, not 41.\nWhy doesn't 100x - x work?\n", + "A": "it depends whether you want to use the repeating section 14 or 41 because the repeating period or cycle is just every two digits. Some numbers like 1/23, 1/4093 or 1/8779 have repeating cycles of 22 digits", + "video_name": "Ihws0d-WLzU", + "timestamps": [ + 160 + ], + "3min_transcript": "So 100x is going to be equal to-- the decimal is going to be there now, so it's going to be 36.363636 on and on and on forever. And then let me rewrite x over here. We're going to subtract that from the 100x. x is equal to 0.363636 repeating on and on forever. And notice when we multiplied by 100x, the 3's and the 6's still line up with each other when we align the decimals. And you want to make sure you get the decimals lined up appropriately. And the reason why this is valuable is now that when we subtract x from 100x, the repeating parts will cancel out. So let's subtract. Let us subtract these two things. So on the left-hand side, we have 100x minus x. So that gives us 99x. And then we get, on the right-hand side, this part cancels out with that part. And we're just left with 36. are left with x is equal to 36 over 99. And both the numerator and the denominator is divisible by 9, so we can reduce this. If we divide the numerator by 9, we get 4. The denominator by 9-- we get 11. So 0.363636 forever and forever repeating is 4/11. Now let's do another interesting one. Let's say we have the number 0.714, and the 14 is repeating. And so this is the same thing. So notice, the 714 isn't going to repeat. Just the 14 is going to repeat. So this is 0.7141414, on and on and on and on. So let's set this equal to x. the decimal all the way clear of 714. But you actually don't want to do that. You want to shift it just enough so that the repeating pattern can be right under itself when you do the subtraction. So again in this situation, even though we have three numbers behind the decimal point, because only two of them are repeating, we only want to multiply by 10 to the second power. So once again, you want to multiply by 100. So you get 100x is equal to-- we're moving the decimal two to the right, one, two-- so it's going to be 71.4141, on and on and on and on. So it's going to be 71.4141414 and on and on and on. And then let me rewrite x right below this. We have x is equal to 0.7141414. And notice, now the 141414's, they're" + }, + { + "Q": "At 5:05 can you just multiply by 10 every time your numerator is a decimal?\n", + "A": "Yes, but only if the decimal stops at the tenths place...if it goes to the hundredths place, multiply by 100. If it goes to the thousandths place, multiply by 1000, and so on...", + "video_name": "Ihws0d-WLzU", + "timestamps": [ + 305 + ], + "3min_transcript": "the decimal all the way clear of 714. But you actually don't want to do that. You want to shift it just enough so that the repeating pattern can be right under itself when you do the subtraction. So again in this situation, even though we have three numbers behind the decimal point, because only two of them are repeating, we only want to multiply by 10 to the second power. So once again, you want to multiply by 100. So you get 100x is equal to-- we're moving the decimal two to the right, one, two-- so it's going to be 71.4141, on and on and on and on. So it's going to be 71.4141414 and on and on and on. And then let me rewrite x right below this. We have x is equal to 0.7141414. And notice, now the 141414's, they're So it's going to work out when we subtract. So let's subtract these things. 100x minus x is 99x. And this is going to be equal to-- these 1414's are going to cancel with those 1414's. And we have 71.4 minus 0.7. And we can do this in our head, or we can borrow if you like. This could be a 14. This is a 0. So you have 14 minus 7 is 7 and then 70 minus 0. So you have 99x is equal to 70.7. And then we can divide both sides by 99. And you could see all of the sudden something strange is happening because we still have a decimal. But we can fix that up at the end. So let's divide both sides by 99. You get x is equal to 70.7 over 99. Now obviously, we haven't converted this into a pure fraction yet. We still have a decimal in the numerator. But that's pretty easy to fix. to get rid of this decimal. So let's multiply the numerator by 10 and the denominator by 10. And so we get 707/990. Let's do one more example over here. So let's say we had something like-- let me write this way-- 3.257 repeating, and we want to convert this into a fraction. So once again, we set this equal to x. And notice, this is going to be 3.257257257. The 257 is going to repeat on and on and on. Since we have three digits that are repeating, we want to think about 1,000x, 10 to the third power times x. And that'll let us shift it just right so that the repeating parts can cancel out. So 1,000x is going to be equal to what?" + }, + { + "Q": "\nAt 4:25 I was confused on why it was changed to p^2-17p+4 why is it plus 4?", + "A": "This is Algebra II. The process of going from equation 1 to equation involves multiplying by 4. This is one of the most common operations in Algebra. If the process confuses you, consider reviewing the courses of Algebra that involve solving quadratic equations.", + "video_name": "GDppV18XDCs", + "timestamps": [ + 265 + ], + "3min_transcript": "in a row, of getting anything else twice in a row? Well, you're just going to multiply this probability times itself. It's going to be 1 minus p times 1 minus p, or we could just write that as 1 minus p squared. So this right over here is the probability of getting a different candy, any other candy, twice in a row. So prob of any other non-\"Honey Bunny\" candy, any other candy, twice in a row. Now, they tell us that this probability needs to be greater than 2 and 1/4 times the probability of getting \"Honey Bunny\" in one try. So is greater than 2 and 1/4 times the probability of getting \"Honey Bunny\" in one So we have just set up the first part. We have written an inequality that models the situation. Now let's actually solve this inequality. And so to do that, I will just expand 1 minus p squared out. 1 minus p squared is the same thing as-- well, I'll just multiply it out. So this is going to be 1 squared minus 2p plus p squared. And that's going to be greater than 2 and 1/4 p. Now let's see. If we subtract 2 and 1/4 p from both sides, we're going to be left with-- and I'm going to reorder this. We're going to get p squared. So you have minus 2p minus 2 and 1/4 p, so that's going to get us minus 4 and 1/4 p. greater than 0. And so let's think about solving this quadratic right over here. And under which circumstances is this greater than 0? To think about it, let's factor it. And actually, before we factor it, let's simplify it a little bit. I don't like having this 17/4 right over here, so let's multiply both sides times 4. And since 4 is a positive number, it's not going to change the sign, the direction of this inequality. So we could rewrite this as 4p squared minus 17p plus 4 is greater than 0. What are the roots of this? And we could use the quadratic formula if we wanted to do it really quick. We could probably do it other ways." + }, + { + "Q": "\nI don't understand the braiding thing Vi does at 1:33. Someone please help!", + "A": "All she does is put 2 like an 8, and weaves the other, whoch she cut in half, into the others.", + "video_name": "4tsjCND2ZfM", + "timestamps": [ + 93 + ], + "3min_transcript": "So say your vector field green bean casserole is in the oven, and now it's time to think about a nice, crispy onion topping. Normal people might just use, for instance, French's French fried onions in a can, put super awesome people use a real French person, and real fresh onions, to make their own fresh onion toroids. And they fry free linked with the Brunnian property to get Borromean onion rings. The Borromean rings show up in many forms, they come flat and in 3D, round, rectangly, triangly. But, the important thing is not the way the rings appear, but the way they are connected to each other. The thing about the Borromean rings is that no two of the rings are actually linked together. Ignore the pink and look at just the green and brown. They're sitting on top of each other, not linked. And if you just look at the green and pink, or pink and brown, it's the same thing. And yet, all three together are linked inseparably. So to make your Borromean rings out of onion rings, you will have to cut one of your rings and then fasten it back together with a toothpick or something, which can be removed after frying. Or you can use the fourth dimension. And luckily I have a four-dimensional guest to help me out. If you're stuck in three dimensions, you can think of it like this. Now, the third ring which I have cut, is going to go outside of the outside ring, but inside of the inside ring. Each ring is wholly out of, and wholly inside of the other two rings so that no two are linked, but all three are. You can also think of laying two on each other flat, one on top of the other. And then having the third weave through them, so that it goes over the one on top, and under the one on bottom. The result can be made to be flatter or more spherical, in some you can see the relationship that Borromean rings have with braids. Sure the orange, yellow, and red ribbons are all twisted together, but no two strands are twisted together. If I pull out the orange one, the other two fall apart. Some people and cultures and stuff think of this togetherness property as a metaphor for unity. So when you eat Borromean onion rings, you get to feel all deep and symbolic. But don't forget to save enough to put on top of your green bean matherole. And there we go. At this point I've got a gelatinous cranberry cylinder, bread spheres with butter prism, masked potatoes, a vector field green bean matherole with Borromean onion rings, All I need is a double helix cut ham, and of course, the crowning glory of this feast which I will tell you about next time." + }, + { + "Q": "who's she cooking with at 0:11?\n", + "A": "Marc ten Bosch, the 4 dimensional Frenchman who invented the Borromean Onion Rings.", + "video_name": "4tsjCND2ZfM", + "timestamps": [ + 11 + ], + "3min_transcript": "So say your vector field green bean casserole is in the oven, and now it's time to think about a nice, crispy onion topping. Normal people might just use, for instance, French's French fried onions in a can, put super awesome people use a real French person, and real fresh onions, to make their own fresh onion toroids. And they fry free linked with the Brunnian property to get Borromean onion rings. The Borromean rings show up in many forms, they come flat and in 3D, round, rectangly, triangly. But, the important thing is not the way the rings appear, but the way they are connected to each other. The thing about the Borromean rings is that no two of the rings are actually linked together. Ignore the pink and look at just the green and brown. They're sitting on top of each other, not linked. And if you just look at the green and pink, or pink and brown, it's the same thing. And yet, all three together are linked inseparably. So to make your Borromean rings out of onion rings, you will have to cut one of your rings and then fasten it back together with a toothpick or something, which can be removed after frying. Or you can use the fourth dimension. And luckily I have a four-dimensional guest to help me out. If you're stuck in three dimensions, you can think of it like this. Now, the third ring which I have cut, is going to go outside of the outside ring, but inside of the inside ring. Each ring is wholly out of, and wholly inside of the other two rings so that no two are linked, but all three are. You can also think of laying two on each other flat, one on top of the other. And then having the third weave through them, so that it goes over the one on top, and under the one on bottom. The result can be made to be flatter or more spherical, in some you can see the relationship that Borromean rings have with braids. Sure the orange, yellow, and red ribbons are all twisted together, but no two strands are twisted together. If I pull out the orange one, the other two fall apart. Some people and cultures and stuff think of this togetherness property as a metaphor for unity. So when you eat Borromean onion rings, you get to feel all deep and symbolic. But don't forget to save enough to put on top of your green bean matherole. And there we go. At this point I've got a gelatinous cranberry cylinder, bread spheres with butter prism, masked potatoes, a vector field green bean matherole with Borromean onion rings, All I need is a double helix cut ham, and of course, the crowning glory of this feast which I will tell you about next time." + }, + { + "Q": "\nat 1:01 , i do not get why it is 20-6, when it is 200-60", + "A": "because you can take away a 0 from the end of the numbers, the you can add a zero to the answer i.e. 200-6 20-6 = 14 which becomes 140", + "video_name": "iivtjjdSu9I", + "timestamps": [ + 61 + ], + "3min_transcript": "What I want to do in this video is show you a way of subtracting numbers that is different than the regrouping technique. And this is closer to what I actually do in my head. And this might not be what you see in school, so be careful while you're looking at this. Some people might not fully approve of what I'm about to show you. But the idea, the reason why I'm showing you this is to understand that there's not just one way to do things. As long as you understand the underlying principles, what these numbers represent, and you do things that make sense, you should be OK. And what's neat about this technique is that we don't have to regroup. We're just going to start with the hundreds place and keep on moving. So, for example, if I think of 301, I first want to subtract 100. So 300 minus 100 is going to get me to 200. Now I need to subtract 60. And so I can think about what is 20-- two, zero-- minus 6. So this is essentially saying what is 200 minus 60? Well, 20 minus 6 is 14. So I just subtract, and now I'm left with 14. So I really just have to think about what-- well, or I could do what 141 minus 9 is. Well, 141, it's a little bit more mental computation than you might be used to. But 141 minus 9 is going to be 132. So we are left with 132. Let's do this one, same technique. And I encourage you to pause the video and try it yourself, this same technique. Well, 9 minus 2, it's really 900 minus 200. You're going to be left with 700. Then 71 minus 8, let's see, 11 minus 8 is 13. So you're going to be left with 63. And now we have to subtract 633 minus 6. 13 minus 6 is 7. So it's going be 627. And once again, try to pause the video and do it on your own. go to 72 minus 8, which is really 720 minus 80. But let's just think in terms of 72 minus 8. Well, 12 minus 8 is 4, so 72 minus 8 is going to be 64. So now this is the same thing as 641 minus 8. Well, 41 minus 8 is going to be 33. I have to do a little bit of mental computation here. So this will result in 633." + }, + { + "Q": "\nat 2:37, why did Sal use the phrase \"Not to beat a dead horse\"? Is it a simile, metaphor, or WHAT? Also, i like horses. I am very much offended.", + "A": "It means not to belabor the point. Yes, it is a figurative phrase.", + "video_name": "AuD2TX-90Cc", + "timestamps": [ + 157 + ], + "3min_transcript": "We could say, and one tenth and five hundredths, or we could just say, look, this is fifteen hundredths. One tenth is ten hundredths. So one tenth and five hundredths is fifteen hundredths. So maybe I can write it like this: sixty-three and fifteen hundredths. Just like that. Now, it might have been a little bit more natural to say, how come I don't say one tenth and then five And you could, but that would just make it a little bit harder for someone's brain to process it when you say it. So it could have been sixty-three-- so let me copy and paste that. It could be sixty-three and, and then you would write, one Sixty-three and one tenth and five hundredths is hard for most people's brains to process. But if you say, fifteen hundredths, people get what you're saying. Not to beat a dead horse, but this right here, this is 1/10 right here and then this is 5/100, 5 over 100. But if you were to add these two, If you were to add 1/10 plus 5/100 -- so let's do that. If you were to add 1/10 plus 5/100, how would you do it? You need a common denominator. numerator and denominator of this character by 10. You get 10 on the top and 100 on the bottom. 1/10 is the same thing as 10 over 100. 10/100 plus 5/100 is equal to 15 over 100, so this piece right here is equal to 15/100. And that's why we say sixty-three and fifteen hundredths." + }, + { + "Q": "What does beat a dead horse mean? Because at 2:36, Sal said, \"Not to beat a dead horse...\"\n", + "A": "Beat a dead horse means, Keep doing what we have done enough of. , or Waste energy doing something that useless.", + "video_name": "AuD2TX-90Cc", + "timestamps": [ + 156 + ], + "3min_transcript": "We could say, and one tenth and five hundredths, or we could just say, look, this is fifteen hundredths. One tenth is ten hundredths. So one tenth and five hundredths is fifteen hundredths. So maybe I can write it like this: sixty-three and fifteen hundredths. Just like that. Now, it might have been a little bit more natural to say, how come I don't say one tenth and then five And you could, but that would just make it a little bit harder for someone's brain to process it when you say it. So it could have been sixty-three-- so let me copy and paste that. It could be sixty-three and, and then you would write, one Sixty-three and one tenth and five hundredths is hard for most people's brains to process. But if you say, fifteen hundredths, people get what you're saying. Not to beat a dead horse, but this right here, this is 1/10 right here and then this is 5/100, 5 over 100. But if you were to add these two, If you were to add 1/10 plus 5/100 -- so let's do that. If you were to add 1/10 plus 5/100, how would you do it? You need a common denominator. numerator and denominator of this character by 10. You get 10 on the top and 100 on the bottom. 1/10 is the same thing as 10 over 100. 10/100 plus 5/100 is equal to 15 over 100, so this piece right here is equal to 15/100. And that's why we say sixty-three and fifteen hundredths." + }, + { + "Q": "what does Sal mean at 2:35 \"not to beat a dead horse\"?\n", + "A": "The phrase means to keep doing or saying something over and over even after it has become pointless .", + "video_name": "AuD2TX-90Cc", + "timestamps": [ + 155 + ], + "3min_transcript": "We could say, and one tenth and five hundredths, or we could just say, look, this is fifteen hundredths. One tenth is ten hundredths. So one tenth and five hundredths is fifteen hundredths. So maybe I can write it like this: sixty-three and fifteen hundredths. Just like that. Now, it might have been a little bit more natural to say, how come I don't say one tenth and then five And you could, but that would just make it a little bit harder for someone's brain to process it when you say it. So it could have been sixty-three-- so let me copy and paste that. It could be sixty-three and, and then you would write, one Sixty-three and one tenth and five hundredths is hard for most people's brains to process. But if you say, fifteen hundredths, people get what you're saying. Not to beat a dead horse, but this right here, this is 1/10 right here and then this is 5/100, 5 over 100. But if you were to add these two, If you were to add 1/10 plus 5/100 -- so let's do that. If you were to add 1/10 plus 5/100, how would you do it? You need a common denominator. numerator and denominator of this character by 10. You get 10 on the top and 100 on the bottom. 1/10 is the same thing as 10 over 100. 10/100 plus 5/100 is equal to 15 over 100, so this piece right here is equal to 15/100. And that's why we say sixty-three and fifteen hundredths." + }, + { + "Q": "I might be asking a question explained in another video, but at 0:47, he said that, \"We have two angles (angle CDE and angle ABC) that are congruent.\" Why is that?\n\n(I am young in learning this)\n", + "A": "In geometry, the alternate interior angle rule says that those angles are congruent. Try searching for that rule.", + "video_name": "R-6CAr_zEEk", + "timestamps": [ + 47 + ], + "3min_transcript": "In this first problem over here, we're asked to find out the length of this segment, segment CE. And we have these two parallel lines. AB is parallel to DE. And then, we have these two essentially transversals that form these two triangles. So let's see what we can do here. So the first thing that might jump out at you is that this angle and this angle are vertical angles. So they are going to be congruent. The other thing that might jump out at you is that angle CDE is an alternate interior angle with CBA. So we have this transversal right over here. And these are alternate interior angles, and they are going to be congruent. Or you could say that, if you continue this transversal, you would have a corresponding angle with CDE right up here and that this one's just vertical. Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles and two of the corresponding angles are the same. And that by itself is enough to establish similarity. We actually could show that this angle and this angle but we don't have to. So we already know that they are similar. And actually, we could just say it. Just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar, even before doing that. So we already know that triangle-- I'll color-code it so that we have the same corresponding vertices. And that's really important-- to know what angles and what sides correspond to what side so that you don't mess up your, I guess, your ratios or so that you do know what's corresponding to what. So we know triangle ABC is similar to triangle-- so this vertex A corresponds to vertex E over here. It's similar to vertex E. And then, vertex B right over here corresponds to vertex D. EDC. Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it in just the way that we've written down the similarity. If this is true, then BC is the corresponding side to DC. So we know that the length of BC over DC right over here is going to be equal to the length of-- well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to-- what's the corresponding side to CE? The corresponding side over here is CA." + }, + { + "Q": "\nat 4:26, wouldn't Aij be the determinate of the submatrix, not the submatrix itself?", + "A": "No, Sal defined Aij to be the submatrix itself.", + "video_name": "H9BWRYJNIv4", + "timestamps": [ + 266 + ], + "3min_transcript": "So this is going to be an n minus 1 by n minus 1 matrix. So if this is 7 by 7, the submatrix is going to be 6 by 6, one less in each direction. So this is going to be the n minus 1 by n minus 1 matrix you get if you essentially ignore or if you take away-- maybe I should say take away. Let's say ignore, like the word ignore. If you ignore the i-th row, this right here is the row, the i-th row and the j-th column of a. So, for example, let's go back to our 3 by 3 right here. This thing could be denoted based on that definition I-- this term right here. We could denote the matrix when you get rid of the first column and the first row or the first row and the first column, we could call this thing right here, we could call that big matrix A11. So this was big matrix A11. This is big matrix A21, or actually, this matrix was called C, so this would be C11 right there. We could call this one, this would be matrix C12. Because if you get rid of the first row, let me get rid of the first row, right? The first term is your row. If you get rid of the first row and the second column, this is the matrix that's left over: 2, 3, 4, 1. So this is this guy and this guy. 2, 3, 4, 1. C, But this one is C12. I know it's a little bit messy there. So that's all we mean by the submatrix. Very similar to what we did in the 3 by 3 case. You essentially get rid of-- so if you want to find out this guy's submatrix, you would call that a11, and you would literally cross out the first row and the first column, and everything left over here would be that submatrix. Now, with that out of the way, we can create a definition, and it might seem a little circular to you at first, and on some level it is. We're going to define the determinant of A to be equal to-- this is interesting. It's actually a recursive definition. I'll talk about that in a second. It's equal to-- we start with a plus. It's equal to a11 times the submatrix if you remove this guy's row and column. So that, by definition, is just A, big capital A11's" + }, + { + "Q": "At 6:26, we get the same answer by treating the question as ratios. That is, 0.6:0.7 as 0.5:x. Is the reasoning sound?\n", + "A": "Yup, because ratios are just fractions, so in the end everything will work out; your reasoning is sound.", + "video_name": "6xPkG2pA-TU", + "timestamps": [ + 386, + 360 + ], + "3min_transcript": "Well, we know a lot of this information. We know the probability of A given B is 0.7. So let me write that, I'll scroll down a little bit. This is 0.7. We know that the probability of B is 0.5. So this is 0.5. So we know that the probability of A and B is the product of these two things. That's going to be 0.35. Seven times five is 35 or, I guess you could say, half of .7 is 0.35. .5 of .7. And that is going to be equal to what we need to figure out. Probability of B given A times probability of A. But we know probability of A. We know that that is 0.6. We know that this is 0.6. So just like that, we've set up a situation, an equation, where we can solve for the probability of B given A. The probability of B given A. Notice, let me just rewrite it right over here. 0.6, 0.6 times the probability of B given A. Times that, right over there. And I'll just copy and paste it so I don't have to keep changing colors. That, over there, is equal to 0.35. Is equal to 0.35. And so to solve for the probability of B given A, we can just divide both sides by 0.6. 0.6, 0.6 and we get the probability of B given A is equal to ... Let me get our calculator out. So 0.35 divided by, divided by 0.6 and we deserve a little bit of a drum roll here, is .5833 ... It keeps going. They tell us to round to the nearest hundredth. So it's 0.58. So notice, this is equal to 0 ... or I'll say approximately equal to 0.58. Once again, verifying that these are dependent. The probability that B happens given A is true, is higher than just the probability that B by itself, or without knowing anything else. Just the probability of B is lower than the probability of B given that you know, given that you know A has happened, or event A is true. And we're done." + }, + { + "Q": "\nAt 1:30 , Sal wrote down cos 58 with no parentheses. Usually he uses parentheses when he has a degree measure. What is the correct way to do it?", + "A": "It s better to write it using parentheses,", + "video_name": "yiH6GoscimY", + "timestamps": [ + 90 + ], + "3min_transcript": "We are told that the cosine of 58 degrees is roughly equal to 0.53. And that's roughly equal to, because it just keeps going on and on. I just rounded it to the nearest hundredth. And then we're asked, what is the sine of 32 degrees? And I encourage you to pause this video and try it on your own. And a hint is to look at this right triangle. One of the angles is already labeled 32 degrees. Figure out what all of the angles are, and then use the fundamental definitions, your sohcahtoa definitions, to see if you can figure out what sine of 32 degrees is. So I'm assuming you've given a go at it. Let's work it through now. So we know that the sum of the angles of a triangle add up to 180. Now in a right angle, one of the angles is 90 degrees. So that means that the other two must add up to 90. These two add up to 90 plus another 90 is going to be 180 degrees. Or another way to think about is that the other two non-right angles are going to be complementary. So what plus 32 is equal to 90? So this right over here is going to be 58 degrees. Well, why is that interesting? Well, we already know what the cosine of 58 degrees is equal to. But let's think about it in terms of ratios of the lengths of sides of this right triangle. Let's just write down sohcahtoa. Soh, sine, is opposite over hypotenuse. Cah, cosine, is adjacent over hypotenuse. Toa, tangent, is opposite over adjacent. So we could write down the cosine of 58 degrees, which we already know. If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse. This is a 58 degree angle. The side that is adjacent to it is-- let me do it in this color-- is side BC right over here. It's one of the sides of the angle, the side of the angle that is not the hypotenuse. The other side, this over here, is a hypotenuse. So this is going to be the adjacent, the length of the adjacent side, BC, over the length of the hypotenuse. Now let's think about what the sine of 32 degrees would be. Well, sine is opposite over hypotenuse. So now we're looking at this 32 degree angle. What side is opposite it? Well, it opens up onto BC. And what's the length of the hypotenuse? It's AB. Notice, the sine of 32 degrees is BC over AB. The cosine of 58 degrees is BC over AB. Or another way of thinking about it, the sine of this angle is the same thing as the cosine of this angle. So we could literally write the sine-- I want to do that in that pink color-- the sine of 32 degrees" + }, + { + "Q": "at 5:04 it confuses me that the two sides of a seemingly isosceles triangle to have different angles at both acute sides is that impossible or am i mistaken\n", + "A": "it may look like an isosceles triangle, but it is not. If the 2 seemingly equal sides were in fact of equal length, the 2 angles would have each been 30\u00c2\u00b0, instead of 25\u00c2\u00b0 and 35\u00c2\u00b0.", + "video_name": "D5lZ3thuEeA", + "timestamps": [ + 304 + ], + "3min_transcript": "I want to make it a little bit more obvious. So let's say a triangle like this. If this angle is 60 degrees, maybe this one right over here is 59 degrees. And then this angle right over here is 61 degrees. Notice they all add up to 180 degrees. This would be an acute triangle. Notice all of the angles are less than 90 degrees. A right triangle is a triangle that has one angle that is exactly 90 degrees. So for example, this right over here would be a right triangle. Maybe this angle or this angle is one that's 90 degrees. And the normal way that this is specified, people wouldn't just do the traditional angle measure and write 90 degrees here. They would draw the angle like this. And that tells you that this angle right over here is 90 degrees. And because this triangle has a 90 degree angle, and it could only have one 90 degree angle, this is a right triangle. So that is equal to 90 degrees. Now you could imagine an obtuse triangle, based on the idea that an obtuse angle is larger than 90 degrees, an obtuse triangle is a triangle that has one angle that is larger than 90 degrees. So let's say that you have a triangle that looks like this. Maybe this is 120 degrees. And then let's see, let me make sure that this would make sense. Maybe this is 25 degrees. Or maybe that is 35 degrees. And this is 25 degrees. Notice, they still add up to 180, or at least they should. But the important point here is that we have an angle that is a larger, that is greater, than 90 degrees. Now, you might be asking yourself, hey Sal, can a triangle be multiple of these things. Can it be a right scalene triangle? Absolutely, you could have a right scalene triangle. In this situation right over here, actually a 3, 4, 5 triangle, a triangle that has lengths of 3, 4, and 5 actually is a right triangle. And this right over here would be a 90 degree angle. You could have an equilateral acute triangle. In fact, all equilateral triangles, because all of the angles are exactly 60 degrees, all equilateral triangles are actually acute. So there's multiple combinations that you could have between these situations and these situations right over here." + }, + { + "Q": "At 3:13 you are multiplying the results by -1.....where did the \"-1\" come from?\n", + "A": "Sal multiplied both sides of the equation by -1 because it was in the form -p = -17. We are trying to solve for p not for negative p. The fastest way to transform one side of an equation from negative to positive or vice versa is to multiply both sides by -1 because only the sign changes and not the value.", + "video_name": "bRwJ-QCz9XU", + "timestamps": [ + 193 + ], + "3min_transcript": "And I'm going to rewrite it over here. So we have 4 over p minus 1 is equal to 5 over p plus 3. So the first thing we could do, especially because we can assume now that neither of these expressions And this is going to be defined, since we've excluded these values of p, is to get the p minus 1 out of the denominator. We can multiply the left hand side by p minus 1. But remember, this is an equation. If you want them to continue to be equal, anything you do left hand side, you have to do to the right hand side. So I'm multiplying by p minus 1. Now I also want to get this p plus 3 out of the denominator here on the right hand side. So the best way to do that is multiply the right hand side by p plus 3. But if I do that to the right hand side, I also have to do that to the left hand side. p plus 3. And so what happens? We have a p minus 1 in the numerator, p minus 1 at the denominator. These cancel out. So you have just a 1 of the denominator, or you have no denominator anymore. And the left hand side simplifies to distribute the 4, 4 times p plus 3. So that is 4p plus 12. And then the right hand side, you have plus 3 canceling with a p plus 3. This is p plus 3 divided by p plus 3. And all you're left with is 5 times p minus 1. If you distribute the 5 you get 5p minus 5. And now this is a pretty straightforward linear equation to solve. We just want to isolate the p's on one side and the constants on the other. So let's subtract 5p from both sides. I'll switch colors. So let's subtract 5p from both sides. And we get on the left hand side, 4p minus 5p is negative p plus 12. Is equal to, these cancel out, is equal to negative 5. And then we could subtract 12 from both sides. negative 5 minus 12 is negative 17. And we're almost done. We can multiply both sides by negative 1 or divide both sides by negative 1 depending on how you want to view it. And we get, negative one times negative p is. So let me just scroll down a little bit so I have a little bit of real estate. That's positive p is equal to 17. p is equal to 17. And let's verify that this really works. Well, it wasn't one of our excluded values, but just in case, let's verify that it really works. If we go, if we have p is 17. We get 4 over 17 minus 1, needs to be equal to 5 over 17 plus 3. I'm just putting 17 in for p, because that's our solution. So this is the same thing as 4 over 16, needs to be the same thing as 5 over 20." + }, + { + "Q": "At 1:15, why does Sal write a 16 instead of a 6?\n", + "A": "That s bracket start (6 months) .", + "video_name": "BKGx8GMVu88", + "timestamps": [ + 75 + ], + "3min_transcript": "Let's say that you are desperate for a dollar. So you come to me the local loan shark, and you say hey I need to borrow a dollar for a year. I tell you I'm in a good mood, I willing to lend you that dollar that you need for a year. I will lend it to you for the low interest of 100% per year. 100% per year. How much would you have to pay me in a year? You're going to have to pay the original principal what I lent you plus 100% of that. Plus one other dollar. Which is clearly going to be equal to $2. You say oh gee, that's a lot to have to pay to pay back twice what I borrowed. There's a possibility that I might have the money in 6 months. What kind of a deal could you get me for that Mr. Loanshark? I say oh gee, if your willing to pay back in 6 months, then I'll just charge you half the You borrow one dollar, so in 6 months, I will charge you 50% interest. 50% interest over 6 months. This, of course, was 1 year. How much would you have to pay? Well, you would have to pay the original principal what you borrowed. The one dollar plus 50% of that one dollar. Plus 0.50, and that of course is equal to 1.5. That is equal to $1.50. I'll just write it like this. $1.50. Now you say well gee that's I guess better. What happens if I don't have the money then? If I still actually need a year. We actually have a system for that. What I'll do is just say that okay, you don't have the money for me yet. I'll essentially ... we could think about it. I will just lend that amount that you need for you for another 6 months. We'll lend that out. same interest rate at 50% for the next 6 months. Then you'll owe me the principal a $1.50 plus 50% of the principal, plus 75 cents. Plus 75 cents, and that gets us to $2.25. That equals $2.25. Another way of thinking about it is to go from $1 over the first period, you just multiply that times 1.5. If your going to grow something by 50%, you just multiply it times 1.5. If your going to grow it by another 50%, you can multiply by 1.5 again. One way of thinking about it that 50% interest is the same thing as multiplying by 1.5. Multiplying by 1.5. If you start with 1 and multiply by 1.5 twice, this is going to be the same thing. $2.25 is going to be 1 multiplied by 1.5 twice." + }, + { + "Q": "\nat 3:00 ,why does the square root of 2 times the square root of 6 equals the square root of 12 ?", + "A": "Because if you multiply 2 and 6, you get 12. You wouldn t be able to combine/simplify them if you were adding, but since you are multiply, it s ok.", + "video_name": "yAH3722GrP8", + "timestamps": [ + 180 + ], + "3min_transcript": "So let's do that. So we get x squared minus the principal square root of 6 times this term-- I'll do it in yellow-- times x squared. And then we have plus this thing again. We're just distributing it. It's just like they say. It's sometimes not that intuitive because this is a big expression, but you can treat it just like you would treat a variable over You're distributing it over this expression over here. And so then we have x squared minus the principal square root of 6 times the principal square root of 2. And now we can do the distributive property again, but what we'll do is we'll distribute this x squared onto each of these terms and distribute the square root of 2 onto each of these terms. It's the exact same thing as here, it's just you could imagine writing it like this. x plus y times a is still going to be ax plus ay. as this up here, we're just switching the order of the multiplication. You can kind of view it as we're distributing from the right. And so if you do this, you get x squared times x squared, which is x to the fourth, that's that times that, and then minus x squared times the principal square root of 6. And then over here you have square root of 2 times x squared, so plus x squared times the square root of 2. And then you have square root of 2 times the square root of 6. And we have a negative sign out here. Now if you take the square root of 2-- let me do this on the side-- square root of 2 times the square root of 6, we know from simplifying radicals that this is the exact same thing as the square root of 2 times 6, or the principal square root of 12. So the square root of 2 times square root of 6, we have a negative sign out here, it becomes minus the square root of 12. You have an x to the fourth term. And then here you have-- well depending on how you want to view it, you could say, look, we have to second degree terms. We have something times x squared, and we have something else times x squared. So if you want, you could simplify these two terms over here. So I have square root of 2 x squareds and then I'm going to subtract from that square root of 6 x squareds. So you could view this as square root of 2 minus the square root of 6, or the principal square root of 2 minus the principal square root of 6, x squared. And then, if you want, square root of 12, you might be able to simplify that. 12 is the same thing as 3 times 4. So the square root of 12 is equal to square root of 3 times square root of 4. And the square root of 4, or the principal square root of 4 I should say, is 2. So the square root of 12 is the same thing as 2 square roots of 3." + }, + { + "Q": "2:10, if |r| > 1, the number will become massively huge. How can we come up with the formula \u00ce\u00a3 = a / (1 - r) if |r| > 1? Please notice that I fully understand \u00ce\u00a3 = a / (1 - r) if 0 < |r| < 1. Did I miss something there?\n", + "A": "You re right--the formula \u00ce\u00a3 = a / (1 - r) does not apply if |r| > 1. So our formula for the sum of a geometric series only applies if |r| < 1, which Sal begins to address around 2:30.", + "video_name": "b-7kCymoUpg", + "timestamps": [ + 130 + ], + "3min_transcript": "In a previous video, we derived the formula for the sum of a finite geometric series where a is the first term and r is our common ratio. What I want to do in this video is now think about the sum of an infinite geometric series. And I've always found this mildly mind blowing because, or actually more than mildly mind blowing, because you're taking the sum of an infinite things but as we see, you can actually get a finite value depending on what your common ratio is. So there's a couple of ways to think about it. One is, you could say that the sum of an infinite geometric series is just a limit of this as n approaches infinity. So we could say, what is the limit as n approaches infinity of this business, of the sum from k equals zero to n of a times r to the k. as n approaches infinity right over here. So that would be the same thing as the limit as n approaches infinity of all of this business. Let me just copy and paste that so I don't have to keep switching colors. So copy and then paste. So what's the limit as n approaches infinity here? Let's think about that for a second. I encourage you to pause the video, and I'll give you one hint. Think about it for r is greater than one, for r is equal to one, and actually let me make it clear-- let's think about it for the absolute values of r is greater than one, the absolute values of r equal to one, and then the absolute value of r less than one. Well, I'm assuming you've given a go at it. So if the absolute value of r is greater than one, as this exponent explodes, as it approaches infinity, this number is just going to become massively, And so the whole thing is just going to become, or at least you could think of the absolute value of the whole thing, is just going to become a very, very, very large number. If r was equal to one, then the denominator is going to become zero. And we're going to be dividing by that denominator, and this formula just breaks down. But where this formula can be helpful, and where we can get this to actually give us a sensical result, is when the absolute value of r is between zero and one. We've already talked about, we're not even dealing with the geometric, we're not even talking about a geometric series if r is equal to zero. So let's think about the case where the absolute value of r is greater than zero, and it is less than one. What's going to happen in that case? Well, the denominator is going to make sense, right over here. And then up here, what's going to happen? Well, if you take something with an absolute value less than one, and you take it to higher and higher and higher" + }, + { + "Q": "At 0:13 Sal talks about a vertical cut. What is a vertical cut?\n", + "A": "It means you re cutting up and down. A horizontal cut would be cutting from side to side.", + "video_name": "hoa1RBk4dTo", + "timestamps": [ + 13 + ], + "3min_transcript": "So I have a three-dimensional solid right over here. And I want to imagine what type of a shape I would get if I were to make a vertical cut. And just to refresh ourselves, or give us a sense of what a vertical cut is, imagine if this was made out of jello or something kind of fairly soft. But it's still a three-dimensional solid. And I were to make a cut-- let's say I were to make a cut right over here. So let's say I had this big, sharp metal thing-- let me draw it like that. So you have this big, sharp metal thing. Let me draw it a little bit neater. So you have this big, sharp metal thing that I'm going to cut right over here. So this is the thing that I'm going to make the cut. And I'm going to go straight down. This is a vertical cut that we're talking about. So this is the thing that I'm going to cut with. Let me make it big enough so that it can capture the shape that will result. So this thing right over here, it's right in front. And I'm going to cut-- I'm going to make it go straight you want to call it, this rectangular pyramid of jello. And what would be the resulting shape of the intersection between the jello and this thing that I'm using to cut it? And now I encourage you to pause your video and think about what the resulting shape would be. And the shape would be in two dimensions. Because this purple surface is a two-dimensional-- you could view it as part of a plane. And so where this intersects when you cut down this rectangular pyramid is the shape we're looking for. So I encourage you to pause the video and think about it or try to come up with it on your own. So let's think about it. And let me draw the rectangular pyramid again. So that's the same one. And now let me see what it would look like once I've done my cut, once I've brought this thing down. So then this is where I cut. So I cut it right over here. It'll cut along this side like that, cut along that side like that. And then it'll exit the bottom right over there. And so let me draw my whole thing. And so once I slice it down, it will look like this. My best shot at drawing it-- it will look like this. This is a vertical cut. So I've brought this thing down. And now the intersection between the thing that I'm cutting with and this pyramid is going to be this shape right over here. It cut into the top right over there. It would get all the way to the bottom right over there. And along this side, it would cut right there. And along that side, it would cut right over there. So what's the resulting shape in two dimensions of essentially the intersection between the slicer" + }, + { + "Q": "\n9:00 Does that mean, for instance, that the range of ground movement in 2011 Japan was 100 times greater than in 1989 Loma Prieta?", + "A": "Yes; you re moving from 7.0 to 9.0, a difference of 2. On the logarithmic scale we use, that s 10^2 or 100.", + "video_name": "RFn-IGlayAg", + "timestamps": [ + 540 + ], + "3min_transcript": "But remember, this is a logarithmic scale. And I encourage you to watch the videos we made on the logarithmic scale. On a logarithmic scale, a fixed distance is not a fixed amount of movement on that scale. Or a change on that scale is not a fixed linear distance. It is actually a scaling factor. And you're not scaling by 1.2 over here. You're scaling by 10 to the 1.2 power. So this is times 10 to the 1.2 power. So I'll get my calculator out right over here. And let's figure what that is. So you could imagine what it's going to be. 10 to the first power is 10. And then you have 0.2. So it's going to be, let's just do it, 10 to the 1.2 power. It's 15.8. So it's roughly 16 times stronger. So whatever shaking that was just felt on the east coast, and maybe some of you all watching this might have felt it, Loma Prieta earthquake was 16 times stronger than the earthquake-- let that we just had on the east coast. So that's a dramatic difference. Even though this caused some damage, and this is kind of shaking on a pretty good scale. Imagine 16 times as much shaking. And how much damage that would cause. I actually just met a reporter who told me that she was in her backyard during the Loma Prieta earthquake, not too far from where I live now. And she says all the cars were like jumping up and down. So it was a massive, massive earthquake. Now let's think about the Japanese earthquake. We could think about how much stronger it was than Loma Prieta. So remember, this isn't, you don't just think of this as just 2 times stronger. It is 10 to the second times stronger. And we know how to figure that out. 10 to the second power is 100. So this right over here. So cars were jumping up and down at the Loma Prieta earthquake. The Japanese earthquake was 100 times stronger. 100 times stronger than Loma Prieta. And if you compare it to the east coast earthquake, that occurred in August of 2011. So massive, massive, massive earthquake. And just to get a sense of how much stronger the Chilean earthquake was in 1960-- and just to-- there's some fascinating outcomes of the Japanese earthquake. It was estimated that Japan, just over the course of the earthquake got 13 feet wider. So this is doing something to the actual shape of a huge island. And on top of that, it's estimated that because of the shaking, and the distortions in earth caused by that shaking, that the day on earth got one millionth of a second shorter. A little over a millionth of a second shorter. So you might say, hey, that's only a millionth of a second. But I'd say, hey, look it actually changed the day of the earth. A very fundamental thing. It actually matters when people send things into space, and probes into Mars, is that they are to be able to know that our day just got a millionth of a second shorter." + }, + { + "Q": "At 2:41 Sal says 1 -3= -1\nIs that a mistake or am I just not understanding something?\n", + "A": "It said in the corner that Sal meant -2, but accidentally said -1. He just made a mistake.", + "video_name": "s4cLM0l1gd4", + "timestamps": [ + 161 + ], + "3min_transcript": "Well, it gets to y equals negative 2. So what's halfway between 4 and negative 2? Well, you could eyeball it, or you could count, or you could, literally, just take the average between 4 and negative 2. So 4-- so the midline is going to be the horizontal line-- y is equal to 4 plus negative 2 over 2. Just literally the mean, the arithmetic mean, between 4 and negative 2. The average of 4 and negative 2, which is just going to be equal to one. So the line y equals 1 is the midline. So that's the midline right over here. And you see that it's kind of cutting the function where you have half of the function is above it, and half of the function is below it. So that's the midline. Now, let's think about the amplitude. Well, the amplitude is how much this function And the midline is in the middle, so it's going to be the same amount whether you go above or below. One way to say it is, well, at this maximum point, right over here, how far above the midline is this? Well, to get from 1 to 4 you have to go-- you're 3 above the midline. Another way of thinking about this maximum point is y equals 4 minus y equals 1. Well, your y can go as much as 3 above the midline. Or you could say your y-value could be as much as 3 below the midline. That's this point right over here, 1 minus 3 is negative 1. So your amplitude right over here is equal to 3. You could vary as much as 3, either above the midline or below the midline. Finally, the period. And when I think about the period I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's a nice-- when x is at negative 2, y is it And so what I want to do is keep traveling along this curve until I get to the same y-value but not just the same y-value but I get the same y-value that I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here because, once again, y is equal to 1? You haven't completed a cycle here because notice over here where our y is increasing as x increases. Well here our y is decreasing as x increases. Our slope is positive here. Our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals 1. Or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. So that gets us to right over there." + }, + { + "Q": "\nat 4:13 in the video when he talks about periods, does the point have to reach zero in order to determine the period, or can you start from anywhere, like start from 4, 6 or anywhere on the graph?", + "A": "Period can be derived from any two points in the domain of the function that have the exact same y value and have identical slopes.", + "video_name": "s4cLM0l1gd4", + "timestamps": [ + 253 + ], + "3min_transcript": "And the midline is in the middle, so it's going to be the same amount whether you go above or below. One way to say it is, well, at this maximum point, right over here, how far above the midline is this? Well, to get from 1 to 4 you have to go-- you're 3 above the midline. Another way of thinking about this maximum point is y equals 4 minus y equals 1. Well, your y can go as much as 3 above the midline. Or you could say your y-value could be as much as 3 below the midline. That's this point right over here, 1 minus 3 is negative 1. So your amplitude right over here is equal to 3. You could vary as much as 3, either above the midline or below the midline. Finally, the period. And when I think about the period I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's a nice-- when x is at negative 2, y is it And so what I want to do is keep traveling along this curve until I get to the same y-value but not just the same y-value but I get the same y-value that I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here because, once again, y is equal to 1? You haven't completed a cycle here because notice over here where our y is increasing as x increases. Well here our y is decreasing as x increases. Our slope is positive here. Our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals 1. Or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. So that gets us to right over there. So the change in x needed to complete one cycle. That is your period. So to go from negative 2 to 0, your period is 2. So your period here is 2. And you could do it again. So we're at that point. Let's see, we want to get back to a point where we're at the midline-- and I just happen to start right over here at the midline. I could have started really at any point. You want to get to the same point but also where the slope is the same. We're at the same point in the cycle once again. So I could go-- so if I travel 1 I'm at the midline again but I'm now going down. So I have to go further. Now I am back at that same point in the cycle. I'm at y equals 1 and the slope is positive. And notice, I traveled. My change in x was the length of the period. It was 2." + }, + { + "Q": "At 4:21, why is the period not 2pi over two (simplifies to pi)? Why is it two?\n", + "A": "Why would it be 2\u00f0\u009d\u009c\u008b/2 = \u00f0\u009d\u009c\u008b? Nothing in the problem suggests that it is! The period of a function is the smallest \u00f0\u009d\u0091\u009d > 0 such that \u00f0\u009d\u0091\u0093(\u00f0\u009d\u0091\u00a5) = \u00f0\u009d\u0091\u0093(\u00f0\u009d\u0091\u00a5 + \u00f0\u009d\u0091\u009d) for all \u00f0\u009d\u0091\u00a5 \u00e2\u0088\u0088 Dom \u00f0\u009d\u0091\u0093. In this case, the function repeats every 2 units away from the point being considered, hence by definition, the period of the function is 2.", + "video_name": "s4cLM0l1gd4", + "timestamps": [ + 261 + ], + "3min_transcript": "And the midline is in the middle, so it's going to be the same amount whether you go above or below. One way to say it is, well, at this maximum point, right over here, how far above the midline is this? Well, to get from 1 to 4 you have to go-- you're 3 above the midline. Another way of thinking about this maximum point is y equals 4 minus y equals 1. Well, your y can go as much as 3 above the midline. Or you could say your y-value could be as much as 3 below the midline. That's this point right over here, 1 minus 3 is negative 1. So your amplitude right over here is equal to 3. You could vary as much as 3, either above the midline or below the midline. Finally, the period. And when I think about the period I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's a nice-- when x is at negative 2, y is it And so what I want to do is keep traveling along this curve until I get to the same y-value but not just the same y-value but I get the same y-value that I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here because, once again, y is equal to 1? You haven't completed a cycle here because notice over here where our y is increasing as x increases. Well here our y is decreasing as x increases. Our slope is positive here. Our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals 1. Or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. So that gets us to right over there. So the change in x needed to complete one cycle. That is your period. So to go from negative 2 to 0, your period is 2. So your period here is 2. And you could do it again. So we're at that point. Let's see, we want to get back to a point where we're at the midline-- and I just happen to start right over here at the midline. I could have started really at any point. You want to get to the same point but also where the slope is the same. We're at the same point in the cycle once again. So I could go-- so if I travel 1 I'm at the midline again but I'm now going down. So I have to go further. Now I am back at that same point in the cycle. I'm at y equals 1 and the slope is positive. And notice, I traveled. My change in x was the length of the period. It was 2." + }, + { + "Q": "How can you derive an already derived function (0:13) ?\nI mean, given that f(x)=2x^2, it's derivative would be f '(x)=4x.\nSo, if i understood correctly, to get the antiderivative, i would have to derive f ' (x), which is F(x)=4\nbut... thats not equal to f(x).\nI'm confused.\n", + "A": "So, if i understood correctly You didnt. If you do a deriavative of a derivative you get the sencond derivative. f(x)=2x^2 f (x)=4x f (x)=4 To get the antiderivative, you have to find a function such that its derivative is your original function. F(x) = 2/3 x^3 Do the derivative and you get back to: F (x) = f(x) = 2 x^2", + "video_name": "61ecnr8m04U", + "timestamps": [ + 13 + ], + "3min_transcript": "In the last video we looked at a function and tried to draw its derivative. Now in this video, we're going to look at a function and try to draw its antiderivative. Which sounds like a very fancy word, but it's just saying the antiderivative of a function is a function whose derivative is that function. So for example, if we have f of x, and let's say that the antiderivative of f of x is capital F of x. And this tends to be the notation, when you're talking about an antiderivative. This just means that the derivative of capital F of x, which is equal to, you could say capital f prime of x, is equal to f of x. So we're going to try to do here is, we have our f of x. And we're going to try to think about what's a possible function that this could be the derivative of? when you start looking at integral calculus. But there's actually many possible functions that this could be the derivative of. And our goal in this video is just to draw a reasonable possibility. So let's think about it a little bit. So let's, over here on the top, draw y is equal to capital F of x. So what we're going to try to draw is a function where its derivative could look like this. So what we're essentially doing is, when we go from what we're draw up here to this, we're taking the derivative. So let's think about what this function could look like. So when we look at this derivative, it says over this interval over this first interval right over here, let me do this in purple, it says over this interval from x is equal to 0 all the way to whatever value of x, this right over here, it says that the slope is a constant positive 1. So let me draw a line with a slope of a constant positive 1. And I could shift that line up and down. I will just pick a reasonable one. So I could have a line that looks something like this. I want to draw a slope of positive 1 as best as I can. So let's say it looks something like this. And I could make the function defined here or undefined here. The derivative is undefined at this point. I could make the function defined or undefined as I see fit. This will probably be a point of discontinuity on the original function. It doesn't have to be, but I'm just trying to draw a possible function. So let's actually just say it actually is defined at that point right over there. But since this is going to be discontinuous, the derivative is going to be undefined at that point. So that's that first interval. Now let's look at the second interval. The second interval, from where that first interval ended, all the way to right over here. The derivative is a constant negative 2." + }, + { + "Q": "at 0:43 sal says there is a positive trend between them, i do not see any?\n", + "A": "The scores appear to be getting generally higher as study time is increased, a trend that teachers expect to happen. (positive trend is as independent increases, dependent also increases). Lowest scores are on the left and highest scores are on the right (even though the max score is not with max study time). What do you see that thinks it is not positive?", + "video_name": "Jpbm5YgciqI", + "timestamps": [ + 43 + ], + "3min_transcript": "The graphs below show the test grades of the students in Dexter's class. The first graph shows the relationship between test grades and the amount of time the students spent studying. So this is study time on this axis and this is the test grade on this axis. And the second graph shows the relationship between test grades and shoes size. So shoe size on this axis and then test grade. Choose the best description of the relationship between the graphs. So first, before looking at the explanations, let's look at the actual graphs. So this one on the left right over here, it looks like there is a positive linear relationship right over here. I could almost fit a line that would go just like that. And it makes sense that there would be, that the more time that you spend studying, the better score that you would get. Now for a certain amount of time studying, some people might do better than others, but it does seem like there's this relationship. really much of a relationship. You see the shoe sizes, for a given shoe size, some people do not so well and some people do very well. Someone with a size 10 and 1/2, it looks like, someone it looks like they flunked the exam. Someone else, looks like they got A minus or a B plus And it really would be hard to somehow fit a line here. No matter how you draw a line, these dots don't seem to form a trend. So let's see which of these choices apply. There's a negative linear relationship between study time and score. No, that's not true. It looks like there's a positive linear relationship. The more you study, the better your score would be. A negative linear relationship would trend downwards like that. There is a non-linear relationship between study time and score and a negative linear relationship between shoe size and score. Well that doesn't seem right either. A non-linear relationship, it would not be easy to fit a line to it. And this one seems like a line would be very reasonable. between shoe size and score. So I wouldn't pick this one either. There's a positive linear relationship between study time and score. That's right. And no relationship between shoe size and score. Well, I'm going to go with that one. Both graphs show positive linear trends of approximately equal strength. No, not at all. This one doesn't show a linear relationship of really any strength." + }, + { + "Q": "\nat 0:36 what is a parabola?", + "A": "the graph of the quadratic function. This ------> y = x^2 is just an example of the parent quadratic function. it could be something as broad as y = 456x^2+567x. The parent parabolic graph is an even graph because it is symmetric over the y-axis.", + "video_name": "0A7RR0oy2ho", + "timestamps": [ + 36 + ], + "3min_transcript": "Let's see if we can learn a thing or two about conic sections. So first of all, what are they and why are they called conic sections? Actually, you probably recognize a few of them already, and I'll write them out. They're the circle, the ellipse, the parabola, and the hyperbola. Hyperbola. And you know what these are already. When I first learned conic sections, I was like, oh, I know what a circle is. I know what a parabola is. And I even know a little bit about ellipses and hyperbolas. Why on earth are they called conic sections? So to put things simply because they're the intersection of a plane and a cone. And I draw you that in a second. But just before I do that it probably makes sense to just draw them by themselves. And I'll switch colors. Circle, we all know what that is. Actually let me see if I can pick a thicker line for my circles. It's all the points that are equidistant from some center, and that distance that they all are that's the radius. So if this is r, and this is the center, the circle is all the points that are exactly r away from this center. We learned that early in our education what a circle is; it makes the world go round, literally. Ellipse in layman's terms is kind of a squished circle. It could look something like this. Let me do an ellipse in another color. So an ellipse could be like that. Could be like that. It's harder to draw using the tool I'm drawing, but it could also be tilted and rotated around. But this is a general sense. And actually, circles are a special case of an ellipse. It's an ellipse where it's not stretched in one dimension It's kind of perfectly symmetric in every way. Parabola. You've learned that if you've taken algebra two and you probably have if you care about conic sections. A parabola looks something like this, kind of a U shape and you know, the classic parabola. I won't go into the equations right now. Well, I will because you're probably familiar with it. y is equal to x squared. And then, you could shift it around and then you can even have a parabola that goes like this. That would be x is equal to y squared. You could rotate these things around, but I think you know the general shape of a parabola. We'll talk more about how do you graph it or how do you know what the interesting points on a parabola actually are. And then the last one, you might have seen this before, is a hyperbola. It almost looks like two parabolas, but not quite, because the curves look a little less U-ish and a little more open. But I'll explain what I mean by that. So a hyperbola usually looks something like this. So if these are the axes, then if I were to draw-- let" + }, + { + "Q": "At 12:10 Sal says that if he has time then he would make a video so I was wondering If he can make a video seeing as to that there are no exercises to make sure that I have learned the material. Where would I go to ask the site staff or Sal to ask him to make a video as a continuation to Parametric equations?\n", + "A": "If you click on ask a question... look on the right side. there is an option to request a feature, rather than post a question.", + "video_name": "IReD6c_njOY", + "timestamps": [ + 730 + ], + "3min_transcript": "So if you have the shape, you don't know what parametric equation you can go back to. You could make up one, but you don't know which one it'll go back to. And just to kind of hit the point home of kind of making up a parametric equation. Sometimes that's asked of you. So let's just do a very simple example. Let's start with a normal function of x. So let's say that we have the equation y is equal to x squared plus x. I just made it up. And let's say that you were asked to turn this into a parametric equation. And this is often a very hard thing for people because there is not one right answer. You can turn into an arbitrary number of parametric equations. I just may do something really crazy and arbitrary. I could say x is equal-- When I have y defined explicitly in terms of x like this, I can make x into anything I could say it's cosine of t minus the natural log of t. That's just a random thing. But then if x is this, then y is going to be equal to-- I just substitute this back in --cosine of t minus ln of t squared plus cosine of t minus ln of t. We just converted this into a parametric equation or a set of parametric equations. I could've also just written x is equal to t. And then what would y equal? y would be equal to t squared plus t. You might say well what's the difference between this parametric equation and this parametric equation? Well, they're both going to have the same shape of their paths. It's going to be something like a parabola. But the rate and the direction with which they progress on those paths will be very very different. It's actually a very interesting thing There are some paths that you could take where-- Let's say Let's say that the shape is-- You know everything we've done so far, we've always been going in one direction, but you could have a-- There are scenarios-- And maybe if I have time, I'll make a video. And this isn't what-- Let's say the shape of the path is some type of, I dunno, let's say it's a circle of some kind. I'm just going to-- That's not this. I'm doing a completely different example right now. I'm just kind of-- Sometimes, with the ellipses we had paths that went counterclockwise and then paths that went clockwise. You can also have paths that kind of isolate in between, move around, move back and forth along the thing. So there's all sorts of parametric equations that you can define. And you can say if t equals from this to this, you'll use this set of parametric equations. If it's another set of t's, use another one. So there's all sorts of crazy things you can do to say what happens as you move along the path. So the difference-- Not that this is the path of that. This is actually more of a parabola. But the difference between this and this is how you move along the shape." + }, + { + "Q": "\nAt 9:00 Sal says that making both negative will make the path reverse. Wouldn't it only reverse if one was negative and one positive?", + "A": "cos -t = cos t so the only thing that really matters is changing the sign of sin t", + "video_name": "IReD6c_njOY", + "timestamps": [ + 540 + ], + "3min_transcript": "We get all the way-- Oh. That's the same color I used before. Let me see this. Let me do this color. We're here. Now notice: in the first one, when we went from t equals 0 to t equals pi over 2, we went from here to there. We went kind of a quarter of the way around the ellipse. But now when we went from t equals zero to pi over 2, where did we go? Went halfway around the ellipse. We went all the way from there, all the way over there. And likewise, when we went from t equals pi over 2 to t equals pi with this set of parametric equations, we went another quarter of the ellipse. We went from there to there. But here, when we go from t equals pi ever 2 to t equals pi, we go all of this way. We go back to the beginning part of our ellipse. has the exact same shape of its path as this set of parametric equations. Except it's going around it at twice as fast of a rate. For every time when t increases by pi over 2 here, we go by-- we kind of go a quarter way around the ellipse. But when t increases by pi over 2 here, we go halfway around the ellipse. So the thing to realize-- and I know I've touched on this before --is that even though both of these sets of parametric equations, when you do the algebra, they can kind of be converted into this shape. You lose the information about where our particle is as it's rotating around the ellipse or how fast it's rotating And that's why you need these parametric equations. We can even set up a parametric equation that goes in the other direction. Instead of having these-- and I encourage you to play with that --but if you instead of this, if you just put a minus sign right here. Instead of going in that direction, it would go in this direction. It would go in a clockwise direction. So one thing that you've probably been thinking from the beginning is OK, I was able to go from my parametric equations to this equation of ellipse in terms of just x and y. Can you go back the other way? Could you go from this to this? And, I think you might realize now, that the answer is no. Because there's no way, just with the information that you're given here, to know that you should go to this parametric equation or this parametric equation or any of an infinite number of parametric equations. I mean anything of the form x is equal to 3 cosine of really anything times t and y is equal to 3 times cosine of-- As long as it's the same anything-- I drew the two squiggly marks the same. --as long as these two things are the same, then you" + }, + { + "Q": "At 0:12 are the numbers on the line tenths or wholes\n", + "A": "The blue lines are whole numbers and the little black lines are tenths.", + "video_name": "qb0QSP7Sfz4", + "timestamps": [ + 12 + ], + "3min_transcript": "- [Voiceover] Where is the point on the number line? Well, here it is, here is the point. But I'm guessing that they're asking not literally just to find it and look at it but what number is this point graphed at. Where is this on the number line? So, one thing we know pretty quickly is the number is between three and four. It's greater than three but it's not quite four. But to figure out how much greater than three we need to know what these black tick marks represent. So, between three and four there is one, two, three, four, five, six, seven, eight, nine, ten equal spaces. So, each of these distances, each of these equal spaces, is one tenth or one tenth of the distance between three and four. It's one out of ten equal spaces. So, if that's one tenth and this next space is another one tenth. And then we have to travel one more tenth So, we went three, we know it's three. Plus, one, two, three tenths. Three and three tenths. Or, let's write this as a decimal, let's look at it as a decimal. If we wanted, we could have our ones place value and then after the ones, the decimal and the tenths. So, for the ones, there's three ones. And how many tenths did we see here? There were three tenths. So, either way we can say three and three tenths or three and three tenths. Our decimal, our point is 3.3 on the number line." + }, + { + "Q": "\nAt 3:24, Sal is just dividing both the sides by tan... Am I right? Thanks in advance :)", + "A": "Yes...you could think about it that way.", + "video_name": "aHzd-u35LuA", + "timestamps": [ + 204 + ], + "3min_transcript": "And they've given us two pieces of information. They gave us the side that is opposite the angle. And they've given us the side that is adjacent to the angle. So what trig function deals with opposite and adjacent? And to remind ourselves, we can write, like I always like to do, soh, cah, toa. And these are really by definition. So you just have to know this, and soh cah toa helps us. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. We can write that the tangent of theta is equal to the length of the opposite side-- 324 meters-- over the length of the adjacent side-- over 54 meters. What angle, when I take its tangent, gives me 324/54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the Inverse Tan Function. So we could rewrite this as we're going to take the inverse tangent-- and sometimes it's written as tangent with this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324/54. And just to be clear, what is this inverse tangent? This just literally says, this will return what is the angle that, when I take the tangent of it, gives me 324/54. This says, what is the angle that, when I take the tangent of it, gives me tangent of theta? Theta is the angle that when you get the tangent of it gets you tangent of theta. And so we get theta is equal to inverse tangent of 324/54. Once again, this inverse tangent thing you might find confusing. But all this is saying is, over here, we're saying tangent of some angle is 324/54. This is just saying my angle is whatever angle I need so that when I take the tangent of it, I get 324/54. It's how we will solve for theta. So let's get our calculator out. And let's say that we want our answer in degrees. Well, I'm just going to assume that they want our answers in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the 2nd mode right over here. And actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees." + }, + { + "Q": "At 3:56, you change the calculator to degree mode. For CC Geometry, should you keep the calculator at degree mode?\n", + "A": "It depends on the question, if its asked to write the answer in Radians, there is no other way!", + "video_name": "aHzd-u35LuA", + "timestamps": [ + 236 + ], + "3min_transcript": "What angle, when I take its tangent, gives me 324/54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the Inverse Tan Function. So we could rewrite this as we're going to take the inverse tangent-- and sometimes it's written as tangent with this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324/54. And just to be clear, what is this inverse tangent? This just literally says, this will return what is the angle that, when I take the tangent of it, gives me 324/54. This says, what is the angle that, when I take the tangent of it, gives me tangent of theta? Theta is the angle that when you get the tangent of it gets you tangent of theta. And so we get theta is equal to inverse tangent of 324/54. Once again, this inverse tangent thing you might find confusing. But all this is saying is, over here, we're saying tangent of some angle is 324/54. This is just saying my angle is whatever angle I need so that when I take the tangent of it, I get 324/54. It's how we will solve for theta. So let's get our calculator out. And let's say that we want our answer in degrees. Well, I'm just going to assume that they want our answers in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the 2nd mode right over here. And actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees. And let me just type in the inverse tangent-- so it's in this yellow color right here-- inverse tangent of 324 divided by 54 is going to be-- and they told us to round to two decimal places-- 80.54 degrees. So theta is equal to 80.54 degrees. That's the angle at which you should shoot the gun to help defeat this horrible alien." + }, + { + "Q": "At 1:41 what does he mean by transitive, communicative and commutative?\n", + "A": "The Commutative Laws say we can swap numbers over and still get the same answer ... a+b = b+a Hope it helps :D Have a great day/night :) :)", + "video_name": "d8lP5tR2R3Q", + "timestamps": [ + 101 + ], + "3min_transcript": "Welcome to the presentation on multiplying and dividing negative numbers. Let's get started. I think you're going to find that multiplying and dividing negative numbers are a lot easier than it might look initially. You just have to remember a couple rules, and I'm going to teach probably in the future like I'm actually going to give you more intuition on why these rules work. But first let me just teach you the basic rules. So the basic rules are when you multiply two negative numbers, so let's say I had negative 2 times negative 2. First you just look at each of the numbers as if there was no negative sign. Well you say well, 2 times 2 that equals 4. And it turns out that if you have a negative times a negative, that that equals a positive. So let's write that first rule down. A negative times a negative equals a positive. What if it was negative 2 times positive 2? two numbers without signs. We know that 2 times 2 is 4. But here we have a negative times a positive 2, and it turns out that when you multiply a negative times a positive you get a negative. So that's another rule. Negative times positive is equal to negative. What happens if you have a positive 2 times a negative 2? I think you'll probably guess this one right, as you can tell that these two are pretty much the same thing by, I believe it's the transitive property -- no, no I think it's the communicative property. I have to remember that. But 2 times negative 2, this also equals negative 4. So we have the final rule that a positive times a negative also equals the negative. of the same thing. A negative times a positive is a negative, or a positive times a negative is negative. You could also say that as when the signs are different and you multiply the two numbers, you get a negative number. And of course, you already know what happens when you have a positive times a positive. Well that's just a positive. So let's review. Negative times a negative is a positive. A negative times a positive is a negative. A positive times a negative is a negative. And positive times each other equals positive. I think that last little bit completely confused you. Maybe I can simplify it for you. What if I just told you if when you're multiplying and they're the same signs that gets you a positive result. And different signs gets you a negative result." + }, + { + "Q": "\nWho is saying this? 0:00", + "A": "sal is doing this.", + "video_name": "d8lP5tR2R3Q", + "timestamps": [ + 0 + ], + "3min_transcript": "Welcome to the presentation on multiplying and dividing negative numbers. Let's get started. I think you're going to find that multiplying and dividing negative numbers are a lot easier than it might look initially. You just have to remember a couple rules, and I'm going to teach probably in the future like I'm actually going to give you more intuition on why these rules work. But first let me just teach you the basic rules. So the basic rules are when you multiply two negative numbers, so let's say I had negative 2 times negative 2. First you just look at each of the numbers as if there was no negative sign. Well you say well, 2 times 2 that equals 4. And it turns out that if you have a negative times a negative, that that equals a positive. So let's write that first rule down. A negative times a negative equals a positive. What if it was negative 2 times positive 2? two numbers without signs. We know that 2 times 2 is 4. But here we have a negative times a positive 2, and it turns out that when you multiply a negative times a positive you get a negative. So that's another rule. Negative times positive is equal to negative. What happens if you have a positive 2 times a negative 2? I think you'll probably guess this one right, as you can tell that these two are pretty much the same thing by, I believe it's the transitive property -- no, no I think it's the communicative property. I have to remember that. But 2 times negative 2, this also equals negative 4. So we have the final rule that a positive times a negative also equals the negative. of the same thing. A negative times a positive is a negative, or a positive times a negative is negative. You could also say that as when the signs are different and you multiply the two numbers, you get a negative number. And of course, you already know what happens when you have a positive times a positive. Well that's just a positive. So let's review. Negative times a negative is a positive. A negative times a positive is a negative. A positive times a negative is a negative. And positive times each other equals positive. I think that last little bit completely confused you. Maybe I can simplify it for you. What if I just told you if when you're multiplying and they're the same signs that gets you a positive result. And different signs gets you a negative result." + }, + { + "Q": "At around 6:30, Sal starts to talk about how the overlap is only supposed to be counted once. But why would it be counted even once if the question is the P (yellow OR cube?) anyways? Doesn't this question not want the area where we can get both outcomes? If that is the case, then is the addition rule really necessary? Couldn't you just simply add together the P(only Yellow), which would be 7, and P(only Cube), which would be 8?\n", + "A": "P(yellow or cube) includes yellow cubes because you re looking for the probability of either. Since yellow cubes have both traits, they automatically have either and are therefore part of the outcome we re looking for. The rule exists so that they re counted, but not double-counted.", + "video_name": "QE2uR6Z-NcU", + "timestamps": [ + 390 + ], + "3min_transcript": "that I've drawn? This Venn diagram is just a way to visualize the different probabilities. And they become interesting when you start thinking about where sets overlap, or even where they don't overlap. So here we are thinking about things that are members of the set yellow. So they're in this set, and they are cubes. So this area right over here-- that's the overlap of these two sets. So this area right over here-- this represents things that are both yellow and cubes, because they are inside both circles. So this right over here-- let me rewrite it right over here. So there's five objects that are both yellow and cubes. Now let's ask-- and this is probably the most interesting thing to ask-- what is the probability of getting something that is yellow or or a cube, a cube of any color? or a cube of any color-- well, we still know that the denominator here is going to be 29. These are all of the equally likely possibilities that might jump out of the bag. But what are the possibilities that meet our conditions? Well, one way to think about it is, well, the probability-- there's 12 things that would meet the yellow condition. So that would be this entire circle right over here-- 12 things that meet the yellow condition. So this right over here is 12. This is the number of yellow. That is 12. And then to that, we can't just add the number of cubes, because if we add the number of cubes, we've already counted these 5. These 5 are counted as part of this 12. One way to think about it is there are 7 yellow objects that are not cubes. Those are the spheres. There are 5 yellow objects that are cubes. And then there are 8 cubes that are not yellow. That's one way to think about. we counted all of this. So we can't just add the number of cubes to it, because then we would count this middle part again. So then we have to essentially count cubes, the number of cubes, which is 13. So the number of cubes, and we'll have to subtract out this middle section right over here. Let me do this. So subtract out the middle section right over here. So minus 5. So this is the number of yellow cubes. It feels weird to write the word yellow in green. The number of yellow cubes-- or another way to think about it-- and you could just do this math right here. 12 plus 13 minus 5 is 20. Did I do that right? 12 minus, yup, it's 20. So that's one way. You just get this is equal to 20 over 29. But the more interesting thing than even the answer of the probability of getting that," + }, + { + "Q": "\nAt 1:37, why do we take half of -6?", + "A": "We are working with completing the square. It starts from the concept that (x + b) ^2 = x(x+b)+b(x+b) = x^2 + bx + bx + b^2 or x^2 + 2bx + b^2. Since the middle has to come from two middle terms, we divide by two. In the specific problem, we have to say that 2b = - 6, thus when we divide both sides, we get b = -3 and b^2 = (-3)^2 = 9.", + "video_name": "4Bx06GFyhUA", + "timestamps": [ + 97 + ], + "3min_transcript": "Which quadratic has the lowest maximum value? So let's figure out the maximum value for each of these-- and they're defined in different ways-- and then see which one is the lowest. And I'll start with the easiest. So h of x. We can just graphically look at it, visually look at it, and say-- what's the maximum point? And the maximum point looks like it's right over here when x is equal to 4. And when x is equal to 4, y or h of x is equal to negative 1. So the maximum for h of x looks like it is negative 1. Now, what's the maximum for g of x? And they've given us some points here and here. Once again, we can just eyeball it, and say-- well, what's the maximum value they gave us? Well, 5 is the largest value. It happens when x is equal to 0. g of 0 is 5. So the maximum value here is 5. Now, f of x. They just give us an expression to define it. And so it's going take a little bit of work to figure out what the maximum value is. The easiest way to do that for a quadratic is to complete the square. And so let's do it. plus 6x minus 1. I never like having this negative here. So I'm going to factor it out. This is the same thing as negative times x squared minus 6x and plus 1. And I'm going to write the plus 1 out here because I'm fixing to complete the square. Now, just as a review of completing the square, we essentially want to add and subtract the same number so that part of this expression is a perfect square. And to figure out what number we want to add and subtract, we look at the coefficient on the x term. It's a negative 6. You take half of that. That's negative 3. And you square it. Negative 3 squared is 9. Now, we can't just add a 9. That would change the actual value of the expression. We have to add a 9 and subtract a 9. And you might say-- well, why are we adding and subtracting the same thing if it doesn't change the value of the expression? And the whole point is so that we can get this first part of the expression to represent a perfect square. So I can rewrite that part as x minus 3 squared and then minus 9-- or negative 9-- plus 1 is negative 8. Let me do that in a different color so we can keep track of things. So this part right over here is negative 8. And we still have the negative out front. And so we can rewrite this as-- if we distribute the negative sign-- negative x minus 3 squared plus 8. Now, let's think about what the maximum value is. And to understand the maximum value, we have to interpret this negative x minus 3 squared. Well, x minus 3 squared-- before we think about the negative-- that is always going to be a positive value. Or it's always going to be non-negative. But then, when we make it negative, it's always going to be non-positive." + }, + { + "Q": "\nOkay I'm having troubles at 2:30.\nI don't understand where you got the 4 from, when I multiply 3x -2^2 I get 12 not 4?", + "A": "He got twelve, but then subtracted it.", + "video_name": "za0QJRZ-yQ4", + "timestamps": [ + 150 + ], + "3min_transcript": "Determine the domain and range of the function f of x is equal to 3x squared plus 6x minus 2. So, the domain of the function is: what is a set of all of the valid inputs, or all of the valid x values for this function? And, I can take any real number, square it, multiply it by 3, then add 6 times that real number and then subtract 2 from it. So essentially any number if we're talking about reals when we talk about any number. So, the domain, the set of valid inputs, the set of inputs over which this function is defined, is all real numbers. So, the domain here is all real numbers. And, for those of you who might say, well, you know, aren't all numbers real? You may or may not know that there is a class of numbers, that are a little bit bizarre when you first learn them, called imaginary numbers and complex numbers. But, I won't go into that right now. But, most of the traditional numbers that you know of, they are part of the set of real numbers. So, you take any real number and you put it here, you can square it, multiply it by 3, then add 6 times it and subtract 2. Now, the range, at least the way we've been thinking about it in this series of videos-- The range is set of possible, outputs of this function. Or if we said y equals f of x on a graph, it's a set of all the possible y values. And, to get a flavor for this, I'm going to try to graph this function right over here. And, if you're familiar with quadratics-- and that's what this function is right over here, it is a quadratic-- you might already know that it has a parabolic shape. And, so its shape might look something like this. And, actually this one will look like this, it's upward opening. But other parabolas have shapes like that. And, you see when a parabola has a shape like this, it won't take on any values below its vertex when it's upward opening, and it won't take on any values above its vertex when it is downward opening. So, let's see if we can graph this and maybe get a sense of its vertex. but let's see how we can think about this problem. So, I'm gonna try some x and y values. There's other ways to directly compute the vertex. Negative b over 2a is the formula for it. It comes straight out of the quadratic formula, which you get from completing the square. Lets try some x values and lets see what f of x is equal to. So, let's try, well this the values we've been trying the last two videos. What happens when x is equal to negative two? Then f of x is 3 times negative 2 squared, which is 4, plus 6 times negative 2, which is 6 times negative 2, so it's minus 12 minus 2. So, this is 12 minus 12 minus 2. So, it's equal to negative 2. Now, what happens when x is equal to negative 1? So, this is going to be 3 times negative 1 squared, which is just 1, minus, or I should say plus 6 times negative 1 which is" + }, + { + "Q": "At 2:33, did Sal just choose -2 randomly? If it wasn't random, why did he choose -2?\n", + "A": "He picked -2 because that is a place that he normally starts his tables and graphs on, it is just habit, there is no specific reason.", + "video_name": "za0QJRZ-yQ4", + "timestamps": [ + 153 + ], + "3min_transcript": "So, you take any real number and you put it here, you can square it, multiply it by 3, then add 6 times it and subtract 2. Now, the range, at least the way we've been thinking about it in this series of videos-- The range is set of possible, outputs of this function. Or if we said y equals f of x on a graph, it's a set of all the possible y values. And, to get a flavor for this, I'm going to try to graph this function right over here. And, if you're familiar with quadratics-- and that's what this function is right over here, it is a quadratic-- you might already know that it has a parabolic shape. And, so its shape might look something like this. And, actually this one will look like this, it's upward opening. But other parabolas have shapes like that. And, you see when a parabola has a shape like this, it won't take on any values below its vertex when it's upward opening, and it won't take on any values above its vertex when it is downward opening. So, let's see if we can graph this and maybe get a sense of its vertex. but let's see how we can think about this problem. So, I'm gonna try some x and y values. There's other ways to directly compute the vertex. Negative b over 2a is the formula for it. It comes straight out of the quadratic formula, which you get from completing the square. Lets try some x values and lets see what f of x is equal to. So, let's try, well this the values we've been trying the last two videos. What happens when x is equal to negative two? Then f of x is 3 times negative 2 squared, which is 4, plus 6 times negative 2, which is 6 times negative 2, so it's minus 12 minus 2. So, this is 12 minus 12 minus 2. So, it's equal to negative 2. Now, what happens when x is equal to negative 1? So, this is going to be 3 times negative 1 squared, which is just 1, minus, or I should say plus 6 times negative 1 which is So, this is 3 minus 6 is negative 3 minus 2 is equal negative 5, and that actually is the vertex. And, you know the formula for the vertex, once again, is negative b over 2 a. So, negative b. That's the coefficient on this term right over here. It's negative 6 over 2 times this one right over here, 2 times 3. 2 times 3, this is equal to negative 1. So, that is the vertex, but let's just keep on going right over here. So, what happens when x is equal to 0? These first two terms are 0, you're just left with a negative 2. When x is equal to positive 1. And, this is where you can see that this is the vertex, and you start seeing the symmetry. If you go one above the vertex, f of x is equal to negative 2. If you go one x value below the vertex, or below the x value of the vertex, f of x is equal to negative 2 again. But, let's just keep going. We could try, let's do one more point over here." + }, + { + "Q": "I Am SOOOO Lost?\nCan Anyone Help Me\n0:00 - 4:17\n", + "A": "ask a question and i think someone will come along.in the meantime watch the math videos and play them more than once if you miss what the person is saying", + "video_name": "tVDslyeLefU", + "timestamps": [ + 0, + 257 + ], + "3min_transcript": "So I have this rectangular prism here. It's kind of the shape of a brick or a fish tank, and it's made up of these unit cubes. And each of these unit cubes we're saying is 1/4 of a foot by 1/4 of a foot by 1/4 of a foot. So you could almost imagine that this is-- so let me write it this way-- a 1/4 of a foot by 1/4 of a foot by 1/4 of a foot. Those are its length, height, and width, or depth, whatever you want to call it. So given that, what is the volume of this entire rectangular prism going to be? So I'm assuming you've given a go at it. So there's a couple of ways to think about it. You could first think about the volume of each unit cube, and then think about how many units cubes there are. So let's do that. The unit cube, its volume is going to be 1/4 of a foot times 1/4 of a foot times 1/4 of a foot. 1/4 times 1/4 cubic feet, which is often written as feet to the third power, cubic feet. So 1/4 times 1/4 is 1/16, times 1/4 is 1/64. So this is going to be 1 over 64 cubic feet, or 1/64 of a cubic foot. That's the volume of each of these. That's the volume of each of these unit cubes. Now, how many of them are there? Well, you could view them as kind of these two layers. The first layer has 1, 2, 3, 4, 5, 6, 7, 8. That's this first layer right over here. And then we have the second layer down here, which would be another 8. So it's going to be 8 plus 8, or 16. So the total volume here is going is going to be equal to 16/64 cubic feet, which is the same thing. 16/64 is the same thing as 1/4. Divide the numerator and the denominator by 16. This is the same thing as 1/4 of a cubic foot. And that's our volume. Now, there's other ways that you could have done this. You could have just thought about the dimensions of the length, the width, and the height. The width right over here is going to be 2 times 1/4 feet, which is equal to 1/2 of a foot. The height here is the same thing. So it's going to be 2 times 1/4 of a foot, which is equal to 2/4, or 1/2 of a foot." + }, + { + "Q": "why would you divide 8:20 to 2:5 when 20 can divide by more then just 5?\n", + "A": "8 : 20 is the ratio and when you simplify a ratio you need to take BOTH values into consideration. If you watch some Algebra/ Pre-Algebra concepts you will see that what you do to one side needs to be done to the other to balance it out. 8 : 20 can be simplified no lower than 2 : 5 because the greatest divisor of 8 and 20 is 4 so it ends up as 2 :5. I hope this helped.", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 500, + 125 + ], + "3min_transcript": "" + }, + { + "Q": "\nHow are fractions and ratios the same? How are they different? Is 2:5 the same as 2/5?", + "A": "Yes, they are the same exact thing.", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 125 + ], + "3min_transcript": "Voiceover:Let's think about another scenario involving ratios. In this case, let's think about the ratio of the number of apples. Number of apples to ... Instead of taking the ratio of the number of apples to the number of oranges, let's take the number of apples to the number of fruit. The number of fruit that we have over here. And I encourage you to pause the video and think about that on your own. Well, how many total apples do we have? We have 2, 4, 6, 8 apples. So we're going to have 8 apples. And then how much total fruit do we have? Well we have 8 apples and we have 3, 6, 9, 12 oranges. So our total fruit is 8 plus 12. We have 20 pieces of fruit. So this ratio is going to be 8 to ... 8 to 20. Or, if we want to write this in a more reduced form, we can divide both of these by 4. 4 is their greatest common divisor. 8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to." + }, + { + "Q": "So the first ratio number goes on top of the fraction so 4:6 is 4 over 6?\n", + "A": "Yes, that is correct.", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 246 + ], + "3min_transcript": "8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to. Or we could say it's 2/5, the fraction 2/5, which would sometimes be read as 2 to 5. This is also, when it's written this way, you could also read that as a ratio, depending on the context. In a sentence like this I would read this as 2/5 of the fruit are apples." + }, + { + "Q": "\nSo can 2:3 become 2/3? I'm taking notes...", + "A": "The bottom number does not (nor should it ever) refer to the whole group. The first and second items (example 2:3) added together will sum to the whole group. Instructions for mixing chemicals are given in this notation.", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 123 + ], + "3min_transcript": "Voiceover:Let's think about another scenario involving ratios. In this case, let's think about the ratio of the number of apples. Number of apples to ... Instead of taking the ratio of the number of apples to the number of oranges, let's take the number of apples to the number of fruit. The number of fruit that we have over here. And I encourage you to pause the video and think about that on your own. Well, how many total apples do we have? We have 2, 4, 6, 8 apples. So we're going to have 8 apples. And then how much total fruit do we have? Well we have 8 apples and we have 3, 6, 9, 12 oranges. So our total fruit is 8 plus 12. We have 20 pieces of fruit. So this ratio is going to be 8 to ... 8 to 20. Or, if we want to write this in a more reduced form, we can divide both of these by 4. 4 is their greatest common divisor. 8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to." + }, + { + "Q": "So for example does 8 to 8 equal to 8:8 and 8/8?\n", + "A": "Yes... and if you reduce it, the ratio is 1 to 1", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 488 + ], + "3min_transcript": "" + }, + { + "Q": "\nIs 2:3, 4:6, and 8:12 equivalent", + "A": "Yes, they are equivalent.", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 123, + 246, + 492 + ], + "3min_transcript": "Voiceover:Let's think about another scenario involving ratios. In this case, let's think about the ratio of the number of apples. Number of apples to ... Instead of taking the ratio of the number of apples to the number of oranges, let's take the number of apples to the number of fruit. The number of fruit that we have over here. And I encourage you to pause the video and think about that on your own. Well, how many total apples do we have? We have 2, 4, 6, 8 apples. So we're going to have 8 apples. And then how much total fruit do we have? Well we have 8 apples and we have 3, 6, 9, 12 oranges. So our total fruit is 8 plus 12. We have 20 pieces of fruit. So this ratio is going to be 8 to ... 8 to 20. Or, if we want to write this in a more reduced form, we can divide both of these by 4. 4 is their greatest common divisor. 8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to." + }, + { + "Q": "\nHow can you get 30:39 equal?", + "A": "If you mean how do you simplify 30:39, then you just divide by three. 30 divided by 3 is 10 and 39 divided by 3 is 13. Your final simplified ratio is 10 to 13. Both 10 to 13 and 30 to 39 are equal though. Hope this helps! :)", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 1839 + ], + "3min_transcript": "" + }, + { + "Q": "\nAt 7:43 in the video, isn't A really A transpose? And what Sal calls r1T, r2T aren't they really the transpose of Column 1 Column2, so maybe they should be labelled c1T, c2T? Because the r1T suggests the transpose of a row to a column.", + "A": "@Jonh I believe you right. r1T is in reality c1T, but as siddhantsabo said, the notation used was to point you re dealing now with rows instead of columns.", + "video_name": "QOTjdgmNqlg", + "timestamps": [ + 463 + ], + "3min_transcript": "And the way that we can write the row space of A, this thing right here, the row space of A, is the same thing as the column space of A transpose. So one way you can rewrite this sentence right here, is that the null space of A is the orthogonal complement of the row space. The row space is the column space of the transpose matrix. And the claim, which I have not proven to you, is that this is the orthogonal complement of this. This is equal to that, the little perpendicular superscript. That's the claim, and at least in the particular example that actually showed you that 2 by 3 matrix. But let's see if this applies generally. So let me write my matrix A like this. So my matrix A, I can write it as just a bunch of row vectors. But just to be consistent with our notation, with vectors we tend to associate as column vectors, so to represent the row vectors here I'm just going to write them as transpose vectors. Because in our reality, vectors will always be column vectors, and row vectors are just transposes of those. r1 transpose, r2 transpose and you go all the way down. We have m rows. So you're going to get rm transpose. Don't let the transpose part confuse you. I'm just saying that these are row vectors. I'm writing transposes there just to say that, look these represent these rows. But if it's helpful for you to imagine them, just imagine this is the first row of the matrix, this is the second row of that matrix, so on and so forth. Now, what is the null space of A? Well that's all of the vectors here. Let me do it like this. The null space of A is all of the vectors x that satisfy the equation that this is going to be equal to the zero vector. Now to solve this equation, what can we do? We've seen this multiple times. This matrix-vector product is essentially the same thing as saying-- let me write it like this-- it's going to be equal to the zero vector in rm. You're going to have m 0's all the way down to the m'th 0. So another way to write this equation, you've seen it before, is when you take the matrix-vector product, you" + }, + { + "Q": "At 1:36, how did you get the side length of 10?\n", + "A": "It s indicated on the diagram along the bottom edge.", + "video_name": "mtMNvnm71Z0", + "timestamps": [ + 96 + ], + "3min_transcript": "- What I want to do in this video is get some practice finding surface areas of figures by opening them up into what's called nets. And one way to think about it is if you had a figure like this, and if it was made out of cardboard, and if you were to cut it, if you were to cut it right where I'm drawing this red, and also right over here and right over there, and right over there and also in the back where you can't see just now, it would open up into something like this. So if you were to open it up, it would open up into something like this. And when you open it up, it's much easier to figure out the surface area. So the surface area of this figure, when we open that up, we can just figure out the surface area of each of these regions. So let's think about it. So what's first of all the surface area, what's the surface area of this, right over here? Well in the net, that corresponds to this area, it's a triangle, it has a base of 12 and height of eight. So this area right over here is going to be one half times the base, so times 12, So this is the same thing as six times eight, which is equal to 48 whatever units, or square units. This is going to be units of area. So that's going to be 48 square units, and up here is the exact same thing. That's the exact same thing. You can't see it in this figure, but if it was transparent, if it was transparent, it would be this backside right over here, but that's also going to be 48. 48 square units. Now we can think about the areas of I guess you can consider them to be the side panels. So that's a side panel right over there. It's 14 high and 10 wide, this is the other side panel. It's also this length over here is the same as this length. It's also 14 high and 10 wide. So this side panel is this one right over here. And then you have one on the other side. And so the area of each of these 14 times 10, they are 140 square units. This one is also 140 square units. the area of I guess you can say the base of the figure, so this whole region right over here, which is this area, which is that area right over there. And that's going to be 12 by 14. So this area is 12 times 14, which is equal to let's see. 12 times 12 is 144 plus another 24, so it's 168. So the total area is going to be, let's see. If you add this one and that one, you get 96. 96 square units. The two magenta, I guess you can say, side panels, 140 plus 140, that's 280. 280. And then you have this base that comes in at 168. We want it to be that same color. 168. One, 68. Add them all together, and we get the surface area for the entire figure. And it was super valuable to open it up into this net" + }, + { + "Q": "\nAt 3:50, is (350/50)=(350/10)/(50/10)=35/5=7 correct?", + "A": "That is right. When you get rid of the zeroes, it is the same as 35/7.", + "video_name": "ccS5Fy5yLjk", + "timestamps": [ + 230 + ], + "3min_transcript": "well this fraction bar right here is the same as the division sign up here. 350 divided by 50 is the same as 350 over 50. And, when we have a fraction like this fraction down here, we can simplify it. In this case, because there's zeroes on the end, we know they're both multiples of 10, so we can divide them both by 10. We can divide our numerator and our denominator by 10. And, when we divide whole numbers by 10, we have a trick we can use, a pattern, really, which is that the whole number, in this case 350, when it's divided by 10 we drop a zero from the end. So, 350 divided by 10 is 35. 350 could be divided into 10 groups of 35. When we divided 50 by 10, we drop that zero off the end, or another way to think about it is 50 divided into groups of 10 would make five groups. And, then we end up with our simplified fraction of 35 fifths or 35 divided by five, which is the same thing here. So, in both of these cases, we can see that 350 divided by 50 is the same as 35 divided by 5. So, when both whole numbers, when we're dividing whole numbers and they both end in zeroes, we can cancel those zeroes. Basically, we're factoring out a 10. We're taking the 10, the divided by 10 out of both of them, out of both numbers. So, we can cancel those zeroes which leaves us with smaller numbers and at least for me, I find division a lot simpler when I'm working with smaller numbers. Let's try one like 420 divided by 70. So, we can see we have two whole numbers, both end in zero. So, we're gonna cancel those zeroes, basically we're dividing a ten out of both numbers and end up with a simpler division problem of 42 divided by seven. 42 divided by seven equals six, therefore 420 divided by 70 also equals six. And, here's one last one. What is we had 5600 divided by 80? So, the first thing I notice is I have zeroes at the end" + }, + { + "Q": "\nSal, at 0:42, how do you know that a-sub2 is equal to -8/5, why is it not a-sub1? Why did you start with 2 instead of 1?", + "A": "Because i = 2 in the Sigma statement.", + "video_name": "yvddTWa9ptU", + "timestamps": [ + 42 + ], + "3min_transcript": "Let's say that we're told that this sum right over here, where our index starts at 2 and we go all the way to infinity, that this infinite series is negative 8/5 plus 16/7 minus 32/9 plus-- and we just keep going on and on forever. And so what I want to do is to explicitly define what a sub n is here. So right now we just say, hey, if you take the sum of a sub n from n equals 2 to infinity, it turns out you get this sum right over here. But let's think about what a sub n-- how we can actually define it in terms of n. And I encourage you to pause the video right now and try it on your own. So the first thing that you might realize is, well, this is the number that we're going to get. Let me write it this way. a sub 2 is equal to negative 8/5. a sub 3 is equal to 16/7. a sub 4 is equal to negative 32/9. Negative 8/5 is the same thing as negative 8 over 5. Let me make that a little bit clearer. So I'll make that a little bit clearer. So this is negative 8/5. Obviously, this is positive, so I don't have to really worry about it too much. And then here, I'm just saying negative 32/9, so it's the same thing as negative 32 over 9. So let's see if we can first find a pattern in the numerator. So when we go from negative 8 to 16, what's happening? Well, we're multiplying by negative 2. Now, to go from 16 to negative 32, we're multiplying by negative 2 again. So you might say, OK, well, whatever we have in the numerator must be a power of negative 2. And, all right, if you say, well, maybe this is negative 2 squared, well, you know that negative 8 isn't negative 2 squared. Negative 2 squared is equal to positive 4. Negative 8, that is equal to negative 2 to the third power. 16 is equal to negative 2 to the fourth power. Negative 32 is equal to negative 2 to the fifth power. So notice, our exponent on the negative 2 is always going to be one more than our index. Our index is 2, our exponent is 3. Our index is 3, our exponent is 4. Our index is 4, our exponent is 5. So that gives a sense that at least the numerator is going to be-- whatever our index is, it's going to be-- so let me write this down. So a sub n is equal to-- well, it's going to be negative 2 to whatever index we're at, to that index plus 1 power. So that's a reasonable way to think about our numerator." + }, + { + "Q": "\nSee 1:47 into the video. Referring to the denominator, why is it 0-(-a) and not (-a)-0. Thanks ;)", + "A": "You could do that, but the answer would be the same. For example, if the 2 points were (9,2) and (7,5), 5-2/7-9 would equal 3/-2, which equals -(3/2.) If you switched the order, it would be 2-5/9-7, which is -3/2. The answers are the same. It does not matter which you put first, as long as when you put the y value of it first, then that point s x value has to be in front too.", + "video_name": "H6ZNLD1AeM8", + "timestamps": [ + 107 + ], + "3min_transcript": "Consider the graph of the function f of x that passes through three points as shown. So these are the three points, and this curve in blue is f of x. Identify which of the statements are true. So they give us these statements. This first one says f of negative a is less than 1 minus f of negative a over a. So this seems like some type of a bizarre statement. How are we able to figure out whether this is true from this right over here? So let's just go piece by piece and see if something starts to make sense. So f of negative a, where do we see that here? Well, this is f of negative a. This is the point x equals negative a. So this is negative a, and this is y is equal to f of negative a. So this is f of negative a right over here. And what we know about f of negative a, based on looking at this graph, is that f of negative a is between 0 and 1. So we can write that. 0 is less than f of negative a, which is less than 1. So that's all I can deduce about f of negative a right Now let's look at this crazy statement, 1 minus f of negative a over a. What is this? Well, let's think about what happens if we take the secant line, if we're trying to find the slope of the secant line, between this point and this point, if we wanted to find the average rate of change between the point negative a, f of negative a, and the point 0, 1. If this is our endpoint, our change in y is going to be 1 minus f of negative a. So 1 minus f of negative a is equal to our change in y. And our change in x, going from negative a to 0, so change in x is going to be equal to 0 minus negative a, which is equal to positive a. So this right over here is essentially our change in x over our change in y from this point to this point. It is our average rate of change from this point to this point. or you could say it's the slope of the secant line. So the secant line would look something like this. Slope of the secant line, so this right over here is slope of secant line between from f of negative a to 0 comma 1. So just looking at this diagram right over here, what do we know about this slope? And in particular, can we make any statements about that slope relative to, say, 0 or 1 or anything like that? Well, let's think about what a line of slope 1 would look like. Well, a line of slope 1, especially one that went through this point right over here, would look something like this. A line of slope 1 would look something like this. So this line right over here that I've just" + }, + { + "Q": "\nIn the example at 5:50 wouldn't the 2a affect the actual slope of the line because it is not a quadratic?", + "A": "I m not quite sure what you mean, but we are only comparing the slope of the secant lines, not the actual slope of the curve. The slope of the secant line is \u00ce\u0094y/\u00ce\u0094x", + "video_name": "H6ZNLD1AeM8", + "timestamps": [ + 350 + ], + "3min_transcript": "that we just looked at. So this is the same value right over here. So we're comparing this slope right over here, to what's this? f of a minus 1 over a. Well, this is the slope of this secant line, this is the slope of this secant line that I'm drawing in this-- let me do it in more contrast. Let me do it in orange. That is the slope of this secant line. So which one has a higher slope? Well, it's pretty clear that the blue secant line has a higher slope than this orange secant line. But here, it's saying that the blue's slope is lower than the orange. So this is not going to be true. So this is not true. Then finally, let's look at this over here-- f of a minus f of negative a over 2a. So this is the slope. Let me draw this. So this right over here, this is the slope of the secant line between this point Our change in y is f of a minus f of negative a. Our change in x is a minus negative a, which is 2a. So this is this secant line right over here. So let me draw it. So this secant line right over here, so they're comparing that slope to this slope. f of a minus 1 is our change in y, over a is our change in x. So we're comparing it to that one right over there. And you could immediately eyeball this kind of brownish, maroon-- I guess it's kind of a brown color-- this secant line that goes all the way from here to here is clearly steeper than this one right over here. And we know that, that the average rate of change from here to here is going to be higher than the average rate of change from here to here, because at least from negative a to 0, we were increasing at a much faster rate. And then we slowed down to this rate. So the average over the entire interval is definitely going to be more than what we get from 0 to a. This has a higher-- we actually know that this is false. Both of these would have been true if we swapped these signs around, if this was a greater than sign, if this was a greater than sign. So this is the only one that applies." + }, + { + "Q": "At 2:59 min, why is the \"divided by 2\" switched with \"times 1/2\"?\n", + "A": "1/2 and divided by 2 are both 0.5 A Half can be written... As a fraction: 1/2 As a decimal: 0.5 As a percentage: 50%", + "video_name": "GwycEivqYYI", + "timestamps": [ + 179 + ], + "3min_transcript": "The figure is 13.09 centimeters. So let's say that this is the figure right over here. So this is the figure that she wants to place in the middle of the page. So this is 13.09 centimeters. So the question is, how much margin should she leave on the left so that the figure is centered? So they're really asking, what should this distance be here so that the figure is centered? And the key thing is, if the figure is centered, this distance is going to be the same as this distance, so that the figure is right in the middle. Whatever this is, it's going to be the same as this. So one way we could think about it-- we could say, OK, let's take 13.09 centimeters from 21.59 And if we do that, then we'll know how much leftover space this plus this combined is. We want half of it to go on the right side, and we want half of it to go on the left side. So let's see how much this left margin and the right margin combined need to be. And that's just 21.59. That's just going to be 21.59 centimeters minus 13.09 centimeters. And this, we get to-- let's see. This is 8.50 centimeters, which is the combined margins. Not just the left or not just the right-- this is the left plus the right margin, is 8.50 centimeters. This is this distance plus this distance. Now, if we want to figure out what the left margin should be, if we want to center it. Well, we just want to split this evenly between the left and the right margin. So we just want to divide this by 2. So if you divide 18.50 by 2-- or multiply it by 1/2, I guess. So if we multiply this by 1/2, you are left with 1/2 of 8.50, is 4.25 centimeters. So if we want it centered, this is going to be 4.25 centimeters. And of course, this is going to be 4.25 centimeters, as well. And you see, when you add 4.25 to 4.25, you get 8.5 centimeters. And then you add that to 13.09. You get the entire width of the piece of paper. So how much margin should she leave on the left? 4.25 centimeters." + }, + { + "Q": "\nAt 0:36 Sal starts to extend the number. Do I need to do that all the time?", + "A": "No, that s only to help you understand the concept. If you get it, that s great. =) You may have to do this when asked to expand the number.", + "video_name": "jxA8MffVmPs", + "timestamps": [ + 36 + ], + "3min_transcript": "Find the place value of 3 in 4,356. Now, whenever I think about place value, and the more you do practice problems on this it'll become a little bit of second nature, but whenever I see a problem like this, I like to expand out what 4,356 really is, so let me rewrite So if I were to write it-- and I'll write it in different colors. So 4,356 is equal to-- and just think about how I just said it. It is equal to 4,000 plus 300 plus 50 plus 6. And you could come up with that just based on how we said it: four thousand, three hundred, and fifty-six. Now another way to think about this is this is just like saying this is 4 thousands plus-- or you could even think as 5 tens plus 6. And instead of 6, we could say plus 6 ones. And so if we go back to the original number 4,356, this is the same thing as 4-- I'll write it down. Let me see how well I can-- I'll write it up like this. This is the same thing is 4 thousands, 3 hundreds, 5 tens and then 6 ones. So when they ask what is the place value of 3 into 4,356, we're concerned with this 3 right here, It's in the hundreds place. If there was a 4 here, that would mean we're dealing with 4 hundreds. If there's a 5, 5 hundreds. It's the third from the right. This is the ones place. That's 6 ones, 5 tens, 3 hundreds. So the answer here is it is in the hundreds place." + }, + { + "Q": "At 3:40 , i didn't understand why sal used that formula ? May i know which video can i look upon for the formula ?\n", + "A": "Finding the sum of n squares part 1/ part 2", + "video_name": "LwhJVURumAA", + "timestamps": [ + 220 + ], + "3min_transcript": "when this is 7, all the way to 14. You could factor out a 2. And so this is going to become 2 times 1 plus 2 plus 3 all the way to 7. And so you can rewrite this piece right over here as 2 times the sum-- so we're essentially just factoring out the 2-- 2 times the sum, which is the sum from n equals 1 to 7 of n. So this is this piece. We still have this 28 that we have to add. So we have this 28. And we draw the parentheses so you don't think that the 28 is part of this right over here. And now we can do the same thing with this. 3 times n-- we're taking from n equals 1 to 7 of 3 n squared. Doing the same exact thing as we just did in magenta, this is going to be equal to 3 times the sum from n We're essentially factoring out the 3. We're factoring out the 2. n squared. And once again, we can put parentheses just to clarify things. Now, at this point, there are formulas to evaluate each of these things. There's a formula to evaluate this thing right over here. There's a formula to evaluate this thing over here. And you can look them up. And actually, I'll give you the formulas, in case you're curious. This formula, one expression of this formula is that this is going to be n to the third over 3 plus n squared over 2 plus n over 6. That's one formula for that. And one formula for this piece right over here, going from n equals 1 to 7-- sorry. Let me make it clear. This n is actually what your terminal value should be. So this should be 7 to the third power over 3-- I was just mindlessly using the formula-- 7 to the third over 3 plus 7 squared over 2 plus 7/6. So that's this sum. And this sum, you could view it as the average of the first and the last terms. So the first term is 1. The last term is 7. So take their average and then multiply it times the number of terms you have. So times-- you have 7 terms. So what is this middle one going to evaluate to? Well, 1 times-- and of course, we have this 2 out front. This green is just this part right over here. So you have 2 times this. And over here, you have 3 times this business right over here. So if we evaluate this one, 2 times-- let's see. 1 plus 7 is 8, divided by 2 is 4. 4 times 2 is 8. Times 7, it's 56. So that becomes 56. Now, this-- let's see." + }, + { + "Q": "at 1:15 you say that the domain is 6. but when you substitute x as 6 in the numerator as well you get\n6^2 - 36/6 - 6\n=36 - 36/6 - 6\n=0/0 = 0\n", + "A": "The domain is not 6. In fact, the domain is all real numbers EXCEPT 6, because 6 doesn t work in the expression (leads to division by zero). In the simplification you just did, at the end, 0/0 = undefined or indeterminate, 0/0 does not equal 0.", + "video_name": "ey_b3aPsRl8", + "timestamps": [ + 75 + ], + "3min_transcript": "Simplify the rational expression and state the domain. Let's see if we can start with the domain part of the question, if we can start with stating the domain. Now, the domain is the set of all of the x values that you can legitimately input into this if you view this as a function, if you said this is f of x is equal to that. The domain is a set of all x values that you could input into this function and get something that is well-defined. The one x value that would make this undefined is the x value that would make the denominator equal 0-- the x value that would make that equal 0. So when does that happen? Six minus x is equal to 0. Let's add x to both sides. We get 6 is equal to x, so the domain of this function is equal to the set of all real numbers except 6. is 6 then you're dividing by 0, and then this expression is undefined. We've stated the domain, now let's do the simplifying the rational expression. Let me rewrite it over here. We have x squared minus 36 over 6 minus x. Now, this might jump out at you immediately, as it's that special type of binomial. It's of the form a squared minus b squared, and we've seen this multiple times. This is equivalent to a plus b times a minus b. And in this case, a is x and b is 6. This top expression right here can be factored as x plus 6 times x minus 6, all of that over 6 minus x. Now, at first you might say, I have a x minus 6 and a 6 minus x. Those aren't quite equal, but what maybe will jump out at Try it out. Let's multiply by negative 1 and then by negative 1 again. Think of it that way. If I multiply by negative 1 times negative 1, obviously, I'm just multiplying the numerator by 1, so I'm not in any way changing the numerator. What happens if we just multiply the x minus 6 by that first negative 1? What happens to that x minus 6? Let me rewrite the whole expression. We have x plus 6, and I'm going to distribute this negative 1. If I distribute the negative 1, I have negative 1 times x is negative x. Negative 1 times negative 6 is plus 6. And then I have a negative 1 out here. I have a negative 1 times negative 1, and all of that is over 6 minus x. Now, negative plus 6. This is the exact same thing as 6 minus x if you just rearrange the two terms. Negative x plus 6 is the same" + }, + { + "Q": "At 1:52 how would you solve the equation if you left in the 75?\n", + "A": "You would factor the expression. You need to find two numbers that add up to 10 and multiply to equal -75. Those two numbers are 15 and -5. Thus, x^2+10x -75 = (x+15)(x-5). Now, solving either bracket for x gives you x = -15 and x = 5.", + "video_name": "TV5kDqiJ1Os", + "timestamps": [ + 112 + ], + "3min_transcript": "We're asked to complete the square to solve 4x squared plus 40x minus 300 is equal to 0. So let me just rewrite it. So 4x squared plus 40x minus 300 is equal to 0. So just as a first step here, I don't like having this 4 out front as a coefficient on the x squared term. I'd prefer if that was a 1. So let's just divide both sides of this equation by 4. So let's just divide everything by 4. So this divided by 4, this divided by 4, that divided by 4, and the 0 divided by 4. Just dividing both sides by 4. So this will simplify to x squared plus 10x. And I can obviously do that, because as long as whatever I do to the left hand side, I also do the right hand side, that will make the equality continue to be valid. So that's why I can do that. So 40 divided by 4 is 10x. And then 300 divided by 4 is what? That is 75. Let me verify that. 7 times 4 is 28. You subtract, you get a remainder of 2. Bring down the 0. 4 goes into 20 five times. 5 times 4 is 20. Subtract zero. So it goes 75 times. This is minus 75 is equal to 0. And right when you look at this, just the way it's written, you might try to factor this in some way. But it's pretty clear this is not a complete square, or this is not a perfect square trinomial. Because if you look at this term right here, this 10, half of this 10 is 5. And 5 squared is not 75. So this is not a perfect square. So what we want to do is somehow turn whatever we have on the left hand side into a perfect square. And I'm going to start out by kind of getting this 75 out You'll sometimes see it where people leave the 75 on the left hand side. I'm going to put on the right hand side just so it kind of clears things up a little bit. So let's add 75 to both sides to get rid of the 75 from the left hand side of the equation. plus 75. Those guys cancel out. And I'm going to leave some space here, because we're going to add something here to complete the square that is equal to 75. So all I did is add 75 to both sides of this equation. Now, in this step, this is really the meat of completing the square. I want to add something to both sides of this equation. I can't add to only one side of the equation. So I want to add something to both sides of this equation so that this left hand side becomes a perfect square. And the way we can do that, and saw this in the last video where we constructed a perfect square trinomial, is that this last term-- or I should say, what we see on the left hand side, not the last term, this expression on the left hand side, it will be a perfect square if we have a constant term that is the square of half of the coefficient on the first degree So the coefficient here is 10. Half of 10 is 5." + }, + { + "Q": "\nAt 3:40, why did he color in 8/15 instead of 10/15 because there is still 2/15 remaining on the side.?", + "A": "He wasn t dealing with the whole 15/15 but rather 4/5 of it = 12/15 2/3 x 4/5 = 8/15", + "video_name": "hr_mTd-oJ-M", + "timestamps": [ + 220 + ], + "3min_transcript": "to represent 2/3 of that 4/5 and see what fraction of the whole you actually have. So pause now. So let's think about this. Let's represent 4/5. So if I have a whole like this, let me try to divide it into 5 equal sections. 5 equal sections, so let's say that is 1 equal section, that is 2 equal sections, that is 3, 4, and 5-- I can do a better This is always the hard part. I'm trying my best to make them look, at least, like equal sections-- 2, 3, 4, and 5. I think you get the point here. I'm trying to make them equal sections. And we want 4/5. So we want 4 of these 5 equal sections. So this would be 1 of the 5 equal sections, 2 of them, 3 of them, and then 4 of them. So that right over there is 4/5. So how can we think about that? Well, we could take this section and divide it into thirds. So let's do that. Divide it into thirds. So we're going divide it into 3 equal sections. So that's 1/3, and then 2/3. So we took each of the 5 equal sections, and we divided them into 3 equal sections. Now what's going to be 2/3 of the 4/5? Well, that's going to be this part right over here. So let me make this clear. This is 1/3 of the 4/5. And then this would be 2/3 of the 4/5. So this right over here, would be 2/3 of the 4/5, or 2/3 times 4/5. But what fraction of the whole does that represent? Well, how many total, how many total equal sections do we now have? Well, we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, So we have 15 equal sections. I'm using a new color. We have 15 equal sections, and that make sense. We started with 5 equal sections, but then we divided each of those into 3 equal sections. So now we have 5 times 3 total equal sections. And then how many of those are now colored in? Well, we see it's 2 times 4. 1, 2, 3, 4, 5, 6, 7, 8. How many of them are in the 2/3 of the 4/5, I should say. And there's 8 of them, 8 of the 15 equals sections. And so there you have it. It should hopefully now make visual sense, or it makes conceptual sense, that 2/3 times 4/5-- you can obviously compute it by just multiplying the numerators, 2 times 4 is 8. And then multiplying the denominators, 3 times 5 is 15-- but hopefully this now makes conceptual sense as 2/3 of 4/5." + } +] \ No newline at end of file diff --git a/MathSc-Timestamp/test.json b/MathSc-Timestamp/test.json new file mode 100644 index 0000000000000000000000000000000000000000..7228c35f414e8552a6089b393d98fa04c728c714 --- /dev/null +++ b/MathSc-Timestamp/test.json @@ -0,0 +1,21347 @@ +[ + { + "Q": "At 7:15,Sal says that PV=k.KE(system).In the last video,he only said that PV=k .So,how come it is k.KE now.I don't get it.Please help.Thanks in advance.", + "A": "Well,I understand your point.I also think of it this way-the units of PV =the units of KE (k is a constant) PV(FV/A)=Nm^3/A=Nm^3/m^2=Nm=kgm^2/s^2 KE(mv^2/2)=kgm^2/s^2 Thus,the units are just the same.The only difference is the constant. Is it because of this that PV=kKE?Am I right?", + "video_name": "x34OTtDE5q8", + "timestamps": [ + 435 + ], + "3min_transcript": "I think you also have the sense that-- what would have more energy? A 100 degree cup of tea, or a 100 degree barrel of tea. I want to make them equivalent in terms of what they're holding. I think you have a sense. Even though they're the same temperature, they're both pretty warm-- let's say this is 100 degrees Celsius, so they're both boiling-- that the barrel, because there's more of it, is going to have more energy. It's equally hot, and there's just more molecules there. That's what temperature is. Temperature, in general, is a measure roughly equal to some energy-- per molecule. So the average kinetic energy of the system divided by the total number of molecules we have. Another way we could talk about is, temperature is essentially energy per molecule. So something that has a lot of molecules, where N is the number of molecules. Another way we could view this is that the kinetic energy of the system is going to be equal to the number of molecules times the temperature. This is just a constant-- times 1 over K, but we don't even know what this is, so we could say that's still a constant-- so the kinetic energy of the system is going to be equal to some constant times the number of particles We don't know what this is, and we're going to figure this out later. This is another interesting concept. We said that pressure times volume is proportional to the kinetic energy of the system-- the aggregate, if you take all of the molecules and combine their kinetic energies. These aren't the same K's-- I could put another constant here and call that K1. And we also know that the kinetic energy of the system is equal to some other constant times the number of molecules I have times the temperature. If you think about it, you could also say that this is proportional to this, and this is proportional to this. You could say that pressure times volume is proportional" + }, + { + "Q": "at 7:00 why is Newtons the number of particles?? isnt newton kg*m/s^2\nat 6:15 Sal say its number of molecules. how is that possible if newton is kg*m/s^2?", + "A": "There are no Newtons in this video. N stands for Number, as in Number of molecules. It s a variable, not a unit.", + "video_name": "x34OTtDE5q8", + "timestamps": [ + 420, + 375 + ], + "3min_transcript": "I think you also have the sense that-- what would have more energy? A 100 degree cup of tea, or a 100 degree barrel of tea. I want to make them equivalent in terms of what they're holding. I think you have a sense. Even though they're the same temperature, they're both pretty warm-- let's say this is 100 degrees Celsius, so they're both boiling-- that the barrel, because there's more of it, is going to have more energy. It's equally hot, and there's just more molecules there. That's what temperature is. Temperature, in general, is a measure roughly equal to some energy-- per molecule. So the average kinetic energy of the system divided by the total number of molecules we have. Another way we could talk about is, temperature is essentially energy per molecule. So something that has a lot of molecules, where N is the number of molecules. Another way we could view this is that the kinetic energy of the system is going to be equal to the number of molecules times the temperature. This is just a constant-- times 1 over K, but we don't even know what this is, so we could say that's still a constant-- so the kinetic energy of the system is going to be equal to some constant times the number of particles We don't know what this is, and we're going to figure this out later. This is another interesting concept. We said that pressure times volume is proportional to the kinetic energy of the system-- the aggregate, if you take all of the molecules and combine their kinetic energies. These aren't the same K's-- I could put another constant here and call that K1. And we also know that the kinetic energy of the system is equal to some other constant times the number of molecules I have times the temperature. If you think about it, you could also say that this is proportional to this, and this is proportional to this. You could say that pressure times volume is proportional" + }, + { + "Q": "Great video, very interesting. When you talk about the Bacteriophages, ( 17:50 ) that inject their DNA though the harder cell walls, how does the DNA then go on to alter the cell if it is just loose genetic material within the cell?", + "A": "Various viruses can do it differently. Sometimes the viral genome comes with proteins that will splice it into the host genome. Sometimes the viral genome is made of RNA that can do the splicing on its own. There are numerous other mechanisms.", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 1070 + ], + "3min_transcript": "So maybe most of it was the viral DNA, but it might have, when it transcribed and translated itself, it might have taken a little bit-- or at least when it translated or replicated itself-- it might take a little bit of the organism's previous DNA. So it's actually cutting parts of DNA from one organism and bringing it to another organism. Taking it from one member of a species to another member of But it can definitely go cross-species. So you have this idea all of a sudden that DNA can jump between species. It really kind of-- I don't know, for me it makes me appreciate how interconnected-- as a species, we kind of imagine that we're by ourselves and can only reproduce with each other and have genetic variation within But viruses introduce this notion of horizontal transfer via transduction. Horizontal transduction is just the idea of, look when I replicate this virus, I might take a little bit of the organism that I'm freeloading off of, I might take a little And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia." + }, + { + "Q": "At 17:45, Sal mentions that the bacteria could be far worse for the virus. Could someone explain how the bacteria could potentially harm the virus?", + "A": "I personally think that the bacteria could be more harm to the virus because the virus simply attaches itself to the bacteria, injects everything inside of it, and then it might just kind of sit there and become inactive. I don t believe that things inside of the bacteria can take some of the bacteria s membrane a form a new bacteria. I could be wrong though.", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 1065 + ], + "3min_transcript": "So maybe most of it was the viral DNA, but it might have, when it transcribed and translated itself, it might have taken a little bit-- or at least when it translated or replicated itself-- it might take a little bit of the organism's previous DNA. So it's actually cutting parts of DNA from one organism and bringing it to another organism. Taking it from one member of a species to another member of But it can definitely go cross-species. So you have this idea all of a sudden that DNA can jump between species. It really kind of-- I don't know, for me it makes me appreciate how interconnected-- as a species, we kind of imagine that we're by ourselves and can only reproduce with each other and have genetic variation within But viruses introduce this notion of horizontal transfer via transduction. Horizontal transduction is just the idea of, look when I replicate this virus, I might take a little bit of the organism that I'm freeloading off of, I might take a little And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia." + }, + { + "Q": "at13:17, Why isn't there any cure/vaccine for AIDS?", + "A": "The virus mutate itself in no time", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 797 + ], + "3min_transcript": "codes it into DNA. So its RNA to DNA. Which when it was first discovered was, kind of, people always thought that you always went from DNA to RNA, but this kind of broke that paradigm. But it codes from RNA to DNA. And if that's not bad enough, it'll incorporate that DNA into the DNA of the host cell. So that DNA will incorporate itself into the DNA of the host cell. Let's say the yellow is the DNA of the host cell. And this is its nucleus. So it actually messes with the genetic makeup of what it's infecting. And when I made the videos on bacteria I said, hey for every one human cell we have twenty bacteria cells. And they live with us and they're useful and they're part of us and they're 10% of our dry mass and all of that. But bacteria are kind of along for the ride. But these retroviruses, they're actually changing our I mean, my genes, I take very personally. They define who I am. But these guys will actually go in and change my genetic makeup. And then once they're part of the DNA, then just the natural DNA to RNA to protein process will code their actual proteins. Or their-- what they need to-- so sometimes they'll lay dormant and do nothing. And sometimes-- let's say sometimes in some type of environmental trigger, they'll start coding for themselves again. And they'll start producing more. But they're producing it directly from the organism's cell's DNA. They become part of the organism. I mean I can't imagine a more intimate way to become part of an organism than to become part of its DNA. I can't imagine any other way to actually define an organism. And if this by itself is not eerie enough, and just so you know, this notion right here, when a virus becomes part of But if this isn't eerie enough, they estimate-- so if this infects a cell in my nose or in my arm, as this cell experiences mitosis, all of its offspring-- but its offspring are genetically identical-- are going to have this viral DNA. And that might be fine, but at least my children won't get it. You know, at least it won't become part of my species. But it doesn't have to just infect somatic cells, it could infect a germ cell. So it could go into a germ cell. And the germ cells, we've learned already, these are the ones that produce gametes. For men, that's sperm and for women it's eggs. But you could imagine, once you've infected a germ cell, once you become part of a germ cell's DNA, then I'm passing on that viral DNA to my son or my daughter." + }, + { + "Q": "At 19:05, I saw this big red thing on the white blood cell. What was it?", + "A": "This question has been asked several times before, and I ll say what I said before - I m not positive, but I believe that it is a platelet.", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 1145 + ], + "3min_transcript": "And infect that DNA into the next organism. So you actually have this DNA, this jumping, from organism to organism. So it kind of unifies all DNA-based life. Which is all the life that we know on the planet. And if all of this isn't creepy enough-- and actually maybe I'll save the creepiest part for the end. But there's a whole-- we could talk all about the different classes of viruses. But just so you're familiar with some of the terminology, when a virus attacks bacteria, which they often do. And we study these the most because this might be a good alternative to antibiotics. Because viruses that attack bacteria might-- sometimes the bacteria is far worse for the virus-- but these are called bacteriaphages. And I've already talked to you about how they have their DNA. But since bacteria have hard walls, they will just inject the DNA inside of the bacteria. And when you talk about DNA, this idea of a provirus. called the lytic cycle. This is just some terminology that's good to know if you're going to take a biology exam about this stuff. And when the virus incorporates it into the DNA and lays dormant, incorporates into the DNA of the host organism and lays dormant for awhile, this is called the lysogenic cycle. And normally, a provirus is essentially experiencing a lysogenic cycle in eurkaryotes, in organisms that have a nuclear membrane. Normally when people talk about the lysogenic cycle, they're talking about viral DNA laying dormant in the DNA of bacteria. Or bacteriophage DNA laying dormant in the DNA of bacteria. But just to kind of give you an idea of what this, quote unquote, looks like, right here. I got these two pictures from Wikipedia. These little green dots you see right here all over the surface, this big thing you see here, this is a white blood cell. Part of the human immune system. This is a white blood cell. And what you see emerging from the surface, essentially budding from the surface of this white blood cell-- and this gives you a sense of scale too-- these are HIV-1 viruses. And so you're familiar with the terminology, the HIV is a virus that infects white blood cells. AIDS is the syndrome you get once your immune system is weakened to the point. And then many people suffer infections that people with a strong immune system normally won't suffer from." + }, + { + "Q": "so from 00:01 to 23:17 he talking about the common cold and flu (influenza)? :|", + "A": "He s talking about Viruses in general and not about a specific one. At the beginning he says that because he has a cold that he s going to talk about Viruses. In between those times he covers how a most viruses interact with living cells and also how a retrovirus might.", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 1, + 1397 + ], + "3min_transcript": "Considering that I have a cold right now, I can't imagine a more appropriate topic to make a video on than a virus. And I didn't want to make it that thick. A virus, or viruses. And in my opinion, viruses are, on some level, the most fascinating thing in all of biology. Because they really blur the boundary between what is an inanimate object and what is life? I mean if we look at ourselves, or life as one of those things that you know it when you see it. If you see something that, it's born, it grows, it's constantly changing. Maybe it moves around. Maybe it doesn't. But it's metabolizing things around itself. It reproduces and then it dies. You say, hey, that's probably life. And in this, we throw most things that we see-- or we throw in, us. We throw in bacteria. We throw in plants. I mean, I could-- I'm kind of butchering the taxonomy system here, but we tend to know life when we see it. information inside of a protein. Inside of a protein capsule. So let me draw. And the genetic information can come in any form. So it can be an RNA, it could be DNA, it could be single-stranded RNA, double-stranded RNA. Sometimes for single stranded they'll write these two little S's in front of it. Let's say they are talking about double stranded DNA, they'll put a ds in front of it. But the general idea-- and viruses can come in all of these forms-- is that they have some genetic information, some chain of nucleic acids. Either as single or double stranded RNA or single or double stranded DNA. And it's just contained inside some type of protein structure, which is called the capsid. And kind of the classic drawing is kind of an icosahedron type looking thing. Let me see if I can do justice to it. It looks something like this. And not all viruses have to look exactly like this. And we're really just scratching the surface and understanding even what viruses are out there and all of the different ways that they can essentially replicate themselves. We'll talk more about that in the future. And I would suspect that pretty much any possible way of replication probably does somehow exist in the virus world. But they really are just these proteins, these protein capsids, are just made up of a bunch of little proteins put together. And inside they have some genetic material, which might be DNA or it might be RNA. So let me draw their genetic material. The protein is not necessarily transparent, but if it was, you would see some genetic material inside of there. So the question is, is this thing life? It seems pretty inanimate. It doesn't grow. It doesn't change. It doesn't metabolize things. This thing, left to its own devices, is just It's just going to sit there the way a book on a table just sits there. It won't change anything." + }, + { + "Q": "At 6:32, is it possible that the ice may slow down due to hydrogen bonding between the molecules?", + "A": "Assume there is no friction, thus now intermolecular bonds between Hs which creates friction", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 392 + ], + "3min_transcript": "We've never seen, at least in our human experience, it looks like everything will eventually stop. So this is a very unintuitive thing to say, that something in motion will just keep going in motion indefinitely. Everything in human intuition says if you want something to keep going in motion, you have to keep putting more force, keep putting more energy into it for it to keep going. Your car won't go forever, unless you keep, unless the engine keeps burning fuel to drive and consuming energy. So what are they talking about? Well, in all of these examples-- and I think this is actually a pretty brilliant insight from all of these fellows is that-- all of these things would have gone on forever. The ball would keep going forever. This ice block would be going on forever, except for the fact that there are unbalanced forces acting on them to stop them. So in the case of ice, even though ice on ice doesn't have a lot of friction, there is some friction between these two. And so you have, in this situation, the force of friction is going to be acting against the direction of the movement of the ice. so if you have the actual water molecules in a lattice structure in the ice cube, and then here are the water molecules in a lattice structure on the ice, on the actual kind of sea of ice that it's traveling on-- they do kind of bump and grind into each other. Although they're both smooth, there are imperfections here. They bump and grind. They generate a little bit of heat. And they'll, essentially, be working against the movement. So there's a force of friction that's being applied to here. And that's why it's stopping. Not only a force of friction, you also have some air resistance. The ice block is going to be bumping into all sorts of air particles. It might not be noticeable at first, but it's definitely going to keep it from going on forever. Same thing with the ball being tossed to the air. Obviously, at some point, it hits the ground because of gravity. So that's one force acting on it. But even once it hits the ground, it doesn't keep rolling forever, once again, because of the friction, especially if there's grass The grass is going to stop it from going. And even while it's in the air, it's going to slow down. It's not going to have a constant velocity. Because you have all of these air particles that are going to bump into it and exert force to slow it down. So what was really brilliant about these guys is that they could imagine a reality where you didn't have gravity, where you did not have air slowing things down. And they could imagine that in that reality, something would just keep persisting in its motion. And the reason why Galileo, frankly, was probably good at thinking about that is that he studied the orbits of planets. And he could, or at least he's probably theorized that, hey, maybe there's no air out there. And that maybe that's why these planets can just keep going round and round in orbit. And I should say their speed, because their direction is changing, but their speed never slows down, because there's nothing in the space to actually slow down those planets." + }, + { + "Q": "At 3:10, the ice has 2 equal forces acting upon it at opposite sides, but wouldn't the ice just go upwards instead of remaining stationary because of friction?", + "A": "No. Friction, in this case, points sideways, so it can t make it go up. The reaction force does point upwards, but it s as big as gravity, so it also can t do that.", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 190 + ], + "3min_transcript": "I can keep observing that rock. And it is unlikely to move, assuming that nothing happens to it. If there's no force applied to that rock, that rock will just stay there. So the first part is pretty obvious. So, \"Every body persists in a state of being at rest\"-- I'm not going to do the second part-- \"except insofar as there's some force being applied to it.\" So clearly a rock will be at rest, unless there's some force applied to it, unless someone here tries to push it or roll it or do something to it. What's less intuitive about the first law is the second part. \"Every body persists in,\" either, \"being in a state of rest or moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.\" So this Newton's first law-- and I think I should do a little aside here, because, this right here is Newton. And if this is Newton's first law, Well, the reason is is because Newton's first law is really just a restatement of this guy's law of inertia. And this guy, another titan of civilization really, this is Galileo Galilei. And he is the first person to formulate the law of inertia. And Newton just rephrased it a little bit and packaged it with his other laws. But he did many, many, many other things. So you really have to give Galileo credit for Newton's first law. So that's why I made him bigger than here. But I was in the midst of a thought. So we understand if something is at rest, it's going to stay at rest, unless there's some force that acts on it. And in some definitions, you'll see unless there's some unbalanced force. And the reason why they say unbalanced is, because you could have two forces that act on something and they might balance out. For example, I could push on this side of the rock with a certain amount of force. And if you push on this side of the rock with the exact same amount of force, the rock won't move. And the only way that it would move if there's a lot more so if you have an unbalanced force. So if you have a ton of-- and maybe the rock is a bad analogy. Let's take ice, because ice is easier to move, or ice on ice. So there's ice right here. And then, I have another block of ice sitting on top of that ice. So once again, we're familiar with the idea, if there's no force acting on it that ice won't move. But what happens if I'm pushing on the ice with a certain amount of force on that side, and you're pushing on the ice on that side with the same amount of force? The ice will still not move. So this right here, this would be a balanced force. So the only way for the ice to change its condition, to change its restful condition is if the force is unbalanced. So if we add a little bit of force on this side, so it more than compensates the force pushing it this way, then you're going to see the ice block start to move, start to really accelerate in that direction." + }, + { + "Q": "What I don't understand, is that wouldn't a body not be in a state of rest because of gravity? Sal uses the rock (1:10) as an example, and he says there is no force applied to that rock. Wouldn't gravity be a force on the rock? I mean, wouldn't the rock fly out of the atmosphere from the smallest amount of gravity somewhere else? What I'm thinking is that Earth's gravity would keeping the rock there, but the rock wouldn't be at \"rest.\" So, in this definition there would rarely/never be rest?", + "A": "Force of gravity is balanced by other forces (by restoring force I think but I m not sure), so rock is in state of rest. Many forces are applied to that rock but they balance each other and that s why rock doesn t move.", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 70 + ], + "3min_transcript": "In this video, I want to talk a little bit about Newton's First Law of Motion. And this is a translation from Newton's Principia from Latin into English. So the First Law, \"Every body persists in a state of being at rest, or moving uniformly straightforward, except insofar as it is compelled to change its state by force impressed.\" So another way to rephrase what they're saying is, that if there's something-- every body persists-- so everything will stay at rest, or moving with a constant velocity, unless it is compelled to change its state by force. Unless it's acted on by a force, especially an unbalanced force. and I'll explain that in a second. So if I have something that's at rest, so completely at rest. So I have-- and this is something that we've seen before. Let's say that I have a rock. Let's say that I have a rock someplace I can keep observing that rock. And it is unlikely to move, assuming that nothing happens to it. If there's no force applied to that rock, that rock will just stay there. So the first part is pretty obvious. So, \"Every body persists in a state of being at rest\"-- I'm not going to do the second part-- \"except insofar as there's some force being applied to it.\" So clearly a rock will be at rest, unless there's some force applied to it, unless someone here tries to push it or roll it or do something to it. What's less intuitive about the first law is the second part. \"Every body persists in,\" either, \"being in a state of rest or moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.\" So this Newton's first law-- and I think I should do a little aside here, because, this right here is Newton. And if this is Newton's first law, Well, the reason is is because Newton's first law is really just a restatement of this guy's law of inertia. And this guy, another titan of civilization really, this is Galileo Galilei. And he is the first person to formulate the law of inertia. And Newton just rephrased it a little bit and packaged it with his other laws. But he did many, many, many other things. So you really have to give Galileo credit for Newton's first law. So that's why I made him bigger than here. But I was in the midst of a thought. So we understand if something is at rest, it's going to stay at rest, unless there's some force that acts on it. And in some definitions, you'll see unless there's some unbalanced force. And the reason why they say unbalanced is, because you could have two forces that act on something and they might balance out. For example, I could push on this side of the rock with a certain amount of force. And if you push on this side of the rock with the exact same amount of force, the rock won't move. And the only way that it would move if there's a lot more" + }, + { + "Q": "3:23 sal says that while ice sits on ice theres no external force acting upon it but as i understand it the very fact that ice sits there is probabaly because gravity acts on the blocks ,could anyone please explain it to me", + "A": "To get the ice moving, you would need an outside force. As said above, gravity pulls downward, but that is an acceleration. If you want to talk about forces, the weight of the ice is what keeps the ice grounded. The normal force pushes upwards to balance the weight force. Static friction is the reason that is stays still side to side.", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 203 + ], + "3min_transcript": "I can keep observing that rock. And it is unlikely to move, assuming that nothing happens to it. If there's no force applied to that rock, that rock will just stay there. So the first part is pretty obvious. So, \"Every body persists in a state of being at rest\"-- I'm not going to do the second part-- \"except insofar as there's some force being applied to it.\" So clearly a rock will be at rest, unless there's some force applied to it, unless someone here tries to push it or roll it or do something to it. What's less intuitive about the first law is the second part. \"Every body persists in,\" either, \"being in a state of rest or moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.\" So this Newton's first law-- and I think I should do a little aside here, because, this right here is Newton. And if this is Newton's first law, Well, the reason is is because Newton's first law is really just a restatement of this guy's law of inertia. And this guy, another titan of civilization really, this is Galileo Galilei. And he is the first person to formulate the law of inertia. And Newton just rephrased it a little bit and packaged it with his other laws. But he did many, many, many other things. So you really have to give Galileo credit for Newton's first law. So that's why I made him bigger than here. But I was in the midst of a thought. So we understand if something is at rest, it's going to stay at rest, unless there's some force that acts on it. And in some definitions, you'll see unless there's some unbalanced force. And the reason why they say unbalanced is, because you could have two forces that act on something and they might balance out. For example, I could push on this side of the rock with a certain amount of force. And if you push on this side of the rock with the exact same amount of force, the rock won't move. And the only way that it would move if there's a lot more so if you have an unbalanced force. So if you have a ton of-- and maybe the rock is a bad analogy. Let's take ice, because ice is easier to move, or ice on ice. So there's ice right here. And then, I have another block of ice sitting on top of that ice. So once again, we're familiar with the idea, if there's no force acting on it that ice won't move. But what happens if I'm pushing on the ice with a certain amount of force on that side, and you're pushing on the ice on that side with the same amount of force? The ice will still not move. So this right here, this would be a balanced force. So the only way for the ice to change its condition, to change its restful condition is if the force is unbalanced. So if we add a little bit of force on this side, so it more than compensates the force pushing it this way, then you're going to see the ice block start to move, start to really accelerate in that direction." + }, + { + "Q": "At 8:12 Sal says \"if gravity disappeared, and you had no air...\". What's the relation between gravity and air?", + "A": "He uses gravity and air as two examples of forces that occur naturally all around us.", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 492 + ], + "3min_transcript": "so if you have the actual water molecules in a lattice structure in the ice cube, and then here are the water molecules in a lattice structure on the ice, on the actual kind of sea of ice that it's traveling on-- they do kind of bump and grind into each other. Although they're both smooth, there are imperfections here. They bump and grind. They generate a little bit of heat. And they'll, essentially, be working against the movement. So there's a force of friction that's being applied to here. And that's why it's stopping. Not only a force of friction, you also have some air resistance. The ice block is going to be bumping into all sorts of air particles. It might not be noticeable at first, but it's definitely going to keep it from going on forever. Same thing with the ball being tossed to the air. Obviously, at some point, it hits the ground because of gravity. So that's one force acting on it. But even once it hits the ground, it doesn't keep rolling forever, once again, because of the friction, especially if there's grass The grass is going to stop it from going. And even while it's in the air, it's going to slow down. It's not going to have a constant velocity. Because you have all of these air particles that are going to bump into it and exert force to slow it down. So what was really brilliant about these guys is that they could imagine a reality where you didn't have gravity, where you did not have air slowing things down. And they could imagine that in that reality, something would just keep persisting in its motion. And the reason why Galileo, frankly, was probably good at thinking about that is that he studied the orbits of planets. And he could, or at least he's probably theorized that, hey, maybe there's no air out there. And that maybe that's why these planets can just keep going round and round in orbit. And I should say their speed, because their direction is changing, but their speed never slows down, because there's nothing in the space to actually slow down those planets. Because on some level, it's super-duper obvious. But on a whole other level, it's completely not obvious, especially this moving uniformly straightforward. And just to make the point clear, if gravity disappeared, and you had no air, and you threw a ball, that ball literally would keep going in that direction forever, unless some other unbalanced force acted to stop it. And another way to think about it-- and this is an example that you might see in everyday life-- is, if I'm in an airplane that's going at a completely constant velocity and there's absolutely no turbulence in the airplane. So if I'm sitting in the airplane right over here. And it's going at a constant velocity, completely smooth, no turbulence. There's really no way for me to tell whether that airplane is moving without looking out the window. Let's assume that there's no windows in that airplane. It's going at a constant velocity." + }, + { + "Q": "At around 5:05 Sal says everything will eventually stop, if so, how is the earth constantly orbiting the sun? What is causing the earth to move?", + "A": "The Earth continues to spin upon its axis because there are no outside forces acting to stop its rotation.", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 305 + ], + "3min_transcript": "so if you have an unbalanced force. So if you have a ton of-- and maybe the rock is a bad analogy. Let's take ice, because ice is easier to move, or ice on ice. So there's ice right here. And then, I have another block of ice sitting on top of that ice. So once again, we're familiar with the idea, if there's no force acting on it that ice won't move. But what happens if I'm pushing on the ice with a certain amount of force on that side, and you're pushing on the ice on that side with the same amount of force? The ice will still not move. So this right here, this would be a balanced force. So the only way for the ice to change its condition, to change its restful condition is if the force is unbalanced. So if we add a little bit of force on this side, so it more than compensates the force pushing it this way, then you're going to see the ice block start to move, start to really accelerate in that direction. This, you know, something that's at rest will stay at rest, unless it's being acted on by an unbalanced force. What's less obvious is the idea that something moving uniformly straightforward, which is another way of saying something having a constant velocity. What he's saying is, is that something that has a constant velocity will continue to have that constant velocity indefinitely, unless it is acted on by an unbalanced force. And that's less intuitive. Because everything in our human experience-- even if I were to push this block of ice, eventually it'll stop. It won't just keep going forever, even assuming that this ice field is infinitely long, that ice will eventually stop. Or if I throw a tennis ball. That tennis ball will eventually stop. It'll eventually grind to a halt. We've never seen, at least in our human experience, it looks like everything will eventually stop. So this is a very unintuitive thing to say, that something in motion will just keep going in motion indefinitely. Everything in human intuition says if you want something to keep going in motion, you have to keep putting more force, keep putting more energy into it for it to keep going. Your car won't go forever, unless you keep, unless the engine keeps burning fuel to drive and consuming energy. So what are they talking about? Well, in all of these examples-- and I think this is actually a pretty brilliant insight from all of these fellows is that-- all of these things would have gone on forever. The ball would keep going forever. This ice block would be going on forever, except for the fact that there are unbalanced forces acting on them to stop them. So in the case of ice, even though ice on ice doesn't have a lot of friction, there is some friction between these two. And so you have, in this situation, the force of friction is going to be acting against the direction of the movement of the ice." + }, + { + "Q": "At 6:20, Sal said that water has a \"lattice structure\". What does that mean?", + "A": "Hi Amol Chavan, Sal refers to a lattice structure or crystal structure is an arrangement of atoms or molecules in a crystalline solid or a liquid. This term is often used to illustrate the bonding of the hydrogen and oxygen atoms in water. Hope that helps! - JK", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 380 + ], + "3min_transcript": "This, you know, something that's at rest will stay at rest, unless it's being acted on by an unbalanced force. What's less obvious is the idea that something moving uniformly straightforward, which is another way of saying something having a constant velocity. What he's saying is, is that something that has a constant velocity will continue to have that constant velocity indefinitely, unless it is acted on by an unbalanced force. And that's less intuitive. Because everything in our human experience-- even if I were to push this block of ice, eventually it'll stop. It won't just keep going forever, even assuming that this ice field is infinitely long, that ice will eventually stop. Or if I throw a tennis ball. That tennis ball will eventually stop. It'll eventually grind to a halt. We've never seen, at least in our human experience, it looks like everything will eventually stop. So this is a very unintuitive thing to say, that something in motion will just keep going in motion indefinitely. Everything in human intuition says if you want something to keep going in motion, you have to keep putting more force, keep putting more energy into it for it to keep going. Your car won't go forever, unless you keep, unless the engine keeps burning fuel to drive and consuming energy. So what are they talking about? Well, in all of these examples-- and I think this is actually a pretty brilliant insight from all of these fellows is that-- all of these things would have gone on forever. The ball would keep going forever. This ice block would be going on forever, except for the fact that there are unbalanced forces acting on them to stop them. So in the case of ice, even though ice on ice doesn't have a lot of friction, there is some friction between these two. And so you have, in this situation, the force of friction is going to be acting against the direction of the movement of the ice. so if you have the actual water molecules in a lattice structure in the ice cube, and then here are the water molecules in a lattice structure on the ice, on the actual kind of sea of ice that it's traveling on-- they do kind of bump and grind into each other. Although they're both smooth, there are imperfections here. They bump and grind. They generate a little bit of heat. And they'll, essentially, be working against the movement. So there's a force of friction that's being applied to here. And that's why it's stopping. Not only a force of friction, you also have some air resistance. The ice block is going to be bumping into all sorts of air particles. It might not be noticeable at first, but it's definitely going to keep it from going on forever. Same thing with the ball being tossed to the air. Obviously, at some point, it hits the ground because of gravity. So that's one force acting on it. But even once it hits the ground, it doesn't keep rolling forever, once again, because of the friction, especially if there's grass" + }, + { + "Q": "at 6:15 he used H2O as a base can we use the HSO4- from the process where we generate electrophile?", + "A": "HSO\u00e2\u0082\u0084\u00e2\u0081\u00bb can act as a base, but H\u00e2\u0082\u0082O is stronger and it is present in much larger amounts.", + "video_name": "rC165FcI4Yg", + "timestamps": [ + 375 + ], + "3min_transcript": "And then over here, we would have an oxygen with three lone pairs of electrons, giving that a negative 1 formal charge. And the nitrogen, of course, is still going to have a plus 1 formal charge like that. All right, let me go ahead and highlight those electrons. So once again, these pi electrons are going to be attracted to the positive charge, nucleophile-electrophile. And those pi electrons are going to form this bond right here to our nitro group. Well, once again, as we've seen several times before, we took away a bond from this carbon. So that's where our plus 1 the formal charge is going to go like that. And so we can draw some resonance structures. So let's go ahead and show a resonance structure for this. We could move these pi electrons in over here. So let's go ahead and draw that. So we had a hydrogen up here. And you could just show a nitro group as NO2. So I'm just going to go ahead and do that to save some time. And I'm saying that those pi electrons moved over to here. So let me go ahead and highlight those. So these pi electrons in blue move over to here, took a bond away from that carbon. So now we can put a plus 1 formal charge at that carbon like that. We can draw yet another resonance structure. So I could show these electrons over here moving to here. So let me go ahead and draw that. So we have our ring. We have our nitro group already on our ring. We have some pi electrons right here. And we have some more pi electrons moving from here to here, which, of course, takes a bond away from this top carbon. So that's where our positive 1 formal charge is now. So now we have our three resonance structures. And remember, once again, that the sigma complex is a hybrid of these three. And we're now ready for our last step, So if we go back up to here, we think, what could function as a base? Well, the water molecule here could function as a base. So a lone pair of electrons on our water molecule are going to take that proton, which would cause these electrons to move in here to reform your aromatic ring. So let's go ahead and show that. So we're going to reform our benzene ring here. And we took off the proton. So deprotonation of the sigma complex yields our product with a nitro group substituted in. So let me go ahead and highlight those electrons again. So this time I'll use green. So these electrons right in here, when that sigma complex is deprotonated, those electrons are going to move in here to restore the aromatic ring, and we have created our product. We have added in our nitro group." + }, + { + "Q": "At 10:30 he says that the Egg gets all the organelles. Does that mean that we get all of the organelles from our mother? (I know about the maternal inheritance of mitochondria.) If so, what happens with the organelles in the sperm cells?", + "A": "No such thing. Organelles are made as instructed by DNA. The part of mitochondrial genome that is still within it is maternally inherited. All others are synthesised using simple molecules, as instructed by maternal and paternal DNA in the nucleus.", + "video_name": "TX7-Kdn6lJQ", + "timestamps": [ + 630 + ], + "3min_transcript": "and were then pulled in half, but not here. In meiosis, each chromosome lines up next to it's homologous pair partner that it's already swapped a few genes with. Now, the homologous pairs get pulled apart and migrate to either end of the cell and that's anaphase I. The final phase of the first round, telophase I rolls out in pretty much the same way as mitosis. The nuclear membrane reforms, the nucleoli form within them, the chromosomes fray out back into chromatin, a crease forms between the two new cells called cleavage and then the two new nuclei move apart from each other, the cells separate in a process called cytokinesis, literally again, cell movement and that is the end of round one. We now have two haploid cells, each with 23 double chromosomes that are new, unique combinations of the original chromosome pairs. In these new cells, the chromosomes are still duplicated and still connected at the centromeres. They still look like X's, but remember, the aim is to end up with four cells. Here, the process is exactly the same as mitosis, except that the aim here isn't to duplicate the double chromosomes, but instead to pull them apart into separate single strand chromosomes. Because of this, there's no DNA replication involved in prophase II. Instead, the DNA just clumps up again into chromosomes and the infrastructure for moving them, the microtubules are put back in place. In metaphase II, the chromosomes are moved into alignment into the middle of the cell and in anaphase II, the chromotids are pulled apart into separate single chromosomes. The chromosomes uncoil into chromatin, the crease form in cleavage and the final separation of cytokinesis then mark the end of telophase II. From one original cell with 46 chromosomes, we now have four new cells with 23 single chromosomes each. If these are sperm, all four of the resulting cells are the same size, but they each have slightly different genetic information and half will be for making girls and half will be for making boys, but if this is the egg making process, and the result is only one egg. To rewind a little, during telophase I, more of the inner goodness of a cell, the cytoplasm, the organelles heads into one of the cells that gets split off then to the other one. In telophase II, when it's time to split again, the same thing happens with more stuff going into one of the cells than the other. This big ol' fat remaining cell becomes the egg with more of the nutrients and cytoplasm and organelles that it will take to make a new embryo. The other three cells that were produced, the little ones, are called polar bodies and they're totally useless in people, though they are useful in plants. In plants, those polar bodies actually, also get fertilized too and they become the endosperm. That's the starchy, protein-ey stuff that we grind into wheat, or pop into popcorn and it's basically the nutrients that feed the plant embryo, the seed. And that's all there is to it. I know you were probably were really excited when I started talking about reproduction, but then I rambled on for a long time" + }, + { + "Q": "At 9:30 he talked about the organ of corti that is comprised of both a basilar and tectorial membrane. What is the functionality of the two membranes and where are they located within the organ of corti? Thank you.", + "A": "the basilar membrane is within the cochlea of the ear and is quite stiff . There are two fluid filled tubes that are in the coil and the function of the membrane is to keep these two fluids away from each other as they are very different. the tectorial membrane is a gel with 97% being water. it is parallel to the basilar membrane. its exact function has not yet been found but it is understood that it is essential for normal functioning of the ear.", + "video_name": "6GB_kcdVMQo", + "timestamps": [ + 570 + ], + "3min_transcript": "This vibration causes three little bones, known as the malleus, incus and stapes to vibrate back and forth accordingly. The next thing that happens is the stapes is attached to this oval window over here. It's known as the elliptical window, which I'm underlining here. It's also known as the oval window. This oval window starts to vibrate back and forth as well. The next thing that happens is there's actually fluid, so this structure that the oval window is attached to is known as the cochlea. This round structure right here is known as the cochlea. Inside the cochlea is a bunch of fluid. As the oval window gets pushed inside and outside of the cochlea by the stapes, it actually pushes the fluid. It causes the fluid to be pushed this way, and causes the fluid to go all the way around the cochlea. It keeps going all the way around the cochlea, until it reaches the tip of the cochlea. what does it do? The only thing it can do is go back. Now the fluid is gonna have to go back. Let's just follow this green line over here. The fluid moves back towards where it came, but it actually doesn't go back to the oval window. It actually goes to this other window known as the circular, or round, window. Let me just fix that, so it goes to this circular or round window. It causes the round window to get pushed out. This basically keeps happening, so the fluid moves all the way to the tip of the cochlea, all the way back out, and back and forth, and back and forth, until the energy of this sound wave - eventually the fluid stops moving - all that energy is dissipated. Meanwhile, hair cells inside the cochlea are being pushed back and forth, and that transmits an electrical impulse The reason that the fluid doesn't move back to the oval window when it goes to the very tip of the cochlea is because in between, in the very middle of the cochlea, is a membrane. Let me use a different color. There's actually a membrane. I'm gonna use this black line to demarcate the membrane that runs along the length of the cochlea. This membrane is something known as the organ of corti. Let me just write that down here - organ of corti. This organ of corti is actually composed of two different things. It's composed of something known as the basilar membrane, and another membrane known as the tectorial membrane. tectorial membrane. One final thing that I just want to touch upon is a general classification of the different parts of the ear." + }, + { + "Q": "At 3:21 he mentioned frequency as how close the peaks are, instead that distance between two peaks is called wavelength. Frequency is just the number of waves in a specific time period. Feel free to comment, I am not 100% sure!", + "A": "Yes ,you are correct.", + "video_name": "6GB_kcdVMQo", + "timestamps": [ + 201 + ], + "3min_transcript": "and it makes a very distinct sound. Let's imagine that these two lines right here are your hands. When you clap your hands, the lines move towards each other, so your hands are moving towards each other. They're moving towards each other fairly quickly. In between your hands are a whole bunch of little air molecules, which I'm drawing. These air molecules, which I'm drawing right now, let's imagine that they're these little purple dots. So, in between your hands are a whole bunch of these air molecules. They're just floating around doing their thing. Then all of a sudden, the hands are moving towards each other, and all of a sudden, this space that these air molecules occupy gets a lot smaller. A little bit later in time, as the hands are moving towards each other - so here we are just drawing the hands almost about to touch. What happened was all these air molecules that are just floating around, they had all this space.. Now all of a sudden, they're really compressed, so they're really, really close together, They're very compacted now. You can imagine that as your hands are even closer together, that the air molecules get even more compacted. Basically, what is effectively going on is the air molecules here are getting pressurized. As you bring your hands together, you're actually adding all the molecules up, and it creates this pressure. This area of pressure actually tries to escape. It tries to escape and it kinda goes this way. It tries to escape out wherever it can. As it's escaping, it creates these areas of high and low pressure. That's what I'm representing here by these lines. These areas of high and low pressure are known as sound waves. We can have different types of sound waves. We can have sound waves that are really, really close together, or really far away from each other. If we draw this graphically, Basically, what I'm drawing here is, up here would be an area of high pressure. Over here would be an area of low pressure. Basically, there are just areas of high and low pressure. How close these peaks are together is the frequency. If I clap my hands even faster together, or if there's something else that's a higher frequency, a higher-pitched sound, the sound waves would be closer to one another, and it would look something like this. Depending on the frequency of the sound wave, it's perceived to be a different noise. Let's imagine that this sound wave right here is F1, and that this one over here is F2. Sound waves of lower frequency actually travel further. This actually happens in the ear, so these lower frequency sound waves actually travel further, and they actually penetrate deeper into the cochlea," + }, + { + "Q": "Does the equation at 4:56 imply that there is no magnetic force when a charge isn't moving? If so, how does a paperclip feel a magnetic force towards a magnet when both objects are held stationary?", + "A": "Go to youtube and search for veritasium how do magnets work and watch the pair of videos.", + "video_name": "NnlAI4ZiUrQ", + "timestamps": [ + 296 + ], + "3min_transcript": "So that's fine, you say, Sal, that's nice. You drew these field lines. And you've probably seen it before if you've ever dropped metal filings on top of a magnet. They kind of arrange themselves But you might say, well, that's kind of useful. But how do we determine the magnitude of a magnetic field at any point? And this is where it gets interesting. The magnitude of a magnetic field is really determined, or it's really defined, in terms of the effect that it has on a moving charge. So this is interesting. I've kind of been telling you that we have this different force called magnetism that is different than the electrostatic force. But we're defining magnetism in terms of the effect that it has on a moving charge. And that's a bit of a clue. And we'll learn later, or hopefully you'll learn later as you advance in physics, that magnetic force or a magnetic field is nothing but an electrostatic field moving at a very high speed. Or you could almost view it as they are the same thing, just from different frames of reference. I don't want to confuse you right now. But anyway, back to what I'll call the basic physics. So if I had to find a magnetic field as B-- so B is a vector and it's a magnetic field-- we know that the force on a moving charge could be an electron, a proton, or some other type of moving charged particle. And actually, this is the basis of how they-- you know, when you have supercolliders-- how they get the particles to go in circles, and how they studied them by based on how they get deflected by the magnetic field. But anyway, the force on a charge is equal to the magnitude of the charge-- of course, this could be positive or negative-- times, and this is where it gets interesting, the velocity of the charge cross the magnetic field. So you take the velocity of the charge, you could either multiply it by the scalar first, or you could take the cross product then multiply it by the scalar. this isn't a vector. But you essentially take the cross product of the velocity and the magnetic field, multiply that times the charge, and then you get the force vector on that particle. Now there's something that should immediately-- if you hopefully got a little bit of intuition about what the cross product was-- there's something interesting going on here. The cross product cares about the vectors that are perpendicular to each other. So for example, if the velocity is exactly perpendicular to the magnetic field, then we'll actually get a number. If they're parallel, then the magnetic field has no impact on the charge. That's one interesting thing. And then the other interesting thing is when you take the cross product of two vectors, the result is perpendicular to both of these vectors. So that's interesting. A magnetic field, in order to have an effect on a charge, has to be perpendicular to its you velocity. And then the force on it is going to be perpendicular to" + }, + { + "Q": "what does Q stand for at 5:03", + "A": "The magnitude of the charge.", + "video_name": "NnlAI4ZiUrQ", + "timestamps": [ + 303 + ], + "3min_transcript": "So that's fine, you say, Sal, that's nice. You drew these field lines. And you've probably seen it before if you've ever dropped metal filings on top of a magnet. They kind of arrange themselves But you might say, well, that's kind of useful. But how do we determine the magnitude of a magnetic field at any point? And this is where it gets interesting. The magnitude of a magnetic field is really determined, or it's really defined, in terms of the effect that it has on a moving charge. So this is interesting. I've kind of been telling you that we have this different force called magnetism that is different than the electrostatic force. But we're defining magnetism in terms of the effect that it has on a moving charge. And that's a bit of a clue. And we'll learn later, or hopefully you'll learn later as you advance in physics, that magnetic force or a magnetic field is nothing but an electrostatic field moving at a very high speed. Or you could almost view it as they are the same thing, just from different frames of reference. I don't want to confuse you right now. But anyway, back to what I'll call the basic physics. So if I had to find a magnetic field as B-- so B is a vector and it's a magnetic field-- we know that the force on a moving charge could be an electron, a proton, or some other type of moving charged particle. And actually, this is the basis of how they-- you know, when you have supercolliders-- how they get the particles to go in circles, and how they studied them by based on how they get deflected by the magnetic field. But anyway, the force on a charge is equal to the magnitude of the charge-- of course, this could be positive or negative-- times, and this is where it gets interesting, the velocity of the charge cross the magnetic field. So you take the velocity of the charge, you could either multiply it by the scalar first, or you could take the cross product then multiply it by the scalar. this isn't a vector. But you essentially take the cross product of the velocity and the magnetic field, multiply that times the charge, and then you get the force vector on that particle. Now there's something that should immediately-- if you hopefully got a little bit of intuition about what the cross product was-- there's something interesting going on here. The cross product cares about the vectors that are perpendicular to each other. So for example, if the velocity is exactly perpendicular to the magnetic field, then we'll actually get a number. If they're parallel, then the magnetic field has no impact on the charge. That's one interesting thing. And then the other interesting thing is when you take the cross product of two vectors, the result is perpendicular to both of these vectors. So that's interesting. A magnetic field, in order to have an effect on a charge, has to be perpendicular to its you velocity. And then the force on it is going to be perpendicular to" + }, + { + "Q": "i dont get what a magnetic mono-pole is ( 0:42 - 0:46 ) ?", + "A": "Thanx Matt :)", + "video_name": "NnlAI4ZiUrQ", + "timestamps": [ + 42, + 46 + ], + "3min_transcript": "We know a little bit about magnets now. Let's see if we can study it further and learn a little bit about magnetic field and actually the effects that they have on moving charges. And that's actually really how we define magnetic field. So first of all, with any field it's good to have a way to visualize it. With the electrostatic fields we drew field lines. So let's try to do the same thing with magnetic fields. Let's say this is my bar magnet. This is the north pole and this is the south pole. Now the convention, when we're drawing magnetic field lines, is to always start at the north pole and go towards the south pole. And you can almost view it as the path that a magnetic north monopole would take. So if it starts here-- if a magnetic north monopole, even though as far as we know they don't exist in nature, although they theoretically could, but let's just say for the sake of argument that we do have a magnetic north monopole. If it started out here, it would want to run away from this north pole and would try to get to the south pole. something like this. If it started here, maybe its path would look something like this. Or if it started here, maybe its path would look something like this. I think you get the point. Another way to visualize it is instead of thinking about a magnetic north monopole and the path it would take, you could think of, well, what if I had a little compass here? Let me draw it in a different color. Let's say I put the compass here. That's not where I want to do it. Let's say I do it here. The compass pointer will actually be tangent to the field line. So the pointer could look something like this at this point. It would look something like this. And this would be the north pole of the pointer and this would be the south pole of the pointer. Or you could-- that's how north and south were defined. People had compasses, they said, oh, this is the north seeking pole, and it points in that direction. of the larger magnet. And that's where we got into that big confusing discussion of that the magnetic geographic north pole that we're used to is actually the south pole of the magnet that we call Earth. And you could view the last video on Introduction to Magnetism to get confused about that. But I think you see what I'm saying. North always seeks south the same way that positive seeks negative, and vice versa. And north runs away from north. And really the main conceptual difference-- although they are kind of very different properties-- although we will see later they actually end up being the same thing, that we have something called an electromagnetic force, once we start learning about Maxwell's equations and relativity and all that. But we don't have to worry about that right now. But in classical electricity and magnetism, they're kind of a different force. And the main difference-- although you know, these field lines, you can kind of view them as being similar-- is that magnetic forces always come in dipoles, soon. while you could have electrostatic forces that are monopoles." + }, + { + "Q": "at2:01 why does sal substitute the value of acceleration due to gravity?", + "A": "Because this is equal to the centripetal acceleration at this point as the only force acting on the object at the top of the circle is Fg (draw the free body diagram)", + "video_name": "4SQDybFjhRE", + "timestamps": [ + 121 + ], + "3min_transcript": "What I want to do now is figure out, what's the minimum speed that the car has to be at the top of this loop de loop in order to stay on the track? In order to stay in a circular motion. In order to not fall down like this. And I think we can all appreciate that is the most difficult part of the loop de loop, at least in the bottom half right over here. The track itself is actually what's providing the centripetal force to keep it going in a circle. But when you get to the top, you now have gravity that is pulling down on the car, almost completely. And the car will have to maintain some minimum speed in order to stay in this circular path. So let's figure out what that minimum speed is. And to help figure that out, we have to figure out what the radius of this loop de loop actually is. And it actually does not look like a perfect circle, based on this little screen shot that I got here. It looks a little bit elliptical. But it looks like the radius of curvature right over here is actually smaller than the radius That if you made this into a circle, it would actually be maybe even a slightly smaller circle. But let's just assume, for the sake of our arguments right over here, that this thing is a perfect circle. And it was a perfect circle, let's think about what that minimum velocity would have to be up here at the top of the loop de loop. So we know that the magnitude of your centripetal acceleration is going to be equal to your speed squared divided by the radius of the circle that you are going around. Now at this point right over here, at the top, which is going to be the hardest point, the magnitude of our acceleration, this is going to be 9.81 meters per second squared. And the radius, we can estimate-- I copied and pasted the car, and it looks like I can get it to stack on itself four times to get And I looked it up on the web, and a car about this size is going to be about 1.5 meters high from the bottom of the tires to the top of the car. And so it looks like-- just eyeballing it based on these copying and pasting of the cars, that the radius of this loop de loop right over here is 6 meters. So this right over here is 6 meters. So you multiply both sides by 6 meters. Or actually, we could keep it just in the variables. So let me just rewrite it-- just to manipulate it so we can solve for v. We have v squared over r is equal to a. And then you multiply both sides by r. You get v squared is equal to a times r. And then you take the principal square root of both sides. You get v is equal to the principal square root of a times r. And then if we plug in these numbers, this velocity that we have to have in order to stay in the circle is going to be the square root of 9.81" + }, + { + "Q": "at 9:00, you said the cyclobutadiene is antiaromatic, but you didn't really mention this molecule is planar.\nonly at the start when you talk about this molecule, you mentioned the p orbitals may overlap each other.\nbut may doesn't mean definitely", + "A": "If a compound doesn t follow Huckel s rule it can t be aromatic. In fact cyclobutadiene has 4n pi electrons which would make it antiaromatic. And yes it is planar. In fact what happens to avoid this issue: cyclobutadiene will distort into a rectangle with 2 long sides and 2 short sides, this will make more sense if you draw the frost circle for the rectangle. Hopefully that makes sense.", + "video_name": "yg0XJWHPqOA", + "timestamps": [ + 540 + ], + "3min_transcript": "are going to give me four molecular orbitals. And I can draw in my frost circle right here. So once again, we're going to draw a line through the center to separate my bonding from my antibonding molecular orbitals. You always start at the bottom of your frost circle. Four-membered ring. So we're going try to draw a four-sided figure in our frost circle. So I'm going to attempt to draw this square in here. And once again, where our polygon intersects with our circle represents the energy level of our molecular orbitals. So I have a total of four molecular orbitals. And if I go over here, I have a molecular orbital below the line. So that's my bonding molecular orbital. I have a molecular orbital above the center line there. So that's my antibonding molecular orbital. And this time I have two molecular orbitals that are right on the line, which represents non-bonding molecular orbitals. When I go ahead and put in my 4 pi electrons, So that takes care of 2 pi electrons. And now I have two more. And so here I have the non-bonding molecular orbitals are on the same energy level. And so if you remember Hund's rule from electron configurations, you can't pair up these last two pi electrons because the orbitals are of equal energy. And so this is the picture that we get. And you can see that I have two unpaired electrons. And two unpaired electrons implies that this molecule is extremely reactive. And experimentally, it is. So cyclobutadiene will actually react with itself. So it's experimentally extremely reactive which tells you that it's not extra stable. It's not aromatic. So you don't have 4n plus 2 pi electrons. And so the second criteria is not fulfilled, but this compound does satisfy the first criteria. And so the term for this compound is anti-aromatic, which means it fulfills the first criteria, but does not satisfy the second criteria. It does not have 4n plus 2 pi electrons. It has 4n pi electrons. And so we say that is anti-aromatic. And there are actually very few examples of anti-aromatic compounds. But cyclobutadiene is considered to be anti-aromatic. In the next video, we're going to look at a few more examples of aromatic stability." + }, + { + "Q": "At 0:43 where was the particle before expansion? What were its surroundings?", + "A": "there re no surroundings. The universe is only expanding on the inside because there is no outside. If there were an outside of the universe, it s the same size it s always been.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 43 + ], + "3min_transcript": "Right now, the prevailing theory of how the universe came about is commonly called the Big Bang theory. And really is just this idea that the universe started as kind of this infinitely small point, this infinitely small singularity. And then it just had a big bang or it just expanded from that state to the universe that we know right now. And when I first imagined this-- and I think if it's also a byproduct of how it's named-- Big Bang, you kind of imagine this type of explosion, that everything was infinitely packed in together and then it exploded. And then it exploded outward. And then as all of the matter exploded outward, it started to condense. And then you have these little galaxies and super clusters of galaxies. And they started to condense. And then within them, planets condensed and stars condensed. And then we have the type of universe that we have right now. has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that" + }, + { + "Q": "At 2:40, if the Universe is finite or has no edge then would you be able to get outside or would it be impossible?", + "A": "it would be impossible", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 160 + ], + "3min_transcript": "has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that Let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitudinal lines on this sphere. On this sphere, all of a sudden-- and I'll shade it in a little bit, make it look nice-- this type of a sphere, you have a finite area. You could imagine the surface of a balloon, or the surface of a bubble, or the surface of the Earth. You have a finite area, but you have no edge. If you keep going forever in one direction, you're going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and-- and I don't want to say finite area anymore, because we're not talking about a three-dimensional space. Let me draw it over here. So let's think about a three-dimensional space, so a three-dimensional space." + }, + { + "Q": "At 10:30 , if three points are in the universe , and after some significant amount of time , the universe expands and the three points get separated farther apart. So , is the EARTH getting any farther apart from the SUN and the MOON getting farther apart from EARTH??", + "A": "No, gravitational forces overpower the force of expansion at closer ranges.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 630 + ], + "3min_transcript": "means that if you go up and you just keep going up, you'll eventually come back from the bottom. So if you keep going all the way up, you'll eventually come back to the point that you were. It might be an unbelievably large distance, but you'll eventually get back where you were. If you go to the right, you'll eventually come back all the way around to the point where you were. And if you were to go into the page-- so if you were to go into the page-- let me draw it that way-- if you go into the page, you would eventually come back from above the page and come back to the point that you are. So that's what this implication would be. That you would eventually get back to where you are. So let's go back to the question of an expanding universe, a expanding universe that's not expanding into any other space. That is all of the space, but it's still expanding. Well, this is the model. So you could imagine shortly after the Big Bang, our four-dimensional sphere looked like this. Maybe right at the Big Bang, it was like this little unbelievably small sphere. Then a little bit later, it's this larger sphere. Let me just shade it in to show you that it's kind of popping out of the page, that's it's a sphere. And then at a later time, the sphere might look like this. The sphere might look like this. Now, your temptation might be to say, wait, Sal, isn't this stuff outside of this sphere, isn't that some type of a space that it's expanding into? Isn't that somehow part of the universe? And I would say if you're talking in three dimensions, no, it's not. The entire universe is this surface. It is this surface of this four-dimensional sphere. If you start talking about more dimensions, then, yes, you could talk about maybe things outside of our three-dimensional universe. So as this expands in space/time-- so one way to view the fourth dimension getting further and further apart. And I'll talk about more evidence in future videos for why the Big Bang is the best theory we have out there right now. But as you could imagine, if you have two points on this sphere that are that far apart, as this sphere expands, this four-dimensional sphere, as this bubble blows up or this balloon blows up, those two points are just-- let me draw three points. Let's say those are three points. Those three points are just going to get further and further apart. And that's actually one of the main points that-- or one of the first reasons why it made sense to believe the Big Bang-- is that everything is expanding, not from some central point. But everything is expanding from everything. That if you go in any direction from any point in the universe, everything else is expanding away. And the further away you go, it looks like the faster it's expanding away from you. So I'll leave you there, something for you to kind of think about a little bit. And then we'll build on some of this to think about what it means to kind of observe the observable universe." + }, + { + "Q": "in 5:52, If time was the 4th dimension, wouldn't the universe experience a cycle through time?", + "A": "The sphere as a fourth dimension would be infinitely expanding, so even time cannot make it back to the point of origin.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 352 + ], + "3min_transcript": "I want to talk about a finite volume and no edge. How do I do that? And when you think about it superficially, well, look, if I have a finite volume, maybe it'll be contained in some type of a cube. And then we clearly have edges in those situations. Or you could even think about a finite volume as being the inside of a sphere. And that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has a finite volume and no edge? And that I'm going to tell you right now, it's very hard for us to visualize it. But in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy, unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere. This is a two-dimensional surface. On the surface of the sphere, you can only move into directions, two perpendicular directions. You could move like that or you could move like that. You could move left and right or you could move up and down. So it's a two-dimensional surface of a three-dimensional sphere. So if we take it by analogy, let's imagine, and it's hard to imagine, a three-dimensional surface. And you can do it mathematically. The math here is actually not that difficult. It's a three-dimensional surface of a four-dimensional sphere. And I'm going to draw it the same way. are just these two dimensions of the surface, the same thing. It's the same thing. And if you imagine that-- I'm not saying that this is actually the shape of the universe. We don't know the actual shape. But we do know that it does have a slight curvature. We don't know the actual shape, but a sphere is the simplest. There's other ones we could do. A toroid would also fit the bill of having a finite volume with no edge. And another thing, I want to make it clear, we actually don't even know whether it has just a finite volume. That's still an open question. But what I want to do is show you that it can have a finite volume and also have no edge. And most people believe-- and I want to say \"believe\" here because we can just go based on evidence and all that-- that we are talking about something with a finite volume, especially when you talk about the Big Bang theory. That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite." + }, + { + "Q": "At 10:14 Sal talks about the sphere getting bigger and the points are getting further away from each other. So, if the universe is a sphere and it is continually expanding, Will we (earth) ever possibly get further away from the sun? Will the sun engulf us all before that happens?", + "A": "Earth is held near the sun by gravity, which keeps them together even as the space they are in stretches.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 614 + ], + "3min_transcript": "means that if you go up and you just keep going up, you'll eventually come back from the bottom. So if you keep going all the way up, you'll eventually come back to the point that you were. It might be an unbelievably large distance, but you'll eventually get back where you were. If you go to the right, you'll eventually come back all the way around to the point where you were. And if you were to go into the page-- so if you were to go into the page-- let me draw it that way-- if you go into the page, you would eventually come back from above the page and come back to the point that you are. So that's what this implication would be. That you would eventually get back to where you are. So let's go back to the question of an expanding universe, a expanding universe that's not expanding into any other space. That is all of the space, but it's still expanding. Well, this is the model. So you could imagine shortly after the Big Bang, our four-dimensional sphere looked like this. Maybe right at the Big Bang, it was like this little unbelievably small sphere. Then a little bit later, it's this larger sphere. Let me just shade it in to show you that it's kind of popping out of the page, that's it's a sphere. And then at a later time, the sphere might look like this. The sphere might look like this. Now, your temptation might be to say, wait, Sal, isn't this stuff outside of this sphere, isn't that some type of a space that it's expanding into? Isn't that somehow part of the universe? And I would say if you're talking in three dimensions, no, it's not. The entire universe is this surface. It is this surface of this four-dimensional sphere. If you start talking about more dimensions, then, yes, you could talk about maybe things outside of our three-dimensional universe. So as this expands in space/time-- so one way to view the fourth dimension getting further and further apart. And I'll talk about more evidence in future videos for why the Big Bang is the best theory we have out there right now. But as you could imagine, if you have two points on this sphere that are that far apart, as this sphere expands, this four-dimensional sphere, as this bubble blows up or this balloon blows up, those two points are just-- let me draw three points. Let's say those are three points. Those three points are just going to get further and further apart. And that's actually one of the main points that-- or one of the first reasons why it made sense to believe the Big Bang-- is that everything is expanding, not from some central point. But everything is expanding from everything. That if you go in any direction from any point in the universe, everything else is expanding away. And the further away you go, it looks like the faster it's expanding away from you. So I'll leave you there, something for you to kind of think about a little bit. And then we'll build on some of this to think about what it means to kind of observe the observable universe." + }, + { + "Q": "Whats finite at 07:12", + "A": "finite means it has limits unlike infinite", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 432 + ], + "3min_transcript": "This is a two-dimensional surface. On the surface of the sphere, you can only move into directions, two perpendicular directions. You could move like that or you could move like that. You could move left and right or you could move up and down. So it's a two-dimensional surface of a three-dimensional sphere. So if we take it by analogy, let's imagine, and it's hard to imagine, a three-dimensional surface. And you can do it mathematically. The math here is actually not that difficult. It's a three-dimensional surface of a four-dimensional sphere. And I'm going to draw it the same way. are just these two dimensions of the surface, the same thing. It's the same thing. And if you imagine that-- I'm not saying that this is actually the shape of the universe. We don't know the actual shape. But we do know that it does have a slight curvature. We don't know the actual shape, but a sphere is the simplest. There's other ones we could do. A toroid would also fit the bill of having a finite volume with no edge. And another thing, I want to make it clear, we actually don't even know whether it has just a finite volume. That's still an open question. But what I want to do is show you that it can have a finite volume and also have no edge. And most people believe-- and I want to say \"believe\" here because we can just go based on evidence and all that-- that we are talking about something with a finite volume, especially when you talk about the Big Bang theory. That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite. Let's imagine this sphere. Once again, if you're on this surface of this four-dimensional sphere-- I obviously can not draw a four-dimensional sphere. But if you're on the surface of this four-dimensional sphere, if you go in any direction, you'll come back out and come back to where you started. If you go that way, you'll come back around here. Now, the universe is super huge. So even light, maybe light itself will take an unbelievable amount of time to traverse it. And if this sphere itself is expanding, it might be expanding so fast that light might not ever be able to come back around it. But in theory, if something were fast enough, if something were to keep going around, it could eventually go back to this point. Now, when we talk about a three-dimensional surface-- it's a three-dimensional surface of a four-dimensional sphere-- that means that any of the three dimensions-- over here, on the surface, I can only draw two. But that means if this is true, if the universe is" + }, + { + "Q": "if the percentage of mercury to chlorine is higher, why is there a 1:2 ratio of mercury?", + "A": "An atom of mercury has a lot more mass than an atom of chlorine. Compare the relative atomic masses from the periodic table, mercury is 200.59 while chlorine is only 35.45...so 1 atom of mercury has the mass of about 5.6 chlorine atoms. This is why we need to use moles!!", + "video_name": "NM0WycKCCDU", + "timestamps": [ + 62 + ], + "3min_transcript": "Sal: What I want to do in this video is start with mass composition and see if we can figure out the empirical formula of the molecule that we're dealing with based on the mass composition. Let's say that we have a bag and we're able to measure that this bag is 73 percent, it's 73 percent mercury and it is, the remainder of the bag, 27 percent chlorine. Based just on this can we figure out the likely empirical formula for the molecule that we have in that bag? I encourage you to pause the video and try to see if you can figure it out on your own. Well one way to think about it is let's just assume a number. This is all the information we have, let's just assume we have 100 grams of it. We could assume a thousand grams or 10,000 grams or 57 grams, but I'll pick 100 grams because it will make the numbers easy to work with in our head. Let's just assume, let me make it clear that I'm assuming this. I'm going to assume that I have 100 grams of this molecule If I assume that that means that the 73 percent that is mercury is going to be 73 grams and the 27 percent that is chlorine is going to be 27 grams of chlorine. Let me make it clear this is mercury and this is chlorine. Now I just need to think about, well how many moles of mercury is 73 grams and how many moles of chlorine is 27 grams. To do that I'll look up the periodic table right here. I have the atomic weight which is the weighted average of the atomic masses that's found in nature. The atomic weight here for mercury is 200.59. That means, let me write this right over here. One mole of, one mole of mercury is, Similarly we could look up the atomic weight for chlorine. Chlorine right over here, 35.453. We could say one mole of chlorine, and once again this is a weighted average of all of the isotopes of chlorine as found in nature. I guess we'll just go with that number. One mole of chlorine is going to be 35.453, 35.453, 35.453 grams. Given this information right over here how many moles of mercury is this, roughly, and how many moles of chlorine is this, roughly? I say roughly because getting an empirical formula from measurements of mass composition is going to be necessarily a messy affair, it's not going to come out completely," + }, + { + "Q": "at 6:16 for calculating the ratio, is the bigger number of moles always divided by the smallest number of moles, regardless of the differences in mass percentages?", + "A": "Yes. Always divide all the moles you have calculated by the smallest moles, that gives you the ratio between the atoms which is essentially the empirical formula.", + "video_name": "NM0WycKCCDU", + "timestamps": [ + 376 + ], + "3min_transcript": "27 divided by 35.453 is equal to .76, I'll just say two. So 0.762 moles of chlorine. What's going to be the ratio of mercury to chlorine? Or I guess we could say since chlorine, there's more of that, chlorine to mercury. Remember, this is just a number. When I say 0.762 moles, this is just 0.762 times Avogadro's number of chlorine atoms. This is 0.364 times Avogadro's number of mercury atoms. We can literally think of this as the ratio. This is a certain number of moles, Well what's the ratio, let's see. What's the ratio of chlorine to mercury? Well you can eyeball it, it looks like it's roughly two to one, and you can verify that if you take that number and you divide it by .3639, and once again I'm just going to get the rough approximate. You can see it's pretty close to two. So when you see something like this the simplest explanation is often the best. Okay, there probably will be some measuring error right over here, but you can say that it looks like roughly, this is what I'm talking about when you're trying to find the empirical formula for mass composition it tends to be a rough science. You can say roughly the ratio of chlorine to mercury is two to one. You have two chlorines for every mercury. And because of that you can say this is likely to be, so likely, for every mercury you have two chlorines, you have two chlorines. very likely that you have mercury two chloride. The reason why it's called mercury two chloride is because, well I won't go into too much detail right over here but chlorine is highly electronegative, it's an oxidizing agent, it likes to take other people's electrons or hog other people electrons. In this case it's hogging, since each of the chlorine likes to hog one electron, this case two chlorines are going to hog two electrons, so it's hogging two electrons from the mercury. When you lose electrons or when your electrons are being hogged you're being oxidized, so the oxidation state on mercury right over here is two. Two of its electrons are being hogged, one by each of the two chlorines. This is mercury two chloride, where the two is the oxidation state of the mercury." + }, + { + "Q": "at 3:57, I thought that each element had a set number of electrons. So is that a hypothetical question orrr...", + "A": "In a NEUTRAL atom the number of protons is equal to the number of electrons. But this does not always have to be the case. Atoms can and do gain or lose electrons. This is the whole point of the video. An ion is an atom that does not have the same number of electrons and protons, so it has a charge.", + "video_name": "zTUnjPALX_U", + "timestamps": [ + 237 + ], + "3min_transcript": "what element you're dealing with, so now if you look at what element has five protons we're dealing with boron. So this is going to be boron. Neutral boron would have five protons and five electrons. But this one has one extra electron, so it has one extra negative charge. So you can write it like this, one minus. Or you could just say it has a negative charge. So this is a boron ion right over here. As soon as you have an imbalance between protons and electrons you no longer would call it an atom, you would call it an actual ion. Now let's do an example question dealing with this. So our question tells us... Our question ... our question tells us ... So let's just look up platinum on our periodic table. Platinum is sitting right over here if you can see it. So an atom of platinum has a mass number of 195. And 195 looks pretty close to that atomic mass we have there. And it contains 74 electrons. 74 electrons. How many protons and neutrons does it contain and what is its charge? Alright, so let's think about this a little bit. So we're dealing with platinum. So by definition platinum has 78 protons, so we know that. It has 78 protons. They're telling us it has 74 electrons. 74 electrons. protons than electrons. So you're going to have a positive four charge. Four more of the positive thing than you have of the negative things. So you could write this as platinum with a plus four charge. This is a platinum ion, a positive platinum ion. The general term when we're talking about a positive ion, we're talking about a cation. That is a positive ion. Up there when we talked about boron being negative, a negative ion, that is an anion. This is just to get ourselves used to some of the terminology. But we're not done answering the question. They say an atom of platinum has a mass number of 195" + }, + { + "Q": "Hey Sal! at 5:03, you mentioned that there's a 'propyl' functional group on Carbon-3...just to clarify, it's an 'ethyl' group, not propyl (C2H5) :)", + "A": "people with exceptional talent are prone to commit minor mistakes more often - my maths tution teacher", + "video_name": "GFiizJ-jGVw", + "timestamps": [ + 303 + ], + "3min_transcript": "So this molecule is zusammen. Which on some levels, you can think of as the same thing as cis, but cis and trans stops applying when you start having more than two functional groups. In this case, we have three. So we would call this Z-4-methylhept-3-ene. And that's because the higher priority functional groups are on the same side of the double bond. Now let's do this one over here. And someone pointed out, rightly, that I had misspelled zusammen in the last video. It's actually spelled like this, zusammen. I had spelled it with two s's and one m. I guess you can forgive me. I don't speak German. But anyway, I thought I would point that out. Now let's try to label this thing right over here. So the first thing, this once again is an alkene. Let's identify the longest carbon chain here. So it looks like one, two, three, four, five, six, seven, eight carbons. Double bonds are closer to the left hand side. One, two, three, four, five, six, seven, eight. So just the main chain is oct-- let me make sure I have some space here-- it is oct-3-ene. And then we have, well we have one functional group sitting off of the main chain. We have this bromine sitting right over there on the third carbon. So we would call this 3-bromooct-3-ene. And now we have to figure out is it entgegen or zusammen. So if we look on the carbon on the right hand side, it's pretty obvious that this is the only functional group. We just have a hydrogen there. So let me circle it in the magenta. And then on the left hand side we have two functional groups. We have this [UNINTELLIGIBLE] bromo or we have a bromine sitting right there. And then we have this propyl group. tempted to say it takes higher priority. But remember, in the Cahn-Ingold-Prelog system, you give higher priority to the atom that has a higher atomic number. Bromine has an atomic number of 35. Carbon has an atomic number of only 6. So Bromine is actually higher priority. So this is the higher priority functional group right over here. So now for deciding whether it's entgegen or zusammen, we see that our higher priority groups are apart. They're on opposite sides of the double bond. This one is on top. This one is below. We are apart. So this is entgegen. Or we would write this is E-3-bromooct-3-ene. And E is for-- just as a bit of a refresher-- it's for entgegen, a word that I enjoy saying, entgegen." + }, + { + "Q": "What would the naming be for the last molecule if there were 2 bromines attached to the left hand side carbon in the last molecule (\"3:52\")? Would the E-Z naming apply at all in this case?", + "A": "If there s two then you won t have to convey which bromine is where using the E-Z convention because you have no choice but to draw both in the correct position.", + "video_name": "GFiizJ-jGVw", + "timestamps": [ + 232 + ], + "3min_transcript": "And what we need to do to identify the highest priority group is to use the Cahn-Ingold-Prelog namings or priority scheme that we learned several videos ago. And there you literally go from this carbon, you look at what it's bonded to, and you compare the atomic numbers. But in both cases it's a carbon to a carbon. This is a carbon to a carbon. So their atomic numbers are the same. So then you go one bond further away and you see which one is bonded to a higher atomic number atom. This carbon bonds to a carbon, which is a higher atomic number. This carbon only bonds to three hydrogens. This one does two hydrogens and one carbon. Because it's bonded to another carbon, it takes priority. This propyl group is a higher priority functional group. So now when we're trying to decide whether it is entgegen or zusammen, we look at these 2 groups. And we see that they are sitting on the same side of the double bond. They are both above the carbons. So this molecule is zusammen. Which on some levels, you can think of as the same thing as cis, but cis and trans stops applying when you start having more than two functional groups. In this case, we have three. So we would call this Z-4-methylhept-3-ene. And that's because the higher priority functional groups are on the same side of the double bond. Now let's do this one over here. And someone pointed out, rightly, that I had misspelled zusammen in the last video. It's actually spelled like this, zusammen. I had spelled it with two s's and one m. I guess you can forgive me. I don't speak German. But anyway, I thought I would point that out. Now let's try to label this thing right over here. So the first thing, this once again is an alkene. Let's identify the longest carbon chain here. So it looks like one, two, three, four, five, six, seven, eight carbons. Double bonds are closer to the left hand side. One, two, three, four, five, six, seven, eight. So just the main chain is oct-- let me make sure I have some space here-- it is oct-3-ene. And then we have, well we have one functional group sitting off of the main chain. We have this bromine sitting right over there on the third carbon. So we would call this 3-bromooct-3-ene. And now we have to figure out is it entgegen or zusammen. So if we look on the carbon on the right hand side, it's pretty obvious that this is the only functional group. We just have a hydrogen there. So let me circle it in the magenta. And then on the left hand side we have two functional groups. We have this [UNINTELLIGIBLE] bromo or we have a bromine sitting right there. And then we have this propyl group." + }, + { + "Q": "At 4:40 what is Thermogenesis?", + "A": "the production of heat, especially in a human or animal body.", + "video_name": "f_Z1zsR9lFM", + "timestamps": [ + 280 + ], + "3min_transcript": "So what tends to make a paracrine hormone work regionally is that the high concentration of the receptors are very close to the site of synthesis. And the same with autocrine, is often they're made, and there's a very high concentration of the receiving end right at that cell, right next to that cell. SALMAN KHAN: And this might be a silly question, but it's called endocrinology. Are there paracrinologists? NEIL GESUNDHEIT: Well, it's a good point. I don't think so. I think we just, perhaps because the paracrine function of hormones was discovered later, we still carry this all under the umbrella of endocrinology. SALMAN KHAN: Right. So all of hormones is endocrinology, even though endocrine hormones are the ones that act at far distances. NEIL GESUNDHEIT: That's right. I think that's a good way to summarize it. Now I like the diagram that you created here because it illustrates some of the major endocrine organs, the ones we'll be focusing on in later lectures. So the first one that you showed very nicely in the head, at the base of the brain, is that orange structure. And that would be the pituitary gland. That's right. because from the pituitary, we make hormones that work on yet other organs. So I'll give you an example. One of the hormones that's made by the pituitary is called thyroid stimulating hormone, or TSH. And after it leaves the pituitary, it goes into the circulation and it acts on the thyroid gland, where there are high receptors for TSH on the surface of the thyroid cells. And it stimulates the thyroid gland to make thyroid hormone, typically thyroxine T4 or triodothyronine, T3. Those would be the two main circulating thyroid hormones. SALMAN KHAN: And what do those do? NEIL GESUNDHEIT: Those regulate metabolism, they regulate appetite, they regulate thermogenesis, they regulate muscle function. They have widespread activities on other parts of the body. SALMAN KHAN: But it kind of upregulates or downregulates the entire body and the metabolism. NEIL GESUNDHEIT: That's right. So someone with hyperthyroidism would have very high metabolism. You may know the classic picture someone with a high heart rate, rapid metabolism, weight loss. That would be someone with excess amounts And then you see pretty much the inverse picture when someone has a deficiency of thyroid hormone and someone with hypothyroidism. So it's critical to maintain just the right amount of almost all of these hormones, and the thyroid hormones are good examples of this. But the ultimate regulation is from that pituitary gland. SALMAN KHAN: This is kind of the master one. It sends a signal there, and then that kind of does the-- NEIL GESUNDHEIT: That's right. And we'll talk later about feedback loops, because how does the pituitary know when to stop making TSH? And basically, like a thermostat, it can sense the levels of thyroid hormone. And when those levels are just at the right level, and not too high, it'll decrease the amount of TSH it makes. If the levels are too low, it'll increase TSH to try to stimulate the thyroid gland to make yet more thyroid hormone. SALMAN KHAN: Very cool. And what else do we have here? So the other hormones, some of the major ones. The pituitary, in addition to making the thyroid stimulating hormones, it makes a hormone called ACTH, adrenal corticotrophic hormone, which acts on the adrenal cortex. And the adrenal is that gland exactly" + }, + { + "Q": "What does that ATTACK word signifies at 5:16 ?\nPlease help. Thank you.", + "A": "Form a bond. The electrons of the O bond to the C.", + "video_name": "Z4F88tTx9-8", + "timestamps": [ + 316 + ], + "3min_transcript": "So let me show the pi bond here, and pi bonds are regions of high electron density so this pi bond can act like a nucleophile in an organic chemistry mechanism. Now let's look at electrophiles. So an electrophile is electron-loving and since electrons are negatively charged we're gonna think about an electrophile as having a region of low electron density so it could have a full positive charge on it because a positive charge would be attracted to electrons, or you could be talking about a partial positive charge. So first let's look at this compound. We know that chlorine is more electronegative than carbon. So chlorine is going to withdraw some electron density and if chlorine is withdrawing electron density away from this carbon, this carbon is partially positive. Next let's look at acetone. So oxygen is more electronegative than carbon so oxygen is going to withdraw some electron density away from this carbon here and this carbon would be partially positive, so this carbon is the electrophilic portion of this compound. Next let's look at a carbocation where there's a full positive charge on this carbon so this carbon has only three bonds to it which gives it a full positive charge. Obviously a full positive charge is going to love electrons. Opposite charges attract, so this carbon is the electrophilic portion of this ion. And finally let's look at this compound, right. We know that oxygen is more electronegative than carbon so oxygen withdraws some electron density away from structure here, so let me take these pi electrons and move them out onto the oxygen, so let's draw a resonance structure so I put in my double bond. Now if I'm showing those pi electrons moving off onto the oxygen I would need three lone pairs of electrons on that top oxygen giving it a negative one formal charge. I took a bond away from this carbon in magenta which is this carbon which gives it a plus one formal charge, so that's one of the possible resonance structures that you can draw and of course we know the carbon in magenta is an electrophilic center, but I could draw another resonance structure so let me go ahead and do that, put in my brackets over here. I could take these pi electrons, I'll show it on this one actually, these pi electrons and move them over to here, so let's draw the resulting resonance structure. So I'd have a double bond here now" + }, + { + "Q": "At 6:17, wouldn't the lack of Hydrogen in the middle Carbon (Carbocation) make it negative instead of positive?", + "A": "No. The middle carbon could be positive, negative, or a radical. It all depends on whether the carbon has lost an H\u00e2\u0081\u00ba, H\u00e2\u0081\u00bb, or H.", + "video_name": "Z4F88tTx9-8", + "timestamps": [ + 377 + ], + "3min_transcript": "Next let's look at acetone. So oxygen is more electronegative than carbon so oxygen is going to withdraw some electron density away from this carbon here and this carbon would be partially positive, so this carbon is the electrophilic portion of this compound. Next let's look at a carbocation where there's a full positive charge on this carbon so this carbon has only three bonds to it which gives it a full positive charge. Obviously a full positive charge is going to love electrons. Opposite charges attract, so this carbon is the electrophilic portion of this ion. And finally let's look at this compound, right. We know that oxygen is more electronegative than carbon so oxygen withdraws some electron density away from structure here, so let me take these pi electrons and move them out onto the oxygen, so let's draw a resonance structure so I put in my double bond. Now if I'm showing those pi electrons moving off onto the oxygen I would need three lone pairs of electrons on that top oxygen giving it a negative one formal charge. I took a bond away from this carbon in magenta which is this carbon which gives it a plus one formal charge, so that's one of the possible resonance structures that you can draw and of course we know the carbon in magenta is an electrophilic center, but I could draw another resonance structure so let me go ahead and do that, put in my brackets over here. I could take these pi electrons, I'll show it on this one actually, these pi electrons and move them over to here, so let's draw the resulting resonance structure. So I'd have a double bond here now so let me put that in here, draw in the hydrogen, put in my brackets, and I removed a bond, we took a bond away from, let me use blue for this, from this carbon, so this carbon now has a plus one formal charge, so the carbon in blue is this carbon over here, so let me draw in a plus one formal charge, so that is also electrophilic, right, a full positive charge is going to be attracted to a negative charge, so this compound actually has, this compound actually has two electrophilic centers, so this carbon here and also this carbon." + }, + { + "Q": "At 11:06, What is Molar Concentration?", + "A": "Molar concentration is defined as the number of moles of solute per liter of solvent. For instance, adding 1 mol of NaCl to make a 1 L solution makes a 1 molar solution of NaCl.", + "video_name": "qbCZbP6_j48", + "timestamps": [ + 666 + ], + "3min_transcript": "through beer. The Beer-Lambert law tells us that the absorbance is proportional-- I should write it like this-- the absorbance is proportional to the path length-- so this would be how far does the light have to go through the solution. So it's proportional to the path length times the concentration. And usually, we use molarity for the concentration. Or another way to say it is that the absorbance is equal to some constant-- it's usually a lowercase epsilon like that-- and this is dependent on the solution, or the solute in question, what we actually have in here, and the temperature, and the pressure, and all of that. Well it's equal to some constant, times the length it has to travel, times the concentration. This thing right here is concentration. And the reason why this is super useful is, you can imagine, if you have something of a known concentration-- let me draw right here. So let's say we have an axis right here, that's axis. And over here I'm measuring concentration. This is our concentration axis. And we're measuring it as molarity. And let's say the molarity starts at 0. It goes, I don't know, 0.1, 0.2, 0.3, so on and so forth. And over here you're measuring absorbance, in the vertical axis you measure absorbance. You measure absorbance just like that. Now let's say you have some solution and you know the So let me write down M for molar. And you measure its absorbance, and you just get some number here. So you measure its absorbance and you get its absorbance. So this is a low concentration, it didn't absorb that much. You get, I don't know, some number here, so let's say it's 0.25. And then, let's say that you then take another known concentration, let's say 0.2 molar. And you say that, oh look, it has an absorbance of 0.5. So let me do that in a different color. It has an absorbance, right here, at 0.5. And I should put a 0 in front of these, 0.5 and 0.25. What this tells you, this is a linear relationship. That for any concentration, the absorbance is going to be on a line. And if you want a little review of algebra, this epsilon is actually going to be the slope of that line. Well actually, the epsilon times the length will be the slope. I don't want to confuse you too much." + }, + { + "Q": "Where does the term \"spectrophotometry\"come from? 0:04", + "A": "Spectro- meaning to look photo- meaning light -metry- meaning measurement", + "video_name": "qbCZbP6_j48", + "timestamps": [ + 4 + ], + "3min_transcript": "What I want to do in this video is to talk a little bit about spectrophotometry. Spectrophotometry sounds fairly sophisticated, but it's really based on a fairly simple principle. So let's say we have two solutions that contain some type of solute. So that is solution one, and then this is solution two. And let's just assume that our beakers have the same width. Now let's say solution 1-- let me put it right here, number 1, and number 2. Now let's say that solution 1 has less of the solute in it. So that's the water line right there. So this guy has less of it. And let's say it's yellow or to our eyes it looks yellow. So this has less of it. Actually, let me do it this way. Let me shade it in like this. So it has less of it. And let's say solution number 2 has more of the solute. So it's more. So I'll just kind of represent that as more So the concentration of the solute is higher here. So let me write higher concentration. And let's say this is a lower concentration. Now let's think about what will happen if we shine some light through each of these beakers. And let's just assume that we are shining at a wavelength of light that is specifically sensitive to the solute that we have dissolved in here. I'll just leave that pretty general right now. So let's say I have some light here of some intensity. So let's just call that the incident intensity. I'll say that's I0. So it's some intensity. What's going to happen as the light exits the other side of this beaker right here? Well, some of it is going to be at absorbed. Some of this light, at certain frequencies, is going to be And so you're actually going to have less light coming out from the other side. Especially less of those specific frequencies that these molecules in here like to absorb. So your're going to have less light come out the other side. I'll call this I1. Now in this situation, if we shine the same amount of light-- so I0-- that's supposed to be an arrow there, but my arrow is kind of degrading. If we shined the same amount of light into this beaker-- so it's the same number, that and that is the same-- the same intensity of light, what's going to happen? Well more of those specific frequencies of light are going to be absorbed as the light travels through this beaker. It's just going to bump into more molecules because it's a higher concentration here." + }, + { + "Q": "Near 6:40, why does he write twiceT=I2/I0?", + "A": "Because he is mentioning about the second case, but it should be noted in order to not to be confused that it is not twice , it is just a subscript 2 identifying the second case.", + "video_name": "qbCZbP6_j48", + "timestamps": [ + 400 + ], + "3min_transcript": "You're getting the most light into your eye. This would be a slightly darker color, and this would be the darkest color. That makes complete sense. If you dissolve something, if you dissolve a little bit of something in water, it will still be pretty transparent. If you dissolve a lot of something in water, it'll be more opaque. And if the cup that you're dissolving in, or the beaker that you're in gets even longer, it'll get even more opaque. So hopefully that gives you the intuition behind spectrophotometry. And so the next question is, well what is it even good for? Why would I even care? Well you could actually use this information. You could see how much light is transmitted versus how much you put in to actually figure out the concentration of a solution. That's why we're even talking about it in a chemistry context. So before we do that-- and I'll show you an example of that in the next video-- let me just define some terms of ways of measuring how concentrated this is. Or ways of measuring how much light is transmitted versus how much was put in. And so when the people who defined it said, well you know, what we care about is how much is transmitted versus how much went in. So let's just define transmittance as that ratio, the amount that gets through. So in this example, the transmittance of number 1 would be the amount that got through over the amount that you put in. Over here, the transmittance would be the amount that you got out over the amount that you put in. And as we see, this one right here will be a lower number. I2 is lower than I1. So this will have a lower transmittance than number 1. So let's call this transmittance 2. This is transmittance 1. And transmittance 3 is the light that comes out, that And this is the smallest number, followed by that, followed by that. So this will have the least transmittance-- it's the most opaque-- followed by that, followed by that. Now another definition-- which was really kind of a derivative of the-- not in the calculus sense, this is just derived from transmittance and we'll see it has pretty neat properties-- is the notion of absorbance. And so here, we're trying to measure how good is it at absorbing? This is measuring how good are you at transmitting? A higher number says your transmitting a lot. But absorbance is how good you're absorbing. So it's kind of the opposite. If you're good at transmitting, that means you're bad at absorbing, you don't have a lot to absorb. If you're good at absorbing, that means you're not transmitting much. So absorbance right here. And that is defined as the negative log of transmittance. And this logarithm is base 10." + }, + { + "Q": "At 3:12, couldn't you just round the numbers like 1.0079 to just 1.01?", + "A": "No, that is not being done correctly. Unlike what Sal does in many of his videos, you cannot round to whatever amount you find convenient. You MUST respect the number of significant digits you have. Almost any chemistry will mark a problem wrong if you do not round in such a way that your number of significant digits requires. You cannot round less than that, you cannot round more than that.", + "video_name": "UPoXG1Z3sI8", + "timestamps": [ + 192 + ], + "3min_transcript": "this is going to be 12.011 atomic mass units, and the contribution from hydrogen, the contribution from hydrogen, let me do that in yellow... The contribution from hydrogen, from the six hydrogens, is going to be six times the atomic weight of hydrogen, which is 1.0079, by this periodic table that I have right here, once again, the weighted average of all the isotopes, et cetera, et cetera, 1.0079. 1.0079, and so, the molecular weight of this whole thing, a typical benzene molecule, if you were to take the weighted average of all of its molecular masses, based on the prevalence of the different isotopes on Earth, well, you would just say it's just going to be the sum of these two things. It is going to be six times 12.011 is that going to give us? Well, let's see, let's get a calculator out, so let me clear that, so 12.011 times six gives us 72.066, so this right over here. is 72.066, and actually, I probably could have done that in my head, well anyway, let's look at what we have from the hydrogen, so the hydrogen are going to be six times 1.0079 gets us to 6.0474, plus 6.0474, but since I only go to the thousandth place in terms of precision here, if I care about significant figures, significant digits, then I'm only going to go three spaces to the right of the decimal, but let's add these two together, so I'm going to get 6.0474, plus 72.066 equals, and I'm just going to go three to the right, so 78.113, so this is equal to, 78.113 atomic mass units, that's the molecular mass of a molecule of benzene, now what percentage is from the carbon? Well, it's going to be equal to the 72.066 over the 78.113 which is equal to, all right, so let me just clear this." + }, + { + "Q": "02:23 - If you caught the flu between Jan 7th and 10th, why did the symptoms show up only on 11th and not as soon as you caught it ?\n\nI do understand that it takes some time for the virus to start affecting the body, but isnt 3-4 days too long?", + "A": "It takes some time for the tiny amount of virus that you actually pick up to infiltrate cells, reproduce, and infiltrate more cells. Some of the symptoms are from the effects of the virus invading the cells, and some are side effects from your immune system attacking the virus.", + "video_name": "6vy5CX6vK0I", + "timestamps": [ + 143 + ], + "3min_transcript": "So today's January 11, 2013. I'm just going to circle the date on my little calendar here. And let's say I came home from work and I just felt really lousy, just awful. Fevers, sore throat, cough, body aches, you name it. This is the first day I've been feeling like this. And up until now, since the new year started, I was feeling really great. I had no symptoms. I was going to work. Feeling myself. So all these days I was feeling good. And then all of a sudden the 11th hit and I just got, all of a sudden, these symptoms. So I'm already suspecting the flu based on what we know. It started abruptly. I've got the symptoms for it. And a few questions start popping up in my head. The first question I want to know is, when can I expect to start feeling better? That's usually the first thing people want to know. So let's think about what we know about the flu in terms of how long the symptoms usually last. Because that's going to help us predict when I can expect to start feeling myself again. So we know that usually symptoms last for three to seven days. going to expect to feel kind of the same maybe. I might start feeling a little bit better by the 16th or 17th. But that would be seven days. So these are the days I can expect to feel kind of lousy. And on average I should start feeling myself again maybe by the 18th and 19th. I should start feeling the way I normally do. So according to this calendar I would start feeling better by January 18. That would be my target date. And this isn't exact. This is just a rough idea. So what's the next thing that people usually try to figure out about the flu? They always want to know, who did they get it from? They always want to figure out who the culprit was. Who gave the flu to them. So I'm no different. I want to know who did I get it from. And so I think back and I say, well, I felt good on the 10th and I felt pretty lousy on the 11th. And your instinct might be to say, well, On Thursday. But, in fact, you have to go a little bit further back. Sometimes you can get it back as far as four days. So I'm going to circle the days that I could've potentially gotten the flu from somebody. And it turns out it could've been any time this week. So I'm going to write that down. So January 7th to January 10th. That's the window in which somebody gave me the flu. Now how do I know that I got it from somebody? Maybe I got it from the doorknob. Or maybe I got it from the remote control that someone was touching. And those kinds of environmental objects, sometimes you can get diseases from there. But with the flu you generally get it from another person. And the reason is, is because you've got this RNA that's protected by an envelope. Remember this green layer here, this double layer, is a lipid or a fat bilayer." + }, + { + "Q": "Based on the pneumonia videos, pneumonia is the infection of the flu virus in the lungs in the situation described at 10:04, and pneumonitis (inflammation of alveolar walls) is secondary to this infection. At 10:04 when he describes inflammation of the alveolar walls, is that an instance where pneumonitis and pneumonia are interchangeable? Or, should inflammation of alveolar walls always be called pneumonia if it is caused by pneumonia?", + "A": "The disease entity is called pneumonia clinically and usually there is further description such as bacterial pneumonia, viral pneumonia, fungal pneumonia, etc. At the microscopic, pathologic level, the changes seen are called pneumonitis. When there is no infection, sometimes the disease is described based upon the pathologic changes hence things like aspiration pneumonitis.", + "video_name": "6vy5CX6vK0I", + "timestamps": [ + 604, + 604 + ], + "3min_transcript": "These are high risk individuals. So why do we care so much about these high risk individuals? Well, it's because they develop complications of flu. And this is what it really boils down to. You remember we initially talked about all the hundreds of thousands of people in the US and around the world that get hospitalized for the flu. And then the numbers of people that die from the flu. Well, overwhelmingly it's people in this group. This high risk group. And the things that they get, the kind of complications they get, are many. Actually, flu leads to many different types of complications. And I'm going to draw out just a few of them for you. I don't want to give you an exhaustive list. But I want you to at least get an appreciation for the kinds of things we're talking about. So, for example, let's say these are your lungs. I'm drawing two branches of your lungs. And this is going to your left lung So this is your trachea splitting up. And you know that the flu, the influenza, is going to affect the cells in your respiratory tree. So it's going to affect these cells and it's going to cause inflammation. You're going to get a big immune response. And if that response is really big, let's say you have a big response, and if it's around these airways here, these bronchioles-- let me actually extend this out a little bit, so you can at least appreciate where the arrow's going. If the response is really strong in the bronchioles, we call that bronchitis. So someone might actually develop bronchitis as a result of getting the flu. Now someone else might actually have a big inflammatory reaction in these little air sacs. Your lungs end in thousands and thousands of little air sacs. And if that happens, then you might call that pneumonia. You might say, well, this person has pneumonia. Actually, lots and lots of it. Smooth muscle that wraps around the bronchioles. And sometimes with the flu you actually can trigger twitichiness of that smooth muscle. It starts spasming. And when that happens we know that sometimes as an asthma attack. So you can actually get an asthma attack related to the flu. So all sorts of things like this can happen. And it's awful. These are things that can actually land you in the hospital. Or can cause death, as well. So these are the kinds of complications. And there are other ones. Things like ear infections and sinus infections and many, many other things as well. But here I just wanted to show you a few of the complications that people get. And show you and remind you that it's usually the high risk people that you have to worry about." + }, + { + "Q": "What would happen if in 1:45 you push on both pistons with the same amount of force? Would the water just displace? What if there is no possibility for the water to displace? Then could you no longer push on the pistons?", + "A": "I think that if you push on both the pistons with the same force, the piston with larger area will displace the liquid as more pressure is applied on it. if there is no possibility for water to displace, then you can no longer push on the pistons.", + "video_name": "lWDtFHDVqqk", + "timestamps": [ + 105 + ], + "3min_transcript": "Welcome back. To just review what I was doing on the last video before I ran out of time, I said that conservation of energy tells us that the work I've put into the system or the energy that I've put into the system-- because they're really the same thing-- is equal to the work that I get out of the system, or the energy that I get out of the system. That means that the input work is equal to the output work, or that the input force times the input distance is equal to the output force times the output distance-- that's just the definition of work. Let me just rewrite this equation here. If I could just rewrite this exact equation, I could say-- the input force, and let me just divide it by this area. The input here-- I'm pressing down this piston that's pressing down on this area of water. So this input force-- times the input area. Let's call the input 1, and call the output 2 for simplicity. Let me do this in a good color-- brown is good color. I have another piston here, and there's going to be some outward force F2. The general notion is that I'm pushing on this water, the water can't be compressed, so the water's going to push up on this end. The input force times the input distance is going to be equal to the output force times the output distance right-- this is just the law of conservation of energy and everything we did with work, et cetera. I'm rewriting this equation, so if I take the input force and divide by the input area-- let me switch back to green-- then I multiply by the area, and then I just multiply times D1. You see what I did here-- I just multiplied and divided by A1, which you can do. You can multiply and divide by any number, and these two cancel out. It's equal to the same thing on the other side, which is over A2 times A2 times D2. Hopefully that makes sense. What's this quantity right here, this F1 divided by A1? Force divided by area, if you haven't been familiar with it already, and if you're just watching my videos there's no reason for you to be, is defined as pressure. Pressure is force in a given area, so this is pressure-- we'll call this the pressure that I'm inputting into the system. What's area 1 times distance 1? That's the area of the tube at this point, the cross-sectional area, times this distance. That's equal to this volume that I calculated in the previous video-- we could say that's the input volume, or V1. Pressure times V1 is equal to the output pressure-- force 2" + }, + { + "Q": "At around 9:30...how do the -1, 0, 1, relate to the px, py, and pz?", + "A": "This is just a convention that is used. However, the three p orbitals are equivalent to one another and only acquire the x, y and z suffixes based on how the three axes are drawn. Therefore, it is arbitrary which p orbital is linked to -1, 0 and 1 and it would be equally valid to, say, relate -1, 0 and 1 to pz, px and py, or to any other permutation. All you need to know is that if l =1 then this means there are three p orbitals, and that these three p orbitals are along different axes.", + "video_name": "KrXE_SzRoqw", + "timestamps": [ + 570 + ], + "3min_transcript": "Let me use a different color here. If l is equal to zero, we know we're talking about an s orbital. When l is equal to zero, we're talking about an s orbital, which is shaped like a sphere. If you think about that, we have only one allowed value for the magnetic quantum number. That tells us the orientation, so there's only one orientation for that orbital around the nucleus. And that makes sense, because a sphere has only one possible orientation. If you think about this as being an xyz axis, (clears throat) excuse me, and if this is a sphere, there's only one way to orient that sphere in space. So that's the idea of the magnetic quantum number. Let's do the same thing for l is equal to one. Let's look at that now. If we're considering l is equal to one ... Let me use a different color here. l is equal to one. If l is equal to one, what are the allowed values for the magnetic quantum number? ml is equal to -- This goes from negative l to positive l, so any integral value from negative l to positive l. Negative l would be negative one, so let's go ahead and write this in here. We have negative one, zero, and positive one. So we have three possible values. When l is equal to one, we have three possible values for the magnetic quantum number, one, two, and three. The magnetic quantum number tells us the orientations, the possible orientations of the orbital or orbitals around the nucleus here. So we have three values for the magnetic quantum number. That means we get three different orientations. We already said that when l is equal to one, we're talking about a p orbital. A p orbital is shaped like a dumbbell here, so we have three possible orientations If we went ahead and mark these axes here, let's just say this is x axis, y axis, and the z axis here. We could put a dumbbell on the x axis like that. Again, imagine this as being a volume. This would be a p orbital. We call this a px orbital. It's a p orbital and it's on the x axis here. We have two more orientations. We could put, again, if this is x, this is y, and this is z, we could put a dumbbell here on the y axis. There's our second possible orientation. Finally, if this is x, this is y, and this is z, of course we could put a dumbbell on the z axis, like that. This would be a pz orbital. We could write a pz orbital here, and then this one right here would be a py orbital." + }, + { + "Q": "At 4:10, if the electron is not found in the sphere of the s orbital, than where else would it be found?", + "A": "That sphere is just the most likely place for an electron to be found, it could be on the other side of the universe but the probability of that is immensely low.", + "video_name": "KrXE_SzRoqw", + "timestamps": [ + 250 + ], + "3min_transcript": "For n is equal to one, let's say the average distance from the nucleus is right about here. Let's compare that with n is equal to two. n is equal to two means a higher energy level, so on average, the electron is further away from the nucleus, and has a higher energy associated with it. That's the idea of the principal quantum number. You're thinking about energy levels or shells, and you're also thinking about average distance from the nucleus. All right, our second quantum number is called the angular momentum quantum number. The angular momentum quantum number is symbolized by l. l indicates the shape of the orbital. This will tell us the shape of the orbital. Values for l are dependent on n, so the values for l go from zero so it could be zero, one, two, or however values there are up to n minus one. For example, let's talk about the first main energy level, or the first shell. n is equal to one. There's only one possible value you could get for the angular momentum quantum number, l. n minus one is equal to zero, so that's the only possible value, the only allowed value of l. When l is equal to zero, we call this an s orbital. This is referring to an s orbital. The shape of an s orbital is a sphere. We've already talked about that with the hydrogen atom. Just imagine this as being a sphere, so a three-dimensional volume here. The angular momentum quantum number, l, since l is equal to zero, that corresponds to an s orbital, so we know that we're talking about an s orbital here So the electron is most likely to be found somewhere in that sphere. Let's do the next shell. n is equal to two. If n is equal to two, what are the allowed values for l? l goes zero, one, and so on all the way up to n minus one. l is equal to zero. Then n minus one would be equal to one. So we have two possible values for l. l could be equal to zero, and l could be equal to one. Notice that the number of allowed values for l is equal to n. So for example, if n is equal to one, we have one allowed value. If n is equal to two, we have two allowed values. We've already talked about what l is equal to zero, what that means. l is equal to zero means an s orbital, shaped like a sphere. Now, in the second main energy level, or the second shell, we have another value for l. l is equal to one." + }, + { + "Q": "At 5:15, why would you not want to take one of these devices apart? What could be in items like this that you would need to be afraid of?", + "A": "Underneath the bottom cover you find the electric heater and water hoses. This is a dangerous combination (water and electricity). The manufacturer wants to keep you away from the chance of messing up the connections and putting a defective coffeemaker back in service. In this demo video, the coffeemaker takes a one-way trip to full disassembly. It will never make another cup.", + "video_name": "XQTIKNXDAao", + "timestamps": [ + 315 + ], + "3min_transcript": "It would make it easy to pull the mold out this way. They probably also had, it was probably a three part mold and there was a section that also came out in this direction. Then this is just another injection molded part that snaps onto this one, as we've seen. This is the part that holds the handle on. Very important part. I think they definitely paid the extra money for a stainless piece there because it's really important that that doesn't come loose and it probably gets fairly wet, so if it was made out of regular steel or another material it might rust and could potentially come apart. We wouldn't want hot coffee on us, now would we. Alright, so that's the coffee kettle. So, inside, here's our coffee maker. We know that hot water... We've got a container here and in this container, is a space where we put our coffee filter and then we put our coffee grounds and we fill this with water and then we close the top and we turn it on and we wait. What happens is that water that we pour in drains down a little hole on the inside there, you can see it right there. Let me point to it with the screw driver. It drains down that hole and it goes down into this underside, so we'll take a look at the underside and see what happens down there. Okay, so, I've modified a screw driver. This was a low-cost screw driver. It was a 99 cent one, so I modified the end of it so I could take out these safety screws. Don't do this at home unless you have a professional with you because this is not meant to be taken apart. That's why they use these special screw heads, so you won't take it apart. There we go. Again, this is an injection molded part. This is a co-molded part, it looks like. Which means that there were two different materials molded together. Let's see if I can knock that screw out. Okay, it wants to stay, that's fine. This material here is, these feet are made out of a softer material and this is a polypropylene material. So it's a plastic, a low-cost plastic. So the mold comes together and they injection mold this material and then once this material has begun to harden, they injection mold the softer material, so it's co-molded or it's a dual molded part. You can see other parts are done like this, like sometimes you'll see toothbrushes that have soft saniprene and then the hard toothbrush and they're molded in one mold. It's a dual shot mold. In any case, so that's the bottom." + }, + { + "Q": "At 2:50, when Sal started talking about velocity, was he actually meaning instantaneous velocity?", + "A": "Yes, he was meaning instantaneous velocity. There are two kinds of velocity: average velocity and instantaneous velocity. By just saying velocity , we have to think about which one it means through the context.", + "video_name": "ITA1rW5UraU", + "timestamps": [ + 170 + ], + "3min_transcript": "" + }, + { + "Q": "At 12:05, what are some other things that can cause Cardiac Arrest?", + "A": "Many. Among others, Brugada syndrome.", + "video_name": "vYnreB1duro", + "timestamps": [ + 725 + ], + "3min_transcript": "It's less likely but sometimes a plaque could also go downstream, kind of form a thromboembolism. It would be this thrombogenic material, the clots around it. It would actually go and block the artery further downstream and be embolism. That can also block the artery and cause tissue to die. But the main cause is this intense clotting that can occur pretty quickly and completely obstruct the artery. There is one last word i want to touch sometimes mixed in with all the other words, that is cardiac arrest. That's because sometimes we use them in the same context. one thing can lead to another. Heart attack is not cardiac arrest. Cardiac arrest is the actual dying of the heart. Some part infarct, that's what they called myocardial infarction. Myocardial means the tissue of heart or the muscle of the heart that's dying. Sometimes it is called myocardial infarction. That is not cardiac arrest. Cause you can have some of your heart tissue die and you can survive. Your heart would be impaired. But you will continue to live. Cardiac arrest is literally your heart stopping. This would obviously cause someone to die. If you have a bad enough heart attack, if you have enough of the tissue get starved of oxygen so that it dies, infarction occurs. Then it could lead to cardiac arrest. It always won't lead to cardiac arrest. Frankly, heart attack is not the only thing that can cause cardiac arrest. Cardiac arrest is heart stopping. Heart failure is essentially just saying that heart can not provide all of the needs for the body." + }, + { + "Q": "At about 3:00 Sal talks about plaque and how it can damage your heart. So, if you do get plaque in your blood vessel, how do you get rid of it?", + "A": "Presumably exercise, a low fat, healthy diet and stopping smoking if one is doing so already.", + "video_name": "vYnreB1duro", + "timestamps": [ + 180 + ], + "3min_transcript": "so this was my ....this is what I thought people were talking about when they were saying clogging of the arteries and maybe when they got clogged enough, the stopped blood flow to the rest of the body somehow and that would actually kill the person. I want to make it very clear right now. Those are not the arteries that people are talking about getting clogged, when people talk about heart disease or heart attacks. The arteries that they are talking about are the arteries that actually provide blood to the heart. Remember the heart itself is a muscle. It itself needs oxygen. So you have these arteries right over here, the red tubes. Those are arteries. and then the blue ones are veins. They're taking the de-oxygenated blood away from the tissue of the heart. And these are called coronary arteries. And this one over here at least from the point of view of me or you looks like it's on the right. This right over here is called the left coronary artery or LCA. And this right over here in red is called the right coronary arteries or the RCA. And so when people talk about arteries getting blocked or getting clogged, they're talking about the coronary arteries. They're talking about the things that supply blood to the heart. So let's zoom in on one of them....Maybe we can zoom in right over here, that part of the artery. That's the tube....clear where I am zooming in. I am zooming in right over here. So over time, I am not going into the details how this happened. It is subject for another video. You can have these plaques build up along the walls of the artery. So over time if a person doesn't have the right diet, or maybe they just have a predisposition to it, And the plaques, the material inside of them are lipids, so things like fat, cholesterol and also dead white blood cells, which is this kind of messy substance right over here. This is what we call a plaque. And the formation of these plaques that obstruct the actual blood vessel, that actually obstruct the artery. We call it.....make it clear you see that. This is kind of tube over here. Let me draw the blood So this formation of these plaques we call atherosclerosis. So you can imagine if you have these things build up," + }, + { + "Q": "at 9:35 why is the carbon cation sp2 hybridized? shouldn't it be sp3?", + "A": "No. A carbocation only has 3 electron domains so it is planar and sp2 hybridized.", + "video_name": "KPh60w6McPI", + "timestamps": [ + 575 + ], + "3min_transcript": "so we form a carbo cation. So this is a stable carbo cation. This is a tertiary carbo cation. So that is why this tertiary alcohol reacts via a SN-one type mechanism. The stability of the carbo cation. And so in our final step, we have the nucleophile is going to attack our electrophile. So the nucleophile attacks our electrophile. The chloride anion attacks our carbo cation, attacks that carbon there and so we form our final product, which is tertbutyl chloride. So let me go ahead and draw in these electrons and lets highlight some again. The electrons in blue. These electrons form the bond, So they bonded right here. And so let me go ahead and highlight this carbon in red. So this carbon in red is this one right here. So we form tertbutyl chloride and we lost water in the process. So this is a very easy reaction to do. It occurs at room temperature and just take and just shake them together and you can form your final product this way. So thinking about stereochemistry and SN-one type mechanism, the carbon in red right here is not a chiral center and so we don't have to worry about what kind of stereochemical outcome that we would predict for the product. So this is our final product. There is no stereochemistry. Just to refresh your memory, for a SN-one type mechanism because this formation of this carbo cation, this carbon and the carbo cation is SP-two hybridized, and so it is planar. And so when we draw out that carbon in red here. Lets say that's that carbon in red. It's SP-two hybridized, which means that the carbons that are directly bonded to it lie on the same plane. So these carbons lie on the same plane. SP-two hybridized with a p orbital. So there is a p orbital. Lets sketch that in. So there is a plus one formal charge in this carbon. So when your nucleophile attacks, your nucleophile could attack from either side of that plane. or it could attack from this side. So if your final product has a chiral center, you need to think about stereochemistry. But not in this case. In this case we don't have one. So we lucked out. This one was a little bit straightforward. So that's a couple of examples of SN-one and SN-two reactions of alcohols." + }, + { + "Q": "at 08:22 , Sal says that the [ln V] is evaluated over final \"velocity\" to starting \"velocity\". I strongly think it should be replaced with \"volume\".\nPls. correct me if I'm wrong.\n\nthank you.", + "A": "He misspoke. It is volume. After all, the video is on volume...", + "video_name": "ixRtSV3CXPA", + "timestamps": [ + 502 + ], + "3min_transcript": "Well, on this term, the n's cancel out, the R cancels out. Over here, this nRT cancels out with this nRT. And what are we left with? We're left with 3/2-- we have this 1 over T left-- times 1 over T delta T plus 1 over V delta V is equal to-- well, zero divided by anything is just equal to 0. Now we're going to integrate over a bunch of really small delta T's and delta V's. So let me just change those to our calculus terminology. We're going to do an infinite sum over infinitesimally small changes in delta T and delta V. So I'll rewrite this as 3/2 1 over T dt plus 1 over V dv is small change in volume. This is a very, very, very, small change, an infinitesimally small change, in temperature. And now I want to do the total change in temperature. I want to integrate over the total change in temperature and the total change in volume. So let's do that. So I want to go from always temperature start to temperature finish. And this will be going from our volume start to volume finish. Fair enough. Let's do these integrals. This tends to show up a lot in thermodynamics, these antiderivatives. The antiderivative of 1 over T is natural log of T. So this is equal to 3/2 times the natural log of T. We're going to evaluate it at the final temperature and then the starting temperature, plus the natural log-- the plus the natural log of V, evaluated from our final velocity, and we're going to subtract out the starting velocity. This is just the calculus here. And this is going to be equal to 0. I mean, we could integrate both sides-- well, if every infinitesimal change is equal to the sum is equal to 0, the sum of all of the infinitesimal changes are also So this is still equal to 0. See what we can do here. So we could rewrite this green part as-- so it's 3/2 times the natural log of TF minus the natural log of TS, which is just, using our log properties, the natural log of TF over the natural log of TS. Right? When you evaluate, you get natural log of TF minus the natural log of TS." + }, + { + "Q": "at 11:16 why is it (Tf/Ts)^2/3(Vf/Vs)=1????\nwhy is it 1???", + "A": "To start with, ln [(Tf/Ts)^3/2(Vf/Vs)]=0. Now, ln (natural log) is logarithm base e (a constant), so log e [(Tf/Ts)^3/2(Vf/Vs)] = 0. From this, and by the property of logarithm, e to the power of 0 (e^0) = [(Tf/Ts)^3/2(Vf/Vs)]. Property of logarithm says whatever number to the 0th power equals 1. So since (e^0) = [(Tf/Ts)^3/2(Vf/Vs)], this whole [(Tf/Ts)^3/2(Vf/Vs)] expression equals 1.", + "video_name": "ixRtSV3CXPA", + "timestamps": [ + 676 + ], + "3min_transcript": "Plus, for the same reason, the natural log of VF over the natural log of VS. When you evaluate this, it's the natural log of VF minus the natural log of VS, which can be simplified this to this, just from our logarithmic properties. So this equals 0. And now we can-- this coefficient out front, we can use our logarithmic properties. Instead of putting a 3/2 natural log of this, we can rewrite this as the natural log of TF over TS to the 3/2. Now we can keep doing our logarithm properties. You take the log of something plus the log of something. That's equal to the log of their product. So this is equal to-- I'll switch colors-- The natural log of TF over TS to the 3/2 power, times the natural log of VF over VS. And this is a fatiguing proof. Now what can we say? Well, we're saying that e to the 0 power-- the natural log is log base e-- e to the 0 power is equal to this thing. So this thing must be 1. E to the 0 power is 1. So we can say-- we're almost there-- that TF, our final temperature over our starting temperature to the 3/2 power, times our final volume over our starting volume is equal to 1. Now let's take this result that we worked reasonably hard to produce. Remember all of this, we just said, we're dealing with an adiabatic process, and we started from the principle of just what the definition of internal energy is. And then we substitute it with our PV equals nRT formulas. point is equal to 3/2 times PV. And then we integrated over all the changes, and we said, look, this is adiabatic. So the total change-- the sum of all of our change in internal energy and work done by the system has to be 0, then we use the property of log to get to this result. Now let's do these for both of these adiabatic processes over here. So the first one we could do is this one where we go from volume B at T1 to volume C at T2. Watch the Carnot cycle video, if you forgot that. This was the VB All of these things up here were at temperature 1. All of the things down here were at temperature 2 So we're at temperature 1 up here, and temperature 2 down here, volume C. So let's look at that. So on that right part, that right process, our final temperature was temperature 2. So let me write it down. Temperature 2." + }, + { + "Q": "Around 2:00, for the equation for the very first question, why is molarity used instead of the number of moles present?", + "A": "HCl will dissociate completely and form 0.500 moles of H3O+. Molarity is the number of moles present, i.e. the concentration.", + "video_name": "JoGQYSTlOKo", + "timestamps": [ + 120 + ], + "3min_transcript": "- [Voiceover] Let's say we're doing a titration and we start with 20 mL of .500 molar HCl. So we're starting with a strong acid, and to the strong acid, we're going to add a solution of a strong base. We're going to add a .500 molar solution of NaOH, and as we add the base, the pH is going to increase, and we can show this on our titration curve. So we put the pH on the y-axis, and on the x-axis we put the volume of base that we are adding. So in part A, our goal is to find the pH before we've added any of our sodium hydroxide, so the addition of 0.0 mL of our base. So if we haven't added any base, the only thing we have present is acid. We know that HCl is a strong acid, so it is going to ionize 100%. So if HCl donates a proton to H2O, then we get H3O plus, and we would get the conjugate base to HCl, which is Cl minus. let's go ahead and write that, .500 molar HCl, and since we know this is 100% ionization because we have a strong acid, that's the same concentration of hydronium ions that we'll have in solution, so we have .500 molar for the concentration of hydronium ions. Now it's easy to find the pH because we know that the pH is equal to the negative log of the concentration of hydronium ions. So let's go ahead and do the calculation. We take the negative log of the concentration of hydronium which is .500, so that gives us .301. So we plug this into here, and we get a pH equal to 0.301. So we can find that point on our titration curve. We've added 0.0 mL of base, and our pH looks like it's just above zero here on our titration curve, and we calculated it to be .301. Let's find another point on our titration curve, this time after we add 10 mL of our base. So we find 10 mL of our base right here, and we're trying to find this point. So what is the pH after we add 10 mL? Well it looks like it's pretty close to one. Let's see if we can calculate what the pH is. So if we're adding base, we know that the base that we're adding, the hydroxide ions that we're adding, are going to neutralize the hydronium ions that are already present. So first, let's calculate how many moles of hydronium ions that we had present here. So the concentration of hydronium ions is .500." + }, + { + "Q": "At 1:20, shouldn't it be hydroxonium not hydronium?", + "A": "Hydroxonium and hydronium mean the same thing and both terms are in use.", + "video_name": "JoGQYSTlOKo", + "timestamps": [ + 80 + ], + "3min_transcript": "- [Voiceover] Let's say we're doing a titration and we start with 20 mL of .500 molar HCl. So we're starting with a strong acid, and to the strong acid, we're going to add a solution of a strong base. We're going to add a .500 molar solution of NaOH, and as we add the base, the pH is going to increase, and we can show this on our titration curve. So we put the pH on the y-axis, and on the x-axis we put the volume of base that we are adding. So in part A, our goal is to find the pH before we've added any of our sodium hydroxide, so the addition of 0.0 mL of our base. So if we haven't added any base, the only thing we have present is acid. We know that HCl is a strong acid, so it is going to ionize 100%. So if HCl donates a proton to H2O, then we get H3O plus, and we would get the conjugate base to HCl, which is Cl minus. let's go ahead and write that, .500 molar HCl, and since we know this is 100% ionization because we have a strong acid, that's the same concentration of hydronium ions that we'll have in solution, so we have .500 molar for the concentration of hydronium ions. Now it's easy to find the pH because we know that the pH is equal to the negative log of the concentration of hydronium ions. So let's go ahead and do the calculation. We take the negative log of the concentration of hydronium which is .500, so that gives us .301. So we plug this into here, and we get a pH equal to 0.301. So we can find that point on our titration curve. We've added 0.0 mL of base, and our pH looks like it's just above zero here on our titration curve, and we calculated it to be .301. Let's find another point on our titration curve, this time after we add 10 mL of our base. So we find 10 mL of our base right here, and we're trying to find this point. So what is the pH after we add 10 mL? Well it looks like it's pretty close to one. Let's see if we can calculate what the pH is. So if we're adding base, we know that the base that we're adding, the hydroxide ions that we're adding, are going to neutralize the hydronium ions that are already present. So first, let's calculate how many moles of hydronium ions that we had present here. So the concentration of hydronium ions is .500." + }, + { + "Q": "3:55 What sets off that movement/What causes it?\n\nThe other two steps in the molozonide breaking apart make sense, but this first one doesn't.", + "A": "The instability of oxygen-oxygen bonds and electrostatic attraction. The bonds between oxygen are very weak and are prone to breaking. Once that bond breaks, the electrons will be attracted to a positive charge. Since there is a slight partial positive charge on the carbon, the electrons move into carbon s valence shell. However, this violates the octet rule and thus breaks an adjacent C-C bond in order to have only 8 electrons associated with carbon, setting off a chain of bonds breaking/electron movement.", + "video_name": "bFj3HpdC4Uk", + "timestamps": [ + 235 + ], + "3min_transcript": "on the oxygen-- this lone pair of electrons here-- is going to attack this carbon which would push these pi electrons off. And those pi electrons are actually going go to this oxygen right here which would push these pi electrons in here off onto this oxygen. So it's a concerted mechanism here. And so let's go ahead and draw the results of those electrons moving. So the oxygen on the left is now bonded to the carbon on the left. The carbon on the left now has a single bond to the carbon on the right. The carbon on the right is now bonded to this oxygen. And then these two oxygens are bonded to an oxygen in the center. For lone pairs of electrons, all of our oxygens are going to have two lone pairs of electrons. Like that. And so we had so many electrons moving, let's say if we can follow them. So let's color coordinate some electrons here. So I'm going to say that these electrons in blue, those between the oxygen and the carbon. And I'm going to say that these pi electrons here in red, those are the ones that formed this bond between this carbon and this oxygen. And then finally, these electrons-- I'm saying these are my pi electrons in here-- are going to move off onto this top oxygen. So you could say that those magenta electrons would be right there. And so now we have this structure. Now oxygen oxygen bonds are relatively weak. So they're unstable and so we have two oxygen oxygen bonds in this molecule. And so one of those oxygen oxygen bonds is going to break in the next step of the mechanism. And so I could pick either one since they're symmetrical. I'm just going to say that--- I'm just going to say that these electrons over here-- so I'm going to say that this oxygen oxygen bond is going to break in the next part of our mechanism. And so, if we think about these electrons and this oxygen moving in here, that would break this bond between the two to the carbon on the right. So this bond is going to break, push those electrons into here, and then these electrons in green are going to come off onto the top oxygen [INAUDIBLE]. So our unstable oxygen oxygen bond breaks. And so let's go ahead and draw what we would get. Well, on the left side, we have a carbon. And now that carbon is going to have two bonds to the oxygen. And the oxygen is going to have two lone pairs of electrons. It would make a carbonyl compound. On the right, the carbon on the right is bonded to two other things and now has a double bond to this oxygen here. And now this oxygen has only one lone pair of electrons on it. And then this oxygen is bonded to the other oxygen, and that oxygen now has three lone pairs of electrons, which would give it a negative 1 formal charge. So we have a negative 1 formal charge here, this oxygen right here gets a positive 1 formal charge, and we form a carbonyl oxide on the right. So this charge compound on the right's" + }, + { + "Q": "Why do seismic waves travel faster through denser material (4:45) ?", + "A": "You can think that in a dense material the molecules are very closer together and when a wave hits one molecule it takes less time for it to reach the next one and the wave then travles faster. In air the molecules are far apart and the wave than travles slower.", + "video_name": "yAQSucmHrAk", + "timestamps": [ + 285 + ], + "3min_transcript": "All of a sudden, the waves were reaching there faster. The slope of this line changed. It took less time for each incremental distance. So for some reason, the waves that we're going at these farther stations, the stations that were more than 200 kilometers away, somehow they were accelerated. Somehow they were able to move faster. And he's the one that realized that this was because the waves that were getting to these further stations must have traveled through a more dense layer of the earth. So let's just think about it. So if we have a more dense layer, it will fit this information right over here. So if we have a layer like this, which we now know to be the crust, and then you have a denser layer, which we now know to be the mantle, then what you would have is-- so you have your earthquake right over here, closer by, while you're still within the crust, it would be proportional. And then let's say that this is exactly, this right here is 200 kilometers away. But then if you go any further, the waves would have to travel. They would travel, so they would go like this. And then they would get refracted even harder. So they would get refracted. So they would be a little bit curved ahead of time. But then they're going to a much denser material. Or it's not gradually dense, it's actually kind of a all of a sudden a considerably more dense material, so it will get refracted even more. And then it'll go over here. And since it was able to travel all of this distance in a denser material, it would have traveled faster along this path. And so it would get to this distance on the surface that's more than 200 kilometers away, it would get there faster. And so he said that there must be a denser layer that those waves are traveling through, which we now know to be the mantle. And the boundary between what we now know to be the crust and this denser layer, It's called the Mohorovicic discontinuity. And sometimes this is called the Moho for short. So that boundary between the crust and the mantle is now named for him. But this was a huge discovery, because not only was he able to tell us, based on the data-- based on, kind of, indirect data, just based on earthquakes happening, and measuring when the earthquakes reach different points of the earth-- that there probably is a denser layer. And if you do the math, under continental crust that denser layer is about 35 kilometers down. He was able to tell us that there is that layer. But even more importantly, he was able to give the clue that just using information from earthquakes, we could essentially figure out the actual composition of the earth. Because no one has ever dug that deep. No one has ever dug into the mantle, much less the outer core or the inner core. In the next few videos, we're going to kind of take this insight, that we can use information from earthquakes, to actually think about how we know that there is an outer liquid core" + }, + { + "Q": "At 2:33 how is the molecule cyclic? Does that mean it can form a ring structure? if so, i'm having trouble picturing how it would do so.", + "A": "The molecule is cyclic because when you follow the bonds from one atom to another, you come back in a loop.to your starting point. The loop is not round like a circle but, because it is a closed loop, it is described as cyclic and as a ring ..", + "video_name": "FaOOx6IZxV8", + "timestamps": [ + 153 + ], + "3min_transcript": "with a partial positive charge. The bottom halogen is now an electrophile, so it wants electrons. It's going to get electrons from those pi electrons here, which are going to move out and nucleophilic attack that partially positively charged halogen atom. And then this lone pair of electrons is going to form a bond with this carbon at the same time these blue electrons move out onto the halogen. So when we draw the result of all those electrons moving around, we're going to form a bond between the carbon on the right and the halogen, and we use the magenta electrons to show that. So there's now a bond there. And so we used red electrons before to show these electrons in here forming a bond with the carbon on the left. That halogen had two lone pairs of electrons still on it, like that, which gives that halogen a plus 1 formal charge. in an earlier video. And if I think about that cyclic halonium ion, I think about the halogen being very electronegative. It's going to attract, I'll say the electrons in magenta again just to be consistent, closer towards it. So it's going to take away a little bit of electron density from this carbon right down here. So I'm going to say this carbon is partially positive. It's going to have some partial carbocationic character. So in the next step of the mechanism, water's going to come along. And water's going to function as a nucleophile. So one of the lone pairs of electrons on water is going to nucleophilic attack our electrophile, which is this carbon right here. And so when that lone pair of electrons on oxygen attacks this carbon, that's going to kick the electrons in magenta off onto your halogen. And so let's go ahead and draw the product. We're going to have, on the left carbon, this halogen now used to have two lone pairs of electrons. so now it looks like that. On the right, we still have the carbon on the right bonded to other things, except now it's bonded to what used to be our water molecule. So the oxygen is now bonded to the carbon. And there's still one lone pair of electrons on that oxygen, giving it a plus 1 formal charge. So let's go ahead and highlight these electrons here in blue. Those electrons in blue are the ones that formed this bond between the carbon and the oxygen. So we're almost done. The last step of the mechanism would just be an acid-base reaction. So another water molecule comes along, and one of the lone pairs of electrons on the water molecule is going to function as a base and take this proton, leaving these two electrons behind on the oxygen. And we are finally done. We have formed our halohydrin, right? So I have my halogen on one side. And then I now have my OH on the opposite side, like that." + }, + { + "Q": "At 6:50, how does the oxygen attach to the tertiary carbocation instead of bromine? Isn't the bromine atom more electronegative than oxygen to hold onto the bond?", + "A": "Water is much more electrophilic than Br\u00e2\u0081\u00bb, and there is much more water available to attack the cyclic bromonium ion.", + "video_name": "FaOOx6IZxV8", + "timestamps": [ + 410 + ], + "3min_transcript": "So I'm going to say that my methyl group is now going down in space. So with the addition of my bromonium ion, that would be my intermediate. And so now, when I think about water coming along and acting as a nucleophile, so here is H2O, and I think about which carbon will the oxygen attack? So I have two options, right? This oxygen could attack the carbon on the left, or it could attack the carbon on the right. It's been proven that the option is going to attack the most substituted carbon. So if I look at the carbon on the left, and if I think about what sort of carbocation would that be, the carbon on the left is bonded to two other carbons. So this would be similar to a secondary carbocation, or a partial carbocation in character. So you could think about it as being if this was a partial positive. Or on the right, if I think about this carbon right here, the one in red. And if I think about that being a carbocation, that would be bonded to one, two, three other carbons. So it's like a tertiary carbocation. And we know that tertiary carbocations are more stable than secondary. So even though this isn't a full carbocation, this carbon in red exhibits some partial carbocation character, and that is where our water is going to attack. So the nucleophile is going to attack the electrophile. And it's more stable for it to attack this one on the right, since it has partial carbocation character similar to a tertiary carbocation. And if it attacks that carbon on the right, these electrons here we kick off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. OK, so what would we have here? And the bromine is going to swing over to the carbon on the left. It's now going to have three lone pairs of electrons around it, like that. And the methyl group that was down relative to the plane is going to be pushed up when that water nucleophilic attacks. And now the methyl group is up, and this oxygen is now going to be bonded to this carbon. And so we still have our two hydrogens attached to it, like that. And there's a lone pair of electrons on this oxygen, giving it a plus 1 formal charge. So once again, let's go ahead and highlight those electrons. I'll draw them in blue here. These electrons right here, those are the ones that formed this bond. So let's go back and let's think about the formation of that cyclic bromonium ion in a different way here. So on the left I showed the bromine adding from the top. If I think about the alkene portion of my starting material, well, there's" + }, + { + "Q": "At 4:27, Jay numbers one of the carbons as Beta Carbon 3. I don't understand why Carbon 3 is a Beta-carbon. It's not connected to the alpha carbon.", + "A": "The \u00ce\u00b1-carbon is the carbon bearing the leaving group (C-2). So the \u00ce\u00b2-carbons are the ones next to it (C-1 and C-3).", + "video_name": "uCW6154hPkc", + "timestamps": [ + 267 + ], + "3min_transcript": "The chlorine has to be up axial, and so if I go around to carbon six, so this would be carbon one here, two, three, four, five, and six, I have a methyl group going away from me in space, so this would be going down, so I'm gonna draw in a Me for a methyl group right here, so it's down axial. So now let's draw in some hydrogens on our beta carbons, so let me highlight our beta carbons here. I'll use red, so this would be, what I've marked is being beta one, so I have two hydrogens on that carbon, so I'll draw those in here, and then my other beta carbon which I called beta two up here, so I only have, so this is beta two, I have only one hydrogen, and it is equatorial. So let's go to a video, so we can analyze which one of these protons will participate in our E two mechanism. Here's our chair conformation, we have the yellow chlorine up and axial. When we go to the beta one carbon, the hydrogen in green is the only one that's anti-periplanar to the halogen. If I turn to the side here, it's easier to see that we have all four of those atoms in the same plane, so the green proton is anti-periplanar to the chlorine. The other hydrogen, the one in white, is not anti-periplanar, so it will not participate in our E two mechanism. We go to the beta two carbon, and this hydrogen in white is not anti-periplanar, and when we look at the down axial position, it's occupied by a methyl group, so that is where a hydrogen would need to be if it were to participate in an E two mechanism. For E two elimination in cyclohexanes, the halogen must be axial, so here is our halogen that's axial, and when the halogen is axial in this chair conformation, is this one in green as we saw in the video, so if a strong base comes along, and takes the proton in green, the electrons in here would move in to form our double bond, and these electrons come off onto the chlorine, so a double bond forms between the alpha and the beta one carbons, which would give us this as our only product, so we don't get a double bond forming between our alpha and our beta two carbon because we would need to have a hydrogen where our methyl group is, so this, if we did have a hydrogen here, this hydrogen would be anti-periplanar to our halogen, but in this case, we get only one product, so this is the only product observed, and we figured that out because we drew our chair conformation. Let's do another E two mechanism for a substituted cyclohexane, and I'll start by numbering my cyclohexane, so that's carbon one, and this is carbon two, this is carbon three, and this is carbon four." + }, + { + "Q": "at 9:52, why does he say that because the iso-propyl group is axial can't participate in the mechanism?", + "A": "In order to get elimination of HCl, the Cl onC2 and the \u00ce\u00b2 H must be in a trans diaxial conformation. If the Cl is axial, the isopropyl group on C1 is also axial and the \u00ce\u00b2 H on C1 is equatorial. There is no axial H on C1, because the isopropyl group has replaced it. The axial \u00ce\u00b2 H must therefore come from C3.", + "video_name": "uCW6154hPkc", + "timestamps": [ + 592 + ], + "3min_transcript": "and next to that would be a beta carbon, and this beta hydrogen is anti-periplanar to this halogen. We have another beta carbon over here with another hydrogen that's anti-periplanar to this halogen. Finally, let's draw our two products, so let's take a proton from the beta one position first, so our base, let me draw it in here, so our strong base is going to take this proton, and so these electrons would move into here to form our double bond, and these electrons come onto the chlorine to form the chloride anion as our leaving group. So this would form a double bond between what I called carbons two and three 'cause this is carbon one, this is carbon two, this is carbon three, and this is carbon four, so we have a methyl group that's up at carbon one, so let me draw in our methyl group up at carbon one, so we put that on a wedge, so if that's carbon one, then this is carbon two, and this is carbon three, and that's where our double bond forms, so the double bond forms between carbons two and three, so let me go ahead and put that in, going away from us, we put that on a dash, so that's one product. If our base took this proton, then the electrons would move into here, and these electrons would come off onto the chlorine, so if we took a proton from the beta two carbon, we would form a double bond between carbons three and four, so here's carbons three and four, so I put a double bond in there. I still have a methyl group that's going up at what I called carbon one, and my isopropyl group is at carbon four, but since now, this carbon is sp two hybridized, I need to draw in this isopropyl group on a straight line, so sometimes, students would put this isopropyl group in on a wedge or a dash, but you're trying to show the planar geometry around this carbon, so a straight line is what you need, and so those are the two products. Let's do one more, and you can see this substrate in the previous example. The only difference is this time, the chlorine is on a wedge instead of a dash, so if I number my ring one, two, three, four, I've already put in both chair conformations to save time, so that's carbon one, this is carbon two, this is carbon three, and this is carbon four. On the other chair conformation, this is carbon one, two, three, and four, and notice for the chair conformation on the left, we have the chlorine in the axial position, so this would be the alpha carbon, and the carbons next to the alpha carbon would be the beta carbons, so this one on the right is a beta carbon, and the one on the left is a beta carbon. We need a proton that's anti-periplanar, and the only one that fits would be this hydrogen right here, so if a base takes that proton, so let me draw in our strong base, taking this proton, these electrons would move into here to form our double bond, the electrons come off onto our chlorine, and a double bond forms between carbons two and three," + }, + { + "Q": "At 1:52 someone mentioned protozoans. What are protozoans", + "A": "In some systems of biological classification, the Protozoa are a diverse group of unicellular eukaryotic organisms. Historically, protozoa were defined as single-celled organisms with animal-like behaviours, such as motility and predation.", + "video_name": "1aJBToJrlvA", + "timestamps": [ + 112 + ], + "3min_transcript": "Man: This is an animal. This is also an animal. Animal. Animal carcass. Animal. Animal carcass again. Animal. The thing that all of these other things have in common is that they're made out of the same basic building block, the animal cell. (music) Animals are made up of your run-of-the-mill eukaryotic cells. These are called eukaryotic because they have a true kernel in the Greek, a good nucleus. That contains the DNA and calls the shots for the rest of the cell, also containing a bunch of organelles. There's a bunch of different kinds of organelles and they all have very specific functions. All of this is surrounded by the cell membrane. Of course, plants are eukaryotic cells too, but theirs are set up a little bit differently. Of course, they have oranelles that allow them to make their own food, which is super nice. We don't have those. Also, their cell membrane is actually a cell wall that's made of cellulose. It's rigid which is why plants can't dance. we did a whole video on it and you can click on it hit here, if it's online yet; it might not be. A lot of the stuff in this video is going to apply to all eukaryotic cells, which includes plants and fungi and protists. Rigid cell walls, that's cool and all, but one of the reasons that animals have been so successful is that their flexible membrane, in addition to allowing them the ability to dance, gives animals the flexibility to create a bunch of different cell types and organ types and tissue types that could never be possible in a plant. Cell walls that protect plants and give them structure prevent them from evolving complicated nerve structures and muscle cells that allow animals to be such a powerful force for eating plants. Animals can move around, find shelter and food, find things to mate with, all that good stuff. In fact, the ability to move oneself around using specialized muscle tissue has been 100% trademarked by kindgom animalia. Voiceover: What about protozoans? Man: Excellent point. What about protozoans? They don't have specialized muscle tissue. They move around with cilia and flagella and that kind of thing. Robert Hooke discovered cells with his kind of crude beta version microscope. He called them cells because they looked like bare, spartan Monk's bedrooms with not much going on inside. Hooke was a smart guy and everything, but he could not have been more wrong about what was going on inside of a cell. There is a whole lot going on inside of a eukaryotic cell. It's more like a city than a Monk's cell. In fact, let's go with that. A cell is like a city. It has defined geographical limits, a ruling government, power plants, roads, waste treatment plants, a police force, industry, all the things a booming metropolis needs to run smoothly. But, this city does not have one of those hippy governments where everybody votes on stuff and talks things out at town hall meetings Nope. Think fascist Italy circa 1938. Think Kim Jong-Il's, I mean Kim Jong-Un's North Korea and you might be getting a closer idea of how eukaryotic cells do their business. Let's start out with city limits. As you approach the city of Eukaryopolis, there's a chance that you will notice something that a traditional city never has, which is either cilia or flagella." + }, + { + "Q": "At 6:17, why is the pz, px, py etc. used and what do the subscripts stand for?", + "A": "In the p subshell there are three p orbitals: the px, py, and pz orbitals. These three orbitals are identical, except that they point in different directions (they are orthogonal to each other). The subscripts distinguish the p orbitals based on their orientation; if you draw an imaginary x-y-z axis with the origin at your atom of interest, then the px orbital points along the x-axis, the py orbital points along the y axis, and the pz orbital points along the z axis.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 377 + ], + "3min_transcript": "it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon." + }, + { + "Q": "At 1:57, the electron went way far out, how far out can that get?", + "A": "There is no limit, but the probability of being very far away is infinitesimal.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 117 + ], + "3min_transcript": "In the last few videos we learned that the configuration of electrons in an atom aren't in a simple, classical, Newtonian orbit configuration. And that's the Bohr model of the electron. And I'll keep reviewing it, just because I think it's an important point. If that's the nucleus, remember, it's just a tiny, tiny, tiny dot if you think about the entire volume of the actual atom. And instead of the electron being in orbits around it, which would be how a planet orbits the sun. Instead of being in orbits around it, it's described by orbitals, which are these probability density functions. So an orbital-- let's say that's the nucleus it would describe, if you took any point in space around the nucleus, the probability of finding the electron. So actually, in any volume of space around the nucleus, it would tell you the probability of finding the electron within that volume. And so if you were to just take a bunch of snapshots of electrons -- let's say in the 1s orbital. You can barely see it there, but it's a sphere around the nucleus, and that's the lowest energy state that an electron can be in. If you were to just take a number of snapshots of electrons. Let's say you were to take a number of snapshots of helium, which has two electrons. Both of them are in the 1s orbital. It would look like this. If you took one snapshot, maybe it'll be there, the next snapshot, maybe the electron is there. Then the electron is there. Then the electron is there. Then it's there. And if you kept doing the snapshots, you would have a bunch of them really close. And then it gets a little bit sparser as you get out, as you get further and further out away from the electron. But as you see, you're much more likely to find the electron close to the center of the atom than further out. Although you might have had an observation with the electron sitting all the way out there, or sitting over here. So it really could have been anywhere, but if you take multiple observations, It's saying look, there's a much lower probability of finding the electron out in this little cube of volume space than it is in this little cube of volume space. And when you see these diagrams that draw this orbital like this. Let's say they draw it like a shell, like a sphere. And I'll try to make it look three-dimensional. So let's say this is the outside of it, and the nucleus is sitting some place on the inside. They're just saying -- they just draw a cut-off -- where can I find the electron 90% of the time? So they're saying, OK, I can find the electron 90% of the time within this circle, if I were to do the cross-section. But every now and then the electron can show up outside of that, right? Because it's all probabilistic. So this can still happen. You can still find the electron if this is the orbital we're talking about out here. Right? And then we, in the last video, we said, OK, the electrons fill up the orbitals from lowest energy state to high energy state." + }, + { + "Q": "In 11:50, he explains about the structure of Nitrogen. It is 1s^2, 2s^2,2p^3. How can there be 3 atoms in p, where it is not possible when its configuration is s?", + "A": "An orbital can have at most 2 electrons. In the s subshell, there is only one orbital, thus an s subshell is full with 2 electrons. In the p subshell, there are three orbitals, each can have at most 2 electrons, for a total maximum of 6 electrons. Similarly, the d subshell has 5 orbitals for a maximum of 10 electrons. The f subshell has 7 orbitals for a maximum of 14 electrons. Thus, in nitrogen each of the three p orbitals in the p subshell has one electron, for a total of 3 electrons in the 2p subshell.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 710 + ], + "3min_transcript": "Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2." + }, + { + "Q": "When you were in the second period at 8:01 to 8:07 you only count B for boron before making it 2p. Why didnt you count Li lithium and Be Berrilium?", + "A": "The valence electrons of Li and Be occupy the 2s orbital, not the 2p. Thus, they don t have anything to do with the 2p.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 481, + 487 + ], + "3min_transcript": "that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon. the first two electrons go into, so, 1s1, 1s2. So then it fills-- sorry, you can't see everything. So it fills the 1s2, so carbon's configuration. It fills 1s1 then 1s2. And this is just the configuration for helium. And then it goes to the second shell, which is the second period, right? That's why it's called the periodic table. We'll talk about periods and groups in the future. And then you go here. So this is filling the 2s. We're in the second period right here. That's the second period. One, two. Have to go off, so you can see everything. So it fills these two. So 2s2. And then it starts filling up the p orbitals. So then it starts filling 1p and then 2p. And we're still on the second shell, so 2s2, 2p2. if we just wanted to visualize this orbital right here, the p orbitals? So we have two electrons. So one electron is going to be in a-- Let's say if this is, I'll try to draw some axes. That's too thin. So if I draw a three-dimensional volume kind of axes. If I were to make a bunch of observations of, say, one of the electrons in the p orbitals, let's say in the pz dimension, sometimes it might be here, sometimes it might be there, sometimes it might be there. And then if you keep taking a bunch of observations, you're going to have something that looks like this bell shape, this barbell shape right there. And then for the other electron that's maybe in the x direction, you make a bunch of observations. Let me do it in a different," + }, + { + "Q": "Around 12:55 it explains silicone and i dont get how you calculate the electron configuration. I find it terrible confusing. how you have added up the numbers. which numbers do i add up to make the electron configuration?", + "A": "Firstly silicon is not the same as silicone (be careful on exams!). For a neutral atom the number of electrons = the number of protons. For silicon that is 14. The fill order is: 1s 2s, 2p 3s, 3p 4s (google orbital fill diagram) An s orbital can take 2 electrons, a p orbital can take 6 electrons. We just keep filling until we get to 14. 1s2 (12 left), 2s2 (10 left), 2p6 (4 left), 3s2 (2 left), 3p2 (none left). Sorry can t do the superscripts here.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 775 + ], + "3min_transcript": "The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons So this is the electron configuration, right here, of nitrogen. And just to make sure you did your configuration right, what you do is you count the number of electrons. So 2 plus 2 is 4 plus 3 is 7. And we're talking about neutral atoms, so the electrons should equal the number of protons. The atomic number is the number of protons. So we're good. Seven protons. So this is, so far, when we're dealing just with the s's and the p's, this is pretty straightforward. And if I wanted to figure out the configuration of silicon, right there, what is it? Well, we're in the third period. One, two, three. That's just the third row. And this is the p-block right here. So this is the second row in the p-block, right? One, two, three, four, five, six. Right. We're in the second row of the p-block, so we start off with 3p2. And then it filled up all of this p-block over here. So it's 2p6. And then here, 2s2. And then, of course, it filled up at the first shell before it could fill up these other shells. So, 1s2. So this is the electron configuration for silicon. And we can confirm that we should have 14 electrons. 2 plus 2 is 4, plus 6 is 10. 10 plus 2 is 12 plus 2 more is 14. So we're good with silicon. I think I'm running low on time right now, so in the next video we'll start addressing what happens when you go to these elements, or the d-block. And you can kind of already guess what happens. We're going to start filling up these d orbitals here that have even more bizarre shapes. And the way I think about this, not to waste too much time, is that as you go further and further out from the nucleus," + }, + { + "Q": "at 05:38 it shows wierd looking things, what are those?", + "A": "Those weird looking things are orbitals, the s, p, d and f orbitals. These orbitals are the mathematical functions of the regions of space about the nucleus in which the electrons can be found, maximum of 2 electrons per orbital.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 338 + ], + "3min_transcript": "it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon." + }, + { + "Q": "What are the diagrams at 04:49 , I don't know them even though Sal mentioned them.", + "A": "These diagrams show the spacial geometry of atomic orbitals. The s orbitals (first column) are actually spheres, but they are shown here as cross-sections to show their nodes (places where the probability of finding an electron is 0). The red shows where the formula describing the orbital (called a wavefunction, from quantum mechanics) has a positive value, and the blue is where it has a negative value.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 289 + ], + "3min_transcript": "well I don't know if Tetris is the thing-- but if I'm stacking cubes, I lay out cubes from low energy, if this is the floor, I put the first cube at the lowest energy state. And let's say I could put the second cube But I only have this much space to work with. So I have to put the third cube at the next highest energy state. In this case our energy would be described as potential energy, right? This is just a classical, Newtonian physics example. But that's the same idea with electrons. Once I have two electrons in this 1s orbital -- so let's say the electron configuration of helium is 1s2 -- the third electron I can't put there anymore, because there's only room for two electrons. The way I think about it is these two electrons are now going to repel the third one I want to add. So then I have to go to the 2s orbital. it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves." + }, + { + "Q": "At 11:17, why is Be 2s one when Li is 1s two? They are right beside each other.", + "A": "He was only doing lithium. The complete electron configuration of Li is 1s2 2s1. The electron configuration of Be is 1s2 2s2.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 677 + ], + "3min_transcript": "It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons" + }, + { + "Q": "In 4:50, does the red and blue represent + 1/2 and -1/2 spin?", + "A": "No, the different colours do not represent the spin as if you look at the 4 lobes of the d orbital (for example the dxz sub-shell), there are two red and two blue sections, but at most it will only contain 2 electrons, which can be found in any of the 4 lobes and not one in each section or even pair of sections.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 290 + ], + "3min_transcript": "well I don't know if Tetris is the thing-- but if I'm stacking cubes, I lay out cubes from low energy, if this is the floor, I put the first cube at the lowest energy state. And let's say I could put the second cube But I only have this much space to work with. So I have to put the third cube at the next highest energy state. In this case our energy would be described as potential energy, right? This is just a classical, Newtonian physics example. But that's the same idea with electrons. Once I have two electrons in this 1s orbital -- so let's say the electron configuration of helium is 1s2 -- the third electron I can't put there anymore, because there's only room for two electrons. The way I think about it is these two electrons are now going to repel the third one I want to add. So then I have to go to the 2s orbital. it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves." + }, + { + "Q": "how can we place helium[inert gas]in the second group \"10:20\"?", + "A": "becouse it has two electrons in its outer shell just like the other elements of the second group. the rest of the noblegasses have 8 electrons in their outer shell", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 620 + ], + "3min_transcript": "if we just wanted to visualize this orbital right here, the p orbitals? So we have two electrons. So one electron is going to be in a-- Let's say if this is, I'll try to draw some axes. That's too thin. So if I draw a three-dimensional volume kind of axes. If I were to make a bunch of observations of, say, one of the electrons in the p orbitals, let's say in the pz dimension, sometimes it might be here, sometimes it might be there, sometimes it might be there. And then if you keep taking a bunch of observations, you're going to have something that looks like this bell shape, this barbell shape right there. And then for the other electron that's maybe in the x direction, you make a bunch of observations. Let me do it in a different, It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2." + }, + { + "Q": "I dont get how to use the chart at 4:46?", + "A": "The chart show the shapes of each of the orbitals in a given energy level. For example, for n = 4, the chart shows in turn the shapes of the orbitals: 4s, 4px, 4py, 4pz, the five 4d orbitals, and some of the 4f orbitals.Except for levels 1 to 3, you do not have to memorize the shapes. It is more of a reference tool when you want to know what the shapes look like.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 286 + ], + "3min_transcript": "well I don't know if Tetris is the thing-- but if I'm stacking cubes, I lay out cubes from low energy, if this is the floor, I put the first cube at the lowest energy state. And let's say I could put the second cube But I only have this much space to work with. So I have to put the third cube at the next highest energy state. In this case our energy would be described as potential energy, right? This is just a classical, Newtonian physics example. But that's the same idea with electrons. Once I have two electrons in this 1s orbital -- so let's say the electron configuration of helium is 1s2 -- the third electron I can't put there anymore, because there's only room for two electrons. The way I think about it is these two electrons are now going to repel the third one I want to add. So then I have to go to the 2s orbital. it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves." + }, + { + "Q": "great vid Sal. But wouldn't @11:11 Lithium be 2s^1 then 2s^2?", + "A": "You would have to fill up the lowest orbital, which means the electron configuration always starts with 1s^1, 1s^2 and so forth. When writing it s like taking the electron configuration of the rightmost element immediately above the period of that element and then writing out the next configuration of the period it is in.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 671 + ], + "3min_transcript": "It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons" + }, + { + "Q": "configuration of berilium is 2s1 or 2s2 at 11:30", + "A": "Beryllium has four electrons, so its configuration is 1s\u00c2\u00b22s\u00c2\u00b2.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 690 + ], + "3min_transcript": "It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons" + }, + { + "Q": "At 00:42 Sal says that the reaction is \"spontaneous\", but I don't exactly understand what that means. How did he figure that out?", + "A": "Spontaneous means that somethings starting energy is higher than its ending, it comes from something called Gibbs free energy. This means it releases energy and doesn t need energy to proceed.", + "video_name": "-KE7jTXwNYs", + "timestamps": [ + 42 + ], + "3min_transcript": "- We've talked a lot about ATP being the energy currency of cells, but I want to dig a little bit deeper into that in this video. And as we'll see when we go from ATP to ADP, ADP + a Phosphate group, we have a release of free energy. If we look at just the system, ATP's free energy is over here, but once hydrolysis has taken place and now it's ADP + a Phosphate group, the free energy has dropped by roughly 30.5 KJ/mol our Delta G is -30.5 Kilojoules per mole and if you watch the videos on Gibbs Free Energy this tells us that this is a spontaneous reaction. Delta, Delta G is less than zero, which tells us that this is going to be spontaneous. Now when I first learned this I was like, well if it's going to be spontaneous why wouldn't all of the ATP just spontaneously and all of the water turn into ADP and just release it's energy as heat or whatever else? And the key is, it has to get over this hump. it has to go uphill a little bit if there's no enzyme to catalyze it. And the reason why we have this uphill hump is the way that ATP gets broken up is that you have to, if we're talking about a water molecule doing the hydrolysis which is typically what people think about hydrolysis, although it can sometimes be done by a different molecule, if you think about the water molecule what needs to happen is, you need this lone pair of electrons on this oxygen to be able to do what we call a nucleophilic attack on this Phosphorous in this phosphate group and if that happens, it forms this bond and then these electrons can be taken back by this oxygen which gives its negative charge right over there. Now you might say, \"Okay, this makes a lot of sense\", but you have to rememeber this, electrons are negative and they're surrounded by these negative charges. So they have to overcome getting close to these things so as they're approaching these negative charges, they want to repel each other, so you have to overcome that. And the way that it's overcome is a class of enzymes And what they do is, remember these enzymes are these big protein complexes and the ATP molecule combined in the right place. And they essentially try and surround the ATP molecule with some positive ions. So let's say there's a positive ion here. So it can keep these electrons busy while the water, or whatever is doing the nucleophilic attack, can not have to worry about these electrons over here. So it might have a positive ion here. And remember, if we think in three dimensions, this thing is all wrapped around in different ways around the ATP molecule, so this is the ATPase. And so by having the enzyme over here, you're lowering the activation energy. And so it might end up looking something more like this. And so the reaction can actually occur. So the reason why you don't just see this happening all the time without an enzyme is why you have to overcome this hump." + }, + { + "Q": "At 2:06 NH4Cl is called an acid, but isn't it a salt?", + "A": "It is a salt, but NH4+ is ammonium, which is the conjugate acid of ammonia (NH3).", + "video_name": "lsHq5aqz4uQ", + "timestamps": [ + 126 + ], + "3min_transcript": "- [Voiceover] Let's do some buffer solution calculations using the Henderson-Hasselbalch equation. So in the last video I showed you how to derive the Henderson-Hasselbalch equation, and it is pH is equal to the pKa plus the log of the concentration of A minus over the concentration of HA. So we're talking about a conjugate acid-base pair here. HA and A minus. And for our problem HA, the acid, would be NH four plus and the base, A minus, would be NH three or ammonia. So the first thing we need to do, if we're gonna calculate the pH of our buffer solution, is to find the pKa, all right, and our acid is NH four plus. So let's say we already know the Ka value for NH four plus and that's 5.6 times 10 to the negative 10. To find the pKa, all we have to do is take the negative log of that. of 5.6 times 10 to the negative 10. So let's get out the calculator and let's do that math. So the negative log of 5.6 times 10 to the negative 10. Is going to give us a pKa value of 9.25 when we round. So pKa is equal to 9.25. So we're gonna plug that into our Henderson-Hasselbalch equation right here. So the pH of our buffer solution is equal to 9.25 plus the log of the concentration of A minus, our base. Our base is ammonia, NH three, and our concentration in our buffer solution is .24 molars. We're gonna write .24 here. And that's over the concentration of our acid, that's NH four plus, and our concentration is .20. So let's find the log, the log of .24 divided by .20. And so that is .080. So 9.25 plus .08 is 9.33. So the final pH, or the pH of our buffer solution, I should say, is equal to 9.33. So remember this number for the pH, because we're going to compare what happens to the pH when you add some acid and when you add some base. And so our next problem is adding base to our buffer solution. And we're gonna see what that does to the pH. So now we've added .005 moles of a strong base to our buffer solution. Let's say the total volume is .50 liters. So what is the resulting pH?" + }, + { + "Q": "how can i identify that solution is buffer solution ? And at 4:35 how does he know that whole of the NH4Cl is going to dissociate into 0.20M of NH4+", + "A": "You need to identify the conjugate acids and bases, and I presume that comes with practice. The same way you know that HCl dissolves to form H+ and Cl-, or H2SO4 form 2H+ and (SO4)2-. In this example with NH4Cl, the conjugate acids and bases are NH4+ and Cl-. Notice how also the way the formula is written will help you identify the conjugate acids and bases (acids come first on the left, bases on the right)", + "video_name": "lsHq5aqz4uQ", + "timestamps": [ + 275 + ], + "3min_transcript": "moles of sodium hydroxide, and our total volume is .50. So if we divide moles by liters, that will give us the concentration of sodium hydroxide. .005 divided by .50 is 0.01 molar. So that's our concentration of sodium hydroxide. And since sodium hydroxide is a strong base, that's also our concentration of hydroxide ions in solution. So this is our concentration of hydroxide ions, .01 molar. So we're adding a base and think about what that's going to react with in our buffer solution. So our buffer solution has NH three and NH four plus. The base is going to react with the acids. So hydroxide is going to react with NH four plus. Let's go ahead and write out the buffer reaction here. ammonium is going to react with hydroxide and this is going to go to completion here. So if NH four plus donates a proton to OH minus, OH minus turns into H 2 O. So we're gonna make water here. And if NH four plus donates a proton, we're left with NH three, so ammonia. Alright, let's think about our concentrations. So we just calculated that we have now .01 molar concentration of sodium hydroxide. For ammonium, that would be .20 molars. So 0.20 molar for our concentration. And for ammonia it was .24. So let's go ahead and write 0.24 over here. So if .01, if we have a concentration of hydroxide ions of .01 molar, all of that is going to react with the ammonium. So we're gonna lose all of this concentration here for hydroxide. the same amount of ammonium over here. So we're gonna be left with, this would give us 0.19 molar for our final concentration of ammonium. Hydroxide we would have zero after it all reacts, And then the ammonium, since the ammonium turns into the ammonia, if we lose this much, we're going to gain the same concentration of ammonia. So over here we put plus 0.01. So the final concentration of ammonia would be 0.25 molar. And now we can use our Henderson-Hasselbalch equation. So let's go ahead and plug everything in. So ph is equal to the pKa. We already calculated the pKa to be 9.25. And then plus, plus the log of the concentration of base, all right, that would be NH three. So the concentration of .25." + }, + { + "Q": "At 8:48, why is it that you first have to react HCL with H2O to get H3O, and in turn, react the with NH3 instead of directly reacting HCL with HN3 as you did (with NH4 and NaOH) in the first example? I would like to know, for future examples, what instances I would have to utilize this method instead of the first method.", + "A": "The HCl exists in water almost exclusively as H\u00e2\u0082\u0083O\u00e2\u0081\u00ba and Cl\u00e2\u0081\u00bb. So it is more correct chemically to write the reaction using H\u00e2\u0082\u0083O\u00e2\u0081\u00ba. In practice, you can do it either way. What matters is the concentration of NH\u00e2\u0082\u0084\u00e2\u0081\u00ba, not the equations you use to produce it.", + "video_name": "lsHq5aqz4uQ", + "timestamps": [ + 528 + ], + "3min_transcript": "So this shows you mathematically how a buffer solution resists drastic changes in the pH. Next we're gonna look at what happens when you add some acid. So we're still dealing with our same buffer solution with ammonia and ammonium, NH four plus. But this time, instead of adding base, we're gonna add acid. So we add .03 moles of HCl and let's just pretend like the total volume is .50 liters. And our goal is to calculate the pH of the final solution here. So the first thing we could do is calculate the concentration of HCl. So that would be moles over liters. So that's 0.03 moles divided by our total volume of .50 liters. And .03 divided by .5 gives us 0.06 molar. And HCl is a strong acid, so you could think about it as being H plus and Cl minus. And since this is all in water, H plus and H two O would give you H three O plus, or hydronium. So .06 molar is really the concentration of hydronium ions in solution. And so the acid that we add is going to react with the base that's present in our buffer solution. So this time our base is going to react and our base is, of course, ammonia. So let's write out the reaction between ammonia, NH3, and then we have hydronium ions in solution, H 3 O plus. So this reaction goes to completion. And if ammonia picks up a proton, it turns into ammonium, NH4 plus. And if H 3 O plus donates a proton, we're left with H 2 O. So we write H 2 O over here. For our concentrations, we're gonna have .06 molar And the concentration of ammonia is .24 to start out with. So we have .24. And for ammonium, it's .20. So we write 0.20 here. So all of the hydronium ion is going to react. So we're gonna lose all of it. So we're left with nothing after it all reacts. So it's the same thing for ammonia. So that we're gonna lose the exact same concentration of ammonia here. So we're gonna lose 0.06 molar of ammonia, 'cause this is reacting with H 3 O plus. And so after neutralization, we're left with 0.18 molar for the concentration of ammonia. And whatever we lose for ammonia, we gain for ammonium since ammonia turns into ammonium. So we're going to gain 0.06 molar" + }, + { + "Q": "At 3:52 what life could survive without nitrogen?", + "A": "Nothing. Nitrogen is one of the building blocks of protein. As all living things need protein, all life would die - plants and animals.", + "video_name": "6rwoktPmqpY", + "timestamps": [ + 232 + ], + "3min_transcript": "you'll notice here that nitrogen gas is made up of two nitrogen atoms stuck together with a triple bound. It's one thing to break apart a single covalent bond, but three? As you can imagine those two nitrogen atoms are a total pain to pry apart, but that molecule has to be split in order for a plant to get at the pieces. In fact, plants can assimilate a bunch of different forms of nitrogen, nitrates, nitrites to a lesser extent and even ammonium, which is what you get when you mix ammonia with water, but all that darn nitrogen gas in the atmosphere is beyond their powers of assimilation. So, plants need help taking advantage of this ocean of nitrogen that we're all swimming in, which is why they need to have that nitrogen fixed so that they can use it. Even though plants aren't wile enough to wrangle those two nitrogen atoms apart, certain nitrogen fixing bacteria are. These bacteria hang out in soil or water or even form symbiotic relationships with the root nodules of some plants, most of which are legumes. That's a pretty big family of plants, soybeans, clover, peanuts, kudzu, all legumes. which then becomes ammonium when it's mixed with water which can be used by plants. They do this with a special enzyme called nitrogenase which is the only biological enzyme that can break that crazy triple bond. Ammonia can also be made by decomposers, fungi, protists; other kinds of bacteria that munch on your proteins and DNA after you die, but they're not picky they like poop and urine too. Then once this has happened, other bacteria known as nitrifying bacteria can take this ammonia and convert it into nitrates; three oxygen atoms attached to a single nitrate atom and nitrite; two oxygens attached to a nitrogen. Those are even easier than ammonium for plants to assimilate. So, the take home here is, if it wasn't for these bacteria there would be a whole lot less of biologically available nitrogen hanging around and as a result there would be a lot fewer living things on the planet. As usual, thanks bacteria we owe you one, but I should mention that it's not just bacteria who can wrangle those two nitrogen atoms apart, lightning of all things has enough energy to break bonds between nitrogens, And in the 20th century, smarty pants humans also figured out various ways to synthetically fix a ton of nitrogen all at once, which is why we have synthetic fertilizers now, there's so much food growing all over the place. Once the atmospheric nitrogen is converted into a form that plants can use to make DNA and RNA and amino acids, organic nitrogen takes off up the food chain, animals eat the plants and use all that sweet, sweet, bio-available nitrogen to make our own amino acid. Then we pee or poop it out or die and the decomposers go to town on it, breaking it down into ammonia and it just keeps going, until one day, that organic nitrogen finds itself in denitrifying bacteria who's job it is to metabolize the nitrogen oxides and turn them back into nitrogen gas using a special enzyme called nitrate reductase. These guys do their business and then release the N2 back in the atmosphere. That my friends is the nitrogen cycle, if you remember nothing else remember that a) you owe bacteria a solid because they were smart enough to make an enzyme" + }, + { + "Q": "I have a problem with how we include friction in this problem. If the entire premise of the easy style of doing these problems is to treat the system as a single object, shouldn't we use 20 kg for the friction component, and not 12?\n\nIt makes sense to me that the box is being pressed down on the table by the additional force, and the normal force pressing back upwards would also be 20 kg * 9.8, rather than 12 kg *9.8.\n\nThis occurs at 4:40", + "A": "How is the box pressing down with 20 * 9.8 N of force? The force from the 3 and 5 kg blocks on the 12 kg block comes from the ropes that are going over the pulleys so it is horizontal and not not down so it doesn t add to the normal force.", + "video_name": "ibdidr-bEvI", + "timestamps": [ + 280 + ], + "3min_transcript": "to reduce the acceleration? Yeah, there's this force of gravity over here. This force of gravity on the three kilogram mass is trying to prevent the acceleration because it's pointing opposite the direction of motion. The motion of this system is upright and down across this direction but this force is pointing opposite that direction. This force of gravity right here. So, I'm gonna have to subtract three kilograms times 9.8 meters per second squared. Am I gonna have any other forces that try to prevent the system from moving? You might think the force of gravity on this 12 kilogram box, but look, that doesn't really, in and of itself, prevent the system from moving or not moving. That's perpendicular to this direction. I've called the direction of motion, this positive direction. If it were a force this way, if it were a force this way or a force that way it'd try to cause acceleration of the system. by the normal force, so I don't even have to worry about that force. So, are there any forces associated with the 12 kilogram box that try to prevent motion? It turns out there is. There is going to be a force of friction between the table because there's this coefficient of kinetic friction. So, I've got a force this way, this kinetic frictional force, that's gonna be, have a size of Mu K times f n. That's how you find the normal force and so this is gonna be minus, the Mu K is 0.1 and the normal force will be the normal force for this 12 kilogram mass. So, I'll use 12 kilograms times 9.8 meters per second squared. You might object, you might say, \"Hey, hold on, 12 times 9.8, that's the force of gravity. \"Why are you using this force? \"I thought you said we didn't use it?\" Well, we don't use this force by itself, but it turns out this force of friction depends on this force. So, we're really using a horizontal force, which is why we've got this negative sign here, but it's a horizontal force. It just so happens that this horizontal force depends on a vertical force, which is the normal force. And so that's why we're multiplying by this .1 that turns this vertical force, which is not propelling the system, or trying to stop it, into a horizontal force which is trying to reduce the acceleration of the system. That's why I subtracted and then I divide by the total mass and my total mass is gonna be three plus 12 plus five is gonna be 20 kilograms. Now, I can just solve. If I solve this, I'll get that the acceleration of this system is gonna be 0.392 meters per second squared. So, this is a very fast way. Look it, this is basically a one-liner. If you could put this together right, it's a one-liner. There's much less chance for error than when you're trying to solve three equations with three unknowns. This is beautiful." + }, + { + "Q": "At 0:13 Sal uses a sign to represent the angle of the launch. I wanted to know what that sign means? Thanks.", + "A": "It s the Greek letter theta, which stands for angle (it s often used in place of an unknown angle). Hope this helps!", + "video_name": "RhUdv0jjfcE", + "timestamps": [ + 13 + ], + "3min_transcript": "Let's say we're going to shoot some object into the air at an angle. Let's say its speed is s and the angle at which we shoot it, the angle above the horizontal is theta. What I want to do in this video is figure out how far this object is going to travel as a function of the angle and as a function of the speed, but we're going to assume that we're given the speed. That that's a bit of a constant. So if this is the ground right here, we want to figure out how far this thing is going to travel. So you can imagine, it's going to travel in this parabolic path and land at some point out there. And so if this is at distance 0, we could call this distance out here distance d. Now whenever you do any problem like this where you're shooting something off at an angle, the best first step is to break down that vector. Remember, a vector is something that has magnitude and direction. The magnitude is s. Maybe feet per second or miles per hour. So if you have s and theta, you're giving me a vector. And so what you want to do is you want to break this vector down into its vertical and horizontal components first and then deal with them separately. One, to help you figure out how long you're in the air. And then, the other to figure out how far you actually travel. So let me make a big version of the vector right there. Once again, the magnitude of the vector is s. So you could imagine that the length of this arrow is s. And this angle right here is theta. And to break it down into its horizontal and vertical components, we just set up a right triangle and just use our basic trig ratios. So let me do that. So this is the ground right there. I can drop a vertical from the tip of that arrow to set up a right triangle. And the length of the-- or the magnitude of the vertical component of our velocity is going to be this That is going to be-- you could imagine, the length of that is going to be our vertical speed. So this is our vertical speed. Maybe I'll just call that the speed sub vertical. And then, this right here, the length of this part of the triangle-- let me do that in a different color. The length of this part of the triangle is going to be our horizontal speed, or the component of this velocity in the horizontal direction. And I use this word velocity when I specify a speed and a direction. Speed is just the magnitude of the velocity. So the magnitude of this side is going to be speed horizontal. And to figure it out, you literally use our basic trig ratios. So we have a right triangle. This is the hypotenuse. And we could write down soh cah toa up here." + }, + { + "Q": "at 3:23 shouldn't the hypothesis be in an if, then statement?", + "A": "You can set up a hypothesis as an if/then statement, but it isn t always required. Hope that helps!", + "video_name": "N6IAzlugWw0", + "timestamps": [ + 203 + ], + "3min_transcript": "in I don't know, northern Canada or something, and let's say that you live near the beach, but there's also a pond near your house, and you notice that the pond, it tends to freeze over sooner in the Winter than the ocean does. It does that faster and even does it at higher temperatures than when the ocean seems to freeze over. So, you could view that as your observation. So, the first step is you're making an observation. Observation. In our particular case is that the pond freezes over at higher temperatures than the ocean does, and it freezes over sooner in the Winter. Well, the next question that you might wanna, or the next step you could view as a scientific method. It doesn't have to be this regimented, but this is a structured way of thinking about it. Well, ask yourself a question. Ask a question. Why does, so in this particular question, or in this particular scenario, why does the pond tend than the ocean does? Well, you then try to answer that question, and this is a key part of the scientific method, is what you do in this third step, is that you try to create an explanation, but what's key is that it is a testable explanation. So, you try to create a testable explanation. Testable explanation, and this is kind of the core, one of the core pillars of the scientific method, and this testable explanation is called your hypotypothesis. Your hypothesis. And so, in this particular case, a testable explanation could be that, well the ocean is made up of salt water, and this pond is fresh water, so your testable explanation could be salt water, salt water has lower freezing point. freezing point. Lower freezing point, so it takes colder temperatures to freeze it than fresh water. Than fresh water. So, this, right over here, this would be a good hypothesis. It doesn't matter whether the hypothesis is actually true or not. We haven't actually run the experiment, but it's a good one, because we can construct an experiment that tests this very well. Now, what would be an example of a bad hypothesis or of something that you couldn't even necessarily consider as part of the scientific method? Well, you could say that there is a fairy that blesses that blesses, let's say that performs magic, performs magic on the pond to freeze it faster." + }, + { + "Q": "oxygen and carbon are having a triple bond at 6:17 the valency of oxygen is 2 how is it possible?", + "A": "I think this is because the new third bond is a dative covalent bond , when both electrons of electron pair shared to form a bond is from ONE atom (in this case, oxygen) only. It can be wrong so plz look it up :)", + "video_name": "vFfriC55fFw", + "timestamps": [ + 377 + ], + "3min_transcript": "" + }, + { + "Q": "at 2:59, what is a photon?", + "A": "Its the basic unit of light. Its the smallest amount of light that you can play around(create, reflect, refract) with.", + "video_name": "y55tzg_jW9I", + "timestamps": [ + 179 + ], + "3min_transcript": "In a vacuum. There's nothing there, no air, no water, no nothing, that's where the light travels the fastest. And let's say that this medium down here, I don't know, let's say it's water. Let's say that this is water. All of this. This was all water over here. This was all vacuum right up here. So what will happen, and actually, that's kind of an unrealistic-- well, just for the sake of argument, let's say we have water going right up against a vacuum. This isn't something you would normally just see in nature but let's just think about it a little bit. Normally, the water, since there's no pressure, it would evaporate and all the rest. But for the sake of argument, let's just say that this is a medium where light will travel slower. What you're going to have is is this ray is actually going to switch direction, it's actually going to bend. Instead of continuing to go in that same direction, it's going to bend a little bit. It's going to go down, in that direction is the refraction. That's the refraction angle. Refraction angle. Or angle of refraction. This is the incident angle, or angle of incidence, and this is the refraction angle. Once again, against that perpendicular. And before I give you the actual equation of how these two things relate and how they're related to the speed of light in these two media-- and just remember, once again, you're never going to have vacuum against water, the water would evaporate because there's no pressure on it and all of that type of thing. But just to--before I go into the math of actually how to figure out these angles relative to the velocities of light in the different media I want to give you an intuitive understanding of not why it bends, 'cause I'm not telling you actually how light works this is really more of an observed property and light, as we'll learn, as we do more and more videos about it, can get pretty confusing. Sometimes you want to treat it as a ray, sometimes you want to treat it as a wave, sometimes you want to treat it as a photon. I actually like to think of it as kind of a, as a bit of a vehicle, and to imagine that, let's imagine that I had a car. So let me draw a car. So we're looking at the top of a car. So this is the passenger compartment, and it has four wheels on the car. We're looking at it from above. And let's say it's traveling on a road. It's traveling on a road. On a road, the tires can get good traction. The car can move pretty efficiently, and it's about to reach an interface it's about to reach an interface where the road ends and it will have to travel on mud. It will have to travel on mud. Now on mud, obviously, the tires' traction will not be as good. The car will not be able to travel as fast. So what's going to happen? Assuming that the car, the steering wheel isn't telling it to turn or anything, the car would just go straight in this direction. But what happens right when--which wheels are going to reach the mud first? Well, this wheel. This wheel is going to reach the mud first." + }, + { + "Q": "At 6:30, don't you need to take the inverse of that expression to get the equivalent resistance of the 2 resistors in parallel?", + "A": "There are two formulas for computing 2 parallel resistors. The one I used is Rp = (R1 R2)/(R1+R2). The other formula is the one with all the reciprocals. That s the one you are thinking of: 1/Rp = 1/R1 + 1/R2. Both give the same answer.", + "video_name": "j-iR7puLj6M", + "timestamps": [ + 390 + ], + "3min_transcript": "If you look here, I have two batteries that are hooked up, their inputs, their positive side is hooked up together and their negative side is hooked up together, so they're actually just acting like one, big battery, so let me draw that. I'm going to draw the circuit again so it looks like this. Here's my combined big battery. And it goes to... Relabel these again so we don't get mixed up. Okay. This is R1. This is R2. So this circuit looks a little simpler, and I'm gonna look at it again, see if I can do any more simplification, so what I recognize right here, right in this area right here, R1 and R2 are in parallel. They have the same voltage on their terminals. That means they're in parallel. I know how to simplify parallel resistors. We'll just use the parallel resistor equation that I have in my head, and that looks like this, let's go to this color here. Okay, so parallel resistors, R1 in parallel with R2. I made up this symbol, two vertical lines, that means they're in parallel, and the formula for two parallel resistors is R1 times R2, over R1 plus R2. Now I'll plug in the values. over here at our schematic, they're the same value, and that has a special thing when in parallel resistors so it's actually R R over 2R. Because those resistors are the same. And you can see I can cancel that and I can cancel that and two parallel resistors, if the resistors are equal, is equal to half the resistance. And let's plug in the real values, 1.4 ohms over 2 equals 0.7 ohms. That's the equivalent resistance of these two resistors in parallel. So this is a good time to redraw this circuit again. Let's do it again. Here's our battery. This time I'm going to draw the equivalent resistance. Then we have R3," + }, + { + "Q": "at 7:30,what are microtubules?", + "A": "mi\u00c2\u00b7cro\u00c2\u00b7tu\u00c2\u00b7bule A microscopic tubular structure present in numbers in the cytoplasm of cells, sometimes aggregating to form more complex structures", + "video_name": "X1bmedVziGw", + "timestamps": [ + 450 + ], + "3min_transcript": "by like, completely disintegrating and the centrosomes then peel away from the nucleus, start heading to the opposite ends of the cell. As they go, they leave behind a wide trail of protein ropes called microtubules running from one centrosome to the other. You might recall from our anatomy of the animal cell the microtubules help provide a kind of structure to the cell and this is exactly what they're doing here. Now we reach the metaphase, which literally means after phase and it's the longest phase of mitosis. It could take up to 20 minutes. During the metaphase, the chromosomes attach to those ropey microtubules right in the middle at their centromeres. The chromosomes then begin to be moved around and this seems to be being done by molecules called motor proteins. And while we don't know too much about how these motors work, we do know, for instance that there are two of them on each side of the centromere. These are called centromere associated protein E. So these motor proteins attached to the microtubule ropes basically serve to spool up the tubule slack. Now at the same time, another protein called dynein near the cell membranes. After being pulled in this direction and that, the chromosomes line up right down the middle of the cell and that brings us to the latest installment of biolo-graphy. (music) So how do chromosomes line up like that? We know that there are motor proteins involved, but like, how? What are they doing? Well, remember when I said earlier that there were a lot of things that we don't totally understand about mitosis? It's sort of weird that we don't because we can literally watch mitosis happening under microscopes, but chromosome alignment is a good example of a small detail that is only very recently been figured out. And it was a revelation like 130 years in the making. Mitosis was first observed by a German biologist by the name of Walther Flemming, who in 1878 was studying the tissue of salamander gills and fins when he saw cell's nuclei split in two and migrate away from each other to form two new cells. He called this process mitosis because of the messy jumble of chromatin, a term he also coined that he saw on the nuclei. But Flemming didn't pick up on the implications of this discovery for genetics, which was still a young discipline and over the next century generations of scientists started piecing together the mitosis puzzle by determining the role of microtubules, say, or identifying motor proteins. And the most recent contribution to this research was made by a postdoctoral student named Tomomi Kiyomitsu at MIT. He watched the same process that Flemming watched and figured out how, at least one, of the motor proteins helps snap the chromosomes into line. He was studying the motor protein called dynein which sits on the inside of the membrane. Think of the microtubules as tug-of-war ropes with the chromosomes as the flag in the middle. What Kiyomitsu discovered was that the dynein plays tug-of-war with itself. Dynein grabs on to one end of the microtubules and pulls the tubules and chromosomes towards one end of the cell. When the ends of the microtubules come too close to the cell membrane they release a chemical signal that punts the dynein" + }, + { + "Q": "At 9:07 does mol mean molecule or is that what it is called.", + "A": "mol means mole (which is Avagadro s number of an atom or molecule).", + "video_name": "-QpkmwIoMaY", + "timestamps": [ + 547 + ], + "3min_transcript": "atmospheres-- let me get my calculator out, get the calculator out, put it right over there-- so it's going to be 23.8 times 1 over 760, or just divided by 760. And we have three significant digits, so it looks like 0.0313. So this is equal to 0.0313 atmospheres. That is our vapor pressure. So let's just deal with this right here. So the number of molecules of water that are going to be in the air in the gaseous state, in the vapor state, is going to be equal to our vapor pressure. That's our equilibrium pressure. If more water molecules evaporate after that point, then we're going to have a higher pressure, which will state, so we'll go kind of past the equilibrium, which is not likely. Or another way to think about it-- more water molecules are not going to of evaporate at a faster rate than they are going to condense beyond that pressure. Anyway, the pressure here is 0.0313 atmospheres. The volume here-- they told us right over here-- so that's the volume-- 4.25. 4.25 times 10 to the fourth liters. And then we want to divide that by-- and you want to make sure that your universal gas constant has the right units, I just looked mine up on Wikipedia-- 0.08-- see everything has three significant digits. So let me just allow that more significant digits and we'll just round at the end. 0.082057, and the units here are liters atmospheres per And this makes sense. This liter will cancel out with that liter. That atmospheres cancels out with that atmospheres. I'm about to multiply it by temperature right here in kelvin. We'll cancel out there. And then we'll have a 1 over moles in the denominator. A 1 over moles in the denominator will just be a moles because you're going to invert it again. So that gives us our answer in moles. And so finally our temperature-- and you've got to remember you've got to do it in kelvin. So 25 degrees Celsius-- let me right it over here-- 25 degrees Celsius is equal to, you just add 273 to it, so this is equal to 298 kelvin. So times 298 kelvin. And now we just have to calculate this. So let's do that." + }, + { + "Q": "At 13:26 I don't understand why Sal multiplied the H2O g with the density of water that's stated in the question above. I would love an explanation.", + "A": "Density is mass per volume of a substance. So Density=Mass/Volume (which is why the unit is grams/mL). We have the mass for water = 980grams and density of water (at the same temp) as 0.997g/ml. so substituting the value we can get the volume of water as 980g/0.997g/mL which gives us the volume in mL.", + "video_name": "-QpkmwIoMaY", + "timestamps": [ + 806 + ], + "3min_transcript": "So this is equal to 54.4 moles. And we could see this liters cancels out with that liters. Kelvin cancels out with kelvin. Atmospheres with atmospheres. You have a 1 over mole in the denominator. So then 1 over 1 over moles is just going to be moles. Now, this is going to be 54.4 moles of water vapor in the room to have our vapor pressure. If more evaporates, then more will condense-- we will be beyond our equilibrium. So we won't ever have more than this amount evaporate in that room. So let's figure out how much liquid water that actually is. Let me do it over here. So 54.4 moles-- let me write it down-- moles of H2O. That's going to be in its vapor form and its going to evaporate. So what is the molar mass of water? Well, it's roughly 18. I actually figured it out exactly. It's actually 18.01 if you actually use the exact numbers on the periodic table, at least one that I used. So we could say that there's 18.01 grams of H2O for every 1 mole of H2O. And obviously, you can just look up the atomic weight of hydrogen, which is a little bit over 1, and the atomic weight of oxygen, which is a little bit below 16. So you have two of these. So 2 plus 16 gives you pretty close to 18. So this right here will tell you the grams of water that can evaporate to get us to that equilibrium pressure. So let's get the calculator out. So we have the 54.4 times 18.01 is equal to 970-- well, this 0.7, it becomes 980. So this is 980 grams of H2O needs to evaporate for us to get to our equilibrium pressure, to our vapor pressure. So let's figure out how many milliliters of water this is. So they tell us the density of water right here. 0.997-- let me do this in a darker color-- 0.997 grams per millileter. Or another way you could view this is for everyone 1 milliliter you have 0.997 grams of water at 25 degrees Celsius. So for every milliliter-- this is grams per milliliter-- we want milliliters per gram because we want this and this to cancel out. So we're essentially just going to divide 980 by 0.997." + }, + { + "Q": "At 8:30, why is the volume of the room used instead of the volume of water?", + "A": "As the liquid water sits in the container, it releases water vapour which spreads throughout the room until it is at an equilibrium with the liquid water. There will be equal amounts of water vapour in every spot of the room and Sal wants to know how many total molecules of vapour there are. So, he takes the entire volume of the room and not just that of the liquid water.", + "video_name": "-QpkmwIoMaY", + "timestamps": [ + 510 + ], + "3min_transcript": "Now, the hardest thing about this is just making sure you have your units right and you're using the right ideal gas constant for the right units, and we'll do that right here. So what I want to do, because the universal gas constant that I have is in terms of atmospheres, we need to figure out this vapor pressuree- this equilibrium pressure between vapor and liquid-- we need to write this down in terms of atmospheres. So let me write this down. So the vapor pressure is equal to 23.8 millimeters of mercury. And you can look it up at a table if you don't have this One atmosphere is equivalent to 760 millimeters of mercury. atmospheres-- let me get my calculator out, get the calculator out, put it right over there-- so it's going to be 23.8 times 1 over 760, or just divided by 760. And we have three significant digits, so it looks like 0.0313. So this is equal to 0.0313 atmospheres. That is our vapor pressure. So let's just deal with this right here. So the number of molecules of water that are going to be in the air in the gaseous state, in the vapor state, is going to be equal to our vapor pressure. That's our equilibrium pressure. If more water molecules evaporate after that point, then we're going to have a higher pressure, which will state, so we'll go kind of past the equilibrium, which is not likely. Or another way to think about it-- more water molecules are not going to of evaporate at a faster rate than they are going to condense beyond that pressure. Anyway, the pressure here is 0.0313 atmospheres. The volume here-- they told us right over here-- so that's the volume-- 4.25. 4.25 times 10 to the fourth liters. And then we want to divide that by-- and you want to make sure that your universal gas constant has the right units, I just looked mine up on Wikipedia-- 0.08-- see everything has three significant digits. So let me just allow that more significant digits and we'll just round at the end. 0.082057, and the units here are liters atmospheres per" + }, + { + "Q": "10:50 \"You could treat a Covalent Bond like an Ionic Bond\", this confuses me. Does an electron actually move into the other atoms shell in an Ionic Bond ? or is the Ionic Bond just a Covelant Bond, where the probability of finding the shared electron is much higher around the more Electroniegative atom ?", + "A": "It s somewhere in between. In a completely ionic bond, the electron would move into the other atom s shell and not be shared at all. In most, if not all, actual ionic bonds, the probability of not finding the shared electron in an orbital around the more electronegative atom is very low, so we just say that the electron moved to the more electronegative atom.", + "video_name": "126N4hox9YA", + "timestamps": [ + 650 + ], + "3min_transcript": "in red so strongly that it completely steals them. So those two electrons in red are going to be stolen by the chlorine, like that. And so the sodium is left over here. And so chlorine has an extra electron, which gives it a negative 1 formal charge. So we're no longer talking about partial charges here. Chlorine gets a full negative 1 formal charge. Sodium lost an electron, so it ends up with a positive formal charge, like that. And so we know this is an ionic bond between these two ions. So this represents an ionic bond. So the difference in electronegativity is somewhere between 1.5 and 2.1, between a polar covalent bond and an ionic bond. So most textbooks we'll see approximately somewhere around 1.7. So if you're higher than 1.7, it's generally considered to be mostly an ionic bond. Lower than 1.7, in the polar covalent range. So we'll come back now to the example between carbon and lithium. So if we go back up here to carbon and lithium, here we treat it like a polar covalent bond. But sometimes you might want to treat the bond in red as being an ionic bond. So let's go ahead and draw a picture of carbon and lithium where we're treating it as an ionic bond. So if carbon is more electronegative than lithium, carbon's going to steal the two electrons in red. So I'll go ahead and show the electrons in red have now moved on to the carbon atom. So it's no longer sharing it with the lithium. Carbon has stolen those electrons. And lithium is over here. So lithium lost one of its electrons, giving it a plus 1 formal charge. Carbon gained an electron, giving it a negative 1 formal charge. And so here, we're treating it like an ionic bond. Full formal charges here. And this is useful for some organic chemistry reactions. And so what I'm trying to point out here is these divisions, 1.7, it's not absolute. You could draw the dot structure above, and this would be considered be correct. You could draw it like this. Or you could treat it like an ionic bond down here. This is relatively close to the cutoff. So this is an overview of electronegativity. And even though we've been dealing with numbers in this video, in future videos, we don't care so much about the numbers. We care about the relative differences in electronegativity. So it's important to understand something as simple as oxygen is more electronegative than carbon. And that's going to help you when you're doing organic chemistry mechanisms." + }, + { + "Q": "At 3:58 he says that carbon and hydrogen have an electromagnetism difference of 0.4 and that still is considered a non polar covalent bond. At what difference would it be considered a polar covalent bond? 0.5, 0.8, 1?", + "A": "You must have meant electronegativity difference The auto-correct changes that to electromagnetism :) generally 0.5 works. Some textbooks say other values, so I don t believe there is a standard value.", + "video_name": "126N4hox9YA", + "timestamps": [ + 238 + ], + "3min_transcript": "Carbon is losing a little bit of negative charge. So carbon used to be neutral, but since it's losing a little bit of negative charge, this carbon will end up being partially positive, like that. So the carbon is partially positive. And the oxygen is partially negative. That's a polarized situation. You have a little bit of negative charge on one side, a little bit of positive charge on the other side. So let's say it's still a covalent bond, but it's a polarized covalent bond due to the differences in electronegativities between those two atoms. Let's do a few more examples here where we show the differences in electronegativity. So if I were thinking about a molecule that has two carbons in it, and I'm thinking about what happens to the electrons in red. Well, for this example, each carbon has the same value for electronegativity. So the carbon on the left has a value of 2.5. The carbon on the right has a value of 2.5. That's a difference in electronegativity of zero. aren't going to move towards one carbon or towards the other carbon. They're going to stay in the middle. They're going to be shared between those two atoms. So this is a covalent bond, and there's no polarity situation created here since there's no difference in electronegativity. So we call this a non-polar covalent bond. This is a non-polar covalent bond, like that. Let's do another example. Let's compare carbon to hydrogen. So if I had a molecule and I have a bond between carbon and hydrogen, and I want to know what happens to the electrons in red between the carbon and hydrogen. We've seen that. Carbon has an electronegativity value of 2.5. And we go up here to hydrogen, which has a value of 2.1. So that's a difference of 0.4. So there is the difference in electronegativity between those two atoms, but it's a very small difference. And so most textbooks would consider the bond between carbon and hydrogen Let's go ahead and put in the example we did above, where we compared the electronegativities of carbon and oxygen, like that. When we looked up the values, we saw that carbon had an electronegativity value of 2.5 and oxygen had a value of 3.5, for difference of 1. And that's enough to have a polar covalent bond. Right? This is a polar covalent bond between the carbon and the oxygen. So when we think about the electrons in red, electrons in red are pulled closer to the oxygen, giving the oxygen a partial negative charge. And since electron density is moving away from the carbon, the carbon gets a partial positive charge. And so we can see that if your difference in electronegativity is 1, it's considered to be a polar covalent bond. And if your difference in electronegativity is 0.4, that's considered to be a non-polar covalent bond. So somewhere in between there must" + }, + { + "Q": "i dont understand 1.7 means what at 9:53", + "A": "If the difference in electronegativities of the two atoms is greater than 1.7, the bond between them is considered to be ionic.", + "video_name": "126N4hox9YA", + "timestamps": [ + 593 + ], + "3min_transcript": "And we'll put in our electrons. And we know that this bond consists of two electrons, like that. Let's look at the differences in electronegativity between sodium and chlorine. All right. So I'm going to go back up here. I'm going to find sodium, which has a value of 0.9, and chlorine which has a value of 3. So 0.9 for sodium and 3 for chlorine. So sodium's value is 0.9. Chlorine's is 3. That's a large difference in electronegativity. That's a difference of 2.1. And so chlorine is much more electronegative than sodium. And it turns out, it's so much more electronegative that it's no longer going to share electrons with sodium. It's going to steal those electrons. So when I redraw it here, I'm going to show chlorine being surrounded by eight electrons. So these two electrons in red-- let me go ahead and show them-- these two electrons in red here between the sodium and the chlorine, since chlorine is so much more in red so strongly that it completely steals them. So those two electrons in red are going to be stolen by the chlorine, like that. And so the sodium is left over here. And so chlorine has an extra electron, which gives it a negative 1 formal charge. So we're no longer talking about partial charges here. Chlorine gets a full negative 1 formal charge. Sodium lost an electron, so it ends up with a positive formal charge, like that. And so we know this is an ionic bond between these two ions. So this represents an ionic bond. So the difference in electronegativity is somewhere between 1.5 and 2.1, between a polar covalent bond and an ionic bond. So most textbooks we'll see approximately somewhere around 1.7. So if you're higher than 1.7, it's generally considered to be mostly an ionic bond. Lower than 1.7, in the polar covalent range. So we'll come back now to the example between carbon and lithium. So if we go back up here to carbon and lithium, here we treat it like a polar covalent bond. But sometimes you might want to treat the bond in red as being an ionic bond. So let's go ahead and draw a picture of carbon and lithium where we're treating it as an ionic bond. So if carbon is more electronegative than lithium, carbon's going to steal the two electrons in red. So I'll go ahead and show the electrons in red have now moved on to the carbon atom. So it's no longer sharing it with the lithium. Carbon has stolen those electrons. And lithium is over here. So lithium lost one of its electrons, giving it a plus 1 formal charge. Carbon gained an electron, giving it a negative 1 formal charge. And so here, we're treating it like an ionic bond. Full formal charges here. And this is useful for some organic chemistry reactions. And so what I'm trying to point out here is these divisions, 1.7, it's not absolute." + }, + { + "Q": "at 10:44, carbon forms an ionic bond with lithium. But as we have seen earlier carbon ALWAYS FORMS COVALENT BONDS. So how come it is forming an ionic bond with lithium?", + "A": "Bonds vary all the way from 100 % ionic to 100 % covalent. The C-Li bond is about 43 % ionic and 57 % covalent. The bond is highly polar covalent. It behaves in many reactions as if it were ionic.", + "video_name": "126N4hox9YA", + "timestamps": [ + 644 + ], + "3min_transcript": "in red so strongly that it completely steals them. So those two electrons in red are going to be stolen by the chlorine, like that. And so the sodium is left over here. And so chlorine has an extra electron, which gives it a negative 1 formal charge. So we're no longer talking about partial charges here. Chlorine gets a full negative 1 formal charge. Sodium lost an electron, so it ends up with a positive formal charge, like that. And so we know this is an ionic bond between these two ions. So this represents an ionic bond. So the difference in electronegativity is somewhere between 1.5 and 2.1, between a polar covalent bond and an ionic bond. So most textbooks we'll see approximately somewhere around 1.7. So if you're higher than 1.7, it's generally considered to be mostly an ionic bond. Lower than 1.7, in the polar covalent range. So we'll come back now to the example between carbon and lithium. So if we go back up here to carbon and lithium, here we treat it like a polar covalent bond. But sometimes you might want to treat the bond in red as being an ionic bond. So let's go ahead and draw a picture of carbon and lithium where we're treating it as an ionic bond. So if carbon is more electronegative than lithium, carbon's going to steal the two electrons in red. So I'll go ahead and show the electrons in red have now moved on to the carbon atom. So it's no longer sharing it with the lithium. Carbon has stolen those electrons. And lithium is over here. So lithium lost one of its electrons, giving it a plus 1 formal charge. Carbon gained an electron, giving it a negative 1 formal charge. And so here, we're treating it like an ionic bond. Full formal charges here. And this is useful for some organic chemistry reactions. And so what I'm trying to point out here is these divisions, 1.7, it's not absolute. You could draw the dot structure above, and this would be considered be correct. You could draw it like this. Or you could treat it like an ionic bond down here. This is relatively close to the cutoff. So this is an overview of electronegativity. And even though we've been dealing with numbers in this video, in future videos, we don't care so much about the numbers. We care about the relative differences in electronegativity. So it's important to understand something as simple as oxygen is more electronegative than carbon. And that's going to help you when you're doing organic chemistry mechanisms." + }, + { + "Q": "At 4:19, Sal started listing momentums and positions for the atoms. But doesn't that violate the Heisenberg Uncertainty Principle?", + "A": "Yes, it does. He can t be certain of both at the same time, however, his point was not to show the momentums and positions but rather to demonstrate the difference between micro and macrostates.", + "video_name": "5EU-y1VF7g4", + "timestamps": [ + 259 + ], + "3min_transcript": "Now we know that that pressure is due to things like, you have a bunch of atoms bumping around. And let's say that this is a gas-- it's a balloon- it's going to be a gas. And we know that the pressure is actually caused-- and I've done several, I think I did the same video in both the chemistry and the physics playlist. I did them a year apart, so you can see if my thinking has evolved at all. But we know that the pressure's really due by the bumps of these particles as they bump into the walls and the side of the balloon. And we have so many particles at any given point of time, some of them are bumping into the wall the balloon, and that's what's essentially keeping the balloon pushed outward, giving it its pressure and its volume. We've talked about temperature, as essentially the average kinetic energy of these-- which is a function of these particles, which could be either the molecules of gas, or if it's an ideal gas, it could be just the atoms of the gas. Maybe it's atoms of helium or neon, or something like that. So for example, I could describe what's going on with I could say, hey, you know, there are-- I could just make up some numbers. The pressure is five newtons per meters squared, or some The units aren't what's important. In this video I really just want to make the differentiation between these two ways of describing what's going on. I could say the temperature is 300 kelvin. I could say that the volume is, I don't know, maybe it's one liter. And I've described a system, but I've described in on a macro level. Now I could get a lot more precise, especially now that we know that things like atoms and molecules exist. What I could do, is I could essentially label every one of contained in the balloon. And I could say, at exactly this moment in time, I could say at time equals 0, atom 1 has-- its momentum is equal to x, and its position, in three-dimensional coordinates, is x, y, and z. And then I could say, atom number 2-- its momentum-- I'm just using rho for momentum-- it's equal to y. And its position is a, b, c. And I could list every atom in this molecule. Obviously we're dealing with a huge number of atoms, on the order of 10 to the 20 something. So it's a massive list I would have to give you, but I could literally give you the state of every atom in this balloon." + }, + { + "Q": "At 9:03 Sal says that it all takes place in space . So if it takes place in space how would the rock move downwards . I mean there is no gravitational force there acting vertically downwards .", + "A": "By space, he meant vacuum. He didn t mean the space where there is no gravity", + "video_name": "5EU-y1VF7g4", + "timestamps": [ + 543 + ], + "3min_transcript": "macro level. That's often a very important thing to think about. And we'll go into concepts like entropy and internal energy, and things like that. And you can rack your brain, how does it relate to atoms? And we will relate them to atoms and molecules. But it's useful to think that the people who first came up with these concepts came up with them not really being sure of what was going on at the micro level. They were just measuring everything at the macro level. Now I want to go back to this idea here, of equilibrium. Because in order for these macrostates to be defined, the system has to be in equilibrium. And let me explain what that means. If I were to take a cylinder. And we will be using this cylinder a lot, so it's good to get used to this cylinder. And it's got a piston in it. move up and down. This is the roof of the cylinder. The cylinder's bigger, but let's say this is a, kind of a roof of the cylinder. And we can move this up and down. And essentially we'll just be changing the volume of the cylinder, right? I could have drawn it this way. I could have drawn it like a cylinder. I could have drawn it like this, and then I could have drawn the piston like this. So there's some depth here that I'm not showing. We're just looking at the cylinder front on. And so, at any point in time, let's say the gas is between the cylinder and the floor of our container. You know, we have a bunch of molecules of gas here, a huge number of molecules. And let's say that we have a rock on the cylinder. We're doing this in space so everything above the piston is a vacuum. Actually just let me erase everything above. Let me just erase this stuff, just so you see. Just let me write that down. So all of this stuff up here is a vacuum, which essentially says there's nothing there. There's no pressure from here, there's no particles here, just empty space. And in order to keep this-- we know already, we've studied it multiple times, that this gas is generating, you know things are bumping into the wall, the floor of this piston all the time. They're bumping into everything, right? We know that's continuously happening. So we would apply some pressure to offset the pressure being generated by the gas. Otherwise the piston would just expand. It would just move up and the whole gas would expand. So let's just say we stick a big rock or a big weight on top of-- let me do it in a different color-- We put a big weight on top of this piston, where the force-- completely offsets the force being applied by the gas." + }, + { + "Q": "At 10:01, if you're doing this in space, the rock on top would not be able to offset the pressure inside of the cylinder, right? Since there's no gravity, it would render the rock weightless. I guess I'm being nit-picky, haha.", + "A": "That bothered me a little too. I believe it s in space so that there s no extra air pressure acting on the system from outside, or anything like that, but he may have forgotten about that when he talked about the rock. In the right setup, you could still experience enough gravity in space to make this work. I settled for, It s on the moon, as my explanation. EDIT: And it seems other possibilities are covered in other people s questions and answers!", + "video_name": "5EU-y1VF7g4", + "timestamps": [ + 601 + ], + "3min_transcript": "move up and down. This is the roof of the cylinder. The cylinder's bigger, but let's say this is a, kind of a roof of the cylinder. And we can move this up and down. And essentially we'll just be changing the volume of the cylinder, right? I could have drawn it this way. I could have drawn it like a cylinder. I could have drawn it like this, and then I could have drawn the piston like this. So there's some depth here that I'm not showing. We're just looking at the cylinder front on. And so, at any point in time, let's say the gas is between the cylinder and the floor of our container. You know, we have a bunch of molecules of gas here, a huge number of molecules. And let's say that we have a rock on the cylinder. We're doing this in space so everything above the piston is a vacuum. Actually just let me erase everything above. Let me just erase this stuff, just so you see. Just let me write that down. So all of this stuff up here is a vacuum, which essentially says there's nothing there. There's no pressure from here, there's no particles here, just empty space. And in order to keep this-- we know already, we've studied it multiple times, that this gas is generating, you know things are bumping into the wall, the floor of this piston all the time. They're bumping into everything, right? We know that's continuously happening. So we would apply some pressure to offset the pressure being generated by the gas. Otherwise the piston would just expand. It would just move up and the whole gas would expand. So let's just say we stick a big rock or a big weight on top of-- let me do it in a different color-- We put a big weight on top of this piston, where the force-- completely offsets the force being applied by the gas. the area of the piston-- over some areas so that we could figure out its pressure. And that pressure will completely offset the pressure But the pressure of the gas, just as a reminder, is going in every direction. The pressure on this plate is the same as the pressure on that side, or on that side, or on the bottom of the container that we're dealing with. Now let's say that we were to just evaporate this-- well let's not say that we evaporate the rock. Let's say that we just evaporate half of the rock immediately. So all of a sudden our weight that's being pushed down, or the force that's being pushed down just goes to half immediately. Let me draw that. So I have-- maybe I would be better off just cut and pasting this right here. So if I copy and paste it. So now I'm going to evaporate half of that rock magically. So let me take my eraser tool. And I just evaporate half of it." + }, + { + "Q": "As at 1:38 minutes it is said that this emission spectrum is unique to hydrogen atom , which means we have different emission spectrum's for different atoms , so does that in turn mean that we have different energies for same energy levels in different atoms ?", + "A": "Yes. Some even swap places.", + "video_name": "Kv-hRvEOjuA", + "timestamps": [ + 98 + ], + "3min_transcript": "- [Voiceover] I'm sure that most of you know the famous story of Isaac Newton where he took a narrow beam of light and he put that narrow beam of light through a prism and the prism separated the white light into all the different colors of the rainbow. And so if you did this experiment, you might see something like this rectangle up here so all of these different colors of the rainbow and I'm gonna call this a continuous spectrum. It's continuous because you see all these colors right next to each other. So they kind of blend together. So that's a continuous spectrum If you did this similar thing with hydrogen, you don't see a continuous spectrum. So, if you passed a current through a tube containing hydrogen gas, the electrons in the hydrogen atoms are going to absorb energy and jump up to a higher energy level. When those electrons fall down to a lower energy level they emit light and so we talked about this in the last video. This is the concept of emission. If you use something like a prism or a defraction grading to separate out the light, for hydrogen, you don't get a continuous spectrum. So, since you see lines, we call this a line spectrum. So this is the line spectrum for hydrogen. So you see one red line and it turns out that that red line has a wave length. That red light has a wave length of 656 nanometers. You'll also see a blue green line and so this has a wave length of 486 nanometers. A blue line, 434 nanometers, and a violet line at 410 nanometers. And so this emission spectrum is unique to hydrogen and so this is one way to identify elements. And so this is a pretty important thing. And since line spectrum are unique, this is pretty important to explain where those wavelengths come from. And we can do that by using the equation we derived in the previous video. So I call this equation the Balmer Rydberg equation. And you can see that one over lamda, is equal to R, which is the Rydberg constant, times one over I squared, where I is talking about the lower energy level, minus one over J squared, where J is referring to the higher energy level. For example, let's say we were considering an excited electron that's falling from a higher energy level n is equal to three. So let me write this here. So we have an electron that's falling from n is equal to three down to a lower energy level, n is equal to two. All right, so it's going to emit light when it undergoes that transition. So let's look at a visual representation of this. Now let's see if we can calculate the wavelength of light that's emitted. All right, so if an electron is falling from n is equal to three to n is equal to two, I'm gonna go ahead and draw an electron here. So an electron is falling from n is equal to three energy level down to n is equal to two, and the difference in those two energy levels" + }, + { + "Q": "He explains the process of the oxidation of water (at about 15:26) but where does the energy for this oxidation come from?", + "A": "From the sun, every process in photosynthesis is fueled indirectly through light. During the light reaction ATP (energy) and NADPH2 (reduction power) is produced which can be later used for reducing CO2 to sugar. The NADPH2 contains the electrons which were taken from the water oxidation.", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 926 + ], + "3min_transcript": "the stroma to the lumen. Then the hydrogen protons want to go back. They want to-- I guess you could call it-- chemiosmosis. They want to go back into the stroma and then that drives ATP synthase. Right here, this is ATP synthase. ATP synthase to essentially jam together ADPs and phosphate groups to produce ATP. Now, when I originally talked about the light reactions and dark reactions I said, well the light reactions have two byproducts. It has ATP and it also has-- actually it has three. It has ATP, and it also has NADPH. NADP is reduced. It gains these electrons and these hydrogens. So where does that show up? Well, if we're talking about non-cyclic oxidative photophosphorylation, or non-cyclic light reactions, the final electron acceptor. energy states, the final electron acceptor is NADP plus. So once it accepts the electrons and a hydrogen proton with it, it becomes NADPH. Now, I also said that part of this process, water-- and this is actually a very interesting thing-- water gets oxidized to molecular oxygen. So where does that happen? So when I said, up here in photosystem I, that we have a chlorophyll molecule that has an electron excited, and it goes into a higher energy state. And then that electron essentially gets passed from one guy to the next, that begs the question, what can we use to replace that electron? And it turns out that we use, we literally use, the electrons in water. So over here you literally have H2O. So you can kind of imagine it donates two hydrogen protons and two electrons to replace the electron that got excited by the photons. Because that electron got passed all the way over to photosystem I and eventually ends up in NADPH. So, you're literally stripping electrons off of water. And when you strip off the electrons and the hydrogens, you're just left with molecular oxygen. Now, the reason why I want to really focus on this is that there's something profound happening here. Or at least on a chemistry level, something profound is happening. You're oxidizing water. And in the entire biological kingdom, the only place where we know something that is strong enough of an oxidizing agent to oxidize water, to literally take away electrons from water. Which means you're really taking electrons away from oxygen. So you're oxidizing oxygen. The only place that we know that an oxidation agent is" + }, + { + "Q": "at 10:04 sal says 'hydrogen protons,what is the difference between an hydrogen proton and a simple proton?", + "A": "protons are protons. He s just explaining how they came to be there.", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 604 + ], + "3min_transcript": "Photosystem II. You have maybe another complex. And these are hugely complicated. I'll do a sneak peek of what photosystem II actually looks like. This is actually what photosystem II looks like. So, as you can see, it truly is a complex. These cylindrical things, these are proteins. These green things are chlorophyll molecules. I mean, there's all sorts of things going here. And they're all jumbled together. I think a complex probably is the best word. It's a bunch of proteins, a bunch of molecules just jumbled together to perform a very particular function. We're going to describe that in a few seconds. So that's what photosystem II looks like. Then you also have photosystem I. And then you have other molecules, other complexes. You have the cytochrome B6F complex and I'll draw this in a different color right here. I don't want to get too much into the weeds. Because the most important thing is just to understand. So you have other protein complexes, protein molecular But the general idea-- I'll tell you the general idea and then we'll go into the specifics-- of what happens during the light reaction, or the light dependent reaction, is you have some photons. Photons from the sun. They've traveled 93 million miles. so you have some photons that go here and they excite electrons in a chlorophyll molecule, in a chlorophyll A molecule. And actually in photosystem II-- well, I won't go into the details just yet-- but they excite a chlorophyll molecule so those electrons enter into a high energy state. Maybe I shouldn't draw it like that. They enter into a high energy state. And then as they go from molecule to molecule they keep going down in energy state. But as they go down in energy state, you have hydrogen the electrons. So you have all of these hydrogen protons. Hydrogen protons get pumped into the lumen. They get pumped into the lumen and so you might remember this from the electron transport chain. In the electron transport chain, as electrons went from a high potential, a high energy state, to a low energy state, that energy was used to pump hydrogens through a membrane. And in that case it was in the mitochondria, here the membrane is the thylakoid membrane. But either case, you're creating this gradient where-- because of the energy from, essentially the photons-- the electrons enter a high energy state, they keep going into a lower energy state. And then they actually go to photosystem I and they get hit by another photon. Well, that's a simplification, but that's how you can think of it. Enter another high energy state, then they go to a lower, lower and lower energy state. But the whole time, that energy from the electrons going from a high energy state to a low energy state is used" + }, + { + "Q": "At 15:30, Sal explains that water is hydrolyzed to resupply photosystem II with electrons it transferred to photosystem I. I always learned this reaction was carried out by a photoactive enzyme on the phospholipid bilayer but Sal seems to indicate its actually carried out by the photosystem. Any ideas?", + "A": "The reaction is carried out by a poorly understood Oxygen Evolving Complex (OEC) which is very integrally attached to the photosystem II and its working is strongly coupled with the working of PSII and therefore it is actually considered to be single complex.", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 930 + ], + "3min_transcript": "the stroma to the lumen. Then the hydrogen protons want to go back. They want to-- I guess you could call it-- chemiosmosis. They want to go back into the stroma and then that drives ATP synthase. Right here, this is ATP synthase. ATP synthase to essentially jam together ADPs and phosphate groups to produce ATP. Now, when I originally talked about the light reactions and dark reactions I said, well the light reactions have two byproducts. It has ATP and it also has-- actually it has three. It has ATP, and it also has NADPH. NADP is reduced. It gains these electrons and these hydrogens. So where does that show up? Well, if we're talking about non-cyclic oxidative photophosphorylation, or non-cyclic light reactions, the final electron acceptor. energy states, the final electron acceptor is NADP plus. So once it accepts the electrons and a hydrogen proton with it, it becomes NADPH. Now, I also said that part of this process, water-- and this is actually a very interesting thing-- water gets oxidized to molecular oxygen. So where does that happen? So when I said, up here in photosystem I, that we have a chlorophyll molecule that has an electron excited, and it goes into a higher energy state. And then that electron essentially gets passed from one guy to the next, that begs the question, what can we use to replace that electron? And it turns out that we use, we literally use, the electrons in water. So over here you literally have H2O. So you can kind of imagine it donates two hydrogen protons and two electrons to replace the electron that got excited by the photons. Because that electron got passed all the way over to photosystem I and eventually ends up in NADPH. So, you're literally stripping electrons off of water. And when you strip off the electrons and the hydrogens, you're just left with molecular oxygen. Now, the reason why I want to really focus on this is that there's something profound happening here. Or at least on a chemistry level, something profound is happening. You're oxidizing water. And in the entire biological kingdom, the only place where we know something that is strong enough of an oxidizing agent to oxidize water, to literally take away electrons from water. Which means you're really taking electrons away from oxygen. So you're oxidizing oxygen. The only place that we know that an oxidation agent is" + }, + { + "Q": "Why is photosystem II before photosystem I? 9:10", + "A": "Photosystem I was discovered before photosystem II, even though photosystem I is the one that comes first. So they re named a bit counterintuitively, but that s only because of the order they were discovered in.", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 550 + ], + "3min_transcript": "The important thing from the photosynthesis point of view is that it's this membrane. And on the outside of the membrane, right here on the outside, you have the fluid that fills up the entire chloroplast. So here you have the stroma. And then this space right here, this is the inside of your thylakoid. So this is the lumen. So if I were to color it pink, right there. This is your lumen. Your thylakoid space. And in this membrane, and this might look a little bit familiar if you think about mitochondria and the electron transport chain. What I'm going to describe in this video actually is an electron transport chain. Many people might not consider it the electron transport chain, but it's the same idea. Same general idea. So on this membrane you have these proteins and these complexes of proteins and molecules that span this membrane. So let me draw a couple of them. So maybe I'll call this one, photosystem II. Photosystem II. You have maybe another complex. And these are hugely complicated. I'll do a sneak peek of what photosystem II actually looks like. This is actually what photosystem II looks like. So, as you can see, it truly is a complex. These cylindrical things, these are proteins. These green things are chlorophyll molecules. I mean, there's all sorts of things going here. And they're all jumbled together. I think a complex probably is the best word. It's a bunch of proteins, a bunch of molecules just jumbled together to perform a very particular function. We're going to describe that in a few seconds. So that's what photosystem II looks like. Then you also have photosystem I. And then you have other molecules, other complexes. You have the cytochrome B6F complex and I'll draw this in a different color right here. I don't want to get too much into the weeds. Because the most important thing is just to understand. So you have other protein complexes, protein molecular But the general idea-- I'll tell you the general idea and then we'll go into the specifics-- of what happens during the light reaction, or the light dependent reaction, is you have some photons. Photons from the sun. They've traveled 93 million miles. so you have some photons that go here and they excite electrons in a chlorophyll molecule, in a chlorophyll A molecule. And actually in photosystem II-- well, I won't go into the details just yet-- but they excite a chlorophyll molecule so those electrons enter into a high energy state. Maybe I shouldn't draw it like that. They enter into a high energy state. And then as they go from molecule to molecule they keep going down in energy state. But as they go down in energy state, you have hydrogen" + }, + { + "Q": "At 12:00, is the ATP synthase drawn backwards? Shouldn't the rotor be on bottom if it is pumping the electrons from the lumen to the stroma and producing ATP?", + "A": "Sal drew the ATP synthase the correct way round - upside down relative to in a mitochondrion. In the thylakoid, protons go from inside (the lumen) to outside (the stroma), and ATP is generated in the stroma. In a mitochondrion, protons go from outside to the inside, and generate ATP within the mitochondrion. In some bacteria, the ATP synthase is used in reverse, so ATP is hydrolysed to pump protons out of the cell, but this is not what s happening here.", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 720 + ], + "3min_transcript": "the electrons. So you have all of these hydrogen protons. Hydrogen protons get pumped into the lumen. They get pumped into the lumen and so you might remember this from the electron transport chain. In the electron transport chain, as electrons went from a high potential, a high energy state, to a low energy state, that energy was used to pump hydrogens through a membrane. And in that case it was in the mitochondria, here the membrane is the thylakoid membrane. But either case, you're creating this gradient where-- because of the energy from, essentially the photons-- the electrons enter a high energy state, they keep going into a lower energy state. And then they actually go to photosystem I and they get hit by another photon. Well, that's a simplification, but that's how you can think of it. Enter another high energy state, then they go to a lower, lower and lower energy state. But the whole time, that energy from the electrons going from a high energy state to a low energy state is used So you have this huge concentration of hydrogen protons. And just like what we saw in the electron transport chain, that concentration is then-- of hydrogen protons-- is then used to drive ATP synthase. So the exact same-- let me see if I can draw that ATP synthase here. You might remember ATP synthase looks something like this. Where literally, so here you have a huge concentration of hydrogen protons. So they'll want to go back into the stroma from the lumen. And they go through the ATP synthase. Let me do it in a new color. So these hydrogen protons are going to make their way back. Go back down the gradient. And as they go down the gradient, they literally-- it's like an engine. And I go into detail on this when I talk about respiration. And that turns, literally mechanically turns, this top And it puts ADP and phosphate groups together. It puts ADP plus phosphate groups together to produce ATP. So that's the general, very high overview. And I'm going to go into more detail in a second. But this process that I just described is called photophosphorylation. Let me do it in a nice color. Why is it called that? Well, because we're using photons. That's the photo part. We're using light. We're using photons to excite electrons in chlorophyll. As those electrons get passed from one molecule, from one electron acceptor to another, they enter into lower and lower energy states. As they go into lower energy states, that's used to drive," + }, + { + "Q": "At 1:26 it says that the red light reflected of the rose enters our eye and falls on one of the red colored cones, we can see that the rose is red. What would happen if this ray of light falls on a blue or green cone, is this situation even possible?", + "A": "Different receptors are built to respond to different stimuli. The other cones would not respond to the light of a different frequency. Think about sound, people lose their high range because these specific receptors are damaged and die. Think about the skin, some receptors respond to touch, different ones respond to heat. It is amazing.", + "video_name": "0ugcw7wOZBg", + "timestamps": [ + 86 + ], + "3min_transcript": "When you're looking at an object, it's necessary to break it down into its component features in order to make sense of what you're looking at. This is known as feature detection. There are many components that make up feature detection. So let's go into these. When you're looking at a rose, not only do you have to look and decide, OK, what color am I looking at, you also have to figure out, OK, what shape is the rose. So you have to get a little bit of information about the form of the rose. You also need to get information about motion. So is the rose moving? Am I throwing it across a room? What's going on? So whenever we look at any object, we need to get information about color, we need to get information about the form of the object, and information about motion. So let's go into each one of these different features. So our ability to sense color actually arises from the presence of cones within our retina. Cones are extremely important, because they're So we have three major types of cones. There are red comes, which make up 60% of the cones in your eye. There are green cones, which make up 30% of the cones in your eye. And there are blue cones, which make up about 10% of the cones in your eye. Now, we divide them into red, green, and blue cones because the red cones are extremely sensitive to red light. So if we're looking at this rose, the petals are actually reflecting red light. And so this is red light enters your eye. And if it happens to hit a red cone, then the cone will activate, and it will fire an action potential. And this action potential will reach your brain, and your brain will recognize that you're looking at something that is red. So this basically happens regardless of what we're looking at. And this has come to be known as the trichromatic theory of color vision. So what other features do we need to take into consideration when we're looking at this rose? we also need to figure out, OK, what are the boundaries of the rose, so the boundaries of the stem, the boundaries of the leaf, the boundaries of the petals, from the background. And this is also really important, because not only do we need to distinguish the boundaries, but we also need to figure out, OK, what shape are the leaves, what shape are the petals. And these are all very important things that your brain ultimately is able to break down. So in order for us to figure out what the form of an object is, we use a very specialized pathway that exists in our brain, which is known as the parvo pathway. So the parvo pathway is responsible for figuring out what the shape of an object is. So another way to say this is that the parvo pathway is really good at spatial resolution. Let me write that down. So spatial resolution." + }, + { + "Q": "At 9:28 why exactly does he write 3d6 when he should be writing 4d6?", + "A": "He writes 3d\u00e2\u0081\u00b6 because the 3d subshell id filled after the 4s subshell.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 568 + ], + "3min_transcript": "This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy." + }, + { + "Q": "At 5:34, why does carbon have 4 valence electrons instead of 2 when the 2s2 shell is filled already?", + "A": "To expand on Just Keith s answer and clarify a bit more, only the level 1 shell is filled in carbon. The 2s sublevel is filled, but the electrons in it are still much closer in energy to the 2p sublevel than they are to those in the filled level 1.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 334 + ], + "3min_transcript": "It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange. Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair" + }, + { + "Q": "At 5:00, why am I not supposed to write the electron confu-thingy as Be 2p2?\nIts confusing.", + "A": "Be is not a noble gas, the square bracket notation is only used with noble gasses", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 300 + ], + "3min_transcript": "but the important thing to realize and actually for the example of hydrogen sodium is that all of these group one elements are going to have one Valence electron. They're going to have one electron that they tend to use when they are either getting lost to form an ion or that they might be able to use to form a covalent bond. Now let's think about helium and helium's an interesting character because all of the rest of the noble gases have eight Valence electrons which makes them very stable but helium only has two Valence electrons. The reason why it's included here is because helium is also very stable because for that first shell, you only need two electrons to fill full, to fill stable. Helium has two Valence electrons, its electron configuration is one S two. Once again the reason why it's out here with the noble gases is because it's very stable and very inert like the noble gases that's why we now use those helium for balloons It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange." + }, + { + "Q": "At 9:38 when Sal says, \"the 4th energy subshell,\" is subshell interchangeable with energy level? I thought the subshell related to the shape of the orbital.", + "A": "He should have said the 4s subshell. Energy levels can split into subshells which are composed of orbitals that correspond with the type of subshell. The way we name these are a bit confusing, because there is a difference between say the 2p subshell and a 2p orbital. It is very easy to use the wrong word when describing this stuff.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 578 + ], + "3min_transcript": "could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy. figuring out the Valence electrons, the electrons that are most slightly right becomes a little bit hard to predict. Some people, some even textbooks will say, \"Oh, all the transition metals\" \"have two Valence electrons\" \"because they all get the four S two\" \"and then they're back filling.\u201d They would say, \"Okay,\" \"these are the two Valence electrons\" \"for all of these transition metals.\" Well that doesn't hold up for all of them because you even have special cases like copper and chromium that only go four S one and then start filling three D depending on the circumstances. Sometimes it does it otherwise but even for the other transition elements like say iron is not necessarily the case so these are the one, the only two electrons that are going to react. You might have some of your D electrons, your three D electrons which are high energy might also be involved in reaction. Might be taken away or might form a bond somehow." + }, + { + "Q": "9:28 Why is it 3d^2 and not 4d^2? Why is it written [Ne] when he's talking about iron (Fe)?", + "A": "Because it is? The 4s and 3d orbitals are similar in energy so they are filled around the same time, but the 4d orbitals are quite a bit higher than those two. Why is what written [Ne]? That element in square brackets notation represents the electron configuration of the noble gas from the row above, it saves time when you get to much heavier elements by removing redundant information. Argon is the noble gas in the row above iron so you use [Ar] to represent the following: 1s^2 2s^2 2p^6 3s^2 3p^6", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 568 + ], + "3min_transcript": "This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy." + }, + { + "Q": "At 5:20, is the esophagus a part of the lower or upper respiratory tract, or is it not included because it is meant for food and water?", + "A": "It is not included because it is meant for food and water, it is part of the digestive tract.", + "video_name": "Z-yv3Yq4Aw4", + "timestamps": [ + 320 + ], + "3min_transcript": "is kind of the more medical word, I guess. And sitting over the larynx is the epiglottis. And the epiglottis is basically like a lid kind of protecting the larynx from making sure that food and water don't go into it. Now, there's another tube I just alluded to, and it's sitting right here, and this purple tube, is our esophagus So the esophagus is basically, it's fantastic for things like food and water. You want food and water to go down the esophagus because it's going to lead to the stomach. So you want food and water to go that way, but you don't want food and water to go into the larynx. And so you want to make sure that the epiglottis, that lid, is working really well. And if you're swallowing food and water, this epiglottis will literally just kind of close up and protect your larynx. But in this case, that's not happening. We're not actually food and water, we're a little molecule of oxygen, Let's see what happens to it. I'm going to drag up the canvas a little bit. Let's make a little bit of space, and I want to just stop it right there because I want to show you that the air molecule, the oxygen molecule has already kind of made an interesting crossroads. It's actually kind of broken an important boundary, and that's this boundary right here. And on the top of this boundary, I've included the larynx and of course, all the other stuff we just talked about-- the mouth and the nose-- and this is considered our upper respiratory tract. So anything above this dashed line is our upper respiratory tract, and then, of course, you can then guess that anything below the line must then be our lower respiratory tract. So this is an important boundary because people will talk about the upper and lower tract, and I want to make sure you know what is on which side. So on the top of it, is the larynx and everything Let me label that here. The trachea is right here, the wind pipe or the trachea, and everything below that, which, of course, mainly includes things like the lungs, but as we'll see a few other structures that we're going to name. So I'm going to keep moving down, but now you know that important boundary exists. So now let me just make a little bit more space you can see the entire lungs. You can see the molecule is going to go through the trachea, and actually, I have my left lung incompletely drawn. Let me just finish it off right there. So we have our right and left lung, right? These are the two lungs, and our air is going to just kind of slowly pass down-- our molecule of oxygen is going to pass down, and it's going to go either into the right lung or the left lung. Now here, I want to make sure I just take a quick pause and show you the naming structure. And the important word of the day" + }, + { + "Q": "At 11:05, Rishi said respiratory bronchials, did he mean respiratory bronchides?", + "A": "I think so since there s alveoli on bottom.", + "video_name": "Z-yv3Yq4Aw4", + "timestamps": [ + 665 + ], + "3min_transcript": "the left and right lung. And the other clue we said was the lobes, so of course, the right one has the upper lobe, the middle lobe, and the lower lobe, and the left lung only has the upper and lower. So that's an important clue. I just want to make sure we don't forget our little tricks that we've learned for telling apart the lungs. So I'm going to take a little pause there, and now, I'm going to show you in a sped up version all of the different branch points. So for example, here we have just a couple of branch points, one, two, and getting into this segmental bronchi would be the third branch point. But I'm going to speed things along and show you how many more branch points there actually are before we get to the final part of our lung where the gas exchange actually happens, so enjoy. [MUSIC - NIKOLAI RIMSKY-KORSAKOV, \"FLIGHT OF THE BUMBLEBEE\"] we start with kind of a bronchi, and we said that there is a naming structure for how to name the bronchi, but that's really just the first three branches. And then after the first three branches, all of the orange stuff, all those branches going from branch point 4 all the way down to about branch point 20, those are the conducting bronchials. So that's the name we give them. They're no longer bronchi, they're bronchials. And so if you see that word just keep that in mind, that we're a little bit further along in the lungs. And then after the conducting bronchials, and we call them the respiratory bronchials. And actually the final I should mention this the final conducting bronchial, sometimes you'll see this called the terminal bronchial. It's that kind of a bad name because terminal sounds like we're done, but actually we're just done with the conducting bronchials, and we're still kind of going into the respiratory bronchials. I guess if I'm only point to one, I should just probably make it singular. And then finally, we get into the alveolar ducts and the alveolar sac, which is kind of a few alveoli put together. And if it's plural as alveoli, then singular just talking about one little part of that sac would be alveolus. So that's where our little molecule of oxygen ends up, and this is kind of where it ends up before it's going to participate in gas exchange. Now, this entire area, going from respiratory bronchials on downwards, this is all called the respiratory zone," + }, + { + "Q": "I'm pretty sure at 2:17 the C has only 2 methyl groups on it; the third one on the right is supposed to be the bond that was connected to X, right?", + "A": "Don t they have to be hydrogens? SN2 can t occur with a tertiary carbon, or occurs so little to be negligible.", + "video_name": "X9ypryY7hrQ", + "timestamps": [ + 137 + ], + "3min_transcript": "One way to make ethers is to use the Williamson ether synthesis, which is where you start with an alcohol, and you add a strong base to deprotonate the alcohol. Once you deprotonate the alcohol, you add an alkyl halide, and primary alkyl halides work the best. We'll talk about why in a minute. And what happens is you end up putting the R prime group from your alkyl halide on to what used to be your alcohol to form your ether like that. So let's look at the mechanism for the Williamson ether synthesis, where you start with your alcohol. We know that alcohols can function as weak acids. So if you react an alcohol with a strong base, something like sodium hydride, we know that the hydride portion of the molecule is going to function as a strong base. This lone pair of electrons is going to take that proton, which is going to kick these electrons off onto the oxygen. So if we're drawing the product of that acid-based reaction, we now have an oxygen with three lone pairs of electrons around it, giving it a negative 1 formal charge. And we call that an alkoxide anion, which charged sodium ion floating around. So there's some electrostatic or ionic interaction between those opposite charges. And here's where you introduce your alkyl halide. So if we draw our alkyl halide, it would look like this. And we know that there's an electronegativity difference between our halogen and our carbon, where our halogen is going to be partially negative, and our carbon is going to be partially positive. Partially positive carbon means that that carbon wants electrons. It's going to function as an electrophile in the next step of the mechanism. And a lone pair of electrons in the oxygen is going to function as a nucleophile. So opposite charges attract. A lone pair of electrons on our nucleophile are going to attack our electrophile, our carbon. At the same time, the electrons in the bond between the carbon the halogen are going to kick off onto the halogen like that. So this is an SN2-type mechanism, because that has the decreased steric hindrance compared to other alkyl halides. So what will happen is, after nucleophilic attack, we're going to attach our oxygen to our carbon like that, and we form our ether. So if we wanted to, we could just rewrite our ether like this to show it as we added on an R prime group like that. Let's look at an example of the Williamson ether synthesis. So if I start with a molecule over here on the left, and it's kind of an interesting-looking molecule. It's called beta-naphthol. And so beta-naphthol has two rings together like this, and then there's an OH coming off of one of the rings, So that's beta-naphthol. And in the first part, we're going to add potassium hydroxide as our base." + }, + { + "Q": "At about 6:28, Sal says that he should only have 3 spots to the right of the decimal, but shouldn't he actually end up expressing his answer as 18.0149 because that still has the same amount of significant figures as the 15.999 because the 0 to the right of the decimal point before the other digits in 18.0149 should not be considered significant.", + "A": "That 0 is significant", + "video_name": "_WXndBGQnyI", + "timestamps": [ + 388 + ], + "3min_transcript": "have an atomic mass of one. If you have two times one, it's just gonna be two atomic mass units. Then 16 from the oxygen. Two plus 16, which is going to get us... Let me do this in another color, since I've been using... It's going to give us 18 atomic mass units. Now, if you wanted to be more precise or if you wanted to say, \"Well, I have this big bag \"of water, I'm not looking exactly at one water molecule. \"What is going to be, on average, \"the mass of those water molecules?\" It might be helpful to get a little it more precise. Then it is helpful, especially if you're talking about a large number of molecules and you really just wanna take the weighted average of all of those molecules. Then it makes sense to say, \"Well, let's \"use the atomic weight.\" So for hydrogen, the atomic weight was 1.0079. We call it atomic weight but it's really just the weighted average, it's not weight in kind of the physics sense of measuring a force. So, 1.0079 atomic mass units and the oxygen is is 15.999 atomic mass units and if you wanted to get a more precise number here, let's get a calculator out. I got my calculator. I'm gonna have two times 1.0079 is equal to, and then to that I'm gonna add, plus 15.999. Gets us to... Let's see, I should go no more than three decimal places to the right. Since I added this, if I don't wanna If you wanna a review of that, you should look at the video on significant figures. Since I added something with just three decimals to to the right I shouldn't have more than three decimals in my answer so, 18.01, I'll round in the thousandths place, 015. So the real one is 18.015 atomic mass units. You could actually consider this the molecular weight. Because once again, we're using atomic weights. We're using weighted average and you could think of, maybe the weighted average of water molecules would be a little bit closer to this. These two numbers are very close. So we might call this molecular weight. Either way, these ideas are very closely related. When we're thinking on the atomic scale, when people use the word weight, they're not using it like a force like you use in physics that you would measure in newtons or pounds." + }, + { + "Q": "9:13, when did the first oceans come from , how did they form", + "A": "We re not sure, but water is fairly common in the solar system. The water on earth might have come mostly from comets striking the earth.", + "video_name": "nYFuxTXDj90", + "timestamps": [ + 553 + ], + "3min_transcript": "And so we have rocks from, that are roughly 3.8 billion years So we kind of put that as the beginning of the Archean Eon. And so there's two things there. One, rocks have survived from the beginning of the Archean And also, that's roughly when we think that the first life existed. And so we're now in the Archean Eon. And you might say, oh, maybe Earth is a more pleasant place now. But it would not be. It still has no to little oxygen in the environment. If you were to go to Earth at that time, it might have looked something like this. It would have a been a reddish sky. You would have had nitrogen and methane and carbon dioxide in the atmosphere. There would have been nothing for you to breathe. There still would have been a lot of volcanic activity. This right here, these are pictures of stromatolites. And these are formed from bacteria And over time, these things get built up. But the most significant event in the Archean period, at least in my humble opinion, was what we believe started to happen about 3.5 billion years ago. And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun, to actually do photosynthesis. And the real fascinating byproduct of that, other than the fact that they can now use energy directly from the sun, is that it started to produce oxygen, so starts to produce oxygen. And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere. Because you had all of this iron that was dissolved in the oceans. And let me be clear. the next several billion years, it all occurred in the ocean. We had no ozone layer now. The land was being irradiated. The land was just a completely inhospitable environment for life. So all of this was occurring in the ocean. And so the first oxygen that actually got produced, it actually, instead of just being released into the atmosphere, it ended up bonding with the iron that was dissolved in the ocean at that time. So it actually didn't have a chance to accumulate in the atmosphere. And when we fast forward past the Archean period, we're going to see, that once a lot of that iron was oxidized and the oxygen really did start to get released in the atmosphere, it actually had-- it's funny to say-- a cataclysmic effect or a catastrophic effect on the other anaerobic life on the planet at the time. And it's funny to say that because it was a catastrophe for them. But it was kind of a necessary thing that had to happen for us to happen. So for us, it was a blessing that this cyanobacteria" + }, + { + "Q": "what does it mean for the orbit to go outward at 1:40 ?", + "A": "They got further away from the Sun.", + "video_name": "nYFuxTXDj90", + "timestamps": [ + 100 + ], + "3min_transcript": "We finished off the last video in the Hadean Eon. It was named for Hades or the ancient Greek underworld. Hades is also the name of the god that ran the Greek underworld, Zeus's oldest brother. And it was an appropriate name, although the idea of, the ancient Greek notion of the underworld isn't exactly the more modern notion of Hell. But it was a hellish environment. You had all this lava flowing around. You had things impacting the Earth from space. And as far as we can tell right now, it was completely inhospitable to life. And to make matters worse, even though the Earth started to cool down a little bit, maybe the crust became a little bit more solid. Maybe the collisions started to happen less and less, as we started to go a few hundred million years fast After the Theia rammed into the early earth and formed the moon, there was something called the Late Heavy Bombardment. And right now, the consensus is that life, whatever we are descended from, would have Because this was a time where so many things from outer space were hitting Earth, that it was so violent, that it might have killed off any kind of primitive, self-replicating organisms or molecules that might have existed before it. And I won't go into the physics of the Late Heavy Bombardment. But we believe that it happened, because Uranus and Neptune-- so if this is the sun right here-- that is the sun. This is the asteroid belt. That's outside the orbits of the inner, rocky planets. That Uranus and Neptune, their orbits moved outward. And I'm not going to go into the physics. But what that caused is, gravitationally, it caused a lot of the asteroids in the asteroid belt to move inward and start impacting the inner planets. And of course, Earth was one of the inner planets. And I should make the sun like orange or something, not blue. I don't want you to think that's Earth. And it's more obvious on the moon, because the moon does not have an atmosphere to kind of smooth over the impact. So the consensus is that only after the Late Heavy Bombardment was Earth kind of ready for life. And we believe that the first life formed 3.824 billion years ago. Remember, g for giga, for billion years ago. And when we talk about life at this period, we're not talking about squirrels or panda bears. We're talking about extremely simple life forms. We're talking about prokaryotes. And let me give you a little primer on that right now, though we go into much more detail in the biology playlist. We're talking about prokaryotes. And I'll compare them to eukaryotes. Prokaryotes are, for the most part, unicellular organisms that have no nucleuses. They also don't have any other membrane-bound" + }, + { + "Q": "At 2:17, if Earth had life only for part of its history (starting in ~3.8 - 4 Ga), then does that mean that other planets might have had life at one point in their histories?", + "A": "Sure, that s possible. There are some tantalizing hints that Mars may have had bacteria-like life at one point, but we don t know.", + "video_name": "nYFuxTXDj90", + "timestamps": [ + 137 + ], + "3min_transcript": "We finished off the last video in the Hadean Eon. It was named for Hades or the ancient Greek underworld. Hades is also the name of the god that ran the Greek underworld, Zeus's oldest brother. And it was an appropriate name, although the idea of, the ancient Greek notion of the underworld isn't exactly the more modern notion of Hell. But it was a hellish environment. You had all this lava flowing around. You had things impacting the Earth from space. And as far as we can tell right now, it was completely inhospitable to life. And to make matters worse, even though the Earth started to cool down a little bit, maybe the crust became a little bit more solid. Maybe the collisions started to happen less and less, as we started to go a few hundred million years fast After the Theia rammed into the early earth and formed the moon, there was something called the Late Heavy Bombardment. And right now, the consensus is that life, whatever we are descended from, would have Because this was a time where so many things from outer space were hitting Earth, that it was so violent, that it might have killed off any kind of primitive, self-replicating organisms or molecules that might have existed before it. And I won't go into the physics of the Late Heavy Bombardment. But we believe that it happened, because Uranus and Neptune-- so if this is the sun right here-- that is the sun. This is the asteroid belt. That's outside the orbits of the inner, rocky planets. That Uranus and Neptune, their orbits moved outward. And I'm not going to go into the physics. But what that caused is, gravitationally, it caused a lot of the asteroids in the asteroid belt to move inward and start impacting the inner planets. And of course, Earth was one of the inner planets. And I should make the sun like orange or something, not blue. I don't want you to think that's Earth. And it's more obvious on the moon, because the moon does not have an atmosphere to kind of smooth over the impact. So the consensus is that only after the Late Heavy Bombardment was Earth kind of ready for life. And we believe that the first life formed 3.824 billion years ago. Remember, g for giga, for billion years ago. And when we talk about life at this period, we're not talking about squirrels or panda bears. We're talking about extremely simple life forms. We're talking about prokaryotes. And let me give you a little primer on that right now, though we go into much more detail in the biology playlist. We're talking about prokaryotes. And I'll compare them to eukaryotes. Prokaryotes are, for the most part, unicellular organisms that have no nucleuses. They also don't have any other membrane-bound" + }, + { + "Q": "What do the arrows below the reaction arrow represent? (At 3:18)", + "A": "They are just lines connecting the Br\u00c3\u00b8nsted acids to their conjugate bases, and the Br\u00c3\u00b8nsted bases to their conjugate a", + "video_name": "jIL333CKE9A", + "timestamps": [ + 198 + ], + "3min_transcript": "And then what do you get if you add an H plus to OH minus? You would get H2O or water. So let me go ahead and draw water in here. And I'll put in my lone pairs of electrons. Let's follow our electrons along so the two electrons, right, this lone pair right here on the hydroxide anion picked up this proton. So let's say those two electrons in magenta are these two electrons, and this was the proton that they picked up. And then we also need to follow the electrons, the electrons in here, I'll make them blue. So these electrons in blue come off onto the oxygen. So let's say those electrons in blue are right here, which gives the oxygen a negative one formal charge. So this is an acid-based reaction. And we can even identify conjugate acid-based pairs here. So on the left, right, on the left this was acetic acid. This was our Bronsted-Lowry acid. What is the conjugate base to acetic acid? Well, that would be over here, right. and this would be the conjugate base. Let me identify this as being the conjugate base. This is the acetate anion, right. So this is our conjugate base. For hydroxide, hydroxide on the left side functioned as a base, right. So the conjugate acid must be on the right side. So if you add a proton to OH minus, you get H2O. So water is the conjugate, oops. I'm writing conjugate base here, but it's really the conjugate acid, right. So we've identified our conjugate acid-base pairs. All right, the biggest mistake that I see when students are drawing acid-base mechanisms is they mess up their curved arrows. So the biggest mistake I see, and I'll do this in red so it'll remind you not to do it, is they show this proton right here moving to the hydroxide anion. And that is incorrect, all right. That's a very common mistake, because curved arrows show the movement of electrons, right. These two electrons up here in magenta are the ones that are taking that acetic proton. So this is incorrect. Don't draw your acid-base mechanisms like this. Let's do one more acid-base mechanism for some extra practice here. So on the left we have acetone and on the right we have the hydronium ion, H3O plus. So the hydronium ion is gonna function as our Bronsted-Lowry acid. It's going to donate a proton to acetone, which is going to be our Bronsted-Lowry base. Remember when you're drawing an acid-base mechanism, your curved arrows show the movement of electrons. So if acetone functions as our base, a lone pair of electrons on this oxygen could take this proton right here and leave these electrons behind on this oxygen. So let's show the results of our acid-base mechanism. So on the left, right, the lone pair on the left of the oxygen didn't do anything." + }, + { + "Q": "At 6:32 why did the empirical formula was H2SO4 and why not H2O4S", + "A": "The sulfate anion is just more formally written as SO\u00e2\u0082\u0084 rather than as O\u00e2\u0082\u0084S. Many other elements form similar ions, and they re also written with the general formula of YO\u00e2\u0082\u0093.", + "video_name": "sXOIIEZh6qg", + "timestamps": [ + 392 + ], + "3min_transcript": "There's a city in Louisiana which we used to drive to all the time -- I think we had some family friends there --called Port Sulfur. And they did a lot of sulfur processing there. And lucky for the residents, at least at the time -- I apologize if they fixed the issue -- it smelled like sulfur, which smells like rotten eggs. But anyway, one mole of sulfur. One mole of sulfur. So sulfur's atomic mass is 32 atomic mass units per sulfur atom. So a whole mole of it is going to have a mass of 32 grams. So a mole of sulfur-- not a mule. Maybe I should invent a new unit called the mule. So a mole of sulfur is 32 grams. So how many moles do we have? We have a little bit more than one, but let's be precise here, because everything else is a little bit of a decimal. So if we have 32.65 grams of sulfur -- divided by 32 grams per mole we have 1.02 moles of sulfur. This was hydrogen up here. So here, you should hopefully see a pretty good ratio, here. For every one sulfur atom-- I mean, the ratio worked exactly out and that's because I did this problem before. I actually made up this problem before we worked, so I made it so the numbers worked out. But one mole of sulfur, for every mole of sulfur, so for every 6.02 times 10 to the 23 sulphur atoms, you have two moles of hydrogen, right? This ratio is 1:2. Two times 1.02 is 2.04. And then, for every one mole of sulfur, you have four moles of oxygen. Right? Literally, if you multiply this times four you get 4.08. So the ratio of hydrogen to sulfur to oxygen we have two hydrogens and we have four oxygens. So the empirical formula of this is H2. And then we have one sulfur. And then we have four oxygens. And this is sulfuric acid, one of the things you would least like poured on you most of the time. Anyway, hope you found that useful." + }, + { + "Q": "0:34: Isn't 2.04% + 65.3% +32.65% = 99.99% and not 100% so wouldn't that statement be incorrect?", + "A": "he rounded a tiny bit, that .01% is not enough to make the statement incorrect", + "video_name": "sXOIIEZh6qg", + "timestamps": [ + 34 + ], + "3min_transcript": "I said I would get you a more interesting mass composition to empirical formula problem, one that doesn't just have a straight-up 2:1 ratio. And so here it is. I have a bag of stuff. Or let's call this a bottle of stuff. Maybe it's in its liquid form. And it happens to be 2.04% hydrogen, 65.3% oxygen, and 32.65% sulfur. What is the empirical formula, what's our best stab at the empirical formula, of this substance? So what we would do, like we do in all these problems, let's just assume we've got 100 grams of the stuff. So we have 100 grams of the stuff. So we assume 100 grams. Let me do that in a good yellow. So let's say, assume I have 100 grams. How many grams of hydrogen do I have? If I have 100 grams total, 2.04% of that is hydrogen, so I have 2.04 grams of hydrogen. I have 65.3 grams of oxygen. Now, what we need to do now is figure out how many moles of hydrogen is this. How many moles of oxygen. And how many moles of sulfur. Then we can compare the ratios and we should be able to know the empirical formula. So how much 1 mole of hydrogen? What is the mass of 1 mole of hydrogen? Let me write that. So 1 mole of hydrogen. Well we know what the mass number for hydrogen is. It's 1. And especially, the atomic weight, also for hydrogen, if we were to take it on Earth. The composition, you pretty much just find hydrogen nucleuses. If it's neutral, it has an electron, but it has no neutrons. So it has an atomic mass of one atomic mass unit. So one mole of hydrogen. If you have a ton of hydrogens together, or a mole of them, not a ton, I shouldn't say, you have 6.02 times 10 to the 23 hydrogens. Then you take hydrogen's atomic mass number in atomic mass units. And you say, well, it'll be that many grams of hydrogen, right? So if you immediately look up here, if we have 2.04 grams of hydrogen, how many moles of hydrogen do we have? Well, one mole is one gram, so we have 2.04 moles of hydrogen. Notice, this said what the mass of the hydrogen is. This tells us how many hydrogen molecules we have. Remember, this is 2.04 times 6.02 times 10 to the 23 hydrogen atoms. Moles of hydrogen. Maybe I should write that down. So one mole of hydrogen. There you go. And then oxygen. One mole of oxygen. Oxygen's mass number, in case you forgot, is 16." + }, + { + "Q": "at 6:31, can I write H2O4S instead of H2SO4 like Sal?", + "A": "yes most people go alphabetically when arranging molecular formulas.", + "video_name": "sXOIIEZh6qg", + "timestamps": [ + 391 + ], + "3min_transcript": "There's a city in Louisiana which we used to drive to all the time -- I think we had some family friends there --called Port Sulfur. And they did a lot of sulfur processing there. And lucky for the residents, at least at the time -- I apologize if they fixed the issue -- it smelled like sulfur, which smells like rotten eggs. But anyway, one mole of sulfur. One mole of sulfur. So sulfur's atomic mass is 32 atomic mass units per sulfur atom. So a whole mole of it is going to have a mass of 32 grams. So a mole of sulfur-- not a mule. Maybe I should invent a new unit called the mule. So a mole of sulfur is 32 grams. So how many moles do we have? We have a little bit more than one, but let's be precise here, because everything else is a little bit of a decimal. So if we have 32.65 grams of sulfur -- divided by 32 grams per mole we have 1.02 moles of sulfur. This was hydrogen up here. So here, you should hopefully see a pretty good ratio, here. For every one sulfur atom-- I mean, the ratio worked exactly out and that's because I did this problem before. I actually made up this problem before we worked, so I made it so the numbers worked out. But one mole of sulfur, for every mole of sulfur, so for every 6.02 times 10 to the 23 sulphur atoms, you have two moles of hydrogen, right? This ratio is 1:2. Two times 1.02 is 2.04. And then, for every one mole of sulfur, you have four moles of oxygen. Right? Literally, if you multiply this times four you get 4.08. So the ratio of hydrogen to sulfur to oxygen we have two hydrogens and we have four oxygens. So the empirical formula of this is H2. And then we have one sulfur. And then we have four oxygens. And this is sulfuric acid, one of the things you would least like poured on you most of the time. Anyway, hope you found that useful." + }, + { + "Q": "At 6:35, Sal writes the empirical formula as H2 S O4. Is there usually a specific order we need to write empirical or molecular formulas?", + "A": "In H2SO4 H2 is the positively charged ion while the SO4 is negatively charged. While writing the molecular formula, the most electropositive element is written first. Hope this helps :)", + "video_name": "sXOIIEZh6qg", + "timestamps": [ + 395 + ], + "3min_transcript": "There's a city in Louisiana which we used to drive to all the time -- I think we had some family friends there --called Port Sulfur. And they did a lot of sulfur processing there. And lucky for the residents, at least at the time -- I apologize if they fixed the issue -- it smelled like sulfur, which smells like rotten eggs. But anyway, one mole of sulfur. One mole of sulfur. So sulfur's atomic mass is 32 atomic mass units per sulfur atom. So a whole mole of it is going to have a mass of 32 grams. So a mole of sulfur-- not a mule. Maybe I should invent a new unit called the mule. So a mole of sulfur is 32 grams. So how many moles do we have? We have a little bit more than one, but let's be precise here, because everything else is a little bit of a decimal. So if we have 32.65 grams of sulfur -- divided by 32 grams per mole we have 1.02 moles of sulfur. This was hydrogen up here. So here, you should hopefully see a pretty good ratio, here. For every one sulfur atom-- I mean, the ratio worked exactly out and that's because I did this problem before. I actually made up this problem before we worked, so I made it so the numbers worked out. But one mole of sulfur, for every mole of sulfur, so for every 6.02 times 10 to the 23 sulphur atoms, you have two moles of hydrogen, right? This ratio is 1:2. Two times 1.02 is 2.04. And then, for every one mole of sulfur, you have four moles of oxygen. Right? Literally, if you multiply this times four you get 4.08. So the ratio of hydrogen to sulfur to oxygen we have two hydrogens and we have four oxygens. So the empirical formula of this is H2. And then we have one sulfur. And then we have four oxygens. And this is sulfuric acid, one of the things you would least like poured on you most of the time. Anyway, hope you found that useful." + }, + { + "Q": "Shouldn't a mole of different gases have different volumes not all the same of 22.4 at 11:50? Is it because in ideal gases we disregard the gas's volume?", + "A": "Most gases approximate ideal gas behavior as long as the pressure is not to high and the temperature is not too low.", + "video_name": "GwoX_BemwHs", + "timestamps": [ + 710 + ], + "3min_transcript": "Pressure is 1 atmosphere, but remember we're dealing with atmospheres. 1 atmosphere times volume-- that's what we're solving for. I'll do that and purple-- is equal to 1 mole times R times temperature, times 273. Now this is in Kelvin; this is in moles. We want our volume in liters. So which version of R should we use? Well, we're dealing with atmospheres. We want our volume in liters, and of course, we have moles and Kelvin, so we'll use this version, 0.082. So this is 1, so we can ignore the 1 there, the 1 there. So the volume is equal to 0.082 times 273 degrees Kelvin, So if I have any ideal gas, and all gases don't behave ideally ideal, but if I have an ideal gas and it's at standard temperature, which is at 0 degrees Celsius, or the freezing point of water, which is also 273 degrees Kelvin, and I have a mole of it, and it's at standard pressure, 1 atmosphere, that gas should take up exactly 22.4 liters. And if you wanted to know how many meters cubed it's going to take up, well, you could just say 22.4 liters times-- now, how many meters cubed are there -- so for every 1 meter cubed, you have 1,000 liters. I know that seems like a lot, but it's true. Just think about how big a meter cubed is. If you have something at 1 atmosphere, a mole of it, and at 0 degrees Celsius. Anyway, this is actually a useful number to know sometimes. They'll often say, you have 2 moles at standard temperature and pressure. How many liters is it going to take up? Well, 1 mole will take up this many, and so 2 moles at standard temperature and pressure will take up twice as much, because you're just taking PV equals nRT and just doubling. Everything else is being held constant. The pressure, everything else is being held constant, so if you double the number of moles, you're going to double the volume it takes up. Or if you half the number of moles, you're going to half the volume it takes up. So it's a useful thing to know that in liters at standard temperature and pressure, where standard temperature and pressure is defined as 1 atmosphere and 273 degrees Kelvin, an ideal gas will take up 22.4 liters of volume." + }, + { + "Q": "At 4:59, there's a model of a molecular structure. What kind of molecule is it modeling?", + "A": "That is nitroglycerin. The black balls represent carbons, white is hydrogen, blue is nitrogen and red is oxygen.", + "video_name": "Rd4a1X3B61w", + "timestamps": [ + 299 + ], + "3min_transcript": "that have all of these different properties. So when you think about chemistry, yes, it might visually look something like this. These are obviously much older pictures. But at its essence, it's how do we create models and understand the models that describe a lot of the complexity in the universe around us? And just to put chemistry in, I guess you could say, in context with some of the other sciences, many people would say at the purest level, you would have mathematics. That math, you're studying ideas, which could even be independent, you're seeing logical ideas that could be even independent of anything that you've ever observed or experienced. And a lot of folks that say if we ever communicate with another intelligent species that could be completely different than us, math might be that common language. Because even if we perceive the world differently math might be that common language. But on top of math, we start to say, well how is our reality actually structured? At the most basic level, what are the constituents of matter and what are the mathematical properties that describe how they react together? And then, or interact with each other? Then you go one level above that, you get to the topic of this video, which is chemistry. Which is very closely related to physics. When we talk about these chemical equations and we create these molecular structures, the interactions between these atoms, these are quantum mechanical interactions which we do not fully understand at the deepest level yet. But with chemistry, we can start to make use of the math and they physics to start to think about how all of these different building blocks can interact to explain all sorts of different phenomena. This chemical equation you see right here, This is hydrogen combusting with oxygen to produce a lot of energy. To produce energy. When we imagine combustion, we think of fire. But what even is fire at its most fundamental level? How do we get, why do we perceive this thing here? And chemistry is super important because on top of that, we build biology. We build biology. And as you'll see as you study all of these things, there's points where these things start to bleed together. But the biology in, say, a human being, or really in any species, it's based on molecular interactions. Interactions between molecules, between atoms, which, at the end of the day, is all about chemistry. As I speak, the only reason why I'm able to speak is because of really, hard to imagine the number of chemical interactions happening in me right now to create this soundness. To create this thing that thinks it exists that wants to make a video about how awesome" + }, + { + "Q": "At 5:26 why did it become 16? and where do 4 come from? why is it plus not minus not like the other example?", + "A": "CO2 C has 4 valence electrons, O has 6 valence electrons 4 + 6 + 6 = 16 valence electrons", + "video_name": "97POZGcfoY8", + "timestamps": [ + 326 + ], + "3min_transcript": "And we'll go ahead and move on to the next step. So let me go ahead and put in my lone pairs of electrons around my chlorine here. So we have our dot structure. Next, we're going to count the number of electron clouds that surround the central atom. And I like to use the term electron cloud. You'll see many different terms for this in different textbooks. You'll see charge clouds, electron groups, electron domains, and they have slightly different definitions depending on which textbook you look in. And really the term of electron cloud helps describe the idea of valence electrons in bonds and in lone pairs of electrons occupying these electron clouds. And you could think about them as regions of electron density. And, since electrons repel each other, those regions of electron density, those clouds, want to be as far apart from each other as they possibly can. And so let's go ahead and analyze our molecule here. So surrounding the central atom. surrounding our central atom. So we could think about those as being an electron cloud. And then over here we have another electron cloud. So we have two electron clouds for this molecule, and those electron clouds are furthest apart when they point in opposite directions. And so the geometry or the shape of the electron clouds around the central atom, if they're pointing in opposite directions, it's going to give you a linear shape here. So this molecule is actually linear because we don't have any lone pairs to worry about here. So we're going to go ahead and predict the geometry of the molecule as being linear. And if that's linear, then we can say the bond angle-- so the angle between the chlorine, the beryllium, and the other chlorine-- is 180 degrees. So just a straight line. All right. So that's how to use VSEPR to predict the shape. Let's do another example. So CO2-- so carbon dioxide. Carbon has four valence electrons. Oxygen has six. And we have two of them. So 6 times 2 gives us 12. 12 plus 4 gives us 16 valence electrons to deal with for our dot structure. The less electronegative atom goes in the center, so carbon is bonded to oxygen, so two oxygens like that. We just represented four valence electrons. Right? So two here and two here, so that's four. So 16 minus 4 gives us 12 valence electrons left. Those electrons are going to go on our terminal atoms, which are oxygens if we are going to follow the octet rule. So each oxygen is surrounded by two electrons. So, therefore, each oxygen needs six more valence electrons. I'll go ahead and put in six more valence electrons on our oxygen. Now you might think we're done, but, of course, we're not because carbon is going to follow the octet rule. Carbon does not have a formal charge of 0 in this dot structure, so even though we've represented all of our valence electrons now," + }, + { + "Q": "At 1:56 how come we don't put 2 more lone pairs of electrons on Be?", + "A": "Total number of electrons in BeCL2 = 16 Which are all used up. ( 6 each to Cl + 4 electrons as bonds)", + "video_name": "97POZGcfoY8", + "timestamps": [ + 116 + ], + "3min_transcript": "This next set of videos, we're going to predict the shapes of molecules and ions by using VSEPR, which is an acronym for valence shell electron pair repulsion. And really all this means is that electrons, being negatively charged, will repel each other. Like charges repel, and so when those electrons around a central atom repel each other, they're going to force the molecule or ion into a particular shape. And so the first step for predicting the shape of a molecule or ion is to draw the dot structure to show your valence electrons. And so let's go ahead and draw a dot structure for BeCl2. So you find beryllium on the periodic table. It's in group 2, so two valence electrons. Chlorine is in group 7, and we have two of them. So 2 times 7 is 14. And 14 plus 2 gives us a total of 16 valence electrons that we need to account for in our dot structure. So you put the less electronegative atom So beryllium goes in the center. We know it is surrounded by two chlorines, And we just represented four valence electrons. So here's two valence electrons. And here's another two for a total of four. So, instead of 16, we just showed four. So now we're down to 12 valence electrons that we need to account for. So 16 minus 4 is 12. We're going to put those left over electrons on our terminal atoms, which are our chlorines. And chlorine is going to follow the octet rule. Each chlorine is already surrounded by two valence electrons, so each chlorine needs six more. So go ahead and put six more valence electrons on each chlorine. And, since I just represented 12 more electrons there, now we're down to 0 valence electron. So this dot structure has all of our electrons in it. And some of you might think, well, why don't you keep going? Why don't you show some of those lone pairs of electrons in chlorine moving in to share them with the beryllium to give it an octet of electrons? And the reason you don't is because of formal charge. So let's go ahead and assign a formal charge So remember each of our covalent bonds consists of two electrons. So I go ahead and put that in. And if I want to find formal charge, I first think about the number of the valence electrons in the free atom. And that would be two, four-- four berylliums. So we have two electrons in the free atom. And then we think about the bonded atom here, so when I look at the covalent bond, I give one of those electrons to chlorine and one of those electrons to beryllium. And I did the same thing for this bond over here, and so you can see that it is surrounded by two valence electrons. 2 minus 2 gives us a formal charge of 0. And so that's one way to think about why you would stop here for the dot structure. So it has only two valence electrons, so even though it's in period 2, it doesn't necessarily have to follow the octet rule. It just has to have less than eight electrons. And so, again, formal charge helps you understand why you can stop here for your dot structure. Let me go ahead and redraw our molecule" + }, + { + "Q": "At 6:45 Sal says that the base must have the same number of moles as the weak acid. Doesn't that statement assume the acid is monoprotic (donates 1 H+)? How would the calculation change for a diprotic acid (donates 2 H+)? The moles for the acid would just be doubled, right?", + "A": "You re right, so the normality of the base must equal the normality of the acid. Therefore, for diprotic molecules you would need twice as much base.", + "video_name": "BBIGR0RAMtY", + "timestamps": [ + 405 + ], + "3min_transcript": "And when we're adding more hydrogens, we're getting really acidic really fast. But we have a lot of the conjugate acid there in the solution already. So we're going to have an acidic equivalence point. Now, let me give you an actual problem, just to hit all the points home. Because everything I've done now has been very hand-wavey, and no numbers. So let me draw one. Let me draw a weak acid. And you'll recognize it because you're good at this now. But I'll deal with some real numbers here. So let's say that's a pH of 7. We're going to titrate it. It starts off at a low pH because it's a weak acid. And as we titrate it, it's pH goes up. And then it hits the equivalence point and it goes like that. The equivalence point is right over here. And let's say our reagent that we were adding is sodium hydroxide. And let's say it's a 0.2 molar solution. I'll use 700 milliliters of sodium hydroxide is our equivalence point. Right there. So the first question is how much of our weak acid did we have? So what was our original concentration of our weak acid? This is just a general placeholder for the acid. So original concentration of our weak acid. Well, we must have added enough moles of OH at the equivalent point to cancel out all of the moles of the weak acid in whatever hydrogen was out there. But the main concentration was from the weak acid. This 700 milliliters of our reagent must have the same number of moles as the number of moles of weak acid we started off with. And let's say our solution at the beginning was 3 liters. 3 liters to begin with, before we started titrating. the solution. But let's just say that in the beginning, we started with 3 liters. So how many moles have we sopped up? Well, how many moles of OH are there in 700 milliliters of our solution? Well, we know that we have 0.2 moles per liter of OH. And then we know that we don't have-- times 0.7 liters, right? 700 milliliters is 0.7 liters. So how many moles have we added to the situation? 2 times 7 is 14. And we have 2 numbers behind the decimal. So it's 0.14. So 700 milliliters of 0.2 molar sodium hydroxide, and we have 700 milliliters of it, or 0.7 liters. We're going to have 0.14 moles of, essentially, OH that we" + }, + { + "Q": "A magnetic field creates a force on a moving charge and a moving charge creates a magnetic field which would then create a force on a moving charge..., right? How does stationary charge create its own static electric charge like Sal is talking about at 9:01? Did I misunderstand him? I am trying to grasp how electricity and magnetism are the same. Thanks for your help! :-)", + "A": "Electric fields originate from any charge (moving or not) and changing magnetic fields. Magnetic fields originate from moving charge (ie. current) and changing electric fields. A charge will not interact with the field it generates itself.", + "video_name": "Ri557hvwhcM", + "timestamps": [ + 541 + ], + "3min_transcript": "It's 3 meters. So 2 pi times 3. So it equals the permeability of free space. The 2 and the 2 cancel out over 3 pi. So how do we calculate that? Well, we get out our trusty TI-85 calculator. And I think you'll be maybe pleasantly surprised or shocked to realize that-- I deleted everything just so you can see how I get there-- that it actually has the permeability of free space stored in it. So what you do is you go to second and you press constant, which is the 4 button. It's in the built-in constants. Let's see, it's not one of those. You press more. It's not one of those, press more. Oh look at that. Mu not. The permeability of free space. That's what I need. And I have to divide it by 3 pi. Divide it by 3-- and then where is pi? There it is. Divided by 3 pi. It equals 1.3 times 10 to the negative seventh. It's going to be teslas. The magnetic field is going to be equal to 1.3 times 10 to the minus seventh teslas. So it's a fairly weak magnetic field. And that's why you don't have metal objects being thrown around by the wires behind your television set. But anyway, hopefully that gives you a little bit-- and just so you know how it all fits together. We're saying that these moving charges, not only can they be affected by a magnetic field, not only can a current be affected by a magnetic field or just a moving charge, it actually creates them. And that kind of creates a little bit of symmetry in your head, hopefully. Because that was also true of electric field. A charge, a stationary charge, is obviously pulled or pushed by a static electric field. And it also creates its own static electric field. Because if you keep studying physics, you're going to actually prove to yourself that electric and magnetic fields are two sides of the same coin. And it just looks like a magnetic field when you're in a different frame of reference, When something is whizzing past you. While if you were whizzing along with it, then that thing would look static. And then it might look a little bit more like an electric field. But anyway, I'll leave you there now. And in the next video I will show you what happens when we have two wires carrying current parallel to each other. And you might guess that they might actually attract or repel each other. Anyway, I'll see you in the next video." + }, + { + "Q": "Why does the acid and base stay liquid and turn pink but not form H20 (water) and NaCl (Salt) as the formula stated at about 1:40 in the video?", + "A": "It turns pink because you add an indicator, which has a pink colour in basic solutions. And they reaction has produced water en NaCl, but the salt stays dissolved in the solution. (NaCl is easily soluble).", + "video_name": "d1XTOsnNlgg", + "timestamps": [ + 100 + ], + "3min_transcript": "Titration is a procedure for determining the concentration of a solution. And so let's say we're starting with an acidic solution. So in here let's say we have some hydrochloric acid. So we have come HCl. And we know the volume of HCL, let's say we're starting with 20.0 milliliters of HCl. But we don't know the concentration right? So question mark here for the concentration of HCl. We can find out that concentration by doing a titration. Next we need to add a few drops of an acid base indicator. So to this flask we're also going to add a few drops of an acid base indicator. We're gonna use phenolphthalein. And phenolphthalein is colorless in acid but turns pink in the presence of base. And since we have our phenolphthalein in acid right now we have a clear solution. There's no color to it. Up here we're gonna have our standard solution right? We're gonna have a known concentration of sodium hydroxide. So let's say we have a solution of sodium hydroxide And we're ready to start our titration. So we allow the sodium hydroxide to drip into our flask containing our HCl and our indicator. And the acid in the base will react, right? So we get an acid base neutralization reaction. HCl plus NaOH right? If we think about the products, this would be OH minus, this would be H plus, H plus and OH minus give us H2O. And our other product we would have Na plus and Cl minus, which give us NaCl, or sodium chloride. So let's say we add a certain volume of base right? So now this would be higher, and we see our solution turn light pink. Alright so let's say we see our solution turn light pink and it stays light pink. That means that all of the acid And we have a tiny amount of excess base present, and that's causing the acid base indicator to remain pink. So a tiny excess of base means we've neutralized all of the acid present. And where the indicator changes color, this is called the end point of a titration, alright? So when our solution changes color, that's the end point of our titration. And here we stop and we check and see the volume of base that we used in our titration. So if we started right here, if we started with that much base, let's say we ended down here, alright? So we still have a little bit of base left. And this would be the volume of base that we used in the titration. Alright so we have a change in volume here, and let's say that it's 48.6 milliliters. So it took 48.6 milliliters of our base to completely neutralize the acid that we had present." + }, + { + "Q": "At 4:53, how do you know that there's a 1-1 mole ratio? And what does he mean by that?", + "A": "From the balanced equation; the unwritten (1) before the element/compound is the balance. (1)NaOH+(1)HCl==>(1)NaCl+(1)H2O", + "video_name": "d1XTOsnNlgg", + "timestamps": [ + 293 + ], + "3min_transcript": "Alright so let's go ahead and do that, and let's start with the concentration of sodium hydroxide. Alright we know that we started with point one zero zero molar solution of sodium hydroxide. So point one zero zero molar. And molarity is equal to mols over liters. Alright so this is equal to mols over liters. And our goal is to figure out how many mols of base that we used to neutralize the acid that was present. Alright so we can take our volume here, 48.6 mililiters and we can convert that into liters. Alright so just move your decimal place three places to the left. So one, two, three. So that's point zero four eight six liters. So this is equal to mols over And so let's get some more space. Alright let me just rewrite this really quickly. Zero point one zero zero is equal to X over zero point zero four eight six. So we're just solving for X, and X represents the mols of sodium hydroxide that were necessary to neutralize the acid that we had present. Alright so when you solve for X, you get zero point zero zero four eight six mols of sodium hydroxide used in our titration. Next you look at the balanced equation for what happened . So if I look at my balanced equation alright there's a one here and there's a one here. So we have a one to one mol ratio. And the equivalence point is where just enough of your standard solution has been added to completely react with the solution that's being titrated. all of the acid has been neutralized. Right? So it's completely reacted. And since we have a one to one mol ratio, if I used this many mols of sodium hydroxide, that must be how many mols of HCl that we had present in our original solution. So therefore, I can go ahead and write that I must have had zero point zero zero four eight six mols of HCl present in the flask before we started our titration. Right and I knew that because of the one to one mol ratio. Remember our goal was to find the concentration of HCl. The original concentration. And concentration, molarity is equal to mols over liters. So now I know how many mols of HCl I had, and my original volume of HCl was 20 milliliters right?" + }, + { + "Q": "At around 3:45, David said that the momentums were positive 5 and negative 10. Could you choose that moving to the left was instead positive and moving to the right was negative? Making the equation: (0.2kg)(-5m/s) - (0.2kg)(10m/s)?", + "A": "Try it and see what happens. Good way to learn.", + "video_name": "uMYAc04D0ak", + "timestamps": [ + 225 + ], + "3min_transcript": "the impulse from all forces on an object, like this ball, that should just equal the change in momentum of that object, like the change in momentum of this ball. So if we can figure out the change in momentum of this ball, we can figure out the net impulse on this ball. And since it's the net impulse, and this formula appears also true, this is equivalent, which is saying that it's the net force, multiplied by the time duration, during which that net force is acting. This is hard for people to remember, sometimes my students like to remember it as Jape Fat. So, if you look at this, it looks like J-A-P, this kinda looks like an E, F-A-T. So if you need a way, a pneumonic device, to remember this, Jape Fat is a way to remember how impulse, change in momentum, force, and time, are all related. So let's do it. We can't use force because we don't know it yet, but I can figure out the change in momentum 'cause I know the velocities. So, we know that the change in momentum is gonna be P final, the final momentum, What's my final momentum? My final momentum is M times V, so it's gonna be mass times V final, minus mass times V initial, and my mass is .2, so I've got a mass of 0.2 kilograms. My final velocity is five, because the ball recoiled to the right with positive five. Positive five 'cause it's moving to the right. I'm gonna assume rightward is positive. Then minus, the mass is .2 again, so 0.2 kilograms. My initial velocity is not 10. This is 10 meters per second to the left, and momentum is a vector, it has direction, so you have to be careful with negative signs here. This is the most common mistake. People just plug in positive 10, then get the wrong answer. But this ball changed directions, so the two velocities here have to have two different sides, so this has to be a negative 10 meters per second, if I'm assuming rightward is positive. This leftward velocity, and this leftward initial velocity, And, if you didn't plug that in, you'd get a different answer, so you gotta be careful. So, what do I get here if I multiply this all out? I'm gonna get zero, no, sorry, I'm gonna get one kilogram meters per second, minus a negative two kilogram meters per second, and that's gonna give me positive three kilogram meters per second is the impulse, and that should make sense. The impulse was positive. The direction of the impulse, which is a vector, is the same direction as the direction of the force. So, which way did our face exert a force on the ball? Our face exerted a force on the ball to the right. That's why the impulse on the ball is to the right. The impulse on this person's face is to the left, but the impulse on the ball is to the right, because the ball was initially going left and it had a force on it to the right that made it recoil and bounce back to the right. That's why this impulse has a positive direction to it." + }, + { + "Q": "how does bacteria get its energy 1:40", + "A": "Bacteria obtain energy by either ingesting other organisms and organic compounds or by producing their own food. The bacteria that produce their own food are called autotrophs. Bacteria that must consume other organic molecules for energy are called heterotrophs.", + "video_name": "dQCsA2cCdvA", + "timestamps": [ + 100 + ], + "3min_transcript": "- [Voiceover] I would like to welcome you to Biology at Khan Academy. And biology, as you might now, is the study of life. And I can't really imagine anything more interesting than the study of life. And when I say \"life,\" I'm not just talking about us, human beings. I'm talking about all animals. I'm talking about plants. I'm talking about bacteria. And it really is fascinating. How do we start off with inanimate molecules and atoms? You know, this right here is a molecule of DNA. How do we start with things like that, and we get the complexity of living things? And you might be saying, well, what makes something living? Well, living things convert energy from one form to another. They use that energy to grow. They use that energy to change. And I guess growth is a form of change. They use that energy to reproduce. And these are all, in and of themselves, How do they do this? You know, we look around us. How do we, you know, eat a muffin? And how does that allow us to move around and think and do all the things we do? Where did the energy from that muffin come from? How are we similar to a plant or an insect? And we are eerily or strangely similar to these things. We actually have a lot more in common with, you know, that tree outside your window, or that insect, that bee, that might be buzzing around, than you realize. Even with the bacteria that you can only even see at a microscopic level. In fact, we have so much bacteria as part of what makes us, us. So these are fascinating questions. How did life even emerge? And so over the course of what you see in Biology on Khan Academy, we're going to answer these fundamental, fascinating questions. We're going to think about things like energy and the role of energy in life. We're going to thing about important molecules in biology. DNA and its role in reproduction and containing information. And we're going to study cells, which are the basic building block of life. And as we'll see, even though we view cells as these super, super small, small things, cells in and of themselves are incredibly complex. And if you compare them to an atomic scale, they're quite large. In fact, this entire blue background that I have there, that's the surface of an immune cell. And what you see here emerging from it, these little yellow things. These are HIV viruses, emerging from an immune cell. So even though you imagine cells as these very, very small microscopic things, this incredible complexity. Even viruses. Viruses are one of these fascinating things that kind of are right on the edge between life and nonlife. They definitely reproduce, and they definitely evolve. But they don't necessarily have a metabolism." + }, + { + "Q": "2:07\nhow much DNA do we have?", + "A": "Humans have 46 chromosomes that contain all of the genetic information, and there are over 25,000 genes in the human genome. Genes are composed of DNA, and it is predicted that there are over 3 billion base pairs in the human genome. Humans have approximately 10 trillion cells, so if you were to line all of the DNA found in every cell of a human body it would stretch from the earth to the sun 100 times!", + "video_name": "dQCsA2cCdvA", + "timestamps": [ + 127 + ], + "3min_transcript": "- [Voiceover] I would like to welcome you to Biology at Khan Academy. And biology, as you might now, is the study of life. And I can't really imagine anything more interesting than the study of life. And when I say \"life,\" I'm not just talking about us, human beings. I'm talking about all animals. I'm talking about plants. I'm talking about bacteria. And it really is fascinating. How do we start off with inanimate molecules and atoms? You know, this right here is a molecule of DNA. How do we start with things like that, and we get the complexity of living things? And you might be saying, well, what makes something living? Well, living things convert energy from one form to another. They use that energy to grow. They use that energy to change. And I guess growth is a form of change. They use that energy to reproduce. And these are all, in and of themselves, How do they do this? You know, we look around us. How do we, you know, eat a muffin? And how does that allow us to move around and think and do all the things we do? Where did the energy from that muffin come from? How are we similar to a plant or an insect? And we are eerily or strangely similar to these things. We actually have a lot more in common with, you know, that tree outside your window, or that insect, that bee, that might be buzzing around, than you realize. Even with the bacteria that you can only even see at a microscopic level. In fact, we have so much bacteria as part of what makes us, us. So these are fascinating questions. How did life even emerge? And so over the course of what you see in Biology on Khan Academy, we're going to answer these fundamental, fascinating questions. We're going to think about things like energy and the role of energy in life. We're going to thing about important molecules in biology. DNA and its role in reproduction and containing information. And we're going to study cells, which are the basic building block of life. And as we'll see, even though we view cells as these super, super small, small things, cells in and of themselves are incredibly complex. And if you compare them to an atomic scale, they're quite large. In fact, this entire blue background that I have there, that's the surface of an immune cell. And what you see here emerging from it, these little yellow things. These are HIV viruses, emerging from an immune cell. So even though you imagine cells as these very, very small microscopic things, this incredible complexity. Even viruses. Viruses are one of these fascinating things that kind of are right on the edge between life and nonlife. They definitely reproduce, and they definitely evolve. But they don't necessarily have a metabolism." + }, + { + "Q": "At 11:20:\n\nWhat did 12 ever do?\nHow was it activated?", + "A": "Factor 12 is the first factor that is activated in the intrinsic pathway. It is activated by a called Kallikrein. Factor 12 then simply becomes a catalyst to convert 11 from its inactive form to its active form.", + "video_name": "FNVvQ788wzk", + "timestamps": [ + 680 + ], + "3min_transcript": "which is actually one of these little yellow guys. And that tissue factor activates VII, which activates X, so you get a shot - a spark that shoots down this way and activates a little bit of X. And then X will activate a little bit of thrombin, and then thrombin will get the intrisic workhorse going. And how will thrombin do that? Well thrombin actually activates a whole bunch of these guys, and to remember the ones that it activates, you just need to take the five odd numbers starting at five. So what is that? That's V, VII, IX, XI, and XIII. Actually, this is just almost right, but it actually turns out that it's not IX, it's VIII because it couldn't be quite that easy. So those are the five that it activates. So let's draw that in here in our drawing. So let's draw that in the form of blue arrows because thrombin is blue. We said it's going to activate VII. We said it's going to activate not IX, but VIII, so this will be an awkward arrow to draw. We said it's going to activate XI, and we said it's going to activate XIII. Where's our XIII? Well, we haven't actually drawn it in yet, so let's quickly chat about that. The end goal of this whole cascade is to get these fibrin molecules, and these fibrin molecules together will form some strands. It actually turns out that there's one more step, which is to connect these strands together. So we're going to want to connect these strands together with some cross links. These cross links will just hold them together so that they actually form a tight mesh. It turns out that it's this step right here, which is enabled by factor XIII. So let's draw the final thrombin activity, which is to activate XIII. it's going to activate all the necessary things in this intrinsic pathway to get it going. You might actually be wondering about XII up there because thrombin is not hitting him, and actually it turns out that if you remove a person's factor XII, they can still clot pretty well. So it's clear that XII is not a totally necessary part of this intrinsic pathway. And to be clear again, with our use of arrows, this green arrow here is different from these white arrows in the sense that here, we are saying that fibrin is going to become fibrin strands, which is going to become interlaced fibrin strands. So if this was all there was to the story, then every time you had a little bit of damage to your endothelium, you would cause the extrinsic pathway to fire. So you'd create a little activated VII. You would activate some X, which would activate some II, which is thrombin, which would start to create fibrin from fibrinogen. And moreover, the thrombin would have" + }, + { + "Q": "At 4:08 why would it be an SN2 if that is a tertiary carbon and SN2 rxn only happens in primary and secondary carbons?", + "A": "its not a tertiary carbon, its given at the start that its a secondary or primary carbon", + "video_name": "LccmkSz-Y-w", + "timestamps": [ + 248 + ], + "3min_transcript": "And these electrons would kick off on to that chlorine. So when we draw the next intermediate here, we would now have our oxygen, still bonded to a hydrogen, still with a plus 1 formal charge like that. And now our sulfur is double bonded to our oxygen again with two lone pairs of electrons on the oxygen. The sulfur is also bonded to one chlorine now, so one of the chlorines left. And we can go ahead and draw in that chlorine. So one of the chlorines left here. It's a negatively charged chloride anion. And then still there's a lone pair of electrons on our sulfur like that. So at this part of the mechanism, the pyridine comes along. So if we go ahead and draw the dot structure for pyridine. It's a base, and so it looks like a benzene ring, except we have a nitrogen here instead. And there'd be a lone pair of electrons on this nitrogen. And so that lone pair of electrons and take this proton here on the oxygen. And that would kick these electrons back off onto this oxygen. So when we go ahead and draw that-- let's go ahead and get some more room here-- so what would we get? We would now have our carbon bonded to our oxygen. Our oxygen now has two lone pairs of electrons around it. And we have our sulfur, and our chlorine, and our lone pair of electrons on the sulfur. And now we've made a better leaving group. So this is a better leaving group than the OH was in the beginning. And if we think about an SN2 type mechanism now, we know that the bond between carbon and oxygen is polarized, right? Oxygen being more electronegative, it will be partially negative. And this carbon here be partially positive. And so now we can think about our SN2 type mechanism. Our nucleophile will be this chloride anion up here that we formed in the mechanism. and it's going to attack our partially positive carbon. An SN2 type mechanism. So as the chloride attacks, this stuff on the right is going to leave. So the electrons in magenta are actually going to move in here, and then these electrons are going to kick off onto that chlorine. So when we draw the product, we can go ahead and show the chlorine has now added on to our carbon on the left. And on the right, if you follow the movement of those electrons, they're going to form sulfur dioxides. So SO2. And also the chloride anions, so the Cl minus, like that. And so we've done it. We've substituted our chlorine atom for the OH and formed an alkyl halide. So this is just a better way of forming an alkyl chloride from an alcohol. So if we look at an example, we'll just take something like ethanol here." + }, + { + "Q": "Hey, at 5:40, is there any reason we use PBr3 instead of H-Br? Thanks.", + "A": "The phosphate ester that is made when you use PBr3 provides a better leaving group than OH on its own. Jay makes this point in the video.", + "video_name": "LccmkSz-Y-w", + "timestamps": [ + 340 + ], + "3min_transcript": "and it's going to attack our partially positive carbon. An SN2 type mechanism. So as the chloride attacks, this stuff on the right is going to leave. So the electrons in magenta are actually going to move in here, and then these electrons are going to kick off onto that chlorine. So when we draw the product, we can go ahead and show the chlorine has now added on to our carbon on the left. And on the right, if you follow the movement of those electrons, they're going to form sulfur dioxides. So SO2. And also the chloride anions, so the Cl minus, like that. And so we've done it. We've substituted our chlorine atom for the OH and formed an alkyl halide. So this is just a better way of forming an alkyl chloride from an alcohol. So if we look at an example, we'll just take something like ethanol here. had some pyridine as our base. We're going to replace the OH with our chlorine like that. And so once again, if we look at our alcohol, this is a primary alcohol. And so primary alcohols work the best because there's decreased steric hindrance. And we don't have to worry about stereochemistry, since we don't have any chirality centers in our product. Let's look at a way to form alkyl bromide. So we just formed an alkyl chloride. Let's look at the general reaction for forming an alkyl bromide here. So I go ahead and have my alcohol. And I react that with phosphorus tribromide, PBr3. The OH group is going to leave and I'm going to put a bromine in its place. And once again, this mechanism is an SN2 type mechanism. So primary or secondary alcohols only. And possible inversion of configuration present or not. So another SN2 mechanism. And again, we need to use phosphorus tribromide because the OH group is not the best leaving group. When we look at this mechanism here, let's go ahead and show that lone pair of electrons better, like that. We have phosphorus tribromide. So I'm going to go to draw the dot structure. So we would have these bromines here with lone pairs of electrons. And there's three of them. So I'll go ahead and put in those bromines. And we still have two more valence electrons to account for, and those would go on our phosphorus like that. So the first step, it's analogous to our previous mechanism. Lone pair of electrons on oxygen are going to for a bonds with phosphorus. And that would kick these electrons off onto one of the bromines, so I just chose that one. It doesn't matter which one you choose. And so when we show the result of that, we would now have our oxygen bonded to a phosphorus." + }, + { + "Q": "at 1:35 you said that a bullet goes as fast as a jet. how fast does a jet go in miles?", + "A": "Legally around 300MPH for the big commercial jets, Regular Jets can go around 600MPH legally, but when we talking Illegally were talking 1000MPH+", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 95 + ], + "3min_transcript": "My goal in this video and the next video is to start giving a sense of the scale of the earth and the solar system. And as we see, as we start getting into to the galaxy and the universe, it just becomes almost impossible to imagine. But we'll at least give our best shot. So I think most of us watching this video know that this right here is earth. And just to get a sense of scale here, I think probably the largest distance that we can somehow relate to is about 100 miles. You can get into a car for an hour, hour and a half, and go about 100 miles. And on the earth that would be about this far. It would be a speck that would look something like that. That is 100 miles. And also to get us a bit of scale, let's think about a speed that at least we can kind of comprehend. And that would be, maybe, the speed of a bullet. Maybe we can't comprehend it, but I'll say this is the fastest thing that we could maybe comprehend. depending on the type of gun and all of that-- about 280 meters per second, which is about 1,000 kilometers per hour. And this is also roughly the speed of a jet. So just to give a sense of scale here, the earth's circumference-- so if you were to go around the planet-- is about 40,000 kilometers. So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the earth. You might have taken a 12- or 15-hour flight that gets you-- not all the way around the earth-- but gets you pretty far. San Francisco to Australia, or something like that. So right now these aren't scales that are too crazy. Although, even for me, the earth itself is a pretty mind-blowingly large object. Now, with that out of the way let's think about the sun. Because the sun starts to approach something far huger. So this obviously here is the sun. And I think most people appreciate that the sun is much larger than the earth, and that it's pretty far away from the earth. But I don't think most people, including myself, fully appreciate how large the sun is or how far it is away from the earth. So just to give you a sense, the sun is 109 times the circumference of the earth." + }, + { + "Q": "At 6:15, when you begin to talk about AUs, the measurement is about equal to the distance between the earth and the sun. Was this done on purpose?", + "A": "yes, that s the definition of the AU", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 375 + ], + "3min_transcript": "So it's 4,360 hours to circumnavigate the sun, going at the speed of a bullet or a jetliner. And so that is-- 24 hours in the day-- that is 181 days. It would take you roughly half a year to go around the sun at the speed of a jetliner. Let me write this down. Half a year. The sun is huge. Now, that by itself may or may not be surprising--and actually let me give you a sense of scale here, because I have this other diagram of a sun. And we'll talk more about the rest of the solar system in the next video. But over here, at this scale, the sun, at least on my screen-- if I were to complete it, it would probably be about 20 inches in diameter. than a raindrop. If I were to draw it on this scale, where the sun is even smaller, the earth would be about that big. Now, what isn't obvious, because we've all done our science projects in third and fourth grade--or we always see these diagrams of the solar system that look something like this-- is that these planets are way further away. Even though these are depicted to scale, they're way further away from the sun than this makes it look. So the earth is 150 million kilometers from the sun. So if this is the sun right here, at this scale you wouldn't even be able to see the earth. It wouldn't even be a pixel. But it would be 150 million kilometers from the earth. and we'll be using that term in the next few videos just because it's an easier way to think about distance-- sometimes abbreviated AU, astronomical unit. And just to give a sense of how far this is, light, which is something that we think is almost infinitely fast and that is something that looks instantaneous, that takes eight minutes to travel from the sun to the earth. If the sun were to disappear, it would take eight minutes for us to know that it disappeared on earth. Or another way, just to put it in the sense of this jet airplane-- let's get the calculator back out. So we're talking about 150 million kilometers." + }, + { + "Q": "At 3:32, how do you know the speed of a bullet.", + "A": "The muzzle velocity of standard rounds for guns are well known. The manufacturing is well controlled so that they will have predictable trajectories when fired.", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 212 + ], + "3min_transcript": "You might have taken a 12- or 15-hour flight that gets you-- not all the way around the earth-- but gets you pretty far. San Francisco to Australia, or something like that. So right now these aren't scales that are too crazy. Although, even for me, the earth itself is a pretty mind-blowingly large object. Now, with that out of the way let's think about the sun. Because the sun starts to approach something far huger. So this obviously here is the sun. And I think most people appreciate that the sun is much larger than the earth, and that it's pretty far away from the earth. But I don't think most people, including myself, fully appreciate how large the sun is or how far it is away from the earth. So just to give you a sense, the sun is 109 times the circumference of the earth. if we said, OK, if I'm traveling at the speed of a bullet or the speed of a jetliner, it would take me 40 hours to go around the earth. Well, how long would it take to go around the sun? So if you were to get on a jet plane and try to go around the sun, or if you were to somehow ride a bullet and try to go around the sun-- do a complete circumnavigation of the sun-- it's going to take you 109 times as long as it would have taken you to do the earth. So it would be 100 times-- I could do 109, but just for approximate-- it's roughly 100 times the circumference of the earth. So 109 times 40 is equal to 4,000 hours. And just to get a sense of what 4,000 is-- actually, since I have the calculator out, let's do the exact calculation. It's 109 times the circumference of the earth times 40 hours. So it's 4,360 hours to circumnavigate the sun, going at the speed of a bullet or a jetliner. And so that is-- 24 hours in the day-- that is 181 days. It would take you roughly half a year to go around the sun at the speed of a jetliner. Let me write this down. Half a year. The sun is huge. Now, that by itself may or may not be surprising--and actually let me give you a sense of scale here, because I have this other diagram of a sun. And we'll talk more about the rest of the solar system in the next video. But over here, at this scale, the sun, at least on my screen-- if I were to complete it, it would probably be about 20 inches in diameter." + }, + { + "Q": "At around 1:21, Sal mentions that the Earth is approximately 40,000 km. About how many miles is that?", + "A": "To convert to miles just divide by 1.6; so 25,000", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 81 + ], + "3min_transcript": "My goal in this video and the next video is to start giving a sense of the scale of the earth and the solar system. And as we see, as we start getting into to the galaxy and the universe, it just becomes almost impossible to imagine. But we'll at least give our best shot. So I think most of us watching this video know that this right here is earth. And just to get a sense of scale here, I think probably the largest distance that we can somehow relate to is about 100 miles. You can get into a car for an hour, hour and a half, and go about 100 miles. And on the earth that would be about this far. It would be a speck that would look something like that. That is 100 miles. And also to get us a bit of scale, let's think about a speed that at least we can kind of comprehend. And that would be, maybe, the speed of a bullet. Maybe we can't comprehend it, but I'll say this is the fastest thing that we could maybe comprehend. depending on the type of gun and all of that-- about 280 meters per second, which is about 1,000 kilometers per hour. And this is also roughly the speed of a jet. So just to give a sense of scale here, the earth's circumference-- so if you were to go around the planet-- is about 40,000 kilometers. So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the earth. You might have taken a 12- or 15-hour flight that gets you-- not all the way around the earth-- but gets you pretty far. San Francisco to Australia, or something like that. So right now these aren't scales that are too crazy. Although, even for me, the earth itself is a pretty mind-blowingly large object. Now, with that out of the way let's think about the sun. Because the sun starts to approach something far huger. So this obviously here is the sun. And I think most people appreciate that the sun is much larger than the earth, and that it's pretty far away from the earth. But I don't think most people, including myself, fully appreciate how large the sun is or how far it is away from the earth. So just to give you a sense, the sun is 109 times the circumference of the earth." + }, + { + "Q": "in 6:10 why does sal multiply both sides of the equation by 2?", + "A": "to remove the 2 times acceleration which is in the denominator in the RHS of the equation.", + "video_name": "2ZgBJxT9pbU", + "timestamps": [ + 370 + ], + "3min_transcript": "that goes by-- this is our change in velocity. So elapsed time is the same thing. I write it over here-- is our change in velocity divided by our acceleration. And just to make sure you understand this, it just comes straight from the idea that acceleration-- or let me write it this way-- that change in velocity is just acceleration times time. Or I should say, acceleration times change in time. So if you divide both sides of this equation by acceleration, you get this right over here. So that is what our displacement-- Remember, I want an expression for displacement in terms of the things we know and the one thing that we want to find out. Well, for this example right over here, we know a couple of things. Well actually, let me take it step by step. We know that our initial velocity is 0. the average velocity is going to be our final velocity divided by 2, since our initial velocity is 0. Our change in velocity is the same thing as final velocity minus initial velocity. And once again, we know that the initial velocity is 0 here. So our change in velocity is the same thing as our final velocity. So once again, this will be times. Instead of writing change in velocity here, we could just write our final velocity because we're starting at 0. Initial velocity is 0. So times our final velocity divided by our acceleration. Final velocity is the same thing as change in velocity because initial velocity was 0. And all of this is going to be our displacement. And now it looks like we have everything written in things we know. or both sides of this equation by 2 times our acceleration on that side. And we multiply the left-hand side by-- I'll do the same colors-- 2 times our acceleration. On the left hand side, we get 2 times our acceleration times our displacement is going to be equal to, on the right hand side, the 2 cancels out with the 2, the acceleration cancels out with the acceleration-- it will be equal to the velocity, our final velocity squared. Final velocity times final velocity. And so we can just solve for final velocity here. So we know our acceleration is negative 9.8 meters" + }, + { + "Q": "At 9:48, wouldn't it be more accurate to express Vf as Vf = - |sqrt(19.6h)| m/s?", + "A": "Yes, but by convention when we use the square root sign we mean the positive root of the value.", + "video_name": "2ZgBJxT9pbU", + "timestamps": [ + 588 + ], + "3min_transcript": "Our convention is very important here. So our displacement over here is going to be negative h meters. So this is the variable, and this is the shorthand for meters. So when you multiply these two things out, lucky for us these negatives cancel out, and you get 19.6h meters squared per second squared is equal to our final velocity squared. And notice, when you square something you lose the sign information. If our final velocity was positive, you square it, you still get a positive value. If it was negative and you square it, you still get a positive value. But remember, in this example, we're going to be moving downward. So we want the negative version of this. So to really figure out our final velocity, of both sides of this equation. So if we were to take the square root of both sides of this, you take the square root of that side, you take the square root of that side, you will get-- and I'll flip them around-- your final velocity, we could say, is equal to the square root of 19.6h. And you can even take the square root of the meter squared per second squared, treat them almost like variables, even though they're units. And then outside of the radical sign, you will get a meters per second. And the thing I want to be careful here is if we just take the principal root here, the principal root here is the positive square root. But we know that our velocity is going to be downwards here, because that is our convention. So we want to make sure we get the negative square root. So let's try it out with some numbers. We've essentially solved what we set out to solve at the beginning of this video, how fast would we be falling, as a function of the height. Well, let's try it out with something. let's say the height is 5 meters, which would be probably jumping off of a or throwing a rock off of a one-story, maybe a commercial one-story building. That's about 5 meters, would be about 15 feet. So yeah, about the roof of a commercial building, give or take. So let's turn it on. And so what do we get? If we put 5 meters in here, we get 19.6 times 5 gives us 98. So almost 100. And then, we want to take the square root of that, so it's going to be almost 10. So the square root of 98 gives us roughly 9.9. And we want the negative square root of that. in that situation, when the height is 5 meters-- So if you jump off of a one-story commercial building, right at the bottom, or if you throw a rock off that, right" + }, + { + "Q": "At 4:00, how does he get (final velocity + initial velocity) / 2 ?", + "A": "That s how you calculate an average of two numbers. You add them up and divide by 2.", + "video_name": "2ZgBJxT9pbU", + "timestamps": [ + 240 + ], + "3min_transcript": "that positive velocity means upwards, or a positive vector means up, a negative vector means down. So we're going to have an initial velocity over here of 0. And then at the bottom we're going to have some final velocity here that is going to be a negative number. So it's going to have some negative value over here. So this is going to be negative. This is going to be a negative number right over there. And we know that the acceleration of gravity for an object on free fall, an object in free fall near the surface of the earth. We know it, and we're going to assume that it's constant. So our constant acceleration is going to be negative 9.8 meters per second squared. and given that their initial velocity is 0 and that our acceleration is negative 9.8 meters per squared, we want to figure out what our final velocity is going to be right before we hit the ground. We're going to assume that this h is given in meters, right over here. And we'll get an answer in meters per second for that final velocity. So let's see how we can figure it out. So we know some basic things. And the whole point of these is to really show you that you can always derive these more interesting questions from very basic things that we know. So we know that displacement is equal to average velocity times change in time. And we know that average velocity-- if we assume acceleration is constant, which we are doing-- average velocity is the final velocity plus the initial velocity over 2. that goes by-- this is our change in velocity. So elapsed time is the same thing. I write it over here-- is our change in velocity divided by our acceleration. And just to make sure you understand this, it just comes straight from the idea that acceleration-- or let me write it this way-- that change in velocity is just acceleration times time. Or I should say, acceleration times change in time. So if you divide both sides of this equation by acceleration, you get this right over here. So that is what our displacement-- Remember, I want an expression for displacement in terms of the things we know and the one thing that we want to find out. Well, for this example right over here, we know a couple of things. Well actually, let me take it step by step. We know that our initial velocity is 0." + }, + { + "Q": "At 9:13 when both sides are square rooted, how come the 19.6 doesnt become 4.42?", + "A": "It is true that you could write 4.42 instead of \u00e2\u0088\u009a19.6, but you would lose the accuracy of the number. 19.6 is not equal to 4.42, rather 4.4271887242357310647984509622058. And yet, though it is more accurate, it is still not. If you get an irrational number, just keep it in the \u00e2\u0088\u009a form.", + "video_name": "2ZgBJxT9pbU", + "timestamps": [ + 553 + ], + "3min_transcript": "So let me write this over here. So this is negative 9.8. So we have 2 times negative 9.8-- let me just multiply that out. So that's negative 19.6 meters per second squared. And then what's our displacement going to be? What's the displacement over the course of dropping this rock off of this ledge or off of this roof? So you might be tempted to say that our displacement is h. But remember, these are vector quantities, so you want to make sure you get the direction right. From where the rock started to where it ends, what's it doing? It's going to go a distance of h, but it's going to go a distance of h downwards. And our convention is down is negative. So in this example, our displacement from when it leaves your hand to when it hits the ground, the displacement is going to be equal to negative h. It's going to travel a distance of h, but it's going to travel that distance downwards. Our convention is very important here. So our displacement over here is going to be negative h meters. So this is the variable, and this is the shorthand for meters. So when you multiply these two things out, lucky for us these negatives cancel out, and you get 19.6h meters squared per second squared is equal to our final velocity squared. And notice, when you square something you lose the sign information. If our final velocity was positive, you square it, you still get a positive value. If it was negative and you square it, you still get a positive value. But remember, in this example, we're going to be moving downward. So we want the negative version of this. So to really figure out our final velocity, of both sides of this equation. So if we were to take the square root of both sides of this, you take the square root of that side, you take the square root of that side, you will get-- and I'll flip them around-- your final velocity, we could say, is equal to the square root of 19.6h. And you can even take the square root of the meter squared per second squared, treat them almost like variables, even though they're units. And then outside of the radical sign, you will get a meters per second. And the thing I want to be careful here is if we just take the principal root here, the principal root here is the positive square root. But we know that our velocity is going to be downwards here, because that is our convention. So we want to make sure we get the negative square root. So let's try it out with some numbers. We've essentially solved what we set out to solve at the beginning of this video, how fast would we be falling, as a function of the height. Well, let's try it out with something." + }, + { + "Q": "at 3:34 how come it is called the swimmers view?", + "A": "Because one arm is raised up by the patient s head as if they were swimming the freestyle stroke.", + "video_name": "cbkTTluHaTw", + "timestamps": [ + 214 + ], + "3min_transcript": "is blocking it. DR. MAHADEVAN: Exactly. You can see that that big white thing there is the shoulder that's gotten in the way. And it's making it hard to see. SAL KHAN: They shouldn't have worn those lead shoulder pads. DR. MAHADEVAN: [LAUGHING] And it's making it hard to see whether there's something going on down there right now. So it's really a mystery, as you've shown. SAL KHAN: So how do you solve this problem? DR. MAHADEVAN: If you look over at the other film, it's what we call a swimmer's view. And what we've asked the patient to do is raise one arm up and lower the other. And in doing so, you kind of clear that lower cervical spine and allow better visualization of the entire spine. SAL KHAN: I see. And you're taking it from the direction of the raised arm, on the side of raised arm. DR. MAHADEVAN: You take it from the side. And you can see. SAL KHAN: This is the raised arm right over here. DR. MAHADEVAN: Exactly, that's the raised arm. SAL KHAN: I see. And the other arm on the further side of the patient is down. And that's what allows us to get to the shoulder in a position, so it doesn't block like it does in this left view. DR. MAHADEVAN: Exactly, exactly. SAL KHAN: I see. And over here, it is much clearer And OK. So let me see. So we can count. This is number one right up here. DR. MAHADEVAN: That's one. SAL KHAN: One, two, three, four, five, six-- yeah, we already got to six. We didn't see six over here. And then we got seven. DR. MAHADEVAN: Exactly. SAL KHAN: OK. And so you would call this is an adequate view for what we're trying-- of the neck, because now we can look at all the way DR. MAHADEVAN: Absolutely. We can get all the way down to seven. And ideally, you want to see the top of one, which comes-- actually, in this counting system, we go one through seven. And then we start back at one again, because we're starting with the thoracic vertebrae. SAL KHAN: Oh, look at that. It's like with those streets, where they restart numbering. And you can't find it. So it becomes one again. DR. MAHADEVAN: Exactly. SAL KHAN: Did I number that right? And again, we're looking more to the front. You've got your numbers perfectly on every spinous process, the little bump that you can feel, if you press on the back of your neck. But what we're really interested is the alignment of the front of the vertebral bodies. SAL KHAN: So this is one. This is two, three, four, five, six, seven. Where's the top of one? DR. MAHADEVAN: If you just continue down right there. And it sometimes is difficult to see. But exactly, you want to see that there's alignment right in front of-- I'll assume that there's something here that I can't really see. But you're an expert. So maybe you see things that I don't. OK, so now what do we do with this? DR. MAHADEVAN: Now, we've shown you that you can get a swimmer's view. And it can show you all the way down to C7, T1. But on the original view, as you've shown, we can't see that. So what we did for this patient was get a swimmer's view. SAL KHAN: I see. So this is adequate. And we have this other slide right over here. We have this other one right over here. And why is this one interesting? DR. MAHADEVAN: This is the same patient. And now we've taken that same view that we talked about before, the swimmer's view, where you've got-- SAL KHAN: This is the same patient as this patient right over here, not this patient over here SAL KHAN: Because that one looked overall pretty healthy. DR. MAHADEVAN: That was a normal swimmer's view. But here is an abnormal swimmer's view." + }, + { + "Q": "3:40 why did carbon have 'too many bonds' isnt it perfectly stable with 4 bonds and 0 formal charge?", + "A": "You re forgetting the implied hydrogen on each carbon, benzene has the formula C6H6, each carbon is bonded to 2 other carbons and 1 hydrogen", + "video_name": "oxf0LMJTklg", + "timestamps": [ + 220 + ], + "3min_transcript": "on this oxygen. I'll make it magenta. That lone pair is next to the pi bond. The one in red and so, we can go ahead and draw a resonance structure and we take these electrons in magenta and move then into here. That would mean too many bonds to this carbon. We take the electrons in red and we push them off onto this carbon. Let's go ahead and draw our resonance structure. We have our ring here and we have now a double bond between the oxygen and the carbon. Only two lone pairs of electrons on this oxygen now. The electrons in magenta move in here to form a pi bond and the electrons in red move off onto this carbon right here. That's gonna give that carbon a -1 formal charge. Let's go ahead and draw a -1 formal charge here. The electrons in blue have not moved and the electrons in green, We put those in there like that. All right, next, we have the exact same pattern that we did before. We have a lone pair of electrons next to a pi bond. The lone pair of electrons are the electrons in red right here. Next to a pi bond, the electrons in blue. Let's go ahead and draw another resonance structure. We could take these electrons in red, push them into here. That would mean too many bonds to this carbon. If you take these electrons in blue and push them off onto this carbon. Let's draw that resonance structure. Once again, we have our carbon double bonded to an oxygen up here. We said that these electrons were the ones in magenta and the electrons in red moving here to form a pi bond. The electrons in blue move off onto this carbon and that gives this carbon a -1 formal charge. This carbon has a -1 formal charge. that has the blue electrons on it. We still have our electrons in green over here and we have the exact same pattern. We have a lone pair next to a pi bond. The lone pair are the ones in blue and this time, the pi bond are the electrons in green here. We can draw yet another resonance structure. We could take the electrons in blue, move them into here. That would mean too many bonds to this carbon. To take the electrons in green and push them off onto that carbon and let's draw that resonance structure. Once again, we have our ring. We draw our ring in here. We have this double bond up here. Put in lone pairs of electrons on the oxygen and these electrons were the ones in magenta and we go around the ring. We had our electrons in red right here. The electrons in blue move into here and finally, the electrons in green move off onto this carbon. This carbon right here in green." + }, + { + "Q": "at 8:38 aren't the both of the products same molecule flipped around ?", + "A": "I don t know which way you want to flip, but the answer is No . They are different molecules. You can flip them any way you want, and you will not be able to make all the bonds coincide with each other.", + "video_name": "fSk1Crn3R2E", + "timestamps": [ + 518 + ], + "3min_transcript": "All I have to do is take away that double bond and I'm done. Well sometimes that's true. But in this case, we actually formed two new chirality centers, right? So this top carbon here is a chirality center, and this bottom carbon here is also a chirality center. So sometimes it's not quite that simple. We need to think about the syn addition of those hydrogens when you think about the possible products that would result. So we're going to get two products here. Let's look at the one on the left. Well, one possibility is I can add those two hydrogens on the same side as a wedge, right? So I have one hydrogen as a wedge, the other hydrogen as a wedge. That's our syn addition. And that means that this top carbon here, this ethyl group, must be going away from me. And down here at the bottom carbon, the methyl group must be going away from me. So that's one possible product. The other possibility, instead of having my two hydrogens So there's a hydrogen and then here's a dash, and there's a hydrogen. So at this top carbon here, now my ethyl group is coming out at me. And at this bottom carbon now my methyl group is coming out at me, like this. So I have two possibilities. And if I look at these two products, I can see that they are enantiomers, they are mirror images of each other. So these two would be my enantiomers, and these would be the products of my reaction. So be very careful when thinking about syn additions here. Let's do one more example of a hydrogenation reaction. Let's do a bridged bicyclic compound. So let's look at a famous bridged bicyclic compound. Let's see if we can draw it here. off like that. And my double bond is going to go right here. And then this is going to be a methyl group. And then up here there are going to be two methyl groups, like that. So this is alpha-pinene, found in turpentine. And you can see there's an alkene on this. So if I took this alpha-pinene molecule and I wanted to hydrogenate it, I could use palladium and charcoal, palladium and carbon. And if I think about what happens in this mechanism, I know that my metal catalyst there, my palladium, is going to be flat, like that. And so, when it has those hydrogens, when the palladium adsorbs those hydrogens, it's going to add those two hydrogens to my double bond, think about this guy over here, think about the alpha-pinene as molecules like a spaceship, And the spaceship is approaching the docking station." + }, + { + "Q": "At 9:32, are we supposed to just know that charcoal (C) is needed with our metal (Pd), or is there some rule being followed?", + "A": "This is just something to know. When the Pd is spread over carbon, it greatly increases the available surface area, and thus the efficiency, of the catalyst. Thus, practically, a palladium hydrogenation is usually palladium on carbon (written as Pd/C).", + "video_name": "fSk1Crn3R2E", + "timestamps": [ + 572 + ], + "3min_transcript": "So there's a hydrogen and then here's a dash, and there's a hydrogen. So at this top carbon here, now my ethyl group is coming out at me. And at this bottom carbon now my methyl group is coming out at me, like this. So I have two possibilities. And if I look at these two products, I can see that they are enantiomers, they are mirror images of each other. So these two would be my enantiomers, and these would be the products of my reaction. So be very careful when thinking about syn additions here. Let's do one more example of a hydrogenation reaction. Let's do a bridged bicyclic compound. So let's look at a famous bridged bicyclic compound. Let's see if we can draw it here. off like that. And my double bond is going to go right here. And then this is going to be a methyl group. And then up here there are going to be two methyl groups, like that. So this is alpha-pinene, found in turpentine. And you can see there's an alkene on this. So if I took this alpha-pinene molecule and I wanted to hydrogenate it, I could use palladium and charcoal, palladium and carbon. And if I think about what happens in this mechanism, I know that my metal catalyst there, my palladium, is going to be flat, like that. And so, when it has those hydrogens, when the palladium adsorbs those hydrogens, it's going to add those two hydrogens to my double bond, think about this guy over here, think about the alpha-pinene as molecules like a spaceship, And the spaceship is approaching the docking station. The spaceship is going to approach the docking station. And there's only one way the spaceship can approach the docking station. And that is the way in which we have drawn it right here. It could not flip upside down and approach it from the top, because of the steric hindrance of these methyl groups. Right? So this is the way that it approaches. In this part of the molecule, your alkene, is the flat part, right? So it's easiest for the molecule to approach in this way. The spaceship analogy always helps my students. So there's only one product for this reaction. And let's see if we can draw it here. And let's see what it would look like. It would look something like this. So we have our two methyl groups right here. So the hydrogens are going to add from below, right? So this hydrogen, let's say it adds right here. That's going to push this methyl group up." + }, + { + "Q": "ok, wait. for statement two, or 3:30, an unbalanced force doesn't necessarily mean tethering the moving object. what if the object hit a wall? then it would slow down, right? or if it rubbed against a surface, it might change direction, but it would also change speed. how is statement 2 wrong?", + "A": "hitting a wall and rubbing against a surface are unbalanced forces and they change the motion of the object", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 210 + ], + "3min_transcript": "its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope, they were traveling in when they let go. They'll keep going on in that direction. And if we assume very, very, very small frictions from the ice skating rink, they'll actually have the same speed. So the force, the inward force, the tension from the rope pulling on the skater in this situation, would have only changed the skater's direction. So and unbalanced force doesn't necessarily have to impact the object's speed. It often does. But in that situation, it would have only impacted the skater's direction. Another situation like this-- and once again, this involves centripetal acceleration, inward forces, inward acceleration-- is a satellite in orbit, or any type of thing in orbit. So if that is some type of planet, and this is one of the planet's moons right over here, the reason why it stays in orbit is because the pull of gravity keeps making the object change its direction, but not its speed." + }, + { + "Q": "At 3:58,Sal said that the satellite changes its direction but not its speed.So if that same planet is moving around the sun, shouldn't the speed of the satellite vary if it has to go around the planet?", + "A": "I think when the satellite is in the geo stationary orbit its velocity is constant other wise the satellite velocity changes like a planet around the sun", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 238 + ], + "3min_transcript": "impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope, they were traveling in when they let go. They'll keep going on in that direction. And if we assume very, very, very small frictions from the ice skating rink, they'll actually have the same speed. So the force, the inward force, the tension from the rope pulling on the skater in this situation, would have only changed the skater's direction. So and unbalanced force doesn't necessarily have to impact the object's speed. It often does. But in that situation, it would have only impacted the skater's direction. Another situation like this-- and once again, this involves centripetal acceleration, inward forces, inward acceleration-- is a satellite in orbit, or any type of thing in orbit. So if that is some type of planet, and this is one of the planet's moons right over here, the reason why it stays in orbit is because the pull of gravity keeps making the object change its direction, but not its speed. So this was its speed right here. If the planet wasn't there, it would just keep going on in that direction forever and forever. But the planet right over here, there's an inward force of gravity. And we'll talk more about the force of gravity in the future. But this inward force of gravity is going to accelerate this object inwards while it travels. And so after some period of time, this object's velocity vector-- if you add the previous velocity with how much it's changed its new velocity vector. Now this is after its traveled a little bit-- its new velocity vector might look something like this. And it's traveling at the exact right speed so that the force of gravity is always at a right angle to its actual trajectory. It's the exact right speed so it doesn't go off into deep space and so it doesn't plummet into the earth. And we'll cover that in much more detail. But the simple answer is, unbalanced force on a body" + }, + { + "Q": "at 2:56, it says \"and they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. wound't they eventually stop from the friction on the ice?", + "A": "yes but he is imagining ice that is so slippery that it has no friction, so that we can get the idea of what it means for an object to have no net force on it", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 176 + ], + "3min_transcript": "its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope, they were traveling in when they let go. They'll keep going on in that direction. And if we assume very, very, very small frictions from the ice skating rink, they'll actually have the same speed. So the force, the inward force, the tension from the rope pulling on the skater in this situation, would have only changed the skater's direction. So and unbalanced force doesn't necessarily have to impact the object's speed. It often does. But in that situation, it would have only impacted the skater's direction. Another situation like this-- and once again, this involves centripetal acceleration, inward forces, inward acceleration-- is a satellite in orbit, or any type of thing in orbit. So if that is some type of planet, and this is one of the planet's moons right over here, the reason why it stays in orbit is because the pull of gravity keeps making the object change its direction, but not its speed." + }, + { + "Q": "At 1:10, why the first statement is true?", + "A": "Why wouldn t it be true? All the statement is doing is rephrasing Newton s First Law.", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 70 + ], + "3min_transcript": "Now that we know a little bit about Newton's First Law, let's give ourselves a little quiz. And what I want you to do is figure out which of these statements are actually true. And our first statement is, \"If the net force on a body is zero, its velocity will not change.\" Interesting. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" Also an interesting statement. Statement number three, \"The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.\" And statement four, \"An unbalanced force on an object will always change the object's direction.\" So I'll let you think about that. So let's think about these statement by statement. So our first statement right over here, its velocity will not change.\" This is absolutely true. This is actually even another way of rephrasing Newton's First Law. If I have some type of object that's just traveling through space with some velocity-- so it has some speed going in some direction, and maybe it's deep space. And we can just, for purity, assume that there's no gravitational interactions. There will always be some minuscule ones, but we'll assume no gravitational interactions. Absolutely no particles that it's bumping into, absolute vacuum of space. This thing will travel on forever. Its velocity will not change. Neither its speed nor its direction will change. So this one is absolutely true. Statement number two, \"An unbalanced force on a body will always impact the object's speed.\" And the key word right over here is \"speed.\" If I had written \"impact the object's velocity,\" then this would be a true statement. impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope," + }, + { + "Q": "Once Sal has simplified the equation (12:51) to be 3PV=mv^2 N he divides the equation by two. This takes on the form 3/2PV=(mv^2/2 N). Why doesn't the number of molecules in the system also get divided by two when he divides the two halves of the equation by two? It seems to be exempt from his division of two on both sides of the equation. Shouldn't the equation look like 3/2PV=(mv^2 N)/2?", + "A": "If you remember order of operations, division and multiplication can be done in either order. He wrote it as (mv^2)/2 *N so that you could see that 1/2 * mv^2 is kinetic energy", + "video_name": "qSFY7GKhSRs", + "timestamps": [ + 771 + ], + "3min_transcript": "This is the pressure from one particle. If I wanted the pressure from all of the particles on that wall-- so the total pressure on that wall is going to be from N over 3 of the particles. The other particles aren't bouncing off that wall. So we don't have to worry about them. So if we want the total pressure on that wall-- I'll just write, pressure sub on the wall. Total pressure on the wall is going to be the pressure from one particle, mv squared, over our volume, times the total number of particles hitting the wall. The total number of particles is N divided by 3, because only 3 will be going in that direction. So, the total pressure on that wall is equal to mv squared, over our volume of our container, times the total particles divided by 3. Let's see if we can manipulate this thing a little bit. So if we multiply both sides by-- let's see what we can do. to mv squared, times N, where N is the number of particles. Let's divide both sides by N. So we get 3pv over-- actually, no, let me leave the N there. Let's divide both sides of this equation by 2. So we get, what do we get? We get 3/2 pv is equal to-- now this is interesting. It's equal to N, the number of particles we have, times mv squared over 2. Remember, I just divided this equation right here by 2 to get this. And I did this for a very particular reason. mv squared over 2 is the kinetic energy of that little particle we started off with. That's the formula for kinetic energy. Kinetic energy is equal to mv squared over 2. So this is the kinetic energy of one particle. Now, we're multiplying that times the total number of particles we have, times N. So N times the kinetic energy of one particle is going to be the kinetic energy of all the particles. And, of course, we also made another assumption. I should state that I assumed that all the particles are moving with the same velocity and have the same mass. In a real situation, the particles might have very different velocities. But this was one of our simplifying assumptions. So, we just assumed they all have that. So, if I multiply N times that-- this statement right here-- is the kinetic energy of the system." + }, + { + "Q": "At 2:35, it is said that experimental studies have shown that the structure of the amide is planar, which means the molecule's sp2 hybridized resonance structure contributes more towards the resonance hybrid. But the stability of the first structure (with sp3 hybrid N) is greater due to zero formal charge. Why is that the stable structure does not contribute more to the resonance hybrid?", + "A": "Although the first structure does not have any formal charges, the electronegativity of the oxygen atom does give it a partial negative charge. The resonance on the second structure is a stabilizing factor, and its sp2 hybridization is more stable than the sp3 hybridization of the first structure. (Covered later in the section on acid-base chemistry.)", + "video_name": "kQCS1AhAnMI", + "timestamps": [ + 155 + ], + "3min_transcript": "force us to push some pi electrons off, onto this oxygen, so let's go ahead and draw the other resonance structure. So this top oxygen would now have three lone pairs of electrons around it, a negative one formal charge, and there'd be a double-bond between this carbon and this nitrogen, so let's go ahead and draw in everything. This nitrogen how has a plus one formal charge, and we can go ahead and complete our resonance bracket here. So let's follow those electrons along, the electrons in magenta, the lone pair of electrons moved in here to form our pi bond, and the pi electrons over here in blue, came off onto the oxygen, to give the oxygen a negative one formal charge. All right, let's now calculate a steric number for this nitrogen, in our second resonance structure. So, steric number is equal to number of sigma bonds: So here's a sigma bond, here's a sigma bond, and our and one of them is a pi bond. So we have a total of three sigma bonds around our nitrogen, zero lone pairs of electrons, so three plus zero gives us a steric number of three; that implies three hybrid orbitals, which means SP two hybridization, and a trigonal planar geometry around that nitrogen, so a planar geometry. And, here we've shown our electrons being de-localized, so the lone pair of electrons are de-localized due to resonance, and so, experimental studies have shown that the amide function group is planar, so these atoms actually are planar here, which means that the electrons in magenta are not localized to that nitrogen; they are actually de-localized. And so, that implies this nitrogen is SP two hybridized, and has a P orbital, and that allow that lone pair of electrons in magenta to be de-localized. And so, here's a situation where drawing actually happening: That lone pair is participating in resonance, which makes this nitrogen SP two hybridized, so it has a P orbital. All right, let's look at this example down here, and let's look first at this left side of the molecule, and so we can see our amide functional group, and if I look at the lone pair of electrons on this nitrogen, we've just talked about the fact that this lone pair of electrons is actually de-localized, so this lone pair is participating in resonance. And so, that affects the geometry, and how you think about the hybridization of this nitrogen, here. So, the electrons in magenta are de-localized because they participate in resonance, and if I think about, let's make this a different color here; let's make these electrons in here blue, so the electrons in blue" + }, + { + "Q": "What is the proper usage of the spelling (5:35), blastocoele vs. blastocoel? My textbook says blastocoele, and I am just curious. Thanks!", + "A": "both are right", + "video_name": "-yCIMk1x0Pk", + "timestamps": [ + 335 + ], + "3min_transcript": "to start filling in some of this gap between the embryoblast and the trophoblast, so you're going to start having some fluid that comes in there, and so the morula will eventually look like this, where the trophoblast, or the outer membrane, is kind of this huge sphere of cells. And this is all happening as they keep replicating. Mitosis is the mechanism, so now my trophoblast is going to look like that, and then my embryoblast is going to look like this. Sometimes the embryoblast-- so this is the embryoblast. Sometimes it's also called the inner cell mass, so let me write that. And this is what's going to turn into the organism. And so, just so you know a couple of the labels that are organism, and we are mammals, we call this thing that the morula turned into is a zygote, then a morula, then the cells of the morula started to differentiate into the trophoblast, or kind of the outside cells, and then the embryoblast. And then you have this space that forms here, and this is just fluid, and it's called the blastocoel. A very non-intuitive spelling of the coel part of blastocoel. But once this is formed, this is called a blastocyst. That's the entire thing right here. Let me scroll down a little bit. This whole thing is called the blastocyst, and this is the case in humans. Now, it can be a very confusing topic, because a lot of times in a lot of books on biology, you'll say, hey, you go from the morula to the blastula or the Let me write those words down. So sometimes you'll say morula, and you go to blastula. Sometimes it's called the blastosphere. And I want to make it very clear that these are essentially the same stages in development. These are just for-- you know, in a lot of books, they'll start talking about frogs or tadpoles or things like that, and this applies to them. While we're talking about mammals, especially the ones that are closely related to us, the stage is the blastocyst stage, and the real differentiator is when people talk about just blastula and blastospheres. There isn't necessarily this differentiation between these outermost cells and these embryonic, or this embryoblast, or this inner cell mass here. But since the focus of this video is humans, and really that's where I wanted to start from, because that's what we are and that's what's interesting, we're going to" + }, + { + "Q": "At 6:00 what is the difference between MORULA and BLASTOCYST??", + "A": "A morula is a special kind of blastocyst, it has 16 cells assembled in a solid block of cells and looks like a mulberry/ Latin: morula. A proper blastocyst is a few days older, has more cell and forms a hollow sphere.", + "video_name": "-yCIMk1x0Pk", + "timestamps": [ + 360 + ], + "3min_transcript": "to start filling in some of this gap between the embryoblast and the trophoblast, so you're going to start having some fluid that comes in there, and so the morula will eventually look like this, where the trophoblast, or the outer membrane, is kind of this huge sphere of cells. And this is all happening as they keep replicating. Mitosis is the mechanism, so now my trophoblast is going to look like that, and then my embryoblast is going to look like this. Sometimes the embryoblast-- so this is the embryoblast. Sometimes it's also called the inner cell mass, so let me write that. And this is what's going to turn into the organism. And so, just so you know a couple of the labels that are organism, and we are mammals, we call this thing that the morula turned into is a zygote, then a morula, then the cells of the morula started to differentiate into the trophoblast, or kind of the outside cells, and then the embryoblast. And then you have this space that forms here, and this is just fluid, and it's called the blastocoel. A very non-intuitive spelling of the coel part of blastocoel. But once this is formed, this is called a blastocyst. That's the entire thing right here. Let me scroll down a little bit. This whole thing is called the blastocyst, and this is the case in humans. Now, it can be a very confusing topic, because a lot of times in a lot of books on biology, you'll say, hey, you go from the morula to the blastula or the Let me write those words down. So sometimes you'll say morula, and you go to blastula. Sometimes it's called the blastosphere. And I want to make it very clear that these are essentially the same stages in development. These are just for-- you know, in a lot of books, they'll start talking about frogs or tadpoles or things like that, and this applies to them. While we're talking about mammals, especially the ones that are closely related to us, the stage is the blastocyst stage, and the real differentiator is when people talk about just blastula and blastospheres. There isn't necessarily this differentiation between these outermost cells and these embryonic, or this embryoblast, or this inner cell mass here. But since the focus of this video is humans, and really that's where I wanted to start from, because that's what we are and that's what's interesting, we're going to" + }, + { + "Q": "at around 3:56 , does this mean that oxygen can have an oxidation state of 1+ when bonded with any other element in it's group ?", + "A": "No, oxygen is more electronegative than the other elements in its group so it will be -2 with all of them.", + "video_name": "R2EtXOoIU-E", + "timestamps": [ + 236 + ], + "3min_transcript": "And if they were hypothetically ionic bonds, what would happen? Well, if you had to give these electrons to somebody, you would give them to the oxygen, the electrons in this period, give them to the oxygen, giving it an oxidation state of negative 1. With the hydrogen having these electrons taken away, it's going to have an oxidation state of positive 1. And the same thing's going to be true for that oxygen and that hydrogen right over there. So this is fascinating, because this is an example where oxygen has an oxidation state not of negative 2, but an oxidation state of negative 1. So this is already kind of interesting. Now it gets even more interesting when we go to oxygen difluoride. Why is this more interesting? Because fluorine is the one thing on this entire table that is more electronegative than oxygen. This is a covalent bond, but in our hypothetical ionic bond, if we had to give these electrons to one So the fluorine, each of them would have an oxidation state of negative 1. And the oxygen here-- now, you could imagine, this is nuts for oxygen. The oxidation state for oxygen, it's giving up these electrons. It would be a positive 2. And we talk about oxidation states when we write this little superscript here. We write the sign after the number. And that's just the convention. But it has an oxidation state of positive 2. Oxygen, the thing that likes to oxidize other things, it itself has been oxidized by fluorine. So this is a pretty dramatic example of how something might stray from what's typical oxidation state or it's typical oxidation number. And in general, oxygen will have an oxidation state or oxidation number in most molecules of negative 2. But unless it's bonded with another oxygen or it's bonded to fluorine, which is a much more but it's the only atom that is more electronegative than-- or the only element is more electronegative than oxygen." + }, + { + "Q": "At 6:33, Sal says vector v is the sum of is the sum the other two component vector and he has even proved it in his previous video. But, doesn't that break down the traditional Pythagorean Theorem?\nEven the result at the end of the video doesn't sum up to 10(value of vector v) but when we use the Pythagorean Theorem, the equation does satisfy. Anyone please explain .", + "A": "When you use the Pythagoras theorem, you find the magnitude of the vector. But, in this video, Sal was talking about the vectors, not just their magnitude. Hope this helps :)", + "video_name": "2QjdcVTgTTA", + "timestamps": [ + 393 + ], + "3min_transcript": "or sometimes it's called a caret character-- that tells us that it is a vector, but it is a unit vector. It has a magnitude of 1. And by definition, the vector j goes and has a magnitude of 1 in the positive y direction. So the y component of this vector, instead of saying it's 5 meters per second in the upwards direction or instead of saying that it's implicitly upwards because it's a vertical vector or it's a vertical component and it's positive, we can now be a little bit more specific about it. We can say that it is equal to 5 times j. Because you see, this magenta vector, it's going the exact same direction as j, it's just 5 times longer. I don't know if it's exactly 5 times. I'm just trying to estimate it right now. It's just 5 times longer. Now what's really cool about this is besides just being able to express the components as now able to do that-- which we did do, we're representing the components as explicit vectors-- we also know that the vector v is the sum of its components. If you start with this green vector right here and you add this vertical component right over here, you have head to tails. You get the blue vector. And so we can actually use the components to represent the vector itself. We don't always have to draw it like this. So we can write that vector v is equal to-- let me write it this way-- it's equal to its x component vector plus the y component vector. And we can write that, the x component vector is 5 square roots of 3 times i. And then it's going to be plus the y component, the vertical component, which is 5j, which is 5 times j. vector in two dimensions by some combination of i's and j's or some scaled up combinations of i's and j's. And if you want to go into three dimensions, and you often will, especially as the physics class moves on through the year, you can introduce a vector in the positive z direction, depending on how you want to do it. Although z is normally up and down. But whatever the next dimension is, you can define a vector k that goes into that third dimension. Here I'll do it in a kind of unconventional way. I'll make k go in that direction. Although the standard convention when you do it in three dimensions is that k is the up and down dimension. But this by itself is already pretty neat because we can now represent any vector through its components and it's also going to make the math much easier." + }, + { + "Q": "At 2:15 Sal mentions an \" important caveat \" , what exactly is that ?", + "A": "A caveat often means an exeption or a problem, ex. It is fun to ride a bike but there s a caveat, you might fall over and hurt yourself", + "video_name": "zA0fvwtvgvA", + "timestamps": [ + 135 + ], + "3min_transcript": "" + }, + { + "Q": "6:35 Isn't the inaccuracy in finding the work in each measured area coming from the fact that you are only measured the area of the rectangle instead of including the triangle above it as well? Assuming that the pressure changes form a straight line, in order to get an accurate value, couldn't you get the area of the rectangle, then use an equation to find the area of a right triangle to find the area above, then add both to find the correct area?", + "A": "I suggest that you watch the calculus videos, it will help to make sense of finding the areas under curves. You are right, when dV is not infinitely small, then it is not exactly a perfect rectangle. But using calculus, we can make it so that we are adding up all the rectangles where dV is so infinitely small that the little triangle at the top doesn t exist. I m sure Sal can explain this a lot better than me, in his Calculus videos :)", + "video_name": "M5uOIy-JTmo", + "timestamps": [ + 395 + ], + "3min_transcript": "This is after removing each of the pebbles, so that our pressure and volume macro states are always well defined. But in state 2, we now have a pressure low and volume is high. The volume is high, you can just see that, because we kept pushing the piston up slowly, slowly, trying to maintain ourselves in equilibrium so our macrostates are always defined. And our pressure is lower just because we could have the same number of particles, but they're just going to bump into the walls a little bit less, because they have a little bit more room to move around. And that's all fair and dandy. So this describes the path of our system as it transitioned or as it experienced this process, which was a quasi-static process. Everything was defined at every point. Now we said that the work done at any given point by the system is the pressure times the change in volume. Now, how does that relate to here? Change in volume is just a certain distance along this x-axis. This is a change in volume. We started off at this volume, and let's say when we removed one pebble we got to this volume. Now, we want to multiply that times our pressure. Since we did it over such a small increment, and we're so close to equilibrium, we could assume that our pressure's is roughly constant over that period of time. So we could say that this is the pressure over that period of time. And so how much work we did, it's this pressure over here, times this volume, which is the area of this rectangle right there. And for any of you all who've seen my calculus videos, this should start looking a little bit familiar. And then what about when we could take our next pebble? Well now our pressure is a little bit lower. This is our new pressure. Our pressure is a little bit lower. And we multiply that times our new change in volume-- times this change in volume-- and we have that increment of work. And if you keep doing that, the amount of work we do is essentially the area of all of these rectangles as we remove each pebble. And now you might say, especially those of you who haven't watched my calculus videos, gee, you know, this might be getting close, but the area of these rectangles isn't exactly the area of this curve. It's a little inexact. And what I would say is, well if you're worried about that, what you should do is use smaller increments of volume. And if you want to have smaller changes in volume along each step, what you do is you remove even smaller pebbles. And this goes back to trying to get to that ideal quasi-static process. So if you did that of, eventually the delta V's would get smaller and smaller and smaller, and the rectangles would get thinner and thinner and thinner. You'd have to do it over more and more steps. But eventually you'll get to a point, if you assume really small changes in our delta V." + }, + { + "Q": "At 13:54 the amount of work needed for the system to reach from state 1 to state 2 is larger than the amount of the whole cycle. Is it because when the system was returning from state 2 to state 1 it was doing work?", + "A": "You have it in reverse. From state 1 to state 2, the system is doing work on the surroundings. (Volume is increasing (work of expansion)). From state 2 to state 1, the surroundings are doing work on the system. (Volume is decreasing (work of compression)). Because the area under 1 to 2, is greater than the area under 2 to 1, the net effect is more work being done on the surroundings than is being done on the system.", + "video_name": "M5uOIy-JTmo", + "timestamps": [ + 834 + ], + "3min_transcript": "State 1. And I do something, you know, I'm in a quasi-static process and it, you know, it's doing something weird, and I get to state 2 here. And it's going in this direction. So my volume is increasing. So in this situation, what is the work done by the system? Easy enough, it's the area under this curve. Now let's say that I keep doing some type of quasi-static process, but it takes a different path. I'm doing something else, other than adding the marbles directly back. So my new path looks something like this to get back to state 1. So these arrows are going back. So now what is the work done to the system? well my volume is decreasing, so it's the area under the The area under the second curve is the work done to the system. So if I want to know what the net work the system did, going from state 1 to state 2, and then going back to state 1-- remember, this is a pressure and volume diagram-- what is it? Well the work that the system did was this whole area under this brown curve. And then it had some work done to it, which is the area under this magenta curve. So the net work it did is essentially the white, the whole area, minus this red area. So the net work it did would be essentially just the area inside this loop. And hopefully you don't have to know calculus to do this, although calculus you would actually use to compute these areas. But I just want to give you that intuition, that the area inside this closed loop is actually the amount of work that our system has done. And what's important is the direction that it's going. of this clockwise motion. This is the work that our system has done, which, I don't know, to me is a pretty interesting thing. And later we can use this notion to come up with some other ideas behind our state variables I'll make one little aside here. Remember, our state variable pressure volume, we did stuff to it then we went back to that state. That stayed the same. And I want to say another thing. For our purposes, when we're dealing with ideal gases, where the internal energy is essentially the kinetic energy of the system, if we go and do all sorts of crazy stuff and come back, our internal energy hasn't changed. So the internal energy is always going to be the same at this point. So if I said, I did all of this stuff and came back here, what is my change in internal energy? It's 0. The change is 0. Now if I said I went from here to here, I would have a different internal energy and my change would be something real. But since this is a state function, it doesn't care how I got there." + }, + { + "Q": "At 8:22 how does thermal radiation interact with our skin and transfer energy?Is it the magnetic and electric field that interacts or the photon of light that transfers heat?", + "A": "I was confused about this as well, seeing as the apparent cold air still exists between you and the fire as you are warming up. I m not sure that is true once the fire is active. You are correct in assuming that the photon transfers the heat. Photons can carry heat energy, even though they have no mass. This phenomenon also explains how you warm up when you step into the sunlight on a chilly day :)", + "video_name": "8GQvMt-ow4w", + "timestamps": [ + 502 + ], + "3min_transcript": "charged particles, are made up of protons and electrons. And so as you accelerate these and the more that you accelerate these the more radiation you are going to release. And you might say, Okay, you said I'm observing that in fire, where am I observing radiation? Well just the very fact that you can see the fire, the light emitted from the fire, that is electromagnetic radiation, that is electromagnetic radiation. It's just the electromagnetic radiation in the wavelengths that your eye considers to be visible light, or that your eye considers to be light. Even the particles up here, so even the particles up here that are still quite hot, they are also emitting electromagnetic radiation because they're all bumping into each other and their electrons and their protons are all getting accelerated in different ways, they're also releasing electromagnetic radiation, but it is at a slower wavelength than your eye is capable of perceiving as light. If you had an infrared camera you would see the flames being much larger. And if you were to look at the flame closely you would see down here right where the combustion reaction is happening, the flame looks blue. And that's because the blue light is higher energy light, and that's because the particles are being accelerated more down here, and then it goes from blue to kind of a white to a yellow, to a red or to an orange, to a red, and then it, to your regular eyes, it disappears. But everything, everything that has some temperature is releasing electromagnetic radiation. And you're like, Okay, well that's all fine, I can see it but how is that a form of energy transfer? Well, if you're ever sat next to a flame you will feel the heat. In fact, even if the air between you and the flame is cold you would still feel like you're getting warm. So if this is a flame right over here, so that is fire, and let's say you have cold air, cold air, let's say you're at a campfire right over here, if you are standing right over here you would still feel heat, you would still feel like you're getting warmed up. And that's because that electromagnetic radiation is being emitted from the air particles that we perceive as fire and then that can actually excite particles on your skin and it will transfer energy to your skin and so you feel like you are actually getting warmed up. I remember once, this is kind of a strange story, but I was on the highway and there was a car on fire and I was literally, we drove to the far lane because it was on fire, we're three lanes away from it and it kind of exploded. I don't think anyone was in it, hopefully no one was in it, but I remember right when it exploded it was an intense, immediate heat that we felt through the window of the car, and that was electromagnetic radiation. That was thermal radiation being released by these accelerated particles in the air around that explosion," + }, + { + "Q": "At 3:20 why 7 times?", + "A": "Just an arbitrary amount of times, he could have done it more or less. The amount of windings would effect the current produced.", + "video_name": "jabo8iTesqQ", + "timestamps": [ + 200 + ], + "3min_transcript": "Of course, the cap is what we're going to make our winding around. So the washers that are going to support the winding need to be pretty close to the cap. And the permanent magnets also need to be pretty close. So we're just making our marks here so that we can position everything. And it's pretty important that things are lined up fairly accurately, otherwise the winding may not turn smoothly. So we determined that we're going to measure off of the center point and about an inch away from the center of the square. So now that we have our intersection points marked, we're going to go ahead and drill our holes. We've taped the drill about 1/2 an inch up, so we want the hole to go in about 1/2 an inch. And we're using a 3/16 inch bit on the sides. And then we'll use a 1/8 inch bit for our-- the other sides. And the other sides are going to hold our bearings. So these are the bearing screws. so that's what these screws are from-- or for. These actually came from an Ikea bed. And then these are just some scrap screws that we had laying around. You can use any screws or nails. What's important is that the screws are lined up right and that they're the same level. So that will improve the efficiency of our motor. So we're just checking the level and making sure everything's lined up correctly. And now we're going to take some hot glue and get our permanent magnets. Make sure that they are oriented so that they are attracted to one another. So we want their opposite poles facing one another. And so we just put some hot glue on the screw and then set the magnet in there. And the main reason we did it this way is that it's really easy to change the position of the magnets if we need to. And we may need to do that. So we've got a washer here. in a vertical orientation. And the hot glue allows for easy adjustments. And so now we're going to take our bottle cap that we had talked about earlier. We're going to wrap our field coil around it. And we're going to go around it seven times. And then once we've got the coil a fairly good length, we have about 3 inches of wire on either side. We'll loop the wire through and tie it off so that it stays in a consistent loop. There we go. And then we'll do the other side as well. I'll just loop the wire through. And that just kind of holds the loop together and gives us something to connect with our washers. All right, now we're just going to tape the edges of the wire to keep it together." + }, + { + "Q": "0:30 How do you know that Co2 is going to be your cation? Sorry, I'm a bit behind. (And confused)", + "A": "In the formula of an ionic compound (at least in one that only has 2 different elements) the first element is the cation and the second is the anion.", + "video_name": "vVTwzjvWySs", + "timestamps": [ + 30 + ], + "3min_transcript": "- [Instructor] So we have the formula for an ionic compound here, and the goal of this video is what do we call this thing? It clearly involves some cobalt and some sulfur, but how do we name it? Well, the convention is, is the first element to be listed is going to be our cation, and if we look at cobalt over here, we see that it is a D-block element and D-block elements are tricky because you don't know exactly how it will ionize. So we know that this is going to be our cation, it's going to be our positive ion, but we don't know what the charge on each of those cobalt is actually going to be. So now let's look at the anion, let's look at the sulfur, or as an anion, the sulfide. So let me underline that. And on the periodic table, we see sulfur is out here that in its group, it would want to gain two electrons in order to have a complete outer shell. So the sulfide anion will look like this. So it will have sulfur when it ionizes will have a two minus charge, just like oxygen, just like everything else in this group. It would want to gain one, two electrons so that its outer shell looks like that of a normal gas, looks like that of argon. We can use this as a clue to figure out what must be the charge on the cobalts because we have three of the sulfides. Each of the sulfides has a two minus charge, and we have three of them, so that's going to give us a six minus charge all in. And then the cobalt, we have two of them. And so these two cobalt have to offset They have to have a six plus charge. Well that means that each of them need to have a three plus charge. If each of these have a three plus charge and you have two of them, then you're gonna have six plus on the positive side and you're gonna have six minus from the sulfides. And the reason why this is useful for us is now we can name this. We would call this ionic compound Cobalt III, cobalt and you would write it with Roman numerals here, Cobalt III Sulfide, Cobalt III Sulfide. Now I know what you might be thinking. Hey, when we looked at other ionic compounds, I didn't have to write the charge of the cation there and the reason why the convention is to do it here, and I don't have to write in upper case there, so let me rewrite it as" + }, + { + "Q": "Around 9:30, Sal is trying to find the legs of the right triangle. Why does he use \"soh\" and \"cah\"? How do either of those relate to the length of the side?\nAt 9:55, he says \"5 sine, of 36.899, is equal to...\" What is \"sine\" and \"cosine\"? And how are they used here?\nThanks in advance.", + "A": "soh, cah and toa are mnemonics (look it up) for Sine = Opposite side / Hypotenuse ==> SOH Cosine = Adjacent side / Hypotenuse ==> CAH Tangent = Opposite side / Adjacent side ==> TOA", + "video_name": "xp6ibuI8UuQ", + "timestamps": [ + 570, + 595 + ], + "3min_transcript": "It's 36.8699 degrees. So I'm picking that particular number for a particular reason. Now what I wanna do is I wanna figure out this vector's horizontal and vertical component. So I wanna break it down into something that's going straight up or down and something that's going straight right or left. So how do I do this? Well, one, I could just draw them, visually, see what they look like. So its vertical component would look like this. It would start... Its vertical component would look like this. And its horizontal component would look like this. Its horizontal component would look like this. The horizontal component, the way I drew it, it would start where vector A starts and go as far in the X direction as vector A's tip, but only in the X direction, and then you need to, to get back to the head of vector A, you need to have its vertical component. we could call the vertical component over here A sub Y, just so that it's moving in the Y direction. And we can call this horizontal component A sub X. Now what I wanna do is I wanna figure out the magnitude of A sub Y and A sub X. So how do we do that? Well, the way we drew this, I've essentially set up a right triangle for us. This is a right triangle. We know the length of this triangle, or the length of this side, or the length of the hypotenuse. That's going to be the magnitude of vector A. And so the magnitude of vector A is equal to five. We already knew that up here. So how do we figure out the sides? Well, we could use a little bit of basic trigonometry. If we know the angle, and we know the hypotenuse, how do we figure out the opposite side to the angle? So this right here, this right here is the opposite side to the angle. And if we forgot some of our basic trigonometry we can relearn it right now. Soh-cah-toa. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we have the angle, we want the opposite, and we have the hypotenuse. So we could say that the sine of our angle, the sine of 36.899 degrees, is going to be equal to the opposite over the hypotenuse. The opposite side of the angle is the magnitude of our Y component. ...is going to be equal to the magnitude of our Y component, the magnitude of our Y component, over the magnitude of the hypotenuse, over this length over here, which we know is going to be equal to five. Or if you multiply both sides by five, you get five sine of 36.899 degrees, is equal to the magnitude of the vertical component" + }, + { + "Q": "In 6:25, Sal writes ||a||. What do the double lines mean?", + "A": "Those lines indicate that we are taking the magnitude of the vector between them. They are sort of like a vector version of the absolute value sign in math.", + "video_name": "xp6ibuI8UuQ", + "timestamps": [ + 385 + ], + "3min_transcript": "this vector right here in green and this vector right here in red. Notice, X starts at the tail of the green vector and goes all the way to the head of the magenta vector. And the magenta vector starts at the head of the green vector and then finishes, I guess, well where it finishes is where vector X finishes. And the reason why I do this... And, you know, hopefully from this comparable explanation right here, says, okay, look, the green vector plus the magenta vector gives us this X vector. That should make sense. I put the head of the green vector to the tail of this magenta vector right over here. But the whole reason why I did this is, if I can express X as a sum of these two vectors, it then breaks down X into its vertical component and its horizontal component. So I could call this or I should say the vertical component. X vertical. And then I could call this over here the X horizontal. Or another way I could draw it, I could shift this X vertical over. Remember, it doesn't matter where I draw it, as long as it has the same magnitude and direction. And I could draw it like this. X vertical. And so what you see is is that you could express this vector X... Let me do it in the same colors. You can express this vector X as the sum of its horizontal and its vertical components. As the sum of its horizontal and its vertical components. Now we're gonna see over and over again that this is super powerful because what it can do is it can turn a two-dimensional problem into two separate one-dimensional problems, one acting in a horizontal direction, one acting in a vertical direction. Now let's do it a little bit more mathematical. I've just been telling you about length and all of that. Let me just show you what this means, to break down the components of a vector. So let's say that I have a vector that looks like this. Let me do my best to... Let's say I have a vector that looks like this. It's length is five. So let me call this vector A. So vector A's length is equal to five. And let's say that its direction... We're gonna give its direction by the angle between the direction its pointing in and the positive X axis. So maybe I'll draw an axis over here. So let's say that this right over here is the positive Y axis going in the vertical direction. This right over here is the positive X axis going in the horizontal direction. And to specify this vector's direction I will give this angle right over here. And I'm gonna give a very peculiar angle, but I picked this for a specific reason, just so things work out neatly in the end." + }, + { + "Q": "at 12:30 What does he mean by saying making this a 2 component velocity", + "A": "Sal said: break that down into two component velocities . These velocities are horizontal and vertical components of the given velocity.", + "video_name": "xp6ibuI8UuQ", + "timestamps": [ + 750 + ], + "3min_transcript": "Now before I take out the calculator and figure out what this is, let me do the same thing for the horizontal component. Over here we know this side is adjacent to the angle. And we know the hypotenuse. And so cosine deals with adjacent and hypotenuse. So we know that the cosine of 36.899 degrees is equal to... Cosine is adjacent over hypotenuse. So it's equal to the magnitude of our X component over the hypotenuse. The hypotenuse here has... Or the magnitude of the hypotenuse, I should say, which has a length of five. Once again, we multiply both sides by five, and we get five times the cosine of 36.899 degrees is equal to the magnitude of our X component. So let's figure out what these are. Let me get the calculator out. I wanna make sure it's in degree mode. So let me check. Yep, we're in degree mode right over there. Don't wanna... Make sure we're not in radian mode. Now let's exit that. And we have the vertical component is equal to five times the sine of 36.899 degrees, which is, if we round it, right at about three. So this is equal to... So the magnitude of our vertical component is equal to three. And then let's do the same thing for our horizontal component. So now we have five times the cosine of 36.899 degrees, is, if once again we round it to, I guess, our hundredths place, we get it to being four. So we see here is a situation where we have... This is a classic three-four-five Pythagorean triangle. The magnitude of our horizontal component is four. The magnitude of our vertical component, right over here, is equal to three. And once again, you might say, Sal, why are we going through all of this trouble? that if we say something has a velocity, in this direction, of five meters per second, we could break that down into two component velocities. We could say that that's going in the upwards direction at three meters per second, and it's also going to the right in the horizontal direction at four meters per second. And it allows us to break up the problem into two simpler problems, into two one-dimensional problems, instead of a bigger two-dimensional one." + }, + { + "Q": "At 3:41, what is a nuetron", + "A": "A nuetron is a sub-atomic particle that is inisde an atom. In a nuclues (which is in an atom) there are two sub-atomic particles - protons and neutrons. Neutrons have a mass of 1 (relative to the mass of a proton) and a charge of 0.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 221 + ], + "3min_transcript": "Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original. results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling." + }, + { + "Q": "at 5:00,it is said that hydrogen will turn to deuterium and then to helium , why cant deuterium change to TRITIUM and then to helium ?", + "A": "You start with a bunch of Hydrogen-1, which is basically free protons. Some of the protons fuse, and immediately decay to form deuterium. Where do the free neutrons, which only have a 15 minute half life, come from to then fuse to form tritium?", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 300 + ], + "3min_transcript": "to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original. results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling. This is still hydrogen because it has one proton and one neutron now. It is not helium yet. This does not have two-- it does not have two protons. But then the deuterium keeps fusing. And then we eventually end up with helium. And we can even see that on the periodic table. Oh, I lost my periodic table. Well, I'll show you the next video. But we know hydrogen in its atomic state has an atomic number of 1. And it also has a mass of 1. It only has one nucleon in its nucleus. But it's being fused. It goes to hydrogen-2, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused-- and I'm not going into the detail of the reaction-- into helium. And by definition, helium has two protons and two neutrons. So it has-- or we're talking about helium-4, in particular, that isotope of helium-- it has an atomic mass of 4." + }, + { + "Q": "I was reading that a neutron is heavier than a proton. Then how could a proton and a neutron have a smaller mass than 2 protons (as Sal mentioned at around 4:00)? To convert a proton into a neutron, one would have to add the mass difference in the form of energy. Then how would the reaction be exothermic?", + "A": "The Deuteron D is a stable particle composed of a proton p and a neutron n. To separate D into its components you have to provide energy in order to overcome the strong binding forces. This energy you have to add corresponds to a mass according to Einsteins formula E=mc^2 Therefore the products (p+n) are heavier than the particle D you startet from.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 240 + ], + "3min_transcript": "Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original. results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling." + }, + { + "Q": "As far as i know a neutron is composed of one proton and one electron but Sal at 3:53 said that when two protons come very very close to each other, a proton degrades into a neutron.\nSo, how is it possible for the creation of a neutron without a proton?", + "A": "A neutron is not composed of a proton and electron. A proton decays into a neutron when one of its constituent up quarks decays into a down quark. A neutron can also decay into a proton via the reverse process.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 233 + ], + "3min_transcript": "Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original. results in a lot of energy-- plus energy. And this energy is why we call it ignition. And so what this energy does is it provides a little bit of outward pressure, so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs. And then that energy provides outward pressure to balance what is now a star. So now we are at where we actually have the ignition at the center. We have-- and we still have all of the other molecules trying to get in providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction-- and I'm just kind of doing the most basic type of fusion that happens in stars-- the hydrogen gets fused into deuterium. I have trouble spelling." + }, + { + "Q": "During the video on the birth of stars, Sal said at about 0:17 that gravity would bring the hydrogen atoms together. But excluding the clouds of hydrogen, there is really nothing in space. Why won't the hydrogen atoms just diffuse and get evenly distributed in the universe?", + "A": "Even individual atoms have a tiny tiny bit of gravitational force that attracts them into forming clouds. If a certain cluster of hydrogen atoms form a dense enough cloud, they attract even more hydrogen atoms until eventually you get the kind of mass associated with stars.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 17 + ], + "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring" + }, + { + "Q": "In the beginning Sal starts with hydrogen atoms...\nQ1.Were did that hydrogen come from in the first place?\nQ2.How did it become a cloud 0:03? (Why were the hydrogen atoms slightly close in the beginning)", + "A": "The hydrogen came from the big bang (supposedly), just like all of the other matter. It became a cloud because the atoms were and are attracted to each other, and they like to form together.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 3 + ], + "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring" + }, + { + "Q": "0:30 wouldn't the hydrogen atoms accelerate since the gravatational force begins acting over a shorter and shorter distance therefore becoming stronger", + "A": "I would imagine that since as they move in more, they heat up more, and so the gas would also try to expand, creating a counter force that would balance it out.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 30 + ], + "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring" + }, + { + "Q": "What's a nuclei? {2:48}", + "A": "More than one nucleus.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 168 + ], + "3min_transcript": "They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring to occur, for fusion ignition. And the reason why-- and I want to be very careful. It's not ignition. It's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion. It's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure-- obviously, this would not happen with just the Coulomb forces-- with enough pressure they get close enough. And then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original." + }, + { + "Q": "7:10- What does he say?", + "A": "At 7:10, he says, ...there s a huge step temperature...", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 430 + ], + "3min_transcript": "This is still hydrogen because it has one proton and one neutron now. It is not helium yet. This does not have two-- it does not have two protons. But then the deuterium keeps fusing. And then we eventually end up with helium. And we can even see that on the periodic table. Oh, I lost my periodic table. Well, I'll show you the next video. But we know hydrogen in its atomic state has an atomic number of 1. And it also has a mass of 1. It only has one nucleon in its nucleus. But it's being fused. It goes to hydrogen-2, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused-- and I'm not going into the detail of the reaction-- into helium. And by definition, helium has two protons and two neutrons. So it has-- or we're talking about helium-4, in particular, that isotope of helium-- it has an atomic mass of 4. Because the atomic mass of the helium that gets produced is slightly lower than four times the atomic mass of each of the constituent hydrogens. So all of this energy, all this energy from the fusion-- but it needs super high pressure, super high temperatures to happen-- keeps the star from collapsing. And once a star is in this stage, once it is using hydrogen-- it is fusing hydrogen in its core, where the pressure and the temperature is the most, to form helium-- it is now in its main sequence. This is now a main sequence star. And that's actually where the sun is right now. Now there's questions of, well, what if there just wasn't enough mass to get to this level over here? And there actually are things that never get to quite that threshold to fuse all the way into helium. There are a few things that don't quite So they are generating some of their heat. Or there are even smaller objects that just get to the point there's a huge temperature and pressure, but fusion is not actually occurring inside of the core. And something like Jupiter would be an example. And you could go several masses above Jupiter where you get something like that. So you have to reach a certain threshold where the mass, where the pressure and the temperature due to the heavy mass, get so large that you start this fusion. And-- but the smaller you are above that threshold, the slower the fusion will occur. But if you're super massive, the fusion will occur really, really fast. So that's the general idea of just how stars get formed and why they don't collapse on themselves and why they are these kind of balls of fusion reactions existing in the universe. In the next few videos, we'll talk about what happens once that hydrogen fuel in the core starts to run out." + }, + { + "Q": "So at 7:16 he mentioned that Jupiter was a example of not fusing into a star. If there was enough mass, could Juipter become a star?", + "A": "Jupiter would have to engulf all of the other planets in the solar system 20 times over in order to be massive enough to star fusing its hydrogen.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 436 + ], + "3min_transcript": "This is still hydrogen because it has one proton and one neutron now. It is not helium yet. This does not have two-- it does not have two protons. But then the deuterium keeps fusing. And then we eventually end up with helium. And we can even see that on the periodic table. Oh, I lost my periodic table. Well, I'll show you the next video. But we know hydrogen in its atomic state has an atomic number of 1. And it also has a mass of 1. It only has one nucleon in its nucleus. But it's being fused. It goes to hydrogen-2, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused-- and I'm not going into the detail of the reaction-- into helium. And by definition, helium has two protons and two neutrons. So it has-- or we're talking about helium-4, in particular, that isotope of helium-- it has an atomic mass of 4. Because the atomic mass of the helium that gets produced is slightly lower than four times the atomic mass of each of the constituent hydrogens. So all of this energy, all this energy from the fusion-- but it needs super high pressure, super high temperatures to happen-- keeps the star from collapsing. And once a star is in this stage, once it is using hydrogen-- it is fusing hydrogen in its core, where the pressure and the temperature is the most, to form helium-- it is now in its main sequence. This is now a main sequence star. And that's actually where the sun is right now. Now there's questions of, well, what if there just wasn't enough mass to get to this level over here? And there actually are things that never get to quite that threshold to fuse all the way into helium. There are a few things that don't quite So they are generating some of their heat. Or there are even smaller objects that just get to the point there's a huge temperature and pressure, but fusion is not actually occurring inside of the core. And something like Jupiter would be an example. And you could go several masses above Jupiter where you get something like that. So you have to reach a certain threshold where the mass, where the pressure and the temperature due to the heavy mass, get so large that you start this fusion. And-- but the smaller you are above that threshold, the slower the fusion will occur. But if you're super massive, the fusion will occur really, really fast. So that's the general idea of just how stars get formed and why they don't collapse on themselves and why they are these kind of balls of fusion reactions existing in the universe. In the next few videos, we'll talk about what happens once that hydrogen fuel in the core starts to run out." + }, + { + "Q": "at 0:30, Sal says that the gravity interacts with \"atoms\" in a huuge distance, Doesn't gravity has a limitation? Sorry for my english.", + "A": "No, as far as we understand it, gravity has unlimited range, although the strength of the field declines with the square of the distance from the mass that generates the field.", + "video_name": "i-NNWI8Ccas", + "timestamps": [ + 30 + ], + "3min_transcript": "Let's imagine we have a huge cloud of hydrogen atoms floating in space. Huge, and when I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. It's-- we normally don't think about the gravity of atoms. But it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. And the hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. And so it's going to get denser and denser and denser. Now remember, it's a huge mass of hydrogen atoms. So the temperature is going up. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really to give you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. And to kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. A hydrogen-- and even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. Atoms are actually-- the nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from our-- from what we learned about the four forces-- that if they did get close enough to each other, that if they did get-- if somehow under huge temperatures and huge pressures you were able to get these two protons close enough to each other, then all of a sudden, the strong force will overtake. It's much stronger than the Coulomb force. And then these two hydrogens will actually-- these nucleuses would actually fuse-- or is it nuclei? Well, anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring" + }, + { + "Q": "At 0:33 , why sal took theta between 0 and 90 degrees ?\nPlease explain .\nThank you .", + "A": "At 0 or 90 degrees, the given equation for distance (d=(2s^2/g)*cosX*sinX) is 0 (sin 90=0, cos 0=0). This can be seen in the graph at 1:26, where 0 and 90 are x-axis intercepts. Since we re looking for optimal distance, these values can be immediately discarded. This makes sense in a real world way, because the horizontal distance traveled if the projectile is shot straight up or straight into the surface will be 0.", + "video_name": "snw0BrCBQYQ", + "timestamps": [ + 33 + ], + "3min_transcript": "Now that we have distance explicitly as a function of the angle that we're shooting the object at, we can use a little bit of calculus to figure out the optimal angle, the angle that's going optimize our distance. And since we only care about angles from 0 degrees to really 90 degrees, let's constrain ourselves. So we're going to optimize things for angles between 0 degrees. So theta is going to be greater than or equal to 0 and less than or equal to 90. So let's see how we can do it. And just to get an idea of what we're even conceptually doing with the calculus, remember when you take a derivative, you are finding the slope of a line, an instantaneous slope of a line. And if you were to graph this-- and I encourage you to graph it on your own, maybe with a graphing calculator-- it will look something like this over the interval. It will look like this where that is the distance as a function of theta axis and then this would And we care about angles between 0 and 90 degrees. So if you were to graph this thing, so this is 0 degrees, this is maybe 90 degrees right here. The graph of this function will look like this. It'll look something like this. It will look something like that. And what we want to do is find the angle, there's some angle here that gives us the optimal distance. So this is, right here, this is the optimal distance. And what we want to do is find that out. And when you look at the graph, and you could do it on a graphing calculator if you like, what happens to the instantaneous slope at that optimal distance? Well it's flat. The slope there is 0. So what we need to do is take the derivative of this function and then just figure out at what angle is the function equal to 0? And then we're done We will know this mystery angle, this optimal angle, to shoot the object at. So let's take the derivative. So the derivative, we'll just use our derivative rules here. The derivative of-- I will call it d prime I guess, or we could say the derivative of the distance with respect to theta is equal to-- we're assuming that s and g are constants, so we don't have to worry about them right now. We could just put them out front since we're assuming they're constants. And then we can do the product rule to take the derivative of this part with respect to theta. In the product rule, we take the derivative of the first function times the second function. So the derivative of cosine of theta is negative sine of theta. And we're going to multiply that times the second function. So that's times the sine of theta." + }, + { + "Q": "At 5:30, in reference to the formula, how can one tell what the graph will look like?", + "A": "Calculate a few points on it, and make a sketch", + "video_name": "2GQTfpDE9DQ", + "timestamps": [ + 330 + ], + "3min_transcript": "Here's why I'm taking the absolute value of the product, well, if they're different charges, this will be a negative number, but we just want the overall magnitude of the force. So we could take, it's proportional to the absolute value of the product of the charges and it's inversely proportional to not just the distance between them, not just to r, but to the square of the distance. The square of the distance between them. And what's pretty neat about this is how close it mirrors Newton's law of gravitation. Newton's law of gravitation, we know that the force, due to gravity between two masses, remember mass is just another property of matter, that we sometimes feel is a little bit more tangible because it feels like we can kind of see weight and volume, but that's not quite the same, or we feel like we can feel or internalize things like weight and volume which are related to mass, but in some ways it is just another property, another property, especially as you get into more of a kind of fancy physics. Our everyday notion of even mass starts to But Newton's law of gravitation says, look the magnitude of the force of gravity between two masses is going to be proportional to, by Newton's, by the gravitational concept, proportional to the product of the two masses. Actually, let me do it in those same colors so you can see the relationship. It's going to be proportional to the product of the two masses, m one m two. And it's going to be inversely proportional to the square of the distance. The square of the distance between two masses. Now these proportional personality constants are very different. Gravitational force, we kind of perceive this is as acting, being strong, it's a weaker force in close range. But we kind of imagine it as kind of what dictates what happens in the, amongst the stars and the planets and moons. While the electrostatic force at close range is a much stronger force. It can overcome the gravitational force very easily. But it's what we consider happening at either an atomic level or kind of at a scale But needless to say, it is very interesting to see how this parallel between these two things, it's kind of these patterns in the universe. But with that said, let's actually apply let's actually apply Coulomb's law, just to make sure we feel comfortable with the mathematics. So let's say that I have a charge here. Let's say that I have a charge here, and it has a positive charge of, I don't know, let's say it is positive five times 10 to the negative three Coulombs. So that's this one right over here. That's its charge. And let's say I have this other charge right over here and this has a negative charge. And it is going to be, it is going to be, let's say it's negative one... Negative one times 10 to the negative one Coulombs." + }, + { + "Q": "At 3:03 how does the OH molecule gets 1+ve formal charge? I'm not able to understand......", + "A": "The O atom in the oxonium ion has two electrons in a lone pair, plus one electron from each of the \u00cf\u0083 bonds (5 valence electrons). That\u00e2\u0080\u0099s one less valence electron than in an isolated O atom. \u00e2\u0088\u00b4 Formal charge = 6 - 5 = +1.", + "video_name": "wCspf85eQQo", + "timestamps": [ + 183 + ], + "3min_transcript": "which is an excellent leaving group. Let's draw what we would form. Let's sketch in our carbon chain here and we know that a bond forms between the oxygen and the carbon in red, so the carbon in red is this carbon and let's make these electrons magenta. So those electrons form a bond between the oxygen and the carbon in red. The oxygen is still attached to this ethyl group here, so let's draw in those two carbons. And the oxygen is still bonded to this hydrogen. So let's put in this hydrogen. We still have a lone pair of electrons left on this oxygen, so I'll put in that lone pair right here. And that gives us a plus one formal charge on the oxygen. Next, we need to make a neutral molecule for our product, so we need to have another molecule of ethanol come along, so let's draw that in here. So, ethanol is our solvent. is going to function as a base. We need to take this proton here and these electrons are left behind on the oxygen. Let's draw our final product. We sketch in our carbon chain. We have our oxygen. We have these two carbons. And now we have two lone pairs of electrons on this oxygen. Let's make these electrons blue. So, our second step is an acid-base reaction where we take a proton and these are, these electrons in blue here, to form our final product which is an ether. Notice we don't have to worry about any stereochemistry for our final product. We don't have any chiral centers to worry about. Let's look at another reaction. For this reaction, we're starting with a secondary alkyl halides. If I look at my summary over here with a secondary substrate, we could have either an S N 2 mechanism or an S N 1 mechanism, First, let's look at the nucleophile. This is N a plus and SH minus, so let me draw in the SH minus here which that is going to be our nucleophile and that's a strong nucleophile. A negative charge on a sulfur would make a strong nucleophile. And for our solvent, we saw in an earlier video, DMSO is a polar A product solvent which favors an S N 2 reaction. So with a strong nucleophile and a polar A product solvent, we need to think about in S N 2 mechanism. So we know our nucleophile attacks at the same time that we get loss of our leaving group, so our nucleophile is going to attack this carbon. So again, I'll make this carbon red. At the same time that we get loss of leaving groups, so these electrons are gonna come off onto the bromine to form the bromide anion. For this reaction, we need to think about the stereochemistry of our S N 2 reaction. Our nucleophile has to attack from the opposite side" + }, + { + "Q": "At 4:32 It says that only one cell will be a successful sex cell while the other three are polar bodies- what do these polar bodies do?", + "A": "In humans, they really don t do anything. However, in plants, they also become sex cells.,", + "video_name": "IQJ4DBkCnco", + "timestamps": [ + 272 + ], + "3min_transcript": "that each have the haploid number that each have the haploid number of chromosomes. And they don't all necessarily have the same genetic informatioin anymore. Because as we go through this first phase, right over here of meiosis, and this first phase here you go from diploid to haploid, right over here, this is called Meiosis One. Meiosis One, you're essentially splitting the homologous pairs and so this one might get some of the ones that you originally got from your father, and some that you originally got from your mother, some that you originally got from your father, some that you originally got from your mother, they split randomly, but each homogolous pair gets split up. And then in this phase, Meiosis Two, so this phase right over here is called Meiosis Two, it's very similar to mitosis, except your now dealing with cells that start off with the haploid number. These cells that you have over here, these are gametes. This are sex cells. These are gametes. This can now be used in fertilization. If we're talking about, if you're male, this is happening in your testes, and these are going to be sperm cells If you are female, this is happening in your ovaries and these are going to be egg cells. If you a tree, this could be pollen or it could be an ovul. But these are used for fertilization. These will fuse together in sexual reproduction to get to a fertilized egg, which then can undergo mitosis to create an entirely new organism. So not a cycle here, although these will find sex cells from another organism and fuse with them and those can turn into another organism. And I guess the whole circle of life starts again. going and going, going. This cell is just like this cell, while these sex cells are differeent than this one right over here. Now, where does this happen in the body? We've talked about this in previous videos. These are your somatic cells right over here. These are the ones that make up the bulk of your body, somatic cells. And where is this happening? Well, this is happening in germ cells, As we mentioned, if you're male it's in your tesis and if you're female it's in your ovaries. And germ cells actually can undergo mitosis to produce other germ cells that have a diploid number of chromosomes, or they can undergo meiosis in order to produce sperm or egg cells in order to produce gametes." + }, + { + "Q": "at 0:19, what is a diploid number?", + "A": "Diploid number means the number of chromosomes in the body cells of a diploid organism. In humans, 46 is the diploid number (body cells) and 23 is the haploid number (egg and sperm). Note: A diploid organism has paired sets of chromosomes in a cell. In humans, there are 23 homologous pairs of chromosomes where one set is inherited from each parent. Hope it helps.", + "video_name": "IQJ4DBkCnco", + "timestamps": [ + 19 + ], + "3min_transcript": "- [Voiceover] Before we go in-depth on meiosis, I want to do a very high level overview comparing mitosis to meiosis. So, in mitosis, this is all a review, if you've watched the mitosis video, in mitosis, we start with a cell, that has a diploid number of chromosomes. I'll just write 2n to show it has a diploid number. For human beings, this would be 46 chromosomes. 46 for humans, you get 23 chromosomes from your mother, 23 chromosomes from your father or you can say you have 23 homologous pairs, which leads to 46 chromosomes. Now after the process of mitosis happens and you have your cytokinesis and all the rest, you end up with two cells that each have the same genetic information as the original. So you now have two cells that each have the diploid number of chromosomes. So, 2n and 2n. are just like this cell was, it can go through interphase again. It grows and it can replicate its DNA and centrosomes and grow some more then each of these can go through mitosis again. And this is actually how most of the cells in your body grow. This is how you turn from a single cell organism into you, or for the most part, into you. So that is mitosis. It's a cycle. After each of these things go through mitosis, they can then go through the entire cell cycle again. Let me write this a little bit neater. Mitosis, that s was a little bit hard to read. Now what happens in meiosis? What happens in meiosis? I'll do that over here. In meiosis, something slightly different happens and it happens in two phases. You will start with a cell that has a diploid number of chromosomes. that has a diploid number of chromosomes. And in it's interphase, it also replicates its DNA. And then it goes through something called Meiosis One. And in Meiosis One, what you end up with is two cells that now have haploid number of chromosomes. So you end up with two cells, You now have two cells that each have a haploid number of chromosomes. So you have n and you have n. So if we're talking about human beings, you have 46 chromosomes here, and now you have 23 chromosomes in this nucleus. And now you have 23 in this nucleus. But you're still not done. Then each of these will go through a phase, which I'll talk about in a second, which is very similar to mitosis, which will duplicate this entire cell into two. So actually, let me do it like this. So now, this one," + }, + { + "Q": "4:37 If Meiosis is occurring in the respective reproductive organs than why does he speak about mixing our parents dna, I know that DNA get's passed on, but how does the whole \"mixing process\" occur in our own bodies? And to what extent?", + "A": "During sexual reproduction, the male and female gametes (The sperm and the egg) fuse, leading to the formation of a zygote. The zygote s DNA is a mixture of both the parents DNA, which comes from the respective gametes.", + "video_name": "IQJ4DBkCnco", + "timestamps": [ + 277 + ], + "3min_transcript": "that each have the haploid number that each have the haploid number of chromosomes. And they don't all necessarily have the same genetic informatioin anymore. Because as we go through this first phase, right over here of meiosis, and this first phase here you go from diploid to haploid, right over here, this is called Meiosis One. Meiosis One, you're essentially splitting the homologous pairs and so this one might get some of the ones that you originally got from your father, and some that you originally got from your mother, some that you originally got from your father, some that you originally got from your mother, they split randomly, but each homogolous pair gets split up. And then in this phase, Meiosis Two, so this phase right over here is called Meiosis Two, it's very similar to mitosis, except your now dealing with cells that start off with the haploid number. These cells that you have over here, these are gametes. This are sex cells. These are gametes. This can now be used in fertilization. If we're talking about, if you're male, this is happening in your testes, and these are going to be sperm cells If you are female, this is happening in your ovaries and these are going to be egg cells. If you a tree, this could be pollen or it could be an ovul. But these are used for fertilization. These will fuse together in sexual reproduction to get to a fertilized egg, which then can undergo mitosis to create an entirely new organism. So not a cycle here, although these will find sex cells from another organism and fuse with them and those can turn into another organism. And I guess the whole circle of life starts again. going and going, going. This cell is just like this cell, while these sex cells are differeent than this one right over here. Now, where does this happen in the body? We've talked about this in previous videos. These are your somatic cells right over here. These are the ones that make up the bulk of your body, somatic cells. And where is this happening? Well, this is happening in germ cells, As we mentioned, if you're male it's in your tesis and if you're female it's in your ovaries. And germ cells actually can undergo mitosis to produce other germ cells that have a diploid number of chromosomes, or they can undergo meiosis in order to produce sperm or egg cells in order to produce gametes." + }, + { + "Q": "Sorry if this seems like an awfully basic question, but why does O get a negative charge at 4:01?", + "A": "Well oxygen is a very electronegative atom hence it can remain stable holding an additional electron/negative charge.", + "video_name": "nv2kfBFkv4s", + "timestamps": [ + 241 + ], + "3min_transcript": "or dehydration synthesis. We saw this type of reaction when we were putting glucoses together, when we were forming carbohydrates. Dehydration synthesis. But whenever I see a reaction like this, it's somewhat satisfying to just be able to do the counting and say, \"All right, this is gonna bond \"with that, we see the bond right over there, \"and I'm gonna let go of an oxygen and two hydrogens, \"which net net equals H2O, equals a water molecule.\" But how can we actually imagine this happening? Can we push the electrons around? Can we do a little bit of high-level organic chemistry to think about how this happens? And that's what I wanna do here. I'm not gonna do a formal reaction mechanism, but really get a sense of what's going on. Well, nitrogen, as we said, has got this lone pair, it's electronegative. And this carbon right over here, it's attached to two oxygens, oxygens are more electronegative. The oxygens might hog those electrons. what we call in organic chemistry a nucleophilic attack on this carbon right over here. And when it does that, if we were doing a more formal reaction mechanism, we could say, \"Hey, well, maybe one \"of the double bonds goes back, \"the electrons in it go back to this oxygen, \"and then that oxygen would have a negative charge.\" But then that lone pair from that double bond could then reform, and as that happens, this oxygen that's in the hydroxyl group will take back both of these electrons. Would take back both of those electrons, and now it's going to have an extra lone pair. Let me do that by erasing this bond and then giving it an extra lone pair. It already had two lone pairs, and then when it took that bond, it's gonna have a third lone pair. And then it's going to have a negative charge. to grab a hydrogen proton someplace. And now it could just grab any hydrogen proton, but probably the most convenient one would be this one, because if this nitrogen is going to use this lone pair to form a bond with carbon, it's going to have a positive charge, and it might wanna take these electrons back. So you could imagine where one of these lone pairs is used to grab this hydrogen proton, and then the nitrogen can take these electrons, can take these electrons back. So hopefully you didn't find this too convoluted, but I always like to think, what could actually happen here? And so you see, this lone pair of electrons from the nitrogen forms this orange bond with the carbon. Let me do that in orange color if I'm going to call it an orange bond. It forms this orange bond. What we call this orange bond, we could call this a peptide bond, or a peptide linkage. Peptide bond, sometimes called a peptide..." + }, + { + "Q": "at 0:44, Sal describes the first law of thermodynamics. however, isn't it the law of conservation of energy?", + "A": "Indeed the first law of thermodynamics is a version of the law of conservation of energy.", + "video_name": "Xb05CaG7TsQ", + "timestamps": [ + 44 + ], + "3min_transcript": "I've now done a bunch of videos on thermodynamics, both in the chemistry and the physics playlist, and I realized that I have yet to give you, or at least if my memory serves me correctly, I have yet to give you the first law of thermodynamics. And I think now is as good a time as any. The first law of thermodynamics. And it's a good one. It tells us that energy-- I'll do it in this magenta color-- energy cannot be created or destroyed, it can only be transformed from one form or another. So energy cannot be created or destroyed, only transformed. So let's think about a couple of examples of this. And we've touched on this when we learned mechanics and bunch of this in the chemistry playlist as well. So let's say I have some rock that I just throw as fast as I can straight up. Maybe it's a ball of some kind. So I throw a ball straight up. That arrow represents its velocity vector, right? it's going to go up in the air. Let me do it here. I throw a ball and it's going to go up in the air. It's going to decelerate due to gravity. And at some point, up here, the ball is not going to have any velocity. So at this point it's going to slow down a little bit, at this point it's going to slow down a little bit more. And at this point it's going to be completely stationary and then it's going to start accelerating downwards. In fact, it was always accelerating downwards. It was decelerating upwards, and then it'll start accelerating downwards. So here its velocity will look like that. And here its velocity will look like that. Then right when it gets back to the ground, if we assume negligible air resistance, its velocity will be the same magnitude as the upward but in the downward direction. tons in the projectile motion videos in the physics playlist, over here we said, look, we have some kinetic And that makes sense. I think, to all of us, energy intuitively means that you're doing something. So kinetic energy. Energy of movement, of kinetics. It's moving, so it has energy. But then as we decelerate up here, we clearly have no kinetic energy, zero kinetic energy. So where did our energy go? I just told you the first law of thermodynamics, that energy cannot be created or destroyed. But I clearly had a lot of kinetic energy over here, and we've seen the formula for that multiple times, and here I have no kinetic energy. So I clearly destroyed kinetic energy, but the first law of thermodynamics tells me that I can't do that. So I must have transformed that kinetic energy. I must have transformed that kinetic energy into something else." + }, + { + "Q": "heat isn't a form of energy is it? @ 7:28 I thought heat was the process or transfer of energy from system to surroundings?", + "A": "Heat is a form of thermal energy.", + "video_name": "Xb05CaG7TsQ", + "timestamps": [ + 448 + ], + "3min_transcript": "And the first law says, oh, Sal, it all turned into potential energy up here. And you saw it turned into potential energy because when the ball accelerated back down, it turned back into But then I say, no, Mr. First Law of Thermodynamics, look, at this point I have no potential energy, and I had all kinetic energy and I had a lot of kinetic energy. Now at this point, I have no potential energy once again, but I have less kinetic energy. My ball has fallen at a slower rate than I threw it to begin with. And the thermodynamics says, oh, well that's because you have air. And I'd say, well I do have air, but where did the energy go? And then the first law of thermodynamics says, oh, when your ball was falling-- let me see, that's the ball. Let me make the ball yellow. So when your ball was falling, it was rubbing up against air particles. It was rubbing up against molecules of air. And right where the molecules bumped into the wall, there's Friction is just essentially, your ball made these molecules that it was bumping into vibrate a little bit faster. And essentially, if you think about it, if you go back to the macrostate/ microstate problem or descriptions that we talked about, this ball is essentially transferring its kinetic energy to the molecules of air that it rubs up against as it falls back down. And actually it was doing it on the way up as well. And so that kinetic energy that you think you lost or you destroyed at the bottom, of here, because your ball's going a lot slower, was actually transferred to a lot of air particles. It was a lot of-- to a bunch of air particles. Now, it's next to impossible to measure exactly the kinetic energy that was done on each individual air particle, because we don't even know what their microstates were to begin with. But what we can say is, in general I transferred some heat to these particles. I raised the temperature of the air particles that the kinetic energy. Remember, temperature is just a measure of kinetic-- and temperature is a macrostate or kind of a gross way or a macro way, of looking at the kinetic energy of the individual molecules. It's very hard to measure each of theirs, but if you say on average their kinetic energy is x, you're essentially giving an indication of temperature. So that's where it went. It went to heat. And heat is another form of energy. So that the first law of thermodynamics says, I still hold. You had a lot of kinetic energy, turned into potential, that turned into less kinetic energy. And where did the remainder go? It turned into heat. Because it transferred that kinetic energy to these air particles in the surrounding medium. Fair enough. So now that we have that out of the way, how do we measure the amount of energy that something contains? And here we have something called the internal energy. The internal energy of a system." + }, + { + "Q": "At 11:06, Sal used the full decimal for the atomic weight of Chlorine, but rounded for the atomic weight of Phosphorus. Won't this throw off some of his data?", + "A": "He will have to round his final answer to the same number of sig figs as his chlorine number.", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 666 + ], + "3min_transcript": "" + }, + { + "Q": "8:00 , how do we know what ratio goes on top or bottom", + "A": "The compound you have goes on the bottom, the one you want goes on top. In that example he needs to convert moles P4 to moles Cl2 so in the ratio moles P4 goes in the denominator so it will cancel, and moles Cl2 goes in the numerator.", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 480 + ], + "3min_transcript": "" + }, + { + "Q": "at 4:05 is there a difference between \"mole\" and \"mol\" ?", + "A": "Using mol for mole is like using m for meter", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 245 + ], + "3min_transcript": "" + }, + { + "Q": "i thought at 3:10 ~ 3:20 sal said atomic weight but ment atomic mass unit\nwhat is a atomic weight? or is this a typo?", + "A": "Atomic weight is an outdated term that we no longer use, as we once did, to refer to atomic mass.", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 190, + 200 + ], + "3min_transcript": "" + }, + { + "Q": "why did sal used 1.45g p4 at time 4:47?", + "A": "If you go back to 00:00 you ll see it s given in the question", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 287 + ], + "3min_transcript": "" + }, + { + "Q": "At 7:04, why 6 moles multiplied in the conversion equation instead of 12 moles?", + "A": "See the balanced equation,in that 6 is the stoichiometric coefficient of cl2 . this meant 6 moles of cl2 . this was why 6 was multiplied.", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 424 + ], + "3min_transcript": "" + }, + { + "Q": "at 9:20, why did he use 35.453 instead of rounding up to 36 or down to 35 like he would have in the other videos?", + "A": "Because Chlorine exists as three different isotopes with atomic weights of 35, 36 and 37, precision chemistry uses an average atomic weight based on the proportion of the isotopes in nature. Early chemistry just uses the Cl 35 for simplicity as it is the most abundant.", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 560 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:24, does the unit like grams or moles effect the problem?", + "A": "Is it different to have 2 kg of socks instead of 2 socks? Yes, right? A mole is a number. Gram is a measure of mass. Not interchangeable.", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 264 + ], + "3min_transcript": "" + }, + { + "Q": "at 3:59 you wrote mol. do you mean mole?", + "A": "mol is the standard abbreviation for mole.", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 239 + ], + "3min_transcript": "" + }, + { + "Q": "At around 9:20, why did Sal not round the atomic weight of chlorine to 35? You can't have a chlorine atom that has 18.453 neutrons right?", + "A": "No but you can have trillions and trillions of atoms that average out to that weight.", + "video_name": "jFv6k2OV7IU", + "timestamps": [ + 560 + ], + "3min_transcript": "" + }, + { + "Q": "About 2:40, when he's describing the configuration for nickel, if the 3d8 is at the end then it's organized by energy state, right? And if the 4s2 is last with the 3d8 behind it then it's organized by distance from the nucleus?", + "A": "yes", + "video_name": "YURReI6OJsg", + "timestamps": [ + 160 + ], + "3min_transcript": "We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that from the nucleus than these right here. Now, another way to figure out the electron configuration for nickel-- and this is covered in some chemistry classes, although I like the way we just did it because you look at the periodic table and you gain a familiarity with it, which is important, because then you'll start having an intuition for how different elements react with each other -- is to just say, oK, nickel has 28 electrons, if it's neutral. It has 28 electrons, because that's the same number of protons, which is the atomic number. Remember, 28 just tells you how many protons there are. This is the number of protons. We're assuming it's neutral. So it has the same number of electrons. That's not always going to be the case. But when you do these electron configurations, that tends to be the case. So if we say nickel has 28, has an atomic number of 28, so it's electron configuration we can do it this way, too. We can write the energy shells. So one, two, three, four." + }, + { + "Q": "is D orbitals always like the chart that khan made in 5:53", + "A": "Yes the d orbitals have more energy, but backfill previous shells.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 353 + ], + "3min_transcript": "Well we're not going to get to f. But you could write f and g and h and keep going. What's going to happen is you're going to fill this one first, then you're going to fill this one, then that one, then this one, then this one. Let me actually draw it. So what you do is, these are the shells that exist, period. These are the shells that exist, in green. What I'm drawing now isn't the order that you fill them. This is just, they exist. So there is a 3d subshell. There's not a 3f subshell. There is a 4f subshell. Let me draw a line here, just so it becomes a little bit neater. And the way you fill them is you make these diagonals. So first you fill this s shell like that, then you fill this one like that. Then you do this diagonal down like that. Then you do this diagonal down like that. And then this diagonal down like that. six in p, in this case, 10 in d. And we can worry about f in the future, but if you look at the f-block on a periodic table, you know how many there are in f. So you fill it like that. So first you just say, OK. For nickel, 28 electrons. So first I fill this one out. So that's 1s2. 1s2. Then I go, there's no 1p, so then I go to 2s2. Let me do this in a different color. So then I go right here, 2s2. That's that right there. Then I go up to this diagonal, and I come back down. And then there's 2p6. And you have to keep track of how many electrons you're dealing with, in this case. So we're up to 10 now. So we used that one up. Then the arrow tells us to go down here, so now we do the third energy shell. So 3s2. And then where do we go next? Then we follow the arrow. We start there, there's nothing there, there's something here. So we go to 3p6. And then the next thing we fill out is 4s2. So then we go to 4s2. And then what's the very next thing we fill out? We have to go back to the top. We come here and then we fill out 3d. And then how many electrons do we have left to fill out? So we're going to be in 3d. So 3d. And how many have we used so far? 2 plus 2 is 4. 4 plus 6 is 10. 10 plus 2 is 12. 18. 20. We've used 20, so we have 8 more electrons to configure. And the 3d subshell can fit the 8 we need, so we have 3d8. And there you go, you've got the exact same answer that we had when we used the first method. Now I like the first method because you're looking at the periodic table the whole time, so you kind of understand an intuition" + }, + { + "Q": "at 7:23, where are noble gases on the periodic table and what are they?", + "A": "They are all the way over on the right side, and they are gases that generally will not engage in chemical reactions. They are inert.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 443 + ], + "3min_transcript": "six in p, in this case, 10 in d. And we can worry about f in the future, but if you look at the f-block on a periodic table, you know how many there are in f. So you fill it like that. So first you just say, OK. For nickel, 28 electrons. So first I fill this one out. So that's 1s2. 1s2. Then I go, there's no 1p, so then I go to 2s2. Let me do this in a different color. So then I go right here, 2s2. That's that right there. Then I go up to this diagonal, and I come back down. And then there's 2p6. And you have to keep track of how many electrons you're dealing with, in this case. So we're up to 10 now. So we used that one up. Then the arrow tells us to go down here, so now we do the third energy shell. So 3s2. And then where do we go next? Then we follow the arrow. We start there, there's nothing there, there's something here. So we go to 3p6. And then the next thing we fill out is 4s2. So then we go to 4s2. And then what's the very next thing we fill out? We have to go back to the top. We come here and then we fill out 3d. And then how many electrons do we have left to fill out? So we're going to be in 3d. So 3d. And how many have we used so far? 2 plus 2 is 4. 4 plus 6 is 10. 10 plus 2 is 12. 18. 20. We've used 20, so we have 8 more electrons to configure. And the 3d subshell can fit the 8 we need, so we have 3d8. And there you go, you've got the exact same answer that we had when we used the first method. Now I like the first method because you're looking at the periodic table the whole time, so you kind of understand an intuition And you also don't have to keep remembering, OK, how many have I used up as I filled the shells? Right? Here you have to say, i used two, then I used two more. And you have to draw this kind of elaborate diagram. Here you can just use the periodic table. And the important thing is you can work backwards. Here there's no way of just eyeballing this and saying, OK, our most energetic electrons are going to be and our highest energy shell is going to be 4s2. There's no way you could get that out of this without going through this fairly involved process. But when do you use this method, you can immediately say, OK, if I'm worried about element Zr, right here. If I'm worried about element Zr. I could go through the whole exercise of filling out the entire electron configuration. But usually the highest shell, or the highest energy electrons, are the ones that matter the most. So you immediately say, OK, I'm filling in 2 d there, but remember, d, you go one period below." + }, + { + "Q": "2:30 , why is isn't 4D 8 , why do we subtract 1 from the D block", + "A": "there is a trend with the respective D and S orbitals once you are referring to Copper and Chromium. The trend is that when 4s orbitals have a lower energy than the 3d orbitals so they give off an electron and have only one electron on its orbital instead of the required electron pair. When referring to 4D and 5s orbitals its the same trend.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that" + }, + { + "Q": "At 02:43, shouldn't the 3d8 come before the 4s2?", + "A": "4s comes before 3d because, before the orbitals are filled, the 4s is lower in energy than the 3d so is occupied first.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 163 + ], + "3min_transcript": "We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that from the nucleus than these right here. Now, another way to figure out the electron configuration for nickel-- and this is covered in some chemistry classes, although I like the way we just did it because you look at the periodic table and you gain a familiarity with it, which is important, because then you'll start having an intuition for how different elements react with each other -- is to just say, oK, nickel has 28 electrons, if it's neutral. It has 28 electrons, because that's the same number of protons, which is the atomic number. Remember, 28 just tells you how many protons there are. This is the number of protons. We're assuming it's neutral. So it has the same number of electrons. That's not always going to be the case. But when you do these electron configurations, that tends to be the case. So if we say nickel has 28, has an atomic number of 28, so it's electron configuration we can do it this way, too. We can write the energy shells. So one, two, three, four." + }, + { + "Q": "In the first video in this series on 'Orbitals' Sal instructed us to visualize that those electrons furthest away from the nucleus posses the greatest potential energy; they are in the highest energy state. On this logic, how is it that at around 3:20 in this video Sal lets us know that within Nickel [Ni] the 3D8 electrons are higher energy than the 4S2; how can this be when the 4S2 electrons are further away from the nucleus than 3D8?", + "A": "The 3d electrons in Ni have to fit in between the negative charges of all of the 3s and 3p electrons. The 4s shell doesn t have as much repulsion, so it takes less energy for an e- to live in a 4s orbital than in a 3d orbital.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 200 + ], + "3min_transcript": "We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that from the nucleus than these right here. Now, another way to figure out the electron configuration for nickel-- and this is covered in some chemistry classes, although I like the way we just did it because you look at the periodic table and you gain a familiarity with it, which is important, because then you'll start having an intuition for how different elements react with each other -- is to just say, oK, nickel has 28 electrons, if it's neutral. It has 28 electrons, because that's the same number of protons, which is the atomic number. Remember, 28 just tells you how many protons there are. This is the number of protons. We're assuming it's neutral. So it has the same number of electrons. That's not always going to be the case. But when you do these electron configurations, that tends to be the case. So if we say nickel has 28, has an atomic number of 28, so it's electron configuration we can do it this way, too. We can write the energy shells. So one, two, three, four." + }, + { + "Q": "At 00:21 Sir said that helium belongs to 's' block but all noble gases belong to 'p' block. This can disturb the configuration of electrons in Ni atom. Why is 'He' considered to be in 's' block?", + "A": "He has no p orbitals at all. It just has a completely full 1s\u00c2\u00b2 orbital and that is all.", + "video_name": "YURReI6OJsg", + "timestamps": [ + 21 + ], + "3min_transcript": "Let's figure out the electron configuration for nickel, right there. 28 electrons. We just have to figure out what shells and orbitals they go in. 28 electrons. So the way we've learned to do it is, we defined this as the s-block. And we can just remember that helium actually belongs here when we talk about orbitals in the s-block. This is the d-block. This is the p-block. And so we could start with the lowest energy electrons. We could either work forward or work backwards. If we work forwards, first we fill up the first two electrons going to 1s2. So remember we're doing nickel. So we fill up 1s2 first with two electrons. Then we go to 2s2. And remember this little small superscript 2 just means we're putting two electrons into that subshell or into that orbital. Actually, let me do each shell in a different color. So 2s2. We fill out all of these, right there. So 2p6. Let's see, so far we've filled out 10 electrons. We've configured 10. You can do it that way. Now we're on the third shell. The third shell. So now we go to 3s2. Remember, we're dealing with nickel, so we go to 3s2. Then we fill out in the third shell the p orbital. So 3p6. We're in the third period, so that's 3p6, right there. There's six of them. And then we go to the fourth shell. I'll do it in yellow. So we do 4s2. 4s2. And now we're in the d-block. And so we're filling in one, two, three, four, five, six, seven, eight in this d-block. So it's going to say d8. And remember, it's not going to be 4d8. So it will be 3d8. So we could write 3d8 here. So this is the order in which we fill, from lowest energy state electrons to highest energy state. But notice the highest energy state electrons, which are these that we filled in, in the end, these eight, these went into the third shell. So when you're filling the d-block, you take the period that you're in minus one. So we were in the fourth period in the periodic table, but we subtracted one, right? This is 4 minus 1. So this is the electron configuration for nickel. And of course if we remember, if we care about the valence electrons, which electrons are in the outermost shell, then you would look at these right here. These are the electrons that will react, although these are in a higher energy state. And these react because they're the furthest. Or at least, the way I visualize them is that" + }, + { + "Q": "At 1:35, are there more than one type of current if there is electron current? I thought that the only thing moving was the electrons moving from atom to atom, or are the atoms actually moving, because protons can't move on their own.", + "A": "Current is a general concept of moving charge. There are two types of charge, but we don t divide that into two types of current. One + charge moving to the left is the same current as one - charge moving to the right.", + "video_name": "17EhKw2tsu4", + "timestamps": [ + 95 + ], + "3min_transcript": "- [Voiceover] When we start to study electricity, we need to get an idea of what is Current and what is Voltage and in two earlier videos, I talked about the idea of current and voltage, current and voltage and, and what they meant. And when we talked about current, it's easiest to describe current when we talk about wires. Let's say we have a copper wire. We talked about a copper wire and inside it was, there was electrons in it, and they have a negative charge, we know they have a negative charge, and if we put a voltage on them, those electrons would move in some direction like that. So if I put a plus voltage over here and a minus voltage over here, the electrons are repelled by the minus voltage and they're attracted to the positive voltage. That is called an Electron Current. in terms of what's actually happening inside a wire makes some sense, it's easier to understand current and that these electrons are moving around. And whenever we talk about this, we'll talk about it specifically that there's an electron current going on here. Now at the same time, what I said in that video, and I'll say again, is the convention for describing current is this. This is called the Conventional Current Direction. The convention we've had for hundreds of years is that current is the direction that a positive charge would move if there was a positive charge there. So, whenever we talk about current from now on, it'll always be conventional current, and in fact, we don't even need to mention conventional any more, it's just current. Current is the direction that positive charges would move. the word, electron current. Now, as a reminder, when we talked about voltage, uh, this was built up by analogy and the analogy was to electrons rolling down a mountaintop, so here's our mountain, remember this? And I built a battery or another voltage source like this, and we said, that what a battery does is it pumps out energetic electrons, and they go down a hill. Roll downhill and go back into the, back into the positive terminal of the battery. And when we design circuits, what we do, is we, we put stuff in the way of this electron on its path, and this is where we build our circuits. So the, the electron current is going in this direction here down the hill. The conventional current direction or the current direction is this way." + }, + { + "Q": "9:50 When you said 1c=6.24x10^18, can you show me step by step to convert it to electrons? how did you get 1.6x10^-19?", + "A": "1 C = 6.24E18 electrons. Divide both sides bye 6.24E18 and you will have coulombs per electron.", + "video_name": "IDQYakHRAG8", + "timestamps": [ + 590 + ], + "3min_transcript": "much, much, much, much smaller mass than a proton, most of the mass of an atom is from the protons and the neutrons. So an electron has a much, much smaller mass than the protons and the neutrons, but it has the same but opposite charge as a proton. So sometimes the convention is to write negative e, or maybe even negative one e, sometimes depending on whether you view this as a kind of the actual charge or whether you view this as a unit, but here I'll view this as the actual charge. You could view negative e as the charge, as the charge of an electron. And something that has no charge, like a neutron, we say they're neutral, and actually that is why they are called neutrons, because they are neutral, they don't have charge. So that right over there, that over there is, is a neutron. Now when we start to get on kind of a larger scale, not on a sub-atomic scale anymore, in general the unit of charge that we typically use is the coulomb, is the coulomb. Coulomb, it's named for Charles Augustin de Coulomb, so if we're talking about the guy, and he was an 18th Century French physicist, we would use capital C, but if we're talking about the units, we would use lowercase c, the coulomb, the coulomb. And the coulomb is defined, so one coulomb, let me write it right over here, one coulomb and it uses the abbreviation uppercase C, is equal, or I'll say approximately equal to, we're going to round here, it's approximately equal to 6.24, 6.24 times 10 to the eighteenth e, or you could say, in magnitude wise, it's equal to the charge of 6.24 times 10 to the eighteenth protons, it would be the opposite if you're talking about electrons, it would be 6.24 times 10 to the eighteenth electrons. Now if you want to go the other way around, what is the charge of, the magnitude of the charge of say a proton in terms of coulombs, well you would just take the inverse of this. So you could say that e is approximately equal to the inverse of this which is 1.60, I guess you could say the reciprocal of this, 1.60 times 10 to the negative 19, times 10 to the negative 19 coulombs. So hopefully this gives you an appreciation for, I guess at a base level, what charge is. And in some ways it's like it's this everyday thing, you're used to it, we're used to dealing with electricity and we'll talk much more about that in depth. But at some levels it is this thing, one of the mysteries of the universe, how did these two particles know to attract each other, you know it looks like they're at a distance, how do they immediately exert a force on each other. how do these particles know immediately to repel each other," + }, + { + "Q": "At 6:55, Sal says that your hair is attracted to the balloon because the negative charge of the balloon has repelled the electrons in your hair, leaving a disproportionate number of protons. But wouldn't it also be the case (and simpler to explain) that you hair is positively charged simply because it's missing the electrons that were \"stolen\" by the balloon?", + "A": "But the balloon attracts hair that it did not rub. How does it do that?", + "video_name": "IDQYakHRAG8", + "timestamps": [ + 415 + ], + "3min_transcript": "We associate mass as just, oh this is just something that we get, we understand it in our everyday life, but even mass, this is just a property of objects, it's just a property of matter, and we feel like we understand it because on our scales we understand notions of things like weight and volume, but even mass can get quite exotic. But anyway, the whole point of this video is not to talk about mass, it's to talk about charge. But all of these things that we talk about in physics, these are just properties that will help us deal with these notions, these behaviors in different frameworks. But anyway, let's get back to this little atom that I was constructing. So this atom, let's say it has two electrons, and obviously this is not drawn to scale, and each of these electrons have a negative charge, and they're kind of jumping around here, buzzing around this nucleus of this atom. And the reason why, this model, even going down to the atomic scale and thinking in protons and electrons is interesting, is that it allows us to start explaining what is happening in the triboelectric effect. rub that balloon on your hair, because of the property of the balloon, the material of the balloon, and the materials of your hair, when they come in contact and they rub, the balloon is grabbing electrons from your hair. So the balloon is grabbing electrons from your hair, and so it is getting more negatively charged, it is getting more negatively charged, and your hair is getting more positively charged, or essentially it's lost these electrons. And so when you put the balloon now close to your hair, remember like charges repel each other, so the electrons in your hair try to move away from these other electrons, the negative charge tries to move away from the negative charge, and I guess you could say that the tips of your hair will then become more positive. Are more positive and they will be attracted, and they will be attracted to the balloon. of transfer of electrons, that's exactly what's happening. And so when you think that way, it's like ok, we are scientists, this is a nice model, we can start to think about what's happening here. This model actually explains a whole ton of behavior that we've observed in the universe, including things like, lightning and whatever else, you know the static shock that you get when you might touch a doorknob after rubbing your shoes along the carpet. But we like to start, we like to quantify things, so we can start seeing how much they repel or how much they attract each other. And so the fundamental unit of charge, or one of the fundamental units of charge, or I guess you could say the elementary unit of charge is defined in terms of the charge of a proton or an electron. So the fundamental, or I guess you could say the elementary unit of charge is denoted by the letter e, and this is the charge of a proton, this is e for elementary, charge of proton." + }, + { + "Q": "At 6:16, why does the balloon grab electrons only? Why not protons?", + "A": "Protons are very heavy and are tightly bound in the nucleus of the atoms. Electrons are light and are loosely bound to their atoms", + "video_name": "IDQYakHRAG8", + "timestamps": [ + 376 + ], + "3min_transcript": "But if we run with this model, we can imagine at the atomic scale, the nuclei of atoms are composed of protons and neutrons. So if you have some protons, and then you have some neutrons, I'll do two of each, you have some neutrons, and based on this framework, protons have a positive charge. Protons have a positive charge. Now once again, this convention of calling them positive and putting a plus on it, it's not like protons have a little plus sign tattooed onto them somehow. We could have called those, we could have said they have a red charge, or we could have even said, we wouldn't of had to even use the word charge, this is just a convention that we have decided to use. And so we say protons have positive charge and then, kind of buzzing around the nucleus of an atom, you often, or usually, or often have electrons. Electrons have a lot less mass. We associate mass as just, oh this is just something that we get, we understand it in our everyday life, but even mass, this is just a property of objects, it's just a property of matter, and we feel like we understand it because on our scales we understand notions of things like weight and volume, but even mass can get quite exotic. But anyway, the whole point of this video is not to talk about mass, it's to talk about charge. But all of these things that we talk about in physics, these are just properties that will help us deal with these notions, these behaviors in different frameworks. But anyway, let's get back to this little atom that I was constructing. So this atom, let's say it has two electrons, and obviously this is not drawn to scale, and each of these electrons have a negative charge, and they're kind of jumping around here, buzzing around this nucleus of this atom. And the reason why, this model, even going down to the atomic scale and thinking in protons and electrons is interesting, is that it allows us to start explaining what is happening in the triboelectric effect. rub that balloon on your hair, because of the property of the balloon, the material of the balloon, and the materials of your hair, when they come in contact and they rub, the balloon is grabbing electrons from your hair. So the balloon is grabbing electrons from your hair, and so it is getting more negatively charged, it is getting more negatively charged, and your hair is getting more positively charged, or essentially it's lost these electrons. And so when you put the balloon now close to your hair, remember like charges repel each other, so the electrons in your hair try to move away from these other electrons, the negative charge tries to move away from the negative charge, and I guess you could say that the tips of your hair will then become more positive. Are more positive and they will be attracted, and they will be attracted to the balloon." + }, + { + "Q": "6:19\nThis might be an arbitrary detail, but why is the balloon grabbing electrons from your hair and not protons? Is there a reason for this?\nThanks.", + "A": "Atoms have electrons on the outside, and nuclei on the inside. The electrons are very small compared to the nucleus.", + "video_name": "IDQYakHRAG8", + "timestamps": [ + 379 + ], + "3min_transcript": "But if we run with this model, we can imagine at the atomic scale, the nuclei of atoms are composed of protons and neutrons. So if you have some protons, and then you have some neutrons, I'll do two of each, you have some neutrons, and based on this framework, protons have a positive charge. Protons have a positive charge. Now once again, this convention of calling them positive and putting a plus on it, it's not like protons have a little plus sign tattooed onto them somehow. We could have called those, we could have said they have a red charge, or we could have even said, we wouldn't of had to even use the word charge, this is just a convention that we have decided to use. And so we say protons have positive charge and then, kind of buzzing around the nucleus of an atom, you often, or usually, or often have electrons. Electrons have a lot less mass. We associate mass as just, oh this is just something that we get, we understand it in our everyday life, but even mass, this is just a property of objects, it's just a property of matter, and we feel like we understand it because on our scales we understand notions of things like weight and volume, but even mass can get quite exotic. But anyway, the whole point of this video is not to talk about mass, it's to talk about charge. But all of these things that we talk about in physics, these are just properties that will help us deal with these notions, these behaviors in different frameworks. But anyway, let's get back to this little atom that I was constructing. So this atom, let's say it has two electrons, and obviously this is not drawn to scale, and each of these electrons have a negative charge, and they're kind of jumping around here, buzzing around this nucleus of this atom. And the reason why, this model, even going down to the atomic scale and thinking in protons and electrons is interesting, is that it allows us to start explaining what is happening in the triboelectric effect. rub that balloon on your hair, because of the property of the balloon, the material of the balloon, and the materials of your hair, when they come in contact and they rub, the balloon is grabbing electrons from your hair. So the balloon is grabbing electrons from your hair, and so it is getting more negatively charged, it is getting more negatively charged, and your hair is getting more positively charged, or essentially it's lost these electrons. And so when you put the balloon now close to your hair, remember like charges repel each other, so the electrons in your hair try to move away from these other electrons, the negative charge tries to move away from the negative charge, and I guess you could say that the tips of your hair will then become more positive. Are more positive and they will be attracted, and they will be attracted to the balloon." + }, + { + "Q": "10:03 i thought coulombs were only for electrons/positrons", + "A": "Coulomb is the unit of charge. Whatever has charge, the charge can be measured in coulombs.", + "video_name": "IDQYakHRAG8", + "timestamps": [ + 603 + ], + "3min_transcript": "much, much, much, much smaller mass than a proton, most of the mass of an atom is from the protons and the neutrons. So an electron has a much, much smaller mass than the protons and the neutrons, but it has the same but opposite charge as a proton. So sometimes the convention is to write negative e, or maybe even negative one e, sometimes depending on whether you view this as a kind of the actual charge or whether you view this as a unit, but here I'll view this as the actual charge. You could view negative e as the charge, as the charge of an electron. And something that has no charge, like a neutron, we say they're neutral, and actually that is why they are called neutrons, because they are neutral, they don't have charge. So that right over there, that over there is, is a neutron. Now when we start to get on kind of a larger scale, not on a sub-atomic scale anymore, in general the unit of charge that we typically use is the coulomb, is the coulomb. Coulomb, it's named for Charles Augustin de Coulomb, so if we're talking about the guy, and he was an 18th Century French physicist, we would use capital C, but if we're talking about the units, we would use lowercase c, the coulomb, the coulomb. And the coulomb is defined, so one coulomb, let me write it right over here, one coulomb and it uses the abbreviation uppercase C, is equal, or I'll say approximately equal to, we're going to round here, it's approximately equal to 6.24, 6.24 times 10 to the eighteenth e, or you could say, in magnitude wise, it's equal to the charge of 6.24 times 10 to the eighteenth protons, it would be the opposite if you're talking about electrons, it would be 6.24 times 10 to the eighteenth electrons. Now if you want to go the other way around, what is the charge of, the magnitude of the charge of say a proton in terms of coulombs, well you would just take the inverse of this. So you could say that e is approximately equal to the inverse of this which is 1.60, I guess you could say the reciprocal of this, 1.60 times 10 to the negative 19, times 10 to the negative 19 coulombs. So hopefully this gives you an appreciation for, I guess at a base level, what charge is. And in some ways it's like it's this everyday thing, you're used to it, we're used to dealing with electricity and we'll talk much more about that in depth. But at some levels it is this thing, one of the mysteries of the universe, how did these two particles know to attract each other, you know it looks like they're at a distance, how do they immediately exert a force on each other. how do these particles know immediately to repel each other," + }, + { + "Q": "At 10:40 Why does he say -19 not -18?", + "A": "As when we divide 1 by 6.24x10 to the power 18,then we get a value 0.1609x10 to the power -18,and shifting one decimal backwards,th value becomes 1.609x 10 to the power -19.", + "video_name": "IDQYakHRAG8", + "timestamps": [ + 640 + ], + "3min_transcript": "in general the unit of charge that we typically use is the coulomb, is the coulomb. Coulomb, it's named for Charles Augustin de Coulomb, so if we're talking about the guy, and he was an 18th Century French physicist, we would use capital C, but if we're talking about the units, we would use lowercase c, the coulomb, the coulomb. And the coulomb is defined, so one coulomb, let me write it right over here, one coulomb and it uses the abbreviation uppercase C, is equal, or I'll say approximately equal to, we're going to round here, it's approximately equal to 6.24, 6.24 times 10 to the eighteenth e, or you could say, in magnitude wise, it's equal to the charge of 6.24 times 10 to the eighteenth protons, it would be the opposite if you're talking about electrons, it would be 6.24 times 10 to the eighteenth electrons. Now if you want to go the other way around, what is the charge of, the magnitude of the charge of say a proton in terms of coulombs, well you would just take the inverse of this. So you could say that e is approximately equal to the inverse of this which is 1.60, I guess you could say the reciprocal of this, 1.60 times 10 to the negative 19, times 10 to the negative 19 coulombs. So hopefully this gives you an appreciation for, I guess at a base level, what charge is. And in some ways it's like it's this everyday thing, you're used to it, we're used to dealing with electricity and we'll talk much more about that in depth. But at some levels it is this thing, one of the mysteries of the universe, how did these two particles know to attract each other, you know it looks like they're at a distance, how do they immediately exert a force on each other. how do these particles know immediately to repel each other, that they're communicating somehow, or I guess once you get to quantum mechanical, an argument can be made that they are communicating somehow. But in our everyday, kind of logical sense, it's like well at a distance, how do these things actually know to repel or attract, and what is this charge anyway? You know we've put all these names around it but to kind of help us think about it and have a framework and predict what will happen. But do we really know what this charge thing is. So on one level it's kind of plain and mundane, and it deals with balloons and hair, but on another level it's this deep thing about this universe, it's a deep property of matter that we can manipulate and we can predict, but it is still this very fundamental and somewhat mysterious thing." + }, + { + "Q": "Starting with 7:45, Sal multiplied the number of the mole( ~6 * 10 ^ 23 molecules / mole) with the 1 * 10 ^ -7 moles / liter.\nI know what the term \"mole\" is.\nHowever, I just don't get the part why would you multiply the mole on the already made(shown) [H3O]?", + "A": "I didn t watch the video, but he must have wanted to know how many molecules there are per liter (molecules / mole) * (moles / liter ) = molecules / liter", + "video_name": "NUyYlRxMtcs", + "timestamps": [ + 465 + ], + "3min_transcript": "giving that a positive charge. And so you might say, \"Well how frequently would \"I find hydronium ions in water?\" Well the concentration, let me actually draw a little tub of water here, let's say this is a liter of water. This is a liter, this is a liter of water. The concentration of hydronium in typical water, the concentration of H three O, the concentration of H three O in typical water, and you put brackets around something to denote \"concentration,\" is one times ten to the negative seven molar. And molar, this just means \"moles per liter.\" This is the same thing as one times ten to the negative seven moles, moles per liter. And now you might be saying, \"Well, what's a mole?\" Well I encourage you to watch the video on what a mole is, but a mole is a quantity. It's like saying, \"a dozen.\" But it's a much larger, a dozen is equal to 12 of something. A mole is roughly equal to, let me write it. A mole is approximately equal to 6.02 times ten to the 23rd. And you're typically talking about molecules. A mole of a substance means approximately 6.022, it actually keeps going, times ten to the 23rd molecules of that thing. So you might say, \"Hey, one times ten to the negative seven \"times 6.02 times ten to the 23rd, \"that would still get us,\" well let's see, this, One times ten to the negative seven moles per liter, times, times, I'll do it this way. Times six, I'll just go with six, since we're gonna go approximately. So approximately, six times ten to the 23rd. Six times 23rd molecules, molecules per mole. Molecules per mole, well these two would cancel out, and you would multiply these two numbers, you would get six times, let's see. that's still gonna be ten to the 16th power, molecules per liter. Molecules per liter, so your first reaction is, \"Oh my God!\" \"I'm gonna have six times,\" or roughly, I'll say roughly. \"Approximately six times ten to the 16th \"molecules of hydronium in this?\" \"That's a lot, we should see it all the time.\" But we have to remind ourselves. There's just a lot of molecules of water in there as well. In fact, a liter of water is roughly, so one liter of H two O, contains, contains approximately 56, 56 moles, moles of H two O. So one way to think about it is, \"I have one, I have one times,\" and if I'm thinking about a liter of water, I have, I'll do it over here, \"I have one times ten to the negative seven \"moles of, moles of H three O for every," + }, + { + "Q": "4:05, what are ions?", + "A": "Ions are atoms that have either lost or gained an electron, giving them an overall charge since they have an unequal number of protons (positive charge) and electrons (negative charge).", + "video_name": "NUyYlRxMtcs", + "timestamps": [ + 245 + ], + "3min_transcript": "and the partial positive charge over here. So these would be attracted to this partial positive charge. Remember there, you have a partial negative charge over here, this is actually what's forming the hydrogen bond. And it actually could bond to the hydrogen proton, while both of these electrons, including one of these electrons that used to be part of this hydrogen, or you could consider used to be part of that hydrogen, are nabbed, are nabbed by this oxygen. And in this circumstance, and I'm not saying that this happens all the time, but under just the right conditions, this actually can happen, and what would result, so let me, what result is, this thing over here, instead of just being a neutral water molecule, would look like this. So you have your oxygen, you have, not only your two hydrogens now, you now have a third hydrogen. You now have a third hydrogen. So you have theses two covalent bonds, these two covalent bonds, this lone pair. And now this lone pair, which I have circled in blue, This electron right over here of the hydrogen got nabbed by this oxygen. So now you've formed another covalent bond. And now this character over here, he's lost the hydrogen proton, but he's kept all of the electrons. So this character over here's gonna look like this. You're gonna have your oxygen, and now it's only going to only be bonded to one hydrogen, only bonded to one hydrogen. Has these two original lone pairs. These two original lone pairs right over here. And then took both of the electrons from this covalent bond. And took both of the electrons from this covalent bond. And so it has another lone pair. So this molecule gained just a proton without getting any electrons. So if you do that, you're now going to have a net positive charge for this one over here. And this molecule over here, actually let me, let me, ugh, let me just write it. I wanna write it a little bit neater. And this molecule over here, so we have this molecule plus this one, this one So it now has a negative charge. So just like that, you went from two neutral water molecules, to two ions. And these ions, this one over here, the one on the left, the one that is now H three O, H three O, H three O, and it now has a positive charge, positive charge, actually I put that O in a different color. H three, H three O. It's a positive charge, this is called the \"hydronium ion.\" Hydronium, hydronium. And this one over here, that is OH minus, so it's OH, O, let me get the colors right. OH minus. This is called the \"hydroxide ion,\" or since it's negative you can just call it an \"anion.\"" + }, + { + "Q": "How can that one Hydrogen at 3:37 take 2 electrons if it only has one proton (+1) charge?", + "A": "No hydrogens took any electrons at any point here Both the electrons that once were in an oxygen-hydrogen bond on the right water molecule have become a third lone pair of electrons on that right product (hydroxide)", + "video_name": "NUyYlRxMtcs", + "timestamps": [ + 217 + ], + "3min_transcript": "and the partial positive charge over here. So these would be attracted to this partial positive charge. Remember there, you have a partial negative charge over here, this is actually what's forming the hydrogen bond. And it actually could bond to the hydrogen proton, while both of these electrons, including one of these electrons that used to be part of this hydrogen, or you could consider used to be part of that hydrogen, are nabbed, are nabbed by this oxygen. And in this circumstance, and I'm not saying that this happens all the time, but under just the right conditions, this actually can happen, and what would result, so let me, what result is, this thing over here, instead of just being a neutral water molecule, would look like this. So you have your oxygen, you have, not only your two hydrogens now, you now have a third hydrogen. You now have a third hydrogen. So you have theses two covalent bonds, these two covalent bonds, this lone pair. And now this lone pair, which I have circled in blue, This electron right over here of the hydrogen got nabbed by this oxygen. So now you've formed another covalent bond. And now this character over here, he's lost the hydrogen proton, but he's kept all of the electrons. So this character over here's gonna look like this. You're gonna have your oxygen, and now it's only going to only be bonded to one hydrogen, only bonded to one hydrogen. Has these two original lone pairs. These two original lone pairs right over here. And then took both of the electrons from this covalent bond. And took both of the electrons from this covalent bond. And so it has another lone pair. So this molecule gained just a proton without getting any electrons. So if you do that, you're now going to have a net positive charge for this one over here. And this molecule over here, actually let me, let me, ugh, let me just write it. I wanna write it a little bit neater. And this molecule over here, so we have this molecule plus this one, this one So it now has a negative charge. So just like that, you went from two neutral water molecules, to two ions. And these ions, this one over here, the one on the left, the one that is now H three O, H three O, H three O, and it now has a positive charge, positive charge, actually I put that O in a different color. H three, H three O. It's a positive charge, this is called the \"hydronium ion.\" Hydronium, hydronium. And this one over here, that is OH minus, so it's OH, O, let me get the colors right. OH minus. This is called the \"hydroxide ion,\" or since it's negative you can just call it an \"anion.\"" + }, + { + "Q": "at 11:13, how are you getting 116 J? what is the conversion?", + "A": "1/2 * 0.145 kg * (40 m/s)\u00c2\u00b2 = 116 kg*m\u00c2\u00b2/s\u00c2\u00b2 = 116 J", + "video_name": "o7_zmuBweHI", + "timestamps": [ + 673 + ], + "3min_transcript": "If an object is translating and rotating and you want to find the total kinetic energy of the entire thing, you can just add these two terms up. If I just take the translational one half M V squared, and this would then be the velocity of the center of mass. So you have to be careful. Let me make some room here, so let me get rid of all this stuff here. If you take one half M, times the speed of the center of mass squared, you'll get the total translational kinetic energy of the baseball. And if we add to that the one half I omega squared, so the omega about the center of mass you'll get the total kinetic energy, both translational and rotational, so this is great, we can determine the total kinetic energy altogether, rotational motion, translational motion, from just taking these two terms added up. So what would an example of this be, let's just get rid of all this. Let's say this baseball, someone pitched this thing, and the radar gun shows that this baseball was hurled So it's heading toward home plate at 40 meters per second. The center of mass of this baseball is going 40 meters per second toward home plate. Let's say it's also, someone really threw the fastball. This thing's rotating with an angular velocity of 50 radians per second. We know the mass of a baseball, I've looked it up. The mass of a baseball is about 0.145 kilograms and the radius of the baseball, so a radius of a baseball is around seven centimeters, so in terms of meters that would be 0.07 meters, so we can figure out what's the total kinetic energy, well there's gonna be a rotational kinetic energy and there's gonna be a translational kinetic energy. The translational kinetic energy, gonna be one half the mass of the baseball times the center of mass speed of the baseball squared which is gonna give us one half. The mass of the baseball was 0.145 and the center of mass speed of the baseball is 40, that's how fast the center of mass of this baseball is traveling. translational kinetic energy. How much rotational kinetic energy is there, so we're gonna have rotational kinetic energy due to the fact that the baseball is also rotating. How much, well we're gonna use one half I omega squared. I'm gonna have one half, what's the I, well the baseball is a sphere, if you look up the moment of inertia of a sphere cause I don't wanna have to do summation of all the M R squareds, if you do that using calculus, you get this formula. That means in an algebra based physics class you just have to look this up, it's either in your book in a chart or a table or you could always look it up online. For a sphere the moment of inertia is two fifths M R squared in other words two fifths the mass of a baseball times the raise of the baseball squared. That's just I, that's the moment of inertia of a sphere. So we're assuming this baseball is a perfect sphere. It's got uniform density, that's not completely true. But it's a pretty good approximation. Then we multiply by this omega squared," + }, + { + "Q": "5:57 When we are numbering the \"shortest path\" as stated in the video, i.e. the carbons between the bridge carbons, do we automatically number away from carbon number 1? So if there were 2 carbons between bridge carbons 1 and 4, how would we number them?", + "A": "If it s symmetrical then it doesn t really matter If it was subsituted then you would number the carbon with the functional group first", + "video_name": "cM-SFbffb7k", + "timestamps": [ + 357 + ], + "3min_transcript": "bicyclo[4.2.0.]octane. Let's do another one following our general formula here for naming a bicyclo compound. So let's take a cyclohexane ring. And let's make this a bridgehead carbon. And we're actually going to put a carbon between our bridgehead carbons like that. Now, this is hard to see when it's drawn this way. So usually, you'll see it drawn a little bit differently. Usually you'll see it drawn with a little bit more three-dimensionality to it. So hopefully we can do that here. So it looks something like this. So let's find our bridgehead carbons here. So we know that this is a bridgehead carbon, this is a bridgehead carbon. Those correspond to these guys over here. And how many cuts would it take to turn this into an open chain alkane? Well, let's go ahead and look at the drawing on the left. we can see it's now an open chain compound. So it took two cuts, so it's bicyclo. Let's go ahead and put those bonds back in there like that. So it's a bicyclo compound. So we can go ahead and start naming it as a bicyclo compound. We need to go ahead and number it, so let's start with our bridgehead carbon. So we'll start with this is our bridgehead carbon. Find the longest path. Well, it doesn't matter if I go left or right here, since it's the same length each way around. So I'll just go to the right. So 1, 2, 3, 4. Second longest path, so I'm going to keep on going this way, so 5 and 6. And then finally, the shortest path, which is the one carbon in between my two bridgehead carbons, which would then get a 7, so seven total carbons. All right, so when I'm naming this, I need to put my brackets in here. And I put the number of carbons in the longest path, So how many carbons were my longest path, excluding Here's 1 and here is 2, so I'm going to go ahead and put a 2 in here. And then I go to my second longest path. Well, that was over on this side. And there are also two carbons on my second longest path. Obvious, it's the same length as the other one. So it's 2, 2. And then my shortest path, how many carbons are there in my shortest path, not counting my bridgehead carbon? There's one. So I go ahead and put a 1 here. And so I have bicyclo[2.2.1]. And then my total number of carbons is 7. So this is a bicycloheptane molecules. So bicyclobicyclo[2.2.1]heptane would be the official IUPAC name for this molecule. This molecule turns up a lot in nature. So much so, that it actually has its own common name. This is also called norborane. So let's go ahead and write that here. So this is norborane, again, a structure" + }, + { + "Q": "9:43. Looking at the bicyclo on the left...why couldn't you start from the bridge carbon and then go left and make the 6th carbon into the second? I'm guessing Counting on the longest carbon chain takes precedence over having the methyl on a lower carbon?", + "A": "Correct. Counting on the longest carbon chain takes precedence over having the methyl on a lower numbered carbon. But if the 6-methyl had been on C-7, the name would still be 6,8-dimethylbicyclo[3.2.1]octane. We would still start numbering by going around the largest bridge, but we would use the numbering in the crossed-out structure in order to get the lowest numbers.", + "video_name": "cM-SFbffb7k", + "timestamps": [ + 583 + ], + "3min_transcript": "I could start numbering with either of these be number one. So let's just make this number 1 here. So if that's number 1, I next go to my longest path. So that would be going to the right here. So this would give me a 2 for this carbon, a 3 for this carbon, a 4 for this carbon, a 5 for this carbon. Next, the second longest path. Well, I can just keep going, make that a 6, make this a 7. And then finally, my shortest path, which is up here at the top. So that would be 8 total carbons for my parent name. On the dot structure on the left, again, the exact same dot structure. This time, I'm going to start with the opposite bridgehead carbons. So I'm going to start with this bridgehead carbon. And then follow the same rule. So the longest path, which would be to the right here. So 2, 3, 4, 5, 6, 7, and then 8. So which one of these numbering systems is the correct one? Remember, our goal is to get the lowest And if I look at the example on the right, I have a methyl group in the 7 position and a methyl group in the 8 position. The example on the left has a methyl group in the 6 position and a methyl group in the 8 position. Since I want to give my lowest number possible to my substituent, I'm going to not number it the way on the right. So I'm going to go with the way on the left here. So let's go ahead and start naming it. We already know this is a bicyclo compound. And we have two methyl groups, one at 6 and one at 8. So we could start naming it by saying 6,8-dimethyl. And then we know this is a bicyclo compound, so bicyclo. And then in the brackets, we're going to do the longest path first, excluding the bridgehead carbon. So the longest path is on the right. How many carbons are there in the longest path? There are three excluding the bridgehead carbon, Next, the second longest path, excluding the bridgehead carbon. Well, that was these two right here. So next, there'll be a 2. And then finally, the shortest path excluding our bridgehead carbon. There's only one carbon in our shortest path. So we put that in there. And then, there were 8 total carbons, so it is octane. So the final IUPAC name for this molecule is 6,8-dimethylbicyclo[3.2.1.]octane." + }, + { + "Q": "At 1:48, how is the OH coming out at us in space", + "A": "That picture of the model with black carbons and a red oxygen is showing how the model looks in 3D. If you put the carbon backbone in the same plane, then the groups coming off the carbon are either coming towards you or going away. It may help if you get your hands on a molecular modelling kit.", + "video_name": "ZAgQH2azx3w", + "timestamps": [ + 108 + ], + "3min_transcript": "- [Voiceover] In the video on bond line structures, we started with this Lewis dot structure on the left, and I showed you how to turn this Lewis dot structure into a bond line structure. So here's the bond line structure that we drew in that video. Bond line structures contain the same information as a Lewis dot structure, but it's obviously much easier, much faster, to draw the bond line structure on the right than the full Lewis dot structure on the left. What about three dimensional bond line structure? So how could you represent this molecule in three dimensions, using a flat sheet of paper? Well, on the left here, is a picture where I made a model of this molecule, and this is going to help us draw this molecule in three dimensions. So we have a flat sheet of paper, how could we represent this picture on our flat sheet of paper? Let's start with the carbon in the center, so that's our carbon in magenta, so that's this one on our Lewis dot structure, this one on our bond line structure. Well, the carbon in magenta is SP3 hybridized, tetrahedral geometry around that carbon. And if you look at that carbon on the picture here, you can see that this bond and this bond are in the same plane. So if you had a flat sheet of paper, you could say those bonds are in the same plane. So a line represents a bond in the plane of the paper, let me go ahead and draw that, so this is the carbon in magenta, and then we have these two bonds here, and those bonds are in the plane of the paper. Next, let's look at what else is connected to the carbon in magenta. Well, obviously, there is an OH, so let me go ahead and circle that. There is an OH we can see there is an OH here, and the OH, the OH in our picture, is coming out at us in space, so hopefully you can visualize that this bond in here is coming towards you in space, which is why this oxygen, this red oxygen atom, looks so big. So this is coming towards you, so let me go ahead and draw a wedge in here, and a wedge means that the bond is in front of your paper, so this means the OH is coming out at you in space, let me draw in the OH like that. Now let's look at what else is connected to that carbon in magenta, we know there's a hydrogen. We didn't draw it over here, but we know there's a hydrogen connected to that carbon, and we can see that this hydrogen, this hydrogen right here, let me go ahead and switch colors, this hydrogen is going away from us in space, so this bond is going away from us in space, or into the paper, or the bond is behind the paper. And we represent that with a dash, so I'm going to draw a dash here, showing that this hydrogen is going away from us. So we're imagining our flat sheet of paper and the OH coming out at us, and that hydrogen going away from us. All right, next, let's look at the carbon on the left here, so this carbon in blue, so that's this carbon, and I'll say that's this carbon over here on the left." + }, + { + "Q": "7:13 - 7:20 | Is there not also some Carbon-13 and Carbon-11?", + "A": "Most carbon is carbon-12, followed by a small about of carbon-13. All other isotopes of carbon are radioactive and only exist in insignificant trace amounts. However, carbon-11 is important for medical purposes.", + "video_name": "NG-rrorZcM8", + "timestamps": [ + 433, + 440 + ], + "3min_transcript": "Atomic mass units. But it's not the mass of just one atom or just one molecule. It's a weighted average across many, many ... of how typically, what you would see, or the makeup of what you would see on Earth. What do I mean by that? Well, on Earth, there are two ... The primary isotope of carbon is carbon-12. Carbon-12, which is defined as having a mass of exactly 12 atomic mass units. But there's also some carbon-14. Carbon-14. What do these numbers mean, just as a reminder? Well, carbon-12 has six protons, and the six protons are what make it carbon. Carbon-14 is also going to have six protons. But carbon-12, carbon-12 also has six neutrons. Six neutrons. I know what you're already thinking. You're, like, \"Well, wait. \"Why don't we say that a proton or a neutron \"weighs one atomic mass unit? \"Because it looks like this is 12, \"and I'm guessing that this, \"that this, the mass of this is going to be \"pretty close to 14.\" If you're thinking that way, that's not an unreasonable way to think. In fact, when I'm kind of just working through chemistry, that is how I think about it. But they don't weigh exactly one atomic mass unit by this definition. Remember, the electron is ever so small, it has very small mass, but it is contributing, or the electrons are contributing, something to the mass. So, a proton or a neutron have very, very, very close ... They are close to one atomic mass unit. Let me write this down. One proton, one proton, or one neutron, one neutron, very close to one atomic mass unit, but not exactly. But anyway, going back to what atomic weight is, the most common isotope of carbon ... Remember, when we're saying \"isotopes,\" we're saying the same element, we have the same number of protons, but we have different number of neutrons. The most common isotope on Earth is carbon-12, but there's also some carbon-14. If you were to take a weighted average, as found on the Earth, of all the carbon-12 and all of the carbon-14, the weighted average of the atomic masses is the atomic weight. And the atomic weight of carbon ... And you'll see this on a periodic table. In fact, I have one right over here. Notice, the six protons, this is what defines it to be carbon. But then they write 12.011, which is the weighted average of the masses of all of the carbons. Now, it's very close to 12, as opposed to being closer to 14, because most of the carbon on Earth is carbon-12. We could write this down. This is the atomic weight. This is the atomic weight of carbon on Earth. This is 12.011." + }, + { + "Q": "At 6:50 Sal says \"one proton or one neutron is very close to 1 amu\". He had said before that the electron has really small mass. Does that mean that 1 amu is essentially the combined mass of 1 proton or neutron plus 1 electron?", + "A": "No. An individual proton is 1.007276466812 u. An individual neutron is 1.00866491600 u. And an individual electron is 0.00054857990946 u. The amu is defined as 1/12th the mass of Carbon-12, which includes 6 protons, neutrons, electrons. It also has binding energy, which converts some of the mass into energy to hold the nucleus together.", + "video_name": "NG-rrorZcM8", + "timestamps": [ + 410 + ], + "3min_transcript": "Atomic mass units. But it's not the mass of just one atom or just one molecule. It's a weighted average across many, many ... of how typically, what you would see, or the makeup of what you would see on Earth. What do I mean by that? Well, on Earth, there are two ... The primary isotope of carbon is carbon-12. Carbon-12, which is defined as having a mass of exactly 12 atomic mass units. But there's also some carbon-14. Carbon-14. What do these numbers mean, just as a reminder? Well, carbon-12 has six protons, and the six protons are what make it carbon. Carbon-14 is also going to have six protons. But carbon-12, carbon-12 also has six neutrons. Six neutrons. I know what you're already thinking. You're, like, \"Well, wait. \"Why don't we say that a proton or a neutron \"weighs one atomic mass unit? \"Because it looks like this is 12, \"and I'm guessing that this, \"that this, the mass of this is going to be \"pretty close to 14.\" If you're thinking that way, that's not an unreasonable way to think. In fact, when I'm kind of just working through chemistry, that is how I think about it. But they don't weigh exactly one atomic mass unit by this definition. Remember, the electron is ever so small, it has very small mass, but it is contributing, or the electrons are contributing, something to the mass. So, a proton or a neutron have very, very, very close ... They are close to one atomic mass unit. Let me write this down. One proton, one proton, or one neutron, one neutron, very close to one atomic mass unit, but not exactly. But anyway, going back to what atomic weight is, the most common isotope of carbon ... Remember, when we're saying \"isotopes,\" we're saying the same element, we have the same number of protons, but we have different number of neutrons. The most common isotope on Earth is carbon-12, but there's also some carbon-14. If you were to take a weighted average, as found on the Earth, of all the carbon-12 and all of the carbon-14, the weighted average of the atomic masses is the atomic weight. And the atomic weight of carbon ... And you'll see this on a periodic table. In fact, I have one right over here. Notice, the six protons, this is what defines it to be carbon. But then they write 12.011, which is the weighted average of the masses of all of the carbons. Now, it's very close to 12, as opposed to being closer to 14, because most of the carbon on Earth is carbon-12. We could write this down. This is the atomic weight. This is the atomic weight of carbon on Earth. This is 12.011." + }, + { + "Q": "At 3:10, can't we write HOH as H2O?", + "A": "You can, but HOH is more commonly used in university or higher level organic chemistry when you deal with reaction mechanism.", + "video_name": "XEPdMvZqCHQ", + "timestamps": [ + 190 + ], + "3min_transcript": "And the carbon in magenta is bonded to three other hydrogens. So we could represent that as a CH three. So I could write CH three here, and the carbon in red is this one and the carbon in magenta is this one. On the left side, the carbon in red is bonded to another carbon in blue and the carbon in blue is bonded to three hydrogens, so there's another CH three on the left side, so let me draw that in, so we have a CH three on the left and the carbon in blue is directly bonded to the carbon in red. So this is called a partially condensed structure so this is a partially condensed, partially condensed structure. We haven't shown all of the bonds here but this structure has the same information as the Lewis structure on the left. it's just a different way to represent that molecule. We could keep going. We could go for a fully condensed structure. So let's do that. Focus in on the carbon in red. So this one right here. So let me draw in that carbon over here. So that's that carbon. That carbon is bonded to two CH three groups. There's a CH three group on the right, so there's the CH three group on the right. And there's a CH three group on the left. So I could write CH three and then I could write a two here which indicates there are two CH three groups bonded to directly bonded to the carbon in red. What else is bonded to the carbon in red? There's a hydrogen, so I'll put that in. So the carbon is bonded to a hydrogen. The carbon is also bonded to an OH, so I'll write in here an OH. This is the fully condensed version, so this is completely condensed and notice there are no bonds shown. you have to infer you have to infer the bonding from the condensed. All right, let's start with the condensed and go all the way to a Lewis structure, so we'll start with a condensed and then we'll do partially condensed structure, and then we'll go to a full Lewis structure. Just to get some more practice here. So I'll draw in a condensed one, so we have CH three, three and then COCH three. All right let's turn that into a partially condensed structure. So this carbon in red right here we're gonna start with that carbon, so I'll start drawing in that carbon right here. What is bonded to that carbon? Well, we have CH three groups and we have three of them. So there are three CH three groups directly bonded to that carbon. So let me draw them in. So here's one CH three group here is another CH three group, and then finally here's the third CH three group." + }, + { + "Q": "At 5:07 (CH)3COCH3, Is the Centered Carbon primary,secondary or tertiary?", + "A": "Given that the definitions for primary through quaternary carbons are based on how many other carbon atoms are attached - what do you think the answer is? Look at the Lewis structure 6:09, there are three bonds to the carbons of CH3 groups and one to oxygen. That means the central carbon is a tertiary carbon. Note that the condensed structure is actually (CH3)3COCH3.", + "video_name": "XEPdMvZqCHQ", + "timestamps": [ + 307 + ], + "3min_transcript": "you have to infer you have to infer the bonding from the condensed. All right, let's start with the condensed and go all the way to a Lewis structure, so we'll start with a condensed and then we'll do partially condensed structure, and then we'll go to a full Lewis structure. Just to get some more practice here. So I'll draw in a condensed one, so we have CH three, three and then COCH three. All right let's turn that into a partially condensed structure. So this carbon in red right here we're gonna start with that carbon, so I'll start drawing in that carbon right here. What is bonded to that carbon? Well, we have CH three groups and we have three of them. So there are three CH three groups directly bonded to that carbon. So let me draw them in. So here's one CH three group here is another CH three group, and then finally here's the third CH three group. is this carbon. The carbon in red is also bonded to an oxygen all right, so we need to draw in an oxygen next. So now we have our oxygen. Notice the carbon in red now has an octet of electrons around it. The oxygen is bonded to another CH three group. So the oxygen is bonded to another CH three and let's draw that in so we have our CH three and since we're doing a partially condensed I won't draw in those bonds. We have CH three like that. I could put in my lone pairs of electrons on the oxygen to give the oxygen an octet of electrons. And now we have our partially condensed structure. If we want to expand it even more, and draw the full Lewis structure, again we start with the carbon in red. So here's the carbon in red, and that carbon is bonded to another carbon and this carbon is bonded to three hydrogens, So this CH three group that I just drew is this one. All right, next, we have a CH three group on the left side, so I need to draw in a CH three on the left, hopefully I have enough room to do that. I'll squeeze it in here, so we have our hydrogens, and that's our second CH three group. So let me circle it in green here. So here's a CH three. And then finally we have let me make this blue down here, we have another CH three group, so I'll draw that one in. So we have another CH three, we'll make room for all these hydrogens here. And that's the one in blue. So we're drawing out all of the bonds now in our full Lewis structure. Next we have an oxygen, so we have an oxygen right in here, with two lone pairs of electrons on the oxygen. The oxygen is bonded to another CH three. So let me let me pick a color here for that one. So we have another CH three on the right," + }, + { + "Q": "At 1:25, instead of having two double bonds in the sulphuric acid shouldn't have two dative bonds?", + "A": "As far as I know, either way, it should mean the same thing.", + "video_name": "dbdVMThH1n8", + "timestamps": [ + 85 + ], + "3min_transcript": "Let's think about what might happen if we had a solution of this carboxylic acid here. We might as well name it just to get some practice. We have one, two, three, four, five, six, seven carbons. So this is heptan-, and then we don't write heptane, because this is a carboxylic acid. It is heptanoic acid. So let's see what happens if we have heptanoic acid reacting with-- this is one, two carbons, and then it has an OH group, so this is ethanol. That's what the OH group does. It makes this an alcohol. And it's in the presence of a sulfuric acid catalyst. This right here is sulfuric acid, one of the stronger acids. I'll actually draw its structure, because I always find it frustrating when people just write just the formula here without the actual structure, because the structure actually shows you why it's so acidic. So it has a double bond to an oxygen, another double bond to an oxygen, and then it has a single bond to an OH group, and then it has another single bond to an OH group. And notice, it has one, two, three, four, five, six valence electrons. Now, the reason why this is such a strong acid, is that if either of these oxygens take the electron from this proton, so actually give away the proton to the solution, there's a ton of resonance structures here. And maybe I'll make a whole video on sulfuric acid, just to show you all of the resonance structures. But, in general, whenever you see a reaction when they say it's catalyzed by an acid, all you have to do is realize it's just going to make the surrounding solution a lot more acidic, just a ton more acidic. Maybe we're in a solution of ethanol, and if we are in a solution of ethanol, it'll just add protons to the ethanol itself. So you can imagine this guy right over here. Let me draw these oxygen-hydrogen bonds, so you have this oxygen and it is bonded to a hydrogen there. This is floating around the solution. You have your ethanol that looks like this, so two carbons and then bonded to an oxygen, and then that oxygen is bonded to a hydrogen. The oxygen has two lone pairs, just like that. And so this guy really is good at getting rid of the protons. So you have a situation where this electron can be taken back by this oxygen, and then it can actually be given here, and there's all these resonance structures. But it's just very good at taking that electron. And that can happen at the exact same time that one of the ethanols capture the hydrogen proton, at the exact same time that this oxygen captures that hydrogen proton. And if you just look at this part right here, this will just result in-- this part alone will just result in a" + }, + { + "Q": "I am actually looking at an updated organic textbook right now and I believe 13:53 should be be two different steps. Deprotanation occurs after the leaving group leaves hope that helps all you guys understand the mechanism better but thanks Kahn for doing most of it right!", + "A": "No. Actually you are mistaken. You are a step premature....there must be a proton transfer between OH and OR before the leaving group leaves. He has the correct mechanism...I have no idea what you re specifically referring to.", + "video_name": "dbdVMThH1n8", + "timestamps": [ + 833 + ], + "3min_transcript": "And so if that happened, then the next step in our reaction-- and remember, these can all go in either direction. The next step in our reaction will look like this. You would have your two, three, four, five, six, seven carbons, single bond to an oxygen, and then you have your bond to this oxygen, which is bound to two carbons, one, two carbons, just like that. And this guy on top is bonded to an oxygen. He's got two lone pairs. And this guy over here grabbed a proton. He gave an electron to a hydrogen proton. He had two, but now he gave one of them to this proton. And so he gave it to that hydrogen. That hydrogen is now neutral. It gained an electron. This oxygen is now positive, because it It still has this other lone pair over here. It is now positive, and frankly, it is now a good leaving group. So, in the next step, you can have someone else. Remember, other people need protons earlier in this reaction. So this proton might get lost, maybe by one of the other ethanol molecules, so let me draw that. Let me do a color I haven't used yet. I'll use orange. So maybe one of the other ethanol molecules, or one of the other intermediaries in this whole reaction. So ethanol, I'll just do ethanol because it's easier to draw. It might give an electron to just the nucleus. And then this guy can take the electron back. This guy could take that hydrogen's electron back and give it to this carbon that was the carbonyl carbon several steps ago. And then since he's got that electron, he can then give back this electron to this, what was an OH group, but now back to him. And then the resulting products will look like this. And this is all in equilibrium. We now have two, three, four, five, six, seven carbons. Now, this guy has a double bond again. So you have an oxygen right there. It is now a double bond. I'll draw this newly formed double bond in magenta. This guy has left as water, so I can just draw that, so now you have this other OH group that is bonded to this hydrogen over here. He has now left as water. And now you have this other thing over here. You have what was that ethanol group has now attached itself. That ethanol has lost its hydrogen. It has attached itself to what was a carboxylic acid, so now it looks like this. It now is bonded to an oxygen and is bonded to one, two carbons, just like that. And this whole reaction that I showed you is called" + }, + { + "Q": "At 3:36, can you give me an example of a good mutation?", + "A": "Bacterial flagella, antibiotic resistance, and the ability to metabolize many sugars for food. hope that helps :)", + "video_name": "tzqZsPjHFVQ", + "timestamps": [ + 216 + ], + "3min_transcript": "Just like Fast and Furious movies, there are five of them. Unlike Fast and Furious movies, they're actually very, very important and are the basic reason why all complex life on earth exists. The main selective pressure is simply natural selection itself, Darwin's sweet little baby which he spent a lot of his career defending from haters. Obviously we know these natural selection makes the alleles that make animals the strongest and most virile and least likely to die more frequent in the population. Most selective pressures are environmental ones like food supply, predators or parasites. At the population level, one of the most important evolutionary forces is sexual selection. Population genetics gets its special attention particularly when it comes to what's called non-random mating which is a lifestyle that I encourage in all of my students, do not mate randomly. Sexual selection is the idea that certain individuals will be more attractive mates than others because of specific traits. This means they'll be chosen to have more sex and therefore offspring. The pop-gen spend on things if that sexual selection There are specific traits that are preferred even though they may not make the animals technically more fit for survival. Sexual selection changes a genetic make up of a population because the alleles of the most successful maters are going to show up more often in the gene-pool. Maters are going to mate. Another important factor here, and another thing that Darwin wished he had understood is mutation. Sometimes when eggs and sperm are formed through meiosis, a mistake happens in the copying process of DNA, that errors in the DNA could result in the death or deformation of offspring. But not all mutations are harmful. Sometimes these mistakes can create new alleles that benefit the individual by making it better at finding food or avoiding predators or finding a mate. These good errors and the alleles they made are then passed to the next generation and into the population. Fourth, we have genetic drift which is when an alleles frequency changes due to random chance. A chance that's greater if the population is small. Those happens much more quickly if the population Genetic drift does not cause individuals to be more fit, just different. Finally, when it comes to allele game changers you got to respect the gene flow which is when individuals with different genes find their way into a population and spread their alleles all over the place. Immigration and emigration are good examples of this. As with genetic drift, its effects are most easily seen in small populations. Again, our factors: Natural selection; alleles for fitter organisms become more frequent. Sexual selection; alleles for more sexually attractive organisms become more frequent. Mutation; new alleles popping up due to mistakes in DNA. Genetic drift; changes an allele frequency due to random chance. Gene flow; changes in allele frequency due to mixing with new genetically different populations. Now that you know all that, in order to explain specifically how these processes influence populations we're going to have to completely forget about them. This is what's called the Hardy-Weinberg principle. Godfrey Hardy and Wilhelm Weinberg were two scientists in 1908" + }, + { + "Q": "around 3:02 John talks about sexual selection choosing a mate. I thought it was endorphins in the brain that caused this. Correct me if i am wrong", + "A": "Half to half. Yes, sexual selection does help choose a mate, but endorphins help.", + "video_name": "tzqZsPjHFVQ", + "timestamps": [ + 182 + ], + "3min_transcript": "it's the study of how populations of a species change genetically overtime leading to species evolving. Let's start up by defining what a population is. It's simply a group of individuals of a species that can interbreed. Because we have a whole bunch of fancy genetic testing gadgets and because unlike Darwin we know a whole lot about heredity. We can now study the genetic change in populations over just a couple of generations. This is really exciting and really fun because it's basically like scientific instant gratification. I can now observe evolution happening within my lifetime. You know, just cross that off the bucket list. Now, part of population genetics or pop-gen and now we've got fancy abbreviations for everything now, involves the study of factors that cause changes and what's called allele frequency. Which is just how often certain alleles turn up within a population. Those changes are at the heart of how and why evolution happens. There are several factors that change allele frequency Just like Fast and Furious movies, there are five of them. Unlike Fast and Furious movies, they're actually very, very important and are the basic reason why all complex life on earth exists. The main selective pressure is simply natural selection itself, Darwin's sweet little baby which he spent a lot of his career defending from haters. Obviously we know these natural selection makes the alleles that make animals the strongest and most virile and least likely to die more frequent in the population. Most selective pressures are environmental ones like food supply, predators or parasites. At the population level, one of the most important evolutionary forces is sexual selection. Population genetics gets its special attention particularly when it comes to what's called non-random mating which is a lifestyle that I encourage in all of my students, do not mate randomly. Sexual selection is the idea that certain individuals will be more attractive mates than others because of specific traits. This means they'll be chosen to have more sex and therefore offspring. The pop-gen spend on things if that sexual selection There are specific traits that are preferred even though they may not make the animals technically more fit for survival. Sexual selection changes a genetic make up of a population because the alleles of the most successful maters are going to show up more often in the gene-pool. Maters are going to mate. Another important factor here, and another thing that Darwin wished he had understood is mutation. Sometimes when eggs and sperm are formed through meiosis, a mistake happens in the copying process of DNA, that errors in the DNA could result in the death or deformation of offspring. But not all mutations are harmful. Sometimes these mistakes can create new alleles that benefit the individual by making it better at finding food or avoiding predators or finding a mate. These good errors and the alleles they made are then passed to the next generation and into the population. Fourth, we have genetic drift which is when an alleles frequency changes due to random chance. A chance that's greater if the population is small. Those happens much more quickly if the population" + }, + { + "Q": "On 6:30, what does he mean by \"to nab on to a hydrogen proton\"?", + "A": "To nab on to a hydrogen proton basically means that the oxygen can bond to a hydrogen proton.", + "video_name": "L677-Fl0joY", + "timestamps": [ + 390 + ], + "3min_transcript": "and then these hydrogen will just be grabbed by another water molecule or something so the proton will be let go. That's why we call it an acid. If it wasn't in a solution it would have the hydrogens but it would be very acidic as soon as you put it into a neutral solution it's going to lose those hydrogens. The phosphate groups are what make it, are what make it an acid but it's confusing sometimes because usually when you see it depicted, you see it with these negative charges and that's because it has already lost its hydrogen proton. You're actually depicting the conjugate base here but that's where it gets its acidic name from because it starts protonated or it gets in this acid form, it's protonated but it readily loses it. And so that's why it has its, that's where it gets the name acid form from. Each of these nucleotides they have a phosphate group. Now the next thing you might notice, the next thing you might notice is. The next thing you might notice is this group right over here. It is a cycle, it is a ring and that's because it is a sugar. This sugar is based on, it's a five-carbon sugar. What I have depicted here, this sugar, this is ribose. This sugar right over here is ribose. This is when it's just as a straight chain and like many sugars, it can take a cyclical form. Actually it can take many different cyclical forms but the one that's most typically described is when you have that. Let me number the carbons because carbon numbering is important when we talk about DNA. But if we start carbonyl group right over here we call that the one carbon or the one prime carbon. One prime, two prime, three prime, four prime and five prime. That's the five prime carbon. You form the cyclical form of ribose as if you have the oxygen. You have the oxygen right over here on the four prime carbon. It uses one of its lone pairs. It uses one of its lone pairs to form a bond. With the one prime carbon and I drew it that way because it kind of does bend. The whole molecule's going to have to bend that way to form this structure. And then when it forms that bond the carbon can let go of one of these double bonds and then that can, then the oxygen, the oxygen can use that. The oxygen can use those electrons to go grab a hydrogen proton from some place. To nab on to a hydrogen proton. When it does that you're in this form and this form, just to be clear of what we're talking about, this is the one prime carbon. One prime, two prime, three prime, four prime and five prime carbon. Where we see this bond, this is the one prime carbon. it was part of a carbonyl. Now it lets go of one of those double bonds so that this oxygen can form a bond with a hydrogen proton. It let go of a double bond there so that this could form a bond with a hydrogen proton." + }, + { + "Q": "At about 7:40 what is hydronium?", + "A": "Hydronium is H30. A water molecule is H20. When some molecular structure release a hydrogen ion (hydroxide), the water molecule, being electronegative (hoging electrons), take the hydroxide, thus forming Hydronium, with three Hydrogen molecules and one Oxygen molecule.", + "video_name": "L677-Fl0joY", + "timestamps": [ + 460 + ], + "3min_transcript": "With the one prime carbon and I drew it that way because it kind of does bend. The whole molecule's going to have to bend that way to form this structure. And then when it forms that bond the carbon can let go of one of these double bonds and then that can, then the oxygen, the oxygen can use that. The oxygen can use those electrons to go grab a hydrogen proton from some place. To nab on to a hydrogen proton. When it does that you're in this form and this form, just to be clear of what we're talking about, this is the one prime carbon. One prime, two prime, three prime, four prime and five prime carbon. Where we see this bond, this is the one prime carbon. it was part of a carbonyl. Now it lets go of one of those double bonds so that this oxygen can form a bond with a hydrogen proton. It let go of a double bond there so that this could form a bond with a hydrogen proton. that hydrogen proton right over there and this green bond that gets formed between the four prime carbon and or between the oxygen that's attached to the four prime carbon and the one prime carbon, that's this. That's this bond right over here. This oxygen is that oxygen right there. Notice, this oxygen is bound to the four prime carbon and now it's also bound to the one prime carbon. It was also attached to a hydrogen. It was also attached to a hydrogen so that hydrogen is there but then that can get nabbed up by another passing water molecule to become hydronium so it can get lost. It grabs up a hydrogen proton right over here and so it can lose a hydrogen proton right there. It's not adding or losing in that net. You form this cyclical form and the cyclical form right over here is very close to what we see in a DNA molecule. It's actually what we would see in an RNA molecule, in a ribonucleic acid. when we say deoxyribonucleic acid. Well, you can start with you have a ribose here but if we got rid of one of the oxygen groups and in particular one of... Well, actually if we just got rid of one of the oxygens we replace a hydroxyl with just a hydrogen, well then you're gonna have deoxyribose and you see that over here. This five-member ring, you have four carbons right over here. it looks just like this. The hydrogens are implicit to the carbons, we've seen this multiple time. The carbons are at where these lines intersect or I guess at the edges or maybe and also where these lines end right over there. But you see this does not have an... This molecule if we compare these two molecules, if we compare these two molecules over here, we see that this guy has an OH, and this guy implicitly just has... This has an OH and an H. This guy implicitly has just two hydrogens over here. He's missing an oxygen. This is deoxyribose." + }, + { + "Q": "At 7:37 Sal says the Hydrogen might have lost an electron. How does that happen?", + "A": "Atoms lose electrons in chemical reactions or when they come in contact with other atoms whether of the same element or another. So when the hydrogen atom comes in contact with another atom, it may lose or also gain electrons (from its outermost shell). Thus an Hydrogen ion is formed.", + "video_name": "TStjgUmL1RQ", + "timestamps": [ + 457 + ], + "3min_transcript": "It just really means that it's very unlikely to go the other way. You have to supply a lot of energy to go the other way. To make this reaction go the other way, you would have to do something called electrolysis, you provide energy, etcetera, etcetera. But in general, the way that this is written, because the arrow is only pointing in one direction, this is implying that it is irreversible. Irreversible. Irreversible. Which probably makes you think, well what about reversible reactions? And I have an example of a reversible reaction, right over here. I have a one bicarbonate ion. And the word ion, that's just used to describe any molecule or atom that has either, has an imbalance of electrons or protons that cause it to have a net charge. So this makes this an ion, and actually right over here, this is a hydrogen, this is a hydrogen ion right over here. Both of these are charged. One has a positive charge, one has a negative charge. And this reactions right over here, you have the bicarbonate ion that looks something like this. This is just my hand-drawing of it. Reacting with a hydrogen ion, it's really a hydrogen atom that has lost it's electron, so some people would even say this is a proton right over here. This is an equilibrium reaction, where it can form carbonic acid. And notice all that's happening is this hydrogen is attaching to one of the oxygens over here. And this is an equilibrium because if in an actual, in an actual solution, it's going back and forth. If you actually provide more reactants, you're gonna go more in that direction. If you provide more of the products over here, then you're gonna go in that direction. And so in an actual, in an actual environment, in an actual system, it's constantly going back and forth between these two things. And different reversible reactions might tend to one side or the other. If you provide more of the stuff on one side, are gonna, they're gonna be more likely to interact, Or if you provide more of this, it might go in the other direction because these might more likely react with their surroundings or disassociate in some way. Now just to get a sense of, you know, it's nice to kind of, you know, are these just some random letters that I wrote here? Carbonic acid is actually an incredibly important molecule, or we could call it a compound because it's made up of two or, two or more elements, in living systems and in fact, you know, even in the environment. And even when you go out to get some fast food. When you have carbonated drinks, it has carbonic acid in it that disassociates into carbon dioxide and that carbon dioxide is what you see bubbling up. Carbonic acid is incredibly important in how your body deals with excess carbon dioxide in its bloodstream. Carbonic acid is involved in the oceans taking up carbon dioxide from the atmosphere. So when you're studying chemistry, especially in the context of biology, these aren't just," + }, + { + "Q": "Around 6:50, what is meant by 'imbalance of electrons or protons'? How does this imbalance occur?", + "A": "Imbalance of protons and electrons means that it will either have more electrons (and less protons) or more protons (and less electrons). This imbalance causes it to have a positive or negative charge.", + "video_name": "TStjgUmL1RQ", + "timestamps": [ + 410 + ], + "3min_transcript": "And just to get an appreciation of how much energy this produces, let me just show you this picture right over here. That's the space shuttle and this, this big tank right over here, let me... This big tank contains a bunch of liquid oxygen and hydrogen. And to create this incredible amount of energy, it actually just... You mix the two together with a little bit of energy and then you produce a ton of energy that makes the rocket, that makes the space shuttle. Well, space shuttle's been discontinued now, but back when they did it, to make it get it's necessary, it's necessary velocity. Now let's talk about the idea. So, you know, this reaction, strongly goes in this, in the direction of going to water. But it can actually go the other way, but it's very, very hard for it to go the other way. So in general we would consider this to be an irreversible reaction, even though it is. You know irreversible sounds like, It just really means that it's very unlikely to go the other way. You have to supply a lot of energy to go the other way. To make this reaction go the other way, you would have to do something called electrolysis, you provide energy, etcetera, etcetera. But in general, the way that this is written, because the arrow is only pointing in one direction, this is implying that it is irreversible. Irreversible. Irreversible. Which probably makes you think, well what about reversible reactions? And I have an example of a reversible reaction, right over here. I have a one bicarbonate ion. And the word ion, that's just used to describe any molecule or atom that has either, has an imbalance of electrons or protons that cause it to have a net charge. So this makes this an ion, and actually right over here, this is a hydrogen, this is a hydrogen ion right over here. Both of these are charged. One has a positive charge, one has a negative charge. And this reactions right over here, you have the bicarbonate ion that looks something like this. This is just my hand-drawing of it. Reacting with a hydrogen ion, it's really a hydrogen atom that has lost it's electron, so some people would even say this is a proton right over here. This is an equilibrium reaction, where it can form carbonic acid. And notice all that's happening is this hydrogen is attaching to one of the oxygens over here. And this is an equilibrium because if in an actual, in an actual solution, it's going back and forth. If you actually provide more reactants, you're gonna go more in that direction. If you provide more of the products over here, then you're gonna go in that direction. And so in an actual, in an actual environment, in an actual system, it's constantly going back and forth between these two things. And different reversible reactions might tend to one side or the other. If you provide more of the stuff on one side," + }, + { + "Q": "At 6:55 Sal talks about a positive \"+\" Hydrogen ion.\nWouldn't that just be a proton, as Hydrogen has only 1 electron and 1 proton as an atom?", + "A": "Yes a hydrogen ion is a proton.", + "video_name": "TStjgUmL1RQ", + "timestamps": [ + 415 + ], + "3min_transcript": "And just to get an appreciation of how much energy this produces, let me just show you this picture right over here. That's the space shuttle and this, this big tank right over here, let me... This big tank contains a bunch of liquid oxygen and hydrogen. And to create this incredible amount of energy, it actually just... You mix the two together with a little bit of energy and then you produce a ton of energy that makes the rocket, that makes the space shuttle. Well, space shuttle's been discontinued now, but back when they did it, to make it get it's necessary, it's necessary velocity. Now let's talk about the idea. So, you know, this reaction, strongly goes in this, in the direction of going to water. But it can actually go the other way, but it's very, very hard for it to go the other way. So in general we would consider this to be an irreversible reaction, even though it is. You know irreversible sounds like, It just really means that it's very unlikely to go the other way. You have to supply a lot of energy to go the other way. To make this reaction go the other way, you would have to do something called electrolysis, you provide energy, etcetera, etcetera. But in general, the way that this is written, because the arrow is only pointing in one direction, this is implying that it is irreversible. Irreversible. Irreversible. Which probably makes you think, well what about reversible reactions? And I have an example of a reversible reaction, right over here. I have a one bicarbonate ion. And the word ion, that's just used to describe any molecule or atom that has either, has an imbalance of electrons or protons that cause it to have a net charge. So this makes this an ion, and actually right over here, this is a hydrogen, this is a hydrogen ion right over here. Both of these are charged. One has a positive charge, one has a negative charge. And this reactions right over here, you have the bicarbonate ion that looks something like this. This is just my hand-drawing of it. Reacting with a hydrogen ion, it's really a hydrogen atom that has lost it's electron, so some people would even say this is a proton right over here. This is an equilibrium reaction, where it can form carbonic acid. And notice all that's happening is this hydrogen is attaching to one of the oxygens over here. And this is an equilibrium because if in an actual, in an actual solution, it's going back and forth. If you actually provide more reactants, you're gonna go more in that direction. If you provide more of the products over here, then you're gonna go in that direction. And so in an actual, in an actual environment, in an actual system, it's constantly going back and forth between these two things. And different reversible reactions might tend to one side or the other. If you provide more of the stuff on one side," + }, + { + "Q": "At 3:51 Sal says the atoms are bouncing around and have a lot of energy,but where does that energy come from", + "A": "Energy comes from the universe. Just like matter, energy can not be created or destroyed, only transferred. Sometimes energy is transferred through heat, or it could be seen as movement. Energy comes from the atoms surroundings, and it will in turn release energy to the surroundings, depending on whether or not the reaction is exothermic or endothermic.", + "video_name": "TStjgUmL1RQ", + "timestamps": [ + 231 + ], + "3min_transcript": "that we have right over here, that tells us that we're dealing with two of those molecules for this reaction to happen, that we need two of these molecules for every, for every molecule of molecular oxygen. And molecular oxygen, once again, this is composed of two oxygen atoms. One, two. So under the right conditions, so you need a little bit of energy to make this happen. If under the right conditions these two things are going to react. And actually it's very, very reactive, molecular hydrogen and molecular oxygen. So much so that it's actually used for rocket fuel. You are going to produce two molecules of water. We see that right over here. And look, I did not create or destroy any atoms. I had one, I had one, I had one oxygen atom here. It was part of the oxygen molecule right here, then I have the second one right over here now. Now they are part of separate molecules. I had, I had a, I had one two, three, four hydrogens. four hydrogens, just like that. And actually this produces a... So we could say some energy, and I'm being inexact right over here. Some energy and then we could say a lot of energy. A lot of energy. So this is a reaction that you just give it a little bit of a kick-start and it really wants to happen. A lot, a lot of energy. So one thing that you might wonder, and this is something that I first wondered when I learned about reactions, well how do, how does this happen? You know, is this a very organized thing? You know, do these molecules somehow know to react with each other? And the answer's no. Chemistry is a incredibly messy thing. You have these things bouncing around, they have energy. They're bouncing around all over the place and actually when you provide energy, they're gonna bounce around even more rigorously, enough so that they collide in the right ways so that they break their old bonds So whenever you see these reactions in biology or chemistry class, keep that in mind. It looks all neat and organized but in a real system, these are all of these things just bouncing around in all different crazy ways. And that's why energy's an important thing here. Because the more energy you apply to the system, the more that they're going to bounce around, the more that they're going to interact with each other. The more reactants you put in, the more chance they're going to bounce around and be able to react with each other. Now I'm gonna introduce another word that you're gonna see in chemistry a lot. This water, these two... We see we have two water molecules here. We could call them molecules, but since they are actually made up of two or more different elements, we could also call this a compound. So water, water is, you could call it a molecule, or you could call it a compound. So this is a molecule or compound, while this molecular hydrogen, you would not call this a compound. And this molecular oxygen, of course it's a molecule," + }, + { + "Q": "Around 3:20, Sal mentions that the formation of water produces a lot of energy, but how can a chemical reaction lead to the production of energy. Doesn't it violates the law of Conservation of Energy?", + "A": "no. Potential energy gets converted to another form (thermal, or KE)", + "video_name": "TStjgUmL1RQ", + "timestamps": [ + 200 + ], + "3min_transcript": "or we could say the products. And so what are the reactants here? Well we have molecular hydrogen and we have molecular oxygen. Now why did I say molecular hydrogen? Because molecular hydrogen, which is the state that you would typically find hydrogen in if you just have it by itself, it is actually made up of two hydrogen atoms. You see it right over here, one, two hydrogen atoms. And what we have in order to have this reaction, you don't just need one molecular hydrogen and one, or one molecule of hydrogen and one molecule of oxygen. For every, for this reaction to happen, you actually have two molecules of molecular hydrogen. So this is actually made up of four hydrogen atoms. So let me make this clear. So this right over here, this is two molecules of molecular hydrogen. And that's why we have the two right out front of the H sub-two. This little subscript two tells us there's two of the hydrogen atoms in this molecule. that we have right over here, that tells us that we're dealing with two of those molecules for this reaction to happen, that we need two of these molecules for every, for every molecule of molecular oxygen. And molecular oxygen, once again, this is composed of two oxygen atoms. One, two. So under the right conditions, so you need a little bit of energy to make this happen. If under the right conditions these two things are going to react. And actually it's very, very reactive, molecular hydrogen and molecular oxygen. So much so that it's actually used for rocket fuel. You are going to produce two molecules of water. We see that right over here. And look, I did not create or destroy any atoms. I had one, I had one, I had one oxygen atom here. It was part of the oxygen molecule right here, then I have the second one right over here now. Now they are part of separate molecules. I had, I had a, I had one two, three, four hydrogens. four hydrogens, just like that. And actually this produces a... So we could say some energy, and I'm being inexact right over here. Some energy and then we could say a lot of energy. A lot of energy. So this is a reaction that you just give it a little bit of a kick-start and it really wants to happen. A lot, a lot of energy. So one thing that you might wonder, and this is something that I first wondered when I learned about reactions, well how do, how does this happen? You know, is this a very organized thing? You know, do these molecules somehow know to react with each other? And the answer's no. Chemistry is a incredibly messy thing. You have these things bouncing around, they have energy. They're bouncing around all over the place and actually when you provide energy, they're gonna bounce around even more rigorously, enough so that they collide in the right ways so that they break their old bonds" + }, + { + "Q": "are the two products that were formed(b1 and b2) at 7:00 exactly the same or are they enantiomers?", + "A": "They are identical, there are no chiral centers on the molecule, and the double bond is in the same spot", + "video_name": "MDh_5n0OO2M", + "timestamps": [ + 420 + ], + "3min_transcript": "And so, let me draw in a proton there. And we think about a weak base coming along and taking that proton, so I'll draw in my weak base here. So it takes this proton and these electrons move into here, so let's draw this product. So we would have our alkene that looks like this. So let's follow those electrons. I'll make them dark blue. Electrons in this bond move into here to form our double bond. And so, now we've gone through the complete mechanism, and we have two products. So let me circle our two products, so this is really just one product, and then this would be our second product. For this reaction, we actually get 90% of the alkene on the right and 10% of the alkene on the left. And so, let's look at the degree of substitution of our two products, So, let me use red for this. If we think about the degree of substitution for the alkene on the right, by drawing my hydrogen right here, it makes it a little bit easier to see we have three alkyl groups, so this one, this one, and this one. So this would be a trisubstituted alkene. So the one on the right is a trisubstituted alkene, and the one on the left, so this one right here, would be a disubstituted alkene. These are the two carbons across our double bond. We have two hydrogens on this carbon, and the carbon on the right has two alkyl groups bonded to it. So this one is a disubstituted alkene. Now we've gone through the whole E1 mechanism, and we've seen that we get a disubstituted product, and a trisubstituted. Now let's think about regiochemistry. For this reaction, it's the region of the molecule where the double bond forms. the double bond formed in this region of the molecule, and for the trisubstituted product, the double bond formed in this region. The trisubstituted product is the major product, and it's also the more stable alkene. So remember, from the video in alkene stability, the more substituted your alkene is, the more stable it is, so this product is more stable, and that's why we form more of it. And the more stable products or the more substituted product is called the Zaitsev product. So we say that this E1 reaction is regioselective because it has a preference to form the more stable product, the more substituted product, which we call the Zaitsev product." + }, + { + "Q": "at 7:33 could these three (two) molecules be considered isoprenes? Does the location of the double bond matter in an isoprene?", + "A": "Isoprene is 2-methylbuta-1,3-diene. These molecules are not isoprenes, although they have an isoprene skeleton (C atoms joined as in 2-methylbutane). When we say that molecules contain isoprene units, we mean that they consist of 5-carbon isoprene units joined together. The location of any double bonds doesn t matter.", + "video_name": "MDh_5n0OO2M", + "timestamps": [ + 453 + ], + "3min_transcript": "So, let me use red for this. If we think about the degree of substitution for the alkene on the right, by drawing my hydrogen right here, it makes it a little bit easier to see we have three alkyl groups, so this one, this one, and this one. So this would be a trisubstituted alkene. So the one on the right is a trisubstituted alkene, and the one on the left, so this one right here, would be a disubstituted alkene. These are the two carbons across our double bond. We have two hydrogens on this carbon, and the carbon on the right has two alkyl groups bonded to it. So this one is a disubstituted alkene. Now we've gone through the whole E1 mechanism, and we've seen that we get a disubstituted product, and a trisubstituted. Now let's think about regiochemistry. For this reaction, it's the region of the molecule where the double bond forms. the double bond formed in this region of the molecule, and for the trisubstituted product, the double bond formed in this region. The trisubstituted product is the major product, and it's also the more stable alkene. So remember, from the video in alkene stability, the more substituted your alkene is, the more stable it is, so this product is more stable, and that's why we form more of it. And the more stable products or the more substituted product is called the Zaitsev product. So we say that this E1 reaction is regioselective because it has a preference to form the more stable product, the more substituted product, which we call the Zaitsev product." + }, + { + "Q": "At 3:26 and 3:42, how does he know that both the products are either trisusbtitual or disubstitual?", + "A": "You count the number of C atoms that are directly attached to the alkene carbons. In 1-methylcyclohexene, the alkene carbons are C1 and C2. The carbon atoms directly attached to C1 are C6 and the methyl carbon. The carbon atom directly attached to C2 is C3. This makes a total of three carbons, so the alkene is trisubstituted. In 3-methylcyclohexene, the carbon atom directly attached to C1 is C6. The carbon atom directly attached to C2 is C3. This makes a total of two carbons, so the alkene is disubstituted.", + "video_name": "MDh_5n0OO2M", + "timestamps": [ + 206, + 222 + ], + "3min_transcript": "and a plus one formal charge on the oxygen. So the loan pair, let's say this lone pair here in magenta, picks up a proton from sulfuric acid to form this bond, and that gives us water as a leaving group. And we know water is a good leaving group. The electrons in this bond can come off onto the oxygen to form H2O. And when that happens, we take a bond away from this carbon in red. So we're gonna form a carbocation. So we take away a bond from the carbon in red. Let me go ahead and draw in our carbocation here. So the carbon in red would be this carbon. That carbon would have a plus one formal charge. This is a tertiary carbocation, because the carbon in red is directly bonded to three other carbons. So this one, this one, and this one. So this is a stable carbocation. Next, let's think about the next step of an E1 mechanism. and take a proton from one of the beta carbons. And let's start with beta two. So let's thinking about a proton on the carbon, so let me draw on in here. And our weak base comes along and takes this proton, which would leave this electrons to move into here to form a double bond. So let's draw that product. So we would have our double bond forming right here. Let me draw in the rest of the molecule. So our electrons in, let me make these blue here. So the electrons in light blue are going to move in to form our double bond, and so we would get this out alkene. So that's beta two. Let's think about what would happen if we took a proton away from beta one. So let's draw that one in next. So I'm gonna draw the carbocation again. Let me draw that in here. And beta one would be up here, so let me put in a proton on beta one. and taking this proton, so our base is probably water. And these electrons would move into here this time. So if that happens, let's draw that product. Our double bond would form up here, and let me draw in the rest of this molecule. So let me use red for those electrons. So electrons in red are going to move into here to form this alkene. Notice that the two alkenes that we just drew is really the same molecule. This is the same compound. So we haven't formed two different products. If you take a proton away from the beta one or the beta two carbon, you're gonna make the same alkene. But what about the beta three carbon? So that's our last example. And let me go ahead. I forgot to in a plus one formal charge on our carbocation. Let me draw one more carbocation, the same one at tertiary carbocation. The difference is this time" + }, + { + "Q": "At 5:49 you named the molecule Cyclohexane Carbaldehyde but you did not explain why that particular compound is called a Carbaldehyde.", + "A": "A carbaldehyde is an aldehyde that is attached to another entity which is often a ring system.", + "video_name": "JMsqu236bZo", + "timestamps": [ + 349 + ], + "3min_transcript": "So if you see benzaldehyde, you can use that in your iupac name. And so if we look at it over here on the right you can see that benzaldehyde can form the base for the name for this molecule. So let's go ahead and write that over here. So we have \"benzaldehyde.\". And then we're going to number to give this carbon on our ring, number one. Right so we have two choices we can go around the ring clockwise or counterclockwise. And we know we want to give the lowest number possible to our substituents and so that would be by going around clockwise. So let's do that, so two, three and then four. So we have two substituents, right. We have a methoxy substituent here at carbon three. We have a hydroxy substituent here at carbon four. And so we need to put those into alphabetical order so H comes before M and so we're going to start with the hydroxies, so let's, hopefully we'll have enough room here. So \"four-hydroxy\" and then we have a \"three-methoxy So vanillin of course is where we get our vanilla flavor from. So a fantastic vanilla smell so these are white crystals which have an amazing vanilla scent to them. So also a lot of fun to do labs with. Alright so a lot of aldehydes have nice smells to them. Let's look at two more aldehydes. Alright so let's look at the one on the left here right. So it's not benzaldehyde, we don't have benzene ring here anymore, we have a cyclohexene, so this one is going to be called \"cyclohexane carbaldehyde.\" So, \"cyclohexane carbaldehyde\" is the iupac name for this. So, \"carbaldehyde.\" Alright, let's look at this molecule over here on the right so this is two aldehydes Alright so let's look at this, so it'll be one, two, three, four and five total carbons. And so it's two aldehydes, so it's going to be \"dial\" and it's five carbons total so it's going to be \"pentane\", so we could write \"pentane.\" We could write \"pentanedial\" here. And if you wanted to put where the numbers are, so it'll be one, five, so you can put that in there. So \"one-five-pentanedial\" for two aldehydes. Alright, let's look at ketone nomenclature next and so just to remind you of the general structure of ketones. So once again you have a carbonyl, a carbon double bonded to an oxygen. This time you have two alkyler groups right, so you have two alkyler groups here and they can be the same or they can be different right. So I can write \"R prime\" here and either one would still be a ketone. So once again we looked at butane earlier, so our four-carbon alkane, \"butane.\"" + }, + { + "Q": "From 6:56-7:08, is there an example that would make what was said about the relationship about mass and acceleration easier to understand? At 7:08, what do you mean by \"the harder it is to change it's constant velocity?\"", + "A": "the harder it is to change it s constant velocity A velocity is changed by accelerating or decelarating. Sal is saying that if you apply the same force to an object with a larger mass, it will accelerate less than an object with lower mass. This happens when you strike a tennis ball and a bowling ball with the same amount of force. The tennis ball will accelerate more than the bowling ball, because the mass of a tennis ball is lower than a bowling ball.", + "video_name": "ou9YMWlJgkE", + "timestamps": [ + 416, + 428, + 428 + ], + "3min_transcript": "divide both sides by 2 kilograms So let's divide the left by 2 kilograms let's divide the right by 2 kilograms that cancels out. The 10 and the 2-- 10 divided by 2 is 5 and then you have kilograms cancelling kilograms. Your left hand side you get 5 metres per second squared and then that's equal to your acceleration. Now just for fun, what happens if I double that force? Well then I have 20Newtons--I'll actually work it out-- 20 kilograms.metres per second squared is equal to --I'll actually color-code this-- 2 kilograms times the acceleration and what do we get? [cancels out] 20 divided by 2 is 10 kilograms cancel kilograms and so we have the acceleration, in this situation is equal to 10 metres per second squared--is equal to the acceleration. So when we doubled the force, we went from 10 Newtons to 20 Newtons, the acceleration doubled. We went from 5 metres per second squared to 10 metres per second squared. So we see that they are directly proportional and the mass is how proportional they are. And so you can imagine what happens if we double the mass. If we double in, let's say in this situation, with 20 Newtons then we won't be dividing by 2 kilograms anymore we'll be dividing by 4 kilograms. And so then we'll have 20 divided by 4 which will be 5 and it'll be metres per second squared. So if you make the mass larger, if you double it then your acceleration would be half as much. you need to accelerate it or for a given force, the less that it will accelerate it. The harder it is to change it's constant velocity." + }, + { + "Q": "at 1:10 Sal says that Force equals mass times acceleration. Isn't \"net force\" equals mass times acceleration?", + "A": "Yes it is, but f=ma is just an easy-to-understand, basic version of the equation. The real equation is: The sum of the forces on an object equals the mass of the object times its acceleration \u00ce\u00a3F = m * a", + "video_name": "ou9YMWlJgkE", + "timestamps": [ + 70 + ], + "3min_transcript": "Newton's first law tells us that an object at rest will stay at rest, and an object with a constant velocity will keep having that constant velocity unless it's affected by some type of net force or you actually can say that an object with constant velocity will stay having a constant velocity unless it's affected by net force because really this takes into consideration the situation where an object is at rest. You could just have a situation where the constant velocity is zero. So Newton's first law-you're gonna have your constant velocity it could be zero, it's going to stay being that constant velocity unless it's affected, unless there's some net force that acts on it. So that leads to the natural question. How does a net force affect the constant velocity or how does it affect the state of an object? And that's what Newton's second law gives us- Newton's Second Law of Motion And this one is maybe the most famous -actually I won't pick favorites here- but this one gives us the famous formula; Force is equal to mass times acceleration And acceleration is a vector quantity and force is a vector quantity. And what it tells us- 'cause we're saying ok if you apply force it might change that constant velocity but how does it change that constant velocity? Well say I have a brick right here and it is floating in space Newton's second law tells us that it's pretty nice for us that the laws of the universe or at least in the classical sense before Einstein showed up The laws of the universe actually dealt with pretty simple mathematics. What it tells us is if you apply a net force let's say on this side of the object and we talk about net force because if you apply two forces that cancel out and that have zero net force then the object won't change it's constant velocity. If you have a net force applied to one side of this object going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff" + }, + { + "Q": "In 5:07 why do we divide both sides by 2kg?", + "A": "We divide both sides by 2kg so that the mass gets neutralized on one side so that we can find the acceleration.", + "video_name": "ou9YMWlJgkE", + "timestamps": [ + 307 + ], + "3min_transcript": "so weight is a force mass is telling you how much stuff there is. And this is really neat, that this formula is so simple because maybe we could have lived in a universe where force is equal to mass squared times acceleration times the square root of acceleration which would have made all of our math much more complicated. But it's nice that it's just this constant of proportionality right over here. It's just this nice simple expression. And just to get our feet wet a little bit with computations involving force mass and acceleration, let's say that I have a force and the unit of force is appropriately called the Newton. So let's say I have a force of 10Newtons - and just to be clear, a Newton is the same thing - so this is the same thing as 10kilogram.metre per seconds squared as kilograms.metres per second square because that's exactly what you get on this side of the formula. So let's say I have a force of 10 Newtons and it is acting on -it is acting on a mass, let's say that the mass is 2 kilograms and I wanna know the acceleration. And once again in this video, these are vector quantities. If I have a positive value here I'm going to--we're going to make the assumption that it's going to the right. If I had a negative value then it would be going to the left. So implicitly I'm giving you not only the magnitude of the force but I'm also giving you the direction. I'm saying it is to the right because it is positive. So what will be the acceleration? Well we just use F=ma You have-on the left hand side 10 - I could write 10 Newtons here or I could write 10kilograms.metres per second squared and that is going to be equal to the mass which is 2 kilograms times the acceleration. divide both sides by 2 kilograms So let's divide the left by 2 kilograms let's divide the right by 2 kilograms that cancels out. The 10 and the 2-- 10 divided by 2 is 5 and then you have kilograms cancelling kilograms. Your left hand side you get 5 metres per second squared and then that's equal to your acceleration. Now just for fun, what happens if I double that force? Well then I have 20Newtons--I'll actually work it out-- 20 kilograms.metres per second squared is equal to --I'll actually color-code this-- 2 kilograms times the acceleration" + }, + { + "Q": "At 1:16, Sal said something about a vector quantity. Um, what is a vector quantity?", + "A": "A quantity which has magnitude and direction both", + "video_name": "ou9YMWlJgkE", + "timestamps": [ + 76 + ], + "3min_transcript": "Newton's first law tells us that an object at rest will stay at rest, and an object with a constant velocity will keep having that constant velocity unless it's affected by some type of net force or you actually can say that an object with constant velocity will stay having a constant velocity unless it's affected by net force because really this takes into consideration the situation where an object is at rest. You could just have a situation where the constant velocity is zero. So Newton's first law-you're gonna have your constant velocity it could be zero, it's going to stay being that constant velocity unless it's affected, unless there's some net force that acts on it. So that leads to the natural question. How does a net force affect the constant velocity or how does it affect the state of an object? And that's what Newton's second law gives us- Newton's Second Law of Motion And this one is maybe the most famous -actually I won't pick favorites here- but this one gives us the famous formula; Force is equal to mass times acceleration And acceleration is a vector quantity and force is a vector quantity. And what it tells us- 'cause we're saying ok if you apply force it might change that constant velocity but how does it change that constant velocity? Well say I have a brick right here and it is floating in space Newton's second law tells us that it's pretty nice for us that the laws of the universe or at least in the classical sense before Einstein showed up The laws of the universe actually dealt with pretty simple mathematics. What it tells us is if you apply a net force let's say on this side of the object and we talk about net force because if you apply two forces that cancel out and that have zero net force then the object won't change it's constant velocity. If you have a net force applied to one side of this object going in the same direction and what Newton's second law of motion tells us is that acceleration is proportional to the force applied or the force applied is proportional to that accleration And the constant of proportionality to figure out what you have to multiply the acceleration by to get the force or what you have to divide the force by to get the acceleration is called mass that is an object's mass. And I'll make a whole video on this you should not confuse mass with weight and I'll make a whole video on the difference between mass and weight. Mass is a measure of how much stuff there is Now that-we'll see in the future there are other things that we don't normallyconsider stuff that does start to have mass but for our classical or at least first year Physics course you could really just imagine how much stuff there is. Weight as we'll see in a future video is how much that stuff" + }, + { + "Q": "At 0:39, Sal says methanol is a protic solvent and releases H+ ions into the solution, but at 7:05, he says that methanol is a weak base. If methanol releases H+, it should be an acid but here it also acts like a base...so which one is it?", + "A": "Methanol is like water. It can act as either an acid or a base.", + "video_name": "MtwvLru62Qw", + "timestamps": [ + 39, + 425 + ], + "3min_transcript": "Let's think about what type of reaction might occur if we have this molecule right over here. I won't go through the trouble of naming it. It would take up too much time in this video. But it's dissolved in methanol. When we talk about what type of reactions, we're going to pick between Sn2, Sn1, E2 and E1 reactions. Now, maybe the first place to start or the place I like to start is to just look at the solvent itself. And when we're trying to decide what type of reaction will occur, the important thing to think about is, is the solvent protic or aprotic? And if you look at this solvent right here, this is methanol. It is protic. And in case you don't remember what protic means, it means that there are protons flying around in the solvent, that they can kind of go loose and jump around from one molecule to another. And the reason why I know that methanol is protic is because you have hydrogen bonded to a very electronegative atom in oxygen. oxygen can steal hydrogen's electron. And then the hydrogen itself, without the electron, the hydrogen proton will be flying around because it doesn't have a neutron. So this is a protic solvent. Now, you might say, well does anything with a hydrogen, would that be protic? And the answer is no. If you have a bunch of hydrogens bonded to just carbons, that is not protic. Carbon is not so electronegative that it could steal a hydrogen's electron and have the hydrogens So a big giveaway is hydrogen bonded to a very electronegative atom like oxygen. So this is protic. And when we think about protic out of all of the reactions we studied, that favors-- well, even a better way to think about it is it disfavors. So it tells us that it's unlikely to have an Sn2 or an E2 reaction. And the logic there is an Sn2 reaction needs a strong An E2 reaction needs a strong base. Now, if you have protons flying around, the nucleophile or the base is likely to react with the proton. It would not be likely to react with the substrate itself. So a protic solution, you're likely to have an Sn2 or E2. What you are likely to have is an Sn1 or an E1 reaction. Both of these need the leaving group to leave on its own, and actually, having protons around might help to stabilize the leaving group to some degree. So it makes Sn2, E2 unlikely, Sn1, E1 a little more likely. So far, these are our good candidates. Now, the next thing to think about is to just look at the leaving group itself, or see if there is even a leaving group. And over here, everything we see on this molecule is either a carbon or a hydrogen, except for this iodine right here." + }, + { + "Q": "At 7:53 he says 'a meagenta electron has been donated to the carbocation', what is a 'magenta electron'?", + "A": "He is simply referring to the colour of the dot that he uses on the screen to make it clear which is the moving electron. All the other dots in the structure of methanol on the screen are blue. There is no such thing as a magenta or a blue electron. Electrons are too small to have any colour.", + "video_name": "MtwvLru62Qw", + "timestamps": [ + 473 + ], + "3min_transcript": "So this step right here is common to both Sn1 and E1 reaction. The leaving group has to leave. Now, after this, they start to diverge. In an Sn1, the leaving group essentially gets substituted with a weak nucleophile. In an E1, a weak base strips off one of the beta hydrogens and forms an alkene. So let's do them separately. So over here, I'm going to do the Sn1. And on the right-hand side, I will do the E1 reaction. So let me start over here. So the Sn1 is starting over here at this step. I'll just redo this step over here. So this has a positive charge. That has a positive charge here. The iodide has left. I don't have to draw all its valence electrons anymore. We're going to get substituted with the weak base, and the weak base here is actually the methanol. The weak base here is the methanol. So let me draw some methanol here. It's got two unbonded pairs of electrons and one of them, it's a weak base. It was willing to give an electron. It has a partial negative charge over here because oxygen is electronegative, but it doesn't have a full negative charge, so it's not a strong nucleophile. But it can donate an electron to this carbocation, and that's what is going to happen. It will donate an electron to this carbocation. And then after that happens, it will look like this. That's our original molecule. Now this magenta electron has been donated to the carbocation. The other end of it is this blue electron right here on the oxygen. That is our oxygen. Here's that other pair of electrons on that oxygen, and it is bonded to a hydrogen and a methyl group. And then the last step of this is another weak base might be able to come and nab off the hydrogen proton right there. Oh, I want to be very clear here. The oxygen was neutral. The methanol here is neutral. It is giving away an electron to the carbocation. The carbocation had a positive charge because it had lost it originally. Now it gets an electron back. It becomes neutral. The methanol, on the other hand, was neutral, gives away an electron, so now it becomes-- it now is positive. So now you might have another methanol. You might have another methanol molecule sitting out here someplace that might also nab the proton off of this positive ion. So this one right here, it would nab it or it It would give the electron to the hydrogen proton, really." + }, + { + "Q": "At 4:00, Sal talks about friction between the 'piston' and the 'wall' of the cylinder. Now my question is more of an observation. Sal doesn't seem to mention anything about the friction generated from the molecules within the cylinder, assuming they have KE during this process. Therefore, wouldn't the heat generated from this also influence the result in this video? Like I said, just an observation and wondered whether there might be a reason for sal not saying anything about it.", + "A": "Ideal gases have no friction between molecules. Most of the time when we are doing introductory level thermo we are assuming an ideal gas, which is not a bad assumption because many gases behave close to ideal as long as the pressure is not too high and the temperature is not too low", + "video_name": "PFcGiMLwjeY", + "timestamps": [ + 240 + ], + "3min_transcript": "how I got there. It should only be dependent on my state variables. So even if I go on some crazy path, at the end of the day, it should get back to 0. But I did something, I guess, a little bit-- what I did wasn't a proof that this is always a valid state variable. It was only a proof that it's a valid state variable if we look at the Carnot cycle. But it turns out that it was only valid because the Carnot cycle was reversible. And this is a subtle but super important point, and I really should've clarified this on the first video. I guess I was too caught up showing the proof of the Carnot cycle to put the reversibility there. And before I even show you why it has to be reversable, let me just review what reversibility means. Now, we know that in order to even define a path here, the system has to be pretty close to equilibrium the whole time. That's the whole reason why throughout these videos, I've here, and then always-- instead of having one big weight on top that I took off or took on, because it would throw the system out of equilibrium-- I did it in really small increments. I just moved grains of sand, so that the system was always really close to equilibrium. And that's called quasistatic. and I've defined that before. And that means that you're always in kind of a quasi-equilibrium. So your state variables are always defined. But that, by itself, does not give you reversability. You have to be quasistatic and frictionless in order to be reversible. Now, what do we mean by frictionless? Well, I think you know what frictionless means. Is that like you see in this system right here, if I make this piston a little bit bigger, that when this piston rubs against the side of this wall, in kind of our real Those molecules start bumping against each other, and then they start making them vibrate, so they transfer some kinetic energy. From just by rubbing into each other, they start generating some kinetic energy, or some heat. So you normally have some heat generated from friction. Now, if you have some heat generated from friction, when I remove a pebble-- first all, when I remove that first pebble, it might not even do anything. Because it might not even overcome-- you can kind of view it as the force of friction. But let's say I remove some pebbles, and this thing moves up a little bit. But because some of the, I guess you could say, the force differential, the pressure differential between the pebbles and the gas inside, and the pressure of the gas, was used to generate heat as opposed to work, when I add the pebbles back, if I have friction, I'm not going to get back to the same point that I was before. Because friction is always resisting the movement." + }, + { + "Q": "At 0:03, Sal said \"nucleuses\". Is it nuclei?", + "A": "Yes, the plural of nucleus is nuclei.", + "video_name": "lJX8DxoPRfk", + "timestamps": [ + 3 + ], + "3min_transcript": "" + }, + { + "Q": "I have a question at 10:44, when Sal starts to draw the pi bonds - those p-Orbitals overlap two times, so in total the C Atoms would have 3 bonds, two pi bonds and including the sigma bond, right? Shouldn't it be only one overlap of the p-Orbitals, as each C Atom has only one more electron to share in the p-Orbital? Or is it irrelevant, how many times this p-Orbital overlaps, because it only has one more electron inside?", + "A": "While the drawing shows two overlaps, of the pi bond it is actually a theoretical drawing of the possibilities of where the electrons can be. The electron can not be in two places at one time, so though it appears to overlap two times, you are correct to think that it is irrelevant how many times it overlaps because there is only one electron inside. Short answer, though the drawing overlaps two times, it is only representing one bond.", + "video_name": "lJX8DxoPRfk", + "timestamps": [ + 644 + ], + "3min_transcript": "" + }, + { + "Q": "This might be a stupid question, but at 11:00 Sal divided the products over the reactants, and this is really the equiiibrium constant formula. But what difference would it make if we took the reactants and divided by the products? (Ok, we would get the inverse value, but what difference would it make?)", + "A": "As long as the inverse of the equilibrium constant is used, there is no difference. It is normally just easier to use the products divided by the reactants because then the inverse is not needed.", + "video_name": "ONBJo7dXJm8", + "timestamps": [ + 660 + ], + "3min_transcript": "-- let's call that K-minus-- the same exact logic holds. We're just going in this direction now. If we look at our original one, we're going in that direction. So for this reaction, we do the same thing. We literally just do different letters, so the reverse reaction is just going to be the concentration of the Y molecule to the c power, because we need c of them there roughly at the same time, times the concentration of the Z molecule to the d power. Now, just at the beginning of the video, we said that equilibrium is when these rates equal each other. I wrote it down right here. So if the reverse rate is equal to some constant times this, and the forward rate is equal to some constant times that, then we reach equilibrium when these two are equal to each other. Let me clear up somespace here. Let me clear this up, too. So when are they going to be equal to each other? -- the forward rate is this. That's our forward constant, which took into account a whole bunch of temperature and molecular structure and all of that-- times the concentration of our V molecule to the a power. You can kind of view that as what's the probability of finding in a certain volume -- and that certain volume can be factored into that K factor as well-- but what's the probability of finding V things, a V molecules in some volume. And it's the concentration of V to the a power times concentration of X to the b power -- that's the forward reaction-- and that has to equal the reverse reactions. So K-minus times the concentration of Y to the c power times the concentration of Z to the d power. Now, if we divide both sides by -- let me erase more space. Nope, not with that. All right. So let's divide both sides by K-minus so you get K-plus over K-minus is equal to that, is equal to Y to the c times Z to the d. All of that over that-- V to the a times the concentration of X to the b. Let me put this in magenta just so you know that this was this K-minus right here. And then, these are just two arbitrary constants, so we could just replace them and call them the equilibrium constant. And we're there where we need to be. We're at the formula for the equilibrium constant. Now, I know this was really hand wavy, but I want you to at least get the sense that this doesn't come from out of the blue, and there is -- at least I think there is-- there's an intuition here. These are really calculating the probabilities of finding -- this is the forward reaction rate probabilities" + }, + { + "Q": "At 2:30, it's being said that energy of photon =E3 - E1\nBut in reality we always do final energy shell - initial energy shell so over here also it should be E1 - E3....?na..??", + "A": "Which way you write it depends on whether you are referring to the excitation of the electron from E1 to E3, or the dropping back down of the electron from E3 to E1. The magnitude of the two will be the same.", + "video_name": "AznXSVx2xX0", + "timestamps": [ + 150 + ], + "3min_transcript": "- We've been talking about the Bohr model for the hydrogen atom, and we know the hydrogen atom has one positive charge in the nucleus, so here's our positively charged nucleus of the hydrogen atom and a negatively charged electron. If you're going by the Bohr model, the negatively charged electron is orbiting the nucleus at a certain distance. So, here I put the negatively charged electron a distance of r1, and so this electron is in the lowest energy level, the ground state. This is the first energy level, e1. We saw in the previous video that if you apply the right amount of energy, you can promote that electron. The electron can jump up to a higher energy level. If we add the right amount of energy, this electron can jump up to a higher energy level. So now this electron is a distance of r3, so we're talking about the third energy level here. This is the process of absorption. The electron absorbs energy and jumps up to a higher energy level. This is only temporary though, It's eventually going to fall back down to the ground state. Let's go ahead and put that on the diagram on the right. Here's our electron, it's at the third energy level. It's eventually going to fall back down to the ground state, the first energy level. Here's the electron going back to the first energy level here. When it does that, it's going to emit a photon. It's going to emit light. When the electron drops from a higher energy level to a lower energy level, it emits light. This is the process of emission. I could represent that photon here. This is how you usually see it in textbooks. We emit a photon, which is going to have a certain wavelength. Lambda is the symbol for wavelength. We need to figure out how to relate lambda to those different energy levels. The energy of the photon is, the energy of the emitted photon is those two energy levels. We have energy with the third energy level and the first energy level. The difference between those... So, the energy of the third energy level minus the energy of the first energy level. That's equal to the energy of the photon. This is equal to the energy of that photon here. We know the energy of a photon is equal to h nu. Let me go ahead and write that over here. Energy of a photon is equal to h nu. H is Planck's constant, this is Planck's constant. Nu is the frequency. We want to think about wavelength. We need to relate the frequency to the wavelength. The equation that does that is of course, C is equal to lambda nu. So, C is the speed of light, lambda is the wavelength, and nu is the frequency." + }, + { + "Q": "At about 4:46, he says that the back panel was injection molded. What exactly does it mean if something is injection molded?", + "A": "It means that a material with a high melting point (usually metal) is used to make a mold in which molten plastic or polypropylene or other materials can be injected. After the material cools off, the mold is opened and the object can be taken out. Injection molding is usually used because it s quite cheap", + "video_name": "qLMsZKx_a8s", + "timestamps": [ + 286 + ], + "3min_transcript": "So it's really a handy way to do it. It used to be you'd have to look it up or just memorize what the different color band codes meant. But this particular resistor, it's got a green band. So we'll put it on green. There we go. And it's got a-- it looks like a navy blue band, and a gold band, green-- oh, wait. Actually, there's a black one. Sorry-- and then the gold one. There we go. So this is a 56 ohm resistor. And that's the amount of resistance that that resistor provides. And the switch right here is just a momentary switch. All right. It's not a momentary switch. It's a continuous switch. So that means when you push it down, it stays down. So the light will stay on after you push it. And the circuit is extremely simple. Basically, you've got the power from the batteries. It comes in through the loop here and the switch basically opens and closes and stops the power flow. Or when you push down on it, it closes and allows the power to flow in the continuous loop there. And so that's what's inside of a tap light. Let's take the batteries out really quick so. These are double A's. And now, it looks like this back panel here was injection molded. And you can see the ejector pins there. Those are the pins that push it out of the mold. And so it looks like it was injection molded. And I would say-- it doesn't have the plastic designation marking on it, but I would guess that it's probably either polypropylene or ABS plastic. are probably made out of-- I initially thought that they were made out of steel. But let's take a look. We've got some magnets here, so we'll take one of our magnets and-- I don't think they are. So they're probably brass contacts, because the magnets are not attracted to them. So it's not a ferrous metal. So then we've got this loop here. And this plastic loop prevents the positive terminal on the battery from slipping below the contacts. So it stays in constant contact and keeps everything together. And you can see there's another one on this side. And that's pretty much it. Oh, and there's also a feature right here so that if you have a screw or a nail on your wall, you can put the tap light in and just hang it like that. But that's pretty much the tap light." + }, + { + "Q": "At 4:06, why doesn't the ethylene glycol react with the carboxylic acid as well as the ketone?", + "A": "It could, but I believe the reaction with the ketone is faster because there is less steric hindrance. Therefore, this will be your major product.", + "video_name": "fY_ejjMRYg0", + "timestamps": [ + 246 + ], + "3min_transcript": "we have the structure of our alcohol product here. So it would be a four-carbon alcohol, so one, two, three, and four, and the exact same thing down here. So our other product would be butanol. And we have two equivalents of it. So that'd be one, two, three, four carbons. So, by looking at our acetal and thinking about hydrolyzing it and thinking about where those portions came from, we can easily come up with the products of this hydrolysis of this acetal. So let's see how we could use this reaction in, if we were trying to synthesize this molecule over here on the right. And so, if we're trying to synthesize this molecule from this molecule, first you might be able to think you could figure it out by looking at the functional groups. Alright, so we have a ketone over here on the left and we want a ketone over here for our product. We also have a carboxylic acid over here to the left, and then we have an alcohol over here on the right. this kind of transformation, you might think to reduce the carboxylic acid to form your alcohol. And so, you could do that with something like lithium aluminum hydride. So you might think, and you could just add some lithium aluminum hydride here, in the first step. And the second step, add a source of protons to protonate your alkoxide anion to form your alcohol as your product. The only problem with trying to do this in one step is lithium aluminum hydride is also going to reduce your ketone here, to form a secondary alcohol. And so, this will not work. The first thing you have to do is protect the ketone, and then you can use your lithium aluminum hydride. So, we're going to use, we're going to protect our ketone using an acetal, of course. And we're going to react it with ethylene glycol. So if we react our starting compound here with ethylene glycol, and we use an acid Alright, so the ethylene glycol is going to react with the ketone portion of the molecule to form an acetal, specifically a cyclic acetal. Alright, so over here on the left we still have our carboxylic acid. And over here on the right, now we're going to have an oxygen bonded to this carbon, another oxygen bonded to this carbon, and they are, of course, connected. And so, that's our cyclic acetal, which came from this oxygen, this carbon, this carbon, and this oxygen. Right, so that's this oxygen, this carbon, this carbon, and this oxygen. So acetals are stable in basic conditions. And so, now we can add our lithium aluminum hydride, so we add our lithium aluminum hydride here in the first step. And that is going to reduce our carboxylic acid. And so in the second step we can add a source of protons, and we can add some excess water." + }, + { + "Q": "At 2:37 isn't the angle supposed to be 60degrees? Because if there's a right angled triangle, then the right angle would be 90degrees and theta would be 30 so 180-90-30=60, no?", + "A": "Since the is an interior angle alternate to the angle theta, it is 30 degrees. Sal does a great job of explaining this in the previous video, Inclined Plane Force Components, starting at about 4:24.", + "video_name": "Mz2nDXElcoM", + "timestamps": [ + 157 + ], + "3min_transcript": "times the gravitational field times 9.8 meters per second squared. So it's going to be 98 newtons downward. So this is 98 newtons downward. I just took 10 kilograms. Let me write it out. So the force due to gravity is going to be equal to 10 kilograms times 9.8 meters per second squared downward. This 9.8 meters per second squared downward, that is the field vector for the gravitational field of the surface of the earth, I guess is one way to think about it. Sometimes you'll see the negative 9.8 meters per second squared. And then that negative is giving you the direction implicitly because the convention is normally that positive is upward and negative is downward. We'll just go with this right over here. So the magnitude of this vector is 10 times 9.8, which is 98 kilogram meters per second squared, which So the magnitude here is 98 newtons and it is pointing downwards. Now what we want to do is break this vector up into the components that are perpendicular and parallel to the surface of this ramp. So let's do that. So first, let's think about perpendicular to the surface of the ramp. So perpendicular to the surface of the ramp. So this right over here is a right angle. And we saw in the last video, that whatever angle this over here is, that is also going to be this angle over here. So this angle over here is also going to be a 30-degree angle. And we can use that information to figure out the magnitude of this orange vector right over here. And remember, this orange vector is the component of the force of gravity that is perpendicular to the plane. And then there's going to be some component that is parallel to the plane. I'll draw that in yellow. Some component of the force of gravity that is parallel to the plane. And clearly this is a right angle, And this is parallel to the plane. If it's perpendicular to the plane, it's also perpendicular to this vector right over here. So we can use some basic trigonometry, like we did in the last video, to figure out the magnitude of this orange and this yellow vector right over here. This orange vector's magnitude over the hypotenuse is going to be equal to the cosine of 30. Or you could say that the magnitude of this is 98 times the cosine of 30 degrees newtons. 98 times the cosine of 30 degrees newtons. And if you want the whole vector, it's in this direction. And the direction going into the surface of the plane. And, based on the simple trigonometry-- and we go into this in a little bit more detail in the last video-- we know that the component of this vector that is parallel to the surface of this plane is going to be 98 sine of 30 degrees." + }, + { + "Q": "At 1:15, Sal says that the force is 98 Newtons. Because it is downward, wouldn't it be -98?", + "A": "Sal is probably referring to the magnitude of the force, but yes, if you define up to be positive direction, then down would be negative.", + "video_name": "Mz2nDXElcoM", + "timestamps": [ + 75 + ], + "3min_transcript": "Let's say that I have a ramp made of ice. Looks like maybe a wedge or some type of an inclined plane made of ice. And we'll make everything of ice in this video so that we have negligible friction. So this right here is my ramp. It's made of ice. And this angle right over here, let's just go with 30 degrees. And let's say on this ramp made of ice, I have another block of ice. So this is a block of ice. It is a block of ice, it's shiny like ice is shiny. And it has a mass of 10 kilograms. And what I want to do is think about what's going to happen to this block of ice. So first of all, what are the forces that we know Well if we're assuming we're on Earth, and we will, and we're near the surface, then there is the force of gravity. There's the force of gravity acting on this block of ice. And the force of gravity is going to be equal to-- it's going to be in the downward direction, times the gravitational field times 9.8 meters per second squared. So it's going to be 98 newtons downward. So this is 98 newtons downward. I just took 10 kilograms. Let me write it out. So the force due to gravity is going to be equal to 10 kilograms times 9.8 meters per second squared downward. This 9.8 meters per second squared downward, that is the field vector for the gravitational field of the surface of the earth, I guess is one way to think about it. Sometimes you'll see the negative 9.8 meters per second squared. And then that negative is giving you the direction implicitly because the convention is normally that positive is upward and negative is downward. We'll just go with this right over here. So the magnitude of this vector is 10 times 9.8, which is 98 kilogram meters per second squared, which So the magnitude here is 98 newtons and it is pointing downwards. Now what we want to do is break this vector up into the components that are perpendicular and parallel to the surface of this ramp. So let's do that. So first, let's think about perpendicular to the surface of the ramp. So perpendicular to the surface of the ramp. So this right over here is a right angle. And we saw in the last video, that whatever angle this over here is, that is also going to be this angle over here. So this angle over here is also going to be a 30-degree angle. And we can use that information to figure out the magnitude of this orange vector right over here. And remember, this orange vector is the component of the force of gravity that is perpendicular to the plane. And then there's going to be some component that is parallel to the plane. I'll draw that in yellow. Some component of the force of gravity that is parallel to the plane. And clearly this is a right angle," + }, + { + "Q": "You found out force acting parallel to the plane is 49N while force acting perpendicular to the plane if 49*(1.732) [49root3]. Therefore force along vertical is more than horizontal. Therefore the body should not move. But the body will move with acceleration 4.9m/s^2 as mentioned in 10:05. Why ?", + "A": "Why do you think the body should not move? There is a force of 49 N pushing it down the plane. Why would that force not cause acceleration? I think maybe you should watch the video again. You might need to also watch some of the earlier videos about Newton s laws", + "video_name": "Mz2nDXElcoM", + "timestamps": [ + 605 + ], + "3min_transcript": "We're not talking about accelerating straight towards the center of the earth. We're talking about accelerating in that direction. We broke up the force into kind of the perpendicular direction and the parallel direction. So you have this counteracting normal force. And that's why you don't have the block plummeting or accelerating into the plane. Now what other forces do we have? Well, we have the force that's parallel to the surface. And if we assume that there's no friction-- and I can assume that there's no friction in this video because we are assuming that it is ice on ice-- what is going to happen? There's no counteracting force to this 49 newtons. 49 newtons parallel downwards, I should say parallel downwards, to the surface of the plane. So what's going to happen? Well, it's going to accelerate in that direction. You have force is equal to mass times acceleration. Or you divide both sides by mass, you get force over mass is equal to acceleration. Over here, our force is 49 newtons in that direction, parallel downwards to the surface of the plane. And so if you divide both by mass, if you divide both of these by mass. So that's the same thing as dividing it by 10 kilograms, dividing by 10 kilograms, that will give you acceleration. That will give you our acceleration. So acceleration is 49 newtons divided by 10 kilograms in that direction, in this direction right over there. And 49 divided by 10 is 4.9, and then newtons divided by kilograms is meters per second squared. So then you get your acceleration. Your acceleration is going to be 4.9 meters per second squared. That's two bars. Or maybe I'll write parallel. Parallel downwards to the surface. Now I'm going to leave you there, and I'll let you think about another thing that I'll address in the next video is, what if you had this just standing still? If it wasn't accelerating downwards, if it wasn't accelerating and sliding down, what would be the force that's keeping it in a kind of a static state? We'll think about that in the next video." + }, + { + "Q": "At 1:05, you talked of the asteroid that wiped off the dinosaurs. What is the name of this asteroid? And what are the other theories of the beginning of the universe except the big bang?", + "A": "The asteroid impact crater linked to the end of the age of dinosaurs is found at Chicxulub, Mexico. The big bang theory is the only currently successful scientific theory for a aging universe. It replaced the solid state theory after the discovery, by Erwin Hubble, of the universe s expansion.", + "video_name": "DRtLXagrMHw", + "timestamps": [ + 65 + ], + "3min_transcript": "What I've done here is I've copied and pasted a bunch of pictures that signify events in our history, when you think about history on a grander scale, that most of us have some relation to or we kind of have heard it talked about a little bit. And the whole point of this is to try to understand, or try to begin to understand, how long 13.7 billion years is. So just to start off, I have here-- this is the best depiction I could find where it didn't have copyrights. This is from NASA-- of the Big Bang. And I've talked about it several times. The Big Bang occurred 13.7 billion years ago. And then if we go a little bit forward, actually a lot forward, we get to the formation of our actual solar system and the Earth. This is kind of the protoplanetary disk or a depiction of a protoplanetary disk forming around our young Sun. And so this right here is 4.5 billion years ago. Now this over here-- once again, these aren't pictures of them. These are just depictions because no one was there This is what we think the asteroid that killed the dinosaurs looked like when it was impacting Earth. And it killed the dinosaurs 65 million years ago. So until then, we had land dinosaurs. And then this, as far as the current theories go, got rid of them. Now, we'll fast forward a little bit more. At about 3 million years ago-- let me do this in a color that you can see-- about 3 million, so three million years ago, our ancestors look like this. This is Australopithecus afarensis. This is I think a depiction of-- this is Lucy. I believe the theory is that all of us have some DNA from her. But this was 3 million years ago. And you fast forward some more and you actually that looked and thought like you and me. This is 200,000 years ago. That's right over here. Obviously, this drawing was done much later. But this is a depiction of a modern human, so 200,000 years ago. And then you fast forward even more. And I don't want to keep picking on Jesus. I did that with him getting on the jet liner. And I genuinely don't mean any offense to anyone. I just keep picking Jesus because frankly our calendar is kind of-- he's a good person that most people know about, 2,000 years ago. And so when we associate kind of a lot of modern history occurring after his birth. So this right here is obviously a painting of the birth of Jesus. And this is 2,000 years ago. And then this might be a little bit American-centric. But the Declaration of Independence, it" + }, + { + "Q": "9:51 wouldn't the line be more of a curve?", + "A": "Hi, The line would be straight because in this we assume the velocity is constant.", + "video_name": "T0zpF_j7Mvo", + "timestamps": [ + 591 + ], + "3min_transcript": "as it went up, but it's actually slowing it down it's pulling it, it's accelerating it in downward direction so that's why you have a negative right over there, that was out convention at the beginning of last video, up is positive, down is negative, so let's focus So this part right over here, negative 4.9 m/s*s times delat t square times delta t, times delta t square, this will make it a little bit easier Although it still, let me get the calculator out So when one second has past, I let my trusty TI-85 out now when 1s past it's 19.6 times 1, well that's just 19.6 minus 4.9 times 1 square So that's just minus 4.9, mius 4.9, gives us 14.7 meters So 14.7 meters So after 1s, the ball has travel 14.7 meters in the air I'll do the same agenda, so after 2s, our velocity is 19.6 minus 9.8 times 2, times 2 this is 2s has gone by well 9.8 times 2, 9.8 times 2 square seconds gives us 19.6 m/s so these just cancel out, so we get out velocity now is zero So after two seconds our velocity now is zero let me make it so this thing should more look like a line I don't get a sense, so this is let me just draw the line like this so our velocity now is zero after 2 seconds what is our displacement? So literally we at the point with the ball has no velocity and right for that exact moment of time it's stationary and then what do we have going on in our displacement? We have 19.6 let me get the calculator out for this We can do it by hand but for the sake of quickness 19.6 times 2 minus 4.9 times 2s squared, this is 2s squared so that's times four, so that gives us 19.6 meters So we have 19, we are at 19.6 meters after 2 seconds we are 19.6 meters in the air now let's go to 3s, so after 3s, our velocity now" + }, + { + "Q": "At 08:10, you state that for the molecule, you couldn't get a tertiary carbocation, and I was wondering why not. Couldn't you move one of the hydrogens from the second carbon?", + "A": "You could, but then the second carbon would still be secondary, not tertiary.", + "video_name": "iEKA0jUstPs", + "timestamps": [ + 490 + ], + "3min_transcript": "to move over here, shift over one carbon, and form a new covalent bond. So what would we get if we get a hydride shift in our mechanism? Well, now our hydride has shifted over here to that carbon. This carbon no longer has a positive charge on it. We took a bond away from this carbon. So now, this is where our positive charge is. So we have a carbocation. How would we classify this carbocation? Well, one, two, three other carbons. So it's tertiary. It's more stable than our secondary carbocation. So in the final step of our mechanism, we had our chloride anion over here from the first step of our mechanism. So a chloride anion, negatively charged nucleophile file. So a nucleophilic attack on our carbocation. So right there. And we're going to form a bond between that halogen and that carbon. So our final product is going to end up Let's do another one. So we have our cyclohexane ring like that. And then, we have our ethyl group attached to this carbon. And then, our chlorine attached to that carbon. So that's going to be our major product. All right. Let's look at the stereochemistry of this reaction really fast. So let's look at what happens if we react this alkene with hydrochloric acid. So what will we get it? All right. First step-- pi electrons take the proton. Kick the electrons off onto the chlorine. So once again, which side do we add our proton? So two possibilities. I could add to the left side of the double bond or add to the right side of the double bond. It makes sense to add it to the right side of the double bond because that gives us a more stable carbocation. So if I add it to the right side of the double bond, this carbon ends up being positively charged. What kind of a carbocation is that? That carbon is bonded to two other carbons. So it's a secondary carbocation here. There's no kind of rearrangement that we could get here to get a tertiary carbocation. So a secondary carbocation is as stable as we're going to get. Now, carbocations are carbons with three bonds to them, meeting that carbon is sp2 hybridized. So let's redraw this carbocation here. So I'm going to say that this carbon right here represents my carbocation. What's bonded to it? Well, on the left side, there is an ethyl group-- CH2, CH3 right here. And I know that there's a methyl group bonded to it. I'm going to put the methyl group going back in space here. And then there's also a hydrogen attached to that carbon. We just didn't draw it in on our carbocation. So like this. I know this carbon is sp2 hybridized, meaning there's an untouched, unhybridized p orbital on this carbon. All right. So let me draw my p orbital in there like that. And let me go ahead and make sure that everyone realizes this is my carbocation. So sp2 hybridized carbon means the atoms" + }, + { + "Q": "At 6:26, I don't understand why that tertiary carbon has a positive charge.. Didn't it lose a proton? Therefore shouldn't it be negatively charged?", + "A": "No it lost a hydride, the H took both electrons that were in the C-H bond. That carbon now has 3 bonds and 0 lone pairs Formal charge = valence electrons - lone pair electrons - bonds 4 - 0 - 3 = +1", + "video_name": "iEKA0jUstPs", + "timestamps": [ + 386 + ], + "3min_transcript": "So let's go ahead and write that. See if we can spell Markovnikov. The halogen adds the more substituted carbon. And the reason it does that is because the more substituted carbon is the one that was the more stable carbocation in the mechanism. So let's do another mechanism here. Whenever you have a carbocation present, you could have rearrangement. So let's do one where there's some rearrangement. So let's start out with this as our alkene and react that with hydrochloric acid once again. First steps-- pi electrons function as a base. These electrons kick off onto your chlorine. So which side do we add the proton to? Right? We could add the proton to the left side of the double bond. We could add the proton to the right side of the double bond. the most stable carbocation that we can. So it makes sense to add the proton to the right side of the double bond right here because that's going to give us this as a carbocation. What kind of carbocation is that? So let's identify this carbon as the one that has our positive charge. That carbon is bonded to two other carbons. So it is a secondary carbocation. If we had added on the proton to the left side of the double bond, we would have a primary carbocation here. So a secondary carbocation is more stable. Can we form a tertiary carbocation? Because we know tertiary carbocations are even more stable than secondary carbocations. And of course, we can. There's a hydrogen attached to this carbon. And we saw-- in our earlier video on carbocations and rearrangements-- we could get a hydride shift here. All right. So the proton and these two electrons here are hydride anion. to move over here, shift over one carbon, and form a new covalent bond. So what would we get if we get a hydride shift in our mechanism? Well, now our hydride has shifted over here to that carbon. This carbon no longer has a positive charge on it. We took a bond away from this carbon. So now, this is where our positive charge is. So we have a carbocation. How would we classify this carbocation? Well, one, two, three other carbons. So it's tertiary. It's more stable than our secondary carbocation. So in the final step of our mechanism, we had our chloride anion over here from the first step of our mechanism. So a chloride anion, negatively charged nucleophile file. So a nucleophilic attack on our carbocation. So right there. And we're going to form a bond between that halogen and that carbon. So our final product is going to end up" + }, + { + "Q": "At 5:52, Isn't it 2-methylbutyl? Shouldn't the methyl get the lowest possible number(Count Carbons from the right)? Oh, maybe when counting the carbons in a substituent, I have to count from the carbon ATTACHED to the main chain? Am I right?", + "A": "Yes when you re numbering something like this you number from where the group attaches to the main chain.", + "video_name": "joQd0qVnX4M", + "timestamps": [ + 352 + ], + "3min_transcript": "Well, I have a methyl group coming off of carbon 2 this time, so it would be 2-methylpropyl for this complex substituent. The common name for this is isobutyl. So butyl again because there are four total carbons in this complex substituent. Iso because, once again, you have these two methyl groups, so they're like the same, so you get that Y formation. So that's isobutyl. The next one, longest carbon chain, there are two carbons in my longest carbon chain so that would be ethyl. And when I number my longest carbon chain, I can see that I have two methyl groups, and each of those methyl groups is coming off of carbon 1. So I would say this would be 1,1-dimethylethyl. So 1,1-dimethylethyl would be the IUPAC name for this complex You will also see tert-butyl. So tert-butyl is probably used even more frequently. Again, butyl because there are a total of four carbons here. So those are the three possibilities for a complex substituent with a total of four carbons. Let's look at just a few of the possibilities for complex substituents that have five carbons. There are actually much more than this, but these are the ones that are most commonly used. So let's just focus in on these two. So once again, we'll draw our zigzag line to represent the fact that this is actually connected to some straight-chain alkane. And once again, we find our longest carbon chain-- 1, 2, 3, 4. So that would be butyl. And when I number that carbon chain-- 1, 2, 3, 4-- I can see that I have a methyl group coming off of carbon 3. So it would be 3-methylbutyl for the IUPAC name. So you could say isopentyl since there are five carbons now, and iso, because again, you have this methyl group and this methyl group looking like a Y. They're like the same thing. Or I've seen this called isoamyl before. So isoamyl or isopentyl are acceptable IUPAC names as well. What about this one on the right? Longest carbon chain-- 1, 2, 3. So that would be propyl. And numbering it 1, 2, 3, immediately it is obvious that you have two methyl groups coming off of carbon 2. So it would be 2,2-dimethylpropyl, otherwise known as neopentyl since once again you have five carbons for these. So again, there are many more, and we'll stop with those. And so that gives you an idea about how to approach naming complex substituents." + }, + { + "Q": "At 6:50, NeoPentyl can also be IsoPentyl, because the C is touching only one other C, am I right?", + "A": "No. Neopentyl has two methyl groups on C2 of a three-carbon chain. Isopentyl has one methyl group on C3 of a four-carbon chain. They are two completely different things.", + "video_name": "joQd0qVnX4M", + "timestamps": [ + 410 + ], + "3min_transcript": "You will also see tert-butyl. So tert-butyl is probably used even more frequently. Again, butyl because there are a total of four carbons here. So those are the three possibilities for a complex substituent with a total of four carbons. Let's look at just a few of the possibilities for complex substituents that have five carbons. There are actually much more than this, but these are the ones that are most commonly used. So let's just focus in on these two. So once again, we'll draw our zigzag line to represent the fact that this is actually connected to some straight-chain alkane. And once again, we find our longest carbon chain-- 1, 2, 3, 4. So that would be butyl. And when I number that carbon chain-- 1, 2, 3, 4-- I can see that I have a methyl group coming off of carbon 3. So it would be 3-methylbutyl for the IUPAC name. So you could say isopentyl since there are five carbons now, and iso, because again, you have this methyl group and this methyl group looking like a Y. They're like the same thing. Or I've seen this called isoamyl before. So isoamyl or isopentyl are acceptable IUPAC names as well. What about this one on the right? Longest carbon chain-- 1, 2, 3. So that would be propyl. And numbering it 1, 2, 3, immediately it is obvious that you have two methyl groups coming off of carbon 2. So it would be 2,2-dimethylpropyl, otherwise known as neopentyl since once again you have five carbons for these. So again, there are many more, and we'll stop with those. And so that gives you an idea about how to approach naming complex substituents. you have to use them when you're naming straight-chain or cycloalkane molecules. So let's look at a cycloalkane molecule, and let's see how to name this guy. Well, I have four carbons in my ring, and I have four carbons in this group. So tie goes to the cycloalkane. So remember from the last video, if you have an equal number of carbons in your ring as with your chain, you're going to name it as an alkyl cycloalkane. The cycloalkane wins the tie. So there are four carbons, so this'll be cyclobutane. So let's go ahead and write cyclobutane here. And once you've determined that you're going to name it as a cycloalkane, then you have to look at this complex substituent and say, OK, well, that's 1, 2, 3, so that would be propyl. And then when you number that complex substituent 1," + }, + { + "Q": "At 1:00 when 1-methylethyl is being named, why is the longest carbon chain ethyl which has two carbons rather than propyl which would be a three carbon chain?", + "A": "This is because you need to be clear which carbon the group of connected by. You CAN call it a butyl group, but you still must specify which carbon the group is attached to, in this case, it is a sec-butyl.", + "video_name": "joQd0qVnX4M", + "timestamps": [ + 60 + ], + "3min_transcript": "So how do we name this molecule? Well, we start with the longest carbon chain. So there are seven carbons in my longest carbon chain. So I would call this heptane. And I number it to give the substituent the lowest number possible. So in this example, it doesn't really matter if I start from the left or from the right. In both examples, you would end up with a 4 for your substituent there. Now, this substituent looks different from ones we've seen before. There are three carbons in it, but those carbons are not in a straight-chain alkyl group. So if I look at it, right there are three carbons, but they're not going in a straight chain. They're branching of branching here. So this is kind of weird. How do we name this substituent? Well, down here, I have the same substituent, and I'm going to draw this little zigzag line to indicate that that substituent is coming off of some straight-chain alkane. And when you're naming a complex substituent like this, you actually use the same rules that you would use for a straight-chain alkane. which in this case is only two carbons. So that would be an ethyl group coming off of my carbon chain. So I'm going to go ahead and name that as an ethyl group. I'm going to go ahead number it to give my branching group there the lowest number possible. So I go 1 and 2. So what is my substituent coming off of my ethyl group? Well, that's a methyl group coming off of carbon 1. So I name it as 1-methylethyl. OK, so now, that complex substituent is named as 1-methylethyl. So I could go ahead and put that into my name. So coming off of carbon 4, I have 1-methylethyl. And I'm going to put that in parentheses. And all of that is coming off of carbon 4 for my molecule. way of naming that molecule. So if your naming your complex substituent as 1-methylethyl, that's the official IUPAC way, but there are also common names for these complex substituents. So the common name for 1-methylethyl is isopropyl. So isopropyl is the common name. And isopropyl is used so frequently that it's perfectly acceptable to use isopropyl for the name of this molecule as well. So you could have said, oh, this is 4-isopropylheptane, and you would have been absolutely correct. So that's yet another IUPAC name. So iso means same, and it probably comes from the fact that you have these two methyl groups giving you this Y shape that are the same. So that's one complex substituent, one that has three carbons on it. Let's look at a bunch of complex substituents" + }, + { + "Q": "At 8:16, why is the carbon compound called 1-methylpropylcyclobutane but not butylcyclobutane?", + "A": "A butyl group consists of a consists of a chain of four carbon atoms. The longest chain (starting from the ring attachment!) is only three carbon atoms long, so it cannot be named butyl, even though it has four carbon atoms in total.", + "video_name": "joQd0qVnX4M", + "timestamps": [ + 496 + ], + "3min_transcript": "So you could say isopentyl since there are five carbons now, and iso, because again, you have this methyl group and this methyl group looking like a Y. They're like the same thing. Or I've seen this called isoamyl before. So isoamyl or isopentyl are acceptable IUPAC names as well. What about this one on the right? Longest carbon chain-- 1, 2, 3. So that would be propyl. And numbering it 1, 2, 3, immediately it is obvious that you have two methyl groups coming off of carbon 2. So it would be 2,2-dimethylpropyl, otherwise known as neopentyl since once again you have five carbons for these. So again, there are many more, and we'll stop with those. And so that gives you an idea about how to approach naming complex substituents. you have to use them when you're naming straight-chain or cycloalkane molecules. So let's look at a cycloalkane molecule, and let's see how to name this guy. Well, I have four carbons in my ring, and I have four carbons in this group. So tie goes to the cycloalkane. So remember from the last video, if you have an equal number of carbons in your ring as with your chain, you're going to name it as an alkyl cycloalkane. The cycloalkane wins the tie. So there are four carbons, so this'll be cyclobutane. So let's go ahead and write cyclobutane here. And once you've determined that you're going to name it as a cycloalkane, then you have to look at this complex substituent and say, OK, well, that's 1, 2, 3, so that would be propyl. And then when you number that complex substituent 1, of carbon 1. So you would write 1-methylpropyl. And if you wanted to, you could identify that 1-methylpropyl as coming off of carbon 1 of your cyclobutane. So you could put this in parentheses and write 1-(1-methylpropyl)cyclobutane. Or you could just leave the one off, and say (1-methylpropyl)cyclobutane, because, again, it is implied. What is the common name for this complex substituent? So 1-methylpropyl, and we go back up here, and we find 1-methylpropyl was also called sec-butyl. So we could also have named this molecule sec-butylcyclobutane. So let's go ahead and write that. So sec-butylcyclobutane is a perfectly acceptable IUPAC name as well." + }, + { + "Q": "At 6:25, where do you get the Oxaloacetic Acid from?", + "A": "At the end of krebs cycle oxalo acetate is formed from malate in the presence of the enzyme malate dehydrogenase. the same oxalo acetic acid is used to combine with Acetyl Co enzyme-A The whole thing s a cycle", + "video_name": "juM2ROSLWfw", + "timestamps": [ + 385 + ], + "3min_transcript": "So we have this kind of preparation step for the Krebs Cycle. We call that pyruvate oxidation. And essentially what it does is it cleaves one of these carbons off of the pyruvate. And so you end up with a 2-carbon compound. You don't have just two carbons, but its backbone of carbons is just two carbons. Called acetyl-CoA. And if these names are confusing, because what is acetyl coenzyme A? These are very bizarre. You could do a web search on them But I'm just going to use the words right now, because it will keep things simple and we'llget the big picture. So it generates acetyl-CoA, which is this 2-carbon compound. And it also reduces some NAD plus to NADH. And this process right here is often given credit-- or the Krebs cycle or the citric acid cycle gets But it's really a preparation step for the Krebs cycle. Now once you have this 2-carbon chain, acetyl-Co-A right here. you are ready to jump into the Krebs cycle. This long talked-about Krebs cycle. And you'll see in a second why it's called a cycle. Acetyl-CoA, and all of this is catalyzed by enzymes. And enzymes are just proteins that bring together the constituent things that need to react in the right way so that they do react. So catalyzed by enzymes. This acetyl-CoA merges with some oxaloacetic acid. A very fancy word. But this is a 4-carbon molecule. These two guys are kind of reacted together, or merged together, depending on how you want to view it. I'll draw it like that. It's all catalyzed by enzymes. And this is important. Some texts will say, is this an enzyme catalyzed reaction? Everything in the Krebs cycle is an And they form citrate, or citric acid. Which is the same stuff in your lemonade or your orange juice. And this is a 6-carbon molecule. You have a 2-carbon and a 4-carbon. You get a 6-carbon molecule. And then the citric acid is then oxidized over a bunch of steps. And this is a huge simplification here. But it's just oxidized over a bunch of steps. Again, the carbons are cleaved off. Both 2-carbons are cleaved off of it to get back to oxaloacetic acid. And you might be saying, when these carbons are cleaved off, like when this carbon is cleaved off, what happens to it? It becomes CO2. It gets put onto some oxygen and leaves the system. So this is where the oxygen or the carbons, or the carbon dioxide actually gets formed. And similarly, when these carbons get cleaved off, it forms CO2." + }, + { + "Q": "@12:27 its shows that 10 NADH molecules are formed from one glucose molecule, however at the end of the video we count 8NADH molecules. Which one is correct?", + "A": "You should have 10NADH at the very end of cellular respiration. 2 from glycolysis, 2 from bridging reaction and 6 from TCA/Kreb/Citric acid cycle.", + "video_name": "juM2ROSLWfw", + "timestamps": [ + 747 + ], + "3min_transcript": "We already accounted for the glycolysis right there. Two net ATPs, two NADHs. Now, in the citric acid cycle, or in the Krebs cycle, well first we have our pyruvate oxidation. That produced one NADH. But remember, if we want to say, what are we producing for every glucose? This is what we produced for each of the pyruvates. This NADH was from just this pyruvate. But glycolysis produced two pyruvates. So everything after this, we're going to multiply by two for every molecule of glucose. So I'll say, for the pyruvate oxidation times two means that we got two NADHs. And then when we look at this side, the formal Krebs cycle, what do we get? We have, how many NADHs? One, two, three NADHs. this cycle for each of the pyruvates produced from glycolysis. So that gives us six NADHs. We have one ATP per turn of the cycle. That's going to happen twice. Once for each pyruvic acid. So we get two ATPs. And then we have one FADH2. But it's good, we're going to do this cycle twice. This is per cycle. So times two. We have two FADHs. Now, sometimes in a lot of books these two NADHs, or per turn of the Krebs cycle, or per pyruvate this one NADH, they'll give credit to the Krebs cycle for that. So sometimes instead of having this intermediate step, they'll just write four NADHs right here. And you'll do it twice. Once for each puruvate. So they'll say eight NADHs get produced from the Krebs cycle. But the reality is, six from the Krebs cycle two from the Now the interesting thing is we can account whether we get to the 38 ATPs promised by cellular respiration. We've directly already produced, for every molecule of glucose, two ATPs and then two more ATPs. So we have four ATPs. Four ATPs. How many NADHs do we have? 2, 4, and then 4 plus 6 10. We have 10 NADHs. And then we have 2 FADH2s. I think in the first video on cellular respiration I said FADH. It should be FADH2, just to be particular about things. And these, so you might say, hey, where are our 38 ATPs? We only have four ATPs right now. But these are actually the inputs in the electron transport chain. These molecules right here get oxidized in the electron transport chain. Every NADH in the electron transport chain produces three ATPs." + }, + { + "Q": "At 2:35 a correction appears saying he meant to say glycolysis occurs in the cytosol not the cytoplasm. I learned in my class it does occur in the cytoplasm, what's the difference between these two terms?", + "A": "Cytosol is a part of cytoplasm. It is the part containing all the water and organic molecules. Cytoplasm consist of the cytosol and the organelles suspended in it.", + "video_name": "juM2ROSLWfw", + "timestamps": [ + 155 + ], + "3min_transcript": "And I always say the net there, because remember, it used two ATPs in that investment stage, and then it generated four. So on a net basis, it generated four, used two, it gave us two ATPs. And it also produced two NADHs. That's what we got out of glycolysis. And just so you can visualize this a little bit better, let me draw a cell right here. Maybe I'll draw it down here. Let's say I have a cell. That's its outer membrane. Maybe its nucleus, we're dealing with a eukaryotic cell. That doesn't have to be the case. It has its DNA and its chromatin form all spread around like that. And then you have mitochondria. And there's a reason why people call it the power We'll look at that in a second. So there's a mitochondria. It has an outer membrane and an inner I'll do more detail on the structure of a mitochondria, maybe later in this video or maybe I'll do a whole video on them. That's another mitochondria right there. And then all of this fluid, this space out here that's between the organelles-- and the organelles, you kind of view them as parts of the cell that do specific things. Kind of like organs do specific things within our own bodies. So this-- so between all of the organelles you have this fluidic space. This is just fluid of the cell. And that's called the cytoplasm. And that's where glycolysis occurs. So glycolysis occurs in the cytoplasm. Now we all know-- in the overview video-- we know what the next step is. The Krebs cycle, or the citric acid cycle. And that actually takes place in the inner membrane, or I should say the inner space of these mitochondria. Let me draw a mitochondria here. So this is a mitochondria. It has an outer membrane. It has an inner membrane. If I have just one inner membrane we call it a crista. If we have many, we call them cristae. This little convoluted inner membrane, let me give it a label. So they are cristae, plural. And then it has two compartments. Because it's divided by these two membranes. This compartment right here is called the outer compartment. This whole thing right there, that's the outer compartment. And then this inner compartment in here, is called the matrix. Now you have these pyruvates, they're not quite just ready for the Krebs cycle, but I guess-- well that's a good intro into how do you make them ready for the Krebs cycle? They actually get oxidized. And I'll just focus on one of these pyruvates. We just have to remember that the pyruvate, that this" + }, + { + "Q": "at 5:23, sal was trying to find out the work required to move the charger a distance of 5 meter closer. shoudnt he have multiplied the force by the distance before integrating?", + "A": "well he already did that when he multiplied it by dr. And he also mentioned it at about 5:50 that it is the work done in moving the particle by a distance dr and he is going to sum up the work done by using integration. She already has multiplied the force by the infinitiesmally small distance and then adding them up using integral.", + "video_name": "CqsYCIjSm9A", + "timestamps": [ + 323 + ], + "3min_transcript": "Because the field is pushing it outward. It takes work to push it inward. So let's say we want to push it in. Let's say it's at 10 meters. Let's say that this distance right here-- let me draw a radial line-- let's say that this distance right here is 10 meters, and I want to push this particle in 5 meters, so it eventually gets right here. This is where I'm eventually going to get it so then it's going to be 5 meters away. So how much work does it take to move it 5 meters towards this charge? Well, the way you think about it is the field keeps changing, right? But we can assume over a very, very, very, very infinitely small distance, and let's call that infinitely small distance dr, change in radius, and as you can see, we're about to If you don't understand what any of this is, you might want to review or learn the calculus in the calculus playlist, but how much work does it require to move this particle a very, very small distance? Well, let's just assume over this very, very, very small distance, that the electric field is roughly constant, and so we can say that the very, very small amount of work to move over that very, very small distance is equal to Coulomb's constant q1 q2 over r squared times dr. Now before we move on, let's think about something for a second. Coulomb's Law tells us that this is the outward force that this charge is exerting on this particle or that the field is exerting on this particle. The force that we have to apply to move the particle from here to here has to be an inward force. to be a negative. And why is that? Because we have to completely offset the force of the field. Maybe if the particle was already moving a little bit, then our force will keep it from decelerating from the field, and if it wasn't already moving, we would have to nudge it just an infinitely small amount just to get it moving, and then our force would completely offset the force of the field, and the particle would neither accelerate nor decelerate. So this is the amount of work, and I just want to explain that we want to put that negative sign there because we going in the opposite direction of the field. So how do we figure out the total amount of work? We figured out the amount of work to get it from here to here, and I even drew it much bigger than it would be. These dr's, this is an infinitely small change in radius. If we want to figure out the total work, then we keep adding them up. We say, OK, what's the work to go from here to here, then the work to go from there to there, then the work to go from there to there, all the way until we get to 5 meters" + }, + { + "Q": "At 7:00 , could anyone do a brief explanation of why integration is important of it works applying it in this case?", + "A": "integration here is important as the electric field is not uniform. it keeps changing with every point, so by applying the formula we do not get a clear idea of the work done throughout, so we do integration of small distance", + "video_name": "CqsYCIjSm9A", + "timestamps": [ + 420 + ], + "3min_transcript": "to be a negative. And why is that? Because we have to completely offset the force of the field. Maybe if the particle was already moving a little bit, then our force will keep it from decelerating from the field, and if it wasn't already moving, we would have to nudge it just an infinitely small amount just to get it moving, and then our force would completely offset the force of the field, and the particle would neither accelerate nor decelerate. So this is the amount of work, and I just want to explain that we want to put that negative sign there because we going in the opposite direction of the field. So how do we figure out the total amount of work? We figured out the amount of work to get it from here to here, and I even drew it much bigger than it would be. These dr's, this is an infinitely small change in radius. If we want to figure out the total work, then we keep adding them up. We say, OK, what's the work to go from here to here, then the work to go from there to there, then the work to go from there to there, all the way until we get to 5 meters And what we do when we take the sum of these, we assume that it's an infinite sum of infinitely small increments. And as you learned, that is nothing but the integral, and so that is the total work is equal to the integral. That's going to be a definite integral because we're starting at this point. We're summing from-- our radius is equal to 10 meters-- that's our starting point-- to radius equals 5 meters. That might be a little unintuitive that we're starting at the higher value and ending at the lower value, We're pushing it inwards. And then we're taking the integral of minus k q1 q2 over r squared dr. All of these are constant terms up here, right? So we could take them out. So this is the same thing-- I don't want to run out of over r squared-- or to the negative 2-- dr. And that equals minus k-- I'm running out of space-- q1 q2. We take the antiderivative. We don't have to worry about plus here because it's a definite integral. r to the negative 2, what's the antiderivative? It's minus r to the negative 1. Well, that minus r, the minus on the minus r will just cancel with this. That becomes a plus r to the negative 1, And you evaluate it at 5 and then subtract it and evaluate it at 10. And then-- let me just go up here. Actually, let me erase some of this. Let me erase this up here." + }, + { + "Q": "in 3:37, Sal says that the arteries are blue. Is that true, because I thought the veins were blue?", + "A": "Sal might have said that arteries are blue but may be it meant pulmonary arteries are blue.", + "video_name": "QhiVnFvshZg", + "timestamps": [ + 217 + ], + "3min_transcript": "So I've been all zoomed in here on the alveolus and these capillaries, these pulmonary capillaries-- let's zoom out a little bit-- or zoom out a lot-- just to understand, how is the blood flowing? And get a better understanding of pulmonary arteries and veins relative to the other arteries and veins that are in the body. So here-- I copied this from Wikipedia, this diagram of the human circulatory system-- and here in the back you can see the lungs. Let me do it in a nice dark color. So we have our lungs here. You can see the heart is sitting right in the middle. And what we learned in the last few videos is that we have our little alveoli and our lungs. Remember, we get to them from our bronchioles, which are branching off of the bronchi, which branch off of the trachea, which connects to our larynx, which connects to our pharynx, which connects to our mouth and nose. we have the capillaries. So when we go away from the heart-- and we're going to delve a little bit into the heart in this video as well-- so when blood travels away from the heart, it's de-oxygenated. It's this blue color. So this right here is blood. This right here is blood traveling away from the heart. It's going behind these two tubes right there. So this is the blood going away from the heart. So this blue that I've been highlighting just now, these are the pulmonary arteries and then they keep splitting into arterials and all of that and eventually we're in capillaries-- super, super small tubes. They run right past the alveoli and then they become oxygenated and now we're going back to the heart. So we're talking about pulmonary veins. So we go back to the heart. Now we're going to go back to the heart. Hope you can see what I'm doing. And we're going to enter the heart on this side. You actually can't even see where we're entering the heart. We're going to enter the heart right over here-- and I'm going to go into more detail on that. Now we have oxygenated blood. And then that gets pumped out to the rest of the body. Now this is the interesting thing. When we're talking about pulmonary arteries and veins-- remember, the pulmonary artery was blue. As we go away from the heart, we have de-oxygenated blood, but it's still an artery. Then as we go towards the heart from the lungs, we have a vein, but it's oxygenated." + }, + { + "Q": "At 9:28, Sal said that Veins carry deoxygenated blood but before it he labeled that veins carry oxygenated blood.", + "A": "The one Sal labelled before, so he was talking about the pulmonary vein. And where he labelled on 9:28 that was not dealing with lungs, thus it was labelled a vein that caries deoxygenated blood. Remember, when one is talking about pulmonary , it refers to that one which which is dealing with the lungs. The lungs have an opposite effect as compared to the other body; the veins carry oxygenated while the arteries carry deoxygenated blood.", + "video_name": "QhiVnFvshZg", + "timestamps": [ + 568 + ], + "3min_transcript": "So you don't see it. I'm going to do a detailed diagram in a second-- into the pulmonary artery. We're going away from the heart. This was a vein, right? This is a vein going to the heart. This is a vein, inferior vena cava vein. This is superior vena cava. They're de-oxygenated. Then I'm pumping this de-oxygenated blood away from the heart to the lungs. Now this de-oxygenated blood, this is in an artery, right? This is in the pulmonary artery. It gets oxygenated and now it's a pulmonary vein. And once it's oxygenated, it shows up here in the left-- let me do a better color than that-- it shows up right here in the left atrium. Atrium, you can imagine-- it's kind of a room with a skylight or that's open to the outside and in both of these cases, things are entering from above-- not sunlight, but blood is entering from above. And in the left atrium, the blood is entering-- and remember, the left atrium is on the right-hand side from our point of view-- on the left atrium, the blood is entering from above from the lungs, from the pulmonary veins. Veins go to the heart. Then it goes into-- and I'll go into more detail-- into the left ventricle and then the left ventricle pumps that oxygenated blood to the rest of the body via the non-pulmonary arteries. So everything pumps out. Let me make it a nice dark, non-blue color. So it pumps it out through there. You don't see it right here, the way it's drawn. It's a little bit of a strange drawing. It's hard to visualize, but I'll show it in more detail and then it goes to the rest of the body. Let me show you that detail right now. So we said, we have de-oxygenated blood. Let's label it right here. This is the superior vena cava. our arms and heads. This is the inferior vena vaca. This is veins from our abdomen and from our legs and the rest of our body. So it it first enters the right atrium. Remember, we call the right atrium because this is someone's heart facing us, even though this is on the left-hand side. It enters through here. It's de-oxygenated blood. It's coming from veins. the body used the oxygen. Then it shows up in the right ventricle, right? These are valves in our heart. And it passively, once the right ventricle pumps and then releases, it has a vacuum and it pulls more blood from the It pumps again and then it pushes it through here. Now this blood right here-- remember, this one still is de-oxygenated blood. De-oxygenated blood goes to the lungs to become oxygenated. So this right here is the pulmonary-- I'm using the word pulmonary because it's going to or from the lungs." + }, + { + "Q": "At 10:18 sal says that the blood from the pulmonary artery goes to the heart but shouldn't it be going to the lungs?", + "A": "Yes, it should go to the lungs. \u00f0\u009f\u0098\u008a", + "video_name": "QhiVnFvshZg", + "timestamps": [ + 618 + ], + "3min_transcript": "And in the left atrium, the blood is entering-- and remember, the left atrium is on the right-hand side from our point of view-- on the left atrium, the blood is entering from above from the lungs, from the pulmonary veins. Veins go to the heart. Then it goes into-- and I'll go into more detail-- into the left ventricle and then the left ventricle pumps that oxygenated blood to the rest of the body via the non-pulmonary arteries. So everything pumps out. Let me make it a nice dark, non-blue color. So it pumps it out through there. You don't see it right here, the way it's drawn. It's a little bit of a strange drawing. It's hard to visualize, but I'll show it in more detail and then it goes to the rest of the body. Let me show you that detail right now. So we said, we have de-oxygenated blood. Let's label it right here. This is the superior vena cava. our arms and heads. This is the inferior vena vaca. This is veins from our abdomen and from our legs and the rest of our body. So it it first enters the right atrium. Remember, we call the right atrium because this is someone's heart facing us, even though this is on the left-hand side. It enters through here. It's de-oxygenated blood. It's coming from veins. the body used the oxygen. Then it shows up in the right ventricle, right? These are valves in our heart. And it passively, once the right ventricle pumps and then releases, it has a vacuum and it pulls more blood from the It pumps again and then it pushes it through here. Now this blood right here-- remember, this one still is de-oxygenated blood. De-oxygenated blood goes to the lungs to become oxygenated. So this right here is the pulmonary-- I'm using the word pulmonary because it's going to or from the lungs. And it's going away from the heart. It's the pulmonary artery and it is de-oxygenated. Then it goes to the heart, rubs up against some alveoli and then gets oxygenated and then it comes right back. Now this right here, we're going to the heart. So that's a vein. It's in the loop with the lungs so it's a pulmonary vein and it rubbed up against the alveoli and got the oxygen diffused into it so it is oxygenated. And then it flows into your left atrium. Now, the left atrium, once again, from our point of view, is on the right-hand side, but from the dude looking at it, it's his left-hand side." + }, + { + "Q": "At 8:52, he mentions something about a hydronium ion being produced. Does this ion affect the solution or organism such that it causes harm to them?", + "A": "The hydronium ion is just a water molecule that has been protonated. Water is H2O and the hydronium ion is H3O+ (ie, water with an extra H, giving rise to a positive charge). It occurs naturally in water, even at neutral pH, and is present at increasingly higher concentrations as the pH of water is reduced by adding acid. It is only harmful if the pH of the solution is very low although bear in mind that the pH of stomach acid is very low (pH 1.5 to 3.5).", + "video_name": "-Aj5BTnz-v0", + "timestamps": [ + 532 + ], + "3min_transcript": "and in the double bond right [over] here it could let go of one of the bonds the electrons in one of the bonds and then that can be taken back by the oxygen or even better that can be used by that oxygen to capture a Hydrogen proton in the solution and actually probably part of a Hydronium Molecule But let me just draw it this way this would just be used to capture a Hydrogen proton that would just be a hydrogen a hydrogen atom without its electron. It's just a hydrogen ion It would just be a hydrogen proton and that would form this bond That would form this bond right over here and let me let me just be very clear this carbon this carbon right over here is This carbon right over there this oxygen this oxygen is This oxygen is that oxygen right over there, and so hopefully you see how it forms a cyclone. You're probably saying Oh, wait wait isn't the way I've drawn it looks like there's an extra hydrogen over here, and then that would leave this guy with a Positive charge we leave with a positive charge, but you can imagine we're in a solution of water then hey I have some I have another water molecule right over here And you know these things are all bouncing around and interacting in different ways But it could use let me do that in the right color it could use So that's oxygen it could use one of its lone pairs instead of this you know this will become positive temporarily But then it can use it can do it can use one of its lone pairs to grab just the hydrogen proton which would allow Which will allow this character to take its to take its Electrons to take these electrons back and turn into this character and just be neutral and then this this guy Would have gained so we have a proton going into the solution you have hi But we took a proton from [the] solution We took a proton we gave a proton to the solution And so you could end up with this so the whole reason I did This is something that's really valuable to get very very familiar [with] because you're going to see Glucose and other sugars in many many many different molecules throughout your academic career" + }, + { + "Q": "When Sal talks about the atoms 'ionizing' (1:30), where do the electrons that were knocked off go? Could there be He-, like there is He+?", + "A": "The electrons just jump off into the space around the atom, but they are no longer orbiting it. Sort of like the nucleus was swinging them around on a string; then the string broke. Another atom might pick up the electron or it could just float around between atoms. And He- can exist but it s not very happy so that extra electron tends to not stick around for very long.", + "video_name": "X_3QAB3o4Vw", + "timestamps": [ + 90 + ], + "3min_transcript": "- [Voiceover] In the last video we learned that there are a class of stars called Cepheid variables. And these are the super giant stars, as much as 30000 times as bright as the sun. A mass, as much as 20 times the mass of the sun. And what's neat about them is, one, because they're so large and so bright, you can see them really really far away. And what's even neater about them is that they're variable, that they pulsate. And because their pulsations are related to their actual luminosity, you know if you see a cepheid variable star in some distant galaxy, you know what it's luminosity actually is if you were kind of at the star, because you know you can see how it's period of pulsation. And so if you know it's actual luminosity, and you know it's obviously apparent luminosity, you know how much it's gotten dimmed. And the more dim it's gotten from its actual state, you know the farther away it is. So that's the real value of them. What I want to do in this video is to try to explain why they're variables. Why they pulsate. And to do that, to do that, is doubly and singly ionized helium. And just to review, helium, so neutral helium, let me draw a neutral helium, neutral helium's got two protons, it's got two protons, two neutrons, two neutrons, and then two electrons and obviously this is not drawn to scale. So this is neutral helium right over here. Now, if you singly ionize helium you knock off one of these electrons. And these type of things happen in stars when you have a lot of heat, easier to ionize things. So singly ionized helium will look like this. It'll have the same nucleus, two protons, two neutrons. One of the electrons gets knocked off so now you only have one electron. And now you have a net positive charge. So here, let me do this in a different color, this helium now has a net charge, we could write one plus here, but if you just write a plus you implicitly mean a positive charge of one. Now you can also double the ionized helium if the environment is hot enough. and doubly ionizing helium is essentially knocking off both of the electrons. So then it's really just a helium nucleus. It's really just a helium nucleus like this. This right here is doubly, doubly ionized helium. Now I just said in order to do this you have to have a hotter environment. There has to be a hotter environment in order to be able to knock off both these, this electron really doesn't want to leave, to take an electron off of something that's already positive is difficult. You have to have a lot of really pressure and temperature. This is cooler. And this is all relative, we're talking about the insides of stars. So, you know, it's hot, this is a hotter part of the star versus a cooler part of the star I guess is a way you think about this, it's still a very hot environment by our traditional, every day standards. Now the other thing about the doubly ionized helium is that it is more opaque. It is more opaque," + }, + { + "Q": "At 2:03, why the voltage drop is +10 V and voltage rise is -10V?\nshouldn't be in a opposite way?", + "A": "Hello Moon, It s a play on words. To drop an object is to give it a negative rise. Same thing in each case. Regards, APD", + "video_name": "Bt6V7D5av9A", + "timestamps": [ + 123 + ], + "3min_transcript": "- [Voiceover] Now we're ready to start hooking up our components into circuits, and one of the two things that are going to be very useful to us are Kirchhoff's laws. In this video we're gonna talk about Kirchhoff's voltage law. If we look at this circuit here, this is a voltage source, let's just say this is 10 volts. We'll put a resistor connected to it and let's say the resistor is 200 ohms. Just for something to talk about. One of the things I can do here is I can label this with voltages on the different nodes. Here's one node down here. I'm going to arbitrarily call this zero volts. Then if I go through this voltage source, this node up here is going to be at 10 volts. 10 volts. So here's a little bit of jargon. We call this voltage here. The voltage goes up as we go through the voltage source, and that's called a voltage rise. at this point in the circuit right here and we went from this node down to this node, like that, the voltage would go from 10 volts down to zero volts in this circuit, and that's called a voltage drop. That's just a little bit of slang, or jargon that we use to talk about changes in voltage. Now I can make an observation about this. If I look at this voltage rise here, it's 10 volts, and if I look at that voltage drop, the drop is 10 volts. I can say the drop is 10 volts, or I could say the rise on this side is minus 10 volts. A rise of minus 10. These two expressions mean exactly the same thing. It meant that the voltage went from 10 volts to zero volts, sort of going through this 200 ohm resistor. which is, v-rise minus v-drop equals what? Equals zero. I went up 10 volts, back down 10 volts, I end up back at zero volts, and that's this right here. This is a form of Kirchhoff's voltage law. It says the voltage rises minus the voltage drops is equal to zero. So if we just plug our actual numbers in here what we get is 10 minus 10 equals zero. I'm gonna draw this circuit again. Let's draw another version of this circuit. This time we'll have two resistors instead of one. We'll make it..." + }, + { + "Q": "0:47 Why is it weak?\nIs it weak because both oxygens have the same electronegitivity?", + "A": "Not because of the electronegativity but because of how alcohols are more stable than cyclic peroxides and can easily form from a reaction with water.", + "video_name": "KfTosrMs5W0", + "timestamps": [ + 47 + ], + "3min_transcript": "If you start with an alkene and add to that alkene a percarboxylic acid, you will get epoxide. So this is an epoxide right here, which is where you have oxygen in a three-membered ring with those two carbons there. You can open up this ring using either acid or base catalyzed, and we're going to talk about an acid catalyzed reaction in this video. And what ends up happening is you get two OH groups that add on anti, so anti to each other across from your double bond. So the net result is you end up oxidizing your alkene. So you could assign some oxidation numbers on an actual problem and find out that this is an oxidation reaction. All right. Let's look at the mechanism to form our epoxide. So we start with our percarboxylic acid here, which looks a lot like a carboxylic acid except it has an extra oxygen. And the bond between these two oxygen atoms is weak, so this bond is going to break in the mechanism. The other important thing to note about the structure of our percarboxylic acid is the particular confirmation that it's in. So this hydrogen ends up being very a source of attraction between those atoms. There's some intramolecular hydrogen bonding that keeps it in this conformation. When the percarboxylic acid approaches the alkene, when it gets close enough in this confirmation, the mechanism will begin. This is a concerted eight electron mechanism, which means that eight electrons are going to move at the same time. So the electrons in this bond between oxygen and hydrogen are going to move down here to form a bond with this carbon. The electrons in this pi bond here are going to move out and grab this oxygen. That's going to break this weak oxygen-oxygen bond, and those electrons move into here. And then finally, the electrons in this pi bond are going to move to here to form an actual bond between that oxygen and that hydrogen. So let's see if we can draw the results of this concerted eight electron mechanism. So, of course, at the bottom here we're going to form our epoxide. So we draw in our carbons, and then we can put in our oxygen And then we show the bond between those like that. And then up at the top here, here's my carbonyl carbon. So now there's only one bond between that carbon and this oxygen. There is a new bond that formed between that oxygen and that hydrogen, and there is an R group over here. And then there used to be only one bond to this oxygen, but another lone pair of electrons moved in to form a carbonyl here. So this is our other product, which you can see is a carboxylic acid. Let's color code these electrons so we can follow them a little bit better. So let's make these electrons in here, those electrons are going to form the bond on the left side between the carbon and the oxygen like that. Let's follow these electrons next. So now let's look at these electrons in here, the electrons in this pi bond. Those are the ones that are going" + }, + { + "Q": "around 7:25 - why would the H2O nucleophilic attack the partially positive carbons when the oxygen in the epoxide has a formal positive charge?", + "A": "Formal charge is not necessarily the same thing as the actual charge. In this case much of the actual positive charge is on the carbons. Also, the result of that interaction would be a peroxide with a highly unstable (high energy) bond between two oxygens.", + "video_name": "KfTosrMs5W0", + "timestamps": [ + 445 + ], + "3min_transcript": "So I'm going to put my oxygen right here, and then that's bonded to our two carbons like this. And then we see if we can draw the rest of the ring. And so in the back here, here is the rest of my cyclohexane ring like that. And we'll go ahead and put in our lone pairs of electrons. So this is the same exact drawing above here. Now I have my H3O plus in here like this, so my hydronium ion is present with a lone pair of electrons, giving us a plus one formal charge like that. So the oxygen on our epoxide is going to act as a base. It's going to take a proton. So this lone pair of electrons is going to take this proton right here, which would kick these electrons in here off onto my oxygen. So let's draw the result of that acid-base reaction. So I'm going to make a protonated epoxide. and it's connected to those carbons down here. So I'll go ahead and draw the rest of my ring in the back here like that. And then one lone pair of electrons didn't do anything, so it's still there. One lone pair of electrons is the one that formed the bond on that proton, so this is my structure now. And this would give this oxygen a plus one formal charge, so it's positively charged now. So this is the same structure that we saw in the earlier videos, like with our cyclic halonium ion. And just like the cyclic halonium ion in those earlier videos-- check out the halohydrin video-- you're going to get a partial carbocation character with these carbons down here. So the resonance hybrid is going to give these carbons some partial positive character. So when water comes along as a nucleophile, are going to be attracted to those carbons. So opposite charges attract. These two blue carbons are partially positive. The negative electrons are attracted to the partially positive carbon, and you're going to get nucleophilic attacks. So let's say this lone pair of electrons attacks right here. Well, that would kick the electrons in this bond off onto your oxygen. So let's go ahead and draw the result of an attack on the carbon on the left. So let's get some more room here. So what would happen in that instance? Well, let's go ahead and draw our cyclohexane ring back here. So here is our cyclohexane ring. The oxygen attacked the carbon on the left. So there is the oxygen that did the nucleophilic attack, so it has two hydrogens on it. It has one lone pair of electrons now, and it formed a plus one formal charge. Our epoxide opened." + }, + { + "Q": "6:58 Where does the water come from?", + "A": "The H2O is deprotentated (minus 1 hydrogen) H3O.", + "video_name": "KfTosrMs5W0", + "timestamps": [ + 418 + ], + "3min_transcript": "So I'm going to put my oxygen right here, and then that's bonded to our two carbons like this. And then we see if we can draw the rest of the ring. And so in the back here, here is the rest of my cyclohexane ring like that. And we'll go ahead and put in our lone pairs of electrons. So this is the same exact drawing above here. Now I have my H3O plus in here like this, so my hydronium ion is present with a lone pair of electrons, giving us a plus one formal charge like that. So the oxygen on our epoxide is going to act as a base. It's going to take a proton. So this lone pair of electrons is going to take this proton right here, which would kick these electrons in here off onto my oxygen. So let's draw the result of that acid-base reaction. So I'm going to make a protonated epoxide. and it's connected to those carbons down here. So I'll go ahead and draw the rest of my ring in the back here like that. And then one lone pair of electrons didn't do anything, so it's still there. One lone pair of electrons is the one that formed the bond on that proton, so this is my structure now. And this would give this oxygen a plus one formal charge, so it's positively charged now. So this is the same structure that we saw in the earlier videos, like with our cyclic halonium ion. And just like the cyclic halonium ion in those earlier videos-- check out the halohydrin video-- you're going to get a partial carbocation character with these carbons down here. So the resonance hybrid is going to give these carbons some partial positive character. So when water comes along as a nucleophile, are going to be attracted to those carbons. So opposite charges attract. These two blue carbons are partially positive. The negative electrons are attracted to the partially positive carbon, and you're going to get nucleophilic attacks. So let's say this lone pair of electrons attacks right here. Well, that would kick the electrons in this bond off onto your oxygen. So let's go ahead and draw the result of an attack on the carbon on the left. So let's get some more room here. So what would happen in that instance? Well, let's go ahead and draw our cyclohexane ring back here. So here is our cyclohexane ring. The oxygen attacked the carbon on the left. So there is the oxygen that did the nucleophilic attack, so it has two hydrogens on it. It has one lone pair of electrons now, and it formed a plus one formal charge. Our epoxide opened." + }, + { + "Q": "so at 9:44 the picture has two purple dun bell shaped p orbtials are they individual or is it just one long p orbital for C2H4", + "A": "At 2:30 he shows the shape of a p orbital. Every p orbital is made out of two halfs. :) In ethin every C atom has ONE p orbital.", + "video_name": "ROzkyTgscGg", + "timestamps": [ + 584 + ], + "3min_transcript": "between the two carbons in the ethane molecule. Remember for ethane, the distance was approximately 1.54 angstroms. A double bond is shorter than a single bond. One way to think about that is the increased S character. This increased S character means electron density is closer to the nucleus and that's going to make this lobe a little bit shorter than before and that's going to decrease the distance between these two carbon atoms here. 1.34 angstroms. Let's look at the dot structure again and see how we can analyze this using the concept of steric number. Let me go ahead and redraw the dot structure. We have our carbon carbon double bond here and our hydrogens like that. If you're approaching this situation using steric number remember to find the hybridization. Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. If my goal was to find the steric number for this carbon. I count up my number of sigma bonds. That's one, two and then I know when I double bond one of those is sigma and one of those is pi. One of those is a sigma bond. A total of three sigma bonds. I have zero lone pairs of electrons around that carbon. Three plus zero, gives me a steric number of three. I need three hybrid orbitals and we've just seen in this video that three SP2 hybrid orbitals form if we're dealing with SP2 hybridization. If we get a steric number of three, you're gonna think about SP2 hybridization. One S orbital and two P orbitals hybridizing. That carbon is SP2 hybridized and of course, this one is too. Both of them are SP2 hybridized. Let's do another example. Let's do boron trifluoride. If you wanna draw the dot structure of BF3, you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine. I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here. What is the hybridization stage of this boron? Let's use the concept of steric number. Once again, let's use steric number. Find the hybridization of this boron. Steric number is equal to number of sigma bonds. That's one, two, three. Three sigma bonds plus lone pairs of electrons. That's zero. Steric number of three tells us this boron is SP2 hybridized. This boron is gonna have three SP2 hybrid orbitals and one P orbital. One unhybridized P orbital. Let's go ahead and draw that." + }, + { + "Q": "@10:55 there are 9 lone pairs right? then y it is taken 0", + "A": "those pairs are of Flourine (F) not of (B) so it doens t count", + "video_name": "ROzkyTgscGg", + "timestamps": [ + 655 + ], + "3min_transcript": "Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. If my goal was to find the steric number for this carbon. I count up my number of sigma bonds. That's one, two and then I know when I double bond one of those is sigma and one of those is pi. One of those is a sigma bond. A total of three sigma bonds. I have zero lone pairs of electrons around that carbon. Three plus zero, gives me a steric number of three. I need three hybrid orbitals and we've just seen in this video that three SP2 hybrid orbitals form if we're dealing with SP2 hybridization. If we get a steric number of three, you're gonna think about SP2 hybridization. One S orbital and two P orbitals hybridizing. That carbon is SP2 hybridized and of course, this one is too. Both of them are SP2 hybridized. Let's do another example. Let's do boron trifluoride. If you wanna draw the dot structure of BF3, you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine. I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here. What is the hybridization stage of this boron? Let's use the concept of steric number. Once again, let's use steric number. Find the hybridization of this boron. Steric number is equal to number of sigma bonds. That's one, two, three. Three sigma bonds plus lone pairs of electrons. That's zero. Steric number of three tells us this boron is SP2 hybridized. This boron is gonna have three SP2 hybrid orbitals and one P orbital. One unhybridized P orbital. Let's go ahead and draw that. and also it's going to have an unhybrized P orbital. Now, remember when you are dealing with Boron, it has one last valance electron and carbon. Carbon have four valance electrons. Boron has only three. When you're thinking about the SP2 hybrid orbitals that you create. SP2 hybrid orbital, SP2, SP2 and then one unhybridized P orbital right here. Boron only has three valance electrons. Let's go ahead and put in those valance electrons. One, two and three. It doesn't have any electrons in its unhybridized P orbital. Over here when we look at the picture, this has an empty orbital and so boron can accept a pair of electrons. We're thinking about its chemical behavior, one of the things that BF3 can do, the Boron can accept an electron pair and function as a lewis acid. That's one way in thinking about how hybridizational" + }, + { + "Q": "How can he tell its going to be a sigma bond at 10:40?", + "A": "Because a single covalent bond is a sigma bond.", + "video_name": "ROzkyTgscGg", + "timestamps": [ + 640 + ], + "3min_transcript": "Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. If my goal was to find the steric number for this carbon. I count up my number of sigma bonds. That's one, two and then I know when I double bond one of those is sigma and one of those is pi. One of those is a sigma bond. A total of three sigma bonds. I have zero lone pairs of electrons around that carbon. Three plus zero, gives me a steric number of three. I need three hybrid orbitals and we've just seen in this video that three SP2 hybrid orbitals form if we're dealing with SP2 hybridization. If we get a steric number of three, you're gonna think about SP2 hybridization. One S orbital and two P orbitals hybridizing. That carbon is SP2 hybridized and of course, this one is too. Both of them are SP2 hybridized. Let's do another example. Let's do boron trifluoride. If you wanna draw the dot structure of BF3, you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine. I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here. What is the hybridization stage of this boron? Let's use the concept of steric number. Once again, let's use steric number. Find the hybridization of this boron. Steric number is equal to number of sigma bonds. That's one, two, three. Three sigma bonds plus lone pairs of electrons. That's zero. Steric number of three tells us this boron is SP2 hybridized. This boron is gonna have three SP2 hybrid orbitals and one P orbital. One unhybridized P orbital. Let's go ahead and draw that. and also it's going to have an unhybrized P orbital. Now, remember when you are dealing with Boron, it has one last valance electron and carbon. Carbon have four valance electrons. Boron has only three. When you're thinking about the SP2 hybrid orbitals that you create. SP2 hybrid orbital, SP2, SP2 and then one unhybridized P orbital right here. Boron only has three valance electrons. Let's go ahead and put in those valance electrons. One, two and three. It doesn't have any electrons in its unhybridized P orbital. Over here when we look at the picture, this has an empty orbital and so boron can accept a pair of electrons. We're thinking about its chemical behavior, one of the things that BF3 can do, the Boron can accept an electron pair and function as a lewis acid. That's one way in thinking about how hybridizational" + }, + { + "Q": "Like it said at 11:40, in overall, is it you always fill valence electrons into sp2 orbitals first then free p orbital right?", + "A": "Yes, however this is a very unusual situation because boron does not usually follow the octet rule. Ordinarily, you would have the sp\u00c2\u00b2 and the p all filled (counting the electrons shared with the other atoms).", + "video_name": "ROzkyTgscGg", + "timestamps": [ + 700 + ], + "3min_transcript": "If you wanna draw the dot structure of BF3, you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine. I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here. What is the hybridization stage of this boron? Let's use the concept of steric number. Once again, let's use steric number. Find the hybridization of this boron. Steric number is equal to number of sigma bonds. That's one, two, three. Three sigma bonds plus lone pairs of electrons. That's zero. Steric number of three tells us this boron is SP2 hybridized. This boron is gonna have three SP2 hybrid orbitals and one P orbital. One unhybridized P orbital. Let's go ahead and draw that. and also it's going to have an unhybrized P orbital. Now, remember when you are dealing with Boron, it has one last valance electron and carbon. Carbon have four valance electrons. Boron has only three. When you're thinking about the SP2 hybrid orbitals that you create. SP2 hybrid orbital, SP2, SP2 and then one unhybridized P orbital right here. Boron only has three valance electrons. Let's go ahead and put in those valance electrons. One, two and three. It doesn't have any electrons in its unhybridized P orbital. Over here when we look at the picture, this has an empty orbital and so boron can accept a pair of electrons. We're thinking about its chemical behavior, one of the things that BF3 can do, the Boron can accept an electron pair and function as a lewis acid. That's one way in thinking about how hybridizational and how something might react. This boron turns out to be SP2 hybridized. This boron here is SP2 hybridized and so we can also talk about the geometry of the molecule. It's planar. Around this boron, it's planar and so therefore, your bond angles are 120 degrees. If you have boron right here and you're thinking about a circle. A circle is 360 degrees. If you divide a 360 by 3, you get 120 degrees for all of these bond angles. In the next video, we'll look at SP hybridization." + }, + { + "Q": "at 10:45 why the number of lone pairs are taken as zero?", + "A": "Because we are only counting the electrons on boron and it has no lone pairs.", + "video_name": "ROzkyTgscGg", + "timestamps": [ + 645 + ], + "3min_transcript": "Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. If my goal was to find the steric number for this carbon. I count up my number of sigma bonds. That's one, two and then I know when I double bond one of those is sigma and one of those is pi. One of those is a sigma bond. A total of three sigma bonds. I have zero lone pairs of electrons around that carbon. Three plus zero, gives me a steric number of three. I need three hybrid orbitals and we've just seen in this video that three SP2 hybrid orbitals form if we're dealing with SP2 hybridization. If we get a steric number of three, you're gonna think about SP2 hybridization. One S orbital and two P orbitals hybridizing. That carbon is SP2 hybridized and of course, this one is too. Both of them are SP2 hybridized. Let's do another example. Let's do boron trifluoride. If you wanna draw the dot structure of BF3, you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine. I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here. What is the hybridization stage of this boron? Let's use the concept of steric number. Once again, let's use steric number. Find the hybridization of this boron. Steric number is equal to number of sigma bonds. That's one, two, three. Three sigma bonds plus lone pairs of electrons. That's zero. Steric number of three tells us this boron is SP2 hybridized. This boron is gonna have three SP2 hybrid orbitals and one P orbital. One unhybridized P orbital. Let's go ahead and draw that. and also it's going to have an unhybrized P orbital. Now, remember when you are dealing with Boron, it has one last valance electron and carbon. Carbon have four valance electrons. Boron has only three. When you're thinking about the SP2 hybrid orbitals that you create. SP2 hybrid orbital, SP2, SP2 and then one unhybridized P orbital right here. Boron only has three valance electrons. Let's go ahead and put in those valance electrons. One, two and three. It doesn't have any electrons in its unhybridized P orbital. Over here when we look at the picture, this has an empty orbital and so boron can accept a pair of electrons. We're thinking about its chemical behavior, one of the things that BF3 can do, the Boron can accept an electron pair and function as a lewis acid. That's one way in thinking about how hybridizational" + }, + { + "Q": "When you squared that negative number in 2:37, does it become non negative?", + "A": "Yes. The square of a negative number is positive. The cube, however, will be negative.", + "video_name": "gluN2wxqES0", + "timestamps": [ + 157 + ], + "3min_transcript": "" + }, + { + "Q": "At 5:20, it is assumed that tension is not equal to the force of gravity on the 3kg box because the 3kg box is accelerating. How can you assume that the 3kg box is accelerating at that point? How can you know that the box isn't standing still?", + "A": "The surface that the 5kg box is sitting on is assumed to be frictionless---meaning any force that is applied to it will move it.", + "video_name": "QKXeZFwFPS0", + "timestamps": [ + 320 + ], + "3min_transcript": "You always draw a force diagram. So what forces do I have on the 5 kg mass? I'm gonna have a force of gravity, I'll draw that straight down. FG, and there's gonna be an equal force, normal force upward. So this normal force up should be equal to the force of gravity and magnitude because this box is probably not gonna be accelerating vertically. There's no real reason why it should be if this table is rigid. And there's one more force on this box. There's a force to the right. That's gonna be the force of tension. And if there's no friction on this table, then I have no leftward forces here. I'm ignoring air resistance since we usually ignore air resistance. So that's is, the only horizontal force I've got is T, tension. And I divide by the mass of the 5 kg box, which is 5 kg. But we got a problem. Look it, we don't know the acceleration of the 5 kg mass, and we don't know the tension. I can't solve this. Normally what you do in this case, is you go to the vertical direction, the other direction in other words. That's just gonna tell me that the normal force is gonna be equal to the force of gravity. And we kind of already knew that. So that doesn't help. So what do we do? Well, you might note, this is only the equation for the 5 kg mass. And so now I have to do this for the 3 kg mass. So let's come over here, let's say that the acceleration of the 3 kg mass is gonna be equal to the net force on the 3 kg mass, divided by the mass of the 3 kg mass. And again, which direction should we pick? Well this acceleration over here is gonna be vertical. So let's solve this for the vertical direction. I'm gonna add one more sub-script, Y, to remind myself. And you should do this too so you remember which direction you're picking. So what forces do I plug in here? You figure that out with a force diagram. I'm gonna have a force of gravity on this 3 kg mass, and then I'm gonna have the same size of friction, or sorry, the same as tension, that I had over here. So the tension on this side of the rope, it's gonna be the same as the tension on this side. or friction. So assuming that its mass is negligible, there's basically no friction, then I'm gonna have a tension. That tension is gonna be the same size. So I'll draw that coming upward. But it's not gonna be as big as the force of gravity is on this 3 kg mass. I've got the force of gravity here. This tension is gonna be smaller, and the reason is, this 3 kg mass is accelerating downwards. So these forces can't be balanced. The upward force of tension has gotta be smaller than the force of gravity on this 3 kg mass. But this tension here should be the same as this tension here. So I'll plug those in. So let's plug this in. A of the 3 kg mass, in the Y direction is gonna be equal to, I've got two vertical forces. I've got tension up, so I'll make that positive, 'cause we usually treat up as positive. I've got gravity down, and so I'm gonna have negative, 'cause it's downward force of 3 kg times the acceleration" + }, + { + "Q": "At 2:53, isn't it 5,5 diethyldecane if you count from the closest sides to each ethyl group?", + "A": "i think you need to review your counting mechanics, because I do not see two adjacent ethyl groups on any 5,5 carbons", + "video_name": "q_Q9C1Ooofc", + "timestamps": [ + 173 + ], + "3min_transcript": "I think we're ready now to tackle some more, or even more complicated examples. So let's draw something crazy here. Let me draw a chain. Let me draw it like that. And so like we've done in all of the examples, you'll want to find the longest chain. We can count from here one, two, three, four, five, six, seven, Or maybe it's one, two, three, four, five, six, seven. Or maybe it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. That is our longest chain. Let me just make that in green. So our longest chain here is in green. So this backbone has 10 carbons in it. The prefix for 10 is dec. It is a alkane since it has all single bonds. So we can write decane for the backbone. And then it has a group right here. And this group consists of one, two carbons attached The prefix for two carbons is eth, so this is an ethyl group. The y-l is because it's a group attached to the main alkane chain. So we call this ethyl decane. But we have to specify where the ethyl group is attached, and we want to give it as low of a number as possible so we start counting on the side closest to it. So it's one, two, three, four, five. So this is 5-ethyldecane. Now let's complicate this a little bit more. So let me just copy and paste this. So I have pasted it there, and let me complicate this molecule Let me add another ethyl group to it. So let's say we have another ethyl group over there. Now what is this going to be? to be that thing in green. So it's still going to be a decane. But now we have two ethyl groups. One on the five carbon, one, two, three, four, five, and then one on the six carbon. You might be tempted to write 5-ethyl 6-ethyldecane, which really wouldn't be wrong, but it would just be maybe more letters than you want to write. Instead you write 5 comma 6-diethyldecane. The 5, 6, tells us the two carbons on the main backbone that the ethyl groups are attached to, and the di says that we have two ethanes, or two ethyl groups I should say, not ethane groups, two ethyl groups, one over here, and one right over here." + }, + { + "Q": "9:05 Positives in left rod attract positives in right rod?", + "A": "Like charges repel, they don t attract.", + "video_name": "ZgDIX2GOaxQ", + "timestamps": [ + 545 + ], + "3min_transcript": "Let's say you have two conducting rods. Say these are made out of metal. One of them has a net amount of negative charge on it which is going to reside on the outside edge because that's what net charge does on a conductor, but this other rod, this other metal conducting rod, does not have any net charge on it. What would happen if I took this first rod touched it to the second rod? You probably guessed, charges want to get as far away from each other as possible so these negatives realize \"Hey, if we spread out, \"some of us go on to this rod and some of us stay here, \"we can spread out even father away from each other.\" That's what they would do. If these rods were the same size, you'd have equal amounts on each. If the second rod was bigger, more of them would go on to this second one because that would allow them to spread out even more. Some would stay on the smaller one. That's charged by just touching something. That's easy. You can charge something also, you can get clever. You can do something called charge, you can charge something by induction it's called. Charge by induction says alright, first imagine I just take this and I bring it nearby but don't touch it. Just bring it near by this other piece of metal and I don't touch it. What would happen? There is negatives in here, I haven't drawn them. There's positives in here. The negatives can move if they wanted to. Do they want to? Yeah, they want to! These negatives are coming nearby, they want to get as far away from them as possible. Even though there are already some negatives here, a net amount of negatives are going to get moved over to this side. They were located with their atom on this side, but they want to get away from this big negative charge so they can move over here, which leaves a total amount of positive charge over here. I.E. There is a deficit of electrons over here, so this side ends up positively charged. You might think, \"Okay, well that's weird. \"Does anything else happen?\" Yeah because now these positives are closer to the negatives and these positives in this charge rod are attracting these positives. These negatives in this conducting rod are attracting these positive charges because like charges repel and opposites attract but they are also repelling. These negatives in this rod are repelling these negatives. Do those forces cancel? They actually don't because the closer you are to the charge the bigger the force. This would cause this rod to get attracted to the other rod. That's kind of cool. If you took a charged rod, brought it to an empty soda can, let that can sit on the table in this orientation so it could roll, if you bring the rod close the can will start moving towards the rod. It's kind of cool, you should try it if you can. But, that's not charge by induction. Charge by induction is something more. It says alright, take this piece of metal and conduct it to ground. What's ground?" + }, + { + "Q": "At 12:33, how do we know that whether it's the ceiling or rather the balloon that reorients or polarise its atoms in order to create an attractive force?", + "A": "We put the charge on the balloon., not on the ceiling, right? Rubber is an insulator, so it is difficult for charge to move around.", + "video_name": "ZgDIX2GOaxQ", + "timestamps": [ + 753 + ], + "3min_transcript": "but some of the electrons will leave which means that this rod, which used to be uncharged now has a net amount of positive charge in it. I've charged this rod without even touching it because I let the negative electrons leave. If I'm clever, what I can do is I can just cut this wire before I take away the thing that induced the charge. If I remove this now and move it far away, what these negatives would have done is they would have said \"Shoot, okay, \"I am glad that that's over. \"Now I can rejoin. \"I'm attracted to this positive again. \"I'm going to rejoin my positives.\" and this thing will become uncharged again but now they can't get back. They're stuck. There's no way for these to get back because you've cut the cord here and you've permanently charged this piece of metal without even touching it. It's called charge by induction. It's a quick way we charge something up. Everyone's tried this. You take a balloon. What happens? How do you charge it up? You rub it against your hair. It steals electrons from your hair and the balloon becomes negatively charged. What do you do with it? You know what you do with it. You take this thing and you put it near a wall or a ceiling and if you're lucky, it sticks there, which is cool! How does it work? Well, remember, this is an insulating material rubber. The ceiling is an insulating material. Electrons aren't getting transferred but even in an insulating material, the atom can reorient or polarize by shifting. The negatives in that atom can shift to one side and the other side becomes a little more positive and what that does, it causes a net force between the ceiling and the balloon because these positives are a little closer. These positives are attracting negatives and the negatives are attracting the positives with a little bit greater a force than these negatives are repelling the other negatives in the ceiling. is also attracting the balloon and the balloon is attracting the ceiling with greater force than the negatives are repelling the balloon, the balloon can stick because of the insulating material's ability to polarize and cause and electric attraction. This is what I said earlier. Even if it's an insulator, sometimes it can interact with something electric because the atom can shift and polarize." + }, + { + "Q": "At 11:24, if the left rod was positively charged, would the positive charges on the right rod leave to the ground instead?", + "A": "The positive charges were the nuclei, and nuclei can t really move in a solid.", + "video_name": "ZgDIX2GOaxQ", + "timestamps": [ + 684 + ], + "3min_transcript": "and these positives in this charge rod are attracting these positives. These negatives in this conducting rod are attracting these positive charges because like charges repel and opposites attract but they are also repelling. These negatives in this rod are repelling these negatives. Do those forces cancel? They actually don't because the closer you are to the charge the bigger the force. This would cause this rod to get attracted to the other rod. That's kind of cool. If you took a charged rod, brought it to an empty soda can, let that can sit on the table in this orientation so it could roll, if you bring the rod close the can will start moving towards the rod. It's kind of cool, you should try it if you can. But, that's not charge by induction. Charge by induction is something more. It says alright, take this piece of metal and conduct it to ground. What's ground? If you took a big metal pipe and stuck it in the ground that would count, or any other huge supply of electron, a place where you can gain, steal, basically take infinitely many electrons or deposit infinitely many electrons and this ground would not care. So the frame of your car, the actual metal, is a good ground because it can supply a ton of electrons or take them. Or a metal pipe in the earth. Some place you can deposit electrons or take them and that thing won't really notice or care. Now what would happen? If I bring this negative rod close to this rod that was originally had no net charge? Now instead of going to the other side of this, they say \"Hey, I can just leave. \"Let me get the heck out of here.\" These negatives can leave. A whole bunch of negatives can start leaving and what happens when that happens is that your rod is no longer uncharged. It has a net amount of charge now. They won't all leave. You're not going to get left with no electrons in here. but some of the electrons will leave which means that this rod, which used to be uncharged now has a net amount of positive charge in it. I've charged this rod without even touching it because I let the negative electrons leave. If I'm clever, what I can do is I can just cut this wire before I take away the thing that induced the charge. If I remove this now and move it far away, what these negatives would have done is they would have said \"Shoot, okay, \"I am glad that that's over. \"Now I can rejoin. \"I'm attracted to this positive again. \"I'm going to rejoin my positives.\" and this thing will become uncharged again but now they can't get back. They're stuck. There's no way for these to get back because you've cut the cord here and you've permanently charged this piece of metal without even touching it. It's called charge by induction. It's a quick way we charge something up." + }, + { + "Q": "At 9:10 why did he assumes protons are attracted by the electron? isn't more easy for the electrons to move to protons?", + "A": "The point is, they are attracted. It does not matter who moves.", + "video_name": "ZgDIX2GOaxQ", + "timestamps": [ + 550 + ], + "3min_transcript": "Let's say you have two conducting rods. Say these are made out of metal. One of them has a net amount of negative charge on it which is going to reside on the outside edge because that's what net charge does on a conductor, but this other rod, this other metal conducting rod, does not have any net charge on it. What would happen if I took this first rod touched it to the second rod? You probably guessed, charges want to get as far away from each other as possible so these negatives realize \"Hey, if we spread out, \"some of us go on to this rod and some of us stay here, \"we can spread out even father away from each other.\" That's what they would do. If these rods were the same size, you'd have equal amounts on each. If the second rod was bigger, more of them would go on to this second one because that would allow them to spread out even more. Some would stay on the smaller one. That's charged by just touching something. That's easy. You can charge something also, you can get clever. You can do something called charge, you can charge something by induction it's called. Charge by induction says alright, first imagine I just take this and I bring it nearby but don't touch it. Just bring it near by this other piece of metal and I don't touch it. What would happen? There is negatives in here, I haven't drawn them. There's positives in here. The negatives can move if they wanted to. Do they want to? Yeah, they want to! These negatives are coming nearby, they want to get as far away from them as possible. Even though there are already some negatives here, a net amount of negatives are going to get moved over to this side. They were located with their atom on this side, but they want to get away from this big negative charge so they can move over here, which leaves a total amount of positive charge over here. I.E. There is a deficit of electrons over here, so this side ends up positively charged. You might think, \"Okay, well that's weird. \"Does anything else happen?\" Yeah because now these positives are closer to the negatives and these positives in this charge rod are attracting these positives. These negatives in this conducting rod are attracting these positive charges because like charges repel and opposites attract but they are also repelling. These negatives in this rod are repelling these negatives. Do those forces cancel? They actually don't because the closer you are to the charge the bigger the force. This would cause this rod to get attracted to the other rod. That's kind of cool. If you took a charged rod, brought it to an empty soda can, let that can sit on the table in this orientation so it could roll, if you bring the rod close the can will start moving towards the rod. It's kind of cool, you should try it if you can. But, that's not charge by induction. Charge by induction is something more. It says alright, take this piece of metal and conduct it to ground. What's ground?" + }, + { + "Q": "At 9:08, he says \"these positives in this charge rod are attracting these positives\" when explaining charge by induction. But since like charges repel, shouldn't they repel instead of attracting?", + "A": "If he said that, he misspoke.", + "video_name": "ZgDIX2GOaxQ", + "timestamps": [ + 548 + ], + "3min_transcript": "Let's say you have two conducting rods. Say these are made out of metal. One of them has a net amount of negative charge on it which is going to reside on the outside edge because that's what net charge does on a conductor, but this other rod, this other metal conducting rod, does not have any net charge on it. What would happen if I took this first rod touched it to the second rod? You probably guessed, charges want to get as far away from each other as possible so these negatives realize \"Hey, if we spread out, \"some of us go on to this rod and some of us stay here, \"we can spread out even father away from each other.\" That's what they would do. If these rods were the same size, you'd have equal amounts on each. If the second rod was bigger, more of them would go on to this second one because that would allow them to spread out even more. Some would stay on the smaller one. That's charged by just touching something. That's easy. You can charge something also, you can get clever. You can do something called charge, you can charge something by induction it's called. Charge by induction says alright, first imagine I just take this and I bring it nearby but don't touch it. Just bring it near by this other piece of metal and I don't touch it. What would happen? There is negatives in here, I haven't drawn them. There's positives in here. The negatives can move if they wanted to. Do they want to? Yeah, they want to! These negatives are coming nearby, they want to get as far away from them as possible. Even though there are already some negatives here, a net amount of negatives are going to get moved over to this side. They were located with their atom on this side, but they want to get away from this big negative charge so they can move over here, which leaves a total amount of positive charge over here. I.E. There is a deficit of electrons over here, so this side ends up positively charged. You might think, \"Okay, well that's weird. \"Does anything else happen?\" Yeah because now these positives are closer to the negatives and these positives in this charge rod are attracting these positives. These negatives in this conducting rod are attracting these positive charges because like charges repel and opposites attract but they are also repelling. These negatives in this rod are repelling these negatives. Do those forces cancel? They actually don't because the closer you are to the charge the bigger the force. This would cause this rod to get attracted to the other rod. That's kind of cool. If you took a charged rod, brought it to an empty soda can, let that can sit on the table in this orientation so it could roll, if you bring the rod close the can will start moving towards the rod. It's kind of cool, you should try it if you can. But, that's not charge by induction. Charge by induction is something more. It says alright, take this piece of metal and conduct it to ground. What's ground?" + }, + { + "Q": "Is displacement and distance traveled the same thing? because I thought distance between two points is displacement. In here, I noticed Mr.Khan using 'distance traveled' and 'displacement' interchangeably. Like on 14:07 he has the equation he used to find 's' set up equaled to 'D'", + "A": "They are not quite the same. Displacement is a vector whereas distance is a scalar. Displacement and distance are often equivalent, but there are many cases where they aren t. For example, after completing one lap around a circular track, the distance would be the circumference of the track whereas the displacement would be 0 (because you ended up back where you started).", + "video_name": "MAS6mBRZZXA", + "timestamps": [ + 847 + ], + "3min_transcript": "so this is interesting, the distance we travel is equal to one half of the initial velocity plus the final velocity so this is really if you just took this quantity right over here it's just the arithmetic, I have trouble saying that word it's the arithmetic mean of these two numbers, so I'm gonna define, this is something new, I'm gonna call this the average velocity we have to be very careful with this this right here is the average velocity but the only reason why I can just take the starting velocity and ending velocity and adding them together and divide them by two since you took an average of two thing it's some place over here and I take that as average velocity it's because my acceleration is constant but it's not always the assumption but if you do have a constant acceleration like this you can assume that the average velocity is gonna be the average of the initial velocity and the final velocity if this is a curve and the acceleration is changing you could not do that but what is useful about this is if you wanna figure out the distance that was travelled, you just need to know the initial velocity and the final velocity, average their two and multiply the times it goes by so in this situation our final velocity is 13m/s our initial velocity is 5m/s so you have 13 plus 5 is equal to 18 you divided that by 2, you average velocity is 9m/s if you take the average of 13 and 5 and 9m/s times 4s gives you 36m so hopefully it doesn't confuse you I just wannt show you but you shouldn't memorize they can all be deduced" + }, + { + "Q": "at 7:30 , why did sal subtracted initial velocity from the final velocity to find height? instead, he could multiply constant acceleration with delta time. ex} 4s * 2m/s^2 = 8 m/s", + "A": "in this case only the Vi-Vf x t = a x t^2 in other cases it may not be the same.", + "video_name": "MAS6mBRZZXA", + "timestamps": [ + 450 + ], + "3min_transcript": "after 3s we will be 11m/s, and after 4s we will be at 13 m/s so you multiply how much time pass times acceleration this is how much faster we are gonna be going, we are already going 5m/s 5 plus how much faster? 13 m/s so this right up here is 13m/s So I will take a little pause here hopefully intuitive and the whole play of that is to show you this formula you will see in many physics book is not something that randomly pop out of there it just make complete common sense Now the next thing I wanna talk about is what is the total distance that we would have travel? and we know from the last video that distance is just the area under this curve right over here, so it's just the area under this curve and we can use a little symbol of geometry to break it down into two different areas, it's very easy to calculate their areas two simple shapes, you can break it down to two, blue part is the rectangle right over here, easy to figure out the area of a rectangle and we can break it down to this purple part, this triangle right here easy to figure out the area of a triangle and that will be the total distance we travel even this will hopefully make some intuition because this blue area is how far we would have travel if we are not accelerated, we just want 5m/s for 4s so you goes 5m/s 1s 2s 3s 4s so you are going from 0 to 4 you change in time is 4s so if you go 5m/s for 4s you are going to go 20 m this right here is 20m this purple or magentic area tells you how furthur than this are you going because you are accelerating because kept going faster and faster and faster it's pretty easy to calculate this area the base here is still 5(4) because that's 5(4) second that's gone by what's the height here? The height here is my final velocity minus my initial velocity minus my initial velocity or it's the change in velocity due to the accleration 13 minus 5 is 8 or this 8 right over here it is 8m/s so this height right over here is 8m/s the base over here is 4s that's the time that past what's this area of the triangle? the area of this triangle is one half times the base which is 4s" + }, + { + "Q": "At 0:30,u assigned right to positive and left to negative.Is the right direction always the positive one and the left always negative.Or is it something we can choose?", + "A": "You can choose it however you want.", + "video_name": "MAS6mBRZZXA", + "timestamps": [ + 30 + ], + "3min_transcript": "The goal of this video is to explore some of the concept of formula you might see in introductional physics class but more importantly to see they are really just common sense ideas So let's just start with a simple example Let's say that and for the sake of this video keep things that magnitudes and velocities that's the direction of velocity etc. let's just assume that if I have a positive number that it means for example postive velocity that it means I'm going to the right let's say I have a negative number we won't see in this video let's assume we are going to the left In that way I can just write a number down only operating in one dimension you know that by specifying the magnitude and the direction if I say velocity is 5m/s that means 5m/s to the right if I say negative 5m/s that means 5m/s to the left let's say just for simplicitiy, say that we start with initial velocity we start with an initial velocity of 5m/s once again I specify the magnitude and the direction because of this let's say we have a constant acceleration we have a constant acceleration 2m/s^2 or 2m per second square and once again since this is positive it is to the right and let's say that we do this for a duration so my change in time, let's say we do this for a duration of 4 I will just use s, second and s different places so s for this video is seconds So I want to do is to think about how far do we travel? and there is two things how fast are we going? after 4 seconds and how far have we travel over the course of those 4 seconds? so let's draw ourselves a little diagram here So this is my velocity axis, and this over here is time axis So that is my time axis, time this is velocity This is my velocity right over there and I'm starting off with 5m/s, so this is 5m/s right over here So vi is equal to 5m/s And every second goes by it goes 2m/s faster that's 2m/s*s every second that goes by So after 1 second when it goes 2m/s faster it will be at 7 another way to think about it is the slope of this velocity line is my constant accleration, my constant slope here so it might look something like that So what has happend after 4s? So 1 2 3 4 this is my delta t So my final velocity is going to be right over there" + }, + { + "Q": "Could you simplify the equation at 12:55 e4ven further and say 1/2t(vi+vf)?", + "A": "Not unless the amount of time between Vi and Vf is 1 unit of time. The average of 2 velocities is still a velocity not a distance.", + "video_name": "MAS6mBRZZXA", + "timestamps": [ + 775 + ], + "3min_transcript": "minus one half of the original something so what is vi minus one half vi? so anything minus its half is just a positive half left so these two terms, this term and this term will simplify to one half vi one half initial velocity plus one half times the final velocity plus one half times the final velocity and all of that is being multiplied with the change in time the time that has gone by and this tells us the distance, the distance that we travel another way to think about it, let's factor out this one half you get distance that is equal to change in time times factoring out the one half so this is interesting, the distance we travel is equal to one half of the initial velocity plus the final velocity so this is really if you just took this quantity right over here it's just the arithmetic, I have trouble saying that word it's the arithmetic mean of these two numbers, so I'm gonna define, this is something new, I'm gonna call this the average velocity we have to be very careful with this this right here is the average velocity but the only reason why I can just take the starting velocity and ending velocity and adding them together and divide them by two since you took an average of two thing it's some place over here and I take that as average velocity it's because my acceleration is constant but it's not always the assumption but if you do have a constant acceleration like this you can assume that the average velocity is gonna be the average of the initial velocity and the final velocity if this is a curve and the acceleration is changing you could not do that but what is useful about this is if you wanna figure out the distance that was travelled, you just need to know the initial velocity and the final velocity, average their two and multiply the times it goes by so in this situation our final velocity is 13m/s our initial velocity is 5m/s so you have 13 plus 5 is equal to 18 you divided that by 2, you average velocity is 9m/s if you take the average of 13 and 5 and 9m/s times 4s gives you 36m so hopefully it doesn't confuse you I just wannt show you" + }, + { + "Q": "At 1:23, he says that ONE photon hits ONE electron. How do you know if it is just one photon with an energy of say, x or multiple photons with a cumulative energy of x which dislodges the electron ? Is it just an assumption Einstein took or is there a scientific reason behind it ?\nAlso we are talking about Classical Mechanics isn't the photons motion supposed to be studied by Quantum Mechanics ?", + "A": "they could have possibly altered the number of photons in experimental apparatus and observed the same result.", + "video_name": "vuGpUFjLaYE", + "timestamps": [ + 83 + ], + "3min_transcript": "- Sometimes light seems to act as a wave, and sometimes light seems to act as a particle. And, an example of this, would be the Photoelectric effect, as described by Einstein. So let's say you had a piece of metal, and we know the metal has electrons. I'm gonna go ahead and draw one electron in here, and this electron is bound to the metal because it's attracted to the positive charges in the nucleus. If you shine a light on the metal, so the right kind of light with the right kind of frequency, you can actually knock some of those electrons loose, which causes a current of electrons to flow. So this is kind of like a collision between two particles, if we think about light as being a particle. So I'm gonna draw in a particle of light which we call a photon, so this is massless, and the photon is going to hit this electron, and if the photon has enough energy, it can free the electron, right? So we can knock it loose, and so let me go ahead and show that. So here, we're showing the electron being knocked loose let's just say, this direction, with some velocity, v, and if the electron has mass, m, we know that there's a kinetic energy. The kinetic energy of the electron would be equal to one half mv squared. This freed electron is usually referred to now as a photoelectron. So one photon creates one photoelectron. So one particle hits another particle. And, if you think about this in terms of classical physics, you could think about energy being conserved. So the energy of the photon, the energy that went in, so let me go ahead and write this here, so the energy of the photon, the energy that went in, what happened to that energy? Some of that energy was needed to free the electron. So the electron was bound, and some of the energy freed the electron. I'm gonna call that E naught, the energy that freed the electron, and then the rest of that energy must have gone into the kinetic energy of the electron, kinetic energy of the photoelectron that was produced. So, kinetic energy of the photoelectron. So let's say you wanted to solve for the kinetic energy of that photoelectron. So that would be very simple, it would just be kinetic energy would be equal to the energy of the photon, energy of the photon, minus the energy that was necessary to free the electron from the metallic surface. And this E naught, here I'm calling it E naught, you might see it written differently, a different symbol, but this is the work function. Let me go ahead and write work function here, and the work function is different for every kind of metal. So, it's the minimum amount of energy that's necessary to free the electron, and so obviously that's going to be different depending on what metal you're talking about. All right, let's do a problem. Now that we understand the general idea of the Photoelectric effect, let's look at what this problem asks us." + }, + { + "Q": "At 12:23 Why is it said that the structure on the left is the right one? ALSO why is it that we have only two resonance structures for the molecule at 12:15?", + "A": "These are isomers, not resonance structures. The five orbitals have a trigonal bipyramidal geometry. There are only two places where the lone pair can go \u00e2\u0080\u0094 the axial ot the equatorial location. The see-saw geometry is more stable because the lone pair in the axial location has less total repulsion from the other electrons.", + "video_name": "0na0xtIHkXA", + "timestamps": [ + 743, + 735 + ], + "3min_transcript": "" + }, + { + "Q": "How do you find the conjugate acid? They were mentioned around 7:55 but it was not explained how he knew those were the conjugate bases.", + "A": "A conjugate acid/base pair are chemicals that are different by a proton or electron pair. For instance, the strong acid HCl has a conjugate base of Cl-. Remember that acids donate protons (H+) and that bases accept protons. So each conjugate pair essentially are different from each other by one proton. There s a lot of info in the acid base section too!", + "video_name": "7BgiKyvviyU", + "timestamps": [ + 475 + ], + "3min_transcript": "this negative-one formal charge is not localized to this oxygen; it's de-localized. And so, because we can spread out some of that negative charge, that increases the stability of the anion here, so this is relatively stable, so increased stability, due to de-localization. So, the fact that we can draw an extra resonance structure, means that the anion has been stabilized. And so, this is called, \"pushing electrons,\" so we're moving electrons around, and it's extremely important to feel comfortable with moving electrons around, and being able to follow them. So, the only way to get good at this is to do a lot of practice problems, so please do that; do lots of practice problems in your textbook. If we compare that to the ethoxide anion, so over here, if we try to do the same thing, if we try to take a lone pair of electrons on this oxygen, and move it into here, already has four bonds; so it's already bonded to two hydrogens, and then we have this bond, and this bond. And so, moving those electrons in, trying to de-localize those electrons, would give us five bonds to carbon, and so we can't do that; we can't draw a resonance structure for the ethoxide anion. So those electrons are localized to this oxygen, and so this oxygen has a full, negative-one formal charge, and since we can't spread out that negative charge, or it's going to destabilize this anion. So this is not as stable, so decreased stability, compared to the anion on the left, because we can't draw a resonance structure. If we think about the conjugate acids to these bases, so the conjugate acid to the acetate anion would be, of course, acetic acid. So we go ahead, and draw in acetic acid, like that. The conjugate acid to the ethoxide anion So we go ahead, and draw in ethanol. And we think about which one of those is more acidic. We know that acetic acid is more acidic; it's more likely to donate a proton, because the conjugate base is more stable, because, you could think about resonance, or de-localization of electrons. If you're looking at ethanol, ethanol's not as likely to donate its proton, because the conjugate base, the ethoxide anion is not as stable, because you can't draw any resonance structures for it. The negative charge is not able to be de-localized; it's localized to that oxygen. So this is just one application of thinking about resonance structures, and, again, do lots of practice. In the next video, we'll talk about different patterns that you can look for, and we talked about one in this video: We took a lone pair of electrons, so right here in green, and we noticed this lone pair of electrons was next to a pi bond, and so we were able to draw another resonance structure for it. We don't have that situation with ethoxide:" + }, + { + "Q": "At 8:36 Jay states that the ethanol molecule is not as likely to donate its proton because its conjugate base is not as stable. He also states the negative charge is localized to the oxygen. What exactly is it that prevents ethanol from donating this proton? Is it the partial negative dipole on the oxygen that causes a strong attraction to the hydrogen?", + "A": "He s comparing ethanol to acetic acid here, that s what we need to keep in mind. In acetic acid the negative charge can be spread over two oxygen atoms whereas in ethanol it s stuck on one. The ethanol isn t being prevented from donating its proton, it still happens somewhat, but when you have a resonance stabilised conjugate base like there is in acetic acid we find they are better able to act as an acid.", + "video_name": "7BgiKyvviyU", + "timestamps": [ + 516 + ], + "3min_transcript": "already has four bonds; so it's already bonded to two hydrogens, and then we have this bond, and this bond. And so, moving those electrons in, trying to de-localize those electrons, would give us five bonds to carbon, and so we can't do that; we can't draw a resonance structure for the ethoxide anion. So those electrons are localized to this oxygen, and so this oxygen has a full, negative-one formal charge, and since we can't spread out that negative charge, or it's going to destabilize this anion. So this is not as stable, so decreased stability, compared to the anion on the left, because we can't draw a resonance structure. If we think about the conjugate acids to these bases, so the conjugate acid to the acetate anion would be, of course, acetic acid. So we go ahead, and draw in acetic acid, like that. The conjugate acid to the ethoxide anion So we go ahead, and draw in ethanol. And we think about which one of those is more acidic. We know that acetic acid is more acidic; it's more likely to donate a proton, because the conjugate base is more stable, because, you could think about resonance, or de-localization of electrons. If you're looking at ethanol, ethanol's not as likely to donate its proton, because the conjugate base, the ethoxide anion is not as stable, because you can't draw any resonance structures for it. The negative charge is not able to be de-localized; it's localized to that oxygen. So this is just one application of thinking about resonance structures, and, again, do lots of practice. In the next video, we'll talk about different patterns that you can look for, and we talked about one in this video: We took a lone pair of electrons, so right here in green, and we noticed this lone pair of electrons was next to a pi bond, and so we were able to draw another resonance structure for it. We don't have that situation with ethoxide: but we don't have a pi bond next to it, And so, more in the next video on that." + }, + { + "Q": "I didnt understan the part when he says that the rate of the reaction is equal to the rate of O2 (time 06:37). How do we know that?\nThanks.", + "A": "The rate of reaction is equal to the, R = rate of formation of any component of the reaction / change in time. Here in this reaction O2 is being formed, so rate of reaction would be the rate by which O2 is formed.", + "video_name": "8wIodo1HD4Y", + "timestamps": [ + 397 + ], + "3min_transcript": "And let's say that oxygen forms at a rate of 9 x 10 to the -6 M/s. So what is the rate of formation of nitrogen dioxide? Well, if you look at the balanced equation, for every one mole of oxygen that forms four moles of nitrogen dioxide form. So we just need to multiply the rate of formation of oxygen by four, and so that gives us, 3.6 x 10 to the -5 Molar per second. So, NO2 forms at four times the rate of O2. What about dinitrogen pentoxide? So, N2O5. Look at your mole ratios. For every one mole of oxygen that forms we're losing two moles of dinitrogen pentoxide. So if we're starting with the rate because our mole ratio is one to two here, we need to multiply this by 2, and since we're losing dinitrogen pentoxide, we put a negative sign here. So this gives us - 1.8 x 10 to the -5 molar per second. So, dinitrogen pentoxide disappears at twice the rate that oxygen appears. All right, let's think about the rate of our reaction. So the rate of our reaction is equal to, well, we could just say it's equal to the appearance of oxygen, right. We could say it's equal to 9.0 x 10 to the -6 molar per second, so we could write that down here. The rate is equal to the change in the concentration of oxygen over the change in time. All right, what about if we wanted to express this in terms of the formation of nitrogen dioxide. was 3.6 x 10 to the -5. All right, so that's 3.6 x 10 to the -5. So you need to think to yourself, what do I need to multiply this number by in order to get this number? Since this number is four times the number on the left, I need to multiply by one fourth. Right, so down here, down here if we're talking about the change in the concentration of nitrogen dioxide over the change in time, to get the rate to be the same, we'd have to multiply this by one fourth. All right, finally, let's think about, let's think about dinitrogen pentoxide. So, we said that that was disappearing at -1.8 x 10 to the -5. So once again, what do I need to multiply this number by in order to get 9.0 x 10 to the -6?" + }, + { + "Q": "At 5:55, We are losing N2O5 because it's a reactant and not a product, right?", + "A": "Sort of. N2O5 [reactant] is being consumed/used up to generate NO2 and O2 [products]. I usually think of it as the number of moles or the concentration of the reactants become less as products are being formed in the process.", + "video_name": "8wIodo1HD4Y", + "timestamps": [ + 355 + ], + "3min_transcript": "and the formation of B, and we could make this a little bit more general. We could say that our rate is equal to, this would be the change in the concentration of A over the change in time, but we need to make sure to put in our negative sign. We put in our negative sign to give us a positive value for the rate. So the rate is equal to the negative change in the concentration of A over the change of time, and that's equal to, right, the change in the concentration of B over the change in time, and we don't need a negative sign because we already saw in the calculation, right, we get a positive value for the rate. So, here's two different ways to express the rate of our reaction. So here, I just wrote it in a little bit more general terms. Let's look at a more complicated reaction. Here, we have the balanced equation for the decomposition of dinitrogen pentoxide And let's say that oxygen forms at a rate of 9 x 10 to the -6 M/s. So what is the rate of formation of nitrogen dioxide? Well, if you look at the balanced equation, for every one mole of oxygen that forms four moles of nitrogen dioxide form. So we just need to multiply the rate of formation of oxygen by four, and so that gives us, 3.6 x 10 to the -5 Molar per second. So, NO2 forms at four times the rate of O2. What about dinitrogen pentoxide? So, N2O5. Look at your mole ratios. For every one mole of oxygen that forms we're losing two moles of dinitrogen pentoxide. So if we're starting with the rate because our mole ratio is one to two here, we need to multiply this by 2, and since we're losing dinitrogen pentoxide, we put a negative sign here. So this gives us - 1.8 x 10 to the -5 molar per second. So, dinitrogen pentoxide disappears at twice the rate that oxygen appears. All right, let's think about the rate of our reaction. So the rate of our reaction is equal to, well, we could just say it's equal to the appearance of oxygen, right. We could say it's equal to 9.0 x 10 to the -6 molar per second, so we could write that down here. The rate is equal to the change in the concentration of oxygen over the change in time. All right, what about if we wanted to express this in terms of the formation of nitrogen dioxide." + }, + { + "Q": "At 07:18 he says that increase in temperature results in increase in solubility. how is that so?", + "A": "the water molecules move around when heated, then gives the salts the opportunity wedge in between them.", + "video_name": "zjIVJh4JLNo", + "timestamps": [ + 438 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:27 Sal says that the NaCl molecule breaks, isn't it that a molecule needs a lot of energy to break? Where does that energy come from?", + "A": "When the NaCl ions dissociate, the attraction between the ions is replaced by the attraction of the ions to the water molecules. Nevertheless, the process is slightly endothermic but the dissolution is favoured by an increase in entropy as the system becomes more disordered. Solutions always have higher entropy than crystalline solids.", + "video_name": "zjIVJh4JLNo", + "timestamps": [ + 267 + ], + "3min_transcript": "" + }, + { + "Q": "Hi :)\n\nAt 7:24, it is mentioned that the molecules have the same kinetic energy. How so? If the volume of the box was decreased then the molecules kinetic energy would increase, wouldn't it?\n\nOr does it mean that all the molecules inside this smaller box all have the same average kinetic energy? But this kinetic energy is still higher than the kinetic energy of the molecules on the bigger box, isn't it?\n\nThank you!", + "A": "Me thinks the average K.E of molecules remains the same but the extra energy they gained by virtue of reduced volume is transferred as increase in heat of collision which is very minute.", + "video_name": "tQcB9BLUoVI", + "timestamps": [ + 444 + ], + "3min_transcript": "With that out of the way, let me give you a formula. I hope by the end of this video you have the intuition for why this formula works. In general, if I have an ideal gas in a container, the pressure exerted on the gas-- on the side of the container, or actually even at any point within the gas, because it will all become homogeneous at some point-- and we'll talk about entropy in future videos-- but the pressure in the container and on its surface, times the volume of the container, is equal to some constant. We'll see in future videos that that constant is actually proportional to the average kinetic energy of the molecules bouncing around. That should make sense to you. If the molecules were moving around a lot faster, then you would have more kinetic energy, and then they would be changing momentum on the sides of the surface a lot more, so you would have more pressure. why pressure times volume is a constant. Let's say I have a container now, and it's got a bunch of molecules of gas in it. Just like I showed you in that last bit right before I erased, these are bouncing off of the sides at a certain rate. Each of the molecules might have a different kinetic energy-- it's always changing, because they're always transferring momentum to each other. But on average, they all have a given kinetic energy, they keep bumping at a certain rate into the wall, and that determines the pressure. What happens if I were able to squeeze the box, and if I were able to decrease the volume of the box? in it, but I squeeze. I make the volume of the box smaller-- what's going to happen? I have the same number of molecules in there, with the same kinetic energy, and on average, they're moving with the same velocities. So now what's going to happen? They're going to be hitting the sides more often-- at the same time here that this particle went bam, bam, now it could go bam, bam, bam. They're going to be hitting the sides more often, so you're going to have more changes in momentum, and so you're actually going to have each particle exert more force on each surface. Because it's going to be hitting them more often in a given amount of time. The surfaces themselves are smaller. You have more force on a surface, and on a smaller surface, you're going to have higher pressure." + }, + { + "Q": "In the 3rd example, while drawing the final products, why arent there any stereochemistry with the left carbon? (5:10 - 5:20)", + "A": "There is no stereochemistry to worry about in the left carbon because two identical groups i.e. methyl groups are attached to the carbon. Hence it is NOT a chiral carbon.", + "video_name": "b_qDLacdkFg", + "timestamps": [ + 310, + 320 + ], + "3min_transcript": "That carbon, that carbon I just circled is a chiral center. So, we do need to worry about stereochemistry. Alright this carbon over here on the left is not a chiral center. So we don't have to worry about the carbon in red. We do have to worry about the carbon in magenta. So let's think about the stereochemistry with a syn addition Adding the H and the OH to the same sides. Alright so we need to think about this alkene here and these carbons that I've marked in yellow, alright those are sp2 hybridized carbons. So the geometry around those carbons is plainer, and we add our H and OH, right. We're turning those into sp3 hybridized carbons. Alright so let's go ahead and sketch in one of our products here. So one of the possible products would be to add the H and the OH on the same side. I'm gonna draw them as wedges. So here's our hydrogen right as a wedge, and then here's the OH as a wedge. because we already figured that out, right. We know the OH adds to this carbon, the one on the right. We know the H added to the one over here in red. Alright, so I'm showing a syn addition of my hydrogen, of my H and of my OH here and so those are wedges, and so let's think about what's attached to those carbons. So let me go back up to here. This is the carbon in magenta. That's this carbon down here. What else is attached to that carbon? Well there's a hydrogen. Alright, and therefore if I think about the stereochemistry that hydrogen must be going away from me now for this product. We're going from an sp2 hybridized carbon in our alkene to a an sp3 hybridized carbon. So that hydrogen is going away from us. Let's think about, let's think about this other carbon. Let me make this other carbon over here red. So this carbon right here. Alright, this carbon. What else is attached to this carbon? Well there's a methyl group. Alright, and if this hydrogen is coming out at me So we have a methyl group going away from us in space, and so that's one of our possible products. Alright, I could draw this out the way we're used to seeing it where we would show the OH coming out at us in space. Alright, the OH is coming out at us. So that's one of our possible products. Alright our other products, I could show the H and the OH adding on the same side but I can show them adding as dashes. Alright so I could draw in our carbons here, and I can show this time, I can show the H adding as a dash, and the OH adding as a dash. Now that means that this hydrogen right here would have to be a wedge, right. So that hydrogen would have to be a wedge. So I draw in the hydrogen here as a wedge, and then for the other carbon, the one in red, so this one. This methyl group would have to be coming out at us now." + }, + { + "Q": "AT 4:10, you said \"now we divide by the mass that's 3kg.\" But in your previous video while doing the chalkboard question you didn't divide it by the mass, even though it was given as 3Kg. Why is that so? Why you didn't divided it in the vertical acceleration while here you did?", + "A": "He does divide by the mass in the chalkboard question, go back and take another look. The only difference is that he just writes the mass as m and doesn t plug in a value for it. In both problems, the next step is to multiply both sides by this mass. Since the left hand side is zero, and zero times anything is zero, the masses (in both problems) vanish anyway.", + "video_name": "EEnzhdQJUYA", + "timestamps": [ + 250 + ], + "3min_transcript": "It's hard to say, we've got forces vertical, we've got forces horizontal. There's only two directions to pick, X or Y in this problem. We're gonna pick the vertical direction, even though it doesn't really matter too much. But because we know one of the forces in the vertical direction, we know the force of gravity. Force of gravity is 30 Newtons. Usually that's a guod strategy, pick the direction that you know something about at least. So we're gonna do that here. We're gonna say that the acceleration vertically equals to the net force vertically over the mass. And so now we plug in. If this can is just sitting here, if there's no acceleration, if this is in not an elevator transporting these peppers up or down, and it's not in a rocket, if it's just sitting here with no acceleration, our acceleration will be zero. That's gonna equal the net force and the vertical direction. So what are we gonna have? So what are the forces in the vertical direction here? One force is this 30 Newton force of gravity. This points down, we're gonna assume upward is positive, that means down in a negative. I could have written -MG, but we already knew it was 30 Newtons, so I'll write -30 Newtons. Then we've got T1 and T2. Both of those point up. But they don't completely point up, they partially point up. So part of them points to the right, part of them points upward. Only this vertical component, we'll call it T1Y, is gonna get included into this calculation, 'cause this calculation only uses Y directed forces. And the reason is only Y directed forces, vertical forces, affect the vertical acceleration. So this T1Y points upward, I'll do plus T1 in the Y direction. And similarly, this T2. It doesn't all point vertically, only part of it points vertically. So I'll write this as T2 in the Y direction. And that's also upward, so since that's up, I'll count it as plus T2 in the Y direction. And that's it, that's all our forces. Notice we can't plug in the total amount T2 in this formula, 'cause only part of it points up. of the T1 force because only part of it points vertically. And then we divide by the mass, the mass is 3 kilograms. But we're gonna multiply both sides by 3 kilograms, and we're gonna get zero equals all of this right here, so I'll just copy this right here. We use this over again, that comes down right there. But now there's nothing on the bottom here. So what do we do at this point? Now you might think we're stuck. I mean, we've got two unknowns in here. I can't solve for either one, I don't know either one of these. I know they have to add up to 30, so I'd do fine, if I added 30 to both sides, I'd realize that these two vertical components of these tension forces added up have to add up to 30, and that makes sense. They have to balance the force downward. But I don't know either of them, so how do I solve here? Well, let's do this. If you ever get stuck on one of the force equations for a single direction, just go to the next equation. Let's try A in the X direction." + }, + { + "Q": "at 2:40 you said there was a primary carbon. Would'nt that be a methyl carbon because there is only one carbon?", + "A": "primary carbon is carbon attached to only one carbon", + "video_name": "_-I3HdmyYfE", + "timestamps": [ + 160 + ], + "3min_transcript": "That gives this carbon a negative one formal charge, So we form a carbanion here, also called an alkynide anion for this portion. So this is an alkynide anion, a carbanion in here. It's a relatively stable conjugate base, because the electrons, these two electrons here, are housed in an SP hybridized orbital, which has a lot of S character to it, so it's relatively small. So those negatively charged electrons are held a little bit more closely to the positively charged nucleus of this carbon here. So that somewhat stabilizes the conjugate base, which is the reason why a terminal alkyne can function as an acid. So once we formed our alkynide anion, we can use that alkynide anion to do an alkylation reaction. So let's go ahead and redraw that alkynide anion here. So I'm going to draw this portion, this R And then we have our carbon triple bonded to another carbon, a negative charge And this can now function as a nucleophile. So a negatively charged anion can function as a nucleophile. And if we react this alkyline anion with an alkyl halide-- let's go ahead and draw an alkyl halide here. So I'm going to put hydrogen there, and I'll put my halogen over here on the right, so putting my lone pairs of electrons. Let's draw in one R group right here and then a hydrogen over here. So here is my alkyl halides. And if you react a strong nucleophile with an alkyl halide that is not very sterically hindered-- this is a primary alkyl halide right here-- you're going to get an SN2 reaction. So think about this being an SN2 reaction. We have an alkyl halide, which has a polarized bond between the carbon and the halogen. So if the halogen is more electronegative, it's going to pull the electrons and the bond between it and carbon closer to itself. So this halogen ends up being partially negative. This carbon, therefore, will be partially positive, like that. This carbon right here is partial positive charge. It wants electrons. Of course, our nucleophile has those electrons. So the lone pair of electrons on our carbon can attack our electrophile, so nucleophile attacks electrophile. And an SN2 mechanism, remember, is a concerted mechanism, meaning the nucleophile attacks the electrophile at the same time you're leaving group is leaving here. So these electrons are going to kick off onto your halogen. So let's go ahead and draw the product. So now we would have an R group, carbon triple bonded to another carbon. And now this is bonded to yet another carbon. So we formed a carbon carbon bond here in this reaction. And then this hydrogen is up here. This R group is still here, coming out at us, and then the hydrogen going away from us like that in space. And then we have our halogen over here with now four lone pairs of electrons, a negative 1 formal charge." + }, + { + "Q": "At around 5:25, David makes a destructive interference wave by moving one wave source a half a wavelength forward. Later he also notes that this placement of the two wave sources will produce no sound because the amplitudes cancel each other out. But what about that one half wavelength of the sound wave at the start, that wave does not get cancelled out right? So technically isn't there a sound being produced for that half of the wavelength?", + "A": "The destructive interference occurs at a particular point in space, not everywhere at the same time.", + "video_name": "oTjTXS40pqs", + "timestamps": [ + 325 + ], + "3min_transcript": "Are there any applications of this? Well yeah. So imagine you're sitting on an airplane and you're listening to the annoying roar of the airplane engine in your ear. It's very loud and it might be annoying. You put on your noise canceling headphones, and what those noise canceling headphones do? They sit on your ear, they listen to the wave coming in. This is what they listen to. This sound wave coming in, and they cancel off that sound by sending in their own sound, but those headphones Pi shift the sound that's going into your ear. So they match that roar of the engine's frequency, but they send in a sound that's Pi shifted so that they cancel and your ear doesn't hear anything. Now it's often now completely silent. They're not perfect, but they work surprisingly well. They're essentially fighting fire with fire. They're fighting sound with more sound, and they rely on this idea of destructive interference. They're not perfectly, totally destructive, but the waves I've drawn here are totally destructive. If they were to perfectly cancel, we'd call that total destructive interference, And it happens because this wave we sent in was Pi shifted compared to what the first wave was. So let me show you something interesting if I get rid of all this. Let me clean up this mess. If I've got wave source one, let me get wave source two back. So this was the wave that was identical to wave source one. We overlap 'em, we get constructive interference because the peaks are lining up perfectly with the peaks, and these valleys or troughs are matching up perfectly with the other valleys or troughs. But as I move this wave source too forward, look at what happens. They start getting out of phase. When they're perfectly lined up we say they're in phase. They're starting to get out of phase, and look at when I move it forward enough what was a constructive situation, becomes destructive. Now all the peaks are lining up with the valleys, they would cancel each other out. And if I move it forward a little more, it lines up perfectly again and you get constructive, move it more I'm gonna get destructive. Keep doing this, I go from constructive to destructive over and over. So in other words, one way to get constructive interference and just put them right next to each other. And a way to get destructive is to take two wave sources that are Pi shifted out of phase, and put them right next to each other, and that'll give you destructive 'cause all the peaks match the valleys. But another way to get constructive or destructive is to start with two waves that are in phase, and make sure one wave gets moved forward compared to the other, but how far forward should we move these in order to get constructive and destructive? Well let's just test it out. When they're right next to each other we get constructive. If I move this second wave source that was initially in phase all the way to here, I get constructive again. How far did I move it? I moved it this far. The front of that speaker moved this far. So how far was that? Let me get rid of this. That was one wavelength. So look at this picture. From peak to peak is exactly one wavelength. We're assuming these waves have the same wavelength. So notice that essentially what we did, we made it so that the wave from wave source two" + }, + { + "Q": "What is that ocean creature called at 9:20?", + "A": "It is an Opabinia. It lived during the Cambrian Era and had five eyes. It s not part of any living phylum today.", + "video_name": "MS7x2hDEhrw", + "timestamps": [ + 560 + ], + "3min_transcript": "believe, a huge rock, a six-mile in diameter rock, colliding with what is now the Yucatan Peninsula in Mexico, or right off the coast of the Yucatan Peninsula. And it destroyed all of the large land life forms, especially the dinosaurs. And to put all of this in perspective-- and actually the thing that really was an aha moment for me-- it's, OK, plants are 450 million years ago. Grass, I kind of view as this fundamental thing in nature. But grass has only been around for about-- I've seen multiple estimates-- 40 to 70 million years. Grass is a relatively new thing on the planet. Flowers have only been around for 130 million years. So there was a time where you had dinosaurs, but you did not have flowers and you did not have grass. And so you fast forward all the way. And so when you look at this scale, it's kind of funny to look at this. This is the time period where the dinosaurs showed up. This whole brown line is where the mammals showed up. And then, of course, the dinosaurs died out here. Our ancestors, when the giant rock hit the Earth, must have been boroughed in holes and were able to stash some food away, or who knows what, and didn't get fully affected. I'm sure most of the large mammals were destroyed. But what's almost-- it's humbling, or almost humorous, or almost ridiculous, when you look at this chart is they put a little dot-- you can't even see it here, They say 2 million years ago, the first humans-- and even this is being pretty generous when they say first humans. These are really the first prehumans. The first humans that are the same as us, if you took one of those babies and your brought them up in the suburbs and gave them haircuts and stuff, they would be the same thing as we are, those didn't exist until 200,000 years ago, give or take. 200,000 to 400,000 years ago, I've seen estimates. So this is actually a very generous period of time to say first humans. It's actually 200,000 years ago. we are and how new evolution is, it was only 5 million years ago-- and I mentioned this in a previous video-- it was only 5 million years ago-- so this is just to get a sense. This is 0 years. Homo sapien sapien, only around for 200,000 years. The Neanderthals, they were cousin species. They weren't our ancestors. Many people think they were. They were a cousin species. We come from the same root. Although there are now theories that they might have remixed in with Homo sapiens. So maybe some of us have some Neanderthal DNA. And it shouldn't be viewed as an insult. They had big brains. Well, they didn't necessarily have big brains. They had big heads. But that seems to imply a big brain. But who knows? We always tend to portray them as somehow inferior. But I don't want to get into the political correctness of how to portray Neanderthals. But anyway, this is a very small period of time, 200,000. If you go 2 million years, then you" + }, + { + "Q": "at 1:40, what does velocity mean?", + "A": "Velocity is a measurement which includes a speed and a direction: (ie: 40 km/h south)", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 100 + ], + "3min_transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time." + }, + { + "Q": "At 8:33, Sal mentions dimensional analysis. What is dimensional analysis?", + "A": "Dimensional analysis is basically just a fancy name for unit conversion by canceling like dimensions...", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 513 + ], + "3min_transcript": "Well, the first step is to think about how many meters we are traveling in an hour. So let's take that 5 kilometers per hour, and we want to convert it to meters. So I put meters in the numerator, and I put kilometers in the denominator. And the reason why I do that is because the kilometers are going to cancel out with the kilometers. And how many meters are there per kilometer? Well, there's 1,000 meters for every 1 kilometer. And I set this up right here so that the kilometers cancel out. So these two characters cancel out. And if you multiply, you get 5,000. So you have 5 times 1,000. So let me write this-- I'll do it in the same color-- 5 times 1,000. So I just multiplied the numbers. When you multiply something, you can switch around the order. Multiplication is commutative-- I always And then in the units, in the numerator, you have meters, and in the denominator, you have hours. Meters per hour. And so this is equal to 5,000 meters per hour. And you might say, hey, Sal, I know that 5 kilometers is the same thing as 5,000 meters. I could do that in my head. And you probably could. But this canceling out dimensions, or what's often called dimensional analysis, can get useful once you start doing really, really complicated things with less intuitive units than something But you should always do an intuitive gut check right here. You know that if you do 5 kilometers in an hour, that's a ton of meters. So you should get a larger number if you're talking about meters per hour. And now when we want to go to seconds, let's do an intuitive gut check. If something is traveling a certain amount in an hour, it should travel a much smaller amount in a second, are in an hour. So that's your gut check. We should get a smaller number than this when we want to say meters per second. But let's actually do it with the dimensional analysis. So we want to cancel out the hours, and we want to be left with seconds in the denominator. So the best way to cancel this hours in the denominator is by having hours in the numerator. So you have hours per second. So how many hours are there per second? Or another way to think about it, 1 hour, think about the larger unit, 1 hour is how many seconds? Well, you have 60 seconds per minute times 60 minutes per hour. The minutes cancel out. 60 times 60 is 3,600 seconds per hour." + }, + { + "Q": "Does anyone have any idea as to why the letter s is used to represent displacement instead of d? (at time 1:51)\nMy best guess is that s stands for Science... Not or that it is used to be the SUM of the displacement vectors...", + "A": "d is delta in calculus", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 111 + ], + "3min_transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time." + }, + { + "Q": "At 0:38 he said that 5km was a magnitude, but then at 0:48 he said that 5km was the distance. Which it is?", + "A": "Both, the distance is the magnitude.", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 38, + 48 + ], + "3min_transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time." + }, + { + "Q": "At 1:20 does the arrow over the V indicate *which* direction, or is it just to show that it has a direction at all?", + "A": "It is to indicate that the value represented by the letter is a vector, not a scalar. I.e. to show that it has a direction at all.", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 80 + ], + "3min_transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time." + }, + { + "Q": "At 1:50, Sal says that the symbol for displacement is S, but I have been learning in my science class that displacement is X. Is one right and one wrong, or are there just other variables?", + "A": "Both x and s are used for representing displacement", + "video_name": "oRKxmXwLvUU", + "timestamps": [ + 110 + ], + "3min_transcript": "Now that we know a little bit about vectors and scalars, let's try to apply what we know about them for some pretty common problems you'd, one, see in a physics class, but they're also common problems you'd see in everyday life, because you're trying to figure out how far you've gone, or how fast you're going, or how long it might take you to get some place. So first I have, if Shantanu was able to travel 5 kilometers north in 1 hour in his car, what was his average velocity? So one, let's just review a little bit about what we know about vectors and scalars. So they're giving us that he was able to travel 5 kilometers to the north. So they gave us a magnitude, that's the 5 kilometers. That's the size of how far he moved. And they also give a direction. So he moved a distance of 5 kilometers. Distance is the scalar. But if you give the direction too, you get the displacement. So this right here is a vector quantity. He was displaced 5 kilometers to the north. What was his average velocity? So velocity, and there's many ways that you might see it defined, but velocity, once again, is a vector quantity. And the way that we differentiate between vector and scalar quantities is we put little arrows on top of vector quantities. Normally they are bolded, if you can have a typeface, and they have an arrow on top of them. But this tells you that not only do I care about the value of this thing, or I care about the size of this thing, I also care about its direction. The arrow isn't necessarily its direction, it just tells you that it is a vector quantity. So the velocity of something is its change in position, including the direction of its change in position. So you could say its displacement, and the letter for displacement is S. And that is a vector quantity, so that is displacement. And you might be wondering, why don't they use D for displacement? That seems like a much more natural first letter. you start using D for something very different. You use it for the derivative operator, and that's so that the D's don't get confused. And that's why we use S for displacement. If someone has a better explanation of that, feel free to comment on this video, and then I'll add another video explaining that better explanation. So velocity is your displacement over time. If I wanted to write an analogous thing for the scalar quantities, I could write that speed, and I'll write out the word so we don't get confused with displacement. Or maybe I'll write \"rate.\" Rate is another way that sometimes people write speed. So this is the vector version, if you care about direction. If you don't care about direction, you would have your rate. So this is rate, or speed, is equal to the distance that you travel over some time." + }, + { + "Q": "What is the sign Sal uses at 4:46 to show a partial positive or negative charge?", + "A": "The symbol is a a lower case delta . Delta (a triangle, related to the letter D) is used for change, so a lower case delta means a small (partial) change.", + "video_name": "Rr7LhdSKMxY", + "timestamps": [ + 286 + ], + "3min_transcript": "it stabilizes the outer shell, or it stabilizes the hydrogen. And likewise, that electron could be, can be shared with the hydrogen, and the hydrogen can kind of feel more like helium. And then this oxygen can feel like it's a quid pro quo, it's getting something in exchange for something else. It's getting the electron, an electron, it's sharing an electron from each of these hydrogens, and so it can feel like it's, that it stabilizes it, similar to a, similar to a neon. But when you have these covalent bonds, only in the case where they are equally electronegative would you have a case where maybe they're sharing, and even there what happens in the rest of the molecule might matter, but when you have something like this, where you have oxygen and hydrogen, they don't have the same electronegativity. Oxygen likes to hog electrons more than hydrogen does. And so these electrons are not gonna spend an even amount of time. Here I did it kind of just drawing these, you know, these valence electrons as these dots. But as we know, the electrons are in this kind of blur around, around the, around the atoms that make up the atoms. And so, in this type of a covalent bond, the electrons, the two electrons that this bond represents, are going to spend more time around the oxygen then they are going to spend around the hydrogen. And these, these two electrons are gonna spend more time around the oxygen, then are going to spend around the hydrogen. And we know that because oxygen is more electronegative, and we'll talk about the trends in a second. This is a really important idea in chemistry, and especially later on as you study organic chemistry. Because, because we know that oxygen is more electronegative, and the electrons spend more time around oxygen then around hydrogen, it creates a partial negative charge on this side, and partial positive charges on this side right over here, which is why water has many of the properties that it does, and we go into much more in depth in that in other videos. And also when you study organic chemistry, a lot of the likely reactions that are or a lot of the likely molecules that form can be predicted based on elecronegativity. And especially when you start going into oxidation numbers and things like that, electronegativity will tell you a lot. So now that we know what electronegativity is, let's think a little bit about what is, as we go through, as we start, as we go through, as we go through a period, as say as we start in group one, and we go to group, and as we go all the way all the way to, let's say the halogens, all the way up to the yellow column right over here, what do you think is going to be the trend for electronegativity? And once again, one way to think about it is to think about the extremes. Think about sodium, and think about chlorine, and I encourage you to pause the video and think about that. Assuming you've had a go at it, and it's in some ways the same idea," + }, + { + "Q": "At 6:45, does this mean that the elements in groups 1 & 2 are more likely to be in ionic bonds than covalent bonds because it's easier to just give an electron and be positively charged and bonded or does it not matter?", + "A": "That is exactly what it means", + "video_name": "Rr7LhdSKMxY", + "timestamps": [ + 405 + ], + "3min_transcript": "or a lot of the likely molecules that form can be predicted based on elecronegativity. And especially when you start going into oxidation numbers and things like that, electronegativity will tell you a lot. So now that we know what electronegativity is, let's think a little bit about what is, as we go through, as we start, as we go through, as we go through a period, as say as we start in group one, and we go to group, and as we go all the way all the way to, let's say the halogens, all the way up to the yellow column right over here, what do you think is going to be the trend for electronegativity? And once again, one way to think about it is to think about the extremes. Think about sodium, and think about chlorine, and I encourage you to pause the video and think about that. Assuming you've had a go at it, and it's in some ways the same idea, Something like sodium has only one electron in it's outer most shell. It'd be hard for it to complete that shell, and so to get to a stable state it's much easier for it to give away that one electron that it has, so it can get to a stable configuration like neon. So this one really wants to give away an electron. And we saw in the video on ionization energy, that's why this has a low ionization energy, it doesn't take much energy, in a gaseous state, to remove an electron from sodium. But chlorine is the opposite. It's only one away from completing it's shell. The last thing it wants to do is give away electron, it wants an electron really, really, really, really badly so it can get to a configuration of argon, so it can complete it's third shell. So the logic here is that sodium wouldn't mind giving away an electron, while chlorine really would love an electron. So chlorine is more likely to hog electrons, while sodium is very unlikely to hog electrons. So this trend right here, when you go from the left to the right, your getting more electronegative. More electro, electronegative, as you, as you go to the right. Now what do you think the trend is going to be as you go down, as you go down in a group? What do you think the trend is going to be as you go down? Well I'll give you a hint. Think about, think about atomic radii, and given that, what do you think the trend is? Are we gonna get more or less electronegative as we move down? So once again I'm assuming you've given a go at it, so as we know, from the video on atomic radii, our atom is getting larger, and larger, and larger, as we add more and more and more shells. And so cesium has one electron in it's outer most shell, in the sixth shell, while, say, lithium has one electron. Everything here, all the group one elements, have one electron in it's outer most shell, but that fifty fifth electron," + }, + { + "Q": "what i the sign made by Sal at 4:53 to denote the partial positivity ?", + "A": "It s a lowercase delta (Greek letter): \u00ce\u00b4", + "video_name": "Rr7LhdSKMxY", + "timestamps": [ + 293 + ], + "3min_transcript": "it stabilizes the outer shell, or it stabilizes the hydrogen. And likewise, that electron could be, can be shared with the hydrogen, and the hydrogen can kind of feel more like helium. And then this oxygen can feel like it's a quid pro quo, it's getting something in exchange for something else. It's getting the electron, an electron, it's sharing an electron from each of these hydrogens, and so it can feel like it's, that it stabilizes it, similar to a, similar to a neon. But when you have these covalent bonds, only in the case where they are equally electronegative would you have a case where maybe they're sharing, and even there what happens in the rest of the molecule might matter, but when you have something like this, where you have oxygen and hydrogen, they don't have the same electronegativity. Oxygen likes to hog electrons more than hydrogen does. And so these electrons are not gonna spend an even amount of time. Here I did it kind of just drawing these, you know, these valence electrons as these dots. But as we know, the electrons are in this kind of blur around, around the, around the atoms that make up the atoms. And so, in this type of a covalent bond, the electrons, the two electrons that this bond represents, are going to spend more time around the oxygen then they are going to spend around the hydrogen. And these, these two electrons are gonna spend more time around the oxygen, then are going to spend around the hydrogen. And we know that because oxygen is more electronegative, and we'll talk about the trends in a second. This is a really important idea in chemistry, and especially later on as you study organic chemistry. Because, because we know that oxygen is more electronegative, and the electrons spend more time around oxygen then around hydrogen, it creates a partial negative charge on this side, and partial positive charges on this side right over here, which is why water has many of the properties that it does, and we go into much more in depth in that in other videos. And also when you study organic chemistry, a lot of the likely reactions that are or a lot of the likely molecules that form can be predicted based on elecronegativity. And especially when you start going into oxidation numbers and things like that, electronegativity will tell you a lot. So now that we know what electronegativity is, let's think a little bit about what is, as we go through, as we start, as we go through, as we go through a period, as say as we start in group one, and we go to group, and as we go all the way all the way to, let's say the halogens, all the way up to the yellow column right over here, what do you think is going to be the trend for electronegativity? And once again, one way to think about it is to think about the extremes. Think about sodium, and think about chlorine, and I encourage you to pause the video and think about that. Assuming you've had a go at it, and it's in some ways the same idea," + }, + { + "Q": "At 13:13, he says that \"x\" represents the concentration of Hydronium ion [H3O]. But the -log of x was pH. I don't understand why. Can someone please explain this for me?\nThanks :3", + "A": "To you figure out the pH of any substance, the formula says: pH = -log([H3O+]) , or pH = -log([H+]) If x = [H3O+], so pH = -log(x); It s all the same thing. Sorry if my english ain t so good, I am brazillian. I hope it helped you.", + "video_name": "XZWoMXVANww", + "timestamps": [ + 793 + ], + "3min_transcript": "So, NH4+ and NH3 are a conjugate acid-base pair. We're trying to find the Ka for NH4+ And again, that's not usually found in most text books, but the Kb value for NH3, is. It's: 1.8 times 10 to the negative five. So for a conjugate acid-base pair, Ka times Kb is equal to Kw. We're trying to find Ka. We know Kb is 1.8 x 10-5 This is equal to: 1.0 times 10 to the negative 14. So we can once again find Ka on our calculator. So, 1.0 x 10-14... We divide that by 1.8 x 10-5 So if we get some room down here, we say: Ka = 5.6 x 10-10 This is equal to: so it'd be X squared over here... And once again, we're going to assume that X is much, much smaller than .050 So we don't have to worry about X right here, but it's an extremely small number, .050 - X is pretty much the same as .050 So we plug this in and we have: .050, here. So we need to solve for X. We get out the calculator, and we're going to take 5.6 x 10-10, and we're going to multiply by .05 and then we're gonna take the square root of that to get us what X is. So: X = 5.3 x 10-6 X represents the concentration of hydronium ions, so this is a concentration, right? This is the concentration of hydronium ions, so to find the pH, all we have to do is take the negative log of that. So, the pH is equal to the negative log of the concentration of hydronium ions. So we can just plug that into here: 5.3 x 10-6, and we can solve; and let's take the - log(5.3 x 10-6) And so we get: 5.28, if we round up, here. So let me get a little more room... So we're rounding up to 5.28 for our final pH." + }, + { + "Q": "hey but in case of CH3OH ( at 5:00 ) the total no of valence electrons dont match with the no of bonds as shown at 6:29 ..??", + "A": "I think you should also count the dots. CH3OH has a total number of 14 valence electrons. At 6:29, the structure shown has a total of 5 bonds, and 4 dots on the Oxygen. It means that the structure has 14 valence electrons and is correct.", + "video_name": "BIZNBfBuu1w", + "timestamps": [ + 300, + 389 + ], + "3min_transcript": "I can represent those valence electrons as one, two, three, four, and five, like that. And I still have two hydrogens to worry about, right? So I have still have these two hydrogens here. And I can see there's a place for them on the nitrogen. So I can go ahead and put a hydrogen in here and a hydrogen in here, and connect the dots. And I have my dot structure. And I can also check on my octet rule. So carbon has an octet. And nitrogen has an octet as well. So let's go ahead and verify that. So there's two electrons here, four, six, and eight. So nitrogen is in the second period. And so nitrogen is also going to follow the octet rule when you're drawing your dot structures. Let's do one with oxygen next. So if I wanted to draw the dot structure for methanol, methanol is CH3 OH. with the four valence electrons. And I have three hydrogens, each one with one valence electron, like that. And so I can go ahead and put in those three hydrogens. Next I have oxygen. So I need to find oxygen on my organic periodic table. And I can see that oxygen is in group VI right here. So oxygen is going to have six valence electrons around it. So I can go ahead and draw in oxygen. And I can put its six valence electrons in-- one, two, three, four, five, and six, like that. And then I'm going to put in the hydrogen, right? So now I have a hydrogen to worry about. And I know that hydrogen has one valence electron. So I can see there's a place for it over here. And once again, I can connect the dots and see all of the single covalent bonds in this molecule. So that's one bond. That's another bond. And the oxygen has bonded to this hydrogen as well. Again, we can check our octet rule. So the carbon has eight electrons around it. And so does the oxygen. So this would be two right here, and then four, and then six, and then eight. So oxygen is going to follow the octet rule. Now when you're drawing dot structures, you don't always have to do this step where you're drawing each individual atom and summing all of your valence electrons that way. You can just start drawing it. So for an example, if I gave you C2 H6, which is ethane, another way to do it would just be starting to draw some bonds here. And so I have two carbons. And it's a pretty good bet those two carbons are going to be connected to each other. And then I have six hydrogens. And if I look at what's possible around those carbons, I could put those six hydrogens around those two carbon atoms, like that. And if I do that, I'll have an octet around each carbon atom." + }, + { + "Q": "I noticed at 2:40 he writes the formula for Methylamine as CH_3NH_2 as opposed to CH_5N, does the expression of the formula vary based on context?", + "A": "There can be variations in a general formula of an organic compound. This happens due to a phenomenon which we have named as isomerism. Simply put, there can be various structures of a general formula if not specified.", + "video_name": "BIZNBfBuu1w", + "timestamps": [ + 160 + ], + "3min_transcript": "There is a single covalent bond. And then I have two more here. So this is my complete dot structure for methane. Now I can see that carbon is surrounded by eight electrons So we can go ahead and highlight those. So if I'm counting the electrons around carbon, it would be two, four, six, and eight, like that. And eight electrons around carbon makes carbon very stable. And if we look at the periodic table, we can see why. So if I look at the second period, I can see that the valence electrons for carbon would be one, two, three, and four. And to get to eight electrons, we would go five, six, seven, eight. So if carbon is surrounded by eight electrons, it's like it has the electron configuration of a noble gas, which makes it very stable, because all of the orbitals in that energy level are now full. So an octet of electrons is the maximum number of electrons for carbon. that each hydrogen is surrounded by two electrons. And so if I find hydrogen here, hydrogen is in the first energy level. And so here's one electron and here's two electrons. So in the first energy level, there is only an s orbital. And so that s orbital holds a maximum of two electrons. And we get to the electron configuration of a noble gas. And so hydrogen is stable with having only two electrons around it. Let's look at another dot structure. And let's do one that has nitrogen in it. So if I look at the molecular formula CH3 NH2, I'm going to once again start with carbon in the center with its four valence electrons around it, like that. And I know that there are three hydrogens on that carbon. So I can go ahead and put in those three hydrogens. Each hydrogen has one valence electron, like that. And then on the right side, I'm going to think about nitrogen. So I need to find nitrogen on my periodic table. Nitrogen is in group V. Therefore, I can represent those valence electrons as one, two, three, four, and five, like that. And I still have two hydrogens to worry about, right? So I have still have these two hydrogens here. And I can see there's a place for them on the nitrogen. So I can go ahead and put a hydrogen in here and a hydrogen in here, and connect the dots. And I have my dot structure. And I can also check on my octet rule. So carbon has an octet. And nitrogen has an octet as well. So let's go ahead and verify that. So there's two electrons here, four, six, and eight. So nitrogen is in the second period. And so nitrogen is also going to follow the octet rule when you're drawing your dot structures. Let's do one with oxygen next. So if I wanted to draw the dot structure for methanol, methanol is CH3 OH." + }, + { + "Q": "When drawing the diagrams, do you choose the element that has the least number of atoms in that bond for the central atom? Secondly, how do you know how to place the atoms in larger bonds (ex. CH3 NH2 at 2:45) (how to configure the diagram)?", + "A": "H atoms must always be external atoms. That leaves C and N as candidates for the central atom. The central atom is the *least electronegative atom:* C. So, C is the central atom. Attach N to the C. Then add H atoms and electrons to give each atom its octet.", + "video_name": "BIZNBfBuu1w", + "timestamps": [ + 165 + ], + "3min_transcript": "There is a single covalent bond. And then I have two more here. So this is my complete dot structure for methane. Now I can see that carbon is surrounded by eight electrons So we can go ahead and highlight those. So if I'm counting the electrons around carbon, it would be two, four, six, and eight, like that. And eight electrons around carbon makes carbon very stable. And if we look at the periodic table, we can see why. So if I look at the second period, I can see that the valence electrons for carbon would be one, two, three, and four. And to get to eight electrons, we would go five, six, seven, eight. So if carbon is surrounded by eight electrons, it's like it has the electron configuration of a noble gas, which makes it very stable, because all of the orbitals in that energy level are now full. So an octet of electrons is the maximum number of electrons for carbon. that each hydrogen is surrounded by two electrons. And so if I find hydrogen here, hydrogen is in the first energy level. And so here's one electron and here's two electrons. So in the first energy level, there is only an s orbital. And so that s orbital holds a maximum of two electrons. And we get to the electron configuration of a noble gas. And so hydrogen is stable with having only two electrons around it. Let's look at another dot structure. And let's do one that has nitrogen in it. So if I look at the molecular formula CH3 NH2, I'm going to once again start with carbon in the center with its four valence electrons around it, like that. And I know that there are three hydrogens on that carbon. So I can go ahead and put in those three hydrogens. Each hydrogen has one valence electron, like that. And then on the right side, I'm going to think about nitrogen. So I need to find nitrogen on my periodic table. Nitrogen is in group V. Therefore, I can represent those valence electrons as one, two, three, four, and five, like that. And I still have two hydrogens to worry about, right? So I have still have these two hydrogens here. And I can see there's a place for them on the nitrogen. So I can go ahead and put a hydrogen in here and a hydrogen in here, and connect the dots. And I have my dot structure. And I can also check on my octet rule. So carbon has an octet. And nitrogen has an octet as well. So let's go ahead and verify that. So there's two electrons here, four, six, and eight. So nitrogen is in the second period. And so nitrogen is also going to follow the octet rule when you're drawing your dot structures. Let's do one with oxygen next. So if I wanted to draw the dot structure for methanol, methanol is CH3 OH." + }, + { + "Q": "At 9:10, it states that there would be a high potential energy assuming that q and Q are positive. Is the same true for when both charges are negative? What if q is positive, and Q is negative (vise versa as well)?", + "A": "The formula is the same and always works, as long as you put all the signs in correctly. So if Q is positive, it creates a positive electric potential V. If Q is neg, it creates a neg electric potential V. But then if you put a positive q in a positive V, it will have positive potential ENERGY (PE = qV). Same for a neg q near a neg Q (thus a neg V). But a positive q in a neg V or a neg q in a positive V will both have NEG potential energy.", + "video_name": "ks1B1_umFk8", + "timestamps": [ + 550 + ], + "3min_transcript": "when you're doing physics with constants. Look at this is in terms of meters, so I've got to use meters here, so nine centimeters is .09 meters. And if I multiply all this out what you'll get is, 10 to the negative ninth cancels this 10 to the ninth, the powers are 10, these just go away, and then I have nine divided by .09, that gets equal 100. So I chose this so that we got the same answer down there. Okay, 100 Joules per Coulomb, you might be like, where the Joules comes from? And how is this Joules per Coulomb? Well let's look at it, if we took, look at, one of these meters cancels one of these meters, and one of these Coulombs cancels one of those Coulombs, what are we left with? We're left with Newton times meter over Coulomb, but Newton times meter, that's force times distance, that's Joules, that's where we get Joules per Coulomb. So this really does give us the number of Joules per Coulomb of charge that you put there. And it works for any point, if I picked a point twice as close, it's half as far away, let's say some point over here, let's say this r value here was only 4.5 centimeters, well I'm dividing this by r, so if the r is half as big this point over here will have a V value of 200 Joules per Coulomb and the closer I get, if I went even closer, if I went to a point that was three centimeters away, well this is a third as much as this other distance, so if I'm only dividing by a third as much distance as you get three times the result 'cause r is not squared, it's just r. So at this point, we'll have a V value of 300 Joules per Coulomb. This tells me, if I wanted to get a charge that have a whole bunch of Potential energy, I should stick it over here, this will give me a lot of Potential energy. Not quite as much, even less, the further I put my charge the less Potential energy it will have. There will be no Potential energy until there is a charge, there'll just be Electric Potential. But once you place another charge in that region to go with the first one, then you'll have Electric Potential energy and this will be a way to find it, Q times the V that you get out of this calculation. You gotta be careful though, sometimes people get sloppy, and V looks, you know, we use V for Electric Potential and we use V for Voltage, what's the difference? Are they the same? Hmmm, not quite. Sometimes you can treat them as the same but sometimes you do and messes you up. Voltage is a, technically a change in Electric Potential between two points," + }, + { + "Q": "At 2:37, how can we even predict that the Eagle Nebula doesn't exist right now if the photons we are seeing currently depict it as it was 7000 years ago, and nothing travel faster than light, i.e., these photons? Shouldn't the 7000 year old photons be our most recent image of this nebula? How could we know of something that happened after this point in time if light can't reach us that fast?", + "A": "Actually, right now we can see a hint of a destructive energy burst (probably from a nearby supernova) going towards the Eagle Nebula. And what we are seeing IS the Eagle Nebula as it was 7,000 years ago. So actually, the Eagle Nebula as we know it does not exist today. It is either gone or completely disfigured.", + "video_name": "w3IKEa_GOYs", + "timestamps": [ + 157 + ], + "3min_transcript": "so this is an enormous amount of distance remember, the distance from earth to the nearest star was about 4 light years it would take voyager, if it were pointed in the right direction moving at 60 thousand kilometers per hour it would take Voyager 80 thousand years to go 4 light years just this pillar is 7 light years but i wanted to show you this because these type of nebulae, the plural of nebula are where stars can form. so this right here, you actually see, is actually a breeding ground for the birth of new stars this gas is condensing, just like we talked about a couple of videos ago. Until it gets to that critical temperature, the critical density, where you can actually get fusion of hydrogen so this is just a huge interstellar cloud of hydrogen gas and over here you can see its just this breeding ground for stars and we don't even, we think that this structure doesn't even exist anymore in fact is, just so you have the number, this thing is 7000 light years away 7000 light years away which means that what we are seeing now, the photons that are reaching our eyes or telescopes right now left this region of space 7000 years ago so we're seeing it as it was 7000 years ago so a lot of this gas, a lot of this hydrogen, may have already condensed into many many more stars so the structure might not be the way it looks right now and actually there was another super nova that happened that might have blown away a lot of this stuff and we won't even be able to see the effects of this super nova for another thousand years but anyway, this is just a pretty amazing photograph in my opinion especially, and its beautiful at any scale and it's even more mindblowing when you think that this is 7, this is a structure that is 7 light years tall one of the pillars of creation this right here is a star field, and this is as we're looking towards the center of our galaxy this is the Sagittarius star field the neat thing here you see is such a diversity in stars this is also kind of mind numbing because every one of these stars, are inside of our galaxy this is looking towards the center of our galaxy this isn't one of those where we're looking beyond our galaxy or looking at clusters of galaxies this is just stars here but the thing here is that you see a huge variety, you see some stars that are shining red, right over here and obviously, the apparent size, you cannot completely tell because the different stars are at different distances and at difference intensities but the redder stars, these are stars in their red giant phase or they're probably at their red giant phase i haven't done specific research on these stars but that's what we suspect those are in their red giant phase the ones that are kind of in the yellowish white part of the spectrum" + }, + { + "Q": "At 7:31 in the video, Sal stated that the chain goes from 5' to 3'. Is it possible for it to go from 3' to 5'? If not, why?", + "A": "DNA polymerase only works in the 5 to 3 direction. This is down to the specificity of this enzyme. If you think about it, DNA replication could result in a real mess if DNA polymerase could work in both directions.", + "video_name": "0CQ5ls3Uc2Q", + "timestamps": [ + 451 + ], + "3min_transcript": "One prime. Two prime. Three prime. Four prime. Five prime. And then when you look at it as a ring, this was the one prime. This is the two prime. This is the three prime. This is the four prime. This is the five prime. Or if you were to number them on this diagram right over here, actually in the DNA molecule, this is the one prime. This is the two prime carbon. This is the three prime carbon. This is the four prime carbon. And this is the five prime carbon. And so one way to think about it is we'll go phosphate group and it's connected with what we call phosphodiester linkages. Phosphodiester linkages, that's what's essentially allowing these backbones to link up. But we're going from phosphate to five prime carbon and then through the sugar we go to the three prime carbon. And then we go to another phosphate. Then we go to the five prime carbon. Let me label that, this is the five prime carbon. And that just comes straight out of just numbering these starting with the carbon that was a number one carbon. When it's straight chain form, it's part of the carbonyl group. But you see we're going from five. We go phosphate, five prime, three prime, phosphate, five prime, three prime, phosphate. So one way to describe the orientation is saying, \"Hey, we're going in the direction \"from five prime to three prime.\" So we could say that we're going from five prime to three prime, that way on the left-hand chain. And what are we doing on the right hand chain? Well, let's number them again. So this is the one prime carbon. Now this thing relative to this is upside down, it's inverted. So one prime. Two prime. Three prime. Four prime. Five prime. I could do it up here. One prime carbon. Two prime carbon. Three prime carbon. Four prime carbon. Five prime carbon. Here you're going from phosphate, three prime, five prime, phosphate, So the way that the sugars are oriented if you're going from top to bottom you're going from three prime to five prime. So on the right hand side, it's three prime, five prime. And so if you wanted to draw an arrow from five prime to three prime, you could look at it like that. And so you could say these are parallel but since they are essentially pointing in different directions even though they are actually parallel, we would call this structure of DNA antiparallel. So this would be an anti... antiparallel structure of DNA. So these two strands, they're complementary. They're defined by each other. The thymine bonds with the adenine, the cytosine bonds with guanine. They are attracted to each other through these hydrogen bonds. But the two backbones, they're pointed in different directions. And now another interesting thing to think about, since we're talking about the molecular structure of DNA, is how do these things form? How did these things know to orient in this way?" + }, + { + "Q": "At 1:53, Why is the fact that nitrogenous bases are forming hydrogen bonds offset their basic property? Somehow making them \"happy\" or something?", + "A": "as he said in 1:35, it is done in order to take more hydrogen protons", + "video_name": "0CQ5ls3Uc2Q", + "timestamps": [ + 113 + ], + "3min_transcript": "- [Voiceover] In the video on the molecular structure of DNA we saw that DNA is typically made up of two strands where the backbone of each of the strands is made up of phosphate alternating between a-- Do some different colors. A phosphate group and then you have a sugar. You have a phosphate group. And then you have a sugar. And then you have a phosphate group. And then you have a sugar. And so I could draw the strand something like this. So phosphate and then we have a sugar. Oops, let me just draw all the phosphates ahead of time. So you have the phosphates on that end and then you have the sugars. And you see the same thing on the other strand as well. Where we have phosphate with a sugar then another phosphate then a sugar then another phosphate. Let me circle the sugars as well. and then you have the sugar there as well. So on the other strand it's also going to look like this. So let me draw the phosphates. I'm just abstracting them now. So the phosphate and then you have the sugars in between the phosphates. And what links them, you can think of them as the rungs on the ladder. These are the complementary nitrogenous bases. And the reason why we call them nitrogenous bases, I actually forgot to talk about it in the last videos, is that these nitrogens are really electronegative and they can take up more hydrogen protons. They have an extra lone pair. The nitrogens have an extra lone pair that can be used up under the right conditions to potentially sop up more hydrogen protons. Now, a lot of people ask, \"Well, if you have these nitrogenous bases here, \"why is DNA called an acid?\" Why is it called an acid?\" Well the first thing is that the basic properties of the nitrogenous base are offset to a good degree based on the fact that they're able to hydrogen bond with each other. And that's what actually forms when these complimentary nitrogenous bases form these hydrogen bonds with each other. But even more, the reason why we call it an acid is the phosphate groups, when they're protonated, are acids. Now the reason why we tend to draw them deprotonated is they're so acidic that if you put them in a neutral solution, they're going to be deprotonated. So this is the form that you're more likely to find it in the nucleus of an actual cell. Once it's actually already deprotonated. But in general, phosphate groups are considered acidic. And if I were to draw kind of a more pure phosphate group, and I talked about this already in the last video, I would have it protonated and so I wouldn't draw that negative charge like that. So that's just a review of last time. Since I already started abstracting it, let's abstract further. So let's draw the nitrogenous bases a little bit. So I have thymine here. And I will do thymine in this green color." + }, + { + "Q": "At 1:32, Sal says that \"as we get more into physics, we'll see that maybe we shouldn't necessarily think of time as driving; maybe position, in some ways, is driving time.\" What is Sal referring to? Is there a relevant Khan Academy video or Wikipedia page?", + "A": "Spacetime diagrams when you study relativity have time on the vertical axis. In relativity you cannot untangle space from time. if you change position very quickly (travel near the speed of light, for example), you move through time more slowly.", + "video_name": "PRx_R9iIWk4", + "timestamps": [ + 92 + ], + "3min_transcript": "- [Voiceover] Let's say this is me right over here, and I'm drifting through space at a constant velocity relative to any other inertial frame of reference, and so I am, I am in an inertial frame of reference myself, and in fact I'm going to define my frame of reference by me, I'm gonna say I'm at the origin of my frame of reference. So at all times, I consider myself to be stationary, and I am at the point, 'x' equals zero, and we're gonna focus on just the 'x' dimension, to simplify our discussion, and I have, you know, my oxygen and everything, and food, so no need to worry about me. Now, what I've drawn here are some axes, so that I can plot the path of things as time progresses in my frame of reference. And one thing that many of ya'll might have noticed is that I have plotted time in seconds on the vertical axis, and our position, our 'x' position, in meters on the horizontal axis. And that might be a little bit counter-intuitive especially with math backgrounds, in fact, it's a little bit uncomfortable for me, we often prefer to put time on the horizontal axis, and position on the vertical axis, so this is a little bit counter-intuitive. But this is, you know, our choice is a little bit arbitrary, in math class, we liked to think of our independent variable on our horizontal axis, and we think of time as somehow driving position, so that's why we tend to put time there than we'd position there, but you can flip them around. And as we'll get more into physics, we'll see, well maybe we shouldn't necessarily think of time always driving position, or maybe position in some ways are driving time. But we'll get into that. But let's just first get a little bit comfortable with this. So what would be my position over time on this diagram right over here? And you might notice these numbers, one second, two second, three second, and then in meters I have three times to the eighth, six times to the eighth, nine times to the eighth, so these are massive, massive numbers, and you could guess where they are coming from. that the speed of light is approximately 3x10 to the eighth meters per second. But we'll get to that in a second. But let's just think about my position over time, in my frame of reference here. So, time equals zero, my 'x' position in this frame of reference is zero, I consider myself to be stationary. After one second, well my position is still zero, after two seconds my position is still 'x' equals zero, after three seconds my position is still 'x' equals zero, so my, I guess you could almost consider my path on this diagram, where I'm plotting time and space, at least in the 'x' direction, is going to look like, is going to look like, whoops, I can do a better job than that. It's going to look, it's going to look like, it's going to look like that. That would be my path on my little time and space axes," + }, + { + "Q": "at 9:00 it is written vt+1/2atsquare .and delta t is a very small quantityand delta t multilied by delta t gives a much more small quantity and should not be taken into consideration.then why it is writtem as t square", + "A": "The t in that equation is not the delta t of calculus, it s the time during which the acceleration took place.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 540 + ], + "3min_transcript": "All of this was another way to write average velocity. the whole reason why I did this is because we don't have final velocity but we have acceleration and we are going to use change in time as our independent variable. We still have to multiply this by this green change in time here. multiply all of this times the green change in time. All of this is what displacement is going to be. This is displacement, and lets see... we can multiply the change in time times all this actually these 2s cancel out and we get (continued over here) We get: displacement is equal to initial velocity times change in time change in time is a little more accurate plus 1/2 (which is the same as dividing by 2) plus one half times the acceleration times the acceleration times (we have a delta t times delta t) change in time times change in time the triangle is delta and it just means \"change in\" so change in time times change in time is just change in times squared. In some classes you will see this written as d is equal to vi times t plus 1/2 a t squared this is the same exact thing they are just using d for displacement and t in place of delta t. The one thing I want you to realize with this video Maybe if you were under time pressure you would want to be able to whip this out, but the important thing, so you remember how to do this when you are 30 or 40 or 50 or when you are an engineer and you are trying to send a rocket into space and you don't have a physics book to look it up, is that it comes from the simple displacement is equal to average velocity times change in time and we assume constant acceleration, and you can just derive the rest of this. I am going to leave you there in this video. Let me erase this part right over here. We are going to leave it right over here. In the next video we are going to use this formula we just derived. We are going to use this to actually plot the displacement vs time because that is interesting and we are going to be thinking about what happens to the velocity and the acceleration as we move further and further in time." + }, + { + "Q": "At 6:05 wouldn't the final velocity be zero after it hits the ground?", + "A": "Final velocity refers to the speed it has when it first hits the ground, not after it lands. It would not be very useful or interesting to say that everything has a final velocity of 0.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 365 + ], + "3min_transcript": "the force of gravity is downwards. So, little g over here, if you want to give it its direction, is negative. Little g is -9.8m/s2. So, we have the acceleration due to gravity. The acceleration due to gravity (ag) is negative 9.8 meters per second squared (9.8m/s^2). Now I want to plot distance relative to time. Let's think about how we can set up a formula, derive a formula that, if we input time as a variable, we can get distance. We can assume these values right over here. Well actually I want to plot displacement over time because that will be more interesting. We know that displacement is the same thing as average velocity times change in time (displacement=Vavg*(t1-t2)). Right now we have but not in terms of initial velocity and acceleration. We know that average velocity is the same thing as initial velocity (vi) plus final velocity (vf) over 2. (Vavg=(vi+vf)/2) If we assume constant acceleration. We can only calculate Vavg this way assuming constant acceleration. Once again when were are dealing with objects not too far from the center of the earth we can make that assumption. Assuming that we have a constant acceleration Once again we don't have what our final velocity is. So, we need to think about this a little more. We can express our final velocity in terms of our initial velocity and time. Just dealing with this part, the average velocity. So we can rewrite this expression as the initial velocity plus something over 2. and what is final velocity? your initial velocity plus your acceleration times change in time. If you are starting at 10m/s and you are accelerated at 1m/s^2 then after 1 second you will be going 1 second faster than that. (11m/s) So this right here is your final velocity. Let me make sure that these are all vector quantities...(draws vector arrows) All of these are vector quantities. Hopefully it is ingrained in you that these are all vector quantities, direction matters. And let's see how we can simplify this Well these two terms (remember we are just dealing with the average velocity here) These two terms if you combine them become 2 times initial velocity (2vi). two times my initial velocity and then divided by this 2 plus all of this business divided by this 2." + }, + { + "Q": "7:11 I'm confused. You are shifting the numbers around, assumably to find the specific variable, but you boxed in blue the Vi+Vf( which has been moved to another part of the equation, extended and broken down in yellow) into a translation that read 2Vi, which is just Vi+Vi, right? Am I missing something, or how did Vf become a second Vi, and where did it come from?", + "A": "Sal re-wrote Vf, which he brackets at 6:25, as initial velocity plus acceleration times change in time. He then collects the two Vi s together (2Vi) and divides them by two, and also divides the acceleration times change in time by two. (Remember, we re just re-writing the V avg. part of the S = V avg. x delta T equation so it s in terms of initial velocity.) Hope that s clear :)", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 431 + ], + "3min_transcript": "but not in terms of initial velocity and acceleration. We know that average velocity is the same thing as initial velocity (vi) plus final velocity (vf) over 2. (Vavg=(vi+vf)/2) If we assume constant acceleration. We can only calculate Vavg this way assuming constant acceleration. Once again when were are dealing with objects not too far from the center of the earth we can make that assumption. Assuming that we have a constant acceleration Once again we don't have what our final velocity is. So, we need to think about this a little more. We can express our final velocity in terms of our initial velocity and time. Just dealing with this part, the average velocity. So we can rewrite this expression as the initial velocity plus something over 2. and what is final velocity? your initial velocity plus your acceleration times change in time. If you are starting at 10m/s and you are accelerated at 1m/s^2 then after 1 second you will be going 1 second faster than that. (11m/s) So this right here is your final velocity. Let me make sure that these are all vector quantities...(draws vector arrows) All of these are vector quantities. Hopefully it is ingrained in you that these are all vector quantities, direction matters. And let's see how we can simplify this Well these two terms (remember we are just dealing with the average velocity here) These two terms if you combine them become 2 times initial velocity (2vi). two times my initial velocity and then divided by this 2 plus all of this business divided by this 2. All of this was another way to write average velocity. the whole reason why I did this is because we don't have final velocity but we have acceleration and we are going to use change in time as our independent variable. We still have to multiply this by this green change in time here. multiply all of this times the green change in time. All of this is what displacement is going to be. This is displacement, and lets see... we can multiply the change in time times all this actually these 2s cancel out and we get (continued over here) We get: displacement is equal to initial velocity times change in time" + }, + { + "Q": "At 3:52 Sal said that if something is moving up, it's given a positive sign, but if it's moving down, it's given a negative sign. What if it's being affected by the moon's gravity? Then, it's moving up relative to earth, but down relative to the moon.", + "A": "positive being up and negative being down is just a conventional way of doing things. We can choose the y-axis to point in any direction, so long as it is perpendicular to the x-axis. It just seems natural to choose y to point upwards. So when the Earth is pulling something down, we give that a negative value. And using that same convention, then the moon above us would pull things up with a positive value.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 232 + ], + "3min_transcript": "The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall the force of gravity is downwards. So, little g over here, if you want to give it its direction, is negative. Little g is -9.8m/s2. So, we have the acceleration due to gravity. The acceleration due to gravity (ag) is negative 9.8 meters per second squared (9.8m/s^2). Now I want to plot distance relative to time. Let's think about how we can set up a formula, derive a formula that, if we input time as a variable, we can get distance. We can assume these values right over here. Well actually I want to plot displacement over time because that will be more interesting. We know that displacement is the same thing as average velocity times change in time (displacement=Vavg*(t1-t2)). Right now we have" + }, + { + "Q": "At 4:03, Sal stated that a vector is positive if an object goes up and negative if it goes down. What about left and right?", + "A": "It can be whatever way you want, just make sure you stick to it for the whole problem. Personally if I m working on a problem and I have a lot of vectors going in a certain direction, I consider that direction positive and the opposite direction negative. Sometime this is right, sometimes it s left.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 243 + ], + "3min_transcript": "The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is If we assume that mass one (m1) is the mass of the earth, and r is the radius of the earth (the distance from the center of the earth) So you would be correct in thinking that it changes a little bit. The force of gravity changes a little bit, but for the sake of throwing things up into our atmosphere we can assume that it is constant. And if we were to calculate it it is 9.8 meters per second squared and I have rounded here to the nearest tenth. I want to be clear these are vector quantities. When we start throwing things up into the air the convention is if something is moving up it is given a positive value, and if it is moving down we give it a negative value. Well, for an object that is in free fall the force of gravity is downwards. So, little g over here, if you want to give it its direction, is negative. Little g is -9.8m/s2. So, we have the acceleration due to gravity. The acceleration due to gravity (ag) is negative 9.8 meters per second squared (9.8m/s^2). Now I want to plot distance relative to time. Let's think about how we can set up a formula, derive a formula that, if we input time as a variable, we can get distance. We can assume these values right over here. Well actually I want to plot displacement over time because that will be more interesting. We know that displacement is the same thing as average velocity times change in time (displacement=Vavg*(t1-t2)). Right now we have" + }, + { + "Q": "In my physics class we use the equation x = xi + (vi)(\u00ce\u0094t) + 1/2 a(\u00ce\u0094t)^2\nBut at the end of the video (by 9:27) I see Sal using the equation x = (vi)(\u00ce\u0094t) + 1/2 a(\u00ce\u0094t)^2\nWhy is the equation used in my class different?", + "A": "xi is just a measure of where you are starting out. Sal s is assigning the starting point a coordinate of zero. But if you decided to say you were starting from, say 10 feet, then you would have to add that 10 feet if you wanted to calculate your distance from zero.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 567 + ], + "3min_transcript": "All of this was another way to write average velocity. the whole reason why I did this is because we don't have final velocity but we have acceleration and we are going to use change in time as our independent variable. We still have to multiply this by this green change in time here. multiply all of this times the green change in time. All of this is what displacement is going to be. This is displacement, and lets see... we can multiply the change in time times all this actually these 2s cancel out and we get (continued over here) We get: displacement is equal to initial velocity times change in time change in time is a little more accurate plus 1/2 (which is the same as dividing by 2) plus one half times the acceleration times the acceleration times (we have a delta t times delta t) change in time times change in time the triangle is delta and it just means \"change in\" so change in time times change in time is just change in times squared. In some classes you will see this written as d is equal to vi times t plus 1/2 a t squared this is the same exact thing they are just using d for displacement and t in place of delta t. The one thing I want you to realize with this video Maybe if you were under time pressure you would want to be able to whip this out, but the important thing, so you remember how to do this when you are 30 or 40 or 50 or when you are an engineer and you are trying to send a rocket into space and you don't have a physics book to look it up, is that it comes from the simple displacement is equal to average velocity times change in time and we assume constant acceleration, and you can just derive the rest of this. I am going to leave you there in this video. Let me erase this part right over here. We are going to leave it right over here. In the next video we are going to use this formula we just derived. We are going to use this to actually plot the displacement vs time because that is interesting and we are going to be thinking about what happens to the velocity and the acceleration as we move further and further in time." + }, + { + "Q": "stupid question, at 8:35 sal multiplies change in time by change in time but the first change in time is placed over 2...so why is it not half change in time multiplied by change in time? does that make sense?", + "A": "That does make sense, but that s because they will both give them same answer. Half of change in time multiplied by time written as (1/2 t)*t will give you the same answer as half of change in time squared ( (1/2 t^2) ). It s just easier to remember 1/2 at^2. You could test that by putting some numbers in for time. example if t = 5: 1/2 * 5 = 5/2 5/2 * 5 = 25/2 or 12.5 5^2 = 25 1/2 * 25 = 25/2 or 12.5", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 515 + ], + "3min_transcript": "All of this was another way to write average velocity. the whole reason why I did this is because we don't have final velocity but we have acceleration and we are going to use change in time as our independent variable. We still have to multiply this by this green change in time here. multiply all of this times the green change in time. All of this is what displacement is going to be. This is displacement, and lets see... we can multiply the change in time times all this actually these 2s cancel out and we get (continued over here) We get: displacement is equal to initial velocity times change in time change in time is a little more accurate plus 1/2 (which is the same as dividing by 2) plus one half times the acceleration times the acceleration times (we have a delta t times delta t) change in time times change in time the triangle is delta and it just means \"change in\" so change in time times change in time is just change in times squared. In some classes you will see this written as d is equal to vi times t plus 1/2 a t squared this is the same exact thing they are just using d for displacement and t in place of delta t. The one thing I want you to realize with this video Maybe if you were under time pressure you would want to be able to whip this out, but the important thing, so you remember how to do this when you are 30 or 40 or 50 or when you are an engineer and you are trying to send a rocket into space and you don't have a physics book to look it up, is that it comes from the simple displacement is equal to average velocity times change in time and we assume constant acceleration, and you can just derive the rest of this. I am going to leave you there in this video. Let me erase this part right over here. We are going to leave it right over here. In the next video we are going to use this formula we just derived. We are going to use this to actually plot the displacement vs time because that is interesting and we are going to be thinking about what happens to the velocity and the acceleration as we move further and further in time." + }, + { + "Q": "Sal shows that a=g (where a is a vector quantity displayed at (1:30)). My question is, would gravity ALSO be a vector quantity? If so, why isn't there an arrow placed above it?\n\nThanks :)", + "A": "Every force is a vector. If you are talking about a force of gravity, then it is a vector.", + "video_name": "wlB0x9W-qBU", + "timestamps": [ + 90 + ], + "3min_transcript": "What I want to do with this video is think about what happens to some type of projectile, maybe a ball or rock, if I were to throw it straight up into the air. To do that I want to plot distance relative to time. There are a few things I am going to tell you about my throwing the rock into the air. The rock will have an initial velocity (Vi) of 19.6 meters per second (19.6m/s) I picked this initial velocity because it will make the math a little bit easier. We also know the acceleration near the surface of the earth. We know the force of gravity near the surface of the earth is the mass of the object times the acceleration. (let me write this down) The force of gravity is going to be the mass of the object times little g. little g is gravity near the surface of the earth g is 9.8 meters per second squared (9.8m/s^2) you just take the force divided by the mass Because we have the general equation Force equals mass times acceleration (F=ma) If you want acceleration divide both sides by mass so you get force over mass So, lets just divide this by mass If you divide both sides by mass, on the left hand side you will get acceleration and on the right hand side you will get the quantity little g. The whole reason why I did this is when we look at the g it really comes from the universal law of gravitation. You can really view g as measuring the gravitational field strength near the surface of the earth. Then that helps us figure out the force when you multiply mass times g. Then you use F=ma, the second law, to come up with g again which is actually the acceleration. The other thing I want to make clear: when you talk about the Force of gravity generally the force of gravity is equal to big G Big G (which is different than little g) times the product of the masses of the two things over the square of the distance between the two things. You might be saying \"Wait, clearly the force of gravity is dependent on the distance. So if I were to throw something up into the air, won't the distance change.\" And you would be right! That is technically right, but the reality is that when you throw something up into the air that change in distance is so small relative to the distance between the object and the center of the earth that to make the math simple, When we are at or near the surface of the earth (including in our atmosphere) we can assume that it is constant. Remember that little g over there is" + }, + { + "Q": "@ 9:20... Shouldn't the drawing only have one methyl group branching off the 1st carbon on each of the ehtyl groups branching off of first and third carbons of the ring? Doesn't the \"di\" in 1,1-dimethylethyl imply that there are only two methyl groups (instead of four like is shown in your drawing)? Such as how you only have two methyl groups in 2,2-dimethyl-hexane. Or is it different because it is an alkyl group branching off another alkyl group instead of the parent chain?", + "A": "It s sort of like an order of operations problem from arithmetic. You have to pay attention to the parenthesis and consider everything within them for each case. Solve 1,1-dimethylethyl first, then put that in place at 1 and 3 on the cyclopentane.", + "video_name": "6BR0Q5e74bs", + "timestamps": [ + 560 + ], + "3min_transcript": "with the core: cyclopentane. That's just a simple five-carbon ring. A five carbon ring that looks like a pentagon: one, two, three, four, five. There you go. That is a five-carbon ring. We can number it however we want, so one, two, three, four, and five. This is telling us at the one and the three position we have-- and the bis- is kind of redundant. This is saying we have two of these things. Obviously, we have two. We have one at the one and one at the three. So you can kind of ignore the bis-. That's just the convention and we've seen that multiple times. But at each of those positions, we have a 1,1-dimethylethyl. So what's a dimethylethyl look like? So let's think about it a little bit. Let's think about it and let me do it orange. They obviously named it using systematic naming and what we have here, we have an ethyl as kind of the core So if an ethyl is equal to two carbons, so this is two carbons right there. So let me draw a two carbon: one, two. That is two carbons right over there. I'm just drawing it at the three spot. I'll draw it also at the one spot, actually. So that is two carbons right there. That's the ethyl part. And then on 1,1, so if we number them, we number where it's connected, so it's one, two. This is saying 1,1-dimethyl. So on this ethyl chain, you have two methyls. Remember, methyl is equal to one, so this is one carbon. You have one carbon. That's what methyl is, but you have two of them. You have dimethyl. You have it twice at the one spot. So you have one methyl here and then you have another methyl there. You have 1-methyl on the one spot and then you have another 1-methyl on the one spot. And then you are connected at positions one and positions three, so you're connected there and you are connected And you're done, That's it. That is our structure. Now, if you did this with common naming, instead of this group being a 1,1-dimethylethyl, you might see that we're connected to a group that has one, two, three, four carbons in it. The carbon that we're connected to branches off to three other carbons. It is a tert-butyl. So you can also call this a 1,3-- let me just write it down. So another name for this would be 1,3-tert-- or sometimes people just write a t there-- t-butylcyclo-- no, actually I should say di-t-butyl, because we have two of them. 1,3-di-t-butylcyclopentane." + }, + { + "Q": "Shouldnt the 3,6,9,9-tetramethyldodecane be named as a 4,4,7,10-tetramathyldodecane\n\n06:37", + "A": "it could be 1,4,7,7-tetramethyldodecane", + "video_name": "6BR0Q5e74bs", + "timestamps": [ + 397 + ], + "3min_transcript": "ropylcyclohexadecane. Let's do another one. I think we're getting the hang of it. So here, maybe we can do this one a little bit faster. Let's see, we have a tetramethyldodecane, so the main root here is the dodecane, do- for two, dec- for ten. This is a 12-carbon chain. It's not in a cycle, so let me just draw it out. We have one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, and so we can just number them arbitrarily, just because I could have drawn this any which way. So it's one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. That's the dodecane, all single bonds. Then we have a 2,6,9,9-tetramethyl. All this is telling us-- remember, meth- is one carbon, nine, so at the three, the six and twice at the nine spot, we have methyl groups, and we have four methyl groups. That's all the tetramethyl is saying so it's We know we have four of them here: 3,6,9,9. We have methyl groups at each of those places. We have one methyl group at three, and then that is bonded with the third carbon on the dodecane chain. We have one at six bonded to the six carbon on the dodecane chain. We have two at nine, so that's one at nine and then we have another one at nine bonded to the nine carbon on the dodecane chain. That's it. That's 3,6,9,9- tetramethyldodecane. Let's do another one. 1,3-bis(1,1-dime thylethyl)cyclopentane. So once again, just kind of breathe slowly. with the core: cyclopentane. That's just a simple five-carbon ring. A five carbon ring that looks like a pentagon: one, two, three, four, five. There you go. That is a five-carbon ring. We can number it however we want, so one, two, three, four, and five. This is telling us at the one and the three position we have-- and the bis- is kind of redundant. This is saying we have two of these things. Obviously, we have two. We have one at the one and one at the three. So you can kind of ignore the bis-. That's just the convention and we've seen that multiple times. But at each of those positions, we have a 1,1-dimethylethyl. So what's a dimethylethyl look like? So let's think about it a little bit. Let's think about it and let me do it orange. They obviously named it using systematic naming and what we have here, we have an ethyl as kind of the core" + }, + { + "Q": "hey i think there is a mistake at 5:26 in the naming of this molecule the sum of the methyl prefix is 3+6+9+9 =27 but if we named the molecule 4,4,7,10-Tetremethylcyclododecane the actual sum is equal to 25 or am i wrong ?", + "A": "You don t sum them, that isn t the rule despite what your teacher may have said. You use the set of numbers that has a lower number at the first point of difference. 3 is less than 4 so the video is correct.", + "video_name": "6BR0Q5e74bs", + "timestamps": [ + 326 + ], + "3min_transcript": "It's three carbons, so it's going to be one, two, three, and the connection point to the main ring in this case is going to be in the middle carbon, so it kind of forms a Y. All of the isos, the isopropyl, isobutyl, they all look like Y's, so it's going to be linked right over here. That's also going to happen at the ninth carbon, so at the ninth carbon we're going to have another isopropyl. We're going to have another isopropyl at the ninth carbon. All right, we've taken care of the 2,9-isopropyl. Then we have the 6-ethyl, which is just a two carbon. Remember, meth- is one, eth- is two, prop- is three. Let me write this down. So this is going to be prop- is equal to three. Isoprop- is equal to that type of shape right over there. So at six we have an ethyl group, so one, two, carbons, and it's connected at the six carbon on the main ring. And then finally we have a cyclopentyl. So if we look at-- let me find a color I haven't used yet-- cyclopentyl. so pent- is five, but it's five in a cycle, so this is a five-carbon ring that's branching off of the main ring. It's at the first spot. Let me draw a five-carbon rings, so pent- is equal to five, so it would look like this, one: two, three, four, five. It looks just like a pentagon. That's a cyclopentyl group and it's attached to the one carbon on my cyclohexadecane, so it is attached just like that. ropylcyclohexadecane. Let's do another one. I think we're getting the hang of it. So here, maybe we can do this one a little bit faster. Let's see, we have a tetramethyldodecane, so the main root here is the dodecane, do- for two, dec- for ten. This is a 12-carbon chain. It's not in a cycle, so let me just draw it out. We have one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, and so we can just number them arbitrarily, just because I could have drawn this any which way. So it's one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. That's the dodecane, all single bonds. Then we have a 2,6,9,9-tetramethyl. All this is telling us-- remember, meth- is one carbon," + }, + { + "Q": "at 6:38, Wouldn't that be called 3,3,6,9 instead of 3,6,9,9? I mean shouldn't we start where there is more branching at a smaller no.?", + "A": "I know this question was asked 2 years ago but for anyone reding though in the future: No. Count again from the right hand side and you will see that would put the methyls on carbon #4 not carbon #3. 3 is a lower number than 4, the name given in the video is correct.", + "video_name": "6BR0Q5e74bs", + "timestamps": [ + 398 + ], + "3min_transcript": "ropylcyclohexadecane. Let's do another one. I think we're getting the hang of it. So here, maybe we can do this one a little bit faster. Let's see, we have a tetramethyldodecane, so the main root here is the dodecane, do- for two, dec- for ten. This is a 12-carbon chain. It's not in a cycle, so let me just draw it out. We have one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, and so we can just number them arbitrarily, just because I could have drawn this any which way. So it's one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. That's the dodecane, all single bonds. Then we have a 2,6,9,9-tetramethyl. All this is telling us-- remember, meth- is one carbon, nine, so at the three, the six and twice at the nine spot, we have methyl groups, and we have four methyl groups. That's all the tetramethyl is saying so it's We know we have four of them here: 3,6,9,9. We have methyl groups at each of those places. We have one methyl group at three, and then that is bonded with the third carbon on the dodecane chain. We have one at six bonded to the six carbon on the dodecane chain. We have two at nine, so that's one at nine and then we have another one at nine bonded to the nine carbon on the dodecane chain. That's it. That's 3,6,9,9- tetramethyldodecane. Let's do another one. 1,3-bis(1,1-dime thylethyl)cyclopentane. So once again, just kind of breathe slowly. with the core: cyclopentane. That's just a simple five-carbon ring. A five carbon ring that looks like a pentagon: one, two, three, four, five. There you go. That is a five-carbon ring. We can number it however we want, so one, two, three, four, and five. This is telling us at the one and the three position we have-- and the bis- is kind of redundant. This is saying we have two of these things. Obviously, we have two. We have one at the one and one at the three. So you can kind of ignore the bis-. That's just the convention and we've seen that multiple times. But at each of those positions, we have a 1,1-dimethylethyl. So what's a dimethylethyl look like? So let's think about it a little bit. Let's think about it and let me do it orange. They obviously named it using systematic naming and what we have here, we have an ethyl as kind of the core" + }, + { + "Q": "At 5:36, if the penny was thrown up at a velocity of 30 m/s, then it would go up, reach a velocity of 0m/s and by the time it comes down to the same point from which it was thrown, wouldn't the velocity at that point then be different than 30 m/s?", + "A": "It would be -30 m/s, since velocity is a vector. The speed at the bottom would be the same on the way down as on the way up. It has to be, because the acceleration is constant the whole time, and the distance up is the same as the distance down. (ignoring air resistance)", + "video_name": "emdHj6WodLw", + "timestamps": [ + 336 + ], + "3min_transcript": "here, and that might simplify things. If we multiply both sides by 2a, we get-- and I'm just going to switch this to distance, if we assume that we always start at distances equal to 0. di, or initial distance, is always at point 0. We could right 2ad-- I'm just multiplying both sides by 2a-- is equal to vf squared minus vi squared, or you could write it as vf squared is equal to vi squared plus 2ad. I don't know what your physics teacher might show you or written in your physics book, but of these variations will show up in your physics book. The reason why I wanted to show you that previous problem first is that I wanted to show you that you could actually figure out these problems without having to always memorize formulas and resort to the formula. With that said, it's probably not bad idea to memorize some form of this formula, although you should understand how it Now that you have memorized it, or I showed you that maybe you don't have to memorize it, let's use this. Let's say I have the same cliff, and it has now turned purple. It was 500 meters high-- it's a 500 meter high cliff. This time, with the penny, instead of just dropping it straight down, I'm going to throw it straight up at positive 30 meters per second. The positive matters, because remember, we said negative is down, positive is up-- that's just the convention we use. Let's use this formula, or any version of this formula, to figure out what our final velocity was when we hit the bottom of the ground. This is probably the easiest formula to use, because it We can say the final velocity vf squared is equal to the initial velocity squared-- so what's our initial velocity? It's plus 30 meters per second, so it's 30 meters per second squared plus 2ad. So, 2a is the acceleration of gravity, which is minus 10, because it's going down, so it's 2a times minus 10-- I'm going to give up the units for a second, just so I don't run out of space-- 2 times minus 10, and what's the height? What's the change in distance? Actually, I should be correct about using change in distance, because it matters for this problem. In this case, the final distance is equal to minus 500, and the initial distance is equal to 0. The change in distance is minus 500." + }, + { + "Q": "What is beta- plus decay ?\nIt was spoken by sal at 1:38 .", + "A": "Beta decay is a kind of radioactive decay, which occurs so that the nucleus becomes more stable. In beta- decay, a neutron emits an electron to become a proton, so that the nucleus becomes more stable. In beta+ decay, also known as positron emission, a proton emits a positron to become a neutron, to make the nucleus more stable.", + "video_name": "FEF6PxWOvsk", + "timestamps": [ + 98 + ], + "3min_transcript": "What I want to do in this video is give a very high-level overview of the four fundamental forces of the universe. And I'm going to start with gravity. And it might surprise some of you that gravity is actually the weakest of the four fundamental forces. And that's surprising because you say, wow, that's what keeps us glued-- not glued-- but it keeps us from jumping off the planet. It's what keeps the Moon in orbit around the Earth, the Earth in orbit around the Sun, the Sun in orbit around the center of the Milky Way galaxy. So if it's a little bit surprising that it's actually the weakest of the forces. And that starts to make sense when you actually think about things on maybe more of a human scale, or a molecular scale, or even atomic scale. Even on a human scale, your computer monitor and you, have some type of gravitational attraction. But you don't notice it. Or your cell phone and your wallet, there's gravitational attraction. But you don't see them being drawn to each other the way you might see two magnets drawn to each other And if you go to even a smaller scale, you'll see the it matters even less. We never even talk about gravity in chemistry, although the gravity is there. But at those scales, the other forces really, really, really start to dominate. So gravity is our weakest. So if we move up a little bit from that, we get-- and this is maybe the hardest force for us to visualize. Or it's, at least, the least intuitive force for me-- is actually the weak force, sometimes called the weak interaction. And it's what's responsible for radioactive decay, in particular beta minus and beta plus decay. And just to give you an example of the actual weak interaction, if I had some cesium-137-- 137 means it has 137 nucleons. A nucleon is either a proton or a neutron. You add up the protons and neutrons of cesium, you get 137. Now, the weak interaction is what's responsible for one of the neutrons-- essentially one of its quarks flipping and turning into a proton. And I'm not going to go into detail of what a quark is and all of that. And the math can get pretty hairy. But I just want to give you an example of what the weak interaction does. So if one of these neutrons turns into a proton, then we're going to have one extra proton. But we're going to have the same number of nucleons. Instead of an extra neutron here, you now have an extra proton here. And so now this is a different atom. It is now barium. And in that flipping, it will actually emit an electron and an anti-electron neutrino. And I'm not going to go into the details of what an anti-electron neutrino is. These are fundamental particles. But this is just what the weak interaction is. It's not something that's completely obvious to us. It's not the kind of this traditional things pulling" + }, + { + "Q": "at 2:47 sal said anti electron neutrino, well i don't understand what neutrino is?", + "A": "Neutrinos are subatomic particles produced by the decay of radioactive elements and are elementary particles that lack an electric charge.", + "video_name": "FEF6PxWOvsk", + "timestamps": [ + 167 + ], + "3min_transcript": "And if you go to even a smaller scale, you'll see the it matters even less. We never even talk about gravity in chemistry, although the gravity is there. But at those scales, the other forces really, really, really start to dominate. So gravity is our weakest. So if we move up a little bit from that, we get-- and this is maybe the hardest force for us to visualize. Or it's, at least, the least intuitive force for me-- is actually the weak force, sometimes called the weak interaction. And it's what's responsible for radioactive decay, in particular beta minus and beta plus decay. And just to give you an example of the actual weak interaction, if I had some cesium-137-- 137 means it has 137 nucleons. A nucleon is either a proton or a neutron. You add up the protons and neutrons of cesium, you get 137. Now, the weak interaction is what's responsible for one of the neutrons-- essentially one of its quarks flipping and turning into a proton. And I'm not going to go into detail of what a quark is and all of that. And the math can get pretty hairy. But I just want to give you an example of what the weak interaction does. So if one of these neutrons turns into a proton, then we're going to have one extra proton. But we're going to have the same number of nucleons. Instead of an extra neutron here, you now have an extra proton here. And so now this is a different atom. It is now barium. And in that flipping, it will actually emit an electron and an anti-electron neutrino. And I'm not going to go into the details of what an anti-electron neutrino is. These are fundamental particles. But this is just what the weak interaction is. It's not something that's completely obvious to us. It's not the kind of this traditional things pulling associate with the other forces. Now, the next strongest force-- and just to give a sense of how weak gravity is even relative to the weak interaction, the weak interaction is 10 to the 25th times the strength of gravity. And you might be saying, if this is so strong, how come this does it operate on planets or us relative to the Earth? Why doesn't this apply to intergalactic distances the way gravity does? And the reason is the weak interaction really applies to very small distances, very, very small distances. So it can be much stronger than gravity, but only over very, very-- and it really only applies on the subatomic scale. You go anything beyond that, it kind of disappears as an actual force, as an actual interaction." + }, + { + "Q": "At 6:03, how do you know which block it is? (ex. D block)", + "A": "Groups 1-2 are the s block Groups 13-18 are the p block Groups 3-12 are the d block The 2 rows under the main periodic table are the f block The block generally (not always) tells you which orbitals electrons are being filled in to", + "video_name": "UXOcWAfBdZg", + "timestamps": [ + 363 + ], + "3min_transcript": "alkaline earth metals. Once again, they have very similar ... They have very similar properties and that's because they have two valence electrons, two electrons in their outermost shell. Also for them, not as quite as reactive as the alkaline metals. Let me write this out, alkaline earth metals. But for them it's easier to lose two electrons than to try to gain six to get to eight. And so these tend to also be reasonably reactive and they react by losing those two outer electrons. Now something interesting happens as you go to the D block. We studied this when we looked at electron configurations, but if you look at the electron configuration for say scandium right over here, the electron, let me do it in magenta, the electron configuration for scandium, so scandium, scandium's electron configuration It's going to be argon. Then you're going to fill it in we're in the one, two, three, fourth period. It's going to be 4s2. Then we start filling the D block. These are the D block elements here. You have to remember, the D block you backfill. In the D block, this is going to be now 3s1. How many electrons does it have in its outermost shell? Once again its outermost shell is its fourth shell, is its fourth shell. These are, you could argue, higher energy electrons that fills this ... These are filled before that, and there are exceptions to this especially that we see a lot in the D block. This is what's, I guess you could say to some degree, is defining its reactivity. Although in the transition metals, the D block elements, I'm sorry, I made a little mistake there. This is 4s2 3d1. We're backfilling the D block. But these, their outermost electrons are in ... They still have two of those outermost electrons. There, once again, are exceptions in these transition metals right here that for the most part are going in backfilling that D block. Once you've kind of backfilled those D blocks then you come over here and you start filling the P block. For example, if you look at the electron configuration for, let's say carbon, carbon is going to have the same electron configuration as helium, as helium. Then you're going to fill your S block, 2s2, and then 2p one two. So 2p2. How many valence electrons does it have? Well, in its second shell, its outermost shell, it has two plus two. It has four valence electrons. That's going to be true for the things in this group." + }, + { + "Q": "At 7:57, how are there 18 noble gases.\nAren't they just 6(+1)?!", + "A": "Sal says So the noble gasses, that s the other name for the group 18 elements, noble gasses , not that there are 18 noble gasses. Groups in the periodic table are the vertical columns that are numbered from 1-18.", + "video_name": "UXOcWAfBdZg", + "timestamps": [ + 477 + ], + "3min_transcript": "We're backfilling the D block. But these, their outermost electrons are in ... They still have two of those outermost electrons. There, once again, are exceptions in these transition metals right here that for the most part are going in backfilling that D block. Once you've kind of backfilled those D blocks then you come over here and you start filling the P block. For example, if you look at the electron configuration for, let's say carbon, carbon is going to have the same electron configuration as helium, as helium. Then you're going to fill your S block, 2s2, and then 2p one two. So 2p2. How many valence electrons does it have? Well, in its second shell, its outermost shell, it has two plus two. It has four valence electrons. That's going to be true for the things in this group. bonding behavior to silicone, to the other things in its group. We could keep going on, for example, oxygen and sulfur. These would both want to take two electrons from someone else because they have six valence electrons and they want to get to eight. They have similar bonding behavior. You go to this yellow group right over here. These are the halogens. There's special name for them. These are the halogens. These are highly reactive because they have seven valence electrons. They would love nothing more than to get one more valence electron. They love to react. In fact, they especially love to react with the alkali metals over here. Then finally you get to kind of your atomic nirvana in the noble gases here. The noble gases, that's the other name for the group, 18 elements, noble gases. of not being reactive. Why don't they react? Because they have eight valence electrons. They have filled their outermost shell. They don't find the need. They're noble. They're kind of above the fray. They don't find the need to have to react with anyone else." + }, + { + "Q": "At around 5:30, why would it be 4s2?", + "A": "b coz 4s has lower energy than 3d as it is inner orbital more close to the nucleus hence it must get fille first", + "video_name": "UXOcWAfBdZg", + "timestamps": [ + 330 + ], + "3min_transcript": "Well, sodium is going to have the same electron configuration as neon. Then it's going to go 3s1. Once again, it has one valence electron, one electron in its outermost shell. All of these elements in orange right over here, they have one valence electron and they're trying to get to the octet rule, this kind of stable nirvana for atoms. You could imagine is that they're very reactive and when they react they tend to lose this electron in their outermost shell. That is the case. These alkali metals are very, very reactive. Actually they have very similar properties. They're shiny and soft. Because they're so reactive it's hard to find them where they haven't reacted with other things. Let's keep looking at the other groups. If we move one over to the right this group two right over here, these are called the alkaline earth metals. alkaline earth metals. Once again, they have very similar ... They have very similar properties and that's because they have two valence electrons, two electrons in their outermost shell. Also for them, not as quite as reactive as the alkaline metals. Let me write this out, alkaline earth metals. But for them it's easier to lose two electrons than to try to gain six to get to eight. And so these tend to also be reasonably reactive and they react by losing those two outer electrons. Now something interesting happens as you go to the D block. We studied this when we looked at electron configurations, but if you look at the electron configuration for say scandium right over here, the electron, let me do it in magenta, the electron configuration for scandium, so scandium, scandium's electron configuration It's going to be argon. Then you're going to fill it in we're in the one, two, three, fourth period. It's going to be 4s2. Then we start filling the D block. These are the D block elements here. You have to remember, the D block you backfill. In the D block, this is going to be now 3s1. How many electrons does it have in its outermost shell? Once again its outermost shell is its fourth shell, is its fourth shell. These are, you could argue, higher energy electrons that fills this ... These are filled before that, and there are exceptions to this especially that we see a lot in the D block. This is what's, I guess you could say to some degree, is defining its reactivity. Although in the transition metals, the D block elements, I'm sorry, I made a little mistake there. This is 4s2 3d1." + }, + { + "Q": "At 5:35, Sc would have 2, not 3 valence electrons? I'm curious as to why you would not add the 2 and 1 together, since you do that with Carbon (and you add the 2s together to get 4 valence electrons).", + "A": "You do not add the 2 and 1 together because the 1 electron is part of the d block and therefore cannot be a valence electron. The d blocks cannot be used as a valence electron because they are not one of the highest energy furthest out electrons, but p blocks can be used as valence electrons. With carbon you get 4 valence electrons because you add the s and p blocks together.", + "video_name": "UXOcWAfBdZg", + "timestamps": [ + 335 + ], + "3min_transcript": "alkaline earth metals. Once again, they have very similar ... They have very similar properties and that's because they have two valence electrons, two electrons in their outermost shell. Also for them, not as quite as reactive as the alkaline metals. Let me write this out, alkaline earth metals. But for them it's easier to lose two electrons than to try to gain six to get to eight. And so these tend to also be reasonably reactive and they react by losing those two outer electrons. Now something interesting happens as you go to the D block. We studied this when we looked at electron configurations, but if you look at the electron configuration for say scandium right over here, the electron, let me do it in magenta, the electron configuration for scandium, so scandium, scandium's electron configuration It's going to be argon. Then you're going to fill it in we're in the one, two, three, fourth period. It's going to be 4s2. Then we start filling the D block. These are the D block elements here. You have to remember, the D block you backfill. In the D block, this is going to be now 3s1. How many electrons does it have in its outermost shell? Once again its outermost shell is its fourth shell, is its fourth shell. These are, you could argue, higher energy electrons that fills this ... These are filled before that, and there are exceptions to this especially that we see a lot in the D block. This is what's, I guess you could say to some degree, is defining its reactivity. Although in the transition metals, the D block elements, I'm sorry, I made a little mistake there. This is 4s2 3d1. We're backfilling the D block. But these, their outermost electrons are in ... They still have two of those outermost electrons. There, once again, are exceptions in these transition metals right here that for the most part are going in backfilling that D block. Once you've kind of backfilled those D blocks then you come over here and you start filling the P block. For example, if you look at the electron configuration for, let's say carbon, carbon is going to have the same electron configuration as helium, as helium. Then you're going to fill your S block, 2s2, and then 2p one two. So 2p2. How many valence electrons does it have? Well, in its second shell, its outermost shell, it has two plus two. It has four valence electrons. That's going to be true for the things in this group." + }, + { + "Q": "at 1:01 Sal marks the 1st column but he doesn't mark HYDROGEN . Can some tell me the reason for this ?", + "A": "This is because H is not an alkali metal and it doesn t share all the properties of the alkali metals. It doesn t perfectly fit anywhere in the periodic table but it s usually put in Group 1.", + "video_name": "UXOcWAfBdZg", + "timestamps": [ + 61 + ], + "3min_transcript": "- Let's talk a little bit about groups of the periodic table. Now, a very simple way to think about groups is that they just are the columns of the periodic table, and the standard convention is to number them. This is the first column, so that's group one, second column, third group, fourth, fifth, sixth, seventh, eighth, group nine, group 10, 11, 12, 13, 14 15, 16, 17, and 18. As some of ya'll might be thinking, what about these F block elements over here? If we were to properly do the periodic table we would shift all of these that everything from the D block and P block all are right words and make room for these F block elements, but the convention is is that we don't number them. But what's interesting? Why do we go to the trouble about calling one of these columns, about calling these columns a group? This is what's interesting about the periodic table is that all of the elements in a column, but for the most part the elements in the column have very, very, very similar properties. That's because the elements in a column, or the elements in a group tend to have the same number of electrons in their outermost shell. They tend to have the same number of valence electrons. And valence electrons are electrons in the outermost shell they tend to coincide, although there's a slightly different variation. The valence electrons, these are the electrons that are going to react, which tend to be the outermost shell electrons, but there are exceptions to that. There's actually a lot of interesting exceptions that happen in the transition metals in the D block. But we're not gonna go into those details. Let's just think a little about some of the groups that you will hear about and why they react in very similar ways. If we go with group one, group one ... And hydrogen is a little bit of a strange character because hydrogen isn't trying to get to eight valence electrons. Hydrogen in that first shell just wants to get to two Hydrogen is kind of ... It doesn't share as much in common with everything else in group one as you might expect for, say, all of the things in group two. Group one, if you put hydrogen aside, these are referred to as the alkali metals. And hydrogen is not considered an alkali metal. These right over here are the alkali. Alkali metals. Now why do all of these have very similar reactions? Why do they have very similar properties? Well, to think about that you just have to think about their electron configurations. For example, the electron configuration for lithium is going to be the same as the electron configuration of helium, of helium. Then you're going to go to your second shell, 2s1. It has one valence electron. It has one electron in its outermost shell." + }, + { + "Q": "At 6:59, Sal says that Carbon has four valence electrons, but the subscript for the p shell is only 2. I'm confused as to why Sal added on the electrons in the 2s shell to the 2p.", + "A": "Because the 2s electrons are also carbon s valence electrons. For main group elements it will be the electrons in the highest energy shell, both 2s and 2p are in carbon s highest energy shell.", + "video_name": "UXOcWAfBdZg", + "timestamps": [ + 419 + ], + "3min_transcript": "It's going to be argon. Then you're going to fill it in we're in the one, two, three, fourth period. It's going to be 4s2. Then we start filling the D block. These are the D block elements here. You have to remember, the D block you backfill. In the D block, this is going to be now 3s1. How many electrons does it have in its outermost shell? Once again its outermost shell is its fourth shell, is its fourth shell. These are, you could argue, higher energy electrons that fills this ... These are filled before that, and there are exceptions to this especially that we see a lot in the D block. This is what's, I guess you could say to some degree, is defining its reactivity. Although in the transition metals, the D block elements, I'm sorry, I made a little mistake there. This is 4s2 3d1. We're backfilling the D block. But these, their outermost electrons are in ... They still have two of those outermost electrons. There, once again, are exceptions in these transition metals right here that for the most part are going in backfilling that D block. Once you've kind of backfilled those D blocks then you come over here and you start filling the P block. For example, if you look at the electron configuration for, let's say carbon, carbon is going to have the same electron configuration as helium, as helium. Then you're going to fill your S block, 2s2, and then 2p one two. So 2p2. How many valence electrons does it have? Well, in its second shell, its outermost shell, it has two plus two. It has four valence electrons. That's going to be true for the things in this group. bonding behavior to silicone, to the other things in its group. We could keep going on, for example, oxygen and sulfur. These would both want to take two electrons from someone else because they have six valence electrons and they want to get to eight. They have similar bonding behavior. You go to this yellow group right over here. These are the halogens. There's special name for them. These are the halogens. These are highly reactive because they have seven valence electrons. They would love nothing more than to get one more valence electron. They love to react. In fact, they especially love to react with the alkali metals over here. Then finally you get to kind of your atomic nirvana in the noble gases here. The noble gases, that's the other name for the group, 18 elements, noble gases." + }, + { + "Q": "At 6:31, for the cycle pentane, shouldn't there be 12 hydrogens, not 10 (because of the alkane formula Cn H2n+2)?", + "A": "For a normal pentane it would be 12 hydrogens, like: CH3CH2CH2CH2CH3 if you would now want to make a ring out of this ordinary pentane, you would have to bond the two CH3 s together. In order to do so, both would hove to lose a bond first, which are both bonds to hydrogen atoms. So that leaves 10 hydrogens for cyclic pentane.", + "video_name": "NRFPvLp3r3g", + "timestamps": [ + 391 + ], + "3min_transcript": "be pentane. But if I have five carbons and they form a ring, so let me draw it. So it's one, two, three, four, five carbons and it forms a ring. Let me make the drawing a little bit better. So it's really, I'm just drawing a pentagon. But notice, this has five carbons on it. I can draw the carbons here. Carbon, carbon, carbon, carbon, carbon. And just as a review, what you don't see is the hydrogens Each of these guys have two bonds, so they must have two bonds with something else and those are going to be with And I'lll draw it here just as a bit of a review, but you notice very quickly, the drawing gets extremely messy when you draw the two hydrogens on each of these carbons. So it's a little bit over-- maybe I shouldn't be doing that. So it becomes very messy when you draw the hydrogens, so it's better to just assume that they're there. If we don't draw all four bonds of the carbon, the other two bonds are going to be with hydrogen. So here, you might say, OK, this is an alkane, because I All of these are single bonds with the carbon. I have five carbons, so you might say this is pentane, but you have to think about one more thing. It's in a ring, so we add the prefix cyclo- to it. So this is, because it's a ring, we write cyclopentane. So let me just break that apart. This tells us that we're dealing with a ring. You see that this is a ring right there. This tells us that we're dealing with five carbons, and then this tells us right here, the -ane part, that tells us that they are all single bonds. All carbon-carbon single bonds. No double or triple bonds. All single bonds. Let's start with the word and let's see if we can figure out what the actual structure would look like. So what is this telling me? This tells me I'm dealing with a ring. That is a ring. It's going to have a ring structure. It's going to have nine carbons, nine C's, and then it's an alkane, so they're all going to be single bonds. So if I want to draw it, I want to draw nine carbons in a ring, it's not a trivial thing to draw. I'll try my best, so let's see, that's one, two, three, four, five, six, seven, eight. Let's see, let me draw it. I'll try a little a better shot at it. So, let's see, you have one, two, three, four, five, six," + }, + { + "Q": "at around 10:00,\nwhy doesn't the fluoride anion react with H+ and form HF??", + "A": "The Fluoride is a very small anion, and indeed, the smallest anion. So, It s electrons will be held very near to it s nucleus. We have to supply high amount of energy to make it bond.It ll be less reactive than Iodide, chloride and bromide which are considerably bigger than Fluoride and they can share their electrons to bond using lesser Energy supply.", + "video_name": "Z9Jh-Q59xso", + "timestamps": [ + 600 + ], + "3min_transcript": "Sn2-type reaction. In a protic solvent, what happens is that the things that are really electronegative and really small, like a fluoride anion-- let me draw a fluoride anion. In a protic solvent, what's going to happen is it's going to be blocked by hydrogen bonds. It's very negative, right? It has a negative charge. And it's also tightly packed. As you can see right here, its electrons are very close, tied in. It's a much smaller atom or ion, in this case. If we looked at iodide, iodide has 53 electrons, many orbitals. Actually, iodide would have 54. It would have the same as iodine plus one. Fluoride will have 10 electrons, nine from fluorine plus it gains another one, so it's a much smaller atom. So when you have water hanging around it, let's say you have That has a negative charge. Water is polar. Actually, both of these are polar, so I should write down polar for both of these. This is a polar protic solvent. This is a polar aprotic solvent. In this case, water is still more electronegative than the carbon, so it still has a partial negative charge. These parts still have a partial negative. Water still has a partial negative charge. The hydrogen has a partial positive charge so it is going to be attracted to the fluorine. This is going to happen all around the fluorine. And if these waters are attracted to the fluorine in kind of forming a tight shell around it, it makes it hard for fluorine to react. So it's a worse nucleophile than, say, iodide or hydroxide in a polar protic solvent. Hydroxide has the same issue. compare them, iodide is much bigger. Maybe I'll draw it like this. I'll draw its valence shell like this. It's a much bigger ion. It has all these electrons in here. And so, it still will form hydrogen bonds with the water. It still will form hydrogen bonds with the hydrogen end of the water because they're partially positive, but it's going to be less tightly packed. And on top of that, iodide is more polarizable, which means that its electron cloud is so big and the valence electrons are so far away from the nucleus that they can be influenced by things and then be more likely to react. So let's say this iodide is getting close to a carbon that has a partial positive charge." + }, + { + "Q": "When you order the halides in order of decreasing nucleophilicity, where would OH- go? 12:33", + "A": "Good question, Sal actually answers this in the next video. In short it depends on whether the solvent is protic or not. In short, it is a stronger nucleophile than all the halides in an aprotic solvent, and not quite as strong as iodide in a protic solvent.", + "video_name": "Z9Jh-Q59xso", + "timestamps": [ + 753 + ], + "3min_transcript": "it's attached to three hydrogens. We've seen that this will have a partial negative charge. It's more electronegative than the carbon, which will have a partial positive charge. When this guy, this big guy with the electrons really far away, gets close to this, more of the electron cloud is going to be attracted to the partial positive charge. It'll get distorted a little bit and so it is more likely to react in a polar protic solvent. Fluoride, on the other hand, is very tightly packed, blocked by the hydrogen bonds. It's less likely to react. If you were to look at the Periodic Table, if you look at just the halogens in a polar protic solvent, the halides-- this would be the ion version of the halogens-- the halide's iodide will be the best nucleophile. let me write this down-- we have a situation where the iodide is the best nucleophile, followed by bromide, followed by chloride, and then last of all is the fluoride. The exact opposite is true in an aprotic solvent. In an aprotic solvent, the fluoride, which is-- fluorine is far more electronegative. Fluoride is more basic. It will be more stable if it is able to form a bond with something than iodide. Iodide is pretty stable. If you look at a hydrogen iodide, it's actually a highly So iodide itself, the conjugate base of hydrogen iodide, is going to be a very bad base. When you're dealing with an aprotic solvent, you go in the direction of basicity. We're going to learn in the next video that actually basicity and nucleophilicity are related, but they aren't the same concept. We're going to talk about that in a little bit. If you're in an aprotic solvent, you're not reacting with the solvent as much. And then in this situation, fluoride is actually the best nucleophile, followed by chloride, followed by bromide, followed by iodide. So here, you're going in the direction of basicity. This is the best. This is the worse in an aprotic solvent. If it was in a protic solvent, this is flipped around. This becomes the best and this becomes the worst." + }, + { + "Q": "At around 12:35 the Professor talks about transposons and conjugation. During which he said trasposons can jump from cell to cell, what I would like to know is if this process has somethig in relation with synapsis.", + "A": "No, this is a direct linking of bacteria through pili to transfer genetic material, while the signals sent across synapsis are chemical. Neuro transmitters are released from the axon terminals (end of the neurons axon) of one neuron, which then travel across the synapse to the dendrites of the recieving neuron.", + "video_name": "TDoGrbpJJ14", + "timestamps": [ + 755 + ], + "3min_transcript": "by the time you have a million bacteria, you'll have a thousand mutations. So they have mutations,but they also have this form. I don't want to call it sexual reproduction, because it's not sexual reproduction. They don't form gametes and the gametes don't fertilize each other and then produces a zygote. But two bacteria can get near each other and then one of their piluses--I'll do that right here. So the piluses are these little structures on the side of the bacteria They're these little tubes,really. One of the piluses can connect from one bacteria to another, and then essentially you have a mixing of what's inside one bacteria with another. So let me draw their nucleoids. And then they have these other pieces of just DNA that hangs out called plasmids. Maybe this guy has got this extra neat plasmid. He got it from someplace, and it's making him able to do things that this guy couldn't do. Maybe this is the R plasmid,which is known for making a bacteria resistant to a lot of antibiotics. And what happens is,that bacteria--and actually, there's mechanisms where the bacteria know that,hey, this guy doesn't have the R plasmid. And we're just beginning to understand how it actually works, but this will actually replicate itself and give this guy a version of the R plasmid. You could also have these things,transposons, and I should make a whole video on this because we have transposons,too. But there's parts of DNA that can jump from one part of a fragment of DNA to another, and these can also end up in the other one. So what you have is kind of--it's not formal sexual reproduction, but what you essentially have is a connection, and these bacteria are just constantly swapping DNA with each other and DNA is jumping back and forth,so you can imagine all sorts of combinations of DNA happen even within what you used to call one bacterial species and become resistant to different things. If this makes it resistant to an antibiotic, then it can kind of spread the information to produce those resistant proteins or whatever to the other bacteria. So this is kind of a form of introducing variation. And so when you transfer stuff via this pilus,or the plural is pili, this is called conjugation,bacterial conjugation. Now,the last thing I want to talk about, because it's something that you've heard a lot about,are antibiotics. A lot of people,they get sick. The first thing they want to get is an antibiotic. And an antibiotic is just a whole class of chemicals and compounds, some of them naturally derived,some of them not,that kill bacteria. So now if someone is undergoing a surgery and they get a cut, instead of them having to worry about getting an infection," + }, + { + "Q": "At 8:25 the Professor talks about Archaea cells, what type of environment are these unique cells found in ?", + "A": "Thank You that makes everything clear", + "video_name": "TDoGrbpJJ14", + "timestamps": [ + 505 + ], + "3min_transcript": "In bacteria,which are what people originally just classify it on whether or not you have a nucleus, in bacteria,there is no membrane surrounding the DNA. So what they have is just a big bundle of DNA. They just have this big bundle of DNA. It's sometimes in a loop all in one circle called a nucleoid. Now,whenever we look at something,and we say,oh, we have this thing;it doesn't;there's this assumption that somehow we're superior or we're more advanced beings. But the reality is that bacteria have infiltrated far more ecosystems in every part of the planet than Eukarya have, and there's far more diversity in bacteria than there is in Eukarya. these are the more successful organisms. If a comet were to hit the Earth--God forbid-- the organisms more likely to survive are going to be the bacteria than the Eukarya,than the ones with the larger--not always larger, but the organisms that do have this nucleus and have membrane-bound organelles like mitochondria and all that. We'll talk more about it in the future. Bacteria,for the most part,are just big bags of cytoplasm. They have their DNA there. They do have ribosomes because they have to code for proteins just like the rest of us do. Some of those proteins,they'll make some from-- bacteria,they'll make these flagella, which are tails that allow them to move around. They also have these things called pili. Pili is plural for pilus or pee-lus,so these pili. And we'll see in a second that the pili are kind of introducing genetic variation into their populations. Actually,I'll take a little side note here. I'm pointing out bacteria as not having a cell wall. There's actually another class that used to be categorized as type of a bacteria,and they're called Archaea. I should give them a little bit of justice. They're always kind of the stepchild. They used to be called Archaea bacteria, but now people realize,they've actually looked at the DNA, because when they originally looked at these,they said,OK, these guys also have no nucleus and a bunch of DNA running around. These must be a form of bacteria. But now that we've actually been able to look into the DNA of the things,we've seen that they're actually quite different. But all of these,both bacteria and Archaea,are considered prokaryotes. And this just means no nucleus." + }, + { + "Q": "Isn't the direction of angular velocity supposed to be perpendicular to both linear velocity and the radius? At 4:10 David says that the direction of angular velocity is anti-clockwise?", + "A": "The angular velocity vector is perpendicular to the linear velocity and the radius But that still leaves two possible directions for it to point. You need to know the direction of the spin in order to know which of those two directions is the one. Specifying the direction of the spin is the same as specifying which direction the velocity vector points.", + "video_name": "garegCgMxxg", + "timestamps": [ + 250 + ], + "3min_transcript": "in a straight line, the regular displacement was a defined b, the final position minus the initial positions, which we called delta x. And that's just usually called the displacement, which is measured in meters. Okay, so now we know how to quantify the amount of angle that this ball has rotated through, but another quantity that might be useful is the rate at which it's traveling through that angle. Just like up here, knowing about the displacement is good, but you might want to know about the rate that it's being displaced. In terms of regular linear quantities that was called the velocity of the ball, and it was defined to be the displacement per time. So down here we'll define a similar quantity, but it's going to be the angular velocity, which is defined analogously to the regular velocity. If regular velocity is displacement per time, the angular velocity is going to be the angular displacement per time. And the symbol we used to represent angular velocity is the Greek letter omega, which looks like a w, but it's really the Greek letter omega. are going to be radians per seconds. Since delta theta, the angular displacement is in radians, and the time is in seconds. Just like how regular velocity had units of meters per second, angular velocity has units of radians per second. What is angular velocity mean? What is this omega? It represents the rate at which an object is changing its angle in time. So let's say the tennis ball starts here, and it's going through a circle at this leisurely rate, that means the rate at which it's changing its angle is very small and it has a very small omega. Whereas if you had this tennis ball going through a circle very fast, the rate at which it's going in a circle would be large and that means the angular velocity and omega would also be large. So the velocity and the angular velocity are related, they're not equal because the velocity gives you how many meters per second something is going through, and the angular velocity gives you how many radians per second it's going through, but if it's got a larger angular velocity, it's going to have a larger velocity as well. angular velocity is also a vector, so I'll put an arrow over this omega. Which way does it point? Technically speaking, you'd use the same right hand rule you use to determine the direction of the angular displacement. But again if it's rotating counter clockwise, we can just consider that to be positive, and if it's rotating clockwise, we can consider that to be a negative omega, or a negative angular velocity. So let me get rid of these, and let's define our last angular motion variable. You can probably guess what it is. There's regular displacement and there's angular displacement. There's regular velocity and there's angular velocity. And then the next logical step in this motion variable sequence would be the acceleration, which was defined for regular variables to be the change in velocity over the change in time. So we'll define an analogous angular quantity that would be the angular acceleration. And it's going to be defined to be, instead of change in velocity over change in time, it's going to be the change in the angular velocity over the change in time. And the letter we use to denote angular acceleration" + }, + { + "Q": "At 6:46 you mentioned that Si can make 5 bonds because it has d orbitals. Being in the 3rd period and above the d block, how does Si have d orbitals?", + "A": "It is because in the 3rd shell there is 3s2 3p6 3d10 in a full shell and even before the transition metals there are d orbitals since while the 3s and 3p orbitals fill the 3d orbitals form from the electrons having more places to go. This makes all elements past neon hypervalent and thus Iron is able to have a +7 oxidation state instead of just +3 like you would expect from an oxide.", + "video_name": "KsdZsWOsB84", + "timestamps": [ + 406 + ], + "3min_transcript": "So I'm going to go ahead and draw in a fluoride anion here, which is normally an extremely poor nucleophile. So it's actually selective for silicon. So if the fluoride functions as a nucleophile, it's going to attack the silicon here. And it could do this for a couple of reasons. So let's talk about those reasons here. So first of all, the silicon is bonded to some carbons. And silicon is bigger than carbon, if you look at where it is in the periodic table. And so the silicon carbon bonds are longer than we're used to seeing. And that means that there's decreased steric hindrance. So the silicon is a little bit more exposed, and that allows the fluoride anion to attack it a little more. So another factor that allows this is silicon is in the third period on the periodic table. So it has vacant d orbitals. between the fluorine and the silicon. So let me go ahead and draw what we would get after the fluoride attacks the silicon. So we would have this portion of the molecule. And we would have our oxygen bonded to our silicon in this intermediate. And now we could show the fluorine bonded to the silicon, And the silicon is still bonded to two methyl groups and also a tert-butyl group, like that. This will give the silicon a negative 1 formal charge. And it looks a little bit weird, because we see silicon has five bonds to it. But that's, again, it's OK because of where silicon is on the periodic table. It has those has those d orbitals, and so forming five bonds for an intermediate is OK. It's OK for it to have an expanded octet. Another reason why fluoride can attack the silicon very well is because the bond that forms between fluorine and silicon happens to be very strong. And we can finish up by kicking these electrons back onto the oxygen and protonating and forming our target compound. So we would go ahead and form our target compound here. So we would get back our alcohol, like that. And we also successfully added on this portion of the molecule on the right. And then we would also form-- we would now have the fluorine bonded to the silicon, like that. So we selectively removed our protecting group and we formed our target compound. And so that's the idea of a protecting group. It allows you to protect one area of the molecule and react with another area of the molecule. And it's also nice to have it easily removed to get back your target molecule." + }, + { + "Q": "at 6:01 , why did sal draw the graph as \"discontinuous function\" ?!\ngenerally , when i draw a graph like this case how can i decide if i'll draw it as continuous or discontinuous function ?", + "A": "The given function was defined piece-wise with a jump discontinuity at t=2. Since it was defined that way (with the jump discontinuity), you can only draw it that way.", + "video_name": "6FTiHeius1c", + "timestamps": [ + 361 + ], + "3min_transcript": "I'll use on this yellow. Let's say on a rate function, that is... Let me make it a little bit interesting. Let's say it's one meter per second, for our time is zero is less than or equal to time which is less than or equal to two seconds. Honestly, this is all in seconds where we're of time. There's two meters per second, for t is greater than two seconds. What's that going to look like? Actually, try to graph it yourself, and just say, \"Well, what is the total change in distance \"over the first, let's say five seconds?\" We want to do delta t over, not the first four seconds but the first five seconds. Well, let's graph it. Let's graph it. So this is one meter per second. One meter per second. That is two meters per second. That's in meters per second. That's my rate axis. One, they're obviously not of the same scale, three, four, five. What is this rate function look like? Well, my rate is one meter per second between time is zero and two, including two seconds, and then the rate jumps. Nothing can accelerate instantly like this. You'd need an infinite force or an infinitely small mass I guess to, or maybe there's something that's thinking about... Anyway, I only get two complex here but this is unrealistic mode. It's not typical for something to just have spontaneous velocity increase like that but let's just go with it. Then after the two seconds, we are at a constant rate of two meters per second. Now, what is our total change in distance over the first five seconds here? or we can break up the problem. We could say, \"Well, over the first two seconds.\" Change in time is two seconds, times our constant rate over those two seconds. It's going to be two seconds times one meter per second. Well, that's going to give us two meters. So this here is going to be, shall we do that in orange color, that's going to give us two meters there, and then we look at the next section. Our change in time here is three seconds, and then we multiply that, times our constant two meters per second. That's going to give us an area of six. If we look at the units, in both cases, we're multiplying seconds times meters per second which is going to give us meters. This is going to be two plus six meters or eight meters. So hopefully this is giving you the intuition that the area under the rate curve or the rate function is" + }, + { + "Q": "At 7:55, Khan says that if the radius goes down, then the tangential velocity goes up.\n\nConsidering a ball on a string wrapping itself around and around, its velocity is increasing even though there is no torque acting upon it. But according to the Law of Conservation of Momentum, the momentum is conserved. But if the tangential velocity increased, then mv increases and so does the linear momentum.\n\nCan someone explain to me why this is? How can the linear momentum change without torque?", + "A": "Torque is rotational and linear momentum is linear, so torque can t change linear momentum, it can only change angular momentum. To apply conservation of momentum you have to consider the system of the ball and the object to which it is tied. The total momentum of those two is conserved. If you ignore the object to which it is tied, then you can consider the string to be exerting an outside force on the ball. The string is accelerating the ball and changing its momentum.", + "video_name": "nFSMu3bxXVA", + "timestamps": [ + 475 + ], + "3min_transcript": "we'll do torque in pink. If torque is equal to zero, if there's no net torque going on here, if the magnitude of torque is equal to zero, then we will have no change. No change in angular momentum. And we will look at that mathematically in a few seconds. But just from this there's a very interesting thing that arises. And something that you might have observed at even the Olympics or in other things. And this is the idea that you can, by changing your radius, you could actually change your tangential velocity. And as we've seen in previous videos, tangential velocity is closely related to angular velocity. So let's explore that a little bit. So when we write it in the world where, well actually you see it straight out of this, if L is constant, if r went down, So let me rewrite it over here. So L, whoops. L is equal to mass times tangential velocity, or actually well yeah, tangential velocity, or the velocity that's perpendicular to the radius, times the radius. Now what happens, if we assume that this is constant, if we assume that there's no torque, so we're in this world. So this over here is going to be constant. So what happens if we were to reduce r? Somehow this wire started to reel in a little bit or started to wrap around here, and that's actually a reasonable thing, you could imagine as it rotates it starts to wrap around this thing so the wire gets shorter. So if r goes down, and this is constant, the mass isn't going to change, Well if L is constant, mass isn't changing, or the velocity that's perpendicular to the radius is going to go up. And if we wanna think about it, we can think about it in terms of angular velocity, we know that angular velocity, which we would measure in radians per second, we would use the symbol omega, omega is defined, and we go into much more depth in this in other videos, as tangential velocity, the magnitude of the velocity that is perpendicular to the radius, divided by the radius. Or if you solve for tangential velocity, you get v is equal to is equal to omega r. And so if you substitute back into this, really this definition for angular momentum," + }, + { + "Q": "does p always mean -log? 12:17", + "A": "Well, when it s in front of something and it s a LOWERCASE p, then yes, it s always -log. Hope I answered your question well! ;D", + "video_name": "LJmFbcaxDPE", + "timestamps": [ + 737 + ], + "3min_transcript": "With hydrogen and its conjugate base. We know that there's an equilibrium constant for this. We've done many videos on that. The equilibrium constant here is equal to the concentration of our hydrogen proton times the concentration of our conjugate base. When I say concentration, I'm talking molarity. Moles per liter divided by the concentration of our weak acid. Now. Let's solve for hydrogen concentration. Because what I want to do is I want to figure out a formula, and we'll call it the Hendersen-Hasselbalch Formula, which a lot of books want you to memorize, which I don't think you should. I think you should always just be able to go from this kind of basic assumption and get to it. But let's solve for the hydrogen so we can figure out a relationship between pH and all the other stuff that's in this formula. we can multiply both sides by the reciprocal of this right here. And you get hydrogen concentration. Ka times --I'm multiplying both sides times a reciprocal of that. So times the concentration of our weak acid divided by the concentration of our weak base is equal to our concentration of our hydrogen. Fair enough. Now. Let's take the negative log of both sides. So the negative log of all of that stuff, of your acidic equilibrium constant, times HA, our weak acid divided by our weak base, our hydrogen concentration. Which is just our pH, right? Negative log of hydrogen concentration is --that's the definition of pH. I'll write the p and the H in different colors. You know a p just means negative log. Minus log. That's all. Base 10. Let's see if we can simplify this any more. So our logarithmic properties. We know that when you take the log of something and you multiply it, that's the same thing as taking the log of this plus the log of that. So this can to be simplified to minus log of our Ka minus the log of our weak acid concentration divided by its conjugate base concentration. Is equal to the pH." + }, + { + "Q": "9:38, in the correction, isn't the conjugate of a weak acid always a weak base? Like HCN (weak acid) has conjugate CN^- (weak base)", + "A": "HCl is a strong acid. The conjugate base, the chloride ion, is a very weak base, so no. However, the ionisation of HCl in water is not a reversible reaction (it goes to completion), so its not appropriate for a discussion of buffers, as they involve equilibrium reactions..", + "video_name": "LJmFbcaxDPE", + "timestamps": [ + 578 + ], + "3min_transcript": "turning into a weak acid and producing more OH. So the pH won't go down as much as you would expect if you just threw this in water. This is going to lower the pH, but then you have more OH that could be produced as this guy grabs more and more hydrogens from the water. So the way to think about it is it's kind of like a cushion or a spring in terms of what a strong acid or base could do to the solution. And that's why it's called a buffer. Because it provides a cushion on acidity. If you add a strong base to water, you immediately increase its pH. Or you decrease its acidity dramatically. But if you add a strong base to a buffer, because of Le Chatelier's Principal, essentially, you're not going to affect the pH as much. Same thing. If you add and acid to that same buffer, it's not going to affect the pH if you had thrown that acid in water because the equilibrium reaction can always kind of refill the amount of OH that you lost if you're adding acid, or it can refill the amount of hydrogen you lost if you're adding a base. And that's why it's called buffer. It provides a cushion. So it give some stability to the solution's pH. The definition of a buffer is just a solution of a weak acid in equilibrium with its conjugate weak base. That's what a buffer is, and it's called a buffer because it provides you this kind of cushion of pH. It's kind of a stress absorber, or a shock absorber for the acidity of a solution. Now, with that said, let's explore a little bit the math of a buffer, which is really just the math of a weak acid. So if we rewrite the equation again, so HA is in equilibrium. With hydrogen and its conjugate base. We know that there's an equilibrium constant for this. We've done many videos on that. The equilibrium constant here is equal to the concentration of our hydrogen proton times the concentration of our conjugate base. When I say concentration, I'm talking molarity. Moles per liter divided by the concentration of our weak acid. Now. Let's solve for hydrogen concentration. Because what I want to do is I want to figure out a formula, and we'll call it the Hendersen-Hasselbalch Formula, which a lot of books want you to memorize, which I don't think you should. I think you should always just be able to go from this kind of basic assumption and get to it. But let's solve for the hydrogen so we can figure out a relationship between pH and all the other stuff that's in this formula." + }, + { + "Q": "At 4:30, why do you decrease the poh when you increase the oh?", + "A": "Because pH + pOH = 14.00 at 25 Celcius", + "video_name": "LJmFbcaxDPE", + "timestamps": [ + 270 + ], + "3min_transcript": "Right? So for example, if you had 1 mole oh hydrogen molecules in your solution right when you do that, all this is going to react with all of that. And the OHs are going to react with the Hs and form water, and they'll both just kind of disappear into the solution. They didn't disappear, they all turned into water. And so all of this hydrogen will go away. Or at least the hydrogen that was initially there. That 1 mole of hydrogens will disappear. So what should happen to this reaction? Well, know this is an equilibrium reaction. So as these hydrogen disappear, because this is an equilibrium reaction or because this is a weak base, more of this is going to be converted into these two products to kind of make up for that loss of hydrogen. So this hydrogen goes down initially, and then it starts getting to equilibrium very fast. But this is going to go down. This is going to go up. And then this is going to go down less. Because sure, when you put the sodium hydroxide there, it just ate up all of the hydrogens. But then you have this -- you can kind of view as the spare hydrogen capacity here to produce hydrogens. And when these disappear, this weak base will disassociate more. The equilibrium we'll move more in this direction. So immediately, this will eat all of that. But then when the equilibrium moves in that direction, a lot of the hydrogen will be replaced. So if you think about what's happening, if I just threw this sodium hydroxide in water. So if I just did NaOH in an aqueous solution so that's just throwing it in water -- that disassociates completely into the sodium cation and hydroxide anion. increase the quantity of OHs by essentially the number of moles of sodium hydroxide you're adding, and you'd immediately increase the pH, right? Remember. When you increase the amount of OH, you would decrease the pOH, right? And that's just because it's the negative log. So if you increase OH, you're decreasing pOH, and you're increasing pH. And just think OH-- you're making it more basic. And a high pH is also very basic. If you have a mole of this, you end up with a pH of 14. And if you had a strong acid, not a strong base, you would end up with a pH of 0. Hopefully you're getting a little bit familiar with that concept right now, but if it confuses you, just play around with the logs a little bit" + }, + { + "Q": "3:15 Sal says a force is needed to be applied in order to move the particle towards the positive side. How would this force be applied?", + "A": "Well, there might be another electrical force being applied from the other side. There might be something pushing it. There is simply some form of a force. It s abstract. If you were supposed to know it, Sal would tell you.If you need to know it in some other scenario, you ll be given some way to find it out. Unless it s in life. That s one game where no variables are necessarily provided.", + "video_name": "zqGvUbvVQXg", + "timestamps": [ + 195 + ], + "3min_transcript": "Only that particle can have energy. Electrical potential, or electric potential, this is associated with a position. So, for example, if I have a charge and I know that it's at some point with a given electric potential, I can figure out the electric potential energy at that point by just multiplying actually this value by the charge. Let me give you some examples. Let's say that I have an infinite uniformly charged plate. So that we don't have to do calculus, we can have a uniform electric field. Let's say that this is the plate. I'll make it vertical just so we get a little bit of change of pace, and let's say it's positively charged plate. And let's say that the electric field is constant, right? It's constant. No matter what point we pick, these field vectors should all in magnitude it's pushing out, because we assume when we draw field lines that we're using a test charge with a positive charge so it's pushing outward. Let's say I have a 1-coulomb charge. Actually, let me make it 2 coulombs just to hit a point home. Say I have a 2-coulomb charge right here, and it's positive. A positive 2-coulomb charge, and it starts off at 3 meters away, and I want to bring it in 2 meters. I want to bring it in 2 meters, so it's 1 meter away. So what is the electric-- or electrical-- potential energy difference between the particle at this point and at this point? amount of work, as we've learned in the previous two videos, we need to apply to this particle to take it from here to here. So how much work do we have to apply? We have to apply a force that directly-- that exactly-- we assume that maybe this is already moving with a constant velocity, or maybe we have to start with a slightly higher force just to get it moving, but we have to apply a force that's exactly opposite the force provided by Coulomb's Law, the electrostatic force. And so what is that force we're going to have to apply? Well, we actually have to know what the electric field is, which I have not told you yet. I just realized that, as you can tell. So let's say all of these electric field lines are 3 newtons per coulomb. So at any point, what is the force being exerted from this" + }, + { + "Q": "at 5:15 you say the charge closer to the plate has a higher PE. Why is this if the E field is uniform? if F= qE then wouldnt the force be the same at both pts?", + "A": "Hi, The force on the charge of course would not vary since F=qE and Sal did not say that the Force is higher. What he said that the P.E is higher. WE know that P.E is the work done in taking the charge to that point. Also Work done is Force times the Distance. Therefore is force is constant the distance will determine the P.E. And from the video it s clear that the particle will be at a higher potential near the plate since it has to travel a greater distance to reach there. Hope it helps Cheers", + "video_name": "zqGvUbvVQXg", + "timestamps": [ + 315 + ], + "3min_transcript": "amount of work, as we've learned in the previous two videos, we need to apply to this particle to take it from here to here. So how much work do we have to apply? We have to apply a force that directly-- that exactly-- we assume that maybe this is already moving with a constant velocity, or maybe we have to start with a slightly higher force just to get it moving, but we have to apply a force that's exactly opposite the force provided by Coulomb's Law, the electrostatic force. And so what is that force we're going to have to apply? Well, we actually have to know what the electric field is, which I have not told you yet. I just realized that, as you can tell. So let's say all of these electric field lines are 3 newtons per coulomb. So at any point, what is the force being exerted from this Well, the electrostatic force on this particle is equal to the electric field times the charge, which is equal to-- I just defined the electric field as being 3 newtons per coulomb times 2 coulombs. It equals 6 newtons. So at any point, the electric field is pushing this way 6 newtons, so in order to push the particle this way, I have to completely offset that, and actually, I have to get it moving initially, and I'll keep saying that. I just want to hit that point home. So I have to apply a force of 6 newtons in the leftward direction and I have to apply it for 2 meters to get the point here. So the total work is equal to 6 newtons times 2 meters, which is equal to 12 newton-meters or 12 joules. So we could say that the electrical potential energy-- The electrical potential energy difference between this point and this point is 12 joules. Or another way to say it is-- and which one has a higher potential? Well, this one does, right? Because at this point, we're closer to the thing that's trying to repel it, so if we were to just let go, it would start accelerating in this direction, and a lot of that energy would be converted to kinetic energy by the time we get to this point, right? So we could also say that the electric potential energy at this point right here is 12 joules higher than the electric potential energy at this point. Now that's potential energy. What is electric potential? Well, electric potential tells us essentially how much work is necessary per unit of charge, right? Electric potential energy was just how much total work is needed to move it from here to here." + }, + { + "Q": "at 4:50, what does meters have to do with electricity?", + "A": "The equation for work is Force times Distance equals work. The 2 meters represents the distance of the charge.", + "video_name": "zqGvUbvVQXg", + "timestamps": [ + 290 + ], + "3min_transcript": "amount of work, as we've learned in the previous two videos, we need to apply to this particle to take it from here to here. So how much work do we have to apply? We have to apply a force that directly-- that exactly-- we assume that maybe this is already moving with a constant velocity, or maybe we have to start with a slightly higher force just to get it moving, but we have to apply a force that's exactly opposite the force provided by Coulomb's Law, the electrostatic force. And so what is that force we're going to have to apply? Well, we actually have to know what the electric field is, which I have not told you yet. I just realized that, as you can tell. So let's say all of these electric field lines are 3 newtons per coulomb. So at any point, what is the force being exerted from this Well, the electrostatic force on this particle is equal to the electric field times the charge, which is equal to-- I just defined the electric field as being 3 newtons per coulomb times 2 coulombs. It equals 6 newtons. So at any point, the electric field is pushing this way 6 newtons, so in order to push the particle this way, I have to completely offset that, and actually, I have to get it moving initially, and I'll keep saying that. I just want to hit that point home. So I have to apply a force of 6 newtons in the leftward direction and I have to apply it for 2 meters to get the point here. So the total work is equal to 6 newtons times 2 meters, which is equal to 12 newton-meters or 12 joules. So we could say that the electrical potential energy-- The electrical potential energy difference between this point and this point is 12 joules. Or another way to say it is-- and which one has a higher potential? Well, this one does, right? Because at this point, we're closer to the thing that's trying to repel it, so if we were to just let go, it would start accelerating in this direction, and a lot of that energy would be converted to kinetic energy by the time we get to this point, right? So we could also say that the electric potential energy at this point right here is 12 joules higher than the electric potential energy at this point. Now that's potential energy. What is electric potential? Well, electric potential tells us essentially how much work is necessary per unit of charge, right? Electric potential energy was just how much total work is needed to move it from here to here." + }, + { + "Q": "At 10:54, why did Linneas develop the division in the first place? What was his purpose?", + "A": "He simply wanted to organize the cluttered jumble of species.", + "video_name": "oHvLlS_Sc54", + "timestamps": [ + 654 + ], + "3min_transcript": "we are in the phylum chordates. And chordates, we're actually in the subphylum, which I didn't write here, vertebrates, which means we have a vertebra. We have a spinal column with a spinal cord in it. Chordates are a little bit more general. Chordates is a phylum where-- kind of the arrangement of where the mouth is, where are the digestive organs, where the anus is, where the spinal column is, where are the brains, where are the eyes, where are the mouth. They're kind of all in the same place. And if you think about it, everything I've listed here kind of has the same general structure. You have a spinal column. You have a brain. You have a mouth. Then the mouth leads to some type digestive column. And at the end of it, you have an anus over there. And you have eyes in front of the brain. And so this is a general way-- and I'm not being very rigorous here, is how you describe a chordate. And to show a chordate that is not a mammal, you would just have to think of a fish or sharks. So this right over here is a non-mammal chordate Now, let's go even broader. As you'll see, now we're going to things that are very, very not human-like. So you go one step broader. Now we're in Animalia, the kingdom of animals. And this is the broadest category that Carl Linnaeus thought about. Well actually, he did go into trees as well. But when you think of kingdom animals and you think of things that aren't chordates, you start going into things like insects. And you start going into things like jellyfish. If you go even broader, now we're talking about the domain. You go to Eukarya. So these are all organisms that have cells. And inside those cells, they have complex structures. So if you're a Eukarya, you have cells with complex structures. If you're a Prokarya, you don't have complex structures inside your cell. But other Eukarya that are not animals include things like plants. And obviously, I'm giving no justice It could be just as rich or richer than everything I've drawn over here. This is just a small fraction of the entire tree of life, but let's go even broader than that. So if you go even broader than that, you say, well, what's a kind of life form that isn't Eukarya, that wouldn't have these more complex cell structures, the mitochondria in the cells, the cell nucleuses? Then you just have to think about something like bacteria. And if you want to go even broader, there's things like viruses that you could even debate whether they really even are life, because they are dependent on other life forms for their actual reproduction. But they do have genetic material, like everything else. And that, to me, is kind of a mind-blowing idea. As different as a plant is-- look at a house plant that is in your house right now or the tree when you walk home or bacteria or this jellyfish. There is a commonality in that we all have DNA. And that DNA, for the most part, replicates in a very, very, very, similar way. So it's actually crazy that we actually even are related or that we even do have a common ancestor with some" + }, + { + "Q": "what is sal saying at 6:21 ? how we can find kinetic energy from position time graph ?", + "A": "He s saying that the weight slows down until it reaches the maximum displacement, where it changes direction, and at that point (instant) the velocity is 0. And since KE depends on velocity, at that point KE is also 0.", + "video_name": "Nk2q-_jkJVs", + "timestamps": [ + 381 + ], + "3min_transcript": "It will have compressed all the way over here. So at T over 2, it'll have been here. And then at the points in between, it will be at x equals 0, right? It'll be there and there. Hopefully that makes sense. So now we know these points. But let's think about what the actual function looks like. Will it just be a straight line down, then a straight line up, and then the straight line down, and then a straight line up. That would imply-- think about it-- if you have a straight line down that whole time, that means that you would have a constant rate of change of your x value. Or another way of thinking about that is that you would have a constant velocity, right? Well do we have a constant velocity this entire time? We know that at this point right here you have a very high velocity, right? You have a very high velocity. We know at this point you have a very low velocity. So you're accelerating this entire time. actually accelerating at a decreasing rate. But you're accelerating the entire time. And then you're accelerating and then you're decelerating this entire time. So your actual rate of change of x is not constant, so you wouldn't have a zigzag pattern, right? And it'll keep going here and then you'll have a point here. So what's happening? When you start off, you're going very slow. Your change of x is very slow. And then you start accelerating. And then, once you get to this point, right here, you start decelerating. Until at this point, your velocity is exactly 0. So your rate of change, or your slope, is going to be 0. And then you're going to start accelerating back. Your velocity is going to get faster, faster, faster. It's going to be really fast at this point. And then you'll start decelerating at that point. So at this point, what does this point correspond to? You're back at A. So at this point your velocity is now 0 again. And now you're going to start accelerating. Your slope increases, increases, increases. This is the point of highest kinetic energy right here. Then your velocity starts slowing down. And notice here, your slope at these points is 0. So that means you have no kinetic energy at those points. And it just keeps on going. On and on and on and on and on. So what does this look like? Well, I haven't proven it to you, but out of all the functions that I have in my repertoire, this looks an awful lot like a trigonometric function. And if I had to pick one, I would pick cosine. Well why? Because when cosine is 0-- I'll write it down here-- cosine of 0 is equal to 1, right? So when t equals 0, this function is equal to A. So this function probably looks something like A cosine of-- and I'll just use the variable omega t-- it probably" + }, + { + "Q": "At 6:25, I now that the spring does not infinitely go on from experience as a kid. Help?", + "A": "That s because of friction. If there were no friction, it would go on forever.", + "video_name": "Nk2q-_jkJVs", + "timestamps": [ + 385 + ], + "3min_transcript": "It will have compressed all the way over here. So at T over 2, it'll have been here. And then at the points in between, it will be at x equals 0, right? It'll be there and there. Hopefully that makes sense. So now we know these points. But let's think about what the actual function looks like. Will it just be a straight line down, then a straight line up, and then the straight line down, and then a straight line up. That would imply-- think about it-- if you have a straight line down that whole time, that means that you would have a constant rate of change of your x value. Or another way of thinking about that is that you would have a constant velocity, right? Well do we have a constant velocity this entire time? We know that at this point right here you have a very high velocity, right? You have a very high velocity. We know at this point you have a very low velocity. So you're accelerating this entire time. actually accelerating at a decreasing rate. But you're accelerating the entire time. And then you're accelerating and then you're decelerating this entire time. So your actual rate of change of x is not constant, so you wouldn't have a zigzag pattern, right? And it'll keep going here and then you'll have a point here. So what's happening? When you start off, you're going very slow. Your change of x is very slow. And then you start accelerating. And then, once you get to this point, right here, you start decelerating. Until at this point, your velocity is exactly 0. So your rate of change, or your slope, is going to be 0. And then you're going to start accelerating back. Your velocity is going to get faster, faster, faster. It's going to be really fast at this point. And then you'll start decelerating at that point. So at this point, what does this point correspond to? You're back at A. So at this point your velocity is now 0 again. And now you're going to start accelerating. Your slope increases, increases, increases. This is the point of highest kinetic energy right here. Then your velocity starts slowing down. And notice here, your slope at these points is 0. So that means you have no kinetic energy at those points. And it just keeps on going. On and on and on and on and on. So what does this look like? Well, I haven't proven it to you, but out of all the functions that I have in my repertoire, this looks an awful lot like a trigonometric function. And if I had to pick one, I would pick cosine. Well why? Because when cosine is 0-- I'll write it down here-- cosine of 0 is equal to 1, right? So when t equals 0, this function is equal to A. So this function probably looks something like A cosine of-- and I'll just use the variable omega t-- it probably" + }, + { + "Q": "At 8:06 we made the carbocation gets attacked by water . why we couldnot we do it in the starting instead of making it atacked by hydronium ion?", + "A": "Water is too weak to attack when there is no positive charge.", + "video_name": "O_yeKo6-qIg", + "timestamps": [ + 486 + ], + "3min_transcript": "And so this is now just neutral water, and we see that we have a conservation of charge here, this was positive in charge, now our original molecule is positively charged. And what feels good about this is we're getting, we're getting close to our end product, at least on our number three carbon, we now have, we now have this hydrogen. Now we need to think about, well how do we get a hydroxyl group added right over here? Well we have all this water, we have all this water floating around, let me, I could use this water molecule but the odds of it being the exact same water molecule, we don't know. But there's all sorts of water molecules, we're in an aqueous solution, so let me draw another water molecule here. So the water molecules are all equivalent, but let me draw another water molecule here. And you can imagine, if they just pump into each other in just the right way, this is, water is a polar molecule, it has a because the oxygen likes to hog the electrons, and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged, so you can imagine the oxygen end might be attracted to this tertiary carbocation, and so just bumping it in just the right way, it might form a bond. So let me say these two electrons right over here, let's say they form a bond with this, with that number two carbon, and then what is going to result? So let me draw, so what is, what is going to result, let me scroll down a little bit, and let me paste, whoops, let me copy and paste our original molecule again. So, here we go. This is the one we constructed actually, so we have the hydrogen there. We have the hydrogen, now this character, so we have the water molecule, so oxygen bonded to two hydrogens, you have this one lone pair that isn't reacting, but then you have the lone pair that does do the reacting. And so it now forms a bond. Woops, let me do it in that orange color. It now, it now forms an actual bond. And we're really close to our final product, we have our hydrogen on the number three carbon, we have more than we want on our number two carbon, we just want a hydroxyl group, now we have a whole water bonded to the carbon. So somehow we have to get one of these other hydrogens swiped off of it, well that could happen with just another water molecule. So let's draw that. So another water molecule someplace, I'll do the different color just to differentiate, although as we know water, well it's hard to see what color is water if you're looking at the molecular scale." + }, + { + "Q": "At 0:36, He says work is energy transferred by force, Using this definition shouldn't the formula be Work = Energy times Force ?", + "A": "No. Work IS energy. when you exert a force over a distance, you transfer energy to the object you are exerting the force on. The amount of energy you transfer is force*distance.", + "video_name": "2WS1sG9fhOk", + "timestamps": [ + 36 + ], + "3min_transcript": "Welcome back. I'm now going to introduce you to the concepts of work and energy. And these are two words that are-- I'm sure you use in your everyday life already and you have some notion of what they mean. But maybe just not in the physics context, although they're not completely unrelated. So work, you know what work is. Work is when you do something. You go to work, you make a living. In physics, work is-- and I'm going to use a lot of words and they actually end up being kind of circular in their definitions. But I think when we start doing the math, you'll start to get at least a slightly more intuitive notion of what they all are. So work is energy transferred by a force. So I'll write that down, energy transferred-- and I got this from Wikipedia because I wanted a good, I guess, relatively intuitive definition. Energy transferred by a force. And that makes reasonable sense to me. But then you're wondering, well, I know what a force is, you know, force is mass times acceleration. But what is energy? And then I looked up energy on Wikipedia and I found this, But it also I think tells you something that these are just concepts that we use to, I guess, work with what we perceive as motion and force and work and all of these types of things. But they really aren't independent notions. They're related. So Wikipedia defines energy as the ability to do work. So they kind of use each other to define each other. Ability to do work. Which is frankly, as good of a definition as I could find. And so, with just the words, these kind of don't give you much information. So what I'm going to do is move onto the equations, and this'll give you a more quantitative feel of what these words mean. So the definition of work in mechanics, work is equal to force times distance. different color just because this yellow might be getting tedious. And I apply a force of-- let's say I apply a force of 10 Newtons. And I move that block by applying a force of 10 Newtons. I move that block, let's say I move it-- I don't know-- 7 meters. So the work that I applied to that block, or the energy that I've transferred to that block, the work is equal to the force, which is 10 Newtons, times the distance, times 7 meters. And that would equal 70-- 10 times 7-- Newton meters. So Newton meters is one, I guess, way of describing work. And this is also defined as one joule." + }, + { + "Q": "When writing the overall formula at 5:55, what if the force that was applied had a anticlockwise torque? (i.e if the force was applied on top of the board, n the same end as worked out in the video) Doesn't the formula change to - Torque = F.r.cos (theta)?", + "A": "Typically, clockwise torques are negative and counterclockwise/anticlockwise torques are positive. This is because clockwise torques would cause an object to rotate in the negative direction, giving theta a negative value. Counterclockwise/anticlockwise torques would be positive for a similar reason. Thus, the formula is still \u00cf\u0084=Fr sin(\u00ce\u00b8)", + "video_name": "ZQnGh-t25tI", + "timestamps": [ + 355 + ], + "3min_transcript": "in order to get the door to rotate. In other words, only perpendicular components of the force, that is perpendicular to the R, are going to exert a torque. So it's only this component of the 10 newtons that's going to contribute toward the torque, and we can find that. If this was 30 degrees, this is an alternate interior angle. That means that this is also 30 degrees. So these angles are identical based on geometry. That means this component that's perpendicular, I can write as well let's see, it's the opposite side. That side is opposite of this 30 degrees, so I can say that it's gonna be 10 newtons, the hypotenuse would be 10 newtons times sine of 30. And 10 newtons times sine of 30 degrees is five newtons. So finally I can say that the torque exerted on this door by this force of 10 newtons at 30 degrees would be the perpendicular component, which is five newtons, times how far that force was applied from the axis, is gonna be 10 newton meters. So at this point I wouldn't blame you if you weren't like, see, this is why I hate torque. I've gotta remember that this two meters is from the axis to the point where the force is applied. I've gotta remember that I'm only supposed to take the perpendicular component, and I'm supposed to remember that perpendicular means perpendicular to this R vector. You might wonder, is there an easier way to do this? Is there a formula that makes it so I don't have to carry so much cognitive load when I'm trying to solve these problems? And there is. Since this force component is always the component that's perpendicular to the R, we can take that into account when writing down the formula. In other words, the way you find that perpendicular component is by taking the magnitude of the total force, the 10 newtons, and you multiply by sine of the angle between the R vector and the F vector. That's what we did to get this five newtons, so why not just write down this formula explicitly in terms of the total force times sine theta? That's gonna be the perpendicular component, So what this represents is this here, this F sine theta is F perpendicular, and then you multiply by R just like we always do. Now in most textbooks you'll see it written like this. They like putting the sine theta at the end. Looks a little cleaner. So if we do F times R times sine theta, now we can just plug in the entire 10 newtons in for the force, the entire two meters in for the R, and this theta would be the angle between the force and the R vector, but that's crucial. You gotta remember if you're gonna use this formula instead of this formula, you've gotta remember that this angle here is always the angle between the force vector and the R vector, which is the vector from the axis to the point where the force is applied. Which in this case, was 30 degrees. So sometimes it's not obvious. How do you figure out the angle between F and R? Well first you identify the direction of F and the direction of R. The safest way to figure it out would be to imagine taking this F vector and just moving it so its tail is at the tail of the R vector." + }, + { + "Q": "At 5:36, why did he move the hydrogen to the other side of the periodic table?", + "A": "Every element wants to satisfy its Electron or Valence Shell. The first shell occupies 2 electrons, so Hydrogen only needs 1 more to satisfy its shell. The same goes for the elements in Group 7, they only need one more electron to satisfy their shells, so he says in theory it could be moved there.", + "video_name": "CCsNJFsYSGs", + "timestamps": [ + 336 + ], + "3min_transcript": "it just needs one electron. So in theory, hydrogen could have been put there. So hydrogen actually could typically could have a positive or a negative 1 oxidation state. And just to see an example of that, let's think about a situation where hydrogen is the oxidizing agent. And an example of that would be lithium hydride right Now, in lithium hydride, you have a situation where hydrogen is more electronegative. A lithium is not too electronegative. It would happily give away an electron. And so in this situation, hydrogen is the one that's oxidizing the lithium. Lithium is reducing the hydrogen. Hydrogen is the one that is hogging the electron. So the oxidation state on the lithium here is a positive 1. And the oxidation state on the hydrogen here is a negative. to make sure we get the notation. Lithium has been oxidized by the hydrogen. Hydrogen has been reduced by the lithium. Now, let's give an example where hydrogen plays the other role. Let's imagine hydroxide. So the hydroxide anion-- so you have a hydrogen and an oxygen. And so essentially, you could think of a water molecule that loses a hydrogen proton but keeps that hydrogen's electron. And this has a negative charge. This has a negative 1 charge. But what's going on right over here? And actually, let me just draw that, because it's fun to think about it. So this is a situation where oxygen typically has-- 1, 2, 3, 4, 5, 6 electrons. And when it's water, you have 2 hydrogens like that. And then you share. over there sharing that pair, covalent bond sharing that right over there. To get to hydroxide, the oxygen essentially nabs both of these electrons to become-- so you get-- that pair, that pair. Now you have-- let me do this in a new color. Now, you have this pair as well. And then you have that other covalent bond to the other hydrogen. And now this hydrogen is now just a hydrogen proton. This one now has a negative charge. So this is hydroxide. And so the whole thing has a negative charge. And oxygen, as we have already talked about, is more electronegative than the hydrogen. So it's hogging the electrons. So when you look at it right over here, you would say, well, look, hydrogen, if we had to, if we were forced to-- remember, oxidation states is just an intellectual tool which we'll find useful. If you had to pretend this wasn't a covalent bond, but an ionic bond, you'd say, OK, then maybe this hydrogen would fully lose an electron," + }, + { + "Q": "If hydroxide is a polyatomic atom that is represented as OH- then how come at the minute 5:20 you write it as HO- ?", + "A": "It doesn t really matter which way around it is written, although OH- is much more common.", + "video_name": "CCsNJFsYSGs", + "timestamps": [ + 320 + ], + "3min_transcript": "These are typically oxidized. Now, we could keep going. If we were to go right over here to the Group 5 elements, typical oxidation state is negative 3. And so you see a general trend here. And that general trend-- and once again, it's not even a hard and fast rule of thumb, even for the extremes, but as you get closer and closer to the middle of the periodic table, you have more variation in what these typical oxidation states could be. Now, I mentioned that I put hydrogen aside. Because if you really think about it, hydrogen, yes, hydrogen only has one electron. And so you could say, well, maybe it wants to give away that electron to get to zero electrons. That could be a reasonable configuration for hydrogen. But you can also view hydrogen kind of like a halogen. So you could kind of view it kind of like an alkali metal. But in theory, it could have been put here on the periodic table as well. You could have put hydrogen here, because hydrogen, it just needs one electron. So in theory, hydrogen could have been put there. So hydrogen actually could typically could have a positive or a negative 1 oxidation state. And just to see an example of that, let's think about a situation where hydrogen is the oxidizing agent. And an example of that would be lithium hydride right Now, in lithium hydride, you have a situation where hydrogen is more electronegative. A lithium is not too electronegative. It would happily give away an electron. And so in this situation, hydrogen is the one that's oxidizing the lithium. Lithium is reducing the hydrogen. Hydrogen is the one that is hogging the electron. So the oxidation state on the lithium here is a positive 1. And the oxidation state on the hydrogen here is a negative. to make sure we get the notation. Lithium has been oxidized by the hydrogen. Hydrogen has been reduced by the lithium. Now, let's give an example where hydrogen plays the other role. Let's imagine hydroxide. So the hydroxide anion-- so you have a hydrogen and an oxygen. And so essentially, you could think of a water molecule that loses a hydrogen proton but keeps that hydrogen's electron. And this has a negative charge. This has a negative 1 charge. But what's going on right over here? And actually, let me just draw that, because it's fun to think about it. So this is a situation where oxygen typically has-- 1, 2, 3, 4, 5, 6 electrons. And when it's water, you have 2 hydrogens like that. And then you share." + }, + { + "Q": "At 2:00, what is the typical oxidation state of a halogen and why?", + "A": "The typical oxidation state, or oxidation number, of a halogen is -1, mainly because halogens are so willing to gain another electron in order to fill out its octet.", + "video_name": "CCsNJFsYSGs", + "timestamps": [ + 120 + ], + "3min_transcript": "Let's see if we can come up with some general rules of thumb or some general trends for oxidation states by looking at the periodic table. So first, let's just focus on the alkali metals. And I'll box them off. We'll think about hydrogen in a second. Well, I'm going to box-- I'm going to separate hydrogen because it's kind of a special case. But if we look at the alkali metals, the Group 1 elements right over here, we've already talked about the fact they're not too electronegative. They have that one valence electron. They wouldn't mind giving away that electron. And so for them, that oxidation state might not even be a hypothetical charge. These are very good candidates for actually forming ionic bonds. And so it's very typical that when these are in a molecule, when these form bonds, that these are the things that are being oxidized. They give away an electron. So they get to-- a typical oxidation state for them would be positive 1. If we go one group over right over here to the alkaline earth So they're likely to fully give or partially give away two electrons. So if you're forced to assign an ionic-- if you were to say, well, none of this partial business, just give it all away or take it, you would say, well, these would typically have an oxidation state of positive 2. In a hypothetical ionic bonding situation, they would be more likely to give the two electrons because they are not too electronegative, and it would take them a lot to complete their valence shell to get all the way to 8. Now, let's go to the other side of the periodic table to Group 7, the halogens. The halogens right over here, they're quite electronegative, sitting on the right-hand side of the periodic table. They're one electron away from being satisfied from a valence electron point of view. So these are typically reduced. And I keep saying typically, because these are not going to always be the case. There are other things that could happen. But this is a typical rule of thumb that they're likely to want to gain an electron. If we move over one group to the left, Group 6-- and that's where the famous oxygen sits-- we already said that oxidizing something is doing to something what oxygen would have done, that oxidation is taking electrons away from it. So these groups are typically oxidized. And oxygen is a very good oxidizing agent. Or another way of thinking about it is oxygen normally takes away electrons. These like to take away electrons, typically two electrons. And so their oxidation state is typically negative 2-- once again, just a rule of thumb-- or that their charge is reduced by two electrons." + }, + { + "Q": "at 5:28 you talk about having a blood vessel break in a certain part of the brain. since we have different sections in the brain for different subjects and topics such as music,personality,vision, etc. would the stroke effect your relationship with that certain subject?", + "A": "If the braintissue dies due to lack of oxygen or due to the compression of the bloodflow in that part of the brain, then those functions are lost or damaged. So it s possible you go blind, you lose the possibility to speak or understand what s bein said et cetera et cetera...", + "video_name": "xbyfeEW56Nc", + "timestamps": [ + 328 + ], + "3min_transcript": "moving through the blood that eventually blocks a blood vessel. So you can actually have a thromboembolism, you can actually have a blood clot that gets broken off -- so let me ignore this for now-- let me paint over it a little bit so that this isn't the main cause of blockage-- but you could actually have a blood clot that breaks off, becomes an embolus, and since it's an embolus due to a blood clot, you call it a thrombembolus -- I always have trouble saying all of these words-- and eventually it blocks an artery over here. So this right here is an embolism, but either way, you're blocking the blood flow further down the brain, [which] could cause infarction, that brain tissue will die, and whatever that brain tissue did for mental function, or whatever, for this person who is experiencing this stroke to do those things. that's called a silent stroke, but damage is occurring. The person experiencing the stroke-- and I'm not a doctor, so take all of this with a grain of salt-- the person experiencing the stroke could be anywhere from - well, one, they may not even notice that damage is occurring, they might have a headache, or it might be more severe, they might actually not be able to properly move a side of their body, or a side of their face, or properly be able to speak, so it really depends on what part of the brain is being damaged. But in either of these situations, an ischemic stroke is caused by some type of restriction or blockage that causes things downstream to not get proper oxygen, and then, so you can imagine, cells over here aren't going to get their oxygen, and then they might actually die. A hemorrhagic stroke - to hemorrhage means to bleed, it's literally just a fancy word for bleeding- and so in a hemorrhagic stroke you have a situation where a blood vessel I'm actually trying to draw the same blood vessel- where it actually breaks. We'll talk more in the future of why a blood vessel might break - strongly related to high blood pressure and other risk factors, but I don't want to get into that right now - but you could imagine if a blood vessel breaks, you have all this blood spewing into the brain in, kind of, an uncontrolled way. So let's say this little diagram [that] I drew right here, that part of the brain, if you have a hemorrhagic stroke, you have all of this blood that's flowing into the brain, and all of that uncontrolled blood will mess up that part of the brain, that causes those neurons and brain tissue to malfunction and maybe causes some of them to die, and it would also cause the blood flow further dowstream to be impaired, so the stuff downstream aren't going to get the blood they need because all the blood is being released everywhere else. And since 87% of strokes are ischemic strokes, the remainder are hemorrhagic, so this is the remaining 13% of strokes." + }, + { + "Q": "At 6:31, is that also called latent heat?", + "A": "Yes, you noticed right! That s precisely what latent heat is.:-)", + "video_name": "lsXcKgjg8Hs", + "timestamps": [ + 391 + ], + "3min_transcript": "times the amount of ice. That's what we're solving for. So times x. Times the change in temperature. So this is a 10 degree change in Celsius degrees, which is also a 10 degree change in Kelvin degrees. We can just do 10 degrees. I could write Kelvin here, just because at least when I wrote the specific heat units, I have a Kelvin in the denominator. It could have been a Celsius, but just to make them cancel out. This is, of course, x grams. So the grams cancel out. So that heat absorbed to go from minus 10 degree ice to zero degree ice is 2.05 times 10 is 20.5. So it's 20.5 times x joules. Now, once we're at zero degrees, the ice can even absorb more energy before increasing in temperature as it melts. Remember, when I drew that phase change diagram. The ice gains some energy and then it levels out as it melts. As the the bonds, the hydrogen bonds start sliding past each other, and the crystalline structure breaks down. So this is the amount of energy the ice can also absorb. Let me do it in a different color. Zero degree ice to zero -- I did it again --to zero degree water. Well the heat absorbed now is going to be the heat of fusion of ice. Or the melting heat, either one. That's 333 joules per gram. It's equal to 333.55 joules per gram Once again, that's x grams. They cancel out. So the ice will absorb 333.55 joules as it goes from zero degree ice to zero degree water. Or 333.55x joules. Let me put the x there, that's key. So the total amount of heat that the ice can absorb without going above zero degrees... Because once it's at zero degree water, as you put more heat into it, it's going to start getting warmer again. If the ice gets above zero degrees, there's no way it's going to bring the water down to zero degrees. The water can't get above zero degrees. So how much total heat can our ice absorb? So heat absorbed is equal to the heat it can absorb when it goes from minus 10 to zero degrees ice. And that's 20.5x." + }, + { + "Q": "9:29 What is a sigma bond? What is a pi bond? Is pi bond in all double bonds or multiple bonds? Is a sigma bond in every single bond?", + "A": "All single bonds are sigma bonds. However, a double bond consists of 1 sigma bond and 1 pi bond. You can rotate around a sigma bond (single bond), but not a pi bond (involved in double and triple bonds), which I m sure will be important in upcoming videos. A tripple bond consists of 1 sigma bond and 2 pi bonds.", + "video_name": "u1eGSL6J6Fo", + "timestamps": [ + 569 + ], + "3min_transcript": "electron would start kind of jumping around or moving around, depending how you want to think about it, in that orbital over there, 2px. Then you'd have the next electron jumping around or moving around in the 2py orbital, so it would be moving around like this. If you went just off of this, you would say, you know what? These guys, this guy over here and that guy over there is lonely. He's looking for a opposite spin partner. This would be the only places that bonds would form. You would expect some type of bonding to form with the x-orbitals or the y-orbitals. Now, that's what you would expect if you just straight-up kind of stayed with this model of how things fill and how orbitals look. The reality of carbon, and I guess the simplest reality of carbon, is if you look at a methane molecule, is very different than what you would expect here. First of all, what you would expect here is that carbon But we know carbon forms four bonds and it wants to pretend like it has eight electrons. Frankly, almost every atom wants to pretend like it has eight electrons. So in order for that to happen, you have to think about a different reality. This isn't really what's happening when carbon bonds, so not what happens when carbon bonds. What's really happening when carbon bonds, and this will kind of go into the discussion of sp3 hybridization, but what you're going to see is it's not that complicated of a topic. It sounds very daunting, but it's actually pretty straightforward. What really happens when carbon bonds, because it wants to form four bonds with things, is its configuration, you could imagine, looks more like this. So you have 1s. We have two electrons there. Then you have your 2s, 2px, 2py and 2pz. It has four electrons that are willing to pair up with electrons from other molecules. In the case of methane, that other molecule is a hydrogen. So what you could imagine is that the electrons actually-- maybe the hydrogen brings this electron right here into a higher energy state and puts it into 2z. That's one way to visualize it. So this other guy here maybe ends up over there, and then these two guys are over there and over there. Now, all of a sudden, it looks like you have four lonely guys and they are ready to bond, and that's actually more accurate of how carbon bonds. It likes to bond with four other people. Now, it's a little bit arbitrary which electron ends up in each of these things, and even if you had this type of bonding, you would expect things to bond along the x, y, and z axis. The reality is, the reality of carbon, is that these four" + }, + { + "Q": "At 5:12, why did Sal say straight VANILLA s and p orbitals??", + "A": "It s American slang for normal or original . It refers to the original flavour of ice cream (vanilla). Since then, many new, more exciting flavours have been invented. The phrase plain vanilla would then refer to the original or unhybridized atomic orbitals.", + "video_name": "u1eGSL6J6Fo", + "timestamps": [ + 312 + ], + "3min_transcript": "Remember, these are really just probability clouds, but it's helpful to kind of visualize them as maybe a little bit more things that we would see in our world, but I think probability cloud is the best way to think about it. So that is the 2px orbital, and then I haven't talked about how they get filled yet, but then you also have your 2py orbital, which'll go in this axis, but same idea, kind of a dumbbell shape in the y-direction, going in both along the y-axis, going in that direction and in that direction. Then, of course, so let me do this 2py, and then you also have your 2pz, and that goes in the z-direction up like that and then downwards like that. So when you keep adding electrons, the first-- so far, we've added four electrons. If you add a fifth electron, you would expect it to go into So even though this 2px orbital can fit two electrons, the first one goes there. The very next one won't go into that one. It actually wants to separate itself within the p orbital, so the very next electron that you add won't go into 2px, it'll go into 2py. And then the one after that won't go into 2py or 2px, it'll go into 2pz. They try to separate themselves. Then if you add another electron, if you add-- let's see, we've added one, two, three, four, five, six, seven. If you add an eighth electron, that will then go into the 2px orbital, so the eighth electron would go there, but it would have the opposite spin. So this is just a little bit of review with a little bit of visualization. Now, given what we just reviewed, let's think about what's happening with carbon. Carbon has six electrons. 1s orbital. Then 2s2, then 2p2, right? It only has two left, because it has a total of six electrons. Two go here, then there, then two are left to fill the p orbitals. If you go based on what we just drew and what we just talked about here, what you would expect for carbon-- let me just draw it out the way I did this. So you have your 1s orbital, your 2s orbital, and then you have your 2px orbital, your 2py orbital, and then you have your 2pz orbital. If you just go straight from the electron configuration, you would expect carbon, so the 1s orbital fills first, so that's our first electron, our second electron, our third electron. Then we go to our 2s orbital, That fills next, third electron, then fourth electron. Then you would expect maybe your fifth electron to go in the 2px." + }, + { + "Q": "At 14:00 methane has been drawn, but I see no 1s orbital drawn...is it there and just not mentioned, or did it get absorbed into the sp3's somehow?", + "A": "I believe that he just did not put the 1s electrons in there since they play no role in the Hydrogens bonding to the Carbon.", + "video_name": "u1eGSL6J6Fo", + "timestamps": [ + 840 + ], + "3min_transcript": "And instead of having one s and three p orbitals, it has four sp3 orbitals. So let me try my best at drawing the four sp3 orbitals. Let's say this is the big lobe that is kind of pointing near us, and then it has a small lobe in the back. Then you have another one that has a big lobe like that and a small lobe in the back. Then you have one that's going back behind the page, so let me draw that. You can kind of imagine a three-legged stool, and then its small lobe will come out like that. And then you have one where the big lobe is pointing straight up, and it has a small lobe going down. You can imagine it as kind of a three-legged stool. One of them is behind like that and it's pointing straight up, So a three-legged stool with something-- it's kind of like a tripod, I guess is the best way to think about it. have the hydrogens, so that's our carbon right there. Then you have your hydrogens. You have a hydrogen here. A hydrogen just has one electron in the 1s orbital, so the hydrogen has a 1s orbital. You have a hydrogen here that just has a 1s orbital. It has a hydrogen here, 1s orbital, hydrogen here, 1s orbital. So this is how the hydrogen orbital and the carbon orbitals get mixed. The hydrogens 1s orbital bonds with-- well, each of the hydrogen's 1s orbital bonds with each of the carbon's sp3 orbitals. Just so you get a little bit more notation, so when people talk about hybridized sp3 orbitals, all they're saying is, look, carbon doesn't bond. Once carbon-- this right here is a molecule of methane, right? This is CH4, or methane, and it doesn't bond like you would vanilla s and p orbitals. If you just went with straight vanilla s and p orbitals, the bonds would form. Maybe the hydrogen might be there and there, and if it had four hydrogens, maybe there and there, depending on how you want to think about it. But the reality is it doesn't look like that. It looks more like a tripod. It has a tetrahedral shape. The best way that that can be explained, I guess the shape of the structure, is if you have four equally-- four of the same types of orbital shapes, and those four types of orbital shapes are hybrids between s's and p's. One other piece of notation to know, sometimes people think it's a very fancy term, but when you have a bond between two molecules, where the orbitals are kind of pointing at each other, so you can imagine right here, this hydrogen orbital is pointing in that direction. This sp3 orbital is pointing that direction, and they're overlapping right around here." + }, + { + "Q": "1:33 what is an internal pressure?", + "A": "I think there is an actual name for the internal pressure of a plant that allows it to stand upright. It s called Turgor Pressure.", + "video_name": "zdvKhaQxvag", + "timestamps": [ + 93 + ], + "3min_transcript": "we've talked a lot about cells in general but what I thought I would do in this video was focus on plant cells and in particular focus on the cell walls love plant cells so this right over here this is a drawing of a plant cell and my jump out at you immediately is instead of drawing it is this kind ever a roundish shape like that but we have drawn a lot of other cells I've drawn this is kind of a cubic structure rectangular prism and that's plant cells can have a structure like that and so the next question is what gives them that shape which allows them to to form the kind of cubic root angular prism and the answer is its the cell wall so this the cell cell wall so let's make sure we can already into ourselves properly in this picture so clearly if I was if I didn't have this cut out all I would be seeing is the outside all I would be seeing is the cell wall but we've cut it out and we can see the different layers we have the cell wall on the outside right below that right below that we have the cellular membrane or the plasma to this the cell cellular cellular membrane cellular membrane right under that and then under that the cell membrane is control is is containing in the cytoplasm inside of the cytoplasm we have all sorts of this big being that is taking up a lot of the volume inside this plant cell that the back you'll which we have described another in other in other videos vacuo as a combination of this internal pressure things like the back you all in just frankly the pressure from from all the fluid inside the cell pushing outwards plus the cell wall kinda holding it all in that's what gives plant their structure that's why plant is able to grow and be a plant is able to grow and be upright to that's my drawing over a plant active a plant in my room that i'm looking at right now in its able to in be upright and so you have the cell wall you the cellular membrane you have the other organelles at Suncorp last year we have our nuclear membrane urgency the shielding the inert nuclear membranes the DNA inside you the and OPA's make particular one kind containing that the rough ER containing the ribald or having the ribosomes on the membrane the smoothie are not having the ribosomes Golgi apparatus with us a little bit overview but our focus here is on the cell wall so let's go back to that so if we zoom in on this if we zoom in on the cell wall right over here we can look at we can look at this diagram and over here it might be a little bit surprising to you because when I've always imagined ace a a wall cell wall I imagine something like a brick wall something that's impenetrable but this drawing shows a something different and just to be clear what's going on here so this is our cellular membrane sorry routes or remembrance of right over here have my lipid by layer and then a right on top of that I have the cell wall but you see it isn't just ate" + }, + { + "Q": "At 2:06, how is it a probabilty of it happening?", + "A": "A probability of it happening is the concentration of the molecules. If you have a high concentration, you will have a high probability. If you have a low concentration, then the probability is low.", + "video_name": "TsXlTWgyItw", + "timestamps": [ + 126 + ], + "3min_transcript": "Let's say we wanted to figure out the equilibrium constant for the reaction boron trifluoride in the gaseous plus 3 -- so for every mole of this, we're going to have 3 moles of H2O in the liquid state -- and that's in equilibrium. It's going forward and backwards with 3 moles of hydrofluoric acid, so it's in the aqueous state. It's been dissolved in the water. If it wasn't dissolved, if it was in the solid state, you would call this hydrogen fluoride. Once it's in water, you call it hydrofluoric acid, and we'll talk more about naming in the future, hopefully. Plus 1 mole of boric acid, also in the aqueous state. It's dissolved in the water. H3BO3 in the aqueous state. So what would the expression for the equilibrium constant So you might be tempted say, OK, that's easy enough, Sal. So the equilibrium constant, you just take the right-hand side. That's just the convention. There's symmetry here. I could've rewritten it either way, but let's just say you take the right-hand side and say, OK, this is dependent on the concentration of the hydrofluoric acid, he concentration of the HF, or the molarity of the HF, to the third power, times the concentration of the boric acid, H3BO3. And remember, this intuition of why you're taking this to the third power is what's the probability -- because in order for the reaction to go this way, you need to have 3 molecules of hydrofluoric acid being very close to 1 molecule of the boric acid. So if you watched the last video I just made about the intuition behind the equilibrium constant, the probability of this reaction happening or the probability of finding all of these molecules in the same place. Of course, you can adjust it with a constant and that's essentially what that does. But that's on the product side, or the reactant, depending on what direction you're viewing this equation, divided by the molarity of the boron trifluoride times -- and I'll do this in a different color-- the molarity of the h2O to the third power. And that's, of course, the H2O liquid. So there you go. We'll just figure this out. And my rebuttal to you is I want you to figure out the molarity of the water. What is the concentration of the water? Remember, the concentration is moles per volume, but in this case, what's happening? I'm putting some boron trifluoride gas" + }, + { + "Q": "at 3:18 whats a solvent. and how is it related to h20", + "A": "Solvent is the liquid thing that is in abundant quantity. Notice how H2O is liquid (there s a small l written next to it, see?) and the others are aqueous. So, that means, water is everywhere. So it is in abundance. And thus, it is a solvent. I can t explain any better than that.", + "video_name": "TsXlTWgyItw", + "timestamps": [ + 198 + ], + "3min_transcript": "So you might be tempted say, OK, that's easy enough, Sal. So the equilibrium constant, you just take the right-hand side. That's just the convention. There's symmetry here. I could've rewritten it either way, but let's just say you take the right-hand side and say, OK, this is dependent on the concentration of the hydrofluoric acid, he concentration of the HF, or the molarity of the HF, to the third power, times the concentration of the boric acid, H3BO3. And remember, this intuition of why you're taking this to the third power is what's the probability -- because in order for the reaction to go this way, you need to have 3 molecules of hydrofluoric acid being very close to 1 molecule of the boric acid. So if you watched the last video I just made about the intuition behind the equilibrium constant, the probability of this reaction happening or the probability of finding all of these molecules in the same place. Of course, you can adjust it with a constant and that's essentially what that does. But that's on the product side, or the reactant, depending on what direction you're viewing this equation, divided by the molarity of the boron trifluoride times -- and I'll do this in a different color-- the molarity of the h2O to the third power. And that's, of course, the H2O liquid. So there you go. We'll just figure this out. And my rebuttal to you is I want you to figure out the molarity of the water. What is the concentration of the water? Remember, the concentration is moles per volume, but in this case, what's happening? I'm putting some boron trifluoride gas and it's creating these aqueous acids. These other molecules are dissolved completely in the water. So what's the solvent here? The solvent is H2O. This might be how the reaction happens, but pretty much, there's water everywhere. The water is in surplus. So if you were to really figure out the concentration of water, it's everywhere. I mean, you could say I mean, you could say everything but the boron trifluoride, but it's a very high number. And if you think about it from the probability point of view, if you say, OK, in order for this reaction to happen forward, I need to figure out the probability of finding a boron trifluoride atom or molecule -- actually, molecule-- in a certain volume, and it also needs 3 moles of water in that certain volume. But you say, hey, there's water everywhere. This is the solvent. There's water everywhere, so I really just need to worry about the concentration of the boron trifluoride." + }, + { + "Q": "what the heck does 1s^22s^22p^63s^1 in the video at 1:23 mean did i miss something because it makes no sense.", + "A": "The superscripts are not exponents. They are the numbers of electrons in each shell or subshell. 1s\u00c2\u00b2 2s\u00c2\u00b22p\u00e2\u0081\u00b6 3s\u00c2\u00b9 means there are 2 electrons in the 1s shell, 2 in the 2s subshell, 6 in the 2p subshell, and 1 in the 3s subshell.", + "video_name": "akm5H2JsccI", + "timestamps": [ + 83 + ], + "3min_transcript": "Now that we've classified our elements into groups on the periodic table, let's see how to determine the number of valence electrons. And so for this video, we're only talking about the valence electrons for elements in the main groups. When we talk about the main groups, you're using the one through eight system for classifying groups. So one, two, three, four, five, six, seven, and eight. So we're going to ignore the other way to number the groups. And so therefore, we're going to ignore groups three through 12 for this video. And so if we're talking about the main groups, the valence electrons are the electrons in the outermost shell or the outermost energy level. And so let's see if we can figure out how many valence electrons sodium has. So for sodium, if I wanted to write an electron configuration for sodium-- I assume you already know how to do these-- so you would say it is 1S2, 2S2, 2P6. And that takes you all the way over here to neon. or the third energy level. And you have one more electron to worry about. And so that electron would go into a 3S orbital. So the full electron configuration is 1S2, 2S2, 2P6, and 3S1. When I want to figure out how many valence electrons sodium has, the number of valence electrons would be equal to the number of electrons in the outermost shell, the outermost energy level. For sodium, sodium has the first energy level, second energy level, and the third energy level. The outermost energy level would, of course, the third energy level. So if I see how many electrons sodium has in its outermost energy level, it's only one this time. So that means that sodium has one valence electron. And that's very convenient, because sodium is found in group one. And so we can say that for main groups, if you want to figure out how many valence electrons you have, it's just equal to the group number. So the group number is equal to the number And so that makes everything really easy. And so if I wanted to represent a neutral atom of sodium with its one valence electron, I could draw sodium here, and I could draw one valence electron next to sodium like that. All right. Let's go ahead and write the electron configuration for chlorine next. So here's chlorine over here. And so if I wanted to write the electron configuration for chlorine, it would be 1s2, 2s2, 2p6, and once again, that takes me all the way to neon. And so now, I'm over here in the third energy level, or the third period. I can see that I would fill 3s2-- so 3s3. And that puts me into my P orbitals. So how many electrons are in my P orbitals? One, two, three, four, five-- so I'm in the third energy level, I'm in P orbitals, and I have five electrons. And so that would be the electron configuration for chlorine." + }, + { + "Q": "At 9:38, when the tension is being found at the bottom of the circle, is the assumption made that the velocity is the same at the top and the bottom? The number 4 was used for both, and I thought the values would be different?", + "A": "He s just showing you how the problem would be different if you have the velocity at the bottom instead of at the top. Consider it two different problems.", + "video_name": "2lcaBPLLoLo", + "timestamps": [ + 578 + ], + "3min_transcript": "Gravity's pointing away from the center, radially away from the center. That means tension still remains a positive force, but the force of gravity now, for this case down here, would have to be considered a negative centripetal force since it's directed away from the center of the circle. So, if we were to calculate the tension at the bottom of the path, the left hand side would still be v squared over r 'cause that's still the centripetal acceleration. The mass on the bottom would still be m 'cause that's the mass of the yo-yo going in a circle. But instead of T plus mg, we'd have T minus mg since gravity's pointing radially out of the center of the circle. And if we solve this expression for the tension in the string, we'd get that the tension equals, we'd have to multiply both sides by m, and then add mg to both sides. And we'd get that the tension's gonna equal m v squared over r, plus m g. This time, we add m g to this m v squared over r term, whereas over here, we had to subtract it. And that should make sense conceptually since before, up here, both tension and gravity to the total centripetal force, so neither one had to be as big as they might have been otherwise. But down here, not only is gravity not helping the tension, gravity's hurting the centripetal cause by pulling this mass out of the center of the circle, so the poor tension in this case not only has to equal the net centripetal force, it has to add up to more than the net centripetal force just to cancel off this negative effect from the force of gravity. And if we plugged in numbers, we'd see that the tension would end up being bigger. We'd actually get the same exact term here, except that instead of subtracting gravity, we have to add gravity to this net centripetal force expression. And we'd get that the tension would be 10.45 Newtons. So recapping, when solving centripetal force problems, we typically write the v squared over r on the left hand side as a positive acceleration, and by doing that, we've selected in toward the center of the circle as positive since that's the direction which means that all forces that are directed in toward the center of the circle also have to be positive. And you have to be careful because that means downward forces can count as a positive centripetal force as long as down corresponds to toward the center of the circle. And just because a force was positive during one portion of the trip, like gravity was at the top of this motion, that does not necessarily mean that that same force is gonna be positive at some other point during the motion." + }, + { + "Q": "at 2:50, why do we have to change degree Celsius to kelvin?", + "A": "Because Kelvin is a measure of temperature in which 0 temperature corresponds to zero energy. C is a measure in which 0 is arbitrarily chosen. The ideal gas equation is therefore written in such a way (check the units) that it is expecting T to be in Kelvin.", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 170 + ], + "3min_transcript": "So each molecule has two hydrogens in it. And let's say I'm measuring it at 30 degrees Celsius. Use different color. 30 degrees Celsius. My brain is really malfunctioning. 30 degrees, not 30 percent, 30 degrees Celsius. And let's say that the pressure on the outside of the balloon, we've measuredat two atmospheres. So my question to you is how many moles of hydrogen do we have? How many moles... So let's apply our ideal gas equation. And since we're dealing with liters and atmospheres, But in general, if we keep pressure. So our pressure is given in atmospheres. Let me write down all the units, actually. So we have 2 atmospheres times our volume is 2 liters, is equal to n. n is the number of particles we care about, and we care about it in moles, but let's just write n there for now. Is equal to n times R. I'll do R in a second times. R times T. Now you might be atempted to just put 30 degrees in there. But in all of these problems-- in fact in general, whenever you're doing any of these gas problems or thermodynamics problems, or any time you're doing math with temperature-- you should always convert into Kelvin. And just as a bit of review as to what Kelvin is, So for example, the lowest possible temperature that can be achieved in the universe, when you think about it in Celsius, let me draw a little temperature scale here. So if that's the temperature scale. I'll draw two, one for Celsius and one for Kelvin. So the lowest possible temperature that can be achieved in the universe, and when we say the lowest possible temperature that means that the average kinetic energy of the molecules or the atoms are zero. They're just not moving. They're just stationary. So in Celsius, it's minus 273.15 degrees Celsius. So zero might be some place over here. Zero, that's where water freezes. And then 100 degrees, that's where water boils. And you can immediately see, the whole Celsius scale" + }, + { + "Q": "Hi, at time 8:21 in video, when doing final calculation, why is the calculation .082 divided by 303? shouldnt it be 4/(.082*303) as in 4 divided by (.082 multiplied by 303) ?", + "A": "What he did is correct. 4/0.82/303 is equal to 4/(0.82*303) the same as 4/2/2 is equal to 4/(2*2)", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 501 + ], + "3min_transcript": "Remember, we're dealing with atmospheres and liters. So I wrote down a couple of versions of R right here. Let's see we're dealing with atmospheres and liters. And in the denominator we're always dealing with mole and Kelvin no matter what. So those are always going to be there. So we should use this proportionality constant. R is equal to 0.082 liter atmospheres per mole Kelvin. Let me write that down. So let me rewrite our whole equation actually. So I have 2 atmospheres times 2 liters is equal to n times, I have a bad memory, 0.082 liter atmospheres per mole Kelvin, times 303 degrees Kelvin. So let's see what we can do. Let's see if all of the units work out. you can treat units like numbers. So if you divide both sides of this equation by atmospheres, the atmospheres cancel out. Divide both sides of this equation by liters, liters cancel out. You have a Kelvin in the numerator, Kelvin in the denominator, that cancels out. And so we have 2 times 2 is equal to n times 0.082 times 303. And then we have just a per mole and a 1 over the mole. So to solve for n, or the number of moles, what we do is we divide both sides of this equation by all of this stuff. So we get 2 times 2 is 4. 4 divided by 0.082 divided by 303. dividing both sides by it. And when you divide by a per mole, if you put a 1 over a mole here, that's the same thing as multiplying by a mole. So it's good, the units all worked out. We're getting n in terms of moles. And so we just have to get the calculator out and figure out how many moles we're dealing with. So we have 4 divided by 0.082 divided by 303 is equal to 0.16. If we wanted to go more digits, 0.161, but we'll just round. So this is equal to 0.16 moles of H2. I am telling you actually here, the exact number of hydrogen molecules. But if you wanted a number, you'd just multiply this times 6.02 times 10 to the 23 and then you would have a number in kind of the traditional sense. And of course, if you wanted to know" + }, + { + "Q": "At 6:14, why is there so many constant, i know in my chemistry class we only learned one. And what is the difference between all of them, is there different uses for them. And how do they come up with the constant reaction formula?", + "A": "PV = nRT only has one constant.", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 374 + ], + "3min_transcript": "and the boiling point of water. Now, Kelvin. So look at this and you say, if I have something that's 5 degrees and I have another thing that's 10 degrees, when you look at the Celsius scale, you're like, oh, maybe the 10 degree thing it has twice as much energy as the 5 degree thing. It has twice the temperature. But when you look at it from the absolute distance to zero. Let me see if I can draw this. So the 10 degree is all the way over here and the 5 degree is almost as far, that far. So the 10 degrees Celsius is only a slight increment over 5 degrees Celsius, if you were to divide the two. It's not twice as hot. And that's why they came up with the Kelvin scale. Because in the Kelvin scale, absolute zero is defined as 0. Zero Kelvin. So this right here is zero degrees Kelvin. And so zero degrees Kelvin is absolute zero. So what is zero degrees Celsius? And the increments are the same. is one degree change in Kelvin. So at least they keep it, it's just a shift. So this is going to be plus 273 degrees Kelvin. And then 5 degrees would be plus 278 10 degrees would be plus 283 Kelvin. And then you see that 5 and 10 degrees really aren't that different from each other. But in general, if you want to convert from Celsius to Kelvin you just add 273 degrees. So 30 degrees Celsius is what? Well, this 5 and 10 I drew too close to 100. But let's say it's sitting here. It would be 303 degrees Kelvin. So this is equal to 303 degrees Kelvin. All right, so now for our temperature, that's what we were worried about. We wanted to put in the temperature there. So now we can put in our 303 degrees Kelvin. Now we have to figure out what constant to use here. Remember, we're dealing with atmospheres and liters. So I wrote down a couple of versions of R right here. Let's see we're dealing with atmospheres and liters. And in the denominator we're always dealing with mole and Kelvin no matter what. So those are always going to be there. So we should use this proportionality constant. R is equal to 0.082 liter atmospheres per mole Kelvin. Let me write that down. So let me rewrite our whole equation actually. So I have 2 atmospheres times 2 liters is equal to n times, I have a bad memory, 0.082 liter atmospheres per mole Kelvin, times 303 degrees Kelvin. So let's see what we can do. Let's see if all of the units work out." + }, + { + "Q": "at 8:21, why does Sal say \"4 divided by .082 divided by 303...\"? Wouldn't it be times 303? He used the multiplication dot when he wrote it out.", + "A": "They both mean the same thing. You can divide 4 by 0.082*303 or you can divide 4 by 0.082 and then divide it by 303. You will get the same answer. Hope this helps :)", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 501 + ], + "3min_transcript": "Remember, we're dealing with atmospheres and liters. So I wrote down a couple of versions of R right here. Let's see we're dealing with atmospheres and liters. And in the denominator we're always dealing with mole and Kelvin no matter what. So those are always going to be there. So we should use this proportionality constant. R is equal to 0.082 liter atmospheres per mole Kelvin. Let me write that down. So let me rewrite our whole equation actually. So I have 2 atmospheres times 2 liters is equal to n times, I have a bad memory, 0.082 liter atmospheres per mole Kelvin, times 303 degrees Kelvin. So let's see what we can do. Let's see if all of the units work out. you can treat units like numbers. So if you divide both sides of this equation by atmospheres, the atmospheres cancel out. Divide both sides of this equation by liters, liters cancel out. You have a Kelvin in the numerator, Kelvin in the denominator, that cancels out. And so we have 2 times 2 is equal to n times 0.082 times 303. And then we have just a per mole and a 1 over the mole. So to solve for n, or the number of moles, what we do is we divide both sides of this equation by all of this stuff. So we get 2 times 2 is 4. 4 divided by 0.082 divided by 303. dividing both sides by it. And when you divide by a per mole, if you put a 1 over a mole here, that's the same thing as multiplying by a mole. So it's good, the units all worked out. We're getting n in terms of moles. And so we just have to get the calculator out and figure out how many moles we're dealing with. So we have 4 divided by 0.082 divided by 303 is equal to 0.16. If we wanted to go more digits, 0.161, but we'll just round. So this is equal to 0.16 moles of H2. I am telling you actually here, the exact number of hydrogen molecules. But if you wanted a number, you'd just multiply this times 6.02 times 10 to the 23 and then you would have a number in kind of the traditional sense. And of course, if you wanted to know" + }, + { + "Q": "at 3:09, what does \"R\" mean in the equation PV = nRT?", + "A": "The R is just a constant number that relates the other four properties, called the universal gas constant. The other four can change depending on the example or situation but R will always have the same value..", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 189 + ], + "3min_transcript": "So each molecule has two hydrogens in it. And let's say I'm measuring it at 30 degrees Celsius. Use different color. 30 degrees Celsius. My brain is really malfunctioning. 30 degrees, not 30 percent, 30 degrees Celsius. And let's say that the pressure on the outside of the balloon, we've measuredat two atmospheres. So my question to you is how many moles of hydrogen do we have? How many moles... So let's apply our ideal gas equation. And since we're dealing with liters and atmospheres, But in general, if we keep pressure. So our pressure is given in atmospheres. Let me write down all the units, actually. So we have 2 atmospheres times our volume is 2 liters, is equal to n. n is the number of particles we care about, and we care about it in moles, but let's just write n there for now. Is equal to n times R. I'll do R in a second times. R times T. Now you might be atempted to just put 30 degrees in there. But in all of these problems-- in fact in general, whenever you're doing any of these gas problems or thermodynamics problems, or any time you're doing math with temperature-- you should always convert into Kelvin. And just as a bit of review as to what Kelvin is, So for example, the lowest possible temperature that can be achieved in the universe, when you think about it in Celsius, let me draw a little temperature scale here. So if that's the temperature scale. I'll draw two, one for Celsius and one for Kelvin. So the lowest possible temperature that can be achieved in the universe, and when we say the lowest possible temperature that means that the average kinetic energy of the molecules or the atoms are zero. They're just not moving. They're just stationary. So in Celsius, it's minus 273.15 degrees Celsius. So zero might be some place over here. Zero, that's where water freezes. And then 100 degrees, that's where water boils. And you can immediately see, the whole Celsius scale" + }, + { + "Q": "at 8:33, how did that work? why not divide 4 by ( 0.082 * 303 ), as in 4/(0.082*303) ?", + "A": "What Sal did and what you ve suggested are mathematically equivalent. Sal effectively took (4/0.082)/303, which is the same as (4/0.082)*(1/303) which is the same as 4/(0.082*303).", + "video_name": "erjMiErRgSQ", + "timestamps": [ + 513 + ], + "3min_transcript": "you can treat units like numbers. So if you divide both sides of this equation by atmospheres, the atmospheres cancel out. Divide both sides of this equation by liters, liters cancel out. You have a Kelvin in the numerator, Kelvin in the denominator, that cancels out. And so we have 2 times 2 is equal to n times 0.082 times 303. And then we have just a per mole and a 1 over the mole. So to solve for n, or the number of moles, what we do is we divide both sides of this equation by all of this stuff. So we get 2 times 2 is 4. 4 divided by 0.082 divided by 303. dividing both sides by it. And when you divide by a per mole, if you put a 1 over a mole here, that's the same thing as multiplying by a mole. So it's good, the units all worked out. We're getting n in terms of moles. And so we just have to get the calculator out and figure out how many moles we're dealing with. So we have 4 divided by 0.082 divided by 303 is equal to 0.16. If we wanted to go more digits, 0.161, but we'll just round. So this is equal to 0.16 moles of H2. I am telling you actually here, the exact number of hydrogen molecules. But if you wanted a number, you'd just multiply this times 6.02 times 10 to the 23 and then you would have a number in kind of the traditional sense. And of course, if you wanted to know You'd say, OK well one mole of H2 has a mass of what? The mass of one hydrogen is one atomic mass unit. The mass of two hydrogen when it's in its molecular form, is two atomic mass units. So a mole of it is going to be 2 grams. So in this case, we have 0.16 moles. So if we wanted to convert that to grams, this in the case of these hydrogen gas molecules would be 0.32 grams. And I just multiplied it by 2 because each mole is 2 grams. Anyway, I hope you found that there's a bunch more of these problems. Because I think the math is pretty straightforward here. The thing that always makes it daunting, I think, is the units and making sure you're using the right units. What is they are using meters cubed instead of liters, or kilopascals instead of atmospheres." + }, + { + "Q": "what did sal mean at 2:55 ? if the mountain in the image is very plane like a plane mirror without any up\nand downs then we will be able to see the image ?", + "A": "Sal meant that mountain cause diffused reflection all around it and does not reflect specularly.", + "video_name": "sd0BOnN6aNY", + "timestamps": [ + 175 + ], + "3min_transcript": "And you can almost imagine that it bounces off at essentially the same angle, but in the other direction. So then it'll hit the surface, and then it'll bounce off, and it'll go just like that. And then we would call this the reflected ray, after it is kind of bounced off of the surface. Reflected ray. And you may have already noticed this if you've played around a lot with mirrors you would see-- and we're going to look at some images. So you can think about it a little better. Next time you're in front of the bathroom mirror you can think about this, and think about the angle of incidence and the angle of reflection. But they're actually equal. So let me define them right here. So if I were to just drop a straight line that is at a 90 degree, or that is perpendicular to the surface of the actual mirror right over here, we would define this, right here, as the angle of incidence. I'll just use theta. That's just a fancy letter to show that the angle at which and the vertical right there, that's the angle of incidence. And then the angle between that vertical and the blue ray right there, we call that the angle of reflection. And it's just a property of especially mirrors when you're having specular reflection. And you can see this for yourself at all the regular mirrors that you might experience is that the angle of incidence is equal to the angle of reflection. And actually we could see that in a couple of images over here. So let me show you some images of specular reflection, just to make it clear here. So you have some light from the sun hitting this mountain. And we're going to talk about diffuse reflection in a little bit, and that's what's happening. It's being reflected diffusely. That's why we don't see the actual image of the sun here. We just see the white. But then those white light rays, and they're actually are hitting the water. I'm going to try to match up parts of the mountain. So you have this part of the mountain. Let me do this in a better color. You have this part of the mountain up here, and the part of the reflection right over there. So what's happening right here is light is coming from that part of the mountain, hitting this part of the surface of the water. Let me see if I can draw this better. It's hitting this part of the surface of the water, and then it's getting reflected, specular reflection, to our eyes. And it's actually coming straight at us, but I'll draw it at a slight angle. And then it's just coming straight to our eyes like this. If our eye was-- Let's say our eye was here. It's actually coming straight out at us so I actually should just draw a vertical line, but hopefully this makes it clear. And what we just said, the angle of incidence is equal to the angle of reflection, so if you were to draw a vertical, and it might not be that obvious here, but this angle right over here-- Let me draw this a little darker color." + }, + { + "Q": "At 0:30, it is stated that particles move in rotation and curved paths but the kinetic molecular theory states that particles move only in straight lines. Error?", + "A": "Definitely an error", + "video_name": "eEJqaNaq9v8", + "timestamps": [ + 30 + ], + "3min_transcript": "Voiceover: All bodies and systems possess a property called temperature. Most commonly temperature is used to refer to how hot or cold something is, but the real sciency definition of temperature is that it's a measure of the average kinetic energy of the particles in a system. I've got a system and I'm filling it with little individual particles. If we think about this microscopically each little particle in the system is moving in some way whether in rotation or in a straight line, or curving, or by kind of a combination of these means. All of these little particles are moving. The energy of motion is called the kinetic energy. All of these moving particles have kinetic energy. The faster those little particles are moving, the greater their kinetic energy. If each of those little particles in the system has greater kinetic energy, that means the system as a whole has a larger amount of total energy, and we would say that it has a greater temperature of the average kinetic energy of those particles. Because knowing the amount of energy in a system can be really useful in chemistry and in physics, we've developed temperature scales to help us quantify or measure the amount of this value, this value of energy. The three scales most widely used are the Kelvin scale, the Celsius scale, and the Fahrenheit scale. For all of these scales I'm going to draw a little thermometer, one for Kelvin. Then we have a thermometer for Celsius, and then another thermometer for the Fahrenheit scale. The two scales used most in the physical sciences are probably the Celsius scale and the Kelvin scale. As a point of comparison here on these thermometers, the freezing point of water occurs at zero degrees Celsius. That's where water freezes. Then the boiling point of water occurs at 100 degrees Celsius. So the boiling point of water occurs at 100 degrees Celsius. That's where water turns into steam. I'm going to write H20 here real quick just so we don't get confused that we're talking about the freezing and boiling point of water. Now when we use the Kelvin scale, we find that water's freezing point is 273.15 Kelvin. Then we find that water boils at 373.15 Kelvin. They differ fundamentally in the zero point. The Celsius and Kelvin scales differ in the zero points that they use, but between water's freezing point, and water's boiling point, we have a span of 100 temperature units for both scales." + }, + { + "Q": "At 5:23, when Mr. Khan says \"slide down the slope\", shouldn't the block go flying after the first hill?", + "A": "It s a possibility. However, energy is still conserved, that is the initial PE + KE still equals the PE + KE at any place along the block s path, whatever the path truly is.", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 323 + ], + "3min_transcript": "into kinetic energy. We had 100 joules of potential energy, so we're still going to have 100 joules, but now all of it's going to be kinetic energy. And kinetic energy is 1/2 mv squared. So we know that 1/2 mv squared, or the kinetic energy, is now going to equal 100 joules. What's the mass? The mass is 1. And we can solve for v now. 1/2 v squared equals 100 joules, and v squared is equal to 200. And then we get v is equal to square root of 200, which is something over 14. We can get the exact number. Let's see, 200 square root, 14.1 roughly. The velocity is going to be 14.1 meters per second squared downwards. Right before the object touches the ground. Right before it touches the ground. And you might say, well Sal that's nice and everything. We learned a little bit about energy. solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it. This is the ground down here. This is the ground. So what's going to happen this time? I'm still 10 meters in the air, so let me draw that. That's still 10 meters. I should switch colors just so not everything is ice. So that's still 10 meters, but instead of the object going straight down now, it's going to go down here and then start It's going to go sliding along this hill. And then at this point it's going to be going really fast in the horizontal direction. And right now we don't know how fast. And just using our kinematics formula, this would have been a really tough formula. This would have been difficult. I mean you could have attempted it and it actually would have taken calculus because the angle of the slope changes continuously. We don't even know the formula for the angle of the slope. You would have had to break it out into vectors. You would have to do all sorts of complicated things. This would have been a nearly impossible problem. But using energy, we can actually figure out what the velocity of this object is at this point. And we use the same idea. Here we have 100 joules of potential energy." + }, + { + "Q": "At 5:50 Sal says that this problem would be extremely difficult using kinematics equations... I do not understand why it is difficult using kinematics. Vi = 0, a = 9.8, (y2-y1) = 10. Use Vi^2 = Vf^2 + 2a(y2-y1). Final velocity comes out to be 14 m/s. Gravity is the only force acting on the object and our displacement is simple. Why should I not use kinematics here?", + "A": "You can, but most people find it easier to use CoE when it applies. Both will give you the same answer. Try it both ways and see what seems quicker and easier to you. Then see if you can think of a more complicated path that might be very difficult to solve with kinematics (like a curvy roller coaster, for example).", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 350 + ], + "3min_transcript": "solved that problem just using your kinematics formula. So what's the whole point of introducing these concepts of energy? And I will now show you. So let's say they have the same 1 kilogram object up here and it's 10 meters in the air, but I'm going to change things a little bit. Let me see if I can competently erase all of this. Nope, that's not what I wanted to do. OK, there you go. I'm trying my best to erase this, all of this stuff. OK. So I have the same object. It's still 10 meters in the air and I'll write that in a second. And I'm just holding it there and I'm still going to drop it, but something interesting is going to happen. Instead of it going straight down, it's actually going to drop on this ramp of ice. The ice has lumps on it. This is the ground down here. This is the ground. So what's going to happen this time? I'm still 10 meters in the air, so let me draw that. That's still 10 meters. I should switch colors just so not everything is ice. So that's still 10 meters, but instead of the object going straight down now, it's going to go down here and then start It's going to go sliding along this hill. And then at this point it's going to be going really fast in the horizontal direction. And right now we don't know how fast. And just using our kinematics formula, this would have been a really tough formula. This would have been difficult. I mean you could have attempted it and it actually would have taken calculus because the angle of the slope changes continuously. We don't even know the formula for the angle of the slope. You would have had to break it out into vectors. You would have to do all sorts of complicated things. This would have been a nearly impossible problem. But using energy, we can actually figure out what the velocity of this object is at this point. And we use the same idea. Here we have 100 joules of potential energy. Down here, what's the height above the ground? Well the height is 0. So all the potential energy has disappeared. And just like in the previous situation, all of the potential energy is now converted into kinetic energy. And so what is that kinetic energy going to equal? It's going to be equal to the initial potential energy. So here the kinetic energy is equal to 100 joules. And that equals 1/2 mv squared, just like we just solved. And if you solve for v, the mass is 1 kilogram. So the velocity in the horizontal direction will be, if you solve for it, 14.1 meters per second. Instead of going straight down, now it's going to be going in the horizontal to the right. And the reason why I said it was ice is because I wanted this to be frictionless and I didn't want any energy lost to heat or anything like that. And you might say OK Sal, that's kind of interesting. And you kind of got the same number for the velocity than if I just dropped the object straight down." + }, + { + "Q": "At minute 3:30 Sal plugs in the mass which is 1 Kg multiplied by (1/2) shouldn't it be .5 and if we divide both sides by .5 to get v^2 alone shouldn't 100/.5=50?", + "A": "100/0.5 is not 50 Try it on your calculator 100 x 0.5 is 50", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 210 + ], + "3min_transcript": "I'm not going to write the units down just to save space, although you should do this when you do it on your test. And then the height is 10 meters. And the units, if you work them all out, it's in newton meters or joules and so it's equal to 100 joules. That's the potential energy when I'm holding it up there. And I asked you, well when I let go, what happens? Well the block obviously will start falling. And not only falling, it will start accelerating to the ground at 10 meters per second squared roughly. And right before it hits the ground-- let me draw that in brown for ground-- right before the object hits the ground or actually right when it hits the ground, what will be the potential energy of the object? Well it has no height, right? Potential energy is mgh. The mass and the acceleration of gravity stay the same, but the height is 0. So they're all multiplied by each other. So down here, the potential energy is going to be equal to 0. And I told you in the last video that we have the law of conservation of energy. It cannot be created or destroyed. It can just be converted from one form to another. But I'm just showing you, this object had 100 joules of energy or, in this case, gravitational potential energy. And down here, it has no energy. Or at least it has no gravitational potential energy, and that's the key. That gravitational potential energy was converted into something else. And that something else it was converted into is kinetic energy. And in this case, since it has no potential energy, all of that previous potential energy, all of this 100 joules that it has up here is now going to be converted into kinetic energy. And we can use that information to figure out its velocity right before it hits the ground. So how do we do that? Well what's the formula for kinetic energy? And we solved it two videos ago, and hopefully it shouldn't be too much of a mystery to you. It's something good to memorize, but it's also good to know how we got it and go back two videos if you forgot. into kinetic energy. We had 100 joules of potential energy, so we're still going to have 100 joules, but now all of it's going to be kinetic energy. And kinetic energy is 1/2 mv squared. So we know that 1/2 mv squared, or the kinetic energy, is now going to equal 100 joules. What's the mass? The mass is 1. And we can solve for v now. 1/2 v squared equals 100 joules, and v squared is equal to 200. And then we get v is equal to square root of 200, which is something over 14. We can get the exact number. Let's see, 200 square root, 14.1 roughly. The velocity is going to be 14.1 meters per second squared downwards. Right before the object touches the ground. Right before it touches the ground. And you might say, well Sal that's nice and everything. We learned a little bit about energy." + }, + { + "Q": "At 1:56 if an object is in the ground it has 0 PE, so does an object on a cliff or mountain top also have 0 PE", + "A": "Potential energy depends on where you re referential is, where you define your 0 to be. You can choose your 0 to be anywhere you want, as long as you stay true to your choice and don t change it in the middle of your calculations. If your 0 is on the ground, an object on the ground has 0 PE but the one on the cliff has not, because it has altitude relative to your 0 potential point.", + "video_name": "kw_4Loo1HR4", + "timestamps": [ + 116 + ], + "3min_transcript": "Welcome back. At the end of the last video, I left you with a bit of a question. We had a situation where we had a 1 kilogram object. This is the 1 kilogram object, which I've drawn neater in this video. That is 1 kilogram. And we're on earth, and I need to mention that because gravity is different from planet to planet. But as I mentioned, I'm holding it. Let's say I'm holding it 10 meters above the ground. So this distance or this height is 10 meters. And we're assuming the acceleration of gravity, which we also write as just g, let's assume it's just 10 meters per second squared just for the simplicity of the math instead of the 9.8. So what we learned in the last video is that the potential energy in this situation, the potential energy, which equals m times g times h is equal to the mass is 1 kilogram times the acceleration of gravity, which is 10 I'm not going to write the units down just to save space, although you should do this when you do it on your test. And then the height is 10 meters. And the units, if you work them all out, it's in newton meters or joules and so it's equal to 100 joules. That's the potential energy when I'm holding it up there. And I asked you, well when I let go, what happens? Well the block obviously will start falling. And not only falling, it will start accelerating to the ground at 10 meters per second squared roughly. And right before it hits the ground-- let me draw that in brown for ground-- right before the object hits the ground or actually right when it hits the ground, what will be the potential energy of the object? Well it has no height, right? Potential energy is mgh. The mass and the acceleration of gravity stay the same, but the height is 0. So they're all multiplied by each other. So down here, the potential energy is going to be equal to 0. And I told you in the last video that we have the law of conservation of energy. It cannot be created or destroyed. It can just be converted from one form to another. But I'm just showing you, this object had 100 joules of energy or, in this case, gravitational potential energy. And down here, it has no energy. Or at least it has no gravitational potential energy, and that's the key. That gravitational potential energy was converted into something else. And that something else it was converted into is kinetic energy. And in this case, since it has no potential energy, all of that previous potential energy, all of this 100 joules that it has up here is now going to be converted into kinetic energy. And we can use that information to figure out its velocity right before it hits the ground. So how do we do that? Well what's the formula for kinetic energy? And we solved it two videos ago, and hopefully it shouldn't be too much of a mystery to you. It's something good to memorize, but it's also good to know how we got it and go back two videos if you forgot." + }, + { + "Q": "at 3:00 he says that this formula only works for a point charge or for a spherical charge. My question is: does this formula work for a wire's cross-section.", + "A": "it does but you have to apply a bit of calculus for it(or maybe Gauss law)", + "video_name": "-LtvW5783zE", + "timestamps": [ + 180 + ], + "3min_transcript": "The way we'll find a formula for the magnitude of the electric field is simply by inserting what we already know is the formula for the electric force. Coulomb's Law gives us the force between two charges, and we're just gonna put that right in here. Coulomb's Law says that the electric force between two charges is gonna be k, the electric constant, which is always nine times 10 to the ninth, multiplied by Q1, the first charge that's interacting, and that'd be this Q1 over here, multiplied by Q2, the other charge interacting, divided by the center to center distance between them squared, and then because we're finding electric field in here, we're dividing by Q2. Notice what happens here. Q2 is canceling, and we get that the magnitude of the electric field is gonna be equal to k, this electric constant, and I'll write that down over here so we know what it is. K is nine times 10 to the ninth, and it's got kinda weird units, but it makes sure that all the units come out okay when you multiply. And then, what do we still have up here? We've still got a Q1 divided by the center and you might be like, well, the other charge went away. We canceled it out. Centered between which two charges? Well, this could be to any point in space, really. So you imagine your test charge at any point you want. I could put it here, I can move it over to here. The r would just be the distance from the first charge, Q1, to wherever I wanna figure out what the electric field would be. But since this Q2 always divides out, we don't even need to talk about that. We can just figure out the electric field that's created by Q1 at any point in space, so this r is just the distance from the center of the charge creating the field to the point in space where you wanna determine the electric field. And now we've got it. This is a formula for the electric field created by a charge Q1. Technically, though, this is only true if this is a point charge. In other words, if it's really, really small compared to the other dimensions in the problem. Or, if this is spherically symmetric, then it doesn't matter. If you're outside of this charge and you've got a spherically symmetrical charge distribution, where all the charges are lumped on one side of this sphere, throughout, then this formula also works just as well when you're outside the sphere. And what's this formula saying? It's saying that the absolute value, or the magnitude of the electric field created at a point in space is equal to k, the electric constant, times the charge creating the field. This is important. This charge, Q1, is creating this electric field. And then you plug in the distance away from that charge that you wanna determine the electric field, r, you square it, and that'll tell you what the magnitude of the electric field is created by Q1 at any point in space around it. Now why are we being so careful, saying that this is just the magnitude? Here's why. Imagine we plugged in this charge as positive because the charge creating it is positive. You'd get a positive value for the electric field, and you might think, oh, that means positive. That means to the right. And in this case, it works out. It does go to the right at this point. But let's say you put those same calculations for a point over here, and you wanna determine what's the value of the electric field at this point? Well, if you plugged in k, it's a positive number, your Q is a positive number, r is gonna" + }, + { + "Q": "01:01 - 01:32 What does he mean by 'moles'?", + "A": "A mole is just a number, like a dozen. But instead of 12, a mole is 6.02 * 10^23. You could use it to count anything, but since it is a big number is is mostly used to count very small things like atoms and molecules.", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 61, + 92 + ], + "3min_transcript": "In my humble opinion, the single most important biochemical reaction, especially to us, is cellular respiration. And the reason why I feel so strongly about that is because this is how we derive energy from what we eat, or from our fuel. Or if we want to be specific, from glucose. At the end of the day, most of what we eat, or at least carbohydrates, end up as glucose. In future videos I'll talk about how we derive energy from fats or proteins. But cellular respiration, let's us go from glucose to energy and some other byproducts. And to be a little bit more specific about it, let me write the chemical reaction right here. So the chemical formula for glucose, you're going to have six carbons, twelve hydrogens and six oxygens. So that's your glucose right there. So if you had one mole of glucose-- let me write that, mole of glucose, if you had six moles of molecular oxygen running around the cell, then-- and this is kind of a gross simplification for cellular respiration. I think you're going to appreciate over the course of the next few videos, that one can get as involved into this mechanism as possible. But I think it's nice to get the big picture. But if you give me some glucose, if you have one mole of glucose and six moles of oxygen, through the process of cellular respiration-- and so I'm just writing it as kind of a big black box right now, let me pick a nice color. So this is cellular respiration. Which we'll see is quite involved. But I guess anything can be, if you want to be particular enough about it. Through cellular respiration we're going to produce six moles of carbon dioxide. Six moles of water. going to produce energy. We're going to produce energy. And this is the energy that can be used to do useful work, to heat our bodies, to provide electrical impulses in our brains. Whatever energy, especially a human body needs, but it's not just humans, is provided by this cellular respiration mechanism. And when you say energy, you might say, hey Sal, on the last video didn't you just-- well, if that was the last video you watched, you probably saw that I said ATP is the energy currency for biological systems. And so you might say, hey, well it looks like glucose is the energy currency for biological systems. And to some degree, both answers would be correct. But to just see how it fits together is that the process of cellular respiration, it does produce energy directly." + }, + { + "Q": "At 3:40 you say that the energy produces 38 ATPs but I heard that it was 34, which one is correct?", + "A": "4 ATP are generated before the electron transport chain. That, in addition to the 34 ATP, makes 38 ATP.", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 220 + ], + "3min_transcript": "going to produce energy. We're going to produce energy. And this is the energy that can be used to do useful work, to heat our bodies, to provide electrical impulses in our brains. Whatever energy, especially a human body needs, but it's not just humans, is provided by this cellular respiration mechanism. And when you say energy, you might say, hey Sal, on the last video didn't you just-- well, if that was the last video you watched, you probably saw that I said ATP is the energy currency for biological systems. And so you might say, hey, well it looks like glucose is the energy currency for biological systems. And to some degree, both answers would be correct. But to just see how it fits together is that the process of cellular respiration, it does produce energy directly. So if I were to break down this energy portion of cellular respiration right there, some of it would just be heat. You know, it just warms up the cell. And then some of it is used-- and this is what the textbooks will tell you. The textbooks will say it produces 38 ATPs. It can be more readily used by cells to contract muscles or to generate nerve impulses or do whatever else-- grow, or divide, or whatever else the cell might need. So really, cellular respiration, to say it produces energy, a little disingenuous. It's really the process of taking glucose and producing ATPs, with maybe heat as a byproduct. But it's probably nice to have that heat around. We need to be reasonably warm in order for our cells to operate correctly. So the whole point is really to go from glucose, from one mole of glucose-- and the textbooks will tell you-- to 38 ATPs. you'll produce 38 ATPs. I was reading up a little bit before doing this video. And the reality is, depending on the efficiency of the cell in performing cellular respiration, it'll probably be more on the order of 29 to 30 ATPs. But there's a huge variation here and people are really still studying this idea. But this is all cellular respiration is. In the next few videos we're going to break it down into its kind of constituent parts. And I'm going to introduce them to you right now, just so you realize that these are parts of cellular respiration. The first stage is called glycolysis. Which literally means breaking up glucose. And just so you know, this part, the glyco for glucose and then lysis means to break up. When you saw hydrolysis, it means using water to break up a molecule. Glycolysis means we're going to be breaking up glucose." + }, + { + "Q": "at 1:54 and a few other places, Sal says a few \"moles\" of oxygen, hydrogen, etc. Bye \"moles\", does he mean molecules?", + "A": "No, he means moles. A mole is a number, like a dozen, but much larger. 6.02*10^23. A mole of hydrogen atoms is 6.02*10^23 of those atoms.", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 114 + ], + "3min_transcript": "In my humble opinion, the single most important biochemical reaction, especially to us, is cellular respiration. And the reason why I feel so strongly about that is because this is how we derive energy from what we eat, or from our fuel. Or if we want to be specific, from glucose. At the end of the day, most of what we eat, or at least carbohydrates, end up as glucose. In future videos I'll talk about how we derive energy from fats or proteins. But cellular respiration, let's us go from glucose to energy and some other byproducts. And to be a little bit more specific about it, let me write the chemical reaction right here. So the chemical formula for glucose, you're going to have six carbons, twelve hydrogens and six oxygens. So that's your glucose right there. So if you had one mole of glucose-- let me write that, mole of glucose, if you had six moles of molecular oxygen running around the cell, then-- and this is kind of a gross simplification for cellular respiration. I think you're going to appreciate over the course of the next few videos, that one can get as involved into this mechanism as possible. But I think it's nice to get the big picture. But if you give me some glucose, if you have one mole of glucose and six moles of oxygen, through the process of cellular respiration-- and so I'm just writing it as kind of a big black box right now, let me pick a nice color. So this is cellular respiration. Which we'll see is quite involved. But I guess anything can be, if you want to be particular enough about it. Through cellular respiration we're going to produce six moles of carbon dioxide. Six moles of water. going to produce energy. We're going to produce energy. And this is the energy that can be used to do useful work, to heat our bodies, to provide electrical impulses in our brains. Whatever energy, especially a human body needs, but it's not just humans, is provided by this cellular respiration mechanism. And when you say energy, you might say, hey Sal, on the last video didn't you just-- well, if that was the last video you watched, you probably saw that I said ATP is the energy currency for biological systems. And so you might say, hey, well it looks like glucose is the energy currency for biological systems. And to some degree, both answers would be correct. But to just see how it fits together is that the process of cellular respiration, it does produce energy directly." + }, + { + "Q": "At 3:22 when Sal says that energy generates 38 ATPs isn't it 36 ATP?", + "A": "A very efficient cell can produce 38 TOTAL ATP from one glucose molecule, but since glycolysis requires 2 ATP, the NET gain would be 36.", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 202 + ], + "3min_transcript": "mole of glucose, if you had six moles of molecular oxygen running around the cell, then-- and this is kind of a gross simplification for cellular respiration. I think you're going to appreciate over the course of the next few videos, that one can get as involved into this mechanism as possible. But I think it's nice to get the big picture. But if you give me some glucose, if you have one mole of glucose and six moles of oxygen, through the process of cellular respiration-- and so I'm just writing it as kind of a big black box right now, let me pick a nice color. So this is cellular respiration. Which we'll see is quite involved. But I guess anything can be, if you want to be particular enough about it. Through cellular respiration we're going to produce six moles of carbon dioxide. Six moles of water. going to produce energy. We're going to produce energy. And this is the energy that can be used to do useful work, to heat our bodies, to provide electrical impulses in our brains. Whatever energy, especially a human body needs, but it's not just humans, is provided by this cellular respiration mechanism. And when you say energy, you might say, hey Sal, on the last video didn't you just-- well, if that was the last video you watched, you probably saw that I said ATP is the energy currency for biological systems. And so you might say, hey, well it looks like glucose is the energy currency for biological systems. And to some degree, both answers would be correct. But to just see how it fits together is that the process of cellular respiration, it does produce energy directly. So if I were to break down this energy portion of cellular respiration right there, some of it would just be heat. You know, it just warms up the cell. And then some of it is used-- and this is what the textbooks will tell you. The textbooks will say it produces 38 ATPs. It can be more readily used by cells to contract muscles or to generate nerve impulses or do whatever else-- grow, or divide, or whatever else the cell might need. So really, cellular respiration, to say it produces energy, a little disingenuous. It's really the process of taking glucose and producing ATPs, with maybe heat as a byproduct. But it's probably nice to have that heat around. We need to be reasonably warm in order for our cells to operate correctly. So the whole point is really to go from glucose, from one mole of glucose-- and the textbooks will tell you-- to 38 ATPs." + }, + { + "Q": "I assumed that Sal was saying 38 ATPs total at first because he was looking at the total number of ATPs not the net. However, in 10:30 , he says the net gain of ATP is 38. My books say 36. Am I understanding this wrong?", + "A": "well the total ATPs produced in aerobic repiration should be 38... However, muscle cells & neurons produce only 36 molecules of ATP per glucose molecule. Its because the 2 molecules of NADH produced during glycolysis in muscle cells & neurons dont enter the ETC directly but through other carriers, which transfer the electrons and H+ to the cytochromes. Therefore, these two NADH molecules produce 2 molecules of ATP only, instead of the usual 3...", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 630 + ], + "3min_transcript": "And then, and this is kind of the interesting point, there's another process that you can say happens after the Krebs cycle. But we're in a cell and everything's bumping into everything all of the time. But it's normally viewed to be after glycolysis And this requires oxygen. So let me be clear, glycolysis, this first step, no oxygen required. Doesn't need oxygen. It can occur with oxygen or without it. Oxygen not needed. Or you could say this is called an anaerobic process. This is the anaerobic part of the respiration. Let me write that down too. Anaerobic. Maybe I'll write that down here. Glycolysis, since it doesn't need oxygen, we You might be familiar with the idea of aerobic exercise. The whole idea of aerobic exercise is to make you breathe hard because you need a lot of oxygen to do aerobic exercise. So anaerobic means you don't need oxygen. Aerobic means it needs oxygen. Anaerobic means the opposite. You don't need oxygen. So, glycolysis anaerobic. And it produces two ATPs net. And then you go to the Krebs cycle, there's a little bit of setup involved here. And we'll do the detail of that in the future. But then you move over to the Krebs cycle, which is aerobic. It is aerobic. It requires oxygen to be around. And then this produces two ATPs. And then this is the part that, frankly, when I first learned it, confused me a lot. But I'll just write it in order the way it's Then you have something called-- we're using the same colors too much-- you have something called the electron And this part gets credit for producing the bulk of the ATPs. 34 ATPs. And this is also aerobic. It requires oxygen. So you can see, if you had no oxygen, if the cells weren't getting enough oxygen, you can produce a little bit of energy. But it's nowhere near as much as you can produce once you have the oxygen. And actually when you start running out of oxygen, this can't proceed forward, so what happens is some of these byproducts of glycolysis, instead of going into the Krebs cycle and the electron transport chain, where they need oxygen, instead they go through a side process called fermentation. For some organisms, this process of fermentation takes your byproducts of glycolysis and literally produces alcohol." + }, + { + "Q": "At 03:55 you mention there are 38 ATP molecules that are being released According to other sources it is 36 why is that ?", + "A": "because they add the number of ATP molecules from the first and the second step of the cellular respiration. Since the first step is produces 2 ATP molecules and the second step produces 36 ATP molecules so when we add them together it is 38 ATP molecules.", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 235 + ], + "3min_transcript": "going to produce energy. We're going to produce energy. And this is the energy that can be used to do useful work, to heat our bodies, to provide electrical impulses in our brains. Whatever energy, especially a human body needs, but it's not just humans, is provided by this cellular respiration mechanism. And when you say energy, you might say, hey Sal, on the last video didn't you just-- well, if that was the last video you watched, you probably saw that I said ATP is the energy currency for biological systems. And so you might say, hey, well it looks like glucose is the energy currency for biological systems. And to some degree, both answers would be correct. But to just see how it fits together is that the process of cellular respiration, it does produce energy directly. So if I were to break down this energy portion of cellular respiration right there, some of it would just be heat. You know, it just warms up the cell. And then some of it is used-- and this is what the textbooks will tell you. The textbooks will say it produces 38 ATPs. It can be more readily used by cells to contract muscles or to generate nerve impulses or do whatever else-- grow, or divide, or whatever else the cell might need. So really, cellular respiration, to say it produces energy, a little disingenuous. It's really the process of taking glucose and producing ATPs, with maybe heat as a byproduct. But it's probably nice to have that heat around. We need to be reasonably warm in order for our cells to operate correctly. So the whole point is really to go from glucose, from one mole of glucose-- and the textbooks will tell you-- to 38 ATPs. you'll produce 38 ATPs. I was reading up a little bit before doing this video. And the reality is, depending on the efficiency of the cell in performing cellular respiration, it'll probably be more on the order of 29 to 30 ATPs. But there's a huge variation here and people are really still studying this idea. But this is all cellular respiration is. In the next few videos we're going to break it down into its kind of constituent parts. And I'm going to introduce them to you right now, just so you realize that these are parts of cellular respiration. The first stage is called glycolysis. Which literally means breaking up glucose. And just so you know, this part, the glyco for glucose and then lysis means to break up. When you saw hydrolysis, it means using water to break up a molecule. Glycolysis means we're going to be breaking up glucose." + }, + { + "Q": "When Sal was talking about a net gain in ATP in the video about 9:25 in glycolysis, are the other processes in cellular respiration also have a net gain? Because Sal didn't write whether or not the 2 ATP in Krebs and the 34 in the transport chain were net gains. Thanks!", + "A": "they are. In the ideal cell you have a net win of 38 ATPs. 2+2+34=38 It s a bit confusing if you look at the pictures of it. Sometimes it just says you ll win NADH or FADH2. But they are used to win ATP.", + "video_name": "2f7YwCtHcgk", + "timestamps": [ + 565 + ], + "3min_transcript": "that's added on to that. You know, these things are all bonded to other things, with oxygens and hydrogens and whatever. But each of these 3-carbon backbone molecules are called pyruvate. We'll go into a lot more detail on that. But glycolysis, it by itself generates-- well, it needs two ATPs. And it generates four ATPs. So on a net basis, it generates two-- let me write this in a different color-- it generates two net ATPs. So that's the first stage. And this can occur completely in the absence of oxygen. I'll do a whole video on glycolysis in the future. Then these byproducts, they get re-engineered a little bit. And then they enter into what's called the Krebs cycle. And then, and this is kind of the interesting point, there's another process that you can say happens after the Krebs cycle. But we're in a cell and everything's bumping into everything all of the time. But it's normally viewed to be after glycolysis And this requires oxygen. So let me be clear, glycolysis, this first step, no oxygen required. Doesn't need oxygen. It can occur with oxygen or without it. Oxygen not needed. Or you could say this is called an anaerobic process. This is the anaerobic part of the respiration. Let me write that down too. Anaerobic. Maybe I'll write that down here. Glycolysis, since it doesn't need oxygen, we You might be familiar with the idea of aerobic exercise. The whole idea of aerobic exercise is to make you breathe hard because you need a lot of oxygen to do aerobic exercise. So anaerobic means you don't need oxygen. Aerobic means it needs oxygen. Anaerobic means the opposite. You don't need oxygen. So, glycolysis anaerobic. And it produces two ATPs net. And then you go to the Krebs cycle, there's a little bit of setup involved here. And we'll do the detail of that in the future. But then you move over to the Krebs cycle, which is aerobic. It is aerobic. It requires oxygen to be around. And then this produces two ATPs. And then this is the part that, frankly, when I first learned it, confused me a lot. But I'll just write it in order the way it's Then you have something called-- we're using the same colors too much-- you have something called the electron" + }, + { + "Q": "At 08:48, why would you look for a resonance structure? Isn't the Oxygen in this case at a formal charge of 0 (6 - (4+2)). It doesn't make sense to me why you would push the electrons off the double bond and end up with a +O and -C as I thought nature disliked charges and prefers neutral atoms.", + "A": "Yes, it does, but the contributor is still a minor contributor to the resonance hybrid.", + "video_name": "UHZHkZ6_H5o", + "timestamps": [ + 528 + ], + "3min_transcript": "so I put that in, and so when you're doing this for cations, you're not gonna move a positive charge, so when you're drawing your arrows, you're showing the movement of electrons, so the arrow that I drew over here, let me go ahead a mark it in magenta. So this arrow in magenta is showing the movement of those electrons in blue, and when those electrons in blue move, that creates a plus-one formal charge on this carbon, and so don't try to move positive charges: Remember, you're always pushing electrons around. Then finally, let's do one more. So, for this situation, this is for acetone, so we have a carbon right here, double-bonded to an oxygen, and we know that there are differences in electronegativity between carbon and oxygen: Oxygen is more electronegative. So what would happen if we took those pi electrons? blue for pi electrons, so these pi electrons right here, and we move those pi electrons off, onto the more electronegative atom, like that, so let's go ahead and draw our resonance structure. So this top oxygen would have three lone pairs of electrons: one of those lone pairs are the ones in blue, those pi electrons; that's gonna give the oxygen a negative-one formal charge, and we took a bond away from this carbon, so we took a bond away from this carbon, and that's going to give that carbon a plus-one formal charge. And so, when you think about your resonance structures, first if all, I should point out that one negative charge and one positive charge give you an overall charge of zero, so charge is conserved, and over here, of course, the charge is zero. So if you're thinking about the resonance hybrid, we know that both structures contribute to the overall hybrid, but the one on the right isn't going to contribute as much, so this one you have a positive and a negative charge, and the goal, of course, is to get to overall neutral. But, what's nice about drawing this resonance structure, and thinking about this resonance structure, is it's emphasizing the difference in electronegativity, so, for this one, you could just say oxygen get a partial negative, and this carbon right here, gets a partial positive. So that's one way of thinking about it, which is very helpful for reactions. But drawing this resonance structure is just another way of thinking about, emphasizing the fact that when you're thinking about the hybrid, you're thinking about a little more electron density on that oxygen. All right, so once again, do lots of practice; the more you do, the better you get at drawing resonance structures, and the more the patterns, the easier the patterns become." + }, + { + "Q": "at 1:59 how is the formal charge of oxygen negative one?", + "A": "Formal charge = valence electrons - lone pair electrons - bonds 6 - 6 - 1 = -1", + "video_name": "UHZHkZ6_H5o", + "timestamps": [ + 119 + ], + "3min_transcript": "Voiceover: Let's look at a few of the patterns for drawing resonance structures, and the first pattern we're gonna look at, is a lone pair of electrons next to a pi bond. And so, here's a lone pair of electrons; I'm gonna highlight it in magenta, that lone pair of electrons is located on this carbon, let me go ahead and put this carbon in green, here. And I'm saying, there's a negative-one formal charge on that carbon in green, and so that carbon in green is also bonded to a hydrogen, so once again, you need to be very familiar with assigning formal charges. So we have a lone pair of electrons next to a pi bond, because over here, we have a double-bond between the carbon and the oxygen, one of those bonds is a sigma bond, and one of those bonds is a pi bond, so I'm just gonna say that these are the pi electrons. So our goal in drawing a resonance structure is to de-localize that negative-one formal charge, so spread out some electron density. And so, we could take the electrons in magenta, and move them into here, to form a double-bond, between the carbon in green and this carbon right here, and that'd be too many bonds to the carbon in yellow, onto this top oxygen here. So we go ahead, and draw in our brackets, and we put our double-headed resonance arrow, and we draw the other resonance structure, so we have our ring, like that, and then we have, now, a double-bond between those two carbons, and then this top oxygen here, now has only one bond to it. The oxygen used to have two lone pairs of electrons, now it has three, because it just picked up a pair of electrons from that pi bond. So let's go ahead, and follow the electrons. The electrons in magenta moved in here, to form our pi bond, like that, and the electrons in the pi bond, in blue, moved off, onto this oxygen, so I'm saying that they are those electrons. That gives the top oxygen a negative-one formal charge, and so we have our two resonance structures for the enalate anion. We know that both resonance structures contribute to the overall hybrid, and if you think about we have had a negative-one formal charge on the carbon in green, so that's a carb anion; and for the resonance structure on the right, we had a negative one formal charge on the oxygen, so that's an oxyanion. Oxygen is more electronegative than carbon, which means it's more likely to support a negative-one formal charge, and so the resonance structure on the right contributes more to the overall hybrid for an enalate anion. All right, let's do another pattern, a lone pair of electrons next to a positive charge, this time. So, let's look at nitromethane, and we could look at this lone pair of electrons here, on this oxygen, and that lone pair of electrons is next to a positive charge; this nitrogen has a plus one formal charge on it. And, so, let's think about drawing the resonance structure, so our goal is to de-localize charge, to spread charge out." + }, + { + "Q": "At 3:49, how come we start numbering at the left side for the second example, whereas Bromine is at 2?", + "A": "Generally, you want the lowest numbers possible for your substituents. There are some exceptions, such as in alcohols, where you want the lowest number on the -OH group (even if it makes big numbers for any other substituents.)", + "video_name": "nQ7QSV4JRSs", + "timestamps": [ + 229 + ], + "3min_transcript": "We have 1, 2, 3, 4, 5 carbons. So it's going to be pent. And there's no double bonds. So I'll just write pentane right then. And we're not going to just write a pentane because actually, the fact that makes it an alcohol, that takes precedence over the fact that it is an alkane. So it actually, the suffix of the word will involve the alcohol part. So it is pentanol. That tells us that's an alcohol. And to know where the OH is grouped, we'll start numbering closest to the OH. So 1, 2, 3, 4, 5. Sometimes it'll be called 2-pentanol. And this is pretty clear because we only have one group here, only one OH. So we know that that is what the 2 applies to. But a lot of times, if people want to be a little bit more particular, they might write pentan-2-ol. multiple functional groups. So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try the name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. We want to call it an octyne. But because we have an alcohol there, we want to call this an octyne-- let me make it very clear. Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that So it is oct-5-yn. That tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-al. And now we have these two bromo groups here on the 7 carbon. So it's 7,7-dibromo oct-5-yn-4-al. And this would all be one word. Let me make sure that you realize that I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex." + }, + { + "Q": "At 4:28, why is it a oct-5-yn-4-ol but not 5-octyn-4-ol?", + "A": "As it was explained to me, both are acceptable, but the first way is less confusing.", + "video_name": "nQ7QSV4JRSs", + "timestamps": [ + 268 + ], + "3min_transcript": "We have 1, 2, 3, 4, 5 carbons. So it's going to be pent. And there's no double bonds. So I'll just write pentane right then. And we're not going to just write a pentane because actually, the fact that makes it an alcohol, that takes precedence over the fact that it is an alkane. So it actually, the suffix of the word will involve the alcohol part. So it is pentanol. That tells us that's an alcohol. And to know where the OH is grouped, we'll start numbering closest to the OH. So 1, 2, 3, 4, 5. Sometimes it'll be called 2-pentanol. And this is pretty clear because we only have one group here, only one OH. So we know that that is what the 2 applies to. But a lot of times, if people want to be a little bit more particular, they might write pentan-2-ol. multiple functional groups. So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try the name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. We want to call it an octyne. But because we have an alcohol there, we want to call this an octyne-- let me make it very clear. Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that So it is oct-5-yn. That tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-al. And now we have these two bromo groups here on the 7 carbon. So it's 7,7-dibromo oct-5-yn-4-al. And this would all be one word. Let me make sure that you realize that I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex." + }, + { + "Q": "At 7:38 Sal says that the area of the piston = area of the base of the container but dosen't the base of container have walls on it's sides? So, the area of the base is not = to the area of piston!\nPlease explain me!", + "A": "The area of the face of the piston is the same as the area of the base of the cylinder; the piston fits snuggly into the cylinder. I an not sure why you think the area of the walls of the cylinder are involved here.", + "video_name": "obeGVTOZyfE", + "timestamps": [ + 458 + ], + "3min_transcript": "system here did some work. So I'm claiming that it applied a force to this piston, and it applied that force to the piston for some distance. So let's figure out what that is, and if we can somehow relate it to other macro properties that we know reasonably well. Well we know the pressure and the volume, right? We know the pressure that's being exerted on the piston, at least at this point in time. Pressure is equal to force per area. Remember, this piston, you're just seeing it from the side, but it's a kind of a flat plate or a flat ceiling on top of this thing. And at what distance did it move it? You know I could blow it up a little bit. It moved it some-- I didn't draw it too big here-- some x, some distance x. So what is the force that it pushed it up? Well, the force, we know its pressure, the pressure's force per area. So if we want to know the force, we have to multiply pressure times area. If we multiply both sides of this times area, we get force. So we're essentially saying the area of this little ceiling to this container right there, you know, it could be, I could draw with some depth, but I think you It has some area. It's probably the same area as the base of the container. So we could say that the force being applied by our system-- let me do it in a new color-- the force is equal to our pressure of the system, times the area of the ceiling of our container of the piston. Now that's the force. Now what's the distance? The distance is-- I'll do it in blue-- it's this change right here. I didn't draw it too big, but that's that x. Now let's see if we can relate this somehow. Let me draw it a little bit bigger. And I'll try to draw in three dimensions. So let me draw the piston. What color did I do it in? I did it in that brown color. So our piston looks something-- I'll draw it as a elipse-- the piston looks like that. And it got pushed up. So it got pushed up some distance x. Let me see how good I can-- whoops. Let me copy and paste that same-- So the piston gets pushed up some distance x. Let me draw that. It got pushed up some distance x. And we're claiming that our-- oh sorry, this is the force. Sorry, let me be clear." + }, + { + "Q": "At 5:34 how does he get \"(1) in the equation.", + "A": "He is showing that six \u00cf\u0080 electrons satisfy H\u00c3\u00bcckel s 4n+2 rule. 4n +2 = 6 4n = 6 \u00e2\u0080\u0093 2 = 4 n = 4/4 = 1 \u00e2\u0088\u00b4 Six \u00cf\u0080 electrons satisfy 4n +2 when n = 1", + "video_name": "oDigu9YxXUg", + "timestamps": [ + 334 + ], + "3min_transcript": "So we're going to start down here. So we're going to inscribe a hexagon. Let's see if we can put a hexagon in here. And so we have a six-sided figure here in our frost circle. The key point about a frost circle is everywhere your polygon intersects with your circle, that represents the energy level of a molecular orbital. And so this intersection right here, this intersection here, and then all the way around. And so we have our six molecular orbitals. And we have the relative energy levels of those six molecular orbitals. So let me go ahead and draw them over here. So we have three molecular orbitals which are above the center line. And those are higher in energy. And we know that those are called antibonding molecular orbitals. So these are antibonding molecular orbitals, which are the highest in energy. If we look down here, there are three molecular orbitals which are below the center line. So those are lower in energy. And if we had some molecular orbitals that were on the center line, those would be non-bonding molecular orbitals. We're going to go ahead and fill our molecular orbitals with our pi electrons. So go back over here. And remember that benzene has 6 pi electrons. And so filling molecular orbitals is analogous to electron configurations. You're going to fill the lowest molecular orbital first. And each orbital can hold two electrons, like electron configurations. And so we're going to go ahead and put two electrons into the lowest bonding molecular orbital. So I have four more pi electrons to worry about. And I go ahead and put those in. And I have filled the bonding molecular orbitals of benzene. So I have represented all 6 pi electrons. If I think about Huckel's rule, 4n plus 2, I have 6 pi electrons. So if n is equal to 1, Huckel's rule is satisfied. Because I would do 4 times 1, plus 2. And so 6 pi electrons follows Huckel's rule. If we look at the frost circle and we look at the molecular orbitals, we can understand Huckel's rule a little bit better visually. So if I think about these two electrons down here, you could think about that's where the two comes from in Huckel's rule. If think about these four electrons up here, that would be four electrons times our positive integer of 1. So 4 times 1, plus 2 gives us six pi electrons. And we have filled the bonding molecular orbitals of benzene, which confers the extra stability that we call aromaticity or aromatic stabilization. And so benzene is aromatic. It follows our different criteria. In the next few videos, we're going to look at several other examples of aromatic compounds and ions." + }, + { + "Q": "From 4:11 does hank mean the actual taxonomic ' class' when he mentions class ?", + "A": "Yes, hank means classes", + "video_name": "c7Yy9v8dH8s", + "timestamps": [ + 251 + ], + "3min_transcript": "In humans, the notochord is reduced to the disks, the cartilage that we have beween our vertebra. Second, we have the nerve cord itself, called the dorsal hollow nerve cord, a tube made of nerve fibers that develops into the central nervous system. This is what makes chordates different from other animal phyla, which have solid ventral nerve cords, meaning they run along the front, or stomach, side. Third, all chordates have pharyngeal slits. In invertebrates like the lancelet here, they function as filters for feeding. In fish and other aquatic animals, they're heel slits, and in land-dwelling vertebrates like us, they disappear before we're born. But, that tissue develops into areas around our jaws, ears, and other structures in the head and neck. Finally, we can't forget our fourth synapomorphy, the post-anal tail. which is exactly what it sounds like. It helps propel aquatic animals through the water, makes our dog look happy when she wags it, and in humans, it shrinks during embryonic development, into what is known as the coccyx, or tailbone. It's right here. These four traits all began to appear during the Cambrian explosion, more than 500 million years ago. Today, they're shared by members of all three chordate sub-phyla, even if the animals in those sub-phyla look pretty much nothing like each other. For instance, our new friends here in cephalochordata are the oldest living sub-phyla, but you can't forget the other invertebrate group of chordates, the urochordata - literally, tail cords. There are over 2,000 species here, including sea squirts. If you're confused about why this ended up in a phylum with us, it's because they have tadpole-like larva with all four chordate characteristics. The adults, which actually have a highly-developed internal structure, with a heart and other organs, retain the pharyngeal slits, but all the other chordate features disappear or reform into other structures. The third and last, and most complex sub-phylum, is the vertebrata, and has the most species in it, because its members have a hard backbone, which has allowed for an explosion in diversity, You can see how fantastic this diversity really is when you break down vertebrata into its many, many classes, from slimy sea snake-y things to us warm and fuzzy mammals. As these classes become more complex, you can identify the traits they each develop that gave them an evolutionary edge over the ones that came before. For example, how's this for an awesome trait - a brain. Vertebrates with a head that contains sensory organs and a brain are called craniates. They also always have a heart with at least two chambers. Since this is science, you're gonna have to know that there's an exception for every rule that you're gonna have to remember. The exception in this case is the myxini, or hagfish, the only vertebrate class that has no vertebra, but is classified with us because it has a skull. This snake-like creature swims by using segmented muscles to exert force against its notochord. Whatever, hagfish. Closely related to it is the class petromyzontida, otherwise known as lampreys, the oldest-living lineage of vertebrates. Now, these have a backbone made of cartilage," + }, + { + "Q": "What he calls a tetrad at 7:00, I thought (from lectures) was a bivalent - are these two names for the same thing, or is there a difference?", + "A": "There both the same because the definition of bivalent is a pair of homologous chromosomes.", + "video_name": "04gQ0bQu6xk", + "timestamps": [ + 420 + ], + "3min_transcript": "A couple of things happen. The nuclear membrane begins to dissolve. This is very similar to prophase when we're looking at mitosis. So the nuclear envelope begins to dissolve. These things start to maybe migrate a little bit. So these characters are trying to go at different ends. And the DNA starts to bunch up into kind of its condensed form. So now I can draw it. So now I can start to draw it as proper. So this is the one from the father right over here. And this is the one from the mother. And I'm drawing, I'm overlapping on purpose because something very interesting happens especially in meiosis. So it's the mother right over here. Let me see. Let's now do the centromere in blue now. That's the centromere. These are the shorter ones from the mother. And actually, let me just do draw them on opposite sides just to show that they don't have to, the ones from the father aren't always on the left hand side. So this is the shorter one from the father. They couldn't be all on the left hand side but doesn't this all they have to be. And this is the shorter one from the mother. And I will draw this overlapping although they could have. Shorter one from the mother. And once again, each of these, this is a homologous pair, that's a homologous pair over there. Now, the DNA has been replicated so in each of the chromosomes in a homologous pair, you have two sister chromatids. And so, in this entire homologous pair, you have four chromatids. And so, this is sometimes called a tetrad. So let me just give ourselves some terminology. So this right over here is called a tetrad or often called a tetrad. Now, the reason why I drew this overlapping Let me label this. This is prophase I. You can get some genetic recombination, some homologous recombination. Once again, this is homologous pair. One chromosome from the father that I've gotten from the father. The species or the cell got it from its father's cell and one from the mother. And they're homologous. They might contain different base pairs, different actual DNA, but they code for the same genes. Over simplification, but in a similar place on each of these it might code for eye color or I don't know, personality. Nothing is that simple in how tall you get and it's not that simple in DNA but just to give you an idea of how it is. And the reason why I overlapped them like this is to show how the recombination can occur. So actually, let me zoom in. So this is the one from the father. Once again, it's on the condensed form. This is one chromosome made up of two sister chromatids" + }, + { + "Q": "At 7:40, why is the object exerting an equal force on the hand? I mean, after all it's the object that is getting accelerated, so the force on the object should be bigger than the force acting back on the hand, right? It seems to me like, if the forces were truly equal, nothing would move, not the hand, not the object...", + "A": "Reaction pair forces are on different objects, not the same object. I push on the chair, the chair pushes on me. How many forces on the chair? Just one. So it accelerates. If you and the chair are floating in outer space, when you push on the chair it will also push on you and you will move in opposite directions. If you masses are equal, you will have equal but opposite acceleration.", + "video_name": "By-ggTfeuJU", + "timestamps": [ + 460 + ], + "3min_transcript": "" + }, + { + "Q": "At 6:30, wouldn't you have to throw an enormous object with A LOT of mass to accelerate backward enough to grab onto the space shuttle?", + "A": "Yeah, that would be ideal, but you probably don t have a massive object on you when you re out in space. I assume it would require more energy to move around.", + "video_name": "By-ggTfeuJU", + "timestamps": [ + 390 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:37, why is Sal saying that L goes in the same direction as current. My intuition tells me that the current is going through the L of the conductor and how can length have a direction?", + "A": "You can consider L to be displacement rather than distance or length . We know displacement is a vector while distance and length are scalars. So L has a direction! :)", + "video_name": "l3hw0twZSCc", + "timestamps": [ + 277 + ], + "3min_transcript": "And then on this side of I1, what happens? Well, on this side, you can see the fingers come back around. So it pops out when it intersects with your video monitor. So on this side, the vectors-- this is the top of an arrow, coming out at you. Fair enough. So I1, by going in this direction, is generating a magnetic field that, at least where I2 is concerned, that magnetic field is going into the page. So what was our formula? And this all came from the first formula we learned about, the effect of a magnetic field soon. on a moving charge. But what was the formula of the net magnetic force on a current carrying wire? It was the force-- I'll do it in blue-- it's a vector, has a magnitude and direction-- is equal to the current. Well, in this case, we want to know the force on this Caused by this magnetic field, by magnetic field 1. So it will be equal to I2, the magnitude of this current, times L-- where L is-- because you can't just say, oh well, what is the effect on this wire? You have to know how much wire is under consideration. So let's say we have a length of wire. And then of course, if you know the length of wire and we knew its mass and we knew the force on it, we could figure out its acceleration in some directions. So let's say that this distance is L, and it's a vector. L goes in the same direction as the current. That's just the convention we're using. It makes things simple. So that's L. So the force on this wire, or at least the length L of this wire, is going to be equal to current 2 times L. We could call that even L2, just so that you know that it deals with wire 2. That's a vector quantity. It's just a notation. I've seen professors do it either way, I've seen it written either way, as well. Cross the magnetic field that it's in. What's the magnetic field that it's in? The magnetic field-- I'll do it in magenta, because it's the magnetic field created by current 1. So it's magnetic field 1, which is this magnetic field. So before going into the math, let's just figure out what direction is this net force going to be in? So here we say, well, the current is a scalar, so that's not going to affect the direction. What's the direction of L2? This is L2. I didn't label it L2 on the diagram. What's the direction of L2? Well, it's up. And then the direction of B1, the magnetic field created by current 1, is going into the page here. So here we just do the standard cross product. Let me see if I can pull this off. This is actually an easy one to draw." + }, + { + "Q": "Why do you use a c instead of an equal sign? Is this intentional or just a fault of drawing?(10:04)", + "A": "It is an equal sign. Just the video has lower quality, so it looks like it is c.", + "video_name": "7vHh1sfZ5KE", + "timestamps": [ + 604 + ], + "3min_transcript": "get here, they've experienced some potential drop. So the electrons here actually are a little bit less eager to get here. And then once they've gone through here, maybe they're just tired of bumping around so much. And once they're here, they're a little bit less eager to get here. So there's a voltage drop across each device, right? So the total voltage is equal to the voltage drop across each of the devices. And now let's go back to the convention, and we'll say that the current is going in that direction. The total voltage drop is equal to V1 plus V2 plus V3, so the total voltage drop is equal to I1 R1 plus I2 R2 plus I3 R3. And what's the total voltage drop? Well, that's equal to the total current through the whole system. I-total, or we just call it I, times the total resistance is Well, we know that all the I's are the same. Hopefully, you can take it as, just conceptually it makes sense to you that the current through the entire circuit will be the same. So all these I's are the same, so we can just cancel them out. Divide both sides by that I. We assume it's non-zero, so I, I, I, I, and then we have that the total resistance of the circuit is equal to R1 plus R2 plus R3. So when you have resistors in series like this, the total resistance, their combined resistance, is just equal to their sum. And that was just a very long-winded way of explaining something very simple, and I'll do an example. Let's say that this voltage is-- I don't know. Let's say it's 20 volts. Let's say resistor 1 is 2 ohms. Let's say resistor 2 is total resistance through this circuit? Well, the total resistance is 2 ohms plus 3 ohms plus 5 ohms, so it's equal to 10 ohms. So total resistance is equal to 10 ohms. So if I were to ask you what is the current going through this circuit? Well, the total resistance is 10 ohms. We know Ohm's law: voltage is equal to current times resistance. The voltage is just equal to 20. 20 is equal to the current times 10 ohms, right? We just added the resistances. Divide both sides by 10. You get the current is equal to 2 amps or 2 coulombs per second. So what seemed like a very long-winded explanation actually results in something that's very, very, very easy to apply." + }, + { + "Q": "How can the voltage be equal at both the points (02:50)? In the previous video, Sal said that voltage is electric field times distance. Clearly, there is a difference in distance between the two points.", + "A": "Yeah that it is quite confusing, the way to think about voltage in a circuit is work being done on charge . In circuits we are assuming that the work to move a charge through a conducting wire is negligible compared to the work getting the electrons through the resistors. Of course that isn t true, work is needed just to move the electrons through a wire even if it doesn t have any resistance but that amount of work is totally negligible.", + "video_name": "7vHh1sfZ5KE", + "timestamps": [ + 170 + ], + "3min_transcript": "And the higher the voltage, the more they really want to get to this positive terminal. So what's going to happen in this circuit? Actually, let me label everything. So let's call this R1, let's call this R2, let's call this R3. The first thing I want you to realize is that between elements that the voltage is always constant. And why is that? Well, we assume that this is a perfect conductor-- let's say this little segment right here, right? And so it's a perfect conductor. Well, let's look at it at this end. So you have all these electrons. This is a perfect conductor, so there's nothing stopping these electrons from just distributing themselves over this wire. Before you encounter an element in the circuit or device or whatever you want to call that, you can view this ideal conducting wire just from a schematic point of view as an extension of the negative terminal. And similarly, you can view this wire right here, this part of the wire, as an extension of And the reason why I want to say that is because it actually turns out that it doesn't matter if you measure the voltage here. So let's say if I take a measure of the voltage across those two terminals using what we call a voltmeter. And I'll later do a whole video on how voltmeters work, but remember, when we measure voltage, we have to measure it at two points. Because voltage is a potential difference. It's not some kind of absolute number. It's a difference between essentially how bad do electrons want to get from here to here. So if we measure the voltage between those two points, it would be the exact same thing as if we measured the voltage between these two points. Theoretically. As we know, no wires really have no resistivity. All wires have a little bit, but when we draw these schematics, we assume that the wires are perfect conductors and all the resistance takes place in the resistor. So that's the first thing I want you to realize, and it makes things very-- so, for example, everywhere along this wire, this part of the wire, the voltage is constant. Everywhere along this wire, the voltage is constant. get too messy. That's a big important realization when you later become an electrical engineer and have much harder problems to solve. Let me erase all of this. Let me erase all of that. Let me redraw that, because we can't have that gap there, because if there was that gap, current wouldn't flow. That's actually-- well, I'll draw later how you can draw a switch, but a switch is essentially a gap. It looks like a gap in the circuit that you can open or Because if you open it, no current will flow. If you close it, current will flow. OK, so you now know that the voltage between devices is constant. The other thing I want to convince you is that the current through this entire circuit is constant, and that applies to any circuit in series. Now, what do I mean by series? Series just means that everything in the circuit is after one another, right? If we take the convention and we say current flows in this direction, it'll hit this resistor, then the next" + }, + { + "Q": "at 1:12, sal says that the electrons are on the negative terminal... dont they repel??", + "A": "yes, thats why current flows when a circuit is made", + "video_name": "7vHh1sfZ5KE", + "timestamps": [ + 72 + ], + "3min_transcript": "Let's make our circle a little bit more complicated now. So let's say I have a battery again, and let me do it in a different color just for variety. That's the positive terminal, that's the negative terminal. Let's say I have this perfect conductor, and let's say I have one resistor and I have another resistor. I don't know, just for fun, let's throw in a third resistor. And we know, of course, that the convention is that the current flows from positive to negative, that that's the flow of the current. And remember, current is just the charge that flows per unit of time or the speed of the charge flow. But we know, of course, that in reality what is happening, if there's any such thing as reality, is that we have a bunch of electrons here that, because of this voltage across the battery terminals, these electrons want to really badly And the higher the voltage, the more they really want to get to this positive terminal. So what's going to happen in this circuit? Actually, let me label everything. So let's call this R1, let's call this R2, let's call this R3. The first thing I want you to realize is that between elements that the voltage is always constant. And why is that? Well, we assume that this is a perfect conductor-- let's say this little segment right here, right? And so it's a perfect conductor. Well, let's look at it at this end. So you have all these electrons. This is a perfect conductor, so there's nothing stopping these electrons from just distributing themselves over this wire. Before you encounter an element in the circuit or device or whatever you want to call that, you can view this ideal conducting wire just from a schematic point of view as an extension of the negative terminal. And similarly, you can view this wire right here, this part of the wire, as an extension of And the reason why I want to say that is because it actually turns out that it doesn't matter if you measure the voltage here. So let's say if I take a measure of the voltage across those two terminals using what we call a voltmeter. And I'll later do a whole video on how voltmeters work, but remember, when we measure voltage, we have to measure it at two points. Because voltage is a potential difference. It's not some kind of absolute number. It's a difference between essentially how bad do electrons want to get from here to here. So if we measure the voltage between those two points, it would be the exact same thing as if we measured the voltage between these two points. Theoretically. As we know, no wires really have no resistivity. All wires have a little bit, but when we draw these schematics, we assume that the wires are perfect conductors and all the resistance takes place in the resistor. So that's the first thing I want you to realize, and it makes things very-- so, for example, everywhere along this wire, this part of the wire, the voltage is constant. Everywhere along this wire, the voltage is constant." + }, + { + "Q": "I might be wrong in my concepts......please correct me if I'm wrong....\n\nAt 1:55 isn't the compound supposed to be called \"2 butyl pentane\"(as the butyl group is linked to the 2nd carbon of the ring) instead of what sal tells us?", + "A": "First you number the atoms in the ring. If there is only one group attached to the ring, that ring carbon is automatically number 1. Thus the names are butylcyclopentane and sec-butylcyclopentene.", + "video_name": "TJUm860AjNw", + "timestamps": [ + 115 + ], + "3min_transcript": "Let's see if we can get the molecular structure for butylcyclopentane. So you just break this up the way we've done it in the last several videos, the suffix is -ane, so it is an alkane, all single bonds. So single bonds. It's pentane, so we're dealing with five carbons on the base, or on the backbone. So this is five carbons and it's a cyclopentane, so it's five carbons in a ring. So its five-carbon ring is the backbone, and then we have a butyl group added to that five-carbon ring. Now, you might say, hey, Sal, how do I know which carbon to add it to? When you're dealing with a ring and you only have one group on the ring, it doesn't matter. Let me just show you what I mean. So let's draw the five-carbon ring. Let's draw the cyclopentane. So it'll just be a pentagon, so one, two, three, four, One, two, three, four, five. Now, it doesn't matter where I draw the butyl group. It's all symmetric around there. We just have a ring and it's connected to a butyl group at some point. It'll start to matter once we add more than one group. So we can just pick any of these carbons to add the butyl group to. Now, just as a review, the but- prefix, that refers to, remember, methyl, ethyl, propyl, or meth-, eth-, prop-, but-. This is four carbons. This is a four-carbon alkyl group. So let me just add it here. I could have added it to any of these carbons around this cyclopentane ring. So if I just add it right here, so I'm going to have four carbons. So one, two, three, four. That is the butyl part of this whole thing. And then let me just attach them up. So you might be tempted to just draw this right there. And actually, this would be right. This is butylcyclopentane. But a question might arise. this first carbon on the butyl right there. I could have just as easily done it like this. I could have just as easily had it like this, where-- let me draw my butyl again, so I have one, two, three, four. So, once again, this is a butyl, but instead of being bonded to the cyclopentane on my first carbon, maybe it's bonded right here. Let me do it with that yellow color. Maybe it's bonded right here. This seems like maybe this could also be butylcyclopentane. It looks like we have a butyl group. This is a butyl right here. I drew a butyl group right over here, and I also drew a butyl group right over here. But these are fundamentally two different molecular structures. I'm touching the first carbon here." + }, + { + "Q": "At 4:23 when discussing naming this structurewould it be appropriate to call it a dibutyl cyclopentane", + "A": "No.The common name is sec-butylcyclopentane. The IUPAC name is (1-methylpropyl)cyclopentane. A dibutylcyclopentane would have two separate butyl groups attached to the cyclopentane ring.", + "video_name": "TJUm860AjNw", + "timestamps": [ + 263 + ], + "3min_transcript": "this first carbon on the butyl right there. I could have just as easily done it like this. I could have just as easily had it like this, where-- let me draw my butyl again, so I have one, two, three, four. So, once again, this is a butyl, but instead of being bonded to the cyclopentane on my first carbon, maybe it's bonded right here. Let me do it with that yellow color. Maybe it's bonded right here. This seems like maybe this could also be butylcyclopentane. It looks like we have a butyl group. This is a butyl right here. I drew a butyl group right over here, and I also drew a butyl group right over here. But these are fundamentally two different molecular structures. I'm touching the first carbon here. Now, there's two ways to differentiate this. One is the common naming and one is the systematic naming. So let me differentiate between the two. So in the common naming, and this can get a little bit involved, and this frankly is probably the most complicated part of naming organic compounds. Systematic is often more complicated, but it's easier to systematically come up with it. So there's a common and then there's a systematic. So the common way of doing it is, if you just say butylcyclopentane, that implies that you are bonding to the first or, depending on how you view it, the last carbon in the chain. So this right here is butylcyclopentane. This right here is not just butylcyclopentane. What you would do is you definitely have a cyclopentane ring, so this would definitely be a cyclopentane. Let me put some space here. And you do have a butyl group on it, so we do have a butyl group, but because we are bonded-- we aren't bonded to the first carbon. We're bonded to a carbon that is bonded to two other carbons. We call this sec-butylcyclopentane, so this is sec-. And everything I'm doing is obviously free-hand. If you were to see this in a book, the sec- would be italicized, or sometimes it would be written as s-butylcyclopentane. And this sec- means that we have attached to a carbon that is touching two other carbons. So you look at the butyl group, and say, well, which of these carbons is attached to two others? It's either that one or that one. And regardless of whether you're attached to this or this, if you think about it, it's fundamentally the same" + }, + { + "Q": "3:10 till 3:20 were a bit difficult to understand for me, can anyone please elaborate?", + "A": "He is saying that there are two different ways of naming the side chains (alkyl groups) \u00e2\u0080\u0094 the common names and the official or IUPAC names. There are four different butyl groups, and their names in the two systems are Common IUPAC n-butyl butyl isobutyl 2-methylpropyl sec-butyl 1-methylpropyl tert-butyl 1,1-dimethylethyl", + "video_name": "TJUm860AjNw", + "timestamps": [ + 190, + 200 + ], + "3min_transcript": "One, two, three, four, five. Now, it doesn't matter where I draw the butyl group. It's all symmetric around there. We just have a ring and it's connected to a butyl group at some point. It'll start to matter once we add more than one group. So we can just pick any of these carbons to add the butyl group to. Now, just as a review, the but- prefix, that refers to, remember, methyl, ethyl, propyl, or meth-, eth-, prop-, but-. This is four carbons. This is a four-carbon alkyl group. So let me just add it here. I could have added it to any of these carbons around this cyclopentane ring. So if I just add it right here, so I'm going to have four carbons. So one, two, three, four. That is the butyl part of this whole thing. And then let me just attach them up. So you might be tempted to just draw this right there. And actually, this would be right. This is butylcyclopentane. But a question might arise. this first carbon on the butyl right there. I could have just as easily done it like this. I could have just as easily had it like this, where-- let me draw my butyl again, so I have one, two, three, four. So, once again, this is a butyl, but instead of being bonded to the cyclopentane on my first carbon, maybe it's bonded right here. Let me do it with that yellow color. Maybe it's bonded right here. This seems like maybe this could also be butylcyclopentane. It looks like we have a butyl group. This is a butyl right here. I drew a butyl group right over here, and I also drew a butyl group right over here. But these are fundamentally two different molecular structures. I'm touching the first carbon here. Now, there's two ways to differentiate this. One is the common naming and one is the systematic naming. So let me differentiate between the two. So in the common naming, and this can get a little bit involved, and this frankly is probably the most complicated part of naming organic compounds. Systematic is often more complicated, but it's easier to systematically come up with it. So there's a common and then there's a systematic. So the common way of doing it is, if you just say butylcyclopentane, that implies that you are bonding to the first or, depending on how you view it, the last carbon in the chain. So this right here is butylcyclopentane. This right here is not just butylcyclopentane. What you would do is you definitely have a cyclopentane ring, so this would definitely be a cyclopentane. Let me put some space here." + }, + { + "Q": "When Sal gives us the (1,1 methyl ethyl)cyclopentane towards the end of the video (12:45) he uses two one's and a di to describe the two methyls on the same carbon. Is it also correct to just say 1-dimethyl? If so, which would be more correct because I feel like 1,1-dimethyl is more referring to two groups of methyls (an ethyl) on each side of that carbon. Slightly confusing. =(", + "A": "We use numbers to locate the substituents only when ther are many possibility for the positions of the substituents. In this case the name could be (dimethylethyl)cyclopentane because there is no other possibilty for the two methyl group to be attach on the ethyl substituent.", + "video_name": "TJUm860AjNw", + "timestamps": [ + 765 + ], + "3min_transcript": "Where we attached is one, two, three carbons. So once again, I'm doing that same one, two, three carbons. So once again, this is a propyl. Prop- is for three, but with a methyl group now is attached to the one, two, the second carbon. So this is 2-methyl. Let me make some space here. This is 2-methyl. So that describes this group right here. That describes this entire group cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as iso-butylcyclopentane or 2-methylpropylcyclopentane. And this is actually a -yl. I spelled it wrong. And then finally, we do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon and the longest chain I can do starting with that carbon is just one chain So we just have a two-carbon chain, right? One, two. The prefix for a two carbon is ethyl, or eth-. Eth-, and since it's a group, ethyl. And then we have two methyl groups attached right over there, and it's attached on the one carbon, right? This is one and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write one comma one to show that we have two groups attached to the first carbon and both of them are methyl. So we write 1,1-dimethyl, di- for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming is 1,1-. We have two groups attached to this first carbon. 1,1-dimethylethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached They're both attached to the one. We have two of them. That's why we wrote di- over there. So it's 1,1-dimethylethyl- and then finally, cyclopentane. So hopefully, that doesn't confuse you too much. I think if you watch the video over and over and try to practice it with your own problems, you'll see that the systematic name way is actually pretty, pretty logical. And actually, if you have more than five or six carbons in the group, they always or they tend to always use the systematic naming." + }, + { + "Q": "i think\nat 5:28 the vector denoted in blue should directed in downword direction", + "A": "The string is pulling in the upward direction", + "video_name": "_UrfHFEBIpU", + "timestamps": [ + 328 + ], + "3min_transcript": "red point, is stationary. It's not accelerating in either the left/right directions and it's not accelerating in the up/down directions. So we know that the net forces in both the x and y dimensions must be 0. My second question to you is, what is going to be the offset? Because we know already that at this point right here, there's going to be a downward force, which is the force of gravity again. The weight of this whole thing. We can assume that the wires have no weight for simplicity. So we know that there's going to be a downward force here, this is the force of gravity, right? The whole weight of this entire object of weight plus wire is pulling down. So what is going to be the upward force here? Well let's look at each of the wires. This second wire, T2, or we could call it w2, I guess. It has no y components. It's not lifting up at all. So it's just pulling to the left. So all of the upward lifting, all of that's going to occur from this first wire, from T1. So we know that the y component of T1, so let's call-- so if we say that this vector here. Let me do it in a different color. Because I know when I draw these diagrams it starts to get confusing. Let me actually use the line tool. So I have this. Let me make a thicker line. So we have this vector here, which is T1. And we would need to figure out what that is. And then we have the other vector, which is its y component, and I'll draw that like here. This is its y component. And then of course, it has an x component too, and I'll do that in-- let's see. I'll do that in red. Once again, this is just breaking up a force into its component vectors like we've-- a vector force into its x and y components like we've been doing in the last several problems. And these are just trigonometry problems, right? We could actually now, visually see that this is T sub 1 x and this is T sub 1 sub y. Oh, and I forgot to give you an important property of this problem that you needed to know before solving it. Is that the angle that the first wire forms with the ceiling, this is 30 degrees. So if that is 30 degrees, we also know that this is a parallel line to this." + }, + { + "Q": "Could someone give me a hard question to this, i am not sure if i have got this or not, thanks! And how did at 9:28, did Mr. Khan get 200N, he said divide both sides by 1/2, that gives 50N, does it not? So what is happening here", + "A": "It would have given 50 N if it were divided by 2. Diving by 1/2 is the same as multiplying by 2.", + "video_name": "_UrfHFEBIpU", + "timestamps": [ + 568 + ], + "3min_transcript": "So if we solve for T1 sub y we get T1 sine of 30 degrees is equal to T1 sub y. And what did we just say before we kind of dived into the math? We said all of the lifting on this point is being done by the y component of T1. Because T2 is not doing any lifting up or down, it's only pulling to the left. So the entire component that's keeping this object up, keeping it from falling is the y component of this tension vector. So that has to equal the force of gravity pulling down. This has to equal the force of gravity. That has to equal this or this point. So that's 100 Newtons. a little confusing to you. We just said, this point is stationery. It's not moving up or down. It's not accelerating up or down. And so we know that there's a downward force of 100 Newtons, so there must be an upward force that's being provided by these two wires. This wire is providing no upward force. So all of the upward force must be the y component or the upward component of this force vector on the first wire. So given that, we can now solve for the tension in this first wire because we have T1-- what's sine of 30? Sine of 30 degrees, in case you haven't memorized it, sine of 30 degrees is 1/2. So T1 times 1/2 is equal to 100 Newtons. Divide both sides by 1/2 and you get T1 is equal to 200 Newtons. second wire is. And we also, there's another clue here. This point isn't moving left or right, it's stationary. So we know that whatever the tension in this wire must be, it must be being offset by a tension or some other force in the opposite direction. And that force in the opposite direction is the x component of the first wire's tension. So it's this. So T2 is equal to the x component of the first wire's tension. And what's the x component? Well, it's going to be the tension in the first wire, 200 Newtons times the cosine of 30 degrees. It's adjacent over hypotenuse. And that's square root of 3 over 2. So it's 200 times the square root of 3 over 2, which equals 100 square root of 3." + }, + { + "Q": "At 11:36 he said \"homologous\".Is it not homozygous?", + "A": "Homozygous is when the two inherited alleles from both parents turn out to be both dominant or recessive at the same time, in the new organism.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 696 + ], + "3min_transcript": "reproduction have this complete set of chromosomes in it, which I find amazing. But only certain chromosomes-- for example, these genes will be completely useless in my fingernails, because all of a sudden, the straight and the curly don't matter that much. And I'm simplifying. Maybe they will on some other dimension. But let's say for simplicity, they won't matter in certain places. So certain genes are expressed in certain parts of the body, but every one of your body cells, and we call those somatic cells, and we'll separate those from the sex sells or the germs that we'll talk about later. So this is my body cells. So this is the great majority of your cells, and this is opposed to your germ cells. And the germ cells-- I'll just write it here, just so you get a clear-- for a male, that's the sperm cells, and for female that's the egg cells, or the ova. and what I want to give you the idea is that for every trait, I essentially have two versions: one from my mother and one from my father. Now, these right here are called homologous chromosomes. What that means is every time you see this prefix homologous or if you see like Homo sapiens or even the word homosexual or homogeneous, it means same, right? You see that all the time. So homologous means that they're almost the same. They're coding for the most part the same set of genes, but they're not identical. They actually might code for slightly different versions of the same gene. So depending on what versions I get, what is actually another word, and I'm overwhelming you with words here. So my genotype is exactly what alleles I have, what versions of the gene. So I got like the fifth version of the curly allele. There could be multiple versions of the curly allele in our gene pool. And maybe I got some version of the straight allele. That is my genotype. My phenotype is what my hair really looks like. So, for example, two people could have different genotypes with the same-- they might code for hair that looks pretty much the same, so it might have a very similar phenotype. So one phenotype can be represented by multiple genotypes. So that's just one thing to think about, and we'll talk a lot about that in the future, but I just wanted to introduce you to that there. Now, I entered this whole discussion because I wanted to" + }, + { + "Q": "at 12:40 what is a genotype?", + "A": "The genotype is the genetic makeup of a person, which phenotype is they physical makeup.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 760 + ], + "3min_transcript": "and what I want to give you the idea is that for every trait, I essentially have two versions: one from my mother and one from my father. Now, these right here are called homologous chromosomes. What that means is every time you see this prefix homologous or if you see like Homo sapiens or even the word homosexual or homogeneous, it means same, right? You see that all the time. So homologous means that they're almost the same. They're coding for the most part the same set of genes, but they're not identical. They actually might code for slightly different versions of the same gene. So depending on what versions I get, what is actually another word, and I'm overwhelming you with words here. So my genotype is exactly what alleles I have, what versions of the gene. So I got like the fifth version of the curly allele. There could be multiple versions of the curly allele in our gene pool. And maybe I got some version of the straight allele. That is my genotype. My phenotype is what my hair really looks like. So, for example, two people could have different genotypes with the same-- they might code for hair that looks pretty much the same, so it might have a very similar phenotype. So one phenotype can be represented by multiple genotypes. So that's just one thing to think about, and we'll talk a lot about that in the future, but I just wanted to introduce you to that there. Now, I entered this whole discussion because I wanted to So how does variation happen? Well, what's going to happen when I-- well, let What's going to happen when I reproduce? And I have. I have a son. Well, my contribution to my son is going to be a random collection of half of these genes. For each homologous pair, I'm either going to contribute the one that I got from my mother or the one that I got from my father, right? So let's say that the sperm cell that went on to fertilize my wife's egg, let's say it happened to have that one, that one, or I could just pick one from each of these 23 sets. And you say, well, how many combinations are there? Well, for every set, I could pick one of the two homologous chromosomes, and I'm going to do that 23 times. 2 times 2 times 2, so that's 2 to the twenty third." + }, + { + "Q": "What did Sal mean when he said noise at around 5:14?", + "A": "He is referring to random changes that can occur during the budding process. The most likely change is a mutation of genes. Instead of an exact replica happening during the copying of the original, there is a error, and the end result is not identical to the original. An environmental event could also account for a change in the expression of the genes that are present, again resulting in a non-identical offspring.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 314 + ], + "3min_transcript": "organism as well as a whole, although if it occurs after the organism has reproduced, it might not be something that selects against the organism and it also wouldn't be passed on. But anyway, I won't go detailed into that. But the whole point is that mutations don't seem to be a satisfying source of variation. They could be a source or kind of contribute on the margin, but there must be something more profound than mutations that's creating the diversity even within, or maybe I should call it the variation, even within a population. And the answer here is really it's kind of right in front of us. It really addresses kind of one of the most fundamental things about biology, and it's so fundamental that a lot of people never even question why it is the way it is. And that is sexual reproduction. And when I mean sexual reproduction, it's this notion that have nucleuses-- and we call those eukaroytes. Maybe I'll do a whole video on eukaryotes versus prokaryotes, but it's the notion that if you look universally all the way from plants-- not universally, but if you look at cells that have nucleuses, they almost universally have this phenomenon that you have males and you have females. In some organisms, an organism can be both a male and a female, but the common idea here is that all organisms kind of produce versions of their genetic material that mix with other organisms' version of their genetic material. If mutations were the only source of variation, then I could just bud off other Sals. Maybe just other Sals would just bud off from me, and then randomly one Sal might be a little bit different and But that would, as we already talked about, most of the time, we would have very little change, very little variation, and whatever variation does occur because of any kind of noise being introduced into this kind of budding process where I just replicate myself identically, most of the time it'll be negative. Most of the time, it'll break the organism. Now, when you have sexual reproduction, what happens? Well, you keep mixing and matching every possible combination of DNA in a kind of species pool of DNA. So let me make this a little bit more concrete for you. So let me erase this horrible drawing I just did. So we all have-- let me stick to humans because that's what we are. We have 23 pairs of chromosomes, and in each pair, we have one chromosome from our mother and one chromosome from our father." + }, + { + "Q": "Isn't the xy pair chromosomes a girl? And the xx is a man? In the picture at 8:55 he is a girl.", + "A": "It actually depends on the species what type of chromosome pairs determine sexual characteristics; there are different sexual determination systems. For humans (and most mammals), the XY chromosomal pair denotes that you are biologically a male while the XX chromosomal pair denotes that you are biologically a female.", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 535 + ], + "3min_transcript": "I have 23 from my mother. I have 23 from my father. Now, each of these chromosomes, and I made them right next to each other. So let me zoom in on one pair of these. So let's say we look at chromosome number-- I'll just call this chromosome number 3. So let me zoom in on chromosome number 3. I have one from my mother right here. Actually, maybe I'll do it this way. Remember, chromosome is just a big-- if you take the DNA, the DNA just keeps wrapping around, and it actually wraps around all these proteins, and it creates this structure, but it's just a big-- when you see it like that, you're like, oh, maybe the DNA-- no, but this could have millions of base pair, so maybe it'll look something like that. It's a densely wrapped version of-- well, it's a long string always the way it is, and we'll talk more about that, they draw it as densely packed like that. So let's say that's from my mother and that's from my father. Now, let's call this chromosome 3. They're both chromosome 3. And what the idea here is that I'm getting different traits from my father and from my mother. And I'm doing a gross oversimplification here, but this is really just to give you the idea of what's going on. This chromosome 3, maybe it contains this trait for hair color. And maybe my father had-- and I'll use my actual example. My father had very straight hair. So someplace on this chromosome, there's a gene for hair straightness. Let's say it's a little thing right there. And remember, that gene could be thousands of base pairs, but let's say this is hair straightness. So my father's version of that gene, he had the allele for I'll call it the allele straight for straight hair. Now, this other chromosome that my mother gave me, this essentially, and there are exceptions, but for the most part, it codes for the same genes, and that's why I put them next to each other. So this will also have the gene for hair straightness or curlyness, but my mom does happen to actually have curly hair. So she has the gene right there for curly hair. The version of the gene here is allele curly. The gene just says, look, this is the gene for whether or not your hair is curly. Each version of the gene is called an allele. Allele curly. Now, when I got both of these in my body or in my cells, and this is in every cell of my body, every cell of my body except for, and we'll talk in a few seconds about my germ" + }, + { + "Q": "At 18:07 when Sal talks about how a new combination of genes may or may not be passed on, what happens if it is not passed on, do an old combination of genes get passed on? or the genes from the other person get passed on?\nAlso can someone give me an exact definition for genotype and phenotype?", + "A": "Genotypes are genes expressed as letters. (Ex. Tt) A phenotype is the appearance expressed in words. (Ex. tall)", + "video_name": "DuArVnT1i-E", + "timestamps": [ + 1087 + ], + "3min_transcript": "So there's a huge amount of variation that even one couple can produce. And if you thought that even that isn't enough, it turns out that amongst these homologous pairs, and we'll talk about when this happens in meiosis, you can actually have DNA recombination. And all that means is when these homologous pairs during meiosis line up near each other, you can have this thing called crossover, where all of this DNA here crosses over and touches over here, and all this DNA crosses over and touches over there. So all of this goes there and all of this goes there. What you end up with after the crossover is that one DNA, the one that came from my mom, or that I thought came from my mom, now has a chunk that came from my dad, and the chunk that came from my dad, now has a chunk that came from my mom. Let me do that in the right color. It came from my mom like that. And so that even increases the amount of variety even more. chromosomes that you're contributing where the chromosomes are each of these collections of DNA, you're now talking about-- you can almost go to the different combinations at the gene level, and now you can think about it in almost infinite form of variation. You can think about all of the variation that might emerge when you start mixing and matching different versions of the same gene in a population. And you don't just look at one gene. I mean, the reality is that genes by themselves very seldom code for a specific-- you can very seldom look for one gene and say, oh, that is brown hair, or look for one gene and say, oh, that's intelligence, or that is how likable someone is. It's usually a whole set of genes interacting in an incredibly complicated way. You know, hair might be coded for by this whole set of genes on multiple chromosomes and this might be coded for a whole set of genes on multiple chromosomes. And so then you can start thinking about all of the different combinations. And then all of a sudden, maybe some combination that never existed before all of a sudden emerges, and that's But I'll leave you to think about it because maybe that combination might be passed on, or it may not be passed on because of this recombination. But we'll talk more about that in the future. But I wanted to introduce this idea of sexual reproduction to you, because this really is the main source of variation within a population. To me, it's kind of a philosophical idea, because we almost take the idea of having males and females for granted because it's this universal idea. But I did a little reading on it, and it turns out that this actually only emerged about 1.4 billion years ago, that this is almost a useful trait, because once you introduce this level of variation, the natural selection can start-- you can kind of say that when you have this more powerful form of variation than just pure mutations, and maybe you might have some primitive form of crossover before, but now that you have this sexual reproduction and you have this variation, natural selection can occur in a" + }, + { + "Q": "Why do you throw e in there without explaining it? Please walk through the steps a little more deliberately. What is happening with e at 5:14?", + "A": "It s assumed knowledge and it should be about the most complex algebra you need to know to get you through most of chemistry. lnK which is in that equation means the natural log of the variable K. To isolate K, we need to use the inverse function on both sides of the equation, which is e. e^(ln K) = K", + "video_name": "U5-3wnY04gU", + "timestamps": [ + 314 + ], + "3min_transcript": "for this balanced equation. So, let's write down our equation that relates delta-G zero to K. Delta-G zero is equal to negative RT, natural log of K. Delta-G zero is negative 33.0 kilojoules, so, let's write in here, negative 33.0, and let's turn that into joules, so times ten to the third, joules. This is equal to the negative, the gas constant is 8.314 joules over moles times K. So, we need to write over here, joules over moles of reaction. So, for this balanced equation, for this reaction, delta-G zero is equal to negative 33.0 kilojoules. So, we say kilojoules, or joules, over moles of reaction just to make our units work out, here. so, we write 298 K in here, Kelvin would cancel out, and then we have the natural log of K, our equilibrium constant, which is what we are trying to find. So, let's get out the calculator and we'll start with the value for delta-G zero which is negative 33.0 times 10 to the third. So, we're going to divide that by negative 8.314, and we'd also need to divide by 298. And so, we get 13.32. So, now we have 13.32, right, so our units cancel out here, and this is equal to the natural log of the equilibrium constant, K. So, how do we solve for K here? Well, we would take E to both sides. and E to the natural log of K on the right, this would cancel out and K would be equal to E to the 13.32, so let's do that. So, let's take E to the 13.32, and that's equal to, this would be 6.1, 6.1 times ten to the one, two, three, four, five. So, 6.1, 6.1 times ten to the fifth. And since we're dealing with gases, if you wanted to put in a KP here, you could. So, now we have an equilibrium constant, K, which is much greater than one. And we got this value from a negative value for delta-G zero. So, let's go back up to here, and we see that delta-G zero, right, is negative. So, when delta-G zero is less than zero," + }, + { + "Q": "at about 6:20 hank mentions a tepane, what in the world is that? I am really curious now about that.", + "A": "Pace or step.", + "video_name": "cstic6WHr2E", + "timestamps": [ + 380 + ], + "3min_transcript": "your sea anemones, your jellyfish, your corals, that have just one hole that serves as both mouth and anus. (laughing) Aren't you glad we're a little bit more complicated than that. It's worth noting that these animals have radial symmetry. All their junk kind of radiates out from a central point. That is their mouth hole/poo hole and that is a little bit more sophisticated than having no symmetry at all, like a sponge, but just barely. I mean, their anus and their mouth are the same thing. But more complex animals with the notable exception of the echinoderms, like starfish and sand dollars, exhibit bilateral symmetry. We have two-sided bodies that look the same on both sides. Something else we have in common is that we have an anus that is, get this, in a different place than our mouth. This separation is pretty key because it means that we, as animals, are basically built around a tube, a digestive tract, with a mouth at one end and an anus at the other. The process of forming this tract is called gastrulation and it's kind of a big deal. So, when we left our little blastula it was still just hanging out a little round hollow ball of cells. starts to form at a single point on the blastula. This place on the blastula that starts to invaginate or fold in on itself is called the blastopore. Now, for animals whose mouth and anus are the same thing, this is where the development stops, which is why they only have one hole for all their business, but in everything else the invagination continues until the indentation makes its way all the way through and opens on the other side creating what is essentially a hollow bead made of cells. Now we have a gastrula. Now two different things can happen at this point, depending on what kind of animal this is going to be. It can either be an animal whose mouth is the orifice that's formed by the blastopore, called a protostome or one whose anus is the structure that's created by the blastopore and that's called a deuterostome. So, guess which one you are? Write it down. I want to see your guesses. Chordates, that is to say, all vertebrates and a couple of our relatives like starfish are deuterostomes. Meaning that we were once just a butt hole and me, congratulations! And hopefully you're getting the idea here the formation of the digestive tract is the first thing that happens in the development of an animal and it happens to every living thing, whether it's going to be a tardigrade or a polar bear or a [tepane]. The miracle of life! Now so far, the little hollow bead of cells is basically two layers of tissue thick. An outer layer called the ectoderm and an inner layer called the endoderm and these are called your germ layers. For those organisms that stop developing at this point with that classy mouth/anus combo, they only get two germ layers. They're called diploblastic and they were born that way. It's totally okay, but for us more complex animals whose mouths are separate from our anuses, yes! We develop a third layer of tissue making us triploblasts. Here the ectoderm is going to end up being the animals skin and nerves and spinal cord and most of it's brain, while the endoderm" + }, + { + "Q": "At 9:06, how can P.E. be 5 Joules.\nWhen h=1/2d, P.E. should be equal to 5 units", + "A": "Are the units Joules? If so, what s the problem?", + "video_name": "vSsK7Rfa3yA", + "timestamps": [ + 546 + ], + "3min_transcript": "of this machine is equal to 10 Newtons. Mechanical advantage is the output over the input, so the mechanical advantage is equal to the force output by the force input, which equals 10/5, which equals 2. And that makes sense, because I have to pull twice as much for this thing to move up half of that distance. Let's see if we can do another mechanical advantage problem. Actually, let's do a really simple one that we've really been working with a long time. Let's say that I have a wedge. A wedge is actually considered a machine, which it took me a little while to get my mind around that, but a wedge is a machine. And why is a wedge a machine? Because it gives you mechanical advantage. So if I have this wedge here. And this is a 30-degree angle, if this distance up here, distance going to be? Well, it's going to be D sine of 30. And we know that the sine of 30 degrees, hopefully by this point, is 1/2, so this is going to be 1/2D. You might want to review the trigonometry a little bit if that doesn't completely ring a bell for you. So if I take an object, if I take a box-- and let's assume it has no friction. We're not going to go into the whole normal force and all that. If I take a box, and I push it with some force all the way up here, what is the mechanical advantage of this system? Well, when the box is up here, we know what its potential energy is. Its potential energy is going to be the weight of the box. So let's say this is a 10-Newton box. The potential energy at this point is going to be 10 Newtons times its height. So potential energy at this point has to equal 10 Newtons And that's also the amount of work one has to put into the system in order to get it into this state, in order to get it this high in the air. So we know that we would have to put 5 joules of work in order to get the box up to this point. So what is the force that we had to apply? Well, it's that force, that input force, times this distance has to equal 5 joules. So this input force-- oh, sorry, this is going to be-- sorry, this isn't 5 joules. It's 10 times 1/2 times the distance. It's 5D joules. This isn't some kind of units. It's 10 Newtons times the distance that we're up, and that's 1/2D, so it's 5D joules. Sorry for confusing you. And so the force I'm pushing here times this distance has to also equal to 5D joules." + }, + { + "Q": "At 3:51 , What do you mean by \"sp3 hybrid orbital gives 4 hybrid orbitals\" (Is this fixed) ?", + "A": "in sp3 orbitals there are 4 orbitals as 1 of s orbital and 3 of p orbitals", + "video_name": "BM-My1AheLw", + "timestamps": [ + 231 + ], + "3min_transcript": "so that's like electron group geometry, you wanna think about the geometry of the entire molecule. I could think about drawing in those electrons, those bonding electrons, like that. So we have a wedge coming out at us in space, a dash going away from us in space, and then, these lines mean, \"in the plane of the page.\" And so, we can go ahead a draw in our hydrogens, and this is just one way to represent the methane molecule, which attempts to show the geometry of the entire molecule. So the arrangement of the atoms turns out to also be tetrahedral, so let's go ahead and write that. So, tetrahedral. And, let's see if we can see that four-sided figure, so a tetrahedron is a four-sided figure, so we can think about this being one face, and then let's go ahead and draw a second face. And if I draw a line back here, that gives us four faces to our tetrahedron. So our electron group geometry is tetrahedral, and then we also have a bond angle, let me go ahead and draw that in, so a bond angle, this hydrogen-carbon-hydrogen bond angle in here, is approximately 109 point five degrees. All right, let's go ahead and do the same type of analysis for a different molecule, here. So let's do it for ammonia, next. So we have NH three, if I want to find the steric number, the steric number is equal to the number of sigma bonds, so that's one, two, three; so three sigma bonds. Plus number of lone pairs of electrons, so I have one lone pair of electrons here, so three plus one gives me a steric number of four. So I need four hybridized orbitals, and once again, when I need four hybridized orbitals, I know that this nitrogen must be SP three hybridized, because SP three hybridization gives us four hybrid orbitals, and so let's go ahead and draw those four hybrid orbitals. So we would have nitrogen, and let's go ahead and draw in all four of those. those are the four hybrid orbitals. When you're drawing the dot structure for nitrogen, you would have one electron, another electron, another electron, and then you'd have two in this one, like that. And then you'd go ahead, and put in your hydrogens, so, once again, each hydrogen has one electron, in a hybridized S orbital, so we go ahead and draw in those hydrogens, so our overlap of orbitals, so here's a sigma bond, here's a sigma bond, and here's a sigma bond; so three sigma bonds in ammonia, and then we have this lone pair up here. So the arrangement of these electron pairs, is just what we talked about before: So we have this tetrahedral arrangement of electron pairs, or electron groups, so the VSEPR theory tells us that's how they're going to repel. However, that's not the shape of the molecule, so if I go ahead and draw in another picture over here, to talk about the molecular geometry, and go ahead and draw in the bonding electrons, like that, and then I'll put in my non-bonding" + }, + { + "Q": "At, 8:40 - 8:42, there is repulsion between the lone pairs & the bond pairs but what about the lone pairs? Don't they repel themselves too?", + "A": "Of course they do. And they repel one another the most.", + "video_name": "BM-My1AheLw", + "timestamps": [ + 520, + 522 + ], + "3min_transcript": "you have four SP three hybridized orbitals. So this oxygen is SP three hybridized, so I'll go ahead and write that in here, so oxygen is SP three hybridized. So we can draw that out, showing oxygen with its four SP three hybrid orbitals; so there's four of them. So I'm gonna go ahead and draw in all four. In terms of electrons, this orbital gets one, this orbital gets one, and these orbitals are going to get two, like that; so that takes care of oxygen's six valence electrons. When you're drawing in your hyrdogens, so let's go ahead and put in the hydrogen here, so, once again, each hydrogen with one electron, in a un-hybridized S orbital, like that. So in terms of overlap of bonds, here's one sigma bond, and here's another sigma bond; so that's our two sigma bonds for water. Once again, the arrangement of these electron pairs is tetrahedral, so VSEPR theory says the electrons repel, is tetrahedral, but that's not the geometry of the entire molecule, 'cause I was just thinking about electron groups, and these hybrid orbitals. The geometry of the molecule is different, so we'll go ahead and draw that over here. So we have our water molecule, and draw in our bonding electrons, and now let's put in our non-bonding electrons, like that, so we have a different situation than with ammonia. With ammonia, we had one lone pair of electrons repelling these bonding electrons up here; for water, we have two lone pairs of electrons repelling these bonding electrons, and so that's going to change the bond angle; it's going to short it even more than in the previous example. So the bond angle decreases, so this bond angle in here decreases to approximately 105 degrees, rounded up a little bit. So, thinking about the molecular geometry, or the shape of the water molecule, so this is, \"bent geometry,\" because you ignore the lone pairs of electrons, and that would just give you this oxygen here, and then this angle; so you could also call this, \"angular.\" So we have this bent molecular geometry, like that, or angular, and once again, for molecular geometry, ignore your lone pairs of electrons. So these are examples of three molecules, and the central atom in all three of these molecules is SP three hybridized, and so, this is one way to figure out your overall molecular geometry, and to think about bond angles, and to think about how those hybrid orbitals affect the structure of these molecules." + }, + { + "Q": "At 1:44 what is VSEPR theory?", + "A": "VSEPR theory is the theory of hybridized orbitals. It is the theory that dictates unhybridized, sp1, sp2, and sp3 hybridizations. Basically the idea is that orbitals will hybridize to form bonds with equal energy.", + "video_name": "BM-My1AheLw", + "timestamps": [ + 104 + ], + "3min_transcript": "Voiceover: The concept of steric number is very useful, because it tells us the number of hybridized orbitals that we have. So to find the steric number, you add up the number of stigma bonds, or single-bonds, and to that, you add the number of lone pairs of electrons. So, let's go ahead and do it for methane. So, if I wanted to find the steric number, the steric number is equal to the number of sigma bonds, so I look around my carbon here, and I see one, two, three, and four sigma, or single-bonds. So I have four sigma bonds; I have zero lone pairs of electrons around that carbon, so four plus zero gives me a steric number of four. In the last video, we saw that SP three hybridized situation, we get four hybrid orbitals, and that's how many we need, the steric number tells us we need four hybridized orbitals, so we took one S orbital, and three P orbitals, and that gave us four, SP three hybrid orbitals, so this carbon must be SP three hybridized. So let's go ahead, and draw that in here. So this carbon is SP three hybridized, and in the last video, we also drew everything out, for that carbon, and we had one valence electron in each of those four, SP three hybrid orbitals, and then hydrogen had one valence electron, in an un-hybridized S orbital, so we drew in our hydrogens, and the one valence electron, like that. This head-on overlap; this is, of course, a sigma bond, so we talked about this in the last video. And so now that we have this picture of the methane molecule, we can think about these electron pairs, so these electron pairs are going to repel each other: like charges repel. And so, the idea of the VSEPR theory, tell us these electron pairs are going to repel, and try to get as far away from each other as they possibly can, in space. And this means that the arrangement of those electron pairs, ends up being tetrahedral. So let's go ahead and write that. So we have a tetrahedral arrangement of electron pairs around our carbon, like that. so that's like electron group geometry, you wanna think about the geometry of the entire molecule. I could think about drawing in those electrons, those bonding electrons, like that. So we have a wedge coming out at us in space, a dash going away from us in space, and then, these lines mean, \"in the plane of the page.\" And so, we can go ahead a draw in our hydrogens, and this is just one way to represent the methane molecule, which attempts to show the geometry of the entire molecule. So the arrangement of the atoms turns out to also be tetrahedral, so let's go ahead and write that. So, tetrahedral. And, let's see if we can see that four-sided figure, so a tetrahedron is a four-sided figure, so we can think about this being one face, and then let's go ahead and draw a second face. And if I draw a line back here, that gives us four faces to our tetrahedron. So our electron group geometry is tetrahedral," + }, + { + "Q": "Around 7:45 Sal said that the radius of the observable universe is equal to 46 billion light years, does that mean the whole universe is 92 billion light years?", + "A": "The diameter of the observable universe is 92 billion ly. We don t know how big the whole universe is. We have good reason to believe it is much, much bigger than the portion we can observe. Maybe infinite, we don t know.", + "video_name": "06z7Q8TWPyU", + "timestamps": [ + 465 + ], + "3min_transcript": "" + }, + { + "Q": "At 9:16 in the video he states that the light from the object that took 13.7 Billion years to reach us is now 46 Billion light years away. If it all began at the big bang 13.7 Billion years ago, how can anything be 46 Billion light years away?", + "A": "Light years are a measure of distance not time", + "video_name": "06z7Q8TWPyU", + "timestamps": [ + 556 + ], + "3min_transcript": "" + }, + { + "Q": "At 5:54, why does he do a mol ratio of MnO4- over Fe2+ and not the other way around? I'm doing similar problems, and there's seemingly no rhyme or reason as to whether it should be 1/5 or 5/1. It varies from problem to problem.", + "A": "That s because it doesn t matter which way he does it. He is trying to get the moles of Fe\u00c2\u00b2\u00e2\u0081\u00ba. If he had written 5/1 = x/0.000 4000, he would still have gotten x = 0.002 000 mol Fe\u00c2\u00b2\u00e2\u0081\u00ba.", + "video_name": "EQJf8Gb8pg4", + "timestamps": [ + 354 + ], + "3min_transcript": "we can see that we've used a certain volume of our solution. Let's say it took 20 milliliters. We used up 20 milliliters of our potassium permanganate solution to completely titrate our iron two plus. Our goal was to find the concentration of iron two plus. If we're going to find the concentration of iron two plus, we could figure out how many moles of permanganate were necessary to completely react with our iron two plus. We could figure out moles from molarity and volume. Let's get some more room down here. We know that molarity is equal to moles over liters. The molarity of permanganate is .02. We have .02 for the concentration of permanganate ions. Moles is what we're solving for. It took us 20 milliliters for our titration, which we move our decimal place one, two, three, So solve for moles. .02 times .02 is equal to .0004. So we have .0004. This is how many moles of permanganate were needed to completely react with all of the iron two plus that we originally had in our solution. It took .0004 moles of permanganate to completely react with our iron. All right. Next, we need to figure out how many moles of iron two plus that we originally started with. To do that, we need to use our balance redox reaction. We're going to look at the coefficients, because the coefficients tell us mole ratios. The coefficient in front of permanganate is a one. The coefficient in front of iron two plus is a five. If we're doing a mole ratio of permanganate to iron two plus, and iron would be a five. So we set up a proportion here. One over five is equal to ... Well, we need to keep permanganate in the numerator here. How many moles of permanganate were necessary to react with the iron two plus? That was .0004. So we have .0004 moles of permanganate. X would represent how many moles of iron two plus we originally started with. We could cross-multiply here to solve for x. Five times .0004 is equal to .002. X is equal to .002. X represents the moles of iron two plus that we originally had present. We're almost done, because our goal was to find the concentration of iron two plus cations. Now we have moles" + }, + { + "Q": "At 3:36 I don't understand, if there are excess MnO4-, then how can this be the end point? Wouldn't it mean that now there are extra ions which would contribute to an extra volume, and therefore become an inaccurate value to calculate the concentration?", + "A": "The MnO4- has a purple color and the Fe2+ is colorless solution. If the Fe2+ solution is excess, it can be colorless unless the MnO4- is enough. If the color turns purple, that means the Fe2+ ion doesn t exist as Fe2+, they all turn into Fe3+.", + "video_name": "EQJf8Gb8pg4", + "timestamps": [ + 216 + ], + "3min_transcript": "so the oxidation state is plus two. Manganese is going from an oxidation state to plus two. That's a decrease or a reduction in the oxidation state. Therefore, manganese is being reduced in our redox reaction. Let's look at iron two plus. We have iron two plus as one of our reactants here. That means the oxidation state is plus two. For our products, we're making iron three plus, so an oxidation state of plus three. Iron is going from plus two to plus three. That's an increase in the oxidation state. Iron two plus is being oxidized in our redox reaction. As we drip in our potassium permanganate, we're forming our products over here. These ions are colorless in solution. As the permanganate reacts, this purple color disappears we should have a colorless solution. Let's say we've added a lot of our permanganate. Everything is colorless. But then we add one more drop, and a light purple color persists. Everything was clear, but then we add one drop of permanganate and then we get this light purple color. This indicates the endpoint of the titration. The reason why this is the endpoint is because our products are colorless. So if we get some purple color, that must mean we have some unreacted, a tiny excess of unreacted permanganate ions in our solution. That means we've completely reacted all the iron two plus that we originally had present. So we stop our titration at this point. We've reached the endpoint. We've used a certain volume of our potassium permanganate solution. Let's say we finished down here. we can see that we've used a certain volume of our solution. Let's say it took 20 milliliters. We used up 20 milliliters of our potassium permanganate solution to completely titrate our iron two plus. Our goal was to find the concentration of iron two plus. If we're going to find the concentration of iron two plus, we could figure out how many moles of permanganate were necessary to completely react with our iron two plus. We could figure out moles from molarity and volume. Let's get some more room down here. We know that molarity is equal to moles over liters. The molarity of permanganate is .02. We have .02 for the concentration of permanganate ions. Moles is what we're solving for. It took us 20 milliliters for our titration, which we move our decimal place one, two, three," + }, + { + "Q": "How can he assume it is an acidic solution at minute 0:25?", + "A": "The equation has H\u00e2\u0081\u00ba in it. That s an acid.", + "video_name": "EQJf8Gb8pg4", + "timestamps": [ + 25 + ], + "3min_transcript": "- [Voiceover] We've already seen how to do an acid-base titration. Now let's look at a redox titration. Let's say we have a solution containing iron two plus cations. We don't know the concentration of the iron two plus cations, but we can figure out the concentration by doing a redox titration. Let's say we have 10 milliliters of our solution, and let's say it's an acidic solution. You could have some sulfuric acid in there. In solution, we have iron two plus cations and a source of protons from our acid. To our iron two plus solution, we're going to add some potassium permanganate. In here, we're going to have some potassium permanganate, KMnO4. Let's say the concentration of our potassium permanganate is .02 molar. That's the concentration that we're starting with. Potassium permanganate is, of course, the source of permanganate anions, and MnO4 minus. Down here, we have a source of permanganate anions. We're going to drip in the potassium permanganate solution. When we do that, we're going to get a redox reaction. Here is the balanced redox reaction. If you're unsure about how to balance a redox reaction, make sure to watch the video on balancing redox reactions in acid. Let's look at some oxidation states really quickly so we can see that this is a redox reaction. For oxygen, it would be negative two. We have four oxygens, so negative two times four is negative eight. Our total has to add up to equal negative one. For manganese, we must have a plus seven, because plus seven and negative eight give us negative one. Manganese has an oxidation state of plus seven. Over here, for our products, we're going to make Mn two plus. so the oxidation state is plus two. Manganese is going from an oxidation state to plus two. That's a decrease or a reduction in the oxidation state. Therefore, manganese is being reduced in our redox reaction. Let's look at iron two plus. We have iron two plus as one of our reactants here. That means the oxidation state is plus two. For our products, we're making iron three plus, so an oxidation state of plus three. Iron is going from plus two to plus three. That's an increase in the oxidation state. Iron two plus is being oxidized in our redox reaction. As we drip in our potassium permanganate, we're forming our products over here. These ions are colorless in solution. As the permanganate reacts, this purple color disappears" + }, + { + "Q": "At 3:48, you said that lions and tigers can make ligers, but can animals of different species make hybrids of themselves?", + "A": "Species are defined as being a distinct group that can breed within itself and produce viable, fertile offspring. Certain species (e.g. lions & tigers -> ligers, horses & donkeys -> mules) are close enough genetically that they can produce hybrids but these hybrids are sterile and not able to breed among themselves and produce offspring. Otherwise they would not represent a separate distinct species but rather a subpopulation or race.", + "video_name": "Tmt4zrDK3dA", + "timestamps": [ + 228 + ], + "3min_transcript": "but are actually closely related. And we'll talk about what it means to be closely related. And then we can see things that look very similar, that they have similar structures or they have similar behavior, like for example bats and birds, but they are actually all very, very distantly related. So we need a more exact definition for species than just things that look like each other, or just things that act like each other. And so the most typical definition for species are animals that can interbreed. And when we say interbreed, literally they can produce offspring with each other, and the offspring are fertile. Which means that the offspring can then further have babies, that they're not sterile, that they're capable of breeding with other animals and producing more offspring. You find a male lion and a female lioness and most of the time they will be able to have offspring, and those offspring can go and mate with other lions or lionesses, depending on their sex, and then they can have viable offspring. So it seems to work out pretty well for lions. Same thing is true of tigers. Now, it does turn out that if you get a male lion and a female tigress they can breed. They can breed and they can produce offspring. And their offspring-- which was made famous by Napoleon Dynamite, he was kind of fascinated by, these are kind of fascinating animals. Their offspring is called a liger. You get a male lion breeding with a female tiger you produce a liger, which is a hybrid, it's a cross between a lion and a tiger. And they're fascinating animals. They're actually larger than either lions or tigers. They are the largest cats that we know of. Or you can't say that lions and tigers are the same species, because even though they are able to interbreed, their offspring, for the most part, is not fertile, is not able to produce offspring. There have been one off stories about ligers being mated with either a lion or a tiger, but those are one off stories. In general, ligers can't interbreed. And in general, this combination isn't going to produce offspring that can keep interbreeding or that are fertile. So that's why we say that lions and tigers are different species. And that liger, we wouldn't even call it as a species at all. We would actually call it a hybrid between two species. Now the same thing is true-- and actually you might be asking yourself, well, this was a male lion and a female tigress, what if we went the other way around? What if we had a female lioness and a male tiger? In that case, you would produce something else called a tiglon or a tiglon, I actually don't" + }, + { + "Q": "At 3:13, Sal decides to integrate with a lower bound of zero and an upper bound of infinity. Could we also have integrated from infinity to zero, thereby summing all of the rings from an infinite distance away up to the point charge at a distance of zero?", + "A": "Of course but you should then multiply the result by -1 as you reversed the limits of the integration.", + "video_name": "TxwE4_dXo8s", + "timestamps": [ + 193 + ], + "3min_transcript": "So at the end, we meticulously calculated what the y-component of the electric field generated by the ring is, at h units above the surface. So with that out of the way, let's see if we can sum up a bunch of rings going from radius infinity to radius zero and figure out the total y-component. Or essentially the total electric field, because we realize that all the x's cancel out anyway, the total electric field at that point, h units above the surface of the plane. So let me erase a lot of this just so I can free it up for some hard-core math. And this is pretty much all calculus at this point. So let me erase all of this. Watch the previous video if you forgot how it was derived. need a lot of space. There you go. OK, so let me redraw a little bit just so we never forget what we're doing here because that happens. So that's my plane that goes off in every direction. I have my point above the plane where we're trying to figure out the electric field. And we've come to the conclusion that the field is going to point upward, so we only care about the y-component. It's h units above the surface, and we're figuring out the electric field generated by a ring around this point of radius r. And what's the y-component of that electric field? We figured out it was this. So now what we're going to do is take the integral. So the total electric field from the plate is going to be the integral from-- that's a really ugly-looking integral-- So we're going to take a sum of all of the rings, starting with a radius of zero all the way to the ring that has a radius of infinity, because it's an infinite plane so we're figuring out the impact of the entire plane. So we're going to take the sum of every ring, so the field generated by every ring, and this is the field generated by each of the rings. Let me do it in a different color. This light blue is getting a little monotonous. Kh 2pi sigma r dr over h squared plus r squared to the 3/2. Now, let's simplify this a little bit. Let's take some constants out of it just so this looks like a slightly simpler equation. So this equals the integral from zero to-- So let's take" + }, + { + "Q": "At 5:10, Sal explains that when you dump (H^+) into the blood, the buffer system prevents the pH from going up by creating an equilibrium that bonds the H+ to the Bicarbonate. Why does that prevent the pH from going up? There is still a larger amount of H+ in the blood.", + "A": "A solution gets more acidic (lower pH) when there are more free H+ ions in it. The bicarbonate takes these free H+ ions out of the solution. H+ that is bonded to something does not affect the pH.", + "video_name": "gjKmQ501sAg", + "timestamps": [ + 310 + ], + "3min_transcript": "But the topic of this video is why this is also useful for maintaining our blood pH in this range. Because these equilibrium reactions between carbon dioxide, carbonic acid, and bicarbonate this is a buffer system. This is a buffer, this is a buffer system. And the word \"buffer,\" in our everyday language, it refers to something that kind of smooths the impact of something, or it reduces the shock of something. And that's exactly what's happening here. Let's think about, remember, these are all equilibrium reactions, this is a weak acid, and you can even look at the different constituents of these molecules and account for them. You have one carbon here, one carbon here, one carbon there. You have one, two, three oxygens there. You have one, two, three oxygens there. One, two, three oxygens there. You have two hydrogens, two hydrogens, two hydrogens. But let's just think about what if you started dumping hydrogen ions in the blood. So, what if you were to dump hydrogen ions, Well, if you dump more hydrogen ions, if this right over here increases. Actually, let me put it this way, if you were to just dump hydrogen ions and if you didn't have this buffer system, then your pH would decrease. Your pH would go down, and if you do it enough, your pH, you would end up with acidosis. But lucky for us, we have this buffer system. And so if you increase your hydrogen ion concentrations, Le Chatelier's principle tells us, \"Hey, these equilibrium reactions are going to move to the left.\" So the more hydrogen ions you have sitting in the blood, the more likely they're gonna bump into the bicarbonate in just the right way to form carbonic acid. And the more carbonic acid that you have in the blood, well, it's the less likely that you're going to have the carbon dioxide reacting with the water to form more carbonic acid. So, as you add more hydrogen ions, by the bicarbonate. So this equilibrium, this set of equilibrium reactions is going to move to the left. So you're not going to have as big effect on pH. And similarly, if you dumped some base, let's say, you dumped some base in your blood right over here, well, instead of it just making your pH go up, and possibly give you alkalosis, well now, the base is going to sop up the hydrogen ions, and typically that would just make your pH go up, but if you have these things going down, well then, you have fewer of these to react and have the equilibrium reaction go to the left and so the reaction is going to move more and more to the right. And so this reaction, you're just gonna have more carbon dioxide being converted to carbonic acid being converted to bicarbonate. This whole thing is going to move to the right. And so it's going to be able, to some degree, replace the lost hydrogen ions. So this right over here's a buffer system." + }, + { + "Q": "I just wanted to thank you for this video. It was fantastic and it was very helpful. Just really quick question. This turtle doesn't have displacement because she is not changing position. She is just changing distance as seen at 3:33 but she doesn't move left or right? Because she is just moving forwards and backwards this is different than if she were moving left and backwards or right and forwards?", + "A": "you can be displaced forward and backward just as well as left or right. any direction will do.", + "video_name": "GtoamALPOP0", + "timestamps": [ + 213 + ], + "3min_transcript": "If this turtle didn't go forward, down, and up, what did this turtle do? We'll start at t equals zero. We'll go up from there. And at t equals zero the value of this graph is three. And the value of this graph is representing the horizontal position, so the value of the graph is giving you the horizontal position. So at t equals zero, the turtle is at three meters. So let's put her over at three meters. She starts over here. Three meters, that's equal zero. Now what happens? So at t equals one second, same thing. We read our graph by going up, hit the graph, then we go left to figure out where we're at. Again turtle's still at three. At two seconds we come up, hit the graph. We come over to the left to figure out where we're at. This turtle's still at three, that's awkward. This turtle didn't even move. For the first two seconds this turtle's just sitting here. So a straight line a horizontal line on a position graph represents no motion whatsoever. This is awkward. Turtle was probably trying to figure out how to turn on her jet pack. Sorry. Now what happens? Turtle at some later time, four seconds, is at negative five meters. That's all the way back here. So between two seconds and four seconds, this turtle rocketed back this way. That's also awkward. Turn down the reverse booster. What a noob, ah turtle. Here we go. Made it all the way back to here. Then what does the turtle do? After that point, turtle rockets forward. Makes it back to zero at this point. And then all the way back to three meters, so this turtle rockets forward back to three meters. That's what the turtle did. That's what this graph is representing, and that's how you read it. But there's more than that in here. I told you there was a lot of information and there is. So one piece of information you can get is the displacement of the turtle. And the displacement I'm gonna represent this with a delta x. And remember the displacement is the final position. Minus the initial position. You can find the displacement between any two times here, for the total time shown on the graph. But I could've found it between zero and like four seconds. Let's just do zero to 10, the whole thing. So what's the final position? The final position would be the position the turtle has. At 10 seconds she was at three meters. At 10 seconds 'cause I read the graph right there. Minus initially, 'cause we're considering the total time, at zero seconds, the turtle was also at three, that means the total displacement was zero. And that makes sense 'cause this turtle started at three. Rocketed back to five, well actually started at three, stood there for a second or two, rocketed back to five, rocketed back to three, ended at the same place she started, no total displacement. What else can we find? We can figure out the total distance. For the total distance traveled remember distance is the sum of all the path links traveled. So for this first path link, there was no distance traveled. That was the awkward part. We're not gonna talk about that." + }, + { + "Q": "5:27, where did you get the 8? I added from three and went down where it stopped and got more than 8.", + "A": "I think he might have forgotten to count 0 when he was counting down from 3.", + "video_name": "GtoamALPOP0", + "timestamps": [ + 327 + ], + "3min_transcript": "Sorry. Now what happens? Turtle at some later time, four seconds, is at negative five meters. That's all the way back here. So between two seconds and four seconds, this turtle rocketed back this way. That's also awkward. Turn down the reverse booster. What a noob, ah turtle. Here we go. Made it all the way back to here. Then what does the turtle do? After that point, turtle rockets forward. Makes it back to zero at this point. And then all the way back to three meters, so this turtle rockets forward back to three meters. That's what the turtle did. That's what this graph is representing, and that's how you read it. But there's more than that in here. I told you there was a lot of information and there is. So one piece of information you can get is the displacement of the turtle. And the displacement I'm gonna represent this with a delta x. And remember the displacement is the final position. Minus the initial position. You can find the displacement between any two times here, for the total time shown on the graph. But I could've found it between zero and like four seconds. Let's just do zero to 10, the whole thing. So what's the final position? The final position would be the position the turtle has. At 10 seconds she was at three meters. At 10 seconds 'cause I read the graph right there. Minus initially, 'cause we're considering the total time, at zero seconds, the turtle was also at three, that means the total displacement was zero. And that makes sense 'cause this turtle started at three. Rocketed back to five, well actually started at three, stood there for a second or two, rocketed back to five, rocketed back to three, ended at the same place she started, no total displacement. What else can we find? We can figure out the total distance. For the total distance traveled remember distance is the sum of all the path links traveled. So for this first path link, there was no distance traveled. That was the awkward part. We're not gonna talk about that. Then, so this is zero meters, plus between two seconds and four seconds, the turtle went from three to five. That's a distance traveled of eight meters. And should we make that negative? Nope. Distance is always positive. We make all these path links positive, we round them all up. So eight meters. Because the turtle went from three all the way back to five. That's the total distance of eight meters traveled. Plus between four seconds and 10 seconds, the turtle made it from negative five meters all the way back to three meters. That means she traveled another eight meters. That means the total distance traveled was 16 meters for the whole trip. Again you could have found this for two points any two points on here. Alright what else can you figure out? You can figure out the say average velocity, sometimes people represent that with a bar. Sometimes they just say the AVG." + }, + { + "Q": "Why he made an assumption that it is moving horizontally not vertical? at 12:58", + "A": "because objects can easily roll along the ground and maintain their speed, but it is not so easy to do that in the vertical direction. Gravity interferes.", + "video_name": "GtoamALPOP0", + "timestamps": [ + 778 + ], + "3min_transcript": "So at four seconds and negative five meters that's our .2. Alright so x two, that would be negative five, 'cause I'm just reading my graph, that .2. That's negative five. So I got negative five meters minus x one that's this. Don't make x one four. That's a time, that's not a position. So point one, the horizontal position was three. So positive three. Put the negative here 'cause the negative's in the formula. And then divide it by time two, that was four seconds. And minus t one was two seconds. And if you saw this thing, negative five and negative three and negative eight divided by two seconds. Oops can't figure out my units. Oh look at that. I got negative four meters per second. That was the instantaneous velocity at three seconds. Negative four meters per second. Negative because the turtle was going backwards. She turned on the reverse booster instead of the forward booster. Negative and four because look it going four meters every second. Made it eight meters and two seconds, that means she was going four meters per second on average. And since it's a straight line, that was the rate she was going at any moment. Beautiful. Alright that would've been if the follow up question is what is it at 2.4 seconds? Don't get concerned. Look it's the same everywhere. It'd be the same answer. Negative four meters per second. For this whole line. What else can we figure out one last thing. Let's say you were asked what's the instantaneous speed at a point? So I'm gonna write that as SINST instantaneous speed, or just s. 'Cause that's usually what we mean by speed. Equals average value, sorry, absolute value of the instantaneous velocity. So now here I've got to make an assumption. This is gonna get a little subtle. If all we're given is a horizontal position graph, This turtle could've gone back and forth, or the turtle could've been like flying upward, as she went back and forth. And if the horizontal location was the same the whole way, this would've looked exactly the same regardless of whether the turtle had any vertical motion at all. So we gotta be careful, 'cause the speed is the magnitude of the total velocity. This is the just the velocity in the x direction. So we're gonna make an assumption. I'm gonna assume this turtle was just moving horizontally. Instead of having the vertical motion. She's not ready for that yet. Alright so how do you get this? Speed is just the absolute value. The magnitude of the instantaneous velocity. And if this is the only component of velocity, then I can just figure this out pretty easy by saying that, oh I gotta give you time, makes no sense to say instantaneous speed. I gotta say instantaneous speed at a given moment. 'Cause the instantaneous speed here was zero." + }, + { + "Q": "At 12:50, doesn't the conjugate base create hydroxide ions in solution, adding to the total moles of OH-, causing [OH-] to be larger and pH to be lower? Or does acetate not affect the pH after the equivalence point?", + "A": "You are correct! It will add to the total amount of OH-, increasing its molarity. But the change is so insignificant that the change is only in the slightest. I did the whole calculation for you and found out that the concentration changes only 9 (!!) places after the decimal. This will cause only a minuscule change in the pH. Also, the higher the concentration of OH-, the higher the pH.", + "video_name": "WbDL7xN-Pn0", + "timestamps": [ + 770 + ], + "3min_transcript": "So right about there, about 8.67. So, that's our equivalence point for a titration of a weak acid with a strong base for this particular example. Finally we're on to Part D, which asks us, what is the pH after the addition of 300.0 mL of a 0.0500 molar solution of sodium hydroxide? So, once again, we need to find the moles of hydroxide ions that we are adding. The concentration would be equal to 0.0500, so 0.0500 molar is our concentration of hydroxide ions. And that's equal to moles over liters. So, 300.0 mL would be 0.3000 liters. So, we have 0.3000 liters here. Multiply 0.0500 by 0.3000, and you get moles. So, 0.0500 times 0.300 moles of hydroxide ions. In Part C, we saw that we needed 0.0100 moles of hydroxide ions to completely neutralize the acid that we originally had present. So, we're going to use up... We're going to use up 0.0100 moles of hydroxide. That's how much was necessary to neutralize our acid. All right, so how many moles of hydroxide are left over after the neutralization? Well, that would just be 0.0050. So, 0.0050 mol of hydroxide ions are left over after all the acid has been neutralized. So, our goal is to find the pH. So, we could find the pOH if we found if we found the concentration of hydroxide. So, what is the concentration of hydroxide ions now after the neutralization has occurred? so it's 0.0050 divided by... What's our total volume? We started with 50.0, and we have now added 300.0 mL more, so 300.0 plus 50.0 is 350.0 mL. Or 0.3500 liters. So, what is our concentration of hydroxide ions? So, this is .005 divided by .35. So our concentration of hydroxide ions is 0.014. So, let's write 0.014 M here. Once we know that, we can calculate the pOH. So the pOH is the negative log of the concentration of hydroxide ions. So, it's the negative log of 0.014. So, we can do that on our calculator. Negative log of .014. And we get a pOH of 1.85." + }, + { + "Q": "at 4:36 why is it called The Bowman's Capsule?", + "A": "A glomerulus is enclosed in the sac. Fluids from blood in the glomerulus are collected in the Bowman s capsule (i.e., glomerular filtrate) and further processed along the nephron to form urine. This process is known as ultrafiltration. The Bowman s capsule is named after Sir William Bowman, who identified it in 1842", + "video_name": "UU366tJPovg", + "timestamps": [ + 276 + ], + "3min_transcript": "And even though they do an amazing job, I'm not badmouthing your kidneys here, the way that they do it is frankly a little bit janky and inefficient. They start out by filtering a bunch of fluid and the stuff dissolved in the fluid out of your blood and then they basically reabsorb 99 percent of it back before sending that one percent on its way in the form of urine. Seriously 99 percent gets reabsorbed. On an average day your kidneys filter out about 180 liters of fluid from your blood but only 1.5 liters of that ends up getting peed out. So most of your excretory system isn't dedicated to excreting, it's dedicated to reabsorbing. But the system works, obviously. I'm still alive so we can't argue with that. Now it is time to get into the nitty-gritty details of how your kidneys do all this and it's pretty cool but there's lots of weird words so get ready. Your kidneys do all this work using a network of tiny filtering structures called nephrons, each one of your mango sized kidneys has about a million of them. If you were, don't do this, but if you were to unravel all of your nephrons and put them end to end they would stretch over 80 kilometers. we're just going to follow the flow from your heart to the toilet. Blood from the heart enters the kidneys through renal arteries and just so you know, whenever you hear the word renal, it means we're dealing with kidney stuff. As the blood enters it's forced into a system of tiny capillaries until it enters a tangle of porous capillaries called the glomerulus. This is the starting point for a single nephron. The pressure in the glomerulus is high enough that it squeezes some of the fluid out of the blood, about 20 percent of it, and into a cup-like sac called the bowman's capsule. The stuff that gets squeezed out is no longer blood, it is now called filtrate. It's made up of water, urea, some smaller ions and molecules like sodium, glucose and amino acids. The bigger stuff in your blood like the red blood cells and the larger proteins, they don't get filtered. Now the filtrate is ready to be processed from the bowman's capsule it flows into a twisted tube called the proximal convoluted tubule which means the tube near the beginning and that is all windy. Anyway, this is the first of two convoluted tubules in the nephron and these along with other tubules we're talking about are where the osmoregulation takes place. With all kinds of tricked out specialized pumps and other kinds of active and passive transport, they reabsorb water and dissolve materials to create whatever balance your body needs at the time. In the proximal tubule it's mainly organic solutes and the filtrate that are reabsorbed like glucose and amino acids and other important stuff that you want to hang onto. But it also helps to recapture some sodium and potassium and water that we're going to want later. From here the filtrate enters the loop of Henle, which is a long hairpin shaped tubule that passes through the two main layers of the kidney. The outermost layer is the renal cortex, that's where the glomerulus and the bowman's capsule and both convoluted tubules are and the layer beneath that is the renal medulla which is the center of the kidney. Cortex, by the way, is Latin for tree bark so whenever you see it in biology you know that it's the outside of something." + }, + { + "Q": "At 4:01, could you also call it a \"vectorial field?\"", + "A": "Hadnt thought about it that way, but yes, I think you could call it a vectoral field Why do you ask? What are you thinking??", + "video_name": "1E3Z_R5AHdg", + "timestamps": [ + 241 + ], + "3min_transcript": "is equal to 0, especially in this vertical direction. And because the net force is equal to 0, I am not accelerating towards the center of the Earth. I am not in free fall. And because this 9.81 meters per second squared still seems relevant to my situation-- I'll talk about that in a second. But I'm not an object in free fall. Another way to interpret this is not as the acceleration due to gravity near Earth's surface for an object in free fall, although it is that-- a maybe more general way to interpret this is the gravitational-- or Earth's gravitational field. Or it's really the average acceleration, or the average, because it actually changes slightly throughout the surface of the Earth. But another way to view this, as the average gravitational field at Earth's surface. Let me write it that way in pink. So the average gravitational field-- and we'll context in a second-- the average gravitational field at Earth's surface. And this is a little bit more of an abstract thing-- we'll talk about that in a second-- but it does help us think about how g is related to this scenario where I am not an object in free fall. A field, when you think of it in the physics context-- slightly more abstract notion when you start thinking about it in the mathematics context-- but in the physics context, a field is just something that associates a quantity with every point in space. So this is just a quantity with every point in space. And it can actually be a scalar quantity, in which case we call it a scalar field, and in which case it would just be a value. Or it could be a vector quantity, which would be a magnitude and a direction associated with every point in space. In which case you are dealing with a vector field. is, because at near Earth's surface, if you give me a mass-- so for example-- actually, I don't know what my mass is in kilograms. But if you're near Earth's surface and you give me a mass-- so let's say that mass right over there is 10 kilograms-- you can use g to figure out the actual force of gravity on that object at that point in space. So for example, if this has a mass of 10 kilograms, then we know-- and this right over here is the surface of the Earth, so that's the center of the Earth. So it actually associates a vector quantity whose magnitude,-- so its direction is towards the center of the Earth, and the magnitude of this vector quantity is going to be the mass times g. And you could take-- since we're already specifying the direction, we could say 9.81 meters per second squared towards the center of the Earth. And so in this situation, it would be 10 kilograms times 9.81 meters per second squared." + }, + { + "Q": "At 0:44, why is an object in free fall accelerating with a negative value since it is going towards the centre of the earth", + "A": "Nature don t know you are positive or negative, you are free to choose sine conventions, either you choose negative or positive the important think is that you make your calculations right", + "video_name": "1E3Z_R5AHdg", + "timestamps": [ + 44 + ], + "3min_transcript": "What I want to do in this video is think about the two different ways of interpreting lowercase g. Which as we've talked about before, many textbooks will give you as either 9.81 meters per second squared downward or towards the Earth's center. Or sometimes it's given with a negative quantity that signifies the direction, which is essentially downwards, negative 9.81 meters per second squared. And probably the most typical way to interpret this value, as the acceleration due to gravity near Earth's surface for an object in free fall. And this is what we're going to focus on this video. And the reason why I'm stressing this last part is because we know of many objects that are not in free fall. For example, I am near the surface of the Earth right now, and I am not in free fall. What's happening to me right now is I'm sitting in a chair. And so this is my chair-- draw a little stick drawing on my chair, and this is me. And let's just say that the chair is supporting all my weight. So I have-- my legs are flying in the air. So this is me. And so what's happening right now? If I were in free fall, I would be accelerating towards the center of the Earth at 9.81 meters per second squared. But what's happening is, all of the force due to gravity is being completely offset by the normal force from the surface of the chair onto my pants, and so this is normal force. And now I'll make them both as vectors. is equal to 0, especially in this vertical direction. And because the net force is equal to 0, I am not accelerating towards the center of the Earth. I am not in free fall. And because this 9.81 meters per second squared still seems relevant to my situation-- I'll talk about that in a second. But I'm not an object in free fall. Another way to interpret this is not as the acceleration due to gravity near Earth's surface for an object in free fall, although it is that-- a maybe more general way to interpret this is the gravitational-- or Earth's gravitational field. Or it's really the average acceleration, or the average, because it actually changes slightly throughout the surface of the Earth. But another way to view this, as the average gravitational field at Earth's surface. Let me write it that way in pink. So the average gravitational field-- and we'll" + }, + { + "Q": "At 0:26 he says \"methanol.\" I know this is a pretty elementary question, but why does methanol end in \"ol\"? As for a question actually pertaining to the video, does anyone have classic examples of molecules being used to teach oxidation states. (like practice problems?)", + "A": "Methanol is a combination of the base methane plus an alcohol group, which tends to be simplified in nomenclature by ol . Thus, we have methanol.", + "video_name": "CuGg-Tf8lPI", + "timestamps": [ + 26 + ], + "3min_transcript": "- Both formal charge and oxidation states are ways of counting electrons, and they're both very useful concepts. Let's start with formal charge. So one definition for formal charge is the hypothetical charge that would result if all bonding electrons are shared equally. So let's go down to the dot structure on the left here, which is a dot structure for methanol, and let's assign a formal charge to carbon. We need to think about the bonding electrons or the electrons in those bonds around carbon, and we know that each bond consists of two electrons. So the bond between oxygen and carbon consists of two electrons. Let me go ahead and draw in those two electrons. Same for the bond between carbon and hydrogen, right? Each bond consists of two electrons, so I can go around and put in all of my bonding electrons. So if we want to assign a formal charge to carbon, we need to think about the number of valence electrons in the free atom or the number of valence electrons that carbon is supposed to have. four valence electrons, so I could put a four here, and from that four we're going to subtract the number of valence electrons in the bonded atom or the number of valence electrons that carbon has around it in our drawing. And since we're doing formal charge, we need to think about all those bonding electrons being shared equally. So we think about a covalent bond. So if we have two electrons and one bond, and those two electrons are shared equally, we could split them up. We could give one electron to oxygen and one electron to carbon in that bond. We go over here to this carbon hydrogen bond, and we could do the same thing. We have two electrons. We could split up those two electrons. We could give one to carbon and one to hydrogen, and we go all the way around, and we do the same thing over here. Split up those electrons and the same thing here. So how many valence electrons do we see around carbon now? So let me go ahead and highlight them. There's one, two, three, and four. around carbon in our drawing. So four minus four is equal to zero. So zero is the formal charge of carbon. So let me go ahead and highlight that here. So in this molecule the formal charge for carbon is zero. Now let's move on to oxidation states, right? So you could also call these oxidation numbers. So one definition for an oxidation state is the hypothetical charge that would result if all those bonding electrons are assigned to the more electronegative atom in the bond. So let's go to the dot structure on the right of methanol and let's assign an oxidation state to that carbon. We need to think about our bonding electrons again, so let's go ahead and put those in, all right? So we know that each bond consists of two electrons." + }, + { + "Q": "at @6:12 why didnt you change the final velocity vector into negative.", + "A": "Because that is the magnitude it was coming down at, he was not trying to point out the direction.", + "video_name": "sTp4cI9VyCU", + "timestamps": [ + 372 + ], + "3min_transcript": "So the vertical component of our velocity is negative 29.06 times .03 in the downward direction. And the horizontal component of our velocity, we know, hadn't changed the entire time. That, we figured out, was 30 cosine of 80 degrees. So that over here, is 30 cosine of 80 degrees. I'll get the calculator out to calculate it. 30 cosine of 80 degrees, which is equal to 5.21. So this is 5.21 meters per second. These are both in meters per second. So what is the total velocity? Well, I can do the head to tails. So I can shift this guy over so that its tail is at the head of the blue vector. So it would look like that. The length of this-- the magnitude And then we could just use the Pythagorean theorem to figure out the magnitude of the total velocity upon impact. So the length of that-- we could just use the Pythagorean theorem. So the magnitude of our total velocity, that's this length right over here. The magnitude of our total velocity, our total final velocity I guess we can say, is going to be equal to-- well that's-- let me write it this way. The magnitude of our total velocity is going to be equal to square root-- this is just straight from the Pythagorean theorem-- of 5.21 squared plus 29.03 squared. And we get it as being the second-- the square root of 5.21 squared plus 29.03 squared This is equal to 29.49 meters per second. That is the magnitude of our final velocity, but we also need to figure out its direction. And so we need to figure out this angle. And now we're talking about an angle below the horizontal. Or, if you wanted to view it in kind of pure terms, it would be a negative angle-- or we could say an angle below the horizontal. So what is this angle right over here? So if we view it as a positive angle just in the traditional trigonometric way, we could say that the-- we could use any of the trig functions, we could even use tangent. Let's use tangent. We could say that the tangent of the angle, is equal to the opposite over the adjacent-- is equal to 29.03 over 5.21. Or that theta is equal to the inverse tangent, or the arctangent of 29.03 over 5.21." + }, + { + "Q": "At 10:17 Sal said that the reason why the car moves through the curve in a curved manner rather than going straight is because of friction. Shouldn't it be because of the change of vector velocity as Sal explained earlier in the satellite, yoyo and the the object travelling in a circular path in space examples?", + "A": "the frictions is the cause of the change in vector velocity", + "video_name": "vZOk8NnjILg", + "timestamps": [ + 617 + ], + "3min_transcript": "Another example that you are probably somewhat familiar with or at least have heard about is if you have something in orbit around the planet So let's say this is Earth right here and you have some type of a satellite that is in orbit around Earth That satellite has some velocity at any given moment in time What's keeping it from not flying out into space and keeping it going in a circle is the force of gravity So in the example of a satellite or anything in the orbit even the moon in orbit around the Earth the thing that's keeping an orbit as opposed to flying out into space is a centripetal force of Earth's gravity Now another example, this is probably the most everyday example because we do it all the time If you imagine a car traveling around the racetrack Before I tell you the answer, I'll have you think about it It's circular. Let's view the racetrack from above If I have a car on a racetrack. I want you to pause it before I tell it to you because I think it's an interesting thing think about It seems like a very obvious thing that's happening We've all experienced; we've all taken turns in cars So we're looking at the top of a car. Tires When you see a car going at a constant speed so on the speedometer, it might say, 60 mph, 40 mph, whatever the constant speed but it's traveling in a circle so what is keeping--what is the centripetal force in that example? There's no obvious string being pulled on the car towards the center There is no some magical gravity pulling it towards the center of the circle There's obviously gravity pulling you down towards the ground but nothing pulling it to the side like this And I encourage you to pause it right now before I tell you the answer Assuming you now unpaused it and I will now tell you the answer The thing that's keeping it going in the circle is actually the force of friction It's actually the force between the resist movement to the side between the tires and the road And a good example of that is if you would remove the friction if you would make the car driving on oil or on ice or if you would shave the treads of the tire or something then the car would not be able to do this So it's actually the force of friction in this example I encourage you to think about that" + }, + { + "Q": "At 0:59, it was specified that the carbon turns into an sp2 hybridized carbon, meaning that all the substituents should be on the same plane, yet there are still wedges and dashes, indicating that the substituents of the carbon are not on the same plane. Can somebody explain this to me?", + "A": "The substituents are still in the same plane. It is just that you are looking at the plane edge-on. This puts one bond closer to you (a wedge), one bond further away (a dashed line), and one that is equidistant at both ends (an ordinary line).", + "video_name": "sDZDgctzbkI", + "timestamps": [ + 59 + ], + "3min_transcript": "- [Narrator] In this video, we're going to look at the stereochemistry of the SN1 reaction. On the left is our alkyl halide, on the right is our nucleophile with a negative charge on the sulfur. We know that the first step of our SN1 mechanism should be loss of a leaving group. So if these electrons come off onto the bromine, we would form the bromide anion. And we're taking a bond away from the carbon in red. So the carbon in red should get a plus one formal charge. So let's draw the resulting carbocation here. So let me sketch that in. The carbon in red is this carbon. So that carbon should have a plus one formal charge. In the next step of our mechanism, our nucleophile will attack, alright? So the nucleophile attacks the electrophile and a bond will form between the sulfur and the carbon in red. But remember the geometry directly around that carbon in red, the carbons that are bonded to it, so this carbon in magenta, this carbon in magenta, and this carbon in magenta and so the nucleophile could attack from either side of that plane. At this point, I think it's really helpful to look at this reaction using the model set. So here's a screenshot from the video I'm going to show you in a second, and in that video I make bromine green, so here you can see this green bromine, this methyl group coming out at us in space is going to be red in the video. On the right side this ethyl group here will be yellow, and finally on the left side this propyl group will be gray. So here's our alkyl halide with our bromine going away from us, our methyl group coming out at us, our ethyl group on the right, and the propyl group on the left. So I'll just turn this a little bit so we get a different viewpoint, and we know that the first step is loss of our leaving group so I'm going to show these electrons coming off onto our bromine and leaving to form a carbocation. We need planar geometry around that central carbon. So here's another model which is more accurate. Now the nucleophile could attack from the left or from the right, and first let's look at what happens when the nucleophile attacks from the left. So we form a bond between the sulfur and the carbon, and let's go ahead and look at a model set of one of our products. So here's the product that results when the nucleophile attacks from the left side of the carbocation. Here's our carbocation again, and this time let's say the nucleophile approaches from the right side. So we're going to form a bond between this sulfur and this carbon. And let's make a model of the product that forms when the nucleophile attacks from the right. So here is that product. And then we hold up the carbocation so we can compare the two. Now let's compare this product with the product when the nucleophile attacked from the left side." + }, + { + "Q": "When Sal mentions that the helium core fuses into heavier elements at 6:02, why only up to iron 26? What about Uranium, gold, etc. ?\nSecondly what happens when A heavier element fuses with a light by one by somehow traveling from its shell or core to another?", + "A": "It has to do with the strength of the Strong Nuclear force. The Strong Nuclear force has a limited distance that is works over. The size of the iron nucleus is the point where the Strong Nuclear force is weakened to the point where to make a larger atom it takes more energy to push the protons/neutrons together than comes out of the process. In systems called heavy ion colliders that they use them to produce heavy elements. So as long as there is enough energy in the collision they can fuse.", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 362 + ], + "3min_transcript": "as the Earth's orbit around the current sun. Or another way to view it is, where we are right now will be on the surface or near the surface or maybe even inside of that future sun. Or another way to put it, when the sun becomes a red giant, the Earth's going to be not even a speck out here. And it will be liquefied and vaporized at that point in time. So this is super, super huge. And we've even thought about it. Just for light to reach the current sun to our point in orbit, it takes eight minutes. So that's how big one of these stars are. To get from one side of the star to another side of the star, it'll take 16 minutes for light to travel, if it was traveling that diameter, and even slightly longer if it was to travel it in a circumference. So these are huge, huge, huge stars. And we'll talk about other stars in the future. They're even bigger than this when they become supergiants. But anyway, we have the hydrogen in the center-- sorry. Let me write this down. We have a helium core in the center. We're fusing faster and faster and faster. We're now a red giant. The core is getting hotter and hotter and hotter until it gets to the temperature for ignition of helium. So until it gets to 100 million Kelvin-- remember the ignition temperature for hydrogen was 10 million Kelvin. So now we're at 100 million Kelvin, factor of 10. And now, all of a sudden in the core, you actually start to have helium fusion. And we touched on this in the last video, but the helium is fusing into heavier elements. And some of those heavier elements, and predominately, it will be carbon and oxygen. And you may suspect this is how heavier and heavier elements form in the universe. They form, literally, due to fusion in the core of stars. But anyway, the core is now experiencing helium fusion. It has a shell around it of helium that is not quite there, does not quite have the pressures and temperatures to fuse yet. So just regular helium. But then outside of that, we do have the pressures and temperatures for hydrogen to continue to fuse. So out here, you do have hydrogen fusion. And then outside over here, you just have the regular hydrogen plasma. So what just happened here? When you have helium fusion all of a sudden-- now this is, once again, providing some type of energetic outward support for the core. So it's going to counteract the ever-increasing contraction of the core as it gets more and more dense, because now we have energy going outward, energy pushing things outward. But at the same time that that is happening," + }, + { + "Q": "At 2:52, he mentioned that the star was starting to collapse. What exactly is meant by collapse?", + "A": "i think he means that a while a red giant is expanding,there is also a white dwarf inside, which when the red giant sheds his outer layers, would reveal the white dwarf. but i m not sure.", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 172 + ], + "3min_transcript": "So core becomes more dense. And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster. Because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing, so it starts to fuse hotter. So let me write this, so the fusion, so hydrogen fuses faster. And actually, we even see this in our sun. Our sun today is brighter and hotter. It's fusing faster than it was when it was born 4.5 or 4.6 billion years ago. But eventually you're going to get to the point so that the core, you only have helium. So there's going to be some point where the entire core is all helium. And it's going to be way denser than this core over here. All of that mass over there has now been turned into helium. But most of it is now in helium, and it's going to be at a much, much smaller volume. And the whole time, the temperature is increasing, the fusion is getting faster and faster. And now there's this dense volume of helium that's not fusing. You do have, and we saw this in this video, a shell around it of hydrogen that is fusing. So this right here is hydrogen fusion going on. And then this over here is just hydrogen plasma. Now the unintuitive thing, or at least this was unintuitive to me at first, is what's going on the core is that the core is getting more and more dense. It's fusing at a faster rate. And so it's getting hotter and hotter. So the core is hotter, fusing faster, getting more and more dense. I kind of imagine it's starting to collapse. Every time it collapses, it's getting hotter and more dense. But at the same time that's happening, the star itself is getting bigger. Red giants are much, much larger than main sequence stars. But the whole time that this is getting more dense, the rest of the star is, you could kind of view it as getting less dense. And that's because this is generating so much energy that it's able to more than offset, or better offset the gravitational pull into it. So even though this is hotter, it's able to disperse the rest of the material in the sun over a larger volume. And so that volume is so big that the surface, and we saw this in the last video, the surface of the red giant is actually cooler-- let me write that a little neater-- is actually cooler than the surface of a main sequence star. This right here is hotter. And just to put things in perspective, when the sun becomes a red giant, and it will become a red giant, its diameter will be 100 times the diameter that it is today." + }, + { + "Q": "Close to the end of a video, (12:21) it said that over many years, a white dwarf would eventually become burned out and turned into a black dwarf. Can you give me a time frame of how long that would take?", + "A": "I have see estimates for around 8 to 11 billion years for a 0.5 to 1 solar mass white dwarf to cool to be a black dwarf.", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 741 + ], + "3min_transcript": "All of this hydrogen, all of this fusing hydrogen will run out. All of this fusion helium will run out. This is the fusing hydrogen. This is the inert helium, which will run out. It'll be used in kind of this core, being fused into the carbon and oxygen, until you get to a point where you literally just have a really hot core of carbon and oxygen. And it's super-dense. This whole time, it will be getting more and more dense as heavier and heavier elements show up in the course. So it gets denser and denser and denser. But the super dense thing will not, in the case of the sun-- and if it was a more massive star, it would get there-- but in the case of the sun, it will not get hot enough for the carbon and the oxygen to form. So it really will just be this super-dense ball of carbon and oxygen and all of the other material in the sun. Remember, it was superenergetic. The more that we progressed down this, the more energy was releasing outward, and the larger the radius of the star became, and the cooler the outside of the star became, until the outside just becomes this kind of cloud, this huge cloud of gas around what once was the star. And in the center-- so I could just draw it as this huge-- this is now way far away from the star, much even bigger than the radius or the diameter of a red giant. And all we'll have left is a mass, a superdense mass of, I would call it, inert carbon or oxygen. This is in the case of the sun. And at first, when it's hot, and it will be releasing radiation because it's so hot. We'll call this a white dwarf. This right here is called a white dwarf. And it'll cool down over many, many, many, many, many, many, many, years, until it becomes, when it's completely it'll just be this superdense ball of carbon and oxygen, at which point, we would call it a black dwarf. And these are obviously very hard to observe because they're not emitting light. And they don't have quite the mass of something like a black hole that isn't even emitting light, but you can see how it's affecting things around it. So that's what's going to happen to the sun. In the next few videos, we're going to talk about what would happen to things less massive than the sun and what would happen to things more massive can imagine the more massive. There would be so much pressure on these things, because you have so much mass around it, that these would begin to fuse into heavier and heavier elements until we get to iron." + }, + { + "Q": "At 1:40, he said our sun was even hotter than it was before because it is fusing faster. Is this causing global warming then?", + "A": "no As compared to4.6 billion years ago it it hotter but global warming is caused by human activities in the mordern time(industrialisation). I would say that the solar output is stable since last 3 billion years and would be same for next 3-4 billion years", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 100 + ], + "3min_transcript": "In the last video, we started with a star in its main sequence, like the sun. And inside the core of that star, you have hydrogen fusion going on. So that is hydrogen fusion, and then outside of the core, you just had hydrogen. You just hydrogen plasma. And when we say plasma, it's the electrons and protons of the individual atoms have been disassociated because the temperatures and pressures are so high. So they're really just kind of like this soup of electrons and protons, as opposed to proper atoms that we associate with at lower temperatures. So this is a main sequence star right over here. And we saw in the last video that this hydrogen is fusing into helium. So we start having more and more helium here. And as we have more and more helium, the core becomes more and more dense, because helium is a more massive atom. It is able to pack more mass in a smaller volume. So core becomes more dense. And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster. Because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing, so it starts to fuse hotter. So let me write this, so the fusion, so hydrogen fuses faster. And actually, we even see this in our sun. Our sun today is brighter and hotter. It's fusing faster than it was when it was born 4.5 or 4.6 billion years ago. But eventually you're going to get to the point so that the core, you only have helium. So there's going to be some point where the entire core is all helium. And it's going to be way denser than this core over here. All of that mass over there has now been turned into helium. But most of it is now in helium, and it's going to be at a much, much smaller volume. And the whole time, the temperature is increasing, the fusion is getting faster and faster. And now there's this dense volume of helium that's not fusing. You do have, and we saw this in this video, a shell around it of hydrogen that is fusing. So this right here is hydrogen fusion going on. And then this over here is just hydrogen plasma. Now the unintuitive thing, or at least this was unintuitive to me at first, is what's going on the core is that the core is getting more and more dense. It's fusing at a faster rate. And so it's getting hotter and hotter. So the core is hotter, fusing faster, getting more and more dense. I kind of imagine it's starting to collapse. Every time it collapses, it's getting hotter and more dense. But at the same time that's happening, the star itself is getting bigger." + }, + { + "Q": "At 7:48 would we be able to see a he flash from Earth?", + "A": "Its not a literal flash. Its just a point in the Suns life cycle.", + "video_name": "EdYyuUUY-nc", + "timestamps": [ + 468 + ], + "3min_transcript": "But anyway, the core is now experiencing helium fusion. It has a shell around it of helium that is not quite there, does not quite have the pressures and temperatures to fuse yet. So just regular helium. But then outside of that, we do have the pressures and temperatures for hydrogen to continue to fuse. So out here, you do have hydrogen fusion. And then outside over here, you just have the regular hydrogen plasma. So what just happened here? When you have helium fusion all of a sudden-- now this is, once again, providing some type of energetic outward support for the core. So it's going to counteract the ever-increasing contraction of the core as it gets more and more dense, because now we have energy going outward, energy pushing things outward. But at the same time that that is happening, is fusing into helium. So it's making this inert part of the helium core even larger and larger and denser, even larger and larger, and putting even more pressure on this inside part. And so what's actually going to happen within a few moments, I guess, especially from a cosmological point of view, this helium fusion is going to be burning super-- I shouldn't use-- igniting or fusing at a super-hot level. But it's contained due to all of this pressure. But at some point, the pressure won't be able to contain it, and the core is going to explode. But it's not going to be one of these catastrophic explosions where the star is going to be destroyed. It's just going to release a lot of energy all of a sudden into the star. And that's called a helium flash. But once that happens, all of a sudden, then now the star is going to be more stable. And I'll use that in quotes without writing it down getting to be less stable than a main sequence star. But once that happens, you now will have a slightly larger volume. So it's not being contained in as small of a tight volume. That helium flash kind of took care of that. So now you have helium fusing into carbon and oxygen. And there's all sorts of other combinations of things. Obviously, there's many elements in between helium and carbon and oxygen. But these are the ones that dominate. And then outside of that, you have helium forming. You have helium that is not fusing. And then outside of that, you have your fusing hydrogen. Over here, you have hydrogen fusing into helium. And then out here in the rest of the radius of our super-huge red giant, you just have your hydrogen plasma out here. Now what's going to happen as this star ages? Well, if we fast forward this a bunch-- and remember," + }, + { + "Q": "At 1:34, how is gas a fluid? I know it fills in the shape of what it is in, but how can it be a fluid? Also, at 6:30 what is that thing?t", + "A": "a fluid is a substance which has the ability to flow. liquids can flow from a state of higher potential energy to lower potential energy. in gasses, the same thing applies. they are also considered as fluids because they flow to the edges of the container or from a hot surface to a cold surface. gasses are fluids because they have the ability to flow", + "video_name": "Pn5YEMwQb4Y", + "timestamps": [ + 94, + 390 + ], + "3min_transcript": "Let's learn a little bit about fluids. You probably have some notion of what a fluid is, but let's talk about it in the physics sense, or maybe even the chemistry sense, depending on in what context you're watching this video. So a fluid is anything that takes the shape of its container. For example, if I had a glass sphere, and let's say that I completely filled this glass sphere with water. I was going to say that we're in a zero gravity environment, but you really don't even need that. Let's say that every cubic centimeter or cubic meter of this glass sphere is filled with water. Let's say that it's not a glass, but a rubber sphere. If I were to change the shape of the sphere, but not really change the volume-- if I were to change the shape of the sphere where it looks like this now-- the water would just change its shape with the container. and in this case, I have green water. The same is also true if that was oxygen, or if that was just some gas. It would fill the container, and in this situation, it would also fill the newly shaped container. A fluid, in general, takes the shape of its container. And I just gave you two examples of fluids-- you have liquids, and you have gases. Those are two types of fluid: both of those things take the shape of the container. What's the difference between a liquid and a gas, then? A gas is compressible, which means that I could actually decrease the volume of this container and the gas will just become denser within the container. could squeeze that balloon a little bit. There's air in there, and at some point the pressure might get high enough to pop the balloon, but you can squeeze it. A liquid is incompressible. How do I know that a liquid is incompressible? Imagine the same balloon filled with water-- completely filled with water. If you squeezed on that balloon from every side-- let me pick a different color-- I have this balloon, and it was filled with water. If you squeezed on this balloon from every side, you would not be able to change the volume of this balloon. No matter what you do, you would not be able to change the volume of this balloon, no matter how much force or pressure you put from any side on it, while if this was filled with gas-- and magenta, blue in for gas-- you actually could decrease the volume by just increasing the pressure on all sides of the balloon. You can actually squeeze it, and make the entire volume smaller." + }, + { + "Q": "At 5:03 wouldn't you feel the G-force because 250 m/s > 9.8 m/s and couldn't you measure that?", + "A": "you are comparing a velocity to an acceleration. It s not 9.8 m/s it s 9.8 m/s^2. That s acceleration. 250 m/s is a velocity.", + "video_name": "3yaZ7lkQPUQ", + "timestamps": [ + 303 + ], + "3min_transcript": "It would look like it's moving behind you or in this case, the way we're looking at it, to the right at 50 meters per second. Now, what would the plane look like? Well, the plane not only would it look like it's moving to the right at 250 meters per second, not only would it be just that 250 meters per second, but relative to you it'd look like it's going even faster 'cause you're going past it, you are going to the left from the stationary, from the ground's point of view at 50 meters per second. So the plane, to you, is gonna look like it's going 250 plus 50 meters per second. So the vector would look like this and so it would look like it's going to the right at 300, let me write that in that orange color, at 300 meters per second. What if we're talking about the plane's frame of reference? Why don't you pause this video and think about what the velocities would be of the plane, the car, and the ground from the plane's point of view. All right, now let's work through this together. So now, we're sitting in the plane and once again we shouldn't be flying the plane, we're letting someone else do that, we have our physics instruments out and we're trying to measure the velocities of these other things from my frame of reference. Well, the plane, first of all, is going to appear to be stationary and that might seem counterintuitive, but if you've ever sat in a plane, especially when there's no turbulence and the plane is already at altitude and it's not taking off or landing, oftentimes if you close your eyes you don't know if you are moving. In fact, if you close all of the windows, it feels like you are in a stationary object, that you might as well be in a house. So from the plane's point of view, you feel like, or from your point of view in the plane, Now the ground, however, looks like it's moving quite quickly. It'll look like it's moving past you at 250 meters per second. Whoops, try and draw a straight line. At 200... At 250. Sometimes my tools act funny. So, at 250 meters per second to the left. And the car, well it's moving to the left even faster. It's going to be moving to the left 50 meters per second faster than the ground is. So the car is gonna look not like it's just going 50 meters per second, it's gonna look like it's going 50 meters plus another 250 meters per second for a total of 300 meters per second to the left. So this gives you an appreciation" + }, + { + "Q": "Why does it take the same amount of time to cycle regardless of the amplitude? (that is to say, if the starting postion is A/2 it still takes the same amount time to go to -A/2 and return as it would if the starting postion were A)\n\nis it because the acceleration is less at A/2 than at A so the lower velocity means it takes longer to cycle? but why the same amount of time?\n\n(around 7:30 onwards)", + "A": "Well from my understanding, the cycles both take the same amount of time because of its velocity, the one that is stretched more goes faster making it quicker, keeping a constant time, and the one that isnt stretched very far goes much slower, draging the time out even though it has less distance to cover. I m having a hard time explaining, it works well if you visualize it, imagine it and if you were to do it yourself.", + "video_name": "oqBHBO8cqLI", + "timestamps": [ + 450 + ], + "3min_transcript": "both sides of this equation times the inverse of the square root of k over m. And you get, t is equal to 2 pi times the square root-- and it's going to be the inverse of this, right? Of m over k. And there we have the period of this function. This is going to be equal to 2 pi times the square root of m over k. So if someone tells you, well I have a spring that I'm going to pull from some-- I'm going to stretch it, or compress it a little bit, then I let go-- what is the period? How long does it take for the spring to go back to its It'll keep doing that, as we have no friction, or no gravity, or any air resistance, or Air resistance really is just a form of friction. You could immediately-- if you memorize this formula, although you should know where it comes from-- you could It's 2 pi times m over k. That's how long it's going to take the spring to get back-- to complete one cycle. And then what about the frequency? If you wanted to know cycles per second, well that's just the inverse of the period, right? So if I wanted to know the frequency, that equals 1 over the period, right? Period is given in seconds per cycle. So frequency is cycles per second, and this is seconds per cycle. So frequency is just going to be 1 over this. Which is 1 over 2 pi times the square root of k over m. That's the frequency. But I have always had trouble memorizing this, and this. k over m, and m over k, and all of that. All you have to really memorize is this. And even that, you might even have an intuition as to why it's true. You can even go to the differential equations if you want to reprove it to yourself. Because if you have this, you really can answer any question The velocity of the mass, at any time, just by taking the Or the period, or the frequency of the function. As long as you know how to take the period and frequency of trig functions. You can watch my videos, and watch my trig videos, to get a refresher on that. One thing that's pretty interesting about this, is notice that the frequency and the period, right? This is the period of the function, that's how long it takes do one cycle. This is how many cycles it does in one second-- both of them are independent of A. So it doesn't matter, I could stretch it only a little bit, like there, and it'll take the same amount of time to go back, and come back like that, as it would if I stretch it a lot. It would just do that. If I stretched it just a little bit, the function would look like this. Make sure I do this right. I'm not doing that right. Edit, undo. If I just do it a little bit, the amplitude is going to be less, but the function is going to essentially do the" + }, + { + "Q": "1:25\nit said that that molecule's longest chain is 9-carbons\nbut i think that's wrong\nit's (i know this is the incorrect name, but i'm using it for an example) 2-propylheptane\n(emphasis on propyl)\npropyl means 3, but you wrote 4\nthe yellow line means it connects to the 9-carbons\nit's not the actual molecule\nso i believe the actual name is not 4-methylnonane, but 4-methyloctane!\nplease tell me if i am wrong or correct\ni'm just a bit confused...", + "A": "The carbon atom that forms part of the long chain is counted as part of part of the alkane group but not the alkyl group. And remember that the lines are the bonds, the important things are the carbon atoms which are the end-points of the lines.", + "video_name": "CFBKfgGTP98", + "timestamps": [ + 85 + ], + "3min_transcript": "In the last video, we tried to draw a 2-Propylheptane. And we did our best attempt at drawing it, but it was pointed out that this wouldn't even be called a 2-Propylheptane to So you actually should never see something called 2-Propylheptane. Let me show you what I'm talking about. So when do you see something like this, you might immediately say-- and the way we drew it was actually correct, it just wouldn't be called 2-Propylheptane. So you say heptane. So that is a seven carbon alkane. No double bond, so one, two, three, four, five, six, seven. And then on the second carbon, so you have one, two, three, four, five, six, seven, we have a propyl group. Propyl, that is three carbons. So on the second carbon we have a propyl group. That's three carbons. So that is one, two, three. And so if someone gave you 2-Propylheptane, this would be what you would draw. But you wouldn't call this 2-Propylheptane. Because remember, if you're given the molecule you look for the longest chain and the longest chain here is not the It is not one, two, three, it is not this thing in magenta-- in this kind of mauve color. It's this chain where you start over here. If you start over here, one, two, three, four, five, six, seven, eight, nine, you actually get a longer chain. So this would actually be the backbone of this molecule right over there. That right over there would be the backbone, and so you would number it. You start numbering closest to the group that's attached, so one, two, three, four, five, six, seven, eight, nine. So you have nine carbons in your backbone, so we're dealing with nonane. We're dealing with nonane. And you have a methyl group: one carbon attached to the So this is going to be four methyl, this is our methyl group right here, 4-Methylnonane. So it was brought up, I think the user name is Minoctu, and they correctly corrected me, that there would never be such a thing as 2-Propylheptane. I just made that up. If someone were to, kind of, label this molecule they would call it 4-Methylnonane and ask you draw it. But either way, both of these would point you in the right direction. This would just be the incorrect name for it, because you'd be looking at-- if someone gave you this molecule and you named it this way, that would be incorrect. So I apologize for this. This is 4-Methylnonane. If you do heptane you're not finding the longest chain." + }, + { + "Q": "So, what would be the name of the actual neurotransmitter that is released as a response to the Glomus cell depolarizing? Would this be acetocholine or epinephrine? (6:24 in video)", + "A": "Neurotransmitters actually known to be used by the glomus cells are : dopamine, noradrenaline, acetylcholine, substance P, vasoactive intestinal peptide and enkephalins.", + "video_name": "cJXY3Cywrnc", + "timestamps": [ + 384 + ], + "3min_transcript": "and these are the chemoreceptors that we're talking about. So these blue cells together make up a body of tissue, and that's where we get the term \"aortic body\" and \"carotid body.\" Now, on the carotid side, one interesting fact is that this body of tissue gets a lot of blood flow, in fact, some of the highest blood flow in the entire human body. It's about 2 liters per minute for 100 grams. And just to put that in perspective for the carotid body, imagine that you have a little 2-liter bottle of soda. I was thinking of something that would be about 2 liters, and soda came to mind. And you can imagine pouring this soda out over something that's about 100 grams-- maybe a tomato. That's about a 100-gram tomato. And if you could do this in one minute, if you could pour out this bottle in one minute, imagine how wet this tomato's going to get, how much profusion, in a sense, this tomato is going to get. That is how much profusion your carotid body gets. how much blood flow's going into that area. So let's now zoom in a little bit further. Let's say I have a capillary. And inside my capillary, I've got a little red blood cell here, floating around. And my red blood cell, of course, has some hemoglobin in it, which is a protein. And this protein has got some oxygen bound to it. I'm going to draw little blue oxygen molecules. And of course, there's some oxygen out here in the plasma itself as well. And if we're in our carotid body or aortic body, you might have these special little blue cells that I've been drawing, our peripheral chemoreceptor cells. And specifically they have a name. These things are called glomus cells. I had initially misstated it as a globus cell. But actually it's an M-- glomus. And these oxygen molecules-- these are oxygen molecules over here-- are going to diffuse down into the tissue and get into our glomus cell. It's going to look something like that. And if you have a lot of oxygen in the blood, of course, But if you don't have too much in here, then not too much is going to make its way into the cell. And that's actually the key point. Because what our cell is going to be able to do is start to detect low oxygen levels. Low oxygen levels in the glomus cell tells this cell that, actually, there are probably low levels in the blood. And when the levels are low, this cell is going to depolarize. Its membrane is going to depolarize. And what it has on the other side are little vesicles that are full of neurotransmitter. And so when these vesicles detect that, hey, there's a depolarization going on, these vesicles are going to dump their neurotransmitter out. And what you have waiting for them is this nice little neuron. So there's a nice little neuron waiting patiently for a signal, and that signal is going to come in the form of a neurotransmitter. So this is how the communication works. There's going to be a depolarization," + }, + { + "Q": "At 09:43, Sal uses Coulomb's Law to calculate the \"force generated by the ring\". Coulomb's Law defines the electrostatic force that exists between two point charges, and not that between a thin ring and a point on it's axis. He is applying the Law to calculate the latter, which, obviously cannot be justified. Has he made a mistake?", + "A": "No. He didn t make a mistake. It s like we assume a small amount of charge say dq and calculate its electric field at the axis of the coil then we tend to integrate the whole thing to obtain Electric field due to charge on entire coil.", + "video_name": "prLfVucoxpw", + "timestamps": [ + 583 + ], + "3min_transcript": "Let's draw a ring, because all of these points are going to be the same distance from our test charge, right? They all are exactly like this one point that I drew here. You could almost view this as a cross-section of this ring that I'm drawing. So let's figure out what the y-component of the electric force from this ring is on our point charge. So to do that, we just have to figure out the area of this ring, multiply it times our charge density, and we'll have the total charge from that ring, and then we can use Coulomb's Law to figure out its force or the field at that point, and then we could use this formula, which we just figured out, to figure out the y-component. I know it's involved, but it'll all be worth it, because you'll know that we have a constant electric field. So let's do that. So first of all, Coulomb's Law tells us-- well, first of all, let's figure out the charge from this ring. So Q of the ring, it equals what? width of the ring. So let's say the circumference is 2 pi r, and let's say it's a really skinny ring. It's really skinny. It's dr. Infinitesimally skinny. So it's width is dr. So that's the area of the ring, and so what's its charge going to be? It's area times the charge density, so times sigma. That is the charge of the ring. And then what is the electric field generated by the ring at this point here where our test charge is? Well, Coulomb's Law tells us that the force generated by the ring is going to be equal to Coulomb's constant times the charge of the ring times our test charge divided by the distance squared, right? Well, what's the distance between really any point on Well, this could be one of the points on the ring and this could be another one, right? And this is like a cross-section. So the distance at any point, this distance right here, is once again by the Pythagorean theorem because this is also r. This distance is the square root of h squared plus r squared. It's the same thing as that. So it's the distance squared and that's equal to k times the charge in the ring times our test charge divided by Well, distance is the square root of h squared plus r squared, so if we square that, it just becomes h squared plus r squared. And if we want to know the electric field created by that ring, the electric field is just the force per test charge, so if we divide both sides by Q, we learned that the electric field of the ring is equal to Coulomb's constant" + }, + { + "Q": "At 0:10 why is the electric field constant? Doesn't it get weaker the further away you get from the electric field?", + "A": "Not with infinite plates.", + "video_name": "prLfVucoxpw", + "timestamps": [ + 10 + ], + "3min_transcript": "In this video, we're going to study the electric field created by an infinite uniformly charged plate. And why are we going to do that? Well, one, because we'll learn that the electric field is constant, which is neat by itself, and then that's kind of an important thing to realize later when we talk about parallel charged plates and capacitors, because our physics book tells them that the field is constant, but they never really prove it. So we will prove it here, and the basis of all of that is to figure out what the electric charge of an infinitely charged plate is. So let's take a side view of the infinitely charged plate and get some intuition. Let's say that's the side view of the plate-- and let's say that this plate has a charge density of sigma. And what's charge density? It just says, well, that's coulombs per area. Charge density is equal to charge per area. That's all sigma is. So we're saying this has a uniform charge density. So before we break into what may be hard-core mathematics, might want to review some of the electrostatics from the physics playlist, and that'll probably be relatively easy for you. If you're watching this from the physics playlist and you haven't done the calculus playlist, you should not watch this video because you will find it overwhelming. But anyway, let's proceed. So let's say that once again this is my infinite so it goes off in every direction and it even comes out of the video, where this is a side view. Let's say I have a point charge up here Q. So let's think a little bit about if I have a point-- let's say I have an area here on my plate. Let's think a little bit about what the net effect of it is going to be on this point charge. Well, first of all, let's say that this point charge is at a height h above the field. Let me draw that. This is a height h, and let's say this is the point directly right here is r. So first of all, what is the distance between this part of our plate and our point charge? What is this distance that I'll draw in magenta? What is this distance? Well, the Pythagorean theorem. This is a right triangle, so it's the square root of this side squared plus this side squared. So this is going to be the square root of h squared plus r squared. So that's the distance between this area and our test charge. Now, let's get a little bit of intuition. So if this is a positive test charge and if this plate is positively charged, the force from just this area on the charge is going to be radially outward from this area, so it's going to be-- let me do it in another color because I don't want to-- it's going to go in that direction, right?" + }, + { + "Q": "At 7:45 and a little after, why are the oxygens positive? I thought when you have an oxygen-hydrogen bond the oxygen becomes partially negative?", + "A": "The oxygen in this case is positive because it is sharing 2 of its non-valence electrons with that hydrogen. Its 2 valence electrons are already tied up in the two covalent bonds with the two carbons, and when it gives up some of its rights to the 2 non-valence electrons to the hydrogen, it reduces its negative charge (making it positive). :)", + "video_name": "OpyTJbzA7Fk", + "timestamps": [ + 465 + ], + "3min_transcript": "and then just as he's leaving, this guy comes back. So there's different ways that all of these could happen but this is the general idea. And then you have it happening a third time down here. One of these lone pairs come and form a bond with this carbon. This carbon in the carbonyl group, part of this carboxyl group. And so once again, this guy can take those two electrons away and maybe share that pair with a hydrogen proton. Once again, this is forming a water molecule, this is forming a water molecule. So three water molecules are going to be produced. This is why we call it dehydration synthesis. We're losing three water molecules in order to form these bonds. So what's it going to look like after this has happened? Well, let me scroll down here. And actually let me just zoom. Actually just let me scroll down here. So this green bond over here is going to now-- is this green bond. And this second green bond is this green bond. And this third green bond is that green bond. The way I've drawn it right now, each of these oxygens haven't let go of its hydrogens. And that could actually happen before or after or all at the same time. Chemistry actually is not a clean thing. But I could, if I want, I could draw the hydrogens here. I could draw the hydrogens. I could draw the hydrogens over here and then these would have a positive charge. These would have a positive charge. But then you could imagine another water molecule coming by and kind of taking one of these hydrogen protons, taking the hydrogen protons away from each of those oxygens. this molecule right over here. And remember we produced three water molecules. So that's one, two and three water molecules. And now this molecule, if you ignore the water molecules out there, this is a triglyceride. Let me write it again. Actually, let me write the slightly more technical term. Sometimes referred to as triacylglycerol. Well we know where the glycerol comes from. It has this glycerine or this glycerol backbone right over here. Now what is \"acyl\" mean? Well acyl is a functional group where you have a carbon that's part of a carbonyl group. It can be bound to a kind of an organic chain" + }, + { + "Q": "At 3:43, Sal says that the nuclear membrane starts to disappear. Does this not break the law of conservation of mass? Where did it go? And how?", + "A": "This just means it breaks down, not that it disappears into thin air.", + "video_name": "TKGcfbyFXsw", + "timestamps": [ + 223 + ], + "3min_transcript": "the magenta stuff was still considered to be one chromosome. And we can draw the blue chromosome. Once again, it's now in the condensed form. That's one sister chromatid right over there. That's another sister chromatid. They are connected at the centromere. So they're condensing now as we enter into mitosis. And the nuclear membrane starts to go away. So the nuclear membrane is starting to go away. And these two centrosomes are starting to migrate to opposite sides of the cell. So one of them is going over here and one of them is maybe going to go over here. So they're migrating to opposite sides of the cell. And it's pretty incredible. It's easy to say, \"Oh, this happens and then that happens.\" Remember, this cell doesn't have a brain. This is all happening through chemical and thermodynamic of where the cell is and its life cycle. It's amazing that this is happening. It's amazing that the structures, and what sometimes we consider to be a simple thing but it's actually incredibly complex thing. It kind of knows what to do even though it has no intelligence here. And a lot of what I'm talking about, the general overview of the process is well understood but scientists are still understanding exactly how do the different things happen when they should happen and by what mechanism and what's actually happening. Sometimes, molecular or atomic basis. But anyway, this first phase of mitosis, the nuclear envelope, the nuclear membrane starts to disappear. The centrosomes migrate to the opposite ends of the cell. And our DNA condenses into kind of the condensed form of the chromosomes. And we call this prophase. Prophase. Prophase of mitosis. In the next phase, let me draw my cell again. Drawing that same green color. In the next phase, your nuclear membrane is now gone. And the chromosomes start lining up in the middle of the cells. So you have the blue one right over here, the blue one that's one sister chromatid. Here's another sister chromatid and they're connected at the centromere, not to be confused with a centrosome. And then, here's the magenta, one of the magenta sister chromatids and another one. And once again, it's not magenta in real life, I'm just making it a magenta because it looks nice. That's the centromere right over there. Our centrosomes are at opposite ends of the cell at this point. At opposite ends of the cell." + }, + { + "Q": "At 4:48, Sal says that the chromosomes are not magenta in real life. What color are they in real life?", + "A": "I believe it depends on the dye used on the cell before viewing it under the microscope. Though I am not fully certain.", + "video_name": "TKGcfbyFXsw", + "timestamps": [ + 288 + ], + "3min_transcript": "of where the cell is and its life cycle. It's amazing that this is happening. It's amazing that the structures, and what sometimes we consider to be a simple thing but it's actually incredibly complex thing. It kind of knows what to do even though it has no intelligence here. And a lot of what I'm talking about, the general overview of the process is well understood but scientists are still understanding exactly how do the different things happen when they should happen and by what mechanism and what's actually happening. Sometimes, molecular or atomic basis. But anyway, this first phase of mitosis, the nuclear envelope, the nuclear membrane starts to disappear. The centrosomes migrate to the opposite ends of the cell. And our DNA condenses into kind of the condensed form of the chromosomes. And we call this prophase. Prophase. Prophase of mitosis. In the next phase, let me draw my cell again. Drawing that same green color. In the next phase, your nuclear membrane is now gone. And the chromosomes start lining up in the middle of the cells. So you have the blue one right over here, the blue one that's one sister chromatid. Here's another sister chromatid and they're connected at the centromere, not to be confused with a centrosome. And then, here's the magenta, one of the magenta sister chromatids and another one. And once again, it's not magenta in real life, I'm just making it a magenta because it looks nice. That's the centromere right over there. Our centrosomes are at opposite ends of the cell at this point. At opposite ends of the cell. let me label this again. I labeled it in a previous video. That's centrosomes. Centrosomes where the two sister chromatids are connected. That's a centromere. Centromere. Now, you might heard of the word centrioles. Centrioles are actually, they help, they are exist inside the centrosomes. They're these two kind of cylindrical looking structures. Each of the centrosomes have two centrioles. But for the sake of this video, you can say, \"Well, the centrioles are just part of the centrosomes.\" But I'm just listing you all the words that involves centri something. Centrioles right over there and you have two of them per centrosome. So hopefully, that helps clarify some confusion. But these things line up. And a lot of what you're about to see, this orchestration, these things moving around in the cell, things lining up and soon things pulling apart, these are all coordinated with actually a fairly sophisticated mechanism of kind of a scaffold of this kind of ropes, these microtubules." + }, + { + "Q": "at 8:43, how can two chromosomes become four chromosomes?", + "A": "The chromosomes (genetic material) duplicates in the previous cell phases (interphase/prophase) that occur before the splitting in mitosis", + "video_name": "TKGcfbyFXsw", + "timestamps": [ + 523 + ], + "3min_transcript": "What's going to happen next? What's going to happen next is ... Let me, I don't wanna draw it too big 'cause I wanna be able to fit it all in one page. What's going to happen next is that those microtubules are going to start pulling on each of the sister chromatids. Let me draw that. So you have this centrosome right over here. You have all these microtubules in your cell. You have the centrosome right over here, all those microtubules. And this one is going to be pulling, is going to be getting one of the blue chromatids to pull towards it or to move towards it. So let me draw that. So this is one blue chromatid right over here. aAd this one is going to be pulling the other blue chromatid towards it. And same thing for the magenta. And same thing for the magenta. And this one is being pulled that way. And just in case you're concerned about the, some of more of the words of vocabulary involved, the point it was this microtubules connect to what used to be sister chromatids, but now that they're apart, we now call them chromosomes. When they emerge, this is one chromosome and they have two sister chromatids. But now they're apart, we would actually now consider this each an independent chromosome. So now, you actually have four chromosomes over here. This point right over here, we call it kinetochore. And even exactly how that interphase works and exactly how things move is not fully understood. Some of this stuff is understood but some of this is still an area of research. So something even as basic as cell division, not so basic after all. So this right over here where you can start to see the DNA kind of migrating to their respective sides of the cell. We call this the anaphase. We call this the anaphase. and that is called telophase. Telophase. And in telophase, I'm gonna do my best shot to draw it. And you could already see I have started to draw that the cellular membrane is starting to pinch in kind of in preparation for cytokineses, in preparation for the cell splitting into two cells. So I'll do it a little bit more. Cytokineses is usually described as kinda being a separate process in mitosis although it's obviously they are, they kind of together help the cell fully turn into two cells. So now in telophase, so you have this, what used to be a sister chromatid, now we can call it a chromosome on its own. And we have this chromosome. And we have this chromosome right over there. We have to do it on both sides. So you're there and you have" + }, + { + "Q": "At 9:05 it is mentioned that if young boys have any precocious puberty it is pathological, but that may not be always the case in young girls. Is there any reason why is it like this?", + "A": "Girls are more prone to environmental triggers for setting off puberty then boys are... Stress, changes in diet and eating habits, body weight and even sex itself can trigger puberty in women.", + "video_name": "XhYQYVQq6K0", + "timestamps": [ + 545 + ], + "3min_transcript": "If a girl comes in with just breast development but none of the body hair or body odor, we think, aha! she's been exposed to estrogens. And so we get a lot from the physical exam in terms of thinking what hormones to go after. okay, so even though the definitions of precocious is just any of these different physical developments happening too soon, the combination that you see is going to direct what yo uthing is going on wit hthem clinically. Exactly. Ok It will make us more or less worried and it will take us down one avenue or another, thinking about what could be wrong, what level of abnormality in the hormone system is at work. Ok. Can I ask you a question, just curious, so what are the most common diagnosis given to people coming in with their kids, that are concerned for precocious puberty? The common diagnosis, is \"variations are normal\". Aha. There is a condition called fancy doctor way of saying the adrenal glands producting their hormones, we see pubic hair, underarm hair, and acne. Uh... We also can see isolated breast development in little toddler girls that's not concerning. So that would just go away, or? Yes. Typically that begins between six months and two years and But even within the more full blown cases, where we're seeing breast and the pubic hair in a girl, or the enlagrement of the testes and the pubic hair in a boy, where's it's full on, full blown puberty, it may be idiopathic, or not pathological. We think that in girls, who develop fully, 95% of the time, even if they have the full blown puberty, it's idiopathic, or not pathological. So the condition is less common on boys, but when it occurs in boys, it's more worrisome. So it's less common for boys to have any signs of precocious puberty, but when it does happen you worry more. Exactly. Because it's more likeyl to be something bad. Exactly. So I'm guessing as an endocrinologist, once you've sort of ruled out the normal things, then you start th then you start thinking more about sort of what's going on in terms of all the different hormones that are controlling the se things, which maybe is something that we'll talk about in a different lecture exactly. is there any other sort of general things that you wanted to adress when we'll just talking about... I think the general advice I would give to a parent if see signs of puberty in their child, before 8 in a girl, before 9 in a boy, they should at least question their pediatrician" + }, + { + "Q": "5:20 why do we only divide 4.9 by 4 and not time?", + "A": "4.9t is one term - the result of multiplying 4.9 and the change in time. If t is one, 4.9t=4.9. If t is two, 4.9t=9.8. If t is one (4.9t=4.9), then 4.9t/4=4.9/4=1.225. Dividing both by 4 would be 4.9/4/4, or 4.9t/16.", + "video_name": "P7LKEkcNibo", + "timestamps": [ + 320 + ], + "3min_transcript": "that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air-- We're talking when it gets to this peak point, right over here-- that's from two videos ago-- that peak point right over there. So our average velocity is just going to be this stuff divided by 2. So it's going to be 4.9 meters per second squared times delta t over 2. So this right here, this is our average velocity. Velocity average. So let's stick that back over here. So our maximum displacement is going to be our average velocity-- so that is 4.9 meters per second squared-- times delta t, all of that over 2. And then we multiply it again times the time up. So times delta t over 2 again. These are the same thing. And then we can simplify it. Our maximum displacement is equal to 4.9 meters per second squared times delta t squared, all of that over 4. And then we can just divide 4.9 divided by 4. 4.9 divided by 4 is-- let me just get the calculator out. I don't want to do that in my head, get this far and make a careless mistake. 4.9 divided by 4 is 1.225. So our maximum displacement is going to be 1.225 times our total time in the air squared, which is a pretty straightforward calculation." + }, + { + "Q": "At 3:14, why is delta-v equal to 0-initial velocity?", + "A": "THat s how you find a change, you take final minus initial", + "video_name": "P7LKEkcNibo", + "timestamps": [ + 194 + ], + "3min_transcript": "we're just talking about half of the path of this ball. So the time that it gets released, and it's going at kind of its maximum upward velocity, and it goes slower, and slower, slower, all the way until it's stationary for just a moment, and then it starts going down again. Now remember, the acceleration is constant downwards this entire time. So what is the final velocity if we just consider half of the time? Well, it's the time. It's 0. So it's going to be 0 minus our initial velocity, when it was taking off. That's our change in velocity. This is our change in velocity, is going to be equal to the acceleration of gravity, negative 9.8 meters per second squared-- or the acceleration due to gravity when an object is in free fall, to be technically correct-- times the time that we are going up. So times delta t up, which is the same thing. Delta t up is the same thing as our total time in the air divided by 2. And so we get negative the initial velocity is equal to-- this thing, when you divide it by 2, is going to be 4.9 meters per second squared-- we still have our negative out front-- times our delta t. Remember, this is our total time in the air, not just the time up. This is our total time in the air. And then we multiply both sides times a negative. We get our initial velocity is just going to be equal to 4.9 meters per second squared times the total time that we are in the air. Or you could say it's going to be 9.8 meters per second squared times half of the time that we're in the air. Either of those would get you the same calculation. that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air--" + }, + { + "Q": "At 5:16, why does Sal square delta t? To cancel out the units?", + "A": "No..not to cancel the units, but to Simplify the calculation.", + "video_name": "P7LKEkcNibo", + "timestamps": [ + 316 + ], + "3min_transcript": "that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air-- We're talking when it gets to this peak point, right over here-- that's from two videos ago-- that peak point right over there. So our average velocity is just going to be this stuff divided by 2. So it's going to be 4.9 meters per second squared times delta t over 2. So this right here, this is our average velocity. Velocity average. So let's stick that back over here. So our maximum displacement is going to be our average velocity-- so that is 4.9 meters per second squared-- times delta t, all of that over 2. And then we multiply it again times the time up. So times delta t over 2 again. These are the same thing. And then we can simplify it. Our maximum displacement is equal to 4.9 meters per second squared times delta t squared, all of that over 4. And then we can just divide 4.9 divided by 4. 4.9 divided by 4 is-- let me just get the calculator out. I don't want to do that in my head, get this far and make a careless mistake. 4.9 divided by 4 is 1.225. So our maximum displacement is going to be 1.225 times our total time in the air squared, which is a pretty straightforward calculation." + }, + { + "Q": "At around 2:50, Sal divided the equation by (-1), but the right side of the equation remained the same, as he wrote it.\n\nIt matters for the direction of the velocity. Did it stay in the same direction or not?\n\nAnd if it did, why did he not put the velocity in the velocity in absolute value?\n\n\nThanks in advance for the helpers! :)", + "A": "there was a negative symbol on both sides before he did that calculation", + "video_name": "P7LKEkcNibo", + "timestamps": [ + 170 + ], + "3min_transcript": "we're just talking about half of the path of this ball. So the time that it gets released, and it's going at kind of its maximum upward velocity, and it goes slower, and slower, slower, all the way until it's stationary for just a moment, and then it starts going down again. Now remember, the acceleration is constant downwards this entire time. So what is the final velocity if we just consider half of the time? Well, it's the time. It's 0. So it's going to be 0 minus our initial velocity, when it was taking off. That's our change in velocity. This is our change in velocity, is going to be equal to the acceleration of gravity, negative 9.8 meters per second squared-- or the acceleration due to gravity when an object is in free fall, to be technically correct-- times the time that we are going up. So times delta t up, which is the same thing. Delta t up is the same thing as our total time in the air divided by 2. And so we get negative the initial velocity is equal to-- this thing, when you divide it by 2, is going to be 4.9 meters per second squared-- we still have our negative out front-- times our delta t. Remember, this is our total time in the air, not just the time up. This is our total time in the air. And then we multiply both sides times a negative. We get our initial velocity is just going to be equal to 4.9 meters per second squared times the total time that we are in the air. Or you could say it's going to be 9.8 meters per second squared times half of the time that we're in the air. Either of those would get you the same calculation. that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air--" + }, + { + "Q": "At 3:14, Sal starts writing the formula for displacement. Why is he using Time Up as the time instead of Total Time? Can't we use Total Time instead?", + "A": "If he found the displacement over the total time, it would be zero. He is showing us how to find the peak height. The object reaches peak height when half of the total time has passed.", + "video_name": "P7LKEkcNibo", + "timestamps": [ + 194 + ], + "3min_transcript": "we're just talking about half of the path of this ball. So the time that it gets released, and it's going at kind of its maximum upward velocity, and it goes slower, and slower, slower, all the way until it's stationary for just a moment, and then it starts going down again. Now remember, the acceleration is constant downwards this entire time. So what is the final velocity if we just consider half of the time? Well, it's the time. It's 0. So it's going to be 0 minus our initial velocity, when it was taking off. That's our change in velocity. This is our change in velocity, is going to be equal to the acceleration of gravity, negative 9.8 meters per second squared-- or the acceleration due to gravity when an object is in free fall, to be technically correct-- times the time that we are going up. So times delta t up, which is the same thing. Delta t up is the same thing as our total time in the air divided by 2. And so we get negative the initial velocity is equal to-- this thing, when you divide it by 2, is going to be 4.9 meters per second squared-- we still have our negative out front-- times our delta t. Remember, this is our total time in the air, not just the time up. This is our total time in the air. And then we multiply both sides times a negative. We get our initial velocity is just going to be equal to 4.9 meters per second squared times the total time that we are in the air. Or you could say it's going to be 9.8 meters per second squared times half of the time that we're in the air. Either of those would get you the same calculation. that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air--" + }, + { + "Q": "at 6:35, will this equation calculate the total displacement or the half of displacement? because it is only using half of the time.", + "A": "I think it will calculate just how high the ball went in the air because if we calculated the whole displacement it would be 0 because it would land exactly from where we threw the ball,so it would be right at the place we started at, but i think we also calculated half the distance traveled", + "video_name": "P7LKEkcNibo", + "timestamps": [ + 395 + ], + "3min_transcript": "These are the same thing. And then we can simplify it. Our maximum displacement is equal to 4.9 meters per second squared times delta t squared, all of that over 4. And then we can just divide 4.9 divided by 4. 4.9 divided by 4 is-- let me just get the calculator out. I don't want to do that in my head, get this far and make a careless mistake. 4.9 divided by 4 is 1.225. So our maximum displacement is going to be 1.225 times our total time in the air squared, which is a pretty straightforward calculation. of how high we get displaced. Right when that ball is stationary, or has no net velocity, just for a moment, and starts decelerating downwards. So we can use that. If a ball is in the air for 5 seconds-- we can verify our computation from the last video-- our maximum displacement, 1.225, times 5 squared, which is 25, will give us 30.625. That's what we got in the last video. If the ball's in the air for, I don't know, 2.3 seconds-- so it's 1.225 times 2.3 squared-- then that means it went 6.48 meters in the air. So anyway, I just wanted to give you a simple expression that gives you the maximum displacement from the ground, assuming air resistance is negligible, as a function of the total time in the air. And it's a neat game to play." + }, + { + "Q": "At 3:05 when he multiplied the entire equation by a negative, shouldn't change of time be negative?", + "A": "That depends on your definition, you can have negative time if you wanted to take X amount of seconds of something. Like how can I cut down the time it takes me to get to work, if it takes me 60 minutes to get to work taking one route then I take another route and it takes off 20 minutes of my time its 60-20 = 40 minutes. depends on the context of the expression.", + "video_name": "P7LKEkcNibo", + "timestamps": [ + 185 + ], + "3min_transcript": "we're just talking about half of the path of this ball. So the time that it gets released, and it's going at kind of its maximum upward velocity, and it goes slower, and slower, slower, all the way until it's stationary for just a moment, and then it starts going down again. Now remember, the acceleration is constant downwards this entire time. So what is the final velocity if we just consider half of the time? Well, it's the time. It's 0. So it's going to be 0 minus our initial velocity, when it was taking off. That's our change in velocity. This is our change in velocity, is going to be equal to the acceleration of gravity, negative 9.8 meters per second squared-- or the acceleration due to gravity when an object is in free fall, to be technically correct-- times the time that we are going up. So times delta t up, which is the same thing. Delta t up is the same thing as our total time in the air divided by 2. And so we get negative the initial velocity is equal to-- this thing, when you divide it by 2, is going to be 4.9 meters per second squared-- we still have our negative out front-- times our delta t. Remember, this is our total time in the air, not just the time up. This is our total time in the air. And then we multiply both sides times a negative. We get our initial velocity is just going to be equal to 4.9 meters per second squared times the total time that we are in the air. Or you could say it's going to be 9.8 meters per second squared times half of the time that we're in the air. Either of those would get you the same calculation. that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air--" + }, + { + "Q": "At 4:13 Sal Khan says that the first part of the equation is the number of stars in the galaxy but my other source of information says that it is the rate that stars form. Which one is right?", + "A": "I think that the average number of stars that are formed is a better value. Stars are born, and stars die. Using the average of stars currently alive at a given time makes more sense than just using the formation rate.", + "video_name": "YYEgq1bweN4", + "timestamps": [ + 253 + ], + "3min_transcript": "number of detectable civilizations in the Milky Way, in our galaxy. and once again, there could be civilizations, looking back in the star field right over here this star right over here, maybe it has a planet that is in the right place that has liquid water and maybe there is intelligent life on that planet, but they might not be detectable, because, they aren't technologically advanced enough that they are using electromagnetic radiation or maybe they just figured out some other way to communicate or maybe they are beyond using electromagnetic radiation, you know, radio waves and all the rest to communicate so we will never be able to detect them. We are talking about civilizations like ours that are, to some degree, using technology not too different than our own. That's what we mean by 'detectable', I like to start with just the total number of stars in our solar system. So let's just start with, I will call it, N*, so this is the number of stars in our galaxy, and our best guess, I said is this's going to be 100-400 billion stars. We don't even know how many there are, some of them are undetectable, and the center of our galaxy is just a big blur to us we don't even know what's on the other side of that, e can't even see all the stars that are packed into the center, so this is our best guess, 100-400 billion stars. Now obviously, there is going to be subset of those stars that even have planets. So, let's multiply the times that subset; so lets multiply times the frequency of having a planet. If you are a star, this is the percent chance or the frequency or so, I'll write this way \"fractions that have planets\", So, if this is a hundred billion, let's say I am making a guess here and we are learning more about this everyday, there are all these discoveries of 'exoplanets'- planets outside our solar systems, maybe this is one fourth. Then you could say, well that means there are 100 billion times one fourth, that means there are 25 billion stars that have planets around them. But that's still not enough to go to civilizations. We also need to think about planets, there could be a planet like Jupiter, and we don't know how life as we know it can survive on a planet like Jupiter or Neptune or Mercury. It has to have planets that are good for sustaining life. Preferably have a rocky core, liquid water on the outside, that's what we think are the ingredients we need for life, maybe we are just not being creative enough, that's what we know as life is being." + }, + { + "Q": "4:52 Sal said planets that are good for sustaining life preferably have a rocky core. Do he mean molten core?", + "A": "Our core isn t molten, it is a solid ball of iron, nickel, and other assorted amounts of heavy metals. The surrounding area is liquid but the core isn t.", + "video_name": "YYEgq1bweN4", + "timestamps": [ + 292 + ], + "3min_transcript": "I like to start with just the total number of stars in our solar system. So let's just start with, I will call it, N*, so this is the number of stars in our galaxy, and our best guess, I said is this's going to be 100-400 billion stars. We don't even know how many there are, some of them are undetectable, and the center of our galaxy is just a big blur to us we don't even know what's on the other side of that, e can't even see all the stars that are packed into the center, so this is our best guess, 100-400 billion stars. Now obviously, there is going to be subset of those stars that even have planets. So, let's multiply the times that subset; so lets multiply times the frequency of having a planet. If you are a star, this is the percent chance or the frequency or so, I'll write this way \"fractions that have planets\", So, if this is a hundred billion, let's say I am making a guess here and we are learning more about this everyday, there are all these discoveries of 'exoplanets'- planets outside our solar systems, maybe this is one fourth. Then you could say, well that means there are 100 billion times one fourth, that means there are 25 billion stars that have planets around them. But that's still not enough to go to civilizations. We also need to think about planets, there could be a planet like Jupiter, and we don't know how life as we know it can survive on a planet like Jupiter or Neptune or Mercury. It has to have planets that are good for sustaining life. Preferably have a rocky core, liquid water on the outside, that's what we think are the ingredients we need for life, maybe we are just not being creative enough, that's what we know as life is being. So we don't necessarily know that they are going to have life, but they seem like they are just the right distance from the star, not too hot, not too cold. They have the right amount of gravity, water, all the other stuff, and we still don't know what this means, but this means average number, so given a number of solar systems with planets, what's the average number of planets capable of sustaining life, and once again, you don't know this answer. maybe it is 0.1, it's probably less than 1. Therefore any given solar system that has planets, the average number capable of sustaining life maybe its 0.1 maybe it's more than 1, I don't know. We don't know the exact answer here, but I will throw an answer in," + }, + { + "Q": "At 3:38, why did Sal say \" really do 'EXCITE' the electrons in the chlorophyll.....\"", + "A": "Because it means that they move faster. People often say excite rather than speed up.", + "video_name": "-rsYk4eCKnA", + "timestamps": [ + 218 + ], + "3min_transcript": "You have these fusion reactions in the sun 93 million miles away, and it's releasing these photons, and some small subset of those photons reach the surface of Earth. They make their way through clouds and whatever else. And then these plants and bacteria and algae are able to harness that somehow and turn them into sugars that we can then eat or maybe the cow eats them and we eat the cow if we're not vegetarians, and we can then use that for energy. Not that the cow is all carbohydrates, but this is essentially what is used as the fuel or the energy for all of the other important compounds that we eat. This is where we get all of our fuel. So this is fuel for animals. Or you know, if you eat a potato directly, you are directly getting your carbohydrates. But anyway, this is a very simple notion of photosynthesis, but it's not incorrect. I mean, if you had to know one thing about photosynthesis, this would be it. But let's delve a little bit deeper and try to get into the how this actually happens. I find it amazing that somehow photons of sunlight are used to create these sugar molecules or these carbohydrates. So let's delve a little bit deeper. So we can write the general equation for photosynthesis. Well, I've almost written it here. But I'll write it a little bit more scientifically specific. You start off with some carbon dioxide. You add to that some water, and you add to that-- instead of sunlight, I'm going to say photons because these are what really do excite the electrons in the chlorophyll that go down, and you'll see this process probably in this video, and we'll go in more detail in the next few videos. But that excited electron goes to a high energy state, and as it goes to a lower energy state, we're able to harness that energy to produce ATPs, and you'll see NADPHs, and those are used to produce carbohydrates. But we'll see that in a little bit. with these constituents, And then you end up with a carbohydrate. And a carbohydrate could be glucose, doesn't have to be glucose. So the general way we can write a carbohydrates is CH2O. And we'll put an n over here, that we could have n multiples of these, and normally, n will be at least three. In the case of glucose, n is 6. You have 6 carbons, 12 hydrogens and 6 oxygens. So this is a general term for carbohydrates, but you could have many multiples of that. You could have these long-chained carbohydrates, so you end up with a carbohydrate and then you end up with some oxygen. So this right here isn't so different than what I wrote up here in my first overview of how we always imagined photosynthesis in our heads. In order to make this equation balance-- let's see, I have n carbons so I need n carbons there. Let's see, I have two n hydrogens here." + }, + { + "Q": "At 2:51, why is Oxygen O2 and not just O. It is like this in my textbook as well and I do not know why.", + "A": "oxygen is actually called oxygen oxide. this is the oxygen we breath", + "video_name": "-rsYk4eCKnA", + "timestamps": [ + 171 + ], + "3min_transcript": "as a living species. One, we need carbohydrates or we need sugars in order to fuel our bodies. You saw that in the cellular respiration videos. We generate all of our ATP by performing cellular respiration on glucose, which is essentially a byproduct, or a broken down carbohydrate. It's the simplest one for us to process in cellular respiration. And the second hugely important part is getting the oxygen. Once again, we need to breathe oxygen in order for us to break down glucose, in order to respire, in order to perform cellar respiration. So these two things are key for life, especially for life that breathes oxygen. So this process, other than the fact that it's interesting, that there are organisms around us, mostly You have these fusion reactions in the sun 93 million miles away, and it's releasing these photons, and some small subset of those photons reach the surface of Earth. They make their way through clouds and whatever else. And then these plants and bacteria and algae are able to harness that somehow and turn them into sugars that we can then eat or maybe the cow eats them and we eat the cow if we're not vegetarians, and we can then use that for energy. Not that the cow is all carbohydrates, but this is essentially what is used as the fuel or the energy for all of the other important compounds that we eat. This is where we get all of our fuel. So this is fuel for animals. Or you know, if you eat a potato directly, you are directly getting your carbohydrates. But anyway, this is a very simple notion of photosynthesis, but it's not incorrect. I mean, if you had to know one thing about photosynthesis, this would be it. But let's delve a little bit deeper and try to get into the how this actually happens. I find it amazing that somehow photons of sunlight are used to create these sugar molecules or these carbohydrates. So let's delve a little bit deeper. So we can write the general equation for photosynthesis. Well, I've almost written it here. But I'll write it a little bit more scientifically specific. You start off with some carbon dioxide. You add to that some water, and you add to that-- instead of sunlight, I'm going to say photons because these are what really do excite the electrons in the chlorophyll that go down, and you'll see this process probably in this video, and we'll go in more detail in the next few videos. But that excited electron goes to a high energy state, and as it goes to a lower energy state, we're able to harness that energy to produce ATPs, and you'll see NADPHs, and those are used to produce carbohydrates. But we'll see that in a little bit." + }, + { + "Q": "@7:42 Shouldn't he do (1.00029*ans)/(1.33*8.1)??", + "A": "It would be the same thing. Apparantly his TI-85 can work by dividing a number, and then diving another number immediately after it, instead of putting parentheses around the items, but all in all, it would equate the same value.", + "video_name": "10LuSfZZa3E", + "timestamps": [ + 462 + ], + "3min_transcript": "And that's going to be equal to the index of refraction of water. So that's index of water is 1.33-- so let me do that in a different color. So that's going to be-- no, I wanted to do a different color. So that's going to be-- let me do it in this dark blue. So that's going to be 1.33 times sine of theta 2. And so, if we want to solve for sine of theta 2, you just divide both sides of this equation by 1.33. So let's do that. So I'll do it over here. So if you divide both sides by 1.33, we get 1.00029 times 7.92 over 8.1, and we're also going to divide by 1.33. So we're also dividing by 1.33. So let's figure what that is. So let's do that. Get the calculator out. So we have 1.00029 times 7.92. Well, actually I could even say times second answer, if we want this exact value. That was the last-- so I'm going to do that-- second answer. So that's the actual precise, not even rounding. And then we want to divide by 1.33, that's this right here. And then we want to divide by 8.1, and we get that. And that's going to be equal to the sine of theta 2. So that's going to be equal to the sine of theta 2. So let me write this down. So we have 0.735 is equal to the sine of theta 2. Now, we could take the inverse sine So we get theta 2 is equal to-- let's just take the inverse sine of this value. So I take the inverse sine of the value that we just had, so answer is just our last answer. And we have theta 2 being 47.3-- let's say rounded-- 47.34 degrees. So this is 47.34 degrees. So we were able to figure out what theta 2 is, 47.34 degrees. So now we just have to use a little bit of trigonometry to actually figure out this distance over here. Now what trig ratio involves-- so we know this angle. We want to figure out its opposite side, to that angle. And we know the adjacent side-- we know that this right here is 3. So what trig identity deals with opposite and adjacent? Well, tangent-- toa. Tangent is opposite over adjacent. So we know that the tangent of this angle right over here" + }, + { + "Q": "At 0:50 how'd do we have a large number of atoms?", + "A": "Even a speck of a substance contains a huge number of atoms.", + "video_name": "9REPnibO4IQ", + "timestamps": [ + 50 + ], + "3min_transcript": "SAL: In the last video we saw all sorts of different types of isotopes of atoms experiencing radioactive decay and turning into other atoms or releasing different types But the question is, when does an atom or nucleus decide to decay? Let's say I have a bunch of, let's say these are all atoms. I have a bunch of atoms here. And let's say we're talking about the type of decay where an atom turns into another atom. So your proton number is going to change. Your atomic number is going to change. So it could either be beta decay, which would release electrons from the neutrons and turn them into protons. Or maybe positron emission turning protons into neutrons. But that's not what's relevant here. Let's say we have a collection of atoms. And normally when we have any small amount of any element, we really have huge amounts of atoms of that element. And we've talked about moles and, you know, one gram of carbon-12-- I'm sorry, 12 grams-- 12 grams of carbon-12 One mole of carbon-12. And what is one mole of carbon-12? That's 6.02 times 10 to the 23rd carbon-12 atoms. This is a ginormous number. This is more than we can, than my head can really grasp around how large of a number this is. And this is only when we have 12 grams. 12 grams is not a large mass. For example, one kilogram is about two pounds. So this is about, what? I want to say [? 1/50 ?] of a pound if I'm doing [? it. ?] But this is not a lot of mass right here. And pounds is obviously force. You get the idea. On Earth, well anywhere, mass is invariant. This is not a tremendous amount. So with that said, let's go back to the question of how do we know if one of these guys are going to decay in some way. And maybe not carbon-12, maybe we're talking about carbon-14 or something. How do we know that they're going to decay? They all have some probability of the decaying. At any given moment, for a certain type of element or a certain type of isotope of an element, there's some probability that one of them will decay. That, you know, maybe this guy will decay this second. And then nothing happens for a long time, a long time, and all of a sudden two more guys decay. And so, like everything in chemistry, and a lot of what we're starting to deal with in physics and quantum mechanics, everything is probabilistic. I mean, maybe if we really got in detail on the configurations of the nucleus, maybe we could get a little bit better in terms of our probabilities, but we don't know what's going on inside of the nucleus, so all we can do is ascribe some probabilities to something reacting. Now you could say, OK, what's the probability of any given molecule reacting in one second? Or you could define it that way. But we're used to dealing with things on the macro level, on dealing with, you know, huge amounts of atoms. So what we do is we come up with terms that help us get our head around this. And one of those terms is the term half-life." + }, + { + "Q": "8:10 - Ok, what is happening here? One equation is being subtracted from another without any substitutions or anything?\n\nI thought I knew my algebra but this has me confused.", + "A": "When you have system of equation in two variables, then you want to cancel out some terms to solve for at least one variable. so he multiplied the equation and then subtracted the two in order to get rid of T2. You could substitute it as well. fell free to contact me for further clarification informjaka@gmail.com", + "video_name": "zwDJ1wVr7Is", + "timestamps": [ + 490 + ], + "3min_transcript": "Actually, let me do it right here. What's the sine of 30 degrees? The sine of 30 degrees is 1/2 so we get 1/2 T1 plus the sine of 60 degrees, which is square root of 3 over 2. Square root of 3 over 2 T2 is equal to 10. And then I don't like this, all these 2's and this 1/2 here. So let's multiply this whole equation by 2. So 2 times 1/2, that's 1. So you get T1 plus the square root of 3 T2 is equal to, 2 times 10 , is 20. Similarly, let's take this equation up here and let's multiply this equation by 2 and bring it down here. So this is the original one that we got. So if we multiply this whole thing by 2-- I'll do it in this color so that you know that it's a different equation. So if you multiply square root of 3 over 2 times 2-- I'm just doing this to get rid of the 2's in the denominator. So you get square root of 3 T1 minus T2 is equal to 0 because And let's see what we could do. What if we take this top equation because we want to start canceling out some terms. Let's take this top equation and let's multiply it by-- oh, I don't know. Let's multiply it by the square root of 3. So you get the square root of 3 T1. I'm taking this top equation multiplied by the square root of 3. This is just a system of equations that I'm solving for. And the square root of 3 times this right here. Square root of 3 times square root of 3 is 3. So plus 3 T2 is equal to 20 square root of 3. And now what I want to do is let's-- I know I'm doing a lot of equation manipulation here. Let's subtract this equation from this equation. So you can also view it as multiplying it by negative 1 and then adding the 2. So when you subtract this from this, these two terms cancel out because they're the same. And so then you're left with minus T2 from here. Minus this, minus 3 T2 is equal to 0 minus 20 square roots of 3. And so this becomes minus 4 T2 is equal to minus 20 square roots of 3. And then, divide both sides by minus 4 and you get T2 is equal to 5 square roots of 3 Newtons. So that's the tension in this wire. And now we can substitute and figure out T1. Let's use this formula right here because it looks suitably simple." + }, + { + "Q": "On 2:40, Sal says that oxygen and 2 hydrogen make up liquid water but i thought H2o makes up water who is right please help me!", + "A": "H2O = water. This means that a molecule of water consists of 2 H (hydrogen) atoms and 1 molecule of O (oxygen).", + "video_name": "Y3ATc9he254", + "timestamps": [ + 160 + ], + "3min_transcript": "but if this ice were more dense, the solid water were more dense than the liquid water, well then the ice would sink. The ice would sink and collect at the bottom. The ice would collect at the bottom right over here. And then the water over here would freeze again. It would freeze again. But if that were more dense, then it would sink and collect at the bottom. And you keep going with this process on and on and on, and eventually the entire body of water, the entire lake would freeze solid. Would freeze... It would freeze solid. Now you can imagine this wouldn't be that good for the animals that are living in the water. If you imagine some fish in here. Those fish would then freeze. And most living things, there are a few simple organisms that can survive being frozen. But most living things would just die. And so this would not be a good environment for animals to live in, or for even biology to take place. it does not follow this pattern. When we're talking about water, when we're talking about water, when we go to the liquid state, when we go from liquid water to solid ice, to solid ice, we actually get, we actually get less dense. So this right over here is less dense. This is why ice floats. This is more dense, more dense. And this is less dense. And to think about why that is, it all goes back to the hydrogen bonding. So we've seen in previous videos. So, I'm gonna do it all in one color. Oxygen, hydrogen, hydrogen. Let's say this is the liquid state that I'm drawing right over here. This is liquid water. Liquid water. So then you have oxygen, and you have oxygen and hydrogen and hydrogen. And you have oxygen and hydrogen and hydrogen. We've already talked multiple times negative charges at the side away from the hydrogens. Partial negative, partial negative because oxygen is so electro-negative. And you have partial positive charges on the hydrogen ends. Partial positive charges at the hydrogens and these partial negatives and partial positives attract each other and this is called hydrogen bonding. Now the liquid state, you have enough energy. The temperature is just really average kinetic energy that these molecules are able to bounce around and flow past each other. These hydrogen bonds get broken and get reformed over and over again. These things flow past each other and also they have enough energy to kind of push even closer to each other than even the hydrogen bonds would dictate. Sometimes they go closer, sometimes they're further, sometimes they're pushing around. But as we get-- As we get cooler, as we get cooler and we lose heat, then they don't have the kinetic energy to kind of-- They get closer and bump up against each other and move right and flow right past each other and they form a lattice structure" + }, + { + "Q": "at 7:20 what does aq stand for?", + "A": "aq stands for aqueous . that means dissolved in water or a solution in which water is the solvent. Hope this will help.............. ; )", + "video_name": "3ROWXs3jtQU", + "timestamps": [ + 440 + ], + "3min_transcript": "to exit the solution in either direction. And so common examples of these -- well, the one I alway think of, for me, the colloid is Jell-O. but gelatin is a colloid. but gelatin is a colloid. The gelatin molecules stay suspended in the -- the gelatin powder stays suspended in the water that you add to it, and you can leave it in the fridge forever and it just won't ever deposit out of it. Other examples, fog. Fog, you have water molecules inside of an air mixture. And then you have smoke. Fog and smoke, these are examples of aerosols. This is an aerosol where you have a liquid in the air. This is an aerosol where you have a solid in the air. Smoke just comes from little dark particles and they'll never come out of the air. They're small enough that they'll always just float around with the air. Now, if you get below 2 nanometers -- maybe I should eliminate my homogenized milk. If you get below 2 nanometers -- I'm trying to draw in black. If you're less than 2 nanometers, you're now in the realm of the solution. And although this is very interesting in the everyday world, a lot of things that we-- and this is a fun thing to think about in your house, or when you encounter things, is this a suspension? Well, first, you should just think is it homogeneous? And then think is it a suspension? Is it eventually going to not be in the state it's in and then I'll have to shake it? Is it a colloid where it will stay in this kind of nice, thick state in the case of Jell-O or fog or smoke where it will really just stay in the state Or is it a solution? And solution is probably the most important in chemistry. 99% of everything we'll talk about in chemistry involves solutions. And in general, it's an aqueous solution, when you stick something in water. So sometimes you'll see something like this. You'll see some compound x in a reaction and right next to it they'll write this aq. They mean that x is dissolved in water. It's a solute with water as the solvent. So actually, let me put that terminology here, just because I used it just now. So you have a solute. This is the thing that's usually whatever you have a smaller amount of, so thing dissolved. And then you have the solvent. This is often water or it's the thing that's in larger quantity. Or you can think of it as the thing that's all around or the thing that's doing the dissolving. Thing dissolving." + }, + { + "Q": "if 110.mg of fluorine-18 is shipped at 10:00 A.M,how many milligrams of the radioisotope are still active when the sample arrives at the radiology laboratory at 5:20 P.M", + "A": "It s half life is every 110 minutes, or 1 hour and fifty minutes. It will go through it s half life a total of ((7 1/3)x60)/110 times, or 4 times. You can divide 110 by 2 four times (55, 27.5, 13.75) and then 6.875 mg will be left.", + "video_name": "dnYyMHSSb8M", + "timestamps": [ + 600, + 320 + ], + "3min_transcript": "" + }, + { + "Q": "At 12:00, if the period is the time taken to complete one cycle then what does it mean the particle will move up and down in one second (frequency)?", + "A": "The period is 1 second. That s not the frequency. The frequency is 1 per second.", + "video_name": "tJW_a6JeXD8", + "timestamps": [ + 720 + ], + "3min_transcript": "vector, but I think you get the general idea. Your velocity-- what's the distance you travel in a period? Well, the distance you travel in a period is your wavelength after one up, down, back again. The wave pulse would have traveled exactly that far. That would be my wavelength. So I've traveled the distance of a wavelength, and how long did it take me to travel that distance? Well, it took me a period to travel that distance. So it's wavelength divided by period. Now I just said that 1 over the period is the same thing as the frequency. So I could rewrite this as wavelength. And actually, I should be clear here. The notation for wavelength tends to be the Greek letter lambda. wavelength over period. Which is the same thing as wavelength times 1 over my period. And we just said that 1 over the period, this is the same thing is your frequency. So velocity is equal to wavelength times your frequency. And if you know this, you can pretty much solve all of the basic problems that you might encounter in waves. So for example, if someone tells you that I have a velocity of-- I don't know-- 100 meters per second to the right, so in that direction-- velocity you have to give a direction-- and they were to tell you that my frequency is equal to-- let's say my frequency is 20 cycles per second, which is the same thing as 20 hertz. only able to observe this part of your wave, you'd only observe that part of my string. If we're talking about 20 hertz, then in 1 second, you would see this go up and down twenty times. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. In exactly one second, you would see this go up and down 20 times. That's what we mean by the frequency being 20 hertz, or 20 cycles per second. So, they gave you the velocity. They gave you the frequency. What's the wavelength here? So the wavelength, in this situation-- you would say the velocity-- your velocity is equal to your wavelength times your frequency. Divide both sides by 20. And actually, let me make sure I get the units right. So this is meters per second, is equal to lambda times 20 cycles per second." + }, + { + "Q": "at 2:01, we suppose that the electrons go straight from one plate to the other, right?", + "A": "The ions you mean. Positive and negative ions are the charge carriers here, not electrons. And no, here the system works differently, the positive ions move to the negative electrode and the negative ions move to the positive electrode. There they get discharged by gaining/losing electrons from/to the electrode. This is how electrolytes conduct electricity.", + "video_name": "uUhBEufepWk", + "timestamps": [ + 121 + ], + "3min_transcript": "- [Voiceover] Most solids offer some amount of resistance to the flow of current through them. This allows us to define things like the resistivity, or the conductivity, but the same is true for liquids. Consider this container full of a liquid. We can measure its resistivity. Now, if I took a battery, and I put one lead here, and one lead here, if there is a voltage and this liquid is able to conduct electricity, then this current should be able to flow through the intervening liquid over to the other side and then back up. Sometimes, this is done with AC current, otherwise, you might get electrolysis and then you get bubbles in here and that changes the liquid in some way. We want to measure the resistance and the resistivity of the liquid, not of some altered liquid. So, sometimes you use AC, but this is the general principal. Send in a voltage, a certain amount of current will flow. How can we use that, to determine the resistivity? Well, we know resistivity is equal to the resistance that we measure times the area divided by the length, I can imagine getting that. This length in here would just be this distance. There's my length, because my \"resistor,\" is this liquid in here. But, what's my area? So, this would be a bad experiment to do. If we want to measure the resistivity, what we really want, is something where we have a well defined area. Let me get rid of this. Imagine you had two plates. Take these two plates. You put them in the solution you want to measure the resistivity of, so, we put them in here. Stick them into there. They have a well defined area. We've got those. We can measure those if we want. We set them apart some known distance between them, L, and you hook them up to battery. So, take this one, hook it up to a known voltage, hook the other side up to the other plate, and if this solution, if this electrolytic solution in here can conduct electricity, current will flow from this side to the other side, and you can measure these quantities. You measure the area. You got that. How do we measure the resistance? Well, we know the voltage. We can have a known voltage of the battery up here, and you can stick ammeter in here to measure the current. If I stick an ammeter, ammeters measure the current. Now, I can just use Ohm's law, and I know that the resistance is just going to be the voltage divided by the current, and if plug all these values into here, I can get an experimental value for the resistivity of this liquid, sometimes it's called the electrolytic resistivity. Or, one over the electrolytic resistivity would be the electrolytic conductivity. So, this would be the electrolytic conductivity. So, this is an experimental way to do it. Honestly, you don't even have to go through all that much trouble. You can just take a solution. First, put a solution in here that has a known resistivity. That way, you can just do this: R equals Rho L over A." + }, + { + "Q": "what is glucose at 2:53?", + "A": "Glucose is a sugar produced by plants in photosynthesis, it has a molecular formula of C6H12O6. In photosynthesis it is the result of a plant using water (H2O) and Carbon Dioxide (CO2) to make Glucose (C6H12O6) and Oxygen gas (O2). 6H2O + 6CO2 = C6H12O6 + 6O2", + "video_name": "lzWUG4H5QBo", + "timestamps": [ + 173 + ], + "3min_transcript": "And the beanstalk was helping him physically, but also was actually providing him with very precious oxygen. In fact, if the beanstalk didn't do that, he may not have even made it. And we also, we don't know for sure, but we think that perhaps some of this story may have taken place during the day. And in fact, we know that sunlight is quite important for this process. And we think that this process, the name that we give it for the beanstalk anyway, is photosynthesis. And so what is really happening-- we're actually going to kind of write it out here-- between Jack and the beanstalk, and really between all plants and animals? What is this process between them? We know that on the one hand, you have beanstalks doing photosynthesis, and on the other hand, you have folks like Jack doing cellular respiration. And there's this really kind of interesting symbiosis. kind of relying on each other to really work. So you kind of need both of them to work well. And so let's actually take a moment to write out these processes that are happening between Jack and the beanstalk. So let's start with the process of photosynthesis, the beanstalk. So on the one hand, you've got what? You've got water because, of course, the beanstalk needs water, and you've got carbon dioxide. And I'm going to do carbon oxide in orange. So it's taking in water and carbon dioxide. And it's going to put out, it's going to actually take these ingredients if you want to think of it as kind of cooking, it's going to take these ingredients and it's going to put out. It's going to put out what? Oxygen and glucose. So I'll put glucose up top and oxygen down below. So these are the inputs and outputs of photosynthesis. You've got inputs. You've got glucose and oxygen going in. You're going to start seeing some serious similarities here. You've got glucose and oxygen going in. So Jack is taking in those two things. And he is again, of course, processing them. And he's putting out water and carbon dioxide. So this looks really, really nice, right? Looks perfect actually. Because everything is nice and balanced. And you can see how it makes perfect sense that, not only did Jack need the beanstalk, but actually it sounds like the beanstalk needed Jack, based on how I've drawn it. Now remember, none of this would even happen if there was no sunlight. So we actually need light energy. In fact, that's the whole purpose of this, right? Getting energy. So you have to have some light energy. I'm going to put a big plus sign, and I might even circle it because it's so important." + }, + { + "Q": "AT 5:45 you said when the volume goes down pressure increases then why does bigger balloon bursts easily as compared to small ballon?", + "A": "because bigger balloons have much more gas inside them which increases pressure significantly.", + "video_name": "WScwPIPqZa0", + "timestamps": [ + 345 + ], + "3min_transcript": "And just since we're dealing on the molecular scale, the number of particles can often be represented as moles. Remember, moles is just a number of particles. So we're saying that that pressure-- well, I'll say it's proportional, so it's equal to some constant, let's call that R. Because we've got to make all the units work out in the end. I mean temperature is in Kelvin but we eventually want to get back to joules. So let's just say it's equal to some constant, or it's proportional to temperature times the number of particles. And we can do that a bunch of ways. But let's think of that in moles. If I say there are 5 mole particles there, you know that's 5 times 6 times 10 to the 23 particles. So, this is the number of particles. This is the temperature. And this is just some constant. We gave these two examples. Obviously, it is dependent on the temperature; the faster each of these particles move, the higher pressure we'll have. It's also dependent on the number of particles, the more particles we have, the more pressure we'll have. What about the size of the container? The volume of the container. If we took this example, but we shrunk the container somehow, maybe by pressing on the outside. So if this container looked like this, but we still had the same four particles in it, with the same average kinetic energy, or the same temperature. So the number of particles stays the same, the temperature is the same, but the volume has gone down. Now, these guys are going to bump into the sides of the container more frequently and there's less area. So at any given moment, you have more force and less area. So when you have more force and less area, your pressure is going to go up. So when the volume went down, your pressure went up. proportional to volume. So let's think about that. Let's put that into our equation. We said that pressure is proportional-- and I'm just saying some proportionality constant, let's call that R, to the number of particles times the temperature, this gives us the total energy. And it's inversely proportional to the volume. And if we multiply both sides of this times the volume, we get the pressure times the volume is proportional to the number of particles times the temperature. So PV is equal to RnT. And just to switch this around a little bit, so it's in a form that you're more likely to see in your chemistry book, if we just switch the n and the R term. You get pressure times volume is equal to n, the number of" + }, + { + "Q": "at 6:53 why would hydrogen get one electron ?", + "A": "He explains it earlier on, at 1:14. It has plus one charge, which means that it lost an electron. Electrons have a -1 charge. Carbon is more electronegative than hydrogen, so it would take four electrons from the four hydrogen atoms. Each hydrogen atom gives an electron to the carbon, causing them to have 1 less electron, giving them 1+ for charge.", + "video_name": "OE0MMIyMTNU", + "timestamps": [ + 413 + ], + "3min_transcript": "to positive 4? Well, the way to increase your charge or your hypothetical charge is to lose electrons. Every time you lose an electron, this becomes less negative. And eventually, it'll become positive. So you have to lose 8 electrons. So I'll write plus 8 electrons right over here. You take these electrons, give it to this carbon. You're going to get to this side of this reaction. And the way I'm writing right now, these are called \"half reactions\" where I'm independently focusing on each of the elemental components of these reactions. So here, you have carbon. In this reaction, carbon-- in our hypothetical oxidation number world-- has lost 8 electrons. What do we call it when you are losing these hypothetical electrons? Well, we can remind ourselves OIL RIG-- oxidation is losing Or LEO-- losing electrons is oxidation-- the lion says GER-- gaining electrons is reduction. So it's clear right over here that carbon is being oxidized. It is losing electrons. Oxidation is losing electrons. Carbon is oxidized. Let's think about the hydrogen. On the left-hand side, you have 4 hydrogens that each have an oxidation number of plus 1. On the right-hand side, you have 4 hydrogens. We're writing it a slightly different way. We could write it like this-- 2H2's that each have an oxidation number of plus 1. The oxidation numbers for the hydrogens has not changed. The hydrogens has neither been oxidized nor reduced. On the left-hand side, you have 2O2's neutral oxidation number. And on the right-hand side, what do you have? You have 4 total oxygens. I'll combine these together. I'll just write this as 4 total oxygens. And what's each of their oxidation numbers? Well, we see it's a negative 2. So what happened to each of these 4 oxygens? I could have written 4O here instead of 2O2. Either way, I'm just really trying to account for the oxygens. Here I have 4 oxygens with a neutral oxidation number-- with an oxidation number of 0. And here I have 4 oxygens with a negative oxidation number. How do you go from 0 to negative? Each of them must have gained 2 electrons. If you have 4 oxygens, each of them gained 2 electrons," + }, + { + "Q": "At 6:15, how is carbon losing electrons therefore being oxidized if it has 8e- on the other side? Or is the 8e- just to balance the charges?", + "A": "when something is oxidized it is losing electrons so the carbon in the reactant side lost 8 electrons to become the carbon on the products side. Now where will you put these 8 electrons? As i ve said they are lost or removed from the reactant so they are in product side. Hope this answered your question :)", + "video_name": "OE0MMIyMTNU", + "timestamps": [ + 375 + ], + "3min_transcript": "with the hydrogens-- in our hypothetical ionic bond world, oxygen is a good bit more electronegative. So we're assuming it's going to take the electrons from the hydrogens. So each of the hydrogens loses an electron, giving it an oxidation number of 1. I could even write it like this, if you like. An oxidation number of positive 1. The oxygen has gained 2 electrons. So that gives it an oxidation number of negative 2. So now that we've done that, let's think about who is getting oxidized and who is being reduced. So let's first focus on the carbon. The carbon starts off at an oxidation number of negative 4. The reaction takes place. And then, carbon now has an oxidation number of positive 4. to positive 4? Well, the way to increase your charge or your hypothetical charge is to lose electrons. Every time you lose an electron, this becomes less negative. And eventually, it'll become positive. So you have to lose 8 electrons. So I'll write plus 8 electrons right over here. You take these electrons, give it to this carbon. You're going to get to this side of this reaction. And the way I'm writing right now, these are called \"half reactions\" where I'm independently focusing on each of the elemental components of these reactions. So here, you have carbon. In this reaction, carbon-- in our hypothetical oxidation number world-- has lost 8 electrons. What do we call it when you are losing these hypothetical electrons? Well, we can remind ourselves OIL RIG-- oxidation is losing Or LEO-- losing electrons is oxidation-- the lion says GER-- gaining electrons is reduction. So it's clear right over here that carbon is being oxidized. It is losing electrons. Oxidation is losing electrons. Carbon is oxidized. Let's think about the hydrogen. On the left-hand side, you have 4 hydrogens that each have an oxidation number of plus 1. On the right-hand side, you have 4 hydrogens. We're writing it a slightly different way. We could write it like this-- 2H2's that each have an oxidation number of plus 1. The oxidation numbers for the hydrogens has not changed. The hydrogens has neither been oxidized nor reduced." + }, + { + "Q": "at 5:37 whats the differnce between a half reaction and a normal reaction", + "A": "Take a normal reaction and now only consider the atoms being oxidized and the ones being reduced. Ignore all other atoms. Now split your equation into two parts: The reduction side: C(4-) ===> C(4+) + 8e- AND The oxidation side: 2 O2 + 8e- ===> 4 O(2-) The half equation is considering only the atoms being oxidized or reduced and where the electrons are flowing from and ending up. In this case carbon loses 8 electrons and oxygen gains 8 electrons. Hope that helps :)", + "video_name": "OE0MMIyMTNU", + "timestamps": [ + 337 + ], + "3min_transcript": "with the hydrogens-- in our hypothetical ionic bond world, oxygen is a good bit more electronegative. So we're assuming it's going to take the electrons from the hydrogens. So each of the hydrogens loses an electron, giving it an oxidation number of 1. I could even write it like this, if you like. An oxidation number of positive 1. The oxygen has gained 2 electrons. So that gives it an oxidation number of negative 2. So now that we've done that, let's think about who is getting oxidized and who is being reduced. So let's first focus on the carbon. The carbon starts off at an oxidation number of negative 4. The reaction takes place. And then, carbon now has an oxidation number of positive 4. to positive 4? Well, the way to increase your charge or your hypothetical charge is to lose electrons. Every time you lose an electron, this becomes less negative. And eventually, it'll become positive. So you have to lose 8 electrons. So I'll write plus 8 electrons right over here. You take these electrons, give it to this carbon. You're going to get to this side of this reaction. And the way I'm writing right now, these are called \"half reactions\" where I'm independently focusing on each of the elemental components of these reactions. So here, you have carbon. In this reaction, carbon-- in our hypothetical oxidation number world-- has lost 8 electrons. What do we call it when you are losing these hypothetical electrons? Well, we can remind ourselves OIL RIG-- oxidation is losing Or LEO-- losing electrons is oxidation-- the lion says GER-- gaining electrons is reduction. So it's clear right over here that carbon is being oxidized. It is losing electrons. Oxidation is losing electrons. Carbon is oxidized. Let's think about the hydrogen. On the left-hand side, you have 4 hydrogens that each have an oxidation number of plus 1. On the right-hand side, you have 4 hydrogens. We're writing it a slightly different way. We could write it like this-- 2H2's that each have an oxidation number of plus 1. The oxidation numbers for the hydrogens has not changed. The hydrogens has neither been oxidized nor reduced." + }, + { + "Q": "at 6:11 what does he mean by the lion goes ger? is it suppose to be a metaphor?", + "A": "LEO (loss electrons is oxidation) goes GER (gaining electrons is reduction) it s just a way to remember whether a half equation is oxidation or reduction", + "video_name": "OE0MMIyMTNU", + "timestamps": [ + 371 + ], + "3min_transcript": "with the hydrogens-- in our hypothetical ionic bond world, oxygen is a good bit more electronegative. So we're assuming it's going to take the electrons from the hydrogens. So each of the hydrogens loses an electron, giving it an oxidation number of 1. I could even write it like this, if you like. An oxidation number of positive 1. The oxygen has gained 2 electrons. So that gives it an oxidation number of negative 2. So now that we've done that, let's think about who is getting oxidized and who is being reduced. So let's first focus on the carbon. The carbon starts off at an oxidation number of negative 4. The reaction takes place. And then, carbon now has an oxidation number of positive 4. to positive 4? Well, the way to increase your charge or your hypothetical charge is to lose electrons. Every time you lose an electron, this becomes less negative. And eventually, it'll become positive. So you have to lose 8 electrons. So I'll write plus 8 electrons right over here. You take these electrons, give it to this carbon. You're going to get to this side of this reaction. And the way I'm writing right now, these are called \"half reactions\" where I'm independently focusing on each of the elemental components of these reactions. So here, you have carbon. In this reaction, carbon-- in our hypothetical oxidation number world-- has lost 8 electrons. What do we call it when you are losing these hypothetical electrons? Well, we can remind ourselves OIL RIG-- oxidation is losing Or LEO-- losing electrons is oxidation-- the lion says GER-- gaining electrons is reduction. So it's clear right over here that carbon is being oxidized. It is losing electrons. Oxidation is losing electrons. Carbon is oxidized. Let's think about the hydrogen. On the left-hand side, you have 4 hydrogens that each have an oxidation number of plus 1. On the right-hand side, you have 4 hydrogens. We're writing it a slightly different way. We could write it like this-- 2H2's that each have an oxidation number of plus 1. The oxidation numbers for the hydrogens has not changed. The hydrogens has neither been oxidized nor reduced." + }, + { + "Q": "At 9:21, how is it that oxygen is reduced only by carbon and not carbon & hydrogen? On the right side of the equation is oxygen not receiving electrons from both hydrogen and carbon?", + "A": "Hydrogen does not change its oxidation state, thus it cannot directly have been part of the oxidation/reduction process.", + "video_name": "OE0MMIyMTNU", + "timestamps": [ + 561 + ], + "3min_transcript": "On the left-hand side, you have 2O2's neutral oxidation number. And on the right-hand side, what do you have? You have 4 total oxygens. I'll combine these together. I'll just write this as 4 total oxygens. And what's each of their oxidation numbers? Well, we see it's a negative 2. So what happened to each of these 4 oxygens? I could have written 4O here instead of 2O2. Either way, I'm just really trying to account for the oxygens. Here I have 4 oxygens with a neutral oxidation number-- with an oxidation number of 0. And here I have 4 oxygens with a negative oxidation number. How do you go from 0 to negative? Each of them must have gained 2 electrons. If you have 4 oxygens, each of them gained 2 electrons, Actually, let me write it like this. Let me move this part. So cut and paste. Let me move it over to the right a little bit because what I want to show is the gaining of the electrons. So plus 8 electrons. So what happened to oxygen? Well, oxygen gained electrons. What is gaining electrons? Reduction is gaining-- RIG. GER-- gaining electrons is reduction. Oxygen has been reduced. Now, what oxidized carbon? Well, carbon lost electrons to the oxygen. So carbon oxidized by the oxygen, which is part of the motivation for calling it \"oxidation.\" And what reduced the oxygen? So oxygen reduced by the carbon. And this type of reaction-- where you have both oxidation and reduction taking place, and really they're two sides of the same coin. One thing is going to be oxidized if another thing is being reduced, and vice versa. We call these oxidation reduction reactions. Or sometimes \"redox\" for short. Take the \"red\" from \"reduction\" and the \"ox\" from \"oxidation,\" and you've got \"redox.\" This is a redox reaction. Something is being oxidized. Something else is being reduced. Not everything is being oxidized or reduced, and we can see that very clearly when we depict it in these half reactions. And one way to check that your half reactions actually makes sense is you can actually sum up the two sides." + }, + { + "Q": "At 8:26 when he is saying that we wish to minimize the formal charge, why is doing so with one double bond preferred to doing so with two double bonds? Don't you get +1 for Sulfur, -1 for the left Oxygen and 0 for the right oxygen for the one double bond, whereas you get 0 for Sulfur and both oxygens for two double bonds? Wouldn't that be 'minimized' for Sulfur?", + "A": "Did everyone see this pop up at 9:00 ? At 9:00, I drew one of the resonance structures for sulfur dioxide. While forming two double bonds would decrease the formal charge, both ways of representing sulfur dioxide are generally accepted.", + "video_name": "3RDytvJYehY", + "timestamps": [ + 506 + ], + "3min_transcript": "Here's two. And here's two more. So 18 minus 4 is 14 valence electrons left. All right, so we're going to start assigning some those leftover electrons to our terminal atoms, which are our oxygens. Oxygen's going to follow the octet rule. And so, therefore, each oxygen gets six more electrons to give each oxygen an octet. So we go ahead and do that. So I just represented 12 more valence electrons, right? Six on each oxygen. So 14 minus 12 is 2 valence electrons. So I have two valence electrons left over. And, remember, when you have leftover valence electrons, you go ahead and put them on your central atom. So we can go ahead and put those two valence electrons here on our central atom like that. Now, we're not quite done with our dot structure because sulfur doesn't have an octet. That's one way of thinking about it. You could also think about formal charge. Sulfur's formal charge is not minimized in this dot structure. So I could take, let's say-- let me go ahead and make these blue here. So I could take a lone pair of electrons from either oxygen. I'm just going to say that we take a lone pair of electrons from that oxygen and move them in here to form a double bond between the sulfur and the oxygen. So if I do that, now I have a double bond between the sulfur and the oxygen. The oxygen on the right now has only two lone pairs of electrons around it. The oxygen on the left still has three lone pairs of electrons around it like that. And the sulfur still has a lone pair of electrons here in the center. All right, so if we assign formal charges now-- let's go ahead and do that really fast. So we know that we have electrons in these bonds here. And so if we assign a formal charge to the oxygen on the left-- let's do that one first, all right, so this oxygen on the left here. All right, oxygen normally has six valence electrons And in our dot structure, we give to the sulfur. So you can see the oxygen is surrounded by seven valence electrons. This one's a little bit harder to see, so let me go ahead and mark it there. So 6 minus 7 gives us a formal charge of negative 1 on this oxygen. And when we do the same formal charge for the sulfur here, we can see that sulfur is surrounded by five valence electrons. Sulfur's in group six. So in the free atom, there's six. 6 minus 5 gives us a formal charge of plus 1. So we have a formal charge of plus 1 in the sulfur, formal charge of negative 1 on this oxygen. And even though we don't have a formal charge of zero on these atoms, this is about as good as we're going to get in terms of this representation of the molecule here. And another thing to think about is the fact that I didn't have to take the lone pair of electrons from this oxygen, right? I could have taken the lone pair of electrons from over here, on that oxygen. And that would just be another resonance structure of this. So I don't want to go too in detail" + }, + { + "Q": "At 3:00 he mentions that Boron doesn't need to follow the octet rule but it can. Does that mean that BF3 would have four resonance structures? The one that doesn't follow the octet rule AND the three that do?", + "A": "It cannot because of the number of total Valence Electrons. Boron can follow the octet rule, but the total number of Valence Electrons prevents this. Boron normally wants 6-8 electrons.", + "video_name": "3RDytvJYehY", + "timestamps": [ + 180 + ], + "3min_transcript": "We're going to put those leftover electrons on a terminal atoms, which are our fluorines in this case. Fluorine follows the octet rule. So each fluorine is now surrounded by two valence electrons. So each fluorine needs six more to have an octet of electrons. So I'll go ahead and put six more valence electrons on each of the three fluorines. And 6 times 3 is 18. So we just represented 18 more valence electrons. And so now we are all set. We've represented all 24 valence electrons in our dot structure. And some of you might think, well, boron is not following the octet rule here. And that is true. It's OK for boron not to follow the octet rule. And to think about why, let's assign a formal charge to our boron atom here. And so, remember, each covalent bond consists of two electrons. Let me go ahead and draw in those electrons in blue here. And when you're assigning formal charge, remember how to do that. You take the number of valence electrons in the free atoms, which is three. of electrons in the bonded atom. And so when you look at the bonds between boron and fluorine, one of those electrons goes to fluorine. And one of those electrons goes to boron here. So we can see that's the same for all three of these bonds. And so now boron is surrounded by three valence electrons in the bonded atom. 3 minus 3 gives us a formal charge of 0. So, remember, the goal is to minimize a formal charge when you're drawing your dot structures. And so this is a completely acceptable dot structure here, even though boron isn't following an octet. Now, boron can be surrounded by eight electrons. And so you'll even see some textbooks say, well, one of these lone pairs electrons on one of these fluorines could actually move in here to surround the boron with eight electrons, giving it an octet. And that's fine. That would give the boron a formal charge. And that might actually contribute to the overall structure of this molecule. But for us, for our purposes we're, just going to stick with this as being our dot structure here. So let me go ahead and redraw that. And I'm going to draw it in a slightly different way when So let me go ahead and put in our lone pairs of electrons here on the fluorine. And let's think about step two. We're going to count the number of electron clouds that surround the central atom here. So, remember, electron clouds are either the bonding electrons or non-bonding electrons-- the valence electrons in bonds or the lone pairs-- just regions of electron density that can repel each other. All right, so if I'm looking at my central atom, which is my boron, I can see that here are some electrons. All right, so that's an electron cloud. These electrons right here occupy an electron cloud, as well. And then I have another electron cloud here. So I have three electron clouds that are going to repel each other. And that allows us to predict the geometry of those electron clouds around that central atom. They're going to try to get as far away from each other as they possibly can. And it turns out, that happens when those electron" + }, + { + "Q": "At 12:04 it says bent, so is it 120 degrees or 104.5 degrees?", + "A": "In this instance, the bond angle will be 120 degrees. If there was a lone pair on the central atom, then the bond angle would be reduced to 104.5 degrees. See the video titled VSEPR for 4 electron clouds for an example of this", + "video_name": "3RDytvJYehY", + "timestamps": [ + 724 + ], + "3min_transcript": "So we have one electron cloud. Over here on the right, this double bond, we can consider it as an electron cloud. We're not worried about numbers of electrons, just regions of them. And then for the first time, we now have a lone pair of electrons, right? And this, we can also think about as occupying an electron cloud. So we have three electron clouds. And we saw in the previous example that when you have three electron clouds, the electron clouds are going to try to adopt a trigonal planar shape. So I could redraw this dot structure and attempt to show it in more of a trigonal planar shape here. So let's go ahead and show it looking like this-- once again, not worried about drawing informal charges here-- so something like this for the structure. Let me go ahead and put those electrons in our orbital here so we can see that electron cloud a little bit better. And so, once again, our electron clouds are in a trigonal planar geometry. to be approximately 120 degrees, right? So let me go ahead and put this in here. So approximately 120 degrees, it's probably slightly less than that. But that is what we would predict the geometry of the electron cloud to be. So let's go back up and look at our rules here, our steps for predicting the shapes of molecules. So we've done step three, right? We have predicted the geometry of the electron clouds around our central atom. And now we go on to step four here. We're going to ignore any lone pairs on our central atom when we predict the geometry of the overall molecule. And so that now pertains to our example here. We're going to ignore that lone pair of electrons on the sulfur. And we're going to ignore this lone pair of electrons when we're talking about the shape of the molecule. So even though the electron clouds have a trigonal planar geometry, we say that the shape of the molecule has a bent or angular shape. And so if you look at that, if you just look at the atoms, you'll see this kind of bent or angular shape here. And so that's what we say is the shape of the molecule. So you could say bent, or you could say angular here. All right, so that's two examples of molecules with three electron clouds. And remember, it's not just the number of electron clouds. You have to ignore lone pairs of electrons to predict the final shape of the molecule." + }, + { + "Q": "From 8:12 on, could someone please simply define and discriminate between transcription and translation?", + "A": "That is a great question! Transcription is when DNA transfers its genetic information to messenger RNA (mRNA), which carries this genetic information to a ribosome, where each codon (a set of three nucleotides) is translated to a specific protein (this part where the mRNA is in the ribosome is the translation step). Keep these in mind, and you ll be able to remember the difference! :) Note: Sal explains transcription around 8:06, and translation around 10:57.", + "video_name": "6gUY5NoX1Lk", + "timestamps": [ + 492 + ], + "3min_transcript": "it might define information for one gene, it could define a protein, this section right over here could be used to define another gene. And genes could be anywhere from several thousand base pairs long, all the way up into the millions. And as we'll see, the way that a gene is expressed, the way we get from the information for that section of DNA into a protein which is really how it's expressed, is through a related molecule to DNA, and that is RNA. Actually let me write this down. RNA. So RNA stands for ribonucleic acid. Ribonucleic acid, let me write that down. deoxyribonucleic acid, so the sugar backbone in RNA is a very similar molecule, well now it's got its oxy, it's not deoxyribonucleic acid, it's ribonucleic acid. The R, let me make it clear where the RNA come from, the R is right over there, then you have the nucleic, that's the n, and then it's a, acid. Same reason why we call the DNA nucleic acid. So you have this RNA. So what role does this play as we are trying to express the information in this DNA? especially if we're talking about cells with nucleii, the DNA sits there but that information has to for the most part get outside of the nucleus in order to be expressed. And one of the functions that RNA plays is to be that messenger, that messenger between a certain section of DNA and kind of what goes on outside of the nucleus, so that that can be translated into an actual protein. messenger RNA, is called transcription. Let me write that down. And what happens in transcription, let's go back to looking at one side of this DNA molecule. So let's say you have that right over there, let me copy and paste it. So there we go, actually I didn't wanna do that. I wanted the other side. So actually I think I'm on the wrong, let me go back here. And so let me copy and then let me paste. There we go. So let's say you have part of this DNA molecule, or you have 1/2 of it just like we did when we replicated it. But now we're not just trying to duplicate the DNA molecule," + }, + { + "Q": "Do genes make DNA molecules or DNA molecules make up a gene (around 6:04)?", + "A": "At around 6:29 , Sal mentions section of DNA . Genes are kind of like parts of DNA that code for the expression of specific traits. Hope this helps, and please let me know if I m wrong!", + "video_name": "6gUY5NoX1Lk", + "timestamps": [ + 364 + ], + "3min_transcript": "so thymine, adenine. Thymine, adenine. Guanine pairs with cytosine. And then cytosine pairs with guanine. So cytosine just like that. And so you can take half of each of this ladder, and then you can use it to construct the other half, and what you've essentially done is you've replicated the actual DNA. And this is actually a kind of conceptual level of how replication is done before a cell divides and replicates, and the entire cell duplicates itself. So that's replication. So the next thing you're probably thinking about, \"Okay, well it's nice to be able to replicate yourself \"but that's kind of useless if that information can't be \"used to define the organism in some way \"to express what's actually happening.\" And so let's think about how the genes in this DNA molecule are actually expressed. So I'll write this as \"expression\". because you hear sometimes the words DNA and chromosome and gene used somewhat interchangeably, and they are clearly related, but it's worth knowing what is what. So when you're talking about DNA you're talking literally about this molecule here that has this sugar phosphate base and it has the sequence of base pairs, it's got this double helix structure, and so this whole thing this could be a DNA molecule. Now when you have a DNA molecule and it's packaged together with other molecules and proteins and kind of given a broader structure, then you're talking about a chromosome. And when you're talking about a gene, you're talking about a section of DNA that's used to express a certain trait. Or actually used to code for a certain type of protein. So for example this could be, this whole thing could be a strand of DNA, but this part right over, let's say in orange I'll do it, it might define information for one gene, it could define a protein, this section right over here could be used to define another gene. And genes could be anywhere from several thousand base pairs long, all the way up into the millions. And as we'll see, the way that a gene is expressed, the way we get from the information for that section of DNA into a protein which is really how it's expressed, is through a related molecule to DNA, and that is RNA. Actually let me write this down. RNA. So RNA stands for ribonucleic acid. Ribonucleic acid, let me write that down." + }, + { + "Q": "At 10:48 does adenine pair with both uracil and thymine?", + "A": "In a DNA molecule adenine pairs with thymine while in a RNA molecule adenine pairs with uracil. There are no uracil in DNA and no thymine in RNA.", + "video_name": "6gUY5NoX1Lk", + "timestamps": [ + 648 + ], + "3min_transcript": "a corresponding mRNA molecule. At least for that section of, at least for that gene. So this might be part of a gene Actually whoops, let me make sure I'm using the right tool. This might be part of a gene that is this section of our DNA molecule right over there. And so transcription is a very similar conceputal idea, where we're now going to construct a strand of RNA and specifically mRNA 'cause it's going to take that information outside of the nucleus. And so it's very similar except for when we're talking about RNA, adenine, instead of pairing with thymine, is now going to pair with uracil. So let me write this down, so now you're gonna have adenine pairs not with thymine but uracil. DNA has uracil instead of the thymine. But you're still going to have cytosine and guanine pairing. So for the RNA and in this case the mRNA that's going to leave the nucleus A is going to pair with U, so uracil, that's the base we're talking about, let me write it down, uracil. Thymine is still going to pair with adenine, just like that. Guanine is gonna pair with cytosine, and cytosine is going to pair with guanine. And so when you do that, now these two characters can detach, and now you have a single strand of RNA and in this case messenger RNA, that has all the information on that section of DNA. And so now that thing can leave the nucleus, go attach to a ribosome, and we'll talk more about that in future videos exactly how that's happened, and then this code can be used to actually code for proteins. Now how does that happen? And that process is called translation. and turning it into an amino acid sequence. Proteins are made up of sequences of amino acids. So translation. So let's take our mRNA or this little section of our mRNA, and actually let me draw it like this. Now let's see, I have it is U A C, so it's gonna be U A C then U U then A C G okay? And then we have an A, let me make sure I change it to the right color. We have an A there, and then we have this U U A, C G, alright, now let me put a C right over there, I'm just taking this and I'm writing it horizontally. I have a C here, not a G, it's a C. And then finally I have a G. And of course it'll keep going on and on and on. And what happens is each sequence of three," + }, + { + "Q": "I was wondering why from 8:57 to 10:00, you multiplied -9.8 by two and took the square root of it all?", + "A": "It was - -30 = 1/2* -9.8 t^2 If we multiply both side by 2 then the half in the RHS will cancel out and we will be left with- -30*2 = -9.8 t^2 minus minus cancels out and we get- (30*2)/9.8 = t^2 t = sq. rt. of [60/9.8] t = 2.47 s", + "video_name": "jmSWImPs6fQ", + "timestamps": [ + 537, + 600 + ], + "3min_transcript": "that means there is no vertical velocity to start with, I'm gonna have to plug this initial velocity in the y direction as zero. So that's the trick. Don't fall for it now you know how to deal with it. So we want to solve for displacement in the x direction, but how many variables we know in the y direction? I mean we know all of this. But we can't use this to solve directly for the displacement in the x direction. We need to use this to solve for the time because the time is gonna be the same for the x direction and the y direction. So I find the time I can plug back in over to there, because think about it, the time it takes for this trip is gonna be the time it takes for this trip. It doesn't matter whether I call it the x direction or y direction, time is the same for both directions. In other words, the time it takes for this displacement of negative 30 is gonna be the time it takes for this displacement of whatever this is that we're gonna find. So let's solve for the time. Now, how will we do that? Think about it. We know the displacement, we know the acceleration, we know the initial velocity, and we know the time. But we don't know the final velocity we don't want to know it. So let's use a formula that doesn't involve the final velocity and that would look like this. So if we use delta y equals v initial in the y direction times time plus one half acceleration in the y direction times time squared. Alright, now we can plug in values. My displacement in the y direction is negative 30. My initial velocity in the y direction is zero. This is where it would happen, this is where the mistake would happen, people just really want to plug that five in over here. But don't do it, it's a trap. So, zero times t is just zero so that whole term is zero. Plus one half, the acceleration is negative 9.8 meters per second squared. And then times t squared, alright, now I can solve for t. I'm gonna solve for t, and then I'd have to take the square root of both sides because it's t squared, and what would I get? I'd have to multiply both sides by two. and then I have to divide both sides by negative 9.8 meters per second squared, equals, notice if you would have forgotten this negative up here for negative 30, you come down here, this would be a positive up top. You'd have a negative on the bottom. You'd have to plug this in, you'd have to try to take the square root of a negative number. Your calculator would have been all like, \"I don't know what that means,\" and you're gonna be like, \"Er, am I stuck?\" So you'd start coming back here probably and be like, \"Let's just make stuff positive and see if that works.\" It would work because look at these negatives canceled but it's best to just know what you're talking about in the first place. So be careful: plug in your negatives and things will work out alright. So if you solve this you get that the time it took is 2.47 seconds. It's actually a long time. It might seem like you're falling for a long time sometimes when you're like jumping off of a table, jumping off of a trampoline, but it's usually like a fraction of a second. This is actually a long time, two and a half seconds" + }, + { + "Q": "At 8:19 why is the 1/2 there?", + "A": "Because that s the formula", + "video_name": "jmSWImPs6fQ", + "timestamps": [ + 499 + ], + "3min_transcript": "was zero, there was no initial vertical velocity. This person was not launched vertically up or vertically down, this person was just launched straight horizontally, and so the initial velocity in the vertical direction is just zero. People don't like that. They're like \"hold on a minute.\" They're like, this person is gonna start gaining, alright, this person is gonna start gaining velocity right when they leave the cliff, this starts getting bigger and bigger and bigger in the downward direction. But that's after you leave the cliff. We're talking about right as you leave the cliff. That moment you left the cliff there was only horizontal velocity, which means you started with no initial vertical velocity. So this is the part people get confused by because this is not given to you explicitly in the problem. The problem won't say, \"Find the distance for a cliff diver \"assuming the initial velocity in the y direction was zero.\" Now, they're just gonna say, \"A cliff diver ran horizontally off of a cliff. \"Find this stuff.\" And you're just gonna have to know that okay, if I run off of a cliff horizontally that means there is no vertical velocity to start with, I'm gonna have to plug this initial velocity in the y direction as zero. So that's the trick. Don't fall for it now you know how to deal with it. So we want to solve for displacement in the x direction, but how many variables we know in the y direction? I mean we know all of this. But we can't use this to solve directly for the displacement in the x direction. We need to use this to solve for the time because the time is gonna be the same for the x direction and the y direction. So I find the time I can plug back in over to there, because think about it, the time it takes for this trip is gonna be the time it takes for this trip. It doesn't matter whether I call it the x direction or y direction, time is the same for both directions. In other words, the time it takes for this displacement of negative 30 is gonna be the time it takes for this displacement of whatever this is that we're gonna find. So let's solve for the time. Now, how will we do that? Think about it. We know the displacement, we know the acceleration, we know the initial velocity, and we know the time. But we don't know the final velocity we don't want to know it. So let's use a formula that doesn't involve the final velocity and that would look like this. So if we use delta y equals v initial in the y direction times time plus one half acceleration in the y direction times time squared. Alright, now we can plug in values. My displacement in the y direction is negative 30. My initial velocity in the y direction is zero. This is where it would happen, this is where the mistake would happen, people just really want to plug that five in over here. But don't do it, it's a trap. So, zero times t is just zero so that whole term is zero. Plus one half, the acceleration is negative 9.8 meters per second squared. And then times t squared, alright, now I can solve for t. I'm gonna solve for t, and then I'd have to take the square root of both sides because it's t squared, and what would I get? I'd have to multiply both sides by two." + }, + { + "Q": "at 5:53 do you not write 1,2 dimethylHept-2-ene because the Methyl at 1 is part of the chain?", + "A": "Yes, the methyl at 1 is part of the chain so you wouldn t write 1,1 dimethylhex-1-ene. Although that name would indicate the molecule we are talking about, it wouldn t be listed in databases that way. 1,2-dimethylhept-2-ene would actually be written 3-methyloct-3-ene.", + "video_name": "KWv5PaoHwPA", + "timestamps": [ + 353 + ], + "3min_transcript": "It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon. We have a double bond starting from the second carbon to the third carbon. So this thing right here, this double bond from the second carbon to the third carbon. So it's hept-2,3-ene-- sorry, not 2,3, 2-ene. You don't write both endpoints. If there was a three, then there would have been another double bond there. It's hept-2-ene. And then we have this methyl group here, which is also sitting on the second carbon. So this methyl group right there on the second carbon. So we would say 2-methyl-hept-2-ene. It's a hept-2-ene, that's all of this part over here, double bonds starting on the two if we're numbering from the right. And then the methyl group is also attached to that second carbon. So we have a cycle here, and once again the root is going to be the largest chain or the largest ring here. Our main ring is the largest one, and we have one, two, three, four, five, six, carbon. So we are dealing with hex as our root for kind of the core of our structure. It's in a cycle, so it's going to be cyclohex. So let me write that. So it's going to be cyclohex. But it has a double bond in it. So it's cyclohex ene, cyclohexene. Let me do this in a different color. So we have this double bond here, and that's why we know it is an ene. Now you're probably saying, Hey Sal, how come we didn't have to number where the ene is? So if you only have one double bond in a ring, it's assumed that one end point of the double bond is your 1-carbon." + }, + { + "Q": "At 7:49 shouldn't Sal start from the place where the substituent is the closest i.e\nthe opposite direction", + "A": "The numbering must include the alkenes as two consecutive numbers, although you write only the lower number in the name. In cyclohexene, the alkene carbons are automatically 1 and 2. C-1 becomes the one with the methyl group, and the numering must then go in the direction Sal used.", + "video_name": "KWv5PaoHwPA", + "timestamps": [ + 469 + ], + "3min_transcript": "So we have a cycle here, and once again the root is going to be the largest chain or the largest ring here. Our main ring is the largest one, and we have one, two, three, four, five, six, carbon. So we are dealing with hex as our root for kind of the core of our structure. It's in a cycle, so it's going to be cyclohex. So let me write that. So it's going to be cyclohex. But it has a double bond in it. So it's cyclohex ene, cyclohexene. Let me do this in a different color. So we have this double bond here, and that's why we know it is an ene. Now you're probably saying, Hey Sal, how come we didn't have to number where the ene is? So if you only have one double bond in a ring, it's assumed that one end point of the double bond is your 1-carbon. cyclohexene would look just like this. Just like that. You don't have to specify where it is. It's just, one of these are going to be the double bond. Now when you have other constituents on it, by definition or I guess the proper naming mechanism, is one of the endpoints of the double bond will be the 1-carbon, and if any of those endpoints have something else on it, that will definitely be the 1-carbon. So these both are kind of the candidates for the 1-carbon, but this point right here also has this methyl group. We will start numbering there, one, and then you want to number in the direction of the other side of the double bond. One, two, three, four, five, six. So we have three methyl groups, one on one. So these are the-- let me circle the methyl groups. That's a methyl group right there. That's a methyl group right there. That's just one carbon. So we have three methyl groups, so this is going to So it is 1, 4, 6. We have three methyl groups, so it's trimethyl cyclohexene. 1, 4, 6-trimethylcyclohexene. That's what that is, hopefully you found that useful." + }, + { + "Q": "4:07 Can it also be named hept-2,4-diene?", + "A": "Close. You have to keep the a of the ane ending when there is a multiplying prefix.. This makes it easier to pronounce. The name is hepta-2,4-diene.", + "video_name": "KWv5PaoHwPA", + "timestamps": [ + 247 + ], + "3min_transcript": "So this tells us that we have a seven carbon chain that has a double bond starting-- the ene tells us a double bond. Let me write that down. So this double bond right there, that's what the ene tells us. Double bond between two carbons, it's an alkene. The double bond starts-- if you start at this point-- the double bond starts at number two carbon, and then it will go to the number three carbon. Now you might be asking, well, what if I had more than one double bond here? So let me draw a quick example of that. Let's say I have something like, one, two, three, four, five, six, seven. So this is the same molecule again. One, two, three, four, five, six, seven. The way we drew it up here, it would look something like this. What if I had another double bond sitting right here? How would we specify this? One, two, three, four, five, six, seven. So we're still going to have a hept here. It's still going to be an alkene, so we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three, so it would be hept-2. And we also have another double bond starting from four and going to five, so hept-2,4-ene. That's what this molecule right there is. Sometimes, this is the-- I guess-- proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene, they'll write that as 2-heptene, probably because it's easier to say 2-heptene. And from this, you would be able to draw this thing over here, so it's giving you the same amount of information. Similarly over here, they might say 2,4-heptene. But this is the specific, this is the correct It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon." + }, + { + "Q": "At 6:00, Sal says it is \"2-methylhept-2ene\". Is it correct if I write \"2-methylheptene\" ?", + "A": "Sal is correct on this one. Since there are two important structures that have to be pointed out (the methyl group and the double bond), you need to indicate these with two separate numbers. You can t just write the 2 once and say that it applies for both of the groups. Another acceptable name for 2-methylhept-2ene is: 2-methyl-2-heptane", + "video_name": "KWv5PaoHwPA", + "timestamps": [ + 360 + ], + "3min_transcript": "It let's you know the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond, and I also have some side chains. Let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. Now we could go in either direction, it doesn't matter. Seven carbons or seven carbons. Let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence-- well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case, we'd want to start numbering at this end. It's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon. We have a double bond starting from the second carbon to the third carbon. So this thing right here, this double bond from the second carbon to the third carbon. So it's hept-2,3-ene-- sorry, not 2,3, 2-ene. You don't write both endpoints. If there was a three, then there would have been another double bond there. It's hept-2-ene. And then we have this methyl group here, which is also sitting on the second carbon. So this methyl group right there on the second carbon. So we would say 2-methyl-hept-2-ene. It's a hept-2-ene, that's all of this part over here, double bonds starting on the two if we're numbering from the right. And then the methyl group is also attached to that second carbon. So we have a cycle here, and once again the root is going to be the largest chain or the largest ring here. Our main ring is the largest one, and we have one, two, three, four, five, six, carbon. So we are dealing with hex as our root for kind of the core of our structure. It's in a cycle, so it's going to be cyclohex. So let me write that. So it's going to be cyclohex. But it has a double bond in it. So it's cyclohex ene, cyclohexene. Let me do this in a different color. So we have this double bond here, and that's why we know it is an ene. Now you're probably saying, Hey Sal, how come we didn't have to number where the ene is? So if you only have one double bond in a ring, it's assumed that one end point of the double bond is your 1-carbon." + }, + { + "Q": "At 4:20, Rishi says the LAIV can't cause you to get sick. If you had a really weak immune system, couldn't you actually get sick? Thanks in advance.", + "A": "There are some groups of people (those with asthma, etc.) who do not qualify for the live vaccination. These people should get the killed vaccine.", + "video_name": "wDghWK_Rr_E", + "timestamps": [ + 260 + ], + "3min_transcript": "get the full protection from the vaccine. So maybe he was in that two-week time span right after he got the vaccination, and maybe he got sick then. And lastly, we have to remember the flu vaccine has three strains in it. So maybe he had a different strain. Maybe it was just one or the other flu virus strains that was out there that year. So who knows exactly why he got sick that year that he says he got sick. But that doesn't change the fact that the flu vaccine is still going to reduce his chances of getting sick in the future. Now, one thing, I'm going to stick with this gentleman here in the bottom left. He might say, well, a different strain? What the heck? If that's the case, then what's the point? I often hear that. He says, what's the point if you could have a flu illness from some other strain? And when people talk about this, I have to remind them that it's not perfect. So it's not an exact science. And what you have to remember is that it goes back to the seat belt issue. It does reduce your chances of getting sick from the flu, Nothing in life is going to reduce your chances to zero, and there's still a possibility that you could get sick. And so our best offer is that you can reduce the chance of getting sick. Let's move on to this gentleman in the middle then. And so he says to me, well, you know what? I think that the flu vaccine made me sick. So this is a little different. He's saying that the flu vaccine itself made him sick. And this is something a lot of people think. They think, well, maybe I was fine, and maybe it was only after getting the flu vaccine that I got sick. And this can be a very frustrating thing to feel. I mean, that's horrible that you were feeling fine, and then a vaccine came along and made you feel awful. But a couple of facts have to come to mind. So one fact, for example, is that we know that the flu vaccine cannot give you flu. So we know flu vaccine does not cause flu. That's a fact. And the reason I can say that with certainty Remember, there's the TIV and then there's the LAIV. The TIV is a dead vaccine, meaning the virus inside of it is dead. And the LAIV-- that's the other option-- this one is alive, but it's very weak. And so usually with this one you might get a runny nose at best. And so, if he says, I think that I got sick from the flu vaccine, I would say, well, I don't think that's possible from this one, right? If you got the injected version, then that's not possible at all. And if you got the live version, then you may have gotten some symptoms, but we wouldn't call that the flu. And I would also say that if that weak vaccine made you sick, can you only imagine how you would have felt if you had the wild virus, the one that circulates and we're trying to protect you from? What's much more likely for this gentleman is that he probably got sick from maybe this copycat virus, one of the other viruses that are circulating, probably around the time that he had the flu vaccine." + }, + { + "Q": "at 1:30, why Na cant react with water?", + "A": "Na+ is a cation, having a positive charge. Needing a negative charge from water, it could potentially react with either H+ or OH-. H+ doesn t work since it has a positive charge. OH- does not work because the supposedly formed substance would be", + "video_name": "HwkEQfsJenk", + "timestamps": [ + 90 + ], + "3min_transcript": "- Let's say we have some hydrochloric acid, and a solution of sodium hydroxide. We know that hydrochloric acid is a strong acid so we can think about it as consisting of H+ and Cl-. Sodium hydroxide is a strong base, so in solution we're going to have sodium ions and hydroxide anions. Let's think about the products for this reaction. One product would be H+, and OH-. If you put H+ and OH- together, you form H20. So water is one of our products, and the other product would be what's leftover. We have Na+ and Cl-, so that gives us NaCl which is sodium chloride. This is an example of an acid base neutralization reaction where an acid reacts with the base to give you water and a salt. In this case, our salt is sodium chloride. Let's think about an aqueous solution of sodium chloride. and you dissolve your sodium chloride in water to make your solution. In solution, you're gonna have sodium cations and chloride anions. Let's think about what those would do with water. We know that the pH of water is seven. The pH of water is equal to seven. Sodium ions don't react with water so they're not going to affect the pH of our solution. You might think that the chloride anion could function as a weak base, and take a proton from water. However, that's not really gonna happen very well because the chloride anion, Cl-, this is the conjugate base to HCl. We know that HCl is a strong acid, and the stronger the acid, the weaker the conjugate base. With a very strong acid, we're gonna get a very weak conjugate base from water very well so the pH is unaffected. The pH of our solution of sodium chloride is equal to seven. When you have a salt that was formed from a strong acid and a strong base, so sodium chloride was formed from a strong acid and a strong base, these salts form neutral solutions. So your pH should be equal to seven. Let's compare this situation to the salt that's formed from a weak acid, and a strong base. Over here we have acetic acid which we know is a weak acid. Then we have sodium hydroxide which is our strong base. In solution, we would have Na+ and OH-. Hydroxide is going to take the acidic proton on acetic acid, and this is the acidic proton on acetic acid. Once again, H+ and OH- give us H20." + }, + { + "Q": "If DMSO is more likely to take the left structure as Jay at 05:50 says, is it probable for it, used as the solvent in a reaction, to take the right, polar aprotic structure and increase the nucleophilicity? And are the polar aprotic solvents generally bulky, so that they, due to steric hindrance, generally do not solvate the anion of the nucleophile?", + "A": "It never has either the structure on the right or the structure on the left. It always is a resonance hybrid of the two structures, with a partial negative charge on the O atom and a partial positive charge on the S. The major requirement for a polar aprotic solvent is that it have a large dipole moment and a large dielectric constant. It does not have to be bulky. Other examples of polar aprotic solvents are acetone, dimethylformamide, and acetonitrile.", + "video_name": "My5SpT9E37c", + "timestamps": [ + 350 + ], + "3min_transcript": "Next let's look at DMF. DMF is the short way of writing this one here. Again no hydrogen directly connected to an electronegative atom. This hydrogen is directly connected to this carbon and then this carbon would have three hydrogens on it and then this carbon would have three hydrogens on it. So DMF is a polar aprotic solvent. And finally let's look at this last one here. So the abbreviation would be HMPA. So let me write that down here. HMPA. Again no hydrogen is directly connected to an electronegative atom. Polar aprotic solvents favor an SN2 mechanism. So let's look at why. Down here I have an SN2 reaction. On the left we have this alkyl halide. Let's say we have sodium hydroxide. We could use DMSO as our solvent So we are gonna use DMSO. And we know in an SN2 mechanism the nucleophile attacks our alkyl halide at the same time our leaving group leaves. So our nucleophile is the hydroxide ion. It is going to attack this carbon and these electrons are gonna come off on to the bromide to form our bromide anion. So our OH replaces our bromine and we can see that over here in our product. In an SN2 mechanism we need a strong nucleophile to attack our alkyl halide. And DMSO is gonna help us increase the effectiveness of our nucleophile which is our hydroxide ion. So let's look at some pictures of how it helps us. So we have sodium hydroxide here. So first let's focus in on the sodium, our cation. So here is the sodium cation. DMSO is a good solvator of cations and that's because oxygen has a partial negative charge. The sulfur has a partial positive charge help to stabilize the positive charge on our sodium. So same thing over here. Partial negative, partial positive and again we are able to solvate our cation. So the fact that our polar aprotic solvent is a good solvator of a cation means we can separate this ion from our nucleophile. That increases the effectiveness of the hydroxide ion. The hydroxide ion itself is not solvated by a polar aprotic solvent. So you might think, okay well if the oxygen is partially negative and the sulfur is partially positive. The partially positive sulfur could interact with our negatively charged nucleophile. But remember we have these bulky methyl groups here. And because of steric hindrance that prevents our hydroxide ion from interacting with DMSO. So the hydroxide ion is all by itself which of course increases its effectiveness as a nucleophile." + }, + { + "Q": "At 9:38, what are dopamine and serotonin?", + "A": "Those are neurotransmitters, chemicals which let neurons communicate with each other.", + "video_name": "TyZODv-UqvU", + "timestamps": [ + 578 + ], + "3min_transcript": "The pump works against both of these conditions, collecting 3 positively charged sodium ions, and pushing them out into the positively charged sodium ion-rich environment. To get the energy to do this, the protein pump breaks up a molecule of ATP. ATP, adenosine tri-phopsphate, an adenosine molecule with 3 phosphate groups attached to it. So, when ATP connects with a protein pump, and enzyme breaks the covalent bond on one of those phophtaes in a bust of excitement and energy. The split releases enough energy to change the shape of the pump so that it opens outward and releases 3 sodium ions. This new shape also makes it a good fit for potassium ions that are outside the cell, so the pump lets 2 of those in. So, what you end up with is a nerve cell that is literally and metaphorically charged. It has all those sodium ions waiting outside with this intense desire to get inside of the cell, and when something triggers the nerve cell, it lets all of those in. And that gives the nerve cell a bunch of electric chemical energy, which you can then use to help you feel things There is still yet another way that stuff gets inside of cells, and this also requires energy, it's also a form of active transport. It's called vesicular transport, and the heavy lifting is done by vesicles, which are tiny sacs made of phohpholipids, just like the cell membrane. This kind of active transport is also called cytosis, from the Greek for cell action. When vesicles transport materials outside of a cell, it's called exocytosis, or outside cell action. A great example of this is going on in your brain right now, it's how your nerve cells release neurotransmitters. You've heard of neurotransmitters, they are very important in helping you feel different ways; they're like dopamine and serotonin. After neurotransmitters are synthesized and packaged into vesicles, they're transported until the vesicle reaches the membrane. When that happens, the 2 bilayers rearrange so that they fuse, and then the neurotransmitter spills out, and now I remember where I left my keys! Now, just play that process in reverse and you'll see how material gets inside the cell, and that's endocytosis. There are 3 different ways that this happens. My personal favorite is phagocytosis. that name itself means means devouring cell action. Check this out. So, this particle outside here is some kind of dangerous bacterium in your body, and this is a white blood cell. Chemical receptors on the blood cell membrane detect this punk invader and attach to it. Actually, reaching out around and engulfing it. Then the membrane forms a vesicle to carry it inside where it lays a total unholy beat down on it with enzymes and other cool weapons. Pinocytosis, or drinking action, is very similar to phagocytosis, except instead of surrounding whole particles, it just surrounds things that have already been dissolved. Here the membrane just folds in a little to form the beginning of a channel, and then pinches off to form a vesicle that holds the fluid. Most of your cells are doing this right now, because it's how our cells absorb nutrients. But what if a cell needs something that only occurs in very small concentrations? That's when cells use clusters of specialized receptor proteins in the membrane, that form a vesicle when receptors connect with the molecule" + }, + { + "Q": "At 8:26 in the breaking of the covalent bond, is the energy produced heat?", + "A": "this energy can be sound, heat, electromagnetic, nuclear and mechanical energy", + "video_name": "TyZODv-UqvU", + "timestamps": [ + 506 + ], + "3min_transcript": "If our bodies were America, ATP would be credit cards. It's such an important form of information currency, that we're going to do an entire separate episode about it, which will be here. Ha, I was going in the wrong direction, but it will be here when we've done it. But for now, here's what you need to know. When a cell requires active transport it basically has to pay a fee, in the form of ATP, to a transport protein. A particularly important kind of frickin' sweet transport protein is called the sodium potassium pump. Most cells have them, but they're especially vital to cells that need lots of energy, like muscle cells and brain cells. (piano instrumental) Ah! (laughs) Biolography, (laughing) it's my favorite part of the show. The sodium potassium pump was discovered in the 1950s by a Danish medical doctor named Jens Christian Skou, who was studying how anesthetics work on membranes. He noticed that there was a protein in cell membranes and the way he got to know this pump was by studying the nerves of crabs, because crab nerves are huge compared to humans' nerves, and are easier to dissect and observe. But crabs are still small, so he need a lot of them. He struck a deal with a local fisherman, and over the years, studied approximately 25,000 crabs, each of which he boiled to study their fresh nerve fibers. He published his findings on the sodium potassium pump in 1957, and in the meantime became known for the distinct odor that filled the halls of the department of physiology at the university where he worked. 40 years after making his discovery, Sku was awarded the Nobel Prize in chemistry. And here's what he taught us. Turns out, these pumps work against 2 gradients at the same time. 1 is the concentration gradient, and the other is the electric chemical gradient. That's the difference in electrical charge on either side of a cell's membrane. So, then our cells that Sku was studying, like nerve cells in your brain, typically have a negative charge inside relative to the outside. The pump works against both of these conditions, collecting 3 positively charged sodium ions, and pushing them out into the positively charged sodium ion-rich environment. To get the energy to do this, the protein pump breaks up a molecule of ATP. ATP, adenosine tri-phopsphate, an adenosine molecule with 3 phosphate groups attached to it. So, when ATP connects with a protein pump, and enzyme breaks the covalent bond on one of those phophtaes in a bust of excitement and energy. The split releases enough energy to change the shape of the pump so that it opens outward and releases 3 sodium ions. This new shape also makes it a good fit for potassium ions that are outside the cell, so the pump lets 2 of those in. So, what you end up with is a nerve cell that is literally and metaphorically charged. It has all those sodium ions waiting outside with this intense desire to get inside of the cell, and when something triggers the nerve cell, it lets all of those in. And that gives the nerve cell a bunch of electric chemical energy, which you can then use to help you feel things" + }, + { + "Q": "At 0:52 why are there 2 hydrogens connected to Carbon and", + "A": "In that structure you can only see 2 bonds to other carbons and there is not a charge indicated. With these structures you assume that the maximum possible bonds are there (4 for carbon), so there are two hydrogens bonded to that carbon.", + "video_name": "qpP8D7yQV50", + "timestamps": [ + 52 + ], + "3min_transcript": "- [Voiceover] I see a lot of mistakes when students draw resonance structures, and so I wanted to make a video on some of the more common mistakes that I've seen. So let's say we wanted to draw a resonance structure for this carbocation. Some students would take these electrons and move them down to here and say, all right, so on the right, now, I would have this, and this is my resonance structure. Let me highlight those electrons in blue here, so these electrons here move down to here. But this is incorrect, so let me write \"no\" here. So the resonance structure on the right, this is an incorrect resonance structure, why is this resonance structure not possible? Well, let's draw in the hydrogens on the carbons, and it will be much more obvious. So this carbon right here has one hydrogen on it, same with this carbon, and this carbon right here has two hydrogens on it, and the carbon with a plus one formal charge must have one hydrogen. So let's put in those hydrogens for the resonance structure on the right, and it should be obvious why this resonance structure Let's focus in on this carbon right here, the one I marked in red. How many bonds are there to that carbon? Well here's one bond, two, three, four, and five, that's five bonds to a carbon, that does not happen, you can't show carbon with five bonds, because that would be 10 electrons around this carbon, and carbon can never exceed an octet of electrons. Because of carbon's position on the periodic table, in the second period, there's four orbitals, and each orbital can hold a maximum of two electrons, which gives us four times two, which is eight. So carbon can never exceed an octet. There's another reason why this is wrong. If we go to this top carbon here, there's only three bonds around that carbon, so that carbon would have a plus one formal charge. So we added another formal charge, so this is incorrect, this is not a correct resonance structure. So what is the proper resonance structure to draw? Well, let's show that down here. You take your electrons, and you move them in the direction of the positive charge, of the positive one formal charge, and so let's show that. The electrons in, let me make them blue again, the electrons in blue move over to here, like that. And that moves the positive formal charge over to this carbon. If we draw in our hydrogens, it'll be clear why this is correct. So we put in a hydrogen here, we put in a hydrogen here, and we put in a hydrogen here. So let me draw in those three hydrogens on the right. Okay, now it's very obvious, let me point this out in red. It's obvious that this carbon here in red has a plus one formal charge, it has three bonds around it." + }, + { + "Q": "at 5:52 is magnesium actually originated in Magnesia?", + "A": "Yes it did from the district of magnesia. It is a place in greece.", + "video_name": "8Y4JSp5U82I", + "timestamps": [ + 352 + ], + "3min_transcript": "So that might be magnetic north. And magnetic south, I don't know exactly where that is. But it can kind of move around a little bit. It's not in the same place. So it's a little bit off the axis of the geographic north pole and the south pole. And this is another slightly confusing thing. Magnetic north is the geographic location, where the north pole of a magnet will point to. But that would actually be the south pole, if you viewed the Earth as a magnet. So if the Earth was a big magnet, you would actually view that as a south pole of the magnet. And the geographic south pole is the north pole of the magnet. You could read more about that on Wikipedia, I know it's a little bit confusing. But in general, when most people refer to magnetic north, or the north pole, they're talking about the geographic north area. And the south pole is the geographic south area. But the reason why I make this distinction is because we know when we deal with magnets, just like electricity, or shortly-- is that opposite poles attract. So if this side of my magnet is attracted to Earth's north pole then Earth's north pole-- or Earth's magnetic north-- actually must be the south pole of that magnet. And vice versa. The south pole of my magnet here is going to be attracted to Earth's magnetic south. Which is actually the north pole of the magnet we call Earth. Anyway, I'll take Earth out of the equation because it gets a little bit confusing. And we'll just stick to bars because that tends to be a little bit more consistent. Let me erase this. There you go. I'll erase my Magnesia. I wonder if the element magnesium was first discovered in Magnesia, as well. Probably. And I actually looked up Milk of Magnesia, which is a laxative. And it was not discovered in Magnesia, but it has So I guess its roots could be in Magnesia if magnesium was discovered in Magnesia. Anyway, enough about Magnesia. Back to the magnets. So if this is a magnet, and let me draw another magnet. Actually, let me erase all of this. All right. So let me draw two more magnets. We know from experimentation when we were all kids, this is the north pole, this is the south pole. That the north pole is going to be attracted to the south pole of another magnet. And that if I were to flip this magnet around, it would actually repel north-- two north facing magnets would repel each other. And so we have this notion, just like we had in electrostatics, that a magnet generates a field." + }, + { + "Q": "At 10:10, how would you use light years to calculate diameter?", + "A": "at center of universe we would use a light in any direction and then when it will reach the end of the universe that its radius and the twice is the diameter", + "video_name": "5FEjrStgcF8", + "timestamps": [ + 610 + ], + "3min_transcript": "and the Sun-- and you multiply that distance by 25,000. So if the Sun is right here, our nearest star will be in that same pixel. They'll actually be within-- you'd actually get a ton of stars within that one pixel, even though they're so far apart. And then this whole thing is 100,000 light years. It's 25,000 times the distance than the distance between the Sun and the nearest star. So we're talking about unimaginable, unfathomable distances, just for a galaxy. And now we're going to get our-- frankly, my brain is already well beyond anything that it can really process. At this point, it almost just becomes abstract thinking. It just becomes playing with numbers and mathematics. But to get a sense of the universe, itself, the observable universe-- and we have to be clear. Because we can only observe light Because that's how old the universe is. The observable universe is about 93 billion light years across. And the reason why it's larger than 13.7 billion is that the points in space that emitted light 13.7 billion years ago, those have been going away from us. So now they're on the order of 40 billion light years away. But this isn't about cosmology. This is just about scale and appreciating how huge the universe is. Just in the part of the universe that we can theoretically observe, you have to get-- and that we can observe, just because we're getting electromagnetic radiation from those parts of the universe-- you would have to multiply this number. So let me make this clear. 100,000 light years-- that's the diameter of the Milky Way. 1,000 would get you to 100 million light years. This is 100,000 times 1,000 is 100 million. You have to multiply by 1,000 again to get to 100 billion light years. And the universe, for all we know, might be much, much, much, much, larger. It might even be infinite. Who knows? But to get from just the diameter of the Milky Way to the observable universe, you have to multiply by a million. And already, this is an unfathomable distance. So in the whole scheme of things, not only are we pretty small, and not only are the things we build pretty small, and not only is our planet ultra small, and not only is our Sun ultra small, and our solar system ultra small, but our galaxy is really nothing compared to the vastness of the universe." + }, + { + "Q": "at 2:39-ish a pop up comes up stating the \"size\" of the earth, is that from pole to pole?", + "A": "It is the diameter of the Earth which is acually long than from pole to pole", + "video_name": "5FEjrStgcF8", + "timestamps": [ + 159 + ], + "3min_transcript": "you'll get about a 60-story skyscraper. Now, if you took that skyscraper and if you were to lie it down 10 times in a row, you'd get something of the length of the Golden Gate Bridge. And once again, I'm not giving you the exact numbers. It's not always going to be exactly 10. But we're now getting to about something that's a little on the order of a mile long. So the Golden Gate Bridge is actually longer than a mile. But if you go within the twin spans, it's roughly about It's actually a little longer than that. But that gives you a sense of a mile. Now, if you multiply that by 10, you get to the size of a large city. And this right here is a satellite photograph of San Francisco. This is the actual Golden Gate Bridge here. And when I copy and pasted this picture, I tried to make it roughly 10 miles by 10 miles just so you appreciate the scale. And what's interesting here-- and this picture's interesting. Because this is the first time we can relate to cities. it's starting to get larger than what we're used to processing on a daily basis. A bridge-- we've been on a bridge. We know what a bridge looks like. We know that a bridge is huge. But it doesn't feel like something that we can't comprehend. Already, a city is something that we can't comprehend all at once. We can drive across a city. We can look at satellite imagery. But if I were to show a human on this, it would be unbelievably, unbelievably small. You wouldn't actually be able to see it. It would be less than a pixel on this image. A house is less than a pixel on this image. But let's keep multiplying by 10. If you multiply by 10 again, you get to something roughly the size of the San Francisco Bay Area. This whole square over here is roughly that square right over there. Let's multiply by 10 again. So this square is about 100 miles by 100 miles. So this one would be about 1,000 miles by 1,000 miles. And now you're including a big part of the Western United States. You Nevada here. You have Arizona and New Mexico-- so a big chunk of a big continent we're already including. And frankly, this is beyond the scale that we're used to operating. We've seen maps, so maybe we're a little used to it. But if you ever had to walk across this type of distance, it would take you a while. To some degree, the fact that planes goes so fast-- almost unimaginably fast for us-- that it's made it feel like things like continents aren't as big. Because you can fly across them in five or six hours. But these are already huge, huge, huge distances. But once again, you take this square that's about 1,000 miles by 1,000 miles, and you multiply that by 10. And you get pretty close-- a little bit over-- the diameter of the Earth-- a little bit over the diameter of the Earth. But once again, we're on the Earth. We kind of relate to the Earth. If you look carefully at the horizon, you might see a little bit of a curvature, especially if you were to get into the plane. So even though this is, frankly, larger" + }, + { + "Q": "At 7:42, you mention how far apart stars are. What stars are the closest together, and how close are they?", + "A": "The two closest stars every found are the stars in HM Cancri system. This system is made up of two white dwarfs co-orbiting each other with a period of about five minutes. Each stars is moving at about 500 kilometers per second around each other and have an average separation of about 80,000 kilometers from each other (a separation of around 7 earths).", + "video_name": "5FEjrStgcF8", + "timestamps": [ + 462 + ], + "3min_transcript": "It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star-- so remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times-- and once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball-- in our part of the galaxy in a volume the size of the Earth-- so if you had a big volume the size of the Earth, if the stars were the sizes of basketballs, in our part of the galaxy, you would only have a handful of basketballs per that volume. So unbelievably sparse. Even though, when you look at the galaxy-- and this is just an artist's depiction of it-- it looks like something that has the spray of stars, and it looks reasonably dense, there is actually a huge amount of space that the great, great, great, great, great majority of the volume in the galaxy is just empty, empty space. There's no stars, no planets, no nothing. I mean, this is a huge jump that I'm talking about. And then if you really want to realize how large a galaxy, itself, can be, you take this distance between the Sun, or between our solar system and the nearest star-- and the Sun-- and you multiply that distance by 25,000. So if the Sun is right here, our nearest star will be in that same pixel. They'll actually be within-- you'd actually get a ton of stars within that one pixel, even though they're so far apart. And then this whole thing is 100,000 light years. It's 25,000 times the distance than the distance between the Sun and the nearest star. So we're talking about unimaginable, unfathomable distances, just for a galaxy. And now we're going to get our-- frankly, my brain is already well beyond anything that it can really process. At this point, it almost just becomes abstract thinking. It just becomes playing with numbers and mathematics. But to get a sense of the universe, itself, the observable universe-- and we have to be clear. Because we can only observe light" + }, + { + "Q": "At about 5:49, what does one AU mean? What is an AU?", + "A": "astronomical unit. Average distance between sun and earth.", + "video_name": "5FEjrStgcF8", + "timestamps": [ + 349 + ], + "3min_transcript": "can kind of relate to the Earth. Now you multiply the diameter of Earth times 10. And you get to the diameter of Jupiter. And so if you were to sit Earth right next to Jupiter-- obviously, they're nowhere near that close. That would destroy both of the planets. Actually, it would definitely destroy Earth. It would probably just be merged into Jupiter. So if you put Earth next to Jupiter, it would look something like that right over there. So I would say that Jupiter is definitely-- on this diagram that I'm drawing here-- is definitely the first thing that I have I can't comprehend. The Earth, itself, is so vastly huge. Jupiter is-- it's 10 times bigger in diameter. It's much larger in terms of mass, and volume, and all the rest. But just in terms of diameter, it is 10 times bigger. But let's keep going. 10 times Jupiter gets us to the sun. So if this is the Sun-- and if I were to draw Jupiter, it would look something like-- I'll do Jupiter in pink-- Jupiter would be around that big. And then the Earth would be around that big if you were to put them all next to each other. So the Sun, once again, is huge. Even though we see it almost every day, it is unimaginably huge. Even the Earth is unimaginably huge. And the Sun is 100 times more unimaginably bigger. Now we're going to start getting really, really, really wacky. You multiply the diameter of the Sun, which is already 100 times the diameter of the Earth-- you multiply that times 100. And that is the distance from the Earth to the Sun. So I've drawn the Sun here as a little pixel. And I didn't even draw the Earth as a pixel. Because a pixel would be way too large. It would have to be a hundredth of a pixel in order to draw the Earth properly. So this is a unbelievable distance It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star-- so remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times-- and once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball--" + }, + { + "Q": "I thought astrophysicists have claimed that for some time after the creation of the universe, the particles that made up the young universe weren't ionized for some length of time. So wouldn't that mean that they couldn't transmit energy (light) across distances and that light from the early star formation couldn't be visible from our home planet?\n\nI probably worded this wrong but oh well.\n\ntime: ~9:00", + "A": "Prior to about 380,000 years after the big bang the universe was too hot (above about 3000K) for electrons to be in atoms so the majority of matter had an electric charge so light could not travel freely. This is so early in the development of the universe that there was very little to see even if you could.", + "video_name": "5FEjrStgcF8", + "timestamps": [ + 540 + ], + "3min_transcript": "in our part of the galaxy in a volume the size of the Earth-- so if you had a big volume the size of the Earth, if the stars were the sizes of basketballs, in our part of the galaxy, you would only have a handful of basketballs per that volume. So unbelievably sparse. Even though, when you look at the galaxy-- and this is just an artist's depiction of it-- it looks like something that has the spray of stars, and it looks reasonably dense, there is actually a huge amount of space that the great, great, great, great, great majority of the volume in the galaxy is just empty, empty space. There's no stars, no planets, no nothing. I mean, this is a huge jump that I'm talking about. And then if you really want to realize how large a galaxy, itself, can be, you take this distance between the Sun, or between our solar system and the nearest star-- and the Sun-- and you multiply that distance by 25,000. So if the Sun is right here, our nearest star will be in that same pixel. They'll actually be within-- you'd actually get a ton of stars within that one pixel, even though they're so far apart. And then this whole thing is 100,000 light years. It's 25,000 times the distance than the distance between the Sun and the nearest star. So we're talking about unimaginable, unfathomable distances, just for a galaxy. And now we're going to get our-- frankly, my brain is already well beyond anything that it can really process. At this point, it almost just becomes abstract thinking. It just becomes playing with numbers and mathematics. But to get a sense of the universe, itself, the observable universe-- and we have to be clear. Because we can only observe light Because that's how old the universe is. The observable universe is about 93 billion light years across. And the reason why it's larger than 13.7 billion is that the points in space that emitted light 13.7 billion years ago, those have been going away from us. So now they're on the order of 40 billion light years away. But this isn't about cosmology. This is just about scale and appreciating how huge the universe is. Just in the part of the universe that we can theoretically observe, you have to get-- and that we can observe, just because we're getting electromagnetic radiation from those parts of the universe-- you would have to multiply this number. So let me make this clear. 100,000 light years-- that's the diameter of the Milky Way." + }, + { + "Q": "At 4:28, Rishi writes pCO2 and pO2. What does p stand for?", + "A": "partial pressure", + "video_name": "QP8ImP6NCk8", + "timestamps": [ + 268 + ], + "3min_transcript": "is kind of a friendly molecule. And so it also likes to sit where or bind where other oxygens have already bound. What are the two, then, major ways, based on this diagram, how I've drawn it. What are the two major ways that oxygen is going to be transported in the blood? One is hemoglobin binding oxygen. And we call that HbO2. Just Hb for hemoglobin, O2 for oxygen. And this molecule, or this enzyme, then, is not really called hemoglobin anymore. Technically, it's called oxyhemoglobin. That's the name for it. And another way that you can actually transport oxygen around is, that some of this oxygen-- I actually underlined it there-- is dissolved, O2 is dissolved in plasma. So some of the oxygen actually just gets dissolved right into the plasma. And that's how it gets moved around. Now, the majority, the vast majority of it is actually going to be moved through binding to hemoglobin. So just a little bit is dissolved in the plasma. So this red blood cell goes off to do its delivery. Let's say, it's delivering some oxygen out here. And there is a tissue cell. And, of course, it doesn't know where it's going to go that day. But it's going to go wherever its blood flow takes it. So let's say, it takes a pass over to this thigh cell in your, let's say, upper thigh. So this thigh cell has been making CO2. And remember, sometimes we think of CO2 as being made only when the muscle has been working. But you could be napping. You could be doing whatever. And this CO2 is still being made because cellular respiration is always happening. So this red blood cell has moved into the capillary right by this thigh cell. So you've got a situation like this where now some of the CO2 is going to diffuse into the red blood cell like that. And what happens once it gets down there? Just so you get a closer view of what's going on. And we're in the thigh and the two big conditions in the thigh that we have to keep in mind. One is that you have a high amount of CO2 or partial pressure of CO2. And this is dissolved in the blood. And the other is that you have a low amount of oxygen, not too much oxygen in those tissues. So let's focus on that second point. If there's not too much oxygen in the tissues, and we know that the hemoglobin is kind of constantly bumping into oxygen molecules and binding them. And they fall off and new ones bind. So it's kind of a dynamic process. Now, when there's not too much oxygen around, these oxygen molecules are going to fall off as they always do in a dynamic situation. Except new ones are not going to bind. Because there's so little oxygen around in the area, that less and less oxygen is free and is available to bump into hemoglobin and bind to it." + }, + { + "Q": "At 4:05, how do you pronounce 'crastholatian' acid metabolism??", + "A": "Crassulacean , after the Crassulaceae family of plants in which it was first discovered. This family includes orchids, bromeliads (of which pineapple is a member), purslane, agave, aloe and the succulent jade plant.", + "video_name": "xp6Zj24h8uA", + "timestamps": [ + 245 + ], + "3min_transcript": "These are plants that are in the desert. Because these stomata, these pores that are on the leaves, they let in air, but they can also let out water. I mean, if I'm in the rainforest, I don't care about that. But if I'm in the middle of the desert, I don't want to let out water vapour through my stomata. So the ideal situation is, I would want my stomata closed during the daytime. This is what I want. So I want, if I'm in the desert, let me make this clear. If I'm in the desert, I want stomata closed during the day. For obvious reasons. I don't want all my water to vaporise out of these holes in my leaves. But at the same time, the problem is photosynthesis can only occur during the daytime. And that includes the dark reactions. Remember, I've said multiple times, the dark reactions are But they both occur simultaneously, the light independent and light dependent, and only during the daytime. And if your stomata is closed, you need to perform photosynthesis, especially the Calvin cycle, you need CO2 (carbon dioxide). So how can you get around this? If I want to close my stomata during the day, but I need CO2 during the day, how can I solve this problem? And what desert plants, or many desert plants, have evolved to do, essentially does photosynthesis, but instead of fixing the carbon in outer cells and then pushing it in to inner cells and then performing the Calvin cycle, instead of outer and inner cells, they do it in the nighttime and in the daytime. So, in CAM plants - and these are called CAM plants because, It stands for crassulacean acid metabolism. And that's because it was first observed in that species of plant, the crassulacean plant. But these are just called, you could call it CAM photosynthesis or CAM plants. They're essentially a subset of C4 plants. But instead of performing C4 photosynthesis, kind of an outside cells and inside cells, they do it at the night time and the day. And what they do is, at night they keep their stomata open. And they perform, and they're able to fix - and everything occurs in the mesophyll cells and the CAM cells, in the CAM plants. So at night time, when they're not afraid of losing water - let's say this is a mesophyll cell right here - my stomata is open." + }, + { + "Q": "At 5:07 sal says that glass is a fluid,but how? in order to become one it should flow, right?", + "A": "Glass is made of really long and hard to move molecules. Even though they want to move, its just too slow. Glass is not really a liquid, but also it isn t a solid, properly saying. Its state is called a supercooled liquid, or an amorphous solid. That happens when a liquid is cooled beyond its freezing point, but it doesn t freeze.", + "video_name": "G4CgOF4ccXk", + "timestamps": [ + 307 + ], + "3min_transcript": "Let me draw the cylinder in a more vibrant color so you can figure out the volume. So it equals this side, the left side of the cylinder, the input area times the length of the cylinder. That's the velocity of the fluid times the time that we're measuring, times the input velocity times time. That's the amount of volume that came in. If that volume came into the pipe-- once again, we learned several videos ago that the definition of a liquid is a fluid that's incompressible. It's not like no fluid could come out of the pipe and all of the fluid just gets squeezed. The same volume of fluid would have to come out of the pipe, Whatever comes into the pipe has to equal the volume coming out of the pipe. One assumption we're assuming in this fraction of time that we're dealing with is also that there's no friction in this liquid or in this fluid, that it actually is not turbulent and it's not viscous. A viscous fluid is really just something that has a lot of friction with itself and that it won't just naturally move without any resistance. When something is not viscous and has no resistance with itself and moves really without any turbulence, that's called laminar flow. That's just a good word to know about and it's the opposite of viscous flow. Different things have different viscosities, and we'll probably do more on that. Like syrup or peanut butter has a very, very high viscosity. Even glass actually is a fluid with a I think there's some kinds of compounds and magnetic fields that you could create that have perfect laminar flow, but this is kind of a perfect situation. In these circumstances, the volume in, because the fluid can't be compressed, it's incompressible, has to equal the volume out. What's the volume out over that period of time? Similarly, we could draw this bigger cylinder-- that's the area out-- and after T seconds, how much water has come out? Whatever water was here at the beginning of our time period will have come out and we can imagine the cylinder here. What is the width of the cylinder? What's going to be the velocity that the liquid is coming out on the right-hand side?" + }, + { + "Q": "At 2:59 when the nitrogen gets sp2 then the carbon with the negative charge gets sp3?.", + "A": "In that particular resonance structure the carbon with the negative charge would be sp3, but the real structure has all the ring atoms sp2 hybridized with the \u00cf\u0080 electrons delocalized (spread over all of the ring atoms).", + "video_name": "wvVdgGTrh-o", + "timestamps": [ + 179 + ], + "3min_transcript": "which would imply an sp three hybridization state for pyrrole. We know that's not the case because pyrrole is an aromatic molecule, and sp three hybridized nitrogen would mean no p orbitals at that nitrogen, which would violate the first criterion for this compound to be aromatic. And so there must be some way to get that nitrogen to be sp two hybridized, and of course, we saw how to do that in the end to the last video. This lone pair of electrons on this nitrogen is actually not localized to this nitrogen. We can take this lone pair of electrons and move them in here so that lone pair of electrons can participate in resonance. If those lone pairs of electrons move into there to form a pi bond, that would kick these electrons off onto this carbon, so the resonance structure will have nitrogen with a pi bond here now and a lone pair of electrons on this carbon, which would give this carbon a negative one formal charge. We still have a pi bond over here like that, Now when we analyze the hybridization state of this nitrogen, we can see that once again, we're going for sigma bonds, so one sigma bond, two sigma bonds, three sigma bonds, so three sigma bonds, this time no lone pairs of electrons because that lone pair of electrons is now de-localized in resonance, and so three plus zero is of course three, meaning that this nitrogen is now sp two hybridized. Since that nitrogen is sp two hybridized, it has a free p orbital, so we can go ahead and draw the p orbital on that nitrogen. You could think about in terms of dot structure, these two electrons over here, these two electrons are actually de-localized and participate in resonance, so that lone pair of electrons you could think about as occupying a p orbital here and they're actually de-localized. We have all these pi electrons de-localized throughout our ring, and so let's go ahead and check the criteria Pyrrole does contain a ring of continuously overlapping p orbitals, and it does have four n plus two pi electrons in that ring, so let's go ahead and highlight those. We had these pi electrons, so that's two, these pi electrons, so that's four, and then these pi electrons here in magenta are actually de-localized in the ring, so that gives us six pi electrons. So if n is equal to one, four times one plus two gives me six pi electrons. Pyrrole has six pi electrons and also has a ring of continuously overlapping p orbitals, so we can say that it is aromatic. Let's go ahead and look at another molecule here so similar to it. This is imidazole. For imidazole, once again, we have the same sort of situation that we had for pyrrole with this nitrogen right here, so at first, it looks like that nitrogen might be sp three hybridized, but we can draw a resonance structure for it." + }, + { + "Q": "At 2:40, I don't really understand where the numerator came from.", + "A": "To figure total parallel resistance, one adds the reciprocal of all the resistors to get the reciprocal of the final resistance of the circuit. (i.e. one adds the conductances). The 16 is the least common denominator of the various fractions he is adding. He needs to convert each fraction to 16ths in order to add the numerators. For example, the 1/4 becomes 4/16, and so on.", + "video_name": "3NcIK0s3IwU", + "timestamps": [ + 160 + ], + "3min_transcript": "that is 3 ohms. And let's say I have a resistor here. Let's just make it simple: 1 ohm. And just to make the numbers reasonably easy-- I am doing this on the fly now-- that's the positive terminal, negative terminal. Let's say that the voltage difference is 20 volts. So what I want us to do is, figure out what is the current flowing through the wire at that point? Obviously, that's going to be different than the current at that point, that point, that point, that point, all of these different points, but it's going to be the same as the current flowing at this point. So the easiest way to do this is try to figure out the equivalent resistance. Because once we know the equivalent resistance of this big hairball, then we can just use Ohm's law and be done. So first of all, let's just start at, I could argue, the simplest part. Let's see if we could figure out the equivalent resistance of these four resistors in parallel. Well, we know that that resistance is going to be equal to 1/4 plus 1/8 plus 1/16 plus 1/16. So that resistance-- and now it's just adding fractions-- over 16. 1/4 is 4/16 plus 2/16 plus 1 plus 1, so 1/R is equal to 4 plus 2 is equal to 8/16-- the numbers are working out-- is equal to 1/2, so that equivalent resistance is 2. So that, quickly, we just said, well, all of these resistors combined is equal to 2 ohms. So let me erase that Simplify it. So that whole thing could now be simplified as 2 ohms. I lost some wire here. I want to make sure that circuit can still flow. So that easily, I turned that big, hairy mess into something that is a lot less hairy. Well, what is the equivalent resistance of this resistor and this resistor? Well, they're in series, and series resistors, they just add up together, right? So the combined resistance of this 2-ohm resistor and this 1-ohm resistor is just a 3-ohm resistor. So let's erase and simplify. So then we get that combined resistor, right?" + }, + { + "Q": "at 3:39 he begins to add up the resistance of the two resistors in series using ohms law by saying R1+R2 which gave him 3. my question is why is R3 on the vertical right hand side line not being added when it is also a resistor in series", + "A": "R3 is not in series with (R1+R2), but rather in series with [(R1+R2) in parallel with R0]. So the total resistance would be (R1+R2)*R0/(R1+R2+R0) + R3 not just R1+R2+R3", + "video_name": "3NcIK0s3IwU", + "timestamps": [ + 219 + ], + "3min_transcript": "So the easiest way to do this is try to figure out the equivalent resistance. Because once we know the equivalent resistance of this big hairball, then we can just use Ohm's law and be done. So first of all, let's just start at, I could argue, the simplest part. Let's see if we could figure out the equivalent resistance of these four resistors in parallel. Well, we know that that resistance is going to be equal to 1/4 plus 1/8 plus 1/16 plus 1/16. So that resistance-- and now it's just adding fractions-- over 16. 1/4 is 4/16 plus 2/16 plus 1 plus 1, so 1/R is equal to 4 plus 2 is equal to 8/16-- the numbers are working out-- is equal to 1/2, so that equivalent resistance is 2. So that, quickly, we just said, well, all of these resistors combined is equal to 2 ohms. So let me erase that Simplify it. So that whole thing could now be simplified as 2 ohms. I lost some wire here. I want to make sure that circuit can still flow. So that easily, I turned that big, hairy mess into something that is a lot less hairy. Well, what is the equivalent resistance of this resistor and this resistor? Well, they're in series, and series resistors, they just add up together, right? So the combined resistance of this 2-ohm resistor and this 1-ohm resistor is just a 3-ohm resistor. So let's erase and simplify. So then we get that combined resistor, right? then we had a 1-ohm. So we had a 2-ohm and a 1-ohm in series, so those simplify to 3 ohms. Well, now this is getting really simple. So what do these two resistors simplify to? Well, 1 over their combined resistance is equal to 1/3 plus 1/3. 2/3. 1/R is equal to 2/3, so R is equal to 3/2, or we could say 1.5, right? So let's erase that and simplify our drawing. So this whole mess, the 3-ohm resistor in parallel with the other 3-ohm resistor is equal to one resistor with a 1.5 resistance." + }, + { + "Q": "at 6:55 the molecule which contains 2 lone pairs and 2 bond pairs\nnitrogen has valency of only 5\nin that 2 lone pair of electrons are there so there is only one e to be shared then how can it form two bonds\nis the one bond a dative bond?", + "A": "Pretend that molecule had an -NH2 group on the end. Then pretend a very strong base came along and removed one of those hydrogens, and the electrons that were in that N-H bond stayed on nitrogen. What you get is nitrogen having 8 electrons still but only 4 of them are in bonds, the other 4 are in the 2 lone pairs.", + "video_name": "5-MM39VCwc0", + "timestamps": [ + 415 + ], + "3min_transcript": "for the formal charge. Alright, let me redraw that. So we have our nitrogen with four bonds to hydrogen and then nitrogen has a plus one formal charge. You should recognize this as being the ammonium ion from general chemistry. So this has a formal charge of plus one, so we have another pattern to think about here. So let's draw that in. We have one, two, three, four bonds and zero lone pairs of electrons. So when nitrogen has four bonds, four bonds and zero lone pairs, zero lone pairs of electrons, we've already seen the formal charge be equal to plus one. So let's look at some examples where nitrogen has a formal charge of plus one. So the example on the left, we can see there are four bonds and there are no lone pairs on that nitrogen, so that's a plus one formal charge. Here's one bond, two bonds, three bonds, and four bonds and no lone pairs, so a plus one formal charge on the nitrogen. Alright, finally, one more nitrogen to assign a formal charge to. So let's look at this one. Let's draw in the electrons in the bond. So here's two electrons and here's two electrons. What is the formal charge on nitrogen? Formal charge is equal to number of valence electrons nitrogen is supposed to have, which we know is five, and from that we subtract the number of valance electrons nitrogen actually has in our dot structure. So again we go over to here and we look at this bond and we give one electron to nitrogen and one electron to the other atom. And over here we give one electron to nitrogen and one electron to the other atom. And now we have two lone pairs of electrons on the nitrogen. So how many is that total? this would be one, two, three, four, five, and six. So five minus six gives us negative one. So a formal charge of negative one. Let me go ahead and redraw that. So I could draw it out here. So nitrogen with two lone pairs of electrons we just found has a formal charge of negative one. If I wanted to leave off the lone pairs of electrons I could do that, I could just write NH here and put a negative one formal charge, and because of this pattern, you should know there are two lone pairs of electrons on that nitrogen. Let me just clarify the pattern here. The pattern for a formal charge of negative one on nitrogen would be two bonds, here are the two bonds, and two lone pairs of electrons. So when nitrogen has two bonds and two lone pairs of electrons, nitrogen should have a formal charge of negative one. Let's look at some examples of that." + }, + { + "Q": "According to wikipedia: A parsec is the distance from the SUN (not from the EARTH) to an astronomical object that has a parallax angle of one arcsecond.\n\nSal said in 7:20, \"It's the distance that an object needs to be from EARTH in order for it to have a parallax angle of one arc second.\n\nYou could argue that it's approximately the same because 1 AU is miniscule compared to distance of a star from the earth. But still, the concept is different.", + "A": "It s not just approximately the same, it s the same to an almost immeasurable degree. We are looking at objects that are light YEARS away. The distance between the sun and the earth is 8 light MINUTES. 8 minutes / 1 year = .0015%. The distance measurements themselves are nowhere near that precise.", + "video_name": "6zV3JEjLoyE", + "timestamps": [ + 440 + ], + "3min_transcript": "And we just use a little bit of trigonometry. The tangent of this angle, the tangent of 90 minus 1/3600, is going to be this distance in astronomical units divided Well, you divide anything by 1, it's just going to be that distance. So that's the distance right over there. So we get our calculator out. And we want to find the tangent of 90 minus 1 divided by 3,600. And we will get our distance in astronomical units, 206,264. We're going to say 265. So this distance over here is going to be equal to 206,265-- I'm just And if we want to convert that into light years, we just divide. So there are 63,115 astronomical units per light year. Let me actually write it down. I don't want to confuse you with the unit cancellation. So we're dealing with 206,265 astronomical units. And we want to multiply that times 1 light year is equal to 63,115 astronomical units. And we want this in the numerator and the denominator to cancel out. And so if you divide 206,265, this number up here, by 63,115, the number of astronomical units in a light year-- let me delete that right over there-- we get 3.2. Well, the way the math worked out here, it rounded to 3.27 light years. So I should just show it's approximate right over there. But that's where the parsec comes from. So hopefully now you just realize it is just a distance. But even more, you actually understand where it comes from. It's the distance that an object needs to be from Earth in order for it to have a parallax angle of one arc second. And that's where the word came from. Parallax arc second." + }, + { + "Q": "Hi all, I have a question. At 1:47, what does Sal mean when he says, \"Distance traveled as a function of theta.\"?? Thank you.", + "A": "at 1:47 sal means that if theta is changed, the distance changes because distance depends upon theta", + "video_name": "-h_x8TwC1ik", + "timestamps": [ + 107 + ], + "3min_transcript": "We now know how long the object is going to be in the air, so we're ready to figure out how far it's going to travel. So we can just go back to kind of the core formula in all of really kinematics, all of kind of projectile motion or mechanical physical problems, and that's distance is equal to rate times time. Now, we're talking about the horizontal distance. So our distance is going to be equal to-- what's our rate in the horizontal direction? We care about horizontal distance traveled, so our rate needs to be the horizontal component of the velocity, or the magnitude of the horizontal component of the velocity. And we figured that out in the first video. That is s cosine of theta. So let's write that down right here. So our rate is s cosine of theta. And how long will we be traveling at Well, we'll be going at that speed as long as we are in the air. So how long are we in the air? Well, we figured that out in the last video. We're going to be in the air this long-- 2 s sine of theta divided by g. So the time is going to be 2 s sine of theta over g. So the total distance we're going to travel, pretty straightforward, rate times time. It's just the product of these two things. And we could put all of the constants out front, so it's a little bit clearer that it's a function of theta. So we can write that the distance traveled-- let me do that same green. The distance traveled as a function of theta is equal to-- I'll do that in this blue. This s times 2s divided by g is-- I'll do it in a neutral color actually. So 2s squared over g times cosine of theta times sine of theta. So now we have a general function. You give me an angle that I'm going to shoot something off at and you give me the magnitude of its velocity, and you give me the acceleration of gravity. I guess if we were on some other planet, who knows? And I will tell you exactly what the horizontal distance is." + }, + { + "Q": "At 0:30 what is a LCD?", + "A": "Liquid Crystal Display. He mentions it in the vid", + "video_name": "PSy6zQsk8z0", + "timestamps": [ + 30 + ], + "3min_transcript": "- [Karl Voiceover] Okay, so here's our Vivitar ViviCam Camera, and we're gonna take it apart and show you how it works. We got the batteries here, and there's the on switch, and now that we have the camera on we're going to take a few photos. You can see the LCD display is showing what's going on there, and letting us see the pictures. Now the LCD is backlit with an LED, which is, LCD stands for liquid crystal display, and LED stands for light emitting diode. So, let's go ahead and get crackin', we'll take it apart. We'll take all the screws out that are holding the two halves of the camera together. So we'll remove the top part, and that's likely made out of ABS, and there's a electroplated bezel, that surrounds the lens of the camera, it's screwed in place. And that's also probably ABS, plated ABS. And now let's go and take a look at what the camera's latch looks like and what it's made out of. We've got some, looks like chrome-plated stamped steel, from the underside of the back side of the batteries, to the board, and so that's where the positive connection on the board is, and the negative connection is up at the top. So we're gonna remove the pin and we can take the latch out now. Now you can slide the latch back and forth, you can see how it moves. Again this is plated steel, you can see there's four little connections on there, we'll talk about those in a second, but let's remove this back piece. This is just a simple insert, and I think it's probably heat staked in place. What that means is really low-cost way to attach things. This metal piece is as well, you can see those four points there. They're heat staked in and it is magnetic, so it's plated steel. And so, now we're gonna remove these screws that are connecting the plastic bezel that surrounds the camera and connects the two sides. And so you can see, it's just an injection molded piece, but again it didn't have a plastic designation on it, so it's not 100%, could be polypropylene or maybe styrene. Probably styrene. So there's the on button and the off button, I should say, the on button and the shutter button, and they push down and trigger the membrane switches, which are mounted on our printed circuit board, or a piece of the printed circuit board that's connected. So now let's take out the screws that are holding the board in place. And we tried to remove the screws for our tripod mount, but one of them is stripped, so we're just going to leave it in place. And I think we'll just pop the back out there and we'll take a look at the two switch covers. These switches are injection molded and then they are painted, and then heat staked in place. And so you can see the silver colors from the paint, and the heat stake process is basically the plastic, just a little tab at the end of the plastic is melted against the outer part of the housing." + }, + { + "Q": "At 7:48, it says there's a decrease in Sn2 mechanism. I understand that, but if a strong nucleophile is put in a polar protic solvent, it becomes a weak nucleophile. Then, shouldn't Sn1 be favoured over Sn2 as a minor product ?", + "A": "A polar protic solvent doesn t make a strong nucleophile into a weak nucleophile.It makes it into a weaker nucleophile. Ethoxide ion is a strong nucleophile in an aprotic solvent. It is weaker in a protic solvent, but it is still a pretty strong nucleophile even there.", + "video_name": "vFSZ5PU0dIY", + "timestamps": [ + 468 + ], + "3min_transcript": "We would have our ring and a double bond forms between carbon one and carbon six. So, that means a double bond forms in here. And then at carbon two, we still have a methyl group going away from us in space. So, let me draw that in like that. So, the electrons in red, hard to see, but if you think about these electrons in red back here, are gonna move in to form our double bond between what I've labeled as carbon one and carbon six. Let me label those again here. So, carbon one and carbon six. Again not IUPAC nomenclature just so we can think about our product compared to our starting material. So, this would be the major product of our reaction which is an E2 reaction. It would also be possible to get some products from an SN2 mechanism, but since heat is here, an elimination reaction is favored over a substitution. with phosphoric acid and heat. And we saw a lot of these types of problems in the videos on elimination reactions. So, it's not gonna be SN1 or SN2 and we don't have a strong base, so don't think E2, think E1. And our first step would be to protonate our alcohol to form a better leaving group. So phosphoric acid is a source of protons and we're going to protonate this oxygen for our first step. So, let's draw in our ring and we protonate our oxygen, so now our oxygen has two bonds to hydrogen, one lone pair of electrons and a plus one formal charge on the oxygen. So, this lone pair of electrons on the oxygen picked up a proton from phosphoric acid to form this bond. And now we have a better leaving group than the hydroxide ion. These electrons come off onto the oxygen and we remove a bond from this carbon in red which would give us a secondary carbocation. and the carbon in red is this one and that carbon would have a plus one formal charge. So, let me draw in a plus one formal charge here. And now we have water which can function as a weak base in our E1 reaction and take a proton from a carbon next to our carbon with a positive charge. So, let's say this carbon right here. It has two hydrogens on it. I'll just draw one hydrogen in and water functions as a base, takes this proton and these electrons move in to form a double bond. So, let's draw our final product here. We would have a ring, we would have a double bond between these two carbons, so our electrons in, let's use magenta, electrons in magenta moved in to form our double bond. So, our product is cyclohexane. So, a secondary alcohol undergoes an E1 reaction if you use something like sulfuric acid" + }, + { + "Q": "At 3:40 he says that the reaction would go by an Sn1 mechanism. Doesn't Sn1 only occurred in tertiary substrates? The one in the example is secondary", + "A": "No, 2\u00c2\u00b0 substrates can react via SN1 or SN2, depending on the conditions. We have two competing processes. If the nucleophile attacks faster than the leaving group spontaneously leaves, the reaction is SN2. If the leaving group leaves before the nucleophile can successfully attack, we have SN1. In this case, the methanoic acid is a very weak nucleophile, so the rate of SN2 attack is slow. Br\u00e2\u0081\u00bb is a good leaving group. so the rate of the SN1 reaction predominates over that of SN2.", + "video_name": "vFSZ5PU0dIY", + "timestamps": [ + 220 + ], + "3min_transcript": "it does not act like a nucleophile. So SN1 and SN2 are out. And a strong base means an E2 reaction. So, E1 is out. Now that we know we're doing an E2 mechanism, let's analyze the structure of our alkyl halide. The carbon that's directly bonded to our halogen is our alpha carbon and the carbons directly bonded to the alpha carbon are the beta carbons. So, I'll just do the beta carbon on the right since they are the same essentially. And we know that our base is gonna take a proton from that beta carbon. So, let me just draw in a hydrogen here. And DBN is a neutral base, so I'll just draw a generic base here. Our base is going to take this proton at the same time these electrons move in to form a double bond and these electrons come off to form our bromide anion. So, our final product is an alkyne and our electrons in magenta in here we have another secondary alkyl halide, so right now all four of these are possible until we look at our reagent which is sodium hydroxide, Na plus, OH minus, and we know that the hydroxide ion can function as a strong nucleophile or a strong base. So a strong nucleophile makes us think an SN2 reaction and not an SN1. The strong base makes us think about an E2 reaction and not an E1 reaction. Since we have heat, heat favors an elimination reaction over a substitution, so E2 should be the major reaction here. So, when we analyze our alkyl halide, the carbon bonded to the halogen is our alpha carbon and the carbons directly bonded to that would be our beta carbons. So, we have two beta carbons here and let me number this ring. I'm gonna say the alpha carbon is carbon one, I'm gonna go round clockwise, so that's one, two, three, carbon four, And next we're going to translate this to our chair confirmation over here. So, carbon one would be this carbon and then carbon two would be this one. This'll be carbon three, four, five and six. The bromine is coming out at us in space at carbon one which means it's going up. So, if I look at carbon one, we would have the bromine going up, which would be up axial. At carbon two, I have a methyl group going away from me in space, so that's going down, so at carbon two we must have a methyl group going down which makes it down axial. So, we care about carbon two. Let me highlight these again. So, we care about carbon two which is a beta carbon. We also care about carbon six which is another beta carbon. So, let's put in the hydrogens on those beta carbons. At carbon two, we would have a hydrogen that's up equatorial and at carbon six we would have" + }, + { + "Q": "around 9:30 in the video he said \"total distance\"... i thought we were finding the total displacement?", + "A": "In this case, the distance and the displacement are the same, because the plane always moves in the same direction. But you are correct, if he was being super-careful, maybe he should have said displacement .", + "video_name": "VYgSXBjEA8I", + "timestamps": [ + 570 + ], + "3min_transcript": "into the sum of the two terms times the difference of the two terms, so that when you multiply these two out you just get that over there, over 2 times the acceleration. Now what's really cool here is we were able to derive a formula that just deals with the displacement, our final velocity, our initial velocity, and the acceleration. And we know all of those things except for the acceleration. We know that our displacement is 80 meters. We know that this is 80 meters. We know that our final velocity, just before we square it, we know that our final velocity is 72 meters per second. And we know that our initial velocity is 0 meters per second. And so we can use all of this information to solve for our acceleration. And you might see this formula, displacement, the scalar version, and really we are thinking only in the scalar, we're thinking about the magnitudes of all of these things for the sake of this video. We're only dealing in one dimension. But sometimes you'll see it written like this, sometimes you'll multiply both sides times the 2 a, and you'll get something like this, where you have 2 times, really the magnitude of the acceleration, times the magnitude of the displacement, which is the same thing as the distance, is equal to the final velocity, the magnitude of the final velocity, squared, minus the initial velocity squared. Or sometimes, in some books, it'll be written as 2 a d is equal to v f squared minus v i squared. And it seems like a super mysterious thing, but it's not that mysterious. We just very simply derived it from displacement, or if you want to say distance, if you're just thinking about the scalar quantity, is equal to average velocity times the change in time. So, so far we've just derived ourselves a kind of a physics class, but let's use it to actually figure out the acceleration that a pilot experiences when they're taking off of a Nimitz class carrier. So we have 2 times the acceleration times the distance, that's 80 meters, times 80 meters, is going to be equal to our final velocity squared. What's our final velocity? 72 meters per second. So 72 meters per second, squared, minus our initial velocity. So our initial velocity in this situation is just 0. So it's just going to be minus 0 squared, which is just going to be 0, so we don't even have to write it down. And so to solve for acceleration, to solve for acceleration, you just divide, so this is the same thing as 160 meters, well, let's just divide both sides by 2 times 80, so we get acceleration is equal to 72 m/s squared" + }, + { + "Q": "At 11:12, Sal uses the previous value stored in the calculator (72.222 m/s) as the final velocity. But if he had just used 72 m/s like he said during the video, then the answer we get would be slightly different, around an acceleration of 32.4 m/s^2. I know this is a very slight difference, but in the second case, the value rounds down to 32 m/s^2. Should we use the more precise value like Sal used or use the value that complies with the significant figures rule?", + "A": "follow sig fig rules. But you don t round intermediate values - you only round the final answer (to the correct number of sig figs)", + "video_name": "VYgSXBjEA8I", + "timestamps": [ + 672 + ], + "3min_transcript": "physics class, but let's use it to actually figure out the acceleration that a pilot experiences when they're taking off of a Nimitz class carrier. So we have 2 times the acceleration times the distance, that's 80 meters, times 80 meters, is going to be equal to our final velocity squared. What's our final velocity? 72 meters per second. So 72 meters per second, squared, minus our initial velocity. So our initial velocity in this situation is just 0. So it's just going to be minus 0 squared, which is just going to be 0, so we don't even have to write it down. And so to solve for acceleration, to solve for acceleration, you just divide, so this is the same thing as 160 meters, well, let's just divide both sides by 2 times 80, so we get acceleration is equal to 72 m/s squared And what we're gonna get is, I'll just write this in one color, it's going to be 72 divided by 160, times, we have in the numerator, meters squared over seconds squared, we're squaring the units, and then we're going to be dividing by meters. So times, I'll do this in blue, times one over meters. Right? Because we have a meters in the denominator. And so what we're going to get is this meters squared divided by meters, that's going to cancel out, we're going to get meters per second squared. Which is cool because that's what acceleration should be in. And so let's just get the calculator out, to calculate this exact acceleration. So we have to take, oh sorry, this is 72 squared, let me write that down. So this is, this is going to be 72 squared, don't want to forget about this part right over here. 72 squared divided by 160. right over here that we calculated, so let's just square that, and then divide that by 160, divided by 160. And if we go to 2 significant digits, we get 33, we get our acceleration is, our acceleration is equal to 33 meters per second squared. And just to give you an idea of how much acceleration that is, is if you are in free fall over Earth, the force of gravity will be accelerating you, so g is going to be equal to 9.8 meters per second squared. So this is accelerating you 3 times more than what Earth is making you accelerate if you were to jump off of a cliff or something. So another way to think about this is that the force, and we haven't done a lot on force yet, we'll talk about this in more depth," + }, + { + "Q": "at 4:24 where does the 2 come from in the denominator? I don't understand...it seems so random", + "A": "The 2 is used because the average (mean) velocity is being found. The mean of a set is found by dividing the sum of the numbers in the set by the number of numbers. So the average mean speed is (V1+V2)/2, Hope this helped.", + "video_name": "VYgSXBjEA8I", + "timestamps": [ + 264 + ], + "3min_transcript": "So if we want to convert this into seconds, we have, we'll put hours in the numerator, 1 hour, so it cancels out with this hour, is equal to 3600 seconds. I'll just write 3600 s. And then if we want to convert it to meters, we have 1000 meters is equal to 1 km, and this 1 km will cancel out with those kms right over there. And whenever you're doing any type of this dimensional analysis, you really should see whether it makes sense. If I'm going 260 km in an hour, I should go much fewer km in a second because a second is so much shorter amount of time, and that's why we're dividing by 3600. If I can go a certain number of km in an hour a second, I should be able to go a lot, many many more meters in that same amount of time, and that's why we're multiplying by 1000. When you multiply these out, the hours cancel out, you have km canceling out, and you have 260 times 1000 So let me get my trusty TI-85 out, and actually calculate that. So I have 260 times 1000 divided by 3600 gets me, I'll just round it to 72, because that's about how many significant digits I can assume here. 72 meters per second. So all I did here is I converted the take-off velocity, so this is 72 m/s, this has to be the final velocity after accelerating. So let's think about what that acceleration could be, given that we know the length of the runway, and we're going to assume constant acceleration here, just to simplify things a little bit. But what does that constant acceleration have to be? So let's think a little bit about it. The total displacement, I'll do that in purple, the total displacement is going to be times the difference in time, or the amount of time it takes us to accelerate. Now, what is the average velocity here? It's going to be our final velocity, plus our initial velocity, over 2. It's just the average of the initial and final. And we can only do that because we are dealing with a constant acceleration. And what is our change in time over here? What is our change in time? Well our change in time is how long does it take us to get to that velocity? Or another way to think about it is: it is our change in velocity divided by our acceleration. If we're trying to get to 10 m/s, or we're trying to get 10 m/s faster, and we're accelerating at 2 m/s squared, it'll take us 5 seconds. Or if you want to see that explicitly written in a formula, we know that acceleration is equal to" + }, + { + "Q": "at 11:40, shouldn't Sal use 72 instead of 72.2222.... because it's only 2 sig figs? that would make 72^2 / 160 = 32.4, which rounded to 2 sig figs is 32, not 33.", + "A": "Sal is often a bit careless about sig figs. The equation that you have written should have 2 sig figs in its answer, if that is the final answer. Remember not to discard extra sig figs until you report your final answer. You use them all during intermediate calculations.", + "video_name": "VYgSXBjEA8I", + "timestamps": [ + 700 + ], + "3min_transcript": "And what we're gonna get is, I'll just write this in one color, it's going to be 72 divided by 160, times, we have in the numerator, meters squared over seconds squared, we're squaring the units, and then we're going to be dividing by meters. So times, I'll do this in blue, times one over meters. Right? Because we have a meters in the denominator. And so what we're going to get is this meters squared divided by meters, that's going to cancel out, we're going to get meters per second squared. Which is cool because that's what acceleration should be in. And so let's just get the calculator out, to calculate this exact acceleration. So we have to take, oh sorry, this is 72 squared, let me write that down. So this is, this is going to be 72 squared, don't want to forget about this part right over here. 72 squared divided by 160. right over here that we calculated, so let's just square that, and then divide that by 160, divided by 160. And if we go to 2 significant digits, we get 33, we get our acceleration is, our acceleration is equal to 33 meters per second squared. And just to give you an idea of how much acceleration that is, is if you are in free fall over Earth, the force of gravity will be accelerating you, so g is going to be equal to 9.8 meters per second squared. So this is accelerating you 3 times more than what Earth is making you accelerate if you were to jump off of a cliff or something. So another way to think about this is that the force, and we haven't done a lot on force yet, we'll talk about this in more depth, more than 3 times the force of gravity, more than 3 g's. 3 g's would be about 30 meters per second squared, this is more than that. So an analogy for how the pilot would feel is when he's, you know, if this is the chair right here, his pilot's chair, that he's in, so this is the chair, and he's sitting on the chair, let me do my best to draw him sitting on the chair, so this is him sitting on the chair, flying the plane, and this is the pilot, the force he would feel, or while this thing is accelerating him forward at 33 meters per second squared, it would feel very much to him like if he was lying down on the surface of the planet, but he was 3 times heavier, or more than 3 times heavier. Or if he was lying down, or if you were lying down, like this, let's say this is you, this is your feet, and this is your face, this is your hands," + }, + { + "Q": "At 6:02, why does the hydrocarbon chain have no polarity?", + "A": "Carbon and Hydrogen have very similar electronegativity, so the elements don t pull each other to a specific direction. Thus the hydrocarbon chain is not polar.", + "video_name": "Pk4d9lY48GI", + "timestamps": [ + 362 + ], + "3min_transcript": "And then in between, we have a carbon, and we call that the alpha carbon. We call that the alpha carbon. Alpha carbon, and that alpha carbon is bonded, it has a covalent bond to the amino group, covalent bond to the carboxyl group, and a covalent bond to a hydrogen. Now, from there, that's where you get the variation in the different amino acids, and actually, there's even some exceptions for how the nitrogen is, but for the most part, the variation between the amino acids is what this fourth covalent bond from the alpha carbon does. So you see in serine, you have this, what you could call it an alcohol. You could have an alcohol side chain. In valine right over here, you have a fairly pure hydrocarbon, hydrocarbon side chain. And so in general, we refer to these side chains as an R group, and it's these R groups and how they interact with their environment and the types of things they can do. And you can even see, just from these examples how these different sides chains might behave differently. This one has an alcohol side chain, and we know that oxygen is electronegative, it likes to hog electrons, it's amazing how much of chemistry or even biology you can deduce from just pure electronegativity. So, oxygen likes to hog electrons, so you're gonna have a partially-negative charge there. Hydrogen has a low electronegativity relative to oxygen, so it's gonna have its electrons hogged, so you're gonna have a partially positive charge, just like that, and so this has a polarity to it, and so it's going to be hydrophilic, it's going to, at least this part of the molecule is going to be able to be attracted and interact with water. And that's in comparison to what we have over here, this hydrocarbon side chain, this has no polarity over here, So this is going to be hydrophobic. And so when we start talking about the structures of proteins, and how the structures of proteins are influenced by its side chains, you could image that parts of proteins that have hydrophobic side chains, those are gonna wanna get onto the inside of the proteins if we're in an aqueous solution, while the ones that are more hydrophilic will wanna go onto the outside, and you might have some side chains that are all big and bulky, and so they might make it hard to tightly pack, and then you might have other side chains that are nice and small that make it very easy to pack, so these things really do help define the shape, and we're gonna talk about that a lot more when we talk about the structure. But how do these things actually connect? And we're gonna go into much more detail in another video, but if you have... If you have serine right over here, and then you have valine right over here, they connect through" + }, + { + "Q": "At 8:50\n\nWhy would Nitrogen grab a proton (H+) from the environment if this atom is very eletronegative?", + "A": "The atom as a whole is not electronegative, if the carboxyl tail becomes negative by losing a H+ the amino head becomes positive by gaining a H+, to make a neutral molecule.", + "video_name": "Pk4d9lY48GI", + "timestamps": [ + 530 + ], + "3min_transcript": "is the term for two or more amino acids connected together, so this would be a dipeptide, and the bond isn't this big, I just, actually let me just, let me draw it a little bit smaller. So... That's serine. This is valine. They can form a peptide bond, and this would be the smallest peptide, this would be a dipeptide right over here. \"Peptide,\" \"peptide bond,\" or sometimes called a peptide linkage. And as this chain forms, that polypeptide, as you add more and more things to it, as you add more and more amino acids, this is going to be, this can be a protein or can be part of a protein that does all of these things. Now one last thing I wanna talk about, this is the way, the way these amino acids have been drawn is a way you'll often see them in a textbook, but at physiological pH's, the pH's inside of your body, which is in that, you know, that low sevens range, What you have is this, the carboxyl group right over here, is likely to be deprotonated, it's likely to have given away its hydrogen, you're gonna find that more likely than when you have... It's gonna be higher concentrations having been deprotonated than being protonated. So, at physiological conditions, it's more likely that this oxygen has taken both of those electrons, and now has a negative charge, so it's given, it's just given away the hydrogen proton but took that hydrogen's electron. So it might be like this, and then the amino group, the amino group at physiological pH's, it's likely to actually grab a proton. So nitrogen has an extra loan pair, so it might use that loan pair to grab a proton, in fact it's physiological pH's, you'll find a higher concentration of it having grabbed a proton than not grabbing a proton. use its loan pairs to grab a proton, and so it is going to have... So it is going to have a... It is going to have a positive charge. And so sometimes you will see amino acids described this way, and this is actually more accurate for what you're likely to find at physiological conditions, and these molecules have an interesting name, a molecule that is neutral even though parts of it have charge, like this, this is called a zwitterion. That's a fun, fun word. Zwitterion. And \"zwitter\" in German means \"hybrid,\" and \"ion\" obviously means that it's going to have charge, and so this has hybrid charge, even though it has charges at these ends, the charges net out to be neutral." + }, + { + "Q": "At 04:07, why is it that at the 3rd carbon it is not 3-dimethyl since you have two methyl groups coming off?", + "A": "That second methyl has already been accounted for in the longest chain (it s the fourth carbon in the butane base), so you only need to name one methyl substituent at the 3 carbon.", + "video_name": "peQsBg9P4ms", + "timestamps": [ + 247 + ], + "3min_transcript": "to use should have as many simple groups attached to it as possible, as opposed to as few complex groups. So if we used this carbon as part of our longest chain, then this will be a group that's attached to it, which would be a bromomethyl group, which is not as simple as maybe it could be. But if we use this carbon in our longest chain, we'll have We'll have a bromo attached, and we'll also have a methyl group. And that's what we want. We want more simple groups attached to the longest chain. So what we're going to do is we're going to use this carbon, this carbon, this carbon, and that carbon as our longest chain. And we want to start from the end that is closest to something being attached to it, and that bromine is right there. So there's going to be our number one carbon, our number two carbon, our number three carbon, and our number four carbon. figure out what order they should be listed in. So this is a 1-bromo and then this will be a 2-methyl right here. And then just a hydrogen. Then three we have a fluoro, so on a carbon three, we have a fluoro, and then on carbon three, we also have a methyl group right here, so we also have a 3-methyl. So when we name it, we put in alphabetical order. Bromo comes first, so this thing right here is 1-bromo. Then alphabetically, fluoro comes next, 1-bromo-3-fluoro. We have two methyls, so it's going to be 2 comma 3-dimethyl. And remember, the D doesn't count in alphabetical order. chain is four carbons. Dimethylbutane. So that's just the standard nomenclature rules. We still haven't used the R-S system. Now we can do that. Now to think about that, we already said that this is our chiral center, so we just have to essentially rank the groups attached to it in order of atomic number and then use the Cahn-Ingold-Prelog rules, and we'll do all that in this example. So let's look at the different groups attached to it. So when you look at it, this guy has three carbons and a hydrogen. Carbon is definitely higher in atomic number on It has an atomic number 6. Hydrogen is 1. You probably know that already. So hydrogen is definitely going to be number four. So let me put number four there next to the hydrogen. And let me find a nice color. I'll do it in white. So hydrogen is definitely the number four group. We have to differentiate between this carbon group," + }, + { + "Q": "Hi, at 9:00 why does the #2 group move but not the #1 group?", + "A": "He is rotating the whole molecule 120\u00c2\u00b0 about the axis joining the C and #1, so those two atoms stay in place while #2, #3, and #4 rotate about that axis.", + "video_name": "peQsBg9P4ms", + "timestamps": [ + 540 + ], + "3min_transcript": "molecule right now the way it's drawn. I'll do that in magenta. So then you have your number three group. It's behind the molecule, so I'll draw it like this. This is our number three group. And then we have our number four group, which is the hydrogen pointing out right now. And I'll just do that in a yellow. We have our number four group pointing out in front right now. So that is number four, just like that. Actually, let me draw it a little bit clearer, so it looks a little bit more like the tripod structure that it's So let me redraw the number three group. The number three group should look like-- so this is our number three group. Let me draw it a little bit more like this. The number three group is behind us. And then finally, you have your number four group in straight out. So that is coming straight out of-- well, not straight out, but at an angle out of the page. So that's our number four group, I'll just label it It really is just a hydrogen, so I really didn't have to simplify it much there. Now by the R-S system, or by the Cahn-Ingold-Prelog system, we want our number four group to be the one furthest back. So we really want it where the number three position is. And so the easiest way I can think of doing that is you can imagine this is a tripod that's leaning upside down. Or another way to view it is you can view it as an umbrella, where this is the handle of the umbrella and that's the top of the umbrella that would block the rain, I guess. But the easiest way to get the number four group that's actually a hydrogen in the number three position would be to rotate it. You could imagine, rotate it around the axis defined by the number one group. So the number one group is just going to stay where it is. The number four is going to rotate to the Number three is going to rotate around to the number two group, and then the number two group is going to rotate to where the number four group is right now. So if we were to redraw that, let's redraw our chiral carbon. So let me scroll over a little bit. So we have our chiral carbon. I put the little asterisk there to say that that's our chiral carbon. The number four group is now behind. I'll do it with the circles. It makes it look a little bit more like atoms. So the number four group is now behind where the number three group used to be, so number four is now there. Number one hasn't changed. That's kind of the axis that we rotated around. So the number one group has not changed. Number one is still there. Number two is now where number four used to be, so number two is now jutting out of the page." + }, + { + "Q": "At around 4:53, why did Sal 4*56+3*16?", + "A": "Each iron has a mass of 56 and each oxygen has a mass of 16. The total mass of Fe2O3 is calculated as 2*56+3*16. (You have a typo; should be 2*56 instead of 4*56)", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 293 + ], + "3min_transcript": "how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that?" + }, + { + "Q": "At 7:00, do you always multiply that number by 2 or did he only do that because it was in front of Al? Would he have not multiplied it at all if it had nothing i front of the Al?", + "A": "If you look at the reaction equation, there is a coefficient in front of Al. You multiply by the coefficient because that is the ratio in which these chemicals react. Of course, if the coefficient is 1 (in which case you don t write it just like you don t write a coefficient of 1 in algebra) there is no need to multiply.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 420 + ], + "3min_transcript": "So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units." + }, + { + "Q": "At about 7:12 sal says \"for every one molecule of this, we need one molecule of that.\" I thought Fe2O3 would have an ionic bond and therefore technically wouldn't be a molecule. Am I mistaken?", + "A": "You are not mistaken, but we are often a little sloppy in terminology. We technically should be saying formula unit when speaking of Fe\u00e2\u0082\u0082O\u00e2\u0082\u0083 or any other ionic compound instead of molecule .", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 432 + ], + "3min_transcript": "So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units." + }, + { + "Q": "I'm still confuse why use the ratio of 1:2 of FE2O3 and 2Al mole ratio , 85g FE2O2 is 0.53 moles right why the aluminum must be 2(0.53)mole?", + "A": "So, the reaction asks for 1 unit of Fe2O3 for every 2 units of Al. Therefore, if you have 1*(0.53) moles of Fe2O3 then you need 2*(0.53) moles of Al", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 62 + ], + "3min_transcript": "We know what a chemical equation is and we've learned how to balance it. Now, we're ready to learn about stoichiometry. And this is an ultra fancy word that often makes people think it's difficult. But it's really just the study or the calculation of the relationships between the different molecules in a reaction. This is the actual definition that Wikipedia gives, stoichiometry is the calculation of quantitative, or measurable, relationships of the reactants and the products. And you're going to see in chemistry, sometimes people use the word reagents. For most of our purposes you can use the word reagents and reactants interchangeably. They're both the reactants in a reaction. The reagents are sometimes for special types of reactions where you want to throw a reagent in and see if something happens. And see if your belief about that substance is true or things like that. But for our purposes a reagent and reactant is the same thing. So it's a relationship between the reactants and the products in a balanced chemical equation. we know how to get to the balanced point. A balanced chemical equation. So let's do some stoichiometry. Just so we get practice balancing equations, I'm always going to start with unbalanced equations. Let's say we have iron three oxide. Two iron atoms with three oxygen atoms. Plus aluminum, Al. And it yields Al2 O3 plus iron. So remember when we're doing stoichiometry, first of all we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation. But I think it's good practice to actually balance the equations ourselves. So let's try to balance this one. We have two iron atoms here in this iron three oxide. How many iron atoms do we have on the right hand side? We only have one. So let's multiply this by 2 right here. We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams." + }, + { + "Q": "At 5:18 is it okay to NOT use PEMDAS?", + "A": "No, you should always compute numbers using the correct order of operations AKA PEMDAS.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 318 + ], + "3min_transcript": "how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that?" + }, + { + "Q": "Why do we round the atomic mass numbers from the periodic table? Like for Iron at 4:43, we use 56, not 55.85", + "A": "I understand that Iron has multiple isotopes and the atomic mass is an average of all those isotopes. 55.85 is the average mass of all Fe isotopes. Using 56 is simplifying a little bit...basically assuming you re using an Fe isotope with a mass of exactly 56 amu.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 283 + ], + "3min_transcript": "how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that?" + }, + { + "Q": "at 6:30 did he round that with the third number?", + "A": "Yes, he did. 85 g Fe\u00e2\u0082\u0082O\u00e2\u0082\u0083 = 0.532 mol Fe\u00e2\u0082\u0082O\u00e2\u0082\u0083. Only 2 significant figures are justified, but this is an intermediate answer, so he should have carried an extra digit and then rounded off the final answer to 2 significant figures.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 390 + ], + "3min_transcript": "So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum." + }, + { + "Q": "At 4:53, Can you explain to me why you multiplied 2 x 56. I am not very clear...", + "A": "you multiply 2X56 because the atomic mass of iron is approximately 56, times two because you have two atoms of Fe. Hope this helps! :)", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 293 + ], + "3min_transcript": "how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that?" + }, + { + "Q": "Things I am confused about:\n- At 5:20 you say that one molecule of iron three oxide is going to be 160 atomic mass units, then\n- At 5:41 you say that one mole of iron three oxide is going to have a mass of 160 grams.\nI am confused, at either 160 is recognized as grams or atomic mass unites?", + "A": "A MOLECULE has mass of 160 amu. One MOLE of those molecules has mass of 160 grams. A mole is a number, and we picked it so that the grams to amu thing would work.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 320, + 341 + ], + "3min_transcript": "how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that?" + }, + { + "Q": "At 1:31 how did you predict the subscripts of the atoms, and how they would arrange themselves?", + "A": "It s something that he just knows from past experience, and it is something that will come to you with time. When a metal is reacted with oxygen a metal oxide is formed. The subscripts are needed to make the formula work: Al^3+ O^2- The simplest way to make the charges cancel is 2x Al^3+ and 3x O^2- So the formula is Al2O3", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 91 + ], + "3min_transcript": "We know what a chemical equation is and we've learned how to balance it. Now, we're ready to learn about stoichiometry. And this is an ultra fancy word that often makes people think it's difficult. But it's really just the study or the calculation of the relationships between the different molecules in a reaction. This is the actual definition that Wikipedia gives, stoichiometry is the calculation of quantitative, or measurable, relationships of the reactants and the products. And you're going to see in chemistry, sometimes people use the word reagents. For most of our purposes you can use the word reagents and reactants interchangeably. They're both the reactants in a reaction. The reagents are sometimes for special types of reactions where you want to throw a reagent in and see if something happens. And see if your belief about that substance is true or things like that. But for our purposes a reagent and reactant is the same thing. So it's a relationship between the reactants and the products in a balanced chemical equation. we know how to get to the balanced point. A balanced chemical equation. So let's do some stoichiometry. Just so we get practice balancing equations, I'm always going to start with unbalanced equations. Let's say we have iron three oxide. Two iron atoms with three oxygen atoms. Plus aluminum, Al. And it yields Al2 O3 plus iron. So remember when we're doing stoichiometry, first of all we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation. But I think it's good practice to actually balance the equations ourselves. So let's try to balance this one. We have two iron atoms here in this iron three oxide. How many iron atoms do we have on the right hand side? We only have one. So let's multiply this by 2 right here. We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams." + }, + { + "Q": "at 1:30 the iron oxide isFE2O3 why not FEO2?", + "A": "Iron (Fe) in the complex will become a (Fe)3+ ion. The oxygens are each (O)2-. Therefore, in order to balance the charges and have a compound with zero net charge, you need 2(Fe)3+ and 3(O)2-", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 90 + ], + "3min_transcript": "We know what a chemical equation is and we've learned how to balance it. Now, we're ready to learn about stoichiometry. And this is an ultra fancy word that often makes people think it's difficult. But it's really just the study or the calculation of the relationships between the different molecules in a reaction. This is the actual definition that Wikipedia gives, stoichiometry is the calculation of quantitative, or measurable, relationships of the reactants and the products. And you're going to see in chemistry, sometimes people use the word reagents. For most of our purposes you can use the word reagents and reactants interchangeably. They're both the reactants in a reaction. The reagents are sometimes for special types of reactions where you want to throw a reagent in and see if something happens. And see if your belief about that substance is true or things like that. But for our purposes a reagent and reactant is the same thing. So it's a relationship between the reactants and the products in a balanced chemical equation. we know how to get to the balanced point. A balanced chemical equation. So let's do some stoichiometry. Just so we get practice balancing equations, I'm always going to start with unbalanced equations. Let's say we have iron three oxide. Two iron atoms with three oxygen atoms. Plus aluminum, Al. And it yields Al2 O3 plus iron. So remember when we're doing stoichiometry, first of all we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation. But I think it's good practice to actually balance the equations ourselves. So let's try to balance this one. We have two iron atoms here in this iron three oxide. How many iron atoms do we have on the right hand side? We only have one. So let's multiply this by 2 right here. We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams." + }, + { + "Q": "At 5:36 Why does Sal add the atomic masses of O and Fe to get the atomic mass of Ironoxide", + "A": "He added them to calculate the moles of Fe2O3 using the formula: moles = given mass/molar mass.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 336 + ], + "3min_transcript": "So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum." + }, + { + "Q": "At 11:27, in our class we say iron feroxide. Is it the same as iron three oxide as Sal says?", + "A": "Do you mean ferric oxide ? If so, that is an older term for Iron (III) oxide. It is still widely used, but it is not the IUAPC approved term. Ferroxide is a brand name of a pigment, so that is not an acceptable scientific name. But, to answer your question, you should use the method demonstrated in the video with the Roman numerals. This is the officially sanctioned term and would be understood by any chemist.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 687 + ], + "3min_transcript": "Or 6.02 times 10 to 23 aluminium atoms is going to be 27 grams. So if we need 1.06 moles, how many is that going to be? So 1.06 moles of aluminium is equal to 1.06 times 27 grams. And what is that? What is that? Equals 28.62. So we need 28.62 grams of aluminium, I won't write the whole thing there, in order to essentially use up our 85 grams of the iron three oxide. And if we had more than 28.62 grams of aluminium, Assuming we keep mixing it nicely and the whole reaction happens all the way. And we'll talk more about that in the future. And in that situation where we have more than 28.63 grams of aluminium, then this molecule will be the limiting reagent. Because we had more than enough of this, so this is what's going to limit the amount of this process from happening. If we have less than 28.63 grams of, I'll start saying aluminum, then the aluminum will be the limiting reagent, because then we wouldn't be able to use all the 85 grams of our iron molecule, or our iron three oxide molecule. Anyway, I don't want to confuse you in the end with that limiting reagents. In the next video, we'll do a whole problem devoted to limiting reagents." + }, + { + "Q": "at 3:36 he says Iron 3 Oxide, he means Iron 2 oxide right?", + "A": "No, Fe\u00e2\u0082\u0082O\u00e2\u0082\u0083 is Iron (III) Oxide. The number after the metal ion references its oxidation state, not how many atoms are present per formula unit.", + "video_name": "SjQG3rKSZUQ", + "timestamps": [ + 216 + ], + "3min_transcript": "We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams. how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out." + }, + { + "Q": "At 6:28 why asymmetric stretch is taking more energy than symmetric stretch ?", + "A": "Here s my guess: Consider a CH\u00e2\u0082\u0082 group. In a symmetric stretch, the two H atoms are going in the same direction. The group dipole moment changes considerably because both bonds are going in and out at the same time. In an asymmetric stretch, they are going in opposite directions. One bond gets longer as the other gets shorter, so the change in dipole moment is much less. It takes energy to separate positive and negative charges from each other, so the symmetric vibration has a higher frequency (energy).", + "video_name": "9GPuoukU8fM", + "timestamps": [ + 388 + ], + "3min_transcript": "Let's compare it to butylamine. So over here, this is a primary amine. The nitrogen is bonded to one carbon, so we're talking about a primary amine now. And let's analyze the IR spectrum. So once again, we're gonna draw a line around 3,000, and we know that this in here is talking about the carbon-hydrogen bond stretch for an SP3 hybridized carbon. Alright, once again, let's look at just past that, right in the bond to hydrogen region, and we get two signals this time, right? So if we look over here, there are two signals. This signal, let's drop down, this is approximately 3,300, so we have one signal approximately 3,300. And then we have another signal. Let me go ahead and make that green here. So we've got another signal right here, which is a little bit higher in terms of the wave number. So we drop down, this signal is approximately 3,400. the nitrogen-hydrogen bond stretch. We get two signals, and we need to figure out what's going on here. Well, this has to do with symmetric and asymmetric stretching. So let's look at two generic amines here, and let's talk about what the difference is between symmetric and asymmetric stretching. If you have symmetric stretching, so these bonds are stretching in phase, if you will. You can think about the hydrogens stretching away from the nitrogen at the same time. So this is called symmetric stretching. This is symmetric stretching. And this one over here, let me go ahead and draw what's happening over here. So this time these two nitrogen-hydrogen bonds are stretching out of phase. So if that hydrogen is stretching this way, this hydrogen might be contracting here, so that's an asymmetric stretch. Let me go ahead and write that. So we're talking about an asymmetric stretch here. This is why we get these two different signals. It turns out it takes less energy to do the symmetric stretching. So if it takes less energy to do the symmetric stretching, this is the one that we find at a lower wave number. Remember, wave numbers correspond to energy. So it takes less energy to do a symmetric stretching, and so that's this signal. It takes a little more energy to do asymmetric stretch. And so that's this signal, right up here. So we get two different signals here for our primary amine. Two signals, right? And it's tempting to say, \"Oh, we get two signals \"because we have two nitrogen-hydrogen bonds. \"So here's a nitrogen-hydrogen bond, \"and here's a nitrogen-hydrogen bond.\" But that's not really what's happening. Some of the molecules are having a symmetric stretch, and some of the molecules are having an asymmetric stretch. And so that's why you see these two different signals. Once again, let's just really quickly compare these two different amines." + }, + { + "Q": "whats a solar mass as said at 1:28", + "A": "One solar mass is not the mass, but the diameter of Sol(the Sun).", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 88 + ], + "3min_transcript": "In the videos on massive stars and on black holes, we learned that if the remnant of a star, of a massive star, is massive enough, the gravitational contraction, the gravitational force, will be stronger than even the electron degeneracy pressure, even stronger than the neutron degeneracy pressure, even stronger than the quark degeneracy pressure. And everything would collapse into a point. And we called these points black holes. And we learned there's an event horizon around these black holes. And if anything gets closer or goes within the boundary of that event horizon, there's no way that it can never escape from the black hole. All it can do is get closer and closer to the black hole. And that includes light. And that's why it's called a black hole. So even though all of the mass is at the central point, this entire area, or the entire surface of the event horizon, this entire surface of the event horizon-- I'll do it in purple because it's It will emit no light. Now these type of black holes that we described, we call those stellar black holes. And that's because they're formed from collapsing massive stars. And the largest stellar black holes that we have observed are on the order of 33 solar masses, give or take. So very massive to begin with, let's just be clear. And this is what the remnant of the star has to be. So a lot more of the original star's mass might have been pushed off in supernovae. That's plural of supernova. Now there's another class of black holes here and these are somewhat mysterious. And they're called supermassive black holes. And to some degree, the word \"super\" isn't big enough, supermassive black holes, than stellar black holes. They're are a lot more massive. They're on the order of hundreds of thousands to billions of solar masses, hundred thousands to billions times the mass of our Sun, solar masses. And what's interesting about these, other than the fact that there are super huge, is that there doesn't seem to be black holes in between or at least we haven't observed black holes in between. The largest stellar black hole is 33 solar masses. And then there are these supermassive black holes that we think exist. And we think they mainly exist in the centers of galaxies. And we think most, if not all, centers of galaxies actually have one of these supermassive black holes. But it's kind of an interesting question, if all black holes were formed from collapsing stars, wouldn't we see things in between? So one theory of how these really massive black holes form" + }, + { + "Q": "What would almost infinite be? (5:16)", + "A": "There is no ALMOST infinite", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 316 + ], + "3min_transcript": "in an area that has a lot of matter that it can accrete around it. So I'll draw the-- this is the event horizon around it. The actual black hole is going to be in the center of it, or rather the mass of the black hole will be in the center of it. And then over time, you have just more and more mass just falling into this black hole. Just more and more stuff just keeps falling into this black hole. And then it just keeps growing. And so this could be a plausible reason, or at least the mass in the center keeps growing and so the event horizon will also keep growing in radius. Now this is a plausible explanation based on our current understanding. But the reason why this one doesn't gel that well is if this was the explanation for supermassive black holes, you expect to see more black holes in between, maybe black holes with 100 solar masses, or a 1,000 solar masses, or 10,000 solar masses. But we're not seeing those right now. We just see the stellar black holes, So another possible explanation-- my inclinations lean towards this one because it kind of explains the gap-- is that these supermassive black holes actually formed shortly after the Big Bang, that these are primordial black holes. These started near the beginning of our universe, primordial black holes. Now remember, what do you need to have a black hole? You need to have an amazingly dense amount of matter or a dense amount of mass. If you have a lot of mass in a very small volume, then their gravitational pull will pull them closer, and closer, and closer together. And they'll be able to overcome all of the electron degeneracy pressures, and the neutron degeneracy pressures, and the quark degeneracy pressures, to really collapse into what we think is a single point. I want to be clear here, too. We don't know it's a single point. We've never gone into the center of a black hole. Just the mathematics of the black holes, or at least into a single point where the math starts to break down. So we're really not sure what happens at that very small center point. But needless to say, it will be an unbelievably, maybe infinite, maybe almost infinitely, dense point in space, or dense amount of matter. And the reason why I kind of favor this primordial black hole and why this would make sense is right after the formation of the universe, all of the matter in the universe was in a much denser space because the universe was smaller. So let's say that this is right after the Big Bang, some period of time after the Big Bang. Now what we've talked about before when we talked about cosmic background is that at that point, the universe was relatively uniform. It was super, super dense but it was relatively uniform. So a universe like this, there's no reason why anything would collapse into black holes. Because if you look at a point here, sure, there's a ton of mass very close to it." + }, + { + "Q": "Mentioned in 8:32, what are those stars in our solar system rapidly orbiting what is believed to be a black hole?", + "A": "the star are called a high velocity star.", + "video_name": "DxkkAHnqlpY", + "timestamps": [ + 512 + ], + "3min_transcript": "when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough, so that you could kind of have this snowballing effect, so that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the Sun. And, this is maybe even the more interesting part, those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going pass it fast enough, it'll just start going in orbit around the black hole. And so you could imagine that this is how the early galaxies or even our galaxy formed. what about the black hole at the center of the Milky Way? And we think there is one. We think there is one because we've observed stars orbiting very quickly around something at the center of the universe-- sorry, at the center of our Milky Way. I want to be very clear, not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the Sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are and somehow they grow into supermassive black holes and that everything in between we just can't observe. Or that they really are a different class of black holes. Maybe they formed near the beginning of the actual universe. When the density of things was a little uniform, things condensed into each other. And what we're going to talk about in the next video is how these supermassive black holes can help generate unbelievable sources of radiation, even though the black holes themselves aren't emitting them. And those are going to be quasars." + }, + { + "Q": "even at 8:15 the structure that is given\nis it benzophenone or dibenzophenone?", + "A": "I wiki d it up It s benzophenone although some part of me really wants it to be called dibenzophenone :P", + "video_name": "wD15pD5pCt4", + "timestamps": [ + 495 + ], + "3min_transcript": "And I'll just name this systematically right here. And the more complicated things get, the more systematic people will want to name it. So if we have six carbons right here, and they're in a chain, so this is cyclohexane. You'd put the \"e\" there if this carbonyl group wasn't there. But since it is, we would call this cyclohexanone, So this right here tells us to name it cyclohexanone. And then in a ring like this, this would implicitly be the number one carbon. So if this is the number one carbon and we want to number in the direction so that the next groups have the lowest possible number, so we want to make this the two carbon. So this is 2,2-dichlorocyclohexanone. And I'll just show these to you because these tend to be referred to by their common names. So I just want to show them to you real fast. One is this molecule right here, where we have a methyl group on this side: CH3. And over here, we have a benzene ring. Now, that first super simple ketone that we saw, we called this acetone. And so the common name here is actually derived from acetone. Instead of calling it acetone, because it doesn't have just a methyl group here, this is called aceto-, and instead of acetone, it's acetophenone, because we have this phenyl group, that benzene ring right there. Acetophenone, which is a pretty common molecule, and you'll see it referred to this way. Now, the other one that you might see every now and then, looks like this, that has two benzene rings on it. It looks like that. And this is benzophenone. These last two I just really wanted to expose you to their But, in general, I think you have a decent idea at this point of how to name at least the simpler chains, either with the common names, for example, propyl, or methyl propyl ketone, or 2-pentanone. And these are the more typical or maybe the easier naming examples." + }, + { + "Q": "At 7:44 the structure that is drawn\nis it acetophenone or acetophenyl?", + "A": "it s acetophenone. Phenyl is only used as a prefix--phenone is used as a suffix. (e.g. 2-phenylbutane)", + "video_name": "wD15pD5pCt4", + "timestamps": [ + 464 + ], + "3min_transcript": "And I'll just name this systematically right here. And the more complicated things get, the more systematic people will want to name it. So if we have six carbons right here, and they're in a chain, so this is cyclohexane. You'd put the \"e\" there if this carbonyl group wasn't there. But since it is, we would call this cyclohexanone, So this right here tells us to name it cyclohexanone. And then in a ring like this, this would implicitly be the number one carbon. So if this is the number one carbon and we want to number in the direction so that the next groups have the lowest possible number, so we want to make this the two carbon. So this is 2,2-dichlorocyclohexanone. And I'll just show these to you because these tend to be referred to by their common names. So I just want to show them to you real fast. One is this molecule right here, where we have a methyl group on this side: CH3. And over here, we have a benzene ring. Now, that first super simple ketone that we saw, we called this acetone. And so the common name here is actually derived from acetone. Instead of calling it acetone, because it doesn't have just a methyl group here, this is called aceto-, and instead of acetone, it's acetophenone, because we have this phenyl group, that benzene ring right there. Acetophenone, which is a pretty common molecule, and you'll see it referred to this way. Now, the other one that you might see every now and then, looks like this, that has two benzene rings on it. It looks like that. And this is benzophenone. These last two I just really wanted to expose you to their But, in general, I think you have a decent idea at this point of how to name at least the simpler chains, either with the common names, for example, propyl, or methyl propyl ketone, or 2-pentanone. And these are the more typical or maybe the easier naming examples." + }, + { + "Q": "When Sal first talks about the 2nd spaceship, he says \"at exactly the same velocity as her\", but then writes and says 1.5C. He loses me at 1:30. If the 2nd ship is traveling at 1.5C and she's at .5C, how do their relative positions stay the same?\nARE we dealing with SR here? It doesn't seem Newtonian to me.\nThank you", + "A": "You are getting confused with the expressions, writing .5C means point five times the speed of light while writing 1.5C would mean one point five times the speed of light. The equation is not 1.5C its 1.5x10 to the power of 8 m/s which is half the speed of light (3x10 to the power of 8 m/s) So all the spaceships are traveling at the same speed, which is half of the speed of light. Hopefully this makes sense, have another look at the equation being written at 1:30 again.", + "video_name": "F7BU1sXtul4", + "timestamps": [ + 90 + ], + "3min_transcript": "- [Voiceover] In the last video, we started to construct a space-time diagram for my frame of reference and I'm just drifting through space and I'm assuming that I'm in an intertial frame of reference which means I'm moving at a constant velocity relative to all other frames of reference and we set up a situation where I emitted a photon right at time zero, so after one second, it would have moved 3 times 10 to the 8th meters. After two seconds, it would have moved 6 times 10 to the 8th meters, and then we had a, we added a little bit of flavor to our little scenario where right at time zero, a friend passes me up in her spaceship and she is traveling relative to me in the positive-x direction at half the speed of light, and so we plotted her path right at time zero. Her spaceship is right there. I could draw the spaceship. It's at the origin. Then after one, after one second, she would have traveled 1 1/2 times 10 to the 8th meters. After two seconds, she would have traveled 3 times 10 to the 8th meters, so this blue line in the last video was her path. Let's assume that we have actually a whole train of spaceships, all traveling in the positive-x direction at the same velocity as her spaceship. So this is her spaceship right over here, at time equals zero, she is exactly where I am, but let's say that 3 times 10 to the 8th meters in front of her, there's another spaceship, traveling at exactly 1 1/2 times 10 to the 8th meters per second, so they're traveling at the same relative velocity to me, but if you think about it, from each of their point of view, they would seem to be stationary, because the distance between them, in this x-direction is going to stay the same. So this person, at time zero, is going to be 3 times 10 to the 8th meters in the positive-x direction from me. Now, if I were to wait two seconds, they are going to be, so if I were to wait two seconds, they are going to be, they are going to be 6 times 10 to the 8th meters away from me, because they're going half and so I could draw their path and I'm going to do it in a slightly more muted, thinner color, so that we, what I'm essentially going to be doing here is I'm setting up gridlines for my friend's alternate frame of reference. I'm going to put that on top of my frame of reference and to be clear, I'm not assuming special relativity. I'm assuming a Newtonian world, classical mechanics. Just to get familiar with these ideas and to see where the Newtonian world is going to break down but let's say that's not the only ship, let's say there was another ship, that at time equals zero is 6 times 10 to the 8th meters away from me in the positive x-direction. Well, where are they going to be? Where are they going to be after two seconds? Well, after, let's see, after one second, they're going to be 1 1/2 times 10 to the 8th meters further. Then after two seconds, they're going to be 3 times 10 to the 8th meters further," + }, + { + "Q": "At 4:10, there are bulges at the cup. Is there any relations with the convex meniscus?", + "A": "Sort of. The bulge that Sal is talking about does have to do with cohesion, but in this situation the bulge is not truly a convex meniscus because it is formed by surface tension, through cohesion, rather than just cohesion.", + "video_name": "_RTF0DAHBBM", + "timestamps": [ + 250 + ], + "3min_transcript": "and that causes a phenomenon known as surface tension. So you have stronger, you have kind of a deeper, and this is still just hydrogen bonds, but since they're not being pulled in other directions by, upwards by the air, they're able to get a little bit more closely packed, a little bit tighter, and this we refer to as surface tension, surface tension. And you have probably observed surface tension many, many, many times in your life in the form of, say, a water droplet. A water droplet, it's able to have this roughly round shape because all the little water molecules on the surface of the water droplet, and here the surface might even be on the bottom of the water droplet. They are more attracted to each other than they are to the surrounding air, so they're able to form this type of a shape. You might've seen it if you go to a pond or a stream sometimes, so you see some still water. And let's say, let me do this in blue. So let's say that this is the surface You might have seen insects that are able to walk on the surface of the water. And I'm not doing a great job at drawing the insects. They don't look exactly like that. But they can walk on the surface of the water. You might've seen or you might've even tried to do something like put a paperclip on the water. And even though this thing is actually more dense than the water and you might expect it to sink, but because of the surface tension, which really forms something of a film on top of the water, the thing won't penetrate the surface, so the paperclip will float, unless you were to push on it a little bit and it allow it to puncture the surface, and then it would actually sink, which is what you would expect because it is actually denser. You'd even see this if you were to take a cup, if you were to take a cup and you were to fill it all the way up to the rim and then a little bit higher, it won't immediately overflow. It won't immediately overflow. If you're very careful, you'll see that you form a bulge here. And that bulges because those individual water molecules are more attracted to each other than they are to the surrounding air. Obviously if you keep pouring water, at some point, they're just gonna start overflowing because gravity's gonna take over there. Gravity's gonna overwhelm the surface tension. But this bulge will actually form. So surface tension, it is really due to the cohesion of the water. Remember, cohesion is when the molecules are attracted to each other. And it definitely, and especially because they're more attracted to each other than the surrounding air." + }, + { + "Q": "At 4:54, Sal says that 'the same average kinetic energy' . What is he referring to ?\n\nThanks in advance !", + "A": "The wood and the metal have the same kinetic energy, or in other words they have the same temperature since temperature is just a mesure of the jiggling of atoms/molecules. Hope this helps ! If not, feel free to contact me", + "video_name": "6f553BGaufI", + "timestamps": [ + 294 + ], + "3min_transcript": "That is wood. And I have the metal surface, I'll do that in white. So I have the metal surface, right over here. And this metal surface, we already talked about, is going to feel colder. Let me draw the rest of my hand, actually. So the rest of my arm, you get the idea. So what's going on here. So let's just think about it at a microscopic level. So the wood, first of all, its surface is going to be uneven. So you're going to have atoms up here, but then you're going to have gaps, there's going to be air here. Let me actually scroll down a little bit. So it's going to be like this, so you're going to have gaps like that. And it also has internal gaps, like that. So this would be the wood, while the metal is much denser. So the metal, let me do the metal in that white color, the metal atoms are much more closely packed. It is much denser, the surface is smoother, it won't have any internal air pockets, it's not going to have any internal air pockets in it. And so what's going to happen? Well, we've always said, you're going to have a transfer of heat from the higher temperature system, or the higher temperature thing, to the lower temperature thing. And so, they're already going to have some kinetic energy, these things are going to have an average kinetic energy that's consistent with 70 degrees fahrenheit. So, let me just draw a couple of these arrows. Same thing over here, they're going to have the same average kinetic energy. So these things are all jostling around, bouncing around and pushing on each other with the electrostatic forces. But, my hand is warmer, my hand has a higher average kinetic energy. And so the atoms and molecules of my hand are going to bounce into the atoms and molecules of the wood, and they're going to transfer the kinetic energy. But we realize in the wood is, I'm making less contact. Because, first of all, the surface of the wood isn't smooth, so I'm making less contact. So this one over here might just bump into another air particle, it actually won't bump into a wood particle. But some of the wood particles will start to take some of the kinetic energy away from me. And I will sense that as being a little bit cool, so maybe that takes a little kinetic energy, that bumps into this guy. So the kinetic energy does get transferred down. But it's going to be transferred down a lot slower than what would happen in the metal. Because, one, I don't have as much surface contact between my hand and the wood, because of these gaps. I also have air pockets in the wood, like this." + }, + { + "Q": "At 1:00 how would the joules cancel out if we multiply them? We are supposed to divide it right? And where did the eV unit pop up from? If the unit of eV is (Joule)^2, then why is it written in the video that 1eV=1.6*10^-19 J?? I am not able to understand this part!", + "A": "I don t think you have followed the units correctly. When you multiply by 1/x that s the same as dividing by x. So when he does: y J * 1 eV / z J, both J cancel out leaving just eV. He is converting an energy value from J to eV. 1 eV = 1.6 x 10^-19 J is the conversion factor between these.", + "video_name": "nJ-PtF14EFw", + "timestamps": [ + 60 + ], + "3min_transcript": "- [Voiceover] So in that last video, I showed you how to get this equation using a lot of Physics, and so it's actually not necessary to watch the previous video, you can just start with this video if you want. And E one, we said, was the energy associated with an electron, and the lowest energy level of hydrogen. And we're using the Bohr model. And we calculated the value for that energy to be equal to negative 2.17 times 10 to the negative 18 joules. And let's go ahead and convert that into electron volts, it just makes the numbers easier to work with. So one electron volt is equal to 1.6 times 10 to the negative 19 joules. So if I take negative 2.17 times 10 to the negative 18 joules, I know that for every one electron volts, right, one electron volt is equal to 1.6 times 10 to the negative 19 joules, and so I have a conversion factor here. And so, if I multiply these two together, the joules would cancel and give me electron volts as my units. you get negative 13.6 electron volts. So once again, that's the energy associated with an electron, the lowest energy level in hydrogen. And so I plug that back into my equation here, and so I can just rewrite it, so this means the energy at any energy level N is equal to E one, which is negative 13.6 electron volts, and we divide that by N squared, where N is an integer, so one, two, three, and so on. So the energy for the first energy level, right, we already know what it is, but let's go ahead and do it so you can see how to use this equation, is equal to negative 13.6 divided by, so we're saying the energy where N is equal to one, so whatever number you have here, you're gonna plug in here. So this would just be one squared, alright? Which is of course just one, and so this is negative 13.6 electron volts, so we already knew that one. so E two, this would just be negative 13.6, and now N is equal to two, so this would be two squared, and when you do that math you get negative 3.4 electron volts. And then let's do one more. So the energy for the third energy level is equal to negative 13.6, now N is equal to three, so this would be three squared, and this gives you negative 1.51 electron volts. So, we have the energies for three different energy levels. The energy for the first energy level is equal to negative 13.6. E two is equal to negative 3.4, and E three is equal to negative 1.51 electron volts. So energy is quantized using the Bohr models, you can't have a value of energy in between those energies. And also note that your energies are negative," + }, + { + "Q": "Why can't we have a value of energy in between the integers which you mentioned at 2:59?", + "A": "Because that s the way nature works. Energy is quantized. It was a surprising discovery, but it is true. Why it is that way, we don t know.", + "video_name": "nJ-PtF14EFw", + "timestamps": [ + 179 + ], + "3min_transcript": "you get negative 13.6 electron volts. So once again, that's the energy associated with an electron, the lowest energy level in hydrogen. And so I plug that back into my equation here, and so I can just rewrite it, so this means the energy at any energy level N is equal to E one, which is negative 13.6 electron volts, and we divide that by N squared, where N is an integer, so one, two, three, and so on. So the energy for the first energy level, right, we already know what it is, but let's go ahead and do it so you can see how to use this equation, is equal to negative 13.6 divided by, so we're saying the energy where N is equal to one, so whatever number you have here, you're gonna plug in here. So this would just be one squared, alright? Which is of course just one, and so this is negative 13.6 electron volts, so we already knew that one. so E two, this would just be negative 13.6, and now N is equal to two, so this would be two squared, and when you do that math you get negative 3.4 electron volts. And then let's do one more. So the energy for the third energy level is equal to negative 13.6, now N is equal to three, so this would be three squared, and this gives you negative 1.51 electron volts. So, we have the energies for three different energy levels. The energy for the first energy level is equal to negative 13.6. E two is equal to negative 3.4, and E three is equal to negative 1.51 electron volts. So energy is quantized using the Bohr models, you can't have a value of energy in between those energies. And also note that your energies are negative, because this is the one that's closest to zero, so E three is the highest energy level out of the three that we're talking about here. Alright, let's talk about the Bohr model of the hydrogen atom really fast. And so, over here on the left, alright, just to remind you, I already showed you how to get these different radii for the Bohr model, so this isn't drawn perfectly to scale. But if we assume that we have a positively charged nucleus, which I just marked in red here, so there's our positively charged nucleus. We know the electron orbits the nucleus in the Bohr model. So I'm gonna draw an electron here, so again, not drawn to scale, orbiting the nucleus. So the positively charged nucleus attracts the negatively charged electron. And I'm saying that electron is orbiting at R one, so that's this first radius right here. So R one is when N is equal to one," + }, + { + "Q": "at 2:24, what is subscripts", + "A": "It literally means written below, in this case it s the smaller letter. At 2:24, he means a smaller letter to distinguish different things. For example, if I m doing a physics problem and I have 2 velocities, of a car and of a train, I could use subscripts to tell them apart.", + "video_name": "CFygKiTB-4A", + "timestamps": [ + 144 + ], + "3min_transcript": "" + }, + { + "Q": "At 5:55, Sal did F*d, with F being 10 kg * 9.8. Why isn't F=0, because as he said, there is no net force because the elevator is going at a constant velocity?", + "A": "The way it is going at constant velocity is it must have an upward force on it equal to its weight. Otherwise it would accelerate downward (fall)", + "video_name": "3mier94pbnU", + "timestamps": [ + 355 + ], + "3min_transcript": "slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it" + }, + { + "Q": "A little before 3:50, sal talks about pushing upwards with the acceleration of gravity while gravity is pulling downwards with the same acceleration, would the block even move then? Because if the block moves then there isnt any work done.", + "A": "There is no net work done on the block and therefore we should not expect the kinetic energy of the block to increase, according to the work-kinetic energy theorem. But there is work done on the earth-block system, and therefore the potential energy of that system increases.", + "video_name": "3mier94pbnU", + "timestamps": [ + 230 + ], + "3min_transcript": "Let's say we didn't know that someone else had applied a force of m for a distance of d to this object, just by calculating its kinetic energy as 125 joules, we immediately know that that's the amount of work that was necessary. And we don't know if this is exactly how this object got to this velocity, but we know that that is the amount of work that was necessary to accelerate the object to this velocity of 7 meters per second. So let's give another example. And instead of this time just pushing something in a horizontal direction and accelerating it, I'm going to show you an example we're going to push something up, but its velocity really isn't going to change. Invert. Let's say I have a different situation, and we're on this planet, we're not in deep space. And I have a mass of m and I were to apply a force. the acceleration of gravity. Mass times-- let's just call that gravity, right? 9.8 meters per second squared. And I were to apply this force for a distance of d upwards. Right? Or instead of d, let's say h. H for height. So in this case, the force times the distance is equal to-- well the force is mass times the acceleration of gravity, right? And remember, I'm pushing with the acceleration of gravity upwards, while the acceleration of gravity is pulling downwards. So the force is mass times gravity, and I'm applying that for a distance of h, right? d is h. So the force is this. This is the force. And then the distance I'm applying is going to be h. And what's interesting is-- I mean if you want to think of an exact situation, imagine an elevator that is already slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if" + }, + { + "Q": "At 5:15, Sal said that there's no net force, so Fnet =0. But why he take 98 N as a force?", + "A": "For Fnet to be zero the upward force must be equal to the weight (which is 98N). He is finding work done by this upward force.", + "video_name": "3mier94pbnU", + "timestamps": [ + 315 + ], + "3min_transcript": "the acceleration of gravity. Mass times-- let's just call that gravity, right? 9.8 meters per second squared. And I were to apply this force for a distance of d upwards. Right? Or instead of d, let's say h. H for height. So in this case, the force times the distance is equal to-- well the force is mass times the acceleration of gravity, right? And remember, I'm pushing with the acceleration of gravity upwards, while the acceleration of gravity is pulling downwards. So the force is mass times gravity, and I'm applying that for a distance of h, right? d is h. So the force is this. This is the force. And then the distance I'm applying is going to be h. And what's interesting is-- I mean if you want to think of an exact situation, imagine an elevator that is already slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it?" + }, + { + "Q": "At 9:30, does the potential energy when it hits the ground get converted to kinetic energy? And would it have velocity 100 when it hits the ground?", + "A": "There is no more PE when you reach the ground. It s all KE at that point. And the KE then gets converted to thermal energy", + "video_name": "3mier94pbnU", + "timestamps": [ + 570 + ], + "3min_transcript": "But it now has a lot of potential to do work. And what do I mean by potential to do work? Well after I move an object up 100 meters into the air, what's its potential to do work? Well, I could just let go of it and have no outside force other than gravity. The gravity will still be there. And because of gravity, the object will come down and be at a very, very fast velocity when it lands. And maybe I could apply this to some machine or something, and this thing could do work. And if that's a little confusing, let me give you an example. It all works together with our-- So let's say I have an object that is-- oh, I don't know-- a 1 kilogram object and we're on earth. And let's say that is 10 meters above the ground. So we know that its potential energy is equal to mass times So mass is 1. Let's just say gravitational acceleration is 10 meters per second squared. Times 10 meters per second squared. Times 10 meters, which is the height. So it's approximately equal to 100 Newton meters, which is the same thing is 100 joules. And what do we know about this? We know that it would take about 100-- or exactly-- 100 joules of work to get this object from the ground to this point up here. Now what we can do now is use our traditional kinematics formulas to figure out, well, if I just let this object go, how fast will it be when it hits the ground? And we could do that, but what I'll show you is even a faster way. And this is where all of the work and energy really becomes useful. We have something called the law of conservation of energy. It's that energy cannot be created or destroyed, it just And there's some minor caveats to that. But for our purposes, we'll just stick with that. So in the situation where I just take the object and I let go up here, up here it has a ton of potential energy. And by the time it's down here, it has no potential energy because the height becomes 0, right? So here, potential energy is equal to 100 and here, potential energy is equal to 0. And so the natural question is-- I just told you the law of conservation of energy, but if you look at this example, all the potential energy just disappeared. And it looks like I'm running out of time, but what I'll show you in the next video is that that potential energy gets converted into another type of energy. And I think you might be able to guess what type that is because this object is going to be moving really fast right before it hits the ground. I'll see you in the next video." + }, + { + "Q": "at 5:39 you said acceleration of gravity is 9.8 meter per second squared why is it not zero when the object is descending with constant velocity. It's not accelerating", + "A": "If the object descends with constant velocity, it is not accelerating. But that s not how objects fall. They accelerate when they fall. That s why you can jump off a short step but you can t jump off a tall building.", + "video_name": "3mier94pbnU", + "timestamps": [ + 339 + ], + "3min_transcript": "slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it" + }, + { + "Q": "At 3:02, why exactly are we pushing the object up with a force mg? Wouldn't it cancel with the gravity?", + "A": "If so happens then the body will stay at equilibrium rather we are pushing it with a bit greater force to overcome gravitational pull.", + "video_name": "3mier94pbnU", + "timestamps": [ + 182 + ], + "3min_transcript": "So its velocity over here. So let's just say I looked at an object here with mass m and it was moving with the velocity v. I would say well, this has a kinetic energy of 1/2 mv squared. And if the numbers are confusing you, let's say the mass was-- I don't know. Let's say this was a 5 kilogram object and it's moving at 7 meters per second. So I would say the kinetic energy of this object is going to be 5-- 1/2 times the mass times 5 times 7 squared, times It's times 49. So let's see. 1/2 times 49, that's a little under 25. So it'll be approximately 125 Newton meters, which is approximately-- and Newton meter is just a joule-- 125 joules. So this is if we actually put numbers in. And so when we immediately know this, even if we didn't Let's say we didn't know that someone else had applied a force of m for a distance of d to this object, just by calculating its kinetic energy as 125 joules, we immediately know that that's the amount of work that was necessary. And we don't know if this is exactly how this object got to this velocity, but we know that that is the amount of work that was necessary to accelerate the object to this velocity of 7 meters per second. So let's give another example. And instead of this time just pushing something in a horizontal direction and accelerating it, I'm going to show you an example we're going to push something up, but its velocity really isn't going to change. Invert. Let's say I have a different situation, and we're on this planet, we're not in deep space. And I have a mass of m and I were to apply a force. the acceleration of gravity. Mass times-- let's just call that gravity, right? 9.8 meters per second squared. And I were to apply this force for a distance of d upwards. Right? Or instead of d, let's say h. H for height. So in this case, the force times the distance is equal to-- well the force is mass times the acceleration of gravity, right? And remember, I'm pushing with the acceleration of gravity upwards, while the acceleration of gravity is pulling downwards. So the force is mass times gravity, and I'm applying that for a distance of h, right? d is h. So the force is this. This is the force. And then the distance I'm applying is going to be h. And what's interesting is-- I mean if you want to think of an exact situation, imagine an elevator that is already" + }, + { + "Q": "why is gravity negative at 5:52?", + "A": "he said that the tension in the string is equal to mg..!! the net force in the vertical direction is equals zero..!! refer the video once more, carefully..!! gravity is not -ve !!", + "video_name": "3mier94pbnU", + "timestamps": [ + 352 + ], + "3min_transcript": "slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it" + }, + { + "Q": "If potential energy is the ability to do work (7:00), what work does the object do when it falls? Isn't this work (falling of the object) done by gravity?", + "A": "Remember, gravity is an acceleration (9.81m/s^2); it s not a force. Therefore, 10kg of an object has potential to do work WITH GRAVITY, if it s located at proper position.", + "video_name": "3mier94pbnU", + "timestamps": [ + 420 + ], + "3min_transcript": "would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it But it now has a lot of potential to do work. And what do I mean by potential to do work? Well after I move an object up 100 meters into the air, what's its potential to do work? Well, I could just let go of it and have no outside force other than gravity. The gravity will still be there. And because of gravity, the object will come down and be at a very, very fast velocity when it lands. And maybe I could apply this to some machine or something, and this thing could do work. And if that's a little confusing, let me give you an example. It all works together with our-- So let's say I have an object that is-- oh, I don't know-- a 1 kilogram object and we're on earth. And let's say that is 10 meters above the ground. So we know that its potential energy is equal to mass times" + }, + { + "Q": "If you times 1/2 x 5 x 49 this is equal to 122.5 which is also 1.23x10 to the 23rd power... How does he get that answer at 1:41 - 1:58", + "A": "Are you referring to his answer of 125 joules? He s using the approximation of 1/2 * 5 * 50 there. Also, 1.23 * 10 to the 23rd power is a huge number, I don t think that is what you meant to type.", + "video_name": "3mier94pbnU", + "timestamps": [ + 101, + 118 + ], + "3min_transcript": "Welcome back. In the last video, I showed you or hopefully, I did show you that if I apply a force of F to a stationary, an initially stationary object with mass m, and I apply that force for distance d, that that force times distance, the force times the distance that I'm pushing the object is equal to 1/2 mv squared, where m is the mass of the object, and v is the velocity of the object after pushing it for a distance of d. And we defined in that last video, we just said this is work. Force times distance by definition, is work. And 1/2 mv squared, I said this is called kinetic energy. And so, by definition, kinetic energy is the amount of work-- and I mean this is the definition right here. It's the amount of work you need to put into an object or apply to an object to get it from rest So its velocity over here. So let's just say I looked at an object here with mass m and it was moving with the velocity v. I would say well, this has a kinetic energy of 1/2 mv squared. And if the numbers are confusing you, let's say the mass was-- I don't know. Let's say this was a 5 kilogram object and it's moving at 7 meters per second. So I would say the kinetic energy of this object is going to be 5-- 1/2 times the mass times 5 times 7 squared, times It's times 49. So let's see. 1/2 times 49, that's a little under 25. So it'll be approximately 125 Newton meters, which is approximately-- and Newton meter is just a joule-- 125 joules. So this is if we actually put numbers in. And so when we immediately know this, even if we didn't Let's say we didn't know that someone else had applied a force of m for a distance of d to this object, just by calculating its kinetic energy as 125 joules, we immediately know that that's the amount of work that was necessary. And we don't know if this is exactly how this object got to this velocity, but we know that that is the amount of work that was necessary to accelerate the object to this velocity of 7 meters per second. So let's give another example. And instead of this time just pushing something in a horizontal direction and accelerating it, I'm going to show you an example we're going to push something up, but its velocity really isn't going to change. Invert. Let's say I have a different situation, and we're on this planet, we're not in deep space. And I have a mass of m and I were to apply a force." + }, + { + "Q": "at 5:40, why is force = 10*9.8? If force = m*a then shouldn't acceleration be -9.8 not 9.8, since it is downwards", + "A": "You can define the signs of your direction however you want; it s arbitrary. If you want to make down be positive, you can. You just have to be consistent throughout the problem.", + "video_name": "3mier94pbnU", + "timestamps": [ + 340 + ], + "3min_transcript": "slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it" + }, + { + "Q": "At about 3:12 Sal says he's applying a force of mg upwards. Is this a net force? I fi t were just a force wouldn't it be counterbalanced by the force of gravity?", + "A": "If he applies mg upward, and gravity pulls downward with a force of mg, then the net force is zero and the object will maintain its velocity.", + "video_name": "3mier94pbnU", + "timestamps": [ + 192 + ], + "3min_transcript": "So its velocity over here. So let's just say I looked at an object here with mass m and it was moving with the velocity v. I would say well, this has a kinetic energy of 1/2 mv squared. And if the numbers are confusing you, let's say the mass was-- I don't know. Let's say this was a 5 kilogram object and it's moving at 7 meters per second. So I would say the kinetic energy of this object is going to be 5-- 1/2 times the mass times 5 times 7 squared, times It's times 49. So let's see. 1/2 times 49, that's a little under 25. So it'll be approximately 125 Newton meters, which is approximately-- and Newton meter is just a joule-- 125 joules. So this is if we actually put numbers in. And so when we immediately know this, even if we didn't Let's say we didn't know that someone else had applied a force of m for a distance of d to this object, just by calculating its kinetic energy as 125 joules, we immediately know that that's the amount of work that was necessary. And we don't know if this is exactly how this object got to this velocity, but we know that that is the amount of work that was necessary to accelerate the object to this velocity of 7 meters per second. So let's give another example. And instead of this time just pushing something in a horizontal direction and accelerating it, I'm going to show you an example we're going to push something up, but its velocity really isn't going to change. Invert. Let's say I have a different situation, and we're on this planet, we're not in deep space. And I have a mass of m and I were to apply a force. the acceleration of gravity. Mass times-- let's just call that gravity, right? 9.8 meters per second squared. And I were to apply this force for a distance of d upwards. Right? Or instead of d, let's say h. H for height. So in this case, the force times the distance is equal to-- well the force is mass times the acceleration of gravity, right? And remember, I'm pushing with the acceleration of gravity upwards, while the acceleration of gravity is pulling downwards. So the force is mass times gravity, and I'm applying that for a distance of h, right? d is h. So the force is this. This is the force. And then the distance I'm applying is going to be h. And what's interesting is-- I mean if you want to think of an exact situation, imagine an elevator that is already" + }, + { + "Q": "At 7:38, Sal explains why the distance between us and the object emitting the photon has expanded due to the universe. Wouldn't the photon emitted from the object still be moving? It wouldn't have stayed in the same spot, right?", + "A": "Right. It s own motion is bringing it closer, and that is offset to some degree by the expansion of space", + "video_name": "6nVysrZQnOQ", + "timestamps": [ + 458 + ], + "3min_transcript": "But I really just want to give you the idea of what's going on here. So let's just say, well, that photon says in about 10 million years, in roughly 10 million years, I should be right about at that coordinate. I should be about one third of the distance. But what happens over the course of those 10 million years? Well, over the course of those 10 million years, the universe has expanded some. The universe has expanded maybe a good deal. So let me draw the expanded universe. So after 10 million years, the universe might look like this. Actually it might even be bigger than that. Let me draw it like this. After 10 million years, the universe might have expanded a good bit. So this is 10 million years into the future. Still on a cosmological time scale, still almost at kind of the infancy of the universe because we're talking about 13.7 billion years. 10 million years go by. The universe has expanded. This coordinate, where we're sitting today at the present time, is now all the way over here. That coordinate where the photon was originally emitted is now going to be sitting right over here. And that photon, it said, OK, after 10 million light years, I'm going to get over there. And I'm approximating. I'm doing it in a very discrete way. But I really just want to give you the idea. So that coordinate, where the photon roughly gets in 10 million light years, is about right over here. The whole universe has expanded. All the coordinates have gotten further away from each other. Now, what just happened here? The universe has expanded. This distance that was 30 million light years now-- and I'm just making rough numbers here. I don't know the actual numbers here. Now, it is actually-- this is really just for the sake of giving you the idea of why-- well, giving you This distance now is no longer 30 million light years. Maybe it's 100 million. So this is now 100 million light years away from each other. The universe is expanding. These coordinates, the space is actually spreading out. You could imagine it's kind of a trampoline or the surface of a balloon. It's getting stretched thin. And so this coordinate where the light happens to be after 10 million years, it has been traveling for 10 million years, but it's gone a much larger distance. That distance now might be on the order of-- maybe it's on the order of 30 million light years. And the math isn't exact here. I haven't done the math to figure it out. So it's done 30 million light years. And actually I shouldn't even make it the same proportion. Because the distance it's gone and the distance it has to go, because of the stretching, it's not going to be completely linear. At least when I'm thinking about it in my head, it shouldn't be," + }, + { + "Q": "in this video, at 13 minutes and 56 seconds (13:56) sal says that the distance between the two points is 46 billion light years. but isnt the universe 13.7 bilion years old, and wouldn't this imply that the distance between the two points is 13.7 light years???? i'm confused...", + "A": "Some parts of the universe may be expanding faster than light.", + "video_name": "6nVysrZQnOQ", + "timestamps": [ + 836 + ], + "3min_transcript": "is a billion light years. So as you can see, the photon might getting frustrated. As it covers more and more distance, it looks back and says, wow, in only 50 million years, I've been able to cover 600 million light years. But it's frustrated because what it thought was it only had to cover 30 million light years in distance. That keeps stretching out because space itself is stretching. So the reality, just going to the original idea, this photon that is just reaching us, that's been traveling for-- let's say it's been traveling for 13.4 billion years. So it's reaching us is just now. So let me just fast forward 13.4 billion years from this point now to get to the present day. So if I draw the whole visible universe right over here, this point right over here is going to be-- where it was emitted from is right over there. We are sitting right over there. If I'm drawing the whole observable universe, the center actually should be where we are. Because we can observe an equal distance. If things aren't really strange, we can observe an equal distance in any direction. So actually maybe we should put us at the center. So if this was the entire observable universe, and the photon was emitted from here 13.4 billion years ago-- so 300,000 years after that initial Big Bang, and it's just getting to us, it is true that the photon has been traveling for 13.7 billion years. But what's kind of nutty about it is this object, since we've been expanding away from each other, this object is now, in our best estimates, this object is going to be about 46 billion light years away from us. This object is now 46 billion light years away from us. When we just use light to observe it, it looks like, just based on light years, hey, this light has been traveling 13.7 billion years to reach us. That's our only way of kind with light to kind of think about the distance. So maybe it's 13.4 or whatever-- I keep changing the decimal-- but 13.4 billion light years away. But the reality is if you had a ruler today, light year rulers, this space here has stretched so much that this is now 46 billion light years. And just to give you a hint of when we talk about the cosmic microwave background radiation, what will this point in space look like, this thing that's actually 46 billion light years away, but the photon only took 13.7 billion years to reach us? What will this look like? Well, when we say look like, it's based on the photons that are reaching us right now." + }, + { + "Q": "At 8:00, there's a point in space which emitted a photon 10mil. yrs ago and space is expanding between us so it takes longer to get to us. At 9:50,the photon isn\u00e2\u0080\u0099t travelling faster than the speed of light, even though it covered a greater distance.\n?\nIf space is expanding, then that photon must be travelling slower than the speed of light towards us, yet faster than the speed of light away from its origin. Ouch, my head! I realize this is getting into relativity, but can anyone expand on this", + "A": "but the light will be red shifted because it s source is moving away.", + "video_name": "6nVysrZQnOQ", + "timestamps": [ + 480, + 590 + ], + "3min_transcript": "But I really just want to give you the idea of what's going on here. So let's just say, well, that photon says in about 10 million years, in roughly 10 million years, I should be right about at that coordinate. I should be about one third of the distance. But what happens over the course of those 10 million years? Well, over the course of those 10 million years, the universe has expanded some. The universe has expanded maybe a good deal. So let me draw the expanded universe. So after 10 million years, the universe might look like this. Actually it might even be bigger than that. Let me draw it like this. After 10 million years, the universe might have expanded a good bit. So this is 10 million years into the future. Still on a cosmological time scale, still almost at kind of the infancy of the universe because we're talking about 13.7 billion years. 10 million years go by. The universe has expanded. This coordinate, where we're sitting today at the present time, is now all the way over here. That coordinate where the photon was originally emitted is now going to be sitting right over here. And that photon, it said, OK, after 10 million light years, I'm going to get over there. And I'm approximating. I'm doing it in a very discrete way. But I really just want to give you the idea. So that coordinate, where the photon roughly gets in 10 million light years, is about right over here. The whole universe has expanded. All the coordinates have gotten further away from each other. Now, what just happened here? The universe has expanded. This distance that was 30 million light years now-- and I'm just making rough numbers here. I don't know the actual numbers here. Now, it is actually-- this is really just for the sake of giving you the idea of why-- well, giving you This distance now is no longer 30 million light years. Maybe it's 100 million. So this is now 100 million light years away from each other. The universe is expanding. These coordinates, the space is actually spreading out. You could imagine it's kind of a trampoline or the surface of a balloon. It's getting stretched thin. And so this coordinate where the light happens to be after 10 million years, it has been traveling for 10 million years, but it's gone a much larger distance. That distance now might be on the order of-- maybe it's on the order of 30 million light years. And the math isn't exact here. I haven't done the math to figure it out. So it's done 30 million light years. And actually I shouldn't even make it the same proportion. Because the distance it's gone and the distance it has to go, because of the stretching, it's not going to be completely linear. At least when I'm thinking about it in my head, it shouldn't be," + }, + { + "Q": "At 16:09, is the \"white hot plasma\" the same thing as the \"cosmic background radiation\"?", + "A": "The cosmic background radiation is the afterglow left over from the big bang. At first it was seen to be constant (the same everywhere). Now the COBE satellite shows that there are tiny fluctuations which are believed to be differences in density. This is what led to the forming of galaxies. But I m not sure what Sal means exactly by white hot plasma .", + "video_name": "6nVysrZQnOQ", + "timestamps": [ + 969 + ], + "3min_transcript": "This object is now 46 billion light years away from us. When we just use light to observe it, it looks like, just based on light years, hey, this light has been traveling 13.7 billion years to reach us. That's our only way of kind with light to kind of think about the distance. So maybe it's 13.4 or whatever-- I keep changing the decimal-- but 13.4 billion light years away. But the reality is if you had a ruler today, light year rulers, this space here has stretched so much that this is now 46 billion light years. And just to give you a hint of when we talk about the cosmic microwave background radiation, what will this point in space look like, this thing that's actually 46 billion light years away, but the photon only took 13.7 billion years to reach us? What will this look like? Well, when we say look like, it's based on the photons that are reaching us right now. So those photons are the photons being emitted from this primitive structure, from this white-hot haze of hydrogen plasma. So what we're going to see is this white-hot haze. So we're going to see this kind of white-hot plasma, white hot, undifferentiated not differentiated into proper stable atoms, much less stars and galaxies, but white hot. We're going to see this white-hot plasma. The reality today is that point in space that's 46 billion years from now, it's probably differentiated into stable atoms, and stars, and planets, and galaxies. And frankly, if that person, that person, if there is a civilization there right now and if they're sitting right there, and they're observing photons being emitted from our coordinate, from our point in space right now, they're not going to see us. They're going to see us 13.4 billion years ago. of our region of space when it really was just a white-hot plasma. And we're going to talk more about this in the next video. But think about it. Any photon that's coming from that period in time, so from any direction, that's been traveling for 13.4 billion years from any direction, it's going to come from that primitive state or it would have been emitted when the universe was in that primitive state, when it was just that white-hot plasma, this undifferentiated mass. And hopefully, that will give you a sense of where the cosmic microwave background radiation comes from." + }, + { + "Q": "At 9:07, Sal said there may be a black hole at the center of the galaxy, why would this be so?", + "A": "We don t know why they are there, but it appears that most or maybe all galaxies have a super massive black hole in the center.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 547 + ], + "3min_transcript": "to give you a sense that when we saw the solar system, it's not just the Sun. There's all this neat, dynamic stuff. And there are planets, and asteroids, and solar winds. And so there's 200 to 400 billion stars and for the most part, 200 and 400 billion solar systems. So it's an unimaginably, I guess, complex or huge place. And just to make it clear, even when we zoom in to this picture right here, and I think it was obvious based on telling you about this, that these little white pockets right here, This isn't two stars. These are thousands of stars here. So when you go over here, each little blotch of white that you see, that's not a star, that's not a thousand stars. We're starting to talk in the millions of stars when you look at certain blotches here and there. I mean, maybe it might be one star that's closer to you or might be a million stars that are far apart And everything has to be used in kind of loose terms here. And we'll talk more about other galaxies. But even this isn't the upper bound of galaxies. People believe the Andromeda Galaxy has a trillion stars in it, a trillion solar systems. We're talking about these huge, huge, immense distances. And so just to give you a sense of where we fit in the picture, this is a rough location of our Sun. And remember, that little dot I drew just now is including millions of stars, millions of solar systems, already unimaginable the distances. But if you really want to get at the sense relative to the whole galaxy, this is an artist's depiction. Once again, we could never obviously get this perspective on the galaxy. It would take us forever to travel this far so that you could see the galaxy from above. But this is our best guess looking at things from our vantage point. And we actually can't even see this whole area over here because it's on the other side of the center of the galaxy, which is super, super dense and super bright. We think-- or actually there's a super massive black hole at the center of the galaxy. And we think that they're at the center of all or most galaxies. But you know the whole point of this video, actually this whole series of videos, this is just kind of-- I don't know-- to put you in awe a little bit of just how huge Because when you really think about the scale-- I don't know-- no words can really describe it. But just to give you a sense, we're about 25,000 light years from the center of the galaxy. So even when we look at things in the center of the galaxy, that's as they were 25,000 years ago. It took 25,000 years for that light to get to us. I mean when that light left the center of the galaxy, I won't even guess to think what humanity was like at that point in time. So it's these huge distances in the whole galaxy over here." + }, + { + "Q": "At 6:36, Sal mentions that our solar system is in the Orion Spur. Also, on a clear night, you sometimes see what looks like an arm/spur of the Milky Way. Which arm/spur is it?", + "A": "That is the entire milky way. Every arm of it is in that same plane.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 396 + ], + "3min_transcript": "looking at an object that's sitting-- let me do this in a darker color-- if we're sitting here on Earth and we're looking at an object out here that's 500 light years away, we're looking at it as it was 500 years ago because the light that is reaching our eyeballs right now, or our telescopes right now, left this guy over here 500 years ago. In fact, he's not going to even be there anymore. He probably has moved around a little bit. So just even on this scale, we're talking about these unimaginably huge distances. And then when we zoom out, this is kind of our local part of the galaxy right over here. This piece right here, this is called the Orion Spur. And people are still trying to work out exactly the details of the actual shape the Milky Way Galaxy, the galaxy that we're in. But we're pretty sure-- actually, we're very sure-- we have these spiral arms But it's actually very hard to come up with the actual shape, especially because you can't see a lot of the galaxy, because it's kind of on the other side, on the other side of the center. But really just to get a sense of something that at least-- I mean it blows my mind if you really think about what it's saying-- these unbelievable distances show up as a little dot here. This whole drawing shows up as a dot here. Now when we zoom out, over here that dot would no longer even show up. It wouldn't even register a pixel on this drawing right over here. And then this whole drawing, this whole thing right over here, this whole picture is this grid right over here. It is this right over here. So hopefully, that gives you a sense of how small even our local neighborhood is relative to the galaxy as a whole. And the galaxy as a whole, just to give you a sense, has 200 to 400 billion stars. to give you a sense that when we saw the solar system, it's not just the Sun. There's all this neat, dynamic stuff. And there are planets, and asteroids, and solar winds. And so there's 200 to 400 billion stars and for the most part, 200 and 400 billion solar systems. So it's an unimaginably, I guess, complex or huge place. And just to make it clear, even when we zoom in to this picture right here, and I think it was obvious based on telling you about this, that these little white pockets right here, This isn't two stars. These are thousands of stars here. So when you go over here, each little blotch of white that you see, that's not a star, that's not a thousand stars. We're starting to talk in the millions of stars when you look at certain blotches here and there. I mean, maybe it might be one star that's closer to you or might be a million stars that are far apart" + }, + { + "Q": "You state at 3:45 that our images of the galaxy are artist representations. What information do they have to draw from and what do they have to make educated guesses about in their representations? What do these artists use to generate these images?", + "A": "The artists have to add the light and color. What our telescopes see is light waves of various frequencies. A lot of the light is not in the visible spectrum. The telescopes use electronic sensors to record the data. Someone has to decide what that data looks like. What wavelength should be what color? How bright is bright ? That s the artist.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 225 + ], + "3min_transcript": "But most of recorded history is in the last 4,000 or 5,000 years. So this is 80,000 years to travel to the nearest star. So it's a huge distance. Another way to think about it is if the Sun were the size of a basketball and you put that basketball in London, if you wanted to do it in scale, the next closest star, which is actually a smaller basketball, right over here, Proxima Centauri, that smaller basketball you would have to put in Kiev, Ukraine in order to have a similar scale. So these are basketballs sitting in these cities. And you would have to travel about 1,200 miles to place the next basketball. And these basketballs are representing these super huge things that we saw in the first video. The Sun, if you actually made the Earth relative to these basketballs, these would be little grains of sand. So there are any little small planets over here, they would have to be grains of sand So this is a massive, massive distance, already, at least in my mind, unimaginable. And when it gets really wacky is when you start really looking at this. Even this is a super, super small distance relative to the galactic scale. So this whole depiction of kind of our neighborhood of stars, this thing over here is about, give or take-- and we're doing rough estimates right here-- it's about 30 light years. I'll just do LY for short. So that's about 30 light years, And once again, you can take pictures of our galaxy from our point of view. But you actually can't take a picture of the whole galaxy from above it. So these are going to be artists' depictions. But if this is 30 light years, this drawing right here of kind of our local neighborhood of the galaxy, this right here is roughly-- and these are all approximations. This is about 1,000 light years. And this is the 1,000 light years of our Sun's neighborhood, if you can even call it a neighborhood anymore. Even this isn't really a neighborhood if it takes you 80,000 years to get to your nearest neighbor. But this whole drawing over here-- now, it would take forever to get anywhere over here-- it would be 1/30 of this. So it would be about that big, this whole drawing. And what's really going to blow your mind is this would be roughly a little bit more than a pixel on this drawing right here, that spans a 1,000 light years. But then when you start to really put it into perspective-- so now, let's zoom out a little bit-- so this drawing right here, this 1,000 light years is now this 1,000 light years over here. So this is the local vicinity of the Sun. And once again, the word \"local\" is used in a very liberal way at this point. So this right here is 1,000 light years." + }, + { + "Q": "In the picture at 12:07, there is a black space that forms a sort of horizontal line. Is that just empty space between stars? Or is it attributed to something else?", + "A": "In these pictures, the whiter parts are usually gas clouds that reflect off the light, while the darkers areas are where dust clouds partially obscure the light of stars behind them.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 727 + ], + "3min_transcript": "it's hard to say the edge of the galaxy, because there's always going to be a few more stars and other things orbiting around the galaxy as you go further and further out, but it gets less dense with stars. But the main density, the main disk, is about 100,000 light years. 100,000 light years is the diameter roughly of the main part of the galaxy. And it's about 1,000 light years thick. So you can kind of imagine it as this disk, this thing that's fairly flat. But it's 1,000 light years thick. It's 1,000 light years thick. You would have to do this distance 250 times just to go from the top part of the galaxy to the bottom part, much less going across the galaxy. So it might seem relatively flat. But it's still immensely, immensely thick. And just as another way to visualize it, if this thing right over here that includes the Oort Cloud, roughly a light year in diameter, is a grain of sand, then the universe as a whole is going to be the diameter of a football field. And that might tell you, OK, those are two tractable things. I can imagine a grain of sand, a millimeter wide grain of sand in a football field. But remember, that grain of sand is still 50,000 or 60,000 times the diameter of Earth's orbit. And Earth's orbit, it would still take a bullet or something traveling as fast as a jet plane 15 hours to just go half of that-- or sorry, not-- 15 years or 17 years, I forgot the exact number. But it was 15, 16, 17 years to even cover half of that distance. So 30 years just to cover the diameter of Earth's orbit. That's 1/60,000 of our little grain of sand in the football And just to kind of really, I don't know, have an appreciation for how mind-blowing this really is, this is actually a picture of the Milky Way Galaxy, our galaxy, from our vantage point. As you can see, we're in the galaxy and this is looking towards the center. And even this picture, you start to appreciate the complexity of what 100 billion stars are. But what I really want to point out is even in this picture, when you're looking at these things, some of these things that look like stars, those aren't stars. those are thousands of stars or millions of stars. Maybe it could be one star closer up. But when we're starting to approach the center of the galaxy, these are thousands and thousands and millions of stars or solar systems that we're actually looking at. So really, it starts to boggle the mind to imagine what might actually be going on over there." + }, + { + "Q": "at 5:49, Sal mentions the Orion Spur. Where is that in the Milky Way?", + "A": "We re in it.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 349 + ], + "3min_transcript": "This is about 1,000 light years. And this is the 1,000 light years of our Sun's neighborhood, if you can even call it a neighborhood anymore. Even this isn't really a neighborhood if it takes you 80,000 years to get to your nearest neighbor. But this whole drawing over here-- now, it would take forever to get anywhere over here-- it would be 1/30 of this. So it would be about that big, this whole drawing. And what's really going to blow your mind is this would be roughly a little bit more than a pixel on this drawing right here, that spans a 1,000 light years. But then when you start to really put it into perspective-- so now, let's zoom out a little bit-- so this drawing right here, this 1,000 light years is now this 1,000 light years over here. So this is the local vicinity of the Sun. And once again, the word \"local\" is used in a very liberal way at this point. So this right here is 1,000 light years. looking at an object that's sitting-- let me do this in a darker color-- if we're sitting here on Earth and we're looking at an object out here that's 500 light years away, we're looking at it as it was 500 years ago because the light that is reaching our eyeballs right now, or our telescopes right now, left this guy over here 500 years ago. In fact, he's not going to even be there anymore. He probably has moved around a little bit. So just even on this scale, we're talking about these unimaginably huge distances. And then when we zoom out, this is kind of our local part of the galaxy right over here. This piece right here, this is called the Orion Spur. And people are still trying to work out exactly the details of the actual shape the Milky Way Galaxy, the galaxy that we're in. But we're pretty sure-- actually, we're very sure-- we have these spiral arms But it's actually very hard to come up with the actual shape, especially because you can't see a lot of the galaxy, because it's kind of on the other side, on the other side of the center. But really just to get a sense of something that at least-- I mean it blows my mind if you really think about what it's saying-- these unbelievable distances show up as a little dot here. This whole drawing shows up as a dot here. Now when we zoom out, over here that dot would no longer even show up. It wouldn't even register a pixel on this drawing right over here. And then this whole drawing, this whole thing right over here, this whole picture is this grid right over here. It is this right over here. So hopefully, that gives you a sense of how small even our local neighborhood is relative to the galaxy as a whole. And the galaxy as a whole, just to give you a sense, has 200 to 400 billion stars." + }, + { + "Q": "At 9:03 Sal said that scientists think there is a super massive black hole in the middle. Would we one day get consumed by this black hole? Is our galaxy spinning into that black hole?", + "A": "No, even a 4 million solar mass black hole is unable to get us at its distance. This black hole has not swallowed anything of significant size in millions of years.", + "video_name": "rcLnMe1ELPA", + "timestamps": [ + 543 + ], + "3min_transcript": "to give you a sense that when we saw the solar system, it's not just the Sun. There's all this neat, dynamic stuff. And there are planets, and asteroids, and solar winds. And so there's 200 to 400 billion stars and for the most part, 200 and 400 billion solar systems. So it's an unimaginably, I guess, complex or huge place. And just to make it clear, even when we zoom in to this picture right here, and I think it was obvious based on telling you about this, that these little white pockets right here, This isn't two stars. These are thousands of stars here. So when you go over here, each little blotch of white that you see, that's not a star, that's not a thousand stars. We're starting to talk in the millions of stars when you look at certain blotches here and there. I mean, maybe it might be one star that's closer to you or might be a million stars that are far apart And everything has to be used in kind of loose terms here. And we'll talk more about other galaxies. But even this isn't the upper bound of galaxies. People believe the Andromeda Galaxy has a trillion stars in it, a trillion solar systems. We're talking about these huge, huge, immense distances. And so just to give you a sense of where we fit in the picture, this is a rough location of our Sun. And remember, that little dot I drew just now is including millions of stars, millions of solar systems, already unimaginable the distances. But if you really want to get at the sense relative to the whole galaxy, this is an artist's depiction. Once again, we could never obviously get this perspective on the galaxy. It would take us forever to travel this far so that you could see the galaxy from above. But this is our best guess looking at things from our vantage point. And we actually can't even see this whole area over here because it's on the other side of the center of the galaxy, which is super, super dense and super bright. We think-- or actually there's a super massive black hole at the center of the galaxy. And we think that they're at the center of all or most galaxies. But you know the whole point of this video, actually this whole series of videos, this is just kind of-- I don't know-- to put you in awe a little bit of just how huge Because when you really think about the scale-- I don't know-- no words can really describe it. But just to give you a sense, we're about 25,000 light years from the center of the galaxy. So even when we look at things in the center of the galaxy, that's as they were 25,000 years ago. It took 25,000 years for that light to get to us. I mean when that light left the center of the galaxy, I won't even guess to think what humanity was like at that point in time. So it's these huge distances in the whole galaxy over here." + }, + { + "Q": "At 7:40, he says that the boiling point for methane is around -164 degrees celsius. So, can methane be in a liquid form, if it is colder than -164 degrees celsius? And if so, is it present in a liquid form naturally anywhere?", + "A": "Yes. Titan, Saturn s largest moon, has clouds, rain, rivers and lakes of liquid methane.", + "video_name": "pBZ-RiT5nEE", + "timestamps": [ + 460 + ], + "3min_transcript": "And even though the methane molecule here, if we look at it, we have a carbon surrounded by four hydrogens for methane. And it's hard to tell in how I've drawn the structure here, but if you go back and you look at the video for the tetrahedral bond angle proof, you can see that in three dimensions, these hydrogens are coming off of the carbon, and they're equivalent in all directions. And there's a very small difference in electronegativity between the carbon and the hydrogen. And that small difference is canceled out in three dimensions. So the methane molecule becomes nonpolar as a result of that. So this one's nonpolar, and, of course, this one's nonpolar. And so there's no dipole-dipole interaction. There's no hydrogen bonding. The only intermolecular force that's holding two methane molecules together would be London dispersion forces. And so once again, you could think about the electrons that are in these bonds moving in those orbitals. And let's say for the molecule on the left, if for a brief transient moment in time you get a little bit of negative charge to be those electrons have a net negative charge on this side. And then for this molecule, the electrons could be moving the opposite direction, giving this a partial positive. And so there could be a very, very small bit of attraction between these two methane molecules. It's very weak, which is why London dispersion forces are the weakest intermolecular forces. But it is there. And that's the only thing that's holding together these methane molecules. And since it's weak, we would expect the boiling point for methane to be extremely low. And, of course, it is. So the boiling point for methane is somewhere around negative 164 degrees Celsius. And so since room temperature is somewhere around 20 to 25, obviously methane has already boiled, if you will, and turned into a gas. So methane is obviously a gas at room temperature and pressure. Now, if you increase the number of carbons, you're going to increase the number of attractive forces And if you do that, you can actually increase the boiling point of other hydrocarbons dramatically. And so even though London dispersion forces are the weakest, if you have larger molecules and you sum up all those extra forces, it can actually turn out to be rather significant when you're working with larger molecules. And so this is just a quick summary of some of the intermolecular forces to show you the application of electronegativity and how important it is." + }, + { + "Q": "2 questions.\n1. In the first example within the first 4:00, both carbons have the same number of substituents. So couldn't you add the proton and the OH to either one of the carbons?\n\n2. at 11:50, why do you not have to worry about sterochem? Just because you are ignoring it for the sake of teaching?", + "A": "1. Correct. The alkene is symmetrical, so you can add the H and the OH to either end. 2. No. You don\u00c3\u00a8t have to worry about stereochemistry, because the product has no chiral centres.", + "video_name": "dJhxphep_gY", + "timestamps": [ + 240, + 710 + ], + "3min_transcript": "So what's left? In this acid base reaction, we took a proton away from H3O+, which leaves us H2O. So here we have H2O over here, so I'll go ahead and put lone pairs of electrons in on our water molecule. And we know that water can act as a nucleophile here. So this lone pair of electrons is going to be attracted to something that's positively charged. So nucleophilic attack on our carbocation. And this is technically at equilibrium as well, depending on the concentrations of your reactants. So let's go ahead and show that water molecule adding on to the carbon on the left. So the carbon on the right already had a hydrogen or proton added onto it, and the carbon on the left is going to have an oxygen now bonded to that carbon. Two hydrogens bonded to that oxygen, and there was a lone pair of electrons on that oxygen that did not participate in any kind of bonding. This gives this oxygen right here a plus 1 formal charge. And we're almost to our product. So we're almost there. We need one more acid base reaction to get rid of that proton on our oxygen. So water can function as a base this time. So water comes along, and this time it's going to act as a Bronsted Lowry base and accept a proton. So let's get those electrons in there. So this lone pair of electrons, let's say it takes that proton, leaving these electrons behind on my oxygen. Once again, I'll draw my equilibrium arrows here, acid base reaction. And I'm going to end up with an OH on the carbon on the left, and the carbon on the right there is a hydrogen, like that. So I added water. I ended up adding water across my double bond. And to be complete, this would regenerate my hydronium ion. I'd get H2O plus H+ would give me H3O+. And so there you go. So remember, a carbocation is present, so you have to think about Markovnikov addition. And since a carbocation is present, you have to think about possible rearrangements. So Markovnikov and rearrangements. Let's take a look at an example where you have a rearrangement here. So let's look at a reaction. So let's look at this as our starting alkene. And let's go ahead and think about the mechanisms. So we know H3O+ is going to be present. So H3O+ right here. So we're adding our alkene to a solution of water and sulfuric acid. And our first step in the mechanism, the pi electrons are going to function as a base and take a proton from our hydronium ion, leaving these electrons in here letting" + }, + { + "Q": "In the previous video it was said that temp is av KE by no of molecules but in this video at 3:20 it is said that temp is total KE by no of molecules. Which one is correct", + "A": "average, he says total divded by n, which is number of molecules", + "video_name": "HvYUKRMT0VI", + "timestamps": [ + 200 + ], + "3min_transcript": "lot of different ways to measure temperature. We know that in Fahrenheit, what's freezing of water? It's 32 degrees Fahrenheit that's freezing, but that's also 0 degrees Celsius-- actually, that's how the Celsius scale was determined. They said, where does water freeze, and then where does water boil? 100 degrees for Celsius is boiling, and that's how they rated it. You could be colder than the freezing of water, and you'd have to go negative in that situation-- Fahrenheit, I'm actually not sure. I need to look that up in Wikipedia, or that might be something for you to do, and tell me how it came out. I think the boiling of water in Fahrenheit is 212 degrees, so it's a little arbitrary. I think Fahrenheit might be somehow related to human body temperature, but I'm just guessing. You can have different scales in this situation, and they were all kind of a bit arbitrary when they were designed. They were just to have some type of relative to scale-- hotter because they have a higher temperature then when things are freezing. You can't divide 100 by zero, but if something is 1 degree, is it necessarily the case that something that is 100 degrees Celsius is a hundred times hotter, or has a hundred times the kinetic energy? Actually, what we'll see is that no, it's actually not the case-- you don't have 100 times the kinetic energy, so this is a bit of an arbitrary scale. The actual interval is arbitrary-- you could pick the 1 degree as being one hundredth of the distance between zero and 100, but where you start-- at least in the Celsius scale-- is a bit arbitrary. They picked the freezing of water. Later on, people figured out that there is an absolute point to start at. And that absolute point to start at is the temperature at which a molecule or an atom has absolutely no kinetic energy. energy of the system, or the total kinetic energy of the system divided by the number of molecules. Or we could also say the average kinetic energy per molecule. The only way to really say that the temperature is zero-- and this is proportional, because the temperature scales are still a little bit arbitrary-- the only way to get to a temperature of zero should be when the kinetic energy of each and every molecule is zero, or the So they're not moving, they're not vibrating, they're not even blinking-- these molecules are stationary. The point at which that occurs is called absolute zero." + }, + { + "Q": "I've noticed this in a few videos but at 8:50 he draws a 'double head' arrow to show the movement of a electron from the Bromide ion to the carbocation but only one electron is moving so shouldn't it be a 'fish hook' arrow?", + "A": "Yes, that s the convention. Most chemists consider this reaction to be a two-electron movement: Both \u00cf\u0080 electrons move to the H, and both of the electrons in the H-Br bond move on to the Br.", + "video_name": "Z_GWBW_GVGA", + "timestamps": [ + 530 + ], + "3min_transcript": "It has six protons, so it has a positive charge. This carbon right here has a positive charge. And another way to think about it is it was completely neutral and then it lost an electron. So now it will have a positive charge right over there. So this is what we are left after that step of the reaction. Oh, and of course, we can't forget. Bromine over here was neutral. It had seven valence electrons, and that's when bromine is neutral. But now it has eight, so now this will have a negative charge. This will have a negative charge, because it gained an electron. And in general, your total charge-- over here our total charge is zero, So? Our total charge will still be zero. We have a negative and we have a positive. They would cancel out, so our total charge is still zero. So what's likely to happen for the next step of our reaction? Well, we have this positive thing here. Maybe bromine just bumped, just the right way, to let go But now you have this guy who's negative and this guy who's positive. Maybe they'll be attracted to each other. Maybe they'll just bump into each other at the exact right way. And if they bump in the exact right way, maybe this guy can swipe the electron from the bromide ion, from this negative ion right here. And you might say, hey, isn't bromine more electronegative than carbon? Well, it might be, but this guy's electron rich. It's not just a regular bromine atom. This is a bromine plus an extra electron. So he's already hogged an electron, so he's electron rich. So in this situation, he's negative. He's positive. He can give this guy an electron. So if they bump in just the right way, this electron can be swiped by this carbon right over here. And this positive carbon, just so it gives you a little terminology, and we'll go over it in more detail in future videos, is actually called a carbocation. It's a positive ion of carbon. But anyway, if this electron gets swiped by this carbon, it will then form a bond. Because remember, this was the electron that was originally in a bond with this hydrogen. It's still going to be, you could imagine, paired up with this other purple or magenta electron right over there. So if that happens, then we're going to be left with-- so the next step is going to be-- so on this end of the molecule we have C, carbon, hydrogen, hydrogen. Then we have this orange hydrogen that we stole from the hydrogen bromide. Then we have this carbon right here. It has a hydrogen. You have the rest of the chain, CH2-CH2-CH3. And now, since this guy stole an electron, a bond will form with the bromine. Let me draw it. So he's going to steal this-- let me draw it this way. So a bond will form. He's stealing this electron, so now this electron is with" + }, + { + "Q": "At 5:23 Sal mentions continuing mountain chains between North America and Europe. What are some examples of this?", + "A": "The Appalachian mountains in the US, the Scottish Highlands, and the Little Atlas Mountains in Morocco were all formed as parts of an ancient Pangean mountain range.", + "video_name": "axB6uhEx628", + "timestamps": [ + 323 + ], + "3min_transcript": "And the last time we had a supercontinent was Pangaea, about 250 million years ago. And now it's broken up into our current day geography. Now, I won't go into all of the detail why we believe that there was a Pangaea about 250 million years ago-- or, this diagram tells us, about 225 million years ago, give or take. But I'll go into some of the interesting evidence. On a very high level, you have a lot of rock commonalities between things that would have had to combine during Pangaea. And probably the most interesting thing is the fossil evidence. There are a whole bunch of fossils. And here are examples of it, from species that were around between 200 and 300 million years ago. And their fossils are found in a very specific place. This animal right here, cynognathus-- I hope I'm pronouncing that right-- cynognathus. This animal's fossils are only found here, and in this part of Africa. So not only does South America look like it fits very nicely into Africa. But the fossil evidence also makes it look like there was a nice clean band where this animal lived and where we find the fossils. So it really makes it seem like these were connected, at least when this animal lived, maybe on the order of 250 million years ago. This species right over here, its fossils are found in this area-- let me do it in a color that has more contrast-- in this area right over here. This plant, its fossils-- now, this starts to connect to a lot of dots between a lot of cont-- its fossils are found in this entire area, across South America, Africa, Antarctica, India, and Australia. And so not only does it look like the continents fit together in a puzzle piece, not only do we get it to a configuration like this if we essentially just rewind to the movement that we're seeing now-- but the fossil evidence also This animal right here, we find fossils on this nice stripe that goes from Africa through India, all the way to Antarctica. Now, this only gives us evidence of the Southern Hemisphere of Pangaea. But there is other evidence. We find kind of continuing mountain chains between North America and Europe. We find rock evidence, where just the way we see the fossils line up nicely. We see common rock that lines up nicely between South America and Africa and other continents that were at once connected. So all the evidence, as far as we can tell now, does make us think that there at one time was a Pangaea. And, for all we know, all the continents are going to keep moving. And maybe in a few hundred million years, we'll have another supercontinent. Who knows?" + }, + { + "Q": "At 7:04, David says that the force of tension is pointing towards the centre. But in previous videos, David said that the force of tension is only a PULL, not a PUSH. But if the force of tension is pointing towards the centre, wouldn't that be contradictory?", + "A": "If the object that is moved around in a circle by a string, suddently lost the string, then the object would keep moving away from the center. This means that the string is pulling the object to stay in the circle. If this makes sense?", + "video_name": "FfNgm-w9Krw", + "timestamps": [ + 424 + ], + "3min_transcript": "for one of the directions. Which direction should we pick? Well, which force do we want to find? We want to find this force of tension, so even though I could if I wanted to use Newton's second law for this vertical direction, the tension doesn't even point that way, so I'm not gonna bother with that direction first. I'm gonna see if I can get by doing this in one step, so I'm gonna use this horizontal direction and that's gonna be the centripetal direction, i.e., into the circle. And when we're dealing with the centripetal force, we're gonna be dealing with the centripetal acceleration, so over here, when I use a and set that equal to the net force over mass, if I'm gonna use the centripetal force, I'm gonna have to use the centripetal acceleration. In other words, I'm gonna only plug forces that go into, radially into the circle here, and I'm gonna have the radial centripetal acceleration right here. And we know the formula for centripetal acceleration, that's v squared over r, so I'm gonna plug v squared over r into the left hand side. That's the thing that's new. When we used Newton's second law for just regular forces, but now, when you're using this law for the particular direction that is the centripetal direction, you're gonna replace a with v squared over r and then I set it equal to the net force in the centripetal direction over the mass. So what am I gonna plug in up here? What forces do I put up here? I mean, I've got normal force, tension, gravity. A common misconception is that people try to put them all into here. People put the gravitational force, the normal force, the tension, why not? But remember, if we've selected the centripetal direction, centripetal just means pointing toward the center of the circle, so I'm only going to plug in forces that are directed in toward the center of the circle, and that's not the normal force or the gravitational force. These forces do not point inward toward the center of the circle. The only force in this case that points toward the center of the circle is the tension force, and like we already said, that is the centripetal force. So over here, I'd have v squared over r, the only force acting as the centripetal force is the tension. Now, should that be positive or negative? Well, we're gonna treat inward as positive, so any forces that point inward are gonna be positive. Is it possible for a centripetal force to be negative? It is. If there was some force that pointed outward, if for some reason there was another string pulling on the ball outward, we would include that force in this calculation, and we would include it with a negative sign, so forces that are directed out of the circle, we're gonna count as negative and forces that are directed into the circle, we're gonna count as positive in here. And if they're not directed into or out, we're not gonna include them in this calculation at all. Now, you might object. You might say, wait a minute. There is a force out of the circle. This ball wants to go out of the circle. There should be a force this way. This is often referred to as the centrifugal force, and that doesn't really exist. So when people say that there's an outward force trying to direct this ball out of the circle," + }, + { + "Q": "At 1:37, what is the angle between the plane containing two C-H bonds in CH4, and the plane containing the other two C-H bonds in the same molecule?", + "A": "90 degrees is the angle", + "video_name": "ka8Yt4bTODs", + "timestamps": [ + 97 + ], + "3min_transcript": "Let's figure out the shape of the methane molecule using VSEPR theory. So the first thing that you do is draw a dot structure to show it the valence electrons. So for methane, carbon is in group four. So 4 valence electrons. Hydrogen is in group one, and I have four of them, so 1 times 4 is 4, plus 4 is 8 valence electrons that we need to show in our dot structure. Carbon goes in the center and carbon is bonded to 4 hydrogens, so I can go ahead and put my hydrogens in there like that. And this is a very simple dot structure. We've already shown all 8 of our valence electrons. Let me go ahead and highlight those here. 2, 4, six, and 8. So carbon has an octet and we are done. The next thing we need to do is count the number of electron clouds that surround our central atom. So remember, electron clouds are regions of electron density, So we can think about these bonding electrons here as being electron clouds. So that's one electron cloud. Here's another one down here. And then finally, here's another one. So we have four electron clouds surrounding our central atom. The next step is to predict the geometry of your electron clouds around your central atom. And so VSEPR theory tells us that those valence electrons are going to repel each other since they are negatively charged. And therefore, they're going to try to get as far away from each other as they can in space. And when you have four electron clouds, the electron clouds are farthest away from each other if they point towards the course of a tetrahedron, which is a four sided figure. So let me go ahead and draw the molecule here, draw the methane molecule. I'm going to attempt to show it in a tetrahedral geometry. And then I'll actually show you what a tetrahedron looks like here. So here's a quick sketch of what the molecule sort of looks like. And let me go ahead and draw tetrahedron over here so you can get a little bit better idea of the shape, all right? So four sided figure. And so there you go. Something like that. So you could think about the corners of your tetrahedron as being approximately where your hydrogens are of that tetrahedron, that four sided figure here. And so we've created the geometry of the electron clouds around our central atom. And in step four, we ignore any lone pairs around our central atom, which we have none this time. And so therefore, the geometry the molecule is the same as the geometry of our electron pairs. So we can say that methane is a tetrahedral molecule like that. All right, in terms of bond angles. So our goal now is to figure out what the bond angles are in a tetrahedral molecule. Turns out to be 109.5 degrees in space. So that's having those bonding electrons as far away from each other as they possibly can using VSEPR theory. So 109.5 degrees turns out to be the ideal bond angle for a tetrahedral molecule. Let's go ahead and do another one. Let's look at ammonia. So we have NH3." + }, + { + "Q": "At 4:15, why is the y-component 120 sin 30 degrees ?", + "A": "The y- and x-component of the velocity is computed with trig. Go back to the 1-dimensional videos for a full explanation of how to use trig to break a vector into it s vertical and horizontal components. Sal does a good job of explaining how to break a vector into it s components, and why it is useful to do so in those earlier videos.", + "video_name": "jl_gQ-eL3xo", + "timestamps": [ + 255 + ], + "3min_transcript": "Its' a 30-degree angle to the horizontal. And there's a fence 350 feet away that's 30 feet high. It's roughly around there. That's 30. And what we need to do is figure out whether the ball can clear the fence. We figured out the last time when we used the unit vector notation that it doesn't clear the fence. But in this problem, or the second part of this problem, they said that there's a 5 meter per second wind gust to the right. So there's a wind gust of 5 meters per second right when I hit the ball. And you could go into the complications of how much does that accelerate the ball? Or what's the air resistance of the ball? I think for the simplicity of the problem, they're just saying that the x-component of the ball's velocity right after you hit it increases by 5 meters per second. I think that's their point. that we did it the last time, but we'll use a different notation. So we can write that equation that I had written before, that the position at any given time as a function of t is equal to the initial position-- that's an i right there-- plus the initial velocity. These are all vectors. Initial velocity times t plus the acceleration vector over 2t squared. So what's the initial position? And now we're going to use some of our new notation. The initial position when I hit the ball, its x-component is 0, right? It's almost like its coordinate, and they're not that different of a notation. And then the y-position is 4. Easy enough. Let me do it. So we can split it up into the x- and the y-components. The y-component is 120 sine of 30 degrees and then the x component is 120 cosine of 30 degrees. That's just the x-component after I hit it. But then they say there's this wind gust so it's going to be plus 5. I think that's their point when they say that there's this wind gust. They say that right when you hit it, for some reason in the x-direction, it accelerates a little bit by 5 meters per second. So the velocity vector. This notation actually is better, because it takes less space up, and you don't have all these i's and j's and pluses confusing everything. So the initial velocity vector, what's its x-component? It's 120 cosine of 30. Cosine of 30 is square root of 3/2 times 120 is 60 square roots of 3, and then you add 5 to it. Let me just solve it right now." + }, + { + "Q": "At 1:20, I believe there is a mistake. RBCs do not contain a nucleus.", + "A": "That is not a RBC and you re correct, they shed their nuclei before they enter circulation. What he is drawing is endothelial cells lining the blood vessel. Hope this helped! :)", + "video_name": "RQpBj8ebbNY", + "timestamps": [ + 80 + ], + "3min_transcript": "Let's look at a blood vessel. A blood vessel is kind of like a tube. I think you'll agree with me. It's a tube through which blood travels. So here I'm drawing a tube, and I'm gonna ask you a question which is what makes up the walls of this tube? Now, for a blood vessel, what makes it up, is something called an endothelial cell. So, actually, the walls are made up of these kind of gooey endothelial cells that are stuck together tightly and that together form a tube through which the blood will travel. So here I'm drawing a bunch of cells tightly stuck together. They're stuck together tightly to prevent blood from coming out, of course. So this is more or less what it looks like. Each of these is a cell, and so, of course, each one has a nucleus, which I'll just quickly draw like that. Now let's delete that and let's change views So now here is the same blood vessel in cross section. And so these are the endothelial cells, which we're seeing in cross section, and maybe here we can draw the nucleus of this one. Maybe it's visible right there. So now we have blood moving through this blood vessel. Here are some red blood cells, and they're moving along, providing the body with oxygen. But a very important question to ask is what happens if this blood vessel gets damaged? So let's say that those two cells right there split open, and they break open. What's gonna happen if we don\u2019t fix this is that all of our blood is just going to rush out here, and we're gonna lose blood. So what is your body gonna do about this? Well, it's going to use a special player that we haven't talked about yet, which is the platelet. So I'm drawing a platelet here. It doesn't have a nucleus or anything. It's a tiny, little piece of a cell that your body uses to block up holes like this. So you have them floating around in your blood all the time. Here I'm drawing a bunch. And what happens is when you have a hole in your blood vessel, they're going to come together. They're going to stick together, and they're going to clog up this hole. And so, basically, they've built a little barrier so that we won't keep losing all our blood. So are you satisfied? Well, you shouldn\u2019t really be satisfied because there's a bit question here which is why are these platelets clumping here, and why aren't they clumping, you know, for example up here? Why don\u2019t they clump there or maybe even just in circulation? Why don\u2019t they clump up like this? Because if what they do is to clump up, how would they know to clump up here and not here? What's telling them? These are things that we don\u2019t want to happen" + }, + { + "Q": "AT 14:13\nIt is said, that the wavelength of that standing wave can be found using the formula\nWavelength (lambda) = 2* (Length of the tube)/ n , where n=1,2,3,.....\nBut, looking at the graph carefully, we can take n= no. of nodes...\nCan'nt we?", + "A": "If we take n as no. of nodes, we just have to be careful! It works for this case, but not for all of them, ex. if you use n as number of nodes for standing waves on strings, it will not work out :)", + "video_name": "BhQUW9s-R8M", + "timestamps": [ + 853 + ], + "3min_transcript": "And so, how much of a wavelength is this? Let's try it out. Let's reference our one wavelength. So it starts at the bottom, and then it goes all the way up to the top, and then it goes all the way down to the bottom, but this one keeps going. This is more than a whole wavelength. Because that's just this part. That's one whole wavelength. Now I got to go all the way back up to the top. So this wave is actually one wavelength and a half. This amount is one extra half of a wavelength, so this was one wavelength and a half. So in this case, L, the total distance of the tube, that's not changing here. The total distance of the tube is L. This time, the wavelength in there is fitting, and one and 1/2 wavelengths fit in there. That's 3/2 of a wavelength. That means the lambda equals two L over three. So in this case, for lambda three, this is going to be called the third harmonic. This should be two L over three. And so, it keeps going. You can have the fourth harmonic, fifth harmonic, every time you add one more node in here, it's always got to be antinode on one end, antinode at the other. These are the possible wavelength, and if you wanted the possible, all of the possible ones, you can probably see the pattern here. Look: two L, and then just L, then two L over three, the next one turns out to be two L over four, and then two L over five, two L over six, and so, if you wanted to just write them all down, shoot ... lambda n equals -- this is all the possible wavelengths -- two L over n, where n equals one, two, three, four, and so on. And so, look at, if I had 'n equals one' in here, I'd have two L. That's the fundamental. You plug in n equals one, you get the fundamental. If I plug in n equals two, I get two L over two. That's just L. That's my second harmonic, because I'm plugging in n equals two. that's my third harmonic. This is telling me all the possible wavelengths that I'm getting for this standing wave. So that's open, open. In the next video, I'm going to show you how to handle open, closed tubes." + }, + { + "Q": "Aren't you supposed to write solute -> lowers FREEZING point at 3:17?", + "A": "Its true, check the clarifications section.", + "video_name": "z9LxdqYntlU", + "timestamps": [ + 197 + ], + "3min_transcript": "they're just vibrating in place. So you have to get a little bit orderly right there, right? And then, obviously, this lattice structure goes on and on with a gazillion water molecules. But the interesting thing is that this somehow has to get organized. And what happens if we start introducing molecules into this water? Let's say the example of sodium-- actually, I won't do any example. Let's just say some arbitrary molecule, if I were to introduce it there, if I were to put something-- let me draw it again. So now I'll just use that same -- I'll introduce some molecules, and let's say they're pretty large, so they push all of these water molecules out of the way. So the water molecules are now on the outside of that, and let's have another one that's over here, some relatively large molecules of solute relative to water, and this is because a water molecule really isn't that big. Now, do you think it's going to be easier Are you going to have to remove more or less energy to get to a frozen state? Well, because these molecules, they're not going to be part of this lattice structure because frankly, they wouldn't even fit into it. They're actually going to make it harder for these water molecules to get organized because to get organized, they have to get at the right distance for the hydrogen bonds to form. But in this case, even as you start removing heat from the system, maybe the ones that aren't near the solute particles, they'll start to organize with each other. But then when you introduce a solute particle, let's say a solute particle is sitting right here. It's going to be very hard for someone to organize with this guy, to get near enough for the hydrogen bond to start taking hold. This distance would make it very difficult. And so the way I think about it is that these solute particles make the structure irregular, or they add more disorder, and we'll eventually talk about entropy and all of that. and it's making it harder to get into a regular form. And so the intuition is that this should lower the boiling point or make it -- oh, sorry, lower the melting point. So solute particles make you have a lower boiling point. Let's say if we're talking about water at standard temperature and pressure or at one atmosphere then instead of going to 0 degrees, you might have to go to negative 1 or negative 2 degrees, and we're going to talk a little bit about what that is. Now, what's the intuition of what this will do when you want to go into a gaseous state, when you want to boil it? So my initial gut was, hey, I'm already in a disordered state, which is closer to what a gas is, so wouldn't that make it easier to boil? But it turns out it also makes it harder to boil, and this is how I think about it. Remember, everything with boiling deals with what's happening at the surface," + }, + { + "Q": "to be clear, when the solute is added, the boiling point is lower, and when the solute is added it creates a lower vapor pressure. What does vapor pressure have to do with the boiling point because that was mentioned at about 5:45", + "A": "The standard boiling point of a substance is the temperature at which the vapour pressure of the substance is equal to the external pressure. The vapour pressure is decreased when a solute is added and thus, a higher temperature is required to bring the vapour pressure to be equal to the external pressure. This, we have a higher boiling point.", + "video_name": "z9LxdqYntlU", + "timestamps": [ + 345 + ], + "3min_transcript": "So at the surface, we said if I have a bunch of water molecules in the liquid state, we knew that although the average temperature might not be high enough for the water molecules to evaporate, that there's a distribution of kinetic energies. And some of these water molecules on the surface because the surface ones might be going fast enough to escape. And when they escape into vapor, then they create a vapor pressure above here. And if that vapor pressure is high enough, you can almost view them as linemen blocking the way for more molecules to kind of run behind them as they block all of the other ambient air pressure above them. So if there's enough of them and they have enough energy, they can start to push back or to push outward is the way I think about it, so that more guys can come in behind them. So I hope that lineman analogy doesn't completely lose you. Now, what happens if you were to introduce solute into it? It probably doesn't have much of an effect down here, but some of it's going to be bouncing on the surface, so they're going to be taking up some of the surface area. And because, and this is at least how I think of it, since they're going to be taking up some of the surface area, you're going to have less surface area exposed to the solvent particle or to the solution or the stuff that'll actually vaporize. You're going to have a lower vapor pressure. And remember, your boiling point is when the vapor pressure, when you have enough particles with enough kinetic energy out here to start pushing against the atmospheric pressure, when the vapor pressure is equal to the atmospheric pressure, you start boiling. But because of these guys, I have a lower vapor pressure. So I'm going to have to add even more kinetic energy, more heat to the system in order to get enough vapor pressure up here to start pushing back the atmospheric pressure. So solute also raises the boiling point. solute, when you add something to a solution, it's going to make it want to be in the liquid state more. Whether you lower the temperature, it's going to want to stay in liquid as opposed to ice, and if you raise the temperature, it's going to want to stay in liquid as opposed to gas. I found this neat -- hopefully, it shows up well on this video. I have to give due credit, this is from chem.purdue.edu/gchelp/solutions/eboil.html, but I thought it was a pretty neat graphic, or at least a visualization. This is just the surface of water molecules, and it gives you a sense of just how things vaporize as well. There's some things on the surface that just bounce off. And here's an example where they visualized sodium chloride at the surface. And because the sodium chloride is kind of bouncing around on the surface with the water molecules, fewer of those water molecules kind of have the room to escape, so the boiling point gets elevated." + }, + { + "Q": "at 6:04, sal says the solute lowers the vapour pressure but my question is if the solute particles occupy both the positions in surface as well as inside ,won't the solvent molecules (H2O) get dis-organised leading to more of the molecules escaping out of the system therby decreasing its boiling point ??", + "A": "Within the solvent and at surface, the solute molecules become surrounded by layers of associated water molecules, or shells of water of solvation. Formation of these shells reduces number of solvent (water) molecules that have enough kinetic energy to escape as a vapor, decreasing vapor pressure and increasing boiling point.", + "video_name": "z9LxdqYntlU", + "timestamps": [ + 364 + ], + "3min_transcript": "So at the surface, we said if I have a bunch of water molecules in the liquid state, we knew that although the average temperature might not be high enough for the water molecules to evaporate, that there's a distribution of kinetic energies. And some of these water molecules on the surface because the surface ones might be going fast enough to escape. And when they escape into vapor, then they create a vapor pressure above here. And if that vapor pressure is high enough, you can almost view them as linemen blocking the way for more molecules to kind of run behind them as they block all of the other ambient air pressure above them. So if there's enough of them and they have enough energy, they can start to push back or to push outward is the way I think about it, so that more guys can come in behind them. So I hope that lineman analogy doesn't completely lose you. Now, what happens if you were to introduce solute into it? It probably doesn't have much of an effect down here, but some of it's going to be bouncing on the surface, so they're going to be taking up some of the surface area. And because, and this is at least how I think of it, since they're going to be taking up some of the surface area, you're going to have less surface area exposed to the solvent particle or to the solution or the stuff that'll actually vaporize. You're going to have a lower vapor pressure. And remember, your boiling point is when the vapor pressure, when you have enough particles with enough kinetic energy out here to start pushing against the atmospheric pressure, when the vapor pressure is equal to the atmospheric pressure, you start boiling. But because of these guys, I have a lower vapor pressure. So I'm going to have to add even more kinetic energy, more heat to the system in order to get enough vapor pressure up here to start pushing back the atmospheric pressure. So solute also raises the boiling point. solute, when you add something to a solution, it's going to make it want to be in the liquid state more. Whether you lower the temperature, it's going to want to stay in liquid as opposed to ice, and if you raise the temperature, it's going to want to stay in liquid as opposed to gas. I found this neat -- hopefully, it shows up well on this video. I have to give due credit, this is from chem.purdue.edu/gchelp/solutions/eboil.html, but I thought it was a pretty neat graphic, or at least a visualization. This is just the surface of water molecules, and it gives you a sense of just how things vaporize as well. There's some things on the surface that just bounce off. And here's an example where they visualized sodium chloride at the surface. And because the sodium chloride is kind of bouncing around on the surface with the water molecules, fewer of those water molecules kind of have the room to escape, so the boiling point gets elevated." + }, + { + "Q": "3:50 (ish) Why would the OCH3 donate the e- to the hydrogen, instead of the e- going directly to the chlorine? Doesn't going to the H before the Cl just make this unnecessarily convoluted?", + "A": "Duh, wow. Sometimes I just miss the simplest things, I can t believe that went over my head haha thanks so much!", + "video_name": "J0gXdEAaSiA", + "timestamps": [ + 230 + ], + "3min_transcript": "And if you use the Lewis definition of a base, that means it really, really, really wants to give away this electron to something else. If you use the Bronsted-Lowry definition, this means that it really, really, really wants to take a proton off of something else. In this situation, that is exactly what it will do. I'll actually give you the most likely reaction to occur here, and we'll talk about other reactions, and why this is the most likely reaction in future videos. So it wants to nab a proton. It is a strong base. It wants to give away this electron. Let's say that it gives away this electron to this hydrogen right over here. Now, this hydrogen already had an electron. methoxide base, then it can give away its electron to the rest of the molecule. It can give away this electron to the rest of the molecule. Now, carbon won't need the electron. Carbon doesn't want to have a negative charge. Maybe simultaneously that electron goes to that carbon right over there. But once again, this carbon doesn't want it. It already has four bonds, but what we see is we have this chloro here. This is a highly electronegative group. The chlorine is very electronegative, so the whole time, the chlorine was already tugging on this electron right here. Now, all of a sudden, it's all happening simultaneously, when an electron becomes available to this carbon, this carbon doesn't need this electron anymore. The chlorine already wanted it, so now the chlorine can take the electron. And just like that, in exactly one step, let's think about If all of these simultaneous reactions occurred, what are we left with? Let me redraw my two chlorobutanes. I'm going to have to change it now. So now, the chlorine has disappeared. So we clear it. The chlorine has now left. This chlorine is now up here. It had this electron right over there, which is right over there. The other electron it was paired with that was forming a bond with is now also on the chlorine. It becomes a chloride anion. Let me draw the rest of the valence electrons. One, two, three, four, five, six, and it has the seventh one right here. One, two, three, four, five, six, seven, eight, so it now" + }, + { + "Q": "At 6:40+ does the Chloride bond with the Sodium at any point?", + "A": "no, remember that in solution the sodium and chloride would exist as ions anyways because of the ionic bonding", + "video_name": "J0gXdEAaSiA", + "timestamps": [ + 400 + ], + "3min_transcript": "This electron up here-- let me clear this part out as well. Now, this electron right here, this magenta electron, this is now given to this carbon. Let me draw it here. That magenta electron is now given to that carbon. And if we look at the other end, the one that it was paired with, that it was bonded with, it still is bonded with it, so that green electron is now still on that carbon and now they are bonded. Now, they're forming a double bond. This all of the sudden has become an alkene and now the methoxide took the hydrogen. Let me redraw the methoxide. OCH3, the oxygen had one, two, three, four, five. It had six. and then all of these six unpaired ones. Neutral oxygen has six. But now it gave one of them away to a hydrogen. It gave that green electron there to the hydrogen-- I'll make this hydrogen the same color-- to this hydrogen right over here, so now this hydrogen is now bonded with it. So what are we left with? We have a chloride anion, so this is chloride. We now have methanol. This was a strong base. Now it has become its conjugate acid. So now we have methanol, which is the same as the solvent, so it's now mixed in. And now we're left off with one, two, We still have four carbons, but now it's an alkene. We have a double bond, so we could call this but-2-ene, or sometimes called 2-butene. Let's think about what happened here. It all happened simultaneously. Both reactants were present. There was actually only one step so this is the rate-determining step. If we would try to name it, it would probably have a 2 in it someplace. And something got eliminated. The chloride, or I guess you could call it the chloro group, got eliminated. It was a leaving group in this situation. So this was eliminated, and this type of reaction where something is eliminated and both of the reactants are participating in the rate-determining step, and we only had one step here so that was the rate-determining step, is called an E2 reaction." + }, + { + "Q": "At 6:43, i'm not exactly sure what he means by \"faster than the speed of light\" because when you think about it, wouldn't the light travel that 1.5x10^8m with the ship and then travel the other 1.5x10^8m to be a total of 3.0x10^8m away from Sal in one second?", + "A": "When you measure the speed of light when traveling through a vacuum it will always travel at 3 * 10^8 m/s regardless of the velocity of the source of the light when it was emitted. So if you have a rocket going 1.5 * 10^8 m/s and they fire a laser at you they see the light as traveling at 3 * 10^8 m/s and you see it traveling at 3 * 10^8 m/s as well not 4.5 * 10^8 m/s.", + "video_name": "OIwp8m3S30c", + "timestamps": [ + 403 + ], + "3min_transcript": "or it looks like the velocity of that photon is one and a half times 10 to the 8th meters per second in the positive x direction. And this should hopefully makes sense from a Newtonian point of view, or a Galilean point of view. These are called Galilean transformations because if I'm in a car and there's another car and you see this on the highway all the time, if I'm in a car going 60 miles per hour, there's another car going 65 miles per hour, from my point of view, it looks like it's only moving forward at five miles per hour. So that photon will look slower to Sally. Similarly, if we assume this Newtonian, this Galilean world, if she had a flashlight, if she had a flashlight right over here and right at time equals zero she turned it on, and that first photon we were to plot it on her frame of reference, well, it should go the speed of light So it starts here at the origin. And then after one second, in the s prime, in the s prime coordinates, it should have gone three times 10 to the 8th meters. After two seconds, it should've gone six times 10 to the 8th meters. And so it's path on her space-time diagram should look like that. That's her photon, that first photon that was emitted from it. So you might be noticing something interesting. That photon from my point of view is going faster than the speed of light. After one second, its x coordinate is 4.5 times 10 to the 8th meters. It's going 4.5 times 10 to the 8th meters per second. It's going faster than the speed of light. It's going faster than my photon, and that might make intuitive sense except it's not what we actually observe in nature. And anytime we try to make a prediction that's not what's observed in nature, it means that our understanding of the universe is not complete inertial reference frame we are in, the speed of light, regardless of the speed or the relative velocity of the source of that light, is always going three times 10 to the 8th meters per second. So we know from observations of the universe that Sally, when she looked at my photon, she wouldn't see it going half the speed of light, she would see it going three times 10 to the 8th meters per second. And we know from observations of the universe that Sally's photon, I would not observe it as moving at 4.5 times 10 to the 8th meters per second, that it would actually still be moving at three times 10 to the 8th meters per second. So something has got to give. This is breaking down our classical, our Newtonian, our Galilean views of the world. It's very exciting. We need to think of some other way to conceptualize things, some other way to visualize these space-time diagrams for the different frames of reference." + }, + { + "Q": "7:57 i thought the OH takes place where the O was originally at , i mean where the C is before it", + "A": "The O hasn t really moved, it s just drawn in a different position. It s still bonded to the same carbon as before.", + "video_name": "rNJPNlgmhbk", + "timestamps": [ + 477 + ], + "3min_transcript": "So this one and this one. So the reaction happened twice. So if you're doing this reaction with a carboxylic acid, it's a similar mechanism. We don't have time to go through it. But you're going to end up with the same product. You're going to add on two hydrogens on to that original carbonyl carbon. Like that. So let's look at the chemoselectivity of this reaction. So now that we've covered sodium borohydride and lithium aluminum hydride, let's see how you can choose which one of those reagents is the best to use. So if I start here with our reactants-- so let's make it a benzene ring. Like that. And let's put stuff on the benzene ring. So let's go ahead and put a double bond here. And then, we'll make this an aldehyde functional group on one end. And then over here on this end, I'm going to put an ester. Like that. transform different parts of this molecule using different reagents. So let's say we were to do a reaction wherein we add on sodium borohydride. And then, the proton source in the second step. So we need to think about what's going to happen. Sodium borohydride is selective for aldehydes and ketones only. It will not reduce carboxylic acids or esters. So it's only going to react with the aldehyde at the top right portion of this molecule. So let's see if we can draw this in here. So it's going to react with the aldehyde in the top right portion. So we are still going to have our double bond here. And the aldehyde's going to go away to form a primary alcohol. So we're going to get primary alcohol where the aldehyde used to be. Sodium borohydride has reduced that carbonyl. forms your primary alcohol as your product. And the rest of the molecule's going to stay the same. So this ester is going to remain untouched down here. So it's the chemoselectivity of that reaction. Let's say we start with the same starting material. And the first step-- this time we add lithium aluminum hydride. Like that. And the second step-- we add some water. Well, lithium aluminum hydride will reduce aldehydes and ketones, and it will also reduce esters. So lithium aluminum hydride in excess-- so let's just assume this is at an excess here-- it's going to react with this aldehyde portion of the molecule. It's also going to react with this ester portion of the molecule. So it's going to reduce both of those and form alcohol. So let's go ahead and try to draw the product here. So we have our benzene ring, which is untouched. And up here, we know that lithium aluminum hydride" + }, + { + "Q": "At 7:26, Sal describes the fluid part as \"Although we can't call it a liquid yet\". Aren't all fluids liquids, or is there a difference?", + "A": "No, not all fluids are liquids, but all liquids are fluids. Gases are fluids. Sand can act like a fluid.", + "video_name": "f2BWsPVN7c4", + "timestamps": [ + 446 + ], + "3min_transcript": "is kind of in this magma, this deformable, somewhat fluid state, and depending on what depth you go into the mantle there are kind of different levels of fluidity. And then the core, the outer level layer of the core, the outer core is liquid, because the temperature is so high. The inner core is made up of the same things, and the temperature is even higher, but since the pressure is so high it's actually solid. So that's why the mantle, crust, and core differentiations don't tell you whether it's solid, whether it's magma, or whether it's really a liquid. It just really tells you what the makeup is. Now, to think about the makeup, and this is important for plate tectonics, because when we talk about these plates we're not talking about just the crust. We're talking about the outer, rigid layer. Let me just zoom in a little bit. Let's say we zoomed in right over there. So now we have the crust zoomed in. This right here is the crust. talking about the upper mantle. We haven't gotten too deep in the mantle right here. So that's why we call it the upper mantle. Now, right below the crust, the mantle is cool enough that it is also in real solid form. So this right here is solid mantle. And when we talk about the plates were actually talking about the outer solid layer. So that includes both the crust and the solid part of the mantle. And we call that the lithosphere. When people talk about plate tectonics, they shouldn't say crustal plates. They should call these lithospheric plates. And then below the lithosphere you have the least viscous part of the mantle, but the pressure isn't so large as what will happen when you go into the lower part of the mantle that the fluid can actually kind of move past each other, although still pretty viscous. This still a magma. So this is still kind of in its magma state. And this fluid part of the mantle, we can't quite call it a liquid yet, but over large periods of time it does have fluid properties. This, that essentially the lithosphere is kind of riding on top of, we call this the asthenosphere. So when we talk about the lithosphere and asthenosphere we're really talking about mechanical layers. The outer layer, the solid layers, the lithosphere sphere. The more fluid layer right below that is the asthenosphere. When we talk about the crust, mantle, and core, we are talking about chemical properties, what are the things actually made up of." + }, + { + "Q": "At around 1:30, Sal said that new land was forming. I don't get what he means by that. And also, he said that the land is pushing the two plates apart. What makes the land push the two plates apart? Is it the inside of the earth?", + "A": "He means that when the plates get pushed apart magma comes up and cools creating more land also pushing the plates apart because of the growing land. The intense heat from the core moves the mantle moving the plates.", + "video_name": "f2BWsPVN7c4", + "timestamps": [ + 90 + ], + "3min_transcript": "What I want to do in this video is talk a little bit about plate tectonics. And you've probably heard the word before, and are probably, or you might be somewhat familiar with what it discusses. And it's really just the idea that the surface of the Earth is made up of a bunch of these rigid plates. So it's broken up into a bunch of rigid plates, and these rigid plates move relative to each other. They move relative to each other and take everything that's on them for a ride. And the things that are on them include the continents. So it literally is talking about the movement of these plates. And over here I have a picture I got off of Wikipedia of the actual plates. And over here you have the Pacific Plate. Let me do that in a darker color. You have the Pacific Plate. You have a Nazca Plate. You have a South American Plate. I could keep going on. You have an Antarctic Plate. It's actually, obviously whenever you do a projection onto two dimensions of a surface of a sphere, the stuff at the bottom and the top look much bigger than they actually are. Antarctica isn't this big relative It's just that we've had to stretch it out to fill up the rectangle. But that's the Antarctic Plate, North American Plate. And you can see that they're actually moving relative to each other. And that's what these arrows are depicting. You see right over here the Nazca Plate and the Pacific Plate are moving away from each other. New land is forming here. We'll talk more about that in other videos. You see right over here in the middle of the Atlantic Ocean the African Plate and the South American Plate meet each other, and they're moving away from each other, which means that new land, more plate material I guess you could say, is somehow being created right here-- we'll talk about that in future videos-- and pushing these two plates apart. Now, before we go into the evidence for plate tectonics or even some of the more details about how plates are created and some theories as to why the plates might move, what I want to do is get a little bit of the terminology of plate tectonics out of the way. and that's not exactly right. And to show you the difference, what I want to do is show you two different ways of classifying the different layers of the Earth and then think about how they might relate to each other. So what you traditionally see, and actually I've made a video that goes into a lot more detail of this, is a breakdown of the chemical layers of the Earth. And when I talk about chemical layers, I'm talking about what are the constituents of the different layers? So when you talk of it in this term, the top most layer, which is the thinnest layer, is the crust. Then below that is the mantle. Actually, let me show you the whole Earth, although I'm not going to draw it to scale. So if I were to draw the crust, the crust is the thinnest outer layer of the Earth. You can imagine the blue line itself is the crust. Then below that, you have the mantle. So everything between the blue and the orange line," + }, + { + "Q": "At 10:07, what does Kw stand for?", + "A": "Kw= (1.0E-14) equilibrium constant for {H3O][OH-] Ka=(1.0E-7) [H3O] + C-base Kb=(1.0E-7) [OH-] + C-acid. These vales are important in determination of PH and POH of a solution", + "video_name": "3Gm4nAAc3zc", + "timestamps": [ + 607 + ], + "3min_transcript": "So we get Ka, our acidic equilibrium concentration, times Kb is equal to our hydrogen concentration times our hydroxide concentration. Remember, this is all in an aqueous solution. What do we know about this? What do we know about our hydrogen times our hydroxide concentration in an aqueous solution? For example, let me review just to make sure I'm jogging your memory properly. We could have H2O. It can autoionize into H plus. Plus OH minus. And this has an equilibrium. You just put the products. So the concentration of the hydrogen protons And you don't divide by this because it's the solvent. And we already figured out what this was. If we have just completely neutral water, this is 10 to the minus 7. And this is 10 to the minus 7. So this is equal to 10 to the minus 14. Now, these two things could change. I can add more hydrogen, I could add more hydroxide. And everything we've talked about so far, that's what we've been doing. That's what acids and bases do. They either increase this or they increase that. But the fact that this is an equilibrium constant means that, look, I don't care what you do to this. At the end of the day, this will adjust for your new reality of hydrogen protons. And this will always be a constant. As long as we're in an aqueous solution, a solution of water where water is a solvent at 25 degrees. I mean, in just pure water it's 10 to the minus 7. in an aqueous solution, the product is always going to be So that's the answer to this question. This is always going to be 10 to the minus 14. If you multiply hydrogen concentration times OH concentration. Now they won't each be 10 to the minus 7 anymore, because we're dealing with a weak acid or a weak base. So they're actually going to change these things. But when you multiply them, you're still going to get 10 to the minus 14. And let's just take the minus log of both sides of that. Let me erase all this stuff I did down here. I'll need the space. Let's say we take the minus logs of both sides of this equation. So you get the-- let me do a different color-- minus log, of course it's base 10, of Ka. Let me do it in the colors. Ka times Kb is going to be equal to" + }, + { + "Q": "At 5:28, how do you know that it can reach octet? Is there a proof or a way to determine? Does it have 8 valence electrons just because it is in the second period? Then what about other periods?", + "A": "Experiments tell us this is how atoms behave", + "video_name": "p7Fsb21B2Xg", + "timestamps": [ + 328 + ], + "3min_transcript": "will have seven valence electrons. And I have four of them. So 7 times 4 gives me 28 valence electrons for my fluorine. The total number valence electrons for my molecule will be 28 plus 4. So I have to account for 32 valence electrons when I draw this dot structure. So let's go ahead and move on to the next step. Let's go back up here and look at our guidelines. So we figured out how many valence electrons we need to account for for our dot structure. We don't have any kind of charges, so we don't need to worry about the rest of step one here. We move on to step two, where we decide on the central atom of our dot structure. And the way to do this is to pick the least electronegative element that we have here, and then draw the bonds. And so for our example, we're working with silicon and fluorine. And so we can go ahead and find those again on our periodic table. Here's fluorine. Fluorine is the most electronegative element, and so therefore, for silicon tetrafluoride, we're going to put the silicon atom at the center of our dot So I'm going to start with silicon here. And I know that silicon has four bonds to fluorine atoms. I'm going to go ahead and put in some fluorines right here. So here's some fluorines like that. So I just drew four covalent bonds, and we know that each covalent bond represents two valence electrons, right? So here's two valence electrons, here's two, so that's a total of four, six, and eight. So we've represented eight valence electrons so far in our dot structure. So we originally had to represent 32. So I'm going to go ahead and subtract 8 from 32. So 32 minus 8 gives me 24. So now I only have to account for 24 valence electrons. Let's go back up and look at our steps again. So let's find out where we are. So we've decided the central atom, and we've drawn the bonds, and we just subtracted the electrons that we used to draw those bonds from the total that we got in step one. So we're on to step three, where we assign the leftover So in this case, the terminal atoms would be the fluorines. So let's go back down here and look at our dot structure. So fluorine would be the terminal atoms, and we're going to assign electrons to those fluorines. But how many do we need to assign? Well, going back to our periodic table over here, so fluorine is in the second period. So pretty good bet it's going to follow the octet rule here. So we need to surround each fluorine atom with eight electrons. Each fluorine already has two electrons around it, so I'm going to go ahead and put six more around each fluorine, like that. So each fluorine get six more valence electrons. And since I'm assigning six valence electrons to four fluorines, 6 times 4 gives me 24. And so therefore we've now accounted for all of the valence electrons. And so this should be the should be the final dot structure here. And so we don't even need to go on to step four for this molecule. This is a very simple molecule to draw." + }, + { + "Q": "At 3:59, I don't exactly understand what you mean when you say Cytosol is the \"Fluid between the organelles.\"", + "A": "Hi, Dan! I get what you mean... So basically, suppose, that all the organelles are swimming in this swimming pool.... And instead of the water, there s something called Cytoplasm. Now, pretend that these organelles decide to leave it and get out of the pool. Now, because the organelles are out, the swimming pool changes its name to Cytosol. In a more realistic world, when all the cell organelles are taken out , the fluid that remains is called Cytosol. Hope that helps! TQ, Cookie!", + "video_name": "6UqtgH_Zy1Y", + "timestamps": [ + 239 + ], + "3min_transcript": "You have the DNA, you have the mRNA. It's all in here, this big jumble of chromatin inside the nucleus. How does it make its way outside of this double bilipid layer? And the way it makes its way out is through nuclear pores. So a nuclear pore is essentially a tunnel. And there are thousands of these. It's a tunnel through this bilipid layer. So the tunnel is made up of a bunch of proteins. So this right over here-- and this is kind of a cross section of it. But you could almost imagine it if you're thinking of it in three dimensions, you would imagine a tunnel. A protein-constructed-- a tunnel made out of proteins that goes through this double bilipid membrane. And so the mRNA can make its way out and get to a free ribosome, But this right over here is not the complete picture. Because when you translate a protein using a free ribosome, this is for proteins that are used inside the cell. So let me draw the entire cell right over here. This is the cell. This right over here is the cytosol of the cell. And you might be sometimes confused with the term cytosol and cytoplasm. Cytosol is all the fluid between the organelles. Cytoplasm is everything that's inside the cell. So it's the cytosol and the organelles and the stuff inside the organelles is the cytoplasm. So cytoplasm is everything inside of the cell. Cytosol is just the fluid that's between the organelles. is good for proteins used within the cell itself. The proteins can then float around the cytosol and used in whichever way is appropriate. But how do you get protein outside of the cell, or even inside the cellular membrane? Not within it, within the cell, but embedded in the cell membrane or outside of the cell itself. And we know that cells communicate in all sorts of different ways and they produce proteins for other cells or for use in the bloodstream, or whatever it might be. And that's what we're going to focus on in this video. So contiguous with this what's called a perinuclear space right over here, so the space between these two membranes-- So you have this perinuclear space between the inner and outer nuclear membrane. Let me just label that. That's the inner nuclear membrane. That's the outer nuclear membrane. You could continue this outer nuclear membrane, and you get into these kind of flaps and folds and bulges." + }, + { + "Q": "At 0:24, are proteins the same thing as amino acids if you were solving codons?", + "A": "Proteins are amino acids. Protein is made up of long chains of amino acid.", + "video_name": "6UqtgH_Zy1Y", + "timestamps": [ + 24 + ], + "3min_transcript": "We've already talked about the process from going from DNA to messenger RNA. And we call that process transcription. And this occurs in the nucleus. And then that messenger RNA makes its way outside of the nucleus, and it attaches to a ribosome. And then it is translated into a protein. And so you could say that this part right over here, this is being facilitated by a ribosome. Or it's happening at a ribosome. With that high-level overview, I now want to think a little bit in more detail about how this actually happens, or the structure of things where this happens inside of a cell. And so I'm going to now draw the nucleus in a little bit more detail so that we can really see what's happening on its membrane. So this right over here is the nucleus. Actually, let me draw it like this. I'm going to draw it with two lines. Because it's actually a double bilipid membrane. So this is one bilipid layer right over here. And then this is another one right over here. And I'm obviously not drawing it to scale. I'm drawing it so you can get a sense of things. So each of these lines that I'm drawing, if I were to zoom in on this-- so if I were to zoom in on each of these lines, so let's zoom in. And if I got a box like that, you would see a bilipid layer. So a bilipid layer looks like this. You have the circle is a hydrophilic end and those lines are the fatty hydrophobic ends. So that's our bilipid layer. So that's each of these lines that I have drawn, each of them are a bilipid layer. So the question is, well, how does the mRNA-- obviously You have the DNA, you have the mRNA. It's all in here, this big jumble of chromatin inside the nucleus. How does it make its way outside of this double bilipid layer? And the way it makes its way out is through nuclear pores. So a nuclear pore is essentially a tunnel. And there are thousands of these. It's a tunnel through this bilipid layer. So the tunnel is made up of a bunch of proteins. So this right over here-- and this is kind of a cross section of it. But you could almost imagine it if you're thinking of it in three dimensions, you would imagine a tunnel. A protein-constructed-- a tunnel made out of proteins that goes through this double bilipid membrane. And so the mRNA can make its way out and get to a free ribosome," + }, + { + "Q": "3:55 If cytoplasm is everything inside cell, can we say that a cell consist of cytoplasm and cell membrane?", + "A": "no..cytoplasm is everything inside cell...everything such as cytosqueleton, cytoplasm, all organelles including nucleus, mitochondria, vesihecles and such... hope it helped", + "video_name": "6UqtgH_Zy1Y", + "timestamps": [ + 235 + ], + "3min_transcript": "You have the DNA, you have the mRNA. It's all in here, this big jumble of chromatin inside the nucleus. How does it make its way outside of this double bilipid layer? And the way it makes its way out is through nuclear pores. So a nuclear pore is essentially a tunnel. And there are thousands of these. It's a tunnel through this bilipid layer. So the tunnel is made up of a bunch of proteins. So this right over here-- and this is kind of a cross section of it. But you could almost imagine it if you're thinking of it in three dimensions, you would imagine a tunnel. A protein-constructed-- a tunnel made out of proteins that goes through this double bilipid membrane. And so the mRNA can make its way out and get to a free ribosome, But this right over here is not the complete picture. Because when you translate a protein using a free ribosome, this is for proteins that are used inside the cell. So let me draw the entire cell right over here. This is the cell. This right over here is the cytosol of the cell. And you might be sometimes confused with the term cytosol and cytoplasm. Cytosol is all the fluid between the organelles. Cytoplasm is everything that's inside the cell. So it's the cytosol and the organelles and the stuff inside the organelles is the cytoplasm. So cytoplasm is everything inside of the cell. Cytosol is just the fluid that's between the organelles. is good for proteins used within the cell itself. The proteins can then float around the cytosol and used in whichever way is appropriate. But how do you get protein outside of the cell, or even inside the cellular membrane? Not within it, within the cell, but embedded in the cell membrane or outside of the cell itself. And we know that cells communicate in all sorts of different ways and they produce proteins for other cells or for use in the bloodstream, or whatever it might be. And that's what we're going to focus on in this video. So contiguous with this what's called a perinuclear space right over here, so the space between these two membranes-- So you have this perinuclear space between the inner and outer nuclear membrane. Let me just label that. That's the inner nuclear membrane. That's the outer nuclear membrane. You could continue this outer nuclear membrane, and you get into these kind of flaps and folds and bulges." + }, + { + "Q": "At 0:28 what is the bronchai", + "A": "the bronchi is the tubes that air goes through either the right or left lung. there are two types of bronchi, primary bronchi which are the first 2 tubes that creates a passageway to the lungs and the secondary bronchi which are the the tubes that branches of the primary bronchi. after the secondary bronchi, the bronchioles. the last one would be arterioles. hoped this helped", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 28 + ], + "3min_transcript": "" + }, + { + "Q": "2:30 what is diffusion", + "A": "Diffusion is the moment of a solute from a region of high concentration to one of a low concentration.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 150 + ], + "3min_transcript": "" + }, + { + "Q": "At 13:15 Sal says that \"They sop up 98.5% of oxygen \" what happens to the remaining 1.5% of oxygen ?", + "A": "They pretty much stay wherever it was the other oxygen was sopped up from. If the oxygen was taken from the blood plasma, which is what I understood from the video, then the 1.5% will just remain in the blood plasma.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 795 + ], + "3min_transcript": "" + }, + { + "Q": "at 1:36 , Sal said that we take in 78% nitrogen , but i know that we cannot breath in nitrogen and we take it from plants because they are the only living organisms that can take in nitrogen ?", + "A": "In fact we do dissolve a very tiny amount of nitrogen in our blood from the air we inhale. And the more the higher pressure around us. This is significant for divers who breathe under Deep Water. If a diver ascends to quickly nitrogen bubbles can be released in the bloodstream because the pressure diminishes too quickly to release the surplus of nitrogen throug normal breathing. This condition is leathal unless the diver very quickly is put in a pressuretank and is de-compressed slowly.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 96 + ], + "3min_transcript": "" + }, + { + "Q": "If red blood cells have DNA then why did he say they didn't. I don't understand. 14:26", + "A": "they do have dna but they push it out to carry more hemoglobin, if you watch the entire video.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 866 + ], + "3min_transcript": "" + }, + { + "Q": "At 14:35, Sal said that red blood cells don't have DNA, so they can't reproduce. At 15:55, Sal also said that they don't live long. I wonder who who makes those RBCs for us.", + "A": "At first, when produced in the bone marrow (most likely from the femur because it s the biggest bone in the body), the RBCs have nuclei. However, when they lose it the metabolism of the cell it s compromised and they live less then other cells.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 875, + 955 + ], + "3min_transcript": "" + }, + { + "Q": "At 10:27, Sal says that iron is the main component. Is that the reason why if you drink blood by accident it tastes like metal?", + "A": "Yes, the hemoglobin molecule has a big-old Iron atom embedded in the complex protein.", + "video_name": "fLKOBQ6cZHA", + "timestamps": [ + 627 + ], + "3min_transcript": "" + }, + { + "Q": "at 3:21 why do you have to do all problems in kelvin units", + "A": "The most common of the universal gas constants is in the units of ((L*atm)/(mol*K)), so in the instance of using that constant, you must be in kelvin. There are different constants for differing units, but Kelvin has the benefit of never being negative.", + "video_name": "69V-60sga3M", + "timestamps": [ + 201 + ], + "3min_transcript": "or force times one dimension of distance, which is joules, so this is in joules. So what the ideal gas constant essentially does is it converts -- we know that this is proportional to this, but it sets the exact constant of proportionality and it also makes sure we get the units right, so that's all it is. It just helps us translate from a world dealing with moles and Kelvins to a world of dealing with, well, in this case, atmospheres and liters, or bars and meters cubed, or kilopascals and meters cubed. But no matter what the units of the pressure and the volume are, the whole unit of pressure times volume is going to be joules. So it's just a translation between the two and we know that they're proportional. Now, with that said, let's do some more problems. So let's say we want to know how many grams of oxygen. So we want to know grams of O2. O2 in grams. that has a pressure of 12 atmospheres and the temperature is 10 degrees Celsius. Well, we've already broken out our ideal gas equation. Let's see, pressure is 12 atmospheres. So we can say 12. I'll keep the units there. 12 atmospheres times the volume. The volume should be in liters, so this is 300 milliliters, or 300 thousandths or 3 tenths. So that's 0.3 liters. 300 one thousandths of a liter is 0.3 liters. And that is equal to the number of moles we have of this, and that's what we need to figure out. If we know the number of moles, we know the number of grams. So this is equal to n times R. We're dealing with liters and atmospheres. We'll deal with this one. Times 0.082. And what's our temperature? It's 10 degrees Celsius. We always have to do everything in Kelvin. So it's 283 degrees Kelvin. So all we have to solve for is n. We just have to divide both sides of this equation by this right here, and so we have n is equal to 12 atmospheres -- I'm just swapping the sides -- times 0.3 liters divided by 0.082 times 283 degrees. Our answer will be in moles. And if you want to verify that, you can plug in all the units and use units for the ideal gas constant. The number of moles we're dealing with," + }, + { + "Q": "At 3:23, why do you have to change the temperature into kelvin?", + "A": "Because 0 on the Kelvin scale corresponds to 0 heat, but 0 on the celsius scale is arbitrary.", + "video_name": "69V-60sga3M", + "timestamps": [ + 203 + ], + "3min_transcript": "or force times one dimension of distance, which is joules, so this is in joules. So what the ideal gas constant essentially does is it converts -- we know that this is proportional to this, but it sets the exact constant of proportionality and it also makes sure we get the units right, so that's all it is. It just helps us translate from a world dealing with moles and Kelvins to a world of dealing with, well, in this case, atmospheres and liters, or bars and meters cubed, or kilopascals and meters cubed. But no matter what the units of the pressure and the volume are, the whole unit of pressure times volume is going to be joules. So it's just a translation between the two and we know that they're proportional. Now, with that said, let's do some more problems. So let's say we want to know how many grams of oxygen. So we want to know grams of O2. O2 in grams. that has a pressure of 12 atmospheres and the temperature is 10 degrees Celsius. Well, we've already broken out our ideal gas equation. Let's see, pressure is 12 atmospheres. So we can say 12. I'll keep the units there. 12 atmospheres times the volume. The volume should be in liters, so this is 300 milliliters, or 300 thousandths or 3 tenths. So that's 0.3 liters. 300 one thousandths of a liter is 0.3 liters. And that is equal to the number of moles we have of this, and that's what we need to figure out. If we know the number of moles, we know the number of grams. So this is equal to n times R. We're dealing with liters and atmospheres. We'll deal with this one. Times 0.082. And what's our temperature? It's 10 degrees Celsius. We always have to do everything in Kelvin. So it's 283 degrees Kelvin. So all we have to solve for is n. We just have to divide both sides of this equation by this right here, and so we have n is equal to 12 atmospheres -- I'm just swapping the sides -- times 0.3 liters divided by 0.082 times 283 degrees. Our answer will be in moles. And if you want to verify that, you can plug in all the units and use units for the ideal gas constant. The number of moles we're dealing with," + }, + { + "Q": "at 3:00, why does \"n\" stand for moles instead of \"m\"?", + "A": "n stands for number of molecules.", + "video_name": "69V-60sga3M", + "timestamps": [ + 180 + ], + "3min_transcript": "or force times one dimension of distance, which is joules, so this is in joules. So what the ideal gas constant essentially does is it converts -- we know that this is proportional to this, but it sets the exact constant of proportionality and it also makes sure we get the units right, so that's all it is. It just helps us translate from a world dealing with moles and Kelvins to a world of dealing with, well, in this case, atmospheres and liters, or bars and meters cubed, or kilopascals and meters cubed. But no matter what the units of the pressure and the volume are, the whole unit of pressure times volume is going to be joules. So it's just a translation between the two and we know that they're proportional. Now, with that said, let's do some more problems. So let's say we want to know how many grams of oxygen. So we want to know grams of O2. O2 in grams. that has a pressure of 12 atmospheres and the temperature is 10 degrees Celsius. Well, we've already broken out our ideal gas equation. Let's see, pressure is 12 atmospheres. So we can say 12. I'll keep the units there. 12 atmospheres times the volume. The volume should be in liters, so this is 300 milliliters, or 300 thousandths or 3 tenths. So that's 0.3 liters. 300 one thousandths of a liter is 0.3 liters. And that is equal to the number of moles we have of this, and that's what we need to figure out. If we know the number of moles, we know the number of grams. So this is equal to n times R. We're dealing with liters and atmospheres. We'll deal with this one. Times 0.082. And what's our temperature? It's 10 degrees Celsius. We always have to do everything in Kelvin. So it's 283 degrees Kelvin. So all we have to solve for is n. We just have to divide both sides of this equation by this right here, and so we have n is equal to 12 atmospheres -- I'm just swapping the sides -- times 0.3 liters divided by 0.082 times 283 degrees. Our answer will be in moles. And if you want to verify that, you can plug in all the units and use units for the ideal gas constant. The number of moles we're dealing with," + }, + { + "Q": "At 3:35 he said that when our cells gets broken by damage we feel pain, but how about slight pain for example if you poke slightly with a toothpick to your hand? There are no visible damage done but it causes slight sensation of pain or it just kills less cells and releases less proteins who causes pain?", + "A": "Dermic and hypodermic receptors, of which there are many, have a big impact on feeling slight damage/discomfort (e.g., toothpick example). There are also different kinds of nerves that carry different sensations (ex., tiny unmyelinated C fibers produce dull, burning ache while large myelinated axons convey quick, sharp pain) I d be happy to elaborate if you like.", + "video_name": "D-oAsFIHqbY", + "timestamps": [ + 215 + ], + "3min_transcript": "So this is the general idea behind a conformational change. So when heat is applied and also when pain applied via a particular molecule, you have a conformational change in the TrpV1 protein. So let's look at a diagram of a hand and go into this in a little bit more detail. OK, so here we have a hand. And as I mentioned before, we have cells located throughout the hand. And these cells are sensitive to temperature and to pain. And within these cells, there are TrpV1 receptors. So let's imagine that each one of these cells sends a little projection to a nerve that eventually reaches the brain. So these cells, whenever they are stimulated by either a change of temperature or the presence of some sort of painful stimulus-- So what can that be? So let's imagine that something pokes your hand. So let's imagine that we have a sharp object, and it pokes your hand. What happens is, the cell, when it gets poked, thousands of cells get broken up. So the cell gets broken up. And when it gets broken up, it releases all kinds of different molecules. And these molecules will travel around. So let's imagine it releases this little green molecule. It will travel around, and it will bind to one of the little TrpV1 receptors. And when it binds to a TrpV1 receptor, it causes the same conformational change that a change in temperature causes. And so that conformational change actually activates the cell, and the cell will send a signal to the brain. So this nerve over here actually contains three different types of fibers. So there are fast, medium, and slow fibers. So fast fibers are really, really fat in diameter. So we have these really big, fat fibers, and they have a lot myelin. So they are covered in myelin. And what myelin is, it's an insulator that basically allows the cell to conduct an action potential very quickly. So as an action potential or as a signal travels down the cell, if we have a lot of myelin surrounding the cell, the signal is able to travel really quickly. And another way that a signal is able to travel quickly is if the cell has a really big diameter. So a big diameter lowers the resistance. So you have less resistance and you have greater conductance because of the myelin. And these two things produce a very fast-- a cell that is able to produce-- send a signal pretty quickly." + }, + { + "Q": "At 5:45, Sal said that the initial velocity is 24.5 but earlier he said that the initial velocity is 0. i got mixed up. Can someone clarify it?", + "A": "Earlier in the video he said final velocity is 0 m/s. I think this must be where you got confused.", + "video_name": "IYS4Bd9F3LA", + "timestamps": [ + 345 + ], + "3min_transcript": "So the change in time is 2.5 s, times 2.5 s So what is our change in velocity which is also the same thing as negative of our initial velocity Get the calculator out, let me get my calculator, bring it on to the screen, so it is negative 9.8 m/s times 2.5 s Times 2.5 s, it gives us negative 24.5, so this gives us I will write it in new color This gives us negative 24.5 m/s, this seconds cancels out With one of these seconds in the denominator we only have one of the denominator out m/s, and this is the same thing as the negative, as the negative initial velocity Negative initial velocity that's the same thing as change in velocity So that simply we are able to figure out what our velocity is So literally you take the time, the total time in the air divide by two And multiply that by acceleration of gravity and if you take I guess you can take the absolute value of that or take the positive version of that And that gives you your initial velocity So your initial velocity here is literally 24.5 m/s Since it's a positive quantity it is upwards in this example So that's my initial velocity, so we already figure out part of this game The initial velocity that threw upward That's also going to be, we gonna have the same magnitude of velocity The balls about to hit the ground although is gonna be in the other direction So what is the distance or let me make it clear what is the displacement of the ball from its lowest point right when it leaves your hand We just have to remember, all of these come from very straight forward ideas Change in velocity is equal to acceleration times change in time And the other simple idea is that displacement is equal to average velocity, average velocity times change in time Now what is our average velocity? Our average velocity is your initial velocity plus your final velocity Divided by 2, or we assume that acceleration is constant So literally just the arithmetic mean of your initial and final velocity So what is that? That's gonna be 24.5 m/s plus our final velocity In this situation we are just going over to the first 2.5 s So our final velocity is once again 0 m/s We are just talking about when we get to his point over here" + }, + { + "Q": "What makes the time up the same as the time down? (Around 1:15)", + "A": "Acceleration remains constant throughout the flight so, when it has a velocity of zero (at the top), you can know it has to be at its halfway point during the flight. Hope that helps : )", + "video_name": "IYS4Bd9F3LA", + "timestamps": [ + 75 + ], + "3min_transcript": "Let's say you and I are playing a game or I'm trying to figure out how high a ball is being thrown in the air How fast would we throwing that ball in the air? And what we do is one of us has a ball and the other one has a stop watch over here So this is my best attempt to make it more like a cat than a stop watch but I think you get the idea And what we do is one of us throw the ball the other one times how long the ball is in the air And what we do is gonna use that time in the air to figure out how fast the ball was thrown straight up and how long it was in the air or how high it got And there is going to be one assumption I make here frankly that's an assumption we are gonna make in all of these projectile motion type problem is that air resistance is negligible And for something like it, this is a baseball or something like that That's a pretty good approximation So when can I get the exact answer, I encourage you experiment it on your own or even to see what air resistance does to your calculations We gonna assume for this projectile motion in future one We gonna assume air resistance is negligible And what that does for us is we can assume that the time up That the time for the ball to go up to its peak height is the same thing as the time that takes it to go down If you look at this previous video, we've plot it displacement verse time You see after 2 seconds the ball went from being on the ground or I guess the thrower's hand all the way to its peak height And then the next 2 seconds it took the same amount of time to go back down to the ground which makes sense whatever the initial velocity is, it take half the time to go to zero and it takes the same amount of time to now be accelerated into downward direction back to that same magnitude of velocity but now in the downward direction So let's play around with some numbers here Just so you get a little bit more of concrete sense So let's say I throw a ball in the air And you measure using the stop watch and the ball is in the air for 5s Well the first thing we could do is we could say look at the total time in the air was 5 seconds that mean the time, let me write it, that means the change in time to go up during the first half, I guess the ball time in the air is going to be 2.5 seconds and which tells us that over this 2.5 seconds we went from our initial velocity, whatever it was We went from our initial velocity to our final velocity which is a velocity of 0 m/s in the 2.5 seconds And this is a graph for that example, This is the graph for the previous one, The previous example we knew the initial velocity but in whatever the time is you are going from you initial velocity to be stationery at the top, right with the ball being stationery and then start getting increasing velocity in the downward direction So it takes 2.5 seconds to go from some initial velocity to 0 seconds" + }, + { + "Q": "is the velocity calculated by Average on 6:25 because both the rise and fall of the object is taken into question?", + "A": "No, Van. We are only dealing with the first part of the projectile-rising part, in which t=2.5s. This is because the falling part is essentially the reverse of rising part. We use average velocity because velocity is not constant, but acceleration is.", + "video_name": "IYS4Bd9F3LA", + "timestamps": [ + 385 + ], + "3min_transcript": "So the change in time is 2.5 s, times 2.5 s So what is our change in velocity which is also the same thing as negative of our initial velocity Get the calculator out, let me get my calculator, bring it on to the screen, so it is negative 9.8 m/s times 2.5 s Times 2.5 s, it gives us negative 24.5, so this gives us I will write it in new color This gives us negative 24.5 m/s, this seconds cancels out With one of these seconds in the denominator we only have one of the denominator out m/s, and this is the same thing as the negative, as the negative initial velocity Negative initial velocity that's the same thing as change in velocity So that simply we are able to figure out what our velocity is So literally you take the time, the total time in the air divide by two And multiply that by acceleration of gravity and if you take I guess you can take the absolute value of that or take the positive version of that And that gives you your initial velocity So your initial velocity here is literally 24.5 m/s Since it's a positive quantity it is upwards in this example So that's my initial velocity, so we already figure out part of this game The initial velocity that threw upward That's also going to be, we gonna have the same magnitude of velocity The balls about to hit the ground although is gonna be in the other direction So what is the distance or let me make it clear what is the displacement of the ball from its lowest point right when it leaves your hand We just have to remember, all of these come from very straight forward ideas Change in velocity is equal to acceleration times change in time And the other simple idea is that displacement is equal to average velocity, average velocity times change in time Now what is our average velocity? Our average velocity is your initial velocity plus your final velocity Divided by 2, or we assume that acceleration is constant So literally just the arithmetic mean of your initial and final velocity So what is that? That's gonna be 24.5 m/s plus our final velocity In this situation we are just going over to the first 2.5 s So our final velocity is once again 0 m/s We are just talking about when we get to his point over here" + }, + { + "Q": "At around 5:10, Sal found the initial velocity. Could you also use s=v_i*t+1/2a*t^2 and plug in 5 for t, -9.8 for a and 0 for s to find v_i? How does these two completely different methods get the same answer?", + "A": "Approaches that are algebraically the same can have different forms", + "video_name": "IYS4Bd9F3LA", + "timestamps": [ + 310 + ], + "3min_transcript": "We know that the acceleration We know the acceleration of gravity here, we are assuming it's constant or slightly not constant but we are gonna assume it's constant We are just dealing close to the surface of the earth is negative 9.8 m/s*s, so let's think about it This change in velocity, are change in velocity Their change in velocity is the final velocity minus the initial velocity which is the same thing as zero minus the initial velocity which is the negative of the initial velocity That's another way to think about change in velocity We just shown the definition of acceleration change in velocity is equal to acceleration, is equal to acceleration negative 9.8 m/s*s times time or times change in time, our change in time, So the change in time is 2.5 s, times 2.5 s So what is our change in velocity which is also the same thing as negative of our initial velocity Get the calculator out, let me get my calculator, bring it on to the screen, so it is negative 9.8 m/s times 2.5 s Times 2.5 s, it gives us negative 24.5, so this gives us I will write it in new color This gives us negative 24.5 m/s, this seconds cancels out With one of these seconds in the denominator we only have one of the denominator out m/s, and this is the same thing as the negative, as the negative initial velocity Negative initial velocity that's the same thing as change in velocity So that simply we are able to figure out what our velocity is So literally you take the time, the total time in the air divide by two And multiply that by acceleration of gravity and if you take I guess you can take the absolute value of that or take the positive version of that And that gives you your initial velocity So your initial velocity here is literally 24.5 m/s Since it's a positive quantity it is upwards in this example So that's my initial velocity, so we already figure out part of this game The initial velocity that threw upward That's also going to be, we gonna have the same magnitude of velocity The balls about to hit the ground although is gonna be in the other direction So what is the distance or let me make it clear what is the displacement of the ball from its lowest point right when it leaves your hand" + }, + { + "Q": "At 6:50 Sal talks about the energy of the system being 0. Under what conditions would the internal energy change?", + "A": "I had a look and could not see where he says the energy of the system is zero... However, the answer to your question is that internal energy = kinetic energy of the particles + potential energy of the particles. In an ideal gas, the potential energy changes are considered to be zero. Temperature is related to kinetic energy of the particles. So, a change in temp = change in KE = change in Internal energy", + "video_name": "aAfBSJObd6Y", + "timestamps": [ + 410 + ], + "3min_transcript": "Now, we've done a bunch of videos now. We said, OK, how much work was done on this system? Well, the work done on the system is the area under this curve. So some positive work was-- not done on the system, sorry. How much work was done by the system? We're moving in this direction. I should put the direction there. We're moving from left to right. The amount of work done by the system is pressure times volume. We've seen that multiple times. So you take this area of the curve, and you have the work done by the system from A to B. So let's call that work from A to B. Now, that's fair and everything, but what I want to think more about, is how much heat was transferred by my reservoir? Remember, we said, if this reservoir wasn't there, the temperature of my canister would have gone down as I expanded its volume, and as the pressure went down. So how much heat came into it? Well, let's go back to our basic internal energy formula. system minus the work done by the system Now, what is the change in internal energy in this scenario? Well, it was at a constant temperature the whole time, right? And since we're dealing with a very simple ideal gas, all of our internal energy is due to kinetic energy, which temperature is a measure of. So, temperature didn't change. Our average kinetic energy didn't change, which means our kinetic energy didn't change. So our internal energy did not change while we moved from left to right along this isotherm. So we could say our internal energy is zero. And that is equal to the heat added to the system minus the work done by the system. Right? So if you just-- we put the work done by the system on the other side, and then switch the sides, you get heat added to the system is equal to the work done by the system. And that makes sense. The system was doing some work this entire time, so it was essentially maybe some potential energy to these rocks. So it was giving energy away. It was giving energy outside of the system. So how did it maintain its internal energy? Well, someone had to give it some energy. And it was given that energy by this reservoir. So let's say, and the convention for doing this is to say, that it was given-- let me write this down. It was given some energy Q1. We just say, we just put this downward arrow to say that some energy went into the system here. Fair enough. Now let's take this state B and remove the reservoir, and completely isolate ourselves. So there's no way that heat can be transferred to and from our system. And let's keep removing some rocks. So if we keep removing some rocks, where do we get to? Let me go down here. So let's say we remove a bunch of more rocks." + }, + { + "Q": "Hi, at 10:02 Sal Khan says that catalysts lower activation energy but on some other websites and in books it is written that the catalysts do not lower the activation energy . What should I follow ?", + "A": "Catalysts provide an alternate reaction pathway with a lower activation energy. The uncatalyzed pathway still has the same activation energy.", + "video_name": "__zy-oOLPug", + "timestamps": [ + 602 + ], + "3min_transcript": "you don't always have to add it, but if it doesn't happen spontaneously you're gonna have to add some energy to the system to get to this activated state. RIght, so this is when we were at this thing right here. We're there. So some energy has to be in the system, and this energy, the difference between the energy we're at when we were just hydrogen molecules and iodine molecules, and the energy we have to get to to get this activated state, this distance right here this is the activation energy. If we're able to get to somehow put enough energy in the system, then this thing will happen, they'll collide with enough energy and bonds will be broken and reformed. Activation energy. Sometimes it's written as Ea energy of activation, and in the future we'll maybe do reactions where we actually measure the activation energy. But the important thing is to conceptually understand that it's there, that things just don't spontaneously go from here to here. but you've probably heard of the word catalyst or something being catalyzed. And that's something, some other agent, some other thing in the reactions. So right now, so right now we're doing, we have H2 plus I2, yielding 2H hydrogen iodides. Now you could have a catalyst, and I'll just say plus C. And I actually don't know what a good catalyst would be for this reaction, and how a catalyst operates is, it can actually operate in many many different ways, so that's why I don't wanna do it in this video. But what a catalyst is, is something that doesn't change. It doesn't get consumed in the reaction. The catalyst was there before the reaction, the catalyst is there after the reaction. But what it does is it makes the reaction happen either faster, or it lowers the amount of energy for the reaction to happen. Which is kind of the same thing. So if you have a catalyst, then this activation energy And what it does is, it makes it, it might easily, it might be some molecule that allows some other transition state that has less of a potential energy so that you require less heat or less concentration of the molecules for them to bump into each other in the right direction, to get to that other state. So you require less energy. So given how we understand how these kinetics occur, these molecules interact with each other, what do you think are the things that will drive whether a reaction happens or not? I mean we already know that if we have a positive catalyst, there's something called a negative catalyst that will actually slow down a reaction. But if we have a positive catalyst, it lowers... Obviously it lowers the activation energy, so this makes reaction faster. More molecules are gonna bump into each other just right to be able to get over this hump 'cause the hump will be lower, when you have a catalyst." + }, + { + "Q": "@5:07 Could you explain the activated complex a little bit clearer? I was confused whether it is the bond between diatomic molecules or whether it was a bond formed between molecules already bonded.", + "A": "It s more like a temporary phase (at a higher energy level) when the bond is first formed between the atoms of the reacting molecules, in this case an H atom of the H2 molecule and an I atom on an Iodine molecule. The activation complex is that moment when these two guys are trying to strengthen their bond and are JuSt breaking away from their original molecules to form HI . Hope I made sense to you", + "video_name": "__zy-oOLPug", + "timestamps": [ + 307 + ], + "3min_transcript": "The iodine might look something like this, it's a much bigger molecule. Where it's bonded together like this, it's also sharing some electrons in a covalent bond, everything's probabilistic. So in order for these two molecules to turn to this, somehow these bonds have to be broken, and new bonds have to be formed. And what has to happen is that these guys, there's a ton of these guys. I could draw a bunch of them. Or I could copy and paste. So there's a bunch of... There's a bunch of hydrogen molecules around, and some of these iodine gas molecules around. So what has to happen in order for us to get the hydrogen iodide is, they have to collide. And they have to collide in exactly the right way. So let's say this guy, wish I could show it. Let's say he's moving, this is neat, I'm just dragging and dropping. But he's moving. He has to hit this hydrogen molecule just right, And maybe just right, if he just happens to hit it then all of the sudden, let's say we get to this point right here, these electrons are gonna say \"Hey, you know, \"it's nice to be shared this way, \"we're in a stable configuration, \"we're filling the 1S shell, but look at this, \"there's this iodine that's close by \"and they really want me, \"they're much more electro-negative \"than me, the hydrogen\". So maybe they're kind of attracted here, they don't know whether they wanna be here between that hydrogen and this right here between that. And so they kind of enter this higher energy state. And similarly, you know these guys they say, \"Hey, wouldn't it be nicer, \"I don't have to be here, I could kind of go back home \"to my home atom if this guy comes in here\". Because then we're gonna have, then we're gonna have eight valence electrons and the same thing's happening here. And this complex right here, this kind of, right when the collision happens, this is actually a state, this is the high energy state of the transition state of the reaction, and this is called an activated complex. Sometimes you know I just drew it kind of visually, but you could draw it like this. So hydrogen has a covalent bond with another hydrogen, and then here comes along some iodine that has a covalent bond with some other iodine,, but all the sudden these guys like to bond as well. So they start forming, so there is kind of a you know, there's a little bit of an attraction on that side too. So this is another way of drawing the activation complex. But this is a high energy state, 'cause in order for the electrons, the way you can think of it, to kind of go from that bond to this bond, or this bond to that bond, or to go back, they have to enter into a higher energy state. A less stable energy state, than they were before. But they do that if there's enough energy, 'cause you can go from, so you're going from both of these things separate, let me just draw them separate," + }, + { + "Q": "At around 8:00 he says that one has to add energy if its not spontaneous, I thought that even spontaneous reactions do have an activation energy so energy has to be added?", + "A": "You re absolutely right that spontaneous reactions have activation energy -- what makes them spontaneous is that there s enough energy in the environment to overcome the activation energy. This is where conditions such as temperature and pressure can effect spontaneity (think delta H as it plays in to delta G, for example). Hope this helps!", + "video_name": "__zy-oOLPug", + "timestamps": [ + 480 + ], + "3min_transcript": "you have the hydrogen separate, plus the iodine separate. They go to this, which is a higher energy state. But if they can get to that higher energy state, if there's enough energy for the collision and they have enough kinetic energy when they hit in the right orientation, then, from this activated complex or this higher energy state, it will then go to the lowest energy state, and the lowest energy state is the hydrogen iodide. I wanna draw the iodide, and then the hydrogen. This is actually, this is actually a lower energy state than this. But in order to get here you have to go through a higher energy state. And I could do that with an energy diagram. So if we say that, let's say the X-axis is the progression of the reaction, we don't know how fast it's progressing, but this you can kind of view it as time on some dimension, and let's say this is the potential energy. I wanna draw thicker lines. See this is the potential energy. Right there, let me make this line thicker as well. So this is the potential energy. So initially, you are at this reality, and we can kind of view it as the combined potential energy, so this is where eventually, we start off here, and this is the H2 plus I2 And a lower potential energy is when we were in the hydrogen iodide. So this is the lower potential energy down here. Lower potential energy down here. This is the 2HI, right? But to get here, we have to enter this higher activation energy, where the electrons have to get, they have to have some energy to kind of be able to at least figure out what they wanna do with their lives. you don't always have to add it, but if it doesn't happen spontaneously you're gonna have to add some energy to the system to get to this activated state. RIght, so this is when we were at this thing right here. We're there. So some energy has to be in the system, and this energy, the difference between the energy we're at when we were just hydrogen molecules and iodine molecules, and the energy we have to get to to get this activated state, this distance right here this is the activation energy. If we're able to get to somehow put enough energy in the system, then this thing will happen, they'll collide with enough energy and bonds will be broken and reformed. Activation energy. Sometimes it's written as Ea energy of activation, and in the future we'll maybe do reactions where we actually measure the activation energy. But the important thing is to conceptually understand that it's there, that things just don't spontaneously go from here to here." + }, + { + "Q": "In 0:26 shouldn't it be one mole of hydrogen and one mole of iodine?", + "A": "Quite correct. It should be 1 mol of H\u00e2\u0082\u0082 and 1 mol of I\u00e2\u0082\u0082.", + "video_name": "__zy-oOLPug", + "timestamps": [ + 26 + ], + "3min_transcript": "- [Voiceover] When you're studying chemistry you'll often see reactions, in fact you always see reactions. For example if you have hydrogen gas it's a diatomic molecule, 'cause hydrogen bonds with itself in the gassy state, plus iodine gas, I2, that's also in the gassy state, it's very easy to just sort of, oh you know, if you put 'em together they're going to react and form the product, if you have two moles of, hydrogen, two moles of iodine, so it's gonna form two moles of hydrogen iodide. That's all nice and neat and it makes it seem like it's a very clean thing that happens without much fuss. But we know that that isn't the reality and we also know that this doesn't happen just instantly, it's not like you can just take some hydrogen, put it with some iodine, and it just magically turns into hydrogen iodide. That there's some process going on, that these gaseous state particles are bouncing around, and somehow they must bounce into each other and break bonds that they were in before, and that's what we're going to study now. This whole study of how the reaction progresses, and the rates of the reactions is called kinetics. Which is a very fancy word, but you're probably familiar with it because we've talked a lot about kinetic energy. Kinetics. Which is just the study of the rate of reactions. How fast do they happen, and how do they happen? So let's just in our minds, come up with a intuitive way that hydrogen and iodine can combine. So let's think about what hydrogen looks like. So if we get our periodic table out, hydrogen's got one valence electron so if they have two hydrogen atoms they can share them with each other. And then iodine, iodine has seven valence electrons, so if they each share one they get complete as well. So let's just review that right now. So hydrogen this hydrogen might have one, well, will have one electron out there. And then you can have another hydrogen that has another electron out there, this hydrogen can pretend like he has this electron, this hydrogen can pretend like she has that electron, and then they're happy. They both feel like they've completed their 1S shell. Same thing on the iodine side. Where you have two iodines, they both have seven valence electrons. They're halogens, you know that already. Halogens are the group seven elements, so they have seven electrons this guy's got one here, this guy's got one here, if this guy can pretend like he's got that electron, he's happy, he has eight valence electrons. If this guy can pretend like he's got that one, same thing. So there's a bond right here, and this is why hydrogen is a diatomic molecular gas, and this is why iodine is the same. Now, when they're in the gaseous state, you have a bunch of these things that are moving around bumping into each other, I'll do it like this. So the hydrogen might look something like this, the hydrogen is these two atomic spheres that are bonded together," + }, + { + "Q": "At time 10:08 when he says 1 to 6 wouldn't be 1 to 12 since 6O2 is 12 oxygen?", + "A": "He means 1 to 6 oxygen molecules", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 608 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:42, why is oxygen in its molecular form?", + "A": "This is a combustion reaction \u00e2\u0080\u0094 a reaction with oxygen \u00e2\u0080\u0094 and oxygen exists as O\u00e2\u0082\u0082 molecules.", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 42 + ], + "3min_transcript": "" + }, + { + "Q": "at 9:00 couldn't Sal have put down 0.14? does it matter how you round the number?", + "A": "Actually, no he could not have written down .14. You have to look at the number with the least number of sig figs in the problem. Then you have to write your answer with that many sig figs and one more for uncertainty purposes. Since two was the least number of sig figs in the problem, Sal rounded his answer to three sig figs. Significant numbers do matter when you are doing stoichiometry problems. Never round midway.", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 540 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:42, how does Sal know that the oxygen is in its molecular state? Is it the convention?", + "A": "Yes that is the convention. If you see reacts with oxygen they mean O2.", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 42 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:00, do you always have to start by balancing the equation?", + "A": "Yes, it is necessary to balance the equation for any stoichiometry problem. Hope this helps :)", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 60 + ], + "3min_transcript": "" + }, + { + "Q": "To find the amount of C02 and H20 we used only glucose in that ratio at 13:00, but what about 02 molecule?", + "A": "There is an excess of oxygen, so only all the glucose will be used up in the reaction. All of the moles of glucose will be used to make your products, carbon dioxide and water. There will be some oxygen left over after the reaction is complete because there are no more glucose for it to react with. Does this make sense? That s why we are only concerned with the LIMITING REACTANT or REAGENT.", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 780 + ], + "3min_transcript": "" + }, + { + "Q": "In 8:39 why does he divide 25 by 180?? soo confused right now", + "A": "There is 25 grams of glucose The molar mass of glucose is approx 180 grams per mole If you divide like so: 25 g / 180 g/mol this will calculate how many moles of glucose there are. We always need to work in moles when using chemical equations. It s important to be comfortable doing this.", + "video_name": "eQf_EAYGo-k", + "timestamps": [ + 519 + ], + "3min_transcript": "" + }, + { + "Q": "0:25 if there is a vacuum everywhere, then wouldn't the water just go up?", + "A": "between sun and earth, there is vacuum so why earth revolves in an elliptical orbit around the sun, by newton law of gravitation, every particle gravitate to other particles the force is the same but the acceleration is different bcoz of there difference in masses.", + "video_name": "HnfBFeLunk4", + "timestamps": [ + 25 + ], + "3min_transcript": "Before we move on, I just wanted to make sure that you understood that last point that I made at the end of that last video. We said that the pressure inputting into this, that we could view this cup with a hole in it as essentially a pipe, where the opening on the top of the cup is the input to the pipe, and this little mini-hole is the output to the pipe, and we said that this is a vacuum. Let's say this is vacuum all around. I know when I drew it last time, I closed it, but we have a vacuum everywhere. Since there's a vacuum everywhere, the pressure at this point P1 is equal to zero. The point I wanted to make is because we have a hole here, the pressure at that point at P2 is also equal to zero. You can almost view it as maybe the atmospheric pressure at that point, but since we're in a vacuum, that pressure is zero. That might have been a little confusing to you, because you said, well, wait, I thought at depth, if I had a point at that same height, that I would actually have a pressure at that point of rho gh. You do have an innate pressure in the liquid at that point of rho gh, and actually, that's what's causing the liquid to come out. But that's actually taken care of in the potential energy part of the equation. Let me rewrite Bernoulli's equation. The input pressure plus rho g h1 plus rho V1 squared over 2 is equal to the output pressure plus rho g h2 plus PV2 squared over 2. I think you understand that this term is pretty close to zero if the rate at which the surface moves is very slow if this surface area is much bigger than this hole. It's like if you poked a hole in Hoover Dam, that whole lake of the speed at which the water's coming out at the other end, so you could ignore this term. We also defined that the hole was at zero, so the height of h2 is zero. It simplified down to the input pressure, the pressure at the top of the pipe, or at the left side of the pipe, plus rho gh1. This isn't potential energy, but this was kind of the potential energy term when we derived Bernoulli's equation, and that equals the output pressure, or the pressure at the output of the hole, at the right side of the hole, plus the kinetic energy PV2. It's the kinetic energy term, because it doesn't actually doesn't add up completely to kinetic energy, because we manipulated it. I just wanted to really make the point that" + }, + { + "Q": "At 10:59 Sal says that the field lines correspond to the trajectory of the test charge but my textbook stated that this is a common misconception. So do they or do they not? If they don't how would the trajectory of the test charge compare to the field line?", + "A": "The field lines are the force that would be exerted on a unit positive charge present in that field. Remember, it s force; not trajectory. If the charge is moving inside the electric field, the trajectory would depend on the force (and some fancy vector mechanics)", + "video_name": "0YOGrTNgGhE", + "timestamps": [ + 659 + ], + "3min_transcript": "a bunch more in between here. And that makes sense, right? Because as you get closer and closer to the source of the electric field, the charge gets stronger. Another way that you could have done this, and this would have actually more clearly shown the magnitude of the field at any point, is you could have-- you could say, OK, if that's my charge Q, you could say, well, really close, the field is strong. So at this point, the vector, the newtons per coulomb, is that strong, that strong, that strong, that strong. We're just taking sample points. You can't possibly draw them at every single point. So at that point, that's the vector. That's the electric field vector. But then if we go a little bit further out, the vector is going to be-- it falls off. This one should be shorter, then this one should be even shorter, right? You could pick any point and you could actually calculate shorter and shorter the electric field vectors get. And so, in general, there's all sorts of things you can draw the electric fields for. Let's say that this is a positive charge and that this is a negative charge. Let me switch colors so I don't have to erase things. If I have to draw the path of a positive test charge, it would go out radially from this charge, right? But then as it goes out, it'll start being attracted to this one the closer it gets to the negative, and then it'll curve in to the negative charge and these arrows go like this. And if I went from here, the positive one will be repelled really strong, really strong, it'll accelerate fast and it's rate of acceleration will slow down, but then as it gets closer to the negative one, it'll speed up again, and then that would be its path. Similarly, if there was a positive test charge here, its path would be like that, right? If it was here, its path would be like that. If it was there, maybe its path is like that, and at some point, its path might never get to that-- this out here might just go straight out that way. That one would just go straight out, and here, the field lines would just come in, right? A positive test charge would just be naturally attracted to that negative charge. So that's, in general, what electric field lines show, and we could use our little area method and see that over here, if we picked a given area, the electric field is much weaker than if we picked that same area right here. We're getting more field lines in than we do right there. So that hopefully gives you a little sense for what an electric field is. It's really just a way of visualizing what the impact would be on a test charge if you bring it close to another charge. And hopefully, you know a little bit about Coulomb's constant. And let's just do a very simple-- I'm getting this out of the AP Physics book, but they say-- let's do a little simple problem: Calculate the static electric force between" + }, + { + "Q": "is Sal saying that electric fields pervade the entire universe At 2:58? How do electric field lines extend infinite distances?", + "A": "Electric fields do indeed pervade the universe, just like gravitational fields do.", + "video_name": "0YOGrTNgGhE", + "timestamps": [ + 178 + ], + "3min_transcript": "affecting the space around it in some way that whenever I put-- it's creating a field that whenever I put another charge in that field, I can predict how the field will affect that charge. So let's put it in a little more quantitative term so I stop confusing you. So Coulomb's Law told us that the force between two charges is going to be equal to Coulomb's constant times-- and in this case, the first charge is big Q. And let's say that the second notional charge that I eventually put in this field is small q, and then you divide by the distance between them. Sometimes it's called r because you can kind of view the distance as the radial distance between the two charges. So sometimes it says r squared, but it's the distance between them. So what we want to do if we want to calculate the field, we want to figure out how much force is there placed per charge at any point around this Q, so, say, at a given At this distance, we want to know, for a given Q, what is the force going to be? So what we can do is we could take this equation up here and divide both sides by this small 1, and say, OK, the force-- and I will arbitrarily switch colors. The force per charge at this point-- let's call that d1-- is equal to Coulomb's constant times the charge of the particle that's creating the field divided by-- well, in this case, it's d1-- d1 squared, right? Or we could say, in general-- and this is the definition of the electric field, right? Well, this is the electric field at the point d1, and if we wanted a more general definition of the electric field, we'll just make this a general variable, so instead of having a particular distance, we'll define the field for all distances away from the point Q. times the charge creating the field divided by the distance squared, the distance we are away from the charge. So essentially, we've defined-- if you give me a force and a point around this charge anywhere, I can now tell you the exact force. For example, if I told you that I have a minus 1 coulomb charge and the distance is equal to-- oh, I don't know. The distance is equal to let's say-- let's make it easy. Let's say 2 meters. So first of all, we can say, in general, what is the electric field 2 meters away from? So what is the electric field out here?" + }, + { + "Q": "At 4:48,how is the force of static friction is going to be 29.4 N? As we got this as a answer of budging force.", + "A": "Budging force is the term Sal uses for the maximum possible force of static friction.", + "video_name": "ZA_D4O6l1lo", + "timestamps": [ + 288 + ], + "3min_transcript": "the direction is straight down towards the center of the earth The normal force, and that force is there because this block is not accelerating downwards So there must be some force that completely balances off the force of gravity And in this example, it is the normal force So it is acting 49 newtons upward and so these net out. And that's why this block does not accelerate upwards or downwards So what we have is the budge the magnitude of the budging force, needs to be equal to, over the magnitude of the normal force well this thing right over here is going to be 49 newtons Is equal to 0.60 Or we could say that the magnitude of the budging force is equal to 49 newtons times the coefficient of static fiction Or that's 49 newtons times 0.60 So the units here are still going to be in newtons So this 49 times .6 gives us 29.4 newtons This is equal to 29.4 newtons So that's the force that's started to overcome static friction which we are applying more than enough of so with a 100 newtons, we would just start to budge it and right when we are in just in that moment where that thing is just starting to move the net force-- so we have a 100 newtons going in that direction and the force of static friction is going to go in this direction-- maybe I could draw it down here to show it's coming from right over here The force of static friction is going to be 29.4 newtons that way and so right when I am just starting to budge this just when that little movement-- and then kinetic friction starts to matter, but just for that moment just for that moment I'll have a net force of 100 - 29.4 to the right, so I have a net force of 70.6 N for just a moment while I budge it So just exactly while I'm budging it While we're overcoming the static friction, we have a 70.6 N net force in the right direction And so just for that moment, you divide it by 5 kg mass So just for that moment, it will be accelerating at 14.12 m/s^2 So you'll have an acceleration of 14.1 m/s^2 to the right but that will just be for that absolute moment, because once I budge it" + }, + { + "Q": "At 5:40, Sal mentions sperm carry either an X or Y chromosome, determining gender. Are the X and Y chromosome not carried in every sperm?", + "A": "Sperm will always carry one or the other unless they undergo nondisjunction in meiotic divisions. Remember that they are haploid, and as such can only carry one copy of each chromosome, including the sex chromosome.", + "video_name": "-ROhfKyxgCo", + "timestamps": [ + 340 + ], + "3min_transcript": "for her sex-determining chromosome. She's got two x's. That's what makes her your mom and not your dad. And then your dad has an x and a y-- I should do it in capital--and has a Y chromosome. And we can do a Punnett square. What are all the different combinations of offspring? Well,your mom could give this X chromosome, in conjunction with this X chromosome from your dad. This would produce a female. Your mom could give this other X chromosome with that X chromosome. That would be a female as well. Well,your mom's always going to be donating an X chromosome. And then your dad is going to donate either the X or the Y. So in this case,it'll be the Y chromosome. So these would be female,and those would be male. And it works out nicely that half are female and half are male. But a very interesting and somewhat ironic fact might pop out at you when you see this. what determines whether someone is or Who determines whether their offspring are male or female? Well,the mom always donates an X chromosome, so in no way does- what the haploid genetic makeup of the mom's eggs of the gamete from the female, in no way does that determine the gender of the offspring. It's all determined by whether--let me just draw a bunch of-- dad's got a lot of sperm,and they're all racing towards the egg. And some of them have an X chromosome in them and some of them have a Y chromosome in them. And obviously they have others. And obviously if this guy up here wins the race. Or maybe I should say this girl.If she wins the race, then the fertilized egg will develop into a female. If this sperm wins the race, then the fertilized egg will develop into a male. And the reason why I said it's ironic is throughout history, I mean it's not just the case with kings. It's probably true,because most of our civilization is male dominated, that you've had these men who are obsessed with producing a male heir to kind of take over the family name. And,in the case of Henry the VIII,take over a country. And they become very disappointed and they tend to blame their wives when the wives keep producing females,but it's all their fault. Henry the VIII,I mean the most famous case was with Ann Boleyn. I'm not an expert here,but the general notion is that he became upset with her that she wasn't producing a male heir. And then he found a reason to get her essentially decapitated, even though it was all his fault. He was maybe producing a lot more sperm" + }, + { + "Q": "Would writing the angle of the second ball as -38 degrees be the same as when Sal says \" 38 degrees below the horizontal\" at 7:47 ?", + "A": "It might be, if you specify that 0 is a horizontal line to the right. You need to be clear.", + "video_name": "leudxqivIJI", + "timestamps": [ + 467 + ], + "3min_transcript": "And maybe you're not familiar with arcsine yet because I don't think I actually have covered yet in the trig modules, although I will eventually. So we know it's just the inverse function of sine. So sine of theta is equal to 0.625. Then we know that theta is equal to the arcsine of 0.625. This is essentially saying, when you say arcsine, this says, tell me the angle whose sine is this number? That's what arcsine is. And we can take out Google because it actually happens that Google has a-- let's see. Google actually-- it's an automatic calculator. So you could type in arcsine on Google of 0.625. Although I think the answer they give you will be in radians. So I'll take that answer that will be in radians and I want to convert to degrees, so I multiply it times 180 over pi. And let's see what I get. So Google, you see, Google says 38.68 degrees. They multiplied the whole thing times 180 and then divided by pi, but that should be the same thing. So roughly 38.7 degrees is theta. Hope you understand that. You could pause it here if you don't, but let me just write that down. So it's 38 degrees. So theta is equal to 38.7 degrees. So then we're done. We figured out that ball B gets hit. This is ball B and it got hit by ball A. Ball A went off in that direction at a 30 degree angle, at a 30 degree angle at 2 meters per second. And now ball B goes at 38.-- or we could say roughly 39 degrees below the horizontal at a velocity of 3.2 meters per second. And does this intuitively make sense to you? Ball A had a mass of 10 kilograms while ball B had a mass of 5 kilograms. So it makes sense. So let's think about just the y direction. Ball A, we figured out, the y component of its velocity was 1 meter per second. And ball B's y component is 2 meters per second downwards. And does that makes sense? Well sure. Because their momentums have to add up to 0. There was no y component of the momentum before they hit each other. And in order for B to have the same momentum going downwards in the y direction as A going upwards, its velocity has to be essentially double, because its mass is half. And a similar logic, although the cosine-- it doesn't work out exactly like that. But a similar logic would mean that its overall velocity is going to be faster than the- than A's velocity. And so what was I just-- oh yeah." + }, + { + "Q": "At 4:13, If Br is so electronegative, why does it give it's electron to Fe in the first place?", + "A": "it have 7 valence electrons and wants to fufill its orbitals with 8, the FeBr3 has an open spot so that both of the compounds have 8 and therefore are satisfied", + "video_name": "K2tIixiXGOM", + "timestamps": [ + 253 + ], + "3min_transcript": "with the right energies it can happen. So this electron-- let me do it in a different color that I haven't used yet, this green color-- let's say this electron right here gets nabbed by that iron. Then what do we have? Well then we have a situation, we have this bromine-- the blue one-- with one, two, three, four, five, six, seven valence electrons. We have the magenta bromine with one, two, three, four, five. Now it only has the sixth valence electron right here. The seventh got nabbed by the iron. So the iron has the seventh valence electron and then you have the rest of the molecule. So then you have your iron and it's attached, of course, to the three bromines. Just like that. And then our bonds, these guys were bonded. They still are bonded. And now these guys are bonded. This electron jumped over to the iron. And now we have another bond. But because this bromine lost an electron-- it was neutral, it lost an electron-- it now has a positive charge. And the iron, now that it gained this electron, now has a negative charge. So let's think about what's going to happen now. Now we're going to bring the benzene into the mix. So let me redraw the benzene. And we have this double bond, that double bond, and then just to make things clear, let me draw this double bond with the two electrons on either end. So we have the orange electron, you have your green electron right over there, and I'll draw the double bond as being green. Now let's think about this molecule right here. We have a bromine with a positive charge. Bromines are really, really, really, electronegative. positive charge, it really wants to grab an electron. And in the right circumstances, you can imagine where it really wants to grab that electron right there. So maybe, if there was just some way it could pull this electron. But the only way it could pull this electron is maybe if this-- because if it just took that electron than this bromine would have a positive charge, which isn't cool. So this bromine maybe would want to pull an electron. If this guy gets an electron, then this guy can get an electron. So you can imagine this thing as a whole really, really wants to grab an electron. It might be very good at doing it. So this is our strong electrophile. So what actually will happen in the bromination of this benzene ring-- let me draw some hydrogens here just to make things clear. We already have hydrogens on all of these carbons. Sometimes it's important to visualize this when we're doing electrophilic aromatic substitutions." + }, + { + "Q": "At 2:39, When do you know when to use negative for tension pulling up or negative for the force of gravity pulling down. I tried another pulley problem using the same concept pulling up 10kg at 2 m/s^2 and the answer was 2T - mg = ma (answer 60N) instead of 2T + mg = ma ( answer 40N upward). These both give two different answers so I am a bit confused.", + "A": "You just have to decide, in each problem, which way do you want to be positive. If you decide up is positive, then it is always positive. If you decide down is positive, then it is always positive.", + "video_name": "52wxpYnS64U", + "timestamps": [ + 159 + ], + "3min_transcript": "per second, the acceleration of gravity. So the wire must be exerting some upward force on the object. And that is the force of tension. That is what's slowing-- that's what's moderating its acceleration from being 9.8 meters per second squared to being 4.13 meters per second squared. So essentially, what is the net force on this object? On just this object? Well the net force is-- and you can ignore what I said before about the net force in all the other places. But we know that the object is accelerating downwards. Well, we know it's 20 kilograms. So that's its mass. And we know that it's accelerating downwards at 4.13 meters per second squared. So the net force, 20 times-- see, times 20 is 82-- let's just say 83 Newtons. We know that the net force is 83 Newtons down. We also know that the tension force plus the force of gravity-- and what's the force of gravity? The force of gravity is just the weight of the object. So the force of tension, which goes up, plus the weight of -- the force of gravity is equal to the net force. And the way I set this up, tension's going to be a negative number. Just because I'm saying positive numbers are downwards, so a negative number would be upwards. So tension will be what is 83 minus 196? Minus 196 is equal to minus 113 Newtons. And the only reason why I got a negative number is because I used positive numbers for downwards. So minus 113 Newtons downwards, which is the same And so that is the tension in the rope. And you could have done the same thing on this side of the problem, although it would have been-- well, yeah. You could have done the exact same thing on this side of the problem. You would've said, well what would it have accelerated naturally if there wasn't some force of tension on this rope going backwards? And then you're saying, oh, well, we know it would have gone in this direction at some acceleration, but instead it's going in the other direction. So you use that. You figure out the net force, and then you say the tension plus all of these forces have to equal the net force. And then you should solve for the tension. And it would be the same tension. Now we will do a fun and somewhat simple, but maybe instructive problem. So I have a pie. This is the pie. This is parallel. And I have my hand. You can tell that my destiny was really to be a great" + }, + { + "Q": "09:00 You're placing the oxygen on the same side as Br but you're saying that it's inversion. How come?\nAlso, I really don't understand that triangle type bond, what's that?", + "A": "See videos on stereochemistry. This video is way too advanced if you don t understand chirality or wedge and dash representations of bonds.", + "video_name": "3LiyCxCTrqo", + "timestamps": [ + 540 + ], + "3min_transcript": "for the side opposite of the leaving group and you can see with the methyl halide there's no steric hindrance. When we move to a primary alkyl halide, the carbon bonded to the halogen has only one alkyl group bonded to it, it's still easy for the nucleophile to approach. When we move to a secondary alkyl halide, so for a secondary you can see that the carbon bonded to the halogen has two methyl groups attached to it now. It gets a little harder for the nucleophile to approach in the proper orientation. These bulky methyl groups make it more difficult for the nucleophile to get close enough to that electrophilic carbon. When we go to a tertiary alkyl halide, so three alkyl groups. There's one, there's two and there's three. There's a lot more steric hindrance and it's even more difficult for our nucleophile to approach. As we saw on the video, for an SN2 reaction we need decreased steric hindrance. So, if we look at this alkyl halide, is attached to only one alkyl group. This is a primary alkyl halide and that makes this a good SN2 reaction. The decreased steric hindrance allows the nucleophile to attack the electrophile." + }, + { + "Q": "At 16:24, it is said that work is done, but deltaU is zero because T does not change, which means Q-W is 0. If work is not zero, than Q must not be zero, which contradicts with a adiabatic process? What's the problem here?", + "A": "I don t recall this process being described as adiabatic. Lots of heat comes from the reservoir to maintain T to offset the drop in T from work being done.", + "video_name": "WLKEVfLFau4", + "timestamps": [ + 984 + ], + "3min_transcript": "Alright. OK. So this is pressure, this is volume. Now. When we started off, before we blew away the wall, we had some pressure and some volume. So this is V1. And then we blew away the wall, and we got to-- Actually, let me do it a little bit differently. I want that to be just right there. Let me make it right there. So that is our V1. This is our original state that we're in. So state initial, or however we want it. That's our initial pressure. And then we blew away the wall, and our volume doubled, right? So we could call this 2V1. Our volume doubled, our pressure would have gone down, and we're here. That's our state 2. blew away the wall. Now, what we did was not a quasistatic process. I can't draw the path here, because right when I blew away the wall, all hell broke loose, and things like pressure and volume weren't well defined. Eventually it got back to an equilibrium where this filled the container, and nothing else was in flux. And we could go back to here, and we could say, OK, now the pressure and the volume is this. But we don't know what happened in between that. So if we wanted to figure out our Q/T, or the heat into the system, we learned in the last video, the heat added to the system is equal to the work done by the system. We'd be at a loss, because the work done by the system is the area under some curve, but there's no curve to speak of here, because our system wasn't defined while all the hell had broke loose. So what can we do? Well, remember, this is a state function. And this is a state function. And I showed that in the last video. So it shouldn't be dependent on how we got from there to there. with my words. This change in s, so s2 minus s1, should be independent of the process that got me from s1 to s2. So this is independent of whatever crazy path-- I mean, I could have taken some crazy, quasistatic path like that, right? So any path that goes from this s1 to this s2 will have the same heat going into the system, or should have the same-- let me take that-- Any system that goes from s1 to s2, regardless of its path, will have the same change in entropy, or their same change in s. Because their s was something here, and it's something different over here. And you just take the difference between the two. So what's a system that we know that can do that? Well, let's say that we did an isothermal. And we know that these are all the same isotherm, right?" + }, + { + "Q": "At around 2:00, Jay says that one oxygen fewer is hypochlorite. Why isn't perchlorate called HYPERchlorate? For example, you can have HYPOthermia and HYPERthermia.", + "A": "Strangely enough, I believe in the past perchlorate was sometimes referred to as hyperchlorate but the name seems to have gone into disuse. Perhaps there was confusion between hypo and hyper so it was safer to drop the hyper.", + "video_name": "DpnUrVXSLaQ", + "timestamps": [ + 120 + ], + "3min_transcript": "- [Voiceover] When you take a general chemistry class, you often have to memorize some of the common polyatomic ions. So let's go through a list of some of the ones that you might see in your class. So we'll start off with Cation here, so a positively charged ion, NH four plus is called the Ammonium ion. And for Anions, there are many Anions that you should know. CH three COO minus is the Acetate ion. CN minus is the Cyanide ion. OH minus is the Hydroxide anion. MnO four minus is the Permanganate ion. And, when you get to NO three minus versus NO two minus, look at the endings. So NO three minus is Nitrate, so we have ate suffix, ate suffix here, which means more Oxygens. Versus the ite suffix, which means fewer Oxygens. So we can see that Nitrate has three Oxygens and Nitrite has two Oxygens. with some of the other polyatomic ions. For example, let's look at this next set here of four. And let's look at Chlorate. So Chlorate has three Oxygens. It's ClO three minus one. And Chlorite has fewer Oxygens, it has two Oxygens here, ClO two minus. So we have ate meaning more and ite meaning fewer here. What about Perchlorate? So here we have Chlorate, but we've added on a prefix this time and the prefix, per, means one more Oxygen. So Perchlorate means one more Oxygen than Chlorate. Chlorate had three Oxygens and for Perchlorate we add one on and we get four. So Perchlorate is ClO four minus. Next, let's look at Hypochlorite. So we talked about Chlorite up here, so here's Chlorite and then we put a prefix, hypo, in front of it. Hypo means one fewer, we take one away and now we have only one Oxygen. So that must be the Hypochlorite ion. We could have done this for a different halogen, here we're dealing with Chlorine, but let's say, instead of ClO three minus, let's do BrO three minus. ClO three minus was Chlorate, here we have Bromine instead of Chlorine, so this would be Bromate. So there's another polyatomic ion and we can do another example. So instead of ClO minus, which is Hypochlorite, we could have had BrO minus, which would therefore be Hypobromite. So this would be Hypobromite. Alright, let's look at our next set of polyatomic ions. Alright, so let's get some space down here. So we have SO four two minus, is called Sulfate." + }, + { + "Q": "4:30 how do we know we should use the second half-reaction?", + "A": "To find a total cell potential we should take into account both oxidation and reduction reactions. It s not enough to know how much an element wants to be reduced, we also should check how strongly the other element wants to be oxidized, as electrons lost by one element are the electrons gained by another.", + "video_name": "fYUwEAPejbY", + "timestamps": [ + 270 + ], + "3min_transcript": "four hydroxide anions. So once again, this is a reduction reaction because those electrons are being incorporated into the molecule. And so you're left with the hydroxide right over here. And if you look at our metal air cell up here, where is that occurring? Well, you need oxygen, you need water and you need electrons. Well, if you look at the cathode right over here, you got your electrons coming in, you got your oxygen coming in, and water, we can assume, is available in this area. The electrolyte paste has water in it. The cathode is porous, it's a porous membrane-type substance. So they're all available over here. So you can assume that, that reaction could happen right over here in the cathode. So let me just write it, 02 in as a gas. So for every molecule of O2, you're gonna have two molecules of H2O in a liquid form. And then you got your electrons coming in on the wire, four hydroxide anions. Four hydroxide anions. And they're in a solution. They're part of that electrolyte paste that's in water. They're an aqueous solution, a water-based solution. And so that's going to happen right over there so the hydroxide anions, I can even say four hydroxide anions are going to be produced every time this reaction is happening. And they tell us the reduction potential. This has a positive potential, has a positive voltage. And notice they say the reduction potential at pH 11. So that's consistent with alkaline conditions because this paste might seep in through here a little bit and obviously you have all this hydroxide being produced. So you're gonna have a higher pH, a more basic pH. And so the fact that this is a positive voltage, so plus 0.34 volts means that the potential is going in this direction. going from left to right. Now they also said they give us the reduction potentials for the cathode, which we just talked about, this is the reduction potential for the cathode, and three possible metal anodes are given in the table below. So here there are three possible metal anodes. So we could have zinc, we could have sodium, we could have calcium, and so let's just go with zinc since it's the first one listed here. So if we assumed that the metal here was zinc, what's going to be going on? Well, you're going to have the zinc reacting with these hydroxides that are being produced over in the cathode, and then you're going to use that to produce zinc oxide water and electrons. So you're actually gonna have the reverse of this reaction. You're going to have, let me write the reverse. You're going to have zinc in the solid state. This whole anode is made out of metal zinc. And then for every molecule of that, you're gonna have two hydroxide anions that are dissolved" + }, + { + "Q": "At 7:04, Sal explained neutron stars by saying that it is just a ball of neutrons. How does this ball stay in place?", + "A": "It dosn t stay in place . All stars - including neutron stars - orbit the super-massive black hole at the center of their galaxy. Furthermore, the galaxy itself is constantly in motion. So, the neutron star does not stay just sit there. It s moving.", + "video_name": "qOwCpnQsDLM", + "timestamps": [ + 424 + ], + "3min_transcript": "it comes from, I believe-- I'm not an expert here-- Latin for \"new.\" And the first time people observed a nova, they thought it was a new star. Because all of a sudden, something they didn't see before, all of a sudden, it looks like a star appeared. Because maybe it wasn't bright enough for us to observe it before. But then when the nova occurred, it did become bright enough. So it comes from the idea of new. But a supernova is when you have a pretty massive star's core collapsing. And that energy is being released to explode the rest of the star out at unbelievable velocities. And just to kind of fathom the amount of energy that's being released in a supernova, it can temporarily outshine an entire galaxy. And in a galaxy, we're talking about hundreds of billions of stars. Or another way to think about it, in that very short period of time, it can release as much energy as the sun will in its entire lifetime. So these are unbelievably energetic events. And so you actually have the material that's not in the core being shot out of the star So we're talking about things being shot out at up to 10% of the speed of light. Now, that's 30,000 kilometers per second. That's almost circumnavigating the earth every second. So that's, I mean, this is unbelievably energetic events that we're talking about here. And so if the original star was-- and these are rough People don't have kind of a hard limit here. If the original star approximately 9 to 20 times the mass of the sun, then it will supernova. And the core will turn into what's called a neutron star. This is a neutron star, which you can imagine is just this dense ball. It's this dense ball of neutrons. it'll be something about maybe two times the mass of the sun, give or take one and a half to three times the mass of the sun. So this is one and a half to three times the mass of the sun in a volume that has a diameter of about 10-- on the order of tens of kilometers. So it's roughly the size of a city, in a diameter of a city. So this is unbelievably dense, diameter of a city. I mean, we know how much larger the sun is relative to the Earth. And we know how much larger the Earth is relative to a city. But this is something large-- more mass than the sun being squeezed into the density, or into the size of a city, so unbelievably dense. Now if the original star is even more massive, if it's more than 20 times the sun-- so let me write it over here. Let me scroll up. If it's greater than 20 times the sun," + }, + { + "Q": "At 2:49, Sal mentioned that the electrons get captured into the nucleus. Well then, wouldn't the atom be annihilated?", + "A": "Why would it be annihilated?", + "video_name": "qOwCpnQsDLM", + "timestamps": [ + 169 + ], + "3min_transcript": "So let me write this here, electron degeneracy pressure. And all this means is we have all of these iron atoms getting really, really, really close to each other. And the only thing that keeps it from collapsing at this earlier stage, the only thing that keeps it from collapsing altogether, is that they have these electrons. You have these electrons, and these are being squeezed together, now. I mean, we're talking about unbelievably dense states of matter. And electron degeneracy pressure is, essentially-- it's saying these electrons don't want to be in the same place at the same time. I won't go into the quantum mechanics of it. But they cannot be squeezed into each other any more. So that, at least temporarily, holds this thing from collapsing even further. in the case of a white dwarf, that's how a white dwarf actually maintains its shape, because of the electron degeneracy pressure. But as this iron core gets even more massive, more dense, and we get more and more gravitational pressure-- so this is our core, now-- even more gravitational pressure, eventually even this electron degeneracy-- I guess we could call it force, or pressure, this outward pressure, this thing that keeps it from collapsing-- even that gives in. And then we have something called electron capture, which is essentially the electrons get captured by protons in the nucleus. They start collapsing into the nucleuses. It's kind of the opposite of beta negative decay, where you have the electrons get captured, protons get turned into neutrons. But you can imagine an enormous amount of energy is also being released. So this is kind of a temporary-- and then all of a sudden, this collapses. This collapses even more until all you have-- and all the protons are turning into neutrons. Because they're capturing electrons. So what you eventually have is this entire core is collapsing into a dense ball of neutrons. You can kind of view them as just one really, really, really, really, really massive atom because it's just a dense ball of neutrons. At the same time, when this collapse happens, you have an enormous amount of energy being released in the form of neutrinos. Did I say that neutrons are being released? No, no, no, the electrons are being captured by the protons, protons turning into neutrons-- this dense ball of neutrons right here-- and in the process, neutrinos get released, these fundamental particles." + }, + { + "Q": "from 4:30 to 5:00\nsure there's loads of pressure on the core of the star, but why exactly does it suddenly go supernovae??\nis there some repulsion between the neutrons??", + "A": "The collapse stops beyond the point when the neutrons are literally touching each other. The neutrons are actually smaller than normal neutrons because of this pressure. The forces trying to decompress the neutrons is incredible and they fuel the supernova.", + "video_name": "qOwCpnQsDLM", + "timestamps": [ + 270, + 300 + ], + "3min_transcript": "in the case of a white dwarf, that's how a white dwarf actually maintains its shape, because of the electron degeneracy pressure. But as this iron core gets even more massive, more dense, and we get more and more gravitational pressure-- so this is our core, now-- even more gravitational pressure, eventually even this electron degeneracy-- I guess we could call it force, or pressure, this outward pressure, this thing that keeps it from collapsing-- even that gives in. And then we have something called electron capture, which is essentially the electrons get captured by protons in the nucleus. They start collapsing into the nucleuses. It's kind of the opposite of beta negative decay, where you have the electrons get captured, protons get turned into neutrons. But you can imagine an enormous amount of energy is also being released. So this is kind of a temporary-- and then all of a sudden, this collapses. This collapses even more until all you have-- and all the protons are turning into neutrons. Because they're capturing electrons. So what you eventually have is this entire core is collapsing into a dense ball of neutrons. You can kind of view them as just one really, really, really, really, really massive atom because it's just a dense ball of neutrons. At the same time, when this collapse happens, you have an enormous amount of energy being released in the form of neutrinos. Did I say that neutrons are being released? No, no, no, the electrons are being captured by the protons, protons turning into neutrons-- this dense ball of neutrons right here-- and in the process, neutrinos get released, these fundamental particles. But it's an enormous amount of energy. And this actually is not really, really well understood, of all of the dynamics here. Because at the same time that this iron core is undergoing through this-- at first it kind of pauses due to the electron degeneracy pressure. And then it finally gives in because it's so massive. And then it collapses into this dense ball of neutrons. But when it does it, all of this energy's released. And it's not clear how-- because it has to be a lot of energy. Because remember, this is a massive star. So you have a lot of mass in this area over here. But it's so much energy that it causes the rest of the star to explode outward in an unbelievable, I guess, unbelievably bright or energetic explosion. And that's called a supernova." + }, + { + "Q": "At 4:00, Sal says \"should.\" Can some cancer cells be so mutated that they do not produce MHC I at all?", + "A": "Yes. Sometimes cancer cells are so mutated or too close to your actual cells that if your cyto cells attack them, they will also attack your healthy cells. That s why to be safe they don t attack them. They also don t recognize them so they don t kill them. That s why all people need help to fight off cancers since your body can t kill most cancers.", + "video_name": "YdBXHm3edL8", + "timestamps": [ + 240 + ], + "3min_transcript": "Once again this is what's occurring outside of cells. When we found stuff outside of cells, we engulf them and then we presented them on MHC two complexes. Now you're probably thinking well, I mean that's the outside of cells but there's MHC two, there's this helper T cells but we've also referred to cytotoxic T cells. What do those do? We've also, if there's MHC two, there's probably an MHC one complex. What does the MHC one complex do? We can recognize shady things that are happening outside of cells but don't shady things sometimes happen inside cells and how does our immune system respond to that? Actually as you can imagine all of those things will be answered in the rest of this video. Let's think about what happens when shady things start to happen inside the cell. For example it might not even be due to a virus could be the cell itself is gone awry. Let's say that this right over here is a cancer cell. It's had some mutations. It's starting to multiply like crazy. This is a cancer cell and a cancer cell because it had mutations are going to produce, it's going to produce some weird proteins. These cancer cells are going to produce some weird proteins. Every cell with a nucleus in your body and that's pretty much every cell except for red blood cells has MHC one complexes. The whole point of the MHC one complex is to bind two shady things that are produced inside of the cell, and then present them to the membrane. Even a malfunctioning cancer cell should be doing this. that are produced by the mutations inside of the nucleus and then it can present them. You could imagine what the appropriate immune response should be. These cancer cells should be killed and actually let me label this properly. That was MHC two, you're presenting an antigen that was found, those initially found outside of the cells engulfing and taken out. MHC one, it's binding to shady things inside the cell and then presenting it out. This thing should be killed. Now as you can imagine, what's going to kill it? Well that's where the cytotoxic T cell comes into the picture. The cytotoxic T cell is going to have, that's a receptor right there." + }, + { + "Q": "Around 6:23 when the Hydrogen breaks off was it in a hybridised sp2 orbital with the C? and when forming the double bond with the other C does it change to a p orbital?", + "A": "Yes, at that point, the C atom is using an sp\u00c2\u00b2 orbital that is parallel to the vacant p orbital on the adjacent carbon. As the H leaves, the back lobe of the orbital becomes larger and the front lobe becomes smaller until, once the H atom is completely free, both lobes are the same size and you have a p orbital that contains two electrons. That p orbital overlaps with the adjacent empty p orbital to form a \u00cf\u0080 bond.", + "video_name": "U9dGHwsewNk", + "timestamps": [ + 383 + ], + "3min_transcript": "doesn't occur nothing else will. But now that this does occur everything else will happen quickly. In our rate-determining step, we only had one of the reactants involved. It's analogous to the SN1 reaction but what we're going to see here is that we're actually eliminating. We're going to call this an E1 reaction. We're going to see that in a second. Actually, elimination is already occurred. The bromide has already left so hopefully you see why this is called an E1 reaction. It's elimination. E for elimination and the rate-determining step only involves one of the reactants right here. It didn't involve in this case the weak base. Now that the bromide has left, let's think about whether this weak base, this ethanol, can actually do anything. It does have a partial negative charge over here. It does have a partial negative charge and on these ends it has partial positive charges, so it is somewhat attracted to hydrogen, or to protons I should say, to positive charges. But not so much that it can swipe it off of things that Now that this guy's a carbocation, this entire molecule actually now becomes pretty acidic, which means it wants to give away protons. Another way you could view it is it wants to take electrons, depending on whether you want to use the Bronsted-Lowry definition of acid, or the Lewis definition. Either way, it wants to give away a proton. It could be that one. It has excess positive charge. It wants to get rid of its excess positive charge. So it's reasonably acidic, enough so that it can react with this weak base. What you have now is the situation, where on this partial negative charge of this oxygen-- let me pick a nice color here-- let's say this purple electron right here, it can be donated, or it will swipe the hydrogen proton. Then hydrogen's electron will be taken In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs? What's our final product? Let me draw it here. This part of the reaction is going to happen fast. The rate-determining step happened slow. The leaving group had to leave. The carbocation had to form. That's not going to happen super fast but once that forms, it's not that stable and then this thing will happen. This is fast. Let me paste everything again. So now we already had the bromide. It had left. Now the hydrogen is gone. The hydrogen from that carbon right there is gone. This electron is still on this carbon but the electron that" + }, + { + "Q": "Would the name of the new organic product formed at 8:39 be: 3-ethyl pentene?\nAlso, how would you indicate in the name where the double bond is?", + "A": "You would name the compound 3-ethyl-pent-2-ene or 3-ethyl-2-pentene. The 2 indicating the location of the double bond.", + "video_name": "U9dGHwsewNk", + "timestamps": [ + 519 + ], + "3min_transcript": "That electron right here is now over here, and now this bond right over here, is this bond. We formed an alkene and now, what was an ethanol took a hydrogen proton and now becomes a positive cation. Let me draw that. So this electron ends up being given. It's no longer with the ethanol. It gets given to this hydrogen right here. That hydrogen right there. And now they have formed a new bond and since this oxygen gave away an electron, it now has a positive charge. And all along, the bromide anion had left in The bromide anion is floating around with its eight valence electrons, one, two, three, four, five, six, seven, and then it has this one right over here. That makes it negative. Then our reaction is done. We had a weak base and a good leaving group, a tertiary carbon, and the leaving group left. We only had one of the reactants involved. It was eliminated. And then once it was eliminated, then the weak base was then able to take a hydrogen off of this molecule, and that allowed this molecule to become an alkene, formed a double bond. This is called, and I already told you, an E1 reaction. E for elimination, in this case of the halide. One, because the rate-determining step only involved one of the molecules. It did not involve the weak base. We'll talk more about this, and especially different circumstances where you might have the different types of E1 off, and all the things like that. And we're going to see with E1, E2, SN1, and SN2, what kind of environments or reactants need to be there for each one of those to occur in different circumstances." + }, + { + "Q": "At 7:45 What if someone who gets blood on their hand washes their hands right after they got the blood on their hand? Would they still get ebola?", + "A": "Not necessarily. The virus would need to enter the body through broken skin or mucous membranes, for example, if the blood on their hands got into a cut or came in contact with the eyes, nose, or mouth. Regular hand washing may remove the visible sign of blood, but the CDC recommends decontamination via an EPA-registered hospital disinfectant.", + "video_name": "xYoQQCO15GE", + "timestamps": [ + 465 + ], + "3min_transcript": "the world is kind of watching and intervening and helping to make sure that we get this under control. So, in addition to the supportive care the other big area that we kind of think about is how do you avoid transmitting the virus from one person to another. So let me just sketch in a second person here. So here I've drawn a second person and the question becomes how does Ebola get from the first person to the second person? How exactly does it move? And the quick answer is that it moves through bodily fluids. So bodily fluids include everything from blood to vomit and diarrhea or stool and these are kind of the big ones especially blood. Very, very infectious. But also stool and vomit because a lot of these patients are sick and bodily fluids can also be things like saliva and sweat. But again, these first few that I've kind of written out are the ones to really keep in mind, and blood in fact is an interesting one. and kind of illustrates the point where they showed that one drop. Just one drop of blood from a monkey that had Ebola was enough to carry with it a million little Ebola virus particles and that was thought to be enough to infect a million monkeys. So kind of the idea that one single drop could be enough to really decimate a population. So it kind of drives on the point that if there's a drop of blood, just a single drop or another bodily fluid that really gets over, let's say over into this person's hand and they can kind of go from there to if they touch their face. They can get entry into their body or through their nose or eyes or mouth or let's say they have a little cut. It could actually make its way into the blood through that cut. So, a few kind of interesting ways that bodily fluid can travel but really then it gets a ticket into the second person and the second person can get sick. with stool or vomit as well. Now another question that I think a lot of people ask is what about airborne? You know, can you breathe this stuff in? And that's a big question and the idea I think is literally is it kind of floating in the air? And the answer is no. You know, it's not really like other viruses like measles and some other ones where it's airborne. So let me just cross that out so it's very clear. It is not airborne. In fact, I think a lot of the confusion comes when you think about things like sneezes. So with a sneeze you can imagine like kind of you know, we've all seen those sneezes or had those sneezes where kind of a big droplet of goo or mucus kind of goes out, and you get this kind of shower, right, of droplets and this kind of right around where you're standing. Just right in your immediate vicinity and that's not gonna affect anyone that's kind of standing away from you, and you could imagine maybe in kind of a strange scenario. Let's say there's a person kind of hanging out right here" + }, + { + "Q": "At 6:00 what would 3.991 be rounded up to if it could only contain two significant figures? Would this result in 4.0? Or would it simply be 4?", + "A": "4.0 since the number of significant digits are preferably preserved.", + "video_name": "xHgPtFUbAeU", + "timestamps": [ + 360 + ], + "3min_transcript": "Let's say we have another block, and this is the other block right over there. We have a, let's say we have an even more precise meter stick, which can measure to the nearest millimeter. And we get this to be 1.901 meters. So measuring to the nearest millimeter. And let's say those measurements were done a long time ago, and we don't have access to measure them any more, but someone says 'How tall is it if I were stack the blue block on the top of the red block - or the orange block, or whatever that color that is?\" So how high would this height be? Well, if you didn't care about significant figures or precision, you would just add them up. You'd add the 1.901 plus the 2.09. So let me add those up: so if you take 1.901 and add that to 2.09, 0 plus 9 is 9, 9 plus 0 is 9, you get the decimal point, 1 plus 2 is 3. So you get 3.991. And the problem with this, the reason why this is a little bit... it's kind of misrepresenting how precise you measurement is. You don't know, if I told you that the tower is 3.991 meters tall, I'm implying that I somehow was able to measure the entire tower to the nearest millimeter. The reality is that I was only be able to measure the part of the tower to the millimeter. This part of the tower I was able to measure to the nearest centimeter. So to make it clear the our measurement is only good to the nearest centimeter, because there is more error here, then... it might overwhelm or whatever the precision we had on the millimeters there. To make that clear, we have to make this only as precise as the least precise thing that we are adding up. So over here, the least precise thing was, we went to the hundredths, so over here we have to round to the hundredths. and so we can only legitimately say, if we want to represent what we did properly that the tower is 3.99 meters. And I also want to make it clear that this doesn't just apply to when there is a decimal point. If I were tell you that... Let's say that I were to measure... I want to measure a building. I was only able to measure the building to the nearest 10 feet. So I tell you that that building is 350 feet tall. So this is the building. This is a building. And let's say there is a manufacturer of radio antennas, so... or radio towers. And the manufacturers has measured their tower to the nearest foot. And they say, their tower is 8 feet tall. So notice: here they measure to the nearest 10 feet, here they measure to the nearest foot. And actually to make it clear, because once again, as I said, this is ambiguous," + }, + { + "Q": "4:45 Sal said \"a mitochondria\" but the correct singular is \"a mitochondrion\", right?", + "A": "Yes, that is correct.", + "video_name": "J30zpvbmw7s", + "timestamps": [ + 285 + ], + "3min_transcript": "from our NADH to the Oxygen, it would release a lot of energy but it would release so much energy that you wouldn't be able to capture most of it. You wouldn't be able to use it to actually do useful work, and so the process of Oxidative Phosphorylation is all about doing this at a series of steps and we do it by transferring these electrons from one electron acceptor to another electron acceptor, and every time we do that, we release some energy, and then that energy can be, in a more controlled way, be used to actually do work, and in this case, that work is pumping hydrogen protons across a membrane, and then that gradient that forms can actually be used to generate ATP, so let's talk through it a little bit more. So we're gonna go, these electrons, they're gonna be transferred, and I won't go into all of the details, this is to just give you a high-level overview of it. They're going to be transferred to different acceptors which then transfer it to another acceptor, so it might go to a Coenzyme, Cytochrome C, and it keeps going to different things, eventually getting to this state right over here, where those electrons can be accepted by the oxygen to actually form the water, and the process, every step of the way, energy is being released. Energy is being released, and this energy, as we will see in a second, is being used to pump hydrogen protons across a membrane, and we're gonna use that gradient to actually drive the production of ATP. So let's think about that a little bit more. So let's zoom in on, on a mitochondria. So this is mitochondria. Let's say that's our mitochondria, and let me draw the inner membrane and then, these folds in the inner membrane, the singular for them is crista. If we're talking about plurals, cristae. So we have these folds in the inner, So just to be clear, what's going on, this is the outer membrane, outer membrane. That is the inner membrane, inner membrane. The space between the outer and the inner membrane, the space right over here, that is the intermembrane space. Intermembrane, membrane space. And then the space inside the inner membrane, let me make that sure you can read that space properly, this space over here, this is the Matrix. This is the Matrix, and that is the location of our Citric acid Cycle or our Krebs Cycle, and I can symbolize that with this little cycle, we have a cycle going on here. And so that's where the bulk of the NADH is being produced. Now we also talked about some other coenzymes. In some books or classes, you might hear about FAD" + }, + { + "Q": "This isn't intuitive at all for me. If at 2:57 you do exactly what he said (index finger points in the direction of b and middle finger for a) with your right hand, i still get my thumb going into the screen, the same result as in the last video. How is he flexing his thumb to get it to point out of the page? I'm really confused.", + "A": "I struggled with this too, but I think if you keep you index finger straight (as in fully extended and in line with your arm), then you should be able to figure it out.", + "video_name": "o_puKe_lTKk", + "timestamps": [ + 177 + ], + "3min_transcript": "Or when I drew it here, it would point into the page. So let's see what happens with b cross a, so I'm just switching the order. b cross a. Well, the magnitude is going to be the same thing, right? Because I'm still going to take the magnitude of b times the magnitude of a times the sine of the angle between them, which was pi over 6 radians and then times some unit vector n. But this is going to be the same. When I multiply scalar quantities, it doesn't matter what order I multiply them in, right? So this is still going to be 25, whatever my units might have been, times some vector n. And we still know that that vector n has to be perpendicular to both a and b, and now we have to figure out, well, is it, in being perpendicular, it can either kind of point into the page here or it could pop out of the page, or point out of the page. So which one is it? So what we do is we take our right hand. I'm actually using my right hand right now, although you can't see it, just to make sure I draw the right thing. So in this example, if I take my right hand, I take the index finger in the direction of b. I take my middle finger in the direction of a, so my middle figure is going to look something like that, right? And then I have two leftover fingers there. Then the thumb goes in the direction of the cross product, right? Because your thumb has a right angle right there. That's the right angle of your thumb. So in this example, that's the direction of a, this is the direction of b, and we're doing b cross a. That's why b gets your index finger. gets the second term, and the thumb gets the direction of the cross product. So in this example, the direction of the cross product is upwards. Or when we're drawing it in two dimensions right here, the cross product would actually pop out of the page for b cross a. So I'll draw it over. It would be the circle with the dot. Or if I were to draw it analogous to this, so this right here, that was a cross b. And then b cross a is the exact same magnitude, but it goes in the other direction. That's b cross a. It just flips in the opposite direction. And that's why you have to use your right hand, because you might know that, oh, something's going to pop in or out of the page, et cetera, et cetera, but you need to know your right hand to know whether it goes in or out of the page. Anyway, let's see if we can get a little bit more intuition of what this is all about because this is all about intuition." + }, + { + "Q": "What happens if you add to much force?\nFor example, from 4:28 on, what would happen to a 42N weight?", + "A": "The weight would accelerate upwards until gravity brings it back down. It would jump , as you can try with any lever in real life.", + "video_name": "DiBXxWBrV24", + "timestamps": [ + 268 + ], + "3min_transcript": "This is getting monotonous. This is the output distance, from here to here. And let's figure out what the ratio has to be, for the ratio of the input distance to the output distance. Well, if we just divide both sides by 10, we get the distance input. It has to be 10 times the distance output, right? 100 divided by 10. So if the distance from the fulcrum to the weight is, I don't know, 5 meters, then the distance from where I'm applying the force to the fulcrum has to be 10 times that. It has to be 50 meters. So no matter what, the ratio of this length to this length has to be 10. And now what would happen? If I design this machine this way, I will be able to apply 10 newtons here, which is my maximum strength, 10 newtons downwards, and I will lift a 100 newton object. And now what's the trade off though? Nothing just pops out of thin air. The trade off is, is that I am going to have to push down for a much longer distance, for actually 10 times the distance as this object is going to move up. equal the work out. I can't through some magical machine-- and if you were able to invent one, you shouldn't watch this video and you should go build it and become a trillionaire-- but a machine can never generate work out of thin air. Or it can never generate energy out of thin air. That energy has to come from some place. Most machines actually you lose energy to friction or whatever else. But in this situation, if I'm putting in 10 newtons of force times some distance, whatever that quantity is of work, the work cannot change. The total work. It can go down if there is some friction in the system. So let's do another problem. And really they're all kind of the same formula. And then I'll move into a few other types of simple systems. I should use the line tool. We'll make this up on the fly. compound it further and et cetera, et cetera, using some of the other concepts we've learned. But I won't worry about that right now. So let's say that I'm going to push up here. Well no let me see what I want to do. I want to push down here with a force of-- let's say that this distance right here is 35 meters, this distance is 5 meters-- and let's say I'm going to push down with the force of 7 newtons, and what I want to figure out is how heavy of an object can I lift here. How heavy of an object. Well, all we have to do is use the same formula. But the moments-- and I know I used that word once before, so you might not know what it is-- but the moments on both sides of the fulcrum have to be the same." + }, + { + "Q": "At 7:28, what does spontaneously mean?", + "A": "With no additional energy / it will naturally just happen. This will also mean that delta G (change in free energy) is negative", + "video_name": "g_snytB7iQ0", + "timestamps": [ + 448 + ], + "3min_transcript": "indicating that a solid formed. This solid is called a precipitate. This solid spontaneously falls out of solution. This is the process of precipitation, which is the opposite of dissolution. In dissolution, we put a solid into water and we formed ions, right? In precipitation, the ions come together to form a solid, and that solid spontaneously falls out of solution. Silver chloride is our precipitate. We would still have some ions in solution. We would still have sodium cations and nitrate anions. In here, we would have sodium cations in solution and nitrate anions in solution. We could add them into here. We could say, NaNo3, aqueous, meaning those ions are present in solution. Let's look in more detail about what's happening We know that we had silver cations in solution. Here's our silver cation over here. We know that this ion is being surrounded by water molecules in the process of hydration, right? So these oxygens are partially negative right here. Since opposites attract, there's a force that's holding that ion in our solution, the forces of hydration. Same thing for the chloride anion, right? The partial positive charge on the hydrogen, opposite charges attract, right? So those water molecules are stabilizing the chloride anion in solution. But when we pour those two solutions together, we form our precipitate. We form silver chloride. We form this ionic crystal down here. Once again, opposite charges attract. So the positively-charged silver cation chloride anion here. Since we notice this solid to form, the electrostatic attractions of our ionic crystal must be stronger than the forces of hydration. This chloride anion would move into here, and then this silver cation would move into here. So the ions come out of solution and a precipitate spontaneously forms. We form our solid, silver chloride. This is one way to represent what's going on here. We could have also drawn out all of the ions, right? Instead of writing that, another way to represent what's happening would be to say, a solution of sodium chloride would be sodium cations in solution, so we write our sodium cations here, chloride anions, so Cl-. We added to that our solution of silver nitrate, which had silver ions in solution," + }, + { + "Q": "At 3:20, would the equation NaCl (s) --H2O--> Na+ (aq) + Cl- (aq) the same as saying NaCl (s) --H2O--> NaCl (aq) ? Or are they structurally different in solution?", + "A": "They mean the same thing, it is just that one is more specific than the other. NaCl (aq) requires knowing that this compound dissociates in aqueous solution. The other version spells that out explicitly.", + "video_name": "g_snytB7iQ0", + "timestamps": [ + 200 + ], + "3min_transcript": "is going to interact with the positive charge on the sodium. Right? Opposite charges attract. We have one water molecule here attracted to this sodium cation, and this water molecule would do the same, right? So partial negative charge attracted to the positive charge on the sodium. So the water molecules are going to pull off the sodium cations and eventually give you this situation over here. We have the partial negative oxygens, right? Partial negative oxygens are going to interact with the sodium cation, right? So the water is a dipole and the sodium cation is an ion. So we could call this an ion-dipole interaction. The water molecules break the ionic bonds, pull off the sodium cations, surround the sodium cation. We call this process hydration. where the ion is surrounded and stabilized by a shell of our solvent molecules. The same thing would happen with the chloride anions. Right up here, the chloride anion is negatively charged. But this time, the negative charge would be attracted to the positive part of our polar molecule, right? The oxygen is partially negative and this hydrogen here would be partially positive. Opposite charges attract, the positive charge is going to interact with the negative charge. Same for this molecule of water, partially positive hydrogen. Once again, this interaction is going to pull off that chloride anion and move it into solution. So we have our partial positive hydrogens interacting with our negatively-charged chloride anions here. Once again, we get ion-dipole interactions. The chloride anion is surrounded by our water molecules, The end result is, each sodium cation is surrounded by water molecules and each chloride anion is surrounded by water molecules. The sodium chloride has dissolved in water. We formed a solution, an aqueous solution, of sodium chloride. The way that you see that written, you would write, here we have solid sodium chloride, which we put into water, our solvent, and the water molecules surrounded our ions. Now we have sodium ions in solution, so we write an aq here for aqueous sodium ions, and we have chloride anions also in solution, so we write an aq here. So we have an aqueous solution of sodium chloride. The sodium chloride has dissolved in water. Now let's say we have one beaker that contains a solution of NaCl, so an aqueous solution of NaCl has sodium cations in solution" + }, + { + "Q": "When you say in 00:40ish \"the formula we've been speaking about\".. which video is this you are referring to? Thanks.", + "A": "That formula is explained in the previous video, Putting it all together: Pressure, flow, and resistance.", + "video_name": "fy_muPF0390", + "timestamps": [ + 40 + ], + "3min_transcript": "Let's imagine that I'm here. And this is me at the age-- actually, I'll do this right now-- so let's say at the age of 31. And then I go into the future. So this is a line for the present. And then I go into the future. And this is me again. And this is now at the age of 71. So I've got 40 extra years. I've even picked up a cane to help me walk around. And finally, I go even further in the future, and I'm being very optimistic. And I'm going to go ahead and say this is at the age of 101. I live to be a centennial. And I'm sitting in a wheelchair, and I'm going to wave out at you. So that's me in a wheelchair at the age of 101. And we're going to apply that formula we've been talking about so much, delta P equals Q times R. So I said Q is cardiac output. And we've got R is resistance. So we've got this formula. And I go today, and I go in 40 years, and I go again when I'm 101. And today, they tell me my blood pressure is 120 over 80. And actually, I went not too long ago, and that's pretty much what they told me it was. And I go in the future, and in 40 years, they tell me it's actually gone up. My blood pressure is now 150 over 90. And in fact, I go again, when I'm 101, and they say it's 180 over, let's say, 105. So the blood pressure is rising, and that's basically what I'm told. And they say, well, you've got to make sure you eat well and exercise, and that should help your blood pressure. And I'm left wondering what the connection is between the two. So let's figure out what that connection is exactly. So my blood pressure, I just said, is 120 over 80. And if I want to figure out my mean arterial pressure, meaning the average pressure in my arteries, to give me a good guess as to what it's going to be. So I know that I spend about 1/3-- my mean arterial pressure is going to equal 1/3 times my systolic, because I know I spend about 1/3 of the time-- my heart does anyway-- spends 1/3 of its time beating. And it spends 2/3 of its time relaxing. And the relaxing pressure is the diastolic pressure, that's 80. And so that works out to about 95. And so that's how I came up with that 95 number. And that's also why it's not exactly 100, which is what you'd think an average would be between two numbers. But it's because we don't spend the exact same amount of time in systole as diastole. So then if I wanted to figure out here, it would be 1/3 times 150 plus 2/3 times 90. And that works out to 110. And if I wanted to do it at the age of 101, my mean arterial pressure would be 1/3 times 180 plus 2/3 times" + }, + { + "Q": "He only mentions \"BB\" as homozygous dominant (@11:10), but I am under the assumption that \"bb\" would be called homozygous recessive. Is this correct?", + "A": "yes you right", + "video_name": "eEUvRrhmcxM", + "timestamps": [ + 670 + ], + "3min_transcript": "just said a brown-eyed gene, but what I should say is the brown-eyed version of the gene, which is the brown allele, or the blue-eyed version of the gene from my mom, which is the blue allele. Since the brown allele is dominant-- I wrote that up here --what's going to be expressed are brown eyes. Now, let's say I had it the other way. Let's say I got a blue-eyed allele from my dad and I get a brown-eyed allele for my mom. Same thing. The phenotype is going to be brown eyes. Now, what if I get a brown-eyed allele from both my mom and my dad? Let me see, I keep changing the shade of brown, but they're all supposed to be the same. So let's say I get two dominant brown-eyed alleles from my mom and my dad. Then what are you going to see? I'm still going to see brown eyes. So there's only one last combination because these are the only two types of alleles we might see in our population, although for most genes, there's more than two types. For example, there's blood types. There's four types of blood. But let's say that I get two blue, one blue allele from each of my parents, one from my dad, one from my mom. Then all of a sudden, this is a recessive trait, but there's nothing to dominate it. So, all of a sudden, the phenotype will be blue eyes. And I want to repeat again, this isn't necessarily how the alleles for eye color work, but it's a nice simplification to maybe understand how heredity works. There are some traits that can be studied in this simple way. But what I wanted to do here is to show you that many different genotypes-- so these are all different genotypes --they all coded for the same phenotype. didn't know exactly whether they were homozygous dominant-- this would be homozygous dominant --or whether they were heterozygotes. This is heterozygous right here. These two right here are heterozygotes. These are also sometimes called hybrids, but the word hybrid is kind of overloaded. It's used a lot, but in this context, it means that you got different versions of the allele for that gene. So let's think a little bit about what's actually happening when my mom and my dad reproduced. Well, let's think of a couple of different scenarios. Let's say that they're both hybrids. My dad has the brown-eyed dominant allele and he also" + }, + { + "Q": "At 3:15, Two oxygens that don't have double bond have 7 electrons? doesn't it need to have 6? so the formal charge is 1?", + "A": "It has 7 electrons because of the one shared by Nitrogen. The formal charge is 1 because it has 7. Had it only 6, the formal charge would be zero.", + "video_name": "bUCu7bPkZeI", + "timestamps": [ + 195 + ], + "3min_transcript": "Therefore, 24 minus 6 gives us 18 valence electrons left over. We're going to put those leftover valence electrons on our terminal atoms, which are our oxygens. And oxygen's going to follow the octet role. Currently, each oxygen has two valence electrons around it, the ones in magenta. So if each oxygen has two, each oxygen needs six more to complete the octet. And so I go ahead and put six more valence electrons on each one of my oxygens. Now each oxygen is surrounded by eight electrons. So the oxygens are happy. We added a total of six valence electrons to three oxygens. So 6 times 3 is 18. So we've used up all of the electrons that we need to represent. And so this dot structure, so far, it has all of our valence electrons here. Oxygen has an octet. So oxygen is happy. But nitrogen does not have an octet. If you look at the electrons in magenta, there are only six electrons around the nitrogen. And there are a couple of different ways that we could give nitrogen an octet. For example, we could take a lone pair of electrons from this top oxygen here and move them into here to share those electrons between that top oxygen and that nitrogen. So let's go ahead and draw that resulting dot structure. So we would have our nitrogen now with a double bond to our top oxygen. Our top oxygen had three lone pairs of electrons. But now it has only two, because electrons in green moved in to form a double bond. This nitrogen is bonded to an oxygen on the bottom left and an oxygen on the bottom right here. So this is a valid dot structure. We followed our steps. And we'll go ahead and put this in brackets and put a negative charge outside of our brackets like that. So that's one possible dot structure. But we didn't have to take a lone pair of electrons from the top oxygen. We could've taken a lone pair of electrons from the oxygen on the bottom left here. So if those electrons in blue moved in here, would have been equally valid. We could have shown this oxygen on the bottom left now bonded to this nitrogen, and it used to have three lone pairs. Now it has only two. And now this top oxygen is still a single bond with three lone pairs around it. And this bottom right oxygen is still a single bond with three lone pairs around it. So this is a valid dot structure as well. So let's go ahead and put our brackets with a negative charge. And then, of course, we could have taken a lone pair of electrons from the oxygen on the bottom right. So I could have moved these in here to form a double bond. And so now, we would have our nitrogen double bonded to an oxygen on the bottom right. The oxygen on the bottom right now has only two lone pairs of electrons. The oxygen at the top, single bond with three lone pairs. And then the same situation for this oxygen on the bottom left. And so this is, once again, another possible dot structure. And so these are considered to be" + }, + { + "Q": "So at 6:59 when the germ cells go through Meiosis, to make the gametes, does that mean the germ cells are haploid?", + "A": "first they are are diploid then after meiosis they become haploid", + "video_name": "PvoigrzODdE", + "timestamps": [ + 419 + ], + "3min_transcript": "These cells here eventually differentiate into the cells into the lungs, and obviously at this scale, the cells are way too small to even see. These cells differentiate into the cells of the heart. Now, I want to draw an important distinction here. Because most of the cells that I've just depicted here that are just a product of mitosis, these are your, I guess you could say these are your body cells, or these are your somatic cells. So these, all of these cells that I'm pointing out in your heart, your lung, your brain, these are somatic cells, or body cells. Somatic cells. And so you're probably wondering, well how do I eventually get these haploid number cells? How do I eventually get, if I'm talking about a male, how do I eventually get these haploid sex cells, these gametes, these sperm cells? I'm talking about a female, how do I eventually get these ova, these egg cells that have a haploid number? And the way that happens is some of these cells So they're going to differentiate into germ cells. In the case of, and they're going to differentiate when I say into germ cells, they're going to differentiate into your gonads. In the case of a female, the gonads are the ovaries. And in the case of a male, the gonads are the testes. The gonads are the testes. And the germ cells in the gonads or the cells that have differentiated into being part of the testes and ovaries, those germ cells. So we differentiate them from somatic cells. So there are germ cells there. Germ cells in your ovaries and testes. They, through the process of meiosis, they through the process of meiosis, So if you're female you're going to produce eggs. If you're male you're going to produce sperm. But this is through the process of meiosis. Meiosis you're going to produce sperm in the case of a man, and you're going to produce ova in the case of a female. And this brings up a really interesting thing, because throughout biology we talk about mutations and natural selection and whatever else. And it's important to realize how mutations may affect you and your offspring. So if you have a mutation in one of the somatic cells here, let's say in a skin cell, or in you brain, or in the heart, that may affect your ability to, you know, especially if God forbid it's a really dangerous thing like cancer, and it happens when you're young, before you've had a chance to reproduce and you're not able to survive," + }, + { + "Q": "At 2:34 What does accreting mean?", + "A": "accretion is the gradual gathering of matter.", + "video_name": "VbNXh0GaLYo", + "timestamps": [ + 154 + ], + "3min_transcript": "imagine is when you have the shockwave traveling out from a supernova, let's say you had a cloud of molecules, a cloud of gas, that before the shockwave came by just wasn't dense enough for gravity to take over, and for it to accrete, essentially, into a solar system. When the shockwave passes by it compresses all of this gas and all of this material and all of these molecules, so it now does have that critical density to form, to accrete into a star and a solar system. We think that's what's happened, and the reason why we feel pretty strongly that it must've been caused by a supernova is that the only way that the really heavy elements can form, or the only way we know that they can form is in kind of the heat of a supernova, and our uranium, the uranium that seems to be in our solar system on Earth, seems to have formed four and a half billion years ago, and we'll talk in a little bit more depth in future videos on exactly how people figure that out, but since the uranium seems about the same age as our solar system, it must've been formed at around the same time, and it must've been formed by a supernova, and it must be coming from a supernova, so a supernova shockwave must've passed through our part of the universe, and that's a good reason for gas to get compressed and begin to accrete. So you fast-forward a few million years. That gas would've accreted into something like this. It would've reached the critical temperature, critical density and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium, and this right here is our early sun. Around the sun you have all of the gases and particles and molecules that had enough angular velocity to not fall into the sun, to go into orbit around the sun. because you can kinda view this as kind of a big cloud of gas, so they're always bumping into each other, but for the most part it was their angular velocity, and over the next tens of millions of years they'll slowly bump into each other and clump into each other. Even small particles have gravity, and they're gonna slowly become rocks and asteroids and, eventually, what we would call \"planetesimals,\" which are, kinda view them as seeds of planets or early planets, and then those would have a reasonable amount of gravity and other things would be attracted to them and slowly clump up to them. This wasn't like a simple process, you know, you could imagine you might have one planetesimal form, and then there's another planetesimal formed, and instead of having a nice, gentle those two guys accreting into each other, they might have huge relative velocities and ram into each other, and then just, you know, shatter, so this wasn't just a nice, gentle process of constant accretion." + }, + { + "Q": "At 4:08 you say that the moon was a part of the earth ere it was hit by a proto planet named theia so if the earth and the moon is made up of same type of stuff why there is neither gravity nor life on the moon ?", + "A": "The Moon does have gravity. It cannot support life because it isn t composed in the same way as the Earth. It is smaller, mostly composed of lighter elements like the Earth s crust, and lacks a significantly active core to generate any substantial magnetic field. As a result, it isn t able to maintain an atmosphere or liquid water. Thus life isn t really able to even get started on the Moon.", + "video_name": "VbNXh0GaLYo", + "timestamps": [ + 248 + ], + "3min_transcript": "four and a half billion years ago, and we'll talk in a little bit more depth in future videos on exactly how people figure that out, but since the uranium seems about the same age as our solar system, it must've been formed at around the same time, and it must've been formed by a supernova, and it must be coming from a supernova, so a supernova shockwave must've passed through our part of the universe, and that's a good reason for gas to get compressed and begin to accrete. So you fast-forward a few million years. That gas would've accreted into something like this. It would've reached the critical temperature, critical density and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium, and this right here is our early sun. Around the sun you have all of the gases and particles and molecules that had enough angular velocity to not fall into the sun, to go into orbit around the sun. because you can kinda view this as kind of a big cloud of gas, so they're always bumping into each other, but for the most part it was their angular velocity, and over the next tens of millions of years they'll slowly bump into each other and clump into each other. Even small particles have gravity, and they're gonna slowly become rocks and asteroids and, eventually, what we would call \"planetesimals,\" which are, kinda view them as seeds of planets or early planets, and then those would have a reasonable amount of gravity and other things would be attracted to them and slowly clump up to them. This wasn't like a simple process, you know, you could imagine you might have one planetesimal form, and then there's another planetesimal formed, and instead of having a nice, gentle those two guys accreting into each other, they might have huge relative velocities and ram into each other, and then just, you know, shatter, so this wasn't just a nice, gentle process of constant accretion. actually happened early in Earth's history, and we actually think this is why the Moon formed, so at some point you fast-forward a little bit from this, Earth would have formed, I should say, the mass that eventually becomes our modern Earth would have been forming. Let me draw it over here. So, let's say that that is our modern Earth, and what we think happened is that another proto-planet or another, it was actually a planet because it was roughly the size of Mars, ran into our, what it is eventually going to become our Earth. This is actually a picture of it. This is an artist's depiction of that collision, where this planet right here is the size of Mars, and it ran into what would eventually become Earth. This we call Theia. This is Theia, and what we believe happened, and if you look up, if you go onto the Internet, you'll see some simulations" + }, + { + "Q": "at 2:35 sal says \"a few million years ago\" I Dont think thats right", + "A": "2:42 corrects that", + "video_name": "VbNXh0GaLYo", + "timestamps": [ + 155 + ], + "3min_transcript": "imagine is when you have the shockwave traveling out from a supernova, let's say you had a cloud of molecules, a cloud of gas, that before the shockwave came by just wasn't dense enough for gravity to take over, and for it to accrete, essentially, into a solar system. When the shockwave passes by it compresses all of this gas and all of this material and all of these molecules, so it now does have that critical density to form, to accrete into a star and a solar system. We think that's what's happened, and the reason why we feel pretty strongly that it must've been caused by a supernova is that the only way that the really heavy elements can form, or the only way we know that they can form is in kind of the heat of a supernova, and our uranium, the uranium that seems to be in our solar system on Earth, seems to have formed four and a half billion years ago, and we'll talk in a little bit more depth in future videos on exactly how people figure that out, but since the uranium seems about the same age as our solar system, it must've been formed at around the same time, and it must've been formed by a supernova, and it must be coming from a supernova, so a supernova shockwave must've passed through our part of the universe, and that's a good reason for gas to get compressed and begin to accrete. So you fast-forward a few million years. That gas would've accreted into something like this. It would've reached the critical temperature, critical density and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium, and this right here is our early sun. Around the sun you have all of the gases and particles and molecules that had enough angular velocity to not fall into the sun, to go into orbit around the sun. because you can kinda view this as kind of a big cloud of gas, so they're always bumping into each other, but for the most part it was their angular velocity, and over the next tens of millions of years they'll slowly bump into each other and clump into each other. Even small particles have gravity, and they're gonna slowly become rocks and asteroids and, eventually, what we would call \"planetesimals,\" which are, kinda view them as seeds of planets or early planets, and then those would have a reasonable amount of gravity and other things would be attracted to them and slowly clump up to them. This wasn't like a simple process, you know, you could imagine you might have one planetesimal form, and then there's another planetesimal formed, and instead of having a nice, gentle those two guys accreting into each other, they might have huge relative velocities and ram into each other, and then just, you know, shatter, so this wasn't just a nice, gentle process of constant accretion." + }, + { + "Q": "At 6:20 should the empirical formula for water be HHO instead of H2O?", + "A": "No H20 is the empirical formula on its own.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 380 + ], + "3min_transcript": "but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have Well, if it's not drawn, then it must be a hydrogen. That's actually the convention that people use in organic chemistry. So there's multiple ways to do a structural formula, but this is a very typical one right over here. As you see, I'm just getting more and more and more information as I go from empirical to molecular to structural formula. Now, I want to make clear, that empirical formulas and molecular formulas aren't always different if the ratios are actually, also show the actual number of each of those elements that you have in a molecule. A good example of that would be water. Let me do water. Let me do this in a different color that I, well, I've pretty much already used every color. Water. So water we all know, for every two hydrogens, for every two hydrogens, and since I already decided to use blue for hydrogen let me use blue again for hydrogen, for every two hydrogens you have an oxygen. It just so happens to be, what I just wrote down I kind of thought of in terms of empirical formula, in terms of ratios, but that's actually the case. A molecule of hydrogen, sorry, a molecule of water has exactly two hydrogens and, and one oxygen. If you want to see the structural formula, you're probably familiar with it or you might be familiar with it. Each of those oxygens in a water molecule are bonded to two hydrogens, are bonded to two hydrogens. So hopefully this at least begins to appreciate different ways of referring to or representing a molecule." + }, + { + "Q": "3:50 what's the meaningo of those double bonds ?", + "A": "A double bond is where there are four electrons shared between two atoms. You will learn more about these in future videos.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 230 + ], + "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have" + }, + { + "Q": "At 4:53, why do you have to draw a structural formula of a carbon with 3 single lines, and 3 other double lines bonding with each other?", + "A": "Because that simply is what the structure of benzene is. Google it and you ll find many pictures that look just like that. You cannot give each carbon an octet of electrons without making those double bonds.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 293 + ], + "3min_transcript": "That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have Well, if it's not drawn, then it must be a hydrogen. That's actually the convention that people use in organic chemistry. So there's multiple ways to do a structural formula, but this is a very typical one right over here. As you see, I'm just getting more and more and more information as I go from empirical to molecular to structural formula. Now, I want to make clear, that empirical formulas and molecular formulas aren't always different if the ratios are actually, also show the actual number of each of those elements that you have in a molecule. A good example of that would be water. Let me do water. Let me do this in a different color that I, well, I've pretty much already used every color. Water. So water we all know, for every two hydrogens, for every two hydrogens, and since I already decided to use blue for hydrogen let me use blue again for hydrogen, for every two hydrogens you have an oxygen." + }, + { + "Q": "At 3:50, why is there a \"double bond\" on the hexagon of carbon atoms? What is a double bond? When should we use double bonds?", + "A": "That molecule is benzene, it has the formula C6H6. The only way for each carbon to have 8 electrons around it is to form double bonds. Double bonds simply are just another bond between the atoms. As for when to use them the dot structure videos go over this. They re often needed to make elements follow the octet rule or to minimise formal charge.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 230 + ], + "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have" + }, + { + "Q": "At 3:45 Sal draws the Benzene molecule, mentioning that every other atom is joined by a double bond, while the rest are single. What is the significance of this and how do you know what kind of bond (single, double, etc) it is?", + "A": "Benzene (a carcinogen) has 2 possible formations. Going clockwise you can start with a double bond, then a single bond, etc. Or you can do the same thing but go anti-clockwise. the 2 structures are very similar. So instead of being one or the other it is actually somewhere between the 2. Each bond is not single or double but 1.5. The 1/2 is normally shown as a circle inside the carbon hexagon.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 225 + ], + "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have" + }, + { + "Q": "At 4:28, there is a structure of benzene shown. What I wanted to know is in the formula are the hydrogens connected to the carbons connected to each other?", + "A": "Hydrogen is only capable of making a single bond, so all of the hydrogen atoms in benzene are bound only to a single atom of carbon.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 268 + ], + "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have" + }, + { + "Q": "At 3:35, Sal draws a structural formula, but some covalent bonds have two lines. What do two lines mean in structural formulas?", + "A": "It means a double bond. It s exactly how it sounds, 2 bonds between the two atoms instead of 1.", + "video_name": "bmjg7lq4m4o", + "timestamps": [ + 215 + ], + "3min_transcript": "Molecular formula. And the molecular formula for benzene, which is now going to give us more information than the empirical formula, tells us that each benzene molecule has six hydrogens, and, sorry, six carbons and six, (laughs) I'm really having trouble today, six hydrogens, (laughs) six carbons, and, six hydrogens. Now, the ratio is still one to one, you get that right over here, it's very easy to go from a molecular formula to an empirical formula. You essentially are losing information. You're just saying the ratio, OK, look, it's a ratio of six to six, which is the same thing as one to one. If we wanted to, we could write this as C one H one just like that to show us that the ratio for every carbon we have a hydrogen. And we see that that's actually the case in one molecule, for every six carbons you have six hydrogens, That may not satisfy you, you might say, well, OK, but how are these six carbons and six hydrogens actually structured? I want more information. And for that, you would wanna go to a structural formula. Structural formula, which will actually give you the structure, or start to give you the structure of a benzene molecule. A benzene molecule would be drawn like... So you would have six carbons in a hexagon. So one, going to write this way, one, two, three, four, five, six carbons in a hexagon just like that. And then you have a double bond, every other of these bonds on the hexagon is a double bond. Each of these carbons are also attached to a hydrogen, also bonded to a hydrogen. Each of these lines that I'm drawing, this is a bond, it's a covalent bond, but it's a sharing of electrons, and that's what keeps these carbons near each other and what keeps the hydrogens kind of tied to each, or, the hydrogens tied to the carbons and the carbons tied to the hydrogens. So let me draw it just like this. And this is only one variant of a structural, it's hard to see this one I just drew, so let me see if I can do a little bit... Oh, that's about as good, hopefully you see there's a hydrogen there, and there's a hydrogen right over there. This is one variant of a structural formula, some structural formulas will actually give you some 3D information, will tell you whether a molecule is kind of popping in or out of the page. Others might not be as explicit, once you go into organic chemistry chains of carbons are just done, they're just ... You might see something like this for benzene, where the carbons are implicit as the vertex of each, there's an implicit carbon at each of these vertices, and then you say, OK, carbon's gotta have, not gotta, but it's typically going to have" + }, + { + "Q": "What happens to / with the positive electron from the battery? 3:51 \"gonna leave a net positive charge\"... Can you explain this? Thanks", + "A": "the copper wire is neutral when no battery is attached when there is a + from the battery, Cu- will be attracted then the area will have less - , thus have a net + charge hope this solve your problem", + "video_name": "ZRLXDiiUv8Q", + "timestamps": [ + 231 + ], + "3min_transcript": "that guy right there. And that's the one that is the easiest to pull away from copper and have it go participate in conduction, in electric current. If I have a chunk of copper, every copper atom will have the opportunity to contribute one, this one lonely electron out here. If we look at another element, like for instance silver, silver has this same kind of electron configuration, where there's just one out here. And that's why silver and copper are such good, good conductors. Now we're gonna build, let's build a copper wire. Here's sort of a copper wire. It's just made of solid copper. It's all full of copper atoms. There's our little battery. This is the minus sign, this is the plus side. And we'll hook up a battery to this. What's going on in here? Inside this copper is a whole bunch of electrons that are associated with atoms. It's a neutral piece of metal. There's the same number of protons as there is electrons. But these electrons are a little bit loose. So if I put a plus over there, that's this situation right here, where a plus is attracting a minus. So an electron is gonna sort of wander over this way and go like that. And that's gonna leave a net positive charge in this region. So these electrons are all gonna start moving in this direction. And down at the end, travel in here, and it's gonna go in there and make up the difference. So if I had a net positive charge here from the electrons leaving and going to the left, this battery would fill those in. And I'm gonna get a net movement of charge, of negative charge, around in this direction, like this. The question is, how do I measure that? How do I measure or give a number to that amount of stuff that's going on? So we wanna quantify that, we wanna assign a number to the amount of current happening here. What we do is, in our heads, we put a boundary across here. So just make that up in your head. And it cuts all the way through the copper. And what we know, we're gonna stand right here. We're gonna keep our eye right on this boundary down in here. As we watch, what we're gonna do is, we're gonna count the number of electrons" + }, + { + "Q": "Why is it -q instead of +q in the copper wire at 5:03 because the net charge is positive since there will be more protons than electrons?", + "A": "In metals, its the electrons that do the moving. Electrons have a negative charge, so they are labeled -q.", + "video_name": "ZRLXDiiUv8Q", + "timestamps": [ + 303 + ], + "3min_transcript": "There's our little battery. This is the minus sign, this is the plus side. And we'll hook up a battery to this. What's going on in here? Inside this copper is a whole bunch of electrons that are associated with atoms. It's a neutral piece of metal. There's the same number of protons as there is electrons. But these electrons are a little bit loose. So if I put a plus over there, that's this situation right here, where a plus is attracting a minus. So an electron is gonna sort of wander over this way and go like that. And that's gonna leave a net positive charge in this region. So these electrons are all gonna start moving in this direction. And down at the end, travel in here, and it's gonna go in there and make up the difference. So if I had a net positive charge here from the electrons leaving and going to the left, this battery would fill those in. And I'm gonna get a net movement of charge, of negative charge, around in this direction, like this. The question is, how do I measure that? How do I measure or give a number to that amount of stuff that's going on? So we wanna quantify that, we wanna assign a number to the amount of current happening here. What we do is, in our heads, we put a boundary across here. So just make that up in your head. And it cuts all the way through the copper. And what we know, we're gonna stand right here. We're gonna keep our eye right on this boundary down in here. As we watch, what we're gonna do is, we're gonna count the number of electrons and we're gonna have a stop watch and we're gonna time that. So we're gonna get, basically, this is charge, it's negative charge, and it's moving to the side. What we're gonna do is, at one little spot right here, we're just gonna count the number that go by in one second. So we're gonna get charge per second. It's gonna be a negative charge moving by. That's what we call current. It's the same as water flowing by in a river. That's the same idea. Now I'm gonna set up a different situation that also produces a current. And this time, we're gonna do it with water, water and salt. Let's build a tube of salt, of salt water, like this. We're gonna pretend this is some tube that's all full of water. I'm also gonna put a battery here. Let's put another battery." + }, + { + "Q": "At 6:15 in the video it discusses the difficulty of eating and digesting cellulose. Are cells in fruit not composed of cellulose, or does the plant do something magically different in the process of fruiting?", + "A": "Fruit, being made of plant cells, definitely does contain cellulose. We can t digest it, but there are so many other good things in fruit (sugars, nutrients, etc.), that we eat it anyways. Cellulose is the fiber part that is advertised as being so healthy for your digestive system, by virtue of the fact that it can t be digested. Tree branches, on the other hand, are basically pure cellulose, with nothing else, so are noticeably less appetizing.", + "video_name": "d9GkH4vpK3w", + "timestamps": [ + 375 + ], + "3min_transcript": "Something plants inherited from their ancestors was a rigid cell wall surrounding the plasma membrane of each cell. This cell wall of plants is mainly made out of cellulose and lignin, which are two really tough compounds. Cellulose is by far the most common and easy to find complex carbohydrate in nature, though if you were to include simple carbohydrates as well, glucose would win that one. This is because, fascinating fact, cellulose is in fact just a chain of glucose molecules. You're welcome. If you want to jog your memory about carbohydrates and other organic molecules, you can watch this episode right here. Anyway, as it happens, you know who needs carbohydrates to live? Animals. But you know what's a real pain in the ass to digest? Cellulose. Plants weren't born yesterday. Cellulose is a far more complex structure than you'll generally find in a prokaryotic cell, and it's also one of the main things that differentiates a plant cell from an animal cell. Animals do not have this rigid cell wall. They have a flexible membrane that frees them up to move around and eat plants and stuff. However, the cell wall gives structure and also protects it, to a degree. Which is why trees aren't squishy, and they don't giggle when you poke them. The combination of lignin and cellulose is what makes trees, for example, able to grow really, really freaking tall. Both of these compounds are extremely strong and resistant to deterioration. When we eat food, lignin and cellulose is what we call roughage because we can't digest it. It's still useful for us on certain aspects of our digestive system, but it's not nutritious. Which is why eating a stick is really unappetizing and like your shirt. This is a 100% plant shirt, but it doesn't taste good. We can't go around eating wood like a beaver, or grass like a cow, because our digestive systems just aren't set up for that. A lot of other animals that don't have access to delicious donut burgers have either developed gigantic stomachs, like sloths, or multipe stomachs, like goats, in order to make a living eating cellulose. These animals have a kind of bacteria in their stomachs that actually does the digestion of the cellulose for it. glucose molecules which can then be used for food. Other animals, like humans, mostly carnivores, don't have any of that kind of bacteria, which is why it's so difficult for us to digest sticks. But there is another reason why cellulose and lignin are very, very useful to us as humans. It burns, my friends. This is basically what would happen in our stomachs. It's oxidizing. It's producing the energy that we would get out of it if we were able to, except it's doing it very, very quickly. This is the kind of energy, like this energy that's coming out of it right now, is the energy that would be useful to us if we were cows. But we're not, so instead, we just use it to keep ourselves warm on the cold winter nights. (blows air) Ow; it's on me; ow. Anyway, while we animals are walking around, spending our lives searching for ever more digestable plant materials, plants don't have to do any of that. They just sit there and they make their own food. We know how they do that." + }, + { + "Q": "@8:33 when the compound reacts with H2 and Pt, since it goes from an alkyne all the way to an alkane, does the same reaction happen twice? syn addition twice with the reagents?", + "A": "Yes, it is effectively the same reaction happening twice.", + "video_name": "RdFfIEDxo18", + "timestamps": [ + 513 + ], + "3min_transcript": "OK, so carbon triple bonded to another carbon, and we'll put a methyl group on each side like that. OK, so let's do a few different reactions with the same substrate here. So our first reaction will just be a normal hydrogenation with hydrogen gas, and let's use platinum as our catalyst. So this is not a poison catalyst. This is a normal catalyst. So what's going to happen is, first you're going to reduce the alkyne to an alkene. And then since there's no way of stopping it, it's going to reduce the alkene to an alkane. So this is going to reduce the alkyne all the way to an alkane. So if we go back up here to beginning, remember, we said that a poison catalyst will stop at the alkene, but if it's not a poison catalyst, down here. So this reaction is going to produce an alkane. Let's go ahead and draw the product. So we know that there are four carbons in my starting materials. There's going to be four carbons when I'm done here, so these two carbons in the center here are going to turn into CH2's. And then on either side, we still have our CH3's. So this is going to form butane as the product. All right, this time let's use a hydrogen gas, and let's use a Lindlar palladium here. This is our poisoned catalyst. So it's going to reduce our alkyne to an alkene, and then it's going to stop. And you have to think, what kind of alkene will you get? You will get a cis-alkene. So if we draw our two hydrogens adding on to the same sides, So our methyl groups will be going-- this and this would be our product, a cis-alkene. All right, let's do one more, same starting material. So this one right here, except this time we're going to add sodium, and we're going to use ammonia as our solvent. And remember, this will reduce our alkyne to an alkene, but it will form a trans-alkene as your product. So when you're drawing your product down here, you want to make sure that your two hydrogens are trans to each other. So they add on the mechanism, and then your two methyl groups would also be on the opposite side like that. So look very closely as to what you are reacting things with. Is it a normal hydrogenation reaction? Is it a hydrogenation reaction with a poison catalyst, which would form a cis-alkene? Or is it reduction with sodium and ammonia, which will give you a trans-alkene." + }, + { + "Q": "at 5:56 you get another Na with one electron, how do you know there are plenty left over after using Na with one electron already in the first step when u started off the reaction? Same with ammonia in the next step following 5:56?", + "A": "There are billions and billions of Na atoms and NH\u00e2\u0082\u0083 molecules in the reaction mixture. The Na and ammonia are always in excess. The alkyne is the limiting reactant.", + "video_name": "RdFfIEDxo18", + "timestamps": [ + 356, + 356 + ], + "3min_transcript": "These electrons are going to repel, and they're going to want to try to be as far away from each other as they possibly can. So what's going to happen is, we have our two carbons right here. And let's say that these two electrons stay over here on this side. This one electron's going to go over to the opposite side. They're going to try to get as far away from each other as they possibly can. And same thing with these R group here, right? So this R group is going to try to get as far away from this R-prime group as it possibly can. So this trans conformation is the more stable one. So this is our negatively charged carbanion right here. So in the next step of the mechanism, we remember ammonia is present. So let's go ahead and draw an ammonia molecule floating around like that. So here is our ammonia molecule. And the carbanion is going to act as a base, and it's going to take a proton from the ammonia molecule. going to form a new bond with this proton, and these electrons are going to kick off onto the nitrogen. So let's go ahead and draw the results of that acid-base reaction. So now we have our two carbons, with an R group right here, R-prime right here. And now this carbon on the right is bonded to a proton. It bonded to a hydrogen like that. And then we still have our radical down here, so there is one electron on that carbon as well. All right, so the next step of our mechanism? Well, there's plenty of sodium present. So here's a sodium atom with one valence electron. The sodium is going to donate this electron to this carbon. So just use a half-headed arrow to show the movement of one electron. So if that sodium atom donates that one valence electron to that carbon, let's go ahead and draw the results of that. So we have two carbons double bonded, an R group over here, And this carbon had one electron around it. It just picked up one more from a sodium atom. So it's like that, which would give it a negative 1 formal charge. So this carbon has a negative 1 formal charge. So let's go ahead and draw that negative 1 formal charge. It's a carbanion. And once again, ammonia is floating around, so let's go ahead and draw ammonia right here, so NH3 like that. And the same thing is going to happen as did before, right? The negative charge is going to grab a proton. It's going to act as a base. And these electrons are going to kick off onto the nitrogen here. And so we protonate our carbanion, and we have completed our mechanism because now we have our two R groups across from each other. And we added on two hydrogens across from each other as well like that. So we formed a trans-alkene. All right, so that's the mechanism to form a trans-alkene. Let's look at a few examples." + }, + { + "Q": "At 9:50 in the video, 3-hexanol is explained to have a dipole-dipole moment between the O and the H, resulting in hydrogen bonding. My question is this: would there also be a dipole-dipole moment between the O and the C in 3-hexanol, and would that dipole-dipole moment be the same strength as that in 3-hexanone?", + "A": "I agree there must be some polarization between the oxygen and the carbon in the alcohol, but I don t think it would be as strong as in the ketone. In the alcohol the oxygen is pulling electron density from both the hydrogen and the carbon, which is more electronegative than the hydrogen so the electron density shift is mostly away from hydrogen. In contrast, in the ketone the oxygen is pulling electron density exclusively from the carbon.", + "video_name": "pILGRZ0nT4o", + "timestamps": [ + 590 + ], + "3min_transcript": "So we have a dipole for this molecule, and we have the same dipole for this molecule of 3-hexanone down here. Partially negative oxygen, partially positive carbon. And since opposites attract, the partially negative oxygen is attracted to the partially positive carbon on the other molecule of 3-hexanone. And so, what intermolecular force is that? We have dipoles interacting with dipoles. So this would be a dipole-dipole interaction. So let me write that down here. So we're talk about a dipole-dipole interaction. Obviously, London dispersion forces would also be present, right? So if we think about this area over here, you could think about London dispersion forces. But dipole-dipole is a stronger intermolecular force compared to London dispersion forces. And therefore, the two molecules here of 3-hexanone are attracted to each other more than the two molecules of hexane. for these molecules to pull apart from each other. And that's why you see the higher temperature for the boiling point. 3-hexanone has a much higher boiling point than hexane. And that's because dipole-dipole interactions, right, are a stronger intermolecular force compared to London dispersion forces. And finally, we have 3-hexanol over here on the right, which also has six carbons. One, two, three, four, five, six. So we're still dealing with six carbons. If I draw in another molecule of 3-hexanol, let me do that up here. So we sketch in the six carbons, and then have our oxygen here, and then the hydrogen, like that. We know that there's opportunity for hydrogen bonding. Oxygen is more electronegative than hydrogen, so the oxygen is partially negative and the hydrogen is partially positive. The same setup over here on this other molecule of 3-hexanol. So partially negative oxygen, partially positive hydrogen. Let me draw that in. So we have a hydrogen bond right here. So there's opportunities for hydrogen bonding between two molecules of 3-hexanol. So let me use, let me use deep blue for that. So now we're talking about hydrogen bonding. And we know that hydrogen bonding, we know the hydrogen bonding is really just a stronger dipole-dipole interaction. So hydrogen bonding is our strongest intermolecular force. And so we have an increased attractive force holding these two molecules of 3-hexanol together. And so therefore, it takes even more energy for these molecules to pull apart from each other. And that's reflected in the higher boiling point for 3-hexanol, right? 3-hexanol has a higher boiling point than 3-hexanone and also more than hexane. So when you're trying to figure out boiling points, think about the intermolecular forces that are present between two molecules. And that will allow you to figure out" + }, + { + "Q": "around 0:34 sal shows a solar flare. what would be the effects if one of these were powerful enough to get to earth?", + "A": "If a powerful solar eruption were to be aimed at earth, humans themselves would not be affected. But it would have a great affect on Earth s electrical grid. A shock like that could short out anything that isn t shielded and it could be years before the grid is fully restored. Such an event has happened in the past around the Civil war where telegraph systems stopped working.", + "video_name": "jEeJkkMXt6c", + "timestamps": [ + 34 + ], + "3min_transcript": "" + }, + { + "Q": "actually gravitational force is independent of mass right and in the video i saw tha sal said at4:57 as mass of the brick is larger force on the brick is also larger am i right in what i have asked", + "A": "No, gravitational force is not independent of mass. That s quite a strange idea. Gravitational force is directly proportional to mass.", + "video_name": "36Rym2q4H94", + "timestamps": [ + 297 + ], + "3min_transcript": "this height and the center of the moon squared so they both have this exact expression on it So let's replace that expression, let's just call that the gravitational field on the moon define any number of the mass will tell you the weight of that object on the moon, or the gravitational force acting downard on that object on moon so this is the gravitational field on the moon it's called g sub m, and all of this is all of this quantity combined so we simplified that way, the force on the brick due to the moon is going to be equal to lower case g on the moon normally we will use this lower case g for the gravitational constant on earth with the gravitational field on earth or sometime the gravitational acceleration on earth but now it's refering to the moon so that's why the lower case subscript m it's equal to that times the mass of the brick for the case of the feather, the force on the feather is equal to all is equal to g sub m times the mass of the feather so now assuming that the mass of the brick is much greater than the mass of the feather, so let's going to assume, a reasonable thing to assume that the mass of the brick is greater than the mass of the feather, than the mass of the feather what's going to be their relative forces? Here you have the greater mass times the same quantity you have the smaller mass times the same quantity so the mass of the brick is greater than the mass of the feather completely reasonable to say that the force of gravity on the brick is going to be greater than the force of gravity on the feather so if you do all of this so every thing we've done to this point is correct, you might say hey it's gonna be more force due to gravity one the brick, and that's why the brick will be accelerated down that's being gravitational force on this brick but it also has greater mass and we remember the larger the mass is, the less acceleration it will experience for a given force So what really determine how quickly either of these things fall is their accelerations and let's figure out their accelerations So we knew that, I will do this on a neutral color we know that force is equal to mass time acceleration So if we wanna figure out the acceleration of the brick we could write it the other way the acceleration, if we divide both side by mass we get acceleration is equal to force divided by mass acceleration is vector quantity and force is also vector quantity and in this situation we will use, I'll use, we're not using any" + }, + { + "Q": "At about 9:00-9:20 Sal says Earth goes into whats called \"Snowball Earth\" and ices over. Is that like an Ice Age, and if not, how are Snowball Earth and an Ice Age different?", + "A": "Snowball Earth is a period in earth s history where the whole surface of the earth was supposedly covered by ice. Snowball Earth periods are extreme ice ages. During most ice ages there was only glaciation in latitudes up to 40\u00c2\u00b0 or 50\u00c2\u00b0.", + "video_name": "E1P79uFLCMc", + "timestamps": [ + 540, + 560 + ], + "3min_transcript": "by with ultraviolet radiation from the Sun, which is very inhospitable to DNA and to life. And so the only life at this point could occur in the ocean, where it was protected to some degree from the ultraviolet radiation. The land was just open to it. Anything on the land would have just gotten irradiated. It's DNA would get mutated. It just would not be able to live. So what happened, and what I guess has to happen, and the reason why we are able to live on land now is that we have an ozone layer. We have an ozone layer up in the upper atmosphere that helps absorb, that blocks most of the UV radiation from the Sun. And now that oxygen began to accumulate, we have the Oxygen Catastrophe. Oxygen accumulates in the atmosphere. Some of that oxygen goes into the upper atmosphere. So we're now in this time period right over here. It goes into the upper atmosphere. to turn into ozone, which then can help actually block the UV light. And I'll do another video maybe on the ozone/oxygen cycle. So this oxygen production, it's crucial, one, to having an ozone layer so that eventually life can exist on the land. And it's also crucial because eukaryotic organisms need that oxygen. Now, the third thing that happened, and this is also pretty significant event, we believe that that oxygen that started to accumulate in the atmosphere, reacted with methane in the atmosphere. So it reacted with methane. And methane is an ozone-- not an ozone. It's a greenhouse gas. It helps retain heat in the atmosphere. And once it reacts with the oxygen and starts dropping out of the atmosphere as methane, we believe the Earth cooled down. And it entered its first, and some people believe it's longest, snowball period. So that's what they talk about right here It's sometimes called the Huronic glaciation. And that happened because we weren't able to retain our heat, if that theory is correct. And so the whole-- as the theory goes-- the whole Earth essentially just iced over. So as we go through the Proterozoic Eon, I guess the big markers of it is it's the first time that we now have an oxygen-rich atmosphere. It's the first time that eukaryotes can now come into existence because they now have oxygen to, I guess we could say, breathe. And the other big thing is now this is where the ozone forms. So this kind of sets the stage for in the next eon, for animals or living things, to eventually get on to the land. And we'll talk about that in the next video." + }, + { + "Q": "I think there's a mistake at 7:00 minutes, how did he get 4. 75m/s?\nI've been calculating it over and over it it keeps appearing to be 3.7m/s.\nWhat's going on?", + "A": "Try changing your calculator to degrees from radians, is that your problem?", + "video_name": "_0nDUXO0k7o", + "timestamps": [ + 420 + ], + "3min_transcript": "in this perpendicular direction. I'm plugging in the kinetic frictional force this 0.2 turns this perpendicular force into this parallel force, so I'm plugging in the force of kinetic friction and it just so happens that it depends on the normal force. That's why I'm plugging that in, I'm gonna need a negative 0.2 times 4 kg times 9.8 meters per second squared. And then I need to multiply by cosine of the angle in this case the angle is 30 degrees. Alright, now finally I divide by my total mass because I have no other forces trying to propel this system or to make it stop and my total mass is going to be 13 kg. You might object and think wait a minute, there's other forces here like this tension going this way, why don't we include that? benefit and appeal of treating this two-mass system as if it were a single mass is that we don't have to worry about these internal forces, it's there but that tension is also over here and on this side it's resisting the motion because it's pointing opposite the directional motion. On this side it's helping the motion, it's an internal force the internal force is canceled that's why we don't care about them, that's what this trick allows us to do by treating this two-mass system as a single object we get to neglect any internal forces because internal forces always cancel on that object. So if we just solve this now and calculate, we get 4.75 meters per second squared is the acceleration of this system. So this 4 kg mass will accelerate up the incline parallel to it with an acceleration of 4.75 meters per second squared. This 9 kg mass will accelerate downward Remember if you're going to then go try to find out what one of these internal forces are, we neglected them because we treated this as a single mass. But you could ask the question, what is the size of this tension? Often that's like a part two because we might want to know what the tension is in this problem, if we do that now we can look at the 9 kg mass individually so I can say for just the 9 kg mass alone, what is the tension on it and what are the force? We can find the forces on it simply by saying the acceleration of the 9 kg mass is the net force on the 9 kg mass divided by the mass of the 9 kg mass. Now this is just for the 9 kg mass since I'm done treating this as a system. This trick of treating this two-mass system as a single object is just a way to quickly get the magnitude of the acceleration. Now that I have that and I want to find an internal force I'm looking at just this 9 kg box." + }, + { + "Q": "at 1:09 the term \"short\" is mentioned, and then again a few seconds later.\n\nplease may I know what is a short?", + "A": "The term short in this context means to short-out; it basically occurs when a conductive element comes into contact with the circuit board, allowing voltage to bypass all resistors and capacitors. When this happens, this is bad. The voltage will rush through the element instead of the board with almost no resistance and destroy many, if not all, components. The solenoid safe-guards from this issue.", + "video_name": "gFFvaLzhYew", + "timestamps": [ + 69 + ], + "3min_transcript": "Alright today we're going to take a look at the Conair 1875 hair dryer.We're going to look at the different systems and functions inside of it, how it was made and how it works. And we're also going to take a look at how they were able to produce a hair dryer for less than $8.00 and still make a profit and still stay in business as a company because that's a very low price and the way low price. And the way they've done that is they've reduced a lot of cost and complexity and we'll take a look at how they've done that. So the first thing I want to take a look at is the plug here. This is called a \"ground fault interruptor circuit plug\" and it has two different sized prongs right here. There's a larger prong and a smaller prong. And that's very important. The larger prong is the neutral prong and that means you can't plug this in incorrectly, it has to go in in only one way. And that means that the power is grounded properly. So the power always goes to ground and that's a critical thing in a circuit like this. So what this plug does it's actually pretty smart it can tell if there's a power difference might occur when the hair dryer was, say, dropped in water or there was some sort of short that happened. Inside the hair dryer there are open electrical contacts that if they're put into water or some other conductive fluid they'll short out. And it will cause the, you know, it'll electrify the fluid. And in the past that was a huge problem because people would get schocked or electrocuted and now it's not as big a deal because we have these ground fault interruptor circuits. So let's take a look at what's inside of that. And I've already popped this apart to some degree. I'm going to see if I can get it the rest of the way here. Now I want to say one thing really quick here from a safety stand point: It's absolutely critical that you DON'T take apart any plugs ever without a professionnal! And if you do have a professional and you do end up taking apart a plug like this make sure that you never ever moulded housing it was injection moulded. There were two pieces of steel that came together and the molten plastic was injected and you can see there are little pin marks right here. And pins came in inside the mold and pushed this part out. And then there's a little plastic piece here with a spring and that's for the test switch. So the test switch pushes on this part right here on the printed circuit board and the reset switch is right here. So you push on the reset switch and it will reset it so if it gets triggered you can still use your hair dryer again later. So one thing I want to take a look at here is the printed circuit board here. So we've got a lot of really cool things happening on this printed circuit board. It is made out of fiberglass. It's got a thin layer of copper applied to it. And then on top of the copper is a layer of lacquer. (The copper) Before they put the copper layer down they actually etch away parts of the copper. So" + }, + { + "Q": "at 9:40, would that tetra molecule be cis or trans?", + "A": "That would be neither. It has the same group on all 4 positions.", + "video_name": "AiGGaJfoQ1Y", + "timestamps": [ + 580 + ], + "3min_transcript": "one more product for this reaction, and let's go back to our secondary carbocation over here. So, if it doesn't rearrange, you could actually take a proton from this secondary carbocation and form our last product. Let's think about the carbon in red right here. So we're looking at the carbons next door, so the carbon in red. Well, the carbon in magenta to the left doesn't have any proton, so we can't take a proton from that one, but the carbon to the right in magenta does. So if I squeeze in a hydrogen in here, finally, our base could come along and take this proton. And if that happen, then these electrons would move into here, so that gives us our final product. Just let me draw in our skeleton here, and then our double bond will form right... Actually, let me just redraw that double bond, so thing we're getting... We just redraw the whole thing. I think I have a little bit more space that I thought I did. our double bonded form here, and then we have our methyl groups coming off of that carbon. So the electrons in light blue moved into form our double bond. So we'd formed three products, three products from this E1 reaction. If we think about which one would be the major product, let's look at degrees of substitution. So let's go over to the alkene on the right. So when we think about the two carbons across our double bonds, there are one, two, three, four alkyl groups. So this is a tetrasubstituted alkene, so this should be the major product. This is the most stable alkene. Next, let's look at this one. So here's a carbon, and here's a carbon across our double bond. We can see this time we have only two alkyl groups. So this is a disubstituted alkene. And then finally, over here on the left, this would be a monosubstituted alkene. We have one alkyl group, so a monosubstituted alkene. And this one came from the secondary carbocation from no rearrangement, so this one is not gonna be a major product. Only a very small percentage of your products would be this monosubstituted alkene. Most of your products is gonna be your di and your tetrasubstituted alkene, So with your tetrasubstituted alkene being your major product since it is the most stable." + }, + { + "Q": "at 1:53 cant the compound be named as 3-Bromo-4,6-dimethylheptane if the functional group is given the first preference ?", + "A": "It is not the functional group, but the rule of lowest numbers, that determines the numbering. In your name, the lowest number is 3. In the other name, the lowest number is 2. The lowest number wins. But the bromo wins in determining the order in which the substituents are listed in the name. So the compound really is 5-bromo-2,4-dimethylheptane.", + "video_name": "aaZ-isZs4ko", + "timestamps": [ + 113 + ], + "3min_transcript": "- [Lecturer] You often see two different ways to name alkele halides. And so we'll start with the common way first. So think about alkele halides. First you wanna think about an alkele group, and this alkele group is an ethyl group, there are two carbons on it. So we write in here ethyl. And then since it's alkele halides, you wanna think about the halogen you have So this is chlorine so it's gonna end in i, so chloride. So ethyl chloride would be the name for this compound. Now let's name the same molecule using IUPAC nomenclature. In this case it's gonna be named as a halo alkane. So for a two carbon alkane that would be ethane. So I write in here ethane. And of course our halogen is chlorine, so this would be chloro. So chloroethane is the name of this molecule. If I had fluorine instead of chlorine, it would be fluoroethane. So let me write in here fluoro, notice the spelling on that. If I had a bromine instead of the chlorine, And finally if I had an iodine instead of the chlorine, it would be iodoethane. So let me write in here ioto. Let's name this compound using our common system. So again think about the alkyl group that is present. So we saw in earlier videos this alkyl group is isopropyl. So I write in here isopropyl. And again we have chlorine attached to that. So it would be isopropyl chloride using the common system. If I'm naming this using the IUPAC system, I look for my longest carbon chain, so that'd be one, two and three. I know that is propane so I write in here propane. And we have a chlorine attached to carbon two. So that would be 2-chloro, 2-chloropropane. Let's look at how to classify alkyl halides. We find the carbon that's directly bonded to our halogen and we see how many alkyl groups There's only one alkyl group, this methyl group here, attached to this carbon so that's called primary. So ethyl chloride is an example of a primary alkyl halide. If you look at isopropyl chloride down here. This is the carbon that's bonded to our halogen and that carbon is bonded to two alkyl groups. So that's said to be a secondary alkyl halide. And let me draw in an example of another one here really fast. So for this compound the carbon that is bonded to our halogen is bonded to three alkyl groups. So three methyl groups here. So that's called a tertiary alkyl halide. And the name of this compound is tert-butyl chloride. So that's the common name for it. And that's the one that you see used most of the time. For larger molecules it's usually easier to use" + }, + { + "Q": "at 2:48 why does it lose both hydrogen protons.", + "A": "Well..oxalic acid is a diprotic/dibasic acid, which means that it has two replaceable H atoms. Oxalic acid when it ionizes loses or liberates 2 hydrogen protons. It does this quite readily as the oxalate anion (acid with 2 H+ ions removed) is stabilized by -I effect and resonance.", + "video_name": "XjFNmfLv9_Q", + "timestamps": [ + 168 + ], + "3min_transcript": "We have that on the other carbon as well. This right here is what oxalic acid is. And it's an interesting acid, because it can actually donate This proton can be nabbed off, and this proton can also be contributed. And it's actually resident stabalized. If that doesn't mean anything to you, don't worry. You'll learn more about that in organic chemistry. But the important thing to realize here is that there's two protons to nab off of it. Now each molecule of sodium hydroxide-- remember when you put it in the water it really just dissolves, and you can really just think of it as hydroxide-- each molecule of hydroxide can nab one of the hydrogen protons. So for every one molecule of oxalic acid, you're going to need two hydroxides-- one to nab this hydrogen proton, and So let's write down the balanced equation that we're dealing with here. So we're going to start off with some oxalic acid. So that has two hydrogens-- so it's H2-- two carbons, and then four oxygens-- O4. It's dissolved in water, so it's an aqueous solution. And to that, we're going to add sodium hydroxide. Now I just told you that you're going to need two of the hydroxides to fully neutralize the oxalic acid. So you're going to need two of them. And this is also in our aqueous solution. And once the reaction happens, this guy will have lost both of the hydrogen protons, so let me draw that. So it will look like this. No more hydrogen, so it's C2O4. It'll have a negative 2 charge. And actually, you could imagine that it might be So this has a negative 2 charge. We could even write it there if you want-- 2 minus. And then you could have the sodiums over here. You have these two sodiums that have two plus. And this entire molecules becomes neutral. They are attracted to each other. They are still in an aqueous solution. And then, the hydroxide nabs the protons, and then you are left with just water. So plus 2 moles-- or 2 molecules depending on how we're viewing this-- plus two waters. I'll just use that same orange color. Plus two H2Os. One of the hydrogens in each of the water molecules are coming from the oxalic acid, and so two of these hydrogens in these two moles of the water are coming from one entire molecule of oxalic acid." + }, + { + "Q": "When writing the chemical name, does it matter the order of the elements, like are H2O and OH2 the same?\nI'm asking this because at 1:10, Sal wrote HO, but I've seen it mostly as OH.", + "A": "technically doesn t matter but it is common to write the acid part on the left and the basic part of the right", + "video_name": "XjFNmfLv9_Q", + "timestamps": [ + 70 + ], + "3min_transcript": "I've taken this problem from Chapter 4 of the Chemistry & Chemical Reactivity book by Kotz, Treichel and Townsend, and I've done it with their permission. So let's do this example. A 1.034 gram sample of impure oxalic acid is dissolved in water and an acid-base indicator added. The sample requires 34.47 milliliters of 0.485 molar sodium hydroxide to reach the equivalence point. What is the mass of oxalic acid, and what is its mass percent in the sample? So before we even break into the math of this, let's just think about what's happening. We have some oxalic acid, which looks like this. It's really two carbolic acid groups joined together, if that means anything to you. Watch the organic chemistry play list if you want to learn more about that. So we have a double bond to one oxygen, and then another We have that on the other carbon as well. This right here is what oxalic acid is. And it's an interesting acid, because it can actually donate This proton can be nabbed off, and this proton can also be contributed. And it's actually resident stabalized. If that doesn't mean anything to you, don't worry. You'll learn more about that in organic chemistry. But the important thing to realize here is that there's two protons to nab off of it. Now each molecule of sodium hydroxide-- remember when you put it in the water it really just dissolves, and you can really just think of it as hydroxide-- each molecule of hydroxide can nab one of the hydrogen protons. So for every one molecule of oxalic acid, you're going to need two hydroxides-- one to nab this hydrogen proton, and So let's write down the balanced equation that we're dealing with here. So we're going to start off with some oxalic acid. So that has two hydrogens-- so it's H2-- two carbons, and then four oxygens-- O4. It's dissolved in water, so it's an aqueous solution. And to that, we're going to add sodium hydroxide. Now I just told you that you're going to need two of the hydroxides to fully neutralize the oxalic acid. So you're going to need two of them. And this is also in our aqueous solution. And once the reaction happens, this guy will have lost both of the hydrogen protons, so let me draw that. So it will look like this. No more hydrogen, so it's C2O4. It'll have a negative 2 charge. And actually, you could imagine that it might be" + }, + { + "Q": "at 2:30 how is one going to know it takes 2naoh to neutralize?", + "A": "Oxalic acid has 2 Hydrogen protons which are formed when it disassociates in water. Thus, 2 OH- ions are required to neutralize them. Every NaOH molecule releases only 1 OH- ions. Thus, 2 NaOH are required to neutralize the oxalic acid. Thus, when there are two H+ ions which are formed on disassociation, two OH_ ions will be required to neutralize it. Hope this helps :)", + "video_name": "XjFNmfLv9_Q", + "timestamps": [ + 150 + ], + "3min_transcript": "I've taken this problem from Chapter 4 of the Chemistry & Chemical Reactivity book by Kotz, Treichel and Townsend, and I've done it with their permission. So let's do this example. A 1.034 gram sample of impure oxalic acid is dissolved in water and an acid-base indicator added. The sample requires 34.47 milliliters of 0.485 molar sodium hydroxide to reach the equivalence point. What is the mass of oxalic acid, and what is its mass percent in the sample? So before we even break into the math of this, let's just think about what's happening. We have some oxalic acid, which looks like this. It's really two carbolic acid groups joined together, if that means anything to you. Watch the organic chemistry play list if you want to learn more about that. So we have a double bond to one oxygen, and then another We have that on the other carbon as well. This right here is what oxalic acid is. And it's an interesting acid, because it can actually donate This proton can be nabbed off, and this proton can also be contributed. And it's actually resident stabalized. If that doesn't mean anything to you, don't worry. You'll learn more about that in organic chemistry. But the important thing to realize here is that there's two protons to nab off of it. Now each molecule of sodium hydroxide-- remember when you put it in the water it really just dissolves, and you can really just think of it as hydroxide-- each molecule of hydroxide can nab one of the hydrogen protons. So for every one molecule of oxalic acid, you're going to need two hydroxides-- one to nab this hydrogen proton, and So let's write down the balanced equation that we're dealing with here. So we're going to start off with some oxalic acid. So that has two hydrogens-- so it's H2-- two carbons, and then four oxygens-- O4. It's dissolved in water, so it's an aqueous solution. And to that, we're going to add sodium hydroxide. Now I just told you that you're going to need two of the hydroxides to fully neutralize the oxalic acid. So you're going to need two of them. And this is also in our aqueous solution. And once the reaction happens, this guy will have lost both of the hydrogen protons, so let me draw that. So it will look like this. No more hydrogen, so it's C2O4. It'll have a negative 2 charge. And actually, you could imagine that it might be" + }, + { + "Q": "At around 1:40 he is drawing the H atoms on the diagram and called them protons. Does he mean to say hydrons?", + "A": "Hydrogen atoms are just protons with one electron.", + "video_name": "XjFNmfLv9_Q", + "timestamps": [ + 100 + ], + "3min_transcript": "I've taken this problem from Chapter 4 of the Chemistry & Chemical Reactivity book by Kotz, Treichel and Townsend, and I've done it with their permission. So let's do this example. A 1.034 gram sample of impure oxalic acid is dissolved in water and an acid-base indicator added. The sample requires 34.47 milliliters of 0.485 molar sodium hydroxide to reach the equivalence point. What is the mass of oxalic acid, and what is its mass percent in the sample? So before we even break into the math of this, let's just think about what's happening. We have some oxalic acid, which looks like this. It's really two carbolic acid groups joined together, if that means anything to you. Watch the organic chemistry play list if you want to learn more about that. So we have a double bond to one oxygen, and then another We have that on the other carbon as well. This right here is what oxalic acid is. And it's an interesting acid, because it can actually donate This proton can be nabbed off, and this proton can also be contributed. And it's actually resident stabalized. If that doesn't mean anything to you, don't worry. You'll learn more about that in organic chemistry. But the important thing to realize here is that there's two protons to nab off of it. Now each molecule of sodium hydroxide-- remember when you put it in the water it really just dissolves, and you can really just think of it as hydroxide-- each molecule of hydroxide can nab one of the hydrogen protons. So for every one molecule of oxalic acid, you're going to need two hydroxides-- one to nab this hydrogen proton, and So let's write down the balanced equation that we're dealing with here. So we're going to start off with some oxalic acid. So that has two hydrogens-- so it's H2-- two carbons, and then four oxygens-- O4. It's dissolved in water, so it's an aqueous solution. And to that, we're going to add sodium hydroxide. Now I just told you that you're going to need two of the hydroxides to fully neutralize the oxalic acid. So you're going to need two of them. And this is also in our aqueous solution. And once the reaction happens, this guy will have lost both of the hydrogen protons, so let me draw that. So it will look like this. No more hydrogen, so it's C2O4. It'll have a negative 2 charge. And actually, you could imagine that it might be" + }, + { + "Q": "AT 3:42 Sal said that the Water Molecule (H_2O) got a \"proton\" from the hydrogen ion .\n\"Can Protons be also added in a atom or molecule\", during a chemical reaction?\nI know that electrons do but can Protons tooo....?", + "A": "When we say proton in chemical reactions it means a hydrogen cation H+, it doesn t mean one of the nuclei is gaining a proton.", + "video_name": "Y4HzGldIAss", + "timestamps": [ + 222 + ], + "3min_transcript": "so the hydrogen is just going to be left as a hydrogen proton. And then the chlorine, the chlorine has just nabbed that electron. It had the electrons it had before, and then it just nabbed an electron from the hydrogen, and so it now has a negative charge, and these are both in aqueous solution still. It's still, they're still both dissolved in the water. And so you see very clearly here, you put this in an aqueous solution, you're going to increase the amount of, you're going to increase the amount of hydrogen ions, the amount of protons in the solution. And we've talked about this before, you'll often see a reaction written like this, but the hydrogen protons, they just don't sit there by themselves in the water. They are going to bond with the water molecules to actually form hydronium. So another way that you'll often see this is like this. You have the hydrochloric acid, hydrochloric acid. It's in an aqueous solution, and then you have the H2O. You have the water molecules, H2O, and you'll sometimes see written, okay, it's in its liquid form, and it's going to yield. Instead of just saying that you have a hydrogen ion right over here, you'll say, \"Okay, that thing, \"the hydrogen is actually gonna get bonded \"to a water molecule.\" And so what you're gonna be left with is actually H3O. Now this thing, this was a water molecule, and all it got was a hydrogen ion. All that is is a proton. It didn't come with any electrons, so now this is going to have a positive charge. It's going to have a positive charge, and we could now say that this is going to be in an aqueous solution, hydronium is going to be in an aqueous solution, and you're going to have plus, now you're going to still have the chloride ion, Chloride, chloride anion, and this is still in an aqueous solution. It is dissolved in water, and remember all that happened here is that the chlorine here took all of the electrons, leaving hydrogen with none. Then that hydrogen proton gets nabbed by a water molecule and becomes hydronium. So even by this definition you might say it increases the concentration of hydrogen protons. You could say it increase the concentration of hydronium, of hydronium right over here. Hydronium ions. So that makes, by the Arrhenius definition, that makes hydrochloric acid a strong acid. That makes it a strong acid. Now what would be a strong base by the Arrhenius definition of acids and bases? Well one would be sodium hydroxide. So let me write that down, so if I have sodium hydroxide," + }, + { + "Q": "1:22 what is aqueous solution??", + "A": "Solution whereby the solute is dissolved in water.", + "video_name": "Y4HzGldIAss", + "timestamps": [ + 82 + ], + "3min_transcript": "- [Voiceover] The first, I guess you could say, modern conception of an acid and base comes from this gentleman right over here, Svante Arrhenius, and he was actually the third recipient of the Nobel Prize in Chemistry in 1903. And his definition of acids, under his definition of acids and bases, an acid is something that increases the concentration, increases the concentration, concentration of Hydrogen protons, and we can say protons when put in an aqueous solution, when in aqueous, aqueous solution, and that's just a water solution. And then you can imagine what a base would be. You could think, oh maybe a base is something that decreases the protons and that's Or you could say, it decreases, or actually let me write this, it increases It increases the hydroxide concentration. when put in aqueous solution. When in aqueous, aqueous solution. So let's make that concrete. Let's look at some examples. So a strong Arrhenius acid, and actually, this would be a strong acid by other definitions as well, would be hydrochloric acid. Hydrochloric acid, you put it in a an aqueous solution. So that's the hydrogen. You have the chlorine. You put it in an aqueous solution. You put it in an aqueous solution, it will readily disassociate. This is a, this reaction occurs, strongly favors moving from the left to right. You're going to have the chlorine strip off the two electrons in the covalent bond with the hydrogen, so the hydrogen is just going to be left as a hydrogen proton. And then the chlorine, the chlorine has just nabbed that electron. It had the electrons it had before, and then it just nabbed an electron from the hydrogen, and so it now has a negative charge, and these are both in aqueous solution still. It's still, they're still both dissolved in the water. And so you see very clearly here, you put this in an aqueous solution, you're going to increase the amount of, you're going to increase the amount of hydrogen ions, the amount of protons in the solution. And we've talked about this before, you'll often see a reaction written like this, but the hydrogen protons, they just don't sit there by themselves in the water. They are going to bond with the water molecules to actually form hydronium. So another way that you'll often see this is like this. You have the hydrochloric acid, hydrochloric acid. It's in an aqueous solution," + }, + { + "Q": "At 9:00, why isn't that molecule designated as \"cis\" or \"trans\"? Can't you have both an R/S designation and a cis/trans designation?", + "A": "There are no double bonds and each of the 4 groups at the chiral centres are different. How could you determine cis/trans here? You can have R/S and cis/trans in cyclic molecules though.", + "video_name": "kFpLDQfEg1E", + "timestamps": [ + 540 + ], + "3min_transcript": "So that would be the lowest priority. So that would get a number four here. So the oxygen has the highest atomic number. So it's going to get a number one. And then we have a longer chain over here for this carbon on the right. So that's going to get a number two, we have a number three. And then the hydrogen would be a number four. So there's a little trick that I covered in an earlier video. So if you ignore the hydrogen, it looks like you're going around this way, it looks like you're going around counterclockwise. So it looks like it's S. But you have this lowest priority group is actually coming out at you. So remember the trick was, if it looks like it's S with those three, just reverse it. And so it must be R. It must have an absolute configuration of R at carbon three. So we'll go ahead and put in here a three R. And then we have to worry about the absolute configuration at carbon six. So at carbon six here, this is another chirality center. There's also a hydrogen attached to this carbon, going away from us like that. And let's think about the highest priority. Well, this chain over here on the left is going to get the highest priority. It has the most carbons, it has an oxygen over here. So it's definitely going to be highest priority There are more carbons this one over here on the right. So again, when you assign priority, this is going to get highest one. And then this is going to get a third up here. So this time, you're going around one, two, three, you're going around counterclockwise. But your lowest priority group, which is this hydrogen back here, is going away from you. So this actually is going to be S. So it's counterclockwise. So it's S. Three R, six S. And I went through those kind of fast. So you need to go back and watch some of the earlier videos on absolute configurations if that was a little bit too fast for you. All right. Let's look at cyclic alcohol. So ring systems. And then there's an O-H coming off of it like that. So six carbons without the O-H, we would call that cyclohexane. And since this is an alcohol, we would just change that to cyclohexanol. So that's very simple nomenclature. You don't really need a number. But you could write a one there. It's implied if you don't put in. So that would be cyclohexanol. What about something that has a ring system with two hydroxyl groups? So let's say we'll put in some stereochemistry, too. So let's say we have an O-H coming out at us. And let's say we have an O-H going away from us like that. So when you have a situation like this, when you have two alcohols in the same molecule, your prefixes be di. So this is actually a diol. And the nomenclature is based off the cyclohexane molecule. So you would write cyclohexane." + }, + { + "Q": "At 3:36, wouldn't the longest carbon chain start at the carbon in the top right (above what Jay designated as the first carbon) and continue down to what he designated at carbon 7? That would make a continuous 8 carbon chain", + "A": "The OH group needs to be included in the main chain as it is the highest priority substituent in this molecule, even though there is a longer carbon chain.", + "video_name": "kFpLDQfEg1E", + "timestamps": [ + 216 + ], + "3min_transcript": "So it's also called propanol. The difference is the hydroxl group is on a different carbon, It's now on carbon two. So we're going to write two-propanol here, which is the IUPAC name. This is also called isopropanol, rubbing alcohol, it's all the same stuff. But two-propanol would be the proper IUPAC nomenclature. How would you classify two-propanol? So once again, we find the carbon attached to the O-H. That's this one. How many carbons is that carbon attached to? It's attached to one and two other carbons. So therefore, this is a secondary alcohol. So we have an example of a primary alcohol, and an example of a secondary alcohol here. Let's do a little bit more complicated nomenclature question. And so let's go ahead and draw out a larger molecule with more substituents. So let's put an O-H here. And then let's go ahead and do that as well. So give the full IUPAC name for this molecule. So you want to find the longest carbon chain that includes the O-H. OK so you have to find the longest carbon chain that includes the O-H, and you want to give the O-H the lowest number possible. So that's going to mean that you're going to start over here. And make this carbon number one like that. So if that's carbon number one, this must be carbon number two, three, four, five, six, and seven. So we have a seven-carbon alcohol. So seven-carbon alcohol would be heptanol. So we can go and start naming this. Make sure to give us plenty of space here. So we have heptanol. And we know that the O-H is coming off of carbon two. So we can go ahead and write two-heptanol like that. Let's look at the other substituents that we have. Well, what do we have right here coming off of our ring? So that would be propyl. So we have three-propyl. So go ahead and write three-propyl in here. And what else do we have? At carbon five, we have two substituents. So we have a chloro group right here. And we have a methyl group right here. And remember your alphabet. Right, so C comes before M. So we can go ahead and put our methyl in there. All right, coming off of carbon five, so that would be five-methyl, like that. And then also coming off five, we have chloro. So five-chloro. Right in here. And that should do it. Everything follows the alphabet rule. So we have five-chloro, five-methyl, three-propyl, two-heptanol for this molecule. What about a problem that includes some stereochemistry? So let's say they give us one where we have to worry about stereochemistry." + }, + { + "Q": "In 4:55, can't I write \"5-chloro-5-methyl-3-propyl-heptan-2-ol\" instead of 5-chloro-5-methyl-3-propyl-2-heptanol\"?\n\nMy chemistry teacher taught me that the first is correct. What's the difference between the names?", + "A": "You can as long as you remove the dash between propyl and heptan. The preferred way to do it is to put the number in the middle like your first way. People don t always learn the new recommendations. It s also much easier to say 2-heptanol out loud than heptan-2-ol", + "video_name": "kFpLDQfEg1E", + "timestamps": [ + 295 + ], + "3min_transcript": "And then let's go ahead and do that as well. So give the full IUPAC name for this molecule. So you want to find the longest carbon chain that includes the O-H. OK so you have to find the longest carbon chain that includes the O-H, and you want to give the O-H the lowest number possible. So that's going to mean that you're going to start over here. And make this carbon number one like that. So if that's carbon number one, this must be carbon number two, three, four, five, six, and seven. So we have a seven-carbon alcohol. So seven-carbon alcohol would be heptanol. So we can go and start naming this. Make sure to give us plenty of space here. So we have heptanol. And we know that the O-H is coming off of carbon two. So we can go ahead and write two-heptanol like that. Let's look at the other substituents that we have. Well, what do we have right here coming off of our ring? So that would be propyl. So we have three-propyl. So go ahead and write three-propyl in here. And what else do we have? At carbon five, we have two substituents. So we have a chloro group right here. And we have a methyl group right here. And remember your alphabet. Right, so C comes before M. So we can go ahead and put our methyl in there. All right, coming off of carbon five, so that would be five-methyl, like that. And then also coming off five, we have chloro. So five-chloro. Right in here. And that should do it. Everything follows the alphabet rule. So we have five-chloro, five-methyl, three-propyl, two-heptanol for this molecule. What about a problem that includes some stereochemistry? So let's say they give us one where we have to worry about stereochemistry. So let's see, something like that. And let's make an O-H group going away from us. And then let's go ahead and make this one coming out at us like that. So give the full IUPAC name for this molecule, and you have to include stereochemistry. So once again, find your longest carbon chain that includes your O-H group. And you want to give that O-H the lowest number possible so it takes precedence over things like alkyl groups, and halogens, and double bonds. So we're going to start from the left. So one, two, three, four, five, six, seven, eight, nine like that. So we have a nine-carbon alcohol. So that would be nonanol. And the alcohol is coming off of carbon three. The O-H is coming off of carbon three. So we have three-nonanol. Like that. So three-nonanol." + }, + { + "Q": "At 1: 26, in my textbook, it shows that the name should be propan-1-ol (or propanol) and similarly the name of 2-propanol should also be propan-2-ol.\nAlso, at 4:48, according to the rules in my textbook, the name should be 5-chloro-5-methyl-3-propylheptan-3-ol.\nI am very confused now because I don't know if should put -ol at first or at last.\nPlease help me!", + "A": "Yeah ! you are correct, in my textbook too it is written as propan -2-ol. And the textbook contains the correct nomenclature", + "video_name": "kFpLDQfEg1E", + "timestamps": [ + 288 + ], + "3min_transcript": "And then let's go ahead and do that as well. So give the full IUPAC name for this molecule. So you want to find the longest carbon chain that includes the O-H. OK so you have to find the longest carbon chain that includes the O-H, and you want to give the O-H the lowest number possible. So that's going to mean that you're going to start over here. And make this carbon number one like that. So if that's carbon number one, this must be carbon number two, three, four, five, six, and seven. So we have a seven-carbon alcohol. So seven-carbon alcohol would be heptanol. So we can go and start naming this. Make sure to give us plenty of space here. So we have heptanol. And we know that the O-H is coming off of carbon two. So we can go ahead and write two-heptanol like that. Let's look at the other substituents that we have. Well, what do we have right here coming off of our ring? So that would be propyl. So we have three-propyl. So go ahead and write three-propyl in here. And what else do we have? At carbon five, we have two substituents. So we have a chloro group right here. And we have a methyl group right here. And remember your alphabet. Right, so C comes before M. So we can go ahead and put our methyl in there. All right, coming off of carbon five, so that would be five-methyl, like that. And then also coming off five, we have chloro. So five-chloro. Right in here. And that should do it. Everything follows the alphabet rule. So we have five-chloro, five-methyl, three-propyl, two-heptanol for this molecule. What about a problem that includes some stereochemistry? So let's say they give us one where we have to worry about stereochemistry. So let's see, something like that. And let's make an O-H group going away from us. And then let's go ahead and make this one coming out at us like that. So give the full IUPAC name for this molecule, and you have to include stereochemistry. So once again, find your longest carbon chain that includes your O-H group. And you want to give that O-H the lowest number possible so it takes precedence over things like alkyl groups, and halogens, and double bonds. So we're going to start from the left. So one, two, three, four, five, six, seven, eight, nine like that. So we have a nine-carbon alcohol. So that would be nonanol. And the alcohol is coming off of carbon three. The O-H is coming off of carbon three. So we have three-nonanol. Like that. So three-nonanol." + }, + { + "Q": "I thought water was already a conductor. 8:20", + "A": "It s actually not the water itself that is the conductor. All water forms hydronium (H3O+) and hydroxyl (OH-) ions, which conduct the electricity. In pure water the concentrations of these ions are very low (about (about 10^-7 mols/liter of each), so you need to add impurities (such as salts or acids) to increase the number of ions and the conductivity.", + "video_name": "Rw_pDVbnfQk", + "timestamps": [ + 500 + ], + "3min_transcript": "So the strength, and I've touched on this, it also goes into the boiling point. So because these bonds are pretty strong, it has a higher boiling point. If you just took salt crystal and tried to boil it, you'd have to add a lot of heat into the system. So this has a higher boiling point than say-- I mean, definitely things that have just van der Waals forces like the noble gases, but it'll also have a higher boiling point than, say, hydrogen fluoride. Hydrogen fluoride, if you remember from the last video, just had dipole-dipole forces. But what's interesting about this is they have a very high boiling point unless they're dissolved in water. So these are very hard, high boiling point, but the ionic crystals can actually be dissolved in water. And when they are dissolved in water, they form ionic dipole bonds. What does that mean? Ionic dipole or ionic polar bonds. -- and this is actually why it dissolves in water. Because the water molecule, we've gone over this tons of times, it has a negative end, because oxygen is hoarding the electrons, and then the hydrogen ends are positive because the electron's pretty stripped of it. So when you put these sodium and chloride ions in the room, or in the water solution, the positive sodiums want to get attracted to the negative side of this dipole, and then the negative chlorides, Cl minus, want to go near the hydrogens. So they kind of get dissolved in this. They don't necessarily want to be-- they still want to be attracted to each other, but they're still also attracted to different sides of the water, so it allows them to get dissolved and go with the flow of the water. So in this case, when you actually dissolve an ionic crystal into water, not a lot of charge that is really movable in this state. But here, all of a sudden, we have these charged particles that can move. And because they can move, all of a sudden, when you put salt, sodium chloride, in water, that does become conductive. So anyway, I wanted you to be at least exposed to all of these different forms of matter. And now, you should at least get a sense when you look at something and you should at least be able to give a pretty good guess at how likely it is to have a high boiling point, a low boiling point, or is it strong or not. And the general way to look at it is just how strong are the intermolecular bonds. Obviously, if the entire structure is all one molecule, it's going to be super-duper strong. And on the other hand, if you're just talking about neon, a bunch of neon molecules, and all they have are the London dispersion forces, this thing's going to have ultra-weak bonds." + }, + { + "Q": "At 1:37ish, wouldn't the carbons be connected slightly differently? More like a single uniform \"diamond\" shape, rather than a spread out tree.", + "A": "actually, it depends on the cooling rate. atleast what i learned when working with metals was when they were melted and cooling, if cooled slowly, they would form their perfect crystaline shapes but if quenched (dipped in water for a quicker cool) they would create many irregular crystal patterns, this irregularity is what would cause quenched metals to bend easier than slowly cooled ones. if theres a big difference in strength of a metal and its ability to bend, same metals different cooling rate.", + "video_name": "Rw_pDVbnfQk", + "timestamps": [ + 97 + ], + "3min_transcript": "In the last video, I talked about some of the weaker intermolecular forces or structures of elements. The weakest, of course, was the London dispersion force. In this video, I'll start with the strongest structure, and that's the covalent network. So if you have a covalent network crystal and let me actually define the word crystal. Crystal is just when you have a solid, where the molecules that make up the solid are in a regular, relatively consistent pattern, and this is versus an amorphous solid, where everything is kind of just a hodge-podge and there's different concentrations of different things, of different ions, and different molecules, and different parts of the solid. So crystal is just a very regular structure. Ice is a crystal, because once you get the temperature low enough in water, the hydrogen bonds form a crystal, a regular structure. And we've talked about that a bunch. But the strongest of all crystal structures And the biggest, or the prime, example of that is carbon when it forms a diamond. So in the covalent network, carbon has four valence electrons, so it always wants four more. So when carbon shares with itself, it's very happy. So what it can do is it can form four bonds to four more carbons, and then each of those carbons can form four more bonds to four more carbons. And this one, 1, 2, 3, and it just keeps going on. This is the structure of a diamond. And the reason why this is such a strong structure is because you can almost view the entire -- in fact, you should view the entire diamond as one molecule, because they all have covalent bonds. These are actual sharing of electrons, and these are actually the strongest of all molecular bonds. So you can imagine if the entire solid you're going to have an extremely strong, extremely high boiling point substance, and that's why a diamond is so strong, and that's why it's so hard to boil a diamond. Now, the next two, and it depends on your special cases of the next most solid version of a solid, and it depends which case you're talking about, one are the ionic crystals, and I'll do them both here, because one isn't necessarily -- ionic crystal-- and the next is the metal. Well, it's not the next. They're kind of the metallic crystal. And these bonds, I mean, let's say the most common ionic molecule or -- that's not exactly the right word, because to some degree, let's say if I had some sodium and some chloride -- and just remember, what happens with sodium chloride is sodium here really has one extra electron" + }, + { + "Q": "At 6:35, isn't HF held together with hydrogen bonding, not dipole-dipole forces?", + "A": "If we re getting technical hydrogen bonding is a strong type of dipole-dipole interaction.", + "video_name": "Rw_pDVbnfQk", + "timestamps": [ + 395 + ], + "3min_transcript": "their electrons to roam, they all become slightly positive. And so they're kind of embedded in this mesh or this sea of electrons. And so the metallic crystals, depending on what cases you look at, sometimes they're harder than the ionic crystals, sometimes not. Obviously, we could list a lot of very hard metals, but we could list a lot of very soft metals. Gold, for example. If you take a screwdriver and a hammer, you know, pure gold, 24-carat gold, if you take a screwdriver and hit it onto the gold, it'll dent it, right? So this one isn't as brittle as the ionic crystal. It'll often mold to what you want to do with it. It's a little bit softer. Even if you talk about very hard metals, they tend to not be as brittle, because the sea of electrons kind of gives you a little give when you're moving around the metal. But that's not to say that it's not hard. In fact, sometimes that give that a metal has, or that ability to bend or flex, is what actually gives it its strength So the strength, and I've touched on this, it also goes into the boiling point. So because these bonds are pretty strong, it has a higher boiling point. If you just took salt crystal and tried to boil it, you'd have to add a lot of heat into the system. So this has a higher boiling point than say-- I mean, definitely things that have just van der Waals forces like the noble gases, but it'll also have a higher boiling point than, say, hydrogen fluoride. Hydrogen fluoride, if you remember from the last video, just had dipole-dipole forces. But what's interesting about this is they have a very high boiling point unless they're dissolved in water. So these are very hard, high boiling point, but the ionic crystals can actually be dissolved in water. And when they are dissolved in water, they form ionic dipole bonds. What does that mean? Ionic dipole or ionic polar bonds. -- and this is actually why it dissolves in water. Because the water molecule, we've gone over this tons of times, it has a negative end, because oxygen is hoarding the electrons, and then the hydrogen ends are positive because the electron's pretty stripped of it. So when you put these sodium and chloride ions in the room, or in the water solution, the positive sodiums want to get attracted to the negative side of this dipole, and then the negative chlorides, Cl minus, want to go near the hydrogens. So they kind of get dissolved in this. They don't necessarily want to be-- they still want to be attracted to each other, but they're still also attracted to different sides of the water, so it allows them to get dissolved and go with the flow of the water. So in this case, when you actually dissolve an ionic crystal into water," + }, + { + "Q": "At 4:20, Sal says that when we cut something, we are breaking atomic bonds. But I thought we were breaking molecular bonds. Or are they the same ?", + "A": "What he means is that we are breaking bonds between atoms. We don t have knives or scissors sharp enough to cut atoms in pieces.", + "video_name": "Rw_pDVbnfQk", + "timestamps": [ + 260 + ], + "3min_transcript": "you're going to have an extremely strong, extremely high boiling point substance, and that's why a diamond is so strong, and that's why it's so hard to boil a diamond. Now, the next two, and it depends on your special cases of the next most solid version of a solid, and it depends which case you're talking about, one are the ionic crystals, and I'll do them both here, because one isn't necessarily -- ionic crystal-- and the next is the metal. Well, it's not the next. They're kind of the metallic crystal. And these bonds, I mean, let's say the most common ionic molecule or -- that's not exactly the right word, because to some degree, let's say if I had some sodium and some chloride -- and just remember, what happens with sodium chloride is sodium here really has one extra electron Chlorine has seven electrons and it's dying to get a new. So sodium essentially donates its electron to chlorine, and then the chlorine becomes negative, the sodium becomes positive, and they want to be near each other, right? So you have a positive sodium ion and a negative chlorine ion, and the structure of this is going to look something like this, where they're all -- so let me do the sodium in green. So you have a bunch of sodium ions that are positive, and then you have a bunch of chlorine ions that are maybe -- this isn't the exact way that they actually are, but I think you get the idea, that one atom is positive and one atom is negative, so they really, really want to be close to each other. And so this is a pretty strong bond, and it has very-- not a very high boiling point. It can have a pretty high boiling point, and this type of structure is actually quite brittle. So if you take some dry table salt, not dissolved in water, and you slam it with a hammer, you'll see that you'll get, a big slice of it. It'll just fall off, right? Because you're essentially just cutting it along one of these lines really fast. That's the interesting thing. Whenever you do something on a macroscale, like cut sth. you really fundamentally are breaking atomic bonds. So the strength of the atomic bonds really do tell you about how hard or strong something is. Now, the metallic crystal we've talked a lot about. Metals, they like to get rid of their electrons, or not get rid of them, they like to share them. So what happens is, let's say in the case of iron, you have a bunch of iron atoms. This is all iron. And their electrons are allowed to roam free in the neighborhood. These are all the electrons. They're allowed to roam free. And because of this, it forms this sea of electrons that are negative, and that makes it a very good conductor of electricity." + }, + { + "Q": "At 1:46 Sal says that dinosaurs may have been smarter than us, but they couldn't of been smarter than us because if they were, they could stopped the asteroid... Right? Or, they were smart enough, but they didn't have what they needed to stop it? What do you think, because that's confusing me. (We'll probably never know)", + "A": "They were never aware of an asteroid. Even if they had tech, it would have struck unexpectedly. Like today, last year we almost got hit with an asteroid 50 feet across. We didnt know until it got close.", + "video_name": "T5DGZIsfK-0", + "timestamps": [ + 106 + ], + "3min_transcript": "I've talked a bunch about the Drake equation, or our own version of the Drake equation that starts with the number of stars in the galaxy, but I haven't given it a shot yet. I haven't tried my own attempt at thinking about how many detectable civilizations there are. So let's actually do that here. So let's just assume that there are 100 billion stars. 100 billion stars. So that's my first term right over there. Let's say that 1/4 will develop planets. And let's say of the solar systems that develop planets, on average let's say that they develop an average of 0.1 planets capable of sustaining life. Or really, that you'll have one planet for every 10 of these solar systems with planets. That's just my assumption there. I don't know if that's right. Now let's multiply that times the fraction of these planets capable of sustaining And I don't know what that is, but I hinted in previous videos that life is one of those things that it seems like if you have all the right ingredients, it's so robust. And you have life it these underwater volcanoes, you have bacteria that can process all sorts of weird things. So let's say that probability is pretty high. Let's say that is 50%, or half of the plans that are capable of getting life actually do have life. I would guess that that might even be higher. But once again, just a guess. Now we have to think about of the life, what fraction becomes intelligent? What becomes intelligent over some point in the history? Well, I'll say it's a tenth. A tenth of all-- maybe if the asteroids didn't kill the dinosaurs, it wouldn't have happened on Earth. Who knows? Or maybe we'd just have some very intelligent dinosaurs We don't know. And maybe there's other forms of intelligent life even on our own planet that we haven't fully appreciated. Dolphins are a good candidate. Some people believe that octopuses, there's a theory that they could develop eventually the ability to kind one day, if their brains mature, and all of the rest, make tools the same way primitive primates eventually were able to have larger brain sizes and actually manipulate things to make tools. I don't want to get into all of that. So there's a 1 in 10 chance that you get intelligent life, and then assuming that intelligent life shows up, what fraction is going to become detectable? I don't know. I don't know whether dolphins will ever communicate via radio or not. So let's just say that is-- I don't know. Let's say that is another 1 in 10 chance, or I'll say 0.1. And then we have to multiply it times the detectable life of the civilization on average. Once again, huge assumptions being here, but the detectable life of a civilization, let me just put it at 10,000 years." + }, + { + "Q": "At 2:47-2:51 where did \"Milkomeda\" come from?", + "A": "To be fair, Sal didn t make it up, it is the already made name for the future galaxy.", + "video_name": "QXYbGZ3T3_k", + "timestamps": [ + 167, + 171 + ], + "3min_transcript": "so this video is actually spanning billions of years but when you actually speed up time like that you'll see that it really gives you the sense of the actual dynamics of these interactions. The other thing that I want to talk about before I start the video is to make you realise that when we talk about galaxies colliding it doesn't mean that the stars are colliding. In fact there are going to be very few stars that actually collide the probability of a star-star collision is very low and that's because we learned when we learned about interstellar scale that there's mostly free space in between stars the closest star to us is 4.2 light years away and that's roughly 30 million times the diameter of the sun so you have a lot more free space than star space or even solar system space. So lets start up this animation it's pretty amazing and what you'll see here so these are just the... obviously... so one rotation is actually 250 million years give or take. But now you see these stars right here are starting to get attracted to this core and then some of the stuff in that core was attracted to those stars and they get pulled away that was the first pass of these 2 galaxies. Some stuff is just being thrown off into intergalactic space and you might worry that might happen to the earth and there's some probability that it would happen to the earth but it really wouldn't affect what happens within those star's solar system this is happening so slow, you wouldn't feel like some type of acceleration or something. And then this is the second pass so they passed one pass and once again this is occuring over hundreds of millions or billions of years and on the second pass they finally are able to merge. And all of these interactions are just through the gravity over interstellar or you could call it intergalactic distances. You can see they merge into what could be called as \"Milkomeda\" or maybe the Andromedy Way I don't know whatever you want to call it but even though they've merged, a lot of the stuff has still been thrown off into intergalactic space but this is a pretty amazing animation to me but it's also pretty neat how a super computer can do all of the computations to figure out what every particle which is really a star or cluster of stars or group of stars is actually doing to actually give us a sense of the actual dynamics here, but this is pretty neat look at that! Every little dot is whole groups of stars, thousands of stars potentially." + }, + { + "Q": "At 2:11; what is the difference between the Hydrogen atom and Deuterium??", + "A": "Deuterium is a hydrogen atom with one neutron. Protium has no neutrons in the atom. Tritium has 2 neutrons.", + "video_name": "I-Or4bUAIfo", + "timestamps": [ + 131 + ], + "3min_transcript": "So the atomic number is symbolized by Z and it refers to the number of protons in a nucleus. And you can find the atomic number on the periodic table. So we're going to talk about hydrogen in this video. So for hydrogen, hydrogen's atomic number is one. So it's right here, so there's one proton in the nucleus of a hydrogen atom. In a neutral atom, the number of protons is equal to the number of electrons, because in a neutral atom there's no overall charge and the positive charges of the protons completely balance with the negative charges of the electrons. So let's go ahead and draw an atom of hydrogen. We know the atomic number of hydrogen is one, so there's one proton in the nucleus. So there's my one proton in the nucleus, and we're talking about a neutral hydrogen atom, so there's one electron. I'm going to draw that one electron somewhere outside the nucleus and I'm going to use the oversimplified Bohr model. So this isn't actually what an atom looks like, So this is one, this one version of hydrogen. This is one isotope of hydrogen. So this is called protium. Let me go ahead and write that here. So this is protium and let's talk about isotopes. An isotope, isotopes are atoms of a single element. So we're talking about hydrogen here. That differ in the number of neutrons in their nuclei. So let's talk about the next isotope of hydrogen. So this is called deuteriums. Let me go ahead and write deuterium here. Deuterium is hydrogen, so it must have one proton in the nucleus and it must have one electron outside the nucleus, but if you look at the definition for isotopes, atoms of a single element that differ in the number of neutrons, protium has zero neutrons in the nucleus. Deuterium has one. So let me go ahead and draw in deuterium's one neutron. So deuterium has one neutron and since neutrons have mass, deuterium has more mass than protium. So isotopes have different masses because they differ in terms of number of neutrons. Notice though, that they have the same atomic number, they have the same number of protons in the nucleus. Right, it's one proton in the nucleus. And that's important because if you change the number of protons, you're changing the element, and that's not what we're doing here. We're talking about atoms of a single element. Deuterium is still hydrogen, it's an isotope. Finally, our last isotope, which is tritium. So tritium has one proton in the nucleus, one electron outside the nucleus, and we draw that in here, and it must differ in terms of number of neutrons, so tritium has two neutrons. Let me go ahead and draw the two neutrons here in the nucleus. And so those are the isotopes of hydrogen." + }, + { + "Q": "At 5:35 we know oxygen has 6 valence electrons it is making a single covalent bond with sulfur so it must be contributing one of its electrons in bonding which leaves 5 lone electrons with it doesn't it then why is he drawing 6 lone electrons with it?", + "A": "Both of the electrons in the S-O bond are coming from the sulfur.", + "video_name": "dNPs-cr_6Bk", + "timestamps": [ + 335 + ], + "3min_transcript": "And once again you ignore hydrogen so it's between sulfur and oxygen. And if you look at the periodic table, you'll see that oxygen is higher in group six than sulfur is. Therefore oxygen is more electronegative. And so therefore, we're going to put sulfur at the center. So we're going to put sulfur right here. Once again, look at the rules from the previous video if that didn't make quite sense to you here. So we have sulfur attached to four oxygens. And I'll go ahead and put my four oxygens in there like that. And then I have two hydrogens. And by experience, you are talking about an acid here. You're going to put your hydrogens on oxygens. And so we're going to go ahead and put our hydrogens here. And let's see how many valence electrons we've used up a drawing this skeleton here. So we have two, four, six, eight, ten, and twelve. So we've used up 12 valence electrons so 32 minus 12 gives us 20 valence electrons left to worry about. to assign some of those left over electrons to some of the terminal atoms. But again, we're not going to assign those electrons to hydrogen because hydrogen's already surrounded by two electrons. And so we're going to try to assign some electrons to oxygen. And oxygen's going to follow the octet rule. So let's examine, let's say the top oxygen here. And we could see the top oxygen is surrounded by two electrons already right there in green. And so if we're going to give it an octet it needs six more. So we have one, two, three, four, five, six. Same thing for this oxygen down here, it needs an octet. So we go ahead and give it six more electrons like that. Right so, we also have these other oxygens over here to worry about. So let me go ahead and use green again. So let's say this oxygen over here on the left, the one bonded to this hydrogen here. Here's two electrons and here's another two for four. So for that option to have an octet, it needs four more. on this oxygen. And then we're going to do the same thing for this oxygen as well. So let's see, how many electrons did we just represent there? Well we had six on the top oxygen, six in the bottom oxygen. That's 12. And then we had four on the left and four more on the right. So that's eight. So 12 plus 8 is 20. So we've now represented all of the valence electrons that we needed to show. And let's think about this as possibly being the final dot structure. So we have an octet around sulfur, an octet around oxygen, and hydrogen's fine. So you might think that we are done here. However, let's go ahead and assign some formal charges. And let's see what that does. So let me go ahead and draw in some electrons here. So we know that each bond consists of two electrons. I'm going to go ahead and make them red here like that. And let's assign a formal charge to the top oxygen here." + }, + { + "Q": "At about 5:30, the K_a of HCl is discussed. Since HCl is a strong acid, so the reaction is irreversible, and K_a is an equilibrium constant, how is this possible? I thought it was only possible to measure the K_a of a weak acid.", + "A": "Is not completely irreversible though very likely to favor porducts. That s why the KB value is so tiny Strong Acid has a very weak base", + "video_name": "DGMs81-Rp1o", + "timestamps": [ + 330 + ], + "3min_transcript": "This is equal to 1.0 times 10 to the negative 14 which is our value for Kw. When you add reactions together to get a net reaction you multiply the equilibrium constants to get the equilibrium constant for the net reaction which in this case is Kw for the auto-ionization of water. Let's go ahead and go in even more detail here. Ka, that's your products over your reactants. That'd be H3O plus, the concentration of H3O plus times the concentration of ammonia. Let's go ahead and do that. The concentration of H3O plus times the concentration of ammonia, and that's all over the concentration of ammonium. This is Ka. This is all over the concentration of ammonium. This represents, let me go ahead and highlight this here. Next let's think about Kb. Over here is Kb, that'd be the concentration of your products, so NH4 plus times OH minus. Let's go ahead and do that. Let's put this in parenthesis here. We have the concentration of ammonium, NH4 plus, times the concentration of OH minus. That's over the concentration of NH3. That's over the concentration of NH3 here. What do we get? The ammonium would cancel out. That cancels here. Then the NH3 cancels out. We're left with H3O plus times OH minus which we know is equal to 1.0 times 10 to the negative 14. Just another way to think about this. This can be important, relating Ka and Kb to Kw. If you know one you can figure out the other. Let's think about HCl. The conjugate base to HCl would be Cl minus, the chloride anion here. Let's think about what this equation means. HCl is a strong acid which means a very high value for Ka. An extremely, extremely high value for Ka. What does that say about Kb for the conjugate base? The conjugate base here is the chloride anion. If Ka is very large then Kb must be very small for this to be equal to Kw. Kb is extremely small here, so a very small value for Kb. This mathematically describes what we talked about earlier the stronger the acid the weaker the conjugate base. HCl is a very strong acid, so it has a very, very high value for Ka. And the conjugate base is the chloride anion," + }, + { + "Q": "At about 0:45 you mention Heat of Fusion. Don't you mean Heat of Vaporization?", + "A": "We re looking at liquid becoming gaseous, in other words vaporization, so I think he meant to say heat of vaporization.", + "video_name": "hA5jddDYcyg", + "timestamps": [ + 45 + ], + "3min_transcript": "We know that when we have some substance in a liquid state, it has enough kinetic energy for the molecules to move past each other, but still not enough energy for the molecules to completely move away from each other. So, for example, this is a liquid. Maybe they're moving in that direction. These guys are moving a little bit slower in that direction so there's a bit of this flow going on, but still there are bonds between them. They kind of switch between different molecules, but they want to stay close to each other. There are these little bonds between them and they want to If you increase the average kinetic energy enough, or essentially increase the temperature enough and then overcome the heat of fusion, we know that, all of a sudden, even these bonds aren't strong enough to even keep them close, and the molecules separate and they get into a gaseous phase. And there they have a lot of kinetic energy, and they're bouncing around, and they take the shape of their container. But there's an interesting thing to think about. Which implies, and it's true, that all of the molecules do not have the same kinetic energy. Let's say even they did. Then these guys would bump into this guy, and you could think of them as billiard balls, and they transfer all of the momentum to this guy. Now this guy has a ton of kinetic energy. These guys have a lot less. This guy has a ton. These guys have a lot less. There's a huge distribution of kinetic energy. If you look at the surface atoms or the surface molecules, and I care about the surface molecules because those are the first ones to vaporize or-- I shouldn't jump the gun. They're the ones capable of leaving if they had enough kinetic energy. If I were to draw a distribution of the surface molecules-- let me draw a little graph here. So in this dimension, I have kinetic energy, and on this And this is just my best estimate, but it should give you the idea. So there's some average kinetic energy at some temperature, right? This is the average kinetic energy. And then the kinetic energy of all the parts, it's going to be a distribution around that, so maybe it looks something like this: a bell curve. You could watch the statistics videos to learn more about the normal distribution, but I think the normal distribution-- this is supposed to be a normal, so it just gets smaller and smaller as you go there. And so at any given time, although the average is here, there's some molecules that have a very low kinetic energy. They're moving slowly or maybe they have-- well, let's just say they're moving slowly. And at any given time, you have some molecules that have a very high kinetic energy, maybe just because of the random bumps that it gets from other molecules. It's accrued a lot of velocity or at least a lot of momentum." + }, + { + "Q": "At 9:27, why do they want to evaporate at a high KE?", + "A": "High KE means they are moving fast. The ones that are moving fast are the ones that might be able to escape the surface and be carried away.", + "video_name": "hA5jddDYcyg", + "timestamps": [ + 567 + ], + "3min_transcript": "And the vapor state will continue to happen until you get to some type of equilibrium. And when you get that equilibrium, we're at some pressure up here. So let me see, some pressure. And the pressure is caused by these vapor particles over here, and that pressure is called the vapor pressure. I want to make sure you understand this. So the vapor pressure is the pressure created, and this is at a given temperature for a given molecule, right? Every molecule or every type of substance will have a different vapor pressure at different temperatures, and obviously every different type of substance will also have different vapor pressures. For a given temperature and a given molecule, it's the pressure at which you have a pressure created by the vapor molecules where you have an equilibrium. going back into the liquid state. And we learned before that the more pressure you have, the harder it is to vaporize even more, right? We learned in the phase state things that if you are at 100 degrees at ultra-high pressure, and you were dealing with water, you would still be in the liquid state. So the vapor creates some pressure and it'll keep happening, depending on how badly this liquid wants to evaporate. But it keeps vaporizing until the point that you have just as much-- I guess you could kind of view it as density up here, but I don't want to think-- you have just as many molecules here converting into this state as molecules here converting into this state. So just to get an intuition of what vapor pressure is or how it goes with different molecules, molecules that really want to evaporate-- and so why would a molecule want to evaporate? It could have high kinetic energy, so this would be at a It could have low intermolecular forces, right? It could be molecular. Obviously, the noble gases have very low molecular forces, but in general, most hydrocarbons or gasoline or methane or all of these things, they really want to evaporate because they have much lower intermolecular forces than, say, water. Or they could just be light molecules. You could look at the physics lectures, but kinetic energy it's a function of mass and velocity. So you could have a pretty respectable kinetic energy because you have a high mass and a low velocity. So if you have a light mass and the same kinetic energy, you're more likely to have a higher velocity. You could watch the kinetic energy videos for that. But something that wants to evaporate, a lot of its molecules-- let me do it in a different color. Something that wants to evaporate really bad, a lot" + }, + { + "Q": "At 9:11, you reference the Ka value of Ammonium (taken from another video). If solving this problem requires this Ka value, would the value normally be displayed as part of the question?", + "A": "The Ka will either be given to you, or it will be possible to calculate it based on what else the equation gives you", + "video_name": "kWucfgOkCIQ", + "timestamps": [ + 551 + ], + "3min_transcript": "does react with water, but the chloride anion does not react appreciably with water. So you don't have to worry about this being present, but the NH four plus will affect the pH, which our goal, of course, is to figure out the pH. Alright, so let's get some more room and let's write in our initial concentration of NH four plus, which is 0.05. So the initial concentration of NH four plus is 0.05. And, once again, for these problems, these weak acid equilibrium problems, we assume that we're starting with zero for the concentration of our products and our change, we're going to lose a certain concentration of ammonium. A certain concentration of ammonium is going to react with water. So we're gonna lose x. And then whatever we lose for ammonium turns into ammonia. So if we lose x for the concentration of ammonium, for NH three, for ammonia. And therefore, that's also the same concentration of NH three O plus. So at equilibrium, we're gonna have 0.05 minus x for the concentration of NH four plus. For the concentration on H three O plus, we have x. And for the concentration of NH thee, we also have x. Next, this is functioning as an acid, alright? NH four plus is donating a proton to water, so for the equilibrium expression, we have to write Ka here. And Ka is equal to concentration of products over reactants, so we have concentration of H three O plus, which is x times concentration of NH three, which is x all over the concentration of NH four plus, which is 0.05 minus x. So 0.05 minus x. Ka for NH four plus, Ka for ammonium, and we found it to be 5.6 times ten to the negative ten. So this is equal to... This is equal to, x times x is x squared and 0.05 minus x, if x is a very small concentration, 0.05 minus x is approximately the same as 0.05. So we write 0.05 in here, to make our lives easier for the calculation and we can get out the calculator and solve for x. 5.6 times ten to the negative ten. Alright, we're gonna multiply that by 0.05 and then we're gonna take the square root of our answer and solve for x. X is equal to 5.3 times ten to the negative six. So let's go ahead and write that down. X is equal to 5.3 times ten to the negative six" + }, + { + "Q": "At 9:20, you said that you can leave out the \"X\" because is \"X\" would have a small amount. How do you know that ? I saw that the result was small, but I just don't understand when you can leave out the\"X\"", + "A": "Most instructors say that x is very small if it is less than 5 % of the initial concentration of the acid or base. A quick test for this is if the initial concentration is greater than or equal to 400 \u00c3\u0097 K. If [Acid]\u00e2\u0082\u0080/K \u00e2\u0089\u00a5 400, x is small enough to ignore. At 9:20, [Acid]\u00e2\u0082\u0080/K = 0.0500/5.6\u00c3\u009710\u00e2\u0081\u00bb\u00c2\u00b9\u00e2\u0081\u00b0 = 8.9 \u00c3\u0097 10\u00e2\u0081\u00b7. This is much larger than 400, so you can safely ignore x in the denominator of the equation.", + "video_name": "kWucfgOkCIQ", + "timestamps": [ + 560 + ], + "3min_transcript": "does react with water, but the chloride anion does not react appreciably with water. So you don't have to worry about this being present, but the NH four plus will affect the pH, which our goal, of course, is to figure out the pH. Alright, so let's get some more room and let's write in our initial concentration of NH four plus, which is 0.05. So the initial concentration of NH four plus is 0.05. And, once again, for these problems, these weak acid equilibrium problems, we assume that we're starting with zero for the concentration of our products and our change, we're going to lose a certain concentration of ammonium. A certain concentration of ammonium is going to react with water. So we're gonna lose x. And then whatever we lose for ammonium turns into ammonia. So if we lose x for the concentration of ammonium, for NH three, for ammonia. And therefore, that's also the same concentration of NH three O plus. So at equilibrium, we're gonna have 0.05 minus x for the concentration of NH four plus. For the concentration on H three O plus, we have x. And for the concentration of NH thee, we also have x. Next, this is functioning as an acid, alright? NH four plus is donating a proton to water, so for the equilibrium expression, we have to write Ka here. And Ka is equal to concentration of products over reactants, so we have concentration of H three O plus, which is x times concentration of NH three, which is x all over the concentration of NH four plus, which is 0.05 minus x. So 0.05 minus x. Ka for NH four plus, Ka for ammonium, and we found it to be 5.6 times ten to the negative ten. So this is equal to... This is equal to, x times x is x squared and 0.05 minus x, if x is a very small concentration, 0.05 minus x is approximately the same as 0.05. So we write 0.05 in here, to make our lives easier for the calculation and we can get out the calculator and solve for x. 5.6 times ten to the negative ten. Alright, we're gonna multiply that by 0.05 and then we're gonna take the square root of our answer and solve for x. X is equal to 5.3 times ten to the negative six. So let's go ahead and write that down. X is equal to 5.3 times ten to the negative six" + }, + { + "Q": "How can we predict that a neutral molecule such as ethanol or water molecule would act as a base as shown at 10:26?", + "A": "Any molecule that contains an atom with a lone pair of electrons, such as the O in ethanol or water, can accept a proton from an acid. For example. H2O: + H-Cl --> [H2O-H]+ + Cl-. The [H2O-H]+ is usually written as H3O+. Since the water is accepting a proton from the HCl, it is behaving as a Br\u00c3\u00b8nsted-Lowry base.", + "video_name": "l-g2xEV-z7o", + "timestamps": [ + 626 + ], + "3min_transcript": "so our weak base comes along, and takes a proton from here, and these electrons have moved into here, that would give us the same product, right? So this would be, let me go and highlight those electrons, so these electrons in dark blue would move in to form our double bond, but this is the same as that product. Alcohols can also react via an E1 mechanism. The carbon that's bonded to the OH would be the alpha carbon, and the carbon next to that would be the beta carbon, so reacting an alcohol with sulfuric acid and heating up your reaction mixture will give you an alkene, and sometimes, phosphoric acid is used instead of sulfuric acid. So we saw the first step of an E1 mechanism was loss of a leaving group, but if that happens here, if these electrons come off onto the oxygen, that would form hydroxide as your leaving group, and the hydroxide anion is a poor leaving group, and we know that by looking at pKa values. it is the conjugate base to water, but water is not a great asset, and we know that from the pKa value here, so water is not great at donating a proton, which means that the hydroxide anion is not that stable, and since the hydroxide anion is not that stable, it's not a great leaving group. So let's go ahead and take off this arrow here, because the first step is not loss of a leaving group, the first step is a proton transfer. We have a strong acid here, sulfuric acid, and the alcohol will act as a base and take a proton from sulfuric acid. And that would form water as your leaving group, and water is a much better leaving group than the hydroxide anion, and again, we know that by pKa values. Water is the conjugate base to the hydronium ion, H3O+, which is much better at donating a proton, the pKa value is much, much lower. And that means that water is stable, when you are doing an E1 mechanism with an alcohol is to protonate the OH group. So here's our alcohol, and the carbon bonded to the OH is our alpha carbon, and then these carbons next to the alpha carbon would all be beta carbons. We just saw the first step is a proton transfer, a lone pair of electrons on the oxygen take a proton from sulfuric acid, so we transfer a proton, and let's go ahead and draw in what we would have now, so there'd be a plus-one formal charge on the oxygen, so let's highlight our electrons in magenta, these electrons took this proton to form this bond, and now we have water as a leaving group, let me just fix this hydrogen here really fast, and these electrons can come off onto our oxygen, so that gives us water as our leaving group, and let me go ahead and draw in the water molecule here, and let me highlight electrons," + }, + { + "Q": "at 1:41, why does the alcohol group act like a base, when oxygen does not like to be positively charged?", + "A": "Bases remove protons, nucleophiles form bonds. Hyroxide removes the proton and so is a base not a nucleophile. If it formed a bond and stayed, then it would be a nucleophile. In step 1 it is a nuc, step 2 it is a base. Overall it is a base I belive, I could be wrong though", + "video_name": "l-g2xEV-z7o", + "timestamps": [ + 101 + ], + "3min_transcript": "- [Instructor] Let's look at the mechanism for an E1 elimination reaction, and we'll start with our substrate, so on the left. Let's say we're dealing with alkyl halide. So the carbon that's bonded to our halogen would be the alpha carbon, and the carbon next to that carbon would be the beta carbon, so we need a beta hydrogen for this reaction. The first step of an E1 elimination mechanism is loss of our leaving group, so loss of leaving group, let me just write that in here really quickly, and in this case, the electrons would come off onto our leaving group in the first step of the mechanism. So we're taking a bond away from this carbon, the one that I've circled in red here, so that carbon is going from being sp3 hybridized to being sp2 hybridized. So now we have a carbocation, and we know that carbocations, sp2 hybridized carbons have planar geometry around them, so I've attempted to show the planar geometry So that's the first step, loss of the leaving group to form a carbocation. In the second step, our base comes along and takes this proton, which leaves these electrons behind, and those electrons move in to form our alkene, so this is the second step of the mechanism, which is the base takes, or abstracts, a proton, so base takes a proton to form our alkene. And let me go ahead and highlight those electrons, so these electrons here in magenta moved in to form our double bond, and we form our product, we form our alkene. So the first step of the mechanism, the loss of the leaving group, this turns out to be the rate determining step, so this is the slowest step of the mechanism. So if you're writing a rate law, the rate of this reaction would be equal to the rate constant k times the concentration of your substrate, so that's what studies have shown, of only your substrate, this over here on the left, so it's first order with respect to the substrate. And that's because of this rate determining step. The loss of the leaving group is the rate determining step, and so the concentration of your substrate, your starting material, that's what matters. Your base can't do anything until you lose your leaving group. And so, since the base does not participate in the rate determining step, it participates in the second step, the concentration of the base has no effect on the rate of the reaction, so it's the concentration of the substrate only, and since it's only dependent on the concentration of the substrate, that's where the one comes from in E1, so I'm gonna go ahead and write this out here, so in E1 mechanism, the one comes from the fact this is a unimolecular, a unimolecular rate law here, and the E comes from the fact that this is an elimination reaction, so when you see E1," + }, + { + "Q": "4:11 I don't understand.... 20mph per second? Can someone please help?", + "A": "If you are in your car and you go from 0 mph to 20 mph, and you do it in 1 second, then your acceleration was 20 mph per second.", + "video_name": "FOkQszg1-j8", + "timestamps": [ + 251 + ], + "3min_transcript": "" + }, + { + "Q": "5:50 If there were no atmospheric pressure it would turn automatically into a gas?", + "A": "Yes (but it will also depend on temperature), because there is nothing holding things together. In fact, if you had liquid water in space it would freeze because it is cold and evaporate because there is no pressure at the same time.", + "video_name": "tvO0358YUYM", + "timestamps": [ + 350 + ], + "3min_transcript": "You see that the units work out. This kelvin is going to cancel out with that kelvin in the denominator. This gram in the numerator will cancel out with that grams. And it makes sense. Specific heat is the amount of energy per mass per degree that is required to push it that 1 degree. So here we're doing 58 degrees, 1,000 grams, you just The units cancel out. So you have kelvin canceling out with kelvin. You have grams canceling out with grams. And we are left with-- take out the calculator, put it on the side here. So we have 58.29 times 1,000-- times one, two, three-- times 2.44 is equal to-- and we only have three So this is going to be 142-- we'll just round down-- 142,000 kelvin. So this is 142,000. Sorry 142,000 joules. Joules is our units. We want energy. So this right here is the amount of energy to take our ethanol, our 1 kilogram of ethanol, from 20 degrees Celsius to 78.29 degrees Celsius. Or you could view this as from 293 kelvin to whatever this number is plus 273, that temperature in kelvin. Either way, we've raised its temperature by 58.29 kelvin. Now, the next step is, it's just a lot warmer ethanol, liquid ethanol. We now have to vaporize it. It has to become vapor at that temperature. So now we have to add the heat of vaporization. So that's right here. We should call it the enthalpy of vaporization. The enthalpy of vaporization, they tell us, is And this is how much energy you have to do to vaporize a certain amount per gram of ethanol. Assuming that it's already at the temperature of vaporization, assuming that it's already at its boiling point, how much extra energy per gram do you have to add to actually make it vaporize? So we have this much. And we know we have 1,000 grams of enthanol. The grams cancel out. 855 times 1,000 is 855,000 joules. So it actually took a lot less energy to make the ethanol go from 20 degrees Celsius to 78.29 degrees Celsius than it took it to stay at 78.29, but go from the liquid form to the This took the bulk of the energy. But if we want to know the total amount of energy, let's" + }, + { + "Q": "what does he mean by sending an axon to the optic nerve at 9:42 in the video? How do you send an axon? Does he mean send a message to an axon in the optic nerve", + "A": "All the axons of ganglion cells compose the optic nerve. send would be just a figure of speech.", + "video_name": "CqN-XIPhMpo", + "timestamps": [ + 582 + ], + "3min_transcript": "and it basically takes the cyclic GMP and converts it into just regular GMP. So this basically reduces the concentration of cyclic GMP and increases the concentration of GMP. And the reason that this is important is because there's another channel over here, so there's a whole bunch of these sodium channels and they're all over the cell. So they're just a whole bunch of them, and basically what they let the cell do is they allow the cell to take in sodium from the outside. So let's just say there's a little sodium ion, and it allows it to come inside the cell. So in order for this sodium channel to be open, it actually needs cyclic GMP to be bound to it. So as long as cyclic GMP is bound, the channel is open. But as the concentration of cyclic GMP decreases because causes sodium channels to close. So now we have a closing of sodium channels, and now we basically have less sodium entering the cell. And as less sodium enters the cell, it actually causes the cell to hyperpolarize and turn off. So as the sodium channels close, it actually causes the rods to turn off. So basically, without light, the rods are on because these sodium channels are open. Sodium is flowing through, and the rods are turned on. They can actually produce an action potential and activate the next cell and so on. But as soon as they're turned off, what happens is very interesting because-- let's just look at this rod over here. So what happens is really interesting, because there's this other cell over here that is called the bipolar cell. And we'll just give it a kind of a neutral base, because it's bipolar. so there are on-center and there are off-center bipolar cells. So the on-center bipolar cells normally are being turned off when this rod cell is turned off. But as we mentioned, due to the phototransduction cascade, the rods turn off, which actually turns on the bipolar cell. So basically, on-center bipolar cells get turned on with light and get turned off when there's no light. So that's how they get their names. So when the bipolar cell gets turned on, it activates a retinal ganglion cell, which then sends an axon to the optic nerve, and then into the brain. And so that process is known as the phototransduction cascade, and it basically allows your brain to recognize that there's light entering the eyeball." + }, + { + "Q": "At 11:30, why do you divide mole on both sides?", + "A": "Review basic algebra - solving simple equations.", + "video_name": "VqAa_cmZ7OY", + "timestamps": [ + 690 + ], + "3min_transcript": "And I'll take out my calculator. And I have 0.162 divided by 0.03 is equal to 5.4. And actually more significant is, we could really say it's 5.40 since we have at least three significant digits in both situations. So 5.40 is our proportionality constant. And you would actually divide by 1 in both cases. We just want the number here. But if you wanted the units, you'd want to divide by that 1 centimeters as well. Now we can use this to figure out the exact answer to our problem without having to eyeball it like we just did. We know that for potassium permanganate at 540 nanometers, the absorbance is going to be equal to 5.4 The units of this proportionality constant right here is liters per centimeter mole. And you'll see it'll just cancel out with the distance which is in centimeters, or the length, and the molarity which is in moles per liter. And it just gives us a dimension list, absorbance. So times-- in our example the length is 1 centimeter-- times 1 centimeter, times the concentration. Now in our example they told us the absorbance was 0.539. That's going to be equal to 5.4 liters per centimeter mole, times 1 centimeter, times our concentration. over there. And then we can just divide both sides by 5.4 liters per mole. So let's do that. Let's divide both sides by 5.4 liters per mole, and what do we have? So on the right-hand side, all of this business is going to We're just going to have this concentration left over. So our concentration is equal to-- let's figure out what this number is. So we have 0.539 divided by 5.4 gives us-- so we only have-- well this is actually 5.40. So we actually have three significant digits. So we could say 0.0998." + }, + { + "Q": "At 9:03, what does the epsilon mean and what does it refer to?", + "A": "It means permittivity.", + "video_name": "VqAa_cmZ7OY", + "timestamps": [ + 543 + ], + "3min_transcript": "So we can figure out-- just based on one of these data points because we know that it's 0-- at 0 concentration the absorbance is going to be 0. So that's our other one. We can figure out what exactly this constant is right here. So we know all of these were measured at the same length, or at least that's what I'm assuming. They're all in a 1 centimeter cell. That's how far the light had to go through the solution. So in this example, our absorbance, our length, is equal to 1 centimeter. So let's see if we can figure out this constant right here for potassium permanganate at-- I guess this is probably standard temperature and pressure right here-- for this frequency of light. Which they told us up here it was 540 nanometers. take the first one, we get-- the absorbance was 0.162. That's going to be equal to this constant of proportionality times 1 centimeter. That's how wide the vial was. Times-- now what is the concentration? Well when the absorbance was 0.162, our concentration was 0.03 times 0.-- actually, I'll write all the significant digits there-- 0.0300. So if we want to solve for this epsilon, we can just divide both sides of this equation by 0.0300. So you divide both sides by 0.0300 and what do we get? These cancel out, this is just a 1. And so you get epsilon is equal to-- let's figure out And I'll take out my calculator. And I have 0.162 divided by 0.03 is equal to 5.4. And actually more significant is, we could really say it's 5.40 since we have at least three significant digits in both situations. So 5.40 is our proportionality constant. And you would actually divide by 1 in both cases. We just want the number here. But if you wanted the units, you'd want to divide by that 1 centimeters as well. Now we can use this to figure out the exact answer to our problem without having to eyeball it like we just did. We know that for potassium permanganate at 540 nanometers, the absorbance is going to be equal to 5.4" + }, + { + "Q": "At 4:57, at the bottom left of the virgo supercluster image, doesn't it say 250 million light years or am I mistaken?", + "A": "oh yeah", + "video_name": "JiE_kNk3ucI", + "timestamps": [ + 297 + ], + "3min_transcript": "This is 4 million light years. Four light years is just the distance from us to the Alpha Centauri. So that was nothing. That would only take that a Voyager 1 80,000 years to get. This is 4 million light years. So 4 million times the distance to the nearest star. But even this is-- I mean I'm starting to stumble on my words because there's really no words to describe it-- even this is small on an intergalactic scale. Because when you zoom out more, you can see our Local Group. Our local group is right over here. And this right over here is the Virgo Super Cluster. And each dot here is at least one galaxy. But it might be more than one galaxy. And the diameter here is 150 million light years. the distance from the Milky Way to Andromeda, which was 2 and 1/2 million light years, which would be just a little dot just like that, that would be the distance between the Milky Way and Andromeda. And now, we're looking at the Virgo Super Cluster that is 150 million light years. But we're not done yet. We can zoom out even more. We can zoom out even more, and over here. So you had your Virgo Super Cluster, 150 million light years was that last diagram, this diagram right over here. I want to keep both of them on the screen if I can. This diagram right here, 150 million light years across. That would fit right about here on this diagram. So this is all of the super clusters that are near us. And once again, \"near\" has to be used very, very, very loosely. A billion light years is-- two, three, four, five-- a billion light years is about from here to there. So we're starting to talk on a fairly massive-- I guess we've always been talking on a massive scale. But now, it's an even more massive scale. But we're still not done. Because this whole diagram-- now these dots that you're seeing now, I want to make it very clear. These aren't stars. These aren't even clusters of stars. These aren't even clusters of millions or even billions of stars. Each of these dots are clusters of galaxies, each of those galaxies having hundreds of billions to trillions of stars. So we're just on an unbelievably massive scale at this point. But we're still not done. We're still not done. This is roughly about a billion light years across. But right here is actually the best estimate of the visible universe. And in future videos, we're going to talk a lot more about what the visible universe means." + }, + { + "Q": "At 8:31 you talk about the universe Expanding. Does that mean the Sun will one Day be out of range for us to even see it? Please answer my question if you understanding, because the Universe is just mind blowing :)", + "A": "No, it doesn t, because gravity keeps earth near the sun. the effect of the expansion of space is only meaningful over intergalactic distances.", + "video_name": "JiE_kNk3ucI", + "timestamps": [ + 511 + ], + "3min_transcript": "to the beginning of the actual universe. And the reason why it's the visible universe is there might have been something a little bit further out. Maybe it's light hasn't reached us yet or maybe the universe itself, and we'll talk more about this, it's expanding so fast that the light will never, ever reach us. So it's actually a huge question mark on how big the actual universe is. And then some people might say, well, does it even matter? Because this by itself is a huge distance. And I want to make it clear, you might say, OK, if this light over here, if this is coming from 13 billion light years away, or if this is 13 billion light years away, then you could say, hey, so everything that we can observe or that we can even observe the past of, the radius is about 26 billion light years. But even there, we have to be careful because remember the universe is expanding. When this light was emitted-- and I'll do a whole video on this because the geometry of it where we are in the Virgo Super Cluster, inside of the Milky Way Galaxy, where we are was much closer to that point. It was on the order of-- and I want to make sure I get this right-- 36 million light years. So we were super close by, I guess, astronomical scales. We were super close, only 36 million light years, to this object, when that light was released. But that light was coming to us and the whole time the universe So we were also moving away from it, if you just think about all of the space, that everything is expanding away from each other, And only 13 billion years later did it finally catch up with us. But the whole time that that was happening, this object has also been moving. This object has also been moving away from us. And so our best estimate of where this object is now, based on how space is expanding, is on the order a 40 or 45 billion light years away. We're just observing where that light And I want to be very clear. What we are observing, this light is coming from something very, very, very primitive. That object or that area of space where that light was emitted from has now condensed into way more, I guess, mature astronomical structures. If you take it from the other point of view, people sitting where in this point of space now, and they've now moved 46 billion light years out, when they observe our region of space, they're not going to see us. They're not going to see Earth as it is now. They're going to see the region of space where Earth is at a super primitive stage, shortly after the Big Bang. And when I use words like \"shortly,\" I use that also loosely. We're still talking about hundreds of thousands or even millions of years. So we'll talk more about that in a future video. But the whole point of this video is it's beyond mind numbing. I would say the last video, about the Milky Way, that alone was mind numbing. But now, we're going in a reality where just the Milky Way becomes something that's" + }, + { + "Q": "8:35 We used to be close but it took the light 13B years to catch up with us. Does this mean the galaxies are moving apart almost at the speed of light? Or some even faster?", + "A": "even faster.", + "video_name": "JiE_kNk3ucI", + "timestamps": [ + 515 + ], + "3min_transcript": "to the beginning of the actual universe. And the reason why it's the visible universe is there might have been something a little bit further out. Maybe it's light hasn't reached us yet or maybe the universe itself, and we'll talk more about this, it's expanding so fast that the light will never, ever reach us. So it's actually a huge question mark on how big the actual universe is. And then some people might say, well, does it even matter? Because this by itself is a huge distance. And I want to make it clear, you might say, OK, if this light over here, if this is coming from 13 billion light years away, or if this is 13 billion light years away, then you could say, hey, so everything that we can observe or that we can even observe the past of, the radius is about 26 billion light years. But even there, we have to be careful because remember the universe is expanding. When this light was emitted-- and I'll do a whole video on this because the geometry of it where we are in the Virgo Super Cluster, inside of the Milky Way Galaxy, where we are was much closer to that point. It was on the order of-- and I want to make sure I get this right-- 36 million light years. So we were super close by, I guess, astronomical scales. We were super close, only 36 million light years, to this object, when that light was released. But that light was coming to us and the whole time the universe So we were also moving away from it, if you just think about all of the space, that everything is expanding away from each other, And only 13 billion years later did it finally catch up with us. But the whole time that that was happening, this object has also been moving. This object has also been moving away from us. And so our best estimate of where this object is now, based on how space is expanding, is on the order a 40 or 45 billion light years away. We're just observing where that light And I want to be very clear. What we are observing, this light is coming from something very, very, very primitive. That object or that area of space where that light was emitted from has now condensed into way more, I guess, mature astronomical structures. If you take it from the other point of view, people sitting where in this point of space now, and they've now moved 46 billion light years out, when they observe our region of space, they're not going to see us. They're not going to see Earth as it is now. They're going to see the region of space where Earth is at a super primitive stage, shortly after the Big Bang. And when I use words like \"shortly,\" I use that also loosely. We're still talking about hundreds of thousands or even millions of years. So we'll talk more about that in a future video. But the whole point of this video is it's beyond mind numbing. I would say the last video, about the Milky Way, that alone was mind numbing. But now, we're going in a reality where just the Milky Way becomes something that's" + }, + { + "Q": "in 8:35 sal said that the universe would keep expanding, if so, in a large amount of time, would we move farther and farther away from the sun or the other planets or galaxies?", + "A": "The expansion of space is only measurable over distances that are much, much greater than even the distance to nearby galaxies. The effect between the earth and the sun is very tiny, and in any event, the sun s gravity offsets the expansion and pulls the earth back even as the expansion of space minutely stretches it away", + "video_name": "JiE_kNk3ucI", + "timestamps": [ + 515 + ], + "3min_transcript": "to the beginning of the actual universe. And the reason why it's the visible universe is there might have been something a little bit further out. Maybe it's light hasn't reached us yet or maybe the universe itself, and we'll talk more about this, it's expanding so fast that the light will never, ever reach us. So it's actually a huge question mark on how big the actual universe is. And then some people might say, well, does it even matter? Because this by itself is a huge distance. And I want to make it clear, you might say, OK, if this light over here, if this is coming from 13 billion light years away, or if this is 13 billion light years away, then you could say, hey, so everything that we can observe or that we can even observe the past of, the radius is about 26 billion light years. But even there, we have to be careful because remember the universe is expanding. When this light was emitted-- and I'll do a whole video on this because the geometry of it where we are in the Virgo Super Cluster, inside of the Milky Way Galaxy, where we are was much closer to that point. It was on the order of-- and I want to make sure I get this right-- 36 million light years. So we were super close by, I guess, astronomical scales. We were super close, only 36 million light years, to this object, when that light was released. But that light was coming to us and the whole time the universe So we were also moving away from it, if you just think about all of the space, that everything is expanding away from each other, And only 13 billion years later did it finally catch up with us. But the whole time that that was happening, this object has also been moving. This object has also been moving away from us. And so our best estimate of where this object is now, based on how space is expanding, is on the order a 40 or 45 billion light years away. We're just observing where that light And I want to be very clear. What we are observing, this light is coming from something very, very, very primitive. That object or that area of space where that light was emitted from has now condensed into way more, I guess, mature astronomical structures. If you take it from the other point of view, people sitting where in this point of space now, and they've now moved 46 billion light years out, when they observe our region of space, they're not going to see us. They're not going to see Earth as it is now. They're going to see the region of space where Earth is at a super primitive stage, shortly after the Big Bang. And when I use words like \"shortly,\" I use that also loosely. We're still talking about hundreds of thousands or even millions of years. So we'll talk more about that in a future video. But the whole point of this video is it's beyond mind numbing. I would say the last video, about the Milky Way, that alone was mind numbing. But now, we're going in a reality where just the Milky Way becomes something that's" + }, + { + "Q": "At 12:30, do the neurotransmitters bind to receptors on sodium ion channels or on the post synaptic membrane?", + "A": "Receptors are sometimes on sodium ion channels, and the ion channels are in the membrane, and sometimes the receptors are in the membrane but connected with the sodium channels with another molecules. I think they will explain us in some other video, probably next one :)", + "video_name": "Tbq-KZaXiL4", + "timestamps": [ + 750 + ], + "3min_transcript": "neurotransmitters. And when the calcium channels-- they're voltage gated-- when it becomes a little more positive, they open calcium floods in and what the calcium does is, it bonds to these proteins that have docked these vesciles. So these little vesicles, they're docked to the presynpatic membrane or to this axon terminal membrane These proteins are actually called SNARE proteins. It's an acronym, but it's also a good word because they've literally snared the vesicles to this membrane. So that's what these proteins are. And when these calcium ions flood in, they bond to these proteins, they attach to these proteins, and they change the confirmation of the proteins just enough that these proteins bring these vesicles closer to the membrane and also kind of pull apart the two membranes so that the Let me do a zoom in of that just to make it clear what's going on. So after they've bonded-- this is kind of before the calcium comes in, bonds to those SNARE proteins, then the SNARE protein will bring the vesicle ultra-close to the presynaptic membrane. So that's the vesicle and then the presynaptic membrane will look like this and then you have your SNARE proteins. And I'm not obviously drawing it exactly how it looks in the cell, but it'll give you the idea of what's going on. Your SNARE proteins have essentially pulled the things together and have pulled them apart so that these two membranes merge. And then the main side effect-- the reason why all this is happening-- is it allows those neurotransmitters to be dumped into the synaptic cleft. So those neurotransmitters that were inside of our vesicle then get dumped into the synaptic cleft. It's exiting the cytoplasm, you could say, of the presynaptic neuron. These neurotransmitters-- and you've probably heard the specific names of many of these-- serotonin, dopamine, epinephrine-- which is also adrenaline, but that's also a hormone, but it also acts as a neurotransmitter. Norepinephrine, also both a hormone and a So these are words that you've probably heard before. But anyway, these enter into the synaptic cleft and then they bond on the surface of the membrane of the post-synaptic neuron or this dendrite. Let's say they bond here, they bond here, and they bond here. So they bond on special proteins on this membrane surface, but the main effect of that is, that will trigger" + }, + { + "Q": "At 3:50, there is just one product drawn. Shouldn't there be one more? The methylgroup is also an ortho- and para-directing activator, which means that there must be an isomer where -SO3H is next to the -CH3.", + "A": "Jay is only showing the major product, which is based on the hydroxyl being a stronger activator.", + "video_name": "iF-f2-KSw6E", + "timestamps": [ + 230 + ], + "3min_transcript": "And first I look at the OH group, which I know is an ortho para director, right. So this is an ortho para director here. And I'd look at where that would be on my ring. Well, once again this would be the ortho position. And again by symmetry, the same ortho position over there. The para position is once again taken up by this methyl group here. When I think about the methyl group-- so let me just use a different color for that one. The methyl group is also an ortho para director. So what's ortho and para to the methyl group. Well this would be ortho, right. These two spots would be ortho, again identical because of symmetry. The para spot is taken up by this OH here. And so now we have a case where we have two ortho para directors directing to different spots. And so the way to figure out which one wins is to think about the activating strength of these two ortho para directors. So the strongest activator is going to be the directing group. So this is a strong activator. And the methyl group we saw is a weak activator from some of the earlier videos here. So the spot that's going to get the SO3H is this carbon, which again by symmetry would be this one right here. So we can go ahead and draw the product of this sulfonation So let me go ahead and sketch in my benzene ring. Let me do another one here. That one wasn't very good. And we have our ring. We have our OH. We have our methyl group. And since the OH is the strongest activator, we're going to go ahead and put SO3H ortho to the OH. And that is our final product. Let's do one more example. So let's see what happens with this reaction here. So once again I can identify this as being a halogenation reaction where And let's go ahead and analyze the substituents once again. So I have a methyl group again, which I know is an ortho para director. So I'll go ahead and mark the spots that are ortho and para to that methyl group. So this would be the ortho position. This would be the ortho position. And then this time the para position is free, so we could possibly put the chlorine there. If I look at what else is on my ring, right-- so I have another substituent here, which is a chlorine. We know that halogens are also ortho para directors because of the lone pair of electrons that are on the chlorine there. So an ortho para director for the chlorine. So what's ortho to this chlorine here. Well this spot is ortho. So is this spot and so is this spot. And so we have a couple of different possible products here. And let's go ahead and start drawing them here. So if I think about the product--" + }, + { + "Q": "ok so on 5:38 he talks about hydrophobic stuff. So is oil hydrophobic?", + "A": "Yes. The reason being that oil cannot be mixed into water, instead is seprerates and that is the meaning of hydrophobic", + "video_name": "QymONNa5C6s", + "timestamps": [ + 338 + ], + "3min_transcript": "Thanks to adhesion, the water molecules are attracted to the molecules in the straw. As the water molecules adhere to the straw, other molecules are drawn in by cohesion following those fellow water molecules. Thank you cohesion. The surface tension created here causes the water to climb up the straw. It will continue to climb until eventually gravity pulling down on the weight of the water in the straw overpowers the surface tension. The fact that water's a polar molecule also makes it really good at dissolving things, which we call it's a good solvent then. Scratch that. Water isn't a good solvent. It's an amazing solvent. There are more substances that can be dissolved in water than in any other liquid on earth. Yes, that includes the strongest acid that we have ever created. These substances that dissolve in water, sugar or salt being ones that we're familiar with, are called hydrophilic, and they are hydrophilic because they are polar. Their polarity is stronger than the cohesive forces of the water. When you get one of these polar substances in water, it's strong enough that it breaks all the little Instead of hydrogen bonding to each other, the water will hydrogen bond around these polar substances. Table salt is ionic, and right now it's being separated into ions as the poles of our water molecules interact with it. What happens when there is a molecule that cannot break the cohesive forces of water? It can't penetrate and come into it. Basically, what happens when that substance can't overcome the strong cohesive forces of water, can't get inside of the water? That's when we get what we call a hydrophobic substance, or something that is fearful of water. These molecules lack charged poles. They are non-polar and are not dissolving in water because essentially they're being pushed out of the water by water's cohesive forces. Water, we may call it the universal solvent, but that does not mean that it dissolves everything. (boppy music) There have been a lot of eccentric scientists throughout history, but all this talk about water of the eccentrics, a man named Henry Cavendish. He communicated with his female servants only via notes, and added a staircase to the back of his house to avoid contact with his housekeeper. Some believe he may have suffered from a form of autism, but just about everyone will admit that he was a scientific genius. He's best remembered as the first person to recognize hydrogen gas as a distinct substance and to determine the composition of water. In the 1700s, most people thought that water itself was an element, but Cavendish observed that hydrogen, which he called inflammable air, reacted with oxygen, known then by the awesome name, dephlogisticated air, to form water. Cavendish didn't totally understand what he'd discovered here, in part because he didn't believe in chemical compounds. He explained his experiments with hydrogen in terms of a fire-like element called phlagiston. Nevertheless, his experiments were groundbreaking. Like his work in determining the specific gravity basically the comparative density of hydrogen" + }, + { + "Q": "why is the water blue at 3:39", + "A": "Blue food coloring/dye was mixed with the water in order to make it more visible on the kitchen counter (since water is clear it is difficult to see).", + "video_name": "QymONNa5C6s", + "timestamps": [ + 219 + ], + "3min_transcript": "This is actually the way that it appears. It is v-shaped. Because this big old oxygen atom is a little bit more greedy for electrons, it has a slight negative charge; whereas, this area here with the hydrogen atoms has a slight positive charge. Thanks to this polarity, all water molecules are attracted to one another; so much so that they actually stick together and these are called hydrogen bonds. We talked about them last time, but essentially what happens is that the positive pole around those hydrogen atoms bonds to the negative pole around the oxygen atoms of a different water molecule. It's a weak bond, but look, they're bonding! Seriously, I cannot overstate the importance of this hydrogen bond. When your teacher asks you, \"What's important about water?\" start out with the hydrogen bonds and you should put it in all caps and maybe some sparkles around it. One of the cool properties that results from these hydrogen bonds is a high cohesion for water which results in high surface tension. Cohesion is the attraction between two like things, like attraction between one molecule of water and another molecule of water. Water has the highest cohesion of any nonmetalic liquid. wax paper or some teflon or something where the water beads up like this. Some leaves of plants do it really well; it's quite cool. Since water adheres weakly to the wax paper or to the plant, but strongly to itself, the water molecules are holding those droplets together in a configuration that creates the least amount of surface area. This is high surface tension that allows some bugs and even I think one lizard and also one Jesus to be able to walk on water. The cohesive force of water does have its limits, of course. There are other substances that water quite likes to stick to. Take glass, for example. This is called adhesion. The water is spreading out here instead of beading up because the adhesive forces between the water and the glass are stronger than the cohesive forces of the individual water molecules in the bead of water. Adhesion is attraction between two different substances, so in this case the water molecules and the glass molecules. These properties lead to one of my favorite things about water; the fact that it can defy gravity. That really cool thing that just happened is called capillary action. Explaining it can be easily done Thanks to adhesion, the water molecules are attracted to the molecules in the straw. As the water molecules adhere to the straw, other molecules are drawn in by cohesion following those fellow water molecules. Thank you cohesion. The surface tension created here causes the water to climb up the straw. It will continue to climb until eventually gravity pulling down on the weight of the water in the straw overpowers the surface tension. The fact that water's a polar molecule also makes it really good at dissolving things, which we call it's a good solvent then. Scratch that. Water isn't a good solvent. It's an amazing solvent. There are more substances that can be dissolved in water than in any other liquid on earth. Yes, that includes the strongest acid that we have ever created. These substances that dissolve in water, sugar or salt being ones that we're familiar with, are called hydrophilic, and they are hydrophilic because they are polar. Their polarity is stronger than the cohesive forces of the water. When you get one of these polar substances in water, it's strong enough that it breaks all the little" + }, + { + "Q": "at 13:50 what happens to the H on the Oxygen that attacks the electrophilic carbon (leading to a cycle product)?", + "A": "During the workup , the reaction mixture is probably treated with dilute sodium carbonate solution. The hydroxide ions would neutralize the H from the oxonium ion.", + "video_name": "8-ccnvn9DxI", + "timestamps": [ + 830 + ], + "3min_transcript": "reacting it with ethylene glycol, and, once again, we use Toluenesulfonic acid, as our catalyst. And this one's a little bit different, because we can see we have a diol, as one of our reactants; up here, we just had butanol, only one OH, but this one has two on it. So, trying to figure out the product here, sometimes it helps just to run through the mechanism really quickly, and so the Toluenesulfonic acid is going to help us to protonate our carbon EEL, and then we have our nucleophile attack, so one of these OHs is going to attack here. And so, without going through all the steps in the mechanism again, that was obviously a pretty complicated mechanism, I'll jump to one of the later steps of the mechanism, where we have already lost water, so minus H two O, so we've already gotten past the dehydration step. And then that would give us this as our intermediate, so there is actually gonna be a plus one formal charge on this oxygen. And then we have these two carbons over here, so let's go ahead, and color-coordinate some of our atoms once again. So, this oxygen has already bonded, we've already lost water, so that oxygen is this oxygen, right here. And then, we still have another OH on this molecule, and that's this one over here, like that. So when we get to this step, we're actually gonna get an intra-molecular, nucleophilic attack. So, these electrons are going to attack this carbon, and kick these electrons off, onto this oxygen. And so, when you think about the final product, you're actually gonna get a cyclic product here, a cyclic acetone. So, we would have our four carbons, and then we would have this oxygen, and then two carbons, and then this oxygen, and they're both bonded to this carbon right here. And so, once again, let's highlight some of those carbons: so this carbon right here, and this carbon right here, or this carbon, and this carbon, and, in our final product, like that. so a little bit trickier than the previous reactions. So in the next video, we'll see a use of cyclic acetals as a protecting group." + }, + { + "Q": "At 8:43. why does ethanol attack the carbonyl carbon and not the oxygen as oxygen possesses the positive charge (3 bonds) while the carbonyl doesn't? Also wouldn't there be less steric hindrance if it attacked the oxygen?", + "A": "That would make the oxygen violate the octet rule", + "video_name": "8-ccnvn9DxI", + "timestamps": [ + 523 + ], + "3min_transcript": "then that would kick these electrons off onto the oxygen, and then we would have water. So, this is the dehydration portion, so we're gonna form water. So let me go ahead, and mark this as being the next step, right? So, in the next step, when those electrons kick in there, so this would be step five, we're going to lose H two O, so the dehydration step. And we would be left with, once again, our ring, and, this time, a double bond to this oxygen, with an ethyl coming off of that oxygen like this. So let's go ahead and make sure we still have a lone pair of electrons on this oxygen, and a plus one formal charge, and the electrons in green, so these electrons in here, moved in here to give us our double bond once again. And, once again, we have a plus one formal charge on the oxygen, so if you drew a resonance structure for this, as being very electrophilic. So, once again, we're going to get a nucleophile attacking our electrophile in the next step, so this would be step six. So, step six would be a nucleophilic attack. So, in step six, a nucleophile comes along, once again, ethanol is our nucleophile, so here is ethanol, so let's go ahead and show ethanol right here, with lone pairs of electrons. And one of these lone pairs of electrons, of course, would attack our electrophile, so nucleophile attacks electrophile, and that would push these electrons in here off onto this oxygen. So let's go ahead, and draw what we have next. Alright, so we now have an oxygen, with still a hydrogen on it, and ethyl right here, a lone pair of electrons, a plus one formal charge on this oxygen. So, let's highlight those electrons: so, in magenta here, these electrons formed a bond, And then over here on the right, we have an oxygen, with an ethyl group, and now there are two lone pairs of electrons on this oxygen. So, we are almost there, right, last step. So, step seven would be a deprotonation step. So in step seven here, all we have to do is take that proton off, and we would form our acetal product. So, once again, we could have a molecule of ethanol come along, and function as a base, and so, a lone pair of electrons take this proton, leaving these electrons behind, on the oxygen, and then finally we are able to draw our acetal products. So we would have, let's go ahead and make this a little bit more angled, so on the left, we would have our oxygen, with an ethyl, and then this carbon is also bonded to another oxygen, with an ethyl coming off of it like that. And so, let's go ahead and show those final electrons here, on our oxygen like this, and, once again, highlight these electrons," + }, + { + "Q": "At 1:50, how was Sal able to infer that acceleration = gravity = -10m/s^2? Was there some kind of formula or rule he used or is that a constant of some sort?", + "A": "That s the acceleration due to gravity on the surface of the earth. We measure it.", + "video_name": "15zliAL4llE", + "timestamps": [ + 110 + ], + "3min_transcript": "" + }, + { + "Q": "at 4:24 the equation that was used was d=v*t . Could we have used the formula vf^2=vi^2+2ad", + "A": "Yes, They ll both give the same answer as long as you know the value of a", + "video_name": "15zliAL4llE", + "timestamps": [ + 264 + ], + "3min_transcript": "" + }, + { + "Q": "At 5:06, Sal gives us the equation delta V=Vf-Vi. Isn't it necessary for (vf-vi) to be divided by time to find the average velocity?", + "A": "No, it isn t. The change in velocity divided by the change in time gives you the acceleration.", + "video_name": "15zliAL4llE", + "timestamps": [ + 306 + ], + "3min_transcript": "" + }, + { + "Q": "how come distance be -500m at 9:08 as told in the video??", + "A": "at 9:08 the distance is -500m because the average velocity is -50m/s the time is 10s and if they both are multiplied the answer would be -500m if you watch the video again you will find how will it come.", + "video_name": "15zliAL4llE", + "timestamps": [ + 548 + ], + "3min_transcript": "" + }, + { + "Q": "At 6:15, how did you get 10? Did you find the square root of 100 to get 10?", + "A": "When you look at the equation, you ll realize that the change in velocity = acceleration X time. We know that the change in velocity (final velocity - initial velocity) is -100 because -100 - 0 = -100. We also know that acceleration is -10 m/s^2 because it s gravity. Then this gives -100 = -10 X t. It s clear now that t=10. Hope that helped!", + "video_name": "15zliAL4llE", + "timestamps": [ + 375 + ], + "3min_transcript": "" + }, + { + "Q": "Sal talks about 7:07 into the video that since momentum is conserved, the speed must be adjusted to the new weight. What happens in an inelastic collision where the two objects stop dead in their tracks? Any mass times 0 velocity always equals zero, so how can momentum be conserved in an inelastic collision?", + "A": "Let s say (to make it easy) that the colliding objects have equal mass and equal speed. Note that equal speed does not mean equal velocity - the objects are headed toward each other, so one of them has velocity exactly equal to the negative of the other one. Before collision, total momentum is: m* v1 + m*v2 = m*(v1 -v2) = 0! So you see now where this is going, right? After the collision, they are stuck together sitting there, so momentum is 0. 0 before, 0 after means momentum was conserved.", + "video_name": "XFhntPxow0U", + "timestamps": [ + 427 + ], + "3min_transcript": "somehow gets stuck in the truck and they just both keep moving together. So they get stuck together. The question is, what is the resulting speed of the combination truck and car after the collision? Well, all we have to do is think about what is the combined momentum before the collision? The momentum of the car is going to be the mass times the car-- mass of the car. Well the total momentum is going to the mass of the car times the velocity of the car plus the mass of the truck times the velocity of the truck. And this is before they hit each other. So what's the mass of the car? That's 1,000. What's the velocity of the car? It's 9 meters per second. So as you can imagine, a unit of momentum would be kilogram meters per second. So it's 1,000 times 9 kilogram meters per second, but I won't save space. And then the mass of the truck is 2,000. And what's its velocity? Well, it's 0. It's stationary initially. So the initial momentum of the system-- this is 2,000 times 0-- is 9,000 plus 0, which equals 9,000 kilogram meters per second. That's the momentum before the car hits the back of the truck. Now what happens after the car hits the back of the truck? So let's go to that situation. So we have the truck. I'll draw it a little less neatly. And then you have the car and it's probably a little bit-- well, I won't go into whether it's banged up and whether it released heat and all of that. Let's assume that there was nothing-- if this is a simple problem that we can do. So if we assume that, there would be no change in momentum. Because we're saying that there's no net forces acting on the system. And when I say system, I mean the combination of the car and the truck. vehicle called a car truck, its momentum will have to be the same as the car and the truck's momentum when they were separate. So what do we know about this car truck object? Well we know its new mass. The car truck object, it will be the combined mass of the two. So it's 1,000 kilograms plus 2,000 kilograms. So it's 3,000 kilograms. And now we can use that information to figure out its velocity. How? Well, its momentum-- this 3,000 kilogram object's momentum-- has to be the same as the momentum of the two objects before the collision. So it still has to be 9,000 kilogram meters per second. So once again, mass times velocity. So mass is 3,000 times the new velocity. So we could call that, I don't know, new velocity, v sub n. That will equal 9,000." + }, + { + "Q": "you said we need the mass of the moving object, but what happens if I am not given the mass of the car? like said at 5:30", + "A": "To calculate momentum, mass is an essential ingredient. However, you might be given the mass indirectly so to say. E.g. One of the cars can move with 40N with an acceleration with 4m/s. This automatically tells you that the mass is 10kg. Thus either way, you must have the mass somehow. Or else it is impossible to calculate the momentum.", + "video_name": "XFhntPxow0U", + "timestamps": [ + 330 + ], + "3min_transcript": "they use for impulse. But another way of viewing impulse is force times change in time. Well that's the same thing as change in momentum over change in time times change in time. Right? Because this is just the same thing as force. And that's just change in momentum, so that's impulse as well. And the unit of impulse is the joule. And we'll go more into the joule when we do work in all of that. And if this confuses you, don't worry about it too much. The main thing about momentum is that you realize it's mass times velocity. And since force is change in momentum per unit of time, if you don't have any external forces on a system or, on say, on a set of objects, their combined, or their net momentum won't change. And that comes from Newton's Laws. The only way you can get a combined change in momentum is if you have some type of net force acting on the system. momentum problems. Whoops. Invert colors. OK. So let's say we have a car. Say it's a car. Let me do some more interesting colors. A car with a magenta bottom. And it is, let's see, what does this problem say? It's 1,000 kilograms. So a little over a ton. And it's moving at 9 meters per second east. So its velocity is equal to 9 meters per second east, or to the right in this example. And it strikes a stationary 2, 000 kilogram truck. So here's my truck. Here's my truck and this is a 2,000 kilogram truck. And it's stationary, so the velocity is 0. somehow gets stuck in the truck and they just both keep moving together. So they get stuck together. The question is, what is the resulting speed of the combination truck and car after the collision? Well, all we have to do is think about what is the combined momentum before the collision? The momentum of the car is going to be the mass times the car-- mass of the car. Well the total momentum is going to the mass of the car times the velocity of the car plus the mass of the truck times the velocity of the truck. And this is before they hit each other. So what's the mass of the car? That's 1,000. What's the velocity of the car? It's 9 meters per second. So as you can imagine, a unit of momentum would be kilogram meters per second. So it's 1,000 times 9 kilogram meters per second, but I won't" + }, + { + "Q": "At 4:54, why can't we assume switching the positions of CH3 and Br? Since they're all single bonds, wouldn't they rotate freely anyways?", + "A": "No, that s the point. No matter how you try you ll find it s impossible to rotate them in such a way that the 2 molecules can fit on top of one another with each atom in the same place. If you cannot visualise it, a molecular model kit will be able to prove it.", + "video_name": "tk-SNvCPLCE", + "timestamps": [ + 294 + ], + "3min_transcript": "should say usually, are carbons, especially when we're dealing in organic chemistry, but they could be phosphoruses or sulfurs, but usually are carbons bonded to four different groups. And I want to emphasize groups, not just four different atoms. And to kind of highlight a molecule that contains a chiral atom or chiral carbon, we can just think of one. So let's say that I have a carbon right here, and I'm going to set this up so this is actually a chiral atom, that the carbon specific is a chiral atom, but it's partly a chiral molecule. And then we'll see examples that one or both of these are true. Let's say it's bonded to a methyl group. Let's say there's a bromine over here. Let's say behind it, there is a hydrogen, and then above it, we have a fluorine. Now if I were to take the mirror image of this thing right here, we have your carbon in the center-- I want to do it in that same blue. You have the carbon in the center and then you have the fluorine above the carbon. You have your bromine now going in this direction. You have this methyl group. It's still popping out of the page, but it's now going to the right instead of to the left, So CH3. And then you have the hydrogen still in the back. These are mirror images, if you view this as kind of the mirror and you can see on both sides of the mirror. Now, why is this chiral? Well, it's a little bit of a visualization challenge, but no matter how you try to rotate this thing right here, you will never make it exactly like this thing. You might try to rotate it around like that and try to So let's try to do that. If we try to get the methyl group over there, what's going to happen to the other groups? Well, then the hydrogen group is going-- or the hydrogen, I The hydrogen atom is going to move there and the bromine is going to move there. So this would be superimposable if this was a hydrogen and this was a bromine, but it's not. You can imagine, the hydrogen and bromine are switched. And you could flip it and do whatever else you want or try to rotate it in any direction, but you're not going to be able to superimpose them. So this molecule right here is a chiral molecule, and this carbon is a chiral center, so this carbon is a chiral carbon, sometimes called an asymmetric carbon or a chiral center. Sometimes you'll hear something called a stereocenter. A stereocenter is a more general term for any point in" + }, + { + "Q": "I actually cannot understand why will the molecule be not superimposble (at 5:36).", + "A": "He is arguing, correctly, that the images are NOT superimposable because the carbon is chiral.", + "video_name": "tk-SNvCPLCE", + "timestamps": [ + 336 + ], + "3min_transcript": "Let's say there's a bromine over here. Let's say behind it, there is a hydrogen, and then above it, we have a fluorine. Now if I were to take the mirror image of this thing right here, we have your carbon in the center-- I want to do it in that same blue. You have the carbon in the center and then you have the fluorine above the carbon. You have your bromine now going in this direction. You have this methyl group. It's still popping out of the page, but it's now going to the right instead of to the left, So CH3. And then you have the hydrogen still in the back. These are mirror images, if you view this as kind of the mirror and you can see on both sides of the mirror. Now, why is this chiral? Well, it's a little bit of a visualization challenge, but no matter how you try to rotate this thing right here, you will never make it exactly like this thing. You might try to rotate it around like that and try to So let's try to do that. If we try to get the methyl group over there, what's going to happen to the other groups? Well, then the hydrogen group is going-- or the hydrogen, I The hydrogen atom is going to move there and the bromine is going to move there. So this would be superimposable if this was a hydrogen and this was a bromine, but it's not. You can imagine, the hydrogen and bromine are switched. And you could flip it and do whatever else you want or try to rotate it in any direction, but you're not going to be able to superimpose them. So this molecule right here is a chiral molecule, and this carbon is a chiral center, so this carbon is a chiral carbon, sometimes called an asymmetric carbon or a chiral center. Sometimes you'll hear something called a stereocenter. A stereocenter is a more general term for any point in groups that it is joined to. But all of these, especially when you're in kind of in introductory organic chemistry class, tends to be a carbon bonded to four different groups. And I want to to stress that it's not four different atoms. You could have had a methyl group here and a propyl group here, and the carbon would still be bonded directly to a carbon in either case, but that would still be a chiral carbon, and this would still actually be a chiral molecule. In the next video, we'll do a bunch of examples. We'll look at molecules, try to identify the chiral carbons, and then try to figure out whether the molecule itself is--" + }, + { + "Q": "At 6:00 Sal says that an uniform electric field can be formed. But,as i have understood, it is uniform only along the plate not perpendicular to it and thus we cant construct one even with an infinitely long charged plate.(the way it is shown in the diagram i.e. perpendicular to the plate as the field is proportional to 1/d where d is the distance from the wire.\nIs this correct?", + "A": "It is close to uniform as long as the distance between the plates is much smaller than the plates. Also, it is not uniform near the edges.", + "video_name": "elJUghWSVh4", + "timestamps": [ + 360 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:58 Sal says that the upward force we would need to apply to push a certain object up is equal to the force of gravity. But wouldn't tht mean tht the net force on the object equals 0 and it would not move upwards?\nPlease help? Sorry if I'm missing something here.", + "A": "Sal is saying that after getting the object to accelerate a little then we have to apply the same force as gravity. As we have already caused the object to accelerate we then only have to keep the object moving to go up. And only by applying the same and opposite force as gravity, we can do that.", + "video_name": "elJUghWSVh4", + "timestamps": [ + 178 + ], + "3min_transcript": "" + }, + { + "Q": "At 7:23,velocity at time t=0 should be 2 m/s right why is it 1 m/s?", + "A": "You probably mixed the concept of velocity and change in velocity (or acceleration). At t=0, the velocity David assume is 1 m/s, while the acceleration is 2 m/s^2.", + "video_name": "DD58B2siDv0", + "timestamps": [ + 443 + ], + "3min_transcript": "by two meters per second during this time. Now you might object. You might say, \"Wait a minute. \"I'll buy this over here because height times width \"is just a times delta t, \"but triangle, that has an extra factor of a half in it, \"and there's no half up here. \"How does this, I mean, how can we still make this claim?\" We can make this claim because we'll do the same thing we always do. We can imagine, all right, imagine a rectangle here. We're gonna estimate the area with a bunch of rectangles. Then this rectangle, and this rectangle in your line like that looks horrible. That doesn't look like the area of a triangle at all. It's got all these extra pieces right here, right? You don't want all of that. And okay, I agree. That didn't work so well. Let's make them even smaller, right? Smaller width. So we'll do a rectangle like that. We'll do this one. You see we're getting better. This is definitely closer. This is not as bad as the other one but it's still not exact. so we'll make it even smaller rectangle and an even smaller rectangle here all of these at the same width but they're even smaller than the ones before. Now we're getting really close. This area is really gonna get close to the area of the triangle. The point is if you make them infinite testable small, they'll exactly represent the area of a triangle. Each one of them can be found with this formula. The delta v for each one will be the area, or sorry, the acceleration of the height of that rectangle times the small infinite testable width and you'll get the total delta v which is so gonna be the total area. Long story short, area on a, acceleration versus time graphs represents the change in velocity. This is one you got to remember. this is the most important aspect of an acceleration graph, oftentimes the most useful aspect of it, the way you analyze it. So why do we care about change in velocity? Because it will allow us to find the velocity. then we can find the velocity at any other point. For instance, let's say I gave you the velocity Daisy had. For some reason I'm gonna stopwatch. I start my stopwatch at right at that moment. At t equals zero, Daisy had a velocity of, let's say positive one meter per second. So Daisy was traveling that fast at t equals zero. That was her velocity at t equals zero seconds. Now I can get the velocity wherever I want. If I want the velocity at four, let's figure this out. To get the velocity at four, I can say that the delta v during this time period right here, this four seconds. I know what that delta v was. That delta v was positive eight. We found that area, height times width. So positive eight is what the delta v is gotta equal. What's delta v? That's v at four seconds minus v at zero seconds." + }, + { + "Q": "at 8:05, how do we know that the velocity at 0(zero) second was 1?", + "A": "It is just an assumption.", + "video_name": "DD58B2siDv0", + "timestamps": [ + 485 + ], + "3min_transcript": "so we'll make it even smaller rectangle and an even smaller rectangle here all of these at the same width but they're even smaller than the ones before. Now we're getting really close. This area is really gonna get close to the area of the triangle. The point is if you make them infinite testable small, they'll exactly represent the area of a triangle. Each one of them can be found with this formula. The delta v for each one will be the area, or sorry, the acceleration of the height of that rectangle times the small infinite testable width and you'll get the total delta v which is so gonna be the total area. Long story short, area on a, acceleration versus time graphs represents the change in velocity. This is one you got to remember. this is the most important aspect of an acceleration graph, oftentimes the most useful aspect of it, the way you analyze it. So why do we care about change in velocity? Because it will allow us to find the velocity. then we can find the velocity at any other point. For instance, let's say I gave you the velocity Daisy had. For some reason I'm gonna stopwatch. I start my stopwatch at right at that moment. At t equals zero, Daisy had a velocity of, let's say positive one meter per second. So Daisy was traveling that fast at t equals zero. That was her velocity at t equals zero seconds. Now I can get the velocity wherever I want. If I want the velocity at four, let's figure this out. To get the velocity at four, I can say that the delta v during this time period right here, this four seconds. I know what that delta v was. That delta v was positive eight. We found that area, height times width. So positive eight is what the delta v is gotta equal. What's delta v? That's v at four seconds minus v at zero seconds. I know what v at zero second was. That was one. So we can get that v at four minus one meter per second is equal to positive eight meters per second. So I get the velocity at four was positive nine meters per second. And you're like, phew, that was hard. I don't wanna do that every time. Yeah, I wouldn't wanna do that every time either so there's a quick way to do it. We can just do this. What's the velocity we had to start with? That was one. What was our change in velocity? That was positive eight. So what's our final velocity? Well, one plus eight gives us our final velocity. It's positive nine. Well it's just gonna take this change in velocity of this area which represents the change in velocity which is gonna add our initial velocity to it when we solve for this final velocity. for instance, if I didn't make sense, for instance, if we want to find the velocity at six, well, we can just say we started at t equals four seconds with a velocity of positive nine. We start here with positive nine." + }, + { + "Q": "At 6:10, shouldn't the 'd' of 'Dicyclobutyl' be in small case, i.e. 'dicyclobutyl' ?", + "A": "The D would be capitalized at the beginning of a sentence, but not anywhere else.", + "video_name": "ygXkdSKXQoA", + "timestamps": [ + 370 + ], + "3min_transcript": "we're trying to minimize the numbers here. We might want to go clockwise so we hit two right over here, three, four, five, six, seven. So in this situation where do we hit interesting things? We hit interesting things at the one, at the two, or where do we have groups attached to the chain? One, two, four, six, and seven. So that's one option. We'll have groups attached at one, two, four, six, and seven if we start right over there, and if we were to go clockwise. Our other option, is to start, and let me erase those, at this other cyclobutyl group here on the left. Let me erase these so I don't mess up the diagram too much. The other option is to start here, make this the one, number this as one, to hit the cyclopropyl, to hit something else as soon as possible. So two, three, four, five, six, seven. So now where we are getting interesting things? We have groups at the one carbon, the two carbon, the four carbon, the five carbon, and at the seven carbon. So actually this second numbering is preferable. Both of them have something at the one, the two, the four, and the seven, but the second one has something at a five while the first one had something at a six. So we would actually want to do this numbering right over here, and this is the numbering that we have now listed right over here. Let's now write what the name of this molecule actually is. So once again, we start first in alphabetical order. So we're going to start with the cyclobutyl groups. And we have two of them, one at the one carbon, So we could say 1 comma 4. And since there are two cyclobutyl groups we would say dicyclobutyl. 1, 4 dicyclobutyl. And then what comes next in alphabetical order is the cyclopropyl. That comes before isobutyl. \"C\" comes before \"I.\" That's at the two carbon. So then we can say 2 cyclopropyl. And now we can get to the two isobutyls. So there's an isobutyl at the five carbon and at the seven carbon. So we can say 5, 7, and since there's two of them we'd say di isobutyl. 5, 7 di isobutyl." + }, + { + "Q": "At 5:40 how would we number it if the numbers were\n1)1,2,4,6,7\n2)1,2,4,5,8\nBecause ive heard we always take the lowest sum, but what if the sum were equal?", + "A": "5 is lower than 6, the sum doesn t actually matter.", + "video_name": "ygXkdSKXQoA", + "timestamps": [ + 340 + ], + "3min_transcript": "we're trying to minimize the numbers here. We might want to go clockwise so we hit two right over here, three, four, five, six, seven. So in this situation where do we hit interesting things? We hit interesting things at the one, at the two, or where do we have groups attached to the chain? One, two, four, six, and seven. So that's one option. We'll have groups attached at one, two, four, six, and seven if we start right over there, and if we were to go clockwise. Our other option, is to start, and let me erase those, at this other cyclobutyl group here on the left. Let me erase these so I don't mess up the diagram too much. The other option is to start here, make this the one, number this as one, to hit the cyclopropyl, to hit something else as soon as possible. So two, three, four, five, six, seven. So now where we are getting interesting things? We have groups at the one carbon, the two carbon, the four carbon, the five carbon, and at the seven carbon. So actually this second numbering is preferable. Both of them have something at the one, the two, the four, and the seven, but the second one has something at a five while the first one had something at a six. So we would actually want to do this numbering right over here, and this is the numbering that we have now listed right over here. Let's now write what the name of this molecule actually is. So once again, we start first in alphabetical order. So we're going to start with the cyclobutyl groups. And we have two of them, one at the one carbon, So we could say 1 comma 4. And since there are two cyclobutyl groups we would say dicyclobutyl. 1, 4 dicyclobutyl. And then what comes next in alphabetical order is the cyclopropyl. That comes before isobutyl. \"C\" comes before \"I.\" That's at the two carbon. So then we can say 2 cyclopropyl. And now we can get to the two isobutyls. So there's an isobutyl at the five carbon and at the seven carbon. So we can say 5, 7, and since there's two of them we'd say di isobutyl. 5, 7 di isobutyl." + }, + { + "Q": "At 9:42 Sal draws a circle and a dot in the centre to say that the vector is pointing out of the page, though what do you draw/signal to say that the vector pointing away from you? Thank you to whoever can answer this.", + "A": "You draw a circle with an x in it. The idea is that the one coming toward you looks like the head of an arrow coming at you and the one going away from you looks like the feathers of an arrow going away.", + "video_name": "s38l6nmTrvM", + "timestamps": [ + 582 + ], + "3min_transcript": "The magnitude of our force vector times sine of theta, that gave us the component of the force vector that is perpendicular to the arm. And we just multiply that times the magnitude of r, and we got the magnitude of the torque vector, which was 15. We can leave out the newton meters for now. 15, and then its direction is this vector that we specified by n. We can call it the normal vector. And what do we know about this vector? It's perpendicular to both r-- this is r, right-- and it's perpendicular to F. And the only way that I can visualize in our three-dimensional universe, a vector that's perpendicular to both this and this is if it pops in or out of this page, right? Because both of these vectors are in the plane that are defined by our video. So if I'm a vector that is perpendicular to your screen, whatever you're watching this on, then it's going to be perpendicular to both of these vectors. into the page? We use the right hand rule, right? In the right hand rule, we take-- r is our index finger, F is our middle finger, and whichever direction our thumb points in tells us whether or not we are-- the direction of the cross product. So let's draw it. Let me see if I can do a good job right here. So if that is my index finger, and you could imagine your hand sitting on top of this screen. So that's my index finger representing r, and this is my Remember, it only works with your right hand. If you do your left hand, it's going to be the opposite. And then my middle finger is going to go in the direction you to draw this. So if I were to draw it-- let me draw my nails just so you know what this is. So this is the nail on my index finger. This is the nail on my middle finger. And so in this situation, where is my thumb going to be? My thumb is going to be popping out. I wish I could-- that's the nail of my thumb. Hopefully, that makes some sense, right? That's the palm of my hand. That's the other side of my-- and I could keep drawing, but hopefully, that makes some sense. This is my index finger. This is the middle finger. My thumb is pointing out of the page, so that tells us that the torque is actually pointing out of the page. So the direction of this unit vector n is going to be out of the page, and we could signify that by a circle with a dot. And I'm almost at my time limit, and so there you have it: the cross product as it is applied to torque. See you in the next video." + }, + { + "Q": "@ 1:35 how do we predict when something will dissociate and when it will not? Why doesn't the Carbon or any other part of the sorbate dissociate?", + "A": "This sort of thing will come to you with time doing practice problems. In general though just remember that water is not strong enough to break C-C or C-H bonds.", + "video_name": "jzcB3faNdq0", + "timestamps": [ + 95 + ], + "3min_transcript": "- [Voiceover] Potassium sorbate, and they give us its formula right over here, has a molar mass of 150 grams per mole. They put this decimal here to show us that these are actually three significant figures. Even the zero is a significant digit here. Is commonly added to diet soft drinks as a preservative. A stock solution of potassium sorbate, dissolvent's an aqueous solution here, of known concentration must be prepared. A student titrates 45 millileters of the stock solution with one point two five molar hydrochloric acid using both an indicator and a pH meter. The value of K A for sorbic acid is one point seven times ten to the negative fifth. All right, so let's tackle this piece by piece. Write the net ionic equation for the reaction between potassium sorbate and hydrochloric acid. All right, so first off, let's write the ionic equation, and then we'll, I'll write the net ionic equation, and hopefully you'll see the difference. So ionic, ionic equation. is we think about well, if these are dissolved in water, it's an aqueous solution, these are going to disassociate into their, into ions. And so we would write that out on both of the, on both the reactant and the product side. So the potassium sorbate, we can write that as, it's gonna be a potassium ion dissolved in an aqueous solution, plus the C six H seven O two, this is also going to be an ion dissolved in the aqueous solution, plus the hydrochloric acid will dissolve, so you have the hydrogen proton dissolved in the aqueous solution plus the chloride ion, or anion I guess we could say it. so that's going to be in our aqueous solution, What happens? Well, you're going to have the C six H seven O two react with the hydrogen proton to get to sorbic acid. So you're gonna have sorbic acid, H C six H seven O two, that's the sorbic acid. It's going to be in an aqueous solution. So I took care so far of that and that, and then you're going to have, and then you're going to have your potassium ions, your potassium ions and your chloride ions. It's going to be just like that. So this right over here is the ionic equation, not the net ionic equation. I have the ions on the reaction, on the reactant side, and then on the product side right over here. And did I, yep, I included everything. Now you do the net ionic reaction, you can imagine what's going to happen here." + }, + { + "Q": "What is coulom force at 6:09?", + "A": "A Coulomb force is the attractive force between positive and negative charges.", + "video_name": "q--2WP8wXtk", + "timestamps": [ + 369 + ], + "3min_transcript": "there the electrons in that bond could spend some of their time on this atom and some of their time on this atom right over here. And so when you have a covalent bond like this, you can then find the distance between the 2 nuclei and take half of that and call that call that the atomic radius. So these are all different ways of thinking about it. Now, with that out of the way, let's think about what the trends for atomic size or atomic radii would be in the periodic table. So the first thing to think about is what do you think will be the trend for atomic radii as we move through a period. So let's say we're in the fourth period and we were to go from potassium to krypton. What do you think is going to be the trend here? And if you want to think about the extremes, how do you think potassium is going to compare to krypton in terms of atomic radius. I encourage you to pause this video and think about that on your own. the outermost electrons are going to be in your fourth shell. Here, you're filling out 4S1, 4S2. Then you start back filling into the 3D subshell and then you start filling again in 4P1 and so forth. You start filling out the P subshell. So as you go from potassium to krypton, you're filling out that outermost fourth shell. Now what's going on there? Well, when you're at potassium, you have 19- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. You have 19 protons and you have 19 electrons. Well I'll just draw those. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, but you only have 1 electron in that outermost, in that fourth shell, so let's just say that's that electron at a moment, just for visually. It doesn't necessarily have to be there but just to visualize that. So that 1 electron right over there, So, you have some, I guess you could say Coulom force that is attracting it, that is keeping it there. But if you go to krypton, all of a sudden you have much more positive charge in the nucleus. So you have 1, 2, 3, 4, 5, 6, 7, 8- I don't have to do them all. You have 36. You have a positive charge of 36. Let me write that, you have plus 36. Here you have plus 19. And you have 36 electrons, you have 36 electrons- I don't know, I've lost track of it, but in your outermost shell, in your fourth, you're going to have the 2S and then you're going to have the 6P. So you have 8 in your outermost shell. So that'd be 1, 2, 3, 4, 5, 6, 7, 8. So one way to think about it, if you have more positive charge in the center, and you have more negative charge on that outer shell, so that's going to bring that outer shell inward. It's going to have more I guess you could imagine one way," + }, + { + "Q": "At 6:38 Sal said \" 2s and 2p\". But aren't those 4s and 4p orbitals?", + "A": "Yes, they are 4s and 4p orbitals", + "video_name": "q--2WP8wXtk", + "timestamps": [ + 398 + ], + "3min_transcript": "the outermost electrons are going to be in your fourth shell. Here, you're filling out 4S1, 4S2. Then you start back filling into the 3D subshell and then you start filling again in 4P1 and so forth. You start filling out the P subshell. So as you go from potassium to krypton, you're filling out that outermost fourth shell. Now what's going on there? Well, when you're at potassium, you have 19- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. You have 19 protons and you have 19 electrons. Well I'll just draw those. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, but you only have 1 electron in that outermost, in that fourth shell, so let's just say that's that electron at a moment, just for visually. It doesn't necessarily have to be there but just to visualize that. So that 1 electron right over there, So, you have some, I guess you could say Coulom force that is attracting it, that is keeping it there. But if you go to krypton, all of a sudden you have much more positive charge in the nucleus. So you have 1, 2, 3, 4, 5, 6, 7, 8- I don't have to do them all. You have 36. You have a positive charge of 36. Let me write that, you have plus 36. Here you have plus 19. And you have 36 electrons, you have 36 electrons- I don't know, I've lost track of it, but in your outermost shell, in your fourth, you're going to have the 2S and then you're going to have the 6P. So you have 8 in your outermost shell. So that'd be 1, 2, 3, 4, 5, 6, 7, 8. So one way to think about it, if you have more positive charge in the center, and you have more negative charge on that outer shell, so that's going to bring that outer shell inward. It's going to have more I guess you could imagine one way, And because of that, that outer most shell is going to drawn in. Krypton is going to be smaller, is going to have a smaller atomic radius than potassium. So the trend, as you go to the right is that you are getting, and the general trend I would say, is that you are getting smaller as you go to the right in a period. That's the reason why the smallest atom of all, the element with the smallest atom is not hydrogen, it's helium. Helium is actually smaller than hydrogen, depending on how you, depending on what technique you use to measure it. That's because, if we take the simplest case, hydrogen, you have 1 proton in the nucleus and then you have 1 electron in that 1S shell, and in helium you have 2, 2 protons in the nucleus and I'm not drawing the neutrons and obviously there's different isotopes," + }, + { + "Q": "At 3:38,sal says atooms are joined by covalent bonds ......so my first question is what are covalent bonds and how can we find radius if atom is not joined to anyone??", + "A": "Covalent bonds are bonds between two atoms sharing electrons.", + "video_name": "q--2WP8wXtk", + "timestamps": [ + 218 + ], + "3min_transcript": "You might say well that's the radius. But in the next moment, there's some probability it might be likely that it ends up here. But there's some probability that it's going to be over there. Then the radius could be there. So electrons, these orbitals, these diffuse probability distributions, they don't have a hard edge, so how can you say what the size of an atom actually is? There's several techniques for thinking about this. One technique for thinking about this is saying, okay, if you have 2 of the same atom, that are- 2 atoms of the same element that are not connected to each other, that are not bonded to each other, that are not part of the same molecule, and you were able to determine somehow the closest that you could get them to each other without them bonding. So, you would kind of see, what's the closest that they can, they can kind of get to each other? So let's say that's one of them and then this is the other one right over here. that closest, that minimum distance, without some type of, you know, really, I guess, strong influence happening here, but just the minimum distance that you might see between these 2 and then you could take half of that. So that's one notion. That's actually called the Van der Waals radius. Another way is well what about if you have 2 atoms, 2 atoms of the same element that are bonded to each other? They're bonded to each other through a covalent bond. So a covalent bond, we've already- we've seen this in the past. The most famous of covalent bonds is well, a covalent bond you essentially have 2 atoms. So that's the nucleus of one. That's the nucleus of the other. And they're sharing electrons. So their electron clouds actually, their electron clouds actually overlap with each other, actually overlap with each other there the electrons in that bond could spend some of their time on this atom and some of their time on this atom right over here. And so when you have a covalent bond like this, you can then find the distance between the 2 nuclei and take half of that and call that call that the atomic radius. So these are all different ways of thinking about it. Now, with that out of the way, let's think about what the trends for atomic size or atomic radii would be in the periodic table. So the first thing to think about is what do you think will be the trend for atomic radii as we move through a period. So let's say we're in the fourth period and we were to go from potassium to krypton. What do you think is going to be the trend here? And if you want to think about the extremes, how do you think potassium is going to compare to krypton in terms of atomic radius. I encourage you to pause this video and think about that on your own." + }, + { + "Q": "At 7:15, do you have to write the CH2 as carbon bonded with 2 hydrogen atoms? Can't you just write it as CH2?", + "A": "The formula with two separate H atoms is a structural formula. The formula with a CH\u00e2\u0082\u0082 group is a condensed structural formula. You decide which one you want to use for your own purposes.", + "video_name": "pMoA65Dj-zk", + "timestamps": [ + 435 + ], + "3min_transcript": "And the important thing is, no matter what the notation, as long as you can figure out the exact molecular structure, as long as you can-- so there's this last CH3. Whether you have this, this, or this, you know what the molecular structure is. You could draw any one of these given any of the others. Now, there's an even simpler way to write this. You could write it just like this. Let me do it in a different color. You literally could write it so we have three carbons. So one, two, three. Now, this seems ridiculously simple and you're like, how can this thing right here give you the same information as all of these more complicated ways to draw it? Well, in chemistry, and in organic chemistry in particular, any of these-- let me call it a line diagram or a line angle diagram. It's the simplest way and it's actually probably the most useful way to show chains of carbons or to show organic molecules. Once they start to get really, really complicated, because something like this, you assume that the end points of any lines have a carbon on it. So if you see something like that, you assume that there's a carbon at that end point, a carbon at that end point, and a carbon at that end point. And then you know that carbon makes four bonds. There are no charges here. All the carbons are going to make four bonds, and each of the carbons here, this carbon has two bonds, so the other two bonds are implicitly going to be with hydrogens. If they don't draw them, you assume that they're going to be with hydrogens. This guy has one bond, so the other three must be with hydrogen. This guy has one bond, so the other three must be hydrogens. So just drawing that little line angle thing right there, I actually did convey the exact same information as this depiction, this depiction, or this depiction. So you're going to see a lot of this. This really simplifies things. And sometimes you see things that are in between. You might see someone draw it like this, where they'll write CH3, and then they'll draw it like that. molecule where you write the CH3's for the end points, but then you implicitly have the CH2 on the inside. You assume that this end point right here is a C and it's bonded to two hydrogens. So these are all completely valid ways of drawing the molecular structures of these carbon chains or of these organic compounds." + }, + { + "Q": "At 1:14 isn't the metals \"magic number\" 18, as they want to fill their d-orbital.", + "A": "Yes, but that isn t something that should come up in organic chemistry as it almost always deals with C H N O (and a few others here and there...)", + "video_name": "pMoA65Dj-zk", + "timestamps": [ + 74 + ], + "3min_transcript": "The one thing that probably causes some of the most pain in chemistry, and in organic chemistry, in particular, is just the notation and the nomenclature or the naming that we use. And what I want to do here in this video and really the next few videos is to just make sure we have a firm grounding in the notation and in the nomenclature or how we name things, and then everything else will hopefully not be too difficult. So just to start off, and this is really a little bit of review of regular chemistry, if I just have a chain of carbons, and organic chemistry is dealing with chains of carbons. Let me just draw a one-carbon chain, so it's really kind of ridiculous to call it a chain, but if we have one carbon over here and it has four valence electrons, it wants to get to eight. That's the magic number we learned in just regular chemistry. For all molecules, that's the stable valence structure, I guess you could say it. A good partner to bond with is hydrogen. So it has four valence electrons and then hydrogen electron with each other and then they both look pretty happy. I said eight's the magic number for everybody except for hydrogen and helium. Both of them are happy because they're only trying to fill their 1s orbital, so the magic number for those two guys is two. So all of the hydrogens now feel like they have two electrons. The carbon feels like it has eight. Now, there's several ways to write this. You could write it just like this and you can see the electrons explicitly, or you can draw little lines here. So I could also write this exact molecule, which is methane, and we'll talk a little bit more about why it's called methane later in this video. I can write this exact structure like this: a carbon bonded to four hydrogens. And the way that I've written these bonds right here you could imagine that each of these bonds consists of two Now let's explore slightly larger chains. So let's say I have a two-carbon chain. Well, let me do a three-carbon chain so it really looks like a chain. So if I were to draw everything explicitly it might So I have a carbon. It has one, two, three, four electrons. Maybe I have another carbon here that has-- let me do the carbons in slightly different shades of yellow. I have another carbon here that has one, two, three, four electrons. And then let me do the other carbon in that first yellow. And then I have another carbon so we're going to have a three-carbon chain. It has one, two, three, four valence electrons. Now, these other guys are unpaired, and if you don't specify it, it's normally going to be hydrogen, so let me draw some hydrogens over here. So you're going to have a hydrogen there, a hydrogen over there, a hydrogen over here, a hydrogen over here, a" + }, + { + "Q": "why is hydrogen the best partner as explained by sal in 0:56?\nwhy can't other atoms be used?", + "A": "other atoms can be used but the main problem is that we have to draw other electrons of the atom also. For eg:in CO2 we have to draw the rest 4 electrons of oxygen & also number compounds of carbon-hydrogen is more than 1000.", + "video_name": "pMoA65Dj-zk", + "timestamps": [ + 56 + ], + "3min_transcript": "The one thing that probably causes some of the most pain in chemistry, and in organic chemistry, in particular, is just the notation and the nomenclature or the naming that we use. And what I want to do here in this video and really the next few videos is to just make sure we have a firm grounding in the notation and in the nomenclature or how we name things, and then everything else will hopefully not be too difficult. So just to start off, and this is really a little bit of review of regular chemistry, if I just have a chain of carbons, and organic chemistry is dealing with chains of carbons. Let me just draw a one-carbon chain, so it's really kind of ridiculous to call it a chain, but if we have one carbon over here and it has four valence electrons, it wants to get to eight. That's the magic number we learned in just regular chemistry. For all molecules, that's the stable valence structure, I guess you could say it. A good partner to bond with is hydrogen. So it has four valence electrons and then hydrogen electron with each other and then they both look pretty happy. I said eight's the magic number for everybody except for hydrogen and helium. Both of them are happy because they're only trying to fill their 1s orbital, so the magic number for those two guys is two. So all of the hydrogens now feel like they have two electrons. The carbon feels like it has eight. Now, there's several ways to write this. You could write it just like this and you can see the electrons explicitly, or you can draw little lines here. So I could also write this exact molecule, which is methane, and we'll talk a little bit more about why it's called methane later in this video. I can write this exact structure like this: a carbon bonded to four hydrogens. And the way that I've written these bonds right here you could imagine that each of these bonds consists of two Now let's explore slightly larger chains. So let's say I have a two-carbon chain. Well, let me do a three-carbon chain so it really looks like a chain. So if I were to draw everything explicitly it might So I have a carbon. It has one, two, three, four electrons. Maybe I have another carbon here that has-- let me do the carbons in slightly different shades of yellow. I have another carbon here that has one, two, three, four electrons. And then let me do the other carbon in that first yellow. And then I have another carbon so we're going to have a three-carbon chain. It has one, two, three, four valence electrons. Now, these other guys are unpaired, and if you don't specify it, it's normally going to be hydrogen, so let me draw some hydrogens over here. So you're going to have a hydrogen there, a hydrogen over there, a hydrogen over here, a hydrogen over here, a" + }, + { + "Q": "at 6:30, why is the final velocity -5m/s? Wouldn't it be 0m/s because the rocket is not moving at all, it is sitting on the ground? I understand that negatives are used when the rocket is going down, but the final velocity would be when the velocity at the very end of the experiment, when the rocket lands and is sitting stationary on the ground.", + "A": "The equation of motion for a projectile does not include the effect of collisions, so the final velocity is the velocity the projectile has the moment it reaches a surface. After that point, the projectile EOM no longer holds since there are additional forces acting on it.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 390 + ], + "3min_transcript": "It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information" + }, + { + "Q": "at 11:41, why is the average velocity in the horizontal direction is 5 square roots of 3 metres per second? I know Sal said it is because it doesn't change, but why does it not change?", + "A": "Gravity only affects the velocity in the vertical direction, and since we are assuming that there is no air resistance, there is nothing to change the horizontal velocity.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 701 + ], + "3min_transcript": "cosine of 30 degrees is equal to the adjacent side. Is equal to the adjacent side, which is the magnitude of our horizontal component, is equal to the adjacent side over the hypotenuse. Over 10 meters per second. multiply both sides by 10 meters per second, you get the magnitude of our adjacent side, color transitioning is difficult, the magnitude of our adjacent side is equal to 10 meters per second. Is equal to 10 meters per second. Times the cosine, times the cosine of 30 degrees. And you might not remember the cosine of 30 degrees, you can use a calculator for this. Or you can just, if you do remember it, you know that it's the square root of three over two. Square root of three over two. So to figure out the actual component, I'll stop to get a calculator out if I want, well I don't have to use it, do it just yet, because I have 10 times Which is going to be 10 divided by two is five. So it's going to be five times the square root of three meters per second. So if I wanna figure out the entire horizontal displacement, so let's think about it this way, the horizontal displacement, we're trying to figure out, the horizontal displacement, a S for displacement, is going to be equal to the average velocity in the x direction, or the horizontal direction. And that's just going to be this five square root of three meters per second because it doesn't change. So it's gonna be five, I don't want to do that same color, is going to be the five square roots of 3 meters per second times the change in time, times how long it is in the air. And we figure that out! Its 1.02 seconds. Times 1.02 seconds. The seconds cancel out with seconds, and we'll get that answers in meters, and now we get our calculator out to figure it out. times 1.02. It gives us 8.83 meters, So this is going to be equal to, this is going to be equal to, this is going to be oh, sorry. this is going to be equal to 8.8, is that the number I got? 8.83, 8.83 meters. And we're done. And the next video, I'm gonna try to, I'll show you another way of solving for this delta t. To show you, really, that there's multiple ways to solve this. It's a little bit more complicated but it's also a little bit more powerful if we don't start and end at the same elevation." + }, + { + "Q": "How can you prove mathematically that the initial velocity will be the same as the final velocity in this situation? ( 5:37 ) Can I calculate this somehow from v = at or s = 1/2(at^2)?", + "A": "There are several approaches. Some have to do with potential and kinetic energy. And it s not v = at, it s (v-u) = at, the formula most appropriate to that is v^2=u^2 + 2as, when the displacement (vertical) is zero then v^2=u^2 and since you know that the object will be moving in the opposite direction (vertically) when it lands than it was when it was launched then it is necessary that v = -u.", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 337 + ], + "3min_transcript": "It is 1/2. So sin of 30 degrees, use a calculator if you don't remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it's five meters per second. Is equal to the magnitude, is equal to the magnitude of our vertical component. Let me get that in the right color. It's equal to the magnitude of our vertical component. So what does that do? What we're, this projectile, because vertical component is five meters per second, it will stay in the air the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the same vertical component. So let's think about how long it will stay in the air. Since were dealing with a situation and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information" + }, + { + "Q": "7:08 shouldn't the 10 be positive since it is actually 5-(-5) = 5+5", + "A": "At 7:08 the equation he just finished was -5 -5 = (-5) + (-5) = -10", + "video_name": "ZZ39o1rAZWY", + "timestamps": [ + 428 + ], + "3min_transcript": "and we're also finishing at the same elevation, and were assuming the air resistance is negligible, we can do a little bit of a simplification here. Although I'll do another version where we're doing the more complicated, but I guess the way that applies to more situations. We could say, we could say \"well what is our \"change in velocity here?\" So if we think about just the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we're talking, let me label all of this. So we're talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing with the vertical here. So our initial velocity, in the vertical direction, our initial velocity in the vertical direction is going to be five meters per second. Is going to be five meters per second. And we're going to use a convention, that up, that up is positive and that down is negative. And now what is going to be our final velocity? We're going to be going up and would be decelerated by gravity, And then were to start accelerating back down. And, if we assume that air resistance is negligible, when we get back to ground level, we will have the same magnitude of velocity but will be going in the opposite direction. So our final velocity, remember, we're just talking about the vertical component right now. We haven't even thought about the horizontal. We're just trying to figure out how long does this thing stay in the air? So its final velocity is going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going to be looking like that. Same magnitude, just in the opposite direction. So what's our change in velocity in the vertical direction? Change in velocity, in the vertical direction, or in the y-direction, is going to be our final velocity, negative five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information Well we know! We know that our vertical, our change our change in our, in our vertical velocity, is going to be the same thing or it's equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What's our acceleration in the vertical direction? What's the acceleration due to gravity, or acceleration that gravity, that the force of gravity has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that's gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air," + }, + { + "Q": "8:46 If the answer to acceleration or velocity is \"0\", is it necessary to include units?", + "A": "It s better if you do. Best ask your teacher if you really feel like being lazy :)", + "video_name": "d-_eqgj5-K8", + "timestamps": [ + 526 + ], + "3min_transcript": "to be the distance traveled. And lucky for us, this is just going to be a triangle, and we know how to figure out the area for triangle. So the area of a triangle is equal to one half times base times height. Which hopefully makes sense to you, because if you just multiply base times height, you get the area for the entire rectangle, and the triangle is exactly half of that. So the distance traveled in this situation, or I should say the displacement, just because we want to make sure we're focused on vectors. The displacement here is going to be-- or I should say the magnitude of the displacement, maybe, which is the same thing as the distance, is going to be one half times the base, which is five seconds, times the height, which is five meters per second. Times five meters. Five meters per second. The seconds cancel out with the seconds. And we're left with one half times five times five meters. So it's one half times 25, which is equal to 12.5 meters. And so there's an interesting thing here, well one, there's a couple of interesting things. Hopefully you'll realize that if you're plotting velocity versus time, the area under the curve, given a certain amount of time, tells you how far you have traveled. The other interesting thing is that the slope of the curve tells you your acceleration. What's the slope over here? Well, It's completely flat. And that's because the velocity isn't changing. So in this situation, we have a constant acceleration. The magnitude of that acceleration is exactly zero. Our velocity is not changing. Here we have an acceleration of one meter per second squared, and that's why the slope of this line right over here is one. The other interesting thing, is, if even if you have constant acceleration, you could still figure out the distance We were able to figure out there we were able to get 12.5 meters. The last thing I want to introduce you to-- actually, let me just do it until next video, and I'll introduce you to the idea of average velocity. Now that we feel comfortable with the idea, that the distance you traveled is the area under the velocity versus time curve." + }, + { + "Q": "At 7:18, the displacement is taken to be the area under the slope i.e., a triangle. why isn't the other side of the slope considered, tat would form a rectangle?", + "A": "If you go at a constant velocity of 10 m/s, the graph of that is a straight horizontal line at 10 m/s, right? If you go that velocity for 10 s, how far did you go? 100 m, right? You got that by doing 10 m/s x 10 s = 100 m, which is the area UNDER that horizontal line. Now, did we need to add the area above the straight line? No, why would we do that, right? It doesn t have to do with anything. The same idea holds when the line is tilted. The area above the line has no signficance. It s infinite, too.", + "video_name": "d-_eqgj5-K8", + "timestamps": [ + 438 + ], + "3min_transcript": "we have a constant slope. So we have just a line here. We don't have a curve. Now what I want to do is think about a situation. Let's say that we accelerate it one meter per second squared. And we do it for-- so the change in time is going to be five seconds. And my question to you is how far have we traveled? Which is a slightly more interesting question than what we've been asking so far. So we start off with an initial velocity of zero. And then for five seconds we accelerate it one meter per second squared. So one, two, three, four, five. So this is where we go. This is where we are. So after five seconds, we know our velocity. Our velocity is now five meters per second. But how far have we traveled? So we could think about it a little bit visually. We could say, look, we could try to draw rectangles over here. of one meter per second. So if I say one meter per second times the second, that'll give me a little bit of distance. And then the next one I have a little bit more of distance, calculated the same way. I could keep drawing these rectangles here, but then you're like, wait, those rectangles are missing, because I wasn't for the whole second, I wasn't only going one meter per second. I kept accelerating. So I actually, I should maybe split up the rectangles. I could split up the rectangles even more. So maybe I go every half second. So on this half-second I was going at this velocity. And I go that velocity for a half-second. Velocity times the time would give me the displacement. And I do it for the next half second. Same exact idea here. Gives me the displacement. So on and so forth. But I think what you see as you're getting-- is the more accurate-- the smaller the rectangles, you try to make here, the closer you're going to get to the area under this curve. And just like the situation here. to be the distance traveled. And lucky for us, this is just going to be a triangle, and we know how to figure out the area for triangle. So the area of a triangle is equal to one half times base times height. Which hopefully makes sense to you, because if you just multiply base times height, you get the area for the entire rectangle, and the triangle is exactly half of that. So the distance traveled in this situation, or I should say the displacement, just because we want to make sure we're focused on vectors. The displacement here is going to be-- or I should say the magnitude of the displacement, maybe, which is the same thing as the distance, is going to be one half times the base, which is five seconds, times the height, which is five meters per second. Times five meters." + }, + { + "Q": "At 9:58, Sal says that He fuses to C and O.Why does He not get fused into other elements like Lithium,Boron or Nitrogen etc.?", + "A": "It does, but lithium, boron, and nitrogen actually have a lower fusion temperature than helium so they fuse instantly.", + "video_name": "kJSOqlcFpJw", + "timestamps": [ + 598 + ], + "3min_transcript": "And the temperature here keeps going up. So we said that the first ignition, the first fusion, occurs at around 10 million Kelvin. This thing will keep heating up until it gets to 100 million Kelvin. And now I'm talking about a star that's about as massive as the sun. Some stars will never even be massive enough to condense the core so that its temperature reaches But let's just talk about the case in which it does. So eventually, you'll get to a point-- so we're still sitting in the red giant phase, so we're this huge star over here. We have this helium core. And that helium core keeps getting condensed and condensed and condensed. And then we have a shell of hydrogen that keeps fusing into helium around it. So this is our hydrogen shell. Hydrogen fusion is occurring in this yellow shell over here that's causing the radius of the star to get bigger and bigger, to expand. But when the temperature get sufficiently hot-- and now I think you're going to get a sense of how heavier and heavier elements form in the universe, and all of the heavy elements that you see around us, including the ones that are in you, were formed it this way from, initially, hydrogen-- when it gets hot enough at 100 million Kelvin, in this core, because of such enormous pressures, then the helium itself will start to fuse. So then we're going to have a core in here where the helium itself will start to fuse. And now we're talking about a situation. You have helium, and you had hydrogen. And all sorts of combinations will form. But in general, the helium is mainly going to fuse into carbon and oxygen. And it'll form into other things. And it becomes much more complicated. But let me just show you a periodic table. I didn't have this in the last one. I had somehow lost it. But we see hydrogen here has one proton. It actually has no neutrons. It was getting fused in the main sequence into helium, two protons, two neutrons. You need four of these to get one of those. Because this actually has an atomic mass of 4 if we're talking about helium-4. And then the helium, once we get to 100 million Kelvin, can start being fused. If you get roughly three of them-- and there's all of these other things that are coming and leaving the reactions-- you can get to a carbon. You get four of them, four of them at least as the starting raw material. You get to an oxygen. So we're starting to fuse heavier and heavier elements. So what happens here is this helium is fusing into carbon and oxygen. So you start building a carbon and oxygen core. So I'm going to leave you there. I realize I'm already past my self-imposed limit of 10 minutes. But what I want you to think about is what is likely to happen. What is likely to happen here if this star will never" + }, + { + "Q": "at 1:12 why is hydrogen fusing into helium?", + "A": "beryllium is also an element ,heavier than hydrogen or helium but lighter than oxygen or carbon", + "video_name": "kJSOqlcFpJw", + "timestamps": [ + 72 + ], + "3min_transcript": "In the last video, we had a large cloud of hydrogen atoms eventually condensing into a high pressure, high mass, I guess you could say, ball of hydrogen atoms. And when the pressure and the temperature got high enough-- and so this is what we saw the last video-- when the pressure and temperature got high enough, we were able to get the hydrogen protons, the hydrogen nucleuses close enough to each other, or hydrogen nuclei close enough to each other, for the strong force to take over and fusion to happen and release energy. And then that real energy begins to offset the actual gravitational force. So the whole star-- what's now a star-- does not collapse on itself. And once we're there, we're now in the main sequence of a star. What I want to do in this video is to take off from that starting point and think about what happens in the star next. So in the main sequence, we have the core of the star. So this is the core-- star's core. And it's releasing just a ton of energy. And that energy is what keeps the core from imploding. It's kind of the outward force to offset the gravitational force that wants to implode everything, that wants to crush everything. And so you have the core of a star, a star like the sun, and that energy then heats up all of the other gas on the outside of the core to create that really bright object that we see as a star, or in our case, in our sun's case, the sun. Now, as the hydrogen is fusing into helium, you could imagine that more and more helium is forming in the core. So I'll do the helium as green. So more, more, and more helium forms in the core. It'll especially form-- the closer you get to the center, the higher the pressures will be, In fact, the bigger the mass of the star, the more the pressure, the faster the fusion occurs. And so you have this helium building up inside of the core as this hydrogen in the core gets fused. Now what's going to happen there? Helium is a more dense atom. It's packing more mass in a smaller space. So as more and more of this hydrogen here turns into helium, what you're going to have is the core itself is going to shrink. So let me draw a smaller core here. So the core itself is going to shrink. And now it has a lot more helium in it. And let's just take it to the extreme point where it's all helium, where it's depleted. But it's much denser. That same amount of mass that was in this sphere" + }, + { + "Q": "In 9:26, How does the Hydrogen heat up? Is it hot in space? Is the star heating it up? If the star is heating it up, whats it's source?", + "A": "As gravity pulls in the atoms, they are confined to a smaller and smaller volume. When you compress a gas, it heats up.", + "video_name": "kJSOqlcFpJw", + "timestamps": [ + 566 + ], + "3min_transcript": "wavelength than this thing over here. This thing, the core, was not burning as furiously as this thing over here. But that energy was being dissipated over a smaller volume. So this has a higher surface temperature. This over here, the core is burning more-- sorry, the core is no longer burning. The core is now helium that's not burning. It's getting denser and denser as the helium packs in on itself. But the hydrogen fusion over here is occurring more intensely. It's occurring in a hotter way. But the surface here is less hot because it's just a larger surface area. So it doesn't make-- the increased heat is more than mitigated by how large the star has become. Now, this is going to keep happening. And this core is keep-- the pressures keep intensifying because more and more helium is getting And the temperature here keeps going up. So we said that the first ignition, the first fusion, occurs at around 10 million Kelvin. This thing will keep heating up until it gets to 100 million Kelvin. And now I'm talking about a star that's about as massive as the sun. Some stars will never even be massive enough to condense the core so that its temperature reaches But let's just talk about the case in which it does. So eventually, you'll get to a point-- so we're still sitting in the red giant phase, so we're this huge star over here. We have this helium core. And that helium core keeps getting condensed and condensed and condensed. And then we have a shell of hydrogen that keeps fusing into helium around it. So this is our hydrogen shell. Hydrogen fusion is occurring in this yellow shell over here that's causing the radius of the star to get bigger and bigger, to expand. But when the temperature get sufficiently hot-- and now I think you're going to get a sense of how heavier and heavier elements form in the universe, and all of the heavy elements that you see around us, including the ones that are in you, were formed it this way from, initially, hydrogen-- when it gets hot enough at 100 million Kelvin, in this core, because of such enormous pressures, then the helium itself will start to fuse. So then we're going to have a core in here where the helium itself will start to fuse. And now we're talking about a situation. You have helium, and you had hydrogen. And all sorts of combinations will form. But in general, the helium is mainly going to fuse into carbon and oxygen. And it'll form into other things. And it becomes much more complicated." + }, + { + "Q": "At 0:50, it is said that the compund breaks into individual ions, when dissolved in water. But, if this happens, they will no longer be compunds. How are they be able to retain the characterictics of the initial compund?", + "A": "When they dissolve, they become a solution of the compound. It is still the same compound, but it is now dissolved.", + "video_name": "BgTpPM9BMuU", + "timestamps": [ + 50 + ], + "3min_transcript": "- [Instructor] What we have here is a molecular equation describing the reaction of some sodium chloride dissolved in water plus some silver nitrate, also dissolved in the water. They're going to react to form sodium nitrate, still dissolved in water, plus solid silver chloride and if you were to look at each of these compounds in their crystalline or solid form before they're dissolved in water, they each look like this. But once you get dissolved in water, and that's what this aqueous form tells us, it tells us that each of these compounds are going to get dissolved in water, they're no longer going to be in that crystalline form, crystalline form. Instead, you're going to have the individual ions disassociating. So for example, in the case of sodium chloride, the sodium is going to disassociate in the water. Sodium is a positive ion, or cation, and so it's going to be attracted to the partially negative oxygen end. That's what makes it such a good solvent. Now, the chloride anions, similarly, are going to dissolve in water 'cause they're going to be attracted to the partially positive hydrogen ends of the water molecules and the same thing is gonna be true of the silver nitrate. Silver ... The silver ion, once it's disassociated, is going to be positive and the nitrate is a negative. It is an anion. Now, in order to appreciate this and write an equation that better conveys the disassociation of the ions, we could instead write the equation like this. This makes it a little bit clearer that look, the sodium and the chloride aren't going to be necessarily together anymore. The sodium is going to dissolve in the water, like we have here. The chloride is gonna dissolve in the water. The silver ions are going to dissolve in the water So this makes it a little bit clearer and similarly on this end with the sodium nitrate stays dissolved so we can write it like this with the individual ions disassociated. But the silver chloride is in solid form. You can think of it as precipitating out of the solution. This does not have a high solubility, so it's not going to get dissolved in the water and so we still have it in solid form. Now you might say, well which of these is better? Well it just depends what you are trying to go for. This form up here, which we see more typically, this is just a standard molecular equation. Molecular ... Molecular equation. It's in balanced form. We always wanna have our equations balanced. This right over here is known as a complete ionic equation. The complete's there because we've put in all of the ions and we're going to compare it to a net ionic equation" + }, + { + "Q": "At 5:31, why can't Carbon have +2 charge?", + "A": "In theory, carbon can have +2 charge, but for only like, i don t know, one millionth of a nanosecond? Carbocations are highly unstable (most of them!) and react quickly. The bigger the charge, the less stable they are.", + "video_name": "7p2qfyqiXHc", + "timestamps": [ + 331 + ], + "3min_transcript": "this carbon will have trigonal planar geometry around it, and again, that's important when you do your organic chemistry mechanics, so carbocations are extremely important to understand. Let's look at some other examples of carbocations and analyze them a little bit too. So let's start with the carbocation on the far left. The carbon with the plus one formal charge is this one, in the center here, and what is this carbon in red bonded to? Well, the carbon in red is bonded to a CH3 group up here, which we call a methyl group in organic chemistry, the carbon in red is bonded to another CH3 group here, and another CH3 group here. So the carbon in red already has three single bonds with zero loan pairs of electrons, and so the carbon in red is a plus one formal charge. Let's look at this carbocation right here, let's highlight the carbon with the plus one formal charge, it's this one, so this carbon in red is bonded to so we only have two bonds here, we only have two bonds at this point, but we know in order for that carbon in red to have a plus one formal charge, we need three bonds, like the example on the left, in the example on the left we have three bonds here to that carbon, and so where is the last bond? The last bond, of course, must be to a hydrogen, so we draw it in here like that, so the carbon in red is bonded to a hydrogen. Usually you leave off your hydrogens when you make these drawings, but it's important to understand what's actually there. Move on to the last example, this time the positive one formal charge is on this carbon in red, and that carbon in red is directly bonded to one other carbon, so that's one bond, but we know we need a total of three bonds, so the carbon in red must have two more bonds, and those two other bonds must be to hydrogen, so we draw in, there's one bond to hydrogen, and there's another bond to hydrogen, Let's do another formal charge, let's assign formal charge to another carbon. Let's put in our electrons in our bonds, so we put those in, and our goal is find the formal charge on carbon, and so the formal charge on carbon is equal to the number of valence electrons that carbon is supposed to have, which we know is four, and from that, we subtract the number of valence electrons that carbon actually has in our drawing. So we divide up these electrons here in these bonds, and this time, carbon has a loan pair of electrons on it, so how many electrons are around carbon in our drawing? This time, there's one, two, three, and then two more from this loan pair, so four and five, so four minus five gives us a formal charge of negative one, so carbon is supposed to have four valence electrons, here it has five, so it's like it's gained an extra electron, which gives it a negative one formal charge." + }, + { + "Q": "At 3:35, does he mean percentage for that one drop, or for the whole plasma part?", + "A": "Those percentages are applicable for both the drop and the entirety of the plasma.", + "video_name": "5MOn8X-tyFw", + "timestamps": [ + 215 + ], + "3min_transcript": "Let me write it out here. Centrifugation. And the machine is called a centrifuge. So it's basically going to spin really quickly, let's say, in one direction or the other. And as a result, what happens is that the blood starts separating out. And the heavy parts of blood kind of go to the tip of the tube. And the less dense part of blood actually rises towards the lid. So after you've centrifuged-- let's say you've actually gone through this process, and you centrifuge the blood. Now you have the same tube, but I'm going to show you kind of an after picture. So let's say this was before I actually spun the tube, and now I've got an after. This is my after picture. So after I spin the tube, what does it look like? Let me draw the tube. And the biggest key difference here is that instead of having one similar looking homogeneous liquid, like we had before, now it actually You've got three different layers, in fact. I'm going to draw all three layers for you. So this is the first layer. And this is the most impressive layer. The largest volume of our blood is going to be in this top layer. So remember, this is the least dense, right? It's not very dense, and that's why it stayed near the lid. And it's actually going to make up about 55% of our total volume. And we call it plasma. So if you've ever heard that word plasma, now you know what it means. So if I was to take a drop of this stuff-- let's say I took a little drop of this plasma, and I wanted to take a good hard look at what was in my drop-- 90% of plasma is going to be nothing more than water. So that's interesting, right, because the major part of blood is plasma, and the major part of plasma is water. So now you're seeing why it is that we always say, well, make sure you drink a lot of water. Make sure you're hydrated. And in fact, that's true for the rest of your body as well. But I want to stress that it's true for blood as well. So that leaves the rest, right? We've got 90%, we have to get to 100%. So what is 8% of this plasma made up of? It's protein. And let me give you some examples of this protein. So one would be, for example, albumin. And albumin, if you're not familiar with it, it's an important protein in your plasma that keeps the liquid from kind of leaking away out of the blood vessels. Another important protein, the antibody. And this, I'm sure you've heard of, but antibodies are basically involved in your immune system, making sure that you stay nice and healthy and don't get sick with infections. And another part of the protein, another type of protein, to kind of keep in mind, would be fibrinogen. And this is one important protein involved in clotting." + }, + { + "Q": "what does Sal mean by culum of water at 6:20 ...is the liquid in the test tube not mercury ?", + "A": "I think this is a mistake because at @6:50 he writes the density of mercury 13600kg/m-3", + "video_name": "i6gz9VFyYks", + "timestamps": [ + 380 + ], + "3min_transcript": "how to do all the math, and all of a sudden you go, what is specific gravity? All specific gravity is, is the ratio of how dense that substance is to water. All that means is that mercury is 13.6 times as dense as water. Hopefully, after the last video-- because I told you to-- you should have memorized the density of water. It's 1,000 kilograms per meter cubed, so the density of mercury-- let's write that down, and that's the rho, or little p, depending on how you want to do it-- is going to be equal to 13.6 times the density of water, or times Let's go back to the problem. What we want to know is how high this column of mercury is. We know that the pressure-- let's consider this point right here, which is essentially the base of this column of mercury. What we're saying is the pressure on the base of this column of mercury right here, or the pressure at this point down, has to be the same thing as the pressure up, because the mercury isn't moving-- we're in a static state. We learned several videos ago that the pressure in is equal to the pressure out on a liquid system. Essentially, I have one atmosphere pushing down here on the outside of the surface, so I must have one atmosphere pushing up here. The pressure pushing up at this point right here-- we again, and just imagine where the pressure is hitting-- is one atmosphere, so the pressure down right here must be one atmosphere. What's creating the pressure down right there? It's essentially this column of water, or it's this formula, which we learned in the last video. What we now know is that the density of the mercury, times the height of the column of water, times the acceleration of gravity on Earth-- which is where we are-- has to equal one atmosphere, because it has to offset the atmosphere that's pushing on the outside and pushing up here. The density of mercury is this: 13.6 thousand, so 13,600 kilogram meters per meter cubed. That's the density times the height-- we don't know what" + }, + { + "Q": "At around 9:37 can some one please explain how 1N is 1kgm^2/s? I thought that force=mass*acceleration leading to the unit kgm/s^2. Thank you in advance!", + "A": "A Newton is a kg*m/s^2 (mass * acceleratin) A Joule is a kg*m^2/s^2 (Force * distance)", + "video_name": "i6gz9VFyYks", + "timestamps": [ + 577 + ], + "3min_transcript": "13,600 kilograms per meter cubed times 9.8 meters per second squared. Make sure you always have the units right-- that's the hardest thing about these problems, just to know that an atmosphere is 103,000 pascals, which is also the same as newtons per meter squared. Let's just do the math, so let me type this in-- 103,000 We were dealing with newtons, so height is equal to 0.77 meters. And you should see that the units actually work, because we have a meters cubed in the denominator up here, we have a meters cubed in the denominator down here, and then we have kilogram meters per second squared here. We have newtons up here, but what's a newton? A newton is a kilogram meter squared per second, so when you divide you have kilogram meters squared per second squared, and here you have kilogram meter per second squared. When you do all the division of the units, all you're left with is meters, so we have 0.77 meters, or roughly 77 centimeters-- is how high this column of mercury is. And you can make a barometer out of it-- you can say, let me make a little notch on this test tube, and that represents one atmosphere. different parts of the globe are. Anyway, I've run out of time. See you in the next video." + }, + { + "Q": "At 2:43, if there is vacuum in the tube, shouldn't mercury climb up the whole tube to compensate for the difference in pressure?", + "A": "it is usual to think that, if there is a vacuum then things will naturally get sucked into that space. But actually, that only happens if there is enough pressure pushing it. The atmospheric pressure is wha pushes it and that is only big enough to get the mercury so far. The weight of mercury is also pushing against the atmospheric pressure hope that makes sense :)", + "video_name": "i6gz9VFyYks", + "timestamps": [ + 163 + ], + "3min_transcript": "We are actually on Earth, or actually in Paris, France, at sea level, because that's what an atmosphere is defined as-- the atmospheric pressure. Essentially, the way you could think about it-- the weight of all of the air above us is pushing down on the surface of this bowl at one atmosphere. An atmosphere is just the pressure of all of the air above you at sea level in Paris, France. And in the bowl, I have mercury. is actually going to go up this column a little bit. We're going to do the math as far as-- one, we'll see why it's going up, and then we'll do the math to figure out how high up does it go. Say the mercury goes up some distance-- this is all still mercury. And this is actually how a barometer works; this is something that measures pressure. Over here at this part, above the mercury, but still within our little test tube, we have a vacuum-- there is no air. Vacuum is one of my favorite words, because it has two u's in a row. We have this set up, and so my question to you is-- how high is this column of mercury going to go? thing is going up to begin with. We have all this pressure from all of the air above us-- I know it's a little un-intuitive for us, because we're used to all of that pressure on our shoulders all of the time, so we don't really imagine it, but there is literally the weight of the atmosphere above us. That's going to be pushing down on the surface of the mercury on the outside of the test tube. Since there's no pressure here, the mercury is going to go upwards here. This state that I've drawn is a static state-- we have assumed that all the motion has stopped. So let's try to solve this problem. Oh, and there are a couple of things we have to know before we do this problem. It's mercury, and we know the specific gravity-- I'm using terminology, because a lot of these problems, the hardest part is the terminology-- of mercury is 13.6." + }, + { + "Q": "At 2:16, Sal calls the water molecule a tetrahedron. What exactly is a tetrahedron?", + "A": "if i remember corectly, a tetrahedron has 6 edges and 4 vertices", + "video_name": "6G1evL7ELwE", + "timestamps": [ + 136 + ], + "3min_transcript": "- [Voiceover] I don't think it's any secret to anyone that water is essential to life. Most of the biological, or actually in fact all of the significant biological processes in your body are dependent on water and are probably occurring inside of water. When you think of cells in your body, the cytoplasm inside of your cells, that is mainly water. In fact, me, who is talking to you right now, I am 60% to 70% water. You could think of me as kind of this big bag of water making a video right now. And it's not just human beings that need water. Life as we know it is dependent on water. That why when we have the search for signs of life on other planets we're always looking for signs of water. Maybe life can occur in other types of substances, but water is essential to life as we know it. And to understand why water is so special let's start to understand the structure of water and how it interacts with itself. And so water, as you probably already know, is made up of one oxygen atom and two hydrogen atoms. And they are bonded with covalent bonds. And covalent bonds, each of these bonds is this pair of electrons that both of these atoms get to pretend like they have. And so you have these two pairs. And you might be saying, \"Well, why did I draw \"the two hydrogens on this end? \"Why didn't I draw them on opposite sides of the oxygen?\" Well that's because oxygen also has two lone electron pairs. Two lone electron pairs. And these things are always repelling each other. The electrons are repelling from each other, and so, in reality if we were looking at it in three dimensions, the oxygen molecule is kind of a tetrahedral shape. I could try to, let me try to draw it a little bit. So if this is the oxygen right over here then you would have, you could have maybe one lone pair of electrons. I'll draw it as a little green circle there. Another lone pair of electrons back here. Then you have the covalent bond. You have the covalent bond to And then you have the covalent bond to the other hydrogen atom. And so you see it forms this tetrahedral shape, It's pretty close to a tetrahedron. Just like this, but the key is that the hydrogens are on one end of the molecule. And this is, we're going to see, very very important to the unique properties, or to the, what gives water its special properties. Now, one thing to realize is, it's very, in chemistry we draw these electrons very neatly, these dots up here. We draw these covalent bonds very neatly. But that's not the way that it actually works. Electrons are jumping around constantly. They're buzzing around, it's actually much more of a, even when you think about electrons, it's more of a probability of where you might find them. And so instead of thinking of these electrons as definitely here or definitely in these bonds, They're actually more of in this cloud around the different atoms. They're in this cloud that kind of describes a probability" + }, + { + "Q": "at 2:58, why is their a bubble around... i din't realy understand.", + "A": "The bubble is supposed to illustrate that the electrons in a water molecule (and any other molecule) aren t actually stationary in one place (like we usually draw them, with the dots and lines), but are constantly buzzing around.", + "video_name": "6G1evL7ELwE", + "timestamps": [ + 178 + ], + "3min_transcript": "And they are bonded with covalent bonds. And covalent bonds, each of these bonds is this pair of electrons that both of these atoms get to pretend like they have. And so you have these two pairs. And you might be saying, \"Well, why did I draw \"the two hydrogens on this end? \"Why didn't I draw them on opposite sides of the oxygen?\" Well that's because oxygen also has two lone electron pairs. Two lone electron pairs. And these things are always repelling each other. The electrons are repelling from each other, and so, in reality if we were looking at it in three dimensions, the oxygen molecule is kind of a tetrahedral shape. I could try to, let me try to draw it a little bit. So if this is the oxygen right over here then you would have, you could have maybe one lone pair of electrons. I'll draw it as a little green circle there. Another lone pair of electrons back here. Then you have the covalent bond. You have the covalent bond to And then you have the covalent bond to the other hydrogen atom. And so you see it forms this tetrahedral shape, It's pretty close to a tetrahedron. Just like this, but the key is that the hydrogens are on one end of the molecule. And this is, we're going to see, very very important to the unique properties, or to the, what gives water its special properties. Now, one thing to realize is, it's very, in chemistry we draw these electrons very neatly, these dots up here. We draw these covalent bonds very neatly. But that's not the way that it actually works. Electrons are jumping around constantly. They're buzzing around, it's actually much more of a, even when you think about electrons, it's more of a probability of where you might find them. And so instead of thinking of these electrons as definitely here or definitely in these bonds, They're actually more of in this cloud around the different atoms. They're in this cloud that kind of describes a probability and they jump around. And what's interesting about water is oxygen is extremely electronegative. So oxygen, that's oxygen and that's oxygen, it is extremely electronegative, it's one of the more electronegative elements we know of. It's definitely way more electronegative than hydrogen. And you might be saying, \"Well, Sal, \"what does it mean to be electronegative?\" Well, electronegative is just a fancy way of saying that it hogs electrons. It likes to keep electrons for itself. Hogs electrons, so that's what's going on. Oxygen like to keep the electrons more around itself than the partners that it's bonding with. So even in these covalent bonds, you say, \"Hey, we're supposed to be sharing these electrons.\" Oxygen says, \"Well I still want them to \"spend a little bit more time with me.\" And so they actually do spend more time on the side without the hydrogens than they do around the hydrogens." + }, + { + "Q": "@2:50 why did he split the cube/pentacyclo ring the way that he did?", + "A": "Because of the symmetry of the molecule, there are many ways in which he could have done it. The point is that he had to draw the largest ring possible (8 C atoms) that contained two bridgehead carbons. The particular one he chose was probably the easiest to draw.", + "video_name": "ayKHmN90ncc", + "timestamps": [ + 170 + ], + "3min_transcript": "on the left, I'm going to start cutting bonds. And let's see how many cuts it takes to get to an open chain alkane. For example, I could start by cutting right here. So we'll say that's our first cut. And then our second cut, we could make a cut right back here like that. So we could make that my second cut here on my cubane. And then for my third cut, I'm going to go for this one right up here. So we'll take care of that one. So that's three cuts so far. And then, if I just go ahead and take out of this one, this bond right here, and then this bond right here, so that's cuts four and five. I now get an open chain alkane. So it took five cuts for us to do that. So there are five rings in cubane, so it's pentacyclo. So that's just not immediately obvious, to me anyway, as to why there are five rings in cubane. to start the IUPAC name here, so pentacyclo, meaning five rings. And then we start our brackets, just like we did in the video on bicyclic nomenclature. And to finish naming cubane, we're going to pretend like it is a bicyclic compound. And the first thing we do is identify our bridgehead carbons. So the carbons that are common to both of the two rings here. So hopefully it's obvious those are two rings and those are the bridgehead carbons that connect those two rings. When you number a bicyclic compound, you start at one of the bridgehead carbons and then you go the longest path first. So I'm going to start at this carbon, and I'm going to go the longest path, which would be up here. So this would be number 2, this would be number 3, carbon number 4, carbon number 5, carbon number 6, which takes me to the other bridgehead carbon. And then you name your next longest path. and then make this one back here carbon 8. So those are my eight carbons of cubane. And so, once again, I can continue to pretend like it's a bicyclic molecule. And the next thing I would do is I would name the number of carbons in my longest path. So the number of carbons in my longest path would be this one, so there'd be 1, 2, 3, 4 carbons. Remember, you exclude the bridgehead carbons when you're doing this, so we're going to start with a 4 right here, like that. Next, you do the number of carbons in your second longest path. So we can see my second longest path would be this one right here. And there are two carbons in my second longest past. So I go ahead and put a 2 over here like that. And then finally, it's the number of carbons between the bridgehead carbons, which in this example, of course, there are no carbons between my two bridgehead carbons. So I would put a 0 here like that. But of course, cubane is not a bicyclic compound," + }, + { + "Q": "At 0:06 Sal refers to the neuron as a cell. However, the neuron has bits that are cells, like the Schwanne cells. Doesn't this make the neuron a tissue? Or is it a cell with add-ons?", + "A": "It s more like a cell with add-ons. The Schwann cells are helper cells/glial cells that specialize in helping a certain type of cell. These cells are just as important as any other cell except not so well known.", + "video_name": "ob5U8zPbAX4", + "timestamps": [ + 6 + ], + "3min_transcript": "We could have a debate about what the most interesting cell in the human body is, but I think easily the neuron would make the top five, and it's not just because the cell itself is interesting. The fact that it essentially makes up our brain and our nervous system and is responsible for the thoughts and our feelings and maybe for all of our sentience, I think, would easily make it the top one or two cells. So what I want to do is first to show you what a neuron looks like. And, of course, this is kind of the perfect example. This isn't what all neurons look like. And then we're going to talk a little bit about how it performs its function, which is essentially communication, essentially transmitting signals across its length, depending on the signals it receives. So if I were to draw a neuron-- let me pick a better color. So let's say I have a neuron. It looks something like this. So in the middle you have your soma and then from the soma-- let me draw the nucleus. This is a nucleus, just like any cell's nucleus. then the neuron has these little things sticking out from it that keep branching off. Maybe they look something like this. I don't want to spend too much time just drawing the neuron, but you've probably seen drawings like this before. And these branches off of the soma of the neuron, off of its body, these are called dendrites. They can keep splitting off like that. I want to do a fairly reasonable drawing so I'll spend a little time doing that. So these right here, these are dendrites. And these tend to be-- and nothing is always the case in biology. Sometimes different parts of different cells perform other functions, but these tend to be where the neuron receives its signal. And we'll talk more about what it means to receive and probably in the next few. So this is where it receives the signal. So this is the dendrite. This right here is the soma. Soma means body. This is the body of the neuron. And then we have kind of a-- you can almost view it as a tail of the neuron. It's called the axon. A neuron can be a reasonably normal sized cell, although there is a huge range, but the axons can be quite long. They could be short. Sometimes in the brain you might have very small axons, but you might have axons that go down the spinal column or that go along one of your limbs-- or if you're talking about one of a dinosaur's limbs. So the axon can actually stretch several feet. Not all neurons' axons are several feet, but they could be. And this is really where a lot of the distance of the signal gets traveled. Let me draw the axon. So the axon will look something like this." + }, + { + "Q": "What is the myelin sheath and what does it do? Mentioned at 4:42.", + "A": "The myelin sheath helps keep the electrical signals insulated and protected; it is the same concept as rubber on a power cord.", + "video_name": "ob5U8zPbAX4", + "timestamps": [ + 282 + ], + "3min_transcript": "connect to other dendrites or maybe to other types of tissue or muscle if the point of this neuron is to tell a muscle to do something. So at the end of the axon, you have the axon terminal right there. I'll do my best to draw it like that. So this is the axon. This is the axon terminal. And you'll sometimes hear the word-- the point at which the soma or the body of the neuron connects to the axon is as often referred to as the axon hillock-- maybe you can kind of view it as kind of a lump. It starts to form the axon. And then we're going to talk about how the impulses travel. And a huge part in what allows them to travel efficiently are these insulating cells around the axon. actually work, but it's good just to have the anatomical structure first. So these are called Schwann cells and they're covering-- they make up the myelin sheath. So this covering, this insulation, at different intervals around the axon, this is called the myelin sheath. So Schwann cells make up the myelin sheath. I'll do one more just like that. And then these little spaces between the myelin sheath-- just so we have all of the terminology from-- so we know the entire anatomy of the neuron-- these are called the nodes of Ranvier. I guess they're named after Ranvier. Maybe he was the guy who looked and saw they had these little slots here where you don't have myelin sheath. So the general idea, as I mentioned, is that you get a signal here. We're going to talk more about what the signal means-- and then that signal gets-- actually, the signals can be summed, so you might have one little signal right there, another signal right there, and then you'll have maybe a larger signal there and there-- and that the combined effects of these signals get summed up and they travel to the hillock and if they're a large enough, they're going to trigger an action potential on the axon, which will cause a signal to travel down the balance of the axon and then over here it might be connected via synapses to other dendrites or muscles. And we'll talk more about synapses and those might help trigger other things. So you're saying, what's triggering these things here? Well, this could be the terminal end of other neurons' axons, like in the brain. This could be some type of sensory neuron. This could be on a taste bud someplace, so a salt molecule" + }, + { + "Q": "At 5:25, it is mentioned that the effects of the impulses are combined. Does that mean that the multiple impulses are transformed into one or does it mean that the effects of each impulse are measured which will then decide whether or not access is granted to the axon. Or, are you saying the impulse effects are measured and then another signal is created and then travels down the axon?", + "A": "Thanks, you really explained it well :)", + "video_name": "ob5U8zPbAX4", + "timestamps": [ + 325 + ], + "3min_transcript": "connect to other dendrites or maybe to other types of tissue or muscle if the point of this neuron is to tell a muscle to do something. So at the end of the axon, you have the axon terminal right there. I'll do my best to draw it like that. So this is the axon. This is the axon terminal. And you'll sometimes hear the word-- the point at which the soma or the body of the neuron connects to the axon is as often referred to as the axon hillock-- maybe you can kind of view it as kind of a lump. It starts to form the axon. And then we're going to talk about how the impulses travel. And a huge part in what allows them to travel efficiently are these insulating cells around the axon. actually work, but it's good just to have the anatomical structure first. So these are called Schwann cells and they're covering-- they make up the myelin sheath. So this covering, this insulation, at different intervals around the axon, this is called the myelin sheath. So Schwann cells make up the myelin sheath. I'll do one more just like that. And then these little spaces between the myelin sheath-- just so we have all of the terminology from-- so we know the entire anatomy of the neuron-- these are called the nodes of Ranvier. I guess they're named after Ranvier. Maybe he was the guy who looked and saw they had these little slots here where you don't have myelin sheath. So the general idea, as I mentioned, is that you get a signal here. We're going to talk more about what the signal means-- and then that signal gets-- actually, the signals can be summed, so you might have one little signal right there, another signal right there, and then you'll have maybe a larger signal there and there-- and that the combined effects of these signals get summed up and they travel to the hillock and if they're a large enough, they're going to trigger an action potential on the axon, which will cause a signal to travel down the balance of the axon and then over here it might be connected via synapses to other dendrites or muscles. And we'll talk more about synapses and those might help trigger other things. So you're saying, what's triggering these things here? Well, this could be the terminal end of other neurons' axons, like in the brain. This could be some type of sensory neuron. This could be on a taste bud someplace, so a salt molecule" + }, + { + "Q": "At 7:40, why is it quasi-static? What does quasi mean?", + "A": "that means almost such that you get almost static", + "video_name": "lKq-10ysDb4", + "timestamps": [ + 460 + ], + "3min_transcript": "small amount of time. And so you wouldn't have thrown that system into this, you know, havoc that I did this last time. Of course, we haven't moved all the way here yet. But what we have done is, we would have moved from that point maybe to this other point right here that's just a little bit closer to there. I've just removed a little bit of the weight. So my pressure went down just a little bit. And my volume went up a just a little bit. Temperature probably went down. And the key here is I'm trying to do it in such small increments that as I do it, my system is pretty much super close to equilibrium. I'm just doing it just slow enough that at every step it achieves equilibrium almost immediately. Or it's almost in equilibrium the whole time I'm doing it. And then I do it again, and do it again. And I'll just draw my drawings a little less neat, just for the sake of time. infinitely small mass. And now my little piston will move just a little bit higher. And I have, remember I have one less sand up here than I had over here. And then my volume in my gas increases a little bit. My pressure goes down a little bit. And I've moved to this point here. What I'm doing here is I'm setting up what's called a quasi-static process. And the reason why it's called that is because it's almost static. It's almost in equilibrium the whole time. Every time I move a grain of sand I'm just moving a little bit closer. And obviously even a grain of sand, the reality is if I were to do this in real life, even a grain of sand on a small scale is going to reek a little bit of havoc on my system. This piston is going to go up a little bit. So say, let me just do even a smaller grain of sand, and do equilibrium. So you can imagine this is kind of a theoretical thing. If I did an infinitely small grains of sand, and did it just slow enough so that it's just gently moved from this point to this point. But we like to think of it theoretically, because it allows us to describe a path. Because remember, why am I being so careful here? Why am I so careful to make sure that the state, the system is in equilibrium the whole time when I get from there to there? Because our macrostates, our macro variables like pressure, volume, and temperature, our only defined when we're in equilibrium. So if I do this process super slowly, in super small increments, it allows me to keep my pressure and volume and actually my temperature of macrostates at any point in time. So I could actually plot a path. So if I keep doing it small, small, small, I could actually plot a path to say, how did I get from state 1 to state 2 on this on this PV diagram." + }, + { + "Q": "At 1:20, how come the force due to gravity is mg? In free fall particle videos earlier, I remember it being just g (-9.81 m/s^2)?", + "A": "You remember wrong. g is acceleration due to gravity, not force due to gravity. (It is also force per unit of mass.) To find the force of gravity on an object you have to multiply m by g.", + "video_name": "TC23wD34C7k", + "timestamps": [ + 80 + ], + "3min_transcript": "Let's say I have some type of a block here. And let's say this block has a mass of m. So the mass of this block is equal to m. And it's sitting on this-- you could view this is an inclined plane, or a ramp, or some type of wedge. And we want to think about what might happen to this block. And we'll start thinking about the different forces that might keep it in place or not keep it in place and all of the rest. So the one thing we do know is if this whole set up is near the surface of the Earth-- and we'll assume that it is for the sake of this video-- that there will be the force of gravity trying to bring or attract this mass towards the center of the Earth, and vice versa, the center of the earth towards this mass. So we're going to have some force of gravity. Let me start right at the center of this mass right over here. And so you're going to have the force of gravity. The force due to gravity is going to be equal to the gravitational field And so we'll call that g. We'll call that g times the mass. Let me just write it. The mass times the gravitational field near the surface of the Earth. And it's going to be downwards, we know that, or at least towards the surface of the Earth. Now, what else is going to be happening here? Well, it gets a little bit confusing, because you can't really say that normal force is acting directly against this force right over here. Because remember, the normal force acts perpendicular to a surface. So over here, the surface is not perpendicular to the force So we have to think about it a little bit differently than we do if this was sitting on level ground. Well, the one thing we can do, and frankly, that we should do, is maybe we can break up this force, the force due to gravity. We can break it up into components that are either perpendicular to the surface or that are parallel to the surface. what's likely to happen. What are potentially the netting forces, or balancing forces, over here? So let's see if we can do that. Let's see if we can break this force vector, the force due to gravity, into a component that is perpendicular to the surface of this ramp. And also another component that is parallel to the surface of this ramp. Let me do that in a different color. That is parallel to the surface of this ramp. And this is a little bit unconventional notation, but I'll call this one over here the force due to gravity that is perpendicular to the ramp. That little upside down t, I'm saying that's perpendicular. Because it shows a line that's perpendicular to, I guess, this bottom line, this horizontal line over there. And this blue thing over here, I'm going to call this the part of force due to gravity that is parallel." + }, + { + "Q": "at 11:59 He said that when \u00ce\u00b8=0 all forces will act perpendicularly to the plane. Then does that not mean that \u00ce\u00b8=90? Furthermore if \u00ce\u00b8=0, then how do you explain the magnitude of the normal force since mg(cos\u00ce\u00b8) = 0 [cos\u00ce\u00b8=0, mg(0)=0], the normal force should be the same as the weight of the object when \u00ce\u00b8 is zero, could someone please clear my doubts. Thank you", + "A": "Theta is the angle between the plane and the ground. When that angle is 0, the force of the weight is perpendicular to the plane. Look at the drawing.", + "video_name": "TC23wD34C7k", + "timestamps": [ + 719 + ], + "3min_transcript": "over the magnitude of the force due to gravity-- which is the magnitude of mg-- that is going to be equal to what? This is the opposite side to the angle. So the blue stuff is the opposite side, or at least its length, is the opposite side of the angle. And then right over here this magnitude of mg, that is the hypotenuse. So you have the opposite over the hypotenuse. Opposite over hypotenuse. Sine of an angle is opposite over hypotenuse. So this is going to be equal to the sine of theta. This is equal to the sine of theta. Or you multiply both sides times the magnitude of the force due to gravity and you get the component of the force due to gravity that is parallel to the ramp is going to be the force due to gravity total times Times sine of theta. And hopefully you should see where this came from. Because if you ever have to derive this again 30 years after you took a physics class, you should be able to do it. But if you know this right here, and this right here, we can all of a sudden start breaking down the forces into things that are useful to us. Because we could say, hey, look, this isn't moving down into this plane. So maybe there's some normal force that's completely netting it out in this example. And maybe if there's nothing to keep it up, and there's no friction, maybe this thing will start accelerating due to the parallel force. And we'll think a lot more about that. And if you ever forget these, think about them intuitively. You don't have to go through this whole parallel line and transversal and all of that. If this angle went down to 0, then we'll be talking about essentially a flat surface. And if this angle goes down to 0, then all of the force should be acting perpendicular to the surface of the plane. So if this going to 0, if the perpendicular force should be the same thing as the total gravitational force. And that's why it's cosine of theta. Because cosine of 0 right now is 1. And so these would equal each other. And if this is equal to 0, then the parallel component of gravity should go to 0. Because gravity will only be acting downwards, and once again, if sine of theta is 0. So the force of gravity that is parallel will go to 0. So if you ever forget, just do that little intuitive thought process and you'll remember which one is sine and which one is cosine." + }, + { + "Q": "Exactly...At 2:37. Mr. Khan literates that the notation that he will be using is unconventional. Then what is a conventional notation?", + "A": "conventional notation is scientific notation", + "video_name": "TC23wD34C7k", + "timestamps": [ + 157 + ], + "3min_transcript": "And so we'll call that g. We'll call that g times the mass. Let me just write it. The mass times the gravitational field near the surface of the Earth. And it's going to be downwards, we know that, or at least towards the surface of the Earth. Now, what else is going to be happening here? Well, it gets a little bit confusing, because you can't really say that normal force is acting directly against this force right over here. Because remember, the normal force acts perpendicular to a surface. So over here, the surface is not perpendicular to the force So we have to think about it a little bit differently than we do if this was sitting on level ground. Well, the one thing we can do, and frankly, that we should do, is maybe we can break up this force, the force due to gravity. We can break it up into components that are either perpendicular to the surface or that are parallel to the surface. what's likely to happen. What are potentially the netting forces, or balancing forces, over here? So let's see if we can do that. Let's see if we can break this force vector, the force due to gravity, into a component that is perpendicular to the surface of this ramp. And also another component that is parallel to the surface of this ramp. Let me do that in a different color. That is parallel to the surface of this ramp. And this is a little bit unconventional notation, but I'll call this one over here the force due to gravity that is perpendicular to the ramp. That little upside down t, I'm saying that's perpendicular. Because it shows a line that's perpendicular to, I guess, this bottom line, this horizontal line over there. And this blue thing over here, I'm going to call this the part of force due to gravity that is parallel. to show something that is parallel to the surface. So this is the component of force due to gravity that's perpendicular, component of force that is parallel. So let's see if we can use a little bit a geometry and trigonometry, given that this wedge is at a theta degree incline relative to the horizontal. If you were to measure this angle right over here, you would get theta. So in future videos we'll make it more concrete, like 30 degrees or 45 degrees or whatever. But let's just keep in general. If this is theta, let's figure out what these components of the gravitational force are going to be. Well, we can break out our geometry over here. This, I'm assuming is a right angle. And so if this is a right angle, we know that the sum of the angles in a triangle add up to 180. So if this angle, and this 90 degrees-- right angle says 90 degrees-- add up to 180, then that" + }, + { + "Q": "At 2:36, I tried searching up the conventional notation for vector magnitudes but came away a little confused, so does anyone know what it is?", + "A": "Unsure what you mean exactly by conventional notation, since mathematicians and physicists can t agree, but the most COMMON notation is as follows: Math: Vector Q is located at <3,4,5> Meaning 3 units in the x-direction, 4 units in the y-direction, and 5 units in the z-direction. Physics: Vector Q is located at 3i+4j+5k, where i is the x-component of Vector Q (Qsubx), j is the y-component (Qsuby), and k is the z component (Qsubz). That s how it s usually written.", + "video_name": "TC23wD34C7k", + "timestamps": [ + 156 + ], + "3min_transcript": "And so we'll call that g. We'll call that g times the mass. Let me just write it. The mass times the gravitational field near the surface of the Earth. And it's going to be downwards, we know that, or at least towards the surface of the Earth. Now, what else is going to be happening here? Well, it gets a little bit confusing, because you can't really say that normal force is acting directly against this force right over here. Because remember, the normal force acts perpendicular to a surface. So over here, the surface is not perpendicular to the force So we have to think about it a little bit differently than we do if this was sitting on level ground. Well, the one thing we can do, and frankly, that we should do, is maybe we can break up this force, the force due to gravity. We can break it up into components that are either perpendicular to the surface or that are parallel to the surface. what's likely to happen. What are potentially the netting forces, or balancing forces, over here? So let's see if we can do that. Let's see if we can break this force vector, the force due to gravity, into a component that is perpendicular to the surface of this ramp. And also another component that is parallel to the surface of this ramp. Let me do that in a different color. That is parallel to the surface of this ramp. And this is a little bit unconventional notation, but I'll call this one over here the force due to gravity that is perpendicular to the ramp. That little upside down t, I'm saying that's perpendicular. Because it shows a line that's perpendicular to, I guess, this bottom line, this horizontal line over there. And this blue thing over here, I'm going to call this the part of force due to gravity that is parallel. to show something that is parallel to the surface. So this is the component of force due to gravity that's perpendicular, component of force that is parallel. So let's see if we can use a little bit a geometry and trigonometry, given that this wedge is at a theta degree incline relative to the horizontal. If you were to measure this angle right over here, you would get theta. So in future videos we'll make it more concrete, like 30 degrees or 45 degrees or whatever. But let's just keep in general. If this is theta, let's figure out what these components of the gravitational force are going to be. Well, we can break out our geometry over here. This, I'm assuming is a right angle. And so if this is a right angle, we know that the sum of the angles in a triangle add up to 180. So if this angle, and this 90 degrees-- right angle says 90 degrees-- add up to 180, then that" + }, + { + "Q": "at 7:30, how can we write V+ as Vin.. isn't Vin = V+ - V-??", + "A": "Hello Karthikeyan, In this case the non-inverting input terminal (V+) is connected directly to the voltage source (Vin). You are also correct that the op-amp amplifies the difference between the input terminals: Vout = Gain * (V+ - V-) Yes these circuits can be tricky. Different things have the same names. You will need to look at the context to figure out the meaning. Regards, APD", + "video_name": "_Ut-nQ535iE", + "timestamps": [ + 450 + ], + "3min_transcript": "Now what else do I know? Let's look at this resistor chain here. This resistor chain actually looks a lot like a voltage divider, and it's actually a very good voltage divider. Remember we said this current here, what is this current here? It's zero. I can use the voltage divider expression that I know. In that case, I know that V minus, this is the voltage divider equation, equals V out Times the bottom resistor remember this? R2 over R1 plus R2, so the voltage divider expression says that when you have a stack of resistors like this, with the voltage on the top and ground on the bottom, Kay, so what I'm going to do next is I'm going to take this expression and stuff it right in there. Let's do that. See if we got enough room, okay now let's go over here. Now I can say that V out equals A times V plus minus V out times R2 over R1 plus R2, alright so far so good. Let's keep going, let's keep working on this. V not equals A times V plus minus A V not, R2 over R1 plus R2. terms over on the left hand side. Let's try that. So that gives me, V not plus A V not, times R2 over R1 plus R2 and that equals A times V plus, and actually I can change that now V plus is what? V plus is V in. Okay let's keep going I can factor out the V not. V not is one plus A R2 over R1 plus R2 and that equals AV in." + }, + { + "Q": "at 10:54, you said ''because x is <<1\", for me to get that I had to calculate it first, so how did you know that it is much smaller than 1 without any calculations?Since when x<<1 we make assumption, that means subtracting/adding it does not make any significant difference, so from that I also wanna know if I would be wrong if I DO NOT make assumptions instead I just work it out the way it is.", + "A": "You can confidently ignore the value of x when the Kb value is of order 10^(-5) or smaller. The reason is that small Ka would mean small dissociation. Also, if you do not ignore x in the equation, you will arrive at almost the same answer (there may be a difference but it is insignificant). But, you may notice that if you do not ignore x, you will arrive at a quadratic equation which increases our effort to solve for the answer(and its going to get you the same answer). Hope this helps!", + "video_name": "223KLPnJCBI", + "timestamps": [ + 654 + ], + "3min_transcript": "And we're saying that we have zero for our initial concentration of hydroxide. So when the reaction comes to equilibrium here, for ammonia we would have 0.15 minus x. For ammonium, we have two sources right? So this is a common ion here. So we have 0.35 plus x. And then for hydroxide we would have just x. So since ammonia is acting as a weak base here, let's go ahead and write our equilibrium expression. And we would write Kb. And Kb for ammonia is 1.8 times 10 to the negative five. So this is equal to concentration of our products over reactants. So we have concentration of NH4+ times the concentration of OH- all over the concentration of our reactants, leaving out water. So we just have ammonia here, So let's go ahead and plug in what we have. For the concentration of ammonium, we have 0.35 plus x. So we put 0.35 plus x. For the concentration of hydroxide, we have x. So we go ahead and put an x in here. And then that's all over the concentration of ammonia at equilibrium, and we go over here, and for ammonia at equilibrium it's 0.15 minus x. So we write over here 0.15 minus x. And we can plug in the Kb value, 1.8 times 10 to the negative five. So let's go ahead and plug in the Kb. So we have 1.8 times 10 to the negative five is equal to, okay once again, we're going to make the assumption. So if we say that x is extremely small number, then we don't have to worry about it And so we just say this is equal to 0.35. So 0.35 plus x is pretty close to 0.35. So this is times x. And make sure you understand this x is this x. And then once again, 0.15 minus x, if x is a very small number, that's approximately equal to 0.15. So now, we would have this. And we need to solve for x. So let's go ahead and do that. Let's get out the calculator here. We need to solve for x. So 1.8 times 10 to the negative five times 0.15, and then we need to divide that by 0.35. And that gives us what x is equal to. And so x is equal to 7.7 times 10 to the negative six. So x is equal to 7.7 times 10 to the negative six. x represents the concentration" + }, + { + "Q": "Forgive my ignorance, but when he speaks of a step-up transformer @ 1:35, what exactly happens to the voltage? I assume it isn't making free energy...?", + "A": "True, Though the voltage is increased in step up transformer, the current is decreased to make Power= Voltage*Current consant.", + "video_name": "xuQcB-oo-4U", + "timestamps": [ + 95 + ], + "3min_transcript": "- [Karl] Today, we're gonna take apart an alarm clock radio and we're gonna see what's inside it and how it works. There's basically four systems that we're going to evaluate. There's the power system. There's the alarm clock or the clock itself. And the structure of the device in the interface. Then we're also gonna take a look at the radio. The first thing is let's take a look at the power system. I've already cut apart this plug here. You can see the prongs. That's where the power comes from. We've got the two wires here. The two wires connect to what's called the transformer. In the transformer, there are three key components. We've got a primary coil, a secondary coil, and an iron core. The primary coil is wound around a certain number of times. The secondary coil is wound around fewer times in this particular transformer. That means it's a step-down transformer. What this transformer does is it converts AC down to nine volt AC. Because the components in the alarm clock need lower voltage. It steps the power down. The way it does it is this coil induces a current flow in this coil, and the iron core helps that to happen. Because this coil has fewer turns, it's a step-down transformer which means that the voltage is less coming out of this part. If there were more turns, it would be a step-up transformer. The iron core, again, facilitates that process. It's called electromagnetic induction. This coil induces the current flow in this coil. Anyway, the power will travel through the cable here, the wire, and it comes to the alarm clock. Let's take a look at the housing, first of all. It's fairly low cost housing. It's made out of injection-molded plastic. Let's see where that power comes in and where it goes. is they basically just only used one fastener, one separate fastener. This is a screw. The more screws you have in things, the easier they are oftentimes to put together and take apart, but they are also more expensive. Every screw requires either a robot or a person to assemble it and it's an expensive cost adder. The more screws you can take out, the more cost you can reduce. All these fasteners in this are actually molded into the body panels. There's a pin or a tab there, so we can pull this top part off. This part right here is the front plate or the front vessel. It's made out of a tinted acrylic and it's injection-molded. There's two parts of the mold that come together and molded plastic is injected, and then this comes out. The reason it's injection-molded is that it creates very precise part. You can get a nice, clean finish." + }, + { + "Q": "At about 1:00 the transformer is for what?", + "A": "the transformer lowers the voltage so that the alarm clock radio doesn t over heat", + "video_name": "xuQcB-oo-4U", + "timestamps": [ + 60 + ], + "3min_transcript": "- [Karl] Today, we're gonna take apart an alarm clock radio and we're gonna see what's inside it and how it works. There's basically four systems that we're going to evaluate. There's the power system. There's the alarm clock or the clock itself. And the structure of the device in the interface. Then we're also gonna take a look at the radio. The first thing is let's take a look at the power system. I've already cut apart this plug here. You can see the prongs. That's where the power comes from. We've got the two wires here. The two wires connect to what's called the transformer. In the transformer, there are three key components. We've got a primary coil, a secondary coil, and an iron core. The primary coil is wound around a certain number of times. The secondary coil is wound around fewer times in this particular transformer. That means it's a step-down transformer. What this transformer does is it converts AC down to nine volt AC. Because the components in the alarm clock need lower voltage. It steps the power down. The way it does it is this coil induces a current flow in this coil, and the iron core helps that to happen. Because this coil has fewer turns, it's a step-down transformer which means that the voltage is less coming out of this part. If there were more turns, it would be a step-up transformer. The iron core, again, facilitates that process. It's called electromagnetic induction. This coil induces the current flow in this coil. Anyway, the power will travel through the cable here, the wire, and it comes to the alarm clock. Let's take a look at the housing, first of all. It's fairly low cost housing. It's made out of injection-molded plastic. Let's see where that power comes in and where it goes. is they basically just only used one fastener, one separate fastener. This is a screw. The more screws you have in things, the easier they are oftentimes to put together and take apart, but they are also more expensive. Every screw requires either a robot or a person to assemble it and it's an expensive cost adder. The more screws you can take out, the more cost you can reduce. All these fasteners in this are actually molded into the body panels. There's a pin or a tab there, so we can pull this top part off. This part right here is the front plate or the front vessel. It's made out of a tinted acrylic and it's injection-molded. There's two parts of the mold that come together and molded plastic is injected, and then this comes out. The reason it's injection-molded is that it creates very precise part. You can get a nice, clean finish." + }, + { + "Q": "@ 9:39, he said that the copper wire works as an antenna . why does that work?", + "A": "Because as the electricity goes through it will make contact with the satellite. Hope that helps. :)", + "video_name": "xuQcB-oo-4U", + "timestamps": [ + 579 + ], + "3min_transcript": "We have a jumper here which helps to, you can use jumpers to alter the functionality of the seven-segment display. This one may be programmed or maybe set up to function in a certain way so that this jumper allows you to transfer power to the different part of the display. Then we have these little white spots here, and what that is is the back of this module, this seven-segment module, has these little pieces of plastic that stick through and there's a hot plate that basically pushes on those pins that stick through and melts them, and it holds the plate against the printed circuit board. This is just a cheap way or an inexpensive way to fasten things together. It works pretty well. That's the clock portion. The buttons, let's talk a little bit about the interface. The buttons, when you push down on the buttons here, they trigger these pins and you can see the pins flex pretty good right here. The pins are connected by these little standoffs. Because the standoffs are really thin, they can flex. When you press the button, the pin moves and the switch gets triggered. When you wanna snooze in the morning, you push this. It shifts the pin and it causes the sleep button to be triggered. That's kinda how that works. This is kind of ingenious in another way too because it holds a bunch of different things together. It's got the pins. It holds the speaker and it also holds the ferrite rod with the copper coil around it which functions as an antenna. That's an antenna for AM/FM radio. The signals come from here. We got a wire broken there. Signals come from here and they go to this thing which is a setup of four variable capacitors and they help to tune out frequencies we don't want. When we turn our dial, we can go right to 101.1 FM or 538 AMR, whichever station we want. This helps us to select those things. Those variable capacitors help us to filter out unwanted frequencies. and they can be used to oscillate at a particular frequency if they're coupled with a capacitor. That can be useful in performing radio functions as well. This guy right here is a, it's a radio chip. It's an IC chip that helps to demodulate or to separate the music or the signal that you want from the actual wave. AM is amplitude modulation so that means that the wave is changed in its height. FM is frequency modulation so that means that the wave is changed in how often it occurs in order to embed the signal that we get to listen to as radio sound. This chip basically decodes that and says, this is the original wave and then this is embedded signal. That's able to be then sent to our speaker" + }, + { + "Q": "@8:39 how does he know that the equilibrium lies to the left? How do you know if you have more reactants than products? I'm having difficulty understanding whether to point the arrow to the left or to the right. Thanks", + "A": "If Ka > 1, the position of equilibrium lies to the right. If Ka < 1, the position of equilibrium lies to the left.", + "video_name": "BeHOvYchtBg", + "timestamps": [ + 519 + ], + "3min_transcript": "so negative one charge on the oxygen. Let me show those electrons. These electrons in green move off onto the oxygen right here, giving it a negative charge. We're also gonna form a hydronium. All right, so H3O plus, so let me go ahead and draw in hydronium. So plus one formal charge on the oxygen and let's show those electrons in red. All right, so this electron pair picks up the acidic proton. Forming this bond that we get H3O plus. So another way to write this acid base reaction would be just to write acetic acid, CH3, COOH plus H2O gives us the acetate anion, CH3COO minus plus H3O plus. Now acetic acid is a weak acid and weak acids don't donate protons very well. So acetic acid is gonna stay mostly protonated. you're gonna have a relatively high concentration of your reactants here. When we write the equilibrium expression, write KA is equal to the concentration of your product so CH3COO minus times the concentration of H3O plus, all over the concentration of acetic acid because we leave water out. So all over the concentration of acetic acid. All right, the equilibrium lies to the left because acetic acid is not good at donating this proton. So we're going to get a very large number for the denominator, for this concentration so this is a very large number and a very small number for the numerator. All right, so this is a very small number. So if you think about what that does to the KA, all right, a very small number divided by an ionization constant much less than one. All right, so this value is going to be much less than one and that's how we recognize, that's one way to recognize a weak acid. Look at the KA value." + }, + { + "Q": "At 7:47, I know it sounds silly but could acetic acid have acted as a base, and ACCEPTED a proton from water? For example, either the double-bonded oxygen or single bonded oxygen could have accepted the proton. Thanks", + "A": "It can happen, but that reaction is not very favourable in water as water is a much better base (which means it s much more likely to accept H+) If you added a strong acid to pure acetic acid it could happen forming CH3COOH2^+", + "video_name": "BeHOvYchtBg", + "timestamps": [ + 467 + ], + "3min_transcript": "competing base strength. All right, so here we have Bronsted-Lowry. Base water is acting as a Bronsted-Lowry base and accepting a proton. And over here if you think about the reverse reaction, the chloride anion would be trying to pick up a proton from hydronium for the reverse reaction here but since HCL is so good at donating protons, that means that the chloride anion is not very good at accepting them. So the stronger the acid, the weaker the conjugate base. Water is a much stronger base than the chloride anion. Finally let's look at acetic acids. Acetic acid is going to be our Bronsted-Lowry acid and this is going to be the acidic proton. Water is gonna function as a Bronsted-Lowry base and a lone pair of electrons in the auction is going to take this acidic proton, leaving these electrons behind on the oxygen. So let's go ahead and draw our products. We would form the acetate anions. so negative one charge on the oxygen. Let me show those electrons. These electrons in green move off onto the oxygen right here, giving it a negative charge. We're also gonna form a hydronium. All right, so H3O plus, so let me go ahead and draw in hydronium. So plus one formal charge on the oxygen and let's show those electrons in red. All right, so this electron pair picks up the acidic proton. Forming this bond that we get H3O plus. So another way to write this acid base reaction would be just to write acetic acid, CH3, COOH plus H2O gives us the acetate anion, CH3COO minus plus H3O plus. Now acetic acid is a weak acid and weak acids don't donate protons very well. So acetic acid is gonna stay mostly protonated. you're gonna have a relatively high concentration of your reactants here. When we write the equilibrium expression, write KA is equal to the concentration of your product so CH3COO minus times the concentration of H3O plus, all over the concentration of acetic acid because we leave water out. So all over the concentration of acetic acid. All right, the equilibrium lies to the left because acetic acid is not good at donating this proton. So we're going to get a very large number for the denominator, for this concentration so this is a very large number and a very small number for the numerator. All right, so this is a very small number. So if you think about what that does to the KA, all right, a very small number divided by" + }, + { + "Q": "At 1:26 why does the Cl ion have a negative charge when it has a full octet?", + "A": "A chlorine atom has 7 valence electrons. If it has 8 (a full octet) then it has to have gained 1 electron which is why it has a -1 charge.", + "video_name": "BeHOvYchtBg", + "timestamps": [ + 86 + ], + "3min_transcript": "- [Voiceover] Let's look at this acid base reaction. So water is gonna function as a base that's gonna take a proton off of a generic acid HA. So lone pair of electrons on the oxygen pick up this proton leaving these electrons behind on the A. Oxygen, oxygen is now bonded to three hydrogens. So it picked up a proton. That's gonna give this oxygen a plus one formal charge and we can follow those electrons. So these two electrons in red here are gonna pick up this proton forming this bond. So we make hydronium H30 plus and these electrons in green right here are going to come off onto the A to make A minus. Let's go ahead and draw that in. So we're gonna make A minus. Let me draw these electrons in green and give this a negative charge like that. Let's analyze what happened. HA donated a proton so this is our Bronsted-Lowry acid. Once HA donates a proton, we're left with the conjugate base Water, H2O accepted a proton, so this is our Bronsted-Lowry base and then once H2O accepts a proton, we turn into hydronium H3O plus. So this is the conjugate acid. So H3O plus, the conjugate acid and then A minus would be a base. If you think about the reverse reaction, H3O plus donating a proton to A minus then you would get back H2O and HA. Once this reaction reaches equilibrium, we can write an equilibrium expression and we're gonna consider the stuff on the left to be the reactants. We're gonna think about the fourth reaction and the stuff on the right to be the products. Let's write our equilibrium expression. And so we write our equilibrium constant and now we're gonna write KA which we call the acid, the acid ionization constant. So this is the acid ionization constant so acid dissociation. So either one is fine. All right and we know when we're writing an equilibrium expression, we're gonna put the concentration of products over the concentration of reactants. Over here for our products we have H3O plus, so let's write the concentration of hydronium H3O plus times the concentration of A minus, so times the concentration of A minus. All over the concentration of our reactant, so we have HA over here, so we have HA. So we could write that in and then for water, we leave water out of our equilibrium expression. It's a pure liquid. Its concentration doesn't change and so we leave, we leave H2O out of our equilibrium expression. All right, so let's use this idea of writing an ionization constant and let's apply this to a strong acid." + }, + { + "Q": "At 2:40 why we do 2^x =4 Then X=? ,what does X means here?", + "A": "To be more specific, x is the order of the reaction in NO (which is the exponent for [NO] in the rate law).", + "video_name": "Ad0aaYixFJg", + "timestamps": [ + 160 + ], + "3min_transcript": "concentration of nitric oxide affects the rate of our reaction. We're going to look at experiments one and two here. The reason why we chose those two experiments is because the concentration of hydrogen is constant in those two experiments. The concentration of hydrogen is point zero zero two molar in both. If we look at what we did to the concentration of nitric oxide, we went from a concentration of point zero zero five to a concentration of point zero one zero. We increased the concentration of nitric oxide by a factor of two. We doubled the concentration. What happened to the initial rate of reaction? Well the rate went from one point two five times 10 to the negative five to five times 10 to the negative five. The rate increased by a factor of four. We increased the rate by a factor of four. you could just use a calculator and say five times 10 to the negative five and if you divide that by one point two five times 10 to the negative five, this would be four over one, or four. This rate is four times this rate up here. Now we know enough to figure out the order for nitric oxide. Remember from the previous video, what we did is we said two to the X is equal to four. Over here, two to the X is equal to four. Obviously X is equal to two, two squared is equal to four. So we can go ahead and put that in for our rate law. Now we know our rate is equal to K times the concentration of nitric oxide this would be to the second power. So the reaction is second order in nitric oxide. So this time we want to choose two experiments where the concentration of nitric oxide is constant. That would be experiment two and three where we can see the concentration of nitric oxide has not changed. It's point zero one molar for both of those experiments. But the concentration of hydrogen has changed. It goes from point zero zero two to point zero zero four. So we've increased the concentration of hydrogen by a factor of 2 and what happened to the rate of reaction? Well it went from five times 10 to the negative five to one times 10 to the negative four so we've doubled the rate. The rate has increased by a factor of two. Sometimes the exponents bother students. How is this doubling the rate? Well, once again, if you can't do that in your head," + }, + { + "Q": "At 5:38, by Time rate of change, does he mean the speed the electrons are moving, and doesn't this relate closely with Voltage?", + "A": "Time rate of change is always of something. Like: time rate of change of voltage, or time rate of change of position. If the time rate of change of voltage is 5 volts/second, then in the capacitor equation (i = C dv/dt) you would substitute 5 for dv/dt.", + "video_name": "l-h72j2-X0o", + "timestamps": [ + 338 + ], + "3min_transcript": "of the voltage, not to the voltage but to the rate of change of the voltage and the way we write that is current equals, C is the proportionality constant, and we write dv, dt so this is the rate of change of voltage with respect to time. We multiply that by this property of this device called capacitance and that gives us the current. This doesn't have a special name but I'm gonna refer to it as the capacitor equation so now we have two equations. Let's do the third equation which is for the inductor. The inductor has the property very similar to the capacitor. It has the property that the voltage across is proportional to the time rate of change of the current flowing through the inductor so this is a similar but opposite of how a capacitor works. The voltage is proportional to the time rate of change is voltage equals L, di, dt. The voltage is proportional. The proportionality constant is the inductants. The inductance of the inductor and this is the time rate of change of voltage, OH sorry, the time rate of change of current flowing through the inductor so this gives us our three equations. Here they are. These are three element equations and we're gonna use these all the time, right there, those three equations. One final point I wanna make is for both these equations of components, these are ideal, ideal components. that we have in our minds that we're gonna try to build in the real world but we'll come close. We'll come very close. We now have a wonderful set of equations: V equals iR, i equals C, dv, dt. v equals L, di, dt. These are gonna be like poetry for you pretty soon and these ideal equations will produce all kinds of really cool circuits for us." + }, + { + "Q": "At 13:20, I am a little confused on how you went from 0 ice to 0 degrees water ?", + "A": "For each kg of ice, you need a certain amount of energy to melt it to water. During the melting, the temperature stays the same. The amount of energy you need is given by L, the latent heat of fusion, which is in units joules per kilogram.", + "video_name": "zz4KbvF_X-0", + "timestamps": [ + 800 + ], + "3min_transcript": "Which is this right here, that distance right here. I forgot to figure out how much energy to turn that 100 degree water into 100 degree vapor. So that's key. So I really should have done that up here before I calculated the vapor. But I'll do it down here. So to do 100 degree water to 100 degree vapor. That's this step right here, this is the phase change. I multiply the heat of vaporization which is 2,257 joules per gram times 200 grams. And this is equal to 451,400. I'll do it in that blue color. for our sample of 200 grams. This piece right here was 83,000 joules. This piece right here was 3,780 joules. So to know the total amount of energy the total amount of heat that we had to put in the system to go from minus 10 degree ice all the way to 110 degree vapor, we just add up all of the energies which we had to do in all of these steps. Let's see. And I'll do them in order this time. So to go from minus 10 degree ice to zero degree ice. Of course we have 200 grams of it. It was 4,100. Plus the 67,000. So plus 67,110. Plus 83,000. That's to go from 0 degree water to 100 degree water. Plus 83,560. So we're at 154,000 right now, and just to get to 100 degree water. And then we need to turn that 100 degree water into 100 degree vapor. So you add the 451,000. So, plus 451,400 is equal to 606. And then finally, we're at 100 degree vapor, and we want to convert that to 110 degree vapor. So it's another 3,700 joules. So plus 3,780 is equal to 609,950 joules. So this whole thing when we're dealing with 200 grams" + }, + { + "Q": "At 9:55 Sal only crossed out the I's to make the equation simpler. why is that and why didn't he cross out the R's?", + "A": "This is because I(current) is the same throughout the circuit. so Sal cancelled the I s. But the resistance keeps on changing after going through each of the resistor. so he simply cannot cancel out the R s.", + "video_name": "ZrMw7P6P2Gw", + "timestamps": [ + 595 + ], + "3min_transcript": "drew in the previous diagram, although now I will assign numbers to it. Let's say that this resistance is 20 ohms and let's say that this resistance is 5 ohms. What I want to know is, what is the current through the system? First, we'll have to figure out what the equivalent resistance is, and then we could just use Ohm's law to figure out the current in the system. So we want to know what the current is, and we know that the convention is that current flows from the positive terminal to the negative terminal. So how do we figure out the equivalent resistance? Well, we know that we just hopefully proved to you that the total resistance is equal to 1 over this resistor plus 1 over this resistor. So 1 over-- I won't keep writing it. What's 1 over 20? Well, actually, let's just make it a fraction. That's 4/20, right? So 1 over our total resistance is equal to 5/20, which is 1/4. So if 1/R is equal to 1/4, R must be equal to 4. R is equal to 4 ohms. So we could redraw this crazy circuit as this. I'll try to draw it small down here. We could redraw this where this resistance is 4 ohms and this is 16 volts. We could say that this whole thing combined is really just a resistor that is 4 ohms. Well, if we have a 16-volt potential difference, current is flowing that way, even though that's not what the electrons are doing. And that's what our resistance is, 4 ohms. What is the current? V equals IR, Ohms law. It equals the current times 4 ohms. So current is equal to 16 divided by 4, is equal to 4 amps. So let's do something interesting. Let's figure out what the current is flowing through. What's this? What's the current I1 and what's this current I2? Well, we know that the potential difference from here to here is also 16 volts, right? Because this whole thing is essentially at the same potential and this whole thing is essentially at the same potential, so you have 16 volts across there. 16 volts divided by 20 ohms, so let's call this I1. So I1 is equal to 16 volts divided by 20 ohms, which is equal to what? 4/5. So it equals 4/5 of an ampere, or 0.8 amperes." + }, + { + "Q": "13:00\n\nIs there a name for this little period where the temperature stays constant although we are adding or removing a certain amount of heat?", + "A": "Heat of vaporazation or the heat of fusion.", + "video_name": "pKvo0XWZtjo", + "timestamps": [ + 780 + ], + "3min_transcript": "add heat my temperature will go up. Temperature is average kinetic energy. Let's say I'm in the solid state here. And I'll do the solid state in purple. No I already was using purple. I'll use magenta. So as I add heat, my temperature will go up. Heat is a form of energy. And when I add it to these molecules, as I did in this example, what did it do? It made them vibrate more. Or it made them have higher kinetic energy, or higher average kinetic engery, and that's what temperature is a measure of; average kinetic energy. So as I add heat in the solid phase, my average kinetic energy will go up. And let me write this down. This is in the solid phase, or the solid state of matter. Now something very interesting happens. So what happens at zero degrees? Which is also 273.15 Kelvin. Let's say that's that line. What happens to a solid? Well, it turns into a liquid. Ice melts. Not all solids, we're talking in particular about water, about H2O. So this is ice in our example. All solids aren't ice. Although, you could think of a rock as solid magma. Because that's what it is. I could take that analogy a bunch of different ways. But the interesting thing that happens at zero degrees. Depending on what direction you're going, either the freezing point of water or the melting point of ice, something interesting happens. As I add more heat, the temperature does not to go up. As I add more heat, the temperature does not go up for Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or" + }, + { + "Q": "In 0:05 he says that there are only 4 truth is there are 5 states of matter.They are Solid,Liquid,Gas,plasma and Bose- Einstine compound.", + "A": "I learned there were five as well.", + "video_name": "pKvo0XWZtjo", + "timestamps": [ + 5 + ], + "3min_transcript": "I think we're all reasonably familiar with the three states of matter in our everyday world. At very high temperatures you get a fourth. But the three ones that we normally deal with are, things could be a solid, a liquid, or it could be a gas. And we have this general notion, and I think water is the example that always comes to at least my mind. Is that solid happens when things are colder, relatively colder. And then as you warm up, you go into a liquid state. And as your warm up even more you go into a gaseous state. So you go from colder to hotter. And in the case of water, when you're a solid, you're ice. When you're a liquid, some people would call ice water, but let's call it liquid water. I think we know what that is. vapor or steam. So let's think a little bit about what, at least in the case of water, and the analogy will extend to other types of molecules. But what is it about water that makes it solid, and when it's colder, what allows it to be liquid. And I'll be frank, liquids are kind of fascinating because you can never nail them down, I guess is the best way to view them. Or a gas. So let's just draw a water molecule. So you have oxygen there. You have some bonds to hydrogen. And then you have two extra pairs of valence electrons in the oxygen. And a couple of videos ago, we said oxygen is a lot more electronegative than the hydrogen. It likes to hog the electrons. So even though this shows that they're sharing At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules." + }, + { + "Q": "Are the polar bonds forming between the H2O molecules @ 4:06 hydrogen bonds? If not what are hydrogen bonds?", + "A": "Yes, in a hydrogen bond hydrogen is bound to a highly electronegative atom, such as nitrogen, oxygen or fluorine.", + "video_name": "pKvo0XWZtjo", + "timestamps": [ + 246 + ], + "3min_transcript": "At both sides of those lines, you can kind of view that hydrogen is contributing an electron and oxygen is contributing an electron on both sides of that line. But we know because of the electronegativity, or the relative electronegativity of oxygen, that it's hogging these electrons. And so the electrons spend a lot more time around the oxygen than they do around the hydrogen. And what that results is that on the oxygen side of the molecule, you end up with a partial negative charge. And we talked about that a little bit. And on the hydrogen side of the molecules, you end up with a slightly positive charge. Now, if these molecules have very little kinetic energy, they're not moving around a whole lot, then the positive sides of the hydrogens are very attracted to the negative sides of oxygen in other molecules. When we talk about the whole state of the whole matter, we actually think about how the molecules are interacting with Not just how the atoms are interacting with each other within a molecule. I just drew one oxygen, let me copy and paste that. But I could do multiple oxygens. And let's say that that hydrogen is going to want to be near this oxygen. Because this has partial negative charge, this has a partial positive charge. And then I could do another one right there. And then maybe we'll have, and just to make the point clear, you have two hydrogens here, maybe an oxygen wants to hang out there. So maybe you have an oxygen that wants to be here because it's got its partial negative here. And it's connected to two hydrogens right there that have their partial positives. But you can kind of see a lattice structure. Let me draw these bonds, these polar bonds that start forming These bonds, they're called polar bonds because the molecules themselves are polar. And you can see it forms this lattice structure. And if each of these molecules don't have a lot of kinetic energy. Or we could say the average kinetic energy of this matter is fairly low. And what do we know is average kinetic energy? Well, that's temperature. Then this lattice structure will be solid. These molecules will not move relative to each other. I could draw a gazillion more, but I think you get the point that we're forming this kind of fixed structure. And while we're in the solid state, as we add kinetic energy, as we add heat, what it does to molecules is, it just makes them vibrate around a little bit. If I was a cartoonist, they way you'd draw a vibration is to put quotation marks there." + }, + { + "Q": "At round 14:02, Sal says that water takes much more time to vaporize as compared to the time period that ice takes to melt.\nWhy does this happen?", + "A": "In order to vaporize water, you have to give a lot of energy to each of the molecules so that they have enough energy to move far away from each other. To melt ice into water, you also have give the molecules energy, but you only have to give enough to break the bonds between the water molecules, and that doesn t require nearly as much energy as it does to actually pull the molecules away from each other.", + "video_name": "pKvo0XWZtjo", + "timestamps": [ + 842 + ], + "3min_transcript": "So what happens at zero degrees? Which is also 273.15 Kelvin. Let's say that's that line. What happens to a solid? Well, it turns into a liquid. Ice melts. Not all solids, we're talking in particular about water, about H2O. So this is ice in our example. All solids aren't ice. Although, you could think of a rock as solid magma. Because that's what it is. I could take that analogy a bunch of different ways. But the interesting thing that happens at zero degrees. Depending on what direction you're going, either the freezing point of water or the melting point of ice, something interesting happens. As I add more heat, the temperature does not to go up. As I add more heat, the temperature does not go up for Let me draw that. For a little period, the temperature stays constant. And then while the temperature is constant, it stays a solid. We're still a solid. And then, we finally turn into a liquid. Let's say right there. So we added a certain amount of heat and it just stayed a solid. But it got us to the point that the ice turned into a liquid. It was kind of melting the entire time. That's the best way to think about it. And then, once we keep adding more and more heat, then the liquid warms up too. Now, we get to, what temperature becomes interesting again for water? Well, obviously 100 degrees Celsius or 373 degrees Kelvin. I'll do it in Celsius because that's what we're familiar with. That's the temperature at which water will vaporize or But something happens. And they're really getting kinetically active. But just like when you went from solid to liquid, there's a certain amount of energy that you have to contribute to the system. And actually, it's a good amount at this point. Where the water is turning into vapor, but it's not getting any hotter. So we have to keep adding heat, but notice that the temperature didn't go up. We'll talk about it in a second what was happening then. And then finally, after that point, we're completely vaporized, or we're completely steam. Then we can start getting hot, the steam can then get hotter as we add more and more heat to the system. So the interesting question, I think it's intuitive, that as you add heat here, our temperature is going to go up." + }, + { + "Q": "At 4:07 when the ring breaks do the electrons in magenta need to rotate in order to form a bond with the other halogen atom(anti addition)?", + "A": "Nothing needs to rotate, the magenta electrons are not forming a new bond they are going to the Br that was in the ring. The electrons in the new C-Br bond comes from the bromide anion. It would help if Jay coloured these in.", + "video_name": "Yiy84xYQ3es", + "timestamps": [ + 247 + ], + "3min_transcript": "on the right and my halogen, like that. And these electrons over here, I'm going to mark in red. So this lone pair of electrons on my halogen are going to form a bond with the carbon on the left, And that halogen still has two lone pairs of electrons on it. So I'm going to put in those two lone pairs of electrons. And that halogen has a plus 1 formal charge. This is called a cyclic halonium ion, and it's been proven to occur in this mechanism. If I think about that positively charged halogen, halogens are very electronegative, and they want electrons. So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here losing a little bit of electron density, giving it a partial positive charge. And so that's going to function as our electrophile in the next step. in the previous step. So we had a halogen that had 3 lone pairs of electrons around it. It picked up the electrons in blue. Right? So now, it has 4 lone pairs of electrons-- 8 total electrons-- giving it a negative 1 formal charge, meaning it can now function as a nucleophile. So if I think about this cyclic halonium ion here, the halogen on top is going to prevent the nucleophile from attacking from the top. It's going to have to attack from below here. So this negatively charged halide anion is going to nucleophilic attack this electrophile here-- this carbon. And that's going to kick these electrons in magenta off onto this halogen here. So let's go ahead and draw the results of that nucleophilic attack. All right. So now, I'm going to have my 2 carbons still bonded to each other like that. And the top halogen has swung over here to the carbon It picked up the electrons in magenta. So that's what the carbon on the left will look like. The carbon on the right is still bonded to 2 other things. And the halide anion had to add from below. So now we're going to have this halogen down here. Like that. And so now we understand why it's an anti addition of my 2 halogen atoms. Let's go ahead and do a reaction. So we're going to start with cyclohexane as our reactant And we're going to react cyclohexane with bromine-- so Br2. Now, if I think about the first step of the mechanism, I know I'm going to form a cyclic halonium ion. So I'm going to draw that ring. And I'm going to show the formation of my cyclic halonium ion. It's called a bromonium ion. So I'm going to form a ring like this." + }, + { + "Q": "At 8:50 in the video the mechanism is described as happening two more times and the result is a trialkylborane molecule. I can't think of where the other two O's have come from. I don't understand why B, with an octet around it has a negative charge.", + "A": "From another two peroxides. Formal charge = valence electrons - lone pair electrons - bonds 3 - 4 = -1", + "video_name": "00qYQahwuSQ", + "timestamps": [ + 530 + ], + "3min_transcript": "So this is the weird part. Alright so these electrons right here are actually going to form a bond with the oxygen. At the same time these electrons are gonna come off on to this oxygen. So, we form hydroxide. Let me go ahead and show that first. That's maybe the easier part to understand of what's going on here. So we have three long pairs of electrons on the oxygen, negative one formal charge, we're gonna make these electrons green. These electrons in green here come off on to the oxygen. So the oxygen oxygen bond is weak. So, it's relatively easy to break this bond, and let me highlight the other electrons here. Well, these electrons in magenta formed this bond between the oxygen and the boron, and now we get our alkyl group migration and I'll show these electrons in the blue. So the electrons in blue are forming a bond between this carbon and this oxygen. It's pretty hard to see. Let me go ahead and draw what we have and then we'll highlight some electrons here. So now we have our oxygen bonded to this group and our boron has two alkyl groups still bonded to it like that. So let's follow those electrons. The electrons in magenta represent the bond between the oxygen and the boron, and the electrons in blue, right, would be these electrons right here. So it's this carbon is now bonded to this oxygen. So the migration of the alkyl group removes the formal charge on the boron and it breaks the weak oxygen oxygen bond and kicks off hydroxide. Alright so now this process happens two more times. Alright cause we have these two other alkyl groups, and so when that happens two more times we get, let me go ahead and draw it in, a trialkylborane. our group coming off of that and then over here we would have our oxygen and then we have our group coming off of that and then also down here. Alright, our next step is the hydroxide anion functions as a nucleophile. So the hydroxide attacks the boron. Once again our boron has an empty orbital. So nucleophilic attack let's draw the results of our nucleophilic attack. Now we have OH bonded to the boron and the boron still has all of these groups around it. So let me sketch all of those in, and you can see why this mechanism is getting rather cumbersome at this point. Right, drawing in all these groups is a little bit annoying here but we still have a negative one. I should say we have a negative one formal charge and a boron and we still have some lone pairs of electrons on this oxygen and I'm putting those in because in the next step to get rid of the negative one formal charge these electrons come off" + }, + { + "Q": "How is Chlorine less electronegative than oxygen? 6:40", + "A": "Sal explains this in the video on trends in the periodic system (chemistry video no. 10). Simply put, chlorine (compared to oxygen) has an additional shell of valence electrons (and a lot more electrons overall) that repels the electrons of other atoms or molecules. Watch the video or look at Wikipedia for a better explanation!", + "video_name": "8qfzpJvsp04", + "timestamps": [ + 400 + ], + "3min_transcript": "I think he was German-American. London dispersion force, and it's the weakest of the van der Waals forces. I'm sure I'm not pronouncing it correctly. And the van der Waals forces are the class of all of the intermolecular, and in this case, neon-- the molecule, is an atom . It's just a one-atom molecule, I guess you could say. The van der Waals forces are the class of all of the intermolecular forces that are not covalent bonds and that aren't ionic bonds like we have in salts, and we'll touch on those in a second. And the weakest of them are the London dispersion forces. So neon, these noble gases, actually, all of these noble gases right here, the only thing that they experience are London dispersion forces, which are the weakest of all of the intermolecular forces. And because of that, it takes very little energy to get them into a gaseous state. So at a very, very low temperature, That's why they're called noble gases, first of all. And they're the most likely to behave like ideal gases because they have very, very small attraction to each other. Fair enough. Now, what happens when we go to situations when we go to molecules that have better attractions or that are a little bit more polar? Let's say I had hydrogen chloride, right? Hydrogen, it's a little bit ambivalent about whether or not it keeps its electrons. Chloride wants to keep the electrons. Chloride's quite electronegative. It's less electronegative than these guys right here. These are kind of the super-duper electron hogs, nitrogen, oxygen, and fluorine, but chlorine is pretty electronegative. So if I have hydrogen chloride, so I have the chlorine atom right here, it has seven electrons and then it shares an electron with the hydrogen. It shares an electron with the hydrogen, Because this is a good bit more electronegative than hydrogen, the electrons spend a lot of time out here. So what you end up having is a partial negative charge on the side, where the electron hog is, and a partial positive side. And this is actually very analogous to the hydrogen bonds. Hydrogen bonds are actually a class of this type of bond, which is called a dipole bond, or dipole-dipole interaction. So if I have one chlorine atom like that and if I have another chlorine atom, the other chlorin eatoms looks like this. If I have the other chlorine atom-- let me copy and paste it-- right there, then you'll have this attraction between them. You'll have this attraction between these two chlorine atoms-- oh, sorry, between these two hydrogen chloride molecules. And the positive side, the positive pole of this dipole is the hydrogen side, because the electrons have kind of left it, will be attracted to the chlorine side" + }, + { + "Q": "9:15 Hydrogen Flourine is HF not H-FL, right?", + "A": "Yes Hydrogen Flourine is HF.", + "video_name": "8qfzpJvsp04", + "timestamps": [ + 555 + ], + "3min_transcript": "Because this is a good bit more electronegative than hydrogen, the electrons spend a lot of time out here. So what you end up having is a partial negative charge on the side, where the electron hog is, and a partial positive side. And this is actually very analogous to the hydrogen bonds. Hydrogen bonds are actually a class of this type of bond, which is called a dipole bond, or dipole-dipole interaction. So if I have one chlorine atom like that and if I have another chlorine atom, the other chlorin eatoms looks like this. If I have the other chlorine atom-- let me copy and paste it-- right there, then you'll have this attraction between them. You'll have this attraction between these two chlorine atoms-- oh, sorry, between these two hydrogen chloride molecules. And the positive side, the positive pole of this dipole is the hydrogen side, because the electrons have kind of left it, will be attracted to the chlorine side And because this van der Waals force, this dipole-dipole interaction is stronger than a London dispersion force. And just to be clear, London dispersion forces occur in all molecular interactions. It's just that it's very weak when you compare it to pretty much anything else. It only becomes relevant when you talk about things with noble gases. Even here, they're also London dispersion forces when the electron distribution just happens to go one way or the other for a single instant of time. But this dipole-dipole interaction is much stronger. And because it's much stronger, hydrogen chloride is going to take more energy to, get into the liquid state, or even more, get into the gaseous state than, say, just a sample of helium gas. Now, when you get even more electronegative, when this guy's even more electronegative when you're dealing with nitrogen, oxygen or fluorine, you get into a special case of dipole-dipole interactions, and that's the hydrogen bond. So it's really the same thing if a bunch of hydrogen fluorides around the place. Maybe I could write fluoride, and I'll write hydrogen fluoride here. Fluoride its ultra-electronegative. It's one of the three most electronegative atoms on the Periodic Table, and so it pretty much hogs all of the electrons. So this is a super-strong case of the dipole-dipole interaction, where here, all of the electrons are going to be hogged around the fluorine side. So you're going to have a partial positive charge, positive, partial negative, partial positive, partial negative and so on. So you're going to have this, which is really a dipole interaction. But it's a very strong dipole interaction, so people call it a hydrogen bond because it's dealing with hydrogen and a very electronegative atom," + }, + { + "Q": "At 7:27 you said the electron from the hydrogen spends most of its time with the chlorine, but why doesn't the chlorine just take the electron permanently, like an ionic bond?", + "A": "hydrogen bonding can also only form if it s with the elements FON. Flourine Oxygen and Nitrogen. and it s also covalently bonded.", + "video_name": "8qfzpJvsp04", + "timestamps": [ + 447 + ], + "3min_transcript": "I think he was German-American. London dispersion force, and it's the weakest of the van der Waals forces. I'm sure I'm not pronouncing it correctly. And the van der Waals forces are the class of all of the intermolecular, and in this case, neon-- the molecule, is an atom . It's just a one-atom molecule, I guess you could say. The van der Waals forces are the class of all of the intermolecular forces that are not covalent bonds and that aren't ionic bonds like we have in salts, and we'll touch on those in a second. And the weakest of them are the London dispersion forces. So neon, these noble gases, actually, all of these noble gases right here, the only thing that they experience are London dispersion forces, which are the weakest of all of the intermolecular forces. And because of that, it takes very little energy to get them into a gaseous state. So at a very, very low temperature, That's why they're called noble gases, first of all. And they're the most likely to behave like ideal gases because they have very, very small attraction to each other. Fair enough. Now, what happens when we go to situations when we go to molecules that have better attractions or that are a little bit more polar? Let's say I had hydrogen chloride, right? Hydrogen, it's a little bit ambivalent about whether or not it keeps its electrons. Chloride wants to keep the electrons. Chloride's quite electronegative. It's less electronegative than these guys right here. These are kind of the super-duper electron hogs, nitrogen, oxygen, and fluorine, but chlorine is pretty electronegative. So if I have hydrogen chloride, so I have the chlorine atom right here, it has seven electrons and then it shares an electron with the hydrogen. It shares an electron with the hydrogen, Because this is a good bit more electronegative than hydrogen, the electrons spend a lot of time out here. So what you end up having is a partial negative charge on the side, where the electron hog is, and a partial positive side. And this is actually very analogous to the hydrogen bonds. Hydrogen bonds are actually a class of this type of bond, which is called a dipole bond, or dipole-dipole interaction. So if I have one chlorine atom like that and if I have another chlorine atom, the other chlorin eatoms looks like this. If I have the other chlorine atom-- let me copy and paste it-- right there, then you'll have this attraction between them. You'll have this attraction between these two chlorine atoms-- oh, sorry, between these two hydrogen chloride molecules. And the positive side, the positive pole of this dipole is the hydrogen side, because the electrons have kind of left it, will be attracted to the chlorine side" + }, + { + "Q": "At 5:19, how doe he say that the lone pair will move up angularly away from the Central Atom O ?", + "A": "The electron pairs in water point towards the corners of a tetrahedron. The bonding pairs are in the plane of the paper. One lone pair is coming angularly out of the paper, and the other lone pair is pointing angularly behind the paper.", + "video_name": "q3g3jsmCOEQ", + "timestamps": [ + 319 + ], + "3min_transcript": "" + }, + { + "Q": "At 8:07 i did not quite understand why the tetrachloromethane molecule will have 0 D??", + "A": "Same case as in CO2. Since the power of Cl in C(CL)4 attracting the electron to themselves have the same magnitude individually, they cancel each other. Hope this helps.", + "video_name": "q3g3jsmCOEQ", + "timestamps": [ + 487 + ], + "3min_transcript": "" + }, + { + "Q": "At 8:00 it is said that we would expect CCl_4 not to be polar, but aren't the individual C-Cl bonds polar making the molecule polar?", + "A": "Yes, the individual bonds are all polar, but they cancel each other out, so the molecule as a whole is nonpolar.", + "video_name": "q3g3jsmCOEQ", + "timestamps": [ + 480 + ], + "3min_transcript": "" + }, + { + "Q": "1. at 12:05 , once we have the moles quantity. cant we use PV=nRT to figure out the pressure for each?\n\n2. For PV= nRT dont we always have to use Liters as the volume?\n\n3. Would the total number of moles always equall 100 moles when we add them up coming from % ?\n\nThanks for helping. and you are changing the world with your videos, not just how people are learning things.", + "A": "for number 1,yes you can , you just have to replace n with the number of moles of each molecule", + "video_name": "d4bqNf37mBY", + "timestamps": [ + 725 + ], + "3min_transcript": "That's 56.746 kilopascals. Or if we wanted it in atmospheres we just take 56,746 divided by 101,325. It equals 0.56 atmospheres. So that's the total pressure being exerted from all of the gases. I deleted that picture. So this is the total pressure. So our question is, what's the partial pressure? We could use either of these numbers, they're just in different units. What's the partial pressure of just the oxygen by itself? Well, you look at the moles, because we don't care about the actual mass. Because we're assuming that they're ideal gases. We want to look at the number of particles. Because remember, we said pressure times volume is proportional to the number of particles times temperature. And they're all at the same temperature. So the number of particles is what matters. So oxygen represents 20% of the particles. So, the partial pressure of oxygen, let me write that as pressure due to oxygen, due to O2. It's going to be 20% of the total pressure. 20% times, let me write 56.746 kilopascals. I just took this pressure measurement. If I wanted atmospheres, times 0.56 atmospheres. So the partial pressure of oxygen -- well, I already have the 0.56 written there. So times 0.2 is equal to 0.112 atmospheres. It's just 20% of this. How did I get 20%? We have a total of 100 mole molecules in our balloon. 20 moles of them are oxygen. So 20% are oxygen. So 20% of the pressure is going to be due to oxygen. If I took 20% times the 56,000. 0.2 times 56, you get roughly 11.2 kilopascals. I'm just multiplying 20% by any of these numbers. And the numbers are going to change depending on the units. So you do the same process. What is the partial pressure due to nitrogen? Well even though 2/3 of the mass is nitrogen, only 50% of the actual particles are nitrogen. So 50% of the pressures is due to the nitrogen particles. Remember, you have to convert everything to moles. Because we only care about the number of particles. So if you want to know the partial pressure due to the nitrogen molecules, it's 50% of these. So it's 28,373 pascals." + }, + { + "Q": "I know this might be a simple math question but dont you have to divide the entire side of the equation by 4 instead of just the 100 @ 8:52 ?", + "A": "this might help: since there are no variables or anything complicated, the entire side is just multiplying numbers. because of the order of operations (BEDMAS, or BEMDAS, or PEMDAS, or however you learned it), it doesn t matter whether you divide the whole thing by at the end, or divide one of the numbers by 4 in the middle. you re just dividing by 4! try it on a calculator, you get the same result :)", + "video_name": "d4bqNf37mBY", + "timestamps": [ + 532 + ], + "3min_transcript": "There you go. And I can erase some stuff that you're not seeing off the screen. And now I'm ready. So we have 100 moles. So we just do our PV is equal to nRT. We're trying to solve for P. P times 4 meters cubed is equal to n. n is the number of moles. We have 100 moles. Is equal to 100 moles times R. I'll put a blank there for R, because we have to figure out which R we want to use. Times temperature, remember we have to do it in Kelvin. So 0 degrees Celsius is 273 Kelvin. And then which R do we use? I always like to write my R's down here. So we're dealing with meters cubed, we're not dealing with liters, so let's use this one. 8.3145 meters cubed pascals per mole Kelvin. these are in meters cubed pascals divided by moles Kelvin. And then our temperature was 273 Kelvin. Now let's do a little dimensional analysis to make sure that we're doing things right. These meters cancel out with those meters. We divide both sides of the equation by meters. These moles cancel with these moles. The moles of the numerator, the moles of the denominator. Kelvin in the numerator, Kelvin in the denominator. And all we're left with is pascals. Which is good because that is a unit of pressure. So if we divide both sides of this equation by 4. I'll just divide the 100 by 4. 25 times 8.3145 times 273. Which is nice, because that's a unit of pressure. So, let's do the math, 25 times 8.3145 times 273 is equal to 56,746 pascals. And that might seem like a crazy number. But the pascal is actually a very small amount of pressure. It actually turns out that 101,325 pascals is equal to one atmosphere. So if we want to figure out how many atmospheres this is, we could just divide that. Let me look it up on this table. Yes, 101,325." + }, + { + "Q": "At 4:28, why did he change oxygen from 16 to 32?", + "A": "The relative atomic mass of one oxygen atom (O) is 16, while the relative atomic mass of one oxygen molecule (O2) is twice that at 32. It depends on if you are talking about the atom or the molecule.", + "video_name": "d4bqNf37mBY", + "timestamps": [ + 268 + ], + "3min_transcript": "So let's do it in grams. Because when we talk about molecular mass it's always in grams. It doesn't have to be. But it makes it a lot simpler to convert between atomic mass units and mass in our world. So this is 2/3 of 2100, that's 1400 grams of N2. Now what's the molar mass of this nitrogen molecule? Well we know that the atomic mass of nitrogen is 14. So this molecule has two nitrogens. So its atomic mass is 28. So one of these molecules will have a mass of 28 atomic mass units. Or one mole of N2 would have a mass of 28 grams. So one mole is 28 grams. We have 1400 grams -- or we say grams per mole, if we want to keep our units right. by 28 grams per mole we should get the number of moles. So 1400 divided by 28 is equal to 50. That worked out nicely. So we have 50 moles of N2. We could write that right there. All right. Let's do oxygen next. So we do the same process over again. 30% is oxygen. So let's do oxygen down here, O2. So we take 30%. Remember, these percentages I gave you, these are the percentages of the total mass, not the percentage of the moles. So we have to figure out what the moles are. So 30.48% of 2100 grams is equal to about 640. So this is equal to 640 grams. And then what is the mass of one mole of the oxygen gas molecule? The atomic mass of one oxygen atom is 16. You can look it up on the periodic table, although you should probably be pretty familiar with it by now. So the atomic mass of this molecule is 32 atomic mass units. So one mole of O2 is going to be 32 grams. We have 640 grams. So how many moles do we have? 640 divided by 32 is equal to 20. We have 20 moles of oxygen. Let me write that down. We have 20 moles. Now we just have to figure out the hydrogen." + }, + { + "Q": "At 7:52 how did you get 3 different R equations? And how can you be sure that you are choosing the right one, is it that you choose which ever one matches your units ?", + "A": "You re right - you always want to choose the R equation that matches the units you re using (or, you have to convert your units to match your R equation!).", + "video_name": "d4bqNf37mBY", + "timestamps": [ + 472 + ], + "3min_transcript": "was a super small fraction of the total mass of the gas that we have inside of the container, we actually have more actual particles, more actual molecules of hydrogen than we do of oxygen. That's because each molecule of hydrogen only has an atomic mass of 2 atomic mass units, while each molecule of oxygen has 32 because there's two oxygen atoms. So already we're seeing we actually have more particles do the hydrogen than do the oxygen. And the particles are what matter, not the mass, when we talk about part pressure and partial pressure. So the first thing we can think about is how many total moles of gas, how many total particles do we have bouncing around? 20 moles of oxygen, 30 moles of hydrogen, 50 moles of nitrogen gas. Add them up. We have 100 moles of gas. So if we want to figure out the total pressure first, we can just apply this 100 moles. Let me erase this. There you go. And I can erase some stuff that you're not seeing off the screen. And now I'm ready. So we have 100 moles. So we just do our PV is equal to nRT. We're trying to solve for P. P times 4 meters cubed is equal to n. n is the number of moles. We have 100 moles. Is equal to 100 moles times R. I'll put a blank there for R, because we have to figure out which R we want to use. Times temperature, remember we have to do it in Kelvin. So 0 degrees Celsius is 273 Kelvin. And then which R do we use? I always like to write my R's down here. So we're dealing with meters cubed, we're not dealing with liters, so let's use this one. 8.3145 meters cubed pascals per mole Kelvin. these are in meters cubed pascals divided by moles Kelvin. And then our temperature was 273 Kelvin. Now let's do a little dimensional analysis to make sure that we're doing things right. These meters cancel out with those meters. We divide both sides of the equation by meters. These moles cancel with these moles. The moles of the numerator, the moles of the denominator. Kelvin in the numerator, Kelvin in the denominator. And all we're left with is pascals. Which is good because that is a unit of pressure. So if we divide both sides of this equation by 4. I'll just divide the 100 by 4. 25 times 8.3145 times 273." + }, + { + "Q": "At 3:43 you say that the star will enlarge its radius like a red giant, but not get to the same size as a sun-sized star becoming a red giant. Does it also a experience a color change toward red, and if so what color was it in the first place? Does the creation of heavier elements in the core affect the color as well?", + "A": "No, as the stars surface gets farther from the core, it gets cooler. Color is entirely related to temperature the cooler surface glows red instead of yellow.", + "video_name": "UhIwMAhZpCo", + "timestamps": [ + 223 + ], + "3min_transcript": "But anyway, let's think about what happens. And so far, just the pattern of what happens, it's going to happen faster because we have more pressure, more gravity, more temperature. But it's going to happen in pretty much the same way as what we saw with a star the mass of the sun. Eventually that helium-- sorry, that hydrogen is going to fuse into a helium core that's going to have a hydrogen shell around it. It's going to have a hydrogen shell around it, hydrogen fusion shell around it. And then you have the rest of the star around that. So let me label it. This right here is our helium core. And more and more helium is going to be built up as this hydrogen in this shell fuses. And in a star the size of our sun or the mass of our sun, this is when it starts to become a red giant. Because this core is getting denser and denser and denser as more and more helium is produced. And as it gets denser and denser and denser, being put on the hydrogen, on this hydrogen shell out here, where we have fusion still happening. And so that's going to release more outward energy to push out the radius of the actual star. So the general process, and we're going to see this as the star gets more and more massive, is we're going to have heavier and heavier elements forming in the core. Those heavier and heavier elements, as the star gets denser and denser, will eventually ignite, kind of supporting the core. But because the core itself is getting denser and denser and denser, material is getting pushed further and further out with more and more energy. Although if the star is massive enough, it's not going to be able to be pushed out as far as you will have in kind of a red giant, with kind of a sun-like star. But let's just think about how this pattern is going to continue. So eventually, that helium, once it gets dense enough, it's going to ignite and it's going to fuse into carbon. And you're going to have a carbon core forming. So that is carbon core. That's a carbon core. Around that, you have a helium core. you have a shell of helium fusion-- that's helium, not hydrogen-- turning into carbon, making that carbon core denser and hotter. And then around that, you have hydrogen fusion. Have to be very careful. You have hydrogen fusion. And then around that, you have the rest of the star. And so this process is just going to keep continuing. Eventually that carbon is going to start fusing. And you're going to have heavier and heavier elements form the core. And so this is a depiction off of Wikipedia of a fairly mature massive star. And you keep forming these shells of heavier and heavier elements, and cores of heavier and heavier elements until eventually, you get to iron. And in particular, we're talking about iron 56. Iron with an atomic mass of 56. Here on this periodic table that 26 is its atomic number. It's how many protons it has. 56, you kind of view it as a count of the protons" + }, + { + "Q": "4:00-4:30 are there others stars that have different elements or have we discovered all of the elements ?", + "A": "we have discovered all the elements that naturally exist, and even if we hadn t there would have to be stars as big as the milky way to fuse elements beyond iron.", + "video_name": "UhIwMAhZpCo", + "timestamps": [ + 240, + 270 + ], + "3min_transcript": "But anyway, let's think about what happens. And so far, just the pattern of what happens, it's going to happen faster because we have more pressure, more gravity, more temperature. But it's going to happen in pretty much the same way as what we saw with a star the mass of the sun. Eventually that helium-- sorry, that hydrogen is going to fuse into a helium core that's going to have a hydrogen shell around it. It's going to have a hydrogen shell around it, hydrogen fusion shell around it. And then you have the rest of the star around that. So let me label it. This right here is our helium core. And more and more helium is going to be built up as this hydrogen in this shell fuses. And in a star the size of our sun or the mass of our sun, this is when it starts to become a red giant. Because this core is getting denser and denser and denser as more and more helium is produced. And as it gets denser and denser and denser, being put on the hydrogen, on this hydrogen shell out here, where we have fusion still happening. And so that's going to release more outward energy to push out the radius of the actual star. So the general process, and we're going to see this as the star gets more and more massive, is we're going to have heavier and heavier elements forming in the core. Those heavier and heavier elements, as the star gets denser and denser, will eventually ignite, kind of supporting the core. But because the core itself is getting denser and denser and denser, material is getting pushed further and further out with more and more energy. Although if the star is massive enough, it's not going to be able to be pushed out as far as you will have in kind of a red giant, with kind of a sun-like star. But let's just think about how this pattern is going to continue. So eventually, that helium, once it gets dense enough, it's going to ignite and it's going to fuse into carbon. And you're going to have a carbon core forming. So that is carbon core. That's a carbon core. Around that, you have a helium core. you have a shell of helium fusion-- that's helium, not hydrogen-- turning into carbon, making that carbon core denser and hotter. And then around that, you have hydrogen fusion. Have to be very careful. You have hydrogen fusion. And then around that, you have the rest of the star. And so this process is just going to keep continuing. Eventually that carbon is going to start fusing. And you're going to have heavier and heavier elements form the core. And so this is a depiction off of Wikipedia of a fairly mature massive star. And you keep forming these shells of heavier and heavier elements, and cores of heavier and heavier elements until eventually, you get to iron. And in particular, we're talking about iron 56. Iron with an atomic mass of 56. Here on this periodic table that 26 is its atomic number. It's how many protons it has. 56, you kind of view it as a count of the protons" + }, + { + "Q": "At 2:10 how do you know it is cosine and not tangent or sine?", + "A": "Think about the problem, if your moving something forward (applied force is greater then friction), and were to use sin0\u00c2\u00b0 then your work F(d)(0)=0 and that doesn t make sense because there is some type of work being done BC the object is moving", + "video_name": "udgMh3Y-dTk", + "timestamps": [ + 130 + ], + "3min_transcript": "I'm going to show you some examples of how to solve problems involving work. Imagine a 4 kilogram trashcan. The trashcan is disgusting. So someone ties a string to it and pulls on the string with a force of 50 newtons. The force of kinetic friction on the trashcan while it slides is 30 newtons. The trash can slides across the ground for a distance of 10 meters. Let's try to find the work done by each force on the trash can as it slides across the ground. To find the work done by each force, we should recall the formula definition of work. Work equals Fd cosine theta, where theta is the angle between the force doing the work and the direction the trashcan is moving. There are four forces involved here-- tension, the normal force, the gravitational force, and the force of kinetic friction. In finding the work done for all of these forces, the displacement is going to be 10 meters. But the value of the force and the angle between that force and the displacement is going to differ for each of the forces. we'll plug in the size of the tension, which is 50 newtons. The displacement is 10 meters. And since the tension force is pointed in the same direction as the displacement, the angle between the force of tension and the displacement is 0 degrees. And since cosine of 0 is 1, the work done by the tension force is 500 joules. To find the work done by friction, we'll plug in the size of the force of friction, which is 30 newtons. The displacement is still 10 meters. And since the force of friction points in the opposite direction as the displacement, the angle between the force of friction and the displacement is 180 degrees. Since cosine of 180 is negative 1, the work done by the force of friction is negative 300 joules. Now let's figure out the work done by the gravitational force. The force of gravity is mg. So the force of gravity is 4 kilograms times 9.8 meters per second squared, which is 39.2 newtons. But the angle between the gravitational force and the direction of the displacement is 90 degrees in this case. And since cosine of 90 is 0, the gravitational force does no work on this trashcan. Similarly, if we were to find the work done by the normal force, the angle between the direction of the displacement and the normal force is 90 degrees. So the normal force also does no work on the trashcan. This makes sense because forces that are perpendicular to the motion can never do any work on that object. So that's how you can find the work done by individual forces. And if we wanted to know the net work done on this trashcan, we could just add up the work done by each individual force. So the net work is going to be 200 joules. Now that we know the net work done on the trashcan, we can use the work-energy principle to figure out the speed of the trashcan after it's slid the 10 meters. The work-energy principle says that the net work done" + }, + { + "Q": "At 7:00, Sal says that astronauts can't tell whether they're in free fall near an object with a gravitational pull, or in deep space without any noticeable gravitational forces. How is this possible? Won't they feel the difference in acceleration?", + "A": "How can they feel the acceleration of free fall? The astronauts on the ISS are in free fall. What do you think they feel?", + "video_name": "oIZV-ixRTcY", + "timestamps": [ + 420 + ], + "3min_transcript": "If they were to just slow themselves down, if they were to just brake relative to the Earth, and if they were to just put their brakes on right over there, they would all just plummet to the Earth. So there's nothing special about going 300 or 400 miles up into space, that all of a sudden gravity disappears. The influence of gravity, actually on some level, it just keeps going. You can't, it might become unnoticeably small at some point, but definitely for only a couple of hundred miles up in the air, there is definitely gravity there. It's just they're in orbit, they're going fast enough. So if they just keep falling, they're never going to hit the Earth. And if you want to simulate gravity, and this is actually how NASA does simulate gravity, is that they will put people in a plane, and they call it the vomit rocket because it's known to make people sick, and they'll make them go in a projectile motion. So if this is the ground, in a projectile path or in a parabolic path I should say, so the plane will take off, and it or in a parabolic path. And so anyone who's sitting in that plane will experience free fall. So if you've ever been in, if you've ever right when you jump off of a or if you've ever bungee jumped or skydived or even the feeling when a roller coaster is going right over the top, and it's pulling you down, and your stomach feels a little ill, that feeling of free fall, that's the exact same feeling that these astronauts feel because they're in a constant state of free fall. But that is an indistinguishable feeling from, if you were just in deep space and you weren't anywhere close any noticeable mass, that is an identical feeling that you would feel to having no gravitational force around you. So hopefully that clarifies things a little bit. To someone who's just sitting in the space shuttle, and if they had no windows, there's no way of them knowing whether they are close to a massive object and they're just in free fall around it, they're in orbit, or whether they're and they really are in a state of or in a place where there's very little gravity." + }, + { + "Q": "At 0:01 what is \"level 4\" multiplication? Does Sal mean grade 4?", + "A": "no its just to show your going in a more complicated type of multplication", + "video_name": "_k3aWF6_b4w", + "timestamps": [ + 1 + ], + "3min_transcript": "Welcome to Level 4 multiplication. Let's do some problems. Let's see, we had 235 times-- I'm going to use two different colors here, so bear with me a second. Let's say times 47. So you start a Level 4 problem just like you would normally do a Level 3 problem. We'll take that 7, and we'll multiply it by 235. So 7 times 5 is 35. 7 times 3 is 21, plus the 3 we just carried is 24. 7 times 2 is 14, plus the 2 we just carried. This is 16. So we're done with the 7. Now we have to deal with this 4. Well, since that 4 is in the tens place, we add a 0 here. You could almost view it as we're multiplying 235, not by we put that 0 there. But once you put the 0 there, you can treat it just like a 4. So you say 4 times 5, well, that's 20. Let's ignore what we had from before. 4 times 3 is 12, plus the 2 we just carried, which is 14. 4 times 2 is 8, plus the 1 we just carried, so that's 9. And now we just add up everything. 5 plus 0 is 5, 4 plus 0 is 4, 6 plus 4 is 10, carry the 1, and 1 plus 9, well, that's 11. So the answer's 11,045. Let's do another problem. Let's say I had 873 times-- and I'm making these numbers up on the fly, so bear with me-- 873 times-- some high numbers-- hopefully get a better understanding of what I'm trying to explain. Let's say 97. No, I just used a 7. Let's make it 98. So just like we did before, we go to the ones place first, and that's where that 8 is, and we multiply that 8 times 873. So 8 times 3 is 24, carry the 2. 8 times 7 is 56, plus 2 is 58, carry the 5. 8 times 8 is 64, plus the 5 we just carried. That's 69. We're done with the 8. Now we have to multiply the 9, or we could just do it as we're multiplying 873 by 90. But multiplying something by 90 is just the same thing as multiplying something by 9 and then adding a 0 at the end, so that's why I put a 0 here." + }, + { + "Q": "at 1:03 where did he get a and b?", + "A": "For Ax^2 +Bx +C a*b = AC a + b = B He set the equality, then factored the resultants to obtain candidates for a and b. This takes experience and trial and error.", + "video_name": "dstNU7It-Ro", + "timestamps": [ + 63 + ], + "3min_transcript": "Simplify the rational expression and state the domain. Once again, we have a trinomial over a trinomial. To see if we can simplify them, we need to factor both of them. That's also going to help us figure out the domain. The domain is essentially figuring out all of the valid x's that we can put into this expression and not get something that's undefined. Let's factor the numerator and the denominator. So let's start with the numerator there, and since we have a 2 out front, factoring by grouping will probably be the best way to go, so let's just rewrite it here. I'm just working on the numerator right now. 2x squared plus 13x plus 20. We need to find two numbers, a and b, that if I multiply them, a times b, needs to be equal to-- let me write it over here on the right. a times b needs to be equal to 2 times 20, so it has to be equal to positive 40. And then a plus b has to be equal to 13. are 5 and 8, right? 5 times 8 is 40. 5 plus 8 is 13. We can break this 13x into a 5x and an 8x, and so we can rewrite this as 2x squared. It'll break up the 13x into-- I'm going to write the 8x first. I'm going to write 8x plus 5x. The reason why I wrote the 8x first is because the 8 shares common factors with the 2, so maybe we can factor out a 2x here. It'll simplify it a little bit. 5 shares factors with the 20, so let's see where this goes. We finally have a plus 20 here, and now we can group them. That's the whole point of factoring by grouping. You group these first two characters right here. Let's factor out a 2x, so this would become 2x times-- well, 2x squared divided by 2x is just going to be x. 8x divided by 2x is going to be plus 4. And if we factor out a 5, what do we get? We get plus 5 times x plus 4. 5x divided by 5 is x, 20 divided by 5 is 4. We have an x plus 4 in both cases, so we can factor that out. We have x plus 4 times two terms. We can undistribute it. This thing over here will be x plus four times-- let me do it in that same color-- 2x plus 5. And we've factored this numerator expression right there. Now, let's do the same thing with the denominator I'll do that in a different-- I don't want to run out of colors. So the denominator is right over here, let's do the same exercise with it. We have 2x squared plus 17x plus 30." + }, + { + "Q": "why we need the slope? he mention it in min 2:34", + "A": "Since the line of reflection is not horizontal or vertical, he noted that it has a slope of 1. We use iy-intercepts and slopes to graph lines.", + "video_name": "3aDV3L8aZtY", + "timestamps": [ + 154 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:30 he takes the derivative of e^(-2x^2) with respect to (-2x^2). Why doesn't the power rule apply here?", + "A": "The power rule applies when the base is a variable and the power is a number. This is an exponential, the base is a number and the variable is in the power. They are very different functions, and therefore have different derivatives.", + "video_name": "MUQfl385Yug", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's say that f of x is equal to x times e to the negative two x squared, and we want to find any critical numbers for f. I encourage you to pause this video and think about, can you find any critical numbers of f. I'm assuming you've given a go at it. Let's just remind ourselves what a critical number is. We would say c is a critical number of f, if and only if. I'll write if with two f's, short for if and only if, f prime of c is equal to zero or f prime of c is undefined. If we look for the critical numbers for f we want to figure out all the places where the derivative of this with Let's think about how we can find the derivative of this. f prime of x is going to be, well let's see. We're going to have to apply some combination of the product rule and the chain rule. It's going to be the derivative with respect to x of x, so it's going to be that, times e to the negative two x squared plus the derivative with respect to x of e to the negative two x squared times x. This is just the product rule right over here. Derivative of the x times e to the negative of two x squared plus the derivative of e to the What is this going to be? Well all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to one. This first part is going to be equal to e to the negative two x squared. Now the derivative of e to the negative two x squared over here. I'll do this in this pink color. This part right over here, that is going to be equal to- We'll just apply the chain rule. Derivative of e to the negative two x squared with respect to negative two x squared, well that's just going to be e to the negative two x squared. We're going to multiply that times the derivative of negative two x squared with respect to x. That's going to be what, negative four x. Times negative four x, and of course we have this x over here. We have that x over there and let's see," + }, + { + "Q": "At 5:15 why is the answer a positive and a negative? e.g. + or - 1/2", + "A": "It s because you are taking the square root of (1/4). When you take the square root you ll have two solutions ( + or - 1/2 in this case ) since when you square each one you ll get (1/4).", + "video_name": "MUQfl385Yug", + "timestamps": [ + 315 + ], + "3min_transcript": "Well obviously both of these terms have an e to the negative two x squared. I'm going to try to figure out where this is either undefined or where this is equal to zero. Let's think about this a little bit. If we factor out e to the negative two x squared, I'll do that in green. We're going to have, this is equal to e to the negative two x squared times, we have here, one minus four x squared. One minus four x squared. This is the derivative of f. Where would this be undefined or equal to zero? e to the negative two x squared, this is going to be defined for any value of x, and this part is also going to be defined for any value of x. There's no point where this is undefined. Let's think about when this is going to be equal to zero. The product of these two expressions equalling zero, e to the negative two x squared, that will never be equal to zero. If you get this exponent to be a really, I guess you could say very negative number, you will approach zero but you will never get it to be zero. This part here can't be zero. If the product of two things are zero at least one of them has to be zero, so the only way we can get f prime of x to be equal to zero is when one minus four x squared is equal to zero. One minus four x squared is equal to zero, let me rewrite that. One minus four x squared is equal to zero, when does that happen? This one we can just solve. Add four x squared to both sides, you get one is equal to four x squared. Divide both sides by four, you get Then what x values is this true at? We just take the plus or minus square root of both sides and you get x is equal to plus or minus one half. Negative one half squared is one fourth, positive one half squared is one fourth. If x equals plus or minus one half f prime, or the derivative, is equal to zero. Let me write it this way. f prime of one half is equal to zero, and you can verify that right over here. And f prime of negative one half is equal to zero. If someone asks what are the critical numbers here, they are one half and negative one half." + }, + { + "Q": "At about 2:29, Sal is saying that 3a(a^5)(a^2) is 3(a^1*a^7). How? I am probably just confused on this topic, though.", + "A": "a = a^1 3\u00e2\u0080\u00a2a^1\u00e2\u0080\u00a2a^5\u00e2\u0080\u00a2a^2 = 3\u00e2\u0080\u00a2a^(1 + 5 + 2) = 3\u00e2\u0080\u00a2a^8", + "video_name": "-TpiL4J_yUA", + "timestamps": [ + 149 + ], + "3min_transcript": "Simplify 3a times a to the fifth times a squared. So the exponent property we can use here is if we have the same base, in this case, it's a. If we have it raised to the x power, we're multiplying it by a to the y power, then this is just going to be equal to a to the x plus y power. And we'll think about why that works in a second. So let's just apply it here. Let's start with the a to the fifth times a squared. So if we just apply this property over here, this will result in a to the fifth plus two-th power. So that's what those guys reduce to, or simplify to. And of course, we still have the 3a out front. Now what I want to do is take a little bit of an aside and realize why this worked. Let's think about what a to the fifth times a squared means. A to the fifth literally means a times a times a times a times a. Now, a squared literally means a times a. So we're multiplying these five a's times these two a's. And what have we just done? We're multiplying a times itself five times, and then another two times. We are multiplying a times itself. So let me make it clear. This over here is a to the fifth. This over here is a squared. When you multiply the two, you're multiplying a by itself itself seven times. 5 plus 2. So this is a to the seventh power. a to the 5 plus 2 power. So this simplifies to 3a times a to the seventh power. Now you might say, how do I apply the property over here? What is the exponent on the a? And remember, if I just have an a over here, this is equivalent to a to the first power. So I can rewrite 3a is 3 times a to the first power. A to the first power-- and the association property of multiplication, I can do the multiplication of the a's before I worry about the 3's. So I can multiply these two guys first. So a to the first times a to the seventh-- I just have to add the exponents because I have the same base and I'm taking the product-- that's going to be a to the eighth power. And I still have this 3 out front. So 3a times a to the fifth times a squared simplifies to 3a to the eighth power." + }, + { + "Q": "At 1:28 why is he making those markings? He draws a line and then crosses it out... as to say, what? If he wants to make it easier for himself to follow along, shouldn't he be crossing out the 2 digit and 3 digit instead?", + "A": "He circles the numbers to see which numbers he s multiplying. He s just trying to help us understand by giving us a visual. :)", + "video_name": "TqRReFvbpXA", + "timestamps": [ + 88 + ], + "3min_transcript": "Let's multiply 7 times 253 and see what we get. So just like in the last example, what I like to do is I like to rewrite the largest number first. So that's 253. And then write the smaller number below it and align the place value, the 7. It only has a ones place, so I'll put the 7 right over here below the ones place in 253. And then put the multiplication symbol right over here. So you could read this as 253 times 7, which we know is the same thing as 7 times 253. And now we are ready to compute. And there are many ways of doing this, but this one you could call the standard way. So what I do is I start with my 7. And I multiply it times each of the numbers up here, and I carry appropriately. So first I start with 7 times 3. Well, 7 times 3 we know is 21. Let me write that down. 7 times 3 is equal to 21. You could do this part in your head, but I just want to make it clear where I'm getting these numbers from. is I would write the 1 into 21 down here, but then carry the 2 to the tens place. Now I want to figure out what 7 times 5 is. We know from our multiplication tables that 7 times 5 is equal to 35. Now, we can't just somehow put the 35 down here. We still have to deal with this 2 that we carried. So we compute 7 times 5 is 35, but then we also add that 2. So it's 35 plus 2 is 37. Now, we write the 7 right over here in the tens place and carry the 3. Now we need to compute what 7 times 2 is. We know that 7 times 2 is 14 from our multiplication tables. We can't just put a 14 down here. We have this 3 to add. So 7 times 2 is 14, plus 3 is 17. because 2 is the last number that we had to deal with. And so we have our answer. 7 times 253 is 1,771." + }, + { + "Q": "At 4:33, how would you find the area of an inscribed polygons?", + "A": "I would break it into triangles, each with a vertex at the center of the circle, then use trigonometry (or special triangles) to solve for the sides and area.", + "video_name": "LrxZMdQ6tiw", + "timestamps": [ + 273 + ], + "3min_transcript": "So we can use that information to figure out what the other angles are. Because these two base angles-- it's an isosceles triangle. The two legs are the same. So our two base angles, this angle is going to be congruent to that angle. If we could call that y right over there. So you have y plus y, which is 2y, plus 60 degrees is going to be equal to 180. Because the interior angles of any triangle-- they add up to 180. And so subtract 60 from both sides. You get 2y is equal to 120. Divide both sides by 2. You get y is equal to 60 degrees. Now, this is interesting. I could have done this with any of these triangles. All of these triangles are 60-60-60 triangles, which tells us-- and we've proven this earlier on when we first started studying equilateral triangles-- we know that all of the angles of a triangle are 60 degrees, then we're dealing with an equilateral triangle, which means that all the sides have the same length. So if this is 2 square roots of 3, then so is this. And this is also 2 square roots of 3. So pretty much all of these green lines are 2 square roots of 3. And we already knew, because it's a regular hexagon, that each side of the hexagon itself is also 2 square roots of 3. So now we can essentially use that information to figure out-- actually, we don't even have to figure this part out. I'll show you in a second-- to figure out the area of any one of these triangles. And then we can just multiply by 6. So let's focus on this triangle right over here and think about how we can find its area. We know that length of DC is 2 square roots of 3. We can drop an altitude over here. We can drop an altitude just like that. And then if we drop an altitude, we know that this is an equilateral triangle. And we can show very easily that these two triangles are symmetric. These are both 90-degree angles. We know that these two are 60-degree angles already. And then if you look at each of these two independent triangles, you'd have to just say, well, So this has to be 30 degrees. This has to be 30 degrees. All the angles are the same. They also share a side in common. So these two are congruent triangles. So if we want to find the area of this little slice of the pie right over here, we can just find the area of this slice, or this sub-slice, and then multiply by 2. Or we could just find this area and multiply by 12 for the entire hexagon. So how do we figure out the area of this thing? Well, this is going to be half of this base length, so this length right over here. Let me call this point H. DH is going to be the square root of 3. And hopefully we've already recognized that this is a 30-60-90 triangle. Let me draw it over here. So this is a 30-60-90 triangle. We know that this length over here is square root of 3. And we already actually did calculate that this is 2 square roots of 3. Although we don't really need it. What we really need to figure out is this altitude height. And from 30-60-90 triangles, we know" + }, + { + "Q": "At 2:10, how ar the triangles congruent by SSS? I thought SSS was a similarity postulate and the triangles are congruent through the SAS Postulate.", + "A": "SSS is both a congruency and a similarity postulate. We cannot use SAS because we haven t proven that all the central angles are congruent.", + "video_name": "LrxZMdQ6tiw", + "timestamps": [ + 130 + ], + "3min_transcript": "We're told that ABCDEF is a regular hexagon. And this regular part-- hexagon obviously tells us that we're dealing with six sides. And you could just count that. You didn't have to be told it's a hexagon. But the regular part lets us know that all of the sides, all six sides, have the same length and all of the interior angles have the same measure. Fair enough. And then they give us the length of one of the sides. And since this is a regular hexagon, they're actually giving us the length of all the sides. They say it's 2 square roots of 3. So this side right over here is 2 square roots of 3. This side over here is 2 square roots of 3. And I could just go around the hexagon. Every one of their sides is 2 square roots of 3. They want us to find the area of this hexagon. Find the area of ABCDEF. And the best way to find the area, especially of regular polygons, is try to split it up into triangles. And hexagons are a bit of a special case. Maybe in future videos, we'll think about the more general case of any polygon. But with a hexagon, what you could think about is if we take this point right over here. And let's call this point G. And let's say it's the center of the hexagon. I'm talking about a point. It can't be equidistant from everything over here, because this isn't a circle. But we could say it's equidistant from all of the vertices, so that GD is the same thing as GC is the same thing as GB, which is the same thing as GA, which is the same thing as GF, which is the same thing as GE. So let me draw some of those that I just talked about. So that is GE. There's GD. There's GC. All of these lengths are going to be the same. So there's a point G which we can call the center of this polygon. And we know that this length is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length. We also know that if we go all the way around the circle like that, we've gone 360 degrees. And we know that these triangles are all going to be congruent to each other. And there's multiple ways that we could show it. But the easiest way is, look, they have two sides. All of them have this side and this side And they all have this third common side of 2 square roots of 3. So all of them, by side-side-side, they are all congruent. What that tells us is, if they're all congruent, then this angle, this interior angle right over here, is going to be the same for all six of these triangles over here. And let me call that x. That's angle x. That's x. That's x. That's x. That's x. And if you add them all up, we've gone around the circle. We've gone 360 degrees. And we have six of these x's. So you get 6x is equal to 360 degrees. You divide both sides by 6. You get x is equal to 60 degrees. All of these are equal to 60 degrees. Now there's something interesting. We know that these triangles-- for example, triangle GBC-- and we could do that for any of these six triangles. It looks kind of like a Trivial Pursuit piece. We know that they're definitely isosceles triangles," + }, + { + "Q": "At around 1:40, couldn't you take the (positive and negative) square roots of both sides?", + "A": "You could. (and then x = -2 and x = 2) Good question!", + "video_name": "bml74_PsfwA", + "timestamps": [ + 100 + ], + "3min_transcript": "Solve and eliminate any extraneous solutions. And what they mean by extraneous solutions are, in the course of solving this rational equation right here, we might get some solutions that if we actually put it into the original problem would give us undefined expressions. And so those solutions are extraneous solutions. They actually don't apply. You actually want to throw them out. And so let's look at this equation. We have x squared over x plus 2 is equal to 4 over x plus 2. So right from the get-go, we don't know if this is going to necessarily be a solution to this equation. But we know, just looking at this, that if x is equal to negative 2, then this denominator and this denominator are going to be 0. And you're dividing by 0. It would be undefined. So we can, right from the get-go, exclude x is equal to negative 2. So x cannot be equal to negative 2. That would make either of these expressions undefined, on either side of the equation. So with that out of the way, let's try to solve it. So as a first step, we want to get the x plus 2 out of the denominator. So let's multiply both sides by x plus 2. And we can assume that x plus 2 isn't 0. So it's going to be defined. x plus 2 divided by x plus 2 is just 1. And so our equation has simplified to x squared is equal to 4. And you could probably do this in your head, but I want to do it properly. So you can write this. You could subtract 4 from both sides. Do it in kind of the proper quadratic equation form. So x squared minus 4 is equal to 0. I just subtracted 4 from both sides over here. And so you could factor this. This is a difference of squares. You get x plus 2 times x minus 2 is equal to 0. And then if this is equal to 0, if the product of two things are equal to 0, that means either one or both of them are equal to 0. So this tells us that x plus 2 is equal to 0 or x minus 2 is equal to 0. If you subtract 2 from both sides of this equation right If you add 2 to both sides of this equation right over here, you get x is equal to 2. And we're saying that either of these would make this last expression 0. Now, we know that we need to exclude one of them. We know that x cannot be equal to negative 2. So x equals negative 2 is an extraneous solution. It's not really a solution for-- it is a solution for this, once we got rid of the rational expressions. But it's not a solution for this original problem up here, because it would make the expressions undefined. It would cause you to divide by 0. So the only solution here is x is equal to 2. And you can check it yourself. If you do 2 squared, you get 4, over 2 plus 2 is 4. And that should be equal to 4 over 2 plus 2, over 4, which it definitely does. 1 is equal to 1." + }, + { + "Q": "At 2:33 Sal says it will make sense. I never really got past the 2 to the 0 power. I don't understand how he took two, put a zero above it, and then turned it into a one?! How does that work?", + "A": "What he showed helped me understand why it does that, so now let me try to explain it for you: He changed the way he did the exponents to multiplying 1 times how many numbers (the number that the exponent is) to that one. So when it is 2\u00e2\u0081\u00b0 = 1 because there aren t any 2s to multiply by. So something as big as 1,000,000\u00e2\u0081\u00b0 = 1 Now let s do it regular: When you have 4\u00e2\u0081\u00b6 = 1 \u00c3\u0097 4 \u00c3\u0097 4 \u00c3\u0097 4 \u00c3\u0097 4 \u00c3\u0097 4 \u00c3\u0097 4 = 4096 Well, that is a bit too big of a number so let s do 3\u00c2\u00b2 = 1 \u00c3\u0097 3 \u00c3\u0097 3 = 9 Ask me to clarify anything.", + "video_name": "dAvosUEUH6I", + "timestamps": [ + 153 + ], + "3min_transcript": "" + }, + { + "Q": "is there other basic rigid motions other than reflect,translate, and rotate?\n\nas said in 1:07", + "A": "Those three translations are the three basic geometric translations besides dilation.", + "video_name": "EDlZAyhWxhk", + "timestamps": [ + 67 + ], + "3min_transcript": "So we have another situation where we want to see whether these two figures are congruent. And the way we're going to test that is by trying to transform this figure by translating it, rotating it, and reflecting it. So the first thing that might-- let me translate it. So if these two things are congruent, looks like point E and the point that I'm touching right now, those would correspond to each other. Let me try to rotate it a little bit. So let me rotate-- whoops, I don't want to rotate there, I want to rotate around point E since I already have those on top of each other. So just like that. And it looks like now, if I reflect it across that line, I'm going to be-- oh no, I'm not there. You see, this is tricky. See, when I reflect it, this point, this point, this point, and this point seem to be in the exact same place, but point C does not correspond with that point right over there. So these two polygons are not congruent. And this is why it's important to do this, to make sure the rotations work out. So these two are not congruent. through translations, rotations, and reflections. So are these polygons congruent? No." + }, + { + "Q": "At 1:09 Sal says that he doesn't like using FOIL. FOIL is really easy and is much less confusing then the way Sal did the distributive property twice even though you get the same answer. Why does Sal not like FOIL?", + "A": "FOIL won t help you if you have to expand a product that isn t two binomials multiplied together; for example, two trinomials multiplied together. It s usually better to understand what you re doing instead of relying on mnemonics. For example: (a + b + c) * (d + e + f) = ad + ae + af + bd + be + bf + cd + ce + cf", + "video_name": "JKvmAexeMgY", + "timestamps": [ + 69 + ], + "3min_transcript": "f of x is equal to 7x minus 5. g of x is equal to x to the third power plus 4x. And then they ask us to find f times g of x So the first thing to realize is that this notation f times g of x is just referring to a function that is a product of f of x and g of x. So by definition, this notation just means f of x times g of x. And then we just have to substitute f of x with this definition, g of x with this definition, and then multiply out these algebraic expressions. f of x is right over there. And g of x, is right over there. So let's do it. So this is going to be equal to-- switch back to the orange color. It's going to be equal to f of x, which is 7x minus 5 times g of x, and g of x is x to the third power plus 4x. that each have two terms. You could use FOIL if you like. I don't like using FOIL because you might forget what it's even about. Foil is really just using the distributive property twice. So for example, you take this expression. Whatever you have out here, if you had a 9 out here, or an a, or an x, or anything. Now you have 7x minus 5. If you're multiplying it times this expression, you would multiply this times each term over here. So when you multiply 7x minus 5 times x to the third, you get-- I'll write it this way. You get x to the third times-- actually, let me write it the other way. You get 7x minus 5 times x to the third. And then you have plus 7x minus 5 times 4x. And now we can do the distributive property again. the things we distribute on the right hand side. It's the same exact idea. We could put the x to the third here as well. And when we distribute, you multiply x to the third times 7x and times negative 5. x to the third times 7x is 7x to the fourth power. X to the third times negative 5 is minus 5x to the third. And then you do it over here. You distribute the 4x over the 7x. 4x times 7x is plus 28x squared. 4x times negative 5 is minus 20x. And let's see if we can simplify this. We only have one fourth degree term, one third degree term, one second degree term, and one first degree term. Actually, we can't simplify this anymore. And we're done. This is the product of those two function definitions. This is f times g of x. It is a new function created by multiplying the other two functions." + }, + { + "Q": "At around 2:00-2:10, why is it when x=1 and when x=3? Why not something else?", + "A": "You can pick any x you like. 0 and 1 are frequently picked because 0*a=0 and 1*a=a. Since the multiplication is easy, you can solve the problem quicker.", + "video_name": "7QMoNY6FzvM", + "timestamps": [ + 120, + 130 + ], + "3min_transcript": "We're asked to graph the equation y is equal to negative 2 times x minus 2 squared plus 5. So let me get by scratch pad out so we could think about this. So y is equal to negative 2 times x minus 2 squared plus 5. So one thing, when you see a quadratic or a parabola graph expressed in this way, the thing that might jump out at you is that this term right over here is always going to be positive because it's some quantity squared. Or I should say, it's always going to be non-negative. It could be equal to 0. So it's always going to be some quantity squared. And then we're multiplying it by a negative. So this whole quantity right over here is going to be non positive. It's always going to be less than or equal to 0. So this thing is always less than or equal to 0, the maximum value that y will take on is when this thing actually does equal 0. So the maximum value for y is at 5. And when does that happen? Well, y hits 5 when this whole thing is 0. And when does this thing equal 0? Well, this whole thing equals 0 when x minus 2 is equal to 0. And x minus 2 is equal to 0 when x is equal to 2. So the point 2 comma 5 is the maximum point for this parabola. And it is actually going to be the vertex. So if we were to graph this, so the point 2 comma 5. So that's my y-axis. This is my x-axis. So this is 1, 2, 1, 2, 3, 4, 5. So this right here is the point 2 comma 5. This is a maximum point, it's a maximum point for this parabola. And now I want to find two more points so that I can really determine the parabola. Three points completely determine a parabola. So that's 1, the vertex, that's interesting. Now, what I'd like to do is just get two points that are equidistant from the vertex. what happens when x is equal to 1 and when x is equal to 3. So I could make a table here actually, let me do that. So I care about x being equal to 1, 2, and 3, and what the corresponding y is. We already know that when x is equal to 2, y is equal to 5. 2 comma 5 is our vertex. When x is equal to 1, 1 minus 2 is negative 1, squared is just 1. So this thing is going to be negative 2 plus 5, so it's going to be 3. And when x is equal to 3, this is 3 minus 2, which is 1 squared is 1 times negative 2 is negative 2 plus 5 is 3 as well. So we have three points. We have the point 1 comma 3, the point 2 comma 5, and the point 3 comma 3 for this parabola. So let me go back to the exercise and actually put those three points in. And so we have the point 1 comma 3, we have the point 2 comma 5," + }, + { + "Q": "1:47\nHow do you determine whether the parabola faces upward \"U\" or downward?\nSOS", + "A": "Clarifiction to the prior response you were given: You need to look at the coefficient of the X^2 terms or the coefficient in front of (X+b)^2. The coefficient of whatever is being squared determines the direction of the parabola. If the number is positive, then the parabola opens upward. If the coefficient is negative, the parabola opens downward.", + "video_name": "7QMoNY6FzvM", + "timestamps": [ + 107 + ], + "3min_transcript": "We're asked to graph the equation y is equal to negative 2 times x minus 2 squared plus 5. So let me get by scratch pad out so we could think about this. So y is equal to negative 2 times x minus 2 squared plus 5. So one thing, when you see a quadratic or a parabola graph expressed in this way, the thing that might jump out at you is that this term right over here is always going to be positive because it's some quantity squared. Or I should say, it's always going to be non-negative. It could be equal to 0. So it's always going to be some quantity squared. And then we're multiplying it by a negative. So this whole quantity right over here is going to be non positive. It's always going to be less than or equal to 0. So this thing is always less than or equal to 0, the maximum value that y will take on is when this thing actually does equal 0. So the maximum value for y is at 5. And when does that happen? Well, y hits 5 when this whole thing is 0. And when does this thing equal 0? Well, this whole thing equals 0 when x minus 2 is equal to 0. And x minus 2 is equal to 0 when x is equal to 2. So the point 2 comma 5 is the maximum point for this parabola. And it is actually going to be the vertex. So if we were to graph this, so the point 2 comma 5. So that's my y-axis. This is my x-axis. So this is 1, 2, 1, 2, 3, 4, 5. So this right here is the point 2 comma 5. This is a maximum point, it's a maximum point for this parabola. And now I want to find two more points so that I can really determine the parabola. Three points completely determine a parabola. So that's 1, the vertex, that's interesting. Now, what I'd like to do is just get two points that are equidistant from the vertex. what happens when x is equal to 1 and when x is equal to 3. So I could make a table here actually, let me do that. So I care about x being equal to 1, 2, and 3, and what the corresponding y is. We already know that when x is equal to 2, y is equal to 5. 2 comma 5 is our vertex. When x is equal to 1, 1 minus 2 is negative 1, squared is just 1. So this thing is going to be negative 2 plus 5, so it's going to be 3. And when x is equal to 3, this is 3 minus 2, which is 1 squared is 1 times negative 2 is negative 2 plus 5 is 3 as well. So we have three points. We have the point 1 comma 3, the point 2 comma 5, and the point 3 comma 3 for this parabola. So let me go back to the exercise and actually put those three points in. And so we have the point 1 comma 3, we have the point 2 comma 5," + }, + { + "Q": "At 5:00, doesn't the \"function\" Sal draws fail the vertical line test? Why does he still call it a function?", + "A": "That s called a piecewise function - they re defined like this: f(x) = { x + 1 for x > 1 x - 1 for x < 1 x for x = 1 }", + "video_name": "8VgmBe3ulb8", + "timestamps": [ + 300 + ], + "3min_transcript": "from the beginning-- because this is really the definition of an even function-- is when you look at this, you're like hey, what does this mean? f of x is equal to f of negative x. And all it does mean is this. It means that if I were to take f of 2, f of 2 is 4. So let me show you with a particular case. f of 2 is equal to f of negative 2. And this particular case for f of x is equal to x squared, they are both equal to 4. So really, this is just another way of saying that the function can be reflected, or the left side of the function is the reflection of the right side of the function across the vertical axis, across the y-axis. Now just to make sure we have a decent understanding here, let me draw a few more even functions. And I'm going to draw some fairly wacky things just so you really kind of learn to visually recognize them. So let's say a function like this, it does something like that. And then on this side, it does the same thing. It's the reflection, so it jumps up here, then it goes like this, then it goes like this. I'm trying to draw it so it's the mirror image of each other. This is an even function. You take what's going on on the right hand side of this function and you literally just reflect it over the y-axis, and you get the left hand side of the function. And you could see that even this holds. If I take some value-- let's say that this value right here is, I don't know, 3. And let's say that f of 3 over here is equal to, let's say, that that is 5. So this is 5. We see that f of negative 3 is also going to be equal to 5. And I can draw, let me just draw one more to really make sure. I'll do the axis in that same green color. Let me do one more like this. And you could have maybe some type of trigonometric looking function that looks like this, that looks like that. And it keeps going in either direction. So something like this would also be even. So all of these are even functions. Now, you are probably thinking, well, what is an odd function? And let me draw an odd function for you. So let me draw the axis once again. x-axis, y-axis, or the f of x-axis. And to show you an odd function, I'll give you a particular odd function, maybe the most famous of the odd functions. This is probably the most famous of the even functions. And it is f of x-- although there are probably" + }, + { + "Q": "at 4:07, when he draws the funny function graph, isn't it true that it isn't a function because it doesn't pass the vertical line test? It appears that the line would go through the open circle and the wing-shaped line on the left side.", + "A": "Perhaps he just didn t draw it clearly. I think he meant to draw an open circle as you suggested. If that was not his intention, then you are correct, it isn t a function.", + "video_name": "8VgmBe3ulb8", + "timestamps": [ + 247 + ], + "3min_transcript": "And this, hopefully, or maybe makes complete sense to you. You're like, well, Sal, obviously if I just reflect this function over the y-axis, that's going to be the case. Whatever function value I get at the positive value of a number, I'm going to get the same function value at the negative value. And this is what kind of leads us to the formal definition. If a function is even-- or I could say a function is even if and only of-- so it's even. And don't get confused between the term even function and the term even number. They're completely different kind of ideas. So there's not, at least an obvious connection that I know of, between even functions and even numbers or odd functions and odd numbers. So you're an even function if and only if, f of x is equal to f of negative x. from the beginning-- because this is really the definition of an even function-- is when you look at this, you're like hey, what does this mean? f of x is equal to f of negative x. And all it does mean is this. It means that if I were to take f of 2, f of 2 is 4. So let me show you with a particular case. f of 2 is equal to f of negative 2. And this particular case for f of x is equal to x squared, they are both equal to 4. So really, this is just another way of saying that the function can be reflected, or the left side of the function is the reflection of the right side of the function across the vertical axis, across the y-axis. Now just to make sure we have a decent understanding here, let me draw a few more even functions. And I'm going to draw some fairly wacky things just so you really kind of learn to visually recognize them. So let's say a function like this, it does something like that. And then on this side, it does the same thing. It's the reflection, so it jumps up here, then it goes like this, then it goes like this. I'm trying to draw it so it's the mirror image of each other. This is an even function. You take what's going on on the right hand side of this function and you literally just reflect it over the y-axis, and you get the left hand side of the function. And you could see that even this holds. If I take some value-- let's say that this value right here is, I don't know, 3. And let's say that f of 3 over here is equal to, let's say, that that is 5. So this is 5. We see that f of negative 3 is also going to be equal to 5." + }, + { + "Q": "AT 9:39, why is there a gap between the reflection about the y-axis and the reflection about the x-axis?\nAt 6:58, there was no gap.", + "A": "It s just a different equation he s graphing (I m not sure what the equation is; if anyone knows I d love to hear). It does look rather strange as a function, but it s still classified as an odd function.", + "video_name": "8VgmBe3ulb8", + "timestamps": [ + 579, + 418 + ], + "3min_transcript": "we figured out f of 2 is 8. 2 to the third power is 8. We know that f of negative 2 is negative 8. Negative 2 to the third power is negative 8. So you have the negative of negative 8, negatives cancel out, and it works out. So in general, you have an odd function. So here's the definition. You are dealing with an odd function if and only if f of x for all the x's that are defined on that function, or for which that function is defined, if f of x is equal to the negative of f of negative x. Or you'll sometimes see it the other way if you multiply both sides of this equation by negative 1, you would get negative f of x is equal to f of negative x. And sometimes you'll see it where it's swapped around where they'll say f of negative x is equal to-- let me write that careful-- I just swapped these two sides. So let me just draw you some more odd functions. So I'll do these visually. So let me draw that a little bit cleaner. So if you have maybe the function does something wacky like this on the right hand side. If it was even, you would reflect it there. But we want to have an odd function, so we're going to reflect it again. So the rest of the function is going to look like this. So what I've drawn in the non-dotted lines, this right here is an odd function. And you could even look at the definition. If you take some value, a, and then you take f of a, which would put you up here. This right here would be f of a. If you take the negative value of that, if you took negative a here, f of negative a is going to be down here. the same distance from the horizontal axis. It's not completely clear the way I drew it just now. So it's maybe going to be like right over here. So this right over here is going to be f of negative a, which is the same distance from the origin as f of a, it's just the negative. I didn't completely draw it to scale. Let me draw one more of these odd functions. I think you might get the point. Actually, I'll draw a very simple odd function, just to show you that it doesn't always have to be something crazy. So a very simple odd function would be y is equal to x, something like this. Whoops. y is equal going through the origin. You reflect what's on the right onto to the left. You get that. And then you reflect it down, you get all of this stuff in the third quadrant. So this is also an odd function. Now, I want to leave you with a few things that are not odd functions and that sometimes might" + }, + { + "Q": "at 0:10, he says rational #s are different from natural numbers. What exactly ARE rational numbers?", + "A": "natural numbers are the numbers you learned for counting: 1, 2, 3 ... whole numbers are the numbers of apples you might have: 0, 1, 2, 3 ... integers are positive or negative numbers that are not fractions: ... -3, -2, -1, 0, 1, 2, 3 ... rational numbers are any number that can be expressed as a fraction of integers.", + "video_name": "i1i2_9wg6N8", + "timestamps": [ + 10 + ], + "3min_transcript": "- [Voiceover] What I'd like to do in this video is order these six numbers from least to greatest. So the least of them being on the left hand side, and the greatest on the right. And I encourage you to pause this video, and see if you can do it on your own, before we work through it together. So assuming you've had a go at it, so let's do it together, and to help us there, let's plot these numbers on a number line, and I have a number line up here, so there you go, there is a handy number line. And let's just take them one by one. So the first number here, we have 7/3. So let's see if we can express that in a different way, if we can write that as a mixed number. So 7/3, how many wholes are here? And the whole is going to be 3/3. So this is going to be 3/3, plus another 3/3, is going to get us to 6/3. And so you're going to have one more third left. So this is 7/3. Three plus three, plus one is seven. So this is 3/3 is one whole, three thirds is one whole, so this is two So seven thirds, same thing as one, two, and you see, between consecutive integers, we have three spaces, so we are essentially marking off thirds, so two and 1/3, is going to be 1/3 of the way between two and three, so it's going to be right over there, so that is 7/3. Then we have negative 5/2, so same logic. Negative, let me do that in that green color. So, I can do it over here. Negative five over two, well that's the same thing as the negative of 5/2, and 5/2 is going to be 2/2 plus another 2/2, plus 1/2 so this is two and 1/2, this is one, this is one, and that's 1/2. So this is going to be one, plus one, plus 1/2, two and 1/2, we have our negative out there, so it's negative two and 1/2. and the negative two and 1/2 is going to be halfway between negative two, and negative three, so it's going to be right over there. So that is negative 5/2. Then we have zero, not too difficult. It's actually labeled on our number line for us. Then we have negative two. Negative two, once again, on our number line for us. Two, two steps, two whole numbers to the left of zero. So negative two is going to put us right over there. Then we have negative 12/4 so it might jump out at you immediately, 12 divided by four is three, so this is going to be the same thing as negative three. So this is negative 12/4 or if you want to use the type of logic that we used in these first two numbers, you could say negative 12/4 is the same thing as the negative of 12/4 which is 4/4 plus another 4/4" + }, + { + "Q": "At 0:21, what does Sal mean by \" arbrutary\"?", + "A": "Any old An arbitrary angle measure is a random angle or just any old angle", + "video_name": "0gzSreH8nUI", + "timestamps": [ + 21 + ], + "3min_transcript": "Thought I would do some more example problems involving triangles. And so this first one, it says the measure of the largest angle in a triangle is 4 times the measure of the second largest angle. The smallest angle is 10 degrees. What are the measures of all the angles? Well, we know one of them. We know it's 10 degrees. Let's draw an arbitrary triangle right over here. So let's say that is our triangle. We know that the smallest angle is going to be 10 degrees. And I'll just say, let's just assume that this right over here is the measure of the smallest angle. It's 10 degrees. Now let's call the second largest angle-- let's call that x. So the second largest angle, let's call that x. So this is going to be x. And then the first sentence, they say the measure of the largest angle in a triangle is 4 times the measure of the second largest angle. So the second largest angle is x. 4 times that measure is going to be 4x. So the largest angle is going to be 4x. of the angles inside of a triangle is that they add up to 180 degrees. So we know that 4x plus x plus 10 degrees is going to be equal to 180 degrees. It's going to be equal to 180. And 4x plus x, that just gives us 5x. And then we have 5x plus 10 is equal to 180 degrees. Subtract 10 from both sides. You get 5x is equal to 170. And so x is equal to 170/5. And let's see, it'll go into it-- what is that, 34 times? Let me verify this. So 5 goes into-- yeah, it should be 34 times because it's going to go into it twice as many times as 10 would go into it. 10 would go into 170 17 times. 5 would go into 170 34 times. So we could verify it. Go into 170. 5 goes into 17 three times. 3 times 5 is 15. Bring down the 0. 5 goes into 20 four times, and then you're not going to have a remainder. 4 times 5 is 20. No remainder. So it's 34 times. So x is equal to 34. So the second largest angle has a measure of 34 degrees. This angle up here is going to be 4 times that. So 4 times 34-- let's see, that's going to be 120 degrees plus 16 degrees. This is going to be 136 degrees. Is that right? 4 times 4 is 16, 4 times 3 is 120, 16 plus 120 is 136 degrees. So we're done. The three measures, or the sizes of the three angles, are 10 degrees, 34 degrees, and 136 degrees. Let's do another one. So let's see. We have a little bit of a drawing here. And what I want to do is-- and we could think about different things. We could say, let's solve for x. I'm assuming that 4x is the measure of this angle. 2x is the measure of that angle right over there." + }, + { + "Q": "Would 3 quadrupled by shown as a radical with a little four in the \"notch\", as a cube root is described at 4:32?", + "A": "4th roots would have a little 4 in the notch of the radical symbol. 5th roots would have a 5 in that position. 6th roots would have a 6 in that position. See the pattern?", + "video_name": "87_qIofPwhg", + "timestamps": [ + 272 + ], + "3min_transcript": "Well the volume is going to be two, times two, times two, which is two to the third power or two cubed. This is two cubed. That's why they use the word cubed because this would be the volume of a cube where each of its sides have length two and this of course is going to be equal to eight. But what if we went the other way around? What if we started with the cube? What if we started with this volume? What if we started with a cube's volume and let's say the volume here is eight cubic units, so volume is equal to eight and we wanted to find the lengths of the sides. So we wanted to figure out what X is cause that's X, that's X, and that's X. It's a cube so all the dimensions have the same length. Well there's two ways that we could express this. We could say that X times X times X or we could use the cube root symbol, which is a radical with a little three in the right place. Or we could write that X is equal to, it's going to look very similar to the square root. This would be the square root of eight, but to make it clear, they were talking about the cube root of eight, we would write a little three over there. In theory for square root, you could put a little two over here, but that'd be redundant. If there's no number here, people just assume that it's the square root. But if you're figuring out the cube root or sometimes you say the third root, well then you have to say, well you have to put this little three right over here in this little notch in the radical symbol right over here. And so this is saying X is going to be some number that if I cube it, I get eight. So with that out of the way, let's do some examples. Let's say that I have... Let's say that I want to calculate the cube root of 27. Well if say that this is going to be equal to X, this is equivalent to saying that X to the third or that 27 is equal to X to the third power. So what is X going to be? Well X times X times X is equal to 27, well the number I can think of is three, so we would say that X, let me scroll down a little bit, X is equal to three. Now let me ask you a question. Can we write something like... Can we pick a new color? The cube root of, let me write negative 64. I already talked about that if we're talking the square root, it's fairly typical that hey you put a negative number in there at least until we learn about imaginary numbers, we don't know what to do with it. But can we do something with this? Well if I cube something, can I get a negative number? Sure. So if I say this is equal to X," + }, + { + "Q": "At 5:02 Sal say's that he is going to calculate the CUBE root of 27.\nI didn't understasnd how it is calculated?", + "A": "So what Sal is saying is he is breaking it down into 3 * 3 = 9 9 * 3 = 27 So, 3 * 3 * 3 = 27. Therefore, 3 is the cube root of 27. Just like cube root of y = x * x * x", + "video_name": "87_qIofPwhg", + "timestamps": [ + 302 + ], + "3min_transcript": "Well the volume is going to be two, times two, times two, which is two to the third power or two cubed. This is two cubed. That's why they use the word cubed because this would be the volume of a cube where each of its sides have length two and this of course is going to be equal to eight. But what if we went the other way around? What if we started with the cube? What if we started with this volume? What if we started with a cube's volume and let's say the volume here is eight cubic units, so volume is equal to eight and we wanted to find the lengths of the sides. So we wanted to figure out what X is cause that's X, that's X, and that's X. It's a cube so all the dimensions have the same length. Well there's two ways that we could express this. We could say that X times X times X or we could use the cube root symbol, which is a radical with a little three in the right place. Or we could write that X is equal to, it's going to look very similar to the square root. This would be the square root of eight, but to make it clear, they were talking about the cube root of eight, we would write a little three over there. In theory for square root, you could put a little two over here, but that'd be redundant. If there's no number here, people just assume that it's the square root. But if you're figuring out the cube root or sometimes you say the third root, well then you have to say, well you have to put this little three right over here in this little notch in the radical symbol right over here. And so this is saying X is going to be some number that if I cube it, I get eight. So with that out of the way, let's do some examples. Let's say that I have... Let's say that I want to calculate the cube root of 27. Well if say that this is going to be equal to X, this is equivalent to saying that X to the third or that 27 is equal to X to the third power. So what is X going to be? Well X times X times X is equal to 27, well the number I can think of is three, so we would say that X, let me scroll down a little bit, X is equal to three. Now let me ask you a question. Can we write something like... Can we pick a new color? The cube root of, let me write negative 64. I already talked about that if we're talking the square root, it's fairly typical that hey you put a negative number in there at least until we learn about imaginary numbers, we don't know what to do with it. But can we do something with this? Well if I cube something, can I get a negative number? Sure. So if I say this is equal to X," + }, + { + "Q": "at 0:31 why when you square 49 the result is 49?", + "A": "He is squaring the result of \u00e2\u0088\u009a49, which is 7. 7 squared is 49, so (\u00e2\u0088\u009a49)^2 = 49.", + "video_name": "87_qIofPwhg", + "timestamps": [ + 31 + ], + "3min_transcript": "- [Voiceover] We already know a little bit about square roots. For example, if I were to tell you that seven squared is equal to 49, that's equivalent to saying that seven is equal to the square root of 49. The square root essentially unwinds taking the square of something. In fact, we could write it like this. We could write the square root of 49, so this is whatever number times itself is equal to 49. If I multiply that number times itself, if I square it, well I'm going to get 49. And that's going to be true for any number, not just 49. If I write the square root of X and if I were to square it, that's going to be equal to X and that's going to be true for any X for which we can evaluate the square root, evaluate the principle root. Now typically and as you advance in math you're going to see that this will change, but typically you say, okay if I'm going to take X has to be non-negative. This is going to change once we start thinking about imaginary and complex numbers, but typically for the principle square root, we assume that whatever's under the radical, whatever's under here, is going to be non-negative because it's hard to square a number at least the numbers that we know about, it's hard to square them and get a negative number. So for this thing to be defined, for it to make sense, it's typical to say that, okay we need to put a non-negative number in here. But anyway, the focus of this video is not on the square root, it's really just to review things so we can start thinking about the cube root. And as you can imagine, where does the whole notion of taking a square of something or a square root come from? Well it comes from the notion of finding the area of a square. If I have a square like this and if this side is seven, well if it's a square, all the sides are going to be seven. it would be seven times seven or seven squared. That would be the area of this. Or if I were to say, well what is if I have a square, if I have, and that doesn't look like a perfect square, but you get the idea, all the sides are the same length. If I have a square with area X. If the area here is X, what are the lengths of the sides going to be? Well it's going to be square root of X. All of the sides are going to be the square root of X, so it's going to be the square root of X by the square root of X and this side is going to be the square root of X as well and that's going to be the square root of X as well. So that's where the term square root comes from, where the square comes from. Now what do you think cube root? Well same idea. If I have a cube. If I have a cube. Let me do my best attempt at drawing a cube really fast. If I have a cube and a cube, all of it's dimensions have the same length so this is a two, by two, by two cube," + }, + { + "Q": "At 0:58, what are Lucas numbers??", + "A": "Lucas numbers are defined in pretty much the same way as the Fibonacci numbers except that the first Lucas number is 2 instead of 1. So, for the Lucas sequence, you get: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, ....", + "video_name": "14-NdQwKz9w", + "timestamps": [ + 58 + ], + "3min_transcript": "So say you're me and you're in math class and you're trying to ignore the teacher and doodle Fibonacci spirals, while simultaneously trying to fend off the local greenery. Only, you become interested in something that the teacher says by accident, and so you draw too many squares to start with. So you cross some out, but cross out too many, and then the teacher gets back on track, and the moment is over. So oh well, might as well try and just do the spiral from here. So you make a three by three square. And here's a four by four, and then seven, and then 11 This works because then you've got a spiral of squares. So you write down the numbers, 1, 3, 4, 7, 11, 18. It's kind of like the Fibonacci series, because 1 plus 3 is 4, 3 plus 4 is 7, and so on. Or maybe it starts at 2 plus 1, or negative 1 plus 2. Either way, it's a perfectly good series. And it's got another similarity with the Fibonacci series. The ratios of consecutive numbers also approach Phi. So, a lot of plants have Fibonacci numbers of spirals, but to understand how they do it, we can learn from the exceptions. This pine cone that has seven spirals one way, and eleven the other, might be showing Lucas numbers. And since Fibonacci numbers and Lucas numbers are related, maybe that explains it. by always growing new parts a Phi-th of a circle around. What angle would give Lucas numbers? In this pine cone, each new pine cone-y thing is about 100 degrees around from the last. We're going to need a Lucas angle-a-tron. It's easy to get a 90 degree angle-a-tron, and if I take a third of a third of that, that's a ninth of 90, which is another 10 degrees. Now you can use it to get spiral patterns like what you see on the Lucas number plants. It's an easy way to grow Lucas spirals, if plants have an internal angle-a-tron. Thing is, 100 is pretty far from 137.5. If plants were somehow measuring angle, you'd think the anomalous ones would show angles close to a Phi-th of a circle. Not jump all the way to 100. Maybe I'd believe different species use different angles, but two pine cones from the same tree? Two spirals on the same cauliflower? And that's not the only exception. A lot of plants don't grow spirally at all. Like this thing. With leaves growing opposite from each other. And some plants have alternating leaves, 180 degrees from each other, which is far from both Phi and Lucas angles. have a fundamentally different growth pattern, and are in a different class of plant or something, but wouldn't it be even better if there were one simple reason for all of these things? These variations are a good clue that maybe these plants get this angle, and Fibonacci numbers, as a consequence of some other process and not just because it mathematically optimizes sunlight exposure if the sun is right overhead, which it pretty much never is, and if the plan are perfectly facing straight up, which they aren't. So how do they do it? Well, you could try observing them. That would be like science. If you zoom in on the tip of a plant, the growing part, there's this part called the meristem. That's where new plant bits form. The biggest plant bits where the first to form off the meristem, and the little ones around the center are newer. As the plants grow, they get pushed away from the meristem, but they all started there. The important part is that a science observer would see the plant bits pushing away, not just from the meristem, but from each other. A couple physicists once tried this thing where they dropped drops of a magnetized liquid in a dish of oil." + }, + { + "Q": "The age of vinod and sanjeev are in the ratio of 5:7. Ten year later the ratio of their age will be 7:9. Find the present age.", + "A": "Let v be vinod age now and s be sajeev age now. 5c= 7v so c = 7/5 v 10 years from now, 7(c+10) = 9(v+10) so 7c + 70 = 9v + 90 or 7c = 9v +20 Substitute into c, so 7(7/5v) = 9v + 20 multiply whole equation by 5 49v = 45v + 100 subtract 45v 4v = 100, v = 25 and c = 7/5(25) = 35 Ratio of 25:35 reduces to 5:7 Ten years, v will be 35 and c will be 45 so ratio of 35:45 or 7:9", + "video_name": "W-5liMGKgHA", + "timestamps": [ + 307, + 429 + ], + "3min_transcript": "And on the right hand side, I can distribute this 3. So 3 times 2 is 6. 3 times y is 3y. 6 plus 3y. And then it's always nice to get all of our constants on one side of the equation, all of our variables on the other side of the equation. So we have a 3y over here. We have more y's on the right hand side than the left hand So let's get rid of the y's on the left hand side. You could do it either way, but you'd end up with negative numbers. So let's subtract a y from each side. And we are left with, on the left hand side, 18. And on the right hand side you have 6 plus 3 y's. Take away one of those y's. You're going to be left with 2 y's. Now we can get rid of the constant term here. So we will subtract 6 from both sides. 18 minus 6 is 12. The whole reason why we subtracted 6 from the right was to get rid of this, 6 minus 6 is 0, so you have 12 Two times the number of years it will take is 12, and you could probably solve this in your head. But if we just want a one-coefficient year, we would divide by 2 on the right. Whatever we do to one side of an equation, we have to do it on the other side. Otherwise, the equation will not still be an equation. So we're left with y is equal to 6, or y is equal to 6. So going back to the question, how many years will it take for Arman to be three times as old as Diya? Well, it's going to take six years. Now, I want you to verify this. Think about it. Is this actually true? Well, in six years, how old is Arman going to be? He's going to be 18 plus 6. We now know that this thing is 6. So in six years, Arman is going to be 18 plus 6, which is 24 years old. How old is Diya going to be? Well, she's going to be 2 plus 6, which is 8 years old. as 8. In 6 years-- Arman is 24, Diya is 8-- Arman is three times as old as Diya, and we are done." + }, + { + "Q": "Okay, quite possibly my concept of integration is wrong but shouldn't increasing the number of trapeziums bring the result closer to the value obtained by integrating f(x)? in this case integrating it brings the value to 7.4, which is close enough to the value obtained at 8:26 , but if i increase the no. of trapeziums to 10, the value of delx becomes 0.5, so the total becomes 3.26. it becomes less accurate.", + "A": "If you increase the number of trapeziums to 10, then \u00ce\u0094x = 0.5, and you got that the area is equal to: A = 0.5/2 [ f(1) + 2f(1.5) + 2f(2) + 2f(2.5) + 2f(3) + 2f(3.5) + 2f(4) + 2f(4.5) + 2f(5) + 2f(5.5) + f(6) ] A = 0.25 [ 0 + 2\u00e2\u0088\u009a0.5 + 2\u00e2\u0088\u009a1 + 2\u00e2\u0088\u009a1.5 + 2\u00e2\u0088\u009a2 + 2\u00e2\u0088\u009a2.5 + 2\u00e2\u0088\u009a3 + 2\u00e2\u0088\u009a3.5 + 2\u00e2\u0088\u009a4 + 2\u00e2\u0088\u009a4.5 + \u00e2\u0088\u009a5 ] A \u00e2\u0089\u0085 7.3847 Which is a much closer result.", + "video_name": "1p0NHR5w0Lc", + "timestamps": [ + 506 + ], + "3min_transcript": "f of 1, let's just remind ourselves what our original function was. Our original function was the square root of x minus 1. So f of 1 is the square root of 1 minus 1, so that is just going to be 0. This expression right over here is going to be 2 times the square root of 2 minus 1. The square root of 2 minus 1 is just 1, so this is just going to be 2. Actually, let me do it in that same-- well, I'm now using the purple for a different purpose than just the first trapezoid. Hopefully, you realized that. I was just sticking with that pen color. Then f of 3. 3 minus 1 is 2-- square root of 2. So the function evaluated at 3 is the square root of 2. So this is going to be 2 times the square root of 2. Then the function evaluated at 4. When you evaluate it at 4, you get the square root of 3. So this is going to be 2 times the square root of 3. And then you get 2 times the square root of 4-- 5 minus 1 2 times the square root of 4 is just four. And then finally, you get f of 6 is square root of 6 minus 1, is the square root of 5. And I think we're now ready to evaluate. So let me get my handy TI-85 out and calculate this. So it's going to be-- well I'm just going to calculate-- well, I'll just multiply. So 0.5 times open parentheses-- well, it's a 0. I'll just write it, just so you know what I'm doing. 0 plus 2 plus-- whoops. Lost my calculator. Plus 2 times the square root of 2 plus 2 times the square root of 3 plus 4-- I'm almost done-- plus the square root of 5-- so let me write that-- gives me-- and I'll just round it-- 7.26. So the area is approximately equal to 7.26 under the curve y is equal to the square root of x minus 1 between x equals 1 and x equals 6. And we did this using trapezoids." + }, + { + "Q": "On the third example, at about 7:25, Sal says that the length marked 2 and the segment parallel to it are equal. He says both are length 2. He says \"We know that these are both length 2 [because] these are all 90 degree angles..\" -- I don't understand how that lets us assume they are equal.", + "A": "Actually, you are right. He can t assume they are equal. BUT, the reason they wouldn t be equal is because the length of either the vertical purple side or the white side (or both) would have been changed. In that case, though, the green side would have changed exactly in the same amount that the other two (white an vertical purple) would have changed, except that if they got shorter, it would have gotten longer, and vice versa.", + "video_name": "vWXMDIazHjA", + "timestamps": [ + 445 + ], + "3min_transcript": "Lets do one more. So here I have a bizarre looking, a bizarre looking shape, and we need to figure out its perimeter. And it it first seems very daunting because they have only given us this side and this side and they have only given us this side right over here. And one thing that we are allowed to assume in this and you don't always have to make you can't always make that assumption and I just didn't draw it here I had time because it would had really crowded out this this diagram. Is it all of the angles in this diagrams are right angles,so i could have drawn a right angle here a right angle here, a right angle there, right angle there, but as you can see it kind of makes things a little bit, it makes things a little bit messy. But how do we figure out the perimeter if we don't know these little distances, if we don't know these little distances here. And the secret here is to kind of shift the sides because all we want to care about is the sum of the sides of the sides. So what I will do is a little exercise in shifting the sides. So this side over here I am going to shift I am going to shift and put it right over there. Then let me keep using different colors, and then this side right over here I am going to shift it and put it right up here. Then finally Iam going to have this side right over here, I can shift it and put it right over there and I think you see what is going on right now. Now all of these sides combined are going to be the same as this side kind of building, even you know this thing was not a rectangle,its its perimeter is going to be a little bit interesting. All we have to think about is this 2 right over here, now lets think about all of these sides that is going up and down. So this side i can shift it all the way to the right and go right over here. Let me make it clear all inside goes all the way to the end, right that it is the exact same all insde. Now this white side I can shift all the way to the right over there, then this green side I can shift right over there and then I have, and then I can shift, and then i can shift this. so I have not, I have not done anything yet, let me be clear I have not done anything yet with that and that I have not shift them over and let me take this side right over here and shift it over. So let me take this entire thing and shift over there and shift it over there. So before I count these two pieces right over here and we know that each have length 2 this 90 degrees angle, so this has link to and this has link to. Before I count these two pieces, I shifted everything else so I was able to form a rectangle. So at least counting everything else I have 7 plus 6, so lets see 7 plus 6 all of these combined are also going to be 7, plus 7, and all of these characters combined are all also going to be 6, plus 6, and then finally I have this 2, right here that I have not counted before, this 2, plus this 2, plus this 2. And then we have our perimeter, so what is this giving us," + }, + { + "Q": "At 3:02 when Sal mentions altitude, what does he mean exactly?", + "A": "Altitude is a geographic term used to describe the height of land.", + "video_name": "vWXMDIazHjA", + "timestamps": [ + 182 + ], + "3min_transcript": "it is going to be equal to the perimeter of the 5 triangles is equal to perimeter of 5 outer triangles. Just call them 5 triangles like this minus their basis, right, if i take the perimeter of all of these sides If i added up the part that should not be part of the perimeter of the star would be this part,that part, that part,that part, that part and that part. those are not the part, those are not the part of the perimeter of the star so should be the perimeter of the 5 triangles minus the links of their bases links of their 5 bases. So what is the perimeter of the 5 triangles? well, the perimeter of each of them is 30, perimeter of 5 of them is going to be 5 times 30 which is 150, now we want to subtract out the links of their 5 bases right over here. So this inner pentagon has a perimeter 50, that is the sum of the 5 bases. So that right over here is 50, so the perimeter of the star is going to be 150 minus 50, or or 100. All we need is to get the perimeter of all triangles, subtracted out these bases which was the perimeter of the inner pentagon and we are done. Now lets do the next problem. What is the area of this this quadrilateral, something that has 4 sides of ABCD? And this is a little bit we have not seen a figure quite like this just yet, it on the right hand side looks like a rectangle, and on the left hand side looks like a triangle and this is actually trapezoid, but we can actually as you could imagine the way we figure out the area of several triangles splitting it up into pieces we can recognise. And the most obvious thing to do here is started A and just drop a rock at 90 degrees and we could call this point E. And what is interesting here is we can split this up into something we recognize a rectangle and a right triangle. But you might say how do, how do we figure out what these you know we have this side and that side, so we can figure out the area of this rectangle pretty straight forwardly. But how would we, how would we figure out the area of this triangle? Well if this side is 6 then that means that this that EC is also going to be 6. If AB is 6, notice we have a rectangle right over her, opposite side of a rectangle are equal. So if AB equals 6, implies that EC is equal to 6, EC is equal to 6, so EC is equal to 6 and if EC is equal to 6 then that tells us that DE is going to be 3. DE is going to be 3, this distance right over here is going to be 3." + }, + { + "Q": "Sal @ 18:00 c2= 1/3 (x2 - 2x1) you forgot the 2 from the equation above. love you Sal. you are the greatest.", + "A": "Sal corrected this error at the end of the video.", + "video_name": "Qm_OS-8COwU", + "timestamps": [ + 1080 + ], + "3min_transcript": "So we have c1 times this vector plus c2 times the b vector 0, 3 should be able to be equal to my x vector, should be able to be equal to my x1 and x2, where these are just arbitrary. So let's see if I can set that to be true. So if this is true, then the following must be true. c1 times 1 plus 0 times c2 must be equal to x1. We just get that from our definition of multiplying vectors times scalars and adding vectors. And then we also know that 2 times c2-- sorry. c1 times 2 plus c2 times 3, 3c2, should be equal to x2. Now, if I can show you that I can always find c1's and c2's any point in R2 using just these two vectors. So let me see if I can do that. So this is just a system of two unknowns. This is just 0. We can ignore it. So let's multiply this equation up here by minus 2 and put it here. So we get minus 2, c1-- I'm just multiplying this times minus 2. We get a 0 here, plus 0 is equal to minus 2x1. And then you add these two. You get 3c2, right? These cancel out. You get 3-- let me write it in a different color. You get 3c2 is equal to x2 minus 2x1. Or divide both sides by 3, you get c2 is equal to 1/3 x2 minus x1. Now we'd have to go substitute back in for c1. this first equation that says c1 plus 0 is equal to x1, so c1 is equal to x1. So that one just gets us there. So c1 is equal to x1. So you give me any point in R2-- these are just two real numbers-- and I can just perform this operation, and I'll tell you what weights to apply to a and b to get to If you say, OK, what combination of a and b can get me to the point-- let's say I want to get to the point-- let me go back up here. Oh, it's way up there. Let's say I'm looking to get to the point 2, 2. So x1 is 2. Let me write it down here. Say I'm trying to get to the point the vector 2, 2. What combinations of a and b can be there? Well, I know that c1 is equal to x1, so that's equal to 2, and c2 is equal to 1/3 times 2 minus 2." + }, + { + "Q": "At 2:21, I noticed the following: 3a+ -2c. This is a bit of a silly question but when I was in middle school we were taught to write the above as 3a+(-2c) as a parenthesis was thought to be needed between the +,- signs. Is the way I was taught to write it wrong/redundant?", + "A": "Either putting parenthesis around the negative number or writing it in the form of 3a-2c is preferred for clarity. In school students are often taught to only do one operation at a time in each math step. However you may choose to combine multiple operations in a single step for speed purposes. Not following these type of conventions just increases the chance of confusion and mistakes but does not invalidate the math. To me it is similar to the difference between formal and informal writing.", + "video_name": "Qm_OS-8COwU", + "timestamps": [ + 141 + ], + "3min_transcript": "One term you are going to hear a lot of in these videos, and in linear algebra in general, is the idea of a linear combination. And all a linear combination of vectors are, they're just a Let me show you what that means. So let's say I have a couple of vectors, v1, v2, and it goes all the way to vn. And they're all in, you know, it can be in R2 or Rn. Let's say that they're all in Rn. They're in some dimension of real space, I guess you could call it, but the idea is fairly simple. A linear combination of these vectors means you just add up the vectors. It's some combination of a sum of the vectors, so v1 plus v2 plus all the way to vn, but you scale them by arbitrary constants. So you scale them by c1, c2, all the way to cn, where member of the real numbers. That's all a linear combination is. Let me show you a concrete example of linear Let me make the vector. Let me define the vector a to be equal to-- and these are all bolded. These purple, these are all bolded, just because those are vectors, but sometimes it's kind of onerous to keep bolding things. So let's just say I define the vector a to be equal to 1, 2. And I define the vector b to be equal to 0, 3. What is the linear combination of a and b? Well, it could be any constant times a plus any constant times b. So it could be 0 times a plus-- well, it could be 0 times a plus 0 times b, which, of course, would be what? That would be 0 times 0, that would be 0, 0. That would be the 0 vector, but this is a completely valid linear combination. big bold 0 like that. I could do 3 times a. I'm just picking these numbers at random. 3 times a plus-- let me do a negative number just for fun. So I'm going to do plus minus 2 times b. What is that equal to? Let's figure it out. Let me write it out. It's 3 minus 2 times 0, so minus 0, and it's 3 times 2 is 6. 6 minus 2 times 3, so minus 6, so it's the vector 3, 0. This is a linear combination of a and b. I can keep putting in a bunch of random real numbers here and here, and I'll just get a bunch of different linear combinations of my vectors a and b. If I had a third vector here, if I had vector c, and maybe" + }, + { + "Q": "At 3:37, he says that (if it's continuous) Psi xy = Psi yx . Is there a video where he proves that?", + "A": "It s proven in all Calc III texts, but most likely in the section that he talks about partial derivatives, it s proven.", + "video_name": "a7wYAtMjORQ", + "timestamps": [ + 217 + ], + "3min_transcript": "So that's one situation to consider. What happens when we take the partial, with respect to x, and then y? So with respect to x, you hold y constant to get just the partial, with respect to x. Ignore the y there. And then you hold the x constant, and you take the partial, with respect to y. So what's the difference between that and if we were to switch the order? So what happens if we were to-- I'll do it in a different color-- if we had psi, and we were to take the partial, with respect to y, first, and then we were to take the partial, with respect to x? So just the notation, just so you're comfortable with it, that would be-- so partial x, partial y. And this is the operator. And it might be a little confusing that here, between these two notations, even though they're the same thing, the order is mixed. That's just because it's just a different way of This says, OK, partial first, with respect to x, then y. This views it more as the operator, so we took the partial of x first, and then we took y, like you're But anyway, so this can also be written as the partial of y, with respect to x-- sorry, the partial of y, and then we took the partial of that with respect to x. Now, I'm going to tell you right now, that if each of the first partials are continuous-- and most of the functions we've dealt with in a normal domain, as long as there aren't any discontinuities, or holes, or something strange in the function definition, they usually are continuous. And especially in a first-year calculus or differential course, we're probably going to be dealing with continuous functions in soon. our domain. If both of these functions are continuous, if both of the first partials are continuous, then these two are going to be equal to each other. So psi of xy is going to be equal to psi of yx. Now, we can use this knowledge, which is the chain knowledge to now solve a certain class of differential equations, first order differential equations, called exact equations. And what does an exact equation look like? An exact equation looks like this. The color picking's the hard part. So let's say this is my differential equation. I have some function of x and y. So I don't know, it could be x squared times cosine of y or something. I don't know, it could be any function of x and y. Plus some function of x and y, we'll call that n, times dy, dx is equal to 0. This is-- well, I don't know if it's an exact equation yet, but if you saw something of this form, your first impulse should be, oh-- well, actually, your very first impulse is, is this separable? And you should try to play around with the algebra a little bit to see if it's separable, because that's always the most straightforward way. If it's not separable, but you can still put it in this form," + }, + { + "Q": "At 3:00, where are the sample sets s1, s2, etc coming from. Sal is just picking out numbers without describing the process.", + "A": "The sets s1, s2 and s3 are outcomes that could be obtained by throwing the unfair dice defined by the distribution in yellow, but they are just examples and could be totally different.", + "video_name": "JNm3M9cqWyc", + "timestamps": [ + 180 + ], + "3min_transcript": "And let's say it's very likely to get a 6 like that. So that's my probability distribution function. If I were to draw a mean-- this the symmetric, so maybe the mean would be something like that. The mean would be halfway. So that would be my mean right there. The standard deviation maybe would look-- it would be that far and that far above and below the mean. But that's my discrete probability distribution function. Now what I'm going to do here, instead of just taking samples of this random variable that's described by this probability distribution function, I'm going to take samples of it. But I'm going to average the samples and then look at those samples and see the frequency of the averages that I get. And when I say average, I mean the mean. Let me define something. Let's say my sample size-- and I could put any number here. But let's say first off we try a sample size of n is equal to 4. And what that means is I'm going to take four samples from this. So let's say the first time I take four samples-- Let's say I get another 1. And let's say I get a 3. And I get a 6. So that right there is my first sample of sample size 4. I know the terminology can get confusing. Because this is the sample that's made up of four samples. But then when we talk about the sample mean and the sampling distribution of the sample mean, which we're going to talk more and more about over the next few videos, normally the sample refers to the set of samples from your distribution. And the sample size tells you how many you actually took from your distribution. But the terminology can be very confusing, because you could easily view one of these as a sample. But we're taking four samples from here. We have a sample size of four. And what I'm going to do is I'm going to average them. So let's say the mean-- I want to be very careful when I say average. The mean of this first sample of size 4 is what? 1 plus 1 is 2. 2 plus 3 is 5. 5 plus 6 is 11. 11 divided by 4 is 2.75. Let me do another one. My second sample of size 4, let's say that I get a 3, a 4. Let's say I get another 3. And let's say I get a 1. I just didn't happen to get a 6 that time. And notice I can't get a 2 or a 5. It's impossible for this distribution. The chance of getting a 2 or 5 is 0. So I can't have any 2s or 5s over here. So for the second sample of sample size 4, my second sample mean is going to be 3 plus 4 is 7. 7 plus 3 is 10 plus 1 is 11. 11 divided by 4, once again, is 2.75. Let me do one more, because I really want to make it clear what we're doing here. Actually, we're going to do a gazillion more. But let me just do one more in detail. So let's say my third sample of sample size 4--" + }, + { + "Q": "At 2:47, he starts using numbers like two and eight that didn't fit with what he was doing a minute before. Are those numbers suppose to be there to help solve the equation or just random numbers he pulled out?", + "A": "They are just numbers he chose to plug in to the equation so we could see that the property was true.", + "video_name": "PupNgv49_WY", + "timestamps": [ + 167 + ], + "3min_transcript": "" + }, + { + "Q": "At 3:07 Why did he multiply instead of adding and where did he get the 16 to add to the 240?", + "A": "He multiplied because when you are adding two logarithms of the same base (in this video, it was base 2) you can rewrite the logarithm as the product of the two numbers that are on the inside. log(8) + log(32) is the same thing as log(8 *32) 8*32 = 256, so we get log(8*32) = log(256) (I didn t write in the base on the logarithm, which would have been base 2.)", + "video_name": "PupNgv49_WY", + "timestamps": [ + 187 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:18 and on, what criteria is used to choose the limiting numbers 1, 2, 3/2, 5/3, 8/5 etc.?", + "A": "Its fibonacci. 1,1,2,3,5,8,13,21 1/1 2/1 3/2 and so on. Its late, but someone will probably like this.", + "video_name": "lOIP_Z_-0Hs", + "timestamps": [ + 138 + ], + "3min_transcript": "Say you're me and you're in math class, and you're doodling flowery petally things. If you want something with lots of overlapping petals, you're probably following a loose sort of rule that goes something like this. Add new petals where there's gaps between old petals. You can try doing this precisely. Start with some number of petals, say five, then add another layer in between. But the next layer, you have to add 10, then the next has 20. The inconvenient part of this is that you have to finish a layer before everything is even. Ideally, you'd have a rule that just lets you add petals until you get bored. Now imagine you're a plant, and you want to grow in a way that spreads out your leaves to catch the most possible sunlight. Unfortunately, and I hope I'm not presuming too much in thinking that, as a plant, you're not very smart. You don't know how to add number to create a series, you don't know geometry and proportions, and can't draw spirals, or rectangles, or slug cats. But maybe you could follow one simple rule. Botanists have noticed that plants seem to be fairly consistent when it comes to the angle between one leaf and the next. So let's see what you could do with that. So you grow your first leaf, and if you didn't change angle would be directly above it. So that's no good, because it blocks all the light or something. You can go 180 degrees, to have the next leaf directly opposite, which seems ideal. Only once you go 180 again, the third leaf is right over the first. In fact, any fraction of a circle with a whole number as a base is going to have complete overlap after that number of turns. And unlike when you're doodling, as a plant you're not smart enough to see you've gone all the way around and now should switch to adding things in between. If you try and postpone the overlap by making the fraction really small, you just get a ton of overlap in the beginning, and waste all this space, which is completely disastrous. Or maybe other fractions are good. The kind that position leaves in a star like pattern. It will be a while before it overlaps, and the leaves will be more evenly spaced in the meantime. But what if there were a fraction that never completely overlapped? For any rational fraction, eventually the star will close. But what if you used an irrational number? The kind of number that can't be expressed as a whole number What if you used the most irrational number? If you think it sounds weird to say one irrational number is might want to become a number theorist. If you are a number theorist, you might tell us that phi is the most irrational number. Or you might say, that's like saying, of all the integers, 1 is the integer-iest. Or you might disagree completely. But anyway, phi. It's more than 1, but less than 2, more than 3/2, less than 5/3. Greater than 8/5, but 13/8 is too big. 21/13 is just a little too small, and 34/21 is even closer, but too big, and so on. Each pair of adjacent Fibonacci numbers creates a ratio that gets closer and closer to Phi as the numbers increase. Those are the same numbers on the sides of these squares. Now stop being a number theorist, and start being a plant again. You put your first leaf somewhere, and the second leaf at an angle, which is one Phi-th of a circle. Which depending on whether you're going one way or the other, could be about 222.5 degrees, or about 137.5. Great, your second leaf is pretty far from the first, gets lots of space in the sun. And now let's add the next one a Phi-th of the circle away." + }, + { + "Q": "At 4:41, in the third line, Sal writes f(x+h) without the denominator of h that it had in line 2. Why isn't this a mistake?\nIf I did the same thing with numbers, it would be as if I rewrote (3 + 5)/2 as 3 + 5/2. That IS a mistake. What's different here?", + "A": "He factored it out, which is not a mistake. This would be like doing the following: (6+10)/5 = 2 \u00e2\u0088\u0099 [(3+5)/5] This is valid because: (ab)/c = a \u00e2\u0088\u0099 (b/c)", + "video_name": "L5ErlC0COxI", + "timestamps": [ + 281 + ], + "3min_transcript": "I just added and subtracted the same thing, but now this thing can be manipulated in interesting algebraic ways to get us to what we all love about the product rule. And at any point you get inspired, I encourage you to pause this video. Well to keep going, let's just keep exploring this expression. So all of this is going to be equal to, it's all going to be equal to the limit as H approaches zero. So the first thing I'm gonna do is I'm gonna look at, I'm gonna look at this part, this part of the expression. And in particular, let's see, I am going to factor out an F of X plus H. So if you factor out an F of X plus H, this part right over here is going to be F of X plus H, F of X plus H, times you're going to be left with G of X plus H. G of X plus H, that's that there, minus G of X, minus G of X, oops, I forgot the parentheses. Oops, it's a different color. I got a new software program and it's making it hard for me to change colors. My apologies, this is not a straightforward proof and the least I could do is change colors more smoothly. Alright, (laughing) G of X plus H minus G of X, that's that one right over there, and then all of that over this H. All of that over H. So that's this part here and then this part over here this part over here, and actually it's still over H, so let me actually circle it like this. So this part over here I can write as. actually here let me, let me factor out a G of X here. So plus G of X plus G of X times this F of X plus H. Times F of X plus H minus this F of X. Minus that F of X. All of that over H. All of that over H. Now we know from our limit properties, the limit of all of this business, well that's just going to be the same thing as the limit of this as H approaches zero plus the limit of this as H approaches zero. And then the limit of the product is going to be the same thing as the product of the limits. So if I used both of those limit properties, I can rewrite this whole thing as the limit, let me give myself some real estate, the limit as H approaches zero of F of X plus H," + }, + { + "Q": "So if we can claim that the the limit as h\u00e2\u0086\u00920 of f(x+h) is f(x), as was stated in the video at 7:30, Why can't you evaluate it as f(x+h)g(x+h)-f(x)g(x)/h = f(x)g(x+h)-f(x)g(x)/h = f(x)((g(x+h)-g(x))/h), which would be equal to f(x)g'(x). This result is clearly wrong, but I can't see where exactly I've made a mistake.", + "A": "You need to put the entirety of the first expression in parentheses as it all must be divided by h", + "video_name": "L5ErlC0COxI", + "timestamps": [ + 450 + ], + "3min_transcript": "actually here let me, let me factor out a G of X here. So plus G of X plus G of X times this F of X plus H. Times F of X plus H minus this F of X. Minus that F of X. All of that over H. All of that over H. Now we know from our limit properties, the limit of all of this business, well that's just going to be the same thing as the limit of this as H approaches zero plus the limit of this as H approaches zero. And then the limit of the product is going to be the same thing as the product of the limits. So if I used both of those limit properties, I can rewrite this whole thing as the limit, let me give myself some real estate, the limit as H approaches zero of F of X plus H, times the limit as H approaches zero, of all of this business, G of X plus H minus G of X, minus G of X, all of that over H, I think you might see where this is going. Very exciting. Plus, plus the limit, let me write that a little bit more clearly. Plus the limit as H approaches zero of G of X, our nice brown colored G of X, times, now that we have our product here, the limit, the limit as H approaches zero of F of X plus H. Of F of X plus H minus F of X, all of that, all of that over H. And let me put the parentheses where they're appropriate. So that, that, that, that. And all I did here, the limit, the limit of this sum, that's gonna be the sum of the limits, that's gonna be the limit of this plus the limit of that, and then the limit of the products is gonna be the same thing as the product of the limits. So I just used those limit properties here. But now let's evaluate them. What's the limit, and I'll do them in different colors, what's this thing right over here? The limit is H approaches zero of F of X plus H. Well that's just going to be F of X. Now, this is the exciting part, what is this? The limit is H approaches zero of G of X plus H minus G of X over H. that's the definition of our derivative. That's the derivative of G. So this is going to be, this is going to be the derivative of G of X, which is going to be G prime of X." + }, + { + "Q": "At 6:00 , how come there are no restrictions on x-3y? How can we tell if there will be restrictions on a certain part or not?", + "A": "If a factor disappears from the denominator, then its restriction needs to be stated. Otherwise there d be no way of knowing that it had ever existed in the original problem. If a factor is still left in the denominator of the simplified fraction, then its restriction does not need to be stated because it is still visible and we all know that division by zero is not allowed.", + "video_name": "e7vA_S7abSY", + "timestamps": [ + 360 + ], + "3min_transcript": "I'm gonna get four y squared and if I add them, I get four y? It looks like two y would do the trick. So it seems like we can rewrite the numerator. This is going to be. So let me draw a little line here to make it clear that this is, this is going to be equal to five times x plus two y times, I could say this is x plus two y squared or I could just say x plus two y times x plus two y. Once again, two y times two y is four y squared. Two y plus two y is four y. And that's all going to be over, that is all going to be over x minus three y times x plus two y. And so now, I have a common factor, x plus two y in both the numerator and the denominator, well that's just going to be one if we assume that x plus two y does not equal zero. And that's actually an important constraint because once we cancel this out, you lose that information. If you want this to be algebraically equivalent, we could say that x plus two y cannot be equal to zero or another way you could say it is that x cannot be equal to, cannot be equal to negative two y. I just subtracted two y from both sides there. And so what you're left with, and we can redistribute this five if we wanna write it out in expanded form. We could rewrite it as, the numerator would be five x. Let me write it over here. Five x plus 10 y. And the denominator is x minus three y. But once again, if we want it to be algebraically equivalent, we would have to say x cannot be equal to, x cannot be equal to negative two y. And now this is algebraically equivalent and you can argue that it's a little bit simpler." + }, + { + "Q": "at 0:45 Sal says that 5^2/3 is equal to 25/9. Shouldn't it be 5^2/3^2 or (5/3)^2 is equal to 25/9?", + "A": "Well, he didn t say that, actually. What he said was that 5/3 squared was equal to 25/9 What he wrote as he segued to the next example was subtly different, and it is true that he should have put his Segue into Park long enough to put the 5/3 into parentheses to emphasize that he was squaring the whole fraction.", + "video_name": "bIFdW0NZ9W4", + "timestamps": [ + 45 + ], + "3min_transcript": "Fractional exponents can be a little daunting at first, so it never hurts to do as many examples as possible. So let's do a few. What if we had 25/9, and we wanted to raise it to the 1/2 power? So we're essentially just saying, well, what is the principal square root of 25/9? So what number times itself is going to be 25/9? Well, we know 5 times 5 is 25, and 3 times 3 is 9. So why don't we just go with 5/3? Because notice, if you have 5/3 times 5/3, that is going to be 25/9. Or another way of saying this, that 5/3 squared is equal to 25/9. So 25/9 to the 1/2 is going to be equal to 5/3. Now let's escalate things a little bit. Let's take a really hairy one. Let's raise 81/256 to the negative 1/4 power. So what's going on here? This negative-- the first thing I always like to do is I want to get rid of this negative in the exponent. So let me just take the reciprocal of this and raise it to the positive. So I could just say that this is equal to 256/81 to the 1/4 power. And so now I can say, well, what number times itself times itself times itself is going to be equal to 256, and what number times itself times itself times itself-- did I say that four times? Well, what number, if I take four of them and multiply, do I get 81? And one way to think about it, this is going to be the same thing-- and we'll talk about this in more depth later on when we talk about exponent properties. to the 1/4 over 81 to the 1/4. You, in fact, saw it over here. This over here was the same thing as the square root of 25 over the square root of 9. Or 25 to the 1/2 over 9 to the 1/2. So we're just doing that over here. So one, we still have to think about what number this is. And this is a little bit of, there's no easy way to do this. You kind of have to just play around a little bit to come up with it. But 4 might jump out at you if you recognize that 16 times 16 is 256. We know that 4 to the fourth power, or you're about to know this, is 4 times 4 times 4 times 4. And 4 times 4 is 16, times 4 is 64, times 4 is equal to 256. So 4 to the fourth is 256, or we could say 4 is equal to 256 to the 1/4 power. Fair enough? Now what about 81? Well, 3 might jump out at you. We know that 3 to the fourth power is equal to 3 times 3 times 3 times 3, which is equal to 81." + }, + { + "Q": "at 1:25 isn't that the transitive property of equality?", + "A": "It can be taken that way. If boys=tall and Bill=tall, that s like saying x=tall and y=tall. Therefore, you can substitute x for tall and say x=y. If you write the equation as boys =tall and tall=bill, then it s like saying x=y and y=z, therefore x=z, that is more the transitive property. This really depends on your perspective, and as I debated today, it can be taken either way.", + "video_name": "GluohfOedQE", + "timestamps": [ + 85 + ], + "3min_transcript": "All right, we're doing the California Standards released questions in geometry now. And here's the first question. It says, which of the following best describes deductive reasoning? And I'm not a huge fan of when they ask essentially definitional questions in math class. But we'll do it and hopefully it will help you understand what deductive reasoning is. Although, I do think that deductive reasoning itself is probably more natural than the definition they'll give here. Well, actually before I even look at the definitions, let me tell you what it is. And then we can see which of these definitions matches it. Deductive reasoning is, if I give you a bunch of statements and then from those statements you deduce, or you come to some conclusion that you know must be true. Like if, I said that all boys are tall. If I told you all boys are tall. And if I told you that Bill is a boy. Bill is a boy. So if you say, OK, if these two statements are true, what can you deduce? Well I say well, Bill is a boy and all boys are tall. Then Bill must be tall. You deduced this last statement from these two other statements that you knew were true. And this one has to be true if those two are true. So Bill must be tall. That is deductive reasoning. Deductive reasoning. Bill must be tall, not Bill must be deductive. So anyway, you have some statements and you deduce other statements that must be true given those. And you often hear another type of reasoning, inductive reasoning. And that's when you're given a couple of examples and you generalize. Well, I don't want to get too complicated here, because this is a question on deductive reasoning. But generalizations often aren't a good thing. But if you see a couple of examples and you see a pattern of broader generalization. That's inductive reasoning. But that's not what they're asking us about this. Let's see if we can find the definition of deductive reasoning in the California Standards language. Use logic to draw conclusions based on accepted statements. Yeah, well actually that sounds about right. That's what we did here. We used logic to draw conclusions based on accepted statements, which were those two. So I'm going to go with A, so far. Accepting the meaning of a term without definition. Well, I don't even know how one can do that. How do you accept the meaning of something without it having being defined? Let's see, so it's not B. I don't think anything is really B. C, defining mathematical terms to correspond with physical objects. No, that's not really anything related to deductive reasoning either. D, inferring a general truth by examining a number of specific examples." + }, + { + "Q": "At 0:57, why did Sal draw those lines over the z? Please click the time to see what I'm talking 'bout", + "A": "Some people draw a line in their z s because it helps other people tell them apart from 2 s, since sloppy z s and 2 s can look like each other.", + "video_name": "cTveNRjWQYo", + "timestamps": [ + 57 + ], + "3min_transcript": "Beatrice is ordering pizzas for a party. Each pizza is cut into 12 pieces. Beatrice wants to have enough pizza so that each person can have 4 pieces. And 16 people will be at the party. A lot of information in this problem. If Beatrice buys 7 pizzas, how many extra pieces will there be after each person eats 4 pieces? So to do this, let's think about how many total pieces she will have. And then let's think about how many total pieces will be eaten by the 16 people. And then we'll figure how much that she had extra. So first, let's think about how many total pieces. And I encourage you to pause the video and try to figure it out yourself. She ordered 7 pizzas. How many total pieces are there going to be? Well, you have 7 pizzas. 7-- that's the number of pizzas. you multiply by 12 to get the total number of pieces. So this is number of pieces per pizza. And 7 times 12 is equal to 84 total pieces. So that's how many pieces she's got. Now let's think about how many get eaten by the 16 people. So you have 16 people. 16-- that's the number of people. And this little symbol, that's shorthand for number. 16 people, and they each eat 4 pieces. So 16 people times 4 pieces per person. So number of pieces per person. So that's 64-- not 84-- 64 pieces are eaten by the 16 people. So how much does she have left over? Well, if you start with 84 pieces and 64 get eaten, 84 minus 64 is 20. We deserve a drum roll now. You have 20 extra pieces after everyone eats their share." + }, + { + "Q": "3:50 wouldn't length 'a' equal length 'b' then?", + "A": "You can t make that assumption. As you shift to a different point on the circle, the length of a and b will vary. If the point was very close to the x-axis, then b would be very short compared to a . If the point was close to the y-axis, then b would be very long compared to a . Hope this makes sense.", + "video_name": "1m9p9iubMLU", + "timestamps": [ + 230 + ], + "3min_transcript": "And then this is the terminal side. So this is a positive angle theta. And what I want to do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates a comma b. The x value where it intersects is a. The y value where it intersects is b. And the whole point of what I'm doing here is I'm going to see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I want to do is I want to make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here. And let me make it clear that this is a 90-degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. you is, what is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of 1. So the hypotenuse has length 1. Now, what is the length of this blue side right over here? You could view this as the opposite side to the angle. Well, this height is the exact same thing as the y-coordinate of this point of intersection. So this height right over here is going to be equal to b. The y-coordinate right over here is b. This height is equal to b. Now, exact same logic-- what is the length of this base going to be? The base just of the right triangle? Well, this is going to be the x-coordinate If you were to drop this down, this is the point x is equal to a. Or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine-- let me use the same green-- what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our soh cah toa definition. That's the only one we have now. We are actually in the process of extending it-- soh cah toa definition of trig functions. And the cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side-- for this angle, the adjacent side has length a. So it's going to be equal to a over-- what's" + }, + { + "Q": "my doubts are in the right triangle in the video.At 3:46, why is the length of the opp side b? and at 4:08, why is the adj side equal to a? what does it have to do with the coordinates of the intersection point?", + "A": "(a,b) in the video is the x,y coordinate of the line from the origin to the perimeter point on the unit circle.", + "video_name": "1m9p9iubMLU", + "timestamps": [ + 226, + 248 + ], + "3min_transcript": "And then this is the terminal side. So this is a positive angle theta. And what I want to do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates a comma b. The x value where it intersects is a. The y value where it intersects is b. And the whole point of what I'm doing here is I'm going to see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I want to do is I want to make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here. And let me make it clear that this is a 90-degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. you is, what is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of 1. So the hypotenuse has length 1. Now, what is the length of this blue side right over here? You could view this as the opposite side to the angle. Well, this height is the exact same thing as the y-coordinate of this point of intersection. So this height right over here is going to be equal to b. The y-coordinate right over here is b. This height is equal to b. Now, exact same logic-- what is the length of this base going to be? The base just of the right triangle? Well, this is going to be the x-coordinate If you were to drop this down, this is the point x is equal to a. Or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine-- let me use the same green-- what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our soh cah toa definition. That's the only one we have now. We are actually in the process of extending it-- soh cah toa definition of trig functions. And the cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side-- for this angle, the adjacent side has length a. So it's going to be equal to a over-- what's" + }, + { + "Q": "At 1:47 .... how is he sure that \" arctan( tan ( g^-1(x) - 3pi/2 ) ) \" will give out ( g^-1(x) - 3pi/2 ) ?\ni mean .... how is he sure , that ( g^-1(x) - 3pi/2 ) is restricted to the domain of tan ?", + "A": "Tangent isn t actually a completely invertible function since it outputs the same value from different inputs. So the fact that the question asks for the inverse function implies it s looking for the simplest inverse function. Since it s just looking for the simplest inverse function, we consider arctangent as if it perfectly cancels tangent.", + "video_name": "QGfdhqbilY8", + "timestamps": [ + 107 + ], + "3min_transcript": "Voiceover:We're told given g of x is equal to ten of x minus three pi over two plus six, find the g inverse of x. They want us to type that in here and then they also want us to figure out what is the domain of g inverse, the domain of g inverse of x. I've got my little scratch pad here to try to work that through. Let's figure out what g inverse of x is. This is g of x, so g inverse of x. Essentially, let me just read this is g of x right over here, g of x is equal to tangent of x minus three pi over two plus six. G inverse of x, I essentially can swap the \u2026 I can replace the x with the g inverse of x and replace to g of x with an x and then solve for g inverse of x. I could write that x is equal to tangent of g inverse of x minus three pi over two plus six. I actually encourage you to pause this video and try to work through this out or work it out on your own. Let's subtract six from both sides to at least get rid of this six here so I'm left with x minus six is equal to the tangent of g inverse of x minus three pi over two. Now let's take the inverse tangent of both sides of this equation so the inverse tangent on the left hand side is the inverse tangent of x minus six and on the right hand side the inverse tangent of tangent. If we restrict the domain in the proper way and we'll talk about that in a little bit is just going to be what the input into the tangent function is. If you restrict the domain in the right way, inverse tangent of the tangent of something, Once again if we restrict the domain, if we restrict what the possible values of theta are in the right way. Let's just assume that we're doing that and so the inverse tangent of the tan, of this is going to be just this stuff right over here. It's just going to be that, it's going to be g inverse of x minus three pi over two. Now we're in the [home] stretch to solve for g inverse of x we could just add three pi over two to both sides so we get and actually let me just swap both sides. We get g inverse of x is equal to the inverse tangent of x minus six and then we're adding three pi over two to both sides so this side is now on this side so plus three pi over two. Let me actually type that and I'm going to see if I can remember it" + }, + { + "Q": "At 0:04 it says two sets of parallel sides. Isn't that the same thing as two sets of parallel lines. Sides on a parallelogram are lines. Even if the question says so, you can also say sets of sides. Not to be critical, just thought I'd point it out.", + "A": "Yes, two sets of parallel sides are the same things as lines. It s just another way of saying it. Just like how you might call a sphere, a ball. There might be many different ways to say that, like two pairs of parallel lines.", + "video_name": "1pHhMX0_4Bw", + "timestamps": [ + 4 + ], + "3min_transcript": "A parallelogram is a blank with two sets of parallel lines. So let's see what the options are. So one option is a quadrilateral. And a parallelogram is definitely a quadrilateral. A quadrilateral is a four-sided figure, and it is definitely a four-sided figure. A parallelogram is not always a rhombus. A rhombus is a special case of a parallelogram where not only do you have to sets of parallel lines as your sides, two sets of parallel sides, but all of the sides are the same length in a rhombus. And a square is a special case of a rhombus where all of the angles are 90 degrees. So here, all we can say is that a parallelogram is a quadrilateral. And so let's check our answer. And it's always a good idea to look at hints. And so it'll kind of say the same thing that we just said, but it would say it for the particular problem that you're actually looking at. Let's do a few more of these. Suzanne is on an expedition to save the universe. Sounds like a reasonable expedition to go on. a game called Find the Rhombuses. A wizard tells her that she has a square, a quadrilateral, and a parallelogram, and she must identify which of the shapes are also rhombuses. Which of these shapes should she pick to save the universe? So a square is a special case of a rhombus. Just to remind ourselves, a rhombus, the opposite sides are parallel to each other. You have two sets of parallel sides. A square has two sets of parallel sides, and it has the extra condition that all of the angles are right angles. So a square is definitely going to be a rhombus. Now, all rhombuses have four sides. So all rhombuses are quadrilaterals. But not all quadrilaterals are rhombuses. You could have a quadrilateral where none of the sides are parallel to each other. So we won't click this one. Once again, a parallelogram. So all rhombuses are parallelograms. two sets of parallel line segments representing their sides. But all parallelograms are not rhombuses. So I would say that if someone gives you square, you can say, look, a square is always going to be a rhombus. A quadrilateral isn't always going to be a rhombus, nor is a parallelogram always going to be a rhombus. We got it right." + }, + { + "Q": "At 3:39, why is it that we minus the dy/dx from the other side? Because it is being multiplied by the (2x-2y) on the left wouldn't we have to divide the other side by dy/dx?", + "A": "... but then we would have an answer that was for (dy/dx)^-1 = dx/dy ... which is not what we want!", + "video_name": "9uxvm-USEYE", + "timestamps": [ + 219 + ], + "3min_transcript": "onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides. 2y from that side. And then we could also subtract a dy dx from both sides, so that all of our dy dx's are on the left hand side, and all of our non dy dx's are on the right hand side. So let's do that. So we're going to subtract a dy dx on the right and a dy dx here on the left. And so what are we left with? Well, on the left hand side, these cancel out. And we're left with 2y minus 2x dy dx minus 1 dy dx, or just minus a dy dx. Let me make it clear. We could write this as a minus 1 dy dx. So this is we can essentially just add these two coefficients. So this simplifies to 2y minus 2x minus 1 times the derivative of y with respect to x, which is going We are left with 1 minus 2x plus 2y. So let me write it that way. Or we could write this as-- so negative, negative 2y is just a positive 2y. And then we have minus 2x. And then we add that 1, plus 1. And now to solve for dy dx, we just have to divide both sides by 2y minus 2x minus 1. And we are left with-- we deserve a little bit of a drum roll at this point. As you can see, the hardest part was really the algebra to solve for dy dx. We get the derivative of y with respect to x is equal to 2y minus 2x plus 1 over 2y minus 2x minus 1." + }, + { + "Q": "At 2:17, Sal distributes the (2x-2y) to the -(dy/dx). I understand that the answer originally is (-2x+2y)(dy/dx), and that he's just rearranging things when he writes it as (2y-2x)(dy/dx). But I don't understand how he gets (-2x+2y)(dy/dx) in the first place! I would have written (2x-2y)(-(dy/dx)). Is he simply applying the negative sign of the (dy/dx) to the (2x-2y)? That's what it looks like, but I don't understand why that is being done. Why can't the negative stay with the (dy/dx)?", + "A": "He is multiplying (2x - 2y)(-(dy/dx)) by one, but by a special form of one which is (-1)(-1). Now he has (-1)(-1)(2x - 2y)(-(dy/dx)). Since we can multiply in any order let s shift things to: (-1)(2x -2y)(-1)(-(dy/dx)). If you multiply the first two terms together, and the 3rd and 4th term together you get: (2y - 2x)(dy/dx).", + "video_name": "9uxvm-USEYE", + "timestamps": [ + 137 + ], + "3min_transcript": "Let's get some more practice doing implicit differentiation. So let's find the derivative of y with respect to x. We're going to assume that y is a function of x. So let's apply our derivative operator to both sides of this equation. So let's apply our derivative operator. And so first, on the left hand side, we essentially are just going to apply the chain rule. First we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1, and the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x. We're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do. Let's make it clear. This right over here, I can rewrite as 2x minus 2y. So let me do that so I can save some space. This is 2x minus 2y If I just distribute the 2. onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides." + }, + { + "Q": "At 1:25ish. Angle BGC and DGC would also be \" Adjacent\", correct?", + "A": "Close. Adjacent angles share a ray, but neither is interior to the other. Each of those angles would be interior to angle BGD. However, the angles you mention are adjacent to each other.", + "video_name": "vAlazPPFlyY", + "timestamps": [ + 85 + ], + "3min_transcript": "We're asked to name an angle adjacent to angle BGD. So angle BGD, let's see if we can pick it out. So here is B, here is G, and here is D, right over here. So angle BGD is this entire angle right over here. So when we talk about adjacent angles, we're talking about an angle that has one of its rays in common. So for example, angle AGB has one of the rays in common, it has GB in common with angle BGD. So we could say angle AGB, which could obviously also be called angle BGA, BGA and AGB are both this angle right over here. You could also go with angle FGB, because that also has GB in common. So you go angle FGB, which could also be written as angle BGF. So you could do this angle right over here, angle EGD. Or you could go all the way out here, angle FGD. These last two sharing ray GD in common. So any one of these responses would satisfy the question of just naming an angle, just naming one. Let's do this next one. Name an angle vertical to angle EGA. So this is this angle right over here. And the way you think about vertical angles is, imagine two lines crossing. So imagine two lines crossing, just like this. And they could literally be lines, and they're intersecting at a point. This is forming four angles, or you could imagine it's forming two sets of vertical angles. it's a vertical angle, it's the one on the opposite side of the intersection. It's one of these angles that it is not adjacent to. So it would be this angle right over here. So going back to the question, a vertical angle to angle EGA, well if you imagine the intersection of line EB and line DA, then the non-adjacent angle formed to angle EGA is angle DGB. Actually, what we already highlighted in magenta right over here. So this is angle DGB. Which could also be called angle BGD. These are obviously both referring to this angle up here. Name an angle that forms a linear pair with the angle DFG. So we'll put this in a new color. Angle DFG. Sorry, DGF, all of these should have G in the middle." + }, + { + "Q": "At 7:27 what about \"x\" does not equal 3?\nx=3 will make the function undefined too right?\nHelp me! I'm a little bit confused.", + "A": "Values that make both the numerator and denominator zero are undefined points (like x = -3 in this example). Values that make only the denominator zero are vertical asymptotes (like x = 3 in this example). Yes, both points are undefined for this function, but in different ways.", + "video_name": "P0ZgqB44Do4", + "timestamps": [ + 447 + ], + "3min_transcript": "but by itself it does not make a vertical asymptote. Let's just think about this denominator right over here so we can factor it out. Actually let's factor out the numerator and the denominator. We can rewrite this as F of X is equal to the numerator is clearly every term is divisible by three so let's factor out three. It's going to be three times X squared minus six X minus 27. All of that over the denominator each term is divisible by six. Six times X squared minus 9 and let's see if we can factor the numerators and denominators out further. This is going to be F of X is equal to three times let's see, two numbers, their product is negative 27, their sum is negative six. Negative nine and three seem to work. You could have X minus nine times X plus three. Just factor the numerator over the denominator. This would be X minus three times X plus three. When does the denominator equal zero? The denominator equals zero when X is equal to positive three or X is equal to negative three. Now I encourage you to pause this video for a second. Think about are both of these vertical asymptotes? Well you might realize that the numerator also equals zero when X is equal to negative three. What we can do is actually simplify this a little bit and then it becomes a little bit clear where our vertical asymptotes are. We could say that F of X, we could essentially divide the numerator and denominator by X plus three and we just have to key, if we want the function to be identical, we have to keep the [caveat] that the function itself is not defined when X is equal to negative three. We have to remember that but that will simplify the expression. This exact same function is going to be if we divide the numerator and denominator by X plus three, it's going to be three times X minus nine over six times X minus three for X does not equal negative three. Notice, this is an identical definition to our original function and I have to put this qualifier right over here for X does not equal negative three because our original function is undefined at X equals negative three. X equals negative three is not a part of the domain of our original function. If we take X plus three out of the numerator and the denominator, we have to remember that. If we just put this right over here, this wouldn't be the same function because this without the qualifier is defined for X equals negative three" + }, + { + "Q": "5:34 So is codomain the same as range?", + "A": "No, range is a subspace (or subset) of codomain. Range is the specific mapping from the elements in set X to elements in set Y, where codomain is every element in set Y. To concretize: Let f map X --> Y Where X = { 0, 1} Where Y = { 2, 3, 4, 5, 6...}. Let f(0) = 2 & f(1) = 3. Clearly, the range of X = [ 1, 2} where as the codomain of X = { 1, 2 ,3, 4, 5, 6..}.", + "video_name": "BQMyeQOLvpg", + "timestamps": [ + 334 + ], + "3min_transcript": "This statement you've probably never seen before, but I like it because it shows the mapping or the association more, while this association I think that you're putting an x into a little meat grinder or some machine that's going to ground up the x or square the x, or do whatever it needs to do to the x. This notation to me implies the actual mapping. You give me an x, and I'm going to associate another number in real numbers called x squared. So it's going to be just another point. And just as a little bit of terminology, and I think you've seen this terminology before, the set that you are mapping from is called the domain and it's part of the function definition. I, as the function creator, have to tell you that every valid input here has to be a set of real numbers. Now the set that I'm mapping to is called the codomain. Sal, when I learned all of this function stuff in algebra II or whenever you first learned it, we never used this codomain word. We have this idea of range, I learned the word range when I was in 9th or 10th grade. How does this codomain relate to range? And it's a very subtle notation. So the codomain is a set that you're mapping to, and in this example this is the codomain. In this example, the real numbers are the domain and the codomain. So the question is how does the range relate to this? So the codomain is the set that can be possibly mapped to. You're not necessarily mapping to every point in the codomain. I'm just saying that this function is generally mapping from members of this set to that set. It could be equal to the codomain. It's some subset. A set is a subset of itself, every member of a set is also a member of itself, so it's a subset of itself. So range is a subset of the codomain which the function actually maps to. So let me give you an example. Let's say I define the function g, and it is a mapping from the set of real numbers. Let me say it's a mapping from R2 to R." + }, + { + "Q": "why is 0 not in the range? it is a real number and the product of any number and zero would be zero, so I don\u00c2\u00b4t get why Sal says in 13:36 that it is not a member of the range?!", + "A": "Elements of the range are not numbers, but triples of numbers. So asking whether or not 0 is in the range makes no sense. Yes it is true that we can make the third coordinate 0, but can we make the first coordinate 5, the second coordinate 1, and the third coordinate 0 all at the same time? The answer is no.", + "video_name": "BQMyeQOLvpg", + "timestamps": [ + 816 + ], + "3min_transcript": "Let's take our h of-- let me use my other notation-- let's say that I said h, and I wanted to find the mapping from the point in R2, let's say the point 2 comma 3. And then my function tells me that this will map to the point in R3. I add the two terms, the 2 plus 3, so it's 5. I'd find the difference between x2 and x1-- so 3 minus 2 is 1-- and then I multiply the two, 6. So clearly this will be in the range, this is a member of the range. So for example the point 2, 3, which might be right there, will be mapped to the three dimensional point, it's kind but I think you get the idea, would be mapped to the three dimensional point 5, 1, 6. So this is definitely a member of the range. Now my question to you, if I have some point in R3, let's say that this is the point 5, 1, 0. Is this point a member of the range? It's definitely a member of the codomain, it's in R3. It's definitely in here, and this by definition is the codomain. But is this in our range? 5 has to be the sum of two numbers, the 1 has to be the difference of two numbers, and then the 0 would have to be the product of two numbers. And clearly we know 5 is the sum, and 1 is the difference, we're dealing with 2 and 3, and there's no way you can get the product of those numbers to be equal to 0. So the range would be the subset of all of these points in R3, so there'd be a ton of points that aren't in the range, and there'll be a smaller subset of R3 that is in the range. Now I want to introduce you to one more piece of terminology when it comes to functions. These functions up here, this function that mapped from points in R2 to R, so its codomain was R. This function up here is probably the most common function you see in mathematics, this is also mapping to R. These functions that map to R are called scalar value or real value, depending on how you want to think about it. But if they map to a one dimensional space, we call them a scalar valued function, or a real valued function." + }, + { + "Q": "What does he mean at 12:42?", + "A": "The function h maps an ordered pair onto an ordered triple. He is observing that the result of the function is an ordered triple of real numbers, an element of the range R^3.", + "video_name": "BQMyeQOLvpg", + "timestamps": [ + 762 + ], + "3min_transcript": "And notice I'm going from a space that has two dimensions to a space it has three dimensions, or three But I can always associate some point with x1, x2 with some point in my R3 there. A slightly trickier question here is, what is the range? Can I always associate every point-- maybe this wasn't the best example because it's not simple enough -- but can I associate every point in R3-- so this is my codomain, my domain was R2, and my function goes from R2 to R3, so that's h. And so my range, as you could see, it's not like every coordinate you can express in this way in some way. Let me give you an example. Let's take our h of-- let me use my other notation-- let's say that I said h, and I wanted to find the mapping from the point in R2, let's say the point 2 comma 3. And then my function tells me that this will map to the point in R3. I add the two terms, the 2 plus 3, so it's 5. I'd find the difference between x2 and x1-- so 3 minus 2 is 1-- and then I multiply the two, 6. So clearly this will be in the range, this is a member of the range. So for example the point 2, 3, which might be right there, will be mapped to the three dimensional point, it's kind but I think you get the idea, would be mapped to the three dimensional point 5, 1, 6. So this is definitely a member of the range. Now my question to you, if I have some point in R3, let's say that this is the point 5, 1, 0. Is this point a member of the range? It's definitely a member of the codomain, it's in R3. It's definitely in here, and this by definition is the codomain. But is this in our range? 5 has to be the sum of two numbers, the 1 has to be the difference of two numbers, and then the 0 would have to be the product of two numbers. And clearly we know 5 is the sum, and 1 is the difference, we're dealing with 2 and 3, and there's no way you can get the product of those numbers to be equal to 0." + }, + { + "Q": "At 2:30, why does a number to the negative power a decimal?", + "A": "When you have a negative exponent, you re essentially dividing the current number, like 5 squared, by the base, or 5. If you keep doing that, you get 1, 0.2, 0.04, and so on. I hope this helped you!", + "video_name": "6phoVfGKKec", + "timestamps": [ + 150 + ], + "3min_transcript": "Express 0.0000000003457 in scientific notation. So let's just remind ourselves what it means to be in scientific notation. Scientific notation will be some number times some power of 10 where this number right here-- let me write it this way. It's going to be greater than or equal to 1, and it's going to be less than 10. So over here, what we want to put here is what that leading number is going to be. And in general, you're going to look for the first non-zero digit. And this is the number that you're going to want to start off with. This is the only number you're going to want to put ahead of or I guess to the left of the decimal point. So we could write 3.457, and it's going to be multiplied by 10 to something. Now let's think about what we're going to have to multiply it by. To go from 3.457 to this very, very small number, to move the decimal to the left a bunch. You have to add a bunch of zeroes to the left of the 3. You have to keep moving the decimal over to the left. To do that, we're essentially making the number much much, much smaller. So we're not going to multiply it by a positive exponent of 10. We're going to multiply it times a negative exponent of 10. The equivalent is you're dividing by a positive exponent of 10. And so the best way to think about it, when you move an exponent one to the left, you're dividing by 10, which is equivalent to multiplying by 10 to the negative 1 power. Let me give you example here. So if I have 1 times 10 is clearly just equal to 10. 1 times 10 to the negative 1, that's equal to 1 times 1/10, which is equal to 1/10. to 0-- let me actually-- I skipped a step right there. Let me add 1 times 10 to the 0, so we have something natural. So this is one times 10 to the first. One times 10 to the 0 is equal to 1 times 1, which is equal to 1. 1 times 10 to the negative 1 is equal to 1/10, which is equal to 0.1. If I do 1 times 10 to the negative 2, 10 to the negative 2 is 1 over 10 squared or 1/100. So this is going to be 1/100, which is 0.01. What's happening here? When I raise it to a negative 1 power, I've essentially moved the decimal from to the right of the 1 to the left of the 1. I've moved it from there to there. When I raise it to the negative 2, I moved it two over to the left. So how many times are we going to have to move it over to the left to get this number right over here?" + }, + { + "Q": "In 9:15 - he simplified 0.1 into 1/10, but it could have been into 10/100 as well... If you do it like that, you have root10 / root100 = root10 / 10\n\nSeems correct to me but has different answer.... why not do it this way?", + "A": "You can do it either way, but depending on what the problem is, it might be easier to do it your way or to do it Sal s way. For instance, if you had 0.2 , then that would simplify to 1/5 Sal s way; but your way, it would simplify to 20/200 , which would turn into root20 / root200 , and that s a little more messy. So, sometimes your way is easier, and sometimes Sal s is. But you can do it whichever way you prefer; they both give correct answers.", + "video_name": "BpBh8gvMifs", + "timestamps": [ + 555 + ], + "3min_transcript": "which is equal to 1/2. Which is clearly rational. It can be expressed as a fraction. So that's clearly rational. Part G is the square root of 9/4. Same logic. This is equal to the square root of 9 over the square root of 4, which is equal to 3/2. Let's do part H. The square root of 0.16. recognize that, gee, if I multiply 0.4 times 0.4, I'll get this. But I'll show you a more systematic way of doing it, if that wasn't obvious to you. So this is the same thing as the square root of 16/100, right? That's what 0.16 is. So this is equal to the square root of 16 over the square root of 100, which is equal to 4/10, which is equal to 0.4. Let's do a couple more like that. Part I was the square root of 0.1, which is equal to the square root of 1/10, which is equal to the square root of 1 over the square root of 10, which is equal to 1 over-- now, the square root of 10-- 10 is just 2 times 5. So that doesn't really help us much. A lot of math teachers don't like you leaving that radical But I can already tell you that this is irrational. You'll just keep getting numbers. You can try it on your calculator, and it will never repeat. Your calculator will just give you an approximation. Because in order to give the exact value, you'd have to have an infinite number of digits. But if you wanted to rationalize this, just to show you. If you want to get rid of the radical in the denominator, you can multiply this times the square root of 10 over the square root of 10, right? This is just 1. So you get the square root of 10/10. These are equivalent statements, but both of them are irrational. You take an irrational number, divide it by 10, you still have an irrational number. Let's do J. We have the square root of 0.01. This is the same thing as the square root of 1/100. Which is equal to the square root of 1 over the square root of 100, which is equal to 1/10, or 0.1." + }, + { + "Q": "at 2:35 shouldn't 2/(square root of) 6 be the other way around", + "A": "Sal had 2*2*2*3 inside his radical. 2*2=4, 2*3=6. So you have 4*6 inside a radical. The sqrt of 4 is a rational number ( a number that can be expressed as a fraction a/b where b doesn t = 0) so we can go ahead and work with it. We remove the sqrt of four from the radical, leaving us with 2 (the sqrt of 4) times the sqrt of 6.", + "video_name": "BpBh8gvMifs", + "timestamps": [ + 155 + ], + "3min_transcript": "square root of 2 times 2 times 2 times 3. That's the same thing as 24. Well, we see here, we have one perfect square right there. So we could rewrite this. This is the same thing as the square root of 2 times 2 times the square root of 2 times 3. Now this is clearly 2. This is the square root of 4. The square root of 4 is 2. And then this we can't simplify anymore. We don't see two numbers multiplied by itself here. So this is going to be times the square root of 6. Or we could even right this as the square root of 2 times the square root of 3. Now I said I would talk about whether things are rational or not. This is rational. This part A can be expressed as the ratio of 2 integers. This is rational. This is irrational. I'm not going to prove it in this video. But anything that is the product of irrational numbers. And the square root of any prime number is irrational. I'm not proving it here. This is the square root of 2 times the square root of 3. That's what the square root of 6 is. And that's what makes this irrational. I cannot express this as any type of fraction. I can't express this as some integer over some other integer like I did there. And I'm not proving it here. I'm just giving you a little bit of practice. And a quicker way to do this. You could say, hey, 4 goes into this. 4 is a perfect square. Let me take a 4 out. This is 4 times 6. The square root of 4 is 2, leave the 6 in, and you would have gotten the 2 square roots of 6. Which you will get the hang of it eventually, but I want to do it systematically first. Square root of 20. Once again, 20 is 2 times 10, which is 2 times 5. So this is the same thing as the square root of 2 times 2, right, times 5. Now, the square root of 2 times 2, that's clearly just going to be 2. It's going to be the square root of this times square root of that. 2 times the square root of 5. And once again, you could probably do that in your head The square root of the 20 is 4 times 5. The square root of 4 is 2. You leave the 5 in the radical. So let's do part D. We have to do the square root of 200. Same process. Let's take the prime factors of it. So it's 2 times 100, which is 2 times 50, which is 2 times 25, which is 5 times 5." + }, + { + "Q": "What does corresponding angles mean? [8:16] and other parts of the video. Thanks!", + "A": "Corresponding angles are created where a transversal crosses other (usually parallel) lines. The corresponding angles are the ones at the same location at each intersection.", + "video_name": "TErJ-Yr67BI", + "timestamps": [ + 496 + ], + "3min_transcript": "that could be parallel, if the alternate interior angles are And we see that they are. These two are kind of candidate alternate interior angles, and they are congruent. So AB must be parallel to CD. Actually, I'll just draw one arrow. AB is parallel to CD by alternate interior angles congruent of parallel lines. I'm just writing in some shorthand. Forgive the cryptic nature of it. I'm saying it out. And so we can then do the exact same-- we've just shown that these two sides are parallel. We could then do the exact same logic to show that these two sides are parallel. And I won't necessarily write it all out, but it's the exact same proof to show that these two. So first of all, we know that this angle Actually, let me write it out. So we know that angle AEC is congruent to angle DEB. They are vertical angles. And that was our reason up here, as well. And then we see the triangle AEC must be congruent to triangle DEB by side-angle-side. So then we have triangle AEC must be congruent to triangle DEB by SAS congruency. Then we know that corresponding angles must be congruent. So for example, angle CAE must be congruent to angle BDE. of congruent triangles. So CAE-- let me do this in a new color-- must be congruent to BDE. And now we have a transversal. The alternate interior angles are congruent. So the two lines that the transversal is intersecting must be parallel. So this must be parallel to that. So then we have AC must be parallel to be BD by alternate interior angles. And we're done. We've just proven that if the diagonals bisect each other, if we start that as a given, then we end at a point where we say, hey, the opposite sides of this quadrilateral must be parallel, or that ABCD is a parallelogram." + }, + { + "Q": "At 16:03, why are you multiplying 1/4sin(t) by 2sintcost? Wasn't it ...+sin2t?", + "A": "At 15:23 Sal reminded you of the trigonometric identity: sin(2\u00ce\u00b8) = 2\u00c2\u00b7sin(\u00ce\u00b8)\u00c2\u00b7cos(\u00ce\u00b8). And even when he s writing it at 16:03, he clearly says this is just a trig identity .", + "video_name": "IW4Reburjpc", + "timestamps": [ + 963 + ], + "3min_transcript": "And we're going to evaluate this whole thing at t, so you get 1/2 sine squared of t minus 1/2 the sine of 0 squared, which is just 0, so that's just minus 0. So far, everything that we have written simplifies to-- let me multiply it all out. So I have 1/2-- let me just pick a good color-- 1/2t sine of t-- I'm just multiplying those out-- plus 1/4 sine of t sine of 2t. And then over here I have minus 1/2 sine squared t times I just took the minus cosine t and multiplied it through here and I got that. Now, this is a valid answer, but I suspect that we can simplify this more, maybe using some more trigonometric identities. And this guy right there looks ripe to simplify. And we know that the sine of 2t-- another trig identity you'll find in the inside cover of any of your books-- is 2 times the sine of t times the cosine of t. So if you substitute that there, what does our whole expression equal? You get this first term. Let me scroll down a little bit. You get 1/2t times the sine of t plus 1/4 sine of t times this thing in here, so times 2 sine of t cosine of t. And then finally I have this minus 1/2 sine squared t cosine of t. No one ever said this was going to be easy, but hopefully it's instructive on some level. At least it shows you that you didn't memorize your trig identities for nothing. So let me rewrite the whole thing, or let me just rewrite this part. So this is equal to 1/4. Now, I have-- well let me see, 1/4 times 2. 1/4 times 2 is 1/2. And then sine squared of t, right? This sine times this sine is sine squared of t cosine of t. And then this one over here is minus 1/2 sine squared of t cosine of t. And luckily for us, or lucky for us, these cancel out. And, of course, we had this guy out in the front. We had this 1/2t sine t out in front. Now, this guy cancels with this guy, and all we're left" + }, + { + "Q": "at 10:00 if we substitute sin\u00cf\u0084.cos\u00cf\u0084 equals to sin2\u00cf\u0084/2 . answer turns out to be different at end. isn't it ? Terms donot cancel out each other in end", + "A": "You do get the same answer. Try again, and make sure that you evaluate it a 0 as well as a t. The integral of \u00c2\u00bdsin(2\u00cf\u0084) becomes -\u00c2\u00bccos(2\u00cf\u0084). That is evaluated at t and at 0, so we are left with [-cos(2(t))/4] - [-cos(2(0))/4] = \u00c2\u00bc - \u00c2\u00bccos(2t). The cos(2t) can be expanded into cos\u00c2\u00b2(t) - sin\u00c2\u00b2(t), so we are left with \u00c2\u00bc + \u00c2\u00bcsin\u00c2\u00b2(t) - \u00c2\u00bccos\u00c2\u00b2(t). All this must be multiplied by the -cos(t) out front, and you are left with \u00c2\u00bccos\u00c2\u00b3(t) - \u00c2\u00bccos(t)sin\u00c2\u00b2(t) - \u00c2\u00bccos(t). You will find that that will cancel with the other integral.", + "video_name": "IW4Reburjpc", + "timestamps": [ + 600 + ], + "3min_transcript": "to the cosine of tau, just the derivative of sine. Or we could write that du is equal to the cosine of tau d tau. We'll undo the substitution before we evaluate the endpoints here. But this was a little bit more of a conundrum. I don't know how to take the antiderivative of cosine squared of tau. It's not obvious what that is. So to do this, we're going to break out some more trigonometric identities. And in a video I just recorded, it might not be the last video in the playlist, I showed that the cosine squared of tau-- I'm just using tau as an example-- is equal to 1/2 And once again, this is just a trig identity that you'll find really in the inside cover of probably your calculus book. So we can make this substitution here, make this substitution right there, and then let's see what our integrals become. So the first one over here, let me just write it here. We get sine of t times the integral from 0 to t of this thing here. Let me just take the 1/2 out, to keep things simple. So I'll put the 1/2 out here. That's this 1/2. So 1 plus cosine of 2 tau and all of that is d tau. And then we have this integral right here, minus cosine of t times the integral from-- let me be very clear. This is tau is equal to 0 to tau is equal to t. And then this thing right here, I did some u subsitution. If u is equal to sine of t, then this becomes u. And we showed that du is equal to cosine-- sorry, u is equal to sine of tau. And then we showed that du is equal to cosine tau d tau, so this thing right here is equal to du. So it's u du, and let's see if we can do anything useful now. So this integral right here, the antiderivative of this is" + }, + { + "Q": "At 6:20, what does Sal mean by the solution satisfying the diff eq for all x's?", + "A": "In this case, it basically means writing the equation in such as a way that you have no differentials remaining and no excess variables. If the differential was in the xy-plane, then x- and y-variables are the intend. In the xyz-space, only x-, y-, and z-variables. Suitable substitutions (like here with m) are allowed.", + "video_name": "zid7J4EhZN8", + "timestamps": [ + 380 + ], + "3min_transcript": "is equal to some coefficient times x, plus some other constant value. Well, in order for a constant value to be equal to a coefficient times x plus some other constant value the coefficient on X must be equal to zero. Another way to think about it is, this should be, you could rewrite the left-hand side here as zero x plus m. So you see, you kind of match the coefficients. So zero must be equal to 3m minus 2, and m is equal to 3b, m is equal to 3b minus 5. m is equal to 3b minus 5. So let's use that knowledge, that information, to solve for m and b. So we could use this first one. So 3m minus 2 must be equal to zero. So let's write that. 3m minus 2 is equal to zero, or 3m is equal to 2, or m is equal to 2/3. So we figured out what m is. because we know that, we know that m is equal to, is equal to 3b minus 5. m is 2/3, so we get 2/3 is equal to 3b minus 5. We could add 5 to both sides, which is the same thing as adding 15/3 to both sides. Is that, did I do that right? Yeah, adding 5 to both sides is the same thing as adding 15/3 to both sides. So let's do that. 15/3 plus 15/3, these cancel out. That's just 5 right over there. On the left-hand side we have 17/3 is equal to 3b, or if you divide both sides by 3 you get b is equal to 17, b is equal to 17/9, and we're done. We just found a particular solution The solution is y is equal to 2/3x plus 17/9. And I encourage you, after watching this video, to verify that this particular solution indeed does satisfy this differential equation for all x's. For all x's." + }, + { + "Q": "At 1:04 is altitude the proper term?", + "A": "YES! According the the definition An altitude is the line segment which join vertex and its opposite sides (known as base) perpendicular to it. And interesting thing about it that it is the the shortest path to reach opposite side from the vertex. A STRAIGHT PATH IS THE SHORTEST PATH!", + "video_name": "APNkWrD-U1k", + "timestamps": [ + 64 + ], + "3min_transcript": "I will now do a proof of the law of sines. So, let's see, let me draw an arbitrary triangle. That's one side right there. And then I've got another side here. I'll try to make it look a little strange so you realize it can apply to any triangle. And let's say we know the following information. We know this angle -- well, actually, I'm not going to say what we know or don't know, but the law of sines is just a relationship between different angles and different sides. Let's say that this angle right here is alpha. This side here is A. The length here is A. Let's say that this side here is beta, and that the length here is B. Beta is just B with a long end there. So let's see if we can find a relationship that connects A and B, and alpha and beta. So what can we do? And hopefully that relationship we find will be the law of sines. So let me draw an altitude here. I think that's the proper term. If I just draw a line from this side coming straight down, and it's going to be perpendicular to this bottom side, which I haven't labeled, but I'll probably, if I have to label it, probably label it C, because that's A and B. And this is going to be a 90 degree angle. I don't know the length of that. I don't know anything about it. All I know is I went from this vertex and I dropped a line that's perpendicular to this other side. So what can we do with this line? Well let me just say that it has length x. The length of this line is x. Can we find a relationship between A, the length of this line x, and beta? Well, sure. Let's see. Let me find an appropriate color. That's, I think, a good color. So what's the relationship? If we look at this angle right here, beta, x is opposite to it and A is the hypotenuse, if we look at this right triangle right here, right? So what deals with opposite and hypotenuse? Whenever we do trigonometry, we should always just right soh cah toa at the top of the page. Soh cah toa. So what deals with opposite of hypotenuse? Soh, and you should probably guess that, because I'm proving the law of sines. So the sine of beta is equal to the opposite over the hypotenuse. It's equal to this opposite, which is x, over the hypotenuse, which is A, in this case. And if we wanted to solve for x, and I'll just do that, because it'll be convenient later, we can multiply both sides of this equation by A and you get A sine of beta is equal to x. Fair enough. That got us someplace." + }, + { + "Q": "At 3:10 you multiply by Y' but why is the reason for this? If the product rule is just f'(x)g(x)+f(x)g'(x) which would mean from my understanding (1)(y^2) + (x)(2y)", + "A": "It s the notation that is confusing. I had a hard time with this as well. If you mentally substitute write in dy/dx here rather than multiply by y I bet you would feel more comfortable", + "video_name": "ZtI94pI4Uzc", + "timestamps": [ + 190 + ], + "3min_transcript": "with different notation. So let's take the derivative of this thing right over here. Well we're going to apply the chain rule. Actually, we're going to apply the chain rule multiple times here. The derivative of e to the something with respect to that something is going to be e to the something times the derivative of that something with respect to x. So times the derivative of xy squared. So that's our left-hand side. We aren't done taking the derivative yet. And on our right-hand side, the derivative of x is just 1. And the derivative with respect to x of y is just going to be minus-- or I could write-- negative dy dx. But instead of writing dy dx, I'm going to write y prime. As you can tell, I like this notation and this notation more because it makes it explicit Here, we just have to assume that we're taking the derivative with respect to x. Here, we have to assume that's the derivative of y with respect to x. But anyway let's stick with this notation right over here. Actually, let me make all of my y primes, all my derivatives of y with respect to x, let me make them pink so I keep track of them. So once again, this is going to be equal to e to the xy squared times the derivative of this. Well the derivative of this, we can just use the product and actually a little bit of the chain rule So the derivative of x is just 1 times the second function. So it's going to be times y squared. And then to that, we're going to add the product of the first function which is this x times the derivative of y squared with respect to x. Well that's going to be the derivative of y squared times the derivative of y with respect to x, which we are now writing as y prime. And then that's going to be equal to 1 minus y prime. And like we've been doing, we now have to just solve for y prime. So let's distribute this exponential, this e to the xy squared. And we get e, or maybe I should say y squared times e to the xy squared. So that's that. Plus 2xye to the xy squared. y prime, the derivative of y with respect to x, is equal to 1 minus the derivative of y with respect" + }, + { + "Q": "4:49 where did the +1 come from?", + "A": "That is basic factoring: 2x + 6x\u00c2\u00b2 Factors to 2x(1+3x) The same thing is happening here, only with slightly more complicated expressions.", + "video_name": "ZtI94pI4Uzc", + "timestamps": [ + 289 + ], + "3min_transcript": "times the derivative of y with respect to x, which we are now writing as y prime. And then that's going to be equal to 1 minus y prime. And like we've been doing, we now have to just solve for y prime. So let's distribute this exponential, this e to the xy squared. And we get e, or maybe I should say y squared times e to the xy squared. So that's that. Plus 2xye to the xy squared. y prime, the derivative of y with respect to x, is equal to 1 minus the derivative of y with respect Now let's get all of our y primes on one side. So let's add y prime to both sides. So let's add-- and just to be clear, I'm adding one y prime to both sides. So let's add a y prime to both sides. And let's subtract this business from both sides. So let's subtract y squared e to the xy squared subtracting from both sides. So we're going to subtract y squared e to the xy squared. And we are left with 2xye to the xy squared plus 1 times y prime. We had this many y primes and then we add another 1y prime so we have this many plus 1 y primes. That's going to be equal to-- well, I purposely added y prime to both sides and so we are left with 1 to the xy squared. And now we just have to divide both sides by this. And we're left with the derivative of y with respect to x is equal to this, which I will just copy and paste. Actually, let me just rewrite it. Scroll down a little bit. It's equal to 1 minus y squared e to the xy squared over this business. Let me get some more space. 2xye to the xy squared plus 1. And we're done. It was kind of crazy, but fundamentally no different than what we've been doing in the last few examples." + }, + { + "Q": "at 4:50 why does he get \"+1)y'\" and not just 2y'? it seems like there is already a y' on that side so now that he added another there should be two right?", + "A": "He just skipped one step, when he added y he got 2xy(e^xy^2)*y +y Now you can t just add this and get 2y but they both have a y so you can do this y (2xy*e^xy^2+1)", + "video_name": "ZtI94pI4Uzc", + "timestamps": [ + 290 + ], + "3min_transcript": "times the derivative of y with respect to x, which we are now writing as y prime. And then that's going to be equal to 1 minus y prime. And like we've been doing, we now have to just solve for y prime. So let's distribute this exponential, this e to the xy squared. And we get e, or maybe I should say y squared times e to the xy squared. So that's that. Plus 2xye to the xy squared. y prime, the derivative of y with respect to x, is equal to 1 minus the derivative of y with respect Now let's get all of our y primes on one side. So let's add y prime to both sides. So let's add-- and just to be clear, I'm adding one y prime to both sides. So let's add a y prime to both sides. And let's subtract this business from both sides. So let's subtract y squared e to the xy squared subtracting from both sides. So we're going to subtract y squared e to the xy squared. And we are left with 2xye to the xy squared plus 1 times y prime. We had this many y primes and then we add another 1y prime so we have this many plus 1 y primes. That's going to be equal to-- well, I purposely added y prime to both sides and so we are left with 1 to the xy squared. And now we just have to divide both sides by this. And we're left with the derivative of y with respect to x is equal to this, which I will just copy and paste. Actually, let me just rewrite it. Scroll down a little bit. It's equal to 1 minus y squared e to the xy squared over this business. Let me get some more space. 2xye to the xy squared plus 1. And we're done. It was kind of crazy, but fundamentally no different than what we've been doing in the last few examples." + }, + { + "Q": "I don't understand how the equation represents a surface (after 7:00). Can anyone recommend any sites that might show an animation to help me visualize it?", + "A": "This is fairly easy. Just imagine now that the only variables we have are m and b. Then you can come up with an equation that looks like this: z=x^2-2*x-2*y+2*x*y+y^2 where x is m, y is b and z is SL error. If you plot that (copy paste it into google search) you ll end up with a surface (3 variables). Here it is drawn as a parabola (z=x^2+y^2), however it resembles only some similarity with the latter. To check it just copy paste provided equations into google search and it will plot it for you.", + "video_name": "f6OnoxctvUk", + "timestamps": [ + 420 + ], + "3min_transcript": "And then we have plus m squared times n times the mean of the x squared values. And then we have-- almost there, home stretch-- we have this over here which is plus 2mb times n times the mean of the x values. And then, finally, we have plus nb squared. So really, in the last two to three videos, all we've done is we simplified the expression for the sum of the squared differences from the those n points to this line, y equals mx plus b. So we're finished with the hard core algebra stage. The next stage, we actually want to optimize this. this expression right over here. We want to find the m and the b values that minimize it. And to help visualize it, we're going to start breaking into a little bit of three-dimensional calculus here. But hopefully it won't be too daunting. If you've done any partial derivatives, it won't be difficult. This is a surface. If you view that you have the x and y data points, everything here is a constant except for the m's and the b's. We're assuming that we have the x's and y's. So we can figure out the mean of the squared values of y, the mean of the xy product, the mean of the y's, the mean of the x squareds. We assume that those are all actual numbers. So this expression right here, it's actually going to be a surface in three dimensions. So you can imagine, this right here, that is the m-axis. And then, you could imagine the vertical axis to be the squared error. This is the squared error of the line axis. So for any combination of m and b, if you're in the mb plane, you pick some combination of m and b. You put it into this expression for the squared It'll give you a point. If you do that for all of the combinations of m's and b's, you're going to get a surface. And the surface is going to look something like this. I'm going to try my best to draw it. It's going to look like this. You could almost imagine it as a kind of a bowl." + }, + { + "Q": "In 6:33 when Sal says that pi is an irrational number, if someone does find out that pi terminates or repeats does that make pi a rational number?", + "A": "Pi cannot be rational. However, if someone can prove that it terminates or repeats, then yes, it would be rational", + "video_name": "qfQv8GzyjB4", + "timestamps": [ + 393 + ], + "3min_transcript": "will cancel out. And we just have to figure out what 34,028 minus 340 is. So let's just figure this out. 8 is larger than 0, so we won't have to do any 2 is less than 4. So we will have to do some regrouping, but we can't borrow yet because we have a 0 over there. And 0 is less than 3, so we have to do some regrouping there or some borrowing. So let's borrow from the 4 first. So if we borrow from the 4, this becomes a 3 and then this becomes a 10. And then the 2 can now borrow from the 10. This becomes a 9 and this becomes a 12. And now we can do the subtraction. 8 minus 0 is 8. 12 minus 4 is 8. 9 minus 3 is 6. 3 minus nothing is 3. 3 minus nothing is 3. So 9,900x is equal to 33,688. So we get 33,688. Now, if we want to solve for x, we just divide both sides by 9,900. Divide the left by 9,900. Divide the right by 9,900. And then, what are we left with? We're left with x is equal to 33,688 over 9,900. Now what's the big deal about this? Well, x was this number. x was this number that we started off with, this number that just kept on repeating. And by doing a little bit of algebraic manipulation and subtracting one multiple of it from another, we're able to express that same exact x as a fraction. Now this isn't in simplest terms. I mean they're both definitely divisible by 2 and it looks like by 4. So you could put this in lowest common form, but we All we care about is the fact that we were able to represent x, we were able to represent this number, as a fraction. As the ratio of two integers. So the number is also rational. It is also rational. And this technique we did, it doesn't only apply to this number. Any time you have a number that has repeating digits, you could do this. So in general, repeating digits are rational. The ones that are irrational are the ones that never, ever, ever repeat, like pi. And so the other things, I think it's pretty obvious, this isn't an integer. The integers are the whole numbers that we're dealing with. So this is someplace in between the integers. It's not a natural number or a whole number, which depending on the context are viewed as subsets of integers. So it's definitely none of those. So it is real and it is rational. That's all we can say about it." + }, + { + "Q": "At 6:31 Sal said that Pi never repeats. however, doesn't Pi always repeat?", + "A": "What he meant was pi doesn t have a pattern. You never see the same set of numbers twice. Unlike a repeating decimal such as 0.3838383838.... and so on.", + "video_name": "qfQv8GzyjB4", + "timestamps": [ + 391 + ], + "3min_transcript": "will cancel out. And we just have to figure out what 34,028 minus 340 is. So let's just figure this out. 8 is larger than 0, so we won't have to do any 2 is less than 4. So we will have to do some regrouping, but we can't borrow yet because we have a 0 over there. And 0 is less than 3, so we have to do some regrouping there or some borrowing. So let's borrow from the 4 first. So if we borrow from the 4, this becomes a 3 and then this becomes a 10. And then the 2 can now borrow from the 10. This becomes a 9 and this becomes a 12. And now we can do the subtraction. 8 minus 0 is 8. 12 minus 4 is 8. 9 minus 3 is 6. 3 minus nothing is 3. 3 minus nothing is 3. So 9,900x is equal to 33,688. So we get 33,688. Now, if we want to solve for x, we just divide both sides by 9,900. Divide the left by 9,900. Divide the right by 9,900. And then, what are we left with? We're left with x is equal to 33,688 over 9,900. Now what's the big deal about this? Well, x was this number. x was this number that we started off with, this number that just kept on repeating. And by doing a little bit of algebraic manipulation and subtracting one multiple of it from another, we're able to express that same exact x as a fraction. Now this isn't in simplest terms. I mean they're both definitely divisible by 2 and it looks like by 4. So you could put this in lowest common form, but we All we care about is the fact that we were able to represent x, we were able to represent this number, as a fraction. As the ratio of two integers. So the number is also rational. It is also rational. And this technique we did, it doesn't only apply to this number. Any time you have a number that has repeating digits, you could do this. So in general, repeating digits are rational. The ones that are irrational are the ones that never, ever, ever repeat, like pi. And so the other things, I think it's pretty obvious, this isn't an integer. The integers are the whole numbers that we're dealing with. So this is someplace in between the integers. It's not a natural number or a whole number, which depending on the context are viewed as subsets of integers. So it's definitely none of those. So it is real and it is rational. That's all we can say about it." + }, + { + "Q": "isn't a 360 degree a cricle i'm confused and whats the difference between supplementary and complemtary at 3:14", + "A": "A supplementary is a pair of angles with 180 degrees total while complementary is a pair of angles that equal 90degrees.", + "video_name": "zrqzG6xKa1A", + "timestamps": [ + 194 + ], + "3min_transcript": "x plus y is equal to 180 degrees. And why does that make sense? Because look, if we add up x plus y we have gone halfway around the circle. So that's 180 degrees, right? So this is part of the way, and this is the rest of the way. So x plus y are going to equal 180 degrees. So hopefully we have learned that. And then let me switch colors for the sake of variety. Let me use my line tool. If I have-- let's see, I'm going to draw perpendicular lines. If I have that line, and then I have that line. And they are perpendicular. And then I have another line. Let's say it goes like that. And then I say that this is angle x. This is angle x. And this is angle y. Well, I said this line and this line are perpendicular, right? So that means that they intersect at a 90 degree angle. So we know that this whole thing is 90 degrees. And so what do we know about x plus y? Well, x plus y is going to equal 90 degrees. Or we could say that x and y are complementary. And I always get confused between supplementary and complementary. You just got to memorize it. I don't know if there's any-- let's see, is there any easy way? 180, supplementary. You could say that 180-- 100 starts with an O, which supplementary does not start with. So there. There's your mnemonic. And 90 starts with an N, and complementary does That's your other mnemonic. Complementary. I don't know if I'm spelling it right. Who cares? Let's move on. So let's learn some more stuff about angles. And what I'm going to do is I'm going to give you an arsenal, and then once you have that arsenal you can just tackle these beastly problems that I'm going to throw at you. So just take these for granted right now, and then in a few videos, probably, we're going to tackle some beastly problems. And you know, I'm using variables here. And if you're not familiar with variables you can put numbers here. If x was 30 degrees, then y is going to be 60 degrees. Or in this case, if x is, I don't know, 45 degrees, then y is going to be 135 degrees. That other way. Let me draw another property of angles of intersecting lines. So if I have two angles, two lines that intersect like this." + }, + { + "Q": "I'm confused about the proof at 7:20. Exactly afterwards when he did 180-x+y=180? Why did he do that?", + "A": "He proved earlier that z = 180 - x , so then he could use this knowledge to replace the meaning of z: (z) + y = 180 (180 - x) + y = 180 The items in parentheses above are the same thing: we know this because it was proven earlier. This is how math proofs happen: you reuse the equations you ve already solved to help rewrite the equation you re trying to solve in a different way. That way, you reduce it into what you know is a fact.", + "video_name": "zrqzG6xKa1A", + "timestamps": [ + 440 + ], + "3min_transcript": "Well what do we know about angle x and angle z? It may not be obvious to you because I've drawn it slightly different, but I'll give you a small hint with an appropriately interesting color. So what angle is this whole thing right here? Well I'm just going along a line, right? That's halfway around a circle. So what angle is that? Well that's 180 degrees. So what does x plus z equal? Well, x plus z is going to equal that larger angle. x plus purple z is going to equal-- I think I'll switch to the blue; maybe it's taking too much time for me to switch-- is equal to 180 degrees. Or x and z are supplementary. So what do we know about z? Well z is equal to 180 minus x. Because x plus z is 180. Fine. Now, what's the relationship between z and y? Well, z and y are also supplementary. Because look, if I drew this angle here. Look at this big angle. Well once again I'm still going halfway around the circle. But now I'm using this line right here. So that's 180 degrees. So we know that angle z plus angle y is also equal to 180 degrees. Or, I don't want to keep writing it, but z and y are also supplementary. But we just figured out that z is 180 minus x. Right? So let's just substitute that back in here. So we get 180 minus x plus y is equal to 180 degrees. Why don't we subtract 180 degrees from both sides of this equation. That cancels out, and we get minus x plus y is equal to 0. And then add x to both sides of this equation, and we get y is equal to x. That was a very long way of showing you something that is fairly simple-- that opposite angles are equal to each other. So x is equal to y. And if you've played around with this, if you just drew a bunch of straight lines and they intersected at different angles, I think when you eyeball it it would make sense." + }, + { + "Q": "At 01:26, why does he put y/x? Would it also work the other way round?", + "A": "He does that because that s how you find the answer to the problem, and no it does not work the other way around.", + "video_name": "Iqws-qzyZwc", + "timestamps": [ + 86 + ], + "3min_transcript": "In this video I'm going to do a bunch of example slope problems. Just as a bit of review, slope is just a way of measuring the inclination of a line. And the definition-- we're going to hopefully get a good working knowledge of it in this video-- the definition of it is a change in y divided by change in x. This may or may not make some sense to you right now, but as we do more and more examples, I think it'll make a good amount of sense. Let's do this first line right here. Line a. Let's figure out its slope. They've actually drawn two points here that we can use as the reference points. So first of all, let's look at the coordinates of those points. So you have this point right here. What's its coordinates? Its x-coordinate is 3. Its y-coordinate is 6. And then down here, this point's x-coordinate is negative 1 and its y-coordinate is negative 6. So there's a couple of ways we can think about slope. We could say change in y-- so slope is change in y over change in x. We can figure it out numerically. I'll in a second draw it graphically. So what's our change in y? Our change in y is literally how much did our y values change going from this point to that point? So how much did our y values change? Our y went from here, y is at negative 6 and it went all the way up to positive 6. So what's this distance right here? It's going to be your end point y value. It's going to be 6 minus your starting point y value. Minus negative 6 or 6 plus 6, which is equal to 12. You say one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. So when we changed our y value by 12, we had to change our x same change in y? Well we went from x is equal to negative 1 to x is equal to 3. Right? x went from negative 1 to 3. So we do the end point, which is 3 minus the starting point, which is negative 1, which is equal to 4. So our change in y over change in x is equal to 12/4 or if we want to write this in simplest form, this is the same thing as 3. Now the interpretation of this means that for every 1 we move over-- we could view this, let me write it this way. Change in y over change in x is equal to-- we could say it's 3 or we could say it's 3/1. Which tells us that for every 1 we move in the positive x-direction, we're going to move up 3 because this is a positive 3 in the y-direction. You can see that. When we moved 1 in the x, we moved up 3 in the y. When we moved 1 in the x, we moved up 3 in the y." + }, + { + "Q": "At 13:41, when Sal says that the slope is undefined, does that just mean that there is no slope for that particular line at all?", + "A": "Not really. Remember, the slope of a line is a fraction: change in y / change in x. If the slope is undefined, it is telling you that the change in x = 0 (denominator = 0). We can t divide by 0, which makes the slope undefined. If you say the line has no slope, this can be confused with a slope = 0. In this case, you have no change in Y (numerator = 0). And, you will have a horizontal line. So, to avoid confusion, it is better to say the slope = 0 or the slope - undefined. Hope this helps.", + "video_name": "Iqws-qzyZwc", + "timestamps": [ + 821 + ], + "3min_transcript": "These are interesting. Let's do the line e right here. Change in y over change in x. So our change in y, when we go from this point to this point-- I'll just count it out. It's one, two, three, four, five, six, seven, eight. It's 8. Or you could even take this y-coordinate 2 minus negative 6 will give you that distance, 8. What's the change in y? Well the y-value here is-- oh sorry what's the change in x? The x-value here is 4. The x-value there is 4. X does not change. So it's 8/0. Well, we don't know. 8/0 is undefined. So in this situation the slope is undefined. When you have a vertical line, you say your slope is undefined. Because you're dividing by 0. But that tells you that you're dealing probably with a vertical line. Now finally let's just do this one. This seems like a pretty straight up vanilla slope You have that point right there, which is the point 3, 1. So this is line f. You have the point 3, 1. Then over here you have the point negative 6, negative 2. So our slope would be equal to change in y. I'll take this as our ending point, just so you can go in different directions. So our change in y-- now we're going to go down in that direction. So it's negative 2 minus 1. That's what this distance is right here. Negative 2 minus 1, which is equal to negative 3. Notice we went down 3. And then what is going to be our change in x? Well, we're going to go back that amount. What is that amount? Well, that is going to be negative 6, that's our end point, minus 3. That gives us that distance, which is negative 9. Which is the same thing as if we go forward 9, we're going to go up 3. All equivalent. And we see these cancel out and you get a slope of 1/3. Positive 1/3. It's an upward sloping line. Every time we run 3, we rise 1. Anyway, hopefully that was a good review of slope for you." + }, + { + "Q": "What is the difference between 80:1 and 1:80? He said it can go the other way around.", + "A": "Both scenarios gives you something that is 80 times as big as another object/line. So changing the order doesn t change much. However, you need to be specific to which object the 1 and the 80 belongs to. Building a miniature ship with a 80:1 ratio is a bit confusing, because the miniature ship is prob 80 times smaller instead of 80 times larger than the real one.", + "video_name": "byjmR7JBXKc", + "timestamps": [ + 4801, + 140 + ], + "3min_transcript": "" + }, + { + "Q": "How did sal in 3:55 get 12.5 from 128,000? -dazed and confused", + "A": "Well, at 3:50 Sal got 12.8 square meters (m^2), in your question you said 12.5 (maybe a typo?), but the conversion is a simple conversion from square centimeters (cm^2) to square meters (m^2).", + "video_name": "byjmR7JBXKc", + "timestamps": [ + 235 + ], + "3min_transcript": "Let me write this down. 400 times 320. Let's think about it. 4 times 32 is going to be 120, plus 8, 128. And I have 1, 2, 3 zeroes. 1, 2, 3. So it's going to be 128,000 centimeters squared. Now that's a lot of square centimeters. What would we do if we wanted to convert it into meters? Well, we just have to figure out how many square centimeters are there in a square meter. So let's think about it this way. A meter is equal to-- 1 meter is equal to 100 centimeters. So a square meter, so that's right over there. is the same thing as 100 centimeters by 100 centimeters. And so if you were to calculate this area in centimeters, 100 times 100 is 10,000, is equal to 10,000 centimeters squared. So you have 10,000 square centimeters for every square meter. And so, if you want to convert 128,000 centimeters squared to meters squared, you would divide by 10,000. So dividing that by 10,000 would give us 12.8 square meters. Now, another way you could've done it, and maybe this would have been easier, is to convert it up here. Instead of saying 400 centimeters times 320 centimeters, you would say, well, And 320 centimeters, well, that's 3.2 meters. And you would say, OK, 4 times 3.2, that is 12.8 square meters. But either way, the area of the living room in the real world in meters squared, or square meters, is 12.8." + }, + { + "Q": "How many more proportion can we make from 2:4::3:6\nPlz answer. My answer is total 8 including ratio given above.", + "A": "Since a ratio is a fraction, we can create a infinite number of proportions from one ratio. Take any ratio, multiply its 2 parts by the same number and you will get an equivalent ratio. Since the 2 ratios are equal, you have a proportion. The number you select to multiply with can be any number. Since there are an infinite set of numbers, you can create an infinite set of ratios.", + "video_name": "qYjiVWwefto", + "timestamps": [ + 124, + 186 + ], + "3min_transcript": "What I want to introduce you to in this video is the notion of a proportional relationship. And a proportional relationship between two variables is just a relationship where the ratio between the two variables is always going to be the same thing. So let's look at an example of that. So let's just say that we want to think about the relationship between x and y. And let's say that when x is one, y is three, and then when x is two, y is six. And when x is nine, y is 27. Now this is a proportional relationship. Why is that? Because the ratio between y and x is always the same thing. And actually the ratio between y and x or, you could say the ratio between x and y, is always the same thing. So, for example-- if we say the ratio y over x-- this is always equal to-- it could be three over one, which is just three. It could be six over two, It could be 27 over nine, which is also just three. So you see that y over x is always going to be equal to three, or at least in this table right over here. And so, or at least based on the data points we have just seen. So based on this, it looks like that we have a proportional relationship between y and x. So this one right over here is proportional. So given that, what's an example of relationships that are not proportional. Well those are fairly easy to construct. So let's say we had-- I'll do it with two different variables. So let's say we have a and b. And let's say when a is one, b is three. And when a is two, b is six. And when a is 10, b is 35. when a is one, b is three so the ratio b to a-- you could say b to a-- you could say well when b is three, a is one. Or when a is one, b is three. So three to one. And that's also the case when b is six, a is two. Or when a is two, b is six. So it's six to two. So these ratios seem to be the same. They're both three. But then all of sudden the ratio is different right over here. This is not equal to 35 over 10. So this is not a proportional relationship. In order to be proportional the ratio between the two variables always has to be the same. So this right over here-- This is not proportional. Not proportional. So the key in identifying a proportional relationship is look at the different values that the variables take on when one variable is one value," + }, + { + "Q": "why in 3:55 the ecuation is equal to 1?", + "A": "try to compare with this simple numeric example 6 / 3 = 2 and then take 2 to other side dividing in the denominator and we have 6 / (3 * 2) = 1", + "video_name": "6YRGEsQWZzY", + "timestamps": [ + 235 + ], + "3min_transcript": "And now, we can substitute the derivative of y with respect to x is just this. And then the right hand side of the equation is this. But we can substitute v for y over x. So let's do that. And so we get v plus x. Instead of dv [? dv ?] x, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared, we're making the substitution v is equal to y over x. All of that over 2v. Now, let's see what we can do. This is where we just get our algebra hat on, and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. So we'll get 2v squared plus 2xv v prime-- 2v times x, yep, Now let's see, let's subtract 2v squared from both sides of this. And we will be left with 2xv v prime is equal to 1 plus-- let's see, we're subtracting 2v squared from both sides. So we're just left with a 1 plus v squared here, right? 3v squared minus 2v squared is just v squared. And let's see, we want it to be separable, so let's put all the v's on the left hand side. So we get 2xv v prime divided by 1 plus v squared is equal to 1. And let's divide both sides by x. So we get the x's on the other side. So then we get 2v-- and I'll now switch back Instead of v prime, I'll write dv dx. 2v times the derivative of v with respect to x divided by 1 plus v squared is equal to-- I'm dividing both sides by x, notice I didn't write the x on this side-- so that is equal to 1 over x. And then, if we just multiply both sides of this times dx, we've separated the two variables and we can integrate So let's do that. Let's go up here. I'll switch to a different color, so you know I'm working on a different column now. So multiply both sides by dx. I get 2v over 1 plus v squared dv is equal to 1 over x dx. And now we can just integrate both sides of this equation. This is a separable equation in terms of v and x. And what's the integral of this? At first, you might think, oh boy, this is complicated. This is difficult, maybe some type of trig function." + }, + { + "Q": "At 3:49, why did he divide 2xvv' by the entire right hand side as opposed to subtracting v^2 from both sides?", + "A": "When solving a separable equation you don t want to have a term you are adding to the dy/dx (or in this case dv/dx) term. The problem is that it will make it harder to separate the variables but lets try: 2x dv/dx - v^2 = 1 2 dv/dx - v^2/x = 1/x 2 dv - ((v^2/x) dx) = 1/x dx So by creating an additional term on the left side of the equation you have a mix of terms and so you can t integrate.", + "video_name": "6YRGEsQWZzY", + "timestamps": [ + 229 + ], + "3min_transcript": "And now, we can substitute the derivative of y with respect to x is just this. And then the right hand side of the equation is this. But we can substitute v for y over x. So let's do that. And so we get v plus x. Instead of dv [? dv ?] x, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared, we're making the substitution v is equal to y over x. All of that over 2v. Now, let's see what we can do. This is where we just get our algebra hat on, and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. So we'll get 2v squared plus 2xv v prime-- 2v times x, yep, Now let's see, let's subtract 2v squared from both sides of this. And we will be left with 2xv v prime is equal to 1 plus-- let's see, we're subtracting 2v squared from both sides. So we're just left with a 1 plus v squared here, right? 3v squared minus 2v squared is just v squared. And let's see, we want it to be separable, so let's put all the v's on the left hand side. So we get 2xv v prime divided by 1 plus v squared is equal to 1. And let's divide both sides by x. So we get the x's on the other side. So then we get 2v-- and I'll now switch back Instead of v prime, I'll write dv dx. 2v times the derivative of v with respect to x divided by 1 plus v squared is equal to-- I'm dividing both sides by x, notice I didn't write the x on this side-- so that is equal to 1 over x. And then, if we just multiply both sides of this times dx, we've separated the two variables and we can integrate So let's do that. Let's go up here. I'll switch to a different color, so you know I'm working on a different column now. So multiply both sides by dx. I get 2v over 1 plus v squared dv is equal to 1 over x dx. And now we can just integrate both sides of this equation. This is a separable equation in terms of v and x. And what's the integral of this? At first, you might think, oh boy, this is complicated. This is difficult, maybe some type of trig function." + }, + { + "Q": "Hi sal, at \"7:49\" you subtract the cx^3 to the left side of the equation. Wouldn't you have to divide that expression to the left side? I was confused.", + "A": "He brought the entire CX^3 term to the left side. Whereas in dividing, he would have to have left either the C or X^3 term on the right side. 1 + 2 = 3 -> 1 + 2 - 3 = 0 hope this helps!", + "video_name": "6YRGEsQWZzY", + "timestamps": [ + 469 + ], + "3min_transcript": "So that's how I knew that this was the antiderivative. And if you don't believe me, use the chain rule and take the derivative of this, and you'll get this. And hopefully, it will make a little bit more sense. But anyway, that's the left hand side, and then that equals-- Well, this one's easy. That's the natural log, the absolute value of x. We could say, plus c, but just so that we can simplify it a little bit, an arbitrary constant c, we can really just write that as the natural log of the absolute value of some constant c. I mean, this is still some arbitrary constant c. So we can rewrite this whole equation as the natural log of 1 plus v squared is equal to-- when you add natural logs, you can essentially just multiply the two numbers that you're taking the natural log of-- the natural log of, we could say, the absolute value of cx. And so the natural log of this is equal to the natural log of this. So we could say that 1 plus v squared is equal to cx. So we know v is equal to y over x, so let's do that. So we get 1 plus y over x squared is equal to cx. Let me scroll this down a little bit. Let's multiply both sides of the equation times x squared. We could rewrite this as y squared over x squared. So we multiply both sides times x squared, you get x squared plus y squared is equal to cx to the third. And we're essentially done. If we want to put all of the variable terms on left hand side, we could say that this is equal to x squared plus y squared minus cx to the third is equal to 0. And this implicitly defined function, or curve, or however you want to call it, is the solution to our original homogeneous first order differential equation. I will see you in the next video. And now, we're actually going to do something. We're going to start embarking on higher order And actually, these are more useful, and in some ways, easier to do than the homogeneous and the exact equations that we've been doing so far. See you in the next video." + }, + { + "Q": "At 6:56 couldn't he have taken everything as a power of e. to cancel the natural log. Or would it have become e^(ln(x) + C) which is the same as e^(ln(x))*e^c which is cx.\n\nSo it would be the same solution, so, nvm. Just figured that in the fly. Was my logic correct?", + "A": "Your logic is correct, I regualary use the approach you suggested in my Differential Equations class and the teachers does it too. I think Sal just took a different approach", + "video_name": "6YRGEsQWZzY", + "timestamps": [ + 416 + ], + "3min_transcript": "reverse chain rule. We have a function here, 1 plus v squared, an expression here. And we have its derivative sitting right there. So the antiderivative of this, and you can make a substitution if you like. You could say u is equal to 1 plus v squared, then du is equal to 2v dv. And then, well, you would end up saying that the antiderivative is just the natural log of u. Or, in this case, the antiderivative of this is just the natural log of 1 plus v squared. We don't even have to write an absolute value there. Because that's always going to be a positive value. So the natural log of 1 plus v squared. And I hope I didn't confuse you. That's how I think about it. I say, if I have an expression, and I have its derivative multiplied there, then I can just take the antiderivative of the whole expression. And I don't have to worry about what's inside of it. So if this was a 1 over an x, or 1 over u, it's just the So that's how I knew that this was the antiderivative. And if you don't believe me, use the chain rule and take the derivative of this, and you'll get this. And hopefully, it will make a little bit more sense. But anyway, that's the left hand side, and then that equals-- Well, this one's easy. That's the natural log, the absolute value of x. We could say, plus c, but just so that we can simplify it a little bit, an arbitrary constant c, we can really just write that as the natural log of the absolute value of some constant c. I mean, this is still some arbitrary constant c. So we can rewrite this whole equation as the natural log of 1 plus v squared is equal to-- when you add natural logs, you can essentially just multiply the two numbers that you're taking the natural log of-- the natural log of, we could say, the absolute value of cx. And so the natural log of this is equal to the natural log of this. So we could say that 1 plus v squared is equal to cx. So we know v is equal to y over x, so let's do that. So we get 1 plus y over x squared is equal to cx. Let me scroll this down a little bit. Let's multiply both sides of the equation times x squared. We could rewrite this as y squared over x squared. So we multiply both sides times x squared, you get x squared plus y squared is equal to cx to the third. And we're essentially done. If we want to put all of the variable terms on left hand side, we could say that this is equal to x squared plus y squared minus cx to the third is equal to 0. And this implicitly defined function, or curve, or however you want to call it, is the solution to our original homogeneous first order differential equation." + }, + { + "Q": "At 3:33, how did Sal get 0.1 from the second equation?", + "A": "The -0.1 is part of the function. It was given as part of the problem definition. Sal didn t create it.", + "video_name": "GA_yxxeFYBU", + "timestamps": [ + 213 + ], + "3min_transcript": "In some ways a recursive function is easier, because you can say okay look. The first term when n is equal to one, if n is equal to one, let me just write it, If n is equal to one, if n is equal to one, what's g of n gonna be? It's gonna be negative 31, negative 31. And if n, if n is greater than one and a whole number, so this is gonna be defined for all positive integers, and whole, and whole number, it's just going to be the previous term, so g of n minus one minus seven, minus seven. We're saying hey if we're just picking an arbitrary term we just have to look at the previous term and then subtract, and then subtract seven. It all works out nice and easy, because you keep looking at previous, previous, previous terms all the way until you get to the base case, which is when n is equal to one, and you can build up back from that. You get this exact same sequence. but let's go the other way around. Here we have a, we have a sequence defined recursively, and I want to create a function that defines a sequence explicitly. Let's think about this. One way to think about it, this sequence, when n is equal to one it starts at 9.6, and then every term is the previous term minus 0.1. The second term is gonna be the previous term minus 0.1, so it's gonna be 9.5. Then you're gonna go to 9.4. Then you're gonna go to 9.3. We could keep going on and on and on. If we want, we could make a little table here, and we could say this is n, this is h of n, and you see when n is equal to one, h of n is 9.6. When n is equal to two, we're now in this case over here, it's gonna h of two minus one, so it's gonna be h of one minus 0.1. which is going to be 9.5. When h is three, it's gonna be h of two, h of two minus 0.1, minus 0.1. H of two is right over here. You subtract a tenth you're gonna get 9.4, exactly what we saw over here. Let's see if we can pause the video now and define this... Create a function that constructs or defines this arithmetic sequence explicitly. Here it was recursively. We wanna define it explicitly. So let's just call it, I don't know, let's just call it f of n. We can say look, it's gonna be 9.6, but we're gonna subtract, we're gonna subtract 0.1 a certain number of times depending on what term we're talking about. We're gonna subtract 0.1, but how many times are we gonna subtract it as a function of n? If we're talking about the first term we subtract zero times." + }, + { + "Q": "At 4:40 when Sal solves the integral and gets 2*ln|x-1|, can I just forget the absolute value of x-1 because of the logarithmic rules? it's the same like ln(x-1)^2 so my argument is always positive and therefore always valid for my ln. Or did I overlook something?", + "A": "Since you took the square out , you must put a modulus there as negative log isn t defined. But if the square is there, its always positive so no worries.", + "video_name": "5j81gyHn9i0", + "timestamps": [ + 280 + ], + "3min_transcript": "but it's actually much more useful for finding the integral. Negative 1/2 plus two over x minus one, d x. Now, how do we evaluate this? Well, the antiderivative of negative 1/2 is pretty straight forward. That's just going to be negative 1/2 x, plus the entire derivative of two over x minus one. You might reduce here ahead. The derivative of x minus one is just one so you could say that the derivative is sitting there. We can essentially use substitution in our heads and say, \"Okay, let's just take the entire derivative, \"we could say with respect to x minus one which will be \"the natural log of the absolute value of x minus one.\" If all of that sounds really confusing, I'll let you do the u-substitution. If I were just trying to evaluate I could see, okay, the derivative of x minus one is just one so I could say u is equal to x minus one, and then d u is going to be equal to d x. This is going to be, we can rewrite in terms of u as two, I'll just take the constant out two times the integral of one over u, d u, which we know as two times the natural log of the absolute value of u plus c. In this case, we know that u is x minus one. This is equal to two times the natural log of x minus one plus c. That's what we're going to have right over here. So plus two times the natural log of the absolute value of x minus one plus c. The plus c doesn't just come from this one, this general overtaking the integral of the whole thing. we take the derivative and the constant will go away. Let me just put the plus c right over there, and we are done." + }, + { + "Q": "Just before 2:31 you got 2*2*x*x*x**y and I did not understand how you got that because dont they factor out or something?", + "A": "That 2*2*x*x*x*y, or 4x^3y, is the greatest common factor of the two polynomials, and he factors it out of both of them.", + "video_name": "_sIuZHYrdWM", + "timestamps": [ + 151 + ], + "3min_transcript": "of its basic constituents. Now let's do the same thing for 8-- I'll color code it-- 8x to the third. Let me do it in similar colors. So in this situation we have 8x to the 1/3 y. So the prime factorization of 8 is 2 times 2 times 2. It's 2 times 2 times 2. Prime, or I should say the factorization of x to the third, or the expansion of it, is just times x times x times x. x multiplied by itself three times. And then we are multiplying everything by a y here, times y. So what factors are common to both of these? And we want to include as many of them as possible to find this greatest common factor. So we have two 2's here, three 2's here. So we only have two 2's in common in both of them. So we only have three x's in common. Three x's and three x's. And we have a y here and a y here. So y is common to both expressions. So the greatest common factor here is going to be 2 times 2. So it's going to be 2 times 2 times x times x times x times Or 4x to the third y. So this is what we want to factor out. So that means we can write this thing as-- if we factor out a 4x to the third y, where essentially we have to divide each of these by 4x to the third. We're factoring it out. So let me rewrite this. So this is 4x to the fourth y plus 8x to the third y. And we're going to divide each of these by 4x to the third y. If we were to multiply this out, we would distribute this 4x to the third y on each terms. And then it would cancel with the denominator. You would have the same thing in the numerator and denominator. And then you would get this expression over here. So hopefully this makes sense that these are the exact same expression. But when you write it this way, then it becomes pretty clear that this is 4x to the third y. And then you just simplify each of these expressions. 4 cancels with 4. x to the fourth divided by x to the third is x. y divided by y is just 1. So you have x plus 8 divided by 4 is 2. x to the third divided by x to the third is 1. Y divided by y is 1. So x plus 2. Another way to see what's left over when you factor it out is if you were to take out the common factor. So we took out this and this. What was left over in 4x to the fourth y when we took this stuff out? When we undistributed it? Well, the only thing that was left was this x right over here. Let me do that in another color." + }, + { + "Q": "say you have 6 peanut cookies and you have 6 friends two of your friends cannot eat nuts. So the ratio is 6:2 (thats how you write a ratio) Now say you baked six more cookies to give to 12 of your friends. For every 6 peanut cookies you have two friends. So what is the ratio if you have 12 cookies?", + "A": "Solving this problem is easier than it seems. The way I do it is that I turn the ratio into a fraction multiply the numerator and denominator by x (in your case,2) then turn it back into a fraction. 6:2 = 6/2 6 X 2 = 12 -------------------- 2 X 2 =4 12/4 = 12:4 By observing the diagram, we can tell the ratio is now 12:4. Hope This Helps!", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 362 + ], + "3min_transcript": "And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples." + }, + { + "Q": "at 2:05 how is 6/9 the same thing as 2/3?", + "A": "6/9 simplifies into 2/3", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 125 + ], + "3min_transcript": "Voiceover:We've got some apples here and we've got some oranges and what I want to think about is, what is the ratio, what is the ratio of apples ... Of apples, to oranges? To oranges. To clarify what we're even talking about, a ratio is giving us the relationship between quantities of 2 different things. So there's a couple of ways that we can specify this. We can literally count the number of apples. 1, 2, 3, 4, 5, 6. So we have 6 apples. And we can say the ratio is going to be 6 to, 6 to ... And then how many oranges do we have? 1, 2, 3, 4, 5, 6, 7, 8, 9. It is 6 to 9. The ratio of apples to oranges is 6 to 9. And you could use a different notation. You could also write it this way. 6 to ... You would still read the ratio as being 6 to 9. because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges" + }, + { + "Q": "kate and nora each have a sum of money. the ratio of the amount of money Kate has to that of Nora is 3:5. After Nora gives $150 to Kate, the ratio of the money is 7:9. Find the sum of money Kate had initially?", + "A": "I think it goes like this Kate to Nora = 3:5 Kate to Total sum = 3:8 Kate+150 to Nora = 7:9 Kate+150 to Total sum = 7: 16 We want Kate initial sum to total moeny. 3/8 and 7/16 have different denominators and we need them the same. 3*2/8*2 = 6/16 Total sum = 150/(1/16) = $2400. Kates initial sum = 2400*(6/16) = $900", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 185, + 429 + ], + "3min_transcript": "because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples." + }, + { + "Q": "At about 1:09 shouldn't he have added fraction form?", + "A": "I think it is covered in the next video.", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 69 + ], + "3min_transcript": "Voiceover:We've got some apples here and we've got some oranges and what I want to think about is, what is the ratio, what is the ratio of apples ... Of apples, to oranges? To oranges. To clarify what we're even talking about, a ratio is giving us the relationship between quantities of 2 different things. So there's a couple of ways that we can specify this. We can literally count the number of apples. 1, 2, 3, 4, 5, 6. So we have 6 apples. And we can say the ratio is going to be 6 to, 6 to ... And then how many oranges do we have? 1, 2, 3, 4, 5, 6, 7, 8, 9. It is 6 to 9. The ratio of apples to oranges is 6 to 9. And you could use a different notation. You could also write it this way. 6 to ... You would still read the ratio as being 6 to 9. because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges" + }, + { + "Q": "Why can\u00e2\u0080\u0099t there be another way to write a ratio, and can there be a negative ratio like -4:10", + "A": "Hello random stranger!! There are 3 different ways to write a ratio. Ex: 2:4 2/4 2 to 4 Hope that helps random stranger! XD", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 250 + ], + "3min_transcript": "because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples." + }, + { + "Q": "Ratio of boys to girls is 3:4 if 252 students attends how many are boys", + "A": "SInce ratios are the same thing as fractions, if the ratio of the boys to girls is 3:4, that means 3/7 children are boys. Therefor, you would have to find 1/7 of 252 (or 36). Then, since 3/7 children are boys, you multiply 1/7x3, or 36+36+36=108. Finally, your answer is 108 boys.", + "video_name": "bIKmw0aTmYc", + "timestamps": [ + 184 + ], + "3min_transcript": "because one way to think about ratios, especially if we're thinking about apples to oranges, is how many apples do we have for a certain number of oranges? When you think about it that way, we can actually reduce these numbers, as you might have already thought about. Both 6 and 9 are divisible by 3. So just like we can reduce fractions, we can also reduce ratios. So if you divide 6 and 9 both by 3. 6 divided by 3 is 2. 6 divided by 3 is 2. And 9 divided by 3 is 3. So we could also say that the ratio of apples to oranges is 2 to 3. Or if we want to use this notation, 2 to 3. 2 to 3. Now, does that make sense? Well look. We divided each of these groups into 3. If you divide this whole total into 3 groups. So 1 group, 1 group. 2 groups, 2 groups. And 3 equal groups. We see that in each of those groups, for every 2 apples, for every 2 apples, we have 3 oranges. For 2 apples we have 3 oranges. For 2 apples we have 3 oranges. So, once again, the ratio of apples to oranges. For every 2 apples we have 3 oranges. But we could think about things the other way around as well. We could also think about what is the ratio ... We could also think about what is the ratio ... Ratio, of oranges to apples? Oranges to apples. And here we would, essentially, switch the numbers. The ratio of oranges to apples. Notice, up here we said apples to oranges And here we're going to say the ratio of oranges to apples, so we've swapped these 2. So we're going to swap the numbers. Here we have 9 oranges for every 6 apples. So we could say the ratio is going to be 9 to 6. The ratio is 9 to 6. Or if we want to reduce it, for every 3 oranges ... So we're going to divide this by 3. So for every 3 oranges we are going to have 2 apples. We are going to have 2 apples. So notice, this is just exactly what we had up here, but when we had apples to oranges it was 6 to 9. 6 apples for every 9 oranges. And now when it's oranges to apples, we say it's 9 to 6. 9 oranges for every 6 apples. Or we could say for every 3 oranges we have exactly 2 apples." + }, + { + "Q": "at 9:46 how is it that you can compare the matrices\n[a] and [b]\n[c] ___ [d]\nto the fractions a/c=b/d ??? I thought matrices were used to notate vectors, not ratios....", + "A": "We are looking at the situation where the determinant is 0. So ad - bc =0 and thus ad = bc ad/cd = bc/cd a/c = b/d", + "video_name": "UqyN7-tRS00", + "timestamps": [ + 586 + ], + "3min_transcript": "And not only would they intersect, they would intersect in an infinite number of places. But still you would have no unique solution. You'd have no one solution to this equation. It would be true at all values of x and y. So you can kind of view it when you apply the matrices to this problem. The matrix is singular, if the two lines that are being represented are either parallel, or they are the exact same line. They're parallel and not intersecting at all. Or they are the exact same line, and they intersect at an infinite number of points. And so it kind of makes sense that the A inverse wasn't defined. So let's think about this in the context of the linear combinations of vectors. That's not what I wanted to use to erase it. So when we think of this problem in terms of linear That this is the same thing as the vector ac times x plus the vector bd times y, is equal to the vector ef. So let's think about it a little bit. We're saying, is there some combination of the vector ac and the vector bd that equals the vector ef. But we just said that if we have no inverse here, we know that because the determinant is 0. And if the determinant is 0, then we know in this situation that a/c must equal b/d. So a/c is equal to b/d. So what does that tell us? Well let me draw it. And maybe numbers would be more helpful here. But I think you'll get the intuition. I'll just assume both sectors are in the first quadrant. Let me draw. The vector ac. Let's say that this is a. Let me do it in a different color. So I'm gonna draw the vector ac. So if this is a, and this is c, then the vector ac looks like that. Let me draw it. I want to make this neat. The vector ac is like that. And then we have the arrow. And what would the vector bd look like? Well the vector bd-- And I could draw it arbitrarily someplace. But we're assuming that there's no derivative-- sorry, no determinant. Have I been saying derivative the whole time?" + }, + { + "Q": "At 0:53, he writes: zero to the second power, is 1*0*0. Why would this equation be 1*0*0? Since any number times one is that number, what would be the difference between 1*0*0 and 0*0? And where did the one come in anyway? Wouldn't zero to the second power just be written as 0*0? Maybe i'm just missing something. Thanks!\n\nLorenzo, age 9", + "A": "I m not exactly sure why he put a one there either. I think it s out of habit from other videos he s done, when taking the powers of non-zero numbers. I think one example is proving that any non-zero number is 1, like this: 5^2 = 1*5*5 5^1 = 1*5 5^0 = 1 His putting a one in front of the equation probably came from that. Perhaps someone might use that to prove that 0^0 is also one if that s his opinion.", + "video_name": "PwDnpb_ZJvc", + "timestamps": [ + 53 + ], + "3min_transcript": "So let's think a little bit about powers of 0. So what do you think 0 to the first power is going to be? And I encourage you to pause this video. Well, let's just think about it. One definition of exponentiation is that you start with a 1, and then, you multiply this number times a 1 one time. So this is literally going to be 1 times-- let me do it in the right color-- it's 1 times 0. You're multiplying the 1 by 0 one time. 1 times 0, well, that's just going to be equal to 0. Now, what do you think 0 squared or 0 to the second power is going to be equal to? Well, once again, one way of thinking about this is that you start with a 1, and we're going to multiply it by 0 two times. So times 0 times 0. Well, what's that going to be? once again, you are going to get 0. And I think you see a pattern here. If I take 0 to any non-zero number-- so to the power of any non-zero, so this is some non-zero number, then this is going to be equal to 0. Now, this raises a very interesting question. What happens at 0 to the 0-th power? So here, 0 to the millionth power is going to be 0. 0 to the trillionth power is going to be 0. Even negative or fractional exponents, which we haven't talked about yet, as long as they're non-zero, this is just going to be equal to 0, kind of makes sense. But now, let's think about what 0 to the 0-th power is, because this is actually a fairly deep question. Well, actually, why don't you pause the video and think a little bit about what 0 to the 0-th power should be. Well, there's two trains of thought here. You could say, look, 0 to some non-zero number is 0. So why don't we just extend this to all numbers and say 0 to any number should be 0. And so maybe you should say that 0 to the 0-th power is 0. But then, there was another train of logic that we've already learned, that any non-zero number, if you take any non-zero number, and you raise it to the 0-th power. We've already established that you start with a 1, and you multiply it times that non-zero number 0 times. So this is always going to be equal to 1 for non-zero numbers. So maybe say, hey, maybe we should extend this to all numbers, including 0. So maybe 0 to the 0-th power should be 1." + }, + { + "Q": "at 2:15 why so many zero and why is he talking about stuff we havent even learned yet", + "A": "Different schools go at different paces and introduce things in different orders. Now expand that idea to all the schools in your country and then expand it again to all the schools in the world. It s not possible to make a website that exactly matches every school and every teacher.", + "video_name": "PwDnpb_ZJvc", + "timestamps": [ + 135 + ], + "3min_transcript": "So let's think a little bit about powers of 0. So what do you think 0 to the first power is going to be? And I encourage you to pause this video. Well, let's just think about it. One definition of exponentiation is that you start with a 1, and then, you multiply this number times a 1 one time. So this is literally going to be 1 times-- let me do it in the right color-- it's 1 times 0. You're multiplying the 1 by 0 one time. 1 times 0, well, that's just going to be equal to 0. Now, what do you think 0 squared or 0 to the second power is going to be equal to? Well, once again, one way of thinking about this is that you start with a 1, and we're going to multiply it by 0 two times. So times 0 times 0. Well, what's that going to be? once again, you are going to get 0. And I think you see a pattern here. If I take 0 to any non-zero number-- so to the power of any non-zero, so this is some non-zero number, then this is going to be equal to 0. Now, this raises a very interesting question. What happens at 0 to the 0-th power? So here, 0 to the millionth power is going to be 0. 0 to the trillionth power is going to be 0. Even negative or fractional exponents, which we haven't talked about yet, as long as they're non-zero, this is just going to be equal to 0, kind of makes sense. But now, let's think about what 0 to the 0-th power is, because this is actually a fairly deep question. Well, actually, why don't you pause the video and think a little bit about what 0 to the 0-th power should be. Well, there's two trains of thought here. You could say, look, 0 to some non-zero number is 0. So why don't we just extend this to all numbers and say 0 to any number should be 0. And so maybe you should say that 0 to the 0-th power is 0. But then, there was another train of logic that we've already learned, that any non-zero number, if you take any non-zero number, and you raise it to the 0-th power. We've already established that you start with a 1, and you multiply it times that non-zero number 0 times. So this is always going to be equal to 1 for non-zero numbers. So maybe say, hey, maybe we should extend this to all numbers, including 0. So maybe 0 to the 0-th power should be 1." + }, + { + "Q": "Wait... at 0:43 Sal says that 0 to the first power is 0*1, but I thought it was 0*0, can anyone explain why? Thanks!", + "A": "Ah, 0^1 = 0 and 0^0 is undefined. Does that help?", + "video_name": "PwDnpb_ZJvc", + "timestamps": [ + 43 + ], + "3min_transcript": "So let's think a little bit about powers of 0. So what do you think 0 to the first power is going to be? And I encourage you to pause this video. Well, let's just think about it. One definition of exponentiation is that you start with a 1, and then, you multiply this number times a 1 one time. So this is literally going to be 1 times-- let me do it in the right color-- it's 1 times 0. You're multiplying the 1 by 0 one time. 1 times 0, well, that's just going to be equal to 0. Now, what do you think 0 squared or 0 to the second power is going to be equal to? Well, once again, one way of thinking about this is that you start with a 1, and we're going to multiply it by 0 two times. So times 0 times 0. Well, what's that going to be? once again, you are going to get 0. And I think you see a pattern here. If I take 0 to any non-zero number-- so to the power of any non-zero, so this is some non-zero number, then this is going to be equal to 0. Now, this raises a very interesting question. What happens at 0 to the 0-th power? So here, 0 to the millionth power is going to be 0. 0 to the trillionth power is going to be 0. Even negative or fractional exponents, which we haven't talked about yet, as long as they're non-zero, this is just going to be equal to 0, kind of makes sense. But now, let's think about what 0 to the 0-th power is, because this is actually a fairly deep question. Well, actually, why don't you pause the video and think a little bit about what 0 to the 0-th power should be. Well, there's two trains of thought here. You could say, look, 0 to some non-zero number is 0. So why don't we just extend this to all numbers and say 0 to any number should be 0. And so maybe you should say that 0 to the 0-th power is 0. But then, there was another train of logic that we've already learned, that any non-zero number, if you take any non-zero number, and you raise it to the 0-th power. We've already established that you start with a 1, and you multiply it times that non-zero number 0 times. So this is always going to be equal to 1 for non-zero numbers. So maybe say, hey, maybe we should extend this to all numbers, including 0. So maybe 0 to the 0-th power should be 1." + }, + { + "Q": "At 1:46 Sal says that this property extends to negative exponents as well. But 0^-1 power would be 1/0^1 = 1/0 which is undefined right?", + "A": "Yes you are correct. By definition, a negative exponent implies a reciprocal. That is, any number, x^(-1) is equivalent to 1/x.", + "video_name": "PwDnpb_ZJvc", + "timestamps": [ + 106 + ], + "3min_transcript": "So let's think a little bit about powers of 0. So what do you think 0 to the first power is going to be? And I encourage you to pause this video. Well, let's just think about it. One definition of exponentiation is that you start with a 1, and then, you multiply this number times a 1 one time. So this is literally going to be 1 times-- let me do it in the right color-- it's 1 times 0. You're multiplying the 1 by 0 one time. 1 times 0, well, that's just going to be equal to 0. Now, what do you think 0 squared or 0 to the second power is going to be equal to? Well, once again, one way of thinking about this is that you start with a 1, and we're going to multiply it by 0 two times. So times 0 times 0. Well, what's that going to be? once again, you are going to get 0. And I think you see a pattern here. If I take 0 to any non-zero number-- so to the power of any non-zero, so this is some non-zero number, then this is going to be equal to 0. Now, this raises a very interesting question. What happens at 0 to the 0-th power? So here, 0 to the millionth power is going to be 0. 0 to the trillionth power is going to be 0. Even negative or fractional exponents, which we haven't talked about yet, as long as they're non-zero, this is just going to be equal to 0, kind of makes sense. But now, let's think about what 0 to the 0-th power is, because this is actually a fairly deep question. Well, actually, why don't you pause the video and think a little bit about what 0 to the 0-th power should be. Well, there's two trains of thought here. You could say, look, 0 to some non-zero number is 0. So why don't we just extend this to all numbers and say 0 to any number should be 0. And so maybe you should say that 0 to the 0-th power is 0. But then, there was another train of logic that we've already learned, that any non-zero number, if you take any non-zero number, and you raise it to the 0-th power. We've already established that you start with a 1, and you multiply it times that non-zero number 0 times. So this is always going to be equal to 1 for non-zero numbers. So maybe say, hey, maybe we should extend this to all numbers, including 0. So maybe 0 to the 0-th power should be 1." + }, + { + "Q": "Why can he switch natural logarithm with the limit at 6:24 ? He explained it but it didn't really convince me. My textbook doesn't seem to mention anything about that kind of limit rule.", + "A": "Think about how we approach analyzing composite functions. For example, if I am looking at the lim as x --> \u00e2\u0088\u009e for sin(1/x), analyzing the sin part of the function doesn t really matter until I determine the behavior of 1/x as x --> \u00e2\u0088\u009e. Therefore, I can rewrite the limit as sin (lim as x --> \u00e2\u0088\u009e of 1/x) and this will not change the answer. Try it and see for yourself.", + "video_name": "3nQejB-XPoY", + "timestamps": [ + 384 + ], + "3min_transcript": "So let's make that substitution. So all of this is going to be equal to the limit as -- now we've gotten rid of our delta x. We're going to say the limit as an approaches infinity of the natural log -- I'll go back to that mauve color -- the natural log of 1 plus -- now, I said that instead of delta x over x, I made the substitution that that is equal to 1 over n. So that's 1 plus 1 over n. And then what's 1 over delta x? Well delta x is equal to x over n, so 1 over delta x is going to be the inverse of this. It's going to be n over x. And then we can re-write this expression right here -- let me re-write it again. This is equal to the limit as n approaches infinity of the natural log of 1 plus 1 over n. I could say this is to the n, and then all of this to the 1 over x. Once again, this is just an exponent property. If I raise something to the n and then to the 1 over x, I could just multiply the exponent to get to the n over x. So these two statements are equivalent. But now we can use logarithm properties to say hey, if this is the exponent, I can just stick it out in front of the coefficient right here. I could put it out right there. And just remember, this was all of the derivative with respect to x of the natural log of x. So what is that equal to? We could put this 1 out of x in the front here. In fact, that 1 out of x term, it has nothing to do with n. It's kind of a constant term when you think of it in terms of n. So we can actually put it all the way out here. We could put it either place. So we could say 1 over x times all that stuff in mauve. The limit as n approaches infinity of the natural log The natural log of all of that stuff. Or, just to make the point clear, we can re-write this part -- let me make a salmon color -- equal to 1 over x times the natural log of the limit as n approaches infinity. I'm just switching places here, because obviously what we care is what happens to this term as it approaches infinity, of 1 plus 1 over n to the n. Well what is -- this should look a little familiar to you on some of the first videos where we talked about e -- this is one of the definitions of e. e is defined. I'm still not using this at all. I'm just stating that the definition of e, e is equal to the limit as n approaches infinity of 1 plus 1 over n to the n." + }, + { + "Q": "At 3:25, why did Sal use the substitution. I don't get how he just defines delta(x) / x equal to 1/n.\nDid I miss something from the previous video? (I just started watching from this video)\n\nPlease help.", + "A": "Because n hasn t appeared anywhere in any equations we are free to use it how we like. we re just making the equation look different to make it easier to recognize what s going on. Sorry if my answer doesn t help.", + "video_name": "3nQejB-XPoY", + "timestamps": [ + 205 + ], + "3min_transcript": "natural log of x plus delta x minus the natural log of x. All of that over delta x. Now we can just use the property of logarithms. If I have the log of a minus the log of b, that's the same thing as a log of a over b. So let me re-write it that way. So this is going to be equal to the limit as delta x approaches 0. I could take this 1 over delta x right here. 1 over delta x times the natural log of x plus delta x divided by this x. Just doing the logarithm properties right there. Then I can re-write this -- first of all, when I have this coefficient in front of a logarithm I can make this the exponent. And then I can simplify this in here. 0 of the natural log -- let me do this in a new color. Let me do it in a completely new color. The natural log of -- the inside here I'll just divide everything by x. So x divided by x is 1. Then plus delta x over x. Then I had this 1 over delta x sitting out here, and I can make that the exponent. That's just an exponent rule right there, or a logarithm property. 1 over delta x. Now I'm going to make a substitution. Remember, all of this, this was all just from my definition of a derivative. This was all equal to the derivative of the natural log of x. I have still yet to in any way use this. And I won't use that until I actually show it to you. I've become very defensive about these claims of circularity. They're my fault because that shows that I wasn't clear to be more clear this time. So let's see if we can simplify this into terms that we recognize. Let's make the substitution so that we can get e in maybe terms that we recognize. Let's make the substitute delta x over x is equal to 1 over n. If we multiply -- this is the same thing. This is the equivalent to substitution. If we multiply both sides of this by x, as saying that delta x is equal to x over n. These are equivalent statements. I just multiplied both sides by x here. Now if we take the limit as n approaches infinity of this term right here, that's equivalent -- that's completely equivalent to taking the limit as delta x approaches 0. If we're defining delta x to be this thing, and we take the limit as its denominator approaches 0, we're going" + }, + { + "Q": "at 2:54, why did sal write 8? wasn't it 7?", + "A": "No it is f(7) = 8, so at x = 7, y reaches it maximum height of 8. Range has to do with possible ys.", + "video_name": "sXP7VhU1gYE", + "timestamps": [ + 174 + ], + "3min_transcript": "is greater than or equal to negative 6. Or we could say negative 6 is less than or equal to x, which is less than or equal to 7. If x satisfies this condition right over here, the function is defined. So that's its domain. So let's check our answer. Let's do a few more of these. The function f of x is graphed. What is its domain? Well, exact similar argument. This function is not defined for x is negative 9, negative 8, all the way down or all the way up I should say to negative 1. At negative 1, it starts getting defined. f of negative 1 is negative 5. So it's defined for negative 1 is less than or equal to x. And it's defined all the way up to x equals 7, including x equals 7. So this right over here, negative 1 is less than or equal to x is less than or equal to 7, the function is defined for any x that satisfies this double inequality right over here. Let's do a few more. What is its range? So now, we're not thinking about the x's for which this function is defined. We're thinking about the set of y values. Where do all of the y values fall into? Well, let's see. The lowest possible y value or the lowest possible value of f of x that we get here looks like it's 0. The function never goes below 0. So f of x-- so 0 is less than or equal to f of x. It does equal 0 right over here. f of negative 4 is 0. And then the highest y value or the highest value that f of x obtains in this function definition is 8. f of 7 is 8. It never gets above 8, but it does equal 8 right over here when x is equal to 7. So 0 is less than f of x, which is less than or equal to 8. So that's its range. Let's do a few more. This is kind of fun. The function f of x is graphed. So once again, this function is defined for negative 2. Negative 2 is less than or equal to x, which is less than or equal to 5. If you give me an x anywhere in between negative 2 and 5, I can look at this graph to see where the function is defined. f of negative 2 is negative 4. f of negative 1 is negative 3. So on and so forth, and I can even pick the values in between these integers. So negative 2 is less than or equal to x, which is less than or equal to 5." + }, + { + "Q": "At 2:13, he says that the formula is clean, what does he mean by that?", + "A": "When he says that the formula of the volume of the cone is clean, he means that the cone is EXACTLY 1/3 of the volume of a cylinder. As you can see in the video, the volume of a cylinder is h(pi)*r^2, and the area of a cylinder is exactly 1/3 of that, which is known as 1/3*h(pi)*r^2. So when he says clean, he means that the volume of a cylinder is exactly 1/3 the volume of a cone.", + "video_name": "hC6zx9WAiC4", + "timestamps": [ + 133 + ], + "3min_transcript": "Let's think a little bit about the volume of a cone. So a cone would have a circular base, or I guess depends on how you want to draw it. If you think of like a conical hat of some kind, it would have a circle as a base. And it would come to some point. So it looks something like that. You could consider this to be a cone, just like that. Or you could make it upside down if you're thinking of an ice cream cone. So it might look something like that. That's the top of it. And then it comes down like this. This also is those disposable cups of water you might see at some water coolers. And the important things that we need to think about when we want to know what the volume of a cone is we definitely want to know the radius of the base. So that's the radius of the base. Or here is the radius of the top part. You definitely want to know that radius. And you want to know the height of the cone. You could call this distance right over here h. And the formula for the volume of a cone-- and it's interesting, because it's close to the formula for the volume of a cylinder in a very clean way, which is somewhat surprising. And that's what's neat about a lot of this three-dimensional geometry is that it's not as messy as you would think it would be. It is the area of the base. Well, what's the area of the base? The area of the base is going to be pi r squared. It's going to be pi r squared times the height. And if you just multiplied the height times pi r squared, that would give you the volume of an entire cylinder that looks something like that. So this would give you this entire volume of the figure that looks like this, where its center of the top is the tip right over here. So if I just left it as pi r squared of this entire can, this entire cylinder. But if you just want the cone, it's 1/3 of that. It is 1/3 of that. And that's what I mean when I say it's surprisingly clean that this cone right over here is 1/3 the volume of this cylinder that is essentially-- you could think of this cylinder as bounding around it. Or if you wanted to rewrite this, you could write this as 1/3 times pi or pi/3 times hr squared. However you want to view it. The easy way I remember it? For me, the volume of a cylinder is very intuitive. You take the area of the base. And then you multiply that times the height. And so the volume of a cone is just 1/3 of that. It's just 1/3 the volume of the bounding cylinder is one way to think about it. But let's just apply these numbers, just to make sure that it makes sense to us. So let's say that this is some type of a conical glass," + }, + { + "Q": "Why is f(x,y)=p(x,y)i+q(x,y)j a function of 't' as Sal mentions around @15:28", + "A": "Well, f(x,y) is a function of x and y. But if we re following the path defined by r(t) = x(t)i + y(t)j then our x and y are themselves just functions of t, so f(x,y) becomes f(x(t),y(t)), in short it depends completely on t (though indirectly through x(t) and y(t)). I.e. f(x(t),y(t)) becomes f(t).", + "video_name": "t3cJYNdQLYg", + "timestamps": [ + 928 + ], + "3min_transcript": "How do we actually calculate something like this? Especially because we have everything parameterized in terms of t. How do we get this in terms of t? And if you just think about it, what is f dot r? Or what is f dot dr? Well, actually, to answer that, let's remember what dr looked like. If you remember, dr/dt is equal to x prime of t, I'm writing it like, I could have written dx dt if I wanted to do, times the i-unit vector, plus y prime of t, times the j-unit vector. And if we just wanted to dr, we could multiply both sides, if we're being a little bit more hand-wavy with the differentials, not too rigorous. We'll get dr is equal to x prime of t dt times the unit vector i plus y prime of t times the differential dt So this is our dr right here. And remember what our vector field was. It was this thing up here. Let me copy and paste it. And we'll see that the dot product is actually not so crazy. So copy, and let me paste it down here. So what's this integral going to look like? This integral right here, that gives the total work done by the field, on the particle, as it moves along that path. Just super fundamental to pretty much any serious physics that you might eventually find yourself doing. So you could say, well gee. It's going to be the integral, let's just say from t is equal to a, to t is equal to b. Right? a is where we started off on the path, t is equal to a to t is equal to b. You can imagine it as being timed, as a particle And then what is f dot dr? Well, if you remember from just what the dot product is, you can essentially just take the product of the corresponding components of your of vector, and add them up. So this is going to be the integral from t equals a to t equals b, of p of p of x, really, instead of writing x, y, it's x of t, right? x as a function of t, y as a function of t. So that's that. Times this thing right here, times this component, right? We're multiplying the i-components. So times x prime of t d t, and then that plus, we're going to do the same thing with the q function. So this is q plus, I'll go to another line. Hopefully you realize I could have just kept writing, but I'm running out of space. Plus q of x of t, y of t, times the component of our dr. Times" + }, + { + "Q": "5:48\nhow does variable summation applies to 1", + "A": "There are a few ways to think about it. One is, when i is 1, then 1 is still 1. When i is 2, then 1 is still 1 right? When i is some number k, then 1 is still 1. When i is N, 1 is still one. So when you sum along the index i, 1 is simply added to itself again and again N times. So the answer is N. The other way to think about it, which is essentially the same is this. Since i is never 0, we can rewrite 1 as i/i. Then Sum(1)=Sum(i/i)=1/1+2/2+3/3+...+(n-2)/(n-2)+(n-1)/(n-1)+n/n=1+1+1+...+1+1+1=N.", + "video_name": "sRVGcYGjUk8", + "timestamps": [ + 348 + ], + "3min_transcript": "this is just a constant term. This is just a constant term. So you can take it out. Times mu squared times the sum from i equals 1 to n. And what's going to be here? It's going to be a 1. We just divided a 1. We just divided this by 1. And took it out of the sigma sign, out of the sum. And you're just left with a 1 there. And actually, we could have just left the mu squared there. But either way, let's just keep simplifying it. So this we can't really do-- well, actually we could. Well, no, we don't know what the x sub i's are. So we just have to leave that the same. So that's the sum. Oh sorry, and this is just the numerator. This whole simplification, we're just simplifying the numerator. And later, we're just going to divide by n. So that is equal to that divided by n, which is equal to this thing divided by n. Because it's the numerator that's the confusing part. We just want to simplify this term up here. So let's keep doing this. So this equals the sum from i equals 1 to n of x sub i squared. And let's see, minus 2 times mu-- sorry, that mu doesn't look good. Edit, Undo, minus 2 times mu times the sum from i is equal to 1 to n of xi. And then, what is this? What is another way to write this? Essentially, we're going to add 1 to itself n times. This is saying, just look, whatever you have here, just iterate through it n times. If you had an x sub i here, you would use the first x term, then the second x term. When you have a 1 here, this is just essentially saying, add one to itself n times, which is the same thing as n. And then see if there's anything else we can do here. Remember, this was just the numerator. So this looks fine. We add up each of those terms. So we just have minus 2 mu from i equals 1 to-- oh well, think about this. What is this? What is this thing right here? Well actually, let's bring back that n. So this simplified to that divided by n, which simplifies to that whole thing, which is simplified to this whole thing, divided by n, which simplifies to this whole thing divided by n, which is the same thing as each of the terms divided by n, which" + }, + { + "Q": "Around 3:30 Sal references the Calculus playlist--I'm not even CLOSE to that playlist yet. Am I watching these videos too soon? It seems like the Statistics playlist is showing up really early on my practice map and I may not have the skills to successfully accomplish the unit. Do you think this could be true? I did okay up through standard deviation, but z-scores, empirical rule and some references are throwing me off!", + "A": "This is like a side tour, sightseeing in a cool neighborhood. You don t need to move into the calculus house to work in statistics. For example, I think the formula for the Standard Deviation of a uniform distribution is (b-a)/sqrt(12). I wanted to know, Why 12? I asked Doctor Math and he (Doctor Anthony) gave me an explanation that I (frankly) didn t understand, but trust. I don t need to know where the 12 came from to use the formula, but I find it comforting to know that someone knows.", + "video_name": "sRVGcYGjUk8", + "timestamps": [ + 210 + ], + "3min_transcript": "This is the same thing as x sub i squared minus-- this is your little algebra going on here. So when you square it-- I mean, we could multiply it out. We could write it. x sub i minus mu times x sub i minus mu. So we have x sub i times x sub i, that's x sub i squared. Then you have x sub i times minus mu. And then you have minus mu times x sub i. So when you add those two together, you get minus 2x sub i mu, because you have it twice. x sub i times mu, that's 1 minus x sub i mu. And then you have another one, minus mu x sub i. When you add them together, you get minus 2x sub i mu. I know it's confusing with me saying sub i and all of that. But it's really no different than when Just the variables look a little bit more complicated. And then the last term is minus mu times minus mu, which is plus mu squared. Fair enough. Let me switch colors just to keep it interesting. Let me cordon that off. The sum of this is the same thing as the sum of-- because if you think about it, we're going to take each x sub i. For each of the numbers in our population, we're going to perform this thing. And we're going to sum it up. But if you think about it, this is the same thing as-- if you're not familiar with sigma notation this is a good thing to know in general, just a little bit of intuition. That this is the same thing as-- I'll do it here to have space. The sum from i is equal to 1 to n of the first term, x sub i squared minus-- and actually, we can bring out the constant terms. is the thing that has the i-th term. So in this case, it's x sub i. So x sub 1, x sub 2. So that's the thing that you have to leave on the right hand side of the sigma notation. And if you've done the calculus playlists already, sigma notation is really like a discrete integral on some level. Because in an integral, you're summing up a bunch of things and you're multiplying them times dx, which is a really small interval. But here you're just taking a sum. And we showed in the calculus playlist that an integral actually is this infinite sum of infinitely small things, but I don't want to digress too much. But this was just a long way of saying that the sum from i equals 1 to n of the second term is the same thing as minus 2 times mu of the sum from i is equal to 1 to n of x sub i." + }, + { + "Q": "At 2:05 Sal says that the line he just drew was called a transversal, because it transversed across the other two lines. But what does it mean to transverse?", + "A": "To transverse means to intercept.", + "video_name": "H-E5rlpCVu4", + "timestamps": [ + 125 + ], + "3min_transcript": "Let's say we have two lines over here. Let's call this line right over here line AB. So A and B both sit on this line. And let's say we have this other line over here. We'll call this line CD. So it goes through point C and it goes through point D. And it just keeps on going forever. And let's say that these lines both sit on the same plane. And in this case, the plane is our screen, or this little piece of paper that we're looking at right over here. And they never intersect. So they're on the same plane, but they never intersect each other. If those two things are true, and when they're not the same line, they never intersect and they can be on the same plane, then we say that these lines are parallel. They're moving in the same general direction, in fact, the exact same general direction. If we were looking at it from an algebraic point of view, we would say that they have the same slope, but they have different y-intercepts. They involve different points. they would intersect that at a different point, but they would have the same exact slope. And what I want to do is think about how angles relate to parallel lines. So right over here, we have these two parallel lines. We can say that line AB is parallel to line CD. Sometimes you'll see it specified on geometric drawings like this. They'll put a little arrow here to show that these two lines are parallel. And if you've already used the single arrow, they might put a double arrow to show that this line is parallel to that line right over there. Now with that out of the way, what I want to do is draw a line that intersects both of these parallel lines. So here's a line that intersects both of them. Let me draw a little bit neater than that. So let me draw that line right over there. Well, actually, I'll do some points over here. Well, I'll just call that line l. And this line that intersects both of these parallel lines, This is a transversal line. It is transversing both of these parallel lines. This is a transversal. And what I want to think about is the angles that are formed, and how they relate to each other. The angles that are formed at the intersection between this transversal line and the two parallel lines. So we could, first of all, start off with this angle right over here. And we could call that angle-- well, if we made some labels here, that would be D, this point, and then something else. But I'll just call it this angle right over here. We know that that's going to be equal to its vertical angles. So this angle is vertical with that one. So it's going to be equal to that angle right over there. We also know that this angle, right over here, is going to be equal to its vertical angle, or the angle that is opposite the intersection. So it's going to be equal to that. And sometimes you'll see it specified like this, where you'll see a double angle mark like that. Or sometimes you'll see someone write" + }, + { + "Q": "At 5:00, if we were given the measure of the yellow angle in the second diagram, could we find out what the measure of the green angle was? If so, how?", + "A": "No, you would not. Since the two angles exist on different lines, and the lines are not parallel, then there is no way to know for certain if the angles would compliment each other or not.", + "video_name": "H-E5rlpCVu4", + "timestamps": [ + 300 + ], + "3min_transcript": "and these two are equal right over here. Now the other thing we know is we could do the exact same exercise up here, that these two are going to be equal to each other and these two are going to be equal to each other. They're all vertical angles. What's interesting here is thinking about the relationship between that angle right over there, and this angle right up over here. And if you just look at it, it is actually obvious what that relationship is-- that they are going to be the same exact angle, that if you put a protractor here and measured it, you would get the exact same measure up here. And if I drew parallel lines-- maybe I'll draw it straight left and right, it might be a little bit more obvious. So if I assume that these two lines are parallel, and I have a transversal here, what I'm saying is that this angle is going to be the exact same measure as that angle there. And to visualize that, just imagine tilting this line. And as you take different-- so it If you take the line like this and you look at it over here, it's clear that this is equal to this. And there's actually no proof for this. This is one of those things that a mathematician would say is intuitively obvious, that if you look at it, as you tilt this line, you would say that these angles are the same. Or think about putting a protractor here to actually measure these angles. If you put a protractor here, you'd have one side of the angle at the zero degree, and the other side would specify that point. And if you put the protractor over here, the exact same thing would happen. One side would be on this parallel line, and the other side would point at the exact same point. So given that, we know that not only is this side equivalent to this side, it is also equivalent to this side over here. And that tells us that that's also equivalent to that side over there. So all of these things in green are equivalent. And by the same exact argument, this angle And that's going to be the same as this angle, because they are opposite, or they're vertical angles. Now the important thing to realize is just what we've deduced here. The vertical angles are equal and the corresponding angles at the same points of intersection are also equal. And so that's a new word that I'm introducing right over here. This angle and this angle are corresponding. They represent kind of the top right corner, in this example, of where we intersected. Here they represent still, I guess, the top or the top right corner of the intersection. This would be the top left corner. They're always going to be equal, corresponding angles. And once again, really, it's, I guess, for lack of a better word, it is a bit obvious. Now on top of that, there are other words that people will see. We've essentially just proven that not only is this angle equivalent to this angle, but it's also equivalent to this angle right over here. And these two angles-- let me label them so that we can make some headway here." + }, + { + "Q": "Hi friends :-) I have a question for you-\n1> at 6:27 \"Sal\" proved us that alternate interior angles are equal , right? so, here is my question ?\nwe know that :- b=c,f=g\nnow b and c are vertical angles and we also know that b = f (corresponding angle) so why can't we say that c = f ,right? and why not alternate exterior angles?", + "A": "Ok, so you said b=c,f=g. Which says b=c and f=g. If you look back at Sals blackboard on the video there is no comma, it simply says b=c=f=g which means they are ALL equal, so yes , c does equal f and c equals g, and b equals g etc. etc.. As for alternate exterior angles, the same rule applies. a=d=e=h. They are all equal.", + "video_name": "H-E5rlpCVu4", + "timestamps": [ + 387 + ], + "3min_transcript": "If you take the line like this and you look at it over here, it's clear that this is equal to this. And there's actually no proof for this. This is one of those things that a mathematician would say is intuitively obvious, that if you look at it, as you tilt this line, you would say that these angles are the same. Or think about putting a protractor here to actually measure these angles. If you put a protractor here, you'd have one side of the angle at the zero degree, and the other side would specify that point. And if you put the protractor over here, the exact same thing would happen. One side would be on this parallel line, and the other side would point at the exact same point. So given that, we know that not only is this side equivalent to this side, it is also equivalent to this side over here. And that tells us that that's also equivalent to that side over there. So all of these things in green are equivalent. And by the same exact argument, this angle And that's going to be the same as this angle, because they are opposite, or they're vertical angles. Now the important thing to realize is just what we've deduced here. The vertical angles are equal and the corresponding angles at the same points of intersection are also equal. And so that's a new word that I'm introducing right over here. This angle and this angle are corresponding. They represent kind of the top right corner, in this example, of where we intersected. Here they represent still, I guess, the top or the top right corner of the intersection. This would be the top left corner. They're always going to be equal, corresponding angles. And once again, really, it's, I guess, for lack of a better word, it is a bit obvious. Now on top of that, there are other words that people will see. We've essentially just proven that not only is this angle equivalent to this angle, but it's also equivalent to this angle right over here. And these two angles-- let me label them so that we can make some headway here. for the angles themselves. So let's call this lowercase a, lowercase b, lowercase c. So lowercase c for the angle, lowercase d, and then let me call this e, f, g, h. So we know from vertical angles that b is equal to c. But we also know that b is equal to f because they are corresponding angles. And that f is equal to g. So vertical angles are equivalent, corresponding angles are equivalent, and so we also know, obviously, that b is equal to g. And so we say that alternate interior angles are equivalent. So you see that they're kind of on the interior of the intersection. They're between the two lines, but they're on all opposite sides of the transversal. Now you don't have to know that fancy word, alternate interior angles, you really just have to deduce what we just saw over here. Know that vertical angles are going to be equal and corresponding angles are going to be equal. And you see it with the other ones, too. We know that a is going to be equal to d, which is going to be equal to h, which is going to be equal to e." + }, + { + "Q": "At 3:11 Sal says that lim kf(x) = K*limf(x)\nx-->c x-->c\nshouldn't it be equal to lim k * limf(x)\nx-->c x-->c", + "A": "At 3:11 if one takes lim k, one will just get K because it is not a function, so one could put it as a function as g(x)=K, but then it would always return K and it would always approach K no matter where you get your limit. so instead of taking lim K, you just use K because K is the limit of K. This is because it is a constant, not a variable. so for example, 2f(2) where f(x)=2x will get you 2(2*2) which is 2*4 which is 8. Hope that made sense.", + "video_name": "lSwsAFgWqR8", + "timestamps": [ + 191 + ], + "3min_transcript": "plus the limit of g of x as x approaches c. Which is equal to, well this right over here is-- let me do that in that same color-- this right here is just equal to L. It's going to be equal to L plus M. This right over here is equal to M. Not too difficult. This is often called the sum rule, or the sum property, of limits. And we could come up with a very similar one with differences. The limit as x approaches c of f of x minus g of x, is just going to be L minus M. It's just the limit of f of x as x approaches c, minus the limit of g of x as x approaches c. So it's just going to be L minus M. And we also often call it the difference rule, or the difference property, of limits. intuitive. Now what happens if you take the product of the functions? The limit of f of x times g of x as x approaches c. Well lucky for us, this is going to be equal to the limit of f of x as x approaches c, times the limit of g of x, as x approaches c. Lucky for us, this is kind of a fairly intuitive property of limits. So in this case, this is just going to be equal to, this is L times M. This is just going to be L times M. Same thing, if instead of having a function here, we had a constant. So if we just had the limit-- let me do it in that same color-- the limit of k times f of x, as x approaches c, where k is just some constant. This is going to be the same thing as k times the limit And that is just equal to L. So this whole thing simplifies to k times L. And we can do the same thing with difference. This is often called the constant multiple property. We can do the same thing with differences. So if we have the limit as x approaches c of f of x divided by g of x. This is the exact same thing as the limit of f of x as x approaches c, divided by the limit of g of x as x approaches c. Which is going to be equal to-- I think you get it now-- this is going to be equal to L over M. And finally-- this is sometimes called the quotient property-- finally we'll look at the exponent property." + }, + { + "Q": "At 2:44 How do I solve this question?\n\n(3r^3-19r^2+27r+4) / (r-4)", + "A": "3r^2+31r-87-344/r-4", + "video_name": "UquFdMg6Z_U", + "timestamps": [ + 164 + ], + "3min_transcript": "The other way to think about it is that we're multiplying this entire expression. So this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you-- they're are all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here, we could just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. And then you have x to the fourth divided by x to the-- well, they don't tell us. But if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the 4 minus 1 power, or x Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1/2. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just-- well, I'll write it. I could write a 1 here. And let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. You could view it two ways-- anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1, Either way, you knew how to do this before you even learned that exponent property. Because x divided by x is 1, and then assuming x does not equal to 0. And we kind of have to assume x doesn't equal 0 in this whole thing. Otherwise, we would be dividing by 0. And then finally, we have 4 over 6x. And there's a couple of ways to think about it. So the simplest way is negative 4 over 6 is the same thing as negative 2/3. Just simplified that fraction. And we're multiplying that times 1/x. So we can view this 4 times 1/x. Another way to think about it is you could have viewed this 4 as being multiplied by x to the 0 power, and this being x to the first power. And then when you tried to simplify it using your exponent properties, you would have-- well, that would be x to the 0 minus 1 power, which is x to the negative 1 power." + }, + { + "Q": "It took me a minute to figure this out, but I wanted to check if I am right. At 4:30, the +1 -2/3x can not be simplified because it can also be stated as + 1x^0 - 2/3x^-1, which makes the x terms different. Is this correct?", + "A": "You are correct. 1 is a constant term (x^0) and is not a like term with (2/3)x^(-1).", + "video_name": "UquFdMg6Z_U", + "timestamps": [ + 270 + ], + "3min_transcript": "Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1/2. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just-- well, I'll write it. I could write a 1 here. And let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. You could view it two ways-- anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1, Either way, you knew how to do this before you even learned that exponent property. Because x divided by x is 1, and then assuming x does not equal to 0. And we kind of have to assume x doesn't equal 0 in this whole thing. Otherwise, we would be dividing by 0. And then finally, we have 4 over 6x. And there's a couple of ways to think about it. So the simplest way is negative 4 over 6 is the same thing as negative 2/3. Just simplified that fraction. And we're multiplying that times 1/x. So we can view this 4 times 1/x. Another way to think about it is you could have viewed this 4 as being multiplied by x to the 0 power, and this being x to the first power. And then when you tried to simplify it using your exponent properties, you would have-- well, that would be x to the 0 minus 1 power, which is x to the negative 1 power. here, but x to the negative 1 is the exact same thing as 1 over x. And so let's just write our answer completely simplified. So it's going to be 3x to the third minus 1/2 x plus 1-- because this thing right here is just 1-- and then minus 2 times 1 in the numerator over 3 times x in the denominator. And we are done. Or we could write this. Depending on what you consider more simplified, this last term right here could also be written in minus 2/3 x to the negative 1. But if you don't want a negative exponent, you could write it like that." + }, + { + "Q": "1:47 why did you add -4x on that side? Not as in addition but you rewrote it as a subtraction sentence using -4x . -8 doesn't have a coefficient so why did Sal do that? I thought it was X minus X or X plus X. Really Confused Here !", + "A": "In order to graph an equation, it is easiest if it is in y=mx+b form. Therefore you want y alone on the left. Since the 4x is hanging out with the 2y, by subtracting it to both sides it helps get y all by itself.", + "video_name": "V6Xynlqc_tc", + "timestamps": [ + 107 + ], + "3min_transcript": "We're asked to convert these linear equations into slope-intercept form and then graph them on a single coordinate plane. We have our coordinate plane over here. And just as a bit of a review, slope-intercept form is a form y is equal to mx plus b, where m is the slope and b is the intercept. That's why it's called slope-intercept form. So we just have to algebraically manipulate these equations into this form. So let's start with line A, so start with a line A. So line A, it's in standard form right now, it's 4x plus 2y is equal to negative 8. The first thing I'd like to do is get rid of this 4x from the left-hand side, and the best way to do that is to subtract 4x from both sides of this equation. So let me subtract 4x from both sides. The left hand side of the equation, these two 4x's cancel out, and I'm just left with 2y is equal to. or negative 8 minus 4, however you want to do it. Now we're almost at slope-intercept form. We just have to get rid of this 2, and the best way to do that that I can think of is divide both sides of this equation by 2. So let's divide both sides by 2. So we divide the left-hand side by 2 and then divide the right-hand side by 2. You have to divide every term by 2. And then we are left with y is equal to negative 4 divided by 2 is negative 2x. Negative 8 divided by 2 is negative 4, negative 2x minus 4. So this is line A, let me graph it right now. So line A, its y-intercept is negative 4. So the point 0, negative 4 on this graph. If x is equal to 0, y is going to be equal to negative 4, you So 0, 1, 2, 3, 4. That's the point 0, negative 4. That's the y-intercept for line A. And then the slope is negative 2x. So that means that if I change x by positive 1 that y goes down by negative 2. So let's do that. So if I go over one in the positive direction, I have to go down 2, that's what a negative slope's going to do, negative 2 slope. If I go over 2, I'm going to have to go down 4. If I go back negative 1, so if I go in the x direction negative 1, that means in the y direction I go positive two, because two divided by negative one is still negative two, so I go over here. If I go back 2, I'm going to go up 4. Let me just do that. Back 2 and then up 4. So this line is going to look like this." + }, + { + "Q": "At 2:50 Sal says that the point at beginning of the interval is a relative maximum point. How do we know that to be true when the points on both sides of the point in question have to have a y-coordinate equal to or less than it, but then everything to the left of the point Sal was referring to was undefined?", + "A": "Well, that means that there is nothing to the left so it only needs to worry about the existing or defined parts of the graph.", + "video_name": "zXyQI4lD4wI", + "timestamps": [ + 170 + ], + "3min_transcript": "They would be the bottom of your valleys. So that's a relative minimum point. This right over here is a relative minimum point, even if there are other parts of the function that are lower. Now there's also an edge case for both relative maxima and relative minima, and that's where the graph is flat. So if you have parts of your function where it's just constant, these points would actually be both. For example, if this is our x-axis right over here, that's our x-axis, if this is our y-axis right over there, and if this is x equals c, if you construct an open interval around c, you notice that the value of our function at c, f of c, is at least as large as the values of the function around it. And it is also at least as small as the values of the function around it, so this point would also be considered But that's an edge case that you won't encounter as often. So with that primer out of the way, let's identify the relative extrema. So first the relative maximum points. Well that's a top of a hill right over there, this is the top of a hill. You might be tempted to look at that point and that point, but notice, at this point right over here, if you go to the right, you have values that are higher than it. So it's really not at the top of a hill. And right over if you go to the left, you have values that are higher than it, so it's also not the top of a hill. And what about the relative minimum points? Well this one right over here is a relative minimum point. This one right over here is a relative minimum point. And this one over here is a relative minimum point. Now let's do an example dealing with absolute extrema. So here we're told to mark the absolute maximum and the absolute minimum points So once again, pause this video and see if you can have a go at this. So you have an absolute maximum point at let's say x equals c if and only if, so I'll write iff for if and only if, f of c is greater than or equal to f of x for all the x's in the domain of the function. And you have an absolute minimum at x equals c if and only if, iff, f of c is less than or equal to f of x for all the x's over the domain. So another way to think about it is, absolute maximum point is the high point. So over here, that is the absolute maximum point. And then the absolute minimum point is interesting because in this case, it would be actually one of, it would happen at one of the endpoints of our domain." + }, + { + "Q": "At 02:52, Sal says that the point (3,-8) would be a relative maximum point but how is that possible? The function is only till -8. How can we assume that the function will have the greatest value considering the points around it? I hope I made my question clear.", + "A": "Sal told us at the beginning of the video that the domain was closed, that is, it included the end points. The domain is [-8,6]. On this particular graph, if we start at x=-8, and move towards the right on the x axis, the next immediate f(x) is less than it was at x=-8. Because we are on a closed interval, that makes the point (-8, 3) a relative maximum. (Make sure you put your x coordinate first when referring to a point on a graph\u00f0\u009f\u0098\u008a.)", + "video_name": "zXyQI4lD4wI", + "timestamps": [ + 172 + ], + "3min_transcript": "They would be the bottom of your valleys. So that's a relative minimum point. This right over here is a relative minimum point, even if there are other parts of the function that are lower. Now there's also an edge case for both relative maxima and relative minima, and that's where the graph is flat. So if you have parts of your function where it's just constant, these points would actually be both. For example, if this is our x-axis right over here, that's our x-axis, if this is our y-axis right over there, and if this is x equals c, if you construct an open interval around c, you notice that the value of our function at c, f of c, is at least as large as the values of the function around it. And it is also at least as small as the values of the function around it, so this point would also be considered But that's an edge case that you won't encounter as often. So with that primer out of the way, let's identify the relative extrema. So first the relative maximum points. Well that's a top of a hill right over there, this is the top of a hill. You might be tempted to look at that point and that point, but notice, at this point right over here, if you go to the right, you have values that are higher than it. So it's really not at the top of a hill. And right over if you go to the left, you have values that are higher than it, so it's also not the top of a hill. And what about the relative minimum points? Well this one right over here is a relative minimum point. This one right over here is a relative minimum point. And this one over here is a relative minimum point. Now let's do an example dealing with absolute extrema. So here we're told to mark the absolute maximum and the absolute minimum points So once again, pause this video and see if you can have a go at this. So you have an absolute maximum point at let's say x equals c if and only if, so I'll write iff for if and only if, f of c is greater than or equal to f of x for all the x's in the domain of the function. And you have an absolute minimum at x equals c if and only if, iff, f of c is less than or equal to f of x for all the x's over the domain. So another way to think about it is, absolute maximum point is the high point. So over here, that is the absolute maximum point. And then the absolute minimum point is interesting because in this case, it would be actually one of, it would happen at one of the endpoints of our domain." + }, + { + "Q": "At 4:06, wouldn't 6/13 plus 6/13 be equal to 12/13? How does it add to zero?", + "A": "The term in the original equation was MINUS 6/13. Sal then added POSITIVE 6/13 to both sides. That s where the zero came from. When combining terms (or numbers) in algebra, you always have to take note of the sign as well as the numerical amount.", + "video_name": "XD-FDGdWnR8", + "timestamps": [ + 246 + ], + "3min_transcript": "1/3 plus 4/3 is indeed equal to 5/3. Let's do another one of these. So let's say that we have the equation K minus eight is equal to 11.8. So once again I wanna solve for K. I wanna have just a K on the left-hand side. I don't want this subtracting this eight right over here. So in order to get rid of this eight, let's add eight And of course, if I do it on the left-hand side, I have to do it on the right-hand side as well. So we're gonna add eight to both sides. The left-hand side, you are substracting eight and then you're adding eight. That's just going to cancel out, and you're just going to be left with K. And on the right-hand side, 11.8 plus eight. Well, 11 plus eight is 19, so it's going to be 19.8. And we're done, and once again, what's neat about equations, 19.8 minus eight is 11.8. Let's do another one, this is too much fun. Alright, so let's say that I had 5/13 is equal to T minus 6/13. Alright, this is interesting 'cause now I have my variable on the right-hand side. But let's just leave it there. Let's just see if we can solve for T by getting rid of everything else on the right-hand side. And like we've done in the past, if I'm subtracting 6/13, so why don't I just add it? Why don't I just add 6/13? I can't just do that on the right-hand side. Then the two sides won't be equal anymore, so I gotta do it on the left-hand side if I wanna hold the equality. So what happens? On the left-hand side I have, let me give myself a little bit more space, I have 5/13 plus 6/13, plus 6/13 are equal to, Well, I was subtracting 6/13, now I add 6/13. Those are just going to add to zero. 6/13 minus 6/13 is just zero, so you're left with T. So T is equal to this. If I have 5/13 and I add to that 6/13, well I'm gonna have 11/13. So this is going to be 11/13 is equal to T, or I could write that the other way around. I could write T is equal to 11/13." + }, + { + "Q": "at 4:45 is Sal writing (e^ln2)^u or (e)^ln2*u? I mean e^ln2 equals 2 and 2 is raised to the power u...so why should it be the latter and not the former?", + "A": "if you remember your logarithmic rules, you will know that log(x^y) = y*log(x)", + "video_name": "C5Lbjbyr1t4", + "timestamps": [ + 285 + ], + "3min_transcript": "So let's see. How can we redefine this right here? Well, 2, 2 is equal to what? 2 is the same thing as e to the natural log of 2, right? The natural log of 2 is the power you have to raise e to to get 2. So if you raise e to that power, you're, of course going to get 2. This is actually the definition of really, the natural log. You raise e to the natural log of 2, you're going to get 2. So let's rewrite this, using this-- I guess we could call this this rewrite or-- I don't want to call it quite a substitution. It's just a different way of writing the number 2. So this will be equal to, instead of writing the number 2, I could write e to the natural log of 2. And all of that to the u du. And now what is this equal to? Well, if I take something to an exponent, and then to another product of those exponents. So this is equal to, let me switch colors, this is equal to the integral of e, to the u, e to the, let me write it this way. e to the natural log of 2 times u. I'm just multiplying these two exponents. I raise something to something, then raise it again, we know from our exponent rules, it's just a product of those two exponents. du. Now, this is just a constant factor, right here. This could be, you know, this could just be some number. We could use a calculator to figure out what this is. We could set this equal to a. But we know in general that the integral, this is pretty straightforward, we've now put it in this form. The antiderivative of e to the au, du, is just 1 over a e to the au. This comes from this definition up here, and of course plus If we take the derivative of this, we take the derivative of the inside, which is just going to be a. We multiply that times the one over a, it cancels out, and we're just left with e to the au. So this definitely works out. So the antiderivative of this thing right here is going to be equal to 1 over our a, it's going to be 1 over our constant term, 1 over the natural log of 2 times our whole expression, e e. And I'm going to do something. This is just some number times u, so I can write it as u times some number. And I'm just doing that to put in a form that might help us simplify it a little bit. So it's e to the u times the natural log of 2, right? All I did, is I swapped this order. I could have written this as e to the natural log of 2 times u. If this an a, a times u is the same thing as u times a. Plus c. So this is our answer, but we have to kind of reverse" + }, + { + "Q": "At 6:01 we have,\n\n= exp [u*ln(2)] / ln(2) + C\n\nWhy can't we substitute u=ln(x) now and use ln(x)ln(2)=ln(x+2) ? i.e.,\n\n= exp [ln(x+2)] / ln(2) + C\n= (x+2) / ln(2) + C\nCan anyone see what I'm doing wrong?", + "A": "You ve confused the property a little bit; ln(x)*ln(2) is not equal to ln(x+2). Rather, ln(x) + ln(2) = ln(2*x), and ln(x)*ln(2) = ln(2^(ln(x))). Hope that helps.", + "video_name": "C5Lbjbyr1t4", + "timestamps": [ + 361 + ], + "3min_transcript": "product of those exponents. So this is equal to, let me switch colors, this is equal to the integral of e, to the u, e to the, let me write it this way. e to the natural log of 2 times u. I'm just multiplying these two exponents. I raise something to something, then raise it again, we know from our exponent rules, it's just a product of those two exponents. du. Now, this is just a constant factor, right here. This could be, you know, this could just be some number. We could use a calculator to figure out what this is. We could set this equal to a. But we know in general that the integral, this is pretty straightforward, we've now put it in this form. The antiderivative of e to the au, du, is just 1 over a e to the au. This comes from this definition up here, and of course plus If we take the derivative of this, we take the derivative of the inside, which is just going to be a. We multiply that times the one over a, it cancels out, and we're just left with e to the au. So this definitely works out. So the antiderivative of this thing right here is going to be equal to 1 over our a, it's going to be 1 over our constant term, 1 over the natural log of 2 times our whole expression, e e. And I'm going to do something. This is just some number times u, so I can write it as u times some number. And I'm just doing that to put in a form that might help us simplify it a little bit. So it's e to the u times the natural log of 2, right? All I did, is I swapped this order. I could have written this as e to the natural log of 2 times u. If this an a, a times u is the same thing as u times a. Plus c. So this is our answer, but we have to kind of reverse the antiderivative with respect to x. But before I do that, let's see if I can simplify this a little bit. What is, if I have, just from our natural log properties, or logarithms, a times the natural log of b. We know this is the same thing as the natural log of b to the a. Let me draw a line here. Right? That this becomes the exponent on whatever we're taking the natural log of. So u, let me write this here, u times the natural log of 2, is the same thing as the natural log of 2 to the u. So we can rewrite our antiderivative as being equal to 1 over the natural log of 2, that's just that part here, times e to the, this can be rewritten based on this logarithm property, as the natural log of 2 to the u, and of course we still have our plus c there. Now, what is e raised to the natural log of 2 to the u? The natural log of 2 to the u is the power that you have to" + }, + { + "Q": "At 5:00, why does the integral of e^(au) become 1/a * e^(au)? Where did the 1/a come from? Thanks!", + "A": "The chain rule, I presume would be the answer, since the expression uses a function of a function ( the thing you wrote up). Hope that helps :)", + "video_name": "C5Lbjbyr1t4", + "timestamps": [ + 300 + ], + "3min_transcript": "So let's see. How can we redefine this right here? Well, 2, 2 is equal to what? 2 is the same thing as e to the natural log of 2, right? The natural log of 2 is the power you have to raise e to to get 2. So if you raise e to that power, you're, of course going to get 2. This is actually the definition of really, the natural log. You raise e to the natural log of 2, you're going to get 2. So let's rewrite this, using this-- I guess we could call this this rewrite or-- I don't want to call it quite a substitution. It's just a different way of writing the number 2. So this will be equal to, instead of writing the number 2, I could write e to the natural log of 2. And all of that to the u du. And now what is this equal to? Well, if I take something to an exponent, and then to another product of those exponents. So this is equal to, let me switch colors, this is equal to the integral of e, to the u, e to the, let me write it this way. e to the natural log of 2 times u. I'm just multiplying these two exponents. I raise something to something, then raise it again, we know from our exponent rules, it's just a product of those two exponents. du. Now, this is just a constant factor, right here. This could be, you know, this could just be some number. We could use a calculator to figure out what this is. We could set this equal to a. But we know in general that the integral, this is pretty straightforward, we've now put it in this form. The antiderivative of e to the au, du, is just 1 over a e to the au. This comes from this definition up here, and of course plus If we take the derivative of this, we take the derivative of the inside, which is just going to be a. We multiply that times the one over a, it cancels out, and we're just left with e to the au. So this definitely works out. So the antiderivative of this thing right here is going to be equal to 1 over our a, it's going to be 1 over our constant term, 1 over the natural log of 2 times our whole expression, e e. And I'm going to do something. This is just some number times u, so I can write it as u times some number. And I'm just doing that to put in a form that might help us simplify it a little bit. So it's e to the u times the natural log of 2, right? All I did, is I swapped this order. I could have written this as e to the natural log of 2 times u. If this an a, a times u is the same thing as u times a. Plus c. So this is our answer, but we have to kind of reverse" + }, + { + "Q": "at 0:42,what does reciprical mean?", + "A": "The reciprocal of a fraction is simply the fraction flipped over. For example, the reciprocal of 6/9 is 9/6.", + "video_name": "yb7lVnY_VCY", + "timestamps": [ + 42 + ], + "3min_transcript": "Tommy is studying for final exams this weekend. He will spend 1/5 of the weekend studying. What fraction of the weekend will he spend studying for each of his 4 subjects if he spends the same amount of time studying for each subject? So the total amount of time he's going to spend studying this weekend is 1/5 of the weekend. And he has to divide that into 4 equal sections. And he's going to spend that much time on each subject. So he's going to divide this by 4. Now, we've already seen that dividing by a number is the same thing as multiplying by its reciprocal. You might say, hey, well, what's the reciprocal of 4? You just have to remind yourself that 4 is the same thing as 4/1. So 1/5 divided by 4/1 is the same thing as 1/5 times 1/4. And you could also view this as 1/4 of 1/5 or 1/5 But here we multiply our numerators to get 1. And then we multiply our denominators, 4 times 5 is 20. So you get 1/20 of the weekend will be spent studying for each subject. Now, let's also try to think about this visually. Let's imagine that this is his entire weekend. And I've divided it into 5 equal sections. And so we already know that the total amount of his weekend spent studying is 1/5. So that's the total amount studying for the weekend is 1/5. Now, he has to divide this into 4 equals section. So let's do that. He's got four subjects, and he's going to spend the same amount of time on each of the 4 subjects. So he's going to divide this into 4 equal sections. So how much time does he spend on one subject? Well, in each subject, that would be this little area that I'm doing in yellow right over here. And what is that? Well, that's 1 over-- how many equal sections are there Well, I've just drawn out the grid. You had 5 rows, and now you have 4 columns. So 5 rows times 4 columns, you have 20 equal sections. So once again, looking at it visually, he's spending 1/20 of his weekend on each of the 4 subjects. And then if you do this for 4 subjects, that means that in this whole weekend, 1/5 will be spent studying. But the question that they're asking, he's spending 1/20 of the weekend on each subject." + }, + { + "Q": "1:15 what does he mean", + "A": "He means that if the weekend was a square split into 20 parts, one of those parts would be spent studying for a single subject (math, science, english, history, etc)", + "video_name": "yb7lVnY_VCY", + "timestamps": [ + 75 + ], + "3min_transcript": "Tommy is studying for final exams this weekend. He will spend 1/5 of the weekend studying. What fraction of the weekend will he spend studying for each of his 4 subjects if he spends the same amount of time studying for each subject? So the total amount of time he's going to spend studying this weekend is 1/5 of the weekend. And he has to divide that into 4 equal sections. And he's going to spend that much time on each subject. So he's going to divide this by 4. Now, we've already seen that dividing by a number is the same thing as multiplying by its reciprocal. You might say, hey, well, what's the reciprocal of 4? You just have to remind yourself that 4 is the same thing as 4/1. So 1/5 divided by 4/1 is the same thing as 1/5 times 1/4. And you could also view this as 1/4 of 1/5 or 1/5 But here we multiply our numerators to get 1. And then we multiply our denominators, 4 times 5 is 20. So you get 1/20 of the weekend will be spent studying for each subject. Now, let's also try to think about this visually. Let's imagine that this is his entire weekend. And I've divided it into 5 equal sections. And so we already know that the total amount of his weekend spent studying is 1/5. So that's the total amount studying for the weekend is 1/5. Now, he has to divide this into 4 equals section. So let's do that. He's got four subjects, and he's going to spend the same amount of time on each of the 4 subjects. So he's going to divide this into 4 equal sections. So how much time does he spend on one subject? Well, in each subject, that would be this little area that I'm doing in yellow right over here. And what is that? Well, that's 1 over-- how many equal sections are there Well, I've just drawn out the grid. You had 5 rows, and now you have 4 columns. So 5 rows times 4 columns, you have 20 equal sections. So once again, looking at it visually, he's spending 1/20 of his weekend on each of the 4 subjects. And then if you do this for 4 subjects, that means that in this whole weekend, 1/5 will be spent studying. But the question that they're asking, he's spending 1/20 of the weekend on each subject." + }, + { + "Q": "At 2:07, why do you have to do:\n(4y^2 -6y) + (10y - 15)\ninstead of\n(4y^2 +10y) + (-6y - 15) <-- it's clearly not the same thing right?\n\nThanks", + "A": "Either way works as long as you can factor it, as both expressions are equivalent. One factors to 2y(2y-3) + 5(2y-3) = (2y+5)(2y-3) The other equals 2y(2y+5) - 3(2y +5) = (2y+5)(2y-3) Hope this helps!", + "video_name": "u1SAo2GiX8A", + "timestamps": [ + 127 + ], + "3min_transcript": "We're asked to factor 4y squared plus 4y, minus 15. And whenever you have an expression like this, where you have a non-one coefficient on the y squared, or on the second degree term-- it could have been an x squared-- the best way to do this is by grouping. And to factor by grouping we need to look for two numbers whose product is equal to 4 times negative 15. So we're looking for two numbers whose product-- let's call those a and b-- is going to be equal to 4 times negative 15, or negative 60. And the sum of those two numbers, a plus b, needs to be equal to this 4 right there. So let's think about all the factors of negative 60, or 60. And we're looking for ones that are essentially 4 apart, because the numbers are going to be of different signs, because their product is negative, so when you take two numbers of different signs and you sum them, you kind of view it as the difference of their absolute values. If that confuses you, don't worry about it. But this tells you that the numbers, since they're going to be of different size, their absolute values are going to So we could try out things like 5 and 12, 5 and negative 12, because one has to be negative. If you add these two you get negative 7, if you did negative 5 and 12 you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10? Then you get a negative 4, if you added these two. But we want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our two numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So this 4y can be rewritten as negative 6y plus 10y, right? Because if you add those you get 4y. And then the other sides of it, you have your 4y squared, your 4y squared and then you have your minus 15. coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the term. Let me do it in a different color. So if I take these two guys, what can I factor out of those two guys? Well, there's a common factor, it looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared, divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. So this group gets factored into 2y times 2y, minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos I've explained why this works. Now here, the greatest common factor is a 5. So we can factor out a 5, so this is equal to plus 5 times 10y, divided by 5 is 2y." + }, + { + "Q": "(9:40 p.m.) Please help me with this problem. I've been out of school for years and I'm preparing the the GMAT. x^2 - 4x + 4 > or = 0", + "A": "Factor it as (x-2)^2, and then you find that it s 0 when x is 2. So you test on either side of 2 to find what the sign will be. If you plug in 1 it s positive, and when you plug in 3 it s positive, so your solutions are all real numbers.", + "video_name": "u1SAo2GiX8A", + "timestamps": [ + 580 + ], + "3min_transcript": "" + }, + { + "Q": "In 4:21 how did he get the m to ?(at 5:13)", + "A": "I use a simple example whenever I can. Say 10>0. If you change the sign on 10, then -10<0. The same exact change happens if there are multiple numbers, as long as you change all of the signs. BTW, if you have numbers on both sides of inequality, say, 10>9, you just change signs on both sides and flip the arrow. -10<-9. Hope this helps.", + "video_name": "xdiBjypYFRQ", + "timestamps": [ + 313 + ], + "3min_transcript": "So just visually looking at it, what x values make this true? Well, this is true whenever x is less than minus 3, right, or whenever x is greater than 2. Because when x is greater than 2, f of x is greater than 0, and when x is less than negative 3, f of x is greater than 0. So we would say the solution to this quadratic inequality, and we pretty much solved this visually, is x is less than minus 3, or x is greater than 2. And you could test it out. You could try out the number minus 4, and you should get f of x being greater than 0. You could try it out here. Or you could try the number 3 and make sure that this works. try out the number 0 and make sure that 0 doesn't work, right, because 0 is between the two roots. It actually turns out that when x is equal to 0, f of x is minus 6, which is definitely less than 0. So I think this will give you a visual intuition of what this quadratic inequality means. Now with that visual intuition in the back of your mind, let's do some more problems and maybe we won't have to go through the exercise of drawing it, but maybe I will draw it just to make sure that the point hits home. Let me give you a slightly trickier problem. Let's say I had minus x squared minus 3x plus 28, let me say, is greater than 0. Well I want to get rid of this negative sign in front of the x squared. I just don't like it there because it makes it look more confusing to factor. I'm going to multiply everything by negative 1. I get x squared plus 3x minus 28, and when you multiply or to swap the sign. So this is now going to be less than 0. And if we were to factor this, we get x plus 7 times x minus 4 is less than 0. So if this was equal to 0, we would know that the two roots of this function -- let's define the function f of x -- let's define the function as f of x is equal to -- well we can define it as this or this because they're the same thing. But for simplicity let's define it as x plus 7 times x minus 4. That's f of x, right? Well, after factoring it, we know that the roots of this, the roots are x is equal to minus 7, and x is equal to 4." + }, + { + "Q": "Hi,\n\nwhen sal took any vector from the original subset V and multiplied it with a vector from the \"orthogonal complement\" subspace at 10:05 , he found that their dot product was equal to 0.\nHow can we be sure that this is the case? I mean, the orthogonal subspace is supposed to have the vectors that are orthogonal to a specific subspace of vectors, not to any in V. I hope I made my question clear...", + "A": "thats the definition of an orthogonal complement... the z axis is the orthogonal complement of the xy plane in a 3 dimensional space. obviously, any vector on the z axis will be orthogonal to vectors in the xy plane, even though the xy plane is a 2 dimensional subspace.", + "video_name": "zlI8mx8Hc8o", + "timestamps": [ + 605 + ], + "3min_transcript": "Let me draw rn again. Let me draw all of rn like that. Now, we have the orthogonal complement. Let me just draw that first. So v perp And then you have the orthogonal complement of the orthogonal complement which could be this set right here. Right? This is v perp. I haven't even drawn the subspace v. All I've shown is, I have some subspace here, which I happen to call v perp. And then I have the orthogonal complement of that subspace. So this means that anything in rn can be represented as the sum of a vector that's here and a vector that's here. So, if I say that w-- let me do it in purple. If I say the vector w-- let me write it this way. The vector v can be represented as the sum of the orthogonal complement of v or v perp And x is a member of its orthogonal complement. Notice, all I'm saying, I could have called this set s. And then this would have been s and its orthogonal complement. And we learned that anything in rn could be represented as the sum of something in a subspace and the subspace's orthogonal complement. So it doesn't matter that v is somehow related to this. It can be represented as a sum of a vector here plus a vector there. Fair enough. Now, what happens if I dot v with w? I'm doing the exact same argument that I did before. Well, if you take anything that's a member of our orthogonal complement, that's going to give us 0. What else is that going to be equal to? If we write v in this way, v dot w is the same thing as this thing dot w. So w plus x dot-- and this is going to be equal to w dot w plus x dot w. And what's x dot w? x is in the orthogonal complement of your orthogonal complement. And w is in the orthogonal complement. So if you take the dot product, you're going to get 0. They're orthogonal to each other. So this is just equal to w dot w or the length of w squared. And since since has to equal 0, we just have a bunch of equals here, that tells us that once again the vector w has to be equal to 0. So that tells us v is equal to w plus x." + }, + { + "Q": "At 4:04 why is the x negative?", + "A": "The x is negative because after going down 4 for the change in y (negative 4), you have to go six to the left to meet up with the line again, (negative six). Going up is positive, going right is positive, left and down are negative.", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 244 + ], + "3min_transcript": "Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6? So it's the same value, you just have to be consistent. If this is my start point, I went down 4, and then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4, so it would be a change in y would be 4, and then my change in x would be 6. And either way, once again, change in y over change in x is going to be 4 over 6, 2/3. So no matter which point you choose, as long as you kind of think about it in a consistent way, you're going to get the same value for slope." + }, + { + "Q": "At about 4:10. Why do the negatives cancel out. Because wouldn't it be -4/6??", + "A": "Review your sign rules. A negative divided by a negative = a positive. So, -4 / (-6) = +4/6", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 250 + ], + "3min_transcript": "Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6? So it's the same value, you just have to be consistent. If this is my start point, I went down 4, and then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4, so it would be a change in y would be 4, and then my change in x would be 6. And either way, once again, change in y over change in x is going to be 4 over 6, 2/3. So no matter which point you choose, as long as you kind of think about it in a consistent way, you're going to get the same value for slope." + }, + { + "Q": "at 3:38 how does he get negative change in y? I get why the x is cause he's moving in the opposite direction but the change in y seems the same as the first example at 2:10. why is it different?", + "A": "In the first example, he was going from a lower point to a higher point, so there was a positive change in y. In the second example, he is going from a higher point to a lower point, so the y is going down. There is a negative change in y. Another way to think about it is this: The y value is going from 1 at the starting point to -3 at the ending point. You get from 1 to -3 by subtracting 4, so the change in y is -4.", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 218, + 130 + ], + "3min_transcript": "Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6? So it's the same value, you just have to be consistent. If this is my start point, I went down 4, and then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4, so it would be a change in y would be 4, and then my change in x would be 6. And either way, once again, change in y over change in x is going to be 4 over 6, 2/3. So no matter which point you choose, as long as you kind of think about it in a consistent way, you're going to get the same value for slope." + }, + { + "Q": "At 2:20 Sal says that the slope is 2/3. I am kinda confused about how you would draw that line using the information found out? Can someone please help me?", + "A": "Okay, using the information Sal mentioned at 2:33 in the video you can t draw a line on a graph. This is because the equation of any linear line on a Cartesian plane can be delfined by the formula y=mx+b. Since we know that m, or the slope equals 2/3 we can add it in the equation. Without any point known on the line, though, we can t solve the equation because any point s coordinates could be plugged into the equation along with the slope to solve for b. Therefore the line is undefined. Hope this helps!", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 140 + ], + "3min_transcript": "Find the slope of the line in the graph. And just as a bit of a review, slope is just telling us how steep a line is. And the best way to view it, slope is equal to change in y over change in x. And for a line, this will always be constant. And sometimes you might see it written like this: you might see this triangle, that's a capital delta, that means change in, change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here-- let me do it in a more vibrant color-- so let's say we start at that point right there. And we want to go to another point that's pretty that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is" + }, + { + "Q": "at 0:55 when you are talking about the starting point, what if a certain graph/problem does not give you a certain starting point. Would you start at (0,0)", + "A": "A problem would not be worded like that, it would be more clear.", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 55 + ], + "3min_transcript": "Find the slope of the line in the graph. And just as a bit of a review, slope is just telling us how steep a line is. And the best way to view it, slope is equal to change in y over change in x. And for a line, this will always be constant. And sometimes you might see it written like this: you might see this triangle, that's a capital delta, that means change in, change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here-- let me do it in a more vibrant color-- so let's say we start at that point right there. And we want to go to another point that's pretty that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is" + }, + { + "Q": "2:28 How the heck did he get a NEGATIVE 32?! What made it negative???? That concept has never been explained to me before!", + "A": "because he multiply 8 * (-4) :) if you have 8(3x - 4), youll have 8 *(3x) + 8*(-4) and this is 24x + (-32), which is basically 24x - 32 :) in this case you have 8(3/8 x - 4 ) so you are left with 3x - 32", + "video_name": "PL9UYj2awDc", + "timestamps": [ + 148 + ], + "3min_transcript": "We have the equation 3/4x plus 2 is equal to 3/8x minus 4. Now, we could just, right from the get go, solve this the way we solved everything else, group the x terms, maybe on the left-hand side, group the constant terms on the right-hand side. But adding and subtracting fractions are messy. So what I'm going to do, right from the start of this video, is to multiply both sides of this equation by some number so I can get rid of the fractions. And the best number to do it by-- what number is the smallest number that if I multiply both of these fractions by it, they won't be fractions anymore, they'll be whole numbers? That smallest number is going to be 8. I'm going to multiply 8 times both sides of this equation. You say, hey, Sal, how did you get 8? And I got 8 because I said, well, what's the least common multiple of 4 and 8? Well, the smallest number that is divisible by 4 and 8 is 8. get rid of the fractions. And so let's see what happens. So 8 times 3/4, that's the same thing as 8 times 3 over 4. Let me do it on the side over here. That's the same thing as 8 times 3 over 4, which is equal to 8 divided by 4 is just 2. So it's 2 times 3, which is 6. So the left-hand side becomes 8 times 3/4x is 6x. And then 8 times 2 is 16. You have to remember, when you multiply both sides, or a side, of an equation by a number, you multiply every term by that number. So you have to distribute the 8. So the left-hand side is 6x plus 16 is going to be equal to-- 8 times 3/8, that's pretty easy, the 8's cancel out and you're just left with 3x. And then 8 times negative 4 is negative 32. And now we've cleaned up the equation a good bit. Now the next thing, let's try to get all the x terms on the left-hand side, and all the constant terms on the right. Let's subtract 3x from both sides to do it. That's the best way I can think of of getting rid of the 3x from the right. The left-hand side of this equation, 6x minus 3x is 3x. 6 minus 3 is 3. And then you have a plus 16 is equal to-- 3x minus 3x, that's the whole point of subtracting 3x, is so they cancel out. So those guys cancel out, and we're just left with a negative 32. Now, let's get rid of the 16 from the left-hand side. So to get rid of it, we're going to subtract 16 from both sides of this equation. Subtract 16 from both sides. The left-hand side of the equation just becomes-- you have this 3x here; these 16's cancel out, you don't have to write anything-- is equal to negative 32 minus 16 is negative 48. So we have 3x is equal to negative 48." + }, + { + "Q": "at 2:06 how do you go from 1 to 3", + "A": "I think Sal is trying to show you that the slope remains constant across the entire line. He started with 2/1 Then, he picked another 2 points and the slope was 6/3. You need to recognize that these are = values. If reduce / divide 6/3, you get 2/1. Same slope. He could have gone up 10 and to the right 5. he would still be on the line. Slope would be: 10/5, which still = 2/1. Hope this helps. Hope that helps.", + "video_name": "MeU-KzdCBps", + "timestamps": [ + 126 + ], + "3min_transcript": "- [Voiceover] As we start to graph lines, we might notice that they're differences between lines. For example, this pink or this magenta line here, it looks steeper than this blue line. And what we'll see is this notion of steepness, how steep a line is, how quickly does it increase or how quickly does it decrease, is a really useful idea in mathematics. So ideally, we'd be able to assign a number to each of these lines or to any lines that describes how steep it is, how quickly does it increase or decrease? So what's a reasonable way to do that? What's a reasonable way to assign a number to these lines that describe their steepness? Well one way to think about it, could say well, how much does a line increase in the vertical direction for a given increase in the horizontal direction? So let's write this down. So let's say if we an increase increase, in vertical, in vertical, for a given increase a given increase in horizontal. So, how can this give us a value? Well let's look at that magenta line again. Now let's just start at an arbituary point in that magenta line. But I'll start at a point where it's going to be easy for me to figure out what point we're at. So if we were to start right here, and if I were to increase in the horizontal direction by one. So I move one to the right. To get back on the line, how much do I have to increase in the vertical direction? Well I have to increase in the vertical direction by two. So at least for this magenta line, it looks like our increase in vertical is two, whenever we have an increase in one in the horizontal direction. Let's see, does that still work if I were to direction by one, if I were increase in the horizontal direction... So let's increase by three. So now, I've gone plus three in the horizontal direction, then to get back on the line, how much do I have to increase in the vertical direction? I have to increase by one, two, three, four, five, six I have to increase by six. So plus six. So when I increase by three in the horizontal direction, I increase by six in the vertical. We were just saying, hey, let's just measure how much to we increase in vertical for a given increase in the horizontal? Well two over one is just two and that's the same thing as six over three. So no matter where I start on this line, no matter where I start on this line, if I take and if I increase in the horizontal direction by a given amount, I'm going to increase twice as much twice as much in the vertical direction. Twice as much in the vertical direction." + }, + { + "Q": "After looking at 5:08, does that mean the slope of a graph basically is a rate of change in a table and graph? In a nutshell, the rate of change and the slope of a graph is the same, right?", + "A": "Yes you are correct. Adding on to that, it can really be applied to Physics. For example, if you found the slope for a velocity over time graph, you get the velocity. See, isnt that cool? like how everything is connected in this universe.", + "video_name": "MeU-KzdCBps", + "timestamps": [ + 308 + ], + "3min_transcript": "divided by increase in horizontal, this is what mathematicians use to describe the steepness of lines. And this is called the slope. So this is called the slope of a line. And you're probably familiar with the notion of the word slope being used for a ski slope, and that's because a ski slope has a certain inclination. It could have a steep slope or a shallow slope. So slope is a measure for how steep something is. And the convention is, is we measure the increase in vertical for a given in increase in horizontal. So six two over one is equal to six over three is equal to two, this is equal to the slope of this magenta line. So let me write this down. So this slope right over here, the slope of that line, is going to be equal to two. And one way to interpret that, for whatever amount you increase in the horizontal direction, you're going to increase twice as much in the vertical direction. What would be the slope of the blue line? Well, let me rewrite another way that you'll typically see the definition of slope. And this is just the convention that mathematicians have defined for slope but it's a valuable one. What is are is our change in vertical for a given change in horizontal? And I'll introduce a new notation for you. So, change in vertical, and in this coordinate, the vertical is our Y coordinate. divided by our change in horizontal. And X is our horizontal coordinate in this coordinate plane right over here. So wait, you said change in but then you drew this triangle. Well this is the Greek letter delta. This is the Greek letter delta. And it's a math symbol used to represent change in. So that's delta, delta. And it literally means, change in Y, change in Y, change in X. So if we want to find the slope of the blue line, we just have to say, well how much does Y change for a given change in X? So, the slope of the blue line. So let's see, let me do it this way. Let's just start at some point here. And let's say my X changes by two so my delta X is equal to positive two. What's my delta Y going to be? What's going to be my change in Y? Well, if I go by the right by two, to get back on the line, I'll have to increase my Y by two. So my change in Y is also going to be plus two. So the slope of this blue line, the slope of the blue line, which is change in Y over change in X. We just saw that when our change in X is positive two, our change in Y is also positive two. So our slope is two divided by two, which is equal to one." + }, + { + "Q": "2:00 What if the line is not strait so the slope cannot only have one number?", + "A": "Good question, you re getting a bit ahead of yourself though. There are other types of equations that will describe curved lines.", + "video_name": "MeU-KzdCBps", + "timestamps": [ + 120 + ], + "3min_transcript": "- [Voiceover] As we start to graph lines, we might notice that they're differences between lines. For example, this pink or this magenta line here, it looks steeper than this blue line. And what we'll see is this notion of steepness, how steep a line is, how quickly does it increase or how quickly does it decrease, is a really useful idea in mathematics. So ideally, we'd be able to assign a number to each of these lines or to any lines that describes how steep it is, how quickly does it increase or decrease? So what's a reasonable way to do that? What's a reasonable way to assign a number to these lines that describe their steepness? Well one way to think about it, could say well, how much does a line increase in the vertical direction for a given increase in the horizontal direction? So let's write this down. So let's say if we an increase increase, in vertical, in vertical, for a given increase a given increase in horizontal. So, how can this give us a value? Well let's look at that magenta line again. Now let's just start at an arbituary point in that magenta line. But I'll start at a point where it's going to be easy for me to figure out what point we're at. So if we were to start right here, and if I were to increase in the horizontal direction by one. So I move one to the right. To get back on the line, how much do I have to increase in the vertical direction? Well I have to increase in the vertical direction by two. So at least for this magenta line, it looks like our increase in vertical is two, whenever we have an increase in one in the horizontal direction. Let's see, does that still work if I were to direction by one, if I were increase in the horizontal direction... So let's increase by three. So now, I've gone plus three in the horizontal direction, then to get back on the line, how much do I have to increase in the vertical direction? I have to increase by one, two, three, four, five, six I have to increase by six. So plus six. So when I increase by three in the horizontal direction, I increase by six in the vertical. We were just saying, hey, let's just measure how much to we increase in vertical for a given increase in the horizontal? Well two over one is just two and that's the same thing as six over three. So no matter where I start on this line, no matter where I start on this line, if I take and if I increase in the horizontal direction by a given amount, I'm going to increase twice as much twice as much in the vertical direction. Twice as much in the vertical direction." + }, + { + "Q": "What does Sal mean at 3:30 when he says \"Let's see if we can break down 115 any further\"? What method does he use, and how does it work?", + "A": "He is literally trying to factorize or break down 115 as much as possible.The method he has showed you is prime factorization . In this you find 2 numbers that when multiplied give you a certain result. you keep doing this until you only have prime numbers.It can also be called a factor tree.", + "video_name": "O64YFlX1_aI", + "timestamps": [ + 210 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:44 how is it a ray without a arrow on the other end?", + "A": "Think of it with everyday things. Take for example sun rays. they go in one direction until they hit an object (us, the atmosphere or a mirror ect.). Do they have arrows? As long as one line looks like it is going to continue on and on, they it should be considered a ray.", + "video_name": "DkZnevdbf0A", + "timestamps": [ + 104 + ], + "3min_transcript": "Any pair of points can be connected by a line segment. That's right. Connect two pairs of black points in a way that creates two parallel line segments. So let's see if we can do that. So I could create one segment that connects this point to this point and then another one that connects this point to this point. And they look pretty parallel. In fact, I think this is the right answer. If we did it another way, if we had connected that point to that point and this point to this point, then it wouldn't look so parallel. These clearly, if they were to keep going, they would intersect at some point. So let me set it back up the way I did it the first time. Let me make these two points parallel. And these are line segments because they have two end points. They each have two end points. And they continue forever. Well they don't continue forever. They continue forever in no directions, in zero directions. If it was a ray, it would continue forever If it's a line, it continues forever In fact, it wouldn't have end points because it would just continue forever in both of these directions. Let's do one more. Drag the ray so it has an endpoint at A, so we want to make its endpoint at A where the ray terminates and goes through one of the other black points. The ray should also be parallel to the pink line. So I have two options. I could make it go through this black point, but it's clearly not parallel. In fact, it looks perpendicular here. So let's try to make it go through this point. Well, yes, when I do that, it does indeed look like my ray is parallel to the pink line. And this is a ray because it has one endpoint. This is where the ray terminates. It's an endpoint. It literally ends there. And it continues forever in one direction. In this case, the direction is to the right. It continues forever to the right. So it continues forever in one-- continues forever in one direction." + }, + { + "Q": "at 5:28 , I dont understand why Sal put +(-14). it is so confusing.", + "A": "He meant 6 plus negative 14, which is 6 minus 14. Hopefully this helped you.", + "video_name": "-4bTgmmWI9k", + "timestamps": [ + 328 + ], + "3min_transcript": "Our goal is four right over here. So this is one, two, three, four, get to positive four. That's positive four right there. We're starting at negative two... Let me do this in a different color. We're starting at negative two. We're saying four is the same thing as negative two, negative two is right over here, negative two plus some amount. And it's clear we're gonna be moving to the right by, let's see, we're gonna move to the right by one, two, three, four, five, six. So we moved to the right by six. So we added six. So negative two plus six is equal to four. This is fascinating. Actually, let's just do several more of these, I can't stop. (laughs) Alright. So let's say we wanted to figure out, Six plus blank is equal to negative eight. Like always, try to pause the video and figure out what this blank is going to be. Let me throw my number line back here. So my number line. And one way to think about it is I am starting at six. So it's five, this is six right over here. And I'm gonna add something to get to negative eight. To get to negative eight this is negative five, negative six, negative seven, negative eight. I want to get right over here, I want to get to negative eight. So what do I have to do to get from six to negative eight? To go from six to negative eight. We're clear I have to go to the left on the number line. And how much do I have to go to the left? Well, let's count it. I have to go one, two, three, four, five, six, seven, eight, nine, So going 14 to the left, you could say that I just subtracted 14. And if we phrase it as six plus what is equal to negative eight? Well, six plus negative 14. If this said six minus something I could have just said six minus 14, but if it's saying six plus what it's going to be six plus negative 14." + }, + { + "Q": "what is a checker board pattern? Sal mentions it at 0:20", + "A": "A checkerboard pattern is when every other square is black or white both in the horizontal and vertical directions. With numbers instead of colors, it could look like this: 1 0 1 0 1 0 1 0 1", + "video_name": "u00I3MCrspU", + "timestamps": [ + 20 + ], + "3min_transcript": "As a hint, I will take the determinant of another 3 by 3 matrix. But it's the exact same process for the 3 by 3 matrix that you're trying to find the determinant of. So here is matrix A. Here, it's these digits. This is a 3 by 3 matrix. And now let's evaluate its determinant. So what we have to remember is a checkerboard pattern when we think of 3 by 3 matrices: positive, negative, positive. So first we're going to take positive 1 times 4. So we could just write plus 4 times 4, the determinant of 4 submatrix. And when you say, what's the submatrix? Well, get rid of the column for that digit, and the row, and then the submatrix is what's left over. So we'll take the determinant of its submatrix. So it's 5, 3, 0, 0. Then we move on to the second item in this row, in this top row. But the checkerboard pattern says we're going to take the negative of it. So it's going to be negative of negative 1-- times the determinant of its submatrix. You get rid of this row, and this column. You're left with 4, 3, negative 2, 0. And then finally, you have positive again. Positive times 1. This 1 right over here. Let me put the positive in that same blue color. So positive 1, or plus 1 or positive 1 times 1. Really the negative is where it got a little confusing on this middle term. But positive 1 times 1 times the determinant of its submatrix. So it's submatrix is this right over here. You get rid of the row, get rid of the column 4, 5, negative 2, 0. So the determinant right over here is going to be 5 times 0 minus 3 times 0. And all of that is going to be multiplied times 4. Well this is going to be 0 minus 0. So this is all just a 0. So 4 times 0 is just a 0. So this all simplifies to 0. Now let's do this term. We get negative negative 1. So that's positive 1. So let me just make these positive. Positive 1, or we could just write plus. Let me just write it here. So positive 1 times 4 times 0 is 0. So 4 times 0 minus 3 times negative 2. 3 times negative 2 is negative 6. So you have 4-- oh, sorry, you have 0 minus negative 6, which is positive 6. Positive 6 times 1 is just 6. So you have plus 6. And then finally you have this last determinant." + }, + { + "Q": "I don't understand the intersect stuff @2:52. Is there a video that talks about this? Or could someone explain it for me? Thx :)", + "A": "What don t you understand about it", + "video_name": "VTlvg4wJ1X0", + "timestamps": [ + 172 + ], + "3min_transcript": "If something-squared is equal to four, that means that the something, that means that this something right over here, is going to be equal to the positive square root of four or the negative square root of four. Or it's gonna be equal to positive or negative two. And so we could write that x plus three could be equal to positive two or x plus three could be equal to negative two. Notice, if x plus three was positive two, two-squared is equal to four. If x plus three was negative two, negative two-squared is equal to four. So, either of these would satisfy our equation. So, if x plus three is equal to two, we could just subtract three from both sides to solve for x and we're left with x is equal to negative one. Or, over here we could subtract three from both sides to solve for x. So, or, x is equal to negative two minus three So, those are the two possible solutions and you can verify that. Take these x-values, substitute it back in, and then you can see when you substitute it back in if you substitute x equals negative one, then x plus three is equal to two, two-squared is four, minus four is zero. And when x is equal to negative five, negative five plus three is negative two, squared is positive four, minus four is also equal to zero. So, these are the two possible x-values that satisfy the equation. Now let's do another one that's presented to us in a slightly different way. So, we are told that f of x is equal to x minus two squared minus nine. And then we're asked at what x-values does the graph of y equals f of x intersect the x-axis. So, if I'm just generally talking about some graph, so I'm not necessarily gonna draw that y equals f of x. So if I'm just, so that's our y-axis, this is our x-axis. And so if I just have the graph of some function. that looks something like that. Let's say that the y is equal to some other function, not necessarily this f of x. Y is equal to g of x. The x-values where you intersect, where you intersect the x-axis. Well, in order to intersect the x-axis, y must be equal to zero. So, y is equal to zero there. Notice our y-coordinate at either of those points are going to be equal to zero. And that means that our function is equal to zero. So, figuring out the x-values where the graph of y equals f of x intersects the x-axis, this is equivalent to saying, \"For what x-values does f of x equal zero?\" So we could just say, \"For what x-values does this thing right over here \"equal zero?\" So, let me just write that down. So we could rewrite this as x, x minus two squared minus nine equals zero." + }, + { + "Q": "4:14, Shouldn't it be the first n+1 squares since we're starting at 0 or do most people start at 1 so it's the first n squares? If the latter, why is Sal starting at 0?", + "A": "I think he started at 0 to eliminate the D term (in the previous video). Our Sigma in this starts at 0, also, so he had to start at 0 to evaluate the formula.", + "video_name": "MkGXR8umLco", + "timestamps": [ + 254 + ], + "3min_transcript": "is equal to negative 9. And now if we add both sides, on the left-hand side, we have-- let's see. 24A minus 18A is 6A-- these cancel-- is equal to 11 minus 9 is 2. Divide both sides by 6. We get A is equal to 2/6, which is the same thing as 1/3. And so now we can substitute back to solve for B. So let's see. We have 6 times 1/3. Our A is 1/3. Plus 2B is equal to 3. 6 times 1/3 is 2. 2 plus 2B is equal to 3. Subtract 2 from both sides. 2B is equal to 1. Divide both sides by 2. So A is 1/3. B is 1/2. Now we just have to solve for C. So we can go back to this original equation right over here. So we have 1/3 plus 1/2 plus C is equal to 1. Actually, let me do that over here so I have some space. So I have some space, let me do it right over here. So I have 1/3 plus 1/2 plus C is equal to 1. Now we find ourselves a common denominator. So let's see. A common multiple of 3, 2, and-- I guess you could say 1-- this is 1 over 1-- is going to be 6. So I can rewrite this as 2/6 plus 3/6 plus C is equal to 6/6. 1 is the same thing as 6/6. So this is 5/6 plus C is equal to 6/6. Subtract 5/6 from both sides. We get C is equal to 1/6. And there we are. We deserve a drum roll now. We've figured out a formula for the sum of the first n squares. So we can rewrite this. This formula is now going to be-- A is 1/3. So it's 1/3 n to the third power plus 1/2 n squared-- let me make sure it doesn't look like that n is a-- 1/2 n squared plus 1/6 n. So this is a pretty handy formula. If now you wanted to find 0 squared plus 1 squared plus 2 squared plus 3 squared, all the way to 100 squared, instead of squaring 100 numbers and then adding them together, you could figure out 1/3 of 100 to the third power plus 1/2 times 100 squared plus 1/6 times 100." + }, + { + "Q": "Okay at 2:13 on question 4 he says Z is 1 but then he says that Z is over 4. I'm really confused about that how can he say one thing but it mean another?", + "A": "The equation is 1/4 * Z So to solve you do: 1/4 * Z/1 (remember Z/1 is still equal to Z) Then you multiply across, first the two numerators (1*Z which equals 1Z or Z) then the two denominators (4*1 which equals 4) And your final answer is Z/4", + "video_name": "aoXUWSwiDzE", + "timestamps": [ + 133 + ], + "3min_transcript": "Let's do some practice problems dealing with variable expressions. So these first problems say write the following in a more condensed form by leaving out the multiplication symbol or leaving out a multiplication symbol. So here we have 2 times 11x, so if we have 11 x's and then we're going to have 2 times those 11 x's, we're going to have 22 x's. So another way you could view this, 2 times 11x, you could view this as being equal to 2 times 11, and all of that times x, and that's going to be equal to 22 x's. You had 11 x's, you're going to have 2 times as many x's, so you're going to have 22 x's. Let's see, you have 1.35 times y. Now here we're just going to do a straight simplifying how we write it. So 1.35 times y-- I'll do it in a different color-- 1.35 times y-- that's a little dot there. If we have a variable following a number, we know that means 1.35 times that variable. So that, we could rewrite as just being equal to 1.35y. We've condensed it by getting rid of the multiplication sign. Let's see, here we have 3 times 1/4. Well, this is just straight up multiplying a fraction. So in problem 3-- this was problem 1, this is problem 2, problem 3-- 3 times 1/4, that's the same thing as 3 over 1 times 1/4. Multiply the numerators, you get 3. Multiply the denominators, 1 times 4, you get 4. So number 3, I got 3/4. And then finally, you have 1/4 times z. We could do the exact same thing we did up here in problem number 2. This was the same thing as 1.35y. That's the same thing as 1.35 times y. 1/4z, or we could view this as being equal to 1 over 4 times z over 1, which is the same thing as z times 1, over 4 times 1, or the same thing as z over 4. So all of these are equivalent. Now, what do they want us to do down here? Evaluate the following expressions for a is equal to 3, b is equal to 2, c is equal to 5, and d is equal to minus 4-- or, actually, I should say negative 4 is the correct terminology. Negative 4. So we just substitute. Every time we see an a, we're going to put a minus 3 there, or a negative 3 there. Every time we see a b, we'll put a positive 2 there." + }, + { + "Q": "@ 4:45, Sal skips from question 6 to question 10.\nCould he possibly either do a re upload where he does all the questions, or maybe someone provide the answers to the rest of the questions so that those who are working directly from the sheet can see if their workings are correct.", + "A": "He is not trying to answer all the questions. The whole point of these videos is to help you understand the concepts. The only reason you could think these videos are not doing their jobs is if, even after you watch the video, you are confused on the topic. If that is true, you might want to try talking to a teacher or parent to help you delve further into the topic. After all, they probably know different ways to help you learn. Not everyone is a visual learner and learns through videos.", + "video_name": "aoXUWSwiDzE", + "timestamps": [ + 285 + ], + "3min_transcript": "And every time we see a d, we'll put a minus 4 there. And I'll do a couple of these. I won't do all of them, just for the sake of time. So let's say problem number 5. They gave us 2 times a plus 3 times b. Well, this is the same thing as 2 times-- instead of an a, we know that a is going to be equal to negative 3. So 2 times minus 3, plus 3 times b-- what's b? They're telling us that b is equal to 2-- so 3 times 2. And what is this equal to? 2 times minus 3-- let me do it in a different color-- 2 times negative 3 is negative 6, plus 3 times 2. 3 times 2 is 6. That's positive 6. So that is equal to 0. And notice the order of operations. before we added the two numbers. Multiplication and division takes precedence over addition and subtraction. Let's do problem 6. I'll do that right here. So you have 4 times c. 4 times-- now what's c equal to? They tell us c is equal to 5. So 4 times 5, that's our c, plus d. d is minus or negative 4. So we have 4 times 5 is 20, plus negative 4-- that's the same thing as minus 4, so that is equal to 16. Problem 6. Now, let's do one of the harder ones down here. This problem 10 looks a little bit more daunting. Problem 10 right there. So we have a minus 4b in the numerator, if you can read it, it's kind of small. a is minus 3. b is 2. So 4 times 2. Remember, this right here is a, that right there is b. They're telling me up here. And then all of that over-- all of that is over 3c plus 2d. So 3 times-- what was c? c is 5 plus 2 times d. What is d? d is negative 4. So let's figure this out. So we have to do order of operations. Multiplication comes first before addition and subtraction. So this is going to be equal to minus 3 minus 4 times 2," + }, + { + "Q": "From 15:32 to 15:43, Sal mentions his wife being a doctor, and if people don't see the decimal point they'll overdose on medication. What does he mean by that?", + "A": "If you are suppose to give a patient 3.2 mg of some medicine And, instead you lose the decimal point and give them 32mg of medicine. You will have given your patient 10 times the amount of medicine they should have. Depending on the medicine, this can have drastic consequences.", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 932, + 943 + ], + "3min_transcript": "the number-- what if we had a 7-- let me think of it this way. Let's say we had a 7, 3 over there. So what would we do? Well, we'd want to go to the first digit right here because this is kind of the largest power of 10 that could go into this thing right here. So if we wanted to represent that thing, let me do another decimal that's like that one. So let's say I did 0.0000516 and I wanted to represent this in scientific notation. I'd go to the first non-digit 0-- the first non-zero digit, not non-digit 0, which is there. And I'm like, OK, what's the largest power of 10 that will fit into that? So I'll go 1, 2, 3, 4, 5. So it's going to be equal to 5.16. So I take 5 there, then everything else is going to be behind the decimal point. So this is going to be the largest power of 10 that fits into this first non-zero number. So it's 1, 2, 3, 4, 5. So 10 to the minus 5 power. Let me do another example. So the point I wanted to make is you just go to the first-- if you're starting at the left, the first non-zero number. That's what you get your power from. That's where i got my 10 to the minus 5 because I counted 1, 2, 3, 4, 5. You got to count that number just like we did over here. And then, everything else will be behind the decimal. Let me do another example. Let's say I had 0.-- and my wife always point out that I have to write a 0 in front of my decimal points because she's a doctor. And if people don't see the decimal point, someone might overdose on some medication. So let's write it her way, 0.0000000008192. Clearly, this is a super cumbersome number to write. And you might forget about a 0 or add too many 0's, which could be costly if you're doing some important scientific research. at this small a dose. Or maybe you would, I don't want to get into that. But how would I write this in scientific notation? So I start off with the first non-zero number, if I'm starting from the left. So it's going to be 8.192. I just put a decimal and write 0.192 times-- times 10 to what? Well, I just count. Times 10 to the 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I have to include that number, 10 to the minus 10. And I think you'll find it reasonably satisfactory that this number is easier to write than that number over there. Now, and this is another powerful thing about scientific notation. Let's say I have these two numbers and I want to multiply them. Let's say I want to multiply the number 0.005 times the number 0.0008. This is actually a fairly straightforward one to do," + }, + { + "Q": "In 6:22, how does it have 23 zeros? 602200000000000000000000 (phew) only has 21 zeros.", + "A": "He was saying 23 digits after the 6.", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 382 + ], + "3min_transcript": "And actually, this might be interesting, just as an aside. You may or may not know what this number is called. This is called a googol. A googol. In the early '90s if someone said, hey, that's a googol, you wouldn't have thought of a search engine. You would have thought of the number 10 to the 100th power, which is a huge number. It's more than the number of atoms, or the estimated number of atoms, in the known universe. In the known universe. It raises the question of what else is there out there. But I was reading up on this not too long ago. And if I remember correctly, the known universe has the order of 10 to the 79th to 10 to the 81 atoms. And this is, of course, rough. No one can really count this. People are just kind of estimating it. Or even better, guesstimating this. But this is a huge number. What may be even more interesting to you a very popular search engine-- Google. Google is essentially just a misspelling of the word \"googol\" with the O-L. And I don't know why they called it Google. Maybe they got the domain name. Maybe they want to hold this much information. Maybe that many bytes of information. Or, it's just a cool word. Whatever it is-- maybe it was the founder's favorite number. But it's a cool thing to know. But anyway, I'm digressing. This is a googol. It's just 1 with a hundred 0's. But I could equivalently have just written that as 10 to 100, which is clearly an easier way. This is an easier way to write this. This is easier. In fact, this is so hard to write that I didn't even take the trouble to write it. It would have taken me forever. This was just twenty 0's right here. A hundred 0's I would have filled up this screen and you have found it boring. So I didn't even write it. So clearly, this is easier to write. But how can we write something that isn't a direct power of 10? How can we use the power of this simplicity? How can we use the power of the simplicity somehow? And to do that, you just need to make the realization. This number, we can write it as-- so this has how many digits in it? It has 1, 2, 3, and then twenty 0's. So it has 23 digits after the 6. 23 digits after the 6. So what happens if I use this-- if I try to get close to it with a power of 10? So what if I were to say 10 to the 23? Do it in this magenta. 10 to the 23rd power. That's equal to what? That equals 1 with 23 0's. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23. You get the idea, that's 10 to the 23rd." + }, + { + "Q": "In the video, at 11:10, Sal said that scientific notation can be used to express any number in an easier way. Does that imply fractions and integers too? Or is it only natural numbers?", + "A": "I m really not sure. I think so, but I don t think that includes repeating decimals. I m somewhat new to scientific notation too.", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 670 + ], + "3min_transcript": "And I wanted to write it in scientific notation. So I guess the best way to think about it is, it's 7,345. So how can I represent a thousand? Well, I wrote it over here, 10 to the third is 1,000. So we know that 10 to the third is equal to 1,000. So that's essentially the largest power of 10 that I can fit into this. This is seven 1,000's. So if this is seven 1,000's, and then it's 0.3 1,00's, then it's 0.4 1,000's-- I don't know if that helps you, we can write this as 7.345 times 10 to the third because it's going to be seven 1,000's plus 0.3 1,000's. What's 0.3 times 1,000? 0.3 times 1,000 is 300. What's 0.04 times 1,000? That's 40. What's 0.05 times 1,000? That's a 5. So 7.345 times 1,000 is equal to 7,345. So if I took 7.345 times 1,000. The way I do it is I ignore the 0's. I essentially multiply 1 times that guy up there. So I get 7, 3, 4, 5. Then I had three 0's here, so I put those on the end. And then I have three decimal places. So 1, 2, 3. Put the decimal right there. And there you have it, 7.345 times 1,000 is indeed 7,345. Let's do a couple of them. Let's say we wanted to write the number 6 in scientific notation. Obviously, there's no need to write in scientific notation. But how would you do it? Well, what's the largest power of 10 that fits into 6? Well, the largest power of 10 that fits into 6 is just 1. So we could write it as something times 10 to the 0. This is just 1, right? That's just 1. Well, it's just 6. So 6 is equal to 6 times 10 to the 0. You wouldn't actually have to write it this way. This is much simpler, but it shows you that you really can express any number in scientific notation. Now, what if we wanted to represent something like this? I had started off the video saying in science you deal with very large and very small numbers. So let's say you had the number-- do it in this color. And you had 1, 2, 3, 4. And then, let's say five 0's. And then you have followed by a 7. Well, once again, this is not an easy number to deal with. But how can we deal with it as a power of 10? As a power of 10? So what's the largest power of 10 that fits into this number, that this number is divisible by? So let's think about it. All the powers of 10 we did before were going to positive or going to-- well, yeah, positive powers of 10. We could also do negative powers of 10. We know that 10 to the 0 is 1." + }, + { + "Q": "at 8:04 Avogadro's number is written as 6.022 x 10^23 but shouldn't it be 6.022 x 10^20 because there is 20 zeroes after the 6022?", + "A": "Everything is fine in the video, notice that there is a dot between 6 and 0. Of course 6.022x 10^23 = 6022x 10^20, but 6022x 10^20 is not a scientific notation, because scientific notation is a number that is at least 1 and less than 10 multiplied by a power of 10.", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 484 + ], + "3min_transcript": "But how can we write something that isn't a direct power of 10? How can we use the power of this simplicity? How can we use the power of the simplicity somehow? And to do that, you just need to make the realization. This number, we can write it as-- so this has how many digits in it? It has 1, 2, 3, and then twenty 0's. So it has 23 digits after the 6. 23 digits after the 6. So what happens if I use this-- if I try to get close to it with a power of 10? So what if I were to say 10 to the 23? Do it in this magenta. 10 to the 23rd power. That's equal to what? That equals 1 with 23 0's. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23. You get the idea, that's 10 to the 23rd. as some multiple of this guy? Because if we multiplied this guy by 6-- if we multiply 6 times 10 to the 23rd, what do we get? Well, we're just going to have a 6 with twenty three 0's. We're going to have a 6, and then you're going to have twenty three 0's. You're going to have twenty three 0's like that. Because all I did, if you take 6 times this. You know how to multiply. You'd have the 6 times this 1. And then all the 6 times the 0's will all be 0. So you'll have 6 followed by twenty three 0's. So that's pretty useful. But still, we're not getting quite to this number. I mean, this had some 2's in there. So how could we do it a little bit better? Well, what if we wrote it as a decimal? This number right here is identical to this number if these 2's were 0's. But if we want to put those 2's there, what can we do? We could put some decimals here. We could say that this is the same thing as 6.022 times 10 And now, this number is identical to this number, but it's a much easier way to write it. And you could verify it, if you like. It will take you a long time. Maybe we should do it with a smaller number first. But if you multiply 6.022 times 10 to the 23rd, and you write it all out, you will get this number right there. You will get Avogadro's number. Avogadro's number. And although this is complicated or it looks a little bit unintuitive to you at first, this was just a number written out. This has a multiplication and then a 10 to a power. You might say, hey, that's not so simple. But it really is. Because you immediately know how many 0's there are. And it's obviously a much shorter way to write this number. Let's do a couple of more. I started with Avogadro's number because it really shows you the need for a scientific notation. So you don't have to write things like that over and over again. So let's do a couple of other numbers. And we'll just write them in scientific notation." + }, + { + "Q": "At 2:00 Sal says Avogadro's number has \" 6 followed by the 23 digits or the 6022 followed by 20 zeros\". Then at 7:15 we multiply 6*10^23, that is 6 followed by 23 digits. At 8:01 we do 6.022*10^23. My question is why does he multiply it by the power of 23 ? Isn't the answer now have 6 and 26 digits now?", + "A": "In both cases, the leading digit before the decimal point is just the 6 . In 6.022 , the other 3 values are to the right of the decimal point. The 10^23 moves the decimal point from where it is currently located in the number. Another way to compare them... You originally stated 6022 followed by 20 zeroes. In this situation, the decimal point is at the end of 6022. To get to 6.022 we only moved the decimal 3 places. So, we need 20+3 = 23 as our exponent. Hope this makes sense.", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 120, + 435, + 481 + ], + "3min_transcript": "I don't think it's any secret that if one were to do any kind of science, they're going to be dealing with a lot of numbers. It doesn't matter whether you do biology, or chemistry, or physics, numbers are involved. And in many cases, the numbers are very large. They are very, very large numbers. Very large numbers. Or, they're very, small, very small numbers. Very small numbers. You could imagine some very large numbers. If I were to ask you, how many atoms are there in the human body? Or how cells are in the human body? Or the mass of the Earth, in kilograms, those are very large numbers. If I were to ask you the mass of an electron, that would be a very, very small number. So any kind of science, you're going to be dealing with these. And just as an example, let me show you one of the most common numbers you're going to see, in especially chemistry. It's called Avogadro's number. Avogadro's number. And if I were write it in just the standard way of writing as-- do it in a new color. It would be 6022-- and then another 20 zeroes. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. And even I were to throw some commas in here, it's not going to really help the situation to make it more readable. Let me throw some commas in here. This is still a huge number. If I have to write this on a piece of paper or if I were to publish some paper on using Avogadro's number, it would take me forever to write this. And even more, it's hard to tell if I forgot to write a zero or if I maybe wrote too many zeroes. So there's a problem here. Is there a better way to write this? So is there a better way to write this than to write it all out like this? To write literally the 6 followed by the 23 digits, or the 6022 followed by the 20 zeroes there? you're curious, Avogadro's number, if you had 12 grams of carbon, especially 12 grams of carbon-12, this is how many atoms you would have in that. And just so you know, 12 grams is like a 50th of a pound. So that just gives you an idea of how many atoms are laying around at any point in time. This is a huge number. The point of here isn't to teach you some chemistry. The point of here is to talk about an easier way to write this. And the easier way to write this we call scientific notation. Scientific notation. And take my word for it, although it might be a little unnatural for you at this video. It really is an easier way to write things like things like that. Before I show you how to do it, let me show you the underlying theory behind scientific notation. If I were to tell you, what is 10 to the 0 power? We know that's equal to 1." + }, + { + "Q": "So, for clarification, scientific notation is an easier way to write 12,000,000,000,000. So how do you know where to put the decimal place in, say, 6,510,000,000,000? Is there a way to know? I'm sorry if its in the video, I'm only at 3:46.", + "A": "The scientific notation formal format is x*10^y, where 1 <= x < 10 and y is an integer. In other words, when writing in scientific notation, the number you multiply the power of ten must be between 1 (included) and 10. Here are the numbers you suggested as an example: 12,000,000,000,000 = 1.2*10^13 6,510,000,000,000 = 6.51*10^12", + "video_name": "trdbaV4TaAo", + "timestamps": [ + 226 + ], + "3min_transcript": "you're curious, Avogadro's number, if you had 12 grams of carbon, especially 12 grams of carbon-12, this is how many atoms you would have in that. And just so you know, 12 grams is like a 50th of a pound. So that just gives you an idea of how many atoms are laying around at any point in time. This is a huge number. The point of here isn't to teach you some chemistry. The point of here is to talk about an easier way to write this. And the easier way to write this we call scientific notation. Scientific notation. And take my word for it, although it might be a little unnatural for you at this video. It really is an easier way to write things like things like that. Before I show you how to do it, let me show you the underlying theory behind scientific notation. If I were to tell you, what is 10 to the 0 power? We know that's equal to 1. That's equal to 10. What's 10 squared? That's 10 times 10. That's 100. What is 10 to the third? 10 to the third is 10 times 10 times 10, which is equal to 1,000. I think you see a general pattern here. 10 to the 0 has no 0's. No 0's in it. 10 to the 1 has one 0. 10 to the second power-- I was going to say the two-th power. 10 to the second power has two 0's. Finally, 10 to the third has three 0's. Don't want to beat a dead horse here, but I think you get the idea. Three 0's. If I were to do 10 to the 100th power, what would that look like? I don't feel like writing it all out here, but it would be 1 followed by-- you could guess it-- a hundred 0's. So it would just be a bunch of 0's. And if we were to count up all of those 0's, you And actually, this might be interesting, just as an aside. You may or may not know what this number is called. This is called a googol. A googol. In the early '90s if someone said, hey, that's a googol, you wouldn't have thought of a search engine. You would have thought of the number 10 to the 100th power, which is a huge number. It's more than the number of atoms, or the estimated number of atoms, in the known universe. In the known universe. It raises the question of what else is there out there. But I was reading up on this not too long ago. And if I remember correctly, the known universe has the order of 10 to the 79th to 10 to the 81 atoms. And this is, of course, rough. No one can really count this. People are just kind of estimating it. Or even better, guesstimating this. But this is a huge number. What may be even more interesting to you" + }, + { + "Q": "I don't get problem 13 at 8:31\nCan anyone help me?", + "A": "For the 2 triangles to be similar the corresponding angles all have to have congruent measurements. For obvious reasons, angle DBE is the same on both tri. For the other 2 corresponding angles to be proven to be congruent, the 2 sides AC and DE must be parallel to eachother. This is true because the transversal of line AB would make the 2 corresponding angles of the triangle congruent ecause of the corresponding angles of the transversal are these angles. This is true for the other transversal as well.", + "video_name": "bWTtHKSEcdI", + "timestamps": [ + 511 + ], + "3min_transcript": "Whatever angle this us, let's call this x. Let's call this angle y and this angle y. We know that x plus 2y is equal to 180, or that 2y is equal to 180 minus x, Or y is equal to 90 minus x over 2. Now if this is x, And let's call these z and z, So we know that x plus 2z is equal to 180. All the angles in a triangle have to add up to 180. Subtract x from both sides, you get 2z is equal to 180 minus x. Divide by 2, you get z is equal to 90 minus x over 2. So z and y are going to be the same angles. So all the angles are the same, so we're dealing with similar triangles. So choice D was definitely correct. 13. Which of the following facts would be sufficient to prove that triangles ABC, that's the big triangle, and triangle DBE, so that's a small one, are similar? So we have to prove that all of their angles are similar. I cannot even look at the choices and I can guess where So we want to prove that those are similar. So first of all, they share the same angle. Angle ABC, this angle, is the same as angle DBE. So they share that same angle. So we got one angle down. Now let's think about it. If we knew that this angle is equal to that angle and that angle is equal to that angle, we'd be done. And the best way to come to that conclusion is if they told us that this and this are parallel. I'm guessing that's where they're going. Now I might have gone on a completely wrong tangent. Because it those two are parallel, then these two lines So that angle and that angle would be corresponding angles, so they would be congruent, and then that angle and that angle would be corresponding angles so they'd also be congruent. So if they told us that these are parallel, we're done. These are definitely similar triangles. And sure enough, choice C, they tell us that AC and DE These are parallel, that's a transversal, this is a corresponding angle of that, so they're congruent. This is a corresponding angle to this, congruent, so all of the angles are congruent. So we have a similar triangle. Problem 14. OK." + }, + { + "Q": "weren't all the sides on the paralellogram on 10:34 the same?", + "A": "There are four paralellograms: Rhombus, Rhomboid, Rectangle, and a Square. The square is the only paralellogram must have four equal sides. While a Rhombus can have 4 equal lengths, for this problem, it is not a given that this is the case. We are looking for proof only that triangles inside the paralellogram are congruent. The answer to your question is the angles could be the same, either way, you should be able to get the right answer even if they were not.", + "video_name": "bWTtHKSEcdI", + "timestamps": [ + 634 + ], + "3min_transcript": "So that angle and that angle would be corresponding angles, so they would be congruent, and then that angle and that angle would be corresponding angles so they'd also be congruent. So if they told us that these are parallel, we're done. These are definitely similar triangles. And sure enough, choice C, they tell us that AC and DE These are parallel, that's a transversal, this is a corresponding angle of that, so they're congruent. This is a corresponding angle to this, congruent, so all of the angles are congruent. So we have a similar triangle. Problem 14. OK. Fair enough. Parallelogram: that tells us that the opposites sides are parallel. That's parallel to that, and then this is parallel to that. And all of the choices got clipped at the bottom, but I'll copy them over. Maybe I'll copy them above the question. Well, let me see what I can do. I think that's good enough. A little unconventional. OK. Parallelogram is shown below. They say which pair of triangles can be established to be congruent to prove that angle DAB is congruent to angle BCD? So DAB is this. Let me do it in another color. DAB is that angle, is congruent to BCD. the same angle measure. OK, and what do we have to show? They say what pair of triangles can be established to be congruent to prove that. OK, if these are both part of two different congruent triangles and they are the corresponding angles, then we know that they're congruent and we'd be done. So let's see what they say. Triangle ADC and BCD. BCD has this angle in it. BCD does help us because it has this angle in it, but triangle ADC does not have this angle in it, right? Triangle ADC has this the smaller angle in it. ADC doesn't involve this whole thing, so that's not going to help us. Triangle AED, once again, does not involve this larger angle, does not involve the angle DAB. It only involves the little smaller angle, so that's not going to help us. Triangle DAB." + }, + { + "Q": "At 5:40, Sal says that if you have two sides and an angle, you can figure everything else out. The problem here is that SSA Congruence is NOT a valid congruence. SSA Congruence includes two sides and an angle. This is because you most likely end up with two solutions for the third side. Does he mean that if you are given two sides and an angle and the angle is between the sides, then you could find everything else?", + "A": "I think when you have a SSA triangle you would just list both possible answers. Like if you were asked to find side C of an SSA triangle you say C=x or C=y", + "video_name": "VjmFKle7xIw", + "timestamps": [ + 340 + ], + "3min_transcript": "that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A. so four times the sine of 105 gives us, it's approximately equal to, let's just round to the nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square root of two over two. Come to think of it, B is four times the sine of 45 degrees. If we wanted actual numerical value, we could just write this as two square roots of two. But let's actually figure out what that is. Two square roots of two is equal to 2.83. So B is approximately equal to 2.83. So [I'm] be clear, this four divided by two is two square roots of two, which is 2.8. Which is approximately equal to 2.83 if we round to the nearest 100th, 2.83, which also seems pretty reasonable here. So the key of the Law of Cosines is if you have two angles and a side, you're able to figure out everything else about it. Or if you actually had two sides and an angle, you also would be able to figure out everything else about the triangle." + }, + { + "Q": "At 2:50, how did you get 1/4?", + "A": "1/2 divided by 2 = 1/4 he got the 1/2 because sin 30 = 1/2", + "video_name": "VjmFKle7xIw", + "timestamps": [ + 170 + ], + "3min_transcript": "And so applying the Law of Sines, actually let me label the different sides. Let's call this side right over here, side A or has length A. And let's call this side, right over here, has length B. So the Law of Sines tells us that the ratio between the sine of an angle, and that the opposite side is going to be constant through this triangle. So it tells us that sine of this angle, sine of 30 degrees over the length of the side opposite, is going to be equal to sine of a 105 degrees, over the length of the side opposite to it. Which is going to be equal to sine of 45 degrees. So sine of 45 degrees over B. And so if we wanted to figure out A, we could solve this equation right over here. And if we wanted to solve for B, we could just set this equal to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you might just remember it from your unit circles or from even 30, 60, 90 triangles and that's 1/2. And if you don't remember it, you can use a calculator to verify that. I have already verified that this is in degree mode, so it's 0.5. So this is going to be equal to 1/2 over two. So another way of thinking about it, that's going to be equal to 1/4, this piece is equal to 1/4 is equal to sine of a 105 degrees over A. Let me write this, this is equal to sine of 105 degrees over A. And actually, we could also say, that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A." + }, + { + "Q": "At around 4:30, why do you need to take the reciprocal of both sides to solve the law of sines?", + "A": "The goal was to isolate the variable. There are several ways of accomplishing this, but since the variable was in the denominator, taking the reciprocal of both sides seemed a useful choice.", + "video_name": "VjmFKle7xIw", + "timestamps": [ + 270 + ], + "3min_transcript": "So sine of 45 degrees over B. And so if we wanted to figure out A, we could solve this equation right over here. And if we wanted to solve for B, we could just set this equal to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you might just remember it from your unit circles or from even 30, 60, 90 triangles and that's 1/2. And if you don't remember it, you can use a calculator to verify that. I have already verified that this is in degree mode, so it's 0.5. So this is going to be equal to 1/2 over two. So another way of thinking about it, that's going to be equal to 1/4, this piece is equal to 1/4 is equal to sine of a 105 degrees over A. Let me write this, this is equal to sine of 105 degrees over A. And actually, we could also say, that this is equal to that. That 1/4 is equal to sine of 45 degrees over B. Actually, sine of 45 degrees is another one of those that is easy to jump out of unit circles. You might remember it's square of two over two. Let's just write, that's square root of two over two. And you can use a calculator, but you'll get some decimal value right over there. But either case, in either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A, we could just multiply both sides times the sine of a 105 degrees. So we get four times the sine of 105 degrees is equal to A. so four times the sine of 105 gives us, it's approximately equal to, let's just round to the nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square root of two over two. Come to think of it, B is four times the sine of 45 degrees." + }, + { + "Q": "at 12:16 why does Sal evaluate one of the Sin(t)'s at Pi all of a sudden (instead of Pi/2)? I got for that problem:\n2Pi + 2 and sal has 4 + 2Pi. Is this a mistake?", + "A": "The second expression with a sine was sin(2t), which, evaluated at pi/2, is sin(2pi/2)=sin(pi)=0", + "video_name": "wyTjyQMVvc4", + "timestamps": [ + 736 + ], + "3min_transcript": "all of that dt. Now this should be reasonably straight forward to get the antiderivative of. Let's just take it. The antiderivative of this is antiderivative of cosine of t; that's a sine of t. The derivative of sine is cosine. So this is going to be 4 sine of t-- the scalars don't affect anything --and then, well let me just distribute this 4. So this is 4 times 1 which is 4 minus 4 cosine of 2t. So the antiderivative of 4 is 4t-- plus 4t --and then the antiderivative of minus 4 cosine of u00b5 t? Let's see it's going to be sine of 2t. The derivative of sine of 2t is 2 cosine of 2t. We're going to have to have a minus sign there, and put a 2 What's the derivative of minus 2 sine of t? Take the derivative of the inside 2 times minus 2 is minus 4. And the derivative of sine of 2t with respect to 2t is cosine of 2t. So there we go; we've figured out our antiderivative. Now we evaluate it from 0 the pi over 2. And what do we get? We get 4 sine-- let me write this down, for I don't want to skip too many --sine of pi over 2 plus 4 times pi over 2-- that's just 2 pi minus 2 sine of 2 times pi over 2 sine of pie, and then all of that minus all this evaluated at 0. That's actually pretty straightforward because sine of 0 is 0. 4 times 0 is 0, and sine of 2 times 0, that's also 0. So everything with the 0's work out nicely. And then what do we have here? Sine of pi over 2-- in my head, I think sine of 90 degrees; And then sine of pi is 0, that's 180 degrees. So this whole thing cancels out. So we're left with 4 plus 2 pi. So just like that we were able to figure out the area of this first curvy wall here, and frankly, that's the hardest part. Now let's figure out the area of this curve. And actually you're going to find out that these other curves as they go along the axes are much, much, much easier, but we're going to have to find different parametrizations for this. So if we take this curve right here, let's do a parametrization for that. Actually, you know what? Let me continue this in the next video because I realize I've been running a little longer. I'll do the next two walls and then we'll sum them all up." + }, + { + "Q": "Wouldn't it be 1/2 +pi^2/8? if not, what happened to the +1 at 10:23 ?could someone please help, because I don't understand where that 1 went.", + "A": "The 1 was multiplied by the sin(t)*cos(t) function inside the integral. The square root of negative sine squared plus cosine squared is one.", + "video_name": "uXjQ8yc9Pdg", + "timestamps": [ + 623 + ], + "3min_transcript": "curtain that has our curve here as kind of its base, and has this function, this surface as it's ceiling. So we go back down here, and let me rewrite this whole thing. So this becomes the integral from t is equal to o to t is equal to pi over 2-- I don't like this color --of cosine of t, sine of t, cosine times sine-- that's just the xy --times ds, which is this expression right here. And now we can write this as-- I'll go switch back to that color I don't like --the derivative of x with respect to t is minus sine of t, and we're going to square it, plus the derivative of y with respect to t, that's cosine of t, and we're going to square it-- let me make my radical a little bit bigger --and then all of that times dt. you realize that this right here, and when you take a negative number and you squared it, this is the same thing. Let me rewrite, do this in the side right here. Minus sine of t squared plus the cosine of t squared, this is equivalent to sine of t squared plus cosine of t squared. You lose the sign information when you square something; it just becomes a positive. So these two things are equivalent. And this is the most basic trig identity. This comes straight out of the unit circle definition: sine squared plus cosine squared, this is just equal to 1. So all this stuff under the radical sign is just equal to 1. And we're taking the square root of 1 which is just 1. So all of this stuff right here will just become 1. And so this whole crazy integral simplifies a good bit pi over 2 of-- and I'm going to switch these around just because it will make it a little easier in the next step --of sine of t times cosine of t, dt. All I did, this whole thing equals 1, got rid of it, and I just switched the order of that. It'll make the next up a little bit easier to explain. Now this integral-- You say sine times cosine, what's the antiderivative of that? And the first thing you should recognize is, hey, I have a function or an expression here, and I have its derivative. The derivative of sine is cosine of t. So you might be able to a u substitution in your head; it's a good skill to be able to do in your head. But I'll do it very explicitly here. So if you have something that's derivative, you define that something as u. So you say u is equal to sine of t and then du, dt, the derivative of u with respect to t is equal to cosine of t." + }, + { + "Q": "At 1:37 why would it be mm squared?", + "A": "when you do an area it is always in mm squared or square mm (or whatever measurement it is in. This is because it would take that many 1mm x 1mm squares to fill up the area of the circle. don t forget a circle is a 2D shape, i may not have used candy for an example on area but just pretend it s the circle bit on one side of the candy lol", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 97 + ], + "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi." + }, + { + "Q": "At 2:47\nDo you get the same anwser if you do 3.14 times R and just add the second power?", + "A": "Not quite. Because of PEMDAS, you square R, then multiply that by pi, or 3.14. Squaring 3.14r would get a completely different answer.", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 167 + ], + "3min_transcript": "Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi. Now, we are using the calculator's internal representation of pi which is going to be more precise than what I had in the last one. And you can 201.06 (to the nearest hundred) So, more precise is 201.06 square millimeters. So, this is closer to the actual answer, because the calculator's representation is more precise than this very rough approximation of what pi is." + }, + { + "Q": "At 1:30, why does Sal say that the answer is still squared, after already squaring the radius?", + "A": "the units are squared, not re-squaring the numbers. Areas are always in square units, Volumes are in cubed units.", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 90 + ], + "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi." + }, + { + "Q": "At 1:53, why after you have squared 8 (64) do you still leave it squared?", + "A": "Because the equation to find the area of a circle is \u00cf\u0080 r2. Therefore to find the area you have to have the radius squared and then you multiply it by \u00cf\u0080. So when you square 8 you get 64 and to find the area you multiply it by \u00cf\u0080 ( \u00cf\u0080 r2 ). He leaves 64 that way because its r2 ( 8*8 ). \u00cf\u0080*64=Area. Hope you got your question answered !!", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 113 + ], + "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi." + }, + { + "Q": "@ 1:18 how did he get 64 mm from 8 mm", + "A": "he found the square root of the radius of the circle", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 78 + ], + "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi." + }, + { + "Q": "2:59 doesnt he mean to say millimeters sq?", + "A": "You can say square millimeters or millimeters squared. Either is correct.", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 179 + ], + "3min_transcript": "Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi. Now, we are using the calculator's internal representation of pi which is going to be more precise than what I had in the last one. And you can 201.06 (to the nearest hundred) So, more precise is 201.06 square millimeters. So, this is closer to the actual answer, because the calculator's representation is more precise than this very rough approximation of what pi is." + }, + { + "Q": "Starting from 0:45, why do we do \"radius^2\", or to be more specific, why \"^2\" to come to the conclusion of the area? any explanation of this somewhere?", + "A": "It comes from a higher branch of mathematics called calculus, which is really good at deriving formulae for areas of shapes. When you try it for a circle, the r^2 pops out of the formula.", + "video_name": "ZyOhRgnFmIY", + "timestamps": [ + 45 + ], + "3min_transcript": "A candy machine creates small chocolate wafers in the shape of circular discs. The Diameter of each wafer is 16 millimeters. Whats it the area of each candy? So, the candy they say is in the shape of circular disc and they tell us that the diameter is 16 millimeters. If I draw a line across the circle, that goes through the center. The length of the line all the way across the circle through the center is 16 millimeters. The Diameter here is 16 millimeters. And they want us to figure out the area of the surface of the candy. Essentially the area of this circle. When we think about area, we know that the area of the circle is pi times the radius of the circle square. They gave us the diameter, what is the radius? Distance from the center of the circle to the outside, to the boundary of the circle. So, it will be this distance over here, which is exactly half of the diameter. So, would be 8 millimeters. So, where we see the radius, we can put 8 millimeters. So, the Area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And, typically this is written as pi after 64. So, you might often see it as 64 pi millimeter squared. Now, this is the answer 64 pi millimeters squared. But sometimes it is not satisfying to leave it as 64 pi millimeter squared. And we could start to use the approximate values of pi. So, the most rough approximate value which is tensed to be used is saying that pi, a very rough approximation, is equal to 3.14. So, in that case we can that this will be equal to 64 times 3.14 millimeters squared. We can get a calculator to figure out what this will be in decimal form. So, we have 64 times 3.14, gives us 200.96 So, we can say that the area is approximately equal to 200.96 square millimeters. Now if we want to get a more accurate representation of this, pi just actually keeps going on and on forever, we could use the calculator's internal representation of pi." + }, + { + "Q": "at 1:35 how does prime factorization work?", + "A": "prime factorization is making a number as small as it can by only using prime numbers", + "video_name": "DKh16Th8x6o", + "timestamps": [ + 95 + ], + "3min_transcript": "We are asked to find the cube root of negative 512. Or another way to think about it is if I have some number, and it is equal to the cube root of negative 512, this just means that if I take that number and I raise it to the third power, then I get negative 512. And if it doesn't jump out at you immediately what this is the cube of, or what we have to raise to the third power to get negative 512, the best thing to do is to just do a prime factorization of it. And before we do a prime factorization of it to see which of these factors show up at least three times, let's at least think about the negative part a little bit. So negative 512, that's the same thing-- so let me rewrite the expression-- this is the same thing as the cube root of negative 1 times 512, which is the same thing as the cube root of negative 1 times the cube root of 512. What number, when I raise it to the third power, do I get negative 1? Well, I get negative 1. This right here is negative 1. Negative 1 to the third power is equal to negative 1 times negative 1 times negative 1, which is equal to negative 1. So the cube root of negative 1 is negative 1. So it becomes negative 1 times this business right here, times the cube root of 512. And let's think what this might be. So let's do the prime factorization. So 512 is 2 times 256. 256 is 2 times 128. 128 is 2 times 64. We already see a 2 three times. 64 is 2 times 32. 32 is 2 times 16. We're getting a lot of twos here. 16 is 2 times 8. And 4 is 2 times 2. So we got a lot of twos. So essentially, if you multiply 2 one, two, three, four, five, six, seven, eight, nine times, you're going to get 512, or 2 to the ninth power is 512. And that by itself should give you a clue of what the cube root is. But another way to think about it is, can we find-- there's definitely three twos here. But can we find three groups of twos, or we could also find-- let me look at it this way. We can find three groups of two twos over here. So that's 2 times 2 is 4. 2 times 2 is 4. So definitely 4 multiplied by itself three times is divisible into this. But even better, it looks like we can get three groups of three twos. So one group, two groups, and three groups." + }, + { + "Q": "When you're drawing a vector with given points (for example, (1,2) for vector v @ 12:57), how do you know what direction the vector is going in? Sal keeps drawing the vector going up but why can't it go down?", + "A": "One Can Draw A Vector In Two Cases. 1)A Point Through Which The Vector Passes And Its Inclination(Angle)With Respect To Any Of The Co-ordinate Axes Are Known Or 2)Minimum Of Two Points Are Known Through Which The Vector Passes. One Can Draw A Vector Through A Single Point In Arbitrary Direction Only.", + "video_name": "r4bH66vYjss", + "timestamps": [ + 777 + ], + "3min_transcript": "we scale our vectors. When we multiply it times some scalar factor. So let me pick new vectors. Those have gotten monotonous. Let me define vector v. v for vector. Let's say that it is equal to 1, 2. So if I just wanted to draw vector v in standard position, I would just go 1 to the horizontal and then 2 to the vertical. That's the vector in standard position. If I wanted to do it in a non standard position, I could do it right here. 1 to the right up 2, just like that. Equally valid way of drawing vector v. Equally valid way of doing it. Now what happens if I multiply vector v. What if I have, I don't know, what if I have 2 times v? 2 times my vector v is now going to be equal to 2 times 2, and then 2 times 2 which is 4. Now what does 2 times vector v look like? Well let me just start from an arbitrary position. Let me just start right over here. So I'm going to go 2 to the right, 1, 2. And I go up 4. 1, 2, 3, 4. So this is what 2 times vector v looks like. This is 2 times my vector v. And if you look at it, it's pointing in the exact same direction but now it's twice as long. And that makes sense because we scaled it by a factor of 2. When you multiply it by a scalar, or you're not changing its direction. Its direction is the exact same thing as it was before. You're just scaling it by that amount. And I could draw this anywhere. I could have drawn it right here. I could have drawn 2v right on top of v. Then you would have seen it, I don't want to cover it. You would have seen that it goes, it's exactly, in this case when I draw it in standard It's along the same line, it's just twice as far. it's just twice as long but they have the exact same direction. Now what happens if I were to multiply minus 4 times our vector v? Well then that will be equal to minus 4 times 1, which is minus 4. And then minus 4 times 2, which is minus 8. So this is on my new vector. Minus 4, minus 8. This is minus 4 times our vector v. So let's just start at some arbitrary point. Let's just do it in standard position. So you go to the right 4. Or you go to the left 4. So so you go to the left 4, 1, 2, 3, 4. And then down 8. Looks like that. So this new vector is going to look like this. Let me try and draw a relatively straight line. There you go. So this is minus 4 times our vector v. I'll draw a little arrow on it to make sure you know it's a vector." + }, + { + "Q": "At 19:40, it is said that \"2x -y -z +3x\" must be equal to 0 in order for b to be valid.\nIsn't it that we have then:\nif b is valid => 5x -y -z = 0\n(i.e. not yet \"<=>\". Necessary condition only).\nThen b is in the plan of equation 5x -y -z = 0, in other words:\n\"the plan of equation 5x -y -z = 0\" is included in C(A). Finally, as C(A) is a plan (dim C(A) = 2), then C(A) = \"the plan of equation 5x -y -z = 0\". (now we have \"<=>\" . b is valid <=> 5x -y -z = 0 )", + "A": "He states Ax = b at 14:00 ish, and asks what are all the possible b s, which form C(A). So if you find out that Ax forms a plane, that equality gives you equivalence .. Ax forms a plane <=> All the b s form a plane, because LHS = RHS (or in other words the validity of b is given). At least that s how I d interpret it.", + "video_name": "EGNlXtjYABw", + "timestamps": [ + 1180 + ], + "3min_transcript": "So let me from the get go try to zero out this third row. And the best way to zero out this third row is to just replace the third row. So the first row-- well, I won't even write the first row. The second row is 0, 1, minus 2, minus 1, and 2x minus y. I'm not even going to worry about the first row right now. But let's replace the third row, just in our attempt to go into reduced row echelon form. Let's replace it with the second row minus the third row. So you get 2x minus y minus z plus 3x. I just took this minus this. So minus z plus 3x. So 0 minus 0 is 0. 1 minus 1 is 0. Minus 2 minus minus 2 is 0, and that's also 0. So we're only going to have a valid solution to Ax equals b What happens if he's not equal to zero? Then we're going to have a bunch of zeroes equaling some number, which tells us that there's no solution. So if I pick a b where this guy does not equal zero, then I'll have no solution. If this guy equals 5, if I pick x, y, and z's such as that this expression is equal to 5, then Ax equal to b will have no solution, because it will have 0 is equal to 5. So this has to equal 0. So 2x minus y minus z plus 3x must be equal to 0 in order for b to be valid, in order for b to be a member of the column space of A, in order for it to be a valid vector that Ax can become, or the product A times x can become for some x. So what does this equal to? If we add the 2x plus the 3x, I get 5x minus y minus z is figured out the basis vectors. We said oh, you know what? The basis vectors, they have to be in the column space themselves by definition. So let me find a normal vector to them both by taking the cross product. I did that, and I said the cross product times any valid vector in our space minus one of the basis vectors has to be equal to zero, and then I got this equation. Or we could have done it the other way. We could've actually literally solved this equation setting our b equal to this. We said what b's will give us a valid solution? And our only valid solution will be obtained when this guy has to be equal to zero, because the rest of his row became zero. And when we set that equal to zero, we got the exact same equation. So, hopefully, you found this to be mildly satisfying, because we were able to tackle the same problem from two" + }, + { + "Q": "I'm so confused.. at about 11:00 Sal decided to factor out 27\u00c2\u00b712x^2 - 4x^6 = 0 to 4x^2(27\u00c2\u00b73-x^4) = 0 .. When I was doing it on my own I multiplied the constants and got 324x^2-4x^6 = 0, factoring out to 4x^2(324-x^4) = 0... Is this an example of where BEDMAS is really important?", + "A": "You forgot to divide the 324 in your final equation by 4. :-)", + "video_name": "zC_dTaEY2AY", + "timestamps": [ + 660 + ], + "3min_transcript": "But if the derivative is equal to 0, the second derivative is equal to 0, you cannot assume that is an inflection point. So what we're going to do is, we're going to find all of the point at which this is true, and then see if we actually do have a sign change in the second derivative of that point, and only if you have a sign change, then you can say it's an inflection point. So let's see if we can do that. So just because a second derivative is 0, that by itself does not tell you it's an inflection point. It has to have a second derivative of 0, and when you go above or below that x, the second derivative has to actually change signs. Only then. So we can say, if f prime changes signs around x, then we can say that x is an inflection. And if it's changing signs around x, then it's definitely if it's negative before x, has to be positive after x,or if it's positive before x, has to be negative after x. So let's test that out. So the first thing we need to do is find these candidate points. Remember, the candidate points are where the second derivative is equal to 0. We're going to find those points, and then see if this is true, that the sign actually changes. We want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where this our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12x squared minus 4x to the sixth is equal to 0. Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So 4x squared. Now we'll have 27 times, if we factor 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the fourth is equal to 0. So the x's that will make this equal to 0 will satisfy either, I'll switch colors, either 4x squared is equal to 0, or, now 27 times 3, I can do that in my head. That's 81. 20 times 3 is 60, 7 times 3 is 21, 60 plus 21is 81. Or 81 minus x to the fourth is equal to 0. Any x that satisfies either of these will make this entire expression equal 0. Because if this thing is 0, the whole thing is If this thing is 0, the whole thing is going to be equal to 0. Let me be clear, this is 81 right there. So let's solve this. This is going to be 0 when x is equal to 0, itself." + }, + { + "Q": "I think at 19:00 the way this notation is formed is as x --> 3(neg. side)\nThe arrow meaning approaching 3 from the negative side.", + "A": "Hay Bennet! Join us as a life-long learner! We can use guys like you in KA. It s been a year since you ve asked this question, & I haven t seen you around since! Miss you around! :)", + "video_name": "zC_dTaEY2AY", + "timestamps": [ + 1140 + ], + "3min_transcript": "a little bit less than 3. If x is a little bit less than 3, if it's like 2.9999, this number is going to be less than 81, so this is also going to be positive. And of course, the denominator is always positive. So as x is less than 3, is approaching from the left, we are concave upwards. This thing's going to be a positive. Then f prime prime is greater than 0. We are upwards, concave upwards. When x is just larger than 3, what's going to happen? Well, this first term is still going to be positive. But if x is just larger than 3, x to the fourth is going to be just larger than 81, and so this second term is going to be negative in that situation. Let me do it ina new color. It's going to be negative when x is larger than 3. Because this is going to be larger than 81. So if this is negative and this is positive, then the whole thing is going to be negative, because this denominator is still going to be positive. going to be concave downwards. One last one. What happens when x is just a greater than minus 3? So just being greater than minus 3, that's like minus 2.99999. So when you take minus 2.99 square it, you're going to get a positive number, so this is going to be positive. And if you take minus 2.99 to the fourth, that's going to be a little bit less than 81, right? Because 2.99 to the fourth is a little bit less than 81, so this is still going to be positive. So you have a positive times a positive divided by a positive, so you're going to be concave upwards, because your second derivative is going to be greater than 0. Concave upwards. And then finally, when x is just, just less than negative 3, remember, when I write this down, I don't mean for all x's larger than negative 3,or all x's smaller than negative 3. There's actually no, well, I can't think of the notation from the left, but what happens if we just go to minus 3.11? Or 3.01, I guess is a better one, or 3.1? Well, this term right here is going to be positive. But if we take minus 3.1 to the fourth, that's going to be larger than positive 81, right? The sign will become positive, it'll be larger than 81, so this'll become negative. So in that case as well, we'll have a positive times a negative divided by a positive, so then our second derivative is going to be negative. And so we're going to be downwards. So I think we're ready to plot. so first of all, is x plus or minus 3 inflection points? As we approach x is equal to 3 from the left, we are concave upwards, and then as we cross 3, the second derivative is 0. The second derivative's 0, I lost it up here. The second derivative is 0. And then, as we go to the right of 3, we become concave downwards." + }, + { + "Q": "At 3:41, I thought that you put dots over the top of repeating decimals. Can you do both?", + "A": "Some students learn to write repeating decimals by putting dots over the repeating numbers, but in this case Sal puts a line over the repeating numbers. However If you want to use Sal s method you can but if you were taught to put dots, its better you do that and yes, you CAN use both lines or dots for your answer. Hope that helped !!", + "video_name": "Y2-tz27nKoQ", + "timestamps": [ + 221 + ], + "3min_transcript": "Well, we could take 100 from the 100's place, and make it 10 10's. And then we could take 1 of those 10's from the 10's place and turn it into 10 1's. And so 9 10's minus 8 10's is equal to 1 10. And then 10 -1 is 9. So it's equal to 19. You probably \u2013 You might have been able to do that in your head. And then we have \u2013 And I see something interesting here \u2013 because when we bring down our next 0, we see 190 again. We saw 190 up here. But let's just keep going. So 27 goes into 190 \u2013 And we already played this game. It goes into it 7 times. 7 \u00d7 27 \u2013 we already figured out \u2013 was 189. We subtracted. We had a remainder of 1. Then we brought down another 0. We said 27 goes into 10 0 times. 0 \u00d7 27 is 0. Subtract. Then you have \u2013 but we've got to bring down another 0. So you have 27, which goes into 100 \u2013 (We've already done this.) \u20133 times. So you see something happening here. It's 0.703703. And we're just going to keep repeating 703. This is going to be equal to 0.703703703703 \u2013 on and on and on forever. So the notation for representing a repeating decimal like this is to write the numbers that repeat \u2013 in this case 7, 0, and 3 \u2013 and then you put a line over all of to indicate that they repeat. So you put a line over the 7, the 0, and the 3, which means that the 703 will keep repeating on and on and on. So let's actually input it into the answer box now. So it's 0.703703. the first six digits of the decimal in your answer. And they don't tell us to round or approximate \u2013 because, obviously, if they said to round to that smallest, sixth decimal place, then you would round up because the next digit is a 7. But they don't ask us to round. They just say, \"Include only the first six digits of the decimal in your answer.\" So that should do the trick. And it did." + }, + { + "Q": "At 1:50 he says that [sq rt]of 3x3x13 = [sq rt] of 3x3x[sq rt] of 13......Why does x13 = [sq rt] of 13? What did I miss?", + "A": "Basically Sal is saying sqrt(3*3*13). Both 3s and the 13 are under the square root symbol. What he is doing is taking out the sqrt(13) , so that sqrt(3*3*13)=sqrt(3*3)*sqrt(13) I hope this helps you.", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 110 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "0:42, 3 doesn't go into 11 3 times. or am i missing something?", + "A": "You are missing something. Since 117 (if I add digits, they =9, so I know that it is divisible by three and 9). When he says 3 goes into 11 3 times, he then multiplies and subtracts (11-9) and drops down the remainder of 2 and divides 27 by 3", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 42 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "Aight a question, Should i watch this after watching the exponent properties playlist?\nBecause im going in order and this one goes first, but sal says at 1:38, and we know this by our exponent properties.", + "A": "then by using Sal s advise watch the exponents video.", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 98 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "Ask a question...I don't know why at 2:04 he crosses out the 3 ?!", + "A": "As he explained in the video, the square root of 3*3 is the same thing as saying the square root of 9. The square root of 9 is 3, so he simplified the square root of 3*3 that he had written down into just 3.", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 124 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "At 1:55 it is explained that 5 sqrt 3*3 (also 5 sqrt 9) equals 15. Sal crosses out one of the 3's and writes a little 3 above the radical. Could someone explain this to me or lead me to a video explaining it?", + "A": "He is just using the basic definition of a square root. Square roots reverse the process of squaring a number. 3^2 = 9 So, sqrt(9) = 3 If you don t understand this concept, you need to go back to the videos that introduce square roots. Use the search bar and search for introduction to square roots to find the video.", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 115 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "Which video (and where) explains why you can add up the digits of a number to see if it's divisible by 3 like at 0:25 - 0:36?", + "A": "go to pre- algabra and in the factors and multiples section you will find divisablity tests at the top of the list and it explains the rule for 3 in the first video", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 25, + 36 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "at 1:48 sal splits the square root.why?", + "A": "He does show to show you that the sqrt(3*3)=3 so that can be taking out of the entire radical.", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 108 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "At 0:09, Sal said that 117 is not a perfect square. What does that mean?", + "A": "It means that a any number times itself won t equal 117", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 9 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can." + }, + { + "Q": "At hte first why did you pick x how do you know that was x 2:35", + "A": "He did not know that it was x (which isn t a number anyway). He just picked x as a variable or as a placeholder for an unknown value. He could have put in T, Delta or Cow, and the results would be the same in the end. The variables would just have wonky names, so he picked x, which sounds sensible.", + "video_name": "1uWZNW5PF-s", + "timestamps": [ + 155 + ], + "3min_transcript": "simplified form if you divide both this height and this width So that's why they're saying greatest common divisor 1. And then they say, find the perimeter of the rectangle. So let's see what we can do here. And I encourage you to pause it and try to do it on your own before I bumble my way through this problem. So let's start in the beginning. Let's start at this square right over here, the center square. And they did tell us that they're all squares. So let's say that that square right over here has a length x and a height x. It's an x by x square. So let me write it. So this is an x, and that is an x. So this is an x by x square right over there. And then you have this square right over here. And we don't know its measurements. So let's say that this square right over here is y by y. So it has y width, and it also has y height. Well, this is an x plus y by an x plus y square, because the width of these two squares combined made the width of this larger square. So what I'm going to do is-- actually, this might be an easier way to write it. Since these are all squares, I'm going to write the dimension of that square inside the square. So this is going to be an x by x square. Kind of a non-conventional notation, but it'll help us keep things a little bit neat. This is going to be a y by y square. So I'm not saying the area is y. I'm saying it's y by y. This over here is going to be an x plus y times-- an x plus y is going to be each of its dimensions. So it's going to be x plus y height and x plus y width. Then this one over here-- well, if this dimension is x plus y and this dimension right over here is x, then this whole side or any of the sides of the square So x plus x plus y is 2x plus y. You can imagine that I'm just labeling the left side of each of these squares. The left side of this square has length y. Left side of this one, x. This one has x plus y. And then this is 2x plus y. And then we can go do this one up here. Well, if this distance right over here is 2x plus y and this distance right over here is x plus y, you add them together to get the entire dimension of one side of the square. So it's going to be 3x plus 2y. I've just added the 2x plus the x and the y plus the y to get 3x plus 2y is the length of one dimension or one side of this square. And they're all the same. Now let's go to this next square. Well, if this length is 3x plus 2y and this length is 2x plus y, then this entire length right over here" + }, + { + "Q": "At 6:13, Sal got 2 differant answers for 2 sides of a square. How did he get the 2 answers?", + "A": "To get the dimension on the left side of the large rectangle, he added the lengths of a side of each square on the left. To get the dimension on the right side of the large rectangle, he added the lengths of a side of each square on the right. Since we know opposite sides of a rectangle have equal length, we can set up an equation (13x+7y=8x+9y) to solve for the ratio of x to y (x=2/5*y).", + "video_name": "1uWZNW5PF-s", + "timestamps": [ + 373 + ], + "3min_transcript": "5x plus 3y is going to be that entire length right over there. And we can also go to this side right over here where we have this length-- let me do that same color. This length is 3x plus 2y. This is x plus y. And this is y. So if you add 3x plus 2y plus x plus y plus y, you get 4x plus-- what is that-- 4y, right? 2y, 3y, 4y. And then we can express this character's dimensions in terms of x and y because this is going to be 5x plus 3y. Then you're going to have 2x plus y. And then you're going to have x. So you add the x's together. 5x plus 2x is 7x, plus x is 8x. And then you add the y's together, 3y plus y, and then you don't have a y there. So that's going to be plus 4y. And then finally, we have this square right over here. Its dimensions are going to be the y plus the 4x plus 4y. So that's 4x plus 5y. And then if we think about the dimensions of this actual rectangle over here, if we think about its height right over there, that's going to be 5x plus 3y plus 8x plus 4y. So 5 plus 8 is 13. So it's 13x plus 3 plus 4 is 7y. So that's its height. But we can also think about its height by going on the other side of it. And maybe this will give us some useful constraints because this is going to have to be the same length as this over here. And so if we add 4x plus 4x, we get 8x. So these are going to have to be equal to each other, so that's an interesting constraint. So we have 13x plus 7y is going to have to equal 8x plus 9y. And we can simplify this. If you subtract 8x from both sides, you get 5x. And if you subtract 7y from both sides, you get 5x is equal to 2y. Or you could say x is equal to 2/5 y. In order for these to show up as integers, we have to pick integers here. But let's see if we have any other interesting constraints if we look at the bottom and the top of this, if this gives us any more information. So if we add 5x plus 3y plus 3x plus 2y plus 4x plus 4y," + }, + { + "Q": "At 3:56 isn't it negative 3/2", + "A": "No: -6/-4 = 3/2 Those minus signs cancel He plugs -2 into 6(x+1)/(x-2) = -6/-4 = 3/2 looks correct", + "video_name": "oUgDaEwMbiU", + "timestamps": [ + 236 + ], + "3min_transcript": "equal to negative 2. So this is equal to 6 times-- we're going divided by x plus 2 in the numerator, x plus 2 in the denominator-- so it's going to be 6 times x plus 1 over x minus 2. And we have to put the constraint here because now we've changed it. Now this expression over here is actually defined at x equals negative 2. But in order to be equivalent to the original function we have to constrain it. So we will say for x not equal to negative 2. And it's also obvious that x can't be equal to 2 here. This one also isn't defined at positive 2 because you're dividing by 0. So you could say, for x does not equal to positive or negative 2 if you want to make it very explicit. But they ask us, what could we assign f of negative 2 to make the function continuous at the point? except that the function is not defined at x equals negative 2. So that's why we have to put that constraint here if we wanted this to be the same thing as our original function. But if we wanted to re-engineer the function so it is continuous at that point then we just have to set f of x equal to whatever this expression would have been when x is equal to negative 2. So let's think about that. Let's think about that. So 6 times negative 2 plus 1 over negative 2 minus 2 is equal to-- this is 6 times negative 1. So it's negative 6 over negative 4, which is equal to 3/2. So if we redefine f of x, if we say f of x is equal to 6x For x not equal positive or negative 2, and it's equal to 3/2 for x equals negative 2. Now this function is going to be the exact same thing as this right over here. This f of x, this new one. This new definition-- this extended definition of our original one-- is now equivalent to this expression, is equal to 6 times x plus 1 over x minus 2. But just to answer their question, what value should be assigned to f of negative 2 to make f of x continuous to that point? Well f of x should be-- or f of negative 2 should be 3/2." + }, + { + "Q": "At 1:10, I don't get why do you start multiplying (-3x -2). It's getting me confused.", + "A": "If you have a variable, like X in a fraction, you usually want to get that variable isolated. So in order to get X by itself you have to multiply both sides by the denominator. When that happens, the side with the fraction has its denominator cancelled out, but what you do to one side, you must do to the other. So that is why he had to multiply both sides by (-3x-2). Hope this helped!!", + "video_name": "PPvd4X3Wv5I", + "timestamps": [ + 70 + ], + "3min_transcript": "So we have 14x plus 4 over negative 3x minus 2 is equal to 8. And I'll give you a few moments to see if you can tackle it on your own. So this equation right here, at first it doesn't look like a straightforward linear equation. We have one expression on top of another expression. But as we'll see, we can simplify this to turn it into a linear equation. So the first thing that I want to do is, I don't like this negative 3x plus 2 sitting here in the denominator, it makes me stressed. So I want to multiply both sides of this equation times negative 3x minus 2. What does that do for us? Well, on the left-hand side, you have this negative 3x minus 2, it's going to be over negative 3x minus 2, they will cancel out. And so you're left with, on the left-hand side, your 14x plus 4. have to multiply 8 times negative 3x minus 2. So you are left with-- well, 8 times negative 3x is negative 24x, and then 8 times negative 2 is negative 16. And there you have it. We have simplified this to just a traditional linear equation. We've got variables on both sides, so we can just keep simplifying. So the first thing I want to do, let's just say we want to put all of our x terms on the left-hand side. So I want to get rid of this negative 24x right over here. So the best way to do that, I'm going to add 24x to the right-hand side. I can't just do it to the right-hand side, I have to also do it to the left-hand side. And so I am left with, on the left-hand side, 14x plus 24x is 38x. And then I have the plus 4-- Is equal to, well, negative 24x plus 24x. are left with just the negative 16. Now we just have to get rid of this 4 here. Let's subtract that 4 from both sides. And we are left with-- and this is the home stretch now-- we are left with 38x is equal to negative 16 minus 4 is negative 20. And so we can divide both sides of this equation by 38. And we are left with x is equal to negative 20 over 38, which can be simplified further. Both the numerator the denominator is divisible by 2. So let's divide the numerator and denominator by 2, and we get negative 10 over 19. x is equal to negative 10 over 19, and we are done. I encourage you to validate this for yourself. It's a little bit of a hairy number right over here," + }, + { + "Q": "At 4:30, the distance formula is mentioned. what is the distance formula? And why is it the same thing as the Pythagorean Theorem? I thought they were two completely different formulas.", + "A": "The distance formula is a formula you can use to find the shortest distance between any 2 points on the coordinate plane. You are correct that the distance formula and Pythagorean theorem are 2 different things but the distance formula is derived from the Pythagorean theorem. The distance formula is: d = \u00e2\u0088\u009a[(({x_1} - {x_2})^(2)) + (({y_1} - {y_2})^(2))]", + "video_name": "iATjsfAX8yc", + "timestamps": [ + 270 + ], + "3min_transcript": "" + }, + { + "Q": "7:10-end you ended up calculating the area of the circle at the start and not the rate of change (I think). r was a function of t, and if you think about the problem just generally thinking about it, the rate of change should not be linear b/c the formula for area is quadratic. In fact, now that I am looking back, I think that this is because you solved for circumference rather than area.", + "A": "A = pi\u00e2\u0080\u00a2r^2 d/dt(A) = d/dt(pi\u00e2\u0080\u00a2r^2) d/dt(A) = pi\u00e2\u0080\u00a2d/dt(r^2) d/dt(A) = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2r\u00e2\u0080\u00a2d/dt(r) dA/dt = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2r\u00e2\u0080\u00a2dr/dt dr/dt = 1 r(0) = 3 dA/dt = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a23\u00e2\u0080\u00a21 dA/dt = 6\u00e2\u0080\u00a2pi C = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2r d/dt(C) = d/dt(2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2r) d/dt(C) = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2d/dt(r) dC/dt = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2dr/dt dr/dt = 1 r(0) = 3 dC/dt = 2\u00e2\u0080\u00a23\u00e2\u0080\u00a21 dC/dt = 6", + "video_name": "kQF9pOqmS0U", + "timestamps": [ + 430 + ], + "3min_transcript": "So the rate at which r of t changes with respect to time? Well, we could just write that as dr dt. These are equivalent expressions. And of course, we have our pi out front. And I just want to emphasize this is just the chain rule right over here. The derivative of something squared with respect to time is going to be the derivative of the something squared with respect to the something. So that's 2 times the something, times the derivative of that something with respect to time. I can't emphasize enough. What we did right over here, this is the chain rule. That is the chain rule. So we're left with pi times this is equal to the derivative of our area with respect to time. Now let me rewrite all this again just so it cleans up a little bit. So we have the derivative of our area with respect to time is equal to pi times-- actually let me put that 2 out front. Is equal to 2 times pi times-- I can now We know that r is a function of t. So I'll just write 2 pi times r times dr dt. Actually, let me make the r in blue. 2 pi r dr dt. Now what do we know? We know what r is. We know that r, at this moment right in time, is 3 centimeters. Right now r is 3 centimeters. We know dr dt right now is 1 centimeter per second. We know this is 1 centimeter per second. So what's da dt going to be equal to? Well, it's going to be equal to-- do that same green-- 2 pi times 3 times 3 times 1 times-- that's purple-- times 1 centimeter per second. And let's make sure we get the units right. So it's going to be centimeters. That's too dark of a color. It's going to be square centimeters, centimeters times centimeters, square centimeter per second, which is the exact units we need for a change in area. So we have da dt is equal to this. da, the rate at which area is changing with respect to time, is equal to 6 pi. So it's going to be a little bit over 18 centimeters squared per second. Right at that moment. Yep, 3 times 2 pi. So 6 pi centimeters squared per second is how fast the area is changing. And we are done." + }, + { + "Q": "At 4:04: Why can you rewrite d/Dt [pi*r^2] as pi*d/dt [r(t)?", + "A": "Because pi is a constant, and you can do that with constants when you are taking derivatives.", + "video_name": "kQF9pOqmS0U", + "timestamps": [ + 244 + ], + "3min_transcript": "Well, they say at what rate is the area of the circle growing? So we need to figure out at what rate is the area of the circle-- where a is the area of the circle-- at what rate is this growing? This is what we need to figure out. So what might be useful here is if we can come up with a relationship between the area of the circle and the radius of the circle and maybe take the derivative with respect to time. And we'll have to use a little bit of the chain rule So what is the relationship at any given point in time between the area of the circle and the radius of the circle? Well, this is elementary geometry. The area of a circle is going to be equal to pi times the radius of the circle squared. Now what we want to do is figure out the rate at which the area is changing with respect to time. So why don't we take the derivative of both sides of this with respect to time? And let me give myself a little more real estate. Actually, let me just rewrite what I just had. So pi r squared. I'm going to take the derivative of both sides of this with respect to time. So the derivative with respect to time. I'm not taking the derivative with respect to r, I'm taking the derivative with respect to time. So on the left-hand side right over here, I'm going to have the derivative of our area. Actually, let me just write it in that green color. I'm going to have the derivative of our area with respect to time on the left-hand side. And on the right-hand side, what do I have? Well, if I'm taking the derivative of a constant times something, I can take the constant out. So let me just do that. Pi times the derivative with respect to time of r squared. And to make it a little bit clearer what I'm about to do, why I'm using the chain rule, we're assuming that r is a function of time. If r wasn't a function of time then area wouldn't be a function of time. So instead of just writing r, let me make it explicit I'll write r of t. So it's r of t, which we're squaring. And we want to find the derivative of this with respect to time. And here we just have to apply the chain rule. We're taking the derivative of something squared with respect to that something. So the derivative of that something squared with respect to the something is going to be 2 times that something to the first power. Let me make it clear. This is the derivative of r of t squared with respect to r of t. The derivative of something squared with respect to that something. If it was a derivative of x squared with respect to x, we'd have 2x. If it was the derivative of r of t squared with respect to r of t, it's 2r of t. But this doesn't get us just the derivative with respect to time. This is just the derivative with respect to r of t. The derivative at which this changes with respect to time, we have to multiply this times the rate at which r of t" + }, + { + "Q": "at 1:36 why does sal multiply a/b with n/n and m/n with b/b", + "A": "I think he did it so he would have the same denominators for both fractions.", + "video_name": "HKUJkMQsGkM", + "timestamps": [ + 96 + ], + "3min_transcript": "What I want to do in this video is think about whether the product or sums of rational numbers are definitely going to be rational. So let's just first think about the product of rational numbers. So if I have one rational number and-- actually, let me instead of writing out the word rational, let me just represent it as a ratio of two integers. So I have one rational number right over there. I can represent it as a/b. And I'm going to multiply it times another rational number, and I can represent that as a ratio of two integers, m and n. And so what is this product going to be? Well, the numerator, I'm going to have am. I'm going to have a times m. And in the denominator, I'm going to have b times n. Well a is an integer, m is an integer. So you have an integer in the numerator. And b is an integer and n is an integer. So you have an integer in the denominator. So now the product is a ratio of two integers right over here, so the product is also rational. So this thing is also rational. you're going to end up with a rational number. Let's see if the same thing is true for the sum of two rational numbers. So let's say my first rational number is a/b, or can be represented as a/b, and my second rational number can be represented as m/n. Well, how would I add these two? Well, I can find a common denominator, and the easiest one is b times n. So let me multiply this fraction. We multiply this one times n in the numerator and n in the denominator. And let me multiply this one times b in the numerator and b in the denominator. Now we've written them so they have a common denominator of bn. And so this is going to be equal to an plus bm, So b times n, we've just talked about. This is definitely going to be an integer right over here. And then what do we have up here? Well, we have a times n, which is an integer. b times m is another integer. The sum of two integers is going to be an integer. So you have an integer over in an integer. You have the ratio of two integers. So the sum of two rational numbers is going to give you another. So this one right over here was rational, and this one is right over here is rational. So you take the product of two rational numbers, You take the sum of two rational numbers, you get a rational number." + }, + { + "Q": "At 0:55 could b^1 also be b^0?", + "A": "Only if b = 1.", + "video_name": "X6zD3SoN3iY", + "timestamps": [ + 55 + ], + "3min_transcript": "Simplify 25 a to the third and a to the third is being raised to the third power, times b squared and all of that over 5 a squared, b times b squared So we can do this in multiple ways, simplify different parts. What I want to do is simplify this part right over here. a to the third power, and we're raising that to the third power. So this is going to be from the power property of exponents, or the power rule this is going to be the same thing as a to the 3 times 3 power So this over here (let me scroll up a little bit) is going to be equal to a the 3 times 3 power, or a to the ninth power. We could also simplify this b times b squared over here. This b times b squared, that is the same thing as b to the first power remember, b is just b to the first power. So it's b to the first power times b to the second power. So b to the first times b to the second power is just equal to b to the one plus two power, which is equal to b to the third power and then last thing we could simplify, just right off the bat just looking at this: or we could say it's going to give us 5 over 1 if you view it as dividing the numerator and the denominator both by 5. So what does our expression simiply to? We have 5a to the ninth, and then we still have this b squared here, b squared. All of that over a squared times b to the third power... times b to the third power. Now, we can use the quotient property of exponents. You have an a to the ninth Let me use a slightly different color. We have an a to the ninth. over a squared. What's that going to simplify to? Well, that's going to simplify to be the same thing, a to the ninth over a squared, the same thing as a to the nine minus two, which is equal to a to the seventh power. and this will get a little bit interesting here So that simplifies too. So b squared over b to the third is equal to b to the two minus three power, which is equal to b to the negative one power. And we'll leave it alone like that right now. So this whole expression simplifies to It simplifies to: 5 times a to the seventh power (because this simplifies to a to the seventh) a to the seventh, times (the bs right here simplify too) b to the negative one. We could leave it like that, you know, that's pretty simple but we may not want a negative exponent there we just have to remeber that b to the negative one power is the same thing as one over b. Now if we remember that, then we can rewrite this entire expression as, the numerator will have a five and will have a to the seventh 5 a to the seventh. And then the denominator will have the b. So we're multiplying this times one over b." + }, + { + "Q": "At 5:34, Sal says that the notation for the length/magnitude of a vector is notated by using double lines around the vector, like this.\n||a||\nCan somebody explain why we use this \"double absolute value\" notation to signify the magnitude of a vector?", + "A": "Well, maybe just as a convention sort of .Someone kept that symbol and maybe we are using it.", + "video_name": "WNuIhXo39_k", + "timestamps": [ + 334 + ], + "3min_transcript": "" + }, + { + "Q": "when he means the length of vector.. does he mean the magnitude of the vector. is it another word for magnitude or is it completely different things. please explain... thanks (4:44 min)", + "A": "i think it s exactly the same thing", + "video_name": "WNuIhXo39_k", + "timestamps": [ + 284 + ], + "3min_transcript": "" + }, + { + "Q": "1:58 Why is 3+5i a complex \"number\"?\nIt consists of 2 numbers... Real and Imaginary.\nCouldn't it be called \"complex numbers\"?", + "A": "Z is the complex number, comprised of a Real Part (5) and an Imaginary Part (3i).", + "video_name": "SP-YJe7Vldo", + "timestamps": [ + 118 + ], + "3min_transcript": "Voiceover:Most of your mathematical lives you've been studying real numbers. Real numbers include things like zero, and one, and zero point three repeating, and pi, and e, and I could keep listing real numbers. These are the numbers that you're kind of familiar with. Then we explored something interesting. We explored the notion of what if there was a number that if I squared it I would get negative one. We defined that thing that if we squared it we got negative one, we defined that thing as i. So we defined a whole new class of numbers which you could really view as multiples of the imaginary unit. So imaginary numbers would be i and negative i, and pi times i, and e times i. This might raise another interesting question. What if I combined imaginary and real numbers? What if I had numbers that were essentially sums or differences of real or imaginary numbers? Let's say I call it z, and z tends to be the most used variable when we're talking about what I'm about to talk about, complex numbers. Let's say that z is equal to, is equal to the real number five plus the imaginary number three times i. So this thing right over here we have a real number plus an imaginary number. You might be tempted to add these two things, but you can't. They won't make any sense. These are kind of going in different, we'll think about it visually in a second, but you can't simplify this anymore. You can't add this real number to this imaginary number. A number like this, let me make it clear, that's real and this is imaginary, imaginary. A number like this we call a complex number, a complex number. It has a real part and an imaginary part. or someone will say what's the real part? What's the real part of our complex number, z? Well, that would be the five right over there. Then they might say, \"Well, what's the imaginary part? \"What's the imaginary part of our complex number, z? And then typically the way that this function is defined they really want to know what multiple of i is this imaginary part right over here. In this case it is going to be, it is going to be three. We can visualize this. We can visualize this in two dimensions. Instead of having the traditional two-dimensional Cartesian plane with real numbers on the horizontal and the vertical axis, what we do to plot complex numbers is we on the vertical axis we plot the imaginary part, so that's the imaginary part. On the horizontal axis we plot the real part. We plot the real part just like that." + }, + { + "Q": "at 3:18 why do we choose y-axis as imaginary and x-axis as real part ,is there any proof for it??", + "A": "It s just a natural representation. The key point here is that the imaginary axis and the real axis must be perpendicular to each other. Imagine rotating your graph by 90 degrees, now the imaginary axis is the horizontal one.", + "video_name": "SP-YJe7Vldo", + "timestamps": [ + 198 + ], + "3min_transcript": "Let's say I call it z, and z tends to be the most used variable when we're talking about what I'm about to talk about, complex numbers. Let's say that z is equal to, is equal to the real number five plus the imaginary number three times i. So this thing right over here we have a real number plus an imaginary number. You might be tempted to add these two things, but you can't. They won't make any sense. These are kind of going in different, we'll think about it visually in a second, but you can't simplify this anymore. You can't add this real number to this imaginary number. A number like this, let me make it clear, that's real and this is imaginary, imaginary. A number like this we call a complex number, a complex number. It has a real part and an imaginary part. or someone will say what's the real part? What's the real part of our complex number, z? Well, that would be the five right over there. Then they might say, \"Well, what's the imaginary part? \"What's the imaginary part of our complex number, z? And then typically the way that this function is defined they really want to know what multiple of i is this imaginary part right over here. In this case it is going to be, it is going to be three. We can visualize this. We can visualize this in two dimensions. Instead of having the traditional two-dimensional Cartesian plane with real numbers on the horizontal and the vertical axis, what we do to plot complex numbers is we on the vertical axis we plot the imaginary part, so that's the imaginary part. On the horizontal axis we plot the real part. We plot the real part just like that. For example, z right over here which is five plus three i, the real part is five so we would go one, two, three, four, five. That's five right over there. The imaginary part is three. One, two, three, and so on the complex plane, on the complex plane we would visualize that number right over here. This right over here is how we would visualize z on the complex plane. It's five, positive five in the real direction, positive three in the imaginary direction. We could plot other complex numbers. Let's say we have the complex number a which is equal to let's say it's negative two plus i. Where would I plot that? Well, the real part is negative two, negative two, and the imaginary part is going to be you could imagine this as plus one i" + }, + { + "Q": "As discussed in 0:57, you divide the numerator and the denominator by the same number. However, for 30/45 and 54/81, do they both have to be divided by 5?", + "A": "No. You see, you could factor out 5 ONLY from 30/45, since both 54 and 81 are not divisible by 5( 54/5=10.8, and 81/5=16.2). You then get 6/9. 3 is divisible by both 6 and 9.You get 2/3. This is can no longer be factored. Now, for 54/81, you see automatically (if you know you re times tables) that 54 and 81 are divisible by 9. You get 6/9.Like 30/45, you see that the numerator and denominator are divisible by 3.You get 2/3. 54/81 and 30/45 are equivalent. I hope this helped.", + "video_name": "Io9i1JkKgN4", + "timestamps": [ + 57 + ], + "3min_transcript": "Determine whether 30/45 and 54/81 are equivalent fractions. Well, the easiest way I can think of doing this is to put both of these fractions into lowest possible terms, and then if they're the same fraction, then they're equivalent. So 30/45, what's the largest factor of both 30 and 45? 15 will go into 30. It'll also go into 45. So this is the same thing. 30 is 2 times 15 and 45 is 3 times 15. So we can divide both the numerator and the denominator by 15. So if we divide both the numerator and the denominator by 15, what happens? Well, this 15 divided by 15, they cancel out, this 15 divided by 15 cancel out, and we'll just be left with 2/3. So 30/45 is the same thing as 2/3. It's equivalent to 2/3. 2/3 is in lowest possible terms, or simplified form, Now, let's try to do 54/81. Now, let's see. Nothing really jumps out at me. Let's see, 9 is divisible into both of these. We could write 54 as being 6 times 9, and 81 is the same thing as 9 times 9. You can divide the numerator and the denominator by 9. So we could divide both of them by 9. 9 divided by 9 is 1, 9 divided by 9 is 1, so we get this as being equal to 6/9. Now, let's see. 6 is the same thing as 2 times 3. 9 is the same thing as 3 times 3. We could just cancel these 3's out, or you could imagine this is the same thing as dividing both the numerator and the denominator by 3, or multiplying both the numerator and the denominator by 1/3. These are all equivalent. I could write divide by 3 or multiply by 1/3. Let me write divide by 3 for now. I don't want to assume you know how to multiply fractions, because we're going to learn that in the future. So we're going to divide by 3. 3 divided by 3 is just 1. 3 divided by 3 is 1, and you're left with 2/3. So both of these fractions, when you simplify them, when you put them in simplified form, both end up being 2/3, so they are equivalent fractions." + }, + { + "Q": "which formula is used here Square root at 6:00", + "A": "the distance formula (:", + "video_name": "GiGLhXFBtRg", + "timestamps": [ + 360 + ], + "3min_transcript": "The centroid of a triangle is just going to be the average of the coordinates of the vertices. Or the coordinate of the centroid here is just going to be the average of the coordinates of the vertices. So this coordinate right over here is going to be-- so for the x-coordinate, we have 0 plus 0 plus a. So we have three coordinates. They add up to a, and we have to divide by 3. So it's a over 3. The y-coordinate is going to b plus 0 plus 0. They add up to b, but we have three of them, so the average is b over 3. And then same thing-- we do it for the z-coordinate. The average is going to be c, is c over 3. And I'm not proving it to you right here. You could verify it for yourself. But it's going to be the average, that if you were to figure out what this line is, this line is, and this line is, this centroid, or this center is just the average of these coordinates. Now, what we want to do is use this information. Let's just use this coordinate right here and then compare just using the distance formula. Let's compare this distance up here in orange to this distance down here in yellow. And remember, this point right over here-- this is the median of this bottom side right over here. It's just going to be the average of these two points. And so the x-coordinate-- 0 plus a over 2 is going to be a over 2. b plus 0 over 2 is going to be b over 2. And then it has no z-coordinates, so it's just going to be 0. 0 plus 0 over 2 is 0. So we know the coordinates for this point that point and that point. So we can calculate the yellow distance and we can calculate the orange distance. So let's calculate the orange distance. So that is going to be equal to the square root of-- of these points squared. So it's a over 3 minus 0 squared. So that's going to be a squared over 9, plus b over 3 minus 0 squared. So that's b squared over 9. Plus c over 3 minus c, which is negative 2/3. And we want to square that. So we're going to have positive 4 over 9c squared. Did I do that right? c over 3, so 1/3 minus 1 is negative 2/3. So this is negative 2/3 c. You're going to get 4/9 c squared. So that's the orange distance. Now, let's calculate-- and if we want to do it, we can express this-- let me express it a little bit simpler than this. This is the same thing as the square root of a squared plus b" + }, + { + "Q": "At 3:59, why is Sal using 3 coordinates? Its only the supposed to be x and y correct?", + "A": "its a triangle so he has to use three points", + "video_name": "GiGLhXFBtRg", + "timestamps": [ + 239 + ], + "3min_transcript": "or around the center of mass. But anyway, the point of this video is not to focus on physics and throwing iron triangles. The point here is I want to show you a neat property of medians. And the property is that if you pick any median, the distance from the centroid to the midpoint of the opposite side-- so this distance-- is going to be half of this distance. So if this distance right here is a, then this distance right here is 2a. Or another way to think about it is this distance is 2/3 of the length of the entire median, and this distance right here is 1/3 of the length of the entire median. And let's just prove it for ourselves just so you don't have to take things on faith. And to do that, I'll draw an arbitrary triangle. I'll do a two-dimensional triangle, and I'll do it in three dimensions because at least in my mind, it makes the math a little bit easier. In general, whenever you have an n-dimensional figure it makes the math a little bit easier. The actual tetrahedron problem that we did, you could actually embed it in four dimensions and it would make the math easier. It's just much harder to visualize, so I didn't do it that way. But let's just have an arbitrary triangle. And let's say it has a vertex and there, a vertex there, and a vertex there. So I'm not making any assumptions about the triangle. I'm not saying it's isosceles, or equilateral or anything. It's just an arbitrary triangle. And so let's say this coordinate right over here is-- I'll call this the x-axis. So, this is the x-axis, the y-axis, and the z-axis. I know some of y'all are used to swapping these two axes, but it doesn't make a difference. So let's call this coordinate right here a, 0, 0. So it's a along the x-axis. Let's call this coordinate 0, b, 0. And let's call this coordinate up here, 0, 0, c. And if you connect the points, you're going to have a triangle just like that. The centroid of a triangle is just going to be the average of the coordinates of the vertices. Or the coordinate of the centroid here is just going to be the average of the coordinates of the vertices. So this coordinate right over here is going to be-- so for the x-coordinate, we have 0 plus 0 plus a. So we have three coordinates. They add up to a, and we have to divide by 3. So it's a over 3. The y-coordinate is going to b plus 0 plus 0. They add up to b, but we have three of them, so the average is b over 3. And then same thing-- we do it for the z-coordinate. The average is going to be c, is c over 3. And I'm not proving it to you right here. You could verify it for yourself. But it's going to be the average, that if you were to figure out what this line is, this line is, and this line is, this centroid, or this center" + }, + { + "Q": "What is tangent? At 0:15.", + "A": "A tangent is a line on the outside of a circle or curve that only touches at one point.", + "video_name": "ZiqHJwzv_HI", + "timestamps": [ + 15 + ], + "3min_transcript": "Angle A is a circumscribed angle on circle O. So this is angle A right over here. Then when they say it's a circumscribed angle, that means that the two sides of the angle are tangent to the circle. So AC is tangent to the circle at point C. AB is tangent to the circle at point B. What is the measure of angle A? Now, I encourage you to pause the video now and to try this out on your own. And I'll give you a hint. It will leverage the fact that this is a circumscribed angle as you could imagine. So I'm assuming you've given a go at it. So the other piece of information they give us is that angle D, which is an inscribed angle, is 48 degrees and it intercepts the same arc-- so this is the arc that it intercepts, arc CB I guess you could call it-- it intercepts this arc right over here. It's the inscribed angle. The central angle that intersects that same arc So this is going to be 96 degrees. I could put three markers here just because we've already used the double marker. Notice, they both intercept arc CB so some people would say the measure of arc CB is 96 degrees, the central angle is 96 degrees, the inscribed angle is going to be half of that, 48 degrees. So how does this help us? Well, a key clue is that angle is a circumscribed angle. So that means AC and AB are each tangent to the circle. Well, a line that is tangent to the circle is going to be perpendicular to the radius of the circle that intersects the circle at the same point. So this right over here is going to be a 90-degree angle, and this right over here is going to be a 90-degree angle. OC is perpendicular to CA. OB, which is a radius, is perpendicular to BA, which is a tangent line, and they both intersect right We have a quadrilateral going on here. ABOC is a quadrilateral, so its sides are going to add up to 360 degrees. So we could know, we could write it this way. We could write the measure of angle A plus 90 degrees plus another 90 degrees plus 96 degrees is going to be equal to 360 degrees. Or another way of thinking about it, if we subtract 180 from both sides, if we subtract that from both sides, we get the measure of angle A plus 96 degrees is going to be equal to 180 degrees." + }, + { + "Q": "I don't understand the end behaviors @4:25", + "A": "End behavior is what the function looks like when it reaches really high or really low values of x.", + "video_name": "tZKzaF28sOk", + "timestamps": [ + 265 + ], + "3min_transcript": "So when we talk about end behavior, we're talking about the idea of what is this function? What does this polynomial do as x becomes really, really, really, really positive and as x becomes really, really, really, really negative? And kind of fully recognizing that some weird things might be happening in the middle. But we just want to think about what happens at extreme values of x. Now obviously for the second degree polynomial nothing really weird happens in the middle. But for a third degree polynomial, we sort of see that some interesting things can start happening in the middle. But the end behavior for third degree polynomial is that if a is greater than 0-- we're starting really small, really low values-- and as a becomes positive, we get to really high values. If a is less than 0 we have the opposite. And these are kind of the two prototypes for polynomials. Because from there we can start thinking about any degree polynomial. So let's just think about the situation of a fourth degree polynomial. plus bx to the third plus cx squared plus dx plus-- I don't want to write e because e has other meanings in mathematics. I'll say plus-- I'm really running out of letters here. I'll just use f, although this isn't the function f. This is just a constant f right over here. So let's just think about what this might look like. Let's think about its end behavior, and we could think about it relative to a second degree polynomial. So its end behavior, if x is really, really, really, really negative, x to the fourth is still going to be positive. And if a is greater than 0 when x is really, really, really negative, we're going to have really, really positive values, just like a second degree. And when x is really positive, same thing. x to the fourth is going to be positive, times a is still going to be positive. So its end behavior is going to look Now, it might do-- in fact it probably will do some funky stuff in between. It might do something that looks kind of like that in between. But we care about the end behavior. I guess you could call the stuff that I've dotted lined in the middle, this is called the non-end behavior, the middle behavior. This will obviously be different than a second degree polynomial. But what happens at the ends will be the same. And the reason why, when you square something, or you raise something to the fourth power, you raise anything to any even power for a very large-- as long as a is greater than 0, for very large positive values, you're going to get positive values. And for very large negative values, you're going to get very large positive values. You take a negative number, raise it to the fourth power, or the second power, you're going to get a positive value. Likewise, if a is less than 0, you're going to have very similar end behavior to this case. For a polynomial where the highest degree" + }, + { + "Q": "At 0:33 what does \"zen\" mean", + "A": "The voice recognition software got a bit confused. I m pretty sure Sal says and we essentially picked no y s then .", + "video_name": "_hrN4rVCOfI", + "timestamps": [ + 33 + ], + "3min_transcript": "Voiceover:What I want to do in this video is hopefully give more intuition as to why the binomial theorem or the binomial formula involves combinatorics. Let's just think about what this expansion would be. I'm just using a particular example that's pretty simple, x plus y to the third power which is x plus y, times x plus y, times x plus y. We already know what the terms look like, we could pick an x from each of these. If we pick exactly an x from each of these that's going to be our x to the third term, and we essentially picked no y [zen] because we picked an x from each of this when we multiplied out to get this term. How many ways are there to construct that? How many ways are there, if we're picking from three things. We're picking from three things, how many ways can we choose exactly zero y's of not picking a y? There's only one way to do that, there's only one possible way of doing that. By not picking a y from each of them The coefficient right over here is one x to the third. There's only one way, when you take this product out, only one of the terms initially is going to have an x to the third in it. Now, what about the next term? So plus. Now we're going to pick from three buckets but we want to pick one y. Another way of thinking about it you have three people or we have three people, this person, this person, and this person. One of them you're going to choose to be I guess be your friend that's another way of saying which of these are you going to pick the y from? This is the exact same thing as a combinatorics problem, from three things you are choosing one of them to be y. Of course three choose one, this is equal to three. If you're picking one y that means you're picking two x's, so it's x squared times y to the one and we could keep doing that logic, so plus. Now we're going to think about the situation where we are picking two y's out of three things. It's x to the first, y squared. We're starting with three things and we're going to pick two to two of this y's so it could be this y and this y to construct the product y. It could be y times y times that x or it could be y times y, times that x or it could be y times x, times y. This is three choose two, out of three things what are the different ways, what are the combinations of picking two things. If you have three friends how many combinations can you put two of them in your two seater car where you don't care about who sits in which seat. You're just thinking about how many different combinations can you pick from that and once again this evaluates to three" + }, + { + "Q": "at 3:04 couldnt he just added 5 to 35 and add 20?", + "A": "Well, he could have, but doing that is basically doing exactly what he did do. He probably chose to do it how he did because if you aren t working with time then you would have to subtract, and he want s us to build good habits in problem solving.", + "video_name": "UhMM68fq9FA", + "timestamps": [ + 184 + ], + "3min_transcript": "We know that there are 60 minutes in an hour. So 1:59 is 1 minute before 2:00 PM. So I'm not drawing it completely it scale, but this right over here is 1:59 PM. So we use 1 minute to get to 2:00 PM. And then, let's see, we have to get up to 96 minutes. So then you get another 60 minutes to go to 3:00 PM. So, so far, we've used a cumulative 61. So this is 1 minute. So this is how long the test has been going on since the beginning. So 1 minute by 2:00 PM, 61 minutes by 3:00 PM. If we go all the way to 4:00 PM, that's going to be another 60. That'd be 121 minutes, which is longer than the actual test. So we know that the entire test is 96 minutes. So we're going to get to some point right over here, some point like this. And what we need to figure out is, is what is left over? What is this distance right over here? Or what is this time, this difference in time right over here? And so to figure that out, we had to figure out the time at which 96 minutes have passed since the beginning of the exam. So at 3 o'clock, 61 minutes have passed. And so if we, if we say well how many more minutes have to go on for the end of the test? It's going to be 96 minus 61. So the amount of time that elapses past 3 minutes is 96 minus 61, which is 35 minutes. So this right over here is going to be 3:35. And so now our question is, how much time is there between 3:35 and 4:00 PM? Well, once again, 60 minutes in an hour. 4:00 PM is essentially the 60th minute. So she has a total of 25 minutes remaining after the test before she has to get to volleyball practice. Just as a review. One minute between 1:59 and 2:00, then another 60. That gets us to 61 total minutes. Then another 35 minutes gets us to 96 minutes have passed. And that also gets us to 3:35, which means we have 25 minutes until 4:00." + }, + { + "Q": "so at 2:32 it says that all four base angles are congruent,but is the angle at the top congruent to them also?", + "A": "Not necessarily, Because the sides of the rhombus do not necessarily have to be equal to one of the diagonals. However, if you have a rhombus with angles 60 and 120 degrees, then the shorter diagonal will split the rhombus into two equilateral triangles and so the vertical angle of the triangle will be also equal to the other two base angles (ie, all the angles will be 60 degrees).", + "video_name": "_QTFeOvPcbY", + "timestamps": [ + 152 + ], + "3min_transcript": "Now, let me draw one of its diagonals. And the way I drew it right here is kind of a diamond. One of its diagonals will be right along the horizontal, right like that. Now, this triangle on the top and the triangle on the bottom both share this side, so that side is obviously going to be the same length for both of these triangles. And then the other two sides of the triangles are also the same thing. They're sides of the actual rhombus. So all three sides of this top triangle and this bottom triangle are the same. So this top triangle and this bottom triangle are congruent. They are congruent triangles. If you go back to your ninth grade geometry, you'd use the side-side-side theorem to prove that. Three sides are congruent, then the triangles themselves are congruent. But that also means that all the angles in the triangle are congruent. So the angle that is opposite this side, this shared side right over here will be congruent to the corresponding angle in the other triangle, the angle So it would be the same thing as that. Now, both of these triangles are also isosceles triangles, so their base angles are going to be the same. So that's one base angle, that's the other base angle, right? This is an upside down isosceles triangle, this is a right side up one. And so if these two are the same, then these are also going to be the same. They're going to be the same to each other, because this is an isosceles triangle. And they're also going to be the same as these other two characters down here, because these are congruent triangles. Now, if we take an altitude, and actually, I didn't even have to talk about that, since actually, I don't think that'll be relevant when we actually want to prove what we want to prove. If we take an altitude from each of these vertices down to this side right over here. So an altitude by definition is going to be perpendicular down here. Now, an isosceles triangle is perfectly symmetrical. If you drop an altitude from the-- I or the unique vertex in an isosceles triangle-- you will split it into two symmetric right triangles. Two right triangles that are essentially the mirror images of each other. You will also bisect the opposite side. This altitude is, in fact, a median of the triangle. Now we could do it on the other side. The same exact thing is going to happen. We are bisecting this side over here. This is a right angle. And so essentially the combination of these two altitudes is really just a diagonal of this rhombus. And it's at a right angle to the other diagonal of the rhombus. And it bisects that other diagonal of the rhombus. And we could make the exact same argument over here. You could think of an isosceles triangle over here. This is an altitude of it." + }, + { + "Q": "At 3:05, why is it just answer 17 but at 3:31, 5 is 25?", + "A": "He is squaring each number. So at 3:05 he squares the squared root of 17, the square root of 17x the square root of 17 equals 17. The square root of 17 is a number slightly bigger than 4, because 4x4 equals 16, so this is just a little bit more than that. At 3:31 he square 5. 5x5=25 The concept is that if you square each number you can compare the numbers without the radical signs........", + "video_name": "KibTbfkoPTs", + "timestamps": [ + 185, + 211 + ], + "3min_transcript": "Because their squares are not going to be irrational numbers. It's going to be much easier to compare, and then we can order them. Because if we order the squares, then they'll tell us what happens if we order their square roots. What am I talking about? Well, I'm just gonna square each of these. So if I take this to the second power, this is going to be four square roots of two, times four square roots of two. You can change the order of multiplication. That's four times four times the square root of two times the square root of two. Now, four times four is 16. Square root of two times square root of two, well, that's just going to be two. So it's gonna be 16 times two which is equal to 32. Now what about two square roots of three? Well, same idea. Let's square it, let's square it. And i'll do this one a little bit faster. So if we square two square roots of three, times square root of three squared. So it's going to be two squared times the square root of three squared. Well, two squared is going to be four. Square root of three squared is going to be three. So this is going to be equal to 12. That's this thing squared. If this step seems a little bit confusing, if you have the product of two things raised to a power, that's the same thing as raising each of them to that power, and then taking the product. And you can actually see, I worked it out here, why that actually makes sense. Notice when I just changed the order of multiplication you had four times four, or four squared, times square root of two squared, which is going to be two. So let's keep doing that. So what is this value squared? It's gonna be three squared, which is nine, times square root of two squared, which is two. What's the square root of 17 squared? That's just going to be seventeen. Do that in blue. This is just going to be 17. What is three square roots of three squared? It's gonna be three squared, which is nine, times square root of three squared. The square root of three times the square root of three is three. So it's gonna be nine times three, or 27. And what is five squared? This is pretty straightforward. That's going to be 25. So let's order them from least to greatest. Which of them, when I square it, gives me the smallest value? Compare 32 to 12 to 18 to 17 to 27 to 25. 12 is the smallest value. So if their square is the smallest, and these are all positive numbers, then this is going to be the smallest value out of all of them. Let me write that first. Two square roots of three. So I've covered that one." + }, + { + "Q": "It states at 4:56 that zero is a positive number. I thought all positive numbers have an opposite. If they do then what is zero's opposite?", + "A": "Yes, all numbers, either positive or negative, should have the opposite, except 0.", + "video_name": "XHHYA2Ug9lk", + "timestamps": [ + 296 + ], + "3min_transcript": "this whole thing is going to be a negative. But let's keep on doing, we said, look if all of them are negative, then this thing would be negative, but that's because I had three numbers here. What if I had four numbers here? What if I had times, What if I had times d here? And if I told you all of these numbers were negative? Let's think about it, if negative, negative, negative, negative. And I can do the multiplication in any order, but I'll just go left to right. A times b, negative times a negative. That would yield a positive. Now if you multiply that product times c, positive times a negative, positive times a negative, positive times a negative that would give you a negative. And then you multiply this negative times this negative, so this whole product, a, b, c, is going to be negative. But then we multiply it times a negative. Well a negative times a negative is going to be a positive. So this whole thing is going to be a positive. And so you're probably seeing a pattern here. if you're multiplying a bunch of numbers, and if you have an odd number of negatives, odd number of negatives being multiplied, and or divided. I just did multiplication here, but this would have also been true if these were all division symbols. If these were all division symbols, we would've been able to say the exact same thing. If you have an odd number of negative numbers in your product or in your quotient, well then you're going to have a negative, you're going to have a negative value for the entire expression. If you have an even, if you have an even number of negative, well then the whole thing is going to be, the whole thing is going to be positive. And so you can view these as generalizations of what we just saw here. Positive times a positive. Zero is actually an even number, so this would be positive. Negative times a negative. Well that's an even number of negatives. That's this case again, so you're going to be positive. Either of these cases, you have one negative. You have one negative in either of these cases, so it's going to be this case, odd number of negatives. So that's going to be a negative. And so, we can use this knowledge to start dealing with negative numbers and exponents. So if I were to say, I would have a to the, let me throw out a wild number here, A to the 101st power, And we know that a is less than zero. What is this going to be? Well this is taking 101 A's and multiplying them together. You have an odd number, an odd number of negatives being multiplied together? Well this whole then is going, this whole thing is going to be less than zero. If we knew what a was, we could calculate it somehow, but we know that this thing is going to be negative. Let's do something, so, and we could do it other ways. We could say something like this." + }, + { + "Q": "At 3:28, Sal says that he can do multiplication in any order, but thought you had to go from left to right.", + "A": "You can do multiplication in any order because of the commutative property of multiplication, which means that you ll get the same answer no matter what order you put it in.", + "video_name": "XHHYA2Ug9lk", + "timestamps": [ + 208 + ], + "3min_transcript": "and now I'm going to write the dot for times. A times b times c, a times b times c. Now if I told you that these were all positive numbers, then you say, \"OK a times b times c is going \"to be positive.\" \"A times b would be positive, and that times c is positive.\" Now what would happen if I were to tell you that they were all negative numbers. What if a, b, and c were all negative? Well if there were all negative, let me write it that way. Let me actually write, let me write, a, b, and c, a, b, and c, they're all going to be negative. So if that's the case, what is this product going to be equal to? Well you're going to have a negative here, times a negative. So a negative times a negative, a times b, if you do that first, and we can when we multiply these numbers. That's going to give you a positive. But then you're going to multiply that times c. You're going to multiply that times c, which is a negative. So you're going to have a positive times a negative, which is going to be a negative. So this one, if a, b, and c are all less than zero, then the product, a, b, c is going to be less than zero as well. This whole thing is going to be negative. Now, if I did something else. If I said, \"There's other ways \"that I can make the product negative.\" If a is, let's say that a, actually let me just write it this way. Let's say that a is positive, b is negative. B is negative, and c is positive. And c is positive. Well here, positive times a negative, if you do this first. Positive times a negative is going to give you a negative. And then a negative times a positive, different signs, is going to give you a negative. this whole thing is going to be a negative. But let's keep on doing, we said, look if all of them are negative, then this thing would be negative, but that's because I had three numbers here. What if I had four numbers here? What if I had times, What if I had times d here? And if I told you all of these numbers were negative? Let's think about it, if negative, negative, negative, negative. And I can do the multiplication in any order, but I'll just go left to right. A times b, negative times a negative. That would yield a positive. Now if you multiply that product times c, positive times a negative, positive times a negative, positive times a negative that would give you a negative. And then you multiply this negative times this negative, so this whole product, a, b, c, is going to be negative. But then we multiply it times a negative. Well a negative times a negative is going to be a positive. So this whole thing is going to be a positive. And so you're probably seeing a pattern here." + }, + { + "Q": "@0:28 Why is it 2cm?? Really confused", + "A": "That is because 2 is the height. When you look at it, two cm is the line coming up that makes the figure 3D, so that is why it is the height. Also, the length and the height make the base and they intersect before the height comes up at the edge.", + "video_name": "feNWZEln6Nc", + "timestamps": [ + 28 + ], + "3min_transcript": "- [Voiceover] Let's see if we can figure out the volume of this figure over here. They've given us some of the dimensions. We see this side over here is two centimeters, this is seven centimeters, this is 12 centimeters, this is five centimeters, this is three centimeters. And so like always, pause this video and see if you can figure it out. Well there's a bunch of ways to do this, but the way I'd like to do it is just to break it up into two rectangular prisms. So what I'm gonna do is, in fact most of the reasonable ways to do this would be to break it up into two rectangular prisms, and the ones that jump out at me is one prism like this that is three centimeters wide, five centimeters high, and then it is seven centimeters long, or seven centimeters deep. So this one right over here. And if this part right over here was transparent you would see it look just like this. You would see it look just like this. And so this one once again, it is three centimeters wide, So this distance right over here is going to be the same as this distance right over here. So seven centimeters long. So the width times the length times the height is five centimeters. Gets us to, let's see. Three times seven is 21, times five is equal to, 20 times five is 100, one times five is five. So it's going to be 105. We can say 105 cubic centimeters, cause you have centimeters times centimeters times centimeters. So this blue part right over here, this blue rectangular prism, has a volume of 105 cubic centimeters. So now we can separately figure out the volume of what I'm now highlighting in this magenta color. What I'm highlighting in this magenta color. If this was transparent, you would see this part back over here and right over here. Well, we know its height is two centimeters, we know that this dimension right over here, I guess you could say its depth, we could call it that, is seven centimeters. But what is this right over here? If we want to consider this, maybe it's length, or maybe it's width, depending on what we want to call it. Well, let's see, this whole thing is 12 centimeters, from here to here is 12 centimeters, and we know that from here to here is three centimeters, so this piece right over here must be nine centimeters. So that must be nine centimeters, is this distance right over here. So the volume of this magenta part is going to be nine centimeters times seven centimers times the height, times two centimeters. Which is going to get us, let's see, nine times seven is 63, 63 times two is equal to," + }, + { + "Q": "At about 1:35, you mention that 5 and 12 are relatively prime. What exactly does that mean?\n\nThe more I think about this, the more puzzled I become. I guess I'm having trouble with how prime-ness can have different degrees. The one solution that I've come up with isn't exactly satisfying: that the two numbers together 'may as well be'/ are 'as good as' prime, since they only have one as a common factor and all primes have 1 as a common factor.", + "A": "12 is not a prime, but if you consider that you are only allowed to divide by 1 and 5 (the factors of 5), then 12 would be a prime. 12 can be divided by 2, 3, 4, 6, but not by any factor of 5. So 15 and 8 are also relatively prime. You could also say: two numbers are relatively prime if they don t share any factors. For example 15=3*5 and 8=2*2*2", + "video_name": "jFd-6EPfnec", + "timestamps": [ + 95 + ], + "3min_transcript": "" + }, + { + "Q": "at 2:1 l don't understand", + "A": "What Sal means is that you need a certain amount of cleaning product (bottles in the numerator) to clean a certain fraction of the bathroom. As in the video, you need 1/3 of a bottle of cleaning product to clean 3/5 of the area of a bathroom.", + "video_name": "2DBBKArGfus", + "timestamps": [ + 121 + ], + "3min_transcript": "It's a little daunting because it has fractions, but then when we work through it step-by-step Hopefully, it'll feel a little bit more intuitive so it says Calvin cleans three-fifths of his bathroom with one third of a bottle of cleaning solution at this rate What fraction of the bottle of cleaning Solution will Calvin used to clean his entire bathroom? And like always encourage you to pause the video and try to take a stab at this yourself So as [I] mentioned it's a little bit You know it's a three-fifths of his bathroom with one third of a bottle How do we [think] about this and what my brain does is I'd like to say well How would we like to answer the question what fraction of the bottle of cleaning solution will Calvin used to clean his entire bathroom? So we want to figure out is we want to figure out how many bottles so?? bottles bottles Per let me write it this way So we want to figure out he uses a certain number up with that? bottles per Bathroom if we knew this then we have the answer to the question if this would this might be I don't know [two-fifths] Bottles two-fifths of a bottle bathroom it might be two Bottles per bathroom, but if we know whatever this? Is and we know the answer How many bottles doesn't need to take to clean a bathroom or what fraction of a bottle? We don't know that they're hinting that it's a fraction of a bottle but how much of a bottle or how many bottles per bathroom or another way to think about this if we want to know if We wanted to express it as a as a rate more mathematically. We could say? bottles bottles per bathroom Bathroom and the Reason why this is helpful It makes it clear that look we want to figure out we want to take our Units tell us that we want to divide the number of bottles or the fraction of a bottle it takes? To clean the number of bathrooms or certain fraction of the bathrooms and they tell us over here they tell us that He's able to take one third of a bottle so I could write it here 13 of a bottle to clean to clean three-fifths of a bathroom three-fifths of a bathroom three-fifths of a bathroom So hopefully it's clear now. Why this was helpful? We say okay? We want to take how many bottles it takes to clean a certain number of bathrooms one third of a bottle take can clean? three-fifths of a bathroom and makes it clear that we need to take the one-third and Divide it by three-fifths because then we're going to get bottles per bathroom. We're going to get the rate We're not going to get bathrooms per bottle we're going to get bottles per bathroom Which is what we care about what fraction of the bottle well? It will it take to clean entire bathroom to clean one entire bathroom? So now we just have to take one-third and divide it by three-fifths" + }, + { + "Q": "At 1:48, isn't the vector drawn vector (a+b), not vector (a-b)?", + "A": "No, it is a - b. To draw a + b, you d put vector b s tail at the head of vector a (i.e. chain them together). Draw some vectors on paper with roughly correct coordinates and give it a go.", + "video_name": "5AWob_z74Ks", + "timestamps": [ + 108 + ], + "3min_transcript": "A couple of videos ago we introduced the idea of the length of a vector. That equals the length. And this was a neat idea because we're used to the length of things in two- or three-dimensional space, but it becomes very abstract when we get to n dimensions. If this has a hundred components, at least for me, it's hard to visualize a hundred dimension vector. But we've actually defined it's notion of length. And we saw that this is actually a scalar value. It's just a number. In this video, I want to attempt to define the notion of an angle between vectors. As you can see, we're building up this mathematics of vectors from the ground up, and we can't just say, oh, I know what an angle is because everything we know about angles and even lengths, it just applies to what we associate with two- or three-dimensional space. But the whole study of linear algebra is abstracting these And I haven't even defined what dimension is yet, but I think you understand that idea to some degree already. When people talk about one or two or three dimensions. So let's say that I have some vector-- let's say I have two vectors, vectors a and b. They're nonzero and they're members of Rn. And I don't have a notion of the angle between them yet, but let me just draw them out. Let me just draw them as if I could draw them in two So that would be vector a right there. Maybe that's vector b right there. And then this vector right there would be the vector a minus b. And you can verify that just the way we've learned to add and subtract vectors. Or you know, this is heads to tails. So b plus a minus b is of course, going to be vector a. To help us define this notion of angle, let me construct another triangle that's going to look a lot like this one. But remember, I'm just doing this for our simple minds to imagine it in two dimensions. But these aren't necessarily two-dimensional beasts. These each could have a hundred components. But let me make another triangle. Well, it should look similar. Say it looks like that. And I'm going to define the sides of the triangles to be the lengths of each of these vectors. Remember, the lengths of each of these vectors, I don't care how many components there are, they're just going to be your numbers. So the length of this side right here is just going to be the length of a. The length of this side right here is just going to be the length of vector a minus vector b. And the length of this side right here is going to be the length of vector b." + }, + { + "Q": "i kind of get but killo means 1,000 or 1,000,000 on 00:14", + "A": "Kilo means 1,000.", + "video_name": "9iulv2QvKwo", + "timestamps": [ + 14 + ], + "3min_transcript": "What I want to do in this video is convert this amount of kilometers into meters and centimeters. So we'll first start with the 2 kilometers. And I encourage you to now pause this video and try to do this on your own. Well, the one thing that we know is that a kilometer literally means 1,000 meters. So you could literally view this is as 2 times 1,000 meters. Let me write that down. So this is going to be equal to 2 times 1,000 meters, which is equal to 2,000 meters. If we wanted to convert the 11 kilometers into meters, it's the same thing. 11 kilometers-- this right over here-- means 1,000 meters. So you could think of it as 11 times 1,000 meters. So 11 times 1,000 is going to be 11,000 meters. Now let's convert these distances into centimeters. And here we just have to remember that 1 meter is equal to 100 centimeters. Let me write that down. And that's because the prefix centi means one hundredth. Another way you can write it is that one centimeter is equal to one hundredth of a meter. But here we have a certain number of meters, and each of those meters are going to be equivalent to 100 centimeters. So if we wanted to write 2,000 meters as centimeters, we could say, well, we have 2,000 meters. Each of those are going to be equivalent to 100 centimeters. And so this is going to be equal to 2,000 times 100. Well, that's going to be 2, and since where we have the three zeroes from the 2,000. And then every time you multiply by 10, you're going to add another zero. Or you're going to have another zero at the end of it. And we're going to be at multiplying by 100 is equivalent to multiplying by 10 twice. So we're going to have two more zeroes. So this is going to be 200,000 centimeters. with the kilometers, the 11 kilometers, which are 11,000 meters. And once again, I encourage you to pause the video and try to convert it into centimeters. Well, same idea. You have 11,000 meters, and each of them are equivalent to 100 centimeters. So this is going to be 11, and let's see, we have one, two, three, four, five zeroes. So this gets us to 1,100,000 centimeters." + }, + { + "Q": "The question is to write an equation, so should we stop at 6:46?", + "A": "We could have stopped but he wanted to solve for the answer.", + "video_name": "jQ15tkoXZoA", + "timestamps": [ + 406 + ], + "3min_transcript": "on the left-hand side just because that might be a little bit more intuitive. So let's subtract 2p squared from both sides. Let's subtract 16p from both sides. We have an 8p and a 12p, and then we're going to subtract a 16p from both sides. And then, actually, let's subtract an 8p as well We have a 16p and an 8p, so that actually works out quite well. So now we've subtracted 8p from both sides, 16p So we've essentially subtracted all of this stuff from both sides. And we are left with-- let's see, I'll do it in degree order. 1.5p squared minus 2p squared is negative 0.5p squared. Now let's see, these cancel out. And then we have plus 64. And then that is going to be equal to 0. And just to simplify this a little bit, or just to make this a little bit cleaner, let's multiply both sides of this equation by negative 2. I want the coefficient over here to be 1. So then we get p squared plus 8p. p squared plus 8p is going to be equal to-- let's see, negative times negative 2. So minus 128-- is going to be equal to 0. So let's see if we can factor this. Can we think of two numbers where if we take their product, we get negative 128? And if we were to add them together, we get positive 8. right over here. So let's see, if we say 12 times-- well, let's see. What numbers could this be? So if we were to think about 128 is the same thing as-- 16, let's see, 16 goes into 128. Let me work through this. 16 goes into 128, does it go 8 times? 8 times 6 is 48. 8 times 10 is 80, plus 40 is 128. Yep, it goes 8 times. So 16 and 8 seem to work. So if you have positive 16 and negative 8, their product would be negative 128. So we can factor this out as p plus 16 times p minus 8 is equal to 0. Now, this is going to be equal to 0 if at least one of these" + }, + { + "Q": "2:32 how is the opposite and adjacent line of theta can become sin theta and cos theta? i thought sin is like opposite over adjacent something like that..", + "A": "In the unit circle, the hypotenuse always equals 1 (it s the radius of the unit circle). Since sin(\u00ce\u00b8)=opposite/hypotenuse, and the hypotenuse equals 1, you can say that sin(\u00ce\u00b8)=opposite/1, or sin(\u00ce\u00b8)=opposite. It s the same idea for cosine. This only works for the unit circle, though. You can watch the videos on the unit circle if you haven t already.", + "video_name": "Idxeo49szW0", + "timestamps": [ + 152 + ], + "3min_transcript": "me-- He's just saying the tangent of some angle is equal to x. And I just need to figure out what that angle is. So let's do an example. So let's say I were walk up to you on the street. There's a lot of a walking up on a lot of streets. I would write -- And I were to say you what is the arctangent of minus 1? Or I could have equivalently asked you, what is the inverse tangent of minus 1? These are equivalent questions. And what you should do is you should, in your head-- If you don't have this memorized, you should draw the unit circle. Actually let me just do a refresher of what tangent is even asking us. The tangent of theta-- this is just the straight-up, vanilla, non-inverse function tangent --that's equal to the sine of theta over the cosine of theta. And the sine of theta is the y-value on the unit function-- on the unit circle. And so if you draw a line-- Let me draw a little unit circle here. So if I have a unit circle like that. And let's say I'm at some angle. Let's say that's my angle theta. And this is my y-- my coordinates x, y. We know already that the y-value, this is the sine of theta. Let me scroll over here. Sine of theta. And we already know that this x-value is the cosine of theta. So what's the tangent going to be? It's going to be this distance divided by this distance. Or from your algebra I, this might ring a bell, because we're starting at the origin from the point 0, 0. This is our change in y over our change in x. Or it's our rise over run. Or you can kind of view the tangent of theta, or it really The slope. So you could write slope is equal to the tangent of theta. So let's just bear that in mind when we go to our example. If I'm asking you-- and I'll rewrite it here --what is the inverse tangent of minus 1? And I'll keep rewriting it. Or the arctangent of minus 1? I'm saying what angle gives me a slope of minus 1 on the unit circle? So let's draw the unit circle. Let's draw the unit circle like that. Then I have my axes like that. And I want a slope of minus 1. A slope of minus 1 looks like this. If it was like that, it would be slope of plus 1. So what angle is this? So in order to have a slope of minus 1, this distance is the same as this distance. And you might already recognize that this is a right angle." + }, + { + "Q": "I really don't get how he looks at the length at about 4:40 and says \"So this is square root of 2 over 2\" is it something he memorized or can you calculate it based on the slope?", + "A": "Sal wasn t looking at the length, really, but rather at the angle Theta (derived from the slope): Theta is -45 degrees. That means that the right triangle he draws is isosceles (with hypotenuse = 1, since this is a unit circle), and a quick calculation (or a good memory) tells us that both of the legs must therefore have a length of sqrt(2)/2.", + "video_name": "Idxeo49szW0", + "timestamps": [ + 280 + ], + "3min_transcript": "The slope. So you could write slope is equal to the tangent of theta. So let's just bear that in mind when we go to our example. If I'm asking you-- and I'll rewrite it here --what is the inverse tangent of minus 1? And I'll keep rewriting it. Or the arctangent of minus 1? I'm saying what angle gives me a slope of minus 1 on the unit circle? So let's draw the unit circle. Let's draw the unit circle like that. Then I have my axes like that. And I want a slope of minus 1. A slope of minus 1 looks like this. If it was like that, it would be slope of plus 1. So what angle is this? So in order to have a slope of minus 1, this distance is the same as this distance. And you might already recognize that this is a right angle. So this has to be a 45 45 90 triangle. This is an isosceles triangle. These two have to add up to 90 and they have to be the same. So this is 45 45 90. And if you know your 45 45 90-- Actually, you don't even have to know the sides of it. In the previous video, we saw that this is going to be-- Right here. This distance is going to be square root of 2 over 2. So this coordinate in the y-direction is minus square root of 2 over 2. And then this coordinate right here on the x-direction is square root of 2 over 2 because this length right there is that. So the square root of 2 over 2 squared plus the square root of 2 over 2 squared is equal to 1 squared. But the important thing to realize is this is a 45 45 90 triangle. So this angle right here is-- Well if you're just looking at the triangle by itself, you would say that this is a 45 degree angle. But since we're going clockwise below the x-axis, we'll call So the tangent of minus 40-- Let me write that down. So if I'm in degrees. And that tends to be how I think. So I could write the tangent of minus 45 degrees it equals this negative value-- minus square root of 2 over 2 over square root of 2 over 2, which is equal to minus 1. Or I could write the arctangent of minus 1 is equal to minus 45 degrees. Now if we're dealing with radians, we just have to convert this to radians. So we multiply that times-- We get pi radians for every 180 degrees. The degrees cancel out. So you have a 45 over 180. This goes four times. So this is equal to-- you have the minus sign-- minus pi over 4 radians." + }, + { + "Q": "At 3:53, what does factorial mean?", + "A": "A FACTORIAL is when a number has the little ! next to it. That means that you multiply that number with every number before it and stop when you get to 1. For example: 5! = 5 x 4 x 3 x 2 x 1 = 120 Hope this helps! :)", + "video_name": "W7DmsJKLoxc", + "timestamps": [ + 233 + ], + "3min_transcript": "" + }, + { + "Q": "How do you know for sure if it's a conic? I mean, what if you try to simplify it and then you can't multiply it by a number so that it has the form ((x+a)^2)/b)+((y+c)^2)/d)=1? For example, what if, at 3:58, the number on the right wasn't 100? What would you do then?", + "A": "I m going to call the number on the right z. If z wasn t a square root, Z= (sqrt(z))^2, right? So you work with that. You would go sqrt(z) in the direction needed.", + "video_name": "cvA4VN1dpuY", + "timestamps": [ + 238 + ], + "3min_transcript": "Is equal to 109. And the things we're going to add, those are what complete the square. Make these things a perfect square. So, if I take this, you have a minus 4 here. I take half of that number. This is just completing the square, I encourage you to watch the video on completing the square where I explain why this works. But I think I have a minus 4. I take half of that, it's minus 2. And then minus 2 squared is plus 4. Now, I can't do one thing to one side of the equation without doing it to the other. And I didn't add a 4 to the left-hand side of the equation. I actually added a 4 times 4, right? Because you have this 4 multiplying it out front. So I added at 16 to the left side of the equation, so I have to also add it to the right-hand side of the equation. This is equivalent to also having a plus 16 here. That might make a little bit clearer, right? When you factor it out, and it becomes a 4. And we would have added a 16 up here as well. Likewise, if we take half of this number here. 1 squared is 1. We didn't add a 1 to the left-hand side of the equation, we added a 1 times minus 25. So we want to put a minus 25 here. And, likewise, this would have been the same thing as adding a minus 25 up here. And you do a minus 25 over here. And now, what does this become? The y terms become 4 times y minus 2 squared. y minus 2 squared. Might want to review factoring a polynomial, if you found that a little confusing, that step. Minus 25 times x plus 1 squared. That's that, right there. x plus 1 squared, is equal to, let's see, 109 plus 16 is 25 minus 25, it equals 100. We're almost there. So we want a 1 here, so let's divide both sides So, you will get y minus 2 squared. 4 divided by 100 is the same thing as 1/25, so this becomes over 25. Minus, let's see, 25/100 is the same thing as 1/4, so this becomes x plus 1 squared over 4 is equal to 1. And there you have it. We have it in standard form and, yes, indeed, we do have a hyperbola. Now, let's graph this hyperbola. So the first thing we know is where the center of this hyperbola is. Is the center of this hyperbola is at the point x is equal to minus one. So it's an x is equal to minus 1. y is equal to 2. And let's figure out the asymptotes of this hyperbola. So if this was -- this is the way I always do it, because I always forget the actual formula. If this was centered at 0 and it looked something like this. y squared over 25 minus x squared over 4 is equal to 1." + }, + { + "Q": "At 0:30 how is 36 the least common multiple of 12 when doesn't it have to not be 1? I don't understand this.", + "A": "It is not the least common multiple (lcm) of 12 but of 12 and 36. 12=12, 24, 36, 48, 60, 72 36=36, 72 36 is the least multiple 12 and 36 have in common. decompositing method. 12=2*2*3 36=2*2*3*3 So the lcm=2*2*3*3=36", + "video_name": "znmPfDfsir8", + "timestamps": [ + 30 + ], + "3min_transcript": "What is the least common multiple of 36 and 12? So another way to say this is LCM, in parentheses, 36 to 12. And this is literally saying what's the least common multiple of 36 and 12? Well, this one might pop out at you, because 36 itself is a multiple of 12. And 36 is also a multiple of 36. It's 1 times 36. So the smallest number that is both a multiple of 36 and 12-- because 36 is a multiple of 12-- is actually 36. There we go. Let's do a couple more of these. That one was too easy. What is the least common multiple of 18 and 12? And they just state this with a different notation. The least common multiple of 18 and 12 is equal to question mark. So let's think about this a little bit. So there's a couple of ways you can think about-- so let's just write down our numbers that we care about. We care about 18, and we care about 12. So there's two ways that we could approach this. One is the prime factorization approach. of these numbers and then construct the smallest number whose prime factorization has all of the ingredients of both of these numbers, and that will be the least common multiple. So let's do that. 18 is 2 times 9, which is the same thing as 2 times 3 times 3, or 18 is 2 times 9. 9 is 3 times 3. So we could write 18 is equal to 2 times 3 times 3. That's its prime factorization. 12 is 2 times 6. 6 is 2 times 3. So 12 is equal to 2 times 2 times 3. Now, the least common multiple of 18 and 12-- let me write this down-- so the least common multiple of 18 and 12 is going to have to have enough prime factors to cover because we want the least common multiple or the smallest common multiple. So let's think about it. Well, it needs to have at least 1, 2, a 3 and a 3 in order to be divisible by 18. So let's write that down. So we have to have a 2 times 3 times 3. This makes it divisible by 18. If you multiply this out, you actually get 18. And now let's look at the 12. So this part right over here-- let me make it clear. This part right over here is the part that makes up 18, makes it divisible by 18. And then let's see. 12, we need two 2's and a 3. Well, we already have one 3, so our 3 is taken care of. We have one 2, so this 2 is taken care of. But we don't have two 2s's. So we need another 2 here. So, notice, now this number right over here has a 2 times 2 times 3 in it, or it has a 12 in it, and it has a 2 times 3 times 3, or an 18 in it. So this right over here is the least common multiple" + }, + { + "Q": "At 1:39 why did you put 2 times 3 times 3", + "A": "Because 2 x 3 x 3 = 18. (In more detail, 2x3 = 6. 6x3= 18.) Same reason with 12. 2x2 = 4 4x3=12. So that s why you see: 2x2x3 Why do we need to do this? Because we need to gather some set of numbers that can be multiplied to get both number 12 and 18, in order to find the Least Common Multiple. LCM: 2x2x3x3 = 36 <--- 2x2x3= 12 and 2x3x3 = 18", + "video_name": "znmPfDfsir8", + "timestamps": [ + 99 + ], + "3min_transcript": "What is the least common multiple of 36 and 12? So another way to say this is LCM, in parentheses, 36 to 12. And this is literally saying what's the least common multiple of 36 and 12? Well, this one might pop out at you, because 36 itself is a multiple of 12. And 36 is also a multiple of 36. It's 1 times 36. So the smallest number that is both a multiple of 36 and 12-- because 36 is a multiple of 12-- is actually 36. There we go. Let's do a couple more of these. That one was too easy. What is the least common multiple of 18 and 12? And they just state this with a different notation. The least common multiple of 18 and 12 is equal to question mark. So let's think about this a little bit. So there's a couple of ways you can think about-- so let's just write down our numbers that we care about. We care about 18, and we care about 12. So there's two ways that we could approach this. One is the prime factorization approach. of these numbers and then construct the smallest number whose prime factorization has all of the ingredients of both of these numbers, and that will be the least common multiple. So let's do that. 18 is 2 times 9, which is the same thing as 2 times 3 times 3, or 18 is 2 times 9. 9 is 3 times 3. So we could write 18 is equal to 2 times 3 times 3. That's its prime factorization. 12 is 2 times 6. 6 is 2 times 3. So 12 is equal to 2 times 2 times 3. Now, the least common multiple of 18 and 12-- let me write this down-- so the least common multiple of 18 and 12 is going to have to have enough prime factors to cover because we want the least common multiple or the smallest common multiple. So let's think about it. Well, it needs to have at least 1, 2, a 3 and a 3 in order to be divisible by 18. So let's write that down. So we have to have a 2 times 3 times 3. This makes it divisible by 18. If you multiply this out, you actually get 18. And now let's look at the 12. So this part right over here-- let me make it clear. This part right over here is the part that makes up 18, makes it divisible by 18. And then let's see. 12, we need two 2's and a 3. Well, we already have one 3, so our 3 is taken care of. We have one 2, so this 2 is taken care of. But we don't have two 2s's. So we need another 2 here. So, notice, now this number right over here has a 2 times 2 times 3 in it, or it has a 12 in it, and it has a 2 times 3 times 3, or an 18 in it. So this right over here is the least common multiple" + }, + { + "Q": "at 1:39 why did you did you put 2 times three times three?", + "A": "2*3*3 is the prime factorization.", + "video_name": "znmPfDfsir8", + "timestamps": [ + 99 + ], + "3min_transcript": "What is the least common multiple of 36 and 12? So another way to say this is LCM, in parentheses, 36 to 12. And this is literally saying what's the least common multiple of 36 and 12? Well, this one might pop out at you, because 36 itself is a multiple of 12. And 36 is also a multiple of 36. It's 1 times 36. So the smallest number that is both a multiple of 36 and 12-- because 36 is a multiple of 12-- is actually 36. There we go. Let's do a couple more of these. That one was too easy. What is the least common multiple of 18 and 12? And they just state this with a different notation. The least common multiple of 18 and 12 is equal to question mark. So let's think about this a little bit. So there's a couple of ways you can think about-- so let's just write down our numbers that we care about. We care about 18, and we care about 12. So there's two ways that we could approach this. One is the prime factorization approach. of these numbers and then construct the smallest number whose prime factorization has all of the ingredients of both of these numbers, and that will be the least common multiple. So let's do that. 18 is 2 times 9, which is the same thing as 2 times 3 times 3, or 18 is 2 times 9. 9 is 3 times 3. So we could write 18 is equal to 2 times 3 times 3. That's its prime factorization. 12 is 2 times 6. 6 is 2 times 3. So 12 is equal to 2 times 2 times 3. Now, the least common multiple of 18 and 12-- let me write this down-- so the least common multiple of 18 and 12 is going to have to have enough prime factors to cover because we want the least common multiple or the smallest common multiple. So let's think about it. Well, it needs to have at least 1, 2, a 3 and a 3 in order to be divisible by 18. So let's write that down. So we have to have a 2 times 3 times 3. This makes it divisible by 18. If you multiply this out, you actually get 18. And now let's look at the 12. So this part right over here-- let me make it clear. This part right over here is the part that makes up 18, makes it divisible by 18. And then let's see. 12, we need two 2's and a 3. Well, we already have one 3, so our 3 is taken care of. We have one 2, so this 2 is taken care of. But we don't have two 2s's. So we need another 2 here. So, notice, now this number right over here has a 2 times 2 times 3 in it, or it has a 12 in it, and it has a 2 times 3 times 3, or an 18 in it. So this right over here is the least common multiple" + }, + { + "Q": "1:24 We're allowed to pick numbers at random? Is that \"legal\"?", + "A": "The idea is to pick points to plot the graph. Which ones you use isn t important as long as you can draw the plot with enough precision. The more points you pick, the more accurate your graph will be around those points.", + "video_name": "rgvysb9emcQ", + "timestamps": [ + 84 + ], + "3min_transcript": "Let's do a couple of problems graphing linear equations. They are a bunch of ways to graph linear equations. What we'll do in this video is the most basic way. Where we will just plot a bunch of values and then connect the dots. I think you'll see what I'm saying. So here I have an equation, a linear equation. I'll rewrite it just in case that was too small. y is equal to 2x plus 7. I want to graph this linear equation. Before I even take out the graph paper, what I could do is set up a table. Where I pick a bunch of x values and then I can figure out what y value would correspond to each of those x values. So for example, if x is equal to-- let me start really low-- if x is equal to minus 2-- or negative 2, I should say-- what is y? Well, you substitute negative 2 up here. It would be 2 times negative 2 plus 7. This is negative 4 plus 7. This is equal to 3. If x is equal to-- I'm just picking x values at random points here. So what happens when x is equal to 0? Then y is going to be equal to 2 times 0 plus 7. Is going to be equal to 7. I just happen to be going up by 2. You could be going up by 1 or you could be picking numbers at random. When x is equal to 2, what is y? It'll be 2 times 2 plus 7. So 4 plus 7 is equal to 11. I could keep plotting points if I like. We should already have enough to graph it. Actually to plot any line, you actually only need two points. So we already have one more than necessary. Actually, let me just do one more just to show you that this really is a line. So what happens when x is equal to 4? Actually, just to not go up by 2, let's do x is equal to 8. Just to pick a random number. Then y is going to be 2 times 8 plus 7, which is-- well this plus 7 is equal to 23. Now let's graph it. Let me do my y-axis right there. That is my y-axis. Let me do my x-axis. I have a lot of positive values here, so a lot of space on the positive y-side. That is my x-axis. And then I use the points x is equal to negative 2. That's negative 1. That's 0, 1, 2, 3, 4, 5, 6, 7, 8. Those are our x values. Then we can go up into the y-axis. I'll do it at a slightly different scale because these numbers get large very quickly. So maybe I'll do it in increments of 2. So this could be 2, 4, 6, 8, 10, 12, 14, 16." + }, + { + "Q": "1:16-1:46 he does the translation visually. Is there any way to do it with an equation.", + "A": "Yes, to translate a figure 2 spots to the right you just add 2 to its x value. So E (3+2,3) -> E (5,3).", + "video_name": "XiAoUDfrar0", + "timestamps": [ + 76, + 106 + ], + "3min_transcript": "- [Voiceover] What I hope to introduce you to in this video is the notion of a transformation in mathematics, and you're probably used to the word in everyday language. Transformation means something is changing, it's transforming from one thing to another. What would transformation mean in a mathematical context? Well, it could mean that you're taking something mathematical and you're changing it into something else mathematical, that's exactly what it is. It's talking about taking a set of coordinates or a set of points, and then changing them into a different set of coordinates or a different set of points. For example, this right over here, this is a quadrilateral we've plotted it on the coordinate plane. This is a set of points, not just the four points that represent the vertices of the quadrilateral, but all the points along the sides too. There's a bunch of points along this. You could argue there's an infinite, or there are an infinite number of points along this quadrilateral. This right over here, the point X equals 0I equals negative four, this is a point on the quadrilateral. Now, we can apply a transformation to this, which just means moving all the points in the same direction, and the same amount in that same direction, and I'm using the Khan Academy translation widget to do it. Let's translate, let's translate this, and I can do it by grabbing onto one of the vertices, and notice I've now shifted it to the right by two. Every point here, not just the orange points has shifted to the right by two. This one has shifted to the right by two, this point right over here has shifted to the right by two, every point has shifted in the same direction by the same amount, that's what a translation is. Now, I've shifted, let's see if I put it here every point has shifted to the right one and up one, they've all shifted by the same amount in the same directions. That is a translation, but you could imagine a translation is not the only kind of transformation. In fact, there is an unlimited variation, there's an unlimited number different transformations. I have another set of points here that's represented by quadrilateral, I guess we could call it CD or BCDE, and I could rotate it, and I rotate it I would rotate it around the point. So for example, I could rotate it around the point D, so this is what I started with, if I, let me see if I can do this, I could rotate it like, actually let me see. So if I start like this I could rotate it 90 degrees, I could rotate 90 degrees, so I could rotate it, I could rotate it like, that looks pretty close to a 90-degree rotation. So, every point that was on the original or in the original set of points I've now shifted it relative to that point that I'm rotating around. I've now rotated it 90 degrees, so this point has now mapped to this point over here. This point has now mapped to this point over here, and I'm just picking the vertices because those are a little bit easier to think about." + }, + { + "Q": "at 5:09, i dont understand how it is the same distance", + "A": "Most transformations, translations, rotations, and reflections all end up with an image that is congruent to the pre-image, so same parts of congruent figures are congruent. The video has the rotation slightly off.", + "video_name": "XiAoUDfrar0", + "timestamps": [ + 309 + ], + "3min_transcript": "The point of rotation, actually, since D is actually the point of rotation that one actually has not shifted, and just 'til you get some terminology, the set of points after you apply the transformation this is called the image of the transformation. So, I had quadrilateral BCDE, I applied a 90-degree counterclockwise rotation around the point D, and so this new set of points this is the image of our original quadrilateral after the transformation. I don't have to just, let me undo this, I don't have to rotate around just one of the points that are on the original set that are on our quadrilateral, I could rotate around, I could rotate around the origin. I could do something like that. Notice it's a different rotation now. It's a different rotation. I could rotate around any point. Now let's look at another transformation, and that would be the notion of a reflection, You imagine the reflection of an image in a mirror or on the water, and that's exactly what we're going to do over here. If we reflect, we reflect across a line, so let me do that. This, what is this one, two, three, four, five, this not-irregular pentagon, let's reflect it. To reflect it, let me actually, let me actually make a line like this. I could reflect it across a whole series of lines. Woops, let me see if I can, so let's reflect it across this. Now, what does it mean to reflect across something? One way I imagine is if this was, we're going to get its mirror image, and you imagine this as the line of symmetry that the image and the original shape they should be mirror images across this line we could see that. Let's do the reflection. There you go, and you see we have a mirror image. This is this far away from the line. This, its corresponding point in the image This point over here is this distance from the line, and this point over here is the same distance but on the other side. Now, all of the transformations that I've just showed you, the translation, the reflection, the rotation, these are called rigid transformations. Once again you could just think about what does rigid mean in everyday life? It means something that's not flexible. It means something that you can't stretch or scale up or scale down it kind of maintains its shape, and that's what rigid transformations are fundamentally about. If you want to think a little bit more mathematically, a rigid transformation is one in which lengths and angles are preserved. You can see in this transformation right over here the distance between this point and this point, between points T and R, and the difference between their corresponding image points, that distance is the same. The angle here, angle R, T, Y, the measure of this angle over here," + }, + { + "Q": "At 4:30 how does Sal go from B/60*-5 into -b/12?", + "A": "Multiply: b/60 * -5/1 = (-5b)/60 Cancel out common factor of 5 and you get: -b/12 Hope this helps.", + "video_name": "Q0tTfe2lKIc", + "timestamps": [ + 270 + ], + "3min_transcript": "is equal to 35 kilometers and just with one equation, we're not going to be able to figure out what W and B are but we have another constraint. We know the total amount of time. So the total amount of time is going to be one and half hours, so we'll just write that over here. This is going to be 1.5, so what's the time traveled by, what's the time he walks? Let me write this over here, time time walking, we'll that's going to be the distance walking divided by the rate walking. So the distance walking is W kilometers W kilometers divided by his rate, the distance divided by your rate is gonna give you your time, so let's see, his rate is five kilometers per hour, five kilometers per hour and so you're gonna and if you divide by or if you have one over hours in the denominator, that's going to be the same thing, this is gonna be W over five hours, so the units work out. So his time walking is W over five, W over five and by that same logic, his time on the bus is going to be the distance on the bus divided by, divided by the average speed of the school bus, so this is going to be 60. This is all going to be in hours and now we can solve this system of equations. We have two linear equations with two unknowns. We should be able to find W and B that satisfy both of these. Now what's an easy thing to do? Let's see, if I can multiply this second equation by negative five, and I'm gonna, this is going to be a negative W here so it'll cancel out with this W up there. So let's do that, let's multiply the second equation by, I'm just gonna switch to one color here, This bottom equation, if I multiply both sides by negative five, so both sides by negative five, I'm going to multiply both sides by negative five I'm going to get negative five times W over five is negative W, negative five times B over 60. Let's see, it's gonna be, it's going to be negative five over 60, that's negative 1/12, so this is negative B over 12 and then it is going to be equal to 1.5 times negative five is negative 7.5, negative 7.5. Now we can just add the left and right hand sides of these two equations. Now let me, I can do this a little bit neater, let me actually delete, let me make these line up a little bit better so that we, delete that, make this, so this first equation was, whoops," + }, + { + "Q": "At around 6:11 , how is it possible that X can be less or equal to -3? Wouldn't that make the square root negative?", + "A": "No, because the absolute value of anything less or equal to -3 is larger than 3, i.e. it is more than 3 removed from zero. Subtracting 3 from something larger than 3 gives a positive outcome, and hence a real solution to the problem. For instance x = -4. Abs value (-4) = 4 4 - 3 = 1 Square root (1) = 1.", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 371 + ], + "3min_transcript": "So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive So, once again, x has to be less than negative 3 or x has to be greater than 3, so we have our domain. So we have it as x is a member of the reals -- I always forget. Is that the line? I forget, it's either a colon or a line. I'm rusty, it's been years since I've done this kind of stuff. But anyway, I think you get the point. It could be any real number here, as long as x is less than negative 3, less than or equal to negative 3, or x is greater than or equal to 3. Let me ask a question now. What if instead of this it was -- that was the denominator, this is all a separate problem up here. So now we have 1 over the square root of the absolute value of x minus 3. So now how does this change the situation? So not only does this expression in the denominator," + }, + { + "Q": "Take the answer of problem at 4:51 . And, let x=12. So, f(x)=(sqrt)9 and\n(sqrt) of 9 has 2 values: 3 and -3. Which means, f(12) = -3 or f(12)= 3 which is not possible according to the definition of a function. How can this be explained??", + "A": "The notation \u00e2\u0088\u009a9 exclusively denotes the POSITIVE value. Its corresponding negative root is denoted as -\u00e2\u0088\u009a9. However, when a positive value does not exist, the radical sign may be used to indicate the negative real value. Eg. ^3\u00e2\u0088\u009a-8 = -2.[The cube root of -8, that can be expressed on a number line, is -2, because (-2)*(-2)*(-2) = -8] That s only when we used symbols; when we say, or write, square root of a certain number, it refers to all applicable values.", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 291 + ], + "3min_transcript": "So here I just made a slight variation on what I had before. Before we said when f of x is equal to x squared that x is just any real number. Now we're saying that x is any real number except for 0. This is just a fancy way of saying it, and then these curly brackets just mean a set. Let's do a couple more ones. Let's say f of x is equal to the square root of x minus 3. So up here we said, well this function isn't defined when we get a 0 in the denominator. But what's interesting about this function? Can we take a square root of a negative number? Well until we learn about imaginary and complex numbers we can't. So here we say well, any x is valid here except for the x's So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive" + }, + { + "Q": "At around 4:25, Sal said the square root couldn't be a negative number, so x would have had to be 3 or more. Wouldn't it have to be 4 or more because 3 - 3 = 0?", + "A": "As Sal says: there is nothing wrong with the expression sqrt(0). Its just zero.", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 265 + ], + "3min_transcript": "And what's 1 over 0? I don't know what it is, so this is undefined. No one ever took the trouble to define what 1 over 0 should be. And they probably didn't do, so some people probably thought about what should be, but they probably couldn't find out with a good definition for 1 over 0 that's consistent with the rest of mathematics. So 1 over 0 stays undefined. So f of 0 is undefined. So we can't put 0 in and get a valid answer for f of 0. So here we say the domain is equal to -- do little brackets, that shows kind of the set of what x's apply. That's those little curly brackets, I didn't draw it that well. x is a member of the real numbers still, such that So here I just made a slight variation on what I had before. Before we said when f of x is equal to x squared that x is just any real number. Now we're saying that x is any real number except for 0. This is just a fancy way of saying it, and then these curly brackets just mean a set. Let's do a couple more ones. Let's say f of x is equal to the square root of x minus 3. So up here we said, well this function isn't defined when we get a 0 in the denominator. But what's interesting about this function? Can we take a square root of a negative number? Well until we learn about imaginary and complex numbers we can't. So here we say well, any x is valid here except for the x's So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the" + }, + { + "Q": "Mr Sal at 3:43 you said\nF(x)=square root of x-3greater or equal to 0\nHow can it be equal to 0, if I plug 0 in place of x i would get -3 which is wrong??", + "A": "Notice the subtle distinction. Sal did not say that x was greater than or equal to 0, but that x-3 was. So x must be greater than or equal to 3.", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 223 + ], + "3min_transcript": "And what's 1 over 0? I don't know what it is, so this is undefined. No one ever took the trouble to define what 1 over 0 should be. And they probably didn't do, so some people probably thought about what should be, but they probably couldn't find out with a good definition for 1 over 0 that's consistent with the rest of mathematics. So 1 over 0 stays undefined. So f of 0 is undefined. So we can't put 0 in and get a valid answer for f of 0. So here we say the domain is equal to -- do little brackets, that shows kind of the set of what x's apply. That's those little curly brackets, I didn't draw it that well. x is a member of the real numbers still, such that So here I just made a slight variation on what I had before. Before we said when f of x is equal to x squared that x is just any real number. Now we're saying that x is any real number except for 0. This is just a fancy way of saying it, and then these curly brackets just mean a set. Let's do a couple more ones. Let's say f of x is equal to the square root of x minus 3. So up here we said, well this function isn't defined when we get a 0 in the denominator. But what's interesting about this function? Can we take a square root of a negative number? Well until we learn about imaginary and complex numbers we can't. So here we say well, any x is valid here except for the x's So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the" + }, + { + "Q": "Why Sal did say at 6:28 the domain of function f(x) = sqrt(abs(x)-3) is x\u00e2\u0082\u00acR | x<=-3 and x>=3 ?\nIts a mistake? We can have a positive square root of 0 ?", + "A": "At this point, Sal says the domain is all real values of x such that x is less than or equal to -3 and greater than or equal to 3. When that value is in the denominator, then the domain is x<=-3 and x>=3 because the denominator cannot be 0. They were two separate equations.", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 388 + ], + "3min_transcript": "So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the So now it's getting a little bit more complicated. Well just like this time around, this expression of the radical still has to be greater than or equal to 0. So you can just say that the absolute value of x minus 3 is greater than or equal to 0. So we have the absolute value of x has to be greater than or equal to 3. And if order for the absolute value of something to be greater than or equal to something, then that means that x has to be less than or equal to negative 3, or x has to be greater than or equal to 3. It makes sense because x can't be negative 2, right? Because negative 2 has an absolute value less than 3. So x has to be less than negative 3. It has to be further in the negative direction than negative 3, or it has to be further in the positive So, once again, x has to be less than negative 3 or x has to be greater than 3, so we have our domain. So we have it as x is a member of the reals -- I always forget. Is that the line? I forget, it's either a colon or a line. I'm rusty, it's been years since I've done this kind of stuff. But anyway, I think you get the point. It could be any real number here, as long as x is less than negative 3, less than or equal to negative 3, or x is greater than or equal to 3. Let me ask a question now. What if instead of this it was -- that was the denominator, this is all a separate problem up here. So now we have 1 over the square root of the absolute value of x minus 3. So now how does this change the situation? So not only does this expression in the denominator," + }, + { + "Q": "So in 0:38, you said that you can add up all the digits (154 adding up to 10 in the video). Does that method always work? I mean, I understand that 5 would not be able to go into 154 because the result always ends in either 0 or 5 as mentioned, but I'm just a bit confused. Could someone possibly explain if the method really works for all numbers?", + "A": "Lets tell parts of division 15\u00c3\u00b75=3 15=dividend formula=d \u00c3\u00b7=division sign 5=divisior formula=D 3=quotient formula=q 0=remainder formula=r Euclid was a great mathmatician who made a division statement (Divisior\u00c3\u0097quotient)+remainder =dividend Eg=lets take 67\u00c3\u00b76=11 (6\u00c3\u009711)+1=67", + "video_name": "5xe-6GPR_qQ", + "timestamps": [ + 38 + ], + "3min_transcript": "Which of the following numbers is a factor of 154? So when a number is going to be a factor of 154 is if we can divide that number into 154 and not have a remainder. Or another way of thinking about it-- a number is a factor of 154 if 154 is a multiple of that number. So let's look at each of these and see which of these we can rule out or say is a factor. So does 3 divide evenly into 154? Or, another way of thinking about it, is 154 a multiple of 3? Well you'll later learn that you could actually test whether something is divisible by 3 by adding up the digits. And if that's divisible by 3, then it's going to be divisible by 3. And so you see here, 1 plus 5 is 6. 6 plus 4 is 10. 10 is not divisible by 3. But if you didn't want to do that little trick-- and we have other videos where we go into more detail about that trick-- you can actually just divide 3 into 154. 3 doesn't go into 1. It does go into 15 five times. Subtract. Then we have 0. Then you bring down a 4. 3 goes into 4 one time. 1 times 3 is 3. Subtract, and you have a remainder of 1. So 3 is clearly not a factor of 154, so we can rule that out. Now what about 5? Well, any multiple of 5 is either going to have 5 or 0 in the ones place. You see that if we write 5 times 1 is 5, 5 times 2 is 10, 5 times 3 is 15, 5 times 4 is 20, you either have a 5 or a 0 in the ones place. This does not have 5 or a 0 the ones place, so it's not going to be divisible by 5. 5 is not a factor. 154 is not a multiple of 5. Now 6 is interesting. You could do the same thing. You could try to divide 6 into 154. But if something is divisible by 6, it's definitely going to be divisible by 3 as well because 6 is divisible by 3. So we can immediately rule this one out as well. it's also not going to be divisible by 6. And you could try it out if you like. And we could make the same argument for 9. If something is divisible by 9, it's going to be divisible by 3 because 9 is divisible by 3. Well, it's not divisible by 3, so we're going to rule out 9 as well. So we've ruled out everything. It looks like 14 is our only option, but let's actually verify it. Let's actually divide 14 into 154. 14 doesn't go into 1. It goes into 15 exactly one time. 1 times 14 is 14. We subtract. We get 1. Bring down the 4. 14 goes into 14 one time. 1 times 14 is 14. And of course, we have no remainder. So 14 goes into 154 exactly 11 times. Or 11 times 14 is 154. 154 is a multiple of 14. Let's do one more. Which of the following numbers is a multiple of 14?" + }, + { + "Q": "I think \"9:50\" does not need a proof as they're just i j k l unit vectors.", + "A": "9:54 A proof may be simple, but still needed. That is the case here.", + "video_name": "JUgrBkPteTg", + "timestamps": [ + 590 + ], + "3min_transcript": "By the same reasoning, no combination of that and that is going to equal this. This is by definition of a pivot entry. When you put it in reduced row echelon form, it's very clear that any pivot column will be the only one to have 1 in that place. So it's very clear that these guys are linearly independent. Now it turns out, and I haven't proven it to you, that the corresponding columns in A-- this is r1, but it's A before we put it in reduced row echelon form-- that these guys right here, so a1, a2, and a4 are also linearly independent. So a1-- let me circle it-- a2, and a4. So if I write it like this, a1, a2, and a4. Let me write it in set notation. These guys are also linearly independant, which I haven't proven. But I think you can kind of get a sense that these row And I'll do a better explanation of this, but I really just wanted you to understand how to develop a basis for the column space. So they're linearly independent. So the next question is do they span our column space? And in order for them to span, obviously all of these 5 vectors, if you have all of them, that's going to span your column space by definition. But if we can show, and I'm not going to show it in this video, but it turns out that you can always represent the non-pivot columns as linear combinations of the pivot columns. And we've kind of touched on that in previous videos where we find a solution for the null space and all that. So these guys can definitely be represented as linear combinations of these guys. I haven't shown you that, but if you take that on faith, then you don't need that column and that column to span. If you did then, or I guess a better way to think it, you are part of the span. Because if you needed this guy, you can just construct him with linear combinations of these guys. So if you wanted to figure out a basis for the column space of A, you literally just take A into reduced row echelon form. You look at the pivot entries in the reduced row echelon form of A, and that's those three. And then you look at the corresponding columns to those pivot columns in your original A. And those form the basis. Because any linear combination of them, or linear combinations of them can be used to construct the non-pivot columns, and they're linearly independant. So I haven't shown you that. But for this case, if you want to know the basis, it's just a1, a2, and a4. And now we can answer another question. So a1, a2, and a4 form a basis for the column space of A," + }, + { + "Q": "at 7:19 sal said it was equal to 112 it is equal to 108 You may do the check if you want 36+108 is 144", + "A": "Well, yeah, I guess. But I think he was a little rushed.", + "video_name": "s9t7rNhaBp8", + "timestamps": [ + 439 + ], + "3min_transcript": "So, we know that C is equal to 5 So, using the Pythagorean Theorem we just figured out that if we know the sides of, one side is 3, the other side is 4 then we can use Pythagorean Theorem for that Hypotenuse of this triangle it has the length of 5 Let's do another example Let's say, once again this is a right angle This side is of length 12, this side is of length 6 and I wanna figure out what this side is So, let's write down the Pythagorean Theorem A squared plus B squared is equal to C squared Where C is the length of the hypotenuse that I just drew Is which side is the hypotenuse? Well this right here, is the right angle So, the hypotenuse is this side right here And we can also eyeball it and say \"oh that's definitely the longest side of this triangle \" So, we know that A squared plus B squared is equal to 12 squared, which is 144 Now, we know that we have one side but we don't have the other side So, I'm gonna ask you as question Does it matter, which side we substitute for A or B? This is because A or B kinda do the same thing in this formula So, we can pick any side to be A other than the hypotenuse And we'll pick the other side to be B So, let's just say that this side is B and let's say this side is A So, we know what A is, so we get 6 squared plus B squared is equal to 144 So, we get 36 plus B squared is equal to 144 Now we got to simplify what the square root of 112 is What we did on those radical modules was probably helpful here So, B is equal to the square root of 112 Let's think about it, how many times does 4 go into a 112? 4 goes into a 125 times, it will go into it 28 times And 4 goes into 28, 7 times I actually think that this is equal to 16 times 7, am I right? 7 times 10 is 70 plus 42 is a 112 So, B equals the square root of 16 times 7 You see, I just factored that as a product of a perfect and a prime number" + }, + { + "Q": "During the video at about 7:10 Sal goes a little to fast,I don't understand the\nrest of the formula. I'm struggling with the last few steps.I know 6 squared is\n36 (side A)and the hypotenuse(side c)12 squared is equal to 144.after this point\ni'm lost. I don't know my square roots real well and I struggle with formulas that\nare more than 3 steps. I know w/lots of practice i will get it.any \"positive\" tips\nmuch appreciated.", + "A": "144-36 = 108, not 112. Other than that, you are correct once you adjust the incorrect values.", + "video_name": "s9t7rNhaBp8", + "timestamps": [ + 430 + ], + "3min_transcript": "So, we know that C is equal to 5 So, using the Pythagorean Theorem we just figured out that if we know the sides of, one side is 3, the other side is 4 then we can use Pythagorean Theorem for that Hypotenuse of this triangle it has the length of 5 Let's do another example Let's say, once again this is a right angle This side is of length 12, this side is of length 6 and I wanna figure out what this side is So, let's write down the Pythagorean Theorem A squared plus B squared is equal to C squared Where C is the length of the hypotenuse that I just drew Is which side is the hypotenuse? Well this right here, is the right angle So, the hypotenuse is this side right here And we can also eyeball it and say \"oh that's definitely the longest side of this triangle \" So, we know that A squared plus B squared is equal to 12 squared, which is 144 Now, we know that we have one side but we don't have the other side So, I'm gonna ask you as question Does it matter, which side we substitute for A or B? This is because A or B kinda do the same thing in this formula So, we can pick any side to be A other than the hypotenuse And we'll pick the other side to be B So, let's just say that this side is B and let's say this side is A So, we know what A is, so we get 6 squared plus B squared is equal to 144 So, we get 36 plus B squared is equal to 144 Now we got to simplify what the square root of 112 is What we did on those radical modules was probably helpful here So, B is equal to the square root of 112 Let's think about it, how many times does 4 go into a 112? 4 goes into a 125 times, it will go into it 28 times And 4 goes into 28, 7 times I actually think that this is equal to 16 times 7, am I right? 7 times 10 is 70 plus 42 is a 112 So, B equals the square root of 16 times 7 You see, I just factored that as a product of a perfect and a prime number" + }, + { + "Q": "From @7:45 to @10:36 , what are you trying to achieve when you convert the matrix to reduced row-echelon form?\n\nI thought you were trying to turn row 2 and 3 to zeros, like you did with the 2x2 matrix.\n\nWhy do you have to convert it to row-echelon form?\n\nThank for the videos btw, they've been really helpful.", + "A": "rref is the solution for the system of equations represented by the augmented matrix. You re finding the vectors v such that Av = lambda*v - i.e., the solution for (lambda*I - A)v = 0; i.e., rref ( [ (lambda*I - A) | 0 ] ).", + "video_name": "3Md5KCCQX-0", + "timestamps": [ + 465, + 636 + ], + "3min_transcript": "And they're the eigenvectors that correspond to eigenvalue lambda is equal to 3. So if you apply the matrix transformation to any of these vectors, you're just going to scale them up by 3. Let me write this way. The eigenspace for lambda is equal to 3, is equal to the span, all of the potential linear combinations of this guy and that guy. So 1/2, 1, 0. And 1/2, 0, 1. So that's only one of the eigenspaces. That's the one that corresponds to lambda is equal to 3. Let's do the one that corresponds to lambda is equal to minus 3. So if lambda is equal to minus 3-- I'll do it up here, I think I have enough space-- lambda is equal to minus 3. This matrix becomes-- I'll do the diagonals-- minus 3 plus 1 is minus 2. Minus 3 minus 2 is minus 5. And all the other things don't change. Minus 2, minus 2, 1. Minus 2, minus 2 and 1. And then that times vectors in the eigenspace that corresponds to lambda is equal to minus 3, is going to be equal to 0. I'm just applying this equation right here which we just derived from that one over there. So, the eigenspace that corresponds to lambda is equal to minus 3, is the null space, this matrix right here, are all the vectors that satisfy this equation. So what is-- the null space of this is the same thing as the null space of this in reduced row echelon form So let's put it in reduced row echelon form. So the first thing I want to do, I'm going to keep my first row the same. I'm going to write a little bit smaller than I normally do because I think I'm going to run out of space. So minus 2, minus 2, minus 2. I will skip some steps. Let's just divide the first row by minus 2. So we get 1, 1, 1. And then let's replace this second row with the second row plus this version of the first row. So this guy plus that guy is 0 minus 5 plus minus-- or let me Let me replace it with the first row minus the second row. So minus 2 minus minus 2 is 0. Minus 2 minus minus 5 is plus 3. And then minus 2 minus 1 is minus 3. And then let me do the last row in a different color for fun. And I'll do the same thing. I'll do this row minus this row. So minus 2 minus minus 2 is a 0. Minus 2 plus 2. Minus 2 minus 1 is minus 3. And then we have minus 2 minus minus 5." + }, + { + "Q": "At about 5:30, Sal has 9x36/2, then he divides the 36 by 2, getting 9x18/1. Why doesn't he divide the 9 by 2 also?", + "A": "36/2 is 36 divided by 2. So then he gets 9 x 18 becuz 36 divided by 2 is 18. Even if you do 9 x 36 and 9 x 2 then you will get 324/18 which is still 18.", + "video_name": "jFSenp9ueaI", + "timestamps": [ + 330 + ], + "3min_transcript": "So let's do it that way. If I have 4 and 1/2 of anything, really-- so let me write 4 and 1/2. I'm trying to find a suitable color. So if I have 4 and 1/2, this is the same thing-- 4 is the same thing as 8/2. This is the same thing-- so let me write it this way. 4 and 1/2 is the same thing as 4 plus 1/2, which is the same thing as-- if we want to have the same denominator as this 2 over here or as this 1/2 over here, this is the same thing as 8/2. Or you could say 4/1 is the same thing as 8/2, if we want to have a common denominator, so 8/2 plus 1/2. Actually, let me write it that way, just so you really understand what we're doing. 4 is the same thing as 4/1. So it's 4/1 plus 1/2. If we want to find a common denominator, it's 2. So 4/1 is the same thing as 8/2 plus 1/2, which is equal to 9/2. Now, I did it this way, which takes longer, why it makes-- hopefully conceptually why it just makes intuitive sense, why 4 and 1/2 is the same thing as 9/2. But if you want a simple process for it, you could just say, look, 4 times 2 is 8. 8 plus 1 is 9. And that gives you that 9 right over there, so 9/2. So we have 9/2 yards that we want to convert to inches. Same process-- times 36 inches per yard. Yard in the numerator, yard in the denominator. We are left with 9/2 times 36. We could say times 36/1 if we like, 36 Inches for every 1 yard. 36-- or the number 36 really is the same as 36/1. And then we're left with just inches in our units. We're just left with inches. And over here there's several ways that we can simplify it. Probably the easiest way to simplify it is we So let me write it this way. I don't want to skip steps. So we have 9 times 36 over 2 times 1, or over 2 inches. And we can divide the numerator and the denominator by 2 to simplify it. They're both divisible by 2. 36 divided by 2 is 18. 2 divided by 2 is 1. So we're really just left with 9 times 18 inches. We can just multiply 9 times 18. Let me do it over here. 18 times 9. 8 times 9 is 72. 1 times 9 is 9, plus 7 is 16, so we get 162 inches. So all of this simplifies to 162 inches, and we are done." + }, + { + "Q": "At 4:21 , the standard deviation of thesample size is given is given as 0.5s why do we have to determine it.", + "A": "because the other one we determined is the standard deviation of the sampling distribution of the sample mean.", + "video_name": "-FtlH4svqx4", + "timestamps": [ + 261 + ], + "3min_transcript": "So how do we think about this? How do we know whether we should accept the alternative hypothesis or whether we should just default to the null hypothesis because the data isn't convincing? And the way we're going to do it in this video, and this is really the way it's done in pretty much all of science, is you say OK, let's assume that the null hypothesis is true. If the null hypothesis was true, what is the probability that we would have gotten these results with the sample? And if that probability is really, really small, then the null hypothesis probably isn't true. We could probably reject the null hypothesis and we'll say well, we kind of believe in the alternative hypothesis. So let's think about that. Let's assume that the null hypothesis is true. So if we assume the null hypothesis is true, let's try gotten this result, that we would have actually gotten a sample mean of 1.05 seconds with a standard deviation of 0.5 seconds. So I want to see if we assumed the null hypothesis is true, I want to figure out the probability-- and actually what we're going to do is not just figure out the probability of this, the probability of getting something like this or even more extreme than this. So how likely of an event is that? To think about that let's just think about the sampling distribution if we assume the null hypothesis. So the sampling distribution is like this. It'll be a normal distribution. We have a good number of samples, we have 100 samples here. So this is the sampling distribution. It will have a mean. Now if we assume the null hypothesis, that the drug has no effect, the mean of our sampling distribution will be distribution, which would be equal to 1.2 seconds. Now, what is the standard deviation of our sampling distribution? The standard deviation of our sampling distribution should be equal to the standard deviation of the population distribution divided by the square root of our sample size, so divided by the square root of 100. We do not know what the standard deviation of the entire population is. So what we're going to do is estimate it with our sample standard deviation. And it's a reasonable thing to do, especially because we have a nice sample size. The sample size is greater than 100. So this is going to be a pretty good approximator. This is going to be a pretty good approximator for this over here. So we could say that this is going to be approximately equal to our sample standard deviation divided by the square root of 100, which is going to be equal to our sample standard deviation is 0.5, 0.5 seconds, and we want" + }, + { + "Q": "At 2:11, how do you know where to subtract the 2 from, like can you subtract it with any of the numbers on the other side of the equation?", + "A": "It s basic equation balancing. To isolate the variable, you have to get rid of the 2 in the equation. To do so but not change the equation, you have to minus 2 from both sides so that technically it s still the same equation just simplified down so the 2 is gone.", + "video_name": "qsL_5Y8uWPU", + "timestamps": [ + 131 + ], + "3min_transcript": "Determine the number of solutions for each of these equations, and they give us three equations right over here. And before I deal with these equations in particular, let's just remind ourselves about when we might have one or infinite or no solutions. You're going to have one solution if you can, by solving the equation, come up with something like x is equal to some number. Let's say x is equal to-- if I want to say the abstract-- x is equal to a. Or if we actually were to solve it, we'd get something like x equals 5 or 10 or negative pi-- whatever it might be. But if you could actually solve for a specific x, then you have one solution. So this is one solution, just like that. Now if you go and you try to manipulate these equations in completely legitimate ways, but you end up with something crazy like 3 equals 5, then you have no solutions. And if you just think about it reasonably, all of these equations are about finding an x that satisfies this. and you were to get something like 3 equals 5, and you were to ask yourself the question is there any x that can somehow magically make 3 equal 5, no. No x can magically make 3 equal 5, so there's no way that you could make this thing be actually true, no matter which x you pick. So if you get something very strange like this, this means there's no solution. On the other hand, if you get something like 5 equals 5-- and I'm just over using the number 5. It didn't have to be the number 5. It could be 7 or 10 or 113, whatever. And actually let me just not use 5, just to make sure that you don't think it's only for 5. If I just get something, that something is equal to itself, which is just going to be true no matter what x you pick, any x you pick, this would be true for. Well, then you have an infinite solutions. So with that as a little bit of a primer, let's try to tackle these three equations. So over here, let's see. Maybe we could subtract. If we want to get rid of this 2 here on the left hand side, If we subtract 2 from both sides, we are going to be left with-- on the left hand side we're going to be left with negative 7x. And on the right hand side, you're going to be left with 2x. This is going to cancel minus 9x. 2x minus 9x, If we simplify that, that's negative 7x. You get negative 7x is equal to negative 7x. And you probably see where this is going. This is already true for any x that you pick. Negative 7 times that x is going to be equal to negative 7 times that x. So we already are going into this scenario. But you're like hey, so I don't see 13 equals 13. Well, what if you did something like you divide both sides by negative 7. At this point, what I'm doing is kind of unnecessary. You already understand that negative 7 times some number is always going to be negative 7 times that number. But if we were to do this, we would get x is equal to x, and then we could subtract x from both sides." + }, + { + "Q": "what is the triangle at 1:00 for?", + "A": "It is the greek letter delta, which is commonly used today in modern applications of science and mathematics to represent a change or difference. Just like how we use pi in geometry, theta in trigonometry, lambda in calculus, phi to represent the golden ration, and more.", + "video_name": "f4MYCepzLyQ", + "timestamps": [ + 60 + ], + "3min_transcript": "definitions for the function D over here over here d of t is equal to 3 t plus 1. We could imagine the d could represent distance as a function of time. Distance measured in meters and time measured in seconds. So over here when time is 0 right when we're starting our distance is going to be 1 meter. After 1 second has gone by our distance is now going to be, so 3 times 1 plus 1 is going to be 4. Our distance now is going to be 4 is going to be 4 meters. After 2 seconds our distance is going to be 3 times two is 6 plus 1 it's going to be 7 it's going to be 7 meters. So given this definition of D of T this function definition. What is the rate of change of distance with respect to time and let me write it this way. What is the change in distance the rate of change of distance with respect to time which people sometimes called speed. Well, what is this going to be? Let's just take two points Let's just say the change in distance over change in time when time goes from time equals 0 to time equals 1 So over here our change in time is equal to 1 our change in time is 1 and what's our change in distance? Well our change in distance when our time increased by 1 our distance increases by 3 it goes from 1 meters to 4 meters. So our change in distance is equal to 3. So it's going to be equal to 3 over 1 or just 3. If we wanted the units it would be 3 meters, every 1 second. Now let's think about it, does that change if we pick any other two points? what if we were to say between 1 second and 2 seconds so between 1 second and 2 seconds our change in time is 1 second and then our change in distance is So once again our change in distance over change in time is 3 meters per second. This is all review and you might recognize, we pick any two points on this line here and we're going to have the same rate of change of distance with respect to time. In fact that's what defines a line. Or one of the ways to think about a line or a linear function is that the rate of change of one variable with respect to the other one, is constant. In this particular one we're talking about the rate of change of the vertical variable with respect to the horizontal one. We're talking about the slope of the line. This is the slope. This line has a slope of 3. That's what defines a line or one of the things that defines a line is the slope between any two points is going to be exactly 3. Just as a little bit of review from other Algebra you've seen before. You can even pick it out in the function definition." + }, + { + "Q": "At 3:02, Sal talks about slope-intercept form. Can anyone give me an explanation of what that is and how it can be used to find average rate of change.", + "A": "y = mx + b is slope-intercept form", + "video_name": "f4MYCepzLyQ", + "timestamps": [ + 182 + ], + "3min_transcript": "Well, what is this going to be? Let's just take two points Let's just say the change in distance over change in time when time goes from time equals 0 to time equals 1 So over here our change in time is equal to 1 our change in time is 1 and what's our change in distance? Well our change in distance when our time increased by 1 our distance increases by 3 it goes from 1 meters to 4 meters. So our change in distance is equal to 3. So it's going to be equal to 3 over 1 or just 3. If we wanted the units it would be 3 meters, every 1 second. Now let's think about it, does that change if we pick any other two points? what if we were to say between 1 second and 2 seconds so between 1 second and 2 seconds our change in time is 1 second and then our change in distance is So once again our change in distance over change in time is 3 meters per second. This is all review and you might recognize, we pick any two points on this line here and we're going to have the same rate of change of distance with respect to time. In fact that's what defines a line. Or one of the ways to think about a line or a linear function is that the rate of change of one variable with respect to the other one, is constant. In this particular one we're talking about the rate of change of the vertical variable with respect to the horizontal one. We're talking about the slope of the line. This is the slope. This line has a slope of 3. That's what defines a line or one of the things that defines a line is the slope between any two points is going to be exactly 3. Just as a little bit of review from other Algebra you've seen before. You can even pick it out in the function definition. our D intercept, when t is equal to 0, is going to be equal to one. All of this is review and if any of this sounds foreign to you, I encourage you to watch the Khan Academy videos on slope and slope intercept form and things like that. But this is all a background to get us to this curve over here because this is interesting. Because here we're no longer dealing with a line. We actually have a curve and this curve right over here. It's a quadratic, it's a parabola. Let's just say distance as a function of time was defined this way instead of it being 3t plus 1 it is t squared plus 1 and what's interesting here is that the rate of change you can have visualized it is it's always changing for example: If you were to pick an arbitrary point on this curve and if you think about a tangent line to it. A tangent line, a line that has the same slope for just that one moment, it just touches on it, over here" + }, + { + "Q": "At 4:00 Sal says 350/360 is = to 35/36 What rules dictate this?", + "A": "That s how you simplify a fraction. Divide the top and bottom numbers by the same factor (in this case 10) to simplify. Another example: 3/6 = 1/2 because you divide the top and bottom numbers by 3.", + "video_name": "tVcasOt55Lc", + "timestamps": [ + 240 + ], + "3min_transcript": "to be pi/2, whatever units we're talking about, long. Now let's think about another scenario. Let's imagine the same circle. So it's the same circle here. Our circumference is still 18 pi. There are people having a conference behind me or something. That's why you might hear those mumbling voices. But this circumference is also 18 pi. But now I'm going to make the central angle an obtuse angle. So let's say we were to start right over here. This is one side of the angle. I'm going to go and make a 350 degree angle. So I'm going to go all the way around like that. So this right over here is a 350 degree angle. And now I'm curious about this arc that subtends this really huge angle. So now I want to figure out this arc length-- so all of this. subtends this really obtuse angle right over here. Well, same exact logic-- the ratio between our arc length, a, and the circumference of the entire circle, 18 pi, should be the same as the ratio between our central angle that the arc subtends, so 350, over the total number of degrees in a circle, over 360. So multiply both sides by 18 pi. We get a is equal to-- this is 35 times 18 over 36 pi. 350 divided by 360 is 35/36. So this is 35 times 18 times pi over 36. so let's divide them both by 18. And so we are left with 35/2 pi. Let me just write it that way-- 35 pi over 2. Or, if you wanted to write it as a decimal, this would be 17.5 pi. Now does this makes sense? This right over here, this other arc length, when our central angle was 10 degrees, this had an arc length of 0.5 pi. So when you add these two together, this arc length and this arc length, 0.5 plus 17.5, you get to 18 pi, which was the circumference, which makes complete sense because if you add these angles, 10 degrees and 350 degrees, you get 360 degrees in a circle." + }, + { + "Q": "Could you cross multiply at 1:32 ?", + "A": "You mean turning it into a = ====1 ------ ------- 1.8pi 360? If so, then yes", + "video_name": "tVcasOt55Lc", + "timestamps": [ + 92 + ], + "3min_transcript": "I have a circle here whose circumference is 18 pi. So if we were to measure all the way around the circle, we would get 18 pi. And we also have a central angle here. So this is the center of the circle. And this central angle that I'm about to draw has a measure of 10 degrees. So this angle right over here is 10 degrees. And what I'm curious about is the length of the arc that subtends that central angle. So what is the length of what I just did in magenta? And one way to think about it, or actually maybe the way to think about it, is that the ratio of this arc length to the entire circumference-- let me write this down-- should be the same as the ratio of the central angle to the total number of angles if you were So let's just think about that. We know the circumference is 18 pi. We're looking for the arc length. I'm just going to call that a. a for arc length. That's what we're going to try to solve for. We know that the central angle is 10 degrees. So you have 10 degrees over 360 degrees. So we could simplify this by multiplying both sides by 18 pi. And we get that our arc length is equal to-- well, 10/360 is the same thing as 1/36. So it's equal to 1/36 times 18 pi, so it's 18 pi over 36, which is the same thing as pi/2. to be pi/2, whatever units we're talking about, long. Now let's think about another scenario. Let's imagine the same circle. So it's the same circle here. Our circumference is still 18 pi. There are people having a conference behind me or something. That's why you might hear those mumbling voices. But this circumference is also 18 pi. But now I'm going to make the central angle an obtuse angle. So let's say we were to start right over here. This is one side of the angle. I'm going to go and make a 350 degree angle. So I'm going to go all the way around like that. So this right over here is a 350 degree angle. And now I'm curious about this arc that subtends this really huge angle. So now I want to figure out this arc length-- so all of this." + }, + { + "Q": "4:25 I don't understand why 5(7k(k+3)-(k+3)) becomes 5(k+3)(7k-1)", + "A": "Try this : If you had for say, 6(k+3) - (k+3), you could clearly say that it is equal to 5(k+3) right? Which is equal to (6-1)(k+3). Same thing is happening in here. You have 7k(k+3)-(k+3) and you are free to transform it into (7k-1)(k+3).", + "video_name": "R-rhSQzFJL0", + "timestamps": [ + 265 + ], + "3min_transcript": "Sorry. If we take 21 and negative 1, their product is negative 21. 21 times negative 1 is negative 21. and if you take their sum, 21 plus negative 1, that is equal to 20. So these two numbers right there fit the bill. Now, let's break up this 20k right here into a 21k and a negative 1k. So let's rewrite the whole thing. We have 5 times 7k squared, and I'm going to break this 20k into a-- let me do it in this color right here-- I'm going to break that 20k into a plus 21k, minus k. Or you could say minus 1k if you want. I'm using those two factors to break it up. And then we finally have the minus 3 right there. Now, the whole point of doing that is so that we can now factor each of the two groups. And so what can we factor out of that group right there? Well, both of these are divisible by 7k, so we can write this as 7k times-- 7k squared divided by 7k, you're just going to have a k left over. And then plus 21k divided by 7k is just going to be a 3. So that factors into that. And then we can look at this group right here. They have a common factor. Well, we can factor out a negative 1 if we like, so this is equal to negative 1 times-- k divided by negative 1 is k. Negative 3 divided by negative 1 is positive 3. And, of course, we have this 5 sitting out there. Now, ignoring that 5 for a second, you see that both of these inside terms have k plus 3 as a factor. So let's ignore this 5 for a second. This inside part right here, the stuff that's inside the parentheses, we can factor k plus 3 out, and it becomes k plus 3, times k plus 3, times 7k minus 1. And if this seems a little bizarre to you, distribute the k plus 3 on to this. K plus 3 times 7k is that term, k plus 3 times negative 1 is that term. And, of course, the whole time you have that 5 sitting outside. You have that 5. We don't even have to put parentheses there. 5 times k plus 3, times 7k minus 1. And we factored it, we're done." + }, + { + "Q": "Hi at 1:03 - 1:06, Sal mentions finding a number whose product is 7 * -3. Why isn't he simply referring to the number -3 as his product?\n\nThanks in advance.", + "A": "The product of two numbers is the number that results from multiplying two numbers together. For example, -3\u00c3\u00977 equals -21. In this instance, -21 is the product. I hope this helps!", + "video_name": "R-rhSQzFJL0", + "timestamps": [ + 63, + 66 + ], + "3min_transcript": "We're asked to factor 35k squared plus 100k, minus 15. And because we have a non-1 coefficient out here, the best thing to do is probably to factor this by grouping. But before we even do that, let's see if there's a common factor across all of these terms, and maybe we can get a 1 coefficient, out there. If we can't get a 1 coefficient, we'll at least have a lower coefficient here. And if we look at all of these numbers, they all look divisible by 5. In fact their greatest common factor is 5. So let's at least factor out a 5. So this is equal to 5 times-- 35k squared divided by 5 is 7k squared. 100k divided by 5 is 20k. And then negative 15 divided by 5 is negative 3. So we were able to factor out a 5, but we still don't have a 1 coefficient here, so we're still going to have to factor by grouping. But at least the numbers here are smaller so it'll be easier to think about it in terms of finding numbers whose product sum is equal to 20. So let's think about that. Let's figure out two numbers that if I were to add them, or even better if I were to take their product, I get 7 times negative 3, which is equal to negative 21. And if I were to take their sum, if I add those two numbers, it needs to be equal to 20. Now, once again, because their product is a negative number, that means they have to be of different signs, so when you add numbers of different signs, you could view it as you're taking the difference of the positive versions. So the difference between the positive versions of the number has to be 20. So the number that immediately jumps out is we're probably going to be dealing with 20 and 21, and 1 will be the negative, because we want to get to a positive 20. So let's think about it. So if we think of 20 and negative 1, their product is Sorry. If we take 21 and negative 1, their product is negative 21. 21 times negative 1 is negative 21. and if you take their sum, 21 plus negative 1, that is equal to 20. So these two numbers right there fit the bill. Now, let's break up this 20k right here into a 21k and a negative 1k. So let's rewrite the whole thing. We have 5 times 7k squared, and I'm going to break this 20k into a-- let me do it in this color right here-- I'm going to break that 20k into a plus 21k, minus k. Or you could say minus 1k if you want. I'm using those two factors to break it up. And then we finally have the minus 3 right there. Now, the whole point of doing that is so that we can now factor each of the two groups." + }, + { + "Q": "At 1:10, why can we not write {(-1)(-52)}^1/2 = {(-1)^1/2}{(-52)^1/2} ?", + "A": "I believe it would be mathematically incorrect, because sqrt(-1 x -52) is sqrt(52). However, sqrt(-1) x sqrt(-52) is not the same as sqrt(-1 x -52) because sqrt(-1) x sqrt(-52) is equal to sqrt(-1) x sqrt(52) x sqrt(-1) = -1 x sqrt(52) which is not the same as sqrt(-52). Sorry you got an answer to your question 2 years after you asked it, hopefully this helps! ;-} By the way, I can do the special effects for the iron-man movies, and am coming out with The last Days . if your interested.", + "video_name": "s03qez-6JMA", + "timestamps": [ + 70 + ], + "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i." + }, + { + "Q": "At 3:35, why is it i*\u00e2\u0088\u009a4*13 and not i^2*\u00e2\u0088\u009a4*13? I thought that by definition i^2= -1", + "A": "i^2 = -1 i = \u00e2\u0088\u009a(-1)", + "video_name": "s03qez-6JMA", + "timestamps": [ + 215 + ], + "3min_transcript": "things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i. And then we're going to multiply that times the square root of 4 times 13. And this is going to be equal to i times the square root of 4. i times the square root of 4, or the principal square root of 4 times the principal square root of 13. The principal square root of 4 is 2. So this all simplifies, and we can switch the order, over here. This is equal to 2 times the square root of 13. 2 times the principal square root of 13, I should say, times i. And I just switched around the order. It makes it a little bit easier to read if I put the i after the numbers over here. But I'm just multiplying i times 2 times the square root of 13. That's the same thing as multiplying 2 times the principal square root of 13 times i. And I think this is about as simplified as we can get here." + }, + { + "Q": "At 2:04, Sal says that I can not split sqrt(-1 x -52) into sqrt(-1) x sqrt(-52). Can I go the opposite way? Would it be mathematically correct to simplify: sqrt(-1) x sqrt(-52) into sqrt(-1 x -52)?", + "A": "I believe it would be mathematically correct, because sqrt(-1 x -52) is sqrt(52). However, sqrt(-1) x sqrt(-52) is not the same as sqrt(-1 x -52) because sqrt(-1) x sqrt(-52) is equal to sqrt(-1) x sqrt(52) x sqrt(-1) = -1 x sqrt(52) which is not the same as sqrt(-52). Sorry you got an answer to your question 2 years after you asked it, hopefully this helps! ;-}", + "video_name": "s03qez-6JMA", + "timestamps": [ + 124 + ], + "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i." + }, + { + "Q": "at 9:01 Sal messed up right? Isn't it supposed to be 365!/362!/(365)to the third??", + "A": "Yes; he did mess up at 9:01", + "video_name": "9G0w61pZPig", + "timestamps": [ + 541 + ], + "3min_transcript": "That'll actually be 30 terms divided by 365 to the 30th power. And you can just type this into your calculator right now. It'll take you a little time to type in 30 numbers, and you'll get the probability that no one shares the same birthday with anyone else. But before we do that let me just show you something that might make it a little bit easier. Is there any way that I can mathematically express this with factorials? Or that I could mathematically express this with factorials? Let's think about it. 365 factorial is what? 365 factorial is equal to 365 times 364 times 363 times-- all the way down to 1. You just keep multiplying. It's a huge number. Now, if I just want the 365 times the 364 in this case, One thing I could do is I could divide this thing by all of these numbers. So 363 times 362-- all the way down to 1. So that's the same thing as dividing by 363 factorial. 365 factorial divided by 363 factorial is essentially this because all of these terms cancel out. So this is equal to 365 factorial over 363 factorial over 365 squared. And of course, for this case, it's almost silly to worry about the factorials, but it becomes useful once we have something larger than two terms up here. So by the same logic, this right here is going to be equal to 365 factorial over 362 factorial over 365 squared. How did we get this 365? Sorry, how did we get this 363 factorial? Well, 365 minus 2 is 363, right? And that makes sense because we only wanted two terms up here. We only wanted two terms right here. So we wanted to divide by a factorial that's two less. And so we'd only get the highest two terms left. This is also equal to-- you could write this as 365 factorial divided by 365 minus 2 factorial 365 minus 2 is 363 factorial and then you just end up with those two terms and that's that there. And then likewise, this right here, this numerator you could rewrite as 365 factorial divided by 365 minus 3-- and we had 3 people-- factorial. And that should hopefully make sense, right? This is the same thing as 365 factorial-- well 365 divided" + }, + { + "Q": "on 0:59 he is a horrible stock investor\nVote up if u agree", + "A": "There isn t enough info to know. -- The problem didn t tell you the time frame. If the portfolio went up in 1 month by 25%, that would be fantastic! -- If the overall market is down for the year by 20% and this person s portfolio went up 25%, that would also be fantastic! -- If it took 20 years for his portfolio to grow 25%, that would not be good, unless his goal was to project his portfolio (low risk). Then, it would be good.", + "video_name": "X2jVap1YgwI", + "timestamps": [ + 59 + ], + "3min_transcript": "Let's do some more percentage problems. Let's say that I start this year in my stock portfolio with $95.00. And I say that my portfolio grows by, let's say, 15%. How much do I have now? I think you might be able to figure this out on your own, but of course we'll do some example problems, just in case it's a little confusing. So I'm starting with $95.00, and I'll get rid of the dollar sign. We know we're working with dollars. 95 dollars, right? And I'm going to earn, or I'm going to grow just because I was an excellent stock investor, that 95 dollars So to that 95 dollars, I'm going to add another 15% of 95. So we know we write 15% as a decimal, as 0.15, so 95 plus 0.15 of 95, so this is times 95-- that dot is just a times sign. It's not a decimal, it's a times, it's a little higher than a decimal-- So 95 plus 0.15 times 95 is what we have now, right? Because we started with 95 dollars, and then we made another 15% times what we started with. Hopefully that make sense. Another way to say it, the 95 dollars has grown by 15%. So let's just work this out. This is the same thing as 95 plus-- what's 0.15 times 95? Let's see. So let me do this, hopefully I'll have enough space here. 95 times 0.15-- I don't want to run out of space. of space-- 95 times 0.15. 5 times 5 is 25, 9 times 5 is 45 plus 2 is 47, 1 times 95 is 95, bring down the 5, 12, carry the 1, 15. And how many decimals do we have? 1, 2. 15.25. Actually, is that right? I think I made a mistake here. See 5 times 5 is 25. 5 times 9 is 45, plus 2 is 47. And we bring the 0 here, it's 95, 1 times 5, 1 times 9, then we add 5 plus 0 is 5, 7 plus 5 is 12-- oh. I made a mistake." + }, + { + "Q": "4:48 Hey, does anyone know why Sal puts the = sign like a smiley face? =D", + "A": "It s not an equal sign, but rather an arrow. He just draws it in a way that it doesn t look quite connected. This is actually a common way to note progression of steps in mathematics.", + "video_name": "X2jVap1YgwI", + "timestamps": [ + 288 + ], + "3min_transcript": "So I'll ask you an interesting question? How did I know that 15.25 was a mistake? Well, I did a reality check. I said, well, I know in my head that 15% of 100 is 15, so if 15% of 100 is 15, how can 15% of 95 be more than 15? I think that might have made sense. The bottom line is 95 is less than 100. So 15% of 95 had to be less than 15, so I knew my answer of 15.25 was wrong. And so it turns out that I actually made an addition error, and the answer is 14.25. So the answer is going to be 95 plus 15% of 95, which is the same thing as 95 plus 14.25, well, that equals what? 109.25. especially this 2 here. 109.25. So if I start off with $95.00 and my portfolio grows-- or the amount of money I have-- grows by 15%, I'll end up with $109.25. Let's do another problem. Let's say I start off with some amount of money, and after a year, let's says my portfolio grows 25%, and after growing 25%, I now have $100. How much did I originally have? Notice I'm not saying that the $100 is growing by 25%. by 25%, and I end up with $100 after it grew by 25%. To solve this one, we might have to break out a little bit of algebra. So let x equal what I start with. So just like the last problem, I start with x and it grows by 25%, so x plus 25% of x is equal to 100, and we know this 25% of x we can just rewrite as x plus 0.25 of x is equal to 100, and now actually we have a level-- actually this might be level 3 system, level 3 linear equation-- but the bottom" + }, + { + "Q": "What's the next step if the number doesn't come out to be exact like 13*16 for 208? (2:19 in video)", + "A": "you go and find the number of ones. example: 114 = 7 * 256 + 16 *13 (D) + 0 * 1 there are 0 1 s. if the number were 115, then it would be 7D1 because there an extra one as supposed to 7D0 = 114. hope this helped", + "video_name": "NFmDz1dQyPU", + "timestamps": [ + 139 + ], + "3min_transcript": "-[Voiceover]Let's now try to convert a number from the decimal system to the hexadecimal system. Let's say we want to convert the number 2,000. This is written in decimal form. Let's say we want to write it in hexadecimal form. Like always, I encourage you to pause this video and try to work through it on your own. The key here is to break this down into multiples of powers of 16. Let's just write down our powers of 16 here. 16 to the zero power is one. 16 to the first power is 16. 16 squared is 256. 16 to the third power is 4,096. We've gotten more than large enough. Let's start decomposing. What's the largest power of 16 that is less than or equal to 2,000? It's going to be 256. How many times does 256 go into 2,000? I'll get a calculator out for that. times 256 goes into it, divided by 256. It goes seven times, plus a little bit. What's going to be left over? 2,000 minus seven times 256 is equal to, you're going to have 208 left over. Let me write that. It's going to be seven times 256. Seven, times 256, plus 208 left over. Now let's see if we can decompose 208 into powers of 16. What's the largest power of 16 that is less than or equal to this? Well it's just going to be 16. So how many 16s go into 208? 208, I'll just take that, divided by 16. 13, exactly 13. We lucked out. So this is exactly 13 times 16. This right over here is exactly 13 times 16. Now, we have broken this down into powers of... We have broken this down into multiples of powers of 16. Now we're ready to write it in hexadecimal. We just have to remind ourselves about the place value. This right over here, this is going to be the ones place. Then we're going to have the 16s place. Then we're going to have the 256s place. We know how many 256s we have. We have seven 256s. We have zero ones. And how many 16s do we have? Well we have 13 16s. Well what's the digit for 13? We can just remind ourselves that we obviously have zero through nine and then we have A is equal to 10, B is equal to 11, C is equal to 12," + }, + { + "Q": "At 4:21, Sal says that that 0i is the same thing as 0 + i. Wouldn't be 0 times i? I am confused.", + "A": "No, he says that i is the same thing as 0 + i , which is true. You are right that 0i is just 0 times i , not 0 + i . So 0i is the same thing as just 0 .", + "video_name": "A_ESfuN1Pkg", + "timestamps": [ + 261 + ], + "3min_transcript": "And 9 is a real number. So we could just add those up. So 2 plus negative 7 would be negative 5. Negative 5 plus 9 would be 4. So the real numbers add up to 4. And now we have these imaginary numbers. So 3 times i minus 5 times i. So if you have 3 of something and then I were to subtract 5 of that same something from it, now you're going to have negative 2 of that something. Or another way of thinking about it is the coefficients. 3 minus 5 is negative 2. So three i's minus five i's, that's going to give you negative 2i. Now you might say, well, can we simplify this any further? Well no, you really can't. This right over here is a real number. 4 is a number that we've been dealing with throughout our mathematical careers. And so what we really consider this is this 4 minus 2i, we can now consider this entire expression to really be a number. So this is a number that has a real part and an imaginary part. And numbers like this we call complex numbers. It is a complex number. Why is it complex? Well, it has a real part and an imaginary part. And you might say, well, gee, can't any real number be considered a complex number? For example, if I have the real number 3, can't I just write the real number 3 as 3 plus 0i? And you would be correct. Any real number is a complex number. You could view this right over here as a complex number. a subset of the complex numbers. Likewise, imaginary numbers are a subset of the complex numbers. For example, you could rewrite i as a real part-- 0 is a real number-- 0 plus i. So the imaginaries are a subset of complex numbers. Real numbers are a subset of complex numbers. And then complex numbers also have all of the sums and differences, or all of the numbers that have both real and imaginary parts." + }, + { + "Q": "at 0:45 why is i2 + to -1?", + "A": "The definition of i is : the number that its square is equal to -1. So: i^2 = -1 by the definition of i itself.", + "video_name": "A_ESfuN1Pkg", + "timestamps": [ + 45 + ], + "3min_transcript": "Now that we know a little bit about the imaginary unit i, let's see if we can simplify more involved expressions, like this one right over here. 2 plus 3i plus 7i squared plus 5i to the third power plus 9i to the fourth power. And I encourage you to pause the video right now and try to simplify this on your own. So as you can see here, we have various powers of i. You could view this as i to the first power. We have i squared here. And we already know that i squared is defined to be negative 1. Then we have i to the third power. I to the third power would just be i times this, or negative i. And we already reviewed this when we first introduced the imaginary unit, i, but I'll do it again. i to the fourth power is just going to be i times this, which is the same thing as negative 1 times i. That's i to the third power times i again. i times i is negative 1. So that's negative 1 times negative 1, which is equal to 1 So we can rewrite this whole thing as 2 plus 3i. 7i squared is going to be the same thing, so i squared is negative 1. So this is the same thing as 7 times negative 1. So that's just going to be minus 7. And then we have 5i to the third power. i to the third power is negative i. So this could be rewritten as negative i. So this term right over here we could write as minus 5i, or negative 5i, depending on how you want to think about it. And then finally, i to the fourth power is just 1. So this is just equal to 1. So this whole term just simplifies to 9. So how could we simplify this more? Well we have several terms that are not imaginary, that they are real numbers. For example, we have this 2 is a real number. And 9 is a real number. So we could just add those up. So 2 plus negative 7 would be negative 5. Negative 5 plus 9 would be 4. So the real numbers add up to 4. And now we have these imaginary numbers. So 3 times i minus 5 times i. So if you have 3 of something and then I were to subtract 5 of that same something from it, now you're going to have negative 2 of that something. Or another way of thinking about it is the coefficients. 3 minus 5 is negative 2. So three i's minus five i's, that's going to give you negative 2i. Now you might say, well, can we simplify this any further? Well no, you really can't. This right over here is a real number. 4 is a number that we've been dealing with throughout our mathematical careers." + }, + { + "Q": "At 5:47 Sal says the x value has to be between 0 and 10. Why isn't it between 0 and 15?", + "A": "Because the piece of card is only 20 wide, and x can be at most half the width.", + "video_name": "MC0tq6fNRwU", + "timestamps": [ + 347 + ], + "3min_transcript": "So what would the volume be as a function of x? Well, the volume as a function of x is going to be equal to the height, which is x, times the width, which is 20 minus x-- sorry, 20 minus 2x times the depth, which is 30 minus 2x. Now, what are possible values of x that give us a valid volume? Well, x can't be less than 0. You can't make a negative cut here. Somehow we would have to add cardboard or something there. So we know that x is going to be greater than or equal to 0. So let me write this down. x is going to be greater than or equal to 0. And what does it have to be less than? Well, I can cut at most-- we can see here the length right over here, this pink color, this mauve color, So this has got to be greater than 0. This is always going to be shorter than the 30 minus 2x, but the 20 minus 2x has to be greater than or equal to 0. You can't cut more cardboard than there is. Or you could say that 20 has to be greater than or equal to 2x, or you could say that 10 is going to be greater than or equal to x, which is another way of saying that x is going to be less than or equal to 10. That's a different shade of yellow. x is going to be less than or equal to 10. So x has got to be between 0 and 10. Otherwise we've cut too much, or we're somehow adding cardboard or something. So first let's think about the volume at the endpoints of our-- essentially of our domain, of what x can be for our volume. Well, our volume when x is equal to 0 is equal to what? We can have 0 times all of this stuff. You're not going to have any height here. So you're not going to have any volume, so our volume would be 0. What is our volume when x is equal to 10? Well, if x equaled 10, then the width here that I've drawn in pink would be 0. So once again, we would have no volume. And this term right over here, if we just look at it algebraically would also be, equal to 0, so this whole thing would be equal to 0. So someplace in between x equals 0 and x equals 10 we should achieve our maximum volume. And before we do it analytically with a bit of calculus, let's do it graphically. So I'll get my handy TI-85 out. So let me get my TI-85 out. And so first let me set my range appropriately before I attempt to graph it. So I'll put my graph function. Let me set my range. So my minimum x-value, let me make that 0. We know that x cannot be less than 0. My maximum x-value, well, 10 seems pretty good. My minimum y-value, this is essentially" + }, + { + "Q": "At 7:40, why did he set a maximum y value? Isn't that already determined since x cannot be greater than 10?", + "A": "Right. He knows that the maximum x-value is 10, but he doesn t know what the maximum y-value is. The y-value represents the volume of the box, which can get pretty big depending on the dimensions. (width, depth, height)", + "video_name": "MC0tq6fNRwU", + "timestamps": [ + 460 + ], + "3min_transcript": "You're not going to have any height here. So you're not going to have any volume, so our volume would be 0. What is our volume when x is equal to 10? Well, if x equaled 10, then the width here that I've drawn in pink would be 0. So once again, we would have no volume. And this term right over here, if we just look at it algebraically would also be, equal to 0, so this whole thing would be equal to 0. So someplace in between x equals 0 and x equals 10 we should achieve our maximum volume. And before we do it analytically with a bit of calculus, let's do it graphically. So I'll get my handy TI-85 out. So let me get my TI-85 out. And so first let me set my range appropriately before I attempt to graph it. So I'll put my graph function. Let me set my range. So my minimum x-value, let me make that 0. We know that x cannot be less than 0. My maximum x-value, well, 10 seems pretty good. My minimum y-value, this is essentially I'm not going to have negative volume, so let me set that equal 0. And my maximum y-value, let's see what would be reasonable here. I'm just going to pick some a random x and see what type of a volume I get. So if x were to be 5, it would be 5 times 20 minus 10, which is 10. So that would be-- did I do that right? Yeah, 20 minus 2 times 5, so that would be 10, and then times 30 minus 2 times 5, which would be 20. So it would be 5 times 10 times 20. So you'd get a volume of 1,000 cubic inches. And I just randomly picked the number 5. So let me give my maximum y-value a little higher than that just in case to that isn't the maximum value. I just randomly picked that. So let's say yMax is 1,500, and if for whatever reason our graph doesn't fit in there, then maybe we can make our yMax even larger. So I think this is going to be a decent range. Now let's actually input the function itself. So our volume is equal to x times 20 And that looks about right. So now I think we can graph it. So 2nd, and I want to select that up there, so I'll do Graph. And it looks like we did get the right range. So this tells us volume is a function of x between x is 0 and x is 10, and it does look like we hit a maximum point right around there. to use the Trace function to figure out roughly what that maximum point is. So let me trace this function. So I can still go higher, higher. So over there my volume is 1,055.5. Then I can get to 1,056. So let's see, this was 1,056.20, this is 1,056.24, then I go back to 1,055. So at least based on the level of zoom that I have my calculator right now," + }, + { + "Q": "At 7:17, is it only a statistical question if you mention \"in 2013\" before you added it in the last example, or can it be statistical either way?", + "A": "Without the in 2013, it is kind of a complicated case. You have to think about what you want to consider. if you say throughout the history of the schools or since 1800, then this does become a statistical question no matter how you look at it. However, considering other things (like in 2013), you have concrete numbers that you can check the difference of, making it not a statistical question", + "video_name": "qyYSQDcSNlY", + "timestamps": [ + 437 + ], + "3min_transcript": "What was the difference in rainfall between Singapore and Seattle in 2013? Well, these two numbers are known. They can be measured. Both the rainfall in Singapore can be measured. The rainfall in Seattle can be measured. And assuming that this has already happened and we can measure them, then we can just find the difference. So you don't need statistics here. You just have to have both of these measurements and subtract the difference. So not a statistical question. In general, will I use less gas driving at 55 miles an hour than 70 miles per hour? This feels statistical, because it probably depends on the circumstance. It might depend on the car. Or even for a given car, when you drive at 55 miles per hour, there's some variation in your gas mileage. It might be how recent an oil change happened, what the wind conditions are like, what the road conditions are like, exactly how you're driving the car. Are you going in a straight line? And same thing for 70 miles an hour. When we're saying in general, there's variation in what the gas mileage is at 55 miles an hour and at 70 miles an hour. What you'd probably want to do is say, well, what's my average mileage when I drive at 55 miles an hour and compare that to the average mileage when I drive at 70. So because we have this variability in each of those cases, this is definitely a statistical question. Do English professors get paid less than math professors? Once again, all English professor don't get paid the same amount, and all math professors don't get paid the same amount. Some English professors might do quite well. Some might make very little. Same thing for math professors. So we'd probably want to find some type of an average to represent the central tendency for each of these. Once again, this is a statistical question. Does the most highly paid English professor at Harvard professor at MIT? Well, now we're talking about two particular individuals. You could go look at their tax forms, see how much each of them get paid, especially if we assume that this is in a particular year. Let's just make it that way, say in 2013, just so that we can remove some variability that they might make from year to year, make it a little bit more concrete. If this was does the most highly paid English professor at Harvard get paid more than the most highly paid math professor at MIT in 2013, then you have an absolute number for each of these people. And then you could just compare them directly. So when we're talking about a particular year, particular people, then it isn't a statistical question." + }, + { + "Q": "There is one point at \"2:25\" that I really want it to be clarified. I think that \"integral f(t) d(t) from a to x\" should be \" F(x) - F(a)\" instead of \"F(x)\". Please help me Im so confused!", + "A": "The integral f(t) d(t) from a to x would equal F(x)-F(a). I believe Sal is simply using F to put the integral as a function of x. Any letter or symbol would do, some calculus books write it as A(x) instead of F(x). They tend to use A because the formula seen in the video is an area accumulation formula (A for area). Whatever x you put into F(x) would give you the amount of area you have accumulated from a to x.", + "video_name": "C7ducZoLKgw", + "timestamps": [ + 145 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:19, Sal explains that every continuous function has an antiderivative, according to the Fundamental Theorem of Calculus. I learned in Calculus A that only continuous functions have derivatives. Is this an inference from the Fundamental Theorem?", + "A": "No. The fact that every continuous function is a derivative (has an antiderivative) has nothing to do with the fact that only continuous functions have derivatives. The FToC is not necessary to show that continuity is a condition for differentiability.", + "video_name": "C7ducZoLKgw", + "timestamps": [ + 259 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:58, it doesn't really matter what size the camel is, since it's infinite, you'll always end up with the whole page covered.", + "A": "No, actually it does matter. If you make the next camel too big, then you will end up needing more than the page to draw all the camels. If you make the next camel too small, then you will not use up all the space on the page.", + "video_name": "DK5Z709J2eo", + "timestamps": [ + 58 + ], + "3min_transcript": "So you mean you're in math class, yet again, because they make you go every single day. And you're learning about, I don't know, the sums of infinite series. That's a high school topic, right? Which is odd, because it's a cool topic. But they somehow manage to ruin it anyway. So I guess that's why they allow infinite serieses in the curriculum. So, in a quite understandable need for distraction, you're doodling and thinking more about what the plural of series should be than about the topic at hand. Serieses, serises, seriesen, seri? Or is it that the singular should be changed? One serie, or seris, or serum? Just like the singular of sheep should be shoop. But the whole concept of things like 1/2 plus 1/4 plus 1/8 plus 1/16, and so on approaches 1 is useful if, say, you want to draw a line of elephants, each holding the tail of the next one. Normal elephant, young elephant, baby elephant, dog-sized elephant, puppy-sized elephant, all the way down to Mr. Tusks, and beyond. Which is at least a tiny bit awesome because you can get an infinite number of elephants in a line and still have it fit across a single notebook page. But there's questions, like what if you started with a camel, which, being smaller than an elephant, only goes across a third of the page. How big should the next camel be in order to properly approach the end of the page? and it's cool that that's possible. But I'm not really interested in doing calculations. So we'll come back to camels. Here's a fractal. You start with these circles in a circle, and then keep drawing the biggest circle that fits in the spaces between. This is called an Apollonian Gasket. And you can choose a different starting set of circles, and it still works nicely. It's well known in some circles because it has some very interesting properties involving the relative curvature of the circles, which is neat, But it also looks cool and suggests an awesome doodle Step 1, draw any shape. Step 2, draw the biggest circle you can within this shape. Step 3, draw the biggest circle you can in the space left. Step 4, see step 3. As long as there is space left over after the first circle, meaning don't start with a circle, this method turns any shape into a fractal. You can do this with triangles. You can do this with stars. And don't forget to embellish. You can do this with elephants, or snakes, or a profile of one I choose Abraham Lincoln. Awesome. OK, but what about other shapes besides circles? For example, equilateral triangles, say, filling this other triangle, which works because the filler triangles, and orientation matters. This yields our friend, Sierpinski's triangle, which, by the way, you can also make out of Abraham Lincoln. But triangles seem to work beautifully in this case. But that's a special case. And the problem with triangles is that they don't always fit snugly. For example, with this blobby shape, the biggest equilateral triangle has this lonely hanging corner. And sure, you don't have to let that stop you, and it's a fun doodle game. But I think it lacks some of the beauty of the circle game. Or what if you could change the orientation of the triangle to get the biggest possible one? What if you didn't have to keep it equilateral? Well, for polygonal shapes, the game runs out pretty quickly, so that's no good. But in curvy, complicated shapes, the process itself becomes difficult. How do you find the biggest triangle? It's not always obvious which triangle has more area, especially when you're starting shape is not very well defined. This is an interesting sort of question because there is a correct answer, but if you were going to write a computer program that filled a given shape with another shape, following even the simpler version of the rules, you might need to learn some computational geometry. And certainly, we can move beyond triangles to squares, or even elephants. But the circle is great because it's just so fantastically round." + }, + { + "Q": "If we do not include two set of numbers can we make the brackets face opposite directions? For example at 5:06 we do not include \"-1 and 4\" Sal wrote them like this (-1,4). But instead of writing parentheses can I write them in such form ]-1,4[", + "A": "no you cannot. If you write it like that, then you re saying that -1 and 4 are not included in the function and the result would be a syntax error. three thumbs up!", + "video_name": "UJQkqV2zGv0", + "timestamps": [ + 306 + ], + "3min_transcript": "So these are all different ways of denoting or depicting the same interval. Let's do some more examples here. So let's-- Let me draw a number line again. So, a number line. And now let me do-- Let me just do an open interval. An open interval just so that we clearly can see the difference. Let's say that I want to talk about the values between negative one and four. Let me use a different color. So the values between negative one and four, but I don't want to include negative one and four. So this is going to be an open interval. So I'm not going to include four, and I'm not going to include negative one. Notice I have open circles here. Over here had closed circles, the closed circles told me that I included negative three and two. it's all the values in between negative one and four. Negative .999999 is going to be included, but negative one is not going to be included. And 3.9999999 is going to be included, but four is not going to be included. So how would we-- What would be the notation for this? Well, here we could say x is going to be a member of the real numbers such that negative one-- I'm not going to say less than or equal to because x can't be equal to negative one, so negative one is strictly less than x, is strictly less than four. Notice not less than or equal, because I can't be equal to four, four is not included. So that's one way to say it. Another way I could write it like this. x is a member of the real numbers such that x is a member of... Now the interval is from negative one to four but I'm not gonna use these brackets. but I'm not going to include them, so I'm going to put the parentheses right over here. Parentheses. So this tells us that we're dealing with an open interval. This right over here, let me make it clear, this is an open interval. Now you're probably wondering, okay, in this case both endpoints were included, it's a closed interval. In this case both endpoints were excluded, it's an open interval. Can you have things that have one endpoint included and one point excluded, and the answer is absolutely. Let's see an example of that. I'll get another number line here. Another number line. And let's say that we want to-- Actually, let me do it the other way around. Let me write it first, and then I'll graph it. So let's say we're thinking about all of the x's that are a member of the real numbers such that let's say negative four is not included, is less than x, is less than or equal to negative one." + }, + { + "Q": "I'm sort of curious about why you would write the inclusion of real all real numbers except for one as (-\u00e2\u0088\u009e,1). If you are including all real numbers except for one, would it not look like [-\u00e2\u0088\u009e,1)? @8:42", + "A": "you can t actually reach infinity, so use ( rather than {", + "video_name": "UJQkqV2zGv0", + "timestamps": [ + 522 + ], + "3min_transcript": "You could say, well hey, everything except for some values. Let me give another example. Let's get another example here. Let's say that we wanna talk about all the real numbers except for one. We want to include all of the real numbers. All of the real numbers except for one. Except for one, so we're gonna exclude one right over here, open circle, but it can be any other real number. So how would we denote this? Well, we could write x is a member of the real numbers such that x does not equal one. So here I'm saying x can be a member of the real numbers but x cannot be equal to one. It can be anything else, but it cannot be equal to one. You could say x is a member of the real numbers such that x is less than one, or x is greater than one. So you could write it just like that. Or you could do something interesting. This is the one that I would use, this is the shortest and it makes it very clear. You say hey, everything except for one. But you could even do something fancy, like you could say x is a member of the real numbers such that x is a member of the set going from negative infinity to one, not including one, or x is a member of the set going from-- or a member of the interval going from one, not including one, all the way to positive, all the way to positive infinity. And when we're talking about negative infinity or positive infinity, you always put a parentheses. all the way up to infinity. It needs to be at least open at that endpoint because infinity just keeps going on and on. So you always want to put a parentheses if you're talking about infinity or negative infinity. It's not really an endpoint, it keeps going on and on forever. So you use the notation for open interval, at least at that end, and notice we're not including, we're not including one either, so if x is a member of this interval or that interval, it essentially could be anything other than one. But this would have been the simplest notation to describe that." + }, + { + "Q": "At 0:50 can anyone tell me why the yellow line's slope is going to be the negative inverse?", + "A": "Because they are both perpendicular lines!!", + "video_name": "0671cRNjeKI", + "timestamps": [ + 50 + ], + "3min_transcript": "We are asked which of these lines are perpendicular. And it has to be perpendicular to one of the other lines, you can't be just perpendicular by yourself. And perpendicular line, just so you have a visualization for what for perpendicular lines look like, two lines are perpendicular if they intersect at right angles. So if this is one line right there, a perpendicular line will look like this. A perpendicular line will intersect it, but it won't just be any intersection, it will intersect at right angles. So these two lines are perpendicular. Now, if two lines are perpendicular, if the slope of this orange line is m-- so let's say its equation is y is equal to mx plus, let's say it's b 1, so it's some y-intercept-- then the equation of this yellow line, its slope is going to be the negative inverse of this guy. This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept. perpendicular, the product of their slopes is going to be negative 1. And so you could write that there. m times negative 1 over m, that's going to be-- these two guys are going to cancel out-- that's going to be equal to negative 1. So let's figure out the slopes of each of these lines and figure out if any of them are the negative inverse of any of the other ones. So line A, the slope is pretty easy to figure out, it's already in slope-intercept form, its slope is 3. So line A has a slope of 3. Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it. So let's do line B over here. Line B, we have x plus 3y is equal to negative 21. Let's subtract x from both sides so that it ends up on the right-hand side. So we end up with 3y is equal to negative x minus 21. and we get y is equal to negative 1/3 x minus 7. So this character's slope is negative 1/3. So here m is equal to negative 1/3. So we already see they are the negative You take the inverse of 3, it's 1/3, and then it's the negative of that. Or you take the inverse of negative 1/3, it's negative 3, and then this is the negative of that. So these two lines are definitely perpendicular. Let's see the third line over here. So line C is 3x plus y is equal to 10. If we subtract 3x from both sides, we get y is equal to" + }, + { + "Q": "at 5:08\nisn't this the formula of Explicit geometric sequence ?", + "A": "In a geometric sequence, the input value is multiplied. Here it is the power.", + "video_name": "G2WybA4Hf7Y", + "timestamps": [ + 308 + ], + "3min_transcript": "two to the first power. So it's going to be one-fourth two. What is h of two going to be equal to? Well, it's going to be one-fourth times two squared, so it's going to be times two times two. Or, we could just view this as this is going to be two times h of one. And actually I should have done this when I wrote this one out, but this we can write as two times h of zero. So notice, if we were to take the ratio between h of two and h of one, it would be two. If we were to take the ratio between h of one it would be two. That is the common ratio between successive whole number inputs into our function. So, h of I could say plus one over h of n is going to be equal to is going to be equal to actually I can work it out mathematically. One-fourth times two to the n plus one one-fourth two to the n. Two to the n plus one, divided by two to the n is just going to be equal to two. That is your common ratio. So for the function h. For the function f, our common ratio is three. If we were to go the other way around, if someone said, hey, I have some function whose initial value, so let's say, I have some function, I'll do this in a new color, I have some function, g, and we know that its initial initial value is five. And someone were to say its common ratio its common ratio is six, what would this exponential function look like? And they're telling you this is an exponential function. Well, g of let's say x is the input, is going to be equal to our initial value, which is five. That's not a negative sign there, Our initial value is five. And then times our common ratio to the x power. So once again, initial value, right over there, that's the five. And then our common ratio is the six, right over there. So hopefully that gets you a little bit familiar with some of the parts of an exponential function, why they are called what they are called." + }, + { + "Q": "sal at 1:08 instead of moving the 5 cant u move the 4 and then add 8 to the other side", + "A": "Yes but the way Sal did it is easier.", + "video_name": "g6nGcnVB8BM", + "timestamps": [ + 68 + ], + "3min_transcript": "We're asked to solve for x. So we have the square root of the entire quantity 5x squared minus 8 is equal to 2x. Now we already have an expression under a radical isolated, so the easiest first step here is really just to square both sides of this equation. So let's just square both sides of that equation. Now the left-hand side, if you square it, the square root of 5x squared minus 8 squared is going to be 5x squared minus 8. This is 5x squared minus 8. And then the right-hand side, 2x squared is the same thing as 2 squared times x squared or 4x squared. Now we have a quadratic. Now let's see what we can do to maybe simplify this a little bit more. Well, we could subtract 4x squared. Or actually, even better, let's subtract 5x squared from both sides so that we just have all our x terms on the right-hand side. So let's subtract 5x squared from both sides. Subtract 5x squared from both sides of the equation. We're just left with negative 8 is equal to 4x squared minus 5x squared, that's negative 1x squared. Or we could just write negative x squared, just like that. And then we could multiply both sides of this equation by negative 1. That'll make it into positive 8. Or I could divide by negative 1, however you want to view it. Negative 1 times that times negative 1. So we get positive 8 is equal to x squared. And now we could take the square root of both sides of this equation. So let's take the square root of both sides of this equation. The principal square root of both sides of this equation. And what do we get? We get, on the right-hand side, x is equal to the square root of 8. And 8 can be rewritten as 2 times 4. And this can be rewritten as the square root of 2 times the I don't like this green color so much. And what's the square root of 4, the principal root of 4? It's 2. So that right there is 2. So this side becomes 2, this 2, times the square root of 2. And that is equal to x. Now let's verify that this is the solution to our original equation. So let's substitute this in, first to the left-hand side. So on the left-hand side, we have 5 times 2 square roots of 2 squared minus 8. And then we're going to have to take the square root of that whole thing. So this is going to be equal to-- we're just focused on the left-hand side right now. This is equal to the square root of 5 times 2 squared, which is 4, times the square root of 2 squared, which is 2. And then minus 8." + }, + { + "Q": "at 3:11 could he use substitution instead of elimination?", + "A": "Of course. He can do anything.", + "video_name": "f7cX-Ar2cEM", + "timestamps": [ + 191 + ], + "3min_transcript": "So 3x times 5 is 15x, y times 5 is plus 5y, and then negative z times 5 is negative 5z-- that's the whole point and why we're multiplying it by 5-- is equal to 3 times 5, which is equal to 15. And so if we add these two equations, we get x plus 15x is 16x, 2y plus 5y is 7y, and 5z minus 5z or plus negative 5z, those are going to cancel out. And that is going to be equal to negative 17 plus 15 is negative 2. So we were able to use the constraints in that equation and that equation, and now we have an equation in just x and y. So let's try to do the same thing. Let's trying to eliminate the z's. But now I'll use this equation and this equation. So this equation-- let me just rewrite it over here. We have 2x minus 3y plus 2z is equal to negative 16. And now, so that this 2z gets eliminated, let's multiply this equation times 2. So let's multiply it times 2, so we'll have a negative 2z here to eliminate with the positive 2z. So 2 times 3x is 6x, 2 times y is plus 2y, and then 2 times negative z is negative 2z is equal to 2 times 3 is equal to 6. And now we can add these two equations. 2x plus 6x is 8x, negative 3y plus 2y is negative y, and then these two guys get canceled out. And then that is equal to negative 16 plus 6 is negative 10. So now we have two equations with two unknowns. We've eliminated the z's. And let's see, if we want to eliminate again, we have a negative y over here. We have a positive 7y. and add the two equations. So let's do that. So let's multiply this times 7. 7 times 8 is 56, so it's 56x minus 7y is equal to 7 times negative 10, is equal to negative 70. And now we can add these two equations. I'm now trying to eliminate the y's. So we have 16x plus 56x. That is 72x. So we have 72x, these guys eliminate, equal to negative 72, negative 2 plus negative 70. Divide both sides by 72, and we get x is equal to negative 1. And now we just have to substitute back to figure out what y and z are equal to." + }, + { + "Q": "what does sal mean when he says at 0:21 when he says \"scaling up\"?", + "A": "By scaling up Sal means multiplying the equation by 2. This scales up all the values in the equation by 2.", + "video_name": "f7cX-Ar2cEM", + "timestamps": [ + 21 + ], + "3min_transcript": "Solve this system. And once again, we have three equations with three unknowns. So this is essentially trying to figure out where three different planes would intersect in three dimensions. And to do this, if we want to do it by elimination, if we want to be able to eliminate variables, it looks like, well, it looks like we have a negative z here. We have a plus 2z. We have a 5z over here. If we were to scale up this third equation by positive 2, then you would have a negative 2z here, and it would cancel out with this 2z there. And then if you were to scale it up by 5, you'd have a negative 5z here, and then that could cancel out with that 5z over there. So let's try to cancel out. Let's try to eliminate the z's first. So let me start with this equation up here. I'll just rewrite it. So we have-- I'll draw an arrow over here-- we have x plus 2y plus 5z is equal to negative 17. And then to cancel out or to eliminate the z's, I'll multiply this equation here times 5. So I'm going to multiply this equation times 5. So 3x times 5 is 15x, y times 5 is plus 5y, and then negative z times 5 is negative 5z-- that's the whole point and why we're multiplying it by 5-- is equal to 3 times 5, which is equal to 15. And so if we add these two equations, we get x plus 15x is 16x, 2y plus 5y is 7y, and 5z minus 5z or plus negative 5z, those are going to cancel out. And that is going to be equal to negative 17 plus 15 is negative 2. So we were able to use the constraints in that equation and that equation, and now we have an equation in just x and y. So let's try to do the same thing. Let's trying to eliminate the z's. But now I'll use this equation and this equation. So this equation-- let me just rewrite it over here. We have 2x minus 3y plus 2z is equal to negative 16. And now, so that this 2z gets eliminated, let's multiply this equation times 2. So let's multiply it times 2, so we'll have a negative 2z here to eliminate with the positive 2z. So 2 times 3x is 6x, 2 times y is plus 2y, and then 2 times negative z is negative 2z is equal to 2 times 3 is equal to 6. And now we can add these two equations. 2x plus 6x is 8x, negative 3y plus 2y is negative y, and then these two guys get canceled out. And then that is equal to negative 16 plus 6 is negative 10. So now we have two equations with two unknowns. We've eliminated the z's. And let's see, if we want to eliminate again, we have a negative y over here. We have a positive 7y." + }, + { + "Q": "at 3:10 isn't pi 3.14 not 3.5 ?", + "A": "correct pi is 3.14 rounded to two decimal places. But Sal is showing that 4*pi is less than 14. Since 4*3.5 is 14, and pi is less than 3.5 then 4*pi must be less than 14. Hope that helps", + "video_name": "EvvxBdNIUeQ", + "timestamps": [ + 190 + ], + "3min_transcript": "Well, area is equal to pi r squared for a circle, where r is a radius. They gave us the diameter. The radius is half of that. So the radius here is going to be half this distance, or 2r. So the area of our circle is going to be pi times 2r, the whole thing squared. This is the radius, right? So we're squaring the entire radius. So this is going to be equal to pi times 4 times r squared. I'm just squaring each of these terms. Or if we were to change the order, the area of the circle is equal to 4 pi r squared. And we want to find the difference. So to find a difference, It's helpful-- just so we don't end up with a negative number-- to figure out which of these two is larger. So they're telling us that p is greater than 7 r. If p is greater than 7r, then 2-- let me write it this way. We know that p is greater than 7r. So if we're going to multiply both sides of this equation by 2rr-- and 2r is positive, we're dealing with positive distances, positive lengths-- so if we multiply both sides of this equation by 2r, it shouldn't change the equation. So multiply that by 2r, and then multiply this by 2r. And then our equation becomes 2rp is greater than 14r squared. Now, why is this interesting? Actually, why did I even multiply this by 2r? Well, that's so that this becomes the same as the area of the rectangle. So this is the area of the rectangle. And what's 14r squared? Well, 4 times pi, is going to get us something less than 14. So this is 4 pi is less than 14. 14 is 4 times 3 and 2-- let me put it this way. 4 times 3.5 is equal to 14. So 4 times pi, which is less than 3.5, is going to be less than 14. So we know that this over here is larger than this quantity over here. It's larger than 4 pi r squared. And so we know that this rectangle has a larger area than the circle. So we can just subtract the circle's area from the rectangle's area to find the difference. So the difference is going to be the area of the rectangle, which we already figured out is 2rp. And we're going to subtract from that the area of the circle. The area of the circle is 4 pi r squared." + }, + { + "Q": "i didnt understand from 1:07-1:47 about the circle", + "A": "it is about pi from 1:07 to 1:47 about the circle", + "video_name": "EvvxBdNIUeQ", + "timestamps": [ + 67, + 107 + ], + "3min_transcript": "Write a binomial to express the difference between the area of a rectangle with length p and width 2r and the area of a circle with diameter 4r. And they tell us that p is greater than 7r. So let's first think about the area of a rectangle with length p and width 2r. So this is our rectangle right here. It has a length of p and it has a width of 2r. So what's its area? Well, it's just going to be the length times the width. So the area here is going to be p-- or maybe I should say 2rp. This is the length times the width, or the width times the length. So area is equal to 2rp for the rectangle. Now, we also want to find the difference between this area and the area of a circle with diameter 4r. So what's the area of the circle going to be? So let me draw our circle over here. So our circle looks like that. Its diameter is 4r. Well, area is equal to pi r squared for a circle, where r is a radius. They gave us the diameter. The radius is half of that. So the radius here is going to be half this distance, or 2r. So the area of our circle is going to be pi times 2r, the whole thing squared. This is the radius, right? So we're squaring the entire radius. So this is going to be equal to pi times 4 times r squared. I'm just squaring each of these terms. Or if we were to change the order, the area of the circle is equal to 4 pi r squared. And we want to find the difference. So to find a difference, It's helpful-- just so we don't end up with a negative number-- to figure out which of these two is larger. So they're telling us that p is greater than 7 r. If p is greater than 7r, then 2-- let me write it this way. We know that p is greater than 7r. So if we're going to multiply both sides of this equation by 2rr-- and 2r is positive, we're dealing with positive distances, positive lengths-- so if we multiply both sides of this equation by 2r, it shouldn't change the equation. So multiply that by 2r, and then multiply this by 2r. And then our equation becomes 2rp is greater than 14r squared. Now, why is this interesting? Actually, why did I even multiply this by 2r? Well, that's so that this becomes the same as the area of the rectangle. So this is the area of the rectangle. And what's 14r squared? Well, 4 times pi, is going to get us something less than 14." + }, + { + "Q": "At 2:42 why is it 14 r squared?", + "A": "Because when you times a r by a r you get r squared and 7 times 2 you get 14 so 14r^2", + "video_name": "EvvxBdNIUeQ", + "timestamps": [ + 162 + ], + "3min_transcript": "Well, area is equal to pi r squared for a circle, where r is a radius. They gave us the diameter. The radius is half of that. So the radius here is going to be half this distance, or 2r. So the area of our circle is going to be pi times 2r, the whole thing squared. This is the radius, right? So we're squaring the entire radius. So this is going to be equal to pi times 4 times r squared. I'm just squaring each of these terms. Or if we were to change the order, the area of the circle is equal to 4 pi r squared. And we want to find the difference. So to find a difference, It's helpful-- just so we don't end up with a negative number-- to figure out which of these two is larger. So they're telling us that p is greater than 7 r. If p is greater than 7r, then 2-- let me write it this way. We know that p is greater than 7r. So if we're going to multiply both sides of this equation by 2rr-- and 2r is positive, we're dealing with positive distances, positive lengths-- so if we multiply both sides of this equation by 2r, it shouldn't change the equation. So multiply that by 2r, and then multiply this by 2r. And then our equation becomes 2rp is greater than 14r squared. Now, why is this interesting? Actually, why did I even multiply this by 2r? Well, that's so that this becomes the same as the area of the rectangle. So this is the area of the rectangle. And what's 14r squared? Well, 4 times pi, is going to get us something less than 14. So this is 4 pi is less than 14. 14 is 4 times 3 and 2-- let me put it this way. 4 times 3.5 is equal to 14. So 4 times pi, which is less than 3.5, is going to be less than 14. So we know that this over here is larger than this quantity over here. It's larger than 4 pi r squared. And so we know that this rectangle has a larger area than the circle. So we can just subtract the circle's area from the rectangle's area to find the difference. So the difference is going to be the area of the rectangle, which we already figured out is 2rp. And we're going to subtract from that the area of the circle. The area of the circle is 4 pi r squared." + }, + { + "Q": "At 3:55 What does he mean by principle root?", + "A": "The principal root is just the positive square root. For instance: \u00e2\u0088\u009a9 = \u00c2\u00b13 But the principal square root of 9 is just 3.", + "video_name": "McINBOFCGH8", + "timestamps": [ + 235 + ], + "3min_transcript": "are equal. So these are the two sides that are equal. And then if the two sides are equal, we have proved to ourselves that the base angles are equal. And if we called the measure of these base angles x, then we know that x plus x plus 90 have to be equal to 180. Or if we subtract 90 from both sides, you get x plus x is equal to 90 or 2x is equal to 90. Or if you divide both sides by 2, you get x is equal to 45 degrees. So a right isosceles triangle can also be called-- and this is the more typical name for it-- it can also be called a 45-45-90 triangle. And what I want to do this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle. Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x. And then we can use the Pythagorean Theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the square of length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. We can take the principal root of both sides of that. I wanted to just change it to yellow. Last, take the principal root of both sides of that. The left-hand side you get, principal root of 2 is just square root of 2, and then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2 So if you have a right isosceles triangle, whatever the two legs are, they're going to have the same length. That's why it's isosceles. The hypotenuse is going to be square root of 2 times that. So c is equal to x times the square root of 2. So for example, if you have a triangle that looks like this. Let me draw it a slightly different way. It's good to have to orient ourselves in different ways every time. So if we see a triangle that's 90 degrees, 45 and 45 like this, and you really just have to know two of these angles to know what the other one is going to be, and if I tell you that this side right over here is 3-- I actually don't even have to tell you that this other side's going to be 3. This is an isosceles triangle, so those two legs are going to be the same. And you won't even have to apply the Pythagorean theorem if you know this-- and this is a good one to know-- that the hypotenuse here, the side opposite the 90 degree side, is just going to be square root of 2 times the length of either of the legs. So it's going to be 3 times the square root of 2." + }, + { + "Q": "At 4:28 wouldn't it be 2x=C", + "A": "You had the equation x^2+x^2 = c^2. Which is 2(x^2)=c^2 which is 2*x*x=c*c When you take the square root of each side, you have to take the square root of both the x*x and the square root of 2. so you get (Square root of 2 * x)=c", + "video_name": "McINBOFCGH8", + "timestamps": [ + 268 + ], + "3min_transcript": "are equal. So these are the two sides that are equal. And then if the two sides are equal, we have proved to ourselves that the base angles are equal. And if we called the measure of these base angles x, then we know that x plus x plus 90 have to be equal to 180. Or if we subtract 90 from both sides, you get x plus x is equal to 90 or 2x is equal to 90. Or if you divide both sides by 2, you get x is equal to 45 degrees. So a right isosceles triangle can also be called-- and this is the more typical name for it-- it can also be called a 45-45-90 triangle. And what I want to do this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle. Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x. And then we can use the Pythagorean Theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the square of length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. We can take the principal root of both sides of that. I wanted to just change it to yellow. Last, take the principal root of both sides of that. The left-hand side you get, principal root of 2 is just square root of 2, and then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2 So if you have a right isosceles triangle, whatever the two legs are, they're going to have the same length. That's why it's isosceles. The hypotenuse is going to be square root of 2 times that. So c is equal to x times the square root of 2. So for example, if you have a triangle that looks like this. Let me draw it a slightly different way. It's good to have to orient ourselves in different ways every time. So if we see a triangle that's 90 degrees, 45 and 45 like this, and you really just have to know two of these angles to know what the other one is going to be, and if I tell you that this side right over here is 3-- I actually don't even have to tell you that this other side's going to be 3. This is an isosceles triangle, so those two legs are going to be the same. And you won't even have to apply the Pythagorean theorem if you know this-- and this is a good one to know-- that the hypotenuse here, the side opposite the 90 degree side, is just going to be square root of 2 times the length of either of the legs. So it's going to be 3 times the square root of 2." + }, + { + "Q": "At 4:59, why is it bn and vn, instead of bk and vk?", + "A": "There s no error here. b belongs to Rn, otherwise original equation Ax=b, where A is nxk matrix would not make any sense. v is a projection of b to the column space, but it still is a member of Rn, hence it also has n components.", + "video_name": "MC7l96tW8V8", + "timestamps": [ + 299 + ], + "3min_transcript": "Or another way to view it, when I say close, I'm talking about length, so I want to minimize the length of-- let me write this down. I want to minimize the length of b minus A times x-star. Now, some of you all might already know where this is going. But when you take the difference between 2 and then take its length, what does that look like? Let me just call Ax. Ax is going to be a member of my column space. Let me just call that v. Ax is equal to v. You multiply any vector in Rk times your matrix A, you're So any Ax is going to be in your column space. And maybe that is the vector v is equal to A times x-star. And we want this vector to get as close as possible to this as long as it stays-- I mean, it has to be in my column space. But we want the distance between this vector and this vector to be minimized. Now, I just want to show you where the terminology for this will come from. I haven't given it its proper title yet. If you were to take this vector-- let just call this vector v for simplicity-- that this is equivalent to the length of the vector. You take the difference between each of the elements. So b1 minus v1, b2 minus v2, all the way to bn minus vn. thing as this. This is going to be equal to the square root. Let me take the length squared, actually. The length squared of this is just going to be b1 minus v1 squared plus b2 minus v2 squared plus all the way to bn minus vn squared. And I want to minimize this. So I want to make this value the least value that it can be possible, or I want to get the least squares estimate here. And that's why, this last minute or two when I was just explaining this, that was just to give you the motivation for why this right here is called the least squares estimate, or the least squares solution, or the least squares approximation for the equation Ax equals b. There is no solution to this, but maybe we can find some" + }, + { + "Q": "@ 4:05 he says it goes down 2/3's then he says its 1 and 1/3. How does he go from 2/3's to 1/3rd? I get the whole thing that 1/ 1/3rd makes 4/3rds, but he went down 2/3rds??", + "A": "When he went down 2/3 it was from 2! The line was 2/3 below 2. So if you look at it in terms of the line being above 1 the line is actually 1/3 above 1. So therefore the line is at 1 1/3.", + "video_name": "9wOalujeZf4", + "timestamps": [ + 245 + ], + "3min_transcript": "That's our end point. That's our starting point. So if you simplify this, b minus b is 0. 1 minus 0 is 1. So you get m/1, or you get it's equal to m. So hopefully you're satisfied and hopefully I didn't confuse you by stating it in the abstract with all of these variables here. But this is definitely going to be the slope and this is definitely going to be the y-intercept. Now given that, what I want to do in this exercise is look at these graphs and then use the already drawn graphs to figure out the equation. So we're going to look at these, figure out the slopes, figure out the y-intercepts and then know the equation. So let's do this line A first. So what is A's slope? Let's start at some arbitrary point. Let's start right over there. We want to get even numbers. If we run one, two, three. One, two, three. Our delta y-- and I'm just doing it because I want to hit an even number here-- our delta y is equal to-- we go down by 2-- it's equal to negative 2. So for A, change in y for change in x. When our change in x is 3, our change in y is negative 2. So our slope is negative 2/3. When we go over by 3, we're going to go down by 2. Or if we go over by 1, we're going to go down by 2/3. You can't exactly see it there, but you definitely see it when you go over by 3. So that's our slope. We've essentially done half of that problem. Now we have to figure out the y-intercept. So that right there is our m. Now what is our b? Our y-intercept. Well where does this intersect the y-axis? Well we already said the slope is 2/3. So this is the point y is equal to 2. gone down by 2/3. So this right here must be the point 1 1/3. Or another way to say it, we could say it's 4/3. That's the point y is equal to 4/3. Right there. A little bit more than 1. About 1 1/3. So we could say b is equal to 4/3. So we'll know that the equation is y is equal to m, negative 2/3, x plus b, plus 4/3. That's equation A. Let's do equation B. Hopefully we won't have to deal with as many fractions here. Equation B. Let's figure out its slope first. Let's start at some reasonable point. We could start at that point. Let me do it right here. B. Equation B. When our delta x is equal to-- let me write it this way, delta x. So our delta x could be 1. When we move over 1 to the right, what happens" + }, + { + "Q": "At about 10:10, he said that the y-intercept was 0. So why does the graph line still have a slope?", + "A": "The y-intercept doesn t have to do with the slope; it just shows where the sloped line crosses the y-axis. So, a graph line can have a slope and y-intercept 0, such as y=x, where the slope is 1 and the y-int 0. The line just crosses the y-intercept at the origin.", + "video_name": "9wOalujeZf4", + "timestamps": [ + 610 + ], + "3min_transcript": "So it's one, two, three, four, five, six. That's our y-intercept when x is equal to 0. This tells us that for every 5 we move to the right, we move down 1. We can view this as negative 1/5. The delta y over delta x is equal to negative 1/5. For every 5 we move to the right, we move down 1. So every 5. One, two, three, four, five. We moved 5 to the right. That means we must move down 1. We move 5 to the right. One, two, three, four, five. We must move down 1. If you go backwards, if you move 5 backwards-- instead of this, if you view this as 1 over negative 5. These are obviously equivalent numbers. If you go back 5-- that's negative 5. One, two, three, four, five. Then you move up 1. If you go back 5-- one, two, three, four, five-- you move up 1. I have to just connect the dots. I think you get the idea. I just have to connect those dots. I could've drawn it a little bit straighter. Now let's do this one, y is equal to negative x. Where's the b term? I don't see any b term. You remember we're saying y is equal to mx plus b. Where is the b? Well, the b is 0. You could view this as plus 0. Here is b is 0. When x is 0, y is 0. That's our y-intercept, right there at the origin. And then the slope-- once again you see a negative sign. You could view that as negative 1x plus 0. So slope is negative 1. When you move to the right by 1, when change in x is 1, change in y is negative 1. When you move up by 1 in x, you go down by 1 in y. Or if you go down by 1 in x, you're going to go up by 1 in y. x and y are going to have opposite signs. They go in opposite directions. So the line is going to look like that. fourth quadrants. Now I'll do one more. Let's do this last one right here. y is equal to 3.75. Now you're saying, gee, we're looking for y is equal to mx plus b. Where is this x term? It's completely gone. Well the reality here is, this could be rewritten as y is equal to 0x plus 3.75. Now it makes sense. The slope is 0. No matter how much we change our x, y does not change. Delta y over delta x is equal to 0. I don't care how much you change your x. Our y-intercept is 3.75. So 1, 2, 3.75 is right around there. You want to get close. 3 3/4. As I change x, y will not change. y is always going to be 3.75. It's just going to be a horizontal line at" + }, + { + "Q": "Hold on.. how did he get 4/3 as the y-intercept in 4:05..?", + "A": "The slope = -2/3. Sal has a point at (-1, 2). He needs to move one uni to the right to get to the y-axis. The slope tells him that as he moves 1 unit to the right, the y-coordinate will decrease by 2/3. Sal finds the y-intercept by doing: 2 - 2/3 = 6/3 - 2/3 = 4/3 Hope this helps.", + "video_name": "9wOalujeZf4", + "timestamps": [ + 245 + ], + "3min_transcript": "That's our end point. That's our starting point. So if you simplify this, b minus b is 0. 1 minus 0 is 1. So you get m/1, or you get it's equal to m. So hopefully you're satisfied and hopefully I didn't confuse you by stating it in the abstract with all of these variables here. But this is definitely going to be the slope and this is definitely going to be the y-intercept. Now given that, what I want to do in this exercise is look at these graphs and then use the already drawn graphs to figure out the equation. So we're going to look at these, figure out the slopes, figure out the y-intercepts and then know the equation. So let's do this line A first. So what is A's slope? Let's start at some arbitrary point. Let's start right over there. We want to get even numbers. If we run one, two, three. One, two, three. Our delta y-- and I'm just doing it because I want to hit an even number here-- our delta y is equal to-- we go down by 2-- it's equal to negative 2. So for A, change in y for change in x. When our change in x is 3, our change in y is negative 2. So our slope is negative 2/3. When we go over by 3, we're going to go down by 2. Or if we go over by 1, we're going to go down by 2/3. You can't exactly see it there, but you definitely see it when you go over by 3. So that's our slope. We've essentially done half of that problem. Now we have to figure out the y-intercept. So that right there is our m. Now what is our b? Our y-intercept. Well where does this intersect the y-axis? Well we already said the slope is 2/3. So this is the point y is equal to 2. gone down by 2/3. So this right here must be the point 1 1/3. Or another way to say it, we could say it's 4/3. That's the point y is equal to 4/3. Right there. A little bit more than 1. About 1 1/3. So we could say b is equal to 4/3. So we'll know that the equation is y is equal to m, negative 2/3, x plus b, plus 4/3. That's equation A. Let's do equation B. Hopefully we won't have to deal with as many fractions here. Equation B. Let's figure out its slope first. Let's start at some reasonable point. We could start at that point. Let me do it right here. B. Equation B. When our delta x is equal to-- let me write it this way, delta x. So our delta x could be 1. When we move over 1 to the right, what happens" + }, + { + "Q": "At 1:10, is that a different way of expressing it using a sigma?", + "A": "In this video, he is discussing sequences, which are just lists of numbers. The sigma, or summation notation comes into play when you are calculating series.", + "video_name": "_cooC3yG_p0", + "timestamps": [ + 70 + ], + "3min_transcript": "What I want to do in this video is familiarize ourselves with a very common class of sequences. And this is arithmetic sequences. And they are usually pretty easy to spot. They are sequences where each term is a fixed number larger than the term before it. So my goal here is to figure out which of these sequences are arithmetic sequences. And then just so that we have some practice with some of the sequence notation, I want to define them either as explicit functions of the term you're looking for, the index you're looking at, or as recursive definitions. So first, given that an arithmetic sequence is one where each successive term is a fixed amount larger than the previous one, which of these are arithmetic sequences? Well let's look at this first one right over here. To go from negative 5 to negative 3, we had to add 2. Then to go from negative 3 to negative 1, you have to add 2. Then to go from negative 1 to 1, you had to add 2. So this is clearly an arithmetic sequence. And there are several ways that we could define the sequence. We could say it's a sub n. And you don't always have to use k. This time I'll use n to denote our index. From n equals 1 to infinity with-- and there's two ways we could define it. We could either define it explicitly, or we could define it recursively. So if we wanted to define it explicitly, we could write a sub n is equal to whatever the first term is. In this case, our first term is negative 5. It's equal to negative 5 plus-- we're going to add 2 one less times than the term we're at. So for the second term, we add 2 once. For the third term, we add 2 twice. For the fourth term, from our base term, we added 2 three times. So we're going to add 2. We're going to add positive 2 one less than the index that we're looking at-- n minus 1 times. So this is an explicit definition of this arithmetic sequence. I could say a sub 1 is equal to negative 5. And then each successive term, for a sub 2 and greater-- so I could say a sub n is equal to a sub n minus 1 plus 3. Each term is equal to the previous term-- oh, not 3-- plus 2. So this is for n is greater than or equal to 2. So either of these are completely legitimate ways of defining the arithmetic sequence that we have here. We can either define it explicitly, or we could define it recursively. Now let's look at this sequence. Is this one arithmetic? Well, we're going from 100. We add 7. 107 to 114, we're adding 7. 114 to 121, we are adding 7. So this is indeed an arithmetic sequence. So just to be clear, this is one," + }, + { + "Q": "I dont understand the part at 6:61", + "A": "Do you mean 7:11? Basically, to prove the quadratic formula Sal is completing the square and then moving everything to one side of the equation.", + "video_name": "r3SEkdtpobo", + "timestamps": [ + 421 + ], + "3min_transcript": "denominator by 4a. In fact, the 4's cancel out and then this a cancels out and you just have a c over a. So these, this and that are equivalent. I just switched which I write first. And you might already be seeing the beginnings of the quadratic formula here. So this I can rewrite. This I can rewrite. The right-hand side, right here, I can rewrite as b squared minus 4ac, all of that over 4a squared. This is looking very close. Notice, b squared minus 4ac, it's already appearing. We don't have a square root yet, but we haven't taken the square root of both sides of this equation, so let's do that. So if you take the square root of both sides, the left-hand side will just become x plus-- let me scroll down a little bit-- x plus b over 2a is going to be equal to the plus or minus square root of this thing. numerator over the square root of the denominator. So it's going to be the plus or minus the square root of b squared minus 4ac over the square root of 4a squared. Now, what is the square root of 4a squared? It is 2a, right? 2a squared is 4a squared. The square root of this is that right here. So to go from here to here, I just took the square root of both sides of this equation. Now, this is looking very close to the quadratic. We have a b squared minus 4ac over 2a, now we just essentially have to subtract this b over 2a from both sides of the equation and we're done. So let's do that. So if you subtract the b over 2a from both sides of this equation, what do you get? You get x is equal to negative b over 2a, plus or minus the square root of b squared minus 4ac over 2a, common Let me do this in a new color. So it's orange. Negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And we are done! By completing the square with just general coefficients in front of our a, b and c, we were able to derive the quadratic formula. Just like that. Hopefully you found that as entertaining as I did." + }, + { + "Q": "At 6:00, why did he simply not just take the square root of b squared and leave it as b?", + "A": "Because you can t distribute square roots like that. You could if the terms in a radical are products or quotients, but when they are sums and differences(i.e. adding and subtracting), you can t distribute a radical.", + "video_name": "r3SEkdtpobo", + "timestamps": [ + 360 + ], + "3min_transcript": "So the left-hand side simplifies to this. The right-hand side, maybe not quite as simple. Maybe we'll leave it the way it is right now. Actually, let's simplify it a little bit. So the right-hand side, we can rewrite it. This is going to be equal to-- well, this is going to be b squared. I'll write that term first. This is b-- let me do it in green so we can follow along. So that term right there can be written as b squared over 4a square. And what's this term? What would that become? This would become-- in order to have 4a squared as the denominator, we have to multiply the numerator and the denominator by 4a. So this term right here will become minus 4ac over 4a squared. And you can verify for yourself that that is the same thing as that. denominator by 4a. In fact, the 4's cancel out and then this a cancels out and you just have a c over a. So these, this and that are equivalent. I just switched which I write first. And you might already be seeing the beginnings of the quadratic formula here. So this I can rewrite. This I can rewrite. The right-hand side, right here, I can rewrite as b squared minus 4ac, all of that over 4a squared. This is looking very close. Notice, b squared minus 4ac, it's already appearing. We don't have a square root yet, but we haven't taken the square root of both sides of this equation, so let's do that. So if you take the square root of both sides, the left-hand side will just become x plus-- let me scroll down a little bit-- x plus b over 2a is going to be equal to the plus or minus square root of this thing. numerator over the square root of the denominator. So it's going to be the plus or minus the square root of b squared minus 4ac over the square root of 4a squared. Now, what is the square root of 4a squared? It is 2a, right? 2a squared is 4a squared. The square root of this is that right here. So to go from here to here, I just took the square root of both sides of this equation. Now, this is looking very close to the quadratic. We have a b squared minus 4ac over 2a, now we just essentially have to subtract this b over 2a from both sides of the equation and we're done. So let's do that. So if you subtract the b over 2a from both sides of this equation, what do you get? You get x is equal to negative b over 2a, plus or minus the square root of b squared minus 4ac over 2a, common" + }, + { + "Q": "At 5:45, when we take square root of both sides, why does the right side end up with a \"plus or minus\" designation while the left side stays only positive?", + "A": "| Because the equation on the left has implied positive domain because it is represented with a (term)^2 and that represents that any term within the brackets which were positive or negative will result in a positive output. While on the right, we do not yet know the domain of the right so it is correct to assume that there could be a negative root. I m sorry if i have confused you. :(", + "video_name": "r3SEkdtpobo", + "timestamps": [ + 345 + ], + "3min_transcript": "So the left-hand side simplifies to this. The right-hand side, maybe not quite as simple. Maybe we'll leave it the way it is right now. Actually, let's simplify it a little bit. So the right-hand side, we can rewrite it. This is going to be equal to-- well, this is going to be b squared. I'll write that term first. This is b-- let me do it in green so we can follow along. So that term right there can be written as b squared over 4a square. And what's this term? What would that become? This would become-- in order to have 4a squared as the denominator, we have to multiply the numerator and the denominator by 4a. So this term right here will become minus 4ac over 4a squared. And you can verify for yourself that that is the same thing as that. denominator by 4a. In fact, the 4's cancel out and then this a cancels out and you just have a c over a. So these, this and that are equivalent. I just switched which I write first. And you might already be seeing the beginnings of the quadratic formula here. So this I can rewrite. This I can rewrite. The right-hand side, right here, I can rewrite as b squared minus 4ac, all of that over 4a squared. This is looking very close. Notice, b squared minus 4ac, it's already appearing. We don't have a square root yet, but we haven't taken the square root of both sides of this equation, so let's do that. So if you take the square root of both sides, the left-hand side will just become x plus-- let me scroll down a little bit-- x plus b over 2a is going to be equal to the plus or minus square root of this thing. numerator over the square root of the denominator. So it's going to be the plus or minus the square root of b squared minus 4ac over the square root of 4a squared. Now, what is the square root of 4a squared? It is 2a, right? 2a squared is 4a squared. The square root of this is that right here. So to go from here to here, I just took the square root of both sides of this equation. Now, this is looking very close to the quadratic. We have a b squared minus 4ac over 2a, now we just essentially have to subtract this b over 2a from both sides of the equation and we're done. So let's do that. So if you subtract the b over 2a from both sides of this equation, what do you get? You get x is equal to negative b over 2a, plus or minus the square root of b squared minus 4ac over 2a, common" + }, + { + "Q": "At 4:00 and 10:00, why is is cos(2a) a minus but sin(2a) is a plus?\n\nAlso which video did I miss? I am so confused here. I don't ever remember learning this in my high school precalc class.", + "A": "You have to review the formula of cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b) From there, you can write cos(2a) as cos(a+a), then go on. You see why the minus in here. That is trigonometry. Go back to the beginning and learn the formula come from.", + "video_name": "a70-dYvDJZY", + "timestamps": [ + 240, + 600 + ], + "3min_transcript": "Because cosine of minus c is the same thing as the cosine of c. Times the cosine of c. And then, minus the sine of c. Instead of writing this, I could write this. Minus the sine of c times the cosine of a. So that we kind of pseudo proved this by knowing this and this ahead of time. Fair enough. And I'm going to use all of these to kind of prove a bunch of more trig identities that I'm going to need. So the other trig identity is that the cosine of a plus b is equal to the cosine of a-- you don't mix up the cosines and the sines in this situation. Cosine of a times the sine of b. And this is minus-- well, sorry. I just said you don't mix it up and then I mixed them up. Times the cosine of b minus sine of a times the sine of b. well, you use these same properties. Cosine of minus b, that's still going to be cosine on b. So that's going to be the cosine of a times the cosine-- cosine of minus b is the same thing as cosine of b. But here you're going to have sine of minus b, which is the same thing as the minus sine of b. And that minus will cancel that out, so it'll be plus sine of a times the sine of b. When you have a plus sign here you get a minus there. When you don't minus sign there, you get a plus sign there. But fair enough. I don't want to dwell on that too much because we have many more identities to show. So what if I wanted an identity for let's say, the cosine of 2a? So the cosine of 2a. Well that's just the same thing as the cosine of a plus a. If my second a is just my b, then this is just equal to cosine of a times the cosine of a minus the sine of a times the sine of a. My b is also an a in this situation, which I could rewrite as, this is equal to the cosine squared of a. I just wrote cosine of a times itself twice or times itself. Minus sine squared of a. This is one I guess identity already. Cosine of 2a is equal to the cosine squared of a minus the sine squared of a. Let me box off my identities that we're showing in this video. So I just showed you that one. What if I'm not satisfied? What if I just want it in terms of cosines? Well, we could break out the unit circle definition of our trig functions. This is kind of the most fundamental identity. The sine squared of a plus the cosine squared" + }, + { + "Q": "at 8:22 how did he get 1-2sin^2(a)?", + "A": "It s one of your identities. cos(2a) = cos^2(a) - sin^2(a) sin^2(a) + cos^2(a) = 1 => cos^2(a) = 1 - sin^2(a) cos(2a) = 1 - sin^2(a) - sin^2(a) cos(2a) = 1 - 2sin^2(a)", + "video_name": "a70-dYvDJZY", + "timestamps": [ + 502 + ], + "3min_transcript": "Now what if I wanted to get an identity that gave me what cosine squared of a is in terms of this? Well we could just solve for that. If we add 1 to both sides of this equation, actually, let me write this. This is one of our other identities. But if we add 1 to both sides of that equation we get 2 times the cosine squared of a is equal to cosine of 2a plus 1. And if we divide both sides of this by 2 we get the cosine squared of a is equal to 1/2-- now we could rearrange these just to do it-- times 1 plus the cosine of 2a. And we're done. And we have another identity. Cosine squared of a, sometimes it's called the power reduction Now what if we wanted something in terms of the sine squared of a? Well then maybe we could go back up here and we know from this identity that the sine squared of a is equal to 1 minus cosine squared of a. Or we could have gone the other way. We could have subtracted sine squared of a from both sides and we could have gotten-- let me go down there. If I subtracted sine squared of a from both sides you could get cosine squared of a is equal to 1 minus sine squared of a. And then we could go back into this formula right up here and we could write down-- and I'll do it in this blue color. We could write down that the cosine of 2a is equal to-- instead of writing a cosine squared of a, I'll write this- is equal to 1 minus sine squared of a minus sine squared of a. So my cosine of 2a is equal to? sine squared of a. So I have 1 minus 2 sine squared of a. So here's another identity. Another way to write my cosine of 2a. We're discovering a lot of ways to write our cosine of 2a. Now if we wanted to solve for sine squared of 2a we could add it to both sides of the equation. So let me do that and I'll just write it here for the sake of saving space. Let me scroll down a little bit. So I'm going to go here. If I just add 2 sine squared of a to both sides of this, I get 2 sine squared of a plus cosine of 2a is equal to 1. Subtract cosine of 2a from both sides. You get 2 sine squared of a is equal to 1 minus cosine of 2a." + }, + { + "Q": "At 4:25 isn't cos(2a) = cos(3a - a) = cos(3a)*(cos)a + sin(3a) * sin(a)", + "A": "You could work it out like that, which would eventually simplify to what he had: cos(2a)=cos^2(a)-sin^2(a); this is one of the double angle formulas", + "video_name": "a70-dYvDJZY", + "timestamps": [ + 265 + ], + "3min_transcript": "Because cosine of minus c is the same thing as the cosine of c. Times the cosine of c. And then, minus the sine of c. Instead of writing this, I could write this. Minus the sine of c times the cosine of a. So that we kind of pseudo proved this by knowing this and this ahead of time. Fair enough. And I'm going to use all of these to kind of prove a bunch of more trig identities that I'm going to need. So the other trig identity is that the cosine of a plus b is equal to the cosine of a-- you don't mix up the cosines and the sines in this situation. Cosine of a times the sine of b. And this is minus-- well, sorry. I just said you don't mix it up and then I mixed them up. Times the cosine of b minus sine of a times the sine of b. well, you use these same properties. Cosine of minus b, that's still going to be cosine on b. So that's going to be the cosine of a times the cosine-- cosine of minus b is the same thing as cosine of b. But here you're going to have sine of minus b, which is the same thing as the minus sine of b. And that minus will cancel that out, so it'll be plus sine of a times the sine of b. When you have a plus sign here you get a minus there. When you don't minus sign there, you get a plus sign there. But fair enough. I don't want to dwell on that too much because we have many more identities to show. So what if I wanted an identity for let's say, the cosine of 2a? So the cosine of 2a. Well that's just the same thing as the cosine of a plus a. If my second a is just my b, then this is just equal to cosine of a times the cosine of a minus the sine of a times the sine of a. My b is also an a in this situation, which I could rewrite as, this is equal to the cosine squared of a. I just wrote cosine of a times itself twice or times itself. Minus sine squared of a. This is one I guess identity already. Cosine of 2a is equal to the cosine squared of a minus the sine squared of a. Let me box off my identities that we're showing in this video. So I just showed you that one. What if I'm not satisfied? What if I just want it in terms of cosines? Well, we could break out the unit circle definition of our trig functions. This is kind of the most fundamental identity. The sine squared of a plus the cosine squared" + }, + { + "Q": "1:25 How are two inscribed angles that are subtended by the same arc equal to each other?", + "A": "For any given arc, there can be any number of inscribed angles that subtend it, but only one central angle will subtend that same arc. Since the inscribed angle theorem tells us that any inscribed angle will be exactly half the measure of the central angle that subtends its arc, it follows that all inscribed angles sharing that arc will be half the measure of the same central angle. Therefore, the inscribed angles must all be congruent. Hope this helps!", + "video_name": "h-_BDon5oes", + "timestamps": [ + 85 + ], + "3min_transcript": "- So what I would like you to do is see if you can figure out the measure of angle DEG here. So try to figure out the measure of this angle. I encourage you to pause the video now and try it on your own. All right, now let's work through this together and the key realization here is to think about this angle, it is an inscribed angle, we see it's vertexes sitting on the circle itself. And then think about the arc that it intercepts. And we see, we see that it intercepts, so let me draw these two sides of the angle, we see that it intercepts arc CD. It intercepts arc CD. And so the measure of this angle, since it's an inscribed angle, is going to be half the measure of arc CD. So if we could figure out the measure of arc CD, then we're going to be in good shape. Because if we figure out the measure of arc CD, then we take half of that and we'll figure out what we care about. inscribed angle that also intercepts arc CD. We have this angle right over here. It also intercepts arc CD. So you could call this angle C ... Whoops. You could call this angle CFD. This also intercepts the same arc. So there's two ways you could think about it. Two inscribed angles that intercept the same arc are going to have the same angle measure so just off of that you could say that this is going to be, that these two angles have the same measure, so you could say this is going to be 50 degrees. Or you could go, you could actually solve what the measure of arc CD is. It's going to be twice the measure of the inscribed angle that intercepts it. So the measure of arc CD is going to be 100 degrees. Twice the 50 degrees. And then you use that and you say, Well, if the measure of that arc is 100 degrees, then an inscribed angle that intercepts it it's going to be 50 degrees. So either way we get to 50 degrees." + }, + { + "Q": "At 1:47, he does 3*-2, and represents it as -2+-2+-2. Can't he also say 3, -2 times?\nIf you took three twice away from zero, it would still equal -6.", + "A": "Yes, you are correct, but in this video, he only shows one way but the way you are thinking is definitely correct, they are the same. It is like saying 3*4 is equal to both 3+3+3+3 and 4+4+4. Hope this helps.", + "video_name": "47wjId9k2Hs", + "timestamps": [ + 107 + ], + "3min_transcript": "We know that if we were to multiply two times three, that would give us positive six. And so we are going to think about negative numbers in this video. One way to think about it, is that I have a positive number times another positive number, and that gives me a positive number. So if I have a positive times a positive, that would give me a positive number. Now it's mixed up a little bit. Introduce some negative numbers. So what happens if I had negative two times three? Negative two times three. Well, one way to think about it-- Now we are talking about intuition in this video and in the future videos. You could view this as negative two repeatedly added three times. So this could be negative two plus negative two plus negative two-- Not negative six. Plus negative two. which would be equal to-- well, negative two plus negative two is negative four, plus another negative two is negative six. This would be equal to negative six. if I had two times three, I would get six. But because one of these two numbers is negative, then my product is going to be negative. So if I multiply, a negative times a positive, I'm going to get a negative. Now what if we swap the order which we multiply? So if we were to multiply three times negative two, it shouldn't matter. The order which we multiply things don't change, or shouldn't change the product. When we multiply two times three, we get six. When we multiply three times two, we will get six. So we should have the same property here. Three times negative two should give us the same result. It's going to be equal to negative six. And once again we say, three times two would be six. One of these two numbers is negative, and so our product is going to be negative. So we could draw a positive times a negative And both of these are just the same thing with the order which we are multiplying switched around. But this is one of the two numbers are negative. Exactly one. So one negative, one positive number is being multiplied. Then you'll get a negative product. Now we'll think about the third circumstance, where both of the numbers are negative. So if I were to multiply--I'll just switch colors for fun here-- If I were to multiply negative two times negative three-- this might be the least intuitive for you of all, and here I'm going to introduce you the rule, in the future I will explore why this is, and why this makes mathematics more--all fit together. But this is going to be, you see, two times three would be six. And I have a negative times a negative, one way you can think about it is that negatives cancel out! So you'll actually end up with a positive six. Actually I don't have to draw a positive here." + }, + { + "Q": "At 1:17, I get confused. Can you help?", + "A": "Ok, so a negative times a negative equals a positive because the negatives cancel out, but if it s a negative times a positive (or vice versa) there s nothing to cancel out that negative so the answer remains negative.", + "video_name": "47wjId9k2Hs", + "timestamps": [ + 77 + ], + "3min_transcript": "We know that if we were to multiply two times three, that would give us positive six. And so we are going to think about negative numbers in this video. One way to think about it, is that I have a positive number times another positive number, and that gives me a positive number. So if I have a positive times a positive, that would give me a positive number. Now it's mixed up a little bit. Introduce some negative numbers. So what happens if I had negative two times three? Negative two times three. Well, one way to think about it-- Now we are talking about intuition in this video and in the future videos. You could view this as negative two repeatedly added three times. So this could be negative two plus negative two plus negative two-- Not negative six. Plus negative two. which would be equal to-- well, negative two plus negative two is negative four, plus another negative two is negative six. This would be equal to negative six. if I had two times three, I would get six. But because one of these two numbers is negative, then my product is going to be negative. So if I multiply, a negative times a positive, I'm going to get a negative. Now what if we swap the order which we multiply? So if we were to multiply three times negative two, it shouldn't matter. The order which we multiply things don't change, or shouldn't change the product. When we multiply two times three, we get six. When we multiply three times two, we will get six. So we should have the same property here. Three times negative two should give us the same result. It's going to be equal to negative six. And once again we say, three times two would be six. One of these two numbers is negative, and so our product is going to be negative. So we could draw a positive times a negative And both of these are just the same thing with the order which we are multiplying switched around. But this is one of the two numbers are negative. Exactly one. So one negative, one positive number is being multiplied. Then you'll get a negative product. Now we'll think about the third circumstance, where both of the numbers are negative. So if I were to multiply--I'll just switch colors for fun here-- If I were to multiply negative two times negative three-- this might be the least intuitive for you of all, and here I'm going to introduce you the rule, in the future I will explore why this is, and why this makes mathematics more--all fit together. But this is going to be, you see, two times three would be six. And I have a negative times a negative, one way you can think about it is that negatives cancel out! So you'll actually end up with a positive six. Actually I don't have to draw a positive here." + }, + { + "Q": "AT 1:39 the regrouping of the problem from -2 x 3 was changed to 3 x -2. Would this be an example of commutative property ?", + "A": "Yes! The communicative property states that the order does not matter for numbers added or multiplied.", + "video_name": "47wjId9k2Hs", + "timestamps": [ + 99 + ], + "3min_transcript": "We know that if we were to multiply two times three, that would give us positive six. And so we are going to think about negative numbers in this video. One way to think about it, is that I have a positive number times another positive number, and that gives me a positive number. So if I have a positive times a positive, that would give me a positive number. Now it's mixed up a little bit. Introduce some negative numbers. So what happens if I had negative two times three? Negative two times three. Well, one way to think about it-- Now we are talking about intuition in this video and in the future videos. You could view this as negative two repeatedly added three times. So this could be negative two plus negative two plus negative two-- Not negative six. Plus negative two. which would be equal to-- well, negative two plus negative two is negative four, plus another negative two is negative six. This would be equal to negative six. if I had two times three, I would get six. But because one of these two numbers is negative, then my product is going to be negative. So if I multiply, a negative times a positive, I'm going to get a negative. Now what if we swap the order which we multiply? So if we were to multiply three times negative two, it shouldn't matter. The order which we multiply things don't change, or shouldn't change the product. When we multiply two times three, we get six. When we multiply three times two, we will get six. So we should have the same property here. Three times negative two should give us the same result. It's going to be equal to negative six. And once again we say, three times two would be six. One of these two numbers is negative, and so our product is going to be negative. So we could draw a positive times a negative And both of these are just the same thing with the order which we are multiplying switched around. But this is one of the two numbers are negative. Exactly one. So one negative, one positive number is being multiplied. Then you'll get a negative product. Now we'll think about the third circumstance, where both of the numbers are negative. So if I were to multiply--I'll just switch colors for fun here-- If I were to multiply negative two times negative three-- this might be the least intuitive for you of all, and here I'm going to introduce you the rule, in the future I will explore why this is, and why this makes mathematics more--all fit together. But this is going to be, you see, two times three would be six. And I have a negative times a negative, one way you can think about it is that negatives cancel out! So you'll actually end up with a positive six. Actually I don't have to draw a positive here." + }, + { + "Q": "At 2:55 he said that the negative cancels out the negative and makes a positive product. Then according to that, wouldn't a positive cancel out a positive, making a negative product.", + "A": "No, that isn t how it works. A negative times a negative is a positive, and a positive times a positive is a positive. But a negative times a positive is a negative, and a positive times negatives is a negative. So, basically, if you are multiplying two of the same thing (like two positives, or two negatives), you get a positive. If you are multiplying two different things (one negative and one positive), you get a negative. So no, a positive does not cancel out another positive. I hope this helps.", + "video_name": "47wjId9k2Hs", + "timestamps": [ + 175 + ], + "3min_transcript": "if I had two times three, I would get six. But because one of these two numbers is negative, then my product is going to be negative. So if I multiply, a negative times a positive, I'm going to get a negative. Now what if we swap the order which we multiply? So if we were to multiply three times negative two, it shouldn't matter. The order which we multiply things don't change, or shouldn't change the product. When we multiply two times three, we get six. When we multiply three times two, we will get six. So we should have the same property here. Three times negative two should give us the same result. It's going to be equal to negative six. And once again we say, three times two would be six. One of these two numbers is negative, and so our product is going to be negative. So we could draw a positive times a negative And both of these are just the same thing with the order which we are multiplying switched around. But this is one of the two numbers are negative. Exactly one. So one negative, one positive number is being multiplied. Then you'll get a negative product. Now we'll think about the third circumstance, where both of the numbers are negative. So if I were to multiply--I'll just switch colors for fun here-- If I were to multiply negative two times negative three-- this might be the least intuitive for you of all, and here I'm going to introduce you the rule, in the future I will explore why this is, and why this makes mathematics more--all fit together. But this is going to be, you see, two times three would be six. And I have a negative times a negative, one way you can think about it is that negatives cancel out! So you'll actually end up with a positive six. Actually I don't have to draw a positive here. This right over here is a positive six. So we have another rule of thumb here. If I have a negative times a negative, the negatives are going to cancel out. And that's going to give me a positive number. Now with these out of the way, let's just do a bunch of examples. I'm encouraging you to try them out before I do them. Pause the video, try them out, and see if you get the same answer. So let's try negative one times negative one. Well, one times one would be one. And we have a negative times a negative. They cancel out. Negative times a negative give me a positive. So this is going to be positive one. I can just write one, or I can literally write a plus sign there to emphasize. This is a positive one. What happened if I did negative one times zero? Now this might seem, this doesn't fit into any of these circumstances, zero is neither positive nor negative. And here you just have to remember anything times zero" + }, + { + "Q": "At 3:41 in the graph we can see a vertical asymptote at approximately x=3, why is this and what is its significance?", + "A": "That is simply due to the fact that 6x^5-100x^2-10 has a zero around 2.57. Because this expression is in the denominator of the rational function, there is a horizontal asymptote (because you can t divide by zero!). However, the vertical asymptotes have nothing to do with the horizontal asymptotes (limits at infinity).", + "video_name": "gv9ogppphso", + "timestamps": [ + 221 + ], + "3min_transcript": "if that actually makes sense. What we're actually saying is that we have a horizontal asymptote at y is equal to 2/3. So lets look at the graph. So right here is the graph. Got it from Wolfram Alpha. And we see, indeed, as x gets larger and larger and larger, f of x seems to be approaching this value that looks right at around 2/3. So it looks like we have a horizontal asymptote right over here. Let me draw that a little bit neater. We have a horizontal asymptote right at 2/3. So let me draw it as neatly as I can. So this right over here is y is equal to 2/3. The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2/3. And when we just look at the graph here, it seems like the same thing is happening from the bottom direction, when x approaches negative infinity. as x approaches negative infinity, that also looks like it's 2/3. And we can use the exact same logic. When x becomes a very, very, very negative number, as it becomes further and further to the left on the number line, the only terms that are going to matter are going to be the 4x to the fifth and the 6x to the fifth. So this is true for very large x's. It's also true for very negative x's. So we could also say, as x approaches negative infinity, this is also true. And then, the x to the fifth over the x to the fifth is going to cancel out. These are the dominant terms. And we're going to get it equaling 2/3. And once again, you see that in the graph here. We have a horizontal asymptote at y is equal to 2/3. We take the limit of f of x as x approaches infinity, we get 2/3. And the limit of f of x as x approaches negative infinity is 2/3. So in general, whenever you do this, you just have to think about what And focus on those." + }, + { + "Q": "why does Sal say at 4:30 \"principal square root\"", + "A": "Because the \u00e2\u0088\u009a means principal square root, not square root.", + "video_name": "egNq4tSfi1I", + "timestamps": [ + 270 + ], + "3min_transcript": "and then we have plus five times, now eight can be written as a product of a perfect square and a not so perfect square, eight can be written as four times two, so lets write it that way so if we view this whole, this is the principle root, the square root of four times two, we can re-write this as the five times the square root of four, or the principle root of four times the principle root of two and what can we simplify here? well we know what the principle root of x squared is, it is the positive square root of x squared, so it is not just x, you might be tempted to say it is x but since we know it is the positive square root we have to say it is the absolute value of x, because what if x was negative? if was x was negative, you'd have , lets say it was negative three, you'd have negative three squared, and so it wouldn't just be x, it wouldnt be negative three, it would be positive three, so you have to take the absolute value, and the other thing that is a perfect square is the four right here, its principle root is two, its principle square root i should say is two, so now you have, if we just change the order we are multiplying right here, you have four, four times the absolute value of x, four times the absolute value of x, times the square root of two, times the square root of two, I want to do that in that same yellow color, times the square root of two, plus plus we have five times two, which is ten, right, this whole thing is simplified to two, so we have plus ten square roots of two, now we could call it a day, and say we are all done adding and simplifying or you could add a little bit more depending on how you wanna view it, because over here you have so you have four absolute value of x of something, and you have ten of that same something, you could add them up, or another way to think about it is, you could factor out a square root of two either one of those works, so you get four times the absolute value of x, plus ten plus ten times times the principle square root of two, so depending on whether you view this of this more simplified, one of those two will will satisfy you" + }, + { + "Q": "So, at 2:50, if we assume the root didn't always indicate the principal square root, could we say that x could be either positive or negative?", + "A": "Yes it can be either or", + "video_name": "egNq4tSfi1I", + "timestamps": [ + 170 + ], + "3min_transcript": "plus three \"a\"'s which will give you four \"a\"'s, in this case \"a\" is all of this business right over here so we added those terms, and then we wanted to think about we have four principle roots of \"a\" and we have one more principle roots of \"a\", so same idea you have four of these things I am circling in magenta and you have one more of these things that I am circling in magenta, that one co-efficient is implicit so if I have four of something plus one more of something it becomes five of that something so plus plus five times the square root, plus five times the square root of eight and now we'll see if we can simplify this anymore, we have four of something and we have five of something else, so you can't just add these two things together, but maybe we can simplify this a little bit so we know that the principle root of two x squared, this is the same thing as, so let me write the four out front, so we have the four, and the principle root of two x squared is the same thing as the principle and then we have plus five times, now eight can be written as a product of a perfect square and a not so perfect square, eight can be written as four times two, so lets write it that way so if we view this whole, this is the principle root, the square root of four times two, we can re-write this as the five times the square root of four, or the principle root of four times the principle root of two and what can we simplify here? well we know what the principle root of x squared is, it is the positive square root of x squared, so it is not just x, you might be tempted to say it is x but since we know it is the positive square root we have to say it is the absolute value of x, because what if x was negative? if was x was negative, you'd have , lets say it was negative three, you'd have negative three squared, and so it wouldn't just be x, it wouldnt be negative three, it would be positive three, so you have to take the absolute value, and the other thing that is a perfect square is the four right here, its principle root is two, its principle square root i should say is two, so now you have, if we just change the order we are multiplying right here, you have four, four times the absolute value of x, four times the absolute value of x, times the square root of two, times the square root of two, I want to do that in that same yellow color, times the square root of two, plus plus we have five times two, which is ten, right, this whole thing is simplified to two, so we have plus ten square roots of two, now we could call it a day, and say we are all done adding and simplifying or you could add a little bit more depending on how you wanna view it, because over here you have" + }, + { + "Q": "Im stuck at 3:10, i dont know how sal has p(x)=f(0)+f\"(0)x.\nIs this like linear approximation or something?", + "A": "First of all, he said p(x)=f(0)+f (0)*x just one prime And yes, this is a linear approximation! Does it remind you of anything? Perhaps point slope form? if y=mx+b, y is p(x), b is f(0), and m is f (0).", + "video_name": "epgwuzzDHsQ", + "timestamps": [ + 190 + ], + "3min_transcript": "So at first, maybe we just want p of 0, where p is the polynomial that we're going to construct, we want p of 0 to be equal to f of 0. So if we want to do that using a polynomial of only one term, of only one constant term, we can just set p of x is equal to f of 0. So if I were to graph it, it would look like this. It would just be a horizontal line at f of 0. And you could say, Sal, that's a horrible approximation. It only approximates the function at this point. Looks like we got lucky at a couple of other points, but it's really bad everywhere else. And now I would tell you, well, try to do any better using a horizontal line. At least we got it right at f of 0. So this is about as good as we can do with just a constant. And even though-- I just want to remind you-- this might not look like a constant, but we're assuming that given the function, will just give us a number. So whatever number that was, we would put it right over here. We'd say p of x is equal to that number. It would just be a horizontal line right there at f of 0. But that obviously is not so great. So let's add some more constraints. Beyond the fact that we want p of 0 to be equal to f of 0, let's say that we also want p prime at 0 to be the same thing as f prime at 0. Let me do this in a new color. So we also want, in the new color, we also want-- that's not a new color. We also want p prime. We want the first derivative of our polynomial, when evaluated at 0, to be the same thing as the first derivative of the function when evaluated at 0. And we don't want to lose this right over here. So what if we set p of x as being equal to f of 0? So we're taking our old p of x, but now we're Plus f prime of times x. So let's think about this a little bit. If we use this as our new polynomial, what happens? What is p is 0? p of 0 is going to be equal to-- you're going to have f of 0 plus whatever this f prime of 0 is times 0. If you put a 0 in for x, this term is just going to be 0. So you're going to be left with p of 0 is equal to f of 0. That's cool. That's just as good as our first version. Now what's the derivative over here? So the derivative is p prime of x is equal to-- you take the derivative of this. This is just a constant, so its derivative is 0. The derivative of a coefficient times x is just going to be the coefficient. So it's going to be f prime of 0. So if you evaluate it at 0-- so p prime of 0." + }, + { + "Q": "Can someone explain the equation for when g(x)?\n5:16", + "A": "are you confused about why the function is called g(x) and not f(x)? if so they mean the same thing . it is like a name for the function", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 316 + ], + "3min_transcript": "you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety, Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this." + }, + { + "Q": "at 8:20, why does G(2) = 1 instead of 4?", + "A": "Because that s part of the function - it says square x unless x = 2 where g(x) gives you a 1", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 500 + ], + "3min_transcript": "let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that. this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1," + }, + { + "Q": "At 1:25, why is f(x) = D?", + "A": "i think you are getting confused by his arrows...", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 85 + ], + "3min_transcript": "In this video, I want to familiarize you with the idea of a limit, which is a super important idea. It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really, really, really simple idea. So let me draw a function here, actually, let me define a function here, a kind of a simple function. So let's define f of x, let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal look, I have the same thing in the numerator and denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true, the difference between f of x equals 1 and this thing right over here, is that this thing can never equal-- this thing is undefined when x is equal to 1. Because if you set, let me define it. Let me write it over here, if you have f of, sorry not f of 0, if you have f of 1, what happens. which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here." + }, + { + "Q": "Wait a minute, at 9:20 Sal says that when x approaches to 2, the value of g(x) would be getting really close of 4, but shouldn't it be 1? Since when x = 2 <=> g(x) = 1?", + "A": "g(2) is in fact 1, given by the dot. However, the limit is different. Remeber g(x) = y. To understand limit, try putting your pencil on the graph and trace it. As you trace it toward x=2, you see that y=g(x) is getting toward 4 and not 1.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 560 + ], + "3min_transcript": "this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1, as we get closer and closer to it. And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1. The limit of g of x as x approaches 2 is equal to 4. And you could even do this numerically using a calculator, and let me do that, because I think that will be interesting. So let me get the calculator out, let me get my trusty TI-85 out. So here is my calculator, and you could numerically say, OK, what's it going to approach as you approach x equals 2. So let's try 1.94, for x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared." + }, + { + "Q": "I don't really get it. At 00:53, why can't you reduce x-1/x-1 to equal to 1?", + "A": "This is only true if x does not equal 1. If x were to equal 1 then you d have 0/0 which is indeterminate. Therefore the division only works for most cases, but it only takes one counterexample to make a conjecture false. So limits will allow for the case up to but not including the problem point.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 53 + ], + "3min_transcript": "In this video, I want to familiarize you with the idea of a limit, which is a super important idea. It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really, really, really simple idea. So let me draw a function here, actually, let me define a function here, a kind of a simple function. So let's define f of x, let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal look, I have the same thing in the numerator and denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true, the difference between f of x equals 1 and this thing right over here, is that this thing can never equal-- this thing is undefined when x is equal to 1. Because if you set, let me define it. Let me write it over here, if you have f of, sorry not f of 0, if you have f of 1, what happens. which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here." + }, + { + "Q": "in the video at 4:24 Sal says that we can get infinitely closer to one. If we can get infinitely closer to one doesn't that mean that we can never approach one?", + "A": "True. Yes, we can always get closer and closer to one but the function actually never reaches one.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 264 + ], + "3min_transcript": "And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here. you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety," + }, + { + "Q": "At 7:30 Sal has drawn a parabola with a gap at the point (2,4).\n\nA point takes up zero space right? It has no size? So how can something have a gap in it if the gap doesn't cover any actual space? How big is the gap? Surely a gap is between two points.\n\nCan someone please help me understand this?", + "A": "At this point it has size 0. The gap is just for understanding that there is no defined point at that x coordinate.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 450 + ], + "3min_transcript": "Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that. this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function" + }, + { + "Q": "9:38 he says its 4 we're getting to but how do you get for without a graph", + "A": "In later lessons you will learn different techniques to do it algebraically. This is just a video introduction of limits, so I recommend you watch other lessons about limits.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 578 + ], + "3min_transcript": "So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1, as we get closer and closer to it. And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1. The limit of g of x as x approaches 2 is equal to 4. And you could even do this numerically using a calculator, and let me do that, because I think that will be interesting. So let me get the calculator out, let me get my trusty TI-85 out. So here is my calculator, and you could numerically say, OK, what's it going to approach as you approach x equals 2. So let's try 1.94, for x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared. so 1.99, and once again, let me square that. Well now I'm at 3.96. What if I do 1.999, and I square that? I'm going to have 3.996. Notice I'm going closer, and closer, and closer to our point. And if I did, if I got really close, 1.9999999999 squared, what am I going to get to. It's not actually going to be exactly 4, this calculator just rounded things up, but going to get to a number really, really, really, really, really, really, really, really, really close to 4. And we can do something from the positive direction too. And it actually has to be the same number when we approach from the below what we're trying to approach, and above what we're trying to approach. So if we try to 2.1 squared, we get 4.4. let me go a couple of steps ahead," + }, + { + "Q": "At 9:20, how is g(x) approaching the value of 4 if it is a hole? Why isn't the limit 1 because it's an actual point?", + "A": "The limit as x approaches 2 of g(x) does not have to equal g(2). The value of this limit (if any) is determined by the behavior of g(x) near, but not at, x = 2. Since g(x) is near 4 when x is near (but not at) 2, the value of this limit is 4. The fact that g(2) = 1 has no effect on the value of this limit. Note that the fact that the value of this limit does not match the value of g(2) indicates that g(x) is discontinuous at x=2. Have a blessed, wonderful day!", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 560 + ], + "3min_transcript": "this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1, as we get closer and closer to it. And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1. The limit of g of x as x approaches 2 is equal to 4. And you could even do this numerically using a calculator, and let me do that, because I think that will be interesting. So let me get the calculator out, let me get my trusty TI-85 out. So here is my calculator, and you could numerically say, OK, what's it going to approach as you approach x equals 2. So let's try 1.94, for x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared." + }, + { + "Q": "At 1:17; what's the difference between infinity and 1/0 ?", + "A": "Infinity (symbol: \u00e2\u0088\u009e) is an abstract concept describing something without any bound or larger than any number. In mathematics, infinity is often treated as a number While the expression a/0 has no meaning, as there is no number which, multiplied by 0, gives a (assuming a\u00e2\u0089\u00a00), and so division by zero is undefined. Since any number multiplied by zero is zero, the expression 0/0 also has no defined value; You can also prove it by the concept of limits. Try it out yourself.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 77 + ], + "3min_transcript": "In this video, I want to familiarize you with the idea of a limit, which is a super important idea. It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really, really, really simple idea. So let me draw a function here, actually, let me define a function here, a kind of a simple function. So let's define f of x, let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal look, I have the same thing in the numerator and denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true, the difference between f of x equals 1 and this thing right over here, is that this thing can never equal-- this thing is undefined when x is equal to 1. Because if you set, let me define it. Let me write it over here, if you have f of, sorry not f of 0, if you have f of 1, what happens. which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here." + }, + { + "Q": "At 1:33, isn't that the Riemann's zeta function?", + "A": "Good question! That would be Riemann s zeta function, evaluated at z = 2. In general, this function is sum n=1 to infinity of 1/(n^z), where z is a complex number (that is, a number of the form a+bi where a and b are real numbers and i is the square root of -1). This function is defined when the real part of z (that is, a) is greater than 1. Have a blessed, wonderful day!", + "video_name": "k9MEOgcc5KY", + "timestamps": [ + 93 + ], + "3min_transcript": "- [Voiceover] Let's say that you have an infinite series, S, which is equal to the sum from n equals one, let me write that a little bit neater. n equals one to infinity of a sub n. This is all a little bit of review. We would say, well this is the same thing as a sub one, plus a sub two, plus a sub three, and we would just keep going on and on and on forever. Now what I want to introduce to you is the idea of a partial sum. This right over here is an infinite series. But we can define a partial sum, so if we say S sub six, this notation says, okay, if S is an infinite series, S sub six is the partial sum of the first six terms. So in this case, this is going to be we're not going to just keep going on forever, this is going to be a sub one, plus a sub two, plus a sub three, plus a sub four, plus a sub five, And I can make this a little bit more tangible if you like. So let's say that S, the infinite series S, is equal to the sum from n equals one, to infinity of one over n squared. In this case it would be one over one squared, plus one over two squared, plus one over three squared, and we would just keep going on and on and on forever. But what would S sub -- I should do that in that same color. What would S -- I said I would change color, and I didn't. What would S sub three be equal to? The partial sum of the first three terms, and I encourage you to pause the video and try to work through it on your own. Well, it's just going to be the first term one, plus the second term, is going to be the sum of the first three terms, and we can figure that out, that's to see if you have a common denominator here, it's going to be 36. It's going to be 36/36, plus 9/36, plus 4/36, so this is going to be 49/36. 49/36. So the whole point of this video, is just to appreciate this idea of a partial sum. And what we'll see is, that you can actually express what a partial sum might be algebraically. So for example, for example, let's give ourselves a little bit more real estate here. Let's say, let's go back to just saying we have an infinite series, S, that is equal to the sum from n equals one to infinity of a sub n. And let's say we know the partial sum, S sub n, so the sum of the first n terms" + }, + { + "Q": "4:06: Wouldn't it be easier to say TRANSPOSE instead of ADJUGATE?! At least would be better to say that T is something that is more used when it comes to studying matrices more? (\u00c3\u008d know its the same thing but might confuce students like me who are having exam about this and never heard of adjugate and instead of TRANSPOSE T)", + "A": "Adjugate of matrix A is Transpose of Cofactor matrix of A", + "video_name": "ArcrdMkEmKo", + "timestamps": [ + 246 + ], + "3min_transcript": "Well that's negative 4 times 5. So that is negative 20. But we're going to subtract negative 20. So that's negative 4 times 5, negative 20, but we're going to subtract negative 20. Obviously that's going to turn into adding positive 20. Then you have negative 1 times 1 times 4, which is negative 4. But we're going to subtract these products. We're going to subtract negative 4. And then you have 2 times 1 times 3, which is 6. But we have to subtract it. So we have subtracting 6. And so this simplifies to negative 5 minus 6 is negative 11, plus 16 gets us to positive 5. So all of this simplifies to positive 5. And then we have plus 20 plus 4. so we don't get confused. So we have plus 20 plus 4 minus 6. So what does this get us? 5 plus 20 is 25, plus 4 is 29, minus 6 gets us to 23. So our determinant right over here is equal to 23. So now we are really in the home stretch. The inverse of this matrix is going to be 1 over our determinant times the transpose of this cofactor matrix. And the transpose of the cofactor matrix is called the adjugate. So let's do that. So let's write the adjugate here. This is the drum roll. We're really in the home stretch. C inverse is equal to 1 over the determinant, And so this is going to be equal to 1/23 times the transpose of our cofactor matrix. So we have our cofactor matrix right over here. So each row now becomes a column. So this row now becomes a column. So it becomes 1, negative 7, 5 becomes the first column. The second row becomes the second column-- 18, negative 11, negative 2. And then finally, the third row becomes the third column. You have negative 4, 5, and 3. And now we just have to multiply, or you could say divide, each of these by 23, and we are there." + }, + { + "Q": "I rally didn't get what you men't by when the video got to 3:29 to 4:00 that was really confusing to be honest", + "A": "I think he was just trying to illustrate what he was multiplying. He shows you closer to 4:30 what he means, and how it all works out.", + "video_name": "j3-XYLnxJDY", + "timestamps": [ + 209, + 240 + ], + "3min_transcript": "has a width of 2 and a height of 3. So you could imagine that being this rectangle right over here. So that is this rectangle right over here. So that's the 2 times 3 rectangle. Now, it looks like the area of the trapezoid should be in between these two numbers. Maybe it should be exactly halfway in between, because when you look at the area difference between the two rectangles-- and let me color that in. So this is the area difference on the left-hand side. And this is the area difference on the right-hand side. If we focus on the trapezoid, you see that if we start with the yellow, the smaller rectangle, it reclaims half of the area, half of the difference between the smaller rectangle It gets exactly half of it on the left-hand side. And it gets half the difference between the smaller and the larger on the right-hand side. So it completely makes sense that the area of the trapezoid, this entire area right over here, should really just be the average. It should exactly be halfway between the areas of the smaller rectangle and the larger rectangle. So let's take the average of those two numbers. It's going to be 6 times 3 plus 2 times 3, all of that over 2. So when you think about an area of a trapezoid, you look at the two bases, the long base and the short base. Multiply each of those times the height, and then you could take the average of them. Or you could also think of it as this is the same thing as 6 plus 2. And I'm just factoring out a 3 here. 6 plus 2 times 3, and then all of that over 2, just writing it in different ways. These are all different ways to think about it-- 6 plus 2 over 2, and then that times 3. So you could view it as the average of the smaller and larger rectangle. So you multiply each of the bases times the height and then take the average. You could view it as-- well, let's just add up the two base lengths, multiply that times the height, and then divide by 2. Or you could say, hey, let's take the average of the two base lengths and multiply that by 3. And that gives you another interesting way If you take the average of these two lengths, 6 plus 2 over 2 is 4. So that would be a width that looks something like-- let me do this in orange. A width of 4 would look something like this." + }, + { + "Q": "At 08:36, why doesn't \"x-1 = +-4\" become \"x = +- 5\"?", + "A": "@James.p.french: He couldn t simplify to x=+-5 because x is not equal to that. He could ve simplified it to the solution x=5 or x=-3.", + "video_name": "lGQw-W1PxBE", + "timestamps": [ + 516 + ], + "3min_transcript": "I know that we're going to be a little bit less than the negative square root, but I'll do it the other way. I'll do it the way I did in the last video. So the other way to think about it is what happens when this term is 0? For this term to be 0, x has to be equal to 1. And does that ever happen? Can x be equal to 1? If x is equal to 1 here this term is 0. And then you have a situation where-- and then you have a minus y squared over 4 would have to equal 1, or this would have to be a negative number. So x could not be equal to 1. So y could be equal to negative 1. Let's try that out. If y is equal to negative 1, this term right here disappears. minus 1 squared over 16 is equal to 1. I just canceled out this term, because I'm saying what happens when y is equal to negative 1. You multiply both sides by 16. Let me do it over here. These get messy. x minus 1 squared is equal to 16. Take the square root of both sides. x minus 1 is equal to positive or negative 4. And so if x is equal to positive 4, if you add 1 to that x would be equal to 5. And then if x minus 1 would be minus 4 and you add 1 to that you will have x is equal to 3. So our 2 points or our 2 points closest to our center are the points 5 comma negative 1 and 3 comma negative 1. Let's plot those 2. So 5, 1 2 3 4 5, negative 1 and 3, negative 1. No, minus 3, because x minus 1 could be minus 4. That's what happens when you skip steps. If you have the minus 4 situation, then x is equal to minus 3. You go 1 2 3 minus 3, minus 1. So those are both points on this hyperbola. And then our intuition was correct, or it was what I said. That-- the positive square root is always going to be slightly below the asymptote, so we get our curve. It's going to look something like this. It's going to get closer and closer, and then here it's going to get closer and closer in that direction. It keeps getting closer and closer to that asymptote. And here, it's going to keep getting closer and closer to" + }, + { + "Q": "At 3:45 isn't it possible that b is -2? I mean if you were to ignore the first table and before knowing c and d. Or am I wrong?", + "A": "Remember b>0 and b=/= 1", + "video_name": "Iz6IVf8frjw", + "timestamps": [ + 225 + ], + "3min_transcript": "Now this is an equivalent statement to saying that b to the a power is equal to ... oh sorry, not b to the a power. This is an equivalent statement to saying b to the 0 power is equal to a. This is saying what exponent do I need to raise b to to get a? You raise it to the 0 power. This is saying b to the 0 power is equal to a. Now what is anything to the 0 power, assuming that it's not 0? If we're assuming that b is not 0, if we're assuming that b is not 0, so we're going to assume that, and we can assume, and I think that's a safe assumption because where we're raising b to all of these other powers, we're getting a non-0 value. Since we know that b is not 0, anything with a 0 power is going to be 1. This tells us that a is equal to 1. We got one figured out. Now let's look at this next piece of information right over here. What does that tell us? That tells us that log base b of 2 is equal to 1. This is equivalent to saying the power that I needed to raise b to get to to 2 is 1. Or if I want to write in exponential form, I could write this as saying that b to the first power is equal to 2. I'm raising something to the first power and I'm getting 2? What is this thing? That means that b must be 2. 2 to the first power is 2. So b is equal to 2. b to the first power is equal to 2. You could say b to the first is equal to 2 to the first. That's also equal to 2. So b must be equal to 2. We've been able to figure that out. This is a 2 right over here. It actually makes sense. 2 to the 1.585 power, yeah, that feels right, that that's about 3. Let's see if we can figure out c. Let's look at this column. Let's see what this column is telling us. That column we could read as log base b. Now our y is 2c. Log base b of 2c is equal to 1.585. Or we could read this as b, if we write in exponential form, b to the 1.585 is equal to 2c. Now what's b to the 1.585? They tell us right over here that b to the 1.585 is 3, is 3, so this right over here is equal to 3. We get 2c is equal to 3, or divide both sides by 2, we would get c is equal to 1.5. This is working out pretty well. Now we have this last column," + }, + { + "Q": "1:00 Is doing this method also the same as using the foil method kind of?", + "A": "Yes, this is basically the intuition behind the foil method. It s showing how the distribution property can be used twice to obtain the quadratic in standard form, which is essentially the foil method :)", + "video_name": "Xy8NKUoyy98", + "timestamps": [ + 60 + ], + "3min_transcript": "- [Voiceover] Let's see if we can figure out the product of x minus four and x plus seven. And we want to write that product in standard quadratic form which is just a fancy way of saying a form where you have some coefficient on the second degree term, a x squared plus some coefficient b on the first degree term plus the constant term. So this right over here would be standard quadratic form. So that's the form that we want to express this product in and encourage you to pause the video and try to work through it on your own. Alright, now let's work through this. And the key when we're multiplying two binomials like this, or actually when you're multiplying any polynomials, is just to remember the distributive property that we all by this point know quite well. So what we could view this is as is we could distribute this x minus four, this entire expression over the x and the seven. So we could say that this is the same thing as x minus four times x plus x minus four times seven. So let's write that. So x minus four times x, or we could write this as x times x minus four. that's right there. Plus seven times x minus four. Times x minus four. Notice all we did is distribute the x minus four. We took this whole thing and we multiplied it by each term over here. We multiplied x by x minus four and we multiplied seven by x minus four. Now, we see that we have these, I guess you can call them two seperate terms. And to simplify each of them, or to multiply them out, we just have to distribute. In this first we're going to have to distribute this blue x. And over here we have to distribute this blue seven. So let's do that. So here we can say x times x is going to be x squared. X times, we have a negative here, so we can say negative four is going to be negative four x. And just like, that we get x squared minus four x. And then over here we have seven times x so that's going to be plus seven x. which is negative 28. And we are almost done. We can simplify it a little bit more. We have two first degree terms here. If I have negative four xs and to that I add seven xs, what is that going to be? Well those two terms together, these two terms together are going to be negative four plus seven xs. Negative four plus, plus seven. Negative four plus seven xs. So all I'm doing here, I'm making it very clear that I'm adding these two coefficients, and then we have all of the other terms. We have the x squared. X squared plus this and then we have, and then we have the minus, and then we have the minus 28. And we're at the home stretch! This would simply to x squared. Now negative four plus seven is three, so this is going to be plus three x. That's what these two middle terms simplify to, to three x." + }, + { + "Q": "At 0:28 Sal Said that an Odd Function Implies j(a) = - j(-a). Is this equivalent to -j(a) = j(-a) the more well known definition of an odd function? Or did Sal make a mistake?", + "A": "Multiply both sides by -1. They re the same.", + "video_name": "zltgXTlUVLw", + "timestamps": [ + 28 + ], + "3min_transcript": "Which of these functions is odd? And so let's remind ourselves what it means for a function to be odd. So I have a function-- well, they've already used f, g, and h, so I'll use j. So function j is odd. If you evaluate j at some value-- so let's say j of a. And if you evaluate that j at the negative of that value, and if these two things are the negative of each other, then my function is odd. If these two things were the same-- if they didn't have this negative here-- then it would be an even function. So let's see which of these meet the criteria of being odd. So let's look at f of x. So we could pick a particular point. So let's say when x is equal to 2. So we get f of 2 is equal to 2. Now, what is f of negative 2? f of negative 2 looks like it is 6. f of negative 2 is equal to 6. So these aren't the negative of each other. In order for this to be odd, f of negative 2 have had to be equal to negative 2. So f of x is definitely not odd. So all I have to do is find even one case that violated this constraint to be odd. And so I can say it's definitely not odd. Now let's look at g of x. So I could use the same-- let's see, when x is equal to 2, we get g of 2 is equal to negative 7. Now let's look at when g is negative 2. So we get g of negative 2 is also equal to negative 7. So here we have a situation-- and it looks like that's the case for any x we pick-- that g of x is going to be equal to g of negative x. So g of x is equal to g of negative x. It's symmetric around the y-- or I should say the vertical axis-- right over here. So which of these functions is odd? Definitely not g of x. So our last hope is h of x. Let's see if h of x seems to meet the criteria. I'll do it in this green color. So if we take h of 1-- and we can look at it even visually. So h of 1 gets us right over here. h of negative 1 seems to get us an equal amount, an equal distance, negative. So it seems to fit for 1. For 2-- well, 2 is at the x-axis. But that's definitely h of 2 is 0. h of negative 2 is 0. But those are the negatives of each other. 0 is equal to negative 0. If we go to, say, h of 4, h of 4 is this negative number. And h of negative 4 seems to be a positive number of the same magnitude. So once again, this is the negative of this. So it looks like this is indeed an odd function." + }, + { + "Q": "At 3:19, why does Sal divide by 7?", + "A": ".7 x 10 will equal 7. .7 represents 79 percent of the full price.", + "video_name": "d1oNF88SAgg", + "timestamps": [ + 199 + ], + "3min_transcript": "That was only that first day that I bought the 6. So how much are those two guavas going to cost me? How much are those two guavas going to cost at full price? At full price? So, a good place to start is, to think about how much would those 6 guavas have cost us at full price? This is the sale price, right here? This is the sale price. How much would those have cost me at full price? So let's do a little bit of algebra here. Pick a suitable color for the algebra. Maybe this grey color. So, let's say that x is equal to the cost of 6 guarvas. 6 guavas, at full price. So, essentially, if we take 30% off of this, we should get $12.60. So let's do that. So if we have the full price of 6 guavas, we're going So that's the same thing as 0.30. Or I could just write 0.3. I could ignore that zero if I like. Actually, let me write it like this. My wife is always bugging me to write zeroes before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. Some I'm just taking 30% off of the full price, off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We could imagine there's a 1 in front -- you know, x is the same thing as 1x. So 1x minus 0.3x is going to be equal to 0.7x. So we get 0.7x, or we could say 0.70 if you like. Same number. Point, or 0.7x, is equal to 12.60. And once you get used to these problems, you might just skip Where you say, 70% of the full price is equal to my sale price, right? I took 30% off. This is 70% of the full price. You might just skip to this step once you get used to these problems in a little bit. And now we just have to solve for x. Divide both sides by 0.7, so you get x is equal to 12.60 divided by 0.7. We could use a calculator, but it's always good to get a little bit of practice dividing decimals. So we get 0.7 goes into 12.60. Let's multiply both of these numbers by 10, which is what we do when we move both of their decimals one to the right. So the 0.7 becomes a 7. Ignore that right there. The 12.60 becomes 126, put the decimal right there. Decimal right there. And we're ready to just do straight up long division. So this is now a 7, not a .7. So 7 goes into 12 1 time." + }, + { + "Q": "At 1:48 Sal states that m=-4w+11 and proceeds to plug that into the other equation that we derived. But couldn't Sal simply plug that right back into the same equation that we got that from. So couldn't Sal use one equation to solve for two unknowns by plugging back in the m in terms of w?", + "A": "Try it. 100*(-4w + 11) + 400*w = 1100 -400*w + 1100 + 400*w = 1100 1100 = 1100 That s called a tautology. It s a true statement, but it s not providing any useful information. When you plug the value of m from one equation into another, your w terms don t cancel each other out, leading to a meaningful result of the form w = value", + "video_name": "2EwPpga_XPw", + "timestamps": [ + 108 + ], + "3min_transcript": "Just as you were solving the potato chip conundrum in the last video, the king's favorite magical bird comes flying along and starts whispering into the king's ear. And this makes you a little bit self-conscious, a little bit insecure, so you tell the king, what is the bird talking about. And the king says, well, the bird says that he thinks that there's another way to do the problem. And you're not used to taking advice from birds. And so being a little bit defensive, you say, well, if the bird thinks he knows so much, let him do this problem. And so the bird whispers a little bit more in the king's ear and says, OK, well I'll have to do the writing because the bird does not have any hands, or at least can't manipulate chalk. And so the bird continues to whisper in the king's ear. And the king translates and says, well, the bird says, let's use one of these equations to solve for a variable. So let's say, let's us this blue equation right over here to solve for a variable. And that's essentially going to be a constraint of one variable in terms of another. So let's see if we can do that. So here, if we want to solve for m, we could subtract 400 w from both sides. If we subtract 400w from the left, this 400w goes away. If we subtract 400w from the right, we have is equal to negative 400w plus 1,100. So what got us from here to here is just subtracting 400w from both sides. And then if we want to solve for m, we just divide both sides by 100. So we just divide all of the terms by 100. And then we get m is equal to negative 400 divided by 100, is negative 4w. 1,100 divided by 100 is 11. Plus 11. So now we've constrained m in terms of w. This is what the bird is saying, using the king as his translator. Why don't we take this constraint and substitute it back for m in the first equation? And then we will have one equation with one unknown. 200, so he's looking at that first equation now, he says 200. Instead of putting an m there, the bird says well, by the second constraint, m is equal to negative 4w plus 11. So instead of writing an m, we substitute for m the expression negative 4w plus 11. And then we have the rest of it, plus 300w, is equal to 1,200. So just to be clear, everywhere we saw an m, we replaced it with this right over here, in that first equation. So the first thing, you start to scratch your head. And you say, is this a legitimate thing to do. Will I get the same answer as I got when I solved the same problem with elimination? And I want you to sit and think about that for a second." + }, + { + "Q": "What does Mister Khan mean when he says, \"open parentheses\", at, 3:48-3:50? What does he mean specifically? Obviously he means that the parentheses are open. but what does the phenomenon of the open parentheses imply?\n\nThat is my question for the day. I hope it was relevant. Goodbye. Reuben.", + "A": "Open the parentheses means to open it up and compute the operations inside and operations having to do what s inside the parentheses. The distributive property is an example. Here s to show: -2 ( 9x+4 ) -- Opening the parentheses would be distributing the two to the things inside. => -2 ( 9x+4 ) = -18x-8 I hope I answered your question.", + "video_name": "rCGHUXSd15s", + "timestamps": [ + 228, + 230 + ], + "3min_transcript": "Now, if you agree to that, now let's work on this a little bit. So we can start with the absolute value of a minus b. And there's a bunch of ways that we could tackle this. But one way we could approach it, and actually let me write a little lower, so I don't bump into this stuff right over here. So let me start with the absolute value of a minus b. Now a, we could rewrite, this is just a, I don't wanna say it's positive or negative, a could be a negative number. This is just a right over here, but we can rewrite this as being the same thing as negative negative a. Or I guess negative negative a plus, minus b. So let's just think about this a little bit, with what I just did. So this a is the same thing as all of this business. The negative negative a. The negative of a negative is going to be a positive, this is positive a. So I haven't changed this in any dramatic way. But what's this going to be the same thing as? Well we can factor out a negative sign. So let's do that, and actually let me, we could factor out, using a color I haven't used yet, we could factor out the negative signs. So this is going to be equal to the absolute value of, I can factor out the negatives, the negative of, and then I have open parentheses, open parentheses negative a. And if you factor out a negative here, this is going to be positive b. So or plus b. Let me just close the parentheses in the same color, Now why is this interesting? Well we just said the absolute value of a negative is the same thing as, the absolute value of negative x is the same thing as the absolute value of x. So the absolute value of the negative of negative a plus b is going to be equal to, this is going to be equal to the absolute value, let me do that yellow color is going to be equal to the absolute value of this thing without the negative right here. You see that right over here. The absolute value of negative x is the same thing as the absolute value of x. So the absolute value of negative negative a plus b, is the same thing as the absolute value of negative a plus b. And negative a plus b, well that's the same thing as b minus a. I'm just swapping the order right over here. B minus a." + }, + { + "Q": "At 2:53, Sal says that the kite has perpendicular lines at an angle of 90 degrees. Do all quadrilaterals with perpendicular lines have to have it at an angle of 90 degrees?", + "A": "There seem to be a couple of questions in your question. First the definition of perpendicular is any 2 lines that form a 90 degree angle. So anything that refers to the word perpendicular is implying that the angle is 90 degrees. For clarification, the video is talking about the diagonals of the quadrilateral intersecting at 90 degrees. This is a property of a kite. Also squares have this property, so they are also kites. However (non-square) rectangles do no have this property, so are not kites. Hope this helps.", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 173 + ], + "3min_transcript": "" + }, + { + "Q": "what did sal mean by the congruent side could be opposite to each other on 1:45", + "A": "This is answered at 2:12 with a color picture of a parallelogram. In a quadrilateral, a shape with four sides, any two sides are either going to be adjacent (share an angle or endpoint) or opposite (do not touch or share an angle or endpoint). We are dealing with 2 pairs of congruent sides. Congruent sides or line segments have the same length. So, if you have two pairs of congruent sides in a quadrilateral, you will either have a parallelogram of a kite.", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 105 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:44, if a pair of congruent sides are adjacent, does it affect the fact that it is congruent?", + "A": "No. Like in triangles, if it is a scalene triangle, all sides will be adjacent, but sense there are no congruent, or equal, sides, then there will be none. Even in a kite, if you think about it, there will be un-congruent ( I just made that word up ) sides that are, in face, adjacent. Please vote for this answer if it was helpful to you! :D", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 104 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:02,can a kite also be a diamond?", + "A": "yes, as it is of the same shape", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 62 + ], + "3min_transcript": "" + }, + { + "Q": "at around 2:00 to 2:11 what is adjacent", + "A": "Next to or adjoining something else.", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 120, + 131 + ], + "3min_transcript": "" + }, + { + "Q": "At 5:19, how is a rhombus also going to be a kite?", + "A": "a kite has two pair of congruent (equal) adjacent (next to) sides. A rhombus has 4 congruent sides, which means each pair of adjacent sides is congruent. Therefore every rhombus is a kite.", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 319 + ], + "3min_transcript": "" + }, + { + "Q": "In 0:39, I don't understand the part where it says that there are a bunch of different Xs. How is that possible?", + "A": "Multiple variables (Same ones) can be in an inequality. Ex: x + x + 1 =23", + "video_name": "UTs4uZhu5t8", + "timestamps": [ + 39 + ], + "3min_transcript": "what i want to do in this video is a handful of fairly simple inequality videos. But the real value of it, I think, will be just to get you warmed up in the notation of inequality. So, let's just start with one. we have x minus 5 is less than 35. So let's see if we can find all of the x's that will satisfy this equation. And that's one of the distinctions of an inequality. In an equation, you typically have one solution, or at least the ones we've solved so far. In the future, we'll see equations where they have more than one solution. But in the ones we've solved so far, you solved for a particular x. In the inequalities, there's a whole set of x's that will satisfy this inequality. So they're saying, what are all the x's, that when you subtract 5 from them, it's going to be less than 35? And we can already think about it. I mean 0 minus 5. That's less than 35. Minus 100 minus 5. That's less than 35. 5 minus 5. That's less than 35. So there's clearly a lot of x's that will satisfy that. essentially encompasses all of the x's. So the way we do that is essentially the same way that we solved any equations. We want to get just the x terms, in this case, on the left-hand side. So I want to get rid of this negative 5, and I can do that by adding 5 to both sides of this equation. So I can add 5 to both sides of this equation. That won't change the inequality. It won't change the less than sign. If something is less than something else, something plus 5 is still going to be less than the other thing plus 5. So on the left-hand side, we just have an x. This negative 5 and this positive 5 cancel out. x is less than 35 plus 5, which is 40. And that's our solution. And to just visualize the set of all numbers that represents, let me draw a number line here. And I'll do it around-- let's say that's 40, And then we could go below 40. 39, 38. You can just keep going below 40. It just keeps going on in both directions. And any x that is less than 40 will satisfy this. So it can't be equal to 40, because if x is equal to 40, 40 minus 5 is 35. It's not less than 35. So x has to be less than 40. And to show this on the number line, we do a circle around 40 to show that we're not including 40. But then we can shade in everything below 40. So everything that's just exactly below 40 is included So everything I've shaded in yellow is included in our solution set. So 39, 39.999999, repeating, which is about as close as you can get to 40 as possible, that's in our solution set. But 40 is not. That's why we put that open circle around it. Let's do another one. Let me do it in another color as well." + }, + { + "Q": "Around 3:00...\nWait, isn't 39.9999(9repeating) equal to 40?\nSince 0.99999(9repeating) is equal to 1,\n39 + 0.99999... =? 39 + 1", + "A": "No it is not one or 40 because it is a decimal. When it is point somethine it is smaller than the rounded number", + "video_name": "UTs4uZhu5t8", + "timestamps": [ + 180 + ], + "3min_transcript": "essentially encompasses all of the x's. So the way we do that is essentially the same way that we solved any equations. We want to get just the x terms, in this case, on the left-hand side. So I want to get rid of this negative 5, and I can do that by adding 5 to both sides of this equation. So I can add 5 to both sides of this equation. That won't change the inequality. It won't change the less than sign. If something is less than something else, something plus 5 is still going to be less than the other thing plus 5. So on the left-hand side, we just have an x. This negative 5 and this positive 5 cancel out. x is less than 35 plus 5, which is 40. And that's our solution. And to just visualize the set of all numbers that represents, let me draw a number line here. And I'll do it around-- let's say that's 40, And then we could go below 40. 39, 38. You can just keep going below 40. It just keeps going on in both directions. And any x that is less than 40 will satisfy this. So it can't be equal to 40, because if x is equal to 40, 40 minus 5 is 35. It's not less than 35. So x has to be less than 40. And to show this on the number line, we do a circle around 40 to show that we're not including 40. But then we can shade in everything below 40. So everything that's just exactly below 40 is included So everything I've shaded in yellow is included in our solution set. So 39, 39.999999, repeating, which is about as close as you can get to 40 as possible, that's in our solution set. But 40 is not. That's why we put that open circle around it. Let's do another one. Let me do it in another color as well. right corner. Say we have x plus 15 is greater than or equal to negative 60. Notice, now we have greater than or equal. So let's solve this the same way we solved We can subtract 15 from both sides. And I like to switch up my notation. Here I added the 5, kind of, on the same line. You could also do your adding or subtracting below the line, like this. So if I subtract 15 from both sides, so I do a minus 15 there, and I do a minus 15 there. Then the left-hand side just becomes an x. Because obviously you have 15 minus 15. That just cancels out. And you get x is greater than or equal to negative 60 minus 15 is negative 75. If something is greater than or equal to something else, if I take 15 away from this and from that, the greater than or" + }, + { + "Q": "4:29 Why did the symbol > did not reversed to < hence you subtracted negative", + "A": "You re right. The symbol only reverses when multiplying or dividing negative numbers.", + "video_name": "UTs4uZhu5t8", + "timestamps": [ + 269 + ], + "3min_transcript": "And then we could go below 40. 39, 38. You can just keep going below 40. It just keeps going on in both directions. And any x that is less than 40 will satisfy this. So it can't be equal to 40, because if x is equal to 40, 40 minus 5 is 35. It's not less than 35. So x has to be less than 40. And to show this on the number line, we do a circle around 40 to show that we're not including 40. But then we can shade in everything below 40. So everything that's just exactly below 40 is included So everything I've shaded in yellow is included in our solution set. So 39, 39.999999, repeating, which is about as close as you can get to 40 as possible, that's in our solution set. But 40 is not. That's why we put that open circle around it. Let's do another one. Let me do it in another color as well. right corner. Say we have x plus 15 is greater than or equal to negative 60. Notice, now we have greater than or equal. So let's solve this the same way we solved We can subtract 15 from both sides. And I like to switch up my notation. Here I added the 5, kind of, on the same line. You could also do your adding or subtracting below the line, like this. So if I subtract 15 from both sides, so I do a minus 15 there, and I do a minus 15 there. Then the left-hand side just becomes an x. Because obviously you have 15 minus 15. That just cancels out. And you get x is greater than or equal to negative 60 minus 15 is negative 75. If something is greater than or equal to something else, if I take 15 away from this and from that, the greater than or So our solution is x is greater than or equal to negative 75. Let's graph it on the number line. So let me draw a number line here. I'll have-- let's say that's negative 75, that's negative 74, that's negative 73, that's negative 76. And so on and so forth. I could keep plotting things. Now, x has to be greater than or equal to negative 75. So x can be equal to negative 75. So we can include the point, because we have this greater than or equal sign. Notice we're not making it hollow like we did there, we're making it filled in because it can equal negative 75, or it needs to be greater than. So greater than or equal. We'll shade in everything above negative 75 as well. So in orange is the solution set. And this obviously, we could keep going to the right. x could be a million, it could be a billion, it could be a googol. It can be an arbitrarily large number as long as it's greater" + }, + { + "Q": "At 4:19, why is the square root of 2 x the square root of 2 equal to 2?", + "A": "Because sqrt(2) * sqrt(2) = sqrt(4) And, sqrt(4) = 2", + "video_name": "s9ppnjgmiyk", + "timestamps": [ + 259 + ], + "3min_transcript": "And to identify the perfect squares you would say, Alright, are there any factors where I have at least two of them? Well I have two times two here. And I also have five times five here. So I can rewrite the square root of 200 as being equal to the square root of two times two. Let me just write it all out. Actually I think I'm going to run out of space. So the square root, give myself more space under the radical, square root of two times two times five times five times two. And I wrote it in this order so you can see the perfect squares here. Well this is going to be the same thing as the square root of two times two. This second method is a little bit more monotonous, (laughing) I guess is one way to think about it. And they really, they boil down to the same method. We're still going to get to the same answer. So square root of two times two times the square root times the square root of five times five, times the square root of two. Well the square root of two times two is just going to be, this is just two. Square root of five times five, well that's just going to be five. So you have two times five times the square root of two, which is 10 times the square root of two. So this right over here, square root of 200, we can rewrite as 10 square roots of two. So this is going to be equal to one over 10 square roots of two. Now some people don't like having a radical in the denominator and if you wanted to get rid of that, you could multiply both 'Cause notice we're just multiplying by one, we're expressing one as square root of two over square root of two, and then what that does is we rewrite this as the square root of two over 10 times the square root of two times the square root of two. Well the square root of two times the square root of two is just going to be two. So it's going to be 10 times two which is 20. So it could also be written like that. So hopefully you found that helpful. In fact, even this one, you could write if you want to visualize it slightly differently, you could view it as one twentieth times the square root of two. So these are all the same thing." + }, + { + "Q": "At around the 6:15 mark, if the interval of convergence included 1 or -1, would the radius of convergence still be 1?", + "A": "To determine the radius of convergence, do not worry about whether the endpoints are included or not. The radius of convergence would be one regardless of whether or not the endpoints were included.", + "video_name": "DlBQcj_zQk0", + "timestamps": [ + 375 + ], + "3min_transcript": "minus our common ratio. What's our common ratio? Our common ratio in this example is x. Going from one term to the next, we're just multiplying by x. We're just multiplying by x right over there. Now, this is pretty neat, because we're going to be able to use this fact to put more traditionally-defined functions into this form, and then try to expand them out using a geometric series. And this whole idea of using power series, or in this special case, geometric series to represent functions, has all sorts of applications in engineering and finance. Using a finite number of terms of these series, you could kind of approximate the functions in a way that's simpler for the human brain to understand, or maybe a simpler way to manipulate in some way. But what's interesting here is instead of just going from the sum to-- instead of going from this expanded-out version to this kind of finite value, in this form and expand it out into a geometric series. But we have to be careful to make sure that we're only doing it over the interval of convergence. This is only going to be true over the interval of convergence. Now, one other term you might see in your mathematical career is a radius. Radius of convergence. And this is how far-- up to what value, but not including this value. So as long as our x value stays less than a certain amount from our c value, then this thing will converge. Now in this case, our c value is 0. So we could ask ourselves a question. As long as x stays within some value of 0, this thing is going to converge. Well, you see it right over here. As long as x stays within one of 0. as it stays less than 1, or as long as it stays greater than negative 1. It can stray anything less than one away from 0, either in the positive direction or the negative direction. Then this thing will still converge. So we could say that our radius of convergence is equal to 1. Another way to think about it, our interval of convergence-- we're going from negative 1 to 1, not including those two boundaries, so our interval is 2. So our radius of convergence is half of that. As long as x stays within one of 0, and that's the same thing as saying this right over here, this series is going to converge." + }, + { + "Q": "In this video Sal uses 6x+2 to get the solution, yet the problem has 6x-2.?? This is at 5:26/10:57 in the video:", + "A": "The quality of those questions weren t the best, if you look closely you can see that the equation really is 6x+2. I had to tilt my screen a little bit before I could tell if it was a plus or a minus.", + "video_name": "_HJljJuVHLw", + "timestamps": [ + 326, + 657 + ], + "3min_transcript": "So a pentagon has three triangles in it. So the sum of its interior angles are going to be 3 times 180, because it has 3 triangles in it. Each triangle has 180 degrees. And I know you can't see what I just wrote. So the sum of all of these angles are going to be the sum of all of the angles in all three triangles. So it's 3 times 180 is equal to 540 degrees. So that's the sum of all of these interior angles. Now, they say that each of them are 2x, 6x, et cetera. So the sum of all of these terms have to be equal 540. I'm going to write them vertically. It makes them easier to add. So if we write there 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. This is going to be the largest, right? So let's add this up. Minus 6, minus 16, that's minus 22. Plus 2 is minus 20. That's right. And 2x plus 6x that is 8x plus 4 is 12, 12 plus 2 is 14, 14 plus 6 is 20. So we have 20x minus 20 is equal to 540 degrees. 20x minus 20 is equal to 540. Let's divide both sides of this equation by 20. So you get x 1 minus 1 is equal to-- it would be 54 divided by 2, which is equal to 27. And they want to know, what is a measure in degrees of the That's going to be this one. That's the largest one. It's 6 times x plus 2. So 6 times 28, that's 48. 2 times 6 is 12 plus 4 is 168. So it's 168 plus 2. It's 170 degrees. Choice C. Problem 48: What is the measure of angle 1? So this we're going into the angle game. And these are fun, because they are kind of these deductive reasoning problems where you just use a couple of simple rules and just fill in the whole thing. So let's think about it. This is 36 degrees. They tell us that this whole angle right here is a right angle. So this angle right here is going to be the complement to 36 degrees." + }, + { + "Q": "8:52, what is equiangular ?", + "A": "Equiangular means that all the angles of the polygon/shape have equal measure. For example an equiangular triangle would have three angles which all have a degree measure of 60.", + "video_name": "_HJljJuVHLw", + "timestamps": [ + 532 + ], + "3min_transcript": "So the answer is A. Problem 49: What is the measure of angle WZX? So they want to know what this angle right here is. Let's do the angle game some more. Let's see, we can immediately figure out what this angle is, because it is the supplement of 132 degrees, so this is going to be 180 minus 132. So this is 48 degrees. This angle plus this angle is going to be equal to this angle. Or this angle plus this angle plus this angle is equal to 180. I don't know what I just said, I think I said something wrong. Write that down. So this angle is going to be equal to 180 minus 52 minus 48. equal to 180 minus 100 which equals 80 degrees. So this angle right here is equal to 80 degrees. And the angle they want us to figure out is the opposite of this angle, or in the U.S., I guess, they say vertical angles. And so opposite or vertical angles are equal or they're congruent, so this is going to be 80 degrees as well. And that is choice A. Problem 50: What is the measure of an exterior angle of a regular hexagon? A regular hexagon tells us that all of the sides are the same, it's equilateral, and all of the angles are the same, equiangular. So if we just knew what's the total degree measure of the interior angles, we could just divide that by 6, and then and then we could use that information to figure out the Let's just do it. So once again, I like to just draw a hexagon. Let's just draw a hexagon and count the triangles in it. Two sides, three sides, four sides, five sides and six sides. And how many triangles do I have here? One, two, three. So I have one, two, three, four triangles. The sum of the interior angles of this hexagon, of any hexagon, whether it's regular or not, are going to be 4 times 180 and that's 720 degrees. And it's a regular hexagon, so all the interior angles are going to be the same. And there's six of them. So each of them are going to be 720 divided by 6." + }, + { + "Q": "At 4:00 and again at 5;10, Sal says that 6x+2 is the largest angle, how can he tell?", + "A": "Among all the angles, ie, 2x, 6x, 4x - 6, 2x - 16 and 6x + 2, In 6x +2 you are multiplying x 6 times and further adding a 2. Try assigning x a value, u will find that 6x + 2 is the largest.", + "video_name": "_HJljJuVHLw", + "timestamps": [ + 240 + ], + "3min_transcript": "OK, so first of all, we have to remember what is the sum of the interior angles of a pentagon? And that's where I always draw an arbitrary pentagon. Let me see if I can do that. Actually, there's a polygon tool here. How does it work? I'm just trying to draw a pentagon. I don't know if that's any different than the line tool, So how many triangles can I draw in a pentagon? And that tells me what my total interior angles are. And there is a formula for that, but I like relying on your reasoning more than the formula, because you might forget the formula, or even worse, you might remember it, but not have the confidence to use it, or you might remember it wrong ten years in the future. So the best thing to do, if you have a polygon, is to count the triangles in it. Straightforward enough. So a pentagon has three triangles in it. So the sum of its interior angles are going to be 3 times 180, because it has 3 triangles in it. Each triangle has 180 degrees. And I know you can't see what I just wrote. So the sum of all of these angles are going to be the sum of all of the angles in all three triangles. So it's 3 times 180 is equal to 540 degrees. So that's the sum of all of these interior angles. Now, they say that each of them are 2x, 6x, et cetera. So the sum of all of these terms have to be equal 540. I'm going to write them vertically. It makes them easier to add. So if we write there 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. This is going to be the largest, right? So let's add this up. Minus 6, minus 16, that's minus 22. Plus 2 is minus 20. That's right. And 2x plus 6x that is 8x plus 4 is 12, 12 plus 2 is 14, 14 plus 6 is 20. So we have 20x minus 20 is equal to 540 degrees. 20x minus 20 is equal to 540. Let's divide both sides of this equation by 20. So you get x 1 minus 1 is equal to-- it would be 54 divided by 2, which is equal to 27." + }, + { + "Q": "At 4:01, isn't it supposed to be 6x-2 and not 6x+2?", + "A": "I thought so too. Probably we can t see it clearly, when I rewatched I thought I saw a hint of the vertical line in the +.", + "video_name": "_HJljJuVHLw", + "timestamps": [ + 241 + ], + "3min_transcript": "OK, so first of all, we have to remember what is the sum of the interior angles of a pentagon? And that's where I always draw an arbitrary pentagon. Let me see if I can do that. Actually, there's a polygon tool here. How does it work? I'm just trying to draw a pentagon. I don't know if that's any different than the line tool, So how many triangles can I draw in a pentagon? And that tells me what my total interior angles are. And there is a formula for that, but I like relying on your reasoning more than the formula, because you might forget the formula, or even worse, you might remember it, but not have the confidence to use it, or you might remember it wrong ten years in the future. So the best thing to do, if you have a polygon, is to count the triangles in it. Straightforward enough. So a pentagon has three triangles in it. So the sum of its interior angles are going to be 3 times 180, because it has 3 triangles in it. Each triangle has 180 degrees. And I know you can't see what I just wrote. So the sum of all of these angles are going to be the sum of all of the angles in all three triangles. So it's 3 times 180 is equal to 540 degrees. So that's the sum of all of these interior angles. Now, they say that each of them are 2x, 6x, et cetera. So the sum of all of these terms have to be equal 540. I'm going to write them vertically. It makes them easier to add. So if we write there 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. This is going to be the largest, right? So let's add this up. Minus 6, minus 16, that's minus 22. Plus 2 is minus 20. That's right. And 2x plus 6x that is 8x plus 4 is 12, 12 plus 2 is 14, 14 plus 6 is 20. So we have 20x minus 20 is equal to 540 degrees. 20x minus 20 is equal to 540. Let's divide both sides of this equation by 20. So you get x 1 minus 1 is equal to-- it would be 54 divided by 2, which is equal to 27." + }, + { + "Q": "at 4:10 Sal mentions that it is ar^k power. Why do the k power instead of nth power?", + "A": "k is the index, where k=0,1,2,3,...n So it would mean ar\u00e2\u0081\u00b0+ar\u00c2\u00b9+ar\u00c2\u00b2+ar\u00c2\u00b3+...+ar^n", + "video_name": "CecgFWTg9pQ", + "timestamps": [ + 250 + ], + "3min_transcript": "Well, we'll start with whatever our first term is. And over here if we want to speak in general terms we could call that a, our first term. So we'll start with our first term, a, and then each successive term that we're going to add is going to be a times our common ratio. And we'll call that common ratio r. So the second term is a times r. Then the third term, we're just going to multiply this one times r. So it's going to be a times r squared. And then we can keep going, plus a times r to the third power. And let's say we're going to do a finite geometric series. So we're not going to just keep on going forever. Let's say we keep going all the way until we get to some a times r to the n. a times r to the n-th power. And I encourage you to pause the video and try it on your own. Well, we could think about it this way. And I'll give you a little hint. You could view this term right over here as a times r to the 0. And let me write it down. This is a times r to the 0. This is a times r to the first, r squared, r third, and now the pattern might be emerging for you. So we can write this as the sum, so capital sigma right over here. We can start our index at 0. So we could say from k equals 0 all the way to k equals n of a times r to the k-th power. a general way to represent a geometric series where r is some non-zero common ratio. It can even be a negative value." + }, + { + "Q": "The example at 8:05 if they asked f(-2) would it be does not exist as the line has an open circle", + "A": "f(-2) would be undefined or does not exist because of the open circle. However, the limit as x->-2 exists and it s 4 as Sal demonstrated.", + "video_name": "nOnd3SiYZqM", + "timestamps": [ + 485 + ], + "3min_transcript": "If x is 7.5, f of 7.5 is here. So it looks like our value of f of x is getting closer and closer and closer to 3. So it looks like the limit of f of x, as x approaches 8 from the negative side, is equal to 3. What about from the positive side? What about the limit of f of x as x approaches 8 from the positive direction or from the right side? Well, here we see as x is 9, this is our f of x. As x is 8.5, this is our f of 8.5. It seems like we're approaching f of x equaling 1. So notice, these two limits are different. So the non-one-sided limit, or the two-sided limit, does not exist at f of x or as we approach 8. So let me write that down. The limit of f of x, as x approaches 8-- this does not exist. Let's do one more example. And here they're actually asking us a question. The function f is graphed below. What appears to be the value of the one-sided limit, the limit of f of x-- this is f of x-- as x approaches negative 2 from the negative direction? So this is the negative 2 from the negative direction. So we care what happens as x approaches negative 2. We see f of x is actually undefined right over there. But let's see what happens as we approach from the negative direction, or as we approach from values less than negative 2, or as we approach from the left. As we approach from the left, f of negative 4 is right over here. So this is f of negative 4. f of negative 3 is right over here. f of negative 2.5 seems to be right over here. We seem to be getting closer and closer So I would say that it looks-- at least, graphically-- the limit of f of x, as x approaches 2 from the negative direction, is equal to 4. Now, if we also asked ourselves the limit of f of x, as x approaches negative 2 from the positive direction, we would get a similar result. Now, we're going to approach from when x is 0, f of x seems to be right over here. When x is 1, f of x is right over here. When x is negative 1, f of x is there. When x is negative 1.9, f of x seems to be right over here. So once again, we seem to be getting closer and closer to 4. Because the left-handed limit and the right-handed limit are the same value. Because both one-sided limits are approaching the same thing, we can say that the limit of f of x, as x approaches" + }, + { + "Q": "At 5:25,when sal wrote the general limit of F(X) as x approaches 4,didn't he forget the minus sign in front of 5?its -5 and not 5", + "A": "Yes he did indeed forget the minus sign although he recognized that by editing in a correction box in the bottom right hand of the screen.", + "video_name": "nOnd3SiYZqM", + "timestamps": [ + 325 + ], + "3min_transcript": "equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that. As x equals 4 from below-- So when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999-- we're getting closer and closer to our function equaling negative 5. So the limit as we approach 4 from below-- this one-sided limit from the left, we could say-- this is going to be equal to negative 5. And if we were to ask ourselves the limit of f of x, as x approaches 4 from the right, from values larger than 4, well, same exercise. f of 5 gets us here. f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of 4 is actually defined, but we're getting closer and closer to it. Even if f of 4 was not defined on either side, we would be approaching negative 5. So this is also approaching negative 5. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say-- so these two things are equal. And because these two things are equal, we know that the limit of f of x, as x approaches 4, is equal to 5. Let's look at a few more examples. So let's ask ourselves the limit of f of x-- now, this is our new f of x depicted here-- as x approaches 8. And let's approach 8 from the left. As x approaches 8 from values less than 8. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to 8." + }, + { + "Q": "I've been thinking about this for a while and I can't figure out how to find the limit of an asymptote. At 6:20 as x=>3- is it negative infinity or undefined?", + "A": "Recall that an asymptote is just a line that a given function or curve tends to (gets closer and closer to). At 6:20, the asymptote is x = 3. However, if you actually meant to find the limit of f(x) as x -> 3\u00e2\u0081\u00bb, it is -\u00e2\u0088\u009e.", + "video_name": "nOnd3SiYZqM", + "timestamps": [ + 380 + ], + "3min_transcript": "As x equals 4 from below-- So when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999-- we're getting closer and closer to our function equaling negative 5. So the limit as we approach 4 from below-- this one-sided limit from the left, we could say-- this is going to be equal to negative 5. And if we were to ask ourselves the limit of f of x, as x approaches 4 from the right, from values larger than 4, well, same exercise. f of 5 gets us here. f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of 4 is actually defined, but we're getting closer and closer to it. Even if f of 4 was not defined on either side, we would be approaching negative 5. So this is also approaching negative 5. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say-- so these two things are equal. And because these two things are equal, we know that the limit of f of x, as x approaches 4, is equal to 5. Let's look at a few more examples. So let's ask ourselves the limit of f of x-- now, this is our new f of x depicted here-- as x approaches 8. And let's approach 8 from the left. As x approaches 8 from values less than 8. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to 8. If x is 7.5, f of 7.5 is here. So it looks like our value of f of x is getting closer and closer and closer to 3. So it looks like the limit of f of x, as x approaches 8 from the negative side, is equal to 3. What about from the positive side? What about the limit of f of x as x approaches 8 from the positive direction or from the right side? Well, here we see as x is 9, this is our f of x. As x is 8.5, this is our f of 8.5. It seems like we're approaching f of x equaling 1. So notice, these two limits are different. So the non-one-sided limit, or the two-sided limit, does not exist at f of x or as we approach 8. So let me write that down. The limit of f of x, as x approaches 8--" + }, + { + "Q": "At 5:29, I thought the limit of f(x) as it approaches 4 would not exist because it is a corner? Aren't corners not differentiable? Does it or does it not exist?", + "A": "The derivative of the function at 4 and limit of the function as x approaches 4 are not the same thing. The derivative does not exist at 4 because of the sharp corner. But the function is continuous at 4, and so the limit is just f(4).", + "video_name": "nOnd3SiYZqM", + "timestamps": [ + 329 + ], + "3min_transcript": "equal the limit of f of x, as x approaches 2 from the positive direction. And since this is the case-- that they're not equal-- the limit does not exist. The limit as x approaches 2 in general of f of x-- so the limit of f of x, as x approaches 2, does not exist. In order for it to have existed, these two things would have had to have been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits-- the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x, as x approaches 4 from below-- so let me draw that. As x equals 4 from below-- So when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999-- we're getting closer and closer to our function equaling negative 5. So the limit as we approach 4 from below-- this one-sided limit from the left, we could say-- this is going to be equal to negative 5. And if we were to ask ourselves the limit of f of x, as x approaches 4 from the right, from values larger than 4, well, same exercise. f of 5 gets us here. f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of 4 is actually defined, but we're getting closer and closer to it. Even if f of 4 was not defined on either side, we would be approaching negative 5. So this is also approaching negative 5. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say-- so these two things are equal. And because these two things are equal, we know that the limit of f of x, as x approaches 4, is equal to 5. Let's look at a few more examples. So let's ask ourselves the limit of f of x-- now, this is our new f of x depicted here-- as x approaches 8. And let's approach 8 from the left. As x approaches 8 from values less than 8. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to 8." + }, + { + "Q": "At 12:22 Sal wrote that N(A) = N(rref(A)). Does that stay true if we exchange rows when we are reducing the matrix to the rref?", + "A": "If you had 2 simultaneous equations (not matrices) in x and y, and you exchanged rows, how would that affect the solution?", + "video_name": "qvyboGryeA8", + "timestamps": [ + 742 + ], + "3min_transcript": "These are just random scalars that are a member of-- We can pick any real number for x3 and we could pick any real number for x4. So our solution set is just a linear combination of those two vectors. What's another way of saying a linear combination of two vectors? Let me write this. The null space of A, which is just a solution set of this equation, it's just all the x's that satisfy this equation, it equals all of the linear combinations of this vector and that vector. What do we call all the linear combinations of two vectors? It's the span of those two vectors. So it equals the span of that vector and that vector. Of the vector 1, minus 2, 1, 0, and the vector 2, minus 3, 0, 1. And this is our null space. Before letting you go, let me just point out one interesting We represented our system of equation like this and we put it into reduced row echelon form, so this is A and this is 0. This right here is, let me make sure I have some space, let me put it right here. That right there is the reduced row echelon form of A. And so where essentially this equation, this is a linear equation that is trying to solve this problem. The reduced row echelon form of A times our vector x is equal to 0. So, all the solutions to this are also the solutions to our original problem, to our original ax is equal to 0. So what's the solution to this? All the x's that satisfy this, these are the null space of the reduced row echelon form of A. Right? So here are all of the x's, this is the null space, this problem, if we find all of the x's here, this is the null But we're saying that this problem is the same problem as this one, right? So we can write that the null space of A is equal to the null space of the reduced row echelon form of A. And that might seem a little bit confusing, hey, why are you even writing this out, but it's the actually very useful when you're trying to calculate null spaces. So we didn't even have to write a big augmented matrix here. We can say, take our matrix A, put it in reduced row echelon form and then figure out it's null space. We would have gone straight to this point right here. This is the reduced row echelon form of A, and then I could have immediately solved these equations, right? I would have just taken the dot product of the reduced row echelon form or, not the dot product, the matrix vector product of the reduced row echelon form of A with this vector, and I would've gotten these equations, and then these equations would immediately, I can just rewrite them in this form, and I would" + }, + { + "Q": "At 5:43 shouldn't the third row be 0 -1 -2 -3? since u r subtracting 4 from each element.", + "A": "A logical thing to do is what you are describing which is to replace the third row by the third row minus four times the first row. What Sal is doing is kind of the reverse but it works too. He is replacing the third row by the first row multiplied by four and then subtracting the third row. Both ways work and we see that they just differ by a factor of -1. I would have done it the way you suggest in your question and then multiplied by -1 but either way you get the same result.", + "video_name": "qvyboGryeA8", + "timestamps": [ + 343 + ], + "3min_transcript": "And then I augment that with the 0 vector. And the immediate thing you should notice is we took the pain of multiplying this times this to equal that, and we wrote this as a system of equations, but now we want to solve the system of equations, we're going back to the augmented matrix world. What does this augmented matrix look like? Well, this is just our matrix A right there. That's just matrix A right there, that's just the 0 vector right there. And to solve this, and we've done this before, we're just going to put this augmented matrix into row echelon form. What you're going to find is when you put it into row echelon form, this right side's not going to change at all, because no matter what you multiply or subtract by, you're just doing it all times 0, so you just keep ending up with 0. So as we put this into reduced row echelon form, were actually just putting matrix A into reduced echelon form. So let me do that, instead of just talking about it. So let me start off by keeping row 1 the same. And then I want to eliminate this 1 right here, so let me replace row 2 with row 2 minus row 1. So 1 minus 1 is 0. 2 minus 1 is 1. 3 minus 1 is 2. 4 minus 1 is 3. 0 minus 0 is 0. You can see the 0's aren't going to change. And then let me replace this guy with 4 times this guy, minus this guy. So I can only get rid of this. So 4 times 1 minus 4 is 0. 4 times 1 minus 3 is 1. 4 times 1 minus 2 is 2. 4 times 1 minus 1 is 3. 4 times 0 minus 0 is 0. Now I want to get rid of, if I want to put this in reduced row echelon form, I want to get rid of that So let me keep my middle row the same. My middle row is 0, 1, 2, 3. So that's 0 on the augmented side of it, although these 0's are never going to change, it's really just a little bit of an exercise just to keep writing them. And my first row, let me replace it with the first row minus the second row, so I can get rid of this 1. So 1 minus 0 is 1. 1 minus 1 is 0. 1 minus 2 is minus 1. 1 minus 3 is minus 2. And 0 minus 0 is 0. And let me replace this last row with the last row minus the middle row. So 0 minus 0 is 0. 1 minus 1 is 0. 2 minus 2 is 0. I think you see where this is going. 3 minus 3 is 0. And obviously 0 minus 0 is 0. So this system of equations has been reduced, just by" + }, + { + "Q": "At 15:00, the equation for projv(x) is\nA(AT A)-1 AT x where AT = A transpose and -1 means inverse\nThis formula seems like it should reduce to x since\n\n(ATA)-1 = (A-1)(AT-1) by the rule for inverse of a product\nso (A A-1) (AT-1 AT) = I I and IIx = x?\nWhat am I missing here?", + "A": "The answer is that A and AT are rectangular, not square so they have no inverse. (ATA) is invertible if A originally had linearly independent columns as per video 106 \u00e2\u0080\u009cLin Alg: Showing that A-transpose x A is invertible\u00e2\u0080\u009d", + "video_name": "cTyNpXB92bQ", + "timestamps": [ + 900 + ], + "3min_transcript": "A transpose A inverse, which'll always exist, times A transpose, times x. Now we said the projection of x onto v is going to be equal to A times y, for some y. Well we just solved for the y using our definition of a projection. We just were able to solve for y. So now, we can define our projection of x onto v as a matrix vector product. So we can write the projection onto v of our vector x is equal to A, times y, and y is just equal to that thing right there. So A times A transpose A inverse-- which always exists transpose, times x. And this thing right here, this long convoluted thing, that's just some matrix, some matrix which always exists for any subspace that has some basis. So we've just been able to express the projection of x onto a subspace as a matrix vector product. So anything that can be any matrix vector product transformation is a linear transformation. And not only did we show that it's a linear transformation, we showed that, look, if you can give me the basis for v, I'm going to make those column vectors equal to the column of some matrix A. And then if I take matrix A, if I take its transpose, if I take A transpose times A, and invert it, and if I multiply them all out in this way, I'm going to get the transformation matrix for the projection. do by hand for many, many projections, but this is super useful if you're going to do some three-dimensional graphical programming. Let's say you have some three-dimensional object, and you want to know what it looks like from the point of view of some observer. So let's say you have some observer. Some observer's point of view is essentially going to be some subspace. You want to see what the projection of this cube onto the subspace, how would it look to the person who's essentially on to this flat screen right there. How would that cube look from this point of view? Well if you know the basis for this subspace, you can just apply this transformation. You can make a matrix whose columns are these basis vectors for this observer's point of view. And then you can apply this to every vector in this cube in" + }, + { + "Q": "At 2:10, why is e^(-st) * e^(at) combine to e^(a-s)t instead of e^(a-s) ?? Wouldn't the -t and t combine to cancel??", + "A": "For multiplying terms which have the same base, (x^a) * (x^b) = x^(a + b). So, e^(-st) * e^(at) = e^(-st + at). Take out the common factor, t from (-st + at) => t(a-s), which gives us e^(a-s)t", + "video_name": "33TYoybjqPg", + "timestamps": [ + 130 + ], + "3min_transcript": "Let's keep doing some Laplace transforms. For one, it's good to see where a lot of those Laplace transform tables you'll see later on actually come from, and it just makes you comfortable with the mathematics. Which is really just kind of your second semester calculus mathematics, but it makes you comfortable with the whole notion of what we're doing. So first of all, let me just rewrite the definition of the So it's the L from Laverne & Shirley. So the Laplace transform of some function of t is equal to the improper integral from 0 to infinity of e to the minus st times our function. Times our function of t, and that's with respect to dt. So let's do another Laplace transform. Let's say that we want to take the Laplace transform-- and now our function f of t, let's say it is e to the at. Well we just substituted it into this definition of the Laplace transform. And this is all going to be really good integration Especially integration by parts. Almost every Laplace transform problem turns into an integration by parts problem. Which, as we learned long ago, integration by parts is just the reverse product rule. This is equal to the integral from 0 to infinity. e to the minus st times e to the at, right? That's our f of t. dt. Well this is equal to just adding the exponents because we have the same base. The integral from 0 to infinity of e to the a minus stdt. Well that's equal to what? With respect to C. So it's equal to-- a minus s, that's just going to be a So we can just leave it out on the outside. 1/a minus s times e to the a minus st. And we're going to evaluate that from t is equal to infinity or the limit as t approaches infinity to t is equal to 0. And I could have put this inside the brackets, but it's just a constant term, right? None of them have t's in them, so I can just pull them out. And so this is equal to 1/a minus s times-- now we essentially have to evaluate t at infinity. So what is the limit at infinity?" + }, + { + "Q": "At 2:50, why is theta taken as the obtuse angle, and not the acute angle?Isn't cos (-60)= -1/2 also?", + "A": "cos(x) = cos(-x)", + "video_name": "eTDaJ4ebK28", + "timestamps": [ + 170 + ], + "3min_transcript": "If I say, you know, what is the inverse cosine of x, my brain says, what angle can I take the cosine of to get x? So with that said, let's try it out on an example. Let's say that I have the arc, I'm told, no, two c's there, I'm told to evaluate the arccosine of minus 1/2. My brain, you know, let's say that this is going to be equal to, it's going to be equal to some angle. And this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2. And as soon as you put it in this way, at least for my brain, it becomes a lot easier to process. So let's draw our unit circle and see if we can make some headway here. So that's my, let me see if I can draw a little straighter. Maybe I could actually draw, put rulers here, and if I put a ruler here, maybe I can draw a straight line. OK, so that is my y-axis, that is my x-axis. Not the most neatly drawn axes ever, but it'll do. Let me draw my unit circle. Looks more like a unit ellipse, but you get the idea. And the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle. So if we have some angle, the x-value is going to be equal a minus 1/2. So we got a minus 1/2 right here. And so the angle that we have to solve for, our theta, is the angle that when we intersect the unit circle, the x-value is minus 1/2. So let me see, this is the angle that we're trying to figure out. This is theta that we need to determine. So how can we do that? So this is minus 1/2 right here. Let's figure out these different angles. out this angle right here. And if I know that angle, I can just subtract that from 180 degrees to get this light blue angle that's kind of the solution to our problem. So let me make this triangle a little bit bigger. So that triangle, let me do it like this. That triangle looks something like this. Where this distance right here is 1/2. That distance right there is 1/2. This distance right here is 1. Hopefully you recognize that this is going to be a 30, 60, 90 triangle. You could actually solve for this other side. You'll get the square root of 3 over 2. And to solve for that other side you just need to do the Pythagorean theorem. Actually, let me just do that. Let me just call this, I don't know, just call this a. So you'd get a squared, plus 1/2 squared, which is 1/4, which is equal to 1 squared, which is 1. You get a squared is equal to 3/4, or a is equal to the" + }, + { + "Q": "At 12:05, what if there is bigger angle like 100 pi instead of 3pi?\nDo you still keep going around the unit circle to find out where you finally land up and then calculate the x coordinate?", + "A": "You could go around the circle 50 times to finally figure out where you land. However, if you understand that 2\u00cf\u0080 radians is a full circle, then you can save yourself a lot of trouble. 100\u00cf\u0080 = 50\u00e2\u0080\u00a22\u00cf\u0080 This means 100\u00cf\u0080 is the same as 50 full circles, and you end up on the same point on the unit circle as 0 radians. When this happens we say 100\u00cf\u0080 is coterminal to 0.", + "video_name": "eTDaJ4ebK28", + "timestamps": [ + 725 + ], + "3min_transcript": "x that you put in here. This is true for any x, any value between negative 1 and 1 including those two endpoints, this is going to be true. Now what if I were ask you what the arccosine of the cosine of theta is? What is this going to be equal to? My answer is, it depends on the theta. So, if theta is in the, if theta is in the range, if theta is between, if theta is between 0 and pi, so it's in our valid a range for, kind of, our range for the product of the arccosine, then this will be equal to theta. If this is true for theta. But what if we take some theta out of that range? Let's try it out. Let's take, so let me do one with theta in that range. one of them that we know. Let's take the cosine of, let's stick with cosine of 2 pi over 3. Cosine of 2 pi over 3 radians, that's the same thing as the arccosine of minus 1/2. Cosine of 2 pi over 3 is minus 1/2. We just saw that in the earlier part of this video. And then we solved this. We said, oh, this is equal to 1 pi over 3. So for in the range of thetas between 0 and pi it worked. And that's because the arccosine function can only produce values between 0 and pi. But what if I were to ask you, what is the arccosine of the cosine of, I don't know, of 3 pi. So if I were to draw the unit circle here, let me draw the unit circle, a real quick one. What's 3 pi? 2 pi is if I go around once. And then I go around another pi, so I end up right here. So I've gone around 1 1/2 times the unit circle. So this is 3 pi. What's the x-coordinate here? It's minus 1. So cosine of 3 pi is minus 1, right? So what's arccosine of minus 1? Arccosine of minus 1. Well remember, the range, or the set of values, that arccosine can evaluate to is in this upper hemisphere. It's between, this can only be between pi and 0. So arccosine of negative 1 is just going to be pi. So this is going to be pi. Arccosine of negative, this is negative 1, arccosine of negative 1 is pi. And that's a reasonable statement, because the difference between 3 pi and pi is just going around the unit" + }, + { + "Q": "When he says that it looks more like an ellipse at 2:20, what is an ellipse, please?", + "A": "Basically, an ellipse is an oval. More technically, here s the definition from wikipedia: In mathematics, an ellipse is a curve on a plane that surrounds two focal points such that the sum of the distances to the two focal points is constant for every point on the curve. As such, it is a generalization of a circle, which is a special type of an ellipse that has both focal points at the same location.", + "video_name": "eTDaJ4ebK28", + "timestamps": [ + 140 + ], + "3min_transcript": "I've already made videos on the arcsine and the arctangent, so I've already made videos on the arcsine and the arctangent, so to kind of complete the trifecta, I might as well make a video on the arccosine. And just like the other inverse trigonometric functions, the arccosine is kind of the same thought process. If I were to tell you the arc, no, I'm doing cosine, if our tell you that arccosine of x is equal to theta. This is an equivalent statement to saying that the inverse cosine of x is equal to theta. These are just two different ways of writing the exact same thing. And as soon as I see either an arc- anything, or an inverse trig function in general, my brain immediately rearranges this. My brain immediately says, this is saying that if I take the cosine of some angle theta, that I'm going to get x. Or that same statement up here. If I say, you know, what is the inverse cosine of x, my brain says, what angle can I take the cosine of to get x? So with that said, let's try it out on an example. Let's say that I have the arc, I'm told, no, two c's there, I'm told to evaluate the arccosine of minus 1/2. My brain, you know, let's say that this is going to be equal to, it's going to be equal to some angle. And this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2. And as soon as you put it in this way, at least for my brain, it becomes a lot easier to process. So let's draw our unit circle and see if we can make some headway here. So that's my, let me see if I can draw a little straighter. Maybe I could actually draw, put rulers here, and if I put a ruler here, maybe I can draw a straight line. OK, so that is my y-axis, that is my x-axis. Not the most neatly drawn axes ever, but it'll do. Let me draw my unit circle. Looks more like a unit ellipse, but you get the idea. And the cosine of an angle as defined on the unit circle definition is the x-value on the unit circle. So if we have some angle, the x-value is going to be equal a minus 1/2. So we got a minus 1/2 right here. And so the angle that we have to solve for, our theta, is the angle that when we intersect the unit circle, the x-value is minus 1/2. So let me see, this is the angle that we're trying to figure out. This is theta that we need to determine. So how can we do that? So this is minus 1/2 right here. Let's figure out these different angles." + }, + { + "Q": "at 9:00, shouldn't we square it and add then take square root?", + "A": "Only if you are working with the magnitude (unit vector etc.). However in this example Sal is purely working with the vector itself in full (not the unit vector) and he s simply adding the two vectors together to get the resultant.", + "video_name": "6Kw2nIwWYL0", + "timestamps": [ + 540 + ], + "3min_transcript": "Scaled up version of the J unit vector and if you add this orange vector to this green vector, you get vector A. Similarly, vector B is equal to the length of the horizontal component and we got to be very careful. It's length is square root of two but it's going in the leftward direction. We're going to put a negative on it times I. If we just squared it, I is doing something like this, square root of two times I would look like that. Negative square root of two would point it to the left. This is negative square root of two times I and then we're going to have plus square root of two times J. Now that we've broken them up in their components, we're ready to figure out what, at least broken up into its components A plus B is equal to, well it's going to be the sum of all of these things. Let me just write that down. It's going to be that, copy and paste. That plus this, let me just copy and paste it and you're going to get that but of course we can simplify this. We can add the I unit vectors to each other. If I have three times the square root of three over two Is and then I have another negative square root of two I, I can add that together. This is going to be equal to three times the square root of three over two minus square root of two times I and then I can add this to this and I'm going to get plus three halves plus square root of two times J. It looks a little bit complicated and get approximations of each of these two values and we essentially have a at least a broken down into its components representation of A plus B. In the next video, we're now going to take this and figure out the actual magnitude and direction of A plus B." + }, + { + "Q": "Why is around 2:30 mins in... (1/3) dividing 3x but on the other side it is multiplying!", + "A": "He s saying that multiplying both sides of the equation by the fraction (1/3) is the same as dividing both sides of the equation by 3.", + "video_name": "kbqO0YTUyAY", + "timestamps": [ + 150 + ], + "3min_transcript": "So once again, we have three equal, or we say three identical objects. They all have the same mass, but we don't know what the mass is of each of them. But what we do know is that if you total up their mass, it's the same exact mass as these nine objects And each of these nine objects have a mass of 1 kilograms. So in total, you have 9 kilograms on this side. And over here, you have three objects. They all have the same mass. And we don't know what it is. We're just calling that mass x. And what I want to do here is try to tackle this a little bit more symbolically. In the last video, we said, hey, why don't we just multiply 1/3 of this and multiply 1/3 of this? And then, essentially, we're going to keep things balanced, because we're taking 1/3 of the same mass. This total is the same as this total. That's why the scale is balanced. Now, let's think about how we can represent this symbolically. So the first thing I want you to think about is, can we set up an equation that expresses that we have these three things of mass x, Can we express that as an equation? And I'll give you a few seconds to do it. Well, let's think about it. Over here, we have three things with mass x. So their total mass, we could write as-- we could write their total mass as x plus x plus x. And over here, we have nine things with mass of 1 kilogram. I guess we could write 1 plus 1 plus 1. That's 3. Plus 1 plus 1 plus 1 plus 1. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9. And actually, this is a mathematical representation. If we set it up as an equation, it's an algebraic representation. It's not the simplest possible way we can do it, but it is a reasonable way to do it. If we want, we can say, well, if I have an x plus another x plus another x, I have three x's. So I could rewrite this as 3x. And 3x will be equal to? Well, if I sum up all of these 1's right over here-- 1 We have 9 of them, so we get 3x is equal to 9. And let me make sure I did that. 1, 2, 3, 4, 5, 6, 7, 8, 9. So that's how we would set it up. And so the next question is, what would we do? What can we do mathematically? Actually, to either one of these equations, but we'll focus on this one right now. What can we do mathematically in order to essentially solve for the x? In order to figure out what that mystery mass actually is? And I'll give you another second or two to think about it. Well, when we did it the last time with just the scales we said, OK, we've got three of these x's here. We want to have just one x here. So we can say, whatever this x is, if the scale stays balanced, it's going to be the same as whatever we have there. There might be a temptation to subtract two of the x's maybe from this side, but that won't help us. And we can even see it mathematically over here. If we subtract two x's from both sides, on the left-hand side you're going to have 3x minus 2x. And on the right-hand side, you're going to have 9 minus 2x." + }, + { + "Q": "I got lost at 3:00 when Sal wrote 9 - 2x", + "A": "pretend 5=x. 9-2x=-1", + "video_name": "kbqO0YTUyAY", + "timestamps": [ + 180 + ], + "3min_transcript": "Can we express that as an equation? And I'll give you a few seconds to do it. Well, let's think about it. Over here, we have three things with mass x. So their total mass, we could write as-- we could write their total mass as x plus x plus x. And over here, we have nine things with mass of 1 kilogram. I guess we could write 1 plus 1 plus 1. That's 3. Plus 1 plus 1 plus 1 plus 1. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9. And actually, this is a mathematical representation. If we set it up as an equation, it's an algebraic representation. It's not the simplest possible way we can do it, but it is a reasonable way to do it. If we want, we can say, well, if I have an x plus another x plus another x, I have three x's. So I could rewrite this as 3x. And 3x will be equal to? Well, if I sum up all of these 1's right over here-- 1 We have 9 of them, so we get 3x is equal to 9. And let me make sure I did that. 1, 2, 3, 4, 5, 6, 7, 8, 9. So that's how we would set it up. And so the next question is, what would we do? What can we do mathematically? Actually, to either one of these equations, but we'll focus on this one right now. What can we do mathematically in order to essentially solve for the x? In order to figure out what that mystery mass actually is? And I'll give you another second or two to think about it. Well, when we did it the last time with just the scales we said, OK, we've got three of these x's here. We want to have just one x here. So we can say, whatever this x is, if the scale stays balanced, it's going to be the same as whatever we have there. There might be a temptation to subtract two of the x's maybe from this side, but that won't help us. And we can even see it mathematically over here. If we subtract two x's from both sides, on the left-hand side you're going to have 3x minus 2x. And on the right-hand side, you're going to have 9 minus 2x. minus 2 something is just 1 of something. So you will just have an x there if you get rid of two of them. But on the right-hand side, you're going to get 9 minus 2 So the x's still didn't help you out. You still have a mystery mass on the right-hand side. So that doesn't help. So instead, what we say is-- and we did this the last time. We said, well, what if we took 1/3 of these things? If we take 1/3 of these things and take 1/3 of these things, we should still get the same mass on both sides because the original things had the same mass. And the equivalent of doing that mathematically is to say, why don't we multiply both sides by 1/3? Or another way to say it is we could divide both sides by 3. Multiplying by 1/3 is the same thing as dividing by 3. So we're going to multiply both sides by 1/3. When you multiply both sides by 1/3-- visually over here, if you had three x's, you multiply it by 1/3, you're only going to have one x left. If you have nine of these one-kilogram boxes, you multiply it by 1/3, you're only going to have three left. And over here, you can even visually-- if you divide by 3, which is the same thing as multiplying" + }, + { + "Q": "6:54 anyone happen to know what the name of the theorem is or where to find out? I seem to remember better if I have a name for these things. Thanks for any help.", + "A": "I think it s called the Base Angle Theorem . Hope that helps a lil... :)", + "video_name": "nMhJLn5ives", + "timestamps": [ + 414 + ], + "3min_transcript": "Well, hopefully you know the angles in a triangle add up to 180 degrees. If you don't it's my fault because I haven't taught you that already. So let's figure out what the angles of this triangle add up to. Well, I mean we know they add up to 180, but using that information, we could figure out what this angle is. Because we know that this angle is 90, this angle is 45. So we say 45-- lets call this angle x; I'm trying to make it messy --45 plus 90-- this [INAUDIBLE] is a 90 degree angle --plus is equal to 180 degrees. And that's because the angles in a triangle always add up to 180 degrees. So if we just solve for x, we get 135 plus x is equal to 180. Subtract 135 from both sides. We get x is equal to 45 degrees. x is also 45 degrees. So we have a 90 degree angle and two 45 degree angles. Now I'm going to give you another theorem that's not named after the head of a religion or the founder of religion. I actually don't think this theorem doesn't have a name at. All It's the fact that if I have another triangle --I'm going to draw another triangle out here --where two of the base angles are the same-- and when I say base angle, I just mean if these two angles are the same, let's call it a. They're both a --then the sides that they don't share-- these angles share this side, right? --but if we look at the sides that they don't share, we know that these sides are equal. I forgot what we call this in geometry class. Maybe I'll look it up in another presentation; But I got this far without knowing what the name of the theorem is. If I were to change one of these angles, the length would also change. Or another way to think about it, the only way-- no, I don't confuse you too much. But you can visually see that if these two sides are the same, then these two angles are going to be the same. If you changed one of these sides' lengths, then the angles will also change, or the angles will not be equal anymore. But I'll leave that for you to think about. But just take my word for it right now that if two angles in a triangle are equivalent, then the sides that they don't share are also equal in length. Make sure you remember: not the side that they share-- because that can't be equal to anything --it's the side that they don't share are equal in length. So here we have an example where we have to equal angles. They're both 45 degrees. So that means that the sides that they don't share-- this is the side they share, right? Both angle share this side --so that means that the side that" + }, + { + "Q": "what 12:00 in night?", + "A": "12:00 at night is also known to many as midnight.", + "video_name": "ftndEjAg6qs", + "timestamps": [ + 720 + ], + "3min_transcript": "" + }, + { + "Q": "Is it correct to say three-oh-five for 3:05", + "A": "I don t know. Though I think it would be five-past-three.", + "video_name": "ftndEjAg6qs", + "timestamps": [ + 185 + ], + "3min_transcript": "We're asked, what time is it? So first, we want to look at the hour hand, which is the shorter hand, and see where it is pointing. So this right over here would have been 12 o'clock, 1 o'clock, 2 o'clock, 3 o'clock, 4 o'clock. And it looks like it's a little bit past 4 o'clock. So we are in the fourth hour. So the hour is 4. And then we have to think about the minutes. The minutes are the longer hand, and every one of these lines represent 5 minutes. We start here. This is 0 minutes past the hour, then 5 minutes past the hour, then 10 minutes past the hour. So the time is-- the minutes are 10, 10 minutes past the hour, and the hour is 4, or it's 4:10. Let's do a few more. What time is it? So first, we want to look at the hour hand. That's the shorter hand right over here. It's at-- let's see. This is 12, 1, 2, 3, 4, 5, 6, 7, 8, 9. It's just past 9. So it's still in the ninth hour. It hasn't gotten to the 10th hour yet. The ninth hour's from starting with 9 all the way until it's right almost before it gets to 10, and then it gets to the 10th hour. So the hour is 9, and then we want the minutes. Well, we can just count from 0 starting at the top of the clock. So 0, 5, 10, 15, 20, 25, 30. It's 9:30. And that also might make sense to you, because we know there are 60 minutes in an hour. And this is exactly halfway around the clock. And so half of 60 is 30. Let's do one more. What time is it? This is 12, 1-- actually, we can even count backwards. We can go 12, 11, 10. So right now we're in the 10th hour. The hour hand has passed 10, but it hasn't gotten to 11 yet. So we are in the 10th hour. So this would be 0, 5, 10, 15, 20 minutes past the hour. That's where the longer hand is pointing. It is 10:20." + }, + { + "Q": "how can i know that xy=5. at 5:53", + "A": "Look at the video. He says, that he has more of these xs and ys. He only gave us a hint how heavy xy is.", + "video_name": "h9ZgZimXn2Q", + "timestamps": [ + 353 + ], + "3min_transcript": "Well think about it this way, we know that X plus Y is equal to 5. So if we were to get rid of an X and a Y on this side,on the left hand side of the equation. What would we have to get rid of on the right hand side of the, or if we know if we get rid of X and Y on the left hand side of the scale What would we get rid of the right hand side of the scale to take away the same mass? Well if we take away the X and Y on the left hand side, we know that an X plus Y is 5kg. So, we'll just have to take 5kg from the right hand side. So, lets think about what that would do. Well then I'll just have an X over here, I'll just have some of these masses left over here.Then I would what X is. Now lets think about how we can represent that algebraically Essentially for taking an X and Y from the left hand side. If I'm taking an X and Y from the left hand side. I'm subtracting an X, and I'm subtracting an X. Actually let me think of it this way. I'm subtracting an X plus Y. I'm subtracting an X and Y on the left hand side. Well an X and a Y we know has a mass of 5. So we can subtract 5 from the right hand side. And the only way I'm gonna be able to do this is because of the information that we got from the second scale. So I can take away 5. So this is going to be equal to taking away 5. Taking away X and a Y is equal to taking away 5. And we know that because an X and a Y is equal to 5kg. And if we take away an X and a Y on the left hand side, what do we left with? Well this is gonna be the same thing. Let me rewrite this part. This, taking away an X and a Y is the same thing if you distribute the negative sign as taking away an X and taking away a Y. And so on the left hand side, we're left with just 2X and we have taken away one of the X's, we're left just an X. We see that visually, we're left with just an X here. And what do we have on the right hand side? We had 8 and we know X and Y is equal to 5, so we took away 5. So to keep the scale balance. And so 8 minus 5 is going to be 3. 8 minus 5 is equal to 3 and just like that using this extra information we're able to figure out that the mass of X is equal to 3. Now, one final question. We're able to figure out the mass of X, can you figure out what the mass of Y is. Well we can go back to either one of these scales. Probably be simpler to go back to this one. We know that the mass of X plus the mass of Y is equal to 5. So we could say, one thing we know that X is now is equal to 3." + }, + { + "Q": "This makes no sense to me once he jumps to the other scale. Essentially what he did was subtract \"x\" from the left side, and ended up with 5 on the right (in which case, the right should be 8 -x, NOT 5.) We don't have a scale in real life to keep \"adding blocks until it equals out\", this is useless, plus this automatically means x = 3 (and thus y = 2.)\n\nCan someone explain the leap in logic starting at 2:20 in the video?", + "A": "The way he put the x and y on the left and the 5kg on the right was NOT derived from previous knowledge. In other words, someone came up to you and told you x+y=5 .", + "video_name": "h9ZgZimXn2Q", + "timestamps": [ + 140 + ], + "3min_transcript": "so now we have a very very very interesting problem on the left hand side of the scale I have two different types of unknown masses one of these X masses and we know that they have the same identical mass, we call that identical each of them having a mass of X But then we have this other blue thing and that has a mass of Y, which isn't necessarily going to be the same as the mass of X. We have two of these X's and a Y.It seems like the total mass or it definitely is the case, their total mass balance it out to these 8 kg right over here. Each of these is 1kg block and balances them out. So the first question I'm going to ask you is, can you express this Mathematically? Can you express what we're seeing here, the fact that this total mass balances out with this total mass. Can you express that mathematically? Well let's just think about our total mass on this side. We have two masses of mass X so those two are gonna total at 2X, and then you have a mass of Y. So,it doesn't get too spread out. On the left hand side, I got 2X plus a mass of Y. That's the total mass. The total mass on the left hand side is 2X plus Y, the total mass on the right hand side is just 8. 1,2,3,4,5,6,7,8. It is equal to 8 And since we see that the scale is balance, this total mass must be equal to this total mass. So, we can write an equal sign there. Now my question to you is there anything we can do just based on the information that we have here to solve for either the mass X or for the mass Y. Is there anything that we can do. Well the simple answer is just with this information here, there's actually very little. You might say that \"Oh well, let me take the Y from both sides\" You might take this Y block up. But if you take this Y block up you have to take away Y from this side and you don't know what Y is. So, you're not gonna get rid of the Y. Same thing with the X's, you actually don't have enough information. Y depends on what X is,and X depends on what Y is. Lucky for us however,we do have some more of these blocks laying around. And what we do is we take one of these X blocks. And I stack it over here,and I also take one of the Y block and I stack it right over there. And then I keep adding all these ones until I balance these things out. So, I keep adding these ones. Obviously if I just place this, this will go down cause there's nothing on that side. But I keep adding these blocks until it all balances out and I find that my scale balances once I have 5 kg on the right hand side So, once again let me ask you this information having X and Y on the left hand side and a 5kg on the right hand side" + }, + { + "Q": "at 1:58 , why did sal divide the numerator by ( x-1 ) , shouldn't it be divided by (x^2 - 1) .\n\nWas this done to actually find factors for which x-1 could be eliminated .", + "A": "Sal divided the numerator by the factor (x-1) to check whether (x-1) was a factor of the polynomial (x^3 - 1), which it is. He was trying to split the numerator into factors so that he could cancel out the factor (x-1) in the main expression. The reason for doing this is because he does not want the denominator to be 0 when he plugs in x=1 to find the limit of the expression algebraically.", + "video_name": "rU222pVq520", + "timestamps": [ + 118 + ], + "3min_transcript": "Let's try to find the limit as x approaches 1 of x to the third minus 1 over x squared minus 1. And at first when you just try to substitute x equals 1, you get 0/0 1 minus 1 over 1 minus 1. So that doesn't help us. So let's see if we can try to simplify this in some way. So you might immediately recognize-- so let's rewrite this expression right over here so it's x to the third minus 1 over x squared minus 1. This on the bottom immediately jumps out as a difference of squares. So we know on the bottom that this could be factored as x minus 1 times x plus 1. And so if somehow this thing on the top also has an x minus 1 as a factor, then that x minus 1 will cancel with this, and then we're not going to have an issue of dividing by 0. The reason why I care about the x minus 1 term is that this is what's making our denominator equal 0. When you say x equals 1, you have 1 minus 1 times 1 plus 1. So if we can have an x minus 1 up here, then we can cancel these out for any x not equal to 1. And then we might have a much simpler thing to find the limit of. So let's think about whether x to the third minus 1 is the product of x minus 1 and something else. And to do that we can do a little bit of algebraic long division. Some of you guys might already recognize a pattern here, but we'll try to do-- well, let's divide x minus 1 into it to see whether it divides evenly into x to the third minus 1. So x minus 1-- we just look at the highest degree term-- x goes into x to the third x squared times. Goes x squared times. Actually, let me do it this way so that way we can keep track of the place. So this would be x-- this would be the second degree place, first degree place, and this would be the constant. So x to the third minus 1. x squared times x is x to the third. x squared times negative 1 is minus x squared. And now we're going to want to subtract this. So we are then left with x squared. x goes into x squared x times plus x. x times x is x squared. x times minus 1 is minus x. And once again we're going to subtract this. We'll swap the signs, negative and positive. And so these cancel out, and we're left with x. And then we bring down a minus 1. x minus 1 goes into x minus 1 exactly one time. 1 times x minus 1 is x minus 1. And then you subtract, and then you have no remainder. So this numerator right over here can be factored as x minus 1 times x squared plus x plus 1. And so we can say that this is the same exact thing." + }, + { + "Q": "And what if at 3:15 x=(-1) ? Is the function undefined there or we can say that as \"x\" does not equal \"1\" it neither equals \"-1\" ?\nThank you.", + "A": "The function is undefined at x = 1 and x = -1. For purposes of this problem we don t care that the function is undefined at x = -1 because we re taking the limit at x = 1. As Sal points out, we also don t have to be concerned that it isn t defined at x = 1 because the operation involved in taking a limit doesn t require a function to be defined at the point where you re taking the limit.", + "video_name": "rU222pVq520", + "timestamps": [ + 195 + ], + "3min_transcript": "So if we can have an x minus 1 up here, then we can cancel these out for any x not equal to 1. And then we might have a much simpler thing to find the limit of. So let's think about whether x to the third minus 1 is the product of x minus 1 and something else. And to do that we can do a little bit of algebraic long division. Some of you guys might already recognize a pattern here, but we'll try to do-- well, let's divide x minus 1 into it to see whether it divides evenly into x to the third minus 1. So x minus 1-- we just look at the highest degree term-- x goes into x to the third x squared times. Goes x squared times. Actually, let me do it this way so that way we can keep track of the place. So this would be x-- this would be the second degree place, first degree place, and this would be the constant. So x to the third minus 1. x squared times x is x to the third. x squared times negative 1 is minus x squared. And now we're going to want to subtract this. So we are then left with x squared. x goes into x squared x times plus x. x times x is x squared. x times minus 1 is minus x. And once again we're going to subtract this. We'll swap the signs, negative and positive. And so these cancel out, and we're left with x. And then we bring down a minus 1. x minus 1 goes into x minus 1 exactly one time. 1 times x minus 1 is x minus 1. And then you subtract, and then you have no remainder. So this numerator right over here can be factored as x minus 1 times x squared plus x plus 1. And so we can say that this is the same exact thing. So that is equal to x squared plus x plus 1 over x plus 1, for x does not equal 1. And that's completely fine, because we're not evaluating x equals 1. We're evaluating as x approaches 1. So this is going to be the same thing as the limit as x approaches 1 of x squared plus x plus 1 over x plus 1. And now this is much easier to find. You could literally just say, well, what happens as we get right to x equals 1? Then you have 1 squared, which is 1 plus 1 plus 1, which is 3, over 1 plus 1, which is 2. So we get that equaling 3/2." + }, + { + "Q": "I lost you at 1:53 when you said x goes in x^2 times, then backtracked and said \"actually lemme do..\" I am so confused -_-. How do you factor this problem.", + "A": "As far as I understood it, he used polynomial division after he had factored the denominator.", + "video_name": "rU222pVq520", + "timestamps": [ + 113 + ], + "3min_transcript": "Let's try to find the limit as x approaches 1 of x to the third minus 1 over x squared minus 1. And at first when you just try to substitute x equals 1, you get 0/0 1 minus 1 over 1 minus 1. So that doesn't help us. So let's see if we can try to simplify this in some way. So you might immediately recognize-- so let's rewrite this expression right over here so it's x to the third minus 1 over x squared minus 1. This on the bottom immediately jumps out as a difference of squares. So we know on the bottom that this could be factored as x minus 1 times x plus 1. And so if somehow this thing on the top also has an x minus 1 as a factor, then that x minus 1 will cancel with this, and then we're not going to have an issue of dividing by 0. The reason why I care about the x minus 1 term is that this is what's making our denominator equal 0. When you say x equals 1, you have 1 minus 1 times 1 plus 1. So if we can have an x minus 1 up here, then we can cancel these out for any x not equal to 1. And then we might have a much simpler thing to find the limit of. So let's think about whether x to the third minus 1 is the product of x minus 1 and something else. And to do that we can do a little bit of algebraic long division. Some of you guys might already recognize a pattern here, but we'll try to do-- well, let's divide x minus 1 into it to see whether it divides evenly into x to the third minus 1. So x minus 1-- we just look at the highest degree term-- x goes into x to the third x squared times. Goes x squared times. Actually, let me do it this way so that way we can keep track of the place. So this would be x-- this would be the second degree place, first degree place, and this would be the constant. So x to the third minus 1. x squared times x is x to the third. x squared times negative 1 is minus x squared. And now we're going to want to subtract this. So we are then left with x squared. x goes into x squared x times plus x. x times x is x squared. x times minus 1 is minus x. And once again we're going to subtract this. We'll swap the signs, negative and positive. And so these cancel out, and we're left with x. And then we bring down a minus 1. x minus 1 goes into x minus 1 exactly one time. 1 times x minus 1 is x minus 1. And then you subtract, and then you have no remainder. So this numerator right over here can be factored as x minus 1 times x squared plus x plus 1. And so we can say that this is the same exact thing." + }, + { + "Q": "In 5:02 he says AB is congruent to segment AC but he wrote AC before AB", + "A": "In Geometry the order only matters when you say AC and AB individually. a is congruent to a, while C is congruent to B. saying CA is congruent to AB would be incorrect. Tell me if it is still unclear, I m not the best explainer", + "video_name": "7UISwx2Mr4c", + "timestamps": [ + 302 + ], + "3min_transcript": "You have two triangles that have three sides that are congruent, or they have the same length. Then the two triangles are congruent. And what's useful about that is if these two triangles are congruent, then their corresponding angles are congruent. And so we've actually now proved our result. Because the corresponding angle to ABC in this triangle is angle ACD in this triangle right over here. So that we then know that angle ABC is congruent to angle ACB. So that's a pretty neat result. If you have an isosceles triangle, a triangle where two of the sides are congruent, then their base angles, these base angles, are also going to be congruent. Now let's think about it the other way. Can we make the other statement? If the base angles are congruent, So let's try to construct a triangle and see if we can prove it the other way. So I'll do another triangle right over here. Let me draw another one just like that. That's not that pretty of a triangle, so let me draw it a little nicer. I'm going to draw it like this. Let me do that in a different color. So I'll call that A. I will call this B. I will call that C right over there. And now we're going to start off with the idea that this angle, angle ABC, is congruent to angle ACB. So they have the same exact measure. And what we want to do in this case-- we want to prove-- so let me draw a little line here to show that we're doing a different idea. Here we're saying if these two sides are the same, then the base angles are going to be the same. We've proved that. Now let's go the other way. If the base angles are the same, do we know that the two sides are the same? So we want to prove that segment AC is congruent to AB. which we would denote that way, is equal to the length of segment AB. These are essentially equivalent statements. Once again in our toolkit, we have our congruency theorems. But in order to apply them, you really do need to have two triangles. So let's construct two triangles here. And this time, instead of defining another point as the midpoint, I'm going to define D this time as the point that if I were to go straight up, the point that is essentially-- if you view BC as straight horizontal, the point that goes straight down from A. And the reason why I say that is there's some point-- you could call it an altitude-- that intersects BC at a right angle." + }, + { + "Q": "Sal uses a calculator throughout the video however, calculators are not allowed in my school so can someone explain how he got 10.7 around the 2:40 mark?", + "A": "Honestly, if you re not allowed calculators, you should probably just leave the answer in terms of tangent, sine, or cosine unless it s an easy value to find. 65\u00c2\u00b0 isn t an easy value to find, so this should be an acceptable answer: a = 5*tan(65\u00c2\u00b0) (this is actually a more exact answer than 10.7)", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 160 + ], + "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse." + }, + { + "Q": "At 4:50 why doesn't he multiply both sides by 5 instead of multiplying them by \"b\"?", + "A": "If you multiply both sides by 5 you would get this: 5cos65 = 25/b Our goal is to get b on its own so multiplying by 5 doesn t help.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 290 + ], + "3min_transcript": "This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse. And so what trigonometric ratios-- or we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that XY is exactly 5 and we don't have to deal with this approximation, let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well we see from SohCahToa cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees is equal to the length of the adjacent side, which is 5 over the length of the hypotenuse, which has a length of b. And then we can try to solve for b. You multiply both sides times b, you're left with b times cosine of 65 degrees is equal to 5. And then to solve for b, you could divide both sides by cosine of 65 degrees. So we're just dividing-- we have to figure it out what our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees. And we're left with b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is 5 divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that 5 squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out" + }, + { + "Q": "At 5:54 he says \" you could of solved this using the Pythagorean theorem... But there is an issue if im not mistaken:\n\n10.7*10.7 + 5*5 does not equal to 11.8?", + "A": "I don t think there is an issue he just wanted to solve using the trigonometric functions because that is what we had just been working on. Just to show, 10.7*10.7 = 114.49 5*5 = 25 114.49+25=139.49 And the square root of 139.49 = 11.8 a^2+b^2=c^2 so don t forget to square root everything.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 354 + ], + "3min_transcript": "And so what trigonometric ratios-- or we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that XY is exactly 5 and we don't have to deal with this approximation, let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well we see from SohCahToa cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees is equal to the length of the adjacent side, which is 5 over the length of the hypotenuse, which has a length of b. And then we can try to solve for b. You multiply both sides times b, you're left with b times cosine of 65 degrees is equal to 5. And then to solve for b, you could divide both sides by cosine of 65 degrees. So we're just dividing-- we have to figure it out what our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees. And we're left with b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is 5 divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that 5 squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out So I'll give you a few seconds to think about what the measure of angle W is. Well here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is-- well we can simplify the left-hand side right over here. 65 plus 90 is 155. So angle W plus 155 degrees is equal to 180 degrees. And then we get angle W-- if we subtract 155 from both sides-- angle W is equal to 25 degrees. And we are done solving the right triangle shown below." + }, + { + "Q": "At 2:13 don't you mean to say 7 7 and 9", + "A": "It s just a minor mistake. You can see that it s 7, 7, and 9. Besides, the videos are still awesome at teaching the topic!", + "video_name": "LEFE1km5ROY", + "timestamps": [ + 133 + ], + "3min_transcript": "A statistician for a basketball team tracked the number of points that each of the 12 players on the team had in one game. And then made a stem-and-leaf plot to show the data. And sometimes it's called a stem-plot. How many points did the team score? And when you first look at this plot right over here, it seems a little hard to understand. Understand we have 0, 1, 2 under leaf you have all of these digits here. How does this relate to the number of points each student, or each player, actually scored? And the way to interpret a stem-and-leaf plot is the leafs contain-- at least the way that this statistician used it-- the leaf contains the smallest digit, or the ones digit, in the number of points that each player scored. And the stem contains the tens digits. And usually the leaf will contain the rightmost digit, or the ones digit, and then the stem will contain all of the other digits. And what's useful about this is it gives kind of a distribution of where the players were. You see that most of the players scored points that started with a 0. And then only one score scored points to started with a 2, and it was actually 20 points. So I'm going to actually write down all of this data in a way that maybe you're a little bit more used to understanding it. So I'm going to write the 0's in purple. So there's, let's see, 1, 2, 3, 4, 5, 6, 7 players had 0 as the first digit. So 1, 2, 3, 4, 5, 6, 7. I wrote seven 0's. And then this player also had a 0 in his ones digit. This player, I'm going to try to do all the colors, this player also had a 0 in his ones digit. This player right here had a 2 in his ones digit, so he scored a total of 2 points. This player, let me do orange, this player had 4 for his ones digit. This player had 7 for his ones digit. Then this player had 7 for his ones digit. this player had 9 for his ones digit. So the way to read this is, you had one player with 0 points. 0, 2, 4, 7, 9 and 9. But you can see, and it's kind of silly saying the zero was a tens digit, you could have even put a blank there. But the 0 lets us know that they didn't score anything in the tens place. But these are the actual scores for those seven players. Now let's go to the next row in the stem-and-leaf plot. So over here, all of the digits start with, or all of the points start with 1, for each of the players. And there's four of them. So 1, 1, 1, and 1. And then we have this player over here, his ones digit, or her ones digit, is a 1. So this player, this represents 11. 1 in the tens place, 1 in the ones place. This player over here also got 11. 1 in the tens place, 1 in the ones place." + }, + { + "Q": "At 0:00, under the Stem column, why is there a 0?", + "A": "Hi Nikki. Good question. When you have a stem and leaf plot, you always have to have something on the stem side in order to have something on the leave side. So, when we want to put numbers such as 3, 7, or 9 (one digit numbers) we put a 0 on the stem side so that the value of the number doesn t change, but there is a stem for the leaves be on. :) Hope that makes sense. Syliva", + "video_name": "LEFE1km5ROY", + "timestamps": [ + 0 + ], + "3min_transcript": "A statistician for a basketball team tracked the number of points that each of the 12 players on the team had in one game. And then made a stem-and-leaf plot to show the data. And sometimes it's called a stem-plot. How many points did the team score? And when you first look at this plot right over here, it seems a little hard to understand. Understand we have 0, 1, 2 under leaf you have all of these digits here. How does this relate to the number of points each student, or each player, actually scored? And the way to interpret a stem-and-leaf plot is the leafs contain-- at least the way that this statistician used it-- the leaf contains the smallest digit, or the ones digit, in the number of points that each player scored. And the stem contains the tens digits. And usually the leaf will contain the rightmost digit, or the ones digit, and then the stem will contain all of the other digits. And what's useful about this is it gives kind of a distribution of where the players were. You see that most of the players scored points that started with a 0. And then only one score scored points to started with a 2, and it was actually 20 points. So I'm going to actually write down all of this data in a way that maybe you're a little bit more used to understanding it. So I'm going to write the 0's in purple. So there's, let's see, 1, 2, 3, 4, 5, 6, 7 players had 0 as the first digit. So 1, 2, 3, 4, 5, 6, 7. I wrote seven 0's. And then this player also had a 0 in his ones digit. This player, I'm going to try to do all the colors, this player also had a 0 in his ones digit. This player right here had a 2 in his ones digit, so he scored a total of 2 points. This player, let me do orange, this player had 4 for his ones digit. This player had 7 for his ones digit. Then this player had 7 for his ones digit. this player had 9 for his ones digit. So the way to read this is, you had one player with 0 points. 0, 2, 4, 7, 9 and 9. But you can see, and it's kind of silly saying the zero was a tens digit, you could have even put a blank there. But the 0 lets us know that they didn't score anything in the tens place. But these are the actual scores for those seven players. Now let's go to the next row in the stem-and-leaf plot. So over here, all of the digits start with, or all of the points start with 1, for each of the players. And there's four of them. So 1, 1, 1, and 1. And then we have this player over here, his ones digit, or her ones digit, is a 1. So this player, this represents 11. 1 in the tens place, 1 in the ones place. This player over here also got 11. 1 in the tens place, 1 in the ones place." + }, + { + "Q": "At 5:03, Sal says we can find the points (the numbers which will make both sides of the equation equal to 0). Why is it so?", + "A": "It s the same point he used to get the formula, P = (4, 9) . He used the 2 points, the variable point Z = (x, y) and a given point P = (4, 9) to get (y-9)/(x-4) = -4; so if he lets Z = (4, 9), then Z and P are the same point, and he gets (9-9)/(4-4) = 0/0, or 9-9 = -4(4-4), or 0 = 0. (Why he s saying this, I don t know).", + "video_name": "LtpXvUCrgrM", + "timestamps": [ + 303 + ], + "3min_transcript": "Again to get to this point? Well, your change in X is positive two. So your change in X is equal to two. And so what's your slope? Change in Y over change in X. Negative eight over two is equal to negative four. So now that we have a, now that we know the slope and we know a point, we know a, we actually know two points on the line, we can express this in point-slope form. And so let's do that. And the way I like to it is I always like to just take it straight from the definition of what slope is. We know that the slope between any two points on this line is going to be negative four. So if we take an arbitrary Y that sits on this line and if we find the difference between that Y and, let's focus on this point up here. So if we find the difference between that Y and this Y, and nine, and it's over the difference between This is going to be the slope between any XY on this line and this point right over here. And the slope between any two points on a line are going to have to be constant. So this is going to be equal to the slope of the line. It's going to be equal to negative four. And we're not in point-slope form or classic point-slope form just yet. To do that, we just multiply both sides times X minus four. So we get Y minus 9, we get Y minus nine is equal to our slope, negative four times X minus four. Time X minus four. And this right over here is our classic, this right over here is our classic point-slope form. We have the point, sometimes they even put parenthesis like this, but we could figure out the point from this point-slope form. The point that sits on this line with things So it would be X equals four, Y equals nine, which we have right up there, and then the slope is right over here, it's negative four. Now from this can we now express this linear equation in y-intercept form? And y-intercept form, just as a bit of a reminder, it's Y is equal to MX plus B. Where this coefficient is our slope and this constant right over here allows us to figure out our y-intercept. And to get this in this form we just have to simplify a little bit of this algebra. So you have Y minus nine. Y minus nine is equal to, well let's distribute this negative four. And I'll just switch some colors. Let's distribute this negative four. Negative four times X is negative four X. Negative four time negative four is plus 16. And now, if we just want to isolate the Y on the left hand side, we can add nine to both sides." + }, + { + "Q": "@1:18 why when Sal convert 6/3 + 1/3 to 2 1/3, he emitted the (+) sign ?", + "A": "because 6/3+1/3 is equal to 2 1/3. 2 1/3 is a mixed number, so even though you say two and one third, the number you write down secludes the and.", + "video_name": "xiIQQNufFuU", + "timestamps": [ + 78 + ], + "3min_transcript": "The graph of the line 2y plus 3x equals 7 is given right over here. Determine its x-intercept. The x-intercept is the x value when y is equal to 0, or it's the x value where our graph actually intersects the x-axis. Notice right over here our y value is exactly 0. We're sitting on the x-axis. So let's think about what this x value must be. Well, just trying to eyeball a little bit, it's a little over 2. It's between 2 and 3. It looks like it's less than 2 and 1/2. But we don't know the exact value. So let's go turn to the equation to figure out the exact value. We essentially have to figure out what x value, when y is equal to 0, will have this equation be true. So we could just say 2 times 0 plus 3x is equal to 7. Well, 2 times 0 is just going to be 0, so we have 3x is equal to 7. Then we can divide both sides by 3 to solve for x, Does that look like 7/3? Well, we just have to remind ourselves that 7/3 is the same thing as 6/3 plus 1/3. And 6/3 is 2. So this is the same thing as 2 and 1/3. Another way you could think about it is 3 goes into 7 two times, and then you have a remainder of 1. So you've still got to divide that 1 by 3. It's 2 full times and then a 1/3, so this looks like 2 and 1/3. And so that's its x-intercept, 7/3. If I was doing this on the exercise on Khan Academy, it's always a little easier to type in the improper fraction, so I would put in 7/3." + }, + { + "Q": "At 1:00, Sal said \"if y is zero..\" he then plugged in zero to the problem to find 2 1/3. How did he know y was zero. I'm lost.\n\nThanks.", + "A": "We re looking for the x-intercept. This is the point where the line crosses the x-axis. What do the coordinates of a point on the x-axis look like? They look like (x , 0). Just like for the y-intercept, the coordinates are (0, y).", + "video_name": "xiIQQNufFuU", + "timestamps": [ + 60 + ], + "3min_transcript": "The graph of the line 2y plus 3x equals 7 is given right over here. Determine its x-intercept. The x-intercept is the x value when y is equal to 0, or it's the x value where our graph actually intersects the x-axis. Notice right over here our y value is exactly 0. We're sitting on the x-axis. So let's think about what this x value must be. Well, just trying to eyeball a little bit, it's a little over 2. It's between 2 and 3. It looks like it's less than 2 and 1/2. But we don't know the exact value. So let's go turn to the equation to figure out the exact value. We essentially have to figure out what x value, when y is equal to 0, will have this equation be true. So we could just say 2 times 0 plus 3x is equal to 7. Well, 2 times 0 is just going to be 0, so we have 3x is equal to 7. Then we can divide both sides by 3 to solve for x, Does that look like 7/3? Well, we just have to remind ourselves that 7/3 is the same thing as 6/3 plus 1/3. And 6/3 is 2. So this is the same thing as 2 and 1/3. Another way you could think about it is 3 goes into 7 two times, and then you have a remainder of 1. So you've still got to divide that 1 by 3. It's 2 full times and then a 1/3, so this looks like 2 and 1/3. And so that's its x-intercept, 7/3. If I was doing this on the exercise on Khan Academy, it's always a little easier to type in the improper fraction, so I would put in 7/3." + }, + { + "Q": "at 3:21, do you always have to flip the numbers over?", + "A": "nope, but it makes it alot easier", + "video_name": "Zm0KaIw-35k", + "timestamps": [ + 201 + ], + "3min_transcript": "Jayda takes 3 hours to deliver 189 newspapers on her paper route. What is the rate per hour at which she delivers the newspaper? So this first sentence tells us that she delivers, or she takes, 3 hours to deliver 189 newspapers. So you have 3 hours for every 189 newspapers. That's what the first sentence told us. But we want to figure out the rate per hour, or the newspapers per hour, so we can really just flip this rate So if we were to just flip it, we would have 189 newspapers for every 3 hours, which is really the same information. We're just flipping what's in the numerator and what's in Now we want to write it in as simple as possible form, and let's see if this top number is divisible by 3. 1 plus 8 is 9, plus 9 is 18. So that is divisible by 3. So let's divide this numerator and this denominator by 3 to simplify things. So if you divide 189 by 3. Let's do it over on the side right here. 3 goes into 189. 3 goes and 18 six times. 6 times 3 is 18. Subtract. Bring down the 9. 18 minus 18 was nothing. 3 goes into 9 exactly three times. 3 times 3 is 9, no remainder. So if you divide 189 by 3, you get 63, and if you divide 3 by 3, you're going to get 1. You have to divide both the numerator and the denominator by the same number. So now we have 63 newspapers for every 1 hour. Or we could write this as 63 over 1 newspapers per hour. thing as 63 newspapers per hour." + }, + { + "Q": "At 2:28, how come he used multiplications to solve pentagon when there was more than 1 of the same numbers, when he didn't use multiplication when he was working out the rectangle", + "A": "For the rectangle, he could ve done 2*5 + 2*3 to get the perimeter of the rectangle. I think he used multiplication for the pentagon because he would have to write 2 five times, which would take too much space. If we understand that we re adding 2 five times, that just 2 multiplied by five.", + "video_name": "9uwLgf84p5w", + "timestamps": [ + 148 + ], + "3min_transcript": "When people use the word \"perimeter\" in everyday language, they're talking about the boundary of some area. And when we talk about perimeter in math, we're talking about a related idea. But now we're not just talking about the boundary. We're actually talking about the length of the boundary. How far do you have to go around the boundary to essentially go completely around the figure, completely go around the area? So let's look at this first triangle right over here. It has three sides. That's why it's a triangle. So what's its perimeter? Well, here, all the sides are the same, so the perimeter for this triangle is going to be 4 plus 4 plus 4, and whatever units this is. If this is 4 feet, 4 feet and 4 feet, then it would be 4 feet plus 4 feet plus 4 feet is equal to 12 feet. Now, I encourage you to now pause the video and figure out the parameters of these three figures. Well, it's the exact same idea. We would just add the lengths of the sides. So let's say that these distances, let's So let's say this is 3 meters, and this is also 3 meters. This is a rectangle here, so this is 5 meters. This is also 5 meters. So what is the perimeter of this rectangle going to be? What is the distance around the rectangle that bounds this area? Well, it's going to be 3 plus 5 plus 3 plus 5, which is equal to-- let's see, that's 3 plus 3 is 6, plus 5 plus 5 is 10. So that is equal to 16. And if we're saying these are all in meters, these are all in meters, then it's going to be 16 meters. Now, what about this pentagon? Let's say that each of these sides are 2-- and I'll make up a unit here. Let's say they're 2 gnus. That's a new unit of distance that I've just invented-- 2 gnus. Well, it's 2 plus 2 plus 2 plus 2 plus 2 gnus. Or we're essentially taking 1, 2, 3, 4, 5 sides. Each have a length of 2 gnus. So the perimeter here, we could add 2 repeatedly five times. Or you could just say this is 5 times 2 gnus, which is equal to 10 gnus, where gnu is a completely made-up unit of length that I just made up. Here we have a more irregular polygon, but same exact idea. How would you figure out its perimeter? Well, you just add up the lengths of its sides. And here I'll just do it unitless. I'll just assume that this is some generic units. And here the perimeter will be 1 plus 4 plus 2 plus 2-- let me scroll over to the right a little bit-- plus 4 plus 6." + }, + { + "Q": "Why doesnt he write 5c in 5:31?", + "A": "It s simpler/cleaner to just call it c. It is a constant of unknown quantity. Sure, it s 5 times something, but it s also 1 times something too.", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 331 + ], + "3min_transcript": "So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1 And then if we want, we can distribute the 5. So this is equal to negative 5x to the negative 1. Now, we could write plus 5 times some constant, but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could [INAUDIBLE] this. If you want it to show that it's a different constant, you could say this is c1, c1, c1. You multiply 5 times c1, you get another constant. We could just call that c, which is equal to 5 times c1. But there you have it. Negative 5x to the negative 1 plus c. And once again, all of these, try to evaluate the derivative, and you will see that you get this business, right over there." + }, + { + "Q": "At 0:24, what does he mean by derivative?", + "A": "At 0:24, Sal refers to the word derivative. A derivative of a function is the slope of the tangent line to a curve at a particular point on the curve. In this video, Sal shows how to find the antiderivative of a function using the power rule. Keep watching the video to find out what the power rule does.", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 24 + ], + "3min_transcript": "Let's take the derivative with respect to x of x to the n plus 1-th power over n plus 1 plus some constant c. And we're going to assume here, because we want this expression to be defined, we're going to assume that n does not equal negative 1. If it equaled negative 1, we'd be dividing by 0, and we haven't defined what that means. So let's take the derivative here. So this is going to be equal to-- well, the derivative of x to the n plus 1 over n plus 1, we can just use the power rule over here. So our exponent is n plus 1. We can bring it out front. So it's going to be n plus 1 times x to the-- I want to use that same color. Colors are the hard part-- times x to the-- instead of n plus 1, we subtract 1 from the exponent. This is just the power rule. So n plus 1 minus 1 is going to be n. And then we can't forget that we were dividing by this n plus 1. And then we have plus c. The derivative of a constant with respect to x-- a constant does not change as x changes, so it is just going to be 0, so plus 0. And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing-- and this is a very general terms-- is equal to x to the n. So given that, what is the antiderivative-- let me switch colors here. What is the antiderivative of x to the n? And remember, this is just the kind of strange-looking notation we use. It'll make more sense when we start doing definite integrals. But what is the antiderivative of x to the n? And we could say the antiderivative with respect to x, if we want to. And another way of calling this is the indefinite integral. is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power" + }, + { + "Q": "At 5:09 how do you transition from x^-1/-1 to -x^-1 ?", + "A": "Dividing by -1 is the same as multiplying by -1, because it only changes the signal of the expression.One easy proof : (-1)^2= 1 =a/a -> {(-1)^2}f(x)=1f(x) -> -f(x).-1=-1f(x)/-1, now divide both sides by -1 and its done -f(x)=f(x)/-1 .", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 309 + ], + "3min_transcript": "So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx. Well, one simplification you can do-- and I haven't rigorously proven it to you just yet-- but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is, indeed, equal to 5 times the antiderivative of x to the negative 2 power, dx. And now we can just use, I guess we could call it this anti-power rule, so this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1, plus some constant. And this is equal to 5 times negative x to the negative 1 And then if we want, we can distribute the 5. So this is equal to negative 5x to the negative 1. Now, we could write plus 5 times some constant, but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could [INAUDIBLE] this. If you want it to show that it's a different constant, you could say this is c1, c1, c1. You multiply 5 times c1, you get another constant. We could just call that c, which is equal to 5 times c1. But there you have it. Negative 5x to the negative 1 plus c. And once again, all of these, try to evaluate the derivative, and you will see that you get this business, right over there." + }, + { + "Q": "At 2:31, Sal writes x^(n+1). Does the denominator \"n+1\" apply to the whole thing or just x? (i.e. does it read ((x^(n+1))/(n+1) or (x/(n+1))^(n+1)?) Does it matter either way? Thanks!", + "A": "The whole thing. It is [(x^(n+1)]/(n+1). And yes it matters!", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 151 + ], + "3min_transcript": "And then we have plus c. The derivative of a constant with respect to x-- a constant does not change as x changes, so it is just going to be 0, so plus 0. And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing-- and this is a very general terms-- is equal to x to the n. So given that, what is the antiderivative-- let me switch colors here. What is the antiderivative of x to the n? And remember, this is just the kind of strange-looking notation we use. It'll make more sense when we start doing definite integrals. But what is the antiderivative of x to the n? And we could say the antiderivative with respect to x, if we want to. And another way of calling this is the indefinite integral. is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx." + }, + { + "Q": "at 0:17, Sal said that if the denominator is the same, you just add the numerator and then the denominator stays the same? I just dont get it. Is it the same on the problems without the same denominators?", + "A": "no you do not add the denominator you add the numerator", + "video_name": "EJjnEau6aeI", + "timestamps": [ + 17 + ], + "3min_transcript": "So we're asked to add 3/15 plus 7/15, and then simplify the answer. So just the process when you add fractions is if they already-- well, first of all, if they're not mixed numbers, and neither of these are, and if they have the same denominator. In this example, the denominators are already the same. The denominator is 15. So if you add these two fractions, your sum is going to have the same denominator, 15, and your numerator is just going to be the sum of the numerator, so it's going to be 3 plus 7, or it's going to be equal to 10/15. Now, if we wanted to simplify this, we'd look for the greatest common factor in both the 10 and the 15, and as far as I can tell, 5 is the largest number that goes into both of them. So divide the 10 by 5 and you divide the 15 by 5, and you get-- 10 divided by 5 is 2 and 15 divided by 5 is 3. You get 2/3. Now, to understand why this works, let's draw it out. So let me split it up into 15 sections. Let me see how well I can do this. Well, actually, even a better way, an easier way might be to draw circles. So let me do the 15 sections. So let me draw. So that is one section right over there. That is one section and then if I copy and paste it, that is a second section, and then a third section, fourth section, and then we have a fifth section. Let me copy and paste this whole thing. So that's five sections right there. Let me copy and then paste that. So that is 10 sections, and then let me do it one more time. So that is 15 sections. So you can imagine this whole thing is like a candy bar or Now, what is 3/15? Well, it's going to be 3 of the 15 sections. So 3/15 is going to be one, two, three: 3/15. Now, to that, were adding 7 of the 1/15 sections, or 7 of the sections. So we're adding 7 of those to it. So that's one, two, three, four, five, six, seven. And you see now, if you take the orange and the blue, you get one, two, three, four, five, six, seven, eight, nine, ten of the sections, or 10 of the 15 sections. And then to see why this is the same thing as 2/3, you can just split this candy bar into thirds, so each third would have five sections in it. So let's do that. One, two, three, four, five, so that is 1/3 right there. One, two, three, four, five, that is" + }, + { + "Q": "Does the endpoint always have to be the first letter when naming a ray?\nFor example at 1:27 the ray is labelled ray JH and J is the endpoint.", + "A": "Yes. Because the arrow always points to the right when naming a ray, the endpoint of the ray must always be the first letter.", + "video_name": "w9jEq6dmqPg", + "timestamps": [ + 87 + ], + "3min_transcript": "Identify all the rays shown in the image below. and this is a reminder what a ray is. A ray start at some point and then goes on forever in some direction. In order to find a ray you need that point that you're starting off on so let's that point over there is called X and then you need another point that sits on the ray and the ray is just keep going beyond, we will name that point as \"Y\" and we will call the ray \"XY\" It starts in \"X\" and keeps going in the direction of \"y\" for ever, it crosses \"y\" and keeps going further we need the second point to specify the direction in which the ray is going. So, lets identify all the rays shown in the image below, we can start anywhere, we will start at is JH going up, goes upto H and keeps on going in that direction beyond H, ray JH, starting from J going through H and going beyond it forever now if we go to H, there is no ray HJ as the line ends in J and does not keep going beyond J, there is no ray H as it is just one point, just usiing one point, we cannot say it as a ray. now looking at our diagram the only ray is JH. Now let us look at the other points. after C to specify it as a ray, we can have a ray CE, starts at C goes through E and goes on for ever, you can also have a ray starting at C, going through F and going on forever, CE & CF are the same rays as F sits on ray CE and E sits of ray CF, so CE & CF are the same rays," + }, + { + "Q": "um on 5:05 when you mentioned about GE as a ray, can GC also be a ray?", + "A": "yeah...... Im pretty sure your right that GC could be a ray too", + "video_name": "w9jEq6dmqPg", + "timestamps": [ + 305 + ], + "3min_transcript": "We can start at point E , go in the direction of C and go beyond C, so it is a ray, ray EC you could start in E and go in the direction of A and go beyond A, so EA is a ray, and we can start at E and go in the direction of F and go beyond F, that is ray EF ray EF and ray CF are different as the starting points are different Now lets go to point F. To the left of point F there is no other point. We can look at the right, we can have ray FE, start at F go through E and go through C and beyond C that is teh same thing as FE ray FE and ray FC are the same as the point E is on ray FC, then finally we have not focussed on point A you may think there is ray AE, but the line does not go beyond E, so it is not a ray, to the top of A there is no other point, so there is no ray there either that is all the rays based on the points specified. If they had given us a point over there, we could have had other rays," + }, + { + "Q": "I don't understand why you need to have two points for it to be a ray. Anyways there are points in between the endpoint A and the arrow (4:41 in the video)", + "A": "The first point serves to show where the ray is starting from. The second point is needed to show in which direction the ray is going. For example, if we just said ray E, it would be confusing. Is it going towards A, towards, F, or towards some other point not even drawn on the graph? Two points are needed in order to show origin and direction.", + "video_name": "w9jEq6dmqPg", + "timestamps": [ + 281 + ], + "3min_transcript": "We can start at point E , go in the direction of C and go beyond C, so it is a ray, ray EC you could start in E and go in the direction of A and go beyond A, so EA is a ray, and we can start at E and go in the direction of F and go beyond F, that is ray EF ray EF and ray CF are different as the starting points are different Now lets go to point F. To the left of point F there is no other point. We can look at the right, we can have ray FE, start at F go through E and go through C and beyond C that is teh same thing as FE ray FE and ray FC are the same as the point E is on ray FC, then finally we have not focussed on point A you may think there is ray AE, but the line does not go beyond E, so it is not a ray, to the top of A there is no other point, so there is no ray there either that is all the rays based on the points specified. If they had given us a point over there, we could have had other rays," + }, + { + "Q": "At 2:40 he mentions CE and CF. Which one would be better to use? or does it matter at all?", + "A": "It depends on the direction that you want to go in.", + "video_name": "w9jEq6dmqPg", + "timestamps": [ + 160 + ], + "3min_transcript": "is JH going up, goes upto H and keeps on going in that direction beyond H, ray JH, starting from J going through H and going beyond it forever now if we go to H, there is no ray HJ as the line ends in J and does not keep going beyond J, there is no ray H as it is just one point, just usiing one point, we cannot say it as a ray. now looking at our diagram the only ray is JH. Now let us look at the other points. after C to specify it as a ray, we can have a ray CE, starts at C goes through E and goes on for ever, you can also have a ray starting at C, going through F and going on forever, CE & CF are the same rays as F sits on ray CE and E sits of ray CF, so CE & CF are the same rays, We can start at point E , go in the direction of C and go beyond C, so it is a ray, ray EC you could start in E and go in the direction of A and go beyond A, so EA is a ray, and we can start at E and go in the direction of F and go beyond F, that is ray EF ray EF and ray CF are different as the starting points are different Now lets go to point F. To the left of point F there is no other point. We can look at the right, we can have ray FE, start at F go through E and" + }, + { + "Q": "0:25\nIs the line of symmetry compulsory to be diagonal or horizontal to the object?", + "A": "Any line where you can fold an object and the two sides match one another is a line of symmetry. Go to Google search images and search for line symmetry .", + "video_name": "LrTn4cvsewk", + "timestamps": [ + 25 + ], + "3min_transcript": "For each of these diagrams, I want to think about whether this blue line represents an axis of symmetry. And the way we can tell is if on both sides of the blue line we essentially have mirror images. Let's take this top part of this polygon, the part that is above this blue line here, and let's reflect it across the blue line-- you could almost imagine that it's a reflection over some type of a lake or something-- and see if we get exactly what we have below. Then this would be an axis of symmetry. So this point right over here, this distance to the blue line, let's go-- the same amount on the other side would get you right there. And so you immediately see we start ending up with a point that is off what's actually here in black, the actual bottom part of the polygon. So this is a pretty good clue that this is not an axis of symmetry. But let's just continue it, just to go through the exercise. So this point, if you reflected it across this blue line, would get you here. This point-- I'll do it in a different color. This point, if you were to reflect it across this blue line, it would get I can do a straighter job than that. So if you go about that distance about it, and I want to go straight down into the blue line, and I'm going to go the same distance on the other side, it gets me to right around there. And then this point over here, if I were to drop it straight down, then if I were to go the same distance on the other side, it gets me right around there. And then finally, this point gets me right around there. So its mirror image of this top part would look something like this. My best attempt to draw it would look something like this, which is very different than the part of the polygon that's actually on the other side of this blue line. So in this case, the blue line is not an axis of symmetry. So this is no. No, this blue line is not an axis of symmetry. Now let's look at it over here. Here you can see that it looks like this blue line really divides this polygon in half. It really does look like mirror images. It really does look like, if you imagine that this is some type of a lake, a still lake, so I shouldn't actually draw waves, but this is some type of a lake, that this is the reflection. And we can even go point by point here. So this point right over here is the same distance from, if we dropped a perpendicular to this point as this one right over here; this one over here, same distance, same distance as this point right over here; and we could do that for all of the points. So in this case, the blue line does represent an axis of symmetry." + }, + { + "Q": "at 0:35 , how does he split 60 into 6 * 10 ?", + "A": "because 6 * 10 is 60, and the order of multiplication doesn t matter.", + "video_name": "jb8mFpA1YI8", + "timestamps": [ + 35 + ], + "3min_transcript": "Let's see if we can figure out 3 times 60. Well, there's a couple of ways you could think about it. You could literally view this as 60 three times. So you could view this as 60 plus 60 plus 60. And you might be able to compute this in your head. 60 plus 60 is 120, plus another 60 is 180. And you'd be done. But another way to think about this is that 3 times 60 is the same thing as 3 times-- instead of thinking of it as 60, you could think of 60 as 6 times 10. 3 times 6 times 10. Now, when you're multiplying three numbers like this, it doesn't matter what order you multiply them in. So we could multiply the 3 times 6 first and get 18 and then multiply that times 10. And 18 times 10 is just going to be 180. It's going to be 18 with another zero. So this is going to be 180. Now, the more practice you get here, you'll realize, but I have to worry about this 0 right over here. So I'm going to put one more zero at the end. It's going to be 180. Same answer that we got right over there. Let's do another one of these. So let's say we want to multiply 50 times 7. And I encourage you to pause the video and think about it yourself, and then unpause it and see what I do. So 50, well, there's a couple of ways you could think about it. One, you could literally try to add 50 seven times. Adding 7 fifty times would take forever, but you could literally say 50 plus 50 plus 50 plus 50-- let's see, that's four-- plus 50 plus 50. Let's see, that is six. I'll do one more right over here. 50 right over here. So this is 50 seven times. If you add together 50 plus 50 is 100, 150, 200, 250, 300, 350. So you could do it that way. You just need to realize that 50 is the same thing as 10 times 5. So we could write this as 10 times 5, and then we're multiplying that times 7. Once again, the order that we multiply does not matter. So we can multiply the 5 times the 7 first. We know that that is 35, and we're going to multiply that times 10. 10 times 35, well, we're just going to stick a zero at the end of the 35. It's going to be equal to 350. Now I want to do that zero in that same color. It's going to be 350. Now, you might realize, hey, look, I could have just looked at this 5 right over here, multiplied the 5 times the 7, and have gotten the 35. And then, not forgetting that it's actually not a 5, it's a 50. So I have to multiply by 10 again, or I have to throw that 0 at the end of it to get that 0 right over here. So 50 times 7 is 350." + }, + { + "Q": "At 10:04, how do you remember which way to move the decimal if it's a negative exponent?", + "A": "okay heres an example -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 Negative is backwards, so just think about it like you have 2^-3 that would be 0.125", + "video_name": "i6lfVUp5RW8", + "timestamps": [ + 604 + ], + "3min_transcript": "And then we're left with this one, times 10 to the negative 3. Now, a very quick way to do it is just to say, look, let me count-- including the leading numeral-- how many spaces I have behind the decimal. 1, 2, 3. So it's going to be 2.81 times 10 to the negative 1, 2, 3 power. Let's do one more like that. Let me actually scroll up here. Let's do one more like that. Let's say I have 1, 2, 3, 4, 5, 6-- how many 0's do I have Well, I'll just make up something. 0, 2, 7. And you wanted to write that in scientific notation. Well, you count all the digits up to the 2, behind the decimal. So 1, 2, 3, 4, 5, 6, 7, 8. So this is going to be 2.7 times 10 to Now let's do another one, where we start with the scientific notation value and we want to go to the numeric value. Just to mix things up. So let's say you have 2.9 times 10 to the negative fifth. So one way to think about is, this leading numeral, plus all 0's to the left of the decimal spot, is going to be five digits. So you have a 2 and a 9, and then you're going to have 4 more 0's. 1, 2, 3, 4. And then you're going to have your decimal. And how did I know 4 0's? Because I'm counting,, this is 1, 2, 3, 4, 5 spaces behind the decimal, including the leading numeral. And so it's 0.000029. And just to verify, do the other technique. How do I write this in scientific notation? I count all of the digits, all of the leading 0's behind the So I have 1, 2, 3, 4, 5 digits. So it's 10 to the negative 5. And so it'll be 2.9 times 10 to the negative 5. And once again, this isn't just some type of black magic here. This actually makes a lot of sense. If I wanted to get this number to 2.9, what I would have to do is move the decimal over 1, 2, 3, 4, 5 spots, like that. And to get the decimal to move over the right by 5 spots-- let's just say with 0, 0, 0, 0, 2, 9. If I multiply it by 10 to the fifth, I'm also going to have to multiply it by 10 to the negative 5. So I don't want to change the number. This right here is just multiplying something by 1. 10 to the fifth times 10 to the negative 5 is 1." + }, + { + "Q": "at 2:11 why is the line dotted instead of solid?", + "A": "A dotted line symbolizes < or > A solid line symbolizes \u00e2\u0089\u00a4 or \u00e2\u0089\u00a5", + "video_name": "CA4S7S-3Lg4", + "timestamps": [ + 131 + ], + "3min_transcript": "Graph the solution set for this system. It's a system of inequalities. We have y is greater than x minus 8, and y is less than 5 minus x. Let's graph the solution set for each of these inequalities, and then essentially where they overlap is the solution set for the system, the set of coordinates that satisfy both. So let me draw a coordinate axes here. So that is my x-axis, and then I have my y-axis. And that is my y-axis. And now let me draw the boundary line, the boundary for this first inequality. So the boundary line is going to look like y is equal to x minus 8. But it's not going to include it, because it's only greater than x minus 8. But let's just graph x minus 8. So the y-intercept here is negative 8. When x is 0, y is going to be negative 8. So just go negative 1, negative 2, 3, 4, 5, 6, 7, 8. So the point 0, negative 8 is on the line. And then it has a slope of 1. You don't see it right there, but I could write it as 1x. So the slope here is going to be 1. I could just draw a line that goes straight up, or you could even say that it'll intersect if y is equal to 0, if y were equal to 0, x would be equal to 8. So 1, 2, 3, 4, 5, 6, 7, 8. And so this is x is equal to 8. If it has a slope of 1, for every time you move to the right 1, you're going to move up 1. So the line is going to look something like this. And actually, let me not draw it as a solid line. If I did it as a solid line, that would actually be this equation right here. But we're not going to include that line. We care about the y values that are greater than that line. So what we want to do is do a dotted line to show that that's just the boundary, that we're not including that in our solution set. Let me do this in a new color. So this will be the color for that line, or for that inequality, I should say. And this says y is greater than x minus 8. So you pick an x, and then x minus 8 would get us on the boundary line. And then y is greater than that. So it's all the y values above the line for any given x. So it'll be this region above the line right over here. And if that confuses you, I mean, in general I like to just think, oh, greater than, it's going to be above the line. If it's less than, it's going to be below a line. But if you want to make sure, you can just test on either side of this line. So you could try the point 0, 0, which should be in our solution set. And if you say, 0 is greater than 0 minus 8, or 0 is greater than negative 8, that works. So this definitely should be part of the solution set. And you could try something out here like 10 comma 0 and see that it doesn't work. Because you would have 10 minus 8, which would be 2, and then you'd have 0. And 0 is not greater than 2. So when you test something out here, you also see that it won't work. But in general, I like to just say, hey look, this is the boundary line, and we're greater than the boundary line for any given x." + }, + { + "Q": "at 1:20 sal says the negative cancel out. What does it mean to cancel out", + "A": "(+)(+)=(+) (-)(-)=(+) (+)(-)=(-) there is two negative, so he cancel out, because negative times or divid negative is equal to positive", + "video_name": "bQ-KR3clFgs", + "timestamps": [ + 80 + ], + "3min_transcript": "Now that we know a little bit about multiplying positive and negative numbers, Let's think about how how we can divide them. Now what you'll see is that it's actually a very similar methodology. That if both are positive, you'll get a positive answer. If one is negative, or the other, but not both, you'll get a negative answer. And if both are negative, they'll cancel out and you'll get a positive answer. But let's apply and I encourage you to pause this video and try these out yourself and then see if you get the same answer that I'm going to get. So eight (8) divided by negative two (-2). So if I just said eight (8) divided by two (2), that would be a positive four (4), but since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative. So eight (8) divided by negative two (-2) is negative four (-4). Now negative sixteen (-16) divided by positive four (4)-- now be very careful here. If I just said positive sixteen (16) divided by positive four (4), that would just be four (4). But because one of these two numbers is negative, then I'm going to get a negative answer. Now I have negative thirty (-30) divided by negative five (-5). If I just said thirty (30) divided by five (5), I'd get a positive six (6). And because I have a negative divided by a negative, the negatives cancel out, so my answer will still be positive six (6)! And I could even write a positive (+) out there, I don't have to, but this is a positive six (6). A negative divided by a negative, just like a negative times a negative, you're gonna get a positive answer. Eighteen (18) divided by two (2)! And this is a little bit of a trick question. This is what you knew how to do before we even talked about negative numbers: This is a positive divided by a positive. Which is going to be a positive. So that is going to be equal to positive nine (9). Now we start doing some interesting things, here's kind of a compound problem. We have some multiplication and some division going on. we're gonna wanna multiply the numerator out, and if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could've written this little \"x\" thing over here but what you're gonna see in Algebra is that the dot become much more common. Because the X becomes used for other-- People don't want to confuse it with the letter X which gets used a lot in Algebra. That's why they used the dot very often. So this just says negative seven (-7) times three (3) in the numerator, and we're gonna take that product and divide it by negative one (-1). So the numerator, negative seven (-7) times three (3), positive seven (7) times three (3) would be twenty-one (21), but since exactly one of these two are negative, this is going to be negative twenty-one (-21), that's gonna be negative twenty-one (-21) over negative one (-1). And so negative twenty-one (-21) divided by negative one (-1), negative divided by a negative is going to be a positive. So this is going to be a positive twenty-one (21). Let me write all these things down. So if I were to take a positive divided by a negative," + }, + { + "Q": "when he takes the two out front of the integral at 2:55, do you have to do that to get the correct answer or does hat just make things easier?", + "A": "It makes things easier since you would not have to factor it out in the end after taking the integration.", + "video_name": "n-iEqLhGfd4", + "timestamps": [ + 175 + ], + "3min_transcript": "and the antiderivative of e to the x is still just e to the x. So let me write this down. So we are saying that f of x-- I'll do it right over here-- f of x is equal to x squared, in which case f prime of x is going to be equal to 2x. And I'm not worrying about the constants right now. We'll just add a constant at the end to make sure that our antiderivative is in the most general form. And then g prime of x is equal to e to the x, which means its antiderivative, g of x, is still equal to e to the x. And now we're ready to apply the right-hand side right over here. So all this thing right over here is going to be equal to f of x, which is x squared-- and let me put it right underneath-- x squared times g of x, which is e to the x, minus. I want to make the colors match up-- minus the antiderivative of f prime of x. Well, f prime of x is 2x, times g of x. g of x is e to the x dx. So you might say, hey, Sal, we're left with another antiderivative, another indefinite integral right over here. How do we solve this one? And as you might guess, the key might be integration by parts again. And we're making progress. This right over here is a simpler expression than this. Notice we were able to reduce the degree of this x squared. It now is just a 2x. And what we can do to simplify this a little bit. Since 2 is just a scalar-- it's a constant, it's multiplying the function-- we can take that out of the integral sign. So let's take it this way. So let me rewrite it this way. that's multiplying the function. So let me put the 2 right out here. And so now what we're concerned about is finding the integral-- let me write it right over here-- the integral of xe to the x dx. And this is another integration by parts problem. And so let's again apply the same principles of integration by parts. What, when I take its derivative, is going to get simpler? Well, x is going to get simpler when I take its derivative. So now, for the purposes of integration by parts, let's redefine f of x to be equal to just x. And then we can still have g prime of x equaling e to the x. And so in this case, let me write it all down. f of x is equal to x. f prime of x is equal to 1. g prime of x is equal to e to the x. g of x, just the antiderivative of this," + }, + { + "Q": "at 6:47\nDose Sal notice this too?: b, h, c, and i are all right angles.", + "A": "If they were, yes. However, it is not given that those angles are right angles. For all we know, they could be 89 or 91 degrees or anything else close to right. Great observation though.", + "video_name": "95logvV8nXY", + "timestamps": [ + 407 + ], + "3min_transcript": "So let's say j is equal to m plus n. And then finally, we can split up h. Remember, it's this whole thing. Let's say that h is the same thing as o plus p plus q. This is o, this is p, this is q. And once again, I wanted to split up these interior angles if they're not already an angle of a triangle. I want to split them up into angles that are parts of these triangles. So we have h is equal to o, plus p, plus q. And the reason why that's interesting is now we can write the sum of these interior angles as the sum of a bunch of angles that are part of these triangles. And then we can use the fact that, for any one triangle, they add up to 180 degrees. So this expression right over her is going to be g. g is that angle right over here. We didn't make any substitutions. let me write the whole thing. So we have 900 minus, and instead of a g, well, actually I'm not making a substitution, so I can write g plus, and instead of an h I can write that h is o plus p plus q. And then plus i. i is sitting right over there. Plus i. And then plus j. And I kind of messed up the colors. The magenta will go with i. And then j is this expression right over here. So j is equal to m plus n instead of writing a j right there. And then finally, we have our f. And f, we've already seen, is equal to k plus l. So plus k plus l. So once again, I just rewrote this part right over here, in terms of these component angles. And now something very interesting is going to happen, because we know what these sums are going to be. They are the measures of the angle for this first triangle over here, for this triangle right over here. So g plus o plus k is 180 degrees. So g-- let me do this in a new color. So for this triangle right over here, we know that g plus o plus k are going to be equal to 180 degrees. So if we cross those out, we can write 180 instead. And then we also know-- let me see, I'm definitely running out of colors-- we know that p, for this middle triangle right over here, we know that p plus l plus m is 180 degrees. So you take those out and you know that sum is going to be equal to 180 degrees. And then finally, this is the home stretch here. We know that q plus n plus i is 180 degrees in this last triangle." + }, + { + "Q": "5:44 into the video, why do you put the 2 on the outside of the square root sign?", + "A": "This problem deals with simplifying a 6th root. So, we have to find factors with an exponent =6. If we find any, then we can find the 6th root for that item. Sal has 6th root( 2^6). This can be simplified to 2. Another way to look at it is to use rational exponents: 6th root( 2^6) = 2^(6/6) = 2 The exponent of 6/6 reduces to 1. 2^1 = 2. Hope this helps.", + "video_name": "iX7ivCww2ws", + "timestamps": [ + 344 + ], + "3min_transcript": "Times 3 to the 1/5. Now I have something that's multiplied. I have 2 multiplied by itself 5 times. And I'm taking that to the 1/5. Well, the 1/5 power of this is going to be 2. Or the fifth root of this is just going to be 2. So this is going to be a 2 right here. And this is going to be 3 to the 1/5 power. 2 times 3 to the 1/5, which is this simplified about as much as you can simplify it. But if we want to keep in radical form, we could write it as 2 times the fifth root 3 just like that. Let's try another one. Let me put some variables in there. Let's say we wanted to simplify the sixth root of 64 times x to the eighth. So let's do 64 first. 64 is equal to 2 times 32, which is 2 times 16. Which is 2 times 4. Which is 2 times 2. So we have 1, 2, 3, 4, 5, 6. So it's essentially 2 to the sixth power. So this is equivalent to the sixth root of 2 to the sixth-- that's what 64 is --times x to the eighth power. Now, the sixth root of 2 to the sixth, that's pretty straightforward. So this part right here is just going to be equal to 2. That's going to be 2 times the sixth root of x to the eighth power. And how can we simplify this? Well, x to the eighth power, that's the same thing as x to the sixth power times x squared. This is the same thing as x to the eighth. So this is going to be equal to 2 times the sixth root of x to the sixth times x squared. And the sixth root, this part right here, the sixth root of x to the sixth, that's just x. So this is going to be equal to 2 times x times the sixth root of x squared. Now, we can simplify this even more if you really think about. Remember, this expression right here, this is the exact same thing as x squared to the 1/6 power. And if you remember your exponent properties, when you raise something to an exponent, and then raise that to an exponent, that's equivalent to x to the 2 times 1/6 power. Or-- let me write this --2 times 1/6 power, which is the same thing-- Let me not forget to write my 2x there." + }, + { + "Q": "At 0:40, why did Sal say to turn the fraction on it's head?", + "A": "First off, notice that multiplying a number by 1/5 gives you the same answer as dividing it by 5. 5 and 1/5 are reciprocals, which means that when you multiply them you get 1 (5/1 is 1/5 flipped over). We use this property to help us divide fractions. We just multiply by the reciprocal instead. The easiest way to do that is to just flip over the fraction.", + "video_name": "tnkPY4UqJ44", + "timestamps": [ + 40 + ], + "3min_transcript": "Divide and write the answer as a mixed number. And we have 3/5 divided by 1/2. Now, whenever you're dividing any fractions, you just have to remember that dividing by a fraction is the same thing as multiplying by its reciprocal. So this thing right here is the same thing as 3/5 times-- so this is our 3/5 right here, and instead of a division sign, you want a multiplication sign, and instead of a 1/2, you want to take the reciprocal of 1/2, which would be 2/1-- so times 2/1. So dividing by 1/2 is the exact same thing as multiplying by 2/1. And we just do this as a straightforward multiplication problem now. 3 times 2 is 6, so our new numerator is 6. 5 times 1 is 5. So 3/5 divided by 1/2 as an improper fraction is 6/5. Now, they want us to write it as at mixed number. many times it goes. That'll be the whole number part of the mixed number. And then whatever's left over will be the remaining numerator over 5. So what we'll do is take 5 into 6. 5 goes into 6 one time. 1 times 5 is 5. Subtract. You have a remainder of 1. So 6/5 is equal to one whole, or 5/5, and 1/5. This 1 comes from whatever is left over. And now we're done! 3/5 divided by 1/2 is 1 and 1/5. Now, the one thing that's not obvious is why did this work? Why is dividing by 1/2 the same thing as multiplying essentially by 2. 2/1 is the same thing as 2. And to do that, I'll do a little side-- fairly simple-- example, but hopefully, it gets the point across. So we have four objects: one, two, three, four. So I have four objects, and if I were to divide into groups of two, so I want to divide it into groups of two. So that is one group of two and then that is another group of two, how many groups do I have? Well, 4 divided by 2, I have two groups of two, so that is equal to 2. Now, what if I took those same four objects: one, two, three, four. So I'm taking those same four objects. Instead of dividing them into groups of two, I want to divide them into groups of 1/2, which means each group will have half of an object in it. So let's say that would be one group right there. That is a second group. That is a third group. I think you see each group has half of a circle in it. That is the fourth. That's the fifth. That's the sixth." + }, + { + "Q": "At 0:16, what does it mean by multiplying its reciprocal?", + "A": "The reciprocal of a fraction is the inverse of it; you basically take a fraction, flip it over, and that s its reciprocal. 3/4 is the reciprocal of 4/3, and 4/3 is the reciprocal of 3/4. Do you get that? When dividing by fractions, we turn the division problem into a multiplication problem: we take the divisor and find it s reciprocal, and then we just multiply. For example 1/2 \u00c3\u00b7 3/4 = 1/2 x 4/3 = 4/6 or 2/3", + "video_name": "tnkPY4UqJ44", + "timestamps": [ + 16 + ], + "3min_transcript": "Divide and write the answer as a mixed number. And we have 3/5 divided by 1/2. Now, whenever you're dividing any fractions, you just have to remember that dividing by a fraction is the same thing as multiplying by its reciprocal. So this thing right here is the same thing as 3/5 times-- so this is our 3/5 right here, and instead of a division sign, you want a multiplication sign, and instead of a 1/2, you want to take the reciprocal of 1/2, which would be 2/1-- so times 2/1. So dividing by 1/2 is the exact same thing as multiplying by 2/1. And we just do this as a straightforward multiplication problem now. 3 times 2 is 6, so our new numerator is 6. 5 times 1 is 5. So 3/5 divided by 1/2 as an improper fraction is 6/5. Now, they want us to write it as at mixed number. many times it goes. That'll be the whole number part of the mixed number. And then whatever's left over will be the remaining numerator over 5. So what we'll do is take 5 into 6. 5 goes into 6 one time. 1 times 5 is 5. Subtract. You have a remainder of 1. So 6/5 is equal to one whole, or 5/5, and 1/5. This 1 comes from whatever is left over. And now we're done! 3/5 divided by 1/2 is 1 and 1/5. Now, the one thing that's not obvious is why did this work? Why is dividing by 1/2 the same thing as multiplying essentially by 2. 2/1 is the same thing as 2. And to do that, I'll do a little side-- fairly simple-- example, but hopefully, it gets the point across. So we have four objects: one, two, three, four. So I have four objects, and if I were to divide into groups of two, so I want to divide it into groups of two. So that is one group of two and then that is another group of two, how many groups do I have? Well, 4 divided by 2, I have two groups of two, so that is equal to 2. Now, what if I took those same four objects: one, two, three, four. So I'm taking those same four objects. Instead of dividing them into groups of two, I want to divide them into groups of 1/2, which means each group will have half of an object in it. So let's say that would be one group right there. That is a second group. That is a third group. I think you see each group has half of a circle in it. That is the fourth. That's the fifth. That's the sixth." + }, + { + "Q": "i am confused at 0:35 i think you switch it to multiplication", + "A": "When you divide by a fraction, it is the same as multiplying by the reciprocal. So if you have 3/8 divided by 1/2, lets say, it would be the same as multiplying 3/8 by 2/1.", + "video_name": "tnkPY4UqJ44", + "timestamps": [ + 35 + ], + "3min_transcript": "Divide and write the answer as a mixed number. And we have 3/5 divided by 1/2. Now, whenever you're dividing any fractions, you just have to remember that dividing by a fraction is the same thing as multiplying by its reciprocal. So this thing right here is the same thing as 3/5 times-- so this is our 3/5 right here, and instead of a division sign, you want a multiplication sign, and instead of a 1/2, you want to take the reciprocal of 1/2, which would be 2/1-- so times 2/1. So dividing by 1/2 is the exact same thing as multiplying by 2/1. And we just do this as a straightforward multiplication problem now. 3 times 2 is 6, so our new numerator is 6. 5 times 1 is 5. So 3/5 divided by 1/2 as an improper fraction is 6/5. Now, they want us to write it as at mixed number. many times it goes. That'll be the whole number part of the mixed number. And then whatever's left over will be the remaining numerator over 5. So what we'll do is take 5 into 6. 5 goes into 6 one time. 1 times 5 is 5. Subtract. You have a remainder of 1. So 6/5 is equal to one whole, or 5/5, and 1/5. This 1 comes from whatever is left over. And now we're done! 3/5 divided by 1/2 is 1 and 1/5. Now, the one thing that's not obvious is why did this work? Why is dividing by 1/2 the same thing as multiplying essentially by 2. 2/1 is the same thing as 2. And to do that, I'll do a little side-- fairly simple-- example, but hopefully, it gets the point across. So we have four objects: one, two, three, four. So I have four objects, and if I were to divide into groups of two, so I want to divide it into groups of two. So that is one group of two and then that is another group of two, how many groups do I have? Well, 4 divided by 2, I have two groups of two, so that is equal to 2. Now, what if I took those same four objects: one, two, three, four. So I'm taking those same four objects. Instead of dividing them into groups of two, I want to divide them into groups of 1/2, which means each group will have half of an object in it. So let's say that would be one group right there. That is a second group. That is a third group. I think you see each group has half of a circle in it. That is the fourth. That's the fifth. That's the sixth." + }, + { + "Q": "At 0:43, is it necessary to multiply each side of the inequality by the reciprocal of -5 or can you just divide each side by -5 and still get the same answer?", + "A": "You can divide both sides by -5 and get the same answer.", + "video_name": "D1cKk48kz-E", + "timestamps": [ + 43 + ], + "3min_transcript": "Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate the c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. We are multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. We would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so it's good that we did not include negative 4. So this is our solution. And this is that solution graphed. And I wanted to do that in that other green color. That's what it looks like." + }, + { + "Q": "The example Sal gives at 5:51,\n(2(x+3)(x-1))/ (x+3)^2(x+1) I don't understand why x at -3 isn't a removable discontinuity. If you put -3 in the original equation without simplying, you get 0/0.\n\nThank you.", + "A": "A removable discontinuity would divide COMPLETELY out of the denominator. In this case, after the division there is still a factor of ( x + 3 ) left in the denominator, so that takes precedence over the fact that a single ( x + 3 ) was able to be divided out and indicates an asymptote at x = -3 instead of a removable discontinuity.", + "video_name": "TX_mx3qULpw", + "timestamps": [ + 351 + ], + "3min_transcript": "So this would be the two zeros. Two zeros. And so let's look at another similar situation. Let's look at a situation where f of x is equal to, you know, the numerator, x plus three times x minus three. And let's say that we do have, let's say that one of those x values, positive or negative three, do make the denominator equal to zero. So let's say, x plus three and then, say times x plus one. Well, you see here, now since x plus three is both the numerator and the denominator, you could divide x plus three divided by x plus three. They cancel out. And here x equals negative three would be a removable discontinuity. So this would have zero at x equals three and a removable discontinuity And so those values that make the numerator equal to zero we now see it could be a zero or it could represent a removable discontinuity. And here I just pick a removable discontinuity be at negative three or it could be the other way around or it could be at both values. If this was x plus three times x minus three over x plus three times x minus three, then you would have a removable discontinuity at both x is positive three and negative three. And then you could go even further. F of x could look like this. It could be two times x plus three times x minus three over x plus three squared and I'll just make up some other, another expression here, x plus one. So, what's going to happen here? Even if you divide the numerator and the denominator by x plus three, you're still going to have one x plus three left over the denominator. That could cancel with one of the x plus threes but you're still going to have an x plus three. And so in this case, you would have a vertical asymptote. at x is equal to three. And you would have a vertical asymptote. I'll just shorten it shorthand. You would have a vertical, I'll just write that out, vertical asymptote at x is equal to negative three. So these particular examples that I just showed you showed you that any value that makes the numerator equals zero aren't necessarily zeros for the function. They could be zeros. They could be removable discontinuities or they could be vertical asymptotes. But they would all occur at x is equal to positive or negative three. So if that lands, now let's look at the choices again. So choice A has a zero at x equals positive three and it has a vertical asymptote at x equals negative three. So that's actually very consistent with this situation that I just described. So choice A actually is looking pretty good. Choice B has a zero at x equals three but its vertical asymptote looks like it's an x equals two" + }, + { + "Q": "At 02:35, Sal says that pi/2 there is equal to 3.5 pi over seven, how does that work out? Really trying to wrap my head around this. Why did he choose the number 3.5?", + "A": "Sal noted that quadrant I contains angles from 0 (X-axis) to pi/2 radians (Y-axis). Because he was discussing the angle of 2pi / 7 radians, he converted pi / 2 to sevenths. Half of 7 is 3.5. So he checked to see if the given angle was between zero radians and 3.5pi / 7. Because the denominators of the angles were the same, it was then easy to compare the numerators and see that 2 pi / 7 radians is less than 3.5pi / 7 radians . Ttherefore the angle of 2 pi / 7 lies in quadrant 1.", + "video_name": "fYQ3GRSu4JU", + "timestamps": [ + 155 + ], + "3min_transcript": "If we go straight up, if we rotate it, essentially, if you want to think in degrees, if you rotate it counterclockwise 90 degrees, that is going to get us to pi over two. That would have been a counterclockwise rotation of pi over two radians. Now is three pi over five greater or less than that? Well, three pi over five, three pi over five is greater than, or I guess another way I can say it is, three pi over six is less than three pi over five. You make the denominator smaller, making the fraction larger. Three pi over six is the same thing as pi over two. So, let me write it this way. Pi over two is less than three pi over five. It's definitely past this. We're gonna go past this. Does that get us all the way over here? If we were to go, essentially, be pointed in the opposite direction. Instead of being pointed to the right, making a full, that would be pi radians. That would be pi radians. But this thing is less than pi. Pi would be five pi over five. This is less than pi radians. We are going to sit, we are going to sit someplace, someplace, and I'm just estimating it. We are gonna sit someplace like that. And so we are going to sit in the second quadrant. Let's think about two pi seven. Two pi over seven, do we even get past pi over two? Pi over two here would be 3.5 pi over seven. We don't even get to pi over two. We're gonna end up, we're gonna end up someplace, someplace over here. This thing is, it's greater than zero, so we're gonna definitely start moving counterclockwise, but we're not even gonna get to... This thing is less than pi over two. This is gonna throw us in the first quadrant. What about three radians? three is a little bit less than pi. Right? Three is less than pi but it's greater than pi over two. How do we know that? Well, pi is approximately 3.14159 and it just keeps going on and on forever. So, three is definitely closer to that than it is to half of that. It's going to be between pi over two, and pi. It's gonna be, if we start with this magenta ray, we rotate counterclockwise by three radians, we are gonna get... Actually, it's probably gonna be, it's gonna look something, it's gonna be something like this. But for the sake of this exercise, we have gotten ourselves, once again, into the second quadrant." + }, + { + "Q": "Infinagons?? 3:05", + "A": "infinigons are polygons that have an infinite number of sides.", + "video_name": "D2xYjiL8yyE", + "timestamps": [ + 185 + ], + "3min_transcript": "is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0." + }, + { + "Q": "At 4:28 Sal finishes the Pascal Expression for (a+b)^4. He starts at a^0. Why not a^1?", + "A": "Because that s the level at which Pascal s triangle starts (2:30)", + "video_name": "v9Evg2tBdRk", + "timestamps": [ + 268 + ], + "3min_transcript": "or you could go like that. Same exact logic: there's three ways to get to this point. And then there's only one way to get to that point right over there. And so let's add a fifth level because this was actually what we care about when we think about something to the fourth power. This is essentially zeroth power-- binomial to zeroth power, first power, second power, third power. So let's go to the fourth power. So how many ways are there to get here? Well I just have to go all the way straight down along this left side to get here, so there's only one way. There's four ways to get here. I could go like that, I could go like that, I could go like that, and I can go like that. There's six ways to go here. Three ways to get to this place, three ways to get to this place. So six ways to get to that and, if you have the time, you could figure that out. There's three plus one-- four ways to get here. And then there's one way to get there. when I'm taking something to the-- if I'm taking something to the zeroth power. This is if I'm taking a binomial to the first power, to the second power. Obviously a binomial to the first power, the coefficients on a and b are just one and one. But when you square it, it would be a squared plus two ab plus b squared. If you take the third power, these are the coefficients-- third power. And to the fourth power, these are the coefficients. So let's write them down. The coefficients, I'm claiming, are going to be one, four, six, four, and one. And how do I know what the powers of a and b are going to be? Well I start a, I start this first term, at the highest power: a to the fourth. And then I go down from there. a to the fourth, a to the third, a squared, a to the first, and I guess I could write a to the zero which of course is just one. And then b to first, b squared, b to the third power, and then b to the fourth, and then I just add those terms together. And there you have it. I have just figured out the expansion of a plus b to the fourth power. It's exactly what I just wrote down. This term right over here, a to the fourth, that's what this term is. One a to the fourth b to the zero: that's just a to the fourth. This term right over here is equivalent to this term right over there. And so I guess you see that this gave me an equivalent result. Now an interesting question is 'why did this work?' And I encourage you to pause this video and think about it on your own. Well, to realize why it works let's just go to these first levels right over here." + }, + { + "Q": "I see many people have questions about this: at approx 4:03, he flipped both sides of the fraction. I understand the concept of doing the same thing to each side of an equation. But what prompted him to do this for the purpose of this problem? I want to understand so I can apply this to other problems.", + "A": "When your variable is on the bottom(of a fraction), you generally want to get it to the top(of the fraction). When he was done simplifying the left side of the equation, he saw that the variable that he was trying to solve for was on the bottom(of the fraction). This probably prompted him to flip both sides of the equation, or to take the reciprocal of both sides of the equation.", + "video_name": "gD7A1LA4jO8", + "timestamps": [ + 243 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:08 Sal defines the range of f(x) as all real number such that f(x) is greater than or equal to 0. Why is it not all real numbers such that x is greater than or equal to 0? I don't really understand in this case the significance of f(x) as a function vs. a variable.", + "A": "X has to do with the domain, not the range, so if you were looking for domain, you would be fine. On the other hand, f(x) has to do with the range (this is functional notation f(x)= mx + b rather than perhaps what you are more used to in the slope intercept form y = mx+b), so that is why he says what he did.", + "video_name": "96uHMcHWD2E", + "timestamps": [ + 248 + ], + "3min_transcript": "and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here. What this looks like. So the graph of \"f(x) is equal to x squared\" is going to look something like this. So, it's gonna look, it's going to look something like this. I'm obviously hand-drawing it, so it's not perfect. It's gonna be a parabola with a, with a vertex right here at the origin. So this is the graph, this is the graph, \"y is equal to f(x),\" this of course is the x-axis, this of course is the y-axis. So let's think about it, what is the set of all possible outputs? Well in this case, the set of all possible outputs is the set of all possible y's here. Well, we see, y can take on any non-negative value. y could be zero, y could one, y could be pi, y could be e, but y cannot be negative. So the range here is, the range... We could, well we could say it a couple of ways, we could say, \"f(x)\", let me write it this way. \"f(x) is a member of the real numbers\" \"or equal to zero.\" We could write it that way, if we wanted to write it in a less mathy notation, we could say that \"f(x) is going to be\" \"greater than or equal to zero.\" f(x) is not going to be negative, so any non-negative number, the set of all non-negative numbers, that is our range. Let's do another example of this, just to make it a little bit, just to make it a little bit, a little bit clearer. Let's say that I had, let's say that I had g(x), let's say I have g(x), I'll do this in white, let's say it's equal to \"x squared over x.\" So we could try to simplify g(x) a little bit, we could say, \"look, if I have x squared\" \"and I divide it by x, that's gonna,\" \"that's the same thing as g(x) being equal to x.\" \"x squared over x\" is x, but we have to be careful." + }, + { + "Q": "At 4:03 Sal places a line after the Real number notation. what does this bar mean?", + "A": "The bar means such that, so it would read all real numbers such that f(x) is greater than or equal to zero. That is also the language he used in the video.", + "video_name": "96uHMcHWD2E", + "timestamps": [ + 243 + ], + "3min_transcript": "and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here. What this looks like. So the graph of \"f(x) is equal to x squared\" is going to look something like this. So, it's gonna look, it's going to look something like this. I'm obviously hand-drawing it, so it's not perfect. It's gonna be a parabola with a, with a vertex right here at the origin. So this is the graph, this is the graph, \"y is equal to f(x),\" this of course is the x-axis, this of course is the y-axis. So let's think about it, what is the set of all possible outputs? Well in this case, the set of all possible outputs is the set of all possible y's here. Well, we see, y can take on any non-negative value. y could be zero, y could one, y could be pi, y could be e, but y cannot be negative. So the range here is, the range... We could, well we could say it a couple of ways, we could say, \"f(x)\", let me write it this way. \"f(x) is a member of the real numbers\" \"or equal to zero.\" We could write it that way, if we wanted to write it in a less mathy notation, we could say that \"f(x) is going to be\" \"greater than or equal to zero.\" f(x) is not going to be negative, so any non-negative number, the set of all non-negative numbers, that is our range. Let's do another example of this, just to make it a little bit, just to make it a little bit, a little bit clearer. Let's say that I had, let's say that I had g(x), let's say I have g(x), I'll do this in white, let's say it's equal to \"x squared over x.\" So we could try to simplify g(x) a little bit, we could say, \"look, if I have x squared\" \"and I divide it by x, that's gonna,\" \"that's the same thing as g(x) being equal to x.\" \"x squared over x\" is x, but we have to be careful." + }, + { + "Q": "At 0:30, Mr.Khan asserts that a domain is \"The set of all inputs over which the function is defined.\"\n\nIf that is the case, then in the last video, at the last example, wouldn't the domain be x = 1 and x = 0?\n\nAm I missing something?", + "A": "No that would have been the range, you put in \u00cf\u0080 and 3, and get out 1 and 0.", + "video_name": "96uHMcHWD2E", + "timestamps": [ + 30 + ], + "3min_transcript": "- As a little bit of a review, we know that if we have some function, let's call it \"f\". We don't have to call it \"f\", but \"f\" is the letter most typically used for functions, that if I give it an input, a valid input, if I give it a valid input, and I use the variable \"x\" for that valid input, it is going to map that to an output. It is going to map that, or produce, given this x, it's going to product an output that we would call \"f(x).\" And we've already talked a little bit about the notion of a domain. A domain is the set of all of the inputs over which the function is defined. So if this the domain here, if this is the domain here, and I take a value here, and I put that in for x, then the function is going to output an f(x). If I take something that's outside of the domain, let me do that in a different color... If I take something that is outside of the domain and try to input it into this function, the function will say, \"hey, wait wait,\" \"I'm not defined for that thing\" \"that's outside of the domain.\" Now another interesting thing to think about, okay, we know the set of all of the valid inputs, that's called the domain, but what about all, the set of all of the outputs that the function could actually produce? And we have a name for that. That is called the range of the function. So the range. The range, and the most typical, there's actually a couple of definitions for range, but the most typical definition for range is \"the set of all possible outputs.\" So you give me, you input something from the domain, it's going to output something, and by definition, because we have outputted it from this function, that thing is going to be in the range, and if we take the set of all of the things that the function could output, that is going to make up the range. So this right over here is the set of all possible, all possible outputs. All possible outputs. So let's make that a little bit more concrete, with an example. So let's say that I have the function f(x) and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here." + }, + { + "Q": "At 3:17, does the term \"parabola\" refer to the shape of the graph, or something else? I have NEVER heard that term before!", + "A": "The word parabola refers to the U-shape of the graph. That is the name of that shape. The equations / functions that create parabolas are quadratics. Hope that helps.", + "video_name": "96uHMcHWD2E", + "timestamps": [ + 197 + ], + "3min_transcript": "okay, we know the set of all of the valid inputs, that's called the domain, but what about all, the set of all of the outputs that the function could actually produce? And we have a name for that. That is called the range of the function. So the range. The range, and the most typical, there's actually a couple of definitions for range, but the most typical definition for range is \"the set of all possible outputs.\" So you give me, you input something from the domain, it's going to output something, and by definition, because we have outputted it from this function, that thing is going to be in the range, and if we take the set of all of the things that the function could output, that is going to make up the range. So this right over here is the set of all possible, all possible outputs. All possible outputs. So let's make that a little bit more concrete, with an example. So let's say that I have the function f(x) and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here. What this looks like. So the graph of \"f(x) is equal to x squared\" is going to look something like this. So, it's gonna look, it's going to look something like this. I'm obviously hand-drawing it, so it's not perfect. It's gonna be a parabola with a, with a vertex right here at the origin. So this is the graph, this is the graph, \"y is equal to f(x),\" this of course is the x-axis, this of course is the y-axis. So let's think about it, what is the set of all possible outputs? Well in this case, the set of all possible outputs is the set of all possible y's here. Well, we see, y can take on any non-negative value. y could be zero, y could one, y could be pi, y could be e, but y cannot be negative. So the range here is, the range... We could, well we could say it a couple of ways, we could say, \"f(x)\", let me write it this way. \"f(x) is a member of the real numbers\"" + }, + { + "Q": "What is a \"porabola\"? 3:20", + "A": "A parabola is a two-dimensional, mirror-symmetrical curve, which is approximately U-shaped.", + "video_name": "96uHMcHWD2E", + "timestamps": [ + 200 + ], + "3min_transcript": "okay, we know the set of all of the valid inputs, that's called the domain, but what about all, the set of all of the outputs that the function could actually produce? And we have a name for that. That is called the range of the function. So the range. The range, and the most typical, there's actually a couple of definitions for range, but the most typical definition for range is \"the set of all possible outputs.\" So you give me, you input something from the domain, it's going to output something, and by definition, because we have outputted it from this function, that thing is going to be in the range, and if we take the set of all of the things that the function could output, that is going to make up the range. So this right over here is the set of all possible, all possible outputs. All possible outputs. So let's make that a little bit more concrete, with an example. So let's say that I have the function f(x) and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here. What this looks like. So the graph of \"f(x) is equal to x squared\" is going to look something like this. So, it's gonna look, it's going to look something like this. I'm obviously hand-drawing it, so it's not perfect. It's gonna be a parabola with a, with a vertex right here at the origin. So this is the graph, this is the graph, \"y is equal to f(x),\" this of course is the x-axis, this of course is the y-axis. So let's think about it, what is the set of all possible outputs? Well in this case, the set of all possible outputs is the set of all possible y's here. Well, we see, y can take on any non-negative value. y could be zero, y could one, y could be pi, y could be e, but y cannot be negative. So the range here is, the range... We could, well we could say it a couple of ways, we could say, \"f(x)\", let me write it this way. \"f(x) is a member of the real numbers\"" + }, + { + "Q": "At 3:48, why can't x be negative? .", + "A": "The domain (values of x) is any real number. It s the range (values of y) that cannot be negative. That s because y = x^2 , and we know that squaring anything (whether x is positive or negative or zero to begin with) cannot produce a negative result.", + "video_name": "96uHMcHWD2E", + "timestamps": [ + 228 + ], + "3min_transcript": "and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here. What this looks like. So the graph of \"f(x) is equal to x squared\" is going to look something like this. So, it's gonna look, it's going to look something like this. I'm obviously hand-drawing it, so it's not perfect. It's gonna be a parabola with a, with a vertex right here at the origin. So this is the graph, this is the graph, \"y is equal to f(x),\" this of course is the x-axis, this of course is the y-axis. So let's think about it, what is the set of all possible outputs? Well in this case, the set of all possible outputs is the set of all possible y's here. Well, we see, y can take on any non-negative value. y could be zero, y could one, y could be pi, y could be e, but y cannot be negative. So the range here is, the range... We could, well we could say it a couple of ways, we could say, \"f(x)\", let me write it this way. \"f(x) is a member of the real numbers\" \"or equal to zero.\" We could write it that way, if we wanted to write it in a less mathy notation, we could say that \"f(x) is going to be\" \"greater than or equal to zero.\" f(x) is not going to be negative, so any non-negative number, the set of all non-negative numbers, that is our range. Let's do another example of this, just to make it a little bit, just to make it a little bit, a little bit clearer. Let's say that I had, let's say that I had g(x), let's say I have g(x), I'll do this in white, let's say it's equal to \"x squared over x.\" So we could try to simplify g(x) a little bit, we could say, \"look, if I have x squared\" \"and I divide it by x, that's gonna,\" \"that's the same thing as g(x) being equal to x.\" \"x squared over x\" is x, but we have to be careful." + }, + { + "Q": "at 7:20 why do you ignore the denominators when solving the equation?", + "A": "Both sides of the equation have the same denominator, so multiplying the numerators on both sides by the denominator cancels them out, which has the same effect as ignoring them. Example: x/3 = y/3 Multiply both sides by the denominator 3: 3 * (x/3) = (y/3) * 3 3x/3 = 3y/3 Simplify: x = y It s the same result as ignoring the denominators.", + "video_name": "S-XKGBesRzk", + "timestamps": [ + 440 + ], + "3min_transcript": "it actually does add up to this, then I'm done and I will have fully decomposed this fraction. I guess is the way-- I don't know if that's the correct terminology. So let's try to do that. So if I were to add these two terms, what do I get? When you add anything, you find the common denominator, and the common denominator, the easiest common denominator, is to multiply the two denominators, so let me write this here. So a over x plus 5 plus b over x minus 8 is equal to-- well, let's get the common denominator-- it's equal to x plus 5 times x minus 8. And then the a term, we would-- a over x plus 5 is the same thing as a times x minus 8 over this whole thing. I mean, if I just wrote this right here, you would just And then you could add that to the common denominator, x plus 5 times x minus 8, and it would be b times x plus 5. Important to realize, that, look. This term is the exact same thing as this term if you just cancel the x minus 8 out, and this term is the exact same thing as this term if you just cancel the x plus 5 out. But now that we have an actual common denominator, we can add them together, so we get-- let me just write the left side here over-- a over x plus 5-- I'm sorry. I want to write this over here. I want to write x plus 3 over plus 5 times x minus 8 is equal to is equal to the sum of these two things on top. a times x minus 8 plus b times x plus 5, all of that over So the denominators are the same, so we know that this, when you add this together, you have to get this. So if we want to solve for a and b, let's just set that equality. We can ignore the denominators. So we can say that x plus 3 is equal to a times x minus 8 plus b times x plus 5. Now, there's two ways to solve for a and b from this point going forward. One is the way that I was actually taught in the seventh or eighth grade, which tends to take a little longer, then there's a fast way to do it and it never hurts to do the fast way first. If you want to solve for a, let's pick an x that'll make this term disappear. So what x would make this term disappear? Well, if I say x is minus 5, then this becomes 0, and" + }, + { + "Q": "4:13 to 4:30 is confusing", + "A": "double lines mean that that angle is the same as the other angle with the same two lines.", + "video_name": "wRBMmiNHQaE", + "timestamps": [ + 253, + 270 + ], + "3min_transcript": "we also know that angle DBA --we know that this is DBA right over here-- we also know that angle DBA and angle DBC are supplementary this angle and this angle are supplementary, their outer sides form a straight angle, they are adjacent so they are supplementary which tells us that angle DBA, this angle right over here, plus angle DBC, this angle over here, is going to be equal to 180 degrees. Now, from this top one, this top statement over here, we can subtract angle DBC from both sides and we get angle CBE is equal to 180 degrees minus angle DBC that's this information right over here, I just put or subtracted it from both sides of the equation and this right over here, if I do the exact same thing, subtract angle DBC from both sides of the equation, I get angle DBA is equal to 180 degrees --let me scroll over to the right a little bit-- is equal to 180 degrees minus angle DBC. So clearly, angle CBE is equal to 180 degrees minus angle DBC angle DBA is equal to 180 degrees minus angle DBC so they are equal to each other! They are both equal to the same thing so we get, which is what we wanted to get, angle CBE is equal to angle DBA. Angle CBE, which is this angle right over here, is equal to angle DBA and sometimes you might see that shown like this; so angle CBE, that's its measure, and you would say that And we have other vertical angles whatever this measure is, and sometimes you will see it with a double line like that, that you can say that THAT is going to be the same as whatever this angle right over here is. You will see it written like that sometimes, I like to use colors but not all books have the luxury of colors, or sometimes you will even see it written like this to show that they are the same angle; this angle and this angle --to show that these are different-- sometimes they will say that they are the same in this way. This angle is equal to this vertical angle, is equal to its vertical angle right over here and that this angle is equal to this angle that is opposite the intersection right over here. What we have proved is the general case because all I did here is I just did two general intersecting lines I picked a random angle, and then I proved that it is equal to the angle that is vertical to it." + }, + { + "Q": "why does he put hash marks on the angle markers ? like at 4:40", + "A": "He puts hash marks on the angle markers to show that the marked angles are congruent. If he didn t put hash marks on the angle markers, the (previously marked) angles would be considered congruent to the other two angles.", + "video_name": "wRBMmiNHQaE", + "timestamps": [ + 280 + ], + "3min_transcript": "we also know that angle DBA --we know that this is DBA right over here-- we also know that angle DBA and angle DBC are supplementary this angle and this angle are supplementary, their outer sides form a straight angle, they are adjacent so they are supplementary which tells us that angle DBA, this angle right over here, plus angle DBC, this angle over here, is going to be equal to 180 degrees. Now, from this top one, this top statement over here, we can subtract angle DBC from both sides and we get angle CBE is equal to 180 degrees minus angle DBC that's this information right over here, I just put or subtracted it from both sides of the equation and this right over here, if I do the exact same thing, subtract angle DBC from both sides of the equation, I get angle DBA is equal to 180 degrees --let me scroll over to the right a little bit-- is equal to 180 degrees minus angle DBC. So clearly, angle CBE is equal to 180 degrees minus angle DBC angle DBA is equal to 180 degrees minus angle DBC so they are equal to each other! They are both equal to the same thing so we get, which is what we wanted to get, angle CBE is equal to angle DBA. Angle CBE, which is this angle right over here, is equal to angle DBA and sometimes you might see that shown like this; so angle CBE, that's its measure, and you would say that And we have other vertical angles whatever this measure is, and sometimes you will see it with a double line like that, that you can say that THAT is going to be the same as whatever this angle right over here is. You will see it written like that sometimes, I like to use colors but not all books have the luxury of colors, or sometimes you will even see it written like this to show that they are the same angle; this angle and this angle --to show that these are different-- sometimes they will say that they are the same in this way. This angle is equal to this vertical angle, is equal to its vertical angle right over here and that this angle is equal to this angle that is opposite the intersection right over here. What we have proved is the general case because all I did here is I just did two general intersecting lines I picked a random angle, and then I proved that it is equal to the angle that is vertical to it." + }, + { + "Q": "at 1:16 Sal put a decimal point and some zeros on the 63 would he be able to put a decimal and zeros on the 35 without changing the question", + "A": "yes, the decimals don t change anything because they have no value. 0.00 = 0 35 + 0.00 = 35 the decimals just help with solving the problem", + "video_name": "xUDlKV8lJbM", + "timestamps": [ + 76 + ], + "3min_transcript": "Let's take 63 and divide it by 35. So the first thing that we might say is, OK, well, 35 doesn't go into 6. It does go into 63. It goes into 63 one time, because 2 times 35 is 70, so that's too big. So it goes one time. So let me write that. 1 times 35 is 35. And then if we were to subtract and we can regroup up here, we can take a 10 from the 60, so it becomes a 50, give that 10 to the 3, so it becomes a 13. 13 minus 5 is 8. 5 minus 3 is 2. So you could just say, hey, 63 divided by 35-- You could say 63 divided by 35 is equal to 1 remainder 28. But this isn't so satisfying. We know that the real answer is going to be one point something, something, something. So what I want to do is keep dividing. I want to divide this thing completely and see what type And to do that, I essentially have to add a decimal here and then just keep bringing down decimal places to the right of the decimal. So 63 is the exact same thing as 63.0, and I could add as many zeroes as I might want to add here. So what we could do is we just make sure that this decimals right over there, and we can now bring down a zero from the tenths place right over here. And you bring down that zero, and now we ask ourselves, how many times does 35 go into 280? And, as always, this is a bit of an art when you're dividing a two-digit number into a three-digit number. So let's see, it's definitely going to be-- if I were to say-- so 40 goes into 280 seven times. 30 goes into 280 about nine times. It's going to be between 7 and 9, so let's try 8. So, let's see what 35 times 8 is. 35 times 8. 5 times 8 is 40, 3 times 8 is 24, plus 4 is 28. So 35 goes into 280 exactly eight times. 8 times 5, we already figured it out. 8 times 35 is exactly 280, and we don't have any remainder now, so we don't have to bring down any more of these zeroes. So now we know exactly that 63 divided by 35 is equal to exactly 1.8." + }, + { + "Q": "Why e^-u at 2:33.I dont get it!", + "A": "Basic property of powers in a fraction: those on the bottom are negatives of those on the top. 1/x = x^(-1) Therefore, 1/e^u = e^(-u) to get rid of the fraction.", + "video_name": "ShpI3gPgLBA", + "timestamps": [ + 153 + ], + "3min_transcript": "So when you look over here, you have a cosine of 5x dx, but we don't have a 5 cosine of 5x dx. But we know how to solve that. We can multiply by 5 and divide by 5. 1/5 times 5 is just going to be 1, so we haven't changed the value of the expression. When we do it this way, we see pretty clearly we have our u and we have our du. Our du is 5. Let me circle that. Let me do that in that blue color-- is 5 cosine of 5x dx. So we can rewrite this entire expression as-- do that 1/5 in purple. This is going to be equal to 1/5-- I hope you don't hear that crow outside. He's getting quite obnoxious. 1/5 times the integral of all the stuff in blue is my du, So how do we take the antiderivative of this? Well, you might be tempted to well, what would you do here? Well, we're still not quite ready to simply take the antiderivative here. If I were to rewrite this, I could rewrite this as this is equal to 1/5 times the integral of e to the negative u du. And so what might jump out at you is maybe we do another substitution. We've already used the letter u, so maybe we'll use w. We'll do some w-substitution. And you might be able to do this in your head, but we'll do w-substitution just to make it a little bit clearer. This would have been really useful if this was just e to the u because we know the antiderivative of e to the u So let's try to get it in terms of the form of e to the something and not e to the negative something. So let's set. And I'm running out of colors here. Let's set w as equal to negative u. In that case, then dw, derivative of w with respect to u, is negative 1. Or if we were to write that statement in differential form, dw is equal to du times negative 1 is negative du. So this right over here would be our w. And do we have a dw here? Well, we have just a du. We don't have a negative du there. But we can create a negative du by multiplying this inside by a negative 1, but then also by multiplying the outside by a negative 1. Negative 1 times negative 1 is positive 1. We haven't changed the value. We have to do both of these in order for it to make sense. Or I could do it like this. So negative 1 over here and a negative 1 right over there." + }, + { + "Q": "Is RSH at 2:39 a real theorem? Or just another name for the HL postulate?", + "A": "RSH is actually the HL congruence Theorem", + "video_name": "q7eF5Ci944U", + "timestamps": [ + 159 + ], + "3min_transcript": "we'll set up some triangles here since we know a lot about triangles now. And we'll set up the triangles by drawing two more radii, radius OC and radius OA. And that's useful for us because we know that they're both radii for the same circle. So they have to be the same length. The radius doesn't change on a circle. So those two things are the same length. And you might already see where this is going. Let me label this point here. Let me call this M because we're hoping that ends up being the midpoint of AC. Triangle AMO is a right triangle. This is its hypotenuse. AO is its hypotenuse. Triangle OMC is a right triangle, and this is its hypotenuse right over there. And so already showed that the hypotenuses have the same length, and both of these right triangles share segment or side OM. So OM is clearly equal to itself. And in a previous video, not the same video where we explained this thing. I think the video is called \"More on why SSA is not a postulate.\" In that video, we say that SSA is not a postulate. So SSA, not a congruency postulate. But we did establish in that video that RSH is a congruency postulate. And RSH tells us that if we have a right triangle-- that's where the R comes from-- if we have a right triangle, and we have one of the sides are congruent, and the hypotenuse is congruent, then we have two congruent triangles. And if you look at this right over here, we have two right triangles. AMO is a right triangle. CMO is a right triangle. They have one leg that's congruent, right over here, MO, and then both of their hypotenuses are congruent to each other. So, by RSH, we know that triangle AMO And so if we know that they're congruent, then their corresponding sides have to be congruent. So based on that, we then know that AM is a corresponding side. Let me do that in a different color. AM is a corresponding side to MC. So we know that AM must be equal to MC because they're corresponding sides. These are corresponding sides. So congruency implies that these are equal. And if those are equal, then we know that OD is bisecting AC. So we've established what we need to do. Another way that we could have proven it without RSH, is just straight up with the Pythagorean theorem. We already know, just by setting up these two radii right over here, we know that OA-- so we draw a little line here." + }, + { + "Q": "At 2:25, why did Sal used inverse of Sine to calculate the Radian angle (am I using that terminology right?) instead of just normal Sine? how do you know when to use the inverse calculation?\n\nALSO, we us 2pi because that'll give us a full circle revolution back to the starting, correct? But why is there a need to multiply it with a n integer? Can the answer just be 0.34 + 2pi?", + "A": "The sine function takes an angle and gives you the ratio of the side lengths. If you have the ratio of the side lengths and you are looking for the angle instead, this is when you would use the inverse of the sine function. And, since sine is a periodic function, the value will repeat infinitely. The 2pi*n is necessary to give ALL of the answers, as opposed to only the answers in the first rotation.", + "video_name": "NC7iWEQ9Kug", + "timestamps": [ + 145 + ], + "3min_transcript": "Voiceover:Which of these are contained in the solution set to sine of X is equal to 1/3? Answers should be rounded to the nearest hundredths. Select all that apply. I encourage you to pause the video right now and work on it on your own. I'm assuming you've given a go at it. Let's think about what this is asking. They're asking what are the X values? What is the solution set? What are the possible X values where sine of X is equal to 1/3? To help us visualize this, let's draw a unit circle. That's my Y axis. This right over here is my X axis. This is X set one. This is Y positive one. Negative one along the X axis, and negative one on the Y axis. The unit circle, I'm going to center it at zero. It's going to have a radius of one, a radius of one and we just have to remind ourselves what the unit circle definition of the sine function is. If we have some angle, one side of the angle if we do this in a color you can see, along the positive X axis. Then the other side, so let's see, this is our angle right over here. Let's say that's some angle theta. The sine of this angle is going to be the Y value of where this ray intersects the unit circle. This right over here, that is going to be sine of theta. With that review out of the way, let's think about what X values, and we're assuming we're dealing in radians. What X values when if I take the sine of it are going to give me 1/3? When does Y equal 1/3 along the unit circle? That's 2/3, 1/3 right over here. We see it equals 1/3 exactly two places, here and here. There's two angles where, or at least two, if we just take one or two on each pass of the unit circle. Then we can keep adding multiples of two pi to get as many as we want. We see just on the unit circle we could have this angle. Or we could go all the way around to that angle right over there. Then we could add any multiple of two pi to those angles to get other angles that would also work where if I took the sine of them I would get 1/3. Now let's think about what these are. Here we can take our calculator out, and we could take the inverse sine of 1/3. Let's do that. The inverse sine of 1 over 3. We have to remember what the range of the inverse sine function is. It's going to give us a value between negative pi over 2 and pi over 2, so a value that sticks us in either the first or the fourth quadrant if we're thinking about the unit circle right over here. We see that gave us zero point, if we round to the nearest hundredth, 34. Essentially they've given us this value. They've given us 0.34. That's this angle right over here." + }, + { + "Q": "At 2:11 in this video, why does Sal use the dot instead of the regular multiplication sign, the regular times symbol?", + "A": "In *Algebra and above, there are things called variables, which are symbols to represent numbers. x is used as one of these variables, so Algebra people use a dot as the multiplication symbol.", + "video_name": "xkg7370cpjs", + "timestamps": [ + 131 + ], + "3min_transcript": "Write 5 and 1/4 as an improper fraction. An improper fraction is just a pure fraction where the numerator is greater than the denominator. This right here, it's not a pure fraction. We have a whole number mixed with a fraction, so we call this a mixed number. So let's think about what 5 and 1/4 represents, and let me rewrite it. So if we're talking about 5 and 1/4, and you can literally think of this as 5 and 1/4 or 5 plus 1/4, that's what 5 and 1/4 represents. So let's think about 5. Five is 5 wholes, or if you're thinking of pie, we could draw literally five pies. Let me just cut up the pies from the get go into four pieces since we're dealing with fourths. So let me just cut up the pies right over here. So that's one pie right over there. Let me copy and paste this. Copy and paste. So I have two pies, and then I have three pies, and then I So this is what the 5 represents. 5 literally represents-- so let me circle all of this together. That is the 5 part right there. That is what 5 literally represents. It represents five whole pies. Now, I have cut up the pies into four pieces, so you can imagine each piece represents a fourth. Now, how many pieces do I have in these five pies? Well, I have four pieces per pie. Let me just right it here. 4 pieces per pie times 5 pies is equal to 20 pieces. this is also equal to 20 times 1/4, or you could just write this as being equal to 20/4. So we have 5 whole pies is equal to 20 fourths. Let me write it like that. 20 fourths. Or we could write it as 20/4. I've kind of done the same thing twice. So that's what the five pies represent. 20/4 or 20 pieces, where each piece is 1/4. Now, the 1/4 right here represents literally one more fourth of a pie or one more piece of a pie, so let me draw another pie here. So that is another pie. Cut it into four pieces. But this 1/4 only represents one of these pieces, right?" + }, + { + "Q": "At around 7:05, he said that 2(pi radius) = 360degree, but afterwards he said that 2(pi RADIANS) = 360degree. Why is this so? I know that radian is an angle, and radius is a length, but how can they be used as the same variable in such an equation?", + "A": "Listen again even more carefully from about 5:40 to 7:15. Sal said that 2pi radiusseses SUBTEND an angle of 360 degrees and that 360 degrees is the SAME as an angle of 2pi radians. Hope this is of use to you!", + "video_name": "EnwWxMZVBeg", + "timestamps": [ + 425 + ], + "3min_transcript": "have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle. Here, the angle in radians tells you exactly the arc length that is subtending the angles. So let's do a couple of thought experiments here. So given that, what would be the angle in radians if we were to go-- so let me draw another circle here. So that's the center, and we'll start right over there. So what would happen if I had an angle-- if I wanted to measure in radians, what angle would this be in radians? And you could almost think of it as radiuses. So what would that angle be? Going one full revolution in degrees, that would be 360 degrees. Based on this definition, what would this be in radians? The arc that subtends this angle is the entire circumference of this circle. Well, what's the circumference of a circle in terms of radiuses? So if this has length r, if the radius is length r, what's the circumference of the circle in terms of r? That's going to be 2 pi r. So going back to this angle, the length of the arc that subtends this angle is how many radiuses? Well, it's 2 pi radiuses. It's 2 pi times r. So this angle right over here, I'll call this a different-- well, let's call this angle x. x in this case is going to be 2 pi radians. If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses. So given that, let's start to think about how we can convert between radians and degrees, and vice versa. If I were to have-- and we can just follow up over here. If we do one full revolution-- that is, 2 pi radians-- how many degrees is this going to be equal to? Well, we already know this. A full revolution in degrees is 360 degrees. Well, I could either write out the word degrees, or I can use this little degree notation there. Actually, let me write out the word degrees. It might make things a little bit clearer that we're kind of using units in both cases. Now, if we wanted to simplify this a little bit, we could divide both sides by 2. In which case, on the left-hand side," + }, + { + "Q": "At 0:07 Sal mentions that Stewart has a negative amount of money in his account; how is that possible?\n\n\nIs it because he's in debt?", + "A": "It is impossible to achieve a negative amount of money in your account unless you take something that isn t yours. So to answer your question, yes, it means he is in debt", + "video_name": "fFdOr8U4mnI", + "timestamps": [ + 7 + ], + "3min_transcript": "At the beginning of the week, Stewart's checking account had a balance of negative $15.08. On Monday morning, he deposited a check for $426.90. On Tuesday morning, he deposited another check for $100. How much was in Stewart's checking account after the second deposit, so after both of these deposits right over here. So he starts off with a negative balance. So a negative balance means that he's overdrawn his checking account. He actually owes the bank money now. Luckily, he's now going to put some money in his bank account. So he'll actually have a positive balance in his checking account. So he's starts off with the negative $15.08. And then to that, he adds $426.90. And then he adds another $100. So he started off with negative $15.08. And then to that, he adds $426.90 and $100. And so how much is going to have in his bank account? He started owing $15.08, and then he's going to add $526.90. So one way to visualize it is, if you think about it on a number line, if this is 0 right over here, he's going to start off at negative $15.08. But then he's going to add $526. So this right over here, this is $15.08 to the left. That's how much he owes. And to that, he's going to add $526. So I'm not drawing this to scale. But to that, he is going to add $526.90. So the amount that he's going to be in the positive is going to be $526.90 minus the $15.08. That's how much he's going to be in the positive. And that's going to be $526.90 minus $15.08. So that's going to be, and we can even just rewrite this so it actually looks exactly like that. That's exactly the same thing as $526.90 minus-- adding a negative is the same thing as subtracting a positive-- minus $15.08. And this is-- I will do this in another color-- $526.90 minus $15.08. Let's see, 0 is less than 8. Let's make that a 10 and borrow from this 9. So that becomes an 8, or I guess you could say we're regrouping. Now, everything up here is larger than everything there. So 10 minus 8 is 2. 8 minus 0 is 8. We have our decimal. 6 minus 5 is 1." + }, + { + "Q": "so at (0:19) twice as much means multiply the age or halve it", + "A": "If the problem says something it s twice as much, it means the amount is two times the value. In this case, Jared is twice as old as Peter. This means he is two times older than peter. If peter is 5, Jared will be 10 since 10 is twice as much as 5.", + "video_name": "bS6EmYzpou4", + "timestamps": [ + 19 + ], + "3min_transcript": "We're told that Jared is twice the age of his brother, Peter. So Jared is twice the age of his brother, Peter. Peter is 4 years old. And Jared is twice the age of his brother, Peter. So Jared is twice Peter's age of 4. And then they tell us, Talia's age is 3 times Jared's age. How old are Talia and Jared? So Talia's age is 3 times Jared's age. So let's write this down. We have Peter is 4 years old. So Peter's age, he's 4 years old. Jared is twice the age of his brother, Peter. Let me do that in yellow. That's not yellow. Jared is twice the age of his brother. You could say Jared is twice the age of his brother or Jared is 2 times the age of his brother. Well, his brother is 4. They told us that right over there. So Jared is going to be 2 times 4 years old. And if you know your multiplication tables, you know that's the same thing as 8. 2 times 4 is 8. 4 times 2 is 8. Or you know this is the same thing as Peter's age twice, which is the same thing as 4 plus 4. That would also get you 8. So Jared's age is equal to 8. And then, finally, they say Talia's age-- so let me write Talia right over here. Talia's age and they tell it to us right over here. It's 3 times Jared's age. So Talia's age is 3 times Jared's age. Well, we were able to figure out Jared's age. It is 3 times Jared's age. If you know your multiplication tables, this should maybe jump out at you. And your multiplication tables really are one of those things in mathematics that you should really just know back and forth just because it will make the rest of your life very, very, very simple. Or you could view 3 times 8 as literally 8 three times. So it's 8 plus 8 plus 8. 8 plus 8 is 16, 16 plus 8 is 24. So 3 times 8 is 24. So Peter is 4, Jared is 8, and Talia is 24." + }, + { + "Q": "You lost me from 0:36", + "A": "It is like that because, let s say if x = 0 and we know that y = x+2 so: y = x+2 y=0+2 y=2 So, there the coordinates will be (0,2) Hope it made sense.", + "video_name": "RLyXTj2j_c4", + "timestamps": [ + 36 + ], + "3min_transcript": "- We're asked to use the reflect tool to find the image of quadrilateral PQRS, that's this quadrilateral right over here, for a reflection over the line y is equal to x plus two. All right, so let's use the reflect tool. So let me scroll down. So let me click on Reflect, it brings up this tool, and I want to reflect across the line y is equal to x plus two. So let me move this so it is the line y equals x plus two and to think about it, let's see, it's going to have a slope of one, the coefficient on the x term is one, so it's going to have a slope of one, so let me see if I can give this a slope of one. Is this a slope of one? Let me put it a little bit -- yep, it looks like a slope of one as the line moves one to the right. We go from one point of the line to the other, you have to go one to the right, and one up or two to the right, and two up, however much you move to the right in the x direction, you have to move that same amount in the vertical direction. So now it has a slope of one, and the y intercept is going to be the point x equals zero, y equals two, When x is equal to zero, y is going to be equal to two, so let me move this. So we see that we now go through that point. When x is equal to zero, y is equal to two. And now, we just need to reflect PQRS, this quadrilateral, over this line, so let's do that. There you go, we did it. The things, the point, like point S right over here that was to the top and left of the line, its reflection, the corresponding point in the image is now to the right and the bottom of the line. The things that were to the right and the bottom of our line, like point P, it's the corresponding point in the reflection is now on the other side of the line. So there you go, I think we're done, and we got it right." + }, + { + "Q": "At 5:02 He says that a 2 - tuple is not a member of 3D real co-ordinate space. Why is that? Can't we just say that a 2 - dimensional vector has zero value for the z-axis? Can't we imagine 3-d space consisting of a lot of planes stacked up?", + "A": "the keyword is 2 tuple . you can t have a 2 tuple that has x y and z components, that s all. but you can represent a 2 dimensional vector as a 3 tuple, it s just that one component will always be zero.", + "video_name": "lCsjJbZHhHU", + "timestamps": [ + 302 + ], + "3min_transcript": "I should make the scale a little bit bigger, so it looks the same-- 1, 2, 3, 4. So it might look something like that. So that would be the vector, negative 3-- let me write a little bit better than that-- negative 3, and negative 4. So if you were to take all of the possible 2-tuples, including the vector 0, 0-- so it has no magnitude, and you could debate what its direction is right over there-- you take all of those combined, and then you have created your two-dimensional real coordinate space. And that is referred to as R2. Now, as you can imagine, the fact that we wrote this 2 here-- we had to specify-- it's like, hey, well, could I put a 3 there? And I would say, absolutely, you could put a 3 there. So R3 would be the three-dimensional real coordinate space. So 3D real coordinate space. the possible real-valued 3-tuples. So, for example, that would be a member of R3. And let me actually label these vectors just so we get in the habit of it. So let's say we call this vector x. Let's say we have a vector b, that looks like this. Negative 1, 5, 3. Both of these would be members of R3. And if you want to see some fancy notation, a member of a set-- so this is a member of R3-- it is a real-valued 3-tuple. Now you say, well, what would not be a member of R3? Well, this right over here isn't a 3-tuple. This right over here is a member of R2. Now, you might be able to extend it in some way, add a zero or something, but formally, Another thing that would not be a member of R3-- let's say someone wanted to make some type of vector that had some imaginary parts in it. So let's say it had i, 0, 1. This is no longer real-valued. We have put an imaginary-- this number up here has an imaginary part. So this is no longer a real-valued 3-tuple. And what's neat about linear algebra is, we don't have to stop there. R3 we can visualize, we can plot these things. In your previous mathematical career, especially if you have some type of a hologram or something, it's not hard to visualize things in three dimensions. But what's neat is that we can keep extending this. We can go into 4, 5, 6, 7, 20, 100 dimensions. And obviously there it becomes much harder, if not impossible, to visualize it. But then we can at least represent it mathematically with an n-tuple of vectors. And so if we were to talk about a real coordinate space generally, you'll often see the notation Rn," + }, + { + "Q": "At 1:55, he says 'neither of these have any imaginary parts'. Does this mean it cannot have x or y or n or any unknown number in it?", + "A": "No, it means that only Real numbers are components of that vector, rather than the option of complex numbers as well, such as the square root of negative 1.", + "video_name": "lCsjJbZHhHU", + "timestamps": [ + 115 + ], + "3min_transcript": "When you get into higher mathematics, you might see a professor write something like this on a board, where it's just this R with this extra backbone right over here. And maybe they write R2. Or if you're looking at it in a book, it might just be a bolded capital R with a 2 superscript like this. And if you see this, they're referring to the two-dimensional real coordinate space, which sounds very fancy. But one way to think about it, it's really just the two-dimensional space that you're used to dealing with in your coordinate plane. To go a little bit more abstract, this isn't necessarily this visual representation. This visual representation is one way to think about this real coordinate space. If we were to think about it a little bit more abstractly, the real R2, the two-dimensional real coordinate space-- let me write this down-- and the two-dimensional real coordinate space. the 2 tells us how many dimensions we're dealing with, and then the R tells us this is a real coordinate space. The two-dimensional real coordinate space is all the possible real-valued 2-tuples. Let me write that down. This is all possible real-valued 2-tuples. So what is a 2-tuple? Well, a tuple is an ordered list of numbers. And since we're talking about real values, it's going to be ordered list of real numbers, and a 2-tuple just says it's an ordered list of 2 numbers. So this is an ordered list of 2 real-valued numbers. Well, that's exactly what we did here when we thought about a two-dimensional vector. This right over here is a 2-tuple, and this is a real-valued 2-tuple. Neither of these have any imaginary parts. So you have a 3 and a 4. Order matters. And even if we were trying to represent them in our axes right over here, this vector, 4, 3, would be 4 along the horizontal axis, and then 3 along the vertical axis. And so it would look something like this. And remember, we don't have to draw it just over there. We just care about its magnitude and direction. We could draw it right over here. This would also be 4, 3, the column vector, 4 3. So when we're talking about R2, we're talking about all of the possible real-valued 2-tuples. So all the possible vectors that you can have, where each of its components-- and the components are these numbers right over here-- where each of its components are a real number. So you might have 3, 4. You could have negative 3, negative 4. So that would be 1, 2, 3, 1, 2, 3, 4," + }, + { + "Q": "8:35 does that also mean that g(x)-h(x) and h(x)-g(x) are also solutions?", + "A": "Yes, it does. 0 - 0 does equal 0. In fact any linear combination of g and h will be solutions. You can do a*g(x) + b*h(x), where a and b are any constants, and that will be a solution.", + "video_name": "UFWAu8Ptth0", + "timestamps": [ + 515 + ], + "3min_transcript": "That's just g prime prime, plus h prime prime, plus B times-- the first derivative of this thing-- g prime plus h prime, plus C times-- this function-- g plus h. And now what can we do? Let's distribute all of these constants. We get A times g prime prime, plus A times h prime prime, plus B times the first derivative of g, plus B times the first derivative of h, plus C times g, plus C times h. And now we can rearrange them. And we get A-- let's take this one; let's take all the g terms-- A times the second derivative of g, plus B times the first derivative, plus C times g-- that's these three terms-- plus A times the second derivative of h, plus B And now we know that both g and h are solutions of the original differential equation. So by definition, if g is a solution of the original differential equation, and this was the left-hand side of that differential equation, this is going to be equal to 0, and so is this going to be equal to 0. So we've shown that this whole expression is equal to 0. So if g is a solution of the differential equation-- of this second order linear homogeneous differential equation-- and h is also a solution, then if you were to add them together, the sum of them is also a solution. So in general, if we show that g is a solution and h is a solution, you can add them. And we showed before that any constant times them is also a solution. So you could also say that some constant times g of x And maybe the constant in one of the cases is 0 or something. I don't know. But anyway, these are useful properties to maybe internalize for second order homogeneous linear differential equations. And in the next video, we're actually going to apply these properties to figure out the solutions for these. And you'll see that they're actually straightforward. I would say a lot easier than what we did in the previous first order homogeneous difference equations, or the exact equations. This is much, much easier. I'll see you in the next video." + }, + { + "Q": "at 5:58, he says that if g(x) is a solution then c1*g(x) is also one, does this mean that 0 is a solution to all of these?", + "A": "Homogeneous systems have a few properties, and one of those properties is that they always have at least one solution, the trivial solution, 0. The goal is to find the non-trivial solutions in most cases.", + "video_name": "UFWAu8Ptth0", + "timestamps": [ + 358 + ], + "3min_transcript": "prime, plus C times g is equal to 0. Right? These mean the same thing. Now, my question to you is, what if I have some constant times g? Is that still a solution? So my question is, let's say some constant c1 gx-- c1 times g-- is this a solution? Well, let's try it out. Let's substitute this into our original equation. So A times the second derivative of this would just be-- and I'll switch colors here; let me switch to brown-- so A times the second derivative of this would be-- the constant, every time you take a derivative, the constant just carries over-- so that'll just be A times c1 g prime prime, plus-- the same thing for the first different than the c1 c-- times g. And let's see whether this is equal to 0. So we could factor out that c1 constant, and we get c1 times Ag prime prime, plus Bg prime, plus Cg. And lo and behold, we already know. Because we know that g of x is a solution, we know that this is true. So this is going to be equal to 0. Because g is a solution. So if this is 0, c1 times 0 is going to be equal to 0. So this expression up here is also equal to 0. Or another way to view it is that if g is a solution to this second order linear homogeneous differential So this is also a solution to the differential equation. And then the next property I want to show you-- and this is all going someplace, don't worry. The next question I want to ask you is, OK, we know that g of x is a solution to the differential equation. What if I were to also tell you that h of x is also a solution? So my question to you is, is g of x plus h of x a solution? If you add these two functions that are both solutions, if you add them together, is that still a solution of our original differential equation? Well, let's substitute this whole thing into our original differential equation, right? So we'll have A times the second derivative of this thing." + }, + { + "Q": "At 0:15, how does the second prize relate to the first prize? Doesn't the first prize have a predetermined ticket, thus making it an independent event?", + "A": "The first prize is a independent event, yes. But the second prize is dependent on the first prize because the ticket drawn for the first prize is not but back in.", + "video_name": "Za7G_eWKiF4", + "timestamps": [ + 15 + ], + "3min_transcript": "The marching band is holding a raffle at a football game with two prizes. After the first ticket is pulled out and the winner determined, the ticket is taped to the prize. The next ticket is pulled out to determine the winner of the second prize. Are the two events independent? Explain. Now before we even think about this exact case, let's think about what it means for events to be independent. It means that the outcome of one event doesn't affect the outcome of the other event. Now in this situation, the first event-- after the first ticket is pulled out and the winner determined-- the ticket is taped to the prize. Then the next ticket is pulled out to determine the winner of the second prize. Now, the winner of the second prize-- the possible winners, the possible outcomes for the second prize, is dependent on who was pulled out for the first prize. You can imagine if there's three tickets, let's say there's tickets A, B, and C in the bag. That's for the first prize. Now, when we think about who could be pulled out for the second prize, it's only going to be tickets B or C. Now the first prize could have gone the other way. It could have been A, B, and C. The first prize could have gone to ticket B. And then the possible outcomes for the second prize would be A or C. So the possible outcomes for the second event, for the second prize, are completely dependent on what happened or what ticket was pulled out for the first prize. So these are not independent events. The second event-- the outcomes for it, are dependent on what happened in the first event. So they are not independent. after the first ticket was pulled out, if they just wrote down the name or something, and then put that ticket back in. Instead they taped it to the prize. But if they put that ticket back in, then the second prize, it would have still had all the tickets there. It wouldn't have mattered who was picked out in the first time because their name was just written down, but their ticket was put back in. And then you would have been independent. So if you had replaced the ticket, you would have been independent. But since they didn't replace the ticket, they taped it to the prize, these are not independent events." + }, + { + "Q": "I know this is simple, but I just wanted to reassure myself. At around 8:27 when Sal multiplies the triangles, the formula is b times hieght divided by 2. Now does that only count for that one half of the triangle? Is that why he had to multiply it once more? Also, hypothetically, if I were to encounter a triangle such as this again, would I do the method shown here to find its area?", + "A": "(base x height)/2 is the formula for the whole triangle, and every other triangle you ll ever see.", + "video_name": "vaOXkt7uuac", + "timestamps": [ + 507 + ], + "3min_transcript": "a squared is equal to 16. a is equal to 4. a is equal to 4. And now we're ready to figure out the area. What's the area of the rectangle? 6 times 6, it's 24. What's the area of each of these triangles? 3 times 4 times 1/2. 3 times 4 is 12 times 1/2 is 6. So the area of that triangle is 6. The area of this triangle is 6. So 6 plus 24 plus 6 is 36. B. Problem 39. What is the area in square inches of the triangle below. Interesting. OK, so this is an equilateral triangle, all the sides are equal. And so we could actually say that since these two triangles are symmetric. That's equal to that. And this comes to a general formula for the area of an equilateral triangle. But let's just figure it all out. So this side is going to be 5. If this is 5 and that's 10, what is this side right here? Let's call it x. Pythagorean theorem. This is the hypotenuse. So x squared plus 5 squared plus 25 is going to be equal to the hypotenuse squared, it's equal to 100. x squared is equal to 100 minus 25, 75. x is equal to the square root of 75. 75 is 25 times 3. So that's equal to the square root of 25 times 3. Which is equal to the square root of 25 times the square root of 3. Which is equal to 5 roots of 3. And now, what's the area of just this right triangle right here? This one on the right side. Well its base is 5, its height is 5 roots of 3. So it's going to be 1/2 times the base, 5, times the height, And that's what? 1/2 times 5 times 5. So it's 25 root 3 over 2 and that's just this triangle right there. Well this triangle's going to have the the exact same area. They are congruent triangles. So the area of the figure is this times 2. So 2 times that is equal to just the 25 root 3. And that's choice B. Next problem, problem 40. The perimeter of two squares are in a ratio of 4:9. What is the ratio between the areas of the two squares? Let me draw two squares. That's one square. Let me draw another square. That's another square. Let's say that the sides of this are x and the sides of this one are y." + }, + { + "Q": "2:02 i was taught like 29+59+?=180\n? = 180 - (29+59)\n? = 180 - 88\n? = 92 degrees.\n\ncan i do it in this way also?", + "A": "Yes, although this method might not work with more complicated multi-step equations. This method does work however for interior triangle angles.", + "video_name": "eTwnt4G5xE4", + "timestamps": [ + 122 + ], + "3min_transcript": "We're given a bunch of lines here that intersect in all different ways and form triangles. And what I want to do in this video, we've been given the measures of some of the angles, this angle, that angle, and that angle. And what we want to do in this video is figure out what the measure of this angle is. And we're going to call that measure x. And so I encourage you to pause the video right now and try it yourself. And then I'm going to give you the solution. So I'm assuming you've unpaused it. And you've solved it or you've given it at least a good shot of it. And what's fun about these is there's multiple ways to solve these. And you kind of just have to keep figuring out what you can figure out. So let's say you start on the left-hand side right over here. If this is 121 degrees, then you'd say, well look, this angle right over here is supplementary to this angle right over there. So this is 121 degrees plus this green angle, that has to be equal to 180 degrees. So this is going to be 180 minus 121. 80 minus 20 would be 60. So that's going to be 59 degrees. So let me write that down. That's going to be 59 degrees. Now we see that we have two angles of a triangle. If you have two angles of a triangle, you can figure out the third angle, because they need to add up to 180. Or you could say that this angle right over here-- so we'll call that question mark-- we know that 59 plus 29 plus question mark needs to be equal to 180 degrees. And if we subtract the 15 out of the 29 from both sides, we get question mark is equal to 180 minus 59 minus 29 degrees. So that is going to be 180 minus 59 minus 29, let's see, 180 minus 59, we already know, is 121. And then 121 minus 29. So if you subtract just 20, you get 101. You subtract another 9, you get 92. This is equal to 92 degrees. Well, this right here is equal to 92 degrees. This angle right here is vertical with that angle. So it is also going to be equal to 92 degrees. And now we're getting pretty close. We can zoom in on this triangle down here. And let me save some space here. So let me just say that that over there is also going to be 92 degrees. And at this triangle down here, we have two of the sides of the triangle. We just have to figure out the third. And actually, we don't even have to do much math here, because we have two of the angles of this triangle. We have to figure out the third angle. So over here, we have one angle that's 92, one angle that's 29. The other one will be 180 minus 92 minus 29. And we don't even have to do any math here, because essentially, this is the exact same angles that we have in this triangle right over here. We have a 92 degree angle, we have a 29 degree angle, and the other one is 59 degrees. So in this case, it has to be also" + }, + { + "Q": "In 1:37 Sal wrote 59+29+x=180, and he wrote 180-59-29=x, why he is subtracting if the original operation is addition? Can you add those two numbers, and subtract the total by 180 and you will get the missing angle?", + "A": "Both ways end up giving you the same answer for x. You can just use the way that is the easiest for you to use.", + "video_name": "eTwnt4G5xE4", + "timestamps": [ + 97 + ], + "3min_transcript": "We're given a bunch of lines here that intersect in all different ways and form triangles. And what I want to do in this video, we've been given the measures of some of the angles, this angle, that angle, and that angle. And what we want to do in this video is figure out what the measure of this angle is. And we're going to call that measure x. And so I encourage you to pause the video right now and try it yourself. And then I'm going to give you the solution. So I'm assuming you've unpaused it. And you've solved it or you've given it at least a good shot of it. And what's fun about these is there's multiple ways to solve these. And you kind of just have to keep figuring out what you can figure out. So let's say you start on the left-hand side right over here. If this is 121 degrees, then you'd say, well look, this angle right over here is supplementary to this angle right over there. So this is 121 degrees plus this green angle, that has to be equal to 180 degrees. So this is going to be 180 minus 121. 80 minus 20 would be 60. So that's going to be 59 degrees. So let me write that down. That's going to be 59 degrees. Now we see that we have two angles of a triangle. If you have two angles of a triangle, you can figure out the third angle, because they need to add up to 180. Or you could say that this angle right over here-- so we'll call that question mark-- we know that 59 plus 29 plus question mark needs to be equal to 180 degrees. And if we subtract the 15 out of the 29 from both sides, we get question mark is equal to 180 minus 59 minus 29 degrees. So that is going to be 180 minus 59 minus 29, let's see, 180 minus 59, we already know, is 121. And then 121 minus 29. So if you subtract just 20, you get 101. You subtract another 9, you get 92. This is equal to 92 degrees. Well, this right here is equal to 92 degrees. This angle right here is vertical with that angle. So it is also going to be equal to 92 degrees. And now we're getting pretty close. We can zoom in on this triangle down here. And let me save some space here. So let me just say that that over there is also going to be 92 degrees. And at this triangle down here, we have two of the sides of the triangle. We just have to figure out the third. And actually, we don't even have to do much math here, because we have two of the angles of this triangle. We have to figure out the third angle. So over here, we have one angle that's 92, one angle that's 29. The other one will be 180 minus 92 minus 29. And we don't even have to do any math here, because essentially, this is the exact same angles that we have in this triangle right over here. We have a 92 degree angle, we have a 29 degree angle, and the other one is 59 degrees. So in this case, it has to be also" + }, + { + "Q": "Is it possible to have a matrix the entries of which are other matrices? Like, in 3:15, when Sal says a matrix can be used to represent the intensity of pixels, maybe you could have a matrix with a cell for each pixel, but the entry is an actual 3x1 matrix that represents the RGB components of this pixel. Does this actually exist?", + "A": "I m also asking this because when I attended some lectures about the Standard Model of Particle Physics, the lecturer kept talking about polymatrices . I cannot say I fully understood the lectures, not even understood, I attended them just for a challenge. Also, my Linear Algebra concepts where almost none (though they are improving now). But despite all this, I deduced he was talking about matrices inside matrices, although might be I deduced wrong. When I try and look up polymatrix , I find no results.", + "video_name": "0oGJTQCy4cQ", + "timestamps": [ + 195 + ], + "3min_transcript": "This is a matrix where 1, 0, negative 7, pi-- each of those are an entry in the matrix. This matrix right over here has two rows. And it has three columns. And because it has two rows and three columns, people will often say that this is a 2 by 3 matrix. Whenever they say it's something by something matrix, they're telling you that it has two rows-- so you see the two rows right over there. And they are telling you that it has three columns. You see the three columns right over there. I could give you other examples of a matrix. So I could have a 1 by 1 matrix. So I could have the matrix 1. This right over here is a 1 by 1 matrix. It has one row, one column. I could have a matrix like this-- 3, 7, and 17. What is this? Well, this has one row. And it has three columns. This is a 1 by 3 matrix. I could have a matrix-- and I think you see where all of this Figuring out the dimensions of a matrix are not too difficult. I could have a matrix that looks like this, where it's 3, 5, 0, 0, negative 1, negative 7. This right over here has three rows. So it's three rows, and it has two columns. So we would call this a 3 by 2. Let me do that in that same color. We would call it a 3 by 2 matrix, three rows and two columns. You know that a matrix is just a rectangular array of numbers. You can say what its dimensions are. You know that each of these numbers that take one of these positions-- we just call those entries. But what are matrices good for? I still might not be clear what the connection is between this and this right over here. this is just a compact representation of a bunch of numbers. It's a way of representing information. They become very valuable in computer graphics because these numbers could represent the color intensity at a certain point. They could represent whether an object is there at a certain point. And as we develop an algebra around matrices, and when we talk about developing an algebra around matrices, we're going to talk about operations that we're going to perform on matrices that we would normally perform with numbers. So we're going to essentially define how to multiply matrices, how to add matrices. We'll learn about taking an inverse of a matrix. And by coming up with an algebra of how we manipulate these things, it'll become very useful in the future when you're trying to write a computer graphics program or you're trying to do an economic simulation or a probability simulation, to say, oh, I have this matrix that represents where different particles are in space. Or I have this matrix that represents the state of some type of a game." + }, + { + "Q": "At 1:08 - 1:13, Sal mentions radians. What are they?", + "A": "Radians are basically units used for measuring angles. In a circle, if finding the radians of a circle, it would be that particular sectors arc length.", + "video_name": "D-EIh7NJvtQ", + "timestamps": [ + 68, + 73 + ], + "3min_transcript": "We already know that an angle is formed when two rays share a common endpoint. So, for example, let's say that this is one ray right over here, and then this is one another ray right over here, and then they would form an angle. And at this point right over here, their common endpoint is called the vertex of that angle. Now, we also know that not all angles seem the same. For example, this is one angle here, and then we could have another angle that looks something like this. And viewed this way, it looks like this one is much more open. So I'll say more open. And this one right over here seems less open. So to avoid having to just say, oh, more open and less open and actually becoming a little bit more exact about it, we'd actually want to measure how open an angle is, or we'd want to have a measure of the angle. Now, the most typical way that angles are measured, The most typical unit is in degrees, but later on in high school, you'll also see the unit of radians being used, especially when you learn trigonometry. But the degrees convention really comes from a circle. So let's draw ourselves a circle right over here, so that's a circle. And the convention is that-- when I say convention, it's just kind of what everyone has been doing. The convention is that you have 360 degrees in a circle. So let me explain that. So if that's the center of the circle, and if we make this ray our starting point or one side of our angle, if you go all the way around the circle, that represents 360 degrees. And the notation is 360, and then this little superscript circle represents degrees. This could be read as 360 degrees. And no one knows for sure, but there's hints in history, and there's hints in just the way that the universe works, or at least the Earth's rotation around the sun. You might recognize or you might already realize that there are 365 days in a non-leap year, 366 in a leap year. And so you can imagine ancient astronomers might have said, well, you know, that's pretty close to 360. And in fact, several ancient calendars, including the Persians and the Mayans, had 360 days in their year. And 360 is also a much neater number than 365. It has many, many more factors. It's another way of saying it's divisible by a bunch of things. But anyway, this has just been the convention, once again, what history has handed us, that a circle is viewed to have 360 degrees." + }, + { + "Q": "At 0:47, when they are talking about more open and less open are they ever going to give names to those angles", + "A": "less open means acute and more open is obtuse, and if the angle forms a right angle it is 90 degrees", + "video_name": "D-EIh7NJvtQ", + "timestamps": [ + 47 + ], + "3min_transcript": "We already know that an angle is formed when two rays share a common endpoint. So, for example, let's say that this is one ray right over here, and then this is one another ray right over here, and then they would form an angle. And at this point right over here, their common endpoint is called the vertex of that angle. Now, we also know that not all angles seem the same. For example, this is one angle here, and then we could have another angle that looks something like this. And viewed this way, it looks like this one is much more open. So I'll say more open. And this one right over here seems less open. So to avoid having to just say, oh, more open and less open and actually becoming a little bit more exact about it, we'd actually want to measure how open an angle is, or we'd want to have a measure of the angle. Now, the most typical way that angles are measured, The most typical unit is in degrees, but later on in high school, you'll also see the unit of radians being used, especially when you learn trigonometry. But the degrees convention really comes from a circle. So let's draw ourselves a circle right over here, so that's a circle. And the convention is that-- when I say convention, it's just kind of what everyone has been doing. The convention is that you have 360 degrees in a circle. So let me explain that. So if that's the center of the circle, and if we make this ray our starting point or one side of our angle, if you go all the way around the circle, that represents 360 degrees. And the notation is 360, and then this little superscript circle represents degrees. This could be read as 360 degrees. And no one knows for sure, but there's hints in history, and there's hints in just the way that the universe works, or at least the Earth's rotation around the sun. You might recognize or you might already realize that there are 365 days in a non-leap year, 366 in a leap year. And so you can imagine ancient astronomers might have said, well, you know, that's pretty close to 360. And in fact, several ancient calendars, including the Persians and the Mayans, had 360 days in their year. And 360 is also a much neater number than 365. It has many, many more factors. It's another way of saying it's divisible by a bunch of things. But anyway, this has just been the convention, once again, what history has handed us, that a circle is viewed to have 360 degrees." + }, + { + "Q": "AT 3:05 and 3:56 what is the difference between the number lines?", + "A": "When the inequality is equal or less than/greater than we use a closed circle (solid dot) as in the first number line. When it is less than/greater than we use an open circle as in the second line.", + "video_name": "ilWDSYnTEFs", + "timestamps": [ + 185, + 236 + ], + "3min_transcript": "Those are all less than 1,500. But what about 1,500 calories? Is it true that 1,500 is less than 1,500? 1,500 is equal to 1,500. So this is not a true statement. But what if I want to eat up to and including 1,500 calories? I want to make sure that I get every calorie in there. How can I express that? How can express that I can eat less than or equal to 1,500 calories, so I can eat up to and including 1,500 calories? Right now, this is only up to but not including 1,500. How could I express that? Well, the way I would do that is to throw this little line under the less than sign. Now, this is not just less than. This is less than or equal to. So this symbol right over here, this So now 1,500 would be a completely legitimate C, a completely legitimate number of calories to have in a day. And if we wanted to visualize this on a number line, the way we would think about it, let's say that this right over here is our number line. I'm not going to count all the way from 0 to 1,500, but let's imagine that this right over here is 0. Let's say this over here is 1,500. How would we display less than or equal to 1,500 a number line? Well, we would say, look, we could be 1,500, so we'll put a little solid circle right over there. And then we can be less than it, so then we would color in everything less than 1,500, is legitimate. And you might say, hey, but what about the situation where it wasn't less than or equal? What about the situation if it was just less than? So let me draw that, too. So going back to where we started, if I were to say C is less than 1,500, the way we would depict that on a number line is-- let's say this is 0, this is 1,500, we want to make it very clear that we're not including the value 1,500. So we would put an open circle around it. Notice, if we're including 1,500, we fill in the circle. If we're not including 1,500, so we're only less than, we were very explicit that we don't color in the circle. But then we show that, look, we can do everything below that. Now, you're probably saying, OK, Sal, you did less than, you did less than or equal, what if you wanted to do it the other way around?" + }, + { + "Q": "at the 9:23 shouldn't it be -1 = f\"(y)", + "A": "He corrects himself at 9:41", + "video_name": "Pb04ntcDJcQ", + "timestamps": [ + 563 + ], + "3min_transcript": "respect to y, is just sine of x. Plus the derivative of e to the y is e to the y. x squared So it's just x squared e to the y, plus-- what's the partial of f of y, with respect to y? It's going to be f prime of y. Well, what did we do? We took M, we integrated with respect to x, and we said, well, we might have lost some function of y, so we added that to it. And then we took the partial of that side that we've almost constructed, and we took the partial of that, with respect to y. Now, we know, since this is exact, that that is going to equal our N. So our N is up there. Cosine of x plus-- So that's going to be equal to-- I want to make sure I can read it up there-- to our N, right? Oh no, sorry. N is up here. Our N is up here. Sine of x-- let me write that-- sine of x plus x So sine of x plus x squared, e to the y, minus 1. That was just our N, from our original differential equation. And now we can solve for f prime of y. So let's see, we get sine of x plus x squared, e to the y, plus f prime of y, is equal to sine of x plus x squared, e to the y, minus 1. So let's see, we can delete sine of x from both sides. We can delete x squared e to the y from both sides. And then what are we left with? We're left with f prime of y is equal to 1. And then we're left with f of y is equal to-- well, it So what is our psi now? We wrote our psi up here, and we had this f of y here, so we So psi is a function of x and y-- we're actually pretty much almost done solving it-- psi is a function of x and y is equal to y sine of x, plus x squared, e to the y, plus y-- oh, sorry, this is f prime of y, minus 1. So this is a minus 1. So this is a minus y plus c. So this is going to be a minus y plus c. So we've solved for psi. And so what does that tell us? Well, we said that original differential equation, up here, using the partial derivative chain rule, that original differential equation, can be rewritten now" + }, + { + "Q": "1:32 sal said 20 when it is actually -20, isn't it?", + "A": "It was just a mistake, and they put an infobox at the bottom-right when the mistake comes along to fix what he says.", + "video_name": "H0q9Fqb8YT4", + "timestamps": [ + 92 + ], + "3min_transcript": "Let's do some examples dividing fractions. Let's say that I have negative 5/6 divided by positive 3/4. Well, we've already talked about when you divide by something, it's the exact same thing as multiplying by its reciprocal. So this is going to be the exact same thing as negative 5/6 times the reciprocal of 3/4, which is 4/3. I'm just swapping the numerator and the denominator. So this is going to be 4/3. And we've already seen lots of examples multiplying fractions. This is going to be the numerators times each other. So we're going to multiply negative 5 times 4. I'll give the negative sign to the 5 there, so negative 5 times 4. Let me do 4 in that yellow color. And then the denominator is 6 times 3. You might already know that 5 times 4 is 20, and you just have to remember that we're multiplying a negative times a positive. We're essentially going to have negative 5 four times. So negative 5 plus negative 5 plus negative 5 plus negative 5 is negative 20. So the numerator here is negative 20. And the denominator here is 18. So we get 20/18, but we can simplify this. Both the numerator and the denominator, they're both divisible by 2. So let's divide them both by 2. Let me give myself a little more space. So if we divide both the numerator and the denominator by 2, just to simplify this-- and I picked 2 because that's the largest number that goes into both of these. It's the greatest common divisor of 20 and 18. 20 divided by 2 is 10, and 18 divided by 2 is 9. So negative 5/6 divided by 3/4 is-- oh, I have to be very careful here. It's negative 10/9, just how we always learned. if the signs are different, then you're going to get a negative value. Let's do another example. Let's say that I have negative 4 divided by negative 1/2. So using the exact logic that we just said, we say, hey look, dividing by something is equivalent to multiplying by its reciprocal. So this is going to be equal to negative 4. And instead of writing it as negative 4, let me just write it as a fraction so that we are clear what its numerator is and what its denominator is. So negative 4 is the exact same thing as negative 4/1. And we're going to multiply that times the reciprocal of negative 1/2. The reciprocal of negative 1/2 is negative 2/1. You could view it as negative 2/1, or you could view it as positive 2 over negative 1, or you could view it as negative 2." + }, + { + "Q": "According to 3:58, is it safe to say that a ray (or 2 rays closed together) has 0 degrees angle and a line has 180 degrees angle?", + "A": "You can do that if you show where the vertex is.", + "video_name": "92aLiyeQj0w", + "timestamps": [ + 238 + ], + "3min_transcript": "we could say the measure of angle XYZ-- sometimes they'll just say angle XYZ is equal to, but this is a little bit more formal-- the measure of angle XYZ is equal to 77. Each of these little sections, we call them \"degrees.\" So it's equal to 77-- sometimes it's written like that, the same way you would write \"degrees\" for the temperature outside. So you could write \"77 degrees\" like that or you could actually write out the word right over there. So each of these sections are degrees, so we're measuring in degrees. And I want to be clear, degrees aren't the only way to measure angles. Really, anything that measures the openness. So when you go into trigonometry, you'll learn that you can measure angles, not only in degrees, but also using something called \"radians.\" But I'll leave that to another day. So let's measure this other angle, angle BAC. So once again, I'll put A at the center, and then AC I'll put along the 0 degree edge of this half-circle And then I'll point AB in the-- well, assuming that I'm drawing it exactly the way that it's Normally, instead of moving the angle, you could actually move the protractor to the angle. So it looks something like that, and you could see that it's pointing to right about the 30 degree mark. So we could say that the measure of angle BAC is equal to 30 degrees. And so you can look just straight up from evaluating these numbers that 77 degrees is clearly larger than 30 degrees, and so it is a larger angle, which makes sense because it is a more open angle. And in general, there's a couple of interesting angles to think about. If you have a 0 degree angle, you actually have something that's just a closed angled. It really is just a ray at that point. As you get larger and larger or as you get more and more open, is completely straight up and down while the other one is left to right. So you could imagine an angle that looks like this where one ray goes straight up down like that and the other ray goes straight right and left. Or you could imagine something like an angle that looks like this where, at least, the way you're looking at it, one doesn't look straight up down or one does it look straight right left. But if you rotate it, it would look just like this thing right over here where one is going straight up and down and one is going straight right and left. And you can see from our measure right over here that that gives us a 90 degree angle. It's a very interesting angle. It shows up many, many times in geometry and trigonometry, and there's a special word for a 90 degree angle. It is called a \"right angle.\" So this right over here, assuming if you rotate it around, would look just like this. We would call this a \"right angle.\"" + }, + { + "Q": "At 3:50 Sal mentioned that we knew b was negative so he changed the inequality. What do we do if we don't know for sure if b is negative or if it could be both?", + "A": "Can I use a condition |A| =/> |B| ? *absolute value of A has to be same or higher than absolute value of B", + "video_name": "0_VaUYoNV7Y", + "timestamps": [ + 230 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:01 in the video, Sal said you need to find the common denominator, which he made 18. It makes sense, but would it also work to make the denominator 6 instead of reducing at it at the end? So instead of having 3/18 and 12/18, you had 1/6 and 4/6?", + "A": "Yes... if you reduce 3/18 before you even start, you get 1/6 Then, you can use a common denominator of 6. Great work in seeing that option!", + "video_name": "8Eb5MWwcMMY", + "timestamps": [ + 61 + ], + "3min_transcript": "Let's add 19 and 3/18 to 18 and 2/3. So I like to separate out the whole number parts from the fraction parts. So 19 and 3/18 is the same thing as 19 plus 3/18. And to that, we are going to add 18 and 2/3, which is the same thing as 18 plus 2/3. Now we can separately add the whole number parts. So we could add the 19 to the 18. So we could do 19 plus 18. And then we can add the fraction parts-- let me do this in green-- plus 3/18 plus 2/3. Now 19/18, pretty straightforward. That is what? Let's see. 19 plus 19 would be 38. So this is going to be 1 less than that. It's going to be 37. So that gives me 37. And then 3/18 plus 2/3, to add them, I need to have the same denominator. So let's convert 2/3 to something over 18. So 2/3, if I want to write it as something over 18, well, I multiplied the denominator by 6, so I'd also have to multiply the numerator by 6. So it's the same thing as 12/18. So I can rewrite 2/3 as 12/18. And now I can add these two things together. That's going to be-- so I have 37 plus-- it's going to be something over 18-- plus something over 18. 3 plus 12 is 15, plus 15/18. And so expressing this as a mixed number, I get 37 and 15/18. And that's the right number. But we can simplify it even more. We can simplify the 15/18. Both the numerator and the denominator are divisible by 3. So let's divide them both by 3. And we're not changing the value because we're And so this gives us, we still have our 37, but the numerator is now 5, and the denominator is now 6. So we get 37 and 5/6. And we're done." + }, + { + "Q": "at 3:10 he breaks 10 down into it's prime numbers. if there is a number all alone do you always have to break it down?", + "A": "I quess you mean break down 10 in its primenumbers like 10=2times5. this is not necessary. remember that the goal was to make the fraction as simple as possible. In that case you make the denomenator as simple as possible. not the numerator. for instance: 2/4 you make it 1/2 3/9 you make it 1/3 6/3 you make it 3/1 so remember 10 is the same as 10/1 which you would simplify to 10. and besides 10 is easier to write than 2times5.", + "video_name": "gcnk8TnzsLc", + "timestamps": [ + 190 + ], + "3min_transcript": "" + }, + { + "Q": "where did the 1/3 come from @4:27 ? is it possible to just reduce the fraction\n27/1 * 2/3--> 9/1 * 2/1 = 18 and 8/1 * 2/3--> becomes 16/3=8 so 18/8 is 9/4 ?? would that work for all other problems?", + "A": "An exponent of 2/3 is not the same as multiplying by 2/3. If you have 9^2, you don t do 9*2. You must do 9*9. An exponent of 2/3 tells you that you need to find the cube root of the base, then square the result. 27^(2/3) = cuberoot(27)^2 = 3^2 = 9 (not the same as your value of 18) 8^(2/3) = cuberoot(8)^2 = 2^2 = 4 (not 16/3, which is your value) Hope this helps.", + "video_name": "S34NM0Po0eA", + "timestamps": [ + 267 + ], + "3min_transcript": "The way I think of it, let me find the cube root of 64, which is 4. And then let me square it. And that is going to get me to 16. Now I'll give you in even hairier problem. And I encourage you to try this one on your own before I work through it. So we're going to work with 8/27. And we're going to raise this thing to the-- and I'll try to color code it-- negative 2 over 3 power, to the negative 2/3 power. I encourage you to pause and try this on your own. Well the first thing I do whenever I see a negative exponent is to say, well, how can I get rid of that negative exponent? And I just remind myself, well, the negative exponent really just says, take the reciprocal of this to the positive exponent. I'm going to use that light mauve color. So this is going to be equal to 27/8. I just took the reciprocal of this right over here. It's equal to 27/8 to the positive 2/3 power. So notice, all I did, I got rid of the exponent and took the reciprocal of the base right over here. 8/27 is the base. Negative 2/3 is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over a denominator to some power-- and this is another very powerful exponent property-- this is going to be the exact same thing as raising 27 to the 2/3 power-- to the 2 over 3 power-- over 8 to the 2/3 power. Notice, if I have something divided by something and I'm raising the whole thing to a power, I can essentially raise the numerator to that power over the denominator raised to that power. Now, let's think about what this is. Well just like we saw before, this is going to be the same thing. This is going to be the same thing as 27 to the 1/3 power and then that squared because 1/3 times 2 is 2/3. So I'm going to raise 27 to the 1/3 power and then square whatever that is. All this color coding is making this have to switch a lot of colors. This is going to be over 8 to the 1/3 power. And then that's going to be raised to the second power. Same thing we were doing in the denominator-- we raise 8 to the 1/3 and then square that. So what's this going to be? Well, 27 to the 1/3 power is the cube root of 27." + }, + { + "Q": "Could what he showed at 1:38 be applied to logarithms in the same way it is showed here?", + "A": "Yes. for example, log(125)25=2/3 because the cube root of the square of 125 (15625) is 25. The cube root of 125 is 5, and 5^2 is 25.", + "video_name": "S34NM0Po0eA", + "timestamps": [ + 98 + ], + "3min_transcript": "We've already seen how to think about something like 64 to the 1/3 power. We saw that this is the exact same thing as taking the cube root of 64. And because we know that 4 times 4 times 4, or 4 to the third power, is equal to 64, if we're looking for the cube root of 64, we're looking for a number that that number times that number times that same number is going to be equal to 64. Well, we know that number is 4, so this thing right over here is going to be 4. Now we're going to think of slightly more complex fractional exponents. The one we see here has a 1 in the numerator. Now we're going to see something different. So what I want to do is think about what 64 to the 2/3 power is. And here I'm going to use a property of exponents that we'll study more later on. But this property of exponents is the idea that-- let's say with a simple number-- if I raise something to the third power and then I were to raise that to, say, the fourth power, to the 3 times 4 power, or 2 to the 12th power, which you could also write as raising it to the fourth power and then the third power. All this is saying is, if I raise something to a power and then raise that whole thing to a power, it's the same thing as multiplying the two exponents. This is the same thing as 2 to the 12th. So we could use that property here to say, well, 2/3 is the same thing as 1/3 times 2. So we could go in the other direction. We could say, hey look, well this is going to be the same thing as 64 to the 1/3 power and then that thing squared. Notice, I'm raising something to a power and then raising that to a power. If I were to multiply these two things, I would get 64 to the 2/3 power. Now, why did I do this? Well, we already know what 64 to the 1/3 power is. We just calculated it. That's equal to 4. So we could say that-- and I'll write it in that same yellow color-- this is equal to 4 squared, which is equal to 16. The way I think of it, let me find the cube root of 64, which is 4. And then let me square it. And that is going to get me to 16. Now I'll give you in even hairier problem. And I encourage you to try this one on your own before I work through it. So we're going to work with 8/27. And we're going to raise this thing to the-- and I'll try to color code it-- negative 2 over 3 power, to the negative 2/3 power. I encourage you to pause and try this on your own. Well the first thing I do whenever I see a negative exponent is to say, well, how can I get rid of that negative exponent? And I just remind myself, well, the negative exponent really just says, take the reciprocal of this to the positive exponent." + }, + { + "Q": "at 4:48 does 3/2=9/2 instead of 9/4?", + "A": "No, 3/2=9/4 because you are squaring both the numerator and the denominator. Therefore: 3^2=9 2^2=4 So, (3/4)^2=9/4", + "video_name": "S34NM0Po0eA", + "timestamps": [ + 288 + ], + "3min_transcript": "I'm going to use that light mauve color. So this is going to be equal to 27/8. I just took the reciprocal of this right over here. It's equal to 27/8 to the positive 2/3 power. So notice, all I did, I got rid of the exponent and took the reciprocal of the base right over here. 8/27 is the base. Negative 2/3 is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over a denominator to some power-- and this is another very powerful exponent property-- this is going to be the exact same thing as raising 27 to the 2/3 power-- to the 2 over 3 power-- over 8 to the 2/3 power. Notice, if I have something divided by something and I'm raising the whole thing to a power, I can essentially raise the numerator to that power over the denominator raised to that power. Now, let's think about what this is. Well just like we saw before, this is going to be the same thing. This is going to be the same thing as 27 to the 1/3 power and then that squared because 1/3 times 2 is 2/3. So I'm going to raise 27 to the 1/3 power and then square whatever that is. All this color coding is making this have to switch a lot of colors. This is going to be over 8 to the 1/3 power. And then that's going to be raised to the second power. Same thing we were doing in the denominator-- we raise 8 to the 1/3 and then square that. So what's this going to be? Well, 27 to the 1/3 power is the cube root of 27. times that same number is going to be equal to 27. Well, it might jump out at you already that 3 to the third is equal to 27 or that 27 to the 1/3 is equal to 3. So the numerator, we're going to end up with 3 squared. And then in the denominator, we are going to end up with-- well, what's 8 to the 1/3 power? Well, 2 times 2 times 2 is 8. So this is 8 to the 1/3 third is 2. Let me do that same orange color. 8 to the 1/3 is 2, and then we're going to square that. So this is going to simplify to 3 squared over 2 squared, which is just going to be equal to 9/4. So if you just break it down step by step, it actually is not too daunting." + }, + { + "Q": "at 0:17 if i flip the sides i mean write 11+a would that still be the same as a+11 is int that the commutative law of addition or it doesn't count when we use a variable?", + "A": "You got it! a + 11 is equivalent to 11 + a via the commutative law.", + "video_name": "640-86yn2wM", + "timestamps": [ + 17 + ], + "3min_transcript": "- [Voiceover] Let's do some examples of the writing expressions with variables exercise. So it says \"Write an expression to represent 11 more than a.\" Well you could just have a but if you want 11 more than a, you would wanna add 11 so you could write that as a plus 11. You could also write that as 11 plus a. Both of them would be 11 more than a. So let's check our answer here. We got it right. Let's do a few more of these. \"Write an expression to represent the sum of d and 9.\" So the sum of d and 9, that means you're gonna add d and 9. So I could write that as d plus 9 or I could write that as 9 plus d. And check our answer. Got that right. Let's do a few more of these. \"Write an expression to represent j minus 15.\" Well, I could just write it with math symbols instead of writing the word minus. Instead of writing M-I-N-U-S, And then I check my answer. Got it right. Let's do a few more of these. This is a lot of fun. \"Write an expression to represent 7 times r.\" There's a couple ways I could do it. I could use this little dot right over here, do 7 times r like that. That would be correct. I could literally just write 7r. If I just wrote 7r that would also count. Let me check my answer. That's right. Let me do a couple of other of these just so you can see that I could've just done 10 and this is not a decimal, it sits a little bit higher than a decimal. It's multiplication and the reason why once you start doing algebra, you use this symbol instead of that kind of cross for multiplication is that x-looking thing gets confused with x when you're using x as a variable so that's why this is a lot more useful. So we wanna write 10 times u, 10 times u, let's check our answer. We got it right. Let's do one more. So we could write it as 8 and then I could write a slash like that, 8 divided by d. And there you go. This is 8 divided by d. Let me check, let me check the answer. I'll do one more of these. Oh, it's 6 divided by b. Alright, same thing. So 6, I could use this tool right over here. It does the same thing as if I were to press the backslash. So 6 divided by b. Check my answer. We got it right." + }, + { + "Q": "In 2:40, Sal said if it goes 6 down but moved his cursor 6 sideways. Why?", + "A": "He said if x goes down by 6. It means 6 in the negative direction of the x axis (left). He should had said decrease instead of down to make it less confusing.", + "video_name": "uk7gS3cZVp4", + "timestamps": [ + 160 + ], + "3min_transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done." + }, + { + "Q": "At 0:50 seconds why did you made the slope zero and cancel it out?", + "A": "He wants to find the y-intercept, which is the point where the line crosses the y-axis. This point obviously has to be exactly on the y-axis. For a point to be exactly on the y-axis its x value has to be exactly zero. So he substitues the 0 for x in the line`s equation to find the y value of this point, which allows him to find the coordinates of the y-intercept.", + "video_name": "uk7gS3cZVp4", + "timestamps": [ + 50 + ], + "3min_transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done." + }, + { + "Q": "At 1:11 how do you know that b is on the y intercept and how do you know y will be on zero?", + "A": "1) What is x where the y axis crosses the x axis? What is the x coordinate of the y axis? 2) What is x where any line intersects the y axis? Now, all the points (x, y) on the line satisfy y = mx + b . What is the x coordinate of the point on the line where the line intersects the y axis? What is the equation for the point on the line at x=0? (y = m*0 + b; or y = b).", + "video_name": "uk7gS3cZVp4", + "timestamps": [ + 71 + ], + "3min_transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done." + }, + { + "Q": "at 8:00 Sal shows that lim u->0 of ln (Z)= ln (lim u->0 of Z). Is this a property of limits that I don't remember?", + "A": "It s the property of limits having to do with continuous functions. If f is continuous, then lim_{x->c}f(g(x)) = f(lim_{x->c}g(x)). basically you can move the limit inside a continuous function.", + "video_name": "yUpDRpkUhf4", + "timestamps": [ + 480 + ], + "3min_transcript": "And we know this is an exponent property, which I'll now do in a different color. We know that a to the bc is equal to a to the b to the c power. So that tells us that this me is equal to the limit as u approaches 0 of the natural log of 1 plus u to the 1/u, because this is one over xu, right? 1/u, and then all of that to the 1/x. And how did I do that? Just from this exponent property, right? If I were to simplify this, I would have 1/x times 1/u, and that's where I get this 1 over xu. If I have b to the a I can put that a out front. So I could take this 1/x and put it in front of the natural log. So now what do I have? We're almost there. We have the limit as u approaches 0. Take that 1/x, put it in front of the natural log sign. 1/x times the natural log of 1 plus u to the 1/u. Fair enough. When we're taking the limit as u approaches 0, x, this term doesn't involve it at all. So we could take this out in front, because the limit doesn't affect this term. And then we're essentially saying what happens to this expression as the limit as u approaches 0. So this thing is equivalent to 1/x times the natural log of And by now hopefully you would recognize that this is the definition. This limit comes to e, if you remember anything from compound interest. You might remember it as the limit-- as n approaches infinity of 1 plus 1 over n to the n. But these things are equivalent. If you just took the substitution u is equal to 1/n, you would get this. You would just get this. So this expression right here is e That expression is e. So we're getting close. So this whole thing is equivalent to 1/x times" + }, + { + "Q": "How did the substitution that u is equal to 1/n get thrown in? At 9:16, Sal gives the example that the limit as n goes to infinity of (1+1/n)n is e, yet the limit being discussed in the problem is as u goes to 0, not infinity.", + "A": "As n goes to infinity, 1/n will go to zero (limit as n->infinity of 1/n = 0) because one over a really big number is very close to zero. Therefore, when we substitute u for n, we change the limit from n->infinity to u->0.", + "video_name": "yUpDRpkUhf4", + "timestamps": [ + 556 + ], + "3min_transcript": "If I have b to the a I can put that a out front. So I could take this 1/x and put it in front of the natural log. So now what do I have? We're almost there. We have the limit as u approaches 0. Take that 1/x, put it in front of the natural log sign. 1/x times the natural log of 1 plus u to the 1/u. Fair enough. When we're taking the limit as u approaches 0, x, this term doesn't involve it at all. So we could take this out in front, because the limit doesn't affect this term. And then we're essentially saying what happens to this expression as the limit as u approaches 0. So this thing is equivalent to 1/x times the natural log of And by now hopefully you would recognize that this is the definition. This limit comes to e, if you remember anything from compound interest. You might remember it as the limit-- as n approaches infinity of 1 plus 1 over n to the n. But these things are equivalent. If you just took the substitution u is equal to 1/n, you would get this. You would just get this. So this expression right here is e That expression is e. So we're getting close. So this whole thing is equivalent to 1/x times the ways to get to e. So the limit as u approaches 0 of 1 plus u to the 1/u. That is e. And what is the natural log? Well it's the log base e. So you know this is equal to 1/x times the log base e of e. So that's saying e to what power is e. Well e to the first power is e, right? This is equal to 1. So 1 times 1/x is equal to 1/x. There we have it. The derivative of the natural log of x is equal to 1/x, which I find kind of neat, because all of the other exponents lead to another exponent. But all of a sudden in the mix here you have the natural log and the derivative of that is equal to x to the negative 1 or 1/x. Fascinating." + }, + { + "Q": "at 3:10 how did we get exactly 1/delta x * ln(1+ delta x/x)??\ni'd like to see more detailed way of getting 1/delta x * ln(1+ delta x/x)", + "A": "a/b is the same thing as 1/b * a. In this case, the a is a complicated looking thing, but the rule still works. the b is \u00ce\u0094x.", + "video_name": "yUpDRpkUhf4", + "timestamps": [ + 190 + ], + "3min_transcript": "So I'm going to take the limit of this whole thing. The natural log ln of x plus delta x-- right, that's like one point that I'm going to take evaluate the function-- minus the ln of x. All of that over delta x. And if you remember from the derivative videos, this is just the slope, and I'm just taking the limit as I find the slope between a smaller and a smaller distance. Hopefully you remember that. So let's see if we can do some logarithm properties to simplify this a little bit. Hopefully you remember-- and if you don't, review the logarithm properties-- but remember that log of a minus log of b is equal to log of a over b, and that comes out of the fact that follow the exponent rules. And if that doesn't make sense to you, you should review those as well. But let's apply this logarithm property to this equation. So let me rewrite the whole thing, and I'm going to keep switching colors to keep it from getting monotonous. So we have the limit as delta x approaches 0 of this big thing. Let's see. So log of a minus b equals log a over b, so this top, the numerator, will equal the natural log of x plus delta x over x. Right? a b a/b, all of that over delta x. And so that equals the limit as delta x approaches 0-- I think it's time to switch colors again-- delta x approaches 0 x out in front. So this is 1 over delta x, and we're going to take the limit of everything. ln x divided by x is 1 plus delta x over x. Fair enough. Now I'm going to throw out another logarithm property, and hopefully you remember that-- and let me put the properties separate so you know it's not part of the proof-- that a log b is equal to log of b to the a. And that comes from when you take something to an exponent, and then to another exponent you just have to multiply those two exponents. I don't want to confuse you, but hopefully you should remember this. So how does apply here? Well this would be a log b. So this expression is the same thing as the limit." + }, + { + "Q": "At 5:27, Sal says \u00ce\u0094x=\u00ce\u00bc, but I thought \u00ce\u0094x=x\u00ce\u00bc.", + "A": "He never said (delta)x = u He said if (delta)x approaches 0, then u also approaches 0, as they are directly proportional.", + "video_name": "yUpDRpkUhf4", + "timestamps": [ + 327 + ], + "3min_transcript": "x out in front. So this is 1 over delta x, and we're going to take the limit of everything. ln x divided by x is 1 plus delta x over x. Fair enough. Now I'm going to throw out another logarithm property, and hopefully you remember that-- and let me put the properties separate so you know it's not part of the proof-- that a log b is equal to log of b to the a. And that comes from when you take something to an exponent, and then to another exponent you just have to multiply those two exponents. I don't want to confuse you, but hopefully you should remember this. So how does apply here? Well this would be a log b. So this expression is the same thing as the limit. delta x over x to the 1 over delta x power. And remember all this is the natural log of this entire thing. And then we're going to take the limit as delta x approaches 0. If you've watched the compound interest problems and you know the definition of e, I think this will start to look familiar. But let me make a substitution that might clean things up a little bit. Let me make the substitution, let me call it n-- no, no, no, let me call u-- is equal to delta x over x. sides by x and we get xu is equal to delta x. Or we would also know that 1 over delta x is equal to 1 over xu. These are all equivalent. So let's make the substitution. So if we're taking the limit is delta x approaches 0, in this expression if delta x approaches 0, what does u approach? u approaches 0. So delta x approaching 0 is the same exact thing as taking the limit as u approaches 0. So we can write this as the limit as u approaches 0 of the natural log of 1 plus-- well we did the substitution, delta x over x is now u-- to the 1 over delta x, and that same substitution told us that's the same thing as one over xu." + }, + { + "Q": "In 0:45. How could we say that we are multiplying dx and the function if dx is just a notation that we are integrating with respect to x?", + "A": "Let me explain dx to you. dx is not a notation. It is used like a notation in evaluating intergals, but it is basically an infinetly small value of x (think of it as limit approaching zero). So, f(x)*dx means a rectangle with height of f(x) and base of an infinetly small number, so that it could be used for intergrating area under curve more accurately than using rectangle with base 1 or 2. Hope this helped you a bit. If not, please check out Sal s lesson about intergration.", + "video_name": "btGaOTXxXs8", + "timestamps": [ + 45 + ], + "3min_transcript": "Over here I've drawn part of the graph of y is equal to x squared. And what we're going to do is use our powers of definite integrals to find volumes instead of just areas. So let's review what we're doing when we take just a regular definite integral. So if we take the definite integral between, say, 0 and 2 of x squared dx, what does that represent? Well, let's look at our endpoints. So this is x is equal to 0. Let's say that this right over here is x is equal to 2. What we're doing is for each x, we're finding a little dx around it-- so this right over here is a little dx. And we're multiplying that dx times our function, times x squared. So what we're doing is we're multiplying this width times this height right over here. The height right over here is x squared. And we're getting the area of this little rectangle. And the integral sign is literally the sum of all of these rectangles for all of the x's But the limit of that as these dx's get smaller and smaller and smaller, get infinitely small, but not being equal to 0. And we have an infinite number of them. That's the whole power of the definite integral. And so you can imagine, as these dx's get smaller and smaller and smaller, these rectangles get narrower and narrower and narrower, and we have more of them, we are getting a better and better approximation of the area under the curve until, at the limit, we are getting the area under the curve. Now we're going to apply that same idea, not to find the area under this curve, but to find the volume if we were to rotate this curve around the x-axis. So this is going to stretch our powers of visualization here. So let's think about what happens when we rotate this thing around the x-axis. So if we were to rotate it, and I'll look at it and say that we're looking it a little bit from the right. So we get kind of a base that looks something like this. So you have a base that looks something like that. And then the rest of the function, if we just think about between 0 and 2, it looks like one of those pieces from-- I don't know if you ever played the game Sorry-- or it looks like a little bit of a weird hat. So it looks like this, and let me shade it in a little bit so it looks something like that. And just so that we're making sure we can visualize this thing that's being rotated. We care about the entire volume of the thing. Let me draw it from a few different angles. So if I drew from the top, it would look something like this. It'll become a little more obvious that it looks something like a hat. It would point up like this, and it goes down like that. It would look something like that. So in this angle, we're not seeing the bottom of it. And if you were to just orient yourself," + }, + { + "Q": "at 16:14, how do we know that m,n,p must be 527, 11, 40. In other word, how do we know for sure that cannot be any other m,n,p that result the same value of area?", + "A": "The question indicates that m and p are relatively prime , which means the fraction m/p is not reducible to any other integers. Also, it says that n is not divisible by the square of any prime, which means the number under the radical cannot have any perfect squares factored out. These instructions rule out any other m, n, or p that can give an equivalent correct answer.", + "video_name": "smtrrefmC40", + "timestamps": [ + 974 + ], + "3min_transcript": "We also know that CM right over here is 25/2. We also know that sine of theta here is equal to 527/625. That's what we just figured out. And we can use that to figure out the height of this triangle because we know that sine is the opposite over the hypotenuse, if we draw a right triangle right here. So sine of theta, which is 527/625, is equal to the opposite, is equal to the height of this triangle, over the hypotenuse, over 5/2 square roots of 11. So we can multiply both sides of this by 5/2 square roots 11. And we get the height of the triangle is equal to 527/625 times 5/2 square roots of 11. So you get a 1 here. And you have a 125 over here. So this is equal to 527 square roots 11 over 125 times 2, which is 250. That's the height. Now, what's the area of the triangle? It's 1/2 base times height. Area is equal to 1/2 base, which is 25/2, times the height, which we just figured out is 527 square roots of 11 over 250. Let's see. Divide the numerator by 25. Divide the denominator by 25. You get a 10 there. So this is equal to 527 square roots of 11 over 2 times So that's our area. And the whole problem-- they didn't want us to find the area. They wanted us to find m plus n plus p, so they want us to find essentially 527 plus 11 plus 40. So 527 plus 11 is-- let me make sure I do this-- 538, and then plus 40 is 578. And we're done." + }, + { + "Q": "At 12:06, how did you get 2 (25/2) ^2 (1 + sine theta) from 2 (25/2) ^2 + 2 (25/2)^2 sine theta?", + "A": "The expression 2(25/2)^2 + 2(25/2)^2sin theta has two terms. Both of these terms has a common factor 2(25/2)^2. Sal pulled this factor out of both terms. Let s make a substitution so it might be easier to see: let x = 2(25/2)^2. Then the expression becomes x + xsin theta. We pull the x out to become x(1 + sin theta). This is simple factoring out, or Sal sometimes calls it undistributing. We can redistribute the x to both terms by multiplying through to get x + xsin theta.", + "video_name": "smtrrefmC40", + "timestamps": [ + 726 + ], + "3min_transcript": "this has it in terms of cosine of theta plus 90 degrees. How can we figure out the sine of theta? That's what we actually care about, to figure out the area of this triangle, or actually to figure out the height of this triangle. And to do that, you just have to make the realization. We know the trig identity that the cosine of theta is equal-- I won't use theta because I don't want to overload theta-- cosine of x is equal to sine of 90 minus x. So the cosine of theta plus 90 degrees is going to be equal to the sine of-- let me put it in parentheses-- 90 minus whatever is here. 90 minus theta minus 90, which is equal to-- the 90's cancel out-- sine of negative theta. And we know sine of negative theta is equal to the negative sine of theta. is the negative sine of theta. So we could write the sine of theta here and then put the negative out here. And this becomes a positive. So what does this simplify to? We have 24 squared, which is 576. 576 is equal to-- let's see. I won't skip any steps here. So we have 25 squared plus 25 squared. This is 2 times 25/2 squared plus 2 times-- this is 25/2 squared again-- times sine of theta. Now we just have to solve for sine of theta. So this is going to be equal to-- well, this is 576 is equal to 2 times 25/2 squared. 1 plus sine of theta. Or we can just divide both sides of the equation by this here. Actually, let me just simplify it. This thing over here is 625/4, but then we're going to multiply that by 2. So this thing over here is 625/2. So let's divide both sides of this by 625/2. And we get 576 times 2 over 625-- just multiplying both sides by the inverse-- is equal to. When you multiply both sides by the inverse [? like this, ?] it actually cancels out-- is equal to 1 plus sine of theta. Or we just subtract 1 from both sides. We get sine of theta is equal to 576 times 2. Let's see, 76 times 2 is 152, plus 1,000. So it's 1,152/625." + }, + { + "Q": "At 5:52 pm what if it ask what is 39 percent of 700 how would you solve that?", + "A": "Best thing for this problem would be to divide 700 by 100 to get the amount of 1% (7) then multiply 7 by 39 to get 39% of 700 which is 273.", + "video_name": "FaDtge_vkbg", + "timestamps": [ + 352 + ], + "3min_transcript": "5 times 16 is 80. You subtract, you have no remainder, and you're done. 4/16 is the same thing as 0.25. Now, 0.25 is the same thing as twenty-five hundredths. Or, this is the same thing as 25/100, which is the same thing as 25%." + }, + { + "Q": "at 7:45, why does the Eigenvector equal span (1/2, 1), not span (1, - 1/2)?", + "A": "these are equivalent, since (1,-1/2) is in the span of (1/2,1) and vice versa. So it doesn t matter which one you choose, both statements are correct.", + "video_name": "3-xfmbdzkqc", + "timestamps": [ + 465 + ], + "3min_transcript": "corresponds to this pivot column, plus or minus 1/2 times my second entry has got to be equal to that 0 right there. Or, v1 is equal to 1/2 v2. And so if I wanted to write all of the eigenvectors that satisfy this, I could write it this way. My eigenspace that corresponds to lambda equals 5. That corresponds to the eigenvalue 5 is equal to the set of all of the vectors, v1, v2, that are equal to some scaling factor. Let's say it's equal to t times what? If we say that v2 is equal to t, so v2 is going to be equal to t times 1. or 1/2 times t. Just like that. For any t is a member of the real numbers. If we wanted to, we could scale this up. We could say any real number times 1, 2. That would also be the span. Let me do that actually. It'll make it a little bit cleaner. Actually, I don't have to do that. So we could write that the eigenspace for the eigenvalue 5 is equal to the span of the vector 1/2 and 1. So it's a line in R2. Those are all of the eigenvectors that satisfy-- that work for the equation where the eigenvalue is equal to 5. Now what about when the eigenvalue is equal to minus 1? So let's do that case. When lambda is equal to minus 1, then we have-- it's going So the eigenspace for lambda is equal to minus 1 is going to be the null space of lambda times our identity matrix, which is going to be minus 1 and 0, 0, minus 1. It's going to be minus 1 times 1, 0, 0, 1, which is just minus 1 there. Minus A. So minus 1, 2, 4, 3. And this is equal to the null space of-- minus 1, minus 1 is minus 2. 0 minus 2 is minus 2. 0 minus 4 is minus 4 and minus 1 minus 3 is minus 4. And that's going to be equal to the null space of the reduced row echelon form of that guy. So we can perform some row operations right here. Let me just put it in reduced row echelon form. So if I replace my second row plus 2 times my first row." + }, + { + "Q": "at 1:24, Where on earth did Sal swipe that 2 from?", + "A": "I think Sal recognized 30 = 6*5 = 2*3*5. That s why he tried multiplying it by 2. Don t wrap your mind around it too much. I tend to like this approach better: Let a = 5x\u00c2\u00b2 The expression would then be a\u00c2\u00b2 - 6a + 9 You now know how to solve it.", + "video_name": "o-ZbdYVGehI", + "timestamps": [ + 84 + ], + "3min_transcript": "We need to factor 25x to the fourth minus 30x squared plus 9. And this looks really daunting because we have something to the fourth power here. And then the middle term is to the second power. But there's something about this that might pop out at you. And the thing that pops out at me at least is that 25 is a perfect square, x to the fourth is a perfect square, so 25x to the fourth is a perfect square. And 9 is also perfect square, so maybe this is the square of some binomial. And to confirm it, this center term has to be two times the product of the terms that you're squaring on either end. Let me explain that a little bit better So, 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now, what is 30x squared? So remember, this needs to be two times the product of what's inside the square, or the square root of this and the square root of that. Given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square. So we can just rewrite this as this is equal to 5x squared-- let me do it in the same color. 5x squared minus 3 times 5x squared minus 3. And if you want to verify it for yourself, multiply this out. You will get 25x to the fourth minus 30x squared plus 9." + }, + { + "Q": "At 1:27 where did the 2 come from?", + "A": "Sal is using the pattern created by squaring a binomial. Here s the pattern: (a+b)^2 = a^2 + 2ab + b^2 Here s where the 2 comes from... use FOIL and multiply (a+b)(a+b) ... (a+b)(a+b) = a^2 + ab + ab + b^2 Notice... the 2 middle terms match. When you add them you get 2ab. That s where the 2 comes from. Hope this helps.", + "video_name": "o-ZbdYVGehI", + "timestamps": [ + 87 + ], + "3min_transcript": "We need to factor 25x to the fourth minus 30x squared plus 9. And this looks really daunting because we have something to the fourth power here. And then the middle term is to the second power. But there's something about this that might pop out at you. And the thing that pops out at me at least is that 25 is a perfect square, x to the fourth is a perfect square, so 25x to the fourth is a perfect square. And 9 is also perfect square, so maybe this is the square of some binomial. And to confirm it, this center term has to be two times the product of the terms that you're squaring on either end. Let me explain that a little bit better So, 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now, what is 30x squared? So remember, this needs to be two times the product of what's inside the square, or the square root of this and the square root of that. Given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square. So we can just rewrite this as this is equal to 5x squared-- let me do it in the same color. 5x squared minus 3 times 5x squared minus 3. And if you want to verify it for yourself, multiply this out. You will get 25x to the fourth minus 30x squared plus 9." + }, + { + "Q": "at 1:00 isn't profit a second variable?", + "A": "Because the problem gives you the value for Profit, a second variable is not needed. You could use a variable, but you would immediately swap it out because you know the value of the variable. Hope this helps.", + "video_name": "roHvNNFXr4k", + "timestamps": [ + 60 + ], + "3min_transcript": "Marcia has just opened her new computer store. She makes $27 on every computer she sells and her monthly expenses are $10,000. What is the minimum number of computer she needs to sell in a month to make a profit? So I'll let you think about that for a second. Well, let's think about what we have to figure out. We have to figure out the minimum number of computers she needs to sell. So let's set that to a variable or set a variable to represent that. So let's let x equal the number of computers she sells. Number of computers sold. Now, let's think about how much net profit she will make in a month. And that's what we're thinking about. How many computers, the minimum number she needs to sell in order to make a net profit? So I'll write her profit is going to be how much money she brings in from selling the computers. And she makes $27 on every computer she sells. So her profit is going to be $27 times the number of computer she sells. But we're not done yet. She still has expenses of $10,000 per month. So we're going to have to subtract out the $10,000. What we care about is making a profit. We want this number right over here to be greater than 0. So let's just think about what number of computers would get us to 0. And then, maybe she needs to sell a little bit more than that. So let's see what gets her to break even. So break even-- that's 0 profit. Neither positive or negative-- is equal to 27 times-- and I'll do it all in one color now. 27x minus 10,000. Well, we've seen equations like this before. We can add 10,000 to both sides. Add 10,000 to both sides, so it's no longer on the right-hand side. And we are left with 10,000. And then to solve for x, we just have to divide both sides by 27. Divide both sides by 27. On our right-hand side, we have x. So let me just write this down. So we have x on our right-hand side is going to be equal to 10,000/27. I switched the right and the left-hand sides here. Now, what is this going to be? Well, we can do a little bit of long division to handle that. So 27 goes into 10,000. So 27 doesn't go into 1. Doesn't go into 10. It goes into 100 three times. 3 times 27 is what? 81. 100 minus 81 is 19. Then we can bring down a 0. 27 goes into 190? It looks like it will go into it about six times." + }, + { + "Q": "At 3:21,how is S the same thing as 9/9?", + "A": "s is not 9/9. if it was, Sal would have gotten rid of the s. any number times 1 is itself, and 9/9 = 1. so 9/9 * s = s * 1", + "video_name": "vBlR2xNAGmo", + "timestamps": [ + 201 + ], + "3min_transcript": "But I'll just do it by hand this time, just so that we don't have to resort to some magical formulas So let's say that we define some sum, this one over here (let's call it S) Let's say that S is equal to what we have in parentheses over here It's going to be equal to four ninths, plus four ninths squared, plus four ninths to the third all the way to infinity Now let's also say that we multiply S by four ninths What's four ninths S going to look like? So then, I'm just essentially multiplying every term here by four ninths So if I take this first term and multiply it by four ninths, what am I going to get? If I take the second term and multiply it by four ninths, I'm going to get four ninths to the third power And we are going to go all the way to infinity So this is interesting When I multiply four ninths times this I get all of the terms here except for this first four ninths Now, this is kind of the magic of how we can actually find the sum of an infinite geometric series We can subtract this term right over here (this pink line) from this green line If we do that, clearly this is equal to that and this is equal to that So if we subtract this from that its equivalent to subtracting the pink from the green So we get S minus four ninths S is equal to... Well, every other term, this guy minus this guy is going to cancel out and on the right hand side you're only going to be left with this four ninths over here Then this four ninths, we can (S is the same thing as nine over nine) write this as nine over nine S minus four ninths S is equal to four ninths So nine over nine minus four over nine of something gives us five over nine So this becomes five ninths S is equal to four ninths Then to solve for S (and this is kind of magical but it's actually quite logical) Multiply both sides times the inverse of this, so times nine fifths on both sides These guys cancel out, and we get S is equal to four fifths That's really neat! We've just shown that this whole thing over here is equal to four fifths" + }, + { + "Q": "at 2:10 sal says between that and that why do we have round down or up for example can 1,251 be rounded to 1,250 rather then 1,300", + "A": "Yes it can! But not if you are rounding to the nearest hundred. If you were rounding to the nearest 50, you WOULD round to 1,250. But 1,250 does not have only zeros after the hundreds place, so in rounding to the nearest hundred that would not be right.", + "video_name": "fh8gkPW_6g4", + "timestamps": [ + 130 + ], + "3min_transcript": "Round 423,275 to the nearest thousand. So let me rewrite it: 423,275. And so the thousands place is the 3 right here, and so if we were round it up to the nearest thousand, we would go to 420-- let me write it so we just focus on the 3-- we would go up to 424,000 if we wanted to round up, 424,000, and if we wanted to round down, we would go to 423,000. We would get rid of the 275. 423,000. So this is our choice. Round up to 424,000 or round down to 423,000. And to figure it out, we just look at the digit one place to If that digit is 5 or greater, you round up. So this is 5. So if this is greater than or equal to 5, 5 or greater, you round up. If it's less than 5, you round down. 2 is definitely less than 5, so we just round down, so it is 423,000. Now just to visualize what this means to the nearest thousand, if I were to do a number line-- and you don't We've gotten the answer, but just to have a little bit better visualization of it, if I were to increment by thousands, you might have 422,000, 423,000. You have 424,000, and then maybe over here, you have 425,000, and you could keep going. And so when we round to the nearest thousand, we have to pick between that and that. We see that it much closer to 423,000 than to 424,000, so we round it right there. But you just use the rules we just came up with, and we rounded down to 423,000." + }, + { + "Q": "Why did Sal draw two lines over the angles at 1:18?", + "A": "To show that both angles became one angle", + "video_name": "jRrRqMJbHKc", + "timestamps": [ + 78 + ], + "3min_transcript": "All right. We're on problem 26. For the quadrilateral shown below, a quadrilateral has four sides, measure of angle A plus the measure of angle C is equal to what? And here, you should know that the sum of all the angles in a quadrilateral are equal to 360 degrees. And you might say, OK, I'll add that to my memory bank of things to memorize. Like the angles in a triangle are equal to 180. And I'll show you no, you don't have to memorize that. Because if you imagine any quadrilateral, let me draw a quadrilateral for you. And this is true of any polygon. So let's say this is some quadrilateral. You don't have to memorize that the sum of the angles is equal to 360. Although it might be useful for a quadrilateral. But I'll show you how to always prove it for any polygon. You just break it up into triangles. Then you only have to memorize one thing. If you break it up into triangles, this angle plus that angle plus that angle has to be equal to 180. equal to 180. So the angles in the quadrilateral itself are this angle and this angle. And then this angle and this angle. Well this one is just the sum of those two, and this one's just the sum of those two. So if these three added up to 180. And these three added up to 180. This plus this plus this, plus this will add up to 360. And you can do that with an arbitrarily shaped polygon. Let's do five sides, let's do a pentagon. So one, two, three, four, five sides. Wow, how many angles are there in a pentagon. Just break it up into triangles. How many triangles can you fit in it? Let's see. One, two. Each of these triangles, their angles, they add up to 180. So if you want to know that, that, that, plus that, that, that, plus that, that, and that. And that also would be the angle measures of the polygon. Because these three angles add up to that angle. That's that. Those angles add up to that one. Those angles add up to that one, and those angles add up So now hopefully, if I gave you a 20 sided polygon, you can figure out how many times can I fit triangles into it. And you'll know how many angles there are. And the sum of all of them. But anyway, back to the quadrilateral. A quadrilateral, the sum of the angles are going to be 360 degrees. So, if we say, measure of angle A, plus measure of angle C, plus these two angles. Let me write it down. Plus 95 plus 32 is going to be equal to 360. So I'll just write A plus C, just a quick notation. Let's see, 95 plus 32 is 127. Plus 127 is equal to 360. A plus C is equal to 360 minus 127." + }, + { + "Q": "At 3:52, why did Sal say 10 to the 3rd power instead of 10 cubed?", + "A": "10 to the 3rd power is the same thing as 10 cubed. They are just two different ways of saying the same thing. Similarly, 10 to the 2nd power is the same thing as 10 squared.", + "video_name": "YJdCw2fK-Og", + "timestamps": [ + 232 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:16 the sign is little bit confusing. More explanation. Thanks", + "A": "Sal wrote b < x < c. Between the points b and c on the x-axis, but not including those points, the function is negative. Notice, as Sal mentions, that this portion of the graph is below the x-axis. That is your first clue that the function is negative at that spot. Hope this helps.", + "video_name": "KxOp3s9ottg", + "timestamps": [ + 136 + ], + "3min_transcript": "- [Voiceover] What I hope to do in this video is look at this graph y is equal to f of x and think about the intervals where this graph is positive or negative and then think about the intervals when this graph is increasing or decreasing. So first let's just think about when is this function, when is this function positive? Well positive means that the value of the function is greater than zero. It means that the value of the function this means that the function is sitting above the x-axis. So it's sitting above the x-axis in this place right over here that I am highlighting in yellow and it is also sitting above the x-axis over here. And if we wanted to, if we wanted to write those intervals mathematically. Well let's see, let's say that this point, let's say that this point right over here is x equals a. Let's say that this right over here is x equals b and this right over here is x equals c. between a and b. At x equals a or at x equals b the value of our function is zero but it's positive when x is between a and b, a and b or if x is greater than c. X is, we could write it there, c is less than x or we could write that x is greater than c. These are the intervals when our function is positive. Let me write this, f of x, f of x positive when x is in this interval or this interval or that interval. So when is f of x negative? Let me do this in another color. F of x is going to be negative. Well, it's gonna be negative if x is less than a. So this is if x is less than a or if x is between b and c F of x is down here so this is where it's negative. So here or, or x is between b or c, x is between b and c. And I'm not saying less than or equal to because at b or c the value of the function f of b is zero, f of c is zero. That's where we are actually intersecting the x-axis. So that was reasonably straightforward. Now let's ask ourselves a different question. When is the function increasing or decreasing? So when is f of x, f of x increasing? Well increasing, one way to think about it is every time that x is increasing then y should be increasing or another way to think about it, you have a, you have a positive rate of change of y with respect to x. We could even think about it as imagine if you had a tangent line at any of these points." + }, + { + "Q": "why is it not factorial at 2:14?", + "A": "Because when you are allowed to use a letter more than once, you do not need to take it out of the possibilities for the next letter. You can use HHH, as Sal said. So you don t need to make it 26x25x24 because the numbers reset every time you add a new slot.", + "video_name": "VYbqG2NuOo8", + "timestamps": [ + 134 + ], + "3min_transcript": "- [Voiceover] So let's ask ourselves some interesting questions about alphabets in the English language. And in case you don't remember and are in the mood to count, there are 26 alphabets. So if you go, \"A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, and Z,\" you'll get, you'll get 26, 26 alphabets. Now let's ask some interesting questions. So given that there are 26 alphabets in the English language, how many possible three letter words are there? And we're not going to be thinking about phonetics or how hard it is to pronounce it. So, for example, the word, the word ZGT would be a legitimate word in this example. Or the word, the word, the word SKJ would be a legitimate word in this example. So how many possible three letter words are there in the English language? I encourage you to pause the video and try to think about it. Alright, I assume you've had a go at it. So let's just think about it, for three letter words possibilities are there for the first one? Well, there's 26 possible letters for the first one. Anything from a to z would be completely fine. Now how many possibilities for the second one? And I intentionally ask this to you to be a distractor because we've seen a lot of examples. We're saying, \"Oh, there's 26 possibilities \"for the first one and maybe there's 25 for the second one, \"and then 24 for the third,\" but that's not the case right over here because we can repeat letters. I didn't say that all of the letters had to be different. So, for example, the word, the word HHH would also be a legitimate word in our example right over here. So we have 26 possibilities for the second letter and we have 26 possibilities for the third letter. So we're going to have, and I don't know what this is, 26 to the third power possibilities, or 26 times 26 time 26 and you can figure out what that is. That is how many possible three letter words we can have for the English language they are, if they meant anything and if we repeated letters. Now let's ask a different question. What if we said, \"How many possible three letter words \"are there if we want all different letters?\" So we want all different letters. So these all have to be different letters. Different, different letters and once again, pause the video and see if you can think it through. Alright, so this is where permutations start to be useful. Although, I think a lot of things like this, it's always best to reason through than try to figure out if some formula applies to it. So in this situation, well, if we went in order, we could have 26 different letters for the first one, 26 different possibilities for the first one. You know, I'm always starting with that one, but there's nothing special about the one on the left. We could say that the one on the right, there's 26 possibilities, well for each of those possibilities, for each of those 26 possibilities, there might be 25 possibilities" + }, + { + "Q": "At 6:00 he swaps a row with another and add a -ve sign. If further down the computation process, if we have to swap another row, do we add another \"-\" sign thus making the Det. positive again..or does the minus stay no matter the no. of row-swaps?\n\nThanks", + "A": "Yes, each row or column swap results in the determinant being multiplied by -1. If you do an odd number of swaps, the determinant is therefore negative, and a positive number of swaps keeps it positive.", + "video_name": "QV0jsTiobU4", + "timestamps": [ + 360 + ], + "3min_transcript": "2 minus 2 is 0. 4 minus 2 is 2. 2 minus 1 is 1. Now let's replace the third row with the third row minus 2 times the second row. So 2 minus 2 times 1 is 0. 7 minus 2 times 2 is 3. 5 minus 2 times 2 is 1. 2 minus 2 times 1 is 0. Let me get a good color here. I'll do pink. Let's replace the last row with the last row, essentially, plus the first row. You could say minus minus 1 times the first row is the same thing as the last row plus the first row. So minus 1 plus 1 is 0. 4 plus 2 is 6. Minus 6 plus 2 is minus 4. So there we have it like that. And this guy has two 0's here, so maybe I want to swap some rows. So let me swap some rows. So if we swap rows, what happens? I'm going to swap the middle two rows just for fun. Well, not just for fun. Because I want a pivot entry right here. I shouldn't say a pivot entry. I want to do it in upper triangular form. So I want a non-zero entry here. This is a 0, so I'm going to move this guy down. So I'm going to keep the top row the same. 1, 2, 2, 1. I'm going to keep the bottom row the same. 0, 0, 6, minus 4, 4. And I'm going to swap these guys right here. So this is going to be 0, 3, 1, 0. And then 0, 0, 2, 1. Now, can I just swap entries like that? Well, I can, but you have to remember that when you swap entries, your resulting determinant is going to be the So if we swap these two guys, the determinant of this is going to be the negative of this determinant. When you swap two rows, you just flip the sign of the determinant. That was one of the first videos we did on these, kind of messing with the determinants. Now, what do we want to do here? To get this guy into upper triangular form, it'd be nice to get this to be a 0. So to get that to be a 0, let me keep everything else the same. So I have a 1, 2, 2, 1. I have a 0, 3, 1, 0. The third row is 0, 0, 2, 1. And now this last row, let me replace it with the last row minus 3 times this row. So let me write it like this. I have to carry that negative sign as well. So I'm going to replace this last row with the last row minus 2 times the second row." + }, + { + "Q": "At 5:54, why exactly did you combine C1 and C2 to have the constant C? Aren't you supposed to solve for C2 and C1 separately?", + "A": "C\u00e2\u0082\u0081 and C\u00e2\u0082\u0082 are just constants, numbers, and they are subtracting each other, so not only it makes sense to combine them into a single constant, but it would be impossible to find their values independently. You would end up with and equation in the form C\u00e2\u0082\u0081 - C\u00e2\u0082\u0082 = 0, and there are an infinite number of possibilities here.", + "video_name": "DL-ozRGDlkY", + "timestamps": [ + 354 + ], + "3min_transcript": "So if that was a two there and if you don't want to change the value of the integral you put the 1/2 right over there. And so now you could either do U substitution explicitly or you could do it in your head where you said U is equal to negative X squared and then DU will be negative to X, DX or you can kind of do this in your head at this point. So I have something and it's derivative so I really could just integrate with respect to that something too with respect to that U. So this is going to be 1/2. This 1/2 right over here. The anti-derivative. This is E to the negative X squared and then of course, I might have some other constant. I'll just call that C two. And once again, if this part over here what I just did seemed strange, the U substitution, you might want to review that piece. Now, what can I do here? We'll have a constant on the left hand side. It's an arbitrary constant. We don't know what it is. we could call it. So, let me just subtract C one from both sides. So if I just subtract C one from both sides I have an arbitrary so this is gonna cancel, and I have C two, sorry. Let me. So, this is C one. So these are going to cancel and C two minus C one. These are both constants, arbitrary constants and we don't know what they are yet. And so, we could just rewrite this as on the left hand side we have Y squared over two is equal to on the right hand side. I'll write 1/2 E. Let me write that in blue just because I wrote it in blue before. 1/2 E to the negative X squared and I'll just say C two minus C one. Let's just call that C. So if you take the sum of those two things let's just call that C. And so now, this is kind of a general solution. We don't know what this constant is but even in this form we can now find a particular solution using this initial condition. Let me separate it out. This was a part of this original expression right over here but using this initial condition. So, it tells us when X is zero, Y needs to be equal to one. So we would have one squared which is just one over two is equal to 1/2. E to the negative zero squared. Well, that's just going to be either the zero is just one. This is gonna be 1/2 plus C and just like that we're able to figure out if you subtract 1/2 from both sides C is equal to zero. So the relationship between Y and X that goes through this point, we could just set C is equal to zero. So that's equal to zero. That's zero right over there. And so we are left with Y squared over two" + }, + { + "Q": "At 4:36 shouldn't the integral be equal to e^ (1 - x^2)/ 2*(1 - x^2) ?", + "A": "No, I m not sure where you re getting (1 - x\u00c2\u00b2). This is an integral best done with u-substition: u = - x\u00c2\u00b2 du = - 2x dx so 1/2\u00e2\u0088\u00ab -2x e^-x\u00c2\u00b2 dx = 1/2 \u00e2\u0088\u00ab e^u du = 1/2 e^u + C = 1/2 e^-x\u00c2\u00b2 + C", + "video_name": "DL-ozRGDlkY", + "timestamps": [ + 276 + ], + "3min_transcript": "a separable differential equation. Differential equation. And it's usually the first technique that you should try. Hey, can I separate the Ys and the Xs and as I said, this is not going to be true of many, if not most differential equations. But now that we did this we can integrate both sides. So let's do that. So, I'll find a nice color to integrate with. So, I'm going to integrate both sides. Now if you integrate the left hand side what do you get? You get and remember, we're integrating with respect to Y here. So this is going to be Y squared over two and we could put some constant there. I could call that plus C one. And if you're integrating that that's going to be equal to. Now the right hand side we're integrating with respect to X. And let's see, you could do U substitution or you could recognize that look, the derivative of negative X squared So if that was a two there and if you don't want to change the value of the integral you put the 1/2 right over there. And so now you could either do U substitution explicitly or you could do it in your head where you said U is equal to negative X squared and then DU will be negative to X, DX or you can kind of do this in your head at this point. So I have something and it's derivative so I really could just integrate with respect to that something too with respect to that U. So this is going to be 1/2. This 1/2 right over here. The anti-derivative. This is E to the negative X squared and then of course, I might have some other constant. I'll just call that C two. And once again, if this part over here what I just did seemed strange, the U substitution, you might want to review that piece. Now, what can I do here? We'll have a constant on the left hand side. It's an arbitrary constant. We don't know what it is. we could call it. So, let me just subtract C one from both sides. So if I just subtract C one from both sides I have an arbitrary so this is gonna cancel, and I have C two, sorry. Let me. So, this is C one. So these are going to cancel and C two minus C one. These are both constants, arbitrary constants and we don't know what they are yet. And so, we could just rewrite this as on the left hand side we have Y squared over two is equal to on the right hand side. I'll write 1/2 E. Let me write that in blue just because I wrote it in blue before. 1/2 E to the negative X squared and I'll just say C two minus C one. Let's just call that C. So if you take the sum of those two things let's just call that C. And so now, this is kind of a general solution. We don't know what this constant is" + }, + { + "Q": "Wait. So are all of the foreheads blue, or not? because if they're all blue it should only take 2 times. (as in open lights shut lights open lights shut and everyone is gone.) At 0:09 that's what it said, but... ok this is getting confusing", + "A": "every person would see a number of blue foreheads. if I HAVE a blue forehead I see one less than the total number of blueforeheads. If I have other than blue forhead I see the total number of blue foreheads. everone would leave on the number of blue foreheads seen plus one. So all blue foreheads would leave on the total number of blue foreheads time the light goes out, while the other foreheads are wating for the total number of blue foreheads plus one.", + "video_name": "-xYkTJFbuM0", + "timestamps": [ + 9 + ], + "3min_transcript": "So we had the hundred logicians. All of their foreheads were painted blue. And before they entered the room, they were told that at least one of you hundred logicians has your forehead painted blue. And then every time that they turned on the lights, so that they could see each other, they said OK, once you've determined that you have a blue forehead, when the lights get turned off again, we want you to leave the room. And then once that's kind of settled down, they'll turn the lights on again. And people will look at each other again. And then they'll turn them off again. And maybe people will leave the room. And so forth and so on. And they're also all told that everyone in the room is a perfect logician. They have infallible logic. So the question was, what happens? And actually maybe an even more interesting question is why does it happen? So I'll answer the first, what happens? And if just take the answer, and you don't know why, it almost seems mystical. That essentially the light gets turned on and off 100 gets turned on, and the lights get turned off again, all of them leave. They all leave. So I mean, it's kind of weird, right? Let's say I'm one of them. Or you're one of them. I go into this room. The lights get turned on. And I see 99 people with blue foreheads. And I can't see my own forehead. They see my forehead, of course. But to any other person, I'm one of the 99, right? But I see 99 blue foreheads. So essentially what happens if we were to watch the show is, the lights get turned on. You see 99 blue foreheads. Then the lights get turned off again. And then the lights get turned on again. And everyone's still sitting there. And I still see 99 blue foreheads. And that happens 100 times. everyone leaves the room. And at first glance, that seems crazy, because nothing changes. Nothing changes between every time we turn on the light. But the way you need to think about this-- and this is what makes it interesting-- is what happens instead of 100, let's say there was one person in the room. So before the show starts-- they never told me that there were going to be 100 people in the room. They just said, at least one of you, at least one of the people in the room, has your forehead painted blue. And as soon as you know that your forehead is painted blue, you leave the room. And that everyone's a perfect logician. So imagine the situation where instead of 100 there's only one perfect logician. Let's say it's me. So that's the room. I walk in. And I sit down. And maybe I should do it with blue." + }, + { + "Q": "At 3:32, is anything better than Sal using the word \"DUDE\" in an educational context?", + "A": "No, no there isn t.", + "video_name": "-xYkTJFbuM0", + "timestamps": [ + 212 + ], + "3min_transcript": "everyone leaves the room. And at first glance, that seems crazy, because nothing changes. Nothing changes between every time we turn on the light. But the way you need to think about this-- and this is what makes it interesting-- is what happens instead of 100, let's say there was one person in the room. So before the show starts-- they never told me that there were going to be 100 people in the room. They just said, at least one of you, at least one of the people in the room, has your forehead painted blue. And as soon as you know that your forehead is painted blue, you leave the room. And that everyone's a perfect logician. So imagine the situation where instead of 100 there's only one perfect logician. Let's say it's me. So that's the room. I walk in. And I sit down. And maybe I should do it with blue. look around the room. And I look around the room, and I see nobody else, right? And remember, even in the case of one, we've painted everyone's forehead blue. So in this case, this one dude, or me or whoever you want to call him. His forehead is painted blue. So he looks around and he sees no one in the room. But he remembers the statement, and maybe it's even written down on a card for him in case he forgets. That at least one of you has your forehead painted blue. So if he looks around the room and he says, well I'm the only dude in the room. And they told me that at least one of the dudes in the room is going to have their foreheads blue. Well, I'm the only dude in the room. So I must have a blue forehead. So as soon as they turn the lights off, he's going to leave. Fair enough. That's almost trivially simple. And you might say, so how does this apply to 100? Well what happens when there are two people. And once again, both of them have their foreheads painted blue. So let me draw another. I don't want to keep drawing the blue forehead room. So let's put ourselves in the head of this guy. Right behind the blue forehead. That's where we're sitting. So when he enters the room. He says, I either have a blue forehead. I either have a blue forehead, or I don't have a blue forehead. No blue. Right? This is what this guy's thinking. Let me draw him. And he has a blue forehead. But he doesn't know it. He can't see it. That's the whole point about painting the forehead blue, as opposed to another part of the body. So he says, I either have a blue forehead or I don't have a blue forehead. He walks in. Let's say this is this guy. He walks in. The first time the lights get turned on, he sees this other dude there who has a blue forehead." + }, + { + "Q": "so does that mean that 1.999....=2? I hope it does, i lost track after 1:26", + "A": "I think this was already answered for you, but I m not sure so I will. Yes, 1.999.... is equal to 2. Think about it as (1 + .999....) = 2.", + "video_name": "TINfzxSnnIE", + "timestamps": [ + 86 + ], + "3min_transcript": "99.9 repeating percent of mathematicians agree, 0.9 repeating equals 1. If because I said so works for you, you can go ahead and do something else now. Maybe you're like, 0.9 repeating equals one, that's this 0.9 repeating-derful! Otherwise, on to reason number 2, or reason 1.9 repeating. See, it's weird, because when we think of the number 1 or 2, in most contexts we mean it as a natural number, like 1, 2, 3, 4, 5. In the sense then, the next number after 1 is 2. 9.9 repeating may equal 10, but you wouldn't say you have 9.9 repeating lords a leaping, in the same way you wouldn't say you had 9.75 lords a leaping plus 1/4 lord a leaping. Lords, leaping or otherwise, come in natural numbers. So what does this statement mean? 0.9 repeating is the same as 1? It looks pretty different, but it equals 1 in the same way that a 1/2 equals 0.5. They have the same value, You can philosophize over whether, if 1 is the loneliest number, 0.78 plus 0.22 is just as lonely, but there's no mathematical doubt that they have the same value, just as 100 years of solitude or 99.9 repeating years of solitude. So reason 2 is not a proof, but a reason to stay open minded. Numbers that look different can have the same value. Another example of this is that, in algebra, 0 equals negative 0. 0.9 repeating is a decimal number, a real number. See, if you want 0.9 repeating to be that number infinitesimally close to 1, but not 1-- and let's face it, some of you do-- then you're writing down the wrong number when you write 0.9 repeating. That number infinitely close to 1, but less than 1, is a number, but it's not 0.9 repeating or any real number. OK, let's do more 3.9 repeating-mal proof for reason 3.9 repeating. According to this 3.9 repeating-mula, 3.9 repeating is 4. First step, say 0.9 repeating equals x. Then multiply each side by 10. Third, subtract 0.9 repeating from this side, which equals x, which we subtract from the other side. And 10x minus x is 9x. Divide by 9, and you get 1 equals x, which you might notice also equals 0.9 repeating. There's no tricks here. It's simple multiplication, subtraction and division by 9, which are all allowed because they are consistent. When something is inconsisten-- 9.9 repeating --t, we just throw it out of algebra altogether. For example, in algebra if you try to divide by 0, you get this problem where anything can equal anything. I mean, if you want to say everything is equal, fine, but your algebra sucks. Normal, everyday elementary algebra, the one they shove down students throats as if it were the only algebra, doesn't allow dividing by 0. So it stays consistent and suspiciously practical. We also could have shifted the decimal point twice, multiplying by 100 to prove that if you have 99.9 repeating bottles of beer on the wall, 99.9 repeating bottles of beer. Take one down, pass it around, 99 bottles of beer on the wall." + }, + { + "Q": "At 07:42: You will never reach _____ steps or 1/3: What is the blank?", + "A": "The blank is infinity", + "video_name": "TINfzxSnnIE", + "timestamps": [ + 462 + ], + "3min_transcript": "Take 0.3 repeating, a repeating decimal equal to 1/3. Multiply it by 3. Obviously, by definition, 3/3 is 1, and 0.3 repeating times 3 is 0.9 repeating, which you might have noticed is also 1. The only assumption here is that 0.3 repeating equals 1/3. Maybe you don't like decimal notation in general, which brings us to reason number 9, this sum of an infinite series thing. 9/10 plus 9/100 plus 9/1000. And we can sum this series and get 1. But I can see why you might be unhappy with this. It recalls Zeno's paradoxes. How can you get across a room, when first you have to walk halfway, and then half of that, and so on. Or, how can you shoot an arrow into a target, when first it needs to go halfway, but before it can get halfway, it needs to go half of halfway, and before that, half of half of halfway, and half of half of half of halfway, and so on. Anyway, it's 1/2 plus 1/4 plus 1/8, dot, dot, dot, dot, dot, to get 1. Each time, you fall short of 1. So how can you ever do anything? Luckily, infinity has got our backs. I mean, that's like the definition of infinity, a numbers so large, you can never get there, no matter how many steps you do, no matter how high you count. This way of writing numbers with this dot, dot, dot business, or with a bar over the repeating part, is a shorthand for an infinite series, whether it be 9/10 plus 9/100, and so on to get 1. Or 3/10 plus 3/100, and so on, to get 1/3. No matter how many 3s you write down, it will always be less than 1/3, but it will also always be less than infinity 3s. Infinity is what gets us there when no real number can. The binary equivalent of 0.9 repeating is 0.1 repeating. That's exactly 1/2, plus 1/4, plus 1/8, and so on. That's how we know a dotted, dotted, dot, dot, dot The ultimate reason that 0.9 repeating equals 1 is because it works. It's consistent, just like 1 plus 1 equals 2 is consistent, and just like 1 divided by 0 equals infinity isn't. Mathematics is about making up rules and seeing what happens. And it takes great creativity to come up with good rules. The only difference between mathematics and art is that if you don't follow your invented rules precisely in mathematics, people have a tendency to tell you you're wrong. Some rules give you elementary algebra and real numbers, and these rules can't tell the difference between 0.9 repeating and 1, just like they can't tell the difference between 0.5 and 1/2, or between 0 and negative 0. I hope you see now that the view that 9.9 repeating does not equal 10 is simply un-- 9.9 repeating --able. If you started this video thinking, I h-- 7.9 repeating that 7.9 repeating is 8, I hope now, you're thinking, oh, sweet, 4.9 repeating is 5? High 4.9 repeating!" + }, + { + "Q": "At 3:24 how does infinity minus 1 still equal infinity?", + "A": "because it was declared as a rule in math. I know that isn t a great answer but it works. The best answer is because its not a real number. When we add something to a real number we get that number plus 1. Since infinity is not bound by the laws of real numbers it doesn t have to behave that way. I suppose you could say it operates above the laws of real numbers in the hyper-real number space.", + "video_name": "TINfzxSnnIE", + "timestamps": [ + 204 + ], + "3min_transcript": "or 99.9 repeating years of solitude. So reason 2 is not a proof, but a reason to stay open minded. Numbers that look different can have the same value. Another example of this is that, in algebra, 0 equals negative 0. 0.9 repeating is a decimal number, a real number. See, if you want 0.9 repeating to be that number infinitesimally close to 1, but not 1-- and let's face it, some of you do-- then you're writing down the wrong number when you write 0.9 repeating. That number infinitely close to 1, but less than 1, is a number, but it's not 0.9 repeating or any real number. OK, let's do more 3.9 repeating-mal proof for reason 3.9 repeating. According to this 3.9 repeating-mula, 3.9 repeating is 4. First step, say 0.9 repeating equals x. Then multiply each side by 10. Third, subtract 0.9 repeating from this side, which equals x, which we subtract from the other side. And 10x minus x is 9x. Divide by 9, and you get 1 equals x, which you might notice also equals 0.9 repeating. There's no tricks here. It's simple multiplication, subtraction and division by 9, which are all allowed because they are consistent. When something is inconsisten-- 9.9 repeating --t, we just throw it out of algebra altogether. For example, in algebra if you try to divide by 0, you get this problem where anything can equal anything. I mean, if you want to say everything is equal, fine, but your algebra sucks. Normal, everyday elementary algebra, the one they shove down students throats as if it were the only algebra, doesn't allow dividing by 0. So it stays consistent and suspiciously practical. We also could have shifted the decimal point twice, multiplying by 100 to prove that if you have 99.9 repeating bottles of beer on the wall, 99.9 repeating bottles of beer. Take one down, pass it around, 99 bottles of beer on the wall. Reason number 5, there's infinite 9s. If anyone ever thinks they have the biggest number, well, they don't, because just add 1, or multiply by 2, or whatever, and it's even bigger. Infinity, though, is not a number you can add 1 to to get a bigger number. Adding 1 is an algebra thing that you do with real numbers. Subtracting doesn't work either. Infinity bottles of beer on the wall minus 1 is still infinity bottles of beer. When we did this decimal shift to multiply by 10, unlike un-infinitely many 9s, there's no last 9 that got shifted over to create a 0. Infinite 9s plus another 9 is still infinite 9s, the kind of infuriating property that makes infinity not a real number and makes that proof work. If you're the type of person who is discon-- 9.9 repeating --t with the idea that 9.9 repeating equals 10, you might also feel that 1 divided by 0 should be infinity. And, as it turns out, there is other systems or calculation besides elementary algebra where it does." + }, + { + "Q": "At 2:35 she did a thing where she said anything equals anything, when all she said was that:\n\nX = 0\n42 * 0 = 0\n42x = x\n0=0\n\nYou should get 0 = 0, which is true.\nCorrect me if I'm wrong please", + "A": "Her last step from 42x=x was to divide both sides by x, getting 42=1. Then you can subtract 1 from both sides to get 41=0. Divide by 41 to get 1=0. And multiply by any number c to get c=0. So any number equals 0. But her work is invalid, because she defined x=0. So when she divided by x, she was dividing by 0, which is undefined precisely because of this issue it raises.", + "video_name": "TINfzxSnnIE", + "timestamps": [ + 155 + ], + "3min_transcript": "or 99.9 repeating years of solitude. So reason 2 is not a proof, but a reason to stay open minded. Numbers that look different can have the same value. Another example of this is that, in algebra, 0 equals negative 0. 0.9 repeating is a decimal number, a real number. See, if you want 0.9 repeating to be that number infinitesimally close to 1, but not 1-- and let's face it, some of you do-- then you're writing down the wrong number when you write 0.9 repeating. That number infinitely close to 1, but less than 1, is a number, but it's not 0.9 repeating or any real number. OK, let's do more 3.9 repeating-mal proof for reason 3.9 repeating. According to this 3.9 repeating-mula, 3.9 repeating is 4. First step, say 0.9 repeating equals x. Then multiply each side by 10. Third, subtract 0.9 repeating from this side, which equals x, which we subtract from the other side. And 10x minus x is 9x. Divide by 9, and you get 1 equals x, which you might notice also equals 0.9 repeating. There's no tricks here. It's simple multiplication, subtraction and division by 9, which are all allowed because they are consistent. When something is inconsisten-- 9.9 repeating --t, we just throw it out of algebra altogether. For example, in algebra if you try to divide by 0, you get this problem where anything can equal anything. I mean, if you want to say everything is equal, fine, but your algebra sucks. Normal, everyday elementary algebra, the one they shove down students throats as if it were the only algebra, doesn't allow dividing by 0. So it stays consistent and suspiciously practical. We also could have shifted the decimal point twice, multiplying by 100 to prove that if you have 99.9 repeating bottles of beer on the wall, 99.9 repeating bottles of beer. Take one down, pass it around, 99 bottles of beer on the wall. Reason number 5, there's infinite 9s. If anyone ever thinks they have the biggest number, well, they don't, because just add 1, or multiply by 2, or whatever, and it's even bigger. Infinity, though, is not a number you can add 1 to to get a bigger number. Adding 1 is an algebra thing that you do with real numbers. Subtracting doesn't work either. Infinity bottles of beer on the wall minus 1 is still infinity bottles of beer. When we did this decimal shift to multiply by 10, unlike un-infinitely many 9s, there's no last 9 that got shifted over to create a 0. Infinite 9s plus another 9 is still infinite 9s, the kind of infuriating property that makes infinity not a real number and makes that proof work. If you're the type of person who is discon-- 9.9 repeating --t with the idea that 9.9 repeating equals 10, you might also feel that 1 divided by 0 should be infinity. And, as it turns out, there is other systems or calculation besides elementary algebra where it does." + }, + { + "Q": "at 3:55, how did he come up with 10-9/12? help?", + "A": "The 10-9/12 came from the right side of the equation: 5/6 - 3/4 and in order to subtract 3/4 from 5/6 we need to have like denominators. The LCM for 5/6 and 3/4 is 12, therefore 5/6 = 10/12 and 3/4 = 9/12 Which gives us 10/12 - 9/12 or 10-9/12 = 1/12", + "video_name": "DopnmxeMt-s", + "timestamps": [ + 235 + ], + "3min_transcript": "And actually, well, if we wanted to check it, we could say, well, the original problem was 2x plus 3 equals minus 15. So we could say 2 times minus 9 plus 3. 2 times minus 9 is minus 18 plus 3. Well, that's equal to minus 15, which is equal to what the original equation said, so we know that's right. That's the neat thing about algebra. You can always check your work. Let's do another problem. I'm going to put some fractions in this time, just to show you that it can get a little bit hairy. So let's say I had minus 1/2x plus 3/4 is equal to 5/6. So we'll do the same thing. First, we just want to get this 3/4 out of the left hand if you want to try working this out yourself, you might want to pause the video and then play it once you're ready to see how I do it. Anyway, let me move forward assuming you haven't paused it. If we want to get rid of this 3/4, all we do is we subtract 3/4 from both sides of this equation. Minus 3/4. Well, the left hand side, the two 3/4 will just cancel. We get minus 1/2x equals, and then on the right hand side, we just have to do this fraction addition or fraction So the least common multiple of 6 and 4 is 12. So this becomes 5/6 6 is 10/12 minus 3/4 is 9/12, so we get minus 1/2x is equal to 1/12. And if that step confused you, I went a little fast, you might just want to review the adding and subtraction of fractions. So going back to where we were. So now all we have to do is, well, the coefficient on the x term is minus 1/2, and this is now a level one problem. So to solve for x, we just multiply both sides by the reciprocal of this minus 1/2x, and that's minus 2/1 times minus 1/2x on that side, and then that's times minus 2/1. The left hand side, and you're used to this by now, simplifies to x. The right hand side becomes minus 2/12, and we could simplify that further to minus 1/6. Well, let's check that just to make sure we got it right. So let's try to remember that minus 1/6. So the original problem was minus 1/2x," + }, + { + "Q": "at 2:19 i didnt hear what sal said what did he say", + "A": "he said ...times minus nine...", + "video_name": "DopnmxeMt-s", + "timestamps": [ + 139 + ], + "3min_transcript": "Welcome to level two linear equations. Let's do a problem. 2x plus 3 is equal to minus 15. Throw the minus in there to make it a little bit tougher. So the first thing we want to do whenever we do any linear equation, is we want to get all of the variable terms on one hand side of the equation and all the constant terms And it doesn't really matter, although I tend to get my variables on the left hand side of the equation. Well, my variables are already on the left hand side of the equation but I have this plus 3 that I somehow want to move to the right hand side of the equation. And the way I can-- you can put it in quotes, move the 3 is I can subtract 3 from both sides of this equation. And look at that carefully as to why you think that works. Because if I subtract 3 from the left hand side, clearly this negative 3 that I'm subtracting and the original 3 will cancel out and become 0. and as long as I do whatever I do on the left hand side, as well, because whatever you do on one side of the equal side, you have to do to the other side, then I'm making a valid operation. So this will simplify to 2x, because the 3's cancel out. They become just 0. Equals minus 15 minus 3. Well, that's minus 18. And now, we're just at a level one problem, and you can just multiply both sides of this equation times the reciprocal on the coefficient of 2x. I mean, some people would just say that we're dividing by 2, which is essentially what we're doing. I like to always go with the reciprocal, because if this 2 was a fraction, it's easier to think about it that way. But either way, you either multiply by the reciprocal, or divide by the number. It's the same thing. So 1/2 times 2x. Well, that's just 1x. So you get x equals, and then minus 18/2. And minus 18/2, well, that just equals minus 9. And actually, well, if we wanted to check it, we could say, well, the original problem was 2x plus 3 equals minus 15. So we could say 2 times minus 9 plus 3. 2 times minus 9 is minus 18 plus 3. Well, that's equal to minus 15, which is equal to what the original equation said, so we know that's right. That's the neat thing about algebra. You can always check your work. Let's do another problem. I'm going to put some fractions in this time, just to show you that it can get a little bit hairy. So let's say I had minus 1/2x plus 3/4 is equal to 5/6. So we'll do the same thing. First, we just want to get this 3/4 out of the left hand" + }, + { + "Q": "At 2:34 couldn't you factor out both the 3 and the 5 to get 15? Wouldn't 15(a+b)=2 be correct?", + "A": "3a+5b=2 To factor, both terms must have a same common factor. Meaning to factor out a 3, the b-term must have a factor of 3 as well. Likewise, to factor out a 5, the a-term must have a factor of 5. Since neither have any common factor, you can t factor it. And certainly, neither of them has a factor of 15.", + "video_name": "CLQRZ2UbQ4Q", + "timestamps": [ + 154 + ], + "3min_transcript": "Let's do a few more examples where we're evaluating expressions with unknown variables. So this first one we're told 3x plus 3y plus 3z is equal to 1, and then we're asked what's 12x plus 12y plus 12z equal to? And I'll give you a few moments to think about that. Well let's rewrite this second expression by factoring out the 12, so we get 12 times x plus y plus z. That's this second expression here, and you can verify that by distributing the 12. You'll get exactly this right up here. Now, what is 12 times x plus y plus z? Well, we don't know yet exactly what x plus y plus z is equal to, but this first equation might help us. This first equation, we can rewrite this left-hand side by factoring out the 3, so we could rewrite this as 3 times x plus y plus z is equal to 1. All I did is I factored the 3 out on the left-hand side. I just divide both sides of this equation by 3, and I'm left with x plus y plus z is equal to 1/3, and so here, instead of x plus y plus z, I can write 1/3. So this whole thing simplified to 12 times 1/3. 12 times 1/3 is the same thing as 12 divided by 3, which is equal to 4. Let's try one more. So here we are told that 3a plus 5b is equal to 2, and then we're asked what's 15a plus 15b going to be equal to? So we might-- let's see. I'll give you a few moments to try to tackle this on your own. Let's see how we might do it. We could approach it the way we've approached the last few problems, trying to rewrite the second expression. We could rewrite it as 15 times a plus b, and so we just have to figure out what a plus b is, And so, it's tempting to look up here, and say maybe we can solve for a plus b somehow, but we really can't. If we divide-- if we try to factor out a 3, we'll get 3 times a plus 5/3b, so this doesn't really simplify things in terms of a plus b. If we try to factor out a 5, we'd get 5 times 3/5a plus b is equal to 2, but neither of these gets us in a form where we can then solve for a plus b. So in this situation, we actually do not have enough information to solve this problem. So it's a little bit of a trick. Not enough info to solve. Anyway, hopefully you enjoyed that." + }, + { + "Q": "WHat does recipocral mean? 3:01", + "A": "the reciprocal means the opposite of a number", + "video_name": "bAerID24QJ0", + "timestamps": [ + 181 + ], + "3min_transcript": "So rewriting it, if I had 5x equals 20, we could do two things and they're essentially the same thing. We could say we just divide both sides of this equation by 5, in which case, the left hand side, those two 5's will cancel out, we'll get x. And the right hand side, 20 divided by 5 is 4, and we would have solved it. Another way to do it, and this is actually the exact same way, we're just phrasing it a little different. If you said 5x equals 20, instead of dividing by 5, we could multiply by 1/5. And if you look at that, you can realize that multiplying by 1/5 is the same thing as dividing by 5, if you know the difference between dividing and multiplying fractions. And then that gets the same thing, 1/5 times 5 is 1, so you're just left with an x equals 4. because when we start having fractions instead of a 5, it's easier just to think about multiplying by the reciprocal. Actually, let's do one of those right now. So let's say I had negative 3/4 times x equals 10/13. Now, this is a harder problem. I can't do this one in my head. We're saying negative 3/4 times some number x is equal to 10/13. If someone came up to you on the street and asked you that, I think you'd be like me, and you'd be pretty stumped. But let's work it out algebraically. Well, we do the same thing. We multiply both sides by the coefficient on x. So the coefficient, all that is, all that fancy word means, is the number that's being multiplied by x. So what's the reciprocal of minus 3/4. to use times, and you're probably wondering why in algebra, there are all these other conventions for doing times as opposed to just the traditional multiplication sign. And the main reason is, I think, just a regular multiplication sign gets confused with the variable x, so they thought of either using a dot if you're multiplying two constants, or just writing it next to a variable to imply you're multiplying a variable. So if we multiply the left hand side by negative 4/3, we also have to do the same thing to the right hand side, minus 4/3. The left hand side, the minus 4/3 and the 3/4, they cancel out. You could work it out on your own to see that they do. They equal 1, so we're just left with x is equal to 10 times minus 4 is minus 40, 13 times 3, well, that's equal to 39." + }, + { + "Q": "At 4:20 how did you multiply the fraction -4/3 to the fraction 10/13.\nI was quite confused there.....", + "A": "To multiply fractions: 1) Multiply the numerators: -4 (10) = -40 2) Multiply the denominators: 3(13) = 39 3) This creates the fraction -40/39. 4) To change to a mixed number, divide -40 by 39. The remainder becomes the new numerator: -40/39 = -1 1/39", + "video_name": "bAerID24QJ0", + "timestamps": [ + 260 + ], + "3min_transcript": "because when we start having fractions instead of a 5, it's easier just to think about multiplying by the reciprocal. Actually, let's do one of those right now. So let's say I had negative 3/4 times x equals 10/13. Now, this is a harder problem. I can't do this one in my head. We're saying negative 3/4 times some number x is equal to 10/13. If someone came up to you on the street and asked you that, I think you'd be like me, and you'd be pretty stumped. But let's work it out algebraically. Well, we do the same thing. We multiply both sides by the coefficient on x. So the coefficient, all that is, all that fancy word means, is the number that's being multiplied by x. So what's the reciprocal of minus 3/4. to use times, and you're probably wondering why in algebra, there are all these other conventions for doing times as opposed to just the traditional multiplication sign. And the main reason is, I think, just a regular multiplication sign gets confused with the variable x, so they thought of either using a dot if you're multiplying two constants, or just writing it next to a variable to imply you're multiplying a variable. So if we multiply the left hand side by negative 4/3, we also have to do the same thing to the right hand side, minus 4/3. The left hand side, the minus 4/3 and the 3/4, they cancel out. You could work it out on your own to see that they do. They equal 1, so we're just left with x is equal to 10 times minus 4 is minus 40, 13 times 3, well, that's equal to 39. And I like to leave my fractions improper because it's easier to deal with them. But you could also view that-- that's minus-- if you wanted to write it as a mixed number, that's minus 1 and 1/39. I tend to keep it like this. Let's check to make sure that's right. The cool thing about algebra is you can always get your answer and put it back into the original equation to make sure you are right. So the original equation was minus 3/4 times x, and here we'll substitute the x back into the equation. Wherever we saw x, we'll now put our answer. So it's minus 40/39, and our original equation said that equals 10/13. Well, and once again, when I just write the 3/4 right next to the parentheses like that, that's just another way of writing times. So minus 3 times minus 40, it is minus 100--" + }, + { + "Q": "At 2:05, how did you get pi. When he said m 10 instead of 8r - 13 < 10. Hope this helped.", + "video_name": "x5EJG_rAtkY", + "timestamps": [ + 213 + ], + "3min_transcript": "This quantity right here has to be between negative 2 and 1/2. It has to be greater than negative 2 and 1/2 right there. And it has to be less than 2 and 1/2, so that's all I wrote there. So let's solve each of these independently. Well, this first went over here, you've learned before that I don't like improper fractions, and I don't like fractions in general. So let's make all of these fractions. Sorry, I don't like mixed numbers. I want them to be improper fractions. So let's turn all of these into improper fractions. So if I were to rewrite it, we get 2r minus 3 and 1/4 is the same thing as 3 times 4 is 12, plus 1 is 13. 2r minus 13/4 is less than-- 2 times 2 is 4, plus 1 is five-- is less than 5/2. So that's the first equation. And then the second question-- and do the same thing here-- we have 2r minus 13 over 4 has to be a greater All right, now let's solve each of these independently. To get rid of the fractions, the easiest thing to do is to multiply both sides of this equation by 4. That'll eliminate all of the fractions, so let's do that. Let's multiply-- let me scroll to the left a little bit-- let's multiply both sides of this equation by 4. 4 times 2r is 8r, 4 times negative 13 over 4 is negative 13, is less than-- and I multiplied by a positive number so I didn't have to worry about swapping the inequality-- is less than 5/2 times 4 is 10, right? You get a 2 and a 1, it's 10. So you get 8r minus 13 is less than 10. Now we can add 13 to both sides of this equation so that we get rid of it on the left-hand side. Add 13 to both sides and we get 8r-- these guys cancel out-- is less than 23, and then we divide And once again, we didn't have to worry about the inequality because we're dividing by a positive number. And we get r is less than 23 over 8. Or, if you want to write that as a mixed number, r is less than-- what is that-- 2 and 7/8. So that's one condition, but we still have to worry about this other condition. There was an and right here. Let's worry about it. So our other condition tells us 2r minus 13 over 4 has to be greater than negative 5/2. Let's multiply both sides of this equation by 4. So 4 times 2r is 8r. 4 times negative 13 over 4 is negative 13, is greater than negative 5/2 times 4 is negative 10. Now we add 13 to both sides of this equation. The left-hand side-- these guys cancel out, you're just" + }, + { + "Q": "I'm curious: In the first problem, why didn't we rewrite y=x^x as \"log base x of y = x\" and take the derivative of that? I tried it, but I get a different answer (I get y*ln(x) ).\n\nThen, another question: at 5:09 Sal says \"If we haven't solved this, you can just keep taking the natural log of this...\" - I tried that ( ln(ln(y))=xln(x)+ln(ln(x)) ), and I get a different answer (I get y*ln(y)*(ln(x)+1+1/(x*ln(x)) ).\n\nWhat am I doing wrong?", + "A": "y = x^x ln(y) = ln(x^x) ln(y) = x\u00e2\u0080\u00a2ln(x) d/dx(ln(y)) = d/dx(x\u00e2\u0080\u00a2ln(x)) d/dx(ln(y)) = d/dx(x)\u00e2\u0080\u00a2ln(x) + x\u00e2\u0080\u00a2d/dx(ln(x)) d/dx(ln(y)) = d/dx(x)\u00e2\u0080\u00a2ln(x) + x\u00e2\u0080\u00a21/x\u00e2\u0080\u00a2d/dx(x) d/dx(ln(y)) = d/dx(x)\u00e2\u0080\u00a2ln(x) + x\u00e2\u0080\u00a21/x\u00e2\u0080\u00a2dx/dx d/dx(ln(y)) = d/dx(x)\u00e2\u0080\u00a2ln(x) + x\u00e2\u0080\u00a21/x d/dx(ln(y)) = d/dx(x)\u00e2\u0080\u00a2ln(x) + 1 d/dx(ln(y)) = dx/dx\u00e2\u0080\u00a2ln(x) + 1 d/dx(ln(y)) = ln(x) + 1 1/y\u00e2\u0080\u00a2d/dx(y) = ln(x) + 1 1/y\u00e2\u0080\u00a2dy/dx = ln(x) + 1 dy/dx = y\u00e2\u0080\u00a2(ln(x) + 1) y = x^x dy/dx = [x^x\u00e2\u0080\u00a2(ln(x) + 1)]", + "video_name": "N5kkwVoAtkc", + "timestamps": [ + 309 + ], + "3min_transcript": "that's what we're going to tackle in this. But it's good to see this problem done first because it gives us the basic tools. So the more difficult problem we're going to deal with is this one. Let me write it down. So the problem is y is equal to x to the-- and here's the twist-- x to the x to the x. And we want to find out dy/dx. We want to find out the derivative of y with respect to x. So to solve this problem we essentially use the same tools. We use the natural log to essentially breakdown this exponent and get it into terms that we can deal with. So we can use the product rule. So let's take the natural log of both sides of this equation like we did last time. You get the natural log of y is equal to the natural log of x to the x to the x. So we can rewrite this as x to the x times the natural log times the natural log of x. So now our expression our equation is simplified to the natural log of y is equal to x to the x times the natural log of x. But we still have this nasty x to the x here. We know no easy way to take the derivative there, although I've actually just shown you what the derivative of this is, so we could actually just apply it right now. I was going to take the natural log again and it would turn into this big, messy, confusing thing but I realized that earlier in this video I just solved for what the derivative of x to the x is. It's this thing right here. It's this crazy expression right here. So we just have to remember that and then apply and then do our problem. So let's do our problem. an unexpected benefit of doing the simpler version of the problem, you could just keep taking the natural log of this, but it'll just get a little bit messier. But since we already know what the derivative of x to the x is, let's just apply it. So we're going to take the derivative of both sides of the equation. Derivative of this is equal to the derivative of this. We'll ignore this for now. Derivative of this with respect to x is the derivative of the natural log of y with respect to y. So that's 1/y times the derivative of y That's just the chain rule. We learned that in implicit differentiation. And so this is equal to the derivative of the first term times the second term, and I'm going to write it out here just because I don't want to skip steps and confuse people. So this is equal to the derivative with respect to x of x to the x times the natural log of x plus the derivative" + }, + { + "Q": "4:24 how does sal equate D/V from V=D/T?", + "A": "V = D/T Multiply both sides by T: VT = D Divide both sides by V: T = D/V", + "video_name": "Uc2Tm4Lr7uI", + "timestamps": [ + 264 + ], + "3min_transcript": "at just 8 miles per hour.\" So we're given a time. And we are given a speed. We should be able to figure out a distance. So let's just do a little bit of aside here. We should be able to figure out the distance to the-- actually let me write it this way. The distance to the store will be equal to-- now we've got to make sure we have our units right. Here they gave it in minutes. Here they have 8 miles per hour. So let's convert this into hours. So 45 minutes in hours, so it's 45 minutes out of 60 minutes per hour. So that's going to give us 45/60. Divide both by 15. That's the same thing as 3/4. So it's going to be 3/4 hours is the time times an average speed So what is the distance to the store? Well, 3/4 times 8. Or you could view it as 3/4 times 8 times 1, is going to be-- well, it's going to be 24 over 4. Let me just write that. That's going to be 24 over 4 which is equal to-- did I get it? Yeah. 24 over 4, which is equal to 6. And units-wise, we're just left with miles. So the distance to the store is 6 miles. 2 times the distance to the gift store, well, this whole thing is going to be 12 miles. 12 miles is the total distance she traveled. Now, what is the time to the gift store? Well, they already told that to us. They already told us that it's 45 minutes. Now, I want to put everything in hours. I'm assuming that they want our average speed in hours. So the time to the gift store was 3/4 of an hour. And what's the time coming back from the gift store? Well, we know her speed. We know her speed coming back. We already know the distance from the gift store. It's the same as the distance to the gift store. So we can take this distance, we can take 6 miles, that's the distance to the gift store, 6 miles divided by her speed coming back, which is 24 miles per hour, so divided by 24 miles per hour. It gives us-- well, let's see. We're going to have 6 over 24 is the same thing as 1/4. It's going to be 1/4. And then miles divided by miles per hour is the same thing as miles times hours per mile. The miles cancel out. And you'll have 1/4 of an hour. So it takes her 1/4 of an hour to get back. And that fits our intuition." + }, + { + "Q": "At 2:58 he says 3/4, Can anyone tell me where he got that from? And also where'd he get 2(distance to gift store) from?\n\nThank you :)", + "A": "45 minutes * 1 hour/60 minutes = 45/60 hours (since minutes cancel out). Both 45 and 60 are divisible by 15 so you can simplify it to 3/4 (hours). The distance to the gift store is the same as the distance from the gift store since he is taking the same path in both directions, so you can view it as 2 * distance to the gift store.", + "video_name": "Uc2Tm4Lr7uI", + "timestamps": [ + 178 + ], + "3min_transcript": "Well, it's time to gift store plus the time coming back from the gift store. Now, we know that the distance to the gift store and the distance back from the gift store is the same. So that's why I just said that the total distance is just going to be two times the distance to the gift store. We don't know-- in fact we know we're going to have different times in terms of times to the gift store and times coming back. How do I know that? Well, she went at different speeds. So it's going to take her-- actually, she went there much slower than she came back. So it would take her longer to get there than it took her to get back. So let's see which of these we can actually-- we already know. So how do we figure out the distance to the gift store? At not point here do they say, hey, the gift store is this far away. But they do tell us, this first sentence right over here, at just 8 miles per hour.\" So we're given a time. And we are given a speed. We should be able to figure out a distance. So let's just do a little bit of aside here. We should be able to figure out the distance to the-- actually let me write it this way. The distance to the store will be equal to-- now we've got to make sure we have our units right. Here they gave it in minutes. Here they have 8 miles per hour. So let's convert this into hours. So 45 minutes in hours, so it's 45 minutes out of 60 minutes per hour. So that's going to give us 45/60. Divide both by 15. That's the same thing as 3/4. So it's going to be 3/4 hours is the time times an average speed So what is the distance to the store? Well, 3/4 times 8. Or you could view it as 3/4 times 8 times 1, is going to be-- well, it's going to be 24 over 4. Let me just write that. That's going to be 24 over 4 which is equal to-- did I get it? Yeah. 24 over 4, which is equal to 6. And units-wise, we're just left with miles. So the distance to the store is 6 miles. 2 times the distance to the gift store, well, this whole thing is going to be 12 miles. 12 miles is the total distance she traveled. Now, what is the time to the gift store? Well, they already told that to us. They already told us that it's 45 minutes. Now, I want to put everything in hours. I'm assuming that they want our average speed in hours." + }, + { + "Q": "At 3:05, how do you convert the minutes into hours?", + "A": "multiply by 60 since there are 60 minutes in an hour. Hope this helps!", + "video_name": "Uc2Tm4Lr7uI", + "timestamps": [ + 185 + ], + "3min_transcript": "Well, it's time to gift store plus the time coming back from the gift store. Now, we know that the distance to the gift store and the distance back from the gift store is the same. So that's why I just said that the total distance is just going to be two times the distance to the gift store. We don't know-- in fact we know we're going to have different times in terms of times to the gift store and times coming back. How do I know that? Well, she went at different speeds. So it's going to take her-- actually, she went there much slower than she came back. So it would take her longer to get there than it took her to get back. So let's see which of these we can actually-- we already know. So how do we figure out the distance to the gift store? At not point here do they say, hey, the gift store is this far away. But they do tell us, this first sentence right over here, at just 8 miles per hour.\" So we're given a time. And we are given a speed. We should be able to figure out a distance. So let's just do a little bit of aside here. We should be able to figure out the distance to the-- actually let me write it this way. The distance to the store will be equal to-- now we've got to make sure we have our units right. Here they gave it in minutes. Here they have 8 miles per hour. So let's convert this into hours. So 45 minutes in hours, so it's 45 minutes out of 60 minutes per hour. So that's going to give us 45/60. Divide both by 15. That's the same thing as 3/4. So it's going to be 3/4 hours is the time times an average speed So what is the distance to the store? Well, 3/4 times 8. Or you could view it as 3/4 times 8 times 1, is going to be-- well, it's going to be 24 over 4. Let me just write that. That's going to be 24 over 4 which is equal to-- did I get it? Yeah. 24 over 4, which is equal to 6. And units-wise, we're just left with miles. So the distance to the store is 6 miles. 2 times the distance to the gift store, well, this whole thing is going to be 12 miles. 12 miles is the total distance she traveled. Now, what is the time to the gift store? Well, they already told that to us. They already told us that it's 45 minutes. Now, I want to put everything in hours. I'm assuming that they want our average speed in hours." + }, + { + "Q": "in the first example, at 2:11 you have simplified it to -sqrt5/sqrt35, the proceed to rewrite this as -sqrt(5/35) and end with the result of -sqrt(1/7).\n\nWhen solving this on my own, after i reached the step you were at at 2:11, I simplified it to:\n-sqrt5 / (sqrt7)(sqrt5) and cancelled the sqrt5 to end with -1/sqrt(7).\n\nI believe these are both correct, as your example of the whole fraction under the radical could simplify to my form, however which would be considered the most simple answer?", + "A": "They are both equivalent. It depends on how your instructor wants the answer to be really. Both are simple enough. Later on you ll learn to rationalize the denominator, so you ll most likely require to do that. You simply multiply - 1/\u00e2\u0088\u009a(7) by \u00e2\u0088\u009a(7) / \u00e2\u0088\u009a(7) which will give -\u00e2\u0088\u009a(7)/7. P.S. He mas an error, forgetting to include his negative sign.", + "video_name": "suwJmCrSDI8", + "timestamps": [ + 131, + 131 + ], + "3min_transcript": "- [Voiceover] We're asked to simplify the expression by removing all factors that are perfect squares from inside the radicals and combining the terms. So, let's see if we can do it and pause the video and give a-go at it before we do it together. Alright, so let's see how we can re-write these radicals. So, four times the square root of 20. Well, thats the same thing as four times the square root of four, times the square root of five, because 20 is the same thing as four times five. And 45, that's the same thing as nine times five. And the reason why I'm thinking about the four and I'm thinking about the nine is because those are perfect squares, so I could write this as, four times the square root of four times the square root of five. And then I could say, this part right over here is minus three, times the square root of nine, times the square root of five. The square root of 45 is the same thing as the square root of nine times five, which is the same thing as square root of nine times the square root of five, and then all of that is going to be over, Now, are there any perfect squares in 35? 35 is seven times five. No, neither of those are perfect squares. So, I could just leave that as square root of 35. And, let's see, the square root of four? Well, thats going to be two. This is the principal root, so we're thinking about the positive square root. The square root of nine is three. And so, this part right over is going to be four times two, times the square root of five, so it's going to be eight square roots of five. And then, this part over here is going to be minus three, times three, times the square root of five. So, minus nine square roots of five. And all of that is going to be over the square root of 35. Square root of 35. And, so let's see, if I have eight of something, and I subtract nine of that something, I'm gonna have negative one of that something. or I could just say negative square root of five. Negative square root of five over the square root of 35. I actually think I could simplify this even more, because this is the same thing. This is equal to the negative of the square root of five over 35. Both the numerator and the denominator are divisible by five. So, we could divide them both by five, and we would get the square root of divide the numerator by five, you get one. Divide the denominator by five, you get seven. So, we can view this as the square root of 1/7th. Square root of 1/7th. And we are all done. Let's do another one of these. These are strangely, strangely fun. And once again, pause it, and see if you can work it out on your own. Perform the indicated operations. Alright, so let's first multiply. So, this essentially is doing the distributive property twice. And, actually let me just do it that way," + }, + { + "Q": "At 5:30 sal says that the limit of f(c) as x approaches from the negative direction does not exist.\nHow is it?\nWhat does the point over the hole mean?", + "A": "Think of a simpler formula, where if x < c, then f(x) = 1; and where x >= c then f(x) = 2. If you are determining the limit from the left (ie the lower direction) your limit would equal 1, but the value for c derived from f(x) is actually 2 - thus the limit does not exist, because: lim f(x) (x->c-) = 1 f(c) = 2 -John", + "video_name": "kdEQGfeC0SE", + "timestamps": [ + 330 + ], + "3min_transcript": "" + }, + { + "Q": "At 3:48 I don't understand how the 2 gets under c^2 and disappears from the other side. Is it being added, subtracted, multiplied, or divided?", + "A": "The 2 dissapears on the left side because the expression is divided by 2 on both sides.", + "video_name": "tSHitjFIjd8", + "timestamps": [ + 228 + ], + "3min_transcript": "So the only right triangle in which the other two angles are equal is a 45-45-90 triangle. So what's interesting about a 45-45-90 triangle? Well other than what I just told you-- let me redraw it. I'll redraw it like this. So we already know this is 90 degrees, this is 45 degrees, this is 45 degrees. And based on what I just told you, we also know that the sides that the 45 degree angles don't share are equal. So this side is equal to this side. And if we're viewing it from a Pythagorean theorem point of view, this tells us that the two sides that are not the hypotenuse are equal. So this is a hypotenuse. We know from the Pythagorean theorem-- let's say the hypotenuse is equal to C-- the Pythagorean theorem tells us that A squared plus B squared is equal to C squared. Right? Well we know that A equals B, because this is a 45-45-90 triangle. So we could substitute A for B or B for A. But let's just substitute B for A. So we could say B squared plus B squared is equal to C squared. Or 2B squared is equal to C squared. Or B squared is equal to C squared over 2. Or B is equal to the square root of C squared over 2. the numerator and the square root of the denominator-- C over the square root of 2. And actually, even though this is a presentation on triangles, I'm going to give you a little bit of actually information on something called rationalizing denominators. So this is perfectly correct. We just arrived at B-- and we also know that A equals B-- but that B is equal to C divided by the square root of 2. It turns out that in most of mathematics, and I never understood quite exactly why this was the case, people don't like square root of 2s in the denominator. Or in general they don't like irrational numbers in the denominator. Irrational numbers are numbers that have decimal places that never repeat and never end. So the way that they get rid of irrational numbers in the denominator is that you do something called rationalizing the denominator. And the way you rationalize a denominator-- let's take our example right now. If we had C over the square root of 2, we just multiply both the numerator and the denominator by the" + }, + { + "Q": "at 8:21 I don't understand wouldn't you divide instead of multiply?", + "A": "b/c you re multiplying it by the inverse", + "video_name": "tSHitjFIjd8", + "timestamps": [ + 501 + ], + "3min_transcript": "So what did we learn? This is equal to B, right? So turns out that B is equal to C times the square root of 2 over 2. So let me write that. So we know that A equals B, right? And that equals the square root of 2 over 2 times C. Now you might want to memorize this, though you can always derive it if you use the Pythagorean theorem and remember that the sides that aren't the hypotenuse in a 45-45-90 triangle are equal to each other. But this is very good to know. Because if, say, you're taking the SAT and you need to solve a problem really fast, and if you have this memorized and someone gives you the hypotenuse, you can figure out what are the sides very fast, or i8f someone gives you one of the sides, you can figure out the hypotenuse very fast. Let's try that out. I'm going to erase everything. So we learned just now that A is equal to B is equal to the So if I were to give you a right triangle, and I were to tell you that this angle is 90 and this angle is 45, and that this side is, let's say this side is 8. I want to figure out what this side is. Well first of all, let's figure out what side is the hypotenuse. Well the hypotenuse is the side opposite the right angle. So we're trying to actually figure out the hypotenuse. Let's call the hypotenuse C. And we also know this is a 45-45-90 triangle, right? Because this angle is 45, so this one also has to be 45, because 45 plus 90 plus 90 is equal to 180. So this is a 45-45-90 triangle, and we know one of the sides-- this side could be A or B-- we know that 8 is equal to the C is what we're trying to figure out. So if we multiply both sides of this equation by 2 times the square root of 2-- I'm just multiplying it by the inverse of the coefficient on C. Because the square root of 2 cancels out that square root of 2, this 2 cancels out with this 2. We get 2 times 8, 16 over the square root of 2 equals C. Which would be correct, but as I just showed you, people don't like having radicals in the denominator. So we can just say C is equal to 16 over the square root of 2 times the square root of 2 over the square root of 2. So this equals 16 square roots of 2 over 2. Which is the same thing as 8 square roots of 2." + }, + { + "Q": "So at 1:16 I understand that the slope is rise over run and that the line has a negative slope because its going down, but why is going right positive for x and left is negative?", + "A": "on a graph onthe x axis the right side is positive while the left is negative", + "video_name": "AqFwKecNaTk", + "timestamps": [ + 76 + ], + "3min_transcript": "A line has a slope of negative 3/4 and goes through the point 0 comma 8. What is the equation of this line in slope-intercept form? So any line can be represented in slope-intercept form, is y is equal to mx plus b, where this m right over here, that is of the slope of the line. And this b over here, this is the y-intercept of the line. Let me draw a quick line here just so that we can visualize that a little bit. So that is my y-axis. And then that is my x-axis. And let me draw a line. And since our line here has a negative slope, I'll draw a downward sloping line. So let's say our line looks something like that. So hopefully, we're a little familiar with the slope already. The slope essentially tells us, look, start at some point on the line, and go how much you had to move in the x direction, that is your run, and then measure how much you had to move in the y direction, that is your rise. And our slope is equal to rise over run. And you can see over here, we'd be downward sloping. Because if you move in the positive x direction, we have to go down. If our run is positive, our rise here is negative. So this would be a negative over a positive, it would give you a negative number. That makes sense, because we're downward sloping. The more we go down in this situation, for every step we move to the right, the more downward sloping will be, the more of a negative slope we'll have. So that's slope right over here. The y-intercept just tells us where we intercept the y-axis. So the y-intercept, this point right over here, this is where the line intersects with the y-axis. This will be the point 0 comma b. And this actually just falls straight out of this equation. this equation, when x is equal to 0. y will be equal to m times 0 plus b. Well, anything times 0 is 0. So y is equal to 0 plus b, or y will be equal to b, when x is equal to 0. So this is the point 0 comma b. Now, they tell us what the slope of this line is. They tell us a line has a slope of negative 3/4. So we know that our slope is negative 3/4, and they tell us that the line goes through the point 0 comma 8. They tell us we go through the-- Let me just, in a new color. I've already used orange, let me use this green color. They tell us what we go through the point 0 comma 8. Notice, x is 0. So we're on the y-axis. When x is 0, we're on the y-axis." + }, + { + "Q": "5:01 what are assymptotes? (I don't think I spelled that right)\nSal keeps mentioning them but I think if I don't know what they are I'm not going anywhere with hyperbolas....", + "A": "an asymptote is a line that the function gets infinitely close to but never touches. You draw them on your graph before you draw your hyperbola (using dotted lines).", + "video_name": "pzSyOTkAsY4", + "timestamps": [ + 301 + ], + "3min_transcript": "you'll probably get confused. Because sometimes they always use the a under the x and the b under the y, or sometimes they always use the a under the positive term and to b under the negative term. So if you just memorize, oh, a divided by b, that's the slope of the asymptote and all of that, you might be using the wrong a and b. So I encourage you to always re-prove it to yourself. And that's what we're going to do right here. It actually doesn't take too long. So these are both hyperbolas. And what I like to do whenever I have a hyperbola is solve for y. So in this case, if I subtract x squared over a squared from both sides, I get-- let me change the color-- I get minus y squared over b squared. That stays there. Is equal to 1 minus x squared over a squared. And then, let's see, I want to get rid of this minus, and I want to get rid of this b squared. So let's multiply both sides of this equation times minus b squared. minus and the b squared go away, and you're just left with y squared is equal to minus b squared. And then minus b squared times a plus, it becomes a plus b squared over a squared x squared. We're almost there. And then you get y is equal to-- and I'm doing this on purpose-- the plus or minus square root, because it can be the plus or minus square root. Of-- and let's switch these around, just so I have the positive term first. b squared over a squared x squared minus b squared. Now you said, Sal, you said this was simple. This looks like a really complicated thing. But remember, we're doing this to figure out asymptotes of the hyperbola, just to kind of give you a sense of where we're going. that other hyperbola. So a hyperbola, if that's the x, that's the y-axis, it has two asymptotes. And the asymptotes, they're these lines that the hyperbola will approach. So if those are the two asymptotes-- and they're always the negative slope of each other-- we know that this hyperbola's is either, and we'll show in a second which one it is, it's either going to look something like this, where as we approach infinity we get closer and closer this line and closer and closer to that line. And here it's either going to look like that-- I didn't draw it perfectly; it never touches the asymptote. It just gets closer and closer and closer, arbitrarily It's either going to look like that, where it opens up to the right and left. Or our hyperbola's going to open up and down. And once again, as you go further and further, and asymptote means it's just going to get closer and closer to one of these lines without ever touching it. It will get infinitely close as you get infinitely far away," + }, + { + "Q": "at 5:15 sal says that the asymptotes are the negative slope of each other. can i say that they are perpendicular lines with the same y-intercept?", + "A": "yes for the perpendicular but no for same y intercept, because in more complex examples there will be horizontal and vertical shifts which will make the y intercepts of each asymptote different. Try searching for examples of hyperbolas with shifts and see their graphs to understand more.", + "video_name": "pzSyOTkAsY4", + "timestamps": [ + 315 + ], + "3min_transcript": "you'll probably get confused. Because sometimes they always use the a under the x and the b under the y, or sometimes they always use the a under the positive term and to b under the negative term. So if you just memorize, oh, a divided by b, that's the slope of the asymptote and all of that, you might be using the wrong a and b. So I encourage you to always re-prove it to yourself. And that's what we're going to do right here. It actually doesn't take too long. So these are both hyperbolas. And what I like to do whenever I have a hyperbola is solve for y. So in this case, if I subtract x squared over a squared from both sides, I get-- let me change the color-- I get minus y squared over b squared. That stays there. Is equal to 1 minus x squared over a squared. And then, let's see, I want to get rid of this minus, and I want to get rid of this b squared. So let's multiply both sides of this equation times minus b squared. minus and the b squared go away, and you're just left with y squared is equal to minus b squared. And then minus b squared times a plus, it becomes a plus b squared over a squared x squared. We're almost there. And then you get y is equal to-- and I'm doing this on purpose-- the plus or minus square root, because it can be the plus or minus square root. Of-- and let's switch these around, just so I have the positive term first. b squared over a squared x squared minus b squared. Now you said, Sal, you said this was simple. This looks like a really complicated thing. But remember, we're doing this to figure out asymptotes of the hyperbola, just to kind of give you a sense of where we're going. that other hyperbola. So a hyperbola, if that's the x, that's the y-axis, it has two asymptotes. And the asymptotes, they're these lines that the hyperbola will approach. So if those are the two asymptotes-- and they're always the negative slope of each other-- we know that this hyperbola's is either, and we'll show in a second which one it is, it's either going to look something like this, where as we approach infinity we get closer and closer this line and closer and closer to that line. And here it's either going to look like that-- I didn't draw it perfectly; it never touches the asymptote. It just gets closer and closer and closer, arbitrarily It's either going to look like that, where it opens up to the right and left. Or our hyperbola's going to open up and down. And once again, as you go further and further, and asymptote means it's just going to get closer and closer to one of these lines without ever touching it. It will get infinitely close as you get infinitely far away," + }, + { + "Q": "At 7:40, Sal got rid of the -b^2, how is that possible?", + "A": "As x approaches infinity, the b\u00c2\u00b2 term becomes less and less significant. For example, consider x = 1,000,000 and b = 5 .", + "video_name": "pzSyOTkAsY4", + "timestamps": [ + 460 + ], + "3min_transcript": "So in order to figure out which one of these this is, let's just think about what happens as x becomes infinitely large. So as x approaches infinity. So as x approaches infinity, or x approaches negative infinity. So I'll say plus or minus infinity, right? It doesn't matter, because when you take a negative, this gets squared. So this number becomes really huge as you approach positive or negative infinity. And you'll learn more about this when we actually do limits, but I think that's intuitive. That this number becomes huge. This number's just a constant. It just stays the same. So as x approaches positive or negative infinity, as it gets really, really large, y is going to be approximately equal to-- actually, I think that's congruent. I always forget notation. Approximately. This just means not exactly but approximately equal to. When x approaches infinity, it's going to be approximately equal to the plus or minus square root of b squared And that is equal to-- now you can take the square root. You couldn't take the square root of this algebraically, but this you can. This is equal to plus or minus b over a x. So that tells us, essentially, what the two asymptotes are. Where the slope of one asymptote will be b over a x. This could give you positive b over a x, and the other one would be minus b over a x. And I'll do this with some example so it makes it a little clearer. But we still know what the asymptotes look like. It's these two lines. Because it's plus b a x is one line, y equals plus b a x. Let's say it's this one. This asymptote right here is y is equal to plus b over a x. I know you can't read that. And then the downward sloping asymptote we could say is y is equal to minus b over a x. So those are two asymptotes. But we still have to figure out whether the hyperbola opens up And there, there's two ways to do this. One, you say, well this is an approximation. This is what you approach as x approaches infinity. But we see here that even when x approaches infinity, we're always going to be a little bit smaller than that number. Because we're subtracting a positive number from this. We're subtracting a positive number, and then we're taking the square root of the whole thing. So we're always going to be a little bit lower than the asymptote, especially when we're in the positive quadrant. So to me, that's how I like to do it. I think, we're always-- at least in the positive quadrant; it gets a little more confusing when you go to the other quadrants-- we're always going to be a little bit lower than the asymptote. So we're going to approach from the bottom there. And since you know you're there, you know it's going to be like this and approach this asymptote. And then since it's opening to the right here, it's also going to open to the left. The other way to test it, and maybe this is more intuitive for you, is to figure out, in the original equation could x or y equal to 0?" + }, + { + "Q": "At 4:25 when multiplying b^2, why does the x^2 get moved from the numerator?", + "A": "x^2 is still part of the numerator - just think of it as x^2/1, multiplied by b^2/a^2. you could also write it as a^2*x^2/b^2, all as one fraction... it means the same thing (multiply x^2 and a^2 and divide by b^2 ->> since multiplication and division occur at the same level of the order of operations, both ways of writing it out are totally equivalent!). hope that helps", + "video_name": "pzSyOTkAsY4", + "timestamps": [ + 265 + ], + "3min_transcript": "minus y squared over b squared is equal to 1. And notice the only difference between this equation and this one is that instead of a plus y squared, we have a minus y squared here. So that would be one hyperbola. The other one would be if the minus sign was the other way around. If it was y squared over b squared minus x squared over a squared is equal to 1. So now the minus is in front of the x squared term instead of the y squared term. And what I want to do now is try to figure out, how do we graph either of these parabolas? Maybe we'll do both cases. And in a lot of text books, or even if you look it up over the web, they'll give you formulas. But I don't like those formulas. One, because I'll always forget it. And you'll forget it immediately after taking the test. You might want to memorize it if you just want to be able to do the test a little bit faster. But you'll forget it. you'll probably get confused. Because sometimes they always use the a under the x and the b under the y, or sometimes they always use the a under the positive term and to b under the negative term. So if you just memorize, oh, a divided by b, that's the slope of the asymptote and all of that, you might be using the wrong a and b. So I encourage you to always re-prove it to yourself. And that's what we're going to do right here. It actually doesn't take too long. So these are both hyperbolas. And what I like to do whenever I have a hyperbola is solve for y. So in this case, if I subtract x squared over a squared from both sides, I get-- let me change the color-- I get minus y squared over b squared. That stays there. Is equal to 1 minus x squared over a squared. And then, let's see, I want to get rid of this minus, and I want to get rid of this b squared. So let's multiply both sides of this equation times minus b squared. minus and the b squared go away, and you're just left with y squared is equal to minus b squared. And then minus b squared times a plus, it becomes a plus b squared over a squared x squared. We're almost there. And then you get y is equal to-- and I'm doing this on purpose-- the plus or minus square root, because it can be the plus or minus square root. Of-- and let's switch these around, just so I have the positive term first. b squared over a squared x squared minus b squared. Now you said, Sal, you said this was simple. This looks like a really complicated thing. But remember, we're doing this to figure out asymptotes of the hyperbola, just to kind of give you a sense of where we're going." + }, + { + "Q": "at 2:30, wouldn't those equations result in the same shape? if not, how are they different?", + "A": "the equation x2/a2-y2/b2 has the x-axis as the major axis while the equation y2/b2-x2/a2 has the y-axis as the major axis.", + "video_name": "pzSyOTkAsY4", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's see if we can learn a thing or two about the hyperbola. And out of all the conic sections, this is probably the one that confuses people the most, because it's not quite as easy to draw as the circle and the ellipse. You have to do a little bit more algebra. But hopefully over the course of this video you'll get pretty comfortable with that, and you'll see that hyperbolas in some way are more fun than any of the other conic sections. So just as a review, I want to do this just so you see the similarity in the formulas or the standard form of the different conic sections. If you have a circle centered at 0, its equation is x squared plus y squared is equal to r squared. And we saw that this could also be written as-- and I'm doing this because I want to show that this is really just the same thing as the standard equation for an ellipse. If you divide both sides of this by r squared, you get x squared over r squared plus y squared over r squared And so this is a circle. And once again, just as review, a circle, all of the points on the circle are equidistant from the center. Or in this case, you can kind of say that the major axis and the minor axis are the same distance, that there isn't any distinction between the two. You're always an equal distance away from the center. So that was a circle. An ellipse was pretty much this, but these two numbers could be different. Because your distance from the center could change. So it's x squared over a squared plus y squared over b squared is equal to 1. That's an ellipse. And now, I'll skip parabola for now, because parabola's kind of an interesting case, and you've already touched on it. So I'll go into more depth in that in a future video. But a hyperbola is very close in formula to this. And so there's two ways that a hyperbola could be written. And I'll do those two ways. minus y squared over b squared is equal to 1. And notice the only difference between this equation and this one is that instead of a plus y squared, we have a minus y squared here. So that would be one hyperbola. The other one would be if the minus sign was the other way around. If it was y squared over b squared minus x squared over a squared is equal to 1. So now the minus is in front of the x squared term instead of the y squared term. And what I want to do now is try to figure out, how do we graph either of these parabolas? Maybe we'll do both cases. And in a lot of text books, or even if you look it up over the web, they'll give you formulas. But I don't like those formulas. One, because I'll always forget it. And you'll forget it immediately after taking the test. You might want to memorize it if you just want to be able to do the test a little bit faster. But you'll forget it." + }, + { + "Q": "But when we are graphing g(x)=2-x at 8:00, should't the line be moved more upwards on a y-axis?\nThank you!", + "A": "If you notice 2 is the y-intercepts in the equation y = -x + 2 in the form y = mx + q", + "video_name": "rOftmuhGLjY", + "timestamps": [ + 480 + ], + "3min_transcript": "that's when x is greater than or equal to zero. When the absolute value is non-negative, if you're taking the absolute value of a non-negative number that is just going to be itself. The absolute value of zero, zero. Absolute value of one is one. The absolute value of a hundred is a hundred. Then you could ignore the absolute value for x is greater than or equal to, not greater than or equal to zero, for x is greater than or equal to one. X is greater than or equal to one, this thing right over here is non-negative. It will just evaluate to x minus one. This is going to be x minus one plus one. Which is the same thing as just x, minus one plus one, they just cancel out. Now when this term right over here is negative and that's going to happen for x is less than one. Well then the absolute value You give me the absolute value of a negative number that's going to be the opposite. Absolute value of negative eight is positive eight. It's going to be that the negative of x minus one is one minus x, plus one. Or we could say two minus x. For x is greater or equal to one, we would look at this expression, now what's the slope of that? Well the slope of that is one. We're going to have a curve that looks like or a line I guess we could say that looks like this. For all x is greater than or equal to one. The important thing, remember, we're going to think about the slope of the tangent line when we think about the derivative of g. Slope is equal to one. For x less than one or our slope now, if we look right over here our slope is negative one. It's going to look like this. For the pointing question, if we're thinking about g prime of nine so nine is some place out here, so what is g prime of nine? G prime of nine, let me make it clear, this graph right over here, this is the graph of g of x or we could say y, this is the graph y equals g of x. Y is equal to g of x. What is g prime of nine? Well that's the slope when x is equal to nine. The slope is going to be equal to one. G prime of nine is one. What does this evaluate to? This is going to be three times three, so this part right over here is nine plus two times one, plus two, which is equal to 11. The slope of the tangent line of h when x is equal to nine is 11." + }, + { + "Q": "At 3:21, how Sal got 1.5 the original population, did he assumed that or there is some logic behind it.", + "A": "He said he wanted the population to increase by 50% over 20 years. So, it ll be 150% of the current population, or 1.5.", + "video_name": "-fIsaqN-aaQ", + "timestamps": [ + 201 + ], + "3min_transcript": "This logistic function. This logistic function is a nonconstant solution, and it's the interesting one we care about if we're going to model population to the logistic differential equation. So now that we've done all that work to come up with this, let's actually apply it. That was the whole goal, was to model population growth. So let's come up with some assumptions. Let's first think about, well let's say that I have an island. So let's say that this is my island, and I start settling it with a 100 people. So I'm essentially saying N naught So I'm saying N naught is equal to 100. Let's say that this environment, given current technology of farming and agriculture, and the availability of water and whatever else, let's say it can only support 1,000 people max. So you get the idea, so we get K is equal to 1000. That's the limit to the population. So now what we have to think about is what is r going to be? So we have to come up with some assumptions. So, let's say in a generation which is about 20 years, well I'll just assume in 20 years, yeah I think it's reasonable that the population grows by, let's say that the population grows by 50%. In 20 years you have 50% growth. 50% increase, increase in the actual population. in order to after 20 years to grow by 50%. Well to think about that I'll get out my calculator. One way to think about it, growing by 50%, that means that you are at 1.5 your original population, and if I take that to the 120th power, and we'll just do 1 divided by 20th. This essentially says how much am I going to grow by or what is going to, this is telling me I'm going to grow by a factor of 1.02 every year, 1.02048. So one way to think about it is if every year I grow by 0.020 I'll just round five then over 20 years" + }, + { + "Q": "Around 4:15 when he found du, why did he keep the ln(2) if it is a constant? Isnt the derivative of a constant 0? Or did he use the product rule?", + "A": "yes he did use the product rule.", + "video_name": "1ct7LUx23io", + "timestamps": [ + 255 + ], + "3min_transcript": "see if we can figure that out. Then we can apply, then we can take, we can evaluate the definite ones. Let's just think about this, let's think about the indefinite integral of x squared times two to the x to the third power d x. I really want to find the anti-derivative of this. Well this is going to be the exact same thing as the integral of, I'll write my x squared still, but instead of two to the x to the third I'm going to write all of this business. Let me just copy and paste that. We already established that this is the same thing as two to the x to the third power. Copy and paste, just like that. Then let me close it with a d x. I was able to get it in terms of e as a base. That makes me a little bit more comfortable but it still seems pretty complicated. You might be saying, \"Okay, look. \"Maybe u substitution could be at play here.\" Because I have this crazy expression, x to the third times Well that's going to be three x squared times the natural log of two, or three times the natural log of two times x squared. That's just a constant times x squared. We already have a x squared here so maybe we can engineer this a little bit to have the constant there as well. Let's think about that. If we made this, if we defined this as u, if we said u is equal to x to the third times the natural log of two, what is du going to be? du is going to be, it's going to be, well natural log of two is just a constant so it's going to be three x squared times the natural log of two. We could actually just change the order we're multiplying a little bit. We could say that this is the same thing as x squared times three natural log of two, which is the same thing just using logarithm properties, as x squared times the natural log of two to the third power. as the natural log of two to the third power. This is equal to x squared times the natural log of eight. Let's see, if this is u, where is du? Oh, and of course we can't forget the dx. This is a dx right over here, dx, dx, dx. Where is the du? Well we have a dx. Let me circle things. You have a dx here, you have a dx there. You have an x squared here, you have an x squared here. So really all we need is, all we need here is the natural log of eight. Ideally we would have the natural log of eight right over here, and we could put it there as long as we also, we could multiply by the natural log of eight as long as we also divide by a natural log of eight. We can do it like right over here, we could divide by natural log of eight. But we know that the anti-derivative of some constant" + }, + { + "Q": "At 5:16, Sal puts a negative sign in front of the inverse of ln8. Is it correct that it should be an equal sign? As far as I can tell, the solution works out only if ln8 is not negative.", + "A": "It actually is an equal sign, but it is very hard to see it the way he wrote it . :)", + "video_name": "1ct7LUx23io", + "timestamps": [ + 316 + ], + "3min_transcript": "Well that's going to be three x squared times the natural log of two, or three times the natural log of two times x squared. That's just a constant times x squared. We already have a x squared here so maybe we can engineer this a little bit to have the constant there as well. Let's think about that. If we made this, if we defined this as u, if we said u is equal to x to the third times the natural log of two, what is du going to be? du is going to be, it's going to be, well natural log of two is just a constant so it's going to be three x squared times the natural log of two. We could actually just change the order we're multiplying a little bit. We could say that this is the same thing as x squared times three natural log of two, which is the same thing just using logarithm properties, as x squared times the natural log of two to the third power. as the natural log of two to the third power. This is equal to x squared times the natural log of eight. Let's see, if this is u, where is du? Oh, and of course we can't forget the dx. This is a dx right over here, dx, dx, dx. Where is the du? Well we have a dx. Let me circle things. You have a dx here, you have a dx there. You have an x squared here, you have an x squared here. So really all we need is, all we need here is the natural log of eight. Ideally we would have the natural log of eight right over here, and we could put it there as long as we also, we could multiply by the natural log of eight as long as we also divide by a natural log of eight. We can do it like right over here, we could divide by natural log of eight. But we know that the anti-derivative of some constant times the anti-derivative of that function. We could just take that on the outside. It's one over the natural log of eight. Let's write this in terms of u and du. This simplifies to one over the natural log of eight times the anti-derivative of e to the u, e to the u, that's the u, du. This times this times that is du, du. And this is straightforward, we know what this is going to be. This is going to be equal to, let me just write the one over natural log of eight out here, one over natural log of eight times e to the u, and of course if we're thinking in terms of just anti-derivative there would be some constant out there. Then we would just reverse the substitution." + }, + { + "Q": "At 0:52 how is their a 2/3 in 5/3 x 2/5? You aren't multiplying anything by 2/3. I understand there is a 2 and a 3 in the equation but that isn't 2/3 like the fraction.", + "A": "With the commutative property for multiplication, you can move the factors around within the numerator and denominator, so you can see the 2/3 fraction that way and factor it out 0:51.", + "video_name": "yUYDhmQsiXY", + "timestamps": [ + 52 + ], + "3min_transcript": "We have three expressions here. This is 2/3 times 7/8. The second expression is 8/7 times 2/3. This third expression is 5 times 2 over 3 times 5. And what I want you to do is pause this video right now and think about which of these expressions is the largest, which one is in the middle in terms of value, and which one is the smallest. And I want you to think about it without actually doing the calculation. If you could just look at them and figure out which of these is the largest, which of these is the smallest, and which of these is in the middle. So pause the video now. Now, you might have taken a shot at it. And I'll give you a little bit of a hint in case you had trouble with it. All of these involve multiplying something by 2/3. And you see a 2/3 here. You see a 2/3 here. And it might not be as obvious, but you also see a 2/3 here. And let me rewrite that to make it a little bit clearer. So this first expression could be rewritten as 7/8 times 2/3. it's already written as 8/7 times 2/3. And then, this last expression, we could write it as, in the numerator, 5 times 2. And then in the denominator, it's over 5 times 3. 5 times 3, which is of course the same thing as 5/5 times 2/3. So you see, all three of these expressions involve something times 2/3. Now, looking at it this way, does it become easier to pick out which of these are the largest, which of these are the smallest, and which of these are someplace in between? I encourage you to pause it again if you haven't thought about it yet. So let's visualize each of these expressions by first trying to visualize 2/3. So let's say the height of what I am drawing right now, let's say the height of this bar right over here is 2/3. The height here is 2/3. So first, let's think about what this one on the right here represents. This is 5/5 times 2/3. Well, what's 5/5? 5/5 is the same thing as 1. This is literally just 1 times 2/3. This whole expression is the same thing as 1 times 2/3, or really, just 2/3. So this, the height here, 2/3, this is the same thing as this thing over here. This is going to be equal to-- this could also be viewed as 5 times 2 over 3 times 5, which was this first expression right over here. Now, let's think about what these would look like. So this is 7/8 times 2/3. So it's less than 8/8 times 2/3. It's less than 1 times 2/3. So we're going to scale 2/3 down." + }, + { + "Q": "At 1:18, are you sure it's 2 by 3 because I'm pretty sure it's 3 by 2?", + "A": "This way of identifying matrices is arbitrary: we all agree (or: most of us do, I think! :) to name the rows first and the columns second. (The first matrix in the video is 2X3, the second is 3X2.) We (or they) could have reversed the order, but since that s the way others do it, so will I (if I can remember it!).", + "video_name": "OMA2Mwo0aZg", + "timestamps": [ + 78 + ], + "3min_transcript": "We're given two matrices over here, matrix E and matrix D. And they ask us, what is ED, which is another way of saying what is the product of matrix E and matrix D? Just so I remember what I'm doing, let me copy and paste this. And then I'm going to get out my little scratch pad. So let me paste that over here. So we have all the information we needed. And so let's try to work this out. So matrix E times matrix D, which is equal to-- matrix E is all of this business. So it is 0, 3, 5, 5, 5, 2 times matrix D, which is all of this. So we're going to multiply it times 3, 3, 4, 4, negative 2, negative 2. Now the first thing that we have to check is whether this is even a valid operation. Now the matrix multiplication is a human-defined operation that happen to have [? new ?] properties. Now the way that us humans have defined matrix multiplication, it only works when we're multiplying our two matrices. So this right over here has two rows and three columns. So it's a 2 by 3 matrix. And this has three rows and two columns, it's 3 by 2. This only works-- we could only multiply this matrix times this matrix, if the number of columns on this matrix is equal to the number of rows on this matrix. And in this situation it is, so I can actually multiply them. If these two numbers weren't equal, if the number of columns here were not equal to the number of rows here, then this would not be a valid operation, at least the way that we have defined matrix multiplication. The other thing you always have to remember is that E times D is not always the same thing as D times E. Order matters when you're multiplying matrices. It doesn't matter if you're multiplying regular numbers, but it matters for matrices. But let's actually work this out. by 2 matrix. But I'm going to create some space here because we're going to have to do some computation. So this is going to be equal to-- I'm going to make a huge 2 by 2 matrix here. So the way we get the top left entry, the top left entry is essentially going to be this row times this product. If you view them each as vectors, and you have some familiarity with the dot product, we're essentially going to take the dot product of that and that. And if you have no idea what that is, I'm about to show you. This entry is going to be 0 times 3, plus 3 times 3, plus 5 times 4. So that is the top left entry. And I already see that I'm going to run out of space here, so let me move this over to the right some space so I have some breathing room." + }, + { + "Q": "At 3:36, Sal mentions that there is a positive derivative and then zero and then a negative derivative after the critical point x=2. But the graph has a negative slope. How is it that it's positive then?", + "A": "The graph is of the derivative not of the original function. This means that when the derivative is positive the function has a positive slope and when the derivative is negative the original function has a negative slope. And the derivative is equal to zero at the function s critical points of +2 and -2.", + "video_name": "pInFesXIfg8", + "timestamps": [ + 216 + ], + "3min_transcript": "But we still don't know whether the function takes on a minimum values at those points, maximum values of those points, or neither. To figure that out we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. I'll draw an axis right over here. I'll do it down here because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2, 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well we have when x is equal to 0 for the derivative we're at negative 12. So this is, we're graphing y is equal to f prime of x. So it looks something like this. These are obviously the 0's of our derivative. So it has to move up to cross the x-axis there and over here. So what is the derivative doing at each of these critical points? Well over here our derivative is crossing from being positive, we have a positive derivative, to being a negative derivative. So we're crossing from being a positive derivative to being a negative derivative, that was our criteria for a critical point to be a maximum point. Over here we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function So a minimum. And I just want to make sure we have the correct intuition. If our function, if some function is increasing going into some point, and at that point we actually have a derivative 0-- the derivative could also be undefined-- but we have a derivative of 0 and then the function begins decreasing, that's why this would be a maximum point. Similarly, if we have a situation where the function is decreasing going into a point, the derivative is negative. Remember this is the graph of the derivative. Let me make this clear. This is the graph of y is equal to not f of x, but f prime of x. So if we have a situation we're going into the point, the function has a negative slope, we see we have a negative slope here-- so the function might look something like this. And then right at this point the function is either undefined or has 0 slope. So in this case it has 0 slope." + }, + { + "Q": "Other than visually (3:44) how do you know if the derivative is positive or negative before the critical point.", + "A": "The simple way to do that is to pick a convenient point just before and just after the critical point and plug those values into the first derivative to see whether it is negative or positive.", + "video_name": "pInFesXIfg8", + "timestamps": [ + 224 + ], + "3min_transcript": "But we still don't know whether the function takes on a minimum values at those points, maximum values of those points, or neither. To figure that out we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. I'll draw an axis right over here. I'll do it down here because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2, 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well we have when x is equal to 0 for the derivative we're at negative 12. So this is, we're graphing y is equal to f prime of x. So it looks something like this. These are obviously the 0's of our derivative. So it has to move up to cross the x-axis there and over here. So what is the derivative doing at each of these critical points? Well over here our derivative is crossing from being positive, we have a positive derivative, to being a negative derivative. So we're crossing from being a positive derivative to being a negative derivative, that was our criteria for a critical point to be a maximum point. Over here we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function So a minimum. And I just want to make sure we have the correct intuition. If our function, if some function is increasing going into some point, and at that point we actually have a derivative 0-- the derivative could also be undefined-- but we have a derivative of 0 and then the function begins decreasing, that's why this would be a maximum point. Similarly, if we have a situation where the function is decreasing going into a point, the derivative is negative. Remember this is the graph of the derivative. Let me make this clear. This is the graph of y is equal to not f of x, but f prime of x. So if we have a situation we're going into the point, the function has a negative slope, we see we have a negative slope here-- so the function might look something like this. And then right at this point the function is either undefined or has 0 slope. So in this case it has 0 slope." + }, + { + "Q": "At 10:40 Sal said that t does not have to be time, so what other things can t be?", + "A": "Maybe you re trying to describe the path of a complicated music box dancer that moves as you turn a crank. In that case, the independent variable might not represent time but the number of rotations that you gave the crank from its resting place, but traditionally in parametric equations we d call it t even though it isn t about time in that case.", + "video_name": "57BiI_iD3-U", + "timestamps": [ + 640 + ], + "3min_transcript": "Cosine of pi over 2 is 0. 0 times 3 is 0. And what's x equal when t is equal to pi? Cosine of pi is minus 1. Minus 1 times 3 is minus 3. Fair enough. Now let's do the y's. When t is 0 what is y? Sine is 0, 0. So 2 times 0 is 0. When t is pi over 2, sine of pi over 2 is 1. 1 times 2 is 2. And when t is pi, sine of pi-- that's sine of 180 degrees-- that's 0. 2 times 0 is 0. So let's plot these points. When time is 0, we're at the point 3, 0. So 3, 0-- 3, 0 is right there. This is t equals 0. When t increases by pi over 2, or if this was seconds, pi over So at t equals pi over 2, we're at the point 0, 2. We're right over here. So this is at t is equal to pi over 2. And then when t increases a little bit more-- when we're at t is equal to pi-- we're at the point minus 3, 0. We're here. So this is t is equal to pi or, you know, we could write 3.14159 seconds. 3.14 seconds. And actually, you know, I want to make the point, t does not have to be time, and we don't have to be dealing with seconds. But I like to think about it that way. I like to think about, maybe this is describing some object in orbit around, I don't know, something else. So now we know the direction. As t increased from 0 to pi over 2 to pi, we went this way. We went counterclockwise. So the direction of t's parametric equations And you might be saying, Sal, you know, why'd we have to do 3 points? We could have just done 2, and made a line. If we just had that point and that point, you might have immediately said, oh, we went from there to there. But that really wouldn't have been enough. Because maybe we got from here to there by going the other way around. So giving that third point lets us know that the direction is definitely counterclockwise. And so what happens if we just take t from 0 to infinity? What happens if we bound t? t is greater than 0 and less than infinity. Well, we're just going to keep going around this ellipse forever. Multiple times. Keep writing over and over, infinite times. If we went from minus infinity to infinity, then we would have always been doing it, I guess is the way to put it. Or if we just wanted to trace this out once, we could go from t is less than or equal to-- or t is greater than or equal to 0. All the way to t is less than or equal to 2 pi." + }, + { + "Q": "in 10:24 when Sal uses pi and pi/2 for his time, is there a reason why he couldn't use 90 or 180?", + "A": "Khan is using angles in radians probably because it is more intuitive rather than in degrees. cos(90\u00c2\u00ba) = 0 = cos(Pi/2) Both expressions are equivalent (just mind the angle units)", + "video_name": "57BiI_iD3-U", + "timestamps": [ + 624 + ], + "3min_transcript": "So to do that, let's make our little table. So let's take some values of t. So we'll make a little table. t, x, and y. It's good to pick values of t. Remember-- let me rewrite the equations again, so we didn't lose it-- x was equal to 3 cosine of t, and y is equal to 2 sine of t. It's good to take values of t where it's easy to figure out what the cosine and sine are, and without using a calculator. We're assuming the t is in radiance, just for simplicity. So let's pick t is equal to 0. t is equal to pi over 2. That's 90 degrees in degrees. And t is equal to pi. And so what is x when t is equal to 0? Well, cosine of 0 is 1 times 3, that's 3. Cosine of pi over 2 is 0. 0 times 3 is 0. And what's x equal when t is equal to pi? Cosine of pi is minus 1. Minus 1 times 3 is minus 3. Fair enough. Now let's do the y's. When t is 0 what is y? Sine is 0, 0. So 2 times 0 is 0. When t is pi over 2, sine of pi over 2 is 1. 1 times 2 is 2. And when t is pi, sine of pi-- that's sine of 180 degrees-- that's 0. 2 times 0 is 0. So let's plot these points. When time is 0, we're at the point 3, 0. So 3, 0-- 3, 0 is right there. This is t equals 0. When t increases by pi over 2, or if this was seconds, pi over So at t equals pi over 2, we're at the point 0, 2. We're right over here. So this is at t is equal to pi over 2. And then when t increases a little bit more-- when we're at t is equal to pi-- we're at the point minus 3, 0. We're here. So this is t is equal to pi or, you know, we could write 3.14159 seconds. 3.14 seconds. And actually, you know, I want to make the point, t does not have to be time, and we don't have to be dealing with seconds. But I like to think about it that way. I like to think about, maybe this is describing some object in orbit around, I don't know, something else. So now we know the direction. As t increased from 0 to pi over 2 to pi, we went this way. We went counterclockwise. So the direction of t's parametric equations" + }, + { + "Q": "3:20 Commutative property of addition 704 = 700 + 4 or 4 + 700\n3:30 Associative property of addition 18 + (4 + 700) = (18 + 4) + 700", + "A": "that is 3th grade math for asossitive property and 6th grade math for comunitive property", + "video_name": "jAfJcgPGqgI", + "timestamps": [ + 200, + 210 + ], + "3min_transcript": "to get to 500? And I could have done that in my head. Okay, I need to add 20. 355 minus 20 is 335. But now this problem is much, much simpler to compute. 335 plus 500, well, it's going to be three hundreds plus five hundreds, it's going to be equal to 800 and then we have 35. 835. Now, to make it a little bit clearer what we did, remember, we wanted to take 20 from here and put it over here, so we could break up 355, we could say that it's going to be equal to, it's going to be equal to, let's break it up into 335 and 20, and remember, the whole reason why I picked 20 is because I'm going to want to add that to 480, but I'm just doing it step-by-step here, so plus 480, and now I can just change the order with which I add, 335 plus 20 plus 480, and instead of adding the 335 and 20 first, I could add the 20 and 480 first, so I could add these two characters first, and so then I'm going to be left with, this is going to be equal to 335 plus, what's 20 plus 480? Well, that was the whole point. I picked 20 so that I can get to 500. 20 plus 480 is going to be 500 and then, now, you can add them. This is going to be 835. So this is just a longer way of saying what I did here. I took 20 from 355, so I can make the 480 into a 500. Let's do a couple more examples of this, and remember, the key is just thinking about how could I add or take away from one of the numbers to make them simpler? So there's a couple ways we could tackle this. One way, we could try to get the 704 to be equal to 700. So, we could say that this is the same thing as 18 plus, I could write 700 plus 4, or I could write it as 4 plus 700, like that, and then I could put the parentheses around this first, and then I could just switch the order in which I add, so this is the same thing as 18 plus 4 plus 700, and now I could add the 18 and the 4 first. Now what's 18 plus 4? It's 22, and then I have plus 700, and now this is pretty easy to compute, and all of these are going to be equal to each other, so let me just write it like that. 22 plus 700, I could do that in my head." + }, + { + "Q": "At 2:18 talking about the passage from the purple curve to the yellow segment of the function, he said a slope is not defined there because we could draw a lot.\nYet a unique limit for that point exists, so it should also exist a derivative right?\n(The same happens between the blue and the orange segments at the end.)", + "A": "just because a limit exists does not mean that a function is differentiable, although it is one of the conditions of that. For a function to be differentiable, the derivative from the left side of the point must be equal to the derivative from the right using one sided limits.", + "video_name": "eVme7kuGyuo", + "timestamps": [ + 138 + ], + "3min_transcript": "So I've got this crazy discontinuous function here, which we'll call f of x. And my goal is to try to draw its derivative right over here. So what I'm going to need to think about is the slope of the tangent line, or the slope at each point in this curve, and then try my best to draw that slope. So let's try to tackle it. So right over here at this point, the slope is positive. And actually, it's a good bit positive. And then as we get larger and larger x's, the slope is still positive, but it's less positive-- and all the way up to this point right over here, where it becomes 0. So let's see how I could draw that over here. So over here we know that the slope must be equal to 0-- right over here. Remember over here, I'm going to try to draw y is equal to f prime of x. And I'm going to assume that this is some type of a parabola. But let's say that, so let's see, here the slope is quite positive. So let's say the slope is right over here. And then it gets less and less and less positive. And I'll assume it does it in a linear fashion. That's why I had to assume that it's some type of a parabola. So it gets less and less and less positive. Notice here, for example, the slope is still positive. And so when you look at the derivative, the slope is still a positive value. But as we get larger and larger x's up to this point, the slope is getting less and less positive, all the way to 0. And then the slope is getting more and more negative. And at this point, it seems like the slope is just as negative as it was positive there. So at this point right over here, the slope is just as negative as it was positive right over there. So it seems like this would be a reasonable view of the slope of the tangent line over this interval. Here the slope seems constant. Our slope is a constant positive value. So once again, our slope here is a constant positive line. Let me be careful here because at this point, our slope won't really be defined, because our slope, you could draw multiple tangent lines at this little pointy point. So let me just draw a circle right over there. But then as we get right over here, the slope seems to be positive. So let's draw that. The slope seems to be positive, although it's not as positive as it was there. So the slope looks like it is-- I'm just trying to eyeball it-- so the slope is a constant positive this entire time. We have a line with a constant positive slope. So it might look something like this." + }, + { + "Q": "at 5:00 how slope of that line is constant, that line rising up so it shouldn't be constant :S", + "A": "The slope of any line as always constant! I can show you algebraically: d/dx (ax+b) = a which is a constant. I can also explain it graphically: The slope of a line is always constant. Therefore, its tangent line is also of constant slope. Even if a line rises, the slope is constant. A rising line simply means a positive slope.", + "video_name": "eVme7kuGyuo", + "timestamps": [ + 300 + ], + "3min_transcript": "Here the slope seems constant. Our slope is a constant positive value. So once again, our slope here is a constant positive line. Let me be careful here because at this point, our slope won't really be defined, because our slope, you could draw multiple tangent lines at this little pointy point. So let me just draw a circle right over there. But then as we get right over here, the slope seems to be positive. So let's draw that. The slope seems to be positive, although it's not as positive as it was there. So the slope looks like it is-- I'm just trying to eyeball it-- so the slope is a constant positive this entire time. We have a line with a constant positive slope. So it might look something like this. I want these things to match up. So let me do my best. So this matches up to that. This matches up over here. And we just said we have a constant positive slope. So let's say it looks something like that over this interval. And then we look at this point right over here. So right at this point, our slope is going to be undefined. There's no way that you could find the slope over-- or this point of discontinuity. But then when we go over here, even though the value of our function has gone down, we still have a constant positive slope. In fact, the slope of this line looks identical to the slope of this line. Let me do that in a different color. The slope of this line looks identical. So we're going to continue at that same slope. It was undefined at that point, but we're going to continue at that same slope. And once again, it's undefined here So the slope will look something like that. And then we go up here. The value of the function goes up, but now the function is flat. So the slope over that interval is 0. The slope over this interval, right over here, is 0. So we could say-- let me make it clear what interval I'm talking about-- the slope over this interval is 0. And then finally, in this last section-- let me do this in orange-- the slope becomes negative. But it's a constant negative. And it seems actually a little bit more negative than these were positive. So I would draw it right over there. So it's a weird looking function. But the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point. And by doing so, we have essentially drawn the derivative over that interval." + }, + { + "Q": "At 3:55, the red colored line is increasing, but is in f(x)<0. So, I get that it will have a constant f'(x) slope, but shouldn't it be in f'(x)<0.", + "A": "Yes, f(x) is negative, but f (x) (or F Prime) will be positive, since it is essentially the slope of the line and the slope at that point is positive.", + "video_name": "eVme7kuGyuo", + "timestamps": [ + 235 + ], + "3min_transcript": "Here the slope seems constant. Our slope is a constant positive value. So once again, our slope here is a constant positive line. Let me be careful here because at this point, our slope won't really be defined, because our slope, you could draw multiple tangent lines at this little pointy point. So let me just draw a circle right over there. But then as we get right over here, the slope seems to be positive. So let's draw that. The slope seems to be positive, although it's not as positive as it was there. So the slope looks like it is-- I'm just trying to eyeball it-- so the slope is a constant positive this entire time. We have a line with a constant positive slope. So it might look something like this. I want these things to match up. So let me do my best. So this matches up to that. This matches up over here. And we just said we have a constant positive slope. So let's say it looks something like that over this interval. And then we look at this point right over here. So right at this point, our slope is going to be undefined. There's no way that you could find the slope over-- or this point of discontinuity. But then when we go over here, even though the value of our function has gone down, we still have a constant positive slope. In fact, the slope of this line looks identical to the slope of this line. Let me do that in a different color. The slope of this line looks identical. So we're going to continue at that same slope. It was undefined at that point, but we're going to continue at that same slope. And once again, it's undefined here So the slope will look something like that. And then we go up here. The value of the function goes up, but now the function is flat. So the slope over that interval is 0. The slope over this interval, right over here, is 0. So we could say-- let me make it clear what interval I'm talking about-- the slope over this interval is 0. And then finally, in this last section-- let me do this in orange-- the slope becomes negative. But it's a constant negative. And it seems actually a little bit more negative than these were positive. So I would draw it right over there. So it's a weird looking function. But the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point. And by doing so, we have essentially drawn the derivative over that interval." + }, + { + "Q": "At 2:15 it says the numbers are equivalent ,how is that?", + "A": "As I understand, it means that both formulas will give you the same number; so it doesn t matter how you go about finding the area, either with formula a or b, both will provide the same result and therefore are equivalent.", + "video_name": "Q3wfb0CPhIY", + "timestamps": [ + 135 + ], + "3min_transcript": "I have this rectangle here, and I want to figure out its area. I want to figure out how much space it is taking up on my screen right over here. And I encourage you to pause this video and try to figure out the area of this entire rectangle. And when you do it, think about the two different ways you could do it. You could either multiply the length of the rectangle times the entire width, so just figure out the area of the entire thing. Or you could separately figure out the area of this red or this purple rectangle and then separately figure out this blue rectangle and realize that their combined areas is the exact same thing as the entire rectangle. So I encourage you to pause the video and try both of those strategies out. So let's just try them out ourselves. First, let's look at the overall dimensions of the larger rectangle. The length is 9, and we're going to multiply that times the width. But what's width here? Well, the width is going to be 8 plus 12. This entire distance right over here is 8 plus 12. This is one way that we could figure out the area of this entire thing. This is just the length times the width. 8 plus 12 is obviously going to be equal to 20. But the other way that we could do it-- and this must be equivalent, because we're figuring out the area of the same thing-- is to separate out the area of these two sub-rectangles. So let's do that. And this must be equal to this thing. So what's the area of this purple rectangle? Well, it's going to be the length. It's going to be 9. Let me do it in that same color. It's going to be 9 times the width, which is 8. It's going to be 9 times 8. And then what's the area of this the blue rectangle? Well, that's going to be 9 times-- so the height here is 9 still. The height is 9. And what's the width? Well, the width is 12. And what's the area of the combined if you wanted to combine the area of the purple rectangle Well, you'd just add these two things together. And of course, when you add these two things together, you get the area of the entire thing. So these things must be equivalent. They are calculating the same area. Now, what's neat about this is we just showed ourselves the distributive property when we're dealing with these numbers. You could try these out for any numbers. They'll work for any numbers, because the distributive property works for any numbers. You see 9 times the sum of 8 plus 12 is equal to 9 times 8 plus 9 times 12. We essentially have distributed the 9-- 9 times 8 plus 9 times 12. And let's actually calculate it just to satisfy ourselves about the area. So if you multiply the length times the entire width, so that's 9 times 8 plus 12, that's the same thing as 9 times 20, which is 180." + }, + { + "Q": "At 2:03, what is the reciprocal?", + "A": "Reciprocal means to turn the fraction upside-down. For example, the reciprocal of 3/4 is 4/3. The reciprocal of 3 is 1/3, because 3 is really 3/1. Also, if you multiply any number by its reciprocal, you will get one (which I think is what nguyensongthienphuc was saying). So 4/3 * 3/4 = 1.", + "video_name": "K2b8iMPY11I", + "timestamps": [ + 123 + ], + "3min_transcript": "- What I hope to do in this video is emphasize the relation, the connection, between fractions and division and then using that knowledge to help us simplify some hairy looking fractions. So, let's say, let's just do a little bit of a review. So, if I say two divided by, two divided by three, two divided by three, I could write this as, I could write this as two over three. Two divided by, two divided by three. So, this expression here, sometimes we might say, \"Hey, this is 2/3. Think of it as a fraction.\" But you can also view this as two divided by three. So, if we went the other way around. If someone were to write four over, actually, let me just do it in a different color, if someone were to say four, 4/9, four, four over nine, we could interpret this, this could be interpreted as four divided by nine. Four divided by nine. this into a decimal that's exactly what we would do. We would calculate what four divided by nine is. We would divide nine into four. Four divided by nine. So, this is all review. So, let's just use this knowledge and I'll think more complex fractions. So, if someone were to say, actually, let me just take this example right over here, if someone were to say two over, instead of saying two over three. If they said two over, I don't know, let me say two over 2/3. Two over 2/3. Well, what would this simplify to? Well, we could go the other way around. This is the same thing as two divided by 2/3. So, let's write it like that. This is the same thing as two divided by, two divided by 2/3, two divided by 2/3, which is the same thing as, dividing by a fraction's the same thing So, this is going to be the same thing as two times, two times the reciprocal of 2/3, which is, we swap the denominator and the numerator, two times three over two. So, two times three halves, if I have three halves twice, well, that's just going to be equal to, that's going to be equal to six halves, and six halves, you can view that as, well, two halves is a whole, so this is gonna be three wholes. Or you can say six divided by two. Well, that's just gonna be equal to three. You can view it either way. And then that is equal to three. Let's do a few more of these. And let's keep making them a little bit more complicated. Just let me get some good practice. And like always, pause the video. You should get excited when you see one of these things. And pause the video and see if you can do it on your own. All right, let's do something really interesting. Let's say negative 16 over nine over, over, I'll do this in a tan color," + }, + { + "Q": "Again 11:53:\nHow do I get from [x]B to [Ax]B ?? that is [x]B = [Ax]B algebraically?", + "A": "Sal is not saying that [Ax]_B = [x]_B. He wrote that D[x]_B = [Ax]_B. We have the rule that some vector v can be expressed in alternative coordinate systems by: C [v]_B = v, and [v]_B = C^-1 v. Ax is some vector. Therefore, we can apply the rule to it. x is also some vector. Therefore, we can apple the rule to it.", + "video_name": "PiuhTj0zCf4", + "timestamps": [ + 713 + ], + "3min_transcript": "Let's see what D times xB is equal to. So let's say if we take D times xB, so this thing right here should be equal to D times the representation or the coordinates of x with respect to the basis B. That's what we're claiming. We're saying that this guy is equal to D times the representation of x with respect to the coordinates with respect to the basis B. Let me write all of this down. I'll do it right here, because I think it's nice to have this graphic up here. So we can say that D times xB is equal to this thing right here. It's the same thing as the transformation of x represented in coordinates with respect to B, or in these nonstandard coordinates. represented in this coordinate system, represented in coordinates with respect to B. We see that right there. But what is the transformation of x? That's the same thing as A times x. That's kind of the standard transformation if x was represented in standard coordinates. So this is equal to x in standard coordinates times the matrix A. Then that will get us to this dot in standard coordinates, but then we want to convert it to these nonstandard coordinates just like that. Now, if we have this, how can we just figure out what the vector Ax should look like? What this vector should look like? Well, we can look at this equation right here. We have this. This is the same thing as this. we want to go the other way. We have this. We have that right there. That's this right there. We want to get just this dot represented in regular standard coordinates. So what do we do? We multiply it by C. Let me write it this way. If we multiply both sides of this equation times C, what do we get? We get this right here. I was looking at the right equation the first time. We have this right here, which is the same-- first intuition is always right. We have this, which is the same thing as this right here. So this can be rewritten. This thing can be rewritten as C inverse-- we don't have an x here. We have an Ax here, so C inverse times Ax. The vector Ax represented in these nonstandard coordinates" + }, + { + "Q": "At 4:31, is it really possible that there are problems like-(-(-2))?", + "A": "Yes it is possible, but you don t see it much.", + "video_name": "3-aryZYsoxU", + "timestamps": [ + 271 + ], + "3min_transcript": "but let's say that A is some number that is right over here. Well negative A is going to be the opposite of A. So if A is three tick marks to the right negative A is going to be three tick marks to the left. One, two, three. And so the opposite of A is going to be this value right over here. And we can write that, we can write opposite of A over here. We could literally write opposite, opposite of A is that number right over there. Or as a shorthand, we can just write-- We can just write negative. We can just write this is negative A right over here. Negative A. So with that in mind, if we literally view this negative symbol as meaning the opposite of whatever this is, let's try something interesting. Let me do this in another color. What would be the negative of negative three? And I encourage you to pause the video and think about it. Well we just said, this negative means the opposite. So you can think about this as meaning this means the opposite... Opposite of negative-- The opposite of negative three. So what is the opposite of negative three? Well negative three is three to the left of zero. One, two, three. So it's opposite is going to be three to the right of zero. One, two, three. So it's going to be positive three. So this is equal to positive three. Or we could just write positive three like that. So hopefully this gives you a better appreciation for what opposite means actual negative symbol. We could keep going. We could do something like what is the negative of the negative of negative-- Let me do a different number. Of negative two. Well, this part right over here the negative, the opposite of negative two, which is really the opposite of the opposite of two, well that's just going to be two. Every time you say opposite you flip over the number line. So this flips you over the number line two to the left and this flips you back two to the right so all of this is going to be two but then you're going to take the opposite of two so that's just going to be negative two. If you threw another negative in front of this, it would be the opposite of all of this. So it would be the opposite of negative two and then all of a sudden it would become positive. So every time you put that negative in front there you're flipping on the other side of the number line." + }, + { + "Q": "what is the angle is an obtuse angle and the protractor isnt big enough to measure it? D:If you dont understand, 3:00\nwell the angle there is acute, but it LOOKS like its obtuse, and then you cant read the angle! D:", + "A": "you measure whats missing instead. then subtract it from 360. use the protractor to extend a ray so you have a strait line and measure the angle you created and subtract that from 360 to find it in degrees", + "video_name": "dw41PMWek6U", + "timestamps": [ + 180 + ], + "3min_transcript": "Let me just keep rotating it. If I could just keep it pressed. That's better. All right. That looks about right. So one side is at the 0 mark. And then my angle, my other side-- or if this was a ray, it points to, looks like, pretty close to the 20 degree mark. So I will type that in off the screen. You don't see that. And that is the right answer. And then we can get another angle. So let's try to measure this one right over here. So once again, place the center of the protractor at the center, at the vertex, of our angle. We can place the 0 degree, the base of the protractor, at this side of the angle. So let's just rotate it a little bit, maybe one more time. That looks about right. And then the angle is now opening up-- let's see, the other side is pointing to 110 degrees. So this is larger than 90 degrees. It's also an obtuse angle. This is obtuse, 110 degrees. More than 90 degrees. So let me type it in. I got the right answer. Let's do a couple more of these. So once again, put the center of the protractor at the vertex of our angle. And now, I want to rotate it. There we go. And this looks like roughly an 80 degree angle, not quite. If I have to be really precise, it looks like it's maybe 81 or 82 degrees. But I'll just go with 80 as my best guess. I got the right answer. Let's do one more of these. So once again, vertex of my angle at the center of my protractor. And then I want to put one side of the angle at the 0 degree. And I want to show you, there's two ways to do that. You could do this. You could do just this. But this isn't too helpful, because the angle is now outside. The other side sits outside of the protractor. So you want the 0 degrees on the side, So let's keep rotating it. There we go. And then our other side opens up or you could say points to 70 degrees. So this is an acute angle right over here. So it is 70 degrees. So I'll leave you with that. Oh look, I'm ready to move on, the exercise tells me. And now we can start talking about more things about angles now that we know how to measure them." + }, + { + "Q": "@ 1:09\nWe get to the expression\n-2.7+ -5\nWhat's one thing that tells us not to solve 7-5 first but multiply the first two values and only then take the difference?\nI'd be very thankful", + "A": "PEMDAS rules always apply. Multiplication is always before subtraction. FYI, your expression as written contains no multiplication. The dot for multiplication has to be raised up: -2 * 7 + (-5). In your expression, you just have a decimal number -2.7 - 5 = -7.7", + "video_name": "uaPm3Tpuxbc", + "timestamps": [ + 69 + ], + "3min_transcript": "We're asked to evaluate negative 2 times f of negative 6 plus g of 1. And they've defined, at least graphically, f of x and g of x here below. So let's see how we can evaluate this. Well, to do this, we first have to figure out what f of negative 6 is. So our input into our function is negative 6. And we'll assume that's along the horizontal axis. So our input is negative 6. And based on our function definition, f of negative 6 is 7. Let me write this down. f of negative 6 is equal to 7. And what is g of 1? Well, once again, here's our input axis. And then the function says that g of 1, which is right over there, is negative 5. g of 1 is equal to negative 5. So this statement simplifies to negative 2 times f of negative 6, which is 7. So times 7 plus g of 1, which is negative 5. Negative 2 times 7 is negative 14 plus negative 5, which is negative 19. And we are done." + }, + { + "Q": "At 2:12 he said that he is going to give it to the 10 hundreds. But are you going to give it to the hundreds or tens?", + "A": "It said he meant 10 tens", + "video_name": "QOtam19NQcQ", + "timestamps": [ + 132 + ], + "3min_transcript": "I've written the same subtraction problem twice. Here we see we're subtracting 172 from 629. And all I did here is I expanded out the numbers. I wrote 629 as 600 plus 20 plus 9, and I rewrote 172, the one is 100. So that's there. This is 7/10. It's in the tens place, so it's 70. And then the 2 is 2 ones, so it just represents 2. And we'll see why this is useful in a second. So let's just start subtracting, and we'll start with the ones place. So we have 9 minus 2. Well, that's clearly just 7. And over here we could also say, well, 9 minus 2, we have the subtraction out front. That is going to be 7. Pretty straightforward. But then something interesting happens when we get to the tens place. We're going to try to subtract 2 minus 7, or we're going to try to subtract 7 from 2. And we haven't learned yet how to do things like negative numbers, which we'll learn in the future, so we have a problem. How do you subtract a larger number from a smaller number? regrouping, sometimes called borrowing. And that's why this is valuable. When we're trying to subtract a 7 from a 2, we're really trying to subtract this 70 from this 20. Well, we can't subtract the 70 from the 20, but we have other value in the number. We have value in the hundreds place. So why don't we take 100 from the 600, so that becomes 500, and give that 100 to the tens place? If we give that 100 to the tens place, what is 100 plus 20? Well, it's going to be 120. So all I did, I didn't change the value of 629. I took 100 from the hundreds place and I gave it to the tens place. Notice 500 plus 120 plus 9 is still 629. We haven't changed the value. So how would we do that right over here? Well, if we take 100 from the hundreds place, give that hundred to the tens place, it's going to be 10 hundreds. So this will now become a 12. This will now become a 12. But notice, this 12 in the tens place represents 12 tens, or 120. So this is just another way of representing what we've done here. There's no magic here. This is often called borrowing, where you say hey, I took a 1 from the 6, and I gave it to the 2. But wait, why did this 2 become a 12? Why was I able to add 10? Well, you've added 10 tens, or 100. You took 100 from here, so 600 became 500, and then 20 became 120. But now we're ready to subtract. 12 tens minus 7 tens is 5 tens. Or you could say 120 minus 70 is 50. And then finally, you have the hundreds place." + }, + { + "Q": "At around 20:08, Sal denotes that P_1 is a vector and only draws a half arrow, (though I assume he meant to draw a full arrow).\nIs there a significant notational different between a full arrow and a half arrow above a variable?", + "A": "There is no difference. Some people, myself included, often find themselves writing half arrows , as you labeled them, instead of full arrows , because they are faster to write.", + "video_name": "hWhs2cIj7Cw", + "timestamps": [ + 1208 + ], + "3min_transcript": "can say, since this is what determines our x-coordinate, we would say that x is equal to 0 plus t times minus 2, or minus 2 times t. And then we can say that y, since this is what determines our y-coordinate, y is equal to 3 plus t times 2 plus 2t. So we could have rewritten that first equation as just x is equal to minus 2t, and y is equal to 2t plus 3. So if you watch the videos on parametric equations, this is just a traditional parametric definition of this line right there. Now, you might have still viewed this as, Sal, this was a waste of time, this was convoluted. You have to define these sets and all that. But now I'm going to show you something that you probably-- that's true of anything. But you probably haven't seen in your traditional algebra class. Let's say I have two points, and now I'm going to deal in three dimensions. So let's say I have one vector. I'll just call it point 1, because these are position vectors. We'll just call it position 1. This is in three dimensions. Just make up some numbers, negative 1, 2, 7. Let's say I have Point 2. Once again, this is in three dimensions, so you have to specify three coordinates. This could be the x, the y, and the z coordinate. Point 2, I don't know. Let's make it 0, 3, and 4. Now, what if I wanted to find the equation of the line that passes through these two points in R3? So this is in R3. Well, I just said that the equation of this line-- so I'll just call that, or the set of this line, let me just call this l. guys, it could be P1, the vector P1, these are all vectors, be careful here. The vector P1 plus some random parameter, t, this t could be time, like you learn when you first learn parametric equations, times the difference of the two vectors, times P1, and it doesn't matter what order you take it. So that's a nice thing too. P1 minus P2. It could be P2 minus P1-- because this can take on any positive or negative value-- where t is a member of the real numbers. So let's apply it to these numbers. Let's apply it right here. What is P1 minus P2? P1 minus P2 is equal to-- let me get some space here. P1 minus P2 is equal, minus 1 minus 0 is minus 1. 2 minus 3 is minus 1." + }, + { + "Q": "At 14:00 why did Sal do b-a instead of b+a ? I thought you had to add vectors together to get the resultant vector. Can someone clarify this for me ?", + "A": "I have the same question. I don t have a good intuitive answer. But if you plot the vectors mentioned in the video we can see that a-b or b-a is the only vector that passes through the tip of the 2 vectors.", + "video_name": "hWhs2cIj7Cw", + "timestamps": [ + 840 + ], + "3min_transcript": "all and I go up. So my vector b will look like that. Now I'm going to say that these are position vectors, that we draw them in standard form. When you draw them in standard form, their endpoints represent some position. So you can almost view these as coordinate points in R2. This is R2. All of these coordinate axes I draw are going be R2. Now what if I asked you, give me a parametrization of the line that goes through these two points. So essentially, I want the equation-- if you're thinking in Algebra 1 terms-- I want the equation for the line that goes through these two points. So the classic way, you would have figured out the slope and all of that, and then you would have substituted back in. But instead, what we can do is, we can say, hey look, this line that goes through both of those points-- you could that's a better-- Both of these vectors lie on this line. Now, what vector can be represented by that line? Or even better, what vector, if I take any arbitrary scalar-- can represent any other vector on that line? Now let me do it this way. What if I were to take-- so this is vector b here-- what if I were to take b minus a? We learned in, I think it was the previous video, that b minus a, you'll get this vector right here. You'll get the difference in the two vectors. This is the vector b minus the vector a. And you just think about it. What do I have to add to a to get to b? I have to add b minus a. So if I can get the vector b minus a-- right, we know how We just subtract the vectors, and then multiply it by any scalar, then we're going to get any point along that line. So what happens if we take t, so some scalar, times our vector, times the vectors b minus a? What will we get then? So b minus a looks like that. But if we were to draw it in standard form-- remember, in standard form b minus a would look something like this. It would start at 0, it would be parallel to this, and then from 0 we would draw its endpoint. So if we just multiplied some scalar times b minus a, we would actually just get points or vectors that lie on this line. Vectors that lie on that line right there. Now, that's not what we set out to do. We wanted to figure out an equation, or parametrization, if you will, of this line, or this set. Let's call this set l. So we want to know what that set is equal to. So in order to get there, we have to start with this, which" + }, + { + "Q": "At 21:30, he mentions that P2 can sub in for P1, and earlier he mentioned that (P1 - P2) can also be (P2-P1). I understand all of that, but can they both sub in at once without any consequences? Can you have (P1 + t(P1 - P2)) and (P1 + t(P2-P1))? And vice versa with P2 as well.", + "A": "Are you asking if P1 + t(P1 - P2) = P1 + t(P2-P1)? Because if so, the answer is no, the two t s would be different at each point on the line it creates depending on which point is in each position in the equation. If you separate it into two equations using the same points and different t s, then the lines will overlap at every point, but having two equations for the same line is redundant.", + "video_name": "hWhs2cIj7Cw", + "timestamps": [ + 1290 + ], + "3min_transcript": "that's true of anything. But you probably haven't seen in your traditional algebra class. Let's say I have two points, and now I'm going to deal in three dimensions. So let's say I have one vector. I'll just call it point 1, because these are position vectors. We'll just call it position 1. This is in three dimensions. Just make up some numbers, negative 1, 2, 7. Let's say I have Point 2. Once again, this is in three dimensions, so you have to specify three coordinates. This could be the x, the y, and the z coordinate. Point 2, I don't know. Let's make it 0, 3, and 4. Now, what if I wanted to find the equation of the line that passes through these two points in R3? So this is in R3. Well, I just said that the equation of this line-- so I'll just call that, or the set of this line, let me just call this l. guys, it could be P1, the vector P1, these are all vectors, be careful here. The vector P1 plus some random parameter, t, this t could be time, like you learn when you first learn parametric equations, times the difference of the two vectors, times P1, and it doesn't matter what order you take it. So that's a nice thing too. P1 minus P2. It could be P2 minus P1-- because this can take on any positive or negative value-- where t is a member of the real numbers. So let's apply it to these numbers. Let's apply it right here. What is P1 minus P2? P1 minus P2 is equal to-- let me get some space here. P1 minus P2 is equal, minus 1 minus 0 is minus 1. 2 minus 3 is minus 1. So that thing is that vector. And so, our line can be described as a set of vectors, that if you were to plot it in standard position, it would be this set of position vectors. It would be P1, it would be-- let me do that in green-- it would be minus 1, 2, 7. I could've put P2 there, just as easily-- plus t times minus 1, minus 1, 3, where, or such that, t is a member of the real numbers. Now, this also might not be satisfying for you. You're like, gee, how do I plot this in three dimensions? Where's my x, y's, and z's? And if you want to care about x, y's, and z's, let's say that this is the z-axis." + }, + { + "Q": "At 0:42why did Sal write 4000+500=3000", + "A": "Look at the original problem at the top of the screen. Sal is not saying 4000+500=3000. He is writing the problem out using numbers. The original problem as + ? that Sal has not yet written in. But, by the end of the video, he does.", + "video_name": "a_mzIWvHx_Y", + "timestamps": [ + 42 + ], + "3min_transcript": "We have 4,5000 equals 3 thousands plus how many hundreds, question mark hundreds? So let's write this left-hand side, but I'm going to write it out in terms of thousands and hundreds. So I'll write the thousands in orange. So this is equal to 4 thousands, which is the same thing as just 4,000, plus 500, which you could also view as 5 hundreds. So this is the left-hand side. Now let's look at the right-hand side. We have 3 thousands. So it's 3 thousands. Now let's not even look at this right now. Let's just think about what do we have to add to this right-hand side in order to get the same thing that we have on the left-hand side? Well, if you compare the 3,000 and the 4,000, you see you have an extra 1,000 over here. So let's add an extra 1,000 on the right. So we're going to add one extra 1,000. And now we just have 3,000 plus 1,000. This makes it the 4,000. But then, of course, we also need another 500. So we're going to need plus a 500 right over here. we need to say 4,000 plus 500 is equal to 3,000 plus 1,500. Now, the way they've set this up, we need to express-- so it almost looks the same. On the left-hand side, this is 4,500. So this right over here, this is the same thing as 4,500. This is this right over here. And on the right-hand side, we have 3 thousands. So that's this right over here. That's the 3 thousands. And then we just need to express this as hundreds. So 1,500, this is the same thing as 15 hundreds. So let's rewrite everything. We can rewrite this as saying 4,500, just to get the exact same form that they wrote it over there. So we could write 4,500 is equal to 3 thousands plus-- now This is 15 hundreds. Literally, if you took 15 times 100, it's going to be equal to 1,500. So this could be viewed as 15 hundreds, so plus 15 hundreds. So in this situation, the question mark is equal to 15." + }, + { + "Q": "@ 7:10. Wouldn't x = 2 also cause the answer to be undefined? (2-2)(2+1)=(0)(3)=0?", + "A": "Yes, but since we re not canceling the (x -2), it s not necessary to include that as a condition.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 430 + ], + "3min_transcript": "That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it." + }, + { + "Q": "At 7:19 how did you come up with negative one? is it opposite from x+1? How do you come up with what is not defined?", + "A": "The fraction is not defined where its denominator goes to zero. You are correct that you set: x+1=0 to see what value of x would cause a problem. x= -1 would make the denominator = 0, so x cannot equal -1.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 439 + ], + "3min_transcript": "That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it." + }, + { + "Q": "could you just divided the 3x^2+3x-18 by 3? I was very confused by Khan's whole \"grouping\" process at 10:48", + "A": "You could divide the whole expression by 3, and then factor from there. However, Sal is trying to explain another way that can be used to factor. On some trinomials, it is difficult to find the correct factors by guessing. A more organized approach is to use grouping.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 648 + ], + "3min_transcript": "Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3. we get negative 6 times x plus 3. And now this is very clear our grouping was successful. This is the same thing as-- we can kind of undistribute this as 3x minus 6 times x plus 3. If we were to multiply this times each of these terms, you get that right there. So the top term, we can rewrite it as 3x minus 6-- let me do it in the same color. So we can rewrite it as 3x minus 6 times x plus 3. That's this term right here. I don't want to make it look like a negative sign. That's that term right there. Now let's factor this bottom part over here. Scroll to the left a little bit. to think of two numbers that when I take their product, I get 2 times 3, which is equal to 6, and they need to add up to be 5. And the two obvious numbers here are 2 and 3. I can rewrite this up here as 2x squared plus 2x plus 3x plus 3, just like that. And then if I put parentheses over here, and I decided to group the 2 with the 2 because they have a common factor of 2, and I grouped the 3 with the 3 because they have a common factor of 3. This right here is 2 and a 3. So here we can factor out a 2x. If you factor out a 2x, you get 2x times x plus 1 plus-- you factor out a 3 here-- plus 3 times x plus 1." + }, + { + "Q": "Why was there no condition in the first problem (at 2:28) ? Shouldn't it have a condition that states x is not equal to -1/3? If x = -1/3, then the denominator would be equal to zero.", + "A": "wait until you finish the video until commenting", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 148 + ], + "3min_transcript": "When we first started learning about fractions or rational numbers, we learned about the idea of putting things in lowest terms. So if we saw something like 3, 6, we knew that 3 and 6 share a common factor. We know that the numerator, well, 3 is just 3, but that 6 could be written as 2 times 3. And since they share a common factor, the 3 in this case, we could divide the numerator by 3 and the denominator by 3, or we could say that this is just 3/3, and they would cancel out. And in lowest terms, this fraction would be 1/2. Or just to kind of hit the point home, if we had 8/24, once again, we know that this is the same thing as 8 over 3 times 8, or this is the same thing as 1 over 3 times 8 over 8. The 8's cancel out and we get this in lowest terms as 1/3. These are rational numbers. Rational expressions are essentially the same thing, but instead of the numerator being an actual number and the denominator be an actual number, they're expressions involving variables. So let me show you what I'm talking about. Let's say that I had 9x plus 3 over 12x plus 4. Now, this numerator up here, we can factor it. We can factor out a 3. This is equal to 3 times 3x plus 1. That's what our numerator is equal to. And our denominator, we can factor out a 4. This is the same thing as 4 times 3x. 12 divided by 4 is 3. 12x over 4 is 3x. So here, just like there, the numerator and the denominator have a common factor. In this case, it's 3x plus 1. In this case, it's a variable expression. It's not an actual number, but we can do the exact same thing. They cancel out. So if we were to write this rational expression in lowest terms, we could say that this is equal to 3/4. Let's do another one. Let's say that we had x squared-- let So let's say we had x squared minus 9 over 5x plus 15. So what is this going to be equal to? So the numerator we can factor. It's a difference of squares." + }, + { + "Q": "at 6:30 what does he mean?", + "A": "He is doing what is called FOIL. in the problem (x+5)(x+1), you first multiply the x in both parentheses, then the x in the first and the 1 in the second, then the 5 and the x and then the 5 and 1. the FOIL means: F-first (the two x s) O-outside (the x and the 1) I-inside (5 and x) L-last (5 and 1) Hope this helps clear things a little!", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 390 + ], + "3min_transcript": "So to make them the same, I also have to add the extra condition that x cannot equal negative 3. So likewise, over here, if this was a function, let's say we wrote y is equal to 9x plus 3 over 12x plus 4 and we wanted to graph it, when we simplify it, the temptation is oh, well, we factored out a 3x plus 1 in the numerator and They cancel out. The temptation is to say, well, this is the same graph as y is equal to the constant 3/4, which is just a horizontal line at y is equal to 3/4. But we have to add one condition. We have to eliminate-- we have to exclude the x-values that would have made this thing right here equal to zero, and that would have been zero if x is equal to negative 1/3. If x is equal to negative 1/3, this or this denominator would be equal to zero. So even over here, we'd have to say x cannot be equal to That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to" + }, + { + "Q": "At 7:28 how come it is x does not equal 1 instead of x does not equal 1 AND 2?", + "A": "You are correct that x\u00e2\u0089\u00a01 and x\u00e2\u0089\u00a02 and you can list that if you like. However, it is traditional not to list any that remain obvious. But, it is not incorrect to list both of them. I personally prefer to list all of them, despite the tradition, just because I might do some later calculations that makes it difficult to tell where the forbidden values of x are.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 448 + ], + "3min_transcript": "That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it." + }, + { + "Q": "At 7:28, (x+5)/(x-2) is not defined for x = -1, but is it defined for x=2?", + "A": "I believe that when we are deciding which term will make a problem undefined we usually go with the one we are dividing by. So if your are canceling out an (x-1) from the numerator and denominator, even if there is another term present, that is the one that cannot be 0 because it is the one that the problem is divisible by.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 448 + ], + "3min_transcript": "That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it." + }, + { + "Q": "at 1:49 how do u get 3(3x+1)", + "A": "Sal starts with 9x+3 in the numerator of the fraction. He used the distributive property to factor out the GCM = 3. 9x + 3 = 3 ( 9x/3 + 3/3) = 3 (3x + 1). Hope this helps.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 109 + ], + "3min_transcript": "When we first started learning about fractions or rational numbers, we learned about the idea of putting things in lowest terms. So if we saw something like 3, 6, we knew that 3 and 6 share a common factor. We know that the numerator, well, 3 is just 3, but that 6 could be written as 2 times 3. And since they share a common factor, the 3 in this case, we could divide the numerator by 3 and the denominator by 3, or we could say that this is just 3/3, and they would cancel out. And in lowest terms, this fraction would be 1/2. Or just to kind of hit the point home, if we had 8/24, once again, we know that this is the same thing as 8 over 3 times 8, or this is the same thing as 1 over 3 times 8 over 8. The 8's cancel out and we get this in lowest terms as 1/3. These are rational numbers. Rational expressions are essentially the same thing, but instead of the numerator being an actual number and the denominator be an actual number, they're expressions involving variables. So let me show you what I'm talking about. Let's say that I had 9x plus 3 over 12x plus 4. Now, this numerator up here, we can factor it. We can factor out a 3. This is equal to 3 times 3x plus 1. That's what our numerator is equal to. And our denominator, we can factor out a 4. This is the same thing as 4 times 3x. 12 divided by 4 is 3. 12x over 4 is 3x. So here, just like there, the numerator and the denominator have a common factor. In this case, it's 3x plus 1. In this case, it's a variable expression. It's not an actual number, but we can do the exact same thing. They cancel out. So if we were to write this rational expression in lowest terms, we could say that this is equal to 3/4. Let's do another one. Let's say that we had x squared-- let So let's say we had x squared minus 9 over 5x plus 15. So what is this going to be equal to? So the numerator we can factor. It's a difference of squares." + }, + { + "Q": "At 7:15, Sal said that x cannot equal -1. However, x also cannot be equal to 2, based off of the answer he got. Why wouldn't you put x cannot equal negative 2 in the final answer, as that would also make the expression undefined?", + "A": "Now that the expression has been simplified to (x+5)(x-2), it is obvious that x cannot be equal to 2, but the fact that x could also not be equal to -1 (otherwise division by 0 resulted) in the original non simplified expression has been lost, so we add the condition as a reminder. You see, we can set x=-1 into (x+5)(x-2) without a problem, but we could not set x=-1 in the original expression - and this simplified expression is based on that.", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 435 + ], + "3min_transcript": "That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1, so we have to add this condition for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared plus 3x minus 18, all of that over 2x squared plus 5x minus 3. So it's always a little bit more painful to factor things that have a non-one coefficient out here, but we've learned how to do that. We can do it by grouping, and this is a good practice for our grouping, so let's do it." + }, + { + "Q": "At 10:45... could you also write (3x-6)(x+3) as 3(x-2)(x+3), or is my thinking quite erred?", + "A": "You sure can. No, you re right. Take the 3 out of the 3x-6 to get 3(x-2)", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 645 + ], + "3min_transcript": "Notice, all I did here is I split this 3x into a 9x minus 6x. The only difference between this expression and this expression is that I split the 3x into a 9x minus 6x. You add these two together, you get 3x. The way I wrote it right here you can actually ignore the parentheses. And the whole reason why I did that is so I can now group it. And normally, I decide which term goes with which based on what's positive or negative or which has common factors. They both have a common factor with 3. Actually, it probably wouldn't matter in this situation, but I like the 9 on this side because they're both positive. So let's factor out a 3x out of this expression on the left. If we factor out a 3x out of this expression this becomes 3x times x plus 3. we get negative 6 times x plus 3. And now this is very clear our grouping was successful. This is the same thing as-- we can kind of undistribute this as 3x minus 6 times x plus 3. If we were to multiply this times each of these terms, you get that right there. So the top term, we can rewrite it as 3x minus 6-- let me do it in the same color. So we can rewrite it as 3x minus 6 times x plus 3. That's this term right here. I don't want to make it look like a negative sign. That's that term right there. Now let's factor this bottom part over here. Scroll to the left a little bit. to think of two numbers that when I take their product, I get 2 times 3, which is equal to 6, and they need to add up to be 5. And the two obvious numbers here are 2 and 3. I can rewrite this up here as 2x squared plus 2x plus 3x plus 3, just like that. And then if I put parentheses over here, and I decided to group the 2 with the 2 because they have a common factor of 2, and I grouped the 3 with the 3 because they have a common factor of 3. This right here is 2 and a 3. So here we can factor out a 2x. If you factor out a 2x, you get 2x times x plus 1 plus-- you factor out a 3 here-- plus 3 times x plus 1." + }, + { + "Q": "How do we do it from slope-intersect form? 00:59", + "A": "Starting from 5x + 3y = 7, you subtract 5x and divide by 3, so 3y = - 5x + 7 or y = -5/3x + 7/3. He graphed 7/3 as the y intercept, then go down 5 over 3 to get second point. These are not very neat numbers to work with.", + "video_name": "MRAIgJmRmag", + "timestamps": [ + 59 + ], + "3min_transcript": "Solve the system of linear equations by graphing, and they give us two equations here. 5x plus 3y is equal to 7, and 3x minus 2y is equal to 8. When they say, \"Solve the system of linear equations,\" they're really just saying find an x and a y that satisfies both of these equations. And when they say to do it by graphing, we're essentially going to graph this first equation. Remember, the graph is really just depicting all of the x's and y's that satisfy this first equation, and then we graph the second equation that's depicting all of the x's and y's that satisfy that one. So if we were looking for an x and a y that satisfies both, that point needs to be on both equations or it has to be on both graphs. So it'll be the intersection of the two graphs. So let's try to see if we can do that. So let's focus on this first equation, and I want to graph it. So I have 5x plus 3y is equal to 7. There's a couple of ways we could graph this. We could put this in slope-intercept form, You just really need two points to graph a line. So let me just set some points over here. Let's say x and y. When x is equal to 0, what does y need to be equal to? So when x is equal to 0, we have-- let me do it over here-- we have 5 times 0, plus 3 times y, is equal to 7. That's just 0 over there. So you have 3y is equal to 7. Divide both sides by 3, you get y is equal to 7/3, which is the same thing as 2 and 1/3 if we want to write it as a mixed number. Now let's set y equal to 0. So if we set y as equal to 0, we get 5x, plus 3 times 0, is equal to 7. Or in this part right over here, it just becomes 0. So we have 5x is equal to 7. Divide both sides by 5, and we get So let's graph both of these points, and then we should be able to graph this line, or at least a pretty good approximation of that line. So we have the point, 0, 2 and 1/3. So that's that point right over there. So I'll call it 0, 7/3 right over there, and then we have the point, 7/5, 0, or 1 and 2/5, 0. So 1 and 2/5. 2/5 is a little less than a half. So 1 and 2/5, 0. So our line is going to look something like this. I just have to connect the dots. It's always hard to draw the straight line. I'll draw it as a dotted line. So it would look something like this. Normally, when you have to solve a system of equations" + }, + { + "Q": "at 1:30 why does 1/2 B become 1B, but H does not become 2H?", + "A": "The commutative law lets us rearrange the right hand side anyway we want. So, it becomes: (2)(1/2)(bh) 2 * 1/2 => 2/2 => 1 So, we end up with: 1*bh which is bh and now we ve got rid of the 1/2 by moving it to the left hand side.", + "video_name": "eTSVTTg_QZ4", + "timestamps": [ + 90 + ], + "3min_transcript": "The formula for the area of a triangle is A is equal to 1/2 b times h, where A is equal to area, b is equal to length of the base, and h is equal to the length of the height. So area is equal to 1/2 times the length of the base times the length of the height. Solve this formula for the height. So just to visualize this a little bit, let me draw a triangle here. Let me draw a triangle just so we know what b and h are. b would be the length of the base. So this distance right over here is b. And then this distance right here is our height. That is the height of the triangle-- let me do that at a lower case h because that's how we wrote it in the formula. Now, they want us to solve this formula for the height. So the formula is area is equal to 1/2 base times height. And we want to solve for h. We essentially want to isolate the h on one side of the equation. It's already on the right-hand side. So let's get rid of everything else on the right-hand side. We could kind of skip steps if we wanted to. But let's see if we can get rid of this 1/2. So the best way to get rid of a 1/2 that's being multiplied by h is if we multiply both sides of the equation by its reciprocal. If we multiply both sides of the equation by 2/1 or by 2. So let's do that. So let's multiply-- remember anything you do to one side of the equation, you also have to do to the other side of the equation. Now, what did this do? Well, the whole point behind multiplying by 2 is 2 times 1/2 is 1. So on the right-hand side of the equation, we're just going to have a bh. And on the left-hand side of the equation, we have a 2A. And we're almost there, we have a b multiplying by an h. If we want to just isolate the h, we could divide both sides of this equation by b. We're just dividing both sides. You can almost view b as the coefficient on the h. We're just dividing both sides by b. And then what do we get? Well, the right-hand side, the b's cancel out. So we get h-- and I'm just swapping the sides here. h is equal to 2A over b. And we're done. We have solved this formula for the height. And I guess this could be useful. If someone just gave you a bunch of areas and a bunch of base lengths, and they said keep giving me the height for those values, or for those triangles." + }, + { + "Q": "At 3:53, Where did you get the y=2(2) from, please explain", + "A": "Because in the Magenta equation we said X=2 you can bring that over into the other equation as X. This only works in systems though", + "video_name": "GWZKz4F9hWM", + "timestamps": [ + 233 + ], + "3min_transcript": "Now to solve for x, we'll subtract 20 from both sides to get rid of the 20 on the left hand side. On the left hand side, we're just left with the -11x and then on the right hand side we are left with -22. Now we can divide both sides by -11. And we are left with x is equal to 22 divided by 11 is 2, and the negatives cancel out. x = 2. So we are not quite done yet. We've done, I guess you can say the hard part, we have solved for x but now we have to solve for y. We could take this x value to either one of these equations and solve for y. But this second one has already explicitly solved for y and instead of x, we now know that the x value where these two intersect, you could view it that way is going to be equal to 2, so 2 * 2 - 5 let's figure out the corresponding y value. So you get y=2(2)-5 and y = 4 - 5 so y = -1. And you can verify that it'll work in this top equation If y = -1 and x=2, this top equation becomes -3(2) which is -6-4(-1) which would be plus 4. And -6+4 is indeed -2. So it satisfies both of these equations and now we can type it in to verify that we got it right, So, let's type it in... x=2 and y=-1. Excellent, now we're much less likely to be embarassed by talking birds." + }, + { + "Q": "When sal added the two 7x at 2:09, together aren't you supposed to add also the exponents?", + "A": "No, he is just combining like terms. If you have 7 apples and add 7 apples, you get 14 apples not 14 squared apples. You add exponents when you multiply variables, so (7x)2 = (7x)(7x) = 49 x^2. or 5x * 3x^2 = 15 x^3.", + "video_name": "xH_GllPuymc", + "timestamps": [ + 129 + ], + "3min_transcript": "- [Voiceover] Let's see if we can figure out what x plus seven, let me write that a little bit neater, x plus seven squared is. And I encourage you to pause the video and work through it on your own. Alright, now let's work through this together. So we just have to remember, we're squaring the entire binomial. So this thing is going to be the same thing as: x plus seven times x plus seven. I'm gonna write the second x plus seven in a different color, which is going to be helpful when we actually multiply things out. When we see it like this, then we can multiply these out the way we would multiply any binomials. And I'll first do it the, I guess you can say, the slower way, but the more intuitive way, applying the distributive property twice. And then we'll think about maybe some shortcuts or some patterns we might be able to recognize, especially when we are squaring binomials. So let's start with just applying the distributive property twice. So let's distribute this yellow x plus seven over this magenta x plus seven. So we can multiply it by the x, this magenta x, So it's going to be magenta x times x plus seven plus magenta seven times yellow x plus seven. X plus seven, and now we can apply the distributive property again. We can take this magenta x and distribute it over the x plus seven. So x times x is x-squared. X times seven is seven x. And then we can do it again over here. This seven, let me do it in a different color, so this seven times that x is going to be plus another seven x and then the seven times the seven is going to be 49. And we're in the home stretch. We can then simplify it. This is going to be x-squared and then these two middle terms we can add together. seven x plus seven x is going to be 14 x plus 14 x plus 49. Plus 49. And we're done. Now the key question is do we see some patterns here? Do we see some patterns that we can generalize and that might help us square binomials a little bit faster in the future? Well, when we first looked at just multiplying binomials, we saw a pattern like x plus a times x plus b is going to be equal to x-squared, let me write it this way, is going to be equal to x-squared plus a plus b x plus b-squared. And so, if both a and b are the same thing, we can say that x plus a" + }, + { + "Q": "At 14:00 could not SQ RT 39/3 be SQ RT 13 ?", + "A": "sqrt(39)/3 is different from sqrt(39/3). Sqrt(39)/3 means square root of 39 is then divided by 3, while sqrt(39/3) means square root of the value that is 39/3, or 13.", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 840 + ], + "3min_transcript": "" + }, + { + "Q": "At 9:45 why can we not square the negative root thereby getting (36 +- 84)/36?", + "A": "That is not the square of (6 \u00c2\u00b1 \u00e2\u0088\u009a\u00e2\u0088\u009284)/6 For reference: (n + \u00e2\u0088\u009ap)\u00c2\u00b2 = n\u00c2\u00b2+ p +2n\u00e2\u0088\u009ap (n \u00e2\u0088\u0092 \u00e2\u0088\u009ap)\u00c2\u00b2 = n\u00c2\u00b2+ p \u00e2\u0088\u0092 2n\u00e2\u0088\u009ap [n + \u00e2\u0088\u009a(\u00e2\u0088\u0092p)]\u00c2\u00b2= n\u00c2\u00b2\u00e2\u0088\u0092 p +2n\u00f0\u009d\u0091\u0096\u00e2\u0088\u009ap [n \u00e2\u0088\u0092 \u00e2\u0088\u009a(\u00e2\u0088\u0092p)]\u00c2\u00b2 = n\u00c2\u00b2\u00e2\u0088\u0092 p\u00e2\u0088\u00922n\u00f0\u009d\u0091\u0096\u00e2\u0088\u009ap", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 585 + ], + "3min_transcript": "" + }, + { + "Q": "At 13:34, Sal simplified the fraction, but I''m not clear how he did it. What happened to the two? Does this simplification leave the square root of 39 alone?\nThanks", + "A": "Yes, just think of the sqrt 39 as some ugly thing multiplied to the 2. When you have (-12 +or- 2*sqrt39) / -6 , there are 3 distinct terms: the neg 12, the 2*sqrt39, and the neg 6. All 3 of those terms have 2 as a factor. That is, neg 12 = 2*neg 6 2*sqrt39 neg 6 = 2*neg 3 So all three terms have a factor of 2, so the 2 can be factored out by dividing each term by 2. That leaves neg 6 sqrt39 neg 3", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 814 + ], + "3min_transcript": "" + }, + { + "Q": "At 13:23, why doesn't he divide the number inside the radical by 2?", + "A": "Terms are things we add or subtract. They are held together by multiplication and division. The numerator only has 2 terms : -12 and 2*sqrt39. Both of those terms were divided by 2 to get -6 and sqrt39. But, if the numerator had been -12 + 2 + sqrt39, (in other words, 3 terms) and then we divided by 2, we would get -6 + 1 + (sqrt39)/2 ( Of course, that would simplify to -5 + (sqrt39)/2. )", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 803 + ], + "3min_transcript": "" + }, + { + "Q": "Isn't there a small mistake at 16:10 ? Sal says \"a little less than one\", where the graph shows a little less than 0 ...", + "A": "he says maybe close to zero but i little bit less than that", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 970 + ], + "3min_transcript": "" + }, + { + "Q": "at 4:16 why did it change to 10 why didn't you put 100?i,m sorry i am so very new to this", + "A": "the 100 was under a square root and then he took the square root of 100 to get 10", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 256 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:05 Sal says that the quadratic formula is at least top 5 most useful formulas. I wonder what other formulas would be as useful as the quadratic formula.", + "A": "off the top of my head: the Pythagorean Theorem <-this is certainly one of the most useful if not the most useful formula, Area of a Triangle (you can usually break other shapes into a series of triangles), Combinations/Permutations, and maybe taking an Average.", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 5 + ], + "3min_transcript": "" + }, + { + "Q": "At 7:11, Sal describes the system as existing in R4, but isn't it also safe to describe this as 4 column vectors in R3?", + "A": "For this system of equations Ax = b , A , not the system, is 4 column vectors. x is an R^4 variable vector, each row is a plane in R^4 , and Ax is constrained to b , an R^3 column vector of constants.", + "video_name": "JVDrlTdzxiI", + "timestamps": [ + 431 + ], + "3min_transcript": "entry is in a lower row than that one. So it's in a column to the right of this one right there. And I just inspected, this looks like a-- this column two looks kind of like a free variable-- there's no pivot entry there, no pivot entry there. But let's see, let's map this back to our system of equations. These are just numbers to me and I just kind of mechanically, almost like a computer, put this in reduced row echelon form. Actually, almost exactly like a computer. But let me put it back to my system of linear equations, to see what our result is. So we get 1 times x1, let me write it in yellow. So I get 1 times x1, plus 2 times x2, plus 0 times x3, plus 3 times x4 is equal to 4. Obviously I could ignore this term right there, I didn't even have to write it. Actually. Then I get 0 times x1, plus 0 times x2, plus 1 times x3, so 1 times x3, minus 2 times x4, is equal to 4. And then this last term, what do I get? I get 0 x1, plus 0 x2 plus 0 x3 plus 0 x4, well, all of that's equal to 0, and I've got to write something on the left-hand side. So let me just write a 0, and that's got to be equal to minus 4. Well this doesn't make any sense whatsoever. 0 equals minus 4. This is this is a nonsensical constraint, this is impossible. 0 can never equal minus 4. This is impossible. Which means that it is essentially impossible to find an intersection of these three systems of equations, or a solution set that satisfies all of them. When we looked at this initially, at the beginning of the of the video, we said there are only three equations, we have four unknowns, maybe there's going But turns out that these three-- I guess you can call them these three surfaces-- don't intersect in r4. These are all four dimensional, we're dealing in r4 right here, because we have-- I guess each vector has four components, or we have four variables, is the way you could think about it. And it's always hard to visualize things in r4. But if we were doing things in r3, we can imagine the situation where, let's say we had two planes in r3. So that's one plane right there, and then I had another completely parallel plane to that one. So I had another completely parallel plane to that first one. Even though these would be two planes in r3, so let me give So let's say that this first plane was represented by the equation 3x plus 6y plus 9z is equal to 5, and the second" + }, + { + "Q": "11:30. To have an infinte number of solutions does one needs to have free variables AND a row of all Zeroes? The video was unclear on this point but alluded to it.", + "A": "row of 0 s is not a necessary condition, e.g. x1 - x2 = 5 x1 - x2 + x3 = 3 reduces to 1 -1 0 | 5 0 0 1 | -2 ` with x2 being free. The solution is (x1, x2, x3) = (5, 0, -2) + x2(1, 1, 0) If you think of a row being a constraint to the solution(s), a row of zero s seems to indicate one that is a linear combination of the remaining one s i.e. it is a superfluous constraint.", + "video_name": "JVDrlTdzxiI", + "timestamps": [ + 690 + ], + "3min_transcript": "of parallel equations, they won't intersect. And you're going to get, when you put it in reduced row echelon form, or you just do basic elimination, or you solve the systems, you're going to get a statement that zero is equal to something, and that means that there are no solutions. So the general take-away, if you have zero equals something, no solutions. If you have the same number of pivot variables, the same number of pivot entries as you do columns, so if you get the situations-- let me write this down, this is good to know. if you have zero is equal to anything, then that means no solution. If you're dealing with r3, then you probably have parallel planes, in r2 you're dealing with parallel lines. If you have the situation where you have the same number of pivot entries as columns, so it's just 1, 1, 1, 1, this I think you get the idea. That equals a, b, c, d. Then you have a unique solution. Now if, you have any free variables-- so free variables look like this, so let's say we have 1, 0, 1, 0, and then I have the entry 1, 1, let me be careful. 0, let me do it like this. 1, 0, 0, and then I have the entry 1, 2, and then I have a bunch of zeroes over here. And then this has to equal zero-- remember, if this was a bunch of zeroes equaling some variable, then I would have no solution, or equalling some constant, let's say this is equal to 5, this is equal to 2. If this is our reduced row echelon form that we eventually get to, then we have a few free variables. This is a free, or I guess we could call this column a free Because it has no pivot entries. These are the pivot entries. So this is variable x2 and that's variable x4. Then these would be free, we can set them equal to anything. So then here we have unlimited solutions, or no unique solutions. And that was actually the first example we saw. And these are really the three cases that you're going to see every time, and it's good to get familiar with them so you're never going to get stumped up when you have something like 0 equals minus 4, or 0 equals 3. Or if you have just a bunch of zeros and a bunch of rows. I want to make that very clear. Sometimes, you see a bunch of zeroes here, on the left-hand side of the augmented divide, and you might say, oh maybe I have no unique solutions, I have an infinite number of solutions. But you have to look at this entry right here. Only if this whole thing is zero and you have free variables, then you have an infinite number of solutions. If you have a statement like, 0 is equal to a, if this is equal to 7 right here, then all of the sudden you would" + }, + { + "Q": "At about 2:08, are we just supposed to assume the solid has the same volume as the shell?", + "A": "Conceptually, we begin by estimating the volume of the solid by adding together the volumes of thin shells that make up the solid. As the shells get thinner and thinner, the estimate of the volume of the solid becomes more accurate. The limit of this sum as the thickness of the shells approaches zero is the integral, which is equal to the actual volume of the solid.", + "video_name": "6Ozz3J-LRrY", + "timestamps": [ + 128 + ], + "3min_transcript": "I've got the function y is equal to x minus 3 squared times x minus 1. And what I want to do is think about rotating the part of this function that sits right over here between x is equal to 1 and x equals 3. And x equals 3 and x equals 1 are clearly the zeroes of this function right over here. And I want to take this region and rotate it around the y-axis. And if I did that, I'd get a shape that looks something like that. And I want to figure out the volume of that shape. And what we're going to do is a new method called the shell method. And the reason we're going to use the shell method-- you might say, hey, in the past, we've rotated things around a vertical line before. We used the disk method. We wrote everything as a function of y, et cetera, et cetera. We created all of these disks. We figured out the volume of each of those disks. But the problem here is this is hard to express as a function of y. How do you solve explicitly for y right over here? So instead, we're going to keep things in terms of x we can come up with the volume. What we're going to imagine instead-- instead of constructing disks, we're going to construct shells. And what do I mean by a shell? So for each x at the interval, on this kind of cut of it, we can construct a rectangle. And what happens if we were to rotate this rectangle? So this is the rectangle right over here. What happens if we rotate this rectangle around the y-axis along with everything else? I'll try my best attempt to draw it. It's going to look something like this. This is challenging my art skills, but I think I can handle it. So it's going to look something not too dissimilar to that right over there. So it looks like a hollowed-out cylinder. I guess that's why we call it a shell. And it's going to have some depth. The depth is going to be dx. going to be the value of my function. The height is f of x. In this case, f of x is x minus 3 squared times x minus 1. How do we figure out the volume of a cylinder like this? Well, if we can figure out the circumference of the cylinder, and then multiply that circumference times the height of the cylinder, we'd essentially figure out the area of the outside surface of our cylinder. And then if we multiply the area of the outside surface of our cylinder by that infinitesimally small depth, then that'll give us the volume-- I shouldn't say cylinder-- of our shell. So let's try to do it. What is the circumference of a shell? What is the circumference of one shell going to be? Well, it's going to be 2 pi times the radius of that shell." + }, + { + "Q": "at 4:32, why doesnt the pi/sqrt(2) distribute to the c? I dont think that it matters, but I would like to know.", + "A": "You re on the right track when you say I don t think it matters. The variable c stands for an arbitrary (unspecified, unknown) constant, which could be any real number. If you multiply an arbitrary constant by some other constant such as \u00cf\u0080/\u00e2\u0088\u009a2, you re still left with an arbitrary constant, so we continue to call it c and don t bother to multiply it by another constant.", + "video_name": "8Yl_u_Otcjg", + "timestamps": [ + 272 + ], + "3min_transcript": "And the square root of cosine squared theta is going to be cosine theta. Now, you might be saying, hey, wait, if I take the square root of something squared, then wouldn't that just be the absolute value of cosine theta? In order to take away the absolute value, I'd have to assume that cosine theta is positive. But we can make that assumption that cosine theta is positive, because if we look right here at this part of our substitution, if we wanted to solve for theta, you'd divide both sides by 2, and you'd get x of 2 is equal to sine of theta. Or we could say that theta is equal to arcsin of x over 2. Now the arcsin function, as it is traditionally defined, will return theta that is between negative pi over 2 and pi over 2. And in that range, cosine of theta is always going to be positive. So we don't have to write the absolute value. We know cosine theta is positive. Cosine theta cancels out with cosine theta. This 2 cancels out with this 2. We can bring this square root of 2 outside. And so we are left with pi over this square root of 2 times the indefinite integral of just d theta. And this is just going to be equal to pi over the square root of 2 times theta plus c. And we're almost done. We just have to rewrite this in terms of x. And we already know that theta is equal to arcsin of x over 2. So we can say that this indefinite integral, or the antiderivative of this expression, is going to be pi over the square root of 2 times arcsin of x over 2 plus c. And we're done. Some people like a square root of 2 in the denominator. If you want to remove it, you can multiply this by square root of 2 over square root of 2, and that will simplify it. But right now, I'll just leave the denominator And this right over here is our antiderivative." + }, + { + "Q": "At 1:15 he changes 25% into a decimal. So does that mean that percent is basically the same thing as decimals?", + "A": "yes, a percentage is always a decimal. You can arrive at the decimal by dividing the number by 100. So 25% = 25/100 = 0.25", + "video_name": "JaScdH47PYg", + "timestamps": [ + 75 + ], + "3min_transcript": "We're asked to identify the percent, amount, and base in this problem. And they ask us 150 is 25% of what number? So another way to think about it is 25% times some number. So I'll do 25% in yellow. And 25% times some number is equal to 150. So the percent is pretty easy to spot. We have a 25% right over here. So this is going to be the percent. And we're multiplying the percent times some base number. So this right over here is the base. So we have the percent times the base. We have the percent times the base is equal to some amount. This is essentially saying 25% of some number, 25% times some number is equal to 150. If it helps, we could rewrite this as 0.25, which is the same thing as 25%. 0.25 times some number is equal to 150. And one interesting thing is just to think about, should that number be larger or smaller than 150? Well, if we only take 25% of that number, if we only take 25 hundredths of that number, if we only take 1/4 of that number, because that's what 25 hundredths is, or that's what 25% is, we get 150. So this number needs to be larger than 150. In fact, it has to be larger than 150 by 4. And to actually figure out what the number is, we can actually multiply. Since this, what's on the left-hand side is equal to the right-hand side, if we want to solve this, we can multiply both sides by 4. If we say, look, we have some value over here, In order for it to still be equal, we have to multiply 150 times 4. 4 times 0.25, or 4 times 25%, or 4 times 1/4, this is just going to be 1. And we're going to get our number is equal to 150 times 4 or it is equal to 600. And that makes sense, 25% of 600 is 150. 1/4 of 600 is 150." + }, + { + "Q": "At 4:30, why does Sal say that you can't simplify -21/20? Can't you simplify it to\n-1 1/20? Or is it -21 over +20? :(", + "A": "First of all you can t simplify one part of the fraction when you want to simplify a fraction you must do it for both numerator and denominator and in this fraction you can t simplify it any further because there isn t any common factor between them (21)=3*7 (20)=4*5 hopefully this helps you", + "video_name": "pi3WWQ0q6Lc", + "timestamps": [ + 270 + ], + "3min_transcript": "you see that there is already common factors in the numerator and the denominator. Both the numerator and the denominator, they're both divisible by 5 and they're both divisible by 3, which essentially tells us that they're divisible by 15. So we can divide the numerator and denominator by 15. So divide the numerator by 15, which is just like dividing by 5 and then dividing by 3. So we'll just divide by 15. Divide by 15. And this is going to be equal to-- well, 5 times 3 is 15. Divided by 15 you get 1 in the numerator. And in the denominator, 9 times 15 divided by 15. Well, that's just going to be 9. So it's equal to 1/9. Let's do another one. What would negative 5/9 times negative 3/15 be? Well, we've already figured out what positive 5/9 times positive 3/15 would be. So now we just have to care about the sign. But now we have to think about the fact that we're multiplying by a negative times a negative. Now, we remember when you multiply a negative times a negative, it's a positive. The only way that you get a negative is if one of those two numbers that you're taking the product of is negative, not two. If both are positive, it's positive. If both are negative, it's positive. Let's do one more example. Let's take 5-- I'm using the number 5 a lot. So let's do 3/2, just to show that this would work with improper fractions. 3/2 times negative 7/10. I'm arbitrarily picking colors. And so our numerator is going to be 3 times negative 7. 3 times negative 7. 2 times 10. So this is going to be the numerator. Positive times a negative is a negative. 3 times negative 7 is negative 21. Negative 21. And the denominator, 2 times 10. Well, that is just 20. So this is negative 21/20. And you really can't simplify this any further." + }, + { + "Q": "In 1:42 I don't get it.", + "A": "this is totally new exercise as 2 fractions times each other that means u can write them in one fraction and the value is still same. in 2 fractions numerators r always times each other and denominators r always time each other but numerators and denominators r always divide that is why Sal wrote 5 x 3 (numerators) and 9 x 15 (denominators). hope this will help.", + "video_name": "pi3WWQ0q6Lc", + "timestamps": [ + 102 + ], + "3min_transcript": "Let's do a few examples multiplying fractions. So let's multiply negative 7 times 3/49. So you might say, I don't see a fraction here. This looks like an integer. But you just to remind yourself that the negative 7 can be rewritten as negative 7/1 times 3/49. Now we can multiply the numerators. So the numerator is going to be negative 7 times 3. And the denominator is going to be 1 times 49. 1 times 49. And this is going to be equal to-- 7 times 3 is 21. And one of their signs is negative, so a negative times a positive is going to be a negative. So this is going to be negative 21. You could view this as negative 7 plus negative 7 plus negative 7. And that's going to be over 49. And this is the correct value, but we can simplify it more because 21 and 49 both share 7 as a factor. So let's divide both the numerator and the denominator by 7. Divide the numerator and the denominator by 7. And so this gets us negative 3 in the numerator. And in the denominator, we have 7. So we could view it as negative 3 over 7. Or, you could even do it as negative 3/7. Let's do another one. Let's take 5/9 times-- I'll switch colors more in this one. That one's a little monotonous going all red there. 5/9 times 3/15. So this is going to be equal to-- we multiply the numerators. So it's going to be 5 times 3. 5 times 3 in the numerator. And the denominator is going to be 9 times 15. 9 times 15. you see that there is already common factors in the numerator and the denominator. Both the numerator and the denominator, they're both divisible by 5 and they're both divisible by 3, which essentially tells us that they're divisible by 15. So we can divide the numerator and denominator by 15. So divide the numerator by 15, which is just like dividing by 5 and then dividing by 3. So we'll just divide by 15. Divide by 15. And this is going to be equal to-- well, 5 times 3 is 15. Divided by 15 you get 1 in the numerator. And in the denominator, 9 times 15 divided by 15. Well, that's just going to be 9. So it's equal to 1/9. Let's do another one. What would negative 5/9 times negative 3/15 be? Well, we've already figured out what positive 5/9 times positive 3/15 would be. So now we just have to care about the sign." + }, + { + "Q": "why at 1:18 does neg. numerator over pos. denominator (-3/+7) become the whole fraction neg. ?", + "A": "You may be aware of the mathematical rule which dictates that if you divide a negative by a positive, you get a negative. A fraction basically means to divide the top number, or the numerator, by three bottom number, the denominator. EXAMPLE: -2/4 We know that +2/+4 is 1/2, so we can simplify -2/4 to -1/2, or -0.5 I hope this this helps!", + "video_name": "pi3WWQ0q6Lc", + "timestamps": [ + 78 + ], + "3min_transcript": "Let's do a few examples multiplying fractions. So let's multiply negative 7 times 3/49. So you might say, I don't see a fraction here. This looks like an integer. But you just to remind yourself that the negative 7 can be rewritten as negative 7/1 times 3/49. Now we can multiply the numerators. So the numerator is going to be negative 7 times 3. And the denominator is going to be 1 times 49. 1 times 49. And this is going to be equal to-- 7 times 3 is 21. And one of their signs is negative, so a negative times a positive is going to be a negative. So this is going to be negative 21. You could view this as negative 7 plus negative 7 plus negative 7. And that's going to be over 49. And this is the correct value, but we can simplify it more because 21 and 49 both share 7 as a factor. So let's divide both the numerator and the denominator by 7. Divide the numerator and the denominator by 7. And so this gets us negative 3 in the numerator. And in the denominator, we have 7. So we could view it as negative 3 over 7. Or, you could even do it as negative 3/7. Let's do another one. Let's take 5/9 times-- I'll switch colors more in this one. That one's a little monotonous going all red there. 5/9 times 3/15. So this is going to be equal to-- we multiply the numerators. So it's going to be 5 times 3. 5 times 3 in the numerator. And the denominator is going to be 9 times 15. 9 times 15. you see that there is already common factors in the numerator and the denominator. Both the numerator and the denominator, they're both divisible by 5 and they're both divisible by 3, which essentially tells us that they're divisible by 15. So we can divide the numerator and denominator by 15. So divide the numerator by 15, which is just like dividing by 5 and then dividing by 3. So we'll just divide by 15. Divide by 15. And this is going to be equal to-- well, 5 times 3 is 15. Divided by 15 you get 1 in the numerator. And in the denominator, 9 times 15 divided by 15. Well, that's just going to be 9. So it's equal to 1/9. Let's do another one. What would negative 5/9 times negative 3/15 be? Well, we've already figured out what positive 5/9 times positive 3/15 would be. So now we just have to care about the sign." + }, + { + "Q": "at 5:05 is he writing out what he did in the model?", + "A": "Yep, that is correct. Sal (the speaker) just replaced the equation with absolute values, which are the lines on both sides of the subtraction.", + "video_name": "DPuK6ZgBGmE", + "timestamps": [ + 305 + ], + "3min_transcript": "Once again, this is two, this is three. She deviates. Her absolute deviation is three. And then we wanna take the mean of the absolute deviation. That's the M in MAD, in Mean Absolute Deviation. This is Manueala's absolute deviation, Sophia's absolute deviation, Jada's absolute deviation, Tara's absolute deviation. We want the mean of those, so we divide by the number of datapoints, and we get zero plus one, plus two, plus three, is six over four. Six over four, which is the same thing as 1 1/2. Or, lemme just write it in all the different ways. We could write it as three halves, or 1 1/2, or 1.5. Which gives us a measure of how much do these datapoints vary from the mean of four. I know what some of you are thinking. \"Wait, I thought there was a formula \"associated with the mean absolute deviation. \"It seems really complex. \"It has all of these absolute-value signs That's all we did. When we write all those absolute-value signs, that's just a fancy way of looking at each datapoint, and thinking about how much does it deviate from the mean, whether it's above or below. That's what the absolute value does. It doesn't matter, if it's three below, we just say three. If it's two above, we just say two. We don't put a positive or negative on. Just so you're comfortable seeing how this is the exact same thing you would've done with the formula, let's do it that way, as well. So the mean absolute deviation is going to be equal to. Well, we'll start with Manueala. How many bubbles did she blow? She blew four. From that you subtract the mean of four, take the absolute value. That's her absolute deviation. Of course, this does evaluate to this zero, to zero here. Then you take the absolute value. Sophia blew five bubbles, and the mean is four. Then you do that for Jada. Jada blew six bubbles; the mean is four. And then you do it for Tara. Then you divide it by the number of datapoints you have. Lemme make it very clear. This right over here, this four, is the mean. This four is the mean. You're taking each of the datapoints, and you're seeing how far it is away from the mean. You're taking the absolute value 'cause you just wanna figure out the absolute distance. Now you see, or maybe you see. Four minus four, this is. Four minus four, that is a zero. That is that zero right over there. Five minus four, absolute value of that? That's going to be. Lemme do this in a new color. This is just going to be one. This thing is the same thing as that over there. We were able to see that just by inspecting this graph, or this chart. And then, six minus four, absolute value of that, that's just going to be two. That two is that two right over here, which is the same thing as this two right over there. And then, finally, our one minus four, this negative three," + }, + { + "Q": "At 4:58, Sal says X squared equals -1. Shouldn't it be 1?", + "A": "No. He has x^2 + 1 = 0 You must subtract 1 from both sides to isolate x^2 (remember, we use the opposite operation to move items to the other side of an equation). This creates Sal s version: x^2 = -1", + "video_name": "uFZvWYPfOmw", + "timestamps": [ + 298 + ], + "3min_transcript": "That would somehow imply that you have only one complex root, which that is not a possibility. Now another way that you could have thought about this-- and this would have been the longer way. But let's say you didn't have the graphs here for you, and someone asked you to just find the roots-- well, you could have attempted to factor this. And this one actually is factorable. y is equal to x to the third plus 3x squared plus x plus 3. As mentioned in previous videos, factoring things of a degree higher than 2, there is something of an art to it. But oftentimes, if someone expects you to, you might be able to group things in interesting ways, especially when you see that several terms have some common factors. So for example, these first two terms right over here have the common factor x squared. So if you were to factor that out, you would get x squared times x plus 3, which is neat because that looks a lot like the second two terms. And then you can factor the x plus 3 out. We could factor the x plus 3 out, and we would get x plus 3 times x squared plus 1. And now, your 0's are going to happen, or this whole y-- remember this is equal to y-- y is going to equal 0 if either one of these factors is equal to 0. So when does x plus 3 equal 0? Well, subtract 3 from both sides. That happens when x is equal to negative 3. When does x squared plus 1 equal 0, I should say? Well, when x squared is equal to negative 1. Well, there's no real x's, no real valued x's. squared is equal to negative 1. x is going to be an imaginary-- or I guess I'll just say it in more general terms-- it's going to be complex valued. So once again, you see you're going to have a pair of complex roots, and you have one real root at x is equal to negative 3." + }, + { + "Q": "At the 1:53 mark he says something like \"you could probably draw a better freehand diagram\". My opinion: I definitely cannot. This guy seems to be good at explaining math in these videos-who is he?", + "A": "He s Richard Rusczyk (yeah, I probably butchered his name), but if you look him up on google you can find more info about him. He s pretty famous.", + "video_name": "rcLw4BlxaRs", + "timestamps": [ + 113 + ], + "3min_transcript": "We've got some 3D geometry here so we're going to have read carefully, visualize what's going on because it's kind of hard to draw in 3D. We got six spheres with a radius 1. Their centers are at the vertices of a regular hexagon that has side length 2. So we're starting with a regular hexagon, and we're going to put spheres centered at each of the vertices. And since the radius of each sphere is 1, side length is 2, that means each of these spheres is going to be tangent to its two neighbors. So we start off with a hexagon, six spheres, each one tangent to each of its two neighbors. And then we're going to have a larger sphere centered at the center of the hexagon such that it's tangent to each of the little spheres. Now, each of the little spheres will touch the inside of this giant sphere. And then we bring out an eighth sphere that's externally tangent to the six little ones. So we got out six little ones down here around the hexagon, and we're going to take this new sphere and just set it right on top of those six. And it's going to touch-- right at the top of it, So we have at least somewhat of a picture of what's going on here, and we want the radius of this last sphere that we dropped in at the top there. And now one thing I like to do with these 3D problems is I like to take 2D cross sections, turn 3D problems into 2D problems. So when I have a problem with a whole bunch of spheres, I like to throw my cross sections through centers of those spheres and through points of tangency whenever I have tangent spheres. Now, a natural place to start here, of course, is the hexagon. We take the cross section with the hexagon, because that's going to go through the centers of seven of these spheres and all kinds of points tangency. So to start off, we'll draw a regular hexagon. And you're going to have to bear with me. On the test, of course, you've got your ruler, you got your protractor, you got your compass so you can draw a perfect diagram. You could probably draw a freehand better diagram better than I can, too. But when we take cross sections of our spheres, we make circles. And we include the points of tangency in this cross section. A cross section of that is a circle that touches each of these little circles. All right, and there we go. This is the cross section through the hexagon. Now we can label some lengths. We know that the radii of the little spheres is 1, and one thing that's really nice about regular hexagons is you can break them up into equilateral triangles. So this is an equilateral triangle. Here's the center of the hexagon center and the big circle, and I can extend this out to the point of tangency of small sphere and the big one. So we know this is 1 because it's a radius of the small sphere. This is 1. This is an equilateral triangle so this side is the same as this side. So it tells us that this is 1, and now we know that the radius of the giant sphere is 3. So we've got the radius of the giant sphere. We got the radii of all these little spheres. All we have left is that eighth sphere we sat on top. And, of course, that sphere's not in this diagram. It's sitting right up here." + }, + { + "Q": "At 1:15... I know this is right, I just need an explanation.\n\nHow does x^2 go into x^3 and x^4?", + "A": "5 goes into 15, 3 times (5*3=15) 8 goes into 16, 2 times (8*2=16) 9 goes into 54, 6 times(9*6=54) x^2 goes into x^3, x times (x^2 * x=x^3) x^2 goes into x^4, x^2 times (x^2 * x^2 = x^4)", + "video_name": "MZl6Mna0leQ", + "timestamps": [ + 75 + ], + "3min_transcript": "We're told to factor 4x to the fourth y, minus 8x to the third y, minus 2x squared. So to factor this, we need to figure out what the greatest common factor of each of these terms are. So let me rewrite it. So we have 4x to the fourth y, and we have minus 8x to the third y, and then we have minus 2x squared. So in the other videos, we looked at it in terms of breaking it down to its simplest parts, but I think we have enough practice now to be able to do a little bit more of it in our heads. So what is the largest number that divides into all of these? When I say number I'm talking about the actual, I guess, coefficients. We have a 4, an 8, a 2. We don't have to worry about the negative signs just yet. And we say, well, the largest, of, the largest common factor of 2, 8 and 4 is 2. 2 goes into all of them, and obviously that's the largest number that can go into 2. So that is the largest number that's going to be part of the greatest common factor. So it's going to be 2. And then what's the greatest, I guess, factor, what's the greatest degree of x that's divisible into all three of these? Well, x squared goes into all three of these, and obviously that's the greatest degree of x that can be divided into this last term. So x squared is going to be the greatest common x degree in all of them. 2x squared. And then what's the largest degree of y that's divisible into all of them? Well, these two guys are divisible by y, but this guy isn't, so there is no degree of y that's divisible into all of them. So the greatest common factor of all three of these guys right here is 2x squared. So what we can do now is we can think about each of these terms as the product of the 2x squared and something else. And to figure that something else we can literally undistribute the 2x squared, say this is the same thing, or even before we undistribute the 2x squared, we could say look, 4x to the fourth y is the same thing as 2x squared, Right? If you just multiply this out, you get 4xy. Similarly, you could say that 8x to the third y-- I'll put the negative out front-- is the same thing as 2x squared, our greatest common factor, times 8x to the third y, over 2x squared. And then finally, 2x squared is the same thing as if we factor out 2x squared-- so we have that negative sign out front-- if we factor out 2x squared, it's the same thing as 2x squared, times 2x squared, over 2x squared. This is almost silly what I'm doing here, but I'm just showing you that I'm just multiplying and dividing both of these terms by 2x squared. Multiplying and dividing. Here it's trivially simple. This just simplifies to 2x squared right there, or this 2x squared times 1." + }, + { + "Q": "At @5:08 you say that the span of T is V but couldn't you also say that it is T itself since a span, unlike a basis, doesn't necessarily need to have linearly independent vectors?", + "A": "T was just S with one extra vector. V in the example was Span(S) = Span(T). V in theory should have infinitely more vectors than T.", + "video_name": "zntNi3-ybfQ", + "timestamps": [ + 308 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:55, it says a sub i is equal to a sub i minus i times 9/10. Is this supposed to be i-1?", + "A": "It s confusing... the dot is actually the multiplication symbol, not a dot on a i . So, Sal does have i-1", + "video_name": "DY9Q3qNmZnw", + "timestamps": [ + 295 + ], + "3min_transcript": "but notice we have a change in sign here, and the key thing is to say well to go from term to the next what are we multiplying by? Well to go from the first term to the second term, we multiply by negative 0.99. And then, so we're multiplying by negative 0.99. Now to go to the next term, we're again multiplying by negative 0.99, so the common ratio is not positive 0.99, but negative zero, negative 0.99, so let me write that, negative 0.99, and of course that is going to be to the 80th power, all over one minus negative 0.99. And so we could simplify this a little bit, this is all going to be equal to, and so this is going to be one minus, so negative 0.99 to the 80th power, I should put parenthesis there to make sure we are taking the negative 0.99 to the 80th power. Well, we're taking it to an even power, so it's going to be positive, so that's going to be the same thing as 0.99 to the 80th power, and all of that over, well subtracting a negative that's just gonna be adding the positive, so all of that over 1.99, and we could attempt to simplify it more but, if we had a calculator we could actually find this exact value or close value actually, most calculators don't give you the exact value when you take something to the 80th power, but this is what that sum is going to be. Let's do one more of these. Alright, so here we have a series defined recursively and so it's useful to just think about So the first term is 10, and then the next term, so the second term A sub two is equal to A sub one times 9/10, alright. So the next term is gonna be the previous term times 9/10, so it's gonna be 10 times nine over 10, and then the next term is gonna be that times, is gonna be the second term, the third term is the second term times 9/10, so 10 times nine over 10, nine over 10 squared. And the way it's written right now, we don't have it written as a finite geometric series, so let's say we wanna take the sum, let's say we want the sum of first, first I don't know 30 terms, sum of first 30 terms. So what will this be? Well we're gonna take S sub one, S sub 30, oh I wrote ten, S sub 30, the sum of the first 30 terms, is gonna be equal to the first term, we've done this before," + }, + { + "Q": "At 3:30 the instructor immediately went to subtracting. But I thought the order of operations was PEDMAS so shouldn't the division go first? The way I did it, was first doing all the divisions then I found the last equation to be (d/ac)-(d/bc)", + "A": "The division bar in the middle of a fraction asks as a grouping symbol. When you have 2 terms in the numerator of a fraction or even 2 terms in the denominator, there are implied sets of parentheses the numbers in the numerator / denominator. So, Sal did the parentheses 1st by subtracting the 2 fractions in the numerator, then doing the division. Sometimes it s easier to see a simpler example: (2+3) / 10 You add 1st = 5 / 10 Then, you divide = 0.5 or 1/2", + "video_name": "_BFaxpf35sY", + "timestamps": [ + 210 + ], + "3min_transcript": "And once again, encourage you, encourage you to pause the video and figure it out on your own. Well, when you divide by a fraction, it is equivalent to multiplying by it's, by it's reciprocal. So this is going to be the same thing as a over b, a over b times, times the reciprocal of this. So times d over, I'm going to use the same color just so I don't confuse you, that d was purple, times d over c, times d over c and then it reduces to a problem like this. You know, and I shouldn't even use this multiplication symbol now that we're in algebra because you might confuse that with an x, so let me write that as times, times d, d over c, times d over c, Well the numerator you're going to have a times d, so it's ad, a, d, over, over bc, over b times c. Now let's do one that's maybe a little bit more involved and see if you can pull it off. So let's say that I had, let's say that I had, I don't know, let me write it as 1 over a, minus 1 over b, all of that over, all of that over c, and let's say, let's also divide that by 1 over d. So this is a more involved expression then what we've seen so far but I think we have all the tools to tackle it so I encourage you to pause the video and see if you can simplify this, if you can actually carry out these operations and come up with a one fraction that represents this. so 1 over a minus 1 over b, let me work through just that part by itself. So 1 over a minus 1 over b, we know how to tackle that, we can find a common denominator, let me write it up here. So 1 over a minus 1 over b, is going to be equal to, we can multiply 1 over a times b over b, so it's going to be b over ba, notice I haven't changed it's value, I just multiplied it times 1, b over b, minus, well I'm going to multiply the numerator and denominator here by a, - a over ab, or I could write that as ba. And the whole reason why I did that is to have the same denominator. So this is going to be equal to b-a, over, I could write is a ba or ab. So this is going to be equal to, this is going to be equal to this numerator right over here, b-a, over ab," + }, + { + "Q": "At 8:16, how did he get A^2=3/4 h^2?", + "A": "Simple algebraic solving techniques. if you take h to be c like it is normally represented, and assume a^2 + b^2 = c^2... and take the 30-60-90 side ratio definition where the side opposite the 30 degree angle is c/2, or hypotenuse/2, then (c/2)^2 + a^2 = c^2. Subtract (c/2)^2 from both sides... you get a^2 = c^2 -(c/2)^2. Fraction stuff... a^2 = (4c/4)^2 - (c/2)^2 so a^2 = (3/4c)^2.", + "video_name": "Qwet4cIpnCM", + "timestamps": [ + 496 + ], + "3min_transcript": "Because that's h over 2, and this is also h over 2. Right over here. So if we go back to our original triangle, and we said that this is 30 degrees and that this is the hypotenuse, because it's opposite the right angle, we know that the side opposite the 30 degree side is 1/2 of the hypotenuse. And just a reminder, how did we do that? Well we doubled the triangle. Turned it into an equilateral triangle. Figured out this whole side has to be the same as the hypotenuse. And this is 1/2 of that whole side. So it's 1/2 of the hypotenuse. So let's remember that. The side opposite the 30 degree side is 1/2 of the hypotenuse. Let me redraw that on another page, because I think this is getting messy. So going back to what I had originally. This is a right angle. This is the hypotenuse-- this side right here. If this is 30 degrees, we just derived that the side opposite that this is equal to 1/2 the hypotenuse. If this is equal to 1/2 the hypotenuse then what is this side equal to? Well, here we can use the Pythagorean theorem again. We know that this side squared plus this side squared-- let's call this side A-- is equal to h squared. So we have 1/2 h squared plus A squared is equal to h squared. This is equal to h squared over 4 plus A squared, is equal to h squared. Well, we subtract h squared from both sides. We get A squared is equal to h squared minus h squared over 4. This is equal to 3/4 h squared. And once going that's equal to A squared. I'm running out of space, so I'm going to go all the way over here. So take the square root of both sides, and we get A is equal to-- the square root of 3/4 is the same thing as the square root of 3 over 2. And then the square root of h squared is just h. And this A-- remember, this is an area. This is what decides the length of the side. I probably shouldn't have used A. But this is equal to the square root of 3 over 2, times h. We've derived what all the sides relative to the hypotenuse are of a 30-60-90 triangle. So if this is a 60 degree side." + }, + { + "Q": "at 0:23, where did you get the radical 2 over 2 from?", + "A": "That was explained in the earlier video about 45-45-90 triangles.", + "video_name": "Qwet4cIpnCM", + "timestamps": [ + 23 + ], + "3min_transcript": "Sorry for starting the presentation with a cough. I think I still have a little bit of a bug going around. But now I want to continue with the 45-45-90 triangles. So in the last presentation we learned that either side of a 45-45-90 triangle that isn't the hypotenuse is equal to the square route of 2 over 2 times the hypotenuse. Let's do a couple of more problems. So if I were to tell you that the hypotenuse of this triangle-- once again, this only works for 45-45-90 triangles. And if I just draw one 45 you know the other angle's got to be 45 as well. If I told you that the hypotenuse here is, let me say, 10. We know this is a hypotenuse because it's opposite the right angle. And then I would ask you what this side is, x. Well we know that x is equal to the square root of 2 over 2 times the hypotenuse. So that's square root of 2 over 2 times 10. 10 divided by 2. So x is equal to 5 square roots of 2. And we know that this side and this side are equal. I guess we know this is an isosceles triangle because these two angles are the same. So we also that this side is 5 over 2. And if you're not sure, try it out. Let's try the Pythagorean theorem. We know from the Pythagorean theorem that 5 root 2 squared, plus 5 root 2 squared is equal to the hypotenuse squared, where the hypotenuse is 10. Is equal to 100. Or this is just 25 times 2. So that's 50. But this is 100 up here. Is equal to 100. And we know, of course, that this is true. So it worked. We proved it using the Pythagorean theorem, and that's actually how we came up with this formula in the first place. Maybe you want to go back to one of those presentations if you forget how we came up with this. type of triangle. And I'm going to do it the same way, by just posing a problem to you and then using the Pythagorean theorem to figure it out. This is another type of triangle called a 30-60-90 triangle. And if I don't have time for this I will do another presentation. Let's say I have a right triangle. That's not a pretty one, but we use what we have. That's a right angle. And if I were to tell you that this is a 30 degree angle. Well we know that the angles in a triangle have to add up to 180. So if this is 30, this is 90, and let's say that this is x. x plus 30 plus 90 is equal to 180, because the angles in" + }, + { + "Q": "At 7:46 how to you get 4?", + "A": "(1/2) squared is (1/4). This multiplied by h is (h/4).", + "video_name": "Qwet4cIpnCM", + "timestamps": [ + 466 + ], + "3min_transcript": "Because that's h over 2, and this is also h over 2. Right over here. So if we go back to our original triangle, and we said that this is 30 degrees and that this is the hypotenuse, because it's opposite the right angle, we know that the side opposite the 30 degree side is 1/2 of the hypotenuse. And just a reminder, how did we do that? Well we doubled the triangle. Turned it into an equilateral triangle. Figured out this whole side has to be the same as the hypotenuse. And this is 1/2 of that whole side. So it's 1/2 of the hypotenuse. So let's remember that. The side opposite the 30 degree side is 1/2 of the hypotenuse. Let me redraw that on another page, because I think this is getting messy. So going back to what I had originally. This is a right angle. This is the hypotenuse-- this side right here. If this is 30 degrees, we just derived that the side opposite that this is equal to 1/2 the hypotenuse. If this is equal to 1/2 the hypotenuse then what is this side equal to? Well, here we can use the Pythagorean theorem again. We know that this side squared plus this side squared-- let's call this side A-- is equal to h squared. So we have 1/2 h squared plus A squared is equal to h squared. This is equal to h squared over 4 plus A squared, is equal to h squared. Well, we subtract h squared from both sides. We get A squared is equal to h squared minus h squared over 4. This is equal to 3/4 h squared. And once going that's equal to A squared. I'm running out of space, so I'm going to go all the way over here. So take the square root of both sides, and we get A is equal to-- the square root of 3/4 is the same thing as the square root of 3 over 2. And then the square root of h squared is just h. And this A-- remember, this is an area. This is what decides the length of the side. I probably shouldn't have used A. But this is equal to the square root of 3 over 2, times h. We've derived what all the sides relative to the hypotenuse are of a 30-60-90 triangle. So if this is a 60 degree side." + }, + { + "Q": "I'm confused about what happens to the negative at 2:37", + "A": "you have (2*(1-3^100))/(-2). then the 2s cancel out: (1-3^100)/(-1). negative from the -1 goes to the top: -(1-3^100)/1. 1 in denominator goes away: -(1-3^100). distribute negative: -1+3^100 rewrite order: 3^100-1", + "video_name": "AXP5PGSaaYk", + "timestamps": [ + 157 + ], + "3min_transcript": "- [Voiceover] Let's do some examples where we're finding sums of finite geometric series. Now let's just remind ourselves in a previous video we derived the formula where the sum of the first n terms is equal to our first term times one minus our common ratio to the nth power all over one minus our common ratio. So let's apply that to this finite geometric series right over here. So what is our first term and what is our common ratio? And what is our n? Well, some of you might just be able to pick it out by inspecting this here, but for the sake of this example, let's expand this out a little bit. This is going to be equal to two times three to the zero, which is just two, plus two times three to the first power, plus two times three to the second power, I can write first power there, plus two times three to the third power, and we're gonna go all the way to two times three to the 99th power. What is our a? Well, a is going to be two. And we see that in all of these terms here. So a is going to be two. What is r? Well, each successive term, as k increases by one, we're multiplying by three again. So, three is our common ratio. So that right over there, that is r. Let me make sure that we, that is a. And now what is n going to be? Well, you might be tempted to say, well, we're going up to k equals 99, maybe n is 99, but we have to realize that we're starting at k equals zero. So there is actually 100 terms here. Notice, when k equals zero, that's our first term, when k equals one, that's our second term, when k equals two, that's our third term, when k equals three, that's our fourth term, when k equals 99, this is our 100th term, 100th term. So what we really want to find is S sub 100. So let's write that down, S sub 100, for this geometric series is going to be equal to to the 100th power, all of that, all of that over, all of that over one minus three. And we could simplify this, I mean at this point it is arithmetic that you'd be dealing with, but down here you would have a negative two, and so you'd have two divided by negative two so that is just a negative. And so negative of one minus three to the 100th, that's the same thing, this is equal to three to the 100th, three to the 100th power minus one. And we're done." + }, + { + "Q": "At 3:10 when Sal factors out the five from the equation, why does he not put the newly factored equation in brackets with a five on the outside?", + "A": "In order to get rid of the 5, Sal divided both sides of the equation by 5, giving (5x^2 - 20x + 15) / 5 = 0/5, which simplifies to x^2 - 4x + 3 = 0. You don t need to include the 5 as a factor in the left-hand side because you divided both sides by it.", + "video_name": "MQtsRYPx3v0", + "timestamps": [ + 190 + ], + "3min_transcript": "when does this equal 0? So I want to figure out those points. And then I also want to figure out the point exactly in between, which is the vertex. And if I can graph those three points then I should be all set with graphing this parabola. So as I just said, we're going to try to solve the equation 5x squared minus 20x plus 15 is equal to 0. Now the first thing I like to do whenever I see a coefficient out here on the x squared term that's not a 1, is to see if I can divide everything by that term to try to simplify this a little bit. And maybe this will get us into a factor-able form. And it does look like every term here is divisible by 5. So I will divide by 5. So I'll divide both sides of this equation by 5. And so that will give me-- these cancel out and I'm left with x squared minus 20 over 5 is 4x. Plus 15 over 5 is 3 is equal to 0 over 5 is just 0. We say are there two numbers whose product is positive 3? The fact that their product is positive tells you they both must be positive. And whose sum is negative 4, which tells you well they both must be negative. If we're getting a negative sum here. And the one that probably jumps out of your mind-- and you might want to review the videos on factoring quadratics if this is not so fresh-- is a negative 3 and negative 1 Negative 3 times negative 1. Negative 3 times negative 1 is 3. Negative 3 plus negative 1 is negative 4. So this will factor out as x minus 3 times x minus 1. And on the right-hand side, we still have that being equal to 0. And now we can think about what x's will make this expression 0, and if they make this expression 0, well they're going to make this expression 0. Which is going to make this expression equal to 0. And so this will be true if either one of these is 0. Or x minus 1 is equal to 0. This is true, and you can add 3 to both sides of this. This is true when x is equal to 3. This is true when x is equal to 1. So we were able to figure out these two points right over here. This is x is equal to 1. This is x is equal to 3. So this is the point 1 comma 0. This is the point 3 comma 0. And so the last one I want to figure out, is this point right over here, the vertex. Now the vertex always sits exactly smack dab between the roots, when you do have roots. Sometimes you might not intersect the x-axis. So we already know what its x-coordinate is going to be. It's going to be 2. And now we just have to substitute back in to figure out its y-coordinate. When x equals 2, y is going to be equal to 5 times 2 squared minus 20 times 2 plus 15, which is equal to-- let's see, this is equal to 2 squared is 4." + }, + { + "Q": "Okay, so at 2:36 when he finds integers that multiplied equal 3 and added equal -4, then moves on using that information? What if you have numbers that don't work? I have the equation x^2 +2x+3=0. Nothing works with this. Is there another way to solve this and graph it as a parabola? Please help.", + "A": "This is because this parabola does not have x-intercepts. You can check it by completing the square: y=(x+1)^2+2 is always positive, which means that every point of the parabola has a positive y value, which means that it is above the x axis. The fastest way to check it though is to evaluate its determinant b^2-4ac and see that it is negative,and therefore has no real zeroes.", + "video_name": "MQtsRYPx3v0", + "timestamps": [ + 156 + ], + "3min_transcript": "when does this equal 0? So I want to figure out those points. And then I also want to figure out the point exactly in between, which is the vertex. And if I can graph those three points then I should be all set with graphing this parabola. So as I just said, we're going to try to solve the equation 5x squared minus 20x plus 15 is equal to 0. Now the first thing I like to do whenever I see a coefficient out here on the x squared term that's not a 1, is to see if I can divide everything by that term to try to simplify this a little bit. And maybe this will get us into a factor-able form. And it does look like every term here is divisible by 5. So I will divide by 5. So I'll divide both sides of this equation by 5. And so that will give me-- these cancel out and I'm left with x squared minus 20 over 5 is 4x. Plus 15 over 5 is 3 is equal to 0 over 5 is just 0. We say are there two numbers whose product is positive 3? The fact that their product is positive tells you they both must be positive. And whose sum is negative 4, which tells you well they both must be negative. If we're getting a negative sum here. And the one that probably jumps out of your mind-- and you might want to review the videos on factoring quadratics if this is not so fresh-- is a negative 3 and negative 1 Negative 3 times negative 1. Negative 3 times negative 1 is 3. Negative 3 plus negative 1 is negative 4. So this will factor out as x minus 3 times x minus 1. And on the right-hand side, we still have that being equal to 0. And now we can think about what x's will make this expression 0, and if they make this expression 0, well they're going to make this expression 0. Which is going to make this expression equal to 0. And so this will be true if either one of these is 0. Or x minus 1 is equal to 0. This is true, and you can add 3 to both sides of this. This is true when x is equal to 3. This is true when x is equal to 1. So we were able to figure out these two points right over here. This is x is equal to 1. This is x is equal to 3. So this is the point 1 comma 0. This is the point 3 comma 0. And so the last one I want to figure out, is this point right over here, the vertex. Now the vertex always sits exactly smack dab between the roots, when you do have roots. Sometimes you might not intersect the x-axis. So we already know what its x-coordinate is going to be. It's going to be 2. And now we just have to substitute back in to figure out its y-coordinate. When x equals 2, y is going to be equal to 5 times 2 squared minus 20 times 2 plus 15, which is equal to-- let's see, this is equal to 2 squared is 4." + }, + { + "Q": "7:30 in the video sal has written -24-10_/5 should the answer be -(-24-12_/5)", + "A": "What Sal wrote is correct. After rationalizing the denominator then simplifying we have: 24 + 12sqrt(5) / -1 Because we are dividing the numerator by negative one, we take the opposite of each of its terms: 24 + 12sqrt(5) / -1 -(24 + 12sqrt(5)) -24 - 12sqrt(5)", + "video_name": "gY5TvlHg4Vk", + "timestamps": [ + 450 + ], + "3min_transcript": "b, negative b squared. These cancel out and you're just left with a squared minus b squared. So 2 minus the square root of 5 times 2 plus the square root of 5 is going to be equal to 2 squared, which is 4. Let me write it that way. It's going to be equal to 2 squared minus the square root of 5 squared, which is just 5. So this would just be equal to 4 minus 5 or negative 1. If you take advantage of the difference of squares of binomials, or the factoring difference of squares, however you want to view it, then you can rationalize this denominator. So let's do that. Let me rewrite the problem. 12 over 2 minus the square root of 5. In this situation, I just multiply the numerator and the denominator by 2 plus the square root of 5 over 2 plus Once again, I'm just multiplying the number by 1. So I'm not changing the fundamental number. I'm just changing how we represent it. So the numerator is going to become 12 times 2, which is 24. Plus 12 times the square root of 5. Once again, this is like a factored difference of squares. This is going to be equal to 2 squared, which is going to be exactly equal to that. Which is 4 minus 1, or we could just-- sorry. 4 minus 5. It's 2 squared minus square root of 5 squared. So it's 4 minus 5. Or we could just write that as minus 1, or negative 1. Or we could put a 1 there and put a negative sign out in front. And then, no point in even putting a 1 in the denominator. We could just say that this is equal to negative 24 minus 12 square roots of 5. So this case, it kind of did simplify it as well. It actually made it look a little bit better. And you know, I don't if I mentioned in the beginning, this is good because it's not obvious. If you and I are both trying to build a rocket and you get this as your answer and I get this as my answer, this isn't obvious, at least to me just by looking at it, that they're the same number. But if we agree to always rationalize our denominators, we're like, oh great. We got the same number. Now we're ready to send our rocket to Mars. Let's do one more of this, one more of these right here. Let's do one with variables in it. So let's say we have 5y over 2 times the square root of y minus 5. So we're going to do this exact same process. We have a binomial with an irrational denominator. It might be a rational. We don't know what y is. But y can take on any value, so at points it's going to be" + }, + { + "Q": "how did he turn 1/4 into a whole number at 3:05", + "A": "he multiplied 12 to 1/4, and that equals 3", + "video_name": "GmD7Czmol0k", + "timestamps": [ + 185 + ], + "3min_transcript": "Then I have plus five. And then I'm gonna subtract. I am subtracting eight times 0.25. 0.25, this is 1/4. I could rewrite this if I want. 0.25, that's the same thing as 1/4. Eight times 1/4, or another way to think about it is eight divided by four is gonna be equal to two. So this whole thing over here is going to be equal to two. So it's gonna be minus, we have this minus out here, so minus two. And what is this going to be? Well, let's think about it. 3.5 plus five is 8.5, minus two is going to be 6.5. So this is equal to this, is equal to 6.5. Let's do another one of these. Alright. And we'll, just like before, try to work through it on your own before we do it together. Alright, now let's do it together. plus eight minus 12 N, when M is equal to 30 and N is equal to 1/4. Alright. So everywhere I see an M I want to replace with a 30. And everywhere I see an N I want to replace with a 1/4. So this is going to be equal to 0.1 times M. M is 30. Times 30 plus eight minus 12 times N, where N is 1/4. N is 1/4. So what is, what is 1/10, This right over here, 0.1, that's the same thing as 1/10 of 30? Well 1/10 of 30, that's going to be three. So this part is three. And we have three plus eight. And then we're gonna have minus. Well what is 12 times 1/4? That's gonna be 12/4, or 12 divided by four, And now when we evaluate this, so that is equal to this, we have three plus eight minus three. Well, threes are going to, you know positive three, then you're gonna subtract three, and you're just going to be left with, you're just going to be left with an eight. And you're done. This expression when M is equal to 30 and N is equal to 1/4 is equal to eight." + }, + { + "Q": "At 2:25, Sal uses the chain rule for the derivative of (2+x^3)^-1. Would it not work if he just used the power rule and left it at that?", + "A": "No, the power rule applies only when you have x to the n, not when you have some function of x raised to the n. You can see this with an example like (x^2)^3. If you just apply the power rule, you get 3(x^2)^2, but we know that s wrong because (x^2)^3 is x^6, so the answer has to be 6x^5 or something equivalent. You get the right result when you apply the chain rule.", + "video_name": "GH8-URjRQpQ", + "timestamps": [ + 145 + ], + "3min_transcript": "We have the curve y is equal to e to the x over 2 plus x to the third power. And what we want to do is find the equation of the tangent line to this curve at the point x equals 1. And when x is equal to 1, y is going to be equal to e over 3. It's going to be e over 3. So let's try to figure out the equation of the tangent line to this curve at this point. And I encourage you to pause this video and try this on your own first. Well, the slope of the tangent line at this point is the same thing as the derivative at this point. So let's try to find the derivative of this or evaluate the derivative of this function right over here at this point. So to do that, first I'm going to rewrite it. You could use the quotient rule if you like, but I always forget the quotient rule. The product rule is much easier for me to remember. So I can rewrite this as y is equal to-- and I might as well color code it-- is equal to e to the x times 2 plus x And so the derivative of this, so let me write it here. So y prime is going to be equal to the derivative of this part of it, e to the x. So the derivative of e to the x is just e to the x. Just let me write that. So we're going to take the derivative of it. And that's what's amazing about e to the x, is that the derivative of e to the x is just e to the x times this thing. So times 2 plus x to the third to the negative 1. And then to that we're going to add this thing. So not its derivative anymore. We're just going to add e to the x times the derivative of this thing right over here. So we're going to take the derivative. So we can do the chain rule. It's going to be the derivative of 2 power with respect to 2 plus x to the third times the derivative of 2 plus x to the third with respect to x. So this is going to be equal to negative-- I'll write it this way-- negative 2 plus x to the third to the negative 2 power. And then we're going to multiply that times the derivative of 2 plus x to the third with respect to x. Well, derivative of this with respect to x is just 3x squared. And of course, we could simplify this a little bit if we like. But the whole point of this is to actually find the value of the derivative at this point. So let's evaluate. Let's evaluate y prime when x is equal to 1. Y prime of 1 when x is equal to 1." + }, + { + "Q": "At 2:25 are you supposed to add the negatives together? Can't you use the old fashioned way too?", + "A": "If you are talking about the negatives of 70, 10, 15, and 21, keep in mind that you can only add them together when they possess the same exponent amounts.", + "video_name": "D6mivA_8L8U", + "timestamps": [ + 145 + ], + "3min_transcript": "We are multiplying 10a minus 3 by the entire polynomial 5a squared plus 7a minus 1. So to do this, we can just do the distributive property. We can distribute this entire polynomial, this entire trinomial, times each of these terms. We could have 5a squared plus 7a minus 1 times 10a. And then 5a squared plus 7a minus 1 times negative 3. So let's just do that. So if we have-- so let me just write it out. Let me write it this way. 10a times 5a squared plus 7a minus 1. That's that right over here. And then we can have minus 3 times 5a squared plus 7a minus 1. And that is this distribution right over here. And then we can simplify it. 10a times 5a squared-- 10 times 5 is 50. 10 times 7 is 70. a times a is a squared. 10a times negative 1 is negative 10a. Then we distribute this negative 3 times all of this. Negative 3 times 5a squared is negative 15a squared. Negative 3 times 7a is negative 21a. Negative 3 times negative 1 is positive 3. And now we can try to merge like terms. This is the only a to the third term here. So this is 50a to the third. I'll just rewrite it. Now we have two a squared terms. We have 70a squared minus 15, or negative 15a squared. So we can add these two terms. 70 of something minus 15 of that something is going to be 55 of that something. So plus 55a squared. We have this negative 10a, and then we have this negative 21a. So if we go negative 10 minus 21, that is negative 31. That is negative 31a. And then finally, we only have one constant term over here. We have this positive 3. So plus 3. And we are done." + }, + { + "Q": "at 1:17, what if there isnt a exact weight given?how would you solve?", + "A": "If there is not an exact weight given then there is to little info to discover an exact answer.", + "video_name": "z1hz8-Kri1E", + "timestamps": [ + 77 + ], + "3min_transcript": "An electronics warehouse ships televisions and DVD players in certain combinations to retailers throughout the country. They tell us that the weight of 3 televisions and 5 DVD players is 62.5 pounds, and the weight of 3 televisions and 2 DVD players-- so they're giving us different combinations-- is 52 pounds. Create a system of equations that represents this situation. Then solve it to find out how much each television and DVD player weighs. Well, the two pieces of information they gave us in each of these statements can be converted into an equation. The first one is is that the weight of 3 televisions and 5 DVD players is 62.5 pounds. Then they told us that the weight of 3 televisions and 2 DVD players is 52 pounds. So we can translate these directly into equations. If we let t to be the weight of a television, and d to be the weight of a DVD player, this first statement up here says that 3 times the weight of a television, or 3 going to be equal to 62.5 pounds. That's exactly what this first statement is telling us. The second statement, the weight of 3 televisions and 2 DVD players, so if I have 3 televisions and 2 DVD players, so the weight of 3 televisions plus the weight of 2 DVD players, they're telling us that that is 52 pounds. And so now we've set up the system of equations. We've done the first part, to create a system that represents the situation. Now we need to solve it. Now, one thing that's especially tempting when you have two systems, and both of them have something where, you know, you have a 3t here and you have a 3t here, what we can do is we can multiply one of the systems by some factor, so that if we were to add this equation to that equation, we would get one of the terms to cancel out. And that's what we're going to do right here. equations to each other, because remember, when we learned this at the beginning of algebra, anything you do to one side of an equation, if I add 5 to one side of an equation, I have to add 5 to another side of the equation. So if I add this business to this side of the equation, if I add this blue stuff to the left side of the equation, I can add this 52 to the right-hand side, because this is saying that 52 is the same thing as this thing over here. This thing is also 52. So if we're adding this to the left-hand side, we're actually adding 52 to it. We're just writing it a different way. Now, before we do that, what I want to do is multiply the second, blue equation by negative 1. And I want to multiply it by negative 1. So negative 3t plus-- I could write negative 2d is equal to negative 52. So I haven't changed the information in this equation. I just multiplied everything by negative 1. The reason why I did that is because now if I add these two equations, these 3t terms are going to cancel out. So let's do that. Let's add these two equations. And remember, all we're doing is we're adding the same thing" + }, + { + "Q": "what is a parabola 0:00 to 2:46", + "A": "It is the line you get when you graph a quadratic.", + "video_name": "v-pyuaThp-c", + "timestamps": [ + 0, + 166 + ], + "3min_transcript": "So you're me, and you're in math class. And your teacher's ranting on and on about this article about whether algebra should be taught in school \u2013 as if he doesn't realize that what he's teaching isn't even algebra \u2013 which could have been interesting \u2013 but how to manipulate symbols and some special cases of elementary algebra \u2013 which isn't. And so, instead of learning about self-consistent systems and logical thought, you spent all week memorizing how to graph parabolas. News flash: No one cares about parabolas. Which is why half the class is playing Angry Birds under their desks. But, since you don't have a smart phone yet, you have to resort to a more noble and outdated form of boredom relief \u2013 that is, doodling. And you've invented a game of your own. A doodle game that connects the dots in ways your math curriculum never will \u2013 except instead of connecting to the closest dots to discover the mysterious hidden picture you've got this precise method of skipping over some number of dots and connecting them that way. In the past you've characterized how this works if your dots are arranged in a circle \u2013 say 11 dots \u2013 and connect one to the dot four dots over, you get these awesome stars. And you can either draw the lines in the order of the dots, or you can just keep going around and maybe it will hit all the dots, or maybe it won't, and how many dots you skip. But then there are other shapes. Circles are good friends with sine waves. And sine waves are good friends with square waves. And let's admit it, that's pretty cool looking. In fact, just two simple straight lines of dots connecting the dots from one line to the other in order somehow gives you this awesome woven curve shape. Another student is asking the teacher when he's ever going to need to know how to graph a parabola \u2013 even as he hides his multi-million dollar enterprise of a parabola graphing game under his desk. If your teacher thought about it, he would probably think shooting birds at things is a great reason to learn about parabolas because he's come to understand that education is about money and prestige and not about becoming a better human able to do great things. You yourself haven't done anything really great yet but you figure the path to your future greatness lies more in inventing awesome new connect-the-dot arrangements than in graphing parabolas or shooting birds at things. And that's when you begin to worry. What if this cool liney curvey thing you drew approximates a parabola? As if your teacher doesn't realize everyone has their phones under their desks, his whole word runs on plausible deniability, so he shouts state-mandated, pass the test, teach-to-the-middle nonsense at students who are not at all fooled by his false enthusiasm or false mathematics and he pretends he's teaching algebra and the students pretend to be taught algebra and everyone else involved in the system is too invested to do anything but pretend to believe them both. You think maybe it's a hyperbola, which is similar to the parabola in that they are both conic sections. A hyperbola is a nice vertical slice of cone, the cone itself being just like a line swirled around in a circle, which is why the cone is like two cones radiating both ways; the lovely hyperbola insecting both parts. Two perfect curves, looking disconnected when seen alone but sharing their common conic heritage. While the boring old parabola is a slice taken at an angle completely meant to miss the top part of the cone and to miss wrapping around the bottom like an ellipse would. And it's such a special, specific case of conic section that all parabolas are exactly the same, just bigger or smaller or moved around. Your teacher could just as well hand you parabolas already drawn and have you draw coordinate grids on parabolas" + }, + { + "Q": "how did she make that shape?! at 4:30", + "A": "She drew a circle out of dots, then drew more circles with the points being their centers and alternated colors between the shapes made by the circles overlapping.", + "video_name": "v-pyuaThp-c", + "timestamps": [ + 270 + ], + "3min_transcript": "his whole word runs on plausible deniability, so he shouts state-mandated, pass the test, teach-to-the-middle nonsense at students who are not at all fooled by his false enthusiasm or false mathematics and he pretends he's teaching algebra and the students pretend to be taught algebra and everyone else involved in the system is too invested to do anything but pretend to believe them both. You think maybe it's a hyperbola, which is similar to the parabola in that they are both conic sections. A hyperbola is a nice vertical slice of cone, the cone itself being just like a line swirled around in a circle, which is why the cone is like two cones radiating both ways; the lovely hyperbola insecting both parts. Two perfect curves, looking disconnected when seen alone but sharing their common conic heritage. While the boring old parabola is a slice taken at an angle completely meant to miss the top part of the cone and to miss wrapping around the bottom like an ellipse would. And it's such a special, specific case of conic section that all parabolas are exactly the same, just bigger or smaller or moved around. Your teacher could just as well hand you parabolas already drawn and have you draw coordinate grids on parabolas And it's stupid, and you hate it, and you don't wanna learn to graph them, even if it means not making a billion dollars from a game about shooting birds at things. Meanwhile anyone who actually learns how to think mathematically can then learn to graph a parabola or anything else they need in like five minutes. But teaching how to think is an individualized process that gives power and responsibility to individuals while teaching what to think can be done with one-size-fits-all bullet points and check-boxes and our culture of excuses demands that we do the latter, keeping ourselves placated in the comforting structure of tautology and clear expectations. Algebra has become a check-box subject and mathematics weeps alone in the top of the ivory tower prison to which she has been condemned. But you're not interested in check-boxes; you're interested in dots, and lines that connect them. Or maybe you could connect them with semicircles, to give visual structure to lines that would otherwise overlap. Or you could say one dot is the center of a circle and another defines a radius and draw the entire circle and do things that way. You could make rules about how every dot is the center of a circle with its neighbor being the radius, and all of the others define radii. But then you just get concentric circles, which I suppose should have been obvious. But what if you did it the other way around and said one dot always stays on the circle and all the other dots are centers, like this. Looks more promising. So you try putting all the dots in a circle and using them as circle centers and choose just one dot for the circles to go through and you get this awesome shape that looks kind of like a heart. So let's call it, oh I don't know, a cardioid. Which happens to be the same curve that you get when parallel lines like rays of light reflect off a circle the same heart of sunshine in a cup. Or maybe instead of circle centers you could have points all on the curve of a circle, which means you need three points to define a circle, maybe just a point and its two closest neighbors to start with. And of course, any collection of circles is two-colorable, which means you can contrast light and dark colors for a classy color scheme. Or maybe you could throw down some random points to make all possible circles. Only that would be a lot of circles, so you choose just ones you like. And then, against your will, you begin to wonder how many points it takes to define the boring old parabola." + }, + { + "Q": "I want so bad to be able to draw the heart thing at 4:29 can someone explain how or make a program maybe??? because I can't understand lol", + "A": "you could take two circles and roll one around the other tracing one point on the edge; OR you could make dots in the shape of a circle ,choose one dot to be a point on the edge of all circles and use the others as centers", + "video_name": "v-pyuaThp-c", + "timestamps": [ + 269 + ], + "3min_transcript": "his whole word runs on plausible deniability, so he shouts state-mandated, pass the test, teach-to-the-middle nonsense at students who are not at all fooled by his false enthusiasm or false mathematics and he pretends he's teaching algebra and the students pretend to be taught algebra and everyone else involved in the system is too invested to do anything but pretend to believe them both. You think maybe it's a hyperbola, which is similar to the parabola in that they are both conic sections. A hyperbola is a nice vertical slice of cone, the cone itself being just like a line swirled around in a circle, which is why the cone is like two cones radiating both ways; the lovely hyperbola insecting both parts. Two perfect curves, looking disconnected when seen alone but sharing their common conic heritage. While the boring old parabola is a slice taken at an angle completely meant to miss the top part of the cone and to miss wrapping around the bottom like an ellipse would. And it's such a special, specific case of conic section that all parabolas are exactly the same, just bigger or smaller or moved around. Your teacher could just as well hand you parabolas already drawn and have you draw coordinate grids on parabolas And it's stupid, and you hate it, and you don't wanna learn to graph them, even if it means not making a billion dollars from a game about shooting birds at things. Meanwhile anyone who actually learns how to think mathematically can then learn to graph a parabola or anything else they need in like five minutes. But teaching how to think is an individualized process that gives power and responsibility to individuals while teaching what to think can be done with one-size-fits-all bullet points and check-boxes and our culture of excuses demands that we do the latter, keeping ourselves placated in the comforting structure of tautology and clear expectations. Algebra has become a check-box subject and mathematics weeps alone in the top of the ivory tower prison to which she has been condemned. But you're not interested in check-boxes; you're interested in dots, and lines that connect them. Or maybe you could connect them with semicircles, to give visual structure to lines that would otherwise overlap. Or you could say one dot is the center of a circle and another defines a radius and draw the entire circle and do things that way. You could make rules about how every dot is the center of a circle with its neighbor being the radius, and all of the others define radii. But then you just get concentric circles, which I suppose should have been obvious. But what if you did it the other way around and said one dot always stays on the circle and all the other dots are centers, like this. Looks more promising. So you try putting all the dots in a circle and using them as circle centers and choose just one dot for the circles to go through and you get this awesome shape that looks kind of like a heart. So let's call it, oh I don't know, a cardioid. Which happens to be the same curve that you get when parallel lines like rays of light reflect off a circle the same heart of sunshine in a cup. Or maybe instead of circle centers you could have points all on the curve of a circle, which means you need three points to define a circle, maybe just a point and its two closest neighbors to start with. And of course, any collection of circles is two-colorable, which means you can contrast light and dark colors for a classy color scheme. Or maybe you could throw down some random points to make all possible circles. Only that would be a lot of circles, so you choose just ones you like. And then, against your will, you begin to wonder how many points it takes to define the boring old parabola." + }, + { + "Q": "At 1:10. why does he make 4.1 to 41?", + "A": "Moving the decimal around is like multiplying or dividing by 10, so he notes that 4.1 hundredths is the same as 41 thousandths (4.1 x 10^-2 = 41 x 10^-3).", + "video_name": "ios3QL9t9LQ", + "timestamps": [ + 70 + ], + "3min_transcript": "- [Voiceover] What I want to do in this video is get a little bit of practice subtracting in scientific notation. So let's say that I have 4.1 x 10 to the -2 power. 4.1 x 10 to the -2 power and from that I want to subtract, I want to subtract 2.6, 2.6 x 10 to the -3 power. Like always, I encourage you to pause this video and see if you can solve this on your own and then we could work through it together. All right, I'm assuming you've had a go it. So the easiest thing that I can think of doing is try to convert one of these numbers so that it has the same, it's being multiplied by the same power of ten as the other one. What I could think about doing, well can we express 4.1 times 10 to the -2? Can we express it as something times 10 to the -3? So we have 4.1 times 10 to the -2. we would divide by 10, but we can't just divide by 10. That would literally change the value of the number. In order to not change it, we want to multiply by 10 as well. So we're multiplying by 10 and dividing by 10. I could have written it like this. I could have written 10/10 times, let me write this a little bit neater. I could have written 10/10 x this and then you take 10 x 4.1, you get 41, and then 10 to the -2 divided by 10 is going to be 10 to the -3. So this right over here, this is equal to 10 x 4.1 is 41 times 10 to the -3. And that makes sense. 41 thousandths is the same thing as 4.1 hundredths and all we did is we multiplied this times 10 and we divided this times 10. So let's rewrite this. We can rewrite it now as 41 X 10 to the -3 So now we have two things. We have 41 X 10 to the -3 - 2.6 x 10 to the -3. Well this is going to be the same thing as 41 - 2.6. - 2.6, let me do it in that same color. That was purple. - 2.6, 10 to the -3. 10, whoops, 10 to the -3. There's 10 to the -3 there, 10 to the -3 there. One way to think about it, I have just factored out a 10 to the -3. Now what's 41 - 2.6? Well 41 - 2 is 39, and then -.6 is going to be 38.4." + }, + { + "Q": "A, B and C are collinear, and B is between A and C. The ratio of AB, B to AC, C is 1:4\nIf A is at (-7,-8) and B is at (-3,-5), what are the coordinates of point C?\nI got the x and y values by subtracting and got 4 and 3. But, when I went to find the ratios of 4 and 3 i'm told to multiply 4 by 1:4 to get 16. And 1:4 by 3 to get 12. But, in the video i'm told to find the ratios by dividing to get 4 1:4=1 so which am I supposed to be doing?", + "A": "I think you are thinking correctly, but I am not completely sure. Setting up proportions and cross multiplying, For x coordinate, AB/AC = 1/4 and if AB is 4, then 4/AC = 1/4 and AC = 16 in x direction For y coordinate, AB/AC = 1/4 and if AB is 3, then 3/AC= 1/4 and AC = 12 in y direction If I start at A and move these distances, then C (-7+16, -8+12) or (9,4)", + "video_name": "lEGS5ECgFxE", + "timestamps": [ + 64, + 64, + 64, + 64 + ], + "3min_transcript": "A, B and C are collinear, and B is between A and C. The ratio of AB to AC is 2 to 5. If is A is at negative 6 comma 9, and B is at negative 2 comma 3, what are the coordinates of point C? And I encourage you to now pause this video and try this on your own. So let's try to visualize this. So A is at negative 6 comma 9. B, let's see, it's less negative in the horizontal direction, it's lower in the vertical direction. So we could put B right over here. B is at the point negative 2 comma 3. And that C, it's going to be collinear, so we're going to go along the same line, let me draw that line right now. So it's going to be collinear and they tell us that the ratio of AB to AC is 2 to 5. So let's say C is-- I'm just trying to eyeball it right now-- let's put C right over here. And we don't know C's coordinates. Well the way we could think about it is to break it up into horizontal change in coordinate and vertical change in coordinate, and apply the same ratio. So for example, what is the horizontal change in coordinates going from A to B? Well let's draw that. Going from A to B, so this is A's x-coordinate, it's at negative 6. B's x-coordinate is at negative 2. So it's this change right over here. This is the horizontal change that we care about. Now what is that? Well if you start at negative 6 and you go to negative 2, you have increased by 4. Another way of thinking about it is negative 2 minus negative 6 Now the ratio between this change and the change of the x-coordinate between A and C is going to be 2 to 5. So let's call that change, this entire change, let's call this x. So we could say that the ratio between 4 and x is equal to-- and notice this is the horizontal change from A to B just if you look on the horizontal axis-- so the ratio of that, which is 4, to the horizontal change between A and C, well that's going to have to be the same ratio. So it's going to be 2/5. Now to solve for x, a fun thing might be to just take the reciprocal of both sides. So x/4 is equal to 5/2. We can multiply both sides times 4 and we are left with x is equal to 5 times 4 divided by 2," + }, + { + "Q": "At 0:03, the word collinear is mentioned. What exactly is collinear?", + "A": "It means you can draw a line through all the points.", + "video_name": "lEGS5ECgFxE", + "timestamps": [ + 3 + ], + "3min_transcript": "A, B and C are collinear, and B is between A and C. The ratio of AB to AC is 2 to 5. If is A is at negative 6 comma 9, and B is at negative 2 comma 3, what are the coordinates of point C? And I encourage you to now pause this video and try this on your own. So let's try to visualize this. So A is at negative 6 comma 9. B, let's see, it's less negative in the horizontal direction, it's lower in the vertical direction. So we could put B right over here. B is at the point negative 2 comma 3. And that C, it's going to be collinear, so we're going to go along the same line, let me draw that line right now. So it's going to be collinear and they tell us that the ratio of AB to AC is 2 to 5. So let's say C is-- I'm just trying to eyeball it right now-- let's put C right over here. And we don't know C's coordinates. Well the way we could think about it is to break it up into horizontal change in coordinate and vertical change in coordinate, and apply the same ratio. So for example, what is the horizontal change in coordinates going from A to B? Well let's draw that. Going from A to B, so this is A's x-coordinate, it's at negative 6. B's x-coordinate is at negative 2. So it's this change right over here. This is the horizontal change that we care about. Now what is that? Well if you start at negative 6 and you go to negative 2, you have increased by 4. Another way of thinking about it is negative 2 minus negative 6 Now the ratio between this change and the change of the x-coordinate between A and C is going to be 2 to 5. So let's call that change, this entire change, let's call this x. So we could say that the ratio between 4 and x is equal to-- and notice this is the horizontal change from A to B just if you look on the horizontal axis-- so the ratio of that, which is 4, to the horizontal change between A and C, well that's going to have to be the same ratio. So it's going to be 2/5. Now to solve for x, a fun thing might be to just take the reciprocal of both sides. So x/4 is equal to 5/2. We can multiply both sides times 4 and we are left with x is equal to 5 times 4 divided by 2," + }, + { + "Q": "I was just wondering, at 1:05, why is point B said to be 2/5 of the way from A? Shouldn't it be that the total distance from A to C is 7 (add both sides of the ratio to find the total number of parts) and then B should be 2 parts out of this total distance? Thus, B is 2/7 of the way? I'm not entirely sure of whether or not I am correct, but I'm assuming I've made some error.", + "A": "Your error was in saying the total distance from A to C is 7. The problem states that AC is 5 when it says the ratio of AB to AC is 2 : 5. In other words, the 2 : 5 ratio is a ratio of a part to the whole, not a ratio of the lengths of 2 portions of the whole.", + "video_name": "lEGS5ECgFxE", + "timestamps": [ + 65 + ], + "3min_transcript": "A, B and C are collinear, and B is between A and C. The ratio of AB to AC is 2 to 5. If is A is at negative 6 comma 9, and B is at negative 2 comma 3, what are the coordinates of point C? And I encourage you to now pause this video and try this on your own. So let's try to visualize this. So A is at negative 6 comma 9. B, let's see, it's less negative in the horizontal direction, it's lower in the vertical direction. So we could put B right over here. B is at the point negative 2 comma 3. And that C, it's going to be collinear, so we're going to go along the same line, let me draw that line right now. So it's going to be collinear and they tell us that the ratio of AB to AC is 2 to 5. So let's say C is-- I'm just trying to eyeball it right now-- let's put C right over here. And we don't know C's coordinates. Well the way we could think about it is to break it up into horizontal change in coordinate and vertical change in coordinate, and apply the same ratio. So for example, what is the horizontal change in coordinates going from A to B? Well let's draw that. Going from A to B, so this is A's x-coordinate, it's at negative 6. B's x-coordinate is at negative 2. So it's this change right over here. This is the horizontal change that we care about. Now what is that? Well if you start at negative 6 and you go to negative 2, you have increased by 4. Another way of thinking about it is negative 2 minus negative 6 Now the ratio between this change and the change of the x-coordinate between A and C is going to be 2 to 5. So let's call that change, this entire change, let's call this x. So we could say that the ratio between 4 and x is equal to-- and notice this is the horizontal change from A to B just if you look on the horizontal axis-- so the ratio of that, which is 4, to the horizontal change between A and C, well that's going to have to be the same ratio. So it's going to be 2/5. Now to solve for x, a fun thing might be to just take the reciprocal of both sides. So x/4 is equal to 5/2. We can multiply both sides times 4 and we are left with x is equal to 5 times 4 divided by 2," + }, + { + "Q": "At 1:35 I solved it as 8^1/4=2, where as in the video its 8^1/3=2. Im Not sure if im missing something of if Sal made a mistake?", + "A": "8^(1/3) is the cube root of 8, which is 2. (2^3)^(1/3)=2^(3/3)=2^1=2 8^(1/4) is the fourth root of 8, which is not 2. 16^(1/4) is the fourth root of 16, which is 2.", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 95 + ], + "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3." + }, + { + "Q": "At the 0:53 problem how does Sal get to an Answer of 1/3 ? What is the work not shown here ?", + "A": "Let s start with 8^x =2. When the variable is in the exponent, it is useful (where possible) to express both sides of the equation using the same base. Since on the righthand side there is a 2 to the first power, ask yourself whether 8 can be expressed as a power of 2? So we end up with (2^3)^x which is the same as 2^3x. And we then have 2^3x = 2^1 as the equation. The bases are the same, so the exponents must be equal. Therefore 3x = 1 so x = 1/3 Hope you find this useful!", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 53 + ], + "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3." + }, + { + "Q": "on 2:37 how does negitive turn into a fraction", + "A": "That is a basic property of exponents. The rule is: a\u00e2\u0081\u00bb\u00e1\u00b5\u0087 = 1/a\u00e1\u00b5\u0087", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 157 + ], + "3min_transcript": "So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3. What would be the log base 8 of 1/2? What does this evaluate to? Let me clean this up so that we have some space to work with. So as always, we're saying, what power do I have to raise 8 to to get to 1/2? So let's think about that a little bit. We already know that 8 to the one-third power is equal to 2. If we want the reciprocal of 2 right over here, we have to just raise 8 to the negative one-third. So let me write that down. 8 to the negative one-third power is going to be equal to 1 over 8 to the one-third power. And we already know the cube root of 8, or 8 to the one-third power, is equal to 2. This is equal to 1/2." + }, + { + "Q": "@1:25, I still dont get it... is something wrong with me? 8^1=8 and 8^0=1, but how do you get.... i dont know; is there another video i can watch to buff up my understanding of this?", + "A": "any number raised to the power 0 is 1, you can think of it like this: x^3 = x*x*x x^2 = x*x which is x^3 divided by x x^1 = x which is x^2 divided by x x^0 = 1 because any number divided by itself is equal to 1. Also, you can think of 8^1 as being 8*1, and 8^2 as being 8*8*1, since any number multiplied by 1 is just itself you don t need to display the 1, but this does explain why 8^1=8 hope this helps :)", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 85 + ], + "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3." + }, + { + "Q": "Don't get the second problem at 0:42.\nAre there like any practice problems for this?", + "A": "2^3 = 8; therefore the cube root of 8 is equal to 2, right? Another way of writing cube root of 8 (remember: that s the square root sign with a little 3 above the check mark) is 8^(1/3). Notice that the exponent is 1/3. That is why the Log (base 8) of 2 is 1/3.", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 42 + ], + "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3." + }, + { + "Q": "The equation is -4x+7. Shortly after the 4:00 mark, Sal replaces the x with -1 and then says, \"4 times -1 = -4\". Shouldn't it be -4 * -1?", + "A": "He misspoke and says 4*-1=4, but what he really meant is -4*-1=4 and he completes the equation as if he had said that correctly. It does not change the problem because he just misspoke and didn t write the incorrect statement down", + "video_name": "nGCW5teACC0", + "timestamps": [ + 240 + ], + "3min_transcript": "all of that over your change in the horizontal axis. Well, your change in the horizontal axis is this distance, which is the same thing as this distance. Notice your change in vertical over change in horizontal, change in vertical over change in horizontal, reviewing the green point as the endpoint. So it's going to be x minus negative 1. And this is the exact same expression. These are the exact same expression. You can simplify them, minus negative 1, and this becomes plus. This could become a plus 1. But these are the exact same expression. So this is the expression, really, for the slope between negative 1 and g of negative 1 and an arbitrary x. Well, we already know that no matter what x you pick, the slope between x, g of , and this point right over here is It's going to be the slope of the line. It's going to be equal to negative 4. This thing is going to be equal to negative 4. It's going to be equal to negative 4. Doesn't matter how close x gets, and weather x comes from the right or whether x comes from the left. So this thing, taking the limit of this, this just gets you to negative 4. It's really just the slope of the line. So even if you were to take the limit as x approaches negative 1, as x gets closer and closer and closer to negative 1, well then, these points are just going to get closer and closer and closer. But every time you calculate the slope, it's just going to be the slope of the line, which Now, you could also do this algebraically. And let's try to do it algebraically. So let's actually just take the limit as x approaches negative 1 of g of x. Well, they already told us what g of x is. It is negative 4x plus 7, minus g of negative 1. Negative 1 times 4 is positive 4. Positive 4 plus 7 is 11. All of that over x plus 1, all of that over x plus 1. And that's really x minus negative 1, is you want to think of it that way. But I'll just write x plus 1 this way here. So this is going to be equal to the limit as x approaches negative 1 of, in our numerator-- let's see. 7 minus 11 is negative 4. We can factor out a negative 4. It's a negative 4 times x plus 1, all of that over x plus 1. And then since we're just trying to find the limit as x approaches negative 1, so we can cancel those out. And this is going to be non-zero for any x value other than negative 1. And so this is going to be equal to negative 4." + }, + { + "Q": "at 1:12, why is the y coordinate of the point g(-1)?", + "A": "At this point Sal has drawn a blue line representing g(x), and wants to plot the point on that line where x = -1, so the y-coordinate has to be the value that the function g assigns to the x-value -1, which is g(-1). Possibly you just lost Sal s train of thought here, but if this doesn t make sense after this explanation, it might help to review function terminology before you proceed, as similar references will come up often.", + "video_name": "nGCW5teACC0", + "timestamps": [ + 72 + ], + "3min_transcript": "Let g of x equal negative 4x plus 7. What is the value of the limit as x approaches negative 1 of all of this? So before we think about this, let's just visualize the line. And then we can think about what they're asking here. So let me draw some axes here. So this is my vertical axis and this is my horizontal axis. And let's say this is my x-axis. We'll label that the x-axis. I'll graph g of x. g of x is going to have a positive-- I guess you would say y-intercept. or vertical axis intercept. It's going to have a slope of negative 4, so it's going to look something like this. Let me draw my best. So it's going to look something like that. And we already know the slope here is going to be negative 4. We get that right from this slope intercept form of the equation, slope is equal to negative 4. And they ask us, what is the limit as x approaches negative 1 of all of this kind of stuff? So let's plot the point negative 1. this point right over here. And this point right over here would be the point negative 1, g of negative 1. Let me label everything else. So I could call this my y-axis. I could call this graph. This is the graph of y is equal to g of x. So what they're doing right over here is they're finding the slope between an arbitrary point x, g of x, and this point right over here. So let's do that. So let's take another x. So let's say this is x. This would be the x, g of x. And this expression right over here, notice it is your change in the vertical axis. That would be your g of x. Let me make it this way. So this would be your change in the vertical axis. That would be g of x minus g of negative 1. And then that's over-- actually, let me write it this way all of that over your change in the horizontal axis. Well, your change in the horizontal axis is this distance, which is the same thing as this distance. Notice your change in vertical over change in horizontal, change in vertical over change in horizontal, reviewing the green point as the endpoint. So it's going to be x minus negative 1. And this is the exact same expression. These are the exact same expression. You can simplify them, minus negative 1, and this becomes plus. This could become a plus 1. But these are the exact same expression. So this is the expression, really, for the slope between negative 1 and g of negative 1 and an arbitrary x. Well, we already know that no matter what x you pick, the slope between x, g of , and this point right over here is" + }, + { + "Q": "Why does Sal draw a triangle in front of X at 2:16?\ndoes that mean change?", + "A": "Yes, it does! The triangle is actually the Greek letter Delta. When used this way, it means change .", + "video_name": "1F7LAJEVp-U", + "timestamps": [ + 136 + ], + "3min_transcript": "Graph the line that represents a proportional relationship between y and x with a unit rate 0.4. That is, a change of one unit in x corresponds to a change of 0.4 units in y. And they also ask us to figure out what the equation of this line actually is. So let me get my scratch pad out and we could think about it. So let's just think about some potential x and y values here. So let's think about some potential x and y values. So when we're thinking about proportional relationships, that means that y is going to be equal to some constant times x. So if we have a proportional relationship, if you have zero x's, it doesn't matter what your constant here is, you're going to have zero y's. So the point 0, 0 should be on your line. So if this is the point 0, 0, this should be on my line right over there. Now, let's think about what happens as we increase x. So if x goes from 0 to 1, we already know that a change of 1 unit in x corresponds to a change of 0.4 units in y. So if x increases by 1, then y is going to increase by 0.4. The 0.4 is hard to graph on this little tool right over here. So let's try to get this to be a whole number. So then when x increases another 1, y is going to increase by 0.4 again. It's going to get to 0.8. When x increases again by 1, then y is going to increase by 0.4 again. It's going to get to 1.2. If x increases again, y is going to increase by 0.4 again. Notice, every time x is increasing by 1, y is increasing by 0.4. That's exactly what they told us here. Now, if x increases by 1 again to 5, then y is going to increase 0.4 to 2. And I like this point because this is nice and easy to graph. So we see that the point 0, 0 and the point 5 comma 2 should be on this graph. And I could draw it. And I'm going to do it on the tool in a second as well. So it'll look something like this. Notice the slope of this actual graph. If our change in x is 5. So notice, here our change in x is 5. Our change in x is 5. You see that as well. When you go from 0 to 5, this change in x is 5. Change in x is equal to 5. What was our corresponding change in y? Well, our corresponding change in y when our change in x was 5, our change in y was equal to 2. And you see that here, when x went from 0 to 5, y went from 0 to 2. So our change in y in this circumstance is equal to 2. So our slope, which is change in y over change in x, is the rate of change of your vertical axis with respect to your horizontal axis, is going to be equal to 2 over 5," + }, + { + "Q": "At 1:57 Why can he take the radical sign off the right and stick it on the left of the equation? I feel like I missed a step or something. I can't figure out why he did that.", + "A": "Oh, right. Thank you. It makes total sense now. I must have been trying too hard and missed it.", + "video_name": "aeyFb2eVH1c", + "timestamps": [ + 117 + ], + "3min_transcript": "We've got the function f of x is equal to x plus 2 squared plus 1, and we've constrained our domain that x has to be greater than or equal to negative 2. That's where we've defined our function. And we want to find its inverse. And I'll leave you to think about why we had to constrain it to x being a greater than or equal to negative 2. Wouldn't it have been possible to find the inverse if we had just left it as the full parabola? I'll leave you -- or maybe I'll make a future video about that. But let's just figure out the inverse here. So, like we've said in the first video, in the introduction to inverses, we're trying to find a mapping. Or, if we were to say that y -- if we were to say that y is equal to x plus 2 squared plus 1. This is the function you give me an x and it maps to y. We want to go the other way. We want to take, I'll give you a y and then map it to an x. So what we do is, we essentially just solve for x in terms of y. So let's do that one step at a time. So, the first thing to do, we could subtract 1 from both sides of this equation. And now to solve here, you might want to take the square root. And that actually will be the correct thing to do. But it's very important to think about whether you want to take the positive or the negative square root at this step. So we've constrained our domain to x is greater than or equal to negative 2. So this value right here, x plus 2, if x is always greater than or equal to negative 2, x plus 2 will always be greater than or equal to 0. So this expression right here, this right here is positive. This is positive. So we have a positive squared. So if we really want to get to the x plus 2 in the appropriate domain, we want to take the positive square root. And in the next video or the video after that, we'll solve an example where you want to take the negative square root. So we're going to take theundefined positive square root, or just the principal root, which is just the square root sign, of both sides. So you get the square root of y minus 1 is equal to x plus 2. the beginning we had a constraint on x. We had for x is greater than or equal to negative 2. But what constraint could we have on y? If you look at the graph right here, x is greater than equal to negative 2. But what's why? What is the range of y-values that we can get here? Well, if you just look at the graph, y will always be greater than or equal to 1. And that just comes from the fact that this term right here is always going to be greater than or equal to 0. So the minimum value that the function could take on is 1. So we could say for x is greater than or equal to negative 2, and we could add that y is always going to be greater than or equal to 1. y is always greater than or equal to 1. The function is always greater than or equal to that right there. To 1. And the reason why I want to write it at the stage is" + }, + { + "Q": "Can someone elaborate why at 1:50 Sal chose to take the positive square root.", + "A": "The values here will be positive. The thing squared is (x+2) and the minimum value possible for x is -2. Therefore, 0 (-2+2) and up is the set of possible values. Since the values are positive (well, technically nonnegative) the principal/positive square root is sufficient.", + "video_name": "aeyFb2eVH1c", + "timestamps": [ + 110 + ], + "3min_transcript": "We've got the function f of x is equal to x plus 2 squared plus 1, and we've constrained our domain that x has to be greater than or equal to negative 2. That's where we've defined our function. And we want to find its inverse. And I'll leave you to think about why we had to constrain it to x being a greater than or equal to negative 2. Wouldn't it have been possible to find the inverse if we had just left it as the full parabola? I'll leave you -- or maybe I'll make a future video about that. But let's just figure out the inverse here. So, like we've said in the first video, in the introduction to inverses, we're trying to find a mapping. Or, if we were to say that y -- if we were to say that y is equal to x plus 2 squared plus 1. This is the function you give me an x and it maps to y. We want to go the other way. We want to take, I'll give you a y and then map it to an x. So what we do is, we essentially just solve for x in terms of y. So let's do that one step at a time. So, the first thing to do, we could subtract 1 from both sides of this equation. And now to solve here, you might want to take the square root. And that actually will be the correct thing to do. But it's very important to think about whether you want to take the positive or the negative square root at this step. So we've constrained our domain to x is greater than or equal to negative 2. So this value right here, x plus 2, if x is always greater than or equal to negative 2, x plus 2 will always be greater than or equal to 0. So this expression right here, this right here is positive. This is positive. So we have a positive squared. So if we really want to get to the x plus 2 in the appropriate domain, we want to take the positive square root. And in the next video or the video after that, we'll solve an example where you want to take the negative square root. So we're going to take theundefined positive square root, or just the principal root, which is just the square root sign, of both sides. So you get the square root of y minus 1 is equal to x plus 2. the beginning we had a constraint on x. We had for x is greater than or equal to negative 2. But what constraint could we have on y? If you look at the graph right here, x is greater than equal to negative 2. But what's why? What is the range of y-values that we can get here? Well, if you just look at the graph, y will always be greater than or equal to 1. And that just comes from the fact that this term right here is always going to be greater than or equal to 0. So the minimum value that the function could take on is 1. So we could say for x is greater than or equal to negative 2, and we could add that y is always going to be greater than or equal to 1. y is always greater than or equal to 1. The function is always greater than or equal to that right there. To 1. And the reason why I want to write it at the stage is" + }, + { + "Q": "At 2:13, Sal makes note of domain and range. I am familiar with the concept, but am unsure of why it is important.", + "A": "A domain is a certain range in which the function remains applicable. In many cases, if the domains are not constrained to a certain range, the function could be meaningless. For example, if S(x) in a function stands for the number of students in a classroom, the domain of x must be greater or equal to 0. Otherwise, this functon is meaningless. I hope you see how it works :)", + "video_name": "aeyFb2eVH1c", + "timestamps": [ + 133 + ], + "3min_transcript": "We've got the function f of x is equal to x plus 2 squared plus 1, and we've constrained our domain that x has to be greater than or equal to negative 2. That's where we've defined our function. And we want to find its inverse. And I'll leave you to think about why we had to constrain it to x being a greater than or equal to negative 2. Wouldn't it have been possible to find the inverse if we had just left it as the full parabola? I'll leave you -- or maybe I'll make a future video about that. But let's just figure out the inverse here. So, like we've said in the first video, in the introduction to inverses, we're trying to find a mapping. Or, if we were to say that y -- if we were to say that y is equal to x plus 2 squared plus 1. This is the function you give me an x and it maps to y. We want to go the other way. We want to take, I'll give you a y and then map it to an x. So what we do is, we essentially just solve for x in terms of y. So let's do that one step at a time. So, the first thing to do, we could subtract 1 from both sides of this equation. And now to solve here, you might want to take the square root. And that actually will be the correct thing to do. But it's very important to think about whether you want to take the positive or the negative square root at this step. So we've constrained our domain to x is greater than or equal to negative 2. So this value right here, x plus 2, if x is always greater than or equal to negative 2, x plus 2 will always be greater than or equal to 0. So this expression right here, this right here is positive. This is positive. So we have a positive squared. So if we really want to get to the x plus 2 in the appropriate domain, we want to take the positive square root. And in the next video or the video after that, we'll solve an example where you want to take the negative square root. So we're going to take theundefined positive square root, or just the principal root, which is just the square root sign, of both sides. So you get the square root of y minus 1 is equal to x plus 2. the beginning we had a constraint on x. We had for x is greater than or equal to negative 2. But what constraint could we have on y? If you look at the graph right here, x is greater than equal to negative 2. But what's why? What is the range of y-values that we can get here? Well, if you just look at the graph, y will always be greater than or equal to 1. And that just comes from the fact that this term right here is always going to be greater than or equal to 0. So the minimum value that the function could take on is 1. So we could say for x is greater than or equal to negative 2, and we could add that y is always going to be greater than or equal to 1. y is always greater than or equal to 1. The function is always greater than or equal to that right there. To 1. And the reason why I want to write it at the stage is" + }, + { + "Q": "why do we ignore the decimals at 0:40?", + "A": "thanks to kwymberry", + "video_name": "JEHejQphIYc", + "timestamps": [ + 40 + ], + "3min_transcript": "We're asked to multiply 32.12, or 32 and 12 hundredths, times 0.5, or just 5 tenths. Now when you multiply decimals, you multiply them the exact same way you would multiply whole numbers, and then you count the number of spaces behind the decimal you have in your two numbers you're multiplying, and you're going to have that many spaces in your product. Let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You could almost ignore the decimal. Right now, you should write the decimal where they belong, but you can almost pretend that this is 3,212 times 5, and then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15, plus 1 is 16. And then we don't have any other places. If we were just doing this as 05, we wouldn't multiply 0 times this whole thing. We would just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal point in the two numbers we're multiplying. So we have one, two, three spaces, or three numbers, to the right of the decimals in the two numbers that we're So we need that many numbers to the right of the decimal in our answer. So we go one, two, three, put the decimal right over there. And this trailing zero right here we can ignore, because it's really not adding any information there. So we could just write this as 16.06. The last thing you want to do is just make sure that this makes sense. You have a number that's almost 32, and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1/2. So we're really multiplying 32.12 times 1/2. We're trying to figure out what one half of 32.12 is. And half of 32 is 16, and half of 0.12 0.06, so this makes complete sense." + }, + { + "Q": "At 0:17, Sal says that multiplying decimals is the same as multiplying whole numbers. Why is this true?", + "A": "He means it s ALMOST the same, but not all the same.", + "video_name": "JEHejQphIYc", + "timestamps": [ + 17 + ], + "3min_transcript": "We're asked to multiply 32.12, or 32 and 12 hundredths, times 0.5, or just 5 tenths. Now when you multiply decimals, you multiply them the exact same way you would multiply whole numbers, and then you count the number of spaces behind the decimal you have in your two numbers you're multiplying, and you're going to have that many spaces in your product. Let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You could almost ignore the decimal. Right now, you should write the decimal where they belong, but you can almost pretend that this is 3,212 times 5, and then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15, plus 1 is 16. And then we don't have any other places. If we were just doing this as 05, we wouldn't multiply 0 times this whole thing. We would just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal point in the two numbers we're multiplying. So we have one, two, three spaces, or three numbers, to the right of the decimals in the two numbers that we're So we need that many numbers to the right of the decimal in our answer. So we go one, two, three, put the decimal right over there. And this trailing zero right here we can ignore, because it's really not adding any information there. So we could just write this as 16.06. The last thing you want to do is just make sure that this makes sense. You have a number that's almost 32, and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1/2. So we're really multiplying 32.12 times 1/2. We're trying to figure out what one half of 32.12 is. And half of 32 is 16, and half of 0.12 0.06, so this makes complete sense." + }, + { + "Q": "At 2:13 - 2:17, How come they remove the 0 at the very end? I know that it has no meaning but why?", + "A": "They remove it because it has no meaning and it is simpler to express the number without it.", + "video_name": "JEHejQphIYc", + "timestamps": [ + 133, + 137 + ], + "3min_transcript": "We're asked to multiply 32.12, or 32 and 12 hundredths, times 0.5, or just 5 tenths. Now when you multiply decimals, you multiply them the exact same way you would multiply whole numbers, and then you count the number of spaces behind the decimal you have in your two numbers you're multiplying, and you're going to have that many spaces in your product. Let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You could almost ignore the decimal. Right now, you should write the decimal where they belong, but you can almost pretend that this is 3,212 times 5, and then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15, plus 1 is 16. And then we don't have any other places. If we were just doing this as 05, we wouldn't multiply 0 times this whole thing. We would just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal point in the two numbers we're multiplying. So we have one, two, three spaces, or three numbers, to the right of the decimals in the two numbers that we're So we need that many numbers to the right of the decimal in our answer. So we go one, two, three, put the decimal right over there. And this trailing zero right here we can ignore, because it's really not adding any information there. So we could just write this as 16.06. The last thing you want to do is just make sure that this makes sense. You have a number that's almost 32, and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1/2. So we're really multiplying 32.12 times 1/2. We're trying to figure out what one half of 32.12 is. And half of 32 is 16, and half of 0.12 0.06, so this makes complete sense." + }, + { + "Q": "Wait is this stuff a joke? Especially 3:42 are there really such diseases?", + "A": "Don t worry, all of them are made up, except maybe the mind-blown syndrom. If you show hexaflexagons to your friends, they could very well be disbelieving at the amazing-ness.", + "video_name": "AmN0YyaTD60", + "timestamps": [ + 222 + ], + "3min_transcript": "clockwise down under the flaps, even if you flip it over or flex it. Well, in this hexaflexagon, it flows counter-clockwise. They're mirror images. The chirality is decided when you fold and tape your triangles into a twisty loop, and once taped, it is impossible to change from one to the other without cutting it apart, at least in three-dimensional euclidean space. A change in chirality could be a sign that your flexagon has been flipped through four-dimensional space and is possibly a highly dangerous multi-dimensional portal. With experience, a hexaflexagon master can construct a hexaflexagon in mere seconds. Some forgo tape and scissors entirely by folding a double strip that's too long and tucking the extra in. This is an advanced technique that should not be attempted without prior training. Beware topological changes. This family seems safe from this philosoraptor, because they live on separate planets with a cold, empty vacuum of space between them. But after a single flex, the unfortunate victims are now doomed, protected only by the inconsequential barrier of their domicile. Your stars might explode, your frowns may become smiles, become the roundest of circles. Perfectly healthy snakes may turn into snake loops, or worse, become decapitated. Either state is fatal for the snake, as having no head can lead to starvation. This can be avoided by simply marking where connections will be across neighboring triangles first. Afterwards, the lines can be filled in however you like. Be aware that with the trihexaflexagon, there are two variations to each face. So you can simply draw one side where triangles connect, and flip and draw the other. But in the hexa-hexaflexagon, the main three faces each appear four different ways. If you use hexaflexagons, keep an eye out for signs of dependency. Overuse can lead to addiction and possibly an overdose. Some users of hexaflexagons report confusion, mind-blown syndrome, hexaflexaperplexia, hexaflexadyslexia, hexaflexaperfectionism, and hexaflexa-Mexican-food-cravings. If you find yourself experiencing any of these symptoms, stop flexagon use immediately, and see the head of your math department. With proper precautions, flexagating can be a great part of your life. Follow these simple safety guidelines, experience." + }, + { + "Q": "At 1:01, Does Vi make two Mobius Strips tied together? Does that mean a hexaflexagon is a mobius strip?", + "A": "Right-o! A hexaflexagon s unique ability to make that weird twist and end up with multiple sides is caused indirectly by the mobius strip it s made out of!", + "video_name": "AmN0YyaTD60", + "timestamps": [ + 61 + ], + "3min_transcript": "Hexaflexagons-- they're cool, hip, and hexa-fun to play with, right? Wrong. Hexaflexagons are not toys. With the increasing number of hexaflexagons finding their way into homes and schools, it's important to be aware of proper flexagation regulations when engaging in flexagon construction and use. Taking proper precautions can help avoid a flexa-catastrophe. Do not wear loose clothing when engaging in flexagation. If you have long hair, tie it back, so it doesn't get caught in a flexagation device. Ties are also a common source of incidents. Stay alert. Never flexagate while under the influence. When using a hexaflexagon, sudden unexpected sides may appear, and drugs like alcohol can slow reaction time. If you aren't sure what kind of flexagon you're dealing with, it's safer to temporarily disable the flexagon. Flexagons can be disarmed by using scissors to cut them apart. You can cut across the original seam where the paper strip was taped together, which may appear on the edge or through the face of the flexagon. In an emergency, however, flexagons can be cut apart right through a triangle, or on three edges if you want to retain symmetry, or into nine separate triangles if you really want to be safe. You can even cut them in half down the length of the paper strip like this, into two separate-- you can figure out what kind it is. If it has nine triangles, that's 18 triangle sides. So at six triangles per hexagon side, that's three sides of trihexaflexagon. Note that some flexagons might be made from a double strip of triangles that have been folded in half, so that marker doesn't bleed through. Don't let yourself be fooled by the extra triangles. Avoid danger during hexaflexagon construction. If you're not working from a printed pattern, you might start your flexagon by picking a point on the edge of a strip of paper, folding that 180 degree angle into thirds to create 360 degree angles, and then using the equilateral triangle that results as a guide to fold the rest of the strip of paper, zigzagging back and forth. Without proper attention and focus, this could easily lead to becoming unreasonably amused with the springy spring of happy triangles that results. Always keep your hexaflexagon in good working order. Pre-creasing all the triangles both ways before configuring them into hexaflexagonal formation will help your flexagon operate properly and avoid accidents. Keep a close watch on the chirality of your hexaflexagon. That is, whether it is right or left handed. clockwise down under the flaps, even if you flip it over or flex it. Well, in this hexaflexagon, it flows counter-clockwise. They're mirror images. The chirality is decided when you fold and tape your triangles into a twisty loop, and once taped, it is impossible to change from one to the other without cutting it apart, at least in three-dimensional euclidean space. A change in chirality could be a sign that your flexagon has been flipped through four-dimensional space and is possibly a highly dangerous multi-dimensional portal. With experience, a hexaflexagon master can construct a hexaflexagon in mere seconds. Some forgo tape and scissors entirely by folding a double strip that's too long and tucking the extra in. This is an advanced technique that should not be attempted without prior training. Beware topological changes. This family seems safe from this philosoraptor, because they live on separate planets with a cold, empty vacuum of space between them. But after a single flex, the unfortunate victims are now doomed, protected only by the inconsequential barrier of their domicile. Your stars might explode, your frowns may become smiles," + }, + { + "Q": "At 1:50 what did you say?", + "A": "At 1:50 the speaker said So we can change the order of the numbers without changing the answer . Hope this helped ! :)", + "video_name": "uHHnwafYivk", + "timestamps": [ + 110 + ], + "3min_transcript": "- [Voiceover] Let's multiply 40 times 70. So 40 times, we have the number 70. So we could actually list that out, the number 70, 40 different times and add it up. But that's clearly a lot of computations to do. And there's gotta be a faster way. So another way is to stick with multiplication, but see if we can break these numbers up, this 40 and this 70, decompose them, break them up in some way to get numbers that might be a little easier to multiply with. For me, multiplying by 10 is the easiest number, because I know the pattern to add a zero. So, I'm gonna break up 40 and say, instead of 40, four times 10. Four times 10 and 40 are equivalent. They're the same thing, so I can replace the 40 with a four times 10. And then for my 70, same thing. I can break this up and write seven times 10. So these two expressions, 40 times 70 and four times 10 times seven times 10, are equal; they're equivalent. So they'll have the same solution. But for me, this one down here is simpler to work out because of these times 10s. So I'll solve this one, knowing that I'll get the same solution as I would have for this top expression. So what we can do is we can re-order these numbers in a different order to, again, continue making this question easier for us to solve. Because in multiplication, the order doesn't matter. If we have five times two, for example, that would be the same as two times five. They're both 10. Five twos or two fives, either way, it's 10. So we can change the order of the numbers without changing the answer. So again, we're going to change our expression a little bit, but what we're not going to change is the solution. So I'm gonna put my one-digit numbers first. And then, I'll put the two-digit numbers, the 10s, times 10 and the other times 10. So we have all the same factors, all the same numbers, in both of these expressions. They've just been re-ordered. And now, I'll solve going across. Four times seven is 28. And now we have 28 times 10, and times another 10. Well, the pattern for times 10 that we know is when we multiply a whole number like 28 times 10, we will add a zero to the end. One zero for that zero in 10, because 28 times 10 is 28 10s, 28 10s, or 280. And that multiplied 28 times 10, and then if we multiply by this other 10, well we have to add another zero." + }, + { + "Q": "At 1:43 what is the difference in tangent line and secant line?", + "A": "A tangent line touches the curve at one point whereas the secant line intersects at two points. The secant slope can be found by simple slope eq: y2 - y1/x2-x1 The tangent slope is found by f(x) - f(a)/ x - a. Hope this helps", + "video_name": "BYTfCnR9Sl0", + "timestamps": [ + 103 + ], + "3min_transcript": "What I want to do in this video is to see whether the power rule is giving us results that at least seem reasonable. This is by no means a proof of the power rule, but at least we'll feel a little bit more comfortable using it. So let's say that f of x is equal to x. The power rule tells us that f prime of x is going to be equal to what? Well, x is the same thing as x to the first power. So n is implicitly 1 right over here. So we bring the 1 out front. It'll be 1 times x to the 1 minus 1 power. So it's going to be 1 times x to the 0 power. x to the 0 So it's just going to be equal to 1. Now, does that makes conceptual sense if we actually try to visualize these functions? So let me actually try to graph these functions. So that's my y-axis. This is my x-axis. And let me graph y equals x. So y is equal to f of x here. So y is equal to x. So it looks something like that. Or this is f of x is equal to x, or y is equal to this f of x right over there. Now, actually, let me just call that f of x just to not confuse you. So this right over here is f of x is equal to x that I graphed right over here. y is equal to f of x, which is equal to x. And now, let me graph the derivative. Let me graph f prime of x. That's saying it's 1. That's saying it's 1 for all x. Regardless of what x is, it's going to be equal to 1. Is this consistent with what we know about derivatives and slopes and all the rest? Well, let's look at our function. What is the slope of the tangent line right at this point? Well, right over here, this has slope 1 continuously. Or it has a constant slope of 1. Slope is equal to 1 no matter what x is. It's a line. And for a line, the slope is constant. So over here, the slope is indeed 1. If you go to this point over here, the slope is indeed 1. If you go over here, the slope is indeed 1. So we've got a pretty valid response there. So let's say I have g of x is equal to x squared. The power rule tells us that g prime of x would be equal to what? Well, n is equal to 2. So it's going to be 2 times x to the 2 minus 1. Or it's going to be equal to 2 x to the first power. It's going to be equal to 2x. So let's see if this makes a reasonable sense. And I'm going to try to graph this one a little bit more precisely. Let's see how precisely I can graph it. So this is the x-axis, y-axis. Let me mark some stuff off here. So this is 1, 2, 3, 4, 5. This is 1, 2, 3, 4. 1, 2, 3, 4. So g of x." + }, + { + "Q": "8:20 but 1/2 squared is 1/4 :'<", + "A": "The ambiguity of not having parenthesis. If you are referring to the blue writing I think it is (1^2)/ 2 and not (1/2)^2 since he substituted in 1 for x in x^2/2.", + "video_name": "vhMl755vR5Q", + "timestamps": [ + 500 + ], + "3min_transcript": "to be equal to the value of this inner function, which is just x. And so this is just y is equal to x. And then we're going to multiply it times the depth, times the depth of each of these disks. And each of these disks are going to have a depth of dx. If you imagine a quarter that has an infinitely-thin depth right over here. So it's going to be dx. And so the volume our, kind of our truffle with a cone carved out, is going to be this integral minus this integral right over here. And we could evaluate it just like that. Or we could even say, OK we could factor out a pi out of both of them. There are actually, there's multiple ways But let's just evaluate it like this, and then I'll generalize it in the next video. So this is going to be equal to the definite integral from 0 to 1. You take the pi outside. Square root of x squared is going to be x dx minus the integral, we can factor the pi out. And we could say this is going to be equal to pi times the antiderivative of x, which is just x squared over 2 evaluated from 0 to 1, minus pi, times the antiderivative of x squared, which is x to the third over 3 evaluated from 0 to 1. This expression is equal to-- and I'm going to arbitrarily switch colors just because the green's getting monotonous-- pi times 1 squared over 2 minus 0 squared over 2. I could write squared. 1 squared over 2 minus 0 squared over 2, minus pi times 1 to the third over 3 minus 0 to the third over 3. me do it in that same blue color-- so this is this simplified. This is just 0 right over here. This is 1 squared over 2, which is just 1/2. So it's just pi over 2, 1/2 times pi minus-- well this is just 0, this is 1/3, minus pi over 3. And then to simplify this, it's just really subtracting fractions. So we can find a common denominator. Common denominator is 6. This is going to be 3 pi over 6. This is 3 pi over 6 minus 2 pi over 6. pi over 3 is 2 pi over 6, pi over 2 is 3 pi over 6. And we end up with, we end up with 3 of something minus 2 of something, you end up with 1 of something. We end up with 1 pi over 6. And we are done. We were able to find the volume of that wacky kind" + }, + { + "Q": "At 1:28 do you have to do it height times width times depth in that order?", + "A": "Yes, the formula for volume in rectangular prisms is: V=B*L*H V=volume B=base L=length H=height.", + "video_name": "I9efKVtLCf4", + "timestamps": [ + 88 + ], + "3min_transcript": "What is the volume of this box? Drag on the box to rotate it. So this is pretty neat. We can actually sit and rotate this box. And here it looks like everything's being measured in meters. So we want to measure our volume in terms of cubic meters. That's going to be our unit cube here. So when we want to think about how many cubic meters could fit in this box, we've already seen examples. You really just have to multiply the three different dimensions of this box. So if you wanted the number of cubic meters that could fit in here, it's going to be six meters times 8 meters times 7 meters which is going to give you something in cubic meters. So let's think about what that is. 6 times 8 is 48. Let me see if I can do this in my head. 48 times 7, that's 40 times 7, which is going to be 280 plus 8 times 7, which is 56, Let's check our answer. Let's do one more of these. So what's the volume of this box? We'll once again, we have its height at six feet. Now everything is being measured in feet. We have it's width being four feet. So we could multiple the height times the width of four feet. And then we can multiply that times its depth of two feet. So 6 times 4 is 24 times 2 is 48 feet. 48, and I should say cubic feet. We're saying how many cubic feet can fit in here? When we multiply the various dimensions measured in feet, we're counting almost how many of those cubic feet can fit into this box." + }, + { + "Q": "0:39 i dont understand the math from here on", + "A": "the video shows how to find the volume from there on for example lxwxh= 2x4x3", + "video_name": "I9efKVtLCf4", + "timestamps": [ + 39 + ], + "3min_transcript": "What is the volume of this box? Drag on the box to rotate it. So this is pretty neat. We can actually sit and rotate this box. And here it looks like everything's being measured in meters. So we want to measure our volume in terms of cubic meters. That's going to be our unit cube here. So when we want to think about how many cubic meters could fit in this box, we've already seen examples. You really just have to multiply the three different dimensions of this box. So if you wanted the number of cubic meters that could fit in here, it's going to be six meters times 8 meters times 7 meters which is going to give you something in cubic meters. So let's think about what that is. 6 times 8 is 48. Let me see if I can do this in my head. 48 times 7, that's 40 times 7, which is going to be 280 plus 8 times 7, which is 56, Let's check our answer. Let's do one more of these. So what's the volume of this box? We'll once again, we have its height at six feet. Now everything is being measured in feet. We have it's width being four feet. So we could multiple the height times the width of four feet. And then we can multiply that times its depth of two feet. So 6 times 4 is 24 times 2 is 48 feet. 48, and I should say cubic feet. We're saying how many cubic feet can fit in here? When we multiply the various dimensions measured in feet, we're counting almost how many of those cubic feet can fit into this box." + }, + { + "Q": "At 1:05, Can we use any symbol (\u00c3\u00ab, \u00c3\u00a6, \u00c2\u00b4\u00c2\u00ac or \u00c2\u00a4) represent an unknown number?", + "A": "I think that s a yes, since Mr. Khan says you could use a smiley face.", + "video_name": "Tm98lnrlbMA", + "timestamps": [ + 65 + ], + "3min_transcript": "I'm here with Jesse Ro, whose a math teacher at Summit San Jose and a Khan Academy teaching fellow and you had some interesting ideas or questions. Yeah, one question that students ask a lot when they start Algebra is why do we need letters, why can't we just use numbers for everything? Why letters? So why do we have all these Xs and Ys and Zs and ABCs when we start dealing with Algebra? Yeah, exactly. That's interesting, well why don't we let people think about that for a second. So Sal, how would you answer this question? Why do we need letters in Algebra? So why letters. So there are a couple of ways I'd think about it. One is if you have an unknown. So if I were to write X plus three is equal to ten the reason why we're doing this is that we don't know what X is It's literally an unknown. And so we're going to solve for it in some way. But it did not have to be the letter X. We could have literally written blank plus three is equal to ten. Or we could have written Question Mark plus three is equal to ten. So it didn't have to be letters, but we needed some type of symbol. But until you know it, you need some type of a symbol to represent whatever that number is. Now we can go and solve this equation and then know what that symbol represents. But if we knew it ahead of time, it wouldn't be an unknown. It wouldn't be something that we didn't know. So that's one reason why I would use letters and where just numbers by itself wouldn't be helpful. The other is when you're describing relationships between numbers. So I could do something like - I could say - that whenever you give me a three, I'm going to give you a four. And I could say, if you give me a five, I'm going to give you a six. And i could keep going on and on forever. If you give me a 7.1, I'm going to give you an 8.1. And I could keep listing this on and on forever. Maybe you could give me any number, and I could tell you what I'm going to give you. But I would obviously run out of space and time if I were to list all of them. And we could do that much more elegantly if we used letters to describe the relationship. And so I say, look, whatever you give me, I'm going to add one to it. And that's what I'm going to give back to you. And so now, this very simple equation here can describe an infinite number of relationships between X or an infinite number of corresponding Ys and Xs. So now someone knows whatever X you give me you give me three, I add one to it, and I'm going to give you four. You give me 7.1, I'm going to add one to it and give you 8.1. So there is no more elegant way that you could've done it than by using symbols. With that said, I didn't have to use Xs and Ys. This is just a convention that kind of comes to use from history. I could've defined what you give me as Star and what I give you as Smiley Face and this also would've been a valid way to express this. So the letters are really just symbols. Nothing more." + }, + { + "Q": "At 2:16, it could have been (Star) + 1 = ... I don't know... Smiley Face). We didn't have to say \"y = x + 1\".", + "A": "It could have been. But people thought letters were more simpler.", + "video_name": "Tm98lnrlbMA", + "timestamps": [ + 136 + ], + "3min_transcript": "I'm here with Jesse Ro, whose a math teacher at Summit San Jose and a Khan Academy teaching fellow and you had some interesting ideas or questions. Yeah, one question that students ask a lot when they start Algebra is why do we need letters, why can't we just use numbers for everything? Why letters? So why do we have all these Xs and Ys and Zs and ABCs when we start dealing with Algebra? Yeah, exactly. That's interesting, well why don't we let people think about that for a second. So Sal, how would you answer this question? Why do we need letters in Algebra? So why letters. So there are a couple of ways I'd think about it. One is if you have an unknown. So if I were to write X plus three is equal to ten the reason why we're doing this is that we don't know what X is It's literally an unknown. And so we're going to solve for it in some way. But it did not have to be the letter X. We could have literally written blank plus three is equal to ten. Or we could have written Question Mark plus three is equal to ten. So it didn't have to be letters, but we needed some type of symbol. But until you know it, you need some type of a symbol to represent whatever that number is. Now we can go and solve this equation and then know what that symbol represents. But if we knew it ahead of time, it wouldn't be an unknown. It wouldn't be something that we didn't know. So that's one reason why I would use letters and where just numbers by itself wouldn't be helpful. The other is when you're describing relationships between numbers. So I could do something like - I could say - that whenever you give me a three, I'm going to give you a four. And I could say, if you give me a five, I'm going to give you a six. And i could keep going on and on forever. If you give me a 7.1, I'm going to give you an 8.1. And I could keep listing this on and on forever. Maybe you could give me any number, and I could tell you what I'm going to give you. But I would obviously run out of space and time if I were to list all of them. And we could do that much more elegantly if we used letters to describe the relationship. And so I say, look, whatever you give me, I'm going to add one to it. And that's what I'm going to give back to you. And so now, this very simple equation here can describe an infinite number of relationships between X or an infinite number of corresponding Ys and Xs. So now someone knows whatever X you give me you give me three, I add one to it, and I'm going to give you four. You give me 7.1, I'm going to add one to it and give you 8.1. So there is no more elegant way that you could've done it than by using symbols. With that said, I didn't have to use Xs and Ys. This is just a convention that kind of comes to use from history. I could've defined what you give me as Star and what I give you as Smiley Face and this also would've been a valid way to express this. So the letters are really just symbols. Nothing more." + }, + { + "Q": "For the graph that Sal draws beginning at 1:35, what does the Y axis represent? I understand how the probability of the event is represented by the area under the curve, but doesn't that mean the Y axis doesn't chart probability, but something different? Is it the probability of the probability? ;)", + "A": "Yes, the y axis charts something different. The y value at each value of x (or possible outcome) is the rate of change of the area under the curve as the interval (range of possible outcomes) increases or decreases. This ensures that the area under the curve in any interval (range of outcomes) is always equal to the probability for that interval.", + "video_name": "Fvi9A_tEmXQ", + "timestamps": [ + 95 + ], + "3min_transcript": "In the last video, I introduced you to the notion of-- well, really we started with the random variable. And then we moved on to the two types of random variables. You had discrete, that took on a finite number of values. And the these, I was going to say that they tend to be integers, but they don't always have to be integers. You have discrete, so finite meaning you can't have an infinite number of values for a discrete random variable. And then we have the continuous, which can take on an infinite number. And the example I gave for continuous is, let's say random variable x. And people do tend to use-- let me change it a little bit, just so you can see it can be something other than an x. Let's have the random variable capital Y. They do tend to be capital letters. Is equal to the exact amount of rain tomorrow. It's actually raining quite hard right now. We're short right now, so that's a positive. We've been having a drought, so that's a good thing. But the exact amount of rain tomorrow. And let's say I don't know what the actual probability distribution function for this is, but I'll draw one and then we'll interpret it. Just so you can kind of think about how you can think about continuous random variables. So let me draw a probability distribution, or they call it its probability density function. And we draw like this. And let's say that there is-- it looks something like this. Like that. All right, and then I don't know what this height is. So the x-axis here is the amount of rain. this is 3 inches, 4 inches. And then this is some height. Let's say it peaks out here at, I don't know, let's say this 0.5. So the way to think about it, if you were to look at this and I were to ask you, what is the probability that Y-- because that's our random variable-- that Y is exactly equal to 2 inches? That Y is exactly equal to two inches. What's the probability of that happening? Well, based on how we thought about the probability distribution functions for the discrete random variable, you'd say OK, let's see. 2 inches, that's the case we care about right now. Let me go up here. You'd say it looks like it's about 0.5." + }, + { + "Q": "At 5:07 Sal says that the two statements P(|Y-2|<.1) and P(1.9= 0 gives Y-2<.1 wich gives Y < 2.1 solving |Y-2|<.1 for (Y-2) < 0 gives -Y+2<.1 wich gives -Y<-1.9 wich gives Y > 1.9 Search for lecture about absolute value for more explanation.", + "video_name": "Fvi9A_tEmXQ", + "timestamps": [ + 307 + ], + "3min_transcript": "And I would say no, it is not a 0.5 chance. And before we even think about how we would interpret it visually, let's just think about it logically. What is the probability that tomorrow we have exactly 2 inches of rain? Not 2.01 inches of rain, not 1.99 inches of rain. Not 1.99999 inches of rain, not 2.000001 inches of rain. Exactly 2 inches of rain. I mean, there's not a single extra atom, water molecule above the 2 inch mark. And not as single water molecule below the 2 inch mark. It's essentially 0, right? It might not be obvious to you, because you've probably heard, oh, we had 2 inches of rain last night. But think about it, exactly 2 inches, right? Normally if it's 2.01 people will say that's 2. But we're saying no, this does not count. We want exactly 2. 1.99 does not count. Normally our measurements, we don't even have tools that can tell us whether it is exactly 2 inches. No ruler you can even say is exactly 2 inches long. At some point, just the way we manufacture things, there's going to be an extra atom on it here or there. So the odds of actually anything being exactly a certain measurement to the exact infinite decimal point is actually 0. The way you would think about a continuous random variable, you could say what is the probability that Y is almost 2? So if we said that the absolute value of Y minus is 2 is less than some tolerance? Is less than 0.1. And if that doesn't make sense to you, this is essentially just saying what is the probability that Y is greater than 1.9 and less than 2.1? I'll let you think about it a little bit. But now this starts to make a little bit of sense. Now we have an interval here. So we want all Y's between 1.9 and 2.1. So we are now talking about this whole area. And area is key. So if you want to know the probability of this occurring, you actually want the area under this curve from this point to this point. And for those of you who have studied your calculus, that would essentially be the definite integral of this probability density function from this point to this point. So from-- let me see, I've run out of space down here. So let's say if this graph-- let me draw it in a different color. If this line was defined by, I'll call it f of x. I could call it p of x or something." + }, + { + "Q": "At 1:23, why is it plus or minus the square root? Why not just square root?", + "A": "A square root undoes a power term, but a power term can hide information. Think what happens if you have x^2 and x = 1, you get 1 right? But if x were to equal -1 you would still get one. So if you were to solve with out putting the + - you only get half of the answer. This will become really big when you start solving quadratic equations.", + "video_name": "RweAgQwLdMs", + "timestamps": [ + 83 + ], + "3min_transcript": "We're asked to solve the equation 2x squared plus 3 is equal to 75. So in this situation, it looks like we might be able to isolate the x squared pretty simply. Because there's only one term that involves an x here. It's only this x squared term. So let's try to do that. So let me just rewrite it. We have 2x squared plus 3 is equal to 75. And we're going to try to isolate this x squared over And the best way to do that, or at least the first step, would be to subtract 3 from both sides of this equation. So let's subtract 3 from both sides. The left hand side, we're just left with 2x squared. That was the whole point of subtracting 3 from both sides. And on the right hand side, 75 minus 3 is 72. Now, I want to isolate this x squared. I have a 2x squared here. So I could have just an x squared here if I divide this side or really both sides by 2. Anything I do to one side, I have to do to the other side if I want to maintain the equality. So the left side, just becomes x squared. So we're left with x squared is equal to 36. And then to solve for x, we can take the positive, the plus or minus square root of both sides. So we could say the plus or-- let me write it this way-- If we take the square root of both sides, we would get x is equal to the plus or minus square root of 36, which is equal to plus or minus 6. Let me just write that on another line. So x is equal to plus or minus 6. And remember here, if something squared is equal to 36, that something could be the negative version or the positive version. It could be the principal root or it could be the negative root. Both negative 6 squared is 36 and positive 6 squared is 36, so both of these work. And you could put them back into the original equation to verify it. Let's do that. If you say 2 times 6 squared plus 3, that's 2 times 36, which is 72 plus 3 is 75. going to get the exact same result. Because negative 6 squared is also 36. 2 times 36 is 72 plus 3 is 75." + }, + { + "Q": "How does Sal know at 6:34, 6:38, and 6:46 that y=x^2, xy=12, and 5/x+y=10 are not linear equations without graphing them first?", + "A": "Linear equations have specific formats. For example, here are some of their formats: 1) Ax + By = C where A, B and C are integers 2) y = mx + b where m and b are numbers None of Sal s equations look like the examples above. y = x^2: this has an exponent on x which makes it non-linear xy = 12: this has x and y being multiplied which doesn t occur in a linear equation 5/x + y = 10: this has x in a denominator which doesn t occur in a linear equation Hope this helps.", + "video_name": "AOxMJRtoR2A", + "timestamps": [ + 394, + 398, + 406 + ], + "3min_transcript": "You can verify that. Four times zero minus three times negative four well that's gonna be equal to positive twelve. And let's see, if y were to equal zero, if y were to equal zero then this is gonna be four times x is equal to twelve, well then x is equal to three. And so you have the point zero comma negative four, zero comma negative four on this line, and you have the point three comma zero on this line. Three comma zero. Did I do that right? So zero comma negative four and then three comma zero. These are going to be on this line. Three comma zero is also on this line. So this is, this line is going to look something like-- something like, I'll just try to hand draw it. Something like that. So once again, all of the xy-- all of the xy pairs that satisfy this, Now what are some examples, maybe you're saying \"Wait, wait, wait, isn't any equation a linear equation?\" And the simple answer is \"No, not any equation is a linear equation.\" I'll give you some examples of non-linear equations. So a non-- non-linear, whoops let me write a little bit neater than that. Non-linear equations. Well, those could include something like y is equal to x-squared. you will see that this is going to be a curve. it could be something like x times y is equal to twelve. This is also not going to be a line. Or it could be something like five over x plus y is equal to ten. This also is not going to be a line. So now, and at some point you could-- I encourage you to try to graph these things, they're actually quite interesting. But given that we've now seen examples of linear equations and non-linear equations, linear equations. One way to think about is it's an equation that if you were to graph all of the x and y pairs that satisfy this equation, you'll get a line. And that's actually literally where the word linear equation comes from. But another way to think about it is it's going to be an equation where every term is either going to be a constant, so for example, twelve is a constant. It's not going to change based on the value of some variable, twelve is twelve. Or negative three is negative three. So every term is either going to be a constant or it's going to be a constant times a variable raised to the first power. So this is the constant two times x to the first power. This is the variable y raised to the first power. You could say that bceause this is just one y. We're not dividing by x or y, we're not multiplying, we don't have a term that has x to the second power, or x to the third power, or y to the fifth power. We just have y to the first power, we have x to the first power. We're not multiplying x and y together like we did over here." + }, + { + "Q": "At 4:55, on the sketched graph I noticed that some values of h(x) were actually smaller than some values of f(x)(check out the minimum of h(x) and the maximum of f(x)), if the inequallity is right, shouldn't the values of h(x) always be bigger than f(x)'s?", + "A": "This means above and below,(Recall [f(x)=y] so its like saying f(x) s y value < g(x) s y value < h(x) s y value Were only comparing the function evaluated at the same x-values (x is the input y [which is equal to f(x)] is the output)", + "video_name": "WvxKwRcHGHg", + "timestamps": [ + 295 + ], + "3min_transcript": "and Diya's calories as a function of the day is always going to be in between those. So now let's make this a little bit more mathematical. So let me clear this out so we can have some space to do some math in. So let's say that we have the same analogy. So let's say that we have three functions. Let's say f of x over some interval is always less than or equal to g of x over that same interval, which is always less than or equal to h of x over that same interval. So let me depict this graphically. So that is my y-axis. This is my x-axis. in the x-axis right over here. So let's say h of x looks something like that. Let me make it more interesting. This is the x-axis. So let's say h of x looks something like this. So that's my h of x. Let's say f of x looks something like this. Maybe it does some interesting things, and then it comes in, and then it goes up like this, so f of x looks something like that. And then g of x, for any x-value, g of x is always in between these two. And I think you see where the squeeze is happening and where the sandwich is happening. If h of x and f of x were bendy pieces of bread, g of x would be the meat of the bread. So it would look something like this. Now, let's say that we know-- this is the analogous thing. On a particular day, Sal and Imran ate the same amount. the limit as f and h approach that x-value, they approach is the same limit. So let's take this x-value right over here. Let's say the x-value is c right over there. And let's say that the limit of f of x as x approaches c is equal to L. And let's say that the limit as x approaches c of h of x is also equal to L. So notice, as x approaches c, h of x approaches L. As x approaches c from either side, f of x approaches L. So these limits have to be defined. Actually, the functions don't have to be defined at x approaches c. Just over this interval, they have to be defined as we approach it." + }, + { + "Q": "What does Sal mean by \"wacky functions\" at 6:54?", + "A": "He means functions that are complex or difficult to manipulate algebraically.", + "video_name": "WvxKwRcHGHg", + "timestamps": [ + 414 + ], + "3min_transcript": "the limit as f and h approach that x-value, they approach is the same limit. So let's take this x-value right over here. Let's say the x-value is c right over there. And let's say that the limit of f of x as x approaches c is equal to L. And let's say that the limit as x approaches c of h of x is also equal to L. So notice, as x approaches c, h of x approaches L. As x approaches c from either side, f of x approaches L. So these limits have to be defined. Actually, the functions don't have to be defined at x approaches c. Just over this interval, they have to be defined as we approach it. And if these limits right over here are defined and because we know that g of x is always sandwiched in between these two functions, therefore, on that day or for that x-value-- I should get out of that food-eating analogy-- this tells us if all of this is true over this interval, this tells us that the limit as x approaches c of g of x must also be equal to L. And once again, this is common sense. f of x is approaching L, h of x is approaching L, g of x is sandwiched in between it. So it also has to be approaching L. And you might say, well, this is common sense. Why is this useful? Well, as you'll see, this is useful for finding the limits of some wacky functions. If you can find a function that's always greater than it and a function that's always less than it, and you can find the limit as they approach some c, know that that wacky function in between is going to approach that same limit." + }, + { + "Q": "Wait, I don't understand something. At 2:50, Sal squared the numbers 15 and 20. Why? What was the need to square them? He then added them, got 625 as a sum, and then found the positive square root of it--25--and deemed that number the length of the rope. Isn't there a more simpler way to solve the problem?", + "A": "Pythagorean theorem: a^2+b^2=c^2 in this case: 15^2+20^2=c^2 you re trying to figure out c (length of rope) so 625 answer -- from that square root of 625 is 25", + "video_name": "JVrkLIcA2qw", + "timestamps": [ + 170 + ], + "3min_transcript": "I can draw it as a pole so it's a little bit clearer. Even shade it in if we like. And then they say a rope attaches to the deck 15 feet away from the base of the mast. So this is the base of the mast. This is the deck right here. The rope attaches 15 feet away from the base of the mast. So if this is the base of the mast, we go 15 feet, might be about that distance right there. Let me mark that. And the rope attaches right here. From the top of the mast all the way that base. So the rope goes like that. And then they ask us, how long is the rope? So there's a few things you might realize. We're dealing with a triangle here. And it's not any triangle. We're assuming that the mast goes straight up and that the deck is straight left and right. So this is a right triangle. This is a 90 degree angle right here. we can always figure out the third side of a right triangle using the Pythagorean theorem. And all that tells us is it the sum of the squares of the shorter sides of the triangle are going to be equal to the square of the longer side. And that longer side is call the hypotenuse. And in all cases, the hypotenuse is the side opposite the 90 degree angle. It is always going to be the longest side of our right triangle. So we need to figure out the hypotenuse here. We know the lengths of the two shorter sides. So we can see that if we take 15 squared, that's one of the short sides, I'm squaring it. And then add that to the square of the other shorter side, to 20 feet squared. hypotenuse. The hypotenuse will always be the longest side. Let's say the hypotenuse is in green just so we get our color coding nice. That is going to be equal to the rope squared. Or the length of the rope. Let's call this distance right here r. r for rope. So 15 squared plus 20 squared is going to be equal to r squared. And what's 15 squared? It's 225. 20 squared is 400. And that's going to be equal to r squared. Now 225 plus 400 is 625. 625 is equal to r squared. And then we can take the principal root of both sides of this equation. Because we're talking about distances, we want the positive square root." + }, + { + "Q": "In 2:05 when Sal said that... when you know two sides of the triangle you calculate the the hypotenuse...My question is that can you calculate the hypotenuse or any one of the sides if you only know ONE side of the triangle?", + "A": "In general, no, unless you know something else about the triangle. There are techniques using trigonometry that will let you find sides if you know one side and one angle (other than the right angle).", + "video_name": "JVrkLIcA2qw", + "timestamps": [ + 125 + ], + "3min_transcript": "The main mast of a fishing boat is supported by a sturdy rope that extends from the top of the mast to the deck. If the mast is 20 feet tall and the rope attaches to the deck 15 feet away from the base of the mast, how long is the rope? So let's draw ourselves a boat and make sure we understand what the deck and the mast and all of that is. So let me draw a boat. I'll start with yellow. So let's say that this is my boat. That is the deck of the boat. And the boat might look something like this. It's a sailing boat. This is the water down here. And then the mast is the thing that holds up the sail. So let me draw ourselves a mast. And they say the mast is 20 feet tall. So this distance right here is 20 feet. I can draw it as a pole so it's a little bit clearer. Even shade it in if we like. And then they say a rope attaches to the deck 15 feet away from the base of the mast. So this is the base of the mast. This is the deck right here. The rope attaches 15 feet away from the base of the mast. So if this is the base of the mast, we go 15 feet, might be about that distance right there. Let me mark that. And the rope attaches right here. From the top of the mast all the way that base. So the rope goes like that. And then they ask us, how long is the rope? So there's a few things you might realize. We're dealing with a triangle here. And it's not any triangle. We're assuming that the mast goes straight up and that the deck is straight left and right. So this is a right triangle. This is a 90 degree angle right here. we can always figure out the third side of a right triangle using the Pythagorean theorem. And all that tells us is it the sum of the squares of the shorter sides of the triangle are going to be equal to the square of the longer side. And that longer side is call the hypotenuse. And in all cases, the hypotenuse is the side opposite the 90 degree angle. It is always going to be the longest side of our right triangle. So we need to figure out the hypotenuse here. We know the lengths of the two shorter sides. So we can see that if we take 15 squared, that's one of the short sides, I'm squaring it. And then add that to the square of the other shorter side, to 20 feet squared." + }, + { + "Q": "at 1:44 he shows that (7) (5) means the same thing is 7x5\nwhat is the reasoning or history for why a number inside a ( ) means to multiply. I would love to understand the thinking behind that better.", + "A": "Like most shortcuts... laziness. I suspect that if you look back far enough, some symbol was regularly used for multiplication... but anything you do that much begs for simplification so at some point someone said how about we call it multiply if there is just no operator and put in the operator if it is anything else ... This works pretty well so we kept it.", + "video_name": "Yw3EoxC_GXU", + "timestamps": [ + 104 + ], + "3min_transcript": "Rewrite 5 plus 5 plus 5 plus 5 plus 5 plus 5 plus 5 as a multiplication expression. And then they want us to write the expression three times using different ways to write multiplication. So let's do the first part. Let's write it as a multiplication expression. So how many times have we added 5 here? Well, we've got it at one, two, three, four, five, six, seven. So one way to think of it, if I just said what is here? How many 5's are there? You'd say, well, I added 5 to itself seven times, right? You could literally say this is 7 times 5. We could literally write, this is 7 times 5, or you could view it as 5 seven times. I'm not even writing it mathematically yet. I'm just saying, look, if I saw seven of something, you would literally say, if these were apples, you would say apples seven times, or you'd say seven times the apple, whatever it is. Now, in this case, we're actually adding the number to each other, and we could figure out what that is, and But the way we would write this mathematically, we would say this is 7 times 5. We could also write it like this. We could write it 7 dot 5. This and this mean the exact same thing. It means we're multiplying 7 times 5 or 5 times 7. You can actually switch the order, and you get the exact same value. You could actually write it 5 times 7. So you could interpret this as 7 five times or 5 seven times, however you like to do it, or 5 seven times. I don't want to confuse you. I just want to show you that these are all equivalent. This is also equivalent. 5 times 7. Same thing. You could write them in parentheses. You could write it like this. This all means the same thing. That's 7 times 5, and so is this. These all evaluate to the same thing: 5 times 7. So these are all equivalent, and since we've worked with it so much, let's just figure out the answer. So if we add up 5 to itself seven times, what do we get? Well, 5 plus 5 is 10. plus 5 is 35. So all of these evaluate to 35, just so you see that they're the same thing. These are all equivalent to 35. And just something to think about, this is also the exact same thing, depending on how you want to interpret this, as 7 five times. They didn't ask us to do it, but I thought I would point it out to you. 7 five times would look like this: 7 plus 7 plus 7 plus 7 plus 7, right? I have 7 five times. I added it to itself five different times. There's five 7's here added to each other. And when you add these up, you'll also get 35. And that's why 5 times 7 and 7 times 5 is the same thing." + }, + { + "Q": "I thought that dx meant that the integral was done relative to the variable x. So at around 3:16 he brings the dx to the middle of the equation. Doesn't that affect the e to the power of .... somehow? Also why is dx being treated as a variable?", + "A": "It doesn t affect the integral at all. dx is essentially meaning a small change in x (think of a minuscule delta-x). It is for all intents and purposes in integration a dummy variable: like any other variable in an expression, you can move it around subject to the laws of multiplicative association. dx is what ultimately creates the +C at the end as well, because of this treatment as a dummy variable. Yet another question in this section that goes to blaming Leibniz.", + "video_name": "b76wePnIBdU", + "timestamps": [ + 196 + ], + "3min_transcript": "But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared, and this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is 3x squared, derivative of x squared is 2x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now, what is going to be the derivative of u with respect to x? du dx. Well, we've done this multiple times. It's going to be 3x squared plus 2x. And now we can write this in differential form. And du dx, this isn't really a fraction of the differential It really is a form of notation, but it is often useful to kind of pretend that it is a fraction, and you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here, how much does u change for a given change in x? You could multiply both sides times a dx. So both sides times a dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to 3x squared plus 2x dx. Now why is this over here? Why did I go to the trouble of doing that? Well we see we have a 3x squared plus 2x, and then it's being multiplied by a dx right over here. I could rewrite this as the integral of-- and let me do it in that color-- of 3x squared plus 2x times dx times e-- let me do that in that other color-- times e to the x to the third plus x squared. Now what's interesting about this? Well the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit, it's going to be equal to-- and what I'm going to do is I'm going to change the order. I'm going to put the du, this entire du, I'm gonna stick it on the other side here, so it looks like more of the standard form that we're used to seeing our indefinite integrals in." + }, + { + "Q": "At 3:26, where did the dx go? Can I just leave it out? And if yes, why?", + "A": "The dx is still there at 3:26. It s between the two functions in the integral.", + "video_name": "b76wePnIBdU", + "timestamps": [ + 206 + ], + "3min_transcript": "But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared, and this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is 3x squared, derivative of x squared is 2x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now, what is going to be the derivative of u with respect to x? du dx. Well, we've done this multiple times. It's going to be 3x squared plus 2x. And now we can write this in differential form. And du dx, this isn't really a fraction of the differential It really is a form of notation, but it is often useful to kind of pretend that it is a fraction, and you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here, how much does u change for a given change in x? You could multiply both sides times a dx. So both sides times a dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to 3x squared plus 2x dx. Now why is this over here? Why did I go to the trouble of doing that? Well we see we have a 3x squared plus 2x, and then it's being multiplied by a dx right over here. I could rewrite this as the integral of-- and let me do it in that color-- of 3x squared plus 2x times dx times e-- let me do that in that other color-- times e to the x to the third plus x squared. Now what's interesting about this? Well the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit, it's going to be equal to-- and what I'm going to do is I'm going to change the order. I'm going to put the du, this entire du, I'm gonna stick it on the other side here, so it looks like more of the standard form that we're used to seeing our indefinite integrals in." + }, + { + "Q": "@2:01 Can you please explain how you used the notation du/dx as a fraction?\nHow can we pretend that it's a fraction?", + "A": "Remember that slope is \u00ce\u0094y/\u00ce\u0094x, change in y over change in x, and that IS a fraction. Well dy/dx (or du/dx) is the same deal, just that we are dealing with infinitesimals, which we call the differential difference, that is dx is what happens to \u00ce\u0094x in the limit when \u00ce\u0094x approaches zero, thus dy/dx, du/dx are fractions.", + "video_name": "b76wePnIBdU", + "timestamps": [ + 121 + ], + "3min_transcript": "Let's say that we have the indefinite integral, and the function is 3x squared plus 2x times e to x to the third plus x squared dx. So how would we go about solving this? So first when you look at it, it seems like a really complicated integral. We have this polynomial right over here being multiplied by this exponential expression, and over here in the exponent, we essentially have another polynomial. It seems kind of crazy. And the key intuition here, the key insight is that you might want to use a technique here called u-substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution. And then over time, you might even be able to do this type of thing in your head. u-substitution is essentially unwinding the chain rule. And the chain rule-- I'll go in more depth in another video, But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared, and this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is 3x squared, derivative of x squared is 2x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now, what is going to be the derivative of u with respect to x? du dx. Well, we've done this multiple times. It's going to be 3x squared plus 2x. And now we can write this in differential form. And du dx, this isn't really a fraction of the differential It really is a form of notation, but it is often useful to kind of pretend that it is a fraction, and you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here, how much does u change for a given change in x? You could multiply both sides times a dx. So both sides times a dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to 3x squared plus 2x dx. Now why is this over here? Why did I go to the trouble of doing that? Well we see we have a 3x squared plus 2x, and then it's being multiplied by a dx right over here." + }, + { + "Q": "i do not understand 4:09 it is a little hazy can someone please explain it to me??", + "A": "it s to easy he seperates the inequality into two peices so it will be more easy to solve", + "video_name": "A3xPhzs-KBI", + "timestamps": [ + 249 + ], + "3min_transcript": "And then the right-hand side, we get 13 plus 14, which is 17. So we get x is less than or equal to 17. So our two conditions, x has to be greater than or equal to negative 1 and less than or equal to 17. So we could write this again as a compound inequality if we want. We can say that the solution set, that x has to be less than or equal to 17 and greater than or equal to negative 1. It has to satisfy both of these conditions. So what would that look like on a number line? So let's put our number line right there. Let's say that this is 17. Maybe that's 18. You keep going down. Maybe this is 0. I'm obviously skipping a bunch of stuff in between. Then we would have a negative 1 right there, maybe a negative 2. So x is greater than or equal to negative 1, so we would We're going to circle it in because we have a greater than or equal to. And then x is greater than that, but it has to be less than or equal to 17. So it could be equal to 17 or less than 17. So this right here is a solution set, everything that I've shaded in orange. And if we wanted to write it in interval notation, it would be x is between negative 1 and 17, and it can also equal negative 1, so we put a bracket, and it can also equal 17. So this is the interval notation for this compound inequality right there. Let's do another one. Let me get a good problem here. Let's say that we have negative 12. I'm going to change the problem a little bit from the one that I've found here. Negative 12 is less than 2 minus 5x, which is less than I want to do a problem that has just the less than and a less than or equal to. The problem in the book that I'm looking at has an equal sign here, but I want to remove that intentionally because I want to show you when you have a hybrid situation, when you have a little bit of both. So first we can separate this into two normal inequalities. You have this inequality right there. We know that negative 12 needs to be less than 2 minus 5x. That has to be satisfied, and-- let me do it in another color-- this inequality also needs to be satisfied. 2 minus 5x has to be less than 7 and greater than 12, less than or equal to 7 and greater than negative 12, so and 2 minus 5x has to be less than or equal to 7." + }, + { + "Q": "at 8:30 i dont get the dif. between \"and\" and \"or\". i went on the practice questions and it said something about \" the first inequality is included........\". im so confused i kept on getting it wrong help!", + "A": "OK. What you have to do is to first solve the first inequality and then the second.", + "video_name": "A3xPhzs-KBI", + "timestamps": [ + 510 + ], + "3min_transcript": "I just wrote this improper fraction as a mixed number. Now let's do the other constraint over here in magenta. So let's subtract 2 from both sides of this equation, just like we did before. And actually, you can do these simultaneously, but it becomes kind of confusing. So to avoid careless mistakes, I encourage you to separate it out like this. So if you subtract 2 from both sides of the equation, the left-hand side becomes negative 5x. The right-hand side, you have less than or equal to. The right-hand side becomes 7 minus 2, becomes 5. Now, you divide both sides by negative 5. On the left-hand side, you get an x. On the right-hand side, 5 divided by negative 5 is negative 1. And since we divided by a negative number, we swap the inequality. It goes from less than or equal to, to greater than or equal to. So we have our two constraints. x has to be less than 2 and 4/5, and it has to be greater So we could write it like this. x has to be greater than or equal to negative 1, so that would be the lower bound on our interval, and it has to be less than 2 and 4/5. And notice, not less than or equal to. That's why I wanted to show you, you have the parentheses there because it can't be equal to 2 and 4/5. x has to be less than 2 and 4/5. Or we could write this way. x has to be less than 2 and 4/5, that's just this inequality, swapping the sides, and it has to be greater than or equal to negative 1. So these two statements are equivalent. And if I were to draw it on a number line, it would look like this. So you have a negative 1, you have 2 and 4/5 over here. Obviously, you'll have stuff in between. We have to be greater than or equal to negative 1, so we can be equal to negative 1. And we're going to be greater than negative 1, but we also have to be less than 2 and 4/5. So we can't include 2 and 4/5 there. We can't be equal to 2 and 4/5, so we can only be less than, so we put a empty circle around 2 and 4/5 and then we fill in everything below that, all the way down to negative 1, and we include negative 1 because we have this less than or equal sign. So the last two problems I did are kind of \"and\" problems. You have to meet both of these constraints. Now, let's do an \"or\" problem. So let's say I have these inequalities. Let's say I'm given-- let's say that 4x minus 1 needs to be greater than or equal to 7, or 9x over 2 needs to be less than 3. So now when we're saying \"or,\" an x that would satisfy these" + }, + { + "Q": "At 1:58, why does he divide by ten?", + "A": "he divided by ten because if you look after you solve the 9x6 which was what was in the parentheses you then solve the 54/10 because in order of operation division is after parentheses, exponents, and multiplication and since there is none of that you divide which comes after multiplication in order of operations.", + "video_name": "STyoP3rCmb0", + "timestamps": [ + 118 + ], + "3min_transcript": "Let's see if we can multiply 9 times 0.6. Or another way to write it, we want to calculate 9 times 0.6. I'll write it like this-- 0.6. We want to figure out what this is equal to. And I encourage you to pause the video and try to figure it out on your own. And I'll give you a little bit of a hint. 0.6 is the same thing as 6 divided by 10. We know that if we start with 6, which we could write as 6.0, and if you were to divide it by 10, dividing by 10 is equivalent to moving the decimal place one place to the left. So 6 divided by 10 is 0.6. We are moving the decimal one place to the left. So I'm assuming you given a go at it. But what I'm going to do is use this that we already know to rewrite what we're trying to multiply. 0.6 is 6 divided by 10. And this expression right over here, we could either do the 6 divided by 10 first, in which case we would get 0.6, and this would turn into this problem. Or we could do the 9 times 6 first. And so let's do 9 times 6, which we know how to calculate, and then divide by 10, which we also know how to do. That. all about just moving the decimal place. So we could write 9 times 6. 9 times 6, we already know, is 54. I'll do that in orange-- is going to be 54. So this right over here is 54. And now to get to this expression, we have to divide by 10. We have to divide by 10. And what happens when we divide something by 10? And we've seen this in previous videos, why this is the case. Each place represents 10 times as much as the place to its right, or each place represents 1/10 of the place So 54 divided by 10, this is going to be-- you could start with 54. And I'll put a 0 here after the decimal. And when you divide by 10, that's equivalent of shifting the decimal one to the left. This is going to be equal to 5.4. And that should make sense to you. 5 times 10 is 50. 0.4 times 10 is 4. So it makes sense that 54 divided by 10-- I shouldn't say equal. I'd write 54 divided by 10 is equal to 5.4. So this right over here is equal to 5.4, and that's what this is. This is equal to 5.4. Notice, 9 times 6 is 54." + }, + { + "Q": "At 1:15 Sal says \"We could either do the 6 divided by the 10 first ... or we could do the 9*6 first\"\n\nWhat happened to BIDMAS?\n\nCan someone explain?\n\nI know the answers will be the same but then where does the principle apply then and how?\n\nThanks", + "A": "Im not familiar with the term BIDMAS but we used PEMDAS meaning parentheses are done first followed by exponents then you can do either division OR multiplication with addition or subtraction as the last operation. I was taught that multiplication and division are equal as far as the sequence of their operation is performed. The same goes for addition and subtraction. It just depends on what operation you are most comfortable performing first.", + "video_name": "STyoP3rCmb0", + "timestamps": [ + 75 + ], + "3min_transcript": "Let's see if we can multiply 9 times 0.6. Or another way to write it, we want to calculate 9 times 0.6. I'll write it like this-- 0.6. We want to figure out what this is equal to. And I encourage you to pause the video and try to figure it out on your own. And I'll give you a little bit of a hint. 0.6 is the same thing as 6 divided by 10. We know that if we start with 6, which we could write as 6.0, and if you were to divide it by 10, dividing by 10 is equivalent to moving the decimal place one place to the left. So 6 divided by 10 is 0.6. We are moving the decimal one place to the left. So I'm assuming you given a go at it. But what I'm going to do is use this that we already know to rewrite what we're trying to multiply. 0.6 is 6 divided by 10. And this expression right over here, we could either do the 6 divided by 10 first, in which case we would get 0.6, and this would turn into this problem. Or we could do the 9 times 6 first. And so let's do 9 times 6, which we know how to calculate, and then divide by 10, which we also know how to do. That. all about just moving the decimal place. So we could write 9 times 6. 9 times 6, we already know, is 54. I'll do that in orange-- is going to be 54. So this right over here is 54. And now to get to this expression, we have to divide by 10. We have to divide by 10. And what happens when we divide something by 10? And we've seen this in previous videos, why this is the case. Each place represents 10 times as much as the place to its right, or each place represents 1/10 of the place So 54 divided by 10, this is going to be-- you could start with 54. And I'll put a 0 here after the decimal. And when you divide by 10, that's equivalent of shifting the decimal one to the left. This is going to be equal to 5.4. And that should make sense to you. 5 times 10 is 50. 0.4 times 10 is 4. So it makes sense that 54 divided by 10-- I shouldn't say equal. I'd write 54 divided by 10 is equal to 5.4. So this right over here is equal to 5.4, and that's what this is. This is equal to 5.4. Notice, 9 times 6 is 54." + }, + { + "Q": "At 3:00 how is how is 9 * 0.6=5.4 . isn't the product of a multiplication problem always bigger then the numbers you are multiplying?", + "A": "not always because when you multiply by a number less than 1 in a multiplication the biggest number gets smaller. 0.6 is 6/10 * 9 is 54/10 and that is 5.4", + "video_name": "STyoP3rCmb0", + "timestamps": [ + 180 + ], + "3min_transcript": "0.6 is 6 divided by 10. And this expression right over here, we could either do the 6 divided by 10 first, in which case we would get 0.6, and this would turn into this problem. Or we could do the 9 times 6 first. And so let's do 9 times 6, which we know how to calculate, and then divide by 10, which we also know how to do. That. all about just moving the decimal place. So we could write 9 times 6. 9 times 6, we already know, is 54. I'll do that in orange-- is going to be 54. So this right over here is 54. And now to get to this expression, we have to divide by 10. We have to divide by 10. And what happens when we divide something by 10? And we've seen this in previous videos, why this is the case. Each place represents 10 times as much as the place to its right, or each place represents 1/10 of the place So 54 divided by 10, this is going to be-- you could start with 54. And I'll put a 0 here after the decimal. And when you divide by 10, that's equivalent of shifting the decimal one to the left. This is going to be equal to 5.4. And that should make sense to you. 5 times 10 is 50. 0.4 times 10 is 4. So it makes sense that 54 divided by 10-- I shouldn't say equal. I'd write 54 divided by 10 is equal to 5.4. So this right over here is equal to 5.4, and that's what this is. This is equal to 5.4. Notice, 9 times 6 is 54. Now you might see a little pattern here. Between these two numbers, I had exactly one number to the right of the decimal. When I take its product, let's say I ignored the decimal. I just said 9 times 6, I would've gotten 54. But then I have to divide by 10 in order to take account of the decimal, take account of the fact this wasn't a 6. This was a 6/10. And so I have one number to the right of the decimal here. And I want to you to think about that whether that's a general principle. Can we just count the total numbers of digits to the right of the decimals and then our product is going to have the same number of digits to right of the decimal? I'll let you to think about that." + }, + { + "Q": "in the video @ 4:54 i do not understand the part where the instructor of the video multiplied 4*84 to get the sum of the previous four tests.\n\n80 + 81 + 87 + 88 = 4 * 84 ? <--- how is this possible, i think its cool but do not understand how that works. thanks!!", + "A": "80+81+87+88=80+80+1+80+7+80+8=(80+80+80+80)+(0+1+7+8)=4*80+16=4*(80+4)=4*84", + "video_name": "9VZsMY15xeU", + "timestamps": [ + 294 + ], + "3min_transcript": "8 plus 8 is 16. I just ran eight miles, so I'm a bit tired. And, 4/8, so that's 32. Plus 1 is 33. And now we divide this number by 4. 4 goes into 336. Goes into 33, 8 times. 8 times 4 is 32. 33 minus 32 is 1, 16. So the average is equal to 84. So depending on what school you go to that's either a B or a C. So, so far my average after the first four exams is an 84. Now let's make this a little bit more difficult. We know that the average after four exams, at four exams, is equal to 84. average an 88, to average an 88 in the class. So let's say that x is what I get on the next test. So now what we can say is, is that the first four exams, I could either list out the first four exams that I took. Or I already know what the average is. So I know the sum of the first four exams is going to 4 times 84. And now I want to add the, what I get on the 5th exam, x. And I'm going to divide that by all five exams. So in other words, this number is the average of my first five exams. We just figured out the average of the first four exams. We add what I got on the fifth exam, and then we divide it by 5, because now we're averaging five exams. And I said that I need to get in an 88 in the class. And now we solve for x. Let me make some space here. So, 5 times 88 is, let's see. 5 times 80 is 400, so it's 440. 440 equals 4 times 84, we just saw that, is 320 plus 16 is 336. 336 plus x is equal to 440. Well, it turns out if you subtract 336 from both sides, you get x is equal to 104. So unless you have a exam that has some bonus problems on it, it's probably impossible for you to get ah an 88 average in the class after just the next exam." + }, + { + "Q": "at 1:27, what in the world is he doing??? I mean, i don't get it. Where did the .25 come from??", + "A": "Did you study long division yet? 4 s into 29 go 7 times with 1 left over. So we do not throw the 1 away, but instead put a decimal point and carry on. We always put a 0 after whatever digit we are left with, so the 1, becomes 10, so how many 4 s go into 10.....? two 4 s go into 10 with 2 left over. Once again put a 0 after the 2 and we get 20, 4 s into 20 go 5 times with none left over. Our final answer is 7.25. :)", + "video_name": "9VZsMY15xeU", + "timestamps": [ + 87 + ], + "3min_transcript": "Welcome to the presentation on averages. Averages is probably a concept that you've already used before, maybe not in a mathematical way. But people will talk in terms of, the average voter wants a politician to do this, or the average student in a class wants to get out early. So you're probably already familiar with the concept of an average. And you probably already intuitively knew that an average is just a number that represents the different values that a group could have. But it can represent that as one number as opposed to giving all the different values. And let's give a couple of examples of how to compute an average, and you might already know how to do this. So let's say I had the numbers 1, 3, 5, and 20. And I asked you, what is the average of these four numbers? Well, what we do is, we literally just add up the numbers. And then divide by the number of numbers we have. So we say 1 plus 3 is 4. 4 plus 5 is 9. 9 plus 20 is 29. And we had 4 numbers; one, two, three, four. So 4 goes into 29. And it goes, 7, 7, 28. And then we have 10, I didn't have to do that decimal there, oh well. 2, 8, 25. So 4 goes into 29 7.25 times. So the average of these four numbers is equal to 7.25. And that might make sense to you because 7.25 is someplace in between these numbers. And we can kind of view this, 7.25, as one way to represent these four numbers without having to list these Like the mode. You'll also the mean, which we'll talk about later, is actually the same thing as the average. But the average is just one number that you can use to represent a set of numbers. So let's do some problems which I think are going to be close to your heart. Let's say on the first four tests of an exam, I got a-- let's see, I got an 80, an 81. An 87, and an 88. What's my average in the class so far? Well, all I have to do is add up these four numbers. So I say, 80 plus 81 plus 87 plus 88. Well, zero plus 1 is 1." + }, + { + "Q": "The ratio of chocolate bars to taffy pieces in a candy shop is 7:3, the total amount of candy is 3000. Then the candy store buys 32 more chocolate for there new daily order. How meny chocolate bars and taffy pieces are there?", + "A": "3000*Candy=3*700*Chocolate+3*300*Taffy=2100*Chocolate+900*Taffy 2100+32=2132 the answer is 2132 chocolate bars and 900 taffy pieces", + "video_name": "MaMk6-f3T9k", + "timestamps": [ + 423 + ], + "3min_transcript": "Table 2-- 11 to 4 and then 12 to 5. Here, it's just incrementing by 1, but the ratios are not the same. 11 to 4 is not the same thing as 12 to 5. So we're not going to be able to-- this right over here is not a legitimate table. Table 3-- so 1 to 1. Then when you double the distance, we double the time. When you triple the distance from 1, you didn't triple the time. So table 3 doesn't seem to make sense, either. Table 4-- so 14 to 10. So that's the same thing as-- let's see, that's the same ratio as, if we were to divide by 2, as 7 to 5 ratio. If we divide both of these by 3, this is also a 7 to 5 ratio. And if you divide both of these by 7, this is also a 7 to 5 So table 4 seems like a completely reasonable scenario." + }, + { + "Q": "At 14:50, you do what you've done many times before and write matrix multiplication as the sum of the each product of the column in the matrix and the row (1 scalar value) in the vector. However, this multiplication is actually defined in the opposite way (first row * first column = entry1,1) etc. Why are you using a different definition for matrix vector multiplication?", + "A": "The two definitions of matrix multiplication are actually equivalent. The definition Sal uses is, however, generally more compact and also carries useful geometric intuition (although not necessarily easier for numerical calculations). From your typical definition, you would find that the first element of the product is v1,1*x1 + v2,1*x2 + ... + vn,1*xn. However, by taking the vector sum in the definition Sal uses, you also obtain the same result.", + "video_name": "ondmopWLiEg", + "timestamps": [ + 890 + ], + "3min_transcript": "So it just equals-- I'll write a little bit lower. That equals c a1 times this column vector, times v1. Plus c a2 times v2 times this guy, all the way to plus c a n times vn. And if you just factor this c out, once again, scalar multiplication times vectors exhibits the distributive property. I believe I've done a video on that, but it's very easy to prove. So this will be equal to c times-- I'll just stay in one color right now-- a1 v1 plus a2 v2 plus all the way to a n vn. And what is this thing equal to? Well that's just our matrix A times our vector-- or our Maybe I'm overloading the letter A. My matrix uppercase A times my vector lowercase a. Where the lowercase a is just this thing right here, a1, a2 This thing up here was the same thing as that. So I just showed you that if I take my matrix and multiply it times some vector that was multiplied by a scalar first, that's equivalent to first multiplying the matrix times a vector and then multiplying by the scalar. So we've shown you that matrix times vector products or matrix vector products satisfied this condition of linear transformations and this condition. So the big takeaway right here is matrix multiplication. And this is a important takeaway. Matrix multiplication or matrix products with vectors And this is a bit of a side note. In the next video I'm going to show you that any linear transformation-- this is incredibly powerful-- can be represented by a matrix product or by-- any transformation on any vector can be equivalently, I guess, written as a product of that vector with a matrix. Has huge repercussions and you know, just as a side note, kind of tying this back to your everyday life. You have your Xbox, your Sony Playstation and you know, you have these 3D graphic programs where you're running around and shooting at things. And the way that the software renders those programs where you can see things from every different angle, you have a cube then if you kind of move this way a little bit, the cube will look more like this and it gets rotated, and you move up and down, these are all transformations of matrices. And we'll do this in more detail." + }, + { + "Q": "Where does the (x-a)^2 come from? [at 1:35]", + "A": "We put it in ourselves to create an equation that could be used to solve for the question mark.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 95 + ], + "3min_transcript": "In this video, I'm going to show you a technique called completing the square. And what's neat about this is that this will work for any quadratic equation, and it's actually the basis for the And in the next video or the video after that I'll prove the quadratic formula using completing the square. But before we do that, we need to understand even what it's all about. And it really just builds off of what we did in the last video, where we solved quadratics using perfect squares. So let's say I have the quadratic equation x squared minus 4x is equal to 5. And I put this big space here for a reason. In the last video, we saw that these can be pretty straightforward to solve if the left-hand side is a perfect square. You see, completing the square is all about making the quadratic equation into a perfect square, engineering So how can we do that? Well, in order for this left-hand side to be a perfect square, there has to be some number here. There has to be some number here that if I have my number squared I get that number, and then if I have two times my number I get negative 4. Remember that, and I think it'll become clear with a few examples. I want x squared minus 4x plus something to be equal to x minus a squared. We don't know what a is just yet, but we know a couple of things. When I square things-- so this is going to be x squared minus 2a plus a squared. So if you look at this pattern right here, that has to be-- sorry, x squared minus 2ax-- this right here has to be 2ax. And this right here would have to be a squared. to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2, and if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2-- same idea there-- and then you square it. You square a, you get positive 4. So we add positive 4 here. Add a 4. Now, from the very first equation we ever did, you should know that you can never do something to just one side You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4" + }, + { + "Q": "At 9:11 why is +9/4 when moved to the right side of the equation not a -9/4?", + "A": "Well, as you probably know from the video, this method is call completing the square . In this step, Sal took the coefficient in front of the x, which is -3. To find the constant we must add to the left-hand side, we do ((coefficient/2)^2), which in this case gives ((-3/2)^2) = 9/4. When adding 9/4 to the left-hand side, in order to keep the equality between both sides of the equal sign, we must thus add 9/4 to the right-hand side of the equation.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 551 + ], + "3min_transcript": "we can get rid of that. We can always turn this into a 1 or a negative 1 if we divide by the right number. But notice, by doing that we got this crazy 4/5 here. So this is super hard to do just using factoring. You'd have to say, what two numbers when I take the product is equal to negative 4/5? It's a fraction and when I take their sum, is equal to negative 3? This is a hard problem with factoring. This is hard using factoring. So, the best thing to do is to use completing the square. So let's think a little bit about how we can turn this into a perfect square. What I like to do-- and you'll see this done some ways and I'll show you both ways because you'll see teachers do it both ways-- I like to get the 4/5 on the other side. So let's add 4/5 to both sides of this equation. You don't have to do it this way, but I like to get the 4/5 out of the way. sides of this equation? The left-hand hand side of the equation just becomes x squared minus 3x, no 4/5 there. I'm going to leave a little bit of space. And that's going to be equal to 4/5. Now, just like the last problem, we want to turn this left-hand side into the perfect square of a binomial. How do we do that? Well, we say, well, what number times 2 is equal to negative 3? So some number times 2 is negative 3. Or we essentially just take negative 3 and divide it by 2, which is negative 3/2. And then we square negative 3/2. So in the example, we'll say a is negative 3/2. And if we square negative 3/2, what do we get? We get positive 9/4. I just took half of this coefficient, squared it, got positive 9/4. The whole purpose of doing that is to turn this left-hand side into a perfect square. got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it." + }, + { + "Q": "at 12:30 wouldn't the answer be 4.55?", + "A": "These are the two values we have to get the answer for x: 3/2 and \u00e2\u0088\u009a(61/20) and we know that x can equal 3/2 plus or minus the value of the square root. so, 3/2 +- \u00e2\u0088\u009a(61/20). We can just get the values of both of the terms and add/subtract appropriately. 3/2 is 1.5 and \u00e2\u0088\u009a(61/20) is 1.74642... So we can solve thusly x = 1.5 + 1.74642 x = 3.246 or x = 1.5 - 1.74642 x = -0.246", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 750 + ], + "3min_transcript": "Crazy number. Now this, at least on the left hand side, is a perfect square. This is the same thing as x minus 3/2 squared. And it was by design. Negative 3/2 times negative 3/2 is positive 9/4. Negative 3/2 plus negative 3/2 is equal to negative 3. So this squared is equal to 61/20. We can take the square root of both sides and we get x minus 3/2 is equal to the positive or the negative square root of 61/20. And now, we can add 3/2 to both sides of this equation and you get x is equal to positive 3/2 plus or minus the And this is a crazy number and it's hopefully obvious you would not have been able to-- at least I would not have been able to-- get to this number just by factoring. And if you want their actual values, you can get your calculator out. And then let me clear all of this. And 3/2-- let's do the plus version first. So we want to do 3 divided by 2 plus the second square root. We want to pick that little yellow square root. So the square root of 61 divided by 20, which is 3.24. This crazy 3.2464, I'll just write 3.246. So this is approximately equal to 3.246, and that was just Let's do the subtraction version. So we can actually put our entry-- if you do second and then entry, that we want that little yellow entry, that's why I pressed the second button. So I press enter, it puts in what we just put, we can just change the positive or the addition to a subtraction and you get negative 0.246. So you get negative 0.246. And you can actually verify that these satisfy our original equation. Our original equation was up here. Let me just verify for one of them. So the second answer on your graphing calculator is the last answer you use. So if you use a variable answer, that's this number right here. So if I have my answer squared-- I'm using answer represents negative 0.24." + }, + { + "Q": "At 4:12, why is the square root of 9 equal to +/- 3?", + "A": "because, nomatter the numbers is positive or nagetive if it be squared, it will become positive. so square +3 =9 square -3 =9 =====\u00e3\u0080\u008b square root of 9 =+3/-3", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 252 + ], + "3min_transcript": "to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2, and if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2-- same idea there-- and then you square it. You square a, you get positive 4. So we add positive 4 here. Add a 4. Now, from the very first equation we ever did, you should know that you can never do something to just one side You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4 It's going to be equal to 5 plus 4. We added 4 on the left-hand side because we wanted this to be a perfect square. But if you add something to the left-hand side, you've got to add it to the right-hand side. And now, we've gotten ourselves to a problem that's just like the problems we did in the last video. What is this left-hand side? Let me rewrite the whole thing. We have x squared minus 4x plus 4 is equal to 9 now. All we did is add 4 to both sides of the equation. But we added 4 on purpose so that this left-hand side becomes a perfect square. What number when I multiply it by itself is equal to 4 and when I add it to itself I'm equal to negative 2? Well, we already answered that question. It's negative 2. So we get x minus 2 times x minus 2 is equal to 9. Or we could have skipped this step and written x minus 2 squared is equal to 9. minus 2 is equal to plus or minus 3. Add 2 to both sides, you get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without completing the square. We could've started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, then their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0." + }, + { + "Q": "at time 10:45, can anyone explain to me how we get the +/- square root of 61/20. specifically, the reason why we have +/- the square root. what rule is that?", + "A": "I don t know if this will help or not but I ll try to explain. So basically you re wondering why should there be a positive + and negative - square root, right? Think about any squared number really does have two possible square roots, the positive and the negative ones. For instance \u00c2\u00b1\u00e2\u0088\u009a9= -3 or 3 Because if we square (-3)^2=9 (\u00e2\u0086\u0092Notice that this different from -3^2 which is equal to =-9) and 3^3=9", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 645 + ], + "3min_transcript": "got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it. Crazy number. Now this, at least on the left hand side, is a perfect square. This is the same thing as x minus 3/2 squared. And it was by design. Negative 3/2 times negative 3/2 is positive 9/4. Negative 3/2 plus negative 3/2 is equal to negative 3. So this squared is equal to 61/20. We can take the square root of both sides and we get x minus 3/2 is equal to the positive or the negative square root of 61/20. And now, we can add 3/2 to both sides of this equation and you get x is equal to positive 3/2 plus or minus the And this is a crazy number and it's hopefully obvious you would not have been able to-- at least I would not have been able to-- get to this number just by factoring. And if you want their actual values, you can get your calculator out. And then let me clear all of this. And 3/2-- let's do the plus version first. So we want to do 3 divided by 2 plus the second square root. We want to pick that little yellow square root. So the square root of 61 divided by 20, which is 3.24. This crazy 3.2464, I'll just write 3.246. So this is approximately equal to 3.246, and that was just" + }, + { + "Q": "At 4:05, why can't you solve the equation as\n(x-2) = 9 (x-2) = 9\nx = 11 x = 11\nWhy do we instead need to take the square root of (x-2)^2 and of 9 to get\n(x-2) = +-3?\n\nIsn't the former how we've been solving these equations up until this point?", + "A": "The ones that you solved before always had the other side was 0 which would allow you to do this. If you have (x-3)(x+4) = 12, you should not have said x - 3 = 12 or x + 4 = 12 which would give you wrong answers. You would have to multiply to get a trinomial, subtract 12, and refactor if you could or use quadratic formula to see if there are factors. Also, if you put your attempted answer back in as a check (x-2)^2 = 9, you would have (11-2)^2 = 9 or 81 = 9 which is an impossibility.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 245 + ], + "3min_transcript": "to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2, and if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2-- same idea there-- and then you square it. You square a, you get positive 4. So we add positive 4 here. Add a 4. Now, from the very first equation we ever did, you should know that you can never do something to just one side You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4 It's going to be equal to 5 plus 4. We added 4 on the left-hand side because we wanted this to be a perfect square. But if you add something to the left-hand side, you've got to add it to the right-hand side. And now, we've gotten ourselves to a problem that's just like the problems we did in the last video. What is this left-hand side? Let me rewrite the whole thing. We have x squared minus 4x plus 4 is equal to 9 now. All we did is add 4 to both sides of the equation. But we added 4 on purpose so that this left-hand side becomes a perfect square. What number when I multiply it by itself is equal to 4 and when I add it to itself I'm equal to negative 2? Well, we already answered that question. It's negative 2. So we get x minus 2 times x minus 2 is equal to 9. Or we could have skipped this step and written x minus 2 squared is equal to 9. minus 2 is equal to plus or minus 3. Add 2 to both sides, you get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without completing the square. We could've started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, then their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0." + }, + { + "Q": "At 10:52, there is a fraction underneath a radical, the square root of 61/20. While Sal uses his calculator to solve it, I want to know how you would simplify a fraction under a radical like that without a calculator. My math teacher takes off points on tests for leaving answers as fractions under radicals, but it's not quite the same as simplifying a radical that is just a whole number, so I'm confused on how to do this. Also, is this process similar if you take the cube of a fraction instead of the square?", + "A": "First of all, 61 is a prime number so in a radical it remains. With 20 you can factor it into 4 and 5, so it can be written as \\ \u00e2\u0088\u009a20 = 2\u00e2\u0088\u009a5 So now we have \u00e2\u0088\u009a61 / 2\u00e2\u0088\u009a5, which is an OK answer as long as radicals in the denominator are permitted. To get rid of the radical in the denominator, multiply \u00e2\u0088\u009a61 / 2\u00e2\u0088\u009a5 by \u00e2\u0088\u009a5/\u00e2\u0088\u009a5 and we have: (\u00e2\u0088\u009a5)(\u00e2\u0088\u009a61)/(2)(\u00e2\u0088\u009a5)(\u00e2\u0088\u009a5), which can be simplified to \u00e2\u0088\u009a(61*5)/10 = \u00e2\u0088\u009a305/10. So now you only have a radical in the numerator.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 652 + ], + "3min_transcript": "got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it. Crazy number. Now this, at least on the left hand side, is a perfect square. This is the same thing as x minus 3/2 squared. And it was by design. Negative 3/2 times negative 3/2 is positive 9/4. Negative 3/2 plus negative 3/2 is equal to negative 3. So this squared is equal to 61/20. We can take the square root of both sides and we get x minus 3/2 is equal to the positive or the negative square root of 61/20. And now, we can add 3/2 to both sides of this equation and you get x is equal to positive 3/2 plus or minus the And this is a crazy number and it's hopefully obvious you would not have been able to-- at least I would not have been able to-- get to this number just by factoring. And if you want their actual values, you can get your calculator out. And then let me clear all of this. And 3/2-- let's do the plus version first. So we want to do 3 divided by 2 plus the second square root. We want to pick that little yellow square root. So the square root of 61 divided by 20, which is 3.24. This crazy 3.2464, I'll just write 3.246. So this is approximately equal to 3.246, and that was just" + }, + { + "Q": "at 5:48 why do they equal zero", + "A": "That s simply the problem statement; at 5:37 Sal starts a new quadratic equation as another example.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 348 + ], + "3min_transcript": "minus 2 is equal to plus or minus 3. Add 2 to both sides, you get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without completing the square. We could've started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, then their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0. negative 1. And in this case, this actually probably would have been a faster way to do the problem. But the neat thing about the completing the square is it will always work. It'll always work no matter what the coefficients are or no matter how crazy the problem is. And let me prove it to you. Let's do one that traditionally would have been a pretty painful problem if we just tried to do it by factoring, especially if we did it using grouping or something like that. Let's say we had 10x squared minus 30x minus 8 is equal to 0. Now, right from the get-go, you could say, hey look, we could maybe divide both sides by 2. That does simplify a little bit. Let's divide both sides by 2. So if you divide everything by 2, what do you get? But once again, now we have this crazy 5 in front of this coefficent and we would have to solve it by grouping which is a reasonably painful process. But we can now go straight to completing the square, and to do that I'm now going to divide by 5 to get a 1 leading coefficient here. And you're going to see why this is different than what we've traditionally done. So if I divide this whole thing by 5, I could have just divided by 10 from the get-go but I wanted to go to this the step first just to show you that this really didn't give us much. Let's divide everything by 5. So if you divide everything by 5, you get x squared minus 3x minus 4/5 is equal to 0. So, you might say, hey, why did we ever do that factoring" + }, + { + "Q": "At 4:10, why does it become +/- 3, instead of just 3?", + "A": "the sqare root of 9 = +/-3 because in algebra, whenever you square a positive OR negative number, the answer is always positive.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 250 + ], + "3min_transcript": "to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2, and if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2-- same idea there-- and then you square it. You square a, you get positive 4. So we add positive 4 here. Add a 4. Now, from the very first equation we ever did, you should know that you can never do something to just one side You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4 It's going to be equal to 5 plus 4. We added 4 on the left-hand side because we wanted this to be a perfect square. But if you add something to the left-hand side, you've got to add it to the right-hand side. And now, we've gotten ourselves to a problem that's just like the problems we did in the last video. What is this left-hand side? Let me rewrite the whole thing. We have x squared minus 4x plus 4 is equal to 9 now. All we did is add 4 to both sides of the equation. But we added 4 on purpose so that this left-hand side becomes a perfect square. What number when I multiply it by itself is equal to 4 and when I add it to itself I'm equal to negative 2? Well, we already answered that question. It's negative 2. So we get x minus 2 times x minus 2 is equal to 9. Or we could have skipped this step and written x minus 2 squared is equal to 9. minus 2 is equal to plus or minus 3. Add 2 to both sides, you get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without completing the square. We could've started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, then their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0." + }, + { + "Q": "At 0:35 why does it have to be greater or equal to? Why not less than or equal to or equal to.", + "A": "Because you can t take the square root of anything less than 0.", + "video_name": "4h54s7BBPpA", + "timestamps": [ + 35 + ], + "3min_transcript": "Find the domain of f of x is equal to the principal square root of 2x minus 8. So the domain of a function is just the set of all of the possible valid inputs into the function, or all of the possible values for which the function is defined. And when we look at how the function is defined, right over here, as the square root, the principal square root of 2x minus 8, it's only going to be defined when it's taking the principal square root of a non-negative number. And so 2x minus 8, it's only going to be defined when 2x minus 8 is greater than or equal to 0. It can be 0, because then you just take the square root of 0 is 0. It can be positive. But if this was negative, then all of a sudden, this principle square root function, which we're assuming is just the plain vanilla one for real numbers, it would not be defined. So this function definition is only defined when 2x minus 8 is greater than or equal to 0. And then we could say if 2x minus 8 has to be greater than or equal to 0, what it's saying about what x has to be. So if we add 8 to both sides of this inequality, you get-- so let me just add 8 to both sides. These 8's cancel out. You get 2x is greater than or equal to 8. 0 plus 8 is 8. And then you divide both sides by 2. Since 2 is a positive number, you don't have to swap the inequality. So you divide both sides by 2. And you get x needs to be greater than or equal to 4. So the domain here is the set of all real numbers that are greater than or equal to 4. x has to be greater than or equal to 4. Or another way of saying it is this function is defined when x is greater than or equal to 4. And we're done." + }, + { + "Q": "i do not understand from 0:22 to 0:56", + "A": "Sal realised that he ll be unable to find the square root of the quantity within the radical if it s negative. So he sets up the inequality to show that that quantity (2x-8) has to be either zero or positive.", + "video_name": "4h54s7BBPpA", + "timestamps": [ + 22, + 56 + ], + "3min_transcript": "Find the domain of f of x is equal to the principal square root of 2x minus 8. So the domain of a function is just the set of all of the possible valid inputs into the function, or all of the possible values for which the function is defined. And when we look at how the function is defined, right over here, as the square root, the principal square root of 2x minus 8, it's only going to be defined when it's taking the principal square root of a non-negative number. And so 2x minus 8, it's only going to be defined when 2x minus 8 is greater than or equal to 0. It can be 0, because then you just take the square root of 0 is 0. It can be positive. But if this was negative, then all of a sudden, this principle square root function, which we're assuming is just the plain vanilla one for real numbers, it would not be defined. So this function definition is only defined when 2x minus 8 is greater than or equal to 0. And then we could say if 2x minus 8 has to be greater than or equal to 0, what it's saying about what x has to be. So if we add 8 to both sides of this inequality, you get-- so let me just add 8 to both sides. These 8's cancel out. You get 2x is greater than or equal to 8. 0 plus 8 is 8. And then you divide both sides by 2. Since 2 is a positive number, you don't have to swap the inequality. So you divide both sides by 2. And you get x needs to be greater than or equal to 4. So the domain here is the set of all real numbers that are greater than or equal to 4. x has to be greater than or equal to 4. Or another way of saying it is this function is defined when x is greater than or equal to 4. And we're done." + }, + { + "Q": "at 1:45 i still do not really understand why after adding 64 you must also subtract 64 from 9. Could someone explain I'm a bit confused.", + "A": "When we make changes to an expression, we need to make sure that the new version is equivalent to the prior version. If you just add 64, you have changed the value of the original expression. The new version would not longer be equal to the original. The identify property of addition says we can add zero to any expression and it doesn t change the value of the expression. Thus, by using: +64 and -64, Sal is adding zero to the expression (+64 - 64 =0). Hope this helps.", + "video_name": "sh-MP-dVhD4", + "timestamps": [ + 105 + ], + "3min_transcript": "- [Voiceover] Let's see if we can take this quadratic expression here, X squared plus 16 X plus nine and write it in this form. You might be saying, hey Sal, why do I even need to worry about this? And one, it is just good algebraic practice to be able to manipulate things, but as we'll see in the future, what we're about to do is called completing the square. It's a really valuable technique for solving quadratics and it's actually the basis for the proof of the quadratic formula which you'll learn in the future. So it's actually a pretty interesting technique. So how do we write this in this form? Well one way to think about is if we expanded this X plus A squared, we know if we square X plus A it would be X squared plus two A X plus A squared, and then you still have that plus B, right over there. So one way to think about it is, let's take this expression, this X squared plus 16 X plus nine, and I'm just gonna write it with a few spaces in it. X squared plus 16 X And so, if we say alright, we have an X squared here. We have an X squared here. If we say that two A X is the same thing as that, then what's A going to be? So this is two A times X. Well, that means two A is 16 or that A is equal to 8. And so if I want to have an A squared over here, well if A is eight, I would add an eight squared which would be a 64. Well I can't just add numbers willy nilly to an expression without changing the value of an expression so if don't want to change the value of the expression, I still need to subtract 64. So notice, all that I have done now, is I just took our original expression and I added 64 and I subtracted 64, so I have not changed the value of that expression. But what was valuable about me doing that, is now this first part of the expression, it fits the pattern of a perfect square quadratic right over here. We have X squared plus two A X, where A is 8, plus A squared, 64. Once again, how did I get 64? I took half of the 16 and I squared it to get to the 64. And so the stuff that I just squared off, this is going to be X plus eight squared. X plus eight, squared. Once again I know that because A is eight, A is eight, so this is X plus eight squared, and then all of this business on the right hand side. What is nine minus 64? Well 64 minus nine is 55, so this is going to be negative 55. So minus 55, and we're done. We've written this expression in this form, and what's also called completing the square." + }, + { + "Q": "In 0:44 , how did Khan get 2ax from (x+a)^2? Can someone explain how and why it works?", + "A": "(x+a)\u00c2\u00b2=(x+a)(x+a). When you multiply it out, you get x\u00c2\u00b2+ax+ax+a\u00c2\u00b2. Simplify it you get x\u00c2\u00b2+2ax+a\u00c2\u00b2", + "video_name": "sh-MP-dVhD4", + "timestamps": [ + 44 + ], + "3min_transcript": "- [Voiceover] Let's see if we can take this quadratic expression here, X squared plus 16 X plus nine and write it in this form. You might be saying, hey Sal, why do I even need to worry about this? And one, it is just good algebraic practice to be able to manipulate things, but as we'll see in the future, what we're about to do is called completing the square. It's a really valuable technique for solving quadratics and it's actually the basis for the proof of the quadratic formula which you'll learn in the future. So it's actually a pretty interesting technique. So how do we write this in this form? Well one way to think about is if we expanded this X plus A squared, we know if we square X plus A it would be X squared plus two A X plus A squared, and then you still have that plus B, right over there. So one way to think about it is, let's take this expression, this X squared plus 16 X plus nine, and I'm just gonna write it with a few spaces in it. X squared plus 16 X And so, if we say alright, we have an X squared here. We have an X squared here. If we say that two A X is the same thing as that, then what's A going to be? So this is two A times X. Well, that means two A is 16 or that A is equal to 8. And so if I want to have an A squared over here, well if A is eight, I would add an eight squared which would be a 64. Well I can't just add numbers willy nilly to an expression without changing the value of an expression so if don't want to change the value of the expression, I still need to subtract 64. So notice, all that I have done now, is I just took our original expression and I added 64 and I subtracted 64, so I have not changed the value of that expression. But what was valuable about me doing that, is now this first part of the expression, it fits the pattern of a perfect square quadratic right over here. We have X squared plus two A X, where A is 8, plus A squared, 64. Once again, how did I get 64? I took half of the 16 and I squared it to get to the 64. And so the stuff that I just squared off, this is going to be X plus eight squared. X plus eight, squared. Once again I know that because A is eight, A is eight, so this is X plus eight squared, and then all of this business on the right hand side. What is nine minus 64? Well 64 minus nine is 55, so this is going to be negative 55. So minus 55, and we're done. We've written this expression in this form, and what's also called completing the square." + }, + { + "Q": "At 3:18 of the video, how does Sal get that x= \u00e2\u0080\u00934 or x= 3 from the \"(x+4)(x-3)=0?\"", + "A": "When we get down to the factored formula (x+4)(x-3)=0, then we use the zero property principle that the others have described to finish the solution. The next step is to set each factor equal to zero by itself and solve. (x+4) = 0 x + 4 -4 = 0 -4 x = -4 First solution for x (x-3) = 0 x-3 +3 = +3 x = 3 Second solution for x", + "video_name": "swFohliPgmQ", + "timestamps": [ + 198 + ], + "3min_transcript": "So this is 1, 2, 3, 4. y-intercept is there and has a slope of 1 so it looks something like this. So when we're looking for the solutions, we're looking for the points that satisfy both. The points that satisfy both are the points that sit on both. So it's that point-- let me do it in green-- It's this point and it's this point right over here. So how do we actually figure that out? Well, the easiest way is to-- well, sometimes the easiest way is to substitute one of these constraints into the other constraint. And since they've already solved for y here, we can substitute y in the blue equation with x plus 1 with this constraint right over here. So instead of saying x squared plus y squared equals 25, we can say x squared plus, and instead of writing a y, we're adding the constraint that y must be x plus 1. So x squared plus x plus 1 squared must be equal to 25. So we get x squared plus-- now we square this. We'll get-- let me write in magenta-- we'll get x squared plus 2x plus 1, and that must be equal to 25. We have 2x squared-- now, I'm just combining these two terms-- 2x squared plus 2x plus 1 is equal to 25. Now, we could just use the quadratic formula to find the-- well, we have to be careful. We have to set this equal to 0, and then use the quadratic formula. So let's subtract 25 from both sides, and you could get 2x squared plus 2x minus 24 is equal to 0. And actually, let's-- just to simplify this-- let's divide both sides by 2, and you get x squared plus x minus 12 is equal to 0. And actually, we don't even have to use the quadratic formula, we can factor this right over here. What are two numbers that when we take their product Well, positive 4 and negative 3 would do the trick. So we have x plus 4 times x minus 3 is equal to 0. So x could be equal to-- well, if x plus 4 is 0 then that would make this whole thing true. So x could be equal to negative 4 or x could be equal to positive 3. So this right over here is a situation where x is negative 4. This right over here is a situation where x is 3. So we're almost done, we just have to find the corresponding y's. And for that, we can just resort to the simplest equation right over here, y is x plus 1. So in this situation when x is negative 4, y is going to be that plus 1. So y is going to be negative 3. This is the point negative 4 comma negative 3. Likewise, when x is 3, y is going to be equal to 4." + }, + { + "Q": "At 0:55, Sal draws a circle, but from an equation like the one in the video, is there any way to figure out exactly how large the circle is?", + "A": "The constant term is 25, so the radius of the circle is sqrt 25 = 5.", + "video_name": "swFohliPgmQ", + "timestamps": [ + 55 + ], + "3min_transcript": "What are the solutions to the system of equations y is equal to x plus 1, and x squared plus y squared is equal to 25? So let's first just visualize what we're trying to do. So let me try to roughly graph these two equations. So my y-axis, this is my x-axis. This right over here, x squared plus y squared is equal to 25, that's going to be a circle centered at 0 with radius 5. You don't have to know that to solve this problem, but it helps to visualize it. So if this is 5, this is 5, 5, 5. This right over here is negative 5. This right over here is negative 5. This equation would be represented by this set of points, or this is a set of points that satisfy this equation. So let me-- there you go. Trying to draw it as close to a perfect circle as I can. And then y equals x plus 1 is a line So this is 1, 2, 3, 4. y-intercept is there and has a slope of 1 so it looks something like this. So when we're looking for the solutions, we're looking for the points that satisfy both. The points that satisfy both are the points that sit on both. So it's that point-- let me do it in green-- It's this point and it's this point right over here. So how do we actually figure that out? Well, the easiest way is to-- well, sometimes the easiest way is to substitute one of these constraints into the other constraint. And since they've already solved for y here, we can substitute y in the blue equation with x plus 1 with this constraint right over here. So instead of saying x squared plus y squared equals 25, we can say x squared plus, and instead of writing a y, we're adding the constraint that y must be x plus 1. So x squared plus x plus 1 squared must be equal to 25. So we get x squared plus-- now we square this. We'll get-- let me write in magenta-- we'll get x squared plus 2x plus 1, and that must be equal to 25. We have 2x squared-- now, I'm just combining these two terms-- 2x squared plus 2x plus 1 is equal to 25. Now, we could just use the quadratic formula to find the-- well, we have to be careful. We have to set this equal to 0, and then use the quadratic formula. So let's subtract 25 from both sides, and you could get 2x squared plus 2x minus 24 is equal to 0. And actually, let's-- just to simplify this-- let's divide both sides by 2, and you get x squared plus x minus 12 is equal to 0. And actually, we don't even have to use the quadratic formula, we can factor this right over here. What are two numbers that when we take their product" + }, + { + "Q": "At 3:27 Sal is telling that angle AEB and angle ECB are congruent because they are corresponding angles but the line l and line m are not parallel. If they are corresponding angles, they have to be parallel. But how he did this?", + "A": "Corresponding angles are actually any two angles in that position. They are only congruent if the lines are parallel, and vice versa. Also, if two corresponding angles are labeled as congruent, the lines are parallel, end of story", + "video_name": "6dIMIBO_2mc", + "timestamps": [ + 207 + ], + "3min_transcript": "let's assume that it is. So let's assume that we do have a quadrilateral where this is indeed the case. Where AEB, this angle right over here, is congruent to angle ECB. We're just going to assume that right from the get-go. Now let's try to visualize this whole thing a little bit differently. DB is a segment, but it's a segment of a larger line. So we could keep extending it off like this. And let's call that larger line, line l. Let me draw it like this. So this right over here is line l. DB is a subset of line l. And CB, which is a segment, is also a subset of a line. And we could call the line, if we were to keep extending it, let's call that line m. So I'll draw it like this. So line m. And then we also have the segment AC. And AC, once again, is a segment but we can view it as a subset of a larger line. And we'll call that larger line, line n. And so let me draw a line n like this. And we see that it intersects both line l and line m. So let me draw it like this, so it looks something like that. That is line n. And we've assumed that angle AEB is congruent to angle ECB. Well point E is just where n and l intersect. So this right over here is point E. And point C is where lines n and m intersect. So this is point C. So this assumption right over here, that we're assuming from the get-go, is saying that this angle AEB-- I'll ECB is congruent to that angle. These angles are these angles right over here. Now what does that tell us about lines m and l? Well the way we've set it up, we have two lines, lines l and m. Line n is a transversal. And now we have two corresponding angles are congruent. We assumed that from the get-go that we could find two quadrilateral, where these two corresponding angles are congruent. But if you have two corresponding angles congruent like this, that means that these two lines must be parallel. So this tells us that line l is parallel to line m. Line l is parallel to line m. Which means that they will never intersect. But we have a contradiction showing up. And I'll let you think about what that is for a second." + }, + { + "Q": "At 7:27, shouldn't it be 1 /\u00e2\u0088\u009a2, rather than \u00e2\u0088\u009a2/2?", + "A": "The second fraction \u00e2\u0088\u009a2/2 is the same as the first fraction 1 /\u00e2\u0088\u009a2, just with a rationalized denominator. The 2 forms are equivalent.", + "video_name": "mSVrqKZDRF4", + "timestamps": [ + 447 + ], + "3min_transcript": "And the derivative of y with respect to x? Well, we don't know what that is. So we're just going to leave that as times the derivative of y with respect to x. So let's just write this down over here. So we have is 2x plus the derivative of something squared, with respect to that something, is 2 times the something. In this case, the something is y, so 2 times y. And then times the derivative of y with respect to x. And this is all going to be equal to 0. Now that was interesting. Now we have an equation that has the derivative of a y with respect to x in it. And this is what we essentially want to solve for. This is the slope of the tangent line at any point. So all we have to do at this point is solve for the derivative of y with respect to x. Solve this equation. And actually just so we can do this whole thing on the same page so we can see where we started, let me copy and paste this up here. This is where we left off. And let's continue there. So let's say let's subtract 2x from both sides. So we're left with 2y times the derivative of y, with respect to x, is equal to-- we're subtracting 2x from both sides-- so it's equal to negative 2x. And if we really want to solve for the derivative of y with respect to x, we can just divide both sides by 2y. And we're left with the derivative of y with respect to x. Let's scroll down a little bit. The derivative of y with respect to x is equal to, well the 2s cancel out. We we're left with negative x over y. So this is interesting. We didn't have to us explicitly define y But we got our derivative in terms of an x and a y. Not just only in terms of an x. But what does this mean? Well, if we wanted to find, let's say we wanted to find the derivative at this point right over here. Which, if you're familiar with the unit circle, so if this was a 45 degree angle, this would be the square root of 2 over 2 comma the square root of 2 over 2. What is the slope of the tangent line there? Well, we figured it out. It's going to be negative x over y. So the slope of the tangent line here is going to be equal to negative x. So negative square root of 2 over 2 over y. Over square root of 2 over 2, which is equal to negative 1. And that looks just about right." + }, + { + "Q": "What if we tried to find dy/dx by explicitly defining y in terms of x first, just like Sal did at 0:47. Would we have to find dy/dx separately for both the positive and negative roots? What would we do afterwards?", + "A": "Yes, since when we define it explicitly we get 2 equations (for a different one, we could get 4, there s an example of this, called a... fleur or something, it s 4 loops, obviously to define it as a correct function you have to limit it, since it loops back on itself). The two functions give you different slopes in terms of which one you re using. When you use y = +sqrt(1 - x^2) you re getting dy/dx for that function, the top half of our circle.", + "video_name": "mSVrqKZDRF4", + "timestamps": [ + 47 + ], + "3min_transcript": "So we've got the equation x squared plus y squared is equal to 1. I guess we could call it a relationship. And if we were to graph all of the points x and y that satisfied this relationship, we get a unit circle like this. And what I'm curious about in this video is how we can figure out the slope of the tangent line at any point of this unit circle. And what immediately might be jumping out in your brain is, well a circle defined this way, this isn't a function. It's not y explicitly defined as a function of x. For any x value we actually have two possible y's that satisfy this relationship right over here. So you might be tempted to maybe split this up into two separate functions of x. You could say y is equal to the positive square root of 1 minus x squared. And you could say y is equal to the negative square root of 1 minus x squared. Take the derivatives of each of these separately. or the derivative of the slope of the tangent line at any point. But what I want to do in this video is literally leverage the chain rule to take the derivative implicitly. So that I don't have to explicitly define y is a function of x either way. And the way we do that is literally just apply the derivative operator to both sides of this equation. And then apply what we know about the chain rule. Because we are not explicitly defining y as a function of x, and explicitly getting y is equal to f prime of x, they call this-- which is really just an application of the chain rule-- we call it implicit differentiation. And what I want you to keep in the back of your mind the entire time is that it's just an application of the chain rule. So let's apply the derivative operator to both sides of this. So it's the derivative with respect to x of x squared plus y squared, on the left hand And then that's going to be equal to the derivative with respect to x on the right hand side. I'm just doing the same exact thing to both sides of this equation. Now if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared, plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. This is going to be x squared, this is going to be y squared. And then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x. So we just get 0. Now this first term right over here, we have done many, many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here. It's going to be 2 times x to the first power." + }, + { + "Q": "at about 4:30, when Sal is taking the derivative of y^2, why is he using the chain rule and not the power rule?", + "A": "He s using both. The chain rule is required here because we re differentiating with respect to x and the expression includes y. We re able to do this by treating y as a function of x. This means that y\u00c2\u00b2 is actually a composition of two functions: the squaring function applies to whatever function would turn x into y. We don t necessarily know what that function is, but that s okay, it s why we re doing implicit differentiation instead of normal (explicit) differentiation.", + "video_name": "mSVrqKZDRF4", + "timestamps": [ + 270 + ], + "3min_transcript": "And then that's going to be equal to the derivative with respect to x on the right hand side. I'm just doing the same exact thing to both sides of this equation. Now if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared, plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. This is going to be x squared, this is going to be y squared. And then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x. So we just get 0. Now this first term right over here, we have done many, many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here. It's going to be 2 times x to the first power. Now what's interesting is what we're doing right over here. The derivative with respect to x of y squared. And the realization here is to just apply the chain rule. If we're taking the derivative with respect to x of this something, we just have to take the derivative-- let me make it clear-- we're just going to take the derivative of our something. The derivative of y squared-- that's what we're taking, you can kind of view that as a function-- with respect to y and then multiply that times the derivative of y with respect to x. We're assuming that y does change with respect to x. y is not some type of a constant that we're writing just an abstract terms. So we're taking the derivative of this whole thing with respect to y. Once again, just the chain rule. And then we're taking the derivative of y It might be a little bit clearer if you kind of thought of it as the derivative with respect to x of y, as a function of x. This might be, or y is a function of x squared, which is essentially another way of writing what we have here. This might be a little bit clearer in terms of the chain rule. The derivative of y is a function of x squared with respect to y of x. So the derivative of something squared with respect to that something, times the derivative of that something, with respect to x. This is just the chain rule. I want to say it over and over again. This is just the chain rule. So let's do that. What do we get on the right hand side over here? And I'll write it over here as well. This would be equal to the derivative of y squared with respect to y, is just going to be 2 times y." + }, + { + "Q": "at 3:14 can someone explain how d/dx (y^2) = dy^2/dy ? how does the denominator of dx turn into dy?", + "A": "First, notice that d/dx (y^2) has a different variable for the function, and the variable we re taking the derivative with respect to. This brings to mind chain rule, that is, if y is a function of x, like y(x). The chain rule says in this example says that d/dx (y(x)^2) = d/dy (y^2) * d/dx (y(x).", + "video_name": "mSVrqKZDRF4", + "timestamps": [ + 194 + ], + "3min_transcript": "or the derivative of the slope of the tangent line at any point. But what I want to do in this video is literally leverage the chain rule to take the derivative implicitly. So that I don't have to explicitly define y is a function of x either way. And the way we do that is literally just apply the derivative operator to both sides of this equation. And then apply what we know about the chain rule. Because we are not explicitly defining y as a function of x, and explicitly getting y is equal to f prime of x, they call this-- which is really just an application of the chain rule-- we call it implicit differentiation. And what I want you to keep in the back of your mind the entire time is that it's just an application of the chain rule. So let's apply the derivative operator to both sides of this. So it's the derivative with respect to x of x squared plus y squared, on the left hand And then that's going to be equal to the derivative with respect to x on the right hand side. I'm just doing the same exact thing to both sides of this equation. Now if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared, plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. This is going to be x squared, this is going to be y squared. And then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x. So we just get 0. Now this first term right over here, we have done many, many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here. It's going to be 2 times x to the first power. Now what's interesting is what we're doing right over here. The derivative with respect to x of y squared. And the realization here is to just apply the chain rule. If we're taking the derivative with respect to x of this something, we just have to take the derivative-- let me make it clear-- we're just going to take the derivative of our something. The derivative of y squared-- that's what we're taking, you can kind of view that as a function-- with respect to y and then multiply that times the derivative of y with respect to x. We're assuming that y does change with respect to x. y is not some type of a constant that we're writing just an abstract terms. So we're taking the derivative of this whole thing with respect to y. Once again, just the chain rule. And then we're taking the derivative of y" + }, + { + "Q": "At 2:20, you said \"the length of u1\" but you just showed u1. Does it matter which way you do it?", + "A": "Here the letter u is meant to stand for unit vector , so you don t need to write ||u1|| .", + "video_name": "rHonltF77zI", + "timestamps": [ + 140 + ], + "3min_transcript": "Let's say I have a set of linearly independent vectors, V1, V2, all the way to Vk, that are a basis for V. We've seen this many times before. That's nice. It's a basis, but we've learned over the last few videos it would be even better to have an orthonormal basis for V. What we're going to address in this video is, is there some way that we can construct an orthonormal basis for V, given that we already have this basis, which I'm assuming isn't orthonormal. It'll work whether or not it's orthonormal. It'll just generate another orthonormal basis. But can we somehow, just given any basis, generate an orthonormal basis for V, and then be able to benefit from all of the properties of having an orthonormal basis? Let's see if we can make some headway here. Let's say that I have a one-dimensional subspace. Let's call that V1, for it's a one-dimensional subspace. I'm going to say it's the span of just this vector V1. Now, or you could say the vector V1 is the basis for the subspace V1. If I had this simple case, how can I ensure that this is orthonormal? What I could do is I could define some vector, let's call it U1. Obviously, this is orthogonal to all of the other guys. There aren't any other guys here. [INAUDIBLE] vectors there, but there aren't any other members of this set, so it's orthogonal to everything else because there isn't anything else. And then to make its length equal one, you can take this vector and divide it by its length. the length of V1, then the length of U1 is going to be what? It's going to be the length of V1 over the length of V1 like this. This is just a constant right here. This is going to be 1 over the length of V1. That could be 5, times length of V1, which is just equal to 1. This one will have a length of 1. So just like that, if we have the set of just U1, we could say that just the set of U1 is an orthonormal basis for V1, for the subspace V1. Now, that was trivially easy." + }, + { + "Q": "what is the thing that sal did 2:43 called?", + "A": "I think improper fraction or mixed number/fraction.", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 163 + ], + "3min_transcript": "Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5. 6 minus 5 is 1. Bring down the 3. 5 goes into 13 two times. And you could have immediately said 5 goes into 63 twelve times, but this way, at least to me, it's a little bit more obvious. And then 2 times 5 is 12, and then we have sorry! 2 times 5 is 10. That tells you not to switch gears in the middle of a math problem. 2 times 5 is 10, and then you subtract, and you have a remainder of 3. So 63/5 is the same thing as 12 wholes and 3 left over, or 3/5 left over. And if you wanted to go back from this to that, just think: 12 is the same thing as 60 fifths, or 60/5." + }, + { + "Q": "at 2:18 what if you have a number that can't be divided normally", + "A": "At 2:18, Sal cancelled out a common factor. If there are no common factors to cancel, then you just multiply. For example: 7/4 * 11/5 = 77/20 = 6 17/20 Hope this helps.", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 138 + ], + "3min_transcript": "Multiply 1 and 3/4 times 7 and 1/5. Simplify your answer and write it as a mixed fraction. So the first thing we want to do is rewrite each of these mixed numbers as improper fractions. It's very difficult, or at least it's not easy for me, to directly multiply mixed numbers. One can do it, but it's much easier if you just make them improper fractions. So let's convert each of them. So 1 and 3/4 is equal to-- it's still going to be over 4, so you're still going to have the same denominator, but your numerator as an improper fraction is going to be 4 times 1 plus 3. And the reason why this makes sense is 1 is 4/4, or 1 is 4 times 1 fourths, right? 1 is the same thing as 4/4, and then you have three more fourths, so 4/4 plus 3/4 will give you 7/4. So that's the same thing as 1 and 3/4. Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5." + }, + { + "Q": "at 3:19 he said 2x5. = 12 and I am pretty sure it = 10 so I'm just confused", + "A": "Yeah, your right, he made a mistake, but if you watch for about two seconds longer, he actually corrects himself. \u00f0\u009f\u0091\u008dgood job on catching his mistake though.", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 199 + ], + "3min_transcript": "Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5. 6 minus 5 is 1. Bring down the 3. 5 goes into 13 two times. And you could have immediately said 5 goes into 63 twelve times, but this way, at least to me, it's a little bit more obvious. And then 2 times 5 is 12, and then we have sorry! 2 times 5 is 10. That tells you not to switch gears in the middle of a math problem. 2 times 5 is 10, and then you subtract, and you have a remainder of 3. So 63/5 is the same thing as 12 wholes and 3 left over, or 3/5 left over. And if you wanted to go back from this to that, just think: 12 is the same thing as 60 fifths, or 60/5." + }, + { + "Q": "At 1:51 you simplify the numerator of one fraction, but then simplify the denominator of the other fraction. How is it that you can do this? I thought you had to simplify the denominator of the same fraction.", + "A": "You can do this if you are multiplying fractions. If you multiply first, you get (7/4)*(36/5)=(7*36)/(4*5). Now you have all the numbers from the original two fractions in the same fraction. So it really doesn t matter at what stage you simplify. I hope this helped you. And remember that this is allowed only in multiplication, not in any other operation.", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 111 + ], + "3min_transcript": "Multiply 1 and 3/4 times 7 and 1/5. Simplify your answer and write it as a mixed fraction. So the first thing we want to do is rewrite each of these mixed numbers as improper fractions. It's very difficult, or at least it's not easy for me, to directly multiply mixed numbers. One can do it, but it's much easier if you just make them improper fractions. So let's convert each of them. So 1 and 3/4 is equal to-- it's still going to be over 4, so you're still going to have the same denominator, but your numerator as an improper fraction is going to be 4 times 1 plus 3. And the reason why this makes sense is 1 is 4/4, or 1 is 4 times 1 fourths, right? 1 is the same thing as 4/4, and then you have three more fourths, so 4/4 plus 3/4 will give you 7/4. So that's the same thing as 1 and 3/4. Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5." + }, + { + "Q": "at 2:37 you only did the denominator for the first fraction and for the second fraction you only did the numerator, why didnt you do all of both fractions?", + "A": "At this point in the video, Sal is cancelling out (dividing out) a common factor. As long as the fractions are being multiplied, we can cancel out a common factor from any numerator with the same factor in any denominator. The only numerator and denominator values that share a common factor are the 36 and the 4. Five (5), the remaining number in a denominator is not a common factor of 7 or of 9. This is why no other numbers were cancelled. out. hope this helps.", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 157 + ], + "3min_transcript": "Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5. 6 minus 5 is 1. Bring down the 3. 5 goes into 13 two times. And you could have immediately said 5 goes into 63 twelve times, but this way, at least to me, it's a little bit more obvious. And then 2 times 5 is 12, and then we have sorry! 2 times 5 is 10. That tells you not to switch gears in the middle of a math problem. 2 times 5 is 10, and then you subtract, and you have a remainder of 3. So 63/5 is the same thing as 12 wholes and 3 left over, or 3/5 left over. And if you wanted to go back from this to that, just think: 12 is the same thing as 60 fifths, or 60/5." + }, + { + "Q": "At 0:34 what do you mean by geometric and arethmatic mean because I'm confused?", + "A": "Geometric and arithmetic mean have different meanings in some areas of math. For now just use the mean you ve learned so far.", + "video_name": "k5EbijWu-Ss", + "timestamps": [ + 34 + ], + "3min_transcript": "Let's say you're trying to design some type of a product for men. One that is somehow based on their height. And the product is for the United States. So ideally, you would like to know the mean height of men in the United States. Let me write this down. So how would you do that? And when I talk about the mean, I'm talking about the arithmetic mean. If I were to talk about some other types of means-- and there are other types of means, like the geometric mean-- I would specify it. But when people just say mean, they're usually talking about the arithmetic mean. So how would you go about finding the mean height of men in the United States? Well, the obvious one is, is you go and ask every or measure every man in the United States. Take their height, add them all together, and then divide by the number of men there are in the United States. is whether that is practical. Because you have on the order-- let's see, there's about 300 million people in the United States. Roughly half of them will be men, or at least they'll be male, and so you will have 150 million, roughly 150 million men in the United States. So if you wanted the true mean height of all of the men United States, you would have to somehow survey-- or not even survey. You would have to be able to go and measure all 150 million men. And even if you did try to do that, by the time you're done, many of them might have passed away, new men will have been born, and so your data will go stale immediately. So it is seemingly impossible, or almost impossible, to get the exact height of every man in the United States in a snapshot of time. And so, instead, what you do is say, well, look, OK, I can't get every man, but maybe I can take a sample. I could take a sample of the men in the United States. And I'm going to make an effort that it's a random sample. happen to play basketball, or played basketball for their college. I don't want to go sample 100 people who are volleyball players. I want to randomly sample. Maybe the first person who comes out of the mall in a random town, or in several towns, or something like that. Something that should not be based in any way, or skewed in any way, by height. So you take a sample and from that sample you can calculate a mean of at least the sample. And you'll hope that that is indicative of-- especially if this was a reasonably random sample-- you'll hope that was indicative of the mean of the entire population. And what you're going to see in much of statistics it is all about using information, using things that we can calculate about a sample, to infer things about a population. Because we can't directly measure the entire population." + }, + { + "Q": "WHY is 5.5 = to 5 feet,6inches? SAL said so at 3:31-3:37", + "A": "Because 1 foot is 12 inches. Compare this to for example hours and minutes. 1.5 hours is one hour plus half an hour, or 1 hour 30 minutes. Likewise, 5.5 feet is 5 feet and half a foot, or 5 feet and 6 inches.", + "video_name": "k5EbijWu-Ss", + "timestamps": [ + 211, + 217 + ], + "3min_transcript": "happen to play basketball, or played basketball for their college. I don't want to go sample 100 people who are volleyball players. I want to randomly sample. Maybe the first person who comes out of the mall in a random town, or in several towns, or something like that. Something that should not be based in any way, or skewed in any way, by height. So you take a sample and from that sample you can calculate a mean of at least the sample. And you'll hope that that is indicative of-- especially if this was a reasonably random sample-- you'll hope that was indicative of the mean of the entire population. And what you're going to see in much of statistics it is all about using information, using things that we can calculate about a sample, to infer things about a population. Because we can't directly measure the entire population. trying to do this, I would recommend doing at least 100 data points, or 1,000, and later on we'll talk about how you can think about whether you've measured enough or how confident you can be. But let's just say you're a little bit lazy, and you just sample five men. And so you get their five heights. Let's say one is 6.2 feet. Let's say one is 5.5 feet-- 5.5 feet would be 5 foot, 6 inches. Let's say one ends up being 5.75 feet. Another one is 6.3 feet. Another is 5.9 feet. Now, if these are the ones that you happen to sample, what would you get for the mean of this sample? Well let's get our calculator out. And we get 6.2 plus 5.5 plus 5.75 plus 6.3 plus 5.9. And then we want to divide by the number of data points we have. So we have five data points. So let's divide 29.65 divided by 5, and we get 5.93 feet. So here, our sample mean-- and I'm going to denote it with an x with a bar over it, is-- and I already forgot the number-- 5.93 feet. This is our sample mean, or, if we want to make it clear, sample arithmetic mean. And when we're taking this calculation based on a sample, and somehow we're trying to estimate it for the entire population, we call this right over here," + }, + { + "Q": "At 4:43, is a capital Sigma only used for sample sums?", + "A": "Per Wiki, a capital Sigma when used by mathematicians indicates summation. A capital Sigma has other uses in other fields like physics and economics. Here s the Wiki link for a more nuanced answer.", + "video_name": "k5EbijWu-Ss", + "timestamps": [ + 283 + ], + "3min_transcript": "trying to do this, I would recommend doing at least 100 data points, or 1,000, and later on we'll talk about how you can think about whether you've measured enough or how confident you can be. But let's just say you're a little bit lazy, and you just sample five men. And so you get their five heights. Let's say one is 6.2 feet. Let's say one is 5.5 feet-- 5.5 feet would be 5 foot, 6 inches. Let's say one ends up being 5.75 feet. Another one is 6.3 feet. Another is 5.9 feet. Now, if these are the ones that you happen to sample, what would you get for the mean of this sample? Well let's get our calculator out. And we get 6.2 plus 5.5 plus 5.75 plus 6.3 plus 5.9. And then we want to divide by the number of data points we have. So we have five data points. So let's divide 29.65 divided by 5, and we get 5.93 feet. So here, our sample mean-- and I'm going to denote it with an x with a bar over it, is-- and I already forgot the number-- 5.93 feet. This is our sample mean, or, if we want to make it clear, sample arithmetic mean. And when we're taking this calculation based on a sample, and somehow we're trying to estimate it for the entire population, we call this right over here, Now, you might be saying, well, what notation do we use if, somehow, we are able to measure it for the population? Let's say we can't even measure it for the population, but we at least want to denote what the population mean is. Well if you want to do that, the population mean is usually denoted by the Greek letter mu. And so in a lot of statistics, it's calculating a sample mean in an attempt to estimate this thing that you might not know, the population mean. And these calculations on the entire population, sometimes you might be able to do it. Oftentimes, you will not be able to do it. These are called parameters. So what you're going to find in much of statistics, it's all about calculating statistics for a sample, finding these sample statistics in order" + }, + { + "Q": "At 1:23 , how many ounces is a ton?", + "A": "There is 32,000 ounces in a ton.", + "video_name": "Dj1rbIP8PHM", + "timestamps": [ + 83 + ], + "3min_transcript": "Let's talk a little bit about the US customary units of weight. So the one that's most typically used is the pound, especially for things of kind of a human scale. And to understand what a pound is, most playing balls are roughly about a pound. So, for example, a soccer ball-- my best attempt to draw a soccer ball. So let's say that this is a soccer ball right over here. And then of course it has some type of pattern on it. So you could imagine a soccer ball is about a pound. So it's roughly one pound. And a pound will often be shorthanded with this \"lb.\" right over here. So it's about a pound. A football, an American football, is also a little under a pound. But we could say it is about a pound, just so we get a sense of what a pound actually represents. Now, if you want to go to scales smaller than a pound, you would think about using the ounce. thinking about weight is that one pound-- let me write this-- is equal to 16 ounces. Or another way of thinking about it is that 1 ounce is equal to 1/16 of a pound. And if you want to visualize things that weigh about an ounce, you could imagine a small box of matches weigh about an ounce. So a small box of matches might weigh about an ounce. Maybe a small AA battery would weigh about an ounce. But that gives you a sense of it. So if you were to take 16 of these together, they would be about the weight of a soccer ball. 16 of these things together, they would be about-- they would be about the weight of a soccer ball or a football. that are larger than a pound, then we would go to the ton. And a ton is equal to-- 1 ton is equal to 2,000 pounds. And you have to be a little bit careful with the ton. We're talking about the US customary units, and this is where we're talking about 2,000 pounds. But when we're talking on a more international level, this is sometimes called the short ton. There's also a long ton. There's also the metric ton. But here we're talking about US customary units, which is the short ton. So one ton is 2,000 pounds. And to get a sense of something that weighs 2,000 pounds, or to get a sense of what 2,000 pounds is like, or what might be measured in tons, a car is a good example. Your average midsize sedan would weigh about a little under to a little over 2 tons, so a little" + }, + { + "Q": "At 6:46, Sal says the function would be defined or all other real numbers except the ones specified. But shouldn't x not be equal to 0 either? Because if x=0, then the denominator would again become 0, hence resulting in an undefined solution. Is that correct?", + "A": "We have h(x)=(x+10)/((x+10)(x-9)(x-5)). In that case, h(0) = (0+10)/((0+10)(0-9)(0-5)) = 10/((10)(-9)(-5) = 1/45", + "video_name": "n17q8CBiMtQ", + "timestamps": [ + 406 + ], + "3min_transcript": "But remember when x is equal 5, we don't look at this part of the compound definition. We look at this part. So it's true that up here you would be dividing by 0 if x is equaling 5 but x equaling 5, you wouldn't even look at -- look there. For input is 5, you use this part of the definition. So you'd divide by 0. Maybe I should write it this way, divide by 0 on, I guess you could say the top -- the top clause or the top part a definition. Part of the definition. If x equals 9, X equals negative 10 or -- and that's it because x equal 5 doesn't apply to this top part. If this clause wasn't here then yes, you would write x equals 5. Now we're almost done, but some of you might say, wait, wait, wait, but look, can't I simplify this? I have x plus 10 in the numerator and x plus 10 in the denominator. Can I just simplify this and that will disappear? And you could except, if you did that, you are now creating a different function definition. Because if you just simplify this, you just said 1 over x minus 9 -- That one actually would be defined at x equals negative 10. But the one that we have started with, this one is not. This is -- you're gonna end up with 0 over 0. You gonna end up with that indeterminate form. So for this function, exactly the way it's written, it's not going to be defined with x is equal to 9 or x is equal to negative 10. So once again if you want a fan -- write in our fancy domain set notation. The domain is going to be x all the x's that are a member of the real such that x does not equal 9 and x does not equal negative 10. Any other real number x -- it's going to work including 5. If x equals 5, h of -- h of 5 is going to be equal to pi, because you default to this one over here. h of 5 so OK x is equal 5, we do that one right over there. Now if you gave x equals 9 by 0, but that's gonna work for anything else. So that right over there is the domain." + }, + { + "Q": "At 1:25, Sal shows us the outer angle. What is the use for these outer angles in the real world?", + "A": "In some math problems, if you are only given the outer angle, you can still solve for the remaining angles. Angles help in many things such as the job of construction workers, carpenters when deciding table lengths, etc.", + "video_name": "QmfoIvgIVlE", + "timestamps": [ + 85 + ], + "3min_transcript": "We're asked to construct a 10 degree angle. So we have this little angle tool here that we can use to construct an angle. So just like that. And then they give us a protractor to actually measure the angle. So let's set it up. Let's put the protractor here. Let's put the vertex of the angle at the center of the protractor, in other words, center the protractor at the center of the vertex. Looks like the protractor's a little bit easier to manipulate. So let's do that there. Now let's put one of the rays here at 0 degrees. And now I'm going to put the other ray at 10 degrees. And it looks like I am done. I have constructed a 10 degree angle. And you want to be careful here when you use this tool because the angle in question-- let me move the protractor show you what I'm talking about-- that this right over here is the angle that we're talking about. If I'd switched these two rays, the way the tool is set up, it might have interpreted the angle as this outer angle right over here. So be careful to look at which angle we're actually measuring. So let me check my answer. And just to be clear what I'm talking about. If you did the 10 degree angle like this, then it definitely would have marked it wrong even though the interior angle right over here or this angle right here might be 10 degrees. The way the tool is set up, you see from this circle, that it thinks that you're looking at this outer angle. So it's important to make sure, at least for the sake of this exercise, that the system, that the computer program knows which angle you're going to talk about. Let's do one more of these. 155 degree angle. And this one's interesting because this is an obtuse angle. So once again, let us put the protractor at the vertex of the angle. So just like that. And now that seems pretty good. Now let's take one ray and put it at 0 degrees, and then let's take the other one and put it at 155. 90-- that gets us to a right angle. Then we'll start getting into obtuse angles, 100, 110, 120, 130, 140, 150. And just to make sure that blue arc is measuring this angle right over here, not the outer one. And let me move the protractor out of the way so we can get a good look at it. And we got it wrong. So let's see what we-- oh, 155 degree angle, not 150 degree angle. Let me fix that. 155 degree angle. Now we got it right." + }, + { + "Q": "What does he mean at 0:31 when he says fair coin?", + "A": "He means a coin with one head and one tail that has an equal chance of flipping one or the other.", + "video_name": "cqK3uRoPtk0", + "timestamps": [ + 31 + ], + "3min_transcript": "Voiceover:Let's say we define the random variable capital X as the number of heads we get after three flips of a fair coin. So given that definition of a random variable, what we're going to try and do in this video is think about the probability distributions. So what is the probability of the different possible outcomes or the different possible values for this random variable. We'll plot them to see how that distribution is spread out amongst those possible outcomes. So let's think about all of the different values that you could get when you flip a fair coin three times. So you could get all heads, heads, heads, heads. You could get heads, heads, tails. You could get heads, tails, heads. You could get heads, tails, tails. You could have tails, heads, heads. You could have tails, head, tails. You could have tails, tails, heads. And then you could have all tails. when you do the actual experiment there's eight equally likely outcomes here. But which of them, how would these relate to the value of this random variable? So let's think about, what's the probability, there is a situation where you have zero heads. So what's the probably that our random variable X is equal to zero? Well, that's this situation right over here where you have zero heads. It's one out of the eight equally likely outcomes. So that is going to be 1/8. What's the probability that our random variable capital X is equal to one? Well, let's see. Which of these outcomes gets us exactly one head? We have this one right over here. We have that one right over there. We have this one right over there. So three out of the eight equally likely outcomes provide us, get us to one head, which is the same thing as saying that our random variable equals one. So this has a 3/8 probability. So what's the probability, I think you're getting, maybe getting the hang of it at this point. What's the probability that the random variable X is going to be equal to two? Well, for X to be equal to two, we must, that means we have two heads when we flip the coins three times. So that's this outcome meets this constraint. This outcome would get our random variable to be equal to two. And this outcome would make our random variable equal to two. And this is three out of the eight equally likely outcomes. So this has a 3/8 probability. And then finally we could say what is the probability that our random variable X is equal to three? Well, how does our random variable X equal three? Well we have to get three heads when we flip the coin. So there's only one out of the eight" + }, + { + "Q": "At 3:28 why is the probability range between 0 and 1? I understand that beyond 1 we have a certainty of something happening, but why 1?", + "A": "In statistics, 1=100%. One hundred percent is is absolute certainty. You can t be more certain than that.", + "video_name": "cqK3uRoPtk0", + "timestamps": [ + 208 + ], + "3min_transcript": "when you do the actual experiment there's eight equally likely outcomes here. But which of them, how would these relate to the value of this random variable? So let's think about, what's the probability, there is a situation where you have zero heads. So what's the probably that our random variable X is equal to zero? Well, that's this situation right over here where you have zero heads. It's one out of the eight equally likely outcomes. So that is going to be 1/8. What's the probability that our random variable capital X is equal to one? Well, let's see. Which of these outcomes gets us exactly one head? We have this one right over here. We have that one right over there. We have this one right over there. So three out of the eight equally likely outcomes provide us, get us to one head, which is the same thing as saying that our random variable equals one. So this has a 3/8 probability. So what's the probability, I think you're getting, maybe getting the hang of it at this point. What's the probability that the random variable X is going to be equal to two? Well, for X to be equal to two, we must, that means we have two heads when we flip the coins three times. So that's this outcome meets this constraint. This outcome would get our random variable to be equal to two. And this outcome would make our random variable equal to two. And this is three out of the eight equally likely outcomes. So this has a 3/8 probability. And then finally we could say what is the probability that our random variable X is equal to three? Well, how does our random variable X equal three? Well we have to get three heads when we flip the coin. So there's only one out of the eight So it's a 1/8 probability. So now we just have to think about how we plot this, to see how this is distributed. So let me draw... So over here on the vertical axis this will be the probability. Probability. And it's going to be between zero and one. You can't have a probability larger than one. So just like this. So let's see, if this is one right over here, and let's see everything here looks like it's in eighths so let's put everything in terms of eighths. So that's half. This is a fourth. That's a fourth. That's not quite a fourth. This is a fourth right over here. And then we can do it in terms of eighths. So that's a pretty good approximation. And then over here we can have the outcomes." + }, + { + "Q": "At 4:36, why didn't Sal distribute the 2 to AC and BD?", + "A": "The Distributive Law only applies to cases of multiplication over addition. The distributive property would not apply in this case, as it is 1/2AC*BD, not 1/2(AC+BD)", + "video_name": "3FManXv4mZM", + "timestamps": [ + 276 + ], + "3min_transcript": "What is the height here? Well we know that this diagonal right over here, that it's a perpendicular bisector. So the height is just the distance from BE. So it's AC times BE, that is the height. This is an altitude. It intersects this base at a 90-degree angle. Or we could say BE is the same thing as 1/2 times BD. So this is-- let me write it. This is equal to, so it's equal to 1/2 times AC, that's our base. And then our height is BE, which we're saying is the same thing as 1/2 times BD. So that's the area of just ABC, that's just the area of this broader triangle right up there, or that larger triangle right up there, that half of the rhombus. But we decided that the area of the whole thing So if we go back, if we use both this information and this information right over here, we have the area of ABCD is going to be equal to 2 times the area of ABC, where the area of ABC is this thing right over here. So 2 times the area of ABC, area of ABC is that right over there. So 1/2 times 1/2 is 1/4 times AC times BD. And then you see where this is going. 2 times 1/4 fourth is 1/2 times AC times BD. Fairly straightforward, which is a neat result. And actually, I haven't done this in a video. I'll do it in the next video. There are other ways of finding the areas of parallelograms, It's essentially base times height, but for a rhombus we could do that because it is a parallelogram, but we also have this other neat little result that we proved in this video. That if we know the lengths of the diagonals, the area of the rhombus is 1/2 times the products" + }, + { + "Q": "0:03 how do you now its a rhombus", + "A": "And it was a given.", + "video_name": "3FManXv4mZM", + "timestamps": [ + 3 + ], + "3min_transcript": "So quadrilateral ABCD, they're telling us it is a rhombus, and what we need to do, we need to prove that the area of this rhombus is equal to 1/2 times AC times BD. So we're essentially proving that the area of a rhombus is 1/2 times the product of the lengths of its diagonals. So let's see what we can do over here. So there's a bunch of things we know about rhombi and all rhombi are parallelograms, so there's tons of things that we know about parallelograms. First of all, if it's a rhombus, we know that all of the sides are congruent. So that side length is equal to that side length is equal to that side length is equal to that side length. Because it's a parallelogram, we know the diagonals bisect each other. So we know that this length-- let me call this point over here B, let's call this E. We know that BE is going to be equal to ED. So that's BE, we know that's going to be equal to ED. And we know that AE is equal to EC. We also know, because this is a rhombus, and we proved this in the last video, but they are also perpendicular. So we know that this is a right angle, this is a right angle, that is a right angle, and then this is a right angle. So the easiest way to think about it is if we can show that this triangle ADC is congruent to triangle ABC, and if we can figure out the area of one of them, we can just double it. So the first part is pretty straightforward. So we can see that triangle ADC is going to be congruent to triangle ABC, and we know that by side-side-side congruency. This side is congruent to that side. This side is congruent to that side, and they both share a C right over here. So this is by side-side-side. And so we can say that the area-- so because of that, we know that the area of ABCD is just we could pick either one of these. We could say 2 times the area of ABC. Because area of ABCD-- actually let me write it this way. The area of ABCD is equal to the area of ADC plus the area of ABC. But since they're congruent, these two are going to be the same thing, so it's just going to be 2 times the area of ABC. Now what is the area of ABC? Well area of a triangle is just 1/2 base times height. So area of ABC is just equal to 1/2 times the base of that triangle times its height, which is equal to 1/2. What is the length of the base? Well the length of the base is AC. So it's 1/2-- I'll color code it. The base is AC." + }, + { + "Q": "In 1:12, How did Sal get 1 from i^25?", + "A": "Sal got one from i^100, not i^25. he said that since 4x25=100, then i^100 is the same thing as (i^4)^25. i^4 is equal to 1, so 1^25=1. Finally, Sal deduced that i^100=1", + "video_name": "QiwfF83NWNA", + "timestamps": [ + 72 + ], + "3min_transcript": "Now that we've seen that as we take i to higher and higher powers, it cycles between 1, i, negative 1, negative i, then back to 1, i, negative 1, and negative i. I want to see if we can tackle some, I guess you could call them, trickier problems. And you might see these surface. And they're also kind of fun to do to realize that you can use the fact that the powers of i cycle through these values. You can use this to really, on a back of an envelope, take arbitrarily high powers of i. So let's try, just for fun, let's see what i to the 100th power is. And the realization here is that 100 is a multiple of 4. So you could say that this is the same thing as i to the 4 times 25th power. And this is the same thing, just from our exponent properties, as i to the fourth power raised to the 25th power. If you have something raised to an exponent, and then that is raised to an exponent, that's the same thing as multiplying the two exponents. that's pretty straightforward. i to the fourth is just 1. i to the fourth is 1, so this is 1. So this is equal to 1 to the 25th power, which is just equal to 1. So once again, we use this kind of cycling ability of i when you take its powers to figure out a very high exponent of i. Now let's say we try something a little bit stranger. Let's try i to the 501st power. Now in this situation, 501, it's not a multiple of 4. So you can't just do that that simply. But what you could do, is you could write this as a product of two numbers, one that is i to a multiple of fourth power. And then one that isn't. And so you could rewrite this. 500 is a multiple of 4. So you could write this as i to the 500th power times i Right? You have the same base. When you multiply, you can add exponents. So this would be i to the 501st power. as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k" + }, + { + "Q": "At 1:12, isn't 100 also divisible by 5?, and since 5 is larger, wouldn't i^100 be i?", + "A": "When working with powers of i, we always use 4 to divide the exponent (regardless of what factors the exponent has) because the values of the powers of i repeat in groups of 4. That way, by using the fact that i^4 is 1, it makes the problem simpler.", + "video_name": "QiwfF83NWNA", + "timestamps": [ + 72 + ], + "3min_transcript": "Now that we've seen that as we take i to higher and higher powers, it cycles between 1, i, negative 1, negative i, then back to 1, i, negative 1, and negative i. I want to see if we can tackle some, I guess you could call them, trickier problems. And you might see these surface. And they're also kind of fun to do to realize that you can use the fact that the powers of i cycle through these values. You can use this to really, on a back of an envelope, take arbitrarily high powers of i. So let's try, just for fun, let's see what i to the 100th power is. And the realization here is that 100 is a multiple of 4. So you could say that this is the same thing as i to the 4 times 25th power. And this is the same thing, just from our exponent properties, as i to the fourth power raised to the 25th power. If you have something raised to an exponent, and then that is raised to an exponent, that's the same thing as multiplying the two exponents. that's pretty straightforward. i to the fourth is just 1. i to the fourth is 1, so this is 1. So this is equal to 1 to the 25th power, which is just equal to 1. So once again, we use this kind of cycling ability of i when you take its powers to figure out a very high exponent of i. Now let's say we try something a little bit stranger. Let's try i to the 501st power. Now in this situation, 501, it's not a multiple of 4. So you can't just do that that simply. But what you could do, is you could write this as a product of two numbers, one that is i to a multiple of fourth power. And then one that isn't. And so you could rewrite this. 500 is a multiple of 4. So you could write this as i to the 500th power times i Right? You have the same base. When you multiply, you can add exponents. So this would be i to the 501st power. as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k" + }, + { + "Q": "At 5:20 Sal says regarding i^96 that \"This is i^4, and then that to the 16th power\". Shouldn't he have said that \"This is i^4, and then that to the 24th power\" instead?", + "A": "Yeah... But the result wasn t wrong.", + "video_name": "QiwfF83NWNA", + "timestamps": [ + 320 + ], + "3min_transcript": "So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over. i to the first power. This is a multiple of 4-- this right here is a multiple of 4-- and I know that because any 1,000 is multiple of 4, any 100 is a multiple of 4, and then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the first power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1, and we're just going to be left with i to the first power, or just i. Let's do another one. i to the 90-- let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99? So this is the same thing as i to the 96th power times i to the third power, right? If you multiply these, same base, add the exponent, you would get i to the 99th power. i to the 96th power, since this is a multiple of 4, this is i to the fourth, and then that to the 16th power. So that's just 1 to the 16th, so this is just 1. And then you're just left with i to the third power. And you could either remember that i to the third power is equal to-- you can just remember that it's equal to negative i. Or if you forget that, you could just say, look, this is the same thing as i squared times i. This is equal to i squared times i. i squared, by definition, is equal to negative 1. So you have negative 1 times i is equal to negative i. Let me do one more just for the fun of it. Let's take i to the 38th power." + }, + { + "Q": "At 5:18 he said 8 games and 16 somgs. How does this meets the first constranit?", + "A": "The first constraint is songs + games must be greater than or equal to 15. 8+16=24 which is greater than 15.", + "video_name": "BUmLw5m6F9s", + "timestamps": [ + 318 + ], + "3min_transcript": "So let's see if we can graph these two constraints. Well, this first constraint, s plus g is going to be greater than or equal to 15. The easiest way to think about this-- or the easiest way to graph this is to really think about the intercepts. If g is 0, what is s? Well, s plus 0 has to be greater than or equal to 15. So if g is 0, s is going to be greater than or equal to 15. Let me put it this way. So if I'm going to graph this one right here. If g is 0, s is greater than or equal to 15. So g is 0, s, 15, let's see, this is 12, 14, 15 is right over there. And s is going to be all of the values equivalent to that or greater than for g equal to 0. If s is equal to 0, g is greater than or equal to 15. So if s is equal to 0, g is greater than or equal to 15. So the boundary line, s plus g is equal to 15, we would just have to connect these two dots. Let me try my best to connect these dots. So it would look something like this. This is always the hardest part. Let me see how well I can connect these two dots. Nope. Let me see. I should get a line tool for this. So that's pretty good. So that's the line s plus g is equal to 15. And we talk about the values greater than 15, we're going And you saw that when g is equal to 0, s is greater than or equal to 15. It's all of these values up here. And when s was 0, g was greater than or equal to 15. So this constraint right here is all of this. All of this area satisfies this. All of this area-- if you pick any coordinate here, it coordinates, because we're not going to buy parts of games. But if you think about all of the integer coordinates here, they represent combinations of s and g, where you're buying at least 15 games. For example here, you're buying 8 games and 16 songs. That's 24. So you're definitely meeting the first constraint. Now the second constraint. 0.89s plus 1.99g is less than or equal to 25. This is a starting point. Let's just draw the line 0.89s plus 1.99 is equal to 25. And then we could think about what region the less than would represent. Oh, 1.99g. And the easiest way to do this, once again, we could do slope y-intercept all that type of thing. But the easiest way is to just find the s- and the g-intercepts. So if s is equal to 0 then we have 1.99g is equal to 25 or g" + }, + { + "Q": "Hay,\nI was just wondering, at 5:24 Khan began to solve for the variables using the standard form. He started to graph it at 7:22. If someone has to graph an inequality converting it to standard form, do you have to use the same interval as in the standard form or can you adjust it due to the converted slope-intercept form?", + "A": "Just use intecepts", + "video_name": "BUmLw5m6F9s", + "timestamps": [ + 324, + 442 + ], + "3min_transcript": "So let's see if we can graph these two constraints. Well, this first constraint, s plus g is going to be greater than or equal to 15. The easiest way to think about this-- or the easiest way to graph this is to really think about the intercepts. If g is 0, what is s? Well, s plus 0 has to be greater than or equal to 15. So if g is 0, s is going to be greater than or equal to 15. Let me put it this way. So if I'm going to graph this one right here. If g is 0, s is greater than or equal to 15. So g is 0, s, 15, let's see, this is 12, 14, 15 is right over there. And s is going to be all of the values equivalent to that or greater than for g equal to 0. If s is equal to 0, g is greater than or equal to 15. So if s is equal to 0, g is greater than or equal to 15. So the boundary line, s plus g is equal to 15, we would just have to connect these two dots. Let me try my best to connect these dots. So it would look something like this. This is always the hardest part. Let me see how well I can connect these two dots. Nope. Let me see. I should get a line tool for this. So that's pretty good. So that's the line s plus g is equal to 15. And we talk about the values greater than 15, we're going And you saw that when g is equal to 0, s is greater than or equal to 15. It's all of these values up here. And when s was 0, g was greater than or equal to 15. So this constraint right here is all of this. All of this area satisfies this. All of this area-- if you pick any coordinate here, it coordinates, because we're not going to buy parts of games. But if you think about all of the integer coordinates here, they represent combinations of s and g, where you're buying at least 15 games. For example here, you're buying 8 games and 16 songs. That's 24. So you're definitely meeting the first constraint. Now the second constraint. 0.89s plus 1.99g is less than or equal to 25. This is a starting point. Let's just draw the line 0.89s plus 1.99 is equal to 25. And then we could think about what region the less than would represent. Oh, 1.99g. And the easiest way to do this, once again, we could do slope y-intercept all that type of thing. But the easiest way is to just find the s- and the g-intercepts. So if s is equal to 0 then we have 1.99g is equal to 25 or g" + }, + { + "Q": "in 9:37 I thought the question said that a total of 15 games should be purchased, i was just wondering if it was relevant?", + "A": "Sal actually said that at least 15 items were purchased. Like English, mathematics is a language. It is important to read the words very carefully, especially in word problems.", + "video_name": "BUmLw5m6F9s", + "timestamps": [ + 577 + ], + "3min_transcript": "Just a little over 28. So 28.08. So that is, g is 0, s is 28. So that is 2, 4, 24, 6, 8. A little over 28. So it's right over there. So this line, 0.89s plus 1.99g is equal to 25 is going to go from this coordinate, which is 0, 28. So that point right there. All the way down to the point 12.56,0. So let me see if I can draw that. It's going to go-- I'll draw up one more attempt. Maybe if I start from the bottom it'll be easier. That was a better attempt. Let me bold that in a little bit, so you can make sure you can see it. So that line represents this right over here. that imply? So if we think about it, when g is equal to 0, 0.89s is less than 25. So when g is equal to 0, if we really wanted the less than there, we could think of it this way. It's less than instead of just doing less than or equal to. So s is less than 28.08. So it'll be the region below. When s is 0, g-- so if we think s is 0, if we use this original equation, 1.99g will be less than or equal to. I use this just to plot the graph, but if we actually care about the actual inequality, we get 1.99g is less than 25. g would be less than or equal to 12.56. So when s is equal to 0, g is less than 12.56. So the area that satisfies this second constraint is everything below this graph. So it's going to be the overlap of the regions that satisfy one of the two. So the overlap is going to be this region right here. Below the orange graph and above the blue graph, including both of them. So if you pick any combination-- so if he buys 4 games and 14 songs, that would work. Or if he bought 2 games and 16 songs, that would work. So you can kind of get the idea. Anything in that region-- and he can only buy integer values-- would satisfy" + }, + { + "Q": "What does Congruent mean? 1:25", + "A": "Congruent means that 2 or more shapes look exactly the same in shape and size.", + "video_name": "tFhBAeZVTMw", + "timestamps": [ + 85 + ], + "3min_transcript": "We know that quadrilateral ABCD over here is a parallelogram. And what I want to discuss in this video is a general way of finding the area of a parallelogram. In the last video, we talked about a particular way of finding the area of a rhombus. You could take half the product of its diagonals. And a rhombus is a parallelogram. But you can't just generally take half the product of the diagonals of any parallelogram. It has to be a rhombus. And now we're just going to talk about parallelograms. So what do we know about parallelograms? Well, we know the opposite sides are parallel. So that side is parallel to that side, and this side is parallel to this side. And we also know that opposite sides are congruent. So this length is equal to this length, and this length is equal to this length over here. Now, if we draw a diagonal-- I'll draw a diagonal AC-- we can split our parallelogram into two triangles. And we've proven this multiple times. These two triangles are congruent. But we can do it in a pretty straightforward way. Obviously, AD is equal to BC. We have DC is equal to AB. And then both of these triangles share this third side right over here. They both share AC. So we can say triangle-- let me write this in yellow-- we could say triangle ADC is congruent to triangle-- let me get this right. So it's going to be congruent to triangle-- so I said A, D, C. So I went along this double magenta slash first, then the pink, and then I went D, and then I went the last one. So I'm going to say CBA. Because I went the double magenta, then pink, then the last one. So CBA, triangle CBA. And this is by side, side, side congruency. All three sides, they have three corresponding sides that are congruent to each other. So the triangles are congruent to each other. And what that tells us is that the areas of these two So if I want to find the area of ABCD, the whole parallelogram, it's going to be equal to the area of triangle-- let me just write it here-- it's equal to the area of ADC plus the area of CBA. But the area of CBA is just the same thing as the area of ADC because they are congruent, by side, side, side. So this is just going to be two times the area of triangle ADC. Which is convenient for us, because we know how to find the areas of triangles. Area of triangles is literally just 1/2 times base times height. So it's 1/2 times base times height of this triangle. And we are given the base of ADC. It is this length right over here. It is DC. You could view it as the base of the entire parallelogram. And if we wanted to figure out the height, we could draw an altitude down like this." + }, + { + "Q": "At 7:46, I would like to make sure that I have the correct answer for the \"cliffhanger\". Is it 20x^9\nThanks", + "A": "Yes, you have the correct answer. Good job. Looks like they cut the video off too quickly.", + "video_name": "iHnzLETGz2I", + "timestamps": [ + 466 + ], + "3min_transcript": "Negative 9x to the fifth power times negative three, use parentheses there, when you have a negative in front, you always wanna use parentheses. Let's do x to the 107th power. If I would have showed you this before this video, you would have said oh my goodness, there's nothing I can do, I'm boxed, there's no way out. But now you know that it's as simple as follow the rules. We're going to multiply the coefficients, negative nine times negative three is 27. Two negatives is a positive and nine times three is 27. I'm gonna add my powers. Five plus 107 is a hundred, ooh, not two, that was almost a mistake I made there. Let's get rid of that, give me a second chance here. Life's all about second chances, And so, this crazy expression, which is two monomials, here's the first, here's the second, when we multiply and simplify we get another monomial, which is 27x to the 112th. I'm gonna leave you on a cliffhanger here. Which, I'm gonna show you a problem. What variable should we use? You notice I've been trying to vary the variables up to show you that it just doesn't matter. That's an ugly five, let's get rid of that. Give me a second chance with that one too. So let's look at 5x to the third power, times 4x to the sixth power. And I'm gonna show you a wrong answer. I had a student that asked to do this, and here's the wrong answer that they gave me. They told me 9x to the 18th power. What did they do wrong? What did they do wrong? I want you to think to yourself, what have we been talking about? What did they do with the five and the four to get the nine? What should they have done? What did they do with the three and the six to get the 18, and what should they have done? That's multiplying monomials by monomials." + }, + { + "Q": "at 2:04 when p^2 = 2p, why wouldn't you solve it as sqrt(p^2) = sqrt(2p)", + "A": "You could do it that way, you d get p = \u00e2\u0088\u009a(2p). The solution would be the same (0 = \u00e2\u0088\u009a0, 2 = \u00e2\u0088\u009a4). It s just easier the way Sal does it, p(p-2) = 0, where you can clearly see the solutions are 0/2.", + "video_name": "ZIqW_sXymrM", + "timestamps": [ + 124 + ], + "3min_transcript": "- [Voiceover] So let's try to find the solutions to this equation right over here. We have the quantity two X minus three squared, and that is equal to four X minus six, and I encourage you to pause the video and give it a shot. And I'll give you a little bit of a hint, you could do this in the traditional way of expanding this out, and then turning it into kind of a classic quadratic form, but there might be a faster or a simpler way to do this if you really pay attention to the structure of both sides of this equation. Well let's look at this, we have two X minus three squared on the left-hand side, on the right-hand side we have four X minus six. Well four X minus six, that's just two times two X minus three, let me be clear there, so this is the same thing as two X minus three squared is equal to, four X minus six, if I factor out a two, that's two times two X minus three. And so this is really interesting, we have something squared is equal to two times that something. let me be very clear here, so the stuff in blue squared is equal to two times the stuff in blue. So if we can solve for what the stuff in blue could be equal to, then we could solve for X, and I'll show you that right now. So let's say, let's just replace two X minus three, we'll do a little bit of a substitution, let's replace that with P. So let's say that P is equal to two X minus three. Well then this equation simplifies quite nicely, the left-hand side becomes P squared, P squared is equal to two times P, 'cause once again two X minus three is P, two times P. And now we just have to solve for P. And I'll switch to just one color now. So we can write this as, if we subtract two P from both sides, we can get P squared minus two P and we can factor out a P, so we get P times P minus two is equal to zero. And we've seen this shown multiple times, if I have the product of two things and they equal to zero, at least one of them needs to be equal to zero, so either P is equal to zero, or P minus two is equal to zero. Well if P minus two is equal to zero, then that means P is equal to two. So either P equals zero, or P equals two. Well we're not quite done yet, because we wanted to solve for X, and not for P. But luckily we know that two X minus three is equal to P. So now we could say either two X minus three is going to be equal to this P value, is going to be equal to zero, or two X minus three is going to be equal to this P value, is going to be equal to two. And so this is pretty straightforward to solve," + }, + { + "Q": "At 4:06, why did Sal multiply 3x and 12 by 1/3?\n\nCouldn't you divide each side by 3 to isolate the variable?\n\nI was taught to use inverse operations. The inverse operation of multiplication is division.\n\nSo why did Sal not use inverse operations to solve 3x=12?", + "A": "Multiplying by 1/3 is the exact same thing as dividing by 3. Remember how to divide fractions? If you have 2/3 divided by 5/6 it is the same as 2/3 times 6/5, right. Remember you can write whole numbers, such as 12 and 3 as 12/1 and 3/1, So 12/3 = 12/1 / 3/1 = 12/1 * 1/3 = 12 * 1/3. So you see, multiplying by 1/3 is the same a dividing by 3. Great Question! Keep Studying! 12/3 = 12/1 * 1/3", + "video_name": "_y_Q3_B2Vh8", + "timestamps": [ + 246 + ], + "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced." + }, + { + "Q": "at 4:05 why didn't he do 3x/3 and 12/3", + "A": "Multiplying by 1/3 and dividing by 3 are the same operation, so your way is the same as Sal s, it just looks a little different", + "video_name": "_y_Q3_B2Vh8", + "timestamps": [ + 245 + ], + "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced." + }, + { + "Q": "At \"2:03\" is the answer Sal has not the correct one, so he has to use another formula to finish the problem?", + "A": "Sal explains earlier (at 0:28) that this problem will have only one solution because there s only one way the absolute value of something can be 0 since the only number with an absolute value of 0 is 0.", + "video_name": "GwjiR2_7A7Y", + "timestamps": [ + 123 + ], + "3min_transcript": "We're told, solve the absolute value of 3x minus 9 is equal to 0, and graph the solution on a number line. So let's just rewrite the absolute value equation. They told us that the absolute value of 3x minus 9 is equal to 0. So we're told that the absolute value of the something-- in this case the something is 3x minus 9-- is equal to 0. If I told you that the absolute value of something is equal to 0, I'm telling you that something has to be exactly 0 away from 0, or 0 away from the origin on the number line. So the only thing that that something could be is 0. If I told you that the absolute value of x is equal to 0, you know that x has to be equal to 0. That's the only value whose absolute value is 0. So if I told you that the absolute value of 3x minus 9 is 0, than we know that 3x minus 9 has to be equal to 0, and that's kind of unique about the 0 is that, it's the absolute value of 0. If you had, say, a 1 here, you could say, oh well, then this thing could be a 1 or a negative 1. But here, if you have a 0, this thing can only be 0. So solving this equation is fairly straightforward. If we want to isolate the 3x, get rid of the negative 9 on the left-hand side, we add 9 to both sides of the equation. Add 9 to both sides of the equation, these 9's cancel out. That's the whole point. On the left-hand side, you're just left with 3x, and on the right-hand side, you are just left with 9. Now we want to solve for x, so we have 3 times x. Let's divide it by 3, because 3 times x divided by 3 is just going to be x. But if we divide the left side by 3, we have to divide the right side by 3. So we are left with-- these guys cancel out. x is equal to 9 over 3, which is 3. And that's our solution. Let's substitute it back into our original equation. So we have the absolute value of 3 times x. Instead of x, I'll just put in our actual answer that we got, 3 times 3 minus 9 has got to be equal to 0. So what's this going to be equal to? 3 times 3 is 9. So it's the absolute value of 9 minus 9, which is the absolute value of 0, which is, indeed, 0. So it does, indeed, equal 0, and we are done." + }, + { + "Q": "How come at 5:37 you had to take the inverse cosine and could not leave the answer as cos(19/20)?", + "A": "In this example we are solving for theta. Notice that theta is stuck inside the COS function when we say: cos(theta) = 19/20 To get theta out in the open we take the inverse cos of both sides: arccos(cos(theta)) = arccos(19/20) theta = acrcos(19/20) theta = 18.19 degrees Notice that arccos and cos are inverse function as they undo each other.", + "video_name": "Ei54NnQ0FKs", + "timestamps": [ + 337 + ], + "3min_transcript": "So let's do that. So this is going to be negative 5,700. Is that right? 5,700 plus... Yes, that is right. Right, because if this was the other way around, if this was 6,100 minus 400 it would be positive 5,700. Alright. And then these two of course cancel out. And this is going to be equal to negative 6,000 times the cosine of theta. Now we can divide both sides by negative 6,000. And we get... I'm just gonna swap the sides. We get cosine of theta is equal to... Let's see we could divide the numerator and the denominator by essentially negative 100. So cosine of theta is equal to 57 over 60. And actually that can be simplified even more. Three goes into 57, is that 19 times? Yep, so this is actually... This could be simplified. This is equal to 19 over 20. We actually didn't have to do that simplification step because we're about to use our calculators, but that makes the math a little more tractable. Right, 3 goes into 57, yeah, 19 times. And so now we can take the inverse cosine of both sides. So we could get theta is equal to the inverse cosine, or the arc cosine, of 19 over 20. So let's get our calculator out and see if we get something that makes sense. So we wanna do the inverse cosine of 19 over 20. We get 18.19 degrees. And I already verified that my calculator is in degree mode. So it gets 18.19 degrees. So if we wanted to round, this is approximately equal to 18.2 degrees, if we wanna round to the nearest tenth. So that essentially gives us a sense of how steep this slope actually is." + }, + { + "Q": "What is an arbitrary triangle? it mentions that term in the video at 1:23. Thank you!", + "A": "It basically means a random triangle. He is paying no attention to side lengths or angle measures. It s not a term you need to know, don t worry.", + "video_name": "Ei54NnQ0FKs", + "timestamps": [ + 83 + ], + "3min_transcript": "Voiceover:Let's say you're studying some type of a little hill or rock formation right over here. And you're able to figure out the dimensions. You know that from this point to this point along the base, straight along level ground, is 60 meters. You know the steeper side, steeper I guess surface or edge of this cliff or whatever you wanna call it, is 20 meters. And then the longer side here, I guess the less steep side, is 50 meters long. So you're able to measure that. But now what you wanna do is use your knowledge of trigonometry, given this information, to figure out how steep is this side. What is the actual inclination relative to level ground? Or another way of thinking about it, what is this angle theta right over there? And I encourage you to pause the video and think about it on your own. Well it might be ringing a bell. Well you know three sides of a triangle and then we want to figure out an angle. And so the thing that jumps out in my head, well maybe the law of cosines could be useful. Let me just write out the law of cosines, before we try to apply it to this So the law of cosines tells us that C-squared is equal to A-squared, plus B-squared, minus two A B, times the cosine of theta. And just to remind ourselves what the A, B's, and C's are, C is the side that's opposite the angle theta. So if I were to draw an arbitrary triangle right over here. And if this is our angle theta, then this determines that C is that side, and then A and B could be either of these two sides. So A could be that one and B could be that one. Or the other way around. As you can see, A and B essentially have the same role in this formula right over here. This could be B or this could be A. So what we wanna do is somehow relate this angle... We wanna figure out what theta is in our little hill example right over here. So if this is going to be theta, what is C going to be? Well C is going to be this 20 meter side. one of these to be A or B. We could say that this A is 50 meters and B is 60 meters. And now we could just apply the law of cosines. So the law of cosines tells us that 20-squared is equal to A-squared, so that's 50 squared, plus B-squared, plus 60 squared, minus two times A B. So minus two times 50, times 60, times 60, times the cosine of theta. This works out well for us because they've given us everything. There's really only one unknown. There's theta here. So let's see if we can solve for theta. So 20 squared, that is 400. 50 squared is 2,500." + }, + { + "Q": "At 2:25, can't you also add 9x to both sides first instead of subtracting 12x?", + "A": "Yes! You can also add 9x to both sides, instead of subtracting 12x. The reason why Sal had subtracted 12x from both sides was that it is often preferred to have x on the left side. He ends up with: -21x - 3 = 18", + "video_name": "YZBStgZGyDY", + "timestamps": [ + 145 + ], + "3min_transcript": "We have the equation negative 9 minus this whole expression, 9x minus 6-- this whole thing is being subtracted from negative 9-- is equal to 3 times this whole expression, 4x plus 6. Now, a good place to start is to just get rid of these parentheses. And the best way to get rid of these parentheses is to kind of multiply them out. This has a negative 1-- you just see a minus here, but it's just really the same thing as having a negative 1-- times this quantity. And here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left-hand side of our equation, we have our negative 9. And then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to-- let's distribute the 3-- 3 times 4x is 12x. Now what we want to do, let's combine our constant terms, if we can. We have a negative 9 and a 6 here, on this side, we've combined all of our like terms. We can't combine a 12x and an 18, so let's combine this. So let's combine the negative 9 and the 6, our two constant terms on the left-hand side of the equation. So we're going to have this negative 9x. So we're going to have negative 9x plus-- let's see, we have a negative 9 and then plus 6-- so negative 9 plus 6 is negative 3. So we're going to have a negative 9x, and then we have a negative 3, so minus 3 right here. That's the negative 9 plus the 6, and that is equal to 12x plus 18. Now, we want to group all the x terms on one side of the equation, and all of the constant terms-- the negative 3 and the positive 18 on the other side-- I like to always You don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right. And the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now, on the left-hand side, I have negative 9x minus 12x. So negative 9 minus 12, that's negative 21. Negative 21x minus 3 is equal to-- 12x minus 12x, well, that's just nothing. So I could write a 0 here, but I don't That was the whole point of subtracting the 12x from the left-hand side. And that is going to be equal to-- so on the right-hand side, we just are left with an 18. We are just left with that 18 here. These guys canceled out. Now, let's get rid of this negative 3 from the left-hand side. So on the left-hand side, we only have x terms, and on the right-hand side, we only have constant terms. So the best" + }, + { + "Q": "At 1:17. Sal says that 12x + 18 cannot happen because x has a coefficient and 18 is a regular number. If this is correct, how did he get 3 + 4x = 12x?", + "A": "He did 3*4x, not 3+4x.", + "video_name": "YZBStgZGyDY", + "timestamps": [ + 77 + ], + "3min_transcript": "We have the equation negative 9 minus this whole expression, 9x minus 6-- this whole thing is being subtracted from negative 9-- is equal to 3 times this whole expression, 4x plus 6. Now, a good place to start is to just get rid of these parentheses. And the best way to get rid of these parentheses is to kind of multiply them out. This has a negative 1-- you just see a minus here, but it's just really the same thing as having a negative 1-- times this quantity. And here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left-hand side of our equation, we have our negative 9. And then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to-- let's distribute the 3-- 3 times 4x is 12x. Now what we want to do, let's combine our constant terms, if we can. We have a negative 9 and a 6 here, on this side, we've combined all of our like terms. We can't combine a 12x and an 18, so let's combine this. So let's combine the negative 9 and the 6, our two constant terms on the left-hand side of the equation. So we're going to have this negative 9x. So we're going to have negative 9x plus-- let's see, we have a negative 9 and then plus 6-- so negative 9 plus 6 is negative 3. So we're going to have a negative 9x, and then we have a negative 3, so minus 3 right here. That's the negative 9 plus the 6, and that is equal to 12x plus 18. Now, we want to group all the x terms on one side of the equation, and all of the constant terms-- the negative 3 and the positive 18 on the other side-- I like to always You don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right. And the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now, on the left-hand side, I have negative 9x minus 12x. So negative 9 minus 12, that's negative 21. Negative 21x minus 3 is equal to-- 12x minus 12x, well, that's just nothing. So I could write a 0 here, but I don't That was the whole point of subtracting the 12x from the left-hand side. And that is going to be equal to-- so on the right-hand side, we just are left with an 18. We are just left with that 18 here. These guys canceled out. Now, let's get rid of this negative 3 from the left-hand side. So on the left-hand side, we only have x terms, and on the right-hand side, we only have constant terms. So the best" + }, + { + "Q": "At 5:10, there should and maybe there is only one x for ( x^2=9 ) as, x = sqrt(9), which is equal to 3, ( by principal squareroot ), not -3!, -3 = -x right, as its the same number 3, & hence, -3 ^ 2 = 9 should always be written as -x ^ 2, for any*, any number even for infinity", + "A": "That is not correct. -3^2 = -9 and you cannot take the square root of a negative number. (-3)^2 = 9. If you have -3 = -x, that is the same as x = 3, so it still is the primary square root. The reason this is one place where we do not just take the primary square root is because we are talking about x intercepts of a quadratic, so the answer is x = +/- 3.", + "video_name": "mbc3_e5lWw0", + "timestamps": [ + 310 + ], + "3min_transcript": "And another way to think about it, it's the positive, this is going to be the positive square root. If someone wants the negative square root of nine, they might say something like this. They might say the negative, let me scroll up a little bit, they might say something like the negative square root of nine. Well, that's going to be equal to negative three. And what's interesting about this is, well, if you square both sides of this, of this equation, if you were to square both sides of this equation, what do you get? Well negative, anything negative squared becomes a positive. And then the square root of nine squared, well, that's just going to be nine. And on the right-hand side, negative three squared, well, negative three times negative three is positive nine. So, it all works out. Nine is equal, nine is equal to nine. And so this is an interesting thing, actually. Let me write this a little bit more algebraically now. the principal root of nine is equal to x. This is, there's only one possible x here that satisfies it, because the standard convention, what most mathematicians have agreed to view this radical symbol as, is that this is a principal square root, this is the positive square root, so there's only one x here. There's only one x that would satisfy this, and that is x is equal to three. Now, if I were to write x squared is equal to nine, now, this is slightly different. X equals three definitely satisfies this. This could be x equals three, but the other thing, the other x that satisfies this is x could also be equal to negative three, 'cause negative three squared is also equal to nine. So, these two things, these two statements, are almost equivalent, although when you're looking at this one, there's two x's that satisfy this one, because this is a positive square root. If people wanted to write something equivalent where you would have two x's that could satisfy it, you might see something like this. Plus or minus square root of nine is equal to x, and now x could take on positive three or negative three." + }, + { + "Q": "at 5:11 why did he put a 1/2 sign as the exponent", + "A": "He used 1/2 as the exponent to explain to us that polynomials can t have fractional or decimal exponents (although other numbers can have them).", + "video_name": "Vm7H0VTlIco", + "timestamps": [ + 311 + ], + "3min_transcript": "So in this first term the coefficient is 10. Lemme write this word down, coefficient. It's another fancy word, but it's just a thing that's multiplied, in this case, times the variable, which is x to seventh power. The first coefficient is 10. The next coefficient. Actually, lemme be careful here, because the second coefficient here is negative nine. We are looking at coefficients. The third coefficient here is 15. You can view this fourth term, or this fourth number, as the coefficient because this could be rewritten as, instead of just writing as nine, you could write it as nine x to the zero power. And then it looks a little bit clearer, like a coefficient. So, in general, a polynomial is the sum of a finite number of terms where each term has a coefficient, which I could represent with the letter A, being multiplied by a variable So, this right over here is a coefficient. It can be, if we're dealing... Well, I don't wanna get too technical. Positive, negative number. Could be any real number. We have our variable. And then the exponent, here, has to be nonnegative. Nonnegative integer. So here, the reason why what I wrote in red is not a polynomial is because here I have an exponent that is a negative integer. Let's give some other examples of things that are not polynomials. So, if I were to change the second one to, instead of nine a squared, if I wrote it as nine a to the one half power minus five, this is not a polynomial because this exponent right over here, it is no longer an integer; it's one half. the square root of a minus five. This also would not be a polynomial. Or, if I were to write nine a to the a power minus five, also not a polynomial because here the exponent is a variable; it's not a nonnegative integer. All of these are examples of polynomials. There's a few more pieces of terminology that are valuable to know. Polynomial is a general term for one of these expression that has multiple terms, a finite number, so not an infinite number, and each of the terms has this form. But there's more specific terms for when you have only one term or two terms or three terms. When you have one term, it's called a monomial. This is a monomial." + }, + { + "Q": "At 0:16, How would you turn that into a linear equation?", + "A": "Good Question. first take the change of your y and x points. which is y( 7) and x(4). now we need to find the slope. to find the slope lest divide our y difference by our x difference: 7/4 now we have our slope! now so far we have Y=7/4x+b at 2:30 Sal was confirming about the dotted line. as we know we are now trying to find the y-Intercept. looking at the graph we can see that the y-Intercept is -7 so now we get!( Drum roll!) Y=7/4X+(-7) or: Y=7/4X-7 Hope This Helps! =)", + "video_name": "wl2iQAuQl7Y", + "timestamps": [ + 16 + ], + "3min_transcript": "f is a linear function whose table of values is shown below. So they give us different values of x and what the function is for each of those x's. Which graphs show functions which are increasing at the same rate as f? So what is the rate at which f is increasing? When x increases by 4, we have our function increasing by 7. So we could just look for which of these lines are increasing at a rate of 7/4, 7 in the vertical direction every time we move 4 in the horizontal direction. And an easy way to eyeball that would actually be just to plot two points for f, and then see what that rate looks like visually. So if we see here when x is 0, f is negative 1. When x is 0, f is negative 1. So when x is 0, f is negative 1. And when x is 4, f is 6, so 1, 2, 3, 4, 5, 6, And two points specify a line. We know that it is a linear function. You can even verify it here. When we increase by 4 again, we increase our function by 7 again. We know that these two points are on f and so we get a sense of the rate of change of f. Now, when you draw it like that, it immediately becomes pretty clear which of these has the same rate of change of f. A is increasing faster than f. C is increasing slower. A is increasing much faster than f. C is increasing slower than f. B is decreasing, so that's not even close. But D seems to have the exact same inclination, the exact same slope, as f. So D is what we would go with. And we could even verify it, even if we didn't draw it in this way. Our change in f for a given change in x our function changed plus 7. It is equal to 7/4. And we can verify that on D, if we increase in the x-direction by 4, so we go from 4 to 8, then in the vertical direction we should increase by 7, so 1, 2, 3, 4, 5, 6, 7. And it, indeed, does increase at the exact same rate." + }, + { + "Q": "At 4:02, we go from d=5m/s * 1h to 5 m*h/s which is cool because we did 5m/s * h/1. But then he just writes 3600/1 s/h without any symbols. Are we just multiplying 5m*h/s * 3600/1 * s/h or what is happening here. Im not really used to adding things into a middle of equations without reasons like simplification etc.", + "A": "Right, there are 3600 seconds in an hour (because there are 60 seconds in a minute and 60 minutes in an hour so 60*60=3600 seconds in an hour). It s 3600 seconds/1 hour, or 3600 s/1 h.", + "video_name": "hIAdCTNi1S8", + "timestamps": [ + 242 + ], + "3min_transcript": "Now you're saying, \"OK, that's cute and everything, \"but this seems like a little bit of too much overhead \"to worry about when I'm just doing \"a simple formula like this.\" But what I want to show you is that even with a simple formula like distance is equal to rate times time, what I just did could actually be quite useful, and this thing that I'm doing is actually called dimensional analysis. It's useful for something as simple as distance equals rate times time, but as you go into physics and chemistry and engineering, you'll see much, much, much more, I would say, hairy formulas. When you do the dimensional analysis, it makes sure that the math is working out right. It makes sure that you're getting the right units. But even with this, let's try a slightly more complicated example. Let's say that our rate is, let's say, let's keep our rate at 5 meters per second, but let's say that someone gave us the time. Instead of giving it in seconds, they give it in hours, so they say the time is equal to 1 hour. We're going to get distance is equal to 5 meters per second, 5 meters per second times time, which is 1 hour, times 1 hour. What's that going to give us? The 5 times the 1, so we multiply the 5 times the 1, that's just going to give us 5. But then remember, we have to treat the units algebraically. We're going to do our dimensional analysis, so it's 5, so we have meters per second times hours, times hours, or you could say 5 meter hours per second. Well, this doesn't look like a ... This isn't a set of units that we know that makes sense to us. This doesn't feel like our traditional units of distance, so we want to cancel this out in some way. It might jump out of you, well, if we can get rid of this hours, if we can express it in terms of seconds, then that would cancel here, and we'd be left with just the meters, which is a unit of distance that we're familiar with. So how do we do that? We'd want to multiply this thing and seconds in the numerator, times essentially seconds per hour. How many seconds are there per hour? Well, there are 3,600 ... Let me do this in a ... I'll do it in this color. There are 3,600 seconds per hour, or you could even say that there are 3,600 seconds for every 1 hour. Now when you multiply, these hours will cancel with these hours, these seconds will cancel with those seconds, and we are left with, we are left with 5 times 3,600. What is that? That's 5 times 3,000 would be 15,000, 5 times 600 is another 3,000, so that is equal to 18,000. The only units that we're left with, we just have the meters there. 18- Oh, it's 18,000, 18,000, 18,000 meters." + }, + { + "Q": "At 2:30 and 2:37, didn't Sal mean a1, a2 and a4 rather than a1, a2 and aN?", + "A": "I believe so. Notice that he switched back. His 4 s do look like n s.", + "video_name": "CkQOCnLWPUA", + "timestamps": [ + 150, + 157 + ], + "3min_transcript": "Two videos ago we asked ourselves if we could find the basis for the columns space of A. And I showed you a method of how to do it. You literally put A in reduced row echelon form, so this matrix R is just a reduced row echelon form of A. And you look at its pivot columns, so this is a pivot column. It has a 1 and all 0's, this is a pivot column, 1 and all 0's, and the 1 is the leading non-zero term in its row. And this is a pivot column, let me circle them, these guys are pivot columns, and this guy's a pivot column right there. You look at those in the reduced row echelon form of the matrix, and the corresponding columns in the original matrix will be your basis. So this guy, this guy, so the first, second, and forth columns. So if we call this a1, this is a2, and let's call this a4, this would be a3, and this is a5. So we could say that a1, a2, and a4 are a basis for the And I didn't show you why two videos ago. I just said this is how you do it. You have to take it as a bit of an article of faith. Now in order for these to be a basis, two things have to be true. They have to be linearly independent, and I showed you in the very last video, the second in our series dealing I showed you that by the fact that this guy is r1, this guy is r2, and this guy is r4, it's clear that these guys are linearly independent. They each have a 1 in a unique entry, and the rest of their entries are 0. We're looking at three pivot columns right now, but it's true if we had n pivot columns. That each pivot column would have a 1 in a unique place, and all the other pivot columns would have 0 in that entry. So there's no way that the other pivot columns, any to each of them. So these are definitely linearly independent. And I showed you in the last video that if we know that these are linearly independent, we do know that they are, given that R has the same null space as A, we know that these guys have to be linearly independant, I did that in the very last video. Now the next requirement for a basis, we checked this one off, is to show that a1 a2 and an, that their span equals the column space of A. Now the column space of A is a span of all five of these vectors, so I had to throw a3 in there and a5. So to show that just these three vectors by themselves span our column space, we just have to show that I can represent a3 and a5 as linear combinations" + }, + { + "Q": "At 3:30 just when I learned that parenthesis most be solved first. But, I also learned about the Distributed Property. SMH", + "A": "Whenever you have parenthesis, do the math inside the parenthesis first, then use the distributive property. 4(11-3)= 4(8)=32 If you want something to help you remember the order of operations, try PEMDAS. Parentheses Exponents (if necessary) Multiplication (this includes the distributive property) Division Addition Subtraction", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 210 + ], + "3min_transcript": "So is equal to 12. You could imagine that I did the reverse distributive property out here. I factored out an x. But the way my head thinks about it is, I have 1.3 of something minus 0.7 of something, that's going to be equal to 1.3 minus 0.7 of those somethings, that x. And of course 1.3 minus 0.7 is 0.6 times x of my somethings is equal to 12. And now, this looks just like one of the problems we did in the last video. We have a coefficient times x is equal to some other number. Well, let's divide both sides of this equation by that coefficient. Divide both sides by 0.6. So the left-hand side will just become an x. X is equal to-- and what is 12 divided by 0.6? 0.6 goes into 12-- let's add some decimal points here-- 6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12. We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2." + }, + { + "Q": "In 9:09, you write an equation which I don't understand. Could you explain it again, please?", + "A": "But why does he put the twelve in brackets then?", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 549 + ], + "3min_transcript": "that I can do it two different ways. But as long as I do legitimate operations, I should get the same answer. So the first way I'm going to do it, is I'm going to multiply both sides of this equation by the inverse of 5/12. So I'm going to multiply both sides by 12 over 5. Because I wanted to get rid of this 5/12 on It makes everything look a little bit messy. And I multiply it by 12 over 5, because these are going to The 5 and the 5 cancel out, the 12 and the 12 cancel out. So the left-hand side of my equation becomes q minus 7 is equal to the right-hand side, 2/3 times 5/12. If you divide the 12 by 3, you get a 4. You divide the 3 by 3, you get a 1. So 2 times 4 is 8 over 5. And now we can add 7 to both sides of this equation. So let's add-- I want to do that in a different color-- add 7 to both sides of this equation. That was the whole point of adding the 7. And you are left with q is equal to 8/5 plus 7. Or we could write 8/5 plus 7 can be written as 35/5. And so this is going to be equal to 8-- well, the denominator is 5. 8 plus 35 is 43. So my answer, going this way, is q is equal to 43/5. And I said I would do it two ways. Let's do it another way. So let me write the same problem down. So I have 5/12-- actually, let me just do it a completely different way. Let me write it the way they wrote it. 5 times q minus 7, over 12 is equal to 2/3. Let me just get rid of the 12 first. Let me multiply both I just don't like that 12 sitting there, so I'm going to multiply both sides by 12. So these are going to cancel out, and you're going to be left with 5 times q minus 7 is equal to 2/3 times 12. That's the same thing as 24 over 3. So this is, let me write this. 2 over 3 times 12 over 1 is equal to-- if you divide that by 3, you get a 4, divide that by 3, you get a 1-- is equal to 8. So you get 5 times q minus 7 is equal to 8. And then instead of dividing both sides by 5, which would get us pretty close to what we were doing over here, let me distribute this 5, I just want to show you, you can do it multiple legitimate ways. So 5 times q is 5q. 5 times negative 7 is minus, or negative 35, is equal to 8. 5q minus 35 is equal to 8. Now, if I want to get rid of that minus 35, or that negative 35, the best way to do it is to" + }, + { + "Q": "8:19 How can 7 be written as 35/5?", + "A": "35/5 is the same thing as 7.", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 499 + ], + "3min_transcript": "inverse of this coefficient. So I multiply 8 over 5 times 5/8s. If I do it to the left-hand side, I have to do it to the right-hand side. 8 over 5. I multiplied by 8 over 5 so that those cancel out and those cancel out. And you are left with s is equal to-- right, this is just a 1-- is equal to-- well, the 5's we can divide. Divide the numerator and the denominator by 5. Divide the numerator by 2 and the denominator by 2. You're left with-- sorry, divide the denominator by 2, you get 6 divided by 2 is 3. You're left with 4/3. s is equal to 4/3. Let's do one more of these. So here I have 5 times q minus 7 over 12 is equal to 2/3. So let me write this. And I could rewrite this as just 5 over 12 times q minus 7 is equal to 2/3. that I can do it two different ways. But as long as I do legitimate operations, I should get the same answer. So the first way I'm going to do it, is I'm going to multiply both sides of this equation by the inverse of 5/12. So I'm going to multiply both sides by 12 over 5. Because I wanted to get rid of this 5/12 on It makes everything look a little bit messy. And I multiply it by 12 over 5, because these are going to The 5 and the 5 cancel out, the 12 and the 12 cancel out. So the left-hand side of my equation becomes q minus 7 is equal to the right-hand side, 2/3 times 5/12. If you divide the 12 by 3, you get a 4. You divide the 3 by 3, you get a 1. So 2 times 4 is 8 over 5. And now we can add 7 to both sides of this equation. So let's add-- I want to do that in a different color-- add 7 to both sides of this equation. That was the whole point of adding the 7. And you are left with q is equal to 8/5 plus 7. Or we could write 8/5 plus 7 can be written as 35/5. And so this is going to be equal to 8-- well, the denominator is 5. 8 plus 35 is 43. So my answer, going this way, is q is equal to 43/5. And I said I would do it two ways. Let's do it another way. So let me write the same problem down. So I have 5/12-- actually, let me just do it a completely different way. Let me write it the way they wrote it. 5 times q minus 7, over 12 is equal to 2/3. Let me just get rid of the 12 first. Let me multiply both" + }, + { + "Q": "@4:42 I didn't understand how you got 3/2. The way i did it was i divided 3 by 2 and got 1.5 .", + "A": "they re really the same thing, 1.5=3/2", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 282 + ], + "3min_transcript": "We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something." + }, + { + "Q": "at 8:19 why is seven written as 35? I thought it would be 8/5 plus 7/1", + "A": "Seven is written as 35/5, not 35. Remember that if I have 35 and divide it by 5 I get 7. It s the same as when you re reducing a fraction: When you reduce a fraction, you divide the top and bottom by the same number. You can also multiply the top and bottom by the same number. In this case, Sal multiplied the top and bottom of 7/1 by 5, which gives (7*5)/(1*5) = 35/5. Hope this helps", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 499 + ], + "3min_transcript": "inverse of this coefficient. So I multiply 8 over 5 times 5/8s. If I do it to the left-hand side, I have to do it to the right-hand side. 8 over 5. I multiplied by 8 over 5 so that those cancel out and those cancel out. And you are left with s is equal to-- right, this is just a 1-- is equal to-- well, the 5's we can divide. Divide the numerator and the denominator by 5. Divide the numerator by 2 and the denominator by 2. You're left with-- sorry, divide the denominator by 2, you get 6 divided by 2 is 3. You're left with 4/3. s is equal to 4/3. Let's do one more of these. So here I have 5 times q minus 7 over 12 is equal to 2/3. So let me write this. And I could rewrite this as just 5 over 12 times q minus 7 is equal to 2/3. that I can do it two different ways. But as long as I do legitimate operations, I should get the same answer. So the first way I'm going to do it, is I'm going to multiply both sides of this equation by the inverse of 5/12. So I'm going to multiply both sides by 12 over 5. Because I wanted to get rid of this 5/12 on It makes everything look a little bit messy. And I multiply it by 12 over 5, because these are going to The 5 and the 5 cancel out, the 12 and the 12 cancel out. So the left-hand side of my equation becomes q minus 7 is equal to the right-hand side, 2/3 times 5/12. If you divide the 12 by 3, you get a 4. You divide the 3 by 3, you get a 1. So 2 times 4 is 8 over 5. And now we can add 7 to both sides of this equation. So let's add-- I want to do that in a different color-- add 7 to both sides of this equation. That was the whole point of adding the 7. And you are left with q is equal to 8/5 plus 7. Or we could write 8/5 plus 7 can be written as 35/5. And so this is going to be equal to 8-- well, the denominator is 5. 8 plus 35 is 43. So my answer, going this way, is q is equal to 43/5. And I said I would do it two ways. Let's do it another way. So let me write the same problem down. So I have 5/12-- actually, let me just do it a completely different way. Let me write it the way they wrote it. 5 times q minus 7, over 12 is equal to 2/3. Let me just get rid of the 12 first. Let me multiply both" + }, + { + "Q": "At 2:11 Khan says that 0.6 goes into 12 two times but doesn't 6.0 not 0.6 go into 12 two times..?", + "A": "he moved the decimal point one place to the right, so instead of it being .6 into 12 it was 6 into 120. He was doing the first part of the problem , 6 into 12 goes twice, so you put a 2, then six goes into 0 zero times so you put a zero. The 2 and the zero together is 20.", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 131 + ], + "3min_transcript": "Let's do a few more examples of solving equations. And I think you're going to see that these equations require a few more steps than the ones we did in the last video. But the fun thing about these is that there's more than one way to do it. But as long as you do legitimate steps, as long as anything you do to the left-hand side, you also do to the right-hand side, you should move in the correct direction, or you shouldn't get the wrong answer. So let's do a couple of these. So the first one says-- I'll rewrite it-- 1.3 times x minus 0.7 times x is equal to 12. Well, here the first thing that my instinct is to do, is to merge these two terms. Because I have 1.3 of something minus 0.7 of that same something. This is the same variable. If I have 1.3 apples minus 0.7 apples, well, why don't I subtract 0.7 from 1.3? And I will get 1.3 minus 0.7 x's, or apples, or whatever So is equal to 12. You could imagine that I did the reverse distributive property out here. I factored out an x. But the way my head thinks about it is, I have 1.3 of something minus 0.7 of something, that's going to be equal to 1.3 minus 0.7 of those somethings, that x. And of course 1.3 minus 0.7 is 0.6 times x of my somethings is equal to 12. And now, this looks just like one of the problems we did in the last video. We have a coefficient times x is equal to some other number. Well, let's divide both sides of this equation by that coefficient. Divide both sides by 0.6. So the left-hand side will just become an x. X is equal to-- and what is 12 divided by 0.6? 0.6 goes into 12-- let's add some decimal points here-- 6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12." + }, + { + "Q": "At 1:40 shouldn't it be (y^3)/2 (instead of (y^3)/3)?", + "A": "Okay, this is corrected later at 2:46", + "video_name": "0pv0QtOi5l8", + "timestamps": [ + 100 + ], + "3min_transcript": "Welcome back. In the last video we were just figuring out the volume under the surface, and we had set up these integral bounds. So let's see how to evaluate it now. And look at this. I actually realized that I can scroll things, which is quite useful because now I have a lot more board space. So how do we evaluate this integral? Well, the first integral I'm integrating with respect to x. I'm adding up the little x sums. So I'm forming this rectangle right here. Or you could kind of view it I'm holding y constant and integrating along the x-axis. I should switch colors. So what's the antiderivative of x y squared with respect to x? Well it's just x squared over 2. And then I have the y squared-- that's just a constant-- all over 2. And I'm going to evaluate that from x is equal to 1 to x is equal to the square root of y, which you might be daunted by. But you'll see that it's actually not that bad once you evaluate them. This is y is equal to 0 to y is equal 1. dy. Now, if x is equal to 1 this expression becomes y squared over 2. Right? y squared over 2, minus-- now if x is equal to square root of y, what does this expression become? If x is equal to the square root of y, then x squared is just y. And then y times y squared is y to the third. Right? So it's y to the third over 3. And now I take the integral with respect to y. So now I sum up all of these rectangles in the y direction. 0, 1. This is with respect to y. And that's cool, right? Because when you take the first integral with respect to x you end up with a function of y anyway, so you might as well have your bounds as functions of y's. It really doesn't make it any more difficult. But anyway, back to the problem. What is the antiderivative of y squared over 2 minus Well the antiderivative of y squared-- and you have to divide by 3, so it's y cubed over 6. Minus y to the fourth-- you have to divide by 4. Minus y to the fourth over-- did I mess up some place? No, I think this is correct. y to the fourth over 12. How did I get a 3 here? That's where I messed up. This is a 2, right? Let's see. x is square root of y. Yeah, this is a 2. I don't know how I ended up. Square root of y squared is y, times y squared y to the third over 2. Right. And then when I take the integral of this it's 4 times 2." + }, + { + "Q": "I'm a little confused about Sal's second table that he draws in green later in the video. Does this signify that the inputs and outputs have changed? At 2:56 you can see the table has written that x is 5, and f \u00e2\u0081\u00bb\u00c2\u00b9(x) is -9. Is 5 now the input, and -9 is the output? Or is x the range and f \u00e2\u0081\u00bb\u00c2\u00b9(x) is the domain?", + "A": "In the inverse function, the domain and the range are switched. domain->range range->domain", + "video_name": "KzaPBzFFLRM", + "timestamps": [ + 176 + ], + "3min_transcript": "With that in mind, let's see if we can evaluate something like f inverse of 8. What is that going to be? I encourage you to pause the video and try to think about it. So f of x, just as a reminder of what functions do, f of x is going to map from this domain, from a value in its domain to a corresponding value in the range. So this is what f does, this is domain... and this right over here is the range. Now f inverse, if you pass it, the value and the range, it'll map it back to the corresponding value in the domain. But how do we think about it like this? so if this was 8, we'd have to say, well, what mapped to 8? We see here f of 9 is 8, so f inverse of 8 is going to be equal to 9. If it makes it easier, we could construct a table, where I could say x and f inverse of x, and what I'd do is swap these two columns. f of x goes from -9 to 5, f inverse of x goes from 5 to -9. All I did was swap these two. Now we're mapping from this to that. So f inverse of x is going to map from 7 to -7. Notice, instead of mapping from this thing to that thing, we're now going to map from that thing to this thing. It's going to map from -7 to 6. It's going to map from 8 to 9, and it's going to map from 12 to 11. Looks like I got all of them, yep. So all I did was swap these columns. The f inverse maps from this column to that column. So I just swapped them out. Now it becomes a little clearer. You see it right here, f inverse of 8, if you input 8 into f inverse, you get 9. Now we can use that to start doing fancier things. We can evaluate something like f of f inverse of 7. f of f inverse of 7." + }, + { + "Q": "At 8:34, she sang something about unary. i dont that that makes sense. help me!", + "A": "Pick a number. 13 for example. In the decimal system (base ten) it is written 13, 1 ten and three 1s. In the binary system (base two) it is written 1101, one 8, one 4, no 2s, one 1. In the unary system (base one) it is written 0000000000000, one 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, and another 1.", + "video_name": "sxnX5_LbBDU", + "timestamps": [ + 514 + ], + "3min_transcript": "backwards E, half a heart on a plate, and a simple, boring, short and straight line. So yeah. If you want to add up all the numbers between 1 and anything, this trick works. For example, all the numbers between 1 and 12. 1 plus 12 is 13. 2 plus 11 is 13. All the way in to 6 plus 7. That's 6 times 13, which I do my head as 60 plus 18 equals 78. So 78 is the 12th triangular number, while 5,050 is the 100th triangular number. At least it's not \"100 Days of Christmas.\" I'll stick with \"Bottles of Beer.\" [SINGING] On the tenth day of Christmas, my true love gave to me, the base of our Arabic numeral system, the base of a nonary numeral and the base of senary, and a number five, quaternary's base, ternary's base also, and binary two, and the base of unary. Here's my favorite way of visualizing the triangular numbers. Say you've got them in this configuration where they make a nice right equilateral triangle, or half a square. Finding the area of a square is easy, because you just square the length of it. In this case, 12 times 12. And the triangle is half of that. Only not really, because half the square means you only get half of this diagonal, so you've got to add back in the other half. But that's easy because there's 12 things in the diagonal, So to get the n-th triangular number, just take n squared over 2 plus n over 2, or n squared plus n over 2. [SINGING] On the eleventh day of Christmas, my true love gave to me the number my amp goes up to, the number of fingers on my hands, the German word for no, what I did after I eat, the number of heads on a hydra, at least until you start cutting them off, the number of strings on a guitar, the number I like to do high, the amount of horsemen of the Apocalypse, the number of notes in a triad, the number of pears in a pair of pears, and the number of partridges in a pear tree. [SINGING] On the twelfth day of Christmas," + }, + { + "Q": "at 0:40 why is y make x =0", + "A": "The y-intercept just means where the line crosses the y axis. Any point on the y-axis has a x value of 0, because you aren t moving left or right, you are just going either up or down. So the y-intercept is where the line crosses the y axis, and that point will have a x value of 0.", + "video_name": "405boztgZig", + "timestamps": [ + 40 + ], + "3min_transcript": "We're told to find the x- and y-intercepts for the graph of this equation: 2 y plus 1/3x is equal to 12. And just as a bit of a refresher, the x-intercept is the point on the graph that intersects the x-axis. So we're not above or below the x-axis, so our y value must be equal to 0. And by the exact same argument, the y-intercept occurs when we're not to the right or the left of the y-axis, so that's when x is equal to 0. So let's set each of these values to 0 and then solve for what the other one has to be at that point. So for the x-intercept, when y is equal to 0, let's solve this. So we get 2 times 0, plus 1/3x is equal to 12. I just set y is equal to 0 right there, right? I put 0 for y. Well, anything times 0 is just 0, so you're just left with 1/3x is equal to 12. sides by 1/3, or we can multiply both sides by the reciprocal of 1/3. And the reciprocal of 1/3 is 3, or you can even think of it as 3 over 1. So times 3 over 1. And so we're left with 3 times 1/3, that just cancels out, so you're left with x is equal to 12 times 3, or x is equal to 36. So when y is equal to 0, x is 36. So the point 36 comma 0 is on the graph of this equation. And this is also the x-intercept. Now, let's do the same thing for the y-intercept. So let's set x equal 0, so you get 2y plus 1/3, times 0 is equal to 12. Once again, anything times 0 is 0. So that's 0, and you're just left with 2y is equal to 12. with y is equal to 12 over 2, is 6. So the y-intercept is when x is equal to 0 and y is equal to 6. So let's plot these two points. I'll just do a little hand-drawn graph, and make it clear what the x- and the y-intercepts are. So let me draw-- that's my vertical axis, and that is my horizontal axis-- and we have the point 36 comma 0. So this is the origin right here, that's the x-axis, that's the y-axis. The point 36 comma 0 might be all the way over here. So that's the point 36 comma 0. And if that's 36, then the point 0, 6 might be right about there. So that's the point 0, 6. And the line will look something like this." + }, + { + "Q": "Why are you multiplying it by 0? 0:22 of the video.", + "A": "I didn t see any multiplication by zero at 0:22. There is some at 0:49. At this point Sal is looking for the x-intercept (where the line crosses the x-axis) which is when y is zero, so he s plugging in 0 for y in the given equation 2y + \u00e2\u0085\u0093x = 12 and solving for x.", + "video_name": "405boztgZig", + "timestamps": [ + 22 + ], + "3min_transcript": "We're told to find the x- and y-intercepts for the graph of this equation: 2 y plus 1/3x is equal to 12. And just as a bit of a refresher, the x-intercept is the point on the graph that intersects the x-axis. So we're not above or below the x-axis, so our y value must be equal to 0. And by the exact same argument, the y-intercept occurs when we're not to the right or the left of the y-axis, so that's when x is equal to 0. So let's set each of these values to 0 and then solve for what the other one has to be at that point. So for the x-intercept, when y is equal to 0, let's solve this. So we get 2 times 0, plus 1/3x is equal to 12. I just set y is equal to 0 right there, right? I put 0 for y. Well, anything times 0 is just 0, so you're just left with 1/3x is equal to 12. sides by 1/3, or we can multiply both sides by the reciprocal of 1/3. And the reciprocal of 1/3 is 3, or you can even think of it as 3 over 1. So times 3 over 1. And so we're left with 3 times 1/3, that just cancels out, so you're left with x is equal to 12 times 3, or x is equal to 36. So when y is equal to 0, x is 36. So the point 36 comma 0 is on the graph of this equation. And this is also the x-intercept. Now, let's do the same thing for the y-intercept. So let's set x equal 0, so you get 2y plus 1/3, times 0 is equal to 12. Once again, anything times 0 is 0. So that's 0, and you're just left with 2y is equal to 12. with y is equal to 12 over 2, is 6. So the y-intercept is when x is equal to 0 and y is equal to 6. So let's plot these two points. I'll just do a little hand-drawn graph, and make it clear what the x- and the y-intercepts are. So let me draw-- that's my vertical axis, and that is my horizontal axis-- and we have the point 36 comma 0. So this is the origin right here, that's the x-axis, that's the y-axis. The point 36 comma 0 might be all the way over here. So that's the point 36 comma 0. And if that's 36, then the point 0, 6 might be right about there. So that's the point 0, 6. And the line will look something like this." + }, + { + "Q": "At around (0:17) he says that the points aren't above or below the x axis....how do you figure out whether or not the point is above or below the x axis?", + "A": "Hi Alice. What Sal means by the x-intercept not being above or below the x-axis is that this point is the point on the line where the line intercepts the x-axis. Finding this means that we have to set y equal to 0, and doing so means that we do not move above or below the x-axis, and that we only move along the x-axis. Hope that helped!", + "video_name": "405boztgZig", + "timestamps": [ + 17 + ], + "3min_transcript": "We're told to find the x- and y-intercepts for the graph of this equation: 2 y plus 1/3x is equal to 12. And just as a bit of a refresher, the x-intercept is the point on the graph that intersects the x-axis. So we're not above or below the x-axis, so our y value must be equal to 0. And by the exact same argument, the y-intercept occurs when we're not to the right or the left of the y-axis, so that's when x is equal to 0. So let's set each of these values to 0 and then solve for what the other one has to be at that point. So for the x-intercept, when y is equal to 0, let's solve this. So we get 2 times 0, plus 1/3x is equal to 12. I just set y is equal to 0 right there, right? I put 0 for y. Well, anything times 0 is just 0, so you're just left with 1/3x is equal to 12. sides by 1/3, or we can multiply both sides by the reciprocal of 1/3. And the reciprocal of 1/3 is 3, or you can even think of it as 3 over 1. So times 3 over 1. And so we're left with 3 times 1/3, that just cancels out, so you're left with x is equal to 12 times 3, or x is equal to 36. So when y is equal to 0, x is 36. So the point 36 comma 0 is on the graph of this equation. And this is also the x-intercept. Now, let's do the same thing for the y-intercept. So let's set x equal 0, so you get 2y plus 1/3, times 0 is equal to 12. Once again, anything times 0 is 0. So that's 0, and you're just left with 2y is equal to 12. with y is equal to 12 over 2, is 6. So the y-intercept is when x is equal to 0 and y is equal to 6. So let's plot these two points. I'll just do a little hand-drawn graph, and make it clear what the x- and the y-intercepts are. So let me draw-- that's my vertical axis, and that is my horizontal axis-- and we have the point 36 comma 0. So this is the origin right here, that's the x-axis, that's the y-axis. The point 36 comma 0 might be all the way over here. So that's the point 36 comma 0. And if that's 36, then the point 0, 6 might be right about there. So that's the point 0, 6. And the line will look something like this." + }, + { + "Q": "7:43. Isn't 1 -3 -2, not 2, or are you saying the absolute value?", + "A": "Sal is using absolute value in this video, and that is why your a little confused. \u00f0\u009f\u0098\u0089", + "video_name": "GdIkEngwGNU", + "timestamps": [ + 463 + ], + "3min_transcript": "It's just gonna be one. And you see that here visually. This point is just one away. It's just one away from three. This point is just one away from three. Four minus three is one. Absolute value of that is one. This point is just one away from three. Four minus three, absolute value. That's another one. every data point was exactly one away from the mean. And we took the absolute value so that we don't have negative ones here. We just care how far it is in absolute terms. So you have four data points. Each of their absolute deviations is four away. So the mean of the absolute deviations are one plus one plus one plus one, which is four, over four. So it's equal to one. One way to think about it is saying, on average, the mean of the distances of these points away from the actual mean is one. And that makes sense because all of these are exactly one away from the mean. Now, let's see how, what results we get for this data set right over here. Let me actually get some space over here. At any point, if you get inspired, I encourage you to calculate the Mean Absolute Deviation on your own. So let's calculate it. The Mean Absolute Deviation here, I'll write MAD, is going to be equal to ... Well, let's figure out the absolute deviation of each of these points from the mean. It's the absolute value of one minus three, that's this first one, plus the absolute deviation, so one minus three, that's the second one, then plus the absolute value of six minus three, that's the six, then we have the four, plus the absolute value of four minus three. Then we have four points. So one minus three is negative two. Absolute value is two. And we see that here. This is two away from three. We just care about absolute deviation. We don't care if it's to the left or to the right. Then we have another one minus three is negative two. It's absolute value, so this is two. That's this. This is two away from the mean. Then we have six minus three. Absolute value of that is going to be three. We see this six is three to the right of the mean. We don't care whether it's to the right or the left. And then four minus three. Four minus three is one, absolute value is one. And we see that. It is one to the right of three. And so what do we have? We have two plus two is four, plus three is seven, plus one is eight, over four, which is equal to two. So the Mean Absolute Deviation ... It fell off over here. Here, for this data set, the Mean Absolute Deviation is equal to two, while for this data set, the Mean Absolute Deviation is equal to one. And that makes sense. They have the exact same means. They both have a mean of three. But this one is more spread out. The one on the right is more spread out because, on average, each of these points are two away from three, while on average, each of these points are one away from three. The means of the absolute deviations on this one is one." + }, + { + "Q": "When Sal say's: du is going to be equal to the derivative of x^4+7 \"with respect to x\" (1:16). What does he mean with the last bit? Would it be possible to have a different variable than x and if so what would happen then?", + "A": "With respect to x means that the equation is being derivated in the form d(u)/d(x), as the equation is written in terms of x. It is possible to have a variable other than x- If we had an equation that was, for example, y=3z-1, we would derivate y with respect to z because the equation is written in terms of z. hence dy/dz = 3. Hope this helps!", + "video_name": "Zp5z0wa0kgo", + "timestamps": [ + 76 + ], + "3min_transcript": "So we want to take the indefinite integral of 4x^3 over x^4 plus 7 dx. So how can we tackle this? It seems like a hairy integral. Now the key inside here is to realize you have this expression x^4 + 7 and you also have its derivative up here. The derivative of x^4 plus 7 is equal to 4x^3. Derivative of x^4 is 4x^3; derivative of 7 is just 0. So that's a big clue that u-substitution might be the tool of choice here. U-sub -- I'll just write u- -- I'll write the whole thing. U-Substitution could be the tool of choice. So given that, what would you want to set your u equal to? And I'll let you think about that 'cause it can figure out this part and the rest will just boil down to a fairly straightforward integral. Well, you want to set u be equal to the expression that you have its derivative laying around. So we could set u equal to x^4 plus 7. Now, what is du going to be equal to? so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du?" + }, + { + "Q": "At 3:10 why does Sal use the \"ln\"? i know ln means natural log but why is this used?", + "A": "All the formulas in calculus involving log use the log with base e or ln. The formula used here is \u00e2\u0088\u00ab (1/x) = ln|x|", + "video_name": "Zp5z0wa0kgo", + "timestamps": [ + 190 + ], + "3min_transcript": "so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du? and we use the absolute value so it'll be defined even for negative use -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!" + }, + { + "Q": "At 2:56, I'm really confused about where the 1 came from. Can someone please explain? Thank you", + "A": "Example: 5/3 = (1/3)*5, right? You had du/u and did the same thing as above, so du/u = (1/u)*du.", + "video_name": "Zp5z0wa0kgo", + "timestamps": [ + 176 + ], + "3min_transcript": "so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du? and we use the absolute value so it'll be defined even for negative use -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!" + }, + { + "Q": "At 3:24 where does the du go? I do understand that 1/u = ln |u| then we put the constant +C . Bu I can not seem to understand what happens to du and where it goes...? Thank you!", + "A": "It disappears the same way dx does when you do a regular integration without a u-substitution. For example, the integral of x dx is (x^2)/2 + C, and the integral of (1/x)dx is ln|x| + C. We re using the same process when we integrate after a u-substitution, but now we re integrating with respect to u, so the integration needs du to work instead of dx.", + "video_name": "Zp5z0wa0kgo", + "timestamps": [ + 204 + ], + "3min_transcript": "so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du? and we use the absolute value so it'll be defined even for negative use -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!" + }, + { + "Q": "At 10:45 he takes the \u00e2\u0088\u009a200 and converts it to 10\u00e2\u0088\u009a2. That lost me completely. What math should I learn so that makes sense? Thank you in advance!", + "A": "The key to simplifying square roots is to take out any perfect squares and leave any non perfect squares inside. So if we notice that 200 = 100 \u00e2\u0080\u00a2 2, we see that 10^2 can be taken out as a 10 and 2 has to stay in. If we prime factor it, we have 2 \u00e2\u0080\u00a2 2 \u00e2\u0080\u00a2 2 \u00e2\u0080\u00a2 5 \u00e2\u0080\u00a2 5, and again \u00e2\u0088\u009a200 = \u00e2\u0088\u009a4 \u00e2\u0080\u00a2 \u00e2\u0088\u009a2 \u00e2\u0080\u00a2 \u00e2\u0088\u009a25 or 2 \u00e2\u0080\u00a2 5 \u00e2\u0088\u009a2", + "video_name": "E4HAYd0QnRc", + "timestamps": [ + 645 + ], + "3min_transcript": "So this is going to be--all right, this is 10/5, which is equal to 2. So the variance here-- let me make sure I got that right. Yes, we have 10/5. So the variance of this less-dispersed data set is a lot smaller. The variance of this data set right here is only 2. So that gave you a sense. That tells you, look, this is definitely a less-dispersed data set then that there. Now, the problem with the variance is you're taking these numbers, you're taking the difference between them and the mean, then you're squaring it. It kind of gives you a bit of an arbitrary number, and if you're dealing with units, let's say if these are distances. So this is negative 10 meters, 0 meters, 10 meters, this is 8 meters, so on and so forth, then when you square it, you get your variance in terms of meters squared. It's kind of an odd set of units. So what people like to do is talk in terms of standard or the square root of sigma squared. And the symbol for the standard deviation is just sigma. So now that we've figured out the variance, it's very easy to figure out the standard deviation of both of these The standard deviation of this first one up here, of this first data set, is going to be the square root of 200. The square root of 200 is what? The square root of 2 times 100. This is equal to 10 square roots of 2. That's that first data set. Now the standard deviation of the second data set is just going to be the square root of its variance, which is just 2. this first data set. This is 10 roots of 2, this is just the root of 2. So this is 10 times the standard deviation. And this, hopefully, will make a little bit more sense. This has 10 times more the standard deviation than this. And let's remember how we calculated it. Variance, we just took each data point, how far it was away from the mean, squared that, took the average of those. Then we took the square root, really just to make the units look nice, but the end result is we said that that first data set has 10 times the standard deviation as the second data set. So let's look at the two data sets. This has 10 times the standard deviation, which makes sense" + }, + { + "Q": "i have doubt about these kind of linear equation which at 7:12 have many solution due to rank of matrix is 2 and order of matrix is 3 * 2", + "A": "No, it s a simple system of two variables. There is one solution for A and another for B.", + "video_name": "hbJ2o9EUmJ0", + "timestamps": [ + 432 + ], + "3min_transcript": "" + }, + { + "Q": "at 5:21 you said that unbiased variance = (n-1)*\u00cf\u0083\u00c2\u00b2/n. we know that (n-1)/n is a increasing function. So how come value decrease as n at 9 compared to at 8", + "A": "That s because the graph is from randomly picked samples and there is some error in this process.", + "video_name": "Cn0skMJ2F3c", + "timestamps": [ + 321 + ], + "3min_transcript": "while the bluer dots are the ones of a larger sample size. You see here these two little, I guess the tails ,so to speak, of this hump, that these ends, are more of a reddish color. that most of the blueish or the purplish dots are focused right in the middle right over here, that they are giving us better estimates. There are some red ones here, and that's why it gives us that purplish color, but out here on these tails, it's almost purely some of these red. Every now and then by happenstance you get a little blue one, but it's disproportionately far more red, which really makes sense when you have a smaller sample size, you are more likely to get a sample mean that is a bad estimate of the population mean, that's far from the population mean, and you're more likely to significantly underestimate the sample variance. Now this next chart really gets to the meat of the issue, that for each of these sample sizes, so this right over here for sample size two, if we keep taking sample size two, and we keep calculating the biased sample variances and dividing that by the population variance, and finding the mean over all of those, you see that over many, many, many trials, and many, many samples of size two, that that biased sample variance over population variance, it's approaching half of the true population variance. When sample size is three, it's approaching 2/3, 66 point six percent, of the true population variance. When sample size is four, it's approaching 3/4 of the true population variance. So we can come up with the general theme that's happening. When we use the biased estimate, we're not approaching the population variance. We're approaching n minus one over n When n was two, this approached 1/2. When n is three, this is 2/3. When n is four, this is 3/4. So this is giving us a biased estimate. So how would we unbias this? Well, if we really want to get our best estimate of the true population variance, not n minus one over n times the population variance, we would want to multiply, I'll do this in a color I haven't used yet, we would want to multiply times n over n minus one. to get an unbiased estimate. Here, these cancel out and you are just left with your population variance. That's what we want to estimate. Over here you are left with our unbiased estimate of population variance," + }, + { + "Q": "Mr. Khan,\nAt 2:52 of the video you mentioned how to write domain. Does {x/x\u00e2\u0082\u00acR} work too to write the domain because I was taught this way?", + "A": "Yes, you could use that when the domain = (-infinity, infinity). The domain and range are sets, so either interval notation is used to describe the set. Or, you could also use set notation which is what you have.", + "video_name": "-DTMakGDZAw", + "timestamps": [ + 172 + ], + "3min_transcript": "over pi. We're able to find the output pretty easily. But I want to do something interesting. Let's attempt to input 0 into the function. If we input 0 then the function tells us what we need to output. Does this definition tell us what we need to output? So if I attempt to put x equal 0, then this definition would say f of 0 be 2 over 0, but 2 over 0 is undefined. Rewrite this -- 2 over 0. This is undefined. This function definition does not tell us what to actually do with 0. It gives us an undefined answer. So this function is not defined here. It gives a question mark. So this gets to the essence of what domain is. Domain is the set of all inputs over which the function is defined. So the domain of this function f would be all real numbers except for x equals 0. So we write down these, these big ideas. This is the domain -- the domain of a function -- A domain of a function is the set of all inputs -- inputs over which the function is defined -- over which the function is defined, or the function has defined outputs over which the function has defined outputs. So the domain for this f in particular -- so the domain for this one -- if I want to say its domain, I could say, look, it's going to be the set of these curly brackets. These are kind of typical mathy set notation. I said OK , it could be the set of -- I gonna put curly brackets like that. Well, x can be a member So this little symbol means a member of the real numbers. But it can't be most of the real numbers except it cannot be 0 because we don't know -- this definition is undefined when you put the input as 0 So x is a member of the real numbers, and we write real numbers -- we write it with this kind of double stroke right over here. That's the set of all real numbers such that -- we have to put the exception. 0 is not a -- x equals to 0 is not a member of that domain -- such that x does not -- does not equal 0. Now let's make this a little bit more concrete by do some more examples So more examples we do, hopefully the clearer this will become. So let's say we have another function. Just be clear, we don't always have to use f's and x's. We could say, let's say we have g of y is equal to the square root of y minus 6. So what is the domain here? What is the set of all inputs over which this function g is defined? So here we are in putting a y it to function g" + }, + { + "Q": "I'm a little confused what the (base) is... 0:53", + "A": "Usually, the bottom side of a parallelogram is thought of as the base, but any side of the parallelogram can be chosen as the base when using the A = bh formula. Once a side is chosen as the base (b), the height (h) must be the perpendicular distance from the side chosen as the base to the side parallel to this base. (Note that the height, h, is usually not the length of a side of the parallelogram.) Have a blessed, wonderful day!", + "video_name": "hm17lVaor0Q", + "timestamps": [ + 53 + ], + "3min_transcript": "- If we have a rectangle with base length b and height length h, we know how to figure out its area. Its area is just going to be the base, is going to be the base times the height. The base times the height. This is just a review of the area of a rectangle. Just multiply the base times the height. Now let's look at a parallelogram. And in this parallelogram, our base still has length b. And we still have a height h. So when we talk about the height, we're not talking about the length of these sides that at least the way I've drawn them, move diagonally. We're talking about if you go from this side up here, and you were to go straight down. If you were to go at a 90 degree angle. If you were to go perpendicularly straight down, you get to this side, that's going to be, that's going to be our height. So in a situation like this when you have a parallelogram, you know its base and its height, what do we think its area is going to be? So at first it might seem well this isn't as obvious But we can do a little visualization that I think will help. So what I'm going to do is I'm going to take a chunk of area from the left-hand side, actually this triangle on the left-hand side that helps make up the parallelogram, and then move it to the right, and then we will see something somewhat amazing. So I'm going to take this, I'm going to take this little chunk right there, Actually let me do it a little bit better. So I'm going to take that chunk right there. And let me cut, and paste it. So it's still the same parallelogram, but I'm just going to move this section of area. Remember we're just thinking about how much space is inside of the parallelogram and I'm going to take this area right over here and I'm going to move it to the right-hand side. And what just happened? What just happened? Let me see if I can move it a little bit better. What just happened when I did that? Well notice it now looks just like my previous rectangle. by taking some of the area from the left and moving it to the right, I have reconstructed this rectangle so they actually have the same area. The area of this parallelogram, or well it used to be this parallelogram, before I moved that triangle from the left to the right, is also going to be the base times the height. So the area here is also the area here, is also base times height. I just took this chunk of area that was over there, and I moved it to the right. So the area of a parallelogram, let me make this looking more like a parallelogram again. The area of a parallelogram is just going to be, if you have the base and the height, it's just going to be the base times the height. So the area for both of these, the area for both of these, are just base times height." + }, + { + "Q": "At 1:05 What does the triangle on x+3 stand for?", + "A": "It s not a triangle, it s delta . In this case it means that you have to find the absolute value of x1 (which is -3) minus x2 (which is 0). The result is 3.", + "video_name": "81SseQCpGws", + "timestamps": [ + 65 + ], + "3min_transcript": "Find the slope of the line pictured on the graph. So the slope of a line is defined to be rise over run. Or you could also view it as change in y over change in x. And let me show you what that means. So let's start at some arbitrary point on this line, and they highlight some of these points. So let's start at one of these points right over here. So if we wanted to start one of these points-- and let's say we want to change our x in the positive direction. So we want to go to the right. So let's say we want to go from this point to this point over here. How much do we have to move in x? So if we want to move in x, we have to go from this point We're going from negative 3 to 0. So our change in x-- and this triangle, that's delta. That means \"change in.\" Our change in x is equal to 3. Well, when we moved from this point to this point, our x-value changed by 3, but what happened to our y-value? Well, our y-value went down. It went from positive 3 to positive 2. Our y-value went down by 1. So our change in y is equal to negative 1. So we rose negative 1. We actually went down. So our rise is negative 1 when our run-- when our change in x-- is 3. So change in y over change in x is negative 1 over 3, or we could say that our slope is negative 1/3. Let me scroll over a little bit. It is negative 1/3. And I want to show you that we can do this with any two points on the line. We could even go further than 3 in the x-direction. Let's start at this point right over here and then move backwards to this point over here, just to show you that we'll still get the same result. So to go from this point to that point, what is our change in x? So our change in x is this right over here. Our change in x is that distance right over there. We started at 3, and we went to negative 3. We went back 6. Over here, our change in x is equal to negative 6. We're starting at this point now. So over here our change in x is negative 6. And then when our change in x is negative 6, when we start at this point and we move 6 back, what is our change of y to get to that point? Well, our y-value went from 1. That was our y-value at this point. And then when we go back to this point, our y-value is 3. So what did we do? We moved up by 2. Our change in y is equal to 2." + }, + { + "Q": "At 5:41 why do we multiply P(A) by P(B)", + "A": "As the events A and B are independent, meaning the outcome of event A does not affect the outcome of event B, then we can calculate the probability of both A and B taking place by multiplying the P(A) with the probability of B. This holds for independent events. But we must be careful about what we are calculating according to the problem we are trying to solve.", + "video_name": "RI874OSJp1U", + "timestamps": [ + 341 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:30 what is sal trying to say??", + "A": "At 0:30 Sal is explaining the significance of the constant of integration added in the end. \u00e2\u0088\u00ab2xdx = x^2 + C The +C is added as the derivative of a constant is 0. So the derivative of x^2, x^2 + 1 or x^2 + e or x^2 + \u00cf\u0080 or in general x^2+C is the same each time.", + "video_name": "MMv-027KEqU", + "timestamps": [ + 30 + ], + "3min_transcript": "We know how to take derivatives of functions. If I apply the derivative operator to x squared, I get 2x. Now, if I also apply the derivative operator to x squared plus 1, I also get 2x. If I apply the derivative operator to x squared plus pi, I also get 2x. The derivative of x squared is 2x. Derivative, with respect to x of pi of a constant, is just 0. Derivative, with respect to x of 1, is just a constant, is just 0. So once again, this is just going to be equal to 2x. In general, the derivative, with respect to x of x squared plus any constant, is going to be equal to 2x. The derivative of x squared, with respect to x, is 2x. Derivative of a constant, with respect to x, a constant does not change with respect to x, so it's just equal to 0. So you have-- You apply the derivative operator Now, let's go the other way around. Let's think about the antiderivative. And one way to think about it is we're doing the opposite of the derivative operator. The derivative operator, you get an expression and you find it's derivative. Now, what we want to do, is given some expression, we want to find what it could be the derivative of. So if someone were to tell-- or give you 2x-- if someone were to say 2x-- let me write this. So if someone were to ask you what is 2x the derivative of? They're essentially asking you for the antiderivative. And so you could say, well, 2x is the derivative of x squared. You could also say that 2x is the derivative of x squared plus pi, I think you get the general idea. So if you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant. So this is what you would consider the antiderivative of 2x. Now, that's all nice, but this is kind of clumsy to have to write a sentence like this, so let's come up with some notation for the antiderivative. And the convention here is to use kind of a strange looking notation, is to use a big elongated s looking thing like that, and a dx around the function that you're trying to take the antiderivative of. So in this case, it would look something like this. This is just saying this is equal to the antiderivative of 2x, and the antiderivative of 2x, we have already seen, is x squared plus c." + }, + { + "Q": "i don't really get what sal says at 4:05 can some one help", + "A": "That order doesn t matter in combinations.", + "video_name": "iKy-d5_erhI", + "timestamps": [ + 245 + ], + "3min_transcript": "Permutations. Now lemme, permutations. Now it's worth thinking about what permutations are counting. Now remember we care, when we're talking about permutations, we care about who's sitting So for example, for example this is one permutation. And this would be counted as another permutation. And this would be counted as another permutation. This would be counted as another permutation. So notice these are all the same three people, but we're putting them in different chairs. And this counted that. That's counted in this 120. I could keep going. We could have that, or we could have that. So our thinking in the permutation world. We would count all of these. Or we would count this as six different permutations. These are going towards this 120. And of course we have other permutations where we would involve other people. Where we have, it could be F, B, C, F, C, B, F, A, C, F, F. I'm going to be a little bit more systematic. F, uh lemme do it. B, B, F, C, B, C, F. And obviously I could keep doing. I can do 120 of these. I'll do two more. You could have C, F, B. And then you could have C, B, F. So in the permutation world, these are, these are literally 12 of the 120 permutations. But what if we, what if all we cared about is the three people we're choosing to sit down, but we don't care in what order that they're sitting, or in which chair they're sitting. So in that world, these would all be one. This is all the same set of three people if we don't care which chair they're sitting in. This would also be the same set of three people. And so this question. If I have six people sitting in three chairs, how many ways can I choose three people out of the six And I encourage you to pause the video, and try to think of what that number would actually be. Well a big clue was when we essentially wrote all of the permutations where we've picked a group of three people. We see that there's six ways of arranging the three people. When you pick a certain group of three people, that turned into six permutations. And so if all you want to do is care about well how many different ways are there to choose three from the six? You would take your whole permutations. You would take your number of permutations. You would take your number of permutations. And then you would divide it by the number of ways to arrange three people. Number of ways to arrange, arrange three people. And we see that you can arrange three people, or even three letters. You can arrange it in six different ways. So this would be equal to 120 divided by six," + }, + { + "Q": "How do I solve this problem?\n: a ski resort has started to keep track of the number of skiers and snowboarders who bought season passes. The ratio of the number of skiers who bought season passes to the number of snowboarders who bought season passes is 1:2. If 1250 more snowboarders bought season passes than skiers, how many snowboarders and how many skiers bought season passes?", + "A": "ratio of skiers to snowboarders 1:2 nbr of skiers = 1 * x = 1x nbr of snowborarders = 2 * x = 2x how many more snow boarders than skiers = 2x -1x = x so x should be equal to 1250 so nbr of skiers = 1x = 1 * 1250 =1250 nbr of snow boarders = 2x = 2 * 1250 = 2450", + "video_name": "tOd2T72eJME", + "timestamps": [ + 62 + ], + "3min_transcript": "The scale on a map is 7 centimeters for every 10 kilometers, or 7 centimeters for 10 kilometers. If the distance between two cities is 60 kilometers-- so that's the actual distance-- how far apart in centimeters are the two cities on the map? Well, they give us the scale. For every 10 kilometers in the real world, the map is going to show 7 centimeters. Or another way to think about it is if you see 7 centimeters on the map, that represents 10 kilometers in the real world. Now, they're saying that the distance between two cities is 60 kilometers. So it's essentially 6 times 10 kilometers, so times 6. So if you have to 6 times 10 kilometers on the map, you're going to have 6 times 7 centimeters, so times 6. And 6 times 7 centimeters gets you to 42 centimeters. Well, 42 centimeters." + }, + { + "Q": "i don't get 0:32", + "A": "Think of it this way: You take a ruler and measure 7cm on the map. That whole 7cm is 10km in the real world. :)", + "video_name": "tOd2T72eJME", + "timestamps": [ + 32 + ], + "3min_transcript": "The scale on a map is 7 centimeters for every 10 kilometers, or 7 centimeters for 10 kilometers. If the distance between two cities is 60 kilometers-- so that's the actual distance-- how far apart in centimeters are the two cities on the map? Well, they give us the scale. For every 10 kilometers in the real world, the map is going to show 7 centimeters. Or another way to think about it is if you see 7 centimeters on the map, that represents 10 kilometers in the real world. Now, they're saying that the distance between two cities is 60 kilometers. So it's essentially 6 times 10 kilometers, so times 6. So if you have to 6 times 10 kilometers on the map, you're going to have 6 times 7 centimeters, so times 6. And 6 times 7 centimeters gets you to 42 centimeters. Well, 42 centimeters." + }, + { + "Q": "What about a ratio like 1/2:3/4", + "A": "A ratio of 1/2:3/4 is a complex fraction (fractions within a fraction). It would need to be simplified by dividing the 2 fractions. 1/2 / 3/4 = 1/2 * 4/3 = 2/3", + "video_name": "tOd2T72eJME", + "timestamps": [ + 123 + ], + "3min_transcript": "The scale on a map is 7 centimeters for every 10 kilometers, or 7 centimeters for 10 kilometers. If the distance between two cities is 60 kilometers-- so that's the actual distance-- how far apart in centimeters are the two cities on the map? Well, they give us the scale. For every 10 kilometers in the real world, the map is going to show 7 centimeters. Or another way to think about it is if you see 7 centimeters on the map, that represents 10 kilometers in the real world. Now, they're saying that the distance between two cities is 60 kilometers. So it's essentially 6 times 10 kilometers, so times 6. So if you have to 6 times 10 kilometers on the map, you're going to have 6 times 7 centimeters, so times 6. And 6 times 7 centimeters gets you to 42 centimeters. Well, 42 centimeters." + }, + { + "Q": "2:22\nWhy can't we count the Probability of \"not catching a sunfish 3 times\" as well 1/2*1/2*1/2?\nAs each time of not catching a sunfish means catching a trout, which has the same probablity.\nThank's!", + "A": "There are 8 equally likely possibilities: SSS, SST, STT, STS, TTT, TTS, TSS, and TST. So the probability you said: 1/2*1/2*1/2 could also be said as: P(not get a sunfish)*P(not get a sunfish)*P(not get a sunfish) which would be the fifth possibility: TTT. Not getting a sunfish three times in a row, however, could be any of the last seven possibilities. I hope this helps! :)", + "video_name": "86nb02Bx_5w", + "timestamps": [ + 142 + ], + "3min_transcript": "You and your friend Jeremy are fishing in a pond that contains ten trout and ten sunfish. Each time one of you catches a fish you release it back into the water. Jeremy offers you the choice of two different bets. Bet number one. We don't encourage betting but I guess Jeremy wants to bet. If the next three fish he catches are all sunfish you will pay him 100 dollars, otherwise he will pay you 20 dollars. Bet two, if you catch at least two sunfish of the next three fish that you catch he will pay you 50 dollars, otherwise you will pay him 25 dollars. What is the expected value from bet one? Round your answer to the nearest cent. I encourage you to pause this video and try to think about it on your own. Let's see. The expected value of bet one. The expected value of bet one where we'll say bet one is -- just to be a little bit better about this. Let's say x is equal what you pay, or I guess you could say , because you might get something, what your profit is from bet one. It's a random variable. The expected value of x is going to be equal to, let's see. What's the probability, it's going to be negative 100 dollars times the probability that he catches three fish. The probably that Jeremy catches three sunfish, the next three fish he catches are Or I should, well you're going to pay that. Since you're paying it we'll put it as negative 100 because we're saying that this is your expected profit, so you're going to lose money there. That's going to be one minus this probability, the probability that Jeremy catches three sunfish. In that situation he'll pay you 20 dollars. You get 20 dollars there. The important thing is to figure out the probability that Jeremy catches three sunfish. Well the sunfish are 10 out of the 20 fish so any given time he's trying to catch fish there's a 10 in 20 chance, or you could say one half probability that it's going to be a sunfish. The probability that you get three sunfish in a row is going to be one half, times one half, times one half." + }, + { + "Q": "At 0:24, why did Sal say 10+10+6? That's 26, instead of 16...", + "A": "Oh, no. Sal said 10+6, not 10+10+6. That s where your trip was.", + "video_name": "vbGwcvXgDlg", + "timestamps": [ + 24 + ], + "3min_transcript": "" + }, + { + "Q": "At 7:35, Sal enters some values into the calculator which are close to -4. What would happen on the calculator if we tried to enter -4 into the function replacing x??", + "A": "When we substitute -4 for x, the equation becomes (-4)^2/(-4)^2-16. We can simplify this expression to 16/16-16 which then becomes 16/0. It s impossible to divide by 0 because nothing times 0 is equal to 16. There is no way to get 16 (or any other number) by multiplying by 0. The expression x^2/x^2-16 becomes undefined when x=-4 and when x=4 because there are no possible outputs from those inputs.", + "video_name": "2N62v_63SBo", + "timestamps": [ + 455 + ], + "3min_transcript": "" + }, + { + "Q": "At 16:59 Sal explains how to get the horizontal asymptote. Is it possible to ever have a negative horizontal asymptote?", + "A": "Yes, a horizontal asymptote can be any y value. It s location is determined by the coefficients of the variables, which can be any number (the coefficients that is).", + "video_name": "2N62v_63SBo", + "timestamps": [ + 1019 + ], + "3min_transcript": "" + }, + { + "Q": "In 3:40, Sal writes |-2-3|, while on the number line, we normally subtract the lesser number from the greater one, i.e. |3-(-2)|, right?", + "A": "As long as you subtract the 2 numbers and take their absolute value, you will create the same result: |-2-3| = |-5| = 5 |3-(-2)| = |5| = 5 So, pick the version you are more comfortable with.", + "video_name": "t4xOkpP8FgE", + "timestamps": [ + 220 + ], + "3min_transcript": "of b minus a, and this is equivalent, either of these expressions is the distance between these numbers. I encourage you to play around with the negatives to see if you can factor out some negatives and think about the absolute value. It will actually make a lot of sense why this is true. In another video, I might do a little bit more of a rigorous justification for it. for this video is to see that this is actually true. So let's say we're in a world, let's get a number line out, and let's look at some examples. So let's say that we want to figure out the distance between, between, let's say negative two, the distance between negative two and positive three. So we can look at the number line and figure out what that distance is. To go from negative two to positive three, or the distance between them, we see is one, two, three, four, five. This distance right over here, this distance right over here is equal to five. One, two, three, four, five. Or you'd have to go five backwards to go from three to negative two. But let's see that what I just wrote actually applies right over here. So if we took negative two to be our a and three to be our b, then we could write this as the absolute value of negative two minus three, what is this going to be equal to? Well this is going to be equal to negative two minus three is negative five, so it's the absolute value of negative five. So this indeed equals five. So notice I subtracted the larger number from the smaller number. I got a negative value, but then I took the absolute value of it. That gave me the actual distance Now what if I did it the other way around? What if I took three minus negative two? So it's going to be the absolute value of three, let me do it in the blue color, the absolute value of three minus, and in parentheses I'll write the negative two. Negative two. Now if you subtract a smaller number from a larger number, you should get a positive value. So the absolute value sign here is just kind of extra. You don't really need it, unless to verify that that's true. This is going to be three minus negative two. That's the same thing as three plus positive two, or five. So this is just going to be the absolute value of five, which of course, is equal to five. So hopefully this makes you feel good that if you want the distance between two numbers, you subtract one from the other, and it doesn't matter which order you do it. You could subtract three from negative two, or negative two from three, be careful with the negative symbols here, and then take the absolute value," + }, + { + "Q": "At around 7:10, why did Sal do 16-2=14 instead of 2+16=18?", + "A": "At 3:00 and at 4:12 or so, he explanes that the change in y is 14 because one point is at +2 and the other is at +16. and if you count that on the graph they are 14 spaces apart.", + "video_name": "WkspBxrzuZo", + "timestamps": [ + 430 + ], + "3min_transcript": "So let me save some space here. So we have 1, 2, 3, 4. It's 4 comma-- 1, 2. So 4 comma 2 is right over here. 4 comma 2. Then we have the point negative 3 comma 16. So let me draw that over here. So we have negative 1, 2, 3. And we have to go up 16. So this is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. So it goes right over here. So this is negative 3 comma 16. Negative 3 comma 16. So the line that goes between them is going to look something like this. Try my best to draw a relatively straight line. That line will keep going. So the line will keep going. So that's my best attempt. And now notice, it's downward sloping. As you increase an x-value, the line goes down. It's going from the top left to the bottom right. As x gets bigger, y gets smaller. And just to visualize our change in x's and our change in y's that we dealt with here, when we started at 4 and we ended at-- or when we started at 4 comma 2 and ended at negative 3 comma 16, that was analogous to starting here and ending over there. And we said our change in x was negative 7. We had to move back. Our run we had to move in the left direction by 7. That's why it was a negative 7. And then we had to move in the y-direction. We had to move in the y-direction positive 14. So that's why our rise was positive. So it's 14 over negative 7, or negative 2. When we did it the other way, we started at this point. We started at this point, and then ended at this point. Started at negative 3, 16 and ended at that point. So in that situation, our run was positive 7. And now we have to go down in the y-direction since we switched the starting and the endpoint. And now we had to go down negative 14. Either way, we got the same slope." + }, + { + "Q": "in 1:16 does m equal slope?", + "A": "In y=mx + b, m is always the slope, and b is always the y-intercept", + "video_name": "WkspBxrzuZo", + "timestamps": [ + 76 + ], + "3min_transcript": "Find the slope of the line that goes through the ordered pairs 4 comma 2 and negative 3 comma 16. So just as a reminder, slope is defined as rise over run. Or, you could view that rise is just change in y and run is just change in x. The triangles here, that's the delta symbol. It literally means \"change in.\" Or another way, and you might see this formula, and it tends to be really complicated. But just remember it's just these two things over here. Sometimes, slope will be specified with the variable m. And they'll say that m is the same thing-- and this is really the same thing as change in y. They'll write y2 minus y1 over x2 minus x1. And this notation tends to be kind of complicated, but all this means is, is you take the y-value of your endpoint and subtract from it the y-value of your starting point. That will essentially give you your change in y. And it says take the x-value of your endpoint And that'll give you change in x. So whatever of these work for you, let's actually figure out the slope of the line that goes through these two points. So we're starting at-- and actually, we could do it both ways. We could start at this point and go to that point and calculate the slope or we could start at this point and go to that point and calculate the slope. So let's do it both ways. So let's say that our starting point is the point 4 comma 2. And let's say that our endpoint is negative 3 comma 16. So what is the change in x over here? What is the change in x in this scenario? So we're going from 4 to negative 3. If something goes from 4 to negative 3, what was it's change? You have to go down 4 to get to 0, and then you have to go down another 3 to get to negative 3. So our change in x here is negative 7. Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. Let's do the same thing for the change in y. And notice, I implicitly use this formula over here. Our change in x was this value, our endpoint, our end x-value minus our starting x-value. Let's do the same thing for our change in y. Our change in y. If we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y-value and subtract from that your starting y-value and you get 14. So what is the slope over here? Well, the slope is just change in y over change in x." + }, + { + "Q": "At 0:32 Sal said \"all the rectangles\" but there was actually one square .", + "A": "A rectangle just means a shape with four sides & four right angles, so technically all squares are rectangles.", + "video_name": "nLY2bzRfQyo", + "timestamps": [ + 32 + ], + "3min_transcript": "So I have this yellow rectangle here, and we know two things about this yellow rectangle. We know that it has a length of 10, that the length of this side right over here is 10. And we also know that this yellow rectangle has an area of 60 square units, whatever units we're measuring this 10 in. So what I want you to do is now pause this video, and based on the information given on these other rectangles-- so in some of them we give you two of their dimensions, in some of them we give you something like the perimeter and one of the dimensions-- I want you to pause the video and think about which of these rectangles, if any of them, have either the same area or the same perimeter as this yellow rectangle. So pause the video right now. Well, the best way to figure out which of these have the same area or perimeter as this original yellow rectangle is to just figure out the area and the perimeter for all of these rectangles and see which ones of them are equivalent. So we already know the area for this one, So how do we figure that out? Well, to figure out perimeter, we would need to know the lengths of all the sides. Well, if the area is 60 square units, that means the length times the width is equal to 60, that 10 times this width right over here is going to be equal to 60. So 10 times what is equal to 60? Well, 10 times 6 is equal to 60. 10 times 6 is equal to 60 square units. 10 units times 6 units is equal to 60 square units. Fair enough. So how do we figure out the perimeter now? Well, this is a rectangle. So we know if this length is 10, then this length must also be 10. And if this width is 6, then this width must be 6 as well. And now we can figure out the perimeter. It's 10 plus 10 plus 6 plus 6, which is 32. So let me write that down. The perimeter of our original yellow rectangle is equal to 32. and figure out what their perimeters and the areas are. We already know the perimeter for this purple or mauve rectangle, but we need to figure out its area. In order to figure out its area, we can't just rely on this one dimension just on its width. We have to figure out its length as well. So how do we figure that out? Well, one way to realize it is that the perimeter is the distance all the way around the rectangle. So what would be the distance halfway around the rectangle? So let me see if I can draw it. What would be the distance of this side, our length, plus this side? Well, it would be half the perimeter. 5 plus something is going to be equal to half the perimeter. Remember, the perimeter is all four sides. If we just took these two sides, which would be half the perimeter. So, these two sides must be equal to, when you take their sum, must be equal to 17, half the perimeter. So 5 plus what is equal to 17? 5 plus this question mark is equal to 17." + }, + { + "Q": "At 1:41, it is 4.", + "A": "Yes, 4 would be the correct opposite for -4.", + "video_name": "2Zk6u7Uk5ow", + "timestamps": [ + 101 + ], + "3min_transcript": "- [Voiceover] What I want to do in this video is think about what it means to have an opposite of a number. Let me draw a number line here. Let's draw a number line. And let's put some numbers on this number line. We can start at zero, and if we go to the right we have positive numbers. One, two, three, four, five. As we go to the left we get more and more negative. So negative one, negative two, negative three, negative four, and I can keep going on and on and on. Let's pick one of these numbers. Let's say that we pick the number three. What is going to be the opposite of the number three? Well the opposite of the number is a number that's the same distance from zero but on the other side. So three is three to the right of zero. One, two, three. So its opposite is going to be three to the left of zero. So the opposite of three is negative three. Let me make a little table here. If we have the number, the number. And then we have its opposite. We have its opposite. So we just figured out that if you have the number three, its opposite is going to be negative three. Now what if your number is negative? What if your number, let's say the number negative four. What's the opposite of that? And I encourage you to pause the video and try to think about it on your own. Well, you say, okay, negative four is right over here. That's negative four. It is four to the left of zero. One, two, three, four to the left of zero. So its opposite is going to be four to the right of it. So one, two, three, four. It's going to be positive. So you're probably starting to see a pattern here. The opposite of a number is going to be the opposite sign of that number. If you have a positive three here, its opposite is going to be negative three. If you start with negative four, its opposite is going to be positive four. One way to think about it, it's going to have the same absolute value but have a different sign. Or another way to think about it is, however if this is three to the right of zero, its opposite is going to be three to the left of zero. Or if the number is four to the left of zero, its opposite is going to be four to the right of zero. So we'll do one last one. What is the opposite of... What's going to be the opposite of one? Well one is one to the right of zero, so its opposite is going to be one to the left of zero, or negative one. Or another way to think about it, one is positive" + }, + { + "Q": "I am confused at 4:35 the video lost me, i am so confused", + "A": "He s basically saying you need all of the factors for both of the numbers and if you leave one out or forget one your answer will be wrong. Hope this helps!", + "video_name": "1Vb8t7Y-pI0", + "timestamps": [ + 275 + ], + "3min_transcript": "find the least common multiple other than just looking at the multiples like this. You could look at it through prime factorization. 30 is 2 times 15, which is 3 times 5. So we could say that 30 is equal to 2 times 3 times 5. And 24-- that's a different color than that blue-- 24 is equal to 2 times 12. 12 is equal to 2 times 6. 6 is equal to 2 times 3. So 24 is equal to 2 times 2 times 2 times 3. So another way to come up with the least common multiple, if we didn't even do this exercise up here, says, look, the number has to be divisible by both 30 and 24. If it's going to be divisible by 30, it's going to have to have 2 times 3 times 5 in its prime factorization. So this makes it divisible by 30. And say, well in order to be divisible by 24, its prime factorization is going to need 3 twos and a 3. Well we already have 1 three. And we already have 1 two, so we just need 2 more twos. So 2 times 2. So this makes it-- let me scroll up a little bit-- this right over here makes it divisible by 24. And so this is essentially the prime factorization of the least common multiple of 30 and 24. You take any one of these numbers away, you are no longer going to be divisible by one of these two If you take a two away, you're not going to be divisible by 24 If you take a two or a three away. If you take a three or a five away, you're not going to be divisible by 30 anymore. And so if you were to multiply all these out, this is 2 times 2 times 2 is 8 times 3 is 24 times 5 is 120. Umama just bought one package of 21 binders. Let me write that number down. 21 binders. She also bought a package of 30 pencils. She wants to use all of the binders and pencils to create identical sets of office supplies for her classmates. What is the greatest number of identical sets Umama can make using all the supplies? So the fact that we're talking about greatest is clue that it's probably going to be dealing with greatest common divisors. And it's also dealing with dividing these things. We want to divide these both into the greatest number of identical sets. So there's a couple of ways we could think about it. Let's think about what the greatest common divisor of both these numbers are. Or I could even say the greatest common factor. The greatest common divisor of 21 and 30. So what's the largest number that divides into both of them?" + }, + { + "Q": "um... in 3:6-11 what do you mean by distribute? that i so confused", + "A": "Distributing is multiplying the number outside the parentheses by everything inside the parentheses 9(1+99) Aka 9*1 and 9*99", + "video_name": "XAzFGx3Ruig", + "timestamps": [ + 186 + ], + "3min_transcript": "or seven, or 11, or 17, why does it work for nine? And actually also works for three, and we'll think about that in the future video. And to realize that, we just have to rewrite 2943. So the two in 2943, it's in the thousands place, so we can literally rewrite it as two times, two times 1000, the nine is in the hundreds place, so we can literally write it as nine times 100, the four is in the tens place, so it's literally the same thing as four times 10, and then finally we have our three in the ones place, we could write it as three times one or just three. This is literally 2000, 900, 40, and three, 2943. But now we can rewrite each of these things, this thousand, this hundred, this ten, as a sum of one plus something that is divisible by nine, so 1000, 1000 I can rewrite as one plus 999, I can rewrite 100 as one plus 99, one plus 99, I can rewrite 10 as one plus 9, and so two times 1000 is the same thing as two times one plus 999, nine times 100 is the same thing as nine times one plus 99, four times 10 is the same thing as four times one plus nine, and then I have this plus three over here. But now I can distribute, I can say, well, this over here is the same thing as two times one, which is just two, plus two times 999, this thing right over here is the same thing as nine times one, just to be clear what I'm doing, I'm distributing the two over the first parenthesis, these first two terms, then the nine, I'm gonna distribute again, so it's gonna be nine, nine times one plus nine times 99, and then over here, I forgot the plus sign right over here, I'm gonna distribute the four, four times one, so plus four, and then four times nine, so plus four times nine, and then finally I have this positive three, or plus three right over here. Now I'm just gonna rearrange this addition, so I'm gonna take all the terms we're multiplying by 999, and I'm going to do that in orange. So I'm gonna take this term, this term, and this term right over here, and so I have two times 999, that's that there, plus nine times 99, plus four times nine, plus four times nine, so that's those three terms, and then I have plus two, plus two, plus nine, plus nine, plus four, plus four, and plus three, plus three. And this is interesting, this is just the sum of our digits," + }, + { + "Q": "At 4:15 mins into the video, Sal says that since 999 is divisible by 9, everything before it is divisible by 9, hence 2.999 is divisible by 9. But 4.9 is not divisible by 9. What am I missing here?", + "A": "let s look here. i think 4*9 is indeed divisible by 9. 4*9 = 36. is 36 divisible by 9? yes, because 36/9 is 4, a whole number. also to do a multiplication sign, use an asterisk (shift-8) on your keyboard instead of a decimal.", + "video_name": "XAzFGx3Ruig", + "timestamps": [ + 255 + ], + "3min_transcript": "I can rewrite 100 as one plus 99, one plus 99, I can rewrite 10 as one plus 9, and so two times 1000 is the same thing as two times one plus 999, nine times 100 is the same thing as nine times one plus 99, four times 10 is the same thing as four times one plus nine, and then I have this plus three over here. But now I can distribute, I can say, well, this over here is the same thing as two times one, which is just two, plus two times 999, this thing right over here is the same thing as nine times one, just to be clear what I'm doing, I'm distributing the two over the first parenthesis, these first two terms, then the nine, I'm gonna distribute again, so it's gonna be nine, nine times one plus nine times 99, and then over here, I forgot the plus sign right over here, I'm gonna distribute the four, four times one, so plus four, and then four times nine, so plus four times nine, and then finally I have this positive three, or plus three right over here. Now I'm just gonna rearrange this addition, so I'm gonna take all the terms we're multiplying by 999, and I'm going to do that in orange. So I'm gonna take this term, this term, and this term right over here, and so I have two times 999, that's that there, plus nine times 99, plus four times nine, plus four times nine, so that's those three terms, and then I have plus two, plus two, plus nine, plus nine, plus four, plus four, and plus three, plus three. And this is interesting, this is just the sum of our digits, and you might see where all of this is going, this orange stuff here, is this divisible by nine? What? Sure, it will definitely, 999, that's divisible by nine, so anything this is multiplying by, it's divisible by nine, so this is divisible by nine, this is definitely divisible by nine, 99, regardless of whether is being multiplied by nine, whatever is multiplying by nine, whatever is multiplying 99 is gonna be divisible by nine, because 99 is divisible by nine, so this works out, and same thing over here, you're always going to be multiplying by a multiple of nine, so all of this business all over here is definitely going to be divisible by nine. And so in order for the whole thing, and all I did is I rewrote 2943 like this right over here, in order for the whole thing to be divisible by nine, this part definitely is divisible by nine, in order for the whole thing does the rest of this sum, it has to be divisible by nine as well. So in order for this whole thing," + }, + { + "Q": "Around the 1:36 mark in your video, I did not understand it when you said that you would multiply the number by 3 jumps. I get the jump part, but i do not get the multiplication part. By that I mean, why would you multiply 8/3 by 3? I would appreciate it very much if anyone could help me figure this out!", + "A": "You have to flip one of the numbers you are dividing and then multiply the flipped number with the other number(in this case 3) to get your answer", + "video_name": "f3ySpxX9oeM", + "timestamps": [ + 96 + ], + "3min_transcript": "Let's think about what it means to take 8/3 and divide it by 1/3. So let me draw a number line here. So there is my number line. This is 0. This is 1. And this is 2. Maybe this is 3 right over here. And let me plot 8/3. So to do that, I just need to break up each whole into thirds. So let's see. That's 1/3, 2/3, 3/3, 4/3, 5/3, 6/3, 7/3, 8/3. So right over here. And then of course, 9/3 would get us to 3. So this right over here is 8/3. Now, one way to think about 8/3 divided by 3 is what if we take this length. And we say, how many jumps would it take to get there, if we're doing it in jumps of 1/3? Or essentially, we're breaking this up. If we were to break up 8/3 into sections of 1/3, how many sections would I have, or how many jumps would I have? If we're trying to take jumps of 1/3, we're going to have to go 1, 2, 3, 4, 5, 6, 7, 8 jumps. So we could view this as-- let me do this in a different color. I'll do it in this orange. So we took these 8 jumps right over here. So we could view 8/3 divided by 1/3 as being equal to 8. Now, why does this actually make sense? Well, when you're dividing things into thirds, for every whole, you're now going to have 3 jumps. So whatever value you're trying to get to, you're going to have that number times 3 jumps. So another way of thinking about it is that 8/3 divided by 1/3 is the same thing as 8/3 times 3. We could write times 3 like that. Or, if we want to write 3 as a fraction, we know that 3 is the same thing as 3/1. And we already know how to multiply fractions. Multiply the numerators. 8 times 3. So you have 8-- let me do that that same color. You have 8 times 3 in the numerator now, 8 times 3. And then you have 3 times 1 in the denominator. Which would give you 24/3, which is the same thing as 24 divided by 3, which once again is equal to 8. Now let's see if this still makes sense. Instead of dividing by 1/3, if we were to divide by 2/3. So let's think about what 8/3 divided by 2/3 is." + }, + { + "Q": "at 2:20 , how are we telling that they are parallel", + "A": "Because of the right angles. It requires a 180 degree angle to make a line fold back over itself, or to be parallel. At 2:20, the top of the square forms a 90 degree angle with the side, and the side forms another 90 degree angle with the bottom. Adding these up, we get 180 degrees, meaning the top and bottom lines are parallel. The same thing is true with the two sides. I hope this answers your question! :)", + "video_name": "wPZIa3SjPF0", + "timestamps": [ + 140 + ], + "3min_transcript": "What is the type of this quadrilateral? Be as specific as possible with the given data. So it clearly is a quadrilateral. We have four sides here. And we see that we have two pairs of parallel sides. Or we could also say there are two pairs of congruent sides here as well. This side is parallel and congruent to this side. This side is parallel and congruent to that side. So we're dealing with a parallelogram. Let's do more of these. So here it looks like a same type of scenario we just saw in the last one. We have two pairs of parallel and congruent sides, but all the sides aren't equal to each other. If they're all equal to each other, we'd be dealing with a rhombus. But here, they're not all equal to each other. This side is congruent to the side opposite. This side is congruent to the side opposite. That's another parallelogram. Now this is interesting. We have two pairs of sides that are parallel to each other, but now all the sides have an equal length. So this would be a parallelogram. And it is a parallelogram, but they're So saying it's a rhombus would be more specific than saying it's a parallelogram. This does satisfy the constraints for being a parallelogram, but saying it's a rhombus tells us even more. Not every parallelogram is a rhombus, but every rhombus is a parallelogram. Here, they have the sides are parallel to the side opposite and all of the sides are equal. Let's do a few more of these. What is the type of this quadrilateral? Be as specific as possible with the given data . So we have two pairs of sides that are parallel, or I should say one pair. We have a pair of sides that are parallel. And then we have another pair of sides that are not. So this is a trapezoid. But then they have two choices here. They have trapezoid and isosceles trapezoid. Now an isosceles trapezoid is a trapezoid where the two non-parallel sides have the same length, just like an isosceles triangle, you have Well we could see these two non-parallel sides do not have the same length. So this is not an isosceles trapezoid. If they did have the same length, then we would pick that because that would be more specific than just trapezoid. But this case right over here, this is just a trapezoid. Let's do one more of these. What is the type of this quadrilateral? Well we could say it's a parallelogram because all of the sides are parallel. But if we wanted to be more specific, you could also see that all the sides are the same. So you could say it's a rhombus, but you could get even more specific than that. You notice that all the sides are intersecting at right angles. So this is-- if we wanted to be as specific as possible-- this is a square. Let me check the answer. Got it right." + }, + { + "Q": "At 0:13, how are you able to tell that the opposite sides are parallel?", + "A": "107+73=180 degrees, so the opp sides are parallel by same-side interior/consecutive interior angles", + "video_name": "wPZIa3SjPF0", + "timestamps": [ + 13 + ], + "3min_transcript": "What is the type of this quadrilateral? Be as specific as possible with the given data. So it clearly is a quadrilateral. We have four sides here. And we see that we have two pairs of parallel sides. Or we could also say there are two pairs of congruent sides here as well. This side is parallel and congruent to this side. This side is parallel and congruent to that side. So we're dealing with a parallelogram. Let's do more of these. So here it looks like a same type of scenario we just saw in the last one. We have two pairs of parallel and congruent sides, but all the sides aren't equal to each other. If they're all equal to each other, we'd be dealing with a rhombus. But here, they're not all equal to each other. This side is congruent to the side opposite. This side is congruent to the side opposite. That's another parallelogram. Now this is interesting. We have two pairs of sides that are parallel to each other, but now all the sides have an equal length. So this would be a parallelogram. And it is a parallelogram, but they're So saying it's a rhombus would be more specific than saying it's a parallelogram. This does satisfy the constraints for being a parallelogram, but saying it's a rhombus tells us even more. Not every parallelogram is a rhombus, but every rhombus is a parallelogram. Here, they have the sides are parallel to the side opposite and all of the sides are equal. Let's do a few more of these. What is the type of this quadrilateral? Be as specific as possible with the given data . So we have two pairs of sides that are parallel, or I should say one pair. We have a pair of sides that are parallel. And then we have another pair of sides that are not. So this is a trapezoid. But then they have two choices here. They have trapezoid and isosceles trapezoid. Now an isosceles trapezoid is a trapezoid where the two non-parallel sides have the same length, just like an isosceles triangle, you have Well we could see these two non-parallel sides do not have the same length. So this is not an isosceles trapezoid. If they did have the same length, then we would pick that because that would be more specific than just trapezoid. But this case right over here, this is just a trapezoid. Let's do one more of these. What is the type of this quadrilateral? Well we could say it's a parallelogram because all of the sides are parallel. But if we wanted to be more specific, you could also see that all the sides are the same. So you could say it's a rhombus, but you could get even more specific than that. You notice that all the sides are intersecting at right angles. So this is-- if we wanted to be as specific as possible-- this is a square. Let me check the answer. Got it right." + }, + { + "Q": "At 1:45, how can we say that two pairs of opposite sides are parallel ?\nIs it just by observation or anything else ?", + "A": "If they ever touch, then it won t be parallel, but they have to continuously have the same space between them.", + "video_name": "wPZIa3SjPF0", + "timestamps": [ + 105 + ], + "3min_transcript": "What is the type of this quadrilateral? Be as specific as possible with the given data. So it clearly is a quadrilateral. We have four sides here. And we see that we have two pairs of parallel sides. Or we could also say there are two pairs of congruent sides here as well. This side is parallel and congruent to this side. This side is parallel and congruent to that side. So we're dealing with a parallelogram. Let's do more of these. So here it looks like a same type of scenario we just saw in the last one. We have two pairs of parallel and congruent sides, but all the sides aren't equal to each other. If they're all equal to each other, we'd be dealing with a rhombus. But here, they're not all equal to each other. This side is congruent to the side opposite. This side is congruent to the side opposite. That's another parallelogram. Now this is interesting. We have two pairs of sides that are parallel to each other, but now all the sides have an equal length. So this would be a parallelogram. And it is a parallelogram, but they're So saying it's a rhombus would be more specific than saying it's a parallelogram. This does satisfy the constraints for being a parallelogram, but saying it's a rhombus tells us even more. Not every parallelogram is a rhombus, but every rhombus is a parallelogram. Here, they have the sides are parallel to the side opposite and all of the sides are equal. Let's do a few more of these. What is the type of this quadrilateral? Be as specific as possible with the given data . So we have two pairs of sides that are parallel, or I should say one pair. We have a pair of sides that are parallel. And then we have another pair of sides that are not. So this is a trapezoid. But then they have two choices here. They have trapezoid and isosceles trapezoid. Now an isosceles trapezoid is a trapezoid where the two non-parallel sides have the same length, just like an isosceles triangle, you have Well we could see these two non-parallel sides do not have the same length. So this is not an isosceles trapezoid. If they did have the same length, then we would pick that because that would be more specific than just trapezoid. But this case right over here, this is just a trapezoid. Let's do one more of these. What is the type of this quadrilateral? Well we could say it's a parallelogram because all of the sides are parallel. But if we wanted to be more specific, you could also see that all the sides are the same. So you could say it's a rhombus, but you could get even more specific than that. You notice that all the sides are intersecting at right angles. So this is-- if we wanted to be as specific as possible-- this is a square. Let me check the answer. Got it right." + }, + { + "Q": "At 1:24, Sal mentions that 2 is a factor of (x^3 - 8). How did he come up with that?", + "A": "If 2 is a root of a polynomial, then (x-2) is a factor. The polynomial is x^3-8. Set that equal to 0 and solve: x^3-8=0 x^3=8 (x^3)^(1/3)=8^(1/3) x=2. So 2 must be a root, and therefore (x-2) is a factor. You can divide x^3-8 by (x-2) and get the quotient of x^2+2x+4. This is how you derive the difference of cubes formula.", + "video_name": "6FrPLJY0rqM", + "timestamps": [ + 84 + ], + "3min_transcript": "Let's see if we can tackle a more complicated partial fraction decomposition problem. I have 10x squared plus 12x plus 20, all of that over x to the third minus 8. The first thing to do with any of these rational expressions that you want to decompose is to just make sure that the numerator is of a lower degree than the denominator, and if it's not, then you just do the algebraic long division like we did in the first video. But here, you can do from [UNINTELLIGIBLE] the highest degree term here is a second-degree term, here it's a third-degree term, so we're cool. This is a lower degree than that one. If it was the same or higher, we would do a little long division. The next thing to do, if we're going to decompose this into its components, we have to figure out the factors of the denominator right here, so that we can use those factors as the denominators in each of the components, and a third-degree polynomial is much, much, much harder to factor than a second-degree polynomial, normally. pop out at you-- if it doesn't immediately, hopefully what I'm about to say will make it pop out at you in the future-- is you should always think about what number, when you substitute into a polynomial, will make it equal to 0. And in this case, what to the third power minus 8 equals 0? And hopefully 2 pops out at you. And this is something you can only do really through inspection or through experience. And you'll immediately see 2 to the third minus 8 is 0, so 2 is a 0 of this, or 2 makes this expression 0, and that tells us that x minus 2 is a factor. So we can rewrite this right here as 10x squared plus 12x plus 20 over x minus 2 times something something We don't know what that something is yet. And I just want to hit the point home of why this is true, or what the intuition vibe has true. expression 0. And we know that 2 would make this factored expression 0, because when you put a 2 right here, this factor become 0, so it'll make the whole thing 0. And so that's why, that's the intuition where, if you substitute a number here and it makes this 0, you do x minus that number here, and we know that that will be a factor of the thing. Well, anyway, the next step if we really want to decompose this rational expression is to figure out what this part of it is, and the way to do that is with algebraic long division. We essentially just divide x minus 2 into x to the third minus 8 to get this, so let's do that. So you get x minus 2 goes into x to the third-- and actually, what I'm going to do is, I'm going to write-- I leave space for the second-degree term, which is 0, the first-degree term, and then minus 8 is the constant term, so minus 8-- I" + }, + { + "Q": "In 2:45, who is blaise pascal?", + "A": "Blaise Pascal (1623-1662) was a French mathematician, physicist, writer, philosopher, and inventor. Pascal s earliest work was in the field of fluids, where the Pascal (Pa), a unit of pressure, was named after him. He then worked to invent the calculator. He became the second inventor of a calculator. Additionally, he basically created the field of projective geometry and worked on probabilities.", + "video_name": "Yhlv5Aeuo_k", + "timestamps": [ + 165 + ], + "3min_transcript": "\"Hmmm what about 29...? pretty sure it's prime.\" But as a number theorist, you'll be shocked to know it takes me a moment to figure these out. But, even though you have your primes memorised up to at least 1000 that doesn't change that primes, in general, are difficult to find. I mean if I ask you to find the highest even number, you'd say, \"that's silly, just give me the number you think is the highest and i'll just add 2.... BAM!!\" But guess what the highest prime number we know is? 2 to the power of 43,112,609 - 1. Just to give you an idea about how big a deal primes are, the guy that found this one won a $100,000 prize for it! We even sent our largest known prime number into space because scientists think aliens will recognise it as something important and not just some arbitrary number. So they will be able to figure out our alien space message... So if you ever think you don't care about prime numbers because they're 'not useful', remember that we use prime numbers to talk to aliens, I'm not even making this up! Anyway, the point is you started doodling because you were bored but ended up discovering some neat patterns. See how the primes tend to line up on the diagonals? Why do they do that?... also this sort of skeletal structure reminds me of bones so lets call these diagonal runs of primes: Prime Ribs! But how do you predict when a Prime Rib will end? I mean, maybe this next number is prime... (but my head is too fuzzy for now this right now so you tell me.) Anyway...Congratulations, You've discovered the Ulam Spiral! So that's a little mathematical doodling history for you. Yyou can stop being Ulam now... or you can continue. Maybe you like being Ulam. (thats fine) However you could also be Blaise Pascal. Here's another number game you can do using Pascal's triangle.(I don't know why I'm so into numbers today but I have a cold so if you'll just indulge my sick predelections maybe I'll manage to infect you with my enthusiasm :D Pascal's Triangle is the one where you get the next row in the triangle by adding two adjacent numbers. Constructing Pascal's Triangle is, in itself a sort of number game because it's not just don't have to do all the adding. I don't know if this was discovered through doodling but it was discovered independantly in: France, Italy, Persia, China and probably other places too so it's possible someone did. Right... so I don't actually care about the individual numbers right now. So, if you still Ulam, you pick a property and highlight it(e.g. if it's even or odd) If you circle all the odd numbers you'll get a form which might be starting to look familiar. And it makes sense you'd get Sierpinski's Triangle because when you add an odd number and an even number, you get an odd number. (odd + odd) = even and (even + even) = even... So it's just like the crash and burn binary tree game. The best part about it is that, if you know these properties, you can forget about the details of the numbers You don't have to know that a space contains a 9 to know that it's going to be odd. Now, instead of two colours, let's try three. we'll colour them depending on what the remainder is when you divide them by three(instead of by two). Here's a chart! :) So, all the multiples of three are coloured red, remainder of one will be coloured black and" + }, + { + "Q": "At 1:34 how come that's the highest prime number known to man isn't there a higher prime number?", + "A": "There are an infinite number of primes, but it is very difficult to prove that a really large number is prime. So, we know that there is a higher prime number, but we don t know what it is. This video is slightly out of date, though, because the highest prime known to man is now 2^57,885,161 \u00e2\u0088\u0092 1.", + "video_name": "Yhlv5Aeuo_k", + "timestamps": [ + 94 + ], + "3min_transcript": "Pretend you're me and you're in math class. Actually... nevermind, I'm sick so I'm staying home today so pretend you are Stanislaw Ulam instead. What I am about to tell you is a true story. So you are Stan Ulam and you're at a meeting but there's this really boring presentation so of course you're doodling and, because you're Ulam and not me, you really like numbers... I mean super like them. So much that what you're doodling is numbers, just counting starting with one and spiralling them around. I'm not too fluent in mathematical notation so so i find things like numbers to be distracting, but you're a number theorist and if you love numbers who am I to judge? Thing is, because you know numbers so intimately, you can see beyond the confusing, squiggly lines you're drawing right into the heart of numbers. And, because you're a number theorist, and everyone knows that number theorists are enamoured with prime numbers( which is probably why they named them \"prime numbers\"), the primes you've doodled suddenly jump out at you like the exotic indivisible beasts they are... So you start drawing a heart around each prime. Well... it was actually boxes but in my version of the story it's hearts because you're not afraid to express your true feelings about prime numbers. You can probably do this instantly but it's going to take me a little longer... I'm all like - \"Hmmm what about 29...? pretty sure it's prime.\" But as a number theorist, you'll be shocked to know it takes me a moment to figure these out. But, even though you have your primes memorised up to at least 1000 that doesn't change that primes, in general, are difficult to find. I mean if I ask you to find the highest even number, you'd say, \"that's silly, just give me the number you think is the highest and i'll just add 2.... BAM!!\" But guess what the highest prime number we know is? 2 to the power of 43,112,609 - 1. Just to give you an idea about how big a deal primes are, the guy that found this one won a $100,000 prize for it! We even sent our largest known prime number into space because scientists think aliens will recognise it as something important and not just some arbitrary number. So they will be able to figure out our alien space message... So if you ever think you don't care about prime numbers because they're 'not useful', remember that we use prime numbers to talk to aliens, I'm not even making this up! Anyway, the point is you started doodling because you were bored but ended up discovering some neat patterns. See how the primes tend to line up on the diagonals? Why do they do that?... also this sort of skeletal structure reminds me of bones so lets call these diagonal runs of primes: Prime Ribs! But how do you predict when a Prime Rib will end? I mean, maybe this next number is prime... (but my head is too fuzzy for now this right now so you tell me.) Anyway...Congratulations, You've discovered the Ulam Spiral! So that's a little mathematical doodling history for you. Yyou can stop being Ulam now... or you can continue. Maybe you like being Ulam. (thats fine) However you could also be Blaise Pascal. Here's another number game you can do using Pascal's triangle.(I don't know why I'm so into numbers today but I have a cold so if you'll just indulge my sick predelections maybe I'll manage to infect you with my enthusiasm :D Pascal's Triangle is the one where you get the next row in the triangle by adding two adjacent numbers. Constructing Pascal's Triangle is, in itself a sort of number game because it's not just" + }, + { + "Q": "At 0:14, did she choose Ulam for any specific reason? Is he just a random person that she chose or does he have something to do with sick number games?", + "A": "Ulam is a famous mathematician", + "video_name": "Yhlv5Aeuo_k", + "timestamps": [ + 14 + ], + "3min_transcript": "Pretend you're me and you're in math class. Actually... nevermind, I'm sick so I'm staying home today so pretend you are Stanislaw Ulam instead. What I am about to tell you is a true story. So you are Stan Ulam and you're at a meeting but there's this really boring presentation so of course you're doodling and, because you're Ulam and not me, you really like numbers... I mean super like them. So much that what you're doodling is numbers, just counting starting with one and spiralling them around. I'm not too fluent in mathematical notation so so i find things like numbers to be distracting, but you're a number theorist and if you love numbers who am I to judge? Thing is, because you know numbers so intimately, you can see beyond the confusing, squiggly lines you're drawing right into the heart of numbers. And, because you're a number theorist, and everyone knows that number theorists are enamoured with prime numbers( which is probably why they named them \"prime numbers\"), the primes you've doodled suddenly jump out at you like the exotic indivisible beasts they are... So you start drawing a heart around each prime. Well... it was actually boxes but in my version of the story it's hearts because you're not afraid to express your true feelings about prime numbers. You can probably do this instantly but it's going to take me a little longer... I'm all like - \"Hmmm what about 29...? pretty sure it's prime.\" But as a number theorist, you'll be shocked to know it takes me a moment to figure these out. But, even though you have your primes memorised up to at least 1000 that doesn't change that primes, in general, are difficult to find. I mean if I ask you to find the highest even number, you'd say, \"that's silly, just give me the number you think is the highest and i'll just add 2.... BAM!!\" But guess what the highest prime number we know is? 2 to the power of 43,112,609 - 1. Just to give you an idea about how big a deal primes are, the guy that found this one won a $100,000 prize for it! We even sent our largest known prime number into space because scientists think aliens will recognise it as something important and not just some arbitrary number. So they will be able to figure out our alien space message... So if you ever think you don't care about prime numbers because they're 'not useful', remember that we use prime numbers to talk to aliens, I'm not even making this up! Anyway, the point is you started doodling because you were bored but ended up discovering some neat patterns. See how the primes tend to line up on the diagonals? Why do they do that?... also this sort of skeletal structure reminds me of bones so lets call these diagonal runs of primes: Prime Ribs! But how do you predict when a Prime Rib will end? I mean, maybe this next number is prime... (but my head is too fuzzy for now this right now so you tell me.) Anyway...Congratulations, You've discovered the Ulam Spiral! So that's a little mathematical doodling history for you. Yyou can stop being Ulam now... or you can continue. Maybe you like being Ulam. (thats fine) However you could also be Blaise Pascal. Here's another number game you can do using Pascal's triangle.(I don't know why I'm so into numbers today but I have a cold so if you'll just indulge my sick predelections maybe I'll manage to infect you with my enthusiasm :D Pascal's Triangle is the one where you get the next row in the triangle by adding two adjacent numbers. Constructing Pascal's Triangle is, in itself a sort of number game because it's not just" + }, + { + "Q": "For 1:40, do you have to represent if it's a natural number, integer, etc..", + "A": "When using sigma notation to express sums, it is implied that we re working with integers (which are a superset of natural numbers). Usually we re only working with the natural numbers, but there s nothing preventing starting at a negative index.", + "video_name": "5jwXThH6fg4", + "timestamps": [ + 100 + ], + "3min_transcript": "What I want to do in this video is introduce you to the idea of Sigma notation, which will be used extensively through your mathematical career. So let's just say you wanted to find a sum of some terms, and these terms have a pattern. So let's say you want to find the sum of the first 10 numbers. So you could say 1 plus 2 plus 3 plus, and you go all the way to plus 9 plus 10. And I clearly could have even written this whole thing out, but you can imagine it becomes a lot harder if you wanted to find the sum of the first 100 numbers. So that would be 1 plus 2 plus 3 plus, and you would go all the way to 99 plus 100. So mathematicians said, well, let's find some notation, instead of having to do this dot dot dot thing-- which you will see sometimes done-- so that we can more cleanly express these types of sums. And that's where Sigma notation comes from. So this sum up here, right over here, this first one, it could be represented as Sigma. And what you do is you define an index. And you could start your index at some value. So let's say your index starts at 1. I'll just use i for index. So let's say that i starts at 1, and I'm going to go to 10. So i starts at 1, and it goes to 10. And I'm going to sum up the i's. So how does this translate into this right over here? Well, what you do is you start wherever the index is. If the index is at 1, set i equal to 1. Write the 1 down, and then you increment the index. And so i will then be equal to 2. i is 2. Put the 2 down. And you're summing each of these terms as you go. And you go all the way until i is equal to 10. So given what I just told you, I encourage you to pause this video and write the Sigma notation for this sum right over here. well, this would be the sum. The first term, well, it might be easy to just say we'll start at i equals 1 again. But now we're not going to stop until i equals 100, and we're going to sum up all of the i's. Let's do another example. Let's imagine the sum from i equals 0 to 50 of-- I don't know, let me say-- pi i squared. What would this sum look like? And once again, I encourage you to pause the video and write it out, expand out this sum. Well, let's just go step by step. When i equals 0, this will be pi times 0 squared. And that's clearly 0, but I'll write it out. pi times 0 squared." + }, + { + "Q": "I don't get the part on 4:34.\n\nOk if you are taking the -sqrt of (x-1)^2 doesn't that equal -(x-1), thus the final value equal (-x+1).\n\nFor example the negative sqrt of 9:\n\n=-sqrt(9)\n=-(3)\n=-3\n\nSomeone tell me if I'm wrong cause I'm confused on that part on how even with negative square root he still gets (x-1).", + "A": "So I read the answer above and yh... I still don t exactly get it. Does it have to do with the fact that x<= 1 ensuring a negative number so when you get a -sqrt of (x-1) you keep it like that so the answer doesn t go positive or is that completely unrelated.", + "video_name": "Bq9cq9FZuNM", + "timestamps": [ + 274 + ], + "3min_transcript": "sides of this equation. And our goal is to get back to negative 3. If we take the positive square root. If we just take the principle root of both sides of that, we would get 3 is equal to 3. But that's not our goal. We want to get back to negative 3. So we want to take the negative square root of our square. So because this expression is negative and we want to get back to this expression, we want to get back to this x minus 1, we need to take the negative square root of both sides. You can always -- every perfect square has a positive or a negative root. The principle root is a positive root. But here we want to take the negative root because this expression right here is going to be negative. And that's what we want to solve for. So let's take the negative root of both sides. So you get the negative square root of y plus 2 is equal to -- and I'll just write this extra step here, just so you Is equal to the negative square root, the negative square root of x minus 1 squared. For y is greater than or equal to negative 2. And x is less than or equal to 1. That's why the whole reason we're going to take the negative square there. And then this expression right here -- so let me just write the left again. Negative square root of y plus 2 is equal to the negative square root of x minus 1 squared is just going to be x minus 1. It's just going to be x minus one. x minus 1 squared is some positive quantity. The negative root is the negative number that you have to square to get it. To get x minus 1 squared. So that just becomes x minus 1. Hopefully that doesn't confuse you too much. We just want to get rid of this squared sign. We want to make sure we get the negative version. We don't want the positive version, which would have been 1 minus x. Don't want to confuse you. So here, we now just have to solve for x. And let me write the 4. y is greater than or equal to negative 2. You get negative square root of y plus 2 plus 1 is equal to x for y is greater than or equal to negative 2. Or, if we want to rewrite it, we could say that x is equal to the negative square root of y plus 2 plus 1 for y is greater than or equal to negative 2. Or if we want to write it in terms, as an inverse function of y, we could say -- so we could say that f inverse of y is equal to this, or f inverse of y is equal to the negative square root of y plus 2 plus 1, for y is greater than or equal to negative 2. And now, if we wanted this in terms of x. If we just want to rename y as x we just replace the y's with x's. So we could write f inverse of x -- I'm just" + }, + { + "Q": "At 4:57 how do you know when to take the negative square root or positive one? I didn't get how in the video. Also if you could tell me, is this level of math algebra 1 or 2? They should do a better job of subdividing the algebra category between the 1 and 2.", + "A": "Whether you take the positive or negative root depends on which side of the graph is present. That s why Sal kept the for y >= -2 and x <= 1 pulled along to every step, to remember which side to use.", + "video_name": "Bq9cq9FZuNM", + "timestamps": [ + 297 + ], + "3min_transcript": "sides of this equation. And our goal is to get back to negative 3. If we take the positive square root. If we just take the principle root of both sides of that, we would get 3 is equal to 3. But that's not our goal. We want to get back to negative 3. So we want to take the negative square root of our square. So because this expression is negative and we want to get back to this expression, we want to get back to this x minus 1, we need to take the negative square root of both sides. You can always -- every perfect square has a positive or a negative root. The principle root is a positive root. But here we want to take the negative root because this expression right here is going to be negative. And that's what we want to solve for. So let's take the negative root of both sides. So you get the negative square root of y plus 2 is equal to -- and I'll just write this extra step here, just so you Is equal to the negative square root, the negative square root of x minus 1 squared. For y is greater than or equal to negative 2. And x is less than or equal to 1. That's why the whole reason we're going to take the negative square there. And then this expression right here -- so let me just write the left again. Negative square root of y plus 2 is equal to the negative square root of x minus 1 squared is just going to be x minus 1. It's just going to be x minus one. x minus 1 squared is some positive quantity. The negative root is the negative number that you have to square to get it. To get x minus 1 squared. So that just becomes x minus 1. Hopefully that doesn't confuse you too much. We just want to get rid of this squared sign. We want to make sure we get the negative version. We don't want the positive version, which would have been 1 minus x. Don't want to confuse you. So here, we now just have to solve for x. And let me write the 4. y is greater than or equal to negative 2. You get negative square root of y plus 2 plus 1 is equal to x for y is greater than or equal to negative 2. Or, if we want to rewrite it, we could say that x is equal to the negative square root of y plus 2 plus 1 for y is greater than or equal to negative 2. Or if we want to write it in terms, as an inverse function of y, we could say -- so we could say that f inverse of y is equal to this, or f inverse of y is equal to the negative square root of y plus 2 plus 1, for y is greater than or equal to negative 2. And now, if we wanted this in terms of x. If we just want to rename y as x we just replace the y's with x's. So we could write f inverse of x -- I'm just" + }, + { + "Q": "I'm confused. Khan says there is only one x term at 1:55. Can someone help me with this?", + "A": "Look at the second line. There is a 3x, a -3x, and there is a x by itself near the middle. Since the -3x cancels out the 3x, the x by itself is left. hope this helps :)", + "video_name": "DMyhUb1pZT0", + "timestamps": [ + 115 + ], + "3min_transcript": "We're asked to simplify this huge, long expression here. x to the third plus 3x minus 6-- that's in parentheses-- plus negative 2x squared plus x minus 2. And then minus the quantity 3x minus 4. So a good place to start, we'll just rewrite this and see if we can eliminate the parentheses in this step. So let's just start at the beginning. We have the x to the third right over there. So x to the third and then plus 3x-- I'll do that in pink-- plus 3x. And then we have a minus 6. And we don't have to put the parentheses around there, those don't really change anything. And we don't have to even write these-- do anything with these parentheses. We can eliminate them. Just because there's a positive sign out here we don't have to distribute anything. Distributing a positive sign doesn't do anything to these numbers. So then plus, we have a negative 2x squared. So this term right here is negative 2x squared, or minus x squared. And then we have a plus x. We have a plus x. Then we have a negative sign times this whole expression. So we're going to have to distribute the negative sign. So it's a positive 3x, but it's being multiplied by negative 1. So it's really a negative 3x. So minus 3x, then you have a negative-- you can imagine this is a negative 1 implicitly out here-- negative 1 times negative 4. That's a positive 4. So plus 4. Now, we could combine terms of similar degree, of the same degree. Now, first we have an x to the third term and I think it's the only third degree term here, because we have x being raised to the third power. So let me just rewrite it here. We have x to the third. And now let's look at our x squared terms. Looks like we only have one. We only have this term right here. So we have minus 2x squared. And then what about our x terms? We have a 3x plus an x minus a 3x again. So that 3x minus the 3x would cancel out, and you're just So plus x. And then finally our constant terms. Negative 6 minus 2 plus 4. Negative 6 minus 2 gets us to negative 8. Plus 4 is negative 4. We have simplified the expression. Now we just have a four term polynomial." + }, + { + "Q": "I understood everything until 1:24. Why does he insert a \"1\" in front of the parenthesis? I mean, where does that come from? He then multiplies \"-1 * -4\", to get a value of 4.... I'm confused why is the 3x ignored? Thanks for any clarification you can provide.", + "A": "He inserted a 1 to show that the ... - ( ... is also equal to ... -1 ( ... The 3x was not ignored because it was already distributed with the - (or -1), making -3x, as shown in 1:20.", + "video_name": "DMyhUb1pZT0", + "timestamps": [ + 84 + ], + "3min_transcript": "We're asked to simplify this huge, long expression here. x to the third plus 3x minus 6-- that's in parentheses-- plus negative 2x squared plus x minus 2. And then minus the quantity 3x minus 4. So a good place to start, we'll just rewrite this and see if we can eliminate the parentheses in this step. So let's just start at the beginning. We have the x to the third right over there. So x to the third and then plus 3x-- I'll do that in pink-- plus 3x. And then we have a minus 6. And we don't have to put the parentheses around there, those don't really change anything. And we don't have to even write these-- do anything with these parentheses. We can eliminate them. Just because there's a positive sign out here we don't have to distribute anything. Distributing a positive sign doesn't do anything to these numbers. So then plus, we have a negative 2x squared. So this term right here is negative 2x squared, or minus x squared. And then we have a plus x. We have a plus x. Then we have a negative sign times this whole expression. So we're going to have to distribute the negative sign. So it's a positive 3x, but it's being multiplied by negative 1. So it's really a negative 3x. So minus 3x, then you have a negative-- you can imagine this is a negative 1 implicitly out here-- negative 1 times negative 4. That's a positive 4. So plus 4. Now, we could combine terms of similar degree, of the same degree. Now, first we have an x to the third term and I think it's the only third degree term here, because we have x being raised to the third power. So let me just rewrite it here. We have x to the third. And now let's look at our x squared terms. Looks like we only have one. We only have this term right here. So we have minus 2x squared. And then what about our x terms? We have a 3x plus an x minus a 3x again. So that 3x minus the 3x would cancel out, and you're just So plus x. And then finally our constant terms. Negative 6 minus 2 plus 4. Negative 6 minus 2 gets us to negative 8. Plus 4 is negative 4. We have simplified the expression. Now we just have a four term polynomial." + }, + { + "Q": "At 5:30 you gave the formula of how to find the area of a quadrilateral but it was quite confusing. Could you please explain it again?", + "A": "Pretty much, with the edges, you form right triangles. Then, you find their area and then add that to the area of the regular quadrilateral. :0)", + "video_name": "gkifo46--JA", + "timestamps": [ + 330 + ], + "3min_transcript": "And then I could drop this down, and then we're done. All of these are pretty straightforward to figure out what their dimensions are. This is 5 by 1. This is 4 by 2. This is 1, 2, 3, 4, 5, 6 by 1, 2. And this is 1 by 1, 2, 3, 4, 5. So what is the area of this figure? And of course, we have this center rectangle right over here. Well, a triangle that is 5 units long and 1 unit high, its area is going to be 1/2 times 1 times 5. Or I could write it 1/2 times 1 times 5, depending on what multiplication symbol you are more comfortable with. Well that's just going to be 1/2 times 5, which is going to be equal to 2.5. So that's 2.5 right over there. This one is going to be 1/2 times 4 times 2. This one is going to be 1/2 times 2 times 1, 2, 3, 4, 5, 6. Well, 1/2 half times 2 is 1 times 6 is just 6. And then this one's going to be 1/2 times 1 times 1, 2, 3, 4, 5. So once again, the area of this one is going to be 2.5. And then finally, this is a 3 by 4 rectangle. And you could even count the unit squares in here. But it has 12 of those unit square, so it has an area of 12. So if we want to find the total area, we just add all of these together. So 2.5 plus 2.5 is 5, plus 4 is 9, plus 6 is 15, plus 12 is 27. So it has a total area of 27." + }, + { + "Q": "At 5:07 Sal says S parameter describes the angle between radius and x-z plane. Has he mistaken it for x-y plane, considering the way he uses the parameter afterwards?", + "A": "Yes, he meant to say x-y plane. A little box pops up in the lower right corner of the screen with the correction.", + "video_name": "owKAHXf1y1A", + "timestamps": [ + 307 + ], + "3min_transcript": "of the torus, or from the z-axis it's a distance of b. It's always going to be a distance of b. It's always, if you imagine the top of the doughnut, let me draw the top of the doughnut. If you're looking down on a doughnut, let me draw a doughnut right here, if you're looking down on a doughnut, it just looks something like that. The z-axis is just going to be popping straight out. The x-axis would come down like this, and then the y-axis would go to the right, like that. So you can imagine, I'm just flying above this. I'm sitting on the z-axis looking down at the doughnut. It will look just like this. And if you imagine the cross section, this circle right here, the top part of the circle if you're looking down, would look just like that. And this distance b is a distance from the z-axis So this distance, let me draw it in the same color, from the center to the center of these circles, that is going to be b. It's just going to keep going to the center of the circles, b. That's going to be b, that's going to be b. That's going to be b. From the center of our torus to the center of our circle that defines the torus, it's a distance of b. So this distance right here, that distance right there is b. And from b, we can imagine we have a radius. A radius of length a. So these circles have radius of length a. So this distance right here is a, this distance right here is a, this distance right there is a, that distance right there is a. If I were look at these circles, these circles have radius a. And what we're going to do is have two parameters. One is the angle that this radius makes with the x-z Let me do that in the same color. You can imagine the x-axis coming out here. So this is the x-z plane. So one parameter is going to be the angle between our radius and the x-z plane. We're going to call that angle, or that parameter, we're going to call that s. And so as s goes between 0 and 2 pi, as s goes between 0 in 2 pi, when the 0 is just going to be at this point right here, and then as it goes to 2 pi, you're going to trace out a circle that looks just like that. Now, we only have one parameter. What we want to do is then spin this circle around. What I just drew is that circle right there. What we want to do is spin the entire circle around. So let's define another parameter. We'll call this one t, and I'll take the top view again." + }, + { + "Q": "at 7:36 how do you get your limits of integration? I understand that you did by inspection, but how would you do it by making the two equations equal to each other ?", + "A": "In that example, Sal is integrating with respect to y. To solve for zero and one analytically, we could make the equations equivalent to each other and solve. Here s the example: sqrt(x) = x^2 x = x^4 This last equation is true only when x = 0 or x = 1.", + "video_name": "WAPZihVUmzE", + "timestamps": [ + 456 + ], + "3min_transcript": "It's going to be 2 minus whatever the x value is right over here. So the x value right over here is going to be y squared. Remember, we want this is a function of y. So our outer radius, this whole distance, is going to be 2 minus the x value here as a function of y. That x value is y squared. So the outer radius is 2 minus x-- sorry. 2 minus y squared. We want it as a function of y. And then our inner radius is going to be equal to what? Well, that's going to be the difference, the horizontal distance, between this vertical line and our inner function, our inner boundary. So it's going to be the horizontal distance between two and whatever x value this is. But this x value as a function of y is just square root of y. So it's going to be 2 minus square root of y. And so now, we can come up with an expression for area. I'll leave the pi there. So it's going to be pi-- right over here-- it's going to be pi times outer radius squared. Well, the outer radius is 2 minus y squared-- and let me just-- well, I'll just write it. 2 minus y squared-- and we're going to square that-- squared, minus pi times the inner radius squared. Well, we already figured that out. The inner radius is 2 minus square root of y. And we're going to square that one, too. So this gives us the area of one of our rings as a function of y, the top of the ring, where I shaded in orange. And now, if we want the volume of one of those rings, we have to multiply it by its depth or its height the way we've drawn it right over here. And its height-- we've done this multiple times already right over here-- is an infinitesimal change in y. So we're going to multiply all that business times dy. This is the volume of one of our rings. So we're going to sum up all of the rings over the interval. And when you take the integral sign, it's a sum where you're taking the limit as you have an infinite number of rings that become infinitesimally small in height or depth, depending on how you view it. And what's our interval? So we've looked at this multiple times. These two graphs-- you could do it by inspection. You could try to solve it in some way, but it's pretty obvious that they intersect at-- remember we care about our y interval. They intersect at y is equal to 0 and y is equal to 1. And there you have it. We've set up our integral for the volume of this shape right over here. I'll leave you there, and in this next video, we will just evaluate this integral." + }, + { + "Q": "At 1:46. I do not understand the purpose of the 1/7 * 7, nor where they even came from, nor what allows him to write them there to begin with.", + "A": "To use u-substitution you have to identify a function u such that a constant multiple of the derivative of u appears in the integrand. In this case the function u is 7x+9. The derivative of 7x+9 is 7. Therefore a constant multiple of 7 must appear in the integrand. Because it does not, Sal multiples the expression by 1 in the form of (1/7 * 7 = 1). In this way he makes sure 7 appears in the integrand therefore facilitating u substitution.", + "video_name": "oqCfqIcbE10", + "timestamps": [ + 106 + ], + "3min_transcript": "Let's take the indefinite integral of the square root of 7x plus 9 dx. So my first question to you is, is this going to be a good case for u-substitution? Well, when you look here, maybe the natural thing to set to be equal to u is 7x plus 9. But do I see its derivative anywhere over here? If we set u to be equal to 7x plus 9, what is the derivative of u with respect to x going to be? Derivative of u with respect to x is just going to be equal to 7. Derivative of 7x is 7. Derivative of 9 is 0. So do we see a 7 lying around anywhere over here? Well, we don't. But what could we do in order to have a 7 lying around, but not change the value of the integral? Well, the neat thing-- and we've seen this multiple times-- is when you're evaluating integrals, scalars can go in and outside of the integral very easily. some scalar a times f of x dx, this is the same thing as a times the integral of f of x dx. The integral of the scalar times a function is equal to the scalar times the integral of the functions. So let me put this aside right over here. So with that in mind, can we multiply and divide by something that will have a 7 showing up? Well, we can multiply and divide by 7. So imagine doing this. Let's rewrite our original integral. So let me draw a little arrow here just to go around that aside. We could rewrite our original integral as being 9 to the integral of times 1/7 times 7 times the square root of 7x plus 9 dx. And if we want to, we could take the 1/7 outside We don't have to, but we can rewrite times the square root of 7x plus 9 dx. So now if we set u equal to 7x plus 9, do we have its derivative laying around? Well, sure. The 7 is right over here. We know that du-- if we want to write it in differential form-- du is equal to 7 times dx. So du is equal to 7 times dx. That part right over there is equal to du. And if we want to care about u, well, that's just going to be the 7x plus 9. That is are u. So let's rewrite this indefinite integral in terms of u. It's going to be equal to 1/7 times the integral of-- and I'll just take the 7 and put it in the back. So we could just write the square root of u du, 7 times dx is du." + }, + { + "Q": "Are du and dx able to just be treated like normal variables?\n@5:30 he treats the dx as another variable, to be multiplied with the 7", + "A": "Yes, when using Leibnitz notation, you can think of them as quantifiable numbers. What they really mean is this - an infinitely small change in the given variable. For example, du is an infinitely small change in u. dx is an infinitely small change in x. Then, if we think of it this way, du/dx as giving the slope makes sense! The change in u over the change in x gives slope! Long story short, you can treat them as actual numbers - infinitely small changes in a variable.", + "video_name": "oqCfqIcbE10", + "timestamps": [ + 330 + ], + "3min_transcript": "It makes it a little bit easier for us to kind of do the reverse power rule here. So we can rewrite this as equal to 1/7 times the integral of u to the 1/2 power du. And let me just make it clear. This u I could have written in white if I want it the same color. And this du is the same du right over here. So what is the antiderivative of u to the 1/2 power? Well, we increment u's power by 1. So this is going to be equal to-- let me not forget this 1/7 out front. So it's going to be 1/7 times-- if we increment the power here, it's going to be u to the 3/2, 1/2 plus 1 is 1 and 1/2 or 3/2. So it's going to be u to the 3/2. times the reciprocal of 3/2, which is 2/3. And I encourage you to verify the derivative of 2/3 u to the 3/2 is indeed u to the 1/2. And so we have that. And since we're multiplying 1/7 times this entire indefinite integral, we could also throw in a plus c right over here. There might have been a constant. And if we want, we can distribute the 1/7. So it would get 1/7 times 2/3 is 2/21 u to the 3/2. And 1/7 times some constant, well, that's just going to be some constant. And so I could write a constant like that. I could call that c1 and then I could call this c2, but it's really just some arbitrary constant. Oh, actually, no we aren't done. We still just have our entire thing in terms of u. So now let's unsubstitute it. So this is going to be equal to 2/21 times u to the 3/2. u is equal to 7x plus 9. Let me put a new color here just to ease the monotony. So it's going to be 2/21 times 7x plus 9 to the 3/2 power plus c. And we are done. We were able to take a kind of hairy looking integral and realize that even though it wasn't completely obvious at first, that u-substitution is applicable." + }, + { + "Q": "You took the average change from 2000-2003 and then 2003-2004 and averaged it out, but why couldn't you just use 2002-2004? 2:27", + "A": "It s mathematically equivalent (and easier) to take the average for 2002 to 2004 (which would be (S(2004) - S(2002))/2) as you suggest. However, the problem specifies that we are to solve by finding the slope of the two secant lines and averaging those slopes, so that is the approach used in the video.", + "video_name": "fI6w2kL295Y", + "timestamps": [ + 147 + ], + "3min_transcript": "The table shows the number of stores of a popular US coffee chain from 2000 to 2006. The number of stores recorded is the number at the start of each year on January 1. So in 2000, there was 1,996 stores, in 2005, 6,177, so on and so forth. Determine a reasonable approximation for the instantaneous rate of change, in coffee stores per year at the beginning of 2003-- so we care about 2003-- by taking the average of two nearby secant slopes. So let's visualize this. So this right over here, I've plotted all of the points. Now let me make sure that the axes are clear. This horizontal axis, this is my t-axis that tells us the year. And then the vertical axis is the number of stores. And we could even say that it is a function of time. So you see in the year 2000, there was 1,996 stores-- 2003, 4,272. 2003, 4,272 stores. Now if you could imagine that they're constantly adding stores, you could even imagine minute by minute they're adding stores. So this is just sampling the number of stores they had on January 1. But if you were to really plot it as a more continuous function, it might look something like this. I'll do my best to approximate it. It might be more of some type of curve that looks something like this. And once again, I'm just approximating what it might actually look like. So when they're saying the instantaneous rate of change in coffee stores per year, so this is the change, the instantaneous rate of change of stores per time. They're really saying, we need to approximate the slope of the tangent line in 2003, when time is 2003. So the tangent line might look something like that. is tangent. Now, they say approximate. We don't have the information to figure it out exactly. But we have some data around it, and we can figure out the slopes of the secant lines between this point and those points. And then we can take the average of the slopes of the secant lines to approximate the slope of this tangent line. So for example, we could find the slope of this secant line right over here as we go from 2002 to 2003. And then we can find the slope of this secant line as we go from 2003 to 2004. And if we average those, that should be a pretty good approximation for the instantaneous rate of change in 2003. So let's do that. So the slope of this pink secant line, as we go from 2002 to 2003, that's going to be the number of stores in 2003 minus the number of stores in 2002. So that's the change in our number of stores over the change in years, or the change in time." + }, + { + "Q": "\"it is a function of time\" can someone rephrase that please, in a very simple way @1:00", + "A": "A function of time meaning that as time passes the number of things change. In this example as the year goes by the number of stores built is increased which can be represented by a function with respect to time. A function being something along the lines of y = x + 1.", + "video_name": "fI6w2kL295Y", + "timestamps": [ + 60 + ], + "3min_transcript": "The table shows the number of stores of a popular US coffee chain from 2000 to 2006. The number of stores recorded is the number at the start of each year on January 1. So in 2000, there was 1,996 stores, in 2005, 6,177, so on and so forth. Determine a reasonable approximation for the instantaneous rate of change, in coffee stores per year at the beginning of 2003-- so we care about 2003-- by taking the average of two nearby secant slopes. So let's visualize this. So this right over here, I've plotted all of the points. Now let me make sure that the axes are clear. This horizontal axis, this is my t-axis that tells us the year. And then the vertical axis is the number of stores. And we could even say that it is a function of time. So you see in the year 2000, there was 1,996 stores-- 2003, 4,272. 2003, 4,272 stores. Now if you could imagine that they're constantly adding stores, you could even imagine minute by minute they're adding stores. So this is just sampling the number of stores they had on January 1. But if you were to really plot it as a more continuous function, it might look something like this. I'll do my best to approximate it. It might be more of some type of curve that looks something like this. And once again, I'm just approximating what it might actually look like. So when they're saying the instantaneous rate of change in coffee stores per year, so this is the change, the instantaneous rate of change of stores per time. They're really saying, we need to approximate the slope of the tangent line in 2003, when time is 2003. So the tangent line might look something like that. is tangent. Now, they say approximate. We don't have the information to figure it out exactly. But we have some data around it, and we can figure out the slopes of the secant lines between this point and those points. And then we can take the average of the slopes of the secant lines to approximate the slope of this tangent line. So for example, we could find the slope of this secant line right over here as we go from 2002 to 2003. And then we can find the slope of this secant line as we go from 2003 to 2004. And if we average those, that should be a pretty good approximation for the instantaneous rate of change in 2003. So let's do that. So the slope of this pink secant line, as we go from 2002 to 2003, that's going to be the number of stores in 2003 minus the number of stores in 2002. So that's the change in our number of stores over the change in years, or the change in time." + }, + { + "Q": "At 0:46 Sal Khan said 30.42 but meant 30.24", + "A": "This is a known error in the video. There is a box that pops up and tells you Sal made an error and provides the correct info.", + "video_name": "Hrjr5f5pZ84", + "timestamps": [ + 46 + ], + "3min_transcript": "Let's see if we can divide 30.24 divided by 0.42. And try pausing the video and solving it on your own before I work through it. So there is a couple of ways you can think about it. We could just write it as 30.24 divided by 0.42. But what do you do now? Well the important realization is, is when you're doing a division problem like this, you will get the same answer as long as you multiply or divide both numbers by the same thing. And to understand that, rewrite this division as 30.42 over 0.42. We could write it really as a fraction. And we know that when we have a fraction like this we're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity. to make it a whole number? Well we can multiply it by 10 and then another 10. So we can multiply it by a 100. If we multiply the denominator by 100 in order to not change the value of this, we also need to multiply the numerator by 100. We are essentially multiplying by 100 over 100, which is just 1. So we're not changing the value of this fraction. Or, you could view this, this division problem. So this is going to be 30.42 times 100. Move the decimal two places to the right, gets you 3,042. The decimal is now there if you care about it. And, 0.42 times 100. Once again move the decimal one, two places to the right, it is now 42. So this is going to be the exact same thing as 3,042 divided by 42. So once again we can move the decimal here, two to the right. then we can move this two to the right. Or we need to move this two to the right. And so this is where, now the decimal place is. You could view this as 3,024. Let me clear that 3024 divided by 42. Let me clear that. And we know how to tackle that already, but lets do it step by step. How many times does 42 go into 3? Well it does not go at all, so we can move on to 30. How many times does 42 go into 30? Well it does not go into 30 so we can move on to 302. How many times does 42 go into 302? And like always this is a bit of an art when your dividing by a two-digit or a multi-digit number, I should say. So lets think about it a little bit. So this is roughly 40. This is roughly 300. So how many times does 40 go into 300? Well how many times does 4 go into 30? Well, it looks like it's about seven times, so I'm going to try out a 7, see if it works out. 7 times 2 is 14." + }, + { + "Q": "At 1:10, Sal says that the perimeter is length+width+length+width, can't you do length*2 and width*2 ?", + "A": "Yup... depends on weather you are faster at addition or multiplication:)", + "video_name": "CDvPPsB3nEM", + "timestamps": [ + 70 + ], + "3min_transcript": "The perimeter of a rectangular fence measures 0.72 kilometers. The length of the fenced area is 160 meters. What is its width? So we have a rectangular fence. Let me draw it. So that is a rectangle. You can imagine we're looking from above. This line is the top of the fence. And if you take its perimeter, the perimeter is the distance around the fence. If you take this distance plus this distance plus that distance plus that distance, it's going to be 0.72 kilometers. That's the total distance of all of the sides. Now, the length of the fenced area is 160 meters. Let's call this the length. So this distance right here is 160 meters. Since it's a rectangle, that distance over there is also going to be 160 meters, and we need to figure out its width. They want us to figure out the width. The width is this distance right over here, which is also Now, what is the perimeter of the fence? The perimeter is the sum of this length plus this length plus that length plus that length. So the perimeter-- let me write it in orange-- is going to be equal to w, the width, plus 160 meters, plus the width, plus 160 meters. And let's assume w is in meters. So we could add it all up. So if you were to add all of these up, you'd get a certain If you were to add all of these up, you have a w plus a w, so you'd be 2w, plus-- 160 plus 160 is 320-- so plus 320. We're assuming that w is in meters. Now, they also told us that the perimeter of the fence is 0.72 kilometers. So the perimeter-- let me do it in that same orange color. The perimeter is also equal to 0.72 kilometers, and we can abbreviate that with km. Now, if we wanted to solve an equation, if we wanted to set this thing equal to the perimeter they gave us, we have to make sure that the units are the same. Here it's in meters. Here it's in kilometers. So let's convert this 0.72 kilometers to meters. And the way to do that, we want the kilometers in the denominator so it cancels out with the kilometers, and you want meters in the numerator. Now, how many meters equal a kilometer? Well, it's 1,000 meters for every 1 kilometer." + }, + { + "Q": "At 8:24 Sal combines X(-b-2a) and X(-a-2b) and is left with only a single X coefficient, I was sure this would be 2X. Can someone please explain why it remains a single X coefficient? Thanks.", + "A": "Because the both 2s in -b-2a and -a-2b would be simplified and that would make it a single X coefficient.", + "video_name": "Vc09LURoMQ8", + "timestamps": [ + 504 + ], + "3min_transcript": "as being equal to that, so let's do that. We get x plus b times c, over a plus 2b is equal to x minus a times c over negative b minus 2a. Well, both sides are divisible by c, we can assume that c is non-zero, so this triangle actually exists, it actually exists in two dimensions, so we can divide both sides by c and we get that. Let's see, now we can cross-multiply. We can multiply x plus b times this quantity right here, and that's going to be equal to a plus 2b times this quantity over here. We can just distribute it, so it's going to be x times all of this business, it's going to be x times negative b minus 2a, plus b times all of this, so I just multiply that times that, is going to equal to this times that, so it's going to be x times all of this business. X times a plus 2b, minus a times that, so distribute the a, minus a squared minus 2ab. Let's see if we can simplify this. We have a negative 2ab on both sides, so let's just subtract that out, so we cancelled things out, let's subtract this from both sides of the equation, so we'll have a minus x times a plus 2b, and I'll subtract that from here, but I'll write it a little different, it'll simplify things, so this will become x times, I'll distribute the negative inside the a plus 2b, so negative a minus 2b, and let's add a b squared to both sides, so plus b squared, plus b squared. and all of the constants on the other. This becomes on the left-hand side, the coefficient on x, I have negative b minus 2b is negative 3b negative 2a minus a is negative 3a, times x is equal to, these guys cancel out, and these guys cancel out, is equal to b squared minus a squared. Let's see if we can factor, this already looks a little suspicious in a good way, something that we should be able to solve. This can be factored. We can factor out a negative three. We can actually factor out a three. We'll get three times b minus a, actually, let's factor out a negative three, so negative three times b plus a times x is equal to, now this is the same thing as b plus a" + }, + { + "Q": "Why does Sal use (1 2, -1 0) matrix at 6:21 ?", + "A": "He s just giving an example of a transformation. Since any matrix can be seen as a transformation, he just choose one at random.", + "video_name": "MIAmN5kgp3k", + "timestamps": [ + 381 + ], + "3min_transcript": "line segment. So that's our L0. It's just a set of vectors. Now we can do the same exercise if we wanted to find out the line, the equation of the line, that goes between x1 and x2. If we wanted to find the equation of that line, we could call this L1. And L1 would be equal to x1 plus t times x2 minus x1 for 0 is less than or equal to t is less than or equal to 1. That's L1. And then finally, if we want to make a triangle out of this, let's define this line right here. Let's define that as L2. L2 would be equal to the set of all of vectors where you start off at x2. Set of all of vectors that are x2 plus some scaled up sum of x0 minus x2 is this vector right here. So x0 minus x2 such that 0 is less than or equal to t is less than or equal to 1. And so if you take the combination, if you were to define kind of a super set-- I could have defined my shape as-- let's say it's the union of all of those guys. Well, let me just write it. L0, L1, and L2. Then you'd have a nice triangle here. If you take the union of all of these three sets, you get that nice triangle there. Now, what I want to do in this video, I think this is all a bit of review for you. But it's maybe a different way of looking at things than we've done in the past. Is I want to understand what happens to this set right here when I take a transformation, a linear transformation, of it? I'll make it a fairly straightforward transformation. Let me define my transformation of x, of any x, to be equal to the matrix 1 minus 1, 2, 0 times whatever vector x. So times x1, x2. And we know that any linear transformation can actually be written as a matrix and vice versa. So you might have said, hey, you know, you're giving an example with the matrix, what about all those other ways to write in your transformation? You can write all of those as a matrix. So let's translate-- let's try to figure out what this is going to look like. What our triangle is going to look like when we transform every point in it. Let me take the transformation first. The transformation of L0 is equal to the transformation of this thing. This is just one of the particular members." + }, + { + "Q": "Would it work if at 1:50, instead of factoring as he did, could we just FOIL? I would only think to do it that way if I knew beforehand that the result should look a certain way.", + "A": "I don t think I understand your question too well... Sal didn t factor at that point in the problem, he just distributed the entire sin expression with the (5 - 3y )", + "video_name": "-EG10aI0rt0", + "timestamps": [ + 110 + ], + "3min_transcript": "Let's say we have the relationship y is equal to cosine of 5x minus 3y. And what I want to find is the rate at which y is changing with respect to x. And we'll assume that y is a function of x. So let's do what we've always been doing. Let's apply the derivative operator to both sides of this equation. On the left-hand side, right over here, we get dy/dx is equal to-- now here on the right hand side, we're going to apply the chain rule. The derivative of the cosine of something, with respect to that something, is going to be equal to negative sine of that something. So negative sine of 5x minus 3y. And then we have to multiply that by the derivative of that something with respect to x. So what's the derivative of the something with respect to x? Well the derivative of 5x with respect to x is just equal to 5. And the derivative of negative 3y with respect to x Negative 3 times the derivative of y with respect to x. And now we just need to solve for dy/dx. And as you can see, with some of these implicit differentiation problems, this is the hard part. And actually, let me make that dy/dx the same color. So that we can keep track of it easier. So this is going to be dy/dx. And then I can close the parentheses. So how can we do it? It's just going to be a little bit of algebra to work through. Well, we can distribute the sine of 5x minus 3y. So let me rewrite everything. We get dy-- whoops, I'm going to do that in the yellow color-- we get dy/dx is equal to-- you distribute the negative sine of 5x minus 3y. You get-- so let me make sure we know what we're doing. It's going to be, we're going to distribute that, and we're going to distribute that. So you're going to have 5 times all of this. So you're going to have negative this 5 times And then you're going to have the negative times a negative, those are going to, you're going to end up with a positive. And so you're going to end up with plus 3 times the sine of 5x minus 3y dy/dx. Now what we can do is subtract 3 sine of 5x minus 3y from both sides. So just to be clear, this is essentially a 1 dy/dx. So if we subtract this from both sides, we are left with-- So on the left-hand side, we're going to have a 1 dy/dx, and we're going to subtract from that 3 sine of 5x minus 3y dy/dx's. So you're going to have 1 minus 3-- I'll keep the color for the 3 for fun--" + }, + { + "Q": "At 2:35, when Sal subtracts the 3Sin(5x-3y), what happened to the dy/dx that was being multiplied to it? Could the 3Sin(5x-3y) even be subtracted without the dy/dx?", + "A": "Yes, you are correct. We are subtracting 3sin(5x-3y)dy/dx. This is exactly what Sal did. He just didn t mention dy/dx out loud.", + "video_name": "-EG10aI0rt0", + "timestamps": [ + 155 + ], + "3min_transcript": "Negative 3 times the derivative of y with respect to x. And now we just need to solve for dy/dx. And as you can see, with some of these implicit differentiation problems, this is the hard part. And actually, let me make that dy/dx the same color. So that we can keep track of it easier. So this is going to be dy/dx. And then I can close the parentheses. So how can we do it? It's just going to be a little bit of algebra to work through. Well, we can distribute the sine of 5x minus 3y. So let me rewrite everything. We get dy-- whoops, I'm going to do that in the yellow color-- we get dy/dx is equal to-- you distribute the negative sine of 5x minus 3y. You get-- so let me make sure we know what we're doing. It's going to be, we're going to distribute that, and we're going to distribute that. So you're going to have 5 times all of this. So you're going to have negative this 5 times And then you're going to have the negative times a negative, those are going to, you're going to end up with a positive. And so you're going to end up with plus 3 times the sine of 5x minus 3y dy/dx. Now what we can do is subtract 3 sine of 5x minus 3y from both sides. So just to be clear, this is essentially a 1 dy/dx. So if we subtract this from both sides, we are left with-- So on the left-hand side, we're going to have a 1 dy/dx, and we're going to subtract from that 3 sine of 5x minus 3y dy/dx's. So you're going to have 1 minus 3-- I'll keep the color for the 3 for fun-- is going to be equal to, well, we subtracted this from both sides. So on the right-hand side, this is going to go away. So we're just going to be left with a negative 5 sine of 5x minus 3y. And we're in the home stretch now. To solve for dy/dx, we just have to divide both sides of the equation by this. And we are left with dy/dx is equal to this thing, negative 5 times the sine of 5x minus 3y. All of that over 1 minus 3 sine of 5x minus 3y." + }, + { + "Q": "at 1:24, how can there be 4 possibilities, one person only have 3 options and not his own self. right?", + "A": "The 4 means that the first person involved in shaking hands can be any of the 4 people. The 3 means that the second person involved in shaking hands can be any of the remaining 3 people not counting the person identified as the first person.", + "video_name": "boH4l1SgJbM", + "timestamps": [ + 84 + ], + "3min_transcript": "- Let's say that there are four people in a room. And you're probably tired of me naming the people with letters, but I'm going to continue doing that. So the four people in the room are people A, B, C, and D. And they are all told, \"You don't know each other. \"So I want you to all meet each other. \"You need to shake the hand, exactly once, \"of every other person in the room so that you all meet.\" So my question to you is, if each of these people need to shake the hand of every other person exactly once, how many handshakes are going to occur? The number of handshakes that are going to occur. So, like always, pause the video and see if you can make sense of this. Alright, I'm assuming you've had a go at it. So one way to think about it is, if you say there's a handshake, two people are party to a handshake. We're not talking about some new three-person handshake or four-person handshake, we're just talking about the traditional, two people shake their right hands. another person in this party. There's four possibilities of one party. And if we assume people aren't shaking their own hands, which we are assuming, they're always going to shake someone else's hand. For each of these four possibilities who's this party, there's three possibilities who's the other party. And so you might say that there's four times three handshakes. Since there's four times three possible handshakes. And what I'd like you to do is think a little bit about whether this is right, whether there would actually be 12 handshakes. You might have thought about it, and you might say, there's four times three, this is actually counting the permutations. This is counting how many ways can you permute four people into two buckets, the two buckets of handshakers, where you care about which bucket they are in. Whether they're handshaker number one, or handshaker number two. and B being the number two handshaker as being different than B being the number one handshaker and A being the number two handshaker. But we don't want both of these things to occur. We don't want A to shake B's hand, where A is facing north and B is facing south. And then another time, they shake hands again where now B is facing north and A is facing south. We only have to do it once. These are actually the same thing, so there's no reason for both of these to occur. So we are going to be double counting. So what we really want to do is think about combinations. One way to think about it is, you have four people. In a world of four people or a pool of four people, how many ways can you choose two?" + }, + { + "Q": "1:43\n\nWe were taught about radial symmetry. Is rotational and radial symmetry the same?", + "A": "rotational and radial symmetry is generally the same, but radial symmetry is usually used in biological contexts", + "video_name": "toKu2-qzJeM", + "timestamps": [ + 103 + ], + "3min_transcript": "So in my last video I joked about folding and cutting spheres instead of paper. But then I thought, why not? I mean, finite symmetry groups on the Euclidean plane are fun and all, but there's really only two types. Some amount of mirror lines around a point, and some amount of rotations around a point. Spherical patterns are much more fun. And I happen to be a huge fan of some of these symmetry groups, maybe just a little bit. Although snowflakes are actually three dimensional, this snowflake doesn't just have lines of mirror symmetry, but planes of mirror symmetry. And there's one more mirror plane. The one going flat through the snowflake, because one side of the paper mirrors the other. And you can imagine that snowflake suspended in a sphere, so that we can draw the mirror lines more easily. Now this sphere has the same symmetry as this 3D paper snowflake. If you're studying group theory, you could label this with group theory stuff, but whatever. I'm going to fold this sphere on these lines, and then cut it, and it will give me something with the same symmetry as a paper snowflake. Except on a sphere, and it's a mess, so let's glue it to another sphere. And now it's perfect and beautiful in every way. But the point is it's equivalent to the snowflake OK, so that's a regular, old 6-fold snowflake, but I've seen pictures of 12-fold snowflakes. How do they work? Sometimes stuff goes a little oddly at the very beginning of snowflake formation and two snowflakes sprout. Basically on top of each other, but turned 30 degrees. If you think of them as one flat thing, it has 12-fold symmetry, but in 3D it's not really true. The layers make it so there's not a plane of symmetry here. See the branch on the left is on top, while in the mirror image, the branch on the right is on top. So is it just the same symmetry as a normal 6-fold snowflake? What about that seventh plane of symmetry? But no, through this plane one side doesn't mirror the other. There's no extra plane of symmetry. But there's something cooler. Rotational symmetry. If you rotate this around this line, you get the same thing. The branch on the left is still on top. If you imagine it floating in a sphere you can draw the mirror lines, and then 12 points of rotational symmetry. So I can fold, then slit it so it can swirl around the rotation point. And cut out a sphereflake with the same symmetry as this. And you can fold spheres other ways to get other patterns. OK what about fancier stuff like this? Well, all I need to do is figure out the symmetry to fold it. So, say we have a cube. What are the planes of symmetry? It's symmetric around this way, and this way, and this way. Anything else? How about diagonally across this way? But in the end, we have all the fold lines. And now we just need to fold a sphere along those lines to get just one little triangle thing. And once we do, we can unfold it to get something with the same symmetry as a cube. And of course, you have to do something with tetrahedral symmetry as long as you're there. And of course, you really want to do icosahedral, but the plastic is thick and imperfect, and a complete mess, so who knows what's going on. But at least you could try some other ones with rotational symmetry. And other stuff and make a mess. And soon you're going to want to fold and cut the very fabric of space itself to get awesome, infinite 3D symmetry groups, such as the one water molecules follow when they pack in together into solid ice crystals. And before you know it, you'll be playing with multidimensional, quasi crystallography, early algebra's, or something. So you should probably just stop now." + }, + { + "Q": "1:23 how did he get 6 as his exponent ?", + "A": "at 1:23 he was subtracting 6 from 2 because there was 2 exponents", + "video_name": "AR1uqNbjM5s", + "timestamps": [ + 83 + ], + "3min_transcript": "Let's do some exponent examples that involve division. Let's say I were to ask you what 5 to the sixth power divided by 5 to the second power is? Well, we can just go to the basic definition of what an exponent represents and say 5 to the sixth power, that's going to be 5 times 5 times 5 times 5 times 5-- one more 5-- times 5. 5 times itself six times. And 5 squared, that's just 5 times itself two times, so it's just going to be 5 times 5. Well, we know how to simplify a fraction or a rational expression like this. We can divide the numerator and the denominator by one 5, and then these will cancel out, and then we can do it by another 5, or this 5 and this 5 will cancel out. And what are we going to be left with? 5 times 5 times 5 times 5 over 1, or you could say that this is just 5 to the fourth power. Essentially we started with six in the numerator, six 5's multiplied by themselves in the numerator, and then we subtracted out. We were able to cancel out the 2 in the denominator. So this really was equal to 5 to the sixth power minus 2. So we were able to subtract the exponent in the denominator from the exponent in the numerator. Let's remember how this relates to multiplication. If I had 5 to the-- let me do this in a different color. 5 to the sixth times 5 to the second power, we saw in the last video that this is equal to 5 to the 6 plus-- I'm trying to make it color coded for you-- 6 plus 2 power. Now, we see a new property. And in the next video, we're going see that these aren't really different properties. They're really kind of same sides of the same coin when we learn about negative exponents. divided by 5 to the second power-- let me do it in a different color-- is going to be equal to 5 to the-- it's time consuming to make it color coded for you-- 6 minus 2 power or 5 to the fourth power. Here it's going to be 5 to the eighth. So when you multiply exponents with the same base, you add When you divide with the same base, you subtract the denominator exponent from the numerator exponent. Let's do a bunch more of these examples right here. What is 6 to the seventh power divided by 6 to the third power?" + }, + { + "Q": "AT 9:20 would 5/4 x^-1 y also be correct", + "A": "It generally would be considered incomplete. Final answers should have positive exponents. And, we would multiply the items together to show them as one fraction. This is why Sal is showing the answer as 5y / (4x)", + "video_name": "AR1uqNbjM5s", + "timestamps": [ + 560 + ], + "3min_transcript": "And we would have simplified this about as far as you can go. Let's do one more of these. I think they're good practice and super-valuable experience later on. Let's say I have 25xy to the sixth over 20y to the fifth x squared. So once again, we can rearrange the numerators and the denominators. So this you could rewrite as 25 over 20 times x over x squared, right? We could have made this bottom 20x squared y to the fifth-- it doesn't matter the order we do it in-- times y to the sixth over y to the fifth. actually just simplify fractions. 25 over 20, if you divide them both by 5, this is equal to 5 over 4. x divided by x squared-- well, there's two ways you could think about it. That you could view as x to the negative 1. You have a first power here. 1 minus 2 is negative 1. So this right here is equal to x to the negative 1 power. Or it could also be equal to 1 over x. These are equivalent. So let's say that this is equal into 1 over x, just like that. And it would be. x over x times x. One of those sets of x's would cancel out and you're just left with 1 over x. And then finally, y to the sixth over y to the fifth, that's y to the 6 minus 5 power, which is just y to the first power, or just y, so times y. rational expression, you have 5 times 1 times y, which would be 5y, all of that over 4 times x, right? This is y over 1, so 4 times x times 1, all of that over 4x, and we have successfully simplified it." + }, + { + "Q": "if at 1:05 L = 1 and R = 0 then that is 01001100110 - a binary number, will someone find what it represents?", + "A": "Bin 01001100110 = Dec 614 If you were to switch them, with L=0, R=1, Bin 10110011001 = Dec 1433", + "video_name": "Gx5D09s5X6U", + "timestamps": [ + 65 + ], + "3min_transcript": "Snakes. Lots and lots of snakes. These snakes are just writhing with potential, similar parts linked together. They move in a specific and limited way. Part of the potential of things is how they break. These snakes break fantastically into these snake modules. You can put them back together too, allowing the existence of the super snake. Super snakes are obviously desirable for many reasons, besides being inherently awesome. You can wear them, and put them on things, and drop them, which I find amusing for some reason. You can arrange them into a space-filling, fractal curve, if that's what you like, which I do. You can even jump snake. But let's not forget. You can make mini-snakes too, which enables the snake stash, and starting from mini-snake gives you room to grow. Snake. Snake. Snake. Snake. They like to bend into this angle. Who cares what it actually is, besides about 90 degrees? But it begs the question, how many ways can I fold this snake if it has 10 segments? I can notate the way it slithers back and forth from tail to head, left, right, right, left, left, right, left. This is a valid slither. This is an invalid one, since anyone who's played snake knows that a snake isn't allowed to run into itself. Given a slither, how can I tell whether it's-- of course, snake lets you go straight too. So you can do another version of this that allows going straight and notate it like this, and wonder whether this snake is a loser snake or not. Snake. Snake. Snake. Snake. Snake. Don't forget to try putting the snake modules together in ways they were never meant to go. You can mix colors. So in theory, I could be hiding a secret message in the color pattern of this snake. But I'm not, because I'm lazy. So it's just the digits of the binary expansion of pi. Even better, you can attach more than one segment at a point. I can have a two-headed serpent. I can play the game where I cut off the heads of a Hydra, adding two more in the old head's place and see how far that gets me. I can put the snake modules on my fingertips and have snaky fingers. That's cool too. I can even do super snaky fingers. Snake. Snake. Snake. Snake. Snake. Snake. Snake. Snake. Snake." + }, + { + "Q": "At 2:06, could we take the y and pull it to the front similarly to how we pulled x to the front earlier in the video since it's an exponent?", + "A": "The Y is not an exponent. We re not raising 5 to the Y power. Rather, it is the result of raising the 5 to some power, i.e., 5 to the what power equals y? So the answer is no.", + "video_name": "RhzXX5PbsuQ", + "timestamps": [ + 126 + ], + "3min_transcript": "We're asked to simplify log base 5 of 25 to the x power over y. So we can use some logarithm properties. And I do agree that this does require some simplification over here, that having this right over here inside of the logarithm is not a pleasant thing to look at. So the first thing that we realize-- and this is one of our logarithm properties-- is logarithm for a given base-- so let's say that the base is x-- of a/b, that is equal to log base x of a minus log base x of b. And here we have 25 to the x over y. So we can simplify. So let me write this down. I'll do this in blue. Log base 5 of 25 to the x over y using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying. is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did it right over there. So this part right over here can be rewritten as x times the logarithm base 5 of 25. And then, of course, we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about. This part right here is asking us, what power do I have to raise 5 to to get to 25? So we have to raise 5 to the second power to get to 25. So this simplifies to 2. So then we are left with, this is equal to-- and I'll write it in front of the x now-- 2 times x minus log base 5 of y." + }, + { + "Q": "At 0:20 can we move the x out in front first, instead of applying the quotient log property first? And then apply the quotient log property?\n\nIf not...is there a certain order in which log properties should be applied (like BEDMAS)??? Thanks in advance :)", + "A": "You can t move the x in front first, because not the whole factor is to the xth power. if we had log(25/y)^x we could. otherwise the whole term, so both log25 and log5 would be times 5, which is wrong. I hope this helps you, English is not my first language, so I don t know if I explained it well enough!", + "video_name": "RhzXX5PbsuQ", + "timestamps": [ + 20 + ], + "3min_transcript": "We're asked to simplify log base 5 of 25 to the x power over y. So we can use some logarithm properties. And I do agree that this does require some simplification over here, that having this right over here inside of the logarithm is not a pleasant thing to look at. So the first thing that we realize-- and this is one of our logarithm properties-- is logarithm for a given base-- so let's say that the base is x-- of a/b, that is equal to log base x of a minus log base x of b. And here we have 25 to the x over y. So we can simplify. So let me write this down. I'll do this in blue. Log base 5 of 25 to the x over y using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying. is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did it right over there. So this part right over here can be rewritten as x times the logarithm base 5 of 25. And then, of course, we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about. This part right here is asking us, what power do I have to raise 5 to to get to 25? So we have to raise 5 to the second power to get to 25. So this simplifies to 2. So then we are left with, this is equal to-- and I'll write it in front of the x now-- 2 times x minus log base 5 of y." + }, + { + "Q": "at 2:00, why x(2 - log_5(y)) = 2x - log_5(y) and not 2x - xlog_5(y)?", + "A": "The 2 resulted from the expression log_5(25) , which was being multiplied by x. There were no parentheses to distribute the x to the second term ( -log_5(y) ), so it was unaffected by it.", + "video_name": "RhzXX5PbsuQ", + "timestamps": [ + 120 + ], + "3min_transcript": "We're asked to simplify log base 5 of 25 to the x power over y. So we can use some logarithm properties. And I do agree that this does require some simplification over here, that having this right over here inside of the logarithm is not a pleasant thing to look at. So the first thing that we realize-- and this is one of our logarithm properties-- is logarithm for a given base-- so let's say that the base is x-- of a/b, that is equal to log base x of a minus log base x of b. And here we have 25 to the x over y. So we can simplify. So let me write this down. I'll do this in blue. Log base 5 of 25 to the x over y using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying. is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did it right over there. So this part right over here can be rewritten as x times the logarithm base 5 of 25. And then, of course, we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about. This part right here is asking us, what power do I have to raise 5 to to get to 25? So we have to raise 5 to the second power to get to 25. So this simplifies to 2. So then we are left with, this is equal to-- and I'll write it in front of the x now-- 2 times x minus log base 5 of y." + }, + { + "Q": "At 1:46, I got confused. If something is \"over\" something else, does that mean divide that number by the bottom number?", + "A": "Yes but do you see that x was being divided by 4 then multiplied by 4 so there is no reason to go through all that math when there opposites it would be like adding 4 to five and getting nine then subtracting $ and getting five again", + "video_name": "p5e5mf_G3FI", + "timestamps": [ + 106 + ], + "3min_transcript": "We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then that is equal to x over 4. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. unnecessary. And so we have negative 18 is equal to x over 4. And our whole goal here is to isolate the x, to solve for the x. And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4, times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4, you're just going to be left with an x. And on the other side, 4 times negative 18. Let's see, that's 40. Well, let's just write it out. So 18 times 4. If we were to multiply 18 times 4, 4 times 8 is 32. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to-- instead of writing x, I'm going to write negative 72-- is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true. This right-hand side, when x is equal to negative 72, does indeed equal negative 16." + }, + { + "Q": "at 2:05 it says add one shouldn't it say add 3?????", + "A": "Hey, you re right! That s funny.... well, he got the answer right anyhow.", + "video_name": "p5e5mf_G3FI", + "timestamps": [ + 125 + ], + "3min_transcript": "We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then that is equal to x over 4. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. unnecessary. And so we have negative 18 is equal to x over 4. And our whole goal here is to isolate the x, to solve for the x. And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4, times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4, you're just going to be left with an x. And on the other side, 4 times negative 18. Let's see, that's 40. Well, let's just write it out. So 18 times 4. If we were to multiply 18 times 4, 4 times 8 is 32. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to-- instead of writing x, I'm going to write negative 72-- is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true. This right-hand side, when x is equal to negative 72, does indeed equal negative 16." + }, + { + "Q": "At 3:30, how does he know that the question is linear?", + "A": "He knows at 3:30 because the equation can be plotted as a line. This means it will be straight no matter how far you extend it, and will not curve or bend anywhere. It is constant and will always remain so. I hope that helps :)", + "video_name": "OWPVZoxNe-U", + "timestamps": [ + 210 + ], + "3min_transcript": "So if we have the minutes they play time playing Bologna and the time playing Cut Your Wire. If I were to add those two together, so if I say x plus y. I'll write that plus in a neutral color, x plus y. What does this need to be equal to? The time I play Bologna Man plus the time I play Have to Cut the Wire. Well if I add them together, they want to spend exactly 45 minutes playing both games. So this is going to be equal to 45 minutes. Now we have set up an equation that relates the time playing Bologna Man and the time playing You Have to Cut the Wire. But now we have to think about is this a linear relationship? for a linear relationship is if you can write it in the traditional form of a line. So if you can write it in the y is equal to mx plus b form, where m is the slope of the line and b is the y-intercept. So let's see if we can do that. Well if we want to do that here, we could just subtract x from both sides. You subtract x from both sides, you get-- so let's subtract an x over here, let's subtract an x over here. So negative x plus this and then subtract an x there. Well, that's going to cancel, and you're going to be left with-- and I'm going try to write it in this form right over here-- y is equal to 45. Let me do that in the same color, just to make it not be confusing. y is equal to, and I'll write the negative x first because we have the x term right over here first. So y is equal to negative x plus 45. But you see here, it has that form. And you might say wait what is m here? I see that b is 45. Well if I write negative x, that's the same thing as writing negative 1x. So this is definitely a line. I was able to write it in this form right over here. So can this relationship be represented using a linear equation? Absolutely, absolutely yes." + }, + { + "Q": "At 4:46, Sal says that there is only one way to get to the cube below the starting point. Isn't that logic flawed because it is a three dimensional object? Couldn't he go to the right an infinite number of times and then go down? Or go to the right one, go down one, and wrap around the cube?", + "A": "You are not travelling along the surface of the cube, you are traveling through the cube itself. Therefore, there is no wrapping around the cube. There is no going right an infinite number of times, because you can only go right twice. You are able to move right twice, down twice, and forward twice. All that matters is the order in which you do those moves.", + "video_name": "wRxzDOloS3o", + "timestamps": [ + 286 + ], + "3min_transcript": "Let me draw some squares in here. It's like that, and like that, and like that, and like that. And then, the middle one was the mauve layer. We'll draw that. The mauve layer looks something like that. And you can imagine I'm slicing it, and just looking at it from above. That's the idea here. And it's going to help us visualize this problem. So the mauve layer looks something like that. And then finally the orange layer. Looks like this. And we're almost ready to actually start doing the problem. Good enough. So just to make sure we understand our visualization, this layer up here-- we call that layer one. We could but this as box one. So I'll put a little two here. And I don't want to get these confused with the paths and all that, so I'm writing it really small. And this is layer three, or level three. And that's right there. And just to make sure you understand, this corner right there, this is our start point. And that's right there. Because this is a whole top. So this is the back left of the top. And are finish point, the bottom right, is right here. So, essentially, our problem goes from, how many ways to get from there to there, to how many ways to get from there to there? So let's just stay within a layer. So how many ways can I get to this point right here? Well I can only go from this point, and go straight in the layer like that. So there's only one way to get there. That movement is the exact same movement as this right here. Going from this box to this box. So there's one way to get there. That's the same thing as there. And similarly, I could go there. And I can just go one more step. So there's only one way to get there. And that's like going there and then there. That's the only way to get there. Or I could go two to the right there. And that's the only way to get there. And now if you watched the two dimensional path counting brain teaser, you know that there's two ways to get here. And the logic is, well you could draw it out. You could go like that. One, two. And that's the same thing as saying, one, two. Though it's easier to visualize here. But the general logic was, well, to know how many ways to get to any square, think about the squares that lead to it, and how many ways can I get to those two squares? And then sum them up. And by the same logic-- so there's two ways to get here. That's that cell. Three ways to get here, right? Two plus one is three. One plus two is three. And three plus three is six. So there were six ways to get to this cube right there, from that one. So this isn't too different from the two dimensional problem so far. But now it gets interesting." + }, + { + "Q": "At 8:43, how does Sal automatically determine that x is 8 or 2. In the equation above, shouldn't x be -8 or -2 since x is getting subtracted by those two numbers?", + "A": "No. Since it is -5, he would have to add 5 to both sides to isolate x. Therefore, it would be 5 plus or minus 3, giving him 8 and 2. If he subtracted 5, the left side would have -10 + x and the right side would have -5 plus or minus 3. This would not be possible. He would have to add 10 to both sides to isolate x, giving him x=5 plus or minus 3.", + "video_name": "55G8037gsKY", + "timestamps": [ + 523 + ], + "3min_transcript": "on factoring so far. We can only do this when this is a perfect square. If you got, like, x minus 3, times x plus 4, and that would be equal to 9, that would be a dead end. You wouldn't be able to really do anything constructive with that. Only because this is a perfect square, can we now say x minus 5 squared is equal to 9, and now we can take the square root of both sides. So we could say that x minus 5 is equal to plus or minus 3. Add 5 to both sides of this equation, you get x is equal to 5 plus or minus 3, or x is equal to-- what's 5 plus 3? Well, x could be 8 or x could be equal to 5 minus 3, or x is equal to 2. Now, we could have done this equation, this quadratic equation, the traditional way, the way you were tempted to do it. What happens if you subtract 9 from both You'll get x squared minus 10x. And what's 25 minus 9? 25 minus 9 is 16, and that would be equal to 0. And here, this would be just a traditional factoring problem, the type that we've seen in the last few videos. What two numbers, when you take their product, you get positive 16, and when you sum them you get negative 10? And maybe negative 8 and negative 2 jump into your brain. So we get x minus 8, times x minus 2 is equal to 0. And so x could be equal to 8 or x could be equal to 2. That's the fun thing about algebra: you can do things in two completely different ways, but as long as you do them in algebraically-valid ways, you're not going to get different answers. And on some level, if you recognize this, this is easier because you didn't have to do that little game in your head, in terms of, oh, what two numbers, when you multiply Here, you just said, OK, this is x minus 5-- oh, I guess you did have to do it. You had to say, oh, 5 times 5 is 25, and negative 10 is negative 5 plus negative 5. So I take that back, you still have to do that little game in your head. So let's do another one. Let's do one more of these, just to really get ourselves nice and warmed up here. So, let's say we have x squared plus 18x, plus 81 is equal to 1. So once again, we can do it in two ways. We could subtract 1 from both sides, or we could recognize that this is x plus 9, times x plus 9. This right here, 9 times 9 is 81, 9 plus 9 is 18. So we can write our equation as x plus 9" + }, + { + "Q": "why do you find the square root of (4x+1 )squared and 8 around 1:30?", + "A": "Since we are trying to solve for x, we want to isolate x on one side of the equation. in order to do that, we have to cancel out as much stuff on the left hand side of the equation as possible. In this example, if we take (4x+1)^2 = 8 and we take the square root of both sides, we are left with 4x+1=sqrt(8), and we are one step closer to isolating x.", + "video_name": "55G8037gsKY", + "timestamps": [ + 90 + ], + "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously." + }, + { + "Q": "At 1:40, what does Sal mean by positive or negative square root. If the square root on one side is positive, shouldn't the square root on the other side be positive, too?", + "A": "Technically, we would get two equations: 4\u00f0\u009d\u0091\u00a5 + 1 = \u00c2\u00b1\u00e2\u0088\u009a8 -(4\u00f0\u009d\u0091\u00a5 + 1) = \u00c2\u00b1\u00e2\u0088\u009a8 Since the two equations are equivalent, we can discard one of them, leaving us with: 4x + 1 = \u00c2\u00b1\u00e2\u0088\u009a8", + "video_name": "55G8037gsKY", + "timestamps": [ + 100 + ], + "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously." + }, + { + "Q": "at 10:09 why is it plus or minus 1, when we square?", + "A": "When you square, you multiply a number by itself. And -3*-3 = 9 as well as 3*3 = 9. So there can be two solutions, when you take the square root of something (9 in my example). It s different when you cube a number. -3*-3*-3 = -27 which is not the same as 3*3*3 = 27.", + "video_name": "55G8037gsKY", + "timestamps": [ + 609 + ], + "3min_transcript": "You'll get x squared minus 10x. And what's 25 minus 9? 25 minus 9 is 16, and that would be equal to 0. And here, this would be just a traditional factoring problem, the type that we've seen in the last few videos. What two numbers, when you take their product, you get positive 16, and when you sum them you get negative 10? And maybe negative 8 and negative 2 jump into your brain. So we get x minus 8, times x minus 2 is equal to 0. And so x could be equal to 8 or x could be equal to 2. That's the fun thing about algebra: you can do things in two completely different ways, but as long as you do them in algebraically-valid ways, you're not going to get different answers. And on some level, if you recognize this, this is easier because you didn't have to do that little game in your head, in terms of, oh, what two numbers, when you multiply Here, you just said, OK, this is x minus 5-- oh, I guess you did have to do it. You had to say, oh, 5 times 5 is 25, and negative 10 is negative 5 plus negative 5. So I take that back, you still have to do that little game in your head. So let's do another one. Let's do one more of these, just to really get ourselves nice and warmed up here. So, let's say we have x squared plus 18x, plus 81 is equal to 1. So once again, we can do it in two ways. We could subtract 1 from both sides, or we could recognize that this is x plus 9, times x plus 9. This right here, 9 times 9 is 81, 9 plus 9 is 18. So we can write our equation as x plus 9 Take the square root of both sides, you get x plus 9 is equal to plus or minus the square root of 1, which is just 1. So x is equal to-- subtract 9 from both sides-- negative 9 plus or minus 1. And that means that x could be equal to-- negative 9 plus 1 is negative 8, or x could be equal to-- negative 9 minus 1, which is negative 10. And once again, you could have done this the traditional way. You could have subtracted 1 from both sides and you would have gotten x squared plus 18x, plus 80 is equal to 0. And you'd say, hey, gee, 8 times 10 is 80, 8 plus 10 is 18, so you get x plus 8, times x plus 10 is equal to 0. And then you'd get x could be equal to negative 8, or x could be equal to negative 10. That was good warm up. Now, I think we're ready to tackle completing the square." + }, + { + "Q": "At 1:50, why does Sal write \u00e2\u0088\u009a8 as \u00e2\u0088\u009a4*2, and how would I know when I should?", + "A": "You can simplify any square roots if they are the product of a perfect square and another number. For example, the number 24. The \u00e2\u0088\u009a24 is the same thing as the (\u00e2\u0088\u009a4*6). We know that \u00e2\u0088\u009a4=2 so, we bring it outside the \u00e2\u0088\u009a sign. We result with 2\u00e2\u0088\u009a6. Same thing Sal did, just with different numbers.", + "video_name": "55G8037gsKY", + "timestamps": [ + 110 + ], + "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously." + }, + { + "Q": "At 5:30 can you simply the fraction even more than that? Can you separate the fraction into 2 parts?", + "A": "It appears you are talking more @4:10 where he ends with the fraction (-1 +/- 2 \u00e2\u0088\u009a2)/4. You cannot simplify the fraction anymore than that, but you could separate it into parts, and you could separate it into two different solutions. Separating into parts to get -1/4 +/- \u00e2\u0088\u009a2/2 does not look as neat and compact as what he has, but it is equivalent.", + "video_name": "55G8037gsKY", + "timestamps": [ + 330 + ], + "3min_transcript": "equation, what do I have? On the left-hand side, I'm just left with 4x. On the right-hand side, I have-- you can't really mathematically, I mean, you could do them if you had a calculator, but I'll just leave it as negative 1 plus or minus the square root, or 2 times the square root of 2. That's what 4x is equal to. If we did it here, as two separate equations, same idea. Subtract 1 from both sides of this equation, you get 4x is equal to negative 1 plus 2, times the square root of 2. This equation, subtract 1 from both sides. 4x is equal to negative 1 minus 2, times the square root of 2. This statement right here is completely equivalent to these two statements. Now, last step, we just have to divide both sides by 4, so you divide both sides by 4, and you get x is equal to root of 2, over 4. Now, this statement is completely equivalent to dividing each of these by 4, and you get x is equal to negative 1 plus 2, times the square root 2, over 4. This is one solution. And then the other solution is x is equal to negative 1 minus 2 roots of 2, all of that over 4. That statement and these two statements are equivalent. And if you want, I encourage you to-- let's substitute one of these back in, just so you feel confident that something as bizarro as one of these expressions can be a solution to a nice, vanilla-looking equation like this. So let's substitute it back in. 4 times x, or 4 times negative 1, plus 2 root 2, over 4, plus 1 squared, minus 8 is equal to 0. plus 2 roots 2, plus 1, squared, minus 8 is equal to 0. This negative 1 and this positive 1 cancel out, so you're left with 2 roots of 2 squared, minus 8 is equal to 0. And then what are you going to have here? So when you square this, you get 4 times 2, minus 8 is equal to 0, which is true. 8 minus 8 is equal to 0. And if you try this one out, you're going to get the exact same answer. Let's do another one like this. And remember, these are special forms where we have squares of binomials in our expression. And we're going to see that the entire quadratic formula is actually derived from a notion like this, because you can actually turn any, you can turn any, quadratic equation into a perfect square equalling something else." + }, + { + "Q": "What does Sal mean when he says, \"vanilla like\"? I am guessing he means not-so-complex-looking, like at 4:48.", + "A": "In the U.S, an object can be described to be as plain as vanilla , which means that something is plain and simple like vanilla ice cream. You are correct in that he means not-so-complex-looking.", + "video_name": "55G8037gsKY", + "timestamps": [ + 288 + ], + "3min_transcript": "equation, what do I have? On the left-hand side, I'm just left with 4x. On the right-hand side, I have-- you can't really mathematically, I mean, you could do them if you had a calculator, but I'll just leave it as negative 1 plus or minus the square root, or 2 times the square root of 2. That's what 4x is equal to. If we did it here, as two separate equations, same idea. Subtract 1 from both sides of this equation, you get 4x is equal to negative 1 plus 2, times the square root of 2. This equation, subtract 1 from both sides. 4x is equal to negative 1 minus 2, times the square root of 2. This statement right here is completely equivalent to these two statements. Now, last step, we just have to divide both sides by 4, so you divide both sides by 4, and you get x is equal to root of 2, over 4. Now, this statement is completely equivalent to dividing each of these by 4, and you get x is equal to negative 1 plus 2, times the square root 2, over 4. This is one solution. And then the other solution is x is equal to negative 1 minus 2 roots of 2, all of that over 4. That statement and these two statements are equivalent. And if you want, I encourage you to-- let's substitute one of these back in, just so you feel confident that something as bizarro as one of these expressions can be a solution to a nice, vanilla-looking equation like this. So let's substitute it back in. 4 times x, or 4 times negative 1, plus 2 root 2, over 4, plus 1 squared, minus 8 is equal to 0. plus 2 roots 2, plus 1, squared, minus 8 is equal to 0. This negative 1 and this positive 1 cancel out, so you're left with 2 roots of 2 squared, minus 8 is equal to 0. And then what are you going to have here? So when you square this, you get 4 times 2, minus 8 is equal to 0, which is true. 8 minus 8 is equal to 0. And if you try this one out, you're going to get the exact same answer. Let's do another one like this. And remember, these are special forms where we have squares of binomials in our expression. And we're going to see that the entire quadratic formula is actually derived from a notion like this, because you can actually turn any, you can turn any, quadratic equation into a perfect square equalling something else." + }, + { + "Q": "At 5:29, how did he get 4 times 2?", + "A": "It s because 2 and sqrt(2) were squared separately, therefore you have 4 times 2. You can also think of 2sqrt(2) as sqrt(8) and when you square it, you ll get 8 which is 4 times 2.", + "video_name": "55G8037gsKY", + "timestamps": [ + 329 + ], + "3min_transcript": "equation, what do I have? On the left-hand side, I'm just left with 4x. On the right-hand side, I have-- you can't really mathematically, I mean, you could do them if you had a calculator, but I'll just leave it as negative 1 plus or minus the square root, or 2 times the square root of 2. That's what 4x is equal to. If we did it here, as two separate equations, same idea. Subtract 1 from both sides of this equation, you get 4x is equal to negative 1 plus 2, times the square root of 2. This equation, subtract 1 from both sides. 4x is equal to negative 1 minus 2, times the square root of 2. This statement right here is completely equivalent to these two statements. Now, last step, we just have to divide both sides by 4, so you divide both sides by 4, and you get x is equal to root of 2, over 4. Now, this statement is completely equivalent to dividing each of these by 4, and you get x is equal to negative 1 plus 2, times the square root 2, over 4. This is one solution. And then the other solution is x is equal to negative 1 minus 2 roots of 2, all of that over 4. That statement and these two statements are equivalent. And if you want, I encourage you to-- let's substitute one of these back in, just so you feel confident that something as bizarro as one of these expressions can be a solution to a nice, vanilla-looking equation like this. So let's substitute it back in. 4 times x, or 4 times negative 1, plus 2 root 2, over 4, plus 1 squared, minus 8 is equal to 0. plus 2 roots 2, plus 1, squared, minus 8 is equal to 0. This negative 1 and this positive 1 cancel out, so you're left with 2 roots of 2 squared, minus 8 is equal to 0. And then what are you going to have here? So when you square this, you get 4 times 2, minus 8 is equal to 0, which is true. 8 minus 8 is equal to 0. And if you try this one out, you're going to get the exact same answer. Let's do another one like this. And remember, these are special forms where we have squares of binomials in our expression. And we're going to see that the entire quadratic formula is actually derived from a notion like this, because you can actually turn any, you can turn any, quadratic equation into a perfect square equalling something else." + }, + { + "Q": "At 2:35, how does drawing a perpendicular line to the base prove that it is an equilateral triangle?", + "A": "I think he started the video already saying that the triangle was equilateral. He is not trying to prove that part.", + "video_name": "SFL4stapeUs", + "timestamps": [ + 155 + ], + "3min_transcript": "So drawing the triangles is always the hard part. This is my best shot at a equilateral triangle. So let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now, what I'm going to do is I'm going So I'm going to drop an altitude right down, and by definition, when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that not only is this an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it, if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. This angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other, So this angle right over here needs to be congruent to that angle right over there. So these two are congruent. And so you can use actually a variety of our congruence postulates. We could say, side-angle-side congruence. We could use angle-side-angle, any of those to show that triangle ABD is congruent to triangle CBD. And what that does for us, and we could use, as I said, we could use angle-side-angle or side-angle-side, whatever we like to use for this. What that does for us is it tells us that the corresponding sides of these triangles are going to be equal. In particular, AD is going to be equal to CD. These are corresponding sides. So these are going to be equal to each other." + }, + { + "Q": "At 0:58 how did he get 90, and if its 90/100 why couldn't the other fraction be 80/100?", + "A": "1st question: he got 90/100 because that = 9/10. he has to + them and you have to have a common denominator. 2nd question: because 8 is in the 100ths place. so you could have it 80/1000. so 9/10 + 8/100 = 90/100 + 8/100 = 98/100 sal makes it easy to understand.", + "video_name": "qbMe4f2yvKs", + "timestamps": [ + 58 + ], + "3min_transcript": "Let's see if we can write 12.98 as a mixed number. So the first thing you might want to realize here is that this is the exact same thing as 12 plus 0.98. And this simplifies it, because then we just have to write it as 12, and to some fraction, that's the same thing as 0.98. So if we could write 0.98 as a fraction, then we're almost done. So let's see if we can do that. So this 9 right over here is in the tenths place. I'll just write it like that. Tenths place. And this right over here is in the hundredths place. So you could view 0.98 two different ways. You could view it as 9/10. That's this part right over here. 9/10 plus 8/100. And if you want to find a common denominator that would be the same thing as 90 over 100 plus 8 over 100, And so 0.98 is 98/100. And another way you could have said that is look, the space right over here is in the hundredths place, and so this is 98/100 or 98/100. So you could have skipped this right over here. So if we just wanted to write it as a mixed number, we could just write it as 12 and, instead of 0.98, 12 and 98/100. Now we haven't reduced this to lowest terms yet. So let's see if we can simplify this. 98 is divisible by 2, and so is 100. So let's divide both of them by 2. They have that common factor. So we're going to divide both of them by 2. And so this is the same thing as 12, and 98 divided by 2 is 49. 100 divided by 2 is 50. And I think that's about as far as we can do. 49 factors, it's divisible by 7, but 50 isn't, so we've So 12.98 can be written as a mixed number, 12 and 49/50." + }, + { + "Q": "At 3:06, he switched but i didn't get why?", + "A": "I believe Sal is trying to show you that you can multiply the 2 numbers ignoring the decimal points. Then, you can set the decimal point in your answer after completing the multiplication.", + "video_name": "4IWfJ7-CYfE", + "timestamps": [ + 186 + ], + "3min_transcript": "Well, I could rewrite 2.91 times 3.2 as being the same thing as. Instead of 2.91, I can write 291 divided by 100. And then times-- instead of writing 3.2, I could write 32 divided by 10. And this can be rewritten as-- this is going to be equal to 291 times 32 divided by 100. I'm just reordering this-- divided by 100, divided by 10. This is equal to 291 times 32. If I divide by 100 and then I divide by 10 again, I'm essentially dividing by 1,000. So this part right over here, I could rewrite as dividing by 1,000. Now, why is this interesting? Well, I already know how to multiply 291 times 32. And then we know how to move the decimal so that when we divide by 1,000. So let's calculate 291 times 32. Let me write it right over here. 291 times 32. Notice, I've just essentially rewritten this without the decimals. So this right over here-- but of course, these are different quantities than this one is right over here. To go from this product to this product, I have to divide by 1,000. But let's just think about this. We already know how to compute this type of thing. 2 times 9 is 18. Carry the 1. 2 times 2 is 4, plus 1 is 5. And now we can think about the 3. 3 times 1-- oh, let me throw a 0 here. Because this isn't a 3. This is now a 30. So this is in the tens place. So that's why I put a 0 there. So 30 times 1 is 30. That's why we say 3 times 1 is 3, but notice, it's in the tens place right now. And then 3 times 9 is 27. Carry the 2. 3 times 2 is 6, plus 2 is 8. And now we can add. And we would get 2. 8 plus 3 is 11. 6 plus 3 is 13. And then you get 9. So you get 9,312. So this is going to be equal to 9,312 divided by 1,000." + }, + { + "Q": "at 0:04 why are the decimals not lined up? aren't they supposed to be lined up when you are multiplying?", + "A": "No, when you multiply, you line it up by how many numbers there are, so in 3.14 and 53.4, you line up the 4 with 4, 1 with 3, and 3 with 5, it doesn t matter where the decimal is. You only line your digits up when you are adding or subtracting.", + "video_name": "4IWfJ7-CYfE", + "timestamps": [ + 4 + ], + "3min_transcript": "Let's see if we can calculate 2.91 times 3.2. And I encourage you to pause this video and try it out on your own. So the way I'm going to think about it is 2.91 is the same thing as 291 divided by 10. Or not divided by 10, divided by 100. And we know that if you divide something by 100, you are going to move the decimal place two places to the left-- one, two. And you would end up at 2.91. It also make sense, if I take 2, and I multiply it by 100, I'd get 200. Or if I take 200 and divided by 100, I would get 2. So it makes sense that 2.91 is the same thing as 291 divided by 100. Similarlarly-- I can never say that word-- 3.2 can be rewritten. It's the same thing as 32 divided by 10. Well, I could rewrite 2.91 times 3.2 as being the same thing as. Instead of 2.91, I can write 291 divided by 100. And then times-- instead of writing 3.2, I could write 32 divided by 10. And this can be rewritten as-- this is going to be equal to 291 times 32 divided by 100. I'm just reordering this-- divided by 100, divided by 10. This is equal to 291 times 32. If I divide by 100 and then I divide by 10 again, I'm essentially dividing by 1,000. So this part right over here, I could rewrite as dividing by 1,000. Now, why is this interesting? Well, I already know how to multiply 291 times 32. And then we know how to move the decimal so that when we divide by 1,000. So let's calculate 291 times 32. Let me write it right over here. 291 times 32. Notice, I've just essentially rewritten this without the decimals. So this right over here-- but of course, these are different quantities than this one is right over here. To go from this product to this product, I have to divide by 1,000. But let's just think about this. We already know how to compute this type of thing." + }, + { + "Q": "I couldn't hear you at 0:25", + "A": "0:20 - 0:30 is What i want to do in this video is test whether any of these four numbers satisfy either of these inequalities. i encourage you to pause the video and try these numbers out, does zero satisfy this inequality? to see what he is saying pause the video and press c on the keyboard for captions or you can scroll down a tiny bit and next to About there is Transcript. Click Transcript to see what he is saying.", + "video_name": "Yh4TXMVq9eg", + "timestamps": [ + 25 + ], + "3min_transcript": "- [Voiceover] We have two inequalities here, the first one says that x plus two is less than or equal to two x. This one over here in I guess this light-purple-mauve color, is three x plus four is greater than five x. Over here we have four numbers and what I want to do in this video is test whether any of these four numbers satisfy either of these inequalities. I encourage you to pause this video and try these numbers out, does zero satisfy this inequality? Does it satisfy this one? Does one satisfy this one? Does it satisfy that one? I encourage you to try these four numbers out on these two inequalities. Assuming you have tried that, let's work through this together. Let's say, if we try out zero on this inequality right over here, let's substitute x with zero. So, we'll have zero plus two needs to be less than or equal to two times zero. Well, on the left hand side, this is two needs to be less than or equal to zero. Is that true, is two less than or equal to zero? No, two is larger than zero. So this is not going to be true, this does not satisfy the left hand side inequality, let's see if it satisfies this inequality over here. In order to satisfy it, three times zero plus four needs to be greater than five times zero. Well three times zero is just zero, five times zero is zero. So four needs to be greater than zero, which is true. So it does satisfy this inequality right over here so zero does satisfy this inequality. Let's try out one. To satisfy this one, one plus two needs to be less than One plus two is three, is three less than or equal than two? No, three is larger than two. This does not satisfy the left hand inequality. What about the right hand inequality right over here? Three times one plus four needs to be greater than five times one. So three times one is three, plus four. So seven needs to be greater than five, well that's true. Both zero and one satisfy three x plus four is greater than five x, neither of them satisfy x plus two is less than or equal to two x. Now let's go to the two. I know it's getting a little bit unaligned, but I'll just do it all in the same color so you can tell. Let's try out two here, two plus two needs to be less than or equal to two times two. Four needs to be less than or equal to four. Well four is equal to four and it just has to be" + }, + { + "Q": "Around 3:10 why not take a proportion like 37% instead of 30%?", + "A": "Look at the null hypothesis. A value of 37% does not satisfy the null hypothesis of p<=0.30. When choosing a value of p, we need to satisfy the null hypothesis, and to choose a conservative number (one which will be less likely to reject the null hypothesis. In this case the largest value allowed by the null, which is 0.30).", + "video_name": "1JT9oODsClE", + "timestamps": [ + 190 + ], + "3min_transcript": "hypothesis for the population. And the given that assumption, what is the probability that 57 out of 150 of our samples actually have internet access. And if that probability is less than 5%, if it's less than our significance level, then we're going to reject the null hypothesis in favor of the alternative one. So let's think about this a little bit. So we're going to start off assuming-- we're going to assume the null hypothesis is true. And in that assumption we're going to have to pick a population proportion or a population mean-- we know that for Bernoulli distributions do the same thing. And what I'm going to do is I'm going to pick a proportion so high so that it maximizes the probability of getting this over here. And we actually don't even know what that number is. And actually so that we can think about a little more proportion even is. We had 57 people out of 150 having internet access. So 57 households out of 150. So our sample proportion is 0.38, so let me write that over here. Our sample proportion is equal to 0.38. So when we assume our null hypothesis to be true, we're going to assume a population proportion that maximizes the probability that we get this over here. So the highest population proportion that's within our null hypothesis that will maximize the probability of getting this is actually if we are right at 30%. So if we say our population proportion, we're going to assume this is true. This is our null hypothesis. We're going to assume that it is 0.3 or 30%. And I want you understand that-- 29% would have been a 28% that would have been a null hypothesis. But for 29% or 28%, the probability of getting this would have been even lower. So it wouldn't have been as strong of a test. If we take the maximum proportion that still satisfies our null hypothesis, we're maximizing the probability that we get this. So if that number is still low, if it's still less than 5%, we can feel pretty good about the alternative hypothesis. So just to refresh ourselves we're going to assume a population proportion of 0.3, and if we just think about the distribution-- sometimes it's helpful to draw these things, so I will draw it. So this is what the population distribution looks like based on our assumption, based on this assumption right over here. Our population distribution has-- or maybe I should write 30% have internet access. And I'll signify that with a 1. And then the rest don't have internet access. 70% do not have internet access." + }, + { + "Q": "at 1:50 how do you divide?", + "A": "You don t, you multiply the improper fraction by the recipricole of 1/5, which is 5/1", + "video_name": "xoXYirs2Mzw", + "timestamps": [ + 110 + ], + "3min_transcript": "Tracy is putting out decorative bowls of potpourri in each room of the hotel where she works. She wants to fill each bowl with 1/5 of a can of potpourri. If Tracy has 4 cans of potpourri, in how many rooms can she place a bowl of potpourri? So she has 4 cans, and she wants to divide this 4 cans into groups of 1/5 of a can. So if you have 4 of something and you're trying to divide it into groups of a certain amount, you would divide by that amount per group. So you want to divide 4 by 1/5. You want to divide 4 cans of potpourri into groups of 1/5. So let's visualize this. Let me draw one can of potpourri right over here. So one can of potpourri can clearly be cut up into 5/5. We have it right over here. 1, 2, 3, 4, 5. So 1 can of potpourri can fill 5 bowls Now, we have 4 cans. So let me paste these. So 2, 3, and 4. So how many total bowls of potpourri can Tracy fill? Well, she's got 4 cans. So this is going to be equal to-- let me do this is the right color-- this is going to be equal to, once again, she has 4 cans. And then for each of those cans, she can fill 5 bowls of potpourri because each bowl only requires a 1/5 of those cans. So this is going to be the same thing as 4 times 5. Or we can even write this as 4 times 5 over 1. 5 is the same thing as 5/1, which is the same thing as 4 times 5, which, of course, is equal to 20. she can fill 20 bowls of potpourri. Now, just as a review here, we've already seen that dividing by a number is equal to multiplying by its reciprocal. And we see that right over here. Dividing by 1/5 is the same thing as multiplying by the reciprocal of 1/5, which is 5/1. So she could fill up 20 bowls of potpourri." + }, + { + "Q": "cane someone pls explain wat sal is trying to say at 3:45 - 4:52?", + "A": "Hi, We need to divide 9.2 by 11.5. As division with decimals get a bit confusing, Sal does the below-mentioned step. 9.2/11.5 = 9.2 *10 / 11.5 *10 ==>92/115 (Multiplying both numerator and denominator by 10, thereby we are not changing the value of the fraction). If you are still confused, take out a calculator and perform both the functions [9.2/11.5 and 92/115]. You will get the same answer. This is the trick in fractions. Hope this helped you.", + "video_name": "EbmgLiSVACU", + "timestamps": [ + 225, + 292 + ], + "3min_transcript": "So here, we have 4.6 times 2. So 4.6 times 2 is 9.2. So that's 9.2. And then 10 to the sixth times 10 to the negative 1-- we have the same base. We're taking the product. We can add the exponents-- is going to be 10 to the 6 minus 1 or 10 to the fifth power. So we've simplified our numerator. And now in our denominator, let's see. 5 times 2.3-- 5 times 2 is 10. 5 times 0.3 is 1.5, so it's going to be 11.5. So this is going to be 11.5. And then if I multiply 10 to the fourth times 10 to the negative 2, that's going to be 10 to the 4 minus 2 or 10 And now I can divide these two things. So this is going to be equal to-- we'll have to think about what 9.2 over 11.5 is. But actually let me just do that right now, get a little practice dividing decimals. Let me get some real estate here. Let me do that in the same color. 9.2 divided by 11.5-- well if we multiply both of these times 10, that's the exact same thing as 92 divided by 115. We're essentially moved the decimal to the right for both of them. And let me add some zeros here because I suspect that I'm going to get a decimal here. So let's think what this is going to be. Let's think about this. Well 115 doesn't go into 9. It doesn't go into 92. It does go into 920. Let's see if that works out. So I have my decimal here. That's a 0. 8 times-- 8 times 5 is 40. 8 times 11 is 88. And then 88 plus 4 is 92. Oh, it went in exactly, very good. So 920, we have no remainder. So 9.2 divided by 11.5 simplified to 0.80. And then 10 to the fifth divided by 10 to the second, we have the same base, and we're dividing. So we can subtract the exponents. That's going to be 10 to the 5 minus 2. So this right over here is going to be 10 to the third power-- so times 10 to the third power. Now, are we done? Well in order to be done, this number right over here needs to be greater than or equal to 1 and less than 10. It is clearly not greater than or equal to 1." + }, + { + "Q": "at 5:07 i got confused.", + "A": "There, 15 goes into 75 ----> 5 times r 1. To review it, when he was writing 5*5 = 25 in the division table, after writing 5, he takes 2 to add with (1*5). But, instead of writing 2 there, he faultily wrote it as 7. Then, after finding the error, he rectified it.", + "video_name": "gHTH6PKfpMc", + "timestamps": [ + 307 + ], + "3min_transcript": "multiply 5 times 5 is 25. 5 times 2 is 10 plus 2, 125. So it goes in exact. So 125 minus 125 is clearly 0. Then we bring down this 0. And 25 goes into 0 zero times. 0 times 25 is 0. Remainder is 0. So we see that 25 goes into 6,250 exactly 250 times. Let's do another problem. Let's say I had-- let me pick an interesting number. Let's say I had 15 and I want to know how many times it goes into 2,265. We say OK, does 15 go into 2? No. So does 15 go into 22? Sure. 15 goes into 22 one time. Notice we wrote the 1 above the 22. If it go had gone into 2 we would've written the 1 here. But 15 goes into 22 one time. 1 times 15 is 15. 22 minus 15-- we could do the whole carrying thing-- 1, 12. 12 minus 5 is 7. 1 minus 1 is 0. 22 minus 15 is 7. Bring down the 6. OK, now how many times does 15 go into 76? Once again, there isn't a real easy mechanical way to do it. You can kind of eyeball it and estimate. Well, 15 times 2 is 30. 15 times 4 is 60. 15 times 5 is 75. So 5 times 5 once again, I already figured it out in my head, but I'll just do it again. 5 times 1 is 5. Plus 7. Oh, sorry. 5 times 5 is 25. 5 times 1 is 5. Plus 2 is 7. Now we just subtract. 76 minus 75 is clearly 1. Bring down that 5. Well, 15 goes into 15 exactly one time. 1 times 15 is 15. Subtract it and we get a remainder of 0. So 15 goes into 2,265 exactly 151 times. So just think about what we're doing here and why it's a" + }, + { + "Q": "at 0:24, he asks what 6250 divided by 25 is. What's the answer?", + "A": "250. @Victoria Fine, you mean pie as in I LIKE PIE or pi as in 3.141592465?", + "video_name": "gHTH6PKfpMc", + "timestamps": [ + 24 + ], + "3min_transcript": "Welcome to the presentation on level 4 division. So what makes level 4 division harder than level 3 division is instead of having a one-digit number being divided into a multi-digit number, we're now going to have a two-addition number divided into a multi-digit number. So let's get started with some practice problems. So let's start with what I would say is a relatively straightforward example. The level 4 problems you'll see are actually a little harder than this. But let's say I had 25 goes into 6,250. So the best way to think about this is you say, OK, I have 25. Does 25 go into 6? Well, no. Clearly 6 is smaller than 25, so 25 does not go into 6. So then ask yourself, well, then if 25 doesn't go into 6, does 25 go into 62? Well, sure. 62 is larger than 25, so 25 will go into 62? 25 times 1 is 25. 25 times 2 is 50. So it goes into 62 at least two times. And 25 times 3 is 75. So that's too much. So 25 goes into 62 two times. And there's really no mechanical way to go about figuring this out. You have to kind of think about, OK, how many times do I think 25 will go into 62? And sometimes you get it wrong. Sometimes you'll put a number here. Say if I didn't know, I would've put a 3 up here and then I would've said 3 times 25 and I would've gotten a 75 here. And then that would have been too large of a number, so I would have gone back and changed it to a 2. Likewise, if I had done a 1 and I had done 1 tmes 25, when I subtracted it out, the difference I would've gotten would be larger than 25. And then I would know that, OK, 1 is too small. I have to increase it to 2. I hope I didn't confuse you too much. I just want you to know that you shouldn't get nervous if you're like, boy, every time I go through the step it's kind of like- I kind of have to guess what the numbers is as And that's true; everyone has to do that. So anyway, so 25 goes into 62 two times. Now let's multiply 2 times 25. Well, 2 times 5 is 10. And then 2 times 2 plus 1 is 5. And we know that 25 times 2 is 50 anyway. Then we subtract. 2 minus 0 is 2. 6 minus 5 is 1. And now we bring down the 5. So the rest of the mechanics are pretty much just like a level 3 division problem. Now we ask ourselves, how many times does 25 go into 125? Well, the way I think about it is 25-- it goes into 100 about four times, so it will go into 125 one more time. It goes into it five times. If you weren't sure you could try 4 and then you would see that you would have too much left over. Or if you tried 6 you would see that you would actually get 6 times 25 is a number larger than 125. So you can't use 6." + }, + { + "Q": "At 2:50 couldnt you multiply both sides by 5, and then subtract by 0.6?", + "A": "Yes, you could. You are right. Good observation. There are multiple ways you can solve this problem.", + "video_name": "BOIA9wsM4ok", + "timestamps": [ + 170 + ], + "3min_transcript": "the negative 10/3 and the positive 10/3, those cancel out to get a zero, and I'm just left with j/4. It's equal to j/4. Now you might recognize 9/3, that's the same thing as nine divided by three. So this is just going to be three. So that simplifies a little bit. Three, let me just rewrite it so you don't get confused. Three is equal to j/4. Now, to solve for j, I could just multiply both sides by four. 'Cause if I divide something by four and then multiply by four, I'm just going to be left with that something. If I start with j and I divide by four, and then I multiply, and then I multiply by four, so I'm just going to multiply by four, then I'm just going to be left with j on the right-hand side. But I can't just multiply the right-hand side by four. I have to do it with the left-hand side So I multiply the left-hand side by four as well. And what I will be left with, four times three is twelve. And then j divided by four times four, So we get j is equal to 12. And the neat thing about equations is you can verify that you indeed got the right answer. You can substitute 12 for j here, and verify that negative 1/3 is equal to 12/4 - 10/3. Does this actually work out? Well 12/4 is the same thing as three, and if I wanted to write that as thirds, this is the same thing as 9/3. And 9/3 - 10/3 is indeed equal to negative 1/3. So we feel very good about that. Let's do another example. So I have n/5 + 0.6 = 2. So let's isolate this term that involves n on the left-hand side. So let's get rid of this 0.6. So let's subtract 0.6 from the left-hand side. But I can't just do it from the left. I have to do it from both sides if I want the equality to hold true. So, subtract 0.6. I'm just going to be left with n/5, and on the right-hand side, 2 - 0.6, that;'s going to be 1.4. And if you don't want to do this in your head, you could work this out It's going to be 2.0 - 0.6. You could say, \"Oh, this is 20/10-6/10\" which is going to be 14/10, which is that there. Or if you want to do it a little bit kind of the traditional method, six from zero, let me re-group.\" That's going to be a 10. I'm going to take from the ones place. If I take a one from the one's place, and that's going to be equal to 10/10. 10/10 - 6/10 is 4/10. And then, bring down one one minus zero ones is just one. So it's 1.4. And now, to solve for n. Well on the left have n being divided by five. If I just want n here, I can just multiply by five. So, if I multiply by five," + }, + { + "Q": "Can someone help me here?\nAt 5:15 rather than distributing the 0.5 to both r and 2.75, he just divided r and 3 by 0.5. If 0.5 can be only divided by r and 3 then why is 2.75 inside the parentheses with r? Wouldn't dividing 0.5 only by r and not by 2.75 create an imbalance?", + "A": "Sal just liked to show you another way to solve that problem, and it is more simple way too.", + "video_name": "BOIA9wsM4ok", + "timestamps": [ + 315 + ], + "3min_transcript": "I'm just going to be left with n/5, and on the right-hand side, 2 - 0.6, that;'s going to be 1.4. And if you don't want to do this in your head, you could work this out It's going to be 2.0 - 0.6. You could say, \"Oh, this is 20/10-6/10\" which is going to be 14/10, which is that there. Or if you want to do it a little bit kind of the traditional method, six from zero, let me re-group.\" That's going to be a 10. I'm going to take from the ones place. If I take a one from the one's place, and that's going to be equal to 10/10. 10/10 - 6/10 is 4/10. And then, bring down one one minus zero ones is just one. So it's 1.4. And now, to solve for n. Well on the left have n being divided by five. If I just want n here, I can just multiply by five. So, if I multiply by five, is going to be just n. But I can't just multiply the left-hand side I have to multiply the right-hand side by five as well. And so what is that going to get us? We are going to get n = 1.4 x 5. 1.4 x 5. Now you might be able to do this in your head, 'cause this is one and 2/5. So this thing should all be equal to seven, but I'll just do it this way as well. Five times four is 20. Re-group the two. One times five is five plus two is seven. And when I look at all the numbers that I'm multiplying, I have one digit to the right of the decimal point. So my answer will have one digit to the right of the decimal point. So it's 7.0, or just 7. n = 7. And you can verify this works, 'cause seven divided by five is going to be equal to 1.4, plus 0.6 is equal to two. Let's do one more example. Alright. 0.5 times the whole quantity (r + 2.75) = 3. Now there's a bunch of ways that you could tackle this. A lot of times when you see something like this, your temptation might be, \"Let's distribute the 0.5.\" But that makes it a little bit hairy, 'cause 0.5 times 2.75. You could calculate that, and you will get the right answer if you do it correctly. But a simpler thing might be, well, let's just divide both sides by 0.5. That way I'm going to get more whole numbers involved. So if I divide. Remember, whatever I do to the left-hand side I have to do to the right-hand side. And the way my brain thought about it was, well, if I divide by 0.5 on the left-hand side I can get rid of this. And if I divide by 0.5 on the right-hand side I'm still going to get an integer. Three divided by 0.5 is six. It's the same thing as three divided by a half. How many halves fit into three? Six halves fit into three. So this going to be six right over here. And then this is going to be equal to six. So the whole thing is simplified now" + }, + { + "Q": "At 0:47, what does it exactly mean to have a slope of 1?", + "A": "A slope of 1 means that for every unit the line goes along the x axis, it goes up the y axis 1 unit.", + "video_name": "5a6zpfl50go", + "timestamps": [ + 47 + ], + "3min_transcript": "Let's say I have the equation y is equal to x plus 3. And I want to graph all of the sets, all of the coordinates x comma y that satisfy this equation right there. And we've done this many times before. So we draw our axis, our axes. That's my y-axis. This is my x-axis. And this is already in mx plus b form, or slope-intercept form. The y-intercept here is y is equal to 3, and the slope here is 1. So this line is going to look like this. We intersect at 0 comma 3-- 1, 2, 3. At 0 comma 3. And we have a slope of 1, so every 1 we go to the right, we go up 1. So the line will look something like that. It's a good enough approximation. So the line will look like this. And remember, when I'm drawing a line, every point on this Or it represents a pair of x and y that satisfy this equation. So maybe when you take x is equal to 5, you go to the line, and you're going to see, gee, when x is equal to 5 on that line, y is equal to 8 is a solution. And it's going to sit on the line. So this represents the solution set to this equation, all of the coordinates that satisfy y is equal to x plus 3. Now let's say we have another equation. Let's say we have an equation y is equal to negative x plus 3. And we want to graph all of the x and y pairs that satisfy this equation. Well, we can do the same thing. This has a y-intercept also at 3, right there. But its slope is negative 1. So it's going to look something like this. to move down 1. Or if you move to the right a bunch, you're going to move down that same bunch. So that's what this equation will look like. Every point on this line represents a x and y pair that will satisfy this equation. Now, what if I were to ask you, is there an x and y pair that satisfies both of these equations? Is there a point or coordinate that satisfies both equations? Well, think about it. Everything that satisfies this first equation is on this green line right here, and everything that satisfies this purple equation is on the purple line right there. So what satisfies both? Well, if there's a point that's on both lines, or essentially, a point of intersection of the lines. So in this situation, this point is on both lines. And that's actually the y-intercept. So the point 0, 3 is on both of these lines. So that coordinate pair, or that x, y pair, must satisfy" + }, + { + "Q": "at 0:06 y=x+3 how do i sketch the graph if x is x square", + "A": "You can t. It s not a linear problem then. The way to graph it is like the quadratics, or other x^2. The y intercept is now the vertex. the line will look like a U.", + "video_name": "5a6zpfl50go", + "timestamps": [ + 6 + ], + "3min_transcript": "Let's say I have the equation y is equal to x plus 3. And I want to graph all of the sets, all of the coordinates x comma y that satisfy this equation right there. And we've done this many times before. So we draw our axis, our axes. That's my y-axis. This is my x-axis. And this is already in mx plus b form, or slope-intercept form. The y-intercept here is y is equal to 3, and the slope here is 1. So this line is going to look like this. We intersect at 0 comma 3-- 1, 2, 3. At 0 comma 3. And we have a slope of 1, so every 1 we go to the right, we go up 1. So the line will look something like that. It's a good enough approximation. So the line will look like this. And remember, when I'm drawing a line, every point on this Or it represents a pair of x and y that satisfy this equation. So maybe when you take x is equal to 5, you go to the line, and you're going to see, gee, when x is equal to 5 on that line, y is equal to 8 is a solution. And it's going to sit on the line. So this represents the solution set to this equation, all of the coordinates that satisfy y is equal to x plus 3. Now let's say we have another equation. Let's say we have an equation y is equal to negative x plus 3. And we want to graph all of the x and y pairs that satisfy this equation. Well, we can do the same thing. This has a y-intercept also at 3, right there. But its slope is negative 1. So it's going to look something like this. to move down 1. Or if you move to the right a bunch, you're going to move down that same bunch. So that's what this equation will look like. Every point on this line represents a x and y pair that will satisfy this equation. Now, what if I were to ask you, is there an x and y pair that satisfies both of these equations? Is there a point or coordinate that satisfies both equations? Well, think about it. Everything that satisfies this first equation is on this green line right here, and everything that satisfies this purple equation is on the purple line right there. So what satisfies both? Well, if there's a point that's on both lines, or essentially, a point of intersection of the lines. So in this situation, this point is on both lines. And that's actually the y-intercept. So the point 0, 3 is on both of these lines. So that coordinate pair, or that x, y pair, must satisfy" + }, + { + "Q": "i don't get the whole rate system. if you have the ratio how do you get the rate? for ex: 8 boxes can hold 48 books the ratio is 48:8 but how do you figure out the rate?", + "A": "Your rate will be 6 because how much does 1 box hold it holds 6 if you divide,if you need any help comment! :D", + "video_name": "qGTYSAeLTOE", + "timestamps": [ + 2888 + ], + "3min_transcript": "" + }, + { + "Q": "By saying that one is going 35 miles per hour, in ratio form would be 35:1, but could it also be expressed the other way around? for example, 1:35, saying that in every hour one is going 35 miles. Is is the same thing?", + "A": "yes it would be the same thing except it has to due with how it is worded. say if the problem said miles per hour it is 29:1 58:2 and so on but if it is hours per amount of miles 1:29 2:29", + "video_name": "qGTYSAeLTOE", + "timestamps": [ + 2101, + 95 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:13, why is theta written in radians?", + "A": "Because it is easier to measure sin cos and tan that way.", + "video_name": "sjUhr0HkLUg", + "timestamps": [ + 73 + ], + "3min_transcript": "" + }, + { + "Q": "At 8:15, Isn't PI transcendental number, so it is not a real number.\nSo is the graph wrong ?", + "A": "\u00cf\u0080 is both transcendental and real. Transcendental just means that the number isn t the root of a polynomial with rational coefficients. \u00e2\u0088\u009a2 is irrational, but not transcendental, since it s the root of x\u00c2\u00b2-2=0.", + "video_name": "sjUhr0HkLUg", + "timestamps": [ + 495 + ], + "3min_transcript": "" + }, + { + "Q": "at 3:18 , Sal goes 3pi/2. Isnt it 3pi/4?\nIf im wrong can someone tell me why?", + "A": "Sal is right, it s 3pi/2. Maybe you got confused because Sal mentioned that we got 3/4 of the way , but you have to remember that the whole way is 2pi. So each quarter is pi/2 and 3*pi/2 is 3pi/2", + "video_name": "sjUhr0HkLUg", + "timestamps": [ + 198 + ], + "3min_transcript": "" + }, + { + "Q": "At 5:50, on the mounting of the graph, why the connection between the points [\u00ce\u00b8,sin(\u00ce\u00b8)] of the table, was made for curved lines instead of straight lines?\nIn other words, why the graph of sin(\u00ce\u00b8) is not like this: /\\/\\/\\/\\/\\/\\/\\/\\/\\/\\ ?", + "A": "You can check this in different ways. If you want to be really confident, just take many points between 0 and pi and take the sine, you will get this smooth line. It s easier to follow the unit circle, as seen from 2:20 - to get from 0 to pi/2 Sal goes along a curved path - this has to be on the graph of the sine als well.", + "video_name": "sjUhr0HkLUg", + "timestamps": [ + 350 + ], + "3min_transcript": "" + }, + { + "Q": "at 6:48 sal says with arcsin we only have to draw the first and fourth quadrants. why is this? I saw somewhere else that sine is between 1 and -1, but I still don't understand this.", + "A": "arc sin is a function, and as such it cannot have different outputs for the same input. If we included the second and third quadrant, we would get two solutions for most input values, so we wouldn t have a function.", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 408 + ], + "3min_transcript": "So let's restrict its range. Actually, just as a side note, what's its domain restricted to? So if I'm taking the arcsine of something. So if I'm taking the arcsine of x, and I'm saying that that is equal to theta, what's the domain restricted to? What are the valid values of x? x could be equal to what? Well if I take the sine of any angle, I can only get values between 1 and negative 1, right? So x is going to be greater than or equal to negative 1 and then less than or equal to 1. That's the domain. Now, in order to make this a valid function, I have to restrict the range. The possible values. I have to restrict the range. Now for arcsine, the convention is to restrict it to the first and fourth quadrants. To restrict the possible angles to this area right here along the unit circle. So theta is restricted to being less than or equal to pi over So given that, we now understand what arcsine is. Let's do another problem. Clear out some space here. Let me do another arcsine. So let's say I were to ask you what the arcsine of minus the square root of 3 over 2 is. Now you might have that memorized. And say, I immediately know that sine of x, or sine of theta is square root of 3 over 2. And you'd be done. But I don't have that memorized. So let me just draw my unit circle. And when I'm dealing with arcsine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I? that means the y-coordinate on the unit circle is minus square root of 3 over 2. So it means we're right about there. So this is minus the square root of 3 over 2. This is where we are. Now what angle gives me that? Let's think about it a little bit. My y-coordinate is minus square root of 3 over 2. This is the angle. It's going to be a negative angle because we're going below the x-axis in the clockwise direction. And to figure out-- Let me just draw a little triangle here. Let me pick a better color than that. That's a triangle. Let me do it in this blue color. So let me zoom up that triangle. Like that. This is theta. That's theta. And what's this length right here? Well that's the same as the y-height, I guess" + }, + { + "Q": "At 7:14, how does Sal know to put the y-axis point in that spot? Is it just an estimate? It seems that the result of the problem would depend on the accuracy of the placement of that point and therefore an estimate would not work.", + "A": "In this example, it is a 30-60-90 triangle. He draws the triangle as a visual aid, not as a means of actually calculating the angle. More generally, you might use a sketch like that to estimate the quadrant but use a calculator if you needed the actual result.", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 434 + ], + "3min_transcript": "So let's restrict its range. Actually, just as a side note, what's its domain restricted to? So if I'm taking the arcsine of something. So if I'm taking the arcsine of x, and I'm saying that that is equal to theta, what's the domain restricted to? What are the valid values of x? x could be equal to what? Well if I take the sine of any angle, I can only get values between 1 and negative 1, right? So x is going to be greater than or equal to negative 1 and then less than or equal to 1. That's the domain. Now, in order to make this a valid function, I have to restrict the range. The possible values. I have to restrict the range. Now for arcsine, the convention is to restrict it to the first and fourth quadrants. To restrict the possible angles to this area right here along the unit circle. So theta is restricted to being less than or equal to pi over So given that, we now understand what arcsine is. Let's do another problem. Clear out some space here. Let me do another arcsine. So let's say I were to ask you what the arcsine of minus the square root of 3 over 2 is. Now you might have that memorized. And say, I immediately know that sine of x, or sine of theta is square root of 3 over 2. And you'd be done. But I don't have that memorized. So let me just draw my unit circle. And when I'm dealing with arcsine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I? that means the y-coordinate on the unit circle is minus square root of 3 over 2. So it means we're right about there. So this is minus the square root of 3 over 2. This is where we are. Now what angle gives me that? Let's think about it a little bit. My y-coordinate is minus square root of 3 over 2. This is the angle. It's going to be a negative angle because we're going below the x-axis in the clockwise direction. And to figure out-- Let me just draw a little triangle here. Let me pick a better color than that. That's a triangle. Let me do it in this blue color. So let me zoom up that triangle. Like that. This is theta. That's theta. And what's this length right here? Well that's the same as the y-height, I guess" + }, + { + "Q": "Hi. In the video you were trying to find the arcsin of rad3/2. Since sin = y/r would not the hypotenuse(r) of the triangle at 8:13 be 2 and not 1? That would make the other sides = rad3 and 1 respectively. 1^2 + rad3^2 = 2^2. I don't understand why one side of the triangle you drew would be rad3/2.", + "A": "the hypotenuse is the radius of the unit circle, which is always 1", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 493 + ], + "3min_transcript": "So given that, we now understand what arcsine is. Let's do another problem. Clear out some space here. Let me do another arcsine. So let's say I were to ask you what the arcsine of minus the square root of 3 over 2 is. Now you might have that memorized. And say, I immediately know that sine of x, or sine of theta is square root of 3 over 2. And you'd be done. But I don't have that memorized. So let me just draw my unit circle. And when I'm dealing with arcsine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I? that means the y-coordinate on the unit circle is minus square root of 3 over 2. So it means we're right about there. So this is minus the square root of 3 over 2. This is where we are. Now what angle gives me that? Let's think about it a little bit. My y-coordinate is minus square root of 3 over 2. This is the angle. It's going to be a negative angle because we're going below the x-axis in the clockwise direction. And to figure out-- Let me just draw a little triangle here. Let me pick a better color than that. That's a triangle. Let me do it in this blue color. So let me zoom up that triangle. Like that. This is theta. That's theta. And what's this length right here? Well that's the same as the y-height, I guess It's a minus because we're going down. But let's just figure out this angle. And we know it's a negative angle. So when you see a square root of 3 over 2, hopefully you recognize this is a 30 60 90 triangle. The square root of 3 over 2. This side is 1/2. And then, of course, this side is 1. Because this is a unit circle. So its radius is 1. So in a 30 60 90 triangle, the side opposite to the square root of 3 over 2 is 60 degrees. This side over here is 30 degrees. So we know that our theta is-- This is 60 degrees. That's its magnitude. But it's going downwards. So it's minus 60 degrees. So theta is equal to minus 60 degrees. But if we're dealing in radians, that's not good enough. So we can multiply that times 100-- sorry --pi radians for every 180 degrees. Degrees cancel out. And we're left with theta is equal to minus pi over 3 radians." + }, + { + "Q": "At 6:02 he writes the range for a general arcsin function, but just to confirm, that's the range for all arcsin functions?? Or am I missing something. Because for the example problem he did at the end, he didn't state the domain and range, but I thought restricting the range was necessary to make the sine function valid. Could someone please clarify?", + "A": "i think you mean all inverse trig functions rather than all arcsin functions. and to answer your question, the range of all inverse trig functions are not the same due to the shape of the graph.", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 362 + ], + "3min_transcript": "Let me go over here. You're like, look pi over 2 worked. 45 degrees worked. But I could just keep adding 360 degrees or I could keep just adding 2 pi. And all of those would work because those would all get me to that same point of the unit circle, right? And you'd be correct. And so all of those values, you would think, would be valid answers for this, right? Because if you take the sine of any of those angles-- You could just keep adding 360 degrees. If you take the sine of any of them, you would get square root of 2 over 2. And that's a problem. You can't have a function where if I take the function-- I can't have a function, f of x, where it maps to multiple values, right? Where it maps to pi over 4, or it maps to pi over 4 plus 2 pi or pi over 4 plus 4 pi. So in order for this to be a valid function-- In order for the inverse sine function to be valid, I have to restrict its range. And the way that-- We'll just restrict its range to So let's restrict its range. Actually, just as a side note, what's its domain restricted to? So if I'm taking the arcsine of something. So if I'm taking the arcsine of x, and I'm saying that that is equal to theta, what's the domain restricted to? What are the valid values of x? x could be equal to what? Well if I take the sine of any angle, I can only get values between 1 and negative 1, right? So x is going to be greater than or equal to negative 1 and then less than or equal to 1. That's the domain. Now, in order to make this a valid function, I have to restrict the range. The possible values. I have to restrict the range. Now for arcsine, the convention is to restrict it to the first and fourth quadrants. To restrict the possible angles to this area right here along the unit circle. So theta is restricted to being less than or equal to pi over So given that, we now understand what arcsine is. Let's do another problem. Clear out some space here. Let me do another arcsine. So let's say I were to ask you what the arcsine of minus the square root of 3 over 2 is. Now you might have that memorized. And say, I immediately know that sine of x, or sine of theta is square root of 3 over 2. And you'd be done. But I don't have that memorized. So let me just draw my unit circle. And when I'm dealing with arcsine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I?" + }, + { + "Q": "What are Radians? 00:14", + "A": "Radians are a different form of measurement of an angle. They act no differently than degrees. They are much like different forms of currency. The Canadian Dollar and the U.S. Dollar serve the same purpose, but have different values. In the same way that a U.S. Dollar has more value than a Canadian Dollar, Radians have more value than degrees. 180 degrees is equal to 1 pi radians.", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 14 + ], + "3min_transcript": "If I were to walk up to you on the street and say you, please tell me what-- so I didn't want to write that thick --please tell me what sine of pi over 4 is. And, obviously, we're assuming we're dealing in radians. You either have that memorized or you would draw the unit circle right there. That's not the best looking unit circle, but you get the idea. You'd go to pi over 4 radians, which is the same thing as 45 degrees. You would draw that unit radius out. And the sine is defined as a y-coordinate on the unit circle. So you would just want to know this value right here. And you would immediately say OK. This is a 45 degrees. Let me draw the triangle a little bit larger. The triangle looks like this. This is 45. That's 45. This is 90. And you can solve a 45 45 90 triangle. The hypotenuse is 1. This is x. They're going to be the same values. Their base angles are the same. So you say, look. x squared plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1/2. x is equal to the square root of 1/2, which is one over the square root of 2. I can put that in rational form by multiplying that by the square root of 2 over 2. And I get x is equal to the square root of 2 over 2. So the height here is square root of 2 over 2. And if you wanted to know this distance too, it would also be the same thing. But we just cared about the height. Because the sine value, the sine of this, is just this height right here. The y-coordinate. And we got that as the square root of 2 over 2. This is all review. We learned this in the unit circle video. But what if someone else-- Let's say on another day, I arcsine of the square root of 2 over 2 is. What is the arcsine? You're like I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize, when they have this word arc in front of it-- This is also sometimes referred to as the inverse sine. This could have just as easily been written as: what is the inverse sine of the square root of 2 over 2? All this is asking is what angle would I have to take the sine of in order to get the value square root of 2 over 2. This is also asking what angle would I have to take the sine of in order to get square root of 2 over 2. I could rewrite either of these statements as saying square-- Let me do it." + }, + { + "Q": "Why didn't Sal round up to 1.4 from 1.47? Wouldn't that be more accurate? 10:01", + "A": "Sal DID round UP. The calculator showed a NEGATIVE 1.047..... which is between - 1.05 and -1.04 The larger number (the one furthest to the right on the number line) is -1.04..", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 601 + ], + "3min_transcript": "It's a minus because we're going down. But let's just figure out this angle. And we know it's a negative angle. So when you see a square root of 3 over 2, hopefully you recognize this is a 30 60 90 triangle. The square root of 3 over 2. This side is 1/2. And then, of course, this side is 1. Because this is a unit circle. So its radius is 1. So in a 30 60 90 triangle, the side opposite to the square root of 3 over 2 is 60 degrees. This side over here is 30 degrees. So we know that our theta is-- This is 60 degrees. That's its magnitude. But it's going downwards. So it's minus 60 degrees. So theta is equal to minus 60 degrees. But if we're dealing in radians, that's not good enough. So we can multiply that times 100-- sorry --pi radians for every 180 degrees. Degrees cancel out. And we're left with theta is equal to minus pi over 3 radians. arcsine of minus square root of 3 over 2 is equal to minus pi over 3 radians. Or we could say the inverse sign of minus square root of 3 over 2 is equal to minus pi over 3 radians. And to confirm this, let's just-- Let me get a little calculator out. I put this in radian mode already. You can just check that. Per second mode. I'm in radian mode. So I know I'm going to get, hopefully, the right answer. And I want to figure out the inverse sign. So the inverse sine-- the second and the sine button --of the minus square root of 3 over 2. It equals minus 1.04. So pi over 3 must be equal to 1.04. Let's see if I can confirm that. So if I were to write minus pi divided by 3, what do I get? I get the exact same value. So my calculator gave me the exact same value, but it might have not been that helpful because my calculator doesn't tell me that this is minus pi over 3." + }, + { + "Q": "At 5:58 it says that the range of theta/arcsin is being less than or equal to pi over 2 and greater than or equal to minus pi over 2. Can it be equal to these numbers? Can the range equal positive pi over 2 or negative pi over 2, or was this a mistake?", + "A": "the range can be pi/2 or -pi/2 because it s still a one to one function", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 358 + ], + "3min_transcript": "Let me go over here. You're like, look pi over 2 worked. 45 degrees worked. But I could just keep adding 360 degrees or I could keep just adding 2 pi. And all of those would work because those would all get me to that same point of the unit circle, right? And you'd be correct. And so all of those values, you would think, would be valid answers for this, right? Because if you take the sine of any of those angles-- You could just keep adding 360 degrees. If you take the sine of any of them, you would get square root of 2 over 2. And that's a problem. You can't have a function where if I take the function-- I can't have a function, f of x, where it maps to multiple values, right? Where it maps to pi over 4, or it maps to pi over 4 plus 2 pi or pi over 4 plus 4 pi. So in order for this to be a valid function-- In order for the inverse sine function to be valid, I have to restrict its range. And the way that-- We'll just restrict its range to So let's restrict its range. Actually, just as a side note, what's its domain restricted to? So if I'm taking the arcsine of something. So if I'm taking the arcsine of x, and I'm saying that that is equal to theta, what's the domain restricted to? What are the valid values of x? x could be equal to what? Well if I take the sine of any angle, I can only get values between 1 and negative 1, right? So x is going to be greater than or equal to negative 1 and then less than or equal to 1. That's the domain. Now, in order to make this a valid function, I have to restrict the range. The possible values. I have to restrict the range. Now for arcsine, the convention is to restrict it to the first and fourth quadrants. To restrict the possible angles to this area right here along the unit circle. So theta is restricted to being less than or equal to pi over So given that, we now understand what arcsine is. Let's do another problem. Clear out some space here. Let me do another arcsine. So let's say I were to ask you what the arcsine of minus the square root of 3 over 2 is. Now you might have that memorized. And say, I immediately know that sine of x, or sine of theta is square root of 3 over 2. And you'd be done. But I don't have that memorized. So let me just draw my unit circle. And when I'm dealing with arcsine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I?" + }, + { + "Q": "At 1:50, Sal says that the sin is the height, but I'm pretty sure that is not usually sin. What special circumstances makes sin the height?", + "A": "Hello Kevin, Height, including negative heights, is an appropriate description for SIN(X). You see, SIN(X) gives the vertical component and COS(X) gives the horizontal component. If this is a fuzzy answer it will make more sense when you get to the unit circle. Regards, APD", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 110 + ], + "3min_transcript": "If I were to walk up to you on the street and say you, please tell me what-- so I didn't want to write that thick --please tell me what sine of pi over 4 is. And, obviously, we're assuming we're dealing in radians. You either have that memorized or you would draw the unit circle right there. That's not the best looking unit circle, but you get the idea. You'd go to pi over 4 radians, which is the same thing as 45 degrees. You would draw that unit radius out. And the sine is defined as a y-coordinate on the unit circle. So you would just want to know this value right here. And you would immediately say OK. This is a 45 degrees. Let me draw the triangle a little bit larger. The triangle looks like this. This is 45. That's 45. This is 90. And you can solve a 45 45 90 triangle. The hypotenuse is 1. This is x. They're going to be the same values. Their base angles are the same. So you say, look. x squared plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1/2. x is equal to the square root of 1/2, which is one over the square root of 2. I can put that in rational form by multiplying that by the square root of 2 over 2. And I get x is equal to the square root of 2 over 2. So the height here is square root of 2 over 2. And if you wanted to know this distance too, it would also be the same thing. But we just cared about the height. Because the sine value, the sine of this, is just this height right here. The y-coordinate. And we got that as the square root of 2 over 2. This is all review. We learned this in the unit circle video. But what if someone else-- Let's say on another day, I arcsine of the square root of 2 over 2 is. What is the arcsine? You're like I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize, when they have this word arc in front of it-- This is also sometimes referred to as the inverse sine. This could have just as easily been written as: what is the inverse sine of the square root of 2 over 2? All this is asking is what angle would I have to take the sine of in order to get the value square root of 2 over 2. This is also asking what angle would I have to take the sine of in order to get square root of 2 over 2. I could rewrite either of these statements as saying square-- Let me do it." + }, + { + "Q": "Don't get 2:55 to 3:55", + "A": "That s an example of using Heron s Formula.", + "video_name": "-YI6UC4qVEY", + "timestamps": [ + 175, + 235 + ], + "3min_transcript": "And to do that we're going to apply something called Heron's Formula. And I'm not going to prove it in this video. I'm going to prove it in a future video. And really to prove it you already probably have the tools necessary. It's really just the Pythagorean theorem and a lot of hairy algebra. But I'm just going to show you the formula now and how to apply it, and then you'll hopefully appreciate that it's pretty simple and pretty easy to remember. And it can be a nice trick to impress people with. So Heron's Formula says first figure out this third variable S, which is essentially the perimeter of this triangle divided by 2. a plus b plus c, divided by 2. Then once you figure out S, the area of your triangle-- of this triangle right there-- is going to be equal to the square root of S-- this variable S right here that you just calculated-- times S minus a, times S minus b, times S minus c. This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18. the square root of S-- 18-- times S minus a-- S minus 9. 18 minus 9, times 18 minus 11, times 18 minus 16. And then this is equal to the square root of 18 times 9 times 7 times 2. Which is equal to-- let's see, 2 times 18 is 36. So I'll just rearrange it a bit. This is equal to the square root of 36 times 9 times 7, which is equal to the square root of 36 times the square" + }, + { + "Q": "At 1:36 what does he mean by \"third variable S\"", + "A": "I think he meant to say the fourth variable (three sides of the triangle being the first three), but the ordinal number used to refer to it is irrelevant. s = the semi-perimeter of the triangle. That would be (a + b + c) / 2.", + "video_name": "-YI6UC4qVEY", + "timestamps": [ + 96 + ], + "3min_transcript": "I think it's pretty common knowledge how to find the area of the triangle if we know the length of its base and its height. So, for example, if that's my triangle, and this length right here-- this base-- is of length b and the height right here is of length h, it's pretty common knowledge that the area of this triangle is going to be equal to 1/2 times the base times the height. So, for example, if the base was equal to 5 and the height was equal to 6, then our area would be 1/2 times 5 times 6, which is 1/2 times 30-- which is equal to 15. Now what is less well-known is how to figure out the area of a triangle when you're only given the sides of the triangle. When you aren't given the height. So, for example, how do you figure out a triangle where I just give you the lengths of the sides. Let's say that's side a, side b, and side c. a, b, and c are the lengths of these sides. And to do that we're going to apply something called Heron's Formula. And I'm not going to prove it in this video. I'm going to prove it in a future video. And really to prove it you already probably have the tools necessary. It's really just the Pythagorean theorem and a lot of hairy algebra. But I'm just going to show you the formula now and how to apply it, and then you'll hopefully appreciate that it's pretty simple and pretty easy to remember. And it can be a nice trick to impress people with. So Heron's Formula says first figure out this third variable S, which is essentially the perimeter of this triangle divided by 2. a plus b plus c, divided by 2. Then once you figure out S, the area of your triangle-- of this triangle right there-- is going to be equal to the square root of S-- this variable S right here that you just calculated-- times S minus a, times S minus b, times S minus c. This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18." + }, + { + "Q": "I was wondering. How come at 4 29 after he has worked it down to the square roots of all of them. How come it is 7? What happens if each number was to not have a perfact sqaure in it? Like a triangle with the sides: 8:10:12? What would happen? What one would you pick?", + "A": "Hi Fuller! If the triangle had sides 8, 10, and 12, then the semiperimeter would equal 15. So, the answer is the square root of 8x10x12 which is the square root of 2x2x2x2x5x2x3x2 which is 2x2x2x the square root of 3x5 which is 8 times the square root of 15.", + "video_name": "-YI6UC4qVEY", + "timestamps": [ + 490 + ], + "3min_transcript": "" + }, + { + "Q": "4:12-4:20 i got confused who can help me ???", + "A": "oh ok thanks for the help ..i really appreciated it", + "video_name": "-YI6UC4qVEY", + "timestamps": [ + 252, + 260 + ], + "3min_transcript": "This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18. the square root of S-- 18-- times S minus a-- S minus 9. 18 minus 9, times 18 minus 11, times 18 minus 16. And then this is equal to the square root of 18 times 9 times 7 times 2. Which is equal to-- let's see, 2 times 18 is 36. So I'll just rearrange it a bit. This is equal to the square root of 36 times 9 times 7, which is equal to the square root of 36 times the square The square root of 36 is just 6. This is just 3. And we don't deal with the negative square roots, because you can't have negative side lengths. And so this is going to be equal to 18 times the square root of 7. So just like that, you saw it, it only took a couple of minutes to apply Heron's Formula, or even less than that, to figure out that the area of this triangle right here is equal to 18 square root of seven. Anyway, hopefully you found that pretty neat." + }, + { + "Q": "at 1:29 what does Sal mean by \"nice trick\"?", + "A": "He s referring to the formula (Heron s formula) itself.", + "video_name": "-YI6UC4qVEY", + "timestamps": [ + 89 + ], + "3min_transcript": "I think it's pretty common knowledge how to find the area of the triangle if we know the length of its base and its height. So, for example, if that's my triangle, and this length right here-- this base-- is of length b and the height right here is of length h, it's pretty common knowledge that the area of this triangle is going to be equal to 1/2 times the base times the height. So, for example, if the base was equal to 5 and the height was equal to 6, then our area would be 1/2 times 5 times 6, which is 1/2 times 30-- which is equal to 15. Now what is less well-known is how to figure out the area of a triangle when you're only given the sides of the triangle. When you aren't given the height. So, for example, how do you figure out a triangle where I just give you the lengths of the sides. Let's say that's side a, side b, and side c. a, b, and c are the lengths of these sides. And to do that we're going to apply something called Heron's Formula. And I'm not going to prove it in this video. I'm going to prove it in a future video. And really to prove it you already probably have the tools necessary. It's really just the Pythagorean theorem and a lot of hairy algebra. But I'm just going to show you the formula now and how to apply it, and then you'll hopefully appreciate that it's pretty simple and pretty easy to remember. And it can be a nice trick to impress people with. So Heron's Formula says first figure out this third variable S, which is essentially the perimeter of this triangle divided by 2. a plus b plus c, divided by 2. Then once you figure out S, the area of your triangle-- of this triangle right there-- is going to be equal to the square root of S-- this variable S right here that you just calculated-- times S minus a, times S minus b, times S minus c. This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18." + }, + { + "Q": "Principal root of a aquare root fnction? What does he mean by this? It is approximately 5:45...", + "A": "Principal root is the positive one of the two roots", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 345 + ], + "3min_transcript": "and complex numbers and all the rest-- you have to expand the definition of what this radical means. So when you are taking the square root of really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function, or this is now defined for complex inputs or the domain, it can also generate imaginary or complex outputs, or I guess you could call that the range. And if you assume that, then really straight from this you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only-- and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true-- So we could apply this when x is greater than or equal to 0. then negative x is clearly a negative number, or I guess it could also be 0. It's a negative number. And then we can apply this right over here. If x was less than 0, then we would be doing all of this nonsense up here. And we start to get nonsensical answers. And if you look at it this way and you say hey, look, i can be the square root of negative one, if it's the principal branch of the complex square root function. Then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really the real fault in this logic when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the complex to include negative numbers in the domain and to include imaginary numbers, then you can do this. You can say the square root of negative x is the square root of negative 1 times-- Or you should say the principal square root of negative x-- I should be particular my words-- is the same thing as the principal square root of negative 1 times the principal square root of x when x is greater than or equal to 0. And I don't want to confuse you, if x is greater than or equal to 0, this negative x, that is clearly a negative, or I guess you should say a non positive number." + }, + { + "Q": "I think Sal make a mistake on (vid @ 5:11) when he write the greater than sign! it should be Less than", + "A": "No, Sal is correct. If he had: i sqrt(x) where X<0, then X is negative. Backup thru Sal steps. If X is negative * (-1) = +X. And he would have started with sqrt(x), not sqrt(-x). He is also trying to highlight that if you had something like: sqrt(12), you would not make this into i sqrt(-12). The imaginary number is not needed if the radical contains a positive number to start with.", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 311 + ], + "3min_transcript": "saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. When we take traditional principal square roots. So when you take the principle square root of 4. We know that this is positive 2, that 4 actually has two square roots. There's negative 2 also is a square root of 4. If you have negative 2 times negative 2 it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non imaginary, non complex numbers, you could really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers-- and complex numbers and all the rest-- you have to expand the definition of what this radical means. So when you are taking the square root of really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function, or this is now defined for complex inputs or the domain, it can also generate imaginary or complex outputs, or I guess you could call that the range. And if you assume that, then really straight from this you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only-- and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true-- So we could apply this when x is greater than or equal to 0. then negative x is clearly a negative number, or I guess it could also be 0. It's a negative number. And then we can apply this right over here. If x was less than 0, then we would be doing all of this nonsense up here. And we start to get nonsensical answers. And if you look at it this way and you say hey, look, i can be the square root of negative one, if it's the principal branch of the complex square root function. Then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really the real fault in this logic when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the complex" + }, + { + "Q": "Does the rule at 2:32 apply if, for example, a is positive and b is negative?", + "A": "Bobo, \u00e2\u0088\u009aa*\u00e2\u0088\u009ab = \u00e2\u0088\u009a(a*b) applies if both a and b are positive or it either a or b are negative. It does not apply if both a and b are negative.", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 152 + ], + "3min_transcript": "the same thing as the square root of a times b. So based on this property of the radical of the principal root, they'll say that this over here is the same thing as the square root of negative 1 times negative 1. If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots. I'm doing this in the other order here. Here I have the principal root of the products, over here I have this on the right. And then from that we all know that negative 1 times negative 1 is 1. So this should be equal to the principal square root of 1. And then the principal square root of 1-- Remember, this radical means principal square root, positive square root, that is just going to be positive 1. And they'll say, this is wrong. Clearly, negative 1 and positive 1 are not the same thing. And they'll argue therefore, you can't make this substitution And what you should then point out is that this was not the incorrect step. That it is true negative 1 is not equal to 1, but the faulty line of reasoning here was in using this property when both a and b are negative. If both a and b are negative, this will never be true. So a and b both cannot be negative. In fact normally when this property is given-- sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time-- but they'll usually give a little bit of a constraint They'll usually say for a and b greater than or equal to 0. So that's where they list this property. This is true for a and b greater than or equal 0. And in particular, it's false if both a and b are both, if they are both negative. saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. When we take traditional principal square roots. So when you take the principle square root of 4. We know that this is positive 2, that 4 actually has two square roots. There's negative 2 also is a square root of 4. If you have negative 2 times negative 2 it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non imaginary, non complex numbers, you could really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers--" + }, + { + "Q": "At , 3:03 Sal says that the people who say that\ni\nisn't the principal square root of -1 are wrong. Is this just Sal's personal thinking, or is it a thing that mathematicians have decided upon? Basically, is it official that\ni\nisn't the principal square root of -1?", + "A": "It honestly depends on who you ask. However, i is, in fact, widely believed to be rad(-1). However, NOT considering i to be such a thing prevents you from solving ( or even understanding ) many problems where using complex numbers is necessary, a lot of which are applicable to real life. But really, like with everything else, it s up to each individual person to decide what they think.", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 183 + ], + "3min_transcript": "the same thing as the square root of a times b. So based on this property of the radical of the principal root, they'll say that this over here is the same thing as the square root of negative 1 times negative 1. If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots. I'm doing this in the other order here. Here I have the principal root of the products, over here I have this on the right. And then from that we all know that negative 1 times negative 1 is 1. So this should be equal to the principal square root of 1. And then the principal square root of 1-- Remember, this radical means principal square root, positive square root, that is just going to be positive 1. And they'll say, this is wrong. Clearly, negative 1 and positive 1 are not the same thing. And they'll argue therefore, you can't make this substitution And what you should then point out is that this was not the incorrect step. That it is true negative 1 is not equal to 1, but the faulty line of reasoning here was in using this property when both a and b are negative. If both a and b are negative, this will never be true. So a and b both cannot be negative. In fact normally when this property is given-- sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time-- but they'll usually give a little bit of a constraint They'll usually say for a and b greater than or equal to 0. So that's where they list this property. This is true for a and b greater than or equal 0. And in particular, it's false if both a and b are both, if they are both negative. saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. When we take traditional principal square roots. So when you take the principle square root of 4. We know that this is positive 2, that 4 actually has two square roots. There's negative 2 also is a square root of 4. If you have negative 2 times negative 2 it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non imaginary, non complex numbers, you could really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers--" + }, + { + "Q": "At 6:41 how can it only apply when x is greater or equal to zero, given that it is a negative number?", + "A": "When the number is positive, it applies as a normal (square) root. For example, the square root of 4 is 2, but the square root of -4 could be seen as 2i, because i=-1. Therefore, x must be greater than or equal to zero in order to have that negative number, and in turn, contain i. Like he says at 5:58, the two negative numbers are where it goes wrong. x cannot be negative.", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 401 + ], + "3min_transcript": "and complex numbers and all the rest-- you have to expand the definition of what this radical means. So when you are taking the square root of really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function, or this is now defined for complex inputs or the domain, it can also generate imaginary or complex outputs, or I guess you could call that the range. And if you assume that, then really straight from this you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only-- and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true-- So we could apply this when x is greater than or equal to 0. then negative x is clearly a negative number, or I guess it could also be 0. It's a negative number. And then we can apply this right over here. If x was less than 0, then we would be doing all of this nonsense up here. And we start to get nonsensical answers. And if you look at it this way and you say hey, look, i can be the square root of negative one, if it's the principal branch of the complex square root function. Then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really the real fault in this logic when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the complex to include negative numbers in the domain and to include imaginary numbers, then you can do this. You can say the square root of negative x is the square root of negative 1 times-- Or you should say the principal square root of negative x-- I should be particular my words-- is the same thing as the principal square root of negative 1 times the principal square root of x when x is greater than or equal to 0. And I don't want to confuse you, if x is greater than or equal to 0, this negative x, that is clearly a negative, or I guess you should say a non positive number." + }, + { + "Q": "At 14:21 Sal points out that S = {[x1, x2] \u00e2\u0088\u0088 R2 | x1 \u00e2\u0089\u00a5 0} is not a subspace because it is not closed under multiplication. Then, at 18:07 he says that the span of any set of vectors is a valid subspace. However, I thought that since the definition of Span is the set of all linear combinations of a set of vectors, there must be a Span ([x1, x2] \u00e2\u0088\u0088 R2 | x1 \u00e2\u0089\u00a5 0). So, there seems to be a contradiction.", + "A": "From what I understand, it must be able to span all R2 vectors in order to be consider a valid subspace of R2 but with the given requirement which is x1\u00e2\u0089\u00a5 0, that is not possible with negative scalar. Thus it is not a span of R2. Am I correct?", + "video_name": "pMFv6liWK4M", + "timestamps": [ + 861, + 1087 + ], + "3min_transcript": "to each other, this thing is also going to be greater than 0. And we don't care what these, these can be anything, I didn't put any constraints on the second component of my vector. So it does seem like it is closed under addition. Now what about scalar multiplication? Let's take a particular case here. Let's take my a, b again. I have my vector a, b. Now I can pick any real scalar. So any real scalar. What if I just multiply it by minus 1? So minus 1. So if I multiply it by minus 1, I get minus a, minus b. If I were to draw it visually, if this is-- let's say a, b was the vector 2, 4. So it's like this. When I multiply it by minus 1, what do I get? I get this vector. Which you can be visually clearly see falls out of, if we view these as kind of position vectors, it falls out of our subspace. Or if you just view it not even visually, if you just do it mathematically, clearly if this is positive then this is going to-- and let's say if we assume this is positive, and definitely not 0. So it's definitely a positive number. So this is definitely going to be a negative number. So when we multiply it by negative 1, for really any element of this that doesn't have a 0 there, you're going to end up with something that falls out of it, right? This is not a member of this set, because to be a member of the set, your first component had to be greater than 0. This first component is less than 0. So this subset that I drew out here, the subset of R2, is not It's not closed under multiplication or scalar multiplication. Now I'll ask you one interesting question. What if I ask you just the span of some set of vectors? Let's say I want to know the span of, I don't know, let's sat I have vector v1, v2, and v3. I'm not even going to tell you how many elements each of these vectors have. Is this a valid subspace of Rn? Where n is the number of elements that each of these have. Let's pick one of the elements." + }, + { + "Q": "At about 5:30 he was talking about the unit circle. How did he come up with 2pi/3 instead of 2pi?", + "A": "Instead of having cos(x), the problem involved cos(3x). In other words, there were 3 repetitions within the usual period of 2\u00cf\u0080. Therefore, each of those periods was of length 2\u00cf\u0080/3.", + "video_name": "SBqnRja4CW4", + "timestamps": [ + 330 + ], + "3min_transcript": "" + }, + { + "Q": "At 5:36 , how did you plot g inverse (x) on the graph, and how did slope come in there ?", + "A": "Well, because y = -x/2 - 1/2 is a linear function, you only need two points to draw it. You can use the y-intercept (x=0): y = - 0/2 - 1/2 = - 1/2. That s the first point. Secondly, you can write -x/2 = -(1*x)/2 = -1/2 x, so -1/2 is the slope of the function. with any x your getting your second point (e.g. x=1 -> y = -1/2 - 1/2 = -1). Draw a line and you re finished.", + "video_name": "wSiamij_i_k", + "timestamps": [ + 336 + ], + "3min_transcript": "like it's supposed to do. Let's do one more of these. So here I have g of x is equal to negative 2x minus 1. So just like the last problem, I like to set y equal to this. So we say y is equal to g of x, which is equal to negative 2x minus 1. Now we just solve for x. y plus 1 is equal to negative 2x. Just added 1 to both sides. Now we can divide both sides of this equation by negative 2, and so you get negative y over 2 minus 1/2 is equal to x, or we could write x is equal to negative y over 2 minus 1/2, or we could write f inverse as a function of y is equal to negative y over 2 minus 1/2, or we can just rename y as x. That shouldn't be an f. The original function was g , so let me be clear. That is g inverse of y is equal to negative y over 2 minus 1/2 because we started with a g of x, not an f of x. Make sure we get our notation right. Or we could just rename the y and say g inverse of x is equal to negative x over 2 minus 1/2. Now, let's graph it. Its y-intercept is negative 1/2. It's right over there. And it has a slope of negative 1/2. Let's see, if we start at negative 1/2, if we move over to 1 in the positive direction, it will go down half. If we move over 1 again, it will go down half again. If we move back-- so it'll go like that. look something like that. It'll just keep going, so it'll look something like that, and it'll keep going in both directions. And now let's see if this really is a reflection over y equals x. y equals x looks like that, and you can see they are a reflection. If you reflect this guy, if you reflect this blue line, it becomes this orange line. But the general idea, you literally just-- a function is originally expressed, is solved for y in terms of x. You just do some algebra. Solve for x in terms of y, and that's essentially your inverse function as a function of y, but then you can rename it as a function of x." + }, + { + "Q": "At 1:30 Sal says that f^-1(x)= -x+4, but in the start f(x) is also equal to -x+4. How can it be the same?\n\nThis may seem silly but i am confused about this question!", + "A": "okay, if we have y=-x + 4, then x = -y +4, that s why the inverse function is same. Another example can be if y=x, then x=y, so the inverse function is same as the function.", + "video_name": "wSiamij_i_k", + "timestamps": [ + 90 + ], + "3min_transcript": "So we have f of x is equal to negative x plus 4, and f of x is graphed right here on our coordinate plane. Let's try to figure out what the inverse of f is. And to figure out the inverse, what I like to do is I set y, I set the variable y, equal to f of x, or we could write that y is equal to negative x plus 4. Right now, we've solved for y in terms of x. To solve for the inverse, we do the opposite. We solve for x in terms of y. So let's subtract 4 from both sides. You get y minus 4 is equal to negative x. And then to solve for x, we can multiply both sides of this equation times negative 1. And so you get negative y plus 4 is equal to x. Or just because we're always used to writing the dependent variable on the left-hand side, we could rewrite this as x is equal to negative y plus 4. Or another way to write it is we could say that f inverse of y is equal to negative y plus 4. it as a function of y, but we can just rename the y as x so it's a function of x. So let's do that. So if we just rename this y as x, we get f inverse of x is equal to the negative x plus 4. These two functions are identical. Here, we just used y as the independent variable, or as the input variable. Here we just use x, but they are identical functions. Now, just out of interest, let's graph the inverse function and see how it might relate to this one right over here. So if you look at it, it actually looks fairly identical. It's a negative x plus 4. It's the exact same function. So let's see, if we have-- the y-intercept is 4, it's going to be the exact same thing. The function is its own inverse. So if we were to graph it, we would put it right on top of this. In the first inverse function video, I talked about how a function and their inverse-- they are the reflection over the line y equals x. So where's the line y equals x here? Well, line y equals x looks like this. And negative x plus 4 is actually perpendicular to y is equal to x, so when you reflect it, you're just kind of flipping it over, but it's going to be the same line. It is its own reflection. Now, let's make sure that that actually makes sense. When we're dealing with the standard function right there, if you input a 2, it gets mapped to a 2. If you input a 4, it gets mapped to 0. What happens if you go the other way? If you input a 2, well, 2 gets mapped to 2 either way, so that makes sense. For the regular function, 4 gets mapped to 0. For the inverse function, 0 gets mapped to 4." + }, + { + "Q": "At 2:41, what does Sal mean by \" And notice, both of these numbers are exactly 10 away from the number 5?\" Why are the numbers 10 units away from 5? Why isn't it 10 away from 0?", + "A": "To be 10 away from zero, the problem would need to be: |x| = 10. Notice the original equation: |x - 5| = 10. You need to take into account the -5 inside the absolute value. That s where the 5 comes from. Hope this helps.", + "video_name": "u6zDpUL5RkU", + "timestamps": [ + 161 + ], + "3min_transcript": "So let's say I have the equation the absolute value of x minus 5 is equal to 10. And one way you can interpret this, and I want you to think about this, this is actually saying that the distance between x and 5 is equal to 10. So how many numbers that are exactly 10 away from 5? And you can already think of the solution to this equation, but I'll show you how to solve it systematically. Now this is going to be true in two situations. Either x minus 5 is equal to positive 10. If this evaluates out to positive 10, then when you take the absolute value of it, you're going to get positive 10. Or x minus 5 might evaluate to negative 10. If x minus 5 evaluated to negative 10, when you take the absolute value of it, you would get 10 again. So x minus 5 could also be equal to negative 10. Now, to solve this one, add 5 to both sides of this equation. You get x is equal to 15. To solve this one, add 5 to both sides of this equation. x is equal to negative 5. So our solution, there's two x's that satisfy this equation. x could be 15. 15 minus 5 is 10, take the absolute value, you're going to get 10, or x could be negative 5. Negative 5 minus 5 is negative 10. Take the absolute value, you get 10. And notice, both of these numbers are exactly 10 away from the number 5. Let's do another one of these. Let's do another one. Let's say we have the absolute value of x plus 2 is equal to 6. So what does that tell us? That tells us that either x plus 2, that the thing inside Or the thing inside of the absolute value sign, the x plus 2, could also be negative 6. If this whole thing evaluated to negative 6, you take the absolute value, you'd get 6. So, or x plus 2 could equal negative 6. And then if you subtract 2 from both sides of this equation, you get x could be equal to 4. If you subtract 2 from both sides of this equation, you get x could be equal to negative 8. So these are the two solutions to the equation. And just to kind of have it gel in your mind, that absolute value, you can kind of view it as a distance, you could rewrite this problem as the absolute value of x minus negative 2 is equal to 6. And so this is asking me, what are the x's that are exactly 6 away from negative 2? Remember, up here we said, what are the x's that are exactly 10 away from positive 5?" + }, + { + "Q": "At about 5:30, Could you write that as -4 1/2? Or is that not allowed?", + "A": "That is allowed, but it just makes it easier to write it like -9/2.", + "video_name": "u6zDpUL5RkU", + "timestamps": [ + 330 + ], + "3min_transcript": "Or the thing inside of the absolute value sign, the x plus 2, could also be negative 6. If this whole thing evaluated to negative 6, you take the absolute value, you'd get 6. So, or x plus 2 could equal negative 6. And then if you subtract 2 from both sides of this equation, you get x could be equal to 4. If you subtract 2 from both sides of this equation, you get x could be equal to negative 8. So these are the two solutions to the equation. And just to kind of have it gel in your mind, that absolute value, you can kind of view it as a distance, you could rewrite this problem as the absolute value of x minus negative 2 is equal to 6. And so this is asking me, what are the x's that are exactly 6 away from negative 2? Remember, up here we said, what are the x's that are exactly 10 away from positive 5? these are both 10 away from positive 5. This is asking, what is exactly 6 away from negative 2? And it's going to be 4, or negative 8. You could try those numbers out for yourself. Let's do another one of these. Let's do another one, and we'll do it in purple. Let's say we have the absolute value of 4x-- I'm going to change this problem up a little bit. 4x minus 1. The absolute value of 4x minus 1, is equal to-- actually, I'll just keep it-- is equal to 19. So, just like the last few problems, 4x minus 1 could be equal to 19. Or 4x minus 1 might evaluate to negative 19. Because then when you take the absolute value, you're going to get 19 again. Or 4x minus 1 could be equal to negative 19. Then you just solve these two equations. Add 1 to both sides of this equation-- we could do them Add 1 to both sides of this equation, you get 4x is equal to negative 18. Divide both sides of this by 4, you get x is equal to 5. Divide both sides of this by 4, you get x is equal to negative 18/4, which is equal to negative 9/2. So both of these x values satisfy the equation. Try it out. Negative 9/2 times 4. This will become a negative 18. Negative 18 minus 1 is negative 19. Take the absolute value, you get 19. You put a 5 here, 4 times 5 is 20. Minus 1 is positive 19. So you take the absolute value. Once again, you'll get a 19. Let's try to graph one of these, just for fun. So let's say I have y is equal to the absolute" + }, + { + "Q": "At 1:39, Sal said that 0 to the zeroth power is undefined. Why?", + "A": "Well, any number to the zeroth power is 1 right? So 0 to the zeroth power would be 1. But zero to any power is 0. So 0 to the zeroth power would be 0. Which is it? It can t be both, so it is undefined.", + "video_name": "NEaLgGi4Vh4", + "timestamps": [ + 99 + ], + "3min_transcript": "Evaluate the expression 4n to the first power minus 2n to the zeroth power for n equals 1 and n equals 5. So, let's do n equals 1 first. So, every place we see an n, let's put a 1 in there. So this is equal to 4 times 1 to the first power minus 2 times 1 to the zeroth power. Every place we saw an n, we have an n there. We have an n to the first, one to the first. So remember, we want to do our order of operations. And I'm going to write it this way just to make it clear that we do parentheses then exponentiation. Then multiplication and division are at the same level, they get the same priority. And then addition and subtraction get the same priority. So, we want to do our exponents before we do anything else. There are no parentheses here. So, we first want to evaluate one to the first power. And anything to the first power is just that anything. If I just have any number to the first power, that is just going to be equal to that number. So, 1 to the first power is just equal to 1. Let me write it. 4 times 1 minus 2 times 1 to the zeroth power. Now, any number other than 0 to the zeroth power is going to be 1. So, if I tell you x to the zeroth power is equal to what? And I tell you that x is not equal to 0. What is x to the zeroth power? You should immediately know that it is equal to 1. I exclude 0 because 0 to the zeroth power is usually undefined. Or for our purposes, you can accept it to be undefined. And maybe in the future, I'll do a video on explaining why it's undefined. There's some good arguments for why this should be 0. Because 0 to pretty much any other power is a 0. But then there's a really good argument for why it should be 1. Because any other number to the zeroth power is 1. So that's why, for the most part, we like to keep this undefined. So with that said, 1 to the zeroth power Anything other than 0 to the zeroth power is 1. So, we have 4 times 1, which is 4, minus 2 times 1, which is 2. So, 4 minus 2 is just equal to 2. Now, let's do the same thing with n equals 5. So it becomes 4 times 5 to the first power minus 2 times 5 to the zeroth power. Same thing. 5 to the first power is just going to be equal to 5. So, I want to do that in a different color. 5 to the first power is just going to be equal to 5. 5 to the zeroth power is just going to be equal to 1. And so, this expression becomes 4 times 5 minus 2 times 1. Or 4 times 5 is 20, minus 2 times 1, that is 2. And this is equal to 18. And we're done." + }, + { + "Q": "At 1:58 why is 0 to the 0 power undefined?", + "A": "it is 1 any number to the power of 0 has to be one", + "video_name": "NEaLgGi4Vh4", + "timestamps": [ + 118 + ], + "3min_transcript": "Evaluate the expression 4n to the first power minus 2n to the zeroth power for n equals 1 and n equals 5. So, let's do n equals 1 first. So, every place we see an n, let's put a 1 in there. So this is equal to 4 times 1 to the first power minus 2 times 1 to the zeroth power. Every place we saw an n, we have an n there. We have an n to the first, one to the first. So remember, we want to do our order of operations. And I'm going to write it this way just to make it clear that we do parentheses then exponentiation. Then multiplication and division are at the same level, they get the same priority. And then addition and subtraction get the same priority. So, we want to do our exponents before we do anything else. There are no parentheses here. So, we first want to evaluate one to the first power. And anything to the first power is just that anything. If I just have any number to the first power, that is just going to be equal to that number. So, 1 to the first power is just equal to 1. Let me write it. 4 times 1 minus 2 times 1 to the zeroth power. Now, any number other than 0 to the zeroth power is going to be 1. So, if I tell you x to the zeroth power is equal to what? And I tell you that x is not equal to 0. What is x to the zeroth power? You should immediately know that it is equal to 1. I exclude 0 because 0 to the zeroth power is usually undefined. Or for our purposes, you can accept it to be undefined. And maybe in the future, I'll do a video on explaining why it's undefined. There's some good arguments for why this should be 0. Because 0 to pretty much any other power is a 0. But then there's a really good argument for why it should be 1. Because any other number to the zeroth power is 1. So that's why, for the most part, we like to keep this undefined. So with that said, 1 to the zeroth power Anything other than 0 to the zeroth power is 1. So, we have 4 times 1, which is 4, minus 2 times 1, which is 2. So, 4 minus 2 is just equal to 2. Now, let's do the same thing with n equals 5. So it becomes 4 times 5 to the first power minus 2 times 5 to the zeroth power. Same thing. 5 to the first power is just going to be equal to 5. So, I want to do that in a different color. 5 to the first power is just going to be equal to 5. 5 to the zeroth power is just going to be equal to 1. And so, this expression becomes 4 times 5 minus 2 times 1. Or 4 times 5 is 20, minus 2 times 1, that is 2. And this is equal to 18. And we're done." + }, + { + "Q": "At 1:17 Sal said \"anything to the zeroth power is equal to 1. What if the number is a negative?\nWouldn't that be -1?", + "A": "It depends on how it s written. If it is expressed as -x^0, or negative x to the power of 0, then it is still equal to 1. If there is a negative sign outside of the term, such as 5 - x^0, you do the exponent first and get 5 - 1.", + "video_name": "NEaLgGi4Vh4", + "timestamps": [ + 77 + ], + "3min_transcript": "Evaluate the expression 4n to the first power minus 2n to the zeroth power for n equals 1 and n equals 5. So, let's do n equals 1 first. So, every place we see an n, let's put a 1 in there. So this is equal to 4 times 1 to the first power minus 2 times 1 to the zeroth power. Every place we saw an n, we have an n there. We have an n to the first, one to the first. So remember, we want to do our order of operations. And I'm going to write it this way just to make it clear that we do parentheses then exponentiation. Then multiplication and division are at the same level, they get the same priority. And then addition and subtraction get the same priority. So, we want to do our exponents before we do anything else. There are no parentheses here. So, we first want to evaluate one to the first power. And anything to the first power is just that anything. If I just have any number to the first power, that is just going to be equal to that number. So, 1 to the first power is just equal to 1. Let me write it. 4 times 1 minus 2 times 1 to the zeroth power. Now, any number other than 0 to the zeroth power is going to be 1. So, if I tell you x to the zeroth power is equal to what? And I tell you that x is not equal to 0. What is x to the zeroth power? You should immediately know that it is equal to 1. I exclude 0 because 0 to the zeroth power is usually undefined. Or for our purposes, you can accept it to be undefined. And maybe in the future, I'll do a video on explaining why it's undefined. There's some good arguments for why this should be 0. Because 0 to pretty much any other power is a 0. But then there's a really good argument for why it should be 1. Because any other number to the zeroth power is 1. So that's why, for the most part, we like to keep this undefined. So with that said, 1 to the zeroth power Anything other than 0 to the zeroth power is 1. So, we have 4 times 1, which is 4, minus 2 times 1, which is 2. So, 4 minus 2 is just equal to 2. Now, let's do the same thing with n equals 5. So it becomes 4 times 5 to the first power minus 2 times 5 to the zeroth power. Same thing. 5 to the first power is just going to be equal to 5. So, I want to do that in a different color. 5 to the first power is just going to be equal to 5. 5 to the zeroth power is just going to be equal to 1. And so, this expression becomes 4 times 5 minus 2 times 1. Or 4 times 5 is 20, minus 2 times 1, that is 2. And this is equal to 18. And we're done." + }, + { + "Q": "I don't understand. At 0:51, Sal says 1/0 - 1/0 is undefined. And then, at 3:11, he says it's indeterminate. Are undefined and indeterminate supposed to be equal to each other? Because the function's the same but it seems to be equaling two different things.", + "A": "This came up in our Calc II class today - undefined and indeterminate are not the same. For example, 1/0 is undefined, however with these limit problems, we know there is a limit (most of the time) even if the Initial Form of the problem is something like 0/0. My understanding is we call this the indeterminate form because with a little manipulation (as Sal demonstrates) there actually is an answer - we just don t see it in the initial form, hence it is indeterminate.", + "video_name": "MeVFZjT-ABM", + "timestamps": [ + 51, + 191 + ], + "3min_transcript": "We want to figure out the limit as x approaches 1 of the expression x over x minus 1 minus 1 over the natural log of x. So let's just see what happens when we just try to plug in the 1. What happens if we evaluate this expression at 1? Well then, we're going to get a one here, over 1 minus 1. So we're going to get something like a 1 over a 0, minus 1 over, and what's the natural log of 1? e to the what power is equal to one? Well, anything to the zeroth power is equal to 1, so e to the zeroth power is going to be equal to 1, so the natural log of 1 is going to be 0. So we get the strange, undefined 1 over 0 minus 1 over 0. It's this bizarre-looking undefined form. But it's not the indeterminate type of form that we looked for in l'Hopital's rule. We're not getting a 0 over a 0, we're not getting an So you might just say, hey, OK, this is a non-l'Hopital's rule problem. We're going to have to figure out this limit some other way. And I would say, well don't give up just yet! Maybe we can manipulate this algebraically somehow so that it will give us the l'Hopital indeterminate form, and then we can just apply the rule. And to do that, let's just see, what happens if we add these two expressions? So if we add them, so this expression, if we add it, it will be, well, the common denominator is going to be x minus 1 times the natural log of x. I just multiplied the denominators. And then the numerator is going to be, well, if I multiply essentially this whole term by natural log of x, so it's going to be x natural log of x, and then this whole term I'm going to multiply by x minus one. So minus x minus 1. And you could break it apart and see that this expression and this expression are the same thing. over x minus 1, because the natural log of x's cancel out. Let me get rid of that. And then this right here is the same thing as 1 over natural log of x, because the x minus 1's cancel out. So hopefully you realize, all I did is I added these two expressions. So given that, let's see what happens if I take the limit as x approaches 1 of this thing. Because these are the same thing. Do we get anything more interesting? So what do we have here? We have one times the natural log of 1. The natural log of 1 is 0, so we have 0 here, so that is a 0. Minus 1 minus 0, so that's going to be another 0, minus 0. So we get a 0 in the numerator. And in the denominator we get a 1 minus 1, which is 0, times the natural log of 1, which is 0, so 0 times 0, that is 0. We have indeterminate form that we need for l'Hopital's rule," + }, + { + "Q": "At 9:00, I am not sure but why didn't Sal solve those 2 equations using matrices? I mean before these vector videos, he was using matrices to solve these kinds of equations. Or is it that these equations can not be solved through matrices? Thanks In Advance.", + "A": "If he used a matrix, it would remove the C constants and sort of defeat the purpose for doing what he was doing. He just wanted to keep the constants intact so you could see what was happening.", + "video_name": "Alhcv5d_XOs", + "timestamps": [ + 540 + ], + "3min_transcript": "In order for them to be linearly dependent, that means that if some constant times 2,1 plus some other constant times this second vector, 3,2 where this should be equal to 0. Where these both aren't necessarily 0. Before I go up for this problem, let's remember what we're going to find out. If either of these are non-zero, if c1 or c2 are non-zero, then this implies that we are dealing with a dependent, linearly dependent set. If c1 and c2 are both 0, if the only way to satisfy this equation -- I mean you can always satisfy it by sitting guys 0, then we're dealing with a linearly independent set. Let's try to do some math. And this'll just take us back to our Algebra 1 days. In order for this to be true, that means 2 times c1 plus 3 times c2 is equal to -- when I say this is equal to 0, it's really the 0 vector. I can rewrite this as 0,0. So 2 times c1 plus 3 times c2 would be equal to that 0 there. And then we'd have 1 times c1 plus 2 times c2 is equal to that 0. And now this is just a system, two equations, two unknowns. A couple of things we could do. Let's just multiply this top equation by 1/2. equal to 0. And then if we subtract the green equation from the red equation this becomes 0. 2 minus 1 and 1/2-- 3/2 is 1 and 1/2 --of this is just 1/2 c2 is equal to 0. And this is easy to solve. c2 is equal to 0. So what's c1? Well, just substitute this back in. c2 is equal to 0. So this is equal to 0. So c1 plus 0 is equal to 0. So c1 is also equal to 0. We could have substituted it back into that top equation as well. So the only solution to this equation involves both c1 and c2 being equal to 0. So they both have to be 0. So this is a linearly independent set of vectors." + }, + { + "Q": "Why do you subtract the green equation from the red one? at 9:17 min", + "A": "That is a good question! This is a system of linear equations having 2 unknown variables and 2 equations with the variables. In order to solve this linear system of equations, we need to eliminate one variable to get an answer for the other variable. So, at 9:17, Sal subtracts the green equation from the red one to eliminate the variable C1 to get the value of C2. After getting C2, you can substitute the value of C2 in any equation to get the value of C1. Sal does this at 9:30. Hope this helped.", + "video_name": "Alhcv5d_XOs", + "timestamps": [ + 557 + ], + "3min_transcript": "In order for them to be linearly dependent, that means that if some constant times 2,1 plus some other constant times this second vector, 3,2 where this should be equal to 0. Where these both aren't necessarily 0. Before I go up for this problem, let's remember what we're going to find out. If either of these are non-zero, if c1 or c2 are non-zero, then this implies that we are dealing with a dependent, linearly dependent set. If c1 and c2 are both 0, if the only way to satisfy this equation -- I mean you can always satisfy it by sitting guys 0, then we're dealing with a linearly independent set. Let's try to do some math. And this'll just take us back to our Algebra 1 days. In order for this to be true, that means 2 times c1 plus 3 times c2 is equal to -- when I say this is equal to 0, it's really the 0 vector. I can rewrite this as 0,0. So 2 times c1 plus 3 times c2 would be equal to that 0 there. And then we'd have 1 times c1 plus 2 times c2 is equal to that 0. And now this is just a system, two equations, two unknowns. A couple of things we could do. Let's just multiply this top equation by 1/2. equal to 0. And then if we subtract the green equation from the red equation this becomes 0. 2 minus 1 and 1/2-- 3/2 is 1 and 1/2 --of this is just 1/2 c2 is equal to 0. And this is easy to solve. c2 is equal to 0. So what's c1? Well, just substitute this back in. c2 is equal to 0. So this is equal to 0. So c1 plus 0 is equal to 0. So c1 is also equal to 0. We could have substituted it back into that top equation as well. So the only solution to this equation involves both c1 and c2 being equal to 0. So they both have to be 0. So this is a linearly independent set of vectors." + }, + { + "Q": "For the final example 13:00:00-end, why does Sal assign a random number for vector c3? How would you solve this to determine dependence without knowing the value(s) of a vector?", + "A": "That s the nice thing about free variables like c3. We get to choose whatever number we want for c3, and no matter what we choose, it will still satisfy the equations. To determine dependence without making a choice for our free variables, all you need to know is that there are free variables. If there are free variables, then the set is linearly dependent.", + "video_name": "Alhcv5d_XOs", + "timestamps": [ + 780 + ], + "3min_transcript": "And I want to know are these linearly dependent or linearly independent. So I go to through the same drill. I use that little theorem that I proved at the beginning of this video. In order for them to be linearly dependent there must be some set of weights that I can multiply these guys. So c1 times this vector plus c2 times this vector plus c3 times that vector, that will equal the 0 vector. And if one of these is non-zero then we're dealing with a linearly dependent set of vectors. And if all of them are 0, then it's independent. Let's just do our linear algebra. So this means that 2 times c1 plus 3 times c2 plus c3 is And then if we do the bottom rows-- Remember when you multiply a scalar times a vector you multiply it by each of these terms. So c1 times 1. 1c1 plus 2c2 plus 2c3 is equal to 0. There's a couple of giveaways on this problem. If you have three two-dimensional vectors, one of them is going to be redundant. Because, in the very best case, even if you assume that that vector and that vector are linearly independent, then these would span r2. Which means that any point, any vector, in your two-dimensional space can be represented by some combination of those two. In which case, this is going to be one of them because this is just a vector in two-dimensional space. So it would be linearly dependent. And then, if you say, well, these aren't linearly In which case, this would definitely be a linearly dependent set. When you see three vectors that are each only vectors in r2, that are each two-dimensional vectors, it's a complete giveaway that this is linearly dependent. But I'm going to show it to you using our dependent, using our little theorem here. So I'm going to show you that I can get non-zero c3's, c2's, and c1's such that I can get a 0 here. If all of these had to be 0-- I mean you can always set them But if they had to be equal to 0, then it would be linearly independent. Let me just show you. I can just pick some random c3. Let me pick c3 to be equal to negative 1. So what would these two equations reduce to? I mean you have just three unknowns and two equations, it means you don't have enough constraints on your system. So if I just set c3-- I just pick that out of a hat. I could have picked c3 to be anything." + }, + { + "Q": "Hey guys I was just wondering about why, at 13:41, can he just pick random values for C3?", + "A": "Because he KNOWS that the vectors are linearly dependent. This is because you only need 2 linearly independent vectors to span R2. Any additional vector MUST be a linear combination of the other two by the very definition of spanning R2.", + "video_name": "Alhcv5d_XOs", + "timestamps": [ + 821 + ], + "3min_transcript": "And then if we do the bottom rows-- Remember when you multiply a scalar times a vector you multiply it by each of these terms. So c1 times 1. 1c1 plus 2c2 plus 2c3 is equal to 0. There's a couple of giveaways on this problem. If you have three two-dimensional vectors, one of them is going to be redundant. Because, in the very best case, even if you assume that that vector and that vector are linearly independent, then these would span r2. Which means that any point, any vector, in your two-dimensional space can be represented by some combination of those two. In which case, this is going to be one of them because this is just a vector in two-dimensional space. So it would be linearly dependent. And then, if you say, well, these aren't linearly In which case, this would definitely be a linearly dependent set. When you see three vectors that are each only vectors in r2, that are each two-dimensional vectors, it's a complete giveaway that this is linearly dependent. But I'm going to show it to you using our dependent, using our little theorem here. So I'm going to show you that I can get non-zero c3's, c2's, and c1's such that I can get a 0 here. If all of these had to be 0-- I mean you can always set them But if they had to be equal to 0, then it would be linearly independent. Let me just show you. I can just pick some random c3. Let me pick c3 to be equal to negative 1. So what would these two equations reduce to? I mean you have just three unknowns and two equations, it means you don't have enough constraints on your system. So if I just set c3-- I just pick that out of a hat. I could have picked c3 to be anything. equations become? You get 2c1 plus 3c2 minus 1 is equal to 0. And you get c1 plus 2c2 minus 2 is equal to 0. 2 times minus 1. What can I do here? If I multiply this second equation by 2, what do I get? I get 2 plus 4c2 minus 4 is equal to 0. And now let's subtract this equation from that equation. So the c1's cancel out. 3c2 minus 4c2 is minus c2. And then minus 1 minus minus 4, so that's minus 1 plus 4. That's plus 3 is equal to 0. And so we get our -- Let me make sure I got that right." + }, + { + "Q": "At 10:44 shouldn't there be a +c for the integration constant?", + "A": "It s a definite integral, so there doesn t need to be a constant of integration.", + "video_name": "AFF8FXxt5os", + "timestamps": [ + 644 + ], + "3min_transcript": "dt, and then you're going to have that plus these two guys multiplied by each other. So that's-- well, there's a minus to sign here so plus. Let me just change this to a minus. Minus cosine squared dt. And if we factor out a minus sign and a dt, what is this going to be equal to? This is going to be equal to the integral from 0 to 2pi of, we could say, sine squared plus-- I want to put the t -- sine squared of t plus cosine squared of t. And actually, let me take the minus sign out to the front. So if we just factor the minus sign, and put a minus there, make this a plus. So the minus sign out there, and then we factor dt out. I did a couple of steps in there, but I think you got it. Now this is just algebra at this point. Factoring out a minus sign, so this becomes positive. And then you have a dt and a dt. Factor that out, and you get this. originally have, if that confuses you at all. And the reason why I did that: we know what sine squared of anything plus cosine squared of that same anything is. That falls right out of the unit circle definition of our trig function, so this is just 1. So our whole integral has been reduced to the minus integral from 0 to 2pi of dt. And this is-- we have seen this before. We can probably say that this is of 1, if you want to put something there. Then the antiderivative of 1 is just-- so this is just going to be equal to minus-- and that minus sign is just the same minus sign that we're carrying forward. The antiderivative of 1 is just t, and we're going to evaluate it from 2pi to 0, or from 0 to 2pi, so this is equal to minus-- that minus sign right there-- 2pi minus t at 0, so minus 0. So this is just equal to minus 2pi. And there you have it. We figured out the work that this field did on the particle, counterclockwise fashion. And our intuition held up. We actually got a negative number for the work done. And that's because, at all times, the field was actually going exactly opposite, or was actually opposing, the movement of, if we think of it as a particle in its counterclockwise direction. Anyway, hopefully, you found that helpful." + }, + { + "Q": "At 1:11, Sal says that the figure he is drawing is a rectangular pyramid. Isn't that a square pyramid?\nThe base side lengths seem equivalent...", + "A": "Rishi@ a rectangle can look like a square because some rectangles are usually very similar looking to them", + "video_name": "ZACf9EecFrY", + "timestamps": [ + 71 + ], + "3min_transcript": "What we're going to explore in this video are polyhedra, which is just the plural of a polyhedron. And a polyhedron is a three-dimensional shape that has flat surfaces and straight edges. So, for example, a cube is a polyhedron. All the surfaces are flat, and all of the edges are straight. So this right over here is a polyhedron. Once again, polyhedra is plural. Polyhedron is when you have one of them. This is a polyhedron. A rectangular pyramid is a polyhedron. So let me draw that. I'll make this one a little bit more transparent. Let me do this in a different color just for fun. I'll make it a magenta rectangular pyramid. So once again, here I have one flat surface. And then I'm going to have four triangular flat surfaces. Now, it clearly looks like a pyramid. Why is it called a rectangular pyramid? Because the base right over here is a rectangle. So these are just a few examples of polyhedra. Now, what I want to think about are nets of polyhedra. And actually, let me draw and make this transparent, too, so we get full appreciation of the entire polyhedron, this entire cube. So now let's think about nets of polyhedron. So what is a net of a polyhedron? Well, one way to think about it is if you kind of viewed this as made up of cardboard, and you were to unfold it in some way so it would become flat, or another way of thinking about it is if you were to cut out some cardboard or some paper, and you wanted to fold it up into one of these figures, how would you go about doing it? And each of these polyhedra has multiple different nets that you could create so that it can be folded up into this three-dimensional figure. So let's take an example. And I'm going to color code it. So let's say that the bottom of this cube was this green color. And so I can represent it like this. That's the bottom of the cube. It's that green color. Now, let's say that this back surface of the cube is orange. Well, I could represent it like this. And notice, I've kind of folded it out. I'm folding it out. And so if I were to flatten it out, it would look like this. It would look like that. Now, this other backside, I'll shade it in yellow. This other backside right over here, I could fold it backwards and keep it connected along this edge, fold it backwards. It would look like this. It would look like that. I think you get the general idea here. And just to be clear, this edge right over here is this edge right over there." + }, + { + "Q": "At 20:00, why does a derivative of 0 mean a maximum/minimum? I get why, but I thought y=x^2 has only one minimum at x=0.", + "A": "Yes and if you let d/dx x^2 = 2x =0, you get x = 0 (and y = 0).", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 1200 + ], + "3min_transcript": "Because the problem wants us to find that point, the maximum point of intersection. They call this the extreme normal line. The extreme normal line is when our second quadrant intersection essentially achieves a maximum point. I know they call it the smallest point, but it's the smallest negative value, so it's really a maximum point. So how do we figure out that maximum point? Well, we have our second quadrant intersection as a function of our first quadrant x. I could rewrite this as, my second quadrant intersection as a function of x0 is equal to minus x minus 1 over 2 x0. So this is going to reach a minimum or a maximum point when its derivative is equal to 0. This is a very unconventional notation, and that's probably the hardest thing about this problem. But let's take this derivative with respect to x0. So my second quadrant intersection, the derivative of straightforward. It's equal to minus 1, and then I have a minus 1/2 times, this is the same thing as x to the minus 1. So it's minus 1 times x0 to the minus 2, right? I could have rewritten this as minus 1/2 times x0 to the minus 1. So you just put its exponent out front and decrement it by 1. And so this is the derivative with respect to my first quadrant intersection. So let me simplify this. So x, my second quadrant intersection, the derivative of it with respect to my first quadrant intersection, is equal to minus 1, the minus 1/2 and the minus 1 become a positive when you multiply them, and so plus 1/2 over x0 squared. Now, this'll reach a maximum or minimum when it equals 0. problem right there. Well, we add one to both sides. We get 1 over 2 x0 squared is equal to 1, or you could just say that that means that 2 x0 squared must be equal to 1, if we just invert both sides of this equation. Or we could say that x0 squared is equal to 1/2, or if we take the square roots of both sides of that equation, we get x0 is equal to 1 over the square root of 2. So we're really, really, really close now. We've just figured out the x0 value that gives us our extreme normal line. This value right here. Let me do it in a nice deeper color. This value right here, that gives us the extreme normal line, that over there is x0 is equal to 1 over the square root of 2. Now, they want us to figure out the equation of the" + }, + { + "Q": "I think there's a mistake right 12:56 when he starts out factoring, or fastforward to 13:46.\nHe's just factoring the equation inside the square root. 1/4xo^2 + (2+4xo^2); o being sub 0, so xo is x sub 0.\nHe factors out 4/xo^2 and gets 4/xo^2(xo^4 + 1/2xo^2 + 1/16)\n\nIn the first part, 4/xo^2 and 1/4xo^2, the 4's cancel and we should be left with 1/xo^4 or xo^-4. Sal just put it as a positive x^4. I might be nitpicking, but I'm just making sure I'm right as there are no annotations correcting this.", + "A": "Yes I do, thank you! It s always the algebra that can be confusing.", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 776, + 826 + ], + "3min_transcript": "So minus b is minus 1 over 2 x0, plus or minus the square root of b squared. So that's that squared. So it's one over four x0 squared minus 4ac. So minus 4 times 1 times this minus thing. So I'm going to have a minus times a minus is a plus, so it's just 4 times this, because there was one there. So plus 4 times this, right here. 4 times this is just 2 plus 4 x0 squared. All I did is, this is 4ac right here. Well, minus 4ac. The minus and the minus canceled out, so There's a 1. So 4 times c is just 2 plus 4x squared. I just multiply this by two, and of course all of this So let's see if I can simplify this. We're just figuring out where the normal line and the parabola intersect. Now, what do we get here. This looks like a little hairy beast here. Let me see if I can simplify this a little bit. So let us factor out-- let me rewrite this. I can just divide everything by 1/2, so this is minus 1 over 4 x0, I just divided this by 2, plus or minus 1/2, that's just this 1/2 right there, times the square root, let me see what I can simplify out of here. So if I factor out a 4 over x0 squared, then what does This term right here will become an x to the fourth, x0 to the fourth, plus, now, what does this term become? This term becomes a 1/2 x0 squared. And just to verify this, multiply 4 times 1/2, you get 2, and then the x0 squares cancel out. So write this term times that, will equal 2, and then you have plus-- now we factored a four out of this and the x0 squared, so plus 1/16. Let me scroll over a little bit. And you can verify that this works out. If you were to multiply this out, you should get this business right here. I see the home stretch here, because this should actually factor out quite neatly. So what does this equal?" + }, + { + "Q": "at 13:20 how did you decide to factor out 4/x^2. It seems like a wild goose chase to me.", + "A": "If you notice that a polynomial has degrees such as -2, then 0, then 2, and you want to turn it into 0, 2, and 4 to make it easier to factor, you can multiply by the second degree, but must make sure to also divide by the second degree to keep it equivalent. Dividing by the second degree is the same as factoring out the reciprocal of the second degree. Example, a = (1/1)a = (n/n)a = (1/n)(n)a = (1/n)(na). I don t know if this helps.", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 800 + ], + "3min_transcript": "So minus b is minus 1 over 2 x0, plus or minus the square root of b squared. So that's that squared. So it's one over four x0 squared minus 4ac. So minus 4 times 1 times this minus thing. So I'm going to have a minus times a minus is a plus, so it's just 4 times this, because there was one there. So plus 4 times this, right here. 4 times this is just 2 plus 4 x0 squared. All I did is, this is 4ac right here. Well, minus 4ac. The minus and the minus canceled out, so There's a 1. So 4 times c is just 2 plus 4x squared. I just multiply this by two, and of course all of this So let's see if I can simplify this. We're just figuring out where the normal line and the parabola intersect. Now, what do we get here. This looks like a little hairy beast here. Let me see if I can simplify this a little bit. So let us factor out-- let me rewrite this. I can just divide everything by 1/2, so this is minus 1 over 4 x0, I just divided this by 2, plus or minus 1/2, that's just this 1/2 right there, times the square root, let me see what I can simplify out of here. So if I factor out a 4 over x0 squared, then what does This term right here will become an x to the fourth, x0 to the fourth, plus, now, what does this term become? This term becomes a 1/2 x0 squared. And just to verify this, multiply 4 times 1/2, you get 2, and then the x0 squares cancel out. So write this term times that, will equal 2, and then you have plus-- now we factored a four out of this and the x0 squared, so plus 1/16. Let me scroll over a little bit. And you can verify that this works out. If you were to multiply this out, you should get this business right here. I see the home stretch here, because this should actually factor out quite neatly. So what does this equal?" + }, + { + "Q": "At 18:45 How does khan figure out that when derivative of -x_0-1/2x_0 is equal to 0 is actually minimum or maximum of the function?", + "A": "derivative = 0 ==> (implies) a maximum or minimum", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 1125 + ], + "3min_transcript": "So minus x0 minus 1 over 4 mine x0. Now what do we have? So let's see. We have a minus 1 over 4 x0, minus 1 over 4 x0. So this is equal to minus x0, minus x0, minus 1 over 2 x0. So if I take minus 1/4 minus 1/4, I get minus 1/2. And so my second quadrant intersection, all this work I did got me this result. My second quadrant intersection, I hope I don't run out of space. My second quadrant intersection, of the normal line and the parabola, is minus x0 minus 1 over 2 x0. Now this by itself is a pretty neat result we just got, but Because the problem wants us to find that point, the maximum point of intersection. They call this the extreme normal line. The extreme normal line is when our second quadrant intersection essentially achieves a maximum point. I know they call it the smallest point, but it's the smallest negative value, so it's really a maximum point. So how do we figure out that maximum point? Well, we have our second quadrant intersection as a function of our first quadrant x. I could rewrite this as, my second quadrant intersection as a function of x0 is equal to minus x minus 1 over 2 x0. So this is going to reach a minimum or a maximum point when its derivative is equal to 0. This is a very unconventional notation, and that's probably the hardest thing about this problem. But let's take this derivative with respect to x0. So my second quadrant intersection, the derivative of straightforward. It's equal to minus 1, and then I have a minus 1/2 times, this is the same thing as x to the minus 1. So it's minus 1 times x0 to the minus 2, right? I could have rewritten this as minus 1/2 times x0 to the minus 1. So you just put its exponent out front and decrement it by 1. And so this is the derivative with respect to my first quadrant intersection. So let me simplify this. So x, my second quadrant intersection, the derivative of it with respect to my first quadrant intersection, is equal to minus 1, the minus 1/2 and the minus 1 become a positive when you multiply them, and so plus 1/2 over x0 squared. Now, this'll reach a maximum or minimum when it equals 0." + }, + { + "Q": "At 14:03, I don't understand what Sal does to get the expression 4/x0(x0^4+1/2(x0^2)+1/16) under the squareroot. It seems to me that he divides the whole expression by 4, but I still can't make sense out of it...", + "A": "He s factoring out 4/(x0)^2 from the three terms. Unfortunately, he turns around the expression so that in the factored form the first and third terms are exchanged, which might be contributing to your confusion (I did a double take as well!). If you try factoring out 4/(x0)^2 from the three terms but keep them in the same order, you end up with: 4/(x0)^2 [1/16 + (1/2)(x0)^2 + (x0)^4]. You can see it s the same as what he has. Hope that helps!", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 843 + ], + "3min_transcript": "So let's see if I can simplify this. We're just figuring out where the normal line and the parabola intersect. Now, what do we get here. This looks like a little hairy beast here. Let me see if I can simplify this a little bit. So let us factor out-- let me rewrite this. I can just divide everything by 1/2, so this is minus 1 over 4 x0, I just divided this by 2, plus or minus 1/2, that's just this 1/2 right there, times the square root, let me see what I can simplify out of here. So if I factor out a 4 over x0 squared, then what does This term right here will become an x to the fourth, x0 to the fourth, plus, now, what does this term become? This term becomes a 1/2 x0 squared. And just to verify this, multiply 4 times 1/2, you get 2, and then the x0 squares cancel out. So write this term times that, will equal 2, and then you have plus-- now we factored a four out of this and the x0 squared, so plus 1/16. Let me scroll over a little bit. And you can verify that this works out. If you were to multiply this out, you should get this business right here. I see the home stretch here, because this should actually factor out quite neatly. So what does this equal? parabola is equal to this. Minus 1 over 4 x0 plus or minus 1/2 times the square root of this business. And the square root, this thing right here is 4 over x0 squared. This is actually, lucky for us, a perfect square. And I won't go into details, because then the video will get too long, but I think you can recognize that this is x0 squared, plus 1/4. If you don't believe me, square this thing right here. You'll get this expression right there. And luckily enough, this is a perfect square, so we can actually take the square root of it. And so we get, the point at which they intersect, our normal line and our parabola, and this is quite a hairy problem. The points where they intersect is minus 1 over 4 x0, plus or minus 1/2 times the square root of this. The square root of this is the square root of this, which is just 2 over x0 times the square root of this, which is" + }, + { + "Q": "At 13:00 Sal begins to factor the expression under the radical sign. This factorization would not have occurred to me. Can anyone comment on the insight that leads to this?", + "A": "First put everything over a common denominator- 4xo^2. I think it becomes much clearer if you do this intermediary step. Impressive nevertheless!", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 780 + ], + "3min_transcript": "So minus b is minus 1 over 2 x0, plus or minus the square root of b squared. So that's that squared. So it's one over four x0 squared minus 4ac. So minus 4 times 1 times this minus thing. So I'm going to have a minus times a minus is a plus, so it's just 4 times this, because there was one there. So plus 4 times this, right here. 4 times this is just 2 plus 4 x0 squared. All I did is, this is 4ac right here. Well, minus 4ac. The minus and the minus canceled out, so There's a 1. So 4 times c is just 2 plus 4x squared. I just multiply this by two, and of course all of this So let's see if I can simplify this. We're just figuring out where the normal line and the parabola intersect. Now, what do we get here. This looks like a little hairy beast here. Let me see if I can simplify this a little bit. So let us factor out-- let me rewrite this. I can just divide everything by 1/2, so this is minus 1 over 4 x0, I just divided this by 2, plus or minus 1/2, that's just this 1/2 right there, times the square root, let me see what I can simplify out of here. So if I factor out a 4 over x0 squared, then what does This term right here will become an x to the fourth, x0 to the fourth, plus, now, what does this term become? This term becomes a 1/2 x0 squared. And just to verify this, multiply 4 times 1/2, you get 2, and then the x0 squares cancel out. So write this term times that, will equal 2, and then you have plus-- now we factored a four out of this and the x0 squared, so plus 1/16. Let me scroll over a little bit. And you can verify that this works out. If you were to multiply this out, you should get this business right here. I see the home stretch here, because this should actually factor out quite neatly. So what does this equal?" + }, + { + "Q": "In the end part of the video, at 7:02, can the -90 degrees be written as 270 degrees? or should it really be -90 degrees?", + "A": "In most cases, they are equivalent. -90 degrees = 270 degrees. There are some contexts when they may not be equivalent. E.g., suppose you are flying a plane, heading North. A -90 degree turn might correspond to a 90 degree LEFT turn (counter-clockwise), to the West. A 270 degree turn might correspond to a 270 degree RIGHT turn (clockwise). You still end up facing the same direction (due West), but you ve taken two very different actions to get there.", + "video_name": "z8vj8tUCkxY", + "timestamps": [ + 422 + ], + "3min_transcript": "for every 180 degrees or we can even write it this way pi radians for every 180 degrees. And here this might be a little less intuitive the degrees cancel out and that's why I usually like to write out the word and you're left with 45 pi/180 radians. Actually let me write this with the words written out for me that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with when you multiply 45 times pi over 180 the degrees have canceled out and you're just left with Which is equal to what? 45 is half of 90 which is half of 180 so this is 1/4 this is equal to pi over 4 radians. Lets do 1 more over here. So lets say that we had negative pi/2 radians. What's that going to be in degrees? Well once again we have to figure out how many degrees are each of these radians. We know that there 180 degrees for every pi radians so we're gonna get the radians cancel out the pi's cancel out and so you have -180/2 this is -90 degrees or we can write it as -90 degrees. and I'll do a couple more example problems here because the more example for this the better and hopefully it will become a little bit intuitive" + }, + { + "Q": "How come at 9:45 he says that if you integrate it you get the same answer? Wouldn't you really get u(x)=x+C not u(x)=x? What happened to the constant of integration?", + "A": "You are allowed to add a constant, but given a constant you have infinitely many solutions so in general we simply take 0 as the constant. And 0 you can easily leave out.", + "video_name": "j511hg7Hlbg", + "timestamps": [ + 585 + ], + "3min_transcript": "of writing it mu prime of x, we could write that as d mu dx. So let's do that. So we could write mu of x is equal to d, the derivative of mu with respect to x, times x. And this is actually a separable differential equation in and of itself. It's kind of a sub-differential equation to solve our broader one. We're just trying to figure out the integrating factor right here. So let's divide both sides by x. So we get mu over x, this is just a separable equation now, is equal to d mu dx. And then, let's divide both sides by mu of x, and we get 1 over x is equal to 1 over mu. That's mu of x, I'll just write 1 over mu right now, for simplicity, times d mu dx. Multiply both sides by dx, you get 1 over x dx is equal to 1 over mu of x d mu. Now, you could integrate both sides of this, and you'll get the natural log of the absolute value of x is equal to the natural log of the absolute value of mu, et cetera, et cetera. But it should be pretty clear from this that x is equal to mu, or mu is equal to x, right? If you look at both sides of this equation there, you can just change x for mu, and it becomes the other side. So, this is obviously telling us that mu of x is equal to x. Or mu is equal to x. So we have our integrating factor. And if you want, you can take the antiderivative of both sides with the natural logs, and all of that. And you'll get the same answer. But this is just, by looking at it, by inspection, you know that mu is equal to x. Because both sides of this equation are completely the same. Anyway, we now have our integrating factor. So in the next video, we're now going to use this integrating factor. Multiply it times our original differential equation. Make it exact. And then solve it as an exact equation. I'll see you in the next video." + }, + { + "Q": "@ 1:25 how do you get 9^2", + "A": "Sal gets the 9^2 because he originally has 9^(t/2 +2) Using the exponent rule: x^a *x^b =X^ab so you get 9^t/2 * 9^2", + "video_name": "Y6wNiYcuCoE", + "timestamps": [ + 85 + ], + "3min_transcript": "- [Voiceover] What I hope to do in this video is get some practice simplifying some fairly hairy exponential expressions. So let's get started. Let's say that I have the expression 10 times nine to the t over two plus two power, times five to the three t. And what I wanna do is simplify this as much as possible, and preferably get it in the form of A times B to the t. And like always, I encourage you to pause this video and see if you can do this on your own using exponent properties, your knowledge, your deep knowledge of exponent properties. All right, so let's work through this together, and it's really just about breaking the pieces up. So 10, I'll just leave that as 10 for now, there doesn't seem to be much to do there. But there's all sorts of interesting things going on here. So nine to the t over two, plus two, so this right over here, I could break this up, write the properties over here. If I have nine to the a plus b power, this is the same thing as nine to the a, times nine to the b power. And over here I have nine to the t over two, plus two, so I could rewrite this as nine to the t over two power, times nine squared. All right, now let's move over to five to the three t. Well, if I have a to the bc, so you could view this as five to the three times t, this is the same thing as a to the b, and then that to the c power. So I could write this as, this is going to be the same thing as five to the third, and then that to the t power. And the whole reason I did that is well this is just going to be a number, then I'm going to have some number to the t power. I want to get as many things just raised just to see if I can simplify this thing. So this character right over here is going to be 81. Nine squared is 81. Five to the third power, 25 times five, that's 125. So we're making good progress and so the only thing we really have to simplify at this point is nine to the t over two. And actually let me do that over here. Nine to the t over two. Well that's the same thing as nine to the one half times t. And by this property right over here, that's the same thing as nine to the one, nine to the one half... And then that to the t power. So what's nine to the one half? Well that's three, so this is going to be equal to three to the t power. So this right over here is three to the t power." + }, + { + "Q": "I like the parentheses at 1:12 rather than multiplying through by -1. Suppose some people might prefer to multiply, though? Seems messier than necessary.", + "A": "It s kind of the same thing.", + "video_name": "8Wxw9bpKEGQ", + "timestamps": [ + 72 + ], + "3min_transcript": "Divide x squared minus 3x plus 2 divided by x minus 2. So we're going to divide this into that. And we can do this really the same way that you first learned long division. So we have x minus 2 being divided into x squared minus 3x plus 2. Another way we could have written the same exact expression is x squared minus 3x plus 2, all of that over x minus 2. That, that, and that are all equivalent expressions. Now, to do this type of long division-- we can call it algebraic long division-- you want to look at the highest degree term on the x minus 2 and the highest degree term on the x squared minus 3x plus 2. And here's the x, and here's the x squared. x goes into x squared how many times? Or x squared divided by x is what? Well, that's just equal to x. So x goes into x squared x times. And I'm going to write it in this column right here above all of the x terms. And then we want to multiply x times x minus 2. That gives us-- x times x is x squared. And just like you first learned in long division, you want to subtract this from that. But that's completely the same as adding the opposite, or multiplying each of these terms by negative 1 and then adding. So let's multiply that times negative 1. And negative 2x times negative 1 is positive 2x. And now let's add. x squared minus x squared-- those cancel out. Negative 3x plus 2x-- that is negative x. And then we can bring down this 2 over here. So it's negative x plus 2 left over, when we only go x times. So then we say, can x minus 2 go into negative x plus 2? Well, x goes into negative x negative one times. You can look at it right here. Negative x divided by x is negative 1. These guys cancel out. Those guys cancel out. So negative 1 times x minus 2-- you have negative 1 times x, which is negative x. Negative 1 times negative 2 is positive 2. just like you do in long division. But that's the same thing as adding the opposite, or multiplying each of these terms by negative 1 So negative x times negative 1 is positive x. Positive 2 times negative 1 is negative 2. These guys cancel out, add up to 0. These guys add up to 0. We have no remainder. So we got this as being equal to x minus 1. And we can verify it. If we multiply x minus 1 times x minus 2, we should get this. So let's actually do that. So let's multiply x minus 1 times x minus 2. So let's multiply negative 2 times negative 1. That gives us positive 2. Negative 2 times x-- that's negative 2x. Let's multiply x times negative 1. That is negative x." + }, + { + "Q": "At 0:21, you mention \"Fibonacci numbers\"; 1,1,2,5,8,13,21,34,...; what are those? And why are they called that?", + "A": "In 1175 A.D., the man who invented the Fibonacci Numbers, Fibonacci, was born. Throughout his life he encountered many circumstances where these numbers appear (like in the spirals of a pineapple or pinecone). This pattern frequently shows up in nature and is extremely fun to find out where it exactly fits in.", + "video_name": "gBxeju8dMho", + "timestamps": [ + 21 + ], + "3min_transcript": "Dear Nickelodeon, I've gotten over how SpongeBob's pants are not actually square. I can ignore most of the time that Gary's shell is not a logarithmic spiral. But what I cannot forgive is that SpongeBob's pineapple house is a mathematical impossibility. There's three easy ways to find spirals on a pineapple. There's the ones that wind up it going right, the ones that spiral up to the left, and the ones that go almost straight up-- keyword almost. If you count the number of spirals going left and the number of spirals going right, they'll be adjacent Fibonacci numbers-- 3 and 5, or 5 and 8, 8 and 13, or 13 and 21. You claim that SpongeBob Squarepants lives in a pineapple under the sea, but does he really? A true pineapple would have Fibonacci spiral, so let's take a look. Because these images of his house don't let us pick it up and turn it around to count the number of spirals going around it, it might be hard to figure out whether it's mathematically a pineapple or not. But there's a huge clue in the third spiral, the one going upwards. In this pineapple there's 8 to the right, 13 to the left. You can add those numbers together to get how many spirals are in the set spiraling In this case, 21. The three sets of spirals in any pineapple are pretty much always adjacent Fibonacci numbers. The rare mutant cases might show Lucas numbers or something, but it will always be three adjacent numbers in a series. What you'll never have is the same number of spirals both ways. Pineapples, unlike people, don't have bilateral symmetry. You'll never have that third spiral be not a spiral, but just a straight line going up a pineapple. Yet, when we look at SpongeBob's supposed pineapple under the sea, it clearly has lines of pineapple things going straight up. It clearly has bilateral symmetry. It clearly is not actually a pineapple at all, because no pineapple could possibly grow that way. Nickelodeon, you need to take a long, hard look in the mirror and think about the way you're misrepresenting the universe to your viewers. This kind of mathematical oversight is simply irresponsible. Sincerely, Vi Hart." + }, + { + "Q": "At 1:30, he said that you would multiply 1 by one half if multiplying by a negative exponent. Am I correct in the sense that if 3^3 is 27, 3^-3 power would be 1/27? Thank you!\n\n-KoKo", + "A": "Yes... 3^(-3) = 1/27. Great job!", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 90 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:04, could you multiply with a decimal? I mean, fractions and decimals are the same things, right?", + "A": "I don t understand your question properly but let say (5.5)^-1 = 1 / 5.5 let say (1/1.3)^-1 = 1.3", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 124 + ], + "3min_transcript": "" + }, + { + "Q": "At 6:25, why was 1/(25/64) changed to it's reciprocal, 64/25?", + "A": "You actually don t have to change the fraction 5/8 to 1/(25/64) or 64/25. Sal states you could just change 5/8 to 8/5 and raise the negative exponent to a positive one. He did this to get the right answer.", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 385 + ], + "3min_transcript": "" + }, + { + "Q": "at 6:41, why does he switch 25/64 to 64/25 ?", + "A": "Multiplication is the inverse operation of division. Instead of dividing by 25/64, you can multiply by 64/25. It is easier to multiply fractions.", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 401 + ], + "3min_transcript": "" + }, + { + "Q": "Did he mis-speak at 0:45-48 seconds? He said to view the top example as multiplying 2 by \"negative 1\"", + "A": "Yes that was a mistake. He meant times 1. :)", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 45 + ], + "3min_transcript": "" + }, + { + "Q": "At 6:24, how did we get 64/25?", + "A": "Sal has: 1 divided by 25/64 Do the division -- change division to multiply by using the reciprocal 1 * 64/25 = 64/25 Hope this helps.", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 384 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:51 why does it change to a positive 3?", + "A": "Because x^-y =1/(x^y). If you don t understand the logic behind that, I can give a more detailed explanation.", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 291 + ], + "3min_transcript": "" + }, + { + "Q": "Why are vectors represented as column form and not row form ? For eg. at 4:00 of this video why wasn't vector V represented as [5,0] in row form.", + "A": "It is simply a convention that most people agree upon, many authors of books write row vectors instead of column vectors. The advantage of the column vector is that it is easy to see how to do matrix multiplication. Further on in linear algebra, one learns about the transpose map and it s relation to functions that operate on vectors to give scalars (linear functionals). Then it becomes a bigger deal whether or not your vector is column or row. For now try not to worry about it too much.", + "video_name": "br7tS1t2SFE", + "timestamps": [ + 240 + ], + "3min_transcript": "And then the direction that the arrow is pointed in specifies it's direction. So this right over here could represent visually this vector. If we say that the horizontal axis is say east, or the positive horizontal direction is moving in the east, this would be west, that would be north, and then that would be south. Now, what's interesting about vectors is that we only care about the magnitude in the direction. We don't necessarily not care where we start, where we place it when we think about it visually like this. So for example, this would be the exact same vector, or be equivalent vector to this. This vector has the same length. So it has the same magnitude. It has a length of 5. And its direction is also due east. So these two vectors are equivalent. Now one thing that you might say is, well, that's fair enough. But how do we represent it with a little bit more mathematical notation? So we don't have to draw it every time. And we could start performing operations on it. want a variable to represent a vector, is usually a lowercase letter. If you're publishing a book, you can bold it. But when you're doing it in your notebook, you would typically put a little arrow on top of it. And there are several ways that you could do it. You could literally say, hey 5 miles per hour east. But that doesn't feel like you can really operate on that easily. The typical way is to specify, if you're in two dimensions, to specify two numbers that tell you how much is this vector moving in each of these dimensions? So for example, this one only moves in the horizontal dimension. And so we'll put our horizontal dimension first. So you might call this vector 5, 0. It's moving 5, positive 5 in the horizontal direction. And it's not moving at all in the vertical direction. And the notation might change. You might also see notation, and actually in the linear algebra context, it's more typical to write it as a column vector like this-- 5, 0. represents how much we're moving in the horizontal direction. And the second coordinate represents how much are we moving in the vertical direction. Now, this one isn't that interesting. You could have other vectors. You could have a vector that looks like this. Let's say it's moving 3 in the horizontal direction. And positive 4. So 1, 2, 3, 4 in the vertical direction. So it might look something like this. So this could be another vector right over here. Maybe we call this vector, vector a. And once again, I want to specify that is a vector. And you see here that if you were to break it down, in the horizontal direction, it's shifting three in the horizontal direction, and it's shifting positive four in the vertical direction." + }, + { + "Q": "5:47 if the magnitude called \"scalar\" on its own, what is the direction on its own?", + "A": "Scalar has no direction.", + "video_name": "br7tS1t2SFE", + "timestamps": [ + 347 + ], + "3min_transcript": "want a variable to represent a vector, is usually a lowercase letter. If you're publishing a book, you can bold it. But when you're doing it in your notebook, you would typically put a little arrow on top of it. And there are several ways that you could do it. You could literally say, hey 5 miles per hour east. But that doesn't feel like you can really operate on that easily. The typical way is to specify, if you're in two dimensions, to specify two numbers that tell you how much is this vector moving in each of these dimensions? So for example, this one only moves in the horizontal dimension. And so we'll put our horizontal dimension first. So you might call this vector 5, 0. It's moving 5, positive 5 in the horizontal direction. And it's not moving at all in the vertical direction. And the notation might change. You might also see notation, and actually in the linear algebra context, it's more typical to write it as a column vector like this-- 5, 0. represents how much we're moving in the horizontal direction. And the second coordinate represents how much are we moving in the vertical direction. Now, this one isn't that interesting. You could have other vectors. You could have a vector that looks like this. Let's say it's moving 3 in the horizontal direction. And positive 4. So 1, 2, 3, 4 in the vertical direction. So it might look something like this. So this could be another vector right over here. Maybe we call this vector, vector a. And once again, I want to specify that is a vector. And you see here that if you were to break it down, in the horizontal direction, it's shifting three in the horizontal direction, and it's shifting positive four in the vertical direction. about how much we're moving up and how much we're moving to the right when we start at the end of the arrow and go to the front of it. So this vector might be specified as 3, 4. 3, 4. And you could use the Pythagorean theorem to figure out the actual length of this vector. And you'll see because this is a 3, 4, 5 triangle, that this actually has a magnitude of 5. And as we study more and more linear algebra, we're going to start extending these to multiple dimensions. Obviously we can visualize up to three dimensions. In four dimensions it becomes more abstract. And that's why this type of a notation is useful. Because it's very hard to draw a 4, 5, or 20 dimensional arrow like this." + }, + { + "Q": "Hi at 1:04 you indicated that speed and direction is velocity, but isn't velocity = change in speed regardless of direction?", + "A": "Velocity is the change in position with regard to direction, Speed is the change in position without regard to direction. A change in either speed or velocity is acceleration.", + "video_name": "br7tS1t2SFE", + "timestamps": [ + 64 + ], + "3min_transcript": "A vector is something that has both magnitude and direction. Magnitude and direction. So let's think of an example of what wouldn't and what would be a vector. So if someone tells you that something is moving at 5 miles per hour, this information by itself is not a vector quantity. It's only specifying a magnitude. We don't know what direction this thing is moving 5 miles per hour in. So this right over here, which is often referred to as a speed, is not a vector quantity just by itself. This is considered to be a scalar quantity. If we want it to be a vector, we would also have to specify the direction. So for example, someone might say it's moving 5 miles per hour east. So let's say it's moving 5 miles per hour due east. So now this combined 5 miles per are due east, this is a vector quantity. And now we wouldn't call it speed anymore. So velocity is a vector. We're specifying the magnitude, 5 miles per hour, and the direction east. But how can we actually visualize this? So let's say we're operating in two dimensions. And what's neat about linear algebra is obviously a lot of what applies in two dimensions will extend to three. And then even four, five, six, as made dimensions as we want. Our brains have trouble visualizing beyond three. But what's neat is we can mathematically deal with beyond three using linear algebra. And we'll see that in future videos. But let's just go back to our straight traditional two-dimensional vector right over here. So one way we could represent it, as an arrow that is 5 units long. We'll assume that each of our units here is miles per hour. And that's pointed to the right, where we'll say the right is east. So for example, I could start an arrow right over here. And I could make its length 5. The length of the arrow specifies the magnitude. And then the direction that the arrow is pointed in specifies it's direction. So this right over here could represent visually this vector. If we say that the horizontal axis is say east, or the positive horizontal direction is moving in the east, this would be west, that would be north, and then that would be south. Now, what's interesting about vectors is that we only care about the magnitude in the direction. We don't necessarily not care where we start, where we place it when we think about it visually like this. So for example, this would be the exact same vector, or be equivalent vector to this. This vector has the same length. So it has the same magnitude. It has a length of 5. And its direction is also due east. So these two vectors are equivalent. Now one thing that you might say is, well, that's fair enough. But how do we represent it with a little bit more mathematical notation? So we don't have to draw it every time. And we could start performing operations on it." + }, + { + "Q": "On 0:04 seconds why does he say number A instead of letter A?", + "A": "In algebra, a letter is a variable that represents a number. When Sal says letter a , a is representing its place on the number line. Hope it helps!", + "video_name": "29P6bar7nHc", + "timestamps": [ + 4 + ], + "3min_transcript": "- [Voiceover] What I have here are three numbers plotted on the number line. We have the number a, the number c, the number b. And then we have three -- (laughs) we have four inequalities, actually. Four inequalities that involve absolute value. And what I want to do is figure out which is these inequalities are true, given where a, c and b are on the number line. And I encourage you to pause the video and try to think through it on your own. All right, let's look at this first one. It says that, \"a is less than b.\" So if we look at a and we look at b, a is clearly to the left of b on the number line. So we know that this is true. Even more we know that a is negative, it's to the left of zero, while b is positive. Which is, if one thing is negative and the other thing is positive, the negative thing is definitely going to be less than the positive thing. But even easier than that, a is to the left of b on the number line. If you're to the left of something else on the number line you're less than that other thing. at least the way we've constructed it, it increases from left to right. All right, the next statement, \"The absolute value of a is greater than the absolute value of b.\" Well, let's just think about where these are on the number line. So we've already said a is three hash marks to the left of zero. That is a. So what is going to be the absolute value of a? Well, the absolute value of a is the distance that a is from zero. So the distance that a is from zero is one, two, three hash marks. So the absolute value of a is just going to be that same distance on the positive side. So the point that we marked as c is also the absolute value of a. So that is also the absolute value of a. The absolute value of a -- sorry, a is three to the left of zero. Absolute value of a is going to be three to the right. It's just a measure of, how many hash marks is it from zero? Well, it's three hash marks from zero So is the absolute value of a greater than the absolute value of b? Or what's the absolute value of b? Well, b is one, two, three, four, five, six, seven, eight hash marks to the right of zero. And so the absolute value of b is going to be on the eighth hash mark. Because it's eight hash marks to the right. So this is also the absolute value of b. And this is consistent with what we've learned about absolute value. Absolute value of a positive number is just going to be that number again. Absolute value of a negative number is going to be the opposite of that number. And absolute value of zero is just going to be zero. So is the absolute value of a greater than the absolute value of b? Well, no. Absolute value of a is to the left of the absolute value of b on our number line. It is less than the absolute value of b. So this is not true. All right, next statement. \"Absolute value of a is less than the absolute value of c.\"" + }, + { + "Q": "at 9:25, why would there be a negative sign?", + "A": "he put a negative sign there because he was trying to subtract the two equations. to do that one of the equations had to be negative. so he multiplied the second equation by -1 to make that possible. so the equation went from:::: 3x+y= 1.79 to:::: -3x-y= -1/79 Hope This Helps You Undertsand", + "video_name": "vA-55wZtLeE", + "timestamps": [ + 565 + ], + "3min_transcript": "And 4 Fruit Roll-Ups. Plus 4 times y, the cost of a Fruit Roll-Up. This is how much Nadia spends. 3 candy bars, 4 Fruit Roll-Ups. And it's going to cost $2.84. That's what this first statement tells us. It translates into that equation. The second statement. Peter also buys 3 candy bars, but could only afford 1 additional Fruit Roll-Up. So plus 1 additional Fruit Roll-Up. His purchase cost is equal to $1.79. What is the cost of each candy bar and each Fruit Roll-Up? And we're going to solve this using elimination. You could solve this using any of the techniques we've seen so far-- substitution, elimination, even graphing, although it's kind of hard to eyeball things with the graphing. So how can we do this? Remember, with elimination, you're going to add-- let's Is there something we could add to both sides of this equation that'll help us eliminate one of the variables? Or let me put it this way, is there something we could add or subtract to both sides of this equation that will help us eliminate one of the variables? Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? And remember, by doing that, I would be subtracting the same thing from both sides of the equation. This is $1.79. Because it says this is equal to $1.79. So if we did that we would be subtracting the same thing from both sides of the equation. So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So let's subtract it. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. I'm just taking the second equation. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. And what do we get? When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. 3x minus 3x is 0x. I won't even write it down. You get 4x minus-- sorry, 4y minus y. That is 3y. And that is going to be equal to $2.84 minus $1.79." + }, + { + "Q": "At 9:04, Sal multiplies the bottom equation by -1. He could have divided by -1, right?", + "A": "Same thing, dividing or multiplying by -1 will give you same equation.", + "video_name": "vA-55wZtLeE", + "timestamps": [ + 544 + ], + "3min_transcript": "And 4 Fruit Roll-Ups. Plus 4 times y, the cost of a Fruit Roll-Up. This is how much Nadia spends. 3 candy bars, 4 Fruit Roll-Ups. And it's going to cost $2.84. That's what this first statement tells us. It translates into that equation. The second statement. Peter also buys 3 candy bars, but could only afford 1 additional Fruit Roll-Up. So plus 1 additional Fruit Roll-Up. His purchase cost is equal to $1.79. What is the cost of each candy bar and each Fruit Roll-Up? And we're going to solve this using elimination. You could solve this using any of the techniques we've seen so far-- substitution, elimination, even graphing, although it's kind of hard to eyeball things with the graphing. So how can we do this? Remember, with elimination, you're going to add-- let's Is there something we could add to both sides of this equation that'll help us eliminate one of the variables? Or let me put it this way, is there something we could add or subtract to both sides of this equation that will help us eliminate one of the variables? Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? And remember, by doing that, I would be subtracting the same thing from both sides of the equation. This is $1.79. Because it says this is equal to $1.79. So if we did that we would be subtracting the same thing from both sides of the equation. So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So let's subtract it. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. I'm just taking the second equation. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. And what do we get? When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. 3x minus 3x is 0x. I won't even write it down. You get 4x minus-- sorry, 4y minus y. That is 3y. And that is going to be equal to $2.84 minus $1.79." + }, + { + "Q": "I have trouble following Sal's explanation from ca. 4:00 to ca. 5:30. d/dx theta = sec^2 theta, but what's the logic behind d(theta)/dt? Did I miss a video demonstrating (or exercises testing how to) differentiate both sides w/ respect to \"something else\". Also, there is no detailed discussion of how/why it is permissible to cancel out terms (i.e. the d(theta)s in this example). Additional chain-rule exercises leading up to these problems would be appreciated.", + "A": "it is not d/dx(theta) it is d(tan theta)/dt = d(tan theta)/d(theta) * d(theta)/dt =sec^2 theta*d(theta)/dt that is how you get the final expression in terms of sec^2theta", + "video_name": "_kbd6troMgA", + "timestamps": [ + 240, + 330 + ], + "3min_transcript": "Well, we're trying to figure out the rate at which the height of the balloon is changing. So if you call this distance right over here h, what we want to figure out is dh dt. That's what we don't know. So what we'd want to come up with is a relationship between dh dt, d theta dt, and maybe theta, if we need it. Or another way to think about it, if we can come up with the relationship between h and theta, then we could take the derivative with respect to t, and we'll probably get a relationship between all of this stuff. So what's the relationship between theta and h? Well, it's a little bit of trigonometry. We know we're trying to figure out h. We already know what this length is right over here. We know opposite over adjacent. That's the definition of tangent. So let's write that down. is equal to the opposite side-- the opposite side is equal to h-- over the adjacent side, which we know is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between d theta dt, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of both sides of this with respect to t implicitly. So let's do that. And actually, let me move over this h over 500 a little bit. So let me move it over a little bit so I have space to show the derivative operator. So let's write it like that. And now let's take the derivative with respect to t. So d dt. I'm going to take the derivative with respect to t on the left. We're going to take the derivative with respect So what's the derivative with respect to t of tangent of theta? Well, we're just going to apply the chain rule here. It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to t, times d theta dt. Once again, this is just the derivative of the tangent, the tangent of something with respect to that something times the derivative of the something with respect to t. Derivative of tangent theta with respect to theta times the derivative theta with respect to t gives us the derivative of tangent of theta with respect to t, which is what we want when we use this type of a derivative operator. We're taking the derivative with respect to t, not just the derivative with respect to theta. So this is the left hand side." + }, + { + "Q": "At 5:20, you take the derivative of h/500. Aren't you supposed to use the quotient rule for that?", + "A": "You don t have a variable in the denominator. That 500 m is a constant. So you would treat it as ( \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0085\u00e2\u0082\u0080\u00e2\u0082\u0080 ) h", + "video_name": "_kbd6troMgA", + "timestamps": [ + 320 + ], + "3min_transcript": "is equal to the opposite side-- the opposite side is equal to h-- over the adjacent side, which we know is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between d theta dt, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of both sides of this with respect to t implicitly. So let's do that. And actually, let me move over this h over 500 a little bit. So let me move it over a little bit so I have space to show the derivative operator. So let's write it like that. And now let's take the derivative with respect to t. So d dt. I'm going to take the derivative with respect to t on the left. We're going to take the derivative with respect So what's the derivative with respect to t of tangent of theta? Well, we're just going to apply the chain rule here. It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to t, times d theta dt. Once again, this is just the derivative of the tangent, the tangent of something with respect to that something times the derivative of the something with respect to t. Derivative of tangent theta with respect to theta times the derivative theta with respect to t gives us the derivative of tangent of theta with respect to t, which is what we want when we use this type of a derivative operator. We're taking the derivative with respect to t, not just the derivative with respect to theta. So this is the left hand side. it's just going to be 1 over 500 dh dt. So 1 over 500 dh dt. We're literally saying it's just 1 over 500 times the derivative of h with respect to t. But now we have our relationship. We have the relationship that we actually care about. We have a relationship between the rate at which the height is changing with respect to time and the rate at which the angle is changing with respect to time and our angle at any moment. So we can just take these values up here, throw it in here, and then solve for the unknown. So let's do that. Let's do that right over here. So we get secant squared of theta. So we get secant squared. Right now our theta is pi over 4. Secant squared of pi over 4. Let me write those colors in to show you that I'm putting these values in. Secant squared of pi over 4. Secant times d theta dt." + }, + { + "Q": "At 4:48, when Sal was writing \"m=2, (-7,5)\" into an equation, why isn't it \"5-b = 2(-7-a)\"? why does Sal replace b with 5, not y with b?", + "A": "In Sal s format of point-slope, you swap out a for the x-value and b for the y-value. You will usually see these written as Xsub1 and Ysub1 where (Xsub1, Ysub1) is the ordered pair. Sal chose to use (a, b) for the ordered pair. The X and Y in the formula stay as X and Y. They become the variables in the equation.", + "video_name": "K_OI9LA54AA", + "timestamps": [ + 288 + ], + "3min_transcript": "And then on the right-hand side, you just have m times x minus a. So this whole thing has simplified to y minus b is equal to m times x minus a. And right here, this is a form that people, that mathematicians, have categorized as point-slope form. So this right over here is the point-slope form of the equation that describes this line. Now, why is it called point-slope form? Well, it's very easy to inspect this and say, OK. Well look, this is the slope of the line in green. That's the slope of the line. And I can put the two points in. If the point a, b is on this line, I'll have the slope times x minus a is equal to y minus b. Now, let's see why this is useful or why people like to use this type of thing. Let's not use just a, b and a slope of m anymore. Let's say that someone tells you that I'm dealing with some line where the slope is equal to 2, and let's say it goes through the point negative 7, 5. So very quickly, you could use this information and your knowledge of point-slope form to write this in this form. You would just say, well, an equation that contains this point and has this slope would be y minus b, which is 5-- y minus the y-coordinate of the point that this line contains-- is equal to my slope times x minus the x-coordinate that this line contains. So x minus negative 7. And just like that, we have written an equation that this point right over here. And if we don't like the x minus negative 7 right over here, we could obviously rewrite that as x plus 7. But this is kind of the purest point-slope form. If you want to simplify it a little bit, you could write it as y minus 5 is equal to 2 times x plus 7. And if you want to see that this is just one way of expressing the equation of this line-- there are many others, and the one that we're most familiar with is y-intercept form-- this can easily be converted to y-intercept form. To do that, we just have to distribute this 2. So we get y minus 5 is equal to 2 times x plus 2 times 7, so that's equal to 14. And then we can get rid of this negative 5 on the left by adding 5 to both sides of this equation. And then we are left with, on the left-hand side, y and, on the right-hand side, 2x plus 19. So this right over here is slope-intercept form. You have your slope and your y-intercept. So this is slope-intercept form." + }, + { + "Q": "I've heard slope-intercept form is y=mx+b. At 6:05, how would that fit in?", + "A": "If you distribute across the parenthesis, and then add/subtract the term on the y side (from both), you should get an answer in slope-intercept form.", + "video_name": "K_OI9LA54AA", + "timestamps": [ + 365 + ], + "3min_transcript": "Let's say that someone tells you that I'm dealing with some line where the slope is equal to 2, and let's say it goes through the point negative 7, 5. So very quickly, you could use this information and your knowledge of point-slope form to write this in this form. You would just say, well, an equation that contains this point and has this slope would be y minus b, which is 5-- y minus the y-coordinate of the point that this line contains-- is equal to my slope times x minus the x-coordinate that this line contains. So x minus negative 7. And just like that, we have written an equation that this point right over here. And if we don't like the x minus negative 7 right over here, we could obviously rewrite that as x plus 7. But this is kind of the purest point-slope form. If you want to simplify it a little bit, you could write it as y minus 5 is equal to 2 times x plus 7. And if you want to see that this is just one way of expressing the equation of this line-- there are many others, and the one that we're most familiar with is y-intercept form-- this can easily be converted to y-intercept form. To do that, we just have to distribute this 2. So we get y minus 5 is equal to 2 times x plus 2 times 7, so that's equal to 14. And then we can get rid of this negative 5 on the left by adding 5 to both sides of this equation. And then we are left with, on the left-hand side, y and, on the right-hand side, 2x plus 19. So this right over here is slope-intercept form. You have your slope and your y-intercept. So this is slope-intercept form." + }, + { + "Q": "At 1:00 in the video, what do the triangles mean when he is talking about what the slope equals?", + "A": "The triangles are supposed to mean The Change in , so it is The Change in Y/ The Change in X , also known as slope, or known as in Slope-intercept form, m.", + "video_name": "K_OI9LA54AA", + "timestamps": [ + 60 + ], + "3min_transcript": "So what I've drawn here in yellow is a line. And let's say we know two things about this line. We know that it has a slope of m, and we know that the point a, b is on this line. And so the question that we're going to try to answer is, can we easily come up with an equation for this line using this information? Well, let's try it out. So any point on this line, or any x, y on this line, would have to satisfy the condition that the slope between that point-- so let's say that this is some point x, y. It's an arbitrary point on the line-- the fact that it's on the line tells us that the slope between a, b and x, y must be equal to m. So let's use that knowledge to actually construct an equation. So what is the slope between a, b and x, y? Well, our change in y-- remember slope is just change in y over change in x. Let me write that. Slope is equal to change in y over change in x. This little triangle character, that's the Greek letter Delta, Our change in y-- well let's see. If we start at y is equal to b, and if we end up at y equals this arbitrary y right over here, this change in y right over here is going to be y minus b. Let me write it in those same colors. So this is going to be y minus my little orange b. And that's going to be over our change in x. And the exact same logic-- we start at x equals a. We finish at x equals this arbitrary x, whatever x we happen to be at. So that change in x is going to be that ending point minus our starting point-- minus a. And we know this is the slope between these two points. That's the slope between any two points on this line. And that's going to be equal to m. So this is going to be equal to m. And so what we've already done here is actually create an equation that describes this line. It might not be in any form that you're used to seeing, any x, y that satisfies this equation right over here will be on the line because any x, y that satisfies this, the slope between that x, y and this point right over here, between the point a, b, is going to be equal to m. So let's actually now convert this into forms that we might recognize more easily. So let me paste that. So to simplify this expression a little bit, or at least to get rid of the x minus a in the denominator, let's multiply both sides by x minus a. So if we multiply both sides by x minus a-- so x minus a on the left-hand side and x minus a on the right. Let me put some parentheses around it. So we're going to multiply both sides by x minus a. The whole point of that is you have x minus a divided by x" + }, + { + "Q": "At 1:20 or so why does he use 2 instead of 1 as the smallest number", + "A": "Starting at 2:03, he notices that he missed the 1 s, and fixes it.", + "video_name": "09Cx7xuIXig", + "timestamps": [ + 80 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:52, why did Sal add 14 and 9? Shouldn't he subtract?", + "A": "He distributed the negative sign for the second polynominal so -9 become +9. If you don t distribute then you would have to do this way, 14 - (-9) which equals to 14+9", + "video_name": "5ZdxnFspyP8", + "timestamps": [ + 112 + ], + "3min_transcript": "Simplify 16x plus 14 minus the entire expression 3x squared plus x minus 9. So when you subtract an entire expression, this is the exact same thing as having 16x plus 14. And then you're adding the opposite of this whole thing. Or you're adding negative 1 times 3x squared plus x minus 9. Or another way to think about it is you can distribute this negative sign along all of those terms. That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part-- I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x because that's positive 1x. Negative 1 times negative 9-- remember, That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second-degree term. We only have one of those. So let me write it over here-- negative 3x squared. And then what do we have in terms of first-degree terms, of just an x, x to the first power? Well, we have a 16x. And then from that, we're going to subtract an x, subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract 1 of them away, you're going to have 15 of that something. And then finally, you have 14. You could view that as 14 times x to the 0 or just 14. 14 plus 9-- they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23." + }, + { + "Q": "At 5:20, couldn't the cube root of a^2b^2 be simplified to a^2/3 b^5/3?", + "A": "Yes, you can use fractional notation as well, in fact I prefer fractional notation in these situations. Sal is just using a different approach. Both are correct. a^2/3 times b^2/3 is not more simplified than the cube root of a^2 times b^2, it is just different. I hope that helps!", + "video_name": "c-wtvEdEoVs", + "timestamps": [ + 320 + ], + "3min_transcript": "to the third times the cube root-- and I'll just group these two guys together just because we're not going to be able to simplify it any more-- times the cube root of a squared b squared. I'll keep the colors consistent while we're trying to figure out what's what. And I could have said that this is times the cube root of a squared times the cube root of b squared, but that won't simplify anything, so I'll just leave these like this. And so we can look at these individually. The cube root of 3 to the third, or the cube root of 27-- well, that's clearly just going to be-- I want to do that in that yellow color-- this is clearly just going to be 3. 3 to the third power is 3 to the third power, or it's equal to 27. This term right over here, the cube root of b to the third-- well, that's just b. And the cube root of c to the third, well, that is clearly-- I want to do So our whole expression has simplified to 3 times b times c times the cube root of a squared b squared. And we're done. And I just want to do one other thing, just because I did mention that I would do it. We could simplify it this way. Or we could recognize that this expression right over here can be written as 3bc to the third power. And if I take three things to the third power, and I'm multiplying it, that's the same thing as multiplying them first and then raising to the third power. It comes straight out of our exponent properties. And so we can rewrite this as the cube root of all of this times the cube root of a squared b And so the cube root of all of this, of 3bc to the third power, well, that's just going to be 3bc, and then multiplied by the cube root of a squared b squared. I didn't take the trouble to color-code it this time, because we already figured out one way to solve it. But hopefully, that also makes sense. We could have done this either way. But the important thing is that we get that same answer." + }, + { + "Q": "I dont understand 1:20, why dont you make a^2 a*a and you make b^5, b^2 and b^3", + "A": "Because he explained that b^5 is not a perfect cube so he had to divide it, thats my explanation to it.", + "video_name": "c-wtvEdEoVs", + "timestamps": [ + 80 + ], + "3min_transcript": "We're asked to simplify the cube root of 27a squared times b to the fifth times c to the third power. And the goal, whenever you try to just simplify a cube root like this, is we want to look at the parts of this expression over here that are perfect cubes, that are something raised to the third power. Then we can take just the cube root of those, essentially taking them out of the radical sign, and then leaving everything else that is not a perfect cube inside of it. So let's see what we can do. So first of all, 27-- you may or may not already recognize this as a perfect cube. If you don't already recognize it, you can actually do a prime factorization and see it's a perfect cube. 27 is 3 times 9, and 9 is 3 times 3. So 27-- its prime factorization is 3 times 3 times 3. So it's the exact same thing as 3 to the third power. So let's rewrite this whole expression down here. But let's write it in terms of things that are perfect cubes and things that aren't. So 27 can be just rewritten as 3 to the third power. a to the third would have been. So we're just going to write this-- let me write it over here. We can switch the order here because we just have a bunch of things being multiplied by each other. So I'll write the a squared over here. b to the fifth is not a perfect cube by itself, but it can be expressed as the product of a perfect cube and another thing. b to the fifth is the exact same thing as b to the third power times b to the second power. If you want to see that explicitly, b to the fifth is b times b times b times b times b. So the first three are clearly b to the third power. And then you have b to the second power after it. So we can rewrite b to the fifth as the product of a perfect cube. So I'll write b to the third-- let me do that in that same purple color. So we have b to the third power over here. And then it's b to the third times b squared. So I'll write the b squared over here. And then finally, we have-- I'll do in blue-- c to the third power. Clearly, this is a perfect cube. It is c cubed. It is c to the third power. So I'll put it over here. So this is c to the third power. And of course, we still have that overarching radical sign. So we're still trying to take the cube root of all of this. And we know from our exponent properties, or we could say from our radical properties, that this is the exact same thing. That taking the cube root of all of these things is the same as taking the cube root of these individual factors and then multiplying them. So this is the same thing as the cube root-- and I could separate them out individually. Or I could say the cube root of 3 to the third b to the third c to the third. Actually, let's do it both ways. So I'll take them out separately. So this is the same thing as the cube root of 3 to the third times the cube root-- I'll write them all in. Let me color-code it so we don't get confused-- times the cube" + }, + { + "Q": "At 1:02 Do we have to do anything else to the numerators and the denominaters other than divide?", + "A": "No, you are only just meant to divide at that point.", + "video_name": "2dbasvm3iG0", + "timestamps": [ + 62 + ], + "3min_transcript": "Use less than, greater than, or equal to compare the two fractions 21/28, or 21 over 28, and 6/9, or 6 over 9. So there's a bunch of ways to do this. The easiest way is if they had the same denominator, you could just compare the numerators. Unlucky for us, we do not have the same denominator. So what we could do is we can find a common denominator for both of them and convert both of these fractions to have the same denominator and then compare the numerators. Or even more simply, we could simplify them first and then So let me do that last one, because I have a feeling that'll be the fastest way to do it. So 21/28-- you can see that they are both divisible by 7. So let's divide both the numerator and the denominator by 7. So we could divide 21 by 7. And we can divide-- so let me make the numerator-- and we can divide the denominator by 7. We're doing the same thing to the numerator and the denominator, so we're not going to change the value of the fraction. So 21 divided by 7 is 3, and 28 divided by 7 is 4. 3/4 is the simplified version of it. Let's do the same thing for 6/9. 6 and 9 are both divisible by 3. So let's divide them both by 3 so we can simplify this fraction. So let's divide both of them by 3. 6 divided by 3 is 2, and 9 divided by 3 is 3. So 21/28 is 3/4. They're the exact same fraction, just written a different way. This is the more simplified version. And 6/9 is the exact same fraction as 2/3. So we really can compare 3/4 and 2/3. So this is really comparing 3/4 and 2/3. And the real benefit of doing this is now this is much easier to find a common denominator for than 28 and 9. Then we would have to multiply big numbers. Here we could do fairly small numbers. The common denominator of 3/4 and 2/3 And 4 and 3 don't share any prime factors with each other. So their least common multiple is really just going to be the product of the two. So we can write 3/4 as something over 12. And we can write 2/3 as something over 12. And I got the 12 by multiplying 3 times 4. They have no common factors. Another way you could think about it is 4, if you do a prime factorization, is 2 times 2. And 3-- it's already a prime number, so you can't prime factorize it any more. So what you want to do is think of a number that has all of the prime factors of 4 and 3. So it needs one 2, another 2, and a 3. Well, 2 times 2 times 3 is 12. And either way you think about it, that's how you would get the least common multiple or the common denominator for 4 and 3. Well, to get from 4 to 12, you've got to multiply by 3." + }, + { + "Q": "At 1:55, Sal says \"their greatest common factor is 3\" what does greatest common factor even mean?", + "A": "The Greatest Common Factor means when you take two numbers and find a factor that both these numbers have in common. For example: What is the greatest common factor of 6 and 9? Well, the first step is to list their factors: The factors for 6 are: 1, 2, 3, and 6. The factors for 9 are: 1, 3, 3, 9. Now what number is the same in both these factor lists? 3! So the Greatest Common Factor will be 3.", + "video_name": "Bt60JVZRVCI", + "timestamps": [ + 115 + ], + "3min_transcript": "Let's think about what fraction of this grid is actually shaded in pink. So the first thing we want to think about is how many equal sections do we have here? Well, this is a 1, 2, 3, 4, 5 by 1, 2, 3 grid. So there's 15 sections here. You could also count it-- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. So there are 15 equal sections here. And how many of those equal sections are actually shaded in this kind of pinkish color? Well, We have 1, 2, 3, 4, 5, 6. So it's 6/15 is shaded in. But I want to simplify this more. I have a feeling that there's some equivalent fractions that represent the exact same thing as 6/15. And to get a sense of that, let me redraw this a little bit, where I still shade in six of these rectangles, but I'll shade them a little bit in one chunk. So let me throw in another grid right over here, and let me attempt to shade in the rectangles So that is 1-- 1 rectangle. I'll even make my thing even bigger. All right, 1 rectangle, 2 rectangles, 3 rectangles-- halfway there-- 4 rectangles, 5 rectangles shaded in and now 6 rectangles shaded in. So this right over here, what I just did, this is still 6 rectangles of the 15 rectangles So this is still 6/15. These are representing the same thing. But how can I simplify this even more? Well, when you look at it numerically, you see that both 6 and 15 are divisible by 3. In fact, their greatest common factor is 3. So what happens if we divide the numerator and denominator by 3? we're not going to be changing the value of the fraction. So let's divide the numerator by 3 and divide the denominator by 3. And what do we get? We get 2 over 5. Now how does this make sense in the context of this diagram right here? Well, we started off with 6 shaded in. You divide by 3, you have 2 shaded in. So you're essentially saying, hey, let's group these into sections of 3. So let's say that this right over here is one section of 3. This is one section of 3 right over here. So that's one section of 3. And then this is another section of 3 right over here. And so you have two sections of 3. And actually let me color it in a little bit better. So you have two sections of 3. And if you were to combine them, it looks just like this." + }, + { + "Q": "I'm confused at 2:10. Sal says that the sequence converges but I learned that when we calculate limits, if you have infinity over infinity, its called a \"indetermination\". Though, what Sal said makes perfectly sense...", + "A": "Recall when we want to find the horizontal asymptote, we take the limit as x approaches infinity. The method was to multiply the numerator and denominator by 1 over the highest term of x on the denominator. Similarly, we find the limit as n approaches infinity to find out if it diverges or converges. So if we multiply by 1/n\u00c2\u00b2 for both the top and bottom, it will converge to 1.", + "video_name": "muqyereWEh4", + "timestamps": [ + 130 + ], + "3min_transcript": "So we've explicitly defined four different sequences here. And what I want you to think about is whether these sequences converge or diverge. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. And diverge means that it's not approaching some value. So let's look at this. And I encourage you to pause this video and try this on your own before I'm about to explain it. So let's look at this first sequence right over here. So the numerator n plus 8 times n plus 1, the denominator n times n minus 10. So one way to think about what's happening as n gets larger and larger is look at the degree of the numerator and the degree of the denominator. And we care about the degree because we want to see, look, is the numerator growing faster than the denominator? In which case this thing is going to go to infinity and this thing's going to diverge. Or is maybe the denominator growing faster, in which case this might converge to 0? Or maybe they're growing at the same level, and maybe it'll converge to a different number. So let's multiply out the numerator and the denominator So n times n is n squared. n times 1 is 1n, plus 8n is 9n. And then 8 times 1 is 8. So the numerator is n squared plus 9n plus 8. The denominator is n squared minus 10n. And one way to think about it is n gets really, really, really, really, really large, what dominates in the numerator-- this term is going to represent most of the value. And this term is going to represent most of the value, as well. These other terms aren't going to grow. Obviously, this 8 doesn't grow at all. But the n terms aren't going to grow anywhere near as fast as the n squared terms, especially for large n's. So for very, very large n's, this is really going to be approaching n squared over n squared, or 1. So it's reasonable to say that this converges. So this one converges. And once again, I'm not vigorously proving it here. Or I should say I'm not rigorously proving it over here. in the numerator and the denominator. So now let's look at this one right over here. So here in the numerator I have e to the n power. And here I have e times n. So this grows much faster. I mean, this is e to the n power. Imagine if when you have this as 100, e to the 100th power is a ginormous number. e times 100-- that's just 100e. Grows much faster than this right over here. So this thing is just going to balloon. This is going to go to infinity. So we could say this diverges. Now let's look at this one right over here. Well, we have a higher degree term. We have a higher degree in the numerator than we have in the denominator. n squared, obviously, is going to grow much faster than n. So for the same reason as the b sub n sequence, this thing is going to diverge. The numerator is going to grow much faster" + }, + { + "Q": "Sal says at 2:00 he isn't 'rigorously proving' his answer, I'm curious, how would one go about rigorously proving the result?", + "A": "One way is to divide the numerator and denominator both by n^2, then take the limit of the result as n approaches infinity. If you haven t seen that process yet, you will soon!", + "video_name": "muqyereWEh4", + "timestamps": [ + 120 + ], + "3min_transcript": "So we've explicitly defined four different sequences here. And what I want you to think about is whether these sequences converge or diverge. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. And diverge means that it's not approaching some value. So let's look at this. And I encourage you to pause this video and try this on your own before I'm about to explain it. So let's look at this first sequence right over here. So the numerator n plus 8 times n plus 1, the denominator n times n minus 10. So one way to think about what's happening as n gets larger and larger is look at the degree of the numerator and the degree of the denominator. And we care about the degree because we want to see, look, is the numerator growing faster than the denominator? In which case this thing is going to go to infinity and this thing's going to diverge. Or is maybe the denominator growing faster, in which case this might converge to 0? Or maybe they're growing at the same level, and maybe it'll converge to a different number. So let's multiply out the numerator and the denominator So n times n is n squared. n times 1 is 1n, plus 8n is 9n. And then 8 times 1 is 8. So the numerator is n squared plus 9n plus 8. The denominator is n squared minus 10n. And one way to think about it is n gets really, really, really, really, really large, what dominates in the numerator-- this term is going to represent most of the value. And this term is going to represent most of the value, as well. These other terms aren't going to grow. Obviously, this 8 doesn't grow at all. But the n terms aren't going to grow anywhere near as fast as the n squared terms, especially for large n's. So for very, very large n's, this is really going to be approaching n squared over n squared, or 1. So it's reasonable to say that this converges. So this one converges. And once again, I'm not vigorously proving it here. Or I should say I'm not rigorously proving it over here. in the numerator and the denominator. So now let's look at this one right over here. So here in the numerator I have e to the n power. And here I have e times n. So this grows much faster. I mean, this is e to the n power. Imagine if when you have this as 100, e to the 100th power is a ginormous number. e times 100-- that's just 100e. Grows much faster than this right over here. So this thing is just going to balloon. This is going to go to infinity. So we could say this diverges. Now let's look at this one right over here. Well, we have a higher degree term. We have a higher degree in the numerator than we have in the denominator. n squared, obviously, is going to grow much faster than n. So for the same reason as the b sub n sequence, this thing is going to diverge. The numerator is going to grow much faster" + }, + { + "Q": "At 2:18 Sal says we should subtract 114 from both the sides.But Instead of that we could transpose it.It is quite easier by transposing", + "A": "Don t ask me why he didn t. It would be quicker.", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 138 + ], + "3min_transcript": "What I want to do now is just a series of problems that really make sure that we know what we're doing with parallel lines and triangles and all the rest. And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here-- so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it just knowing what you know about the sums of the measures of the angles inside of a triangle, and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself because I'm about to give you the solution. So the first thing you might say-- and this is a general way to think about a lot of these problems where they give you some angles and you have to figure out some other angles based on the sum of angles and a triangle equaling 180, or this one doesn't have parallel lines on it. and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114." + }, + { + "Q": "At 3:40 when sal says that to get the last angle in a triangle you have to do (180- a- b) but shouldn't it rather be 180 -(a+b)? Pls clarify my doubt\u00f0\u009f\u0099\u008f\u00f0\u009f\u008f\u00bc", + "A": "Either way works, since the first way you would use GEMDAS (Grouping Symbols, Exponets, Multiplication/Division, and then Addition/Subtraction) and thus subtract a from 180 and then b from whatever is left. You could also see 180 -(a+b) as 180+ -1(a+b)", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 220 + ], + "3min_transcript": "So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114. Notice, this 114 was the exact same sum of these 2 angles over here. And that's actually a general idea, and I'll do it on the side here just to prove it to you. If I have, let's say that these 2 angles-- let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle. So in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You could subtract 180 from both sides. You could add a plus b to both sides. Running out of space on the right hand side. And then you're left with-- these cancel out. On the left hand side, you're left with y. On the right hand side is equal to a plus b. So this is just a general property. You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees, and then you have a supplementary angles Or you could just say, look, if I have the exterior angles right over here, it's equal to the sum of the remote interior angles. That's just a little terminology you could see there. So y is equal to a plus b. 114 degrees, we've already shown to ourselves, is equal to 64 plus 50 degrees. But anyway, regardless of how we do it, if we just reason it out step by step or if we just knew this property from the get go, if we know that y is equal to 114 degrees-- and I like to reason it out every time just to make sure I'm not jumping to conclusions. So if y is 114 degrees, now we know this angle." + }, + { + "Q": "At 00:34, couldn't I just do 64+31+50+x=180?", + "A": "You sure can! There are many ways to figure out each of the angles.", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 34 + ], + "3min_transcript": "What I want to do now is just a series of problems that really make sure that we know what we're doing with parallel lines and triangles and all the rest. And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here-- so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it just knowing what you know about the sums of the measures of the angles inside of a triangle, and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself because I'm about to give you the solution. So the first thing you might say-- and this is a general way to think about a lot of these problems where they give you some angles and you have to figure out some other angles based on the sum of angles and a triangle equaling 180, or this one doesn't have parallel lines on it. and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114." + }, + { + "Q": "Why do we use a question mark in the video at 0:55 seconds in the video?", + "A": "Th question mark represents the measure in degrees of the angle that the question is asking about. At 5:05 he labels the question mark z .", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 55 + ], + "3min_transcript": "What I want to do now is just a series of problems that really make sure that we know what we're doing with parallel lines and triangles and all the rest. And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here-- so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it just knowing what you know about the sums of the measures of the angles inside of a triangle, and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself because I'm about to give you the solution. So the first thing you might say-- and this is a general way to think about a lot of these problems where they give you some angles and you have to figure out some other angles based on the sum of angles and a triangle equaling 180, or this one doesn't have parallel lines on it. and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114." + }, + { + "Q": "At 1:04 you drew a =D, which looks like a smiley face. What is it?", + "A": "It s just a sloppy arrow.", + "video_name": "T4JKO0OGjpQ", + "timestamps": [ + 64 + ], + "3min_transcript": "Let's write 113.9% as a decimal. So percent, this symbol right over here, literally means per hundred. So this is the same thing as 113.9 per 100, which is the same thing as 113.9 divided by 100. So essentially, if we want to write this as a decimal, we just take a 113.9, and we divide it by 100. And to do that, 113.9 divided by 100. If you divide by 10, you move the decimal once to the left. If you divide by 100, you're moving the decimal twice If we're doing it by 1,000, we would go to the left once more. 10,000, each time you divide by 10, you'd go one step to the left, which hopefully makes sense. If we were to multiply by 10, we would be moving the decimal over to the right, and there are other videos that go into far more depth on the intuition behind that. But anyway, we're dividing 113.9 by 100, so move the decimal two spaces to the left. So the decimal will end up right over here. And we are done." + }, + { + "Q": "At 3:00 they wrote two to the sixty third power. Couldn't you also write it as 2^63?", + "A": "Yes, the two symbols mean the same thing", + "video_name": "UCCNoXqCGZQ", + "timestamps": [ + 180 + ], + "3min_transcript": "8, 2's multiplied together. 9, 2's. 10, 11, 12, 13. So all of this stuff multiplied together. 8,192 grains of rice is what we should see right over here. Voiceover:And you know, I had fun last night and I was up late, but there you go. Voiceover:Did you really count out 8,192 grains of rice? Voiceover:More or less. Voiceover:Okay. Let's just say you did. Voiceover:What if we just went, you know, 4 steps ahead. How much rice would be here? Voiceover:4 steps ahead, so we're going to multiple by 2, then multiple by 2 again, then multiply by 2 again, the multiply by 2 again. So it's this number times ... Let's see, 2 times 2 is 4. Times 2 is 8, times 2 is 16. So it's going to get us like 120, like 130,000 or around there. Voiceover:131,672. Voiceover:You had a lot of time last night. We're not even halfway across the board yet. Voiceover:We're not. Voiceover:This is a lot of ... You could throw a party. Voiceover:What about the last square? This is 63 steps. Voiceover:We're going to take 2 times 2 and we're going to do 63 of those. So this is going to be a huge number. And actually, it would be neat if there was a notation for that. Voiceover:I didn't count this one out but it is the size of Mount Everest, the pile of rice. And it would feed 485 trillion people. I mean, you know, this was a little bit of a pain for me to write all of these 2's. Voiceover:So was this. Voiceover:If I were the mathematical community I would want some type of notation. Voiceover:You kind of got on it here. I like this dot, dot, dot and the 63. This I understand this. Voiceover:Yeah, you could understand this but this is still a little bit ... This is a little bit too much. What if, instead, we just wrote ... Voiceover:Mathematicians love being efficient, right? They're lazy. Voiceover:Yeah, they have things to do. They have to go home and count grains of rice. (laughter) Voiceover:Yeah. So that is, take 63, 2's and multiply them Voiceover:This is the first square on our board. We have 1 grain of rice. And when we double it we have 2 grains of rice. Voiceover:Yup. Voiceover:And we double it again we have 4. I'm thinking this is similar to what we were doing, it's just represented differently. Voiceover:Yeah, well, I mean, this one, the one you were making, right, every time you were kind of adding these popsicle sticks, you're kind of branching out. 1 popsicle stick now becomes 2 popsicles sticks. Then you keep doing that. 1 popsicle stick becomes 2 but now you have 2 of them. So here you have 1, now you have 1 times 2. Now each of these 2 branch into 2, so now you have 2 times 2, or you have 4 popsicle sticks. Every stage, every branch, you're multiplying by 2 again. Voiceover:I basically just continue splitting just like a tree does. Voiceover:Yup. Voiceover:Now I can really see what 2 to the power of 3 looks like. Voiceover:And that's what we have here. 1 times 2 times 2 times 2, which is 8. This is 2 to the third power." + }, + { + "Q": "At 1:24 he said line uv and he wrote,it but when he wrote it couldn't he have done another letter because uv looked like w which would get heeps confusing right", + "A": "Most likely it was due that each line contained 2 points. A third letter was not needed. Furthermore, he made up a random problem with letters on each line in ABC order. Hope that sorts out the confusion. :)", + "video_name": "aq_XL6FrmGs", + "timestamps": [ + 84 + ], + "3min_transcript": "Identify all sets of parallel and perpendicular lines in the image below. So let's start with the parallel lines. And just as a reminder, two lines are parallel if they're in the same plane, and all of these lines are clearly in the same plane. They're in the plane of the screen you're viewing right now. But they are two lines that are in the same plane that never intersect. And one way to verify, because you can sometimes-- it looks like two lines won't intersect, but you can't just always assume based on how it looks. You really have to have some information given in the diagram or the problem that tells you that they are definitely parallel, that they're definitely never going to intersect. And one of those pieces of information which they give right over here is that they show that line ST and line UV, they both intersect line CD at the exact same angle, at this angle right here. And in particular, it's at a right angle. And if you have two lines that intersect a third line at the same angle-- so these are actually called corresponding angles and they're the same-- then these two lines are parallel. So line ST is parallel to line UV. And we can write it like this. Line ST, we put the arrows on each end of that top bar to say that this is a line, not just a line segment. Line ST is parallel to line UV. And I think that's the only set of parallel lines in this diagram. Yep. Now let's think about perpendicular lines. Perpendicular lines are lines that intersect at a 90-degree angle. So, for example, line ST is perpendicular to line CD. So line ST is perpendicular to line CD. And we know that they intersect at a right angle or at a 90-degree angle because they gave us this little box here which literally means that the measure of this angle is 90 degrees. Let me make sure I specified these as lines. Line UV is perpendicular to CD. So I did UV, ST, they're perpendicular to CD. And then after that, the only other information where they definitely tell us that two lines are intersecting at right angles are line AB and WX. So AB is definitely perpendicular to WX, line WX. And I think we are done. And one thing to think about, AB and CD, well, they don't even intersect in this diagram. So you can't make any comment about perpendicular, but they're definitely not parallel. You could even imagine that it looks like they're about to intersect. And they give us no information that they intersect the same lines at the same angle. So if somehow they told us that this is a right angle, even" + }, + { + "Q": "At 4:10 why y-4=2x is equals to y/2-2=x? Sorry for my question. Hope somebody could help me with this thing.", + "A": "Sal just divided the entire equation by 2: y/2 - 4/2 =2x/2 Reduce each term and you get: y/2 - 2 = x Hope this helps.", + "video_name": "W84lObmOp8M", + "timestamps": [ + 250 + ], + "3min_transcript": "This right here, that is equal to f of 2. Same idea. You start with 3, 3 is being mapped by the function to 10. It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2, or is there a way to go back from the 10 to the 3? Or is there some other function? Is there some other function, we can call that the inverse of f, that'll take us back? Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as notation, and it'll take us back from 10 to 3. Is there a way to do that? Will that same inverse of f, will it take us back from-- if we apply 8 to it-- will that take us back to 2? What you'll find is it's actually very easy to solve for this inverse of f, and I think once we solve for it, it'll make it clear what I'm talking about. That the function takes you from 2 to 8, the inverse will take us back from 8 to 2. So to think about that, let's just define-- let's just say y is equal to f of x. So y is equal to f of x, is equal to 2x plus 4. So I can write just y is equal to 2x plus 4, and this once again, this is our function. You give me an x, it'll give me a y. But we want to go the other way around. We want to give you a y and get an x. So all we have to do is solve for x in terms of y. So let's do that. If we subtract 4 from both sides of this equation-- let me switch colors-- if we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x, and then if we minus 2-- 4 divided by 2 is 2-- is equal to x. Or if we just want to write it that way, we can just swap the sides, we get x is equal to 1/2y-- same thing as y over 2-- minus 2. So what we have here is a function of y that gives us an x, which is exactly what we wanted. We want a function of these values that map back to an x. So we can call this-- we could say that this is equal to-- I'll do it in the same color-- this is equal to f inverse as a function of y. Or let me just write it a little bit cleaner. We could say f inverse as a function of y-- so we can have 10 or 8-- so now the range is now the domain for f inverse. f inverse as a function of y is equal to 1/2y minus 2. So all we did is we started with our original function, y is equal to 2x plus 4, we solved for-- over here, we've" + }, + { + "Q": "At 4:23, isn't it suppose to be (y-4)/2, not y/2-2?", + "A": "(y-4)/2, and y/2 - 2 are equivalent. So either version will work.", + "video_name": "W84lObmOp8M", + "timestamps": [ + 263 + ], + "3min_transcript": "This right here, that is equal to f of 2. Same idea. You start with 3, 3 is being mapped by the function to 10. It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2, or is there a way to go back from the 10 to the 3? Or is there some other function? Is there some other function, we can call that the inverse of f, that'll take us back? Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as notation, and it'll take us back from 10 to 3. Is there a way to do that? Will that same inverse of f, will it take us back from-- if we apply 8 to it-- will that take us back to 2? What you'll find is it's actually very easy to solve for this inverse of f, and I think once we solve for it, it'll make it clear what I'm talking about. That the function takes you from 2 to 8, the inverse will take us back from 8 to 2. So to think about that, let's just define-- let's just say y is equal to f of x. So y is equal to f of x, is equal to 2x plus 4. So I can write just y is equal to 2x plus 4, and this once again, this is our function. You give me an x, it'll give me a y. But we want to go the other way around. We want to give you a y and get an x. So all we have to do is solve for x in terms of y. So let's do that. If we subtract 4 from both sides of this equation-- let me switch colors-- if we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x, and then if we minus 2-- 4 divided by 2 is 2-- is equal to x. Or if we just want to write it that way, we can just swap the sides, we get x is equal to 1/2y-- same thing as y over 2-- minus 2. So what we have here is a function of y that gives us an x, which is exactly what we wanted. We want a function of these values that map back to an x. So we can call this-- we could say that this is equal to-- I'll do it in the same color-- this is equal to f inverse as a function of y. Or let me just write it a little bit cleaner. We could say f inverse as a function of y-- so we can have 10 or 8-- so now the range is now the domain for f inverse. f inverse as a function of y is equal to 1/2y minus 2. So all we did is we started with our original function, y is equal to 2x plus 4, we solved for-- over here, we've" + }, + { + "Q": "At around 4:10, when Sla carries the 2 from 2x he divides. I understand that but why is it that he does (y/2) -2 instead of (y-4)/2? do either work or is the way Sal does it the only correct way and why?", + "A": "Since your fraction is already in simplest terms, it would also be an acceptable answer.", + "video_name": "W84lObmOp8M", + "timestamps": [ + 250 + ], + "3min_transcript": "This right here, that is equal to f of 2. Same idea. You start with 3, 3 is being mapped by the function to 10. It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2, or is there a way to go back from the 10 to the 3? Or is there some other function? Is there some other function, we can call that the inverse of f, that'll take us back? Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as notation, and it'll take us back from 10 to 3. Is there a way to do that? Will that same inverse of f, will it take us back from-- if we apply 8 to it-- will that take us back to 2? What you'll find is it's actually very easy to solve for this inverse of f, and I think once we solve for it, it'll make it clear what I'm talking about. That the function takes you from 2 to 8, the inverse will take us back from 8 to 2. So to think about that, let's just define-- let's just say y is equal to f of x. So y is equal to f of x, is equal to 2x plus 4. So I can write just y is equal to 2x plus 4, and this once again, this is our function. You give me an x, it'll give me a y. But we want to go the other way around. We want to give you a y and get an x. So all we have to do is solve for x in terms of y. So let's do that. If we subtract 4 from both sides of this equation-- let me switch colors-- if we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x, and then if we minus 2-- 4 divided by 2 is 2-- is equal to x. Or if we just want to write it that way, we can just swap the sides, we get x is equal to 1/2y-- same thing as y over 2-- minus 2. So what we have here is a function of y that gives us an x, which is exactly what we wanted. We want a function of these values that map back to an x. So we can call this-- we could say that this is equal to-- I'll do it in the same color-- this is equal to f inverse as a function of y. Or let me just write it a little bit cleaner. We could say f inverse as a function of y-- so we can have 10 or 8-- so now the range is now the domain for f inverse. f inverse as a function of y is equal to 1/2y minus 2. So all we did is we started with our original function, y is equal to 2x plus 4, we solved for-- over here, we've" + }, + { + "Q": "In 3:55, why couldn't you do the same method for the first example?", + "A": "Sal could have, he just skipped that step. We can apply the method like this: 2 is 2 * 1. 4x is 2 * 2x. We have a common factor here - 2. So we pull it out to get: 2 ( 1 + 2x).", + "video_name": "I6TBBzIvgB8", + "timestamps": [ + 235 + ], + "3min_transcript": "is equal to four X, so plus four X. So in our algebra brains, this will often be reviewed as or referred to as this expression factored or in a factored form. Sometimes people would say that we have factored out the two. You could just as easily say that you have factored out a one plus two X. You have broken this thing up into two of its factors. So let's do a couple of examples of this and then we'll think about, you know, I just told you that we could write it this way but how do you actually figure that out? So let's do another one. Let's say that you had, I don't know, let's say you had, six, let me just in a different color, let's say you had six X six X plus three, six X plus 30, that's interesting. So one way to think about it is can we break up each of these terms so that they have a common factor? Well, this one over here, six X literally represents six times X, and then 30, if I want to break out a six, 30 is divisible by six, so I could write this as six times five, 30 is the same thing as six times five. And when you write it this way, you see, \"Hey, I can factor out a six!\" Essentially, this is the reverse of the distributive property! So I'm essentially undoing the distributive property, taking out the six, and you are going to end up with, so if you take out the six, you end up with six times, so if you take out the six here, you have an X, and you take out the six here, you have plus five. So six X plus 30, if you factor it, we could write it as six times X plus five. And you can verify with the distributive property. six X + five times six or six X + 30. more interesting where we might want to factor out a fraction. So let's say we had the situation ... Let me get a new color here. So let's say we had 1/2 minus 3/2, minus 3/2 X. How could we write this in a, I guess you could say, in a factored form, or if we wanted to factor out something? I encourage you to pause the video and try to figure it out, and I'll give you a hint. See if you can factor out 1/2. Let's write it that way. If we're trying to factor out 1/2, we can write this first term as 1/2 times one and this second one we could write as minus 1/2 times three X. That's what this is, 3/2 X is the same thing as three X divided by two or 1/2 times three X. And then here we can see that we can just factor" + }, + { + "Q": "So with the 6x + 30 I paused the video and did it. His answer was 6( x + 5) I came up with the answer 2( 3x + 15). Math can be done many ways so I was thinking can both our answers be right. Or does this specific thing have to be done a certain way? This was at 4:00 minute", + "A": "Sal s answer is more simplified because you can factor the 3 out of your answer. Usually math problems ask for the most simplified answer.", + "video_name": "I6TBBzIvgB8", + "timestamps": [ + 240 + ], + "3min_transcript": "is equal to four X, so plus four X. So in our algebra brains, this will often be reviewed as or referred to as this expression factored or in a factored form. Sometimes people would say that we have factored out the two. You could just as easily say that you have factored out a one plus two X. You have broken this thing up into two of its factors. So let's do a couple of examples of this and then we'll think about, you know, I just told you that we could write it this way but how do you actually figure that out? So let's do another one. Let's say that you had, I don't know, let's say you had, six, let me just in a different color, let's say you had six X six X plus three, six X plus 30, that's interesting. So one way to think about it is can we break up each of these terms so that they have a common factor? Well, this one over here, six X literally represents six times X, and then 30, if I want to break out a six, 30 is divisible by six, so I could write this as six times five, 30 is the same thing as six times five. And when you write it this way, you see, \"Hey, I can factor out a six!\" Essentially, this is the reverse of the distributive property! So I'm essentially undoing the distributive property, taking out the six, and you are going to end up with, so if you take out the six, you end up with six times, so if you take out the six here, you have an X, and you take out the six here, you have plus five. So six X plus 30, if you factor it, we could write it as six times X plus five. And you can verify with the distributive property. six X + five times six or six X + 30. more interesting where we might want to factor out a fraction. So let's say we had the situation ... Let me get a new color here. So let's say we had 1/2 minus 3/2, minus 3/2 X. How could we write this in a, I guess you could say, in a factored form, or if we wanted to factor out something? I encourage you to pause the video and try to figure it out, and I'll give you a hint. See if you can factor out 1/2. Let's write it that way. If we're trying to factor out 1/2, we can write this first term as 1/2 times one and this second one we could write as minus 1/2 times three X. That's what this is, 3/2 X is the same thing as three X divided by two or 1/2 times three X. And then here we can see that we can just factor" + }, + { + "Q": "At 1:03, for the prime factors of 12, instead of saying 2 x 2 x 3, could you instead say\n2^2 x 3? Would it still be okay?", + "A": "Yes that is another way of writing it, just simplified. Though I am not so sure if the question will allow it. If the question says about anything to simplify, then simplify, if not then don t. ( But in real life, YOU SHOOULD ALWAYS for CLASS unless not allowed for some reason)", + "video_name": "I6TBBzIvgB8", + "timestamps": [ + 63 + ], + "3min_transcript": "- In earlier mathematics that you may have done, you probably got familiar with the idea of a factor. So for example, let me just pick an arbitrary number, the number 12. We could say that the number 12 is the product of say two and six; two times six is equal to 12. So because if you take the product of two and six, you get 12, we could say that two is a factor of 12, we could also say that six is a factor of 12. You take the product of these things and you get 12! You could even say that this is 12 in factored form. People don't really talk that way but you could think of it that way. We broke 12 into the things that we could use to multiply. And you probably remember from earlier mathematics the notion of prime factorization, where you break it up into all of the prime factors. So in that case you could break the six into a two and a three, and you have two times two times three is equal to 12. And you'd say, \"Well, this would be 12 \"in prime factored form or the prime factorization of 12,\" so these are the prime factors. is things that you can multiply together to get your original thing. Or if you're talking about factored form, you're essentially taking the number and you're breaking it up into the things that when you multiply them together, you get your original number. What we're going to do now is extend this idea into the algebraic domain. So if we start with an expression, let's say the expression is two plus four X, can we break this up into the product of two either numbers or two expressions or the product of a number and an expression? Well, one thing that might jump out at you is we can write this as two times one plus two X. And you can verify if you like that this does indeed equal two plus four X. We're just going to distribute the two. is equal to four X, so plus four X. So in our algebra brains, this will often be reviewed as or referred to as this expression factored or in a factored form. Sometimes people would say that we have factored out the two. You could just as easily say that you have factored out a one plus two X. You have broken this thing up into two of its factors. So let's do a couple of examples of this and then we'll think about, you know, I just told you that we could write it this way but how do you actually figure that out? So let's do another one. Let's say that you had, I don't know, let's say you had, six, let me just in a different color, let's say you had six X six X plus three," + }, + { + "Q": "At 1:55 How did Sal get 2(1+2x) instead of 2(1+3x) or 2(2+2x)? ;)", + "A": "You might have wanted to add the terms inside the parentheses but remember that you cannot add 1+3x or 2+2x ...etc because they are not like terms. you multiply them by their common factor separately. thats why the 2(1+2x). try to multiply the 2 by both terms and see what you get.", + "video_name": "I6TBBzIvgB8", + "timestamps": [ + 115 + ], + "3min_transcript": "- In earlier mathematics that you may have done, you probably got familiar with the idea of a factor. So for example, let me just pick an arbitrary number, the number 12. We could say that the number 12 is the product of say two and six; two times six is equal to 12. So because if you take the product of two and six, you get 12, we could say that two is a factor of 12, we could also say that six is a factor of 12. You take the product of these things and you get 12! You could even say that this is 12 in factored form. People don't really talk that way but you could think of it that way. We broke 12 into the things that we could use to multiply. And you probably remember from earlier mathematics the notion of prime factorization, where you break it up into all of the prime factors. So in that case you could break the six into a two and a three, and you have two times two times three is equal to 12. And you'd say, \"Well, this would be 12 \"in prime factored form or the prime factorization of 12,\" so these are the prime factors. is things that you can multiply together to get your original thing. Or if you're talking about factored form, you're essentially taking the number and you're breaking it up into the things that when you multiply them together, you get your original number. What we're going to do now is extend this idea into the algebraic domain. So if we start with an expression, let's say the expression is two plus four X, can we break this up into the product of two either numbers or two expressions or the product of a number and an expression? Well, one thing that might jump out at you is we can write this as two times one plus two X. And you can verify if you like that this does indeed equal two plus four X. We're just going to distribute the two. is equal to four X, so plus four X. So in our algebra brains, this will often be reviewed as or referred to as this expression factored or in a factored form. Sometimes people would say that we have factored out the two. You could just as easily say that you have factored out a one plus two X. You have broken this thing up into two of its factors. So let's do a couple of examples of this and then we'll think about, you know, I just told you that we could write it this way but how do you actually figure that out? So let's do another one. Let's say that you had, I don't know, let's say you had, six, let me just in a different color, let's say you had six X six X plus three," + }, + { + "Q": "At 2:20, He could have continued that lin and made 2(1+[2 {1x}])", + "A": "True, but it is less complicated to do 2(1*2x).", + "video_name": "I6TBBzIvgB8", + "timestamps": [ + 140 + ], + "3min_transcript": "- In earlier mathematics that you may have done, you probably got familiar with the idea of a factor. So for example, let me just pick an arbitrary number, the number 12. We could say that the number 12 is the product of say two and six; two times six is equal to 12. So because if you take the product of two and six, you get 12, we could say that two is a factor of 12, we could also say that six is a factor of 12. You take the product of these things and you get 12! You could even say that this is 12 in factored form. People don't really talk that way but you could think of it that way. We broke 12 into the things that we could use to multiply. And you probably remember from earlier mathematics the notion of prime factorization, where you break it up into all of the prime factors. So in that case you could break the six into a two and a three, and you have two times two times three is equal to 12. And you'd say, \"Well, this would be 12 \"in prime factored form or the prime factorization of 12,\" so these are the prime factors. is things that you can multiply together to get your original thing. Or if you're talking about factored form, you're essentially taking the number and you're breaking it up into the things that when you multiply them together, you get your original number. What we're going to do now is extend this idea into the algebraic domain. So if we start with an expression, let's say the expression is two plus four X, can we break this up into the product of two either numbers or two expressions or the product of a number and an expression? Well, one thing that might jump out at you is we can write this as two times one plus two X. And you can verify if you like that this does indeed equal two plus four X. We're just going to distribute the two. is equal to four X, so plus four X. So in our algebra brains, this will often be reviewed as or referred to as this expression factored or in a factored form. Sometimes people would say that we have factored out the two. You could just as easily say that you have factored out a one plus two X. You have broken this thing up into two of its factors. So let's do a couple of examples of this and then we'll think about, you know, I just told you that we could write it this way but how do you actually figure that out? So let's do another one. Let's say that you had, I don't know, let's say you had, six, let me just in a different color, let's say you had six X six X plus three," + }, + { + "Q": "At 4:48 couldn't you just have figured that angle DBE was 45 because it is the half of 90? I mean if I do it always with that logic can it go wrong?", + "A": "Always prove everything..proving is better.", + "video_name": "7FTNWE7RTfQ", + "timestamps": [ + 288 + ], + "3min_transcript": "Now, we could do either of these. Let's do this one right over here. So what is the measure of angle ABE? So they haven't even drawn segment BE here. So let me draw that for us. And so we have to figure out the measure of angle ABE. So we have a bunch of congruent segments here. And in particular, we see that triangle ABD, all of its sides are equal. So it's an equilateral triangle, which means all of the angles are equal. And if all of the angles are equal in a triangle, they all have to be 60 degrees. So all of these characters are going to be 60 degrees. Well, that's part of angle ABE, but we have to figure out this other part right over here. And to do that, we can see that we're actually dealing with an isosceles triangle kind of tipped over to the left. This is the vertex angle. This is one base angle. This is the other base angle. And the vertex angle right here is 90 degrees. And once again, we know it's isosceles And once again, these two angles plus this angle right over here are going to have to add up to 180 degrees. So you call that an x. You call that an x. You've got x plus x plus 90 is going to be 180 degrees. So you get 2x plus-- let me just write it out. Don't want to skip steps here. We have x plus x plus 90 is going to be equal to 180 degrees. x plus x is the same thing as 2x, plus 90 is equal to 180. And then we can subtract 90 from both sides. You get 2x is equal to 90. Or divide both sides by 2. You get x is equal to 45 degrees. And then we're done because angle ABE is going to be equal to the 60 degrees plus the 45 degrees. So it's going to be this whole angle, which is what we care about. Angle ABE is going to be 60 plus 45, which is 105 degrees. This one looks a little bit simpler. I have an isosceles triangle. This leg is equal to that leg. This is the vertex angle. I have to figure out B. And the trick here is like, wait, how do I figure out one side of a triangle if I only know one other side? Don't I need to know two other sides? And we'll do it the exact same way we just did that second part of that problem. If this is an isosceles triangle, which we know it is, then this angle is going to be equal to that angle there. And so if we call this x, then this is x as well. And we get x plus x plus 36 degrees is equal to 180. The two x's, when you add them up, you get 2x. And then-- I won't skip steps here. 2x plus 36 is equal to 180. Subtract 36 from both sides, we get 2x-- that 2 looks a little bit funny. We get 2x is equal to-- 180 minus 30 is 150." + }, + { + "Q": "At 1:19- 1:24 the video says that the ft would cancel out. I don't see how they cancel out. In class I'm doing this and my teacher says use a chart to do this, but how do i get things to cancel out. i really want to pass the 9th eoct.", + "A": "It is known as simplifying the equation . Suppose you are given 9/3*27/6. You could simplify it and you will get 3/1*9/2, that is 27/2. If you do it the long way you will the same answer. Sal did the same thing.", + "video_name": "F0LLR7bs7Qo", + "timestamps": [ + 79, + 84 + ], + "3min_transcript": "A squirrel is running across the road at 12 feet per second. It needs to run 9 feet to get across the road. How long will it take the squirrel to run 9 feet? Round to the nearest hundredth of a second. Fair enough. A car is 50 feet away from the squirrel-- OK, this is a high-stakes word problem-- driving toward it at a speed of 100 feet per second. How long will it take the car to drive 50 feet? Round to the nearest hundredth of a second. Will the squirrel make it 9 feet across the road before the car gets there? So this definitely is high stakes, at least for the squirrel. So let's answer the first question. Let's figure out how long will it take the squirrel to run 9 feet. So let's think about it. So the squirrel's got to go 9 feet, and we want to figure out how many seconds it's So would we divide or multiply this by 12? Well, to think about that, you could think about the units where we want to get an answer in terms of seconds. We want to figure out time, so it'd be great if we could multiply this times seconds per foot. Then the feet will cancel out, and I'll be left with seconds. Now, right over here, we're told that the squirrel can run at 12 feet per second, but we want seconds per foot. So the squirrel, every second, so they go 12 feet per second, then we could also say 1 second per every 12 feet. So let's write it that way. So it's essentially the reciprocal of this because the units are the reciprocal of this. So, it's 1 second for every 12 feet. Notice, all I did is I took this information right over here, 12 feet per second, and I wrote it as second per foot-- 12 feet for every 1 second, 1 second for every 12 feet. it takes for the squirrel in seconds. So the feet cancel out with the feet, and I am left with 9 times 1/12, which is 9/12 seconds. And 9/12 seconds is the same thing as 3/4 seconds, which is the same thing as 0.75 seconds for the squirrel to cross the street. Now let's think about the car. So now let's think about the car. And it's the exact same logic. They tell us that the car is 50 feet away. So the squirrel is trying to cross the road like that, and the car is 50 feet away coming in like that, and we want to figure out if the squirrel will survive. So the car is 50 feet away. So it's 50 feet away. We want to figure out the time it'll take to travel that 50 feet. Once again, we would want it in seconds. So we would want seconds per feet." + }, + { + "Q": "At 0:30, what are the indices of a sequence?", + "A": "The index is the counting number n (or k, or i or whatever). What he says is that we often view a sequence as a function of the indices. In other words, for each value of n, there is a specified value of the sequence based on the definition in terms of n. If the index is n, and the sequence is defined as starting at n= 0 or n = 1, then for every value of n, we can generate a new term of the sequence. If the sequence is (-1)\u00e2\u0081\u00bf\u00e2\u0081\u00ba\u00c2\u00b9 \u00e2\u0088\u0099 1/n\u00c2\u00b2 when the index = 2, the term is -\u00c2\u00bc Hope that helps", + "video_name": "wzw9ll80Zbc", + "timestamps": [ + 30 + ], + "3min_transcript": "what i want to do in this video is to provide ourselves with a rigorous definition of what it means to take the limit of a sequence as n approaches infinity and what we'll see is actually very similar to the definition of any function as a limit approaches infinity and this is because the sequences really can be just viewed as a function of their indices, so let's say let me draw an arbitrary sequence right over here so actually let me draw like this just to make it clear but the limit is approaching so let me draw a sequence let me draw a sequence that is jumping around little bit, so lets say when n=1, a(1) is there, when n = 2, a(2) is there, when n = 3, a(3) is over there when n=4, a(4) is over here, when n=5 a(5) is over here and it looks like is n is so this is 1 2 3 4 5 a(n) seems to be approaching, seems to be approaching some value it seems to be getting closer and closer, seems to be converging to some value L right over here. What we need to do is come up with a definition of what is it really mean to converge to L. So let's say for any, so we're gonna say that you converge to L for any, for any \u03b5 > 0, for any positive epsilon, you can, you can come up, you can get or you can, there is let me rewrite it this way, for any positive epsilon there is a positive, positive M, capital M, such that, such that if, if, lower case n is greater than capital M, the distance between those two points is less than epsilon. If you can do this for any epsilon, for any epsilon, greater than 0, there is a positive M, such that if n is greater than M, the distance between a(n) and our limit is less than epsilon then we can say, then we can say that the limit of a(n) as n approaches infinity is equal to L and we can say that a(n) converges, converges, converges to L. So let's, let's, let's parse this, so here I was making the claim that a(n) is approaching this L right over here, I tried to draw it as a horizontal line." + }, + { + "Q": "what does mean looking at the frequency of scores at 4:42 and 4:43", + "A": "Frequency means how many times it occurs. So the frequency of scores means how many times each score occurred.", + "video_name": "0ZKtsUkrgFQ", + "timestamps": [ + 282, + 283 + ], + "3min_transcript": "so her score is going to be, let me do that in Efra's color, and that's Efra's score right there. She also got 100. So, Efra... Efra, and then finally, Farah got an 80, so 60, 70, 80, so Farah got an 80. So this is Farah's score right over here. So this is another way of representing the data, and here we see it in visual form, but it has the same information. You can look up someone's name, and then figure out their score. Amy scored a 90, Bill scored a 95, Cam scored 100, Efra also scored 100, Farah scored an 80. And there's even other ways you can have some of this information. In fact sometimes, you might not even know their names, and so then it would be less information but (mumbles) a list of scores. The professor might say, \"Hey, here are the five scores \"that people got on the exam.\" 100... 100, and 80, now, if it was listed, this was all the data you got, this is less information than the data that's in this bar graph, or this histogram, or the data that's given in this table right over here, because here, not only do we know the scores, but we know who got what score. Here, we only know the list of scores, and this is not an exhaustive video of all of the different ways you can represent data. You can also represent data by looking at the frequency of scores. So, the frequency of scores right over here, so instead of writing the people, you could write the scores. So let's see, you could say this is 80, 85, 90, 95, and 100, and then you could record So how many times do we have a score of an 80? Well, Farah is the only person with a score of 80, so you put one data point there. No one got an 85, one person got a 90, so you put a data point there. One person got a 95, so you can put that data point right over there. And then two people got 100. So this is one and two. Let's see the other 100 is in this color, so I'll just do it in the color. You wouldn't necessarily have to color code it like this. So this is another way to represent, and this axis, you could just view it as the number. So this tells you how many 80s there were, how many 90s there are, how many 95s, and how many 100s. So this right over here, has the same data as this list of numbers. It's just another way of looking at it. And once you have your data arranged in any of these ways, we can start to ask interesting questions. We can ask ourselves things like," + }, + { + "Q": "I dont understand what ur saying at 5:30", + "A": "at 5:30, he is combining like terms to make the equation simpler and easier to work with.", + "video_name": "cNlwi6lUCEM", + "timestamps": [ + 330 + ], + "3min_transcript": "entire 12 feet of wood? So the length of all of the shelves have to add up to 12 feet. She's using all of it. So t plus m, plus b needs to be equal to 12 feet. That's the length of each of them. She's using all 12 feet of the wood. So the lengths have to add to 12. So what can we do here? Well, we can get everything here in terms of one variable, maybe we'll do it in terms of m, and then substitute. So we already have t in terms of m. We could, everywhere we see a t, we could substitute with m minus 1/2. But here we have b in terms of t. So how can we put this in terms of m? Well, we know that t is equal to m minus 1/2. So let's take, everywhere we see a t, let's substitute it with this thing right here. That is what t is equal to. So we can rewrite this blue equation as, the length of the but we know that t is equal to m minus 1/2. And if we wanted to simplify that a little bit, this would be that the bottom shelf is equal to-- let's distribute the 2-- 2 times m is 2m. 2 times negative 1/2 is negative 1. And then minus another 1/2. Or, we could rewrite this as b is equal to 2 times the middle shelf minus 3/2. 1/2 is 2/2 minus another 1/2 is negative 3/2, just like that. So now we have everything in terms of m, and we can substitute back here. So the top shelf-- instead of putting a t there, we could put m minus 1/2. So we put m minus 1/2, plus the length of the middle Well, we already put that in terms of m. That's what we just did. This is the length of the bottom shelf in terms of m. So instead of writing b there, we could write 2m minus 3/2. Plus 2m minus 3/2, and that is equal to 12. All we did is substitute for t. We wrote t in terms of m, and we wrote b in terms of m. Now let's combine the m terms and the constant terms. So if we have, we have one m here, we have another m there, and then we have a 2m there. They're all positive. So 1 plus 1, plus 2 is 4m. So we have 4m. And then what do our constant terms tell us? We have a negative 1/2, and then we have a negative 3/2. So negative 1/2 minus 3/2, that is negative 4/2 or negative 2. So we have 4m minus 2." + }, + { + "Q": "Can you explain a little more on why ln(x) + c (@ 1:10) has to be greater than 0?", + "A": "The input for logarithmic equation has to be greater than 0. This is because y=ln(x) <==> e^y=x , but e^y is never 0 or negative. So x>0", + "video_name": "sPPjk4aXzmQ", + "timestamps": [ + 70 + ], + "3min_transcript": "What I want to do in this video is think about the antiderivative of 1/x. Or another way of thinking about it, another way of writing it , is the antiderivative of x to the negative 1 power. And we already know, if we somehow try to apply that anti-power rule, that inverse power rule over here, we would get something that's not defined. We would get x to the 0 over 0, doesn't make any sense. And you might have been saying, OK, well, I know what to do in this case. When we first learned about derivatives, we know that the derivative-- let me do this in yellow-- the derivative with respect to x of the natural log of x is equal to 1 over x. So why can't we just say that the antiderivative of this right over here is equal to the natural log of x plus c? And this isn't necessarily wrong. The problem here is that it's not broad enough. When I say it's not broad enough, is that the domain over here, for our original function that we're taking the antiderivative of, is all real numbers except for x equals 0. So over here, x cannot be equal to 0. So over here, x, so for this expression, x has to be greater than 0. So it would be nice if we could come up with an antiderivative that has the same domain as the function that we're taking the antiderivative of. So it would be nice if we could find an antiderivative that is defined everywhere that our original function is. So pretty much everywhere except for x equaling 0. So how can we rearrange this a little bit so that it could be defined for negative values as well? Well, one one possibility is to think about the natural log of the absolute value of x. So I'll put little question mark here, just because we don't really know what the derivative of this thing is going to be. And I'm not going to rigorously prove it here, but I'll I will give you kind of the conceptual understanding. So to understand it, let's plot the natural log of x. So that right over there is roughly what the graph of the natural log of x looks like. So what would the natural log of the absolute value of x is going to look like? Well, for positive x's, it's going to look just like this. For positive x's you take the absolute value of it, it's just the same thing as taking that original value. So it's going to look just like that for positive x's. But now this is also going to be defined for negative x's. If you're taking the absolute value of negative 1, that evaluates to just 1. So it's the natural log of 1, so you're going to be right there. As you get closer and closer and closer to 0 from the negative side, you're just going to take the absolute value. So it's essentially going to be exactly this curve for the natural log of x, but the left side of the natural log of the absolute value of x is going to be its mirror image, if you were to reflect around It's going to look something like this. So what's nice about this function" + }, + { + "Q": "5:61 Is there something wrong here?", + "A": "if you are writing in decimals it is a decimal dot (looks like a period) not a colon. If it is time, the time is impossible because there can be only 60 minutes per hour actually 59 before the next hour. Good Question! Keep it Up!", + "video_name": "AGFO-ROxH_I", + "timestamps": [ + 361 + ], + "3min_transcript": "So 3 goes into 10,560. It doesn't go into 1. It goes into 10 three times. 3 times 3 is 9. And we subtract. We get 1. Bring down this 5. It becomes a 15. 3 goes into 15 five times. 5 times 3 is 15. We have no remainder, or 0. You bring down the 6. 3 goes into 6 two times. Let me scroll down a little bit. 2 times 3 is 6. Subtract. No remainder. Bring down this last 0. 3 goes into 0 zero times. 0 times 3 is 0. And we have no remainder. So 2 miles is the equivalent to 3,520 yards. That's the total distance he has to travel. Now we want to figure out how many laps there are. We want this in terms of laps, not in terms of yards. So we want the yards to cancel out. And we want laps in the numerator, right? Because when you multiply, the yards will cancel out, and we'll just be left with laps. Now, how many laps are there per yard or yards per lap? Well, they say the distance around the field is 300 yards. So we have 300 yards for every 1 lap. So now, multiply this right here. The yards will cancel out, and we will get 3,520. Let me do that in a different color. We will get 3,520, that right there, times 1/300. When you multiply it times 1, it just becomes 3,520 divided by 300. And in terms of the units, the yards canceled out. We're just left with the laps. So 3,520 divided by 300. Well, we can eyeball this right here. What is 11 times 300? Let's just approximate this right here. So if we did 11 times 300, what is that going to be equal to? Well, 11 times 3 is 33, and then we have two zeroes here. So this will be 3,300. So it's a little bit smaller than that. If we have 12 times 300, what is that going to be? 12 times 3 is 36, and then we have these two zeroes, so it's equal to 3,600. So this is going to be 11 point something. It's larger than 11, right? 3,520 is larger than 3,300. So when you divide by 300 you're going to get something larger than 11. But this number right here is smaller than 3,600 so when you divide it by 300, you're going to get something a little bit smaller than 12." + }, + { + "Q": "On 7:34, it says the first number to the right is the ones place. How about decimal numbers?", + "A": "Decimal numbers have place values below the ones place.", + "video_name": "wx2gI8iwMCA", + "timestamps": [ + 454 + ], + "3min_transcript": "So we could rewrite this. This is equal to, this is equal to three 10s three 10s plus, plus seven ones. Or another way to think about it what are the three 10s? Well if use the same number system to represent three 10s you would write that down as 30. And then seven ones. Once again if you use our same number system you would represent that as seven. So these are all different ways of representing 37. And hopefully this allows you to appreciate how neat our number system is. Where even a number like 37 as soon as you just write scratches on a wall, it becomes pretty hard to read. And you can imagine when you get to much much larger numbers like 1,052 to have to count that many marks every time. But our number system gives us a way of dealing with it." + }, + { + "Q": "At 1:00 how did he know to put t-5 into parentheses instead of 210t-5=41790?", + "A": "t represents the number of trees. t is the x , or independent, or domain variable. The y variable in this problem is the # of oranges (the range, or the dependent variable) When the dude cut down 5 trees, he messed with the domain, t. Since he only messed with the domain, and not with the range (the number of oranges), we want to isolate that -5 stuff with the t. If he didn t put the parentheses, it would be like subtracting 5 oranges, not subtracting 5 trees.", + "video_name": "xKH1Evwu150", + "timestamps": [ + 60 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:10, Why you place 210 on the other side instead of distributing? Isn't it distribution next then you eliminate and carry over to the other side?", + "A": "You can do it both ways. Whether you divide both sides by 210 first or distribute the 210 across the (t-5), you get the same results.", + "video_name": "xKH1Evwu150", + "timestamps": [ + 130 + ], + "3min_transcript": "" + }, + { + "Q": "I watched the videos on the binomial theorem but they didn't show what he is doing here... I can't follow his thinking process at around 2:00 when he keeps expanding the binomial. Any help for me? I wouldn't be able to write out the whole thing by myself is what I mean.", + "A": "The expansion for (a + b)^n always starts with a^n and always ends with b^n. The stuff in the middle comes from the binomial expansion. Check it out in the precalc section! It is most excellent. :)", + "video_name": "dZnc3PtNaN4", + "timestamps": [ + 120 + ], + "3min_transcript": "" + }, + { + "Q": "Since C can be negative (when pi/2 < theta < pi) how do we know the value in the radical at 4:45 will remain positive?\n\nEdit: Nevermind, I see my error. C can never be less than -1 on the unit circle, so that radical will never be negative.", + "A": "you made a small mistake in your edit: if C<-1, the radical becomes more positive, as the formula has 1-C, not 1+C in it, so C would have to become larger than 1 for the radical to become negative.", + "video_name": "yV4Xa8Xtmrc", + "timestamps": [ + 285 + ], + "3min_transcript": "this yellow sine squared theta, and all of this is equal to C or we could get that C is equal to one minus two sine squared theta. And what's useful about this is we just have to solve for sine of theta. So let's see, I could multiply both sides by a negative just so I can switch the order over here. So I could write this as negative C is equal to two sine squared theta minus one and just multiply both sides by a negative and then let's see I could add one to both sides, if I add one to both sides, and I'll go over here, if I add one to both sides, I could divide both sides by two, and then so I get sine squared theta is equal to one minus C over two, or I could write that sine of theta is equal to the plus or minus square root of one minus C over two. So that leads to a question, Is it both? Is it the plus and minus square root? Or is it just one of those? And I encourage you to pause the video again in case you haven't already figured it out, and look at the information here, and think about whether they give us the information of whether we should be looking at the positive or negative sine. Well they tell us that theta is between zero and pi. So if I were to draw a unit circle here, between zero and pi radians. and pi is going all the way over here. So this angle places its terminal ray either in the first or second quadrants. So, it could be an angle like this, it could be an angle like this, it cannot be an angle like this, and we know that the sine of an angle is the Y coordinate, and so we know that for the first of second quadrant the Y coordinate is going to be non-negative. So we would want to take the positive square root, so we would get sine of theta, is equal to the principal root, or we could even think of it as the positive square root of one minus C over two. So let's go back to our... Make sure we can check our answers. So sine of theta is equal to the square root of one minus capital C, all of that over two," + }, + { + "Q": "At 3:00 isn't that called the Reflexive Property?", + "A": "In statement 3 (CA=CA) this is the reflexive property of congruency.", + "video_name": "fSu1LKnhM5Q", + "timestamps": [ + 180 + ], + "3min_transcript": "And you don't have to do something as a two column proof, but this is what you normally see in a normal introductory geometry class. So I thought I would expose you to it. It's a pretty basic idea is that you make a statement, and you just have to give the reason for your statement. Which is what we've been doing with any proof, but we haven't always put it in a very structured way. So I'm just going to do it like this. I'll have two columns like that. And I'll have a statement, and then I will give the reason for the statement. And so the strategy that I'm going to try to do is it looks like, right off the bat, it seems like I can prove the triangle CDA is congruent to triangle CBA based on side side side. And that's a pretty good starting point because once I can base congruency, then I can start to have angles be the same. And the reason why I can do that is because this side is the same as that side, this side the same as that side, and they both share that side. I want to write it out properly early in this two column proof. So we have CD, we had the length of segment CD is equal to the length of CB. CD is equal to CB, and that is given. So these two characters have the same length. We also know that DA, the length of segment DA, is the same as the length of segment BA. So DA is equal to BA, that's also given in the diagram. And then we also know that CA is equal to CA, I guess we could say. So CA is equal to itself. And it's obviously in both triangles. So this is also given, or it's obvious from the diagram. It's a bit obvious. Both triangles share that side. Their corresponding sides have the same length, and so we know that they're congruent. So we know that triangle CDA is congruent to triangle CBA. And we know that by the side side side postulate and the statements given up here. Actually, let me number our statements just so we can refer back to this 1, 2, 3, and 4. And so side side side postulate and 1, 2, and 3-- statements 1, 2, and 3. So statements 1, 2, and 3 and the side side postulate let us know that these two triangles are congruent. And then if these are congruent, then we know, for example, we know that all of their corresponding angles are equivalent. So for example, this angle is going to be equal to that angle. So let's make that statement right over there." + }, + { + "Q": "at 0:33 thers a arc on the outer side what is that called?", + "A": "Even about the reflex angle, it doesn t even count as still an obtuse angle, so the reflex angle is 2/3 of a circle. \u00f0\u009f\u0099\u0082", + "video_name": "4ZyTVTGVPgE", + "timestamps": [ + 33 + ], + "3min_transcript": "Put the vertex of the angle at point A. So let me do that. Make one of the rays go through point B. Make the other ray go through one of the other points to make an acute angle. So an acute angle is an angle less than 90 degrees. So we could make one of these rays go through point B. And then we have to pick where to put the other ray to make it go to one of the other points. And we have to be very careful here because we have to look at this arc that shows which angle the tool is actually measuring. Because we might be tempted to do something like this, thinking that, hey, maybe this is the angle that we're thinking about. But the tool thinks we're referring to this outer angle right over here, this larger huge angle. This angle right over here is well over 180 degrees. So we have to pay attention to this arc to make sure that the tool is looking at the same angle that we are. So once again, we want an acute angle. So this right over here looks like an acute angle. It looks like it is less than 90 degrees. And we have to be very careful that we go exactly So that looks about right. This is an acute angle because its measure is less than 90 degrees. Let's do a few more of these. Make an obtuse angle using the black points. Choose one of the points as the vertex and make the rays go through the other two points. The angle should also be less than 180 degrees. So you could think of it several ways. You could just try to pick that point. But then when you go through these other two points, this right over here is an acute angle. This is less than 90 degrees. You could do something like this where now when I switch the rays, the tool is now thinking about this angle, not this outer angle right over here. But this is larger than 180 degrees. So this also doesn't apply. So we really picked the wrong point for the vertex. If you just move this out of the way and eyeball it, it looks like you get the largest angle that's less than 180 degrees if you put the vertex right over here. So let me do that. And now it looks like I have constructed a 180-degree angle. And we have to be very careful that we go right through those points because otherwise it might mark us wrong. So that looks pretty good. That's an obtuse angle because its measure is greater than 90 degrees. Let's do one more of these. So another obtuse angle using the points. Same general idea so that would be an exactly 180-degree angle. And there we go. And this is an obtuse angle because it is greater than 90 degrees." + }, + { + "Q": "At 6:01\nWhat I don't understand is that if the sign is the same, why don't you add or subtract\nEX#1:\n-15+(-16)\nisn't that technically subtraction?", + "A": "It can be both. You can do it whatever way is easiest to you. That s why he is teaching you it! Hope this helps! :)", + "video_name": "Oo2vGhVkvDo", + "timestamps": [ + 361 + ], + "3min_transcript": "to be the absolute value of 134. It's going to be this distance right over here, which is just 134-- which is just that right over there-- plus this distance right over here. Now, what is this distance? Well, it's the absolute value of negative 128. It's just 128. So it's going to be that distance, 134, plus 128. And that's why it made sense. This way, you're thinking of what's the difference between a larger number and a smaller number. But since it's a smaller number and you're subtracting a negative, it's the same thing as adding a positive. And hopefully this gives you a little bit of that intuition. But needless to say, we can now figure out what's going to be. And this is going to be equal to-- let me figure this out separately over here. So if I were to add 134 plus 128, I get 4 plus 8 It's 262. This right over here is equal to 262. How many degrees difference are there between the coldest and warmest recorded outside temperature? 262 degrees Fahrenheit difference." + }, + { + "Q": "at 3:37, he explains how absolute value is always positive. how come it can't be negative?", + "A": "The mathematical definition of absolute value says that it the magnitude of a real number without regard to its sign. Think about distance. You would use absolute value to calculate this because you can t have a negative distance.", + "video_name": "Oo2vGhVkvDo", + "timestamps": [ + 217 + ], + "3min_transcript": "And from that, you could subtract the smaller number, which is negative 128. So this essentially saying what's the difference between these two numbers? It's going to be positive, because we're subtracting the smaller one from the larger one. This is going to give you the exact same thing as this. Now, there's several ways to think about it. One is we know that if you subtract a negative number, that's the same thing as adding the positive of that number, or adding the absolute value. So this is the same thing. This is going to be equal to 134 plus positive 128 degrees. And what's the intuition behind that? Why does this happen? Well, look at this right over here. We're trying to figure out this distance. This distance is 134 minus negative 128. to be the absolute value of 134. It's going to be this distance right over here, which is just 134-- which is just that right over there-- plus this distance right over here. Now, what is this distance? Well, it's the absolute value of negative 128. It's just 128. So it's going to be that distance, 134, plus 128. And that's why it made sense. This way, you're thinking of what's the difference between a larger number and a smaller number. But since it's a smaller number and you're subtracting a negative, it's the same thing as adding a positive. And hopefully this gives you a little bit of that intuition. But needless to say, we can now figure out what's going to be. And this is going to be equal to-- let me figure this out separately over here. So if I were to add 134 plus 128, I get 4 plus 8 It's 262. This right over here is equal to 262. How many degrees difference are there between the coldest and warmest recorded outside temperature? 262 degrees Fahrenheit difference." + }, + { + "Q": "whats AA at 4:45", + "A": "AA is short for Angle-Angle similarity. This tells us that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. This also means the third angle of the first triangle is congruent with the third angle of the second triangle.", + "video_name": "7bO0TmJ6Ba4", + "timestamps": [ + 285 + ], + "3min_transcript": "is that the ratio between all of the sides are going to be the same. So for example, if we have another triangle right over here-- let me draw another triangle-- I'll call this triangle X, Y, and Z. And let's say that we know that the ratio between AB and XY, we know that AB over XY-- so the ratio between this side and this side-- notice we're not saying that they're congruent. We're looking at their ratio now. We're saying AB over XY, let's say that that is equal to BC over YZ. That is equal to BC over YZ. And that is equal to AC over XZ. that we say, hey, this means similarity. So if you have all three corresponding sides, the ratio between all three corresponding sides are the same, then we know we are dealing with similar triangles. So this is what we call side-side-side similarity. And you don't want to get these confused with side-side-side congruence. So these are all of our similarity postulates or axioms or things that we're going to assume and then we're going to build off of them to solve problems and prove other things. Side-side-side, when we're talking about congruence, means that the corresponding sides are congruent. Side-side-side for similarity, we're saying that the ratio between corresponding sides are going to be the same. So for example, let's say this right over here is 10. Let me think of a bigger number. Let's say this is 60, this right over here is 30, and I just made those numbers because we will soon learn what typical ratios are of the sides of 30-60-90 triangles. And let's say this one over here is 6, 3, and 3 square roots of 3. Notice AB over XY 30 square roots of 3 over 3 square roots of 3, this will be 10. What is BC over XY? 30 divided by 3 is 10. And what is 60 divided by 6 or AC over XZ? Well, that's going to be 10. So in general, to go from the corresponding side here to the corresponding side there, we always multiply by 10 on every side. So we're not saying they're congruent or we're not saying the sides are the same for this side-side-side for similarity. We're saying that we're really just scaling them up by the same amount, or another way to think about it, the ratio between corresponding sides are the same. Now, what about if we had-- let's" + }, + { + "Q": "At around 7:53, I noticed Sal wrote XY/AB = k = BC/YZ. Shouldn't the \"BC/YZ\" be \"YZ/BC\" instead? As if XY/AB = k, then BC/YZ must equal to 1/k.", + "A": "Yes, nice catch! Sal only messes up once in a while. Just submit that in the Report a mistake In the video , and just say, at 7:53, Sal says XY/AB = k = BC/YZ, but should say BC/YZ be YZ/BC, if you are really worried about it that much!", + "video_name": "7bO0TmJ6Ba4", + "timestamps": [ + 473 + ], + "3min_transcript": "Let me draw it like this. Actually, I want to leave this here so we can have our list. So let's draw another triangle ABC. So this is A, B, and C. And let's say that we know that this side, when we go to another triangle, we know that XY is AB multiplied by some constant. So I can write it over here. XY is equal to some constant times AB. Actually, let me make XY bigger, so actually, it That constant could be less than 1 in which case it would be a smaller value. But let me just do it that way. So let me just make XY look a little bit bigger. So let's say that this is X and that is Y. So let's say that we know that XY over AB Or if you multiply both sides by AB, you would get XY is some scaled up version of AB. So maybe AB is 5, XY is 10, then our constant would be 2. We scaled it up by a factor of 2. And let's say we also know that angle ABC is congruent to angle XYZ. I'll add another point over here. So let me draw another side right over here. So this is Z. So let's say we also know that angle ABC is congruent to XYZ, and let's say we know that the ratio between BC and YZ is also this constant. The ratio between BC and YZ is also equal to the same constant. So an example where this 5 and 10, maybe this is 3 and 6. The constant we're kind of doubling So is this triangle XYZ going to be similar? Well, if you think about it, if XY is the same multiple of AB as YZ is a multiple of BC, and the angle in between is congruent, there's only one triangle we can set up over here. We're only constrained to one triangle right over here, and so we're completely constraining the length of this side, and the length of this side is going to have to be that same scale as that over there. And so we call that side-angle-side similarity. So once again, we saw SSS and SAS in our congruence postulates, but we're saying something very different here. We're saying that in SAS, if the ratio between corresponding sides of the true triangle are the same, so AB and XY of one corresponding side and then" + }, + { + "Q": "At 11:00 and 11:19, when he was explaining with two non-similar triagles, why did he say \"unnecessarily similar\"?", + "A": "He said necessarily similar.", + "video_name": "7bO0TmJ6Ba4", + "timestamps": [ + 660, + 679 + ], + "3min_transcript": "so that's between BC and YZ, and the angle between them are congruent, then we're saying it's similar. For SAS for congruency, we said that the sides actually had to be congruent. Here we're saying that the ratio between the corresponding sides just has to be the same. So for example SAS, just to apply it, if I have-- let me just show some examples here. So let's say I have a triangle here that is 3, 2, 4, and let's say we have another triangle here that has length 9, 6, and we also know that the angle in between are congruent so that that angle is equal to that angle. What SAS in the similarity world tells you is that these triangles are definitely going to be similar triangles, that we're actually constraining because there's actually only one triangle we can draw a right over here. It's the triangle where all the sides So there's only one long side right here that we could actually draw, and that's going to have to be scaled up by 3 as well. This is the only possible triangle. If you constrain this side you're saying, look, this is 3 times that side, this is 3 three times that side, and the angle between them is congruent, there's only one triangle we could make. And we know there is a similar triangle there where everything is scaled up by a factor of 3, so that one triangle we could draw has to be that one similar triangle. So this is what we're talking about SAS. We're not saying that this side is congruent to that side or that side is congruent to that side, we're saying that they're scaled up by the same factor. If we had another triangle that looked like this, so maybe this is 9, this is 4, and the angle between them were congruent, you couldn't say that they're similar because this side is scaled up by a factor of 3. This side is only scaled up by a factor of 2. So this one right over there you could not say that it is necessarily similar. and length 6 there, but you did not know that these two angles are the same, once again, you're not constraining this enough, and you would not know that those two triangles are necessarily similar because you don't know that middle angle is the same. Now, you might be saying, well there was a few other postulates that we had. We had AAS when we dealt with congruency, but if you think about it, we've already shown that two angles by themselves are enough to show similarity. So why worry about an angle, an angle, and a side or the ratio between a side? So why even worry about that? And we also had angle-side-angle in congruence, but once again, we already know the two angles are enough, so we don't need to throw in this extra side, so we don't even need this right over here. So these are going to be our similarity postulates, and I want to remind you, side-side-side, this is different than the side-side-side for congruence. We're talking about the ratio between corresponding sides." + }, + { + "Q": "I'm at 2:45, I realized that you also could prove similarity if one angle and two sides are congruent, right ?", + "A": "I believe so, yes. Later on in the video, Sal describes that as another similarity theorem: The SAS (Side Angle Side) way of determining similar triangles.", + "video_name": "7bO0TmJ6Ba4", + "timestamps": [ + 165 + ], + "3min_transcript": "And you've got to get the order right to make sure that you have the right corresponding angles. Y corresponds to the 90-degree angle. X corresponds to the 30-degree angle. A corresponds to the 30-degree angle. So A and X are the first two things. B and Y, which are the 90 degrees, are the second two, and then Z is the last one. So that's what we know already, if you have three angles. But do you need three angles? If we only knew two of the angles, would that be enough? Well, sure because if you know two angles for a triangle, you know the third. So for example, if I have another triangle that looks like this-- let me draw it like this-- and if I told you that only two of the corresponding angles are congruent. So maybe this angle right here is congruent to this angle, and that angle right there is congruent to that angle. Is that enough to say that these two triangles are similar? Because in a triangle, if you know two of the angles, If you know that this is 30 and you know that that is 90, then you know that this angle has to be 60 degrees. Whatever these two angles are, subtract them from 180, and that's going to be this angle. So in general, in order to show similarity, you don't have to show three corresponding angles are congruent, you really just have to show two. So this will be the first of our similarity postulates. We call it angle-angle. If you could show that two corresponding angles are congruent, then we're dealing with similar triangles. So for example, just to put some numbers here, if this was 30 degrees, and we know that on this triangle, this is 90 degrees right over here, we know that this triangle right over here is similar to that one there. And you can really just go to the third angle in this pretty straightforward way. You say this third angle is 60 degrees, so all three angles are the same. That's one of our constraints for similarity. is that the ratio between all of the sides are going to be the same. So for example, if we have another triangle right over here-- let me draw another triangle-- I'll call this triangle X, Y, and Z. And let's say that we know that the ratio between AB and XY, we know that AB over XY-- so the ratio between this side and this side-- notice we're not saying that they're congruent. We're looking at their ratio now. We're saying AB over XY, let's say that that is equal to BC over YZ. That is equal to BC over YZ. And that is equal to AC over XZ." + }, + { + "Q": "So at 9:00, if you have log base 2 (32/sqrt 8), you're NOT supposed to simplify (32/sqrt 8) into (4*sqrt 8)?", + "A": "Yes, you could do that and still get to the same answer, but sal just skipped it", + "video_name": "TMmxKZaCqe0", + "timestamps": [ + 540 + ], + "3min_transcript": "Just so you know logarithms are useful. Let's do some examples. Let's just let's just rewrite a bunch of things in simpler forms. So if I wanted to rewrite the log base 2 of the square root of-- let me think of something. Of 32 divided by the cube-- no, I'll just take the square root. Divided by the square root of 8. How can I rewrite this so it's reasonably not messy? Well let's think about this. This is the same thing, this is equal to-- I don't know I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right? Minus the logarithm base 2 of the square root of 8. Right? Let's see. Well here once again we have a square root here, so we could say this is equal to 1/2 times log base 2 of 32. Minus this 8 to the 1/2, which is the same thing is 1/2 log base 2 of 8. We learned that property in the beginning of this presentation. And then if we want, we can distribute this original 1/2. This equals 1/2 log base 2 of 32 minus 1/4-- because we have to distribute that 1/2-- minus 1/4 log base 2 of 8. This is 5/2 minus, this is 3. 3 times 1/4 minus 3/4. Or 10/4 minus 3/4 is equal to 7/4. I probably made some arithmetic errors, but you get the point." + }, + { + "Q": "At 8:10, how is Log2 sqrt of 32/sqrt8 converted into Log2 (32/sqrt8)^1/2?", + "A": "The square root is the same thing as an exponent of 1/2. Likewise, the cube root is the same thing as an exponent of 1/3. So, this isn t even a computation or an operation, it is just a different way of writing exactly the same thing.", + "video_name": "TMmxKZaCqe0", + "timestamps": [ + 490 + ], + "3min_transcript": "We could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just type in 357 in your calculator and press the log button and you're going to get bada bada bam. Then you can clear it, or if you know how to use the parentheses on your calculator, you could do that. But then you type 17 on your calculator, press the log button, go to bada bada bam. And then you just divide them, and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth. This one, to me it's the most useful, but it doesn't completely-- it does fall out of, obviously, the exponent properties. But it's hard for me to describe the intuition simply, so you probably want to watch the proof on it, if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. Just so you know logarithms are useful. Let's do some examples. Let's just let's just rewrite a bunch of things in simpler forms. So if I wanted to rewrite the log base 2 of the square root of-- let me think of something. Of 32 divided by the cube-- no, I'll just take the square root. Divided by the square root of 8. How can I rewrite this so it's reasonably not messy? Well let's think about this. This is the same thing, this is equal to-- I don't know I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right?" + }, + { + "Q": "I don't get why he corrected from -1/2 to -1/4 in the very end (9:48)? Can someone please explain?", + "A": "He did not distribute the 1/2 to the -1/2 log\u00e2\u0082\u00828 term (he wrote -1/2 log\u00e2\u0082\u00828, which was unchanged from the original term), so he corrected it to -1/4 log\u00e2\u0082\u00828. Hope this helps!", + "video_name": "TMmxKZaCqe0", + "timestamps": [ + 588 + ], + "3min_transcript": "I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right? Minus the logarithm base 2 of the square root of 8. Right? Let's see. Well here once again we have a square root here, so we could say this is equal to 1/2 times log base 2 of 32. Minus this 8 to the 1/2, which is the same thing is 1/2 log base 2 of 8. We learned that property in the beginning of this presentation. And then if we want, we can distribute this original 1/2. This equals 1/2 log base 2 of 32 minus 1/4-- because we have to distribute that 1/2-- minus 1/4 log base 2 of 8. This is 5/2 minus, this is 3. 3 times 1/4 minus 3/4. Or 10/4 minus 3/4 is equal to 7/4. I probably made some arithmetic errors, but you get the point." + }, + { + "Q": "At 8:25-8:38 why did he remove the exponent 1/2 and put it on the left so that it turns into 1/2_log(32/sqrt(8))? He mentioned a property but which one is it?", + "A": "Because, one of the property of logarithm is: log of a^b = b log a (log of a power b equals b log of a) in this video, log of some thing power to 1/2, then equals to 1/2 log of some thing. Hope it helps.", + "video_name": "TMmxKZaCqe0", + "timestamps": [ + 505, + 518 + ], + "3min_transcript": "We could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just type in 357 in your calculator and press the log button and you're going to get bada bada bam. Then you can clear it, or if you know how to use the parentheses on your calculator, you could do that. But then you type 17 on your calculator, press the log button, go to bada bada bam. And then you just divide them, and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth. This one, to me it's the most useful, but it doesn't completely-- it does fall out of, obviously, the exponent properties. But it's hard for me to describe the intuition simply, so you probably want to watch the proof on it, if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. Just so you know logarithms are useful. Let's do some examples. Let's just let's just rewrite a bunch of things in simpler forms. So if I wanted to rewrite the log base 2 of the square root of-- let me think of something. Of 32 divided by the cube-- no, I'll just take the square root. Divided by the square root of 8. How can I rewrite this so it's reasonably not messy? Well let's think about this. This is the same thing, this is equal to-- I don't know I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right?" + }, + { + "Q": "How does Sal cancel out the 2 a's at about 2:30?", + "A": "Remember that multiplying a number by 1 does not change it. So what Sal did was multiply the expression by (1/a*a)/(1/a*a).", + "video_name": "tvj42WdKlH4", + "timestamps": [ + 150 + ], + "3min_transcript": "Simplify 3a to the fifth over 9a squared times a to the fourth over a to the third. So before we even worry about the a's, we can actually simplify the 3 and the 9. They're both divisible by 3. So let's divide the numerator and the denominator here by 3. So if we divide the numerator by 3, the 3 becomes a 1. If we divide the denominator by 3, the 9 becomes a 3. So this reduces to, or simplifies to 1a to the fifth times a to the fourth over-- or maybe I should say, a to the fifth over 3a squared times a to the fourth over a to the third. Now this, if we just multiply the two expressions, this would be equal to 1a to the fifth times a to the fourth in the numerator, and we don't have to worry about the one, it doesn't change the value. So it's a to the fifth times a to the fourth in the numerator. And then we have 3a-- let me write the 3 like this-- and then we have 3 times a squared times a to the third in the denominator. can simplify this from here. One sometimes is called the quotient rule. And that's just the idea that if you have a to the x over a to the y, that this is going to be equal to a to the x minus y. And just to understand why that works, let's think about a to the fifth over a squared. So a to the fifth is literally a times a times a times a times a. That right there is a to the fifth. And we have that over a squared. And I'm just thinking about the a squared right over here, which is literally just a times a. That is a squared. Now, clearly, both the numerator and denominator are both divisible by a times a. We can divide them both by a times a. So we can get rid of-- if we divide the numerator And if we divide the denominator by a times a, we just get a 1. So what are we just left with? We are left with just a times a times a over 1, which is just a times a times a. But what is this? This is a to the third power, or a to the 5 minus 2 power. We had 5, we were able to cancel out 2, that gave us 3. So we could do the same thing over here. We can apply the quotient rule. And I'll do two ways of actually doing this. So let's apply the quotient rule with the a to the fifth and the a squared. So let me do it this way. So let's apply with these two guys, and then let's apply it with these two guys. And of course, we have the 1/3 out front. So this can be reduced to 1/3 times-- if we apply the quotient rule with a" + }, + { + "Q": "At 1:51 Where does the aa being a divisible of both come from? That was confusing, how would you even know to use that formula?", + "A": "This is basically factoring and cancelling but with letters instead of numbers. It s like saying what s 32/4? well, since the numerator is actually divisible by four (2*2) we can simplify by dividing the numerator and denominator by 4 giving us 8/1 or 8. (this is the same problem that Sal shows in the video but with the number 2 in place of a). Does that help?", + "video_name": "tvj42WdKlH4", + "timestamps": [ + 111 + ], + "3min_transcript": "Simplify 3a to the fifth over 9a squared times a to the fourth over a to the third. So before we even worry about the a's, we can actually simplify the 3 and the 9. They're both divisible by 3. So let's divide the numerator and the denominator here by 3. So if we divide the numerator by 3, the 3 becomes a 1. If we divide the denominator by 3, the 9 becomes a 3. So this reduces to, or simplifies to 1a to the fifth times a to the fourth over-- or maybe I should say, a to the fifth over 3a squared times a to the fourth over a to the third. Now this, if we just multiply the two expressions, this would be equal to 1a to the fifth times a to the fourth in the numerator, and we don't have to worry about the one, it doesn't change the value. So it's a to the fifth times a to the fourth in the numerator. And then we have 3a-- let me write the 3 like this-- and then we have 3 times a squared times a to the third in the denominator. can simplify this from here. One sometimes is called the quotient rule. And that's just the idea that if you have a to the x over a to the y, that this is going to be equal to a to the x minus y. And just to understand why that works, let's think about a to the fifth over a squared. So a to the fifth is literally a times a times a times a times a. That right there is a to the fifth. And we have that over a squared. And I'm just thinking about the a squared right over here, which is literally just a times a. That is a squared. Now, clearly, both the numerator and denominator are both divisible by a times a. We can divide them both by a times a. So we can get rid of-- if we divide the numerator And if we divide the denominator by a times a, we just get a 1. So what are we just left with? We are left with just a times a times a over 1, which is just a times a times a. But what is this? This is a to the third power, or a to the 5 minus 2 power. We had 5, we were able to cancel out 2, that gave us 3. So we could do the same thing over here. We can apply the quotient rule. And I'll do two ways of actually doing this. So let's apply the quotient rule with the a to the fifth and the a squared. So let me do it this way. So let's apply with these two guys, and then let's apply it with these two guys. And of course, we have the 1/3 out front. So this can be reduced to 1/3 times-- if we apply the quotient rule with a" + }, + { + "Q": "If you can have the same two terms, such as the time at 6:22, forever and ever in one sequence, can you have one term for a whole sequence? What would an example be?", + "A": "Yes, but it s not useful: a(1) = a1 a_i = a(i-1) + 0 (i > 1)", + "video_name": "Kjli0Gunkds", + "timestamps": [ + 382 + ], + "3min_transcript": "Well, then we're gonna have to figure out what the previous term is. G of five is going to be g of four, g of four plus 3.2, and you would keep going back and back and back, but we've already figured out what g of four is. It's 13.6, so this is 16.8, and then if g of five is 16.8, 16.8, you add 3.2 there, you would get 20. So you could start at g of six and keep backing up, all the way until you get to g of one, and then you figure out what that is. You recurse back to your base case, and then you're able to fill in all of the blanks. Let's do a few more examples of this. So we have this function here. So let's say that this defines a sequence. Let's think about what the first four terms of that sequence are, and once again, I encourage you to pause the video and figure that out. All right, let's work through it. So h of one is, well, they very clearly tell us If n is equal to one, h is 14. H of two, well, now we're falling into this case, 'cause two is greater than one and it's a whole number, so it's gonna be 28 over h of one, over h of one. Well, we know h of one is 14, so it's gonna be 28 over 14, which is equal to two. Now h of three. H of three, we're gonna fall into this case again. It's gonna be 28, 28 divided by h of two, if we're thinking of this as a sequence, divided by the previous term in the sequence. So 28 divided by h of two, we know that h of two is equal to, is equal to two. We just figured that out. So we go back to 14, something very interesting. I think you see where this is going. H of four is gonna be 28 divided by h of three, 28 divided by h of three, this is h of three right over here, we just figured that out, divided by 14, which is back to two. If we were to think of this as a sequence, we'd say, \"All right, let's see, the first term is 14, \"then we get to two, \"then we go to 14, \"then we get to two.\" So one way to think about this sequence is that we just keep alternating between 14 and twos. All of the odd terms of the sequence are 14, all of the even terms of the sequence are two. That's one way to think about it. Or another way to think about it is, we're starting with 14, and each successive term is the previous term divided, is 28 divided by the previous term. So here, 28 divided by 14 is two. 28 divided by two is 14. 28 divided by 14 is two. And we keep going on and on and on, and that's what was actually going on right over there. Let's do one more of these. And this one is interesting, because we now have," + }, + { + "Q": "At 2:11 did it have to be a fraction?", + "A": "Well, you could try to divide and write your answer as a decimal, but seeing as how the answer would probably be a decimal that stretches on for a really long time, using the fraction will be much easier.", + "video_name": "_ETPMszULXc", + "timestamps": [ + 131 + ], + "3min_transcript": "- [Voiceover] The two-way frequency table below shows data on type of vehicle driven. So, this is type of vehicle driven, and whether there was an accident the last year. So, whether there was an accident in the last year, for customers of All American Auto Insurance. Complete the following two-way table of column relative frequencies. So that's what they're talking here, this is a two-way table of column relative frequencies. If necessary, round your answers to the nearest hundredth. So, let's see what they're saying. They're saying, let's see... Of the accidents within the last year, 28 were the people were driving an SUV, a Sport Utility Vehicle and 35 were in a Sports car. Of the No accidents in the last year, 97 were in SUV and a 104 were in sports cars. Another way you could think of it, of the Sport utility vehicles that were driven and the total, let's see it's 28 plus 97 which is going to be 125. Of that 125, 28 had an accident within the last year Similarly, you could say of the 139 Sports cars 35 had an accident in the last year, 104 did not have an accident in the last year. So what they want us to do is put those relative frequencies in here. So the way we could think about it. One right over here, this represents all the Sport utility vehicles. So one way you could think about, that represents the whole universe of the Sport Utility Vehicles, at least the universe that this table shows. So, that's really representative of the 28 plus the 97. And so, in each of these we want to put the relative frequencies. So this right over here is going to be 28 divided by the total. So over here is 28, but we want this number to be a fraction of the total. Well the fraction of the total is gonna be, 28 over 97 plus 28. Which of course is going to be 125. This one right over here is going to be 97 over 125. And of course, when you add this one and this one, it should add up to one. Likewise, this one's going to be 35 over 139. 35 plus 104. So, 139. And this is going to be 104 over 104 plus 35. Which is 139. So, let me just calculate each of them using this calculator. Let me scroll down a little bit. And so, if I do 28 divided by 125, I get 0.224. They said round your answers to the nearest hundredth. So this is 0.22. No accident within the last year, 97 divided by 125. So 97 divided by 125 is equal to, see here if I rounded" + }, + { + "Q": "i dont think the y-intercept is 4/5 at 1:58", + "A": "but that is the xintercept since the y is 0", + "video_name": "5fkh01mClLU", + "timestamps": [ + 118 + ], + "3min_transcript": "In this video I'm going to do a bunch of examples of finding the equations of lines in slope-intercept form. Just as a bit of a review, that means equations of lines in the form of y is equal to mx plus b where m is the slope and b is the y-intercept. So let's just do a bunch of these problems. So here they tell us that a line has a slope of negative 5, so m is equal to negative 5. And it has a y-intercept of 6. So b is equal to 6. So this is pretty straightforward. The equation of this line is y is equal to negative 5x plus 6. That wasn't too bad. Let's do this next one over here. The line has a slope of negative 1 and contains the point 4/5 comma 0. So they're telling us the slope, slope of negative 1. So we know that m is equal to negative 1, but we're not 100% So we know that this equation is going to be of the form y is equal to the slope negative 1x plus b, where b is the y-intercept. Now, we can use this coordinate information, the fact that it contains this point, we can use that information to solve for b. The fact that the line contains this point means that the value x is equal to 4/5, y is equal to 0 must satisfy this equation. So let's substitute those in. y is equal to 0 when x is equal to 4/5. So 0 is equal to negative 1 times 4/5 plus b. I'll scroll down a little bit. So let's see, we get a 0 is equal to negative 4/5 plus b. We can add 4/5 to both sides of this equation. We could add a 4/5 to that side as well. The whole reason I did that is so that cancels out with that. You get b is equal to 4/5. So we now have the equation of the line. y is equal to negative 1 times x, which we write as negative x, plus b, which is 4/5, just like that. Now we have this one. The line contains the point 2 comma 6 and 5 comma 0. So they haven't given us the slope or the y-intercept explicitly. But we could figure out both of them from these So the first thing we can do is figure out the slope. So we know that the slope m is equal to change in y over change in x, which is equal to-- What is the change in y? Let's start with this one right here. So we do 6 minus 0." + }, + { + "Q": "At 2:00, shouldn't Sal add -4/5 to b and 0, resulting in our final answer being negative?", + "A": "Sal ha 0 = -4/5 + b He needs to move the -4/5 to the other side. Remember, we always use the opposite sign, or the opposite operation to move items to the other side of an equation. The 4/5 is negative. This means it is being subtracted from b . So, to move it, we do the opposite, we add 4/5 to both sides. -4/5 + 4/5 will cancel out like Sal shows in the video. If you did: -4/5 +(-4/5), you get -4/5 -4/5 = -9/5. So, the b would not be by itself. Hope this helps.", + "video_name": "5fkh01mClLU", + "timestamps": [ + 120 + ], + "3min_transcript": "In this video I'm going to do a bunch of examples of finding the equations of lines in slope-intercept form. Just as a bit of a review, that means equations of lines in the form of y is equal to mx plus b where m is the slope and b is the y-intercept. So let's just do a bunch of these problems. So here they tell us that a line has a slope of negative 5, so m is equal to negative 5. And it has a y-intercept of 6. So b is equal to 6. So this is pretty straightforward. The equation of this line is y is equal to negative 5x plus 6. That wasn't too bad. Let's do this next one over here. The line has a slope of negative 1 and contains the point 4/5 comma 0. So they're telling us the slope, slope of negative 1. So we know that m is equal to negative 1, but we're not 100% So we know that this equation is going to be of the form y is equal to the slope negative 1x plus b, where b is the y-intercept. Now, we can use this coordinate information, the fact that it contains this point, we can use that information to solve for b. The fact that the line contains this point means that the value x is equal to 4/5, y is equal to 0 must satisfy this equation. So let's substitute those in. y is equal to 0 when x is equal to 4/5. So 0 is equal to negative 1 times 4/5 plus b. I'll scroll down a little bit. So let's see, we get a 0 is equal to negative 4/5 plus b. We can add 4/5 to both sides of this equation. We could add a 4/5 to that side as well. The whole reason I did that is so that cancels out with that. You get b is equal to 4/5. So we now have the equation of the line. y is equal to negative 1 times x, which we write as negative x, plus b, which is 4/5, just like that. Now we have this one. The line contains the point 2 comma 6 and 5 comma 0. So they haven't given us the slope or the y-intercept explicitly. But we could figure out both of them from these So the first thing we can do is figure out the slope. So we know that the slope m is equal to change in y over change in x, which is equal to-- What is the change in y? Let's start with this one right here. So we do 6 minus 0." + }, + { + "Q": "10:20 What is the purpose of functions in a practical real life scenario?", + "A": "Functions are used all the time. -- Computer programs are functions -- the cash register at a store uses functions to determine what you owe -- when you calculate the tip on a bill in a restaurant, you are using a function. Those a just a few. There are many more.", + "video_name": "5fkh01mClLU", + "timestamps": [ + 620 + ], + "3min_transcript": "plus 5/2, plus 5/2. I like to change my notation just so you get familiar with both. So the equation becomes 5/2 is equal to-- that's a 0-- is equal to b. b is 5/2. So the equation of our line is y is equal to 5/6 x plus b, which we just figured out is 5/2, plus 5/2. We are done. Let's do another one. We have a graph here. Let's figure out the equation of this graph. This is actually, on some level, a little bit easier. What's the slope? Slope is change in y over change it x. So let's see what happens. When we move in x, when our change in x is 1, so that is our change in x. So change in x is 1. I'm just deciding to change my x by 1, increment by 1. It looks like y changes exactly by 4. It looks like my delta y, my change in y, is equal to 4 when my delta x is equal to 1. So change in y over change in x, change in y is 4 when change in x is 1. So the slope is equal to 4. Now what's its y-intercept? Well here we can just look at the graph. It looks like it intersects y-axis at y is equal to negative 6, or at the point 0, negative 6. So we know that b is equal to negative 6. So we know the equation of the line. The equation of the line is y is equal to the slope times x plus the y-intercept. I should write that. equation of our line. Let's do one more of these. So they tell us that f of 1.5 is negative 3, f of negative 1 is 2. What is that? Well, all this is just a fancy way of telling you that the point when x is 1.5, when you put 1.5 into the function, the function evaluates as negative 3. So this tells us that the coordinate 1.5, negative 3 is on the line. Then this tells us that the point when x is negative 1, f of x is equal to 2. This is just a fancy way of saying that both of these two points are on the line, nothing unusual. I think the point of this problem is to get you familiar with function notation, for you to not get intimidated if you see something like this. If you evaluate the function at 1.5, you get negative 3." + }, + { + "Q": "At 6:11, how come its 0-5 instead of 5-0? Is that crucial to get the correct answer or does it not matter, because you are still finding the change?", + "A": "(y2 - y1)/(x2-x1) works either way. Just make sure that whatever point you started with for the difference in y in the numerator is the same point you start with for the difference of x in the denominator.", + "video_name": "5fkh01mClLU", + "timestamps": [ + 371 + ], + "3min_transcript": "So, so far we know that the line must be, y is equal to the slope-- I'll do that in orange-- negative 2 times x plus our y-intercept. Now we can do exactly what we did in the last problem. We can use one of these points to solve for b. We can use either one. Both of these are on the line, so both of these must satisfy this equation. I'll use the 5 comma 0 because it's always nice when you have a 0 there. The math is a little bit easier. So let's put the 5 comma 0 there. So y is equal to 0 when x is equal to 5. So y is equal to 0 when you have negative 2 times 5, when x is equal to 5 plus b. So you get 0 is equal to -10 plus b. If you add 10 to both sides of this equation, let's add 10 to both sides, these two cancel out. You get b is equal to 10 plus 0 or 10. Now we know the equation for the line. The equation is y-- let me do it in a new color-- y is equal to negative 2x plus b plus 10. We are done. Let's do another one of these. All right, the line contains the points 3 comma 5 and negative 3 comma 0. Just like the last problem, we start by figuring out the slope, which we will call m. It's the same thing as the rise over the run, which is the same thing as the change in y over the change in x. If you were doing this for your homework, you wouldn't I just want to make sure that you understand that these are all the same things. Then what is our change in y over our change in x? This is equal to, let's start with the side first. It's just So let's say it's 0 minus 5 just like that. So I'm using this coordinate first. I'm kind of viewing it as the endpoint. Remember when I first learned this, I would always be tempted to do the x in the numerator. No, you use the y's in the numerator. So that's the second of the coordinates. That is going to be over negative 3 minus 3. This is the coordinate negative 3, 0. This is the coordinate 3, 5. We're subtracting that. So what are we going to get? This is going to be equal to-- I'll do it in a neutral color-- this is going to be equal to the numerator is negative 5 over negative 3 minus 3 is negative 6." + }, + { + "Q": "at 2:49 how does x squared divided by x equal x?", + "A": "x squared is the same as x * x. If you divide x * x by x you get x.", + "video_name": "FXgV9ySNusc", + "timestamps": [ + 169 + ], + "3min_transcript": "for more complicated problems. So you could have also written this as 2 goes into 2x plus 4 how many times? And you would perform this the same way you would do traditional long division. You'd say 2-- you always start with the highest degree term. 2 goes into the highest degree term. You would ignore the 4. 2 goes into 2x how many times? Well, it goes into 2x x times and you put the x in the x place. x times 2 is 2x. And just like traditional long division, you now subtract. So 2x plus 4 minus 2x is what? It's 4, right? And then 2 goes into 4 how many times? It goes into it two times, a positive two times. Put that in the constants place. 2 times 2 is 4. You subtract, remainder 0. So this might seem overkill for what was probably a do it in a few steps. We're now going to see that this is a very generalizable process. You can do this really for any degree polynomial dividing into any other degree polynomial. Let me show you what I'm talking about. So let's say we wanted to divide x plus 1 into x squared plus 3x plus 6. So what do we do here? So you look at the highest degree term here, which is an x, and you look at the highest degree term here, which is an x squared. So you can ignore everything else. And that really simplifies the process. You say x goes into x squared how many times? Well, x squared divided by x is just x, right? x goes into x squared x times. You put it in the x place. This is the x place right here or the x to So x times x plus 1 is what? x times x is x squared. x times 1 is x, so it's x squared plus x. And just like we did over here, we now subtract. And what do we get? x squared plus 3x plus 6 minus x squared-- let me be very careful-- this is minus x squared plus x. I want to make sure that negative sign only-- it applies to this whole thing. So x squared minus x squared, those cancel out. 3x, this is going to be a minus x. Let me put that sign there. So this is minus x squared minus x, just to be clear. We're subtracting the whole thing. 3x minus x is 2x. And then you bring down the 6, or 6 minus 0 is nothing. So 2x plus 6. Now, you look at the highest degree term, an x and a 2x. How many times does x go into 2x?" + }, + { + "Q": "At 3:00, why did Sal put the answer to x^2 divided by x above the 3x? Why not above the x^2?", + "A": "I agree with you at 3:00. It would be better and less confused, if Sal put the x right on top of x^2 , even though it is not a require. In fact, x can be put any where within the line.", + "video_name": "FXgV9ySNusc", + "timestamps": [ + 180 + ], + "3min_transcript": "for more complicated problems. So you could have also written this as 2 goes into 2x plus 4 how many times? And you would perform this the same way you would do traditional long division. You'd say 2-- you always start with the highest degree term. 2 goes into the highest degree term. You would ignore the 4. 2 goes into 2x how many times? Well, it goes into 2x x times and you put the x in the x place. x times 2 is 2x. And just like traditional long division, you now subtract. So 2x plus 4 minus 2x is what? It's 4, right? And then 2 goes into 4 how many times? It goes into it two times, a positive two times. Put that in the constants place. 2 times 2 is 4. You subtract, remainder 0. So this might seem overkill for what was probably a do it in a few steps. We're now going to see that this is a very generalizable process. You can do this really for any degree polynomial dividing into any other degree polynomial. Let me show you what I'm talking about. So let's say we wanted to divide x plus 1 into x squared plus 3x plus 6. So what do we do here? So you look at the highest degree term here, which is an x, and you look at the highest degree term here, which is an x squared. So you can ignore everything else. And that really simplifies the process. You say x goes into x squared how many times? Well, x squared divided by x is just x, right? x goes into x squared x times. You put it in the x place. This is the x place right here or the x to So x times x plus 1 is what? x times x is x squared. x times 1 is x, so it's x squared plus x. And just like we did over here, we now subtract. And what do we get? x squared plus 3x plus 6 minus x squared-- let me be very careful-- this is minus x squared plus x. I want to make sure that negative sign only-- it applies to this whole thing. So x squared minus x squared, those cancel out. 3x, this is going to be a minus x. Let me put that sign there. So this is minus x squared minus x, just to be clear. We're subtracting the whole thing. 3x minus x is 2x. And then you bring down the 6, or 6 minus 0 is nothing. So 2x plus 6. Now, you look at the highest degree term, an x and a 2x. How many times does x go into 2x?" + }, + { + "Q": "At around 8:00, you need to restrict x=-4 to be equal to the first expression. So if your long division works out with no remainder, you have to make restrictions for your original denominator, or am I missing something?", + "A": "You are correct. The original expression is undefined at x = -4, so the correct answer would be: (x\u00c2\u00b2 +5x + 4) / (x + 4) = x + 1; x \u00e2\u0089\u00a0 -4", + "video_name": "FXgV9ySNusc", + "timestamps": [ + 480 + ], + "3min_transcript": "Let's do another one of these. They're kind of fun. So let's say that we have-- we want to simplify x squared plus 5x plus 4 over x plus 4. So once again, we can do our algebraic long division. We can divide x plus 4 into x squared plus 5x plus 4. And once again, same exact process. Look at the highest degree terms in both of them. x goes into x squared how many times? It goes into it x times. Put it in the x place. This is our x place right here. X times x is x squared. x times 4 is 4x. So let me just put a negative sign there. And then these cancel out. 5x minus 4x is x. 4 minus 0 is plus 4. x plus 4, and then you could even see this coming. You could say x plus 4 goes into x plus 4 obviously one time, or if you were not looking at the constant terms, you would completely just say, well, x goes into x how many times? Well, one time. Plus 1. 1 times x is x. 1 times 4 is 4. We're going to subtract them from up here, so it cancels out, so we have no remainder. So this right here simplifies to-- this is equal to x plus 1. And there's other ways you could have done this. We could have tried to factor this numerator. x squared plus 5x plus 4 over x plus 4. This is the same thing as what? We could have factored this numerator as x plus 4 4 times 1 is 4. 4 plus 1 is 5, all of that over x plus 4. That cancels out and you're left just with x plus 1. Either way would have worked, but the algebraic long division will always work, even if you can't cancel out factors like that, even if you did have a remainder. In this situation, you didn't. So this was equal to x plus 1. Let's do another one of these just to make sure that you really-- because this is actually a very, very useful skill to have in your toolkit. So let's say we have x squared-- let me Let's say we had 2x squared-- I could really make these numbers up on the fly. 2x squared minus 20x plus 12 divided by-- actually, let's" + }, + { + "Q": "It seems like around 8:24 or so, something important got lost. The original formula was undefined when x was -4, because at that x-value the denominator was 0. The new formula of simply x + 1 is defined everywhere. What happened there?", + "A": "The new formula is x + 1, x =/= -4, so the expressions are equivalent. Sal just forgot to mention that.", + "video_name": "FXgV9ySNusc", + "timestamps": [ + 504 + ], + "3min_transcript": "Let's do another one of these. They're kind of fun. So let's say that we have-- we want to simplify x squared plus 5x plus 4 over x plus 4. So once again, we can do our algebraic long division. We can divide x plus 4 into x squared plus 5x plus 4. And once again, same exact process. Look at the highest degree terms in both of them. x goes into x squared how many times? It goes into it x times. Put it in the x place. This is our x place right here. X times x is x squared. x times 4 is 4x. So let me just put a negative sign there. And then these cancel out. 5x minus 4x is x. 4 minus 0 is plus 4. x plus 4, and then you could even see this coming. You could say x plus 4 goes into x plus 4 obviously one time, or if you were not looking at the constant terms, you would completely just say, well, x goes into x how many times? Well, one time. Plus 1. 1 times x is x. 1 times 4 is 4. We're going to subtract them from up here, so it cancels out, so we have no remainder. So this right here simplifies to-- this is equal to x plus 1. And there's other ways you could have done this. We could have tried to factor this numerator. x squared plus 5x plus 4 over x plus 4. This is the same thing as what? We could have factored this numerator as x plus 4 4 times 1 is 4. 4 plus 1 is 5, all of that over x plus 4. That cancels out and you're left just with x plus 1. Either way would have worked, but the algebraic long division will always work, even if you can't cancel out factors like that, even if you did have a remainder. In this situation, you didn't. So this was equal to x plus 1. Let's do another one of these just to make sure that you really-- because this is actually a very, very useful skill to have in your toolkit. So let's say we have x squared-- let me Let's say we had 2x squared-- I could really make these numbers up on the fly. 2x squared minus 20x plus 12 divided by-- actually, let's" + }, + { + "Q": "At 3:30 minutes into the video it says a histogram is really a plot, kind of bar graph... How is a histogram related to a bar graph in anyway?", + "A": "It has the same formation.", + "video_name": "4eLJGG2Ad30", + "timestamps": [ + 210 + ], + "3min_transcript": "I want to put them in order. I have a 2, another 2. Let me stack that above my first 2. I have another 1. Let me stack that above my other two 1's. I have another 0. Let me stack it there. I have another 1. Then I have another 2. Another 1. I have two more 0's. 0, 0. I have two more 2's. I have a 3. I have two more 1's. Another 3. And then I have a 6. So no 5's, and then I have a 6. And that space right there was unnecessary. But right here I've listed-- I've just rewritten those numbers and I've essentially categorized them. Now what I want to do is calculate how many of each of So I want to look at the frequency of each of these numbers. So I have one, two, three, four 0's. I have one, two, three, four, five, six, seven 1's. I have one, two, three, four, five 2's. I have one, two 3's. I have one 4, and one 6. So we could write it this way. We could write the number, and then we can have the frequency. So I have the numbers 0, 1, 2, 3, 4-- we could even throw 5 in there, although 5 has a frequency of 0. And we have a 6. So the 0 showed up four times in this data set. 1 showed up seven times in this data set. showed up one time, 5 didn't show up, and 6 showed up one time. All I did is I counted this data set, and I did this first. But you could say how many times do I see a 0? I see it one, two, three, four times. How many times do I see a 1? One, two, three, four, five, six, seven times. That's what we mean by frequency. Now a histogram is really just a plot, kind of a bar graph, plotting the frequency of each of these numbers. It's going to look a lot like this original thing that I drew. So let me draw some axes here. So the different buckets here are the numbers. And that worked out because we're dealing with very clean integers that tend to repeat. If you're dealing with things that the exact number doesn't repeat, oftentimes people will put the numbers into buckets or ranges. But here they fit into nice little buckets. You have the numbers 0, 1, 2, 3, 4, 5, and 6." + }, + { + "Q": "I dont understand the time 2:50", + "A": "He added the 5 just to go in order, the 5 shows up 0 times.", + "video_name": "4eLJGG2Ad30", + "timestamps": [ + 170 + ], + "3min_transcript": "I want to put them in order. I have a 2, another 2. Let me stack that above my first 2. I have another 1. Let me stack that above my other two 1's. I have another 0. Let me stack it there. I have another 1. Then I have another 2. Another 1. I have two more 0's. 0, 0. I have two more 2's. I have a 3. I have two more 1's. Another 3. And then I have a 6. So no 5's, and then I have a 6. And that space right there was unnecessary. But right here I've listed-- I've just rewritten those numbers and I've essentially categorized them. Now what I want to do is calculate how many of each of So I want to look at the frequency of each of these numbers. So I have one, two, three, four 0's. I have one, two, three, four, five, six, seven 1's. I have one, two, three, four, five 2's. I have one, two 3's. I have one 4, and one 6. So we could write it this way. We could write the number, and then we can have the frequency. So I have the numbers 0, 1, 2, 3, 4-- we could even throw 5 in there, although 5 has a frequency of 0. And we have a 6. So the 0 showed up four times in this data set. 1 showed up seven times in this data set. showed up one time, 5 didn't show up, and 6 showed up one time. All I did is I counted this data set, and I did this first. But you could say how many times do I see a 0? I see it one, two, three, four times. How many times do I see a 1? One, two, three, four, five, six, seven times. That's what we mean by frequency. Now a histogram is really just a plot, kind of a bar graph, plotting the frequency of each of these numbers. It's going to look a lot like this original thing that I drew. So let me draw some axes here. So the different buckets here are the numbers. And that worked out because we're dealing with very clean integers that tend to repeat. If you're dealing with things that the exact number doesn't repeat, oftentimes people will put the numbers into buckets or ranges. But here they fit into nice little buckets. You have the numbers 0, 1, 2, 3, 4, 5, and 6." + }, + { + "Q": "What are the \"buckets' Sal talks about at 3:41?", + "A": "He uses the term bucket to represent a range of data. The example that he uses, he lists all numbers 0,1,2,3,4,5. You might not have a lot of numbers, so you might want to clump, say 0-1, 2-3, 4-5. So you have 3 buckets.", + "video_name": "4eLJGG2Ad30", + "timestamps": [ + 221 + ], + "3min_transcript": "So I want to look at the frequency of each of these numbers. So I have one, two, three, four 0's. I have one, two, three, four, five, six, seven 1's. I have one, two, three, four, five 2's. I have one, two 3's. I have one 4, and one 6. So we could write it this way. We could write the number, and then we can have the frequency. So I have the numbers 0, 1, 2, 3, 4-- we could even throw 5 in there, although 5 has a frequency of 0. And we have a 6. So the 0 showed up four times in this data set. 1 showed up seven times in this data set. showed up one time, 5 didn't show up, and 6 showed up one time. All I did is I counted this data set, and I did this first. But you could say how many times do I see a 0? I see it one, two, three, four times. How many times do I see a 1? One, two, three, four, five, six, seven times. That's what we mean by frequency. Now a histogram is really just a plot, kind of a bar graph, plotting the frequency of each of these numbers. It's going to look a lot like this original thing that I drew. So let me draw some axes here. So the different buckets here are the numbers. And that worked out because we're dealing with very clean integers that tend to repeat. If you're dealing with things that the exact number doesn't repeat, oftentimes people will put the numbers into buckets or ranges. But here they fit into nice little buckets. You have the numbers 0, 1, 2, 3, 4, 5, and 6. And then on the vertical axis we're going to plot the frequency. So one, two, three, four, five, six, seven. So that's 7, 6, 5, 4, 3, 2, 1. So 0 shows up four times. So we'll draw a little bar graph here. 0 shows up four times. Draw it just like that. 0 shows up four times. That is that information right there. 1 shows up seven times. So I'll do a little bar graph. 1 shows up seven times. Just like that. I want to make it a little bit straighter than that-- 1 shows up seven times. 2-- I'll do it in a different color-- 2 shows up five times." + }, + { + "Q": "about 04:00, does this mean that every sum of normal distributed, independent, and squared variables will have the chi-square distribuition?", + "A": "Indeed it does. This is mainly because the sum of normally-distributed, independent, and squared random variables is the very definition of the chi-square distribution.", + "video_name": "dXB3cUGnaxQ", + "timestamps": [ + 240 + ], + "3min_transcript": "here is going to be an example of the chi-square distribution. Actually what we're going to see in this video is that the chi-square, or the chi-squared distribution is actually a set of distributions depending on how many sums you have. Right now, we only have one random variable that we're squaring. So this is just one of the examples. And we'll talk more about them in a second. So this right here, this we could write that Q is a chi-squared distributed random variable. Or that we could use this notation right here. Q is-- we could write it like this. So this isn't an X anymore. This is the Greek letter chi, although it looks a lot like a curvy X. So it's a member of chi-squared. And since we're only taking one sum over here-- we're only taking the sum of one independent, normally distributed, standard or normally distributed variable, we say that this only has 1 degree of freedom. So this right here is our degree of freedom. We have 1 degree of freedom right over there. So let's call this Q1. Let's say I have another random variable. Let's call this Q-- let me do it in a different color. Let me do Q2 in blue. Let's say I have another random variable, Q2, that is defined as-- let's say I have one independent, standard, normally distributed variable. I'll call that X1. And I square it. And then I have another independent, standard, normally distributed variable, X2. And I square it. So you could imagine both of these guys have distributions like this. And they're independent. So get to sample Q2, you essentially sample X1 from this distribution, square that value, sample X2 and then add the two. And you're going to get Q2. This over here-- here we would write-- so this is Q1. Q2 here, Q2 we would write is a chi-squared, distributed random variable with 2 degrees of freedom. Right here. 2 degrees of freedom. And just to visualize kind of the set of chi-squared distributions, let's look at this over here. So this, I got this off of Wikipedia. This shows us some of the probability density functions for some of the chi-square distributions. This first one over here, for k of equal to 1, that's the degrees of freedom. So this is essentially our Q1. This is our probability density function for Q1. And notice it really spikes close to 0. And that makes sense. Because if you are sampling just once from this standard normal distribution," + }, + { + "Q": "At 6:59, how does Sal make the logical leap that the top angle is also Y, given that he has a 90 degree angle and an angle of 90-y?", + "A": "The two acute angles in an right triangle must add up to 90. Therefore, if one value is being subtracted from 90 to represent an angle, that value must represent the other angle. When you add them up, they will add up to 90 (90 - y + y = 90).", + "video_name": "R0EQg9vgbQw", + "timestamps": [ + 419 + ], + "3min_transcript": "At any point if you get excited pause the video and try to finish the proof on your own. The length of segment CB if we just multiply both sides by cosine of x, the length of segment CB is equal to cosine of x times sine of y. Which is neat because we just showed that this thing right over here is equal to this thing right over here. To complete our proof we just need to prove that this thing is equal to this thing right over there. If that's equal to that and that's equal to that well we already know that the sum of these is equal to the length of DF which is sine of x plus y. Let's see if we can figure out, if we can express DE somehow. What angle would be useful? If somehow we could figure If we could figure out this angle then DE we could express in terms of this angle and sine of x. Let's see if we can figure out that angle. We know this is angle y over here and we also know that this is a right angle. EC is parallel to AB so you could view AC as a transversal. If this is angle y right over here then we know this is also angle y. These are once again, notice. If AC is a transversal here and EC and AB are parallel then if this is y then that is y. If that's y then this is 90 minus y. If this is 90 degrees and this is 90 minus y then these two angles combined add up to 180 minus y, then this thing up here must be equal to y. Validate that. y plus 90 minus y plus 90 is 180 degrees, and that is useful for us because now we can express segment DE in terms of y and sine of x. What is DE to y? It's the adjacent angle, so we can think of cosine. We know that the cosine of angle y, if we look at triangle DEC right over here, we know that the cosine of y is equal to segment DE over its hypotenuse, over sine of x. You should be getting excited right about now because we've just shown, if we multiply both sides by sine of x, we've just shown that DE is equal to sine of x times cosine of y." + }, + { + "Q": "So at 2:05, Sal put -2y after the -6x^2. But when I was doing the problem myself, I put 8xy first and then -2y. Is the way I ordered the expression wrong and if it is...why? This was my answer:\n( 6x^2y - 6x^2 + 8xy - 2y +4 ) This was Sal's answer:\n( 6x^2y - 6x^2 - 2y + 8xy +4 )", + "A": "Hey Tushar Gaddi!, it does not matter in what order you put it in as long as you dont change any of the negative( - ) or posiive( + ) signs. the way you wrote it was just fine. for problems like this this is why the Order Of Operations comes into play. no matter what order you put it in you will get the same answer. Hope This Helps! Good Luck and Have Fun! :)", + "video_name": "jroamh6SIo0", + "timestamps": [ + 125 + ], + "3min_transcript": "So let's get some practice simplifying polynomials, especially in the case where we have more than one variable over here. So I have 4x squared y minus 3x squared minus 2y. So that entire expression plus the entire expression 8xy minus 3x squared plus 2x squared y plus 4. So the first thing that jumps out at me is that I'm just adding this expression to this expression. So to a large degree, these parentheses don't matter. So I can just rewrite it as 4x squared y minus 3x squared minus 2y plus 8xy minus 3x squared plus 2x squared y plus 4. Now we can try to group similar terms or like terms. So let's think about what we have over here. So this first term right here is a 4x squared y. So can I add this to any of the other terms here? Do we have any other x squared y terms? is another x squared y term. If I have 4 of something-- in this case, I have 4x squared y's and I add 2x squared y's to it, how many x squared y's do I now have? Well, 4 plus 2-- I now have 6x squared y's. Now let's move on to this term. So I have negative 3x squareds. Do I have any more x squareds in this expression right over here? Well, sure, I have another negative 3x squared. So if I have negative 3 of something and then I have another negative 3, I end up with negative 6 So it's negative 6x squared. Now let's think about this negative 2y term. Are there any other y's over here? Well, it doesn't look like there are. This is an 8xy. This is a 4. There's no just y's. So I'll just rewrite it, negative 2y. And then 8xy-- well, once again, it doesn't seem like that can be added to anything else. So let's just write that over again. And then finally, we just have the constant term plus 4. And it pretty much looks like we're done. We have simplified this as much as we can." + }, + { + "Q": "At 1:11 you go through this really calculated sum but you could just find 25% of 66 and that would be your answer 66!", + "A": "You cant just do 25% of 66 because you didn t know the 66. The only 2 numbers that you know are that you have $50 and the sale is going to take 25% off. you are calculating the $66. Hope this helps.", + "video_name": "4oeoIOan_h4", + "timestamps": [ + 71 + ], + "3min_transcript": "Let's say I go to a store and I have $50 in my pocket. $50 in my wallet. And at the store that day they say it is a 25% off marked price sale. So 25% off marked price means that if the marked price is $100 the price I'm going to pay is going to be 25% less than $100. So my question to you is if I have $50, what is the highest marked price I can afford? Because I need to know that before I go finding something that I might like. So let's do a little bit of algebra. So let x be the highest marked price that I can afford. price, the sale price will be x minus 25% of x is equal to the sale price. And I'm assuming that I'm in a state without sales tax. Whatever the sale price is, is what I have to pay in cash. So x minus 25% x is equal to the sale price. The discount is going to be 25% of x. But we know that this is the same thing as x minus 0.25x. And we know that that's the same thing as-- well, because we know this is 1x, x is the same thing is 1x. 1x minus 0.25x. Well, that means that 0.75x is equal to the sale price, right? All I did is I rewrote x minus 25% of x as 1x minus 0.25x. Because 1 minus 0.25 is 0.75. So 0.75x is going to be the sale price. Well, what's the sale price that I can afford? Well, the sale price I can afford is $50. So 0.75x is going to equal $50. If x is any larger number than the number I'm solving for, then the sale price is going to be more than $50 and I won't be able to afford it. So that's how we set the-- the highest I can pay is $50 and that's the sale price. So going back to how we did these problems before. We just divide both sides by 0.75. And we say that the highest marked price that I can afford is $50 divided by 0.75. And let's figure out what that is. 0.75 goes into 50-- let's add some 0's in the back." + }, + { + "Q": "in 5:31 Sal says mathleat what is that???", + "A": "An athlete is someone who does physical activity and is good at it. A mathlete is some who performs math and is good at it, usually competing against other people.", + "video_name": "xO_1bYgoQvA", + "timestamps": [ + 331 + ], + "3min_transcript": "" + }, + { + "Q": "At 3:35, how is x to the a over x to the b equal to x to the a-b? I know he explains it, but it still doesn't make that much sense.", + "A": "Try a numeric example: x^5 / x^2 This is a fraction and must be reduced. Cancel out the x^2 and you get x^3. So, what happened to the exponents? They were subtracted. x^5 / x^2 = x^(5-2) = x^3 Sal did the same exact thing, except he has variables instead of numeric exponents. Hope this helps.", + "video_name": "0z-yIFzpunM", + "timestamps": [ + 215 + ], + "3min_transcript": "as x to the negative b, which is going to be the same thing as... If I have a base to one exponent times the same base to another exponent, that's the same thing as that base to the sum of the exponents, a plus negative b which is just gonna be a minus b. So, we got to the same place. So, we can re-write this as... So, we can re-write this part as being equal to m to the 7/9 power minus 1/3 power is equal to, is equal to m to the k over nine. And I think you see where this is going. What is 7/9 minus 1/3? Well, 1/3 is the same thing, if we want to have a common denominator, 1/3 is the same thing as 3/9. So, I can re-write this as 3/9. So 7/9 minus 3/9 is going to be 4/9. So, this is the same thing as m to the... is going to be equal to m to the k-ninths power. So, 4/9 must be the same thing as k-ninths. So, we can say 4/9 is equal to k-ninths. Four over nine is equal to k over nine, which tells us that k must be equal to four, and we're all done." + }, + { + "Q": "What was the point in him specifying what the f vetor was at 5:40? He never used it to solve his problem.", + "A": "He uses it to remind us where he gets his P(x,y) and Q(x,y). This have been explained through many of the previous videos, so i just a remainder, and not a explanation.", + "video_name": "gGXnILbrhsM", + "timestamps": [ + 340 + ], + "3min_transcript": "but we're saying that the curve is the boundary of this region and we're going to go in I counterclockwise direction. I have to specify that. So our curve, we could start at any point really, but we're going to go like that. Then get to that point and then come back down along that top curve just like that. And so this met the condition that the inside of the region is always going to be our left, so we can just do the straight up Green's theorem, we don't have to do the negative of it. And let's define our region. Let's just do our region. This integral right here is going to go-- I'll just do it the way-- y varies from y is equal to 2x squared to And so maybe we'll integrate with respect to y first. And then x, I'll do the outside. The boundary of x goes from 0 to 1. So they set us up well to do an indefinite integral. Now we just have to figure out what goes over here-- Green's theorem. Our f would look like this in this situation. f is f of xy is going to be equal to x squared minus y squared i plus 2xy j. We've seen this in multiple videos. You take the dot product of this with dr, you're going to get this thing right here. So this expression right here is our P of x y. And this expression right here is our Q of xy. So inside of here we're just going to apply Green's So the partial of Q with respect to x-- so take the derivative of this with respect to x. We're just going to end up with the 2y. And then from that we're going to subtract the partial of P with respect to y. So if you take the derivative of this with respect to y that becomes 0 and then, here you have-- the derivative with respect to y here is minus 2y. Just like that. And so this simplifies to 2y minus minus 2y. That's 2y plus plus 2y. I'm just subtracting a negative. Or this inside, and just to save space, this inside-- that's just 4y. I don't want to have to rewrite the boundaries. That right there is the same thing as 4y." + }, + { + "Q": "At 5:20, shouldn't the da be a dx?", + "A": "Yes. He writes that originally but the step after that he rewrites it as dx.", + "video_name": "gGXnILbrhsM", + "timestamps": [ + 320 + ], + "3min_transcript": "And let's see. x goes from 0 to 1, so if we make-- that's obviously 0. Let's say that that is x is equal to 1, so that's all the x values. And y varies, it's above 2x squared and below 2x. Normally, if you get to large enough numbers, 2x squared is larger, but if you're below 1 this is actually going to be smaller than that. So the upper boundary is 2x, so there's 1 comma 2. This is a line y is equal to 2x, so that is the line y is-- let me draw a straighter line than that. The line y is equal to 2x looks something like that. That right there is y is equal to 2x. Maybe I'll do that in yellow. And then the bottom curve right here is y is going to be greater than 2x squared. It might look something like this. but we're saying that the curve is the boundary of this region and we're going to go in I counterclockwise direction. I have to specify that. So our curve, we could start at any point really, but we're going to go like that. Then get to that point and then come back down along that top curve just like that. And so this met the condition that the inside of the region is always going to be our left, so we can just do the straight up Green's theorem, we don't have to do the negative of it. And let's define our region. Let's just do our region. This integral right here is going to go-- I'll just do it the way-- y varies from y is equal to 2x squared to And so maybe we'll integrate with respect to y first. And then x, I'll do the outside. The boundary of x goes from 0 to 1. So they set us up well to do an indefinite integral. Now we just have to figure out what goes over here-- Green's theorem. Our f would look like this in this situation. f is f of xy is going to be equal to x squared minus y squared i plus 2xy j. We've seen this in multiple videos. You take the dot product of this with dr, you're going to get this thing right here. So this expression right here is our P of x y. And this expression right here is our Q of xy. So inside of here we're just going to apply Green's" + }, + { + "Q": "At 4:15 of the movie above suddenly you drop the unit vectors(i,j,k) from the result of bxc. Is it ok to do that? If it is ok, then I want to know why is it possible to do that.", + "A": "i, j, and k are (1, 0, 0), (0. 1, 0), and (0, 0, 1). a = (a1, a2, a3) = a1i + a2j + a3k (the resultant of the components of a).", + "video_name": "b7JTVLc_aMk", + "timestamps": [ + 255 + ], + "3min_transcript": "of three-by-threes. And we'll put b's x term, b's y coefficient, and b's z component. And then you do the same thing for the c, cx, cy, cz. And then this is going to be equal to-- so first you have the i component. So it's going to be the i component times b. So you ignore this column and this row. So bycz minus bzcy. So I'm just ignoring all of this. And I'm looking at this two-by-two over here, minus bzcy. And then we want to subtract the j component. Remember, we alternate signs when we take our determinant. Subtract that. And then we take out that column and that row, so it's going to be bxcz-- this is a little monotonous, bxcz minus bzcx. And then finally, plus the k component. OK, we're going to have bx times cy minus bycx. We just did the dot product, and now we want to take the-- oh, sorry, we just did the cross product. I don't want to get you confused. We just took the cross product of b and c. And now we need take the cross product of that with a, or the cross product of a with this thing right over here. Instead of rewriting the vector, let me just set up another matrix here. So let me write my i j k up here. And then let me write a's components. And then let's clean this up a little bit. We're just looking at-- no, I want to do that in black. Let's do this in black, so that we can kind of erase that. Now this is a minus j times that. So what I'm going to do is I'm going to get rid of the minus and the j, but I am going to rewrite this with the signs swapped. So if you swap the signs, it's actually bzcx minus bxcz. So let me delete everything else. So I just took the negative and I multiplied it by this. I hope I'm not making any careless mistakes here, so let me just check and make my brush size little bit bigger, so I can erase that a little more efficiently. And then we also want to get rid of that right over there. Now let me get my brush size back down to normal size." + }, + { + "Q": "When Sal says multiply b and c by the dot products of a and c and a and b at 13:31 (b(a dot c) - c(a dot b)) what does he mean by multiply?", + "A": "Scalar multiplication of the vector b with the scalar (a dot c) and so on", + "video_name": "b7JTVLc_aMk", + "timestamps": [ + 811 + ], + "3min_transcript": "plus by times j, plus bz times k. And then, from that, we're going to subtract all of this, a dot b. We're going to subtract a dot b times the exact same thing. And you're going to notice, this right here is the same thing as vector b. That is vector b. When you do it over here, you're going to get vector c. So I'll just write it over here. You're just going to get vector c. So just like that, we have a simplification for our triple product. I know it took us a long time to get here, but this is a simplification. It might not look like one, but computationally it is. It's easier to do. If I have-- I'll try to color-code it-- a cross b we just saw that this is going to be equivalent to-- and one way to think about it is, it's going to be, you take the first vector times the dot product of-- the first vector in this second dot product, the one that we have our parentheses around, the one we would have to do first-- you take your first vector there. So it's vector b. And you multiply that times the dot product of the other two vectors, so a dot c. And from that, you subtract the second vector multiplied by the dot product of the other two vectors, of a dot b. And we're done. This is our triple product expansion. Now, once again, this isn't something You could always, obviously, multiply it. You could actually do it by hand. You don't have to know this. But if you have really hairy vectors, or if this was some type of math competition, and sometimes it simplifies real fast when you reduce it to dot products, this is a useful thing to know, Lagrange's formula, or the triple product expansion." + }, + { + "Q": "Doesn't he mean cross product at 2:00?", + "A": "Yes, he meant cross product.", + "video_name": "b7JTVLc_aMk", + "timestamps": [ + 120 + ], + "3min_transcript": "What I want to do with this video is cover something called the triple product expansion-- or Lagrange's formula, sometimes. And it's really just a simplification of the cross product of three vectors, so if I take the cross product of a, and then b cross c. And what we're going to do is, we can express this really as sum and differences of dot products. Well, not just dot products-- dot products scaling different vectors. You're going to see what I mean. But it simplifies this expression a good bit, because cross products are hard to take. They're computationally intensive and, at least in my mind, they're confusing. Now this isn't something you have to know if you're going to be dealing with vectors, but it's useful to know. My motivation for actually doing this video is I saw some problems for the Indian Institute of Technology entrance exam that seems to expect that you know Lagrange's formula, or the triple product expansion. So let's see how we can simplify this. So to do that, let's start taking the cross product of b and c. just going to assume-- let's say I have vector a. That's going to be a, the x component of vector a times the unit of vector i plus the y component of vector a times the unit vector j plus the z component of vector a times unit vector k. And I could do the same things for b and c. So if I say b sub y, I'm talking about what's scaling the j component in the b vector. So let's first take this cross product over here. And if you've seen me take cross products, you know that I like to do these little determinants. Let me just take it over here. So b cross c is going to be equal to the determinant. And I put an i, j, k up here. This is actually the definition of the cross product, so no proof necessary to show you why this is true. This is just one way to remember the dot product, of three-by-threes. And we'll put b's x term, b's y coefficient, and b's z component. And then you do the same thing for the c, cx, cy, cz. And then this is going to be equal to-- so first you have the i component. So it's going to be the i component times b. So you ignore this column and this row. So bycz minus bzcy. So I'm just ignoring all of this. And I'm looking at this two-by-two over here, minus bzcy. And then we want to subtract the j component. Remember, we alternate signs when we take our determinant. Subtract that. And then we take out that column and that row, so it's going to be bxcz-- this is a little monotonous," + }, + { + "Q": "At 20:00 couldn't the graph left of zero be the right half of upward concavity ( / )?", + "A": "No, because slope is zero at 0. Isn t it?", + "video_name": "hIgnece9ins", + "timestamps": [ + 1200 + ], + "3min_transcript": "" + }, + { + "Q": "I don't understand why at 8:45 we started looking at x > 2/3.", + "A": "He was attempting to determine concavity. x=2/3 is a point of inflection. It is where the graph changes concavity. You need to determine the sign of the 2nd derivative at points both less than and greater than your point of inflection to determine the concavity of the function along those intervals.", + "video_name": "hIgnece9ins", + "timestamps": [ + 525 + ], + "3min_transcript": "" + }, + { + "Q": "I'm wondering about the result at the end of the video (8:07) : x^3y+1/2(x^2y^2) = c\nIf that were a diffrential equation describing a physic phenomenon, how would you turn this solution into a function you can actually use to describe the physical system you took the diffrential equation from ?\nAnother way of saying this would be : How do you turn x^3y+1/2(x^2y^2) into\ny = some x business ?", + "A": "I believe you can use the quadratic formula to get the explicit solution in terms of y. Just move the c to the left side so you have ay^2 + ay + c = 0 ps I thought you were talking about psychic phenomena for a minute.", + "video_name": "0NyeDUhKwBE", + "timestamps": [ + 487 + ], + "3min_transcript": "I just took the antiderivative of both sides. So, a solution to the differential equation is psi is equal to c. So psi is equal to x to the third y plus 1/2 x squared y squared. And we could have said plus c here, but we know the solution is that psi is equal to c, so we'll just write that there. I could have written a plus c here, but then you have a plus c here. You have another constant there. And you can just subtract them from both sides. And they just merge into another arbitrary constant. But anyway, there we have it. We had a differential equation that, at least superficially, looked exact. It looked exact, but then, when we tested the exactness of it, it was not exact. But we multiplied it by an integrating factor. And in the previous video, we figured out that a possible integrating factor is that we could just multiply both sides by x. And when we did that, we tested it. And true enough, it was exact. would exist where the derivative of psi with respect to x would be equal to this entire expression. So we could rewrite our differential equation like this. And we'd know that a solution is psi is equal to c. And to solve for psi, we just say, OK, the partial derivative of psi with respect to x is going to be this thing. Antiderivative of both sides, and there's some constant h of y-- not constant, there's some function of y-- h of y that we might have lost when we took the partial with respect to x. So to figure that out, we take this expression. Take the partial with respect to y, and set that equal to our N expression. And by doing that, we figured out that that function of y is really just some constant. And we could have written that here. We could have written that plus c. We could call that c1 or something. But we know that the solution of our original differential equation is psi is equal to c. So the solution of our differential equation is psi x to the third y plus 1/2 x squared y We could have had this plus c1 here, then subtracted both sides. But I think I've said it so many times that you understand, why if h of y is just a c, you can kind just ignore it. Anyway, that's all for now, and I will see you in the next video. You now know a little bit about integrating factors. See you soon." + }, + { + "Q": "At 0:44, he said it would have been a function of y that is integrating factor. i tried u(y) instead of u(x) but I can't solve for u(y). I just curious that is it possible to use u(y) for this question too? And how do we know which one would we use? u(x) or u(y), judging for what or trial and error?", + "A": "I have not tried to solve for the integrating factor of u(y) but i do know that there may or may not be an integrating factor for u(y). Like you said it is kind of trial and error. All of these exact equations that need an integrating factor COULD have a u(x), u(y), u(x,y) or any combination of them but it will be up to you to figure out which one it has.", + "video_name": "0NyeDUhKwBE", + "timestamps": [ + 44 + ], + "3min_transcript": "And, in the last video, we had this differential equation. And it at least looked like it could be exact. But when we took the partial derivative of this expression, which we could call M with respect to y, it was different than the partial derivative of this expression, which is N in the exact differential equations world. It was different than N with respect to x. And we said, oh boy, it's not exact. But we said, what if we could multiply both sides of this equation by some function that would make it exact? And we called that mu. And in the last video, we actually solved for mu. We said, well, if we multiply both sides of this equation by mu of x is equal to x, it should make this into an exact differential equation. It's important to note, there might have been a function of y that if I multiplied by both sides it would also make it exact. There might have been a function of x and y that would have done the trick. But our whole goal is just to make this exact. It doesn't matter which one we pick, which integrating factor-- this is called the integrating factor-- which integrating factor we pick. Let's solve the problem. Let's multiply both sides of this equation by mu, and mu of x is just x. We multiply both sides by x. So see, if you multiply this term by x, you get 3x squared y plus xy squared, we're multiplying these terms by x now, plus x to the third plus x squared y, y prime is equal to 0. Well now, first of all, just as a reality check, let's make sure that this is now an exact equation. So what's the partial of this expression, or this kind of sub-function, with respect to y? Well, it's 3x squared, that's just kind of a constant coefficient of y, plus 2xy, that's the partial with respect to y of that expression. So we get 3x squared plus 2xy. And there we have it. The partial of this with respect to y is equal to the partial of this with respect to N. So we now have an exact equation whose solution should be the same as this. All we did is we multiplied both sides of this equation by x. So it really shouldn't change the solution of that equation, or that differential equation. So it's exact. Let's solve it. So how do we do that? Well, what we say is, since we've shown this exact, we know that there's some function psi where the partial derivative of psi with respect to x is equal to this expression right here. So it's equal to 3x squared y plus xy squared. Let's take the antiderivative of both sides with respect to x, and we'll get psi is equal to what?" + }, + { + "Q": "At 7:05, he begins to explain a simpler way to find the value of cesium-137 in the sample after 150 days. Though, because 150 is divisible by 30, isn't there an even simpler way? You could essentially just divide 8 in half 5 times and reach the same answer of .25. Is this valid?", + "A": "yes, that is how I learned it in science class. Just don t forget to include year 0!!", + "video_name": "polop-89aIA", + "timestamps": [ + 425 + ], + "3min_transcript": "And then, they finally say-- how many becquerels of caesium-137 remain in our sample 150 days after its release in the soil? Use the rounded value of r, and round this number to the nearest hundredth. So just to be clear, we already know c and r. We know that the amount of caesium-137 in becquerels-- as a function of time in days-- is going to be equal to 8 times 0.977 to the t power-- where t is the number of days that have passed. And they're essentially saying, well, how much do we have left after 150 days? So they want us to calculate what is A of 150? Well, that's going to be 8 times 0.977 to the 150th power. So let's calculate that. So it's going to be 8 times-- and they tell us to use our rounded value of r, not the exact value of r. So it's going to be 8 times 0.977 to the 150th power. And they want us to round to the nearest hundredth. 0.24. 0.24 becquerels is kind of the radioactivity level of the caesium-137 that we have left over. Now, one interesting thing is they asked us to use the rounded value of r. So we used the rounded value of r. Because this right over here is a multiple of 30, you could actually-- in not too difficult of a way-- find out the exact value that's left over. And actually, you don't even need a calculator for it. I encourage you to pause your video Find the exact value. Well, instead of writing 0.977, let's write A of t as being equal to 8 times our r. This is an approximate value for r. If we wanted to be a little more exact, we can say that our r is one half to the 1/30th power. And we're going to raise that to the t power. Or we could say A of t is equal to 8 times one half to the t/30 power. If we raise something to an exponent and then raise that to an exponent, we can take the product of those exponents. So that's one half to the t/30 power. Let me actually do that in another color. Let me do that in yellow. So that's 8 times one half to the t/30 power. And actually, I don't need this parentheses right over here. This is another way to describe A of t." + }, + { + "Q": "hello there its your friendly neighborhood precal student here again\n\ni have a synthetic division question in my summer homework (yipee!)\nthe denominator is 3x-2\nwould i still put positive 2 at the same place where Sal puts the 3 at 2:00 in the video\nor does the 3x affect this?\n\nThanks for your help :)", + "A": "It depends on the question. You look for the highest polynomial which you want to subtract away. If you re dividing the same polynomial, but now by 3x-2, yes, you would still put a 2 down there, but note that you multiplied it by 3/2. Remember to upvote good questions and helpful answers", + "video_name": "3Ee_huKclEQ", + "timestamps": [ + 120 + ], + "3min_transcript": "Let's do another synthetic division example. And in another video, we actually have the why this works relative to algebraic long division. But here it's going to be another just, let's go through the process of it just so that you get comfortable with it. And now is a good chance to give it a shot, to actually try to simplify this rational expression. So let's think about this step by step. So the first thing I want to do is write all of the coefficients of the numerator. So I have a 2. Oh, I have to be careful here. Because the 2 is the coefficient for x to the fifth, I have no x to the fourth term. Let me start over. So I have the 2 from 2x to the fifth. And then I have no x to the fourth. So it's really 0x to the fourth. So I'll put a 0 as the coefficient for the x to the fourth term. And then I have a negative 1 times x to the third. And then I have a positive 3 times x squared. And then I have a constant term, or zero degree term, of 7. I just have a positive 7. And now let me just draw my little funky synthetic division operator-looking symbol. And remember, the type of synthetic division we're doing, it only applies when we are dividing by an x plus or minus something. There's a slightly different process you would have to do if it was 3x or if was negative 1x or if it was 5x squared. This only works when we have x plus or minus something. In this case we have x minus 3. So we have the negative 3 here. And the process we show-- there's other ways of doing it-- is you take the negative of this. So the negative of negative 3 is positive 3. And now we're ready to perform our synthetic division. So we'll bring down this 2 and then multiply 2 time 3 gives us 6. 0 plus 6 is 6. And then we multiply that times the 3, and we get positive 18. Negative 1 plus 18 is 17. Multiply that times the 3. 17 times 3 is 51. 3 plus 51 is 54. Multiply that times 3. The numbers are getting kind of large now. So that's going to be what? 50 times 3 is 150. 4 times 3 is 12. So this is going to be 162. Negative 2 plus 162 is 160." + }, + { + "Q": "Wait at 4:57, Sal doesn't put the x^5 term. He puts the highest degree as x^4. Am I missing something? I thought that 487 would be the constant and that perhaps that meant the equation was even or something. What happened to the x^5 term?", + "A": "Remember, when you are dividing with terms containing variables that instead of adding the exponents (i.e., like you do when you re multiplying them), you subtract the exponents. So in the example (e.g., 2x^5/x), the exponent subtraction would be 5-1=4 or your x^4 term, just as in the next term : x^3/x would yield your 3-1=2 or x^2 term, etc. Hope this helps.", + "video_name": "3Ee_huKclEQ", + "timestamps": [ + 297 + ], + "3min_transcript": "2 time 3 gives us 6. 0 plus 6 is 6. And then we multiply that times the 3, and we get positive 18. Negative 1 plus 18 is 17. Multiply that times the 3. 17 times 3 is 51. 3 plus 51 is 54. Multiply that times 3. The numbers are getting kind of large now. So that's going to be what? 50 times 3 is 150. 4 times 3 is 12. So this is going to be 162. Negative 2 plus 162 is 160. And you add 480 to 7, and you get 487. And you can think of it, I only have one term or one number to the left-hand side of this bar here. Or I'm just doing the standard, traditional x plus or minus something version of synthetic division, I should say. So I can separate this out, and now I've essentially gotten my answer. And it looks like voodoo, and it kind of is voodoo. And that's why I don't like to do it, because you're just memorizing an algorithm. But there are other videos why we explain why. And it can be fast and convenient and paper saving very often, like you see right here. But then we have our final answer. It's going to be-- and let me work backwards. So I'll start with our remainder. So our remainder is 487. And it's going to be 487 over x minus 3. And so you're going to have plus 160 plus 487 over x minus 3. Now this is our x term. So it's going to be 54x plus all of this. This is going to be our x squared term. So this is going to be 17x squared plus 54x plus 160 and all of that. Then this is going to be x to the third term. So this is going to be 6x to the third plus all of that. And then finally, this is our x to the fourth term-- 2x to the fourth. And let me erase this. So then I have my x to the fourth term. So it is 2x to the fourth. And we are done. This thing simplifies to this right over here. And I encourage you to verify it with traditional algebraic long division." + }, + { + "Q": "At 4:07 why is it h over two? I am still confused on how he exactly got that.", + "A": "The area of a circle is \u00e1\u00b4\u00a8r\u00c2\u00b2, where r is the radius, right? We are told the diameter of the circle in this problems is h, right? What is the relationship between the radius and diameter of a circle? 2r=D, here D=h so 2r=h which means that, the radius r = h/2. You may want to review similar triangles.", + "video_name": "Xe6YlrCgkIo", + "timestamps": [ + 247 + ], + "3min_transcript": "of that relationship, possibly using the chain rule, to come up with a relationship between the rate at which the volume is changing and the rate at which the height is changing. So let's try to do it step by step. So first of all, can we come up with a relationship between the volume and the height at any given moment? Well we have also been given the formula for the volume of a cone right over here. The volume of a cone is 1/3 times the area of the base of the cone, times the height. And we won't prove it here, although we could prove it later on. Especially when we start doing solids of revolutions within in integral calculus. But we'll just take it on faith right now, that this is how we can figure out the volume of a cone. So given this can we figure out volume-- can we figure out an expression that relates volume to the height of the cone? Well we could say that volume-- and I'll do it in this blue color-- the volume of water is what we really care about. The volume of water is going to be equal to 1/3 times the area of the surface of the water-- So times h. So how can we figure out the area of the water surface, preferably in terms of h? Well we see right over here, the diameter across the top of the cone is 4 centimeters. And the height of the whole cup is 4 centimeters. And so that ratio is going to be true of any-- at any depth of water. It's always going to have the same ratio between the diameter across the top and the height. Because these are lines right over here. So at any given point, the ratio between this and this is going to be the same. So at any given point, the diameter across the surface of the water-- if the depth is h, the diameter across the surface of the water is also going to be h. And so from that we can figure out what are the radius is going to be. The radius is going to be h over 2. is going to be pi r squared, pi times the radius squared. h over 2 squared. That's the area of the surface of the water. And of course we still have the 1/3 out here. And we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1/3 times pi h squared over 4 times another h, which is equal to-- we have pi, h to the third power over 12. So that is our volume. Now what we want to do is relate the volume, how fast the volume is changing with respect to time, and how fast the height is changing with respect to time. So we care with respect to time, since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time." + }, + { + "Q": "At 5:20, what is he talking about when he says \"This business\"?", + "A": "Occasionally in these videos Sal saves a little time by using the words this business as a way to refer to some expression he s manipulating on the screen. In this video, as he says that, Sal is putting brackets around the expression \u00cf\u0080h^3/12, so that s what he means here by this business.", + "video_name": "Xe6YlrCgkIo", + "timestamps": [ + 320 + ], + "3min_transcript": "So times h. So how can we figure out the area of the water surface, preferably in terms of h? Well we see right over here, the diameter across the top of the cone is 4 centimeters. And the height of the whole cup is 4 centimeters. And so that ratio is going to be true of any-- at any depth of water. It's always going to have the same ratio between the diameter across the top and the height. Because these are lines right over here. So at any given point, the ratio between this and this is going to be the same. So at any given point, the diameter across the surface of the water-- if the depth is h, the diameter across the surface of the water is also going to be h. And so from that we can figure out what are the radius is going to be. The radius is going to be h over 2. is going to be pi r squared, pi times the radius squared. h over 2 squared. That's the area of the surface of the water. And of course we still have the 1/3 out here. And we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1/3 times pi h squared over 4 times another h, which is equal to-- we have pi, h to the third power over 12. So that is our volume. Now what we want to do is relate the volume, how fast the volume is changing with respect to time, and how fast the height is changing with respect to time. So we care with respect to time, since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time. let me move this over. Let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well the derivative with respect to time of our volume, we could just rewrite that as dV dt, this thing right over here. This is dV dt, and this is going to be equal to-- well we could take the constants out of this-- this is going to be equal to pi over 12 times the derivative with respect to t of h, of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time." + }, + { + "Q": "At 5:50 how did he get the derivative to be just pi/12?", + "A": "That s not what happened here. For convenience, Sal moved the constant (pi/12) outside the derivative. He had the derivative of (pi*h^3)/12, which is the same as (pi/12)*h^3, and he rewrote it as (pi/12) times the derivative of h^3.", + "video_name": "Xe6YlrCgkIo", + "timestamps": [ + 350 + ], + "3min_transcript": "is going to be pi r squared, pi times the radius squared. h over 2 squared. That's the area of the surface of the water. And of course we still have the 1/3 out here. And we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1/3 times pi h squared over 4 times another h, which is equal to-- we have pi, h to the third power over 12. So that is our volume. Now what we want to do is relate the volume, how fast the volume is changing with respect to time, and how fast the height is changing with respect to time. So we care with respect to time, since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time. let me move this over. Let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well the derivative with respect to time of our volume, we could just rewrite that as dV dt, this thing right over here. This is dV dt, and this is going to be equal to-- well we could take the constants out of this-- this is going to be equal to pi over 12 times the derivative with respect to t of h, of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time. As time goes on, the height will change. Because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power. Just to make it clear that this is a function of t. h of t to the third power. Now what is the derivative with respect to t, of h of t to the third power. Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule. The chain rule tells us-- let me rewrite everything else. dV with respect to t, is going to be equal to pi over 12, times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something. So that's going to be-- let me write this in a different color, maybe in orange-- so that's going to be 3 times our something squared," + }, + { + "Q": "what is the meaning of 'inx' at 1:31.", + "A": "That is an l, not an i. The ln stands for the natural logarithm, or the logarithm with base e. The number e is called Euler s number (pronounced oiler ), or the natural base. It is the base of an exponential that is its own derivative, which is a handy thing in calculus and differential equations.", + "video_name": "OkFdDqW9xxM", + "timestamps": [ + 91 + ], + "3min_transcript": "Use the change of base formula to find log base 5 of 100 to the nearest thousandth. So the change of base formula is a useful formula, especially when you're going to use a calculator, because most calculators don't allow you to arbitrarily change the base of your logarithm. They have functions for log base e, which is a natural logarithm, and log base 10. So you generally have to change your base. And that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense, or how we can derive it. So the change of base formula just tells us that log-- let me do some colors here-- log base a of b is the exact same thing as log base x, where x is an arbitrary base of b, over log base, that same base, base x over a. is that we can change our base. Here are our bases, a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying-- this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x, and e is obviously the number 2.71, keeps going on and on and on forever. Now let's apply it to this problem. We need to figure out the logarithm-- and I'll use colors-- base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100-- what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2. So it simplifies to 2 over log base 10 of 5. And we can now use our calculator, because the log function on a calculator is log base 10. So let's get our calculator out. We want to clear this. 2 divided by-- When someone just writes log, they mean base 10. If they press LN, that means base e. So log without any other information is log base 10. So this is log base 10 of 5 is equal to 2 point--" + }, + { + "Q": "At 6:08 in the video, why does the \"log a\" become \"log x\"?\nAlso, why is the \"a\" taken away from the subscript of the log?", + "A": "It is not taken away, but the log form is re-written into exponential form, where a is the base and the exponent is written in superscript. log a does not become log x Sal takes the log base x of both sides (of the exponential equation a^c=b) because it is essentially the same thing by comparison and he is not changing the value of the equation.", + "video_name": "OkFdDqW9xxM", + "timestamps": [ + 368 + ], + "3min_transcript": "think about why this property, why this thing right over here makes sense. So if I write log base a-- I'll try to be fair to the colors-- log base a of b. Let's say I set that to be equal to some number. Let's call that equal to c, or I could call it e for-- Well, I'll say that's equal to c. So that means that a to the c-th power is equal to b. This is an exponential way of writing this truth. This is a logarithmic way of writing this truth. This is equal to b. Now, we can take the logarithm of any base of both sides of this. Anything you do, if you say 10 to the what power equals this? because these two things are equal to each other. So let's take the same logarithm of both sides of this, the logarithm with the same base. And I'll actually do log base x to prove the general case, So I'm going to take log of base x of both sides of this. So this is log base x of a to the c power-- I try to be faithful to the colors-- is equal to log base x of b. And let me close it off with orange, as well. And we know from our logarithm properties, log of a to the c is the same thing as c times the logarithm of whatever base we are of a. Let me put-- I can just write a b, right over there. And if we wanted to solve for c, you just divide both sides by log base x of a. So you would get c is equal to-- and I'll stick to the color-- so it's log base x of b, which is this, over log base x of a. And this was what c was. c was log base a of b. It's equal to log base a of b. Let me write it this way. Let me write it-- Well, let me do the original color codes just so it becomes very clear what I'm doing. I think you know where this is going, but I want to be fair to the colors. So c is equal to log base x of b over-- let me scroll down a little bit-- log base x, dividing both sides by that, of a." + }, + { + "Q": "At 0:49, he says x is an arbitory base. What is that?", + "A": "I m assuming that he means that x is just a letter. Your variable could be almost any letter of the alphabet and it still means the same thing.", + "video_name": "OkFdDqW9xxM", + "timestamps": [ + 49 + ], + "3min_transcript": "Use the change of base formula to find log base 5 of 100 to the nearest thousandth. So the change of base formula is a useful formula, especially when you're going to use a calculator, because most calculators don't allow you to arbitrarily change the base of your logarithm. They have functions for log base e, which is a natural logarithm, and log base 10. So you generally have to change your base. And that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense, or how we can derive it. So the change of base formula just tells us that log-- let me do some colors here-- log base a of b is the exact same thing as log base x, where x is an arbitrary base of b, over log base, that same base, base x over a. is that we can change our base. Here are our bases, a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying-- this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x, and e is obviously the number 2.71, keeps going on and on and on forever. Now let's apply it to this problem. We need to figure out the logarithm-- and I'll use colors-- base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100-- what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2. So it simplifies to 2 over log base 10 of 5. And we can now use our calculator, because the log function on a calculator is log base 10. So let's get our calculator out. We want to clear this. 2 divided by-- When someone just writes log, they mean base 10. If they press LN, that means base e. So log without any other information is log base 10. So this is log base 10 of 5 is equal to 2 point--" + }, + { + "Q": "At 0:16, what is theoretical probability?", + "A": "Theoretical probability is basically calculating a probability for something without doing the actual experiment. It is what should happen in theory, thus theoretical probability.", + "video_name": "RdehfQJ8i_0", + "timestamps": [ + 16 + ], + "3min_transcript": "- [Voiceover] There's a lot of times, there's a lot of situations in which we're studying something pretty straightforward and we can find an exact theoretical probability. So what am I talking about? Just let me write that down. Theoretical probabiity. Well, maybe the simplest example, or one of the simplest examples is if you're flipping a coin. And let's say in theory you're flipping a completely fair coin and you're flipping it in a way that is completely fair. Well, there you know you have two outcomes. Either heads will be on top or tails will be on top. So theoretically you say, \"well, look, \"if I want to figure out the probability \"of getting a heads, in theory I have two \"equally likely possibilities, and heads \"is one of those two equally likely possibilities.\" So you have a 1/2 probability. Once again, if in theory the coin is definitely fair, it's a fair coin and it's flipped in a very fair way, then this is true. You have a 1/2 probability. A fair six-sided die is going to have six possible outcomes: one, two, three, four, five and six. And if you said \"what is the probability of getting \"a result that is greater than or equal to three?\" Well, we have six equally likely possibilities. You see them there. In theory, if they're all equally likely, four of these possibilities meet our constraint of being greater than or equal to three. We have four out of the six of these possibilities meet our constraints. So we have a 2/3, 4/6 is the same thing as 2/3, probability of it happening. Now these are for simple things, like die or flipping a coin. And if you have fancy computers or spreadsheets you can even say \"hey, \"I'm gonna flip a coin a bunch of times \"and do all the combinatorics\" and all that. But there are things that are even beyond what a computer can find the exact theoretical probability for. Let's say you are playing a game, say football, American football, of scoring a certain number of points. Well that isn't very simple because that's going to involve what human beings are doing. Minds are very unpredictable, how people will respond to things. The weather might get involved. Someone might fall sick. The ball might be wet, or just how the ball might interact with some player's jersey. Who knows what might actually result in the score being one point this way, or seven points this way, or seven points that way. So for situations like that, it makes more sense to think more in terms of experimental probability. In experimental probability, we're really just trying to get an estimate of something happening, based on data and experience that we've had in the past. For example, let's say you had data from your football team and it's many games into the season." + }, + { + "Q": "At 3:20, why is it 1/15", + "A": "To change 16/15 into a mixed number, you divide. 15 goes into 16 once (that s the whole number) Subtract 16 - 15 = 1. We have a remainder of 1 that becomes the new numerator Thus, you get 1 1/15 Hope this helps.", + "video_name": "bcCLKACsYJ0", + "timestamps": [ + 200 + ], + "3min_transcript": "as something over 30. So nine over 10. How would I write that as something over 30? Well I multiply the denominator, I'm multiplying the denominator by three. So I've just multiplied the denominator by three. So if I don't want to change the value of the fraction, I have to do the same thing to the numerator. I have to multiply that by three as well because now I'm just multiplying the numerator by three and the denominator by three, and that doesn't change the value of the fraction. So nine times three is 27. So once again, 9/10 and 27/30 represent the same number. I've just written it now with a denominator of 30, and that's useful because I can also write 1/6 with a denominator of 30. Let's do that. So 1/6 is what over 30? I encourage you to pause the video and try to think about it. So what did we do go from six to 30? We had to multiply by five. So if we multiply the denominator by five, so one times five, one times five is five. So 9/10 is the same thing as 27/30, and 1/6 is the same thing as 5/30. And now we can add, now we can add and it's fairly straightforward. We have a certain number of 30ths, added to another number of 30ths, so 27/30 + 5/30, well that's going to be 27, that's going to be 27 plus five, plus five, plus 5/30, plus 5/30, which of course going to be equal to 32/30. 32 over 30, and if we want, we could try to reduce this fraction. We have a common factor of 32 and 30, they're both divisible by two. So if we divide the numerator and the denominator by two, denominator divided by two is 15. So, this is the same thing as 16/15, and if I wanted to write this as a mixed number, 15 goes into 16 one time with a remainder one. So this is the same thing as 1 1/15. Let's do another example. Let's say that we wanted to add, we wanted to add 1/2 to to 11/12, to 11 over 12. And I encourage you to pause the video and see if you could work this out. Well like we saw before, we wanna find a common denominator. If these had the same denominator, we could just add them immediately, but we wanna find a common denominator because right now they're not the same. Well what we wanna find is a multiple, a common multiple of two and 12, and ideally we'll find the lowest common multiple of two and 12, and just like we did before, let's start with the larger of the two numbers, 12." + }, + { + "Q": "At 9:22 How do you know to put the number before the decimal, or at 5:06 the zero?", + "A": "It depends if the denominator divides before the decimal point or after. For instance, if you had 31/15, it would come out to 2.1. I hope that this was helpful! :)", + "video_name": "Gn2pdkvdbGQ", + "timestamps": [ + 562, + 306 + ], + "3min_transcript": "That's the same thing as 35/1,000. And you're probably saying, Sal, how did you know it's 35/1000? Well because we went to 3-- this is the 10's place. Tenths not 10's. This is hundreths. This is the thousandths place. So we went to 3 decimals of significance. So this is 35 thousandths. If the decimal was let's say, if it was 0.030. There's a couple of ways we could say this. Well, we could say, oh well we got to 3-- we went to the thousandths Place. So this is the same thing as 30/1,000. We could have also said, well, 0.030 is the same thing as 0.03 because this 0 really doesn't add any value. So this is the same thing as 3/100. So let me ask you, are these two the same? Sure they are. If we divide both the numerator and the denominator of both of these expressions by 10 we get 3/100. Let's go back to this case. Are we done with this? Is 35/1,000-- I mean, it's right. That is a fraction. 35/1,000. But if we wanted to simplify it even more looks like we could divide both the numerator and the denominator by 5. And then, just to get it into simplest form, that equals 7/200. And if we wanted to convert 7/200 into a decimal using the technique we just did, so we would do 200 goes into 7 and figure it out. We should get 0.035. I'll leave that up to you as an exercise. to convert a fraction into a decimal and maybe vice versa. And if you don't, just do some of the practices. And I will also try to record another module on this or another presentation. Have fun with the exercises." + }, + { + "Q": "In 1:40 how come we don't multiply by -1 to make -3p a positive and flip the inequality? I've seen it in other problems before.", + "A": "You can either multiply with -1, divide by -3, or just swap the numbers (the left goes to the right and vice versa), do as you like.", + "video_name": "0YErxSShF0A", + "timestamps": [ + 100 + ], + "3min_transcript": "Solve for z. 5z plus 7 is less than 27 or negative 3z is less than or equal to 18. So this is a compound inequality. We have two conditions here. So z can satisfy this or z can satisfy this over here. So let's just solve each of these inequalities. And just know that z can satisfy either of them. So let's just look at this. So if we look at just this one over here, we have 5z plus 7 is less than 27. Let's isolate the z's on the left-hand side. So let's subtract 7 from both sides to get rid of this 7 on the left-hand side. And so our left-hand side is just going to be 5z. Plus 7, minus 7-- those cancel out. 5z is less than 27 minus 7, is 20. So we have 5z is less than 20. Now we can divide both sides of this inequality by 5. And we don't have to swap the inequality because we're And so we get z is less than 20/5. z is less than 4. Now, this was only one of the conditions. Let's [? look at ?] the other one over here. We have negative 3z is less than or equal to 18. Now, to isolate the z, we could just divide both sides of this inequality by negative 3. But remember, when you divide or multiply both sides of an inequality by a negative number, you have to swap the inequality. So we could write negative 3z. We're going to divide it by negative 3. And then you have 18. We're going to divide it by negative 3. But we're going to swap the inequality. So the less than or equal will become greater than or equal to. And so these guys cancel out. Negative 3 divided by negative 3 is 1. So we have z is greater than or equal to 18 over negative 3 is negative 6. And remember, it's this constraint or this constraint. And this constraint right over here boils down to this. So our solution set-- z is less than 4 or z is greater than or equal to negative 6. So let me make this clear. Let me rewrite it. So z could be less than 4 or z is greater than or equal to negative 6. It can satisfy either one of these. And this is kind of interesting here. Let's plot these. So there's a number line right over there. Let's say that 0 is over here. We have 1, 2, 3, 4 is right over there. And then negative 6. We have 1, 2, 3, 4, 5, 6. That's negative 6 over there. Now, let's think about z being less than 4. We would put a circle around 4, since we're not including 4. And it'd be everything less than 4." + }, + { + "Q": "at 1:48, i didnt under stand why -32/-3 was crossed out can some one\nexplain it to me", + "A": "Watch the video more carefully. It was not -32/-3, it was -3*z/-3 since you have a -3 on the numerator and denominator they cancel out, so that just leaves you with z.", + "video_name": "0YErxSShF0A", + "timestamps": [ + 108 + ], + "3min_transcript": "Solve for z. 5z plus 7 is less than 27 or negative 3z is less than or equal to 18. So this is a compound inequality. We have two conditions here. So z can satisfy this or z can satisfy this over here. So let's just solve each of these inequalities. And just know that z can satisfy either of them. So let's just look at this. So if we look at just this one over here, we have 5z plus 7 is less than 27. Let's isolate the z's on the left-hand side. So let's subtract 7 from both sides to get rid of this 7 on the left-hand side. And so our left-hand side is just going to be 5z. Plus 7, minus 7-- those cancel out. 5z is less than 27 minus 7, is 20. So we have 5z is less than 20. Now we can divide both sides of this inequality by 5. And we don't have to swap the inequality because we're And so we get z is less than 20/5. z is less than 4. Now, this was only one of the conditions. Let's [? look at ?] the other one over here. We have negative 3z is less than or equal to 18. Now, to isolate the z, we could just divide both sides of this inequality by negative 3. But remember, when you divide or multiply both sides of an inequality by a negative number, you have to swap the inequality. So we could write negative 3z. We're going to divide it by negative 3. And then you have 18. We're going to divide it by negative 3. But we're going to swap the inequality. So the less than or equal will become greater than or equal to. And so these guys cancel out. Negative 3 divided by negative 3 is 1. So we have z is greater than or equal to 18 over negative 3 is negative 6. And remember, it's this constraint or this constraint. And this constraint right over here boils down to this. So our solution set-- z is less than 4 or z is greater than or equal to negative 6. So let me make this clear. Let me rewrite it. So z could be less than 4 or z is greater than or equal to negative 6. It can satisfy either one of these. And this is kind of interesting here. Let's plot these. So there's a number line right over there. Let's say that 0 is over here. We have 1, 2, 3, 4 is right over there. And then negative 6. We have 1, 2, 3, 4, 5, 6. That's negative 6 over there. Now, let's think about z being less than 4. We would put a circle around 4, since we're not including 4. And it'd be everything less than 4." + }, + { + "Q": "At 2:21, could you explain how to determine the greatest integer between an interval for example [0,1]?", + "A": "firstly, integers are all positive and negative whole numbers including 0: ...-3,-2,-1,0,1,2,3... so what are the integers between the interval [0,1]? 0 and 1 which is greater 0 or 1? 1 So the answer to your question is 1.", + "video_name": "CZdziIlYIfI", + "timestamps": [ + 141 + ], + "3min_transcript": "For any real number x, let brackets around x denote the largest integer less than or equal to x, often known as the greatest integer function. Let f be a real valued function defined on the interval negative 10 to 10, including the boundaries by f of x is equal to x minus the greatest integer of x, if the greatest integer of x is odd, and 1 plus the greatest integer of x minus x, if the greatest integer of x is even. Then the value of pi squared over 10 times a definite integral from negative 10 to 10 of f of x cosine of pi of x dx is-- so before even try to attempt to evaluate this integral, let's see if we can at least visualize this function, f of x, right over here. So let's do our best to visualize it. So let me draw my x-axis. And let me draw my y-axis. So let me draw my y-axis. And then let's think about what this function will look like. this is x is equal to 1, x is equal to 2, x is equal to 3. We could go down to negative 1, negative 2. We could just keep going, if we like. Hopefully we'll see some type of pattern, because it seems to change from odd to even. So between 0 and 1, what is the greatest integer of x? So let me just write it over here. So between 0 and 1, until you get to 1-- so maybe I should do this-- from including 0 until 1, the greatest integer of x is equal to 0. If I'm at 0.5, the greatest integer below 0.5 is 0. As I go from 1 to 2, this brackets around x It's the greatest integer. If I'm at 1.9, the greatest integer is 1. And then if I go to above from between 2 and 3, then the greatest integer is going to be 2. If I'm at 2.5, greatest integer is going to be 2. So with that, let's try to at least draw this function over these intervals. So between 0 and 1, the greatest integer is 0. 0 we can consider to be even. 0 is even, especially if we're alternating. 1 is odd, 2 is even, 3 is odd. So 0 is even. So we would look at this circumstance right over here, if x is even. And then over this time frame or over this part of the x-axis, the greatest integer of x is just 0. So the equation or the line or the function is just going to be 1 minus x over this interval, because the greatest integer is 0. So 1 minus x will look like this. If this is 1, 1 minus x just goes down like that." + }, + { + "Q": "At 7:24, I still do not quite understand why the integral of f(x) between 0 and 1 is the same as the integral of f(x) between 1 and 2? I mean I can kind of visualise it, but how can I prove this is true? Thanks a lot:)", + "A": "That is the case because of symmetry. The area between 0 and 1 is the mirror image (along y = 1) of the area between 1 and 2. The easiest way to prove it is to calculate both integrals and see that the result is the same. Int from 0 to 1 (1-x)*cos(pi*x) = Int from 1 to 2 (x-1)*cos(pi*x) = 2/pi^2", + "video_name": "CZdziIlYIfI", + "timestamps": [ + 444 + ], + "3min_transcript": "Don't want you to get that wrong. Cosine pi 0 is cosine of 0. So that's 1. Cosine of pi is negative 1. So when x is equal to 1, this becomes cosine of pi. So then the value of the function is negative 1. It'll be over here. And then cosine of 2 pi, 2 times pi, is then 1 again. So it'll look like this. This is at 1/2. When you put it over here, it'll become pi over 2. Cosine of pi over 2 is 0. So it'll look like this. Let me draw it as neatly as possible. So it will look like this. Cosine, and then it'll keep doing that, and then it'll go like this. So it is also periodic. So if we wanted to figure out the integral of the product from negative 10 to 10, can we simplify that? And it looks like it would just be, because we have this interval, let's look at this interval over here. Let's look at just from 0 to 1. So just from 0 to 1, we're going to take this function and take the product of this cosine times essentially 1 minus x, and then find the area under that curve, whatever it might be. Then when we go from 1 to 2, when we take the product of this and x minus 1, it's actually going to be the same area, because these two, going from 0 to 1 and going from 1 to 2, it's completely symmetric. You can flip it over this line of symmetry, and both functions are completely symmetric. So you're going to have the same area when you take their product. So what we see is, over every interval, from 2 to 3 is clearly the same thing as the integral from 0 to 1. Both functions look identical over that interval. But it will also be the same as going from 1 to 2, because it's completely symmetric. When you take the products of the function, that function will be completely symmetric around this axis. So the integral from here to here will be the same as the integral from there to there. So with that said, we can rewrite this thing over here. So what we want to evaluate, pi squared over 10 times the integral from negative 10 to 10 of f of x cosine of pi x, using the logic we just talked about. This is going to be the same thing as being equal to pi squared over 10 times the integral from 0 to 1, but 20 times that, because we have 20 integers between negative 10 and 10." + }, + { + "Q": "at 10:50 it says du is equal to dx. why he doesn't write 1? and at 12:02 it write again dx, instead of 1.\nCould anyone make some sense of this?\nI guess it is related to the dx at the end of the \"\u00e2\u008c\u00a0f(x) dx \" formula, but still not sure what that dx represent, and why it is supposed to be 1 here... i'm confused", + "A": "1 is there but it is noted implicitly. In calculus, 1 is unusually implied in problems. Dx less us to do the substitution and it essentially means(in that context) that we are differentiating. When I do these kinds of problems, I don t really think about the formal definition of dx.", + "video_name": "CZdziIlYIfI", + "timestamps": [ + 650, + 722 + ], + "3min_transcript": "Well, we just figured out, from 0 to 1, f of x is just 1 minus x. f of x is just 1 minus x from 0 to 1 times cosine of pi x, cosine of pi x dx. And now we just have to evaluate this integral right over here. So let's do that. So 1 minus x times cosine of pi x is the same thing as cosine of pi x minus x cosine of pi x. Now, this right here, well, let's just focus on taking the antiderivative. This is pretty easy. But let's try to do this one, because it seems a little bit more complicated. So let's take the antiderivative of x cosine of pi x dx. And what should jump in your mind is, well, this isn't that simple. But if I were able to take the derivative of x, that would simplify. It's very easy to take the antiderivative of cosine of pi x without making it more complicated. And remember, integration by parts tells us that the integral-- I'll write it up here-- the integral of udv is equal to uv minus the integral of vdu. And we'll apply that here. But I've done many, many videos where I prove this and show examples of exactly what that means. But let's apply it right over here. And in general, we're going to take the derivative of whatever the u thing is. So we want u to be something that's simpler when I take the derivative. And then we're going to take the antiderivative of dv. So we want something that does not become more complicated when I take the antiderivative. So the thing that becomes simpler when I take this derivative is x. So if I set u is equal to x, then clearly du is equal to just dx. Or you say du dx is equal to 1. So du is equal to dx. And then dv is going to be the rest of this. dv is equal to cosine pi x dx. And so v would just be the antiderivative of this with respect to x. v is going to be equal to 1 over pi sine of pi x. If I took the derivative here, derivative of the inside, you get a pi, times 1 over pi, cancels out. Derivative of sine of pi x becomes cosine of pi x. So that's our u, that's our v. This is going to be equal to u times v. So it's equal to x, this x times this. So x over pi sine of pi x minus the integral of v, which is 1 over pi sine of pi x du." + }, + { + "Q": "At about 1:49 Sal is explaining that if x=1 than [x]=0, because it is the largest integer, but it says that [x] must be lesser or EQUAL to x! Why is that?", + "A": "In real life, you can t deal, in decimal number, with how many siblings do you have. Anyway, the greatest integral function is, just, a function to create, only, an integer value. So, why does define as must be lesser or equal to x is because when you evaluate x = 0.7, you notice that 0.7 is not completing any integer number, therefore, it is f([0.7]) = 0. This f(x)= [x] looks, graphically, for the greatest integer function of [x], like stairs. Take off the decimal of any x, make it an integer.", + "video_name": "CZdziIlYIfI", + "timestamps": [ + 109 + ], + "3min_transcript": "For any real number x, let brackets around x denote the largest integer less than or equal to x, often known as the greatest integer function. Let f be a real valued function defined on the interval negative 10 to 10, including the boundaries by f of x is equal to x minus the greatest integer of x, if the greatest integer of x is odd, and 1 plus the greatest integer of x minus x, if the greatest integer of x is even. Then the value of pi squared over 10 times a definite integral from negative 10 to 10 of f of x cosine of pi of x dx is-- so before even try to attempt to evaluate this integral, let's see if we can at least visualize this function, f of x, right over here. So let's do our best to visualize it. So let me draw my x-axis. And let me draw my y-axis. So let me draw my y-axis. And then let's think about what this function will look like. this is x is equal to 1, x is equal to 2, x is equal to 3. We could go down to negative 1, negative 2. We could just keep going, if we like. Hopefully we'll see some type of pattern, because it seems to change from odd to even. So between 0 and 1, what is the greatest integer of x? So let me just write it over here. So between 0 and 1, until you get to 1-- so maybe I should do this-- from including 0 until 1, the greatest integer of x is equal to 0. If I'm at 0.5, the greatest integer below 0.5 is 0. As I go from 1 to 2, this brackets around x It's the greatest integer. If I'm at 1.9, the greatest integer is 1. And then if I go to above from between 2 and 3, then the greatest integer is going to be 2. If I'm at 2.5, greatest integer is going to be 2. So with that, let's try to at least draw this function over these intervals. So between 0 and 1, the greatest integer is 0. 0 we can consider to be even. 0 is even, especially if we're alternating. 1 is odd, 2 is even, 3 is odd. So 0 is even. So we would look at this circumstance right over here, if x is even. And then over this time frame or over this part of the x-axis, the greatest integer of x is just 0. So the equation or the line or the function is just going to be 1 minus x over this interval, because the greatest integer is 0. So 1 minus x will look like this. If this is 1, 1 minus x just goes down like that." + }, + { + "Q": "In 1:03 can a radius come from outside of the circle also?", + "A": "A radius is a line from the center of a circle to the circumference.", + "video_name": "04N79tItPEA", + "timestamps": [ + 63 + ], + "3min_transcript": "Draw a circle and label the radius, diameter, center and the circumference. Let me draw a circle, it won't be that well drawn of a circle but I think you get the idea so that is my circle. I'm going to label the center..over here. So I'll do the center I'll call it 'c'. So that is my center....and I'll draw an arrow there that is the center..of the circle and actually the circle itself is the set of all points that are a fixed distance away from that center and that fixed distance away, they're all from that center..that is the radius. So let me draw..the radius. So this distance right over here is the radius That is the radius and..that's going to be the same as this distance! ..which is the same as that distance! I can draw multiple radii. All of these are radii The distance between the center and any point on the circle now a diameter just goes straight across the circle and it's essentially two radii put toghether. So for example this would be a diameter.. that would be a diameter, you have one radii and another radii all on one line going from one side of the circle to another going through the center. So that is a diameter! ^_^........ that is a diameter. and I could have drawn it other ways, I could have drawn it like this that would be another diameter it would have the exact same length and finally, we have to think about the circumference, and the circumference is really just how far you have to go to go around the circle or if you put a string on that circle, how long would the string have to be? so what I'm tracing out in blue right now the length of what im tracing out is the circumference so....right over here..that is the circumference cir-cum-ference. Cir-cum-ference. And we're done!" + }, + { + "Q": "At 5:34, how did Sal Khan get a rise over run of -5/6 if you're supposed to subtract the 5 from 5x + 6y = 30 and then divide 6y by 6? I thought you get a slope and y-intercept of 5/6 + 5.", + "A": "As you said... you subtract 5x . You do this on both sides of the equation. The 5x is now on the other side with a minus in front of it. Thus it is now -5x . Divide by 6 and you get: -5/6 (x).", + "video_name": "LNSB0N6esPU", + "timestamps": [ + 334 + ], + "3min_transcript": "So let's say that I had the linear equation. Let's say that I have 5x + 6y = 30. I encourage you to pause this video, and figure out what are the x and y-intercepts for the graph that represents the solutions, all the xy pairs that satisfy this equation. Well the easiest thing to do here, let's see what the y value is when x = 0 and what x value is when y = 0. When x = 0 this becomes 6y = 30. So 6 times what is 30? Well y would be equal to 5 here. So when x is 0, y is 5. What about when y is 0? Well when y is 0, that's going to be 0, and you have 5x = 30. Well then x would be equal to 6. So we could plot those points, 0, 5. When x is 0, y is 5. When x is 6, y is 0. So those are both points on this graph and then the actual graph is going to, or the actual line that represents the x and y pairs that satisfy this equation is going to look like, it's going to look like this. I'll just try. So I can make it go, it's going to look like... It's going to go through those two points. So it going to...I can make it go the other way too. Let me see. It's going to go through those two points and so it's going to look something like that. Now what are its' x and y-intercepts? Well, we already kind of figured it out but the intercepts themselves, these are the points on the graph where they intersect the axes. That point is the y-intercept and it happens, it's always going to happen when x = 0, and when x = 0 we know that y = 5. It's that point, the point 0, 5. And what is the y inter...what is the x-intercept? The x-intercept is the point, it's actually the same x-intercept for this equation right over here. It's the point 6, 0. That point right over there." + }, + { + "Q": "At 0:36, why would you add -3/4? Shouldn't it be subtracted?", + "A": "Adding a negative is equal to subtracting its positive! For example, 2-1 = 2+-1 =-1+2, and all of these are correct. I think the -3/4 was placed before -10/6 so that this equation can look like the previous equation dragged down. :)", + "video_name": "9tmtDBpqq9s", + "timestamps": [ + 36 + ], + "3min_transcript": "We have negative 3/4 minus 7/6 minus 3/6. And there's many ways to do this. But it immediately jumps out at me that these last two numbers have a 6 in the denominator. So I'm going to worry about these first. I'm going to view this as negative 7/6 minus 3/6. So if we have negative 7/6 minus 3/6, that's going to be the same thing as negative 7 minus 3 over 6. And of course, we have this negative 3/4 out front that we're going to add to whatever we get over here. So this is these two terms that I'm adding together. Negative 7 minus 3 is negative 10. So it's negative 10 over 6. And then I'm going to have to add that to negative 3/4. And now I have to worry about finding a common denominator. Let me write that so they have a similar size. What is the smallest number that is a multiple of both 4 and 6? Well, it might jump out at you that it's 12. You can literally just go through the multiples of 4. Or you could look at the prime factorization of both of these numbers. And what's the smallest number that has all of the prime factors of both of these? So you need two 2s, and you need a 2 and a 3. So if you have two 2s and a 3, that's 4 times 3 is 12. So let's rewrite this as something over 12 plus something over 12. Well, to get your denominator from 4 to 12, you have to multiply by 3. So let's multiply our numerator by 3 as well. So if we multiply negative 3 times 3, you're going to have negative 9. And to get your denominator from 6 to 12, you have to multiply by 2. So let's multiply our numerator by 2 as well so that we don't change the value of the fraction. So that's going to be negative 20. Our common denominator is 12. And so this is going to be negative 9 plus negative 20, or we could even write that as minus 20, over 12, which is equal to-- and we deserve a drum roll now. This is negative 29 over 12. And 29 is a prime number, so it's not going to share any common factors other than 1 with 12. So we also have this in the most simplified form." + }, + { + "Q": "At 0:14 how does he turn the seven-sixths to negative.", + "A": "Since there is a subtraction sign after the - 3/4, this could be viewed as adding negatives. Since 7/6 ad 3/6 have common denominators already there, we could block them up to solve the problem easier. -7/6 + -3/6 = -10/6 Then you can find the solution by scaling the numerator and denominators of -10/6 and -3/4 and then add the two together.", + "video_name": "9tmtDBpqq9s", + "timestamps": [ + 14 + ], + "3min_transcript": "We have negative 3/4 minus 7/6 minus 3/6. And there's many ways to do this. But it immediately jumps out at me that these last two numbers have a 6 in the denominator. So I'm going to worry about these first. I'm going to view this as negative 7/6 minus 3/6. So if we have negative 7/6 minus 3/6, that's going to be the same thing as negative 7 minus 3 over 6. And of course, we have this negative 3/4 out front that we're going to add to whatever we get over here. So this is these two terms that I'm adding together. Negative 7 minus 3 is negative 10. So it's negative 10 over 6. And then I'm going to have to add that to negative 3/4. And now I have to worry about finding a common denominator. Let me write that so they have a similar size. What is the smallest number that is a multiple of both 4 and 6? Well, it might jump out at you that it's 12. You can literally just go through the multiples of 4. Or you could look at the prime factorization of both of these numbers. And what's the smallest number that has all of the prime factors of both of these? So you need two 2s, and you need a 2 and a 3. So if you have two 2s and a 3, that's 4 times 3 is 12. So let's rewrite this as something over 12 plus something over 12. Well, to get your denominator from 4 to 12, you have to multiply by 3. So let's multiply our numerator by 3 as well. So if we multiply negative 3 times 3, you're going to have negative 9. And to get your denominator from 6 to 12, you have to multiply by 2. So let's multiply our numerator by 2 as well so that we don't change the value of the fraction. So that's going to be negative 20. Our common denominator is 12. And so this is going to be negative 9 plus negative 20, or we could even write that as minus 20, over 12, which is equal to-- and we deserve a drum roll now. This is negative 29 over 12. And 29 is a prime number, so it's not going to share any common factors other than 1 with 12. So we also have this in the most simplified form." + }, + { + "Q": "At 1:30, why does he say \"the negative direction\"?", + "A": "lim : Means you are inputting smaller negative values. x\u00e2\u0086\u00920- Relating the above limt to the example, ln(n \u00e2\u0089\u00a4 0) is undifined. Hence the given limit condition: lim x\u00e2\u0086\u00920+", + "video_name": "CDf_aE5yg3A", + "timestamps": [ + 90 + ], + "3min_transcript": "- [Voiceover] What I would like to tackle in this video is what I consider to be a particularly interesting limits problem. Let's say we want to figure out the limit as X approaches zero from the positive direction of sine of X. This is where it's about to get interesting. Sine of X to the one over the natural log of X power and I encourage you to pause this video and see if you can have a go at it fully knowing that this is a little bit of a tricky exercise. I'm assuming you have attempted. Some of you might have been able to figure out on the first pass. I will tell you that the first time that I encountered something like this, I did not figure it out at the first pass so definitely do not feel bad if you fall into that second category. What many of you all probably did is you said okay, let me think about it. Let me just think about the components here. If I were to think about the limit, if I were to think about the limit as X approaches zero from the positive direction of sine of X, well that's pretty straightforward. That's going to be zero, is going to approach zero but then if you say, and you could say, I guess I should say. The limit as X approaches zero from the positive direction of one over natural log of X and this is why we have to think about it from the positive direction. It doesn't make sense to approach it from the negative direction. You can't take the natural log of a negative number. That's not in the domain for the natural log but as you get closer and closer to zero from the negative direction, the natural log of those values, you have to raise E to more larger and larger negative values. This part over here is going to approach negative infinity. It's going to go to negative infinity. One over negative infinity, one divided by super large or large magnitude negative numbers, well, that's just going to approach zero. You could say that this right over here is also going to be, is also going to be equal to zero. That doesn't seem to help us much because if this thing is going to zero and that thing is going to zero, it's kind of an implication that well to the zero power but we don't really know what zero, let me do the some, those color. Zero to the zero power but this is one of those great fun things to think about in mathematics. There's justifications why this could be zero, justifications why this could be one. We don't really know what to make of this. This isn't really a satisfying answer. Something at this point might be going into your brain. We have this thing that we've been exposed to called L'Hopital's rule. If you have not been exposed to it, I encourage you to watch the video, the introductory video on L'Hopital's rule. In L'Hopital's rule, let me just write it down. L'Hopital's rule helps us out with situations where when we try to superficially evaluate the limit, we get indeterminate forms things like zero over zero. We get infinity over infinity. We get negative infinity over negative infinity and we go into much more detail into that video." + }, + { + "Q": "At 4:30 he says there's no evidence that DC is equivalent to AC. Are we sure? How do we know DC is not AC? Is it bigger or smaller or what?", + "A": "We know DC is not equivalent to AC because if it was the case, then we would have an isosceles triangle. A theorem associated with isosceles triangles is that the two angles opposite the equal lengths must be equal. That is not the case, since 31 degrees is opposite to AC and 59 is opposite to DC.", + "video_name": "TugWqiUjOU4", + "timestamps": [ + 270 + ], + "3min_transcript": "Now is that what they wrote here? No. They wrote AC over EF. Well, where's EF? EF is nowhere to be seen either in this triangle, or even in this figure. EF is this thing right over here. EF is this business right over here. That's EF. It's in a completely different triangle in a completely different figure. We don't even know what scale this is drawn at. There's no way the tangent of this angle is related to this somewhat arbitrary number that's over here. They haven't labelled it. This thing might be a million miles long for all we know. This thing really could be any number. So this isn't the case. We would have to relate it to something within this triangle, or something that's the same length. So if somehow we could prove that EF is the same length as DC, then we could go with that. But there's no way. This is a completely different figure, a completely different diagram. This is a similar triangle to this, but we don't know anything about the lengths. A similar triangle just lets us know that the angles are all might be the same, but it doesn't tell us what this number right over here, doesn't tell us that this side is somehow congruent to DC. So we can't go with this one. Now let's think about the sine of CBA. So the sine-- let me do this in a different color. So the sine of angle CBA. So that's this angle right over here, CBA. Well, sine is opposite over hypotenuse. I guess let me make it clear which I'll do this in yellow. We're now looking at this triangle right over here. The opposite side is AC. That's what the angle opens up into. So it's going to be equal to AC. And what is the hypotenuse? What is the hypotenuse here? Well, the hypotenuse-- so let me see, It's the side opposite the 90 degree side. So this, it's BC. Sine is opposite over hypotenuse, so over BC. Is that what they wrote over here? No. They have DC over BC. Now what is DC equal to? Well, DC is this. And DC is not-- there's no evidence on this drawing right over here that DC is somehow equivalent to AC. So given this information right over here, we can't make this statement, either. So neither of these are true. So let's make sure we got this right. We can go back to our actual exercise, and we get-- oh, that's not the actual exercise. Let me minimize that." + }, + { + "Q": "I'm a bit confused, I thought DC and EF were the same because both triangles are similar 3:30. Can someone explain?", + "A": "If the triangles are congruent, then corresponding parts of corresponding triangles are congruent, however, if triangles are similar (AA is one method to show this), then the corresponding sides are proportional ( a common scale factor to get from each side to its corresponding side on the other triangle).", + "video_name": "TugWqiUjOU4", + "timestamps": [ + 210 + ], + "3min_transcript": "we're dealing with this right triangle right over here. That's the only right triangle that angle ADC is part of. And so what side is opposite angle ADC? Well, it's side CA, or I guess I say AC, side AC. So that is opposite. And what side is adjacent? Well, this side, CD. CD, or I guess I could call it DC, whatever I want to call it. DC, or CD, is adjacent. Now how did I know that this side is adjacent and not side DA? Because DA is the hypotenuse. They both, together, make up the two sides of this angle. But the adjacent side is one of the sides of the angle that is not the hypotenuse. AD or DA in the sohcahtoa context we would consider to be the hypotenuse. For this angle, this is opposite, this is adjacent, this is hypotenuse. Tangent of this angle is opposite over adjacent-- AC Now is that what they wrote here? No. They wrote AC over EF. Well, where's EF? EF is nowhere to be seen either in this triangle, or even in this figure. EF is this thing right over here. EF is this business right over here. That's EF. It's in a completely different triangle in a completely different figure. We don't even know what scale this is drawn at. There's no way the tangent of this angle is related to this somewhat arbitrary number that's over here. They haven't labelled it. This thing might be a million miles long for all we know. This thing really could be any number. So this isn't the case. We would have to relate it to something within this triangle, or something that's the same length. So if somehow we could prove that EF is the same length as DC, then we could go with that. But there's no way. This is a completely different figure, a completely different diagram. This is a similar triangle to this, but we don't know anything about the lengths. A similar triangle just lets us know that the angles are all might be the same, but it doesn't tell us what this number right over here, doesn't tell us that this side is somehow congruent to DC. So we can't go with this one. Now let's think about the sine of CBA. So the sine-- let me do this in a different color. So the sine of angle CBA. So that's this angle right over here, CBA. Well, sine is opposite over hypotenuse. I guess let me make it clear which I'll do this in yellow. We're now looking at this triangle right over here. The opposite side is AC. That's what the angle opens up into. So it's going to be equal to AC. And what is the hypotenuse? What is the hypotenuse here? Well, the hypotenuse-- so let me see," + }, + { + "Q": "At 4:14 where did you get the plus sign from? Wouldn't you put a subtraction sign there? (because the previous color coded terms had a subtraction sign at the beginning?)", + "A": "At that point in the video, Sal is combining: -2x^2 + 3x^2. He does this by adding/subtracting the coefficients: -2 + 3 = +1, not a -1. So, once combined, the 2 terms creates +x^2. Hope this helps.", + "video_name": "FNnmseBlvaY", + "timestamps": [ + 254 + ], + "3min_transcript": "Do I have any other x squared? Yes, I do. I have this 3x squared right over there. So plus 3x squared. And then let's see, I have an x term right over here, and that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term-- I'll circle that in orange-- so plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have two of something and I subtract 3 of that, what am I left with? Well, I'm left with negative 1 of that something. So I could write negative 1y, or I could just write negative y. And another way you could think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. It's 2. Where obviously both are dealing-- they're both y terms, not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1, or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms. If I have 4 of this, 4 xy's and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say add the coefficients, 4 plus negative 4, gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. And so I'm left with no xy's. And then I have right over here-- I could have written 0xy, but that seems unnecessary-- then right over here I have my x squared terms. Negative 2 plus 3 is 1. Or another way of saying it, if I have 3x squared squared, so I'm left with the 1x squared. So this right over here simplifies to 1x squared. Or I could literally just write x squared. 1x squared is the same thing as x squared. So plus x squared, and then these there's nothing really left to simplify. So plus 2x plus y squared. And obviously you might have gotten an answer in some other order, but the order in which I write these terms don't matter. It just matters that you were able to simplify it to these four terms." + }, + { + "Q": "at 0:48 isn't xy a 5 not a 6? just asking :D", + "A": "\u00f0\u009d\u0091\u00a5\u00f0\u009d\u0091\u00a6 = \u00f0\u009d\u0091\u00a5 \u00e2\u0088\u0099 \u00f0\u009d\u0091\u00a6 \u00f0\u009d\u0091\u00a5 = 3, \u00f0\u009d\u0091\u00a6 = 2 \u00e2\u0087\u0092 \u00f0\u009d\u0091\u00a5\u00f0\u009d\u0091\u00a6 = 3 \u00e2\u0088\u0099 2 = 6", + "video_name": "FNnmseBlvaY", + "timestamps": [ + 48 + ], + "3min_transcript": "Now we have a very, very, very hairy expression. And once again, I'm going to see if you can simplify this. And I'll give you little time to do it. So this one is even crazier than the last few we've looked at. We've got y's and xy's, and x squared and x's, well more just xy's and y squared and on and on and on. And there will be a temptation, because you see a y here and a y here to say, oh, maybe I can add this negative 3y plus this 4xy somehow since I see a y and a y. But the important thing to realize here is that a y is different than an xy. Think about it they were numbers. If y was 3 and an x was a 2, then a y would be a 3 while an xy would have been a 6. And a y is very different than a y squared. Once again, if the why it took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same-- I guess you cannot add these two A y is different than a y squared, is different than an xy. Now with that said, let's see if there is anything that we can simplify. So first, let's think about the y terms. So you have a negative 3y there. Do we have any more y term? We have this 2y right over there. So I'll just write it out-- I'll just reorder it. So we have negative 3y plus 2y. Now, let's think about-- and I'm just going in an arbitrary order, but since our next term is an xy term-- let's think about all of the xy terms. So we have plus 4xy right over here. So let me just write it down-- I'm just reordering the whole expression-- plus 4xy. And then I have minus 4xy right over here. Then let's go to the x squared terms. I have negative 2 times x squared, or minus 2x squared. So let's look at this. Do I have any other x squared? Yes, I do. I have this 3x squared right over there. So plus 3x squared. And then let's see, I have an x term right over here, and that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term-- I'll circle that in orange-- so plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have two of something and I subtract 3 of that, what am I left with? Well, I'm left with negative 1 of that something. So I could write negative 1y, or I could just write negative y. And another way you could think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3." + }, + { + "Q": "At 0:02, how is it possible to have a term like 4xy?", + "A": "If I square (2x + y)^2, then (2x+y)(2x+y) = 4x^2 + 4xy + y^2. We can combine variables, and this will we a term that could be combined with any other xy terms we might have.", + "video_name": "FNnmseBlvaY", + "timestamps": [ + 2 + ], + "3min_transcript": "Now we have a very, very, very hairy expression. And once again, I'm going to see if you can simplify this. And I'll give you little time to do it. So this one is even crazier than the last few we've looked at. We've got y's and xy's, and x squared and x's, well more just xy's and y squared and on and on and on. And there will be a temptation, because you see a y here and a y here to say, oh, maybe I can add this negative 3y plus this 4xy somehow since I see a y and a y. But the important thing to realize here is that a y is different than an xy. Think about it they were numbers. If y was 3 and an x was a 2, then a y would be a 3 while an xy would have been a 6. And a y is very different than a y squared. Once again, if the why it took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same-- I guess you cannot add these two A y is different than a y squared, is different than an xy. Now with that said, let's see if there is anything that we can simplify. So first, let's think about the y terms. So you have a negative 3y there. Do we have any more y term? We have this 2y right over there. So I'll just write it out-- I'll just reorder it. So we have negative 3y plus 2y. Now, let's think about-- and I'm just going in an arbitrary order, but since our next term is an xy term-- let's think about all of the xy terms. So we have plus 4xy right over here. So let me just write it down-- I'm just reordering the whole expression-- plus 4xy. And then I have minus 4xy right over here. Then let's go to the x squared terms. I have negative 2 times x squared, or minus 2x squared. So let's look at this. Do I have any other x squared? Yes, I do. I have this 3x squared right over there. So plus 3x squared. And then let's see, I have an x term right over here, and that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term-- I'll circle that in orange-- so plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have two of something and I subtract 3 of that, what am I left with? Well, I'm left with negative 1 of that something. So I could write negative 1y, or I could just write negative y. And another way you could think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3." + }, + { + "Q": "At approximately 8:59, why are we allowed to divide the \"outside bit\" by four, and multiply the \"inside bit\" by four?", + "A": "You can multiply both the outside and the inside because any number is itself multiplied by 1, and 1 = 4 * (1/4) You can hence multiply an expression by 4 * (1/4) while keeping its value. Using the property of multiplication m * (a * b) = (m * a) * b = a * (m * b) you can write that as 1/4 times the outside multiplied by 4 times the inside. I hope this helps. --Phi \u00cf\u0086", + "video_name": "64bH_27Ehoc", + "timestamps": [ + 539 + ], + "3min_transcript": "And so we're going to be left with 3 squared. That's what this is down here, squared. So this is 1/3 squared. And then that, squared. So that's what we're left with, with that orange term. And then we go to the pink term. This pink term. 48 is just 3 times 4 times 4. Three times 4. I'll write 4 squared here, because each time we're going to multiply it times 4 again. So the next one's going to be 4 to the third, because we're really-- each side turns into four sides. That's where that came from. 4 squared, we're factoring out the square root of 3, we're factoring out the four, we're factoring out the s squared. And all we're left is 1 over 3 to the third power, squared. So times 1 over 3 to the third power, squared. And we're just going to keep going like that forever. So on each step, we're multiplying by 4, the power of this 4 is incrementing. So there's really 4 to the 0-th power here. We have a 1 here you can imagine, implicitly. The 4 to the first power, 4 squared, then it'll go 4 to the third. This power is also incrementing-- 3 to the first, 3 to the second, 3 to the third. But we see that this power is always one more than that. And it'll be much easier to calculate this infinite-- what's going to turn into an infinite-- geometric series, if those are actually the same power. So what I want to do is I want to increase the power of 4 in all of those. But I can't just willy-nilly multiply everything by 4. If I'm going to multiply everything by 4, I also need to divide everything by 4. So what I'm going to do in this step right over here is I'm going to multiply and divide everything by 4. So if we divide by 4, I can do that on the outside. So let me multiply 1/4 times this right over here. And so I'm dividing by 4 out here. And then I'm going to multiply this by 4. And so I'm not going to be changing This is going to be 4 plus 3 times 4 plus 3 times 4 to the third. And so what was cool about this is now that the power of 4 and the power of this 3 down here are going to be the same power. But it still seems a little weird because we're taking this 1 over 3 squared and then we're squaring it. And here we just have to realize-- so this is always going to be squared. And this is the thing that's incrementing. But in general, if I have 1 over 3 to the n, and I'm squaring it, this is equal to 1 over 3 to the 2n power. So I'm just multiplying it by 2. If I'm raising something to the exponent, then raising that to an exponent, that's just raising it to the n times 2 exponent. And this is the exact same thing as 1 over 3 squared, raised to the nth power. So we can actually switch these two exponents" + }, + { + "Q": "When he starts factoring at 6:29, why doesn't the square (the 2 on the blue section) get factored out with the rest of the equation? Is it because it's not exactly s^2 like the original yellow equation?", + "A": "sqrt(3)\u00e2\u0080\u00a2s^2/4 + 3\u00e2\u0080\u00a2sqrt(3)\u00e2\u0080\u00a2(s/3)^2/4 + 12\u00e2\u0080\u00a2sqrt(3)\u00e2\u0080\u00a2(s/9)^2/4 + 48\u00e2\u0080\u00a2sqrt(3)\u00e2\u0080\u00a2(s/27)^2/4 ... sqrt(3)\u00e2\u0080\u00a2(s^2/4 + 3\u00e2\u0080\u00a2(s/3)^2/4 + 12\u00e2\u0080\u00a2(s/9)^2/4 + 48\u00e2\u0080\u00a2(s/27)^2/4 ...) sqrt(3)/4\u00e2\u0080\u00a2(s^2 + 3\u00e2\u0080\u00a2(s/3)^2 + 12\u00e2\u0080\u00a2(s/9)^2 + 48\u00e2\u0080\u00a2(s/27)^2 ...) sqrt(3)/4\u00e2\u0080\u00a2(s^2 + 3\u00e2\u0080\u00a2s^2\u00e2\u0080\u00a2(1/3)^2 + 12\u00e2\u0080\u00a2s^2\u00e2\u0080\u00a2(1/9)^2 + 48\u00e2\u0080\u00a2s^2\u00e2\u0080\u00a2(1/27)^2 ...) sqrt(3)\u00e2\u0080\u00a2s^2/4\u00e2\u0080\u00a2(1 + 3\u00e2\u0080\u00a2(1/3)^2 + 12\u00e2\u0080\u00a2(1/9)^2 + 48\u00e2\u0080\u00a2(1/27)^2 ...)", + "video_name": "64bH_27Ehoc", + "timestamps": [ + 389 + ], + "3min_transcript": "first of all, after I do another pass? Well, the previous pass, I had 12 sides. Each of those 12 sides are now going to turn to 4 new sides when I add these little orange bumps there. So I'm going to multiply it times 4 again. I'm going to multiply it times 4. So now I'm going to have 48 sides. And how many new triangles? Well it's going to be the yellow area plus the blue area plus the orange area. So how many new orange triangles do I have.? Well I'm adding a new orange triangle to each of the sides for the previous pass. In the previous pass I had 12 sides. So now I'm going to add 12 orange triangles. And actually let me write that. I'll just write 12 orange triangles. But it's really I just multiplied it times 4. And then I'm going to have times the square root of 3. And now this isn't going to be s over 3 any more. These are now going to be s over 9. These have 1/3 of the dimensions of these blue triangle. s over 9 squared over 4. And so I think you might start to see the pattern building if we do another pass after this one. Move to the right a little bit. What will that look like? Let me do this in a different color that I haven't used yet. Let me see I haven't used this pink, yet. So now I'm going to have the previous number of sides, that's my number of new triangles, 48 times the square root of 3 times s over-- now these are going to be even 1/3 of this-- s over 27 to the second power. All of that over 4. And I'm going to keep adding an infinite number of terms of this to get the area of a true koch snowflake. So I'm just going to keep doing this over, and over again. So the trick really is finding this infinite sum and seeing if we get a finite number over here. So the first thing I want to do, just to simplify, well, let me just rewrite it a little bit differently over here. is that we can factor out a square root of 3s squared over 4. So let me factor that out. So if we factor out a square root of 3s squared over 4 from all of the terms. Then this term right over here will become a 1. This term right over here is going to become a 3. Let's see, we factored out a square root of 3. We factor out a four. And we factored out the s squared. We factor out only the s squared. So now it's going to have plus 3 times 1/3 squared. That's all we have left here. We have the 1/3 squared. And then we have this 3. And I'm not simplifying this on purpose, so that we see a pattern emerge. And then this next term, right over here, plus-- so this 12 is still going to be there. But I'm going to write that as 3 times 4. Let's see, we're factoring out the square root of 3," + }, + { + "Q": "I lost at 0:43 until the video is over. I didnt understand anything.. pls help", + "A": "What did you not understand? The video is about finding a Common Denominator. Is that the part you didn t understand or how he changed the 1/10 into 10/100?", + "video_name": "DR2DYe7PI74", + "timestamps": [ + 43 + ], + "3min_transcript": "Let's see if we can write 0.15 as a fraction. So the important thing here is to look at what place these digits are in. So this 1 right over here, this is in the tenths place, so you could view that as 1 times 1/10. This 5 right over here is in the hundredths place, so you could view that as 5 times 1/100. So if I were to rewrite this, I can rewrite this as the sum of-- this 1 represents 1 times 1/10, so that would literally be 1/10 plus-- and this 5 represents 5 times 1/100, so it would be plus 5/100. And if we want to add them up, we want to find a common denominator. The common denominator is 100. Both 10 and-- the least common multiple. 100 is a multiple of both 10 and 100. So we can rewrite this as something over 100 plus something over 100. This isn't going to change. This was already 5/100. by 10-- that's what we did; we multiplied it by 10-- then we're going to have to multiply this numerator by 10. And so this is the same thing as 10/100. And now we're ready to add. This is the same thing as-- 10 plus 5 is 15/100. And you could have done that a little bit quicker just You would say, look, my smallest place right over here is in the hundredths place. Instead of calling this 1/10, I could call this literally 10/100. Or I could say this whole thing is 15/100. And now if I want to reduce this to lowest terms, we can-- let's see, both the numerator and the denominator are divisible by 5. So let's divide them both by 5. And so the numerator, 15 divided by 5, is 3. The denominator, 100 divided by 5, is 20. And that's about as simplified as we can get." + }, + { + "Q": "What does R^2 (at 0:59) mean?", + "A": "Basically, it s a coordinate space analogous to the xy plane that we all know and love from graphing functions in algebra 2.", + "video_name": "8QihetGj3pg", + "timestamps": [ + 59 + ], + "3min_transcript": "So I have two 2-dimensional vectors right over here, vector a and vector b. And what I want to think about is how can we define or what would be a reasonable way to define the sum of vector a plus vector b? Well, one thing that might jump at your mind is, look, well, each of these are two dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just a sum of the first two components of these two vectors. So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two 2-dimensional vectors. You add them together, you get another two 2-dimensional vectors. If you think about it in terms of real coordinates bases, both of these are members of R2-- I'll write this down here just so we get used to the notation. which is just another way of saying that these are both two tuples. They are both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it, but how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector a, we could visualize, this tells us how far this vector moves in each of these directions-- horizontal direction and vertical direction. So if we put the, I guess you could say the tail of the vector at the origin-- remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction. 1, 2, 3, 4, 5, 6. And then negative 2 in the vertical. So negative 2. So vector a could look like this. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector. And the direction is the direction that it is pointed in. And also just to emphasize, I could have drawn vector a like that or I could have put it over here. These are all equivalent vectors. These are all equal to vector a. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector b. Vector b in the horizontal direction goes negative 4-- 1, 2, 3, 4, and in the vertical direction goes 4-- 1, 2, 3, 4. So its tail if we start at the origin, if its tail is at the origin, its head would be at negative 4, 4. So let me draw that just like that." + }, + { + "Q": "what is that fancy bracket that Sal draws around the 2*3*5 at 1:18 called, if it is called anything other than a bracket?", + "A": "A bracket is [ or ] . A brace is { or } . I guess you can call it a squiggly line that connects the selected objects and relates them to something else in the part that sticks out in the middle.", + "video_name": "QUem_2dkB9I", + "timestamps": [ + 78 + ], + "3min_transcript": "We need to figure out the least common multiple of 30 and 25. So let's get our little scratch pad out here. And we care about 30 and we care about 25. And I'm going to do this using the prime factorization method which I just like more. Let's find the prime factorization of both of these numbers. So 30, it's divisible by 2. It's 2 times 15. 15 is 3 times 5. And now we've expressed 30 as the product of only prime numbers, 2 times 3 times 5. Now let's do the same thing for 25. 25 is-- well that's just 5 times 5. So let me write that down. 25 is equal to 5 times 5. Now to find the least common multiple, let me write this down, the least common multiple of 30 and 25 is going to have a number whose prime factorization is a super set of both of these as we have in any one of these. So it's the least common multiple. Well it has to be divisible by 30. So it's going to need a 2 times a 3 times a 5. This is what makes it divisible by 30. But it needs to also be divisible by 25. And in order to be divisible by 25, you need to have two 5s in your prime factorization. Right now our prime factorization only has one 5. So we have one 5 right over here. We need another 5. So let's throw another 5 right over here. So now this thing clearly has a 25 in it. It's clearly divisible by 25. And this is the least common multiple. I could have, if we just wanted a common multiple, we could have thrown more factors here and it would have definitely been divisible by 30 or 25, but this has the bare minimum of prime factors necessary to be divisible by 30 and 25. I wouldn't be divisible by both anymore. If I got rid of this 2, I wouldn't be divisible by 30 If I got rid of one of the 5s, I wouldn't be divisible by 25 anymore. So let's just multiply it out. This is essentially the prime factorization of our least common multiple. And this is equal to 2 times 3 is 6, 6 times 5 is 30, 30 times 5 is equal to 150. And of course, we can check our answer, 150. Check it, and we got it right." + }, + { + "Q": "But at 4:35, why is it an x minus 2. He is shifting from g(x) to the LEFT, to the f(x). Shouldn't it be a PLUS sign, since he is going to the LEFT? Or is g(x) your end point, and f(x) is your starting point?", + "A": "Think of it this way: make an x/y table for your points. You plug in an x-value to the function and apply the operations to it to solve for the y value, correct? It s reverse order of operations when it is directly attached to the x. (Another way of imagining this would be to think about switching the -2 to the left side of the function: going over the equals sign would change the -2 to a positive number, 2)", + "video_name": "ENFNyNPYfZU", + "timestamps": [ + 275 + ], + "3min_transcript": "equivalent to f of negative 2. So let me write that down. g of 0 is equal to f of negative 2. We could keep doing that. We could say g of 1, which is right over here. This is 1. g of 1 is equal to f of negative 1. g of 1 is equal to f of negative 1. So I think you see the pattern here. g of whatever is equal to the function evaluated at 2 less than whatever is here. So we could say that g of x is equal to f of-- well So f of x minus 2. So this is the relationship. g of x is equal to f of x minus 2. And it's important to realize here. When I get f of x minus 2 here-- and remember the function is being evaluated, this is the input. x minus 2 is the input. When I subtract the 2, this is shifting the function to the right, which is a little bit counter-intuitive unless you go through this exercise right over here. So g of x is equal to f of x minus 2. If it was f of x plus 2 we would have actually shifted f to the left. Now let's think about this one. This one seems kind of wacky. So first of all, g of x, it almost looks like a mirror image but it looks like it's been flattened out. So let's think of it this way. Let's take the mirror image of what g of x is. So I'm going to try my best to take the mirror image of it. So let's see... It gets to about 2 there, then it gets pretty close to 1 right over there. So if I were to take its mirror image, it looks something like this. Its mirror image if I were to reflect it across the x-axis. It looks something like this. So this right over here we would call-- so if this is g of x, when we flip it that way, this is the negative g of x. When x equals 4, g of x looks like it's about negative 3 and 1/2. You take the negative of that, you get positive. I guess it should be closer to here-- You get positive 3 and 1/2 if you were to take the exact mirror image. So that's negative g of x. But that still doesn't get us. It looks like we actually have to triple this value for any point. And you see it here. This gets to 2, but we need to get to 6. This gets to 1, but we need to get to 3." + }, + { + "Q": "at 4:00, how is it positive? I don't understand Sal's explanation. Can someone explain it please? and plz explain how square root of -1 exsists...", + "A": "\u00f0\u009d\u0091\u0096\u00e2\u0081\u00b4 = \u00f0\u009d\u0091\u0096\u00c2\u00b2 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u0096\u00c2\u00b2 = (-1)(-1) = 1 The principal square root of -1 is \u00f0\u009d\u0091\u0096 simply because we define it to be so. In other words, we define \u00f0\u009d\u0091\u0096\u00c2\u00b2 = -1. Mathematics doesn t bother with does or does not exist because it simply deals with logical extensions of definitions of axioms we set.", + "video_name": "ysVcAYo7UPI", + "timestamps": [ + 240 + ], + "3min_transcript": "well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power, Well once again this is going to be \"i\" times \"i\" to the third power. So that's \"i\" times \"i\" to the third power. \"i\" times \"i\" to the third power Well what was \"i\" to the third power? \"i\" to the third power was negative \"i\" This over here is negative \"i\". And so \"i\" times \"i\" would get negative one, but you have a negative out here, so its \"i\" times \"i\" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is \"i\" times negative \"i\", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times \"i\" times \"i\". \"i\" times \"i\", by definition, is negative one. Negative one times negative one is equal to positive one. So \"i\" to the fourth is the same thing as \"i\" to the zeroth power. \"i\" to the fifth power. Well that's just going to be \"i\" to to the fourth times \"i\". And we know what \"i\" to the fourth is. It is one. So its one times \"i\", or it is one times \"i\", or it is just \"i\" again. So once again it is exactly the same thing as \"i\" to the first power. Lets try again just to see the pattern keep going. Lets try \"i\" to the seventh power. Sorry, \"i\" to the sixth power. Well that's \"i\" times \"i\" to the fifth power, that's \"i\" times \"i\" to the fifth, \"i\" to the fifth we already established as just \"i\", so its \"i\" times \"i\", it is equal to, by definition,\"i\" times \"i\" is negative one. And then lets finish off, well we could keep going on this way We can keep putting high and higher powers of \"i\" here. An we'll see that it keeps cycling back. In the next video I'll teach you how taking an arbitrarily high power of \"i\", how you can figure out what that's going to be. But lets just verify that this cycle keeps going. \"i\" to the seventh power is equal to \"i\" times \"i\" to the sixth power." + }, + { + "Q": "Isn't it supposed to be i=\u00c2\u00b1sqrt(-1) at \"0:56\"???", + "A": "But isn t (+sqrt(-1))^2=(-sqrt(-1))^2 or is it just defined and can t be changed???", + "video_name": "ysVcAYo7UPI", + "timestamps": [ + 56 + ], + "3min_transcript": "In this video, I want to introduce you to the number i, which is sometimes called the imaginary, imaginary unit What you're gonna see here, and it might be a little bit difficult, to fully appreciate, is that its a more bizzare number than some of the other wacky numbers we learn in mathematics, like pi, or e. And its more bizzare because it doesnt have a tangible value in the sense that we normally, or are used to defining numbers. \"i\" is defined as the number whose square is equal to negative 1. This is the definition of \"i\", and it leads to all sorts of interesting things. Now some places you will see \"i\" defined this way; \"i\" as being equal to the principle square root of negative one. I want to just point out to you that this is not wrong, it might make sense to you, you know something squared is negative one, then maybe its the principle square root of negative one. And so these seem to be almost the same statement, some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of \"i\" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of \"i\" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with \"i\" to the zeroth power. And so you might say, anything to the zeroth power is one, so \"i\" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including \"i\" is one. well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power," + }, + { + "Q": "At 3:15, Sal said that i^4=i*i^3 and resulted with positive 1. Could you just simply do i^4=i^2*i^2 and get -1*-1 just resulting with positive 1 as well. Is this right or it doesn't work all the time? Thanks for the video!", + "A": "Your technique works. Remember, we can regroup or multiply in any order due to the associative and commutative properties of multiplication.", + "video_name": "ysVcAYo7UPI", + "timestamps": [ + 195 + ], + "3min_transcript": "some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of \"i\" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of \"i\" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with \"i\" to the zeroth power. And so you might say, anything to the zeroth power is one, so \"i\" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including \"i\" is one. well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power, Well once again this is going to be \"i\" times \"i\" to the third power. So that's \"i\" times \"i\" to the third power. \"i\" times \"i\" to the third power Well what was \"i\" to the third power? \"i\" to the third power was negative \"i\" This over here is negative \"i\". And so \"i\" times \"i\" would get negative one, but you have a negative out here, so its \"i\" times \"i\" is negative one, and you have a negative, that gives you positive one. Let me write it out. This is the same thing as, so this is \"i\" times negative \"i\", which is the same thing as negative one times, remember multiplication is commutative, if you're multiplying a bunch of numbers you can just switch the order. This is the same thing as negative one times \"i\" times \"i\". \"i\" times \"i\", by definition, is negative one. Negative one times negative one is equal to positive one. So \"i\" to the fourth is the same thing as \"i\" to the zeroth power." + }, + { + "Q": "A 1:13, how can the square of any no. be negative ? Given that i = under root -1 and square of i = -1 ?", + "A": "You are right. The square of any number either positive or negative, cannot be a negative. i is not any number, it is an imaginary number. You might wonder about the purpose of this. The purpose of i is to compensate for the fact that normal numbers cannot be negative when squared. and square root of -1 would make no sense.", + "video_name": "ysVcAYo7UPI", + "timestamps": [ + 73 + ], + "3min_transcript": "In this video, I want to introduce you to the number i, which is sometimes called the imaginary, imaginary unit What you're gonna see here, and it might be a little bit difficult, to fully appreciate, is that its a more bizzare number than some of the other wacky numbers we learn in mathematics, like pi, or e. And its more bizzare because it doesnt have a tangible value in the sense that we normally, or are used to defining numbers. \"i\" is defined as the number whose square is equal to negative 1. This is the definition of \"i\", and it leads to all sorts of interesting things. Now some places you will see \"i\" defined this way; \"i\" as being equal to the principle square root of negative one. I want to just point out to you that this is not wrong, it might make sense to you, you know something squared is negative one, then maybe its the principle square root of negative one. And so these seem to be almost the same statement, some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of \"i\" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of \"i\" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with \"i\" to the zeroth power. And so you might say, anything to the zeroth power is one, so \"i\" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including \"i\" is one. well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power," + }, + { + "Q": "At time 6:08 it says that even if a is less than 0 it is still going to be positive? So my question is how do you know what to go by ? A is less than 0 so it is negative or a to the 4rth power which indicates it is positive?", + "A": "You would go by the order of operations to figure out the sign (parenthesis, exponents, multiplication, division, addition, subtraction). So since -a^4/3 is the same as -1*(a^4/3) then you would resolve the exponent before multiplying by -1. The exponent is even which means that you will always get a positive number in the numerator giving you a positive fraction which you THEN multiply by -1 making it a negative number. Hope this helps! :)", + "video_name": "Pms4cBWwPSU", + "timestamps": [ + 368 + ], + "3min_transcript": "in less than a second? Well if you just have a bunch of numbers, and then you multiply them times zero at some point in that, the whole thing is just going to be equal to zero. I could literally have a times b times c times d, I could multiply a bunch of numbers. And if I just knew that one of these numbers is zero, then the whole product is going to be equal to zero. You know if c is equal to zero, I could multiply a times b times d times e and get some number, and then multiply that times c, it's gonna be anything times zero is zero. Now notice, q is equal to zero, and q is right over here. So I could take p divided by p, that's gonna be one. One times q, that's just gonna be zero, as one times zero is zero, then you're gonna have zero times four and 2/7, it's just all gonna be zero. And the key is, there's a zero right over here. At some point you're multiplying this entire product times zero, so the whole product is going, or you're multiplying these numbers times zero, so the whole product is going to be zero. So this side isn't positive or negative, it's zero. It's neither positive nor negative. All right, let's do another one. What is the sign of negative 3/4 times negative a to the fourth over three, when a is less than zero? So they're telling us that a is negative. So a is negative. So let me use a more dramatic color. So right over here, I have a negative number to the fourth power. So it's a times a times a times a. And we've already said if you have an even number of negatives being multiplied by each other, then that's going to be a positive. You could even see it over here, I mean they're all negative, but a negative times a negative is going to be a positive. A negative times the negative, I take the third and the fourth one here, that's gonna be a positive. And then a positive times a positive, the whole thing is going to be positive, even though each of these are negative. Negative times negative, or another way you could say it, a times a is going to be positive, then times a is going to be negative, it's going to be positive again. So a to the fourth power is going to be positive. But then you have a negative in front of it. So you're taking the negative of a positive divided by a positive, Well, so this, all of this stuff right over here is all gonna be positive, but then you have this negative in front. So this entire thing inside the parentheses is going to be a negative. And then you have a negative, negative 3/4, times a negative, well a negative times a negative, the whole thing is going to be positive. So what's the sign of this? It's going to be positive. Let's do another one, I'm having too much fun. All right, what is the sign of x to the 59th power, divided by 2.3x times 4/5, when x is less than zero? So x is once again a negative number. So x to the 59th power, that's an odd number of negatives being multiplied together. So this thing right over here is going to be negative." + }, + { + "Q": "At 0:40, why is the volume (4/3)*(pi)*(r^2)? Why is it 4/3? I'm in middle school, so I don't understand why it's 4/3.", + "A": "If you want to find out why it is 4/3 you can look up who created the equation for a sphere. It may explain how that particular person got the equation.", + "video_name": "IelS2vg7JO8", + "timestamps": [ + 40 + ], + "3min_transcript": "Find the volume of a sphere with a diameter of 14 centimeters. So if I have a sphere-- so this isn't just a circle, this is a sphere. You could view it as a globe of some kind. So I'm going to shade it a little bit so you can tell that it's three-dimensional. They're giving us the diameter. So if we go from one side of the sphere straight through the center of it. So we're imagining that we can see through the sphere. And we go straight through the centimeter, that distance right over there is 14 centimeters. Now, to find the volume of a sphere-- and we've proved this, or you will see a proof for this later when you learn calculus. But the formula for the volume of a sphere is volume is equal to 4/3 pi r cubed, where r is the radius of the sphere. So they've given us the diameter. And just like for circles, the radius of the sphere is half of the diameter. So in this example, our radius is going to be 7 centimeters. And in fact, the sphere itself is the set the radius away from the center. But with that out of the way, let's just apply this radius being 7 centimeters to this formula right over here. So we're going to have a volume is equal to 4/3 pi times 7 centimeters to the third power. So I'll do that in that pink color. So times 7 centimeters to the third power. And since it already involves pi, and you could approximate pi with 3.14. Some people even approximate it with 22/7. But we'll actually just get the calculator out to get the exact value for this volume. So this is going to be-- so my volume is going to be 4 divided by 3. And then I don't want to just put a pi there, because that might interpret it as 4 divided by 3 pi. So 4 divided by 3 times pi, times 7 to the third power. before it does the multiplication, so this should work out. And the units are going to be in centimeters cubed or cubic centimeters. So we get 1,436. They don't tell us what to round it to. So I'll just round it to the nearest 10th-- 1,436.8. So this is equal to 1,436.8 centimeters cubed. And we're done." + }, + { + "Q": "Um I don't get \"7:20\"", + "A": "Given: n = l - m n = log_x (A/B) l = log_x (A) m = log_x (B) Therefore (plugging in to the first equation): log_x (A/B) = [log_x (A)] - [log_x (B)] Does that answer your question?", + "video_name": "yEAxG_D1HDw", + "timestamps": [ + 440 + ], + "3min_transcript": "How can we write all of these expressions as exponents? Well, this just says that x to the l is equal to A. Let me switch colors. That keeps it interesting. This is just saying that x to the m is equal to B. And this is just saying that x to the n is equal to A/B. So what can we do here? Well what's another way of writing A/B? Well, that's just the same thing as writing x to the l because that's A, over x to the m. That's B. also be written as x to the l, x to the negative m. Or that also equals x to the l minus m. So what do we know? We know that x to the n is equal to x to the l minus m. Those equal each other. I just made a big equal chain here. So we know that n is equal to l minus m. Well, what does that do for us? Well, what's another way of writing n? I'm going to do it up here because I think we have stumbled upon another logarithm rule. What's another way of writing n? Well, I did it right here. This is another way of writing n. So logarithm base x of A/B-- this is an x over l is this right here. Log base x of A is equal to l. The log base x of A minus m. I wrote m right here. That's log base x of B. There you go. I probably didn't have to prove it. You could've probably tried it out with dividing it, but whatever. But you know are hopefully satisfied that we have this new logarithm property right there. Now I have one more logarithm property to show you, but I don't think I have time to show it in this video. So I will do it in the next video. I'll see you soon." + }, + { + "Q": "At 3:37 Sal says that the range of x values can be narrower than the maximum possible range. Does this mean that delta can vary?", + "A": "Yes. Delta will change depending on the epsilon chosen.", + "video_name": "w70af5Ou70M", + "timestamps": [ + 217 + ], + "3min_transcript": "In this yellow definition right over here, we said you can get f of x as close as you want to L by making x sufficiently close to C. This second definition, which I kind of made as a little bit more of a game, is doing the same thing. Someone is saying how close they want f of x to be to L and the burden is then to find a delta where as long as x is within delta of C, then f of x will be within epsilon of the limit. So that is doing it. It's saying look, if we are constraining x in such a way that if x is in that range to C, then f of x will be as close as you want. So let's make this a little bit clearer by diagramming right over here. You show up and you say well, I want f of x to be within epsilon of our limit. And this right over here might be our limit minus epsilon. And you say, OK, sure. I think I can get your f of x within this range of our limit. And I can do that by defining a range around C. And I could visually look at this boundary. But I could even go narrower than that boundary. I could go right over here. Says OK, I meet your challenge. I will find another number delta. So this right over here is C plus delta. This right over here is C minus-- let me write this down-- is C minus delta. So I'll find you some delta so that if you take any x in the range C minus delta to C defined at C, so we think of ones that maybe aren't C, but are getting very close. If you find any x in that range, f of those x's are going to be as close as you want to your limit. They're going to be within the range L plus epsilon or L minus epsilon. So what's another way of saying this? Another way of saying this is you give me an epsilon, then I will find you a delta. So let me write this in a little bit more math notation. So I'll write the same exact statements with a little bit more math here. But it's the exact same thing. Let me write it this way. Given an epsilon greater than 0-- so that's kind of the first part of the game-- we can find a delta greater than 0, such" + }, + { + "Q": "At 5:28 Sal made the mistake of not making the distance between x and c greater than 0. Is that the same case as with the distance between f(x) and the limit? I feel like because it's distance as well it would need to be greater than 0 too.", + "A": "Think about this. If you are at school (assuming you are not home schooled), the distance between you and your home is some positive number. If you are at home, then the distance between you and your home is zero.", + "video_name": "w70af5Ou70M", + "timestamps": [ + 328 + ], + "3min_transcript": "And this right over here might be our limit minus epsilon. And you say, OK, sure. I think I can get your f of x within this range of our limit. And I can do that by defining a range around C. And I could visually look at this boundary. But I could even go narrower than that boundary. I could go right over here. Says OK, I meet your challenge. I will find another number delta. So this right over here is C plus delta. This right over here is C minus-- let me write this down-- is C minus delta. So I'll find you some delta so that if you take any x in the range C minus delta to C defined at C, so we think of ones that maybe aren't C, but are getting very close. If you find any x in that range, f of those x's are going to be as close as you want to your limit. They're going to be within the range L plus epsilon or L minus epsilon. So what's another way of saying this? Another way of saying this is you give me an epsilon, then I will find you a delta. So let me write this in a little bit more math notation. So I'll write the same exact statements with a little bit more math here. But it's the exact same thing. Let me write it this way. Given an epsilon greater than 0-- so that's kind of the first part of the game-- we can find a delta greater than 0, such So what's another way of saying that x is within delta of C? Well, one way you could say, well, what's the distance between x and C is going to be less than delta. This statement is true for any x that's within delta of C. The difference between the two is going to be less than delta. So that if you pick an x that is in this range between C minus delta and C plus delta, and these are the x's that satisfy that right over here, then-- and I'll do this in a new color-- then the distance between your f of x and your limit-- and this is just the distance between the f of x and the limit, it's going to be less than epsilon. So all this is saying is, if the limit truly does exist," + }, + { + "Q": "At 7:30, is sss the only kind of way to find congruence in a triangle?", + "A": "The congruence postulates are: SSS, AAS, ASA, SAS. AAA shows similarity only, not necessarily congruence. SSA may or may not show congruence, depending on the details. But generally SSA does not show congruence by itself -- you need some other bit of information to establish congruence. For example, if the known angle is right or obtuse, then SSA proves congruence; but, if the known angle is acute, then you would still need more information to establish congruence.", + "video_name": "CJrVOf_3dN0", + "timestamps": [ + 450 + ], + "3min_transcript": "have the same measure, they're congruent We also know that these two corresponding angles I'll use a double arch to specify that this has the same measure as that So, we also know the measure of angle ABC is equal to the measure of angle XYZ And then finally we know that this angle, if we know that these two characters are congruent, then this angle is gonna have the same measure as this angle as a corresponding angle So, we know that the measure of angle ACB is gonna be equal to the measure of angle XZY Now what we're gonna concern ourselves a lot with is how do we prove congruence? 'Cause it's cool, 'cause if you can prove congruence of 2 triangles then all of the sudden you can make all of these assumptions we're gonna assume it for the sake of introductory geometry course This is an axiom or a postulate or just something you assume So, an axiom, very fancy word Postulate, also a very fancy word It really just means things we are gonna assume are true An axiom is sometimes, there's a little bit of distinction sometimes where someone would say \"an axiom is something that is self-evident\" or it seems like a universal truth that is definitely true and we just take it for granted You can't prove an axiom A postulate kinda has that same role but sometimes let's just assume this is true and see if we assume that it's true what can we derive from it, what we can prove if we assume its true But for the sake of introductory geometry class and really most in mathematics today, these two words are use interchangeably An axiom or a postulate, just very fancy words that things we take as a given Things that we'll just assume, we won't prove them, and then we're just gonna build up from there And one of the core ones that we'll see in geometry is the axiom or the postulate That if all of the sides are congruent, if the length of all the sides of the triangle are congruent, then we are dealing with congruent triangles So, sometimes called side, side, side postulate or axiom We're not gonna prove it here, we're just gonna take it as a given So this literally stands for side, side, side And what it tells is, if we have two triangles and So I say that's another triangle right over there And we know that corresponding sides are equal So, we know that this side right over here is equal into, like, that side right over there" + }, + { + "Q": "can you write that congruency sign (at 1:39) on a keyboard?", + "A": "In Windows, like in Microsoft Word or Microsoft Excel, you can your font to Symbol. Then, that symbol is SHIFT+2 (or the @ key on your keyboard.). As a side note, with the Symbol font you get most of the Greek letters..alpha, beta, delta, pi, etc.", + "video_name": "CJrVOf_3dN0", + "timestamps": [ + 99 + ], + "3min_transcript": "Let's talk a little bit about congruence, congruence And one to think about congruence, it's really kind of equivalence for shapes So, when in algebra when something is equal to another thing it means that their quantities are the same But when we're all of the sudden talking about shapes and we say that those shapes are the same, the shapes are the same size and shape then we say that they're congruent And just to see a simple example here: I have this triangle, right over there and let's say I have this triangle right over here And if you are able to shift, you are able to shift this triangle and flip this triangle, you can make it look exactly like this triangle As long as you're not changing the lengths of any of the sides or the angles here But you can flip it, you can shift it, you can rotate it So you can shift, let me write this, you can shift it, you can flip it and you can rotate If you can do those three procedures to make these And if you say that a triangle is congruent, let me label this So, let's call this triangle ABC Now let's call this D, let me call it XYZ XY and Z So, if we were to say, if we make the claim that both of these triangles are congruent So, if we say triangle ABC is congruent And the way you specify it, it almost look like an equal sign But it's equal sign with a curly thing on top Let me write it a little bit either So, we would write it like this If we know that triangle ABC is congruent to triangle XYZ That means their corresponding sides have the same length And their corresponding angles have the same measure So, if we make this assumption or someone tells us that this is true then we know, for example, that AB is going to equal to XY And we could do this like this, and I'm assuming this are the corresponding sides And you can see that actually we've defined these triangles A corresponds to X, B corresponds to Y and C corresponds to Z right over there So, side AB is gonna have the same length as XY Then you can sometimes if you don't have the colors you can denote it just like that These two length are- or this two lines segments have the same length And you can actually say this, you don't always see this written this way You could also make the statement that line segment AB is congruent to line segment XY But congruence of line segments really just means that their lengths are equivalent So, these two things mean the same thing" + }, + { + "Q": "At 1:20 can you reflect a triangle also?", + "A": "yes. that is what flip means.", + "video_name": "CJrVOf_3dN0", + "timestamps": [ + 80 + ], + "3min_transcript": "Let's talk a little bit about congruence, congruence And one to think about congruence, it's really kind of equivalence for shapes So, when in algebra when something is equal to another thing it means that their quantities are the same But when we're all of the sudden talking about shapes and we say that those shapes are the same, the shapes are the same size and shape then we say that they're congruent And just to see a simple example here: I have this triangle, right over there and let's say I have this triangle right over here And if you are able to shift, you are able to shift this triangle and flip this triangle, you can make it look exactly like this triangle As long as you're not changing the lengths of any of the sides or the angles here But you can flip it, you can shift it, you can rotate it So you can shift, let me write this, you can shift it, you can flip it and you can rotate If you can do those three procedures to make these And if you say that a triangle is congruent, let me label this So, let's call this triangle ABC Now let's call this D, let me call it XYZ XY and Z So, if we were to say, if we make the claim that both of these triangles are congruent So, if we say triangle ABC is congruent And the way you specify it, it almost look like an equal sign But it's equal sign with a curly thing on top Let me write it a little bit either So, we would write it like this If we know that triangle ABC is congruent to triangle XYZ That means their corresponding sides have the same length And their corresponding angles have the same measure So, if we make this assumption or someone tells us that this is true then we know, for example, that AB is going to equal to XY And we could do this like this, and I'm assuming this are the corresponding sides And you can see that actually we've defined these triangles A corresponds to X, B corresponds to Y and C corresponds to Z right over there So, side AB is gonna have the same length as XY Then you can sometimes if you don't have the colors you can denote it just like that These two length are- or this two lines segments have the same length And you can actually say this, you don't always see this written this way You could also make the statement that line segment AB is congruent to line segment XY But congruence of line segments really just means that their lengths are equivalent So, these two things mean the same thing" + }, + { + "Q": "At 2:27, how are 1, 3, 9, and 27 [[\"\"positive\"\"]] divisors of 27,000", + "A": "They are all positive numbers that 27 000 can be equally divided into. 27 000/1=1, 27 000/3=9000, 27 000/9=3000, 27 000/27= 1000. These are actually all the positive divisors of 27. He knows that every single divisor of 27 would be a divisor of 27 000 =)", + "video_name": "17st-s5gg10", + "timestamps": [ + 147 + ], + "3min_transcript": "Calculate the sum of all positive divisors of 27,000. The easiest thing that I can think of doing is first take the prime factorization of 27,000, and then that will help us kind of structure our thought of what all of the different divisors of 27,000 would have to look like. So 27,000 is the same thing as 27 times 1,000, which is the same thing as 3 to the third times 10 to the third, and 10 is, of course, the same thing as 2 times 5. So this is the same thing as 2 times 5 to the third, or it's the same thing as 2 to the third times 5 to the third. So 27,000 is equal to 2 to the third times 3 to the third times 5 to the third. So any divisor of 27,000 is going to have to be made up of the product of up to three 2's, up to three 3's, and up So let's try to look at all the combinations and think of a fast way of summing them. So let's just say it has no fives in it. It has no fives in a divisor. So if it has no fives, then it could have up to three 2's, so let's say it has zero 2's. So I'm just going to take the powers of 2, so if it has zero 2's, then we'll put a 1 here, if it has two 2's, it has to be divisible by 4. If it has three 2's, it's going to be divisible by 8. When I say three 2's, I mean 2 times 2 times 2. Now, let's do it with the 3's. If you have, oh wait, I forgot a power. If you have zero 2's, that means it's just divisible by 1 from looking at the 2's. If you have one 2, it's divisible by just 2. If you have two 2's, you're divisible by 4. And if you have three 2's, and when I mean that I'm saying 2 times 2 times 2, that means you're divisible by 8. Let's do the same thing with 3. From the point of view of the 3, if you have no 3's, that means at least you're divisible by 1. If you have one 3, that means you're divisible by 3. If you have three 3's, it means you're divisible by 27. So let's look at all of the possible combinations. And for this grid that I'm going to generate right here, we assume that you're not divisible by 5, or you're only divisible by 5 to the zero power. So what are all the possible numbers here? Well, you have 1 times 1 is 1. That's divisible by 1 and 1. You have 1 times 3, which is 3, 1 times 9 which is 9, 1 times 27 which is 27. So these are all the numbers that are divisible by that have up to three 3's in them, from zero to three 3's in them, and they have no twos in them. If you throw another two in here, you're essentially going to multiply all of these numbers by two. If you throw another two in here, you're going to multiply all of these numbers by 2. Now, before I do this, because I want to do this as fast as possible. I could figure out what these numbers are, I could multiply them. But instead, let's just take the sum. Let's just take the sum here of this row," + }, + { + "Q": "At 0:56, when Sal is graphing y-k=x^2, he puts the vertex higher on the graph. I don't understand why it's higher on the graph if x^2 is k LESS than y. Shouldn't it be lower on the graph?", + "A": "Think of it this way: you would have to decrease the y value by k units to get down to x^2. That must mean that y is higher than it would be on the x^2 curve.", + "video_name": "99v51U3HSCU", + "timestamps": [ + 56 + ], + "3min_transcript": "Here I've drawn the most classic parabola, y is equal to x squared. And what I want to do is think about what happens-- or how can I go about shifting this parabola. And so let's think about a couple of examples. So let's think about the graph of the curve. This is y is equal to x squared. Let's think about what the curve of y minus k is equal to x squared. What would this look like? Well, right over here, we see when x is equal to 0, x squared is equal to 0. That's this yellow curve. So x squared is equal to y, or y is equal to x squared. But for this one, x squared isn't equal to y. It's equal to y minus k. So when x equals a 0, and we square it, 0 squared doesn't get us to y. It gets us to y minus k. So this is going to be k less than y. Or another way of thinking about it, this is 0. If it's k less than y, y must be at k, wherever k might be. So y must be at k, right over there. So at least for this point, it had the effect And that's actually true for any of these values. So let's think about x being right over here. For this yellow curve, you square this x value, and you get it there. And it's clearly not drawn to scale the way that I've done it right over here. But now for this curve right over here, x squared doesn't cut it. It only gets you to y minus k. So y must be k higher than this. So this is y minus k. y must be k higher than this. So y must be right over here. So this curve is essentially this blue curve shifted up by k. So making it y minus k is equal to x squared shifted it up by k. Whatever value this is, shift it up by k. This distance is a constant k, the vertical distance between these two parabolas. And I'll try to draw it as cleanly as I can. Now let's think about shifting in the horizontal direction. Let's think about what happens if I were to say y is equal to, not x squared, but x minus h squared. So let's think about it. This is the value you would get for y when you just square 0. You get y is equal to 0. How do we get y equals 0 over here? Well, this quantity right over here has to be 0. So x minus h has to be 0, or x has to be equal to h. So let's say that h is right over here. So x has to be equal to h. So one way to think about it is, whatever value you were squaring here to get your y, you now have to have an h higher value to square that same thing. Because you're going to subtract h from it. Just to get to 0, x has to equal h. Here, if you wanted to square 1, x just had to be equal to 1." + }, + { + "Q": "At 5:10, how did Sal come to know that (23/4) is (5 3/4) ?", + "A": "Divide it by 4; you have 20 as the nearest multiple so 3 remains giving 5 3/4 With division practice you become better", + "video_name": "w56Vuf9tHfA", + "timestamps": [ + 310 + ], + "3min_transcript": "that really uses our knowledge of the vertex of a parabola to be able to figure out where the focus and the directrix is going to be. So let's think about the vertex of this parabola right over here. Remember, the vertex, if the parabola is upward opening like this, the vertex is this minimum point. If it is downward opening, it's going to be this maximum point. And so when you look over here, you see that you have a negative one-third in front of the x minus one squared. So this quantity over here is either going to be zero or negative. It's not going to add to 23 over four, it's either gonna add nothing or take away from it. So this thing's going to hit a maximum point, when this thing is zero, when this thing is zero, and that's just gonna go down from there and when this thing is zero, y is going to be equal to 23 over four. So our vertex is going to be that maximum point. Well, when x equals one. When x equals one, you get one minus one squared. So zero squared times negative one-third, this is zero. So when x is equal to one, we're at our maximum y value of 23 over four which five and three-fourths. Actually, let me write that as a . Actually, I'll leave just that's our vertex. and it is a downward opening parabola. So actually, let me start to draw this. So we'd get some axis here. So we have to go all the way up to five and three-fourths. So. Let's make this our y, this is our y axis. This is the x axis. That's the x axis. We're gonna see, we're gonna go to one. Let's call that one. Let's call that two. And then I wanna get, let's see, if I go to five and three-fourths, let's go up to, let's see one, two, three, four We can label 'em. One, two, three, four five, six and seven and so our vertex is right over here. One comma 23 over four, so that's five and three-fourths. So it's gonna be right around right around there and as we said, since we have a negative value in front of this x minus one squared term, I guess we could call it, this is going to be a downward opening parabola. This is going to be a maximum point. So our actual parabola is going to look is going to look something it's gonna look something like this. It's gonna look something like this and we could, obviously, I'm hand drawing it, so it's not going to be exactly perfect, but hopefully you get the general idea of what the parabola is going look like and actually, let me just do part of it," + }, + { + "Q": "at 4:58 Sal got c=1 but why I get C=1/2 ?\nhere my solution\nx=-1/2y^-1 +c\nafter subst (1,-1)\nc= 1/2\nNOT: it seems like when I put C in x part like x+c=... and when I put it in y part like x= ....+c I get different values of C. So Do all of C's are correct?", + "A": "Yes, both Cs are correct, because they don t represent the same thing. In Sal s case, his C is on the x side of the equation (plus he swallowed the 2 into his C when he solved for y), and in your case, your C in on the y side of the equation. Still, if you use your value of C = 1/2 in your equation and solve for y, you get the exact same result that Sal did: x = -1/2 y\u00e2\u0081\u00bb\u00c2\u00b9 + C x = -1/2 y\u00e2\u0081\u00bb\u00c2\u00b9 + 1/2 x - 1/2 = -1/2 y\u00e2\u0081\u00bb\u00c2\u00b9 -2(x - 1/2) = y\u00e2\u0081\u00bb\u00c2\u00b9 -2x + 1 = y\u00e2\u0081\u00bb\u00c2\u00b9 y\u00e2\u0081\u00bb\u00c2\u00b9 = -2x + 1 y = 1/(-2x + 1)", + "video_name": "E444KhRcWSk", + "timestamps": [ + 298 + ], + "3min_transcript": "let's see, I can multiply both sides by negative two, and then I'm gonna have, the left hand side you're just gonna have Y to the negative one, or 1/Y is equal to, if I multiply the right hand side times negative two, I'm gonna have negative two times X plus, well it's some arbitrary constant, it's still going to, it's gonna be negative two times this arbitrary constant but I could still just call it some arbitrary constant, and then if we want we can take the reciprocal of both sides, and so we will get Y is equal to, is equal to 1/-2X+C. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for C. Oh sorry, when X is one, when X is one, Y is negative one, so we get negative one is equal to 1/-2+C, or we could say C minus two, we could multiply both sides times C minus two, if then we will get, actually let me just scroll down a little bit, so if you multiply both sides times C minus two, negative one times C minus two is going to be negative C plus two or two minus C is equal to one. All I did is I multiplied C minus two times both sides, and then, let's see, I can subtract two from both sides, so negative C is equal to negative one, and then if I multiply both sides by negative one, we get C is equal to one. So our particular solution is Y is equal to 1/-2X+1. we didn't just ask for the particular solution, we asked, what is Y when X is equal to three. So Y is going to be equal to one over, three times negative two is negative six plus one, which is equal to negative, is going to be equal to 1/-5, or -1/5. And we are done." + }, + { + "Q": "At 1:45, how can we say that sample mean=p (i.e. the proportion of teachers who think computers is a good tool) ? Is there a rule or something to take that value which I've missed?\nBecause for a binomial distribution E(X)= np where n is the number of trials and p is the success proportion.", + "A": "Just use the same old formula (sum x(i))/250, with 108 of the x = 0, and 142 of them = 1. So the mean is 142/250 Now: How many teachers think computers are a good tool? How many teachers are there? What proportion think computers are a good tool?", + "video_name": "SeQeYVJZ2gE", + "timestamps": [ + 105 + ], + "3min_transcript": "In a local teaching district, a technology grant is available to teachers in order to install a cluster of four computers in their classroom. From the 6,250 teachers in the district, 250 were randomly selected and asked if they felt that computers were an essential teaching tool for their classroom. Of those selected, 142 teachers felt that the computers were an essential teaching tool. And then they ask us, calculate a 99% confidence interval for the proportion of teachers who felt that the computers are an essential teaching tool. So let's just think about the entire population. We weren't able to survey all of them, but the entire population, some of them fall in the bucket, and we'll define that as 1, they thought it was a good tool. They thought that the computers were a good tool. And we'll just define a 0 value as a teacher And some proportion of the total teachers think that it is a good learning tool. So that proportion is p. And then the rest of them think it's a bad learning tool, 1 minus p. We have a Bernoulli Distribution right over here, and we know that the mean of this distribution or the expected value of this distribution is actually going to be p. So it's actually going to be a value, it's neither 0 or 1, so not an actual value that you could actually get out of a teacher if you were to ask them. They cannot say something in between good and not good. The actual expected value is something in between. It is p. Now what we do is we're taking a sample of those 250 teachers, and we got that 142 felt that the computers were an essential teaching tool. So in our survey, so we had 250 sampled, and we got 142 So we got 142 1's, or we sampled 1, 142 times from this And then the rest of the time, so what's left over? There's another 108 who said that it's not good. So 108 said not good, or you could view them as you were sampling a 0, right? 108 plus 142 is 250. So what is our sample mean here? We have 1 times 142, plus 0 times 108 divided by our total number of samples, divided by 250. It is equal to 142 over 250. You could even view this as the sample proportion of teachers who thought that the computers were a good teaching tool. Now let me get a calculator out to calculate this." + }, + { + "Q": "at 18:12, shouldn't it be \"take larger samples\" instead of MORE samples? 2 different things", + "A": "When he says take more samples , he means take larger samples or make more observations . People use the word sample ambiguously - as in: a sample of 100 samples , instead of a sample of 100 observations .", + "video_name": "SeQeYVJZ2gE", + "timestamps": [ + 1092 + ], + "3min_transcript": "I can delete this right here. Let me clear it. I can replace this, because we actually did take a sample. So I can replace this with 0.568. So we could be confident that there's a 99% chance that 0.568 is within 0.08 of the population proportion, which is the same thing as the population mean, which is the same thing as the mean of the sampling distribution of the sample mean, so forth and so on. And just to make it clear we can actually swap these two. It wouldn't change the meaning. If this is within 0.08 of that, then that is within 0.08 of this. So let me switch this up a little bit. So we could put a p is within of-- let me switch this up-- of 0.568. And now linguistically it sounds a little bit more like a confidence interval. We are confident that there's a 99% chance that p is within So what would be our confidence interval? It will be 0.568 plus or minus 0.08. And what would that be? If you add 0.08 to this right over here, at the upper end you're going to have 0.648. And at the lower end of our range, so this is the upper end, the lower end. If we subtract 8 from this we get 0.488. So we are 99% confident that the true population proportion is between these two numbers. Or another way, that the true percentage of teachers who think those computers are good ideas is between-- we're 99% confident-- we're confident that there's a 99% chance that the true percentage of teachers that like the computers is between 48.8% and 64.8%. Now we answered the first part of the question. The second part, how could the survey be changed to narrow 99% confidence interval? Well, you could just take more samples. If you take more samples than our estimate of the standard deviation of this distribution will go down because this denominator will be higher. If the denominator is higher then this whole thing will go down. So if the standard deviations go down here, then when we count the standard deviations, when we do the plus or minus on the range, this value will go down and So you just take more samples." + }, + { + "Q": "At 11:01 you just end saying that it is .5 + .495 to get .995 to look up on the z table. Why .995? Just a few seconds before that you said the interval is symmetric about the mu and the right half was .495. Since that is .495 and .495 + .495 = .99 which is the confidence level we want, why do .5 + .495? That lost me.", + "A": "For z = 2.58, probability (area) is .9951. But this is the area from minus infinity to +2.58 SDs. We want the area from the 0 to 2.58 SDs (so we can double it), so we subtract .5. Then we have .4951*2 = .9902, approximately .99 = 99%", + "video_name": "SeQeYVJZ2gE", + "timestamps": [ + 661 + ], + "3min_transcript": "chance, or how many-- let me think of it this way. How many standard deviations away from the mean do we have to be that we can be 99% confident that any sample from the sampling distribution will be in that interval? So another way to think about think it, think about how many standard deviations we need to be away from the mean, so we're going to be a certain number of standard deviations away from the mean such that any sample, any mean that we sample from here, any sample from this distribution has a 99% chance of being plus or minus that many standard So it might be from there to there. So that's what we want. We want a 99% chance that if we pick a sample from the sampling distribution of the sample mean, it will be within this many standard deviations of the actual mean. And to figure that out let's look at an actual Z-table. So another way to think about it if we want 99% confidence, if we just look at the upper half right over here, that orange area should be 0.475, because if this is 0.475 then this other part's going to be 0.475, and we will get to our-- oh sorry, we want to get to 99%, so it's not going to be 0.475. We're going to have to go to 0.495 if we want 99% So this area has to be 0.495 over here, because if that is, that over here will also be. So that their sum will be 99% of the area. Now if this is 0.495, this value on the z table right here will have to be 0.5, because all of this area, if you include all of this is going to be 0.5. So it's going to be 0.5 plus 0.495. Let me make sure I got that right. 0.995. So let's look at our Z-table. So where do we get 0.995. on our z table? 0.995. is pretty close, just to have a little error, it will be right over here-- this is 0.9951. So another way to think about it is 99-- so this value right here gives us the whole cumulative area up to that, up to our mean. So if you look at the entire distribution like this, this is the mean right over here. This tells us that at 2.5 standard deviations above the mean, so this is 2.5 standard deviations above the mean. So this is 2.5 times the standard deviation of the sampling distribution." + }, + { + "Q": "But why wait to round it at 5:30", + "A": "Remember, mathematicians don t like to play with long numbers as shown in scientific notation. Besides, this is just an example, so we want to work the easy way, as long as his watchers understand.", + "video_name": "XJBwJjP2_hM", + "timestamps": [ + 330 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:30, couldn't I multiply 0.3979 and 10^5 BOTH by 10? Wouldn't the value be kept equivalent? Or no?", + "A": "No.It wouldn t since it ll make it 3.979*10^6=3979000 but the original answer is 39790. Hope that helps!", + "video_name": "XJBwJjP2_hM", + "timestamps": [ + 270 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:10, How come the inches in the water increase?", + "A": "the ring s volume is 1.5, so it displaces the water so it stays even, Because atoms cant go inside of each other...", + "video_name": "ViFLPsLTO1k", + "timestamps": [ + 10 + ], + "3min_transcript": "Jamie wants to know the volume of his gold ring in cubic inches. He gets a rectangular glass with base 3 inches by 2 inches. So you see that here, the base is 3 inches by 2 inches. And he fills the glass 4 inches high with water. So you see that over here, 4 inches high with water. Jamie drops his gold ring in the glass and measures the new height of the water to be 4.25 inches. So this is after the gold ring is dropped. What is the volume of Jamie's ring in cubic inches? Well when you start with this water right over here and you add his ring, whatever that volume is of his ring is going to displace an equal volume of water and push it up. And so the incremental volume that you now have is essentially going to be the volume of his ring. Well what is the incremental volume here? Well it's going to be the volume. If you think about going from this before volume to the after volume, the difference is the base stays the same. It's 3 inches by 2 inches, the difference is-- to make it a little bit neater-- the base is the same. The height now is 4.25 inches after dropping in the ring So the water went up by 0.25 inches. Let me write that, 0.25 inches is what the water went up by. So we could just think about, what is this incremental volume going to be? So this incremental volume right over here, that I'm shading in with purple. Well to figure that out we just have to measure. We just have to multiply the length times the width times the height times 0.25. So it's just going to be 3 times 2 times 0.25. 3 times 2 is 6, times 0.25, and you could do that either on paper or you might be able do that in your head. 4 times 0.25 is going to be 1, and you have 2 more So this is going to be 1.50. And we multiply it inches times inches times inches. So this is going to be in terms of cubic inches. 1.5 cubic inches is the volume of Jamie's ring, which is actually a pretty sizable volume for a gold ring. Maybe he has a very big finger or he just likes to spend, or I guess is his, whoever bought him the ring likes to spend a lot on gold." + }, + { + "Q": "at 0:55 what does sal mean by the leibniz notation", + "A": "Gottfried Wilhelm Leibniz, used the symbols dx and dy to represent infinitely small (or infinitesimal) increments of x and y. So f (x)= dy/dx, this was named in honor of Leibniz.", + "video_name": "6o7b9yyhH7k", + "timestamps": [ + 55 + ], + "3min_transcript": "- [Voiceover] Let's now introduce ourselves to the idea of a differential equation. And as we'll see, differential equations are super useful for modeling and simulating phenomena and understanding how they operate. But we'll get into that later. For now let's just think about or at least look at what a differential equation actually is. So if I were to write, so let's see here is an example of differential equation, if I were to write that the second derivative of y plus two times the first derivative of y is equal to three times y, this right over here is a differential equation. Another way we could write it if we said that y is a function of x, we could write this in function notation. We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function. Or if we wanted to use the Leibniz notation, we could also write, the second derivative of y of y with respect to x is equal to three times y. All three of these equations are really representing the same thing, they're saying OK, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself. So just to be clear, these are all essentially saying the same thing. And you might have just caught from how I described it that the solution to a differential equation is a function, or a class of functions. It's not just a value or a set of values. So the solution here, so the solution to a differential equation is a function, or a set of functions, or a class of functions. It's important to contrast this relative to a traditional equation. So let me write that down. traditional equation, differential equations have been around for a while. So let me write this as maybe an algebraic equation that you're familiar with. An algebraic equation might look something like, and I'll just write up a simple quadratic. Say x squared plus three x plus two is equal to zero. The solutions to this algebraic equation are going to be numbers, or a set of numbers. We can solve this, it's going to be x plus two times x plus one is equal to zero. So x could be equal to negative two or x could be equal to negative one. The solutions here are numbers, or a set of values that satisfy the equation. Here it's a relationship between a function and its derivatives. And so the solutions, or the solution," + }, + { + "Q": "At 6:53 Why did Sal say that second deravative of why is also same as the first one ?", + "A": "If y = e^x, then the first derivative y is also equal to e^x (e^x is its own derivative). The derivative of the first derivative, known as the second derivative y , is therefore also equal to e^x. Thus, the first derivative of y is equal to the second derivative of y. Also why is how we pronounce the letter y.", + "video_name": "6o7b9yyhH7k", + "timestamps": [ + 413 + ], + "3min_transcript": "So let's verify that. So we first have the second derivative of y. So that's that term right over there. So we have nine e to the negative three x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative three x. Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative three x. Three e to the negative three x. So these two terms right over here, nine e to the negative three x, essentially minus six e to the negative three x, that's gonna be three e to the negative three x. Which is indeed equal to three e to the negative three x. So y one is indeed a solution to this differential equation. But as we'll see, it is not the only solution to this differential equation. For example, let's say y two is equal to e to the x is also a solution to this differential equation. And I encourage you to pause the video again and verify that it's a solution. So assuming you've had a go at it. So the first derivative of this is pretty straight-forward, is e to the x. Second derivative, one of the profound things of the exponential function, the second derivative here is also e to the x. in those same colors. So the second derivative is going to be e to the x plus two times e to the x is indeed going to be equal to three times e to the x. This is absolutely going to be true. E to the x plus two e to the x is three e to the x. So y two is also a solution to this differential equation. So that's a start. In the next few videos, we'll explore this more. We'll start to see what the solutions look like, what classes of solutions are, techniques for solving them, visualizing solutions to differential equations, and a whole toolkit for kind of digging in deeper." + }, + { + "Q": "At 13:01, Sal says that any non-zero number is equal to 1. Why is zero not included?", + "A": "0^0 is 1 in certain contexts and indeterminate in other contexts.", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 781 + ], + "3min_transcript": "x squared times y squared. So I can rearrange this, and I will rearrange it so that it's in a way that's easy to simplify. So I can multiply 2 times 3, and then I can worry about the x terms. Let me do it in this color. Then I have times x times x to the fourth times x squared. And then I have to worry about the y terms, times y squared times another y squared times another y squared. And now what are these equal to? Well, 2 times 3. You knew how to do that. That's equal to 6. And what is x times x to the fourth times x squared. Well, one thing to remember is x is the same thing as x to the first power. Anything to the first power is just that number. So you know, 2 to the first power is just 2. So what is this going to be equal to? This is going to be equal to-- we have the same base, x. We can add the exponents, x to the 1 plus 4 plus 2 power, and I'll add it in the next step. And then on the y's, this is times y to the 2 plus 2 plus 2 power. And what does that give us? That gives us 6 x to the seventh power, y to the sixth power. And I'll just leave you with some thing that you might already know, but it's pretty interesting. And that's the question of what happens when you take something to the zeroth power? So if I say 7 to the zeroth power, What does that equal? And I'll tell you right now-- and this might seem very counterintuitive-- this is equal to 1, or 1 to the zeroth power is also equal to 1. Anything that the zeroth power, any non-zero number to And just to give you a little bit of intuition on why that is. Think about it this way. 3 to the first power-- let me write the powers-- 3 to the first, second, third. We'll just do it the with the number 3. So 3 to the first power is 3. I think that makes sense. 3 to the second power is 9. 3 to the third power is 27. And of course, we're trying to figure out what should 3 to the zeroth power be? Well, think about it. Every time you decrement the exponent. Every time you take the exponent down by 1, you are dividing by 3. To go from 27 to 9, you divide by 3. To go from 9 to 3, you divide by 3. So to go from this exponent to that exponent, maybe we should divide by 3 again. And that's why, anything to the zeroth power, in this case, 3 to the zeroth power is 1. See you in the next video." + }, + { + "Q": "At 12:57, Sal says that 1^0 = 1. Why is that? Why isn't it undefined?\n\nAlso, why is any number to the 0th power 1?", + "A": "lets take an exponent series of x x^0 x^1 x^2 x^3 Now lets write what we know: x^0 = ? x^1 = x x^2 = x * x x^3 = x * x * x Now to get to x^3 to x^2 what do we do? We divide x^3 by x: x^3/x = x^2 because x * x * x/x = x * x = x^2 This logic works for getting from x^2 to x^1 x * x/x = x = x^1 Now extrapolating backwards, wouldn t going from x^1 to x^0 be the same as just x^1 divided by x? So: x^1/x = x^0 = x/x = 1 So any number to the zeroth power is equal to 1, except for 0 itself.", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 777 + ], + "3min_transcript": "x squared times y squared. So I can rearrange this, and I will rearrange it so that it's in a way that's easy to simplify. So I can multiply 2 times 3, and then I can worry about the x terms. Let me do it in this color. Then I have times x times x to the fourth times x squared. And then I have to worry about the y terms, times y squared times another y squared times another y squared. And now what are these equal to? Well, 2 times 3. You knew how to do that. That's equal to 6. And what is x times x to the fourth times x squared. Well, one thing to remember is x is the same thing as x to the first power. Anything to the first power is just that number. So you know, 2 to the first power is just 2. So what is this going to be equal to? This is going to be equal to-- we have the same base, x. We can add the exponents, x to the 1 plus 4 plus 2 power, and I'll add it in the next step. And then on the y's, this is times y to the 2 plus 2 plus 2 power. And what does that give us? That gives us 6 x to the seventh power, y to the sixth power. And I'll just leave you with some thing that you might already know, but it's pretty interesting. And that's the question of what happens when you take something to the zeroth power? So if I say 7 to the zeroth power, What does that equal? And I'll tell you right now-- and this might seem very counterintuitive-- this is equal to 1, or 1 to the zeroth power is also equal to 1. Anything that the zeroth power, any non-zero number to And just to give you a little bit of intuition on why that is. Think about it this way. 3 to the first power-- let me write the powers-- 3 to the first, second, third. We'll just do it the with the number 3. So 3 to the first power is 3. I think that makes sense. 3 to the second power is 9. 3 to the third power is 27. And of course, we're trying to figure out what should 3 to the zeroth power be? Well, think about it. Every time you decrement the exponent. Every time you take the exponent down by 1, you are dividing by 3. To go from 27 to 9, you divide by 3. To go from 9 to 3, you divide by 3. So to go from this exponent to that exponent, maybe we should divide by 3 again. And that's why, anything to the zeroth power, in this case, 3 to the zeroth power is 1. See you in the next video." + }, + { + "Q": "At 3:13, how does 3x^2 equal 27x^2? They seem like two different numbers.", + "A": "it was actually cubed not squared. On the video it says: 3x . 3x . 3x = (3 . 3 . 3)(x . x . x) (3x)^3 = 27x^3 you can test it by substituting say a 2 for the x value, then you would get 216 = 216", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 193 + ], + "3min_transcript": "5 times 5 times 5 times 5 times 5 times 5 times 5. One, two, three, four, five, six, seven. This is going to be a really, really, really, really, large number and I'm not going to calculate it right now. If you want to do it by hand, feel free to do so. Or use a calculator, but this is a really, really, really, large number. So one thing that you might appreciate very quickly is that exponents increase very rapidly. 5 to the 17th would be even a way, way more massive number. But anyway, that's a review of exponents. Let's get a little bit steeped in algebra, using exponents. So what would 3x-- let me do this in a different color-- what would 3x times 3x times 3x be? Well, one thing you need to remember about multiplication multiplication in. So this is going to be the same thing as 3 times 3 times 3 times x times x times x. And just based on what we reviewed just here, that part right there, 3 times 3, three times, that's 3 to the third power. And this right here, x times itself three times. that's x to the third power. So this whole thing can be rewritten as 3 to the third times x to the third. Or if you know what 3 to the third is, this is 9 times 3, which is 27. This is 27 x to the third power. Now you might have said, hey, wasn't 3x times 3x times 3x. Wasn't that 3x to the third power? Right? You're multiplying 3x times itself three times. And I would say, yes it is. So this, right here, you could interpret that as 3x to the And just like that, we stumbled on one of our exponent properties. Notice this. When I have something times something, and the whole thing is to the third power, that equals each of those things to the third power times each other. So 3x to the third is the same thing is 3 to the third times x to the third, which is 27 to the third power. Let's do a couple more examples. What if I were to ask you what 6 to the third times 6 to the sixth power is? And this is going to be a really huge number, but I want to write it as a power of 6. Let me write the 6 to the sixth in a different color. 6 to the third times 6 to the sixth power, what is this going to be equal to? Well, 6 to the third, we know that's 6 times itself three times. So it's 6 times 6 times 6." + }, + { + "Q": "Is there a test anywhere with questions similar to 10:08? I enjoyed simplifying the expression but I want more challenges like that to see if I can do it again.\n\nThe test \"Practice Multiply Powers\" only has simple questions", + "A": "Keep working thru this section. There are later exercise sets that combine 2 or more of the exponent properties which makes them more challenging.", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 608 + ], + "3min_transcript": "So just to review the properties we've learned so far in this video, besides just a review of what an exponent is, if I have x to the a power times x to the b power, this is going to be equal to x to the a plus b power. We saw that right here. x squared times x to the fourth is equal to x to the sixth, 2 plus 4. We also saw that if I have x times y to the a power, this is the same thing is x to the a power times y to the a power. We saw that early on in this video. We saw that over here. 3x to the third is the same thing as 3 to the third times x to the third. That's what this is saying right here. 3x to the third is the same thing is 3 to the third times x to the third. is, if you have x to the a and then you raise that to the bth power, that's equal to x to the a times b. And we saw that right there. a to the third and then raise that to the fourth power is the same thing is a to the 3 times 4 or a to the 12th power. So let's use these properties to do a handful of more complex problems. Let's say we have 2xy squared times negative x squared y squared times three x squared y squared. And we wanted to simplify this. squared times y squared. So if we take this whole thing to the squared power, this is like raising each of these to the second power. So this part right here could be simplified as negative 1 squared times x squared squared, times y squared. And then if we were to simplify that, negative 1 squared is just 1, x squared squared-- remember you can just multiply the exponents-- so that's going to be x to the fourth y squared. That's what this middle part simplifies to. And let's see if we can merge it with the other parts. The other parts, just to remember, were 2 xy squared, and then 3x squared y squared. Well now we're just going ahead and just straight up multiplying everything. And we learned in multiplication that it doesn't matter which order you multiply things in. So I can just rearrange. We're just going and multiplying 2 times x times y" + }, + { + "Q": "at 11:12, Sal is rearranging the problem. He begins with 2x3. Where is the 3 coming from? Please enlighten me on this.", + "A": "The 3 is coming from that last term in parentheses (3x^2y^2). First he multiplied all the numbers together (the middle term didn t have a number), then the letters.........", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 672 + ], + "3min_transcript": "is, if you have x to the a and then you raise that to the bth power, that's equal to x to the a times b. And we saw that right there. a to the third and then raise that to the fourth power is the same thing is a to the 3 times 4 or a to the 12th power. So let's use these properties to do a handful of more complex problems. Let's say we have 2xy squared times negative x squared y squared times three x squared y squared. And we wanted to simplify this. squared times y squared. So if we take this whole thing to the squared power, this is like raising each of these to the second power. So this part right here could be simplified as negative 1 squared times x squared squared, times y squared. And then if we were to simplify that, negative 1 squared is just 1, x squared squared-- remember you can just multiply the exponents-- so that's going to be x to the fourth y squared. That's what this middle part simplifies to. And let's see if we can merge it with the other parts. The other parts, just to remember, were 2 xy squared, and then 3x squared y squared. Well now we're just going ahead and just straight up multiplying everything. And we learned in multiplication that it doesn't matter which order you multiply things in. So I can just rearrange. We're just going and multiplying 2 times x times y x squared times y squared. So I can rearrange this, and I will rearrange it so that it's in a way that's easy to simplify. So I can multiply 2 times 3, and then I can worry about the x terms. Let me do it in this color. Then I have times x times x to the fourth times x squared. And then I have to worry about the y terms, times y squared times another y squared times another y squared. And now what are these equal to? Well, 2 times 3. You knew how to do that. That's equal to 6. And what is x times x to the fourth times x squared. Well, one thing to remember is x is the same thing as x to the first power. Anything to the first power is just that number. So you know, 2 to the first power is just 2." + }, + { + "Q": "On the second problem (7:23), why is '8' divisible by both '9' and '24'? I honestly don't get it because I'm thinking of it as '9/8' or something. I get the part of 2.2.2, though.\nCould somebody please point out what's wrong here? thanks in advance.", + "A": "The point is that any number divisible by 9 and 24, is also divisible by 8. Why?! because the prime factorization of 9 and 24 contains the prime factorization of 8. by the way, at first that confuse me :)", + "video_name": "zWcfVC-oCNw", + "timestamps": [ + 443 + ], + "3min_transcript": "And once again we just do the prime factorization. We essentially think about the least common multiple of 9 and 24. You take the prime factorization of 9, it's 3 times 3 and we're done. Prime factorization of 24 is 2 times 12. 12 is 2 times 6, 6 is 2 times 3. So anything that's divisible by 9 has to have a 9 in it's factorization or if you did its prime factorization would have to be a 3 times 3, anything divisible by 24 has got to have three 2's in it, so it's gotta have a 2 times a 2 times a 2, and it's got to have at lease one 3 here. And we already have at least one 3 from the 9, so we have that. So this number right here is divisible by both 9 and 24. And this number right here is actually 72. This is 8 times 9, which is 72. let's assume that it was multiple choice. Let's say the choices here were 16 27 5 11 and 9. So 16, if you were to do its prime factorization, is 2 times 2 times 2 times 2. It's 2 to the 4th power. So you would need four 2's here. We don't have four 2's over here. I mean there could be some other numbers here but we don't know what they are. These are the only numbers that we can assume are in the prime factorization of something divisible by both 9 and 24. So we can rule out 16. We don't have four 2's here. 27 is equal to 3 times 3 times 3, so you need three 3's in the prime factorization. We don't have three 3's. So once again, cancel that out. 5's a prime number. There are no 5's here. Rule that out. No 11's here. Rule that out. 9 is equal to 3 times 3. And actually I just realized that this is a silly answer because obviously all numbers divisible by 9 and 24 are also divisible by 9. So obviously 9 is going to work but I shouldn't have made that a choice cuz that's in the problem, but 9 would work, and what also would work if we had a, if 8 was one of the choices, because 8 is equal to 2 times 2 times 2, and we have a 2 times 2 times 2 here. 4 would also work. That's 2 times 2. That's 2 times 2. 6 would work since that's 2 times 3. 18 would work cuz that's 2 times 3 times 3. So anything that's made up of a combination of these prime factors will be divisible into something divisible by both 9 and 24. Hopefully that doesn't confuse you too much." + }, + { + "Q": "4:05 12 is a number that is in the problem!\n\nAll numbers divisible by both 12 and 20 are also divisible by 12?", + "A": "Yup, and all numbers divisible by both 12 and 20 are also divisible by 20.", + "video_name": "zWcfVC-oCNw", + "timestamps": [ + 245 + ], + "3min_transcript": "of 12 and 20. Now this isn't the only number that's divisible by both 12 and 20. You could multiply this number right here by a whole bunch of other factors. I could call them a, b, and c, but this is kind of the smallest number that's divisible by 12 and 20. Any larger number will also be divisible by the same things as this smaller number. Now with that said, let's answer the questions. All numbers divisible by both 12 and 20 are also divisible by: Well we don't know what these numbers are so we can't really address it. They might just be 1's or they might not exist because the number might be 60. It might be 120. Who knows what this number is? So the only numbers that we know can be divided into this number, well we know 2 can be, we know that 2 is a legitimate answer. 2 is obviously divisible into 2 times 2 times 3 times 5. We know that 2 times 2 is divisible into it, cuz we have the 2 times 2 over there. We know that 2 times 3 is divisible into it. So that's 6. Let me write these. This is 4. This is 6. We know that 2 times 2 times 3 is divisible into it. I could go through every combination of these numbers right here. We know that 3 times 5 is divisible into it. We know that 2 times 3 times 5 is divisible into it. So in general you can look at these prime factors and any combination of these prime factors is divisible into any number that's divisible by both 12 and 20, so if this was a multiple choice question, and the choices were 7 and 9 and 12 and 8. You would say, well let's see, 7 is not one of these prime factors over here. 9 is 3 times 3 so I need to have two 3's here. I only have one 3 here so 9 doesn't work. 7 doesn't work, 9 doesn't work. or another way to think about it, 12 is 2 times 2 times 3. Well there is a 2 times 2 times 3 in the prime factorization, of this least common multiple of these two numbers, so this is a 12 so 12 would work. 8 is 2 times 2 times 2. You would need three 2's in the prime factorization. We don't have three 2's, so this doesn't work. Let's try another example just so that we make sure that we understand this fairly well. So let's say we wanna know, we ask the same question. All numbers divisible by and let me think of two interesting numbers, all numbers divisible by 12 and let's say 9, and I don't know, let's make it more interesting, 9 and 24 are also divisible by" + }, + { + "Q": "At 1:18, he says that in order to be divisible by both a number has to have 2 2s, a 3 etc... but why 2 2s and just 1 3?", + "A": "12 needed 2 2s and one 3 and 20 needed a 5 and 2 2s but the 2s were thare and he use prime numbers and 2, 3, and 5s are prime", + "video_name": "zWcfVC-oCNw", + "timestamps": [ + 78 + ], + "3min_transcript": "- [Voiceover] In this video I wanna do a bunch of example problems that show up on standardized exams and definitely will help you with our divisibility module because it's asking you questions like this. And this is just one of the examples. All numbers divisible by both 12 and 20 are also divisible by: And the trick here is to realize that if a number is divisible by both 12 and 20, it has to be divisible by each of these guy's prime factors. So let's take the prime factorization, the prime factorization of 12, let's see, 12 is 2 times 6. 6 isn't prime yet so 6 is 2 times 3. So that is prime. So any number divisible by 12 needs to be divisible by 2 times 2 times 3. So its prime factorization needs to have a 2 times a 2 times a 3 in it, any number that's divisible by 12. Now any number that's divisible by 20 needs to be divisible by, let's take it's prime factorization. 2 times 10 10 is 2 times 5. So any number divisible by 20 2 times 2 times 5. Or another way of thinking about it, it needs to have two 2's and a 5 in its prime factorization. If you're divisible by both, you have to have two 2's, a 3, and a 5, two 2's and a 3 for 12, and then two 2's and a 5 for 20, and you can verify this for yourself that this is divisible by both. Obviously if you divide it by 20, let me do it this way. Dividing it by 20 is the same thing as dividing by 2 times 2 times 5, so you're going to have the 2's are going to cancel out, the 5's are going to cancel out. You're just going to have a 3 left over. So it's clearly divisible by 20, and if you were to divide it by 12, you'd divide it by 2 times 2 times 3. This is the same thing as 12. And so these guys would cancel out and you would just have a 5 left over so it's clearly divisible by both, and this number right here is 60. It's 4 times 3, which is 12, times 5, it's 60. of 12 and 20. Now this isn't the only number that's divisible by both 12 and 20. You could multiply this number right here by a whole bunch of other factors. I could call them a, b, and c, but this is kind of the smallest number that's divisible by 12 and 20. Any larger number will also be divisible by the same things as this smaller number. Now with that said, let's answer the questions. All numbers divisible by both 12 and 20 are also divisible by: Well we don't know what these numbers are so we can't really address it. They might just be 1's or they might not exist because the number might be 60. It might be 120. Who knows what this number is? So the only numbers that we know can be divided into this number, well we know 2 can be, we know that 2 is a legitimate answer. 2 is obviously divisible into 2 times 2 times 3 times 5. We know that 2 times 2 is divisible into it, cuz we have the 2 times 2 over there." + }, + { + "Q": "at 2:42, to isolate the -2, why did you have to divide and not subtract?", + "A": "be careful with leading negative numbers. Try to think of this as a negative number, not subtraction. The negative two is being multiplied by the absolute value. In order to cancel this number we do the opposite of multiplying by -2 which is dividing by -2.", + "video_name": "15s6B7K9paA", + "timestamps": [ + 162 + ], + "3min_transcript": "If these two things are equal, and if I want to keep them equal, if I subtract 6 from the right-hand side, I've got to subtract-- or if I subtract 6 times the absolute value of x plus 10 from the right-hand side, I have to subtract the same thing from the left-hand side. So we're going to have minus 6 times the absolute value of x plus 10. And likewise, I want to get all my constant terms, I want to get this 4 out of the left-hand side. So let me subtract 4 from the left, and then I have to also do it on the right, otherwise my equality wouldn't hold. And now let's see what we end up with. So on the left-hand side, the 4 minus 4, that's 0. You have 4 of something minus 6 of something, that means you're going to end up with negative 2 of that something. Negative 2 of the absolute value of x plus 10. Remember, this might seem a little confusing, but remember, if you had 4 apples and you subtract 6 apples, you now have negative 2 apples, Same way, you have 4 of this expression, you take away 6 of this expression, you now have negative 2 of this expression. Let me write it a little bit neater. So it's negative 2 times the absolute value of x plus 10 is equal to, well the whole point of this, of the 6 times the absolute value of x plus 10 minus 6 times the absolute value of x plus 10 is to make those cancel out, and then you have 10 minus 4, which is equal to 6. Now, we want to solve for the absolute value of x plus 10. So let's get rid of this negative 2, and we can do that by dividing both sides by negative 2. You might realize, everything we've done so far is just treating this red expression as almost just like a variable, and we're going to solve for that red expression and then take it from there. So negative 2 divided by negative 2 is 1. 6 divided by negative 2 is negative 3. So we get the absolute value of x plus 10 is equal to negative 3. You might say maybe this could be the positive version or the negative, but remember, absolute value is always non-negative. If you took the absolute value of 0, you would get 0. But the absolute value of anything else is going to be positive. So this thing right over here is definitely going to be greater than or equal to 0. Doesn't matter what x you put in there, when you take its absolute value, you're going to get a value that's greater than or equal to 0. So there's no x that you could find that's somehow-- you put it there, you add 10, you take the absolute value of it, you're actually getting a negative value. So this right over here has absolutely no solution. And I'll put some exclamation marks there for emphasis." + }, + { + "Q": "At 6:10 can it be y=k/x since it is y=k*1/x", + "A": "Certainly! That is what inverse variations generally look like. People don t usually write y = 8 * 1/x, they usually write it as y = 8/x.", + "video_name": "92U67CUy9Gc", + "timestamps": [ + 370 + ], + "3min_transcript": "That's what it means to vary directly. Now, it's not always so clear. Sometimes it will be obfuscated. So let's take this example right over here. y is equal to negative 3x. And I'm saving this real estate for inverse variation in a second. You could write it like this, or you could algebraically manipulate it. You could maybe divide both sides of this equation by x, and then you would get y/x is equal to negative 3. Or maybe you divide both sides by x, and then you divide both sides by y. So from this, so if you divide both sides by y now, you could get 1/x is equal to negative 3 times 1/y. These three statements, these three equations, are all saying the same thing. So sometimes the direct variation isn't quite in your face. But if you do this, what I did right here with any of these, you will get the exact same result. to this form over here. And there's other ways we could do it. We could divide both sides of this equation by negative 3. And then you would get negative 1/3 y is equal to x. And now, this is kind of an interesting case here because here, this is x varies directly with y. Or we could say x is equal to some k times y. And in general, that's true. If y varies directly with x, then we can also say that x varies directly with y. It's not going to be the same constant. It's going to be essentially the inverse of that constant, but they're still directly varying. Now with that said, so much said, about direct variation, let's explore inverse variation a little bit. Inverse variation-- the general form, if we use the same variables. And it always doesn't have to be y and x. It could be an a and a b. It could be a m and an n. If I said m varies directly with n, we would say m is equal to some constant times n. So if I did it with y's and x's, this would be y is equal to some constant times 1/x. So instead of being some constant times x, it's some constant times 1/x. So let me draw you a bunch of examples. It could be y is equal to 1/x. It could be y is equal to 2 times 1/x, which is clearly the same thing as 2/x. It could be y is equal to 1/3 times 1/x, which is the same thing as 1 over 3x. it could be y is equal to negative 2 over x. And let's explore this, the inverse variation, the same way that we explored the direct variation. So let's pick-- I don't know/ let's pick y is equal to 2/x. And let me do that same table over here. So I have my table. I have my x values and my y values. If x is 1, then y is 2." + }, + { + "Q": "At 4:20,how can we assume it is an isosceles pyramid? We don't know for sure,do we? That question might have been a trick question.", + "A": "I guess because we are using the actual Great Pyramid of Giza for the problem, and it is in real life isosceles, we can assume this.", + "video_name": "Z5EnuVJawmY", + "timestamps": [ + 260 + ], + "3min_transcript": "And what else do we know? Well, we know this is 72. We know that the whole thing is 180. So this is 72, and the whole thing is 180. The part of this edge that's below the water, this distance Let me draw it without cluttering the picture too much. I'll do it in that black color. This distance right over here is going to be 108. 108 plus 72 is 180. So what does this do for us? We need to figure out this height. We know that this right over here is a right triangle. I could color this in just to make it a little bit clearer. This thing in yellow, right over here is a right triangle. If we look at that right triangle, and if we wanted to solve for h and solve for h using a trig ratio based on this angle theta right over here, we know that relative to this angle theta, And this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? Well, we just write SOHCAHTOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse. So we get the cosine of theta is going to be equal to the height that we care about. That's the adjacent side of this right triangle over the length of the hypotenuse, OVER 108. Well, that doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here. Theta is also sitting up here. So maybe if we can figure out what cosine of theta is based up here, then we can solve for h. So if we look at this data, what is the cosine of theta? And now we're looking at a different right triangle. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here. We already know that's 139 meters. So it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here. It's 72 plus 108. Oh, we already have it labeled here. It's 180. We can assume that this is an isosceles-- that this pyramid is an isosceles triangle. So 180 on that side and 180 on that side. So the cosine is adjacent-- 139-- over the hypotenuse, which is 180, over 180. And these data are the same data. We just showed that. So now we have cosine of theta is h/108. Cosine of theta is 139/180. Or we could say that h/108, which is cosine of theta," + }, + { + "Q": "At 1:42 Sal says to move the 0 degrees on the protractor to one side of the angle. How do you know which side to use?", + "A": "One side is in the angle, if the angle is right or acute, than use the side that flows into the actual angle, obtuse angles can vary.", + "video_name": "wJ37GJyViU8", + "timestamps": [ + 102 + ], + "3min_transcript": "This is the video for the measuring angles module because, clearly, at the time that I'm doing this video, there is no video for the measuring angles module. And this is a pretty neat module. This was made by Omar Rizwan, one of our amazing high school interns that we had this past summer. This is the summer of 2011. And what it really is, is it makes you measure angles. And he made this really cool protractor tool here so that you actually use this protractor to measure the angles there. And so the trick here is you would actually measure it the way you would measure any angle using an actual physical protractor. You'd want to put the center of the protractor right at the vertex of where the two lines intersect. You can view it as the vertex of the angle. And then you'd want to rotate it so that, preferably, this edge, this edge at 0 degrees, is at one of these sides. So let's do it so that this edge right over here is right along this line. So let me rotate it. So then-- I've got to rotate it a little bit further, maybe So that looks about right. And then if you look at it this way, you can see that the angle-- and I don't have my Pen tool here. I'm just using my regular web browser-- if you look at the angle here, you see that the other line goes to 130 degrees. So this angle that we need to measure here is 130 degrees, assuming you can read sideways. So that is 130 degrees. Let me check my answer. Very good, I got it right. It would have been embarrassing if I didn't. Let's do the next question. I'll do a couple of examples like this. So once again, let us put the center of the protractor right at the vertex right over there. And let's get this 0 degrees side to be on one of these sides so that this angle will be within the protractor. So let me rotate it this way. And this really is pretty cool what Omar did with this module. So let's see. Let's do it one more time. That's too far. And then you can see that the angle right over here, if we look at where the other line points to, it is 40 degrees. Check answer-- very good. Let's do another one. This is fun. So let's get our protractor right over there. And you don't always have to do it in that same order. You could rotate it first so that the 0 degrees is-- and what you want to do is you want to rotate the 0 degrees to one of the sides so that the angle is still within the protractor. So let's rotate it around. So if you did it like that-- so you don't always have to do it in that same order. Although I think it's easier to rotate it when you have the center of the protractor at the vertex of the angle. So we have to rotate it a little bit more. So 0 degrees is this line. And then as we go further and further up, I guess, since this is on its side, it looks like this other line gets us to 150 degrees. And hopefully you're noticing that the higher the degrees, the more open this angle is." + }, + { + "Q": "at 1:30 \u00c2\u00a8\u00c3\u00aft would be embarrasing if i didnt\u00c2\u00a8 no sal it would be kinda histerical (is that how you spell it?). and show your human", + "A": "*hysterical, that s how it s spelt :) And seeing as angle measuring is quite simple for a mathematician, it would be pretty embarrassing for him.", + "video_name": "wJ37GJyViU8", + "timestamps": [ + 90 + ], + "3min_transcript": "This is the video for the measuring angles module because, clearly, at the time that I'm doing this video, there is no video for the measuring angles module. And this is a pretty neat module. This was made by Omar Rizwan, one of our amazing high school interns that we had this past summer. This is the summer of 2011. And what it really is, is it makes you measure angles. And he made this really cool protractor tool here so that you actually use this protractor to measure the angles there. And so the trick here is you would actually measure it the way you would measure any angle using an actual physical protractor. You'd want to put the center of the protractor right at the vertex of where the two lines intersect. You can view it as the vertex of the angle. And then you'd want to rotate it so that, preferably, this edge, this edge at 0 degrees, is at one of these sides. So let's do it so that this edge right over here is right along this line. So let me rotate it. So then-- I've got to rotate it a little bit further, maybe So that looks about right. And then if you look at it this way, you can see that the angle-- and I don't have my Pen tool here. I'm just using my regular web browser-- if you look at the angle here, you see that the other line goes to 130 degrees. So this angle that we need to measure here is 130 degrees, assuming you can read sideways. So that is 130 degrees. Let me check my answer. Very good, I got it right. It would have been embarrassing if I didn't. Let's do the next question. I'll do a couple of examples like this. So once again, let us put the center of the protractor right at the vertex right over there. And let's get this 0 degrees side to be on one of these sides so that this angle will be within the protractor. So let me rotate it this way. And this really is pretty cool what Omar did with this module. So let's see. Let's do it one more time. That's too far. And then you can see that the angle right over here, if we look at where the other line points to, it is 40 degrees. Check answer-- very good. Let's do another one. This is fun. So let's get our protractor right over there. And you don't always have to do it in that same order. You could rotate it first so that the 0 degrees is-- and what you want to do is you want to rotate the 0 degrees to one of the sides so that the angle is still within the protractor. So let's rotate it around. So if you did it like that-- so you don't always have to do it in that same order. Although I think it's easier to rotate it when you have the center of the protractor at the vertex of the angle. So we have to rotate it a little bit more. So 0 degrees is this line. And then as we go further and further up, I guess, since this is on its side, it looks like this other line gets us to 150 degrees. And hopefully you're noticing that the higher the degrees, the more open this angle is." + }, + { + "Q": "i am confused:\nat 3:20 how does Sal go from =44 to 00000000+000", + "A": "Well he took the commutative property and broke it down. 4*8+4*3 4*8 is equal to 32. 4*3 = 12. I ll break the equation cause it s just how i work out problems. (4^2+4)+(8*3). 4 times itself is a product of 16.16+4 equals 20. 8*3 =24. 24+20=44.", + "video_name": "gl_-E6iVAg4", + "timestamps": [ + 200 + ], + "3min_transcript": "evaluate it that way. But they want us to use the distributive law of multiplication. We did not use the distributive law just now. We just evaluated the expression. We used the parentheses first, then multiplied by 4. In the distributive law, we multiply by 4 first. And it's called the distributive law because you distribute the 4, and we're going to think about what that means. So in the distributive law, what this will become, it'll become 4 times 8 plus 4 times 3, and we're going to think about why that is in a second. So this is going to be equal to 4 times 8 plus 4 times 3. A lot of people's first instinct is just to multiply the 4 times the 8, but no! You have to distribute the 4. You have to multiply it times the 8 and times the 3. This is the distributive property in action right here. Distributive property in action. And then when you evaluate it-- and I'm going to show you in kind of a visual way why this works. But then when you evaluate it, 4 times 8-- I'll do this in a different color-- 4 times 8 is 32, and then so we have 32 plus 4 times 3. 4 times 3 is 12 and 32 plus 12 is equal to 44. That is also equal to 44, so you can get it either way. But when they want us to use the distributive law, you'd distribute the 4 first. Now let's think about why that happens. Let's visualize just what 8 plus 3 is. Let me draw eight of something. So one, two, three, four, five, six, seven, eight, right? And then we're going to add to that three of something, of One, two, three. So you can imagine this is what we have inside of the parentheses. We have 8 circles plus 3 circles. Now, when we're multiplying this whole thing, this whole thing times 4, what does that mean? Well, that means we're just going to add this to itself four times. Let me do that with a copy and paste. Copy and paste. Let me copy and then let me paste. There you go. That's two. That's one, two, three, and then we have four, and we're going to add them all together. So this is literally what? Four times, right? Let me go back to the drawing tool. We have it one, two, three, four times this expression, which is 8 plus 3. Now, what is this thing over here? If you were to count all of this stuff, you would get 44." + }, + { + "Q": "At 1:22 in the video do you do four times eleven?", + "A": "Yep! 4(11) = 44", + "video_name": "gl_-E6iVAg4", + "timestamps": [ + 82 + ], + "3min_transcript": "Rewrite the expression 4 times, and then in parentheses we have 8 plus 3, using the distributive law of multiplication over addition. Then simplify the expression. So let's just try to solve this or evaluate this expression, then we'll talk a little bit about the distributive law of multiplication over addition, usually just called the distributive law. So we have 4 times 8 plus 8 plus 3. Now there's two ways to do it. Normally, when you have parentheses, your inclination is, well, let me just evaluate what's in the parentheses first and then worry about what's outside of the parentheses, and we can do that fairly easily here. We can evaluate what 8 plus 3 is. 8 plus 3 is 11. So if we do that-- let me do that in this direction. So if we do that, we get 4 times, and in parentheses we have an 11. 8 plus 3 is 11, and then this is going to be equal to-- evaluate it that way. But they want us to use the distributive law of multiplication. We did not use the distributive law just now. We just evaluated the expression. We used the parentheses first, then multiplied by 4. In the distributive law, we multiply by 4 first. And it's called the distributive law because you distribute the 4, and we're going to think about what that means. So in the distributive law, what this will become, it'll become 4 times 8 plus 4 times 3, and we're going to think about why that is in a second. So this is going to be equal to 4 times 8 plus 4 times 3. A lot of people's first instinct is just to multiply the 4 times the 8, but no! You have to distribute the 4. You have to multiply it times the 8 and times the 3. This is the distributive property in action right here. Distributive property in action. And then when you evaluate it-- and I'm going to show you in kind of a visual way why this works. But then when you evaluate it, 4 times 8-- I'll do this in a different color-- 4 times 8 is 32, and then so we have 32 plus 4 times 3. 4 times 3 is 12 and 32 plus 12 is equal to 44. That is also equal to 44, so you can get it either way. But when they want us to use the distributive law, you'd distribute the 4 first. Now let's think about why that happens. Let's visualize just what 8 plus 3 is. Let me draw eight of something. So one, two, three, four, five, six, seven, eight, right? And then we're going to add to that three of something, of" + }, + { + "Q": "At 1:56 how did you get x = 4.? And one ore thing I am still not getting what does this theorem states?", + "A": "The theorem says that the third side of a triangle has to be LESS than the sum of the other two sides AND MORE than the rest of the two sides. So when you subtract 6 from 10 you get 4.", + "video_name": "KlKYvbigBqs", + "timestamps": [ + 116 + ], + "3min_transcript": "Let's draw ourselves a triangle. Let's say this side has length 6. Let's say this side right over here has length 10. And let's say that this side right over here has length x. And what I'm going to think about is how large or how small that value x can be. How large or small can this side be? So the first question is how small can it get? Well, if we want to make this small, we would just literally have to look at this angle right over here. So let me take a look at this angle and make it smaller. So let's try to make that angle as small as possible. So we have our 10 side. Actually let me do it down here. So you have your 10 side, the side of length 10, and I'm going to make this angle really, really, really small, approaching 0. If that angle becomes 0, we end up with a degenerate triangle. It essentially becomes one dimension. But as we approach 0, this side starts to coincide or get closer and closer to the 10 side. And you could imagine the case where it actually coincides with it and you actually get the degenerate. So if want this point right over here to get as close as possible to that point over there, essentially minimizing your distance x, the closest way is if you make the angle the way equal to 0, all the way. So let's actually-- let me draw a progression. So now the angle is getting smaller. This is length 6. x is getting smaller. Then we keep making that angle smaller and smaller and smaller all the way until we get a degenerate triangle. So let me draw that pink side. So you have the side of length 10. Now the angle is essentially 0, this angle that we care about. So this side is length 6. And so what is the distance between this point and this point? And that distance is length x. We know that 6 plus x is going to be equal to 10. So in this degenerate case, x is going to be equal to 4. So if you want this to be a real triangle, at x equals 4 you've got these points as close as possible. It's degenerated into a line, into a line segment. If you want this to be a triangle, x has to be greater than 4. Now let's think about it the other way. How large can x be? Well to think about larger and larger x's, we need to make this angle bigger. So let's try to do that. So let's draw my 10 side again. So this is my 10 side. I'm going to make that angle bigger and bigger. So now let me take my 6 side and put it like that. And so now our angle is getting bigger and bigger and bigger. It's approaching 180 degrees. At 180 degrees, our triangle once again" + }, + { + "Q": "I don't understand, at around 8:00 when Sal is explaining vector [c1 c2 0 c4 0] he said that c1,c2,c4 must be zero showing linear independence, now aren't the third and fifth term also zero? If c1 c2 c4 are zero then the solution would be [0 0 0 0 0], doesn't that imply all column vectors to be linearly independent?", + "A": "He says that the 1st, 2nd and 4th c must be 0 for the result to be the 0 vector, and hence are linearly independent. Whereas, the 3rd and 5th just happen to be zero.. but since these vectors can be made from the other vectors, you would also be able to find a way to get back to 0 even if these happened not to be. (He mentions showing this in the next video).", + "video_name": "BfVjTOjvI30", + "timestamps": [ + 480 + ], + "3min_transcript": "do it over here, let me do it in blue-- you get c1 times a1 plus c2 times a2, and then 0 times a3, plus c4 times a4 is equal to 0. Now these guys are going to be linearly independent, if and only if the only solution to this equation is they all equal to 0. Well we know that the only solution to this is that they all equal 0 because anything that's a solution to this is a solution to this. And the only solution to this was, if I go ahead and I constrain these two terms to being equal to 0, the only solution to this is all of these c's have to be 0. So likewise, if I constrain these to be 0, the only solution to this is that c1, c2, and c4 have to be 0. vectors, a1, a2, and a4, so that implies that the set a1, a2, and a4 are linearly independent. So we're halfway there. We've shown that because the pivot columns here are linearly independent. We can show and they have the same solution set. The null space of the reduced row echelon form is the same as the null space of our original matrix. We were able to show that the only solution to c1 times this plus c2 times this plus c4 times this is when all the constants are 0, which shows that these three vectors or a set of those three vectors are definitely linearly independent. Now, the next thing to prove that they are a basis, is to show that all of the other column vectors can be represented as multiples of these three guys. boring you too much, I'll do that in the next video. So in this one we saw that if the pivot columns are linearly independent, they always are. All pivot columns, by definition are linearly Or the set of pivot columns are always linearly independent when you take away the non-pivot columns, then the corresponding columns in your original vector are also In the next one we'll show that these three guys also span your column space." + }, + { + "Q": "At 8:16, when you have the area of 324 pi. Why can't you substitute pi with 3 because 3.14 rounded is 3 and multiply 3 by 324?", + "A": "The answer will not be accurate", + "video_name": "mLE-SlOZToc", + "timestamps": [ + 496 + ], + "3min_transcript": "There's actually an infinite number of points you could pick here. And so, when we talk about the probability that the point also lies in the smaller circle, we're really thinking about the percentage of the points in the larger circle that are also in the smaller circle. Or another way to think about it is the probability that if we pick a point from this larger circle, the probably that it's also in the smaller circle is really just going to be the percentage of the larger circle that is the smaller circle. I know that might sound confusing, but we're really just have to figure out the areas for both of them, and it's really just going to be the ratios so let's think about that. So there's a temptation to just use this 36 pi up here, but we have to remember, this was the circumference, and we need to figure out the area of both of these circles. And so for area, we need to know the radius, because area is pi r squared. So we can figure out the radius from the circumference by saying, well, circumference is equal to 2 times pi times the radius of the circle. is equal to 2 times pi times the radius, we can divide both sides by 2 pi, and on the left hand side, 36 divided by 2 is 18 the pi's cancel out, we get our radius as being equal to 18 for this larger circle. So if we want to know its area, its area is going to be pi r squared, which is equal to pi times 18 squared. And let's figure out what 18 squared is. 18 times 18, 8 times 8 is 64, eight times 1 is 8 plus 6 is 14, and then we put that 0 there because we're now in the tens place, 1 times 8 is 8, 1 times 1 is 1. And really, this is a 10 times the 10, and that's why it gives us 100. Anyway, 4 plus 0 is a 4, 4 plus 8 is a 12, So the area here is equal to pi times 324, or we could say 324 pi. So the area of the entire larger circle, the part that I shaded in yellow, including what's kind of under this orange circle, if you want to view it that way, this area right over here is equal to 324 pi. So the probability that a point that we select from this larger circle is also in the smaller circle is really just a percentage of the larger circle that is the smaller circle. So our probability-- I'll just write it like this-- the probability that the point also lies in the smaller circle-- so all of that stuff The probability of that is going to be equal to the percentage of this larger circle that is this smaller one, and that's going to be--" + }, + { + "Q": "At 5:15 , what is a circumference ?", + "A": "The circumference of a circle is the length of the perimeter of the circle, or the distance all the way around. It is equal to pi times the diameter of the circle or pi times 2 times the radius.", + "video_name": "mLE-SlOZToc", + "timestamps": [ + 315 + ], + "3min_transcript": "of the total possibilities that meet our constraint, and our constraint is being a multiple of 5. So how many total possibilities are there? Let's think about that. How many do we have? Well that's just the total number of numbers we have to pick from, so 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 possibilities. We have an equal chance of picking any one of these 12. Now which of these 12 are a multiple of 5? So let's do this in a different color. So let me pick out the multiples of 5. 32 is not a multiple of 5, 49 is not a multiple of 5. 55 is a multiple of 5. Really, we're just looking for the numbers that in the ones place you either have a 5 or a 0. 55 is a multiple of 5, 30 is a multiple 55 is 11 times 5. Not 56, not 28. This is clearly 5 times 10, this is 8 times 5, this is the same number again, also 8 times 5. So all of these are multiples of 5. 45, that's 9 times 5. 3 is not a multiple of 5. 25, clearly 5 times 5. So I've circled all the multiples of 5. So of all the possibilities, the ones that meet our constraint of being a multiple of 5, there are 1, 2, 3, 4, 5, 6, 7 possibilities. So 7 meet our constraint. So in this example, the probability of a selecting a number that is a multiple of 5 is 7/12. Let's do another one. the circumference of a circle is 36 pi. Let's draw this circle. so let's say the circle looks-- I can draw a neater circle than that. So let's say the circle looks something like that. And its circumference-- we have to be careful here, they're giving us interesting-- the circumference is 36 pi. Then they tell us that contained in that circle is a smaller circle with area 16 pi. So inside the bigger circle, we have a smaller circle that has an area of 16 pi. A point is selected at random from inside the larger circle, so we're going to randomly select some point in this larger circle. What is the probability that the point also lies in this smaller circle? So here's a little bit interesting, because you actually have an infinite number of points in both of these circles, because it's not kind of a separate balls or marbles," + }, + { + "Q": "Before the time of 3:04, shouldn't the non green marbles be 11/14?", + "A": "No, we are trying to find the probability of picking a NON-BLUE marble. There are 14 marbles total, two of which are blue. Thus the number of non-blue marbles is 12/14, which simplifies down to 6/7.", + "video_name": "mLE-SlOZToc", + "timestamps": [ + 184 + ], + "3min_transcript": "And the way you just think about it is what fraction of all of the possible events meet our constraint? So let's just think about all of the possible events first. How many different possible marbles can we take out? Well that's just the total number of marbles there are. So are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 possible marbles. So this is the number of possibilities. And then we just have to think what fraction of those possibilities meet our constraints. And the other way you could have gotten 14 is just taking 9 plus 2 plus 3. So what number of those possibilities meet our constraints? And remember, our constraint is selecting a non-blue marble Another way to think about it is a red or green marble, because the only other two colors we have are red and green. So how many non-blue marbles are there? Well, there's a couple ways to think about it. 2 are blue. So there are going to be 14 minus 2, which is 12 non-blue marbles. Or you could just count them. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 non-blue marbles. So these are the possibilities that meet our constraints over all of the possibilities. And then if we want to-- this isn't in simplified form right here, since both 12 and 14 are divisible by 2. So let's divide both the numerator and the denominator by 2, and you get 6 over 7. So we have a 6/7 chance of selecting a non-blue marble from the bag. Let's do another one. If a number is randomly chosen from the following list, what is the probability that the number is a multiple of 5? of the total possibilities that meet our constraint, and our constraint is being a multiple of 5. So how many total possibilities are there? Let's think about that. How many do we have? Well that's just the total number of numbers we have to pick from, so 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 possibilities. We have an equal chance of picking any one of these 12. Now which of these 12 are a multiple of 5? So let's do this in a different color. So let me pick out the multiples of 5. 32 is not a multiple of 5, 49 is not a multiple of 5. 55 is a multiple of 5. Really, we're just looking for the numbers that in the ones place you either have a 5 or a 0. 55 is a multiple of 5, 30 is a multiple" + }, + { + "Q": "At 1:31, isn't there an easier way of solving?", + "A": "I m sure there is! But Sal is not trying to show us the easiest way, here: he s trying to explain the principles of probabilities and choice. Understanding and explaining these principles is far more important for Sal than showing us the easiest way to find a solution.", + "video_name": "mLE-SlOZToc", + "timestamps": [ + 91 + ], + "3min_transcript": "Let's do a couple of exercises from our probability one module. So we have a bag with 9 red marbles, 2 blue marbles, and 3 green marbles in it. What is the probability of randomly selecting a non-blue marble from of the bag? So let's draw this bag here. So that's my bag, and we're going to assume that it's a transparent bag, so it looks like a vase. But we have 9 red marbles, so let me draw 9 red marbles. 1, 2, 3, 4, 5, 6, 7, 8, 9 red marbles. They're kind of orange-ish, but it does the job. 2 blue marbles, so we have 1 blue marble, 2 blue marbles. And then we have 3 green marbles, let me draw those 3, so 1, 2, 3. What is the probability of randomly selecting a non-blue marble from the bag? So maybe we mix them all up, and we have an equal probability And the way you just think about it is what fraction of all of the possible events meet our constraint? So let's just think about all of the possible events first. How many different possible marbles can we take out? Well that's just the total number of marbles there are. So are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 possible marbles. So this is the number of possibilities. And then we just have to think what fraction of those possibilities meet our constraints. And the other way you could have gotten 14 is just taking 9 plus 2 plus 3. So what number of those possibilities meet our constraints? And remember, our constraint is selecting a non-blue marble Another way to think about it is a red or green marble, because the only other two colors we have are red and green. So how many non-blue marbles are there? Well, there's a couple ways to think about it. 2 are blue. So there are going to be 14 minus 2, which is 12 non-blue marbles. Or you could just count them. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 non-blue marbles. So these are the possibilities that meet our constraints over all of the possibilities. And then if we want to-- this isn't in simplified form right here, since both 12 and 14 are divisible by 2. So let's divide both the numerator and the denominator by 2, and you get 6 over 7. So we have a 6/7 chance of selecting a non-blue marble from the bag. Let's do another one. If a number is randomly chosen from the following list, what is the probability that the number is a multiple of 5?" + }, + { + "Q": "at 5:30 sal says area of a circle. however,circles dont have area,circles have circumfrence", + "A": "All two-dimensional shapes have a perimeter and an area. Perimeter is a distance, the length of the bounding line(s). Area is ... well ... an area. It measures the amount of a flat surface that is contained within the perimeter. For circles, the perimeter is called the circumference.", + "video_name": "mLE-SlOZToc", + "timestamps": [ + 330 + ], + "3min_transcript": "of the total possibilities that meet our constraint, and our constraint is being a multiple of 5. So how many total possibilities are there? Let's think about that. How many do we have? Well that's just the total number of numbers we have to pick from, so 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 possibilities. We have an equal chance of picking any one of these 12. Now which of these 12 are a multiple of 5? So let's do this in a different color. So let me pick out the multiples of 5. 32 is not a multiple of 5, 49 is not a multiple of 5. 55 is a multiple of 5. Really, we're just looking for the numbers that in the ones place you either have a 5 or a 0. 55 is a multiple of 5, 30 is a multiple 55 is 11 times 5. Not 56, not 28. This is clearly 5 times 10, this is 8 times 5, this is the same number again, also 8 times 5. So all of these are multiples of 5. 45, that's 9 times 5. 3 is not a multiple of 5. 25, clearly 5 times 5. So I've circled all the multiples of 5. So of all the possibilities, the ones that meet our constraint of being a multiple of 5, there are 1, 2, 3, 4, 5, 6, 7 possibilities. So 7 meet our constraint. So in this example, the probability of a selecting a number that is a multiple of 5 is 7/12. Let's do another one. the circumference of a circle is 36 pi. Let's draw this circle. so let's say the circle looks-- I can draw a neater circle than that. So let's say the circle looks something like that. And its circumference-- we have to be careful here, they're giving us interesting-- the circumference is 36 pi. Then they tell us that contained in that circle is a smaller circle with area 16 pi. So inside the bigger circle, we have a smaller circle that has an area of 16 pi. A point is selected at random from inside the larger circle, so we're going to randomly select some point in this larger circle. What is the probability that the point also lies in this smaller circle? So here's a little bit interesting, because you actually have an infinite number of points in both of these circles, because it's not kind of a separate balls or marbles," + }, + { + "Q": "At around 2:15 , why does (1-cos(2x))\u00c2\u00b2 turn into 1-2cos(2x)+cos\u00c2\u00b2(2x) ? Where does the 2cos(2x) come from ?", + "A": "Sal was simply expanding the expression (1/2 (1-cos2x))^2. 1/2 squared is 1/4 and (1-cos2x) squared is (1-cos2x) times (1-cos2x). If you recall from Algebra that (a-b)(a-b) = a^2 -2ab + b^2, then let a = 1 and let b = cos2x and multiply it out. a^2 = (1)(1) = 1, -2ab = -2(1)(cos2x) = -2cos2x. b^2 = (cos2x)(cos2x) = cos^2 2x. Hope this helps. Good luck.", + "video_name": "n34jx1FIN8M", + "timestamps": [ + 135 + ], + "3min_transcript": "- [Voiceover] Let's see if we can take the indefinite integral of sine of x to the fourth dx. And like always, pause the video and see if you can work through it on your own. So if we had an odd exponent up here, whether it was a sine or a cosine, then the technique I would use, so if this was sine to the third of x, I would separate one of the sine of xs out, so I would rewrite it as sine squared x times sine of x, and then I would convert this using the Pythagorean Identity, and then when I distribute the sine of x, I'd be able to use u-substitution. We've done that in previous example videos. You could have done this if it was cosine to the third of x as well, or to the fifth, or to the seventh, if you had an odd exponent. But here we have an even exponent. So what do we do? So the technique we will use, and I guess you could call it a trick, the technique or trick to use is once again you want to algebraically manipulate this so you can use integration techniques that we are familiar with. the Double Angle Identity. The Double Angle Identity tells us that sine squared of x is equal to 1/2 times one minus cosine of two x. So how can I apply this over here? Our original integral is just the same thing as, this is going to be the same thing as the integral of sine of x squared, all of that squared, dx. Now I can make this substitution. So this is the same thing as this, which is of course the same thing as this by the Double Angle Identity. So I could rewrite it as the integral of 1/2 times one minus cosine of two x, and then all of that squared dx. That's going to be equal to, let's see. 1/2 squared is 1/4, so I can take that out. So we get 1/4 times the integral of... I'm just going to square all this business. One squared, which is one, minus two cosine of two x plus cosine squared of two x dx. Fairly straightforward to take the indefinite integral, or to take the antiderivative of these two pieces. But what do I do here? Once again I've got an even exponent. Let's apply the Double Angle Identity for cosine. We know that cosine squared of two x is going to be equal to 1/2 times one plus cosine of double this angle, so cosine of four x. Once again just make the substitution." + }, + { + "Q": "towards the end of the video at 10:45 sal says that the cut off for staticians is 5% or less. why is it 5% and less and not 5% and more. can i please have an example as well as to why it is less than 5%", + "A": "We start with an assumption (null hypothesis), and calculate the probability of the observed result if that assumption is actually true. Hence, small probability will make us say Hmm, these results don t match up with our assumption very well, the assumption is probably wrong. Large probability does not make us question our assumption: it says the data are fairly compatible with the assumption, so it might be okay.", + "video_name": "W3C07uH-b9o", + "timestamps": [ + 645 + ], + "3min_transcript": "is this occurring? Well, this one occurs 85 times, this one occurs eight. If you add these two together, 93 out of the thousand times, out of her re-randomization or I guess you could say 9.3 percent of the time, the data... 9.3 percent of the randomized, the 1000 re-randomizations, 9.3 percent of the time she got data that was as validating of a hypothesis or more than the actual experiment. One way to think about this is, the probability of randomly getting the results from her experiment or better results from her experiment are 9.3 percent. They're low, it's a reasonably low probability that this happened purely by chance. Now, a question is, \"What's the threshold?\" If it was a 50 percent you say, \"Okay, this was very If this was a 25 percent you're like, \"Okay, it's less \"likely to happen by chance but it could happen.\" 9.3 percent, it's roughly 10 percent. For every 10 people who do an experiment like she did, even if it was random, one person would get data like this? What typically happens amongst statisticians is they draw a threshold and the threshold for statistical significance is usually five percent. One way to think about it, the probability of her getting this result by chance, this result or a more extreme result? One that more confirms her hypothesis by chance is 9.3 percent. If you're cut-off for significance is five percent. If you said, \"Okay, this has to be five percent or less.\" Then you say, \"Okay, this is not statistically significant.\" There's more than a five percent chance that I could have gotten this result purely through random chance. Once again, that just depends on where you have that threshold. When we go back, I think we've already answered the final question, \"According to the simulations, \"being lower than the control group's median \"by eight minutes or more?\" Which once again, eight minutes or more, that would be negative eight and negative 10. We just figured that out, that was 93 out of the 1000 re-randomizations, so it's a 9.3 percent chance. If you set five percent as your cut-off for statistical significance, you say, \"Okay, this doesn't quite meet my \"cut-off so maybe this is not a statistically \"significant result.\"" + }, + { + "Q": "is there a way to convert the remainder into a decimal on 2:06", + "A": "There sure is. Think of 15/4 (15 divided by 4) as how many groups of 4 can 15 be divided into. The answer of 3r3 means 3 full groups of 4 and a partially completed group, the remained, of 3. That partially completed group holds 4 but only has 3 in it. We call this 3/4. So we can rewrite 3r3 as 3 3/4. We can do our division on 3/4 and convert it into a decimal, which equals 0.75. So 15/4 = 3r3 = 3 3/4 = 3.75", + "video_name": "BIGX05Mp5nw", + "timestamps": [ + 126 + ], + "3min_transcript": "Let's take the number 7 and divide it by 3. And I'm going to conceptualize dividing by 3 as let me see how many groups of 3 I can make out of the 7. So let me draw 7 things-- 1, 2, 3, 4, 5, 6, 7. So let me try to create groups of 3. So I can definitely create one group of 3 right over here. I can definitely create another group of 3. So I'm able to create two groups of 3. And then I can't create any more full groups of 3. I have essentially this thing right over here left over. So this right over here, I have this thing remaining. This right over here is my remainder after creating as many groups of 3 as I can. And so when you see something like this, people will often say 7 divided by 3. Well, I can create two groups of 3. But it doesn't divide evenly, or 3 doesn't divide evenly into 7. I end up with something left over. I have a leftover. I have a remainder of 1. And that makes sense. 2 times 3 is 6. So it doesn't get you all the way to 7. But then if you have your extra remainder, 6 plus that 1 remainder gets you all the way to 7. Let's do another one. Let's imagine 15 divided by 4. Let me draw 15 objects-- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. Now, let me try to divide it into groups of 4. So let's see, that's one group of 4. That's another group of 4. And then that's another group of 4. So I'm able to create three groups of 4. But then I can't create a fourth full group of 4. I am then left with this remainder right over here. I have a remainder right over here of 3. So we could say that 15 divided by 4 is 3 remainder 3. 4 goes into 15 three times. But that only gets us to 12. 4 times 3 is 12. To get all the way to 15, we need to use our remainder. We have to get 3 more. So 15 divided by 4, I have 3 left over. Now, let's try to think about this doing a little bit of our long division techniques. So let's say that I have 4. Let's say I want to divide 75 by 4. Well, traditional long division techniques. 4 goes into 7 one time. And If you're looking at place value, we're really saying the 4 is going into 70 ten times, because we're putting this in the tens place. And then we say, 1 times 4 is 4. But really, once again, since it's in the tens place, this is really representing a 40." + }, + { + "Q": "At 1:28 Sal says that we must divide both the sides with 5 but instead of that we can transpose 5.It is much easier that way!", + "A": "fair enough, but I think that dividing both the sides by 5 is good for beginners because the fundamental belief of algebra, that if you do the same thing to both the sides of an equation,you won t change anything seems much more prominent in dividing both sides by 5 than in transposing.", + "video_name": "c6-FJRda_Vc", + "timestamps": [ + 88 + ], + "3min_transcript": "y is directly proportional to x. If y equals 30 when x is equal to 6, find the value of x when y is 45. So let's just take this each statement at a time. y is directly proportional to x. That's literally just saying that y is equal to some constant times x. This statement can literally be translated to y is equal to some constant times x. y is directly proportional to x. Now, they tell us, if y is 30 when x is 6-- and we have this constant of proportionality-- this second statement right over here allows us to solve for this constant. When x is 6, they tell us y is 30 so we can figure out what this constant is. We can divide both sides by 6 and we get this left-hand side is 5-- 30 divided by 6 is 5. 5 is equal to k or k is equal to 5. So the second sentence tells us, this gives us the information that y is equal to 5 times x. y is 30 when x is 6. And then finally, they say, find the value of x when y is 45. So when y is 45 is equal to-- so we're just putting in 45 for y-- 45 is equal to 5x. Divide both sides by 5 to solve for x. We get 45 over 5 is 9, and 5x divided by 5 is just x. So x is equal to 9 when y is 45." + }, + { + "Q": "I calculated M by subtracting the first formula [(mx^2)+bx=xy] from the second (y=mx+b). Which is the opposite of what sal does @0:50. I get a different formula for M, is this OK? I can't equate the two formulas. Here is the formula I get (all the x and y should have the mean sign. M= [(xy/x)-y]/[(x^2/x)-x)", + "A": "Multiply both top and bottom by -1, and the result is [(-xy/x)+y]/[(-x^2/x)+x] where all the x, x^2, and y should have the mean sign. This is equivalent to [y-(xy/x)]/[x-(x^2/x)] simply by switching the order of the terms in the numerator and denominator, respectively. This is Sal s answer.", + "video_name": "8RSTQl0bQuw", + "timestamps": [ + 50 + ], + "3min_transcript": "So if you've gotten this far, you've been waiting for several videos to get to the optimal line that minimizes the squared distance to all of those points. So let's just get to the punch line. Let's solve for the optimal m and b. And just based on what we did in the last videos, there's We actually now know two points that lie on that line. So we can literally find the slope of that line and then the the y intercept, the b there. Or, we could just say it's the solution to this system of equations. And they're actually mathematically equivalent. So let's solve for m first. And if we want to solve for m, we want to cancel out the b's. So let me rewrite this top equation just the way it's written over here. We have m times the mean of the x squareds plus b times the mean of-- Actually, we could even do it better than that. One step better than that is to, based on the work we did in the last video, we can just subtract this bottom equation from this top equation. So let me subtract it. Or let's add the negatives. So if I make this negative, this is negative. What do we get? We get m times the mean of the x's minus the mean of the x squareds over the mean of x. The plus b and the negative b cancel out. Is equal to the mean of the y's minus the mean of the xy's over the mean of the x's. And then, we can divide both sides of the equation by this. And so we get m is equal to the mean of the y's minus the mean of the xy's over the mean of the x's over this. The mean of the x's minus the mean of the x squareds over the mean of the x's. Now notice, this is the exact same thing that you would get if you found the slope between these two points over here. that right over there. Over the change in x's. The change in that x minus that x is exactly this over here. Now, to simplify it, we can multiply both the numerator and the denominator by the mean of the x's. And I do that just so we don't have this in the denominator both places. So if we multiply the numerator by the mean of the x's, we get the mean of the x's times the mean of the y's minus, this and this will cancel out, minus the mean of the xy's. All of that over, mean of the x's times the mean of the x's is just going to be the mean of the x's squared, minus over here you have the mean of the x squared. And that's what we get for m." + }, + { + "Q": "So, I now understand that you can multiply a real number (or scalar, which was defined at 0:30) by a matrix, but can you multiply a Matrix by another Matrix? Wouldn't it be just like adding or subtracting, except you multiply instead?\nFor example:\n\n[1 8 3] * [2 9 4] = [2 72 12]\n\nWould that be correct?", + "A": "No and it is more complicated than that. You can watch Dr. Khan s video on this but I ll give you the quick version. For 2 matrices to be multiplied, the number of rows in the first matrix must be equal to the number of columns in the second matrix (remember matrix multiplication isn t always commutative as AB isn t necessarily BA). In your case, those two matrices cannot be multiplied.", + "video_name": "TbaltFbJ3wE", + "timestamps": [ + 30 + ], + "3min_transcript": "Now that we know what a matrix is, let's see if we can start to define some operations on matrices. So let's say I have the 2 by 3 matrix, so two rows and three columns, and the entries are 7, 5, negative 10, 3, 8, and 0. And I want to define what happens when I multiply 3 times this whole thing. So first of all, let's get a little terminology out of the way. The number three, in just the everyday world, if you weren't dealing with matrices or vectors, and if you don't know what vectors are, don't worry about them just now, you would just call that a number. You would call this a real number. It's just a regular number sitting out there. But now in the world where we have these new structured things, these matrices, these arrays of numbers, we will refer to these just plain old real numbers that aren't part of some type of an array here, we call these scalars. So essentially what we're defining here, we don't know-- I haven't said what this is actually going to turn out to be, but whatever this turns out where we're multiplying a scalar times a matrix. And so how would you define this? What do you think this should be? 3 times this stuff right over here. Well, the world could have defined scalar multiplication however it saw fit, but one way that we find, perhaps, the most obvious and the most useful, is to multiply this scalar quantity times each of the entries. So this is going to be equal to 3 times 7 in the top left, 3 times 5, 3 times negative 10, 3 times 3, 3 times 8, and 3 times 0, which will give us-- it didn't change the dimensions of the matrix. It didn't change, I guess you could say, the structure of the matrix, it just multiplied each of the entries times 3. So the top left entry is now going to be 21, the entry in the middle row, So when you multiply a matrix times a scalar, you just multiply each of those entries times that scalar quantity." + }, + { + "Q": "at 2:20 I have no idea whats going on", + "A": "He is reversing the multiplication he did to make it a whole number so that the decimal would be in the correct place.", + "video_name": "D5fmcpNygQk", + "timestamps": [ + 140 + ], + "3min_transcript": "Let's multiply 1.21, or 1 and 21 hundredths, times 43 thousandths, or 0.043. And I encourage you to pause this video and try it on your own. So let's just think about a very similar problem but one where essentially we don't write the decimals. Let's just think about multiplying 121 times 43, which we know how to do. So let's just think about this problem first as kind of a simplification, and then we'll think about how to get from this product to this product. So we can start with-- so we're going to say 3 times 1 is 3. 3 times 2 is 6. 3 times 1 is 3. 3 times 121 is 363. And now we're going to go to the tens place, so this is a 40 right over here. So since we're in the tens place, let's put a 0 there. 40 times 1 is 40. 40 times 20 is 800. 40 times 100 is 4,000. and now we can just add all of this together. And we get-- let me do a new color here-- 3 plus 0 is 3. 6 plus 4 is 10. 1 plus 3 plus 8 is 12. 1 plus 4 is 5. So 121 times 43 is 5,203. Now, how is this useful for figuring out this product? Well, to go from 1.21 to 121, we're essentially multiplying by 100. We're moving the decimal two places over to the right. And to go from 0.043 to 43, what are we doing? We're removing the decimal, so we're multiplying by ten, hundred, thousand. We're multiplying by 1,000. So to go from this product to this product and we multiplied by 1,000. So then to go back to this product, we have to divide. We should divide by 100 and then divide by 1,000, which is equivalent to dividing by 100,000. But let's do that. So let's rewrite this number here, so 5,203. Actually let me write it like this just so it's a little bit more aligned, 5,203. And we could imagine a decimal point right over here. If we divide by 100-- so you divide by 10, divide by 100-- and then we want to divide by another 1,000. So divide by 10, divide by 100, divide by 1,000. So our decimal point is going to go right over there, and we're done. 1.21 times 0.043 is 0.05203." + }, + { + "Q": "Can someone explain to me how he got the 6i in (9+6i-1) at 5:50? Thanks!", + "A": "Shivanie, He was multiplying (3+i)(3+i) Using FOIL you get First 3*3 = 9 Outside 3*i = 3i Inside i*3 = 3i Last i*i = -1 So you get 9+3i+3i-1 And the 3i+3i = 6i so you get 9+6i-1 I hope that is of help to you.", + "video_name": "dnjK4DPqh0k", + "timestamps": [ + 350 + ], + "3min_transcript": "Those are the two roots. Now what I want to do is a verify that these work. Verify these two roots. So this one I can rewrite as 3 plus i over 2. These are equivalent. All I did-- you can see that this is just dividing both of these by 2. Or if you were to essentially factor out the 1/2, you could go either way on this expression. And this one over here is going to be 3 minus i over 2. Or you could go directly from this. This is 3 plus or minus i over 2. So 3 plus i over 2. Or 3 minus i over 2. This and this or this and this, or this. These are all equal representations of both of the roots. But let's see if they work. So I'm first going to try this character right over here. It's going to get a little bit hairy, because we're going to have to square it and all the rest. But let's see if we can do it. 2 times this quantity squared. So 2 times 3 plus i over 2 squared plus 5. And we want to verify that that's the same thing as 6 times this quantity, as 6 times 3 plus i over 2. So what is 3 plus i squared? So this is 2 times-- let me just square this. So 3 plus i, that's going to be 3 squared, which is 9, plus 2 times the product of three and i. So 3 times i is 3i, times 2 is 6i. So plus 6i. And if that doesn't make sense to you, I encourage you to kind of multiply it out either with the distributive property or FOIL it out, and you'll get the middle term. You'll get 3i twice. When you add them, you get 6i. I And then plus i squared, and i squared is negative 1. Minus 1. All of that over 4, plus 5, is equal to-- well, you get a 3 here and you get a 1 here. And 3 distributed on 3 plus i is equal to 9 plus 3i. And what we have over here, we can simplify it just to save some screen real estate. 9 minus 1 is 8. So if I get rid of this, this is just 8 plus 6i. We can divide the numerator and the denominator right here by 2. So the numerator would become 4 plus 3i, if we divided it by 2, and the denominator here is just going to be 2. This 2 and this 2 are going to cancel out. So on the left hand side, we're left with 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works." + }, + { + "Q": "Can someone explain to me what Sal is doing at 5:29 onwards?", + "A": "Sal is using the same FOIL technique except now there are complex numbers. ((3 + i) / 2)^2 can also be written as ((3 + i)*(3 + i)) / (2*2). By using FOIL the numerator will become... F: 3*3 = 9 O: 3*i = 3i I: 3*i = 3i L: i*i = -1 (3 + i)(3 + i) ----> 9 + 3i + 3i - 1 ----> 8 + 6i 2(8 + 6i) / 4 ----> 4(4 + 3i) / 4 ----> 4 + 3i I hope this helps", + "video_name": "dnjK4DPqh0k", + "timestamps": [ + 329 + ], + "3min_transcript": "I could even do it one step-- that's the same thing as negative 1 times 4 under the radical, which is the same thing as the square root of negative 1 times the square root of 4. And the principal square root of negative 1 is i times the principal square root of 4 is 2. So this is 2i, or i times 2. So this right over here is going to be 2i. So we are left with x is equal to 6 plus or minus 2i over 4. And if we were to simplify it, we could divide the numerator and the denominator by 2. And so that would be the same thing as 3 plus or minus i over 2. Or if you want to write them as two distinct complex numbers, you could write this as 3 plus i over 2, or 3/2 plus 1/2i. That's if I take the positive version of the i there. Or we could view this as 3/2 minus 1/2i. Those are the two roots. Now what I want to do is a verify that these work. Verify these two roots. So this one I can rewrite as 3 plus i over 2. These are equivalent. All I did-- you can see that this is just dividing both of these by 2. Or if you were to essentially factor out the 1/2, you could go either way on this expression. And this one over here is going to be 3 minus i over 2. Or you could go directly from this. This is 3 plus or minus i over 2. So 3 plus i over 2. Or 3 minus i over 2. This and this or this and this, or this. These are all equal representations of both of the roots. But let's see if they work. So I'm first going to try this character right over here. It's going to get a little bit hairy, because we're going to have to square it and all the rest. But let's see if we can do it. 2 times this quantity squared. So 2 times 3 plus i over 2 squared plus 5. And we want to verify that that's the same thing as 6 times this quantity, as 6 times 3 plus i over 2. So what is 3 plus i squared? So this is 2 times-- let me just square this. So 3 plus i, that's going to be 3 squared, which is 9, plus 2 times the product of three and i. So 3 times i is 3i, times 2 is 6i. So plus 6i. And if that doesn't make sense to you, I encourage you to kind of multiply it out either with the distributive property or FOIL it out, and you'll get the middle term. You'll get 3i twice. When you add them, you get 6i. I And then plus i squared, and i squared is negative 1. Minus 1. All of that over 4, plus 5, is equal to-- well," + }, + { + "Q": "At 2:34 Sal says that any percent is that percent over 100. But however what if your number was greater than 100 say like 109%?", + "A": "109% is 109/100, which is 1.09.", + "video_name": "-gB1y-PMWfs", + "timestamps": [ + 154 + ], + "3min_transcript": "It's 0.18. You could view this as 1 tenth and 8 hundredths, which is the same thing, or 10 hundredths and 8 hundredths, which is 18 hundredths. So this is written in decimal form. And if we write it as a simplified fraction, we need to see if there is a common factor for 18 and 100. And they're both even numbers, so we know they're both divisible by 2, so let's divide both the numerator and the denominator by 2. So we have 18 divided by 2 over 100 divided by 2. And we're going to get 18 divided by 2 is 9. 100 divided by 2 is 50. And I don't think these guys share any common factors. 50 is not divisible by 3. 9 is only divisible by 3 and 1 and 9. So this is the fraction in simplest form. So we have 18% is the same thing as 0.18, which is the Now, I went through a lot of pain here to show you that this really just comes from the word, from percent, from per 100. But if you ever were to see this in a problem, the fast way to do this is to immediately say, OK, if I have 18%, you should immediately say, anything in front of the percent-- that's that anything, whatever this anything is-- it should be equal to that anything. In this case it's 18/100. And another way to think about it, you could view this as 18.0%. I just added a trailing zero there, just so that you see the decimal, really. But if you want to express this as a decimal without the percent, you just move the decimal to the left two spaces. this becomes 0.18. Or you could immediately say that 18% as a fraction is 18/100. When you put it in simplified form, it's 9/50. But you should also see that 18/100, and we have seen this, is the exact same thing as 18 hundredths, or 0.18. Hopefully, this made some connections for you and didn't confuse you." + }, + { + "Q": "he lost me at 2:32, what is he saying. Please help", + "A": "(anything)%=(anything)/100 I hope you understand now!", + "video_name": "-gB1y-PMWfs", + "timestamps": [ + 152 + ], + "3min_transcript": "It's 0.18. You could view this as 1 tenth and 8 hundredths, which is the same thing, or 10 hundredths and 8 hundredths, which is 18 hundredths. So this is written in decimal form. And if we write it as a simplified fraction, we need to see if there is a common factor for 18 and 100. And they're both even numbers, so we know they're both divisible by 2, so let's divide both the numerator and the denominator by 2. So we have 18 divided by 2 over 100 divided by 2. And we're going to get 18 divided by 2 is 9. 100 divided by 2 is 50. And I don't think these guys share any common factors. 50 is not divisible by 3. 9 is only divisible by 3 and 1 and 9. So this is the fraction in simplest form. So we have 18% is the same thing as 0.18, which is the Now, I went through a lot of pain here to show you that this really just comes from the word, from percent, from per 100. But if you ever were to see this in a problem, the fast way to do this is to immediately say, OK, if I have 18%, you should immediately say, anything in front of the percent-- that's that anything, whatever this anything is-- it should be equal to that anything. In this case it's 18/100. And another way to think about it, you could view this as 18.0%. I just added a trailing zero there, just so that you see the decimal, really. But if you want to express this as a decimal without the percent, you just move the decimal to the left two spaces. this becomes 0.18. Or you could immediately say that 18% as a fraction is 18/100. When you put it in simplified form, it's 9/50. But you should also see that 18/100, and we have seen this, is the exact same thing as 18 hundredths, or 0.18. Hopefully, this made some connections for you and didn't confuse you." + }, + { + "Q": "How did Sal decide how big the blue circles were at 1:00 to 1:10", + "A": "The first circle needs to be at least half the diameter of the circle win the problem. Then, with the second circle, Sal simply placed the midpoint of the new compass on the midpoint of the first circle, made the circles the same size by adjusting the new compass, and then placed the new circle s midpoint on the opposite side of the circle in the problem. Hope this helps :)", + "video_name": "-gWtl6mdpeY", + "timestamps": [ + 60, + 70 + ], + "3min_transcript": "Construct a square inscribed inside the circle. And in order to do this, we just have to remember that a square, what we know of a square is all four sides are congruent and they intersect at right angles. And we also have to remember that the two diagonals of the square are going to be perpendicular bisectors of each other. So let's see if we can construct two lines that are perpendicular bisectors of each other. And essentially, where those two lines intersect our bigger circle, those are going to be the vertices of our square. So let's throw a straight edge right over here. And let's make a diameter. So that's a diameter right over here. It just goes through the circle, goes through the center of the circle, to two sides of the circle. And now, let's think about how we can construct a perpendicular bisector of this. And we've done this in other compass construction or construction videos. But what we can do is we can put a circle-- let's throw a circle right over here. And what we're going to do is we're going to reuse this. We're going to make another circle that's the exact same size. Put it there. And where they intersect is going to be exactly along-- those two points of intersection are going to be along a perpendicular bisector. So that's one of them. Let's do another one. I want a circle of the exact same dimensions. So I'll center it at the same place. I'll drag it out there. That looks pretty good. I'll move it on to this side, the other side of my diameter. So that looks pretty good. And notice, if I connect that point to that point, I will have constructed a perpendicular bisector of this original segment. So let's do that. Let's connect those two points. So that point and that point. And then, we could just keep going all the way to the end of the circle. Go all the way over there. That looks pretty good. And now, we just have to connect these four So let's do that. So I'll connect to that and that. And then I will connect, throw another straight edge there. I will connect that with that. And then, two more to go. I'll connect this with that, and then one more. I can connect this with that, and there you go. I have a shape whose vertices intersect the circle. And its diagonals, this diagonal and this diagonal, these are perpendicular bisectors." + }, + { + "Q": "so should i have studied quadratic equations before i got to this? it looks like that's a later topic in the algebra play list. and where do i find the vertex form/formula?\n\nfor instance at 5:02 sal says \"this is in vertex form\" referring to y = 2 (x-4)^2 + 3 \"where x = 4 and y = 3\"... and then he plots that point. i don't knw what that means or where i can look for it in the playlist.\n\nat 5:57 i think this might be completing the square?", + "A": "Yes, you need to know quadratics before doing these videos -- they seem to have been put too early on this list. It is possible to do some simple non-linear equations before mastering quadratics, but I would not recommend it. Try going to the Functions videos and then coming back to these videos once you re completely done with quadratics.", + "video_name": "FksgVpM_iXs", + "timestamps": [ + 302, + 357 + ], + "3min_transcript": "Over here, we have 18 plus or minus the square root of-- let's just use a calculator. I could multiply it out but I think-- we have 18 squared minus 4 times 3 times 37, which is negative 120. It's 18 plus or minus the square root of negative 120. You might have even been able to figure out that this is negative. 4 times 3 is 12. 12 times 37 is going to be a bigger number than 18. Although it's not 100% obvious, but you might be able to just get the intuition there. We definitely end up with a negative number under the radical here. Now, if we're dealing with real numbers, there is no square root of negative 120. So there is no solution to this quadratic equation. There is no solution. If we wanted to, we could have just looked at the The discriminant is this part-- b squared minus 4ac. We see the discriminant is negative, there's no solution, which means that these two guys-- these two equations-- never intersect. There is no solution to the system. There are no x values that when you put into both of these equations give you the exact same y value. Let's think a little bit about why that happened. This one is already in kind of our y-intercept form. It's an upward opening parabola, so it looks something like this. I'll do my best to draw it-- just a quick and dirty version of it. Let me draw my axes in a neutral color. Let's say that this right here is my y-axis, that right there is my x-axis. x and y. This vertex-- it's in the vertex form-- occurs when x is equal to 4 and y is equal to 3. So x is equal to 4 and y is equal to 3. It's an upward opening parabola. So this will look something like this. I don't know the exact thing, but that's close enough. Now, what will this thing look like? It's a downward opening parabola and we can actually put this in vertex form. Let me put the second equation in vertex form, just so we have it. So we have a good sense. So, y is equal to-- we could factor in a negative 1-- negative x squared minus 2x plus 2. Actually, let me put the plus 2 further out-- plus 2, all the way up out there. Then we could say, half of negative 2 is negative 1. You square it, so you have a plus 1 and then a minus 1 there. This part right over here, we can rewrite as x minus 1 squared, so it becomes negative x minus 1 squared." + }, + { + "Q": "What does Sal mean @16:38 when he said usually they are the negative of each other?", + "A": "You normally see vector fields pointing to decreasing the scalar, not increasing the scalar. For example, the force on a particle at a certain point is equivalent to the negative of the gradient of the potential energy at that point.", + "video_name": "K_fgnCJOI8I", + "timestamps": [ + 998 + ], + "3min_transcript": "This associates a value with every point on the xy plane. But this whole exercise, remember this is the same thing as that. This is our whole thing that we were trying to prove: that is equal to f dot dr. f dot dr, our vector field, which is the gradient of the capital F-- remember F was equal to the gradient of F, we assume that it's the gradient of some function capital F, if that is the case, then we just did a little bit of calculus or algebra, whatever you want to call it, and we found that we can evaluate this integral by evaluating capital F at t is equal to b, and then subtracting from that capital F at t is equal to a. But what that tells you is that this integral, the value of this integral, is only dependent at our starting point, t is equal to a, this is the point x of a, y of a, and That integral is only dependent on these two values. How do I know that? Because to solve it-- because I'm saying that this thing exists --I just had to evaluate that thing at those two points; I didn't care about the curve in between. So this shows that if F is equal to the gradient-- this is often called a potential function of capital F, although they're usually the negative each other, but it's the same idea --if the vector field f is the gradient of some scale or field upper-case F, then we can say that f is conservative or that the line integral of f dot dr is path independent. It doesn't matter what path we go on as long as our starting Hopefully found that useful. And we'll some examples with that. And actually in the next video I'll prove another interesting outcome based on this one." + }, + { + "Q": "What does he mean at 3:52", + "A": "He means that to determine whether a point is on a line or not you can input the coordinates into the given equation and if the equation is true, than the point is on the line, but if the equation false, than the point isn t on the line.", + "video_name": "SSNA9gaAOVc", + "timestamps": [ + 232 + ], + "3min_transcript": "to be-- so you're going to have 0 plus 2y is equal to 7. y is going to be equal to 3.5. When x is equal to 1, you have 5 plus 2y is equal to 7. If you subtract 5 from both sides, you get 2y is equal to 2. You get y is equal to 1. So when x is 1, y is 1, and when x is-- well let's try-- well that's actually enough for us to graph. We could keep doing more points. We could even put the point 3, negative 4 there, but let's just try to graph it in this very rough sense right here. So let me draw my x-axis, and then this right over here is my y-axis. So let's say that this is y is 1, 2, 3, 4. This is negative 1, negative 2, negative 3, negative 4. I could keep going down in that direction. This is 1. Let me do it a little bit-- 1, 2, 3, 4, and I could just keep going on and on in the positive x direction. So let's plot these points. I have 0, 3.5. When x is 0, y is 1, 2, 3.5. When x is 1, y is 1. And so if we were to draw this line-- I'll do it as a dotted line, just so that I can make sure I connect the dots. I can do a better job than that though. So it will look something like that. And so if someone gave you this line, you'd say oh, well it's 3, negative 4 on this line, and let's assume that we drew it really nicely Let me try one last attempt at it. So it's going to look something like that. And If someone asked is 3, negative 4 on it, you could visually do it, but it's always hard when you actually don't substitute it, Maybe you're a little bit off. But if you look at it over here, you say when x is equal to 3, what is y? Well, you go down here, and it looks like y is equal to negative 4. So this is a point 3 comma negative 4. Obviously, in general, you don't want to just rely off of inspecting graphs. Maybe this was 3, negative 3.9999, and you just couldn't tell looking at the graph. That's why you always want to just the substitute and make sure that it really does equal, that this equality really does hold true at that point, not just looking at the graph. But it's important to realize that the graph really is another representation to all of the solutions" + }, + { + "Q": "at 1:14 what is the operation?", + "A": "He just multiplies 5 and 3 and adds it to 2 multiplied by -4", + "video_name": "SSNA9gaAOVc", + "timestamps": [ + 74 + ], + "3min_transcript": "Is 3 comma negative 4 a solution to the equation 5x plus 2y is equal to 7? So there's two ways to think about it. One, you could just substitute this x and y value into this equation to see if it satisfies-- and then we'll do that way first-- and the other way is if you had a graph of this equation, you could see if this point sits on that graph, which would also mean that it is a solution to this equation. So let's do it the first way. So we have 5x plus 2y is equal to 7, so let's substitute. Instead of x, let us put in 3 for x. So 5 times 3 plus 2 times y-- so y is negative 4-- plus 2 times negative 4 needs to be equal to 7. I'll put a question mark here, because we're not sure yet if it does. So 5 times 3 is 15, and then 2 times negative 4 is negative 8. and this needs to be equal to 7. And of course, 15 minus 8 does equal 7, so this all works out. This is a solution, so we've answered the question. But I also want to show you, this way we just did it by substitution. If we had the graph of this equation, we could also do it graphically. So let's give ourselves the graph of this equation, and I'll do that by setting up a table. There's multiple ways to graph this. You could put it in a slope-intercept form and all of the rest, but I'll just set up a table of x and y values. And I'll graph it, and then given the graph, I want to see if this actually sits on it. And obviously it will, because we've already shown that this works. In fact, we could try the point 3, negative 4, and that actually is on the graph. We could do it on our table, but I won't do that just yet. I'm just going to do this to give ourselves a graph. So let's try it when x is equal to 0. We have 5 times 0 plus 2 times y is equal to 7. to be-- so you're going to have 0 plus 2y is equal to 7. y is going to be equal to 3.5. When x is equal to 1, you have 5 plus 2y is equal to 7. If you subtract 5 from both sides, you get 2y is equal to 2. You get y is equal to 1. So when x is 1, y is 1, and when x is-- well let's try-- well that's actually enough for us to graph. We could keep doing more points. We could even put the point 3, negative 4 there, but let's just try to graph it in this very rough sense right here. So let me draw my x-axis, and then this right over here is my y-axis." + }, + { + "Q": "Around 5:50, wouldn't you also divide by 2x1 to remove the repeats for tails? Or would that cancel out because tails was not included in the total probability in the first place?", + "A": "I think tails don t factor in because our question didn t address them as something to calculate. If, say, we were asking what the probability would be of getting 3 heads and 2 tails exactly in six flips (assuming that, in this scenario, it is possible to get a coin on its side), then, yes, we would have to factor in tails. Also, tails were included in our total probability (all possible sequences of heads and tails), but again, their chance of occurring wasn t asked for in the question; so we ignore them.", + "video_name": "udG9KhNMKJw", + "timestamps": [ + 350 + ], + "3min_transcript": "" + }, + { + "Q": "At about 1:17, he showed that he was converting the decimals to whole numbers. That method really confuses me. Can somebody please show me a way to solve the problem while keeping the decimals and solving it that way? would it be possible to instead of changing the decimals, keep the decimals and work out the problem with the decimals?", + "A": "You can do 0.6 / 1.2 But, the 1st step in decimal division, is to change the 1.2 into a whole number. Shift the decimal places one place to right 6 / 12. Then, do the long division. You will get c = 0.5", + "video_name": "a3acutLstF8", + "timestamps": [ + 77 + ], + "3min_transcript": "Let's get some practice solving some equations, and we're gonna set up some equations that are a little bit hairier than normal, they're gonna have some decimals and fractions in them. So let's say I had the equation 1.2 times c is equal to 0.6. So what do I have to multiply times 1.2 to get 0.6? And it might not jump out immediately in your brain but lucky for us we can think about this a little bit methodically. So one thing I like to do is say okay, I have the c on the left hand side, and I'm just multiplying it by 1.2, it would be great if this just said c. If this just said c instead of 1.2c. So what can I do there? Well I could just divide by 1.2 but as we've seen multiple times, you can't just do that to the left hand side, that would change, you no longer could say that this is equal to that if you only operate on one side. So you have to divide by 1.2 on both sides. So on your left hand side, 1.2c divided by 1.2, well that's just going to be c. You're just going to be left with c, Now what is that equal to? There's a bunch of ways you could approach it. The way I like to do it is, well let's just, let's just get rid of the decimals. Let's just multiply the numerator and denominator by a large enough number so that the decimals go away. So what happens if we multiply the numerator and the denominator by... Let's see if we multiply them by 10, you're gonna have a 6 in the numerator and 12 in the denominator, actually let's do that. Let's multiply the numerator and denominator by 10. So once again, this is the same thing as multiplying by 10 over 10, it's not changing the value of the fraction. So 0.6 times 10 is 6, and 1.2 times 10 is 12. So it's equal to six twelfths, and if we want we can write that in a little bit of a simpler way. We could rewrite that as, divide the numerator and denominator by 6, you get 1 over 2, And if you look back at the original equation, 1.2 times one half, you could view this as twelve tenths. Twelve tenths times one half is going to be equal to six tenths, so we can feel pretty good that c is equal to one half. Let's do another one. Let's say that we have 1 over 4 is equal to y over 12. So how do we solve for y here? So we have a y on the right hand side, and it's being divided by 12. Well the best way I can think of of getting rid of this 12 and just having a y on the right hand side is multiplying both sides by 12. We do that in yellow. So if I multiply the right hand side by 12, I have to multiply the left hand side by 12. And once again, why did I pick 12? Well I wanted to multiply by some number, that when I multiply it by y over 12" + }, + { + "Q": "At 1:00 Sal tells us about parallel lines. Are they related in any way", + "A": "Yes. They never intersect. If you have 2 unparalleled lines, then somewhere they will intersect and they are therefore not parallel.", + "video_name": "V0xounKGEXs", + "timestamps": [ + 60 + ], + "3min_transcript": "Let's think a little bit about two terms that you'll see throughout your geometry, and really, mathematical career. One is the idea of things being perpendicular. And usually, people are talking about perpendicular. Actually I'm misspelling it-- perpendicular lines, and the idea of parallel lines. So perpendicular lines are two lines that intersect at a right angle. So what am I talking about? So let's say that this is one line right over here and that this is another line right over here. We would say these two lines are perpendicular if they intersect at a right angle. So they clearly intersect. In order for them to intersect at a right angle, the angle formed between these two lines needs to be 90 degrees. And if any one of these angles is 90 degrees, the rest of them are going to be 90 degrees. And if that's 90 degrees, then that's going to be 90 degrees, that's going to be 90 degrees, and that's going to be 90 degrees. So if any of them are 90 degrees, the rest of them are 90 degrees, and we have perpendicular lines. If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines. So this line right over here and this line right over here, the way I've drawn them, are parallel lines. They aren't intersecting. They're both kind of going in the same direction, but they're kind of shifted versions of each other. They will never intersect with each other. So these two are parallel. If we have two lines that, let's say, they intersect, but they don't intersect at a right angle, so let's say we have that line and we have this line right over here, and they're clearly not intersecting at a right angle, then we call these neither perpendicular These lines just intersect." + }, + { + "Q": "I still don't understand how does the computer program calculate the \"pseudo-sample variance\" @4:10 if we don't know mu's value. Can someone please explain?", + "A": "In real life we generally don t know the value of \u00ce\u00bc. However, in a simulation, we are making up the data, and we do in fact know \u00ce\u00bc. What were doing is: 1. Set \u00ce\u00bc and create some data from a distribution with that mean. 2. Pretend that we don t know \u00ce\u00bc, and calculate the mean and standard deviation. 3. Remember that we know \u00ce\u00bc, and perform the calculations shown in the video.", + "video_name": "F2mfEldxsPI", + "timestamps": [ + 250 + ], + "3min_transcript": "In the vertical axis, using this denominator, dividing by n, we calculate two different variances. One variance, we use the sample mean. The other variance, we use the population mean. And this, in the vertical axis, we compare the difference between the mean calculated with the sample mean versus the mean calculated with the population mean. So for example, this point right over here, when we calculate our mean with our sample mean, which is the normal way we do it, it significantly underestimates what the mean would have been if somehow we knew what the population mean was and we could calculate it that way. And you get this really interesting shape. And it's something to think about. And he recommends some thinking about why or what kind of a shape this actually is. The other interesting thing is when you look at it this way, it's pretty clear this entire graph is sitting below the horizontal axis. So we're always, when we calculate our sample variance which we typically do, we're always getting a lower variance than when we use the population mean. Now this over here, when we divide by n minus 1, we're not always underestimating. Sometimes we are overestimating it. And when you take the mean of all of these variances, And here we're overestimating it a little bit more. And just to be clear what we're talking about in these three graphs, let me take a screen shot of it and explain it in a little bit more depth. So just to be clear, in this red graph right over here, let me do this. A color close to at least. So this orange, what this distance is for each of these samples, we're calculating the sample variance using, so let me, using the sample mean. And in this case, we are using n as our denominator. In this case right over here. or I guess you could call this some kind of pseudo sample variance, if we somehow knew the population mean. This isn't something that you see a lot in statistics. But it's a gauge of how much we are underestimating our sample variance given that we don't have the true population mean at our disposal. And so this is the distance. This is the distance we're calculating. And you see we're always underestimating. Here we overestimate a little bit. And we also underestimate. But when you take the mean, when you average them all out, it converges to the actual value. So here we're dividing by n minus 1, here we're dividing by n minus 2." + }, + { + "Q": "Shouldn't the unit vector i go into the other direction of the x-axis? (Minute 12:44) Otherwise we set up a left-handed coordinate system, didn't we?", + "A": "Yes, Sal decided to change the sign of the x coordinate on minute 9:58, turning the system into a left-handed coordinate system.", + "video_name": "bJ_09eoCmag", + "timestamps": [ + 764 + ], + "3min_transcript": "And so our parameterization, and you know, just play with this triangle, and hopefully it'll make sense. I mean, if you say that this is our y-coordinate right here, you just do SOCATOA, cosine of t, CA is equal to adjacent, which is y, right, this is the angle right here, over the hypotenuse. Over b plus a cosine of s. Multiply both sides of the equation times this, and you get y of s of t is equal to cosine of t times this thing, right there. Let me copy and paste all of our takeaways. And we're done with our parameterization. We could leave it just like this, but if we want to represent it as a position vector-valued function, we can define it like this. So let's say our position vector-valued function is r. It's going to be a function of two parameters, s and t, and it's going to be equal to its x-value. Let me do that in the same color. So it's going to be, I'll do this part first. b plus a cosine of s times sine of t, and that's going to go in the x-direction, so we'll say that's times i. And this case, remember, the way I defined it, the positive x-direction is going to be here. So the i-unit vector will look like that. i will go in that direction, the way I've defined things. And then plus our y-value is going to be b plus a cosine of s times cosine of t in the y-unit vector direction. That's our j-unit vector. And then, finally, we'll throw in the z, which was actually the most straightforward. plus a sine of s times the k-unit vector, which is the unit vector in the z-direction. So times the k-unit vector. And so you give me, now, any s and t within this domain right here, and you put it into this position vector-valued function, it'll give you the exact position vector that specifies the appropriate point on the torus. So if you pick, let's just make sure we understand what we're doing. If you pick that point right there, where s and t are both equal to pi over 2, and you might even want to go through the exercise. Take pi over 2 in all of these." + }, + { + "Q": "at 0:03, some people find it offensive to have people ask their ages. and at 0:34, how can you throw ages into a bucket? They aren't physical objects.", + "A": "It is metaphoric. He is putting the ages into categories. The categories are being compared to buckets. They are not actually buckets. Also, due to the scenario, probably no one was offended about the ages, and Sal never asked anyone. He said if you were to go and ask the people at the restaurant, this is what you would get.", + "video_name": "gSEYtAjuZ-Y", + "timestamps": [ + 3, + 34 + ], + "3min_transcript": "- [Voiceover] So let's say you were to go to a restaurant and just out of curiosity you want to see what the makeup of the ages at the restaurant are. So you go around the restaurant and you write down everyone's age. And so these are the ages of everyone in the restaurant at that moment. And so you're interested in somehow presenting this, somehow visualizing the distribution of the ages, because you want just say, well, are there more young people? Are there more teenagers? Are there more middle-aged people? Are there more seniors here? And so when you just look at these numbers it really doesn't give you a good sense of it. It's just a bunch of numbers. And so how could you do that? Well one way to think about it, is to put these ages into different buckets, and then to think about how many people are there in each of those buckets? Or sometimes someone might say how many in each of those bins? So let's do that. So let's do buckets or categories. So, I like, sometimes it's called a bin. So the bucket, I like to think of it more of as a bucket, the bucket and then the number in the bucket. The number in the bucket. It's the, oops. It's the number (laughing), it's the number in the bucket. Alright. So let's just make buckets. Let's make them 10 year ranges. So let's say the first one is ages zero to nine. So how many people... Why don't we just define all of the buckets here? So the next one is ages 10 to 19, then 20 to 29, then 30 to 39, and 40 to 49, 50 to 59, let me make sure you can read that properly, then you have 60 to 69. And I think that covers everyone. I don't see anyone 70 years old or older here. So then how many people fall into the zero to nine-year-old bucket? Well it's gonna be one, two, three, four, five, six people fall into that bucket. How many people fall into the... How many people fall into the 10 to 19-year-old bucket? One, two, three. Three people. And I think you see where this is going. What about 20 to 29? So that's one, two, three, four, five people. Five people fall into that bucket. Alright, what about 30 to 39? We have one, and that's it. Only one person in that 30 to 39 bin or bucket or category. Alright, what about 40 to 49? We have one, two people. Two people are in that bucket. And then 50 to 59. Let's see, you have one, two people. Two people. And then finally, finally, ages 60-69. Let me do that in a different color. 60 to 69. There is one person, right over there." + }, + { + "Q": "At 2:46, how can one list the factors of \"a\" if it has already been declared prime?", + "A": "It hasn t been declared prime. a/b is reduced to lowest terms. 27/32 is reduced to lowest terms. It s (3*3*3)/(2*2*2*2*2).", + "video_name": "W-Nio466Ek4", + "timestamps": [ + 166 + ], + "3min_transcript": "Well, this being rational says I can represent the square root of p as some fraction, as some ratio of two integers. And if I can represent anything as a ratio of two integers, I can keep dividing both the numerator and the denominator by the common factors until I eventually get to an irreducible fraction. So I'm assuming that's where we are right here. So this cannot be reduced. And this is important for our proof-- cannot be reduced, which is another way of saying that a and b are co-prime, which is another way of saying that a and b share no common factors other than 1. So let's see if we can manipulate this a little bit. Let's take the square of both sides. We get p is equal to-- well, a/b, the whole thing squared, that's the same thing as a squared over b squared. We can multiply both sides by b squared, and we get b squared times p is equal to a squared. Well, b is an integer, so b squared must be an integer. So an integer times p is equal to a squared. Well, that means that p must be a factor of a squared. Let me write this down. So a squared is a multiple of p. Now, what does that tell us about a? Does that tell us that a must also be a multiple of p? Well, to think about that, let's think about the prime factorization of a. Let's say that a can be-- and any number-- can be rewritten as a product of primes. Or any integer, I should say. So let's write this out as a product of primes right over here. So let's say that I have my first prime factor times my second prime factor, all the way to my nth prime factor. I'm just saying that a is some integer right over here. So that's the prime factorization of a. What is the prime factorization of a squared going to be? Well, a squared is just a times a. Its prime factorization is going to be f1 times f2, all the way to fn. And then that times f1 times f2 times, all the way to fn. Or I could rearrange them if I want. f1 times f1 times f2 times f2, all the way to fn times fn. Now, we know that a squared is a multiple of p. p is a prime number, so p must be one of these numbers in the prime factorization. p could be f2, or p could be f1, but p needs to be one of these numbers in the prime factorization. So p needs to be one of these factors." + }, + { + "Q": "At 4:53, i don't understand why f2=p, because a^2=p.b^2 so p is a factor of a^2 so i think p=f2.f2 not p=f2. Can you please explain? Thank you", + "A": "We know that P is a factor of a^2, because a^2 = P*b^2. P is a prime number, so it can t be the product f2*f2. P has to be one of the prime factors of a^2. These prime factors come in pairs, as they do in all perfect squares (as Sal shows in the video). Sal picked f2 as a possible example.", + "video_name": "W-Nio466Ek4", + "timestamps": [ + 293 + ], + "3min_transcript": "I'm just saying that a is some integer right over here. So that's the prime factorization of a. What is the prime factorization of a squared going to be? Well, a squared is just a times a. Its prime factorization is going to be f1 times f2, all the way to fn. And then that times f1 times f2 times, all the way to fn. Or I could rearrange them if I want. f1 times f1 times f2 times f2, all the way to fn times fn. Now, we know that a squared is a multiple of p. p is a prime number, so p must be one of these numbers in the prime factorization. p could be f2, or p could be f1, but p needs to be one of these numbers in the prime factorization. So p needs to be one of these factors. pick one of these arbitrarily. Let's say that p is f2. If p is f2, then that means that p is also a factor of a. So this allows us to deduce that a is a multiple of p. Or another way of saying that is that we can represent a as being some integer times p. Now, why is that interesting? And actually, let me box this off, because we're going to reuse this part later. But how can we use this? Well, just like we did in the proof of the square root of 2 being irrational, let's now substitute this back into this equation right over here. So we get b squared times p. We have b squared times p is equal to a squared. that as some integer k times p. So we can rewrite that as some integer k times p. And so, let's see, if we were to multiply this out. So we get b squared times p-- and you probably see where this is going-- is equal to k squared times p squared. We can divide both sides by p, and we get b squared is equal to p times k squared. Or k squared times p. Well, the same argument that we used, if a squared is equal to b squared times p, that let us know that a squared is a multiple of p. So now we have it the other way around." + }, + { + "Q": "At 3:20, why does Sal write -10, is this because the ball is going down to the ground according to physics or he just make typo?", + "A": "It depends how you want to formulate your series. In the video Sal starts the series at n = 0. What is 20 * (1/2)^0? It is 20. However, the first bounce is only 10m, not 20m! Therefore, we need to deduct 10. In other words, after the ball has been dropped (i.e., the 0th bounce) it has travelled a distance of: -10 + 20(1/2)^0 = 10m", + "video_name": "tqTJZEglrvc", + "timestamps": [ + 200 + ], + "3min_transcript": "It's going to go up 5 meters, up half of 10 meters, and then down half of 10 meters. Let me put it this way. So each of these is going to be 10 meters. Actually, I don't have to write the units here. Let me take the units out of the way. Let me write that clear. So the first bounce, once again, it goes straight down 10 meters. Then on the next bounce it's going to go up 10 times 1/2. And then it's going to go down 10 times 1/2. Notice we just care about the total vertical distance. We don't care about the direction. So it's going to go up 10 times 1/2, up 5 meters, and then it's going to go down 5 meters. So it's going to travel a total vertical distance of 10 meters, Now what about on this jump, or on this bounce, I should say. Well here it's going to go half as far as it went there. So it's going to go 10 times 1/2 squared up, and then 10 times 1/2 squared down. And I think you see a pattern here. This looks an awful lot like a geometric series, an infinite geometric series. It's going to just keep on going like that forever and ever. So let's try to clean this up a little bit so it looks a little bit more like a traditional geometric series. So if we were to simplify this a little bit we could rewrite this as 10 plus 20. 20 times 1/2 to the first power, plus 10 1/2 times 1/2 squared plus 10 times 1/2 squared is going to be 20 times 1/2 So this would be a little bit clearer if this were a 20 right over here. But we could do that. We could write 10 as negative 10 plus 20, and then we have plus all of this stuff right over here. Let me just copy and paste that. So plus all of this right over here. And we can even write this first. We can even write this 20 right over here is 20 times 1/2 to the 0 power plus all of this. So now it very clearly looks like an infinite geometric series. We can write our entire sum, and maybe I'll write it up here since I don't want to lose the diagram. We could write it as negative 10. That's that negative 10 right over here." + }, + { + "Q": "At 1:08, when sal divides both sides of 24x/24x wouldn't you be left with 1x? so then you move the variables to one side making 1x-1x making it 0x. ahh i think i just answered my own question. 0xanything is 0. right?", + "A": "he s not dividing though, he s subtracting", + "video_name": "zKotuhQWIRg", + "timestamps": [ + 68 + ], + "3min_transcript": "Solve for x. We have 8 times the quantity 3x plus 10 is equal to 28x minus 14 minus 4x. So like every equation we've done so far, we just want to isolate all of the x's on one side of this equation. But before we do that, we can actually simplify each of these sides. On the left-hand side, we can multiply the quantity 3x plus 10 times 8. So we're essentially just distributing the 8, the distributive property right here. So this is the same thing as 8 times 3x, which is 24x, plus 8 times 10, which is 80, is equal to-- and over here, we have 28x minus 14 minus 4x. So we can combine the 28x and the minus 4x. If we have 28x minus 4x, that is 24x And then you have the minus 14 right over here. Now, the next thing we could-- and it's already looking a little bit suspicious, but just to confirm that it's as suspicious as it looks, let's try to subtract 24x from both sides of this equation. And if we do that, we see that we actually remove the x's and we have a 24x there. You might say, hey, let's put all the x's on the left-hand side. So let's get rid of this 24x. So you subtract 24x right over there, but you have to do it to the left-hand side as well. On the left-hand side, these guys cancel out, and you're left with just 80-- these guys cancel out as well-- is equal to a negative 14. Now, this looks very bizarre. It's making a statement that 80 is equal to negative 14, which we know is not true. This does not happen. 80 is never equal to negative 14. They're just inherently inequal. So this equation right here actually has no solution. This has no solution. There is a no x-value that will make 80 equal to negative 14." + }, + { + "Q": "in 5:31 whats that division symbol really sorry if it a silly doubt", + "A": "At 3:46 he explains what that symbol is. It is a subtraction sign. A division sign is the opposite, like / .", + "video_name": "2B4EBvVvf9w", + "timestamps": [ + 331 + ], + "3min_transcript": "Or the relative complement of B in A. Now, with that out of the way, let's think about things the other way around. What would B slash-- I'll just call it a slash right over here. What would B minus A be? So what would be B minus A? Which we could also write it as B minus A. What would this be equal to? Well, just going back, we could view this as all of the things in B with all of the things in A taken out of it. Or all of the things-- the complement of A that happens to be in B. So let's think of it as the set B with all of the things in A taken out of it. So if we start with set B, we have a 17. Then we have a 19. But there's a 19 in set A, so we have to take the 19 out. Then we have a 6. Oh, well, we don't have to take a 6 out of B because the 6 is not in set A. So we're left with just the 6. So this would be just the set with a single element in it, set 6. Now let me ask another question. What would the relative complement of A in A be? Well, this is the same thing as A minus A. And this is literally saying, let's take set A and then take all of the things that are in set A out of it. Well, I start with the 5. Oh, but there's already a 5. There's a 5 in set A. So I have to take the 5 out. Well, there's a 3, but there's a 3 in set A, so I have to take a 3 out. So I'm going to take all of these things out. And so I'm just going to be left with the empty set, often called the null set. will look like this, the null set, the empty set. There's a set that has absolutely no objects in it." + }, + { + "Q": "(about 3:30 minutes into the video) Why is f(c) greater than or equal to f(x)?\nThe few seconds after that are also not quite clear to me. Help please!", + "A": "Essentially, f(c) is a random point representing a relative max. f(x) represents the rest of the graph within the domain (c-h,c+h). So, if f(c), the point, is higher than the rest of the graph,f(x) in the selected interval, then it must be a relative max. Hopefully that helped.", + "video_name": "Hoyv3-BMAGc", + "timestamps": [ + 210 + ], + "3min_transcript": "It's larger than the other ones. Locally, it looks like a little bit of a maximum. And so that's why this value right over here would be called-- let's say this right over here c. This is c, so this is f of c-- we would call f of c is a relative maximum value. And we're saying relative because obviously the function takes on the other values that are larger than it. But for the x values near c, f of c is larger than all of those. Similarly-- I can never say that word. Similarly, if this point right over here is d, f of d looks like a relative minimum point or a relative minimum value. f of d is a relative minimum or a local minimum value. there's definitely points that are lower. And we hit an absolute minimum for the interval at x is equal to b. But this is a relative minimum or a local minimum because it's lower than the-- if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values. And you're at a minimum if you're at a smaller value than any of the surrounding areas. But how could we write that mathematically? So here I'll just give you the definition that really is just a more formal way of saying what we just said. So we say that f of c is a relative max, relative maximum value, if f of c x that-- we could say in a casual way, for all x near c. So we could write it like that. But that's not too rigorous because what does it mean to be near c? And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0. So does that make sense? Well, let's look at it. So let's construct an open interval. So it looks like for all of the x values in-- and you just have to find one open interval. There might be many open intervals where this is true. But if we construct an open interval that" + }, + { + "Q": "3:25 why bigger or equal to? if it's equal how can it be a maximum?", + "A": "The maximum value a function gets is still the largest value whether the function reaches it one times, five times, or infinitely many times. For a local min that is not a global min, you would typically only have the equal scenario come up if there is a horizontal region in the function.", + "video_name": "Hoyv3-BMAGc", + "timestamps": [ + 205 + ], + "3min_transcript": "It's larger than the other ones. Locally, it looks like a little bit of a maximum. And so that's why this value right over here would be called-- let's say this right over here c. This is c, so this is f of c-- we would call f of c is a relative maximum value. And we're saying relative because obviously the function takes on the other values that are larger than it. But for the x values near c, f of c is larger than all of those. Similarly-- I can never say that word. Similarly, if this point right over here is d, f of d looks like a relative minimum point or a relative minimum value. f of d is a relative minimum or a local minimum value. there's definitely points that are lower. And we hit an absolute minimum for the interval at x is equal to b. But this is a relative minimum or a local minimum because it's lower than the-- if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values. And you're at a minimum if you're at a smaller value than any of the surrounding areas. But how could we write that mathematically? So here I'll just give you the definition that really is just a more formal way of saying what we just said. So we say that f of c is a relative max, relative maximum value, if f of c x that-- we could say in a casual way, for all x near c. So we could write it like that. But that's not too rigorous because what does it mean to be near c? And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0. So does that make sense? Well, let's look at it. So let's construct an open interval. So it looks like for all of the x values in-- and you just have to find one open interval. There might be many open intervals where this is true. But if we construct an open interval that" + }, + { + "Q": "At 4:06 he says that the denominator is zero but isn't the square root of 0 + 1 just equal one. would the graph be different?", + "A": "Yes, but he was finding the limit of 0/(sqrt(0+1)), which is 0.", + "video_name": "xks4cETlN58", + "timestamps": [ + 246 + ], + "3min_transcript": "Now, for positive x'es the absolute value of x is just going to be x. This is going to be x divided by x, so this is just going to be 1. Similarly, right over here, we take the limit as we go to negative infinity, this is going to be the limit of x over the absolute value of x as x approaches negative infinity. Remember, the only reason I was able to make this statement is that f(x) and this thing right over here become very very similar, you can kind of say converge to each other, as x gets very very very large or x gets very very very very negative. Now, for negative values of x the absolute value of x is going to be positive, x is obviously going to be negative and we're just going to get negative 1. And so using this, we can actually try to graph our function. So let's say, that is my y axis, this is my x axis, and we see that we have 2 horizontal asymptotes. We have 1 horizontal asymptote at y=1, so let's say this right over here is y=1, let me draw that line as dotted line, we're going to approach this thing, and then we have another horizontal asymptote at y=-1. So that might be right over there, y=-1. And if we want to plot at least 1 point we can think about what does f(0) equal. So, f(0) is going to be equal to 0 over the square root of 0+1, or 0 squared plus 1. Well that's all just going to be equal to zero. So we have this point, right over here, and we know that as x approaches infinity, we're approaching this blue, horizontal asymptote, Let me do it a little bit differently. There you go. I'll clean this up. So it might look something like this. That's not the color I wanted to use. So it might look something like that. We get closer and closer to that asymptote as x gets larger and larger and then like this -- we get closer and closer to this asymptote as x approaches negative infinity. I'm not drawing it so well. So that right over there is y=f(x). And you can verify this by taking a calculator, trying to plot more points or using some type of graphing calculator or something. But anyway, I just wanted to tackle another situation we're approaching infinity and or negative infinity and we're trying to determine the horizontal asymptotes. And remember, the key is just to say what terms dominate" + }, + { + "Q": "At 9:23 pm, how do you factor problems when the coefficient does not share a common factor with the other numbers?\nFor example:\n2x2 - 3x + 2", + "A": "Your example you gave here riverav is not factorable. To answer your question, however, you do the regular factoring polynomials method sal taught us to get your answer, or if you have exhausted all factors, the problem is not factorable.", + "video_name": "GMoqg_s4Dl4", + "timestamps": [ + 563 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:54, why did you add 3k and -3k? I can't understand why and extra coefficient was added.", + "A": "was the k squared? because that might be why...", + "video_name": "GMoqg_s4Dl4", + "timestamps": [ + 174 + ], + "3min_transcript": "And remember if anything has the form x squared plus bx plus c, where you have a leading 1 coefficient-- this is implicitly a one-- we have that here in this expression in parentheses. Then we literally just need to-- and we can do this multiple ways-- but we need to find two numbers whose sum is equal to the coefficient on x. So two numbers whose sum is equal to negative 3 and whose product is equal to the constant term. And whose product is equal to negative 18. So let's just think about the factors of negative 18 here. Let's see if we can do something interesting. So it could be 1. And since it's negative, one of the numbers has to be positive, one has to be negative 1 and 18 is if it was positive. And then one of these could be positive and then one of these could be negative. But no matter what if this is negative If you switch them, then they add up to negative 17. So those won't work. So either we could write it this way, positive or negative 1, and then negative or positive 18 to show that they have to be different signs. So those don't work. Then you have positive or negative 3. And then negative or positive 6, just to know that they are different signs. So if you have positive 3 and negative 6, they add up to negative 3 which is what we need them to add up to. And clearly, positive 3 and negative 6, their product is negative 18. So it works. So we're going to go with positive 3 and negative 6 as our two numbers. Now, for this example-- just for the sake of this example-- We'll do this by grouping. So what we can do is we can separate this middle term right here as the sum of 3k negative 6k. So I could write the negative 3k as plus 3k minus 6k. And then let me write the rest of it. which is the same thing as this over here. And then we have minus 18. And then all of that's being multiplied by 8. Now we're ready to group this thing. We can group these first two terms, they're both divisible by k. And then we can group-- let me put a positive sign-- let's group these second two terms. So then we have 8 times-- I'll write brackets here instead of drawing double parentheses. Brackets are really just parentheses that look a little bit more serious. Now let's factor out a k from this term right here. I'm going to do this in a different color. Let's factor out the k here. So this is k times k plus 3. And then we have plus. And then over here it looks like we could factor out a negative 6. So let's factor out-- I'm going to do this in a different color-- let's factor out a negative 6 So plus negative 6 times k plus 3." + }, + { + "Q": "Wait, at 0:37 Sal says that for the first flip there's 2 possibilities, same on the second and the third. So 2 and 2 and 2 should be 6, right?", + "A": "The 2 s should be multiplied instead of added together. This should be seen in layers like this: There are 2 possibility for the first flip. For every possibility of the first flip, there are 2 possibility for the second flip. So there are a total of 2 * 2 possibilities for the first two flips. For every possibility of the first two flips, there are 2 possibility for the third flip. So there are a total of (2 * 2) * 2 possibilities for the three flips.", + "video_name": "mkyZ45KQYi4", + "timestamps": [ + 37 + ], + "3min_transcript": "" + }, + { + "Q": "At 6:33, can we also write:\n= 1 - ( 1 / ( 2 ^ 20 ) )\n?", + "A": "No, not quite. There are only 10 2s, so it would be 1 - ( 1 / ( 2 ^ 10 ) ).", + "video_name": "mkyZ45KQYi4", + "timestamps": [ + 393 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:03, how do you add 7 + 1/100?", + "A": "He isn t. He s multiplying them. But if you want to add them, you d get 7 1/100 (seven and one hundredth).", + "video_name": "he4kcTujy30", + "timestamps": [ + 123 + ], + "3min_transcript": "Let's say I have the number 905.074. So how could I expand this out? And what does this actually represent? So let's just think about each of the place values here. The 9 right over here, this is in the hundreds place. This literally represents nine hundreds. So we could rewrite that 9 as nine hundreds. Let me write it two ways. We could write it as 900, which is the same thing as 9 times 100. Now, there's a 0. That's just going to represent zero tens. But zero tens is still just 0. So we don't have to really worry about that. It's not adding any value to our expression or to our number. Now we have this 5. This 5 is in the ones place. It literally represents five ones, or you could just say it represents 5. Now, if we wanted to write it as five ones, we could say well, that's going to be 5 times 1. 100 plus 5 times 1. And you might say hey, how do I know whether I should multiply or add first? Should I do this addition before I do this multiplication? And I'll always remind you, order of operations. In this scenario, you would do your multiplication before you do your addition. So you would multiply your 5 times 1 and your 9 times 100 before adding these two things together. But let's move on. You have another 0. This 0 is in the tenths place. This is telling us the number of tenths we're going to have. This is zero tenths, so it's really not adding much, or it's not adding anything. Now we go to the hundredths place. So this literally represents seven hundredths. So we could write this as 7/100, or 7 times 1/100. So we go to the thousandths place. And we have four thousandths. So that literally represents 4 over 1,000, or 4 times 1/1000. Notice this is coming from the hundreds place. You have zero tens, but I'll write the tens place there just so you see it. So it's zero tens, so I didn't even bother to write that down. Then you have your ones place. You have five ones. Then you have zero tenths. So I didn't write that down. Then you have seven hundredths and then you have four thousandths. We've written this out, really just understanding what this number represents." + }, + { + "Q": "At 3:44, how did we get cos^2(theta) \u00e2\u0080\u0093 sin^2(theta) to equal to 1 \u00e2\u0080\u0093 2 sin^2(theta) and to 2cos^2(theta) \u00e2\u0080\u0093 1. Thank you!", + "A": "First, cos^2 (theta) = 1 - sin^2 (theta). This is from the Pythagorean identity sin^2 (x) + cos^2 (x) = 1. Therefore, cos^2 (theta) - sin^2 (theta) = 1 - sin^2 (theta) - sin^2 (theta) = 1 - 2 sin^2 (theta). Second, we flip the Pythagorean identity around and replace sin^2 (theta) with 1 - cos^2 (theta). We get 1 - 2(1 - cos^2 (theta)) = 1 - 2 + 2 cos^2 (theta) = 2 cos^2 (theta) - 1.", + "video_name": "lXShNH1G6Pk", + "timestamps": [ + 224 + ], + "3min_transcript": "this is going to be cosine of negative pi over 2, right? This is negative pi over 2, cosine of negative pi over 2, if you thought in degrees, that's going to be negative 90 degrees. Well, cosine of that is just going to be zero, so what we end up with is equal to zero over zero, and as we've talked about before, if we had something non-zero divided by zero, we'd say, okay, that's undefined. We might as well give up, but we have this indeterminate form, it does not mean the limit does not exist. It's usually a clue that we should use some tools in our toolkit, one of which is to do some manipulation here to get an expression that maybe is defined at theta is equal to, or does not, is not an indeterminate form, that theta is equal to pi over 4 and we'll see other tools in our toolkit in the future. So let me algebraically manipulate this a little bit. So if I have 1 plus the square root of 2, sine theta, over cosine 2 theta, the things that might be useful here are our trig identities and in particular, cosine of 2 theta seems interesting. Let me write some trig identities involving cosine of 2 theta. I'll write it over here. So we know that cosine of 2 theta is equal to cosine squared of theta minus sine squared of theta which is equal to 1 minus 2 sine squared of theta which is equal to 2 cosine squared theta minus 1, and you can go from this one to this one to this one just using the Pythagorean identity. We proved that in earlier videos in trigonometry on Khan Academy. Now, do any of these look useful? Well, all of these three are going to be differences of squares, so we can factor them in interesting ways, is maybe cancel things out that are making us get this zero over zero, and if I could factor this into something that involved a 1 plus square root of 2 sine theta, then I'm going to be in business, and it looks like, it looks like this right over here, that can be factored as 1 plus square root of 2 sine theta times 1 minus square root of 2 sine theta, so let me use this. Cosine of 2 theta is the same thing, cosine of 2 theta is the same thing as 1 minus 2 sine squared theta, which is just a difference of squares. We can rewrite that as, this is a-squared minus b-squared, this is a plus b times a minus b, so I can just replace this with 1 plus square root of 2 sine theta times 1 minus square root of 2 sine theta," + }, + { + "Q": "at 1:49, why don't we plug in 8 for x^2?", + "A": "we only plug 8 into x^2 to get h (height) at that instant. To get dh/dt we have to differentiate x^2+h^2=10^2 to get relation between dx/dt, dh/dt, x(for that instant) and h(for that instant). Now we will solve dh/dt using simple algebra by putting known values into it.", + "video_name": "kBVDSu7v8os", + "timestamps": [ + 109 + ], + "3min_transcript": "" + }, + { + "Q": "at ~13:20, sal is writing out 3rd/final term of the 3x3 determinant |A|...but shouldn't it be \"-f(ah-bg)\" vs \"-f(dh-eg)\" ?", + "A": "Yes. Transcription error from a few minutes in. [I m writing words to pass the comment screener.]", + "video_name": "32rdijPB-rA", + "timestamps": [ + 800 + ], + "3min_transcript": "have a k aij. So this is the determinant of A prime. We could just take out this constant right here. It has no i's or j's in it. I has no j's in it in particular, so we can just take it out. So it's equal to k times the sum from j is equal to 1 to j is equal to n of minus 1 to the i plus j times aij. This is the coefficient. This is the submatrix for each of those coefficients, aij. That's a matrix right there, an n minus 1 by n minus 1 matrix. Then you immediately recognize-- I think you saw where this was going-- this right here is just the determinant of A. So we get the result that the determinant of A prime is equal to k times the determinant of A. n matrix, if you multiply only one row, not the whole matrix, only one row by some scalar multiple k, the resulting determinant will be your original determinant times k. Now I touched on this in the original video. What is the determinant of k times A? So now we're multiplying every row times k. Or another way to think about is you're multiplying n rows times k. So you're doing this n times. So if you multiply k times itself n times, what do you get? You get k to the n. So this is going to be equal to k to the n times the determinant of A. If you just do it once, you get k times the determinant of A. Now if you do a second row, you're going to get k times k times the determinant of A. times the determinant of A. The fourth row, k to the fourth times the If you do them all, all n rows, you're going to have k to the n times the determinant of A. Anyway, hopefully you found that interesting. I encourage you to experiment with these other ways. Try going down a column and seeing what happens." + }, + { + "Q": "at 0:49 isnt it the other way because it would be 1.5 in real life problem", + "A": "Do you mean 1*5? 1.5 is a decimal number for 1 1/2. Any dot used for multiplication must be raised. Anyway... I think Sal is using the bar under each number to separate the digit from its place value. I don t believe he is using it for division if that is what you were thinking.", + "video_name": "BItpeFXC4vA", + "timestamps": [ + 49 + ], + "3min_transcript": "So I have a number written here. It's a 2, a 3, and a 5. And we already have some experience with numbers like this. We can think about 'what does it represent'. And to think about that we just have to look at the actual place values. So this right-most place right over here. This is the ones place. So this 5 represents five ones, or I guess you could say that's just going to be 5. This 3, this is in the tens place. This is the tens place, so we have three tens. So that's just going to be 30. And the 2 is in the hundreds place. So putting a 2 there means that we have two hundreds. So this number we can view as two hundred, thirty, five. Or you could view it as two hundred plus thirty plus five. Now what I want to do in this video is think about place values to the right of the ones place. And you might say 'wait, wait, I always thought that the ones place was the place furthest to the right.' Well everything that we've done so far, it has been. But to show that you can go even further to the right We call that a 'decimal point'. And that dot means that anything to the right of this is going to be place values that are smaller, I guess you could say, than the ones place. So right to the left you have the ones place and the tens place and the hundreds place, and if you were to keep going you'd go to the thousands place and the ten thousands place. But then if you go to the right of the decimal point now you're going to divide by 10. So what am I talking about? Well, right to the right of the decimal point you are going to have-- find a new color-- this is going to be the tenths place. Well what does that mean? Well whatever number I write here that tells us how many tenths we're dealing with. So if I were to write the number 4 right over here, now my number is 2 hundreds plus 3 tens plus 5 ones plus 4 tenths. Or you could write this as 4 tenths. Not tens, 4 tenths. Or 4 tenths is the same thing as this right over here. So this is a super important idea in mathematics. I can now use our place values to represent fractions. So this right over here, this 'point 4', this is 4/10. So another way to write this number-- I could write it this way, I could write it as two hundred, thirty-- let me do the thirty in blue-- two hundred and thirty five and four tenths. So I could write it like this, as a mixed number. So this up here would be a decimal representation: 235.4 And this right over here would be a mixed number representation: 235 and 4/10 but they all represent 200 plus 30 plus 5 plus 4/10." + }, + { + "Q": "in 4:24 why he drew a point outside the parabola? that point is undefined according to y=x^2, the function f(x)=x^2 but its f(x)=3 when x=2:- this is a wrong statement. And what is the meaning of limit? the limit going to tell u the defined function like in parabola x^2=4 when x=2. what is the meaning of limits?? :S", + "A": "The function he used in this example is a piecewise function which forces the value of f(x) to equal 3 when x=2. A limit is used to describe the value of f(x) as it starts to approach a number. This becomes a key component in many aspects of calculus.", + "video_name": "W0VWO4asgmk", + "timestamps": [ + 264 + ], + "3min_transcript": "The way I think about it is as you move on the curve closer and closer to the expression's value, what does the expression equal? In this case, it equals 4. You're probably saying, Sal, this seems like a useless concept because I could have just stuck 2 in there, and I know that if this is-- say this is f of x, that if f of x is equal to x squared, that f of 2 is equal to 4, and that would have been a no-brainer. Well, let me maybe give you one wrinkle on that, and hopefully now you'll start to see what the use of a limit is. Let me to define-- let me say f of x is equal to x squared when, if x does not equal 2, and let's say it equals 3 when x equals 2. So this is our new f of x. So let me ask you a question. What is-- my pen still works-- what is the limit-- I used cursive this time-- what is the limit as x-- that's an x-- as x approaches 2 of f of x? That's an x. It says x approaches 2. I just-- I don't know. For some reason, my brain is working functionally. OK, so let me graph this now. So that's an equally neat-looking graph as the one I just drew. So now it's almost the same as this curve, except something interesting happens at x equals 2. So it's just like this. It's like an x squared curve like that. But at x equals 2 and f of x equals 4, we We draw a hole because it's not defined at x equals 2. This is x equals 2. This is 2. This is 4. This is the f of x axis, of course. And when x is equal to 2-- let's say this is 3. When x is equal to 2, f of x is equal to 3. This is actually right below this. I should-- it doesn't look completely right below it, but I think you got to get the picture. See, this graph is x squared. It's exactly x squared until we get to x equals 2. At x equals 2, We have a grap-- No, not a grap. We have a gap in the graph, which maybe could be called a grap. We have a gap in the graph, and then we keep-- and then after x equals 2, we keep moving on. And that gap, and that gap is defined right here, what happens when x equals 2? Well, then f of x is equal to 3. So this graph kind of goes-- it's just like x squared, but" + }, + { + "Q": "Could someone explain how Sal solved 21 times 2/3 in his head so quickly? It happened around 2:57 in the video. I would like to understand his method.", + "A": "Its called cross products or cross multiplication.", + "video_name": "XOIhNVeLfWs", + "timestamps": [ + 177 + ], + "3min_transcript": "that do not share a common factor other than one. That just means that A, B, and C in the standard form they want need to be integers. And they want them to not have any common factors. So if we got to the point of say 4x plus 2y is equal to 10, well this number, and this number, and this number are all divisible by two. They all have the common factor of two. So we would want to simplify it more. Divide them all by two and then you would get two so you divide this by two. You get 2x, divide this by two you'll get plus y is equal to 5. So this is the form that they're asking for and probably because it's just easier for the site to know that this is the right answer. Because there's obviously a bunch of forms in this way. So let's see if we can do that. Let's see if we can write it in standard form. So the first thing I would wanna do is, well there's a bunch of ways that you could approach it. well let's get rid of all of these fractions. And the best way to get rid of the fractions is to multiple by three and to multiply by seven. If you multiply by three you get rid of this fraction. If you multiply by seven you get rid of this fraction. So if you multiply three and you multiply by seven. Let me just rewrite it over here. There's actually also a couple of ways that we can do this. So if you multiply. So one way to do it. So we start with Y is equal to 2/3x plus 4/7. So if I multiply this side by three and I multiply by seven, I have to do that to this side as well. So this is going to be multiplied by three and multiplied by seven. So the left hand side becomes 21y. 21y. Three times seven of course is 21, we just figured that out. We would distribute the 21. 21 times 2/3, well let's see. 21 divided by three is seven, times two is 14. And then 21 divided by seven is three times four is 12. So just like that I was able to get rid of the fractions. And now I wanna get all the X's and Y's on one side. So I wanna get this 14x onto the left side. So let's see if I can do that. So I'm gonna do that by. To get rid of this I would want to subtract 14x. I can't just do it on the right hand side I have to do it on the left hand side as well. So I wanna subtract 14x, and then what am I left with? Let me give myself a little bit more space. So on the left hand side I have negative 14x plus 21y. Plus 21y is equal to. Let's see and I subtracted 14x to get rid of this. And then I have this is equal to 12. Now let's see, am I done? Do these share any, do 14, 21, and 12" + }, + { + "Q": "at 0:12 what is o-blood", + "A": "There are different types of bloods, that is just one of them", + "video_name": "qrVvpYt3Vl0", + "timestamps": [ + 12 + ], + "3min_transcript": "According to the pictograph below, how many survey respondents have type O positive blood? How many have O negative blood? So a pictograph is really just a way of representing data with pictures that are somehow related to the data. So in this case, they gave us little pictures of, I'm assuming, blood drops right over here. And then they tell us that each blood drop in this pictograph represents 8 people. So you could kind of view that as a scale of these graphs. Each of these say 8 people. So, for example, if you say how many people have A positive? It would be 1, 2, 3, 4, 5, 6, 7 blood drops. But each of those blood drops represent 8 people, so it would be 56 people have type A positive. But let's answer the actual question that they're asking us. How many survey respondents have type O positive? So this is O, and then we care about O positive. So we have 1 blood drop, 2, 3. I'm going to do this in a new, different color. So we have 8 drops. I'll put those in quotes because they're pictures of drops. And then the scale is 8 people. Let me write it this way. Times 8 people per drop. And so 8 times 8-- and actually even the drops, you could view them as canceling out if you view them as units, so drops, drops. 8 times 8 is equal to 64 people. So they could have written literally the number 64 right over here. 64 people have type O positive blood. Now let's think about the O negative case, O negative blood. Well, this is O. And then within the blood group O, this is O negative. And we have 1 drops, 2 drops. So we have 2 drops times 8 people per drop. So 8 and then 16, or 2 times 8 is equal to 16 people. So 16 of the survey respondents, 16 have type O negative. 64 have type O positive." + }, + { + "Q": "Instead of using his \"aside\" with the triangle beginning at 1:49, I did something similar but using x/3 instead of 'S'. In other words, I said:\nUsing the Pythagorean Theorem: ((x/3)/2)^2+h^2 = (x/3)^2\nh^2 = (x/3)^2-(x/6)^2\nh = x/3-x/6\nh=x/6\n\nObviously this differs from Sal's answer but I can't figure out why?", + "A": "Your mistake was when you square rooted, \u00e2\u0088\u009a(a\u00c2\u00b2+b\u00c2\u00b2) is NOT \u00e2\u0088\u009aa\u00c2\u00b2 + \u00e2\u0088\u009ab\u00c2\u00b2 and is not a+b Here is the correct way to do it: h\u00c2\u00b2 = (\u00e2\u0085\u0093x)\u00c2\u00b2- (\u00e2\u0085\u0099x)\u00c2\u00b2 h\u00c2\u00b2 = \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0089 x\u00c2\u00b2 - \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0083\u00e2\u0082\u0086 x\u00c2\u00b2 h\u00c2\u00b2 = \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0081\u00e2\u0082\u0082 x\u00c2\u00b2 h = x / \u00e2\u0088\u009a12 h = x / (2\u00e2\u0088\u009a3) h = \u00e2\u0085\u0099 x\u00e2\u0088\u009a3", + "video_name": "IFU7Go6Qg6E", + "timestamps": [ + 109 + ], + "3min_transcript": "Let's say that I have a 100 meter long wire. So that is my wire right over there. And it is 100 meters. And I'm going to make a cut someplace on this wire. And so let's say I make the cut right over there. With the left section of wire-- I'm going to obviously cut it in two-- with the left section, I'm going to construct an equilateral triangle. And with the right section, I'm going to construct a square. And my question for you and for me is, where do we make this cut in order to minimize the combined areas of this triangle and this square? Well, let's figure out. Let's define a variable that we're trying to minimize, or that we're trying to optimize with respect to. So let's say that the variable x is the number of meters that we decide to cut from the left. So if we did that, then this length for the triangle would be, well, if we use x up for the left hand side, we're going to have 100 minus x for the right hand side. And so what would the dimensions of the triangle and the square Well, the triangle sides are going to be x over 3, x over 3, and x over 3 as an equilateral triangle. And the square is going to be 100 minus x over 4 by 100 minus x over 4. Now it's easy to figure out an expression for the area of the square in terms of x. But let's think about what the area of an equilateral triangle might be as a function of the length of its sides. So let me do a little bit of an aside right over here. So let's say we have an equilateral triangle. Just like that. And its sides are length s, s, and s. is 1/2 times the base times the height. So in this case, the height we could consider to be altitude, if we were to drop an altitude just like this. This length right over here, this is the height. And this would be perpendicular, just like that. So our area is going to be equal to one half times our base is s. 1/2 times s times whatever our height is, times our height. Now how can we express h as a function of s? Well, to do that we just have to remind ourselves that what we've drawn over here is a right triangle. It's the left half of this equilateral triangle. And we know what this bottom side of this right triangle is. This altitude splits this side exactly into two." + }, + { + "Q": "At 4:38 Sal says the word reciprocal.\nWhat is a reciprocal?", + "A": "The easiest way to find the reciprocal of a number is to think what should I multiply this number with to make it equal 1? That s basically all there is to reciprocals.", + "video_name": "a_Wi-6SRBTc", + "timestamps": [ + 278 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:18, Sal mentions \"interval notation\". I watched all the videos in order in \"Creating and solving linear equations\" but did not see anything about interval notation. Is there a video for this?", + "A": "At 1:18 Sal is introducing interval notation in this video.", + "video_name": "xOxvyeSl0uA", + "timestamps": [ + 78 + ], + "3min_transcript": "Let's do a few more problems that bring together the concepts that we learned in the last two videos. So let's say we have the inequality 4x plus 3 is less than negative 1. So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a zero. No reason to change the inequality just yet. We're just adding and subtracting from both sides, in this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to-- let's see, we can divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1, so we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. So let's get all our x's on the left-hand side, and the best way to do that is subtract 8x from both sides. So you subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantities on both sides. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by" + }, + { + "Q": "At 5:00, how does 8x-5 turn into 8x-20x??", + "A": "8x - 5(4x + 1) Is the right side of the inequality. So, to simplify the right hand side of the inequality, Sal distributes the -5 to (4x + 1), which is -5(4x) + -5(1) = -20x -5. So, it is 8x - 20x - 5. Hope this helped!", + "video_name": "xOxvyeSl0uA", + "timestamps": [ + 300 + ], + "3min_transcript": "to less than. And the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9, maybe this would be negative 8, maybe this would be negative 10. You would start at negative 9, not included, because we don't have an equal sign here, and you go everything less than that, all the way down, as we see, to negative infinity. Let's do a nice, hairy problem. So let's say we have 8x minus 5 times 4x plus 1 is greater Now, this might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. So let's just simplify this. You get 8x minus-- let's distribute this negative 5. So let me say 8x, and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5-- when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5, and then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. And now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to-- we that's negative 7, and then we have this plus 8x left over. Now, I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides, or adding and subtracting from them. The right-hand side becomes-- this thing cancels out, 8x minus 8x, that's 0. So you're just left with a negative 7. And now I want to get rid of this negative 5. So let's add 5 to both sides of this equation." + }, + { + "Q": "at 1:19, if x can be from negative infinity to -1, but not including negative 1, so why does he include -1 in the partenthases", + "A": "The symbols \u00e2\u0080\u0098(\u00e2\u0080\u0098 and \u00e2\u0080\u0098)\u00e2\u0080\u0099 indicate exclusive values, whereas the symbols \u00e2\u0080\u0098[\u00e2\u0080\u0098 and \u00e2\u0080\u0098]\u00e2\u0080\u0099 indicate inclusive values. In the example x<-1 which is represented as (-\u00e2\u0088\u009e,-1) because -1 is not inclusive. If the example had been x\u00e2\u0089\u00a4-1 then -1 would have been included and could be represented as (-\u00e2\u0088\u009e,-1]", + "video_name": "xOxvyeSl0uA", + "timestamps": [ + 79 + ], + "3min_transcript": "Let's do a few more problems that bring together the concepts that we learned in the last two videos. So let's say we have the inequality 4x plus 3 is less than negative 1. So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a zero. No reason to change the inequality just yet. We're just adding and subtracting from both sides, in this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to-- let's see, we can divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1, so we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. So let's get all our x's on the left-hand side, and the best way to do that is subtract 8x from both sides. So you subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantities on both sides. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So now that we divided both sides by a negative number, by" + }, + { + "Q": "At 4:28, what is a basis?", + "A": "At 4:28, A basis is a System which consist of the MINIMAL amount of VECTORS which are needed to define a room (space, coordinate system). All VECTORS in from a BASIS must be liear dependant from eachother.", + "video_name": "C2PC9185gIw", + "timestamps": [ + 268 + ], + "3min_transcript": "We're just assuming that A has at least n linearly independent eigenvectors. In general, you could take scaled up versions of these and they'll also be eigenvectors. Let's see, so the transformation of vn is going to be equal to A times vn. And because these are all eigenvectors, A times vn is just going to be lambda n, some eigenvalue times the vector, vn. Now, what are these also equal to? Well, this is equal to, and this is probably going to be unbelievably obvious to you, but this is the same thing as lambda 1 times vn plus 0 times v2 plus all the way to 0 times vn. And this right here is going to be 0 times v1 plus lambda 2 times v2 plus all the way, 0 times all of the other vectors vn. And then this guy down here, this is going to be 0 times v1 these eigenvectors, but lambda n times vn. This is almost stunningly obvious, right? I just rewrote this as this plus a bunch of zero vectors. But the reason why I wrote that is, because in a second, we're going to take this as a basis and we're going to find coordinates with respect to that basis, and so this guy's coordinates will be lambda 1, 0, 0, because that's the coefficients on our basis vectors. So let's do that. So let's say that we define this as some basis. So B is equal to the set of-- actually, I don't even have to write it that way. Let's say I say that B, I have some basis B, that's equal to that. What I want to show you is that when I do a change of basis-- we've seen this before-- in my standard coordinates or in coordinates with respect to the standard basis, you give me some vector in Rn, I'm going to multiply it times A, and you're going to have the It's also going to be in Rn. Now, we know we can do a change of basis. And in a change of basis, if you want to go that way, you multiply by C inverse, which is-- remember, the change of basis matrix C, if you want to go in this direction, you multiply by C. The change of basis matrix is just a matrix with all of these vectors as columns. It's very easy to construct. But if you change your basis from x to our new basis, you multiply it by the inverse of that. We've seen that multiple times. If they're all orthonormal, then this is the same thing as We can't assume that, though. And so this is going to be x in our new basis. And if we want to find some transformation, if we want to find the transformation matrix for T with respect to our new basis, it's going to be some matrix D. And if you multiply D times x, you're going to get this guy, but you're going to get the B representation of that guy. The transformation of the vector x is B representation." + }, + { + "Q": "At 4:15 is there an easier way to find that number that if given the certain exponent (5) makes it into the 32? (how did he know it was a 2?)", + "A": "With enough practices you ll be able to recognize certain numbers or quickly calculate in your head. Also he was using it as an example so he would pick something he already knows and easy for us to recognize.", + "video_name": "lZfXc4nHooo", + "timestamps": [ + 255 + ], + "3min_transcript": "just so you make sure you get what's going on. And I encourage you to pause it as much as necessary and try to figure it out yourself. So based on what I just told you, what do you think 9 to the 1/2 power is going to be? Well, that's just the square root of 9. The principal root of 9, that's just going to be equal to 3. And likewise, we could've also said that 3 squared is, or let me write it this way, that 9 is equal to 3 squared. These are both true statements. Let's do one more like this. What is 25 to the 1/2 going to be? Well, this is just going to be 5. 5 times 5 is 25. Or you could say, 25 is equal to 5 squared. Now, let's think about what happens when you take something to the 1/3 power. So let's imagine taking 8 to the 1/3 power. So the definition here is that taking something to the 1/3 power is the same thing And the cube root is just saying, well what number, if I had three of that number, and I multiply them, that I'm going to get 8. So something, times something, times something, is 8. Well, we already know that 8 is equal to 2 to the third power. So the cube root of 8, or 8 to the 1/3, is just going to be equal to 2. This says hey, give me the number that if I say that number, times that number, times that number, I'm going to get 8. Well, that number is 2 because 2 to the third power is 8. Do a few more examples of that. What is 64 to the 1/3 power? Well, we already know that 4 times 4 times 4 is 64. So this is going to be 4. And we already wrote over here that 64 is the same thing as 4 to the third. I think you're starting to see a little bit of a pattern here, a little bit of symmetry here. And we can extend this idea to arbitrary rational exponents. let me think of a good number here-- so let's say I have 32. I have the number 32, and I raise it to the 1/5 power. So this says hey, give me the number that if I were to multiply that number, or I were to repeatedly multiply that number five times, what is that, I would get 32. Well, 32 is the same thing as 2 times 2 times 2 times 2 times 2. So 2 is that number, that if I were to multiply it five times, then I'm going to get 32. So this right over here is 2, or another way of saying this kind of same statement about the world is that 32 is equal to 2 to the fifth power." + }, + { + "Q": "Isn't a line segment that has no length a point? Refering to 4:50", + "A": "Yes. A point really has no size, and since lines have no width, if a line also had no length, it would not be a line, it would be a point.", + "video_name": "Oc8sWN_jNF4", + "timestamps": [ + 290 + ], + "3min_transcript": "If it were a regular line, not an infinitely spiked one, scaling up by three would make it three times as much drawing, as expected. But if that spiky line were supposed to represent an infinitely spiked magical fortress city of dragon dungeon doom, by scaling it up in this way, you'd be losing details, making these long lines that should have had spiky bumps in them. Theoretically, no matter how much you scale up the city or no matter how finely you look at it, you'll never get any flat sections. This whole thing scaled down is the same as this section, which is the same as this section, which is the same as this. Three times as big is four times as much stuff. Not three, like if it were a normal 1D line, and certainly not nine, like that 2D area on the inside. Somehow, the infinity fractal-ness of the thing makes it behave differently from all 1D things and all 2D things. You convince yourself that all 1D things got twice as big when you make them twice as big, because you could think And you know how line segments behave. And you convince yourself all 2D things scaled up by two get four times as much stuff, because 2D things can be thought of as being mad of squares, and you know how squares behave. But then there is this which has no straight lines in it. And there's no square areas in it, either. More than three to the one, less than three to the two. It behaves as if it's between one and two dimensions. You think back to Sierpinski's triangle. Maybe it can be thought of as being made out of straight line segments, though there's an infinite amount of them and they get infinitely small. When you make it twice as tall, if you just make all the lines of this drawing twice as long, you're missing detail again. But the tiny lines too small to draw are also twice as long and now visible. And so on, all the way down to the infinitely small line segments. You wonder if your similar line thing works on lines that don't actually have length. Wait. Lines that don't have length? Is that a thing? First, though, you figure out that when you make it twice as Not two, like a 1D triangle outline. Not four, like a solid 2D triangle. But somewhere in between. And the in between-ness seems to be true, no matter which way you make it-- out of lines, or by subtracting 2D triangles, or with squiggles. They all end up the same. An object in fractional dimension. No longer 1D because of infinity infinitely small lines. Or no longer 2D because of subtracting out all the area. Or being an infinitely squiggled up line that's too infinate and squiggled to be a line anymore, but doesn't snuggle into itself enough to have any 2D area, either. Though in the dragon curve, it does seem to snuggle up into itself. Hm. If you pretend this is the complete dragon curve and iterate this way, there's twice as much stuff. That's what you'd expect from a 1D line if it were scaling it by two. But let's see, this is scaling up by, well, not quite two. Let's see. I suppose if you did it perfectly, it's supposed to be an equilateral right triangle. So square root 2. If it were two dimensional, you'd" + }, + { + "Q": "At 5:00, Vi mentions a line that has no length. Wouldn't that be resembling a point? A point has no length whatsoever, or width, hence only trapped in 1st dimension. So does a potential line with no length. So... that means it's a case of a=b and b=c so a=c, right?", + "A": "A point is zero-D it has no length", + "video_name": "Oc8sWN_jNF4", + "timestamps": [ + 300 + ], + "3min_transcript": "If it were a regular line, not an infinitely spiked one, scaling up by three would make it three times as much drawing, as expected. But if that spiky line were supposed to represent an infinitely spiked magical fortress city of dragon dungeon doom, by scaling it up in this way, you'd be losing details, making these long lines that should have had spiky bumps in them. Theoretically, no matter how much you scale up the city or no matter how finely you look at it, you'll never get any flat sections. This whole thing scaled down is the same as this section, which is the same as this section, which is the same as this. Three times as big is four times as much stuff. Not three, like if it were a normal 1D line, and certainly not nine, like that 2D area on the inside. Somehow, the infinity fractal-ness of the thing makes it behave differently from all 1D things and all 2D things. You convince yourself that all 1D things got twice as big when you make them twice as big, because you could think And you know how line segments behave. And you convince yourself all 2D things scaled up by two get four times as much stuff, because 2D things can be thought of as being mad of squares, and you know how squares behave. But then there is this which has no straight lines in it. And there's no square areas in it, either. More than three to the one, less than three to the two. It behaves as if it's between one and two dimensions. You think back to Sierpinski's triangle. Maybe it can be thought of as being made out of straight line segments, though there's an infinite amount of them and they get infinitely small. When you make it twice as tall, if you just make all the lines of this drawing twice as long, you're missing detail again. But the tiny lines too small to draw are also twice as long and now visible. And so on, all the way down to the infinitely small line segments. You wonder if your similar line thing works on lines that don't actually have length. Wait. Lines that don't have length? Is that a thing? First, though, you figure out that when you make it twice as Not two, like a 1D triangle outline. Not four, like a solid 2D triangle. But somewhere in between. And the in between-ness seems to be true, no matter which way you make it-- out of lines, or by subtracting 2D triangles, or with squiggles. They all end up the same. An object in fractional dimension. No longer 1D because of infinity infinitely small lines. Or no longer 2D because of subtracting out all the area. Or being an infinitely squiggled up line that's too infinate and squiggled to be a line anymore, but doesn't snuggle into itself enough to have any 2D area, either. Though in the dragon curve, it does seem to snuggle up into itself. Hm. If you pretend this is the complete dragon curve and iterate this way, there's twice as much stuff. That's what you'd expect from a 1D line if it were scaling it by two. But let's see, this is scaling up by, well, not quite two. Let's see. I suppose if you did it perfectly, it's supposed to be an equilateral right triangle. So square root 2. If it were two dimensional, you'd" + }, + { + "Q": "At 2:29 why does Vi draw a man with 3 \u00f0\u009f\u0091\u0080", + "A": "That man lives in 4 dimensional space, meaning that a person needs three eyes to see properly.", + "video_name": "Oc8sWN_jNF4", + "timestamps": [ + 149 + ], + "3min_transcript": "So you're back in math class again. It just-- it never stops. Day after day you find yourself trapped behind this desk with only your notebook for company, that pale mirror that reflects your thoughts on a comforting simulacrum of shared ideas. You're wasting another irreplaceable hour of your finite life not even pretending to listen to your teacher talk about logarithms. Or, at least, you think it's logarithms that she's trying to teach you. You haven't exactly been paying attention as you sit there casually, disposing of this only this moment you'll ever have, and-- oops, there goes another one. But either way, it's definitely logarithms in particular that you're not even pretending to learn, as opposed to, say, calculus, which your teacher wouldn't even expect you to be pretending to pay attention to. So instead, you're amusing yourself by doodling. Two days ago, you discovered the infinitely self-similar beast that is the fractal. And yesterday, you discovered that when you scale a two-dimensional thing up by two, it grows by a factor of four. And when you scale it by any amount, the area grows by that amount squared, unlike 1D where it's just that amount, and in 3D it's that amount cubed. And in n dimensions, it's that amount to the n. So when you put it like that, you're actually making pretty good use of your limited time And by limited time on this earth, I mean that we're all going to become immortal space robots. Anyway, you're continuing your plans for a fractal city of infinite dragon dungeons, triangles upon triangles upon triangles, each next set scaled down by a factor of 3. That's 1/9 the area. But there's four times as many. And the next set has four times as many as the last, but 1/9 times the size of the last. So the total weight of steel is 1 plus 4/9 plus 4 squared over 9 squared plus 4 to the 3 over 9 to the 3, dot, dot, dot, plus 40 to the n over 9 to the n. And maybe could learn how to add up an infinite series of numbers if your teacher would ever get past logarithms. But at least you know how to create the perfect fractal city, which is good, because understanding scale factors and city planning seems like the kind of thing that might come in handy if you want to help the species on our journey towards becoming immortal space robots. Really, the only thing that could make the city better Or how about three times as big? So you can keep this part of the design and just draw the next iteration. Three times the scale in two dimensions means technically you'll need nine times as much steel. Though, as far as drawing these plans go, it's not nine times as difficult, but only four, since the hard part is the outside spiky part. And that's just copied four times, in order to scale up by three. Weird thing number one. Scaling up by three makes nine times as much steel. But it's the same thing copied four times, plus filling in this middle triangle. So it's also four times the steel plus 9. So if this mystery sum of an infinite series, total weight of steel were x, and 9x equals 4x plus 9, 5x equals 9, x equals 9/5 exactly. Take that, infinity! OK. Now weird thing number two. Look at just the edge, the actual Koch Curve" + }, + { + "Q": "At 5:17, how did you get those numbers if young are adding 2, then, then 6.", + "A": "Each number is 2 more than the previous number. If x is the first number, then the 2nd number is x+2. The third number is 2 more than the previous number, which is the 2nd number, which also is x+2. So the 3rd number is (x+2)+2 = x+3. The 4th number is the 3rd + 2, which is (x+4)+2 = x+6", + "video_name": "8CJ6Qdcoxsc", + "timestamps": [ + 317 + ], + "3min_transcript": "So we can rewrite those literally as 4x. And then we have 2 plus 4, which is 6, plus another 6 is 12. 4x plus 12 is equal to 136. So to solve for x, a good starting point would be to just to isolate the x terms on one side of the equation or try to get rid of this 12. Well to get rid of that 12, we'd want to subtract 12 from the left-hand side. But we can't just do it from the left-hand side. Then this equality wouldn't hold anymore. If these two things were equal before subtracting the 12, well then if we want to keep them equal, if we want the left and the right to stay equal, we've got to subtract 12 from both sides. So subtracting 12 from both sides gives us, well on the left-hand side, we're just left with 4x. And on the right-hand side, we are left with 136 minus 12 is 124. Yeah, 124. Well, we just divide both sides by 4 to solve for x. And we get-- do that in the same, original color-- x is equal to 124 divided by 4. 100 divided by 4 is 25. 24 divided by 4 is 6. 25 plus 6 is 31. And if you don't feel like doing that in your head, you could also, of course, do traditional long division. Goes into 124-- 4 doesn't go into 1. 4 goes into 12 three times. 3 times 4 is 12. You subtract, bring down the next 4. 4 goes into 4 one time. You get no remainder. So x is equal to 31. So x is the smallest of the four integers. So this right over here, x is 31. x plus 2 is going to be 33. And x plus 6 is going to be 37. So our four consecutive odd integers are 31, 33, 35, and 37." + }, + { + "Q": "at 4:45 why is the second derivative negative? i thought if it is going up, than it is positive both below and above the x axis?", + "A": "The second derivative is negative wherever the first derivative is decreasing.", + "video_name": "LcEqOzNov4E", + "timestamps": [ + 285 + ], + "3min_transcript": "" + }, + { + "Q": "Question about 8:37 onwards: so, what if the critical points are undefined for f(x)? i.e.: the \"maximum\" point of the downward concave area doesn't have a value, is discontinuous -- is this point still considered a maximum, even though there's no value for the point, or do maximum and minimum points only apply if the interval is continuous?", + "A": "We can check the function, however, discontinuities in the derivative sometimes result in discontinuities in the original function (such as the derivatives of tanx, 1/x, and 1/x\u00e2\u0081\u00bf).", + "video_name": "LcEqOzNov4E", + "timestamps": [ + 517 + ], + "3min_transcript": "" + }, + { + "Q": "at 7:36 you said slope is increasing. Does that mean slope is becoming more positive?", + "A": "Increasing slope can mean one of two things: more positive or less negative. Whichever situation you have, increasing slope always implies concave up.", + "video_name": "LcEqOzNov4E", + "timestamps": [ + 456 + ], + "3min_transcript": "" + }, + { + "Q": "At 5:40, why is it true that if the first derivative/slope of the line is increasing, then the second derivative must be positive? (and vice versa for decreasing/negative)", + "A": "Consider that when the original function f(x) is increasing/decreasing, then the first derivative f (x) is positive/negative. Similarly, when f (x) is increasing/decreasing, then its derivative f (x) is positive/negative. However, the main idea of this video is to help illustrate that when the second derivative is positive/negative, then the original function is concave up/down. The second derivative is useful for classifying extrema and for identifying concavity of a function.", + "video_name": "LcEqOzNov4E", + "timestamps": [ + 340 + ], + "3min_transcript": "" + }, + { + "Q": "So there is the (1:00):\nf(g(x)) = sqrt( g(x)^2 - 1 )\nI was recently watching video series on complex numbers and I am just curious why can't we rewrite that thing as\ng(x) * sqrt(-1)\nand so\ng(x)*sqrt(i^2)\nto\ng(x)*i hence state the solution in terms of complex numbers? I know it is a boo-boo *but why?*", + "A": "Sorry, your method doesn t work. You changed subtraction: sqrt( g(x)^2 - 1 ) into multiplication: g(x) * sqrt(-1). This is not a valid operation. Also, there is no property of square roots that lets you split the square root by terms (items being added/subtracted). You can only split square roots by factors (items being multiplied/divided). Hope this helps.", + "video_name": "_b-2rZpX5z4", + "timestamps": [ + 60 + ], + "3min_transcript": "Voiceover:When we first got introduced to function composition, we looked at actually evaluating functions at a point, or compositions of functions at a point. What I wanna do in this video is come up with expressions that define a function composition. So, for example, I wanna figure out, what is, f of, g of x? f of, g of x. And I encourage you to pause the video, and try to think about it on your own. Well, g of x in this case, is the input to f of x. So, wherever we see the x in this definition, that's the input. So we're going to replace the input with g of x. We're going to replace the x with g of x. So, f of g of x is going to be equal to the square root of- Well instead of an x, we would write a g of x. g of x, g of x squared. g of x squared, minus one. Now what is g of x equal to? So this is going to be equal to the square root of, g of x, is x over 1 plus x. We're going to square that. We're going to square that, minus 1. So f of g of x, is also a function of x. So f of g of x is a square root of, and we could write this as x squared over 1 plus x squared, but we could just leave it like this. It's equal to the square root of this whole thing, x over 1 plus x, squared, minus one. Now let's go the other way round. What is g of f of x? What is g of f of x? And once again, I encourage you to pause the video, and try to think about it on your own. Well, f of x is now the input into g of x. So everywhere we see the x here, we'll replace it with f of x. So this is going to be equal to, this is going to be equal to, f of x, over- so you can appreciate it better. f of x over, one plus f of x. One plus f of x. And what's that equal to? Well, f of x is equal to the square root, of x squared minus one. x squared minus one. So it's gonna be that over 1, plus the square root. One plus the square root of x squared minus one. So this is a composition f of g of x, you get this thing. This is g of f of x, where you get this thing. And to be clear, these are very different expressions. So typically, you want the composition one way. This isn't gonna be the same as the composition the other way, unless the functions are designed in a fairly special way." + }, + { + "Q": "I don't understand what he says at 6:09, please explain!", + "A": "The length of one cycle of the standard cosine function is 2pi. So we need to ask ourselves: How does 2pi compare to the length of one cycle in the problem? That s why we set up the ratio of 2pi to 365, or written in fractional terms, 2pi/365.", + "video_name": "mVlCXkht6hg", + "timestamps": [ + 369 + ], + "3min_transcript": "either actual degrees or radians, which trigonometric function starts at your maximum point? Well cosine of zero is one. The cosine starts at your maximum point. Sine of zero is zero, so I'm going to use cosine here. I'm going to use a cosine function. So, temperature as a function of days. There's going to be some amplitude times our cosine function and we're going to have some argument to our cosine function and then I'm probably going to have to shift it. So let's think about how we would do that. Well, what's the mid line here? The mid line is the halfway point between our high and our low. So our midpoint, if we were to visualize it, looks just like so. That is our mid line right over there. And what value is this? Well what's the average of 29 and 14? 29 plus 14 is 43 divided by two is 21.5 degrees Celsius. shifted up our function by that amount. If we just had a regular cosine function our mid line would be at zero, but now we're at 21.5 degrees Celsius. I'll just write plus 21.5, that's how much we've shifted it up. Now, what's the amplitude? Well our amplitude is how much we diverge from the mid line. Over here we're 7.5 above the mid line so that's plus 7.5. Here we're 7.5 below the mid line, so minus 7.5. So our amplitude is 7.5, the maximum amount we go away from the mid line is 7.5. So that's our amplitude. And now let's think about our argument to the cosine function right over here. It's going to be a function of the days. And what do we want? When 365 days have gone by, we want this entire argument to be two pi. whole thing to evaluate to two pi. We could put two pi over 365 in here. You might remember your formulas, I always forget them that's why I always try to reason through them again. The formulas, you want two pi divided by your period and all the rest, but I just like to think, \"Okay, look. \"After one period, which is 365 days, I want the whole \"argument over here to be two pi. \"I want to go around the unit circle once and so if this \"is two pi over 365, when you multiply it by 365 \"your argument here is going to be two pi.\" Just like that we've done the first part of this question. We have modeled the average high temperature in Santiago as a function of days after January seventh. In the next video we'll answer this second question. I encourage you to do it ahead of time before watching that next video and I'll give you one clue." + }, + { + "Q": "At 2:28 why does -24x become -25x when you subtract x", + "A": "Because a negative minus a positive is basically adding a negative to a negative.", + "video_name": "711pdW8TbbY", + "timestamps": [ + 148 + ], + "3min_transcript": "In this video, we're going to start having some experience solving radical equations or equations that involve square roots or maybe even higher-power roots, but we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's say I have the equation the square root of x is equal to 2 times x minus 6. Now, one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that, I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8." + }, + { + "Q": "At 5:40 Sal says that the square root of 2.25 could be 1.5 or -1.5. At 5:57, Sal says that 1.5 \u00e2\u0089\u00a0 -1.5. However, at 5:52, couldn't he have as easily used -1.5 as the square root of 2.25 instead of 1.5? If he had used -1.5, the solution 2.25 would have been valid. Is it because you can only principal square roots? If so, why?", + "A": "No. all square roots, even non-perfect squares will always have a positive or negative value, even square root of is +2 or -2. Why? If you might have forgotten; two negatives, when multiplied together, will always become positive. However, cube roots only have a positive root if the value inside the cube root sign is positive and negative if negative. And yes, 1.5 is not equal to -1.5 but their square is equal.", + "video_name": "711pdW8TbbY", + "timestamps": [ + 340, + 357, + 352 + ], + "3min_transcript": "8 goes into 18 two times, remainder 2, so this is equal to 2 and 2/8 or 2 and 1/4 or 2.25, just like that. Now, I'm going to show you an interesting phenomena that occurs. And maybe you might want to pause it after I show you this conundrum, although I'm going to tell you why this conundrum pops up. Let's try out to see if our solutions actually work. So let's try x is equal to 4. If x is equal to 4 works, we get the principal root of 4 should be equal to 2 times 4 minus 6. The principal root of 4 is positive 2. Positive 2 should be equal to 2 times 4, which is 8 minus 6, which it is. This is true. So 4 works. Now, let's try to do the same with 2.25. According to this, we should be able to take the square root, the principal root of 2.2-- let me make my radical a The principal root of 2.25 should be equal to 2 times 2.25 minus 6. Now, you may or may not be able to do this in your head. You might know that the square root of 225 is 15. And then from that, you might be able to figure out the square root of 2.25 is 1.5. Let me just use the calculator to verify that for you. So 2.25, take the square root. It's 1.5. The principal root is 1.5. Another square root is negative 1.5. So it's 1.5. And then, according to this, this should be equal to 2 times 2.25 is 4.5 minus 6. Now, is this true? This is telling us that 1.5 is equal to negative 1.5. This is not true. 2.5 did not work for this radical equation. So 2.25 is an extraneous solution. Now, here's the conundrum: Why did we get 2.25 as an answer? It looks like we did very valid things the whole way down, and we got a quadratic, and we got 2.25. And there's a hint here. When we substitute 2.25, we get 1.5 is equal to negative 1.5. So there's something here, something we did gave us this solution that doesn't quite apply over here. And I'll give you another hint. Let's try it at this step. If you look at this step, you're going to see that both solutions actually work. So you could try it out if you like. Actually, try it out on your own time. Put in 2.25 for x here. You're going to see that it works. Put in 4 for x here and you see that they both work here. So they're both valid solutions to that. So something happened when we squared that made the equation" + }, + { + "Q": "At 1:34, you did some factoring in your head...hard to follow to get the 24x. Could you expound upon that? I know - watch the factoring videos - but that one lost me. It wasn't quite clear WHY you multiply the 2x and the 6 to get the 24.", + "A": "Because the square factor form is as following: (ax-b)^2 ax-b x ax-b _________ ax^2 -abx -bax +b^2 Focus in the part (-abx -bax), since multiplication scalar is commutative, they are twice themselves. Rewrite them in order, (-abx -abx), which is equal to 2(-abx). If a = 2 and b = 6, representing them in the expression 2(-2*6x), 2(-12x) that is equal to -24x.", + "video_name": "711pdW8TbbY", + "timestamps": [ + 94 + ], + "3min_transcript": "In this video, we're going to start having some experience solving radical equations or equations that involve square roots or maybe even higher-power roots, but we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's say I have the equation the square root of x is equal to 2 times x minus 6. Now, one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that, I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8." + }, + { + "Q": "Ok, I know what PEMDAS is, what do you do if you get, say, 3*6:9? M and D are on the same level!", + "A": "work from left to right!", + "video_name": "0uCslW40VHQ", + "timestamps": [ + 369 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:43, the speaker explains PEMDAS . Is there another way to solve expressions?", + "A": "Not in my knowledge. to me, PEMDAS is the best way to do order of operations.", + "video_name": "0uCslW40VHQ", + "timestamps": [ + 43 + ], + "3min_transcript": "Evaluate the expression 5y to the fourth minus y squared when y is equal to 3. So every place we see a y here, we could just replace it with a 3 to evaluate it. So it becomes 5 times 3 to the fourth power minus 3 squared. All I did is every time we saw a y here, I put a 3 there. Every time we saw a y, I put a 3. So what does this evaluate to? And we have to remember our order of operations. Remember, parentheses comes first. Sometimes it's referred to as PEMDAS. Let me write that down. PEMDAS, PEMDAS. P is for parentheses. E is for exponents. M and D are for Multiplication and Division. They're really at the same level of priority. And then addition and subtraction If you really want to do it properly, it should be P-E, and then multiplication and division are really at the same level. And addition and subtraction are at the same level. But what this tells us is that we do parentheses first. But then after that, exponentiation takes priority over everything else here. before we multiply anything or before we subtract anything. So the one exponent we'd have to evaluate is 3 squared. So let's remember. 3 to the first is just 3. It's just 3 times itself once. 3 squared is equal to 3 times 3, 3 multiplied by itself twice. That's equal to 9. 3 to the third power is equal to 3 times 3 times 3. Or you could view it as 3 squared times 3. So it'll be 9. 3 times 3 is 9. 9 times 3 is equal to 27. 3 to the fourth is equal to 3 times 3 times 3 times 3. So 3 times 3 is 9. 3 times 3 is 9. So it's going to be the same thing as 9 times 9. So this is going to be equal to 81. So we now know what 3 to the fourth is. We know what 3 squared is. Let's just put it in the expression. So this is going to be equal to 5 times 3 to the fourth. 3 to the fourth is 81. And we have 3 squared right over here. It is equal to 9. 5 times 81 minus 9. Let's figure out what 5 times 81 is. So 81 times 5. 1 times 5 is 5. 8 times 5 is 40. So this right over here is 405. So it becomes 405 minus 9. So that is going to be equal to-- if we were subtracting 10, it would be 395. But we're subtracting one less than that. So it's 396. And we're done." + }, + { + "Q": "Around 4:50 when Sal is subtracting \"all of this business\", i.e. -e^xcosx+Se^x cosx dx, why does the plus sign become a minus sign? Its e^x sinx - -e^xcosx... so that becomes plus. But why does the plus sign before the antiderivative become a minus sign??", + "A": "When he was taking e^x(sin(x)) - \u00e2\u0088\u00ab e^x(sin(x))dx, the minus sign acts as a negative sign so it would end up something like this in his thinking: e^x(sin(x)) + [-( \u00e2\u0088\u00ab e^x(sin(x))dx)]. You would take the integral of e^x(sin(x))dx to get -e^xcos(x) + \u00e2\u0088\u00ab (e^xcos(x))dx and factor the negative into the whole equation to get: e^x(sin(x)) + e^xcos(x) - \u00e2\u0088\u00ab (e^xcos(x))dx", + "video_name": "LJqNdG6Y2cM", + "timestamps": [ + 290 + ], + "3min_transcript": "prime of xg of x. F prime of x is e to x. And then g of x is negative cosine of x. So I'll put the cosine of x right over here, and then the negative, we can take it out of the integral sign. And so we're subtracting a negative. That becomes a positive. And of course, we have our dx right over there. And you might say, Sal, we're not making any progress. This thing right over here, we now expressed in terms of an integral that was our original integral. We've come back full circle. But let's try to do something interesting. Let's substitute back this-- all right, let me write it this way. Let's substitute back this thing up here. Let's substitute this for this in our original equation. And let's see if we got anything interesting. So what we'll get is our original integral, on the left hand side here. The indefinite integral or the antiderivative of e to the x cosine of x dx is equal to e to the x sine of x, minus all of this business. So let's just subtract all of this business. We're subtracting all of this. So if you subtract negative e to the x cosine of x, it's going to be positive. It's going to be positive e to the x, cosine of x. And then remember, we're subtracting all of this. So then we're going to subtract. So then we have minus the antiderivative of e to the x, Now this is interesting. Just remember all we did is, we took this part right over here. We said, we used integration by parts to figure out that it's the same thing as this. So we substituted this back in. When you subtracted it. When you subtracted this from this, we got this business right over here. Now what's interesting here is we have essentially an equation where we have our expression, our original expression, twice. We could even assign this to a variable and essentially solve for that variable. So why don't we just add this thing to both sides of the equation? Let's just add the integral of e to the x cosine of x dx to both sides. e to the x, cosine of x, dx. And what do you get? Well, on the left hand side, you have two times our original integral. e to the x, cosine of x, dx is equal to all of this business." + }, + { + "Q": "Around 2:00 I don't understand what he is saying about the point in the middle being an equidistant away from each angle. Please explain.", + "A": "He was referencing a video about perpendicular bisectors. Basically, Sal is saying that if you have one point that is equidistant from two other points on a line, but said point ISNT on that line, then it lies on the perpendicular bisector of that line. so since the distance from I is the same distance from AB, and I ISNT on AB, then I is on the perpendicular bisector of AB. Sorry if Im not much help, watch the video!", + "video_name": "21vbBiCVijE", + "timestamps": [ + 120 + ], + "3min_transcript": "I have triangle ABC here. And in the last video, we started to explore some of the properties of points that are on angle bisectors. And now, what I want to do in this video is just see what happens when we apply some of those ideas to triangles or the angles in triangles. So let's bisect this angle right over here-- angle BAC. And let me draw an angle bisector. So the angle bisector might look something-- I want to make sure I get that angle right in two. Pretty close. So that looks pretty close. So that's the angle bisector. Let me call this point right over here-- I don't know-- I could call this point D. And then, let me draw another angle bisector, the one that bisects angle ABC. So let me just draw this one. It might look something like that right over there. And I could maybe call this point E. So AD bisects angle BAC, and BE bisects angle ABC. So the fact that this green line-- AD bisects this angle right over here-- be equal to that angle right over there. They must have the same measures. And the fact that this bisects this angle-- angle ABC-- tells us that the measure of this angle-- angle ABE-- must be equal to the measure of angle EBC. Now, we see clearly that they have intersected at a point inside of the triangle right over there. So let's call that point I just for fun. I'm skipping a few letters, but it's a useful letter based on what we are going to call this in very short order. And there's some interesting things that we know about I. I sits on both of these angle bisectors. And we saw in the previous video that any point that sits on an angle bisector is equidistant from the two sides of that angle. So for example, I sits on AD. So it's going to be equidistant from the two sides of angle BAC. So this is one side right over here. And then this is the other side right over there. So because I sits on AD, we know that these two distances are going to be the same, assuming that this is the shortest distance between I and the sides. And then, we've also shown in that previous video that, when we talk about the distance between a point and a line, we're talking about the shortest distance, which is the distance you get if you drop a perpendicular. So that's why I drew the perpendiculars right And let's label these. This could be point F. This could be point G right over here. So because I sits on AD, sits on this angle bisector, we know that IF is going to be equal to IG. Fair enough. Now, I also sits on this angle bisector. It also sits on BE, which says that it must be equidistant. The distance to AB must be the same as I's distance to BC. I's distance to AB we already just said is this right over here." + }, + { + "Q": "at 3:27,a part of the equation does not show up. why is that?", + "A": "you know how sometimes he slides the screen down I think its cause his screen is bigger than whatever were watching it on.", + "video_name": "CLrImGKeuEI", + "timestamps": [ + 207 + ], + "3min_transcript": "Now we have it in that form. We have ax squared a is negative 1. So let me write this down. a is equal to negative 1. a is equal to negative 1. It's implicit there, you could put a 1 here if you like. A negative 1. Negative x squared is the same thing as negative 1x squared. b is equal to 8. So b is equal to 8, that's the 8 right there. And c is equal to negative 1. That's the negative 1 right there. So now we can just apply the quadratic formula. The solutions to this equation are x is equal to negative b. Plus or minus the square root of b squared, of 8 squared, the green is the part of the formula. The colored parts are the things that we're substituting into the formula. Minus 4 times a, which is negative 1, times negative 1, times c, which is also negative 1. And then all of that-- let me extend the square root sign a little bit further --all of that is going to be over 2 times a. In this case a is negative 1. So let's simplify this. So this becomes negative 8, this is negative 8, plus or minus the square root of 8 squared is 64. And then you have a negative 1 times a negative 1, these just cancel out just to be a 1. So it's 64 minus is 4. That's just that 4 over there. All of that over negative 2. So this is equal to negative 8 plus or minus the All of that over negative 2. And let's see if we can simplify the radical expression here, the square root of 60. Let's see, 60 is equal to 2 times 30. 30 is equal to 2 times 15. And then 15 is 3 times 5. So we do have a perfect square here. We do have a 2 times 2 in there. It is 2 times 2 times 15, or 4 times 15. So we could write, the square root of 60 is equal to the square root of 4 times the square root of 15, right? The square root of 4 times the square root of 15, that's what 60 is. 4 times 15. And so this is equal to-- square root of 4 is 2 times the square of 15. So we can rewrite this expression, right here, as being equal to negative 8 plus or minus 2 times the square" + }, + { + "Q": "Starting at about 3:35, Sal write the equation for L(x). Where does he get f(4) from? Also, is L(x) just the tangent line to f(x) at x = 4?", + "A": "Yes, L(x) is the tangent line to f(x) at x=4. Sal uses f(4) because he is writing the equation of the line tangent to f(x) at the point (4, f(4)) (if the x-value is 4, then the y-value is f(4), right?). You might be more used to always writing y = mx +b, where you put in the slope (which in this case would be f (4)), and then plug in a point to find b, but Sal s method is faster and easier. Basically, he is using the point-slope form instead of the slope-intercept form.", + "video_name": "u7dhn-hBHzQ", + "timestamps": [ + 215 + ], + "3min_transcript": "So that right over here is going to be 2. That's f of 4. And what we wanna approximate is f of 4.36, so 4.36 might be right around, right around there, and so we want to approximate, we wanna approximate this y value right over here. We want to approximate that. Right over here is f of 4.36, and, once again, we're assuming we don't have a calculator at hand. So, how can we do that using what we know about derivatives? Well, what if we were to figure out an equation for the line that is tangent to the point, to tangent to this point right over here. So the equation of the tangent line at x is equal to 4, and then we use that linearization, that linearization defined to approximate values local So what I'm saying is, let's figure out what this, the equation of this line is. Let's call that l of x. And then we can use that to appro, and then we can evaluate that at 4.36, and hopefully that will be a little bit easier to do than to try and figure out this right over here. So how would we do that? Well, one way to think about it, and obviously, there are many ways to express a line, but one way to think about it is, okay, it's going to l of x is going to be f of 4, which is 2. It's going to be f of 4 plus the slope, the slope at, at x equals 4, which is, of course, the derivative f prime of 4, so that's going to be the slope of this line of l of x is f prime of 4. Let me make that clear. So this right over here is the slope. The slope when x is at, at x equals 4, so other point on this is gonna be f' of 4 plus the slope times how far you are away from x equals 4. So it's going to be times x minus 4. Let's just, let's just validate that this makes sense. When we put 4.36 here, when we put 4.36 here, actually let me zoom in on this graph just to make things a little bit clearer. So, if this is, so I'm gonna do a zoom in. I'm gonna do a zoom in. I'm gonna try to zoom in into this region right over here. So, this is the point. This is the point (4, f of 4), and we are going to graph l of x. So let me do that. So this right over here is l of x. And let's say this, right over here, this right over" + }, + { + "Q": "At 0:36 when did \"Deca-\" prefix become \"Deka-\"?", + "A": "Both deka- and deca- are used alternatively and mean the exact same thing. I always used deca-, but apparently many people use the alternative deka-. Doing a web search indicated deca- was used about three times as often as deka-. But using deka- would probably result in less people misreading it as deci-. If I was King, I would declare deka- should always be used instead of deca-.", + "video_name": "SYkmadc2wOI", + "timestamps": [ + 36 + ], + "3min_transcript": "We're asked how many centiliters are in one dekaliter? So the first thing we want to do is just think about how much is a centiliter relative to a liter, and how much is a dekaliter relative to a liter? And I'll write the prefixes down. And really, you should have these memorized because you're going to see these prefixes over and over again for different types of units. So the prefix, kilo, sometimes [? ki-lo, ?] this means 1,000. If you see hecto, hecto means 100. Deka means 10. If you have nothing, then that just means 1. Let's put that there. Then if you have deci, this means 1/10. If you have centi, this means 1/100. If you have milli, this means 1/1,000. So let's go back to what we have. We have centiliters. If you have a centiliter, this is equal to 1/100 of a liter, Or you could say 1 liter for every 100 centiliters, so you could also write it like this: 1 liter for every 100, or per every 100, centiliters. So we got the centi, now let's think about the dekaliter. So the deka is right over here. So a dekaliter means 10 liters. Or another way to say it is for every 10 liters, you will Now, before I actually work out the problem, what's going on here? We're going from a smaller unit to a larger unit, so there are going to be many of the smaller units in one of the larger ones. And we can do it multiple ways. So we want to essentially convert 1 dekaliter into centiliters. Now, we could just do it by looking at this chart, or we could do it with the dimensional analysis, making sure the dimensions work out. Let's do it the first way. So if you have one dekaliter, how many liters is that? 1 dekaliter over here would be the same thing as 10 liters. That's liters. We're assuming that our unit is liters here. And then 10 liters is going to be how many deciliters? It's going to be 100 deciliters, right? Because each of these is 10 deciliters, and you have 10 of them. So every time you go down, you're going to be multiplying" + }, + { + "Q": "At 2:10 could't you just do 17-5 and then get 12 and x=4", + "A": "he should of clarified it simpler", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 130 + ], + "3min_transcript": "" + }, + { + "Q": "at 4:16 why dose it not excist any more?", + "A": "Hah that s a bit funny to read. It still exists but Sal removed it because it was subtracted", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 256 + ], + "3min_transcript": "" + }, + { + "Q": "Why didn't Sal explain why -8/7 is the same that -(8/7) at 8:30. It isn't so simple to understand.", + "A": "just think about it. -8/7 is equal to 8/-7 and also equal to the negative value of 8/7. so -8/7 = 8/-7 = -(8/7)", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 510 + ], + "3min_transcript": "" + }, + { + "Q": "at 7:41 you were saying -10 + 2 does not = -12. i have an easy way to understand it. imagine you are paying a debt of 10 bucks. then, you win 2 bucks in a competition. so when you give the person you owe your 2 bucks,you would only owe 8 bucks, as -10 + 2=8. all those who think my method is practical can answer this.", + "A": "sorry i was supposed to type -8 but i accidentally typed8", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 461 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:14, EX1. y/2-8=10;", + "A": "y/2 - 8 = 10 Add 8 to both sides y/2 = 18 Multiply both sides by 2 y = 36", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 254 + ], + "3min_transcript": "" + }, + { + "Q": "in 1:20-1:25 i dont get it how can 3x+5=17", + "A": "Put, x = 4 and see. The equation holds!", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 80, + 85 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:41, is 5. (1) not the same as 5 x x 1 ?", + "A": "If another operation is before the parentheses, then you don t multiply. If you have something like 3- (-3), you just subtract the -3. If yo have 4*(5), you only multiply once.", + "video_name": "AJNDeVt9UOo", + "timestamps": [ + 41 + ], + "3min_transcript": "A local hospital is holding a raffle as a fundraiser. The individual cost of participating in the raffle is given by the following expression-- 5t plus 3, or 5 times t plus 3, where t represents the number of tickets someone purchases. Evaluate the expression when t is equal to 1, t is equal to 8, and t is equal to 10. So let's first take the situation where t is equal to 1. Then this expression right over here becomes-- and I'll use that same color-- becomes 5 times 1. 5 times 1 plus 3. 5 times 1 plus 3, and we know from order of operations, you do the multiplication before you do the addition. So this will be 5 times 1 is 5 plus 3, and then this is clearly equal to 8. Now let's do it when t is equal to 8. and I'll do the same colors again-- 5 times 8 plus 3. Same color of green. And once again, 5 times 8 is 40, and then we have the plus 3, there so this is equal to 43. And so we have the last situation, with t is equal to 10. I'll do that in blue. So we have 5 times 10. So 5t is 5 times 10. Instead of a t, put a 10 there. 5 times 10 plus 3. That's a slightly different shade of green, but I think you get the idea. 5 times 10 is 50. We do 50, and then we're going to have to add 3 to that, and that is equal to 53. And we're done." + }, + { + "Q": "where does he get 104 at 1:26? 80-24? where does he even get that.", + "A": "to add or subtract the fractions like 85/13 and 8/1 are we take the common denominators of both of the fractions and to maintain the equality the changes which we make in the denominator like to make the denominator of fraction 8/1 we multiplied the denominator by 13 so we do the same thing in the numerator thats how sal got 104", + "video_name": "KV_XLL4K2Fw", + "timestamps": [ + 86 + ], + "3min_transcript": "The following line passes through the point 5 comma 8, and the equation of the line is y is equal to 17/13x plus b. What is the value of the y-intercept b? So we know that this point, this x and y value must satisfy this equation, so we know that when x is equal to 5, y is equal to 8. So we can say-- so when x is equal to 5, y is equal to 8. So we can say 8 must be equal to 17/13 times x times 5 plus b, and then we can solve for b. So if we simplify this a little bit, we get 8 is equal to-- let's see, 5 times 17 is 50, plus 35 is 85-- is 85/13 plus b. Then to solve for b, we just subtract 85/13 from both sides. 85/13 Is equal to b. And now we just have to subtract these two numbers. So 8 is the same thing as-- let's see. 80 plus 24 is 104, so it is 104/13-- this is the same thing as 8-- minus 85/13, which is going to be-- let's see. This is 19/13. Is that right? Yes, if this was 105, then it would be 20/13. So this is 19/13, so it's equal to 19/13, which is equal to b. So the equation of this line is going to be y is equal to 17/13x plus 19/13." + }, + { + "Q": "Why is it (1+ the square root of 5,-2)[at 12:48]", + "A": "adding x coordinate of center point with focal distance y coordinate stays the same", + "video_name": "QR2vxfwiHAU", + "timestamps": [ + 768 + ], + "3min_transcript": "y is equal to minus 2. That's the center. And then, the major axis is the x-axis, because this is larger. And so, b squared is -- or a squared, is equal to 9. And the semi-minor radius is going to be equal to 3. So, if you go 1, 2, 3. Go there. Than you have 1, 2, 3. No. 1, 2, 3. I think this -- let's see. 1, 2, 3. You go there, roughly. And then in the y direction, the semi-minor radius is going to be 2, right? The square root of that. So b is equal to 2. So you go up 2, then you go down 2. And this ellipse is going to look something like -- pick a good color. It's going to look something like this. And what we want to do is, we want to find out the coordinates of the focal points. the semi-major axis. And we need to figure out these focal distances. And then we can essentially just add and subtract them from the center. And then we'll have the coordinates. What we just showed you, or hopefully I showed you, that the the focal length or this distance, f, the focal length is just equal to the square root of the difference between these two numbers, right? It's just the square root of 9 minus 4. So the focal length is equal to the square root of 5. So, if this point right here is the point, and we already showed that, this is the point -- the center of the ellipse is the point 1, minus 2. The coordinate of this focus right there is going to be 1 plus the square root of 5, minus 2. And the coordinate of this focus right there is going to be 1 minus the square root of 5, minus 2. -- since we're along the major axes, or the x axis, I just add and subtract this from the x coordinate to get these two coordinates right there. So, anyway, this is the really neat thing about conic sections, is they have these interesting properties in relation to these foci or in relation to these focus points. And in future videos I'll show you the foci of a hyperbola or the the foci of a -- well, it only has one focus of a parabola. But this is really starting to get into what makes conic sections neat. Everything we've done up to this point has been much more about the mechanics of graphing and plotting and figuring out the centers of conic sections. But now we're getting into a little bit of the the mathematical interesting parts of conic sections. See you in the next video." + }, + { + "Q": "At 0:35, how can Sal draw the ellipse if he doesn't know A and B?", + "A": "At the beginning of the video, Sal isn t trying to measure a specific ellipse. Rather, he s trying to deduce information about ellipsis in general. He could have drawn any size of ellipse and the conclusions he came to would have been the same. So he could draw the ellipse without knowing A and B because ANY ellipse with ANY size A and B would work.", + "video_name": "QR2vxfwiHAU", + "timestamps": [ + 35 + ], + "3min_transcript": "Let's say we have an ellipse formula, x squared over a squared plus y squared over b squared is equal to 1. And for the sake of our discussion, we'll assume that a is greater than b. And all that does for us is, it lets us so this is going to be kind of a short and fat ellipse. Or that the semi-major axis, or, the major axis, is going to be along the horizontal. And the minor axis is along the vertical. And let's draw that. Draw this ellipse. I want to draw a thicker ellipse. Let's say, that's my ellipse, and then let me draw my axes. OK, this is the horizontal right there. And there we have the vertical. And we've studied an ellipse in pretty good detail so far. We know how to figure out semi-minor radius, which in this case we know is b. And that's only the semi-minor radius. Because b is smaller than a. If b was greater, it would be the major radius. And then, of course, the major radius is a. And that distance is this right here. Now, another super-interesting, and perhaps the most interesting property of an ellipse, is that if you take any point on the an ellipse, and measure the distance from that point to two special points which we, for the sake of this discussion, and not just for the sake of this discussion, for pretty much forever, we will call the focuses, or the foci, of this ellipse. And these two points, they always sit along the major axis. So, in this case, it's the horizontal axis. And they're symmetric around the center of the ellipse. So let's just call these points, let me call this one f1. And this is f2. And it's for focus. Focuses. f2. So the super-interesting, fascinating property of an ellipse. you take any point on this ellipse, and measure its distance to each of these two points. So, let's say that I have this distance right here. Let's call this distance d1. And then I have this distance over here, so I'm taking any point on that ellipse, or this particular point, and I'm measuring the distance to each of these two foci. And this is d2. We'll do it in a different color. So this is d2. This whole line right here. That's d2. So when you find these two distances, you sum of them up. So this d2 plus d1, this is going to be a constant that it actually turns out is equal to 2a. But it turns out that it's true anywhere you go on the ellipse. Let me make that point clear. And I'm actually going to prove to you that this constant" + }, + { + "Q": "Around 9:50 he mentions that v3 is a linear combination because it is v1 and v2 added together which gives a vector in between the angle of the two. If I took v1 - v2 is it still a linear combination because it is outside the angle? Or is it still a linear combination as it is in R^2", + "A": "It is a linear combination: the constant you multiplied v2 by was -1.", + "video_name": "CrV1xCWdY-g", + "timestamps": [ + 590 + ], + "3min_transcript": "call this vector 2, is equal to vector 3. So vector 3 is a linear combination of these other two vectors. So this is a linearly dependent set. And if we were to show it, draw it in kind of two space, and it's just a general idea that-- well, let me see. Let me draw it in R2. There's a general idea that if you have three two-dimensional vectors, one of them is going to be redundant. Well, one of them definitely will be redundant. For example, if we do 2, 3, if we do the vector 2, 3, that's the first one right there. I draw it in the standard position. And I draw the vector 7, 2 right there, I could show you combination of these two vectors. We can even do a kind of a graphical representation. I've done that in the previous video, so I could write that the span of v1 and v2 is equal to R2. That means that every vector, every position here can be represented by some linear combination of these two guys. Now, the vector 9, 5, it is in R2. It is in R2, right? Clearly. I just graphed it on this plane. It's in our two-dimensional, real number space. Or I guess we could call it a space or in our set R2. It's right there. So we just said that anything in R2 can be represented by a linear combination of those two guys. So clearly, this is in R2, so it can be represented as a linear combination. So hopefully, you're starting to see the relationship between span and linear independence or linear dependence. Let me do another example. Let's say I have the vectors-- let me do a new color. Let's say I have the vector-- and this one will be a little bit obvious-- 7, 0, so that's my v1, and then I have my second vector, which is 0, minus 1. That's v2. Now, is this set linearly independent? Is it linearly independent? Well, can I represent either of these as a combination of the other? And really when I say as a combination, you'd have to scale up one to get the other, because there's only two vectors here. If I am trying to add up to this vector, the only thing I" + }, + { + "Q": "At 7:45 when Sal asks if the vectors are dependent or independent, don't we know that they can't be independent solely based on the fact that they are 3 vectors that are only written in two dimensions (a 2x1 matrix)?", + "A": "Yes, you can say that. But Sal hasn t proved that that is the case yet, and he is just trying to introduce linear dependence right now.", + "video_name": "CrV1xCWdY-g", + "timestamps": [ + 465 + ], + "3min_transcript": "Anything in this plane going in any direction can be-- any vector in this plane, when we say span it, that means that any vector can be represented by a linear combination of this vector and this vector, which means that if this vector is on that plane, it can be represented as a linear combination of that vector and that vector. So this green vector I added isn't going to add anything to the span of our set of vectors and that's because this is a linearly dependent set. This one can be represented by a sum of that one and that one because this one and this one span this plane. In order for the span of these three vectors to kind of get more dimensionality or start representing R3, the third vector will have to break out of that plane. It would have to break out of that plane. And if a vector is breaking out of that plane, that means it's a vector that can't be represented anywhere on that Where it's outside, it can't be represented by a linear combination of this one and this one. So if you had a vector of this one, this one, and this one, and just those three, none of these other things that I drew, that would be linearly independent. Let me draw a couple more examples for you. That one might have been a little too abstract. So, for example, if I have the vectors 2, 3 and I have the vector 7, 2, and I have the vector 9, 5, and I were to ask you, are these linearly dependent or independent? So at first you say, well, you know, it's not trivial. Let's see, this isn't a scalar multiple of that. That doesn't look like a scalar multiple of either of Maybe they're linearly independent. But then, if you kind of inspect them, you kind of see call this vector 2, is equal to vector 3. So vector 3 is a linear combination of these other two vectors. So this is a linearly dependent set. And if we were to show it, draw it in kind of two space, and it's just a general idea that-- well, let me see. Let me draw it in R2. There's a general idea that if you have three two-dimensional vectors, one of them is going to be redundant. Well, one of them definitely will be redundant. For example, if we do 2, 3, if we do the vector 2, 3, that's the first one right there. I draw it in the standard position. And I draw the vector 7, 2 right there, I could show you" + }, + { + "Q": "At 3:33, it says \"if I draw a vertical line\". Is it the same with a horizontal line?", + "A": "No, you can t-do that with a horizontal line. A function can have multiple x s leading to 1 y, but a function can t have 1 x going to several y s. For that reason, you must have horizontal lines", + "video_name": "qGmJ4F3b5W8", + "timestamps": [ + 213 + ], + "3min_transcript": "It's not defined for 1. We don't know what our function is equal to at 1. So it's not defined there. So 1 isn't part of the domain. It tells us when x is 2, then y is going to be equal to negative 2. So it maps it or associates it with negative 2. That doesn't seem too troublesome just yet. Now, let's look over here. Our function is also defined at x is equal to 3. Our function associates or maps 3 to the value y is equal to 2. That seems pretty straightforward. And then we get to x is equal to 4, where it seems like this thing that could be a function is somewhat defined. It does try to associate 4 with things. But what's interesting here is it tries to associate 4 with two different things. All of a sudden in this thing that we think might have been a function, but it looks like it might not be, we don't know. Do we associate 4 with 5? So this thing right over here is actually a relation. You can have one member of the domain being related to multiple members of the range. But if you do have that, then you're not dealing with a function. So once again, because of this, this is not a function. It's not clear that when you input 4 into it, should you output 5? Or should you output negative 1? And sometimes there's something called the vertical line test that tells you whether something is a function. When it's graphically defined like this, you literally say, OK, when x is 4, if I draw a vertical line, do I intersect the function at two places or more? It could be two or more places. And if you do, that means that there's two or more values that are related to that value in the domain. There's two or more outputs for the input 4. And if there are two or more outputs for that one input, then you're not dealing with a function. You're just dealing with a relation. A function is a special case of a relation." + }, + { + "Q": "I'm confused with these factorials (!) In the problem around 6:12 how does 5!/4!=5?", + "A": "5! = 5\u00e2\u0080\u00a24\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 4! = 4\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 5! 5\u00e2\u0080\u00a24\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 \u00e2\u0094\u0080\u00e2\u0094\u0080 = \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 4! 4\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 5! \u00e2\u0094\u0080\u00e2\u0094\u0080 = 5 4!", + "video_name": "WWv0RUxDfbs", + "timestamps": [ + 372 + ], + "3min_transcript": "Well actually over zero factorial times five minus zero factorial. Well zero factorial is one, by definition, so this is going to be five factorial, over five factorial, which is going to be equal to one. Once again I like reasoning through it instead of blindly applying a formula, but I just wanted to show you that these two ideas are consistent. Let's keep going. I'm going to do x equals one all the way up to x equals five. If you are inspired, and I encourage you to be inspired, try to fill out the whole thing, what's the probability that x equals one, two, three, four or five. So let's go to the probability that x equals two. Or sorry, that x equals one. The probability that x equals one is going to be equal to... Well how do you get one head? It could be, the first one could be head The second one could be head and then the rest of them are gonna be tails. I could write them all out but you can see that there's five different places to have that one head. So five out of the 32 equally likely outcomes involve one head. Let me write that down. This is going to be equal to five out of 32 equally likely outcomes. Which of course is the same thing, this is going to be the same thing as saying I got five flips, and I'm choosing one of them to be heads. So that over 32. You could verify that five factorial over one factorial times five minus-- Actually let me just do it just so that you don't have to take my word for it. So five choose one is equal to five factorial over one factorial, which is just one, times five minus four-- Sorry, five minus one factorial. which is just going to be equal to five. All right, we're making good progress. Now in purple let's think about the probability that our random variable x is equal to two. Well this is going to be equal to, and now I'll actually resort to the combinatorics. You have five flips and you're choosing two of them to be heads. Over 32 equally likely possibilities. This is the number of possibilities that result in two heads. Two of the five flips have chosen to be heads, I guess you can think of it that way, by the random gods, or whatever you want to say. This is the fraction of the 32 equally likely possibilities, so this is the probability that x equals two. What's this going to be? I'll do it right over here. And actually no reason for me to have to keep switching colors." + }, + { + "Q": "@0:34\n\nI understand that it doesn't produce the correct answer, in fact it seems to produce the reciprocal of the correct answer, but why doesn't it work to multiply the left and right sides by the fraction x/x (which would be equal to 1) which would leave us with 10x on the left and 15x on the right?\n\nI'm sure that I'm missing something simple, as usual.", + "A": "If you multiply both sides by x/x, here s what you would get: 7x/x - 10x/x^2 = 2x/x + 15x/x^2 This just gives you a much more complicated equation to try and solve. You need to multiply by x to eliminate the fractions and get the variable into the numerator.", + "video_name": "Z7C69xP08d8", + "timestamps": [ + 34 + ], + "3min_transcript": "So I have the equation 7 minus 10/x is equal to 2 plus 15/x. And so this isn't the type of equation that you might think that you're used to solving. But I'll give you a few moments to see if you can solve it on your own. Well, what we'll see is we can do a quick multiplication of both sides to actually simplify this to a form that we are more used to looking at. So what's probably bothering you, because it's bothering me, is these x's that we have in the denominators right over here. We're like, well, how do we deal with that? Well, whenever we see an x in the denominator, the temptation is to multiply it by x. But we can't just multiply one of the terms by x. We have to multiply the entire side by x. So we could multiply this entire side by x. But we can't just multiply the left-hand side by x. We'd also want to multiply the right-hand side by x. And so what will that give us? Well, we distribute the x. We get x times 7 is 7x. And then x times negative 10/x, well, So you get negative 10 right over there. So the left-hand side simplifies to 7x minus 10. And then your right-hand side, once again, distribute the x. x times 2 is 2x. x times 15/x, well, x times something over x is just going to be the something. x times 15/x is just going to be 15-- plus 15. So now we've simplified this to a linear equation. We have the variable on both sides. So we just have to do some of the techniques that we already know. So the first thing that I like to do is maybe get all my x's on the left-hand side. So I want to get rid of this 2x right over here. So I subtract 2x from the right-hand side. Now, and I always remind you, I can't do that just to the right-hand side. If I did it just to the right-hand side, it wouldn't be an equality anymore. You have to do that to the left-hand side as well. And so we are left with-- let me get that pink color again. well, you're going to have 5 of that something, minus 10. These two x's negate each other. And you're left with equals 15. Now we can get rid of this negative 10 by adding 10 to both sides. You know, I like that green color when I do stuff to both sides. So I can add 10 to both sides. And I'm left with 5x-- these negate each other-- is equal to 25. And this is the home stretch. You see where this is going. We can divide both sides by 5. And we are left with x is equal to 5. Now let's verify that this actually worked. So let's go back to the original equation. We have 7 minus 10/5. This needs to be equal to-- I'm just taking our 5 and substituting it back here. This needs to be equal to 2 plus 15/5." + }, + { + "Q": "At 2:30, Sal crossed out -10 & +10. He said \"these negate each other\". What does negate mean?", + "A": "It means they cancel each other out to make 0.", + "video_name": "Z7C69xP08d8", + "timestamps": [ + 150 + ], + "3min_transcript": "So I have the equation 7 minus 10/x is equal to 2 plus 15/x. And so this isn't the type of equation that you might think that you're used to solving. But I'll give you a few moments to see if you can solve it on your own. Well, what we'll see is we can do a quick multiplication of both sides to actually simplify this to a form that we are more used to looking at. So what's probably bothering you, because it's bothering me, is these x's that we have in the denominators right over here. We're like, well, how do we deal with that? Well, whenever we see an x in the denominator, the temptation is to multiply it by x. But we can't just multiply one of the terms by x. We have to multiply the entire side by x. So we could multiply this entire side by x. But we can't just multiply the left-hand side by x. We'd also want to multiply the right-hand side by x. And so what will that give us? Well, we distribute the x. We get x times 7 is 7x. And then x times negative 10/x, well, So you get negative 10 right over there. So the left-hand side simplifies to 7x minus 10. And then your right-hand side, once again, distribute the x. x times 2 is 2x. x times 15/x, well, x times something over x is just going to be the something. x times 15/x is just going to be 15-- plus 15. So now we've simplified this to a linear equation. We have the variable on both sides. So we just have to do some of the techniques that we already know. So the first thing that I like to do is maybe get all my x's on the left-hand side. So I want to get rid of this 2x right over here. So I subtract 2x from the right-hand side. Now, and I always remind you, I can't do that just to the right-hand side. If I did it just to the right-hand side, it wouldn't be an equality anymore. You have to do that to the left-hand side as well. And so we are left with-- let me get that pink color again. well, you're going to have 5 of that something, minus 10. These two x's negate each other. And you're left with equals 15. Now we can get rid of this negative 10 by adding 10 to both sides. You know, I like that green color when I do stuff to both sides. So I can add 10 to both sides. And I'm left with 5x-- these negate each other-- is equal to 25. And this is the home stretch. You see where this is going. We can divide both sides by 5. And we are left with x is equal to 5. Now let's verify that this actually worked. So let's go back to the original equation. We have 7 minus 10/5. This needs to be equal to-- I'm just taking our 5 and substituting it back here. This needs to be equal to 2 plus 15/5." + }, + { + "Q": "at 3:24 he started say that they are both independent. It was not clear to me if the first choice was dependent or not? Maybe I missed it.", + "A": "Yes, that is correct!", + "video_name": "VjLEoo3hIoM", + "timestamps": [ + 204 + ], + "3min_transcript": "You say, well, this sounds like an interesting game. How much does it cost to play? And the guy tells you, it is $0.35 to play, so obviously a fairly low stakes casino. So my question to you is, would you want to play this game? And don't put the fun factor into it. Just economically, does it make sense for you to actually play this game? Well, let's think through the probabilities a little bit. So, first of all, what's the probability that the first marble you pick is green? Actually, let me just write, first green, probability first green. Well, the total possible outcomes-- there's 5 marbles here, all equally likely. So there's 5 possible outcomes. So there's a 3/5 probability that the first is green. So you have a 3/5 chance, 3/5 probability I should say, that after that first pick you're kind of still in the game. Now, what we really care about is your probability of winning the game. You want the first to be green, and the second green. Well, let's think about this a little bit. What is the probability that the first is green-- first, I'll just write g for green-- and the second is green? Now, you might be tempted to say, oh, well, maybe the second being green is the same probability. It's 3/5. I can just multiply 3/5 times 3/5, and I'll get 9/25, seems like a pretty straightforward thing. But the realization here is what you do with that first green marble. You don't take that first green marble out, look at it, and put it back in the bag. So when you take that second pick, depends on what you got on the first pick. Remember, we take the marble out. If it's a green marble, whatever marble it is, at whatever after the first pick, we leave it on the table. We are not replacing it. So there's not any replacement here. So these are not independent events. Let me make this clear, not independent. Or in particular, the second pick is dependent on the first. Dependent on the first pick. If the first pick is green, then you don't have 3 green marbles in a bag of 5. If the first pick is green, you now have 2 green marbles in a bag of 4. So the way that we would refer to this, the probability of both of these happening, yes, it's definitely equal to the probability of the first green times-- now, this is kind of the new idea--" + }, + { + "Q": "What is that weird N-shaped symbol that Sal drew at 2:32?\nI assume it's some sort of symbol meaning and.", + "A": "The \u00e2\u0088\u00a9 symbol Sal wrote in 2:34 stands for intersection, which you have probably encountered in basic statistics. For example, if you let X and Y be arbitrary sets, X \u00e2\u0088\u00a9 Y would be classified as the set containing the elements that are in Set X AND Set Y.", + "video_name": "VjLEoo3hIoM", + "timestamps": [ + 152 + ], + "3min_transcript": "You say, well, this sounds like an interesting game. How much does it cost to play? And the guy tells you, it is $0.35 to play, so obviously a fairly low stakes casino. So my question to you is, would you want to play this game? And don't put the fun factor into it. Just economically, does it make sense for you to actually play this game? Well, let's think through the probabilities a little bit. So, first of all, what's the probability that the first marble you pick is green? Actually, let me just write, first green, probability first green. Well, the total possible outcomes-- there's 5 marbles here, all equally likely. So there's 5 possible outcomes. So there's a 3/5 probability that the first is green. So you have a 3/5 chance, 3/5 probability I should say, that after that first pick you're kind of still in the game. Now, what we really care about is your probability of winning the game. You want the first to be green, and the second green. Well, let's think about this a little bit. What is the probability that the first is green-- first, I'll just write g for green-- and the second is green? Now, you might be tempted to say, oh, well, maybe the second being green is the same probability. It's 3/5. I can just multiply 3/5 times 3/5, and I'll get 9/25, seems like a pretty straightforward thing. But the realization here is what you do with that first green marble. You don't take that first green marble out, look at it, and put it back in the bag. So when you take that second pick, depends on what you got on the first pick. Remember, we take the marble out. If it's a green marble, whatever marble it is, at whatever after the first pick, we leave it on the table. We are not replacing it. So there's not any replacement here. So these are not independent events. Let me make this clear, not independent. Or in particular, the second pick is dependent on the first. Dependent on the first pick. If the first pick is green, then you don't have 3 green marbles in a bag of 5. If the first pick is green, you now have 2 green marbles in a bag of 4. So the way that we would refer to this, the probability of both of these happening, yes, it's definitely equal to the probability of the first green times-- now, this is kind of the new idea--" + }, + { + "Q": "At 1:43, when he says that the {Ak} equation means that k=1 from the first to the last term, does that mean if he had 5 numbers that the \"4\" at the top of the equation would be a \"5\" instead?", + "A": "You are correct. That s why in the infinite equation, the value at the top (AKA the value the sequence goes to) is infinity.", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 103 + ], + "3min_transcript": "What I want to do in this video is familiarize ourselves with the notion of a sequence. And all a sequence is is an ordered list of numbers. So for example, I could have a finite sequence-- that means I don't have an infinite number of numbers in it-- where, let's say, I start at 1 and I keep adding 3. So 1 plus 3 is 4. 4 plus 3 is 7. 7 plus 3 is 10. And let's say I only have these four terms right over here. So this one we would call a finite sequence. I could also have an infinite sequence. So an example of an infinite sequence-- let's say we start at 3, and we keep adding 4. So we go to 3, to 7, to 11, 15. And you don't always have to add the same thing. We'll explore fancier sequences. The sequences where you keep adding the same amount, we call these arithmetic sequences, which we will also explore in more detail. But to show that this is infinite, to show that we keep this pattern going on and on and on, I'll put three dots. This just means we're going to keep going on and on and on. Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1" + }, + { + "Q": "At 1:49, Sal stops at \"a sub-4\". In this finite sequence, is it possible to have an \"a sub-5\"?", + "A": "Hi shanzi11, In that specific sequence no. This is because he states that it is from term 1 to 4 and therefore in this sequence only, subs greater than 4 do not exist. Hope that helps! - JK", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 109 + ], + "3min_transcript": "What I want to do in this video is familiarize ourselves with the notion of a sequence. And all a sequence is is an ordered list of numbers. So for example, I could have a finite sequence-- that means I don't have an infinite number of numbers in it-- where, let's say, I start at 1 and I keep adding 3. So 1 plus 3 is 4. 4 plus 3 is 7. 7 plus 3 is 10. And let's say I only have these four terms right over here. So this one we would call a finite sequence. I could also have an infinite sequence. So an example of an infinite sequence-- let's say we start at 3, and we keep adding 4. So we go to 3, to 7, to 11, 15. And you don't always have to add the same thing. We'll explore fancier sequences. The sequences where you keep adding the same amount, we call these arithmetic sequences, which we will also explore in more detail. But to show that this is infinite, to show that we keep this pattern going on and on and on, I'll put three dots. This just means we're going to keep going on and on and on. Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1" + }, + { + "Q": "I can't understand the part where he creates the function for the first sequence at around 2:50. How did he come up with 1+3(k-1)", + "A": "k a_k 1 + 3\u00e2\u0080\u00a2(k - 1) 1 1 1 + 3\u00e2\u0080\u00a2(1 - 1) = 1 + 3\u00e2\u0080\u00a20 = 1 + 0 = 1 2 4 1 + 3\u00e2\u0080\u00a2(2 - 1) = 1 + 3\u00e2\u0080\u00a21 = 1 + 3 = 4 3 7 1 + 3\u00e2\u0080\u00a2(3 - 1) = 1 + 3\u00e2\u0080\u00a22 = 1 + 6 = 7 4 10 1 + 3\u00e2\u0080\u00a2(4 - 1) = 1 + 3\u00e2\u0080\u00a23 = 1 + 9 = 10", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 170 + ], + "3min_transcript": "Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1 should write 3 times k minus 1-- same thing. And you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with kind of this function notation. I want to make it clear-- I have essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about. a of k is equal to 1 plus 3 times k minus 1." + }, + { + "Q": "Around 7:09. As a recursive function, Sal said that A sub 2 is:\n\nA(2-1)+3 Which is 1+3 which the answer is 4.\nBut why doesn't it work for A sub 3?\n\nA(3-1)+3 --> 2+3 --> 5\n\nI think I'm making a mistake but I don't know where the mistake is. Please help! Thanks!", + "A": "A(3-1) is not 2, its A(n) when n=2, its the second element\\member of the sequence, indexed at 2, and in this case A(2) = 4, and not 2. the fact that A(1) = 1 is sheer coincidence i m afraid :)", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 429 + ], + "3min_transcript": "So we're adding 4 one less than the term that we're at. So it's going to be plus 4 times k minus 1. So this is another way of defining this infinite sequence. Now, in both of these cases, I defined it as an explicit function. So this right over here is explicit. That's not an attractive color. Let me write this in. This is an explicit function. And so you might say, well, what's another way of defining these functions? Well, we can also define it, especially something like an arithmetic sequence, we can also define it recursively. And I want to be clear-- not every sequence can be defined as either an explicit function like this, or as a recursive function. But many can, including this, which is an arithmetic sequence, where we keep adding the same quantity over and over again. So how would we do that? Well, we could also-- another way of defining starting at k equals 1 and going to 4 with. And when you define a sequence recursively, you want to define what your first term is, with a sub 1 equaling 1. You can define every other term in terms of the term before it. And so then we could write a sub k is equal to the previous term. So this is a sub k minus 1. So a given term is equal to the previous term. Let me make it clear-- this is the previous term, plus-- in this case, we're adding 3 every time. Now, how does this make sense? Well, we're defining what a sub 1 is. And if someone says, well, what happens when k equals 2? Well, they're saying, well, it's going to be a sub 2 minus 1. Well, we know a sub 1 is 1. So it's going to be 1 plus 3, which is 4. Well, what about a sub 3? Well, it's going to be a sub 2 plus 3. a sub 2, we just calculated as 4. You add 3. It's going to be 7. This is essentially what we mentally did when I first wrote out the sequence, when I said, hey, I'm just going to start with 1. And I'm just going to add 3 for every successive term. So how would we do this one? Well, once again, we could write this as a sub k. Starting at k, the first term, going to infinity with-- our first term, a sub 1, is going to be 3, now. And every successive term, a sub k, is going to be the previous term, a sub k minus 1, plus 4. And once again, you start at 3. And then if you want the second term, it's going to be the first term plus 4. It's going to be 3 plus 4. You get to 7. And you keep adding 4. So both of these, this right over here" + }, + { + "Q": "@ 2:56 what does this man mean by we added 3 times one less than the k term times? I don't understand", + "A": "Hey ISAIAH, This man is trying to show that a(k) = 1 + 3(k-1) Since the k term is what term it is (1,2, etc.) It could be referred to as how many times. That is what this man is referring to as the k term times . Hope that helps! - JK", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 176 + ], + "3min_transcript": "Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1 should write 3 times k minus 1-- same thing. And you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with kind of this function notation. I want to make it clear-- I have essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about. a of k is equal to 1 plus 3 times k minus 1." + }, + { + "Q": "Having trouble understanding how to explicitly define a sequence. Can anyone clarify what Sal means at 4:56 when he says \"we are adding 4, one less time\"?", + "A": "Sal is talking us though his calculations, and so he isn t being as clear as he could be, here. One less time means that we do not add four to obtain the first term. That means that with term #2, we add one 4 to the first term, with term #3, we add two fours to the first term, and so forth: we add one less 4 than the number of the term. (Thanks for the time-stamp: they really help with a question like this :-)", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 296 + ], + "3min_transcript": "should write 3 times k minus 1-- same thing. And you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with kind of this function notation. I want to make it clear-- I have essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about. a of k is equal to 1 plus 3 times k minus 1. an allowable input, the domain, is restricted to positive integers. Now, how would I denote this business right over here? Well, I could say that this is equal to-- and people tend to use a. But I could use the notation b sub k or anything else. But I'll do a again-- a sub k. And here, we're going from our first term-- so this is a sub 1, this is a sub 2-- all the way to infinity. Or we could define it-- if we wanted to define it explicitly as a function-- we could write this sequence as a sub k, where k starts at the first term and goes to infinity, with a sub k is equaling-- so we're starting at 3. And we are adding 4 one less time. For the second term, we added 4 once. For the third term, we add 4 twice. So we're adding 4 one less than the term that we're at. So it's going to be plus 4 times k minus 1. So this is another way of defining this infinite sequence. Now, in both of these cases, I defined it as an explicit function. So this right over here is explicit. That's not an attractive color. Let me write this in. This is an explicit function. And so you might say, well, what's another way of defining these functions? Well, we can also define it, especially something like an arithmetic sequence, we can also define it recursively. And I want to be clear-- not every sequence can be defined as either an explicit function like this, or as a recursive function. But many can, including this, which is an arithmetic sequence, where we keep adding the same quantity over and over again. So how would we do that? Well, we could also-- another way of defining" + }, + { + "Q": "at 3:14 why is the depth of the cylinder dy.", + "A": "dy is an extremely small change in y, it is infinitesimal. We set it as this so that we can use the sum of infinitely thin cylinders to take an integral. Imagine if y was a large number. There would be a large volume unaccounted for, and the integral would not be accurate.", + "video_name": "43AS7bPUORc", + "timestamps": [ + 194 + ], + "3min_transcript": "So we care about this part right over here, not the very bottom of it. And let me shade it in a little bit. So it would look something like that. So let me draw it separately, just so we can visualize it. So I'll draw it at different angles. So if I were to draw it with the y-axis kind of coming out the back, it would look something like this. It would look something like-- it gets a little bit smaller like that. And then it gets cut off right over here, right over here like that. So it looks-- I don't know what shape you could call it. But I think hopefully you're conceptualizing this. Let me do it in that same yellow color. The visual-- that's not yellow. The visualization here is probably the hardest part. But as we can see it's not too bad. So it looks something like this. It looks like maybe a truffle, an upside-down truffle. So this right here, let me draw the y-axis just to show how we're oriented. So the y-axis is popping out in this example like that. And then the x-axis is going like this. So I just tilted this over. I tilted it over a little bit to be able to view it at a different angle. This top right over here is this top right over there. So that gives you an idea of what it looks like. But we still haven't thought about how do we actually find the volume of this thing? Well, what we can do, instead of creating discs where the depth is in little dx's, what if we created discs where the depth is in dy? So let's think about that a little bit. So let's create-- let's think about constructing a disc at a certain y-value. So let's think about a certain y-value, and we're going to construct a disc right over there that has the same radius of the shape at that point. So that's our disc. That's our disc right over here. of saying it has a depth of dx, let's say it has a depth of dy. So this depth right over here is dy. So what is the volume of this disc in terms of y? And as you could imagine, we're going to do this definite integral, and it is a definite integral, with respect to y. So what's the volume of this thing? Well, like we did in the last video, we have to figure out the area of the top of each of these discs. Or I guess you could say the face of this coin. Well, to find that area it's pi r squared. If we can figure out this radius right over here, we know the area. So what's that radius? So to think about that radius in terms of y, we just have to solve this equation explicitly in terms of y. So instead of saying it's y is equal to x squared, we can take the principal root of both sides, and we could say that the square root of y is equal to x. And this right over here is only defined for non-negative y's," + }, + { + "Q": "When finding the length of the B side at approx 1:17 do you add 4^2 to 3^2?", + "A": "Yes, you add 4^2 to 3^2 and then taking the square root of this sum. Just pythagorean theorem. Hope this helps.", + "video_name": "2yjSAarzWF8", + "timestamps": [ + 77 + ], + "3min_transcript": "The graph below contains triangle ABC and the point P. Draw the image of triangle ABC under a dilation whose center is P and a scale factor of 2. So essentially, we want to scale this so that every point is going to be twice as far away from P. So for example, B right over here has the same y-coordinate as P, but its x-coordinate is three more. So we want to be twice as far. So if this maps to point B, we just want to go twice as far. So we're at 3 away, we want to go 6 away. So point P's x-coordinate is at 3, now we're at 9. Likewise, point C is 3 below P. Well we want to go twice as far, so we'll go 3 more. And point A is 4 above P. Well we want to go 4 more. We want to go twice as far-- one, two, three, four. And we get right over there. Then they ask us, what are the lengths of side AB and its image? might have to apply the distance formula. Let's see, it's the base right over here. The change in x between the two is 3 and the change in y is 4, so this is actually a 3, 4, 5 right triangle. 3 squared plus 4 squared is equal to 5 squared. So AB is 5 units long. Essentially just using the Pythagorean theorem to figure that out. And its image, well it's image should be twice as long. And let's see whether that actually is the case. So this is a base right over here that's of length 6. This has a height, or this change in y, I could say. Because I'm really just trying to figure out this length, which is the hypotenuse of this right triangle. I don't have my drawing tool, so I apologize. But this height right here is 8. So 8 squared is 64, plus 6 squared is 36, that's 100, which is 10 squared. So notice, our scale factor of 2, the corresponding side Each of these points got twice as far away from our center of dilation." + }, + { + "Q": "At 0:50, why did we go 4 points above?", + "A": "It went 4 points above because the distance between point A and point P is 4, and as the problem states, it wants you to dilate from point P with a scale factor of 2. 2*4=8. And so point A2 is 8 points away from point P.", + "video_name": "2yjSAarzWF8", + "timestamps": [ + 50 + ], + "3min_transcript": "The graph below contains triangle ABC and the point P. Draw the image of triangle ABC under a dilation whose center is P and a scale factor of 2. So essentially, we want to scale this so that every point is going to be twice as far away from P. So for example, B right over here has the same y-coordinate as P, but its x-coordinate is three more. So we want to be twice as far. So if this maps to point B, we just want to go twice as far. So we're at 3 away, we want to go 6 away. So point P's x-coordinate is at 3, now we're at 9. Likewise, point C is 3 below P. Well we want to go twice as far, so we'll go 3 more. And point A is 4 above P. Well we want to go 4 more. We want to go twice as far-- one, two, three, four. And we get right over there. Then they ask us, what are the lengths of side AB and its image? might have to apply the distance formula. Let's see, it's the base right over here. The change in x between the two is 3 and the change in y is 4, so this is actually a 3, 4, 5 right triangle. 3 squared plus 4 squared is equal to 5 squared. So AB is 5 units long. Essentially just using the Pythagorean theorem to figure that out. And its image, well it's image should be twice as long. And let's see whether that actually is the case. So this is a base right over here that's of length 6. This has a height, or this change in y, I could say. Because I'm really just trying to figure out this length, which is the hypotenuse of this right triangle. I don't have my drawing tool, so I apologize. But this height right here is 8. So 8 squared is 64, plus 6 squared is 36, that's 100, which is 10 squared. So notice, our scale factor of 2, the corresponding side Each of these points got twice as far away from our center of dilation." + }, + { + "Q": "At 6:29 I found that the line went the other way. Are u guys sure that is the way the line goes?", + "A": "How did you get it to go the other way? It has a negative slope, so it should go down from left to right (or someone could slide down the hill). With a slope of -1/2, this says go down 1(change in y is -1) and right 2 (change in y is 2). Figure out where you went wrong to get line to go other way.", + "video_name": "unSBFwK881s", + "timestamps": [ + 389 + ], + "3min_transcript": "So if you were to do this for all the possible x's, you would not only get all the points on this line which we've drawn, you would get all the points below the line. So now we have graphed this inequality. It's essentially this line, 4x plus 3, with all of the area below it shaded. Now, if this was just a less than, not less than or equal sign, we would not include the actual line. And the convention to do that is to actually make the line a dashed line. This is the situation if we were dealing with just less than 4x plus 3. Because in that situation, this wouldn't apply, and we would just have that. So the line itself wouldn't have satisfied it, just the area below it. Let's do one like that. So let's say we have y is greater than negative x So a good way to start-- the way I like to start these problems-- is to just graph this equation right here. So let me just graph-- just for fun-- let me graph y is equal to-- this is the same thing as negative 1/2 minus 6. So if we were to graph it, that is my vertical axis, that is my horizontal axis. And our y-intercept is negative 6. So 1, 2, 3, 4, 5, 6. So that's my y-intercept. And my slope is negative 1/2. Oh, that should be an x there, negative 1/2 x minus 6. So my slope is negative 1/2, which means when I go 2 to the right, I go down 1. So if I go 2 to the right, I'm going to go down 1. going to go up 1. So negative 2, up 1. So my line is going to look like this. My line is going to look like that. That's my best attempt at drawing the line. So that's the line of y is equal to negative 1/2 x minus 6. Now, our inequality is not greater than or equal, it's just greater than negative x over 2 minus 6, or greater than negative 1/2 x minus 6. So using the same logic as before, for any x-- so if you take any x, let's say that's our particular x we want to pick-- if you evaluate negative x over 2 minus 6, you're going to get that point right there. You're going to get the point on the line. But the y's that satisfy this inequality are the y's greater than that. So it's going to be not that point-- in fact, you draw an open circle there-- because you can't include the point of negative 1/2 x minus 6. But it's going to be all the y's greater than that." + }, + { + "Q": "why did you go back 4 at time stamp 1:34?", + "A": "He didn t. You have the y-intercept graphed. From there you know that the slope is 4, so up 4 and right 1. But that will only let you graph to the right. To graph to the left, you need to go down 4 and left 1. The complete opposite.", + "video_name": "unSBFwK881s", + "timestamps": [ + 94 + ], + "3min_transcript": "Let's graph ourselves some inequalities. So let's say I had the inequality y is less than or equal to 4x plus 3. On our xy coordinate plane, we want to show all the x and y points that satisfy this condition right here. So a good starting point might be to break up this less than or equal to, because we know how to graph y is equal to 4x plus 3. So this thing is the same thing as y could be less than 4x plus 3, or y could be equal to 4x plus 3. That's what less than or equal means. It could be less than or equal. And the reason why I did that on this first example problem is because we know how to graph that. So let's graph that. Try to draw a little bit neater than that. So that is-- no, that's not good. So that is my vertical axis, my y-axis. And then we know the y-intercept, the y-intercept is 3. So the point 0, 3-- 1, 2, 3-- is on the line. And we know we have a slope of 4. Which means if we go 1 in the x-direction, we're going to go up 4 in the y. So 1, 2, 3, 4. So it's going to be right here. And that's enough to draw a line. We could even go back in the x-direction. If we go 1 back in the x-direction, we're going to go down 4. 1, 2, 3, 4. So that's also going to be a point on the line. So my best attempt at drawing this line is going to look something like-- this is the hardest part. It's going to look something like that. That is a line. It should be straight. I think you get the idea. That right there is the graph of y is equal to 4x plus 3. So all of these points satisfy this inequality, but we have more. This is just these points over here. What about all these where y ix less than 4x plus 3? So let's think about what this means. Let's pick up some values for x. When x is equal to 0, what does this say? When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3. When x is equal to negative 1, what is this telling us? 4 times negative 1 is negative 4, plus 3 is negative 1. y would be less than negative 1. When x is equal to 1, what is this telling us? 4 times 1 is 4, plus 3 is 7. So y is going to be less than 7. So let's at least try to plot these. So when x is equal to-- let's plot this one first. When x is" + }, + { + "Q": "At 5:30 didn't he mean the equation would be y=-1/2x-6? He said it would be y=-1/2-6 and I was wonder where the variable would be. I'm not sure if I'm correct or not, please let me know.", + "A": "Yes, you re correct. Initially he forgot to add the x. About a minute afterwards, though, he saw his mistake and fixed it. :)", + "video_name": "unSBFwK881s", + "timestamps": [ + 330 + ], + "3min_transcript": "So it's all of these points here-- that I'm shading in in green-- satisfy that right there. If I were to look at this one over here, when x is negative 1, y is less than negative 1. So y has to be all of these points down here. When x is equal to 1, y is less than 7. So it's all of these points down here. And in general, you take any point x-- let's say you take this point x right there. If you evaluate 4x plus 3, you're going to get the point on the line. That is that x times 4 plus 3. Now the y's that satisfy it, it could be equal to that point on the line, or it could be less than. So if you were to do this for all the possible x's, you would not only get all the points on this line which we've drawn, you would get all the points below the line. So now we have graphed this inequality. It's essentially this line, 4x plus 3, with all of the area below it shaded. Now, if this was just a less than, not less than or equal sign, we would not include the actual line. And the convention to do that is to actually make the line a dashed line. This is the situation if we were dealing with just less than 4x plus 3. Because in that situation, this wouldn't apply, and we would just have that. So the line itself wouldn't have satisfied it, just the area below it. Let's do one like that. So let's say we have y is greater than negative x So a good way to start-- the way I like to start these problems-- is to just graph this equation right here. So let me just graph-- just for fun-- let me graph y is equal to-- this is the same thing as negative 1/2 minus 6. So if we were to graph it, that is my vertical axis, that is my horizontal axis. And our y-intercept is negative 6. So 1, 2, 3, 4, 5, 6. So that's my y-intercept. And my slope is negative 1/2. Oh, that should be an x there, negative 1/2 x minus 6. So my slope is negative 1/2, which means when I go 2 to the right, I go down 1. So if I go 2 to the right, I'm going to go down 1." + }, + { + "Q": "At 1:31 Sal says that the point is here. Why cannot it be on the left hand side?", + "A": "because it isn t a negative 1 so it is a positive on the graph hope this helps u", + "video_name": "unSBFwK881s", + "timestamps": [ + 91 + ], + "3min_transcript": "Let's graph ourselves some inequalities. So let's say I had the inequality y is less than or equal to 4x plus 3. On our xy coordinate plane, we want to show all the x and y points that satisfy this condition right here. So a good starting point might be to break up this less than or equal to, because we know how to graph y is equal to 4x plus 3. So this thing is the same thing as y could be less than 4x plus 3, or y could be equal to 4x plus 3. That's what less than or equal means. It could be less than or equal. And the reason why I did that on this first example problem is because we know how to graph that. So let's graph that. Try to draw a little bit neater than that. So that is-- no, that's not good. So that is my vertical axis, my y-axis. And then we know the y-intercept, the y-intercept is 3. So the point 0, 3-- 1, 2, 3-- is on the line. And we know we have a slope of 4. Which means if we go 1 in the x-direction, we're going to go up 4 in the y. So 1, 2, 3, 4. So it's going to be right here. And that's enough to draw a line. We could even go back in the x-direction. If we go 1 back in the x-direction, we're going to go down 4. 1, 2, 3, 4. So that's also going to be a point on the line. So my best attempt at drawing this line is going to look something like-- this is the hardest part. It's going to look something like that. That is a line. It should be straight. I think you get the idea. That right there is the graph of y is equal to 4x plus 3. So all of these points satisfy this inequality, but we have more. This is just these points over here. What about all these where y ix less than 4x plus 3? So let's think about what this means. Let's pick up some values for x. When x is equal to 0, what does this say? When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3. When x is equal to negative 1, what is this telling us? 4 times negative 1 is negative 4, plus 3 is negative 1. y would be less than negative 1. When x is equal to 1, what is this telling us? 4 times 1 is 4, plus 3 is 7. So y is going to be less than 7. So let's at least try to plot these. So when x is equal to-- let's plot this one first. When x is" + }, + { + "Q": "1:01 What are the other equation forms useful for?", + "A": "Standard form: Ax+By = C is used frequently when solving systems of equations. Point-Slope form: Y - y1 = m(X - x1) will accept the slope and any point from the line. So, it is easy to use to create the equation of a line. Slope-Intercept form: y = mx + b is used to quickly obtain info need to graph the line. It is also used to change a linear equation into function notation. And, it can be used to create the equation of the line. Hope this helps.", + "video_name": "IL3UCuXrUzE", + "timestamps": [ + 61 + ], + "3min_transcript": "- [Voiceover] There's a lot of different ways that you could represent a linear equation. So for example, if you had the linear equation y is equal to 2x plus three, that's one way to represent it, but I could represent this in an infinite number of ways. I could, let's see, I could subtract 2x from both sides, I could write this as negative 2x plus y is equal to three. I could manipulate it in ways where I get it to, and I'm gonna do it right now, but this is another way of writing that same thing. y minus five is equal to two times x minus one. You could actually simplify this and you could get either this equation here or that equation up on top. These are all equivalent, you can get from one to the other with logical algebraic operations. So there's an infinite number of ways to represent a given linear equation, but I what I wanna focus on in this video is this representation in particular, because this one is a very useful representation this one and this one can also be useful, depending on what you are looking for, but we're gonna focus on this one, and this one right over here is often called slope-intercept form. Slope-intercept form. And hopefully in a few minutes, it will be obvious why it called slop-intercept form. And before I explain that to you, let's just try to graph this thing. I'm gonna try to graph it, I'm just gonna plot some points here, so x comma y, and I'm gonna pick some x values where it's easy to calculate the y values. So maybe the easiest is if x is equal to zero. If x is equal to zero, then two times zero is zero, that term goes away, and you're only left with this term right over here, y is equal to three. Y is equal to three. And so if we were to plot this. Actually let me start plotting it, so that is my y axis, and let me do the x axis, so as straight as I would like it. So that looks pretty good, alright. That is my x axis and let me mark off some hash marks here, so this is x equals one, x equals two, x equals three, this is y equals one, y equals two, y equals three, and obviously I could keep going and keep going, this would be y is equal to negative one, this would be x is equal to negative one, negative two, negative three, so on and so forth. So this point right over here, zero comma three, this is x is zero, y is three. Well, the point that represents when x is equal to zero and y equals three, this is, we're right on the y axis. If they have a line going through it and this line" + }, + { + "Q": "I have no idea how to simplified (y-5)=2(x-1) at 00:43", + "A": "the word simplify in this context means to remove the parentheses using the distributive property and then use the addition property of equality to combine the constant terms. thus you first get y - 5 = 2x - 2. Then adding 5 to both sides you get y=2x+3. This final form is unique for any given line and is called slope-intercept form. It is the function form of the line and it is usually the form that you need if you are going to use a calculator or other graphing software to graph your line.", + "video_name": "IL3UCuXrUzE", + "timestamps": [ + 43 + ], + "3min_transcript": "- [Voiceover] There's a lot of different ways that you could represent a linear equation. So for example, if you had the linear equation y is equal to 2x plus three, that's one way to represent it, but I could represent this in an infinite number of ways. I could, let's see, I could subtract 2x from both sides, I could write this as negative 2x plus y is equal to three. I could manipulate it in ways where I get it to, and I'm gonna do it right now, but this is another way of writing that same thing. y minus five is equal to two times x minus one. You could actually simplify this and you could get either this equation here or that equation up on top. These are all equivalent, you can get from one to the other with logical algebraic operations. So there's an infinite number of ways to represent a given linear equation, but I what I wanna focus on in this video is this representation in particular, because this one is a very useful representation this one and this one can also be useful, depending on what you are looking for, but we're gonna focus on this one, and this one right over here is often called slope-intercept form. Slope-intercept form. And hopefully in a few minutes, it will be obvious why it called slop-intercept form. And before I explain that to you, let's just try to graph this thing. I'm gonna try to graph it, I'm just gonna plot some points here, so x comma y, and I'm gonna pick some x values where it's easy to calculate the y values. So maybe the easiest is if x is equal to zero. If x is equal to zero, then two times zero is zero, that term goes away, and you're only left with this term right over here, y is equal to three. Y is equal to three. And so if we were to plot this. Actually let me start plotting it, so that is my y axis, and let me do the x axis, so as straight as I would like it. So that looks pretty good, alright. That is my x axis and let me mark off some hash marks here, so this is x equals one, x equals two, x equals three, this is y equals one, y equals two, y equals three, and obviously I could keep going and keep going, this would be y is equal to negative one, this would be x is equal to negative one, negative two, negative three, so on and so forth. So this point right over here, zero comma three, this is x is zero, y is three. Well, the point that represents when x is equal to zero and y equals three, this is, we're right on the y axis. If they have a line going through it and this line" + }, + { + "Q": "At 7:42, Sal mentions that we could of taken the Absolute value of the difference between the measurements and the mean instead of squaring them. Why don't we do that, it seems easier?", + "A": "A few reasons. 1. The absolute value function is much harder to deal with mathematically, because the derivative isn t nearly so nice as that of the square function. 2a. Squaring works very well with the Normal distribution. 2b. The sample mean is a natural estimate of location/center, and the Sampling Distribution of the sample mean is Normal, so we d like to use that. Hence, item 2a.", + "video_name": "PWiWkqHmum0", + "timestamps": [ + 462 + ], + "3min_transcript": "And in this case, what's it going to be? It's going to be the square root of 0.316. And then, what are the units going to be? It's going to be just meters. And we end up with-- so let me take the square root of 0.316. And I get 0.56-- I'll just round to the nearest thousandth-- 0.562. So this is approximately 0.562 meters. So you might be saying, Sal, what do we call this thing that we just did? The square root of the variance. And here we're dealing with the population. We haven't thought about sampling yet. The square root of the population variance, what do we call this thing right over here? And this is a very familiar term. Oftentimes, when you take an exam, this is calculated for the scores on the exam. I'm using that yellow a little bit too much. This is the population standard deviation. It is a measure of how much the data is varying from the mean. In general, the larger this value, that means that the data is more varied from the population mean. The smaller, it's less varied. And these are all somewhat arbitrary definitions of how we've defined variance. We could have taken things to the fourth power. We could have done other things. We could have not taken them to a power but taking the absolute value here. The reason why we do it this way is it has neat statistical properties as we try to build on it. But that's the population standard deviation, which gives us nice units-- meters. In the next video, we'll think about the sample standard" + }, + { + "Q": "I wish Khan Academy had a video where Sal explains all the special symbols that are used in math and what they stand for. For example, at 1:02 Sal says \"We're going to use 'Mu'\" what's 'Mu'?", + "A": "Mu = \u00c2\u00b5 = population mean. Population symbols are always Greek.", + "video_name": "PWiWkqHmum0", + "timestamps": [ + 62 + ], + "3min_transcript": "Let's say that you're curious about studying the dimensions of the cars that happen to sit in the parking lot. And so you measure their lengths. Let's just make the computation simple. Let's say that there are five cars in the parking lot. The entire size of the population that we care about is 5. And you go and measure their lengths-- one car is 4 meters long, another car is 4.2 meters long, another car is 5 meters long, the fourth car is 4.3 meters long, and then, let's say the fifth car is 5.5 meters long. So let's come up with some parameters for this population. So the first one that you might want to figure out is a measure of central tendency. And probably the most popular one is the arithmetic mean. So let's calculate that first. So we're going to do that for the population. So we're going to use mu. Well, we just have to add all of these data points up and divide by 5. And I'll just get the calculator out just so it's a little bit quicker. This is going to be for 4 plus 4.2 plus 5 plus 4.3 plus 5.5. And then, I'm going to take that sum and then divide by 5. And I get an arithmetic mean for my population of 4.6. So that's fine. And if we want to put some units there, it's 4.6 meters. Now, that's the central tendency or measure of central tendency. We also might be curious about how dispersed is the data, especially from that central tendency. So what would we use? the population variance. And the population variance is one of many ways of measuring dispersion. It has some very neat properties the way we've defined it as the mean of the squared distances from the mean. It tends to be a useful way of doing it. So let's just a bit. Let's actually calculate the population variance for this population right over here. Well, all we need to do is find the distance from each of these points to our mean right over here. And then, square them. And then, take the mean of those two squared distances. So let's do that. So it's going to be 4 minus 4.6 squared plus 4.2 minus 4.6 squared plus 5 minus 4.6 squared" + }, + { + "Q": "so does phi work for any ratio? even 7:1?", + "A": "No. It only works for the golden ratio.", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 421 + ], + "3min_transcript": "or it's really the constant term right over there. So the solutions to this, phi-- and we're actually only going to care about the positive solution because we're thinking about a positive-- when we go to our original problem here, we're assuming that these are both positive distances, so we care about a positive value right over here. We get phi is equal to-- do it in orange-- negative b. Well negative negative 1 is 1 plus or minus the square root of b squared. b squared is going to be 1 minus 4ac. a is 1, c is negative 1. So negative 4 times negative 1 is positive 4. So 1 plus 4, all of that over 2a. So a is 1, so all of that over 2. So phi is equal to 1. And once again, we only care about the positive solution This is going to be the square root of 5. If you have 1 minus the square root of 5, you're going to get a negative in the numerator. So we only care about the positive solution. 1 plus the square root of 5 over 2. Let's actually take a calculator out and see if we can get the first few places of this magic number phi. So let me get my calculator out. Let's just actually evaluate it. And you might recognize that square root of 5 is an irrational number. And so this whole thing is going to be an irrational number, but I'll prove that in another video, which means it never repeats. It goes on and on and on forever. But let's actually evaluate it. So it's 1 plus the square root of 5 divided by 2. So it says 1.6180339. So let me put that aside. Let me write it down. And this is where it starts to get really interesting and mysterious. So this number right over here is 1.618033988 on never terminating, never repeating. So that by itself, it's this cool number. It's this ratio that has all of these neat properties, which are pretty crazy anyway that you express it. But what's really neat is if we revisit this thing right over here. Because what is 1 over phi going to be? So 1 over phi, which we sometimes denote with a capital phi. We already know 1 over phi is just phi minus 1. So we actually can do this in our heads. 1 over this is just going to be 0.618033988. There's just something wacky about that, that the inverse of the number is really just the decimals left over after you get rid of the 1. That, by itself, is kind of a crazy idea. But it gets even crazier because this number is showing up everywhere. And as you might imagine from the title of this video, this phi right over here, this is called the golden ratio." + }, + { + "Q": "Isn't the number at 7:04 kind of like pi? It keeps on going forever but never repeats.", + "A": "Yes, the number(or Phi) is an irrational number like pi since the numbers or never terminating.", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 424 + ], + "3min_transcript": "or it's really the constant term right over there. So the solutions to this, phi-- and we're actually only going to care about the positive solution because we're thinking about a positive-- when we go to our original problem here, we're assuming that these are both positive distances, so we care about a positive value right over here. We get phi is equal to-- do it in orange-- negative b. Well negative negative 1 is 1 plus or minus the square root of b squared. b squared is going to be 1 minus 4ac. a is 1, c is negative 1. So negative 4 times negative 1 is positive 4. So 1 plus 4, all of that over 2a. So a is 1, so all of that over 2. So phi is equal to 1. And once again, we only care about the positive solution This is going to be the square root of 5. If you have 1 minus the square root of 5, you're going to get a negative in the numerator. So we only care about the positive solution. 1 plus the square root of 5 over 2. Let's actually take a calculator out and see if we can get the first few places of this magic number phi. So let me get my calculator out. Let's just actually evaluate it. And you might recognize that square root of 5 is an irrational number. And so this whole thing is going to be an irrational number, but I'll prove that in another video, which means it never repeats. It goes on and on and on forever. But let's actually evaluate it. So it's 1 plus the square root of 5 divided by 2. So it says 1.6180339. So let me put that aside. Let me write it down. And this is where it starts to get really interesting and mysterious. So this number right over here is 1.618033988 on never terminating, never repeating. So that by itself, it's this cool number. It's this ratio that has all of these neat properties, which are pretty crazy anyway that you express it. But what's really neat is if we revisit this thing right over here. Because what is 1 over phi going to be? So 1 over phi, which we sometimes denote with a capital phi. We already know 1 over phi is just phi minus 1. So we actually can do this in our heads. 1 over this is just going to be 0.618033988. There's just something wacky about that, that the inverse of the number is really just the decimals left over after you get rid of the 1. That, by itself, is kind of a crazy idea. But it gets even crazier because this number is showing up everywhere. And as you might imagine from the title of this video, this phi right over here, this is called the golden ratio." + }, + { + "Q": "At 3:57 how is phi always equal to only part of phi, shouldn't Phi=Sq. root of 1+Phi equal the square root of 1+Phi, if 1+Phi is Phi there SHOULD be no solution", + "A": "At 6:15 Sal talks about how the \u00e2\u0088\u009a5 is an irrational number and so repeats forever like \u00cf\u0080. An equation has no solutions when one side doesn t equal the other not when one sides repeats forever I think. I m twice your age and struggling with this so it s awesome you re on it already.", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 237 + ], + "3min_transcript": "wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let equal to phi plus 1. phi squared is equal to phi plus 1. And then, actually, I'm going to take a little bit of a side But even this is interesting, because then if we take the square root of both sides of this, you get-- let me scroll down a little bit-- you get phi is equal to the square root of-- and I'll just switch the order here-- the square root of 1 plus phi. So once again, we can set up another recursive definition. phi is equal to the square root of 1 plus phi. And I could write phi there, but hey, phi is equal to the square root of 1 plus and I could write phi there, but hey, phi is just equal to the square root of 1 plus, the square root of 1 plus. And we could just keep going on and on like this forever. So even this is neat. The same number that can be expressed this way, the same number where if I just subtract 1 from it, It can also be expressed in these kind of recursive square roots underneath each other. So this is already starting to get very, very, very intriguing, but let's get back to business. Let's actually solve for this magic number, this magic ratio that we started thinking about. And really from a very simple idea, that the ratio of the longer side to the shorter side is equal to the ratio of the sum of the two to the longer side. So let's just solve this as a traditional quadratic. Let's get everything on the left-hand side. So we're going to subtract phi plus 1 from both sides. And we get phi squared minus phi minus 1 is equal to 0. And we can solve for phi now using the quadratic formula, which we've proven in other videos. You can prove using completing the square. But the quadratic formula you say, negative b. Negative b is the coefficient on this term right here. So let me just write it down, a is equal to 1, that's the coefficient on this term. b is equal to negative 1, that's the coefficient on this term." + }, + { + "Q": "At 3:20 he multiplied each side of the equation by phi, wouldnt that make it\nPhi^2=1+1", + "A": "(phi - 1 = 1/phi)phi phi^2 - phi = 1", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 200 + ], + "3min_transcript": "And b over a is just the inverse of this statement right over here. So b over a-- this thing right here over here is phi-- so b over a is going to be 1 over phi. This is going to be 1 over phi. So this is interesting. We've now set up a number, which we're going to call this special ratio, phi is equal to 1 plus 1 over phi. Well just that is kind of a neat statement right over there. First of all, you could, if you subtract 1 from both sides of this, you get phi minus 1 is equal to its inverse. That seems to be a pretty neat property of any number that if I just subtract 1 from it, I get its multiplicative inverse. And so that, already that seems kind of intriguing. But then even this statement over here is kind of interesting because we've defined phi in terms of 1 plus 1 over phi. So we can actually think of it this way. We could say that phi is equal to 1 plus 1 over phi. wait, phi is just 1 plus 1 over-- instead of saying phi-- I could say, well, that's just 1 plus 1 over and I could just write phi again or I could just keep on going. I could just keep on going like this forever. I could say that's 1 over 1 plus 1 over and just keep on going on and on and on, forever. And this is a recursive definition of a function, or a recursive definition of a variable, where it's defined in terms of itself. But even this seems like a pretty neat property. But we want to get a little bit further into it. We actually eventually want to figure out what phi is. What is the value of phi, this weird number, this weird ratio that we're beginning to explore? So let's see if we can turn it into a quadratic equation that we can solve using fairly traditional methods. And the easiest way to do that is to multiply both sides of this equation by phi. And then you get phi squared-- let equal to phi plus 1. phi squared is equal to phi plus 1. And then, actually, I'm going to take a little bit of a side But even this is interesting, because then if we take the square root of both sides of this, you get-- let me scroll down a little bit-- you get phi is equal to the square root of-- and I'll just switch the order here-- the square root of 1 plus phi. So once again, we can set up another recursive definition. phi is equal to the square root of 1 plus phi. And I could write phi there, but hey, phi is equal to the square root of 1 plus and I could write phi there, but hey, phi is just equal to the square root of 1 plus, the square root of 1 plus. And we could just keep going on and on like this forever. So even this is neat. The same number that can be expressed this way, the same number where if I just subtract 1 from it," + }, + { + "Q": "At 12:00, how did he get from (a-b)/b to (a/b)-1?", + "A": "(a-b)/b=(a/b)-(b/b)=(a/b)-1", + "video_name": "5zosU6XTgSY", + "timestamps": [ + 720 + ], + "3min_transcript": "That 1.61 so on and so forth. Let me scroll down a little bit. So that is going to be equal to phi. So that's something interesting to do. Maybe that's a nice looking rectangle of some sort. But let me put out a square here. So let me separate this into a b by b square. So this is a b by b square right over here. And then-- actually let me do it a little bit, let me draw it a little bit differently, this rectangle actually isn't exactly the way I would want to draw it-- so the ratio might look a little bit like this. So the ratio of the width to the length, or the width to the height, is going to be the golden ratio. So a over b is going to be that golden ratio. And let me separate out a little b by b square over here. So this has width b as well. And so this distance right over here is going to be a minus b. Actually, I should say, we have a b by b square, right over here. This is b by b. And then we're left with a b by a minus b rectangle. Now wouldn't it be cool if this was also the golden ratio? And so let's try it out. Let's find the ratio of b to a minus b. So the ratio of b to a minus b. Well, that's going to be equal to 1 over the ratio of a minus b to b. I just took the reciprocal of this right over here. And this is just going to be equal to 1 over a over b. Let me write this, a over b minus 1. I just rewrote this right there. And that's just going to be equal to 1 over phi. The ratio of a to b, we said, by definition was phi minus 1. But what is phi minus 1? Well phi minus 1 is 1 over phi. It's this cool number. So it's equal to 1 over 1 over 1 over phi, which is once again, So once again, the ratio of this smaller rectangle, of its height to its width, is once again this golden ratio, this number that keeps showing up. And then we could do the same thing again. We could separate this into an a minus b by a minus b square. Just like that. And then we'll have another golden rectangle, sometimes it's called, right over there. And then we could separate that into a square and another golden rectangle. Then we could separate that into a square and then another golden rectangle. Then another golden rectangle. Actually let me do it like this. This would be better. So let me separate. Let me do the square up here. So this is an a minus b by a minus b square and then we have another golden rectangle right over here. I could put a square right in there. Then we'll have another golden rectangle. Then we could put another square right over there, you have another golden rectangle. I think you see where this is going." + }, + { + "Q": "at 9:07 how did minus 8 by 24?", + "A": "Because it will go into the negatives.", + "video_name": "-rxUip6Ulnw", + "timestamps": [ + 547 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:11 I really got confused... Like what? I don't get what she means? Like how?", + "A": "9x10 is basically 9 ten times. You move nine to the tens place, and zero is in the ones place since there are no ones left. 9 times 10 is 9, 10 times or 10, 9 times", + "video_name": "Ehd3cgRBvl0", + "timestamps": [ + 131 + ], + "3min_transcript": "- [Voiceover] What is 7 100s times 10? Well, let's focus first on this times 10 part of our expression. Because multiplying by 10 has some patterns in math that we can use to help us solve. One pattern we can think of when we multiply by 10 is if we take a whole number and multiply it by 10, we'll simply add a zero to the end of our whole number. So, for example, if we have a whole number like nine, and we multiply by 10, our solution will be a nine with one zero at the end. Or 90. Because nine times 10 is the same as nine 10s, and nine 10s is ninety. So let's use that pattern first to try to solve. Here we have seven 100s. So, seven times we have 100, or 700, and we're multiplying again times 10. Our solution will add a zero at the end. So if we had 700, 10 times, we would have 700 with a zero on the end. Or 7,000. So seven 100s times ten, is equal to 7,000. But there's another pattern we could use, here. Another pattern to think about when we multiply by 10. And that is that when we multiply by 10, we move every digit one place value, one place value, left. Or one place value greater. So let's look at that one on a place value chart. Here we have a place value chart. To use that earlier example when we had nine ones, and we multiplied it by 10, It moved up to the 10s. Now, we had nine 10s. And we filled in a zero here, because there were no ones left; there were zero ones left. And so, we saw that nine times 10 was equal to 90. So again, it's the same as adding a zero at the end, but we're looking at it another way. We're looking at it in terms of place value and multiplying by 10 moved every digit one place value to the left. So, if we do that with this same question, seven 100s, seven 100s, if we move 100s one place value to the left, we'll end up with 1,000s. So, 700 times 10 is seven 1,000s. Or, as we saw earlier, 7,000. So either one of these is a correct answer." + }, + { + "Q": "At 2:32 , I don't understand how he got that (s-7). Could someone explain it?", + "A": "Ah I see your confusion. Here is what he did: It started as S(S+5) - 7(S+5) What if we look at it all in parenthesis? (S(S+5) - 7(S+5)) What could you factor out of this big mess? Both terms have (S+5) in common right? So pull out S+5 and you get: (S+5)(whatever is not yet factored out) But, the only terms remaining in the original parenthesis was an S-7. So there you go! The final terms are (S+5)(S+7)", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 152 + ], + "3min_transcript": "And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5." + }, + { + "Q": "in 1:17 how do you get 5+-7=-3", + "A": "He got 5 + (-7) = -2, not 5 + (-7) = -3 He also had the constraint that a \u00c3\u0097 b = -35, and 5 \u00c3\u0097 -7 = -35", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 77 + ], + "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers" + }, + { + "Q": "At 1:59 how did Sal get s(s+5)?", + "A": "Arnav, At 1:59, Sal took the part of the expression (s\u00c2\u00b2 + 5s) and using the distribtive property in reverse, he factored out an s Like this (s\u00c2\u00b2 + 5s) is (s*s + 5*s) so factor out (undistribute) an s s*(s+5) and rewrite as s(s+5) And in case you still don t understand the reverse distributive property, just use the distributive property on the answer and see if that helps. s(s+5) dstribute the s s*s + 5*s and rewrite as s\u00c2\u00b2 + 5s I hope that helps make it click for you.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 119 + ], + "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers" + }, + { + "Q": "At 1:41, can you factor out the quadratic equation into 2 binomials? Does it affect the answer?", + "A": "That is exactly what Sal did. He factored the quadratic into the binomials: (s-7)(s+5)=0 So, I m sure what you mean by your questions. If you can clarify, I ll try to help.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 101 + ], + "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers" + }, + { + "Q": "where did the 35 go at 2:11?", + "A": "Sal just factored a -7 out of that part of the polynomial. He divided the two terms by -7 to get -7(s+5). Hope this helps! :D", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 131 + ], + "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers" + }, + { + "Q": "At 2:50, how does s(s+5)-7(s+5)=0 factor to (s+5)(s-7)?", + "A": "s(s+5)-7(s+5) factors into (s+5)(s-7) because s has been factored out of (s^2+5s) and -7 has been factored out of (-7s-35) both of the factored out forms are (s+5) you combine what you factored out of both sides and you get (s-7) leaving you with the factored form (s+5)(s-7).", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 170 + ], + "3min_transcript": "And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5." + }, + { + "Q": "At 2:16, couldn't you do s^2-5s+7s-35? It would mean the same thing right?", + "A": "I don t think you can factor the equation in thay form, but mathematically it means the same thing.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 136 + ], + "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers" + }, + { + "Q": "When solving a quadratic equation by factoring, if both equations could equal zero how come this is not included in the answer? Sal mentions around 3:45 that both could equal zero. In the above example the answer is given as s=-5 or s=7 but not s=-5 and s=7. Why is this? In the equation it makes sense that both could equal 0 as 0x0=0 but how can the answer be s=-5 and s=7?\n\nThanks!", + "A": "s=-5 makes one factor 0. s= 7 makes the other factor 0. Those are two different solutions to making the equation 0. But we didn t know until we did the factoring that the two factors would lead to two different zeros. It could have some out, for example, like this: (s+5)(s+5)=0. Then s= -5 would make both factors zero, and that would be ok because 0*0 = 0.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 225 + ], + "3min_transcript": "So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5. both sides of that equation, and you get s is equal to 7. So if s is equal to negative 5, or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is minus 35. That does equal zero. If you have 7, 49 minus 14 minus 35 does equal zero. So we've solved for s. Now, I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what is that equal to? x times x is x squared, x times b is bx." + }, + { + "Q": "Wouldn't the (s+5), in about 2:45, be squared because there is two of them? I didn't think that you could just get rid of it or ignore it..", + "A": "Emily, We had s(s+5) - 7(s+5) We are combining like terms. The second (s+5) doesn t just disappear it is just combined. Think of the s+5 as being apples . s apples - 7 apples = (s-7) apples. We can do the same thing with (s+5) instead of apples s (s+5) - 7 (s+5) = (s-7) (s+5) I hope that helps make it click for you.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 165 + ], + "3min_transcript": "And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5." + }, + { + "Q": "At 2:51 Sal wrote the equation as (s+5)(s-7)=0. When factoring quadratics, how do you know which constants are supposed to come first, like, in this case, 5 is the first constant and -7 is the second constant? That usually gets me when solving quadratic equations by factoring.", + "A": "It doesn t matter which comes first. The commutative property of multiplication tells use that the order we multiply in doesn t matter. Example 2x3 = 3x2. Apply the same property to the factors: (s+5)(s-7) = (s-7)(s+5). The order of the factors does not matter. Hope this helps.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 171 + ], + "3min_transcript": "And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5." + }, + { + "Q": "at 2:09 why does he put the plus or minus sign?", + "A": "When we multiply +9 x +9 we get 81 also when we multiply -9 x -9 we still get 81. so the square root can be + or - 9. therefore he writes + and - as the root can be either in + or -.", + "video_name": "tRHLEWSUjrQ", + "timestamps": [ + 129 + ], + "3min_transcript": "- Let's see if we can solve the equation P squared is equal to 0.81. So how could we think about this? Well one thing we could do is we could say, look if P squared is equal to 0.81, another way of expressing this is, that well, that means that P is going to be equal to the positive or negative square root of 0.81. Remember if we just wrote the square root symbol here, that means the principal root, or just the positive square root. But here P could be positive or negative, because if you square it, if you square even a negative number, you're still going to get a positive value. So we could write that P is equal to the plus or minus square root of 0.81, which kind of helps us, it's another way of expressing the same, the same, equation. But still, what could P be? In your brain, you might immediately say, well okay, you know if this was P squared is equal to 81, I kinda know what's going on. Because I know that nine times nine is equal to 81. Or we could write that nine squared is equal to 81, to the principal root of 81. These are all, I guess, saying the same truth about the universe, but what about 0.81? Well 0.81 has two digits behind, to the right of the decimal and so if I were to multiply something that has one digit to the right of the decimal times itself, I'm gonna have something with two digits to the right of the decimal. And so what happens if I take, instead of nine squared, what happens if I take 0.9 squared? Let me try that out. Zero, I'm gonna use a different color. So let's say I took 0.9 squared. 0.9 squared, well that's going to be 0.9 times 0.9, which is going to be equal to? Well nine times nine is 81, and I have one, two, numbers to the right of the decimal, so I'm gonna have two numbers to the right of the decimal in the product. So one, two. So that indeed is equal to 0.81. In fact we could write 0.81 as 0.9 squared. is equal to the plus or minus, the square root of, instead of writing 0.81, I could write that as 0.9 squared. In fact I could also write that as negative 0.9 squared. Cause if you put a negative here and a negative here, it's still not going to change the value. A negative times a negative is going to be a positive. I could, actually I would have put a negative there, which would have implied a negative here and a negative there. So either of those are going to be true. But it's going to work out for us because we are taking the positive and negative square root. So this is going to be, P is going to be equal to plus or minus 0.9. Plus or minus 0.9, or we could write it that P is equal to 0.9, or P could be equal to negative 0.9. And you can verify that, you would square either of these things, you get 0.81." + }, + { + "Q": "at 5:46 how did he get b+6", + "A": "Because, the sum of a+b must equal -4 and the product of a*b must equal -60. He just brute force went thru all the combinations possible until finding that +6 and -10 satisfy this. (a+6)(a-10) = b^2-4b-60", + "video_name": "STcsaKuW-24", + "timestamps": [ + 346 + ], + "3min_transcript": "Then you could have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and a 12, still seems too far apart One of them is negative, then you either have their sum being positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They are four apart. So if we make-- and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10 their sum will be negative 4, and their product is negative 60. So that works. So you could literally say that this is equal to b plus 6, times b, minus 10. b plus the a, plus b minus the b. And let me be very careful here. This b over here, I want to make it very clear, I just used this b here to say, look, we're looking for two numbers that add up to this second term It's a different b. I could have said x plus y is equal to negative 4, and x times y is equal to negative 60. In fact, let me do it that way just so we don't get confused. So we could write x plus y is equal to negative 4. And then we have x times y is equal to negative 60. So we have b plus 6, times b plus y. x is 6, y is negative 10. And that is equal to 0. Let's just solve this right here. And then we'll go back and show you. You could also factor this by grouping. But just from this, we know that either one of these is equal to zero. Either b plus 6 is equal to 0, or b minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get b is equal to negative 6. you get b is equal to 10. And those are our two solutions. You could put them back in and verify that they satisfy our constraints. Now the other way that you could solve this, and we're going to get exact same answer. Is you could just break up this negative 4b into its constituents. So you could have broken this up into 0 is equal to b squared. And then you could have broken it up into plus 6b, minus 10b, minus 60. And then factor it by grouping. Group these first two terms. Group these second two terms. Just going to add them together. The first one you could factor out a b. So you have b times b, plus 6. The second one you can factor out a negative 10. So minus 10 times b, plus 6. All that's equal to 0. And now you can factor out a b plus 6. So if you factor out a b plus 6 here," + }, + { + "Q": "i didnt get what it meant in 5:00 could someone explain it to me??", + "A": "He is saying that the b s are not the same he just used b and a instead of x and y", + "video_name": "STcsaKuW-24", + "timestamps": [ + 300 + ], + "3min_transcript": "So we need to factor b squared, minus 4b, minus 60. So what we want to do, we want to find two numbers whose sum is negative 4 and whose product is negative 60. Now, given that the product is negative, we know there are different signs. And this tells us that their absolute values are going to be four apart. That one is going to be four less than the others. So you could look at the products of the factors of 60. 1 and 60 are too far apart. Even if you made one of the negative, you would either get positive 59 as the sum or negative 59 as the sum. 2 and 30, still too far apart. 3 and 20, still too far apart. If you had made one negative you'd Then you could have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and a 12, still seems too far apart One of them is negative, then you either have their sum being positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They are four apart. So if we make-- and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10 their sum will be negative 4, and their product is negative 60. So that works. So you could literally say that this is equal to b plus 6, times b, minus 10. b plus the a, plus b minus the b. And let me be very careful here. This b over here, I want to make it very clear, I just used this b here to say, look, we're looking for two numbers that add up to this second term It's a different b. I could have said x plus y is equal to negative 4, and x times y is equal to negative 60. In fact, let me do it that way just so we don't get confused. So we could write x plus y is equal to negative 4. And then we have x times y is equal to negative 60. So we have b plus 6, times b plus y. x is 6, y is negative 10. And that is equal to 0. Let's just solve this right here. And then we'll go back and show you. You could also factor this by grouping. But just from this, we know that either one of these is equal to zero. Either b plus 6 is equal to 0, or b minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get b is equal to negative 6." + }, + { + "Q": "i notice at 0:29 sal factor the number with -1 ? why he do that? can someone explain me?", + "A": "If you factor a negative number, -1 is a factor. and then you can focus on factoring the rest which is now positive. Also, -1 is a perfect cube, so you can do the cuberoot (-1) = -1.", + "video_name": "drhoIgAhlQM", + "timestamps": [ + 29 + ], + "3min_transcript": "We're are asked to find the cube root of negative 343. Or another way to think about it is some number that when I multiply it by itself three times, I'm going to get negative 343. Or another way to view it-- this is the same thing as negative 343 to the 1/3 power. And the best way to do this is to really just try to factor this out. So the first thing that we could do-- so let me just factor negative 343. So the first thing I'd like to do is just factor out the negative 1. So this is the same thing as negative 1 times 343. And let's think about this. Is this divisible by 2? No. Is it divisible by 3? Let's see-- the digits do not add up to a multiple of 3. They add up to 10. So it's not divisible by 3. Not divisible by 4 since it's odd. Not divisible by 5 because it doesn't end with a 5 or a 0. It's not divisible by 6, because it's not divisible by 2 or 3. Is it divisible by 7? Let's check this out. So 7 goes into 343-- 7 goes into 34 four times. 34 minus 28 is going to be 6. Bring down the 3. 7 goes into 63 exactly nine times. So then we end up-- so 9 times 7 is 63, no remainder. So this is going to be 7 times 49. And we know that 49 is the same thing as 7 times 7. So how can we rewrite this? This is the same thing as taking the cube root of negative 1 times 7 times 7 times 7, which is the same thing as taking the cube root of negative 1 times the cube root of 7 times 7 times 7. Now, what's the cube root of negative 1? Well, negative 1 times itself three times is negative 1. You could verify it. Negative 1 times negative 1 times negative 1 is indeed negative 1. This becomes positive 1. Multiply by negative 1 again, you get negative 1. So this is negative 1. And then this over here, the cube root of 7 times 7 times 7-- well, that's just going to be 7. 7 multiplied by itself three times gives us 7 times 7 times 7 or 343. So it's going to be negative 1 times 7, which is the same thing as negative 7. So our answer is negative 7. And we're done." + }, + { + "Q": "At 0:31 he says something about finding what 0 is equal to.", + "A": "He s not finding what zero equals, he said setting your to zero to solve for x. It is mathematically equivalent to frame the equation as 0 = x as it is to state x = 0.", + "video_name": "6agzj3A9IgA", + "timestamps": [ + 31 + ], + "3min_transcript": "Use completing the square to find the roots of the quadratic equation right here. And when anyone talks about roots, this just means find the x's where y is equal to 0. That's what a root is. A root is an x value that will make this quadratic function equal 0, that will make y equal 0. So to find the x's, let's just make y equal 0 and then solve for x. So we get 0 is equal to 4x squared plus 40x, plus 280. Now, the first step that we might want to do, just because it looks like all three of these terms are divisible by 4, is just divide both sides of this equation by 4. That'll make our math a little bit simpler. So let's just divide everything by 4 here. If we just divide everything by 4, we get 0 is equal to x squared plus 10x, plus-- 280 divided by 4 is 70-- plus 70. me write that 70 a little bit further out, and you'll see why I did that in a second. So let me just write a plus 70 over here, just to have kind of an awkward space here. And you'll see what I'm about to do with this space, that has everything to do with completing the square. So they say use completing the square, which means, turn this, if you can, into a perfect square. Turn at least part of this expression into a perfect square, and then we can use that to actually solve for x. So how do we turn this into a perfect square? Well, we have a 10x here. And we know that we can turn this into a perfect square trinomial if we take 1/2 of the 10, which is 5, and then we square that. So 1/2 of 10 is 5, you square it, you add a 25. Now, you can't just willy-nilly add a 25 to one side of the equation without doing something to the other, or without just subtracting the 25 right here. Think about it, I have not changed the equation. I've added 25 and I've subtracted 25. So I've added nothing to the right-hand side. I could add a billion and subtract a billion and not change the equation. So I have not changed the equation at all right here. But what I have done is I've made it possible to express these three terms as a perfect square. That right there, 2 times 5 is 10. 5 squared is 25. So that is x plus 5 squared. And if you don't believe me, multiply it out. You're going to have an x squared plus 5x, plus 5x, which will give you 10x, plus 5 squared, which is 25. So those first three terms become that, and then the second two terms, right there, you just add them. Let's see, negative 25 plus 70. Let's see, negative 20 plus 70 would be positive 50, and then you have another 5, so it's plus 45." + }, + { + "Q": "1:53 what is the meaning of willy-nilly? without-worry?", + "A": "When I ve heard this used, it has always meant to do something in a random manner, haphazardly, without any method or planning, in any way you please without thinking.", + "video_name": "6agzj3A9IgA", + "timestamps": [ + 113 + ], + "3min_transcript": "Use completing the square to find the roots of the quadratic equation right here. And when anyone talks about roots, this just means find the x's where y is equal to 0. That's what a root is. A root is an x value that will make this quadratic function equal 0, that will make y equal 0. So to find the x's, let's just make y equal 0 and then solve for x. So we get 0 is equal to 4x squared plus 40x, plus 280. Now, the first step that we might want to do, just because it looks like all three of these terms are divisible by 4, is just divide both sides of this equation by 4. That'll make our math a little bit simpler. So let's just divide everything by 4 here. If we just divide everything by 4, we get 0 is equal to x squared plus 10x, plus-- 280 divided by 4 is 70-- plus 70. me write that 70 a little bit further out, and you'll see why I did that in a second. So let me just write a plus 70 over here, just to have kind of an awkward space here. And you'll see what I'm about to do with this space, that has everything to do with completing the square. So they say use completing the square, which means, turn this, if you can, into a perfect square. Turn at least part of this expression into a perfect square, and then we can use that to actually solve for x. So how do we turn this into a perfect square? Well, we have a 10x here. And we know that we can turn this into a perfect square trinomial if we take 1/2 of the 10, which is 5, and then we square that. So 1/2 of 10 is 5, you square it, you add a 25. Now, you can't just willy-nilly add a 25 to one side of the equation without doing something to the other, or without just subtracting the 25 right here. Think about it, I have not changed the equation. I've added 25 and I've subtracted 25. So I've added nothing to the right-hand side. I could add a billion and subtract a billion and not change the equation. So I have not changed the equation at all right here. But what I have done is I've made it possible to express these three terms as a perfect square. That right there, 2 times 5 is 10. 5 squared is 25. So that is x plus 5 squared. And if you don't believe me, multiply it out. You're going to have an x squared plus 5x, plus 5x, which will give you 10x, plus 5 squared, which is 25. So those first three terms become that, and then the second two terms, right there, you just add them. Let's see, negative 25 plus 70. Let's see, negative 20 plus 70 would be positive 50, and then you have another 5, so it's plus 45." + }, + { + "Q": "At around 5:15, Sal says that the t-distribution has fatter tails because the small sample size causes an underestimation of the standard deviation of the sampling distribution of the sample mean.\n\nWouldn't a smaller sample size make you overestimate the standard deviation? And that is what leads to fatter tails? Could someone clarify this for me? Thank you!", + "A": "Too large an SEM would give too large an interval; too small too small. Fatter tails means a larger percentage of the area (probability) at higher SD. So you would need more SDs to get the same probability. So it looks to me like the fatter tails would tend to compensate for underestimating SEM. Why do you think that smaller sample size would overestimate the SEM? It gives a larger SEM, yes (the smaller the sample, the less accurate the sample mean is likely to be). Is it enough larger?", + "video_name": "K4KDLWENXm0", + "timestamps": [ + 315 + ], + "3min_transcript": "we're going to tweak the sampling distribution. We're not going to assume it's a normal distribution because this is a bad estimate. We're going to assume that it's something called a t-distribution. And a t-distribution is essentially, the best way to think about is it's almost engineered so it gives a better estimate of your confidence intervals and all of that when you do have a small sample size. It looks very similar to a normal distribution. It has some mean, so this is your mean of your sampling distribution still. But it also has fatter tails. And the way I think about why it has fatter tails is when you make an assumption that this is a standard deviation for-- let me take one more step. So normally what we do is we find the estimate of the true standard deviation, and then we say that the standard true standard deviation of our population divided by the square root of n. In this case, n is equal to 7. And then we say OK, we never know the true standard, or we seldom know-- sometimes you do know-- we seldom know the true standard deviation. So if we don't know that the best thing we can put in there is our sample standard deviation. And this right here, this is the whole reason why we don't say that this is just a 95 probability interval. This is the whole reason why we call it a confidence interval because we're making some assumptions. This thing is going to change from sample to sample. And in particular, this is going to be a particularly bad estimate when we have a small sample size, a size less than 30. So when you are estimating the standard deviation where you don't know it, you're estimating it with your sample standard deviation, and your sample size is small, and deviation of your sampling distribution, you don't assume your sampling distribution is a normal distribution. You assume it has fatter tails. And it has fatter tails because you're essentially underestimating-- you're underestimating the standard deviation over here. Anyway, with all of that said, let's just actually go through this problem. So we need to think about a 95% confidence interval around this mean right over here. So a 95% confidence interval, if this was a normal distribution you would just look it up in a Z-table. But it's not, this is a t-distribution. We're looking for a 95% confidence interval. So some interval around the mean that encapsulates 95% of the area. For a t-distribution you use t-table, and I have a t-table ahead of time right over here. And what you want to do is use the two-sided row for what" + }, + { + "Q": "at 1:37, how could 10 hundreths and 7 hundreths make sense", + "A": "10 hundredths + 7 hundredths = 0.17 = 1 tenth + 7 hundredths. 1 hundredth = 0.01", + "video_name": "qSPwUDmpnJ4", + "timestamps": [ + 97 + ], + "3min_transcript": "- [Voiceover] Let's say that I had the number zero point one seven. How could I say this number? I said it one way, I said zero point one seven, but what are other ways that I could say it, especially if I wanted to express it in terms of tenths or hundredths or other places? And like always, try to pause the video and try think about it on your own. Alright, so there's actually a couple of ways that we could say this number. One is just to say zero point one seven. Other ways are to say look, I have a one in the tenths place, so that's going to be one tenth, one tenth and one tenth and I have a seven in the hundredths place, so this is a seven right over here in the hundredths place, so I can say one tenth and seven hundredths. Hun- Hundredths. And there you go. Now another to think about it is just say the whole thing in terms of hundredths. So a tenth is how many hundredths? Well a tenth is the same thing as 10 hundredths, so you could say, you could say instead of a tenth, you could say this is 10 hundredths, and the way I'm writing it right now, very few people would actually do it this way. 10 hundredths and and seven hundredths. And seven hundredths. Well not I could just add these hundredths, if I have 10 hundredths and I have another seven hundredths, that's going to be 17 hundredths. So I could just write this down as 17 hundredths. Hundredths. And to make that intuition of how we could just call this 17 hundredths instead of just calling it one tenth and seven hundredths, let's actually count by hundredths. So that is one hundredth, and actually, let me just go straight to nine hundredths. So I skipped a bunch right over here. And what would be the next, how would I say 10 hundredths? Well 10 hundredths, let me write it this way, 10 hundredths is the same thing as one tenth. So if we go from nine hundredths, the next, if I'm counting by hundredths, the next one's going to be 10 hundredths. Now once again, 10 hundredths is the same thing as one tenth, just the same way that 10 ones is the same thing as one 10. I hope that doesn't confuse you, but we could keep counting. 10 hundredths, 11 hundredths, 12 hundredths, 13 hundredths, 14 hundredths, 15 hundredths, 16 hundredths, and then finally 17 hundredths. So hopefully that gives you a little intuition for why we can call this number, instead of just calling it zero point one seven, or one tenth and seven hundredths, we could call this 17 hundredths." + }, + { + "Q": "at 5:41 Sal mentions a \"rule of thumb\". I'm new to statistics, but I can't seem to find it in my books or these videos. What is the \"Rule of Thumb\" and is there a video I'm missing?", + "A": "Rule of thumb is just an expression, it means a good generalization or a simple way to remember something. There is no single Rule of Thumb. :P", + "video_name": "5ABpqVSx33I", + "timestamps": [ + 341 + ], + "3min_transcript": "distributed if our sample size is greater than 30. Even this approximation will be approximately normally distributed. Now, if your sample size is less than 30, especially if it's a good bit less than 30, all of a sudden this expression will not be normally distributed. So let me re-write the expression over here. Sample mean minus the mean of your sampling distribution of the sample mean divided by your sample standard deviation over the square root of your sample size. We just said if this thing is well over 30, or at least 30, then this value right here, this statistic, is going to be normally distributed. If it's not, if this is small, then this is going to have a T-distribution. And then you're going to do the exact same thing you did normal distribution, so this example it was normal. All of Z's are normally distributed. Over here in a T-distribution, and this will actually be a normalized T-distribution right here because we subtracted out the mean. So in a normalized T-distribution, you're going to have a mean of 0. And what you're going to do is you want to figure out the probability of getting a T-value at least this extreme. So this is your T-value you would get, and then you essentially figure out the area under the curve right over there. So a very easy rule of thumb is calculate this quantity either way. Calculate this quantity either way. If you will have more than 30 samples, if your sample size is more than 30, your sample standard deviation is going to be a good approximator for your population standard deviation. And so this whole thing is going to be approximately figure out the probability of getting a result at least that extreme. If your sample size is small, then this statistic, this quantity, is going to have a T-distribution, and then you're going to have to use a T-table to figure out the probability of getting a T-value at least this extreme. And we're going to see this in an example a couple of videos from now. Anyway, hopefully that helped clarify some things in your head about when to use a Z-statistic or when to use a T-statistic." + }, + { + "Q": "At 5:54, how did Sal already find the b for Slope-Intercept Form?", + "A": "He distributed the -2/3(x+3) in the point-slope form.", + "video_name": "-6Fu2T_RSGM", + "timestamps": [ + 354 + ], + "3min_transcript": "So that is our slope, negative 2/3. So we're pretty much ready to use point slope form. We have a point, we could pick one of these points, I'll just go with the negative 3, 6. And we have our slope. So let's put it in point slope form. All we have to do is we say y minus-- now we could have taken either of these points, I'll take this one-- so y minus the y value over here, so y minus 6 is equal to our slope, which is negative 2/3 times x minus our Well, our x-coordinate, so x minus our x-coordinate is negative 3, x minus negative 3, and we're done. We can simplify it a little bit. x minus negative 3 is the same thing as x plus 3. This is our point slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope intercept form. Let's do that. So let's do slope intercept in orange. So we have slope intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2/3, so you get y minus 6 is equal to-- I'm just distributing the negative 2/3-- so negative 2/3 times x is negative 2/3 x. And then negative 2/3 times 3 is negative 2. And now to get it in slope intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side, so let's add 6 to both Left-hand side of the equation, we're just left with a y, these guys cancel out. You get a y is equal to negative 2/3 x. Negative 2 plus 6 is plus 4. So there you have it, that is our slope intercept form, mx plus b, that's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2/3 x to both sides of this equation. So I'll start it here. So we have y is equal to negative 2/3 x plus 4, that's slope intercept form. Let's added 2/3 x, so plus 2/3 x to both sides of this equation. I'm doing that so it I don't have this 2/3 x on the right-hand side, this negative 2/3 x." + }, + { + "Q": "7:01 if -2/3X was positive, would the answer be 2/3x-y=4? or 2/3x+y=4?", + "A": "yes.", + "video_name": "-6Fu2T_RSGM", + "timestamps": [ + 421 + ], + "3min_transcript": "x minus negative 3 is the same thing as x plus 3. This is our point slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope intercept form. Let's do that. So let's do slope intercept in orange. So we have slope intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2/3, so you get y minus 6 is equal to-- I'm just distributing the negative 2/3-- so negative 2/3 times x is negative 2/3 x. And then negative 2/3 times 3 is negative 2. And now to get it in slope intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side, so let's add 6 to both Left-hand side of the equation, we're just left with a y, these guys cancel out. You get a y is equal to negative 2/3 x. Negative 2 plus 6 is plus 4. So there you have it, that is our slope intercept form, mx plus b, that's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2/3 x to both sides of this equation. So I'll start it here. So we have y is equal to negative 2/3 x plus 4, that's slope intercept form. Let's added 2/3 x, so plus 2/3 x to both sides of this equation. I'm doing that so it I don't have this 2/3 x on the right-hand side, this negative 2/3 x. little bit, maybe more than I should have-- the left-hand side of this equation is what? It is 2/3 x, because 2 over 3x, plus this y, that's my left-hand side, is equal to-- these guys cancel out-- is equal to 4. So this, by itself, we are in standard form, this is the standard form of the equation. If we want it to look, make it look extra clean and have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get? 2/3 x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12. These are the same equations, I just multiplied every term by 3. If you do it to the left-hand side, you can do to the right-hand side-- or you have to do to the right-hand side-- and we are in standard form." + }, + { + "Q": "at 7:00, can there be a standard form without an A?", + "A": "There will always be an A, it s just a coefficient. It can be 0 though, in which case the term does not exist.", + "video_name": "-6Fu2T_RSGM", + "timestamps": [ + 420 + ], + "3min_transcript": "x minus negative 3 is the same thing as x plus 3. This is our point slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope intercept form. Let's do that. So let's do slope intercept in orange. So we have slope intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2/3, so you get y minus 6 is equal to-- I'm just distributing the negative 2/3-- so negative 2/3 times x is negative 2/3 x. And then negative 2/3 times 3 is negative 2. And now to get it in slope intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side, so let's add 6 to both Left-hand side of the equation, we're just left with a y, these guys cancel out. You get a y is equal to negative 2/3 x. Negative 2 plus 6 is plus 4. So there you have it, that is our slope intercept form, mx plus b, that's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2/3 x to both sides of this equation. So I'll start it here. So we have y is equal to negative 2/3 x plus 4, that's slope intercept form. Let's added 2/3 x, so plus 2/3 x to both sides of this equation. I'm doing that so it I don't have this 2/3 x on the right-hand side, this negative 2/3 x. little bit, maybe more than I should have-- the left-hand side of this equation is what? It is 2/3 x, because 2 over 3x, plus this y, that's my left-hand side, is equal to-- these guys cancel out-- is equal to 4. So this, by itself, we are in standard form, this is the standard form of the equation. If we want it to look, make it look extra clean and have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get? 2/3 x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12. These are the same equations, I just multiplied every term by 3. If you do it to the left-hand side, you can do to the right-hand side-- or you have to do to the right-hand side-- and we are in standard form." + }, + { + "Q": "at 2:24 how did you get the slope? that is what confused me because when i tried it it didnt work", + "A": "0-6 over 6+3 = -6 over 9 Simplify -2 over 3", + "video_name": "-6Fu2T_RSGM", + "timestamps": [ + 144 + ], + "3min_transcript": "A line passes through the points negative 3, 6 and 6, 0. Find the equation of this line in point slope form, slope intercept form, standard form. And the way to think about these, these are just three different ways of writing the same equation. So if you give me one of them, we can manipulate it to get any of the other ones. But just so you know what these are, point slope form, let's say the point x1, y1 are, let's say that that is a point on the line. And when someone puts this little subscript here, so if they just write an x, that means we're talking about a variable that can take on any value. If someone writes x with a subscript 1 and a y with a subscript 1, that's like saying a particular value x and a particular value of y, or a particular coordinate. And you'll see that when we do the example. But point slope form says that, look, if I know a particular point, and if I know the slope of the line, then putting that line in point slope form would be y minus y1 is equal to m times x minus x1. point negative 3 comma 6 is on the line, then we'd say y minus 6 is equal to m times x minus negative 3, so it'll end up becoming x plus 3. So this is a particular x, and a particular y. It could be a negative 3 and 6. So that's point slope form. Slope intercept form is y is equal to mx plus b, where once again m is the slope, b is the y-intercept-- where does the line intersect the y-axis-- what value does y take on when x is 0? And then standard form is the form ax plus by is equal to c, where these are just two numbers, essentially. They really don't have any interpretation directly on the graph. So let's do this, let's figure out all of these forms. So the first thing we want to do is figure out the slope. Once we figure out the slope, then point slope form is So, just to remind ourselves, slope, which is equal to m, which is going to be equal to the change in y over the change in x. Now what is the change in y? If we view this as our end point, if we imagine that we are going from here to that point, what is the change in y? Well, we have our end point, which is 0, y ends up at the 0, and y was at 6. So, our finishing y point is 0, our starting y point is 6. What was our finishing x point, or x-coordinate? Our finishing x-coordinate was 6. Let me make this very clear, I don't want to confuse you. So this 0, we have that 0, that is that 0 right there. And then we have this 6, which was our starting y point, that is that 6 right there." + }, + { + "Q": "At 6:48 where did you get the 4 from?", + "A": "It is simply the same equation as the one underneath slope-intercept form, because -2+6=4.", + "video_name": "-6Fu2T_RSGM", + "timestamps": [ + 408 + ], + "3min_transcript": "x minus negative 3 is the same thing as x plus 3. This is our point slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope intercept form. Let's do that. So let's do slope intercept in orange. So we have slope intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2/3, so you get y minus 6 is equal to-- I'm just distributing the negative 2/3-- so negative 2/3 times x is negative 2/3 x. And then negative 2/3 times 3 is negative 2. And now to get it in slope intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side, so let's add 6 to both Left-hand side of the equation, we're just left with a y, these guys cancel out. You get a y is equal to negative 2/3 x. Negative 2 plus 6 is plus 4. So there you have it, that is our slope intercept form, mx plus b, that's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2/3 x to both sides of this equation. So I'll start it here. So we have y is equal to negative 2/3 x plus 4, that's slope intercept form. Let's added 2/3 x, so plus 2/3 x to both sides of this equation. I'm doing that so it I don't have this 2/3 x on the right-hand side, this negative 2/3 x. little bit, maybe more than I should have-- the left-hand side of this equation is what? It is 2/3 x, because 2 over 3x, plus this y, that's my left-hand side, is equal to-- these guys cancel out-- is equal to 4. So this, by itself, we are in standard form, this is the standard form of the equation. If we want it to look, make it look extra clean and have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get? 2/3 x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12. These are the same equations, I just multiplied every term by 3. If you do it to the left-hand side, you can do to the right-hand side-- or you have to do to the right-hand side-- and we are in standard form." + }, + { + "Q": "3:00 of the video; I thought 4 to the exponent of 3 times 5 to the exponent of 3 would give you a different answer then then 4x5 to the exponent of 3. So I do not understand the logic of this.", + "A": "(4x5) to the third is (4x5)x(4x5)x(4x5). Because it is multiplication, we can move the numbers around, getting 4x4x4x5x5x5. 4x4x4 is 4 to the third, and 5x5x5 is 5 to the third. So, (4x5)^3 = 4^3 x 5^3.", + "video_name": "rEtuPhl6930", + "timestamps": [ + 180 + ], + "3min_transcript": "When I raise something to the fifth power, that's just like saying 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth, right? All I did is I took 2 to the tenth and I multiplied it by itself five times. That's the fifth power. Well, we know from this rule up here that we can add these exponents because they're all the same base. So if we add 10 plus 10 plus 10 plus 10 plus 10, what do we get? Right, we get 2 to the fiftieth power. So essentially, what did we do here? All we did is we multiplied 10 times 5 to get 50. So that's our third exponent rule, that when I raise an exponent to a power and then I raise that whole expression to another power, I can multiply those two exponents. So let me give you another example. again, all I do is I multiply the 7 and the negative 9, and I get 3 to the minus 63. So, you see, it works just as easily with negative numbers. So now, I'm going to teach you one final exponent property. Let's say I have 2 times 9, and I raise that whole thing to the hundredth power. It turns out of this is equal to 2 to the hundredth power times 9 to the hundredth power. Now let's make sure that that makes sense. Let's do it with a smaller example. What if it was 4 times 5 to the third power? times 4 times 5, right, which is the same thing as 4 times 4 times 4 times 5 times 5 times 5, right? I just switched the order in which I'm multiplying, which you can do with multiplication. Well, 4 times 4 times 4, well, that's just equal to 4 to the third. And 5 times 5 times 5 is equal to 5 to the third. Hope that gives you a good intuition of why this property here is true. And actually, when I had first learned exponent rules, I would always forget the rules, and I would always do this proof myself, or the other proofs. And a proof is just an explanation of why the rule works, just to make sure that I was doing it right. So given everything that we've learned to now-- actually, let me review all of the rules again." + }, + { + "Q": "At the end of the video, when he gives the problem, he says at 5:45 that the problem can be written as (3^2 * (3^2)^8)^-2.... but, it couldn't be (3^2 * 3 * 3^8)^-2? It would be a different answer, but it is following the rules in the same way...how could this be? Someone please?", + "A": "In (3^2 * (3^2)^8)^-2 , the second 3^2 is the expanded form of 9... thus 9 becomes 3^2 (3^2)^8 is 9^8 3 * 3^8 is 3^1+8= 3^9 It is pretty clear here that 9^8 and 3^9 is not the same.. And I don t see what rule you re saying you re following..could you explain that part?", + "video_name": "rEtuPhl6930", + "timestamps": [ + 345 + ], + "3min_transcript": "well, then I can add the exponents, 2 to the tenth. If I have 2 the seventh over 2 the third, well, here I subtract the exponents, and I get 2 to the fourth. If I have 2 to the seventh to the third power, well, here I multiplied the exponents. That gives you 2 to the 21. And if I had 2 times 7 to the third power, well, that equals 2 to the third times 7 to the third. Now, let's use all of these rules we've learned to actually try to do some, what I would call, composite problems that involve you using multiple rules at the same time. And a good composite problem was that problem that I had introduced you to at the end of that last seminar. and all of that I'm going to raise to the negative 2 power. So what can I do here? Well, 3 and 9 are two separate bases, but 9 can actually be expressed as an exponent of 3, right? 9 is the same thing as 3 squared, so let's rewrite 9 like that. That's equivalent to 3 squared times-- 9 is the same thing as 3 squared to the eighth power, and then all of that to the negative 2 power, right? All I did is I replaced 9 with 3 squared because we know 3 times 3 is 9. Well, now we can use the multiplication rule on this to simplify it. which is 16, and all of that to the negative 2. Now, we can use the first rule. We have the same base, so we can add the exponents, and we're multiplying them, so that equals 3 to the eighteen power, right, 2 plus 16, and all that to the negative 2. And now we're almost done. We can once again use this multiplication rule, and we could say 3-- this is equal to 3 to the eighteenth times negative 2, so that's 3 to the minus 36. So this problem might have seemed pretty daunting at first, but there aren't that many rules, and all you have to do is keep seeing, oh, wow, that little part of the problem, I can simplify it. Then you simplify it, and you'll see that you can keep using rules until you get to a much simpler answer. And actually the Level 1 problems don't even involve" + }, + { + "Q": "At 2:56, why isn't (2x9)^100 =18^100? why is it 2^100 x 9^100?", + "A": "They are the same thing. See the example at 3:00.", + "video_name": "rEtuPhl6930", + "timestamps": [ + 176 + ], + "3min_transcript": "When I raise something to the fifth power, that's just like saying 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth, right? All I did is I took 2 to the tenth and I multiplied it by itself five times. That's the fifth power. Well, we know from this rule up here that we can add these exponents because they're all the same base. So if we add 10 plus 10 plus 10 plus 10 plus 10, what do we get? Right, we get 2 to the fiftieth power. So essentially, what did we do here? All we did is we multiplied 10 times 5 to get 50. So that's our third exponent rule, that when I raise an exponent to a power and then I raise that whole expression to another power, I can multiply those two exponents. So let me give you another example. again, all I do is I multiply the 7 and the negative 9, and I get 3 to the minus 63. So, you see, it works just as easily with negative numbers. So now, I'm going to teach you one final exponent property. Let's say I have 2 times 9, and I raise that whole thing to the hundredth power. It turns out of this is equal to 2 to the hundredth power times 9 to the hundredth power. Now let's make sure that that makes sense. Let's do it with a smaller example. What if it was 4 times 5 to the third power? times 4 times 5, right, which is the same thing as 4 times 4 times 4 times 5 times 5 times 5, right? I just switched the order in which I'm multiplying, which you can do with multiplication. Well, 4 times 4 times 4, well, that's just equal to 4 to the third. And 5 times 5 times 5 is equal to 5 to the third. Hope that gives you a good intuition of why this property here is true. And actually, when I had first learned exponent rules, I would always forget the rules, and I would always do this proof myself, or the other proofs. And a proof is just an explanation of why the rule works, just to make sure that I was doing it right. So given everything that we've learned to now-- actually, let me review all of the rules again." + }, + { + "Q": "2:59 of the this video. Doesn't this break the \"order of operation\" because you do what is in the parenthesis first then work your way out?\n\nalso in ths video and the previous video before this one, the subtitle at the bottom seem to flash a lot or something.", + "A": "Because they are all multiplication, we can rearrange the numbers any way we want, using the distributive property.", + "video_name": "rEtuPhl6930", + "timestamps": [ + 179 + ], + "3min_transcript": "When I raise something to the fifth power, that's just like saying 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth, right? All I did is I took 2 to the tenth and I multiplied it by itself five times. That's the fifth power. Well, we know from this rule up here that we can add these exponents because they're all the same base. So if we add 10 plus 10 plus 10 plus 10 plus 10, what do we get? Right, we get 2 to the fiftieth power. So essentially, what did we do here? All we did is we multiplied 10 times 5 to get 50. So that's our third exponent rule, that when I raise an exponent to a power and then I raise that whole expression to another power, I can multiply those two exponents. So let me give you another example. again, all I do is I multiply the 7 and the negative 9, and I get 3 to the minus 63. So, you see, it works just as easily with negative numbers. So now, I'm going to teach you one final exponent property. Let's say I have 2 times 9, and I raise that whole thing to the hundredth power. It turns out of this is equal to 2 to the hundredth power times 9 to the hundredth power. Now let's make sure that that makes sense. Let's do it with a smaller example. What if it was 4 times 5 to the third power? times 4 times 5, right, which is the same thing as 4 times 4 times 4 times 5 times 5 times 5, right? I just switched the order in which I'm multiplying, which you can do with multiplication. Well, 4 times 4 times 4, well, that's just equal to 4 to the third. And 5 times 5 times 5 is equal to 5 to the third. Hope that gives you a good intuition of why this property here is true. And actually, when I had first learned exponent rules, I would always forget the rules, and I would always do this proof myself, or the other proofs. And a proof is just an explanation of why the rule works, just to make sure that I was doing it right. So given everything that we've learned to now-- actually, let me review all of the rules again." + }, + { + "Q": "at about 6:10, when Sal has (3x2x(3x2)to the 8th), why does he only multiply the 8th to the second 3x2, and not both, because they are all within the larger parentheses?", + "A": "First of all it is not multiplication, but power (3^2*(3^2)^8) Power always applies to item it is after. 3^2 means 3 in power 2. The second power applies to 3. The power of 8 is after the closing parentheses, so it only applies to items in parentheses. (3^2)^8 means 3 to the power of 2 and all of that is to the power of 8.", + "video_name": "rEtuPhl6930", + "timestamps": [ + 370 + ], + "3min_transcript": "well, then I can add the exponents, 2 to the tenth. If I have 2 the seventh over 2 the third, well, here I subtract the exponents, and I get 2 to the fourth. If I have 2 to the seventh to the third power, well, here I multiplied the exponents. That gives you 2 to the 21. And if I had 2 times 7 to the third power, well, that equals 2 to the third times 7 to the third. Now, let's use all of these rules we've learned to actually try to do some, what I would call, composite problems that involve you using multiple rules at the same time. And a good composite problem was that problem that I had introduced you to at the end of that last seminar. and all of that I'm going to raise to the negative 2 power. So what can I do here? Well, 3 and 9 are two separate bases, but 9 can actually be expressed as an exponent of 3, right? 9 is the same thing as 3 squared, so let's rewrite 9 like that. That's equivalent to 3 squared times-- 9 is the same thing as 3 squared to the eighth power, and then all of that to the negative 2 power, right? All I did is I replaced 9 with 3 squared because we know 3 times 3 is 9. Well, now we can use the multiplication rule on this to simplify it. which is 16, and all of that to the negative 2. Now, we can use the first rule. We have the same base, so we can add the exponents, and we're multiplying them, so that equals 3 to the eighteen power, right, 2 plus 16, and all that to the negative 2. And now we're almost done. We can once again use this multiplication rule, and we could say 3-- this is equal to 3 to the eighteenth times negative 2, so that's 3 to the minus 36. So this problem might have seemed pretty daunting at first, but there aren't that many rules, and all you have to do is keep seeing, oh, wow, that little part of the problem, I can simplify it. Then you simplify it, and you'll see that you can keep using rules until you get to a much simpler answer. And actually the Level 1 problems don't even involve" + }, + { + "Q": "At 5:32 Sal expresses 9^8 as 3^16 to be able to calculate the product of two exponents with the same base. It looks like you could instead express 3^2 as 9^1 and come up with a final answer of 9^-18 instead of 3^-36. The numbers are equivalent but is one answer better by convention?", + "A": "You are absolutely correct. It just depends if you want your base to be a prime number. If that is your goal you will always come up with the same answer, which has some elegance.", + "video_name": "rEtuPhl6930", + "timestamps": [ + 332 + ], + "3min_transcript": "well, then I can add the exponents, 2 to the tenth. If I have 2 the seventh over 2 the third, well, here I subtract the exponents, and I get 2 to the fourth. If I have 2 to the seventh to the third power, well, here I multiplied the exponents. That gives you 2 to the 21. And if I had 2 times 7 to the third power, well, that equals 2 to the third times 7 to the third. Now, let's use all of these rules we've learned to actually try to do some, what I would call, composite problems that involve you using multiple rules at the same time. And a good composite problem was that problem that I had introduced you to at the end of that last seminar. and all of that I'm going to raise to the negative 2 power. So what can I do here? Well, 3 and 9 are two separate bases, but 9 can actually be expressed as an exponent of 3, right? 9 is the same thing as 3 squared, so let's rewrite 9 like that. That's equivalent to 3 squared times-- 9 is the same thing as 3 squared to the eighth power, and then all of that to the negative 2 power, right? All I did is I replaced 9 with 3 squared because we know 3 times 3 is 9. Well, now we can use the multiplication rule on this to simplify it. which is 16, and all of that to the negative 2. Now, we can use the first rule. We have the same base, so we can add the exponents, and we're multiplying them, so that equals 3 to the eighteen power, right, 2 plus 16, and all that to the negative 2. And now we're almost done. We can once again use this multiplication rule, and we could say 3-- this is equal to 3 to the eighteenth times negative 2, so that's 3 to the minus 36. So this problem might have seemed pretty daunting at first, but there aren't that many rules, and all you have to do is keep seeing, oh, wow, that little part of the problem, I can simplify it. Then you simplify it, and you'll see that you can keep using rules until you get to a much simpler answer. And actually the Level 1 problems don't even involve" + }, + { + "Q": "What does si mean and theda.2:20", + "A": "Psi and theta are Greek letters that usually denote angles.", + "video_name": "MyzGVbCHh5M", + "timestamps": [ + 140 + ], + "3min_transcript": "" + }, + { + "Q": "at 2:01 what does si has mean?", + "A": "Psi (not si) is a letter in Greek alphabet (\u00ce\u00a8). Mathematicians use Greek letters to write angles. They often use other letters : \u00ce\u00b1(alpha), \u00ce\u00b2(beta), \u00ce\u00b3(gamma), \u00ce\u00b8(theta)", + "video_name": "MyzGVbCHh5M", + "timestamps": [ + 121 + ], + "3min_transcript": "" + }, + { + "Q": "how does he know psi 1 equal 1/2 theta 1 at 8:15 in the vidio? i didn't undrstand the proof. i don't think i saw a proof. like wise for scy 2 and thaita 2 at 8:20", + "A": "Sal wasnt really proving anything in particular. All he wanted to show was that the centre neednt be within the arc being subtended. And he just did that using stuff that he had taught before", + "video_name": "MyzGVbCHh5M", + "timestamps": [ + 495, + 500 + ], + "3min_transcript": "" + }, + { + "Q": "At 0:32 what does transversal mean?", + "A": "A transversal is a line that intersects two other lines. A transversal can be useful in determining whether the two lines it intersects are parallel.", + "video_name": "LhrGS4-Dd9I", + "timestamps": [ + 32 + ], + "3min_transcript": "What we're going to prove in this video is a couple of fairly straightforward parallelogram-related proofs. And this first one, we're going to say, hey, if we have this parallelogram ABCD, let's prove that the opposite sides have the same length. So prove that AB is equal to DC and that AD is equal to BC. So let me draw a diagonal here. And this diagonal, depending on how you view it, is intersecting two sets of parallel lines. So you could also consider it to be a transversal. Actually, let me draw it a little bit neater than that. I can do a better job. Nope. That's not any better. That is about as good as I can do. So if we view DB, this diagonal DB-- we can view it as a transversal for the parallel lines AB and DC. And if you view it that way, you can pick out that angle ABD is going to be congruent-- so angle ABD. That's that angle right there-- is going to be congruent to angle BDC, because they are alternate interior angles. So we know that angle ABD is going to be congruent to angle BDC. Now, you could also view this diagonal, DB-- you could view it as a transversal of these two parallel lines, of the other pair of parallel lines, AD and BC. And if you look at it that way, then you immediately see that angle DBC right over here is going to be congruent to angle ADB for the exact same reason. They are alternate interior angles of a transversal intersecting these two parallel lines. So I could write this. This is alternate interior angles are congruent when you have a transversal intersecting And we also see that both of these triangles, triangle ADB and triangle CDB, both share this side over here. It's obviously equal to itself. Now, why is this useful? Well, you might realize that we've just shown that both of these triangles, they have this pink angle. Then they have this side in common. And then they have the green angle. Pink angle, side in common, and then the green angle. So we've just shown by angle-side-angle that these two triangles are congruent. So let me write this down. We have shown that triangle-- I'll go from non-labeled to pink to green-- ADB is congruent to triangle-- non-labeled to pink to green-- CBD. And this comes out of angle-side-angle congruency." + }, + { + "Q": "At 2:12, if both are increasing then don't the negatives cancel out and become positive?", + "A": "You are comparing the slopes, not multiplying the slopes. Both lines are negative, so both lines slant down from left to right. The slope of line F is decreasing faster because its slope is more negative than the slope of line G. Hope this helps.", + "video_name": "fZO-JylMFqY", + "timestamps": [ + 132 + ], + "3min_transcript": "Two functions, f and g, are described below. Which of these statements about f and g is true? So they defined function f as kind of a traditional linear equation right over here. And this right over here is g. So this right over here is g of x. And that also looks like a linear function. We see it's a kind of a downward sloping line. So let's look at our choices and see which of these are true. f and g are both increasing, and f is increasing faster than g. Well, when I look at g-- Well, first of all, g is definitely decreasing. So we already know that that's false. And f is also decreasing. We see here it has a negative slope. Every time we move forward 3 in the x direction, we're going to move down 7 in the vertical direction. So neither of these are increasing so that's definitely not right. f and g are both increasing. Well, that's definitely not right. So we know that both f and g are decreasing. So this first choice says they're both decreasing, and g So let's see what the slope on g is. So the slope on g is every time we move 1 in the x direction, positive 1 in the x direction, we move down 2 in the y direction. So for g of x, if we were to write our change in y over our change in x-- which is our slope-- our change in y over change in x, when we move one in the x direction, positive 1 in the x direction, we move down 2 in the y direction. So our change in y over change in x is negative 2. So g has a slope of negative 2. f has a slope of negative 7/3. Negative 7/3 is the same thing as negative 2 and 1/3. So f's slope is more negative. So it is decreasing faster. than g. So this is not right. And then we have this choice-- f and g are both decreasing, and f is decreasing faster than g. This is right, right over here. We have this last choice-- g is increasing but f is decreasing. We know that's not true. g is actually decreasing." + }, + { + "Q": "At somewhere around 8:00 (it's hard to tell; for some reason this video doesn't show me the time) Sal adds a squared and b squared. I don't understand how he was able to do this. Wasn't he talking about two different triangles?", + "A": "both of the triangles add up to make the larger one; he s trying to get a math statement that applies to the larger one.", + "video_name": "LrS5_l-gk94", + "timestamps": [ + 480 + ], + "3min_transcript": "are going to be similar So, we can say triangle BDC, we went from pink to right, to not labeled So, triangle BDC, triangle BDC is similar to triangle, now we're gonna look at the larger triangle, now we're gonna start the pink angle B, now we go to the right angle CA BCA >From pink angle to right angle to non-labeled angle, at least from the point of view here before the blue Now we setup some type of relationship here We can say that the ratio on the smaller triangle BC, side BC over BA BC over BA Once again we're taking the hypotenuses of both of them So, BC over BA is going to be equal to BD Here's another color, BD, so this one of the legs BD over BC, I'm just taking the corresponding vertices, over BC And once again, we know, BC is the same as lowercase \"b,\" BC is lowercase \"b \" BA is lower case \"c \" And then BD we defined as lower case \"e \" So, this is lowercase \"e \" We can cross multiply here and we b times b Which and I mentioned this in many videos cross multiplying both sides by both denominators b times b is equal to ce And now we can do something kind of interesting We can add these two statements down here Let me rewrite this statement down here So, b squared is equal to ce So, if we add the left hand sides, we get a squared plus b squared, a squared plus b squared is equal to cd, is eqaual to cd And then we have a ce in both of these terms so we can factor it out So, this is gonna be equal to, we can factor out the c, it's gonna be c times d plus e c times d plus e, and close the parenthesis Now what is d plus e? d is this length e is this length So, d plus e is actually gonna be c as well So, this is gonna be c So, if c times c is the same thing as c squared So, now we have an interesting relationship, we have that a squared plus b squared is equal to c squared Let me rewrite that a squared, I'll do that- well let me just arbitrary new color I deleted that by accident, so let me rewrite it" + }, + { + "Q": "how does he figure out for what n means at 6:20?? i am very confused too!!!", + "A": "Since he is trying to get n by itself, he multiplies 8/36 with 36/8. They cancel out, and you are left with n. What you do to one side you do to the other. 10 x 36/8=360/8. There is one n left, so n=360/8. That s your answer.", + "video_name": "GO5ajwbFqVQ", + "timestamps": [ + 380 + ], + "3min_transcript": "Or to figure out what that times what is, you divide 360 divided by 8. So we could divide, and this is a little bit of algebra here, we're dividing both sides of the equation by 8. And we're getting n is equal to 360 divided by 8. You could do that without thinking in strict algebraic terms. You could say 8 times what is 360. Well 8 times 360/8. If I write 8 times question mark is equal to 360, well, question mark could definitely be 360/8. If I multiply these out, this guy and that guy cancel out, and it's definitely 360. And that's why it's 360/8. But now we want to actually divide this to actually get our right answer, or a simplified answer. 8 goes into 360, 8 goes into 36 4 times, 4 times 8 is 32. You have a remainder of 4. Bring down the 0. 8 goes into 40 5 times. 5 times 8 is 40. And you're done. Once again, we got n is equal to 45. Now the last thing I'm going to show you involves a little bit of algebra. If any of the ways before this worked, that's fine. And where this is sitting in the playlist, you're not expected to know the algebra. But I want to show you the algebra just because I wanted to show you that this cross-multiplication isn't some magic, that using algebra, we will get this exact same thing. But you could stop watching this, if you'll find this part confusing. So let's rewrite our proportion, 8/36 is equal to 10/n. And we want to solve for n. Well the easiest way to solve for n is maybe multiply both-- this thing on the left is equal to this thing on the right. So we can multiply them both by the same thing. And the equality will still hold. So we could multiply both of them by n. On the right-hand side, the n's cancel out. On the left-hand side, we have 8/36 times n is equal to 10. If we want just an n here, we would want to multiply this side times 36-- I'll do that in a different color-- we'd want to multiply this side times 36 times 8, because if you multiply these guys out, you get 1. And you just have an n. But since we're doing it to the left-hand side, we also have to do it to the right-hand side, so times 36/8. These guys cancel out and we're left with n is equal to 10 times 36 is 360/8. And notice, we're getting the exact same value that we got with cross-multiplying. And with cross-multiplying, you're actually doing two steps. Actually, you're doing an extra step here. You're multiplying both sides by n, so that you had your 8n. And then you're multiplying both sides by 36, so that you get your 36 on both sides. And you get this value here. But at the end, when you simplify it," + }, + { + "Q": "Couldn't Sal have just converted 5/4 into 1.25 at 0:60 instead of going through all of the 36*5/4 stuff?", + "A": "5/4 = 1.25 and 1.25 * 36 = ?? it is difficult to calculate However if we solve (5/4) * 36 it is equal to (36/4) * 5 = 8 * 5 = 40.Thus we neednt get answer in decimals if it can be cut. This method is fast and easy and removes converting into decimal part.", + "video_name": "GO5ajwbFqVQ", + "timestamps": [ + 60 + ], + "3min_transcript": "We're asked to solve the proportion. We have 8 36ths is equal to 10 over what. Or the ratio of 8/36 is equal to the ratio of 10 to what. And there's a bunch of different ways to solve this. And I'll explore really all of them, or a good selection of them. So one way to think about it is, these two need to be equivalent ratios, or really, equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10? Well you could multiply 8 times 10/8. It will definitely give you 10. So we're multiplying by 10/8 over here. Or another way to write 10/8, 10/8 is the same thing as 5/4. So we're multiplying by 5/4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it. You have to multiply it times 5/4. And so we could say this n, this thing that we just solved for, Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We could divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9. Divide the denominator by 4 you get 1. You get 45. So that's one way to think about it. 8/36 is equal to 10/45. Another way to think about it is, what do we have to multiply 8 by to get its denominator. How much larger is the denominator 36 than 8? Well let's just divide 36/8. So 36/8 is the same thing as-- so we can simplify, dividing the numerator and the denominator by 4. That's the greatest common divisor. That's the same thing as 9/2. you get the denominator. So we're multiplying by 9/2 to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9/2 times 8, let me write this. 8 times 9/2 is equal to 36. That's how we go from the numerator to the denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9/2 again. So then we'll get 10 times 9/2 is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9/2. Divide the numerator and the denominator by 2, you get 5/1, which is 45. So 45 is equal to n. Once again, we got the same way, completely legitimate way, to solve it. Now sometimes when you see proportion like this," + }, + { + "Q": "The point of the video is to Multiply and Divide Scientific Notation, right? So how come at 2:19 you add? Is it something about exponent rules? If so, can someone please explain it, and, if not, can someone please tell me how? Thanks. And also, why do you have to change the number at 4:01? Thanks in advance for the help.", + "A": "Its an exponent rule. 10^2 x 10^3 is equal to 10^(2+3) because (10 x 10)x(10 x 10 x 10)= 10^5", + "video_name": "xxAFh-qHPPA", + "timestamps": [ + 139, + 241 + ], + "3min_transcript": "Multiple, expressing the product in scientific notation. So lets multiply first and then lets try and get what we have into scientific notation. Actually before we do that lets just try and remember what it means to be in scientifc notation. So to be in scientific notation and actually each of these numbers here are in scientific notation. It is going to be the form: a times ten to some power where a can be greater than of equal to one and is going to be less than ten. So both of these numbers are greater than or equal to one and less than ten, and are being multiplied by some power of 10 so lets see how we can multiply this. so this here is the exact same thing. I am doing this in magenta 9.1 times 10 to the 6th. times, times 3.2 times. Actually, lets use dot notation, it will be a little bit more straightforward. So this is equal to 9.1 times 10 to the 6th. Now in multiplication, this comes from the associative property allows us to remove these brackets and says you can multiply that first or these guys first. and you can re-associate them and the communitive property tells us that we can re-arrange these here. and what I want to rearrange is I want to multiply the 9.1 by the 3.2 first. And then multiply that by the ten to the 6 times the ten to the -5. So I am going to rearrange this So I am going to rearrange this using the cumulative property. So this is the same thing as 9.1 x 3.2. and I am going to re-associate so I am going to do these first. Now that times 10^6 times 10^-5. We have the same base here, base 10 and we are taking the product so we can add the exponents. so this part right over here is going to be 10 to the ( 6 - 5 ) or just 10 to the 1st power. which is really just equal to 10. and that is going to be multiplied by 9.1 x 3.2 lets do that over here, so we have 9.1 x 3.2 At first I am going to ignore the decimals so I have 91 by 32 So to have 2 by 1 is 2 2 by 9 is 18 have a zero here as I am in the tens place now and am multiplying everything by 30 and not just 3 thats why my zero is there and I multiply 3 by 1 to get 3 and then 3 by 9 is 27" + }, + { + "Q": "At 5:17 what does adjacent mean? I forget. Thanks", + "A": "Adjacent means next to .", + "video_name": "TgDk06Qayxw", + "timestamps": [ + 317 + ], + "3min_transcript": "it splits that line segment in half. So what it tells is, is that the length of this segment right over here is going to be equivalent to the length of this segment right over there. I have a circle. This radius bisects this chord right over here. And the goal here is to prove that it bisects this chord at a right angle. Or another way to say it-- let me add some points here. Let's call this B. Let's call this C, And let's call this D. I want to prove that segment AB is perpendicular. It intersects it at a right angle. It is perpendicular to segment CD. And as you could imagine, I'm going to prove it pretty much using the side-side-side whatever you want to call it, side-side-side theorem, postulate, or axiom. So let's do it. Let's think about it this way. I need to have some triangles. There's no triangles here right now. But I can construct triangles, and I can construct triangles based on things I know. For example, I can construct-- this has some radius. That's a radius right over here. The length of that is just going to be the radius of the circle. But I can also do it right over here. The length of AC is also going to be the radius of the circle. So we know that these two lines have the same length, which is the radius of the circle. Or we could say that AD is congruent to AC, or they have the exact same lengths. We know from the set-up in the problem that this segment is equal in length to this segment over here. Let me add a point here so I can refer to it. So if I call that point E, we know from the set-up in the problem, that CE is congruent to ED, or they have the same lengths. CE has the same length as ED. And we also know that both of these triangles, the one here the side EA. So EA is clearly equal to EA. So this is clearly equal to itself. It's the same side. The same side is being used for both triangles. The triangles are adjacent to each other. And so we see a situation where we have two different triangles that have corresponding sides being equal. This side is equivalent to this side right over here. This side is equal in length to that side over there. And then, obviously, AE is equivalent to itself. It's a side on both of them. It's the corresponding side on both of these triangles. And so by side-side-side, AEC." + }, + { + "Q": "at 5:57 does khan mean 0, cos(2x)>1 this is how it works: as x>o; cos(2*0)=cos(0)=1 cos(0)=1 so it is 8*1 which is 8 hope that helped", + "video_name": "BiVOC3WocXs", + "timestamps": [ + 415 + ], + "3min_transcript": "Let's see what happens in the numerator. Negative 2 times sine of 0. That's going to be 0. And then plus 4 times sine of 2 times 0. Well, that's still sine of 0, so that's still going to be 0. So once again, we got indeterminate form again. Do we give up? Do we say that L'Hopital's rule didn't work? No, because this could have been our first limit problem. And if this is our first limit problem we say, hey, maybe we could use L'Hopital's rule here because we got an indeterminate form. Both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to-- if the limit exist, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then, plus the derivative of 4 sine of 2x. Well, it's 2 times 4, which is 8. Derivative of sine of 2x is 2 cosine of 2x. And that first 2 gets multiplied by the 4 to get the 8. And then the derivative of the denominator, derivative of sine of x is just cosine of x. So let's evaluate this character. So it looks like we've made some headway or maybe L'Hopital's rule stop applying here because we take the limit as x approaches 0 of cosine of x. That is 1. So we're definitely not going to get that indeterminate form, that 0/0 on this iteration. Let's see what happens to the numerator. We get negative 2 times cosine of 0. Well that's just negative 2 because cosine of 0 is 1. Plus 8 times cosine of 2x. Well, if x is 0, so it's going to be cosine of 0, which is 1. So it's just going to be an 8. So negative 2 plus 8. Well this thing right here, negative 2 plus 8 is 6. 6 over 1. This whole thing is equal to 6. If this was the problem we were given and we said, hey, when we tried to apply the limit we get the limit as this numerator approaches 0 is 0. Limit as this denominator approaches 0 is 0. As the derivative of the numerator over the derivative of the denominator, that exists and it equals 6. So this limit must be equal to 6. Well if this limit is equal to 6, by the same argument, this limit is also going to be equal to 6. And by the same argument, this limit has got to also be equal to 6. And we're done." + }, + { + "Q": "@12:25, when Sal multiplied 10^17 by 10^-1, by the rules of multiplication, shouldn't the answer have been 10^-16, rather than 10^16? Thanks, whoever answers.", + "A": "The rules of exponents state that x^a * x^b = x^(a+b) we are adding, not multiplying, so the exponent in Sal s case stays positive as 17 > 1.", + "video_name": "0Dd-y_apbRw", + "timestamps": [ + 745 + ], + "3min_transcript": "Well, this is equal to 3.2 over 6.4. We can just separate them out because it's associative. So, it's this times 10 to the 11th over 10 to the minus six, right? If you multiply these two things, you'll get that right there. So 3.2 over 6.4. This is just equal to 0.5, right? 32 is half of 64 or 3.2 is half of 6.4, so this is 0.5 right there. And what is this? This is 10 to the 11th over 10 to the minus 6. So when you have something in the denominator, you could write it this way. This is equivalent to 10 to the 11th over 10 to the minus 6. It's equal to 10 to the 11th times 10 to the minus 6 to the minus 1. Or this is equal to 10 to the 11th times 10 to the sixth. This is 1 over 10 to the minus 6. So 1 over something is just that something to the negative 1 power. And then I multiplied the exponents. You can think of it that way and so this would be equal to 10 to the 17th power. Or another way to think about it is if you have 1 -- you have the same bases, 10 in this case, and you're dividing them, you just take the 1 the numerator and you subtract the exponent in the denominator. So it's 11 minus minus 6, which is 11 plus 6, which is equal to 17. So this division problem ended up being equal to 0.5 times 10 to the 17th. Which is the correct answer, but if you wanted to be a stickler and put it into scientific notation, we want something maybe greater than 1 right here. So the way we can do that, let's multiply it by 10 on this side. And divide by 10 on this side or multiply by 1/10. by 10 and divide by 10. We're just doing it to different parts of the product. So this side is going to become 5 -- I'll do it in pink -- 10 times 0.5 is 5, times 10 to the 17th divided by 10. That's the same thing as 10 to the 17th times 10 to the minus 1, right? That's 10 to the minus 1. So it's equal to 10 to the 16th power. Which is the answer when you divide these two guys right there. So hopefully these examples have filled in all of the gaps or the uncertain scenarios dealing with scientific notation. If I haven't covered something, feel free to write a comment on this video or pop me an e-mail." + }, + { + "Q": "At 6:25 for the partial derivative with respect to (x) he did:\n(x^2).2+sin2\n= (2x).2+0\n= (4x)+0\nand I'm ok so far;\nthen he derived (4x) as equal to (4) - yes because (x) becomes 1.. But I thought we already derived x^2 as 2x :-/\nWhat i'm missing?", + "A": "The last line from 4x+0 to 4 is not a derivation, he just replaces x with its value, which is one ^^", + "video_name": "AXqhWeUEtQU", + "timestamps": [ + 385 + ], + "3min_transcript": "Again, it depends on the function. And I'll show you how you can compute something like this in just a moment here. But, first there's kind of an annoying thing associated with partial derivatives, where we don't write them with D's in DX/DF. People came up with this new notation, mostly just to emphasize to the reader of your equation that it's a multi-variable function involved. And what you do, is you say, you write a D, but it's got kind of a curl at the top. It's this new symbol and people will often read it as partial. So, you might read like partial F, partial Y. If you're wondering, by the way, why we call these partial derivatives, it's sort of like, this doesn't tell the full story of how F changes 'cause it only cares about the X direction. Neither does this, this only cares about the Y direction. So, each one is only a small part of the story. So, let's actually evaluate something like this. I'm gonna go ahead and clear the board over here. I think the one-dimensional analogy is something we probably have already. So, if you're actually evaluating something like this, here, I'll write it up here again up here. Partial derivative of F, with respect to X, and we're doing it at one, two. It only cares about movement in the X direction, so it's treating Y as a constant. It doesn't even care about the fact that Y changes. As far as it's concerned, Y is always equal to two. So, we can just plug that in ahead of time. So, I'm gonna say partial, partial X, this is another way you might write it, put the expression in here. And I'll say X squared, but instead of writing Y, I'm just gonna plug in that constant ahead of time. 'Cause when you're only moving in the X direction, this is kind of how the multi-variable function sees the world. And I'll just keep a little note that we're evaluating this whole thing at X equals one. And here, this is actually just an ordinary derivative. This is an expression that's an X, you're asking how it changes as you shift around X and you know how to do this. This is just taking the derivative is gonna be 4x 'cause X squared goes to 2x. And then the derivative of a constant, sin of two is just a constant, is zero. And of course we're evaluating this at X equals one, so your overall answer is gonna be four. And as for practice, let's also do that with derivative with respect to Y. So, we look over here, I'm gonna write the same thing. You're taking the partial derivative of F with respect to Y. We're evaluating it at the same point one, two. This time it doesn't care about movement in the X direction. So, as far as it's concerned, that X just stays constant at one. So, we'd write one squared times Y, plus sin(Y). Sin(Y). And you're saying, oh, I'm keeping track of this at Y=2. So, it's kind of, you're evaluating at Y=2. When you take the derivative, this is just 1xY." + }, + { + "Q": "at 1:30, how did you come up with 50 divided by 4 when you said 50 times 25?", + "A": "True. Sal said 50 times 25 percent. 25% is equivalent to 1/4. So that s why he divided 50 by 4.", + "video_name": "OBVGQt1Eeug", + "timestamps": [ + 90 + ], + "3min_transcript": "We're told to make a table and solve. So they tell us that we have 50 ounces of a 25% saline solution, a mixture of water and salt. How many ounces of a 10% saline solution must you add to make a new solution that is 15% saline? So let's make this table that they're talking about. Let's write amount of solution. Let me write total amount of solution or maybe I should say total volume of solution. And then the next column I'll say percent saline. And then we can use this information to figure out total amount of saline. And let's list it for each of the two solutions that they talk about. We're starting with 50 ounces of a 25% saline solution. So this is what we start with. We'll assume everything is in ounces. It is 25% saline. So if we wanted to figure out the total ounces of saline, we say, well, we have 50 ounces. Multiply that by 25% and we have the total amount of saline in this solution. So 50 times 25%, that's the same thing as 50 divided by 4, so that's 12.5 ounces of saline in this 50 total ounces. It's 25% saline. Now, let's talk about what we're going to add to it, so solution added. Now, they say how many ounces of a 10% solution? So we don't even know how many ounces we're going to add. That's what we have to actually solve for. So we don't know how much we're going to add, but we do know that it is a 10% saline solution. And if we know what x is, we know the total amount of saline is going to be 10% of x. If we had 50 here, it would be 10% of 50. If we had 10 here, it would be 10% of 10. So the amount of saline we have in this solution, in x ounces of this solution, is going to be 0.1x, or 10% percent of x. That's what 10% of the solution being saline means. Now, when we add it, what do we end up with? So let me do this in a different color. Resulting solution. Well, if we started with 50 ounces and we add x ounces, we're going to end up with 50 plus x ounces." + }, + { + "Q": "At 7:53, how is -5 the square root of -\u00e2\u0088\u009a25? I understand that -5*-5 is equal to 25, since the negatives cancel out.\n*Note: I inserted the square root symbol by holding down on ALT and then typing 251 on the keypad.", + "A": "Since the negative sign is out of the parenthesis, it is negative, like if you learned what absolute value is, you ll know -I-5I is -5, since the negative is out of the absolute value signs, but if you are talking about the square root of negative 25, the answer is 5i, which you ll learn about later", + "video_name": "-QHff5pRdM8", + "timestamps": [ + 473 + ], + "3min_transcript": "So, this, right over here, is an irrational number. It's not rational. It cannot be represented as the ratio of two integers. All right, 14 over seven. This is the ratio of two integers. So, this, for sure, is rational. But if you think about it, 14 over seven, that's another way of saying, 14 over seven is the same thing as two. These two things are equivalent. So, 14 over seven is the same thing as two. So, this is actually a whole number. It doesn't look like a whole number, but, remember, a whole number is a non-negative number that doesn't need to be represented as the ratio of two integers. And this one, even though we did represent it as the ratio of two integers, it doesn't need to be represented as the ratio of two integers. You could have represent this as just two. So, that's going to be a whole number. 14 over seven, which is the same thing as two, that is a whole number. Now, two-pi. Now pi is an irrational... Pi is an irrational number. if we just take a integer multiple of pi, like that, this is also going to be an irrational number. If you looked at its decimal representation, it will never repeat. So that's two-pi, right over there. Now what about... Let me do that same, since I've been consistent, relatively consistent, with the colors. So, this is two-pi right over there. Now, what about the negative square root of 25. Well, 25's a perfect square. Square root of that's just gonna be five. So, this thing is going to be, this thing is going to be equivalent to negative five. So, this is just another representation of this, right over here. So, it is an integer. It's not a whole number because it's negative, but it's an integer. Negative square root of 25. These two things are actually... These two things are actually the same number, just different ways of representing them. And then you have, let's see, you have the square root of nine over... The square root of nine over seven. This thing is gonna be the same thing, this thing is the same... Let me do this in a different color. This is the same thing as, square root of nine is three, it's the principal root of nine, so it's three-sevenths. So, this is a ratio of two integers. This is a rational number. Square root of nine over seven is the same thing as three-sevenths. Now, let me just give you one more just for the road. What about pi over pi? What is that going to be? Well, pi divided by pi is going to be equal to one. So, this is actually a whole number. So I could write pi over pi, right over there. That's just a very fancy way of saying one." + }, + { + "Q": "At 4:20, Sal said 22/7 is a rational number. However, 22/7 is a irrational number, it keeps on repeating. Please help me understand.", + "A": "22/7 is definitely a rational number. As you can see it is the ratio of two integers - and that s the definition of rational. You re right though its decimal expansion does go on forever, but the crucial point is that it does repeat. 22/7 = 3.142857 142857 ... Repeating decimals can always be converted back to a fraction. 22/7 is often used as an approximation to \u00cf\u0080, but \u00cf\u0080 is irrational and its decimal expansion never ends and never repeats.", + "video_name": "-QHff5pRdM8", + "timestamps": [ + 260 + ], + "3min_transcript": "Irrational numbers. An integer. Well, if I could say, \"Look, that is an integer. \"Let's think about the integers.\" But I wouldn't say, \"Let's just think about the rational.\" I'd say, \"Let's think about the rational numbers.\" All right, now that we have these categories in place, let's categorize them. Like always, pause the video. See if you can figure out what category these numbers fall into. Where would you put them on this diagram? So, let's start off with three. This is positive three. It can be definitely represented as a fraction. You can represent it as three over one. But, it doesn't have to be represented as a fraction. It, literally, could be just a three, right over there, but it's also non-negative. So three is a whole number. So three, and maybe I'll do it in the color of the category. So, three is a whole number. So, it's a member of that set. But if you're a whole number, you're also an integer, and you're also a rational number. So, three is a whole number, it's an integer, Now, let's think about negative five. Now, negative five, once again, it can be represented as a fraction, but it doesn't have to be, but it is negative. So, it's not gonna be a whole number. So, negative five is going to sit right over here. It's an integer, and if you're an integer, you're definitely going to be a rational number, but it's not a whole number because it is negative. Now we have 0.25. Well, this, for sure, can be represented as a fraction. This is 25-hundreths, right over here. So, we can represent that as a fraction of two integers, I should say. It's 25-hundredths. But there's no way to represent this except using a fraction of two integers. So, 0.25 is a rational number, but it's not an integer and not a whole number. Now what about 22 over seven. Well, here it's clearly represented, already, as a fraction of two integers, except as a fraction of two integers. I can't somehow make this without using a fraction or some type of decimal that might repeat. So, this, right over here, this would also be a rational number, but it's not an integer, not a whole number. Now this over here. 0.2713. Now the 13 repeats. This is the same thing as 0.27131313, that's what line up there represents. Now, you might not realize it yet, but any number that repeats eventually, this one does repeat eventually, you have the .1313, or you have the 0.27131313, any number like this can be represented as a fraction. For example, and I'm not going to do it here, just for the sake of time, but, for example, 0.3, repeating, that's the same thing as one-third. And later on, we're gonna see techniques" + }, + { + "Q": "From 3:53 to 4:23 you said that 22/7 is rational number but, from 6:54 to 6:58 you said that pi is irrational. But, pi=22/7 and 22/7 is not an irrational number. So, how can you say that pi is irrational?", + "A": "22/7 does not = Pi. It is only an approximation for Pi, just like 3.14 is an approximation of Pi. Compare the numbers: Pi = 3.141592653589793238462643383... 22/7 = 3.142857142857142857142857... This is a repeating decimal, the digits 142857 repeat. The decimal values in Pi never repeat and never terminate. So, Pi is an irrational number. 22/7 is the ratio of 2 integers, so it is a rational number. Hope this helps.", + "video_name": "-QHff5pRdM8", + "timestamps": [ + 233, + 263, + 414, + 418 + ], + "3min_transcript": "Irrational numbers. An integer. Well, if I could say, \"Look, that is an integer. \"Let's think about the integers.\" But I wouldn't say, \"Let's just think about the rational.\" I'd say, \"Let's think about the rational numbers.\" All right, now that we have these categories in place, let's categorize them. Like always, pause the video. See if you can figure out what category these numbers fall into. Where would you put them on this diagram? So, let's start off with three. This is positive three. It can be definitely represented as a fraction. You can represent it as three over one. But, it doesn't have to be represented as a fraction. It, literally, could be just a three, right over there, but it's also non-negative. So three is a whole number. So three, and maybe I'll do it in the color of the category. So, three is a whole number. So, it's a member of that set. But if you're a whole number, you're also an integer, and you're also a rational number. So, three is a whole number, it's an integer, Now, let's think about negative five. Now, negative five, once again, it can be represented as a fraction, but it doesn't have to be, but it is negative. So, it's not gonna be a whole number. So, negative five is going to sit right over here. It's an integer, and if you're an integer, you're definitely going to be a rational number, but it's not a whole number because it is negative. Now we have 0.25. Well, this, for sure, can be represented as a fraction. This is 25-hundreths, right over here. So, we can represent that as a fraction of two integers, I should say. It's 25-hundredths. But there's no way to represent this except using a fraction of two integers. So, 0.25 is a rational number, but it's not an integer and not a whole number. Now what about 22 over seven. Well, here it's clearly represented, already, as a fraction of two integers, except as a fraction of two integers. I can't somehow make this without using a fraction or some type of decimal that might repeat. So, this, right over here, this would also be a rational number, but it's not an integer, not a whole number. Now this over here. 0.2713. Now the 13 repeats. This is the same thing as 0.27131313, that's what line up there represents. Now, you might not realize it yet, but any number that repeats eventually, this one does repeat eventually, you have the .1313, or you have the 0.27131313, any number like this can be represented as a fraction. For example, and I'm not going to do it here, just for the sake of time, but, for example, 0.3, repeating, that's the same thing as one-third. And later on, we're gonna see techniques" + }, + { + "Q": "At 4:51, how would do solve a system of equation using substitution with 3 variables and it is a fraction?\nexample: x/3+y/4-z/2=24\nx/2+y/3+z/4=20\nx/4+y/2+z/3=25 ( these are made up equations)", + "A": "Really, nothing has changed. You just solve for a variable and substitute it into another equation. If you have trouble with it, I recommend you look at easier examples of a system with 3 equations, and look into rational expressions. I m afraid I m too lazy to solve the example problem for you since it s quite a pain to simplify fractions that many times.", + "video_name": "u5dPUHjagSI", + "timestamps": [ + 291 + ], + "3min_transcript": "that we saw when we first started looking to algebra, that you can maintain your equality as long as you add the same thing. On the left-hand side, we're going to add this. And on the right-hand side, we're going to add this. But this second equation tells us that those two things are equal. So we can maintain our equality. So let's do that. What do we get on the left-hand side? Well, you have a positive x and a negative x. They cancel out. That was the whole point behind manipulating them in this way. And then you have negative y minus y, which is negative 2y. And then on the right-hand side, you have negative 4 minus 12, which is negative 16. And these are going to be equal to each other. Once again, we're adding the same thing to both sides. To solve for y, we can divide both sides by negative 2. And we are left with y is equal to positive 8. But we are not done yet. We want to go and substitute back into one of the equations. And we can substitute back into this one and to this one, or this one and this one. The solutions need to satisfy all of these essentially. This green equation is another way of expressing this green equation. So I'll go for whichever one seems to be the simplest. And this one seems to be pretty simple right over here. So let's take x minus y-- we just solved that y would be positive 8-- is equal to negative 4. And now to solve for x, we just have to add 8 to both sides. And we are left with, on the left-hand side, negative 8 plus 8 cancels out. You're just left with an x. And negative 4 plus 8 is equal to positive 4. So you get x is equal to 4, y is equal to 8. And you can verify that it would work with either one of these equations. 6 times 4 is 24 minus 6 times 8-- so it's 24 minus 48-- is, indeed, negative 24. Negative 5 times 4 is negative 20, minus negative 40, So it works out for both of these. And we can try it out by inputting our answers. So x is 4, y is 8. So let's do that. So let me type this in. x is going to be equal to 4. y is going to be equal to 8. And let's check our answer. It is correct. Very good." + }, + { + "Q": "I don't get the part @ 1:15. Can someone explain", + "A": "The problem is x-4y=-18, -x+3y=11.At 1:15, they are just combining like terms from both equations.They re adding x and -x and which the x and the -x cancel out because it s 1x and -1x. They re adding -4y and 3y and they get -y.They add -18 and 11 and get -7. -y=-7 Divide both sides by -1 and get y=7 -7=-7 Then substitute the y values into the equations.x-4(7)=-18, - x+3(7)=11 x-28=-18 Add 28 to both sides and get x=10 -x+21=11 Subract 21 from both sides -x=-10 Divide both sides by -1 and get x=10 x=10,y=7", + "video_name": "NPXTkj75-AM", + "timestamps": [ + 75 + ], + "3min_transcript": "- [Instructor] So we have a system of two linear equations here. This first equation, X minus four Y is equal to negative 18, and the second equation, negative X plus three Y is equal to 11. Now what we're gonna do is find an X and Y pair that satisfies both of these equations. That's what solving the system actually means. As you might already have seen, there's a bunch of X and Y pairs that satisfy this first equation. In fact, if you were to graph them, they would form a line, and there's a bunch of other X and Y pairs that satisfy this other equation, the second equation, and if you were to graph them, it would form a line. And so if you find the X and Y pair that satisfy both, that would be the intersection of the lines, so let's do that. So actually, I'm just gonna rewrite the first equation over here, so I'm gonna write X minus four Y is equal to negative 18. So, we've already seen in algebra that as long as we do the same thing to both sides of the equation, we can maintain our equality. So what if we were to add, so we have one equation with one unknown, so what if we were to add this negative X plus three Y to the left hand side here? So negative X plus three Y, well, that looks pretty good because an X and a negative X are going to cancel out, and we are going to be left with negative four Y, plus three Y. Well, that's just going to be negative Y. So by adding the left hand side of this bottom equation to the left hand side of the top equation, we were able to cancel out the Xs. We had X, and we had a negative X. That was very nice for us. So what do we do on the right hand side? We've already said that we have to add the same thing to both sides of an equation. We might be tempted to just say, well, if I have to add the same thing to both sides, well, maybe I have to add a negative X plus three Y to that side. But that's not going to help us much. We're gonna have negative 18 minus X plus three Y. We would have introduced an X but what if we could add something that's equivalent to negative X plus three Y that does not introduce the X variable? Well, we know that the number 11 is equivalent to negative X plus three Y. How do we know that? Well, that second equation tells us that. So once again, all I'm doing is I'm adding the same thing to both sides of that top equation. On the left, I'm expressing it as negative X plus three Y, but the second equation tells us that negative X plus three Y is going to be equal to 11. It's introducing that second constraint, and so let's add 11 to the right hand side, which is, once again, I know I keep repeating it, it's the same thing as negative X plus three Y. So negative 18 plus 11 is negative seven, and since we added the same thing to both sides, the equality still holds, and we get negative Y is equal to negative seven, or divide both sides by negative one or multiply both sides by negative one." + }, + { + "Q": "i dont get the part at 0:44 . how does he know what those little lines are if there not labled?", + "A": "It s okay @supergirlygamer he knows about those because no s are written at the interval of 5 and there are 4 little lines between those intervals therefore, every little line represents 1 unit for e.g. -5 | | | | 0 | | | | 5 above there are 4 lines between the interval so, filling those gaps -5 -4 -3 -2 -1 0 1 2 3 4 5 I hope this clears your doubt :)", + "video_name": "Ddvw2wEBfpc", + "timestamps": [ + 44 + ], + "3min_transcript": "- [Voiceover] We're told to fill in the blanks to complete the equation that describes the diagram. So let's think about what's going on over here. If we start at zero, and we move one, two, three, four spaces to the right of zero, this arrow right over here represents positive four. We already see that right over here in the equation. Then from positive four, from the tip of this arrow, we then go one, two, three, four, five, six spaces to the left. So what we just did here is we just added a negative six to the positive four. Positive four plus negative six. Where does that put us? Well we see it puts us one, two spaces to the left of zero and each of these spaces in this diagram are one. So two spaces to the left of zero is going to be negative two. This is fun. Let's keep doing more examples. your choice, so they're giving us some choice, to describe the diagram. Alright, let's see what's going on here. We're starting at zero and we're going one, two, three, four to the left. So if we're going four to the left or so we can say negative four, -4. And then we're going to go another one, two, three, four, five, six, seven, eight, nine to the left. So we could write this as negative four minus nine is equal to. And when you go four to the left and then you go another nine to the left, you end up 13 to the left of zero which is negative 13. Equals negative 13. So this way I've written it as a subtraction equation I guess you could say. Negative four minus nine, is equal to negative 13. Now another way I could have done it, I could have said negative four plus negative nine Either of those would have been legitimate. Now I've written it as an addition equation. Let's keep going. Fill in the blanks to complete the equation that describes the diagram. So we're starting at zero, we go three to the left of zero, that's negative three. Then we go another three to the left of that. So we're going to add another negative three. We're going to add another negative three and that puts us six to the left of zero. Well six to the left of zero is negative six. And we're done." + }, + { + "Q": "At around 2:55 in the video, is Sal essential combining the operations of -1R2 and R2 + 2R1? That confused me at first...", + "A": "He s doing 2R1 - R2. Your suggestion that he s combining the operations of -1R2 and R2 + 2R1 doesn t look right because the R2s would cancel out leaving 2R1. I might be reading your question wrong though...", + "video_name": "_uTAdf_AsfQ", + "timestamps": [ + 175 + ], + "3min_transcript": "Is this a basis for the space, for example? Is this a linear independent set of vectors? How can we visualize this space? And I haven't answered any of those yet. But if someone just says, hey what's the column space of A? This is the column space of A. And then we can answer some of those other questions. If this is a linearly independent set of vectors, then these vectors would be a basis for the column space of A. We don't know that yet. We don't know whether these are linearly independent. But we can figure out if they're linearly independent by looking at the null space of A. Remember these are linearly independent if the null space of A only contains the 0 vector. So let's figure out what the null space of A is. And remember, we can do a little shortcut here. The null space of A is equal to the null space of the row, the reduced row echelon form of A. And I showed you that when we first calculated the null essentially if you want to solve for the null space of A, you create an augmented matrix. And you put the augmented matrix in reduced row echelon form, but the 0's never change. So essentially you're just taking A and putting it in reduced row echelon form. Let's do that. So I'll keep row one the same, 1, 1, 1, 1. And then let me replace row two with, row two minus row one. So what do I get? No, actually I want to zero this out here. So row two minus, 2 times row one. Actually even better because I eventually So let me do 2 times row one, minus row two. So let me say 2 times row one, and I'm going to minus row two. So 2 times 1 minus 2 is 0, which is exactly what I wanted there. 2 times 1 minus 1 is 1. 2 times 1 minus 4 is minus 2. 2 times 1 minus 3 is minus 1. All right, now let me see if I can zero out this guy here. So what can I do? I could do any combination, anything that essentially zeroes this guy out. But I want to minimize my number of negative numbers. So let me take this third row, minus 3 times this first row. So I'm going take minus 3 times that first row and add it to this third row. So 3 minus 3 times 1 is 0. These are just going to be a bunch of 3's. 4 minus 3 times 1 is 1. 1 minus 3 times 1 is minus 2. And 2 minus 3 times 1 is minus 1." + }, + { + "Q": "At 1:11 Sal said the word \" trend\". What does trend mean?", + "A": "trend means pattern basicly", + "video_name": "mFftY8Y_pyY", + "timestamps": [ + 71 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:10, is Sal explaining the Triangle Inequality Theorem from geometry?", + "A": "Well, sort of. Maybe the opposite, because he is explaining WHY the sum of two vectors can t be greater than the sum of their magnitudes, using that Theorem, not trying to prove it.", + "video_name": "0t8W4JFpP2M", + "timestamps": [ + 130 + ], + "3min_transcript": "Sal:Let's say that we have three vectors, vectors A, B and C, and we know that vector A plus vector B is equal to vector C. Now given this I have some interesting questions. Can you construct a scenario where the magnitude of vector C is equal to the magnitude of vector A plus the magnitude of vector B? And can you also, using potentially different vectors A and B, construct a scenario where the magnitude of vector C is greater than the magnitude of vector A plus the magnitude of vector B? I encourage you to pause this video right now and try to do that. Try to come up with some vectors A and B so that sum is equal to the sum of the magnitudes. And also see if you can come up with some vectors A and B so that if you take the sum of the vectors that the magnitude of the sum is actually greater than the sum of the magnitude. see if you can come up with that. I'm assuming you've given a go at it, and potentially you've gotten a little bit frustrated, especially with the second one. The only way- Let's actually just draw some vectors. If you have vector A like this, and let's say vector B looks something like that, then A plus B ... We can just shift this over, copy and paste. A plus B is going to look like this. A plus B, or vector C I guess we could say, is going to look like that. If you have a triangle, one side cannot be longer than the sum of the other two sides. Think about it. If you wanted this to be longer what you could try to do is maybe change vector B in a way so you're pushing it further and further out. Maybe if you change your vector B a little bit you could get this vector C to be longer and longer. Maybe if you made your vector B like this. Maybe your vector B would look something like this. Now your vector C is getting pretty long, but it's still shorter than the sum of these two sides. To make it equal to the sum of these two sides you essentially have to make these two vectors go in the exact same direction. To make it equal you have to have vector A looking like this." + }, + { + "Q": "Can we write that A is a subset of B? This is at 2:25 in the video.", + "A": "No. A has more elements than set B. This is denoted by other term called Superset . A is a superset of B.", + "video_name": "1wsF9GpGd00", + "timestamps": [ + 145 + ], + "3min_transcript": "Let's define ourselves some sets. So let's say the set A is composed of the numbers 1. 3. 5, 7, and 18. Let's say that the set B-- let me do this in a different color-- let's say that the set B is composed of 1, 7, and 18. And let's say that the set C is composed of 18, 7, 1, and 19. Now what I want to start thinking about in this video is the notion of a subset. So the first question is, is B a subset of A? And there you might say, well, what does subset mean? Well, you're a subset if every member of your set is also a member of the other set. So we actually can write that B is a subset-- this is a subset-- B is a subset of A. B is a subset. So let me write that down. B is subset of A. Every element in B is a member of A. Now we can go even further. We can say that B is a strict subset of A, because B is a subset of A, but it does not equal A, which means that there are things in A that are not in B. So we could even go further and we could say that B is a strict or sometimes said a proper subset of A. And the way you do that is, you could almost imagine that this is kind of a less than or equal sign, and then you kind of cross out this equal part of the less than or equal sign. So this means a strict subset, which means everything that is in B is a member A, but everything that's in A is not a member of B. So let me write this. This is B. B is a strict or proper subset. In fact, every set is a subset of itself, because every one of its members is a member of A. We cannot write that A is a strict subset of A. This right over here is false. So let's give ourselves a little bit more practice. Can we write that B is a subset of C? Well, let's see. C contains a 1, it contains a 7, it contains an 18. So every member of B is indeed a member C. So this right over here is true. Now, can we write that C is a subset? Can we write that C is a subset of A? Can we write C is a subset of A?" + }, + { + "Q": "At 3:36, Sal said (x-2)^2 is \"always positive\" but I think it could be zero too. So shouldn't it be always non-negative?", + "A": "Yes, it should be always non-negative .", + "video_name": "dfoXtodyiIA", + "timestamps": [ + 216 + ], + "3min_transcript": "" + }, + { + "Q": "At 2:45 how was x^2-4x+4 equal to (x-2)^2? Please answer. Thanks.", + "A": "To factor the quadratic, you need to find 2 factors of +4 (the last term) that also add to -4. The 2 factors are: -2 (-2) This creates the factors of (x-2)(x-2) or (x-2)^2 If this doesn t make any sense, then I recommend you go to Algebra 1 and review the section of lessons on Polynomial Factoring. Hope this helps.", + "video_name": "dfoXtodyiIA", + "timestamps": [ + 165 + ], + "3min_transcript": "" + }, + { + "Q": "At about 4:52, why does he use a fraction and a number for r?", + "A": "When you multiply same bases, you add exponents, so 4/3 + (4)1/2 = 10/3 which is an improper fraction, but to make it a proper fraction, we get 3 1/3. So 3 sets of 3 rs come out of the cubed root and one r stays in which he does not get to until the very end of the video. He does the same thing with s when he gets 8.5 which is the same as 8 1/2.", + "video_name": "4F6cFLnAAFc", + "timestamps": [ + 292 + ], + "3min_transcript": "We're taking their square roots, so let's simplify those. So this 4 to the 1/2, that's the same thing as 2. We're taking the principal root of 4. 5 to the 1/2? Well, we can't take the square root of that, so let's just write that as the square root of 5. r to the fourth to the 1/2. There's two ways you can think about it. 4 times 1/2 is 2. So this is r squared. Or you could say the square root of r to the fourth is r squared. So this is r squared. Similarly, the square root of s to the fourth or s to the 1/2 is also s squared. And then this s to the 1/2, let's just write that as the square root of s. Just like that. Let's see what else we can do here. Let me write these other terms. We have an r to the 4/3 squared times s squared times the square root of s. Now, a couple of things we can do here. We could combine these s terms. Let's do that. Actually, just write the 2 out front first. So let's write the 2 out front first. So you have 2 times. Now let's look at these two s terms over here. We have s to the sixth times s squared. When someone says to simplify it, there's multiple interpretations for it. But we'll just say s to the sixth times s squared. That's s to the eighth. 6 plus 2. Times s to the eighth power. Times-- now this one's interesting and we might want to break it up depending on what we consider to be truly We have r to the 4/3 times r squared. r to the 4/3 is the same thing as r to the 1 and 1/3. That's what 4/3 is. So 1 and 1/3 plus 2 is 3 and 1/3. That's a little inconsistent. Over here I'm adding a fraction. Over here with the s I kind of left out the s to the 1/2 from the s's here. But we could play around with it and all of those would be valid expressions. So we've already dealt with the 2. We've already dealt with these two s's. We've already dealt with these r's. And then you have the square root of 5 times the And we could merge them if we want, but I won't do it just yet. Times the square root of 5 times the square root of s. Now there's two ways we could do it. We might not like having a fractional exponent here. And then we could break it out. Or we might want to take this guy and merge it with the eighth power. Because you know that this is the same thing as s to the 1/2. So let's do it both ways. So if we wanted to merge all of the exponents, we could write this as 2 times s to the eighth times s to the 1/2. So s to the eighth and s to the 1/2. That would be 2 times s to the 8-- I can even" + }, + { + "Q": "At 1:20, why does sal use theta instead of a, b, c, etc?", + "A": "We use theta for angles in math. It is not so important now, but when you take trigonometry, you will use it all the time.", + "video_name": "b0U1NxbRU4w", + "timestamps": [ + 80 + ], + "3min_transcript": "Let's say we have a circle, and then we have a diameter of the circle. Let me draw my best diameter. This right here is the diameter of the circle or it's a diameter of the circle. That's a diameter. Let's say I have a triangle where the diameter is one side of the triangle, and the angle opposite that side, it's vertex, sits some place on the circumference. So, let's say, the angle or the angle opposite of this diameter sits on that circumference. So the triangle looks like this. The triangle looks like that. What I'm going to show you in this video is that this triangle is going to be a right triangle. The 90 degree side is going to be the side that is opposite this diameter. I don't want to label it just yet because that would ruin the fun of the proof. Well, we have in our tool kit the notion of an inscribed angle, it's relation to a central angle that subtends the same arc. So let's look at that. So let's say that this is an inscribed angle right here. Let's call this theta. Now let's say that that's the center of my circle right there. Then this angle right here would be a central angle. Let me draw another triangle right here, another line right there. This is a central angle right here. This is a radius. This is the same radius -- actually this distance is the same. But we've learned several videos ago that look, this angle, this inscribed angle, it subtends this arc up here. The central angle that subtends that same arc is going to be twice this angle. We proved that several videos ago. So this is going to be 2theta. Now, this triangle right here, this one right here, this is an isosceles triangle. I could rotate it and draw it like this. If I flipped it over it would look like that, that, and then the green side would be down like that. And both of these sides are of length r. This top angle is 2theta. So all I did is I took it and I rotated it around to draw it for you this way. This side is that side right there. Since its two sides are equal, this is isosceles, so these to base angles must be the same. That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle. Now let me see, I already used theta, maybe I'll use x for these angles. So this has to be x, and that has to be x." + }, + { + "Q": "at 3:48 how did he get ninety out of 180-2 theta? im confused!!", + "A": "thank you! that just made it alot easier! :)", + "video_name": "b0U1NxbRU4w", + "timestamps": [ + 228 + ], + "3min_transcript": "Now, this triangle right here, this one right here, this is an isosceles triangle. I could rotate it and draw it like this. If I flipped it over it would look like that, that, and then the green side would be down like that. And both of these sides are of length r. This top angle is 2theta. So all I did is I took it and I rotated it around to draw it for you this way. This side is that side right there. Since its two sides are equal, this is isosceles, so these to base angles must be the same. That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle. Now let me see, I already used theta, maybe I'll use x for these angles. So this has to be x, and that has to be x. Well, x plus x plus 2theta have to equal 180 degrees. They're all in the same triangle. So let me write that down. We get x plus x plus 2theta, all have to be equal to 180 degrees, or we get 2x plus 2theta is equal to 180 degrees, or we get 2x is equal to 180 minus 2theta. Divide both sides by 2, you get x is equal to 90 minus theta. So x is equal to 90 minus theta. Now let's see what else we could do with this. Well we could look at this triangle right here. This triangle, this side over here also has this distance right here is also a radius of the circle. a radius of a circle. So once again, this is also an isosceles triangle. These two sides are equal, so these two base angles have to be equal. So if this is theta, this is also going to And actually, we use that information, we use to actually show that first result about inscribed angles and the relation between them and central angles subtending So if this is theta, that's theta because this is an isosceles triangle. So what is this whole angle over here? Well it's going to be theta plus 90 minus theta. That angle right there's going to be theta plus 90 minus theta. Well, the thetas cancel out. So no matter what, as long as one side of my triangle is the diameter, and then the angle or the vertex of the angle opposite sits opposite of that side, sits on the circumference, then this angle right here is going to be a" + }, + { + "Q": "0:52 why do you factor", + "A": "So that we can manipulate all the other variables in an equation to find the value of a missing variable.", + "video_name": "NuccqpiUHrk", + "timestamps": [ + 52 + ], + "3min_transcript": "Simplify the cube root of 125 x to the sixth y to the third power. So taking the cube root of something is the same thing as raising that something to the 1/3 power. So this is equal to 125 x to the sixth y to the third power raised to the 1/3 power. And if we take a product of a bunch of stuff and raise that to the 1/3 power, that's the same thing as individually raising each of the things to the 1/3 power and then taking the product. So this is going to be equal to 125 to the 1/3 power times x to the sixth to the 1/3 power times y to the third to the 1/3 power. And then we can think about how we can simplify each of these. What's 125 to the 1/3? Well, let's just factor and see if we can have at least three prime factors of something and maybe more than one prime factor that shows up three times. So 125 is 5 times 25. So 125 really is 5 times 5 times 5. So if you multiply 5 times itself three times you get 125. 125 to the 1/3 power is going to be 5. So this is going to simplify to 5 times. And then x to the sixth to the 1/3 power-- we saw this in a previous example-- if you raise a base to an exponent and then raise that whole thing to another exponent, you can take the product of the two exponents. So 6 times 1/3 is 6/3 or 2. So this part right over here simplifies to x to the sixth divided by 3 power or x squared. And then finally over here, same principle. Raising y to the third power, and then that to the 1/3 power. So that's going to be y to the 3 times 1/3 power, or y to the first power. And then times y. And if you don't want to write this little multiplication here, you could just write this as 5x squared y. And we have simplified." + }, + { + "Q": "At around 12:00 , Sal said that the functions are the inverse of each other. I understand what inverses are, since I've learned it in Algebra videos, but how can I tell that they're inverses? Is there any way to prove it? Thanks.", + "A": "We define arcsin, arccos, and arctan (also known as sin\u00e2\u0081\u00bb\u00c2\u00b9, cos\u00e2\u0081\u00bb\u00c2\u00b9, and tan\u00e2\u0081\u00bb\u00c2\u00b9) to be the inverses of sine, cosine, and tangent functions respectively. There is no proof for this as it is something we defined. It s like asking for a proof that a square has 4 sides.", + "video_name": "G-T_6hCdMQc", + "timestamps": [ + 720 + ], + "3min_transcript": "the denominator right over here is just going to be three. So that we've rationalized a square root of three over three. Fair enough. Now lets use the same triangle to figure out the trig ratios for the sixty degrees, since we've already drawn it. so what is... what is the sine of the sixty degrees? and i think you're hopefully getting the hang of it now. Sine is opposite over adjacent. soh from the \"soh cah toa\". for the sixty degree angle what side is opposite? what opens out into the two square roots of three, so the opposite side is two square roots of three, and from the sixty degree angle the adj-oh sorry its the opposite over hypotenuse, don't want to confuse you. so it is opposite over hypotenuse so it's two square roots of three over four. four is the hypotenuse. so it is equal to, this simplifies to square root of three over two. What is the cosine of sixty degrees? cosine of sixty degrees. adjacent is the two sides, right next to the sixty degree angle. So it's two over the hypotenuse which is four. So this is equal to one-half and then finally, what is the tangent? what is the tangent of sixty degrees? Well tangent, \"soh cah toa\". Tangent is opposite over adjacent opposite the sixty degrees is two square roots of three two square roots of three and adjacent to that adjacent to that is two. Adjacent to sixty degrees is two. So its opposite over adjacent, two square roots of three over two which is just equal to the square root of three. And I just wanted to -look how these are related- the sine of thirty degrees is the same as the cosine of sixty degrees. The cosine of 30 degrees is the same thing as the sine of 60 degrees and i think if you think a little bit about this triangle it will start to make sense why. we'll keep extending this and give you a lot more practice in the next few videos." + }, + { + "Q": "At 2:04 cant 12/45 also be divided by 2?", + "A": "No. 45 is a odd number, therefore if you did 45/2 you would get 22 with a remainder of 1", + "video_name": "_btQus9HV_I", + "timestamps": [ + 124 + ], + "3min_transcript": "Let's try to evaluate 7 and 6/9 minus 3 and 2/5. So like always, I like to separate out the whole number parts from the fractional parts. This is the same thing as 7 plus 6/9 minus 3 minus 2/5. And the reason why I'm saying minus 3 minus 2/5 is this is the same thing as minus 3 plus 2/5. And so you distribute the negative sign. You're subtracting a 3, and then you're subtracting the 2/5. And so now we can worry about the whole number parts, 7 minus 3. Well, 7 minus 3 is going to give us 4. So that's going to give us 4. And then we're going to have 6/9 minus 2/5. So let me think about what 6/9 minus 2/5 are. 6/9 minus 2/5, well, we're going to have to find a common denominator. So this is going to be the same thing. And I think the least common multiple of 9 and 5 is going to be 45. They have no common factors. So it's going to be over 45. To go from 9 to 45, I have to multiply by 5. So I'm going to have to multiply the numerator by 5. So 6 times 5 is 30. Then I'm going to subtract. To go from 5 to 45, I had to multiply by 9. So I have to multiply the numerator by 9 if I don't want to change the value. So 2 times 9 is 18. And 30/45 minus 18/45 is going to be something over 45. 30 minus 18 is 12. If I subtract these two fractions right over here, I get 12/45. So it's 4 plus 12/45. Or if we wanted to write it as a mixed number, this is equal to 4 and 12/45. But we're not done yet. We can simplify this further. 12 and 45 have common factors. They're both divisible by 3. I think we can divide more after that. If we divide the numerator by 3 and the denominator by 3, we end up with 4. And 12 divided by 3 is 4. And 45 divided by 3 is 15. 4 and 4/15. And actually, we're done. These two can't be simplified anymore. 4 and 4/15." + }, + { + "Q": "At 2:25 , 8-3=5. How is it negative? Is it because we're moving towards left and conventionally, it is negative. I need a conceptual clarity there. Thank you :)", + "A": "Yes, it is because in order to move from 8 to 3 on the x-axis, you have to go backwards on the coordinate plane, which means that it will be negative.", + "video_name": "v_W-aaB1irs", + "timestamps": [ + 145 + ], + "3min_transcript": "- [Voiceover] I have some example problems here from our equivalent vectors exercise on Khan Academy, so let's go through these and like always, pause the video and see if you can work through them on your own. So this first one says, \"Are vectors u and w equivalent?\" And so we can see vector u here in blue and vector w right over here. And we have to remember a vector is defined by both having a magnitude and a direction. And so for two vectors to be equivalent they have to have the same magnitude and the same direction. So when I look at these two, they are clearly pointing in different directions. Vector u is pointing to the bottom right, that's the direction it's pointing in, and vector w is pointing to the bottom left, so they definitely aren't equivalent. So they are not equivalent, scratch that out. So they have different directions, different directions. Different directions. If I just look at the length of the arrows, just eyeballing it, they look pretty close. Let me verify that. So if I'm starting at the initial point for vector u, how much do I move in the x direction? Well in the x direction I go from negative eight to negative three, so I could say my change in x is positive five, my x increases by five as I go from the initial point, from the x coordinate of the initial point to the x coordinate of the terminal point, so that length. The magnitude of just the x component is five. And let's see what happens in the y direction. So in the y direction, I start at y equals negative two, right over here, and then I go down to y equals negative eight. So my change in y, change in y, Also think about this one over here. What's my change in x? What's my change in x? Well I'm starting at x equals eight, and I am going to x equals three. So my change in x is negative five. I'm gonna write that, change in x is equal to, this is my change in x, change in x. And then what's my change in y? Well I start at y is equal to eight and I go down y is equal to two. Do it just like that. So my change in y is equal to negative six. Now based on the changes in x's and y's, I can figure out the magnitude of each of these vectors." + }, + { + "Q": "In 4:06 Khan draws his graph from the bottom to the top. Wouldn't it make more sense if he were to draw it from the positive region to the negative region? Considering the fact that Theta is greater than -3pi/2 and has a less obtuse angle measurement than -pi. Meaning if you draw Theta counterclockwise, wouldn't this be a more reasonable array?", + "A": "Since the angle is negative, it makes sense to me to go in the negative direction, clockwise. There is no negative region , but an angle can be in a negative or positive direction, regardless of where it is on the unit circle.", + "video_name": "soIt2TwV6Xk", + "timestamps": [ + 246 + ], + "3min_transcript": "So it's the positive or negative square root of 3 over 2. But how do we know which one of these it actually is? Well, that's where this information becomes useful. Let's draw our unit circle. If you're saying, well, why am I even worried about cosine of theta? Well, if you know sine of theta you know cosine of theta. Tangent of theta is just sine of theta over cosine theta. So then you will know the tangent of theta. But let's look at the unit circle to figure out which value of cosine we should use. So let me draw it, the unit circle. That's my y-axis. That is my x-axis. And I will draw the unit circle in pink. So that's my best attempt at drawing a circle. Please forgive me for its lack of perfect roundness. And it says theta is greater than negative 3 pi over 2. So let's see. This is negative pi over 2. So this is one side of the angle. Let me do this in a color. So this one side of the angle is going to be along the positive x-axis. And we want to figure out where the other side is. So this right over here that's negative pi over 2. This is negative pi. So it's between negative pi, which is right over here. So let me make that clear. Negative pi is right over here. It's between negative pi and negative 3 pi over 2. Negative 3 pi over 2 is right over here. So our angle theta is going to put us someplace over here. And the whole reason I did this-- so this whole arc right here-- you could think of this And the whole reason I did that is to think about whether the cosine of theta is going to be positive or negative. We clearly see it's in the second quadrant. The cosine of theta is the x-coordinate of this point where our angle intersects the unit circle. So this point right over here-- actually let me do it in that orange color again-- this right over here, that is the cosine of theta. Now is that a positive or negative value? Well it's clearly a negative value. So for the sake of this example, our cosine theta is not a positive 1. It is a negative 1 So we could write the cosine theta is equal to the negative square root of 3 over 2. So we figured out cosine theta, but we still have to figure out tangent of theta. And we just have to remind ourselves that the tangent of theta is going to be equal to the sine of theta over the cosine of theta." + }, + { + "Q": "Can someone explain what \"sin\" is at around 9:40 -- ish?\nIs there a video on \"sin\"? If there isn't it would be helpful\nto add one. Thanks!", + "A": "sin stands for the sine function, which is an elementary trigonometric function. It s sort of hard to explain here, but basically: Given a right triangle with one of the acute angles as a , sin(a) would be the ratio of the side opposite to a over the hypotenuse. There are several videos on trigonometry, I think.", + "video_name": "UmiZK6Hgm6c", + "timestamps": [ + 580 + ], + "3min_transcript": "So that is equal to 2. So what trig ratio is the ratio of an angle's opposite side to hypotenuse? So some of you all might get tired of me doing this all the time, but SOH CAH TOA. SOH-- sin of an angle is equal to the opposite over the hypotenuse. So let me scroll down a little bit. I'm running out of space. So the sin of this angle right here, the sin of 60 degrees, is going to be equal to the opposite side, is going to be equal to a/2, over the hypotenuse, which is our radius-- over 2. Which is equal to a/2 divided by 2 is a/4. And what is sin of 60 degrees? And if the word \"sin\" looks completely foreign to you, watch the first several videos on the trigonometry playlist. sin of 60 degrees you might remember from your 30-60-90 triangles. So let me draw one right there. So that is a 30-60-90 triangle. If this is 60 degrees, that is 30 degrees, that is 90. You might remember that this is of length 1, this is going to be of length 1/2, and this is going to be of length square root of 3 over 2. So the sin of 60 degrees is opposite over hypotenuse. Square root of 3 over 2 over 1. sin of 60 degrees. If you don't have a calculator, you could just use this-- is square root of 3 over 2. So this right here is square root of 3 over 2. Now we can solve for a. Square root of 3 over 2 is equal to a/4. Let's multiply both sides by 4. So you get this 4 cancels out. This becomes a 2. This becomes a 1. You get a is equal to 2 square roots of 3. We're in the home stretch. We just figured out the length of each of these sides. We used Heron's formula to figure out the area of the triangle in terms of those lengths. So we just substitute this value of a into there to get our actual area. So our triangle's area is equal to a squared. What's a squared? That is 2 square roots of 3 squared, times the square root of 3 over 4. We just did a squared times the square root of 3 over 4. This is going to be equal to 4 times 3 times the square of 3 over 4. These 4's cancel. So the area of our triangle we got is 3 times the square root of 3. So the area here is 3 square roots of 3." + }, + { + "Q": "At 7:50 how can you know that it is a right triangle? How do you know that the bisector of the obtuse angle is perpendicular to the base?", + "A": "Starting from the 60-60-60 triangle, when you bisect one of the angles, you end up with two 30-60-90 triangles because you created the 30 and the 60 stays the same which means the other has to be 90. In the case that he uses, he has bisected two of the sixty degree angles to create a 30-30-120 angle. When you bisect the 120 degree angle, you end up with two 60 degree angle, and the 30 degree angle does not change, so you still end up with a 30-60-90 triangle.", + "video_name": "UmiZK6Hgm6c", + "timestamps": [ + 470 + ], + "3min_transcript": "Well, they're going to be 60 degrees. I'm bisecting that angle. That is 60 degrees, and that is 60 degrees right there. And we know that I'm splitting this side in two. This is an isosceles triangle. This is a radius right here. Radius r is equal to 2. This is a radius right here of r is equal to 2. So this whole triangle is symmetric. If I go straight down the middle, this length right here is going to be that side divided by 2. That side right there is going to be that side divided by 2. Let me draw that over here. If I just take an isosceles triangle, any isosceles triangle, where this side is equivalent to that side. Those are our radiuses in this example. And this angle is going to be equal to that angle. If I were to just go straight down this angle right here, I would split that opposite side in two. So these two lengths are going to be equal. In this case if the whole thing is a, each of these are going to be a/2. trigonometry to find the relationship between a and r. Because if we're able to solve for a using r, then we can then put that value of a in here and we'll get the area of our triangle. And then we could subtract that from the area of the circle, and we're done. We will have solved the problem. So let's see if we can do that. So we have an angle here of 60 degrees. Half of this whole central angle right there. If this angle is 60 degrees, we have a/2 that's opposite to this angle. So we have an opposite is equal to a/2. And we also have the hypotenuse. Right? This is a right triangle right here. You're just going straight down, and you're bisecting that opposite side. This is a right triangle. So we can do a little trigonometry. Our opposite is a/2, the hypotenuse is equal to r. So that is equal to 2. So what trig ratio is the ratio of an angle's opposite side to hypotenuse? So some of you all might get tired of me doing this all the time, but SOH CAH TOA. SOH-- sin of an angle is equal to the opposite over the hypotenuse. So let me scroll down a little bit. I'm running out of space. So the sin of this angle right here, the sin of 60 degrees, is going to be equal to the opposite side, is going to be equal to a/2, over the hypotenuse, which is our radius-- over 2. Which is equal to a/2 divided by 2 is a/4. And what is sin of 60 degrees? And if the word \"sin\" looks completely foreign to you, watch the first several videos on the trigonometry playlist." + }, + { + "Q": "at 6:55, Sal says that when it biomes pi/2, the slope becomes infinity. But, tan(\u00ce\u00b8) is always one, since its on the unit circle. so even if it does reach pi/2 or beyond that, wouldn't it still be equal to one? I don't understand why the graph does not look like a regular sine or cosine graph. Does that mean a tangent graph cannot go beyond pi/2? What does he mean by reaching infinity?", + "A": "tan(\u00ce\u00b8) is equal to sin(\u00ce\u00b8)/cos(\u00ce\u00b8), when the angle approaches \u00cf\u0080/2: - sin(\u00ce\u00b8) starts to approach 1. - cos(\u00ce\u00b8) starts to approach zero. Since the denominator goes to zero, the function goes to infinity. The tangent line is 1 (i.e., equal to the radius of the unit circle) just when sin(\u00ce\u00b8) and cos(\u00ce\u00b8) have the same magnitude (the sides of the triangle have the same length). That happens at \u00ce\u00b8 = \u00cf\u0080/4. At \u00ce\u00b8 = \u00cf\u0080/2 it s undefined (because it never reaches infinity) and so at \u00ce\u00b8 = -\u00cf\u0080/2, when it goes to negative infinity.", + "video_name": "FK6-tZ5D7xM", + "timestamps": [ + 415 + ], + "3min_transcript": "Tangent of negative pi over four is negative one. Now let's think of it. Right now, if you just saw that, you might say, \"Oh, maybe this is some type of a line,\" but we'll see very clearly it's not a line because what happens as our angle gets closer and closer to, as our angle gets closer and closer to pi over two, what happens to the slope of this line? So that is theta. We're getting closer and closer to pi over two. This ray, I guess I should say, is getting closer and closer to approaching the vertical, so its slope is getting more and more and more positive, and if you go all the way to pi over two, the slope at that point is really undefined but it's approaching, one way to think about it is it is approaching infinity. So as you get closer and closer to pi over two, so I'm going to make a ... I'm going to draw essentially a vertical asymptote right over here at pi over two I guess one way we could think about it, it's approaching infinity there, so this is going to be looking something like this. It's going to be looking something like this. The slope of the ray as you get closer and closer to pi over two is getting closer and closer to infinity. What happens when the angle is getting closer and closer to negative pi over two? Is getting closer and closer to negative pi over two? Well, then, the slope is getting more and more and more negative. It's really approaching negative. It's approaching negative infinity. So let me draw that. Once again, not quite defined right over there, we have a vertical asymptote, and we are approaching negative infinity. We are approaching negative infinity. That's what the graph of tangent of theta looks just over this section of, I guess we could say the theta axis, but then we could keep going. Then we could keep going because so let's say we've just crossed pi over two, so we went right across it, now what is the slope? What is the slope of this thing? Well, the slope of this thing is hugely negative. It looks almost like what I just drew down here. It's hugely negative. So then the graph jumps back down here, and it's hugely negative again. It's hugely negative. And then as we increase our theta, as we increase our theta, it becomes less and less and less negative all the way to when we go to, what is this ... all the way until we go to ... let me plot this ... this angle right over here. Now what is this angle? This, well, I haven't told you yet. Let's say that this angle right over here is three pi over four. Now why did I pick three pi over four? Because that is pi over two plus pi over four. Or you could say two pis over for plus another pi over four" + }, + { + "Q": "in 0:10 what if the denominators were not the same? could that even happen?", + "A": "Later you will learn to make denominators that are not the same the same. You will multiply the fractions to make them equal in order to subtract. For example: 3/4 - 1/2 The denominators are not the same, so you must make them equal by multiplying 1/2 by 2/2 to end up with: 3/4 - 2/4 And now you can subtract and get: 1/4", + "video_name": "UbUdyE1_b9g", + "timestamps": [ + 10 + ], + "3min_transcript": "We're asked to subtract and simplify the answer, and we have 8/18 minus 5/18. So subtracting fractions is very similar to adding fractions. If we have the same denominator, the denominator in the difference is going to be the same as the denominators in the two numbers that we're subtracting, so it's going to be 18. And our numerator is going to be the difference between the numerators. So in this case, it is 8 minus 5, and this will be equal to 3 over 18, which is the answer, but it's not completely simplified, because both 3 and 18 are divisible by 3. So let's divide them both by 3. So you divide 3 by 3, you divide 18 by 3, and you get 3 divided by 3 is 1. 18 divided by 3 is 6, so you get 1/6. And just to see this visually, let me draw 18 parts. Let me draw 18 parts here. So it might be a little bit of a messy drawing. I'll try the best I can. So let me draw six in this direction. We have another three, so that's six parts. And then let me split this into three columns. So there we go. We have 18 parts. Now 8/18 is equal to one, two, three, four, five, six, seven, eight. That's 8/18. And now we want to subtract five of the eighteenths, so we subtract one, two, three, four, five. Now, what do we have left over? Well, we have three of the eighteenths left over, so you have that right there. You have three of the eighteenths left over. Now, if you turn three of the eighteenths into one piece, how many of those bigger pieces do you have? This is one of those big pieces. Now, where are the other ones? Well, this is another big piece right here. This is another big piece right here, another one, If you had 18 pieces and you merged three of the pieces into one, then you actually end up with only six pieces. You end up with six pieces. Hopefully, you see that each row is one of the pieces now, and the blue is exactly one of the six, so 3/18 is the same as 1/6." + }, + { + "Q": "At 2:46, what are those two arrows doing on the lines?Also, how come the shape isn't a irregular quadrilateral?", + "A": "Hi Jonathan! The arrows signify that the two lines that Sal drew are parallel. The figure drawn has 1 pair of parallel sides, therefore it is a trapezoid. Yes, it is irregular, because the sides and angles are not all equal.", + "video_name": "-nufZ41Kg5c", + "timestamps": [ + 166 + ], + "3min_transcript": "So the line looks like this. So every time we increase our x by 1, we decrease our y by 1. So the line looks something like this. y is equal to 3 minus x. Try to draw it relatively, pretty carefully. So that's what it looks like. y is equal to 3 minus x. So that's my best attempt at drawing it. y is equal to 3 minus x. So the quadrilateral is left unchanged by reflection over this. So that means if I were to reflect each of these vertices, I would, essentially, end up with one of the other vertices on it, and if those get reflected you're going to end up with one of these so the thing is not going to be different. So let's think about where these other two vertices of this quadrilateral need to be. So this point, let's just reflect it over this line, over y is equal to 3 minus x. So if we were to try to drop a perpendicular to this line-- notice, we have gone diagonally across one, need to go diagonally across three of them on the left-hand side. So one, two, three gets us right over there. This is the reflection of this point across that line. Now, let's do the same thing for this blue point. To drop a perpendicular to this line, we have to go diagonally across two of these squares. So let's go diagonally across two more of these squares just like that to get to that point right over there. And now we've defined our quadrilateral. Our quadrilateral looks like this. Both of these lines are perpendicular to that original line, so they're going to have the same slope. So that line is parallel to that line over there. And then we have this line and then we have this line. So what type of quadrilateral is this? Well, I have one pair of parallel sides," + }, + { + "Q": "at 4:55 it gets cofusing what does he meen?", + "A": "If you look on the right hand side you ll see that he has written 291 x 6 = 1746 So if 291 x 6 = 1746 Then just add a zero to both sides, and you get 261 x 60 = 17460 and then...add another zero to both sides and it becomes 291 x 600 = 174600", + "video_name": "omFelSZvaJc", + "timestamps": [ + 295 + ], + "3min_transcript": "with one, two, three, four zeros to get this number, 8.73 million. So I have to multiply it by 30,000. But I got that straight from this idea, that 291 times 3 is 873. So let's subtract this right over here. Let's subtract this, 2 minus 0 is 2. 5, 9, 3, 7 minus 3 is 4. 8 minus 7 is 1. 9 minus 8 is 1. So now we're left with 1,143,952. So which of these just gets us right under that? So let's see. If we want to go to-- we can't go straight to 1,746, that will be too big over here. We might want to do 873 again. But this time, we're going to do it 873,000. That is equal to 3-- and then you have one, two, three zeros, 3 times 291 is 873. 3,000 is 873,000. Let me write this a little bit neater. My handwriting is-- so this is going to be 3,000 times 291. And just let me make sure. This is a 2 right over here. 2 minus 0 is 2. And then you subtract again. 2 minus 0 is 2. 5 minus 0 is 5. 9 minus 0 is 9. 3 minus 3 is 0. And then you have 4 minus 7. So the way I like to do it when I have to start regrouping and borrowing is making sure I go from the left. So this 1, I could borrow from there, so that this becomes an 11. And then the 4, I can borrow 1 from here, so that becomes a 10. And then this becomes a 14. So 14 minus 7 is 7. 10 minus 8 is 2. So I'm down to 270,952. So it seems that we can get pretty close if we do 291 times 6, so if you do a 1,746 and then add two zeros to it. This is going to be times 6 with two zeros, so this is times 600. Once again, you subtract. And let's say I'm only using the sixes and the threes, because I figured those out ahead of time, so I didn't have to do any extra math. So 2 minus 0 is 2. 5 minus 0 is 5. 9 minus 6 is 3. 0 minus 4-- well, there's a couple of ways you could think about doing this. You could borrow from here. That will become a 6. This becomes a 10. 10 minus 4 is 6. Now this one's lower, so it has to borrow as well. Make this into a 16. 16 minus 7-- and I have multiple videos on how to borrow, if I'm doing that part too fast. But the idea here is to show you a different way of long division. So 16 minus 7 is 9. So now we're at 96,352." + }, + { + "Q": "2:30 what is the difference that the brackets made does it mean anything", + "A": "The brackets basically mean that you other parentheses inside of them, and so you solve whats inside of them. They are the same thing as parentheses, they just look different, so when you are solving the equation you don t get all of the parentheses mixed up. A case to use brackets would be here: a((-b+c)(d-e)(f+g)) Instead you would have: a[(-b+c)(d-e)(f+g)]", + "video_name": "gjrGd9TjjnY", + "timestamps": [ + 150 + ], + "3min_transcript": "Simplify negative 1 times this expression in brackets, negative 7 plus 2 times 3 plus 2 minus 5, in parentheses, squared. So this is an order of operations problem. And remember, order of operations, you always want to do parentheses first. Parentheses first. Then you do exponents. Exponents. And there is an exponent in this problem right over here. Then you want to do multiplication. Multiplication and division. And then finally, you do addition and subtraction. So let's just try to tackle this as best we can. So first, let's do the parentheses. We have a 3 plus 2 here in parentheses, so we can evaluate that to be equal to 5. And let's see, we could do other things in other parts of this expression that won't affect what's going on right here in the parentheses. We have this negative 5 squared. We want to do the exponent before we worry about it being subtracted. So this 5 squared over here we can rewrite as 25. And so let's not do too many steps at once. So this whole thing will simplify to negative 1. And then in brackets, we have negative 7 plus 2 times 5. And then, 2 times 5. And then close brackets. Minus 25. Now, this thing-- we want to do multiplication. You could say, hey, wait. I still have a parentheses here. Why don't I do that first? But when you just evaluate what's inside of this parentheses, you just get a negative 7. It doesn't really change anything. So we can just leave this here as a negative 7. And this expression. We do want to evaluate this whole expression before we I mean, we could distribute this negative 1 and all of that, but let's just do straight up order of operations here. So let's evaluate this expression. So we get 2 times 5 right over there. 2 times 5 is 10. That is 10. So our whole expression becomes-- and normally, you wouldn't have to rewrite the expression this many times. But we're going to do it this time just to make sure no one gets confused. So it becomes negative 1 times negative 7 plus 10. Plus 10. And we close our brackets. Minus 25. Now, we can evaluate this pretty easily. Negative 7 plus 10. We're starting at negative 7. So I was going to draw a number line there. So we're starting-- let me draw a number line. So we're starting at negative 7. So the length of this line is negative 7. And then, we're adding 10 to it. We're adding 10 to it. So we're going to move 10 to the right. If we move 7 to the right, we get back to 0. And then we're going to go another 3 after that." + }, + { + "Q": "At 4:24, isnt -3-24 just negative three beside negative twenty-four?", + "A": "-3-24 can be rewritten as -3 + (-24).", + "video_name": "gjrGd9TjjnY", + "timestamps": [ + 264 + ], + "3min_transcript": "So we get 2 times 5 right over there. 2 times 5 is 10. That is 10. So our whole expression becomes-- and normally, you wouldn't have to rewrite the expression this many times. But we're going to do it this time just to make sure no one gets confused. So it becomes negative 1 times negative 7 plus 10. Plus 10. And we close our brackets. Minus 25. Now, we can evaluate this pretty easily. Negative 7 plus 10. We're starting at negative 7. So I was going to draw a number line there. So we're starting-- let me draw a number line. So we're starting at negative 7. So the length of this line is negative 7. And then, we're adding 10 to it. We're adding 10 to it. So we're going to move 10 to the right. If we move 7 to the right, we get back to 0. And then we're going to go another 3 after that. So that gets us to positive 3. Another way to think about it is we are adding integers of different signs. We can view the sum as going to be the difference of the integers. And since the larger integer is positive, our answer will be positive. So you could literally just view this as 10 minus 7. 10 minus 7 is 3. So this becomes a 3. And so our entire expression becomes negative 1. Negative 1 times. And just to be clear, brackets and parentheses are really the same thing. Sometimes people will write brackets around a lot of parentheses just to make it a little bit easier to read. But they're really just the same thing as parentheses. So these brackets out here, I could just literally write them like that. And then I have a minus 25 out over here. Now, once again, you want to do multiplication or division before we do addition and subtraction. So let's multiply the negative 1 times 3 is negative 3. So negative 3 minus 25. We are adding two integers of the same sign. We're already at negative 3 and we're going to become 25 more negative than that. So you can view this as we're moving 25 more in the negative direction. Or you could view it as 3 plus 25 is 28. But we're doing it in the negative direction, so it's negative 28. So this is equal to negative 28. And we are done." + }, + { + "Q": "At 6:55 he missed a traingle", + "A": "He figured it out, just keep watching.", + "video_name": "qG3HnRccrQU", + "timestamps": [ + 415 + ], + "3min_transcript": "So we can assume that s is greater than 4 sides. Let's say I have an s-sided polygon, and I want to figure out how many non-overlapping triangles will perfectly cover that polygon. How many can I fit inside of it? And then I just have to multiply the number of triangles times 180 degrees to figure out what are the sum of the interior angles of that polygon. So let's figure out the number of triangles as a function of the number of sides. So once again, four of the sides are going to be used to make two triangles. So those two sides right over there. And then we have two sides right over there. I can draw one triangle over-- and I'm not even going to talk about what happens on the rest of the sides of the polygon. You could imagine putting a big black piece of construction paper. There might be other sides here. I'm not going to even worry about them right now. So out of these two sides I can draw one triangle, just like that. Out of these two sides, I can draw another triangle right over there. So four sides used for two triangles. I've already used four of the sides, but after that, if I have all sorts of craziness here. I could have all sorts of craziness here. Let me draw it a little bit neater than that. So I could have all sorts of craziness right over here. It looks like every other incremental side I can get another triangle out of it. So that's one triangle out of there, one triangle out of that side, one triangle out of that side, one triangle out of that side, and then one triangle out of this side. So for example, this figure that I've drawn is a very irregular-- one, two, three, four, five, six, seven, eight, nine, 10. Is that right? One, two, three, four, five, six, seven, eight, nine, 10. It is a decagon. And in this decagon, four of the sides were used for two triangles. So I got two triangles out of four of the sides. And out of the other six sides I was These are six. This is one, two, three, four, five. Actually, let me make sure I'm counting the number of sides right. So I have one, two, three, four, five, six, seven, eight, nine, 10. So let me make sure. Did I count-- am I just not seeing something? Oh, I see. I actually didn't-- I have to draw another line right over These are two different sides, and so I have to draw another line right over here. I can get another triangle out of that right over there. And so there you have it. I have these two triangles out of four sides. And out of the other six remaining sides I get a triangle each. So plus six triangles. I got a total of eight triangles. And so we can generally think about it. The first four, sides we're going to get two triangles. So let me write this down. So our number of triangles is going to be equal to 2." + }, + { + "Q": "sooo at 3:13 does the mean that the sum of interior angles are just how many triangles are in the polygon or hexagon?", + "A": "Yes, the sum of the interior angles is = 180 * (number of triangles polygon is divided into)", + "video_name": "qG3HnRccrQU", + "timestamps": [ + 193 + ], + "3min_transcript": "that a plus b plus c is equal to 180 degrees. And then if we call this over here x, this over here y, and that z, those are the measures of those angles. We know that x plus y plus z is equal to 180 degrees. And so if we want the measure of the sum of all of the interior angles, all of the interior angles are going to be b plus z-- that's two of the interior angles of this polygon-- plus this angle, which is just going to be a plus x. a plus x is that whole angle. The whole angle for the quadrilateral. Plus this whole angle, which is going to be c plus y. And we already know a plus b plus c is 180 degrees. And we know that z plus x plus y is equal to 180 degrees. So I think you see the general idea here. We just have to figure out how many triangles we can divide something into, and then we just multiply by 180 degrees since each of those triangles will have 180 degrees. Let's do one more particular example. And then we'll try to do a general version where we're just trying to figure out how many triangles can we fit into that thing. So let me draw an irregular pentagon. So one, two, three, four, five. So it looks like a little bit of a sideways house there. Once again, we can draw our triangles inside of this pentagon. So that would be one triangle there. That would be another triangle. So I'm able to draw three non-overlapping triangles that perfectly cover this pentagon. This is one triangle, the other triangle, and the other one. And we know each of those will have 180 degrees if we take the sum of their angles. And we also know that the sum of all of those interior angles of the polygon as a whole. And to see that, clearly, this interior angle is one of the angles of the polygon. This is as well. But when you take the sum of this one and this one, then you're going to get that whole interior angle of the polygon. And when you take the sum of that one and that one, you get that entire one. And then when you take the sum of that one plus that one plus that one, you get that entire interior angle. So if you take the sum of all of the interior angles of all of these triangles, you're actually just finding the sum of all of the interior angles of the polygon. So in this case, you have one, two, three triangles. So three times 180 degrees is equal to what? 300 plus 240 is equal to 540 degrees. Now let's generalize it. And to generalize it, let's realize that just to get our first two triangles, we have to use up four sides. We have to use up all the four sides in this quadrilateral. We had to use up four of the five sides--" + }, + { + "Q": "At 0:03, how could there be negative degrees?", + "A": "Instead of the degrees going counter-clockwise (for positive), the degrees go clockwise from 0 degrees to get negative degrees (i.e reverse from positive).", + "video_name": "O3jvUZ8wvZs", + "timestamps": [ + 3 + ], + "3min_transcript": "- [Instructor] We're asked to convert 150 degrees and negative 45 degrees to radians. Let's think about the relationship between degrees and radians, and to do that, let me just draw a little circle here. So that's the center of the circle, and then do my best shot, best attempt to freehand draw a reasonable-looking circle. That's not, I've done worse than that. Alright, now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that? We know that that would be 360 degrees. If we did the same thing, how many radians is that, if we were to go all the way around the circle? We just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle. So if you go all the way around, you're really talking about the arc length the circumference of the circle. And you're essentially saying, how many radius's this is, or radii, or how many radii is the circumference of the circle. You know a circumference of a circle is two pi times the radius, or you could say that the length of the circumference of the circle is two pi radii. If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi. That just comes from the, really, actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle. If we were to go all the way around this, this is also two pi radians. That tells us that two pi radians, as an angle measure, is the exact same thing, 360 degrees. And then we can take all of this relationship and manipulate it in different ways. If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case, you are left with, if you divide both sides by two, you are left with pi radians is equal to 180 degrees. How can we use this relationship now to figure out what 150 degrees is? Well, this relationship, we could write it in different ways. We could divide both sides by 180 degrees, and we could get pi radians over 180 degrees" + }, + { + "Q": "I don't get what he starts doing at 1:05, can someone explain please?", + "A": "The 1, the 3 and the 9 are all digits in the tens column. They are actually counting tens. 1 ten is 10, 3 tens are 30 and 9 tens are 90. He is just explaining what the numbers are actually representing. Comment again if this doesn t make it clearer for you.", + "video_name": "Wm0zq-NqEFs", + "timestamps": [ + 65 + ], + "3min_transcript": "Let's add 536 to 398. And we're going to do it two different ways so that we really understand what this carrying is all about. So first, we'll do it in the more traditional way. We start in the ones place. We say, \"Well, what's 6 + 8?\" Well, we know that 6 + 8 is equal to 14. And so when we write it down here in the sum, we could say, \u201c \"Well look. The 4 is in the ones place.\u201d So it's equal to 4 + 1 ten.\" So let's write that 1 ten in the tens place. And now we focus on the tens place. We have 1 ten + 3 tens + 9 tens. So, what's that going to get us? 1 + 3 + 9 is equal to 13. Now we have to remind ourselves that this is 13 tens. Or another way of thinking about it, this is 3 tens and 1 hundred. You might say, \"Wait, wait! How does that make sense?\" When we're adding 1 ten + 3 tens + 9 tens, we're actually adding 10 + 30 + 90, and we're getting 130. And so we're putting the 30 (the 3 in the tens place represents the 30) \u2013 So this is the 3. The 3 represents the 30. And then we're placing this 1 in the hundreds place. 10 tens is equal to 100. And now we're adding up the numbers in the hundreds place. 1 + 5 + 3 is equal to \u2013 let's see. 1 + 5 is equal to 6, + 3 is equal to 9. But we have to remind ourselves: this is 9 hundreds. This is in the hundreds place. So this is actually 1 hundred. So this is actually 1 hundred + 5 hundreds + 3 hundreds, is equal to 9 hundreds. 100 + 500 + 300 is equal to 900. And we're done. This is equal to 934." + }, + { + "Q": "Adding three digit numbers like Sall (0:01) is simple, but is there a trick to adding faster in your mind? Thanks.", + "A": "Watch the videos for Regrouping.", + "video_name": "Wm0zq-NqEFs", + "timestamps": [ + 1 + ], + "3min_transcript": "Let's add 536 to 398. And we're going to do it two different ways so that we really understand what this carrying is all about. So first, we'll do it in the more traditional way. We start in the ones place. We say, \"Well, what's 6 + 8?\" Well, we know that 6 + 8 is equal to 14. And so when we write it down here in the sum, we could say, \u201c \"Well look. The 4 is in the ones place.\u201d So it's equal to 4 + 1 ten.\" So let's write that 1 ten in the tens place. And now we focus on the tens place. We have 1 ten + 3 tens + 9 tens. So, what's that going to get us? 1 + 3 + 9 is equal to 13. Now we have to remind ourselves that this is 13 tens. Or another way of thinking about it, this is 3 tens and 1 hundred. You might say, \"Wait, wait! How does that make sense?\" When we're adding 1 ten + 3 tens + 9 tens, we're actually adding 10 + 30 + 90, and we're getting 130. And so we're putting the 30 (the 3 in the tens place represents the 30) \u2013 So this is the 3. The 3 represents the 30. And then we're placing this 1 in the hundreds place. 10 tens is equal to 100. And now we're adding up the numbers in the hundreds place. 1 + 5 + 3 is equal to \u2013 let's see. 1 + 5 is equal to 6, + 3 is equal to 9. But we have to remind ourselves: this is 9 hundreds. This is in the hundreds place. So this is actually 1 hundred. So this is actually 1 hundred + 5 hundreds + 3 hundreds, is equal to 9 hundreds. 100 + 500 + 300 is equal to 900. And we're done. This is equal to 934." + }, + { + "Q": "What does carry mean, at 0:16?", + "A": "When you add, you carry by putting numbers more than 10 to the top so it can be easier to solve.", + "video_name": "Wm0zq-NqEFs", + "timestamps": [ + 16 + ], + "3min_transcript": "Let's add 536 to 398. And we're going to do it two different ways so that we really understand what this carrying is all about. So first, we'll do it in the more traditional way. We start in the ones place. We say, \"Well, what's 6 + 8?\" Well, we know that 6 + 8 is equal to 14. And so when we write it down here in the sum, we could say, \u201c \"Well look. The 4 is in the ones place.\u201d So it's equal to 4 + 1 ten.\" So let's write that 1 ten in the tens place. And now we focus on the tens place. We have 1 ten + 3 tens + 9 tens. So, what's that going to get us? 1 + 3 + 9 is equal to 13. Now we have to remind ourselves that this is 13 tens. Or another way of thinking about it, this is 3 tens and 1 hundred. You might say, \"Wait, wait! How does that make sense?\" When we're adding 1 ten + 3 tens + 9 tens, we're actually adding 10 + 30 + 90, and we're getting 130. And so we're putting the 30 (the 3 in the tens place represents the 30) \u2013 So this is the 3. The 3 represents the 30. And then we're placing this 1 in the hundreds place. 10 tens is equal to 100. And now we're adding up the numbers in the hundreds place. 1 + 5 + 3 is equal to \u2013 let's see. 1 + 5 is equal to 6, + 3 is equal to 9. But we have to remind ourselves: this is 9 hundreds. This is in the hundreds place. So this is actually 1 hundred. So this is actually 1 hundred + 5 hundreds + 3 hundreds, is equal to 9 hundreds. 100 + 500 + 300 is equal to 900. And we're done. This is equal to 934." + }, + { + "Q": "I'm sorry, I still don't get how Sal solved the problem around 5:00.\nThey are \"fundamentally different ratios\"...what does that mean?", + "A": "He s comparing the 5 to 1 and 4 to 1 ratios of y to x, and saying that they have different slopes. Therefore, the two lines must intersect somewhere at one point. If you ve watched enough videos on here, you ll notice that Sal frequently (over)uses the word fundamentally, to just mean certainly or definitely. He didn t mean anything special by the use of the word fundamentally here.", + "video_name": "SuB1gkto9LU", + "timestamps": [ + 300 + ], + "3min_transcript": "let's see if we can think about what types of solutions we might find. So let's take this down. So they say determine how many solutions exist for the system of equations. So you have 10x minus 2y is equal to 4, and 10x minus 2y is equal to 16. So just based on what we just talked about the x's and the y's are on the same side of the equation and the ratio is 10 to negative 2. Same ratio. So something strange is going to happen here. But when we have the same kind of combination of x's and y's in the first one we get 4, and on the second one we get 16. So that seems a little bit bizarre. Another way to think about it, we have the same number of x's, the same number of y's but we got a different number on the right hand side. So if you were to simplify this, and we could even look at the hints to see what it says, you'll see that you're going to end up with the same slope but different y-intercepts. So we convert both the slope intercept form right over here and the green one is y is equal to 5x minus 8. Same slope, same ratio between the x's and the Y's, but you have different values right over here. You have different y-intercepts. So here you have no solutions. That is this scenario right over here if you were to graph it. So no solutions, check our answer. Let's go to the next question. So let's look at this one right over here. So we have negative 5 times x and negative 1 times y. We have 4 times x and 1 times y. So it looks like the ratio if then we're looking at the x's and y's always on the left hand side right over here, it looks like the ratios of x's and y's are different. You have essentially 5 x's for every one y, or you could say negative 5 x's for every negative 1 y, and here you have 4 x's for every 1 y. So this is fundamentally a different ratio. are going to intersect in exactly one place. If you were to put this into slope intercept form, you will see that they have different slopes. So you could say this has one solution and you can check your answer. And you could look at the solution just to verify. And I encourage you to do this. So you see the blue one if you put in the slope intercept form negative 5x plus 10 and you take the green one into slope intercept form negative 4x minus 8. So different slopes, they're definitely going to intersect in exactly one place. You're going to have one solution. Let's try another one. So here we have 2x plus y is equal to negative 3. And this is pretty clear, you have 2x plus y is equal to negative 3. These are the exact same equations. So it's consistent information, there's definitely solutions. But there's an infinite number of solutions right over here. This is a dependent system. So there are infinite number of solutions here and we can check our answer. Let's do one more because that was a little bit too easy. OK so this is interesting right over here, we have it in different forms. 2x plus y is equal to negative 4," + }, + { + "Q": "As stupid as this question might sound, I don't exactly understand why at 1:54 Sal says that the angle is theta as well. Like how? (I understand the parts after that, but I can't seem to figure out why we are taking it as theta. Forgive me, if I've missed something and hence have the doubt; I like thorough explanations of even the simplest things because I tend to get muddled up quite often). Thanks!", + "A": "The yellow ray is the reflection of the green ray over the \u00f0\u009d\u0091\u00a6-axis, which means that if the green ray forms the angle \u00f0\u009d\u009c\u0083 with the positive \u00f0\u009d\u0091\u00a5-axis, then the yellow ray forms the angle \u00f0\u009d\u009c\u0083 with the negative \u00f0\u009d\u0091\u00a5-axis.", + "video_name": "tzQ7arA917E", + "timestamps": [ + 114 + ], + "3min_transcript": "Voiceover:Let's explore the unit circle a little bit more in depth. Let's just start with some angle theta, and for the sake of this video, we'll assume everything is in radians. This angle right over here, we would call this theta. Now let's flip this, I guess we could say, the terminal ray of this angle. Let's flip it over the X and Y-axis. Let's just make sure we have labeled our axes. Let's flip it over the positive X-axis. If you flip it over the positive X-axis, you just go straight down, and then you go the same distance on the other side. You get to that point right over there, and so you would get this ray. You would get this ray that I'm attempting to draw in blue. You would get that ray right over there. Now what is the angle between this ray and the positive X-axis if you start at the positive X-axis? Well, just using our conventions that counterclockwise from the X-axis is a positive angle, this is clockwise. Instead of going theta above the X-axis, so we would call this, by our convention, an angle of negative theta. Now let's flip our original green ray. Let's flip it over the positive Y-axis. If you flip it over the positive Y-axis, we're going to go from there all the way to right over there then we can draw ourselves a ray. My best attempt at that is right over there. What would be the measure of this angle right over here? What was the measure of that angle in radians? We know if we were to go all the way from the positive X-axis to the negative X-axis, that would be pi radians because that's halfway around the circle. This angle, since we know that that's theta, this is theta right over here, the angle that we want to figure out, this is going to be all the way around. It's going to be pi minus, Notice, pi minus theta plus theta, these two are supplementary, and they add up to pi radians or 180 degrees. Now let's flip this one over the negative X-axis. If we flip this one over the negative X-axis, you're going to get right over there, and so you're going to get an angle that looks like this, that looks like this. Now what is going to be the measure of this angle? If we go all the way around like that, what is the measure of that angle? To go this far is pi, and then you're going another theta. This angle right over here is theta, so you're going pi plus another theta. This whole angle right over here, this whole thing, this whole thing is pi plus theta radians. Pi plus theta, let me just write that down. This is pi plus theta. Now that we've figured out" + }, + { + "Q": "At 3:56 why isn't the (x,y ) coordinates are (cos theta, sin theta)? Why is the angle taken as (pi - theta) ?", + "A": "You are almost right. Where the yellow ray hits the circle, the ( x, y ) co-ordinates could either be labelled as Sal does or as ( - cos theta, sin theta). Since the x-co-ordinate is in a negative direction, cosine theta has to be negative. This gives us two of the many trig identities : cos ( pi - theta ) = - cos theta sin ( pi - theta ) = sin theta", + "video_name": "tzQ7arA917E", + "timestamps": [ + 236 + ], + "3min_transcript": "Notice, pi minus theta plus theta, these two are supplementary, and they add up to pi radians or 180 degrees. Now let's flip this one over the negative X-axis. If we flip this one over the negative X-axis, you're going to get right over there, and so you're going to get an angle that looks like this, that looks like this. Now what is going to be the measure of this angle? If we go all the way around like that, what is the measure of that angle? To go this far is pi, and then you're going another theta. This angle right over here is theta, so you're going pi plus another theta. This whole angle right over here, this whole thing, this whole thing is pi plus theta radians. Pi plus theta, let me just write that down. This is pi plus theta. Now that we've figured out let's think about how the sines and cosines of these different angles relate to each other. We already know that this coordinate right over here, that is sine of theta, sorry, the X-coordinate is cosine of theta. The X-coordinate is cosine of theta, and the Y-coordinate is sine of theta. Or another way of thinking about it is this value on the X-axis is cosine of theta, and this value right over here on the Y-axis is sine of theta. Now let's think about this one down over here. By the same convention, this point, this is really the unit circle definition of our trig functions. This point, since our angle is negative theta now, this point would be cosine of negative theta, comma, sine of negative theta. And we can apply the same thing over here. This point right over here, the X-coordinate is cosine of pi minus theta. from the positive X-axis. This is cosine of pi minus theta. And the Y-coordinate is the sine of pi minus theta. Then we could go all the way around to this point. I think you see where this is going. This is cosine of, I guess we could say theta plus pi or pi plus theta. Let's write pi plus data and sine of pi plus theta. Now how do these all relate to each other? Notice, over here, out here on the right-hand side, our X-coordinates are the exact same value. It's this value right over here. So we know that cosine of theta must be equal to the cosine of negative theta. That's pretty interesting. Let's write that down. Cosine of theta is equal to ... let me do it in this blue color," + }, + { + "Q": "At 4:34, is there no horizontal asymptote in this expression? Can anyone explain why? I thought a rational expression must have an asymptote.", + "A": "Why would a rational expression have to have an asymptote? Do you have a reason for thinking this? Let s say x=a, a being any possible real number. What s (x+1)/(x+1)? Try to get the answer to this by yourself. (x+1)/(x+1) is always = 1 for any possible x, so it can t have a vertical asymptote. Does it have a horizontal one? Well, it always = 1, but that s not an asymptote, because the graph doesn t approach f(x) = 1, because it s already 1 everywhere.", + "video_name": "ReEMqdZEEX0", + "timestamps": [ + 274 + ], + "3min_transcript": "would be equal to 1. You would say this would be equal to 1 when x does not equal negative 1 or when these terms don't equal zero. It equals 0/0, which we don't know what that is, when x is equal to negative 1. So in this situation, you would not have a vertical asymptote. So this graph right here, no vertical asymptote. And actually, you're probably curious, what does this graph look like? I'll take a little aside here to draw it for you. This graph right here, if I had to graph this right there, what this would be is this would be y is equal to 1 for all the values except for x is equal to negative 1. So in this situation the graph, it would be y is equal to 1 everywhere, except for y is equal to negative 1. So we actually have a hole there. We actually draw a little circle around there, a little hollowed-out circle, so that we don't know what y is when x is equal to negative 1. So this looks like that right there. It looks like that horizontal line. No vertical asymptote. And that's because this term and that term cancel out when they're not equal to zero, when x is not equal to negative 1. So when your identifying vertical asymptotes-- let me clear this out a little bit. when you're identifying vertical asymptotes, you want to be sure that this expression right here isn't canceling out with something in the numerator. And in this case, it's not. In this case, it did, so you don't have a vertical asymptote. In this case, you aren't canceling out, so this will define a vertical asymptote. x is equal to negative 1 is a vertical asymptote for this graph right here. So x is equal to negative 1-- let me draw the vertical And then to figure out what the graph is doing, we could try out a couple of values. So what happens when x is equal to 0? So when x is equal to 0 we have 2 times 0, which is 0 over 0 plus 1. So it's 0/1, which is 0. So the point 0, 0 is on our curve. What happens when x is equal to 1? We have 2 times 1, which is 2 over 1 plus 1. So it's 2/2. So it's 1, 1 is also on our curve. So that's on our curve right there. So we could keep plotting points, but the curve is going to look something like this. It looks like it's going approach negative infinity as it approaches the vertical asymptote from the right. So as you go this way it, goes to negative infinity. And then it'll approach our horizontal asymptote from the" + }, + { + "Q": "how did sal get rid of the cube root and the ^3? do they cancel each other out? 4:27\nis there a video explaining ^^^?", + "A": "The cube root of a number is another number that, when you multiply it by itself three times, gives you the original number. For example: The cube root of 27 is 3, because 3*3*3 = 27 if a\u00c2\u00b3 = x, then \u00e2\u0088\u009bx = a If we substitute, we have \u00e2\u0088\u009ba\u00c2\u00b3, which is just a. Sal s example had \u00e2\u0088\u009b7\u00c2\u00b3 which is just 7 (the number he used in the next step of the problem).", + "video_name": "8y7xP4zz0UY", + "timestamps": [ + 267 + ], + "3min_transcript": "like this that doesn't seem to be divisible by a lot of things, it's always a good idea to try things like 7, 11, 13. Because those tend to construct very interesting numbers. So let's see if this is divisible by 7. So if I take 343 and if I want to divide it by 7, 7 goes into 30-- it doesn't go into 3-- 7 goes into 34 four times. 4 times 7 is 28. Subtract, 34 minus 28 is 6. Bring down a 3. 7 goes into 63 nine times. 9 times 7 is 63. Subtract. We don't have any remainder. And I forgot to do that last step up here. 3 times 15 is 15. Subtract, no remainder. It went in exactly. So here, 343 can be factored into 7 and 49. And 49 might jump out at you. It can be factored into 7 times 7. I can rewrite all of this here-- the cube root of 3,430-- now as the cube root of-- I'm just going to write it in its factored form-- 2 times 5 times-- I could write 7 times 7 times 7, or I could write times 7 to the third power. That captures these three 7's right over here. I have three 7's, and then I'm multiplying them together. So that's 7 to the third power. And from our exponent properties, we know that this is the exact same thing as the cube root of 2 times 5 times the cube root-- so let me do that in that same, just so we see what colors we're dealing with. So the cube root of 2 times 5, which is the cube root of 10, times the cube root-- and I think Keeping track of the colors is the hard part. And the cube root of 10, we just leave it as 10. We know the prime factorization of 10 is 2 times 5, so you're not going to just get a very simple integer You would get some decimal answer here, but here you get a very clear integer answer. The cube root of 7 to the third, well, that's just going to be 7. So this is just going to be 7. So our entire thing simplifies. This is equal to 7 times the cube root of 10. And this is about as simplified as we can get just using hand arithmetic. If you want to get the exact number here, you're probably best off using a calculator." + }, + { + "Q": "why does he say at 2:06 that 343 isn't divisible by 2 when the sum of 3+4+3 is 10 which is divisible by 2?", + "A": "adding the digits is only a test for divisibility for 3 and 9", + "video_name": "8y7xP4zz0UY", + "timestamps": [ + 126 + ], + "3min_transcript": "Let's see if we can find the cube root of 3,430. And if you're like me, it doesn't jump out of your mind what number times that same number times that same number-- if you have three of those numbers and you were to multiply them together-- would be equal to 3,430. So what I'm going to do is to try to prime factorize this to find all the prime factors of 3,430 and see if any of those prime factors show up at least three times. And that'll help us with this. So 3,430-- it's clearly divisible by 5 and 2, or it's divisible by 10. So let's do that. So first we can divide it by 2. It's 2 times-- let's see. 3,430 divided by 2 is 1,715. Then we can divide it by 5, as well. We can factor 1,715 into 5 and-- let me do a little bit of long division on the side here. So if I have 1,715, and I'm going to divide it by 5. It goes into 17 three times. 3 times 5 is 15. Subtract, you get 2, and then you bring down a 1. 5 goes into 21 four times. 4 times 5 is 20. Subtract. Bring down the 5. 5 goes into 15 three times, so it goes exactly 343 times. So 1,715 can be factored into 5 times 343. Now, 343 might not jump out at you as a number that is easy to factor. It's clearly an odd number, so it won't be divisible by 2. Its digits add up to 10, which is not divisible by 3. So this isn't going to be divisible by 3. It's not going to be divisible by 4, because it's not divisible by 2. It's not going to be divisible by 5. If it wasn't divisible by 3 or 2, it's not going to be divisible by 6. And now we get to 7. like this that doesn't seem to be divisible by a lot of things, it's always a good idea to try things like 7, 11, 13. Because those tend to construct very interesting numbers. So let's see if this is divisible by 7. So if I take 343 and if I want to divide it by 7, 7 goes into 30-- it doesn't go into 3-- 7 goes into 34 four times. 4 times 7 is 28. Subtract, 34 minus 28 is 6. Bring down a 3. 7 goes into 63 nine times. 9 times 7 is 63. Subtract. We don't have any remainder. And I forgot to do that last step up here. 3 times 15 is 15. Subtract, no remainder. It went in exactly. So here, 343 can be factored into 7 and 49. And 49 might jump out at you. It can be factored into 7 times 7." + }, + { + "Q": "at 1:01 why are the fours at the bottom of the denominator not negative I am confused :/", + "A": "a negative exponent is not applied to the coefficient, it just flips the exponential with a negative exponent to the other side of the divide line and thus makes the exponent positive. It is not (-4)^-3 power which would have three negative fours.", + "video_name": "CZ5ne_mX5_I", + "timestamps": [ + 61 + ], + "3min_transcript": "- [Narrator] Let's get some practice with our exponent properties, especially when we have integer exponents. So, let's think about what four to the negative three times four to the fifth power is going to be equal to. And I encourage you to pause the video and think about it on your own. Well there's a couple of ways to do this. See look, I'm multiplying two things that have the same base, so this is going to be that base, four. And then I add the exponents. Four to the negative three plus five power which is equal to four to the second power. And that's just a straight forward exponent property, but you can also think about why does that actually make sense. Four to the negative 3 power, that is one over four to the third power, or you could view that as one over four times four times four. And then four to the fifth, that's five fours being multiplied together. So it's times four times four times four times four times four. And so notice, when you multiply this out, and three fours in the denominator. And so, three of these in the denominator with three of these in the numerator. And so you're going to be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have A to the negative fourth power times A to the, let's say, A squared. What is that going to be? Well once again, you have the same base, in this case it's A, and so since I'm multiplying them, you can just add the exponents. So it's going to be A to the negative four plus two power. Which is equal to A to the negative two power. And once again, it should make sense. This right over here, that is one over A times A times A times A times A times A, so that cancels with that, that cancels with that, and you're still left with one over A times A, which is the same thing as A to the negative two power. Now, let's do it with some quotients. So, what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So, this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent and so, this is going to be equal to 12 to the, subtracting a negative is the same thing as adding the positive," + }, + { + "Q": "( 3:37 ) Could (12^-7) / (12^-5) be written as (12^5) / (12^7) ?", + "A": "Yes it could because the exponents are negative. Correct!", + "video_name": "CZ5ne_mX5_I", + "timestamps": [ + 217 + ], + "3min_transcript": "times A times A, so that cancels with that, that cancels with that, and you're still left with one over A times A, which is the same thing as A to the negative two power. Now, let's do it with some quotients. So, what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So, this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent and so, this is going to be equal to 12 to the, subtracting a negative is the same thing as adding the positive, And once again, we just have to think about, why does this actually make sense? Well, you could actually rewrite this. 12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power. If we take the reciprocal of this right over here, you would make exponent positive and then you would get exactly what we were doing in those previous examples with products. And so, let's just do one more with variables for good measure. Let's say I have X to the negative twentieth power divided by X to the fifth power. Well once again, we have the same base and we're taking a quotient. So, this is going to be X to the negative 20 minus five cause we have this one right over here in the denominator. So, this is going to be equal to X to the negative twenty-fifth power. as X to the negative twentieth and having an X to the fifth in the denominator dividing by X to the fifth is the same thing as multiplying by X to the negative five. So here you just add the exponents and once again you would get X to the negative twenty-fifth power." + }, + { + "Q": "In 1:35 how does he get 8 1/3?", + "A": "He divided 100 by 12, which is 8 with a remainder of 4. The remainder can be written as 4/12, which can be reduced to 1/3, so 8 1/3", + "video_name": "jOZ98FDyl2E", + "timestamps": [ + 95 + ], + "3min_transcript": "- [Voiceover] Let's get some practice comparing and computing rates. So they tell us the pet store has three fish tanks, each holding a different volume of water and a different number of fish. So Tank A has five fish, and it has 40 liters of water, Tank B, 12 fish, 100 liters of water, and Tank C, 23 fish, and it has 180 liters of water. Order the tanks by volume per fish from least to greatest. So let's think about what volume per fish, and we could think about this as volume divided by fish. Volume per fish. All right, so here for Tank A, it's going to be 40 liters for every five fish. 40 for every five fish, and let's see, 40 over five is eight, so you have eight liters per fish, is the rate at which they have to add water per fish for Tank A. Now, Tank B, you have 100 liters so what is this going to be? This is going to be, 12 goes into 100 eight times, so eight times 12 is 96, and then you have four left over. So this is going to be eight and 4/12, or eight and 1/3 liters per fish. And all I did is I converted this improper fraction, 100 over 12, to a mixed number, and I simplified it. 12 goes into 100 eight times with a remainder of four, so it's eight and 4/12, which is the same thing as eight and 1/3. And then finally in Tank C, I have 180 liters for 23 fish. So what is this going to be? 23 goes into 180. Let me try to calculate this. 23 goes into 180, it looks like it's going to be less than nine times. Is it eight times? Eight times, no, not eight times. Seven times three is 21, seven times two is 14, plus two is 16. When you subtract you get a remainder of 19. So this is going to be seven with a remainder of 19, or you could say this is seven and 19/23 liters per fish. So which one has, we're going to order the tanks by volume per fish, from least to greatest, so Tank B has the largest volume per fish, has eight and 1/3 liters per fish, so this is in first place. And then Tank A is in second place. Tank A is in second place. And then Tank C is in third place. Oh, actually we wanna go from least to greatest, so this is, let me write it this way. This is Tank C is the least, and Tank A is the greatest. So we really have to swap this order around. Now, I just copied and pasted this from the Khan Academy exercises. Let me actually bring the actual exercise up," + }, + { + "Q": "at 0:41 he's dividing them but couldn't you just write out the problem in long division too?", + "A": "Sal is working with ratios and rates which are types of fractions. This is why he is writing them initially as fractions. If you wrote the long division form first, it would work provided you got to the answers Sal has written out.", + "video_name": "jOZ98FDyl2E", + "timestamps": [ + 41 + ], + "3min_transcript": "- [Voiceover] Let's get some practice comparing and computing rates. So they tell us the pet store has three fish tanks, each holding a different volume of water and a different number of fish. So Tank A has five fish, and it has 40 liters of water, Tank B, 12 fish, 100 liters of water, and Tank C, 23 fish, and it has 180 liters of water. Order the tanks by volume per fish from least to greatest. So let's think about what volume per fish, and we could think about this as volume divided by fish. Volume per fish. All right, so here for Tank A, it's going to be 40 liters for every five fish. 40 for every five fish, and let's see, 40 over five is eight, so you have eight liters per fish, is the rate at which they have to add water per fish for Tank A. Now, Tank B, you have 100 liters so what is this going to be? This is going to be, 12 goes into 100 eight times, so eight times 12 is 96, and then you have four left over. So this is going to be eight and 4/12, or eight and 1/3 liters per fish. And all I did is I converted this improper fraction, 100 over 12, to a mixed number, and I simplified it. 12 goes into 100 eight times with a remainder of four, so it's eight and 4/12, which is the same thing as eight and 1/3. And then finally in Tank C, I have 180 liters for 23 fish. So what is this going to be? 23 goes into 180. Let me try to calculate this. 23 goes into 180, it looks like it's going to be less than nine times. Is it eight times? Eight times, no, not eight times. Seven times three is 21, seven times two is 14, plus two is 16. When you subtract you get a remainder of 19. So this is going to be seven with a remainder of 19, or you could say this is seven and 19/23 liters per fish. So which one has, we're going to order the tanks by volume per fish, from least to greatest, so Tank B has the largest volume per fish, has eight and 1/3 liters per fish, so this is in first place. And then Tank A is in second place. Tank A is in second place. And then Tank C is in third place. Oh, actually we wanna go from least to greatest, so this is, let me write it this way. This is Tank C is the least, and Tank A is the greatest. So we really have to swap this order around. Now, I just copied and pasted this from the Khan Academy exercises. Let me actually bring the actual exercise up," + }, + { + "Q": "At 1:00, where does the formula come from?", + "A": "One place it comes from is after deriving the quadratic formula, you end up with (-b \u00c2\u00b1 \u00e2\u0088\u009a(b^2-4ac))/(2a), so the -b/2a where the line of symmetry is and thus the x coordinate of the vertex and the (\u00c2\u00b1 \u00e2\u0088\u009a(b^2-4ac))/2a is the distance away from the line of symmetry of the zeroes.", + "video_name": "IbI-l7mbKO4", + "timestamps": [ + 60 + ], + "3min_transcript": "I have an equation right here. It's a second degree equation. It's a quadratic. And I know its graph is going to be a parabola. Just as a review, that means it looks something like this or it looks something like that. Because the coefficient on the x squared term here is positive, I know it's going to be an upward opening parabola. And I am curious about the vertex of this parabola. And if I have an upward opening parabola, the vertex is going to be the minimum point. If I had a downward opening parabola, then the vertex would be the maximum point. So I'm really trying to find the x value. I don't know actually where this does intersect the x-axis or if it does it all. But I want to find the x value where this function takes on a minimum value. Now, there's many ways to find a vertex. Probably the easiest, there's a formula for it. And we talk about where that comes from in multiple videos, where the vertex of a parabola or the x-coordinate of the vertex of the parabola. So the x-coordinate of the vertex And the negative b, you're just talking about the coefficient, or b is the coefficient on the first degree term, is on the coefficient on the x term. And a is the coefficient on the x squared term. So this is going to be equal to b is negative 20. So it's negative 20 over 2 times 5. Well, this is going to be equal to positive 20 over 10, which is equal to 2. And so to find the y value of the vertex, we just substitute back into the equation. The y value is going to be 5 times 2 squared minus 20 times 2 plus 15, which is equal to let's see. This is 5 times 4, which is 20, minus 40, which is negative 20, So just like that, we're able to figure out the coordinate. This coordinate right over here is the point 2, negative 5. Now it's not so satisfying just to plug and chug a formula like this. And we'll see where this comes from when you look at the quadratic formula. This is the first term. It's the x value that's halfway in between the roots. So that's one way to think about it. But another way to do it, and this probably will be of more lasting help for you in your life, because you might forget this formula. It's really just try to re-manipulate this equation so you can spot its minimum point. And we're going to do that by completing the square. So let me rewrite that. And what I'll do is out of these first two terms, I'll factor out a 5, because I want to complete a square here and I'm going to leave this 15 out to the right, because I'm going to have to manipulate that as well. So it is 5 times x squared minus 4x. And then I have this 15 out here." + }, + { + "Q": "At 2:00, what is Sal doing? I don't get how or why he adds 16...", + "A": "Sal is using a method called Completing the square . it involves an equation (M*1/2x)^2. in his example if you take M which is the Middle term(8) and divide by 2 you would get 4. then squar it and you get 16. so that is why he adds and subtracts 16. this wont change the answer at all because +16-16=0. so your adding a fancy way of saying +0", + "video_name": "IbI-l7mbKO4", + "timestamps": [ + 120 + ], + "3min_transcript": "I have an equation right here. It's a second degree equation. It's a quadratic. And I know its graph is going to be a parabola. Just as a review, that means it looks something like this or it looks something like that. Because the coefficient on the x squared term here is positive, I know it's going to be an upward opening parabola. And I am curious about the vertex of this parabola. And if I have an upward opening parabola, the vertex is going to be the minimum point. If I had a downward opening parabola, then the vertex would be the maximum point. So I'm really trying to find the x value. I don't know actually where this does intersect the x-axis or if it does it all. But I want to find the x value where this function takes on a minimum value. Now, there's many ways to find a vertex. Probably the easiest, there's a formula for it. And we talk about where that comes from in multiple videos, where the vertex of a parabola or the x-coordinate of the vertex of the parabola. So the x-coordinate of the vertex And the negative b, you're just talking about the coefficient, or b is the coefficient on the first degree term, is on the coefficient on the x term. And a is the coefficient on the x squared term. So this is going to be equal to b is negative 20. So it's negative 20 over 2 times 5. Well, this is going to be equal to positive 20 over 10, which is equal to 2. And so to find the y value of the vertex, we just substitute back into the equation. The y value is going to be 5 times 2 squared minus 20 times 2 plus 15, which is equal to let's see. This is 5 times 4, which is 20, minus 40, which is negative 20, So just like that, we're able to figure out the coordinate. This coordinate right over here is the point 2, negative 5. Now it's not so satisfying just to plug and chug a formula like this. And we'll see where this comes from when you look at the quadratic formula. This is the first term. It's the x value that's halfway in between the roots. So that's one way to think about it. But another way to do it, and this probably will be of more lasting help for you in your life, because you might forget this formula. It's really just try to re-manipulate this equation so you can spot its minimum point. And we're going to do that by completing the square. So let me rewrite that. And what I'll do is out of these first two terms, I'll factor out a 5, because I want to complete a square here and I'm going to leave this 15 out to the right, because I'm going to have to manipulate that as well. So it is 5 times x squared minus 4x. And then I have this 15 out here." + }, + { + "Q": "At 6:50, he says that a light bulb is on if it has an odd number of factors, so if the bulb number is prime, is it always off(except for 1)?", + "A": "Yes, every prime number, by definition, has exactly two factors, meaning that it will always be off.", + "video_name": "WNhxkpmVQYw", + "timestamps": [ + 410 + ], + "3min_transcript": "And that's just a fancy way of saying, look, if I'm on pass 17, I'm going to switch all the multiples of 17. Or I could say, I know that I'm going to switch light bulb 51, because 17 is a factor of 51. So that tells you that we're always going to be switching one of these light bulbs on or off when one of its factors is our pass. So for example, if we're looking at light bulb eight. This is light bulb eight. So when will it be switched? So on pass one, we're definitely going So pass one, it's going to be switched on. Pass two, it'll be switched off. Pass three, nothing's going to happen. On pass three, nothing's going to happen because this isn't a multiple of three. Pass four, what'll happen? It will be switched. It'll be switched back on. And then pass eight is the next time we'll touch this light bulb, and it'll be switched back. So every time one of its factors go by, we're going to switch this thing. And as you can see, in order for it to be on at the end, you have to have an odd number of factors. So that's an interesting thing. So in order for a light bulb to be on, it has to have odd number of factors. Now that's an interesting question. What numbers have an odd number of factors. And this is something that I think they should teach you in grade school, and they never do. But it's a really interesting kind of number theory. So what numbers are true? Let's do all the factors for some of the starting numbers. So all the factors of one. Well one, the only factor is just one. So one works? One has an odd number of factors. So that means that one will remain on. Because you're only going to turn it on in the first pass. Makes sense. Two. What are all the factors of two. Well you have one and two. So two has an even number. You're going to switch it on the first time, then off the second time. Then you're never going to touch it again. So this is going to stay off. Your factors are one and three. Four. Your factors are one, two and four. Interesting. Here we have three factors. We have an odd number of factors. So four is going to stay on. We're going to turn it on in our first pass, we're going to turn it off on our second pass. And we're going to turn it on again in our fourth pass. Let's keep going. So five." + }, + { + "Q": "8:00 is the answer all square numbers will be on?", + "A": "Yes pretty much the square numbers that are smaller than 100.", + "video_name": "WNhxkpmVQYw", + "timestamps": [ + 480 + ], + "3min_transcript": "Pass three, nothing's going to happen. On pass three, nothing's going to happen because this isn't a multiple of three. Pass four, what'll happen? It will be switched. It'll be switched back on. And then pass eight is the next time we'll touch this light bulb, and it'll be switched back. So every time one of its factors go by, we're going to switch this thing. And as you can see, in order for it to be on at the end, you have to have an odd number of factors. So that's an interesting thing. So in order for a light bulb to be on, it has to have odd number of factors. Now that's an interesting question. What numbers have an odd number of factors. And this is something that I think they should teach you in grade school, and they never do. But it's a really interesting kind of number theory. So what numbers are true? Let's do all the factors for some of the starting numbers. So all the factors of one. Well one, the only factor is just one. So one works? One has an odd number of factors. So that means that one will remain on. Because you're only going to turn it on in the first pass. Makes sense. Two. What are all the factors of two. Well you have one and two. So two has an even number. You're going to switch it on the first time, then off the second time. Then you're never going to touch it again. So this is going to stay off. Your factors are one and three. Four. Your factors are one, two and four. Interesting. Here we have three factors. We have an odd number of factors. So four is going to stay on. We're going to turn it on in our first pass, we're going to turn it off on our second pass. And we're going to turn it on again in our fourth pass. Let's keep going. So five. The factors are one, two, three and six. It's an even number, so they're going to be off when we're done with it. Seven. it's one and seven. We just did that. It's one, two, four and eight. Still going to be off. Nine. Let's see. The factors are one, three and nine. Interesting. Once again we have an odd number of factors. So the light bulb number nine is also going to be on when everything's done. Let's keep going. I don't know, I actually did this at our mental boot camp with some of the kids. And they immediately said, the distance between one and four is three. The distance between four and nine is five. And maybe the distance between nine and the next number is going to be seven. It increases by odd numbers. What's nine plus seven? 16. What are the factors of 16? They're one, two, four, eight and 16." + }, + { + "Q": "At 8:11, what is a hypersphere?", + "A": "Hypersphere is a generalization for any sphere that is more than 3D. Normally we define a sphere in 3D as x\u00c2\u00b2+y\u00c2\u00b2+z\u00c2\u00b2=1. But you can generalize this to as many dimensions as you want. You could have a 4D sphere or a 500D sphere, but in general, they are referred to as hyperspheres.", + "video_name": "iDQ1foxYf0o", + "timestamps": [ + 491 + ], + "3min_transcript": "So \"Three students attempt to define \"what a line segment is.\" And we have a depiction of a line segment right over here. We have point P, point Q, and the line segment is all the points in between P and Q. So, so let's match the teacher's comments to the definitions. Ivy's definition: \"All of the points \"in line with P and Q, extending infinitely \"in both directions.\" Well, that would be the definition of a line. That would be the line P, Q. That would be, if you're extending infinitely in both directions, so ... I would say, \"Are you thinking of a line \"instead of a line segment?\" Ethan's definition: \"The exact distance from P to Q. Well, that's just a ... that's the length of a line segment. That's not exactly what a line segment is. And see, Ebuka's definition. \"The points P and Q, which are called endpoints, \"and all of the points in a straight line Yep. That looks like a good definition for a line segment. So we can just check our, we can just check our answer. So, looking good. Let's do one more of this. I'm just really enjoying pretending to be a teacher. All right. \"Three students attempt to define \"what a circle is.\" Define what a circle is. \"Can you match the teacher's comments to the definitions?\" Duru. \"The set of all points in a plane \"that are the same distance away from some given point, \"which we call the center.\" That just seems like a pretty good definition of a circle. So, I would, you know, \"Stupendous! Well done.\" Oliver's definition. \"The set of all points in 3D space \"that are the same distance from a center point.\" If we're talking about 3D space and the set of all points that are equidistant from that point in 3D space, now we're talking about a sphere, not a circle. And so, \"You seem to be confusing \"a circle with a sphere.\" And then, finally, \"A perfectly round shape.\" But if you're talk about three dimensions, you could be talking about a sphere. If you're talking about, if you go beyond three dimensions, hypersphere, whatever else. In two dimensions, yeah, perfectly round shape, most people would call it a circle. But that doesn't have a lot of precision to it. It doesn't have, it doesn't give us a lot that we can work with from a mathematical point of view. So I would say, actually, what the teacher's saying: \"Your definition needs to be much more precise.\" Duru's definition is much, much more precise. The set of all points that are equidistant from ... in a plane, that are equidistant away from a given point, which we call the center. So yep. Carlos could use a little bit more precision. We're all done." + }, + { + "Q": "At 0:32, how does one know when to write x/y and when to write y/x?", + "A": "its just another way to write division 4/2=2/4 and here sal is just saying 14/2 =/= 2/14 so in proportions you need bigger number on top...", + "video_name": "qcz1Cm_-l50", + "timestamps": [ + 32 + ], + "3min_transcript": "- So, let's set up a relationship between the variables x and y. So, let's say, so this is x and this is y, and when x is one, y is four, and when x is two, y is eight, and when x is three, y is 12. Now, you might immediately recognize that this is a proportional relationship. And remember, in order for it to be a proportional relationship, the ratio between the two variables is always constant. So, for example, if I look at y over x here, we see that y over x, here it's four over one, which is just four. Eight over two is just four. Eight halves is the same thing as four. 12 over three it's the same thing as four. Y over x is always equal to four. In fact, I can make another column here. I can make another column here where I have y over x, here it's four over one, which is equal to four. Here it's eight over two, which is equal to four. Here it's 12 over three, which is equal to four. the ratio, the ratio between y and x is this constant four, to express the relationship between y and x as an equation. In fact, in some ways this is, or in a lot of ways, this is already an equation, but I can make it a little bit clearer, if I multiply both sides by x. If I multiply both sides by x, if I multiply both sides by x, I am left with, well, x divided by x, you'd just have y on the left hand side. Y is equal to 4x and you see that's the case. X is one, four times that is four. X is two, four times that is eight. So, here you go, we're multiplying by four. We are multiplying by four, we are multiplying by four. And so, four, in this case, four, in this case, in this situation, this is our constant of proportionality. Constant, constant, sometimes people will say proportionality constant. Now sometimes, it might even be described as a rate of change and you're like well, Sal, how is this a, how would four be a rate of change? And, to make that a little bit clearer, let me actually do another example, but this time, I'll actually put some units there. So let's say that, let's say that I have, let's say that x-- Let me do this, I already used yellow, let me use blue. So let's x, let's say that's a measure of time and y is a measure of distance. Or, let me put it this way, x is time in terms of seconds. Let me write it this way. So, x, x is going to be in seconds and then, y is going to be in meters. So, this is meters, the units, and this right over here is seconds. So, after one second, we have traveled, oh, I don't know, seven meters." + }, + { + "Q": "at 2:05, does the table of values always has to start with 0 ?", + "A": "No, sometimes the value of x and never even be 0. However, it s often easier for to start at zero if x can indeed be 0 because that way you can find the y-intercept.", + "video_name": "86NwKBcOlow", + "timestamps": [ + 125 + ], + "3min_transcript": "Create a graph of the linear equation 5x plus 2y is equal to 20. So the line is essentially the set of all coordinate, all x's and y's, that satisfy this relationship right over here. To make things simpler, what we're going to do is set up a table where we're going to put a bunch of x values in and then figure out the corresponding y value based on this relationship. But to make it a little bit simpler, I'm going to solve for y here. So it becomes easier to solve for y for any given x. So we have 5x plus 2y is equal to 20. If we want to solve for y, let's just get rid of the 5x on the left-hand side. So let's subtract 5x from both sides of this equation. The left-hand side, these guys cancel out, so we get 2y is equal to the right hand side, you have 20 minus 5x. And then you can divide both sides of this equation by 2. So you divide both sides by 2. The left-hand side, we just have a y, and then the right-hand side, we could leave it that way. That actually would be a pretty straightforward way is 10 minus 5x over 2 or minus 5/2 times x. And so now using this, let's just come up with a bunch of x values and see what the corresponding y values are, and then just plot them. So let me do this in a new color. So let me-- a slightly different shade of yellow. So we have x values, and then let's think about what the corresponding y value is going to be. So I'll start, well, I could start anywhere. I'll start at x is equal to 0, just because that tends to keep things pretty simple. If x is 0, then y is equal to 10 minus 5/2 times 0, which is equal to 5/2 times 0 is just a 0. So it's just 10 minus 0 or 10. So that gives us the coordinate, the point, 0 comma 10. When x is 0, y is 10. So x is 0. So it's going to be right here at the middle of the x-axis. And you go up 10 for the y-coordinate. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So that's the point 0 comma 10. Let's do another point. Let's say that x is 2. I'm going to pick multiples of 2 here just so that I get a nice clean answer here. So when x is 2, then y is equal to 10 minus 5/2 times 2, and the 2 in the denominator cancels out with this 2 in the numerator. So it simplifies to 10 minus 5, or just 5. So that tells us the point x equals 2, y is equal to 5, is on the line. So 2x is equal to 1, 2 right over here. And then y is equal to 5. You go up 5. 1, 2, 3, 4, 5, just like that. So that's the point 2, 5. And when you're drawing a line you actually just need two points. If you have a ruler or any kind of straight edge, we could just connect these two points." + }, + { + "Q": "0:15 why he multiple for 6?? not for 3 i got confused.", + "A": "least common multiple of x/3 and 1/6 is 6. *if you want to add to x/3 and 1/6 you need the denominator be the same.", + "video_name": "CJyVct57-9s", + "timestamps": [ + 15 + ], + "3min_transcript": "Solve for x. And we have x minus 8 is equal to x/3 plus 1/6. Now the first thing I want to do here-- and there's multiple ways to do this problem-- but what I want to do is just to simplify the fraction. I'm going to multiply everything times the least common multiple of all of these guys' denominators. This is essentially x/1. This is 8/1, x/3, 1/6. The least common multiple of 1, 3, and 6 is 6. So if I multiply everything times 6, what that's going to do is going to clear out these fractions. So these weren't fractions to begin with, so we're just multiplying them by 6. So it becomes 6x minus 6 times negative 8, or 6 times 8 is 48. And we're subtracting it right over there. And then we have x/3 times 6. Let me just write it out here. So that's going to be 6 times x/3 plus 6 times 1/6. Or we get 6x minus 48 is equal to 6 times something divided That's the same thing as 6 divided by 3 times that something. That's just going to be equal to 2x plus 6 times 1/6 or 6 divided by 6 is just going to be 1. So that first step cleared out all of the fractions and now this is just a straightforward problem with all integer coefficients or integers on either side of the equation. And what we want to do is we want to isolate all of the x's on one side or the other. And we might as well isolate them all on the left hand side. So let's subtract 2x from both sides. We want to get rid of this 2x here. That's why I'm subtracting the 2x. So let's subtract 2x from both sides. And on the right hand side, I have 2x plus 1 minus 2x. Those cancel out. That was the whole point. So I'm left with just this 1 over here. On the left hand side I have 6x minus 2x. Well, that's just going to be 4x. If I have 6 of something minus 2 of that something, I have 4 of that something. Minus 48. And now I can-- let's see, I want to get rid of this 48 So let me add 48 to both sides of the equation. I'll do this in a new color. So let me add 48 to both sides of this equation. And on the left hand side 4x minus 48 plus 48, I'm left with just a 4x. And on the right hand side, 1 plus 48 is going to be 49. And now I've isolated the x but it's still multiplied by a 4. So to make that a 1 coefficient, let's multiply both sides by 1/4. Or you could also say, let's divide both sides by 4. Anything you do to one side you have to do to the other. And so you have-- what do we have over here? 4x/4 is just x. x is equal to 49 over 4. And that's about as far as we can simplify it because these don't have any common factors, 49 and 4. Let's check to see whether 49/4 is indeed the answer. So let's put it into the original equation. Remember, the original equation is" + }, + { + "Q": "At 7:31 he did 10^200 divided by 10^50. How did he get 150 if 200 + 50 gives you 250.", + "A": "10^200/10^50 the exponent is subtracted 200-50, only multiplication is it added 200+50.", + "video_name": "kITJ6qH7jS0", + "timestamps": [ + 451 + ], + "3min_transcript": "" + }, + { + "Q": "4:16 of the video to 5:00. How does 2 with the exponent of 8 + 2 with the exponent of 8 eventually get broken down into 2 with the exponent of 9; it doesn't seem to make any since because they both would have different equations?", + "A": "this happens as 2^8 + 2^8 = 2 * 2^8 which can be simplified as 2^9", + "video_name": "kITJ6qH7jS0", + "timestamps": [ + 256, + 300 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:35, why does Sal put parentasis around the 6?", + "A": "He puts parenthesis around the 6 because that mean multiplication or simply a form of it", + "video_name": "Rcb7ZUTOQ1I", + "timestamps": [ + 95 + ], + "3min_transcript": "If we know some circle has an area of 36pi-- so it has an area of 36pi-- can we figure out what the circumference of this circle And I encourage you to pause this video, and try to think about that question. Well, from the area, we could figure out what the radius is, and then from that radius, we can figure out what its circumference is. So we know that the area, which is 36pi, is equal to pi r squared. And so if you look at it on both sides of this equation, if we divide-- let me rewrite it so it's a little bit clearer in a different color. So we could set up an equation pi r squared is equal to 36pi. Now, if we want to solve for the radius the first thing that we might want to do is divide both sides by pi. Then, we're left with r squared is equal to 36. Now, if we just solve this as a pure math equation, you might say, OK, we could take the positive and negative r could be plus or minus 6, but we need to remember that r is a distance, so we only care about the positive. So if we take the principal root of 36, we get r is equal to 6. From there, we can use this to figure out the circumference. So the circumference is equal to 2 pi r. Circumference is equal to 2 pi r. And in this case, r is equal to 6. So it's equal to 2 pi times 6, which is going to be equal to 12pi. So that's straightforward, area 36pi, we leverage pi r squared to figure out that the radius was 6, and then from that we were able to figure out that the circumference was 12pi." + }, + { + "Q": "At 6:30 Sal write dt, but I cant see where he get that from. Anyone can help? Thanks", + "A": "dt is the differential, you put it at the end of the integration expression. If you re wondering why its dt and not dx ; its because it parametric, so your functions are functions of t - f(t) - and not like the usual functions of x - f(x). Hope i could help!", + "video_name": "99pD1-6ZpuM", + "timestamps": [ + 390 + ], + "3min_transcript": "And the reason why this is the case, is if you imagine this is a, this is b, that is my f of x. When you do it this way, your dx's are always going to be positive. When you go in that direction, your dx's are always going to be positive, right? Each increment, the right boundary is going to be higher So your dx's are positive. In this situation, your dx's are negative. The heights are always going to be the same, they're always going to be f of x, but here your change in x is a negative change in x, when you go from b to a. And that's why you get a negative integral. In either case here, our path changes, but our ds's are going to be positive. And the way I've drawn this surface, it's above the x-y plane, the f of xy is also going to be positive. So that also kind of gives the same intuition that this should be the exact same area. But let's prove it to ourselves. So let's start off with our first parameterization, just We have x is equal to x of t, y is equal to y of t, and we're dealing with this from, t goes from a to b. And we know we're going to need the derivatives of these, so let write that down right now. We can write dx dt is equal to x prime of t, and dy dt, let me write that a little bit neater, dy dt is equal to y prime of t. This is nothing groundbreaking I've done so far. But we know the integral over c of f of xy. f is a scalar field, not a vector field. ds is equal to the integral from t is equal to a, to t is dt squared, which is the same thing as x prime of t squared, plus dy dt squared, the same thing as y prime of t squared. All that under the radical, times dt. This integral is exactly that, given this parameterization. Now let's do the minus c version. I'll do that in this orange color. Actually, let me do the minus c version down here. The minus the c version, we have x is equal to, you remember this, actually, just from up here, this was from the last video." + }, + { + "Q": "At around 3:19, doesn't point D also have a y value of 6?", + "A": "If you look carefully, point D is slightly below 6 and probably has a y-value of 5.5. This is later mentioned at 5:44.", + "video_name": "P3IlneCNm8A", + "timestamps": [ + 199 + ], + "3min_transcript": "So for example, f prime of 0-- which is the x value for this point right over here-- is going to be some negative value. It's the slope of the tangent line. Similarly, f prime of x, when x is equal to 4-- that's what's going on right over here-- that's going to be the slope of the tangent line. That's going to be a positive value. So if you look at all of these, where is the slope of the tangent line 0? And what does a 0 slope look like? Well, it looks like a horizontal line. So where is the slope of the tangent line here horizontal? Well, the only one that jumps out at me is point B right over here. It looks like the slope of the tangent line would indeed be horizontal right over here. Or another way you could think of it is the instantaneous rate of change of the function, right at x equals 2, looks like it's pretty close to-- if this So out of all of the choices here, I would say only B looks like the derivative at x equals 2. Or the slope of the tangent line at B, it looks like it's 0. So I'll say B right over here. And then they had this kind of crazy, wacky expression here. f of x minus 6 over x. What is that greatest in value? And we have to interpret this. We have to think about what does f of x minus 6 over x actually mean? Whenever I see expressions like this, especially if I'm taking a differential calculus class, I would say well, this looks kind of like finding the slope of a secant line. In fact, all of what we know about derivatives is finding the limiting value of the slope of a secant line. And this looks kind of like that, especially if at some point, my y value is a 6 here. And this could be the change in y value. And if the corresponding x value is 0, then this would be f of x minus 6 over x minus 0. Well, sure. When x is equal to 0, we see that f of x is equal to 6. So what this is right over here-- let me rewrite this. This we could rewrite as f of x minus 6 over x minus 0. So what is this? What does this represent? Well, this is equal to the slope-- let me do some of that color-- this is equal to the slope of the secant line between the points, x, f of x, x, and whatever the corresponding f of x is." + }, + { + "Q": "on the 30:60:90 triangle why do you square root it by three? thats the only part that lost me (:\n\nThanks,\nAshley", + "A": "It s a right triangle, so we can apply Pythagoras (a\u00c2\u00b2 + b\u00c2\u00b2 = c\u00c2\u00b2). The hypotenuse is c = 2, a = 1, and b = \u00e2\u0088\u009a3: a\u00c2\u00b2 + b\u00c2\u00b2 = c\u00c2\u00b2 1\u00c2\u00b2 + (\u00e2\u0088\u009a3)\u00c2\u00b2 = 2\u00c2\u00b2 1 + 3 = 4", + "video_name": "UKQ65tiIQ6o", + "timestamps": [ + 1860 + ], + "3min_transcript": "" + }, + { + "Q": "What do you mean of function of s in 0:16?", + "A": "He means that we will find the area of a triangle that has a side length of s, not of a specific triangle, like one with sides of length 4 or 7 or something. Working this way, the equation he finds will work for any equilateral triangle, and you just have to substitute the value of s (side length) into the equation to get the area.", + "video_name": "UKQ65tiIQ6o", + "timestamps": [ + 16 + ], + "3min_transcript": "Let's say that this triangle right over here is equilateral, which means all of its sides have the same length. And let's say that that length is s. What I want to do in this video is come up with a way of figuring out the area of this equilateral triangle, as a function of s. And to do that, I'm just going to split this equilateral in two. I'm just going to drop an altitude from this top vertex right over here. This is going to be perpendicular to the base. And it's also going to bisect this top angle. So this angle is going to be equal to that angle. And we showed all of this in the video where we proved the relationships between the sides of a 30-60-90 triangle. Well, in a regular equilateral triangle, all of the angles are 60 degrees. So this one right over here is going to be 60 degrees, let me do that in a different color. This one down here is going to be 60 degrees. This one down here is going to be 60 degrees. And then this one up here is 60 degrees, but we just split it in two. So this angle is going to be 30 degrees. And then the other thing that we know is that this altitude right over here also will bisect this side down here. So that this length is equal to that length. And we showed all of this a little bit more rigorously on that 30-60-90 triangle video. But what this tells us is well, if this entire length was s, because all three sides are going to be s, it's an equilateral triangle, then each of these, so this part right over here, is going to be s/2. And if this length is s/2, we can use what we know about 30-60-90 triangles to figure out this side right over here. So to figure out what the actual altitude is. And the reason why I care about the altitude is because the area of a triangle is 1/2 times the base times the height, or times the altitude. So this is s/2, the shortest side. The side opposite the 30 degree angle is s/2. Then the side opposite the 60 degree angle is going to be square root of 3 times that. And we know that because the ratio of the sides of a 30-60-90 triangle, if the side opposite the 30 degree side is 1, then the side opposite the 60 degree side is going to be square root of 3 times that. And the side opposite the 90 degree side, or the hypotenuse, is going to be 2 times that. So it's 1 to square root of 3 to 2. So this is the shortest side right over here. That's the side opposite the 30 degree side. The side opposite the 60 degree side is going to be square root of 3 times this. So square root of 3 s over 2. So now we just need to figure out what the area of this triangle is, using area of our triangle is equal to 1/2 times the base, times the height of the triangle. Well, what is the base of the triangle? Well, the entire base of the triangle right over here is s. So that is going to be s. And what is the height of the triangle?" + }, + { + "Q": "I am really puzzeled about the 7:26-7:35 portion of the video where Sal goes from a^2 =1/2 to\na = 1/sqrt 2....how did he do that?", + "A": "2a^2=1 --------- 2 ......2 The first 2 s cancel out and your left with: a^2 =1/2", + "video_name": "fp9DZYmiSC4", + "timestamps": [ + 446, + 455 + ], + "3min_transcript": "Well, you just look at this graph. You see there's two points of intersection. This point right over here and this point right over here. Just between 0 and 2 pi. These are cyclical graphs. If we kept going, they would keep intersecting with each other. But just over this 2 pi range for theta, you get two points of intersection. Now let's think about what they are, because they look to be pretty close between 0 and pi over 2. And right between pi and 3 pi over 2. So let's look at our unit circle if we can figure out what those values are. It looks like this is at pi over 4. So let's verify that. So let's think about what these values are at pi over 4. So pi over 4 is that angle, or that's the terminal side of it. So this is pi over 4. Pi over 4 is the exact same thing as a 45 degree angle. So we have to figure out what this point is what. What the coordinates are. So let's make this a right triangle. And so what do we know about this right triangle? And I'm going to draw it right over here, to make it a little clear. This is a typical type of right triangle. So it's good to get some familiarity with it. So let me draw my best attempt. Alright. So we know it's a right triangle. We know that this is 45 degrees. What is the length of the hypotenuse? Well this is a unit circle. It has radius 1. So the length of the hypotenuse here is 1. And what do we know about this angle right over here? Well, we know that it too must be 45 degrees, because all of these angles have to add up to 180. And since these two angles are the same, we know that these two sides are going to be the same. And then we could use the Pythagorean Theorem So using the Pythagorean Theorem, knowing that these two sides are equal, what do we get for the length of those sides? Well, if this has length a, well then this also has length a. And we can use the Pythagorean Theorem. And we could say a squared plus a squared is equal to the hypotenuse squared. Is equal to 1. Or 2a squared is equal to 1a squared, is equal to 1/2. Take the principal root of both sides. a is equal to the square root of 1/2 which is the square root of 1, which is 1, over the square root of 2. We can rationalize the denominator here by multiplying by square root of 2 over square root of 2, which gives us a is equal to-- in the numerator-- square root of 2. And in the denominator, square root of 2 times square root of 2 is 2. So this length is the square root of 2. And this length is the same thing. So this length right over here is square root of 2 over 2. And this height right over here is also square root of 2 over 2." + }, + { + "Q": "At 03:36 isnt he marking wrong? He is marking cosine on the Y axis, but on the begining of the video he said cosine of data is the X, axis! And sine was the Y axis! Now im really confused!", + "A": "I understand the confusion. When he said cosine of theta is the x-axis, he was basically saying Instead of the x-axis and integers (i.e. -2, -1, 0, 1, 2 ect...) we are going to call it the theta-axis and use radians. Also, instead of saying cos(theta) = y-axis he mixed up the measurement for the two axis. One axis has been turned into radians (theta, cis(theta)), the other has been left in integer number form. (x/y form).", + "video_name": "fp9DZYmiSC4", + "timestamps": [ + 216 + ], + "3min_transcript": "So what is cosine of theta? What's the x-coordinate here? Which is negative 1. And sine of theta is going to be the y-coordinate, which is 0. Now let's keep going. Now we're down here at 3 pi over 2. If we go all the way around to 3 pi over 2, what is this coordinate? Well this is 0, negative 1. Cosine of theta is the x-coordinate here. So cosine of theta is going to be 0. And what is sine of theta going to be? Well it's going to be negative 1. And then finally we go back to 2 pi, which is making a full revolution around the circle. We went all the way around and we're back to this point right over here. So the coordinate is the exact same thing as when the angle equals 0 radians. And so what is cosine of theta? Well that's 1. And sine of theta is 0. and think about where they might intersect. So first let's do cosine of theta. When theta is 0-- and let me mark this off. So this is going to be when y is equal to 1. And this is when y is equal to negative 1. So y equals cosine of theta. theta equals 0. Cosine of theta equals 1. So cosine of theta is equal to 1. When theta is equal to pi 2, cosine of theta is 0. When theta is equal to pi, cosine of theta is negative 1. When theta is equal to 3 pi over 2, cosine of theta is equal to 0. That's this right over here. And then finally when theta is 2 pi, cosine of theta is 1 again. And the curve will look something like this. My best attempt to draw it. Make it a nice smooth curve. The look of these curves should look somewhat familiar at this point. So this is the graph of y is equal to cosine of theta. Now let's do the same thing for sine theta. When theta is equal to 0, sine theta is 0. When theta is pi over 2, sine of theta is 1. When theta is equal to pi, sine of theta is 0. When theta is equal to 3 pi over 2, sine of theta is negative 1. When theta is equal to 2 pi, sine of theta is equal to 0. And so the graph of sine of theta is going to look something like this. My best attempt at drawing it. So just visually, we can think about the question. At how many points do the graphs of y equals sine of theta and y equals cosine of theta intersect for this range for theta?" + }, + { + "Q": "At 0:25 is one fourth just like a quarter?", + "A": "Think of one whole as 100. If you think of money, one dollar is 100 cents. One quarter is 25 cents out of the whole 100 cents. So a quarter is one fourth(a quarter) of a dollar(a whole).", + "video_name": "gEE6yIObbmg", + "timestamps": [ + 25 + ], + "3min_transcript": "- [Voiceover] Is each piece equal to 1/4 of the area of the pie? So we have a pie, and it has one, two, three, four pieces. So it does have four pieces. So is one of those pieces equal to 1/4 of the pie? Well let's talk about what we mean when we have a fraction like 1/4. The one in the fraction, the numerator, represents a number of pieces. So here, one piece. One piece of pie. And then the four, when we're talking about fractions is always talking about the number of equal size. Equal size pieces. So in this case four equal size pieces. So the question is, is each piece one of four equal size pieces? Let's look at the pie. I think it's pretty clear that these pieces on the end are not equal, they are smaller If you love cherry pie, you are not happy about getting this end piece. Because it is smaller. It is not an equal size piece. So yes, each piece is one out of four pieces. But it is not one of four equal size pieces. Therefore it is not 1/4. So our answer is no. No, no, no. Each piece is not 1/4 or an equal share of the pie." + }, + { + "Q": "At 7:27, Sal said that if r= -1, then the values would keep on oscillating. However, if you actually work it out, the series would converge to a/2. If we say that the series, is S, than S= a-a+a-a+a-a+a-a+a... . Also, a-S would equal a-a+a-a+a-a+a... . This is equal to the original series S. So, we can say a-S=S. Then we get a= 2S. Therefore, S=a/2. Is there something I am doing I am not disregarding in my calculuations?", + "A": "Your logic seems plausible but fails the epsilon-delta test, which is the ultimate test for whether a series converges. There is no delta for which larger values of n will produce S values closer than, say, a/4 (a possible epsilon). We know this because it s clear that there is no point beyond which the sum stops oscillating between a and 0. Your result of a/2 is the average of the two values between which the sum oscillates, not the value to which the sum converges.", + "video_name": "wqnpSzEzq1w", + "timestamps": [ + 447 + ], + "3min_transcript": "My brain isn't working right! 5 times 3/5 is going to be 3 times 3/5. Is going to be-- 3 times this is going to be 9/5-- actually that was right. My brain is working right. Times 3/5 is going to be 27 over 25. Times 3/5 is going to be 81/125. And we keep on going on and on and on forever. And notice these terms are starting to get smaller and smaller and smaller. Well actually all of them are getting smaller and smaller and smaller. We're multiplying by 3/5 every time. We now know what the sum is going to be. It's going to be our first term-- it's going to be 5-- over 1 minus our common ratio. And our common ratio in this case is 3/5. So this is going to be equal to 5 over 2/5, which is 25/2 which is equal to 12 and 1/2, or 12.5. Once again, amazing result. I'm taking a sum of infinite terms here, and I was able to get a finite result. And once again, when does this happen? Well, if our common ratio-- if the absolute value of our common ratio-- is less than 1, then these terms are going to get smaller and smaller and smaller. And you'll even see here it even works out mathematically in this denominator that you are going to get a reasonable answer. And it makes sense because these terms are getting smaller and smaller and smaller that this thing will converge. If r is 0, we're still not dealing strictly with a geometric series anymore, but obviously if r was 0, then you're really only going to have this-- well, even depending on how you define what 0 to 0 is. But if your first term you just said would be a, then clearly you'd just be left with a is the sum, and a over 1 minus 0 is still a. So this formula that we just derived does hold up for that. It does start to break down if r is equal to 1 or negative 1. If r is equal to 1 then as you imagine here, you just have a plus a plus a plus a, going on and on forever. If r is equal to negative 1 you just keep oscillating. a, minus a, plus a, minus a. And so the sum's value keeps oscillating between two values. So in general this infinite geometric series is going to converge if the absolute value of your common ratio is less than 1. Or another way of saying that, if your common ratio is between 1 and negative 1." + }, + { + "Q": "At 7:50 Sal said that the boundaries are -1 < r < 1 ,but why not between 0 < r < 1 ?", + "A": "The problem is that using 0 sign into a < sign or when to leave it the way it is? can someone explain? at 2:58 on the video...", + "A": "You change the sign whenever you divide or multiply by a negative number. And only that condition, anything else you keep the sign as it was.", + "video_name": "y7QLay8wrW8", + "timestamps": [ + 178 + ], + "3min_transcript": "Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12. negative number, I swap the inequality, the greater than becomes a less than. When it was positive, I didn't have to swap it. So 27 divided by negative 12, well, they're both divisible by 3. So we're going to get, if we divide the numerator and the denominator by 3, we get negative 9 over 4 is less than-- these cancel out-- y. So y is greater than negative 9/4, or negative 9/4 is less than y. And if you wanted to write that-- just let me write this-- our answer is y is greater than negative 9/4. I just swapped the order, you could say negative 9/4 is less than y. Or if you want to visualize that a little bit better, 9/4 is 2 and 1/4, so we could also say y is greater than negative 2 and 1/4 if we want to put it as a mixed number. And if we wanted to graph it on the number line-- let me" + }, + { + "Q": "At 0:52 when he multiplied 3 by both sides, that same number has to correspond to the denominator for example if the dominator was -3, you multiply by -3 yes?", + "A": "Correct. This is because we need to get rid of the denominator. Does that make sense? And most don t know why you switch the sign, but I believe you do - otherwise you would have asked to know. I hope I helped!", + "video_name": "y7QLay8wrW8", + "timestamps": [ + 52 + ], + "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12." + }, + { + "Q": "2:50, why do you swap numbers?", + "A": "He doesn t swap, he flips the inequality around because you are dividing by a negative number.", + "video_name": "y7QLay8wrW8", + "timestamps": [ + 170 + ], + "3min_transcript": "Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12. negative number, I swap the inequality, the greater than becomes a less than. When it was positive, I didn't have to swap it. So 27 divided by negative 12, well, they're both divisible by 3. So we're going to get, if we divide the numerator and the denominator by 3, we get negative 9 over 4 is less than-- these cancel out-- y. So y is greater than negative 9/4, or negative 9/4 is less than y. And if you wanted to write that-- just let me write this-- our answer is y is greater than negative 9/4. I just swapped the order, you could say negative 9/4 is less than y. Or if you want to visualize that a little bit better, 9/4 is 2 and 1/4, so we could also say y is greater than negative 2 and 1/4 if we want to put it as a mixed number. And if we wanted to graph it on the number line-- let me" + }, + { + "Q": "At 1:28, Sal simplifies 2/3 by multiplying it by three(because he's doing that to the other side of the inequality sign). He then crosses out both threes and leaves the two. Why? How does that work? It's probably really obvious, but can someone answer this?", + "A": "Sure!, A fraction is just a number divided by another number, therefore to undo the division you multiply. Sal crossed them out because they cancel out each other. For example, if you have 6/3 that will equal 2. then multiply by three to get 6. But you should just multiply by the denominator and cancel out everything at the beginning to get 6! Hope this helps!", + "video_name": "y7QLay8wrW8", + "timestamps": [ + 88 + ], + "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12." + }, + { + "Q": "At 2:47, I'm a little confused .. Why the limit isn't (7) as we are approaching point (8) as it's part of the function at this point? &if so how is the limit exist as we are approaching different points? Please somebody enlightens me!", + "A": "You can say that f (8) = 7 but that s not the limit. The limit is concerned with the value NEAR a point but not AT the point. And notice that between 2:50 and around 3:15 or so Sal checks the values on both sides of 8 and finds the limit to be 1. It s ok that the limit (1) does not match the value of f (8) = 7. It just shows that curve is discontinuous at that point.", + "video_name": "_bBAiZhfH_4", + "timestamps": [ + 167 + ], + "3min_transcript": "But then we jump right at x equals 8. And then we continue from 1 again. So this is our other candidate. So these are the three candidates where the function is not continuous. Now let's think about which of these points, which of these x values, does f of k exist. So if one of these is k, does f of k exist? Well f of negative 2 exists. f of 3 exists, right over here. That's f of 3. This is f of negative 2. And f of 8, all exist. So all of these potential k's meet this constraint-- f of k exists, and f is not continuous at k. So that's true for x equals 8, 3, or negative 2. Now let's look at this first constraint. The limit of f of x as x approaches k needs to exist. Well, if we tried to look at x equals 2, the limit of f as x approaches negative 2 here-- the limit from the left, the limit from values lower than negative 2, it looks like our function is approaching something a little higher. It looks like it's a little higher than 3. And the limit from the right, it looks like our function is approaching negative 3. So this one, the limit does not exist. You get a different limit from the left and from the right. Same thing for x equals positive 3. The limit from the left seems like it's approaching 4 and 1/2, while the limit from the right looks like it's approaching negative 4. So this is also not a candidate. So we only have one left. So for this one, the limit should exist. And we see the limit as f of x as x approaches 8 from the negative direction, it looks like f of x is approaching 1. And it looks like, as we approach 8 from the positive direction, the limit of f It's also equal to 1. So your left- and your right-sided limits approach the same value. So the limit of f of x as x approaches 8 is equal to 1. This limit exists. Now, the reason why the function isn't continuous there is that the limit of f of x as x approaches 8, which is equal to 1, it does not equal the value of f of 8. f of 8, we're seeing, is equal to 7. So that's why it meets the last constraint. The function is not continuous there. The function exists. It's defined, f of 8 is equal to 7. And the limit exists. But the limit of f of x as x approaches k is not the same thing, or is not the same as the value of the function evaluated at that point. And so x equals 8 meets all of our constraints. So we could say k is equal to 8." + }, + { + "Q": "at 4:24 -- what makes Sal determine it's t2/2? Is this a formula?", + "A": "Well, I think the deduction of this equation comes out here: d=Va*t, where d is the distance,and Va means the average velocity. while Va=(Vf+Vi)/2, where Vf is the final velocity and Vi is the initial velocity (in this case Vi=0). In addition,we know that the difference of velocity Vdelta=Vf-Vi=g*t. So,Vf=g*t+Vi,since Vi=0, so Vf=g*t+Vi=g*t+0=g*t. Now replace Vf by g*t: d=Va*t=(Vf+Vi)/2*t=(g*t+Vi)/2*t=(g*t+0)/2*t=g*t/2*t=g*t^2/2.", + "video_name": "m6c6dlmUT1c", + "timestamps": [ + 264 + ], + "3min_transcript": "" + }, + { + "Q": "on 4:18 Hilbert who?", + "A": "David Hilbert. He was one of the leading number theorists of his time.", + "video_name": "ik2CZqsAw28", + "timestamps": [ + 258 + ], + "3min_transcript": "down a squiggle, up, wop, all the way over here. OK, but say you're me and you're in math class. This mean that you have graph paper. Opportunity for precision. You could draw that first curve like this. Squig-a, squig-a, squig-a, squig-a, squig-a, squig-a, squig-a, squig-a. The second iteration to fit squiggles going up and down will have a line three boxes across on top and bottom, if you want the squiggles as close on the grid as possible without touching. You might remind yourself by saying, three a-squig, a-squig, a-squiggle, three, a-squig, a-squig, a-squiggle. The next iteration has a woop, and you have to figure out how long that's going to be. Meanwhile, other lengths change to keep everything close. And, two a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. Two, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle, two, nine. We could write the pattern down like this. So what would the next pattern be? Five. Two a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. Two a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle, two. Nine. Two a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. And 15 all the way over to here. And now, Yeah. I can talk that fast totally. OK, But let's not get too far from your original purpose, which was to nicely fill a page with this squiggle. The nicest page filling squiggles have kind of the same density of squiggle everywhere. You don't want to be clumped up here, but have left over space there, because monsters may start growing in the left over space. On graph paper, you can be kind of precise about it. Say you want a squiggle that goes through every box exactly once, and can be extended infinitely. So you try some of those, and decide that, since the point of them is to fill up all the space, you call them space filling curves. Yeah, that's actually a technical term, but be careful because your curve might actually be a snake, snake, snake, snake, snake, snake, snake, snake, snake, snake-- Also, to make it neater, you draw the lines on the lines, and shift the rules so that you go through each intersection on the graph paper exactly once. Which is the same thing, as far as space is concerned. Here's a space filling curve that a guy named Hilbert made up, because Hilbert was awesome, but he's dead now. Here's the first iteration. For the second one, we're going to build it piece-by-piece by connecting four copies of the first. So here's one. Put the second space away next to it, and connect those. Then turn the page to put the third sideways under the first, and connect those. And then the fourth will be the mirror image of that on the other side. Now you've got one nice curve. The third iteration will be made out of four copies of the second iteration. So first build another second iteration curve out of four copies of the first iteration-- one, two, three, four-- then put another next to it, then two sideways on the bottom. Connect them all up. There you go. The fourth iteration is made of four copies of the third iteration, the same way. If you learn to do the second iteration in one piece, it'll make this go faster. Then build two third iterations facing up next to each other," + }, + { + "Q": "At 5:15 what did she use to make the squiggle?", + "A": "She uses a Pipe Cleaner to make the squiggle. The brand is Dill s", + "video_name": "ik2CZqsAw28", + "timestamps": [ + 315 + ], + "3min_transcript": "Three a-squig, a-squig, a-squiggle. Two, nine. Two a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle, two. Nine. Two a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. And 15 all the way over to here. And now, Yeah. I can talk that fast totally. OK, But let's not get too far from your original purpose, which was to nicely fill a page with this squiggle. The nicest page filling squiggles have kind of the same density of squiggle everywhere. You don't want to be clumped up here, but have left over space there, because monsters may start growing in the left over space. On graph paper, you can be kind of precise about it. Say you want a squiggle that goes through every box exactly once, and can be extended infinitely. So you try some of those, and decide that, since the point of them is to fill up all the space, you call them space filling curves. Yeah, that's actually a technical term, but be careful because your curve might actually be a snake, snake, snake, snake, snake, snake, snake, snake, snake, snake-- Also, to make it neater, you draw the lines on the lines, and shift the rules so that you go through each intersection on the graph paper exactly once. Which is the same thing, as far as space is concerned. Here's a space filling curve that a guy named Hilbert made up, because Hilbert was awesome, but he's dead now. Here's the first iteration. For the second one, we're going to build it piece-by-piece by connecting four copies of the first. So here's one. Put the second space away next to it, and connect those. Then turn the page to put the third sideways under the first, and connect those. And then the fourth will be the mirror image of that on the other side. Now you've got one nice curve. The third iteration will be made out of four copies of the second iteration. So first build another second iteration curve out of four copies of the first iteration-- one, two, three, four-- then put another next to it, then two sideways on the bottom. Connect them all up. There you go. The fourth iteration is made of four copies of the third iteration, the same way. If you learn to do the second iteration in one piece, it'll make this go faster. Then build two third iterations facing up next to each other, You can keep going until you run out of room, or you can make each new version the same size by making each line half the length. Or you can make it out of snakes. Or if you have friends, you can each make an iteration of the same size, and put them together. Or invent your own fractal curve so that you could be cool like Hilbert. Who was like, mathematics? I'm going to invent meta-mathematics like a boss." + }, + { + "Q": "at 0:46 to 0:53 he said 9 is > or = to 5 i don't get that", + "A": "With rounding if the number is 5 or more than 5 than you round it up. For example you have 1.5, you d round it up to 2.", + "video_name": "tx2Niw7aJJ8", + "timestamps": [ + 46, + 53 + ], + "3min_transcript": "A ticket agent sells 42 tickets to a play. The tickets cost $29 each. Use rounding to estimate the total dollars taken in from the sale of the tickets. Now if we wanted the exact number, we could say 42 times 29, and we could work out the multiplication, but they essentially want us to be able to do it in our head. We want to round the numbers first and then multiply. So if we want to round, and really we just have two places here, so if we're going to round anything, it's going to be to the nearest ten because that's the largest place we have. So if we round 42 to the nearest ten-- we've done this drill many times-- 2 in the ones place is the less than 5, so we're going to round down. The nearest ten is 40. We're going to round down to 40. 29, if we round to the nearest ten, 9 in the ones place is greater than or equal to 5, so we round up. The nearest ten is 30. And another way to think about it. Just say, well, you know, 42, that's pretty close to 40. 29 is pretty close to 30. I can figure out, so now I can multiply. And here, once again, we can use-- you could call it a trick, but hopefully, you understand why it works. But 30 times 40, instead of you saying, well, this is going to be the same thing as 3 times 4, but we're going to put two zeroes at the end of it. 30 times 40 is the same thing as 3 times 4 with two zeroes, So you have 3 times 4 is 12, which we know, and then we have two zeroes. We got that zero, so let's stick that zero there, and then we got that blue zero there, so let's put that over there. So they're going to have roughly $1,200 taken it from sales of the tickets. That is our estimate." + }, + { + "Q": "At 09:39 in the video I thought there were 16 different scenarios.(Is it possible?)", + "A": "is what possible? It s written in red at the top of the video: 6/16 = 3/8 .. 6 scenarios for 2 heads are possible and there are 16 different total scenarios for heads or tails.", + "video_name": "8TIben0bJpU", + "timestamps": [ + 579 + ], + "3min_transcript": "then how many different places can that second head show up in? Well, if that first head is in one of the four places, then that second head can only be in three different places. So that second head can only be-- I'm picking a nice color here-- can only be in three different places. And so, you know, it could be in any one of these. It could maybe be right over there. Any one of those three places. And so, when you think about it in terms of the first, and I don't want to say the first head, head one. Actually, let me call it this way. Let me call it head A and head B. That way you won't think that I'm talking about the first flip or the second flip. So this is head A, and this right over there is head B. So if you had a particular, I mean, these heads are identical. These outcomes aren't different, but the way we talk about it right now, it looks like there's four places that we could get this head in, and there's three places where we could get this head in. And so if you were to multiply all of the different ways where this is in four different places, and then this is in one of the three left over places, you get 12 different scenarios. But there would only be 12 different scenarios if you viewed this as being different than this. And let me rewrite it with our new-- So this is head A, this is head B, this is head B, this is head A. There would only be 12 different scenarios if you viewed these two things as fundamentally different. But we don't. We're actually double counting. Because we can always swap these two heads and have the exact same outcome. So what you want to do is actually divide it by two. So you want to divide it by all of the different ways that you can swap two different things. If we had three heads here, you would think about all of the different ways you could swap three different things. If you had four heads here, it would be all the different ways you could swap four different things. So there's 12 different scenarios want to divide it by all of the different ways that you can swap two things. So 12 divided by 2 is equal to 6. Six different scenarios, fundamentally different scenarios, considering that you can swap them. If you assume that head A and head B can be interchangeable. But it's a completely identical outcome for us, because they're really just heads. So there's six different scenarios, and we know that there's a total of 16 equally likely scenarios. So we could say that the probability of getting exactly two heads is 6 times, six scenarios and-- Or there's a couple of ways. You could say there are six scenarios that give us two heads, of a possible 16. Or you could say there are six possible scenarios, and the probability of each of those scenarios is 1/16. But either way, you'll get the same answer." + }, + { + "Q": "at 4:04 he sal listed outcomes for exactly one head... I think he forgot to list [TTTH]. Am I right?", + "A": "He accidentally wrote TTHT twice in his equation, however [TTTH} is in row 3, column 4 of his table.", + "video_name": "8TIben0bJpU", + "timestamps": [ + 244 + ], + "3min_transcript": "possibilities, involve getting exactly one heads? Well, we could list them. You could get your heads. So this is equal to the probability of getting the heads in the first flip, plus the probability of getting the heads in the second flip, plus the probability of getting the heads in the third flip. Remember, exactly one heads. We're not saying at least one, exactly one heads. So the probability in the third flip, and then, or the possibility that you get heads in the fourth flip. Tails, heads, and tails. And we know already what the probability of each of these things are. There are 16 possible events, and each of these are one of those 16 possible events. So this is going to be 1 over 16, 1 over 16, 1 over 16, And so we're really saying the probability of getting exactly one heads is the same thing as the probability of getting heads in the first flip, or the probability of getting heads-- or I should say the probability of getting heads in the first flip, or heads in the second flip, or heads in the third flip, or heads in the fourth flip. And we can add the probabilities of these different things, because they are mutually exclusive. Any two of these things cannot happen at the same time. You have to pick one of these scenarios. And so we can add the probabilities. 1/16 plus 1/16 plus 1/16 plus 1/16. Did I say that four times? Well, assume that I did. And so you would get 4/16, which is equal to 1/4. Fair enough. Now let's ask a slightly more interesting question. Let's ask ourselves the probability of getting exactly two heads. And there's a couple of ways we can think about it. [? We ?] know the number of possibilities and of those equally likely possibilities. And we can only use this methodology because it's a fair coin. So, how many of the total possibilities have two heads of the total of equally likely possibilities? So we know there are 16 equally likely possibilities. How many of those have two heads? So I've actually, ahead of time so we save time, I've drawn all of the 16 equally likely possibilities. And how many of these involve two heads? Well, let's see. This one over here has two heads, this one over here has two heads, this one over here has two heads. Let's see, this one over here has two heads, and this one over here has two heads. And then this one over here has two heads, and I believe we are done after that. So if we count them, one, two, three, four, five, six of the possibilities have exactly two heads." + }, + { + "Q": "I'm baffled as to why we keep using our original sample standard deviations as estimates for the population SDs (c. 6:50) once we're assuming the null hypothesis. If (and I might be barking up the wrong tree here) the hypothesis is that there's no meaningful difference whatsoever in weight loss effect between the two diets, why should their SDs remain distinct when imagined across the whole population? If the two groups' data are basically identical when viewed globally, shouldn't their SDs be identical too?", + "A": "because it would lead to same answer. if you sample twice from the same population then the best variance estimator is ((n1-1)var(x1) + (n2-1)var(x2))/(n1+n2-2) ... i know you understand which symbol means what here .. now calculate for variance of difference of means of two iid samples from this population using the just calculated estimate of variance. It is the same thing as what sal does", + "video_name": "N984XGLjQfs", + "timestamps": [ + 410 + ], + "3min_transcript": "We could reject the null hypothesis and go with the alternative hypothesis. Remember, once again, we can use Z-scores, and we can assume this is a normal distribution because our sample size is large for either of those samples. We have a sample size of 100. And to figure that out, the first step, if we just look at a normalized normal distribution like this, what is your critical Z value? We're getting a result above that Z value, only has a 5% chance. So this is actually cumulative. So this whole area right over here is going to be 95% chance. We can just look at the Z table. We're looking for 95% percent. We're looking at the one tailed case. So let's look for 95%. This is the closest thing. We want to err on the side of being a little bit maybe to So let's say 95.05 is pretty good. So that's 1.65. So this critical Z value is equal to 1.65. Or another way to view it is, this distance right here is going to be 1.65 standard deviations. I know my writing is really small. I'm just saying the standard deviation of that distribution. So what is the standard deviation of that We actually calculated it in the last video, and I'll recalculate it here. The standard deviation of our distribution of the difference of the sample means is going to be equal to the square root of the variance of our first population. Now, the variance of our first population, we don't know it. But we could estimate it with our sample standard deviation. If you take your sample standard deviation, 4.67 and And so this is the variance. This is our best estimate of the variance of the population. And we want to divide that by the sample size. And then plus our best estimate of the variance of the population of group two, which is 4.04 squared. The sample standard deviation of group two squared. That gives us variance divided by 100. I did before in the last. Maybe it's still sitting on my calculator. Yes, it's still sitting on the calculator. It's this quantity right up here. 4.67 squared divided by 100 plus 4.04 squared divided by 100. So it's 0.617. So this right here is going to be 0.617. So this distance right here, is going to" + }, + { + "Q": "Whats a \"reciprocal\"? (4:07)", + "A": "For any fraction, its reciprocal is created by flipping the fraction. Example: 3/4: its reciprocal is 4/3 -5/2: its reciprocal is -2/5 6: Note 6 as a fraction is 6/1. Its reciprocal = 1/6 In the video, Sal is using the reciprocal of (5x^4)/4, which would be 4/(5x^4). Hope this helps.", + "video_name": "6nALFmvvgds", + "timestamps": [ + 247 + ], + "3min_transcript": "defined this way, if you said, if you said f of x is equal to six x to the third over five times two over, times two over three x; and if someone said, well what is f of zero, you would say f of zero is undefined. Undefined. Why is that? Because you put x equals zero there, you're going to get two divided by zero and it's undefined. But if you said, okay, well, can I simplify this a little bit to get the exact same function? Well, we're saying you can say f of x is equal to 4/5 times x squared. But if you just left it at that, you would get f of zero is equal to zero. So now it would be defined at zero, but then this would make it a different function. These are two different functions the way they're written right over here. Instead, to make them, to make it clear that this is equivalent to that one, you would have to say x cannot be equal to zero. if u said f of zero, you'd say all right, x cannot be equal to zero, you know? This would be the case if x is anything other than zero and it's not defined for zero, and so you would say f or zero is undefined. So now, these two functions are equivalent, or these two expressions are algebraically equivalent. So thinking about that, let's tackle this division situation here. So immediately, when you look at this, you say, woah, what are constrains here? Well, x cannot be equal to zero because if x was a zero, this second, this five x to the fourth over four would be zero and you'd be dividing, you'd be dividing by zero. So we can explicitly call out that x cannot be equal to zero. And so if x cannot be equal to zero in the original expression, if the result, whatever we get for the resulting expression, in order for it to be algebraically equivalent, we have to give this same constraint. So let's multiply this, or let's do the division. So this is going to be the same thing as two x times... The reciprocal of this is going to be four over five x to the fourth, which is going to be equal to in the numerator, we're going to have eight x to the fourth. So we're going to have eight x to the fourth, four times two x to the fourth, over seven times five x to the fourth is 35 x to the fourth. And now, there's something. We can do a little bit of simplification here, both the numerator and the denominator are divisible by x to the fourth, so let's divide by x to the fourth and we get eight over 35. So once again, you just look at eight 30, Well, this is going to be defined for any x. X isn't even involved in the expression. But if we want this to be algebraically equivalent to this first expression, then we have to make the same constraint, x does not, cannot be equal to zero. And to see, you know, this even seems a little bit more nonsensical to say x cannot be equal to zero" + }, + { + "Q": "at 1:10 Vi say says twogons. instead of twogons could you use biagons or bigons?", + "A": "Yes, you could use biagons, but that sounds ridiculous, so most people just stick with 2-gons.", + "video_name": "CfJzrmS9UfY", + "timestamps": [ + 70 + ], + "3min_transcript": "Let's say you're me, and you're in math class, and you're supposed to be learning about factoring. Trouble is, your teacher is too busy trying to convince you that factoring is a useful skill for the average person to know, with real-world applications ranging from passing your state exams all the way to getting a higher SAT score. And unfortunately, does not have the time to show you why factoring is actually interesting. It's perfectly reasonable for you to get bored in this situation. So like any reasonable person, you start doodling. Maybe it's because your teacher's soporific voice reminds you of a lullaby, but you're drawing stars. And because you're me, you quickly get bored of the usual five-pointed star and get to wondering, why five? So you start exploring. It seems obvious that a five-pointed star is the simplest one, the one that takes the least number of strokes to draw. Sure, you can make a start with four points, but that's not really a star the way you're defining stars. Then there's a six-pointed star, which is also pretty familiar, but totally different from the five-pointed star because it takes two separate lines to make. And then you're thinking about how, much like you can put two triangles together to make a six-pointed star, you can put two squares together to make an eight-pointed star. And any even-numbered star with p points can be made out of two p/2-gons. It is at this point that you realize maybe drawing stars was not the greatest idea. But wait, four would be an even number of points, but that would mean you could make it out of two 2-gons. Maybe you were taught polygons with only two sides But for the purposes of drawing stars, it works out rather well. Sure, the four-pointed star doesn't look too star-like. But then you realize you can make the six-pointed star out of three of these things, and you've got an asterisk, which is definitely a legitimate star. In fact, for any star where the number of points is divisible by 2, you can draw it asterisk style. But that's not quite what you're looking for. What you want is a doodle game, and here it is. Draw p points in a circle, evenly spaced. Pick a number Q. Starting at one point, go around the circle and connect to the point two places over. Repeat. If you get to the starting place before you've covered all the points, jump to a lonely point, and keep going. That's how you draw stars. And it's a successful game, in that previously you were considering running screaming from the room. Or the window was open, so that's an option, too. But now, you're not only entertained but beginning to become curious about the nature of this game. The interesting thing is that the more points you have, the more different ways there is to draw the star. two really good ways to draw them, but they're still simple. I would like to note here that I've never actually left a math class by the window, not that I can say the same for other subjects. Eight is interesting, too, because not only are there a couple nice ways to draw it, but one's a composite of two polygons, while another can be drawn without picking up the pencil. Then there's nine, which, in addition to a couple of other nice versions, you can make out of three triangles. And because you're me, and you're a nerd, and you like to amuse yourself, you decide to call this kind of star a square star because that's kind of a funny name. So you start drawing other square stars. Four 4-gons, two 2-gons, even the completely degenerate case of one 1-gon. Unfortunately, five pentagons is already difficult to discern. And beyond that, it's very hard to see and appreciate the structure of square stars. So you get bored and move on to 10 dots in a circle, which is interesting because this is the first number where you can make a star as a composite of smaller stars-- that is, two boring old five-pointed stars. Unless you count asterisk stars, in which case 8 was two 4s's or four 2's or two 2's and a 4. But 10 is interesting because you can make it as a composite in more than one way because it's divisible by 5, which itself can be made in two ways." + }, + { + "Q": "In the video, at 00:40 she says 'you can draw a four pointed star, but that's not really a star by the way you're defining your stars' -why not?", + "A": "Well, later on in the video she explains the whole star game. You can t really use that method with four points.", + "video_name": "CfJzrmS9UfY", + "timestamps": [ + 40 + ], + "3min_transcript": "Let's say you're me, and you're in math class, and you're supposed to be learning about factoring. Trouble is, your teacher is too busy trying to convince you that factoring is a useful skill for the average person to know, with real-world applications ranging from passing your state exams all the way to getting a higher SAT score. And unfortunately, does not have the time to show you why factoring is actually interesting. It's perfectly reasonable for you to get bored in this situation. So like any reasonable person, you start doodling. Maybe it's because your teacher's soporific voice reminds you of a lullaby, but you're drawing stars. And because you're me, you quickly get bored of the usual five-pointed star and get to wondering, why five? So you start exploring. It seems obvious that a five-pointed star is the simplest one, the one that takes the least number of strokes to draw. Sure, you can make a start with four points, but that's not really a star the way you're defining stars. Then there's a six-pointed star, which is also pretty familiar, but totally different from the five-pointed star because it takes two separate lines to make. And then you're thinking about how, much like you can put two triangles together to make a six-pointed star, you can put two squares together to make an eight-pointed star. And any even-numbered star with p points can be made out of two p/2-gons. It is at this point that you realize maybe drawing stars was not the greatest idea. But wait, four would be an even number of points, but that would mean you could make it out of two 2-gons. Maybe you were taught polygons with only two sides But for the purposes of drawing stars, it works out rather well. Sure, the four-pointed star doesn't look too star-like. But then you realize you can make the six-pointed star out of three of these things, and you've got an asterisk, which is definitely a legitimate star. In fact, for any star where the number of points is divisible by 2, you can draw it asterisk style. But that's not quite what you're looking for. What you want is a doodle game, and here it is. Draw p points in a circle, evenly spaced. Pick a number Q. Starting at one point, go around the circle and connect to the point two places over. Repeat. If you get to the starting place before you've covered all the points, jump to a lonely point, and keep going. That's how you draw stars. And it's a successful game, in that previously you were considering running screaming from the room. Or the window was open, so that's an option, too. But now, you're not only entertained but beginning to become curious about the nature of this game. The interesting thing is that the more points you have, the more different ways there is to draw the star. two really good ways to draw them, but they're still simple. I would like to note here that I've never actually left a math class by the window, not that I can say the same for other subjects. Eight is interesting, too, because not only are there a couple nice ways to draw it, but one's a composite of two polygons, while another can be drawn without picking up the pencil. Then there's nine, which, in addition to a couple of other nice versions, you can make out of three triangles. And because you're me, and you're a nerd, and you like to amuse yourself, you decide to call this kind of star a square star because that's kind of a funny name. So you start drawing other square stars. Four 4-gons, two 2-gons, even the completely degenerate case of one 1-gon. Unfortunately, five pentagons is already difficult to discern. And beyond that, it's very hard to see and appreciate the structure of square stars. So you get bored and move on to 10 dots in a circle, which is interesting because this is the first number where you can make a star as a composite of smaller stars-- that is, two boring old five-pointed stars. Unless you count asterisk stars, in which case 8 was two 4s's or four 2's or two 2's and a 4. But 10 is interesting because you can make it as a composite in more than one way because it's divisible by 5, which itself can be made in two ways." + }, + { + "Q": "at about 3:25 why does it bother you that a 25 pointed star a square star?", + "A": "It doesn t bother her that a 25 pointed star is a square star. What bothers her is the method by which the 25 pointed star is made. If it is made from 5 pentagons, then that is clearly a square star. But what if it is made from 5 5 pointed stars? Then, although P=25 still, Q=10 instead of 5.", + "video_name": "CfJzrmS9UfY", + "timestamps": [ + 205 + ], + "3min_transcript": "maybe drawing stars was not the greatest idea. But wait, four would be an even number of points, but that would mean you could make it out of two 2-gons. Maybe you were taught polygons with only two sides But for the purposes of drawing stars, it works out rather well. Sure, the four-pointed star doesn't look too star-like. But then you realize you can make the six-pointed star out of three of these things, and you've got an asterisk, which is definitely a legitimate star. In fact, for any star where the number of points is divisible by 2, you can draw it asterisk style. But that's not quite what you're looking for. What you want is a doodle game, and here it is. Draw p points in a circle, evenly spaced. Pick a number Q. Starting at one point, go around the circle and connect to the point two places over. Repeat. If you get to the starting place before you've covered all the points, jump to a lonely point, and keep going. That's how you draw stars. And it's a successful game, in that previously you were considering running screaming from the room. Or the window was open, so that's an option, too. But now, you're not only entertained but beginning to become curious about the nature of this game. The interesting thing is that the more points you have, the more different ways there is to draw the star. two really good ways to draw them, but they're still simple. I would like to note here that I've never actually left a math class by the window, not that I can say the same for other subjects. Eight is interesting, too, because not only are there a couple nice ways to draw it, but one's a composite of two polygons, while another can be drawn without picking up the pencil. Then there's nine, which, in addition to a couple of other nice versions, you can make out of three triangles. And because you're me, and you're a nerd, and you like to amuse yourself, you decide to call this kind of star a square star because that's kind of a funny name. So you start drawing other square stars. Four 4-gons, two 2-gons, even the completely degenerate case of one 1-gon. Unfortunately, five pentagons is already difficult to discern. And beyond that, it's very hard to see and appreciate the structure of square stars. So you get bored and move on to 10 dots in a circle, which is interesting because this is the first number where you can make a star as a composite of smaller stars-- that is, two boring old five-pointed stars. Unless you count asterisk stars, in which case 8 was two 4s's or four 2's or two 2's and a 4. But 10 is interesting because you can make it as a composite in more than one way because it's divisible by 5, which itself can be made in two ways. at all because 11 is prime. Though here you start to wonder how to predict how many times around the circle we'll go before getting back to start. But instead of exploring the exciting world of modular arithmetic, you move on to 12, which is a really cool number because it has a whole bunch of factors. And then something starts to bother you. Is a 25-pointed star composite made of five five-pointed stars a square star? You had been thinking only of pentagons because the lower numbers didn't have this question. How could you have missed that? Maybe your teacher said something interesting about prime numbers, and you accidentally lost focus for a moment. I don't know. It gets even worse. 6 squared would be a 36-pointed star made of six hexagons. But if you allow use of six-pointed stars, then it's the same as a composite of 12 triangles. And that doesn't seem in keeping with the spirit of square stars. You'll have to define square stars more strictly. But you do like the idea that there's three ways to make the seventh square star. Anyway, the whole theory of what kind of stars can be made with what numbers is quite interesting. And I encourage you to explore this during your math class." + }, + { + "Q": "at 1:25 why is 9x^2 not equal to 3x^2 ?", + "A": "We have 9x^2. We want to make it into (ax)^2. 3x^2 does NOT equal 9x^2. However, (3x)^2 does equal 9x^2. Why? When we have (3x)^2 the exponent distributes to both terms (the 3 and the x) so we have: (3x)^2 = 3^2x^2 = 9x^2.", + "video_name": "jmbg-DKWuc4", + "timestamps": [ + 85 + ], + "3min_transcript": "Let's see if we can factor the expression 45x squared minus 125. So whenever I see something like this-- I have a second-degree term here, I have a subtraction sign-- my temptation is to look at this as a difference of squares. We've already seen this multiple times. We've already seen that if we have something of the form a squared minus b squared, that this can be factored as a plus b times a minus b. So let's look over here. Well, over here, it's not obvious that this right over here is a perfect square. Neither is it obvious that this right over here So it's not clear to me that this is a difference of squares. But what is interesting is that both 45 and 125 have some factors in common. And the one that jumps out at me is 5. So let's see if we can factor out a 5, and by doing that, whether we can get something that's a little bit closer to this pattern right over here. 45x squared divided by 5 is going to be 9x squared. And then 125 divided by 5 is 25. Now, this is interesting. 9x squared-- that's a perfect square. If we call this a squared, then that tells us that a would be equal to 3x. 3x-- the whole thing squared is 9x squared. Similarly-- I can never say similarly correctly-- 25 is clearly just 5 squared. So in this case, if we're looking at this template, b would be equal to 5. So now this is a difference of squares, and we can factor it completely. So we can't forget our 5 out front that we factored out. So it's going to be 5 times a plus b. So let me write this. So it's going to be 5 times a plus b times a minus b. 5 times 3x plus 5 times 3x minus 5 is 45x squared minus 125 factored out." + }, + { + "Q": "At 5:25, You said that the median is the middle number, but at 2:25 you said that that average is the typical or the middle as well, but my teachers say to associate average with mean, so are there many averages or if not, whats the deal?", + "A": "Yes, there are many averages. The mean is the most common average, so it s often used as a synonym even though it properly should not be.", + "video_name": "h8EYEJ32oQ8", + "timestamps": [ + 325, + 145 + ], + "3min_transcript": "I'll write in yellow, arithmetic mean. When arithmetic is a noun, we call it arithmetic. When it's an adjective like this, we call it arithmetic, arithmetic mean. And this is really just the sum of all the numbers divided by-- this is a human-constructed definition that we've found useful-- the sum of all these numbers divided by the number of numbers we have. So given that, what is the arithmetic mean of this data set? Well, let's just compute it. It's going to be 4 plus 3 plus 1 plus 6 plus 1 plus 7 over the number of data points we have. So we have six data points. So we're going to divide by 6. And we get 4 plus 3 is 7, plus 1 is 8, plus 6 is 14, 15 plus 7 is 22. Let me do that one more time. You have 7, 8, 14, 15, 22, all of that over 6. And we could write this as a mixed number. 6 goes into 22 three times with a remainder of 4. So it's 3 and 4/6, which is the same thing as 3 and 2/3. We could write this as a decimal with 3.6 repeating. So this is also 3.6 repeating. We could write it any one of those ways. But this is kind of a representative number. This is trying to get at a central tendency. Once again, these are human-constructed. No one ever-- it's not like someone just found some religious document that said, this is the way that the arithmetic mean must be defined. It's not as pure of a computation as, say, finding the circumference of the circle, which there really is-- that was kind of-- we studied the universe. And that just fell out of our study of the universe. It's a human-constructed definition Now there are other ways to measure the average or find a typical or middle value. The other very typical way is the median. And I will write median. I'm running out of colors. I will write median in pink. So there is the median. And the median is literally looking for the middle number. So if you were to order all the numbers in your set and find the middle one, then that is your median. So given that, what's the median of this set of numbers going to be? Let's try to figure it out. Let's try to order it. So we have 1. Then we have another 1. Then we have a 3. Then we have a 4, a 6, and a 7. So all I did is I reordered this. And so what's the middle number? Well, you look here. Since we have an even number of numbers, we have six numbers, there's not one middle number. You actually have two middle numbers here. You have two middle numbers right over here." + }, + { + "Q": "Isn't the part in (3:53) where he finds the arithmetic mean, when he says to add and find the sum then divide, that also I think is one way how to find the average, right?", + "A": "to find Mean: add up all of the numbers and then divide by how many numbers there are Median: The middle number. if there are two in the middle, add them up and divide by two to find the middle of those two numbers Mode: Whatever number is the most common (there can t be two modes. The number has to be repeated the MOST. There cant be two highest repeated numbers.)", + "video_name": "h8EYEJ32oQ8", + "timestamps": [ + 233 + ], + "3min_transcript": "has a very particular meaning, as we'll see. When many people talk about average, they're talking about the arithmetic mean, which we'll see shortly. But in statistics, average means something more general. It really means give me a typical, or give me a middle number, or-- and these are or's. And really it's an attempt to find a measure of central tendency. So once again, you have a bunch of numbers. You're somehow trying to represent these with one number we'll call the average, that's somehow typical, or middle, or the center somehow of these numbers. And as we'll see, there's many types of averages. The first is the one that you're probably most familiar with. It's the one-- and people talk about hey, the average on this exam or the average height. And that's the arithmetic mean. I'll write in yellow, arithmetic mean. When arithmetic is a noun, we call it arithmetic. When it's an adjective like this, we call it arithmetic, arithmetic mean. And this is really just the sum of all the numbers divided by-- this is a human-constructed definition that we've found useful-- the sum of all these numbers divided by the number of numbers we have. So given that, what is the arithmetic mean of this data set? Well, let's just compute it. It's going to be 4 plus 3 plus 1 plus 6 plus 1 plus 7 over the number of data points we have. So we have six data points. So we're going to divide by 6. And we get 4 plus 3 is 7, plus 1 is 8, plus 6 is 14, 15 plus 7 is 22. Let me do that one more time. You have 7, 8, 14, 15, 22, all of that over 6. And we could write this as a mixed number. 6 goes into 22 three times with a remainder of 4. So it's 3 and 4/6, which is the same thing as 3 and 2/3. We could write this as a decimal with 3.6 repeating. So this is also 3.6 repeating. We could write it any one of those ways. But this is kind of a representative number. This is trying to get at a central tendency. Once again, these are human-constructed. No one ever-- it's not like someone just found some religious document that said, this is the way that the arithmetic mean must be defined. It's not as pure of a computation as, say, finding the circumference of the circle, which there really is-- that was kind of-- we studied the universe. And that just fell out of our study of the universe. It's a human-constructed definition" + }, + { + "Q": "At 2:08 Sal started explaining average. Are average and mean the same thing, because Sal never said anything about mean", + "A": "Yes! Average and mean are synonyms. They both involve the total sum of a set of numbers and dividing it by how many numbers given.", + "video_name": "h8EYEJ32oQ8", + "timestamps": [ + 128 + ], + "3min_transcript": "We will now begin our journey into the world of statistics, which is really a way to understand or get our head around data. So statistics is all about data. And as we begin our journey into the world of statistics, we will be doing a lot of what we can call descriptive statistics. So if we have a bunch of data, and if we want to tell something about all of that data without giving them all of the data, can we somehow describe it with a smaller set of numbers? So that's what we're going to focus on. And then once we build our toolkit on the descriptive statistics, then we can start to make inferences about that data, start to make conclusions, start to make judgments. And we'll start to do a lot of inferential statistics, make inferences. So with that out of the way, let's think about how we can describe data. So let's say we have a set of numbers. We can consider this to be data. in our garden. And let's say we have six plants. And the heights are 4 inches, 3 inches, 1 inch, 6 inches, and another one's 1 inch, and another one is 7 inches. And let's say someone just said-- in another room, not looking at your plants, just said, well, you know, how tall are your plants? And they only want to hear one number. They want to somehow have one number that represents all of these different heights of plants. How would you do that? Well, you'd say, well, how can I find something that-- maybe I want a typical number. Maybe I want some number that somehow represents the middle. Maybe I want the most frequent number. Maybe I want the number that somehow represents the center of all of these numbers. And if you said any of those things, you would actually have done the same things that the people who first came up with descriptive statistics said. They said, well, how can we do it? And we'll start by thinking of the idea of average. has a very particular meaning, as we'll see. When many people talk about average, they're talking about the arithmetic mean, which we'll see shortly. But in statistics, average means something more general. It really means give me a typical, or give me a middle number, or-- and these are or's. And really it's an attempt to find a measure of central tendency. So once again, you have a bunch of numbers. You're somehow trying to represent these with one number we'll call the average, that's somehow typical, or middle, or the center somehow of these numbers. And as we'll see, there's many types of averages. The first is the one that you're probably most familiar with. It's the one-- and people talk about hey, the average on this exam or the average height. And that's the arithmetic mean." + }, + { + "Q": "If 6 5/x = 14, what does x equal to?\nFor Example:\n6 multiplied by x = 6x\n6x + 5= 8.75/x\n8.75/x = 14\non 7:38", + "A": "x = 0.625", + "video_name": "9IUEk9fn2Vs", + "timestamps": [ + 458 + ], + "3min_transcript": "this equation by x plus 5. You can say x plus 5 over 1. Times x plus 5 over 1. On the left-hand side, they get canceled out. So we're left with 3 is equal to 8 times x plus five. All of that over x plus 2. Now, on the top, just to simplify, we once again just multiply the 8 times the whole expression. So it's 8x plus 40 over x plus 2. Now, we want to get rid of this x plus 2. So we can do it the same way. We can multiply both sides of this equation by x plus 2 over 1. x plus 2. We could just say we're multiplying both sides by x plus 2. The 1 is little unnecessary. So the left-hand side becomes 3x plus 6. multiplying it times the whole expression. x plus 2. And on the right-hand side. Well, this x plus 2 and this x plus 2 will cancel out. And we're left with 8x plus 40. And this is now a level three problem. Well, if we subtract 8x from both sides, minus 8x, plus-- I think I'm running out of space. Minus 8x. Well, on the right-hand side the 8x's cancel out. On the left-hand side we have minus 5x plus 6 is equal to, on the right-hand side all we have left is 40. Now we can subtract 6 from both sides of this equation. Let me just write out here. Minus 6 plus minus 6. Now I'm going to, hope I don't lose you guys by trying to go up here. But if we subtract minus 6 from both sides, on the left-hand side we're just left with minus 5x equals, and on the Now it's a level one problem. We just multiply both sides times negative 1/5. Negative 1/5. On the left-hand side we have x. And on the right-hand side we have negative 34/5. Unless I made some careless mistakes, I think that's right. And I think if you understood what we just did here, you're ready to tackle some level four linear equations. Have fun." + }, + { + "Q": "At 2:49, why is it 7x + 7 ?", + "A": "With the distributive property, you have to multiply each term in the brackets by the 7, so multiply 7 by x, and 7 by 1 to give you 7x + 7.", + "video_name": "9IUEk9fn2Vs", + "timestamps": [ + 169 + ], + "3min_transcript": "We either just multiply both sides by 1/5, or you could just do that as dividing by 5. If you multiply both sides by 1/5. The left-hand side becomes x. And the right-hand side, 3 times 1/5, is equal to 3/5. So what did we do here? This is just like, this actually turned into a level two problem, or actually a level one problem, very quickly. All we had to do is multiply both sides of this equation by x. And we got the x's out of the denominator. Let's do another problem. Let's have -- let me say, x plus 2 over x plus 1 is equal to, let's say, 7. So, here, instead of having just an x in the denominator, we have a whole x plus 1 in the denominator. To get that x plus 1 out of the denominator, we multiply both sides of this equation times x plus 1 over 1 times this side. Since we did it on the left-hand side we also have to do it on the right-hand side, and this is just 7/1, times x plus 1 over 1. On the left-hand side, the x plus 1's cancel out. And you're just left with x plus 2. It's over 1, but we can just ignore the 1. And that equals 7 times x plus 1. And that's the same thing as x plus 2. And, remember, it's 7 times the whole thing, x plus 1. So we actually have to use the distributive property. And that equals 7x plus 7. So now it's turned into a, I think this is a level three linear equation. And now all we do is, we say well let's get all the x's on the other side of the equation. So I'm going to choose to get the x's on the left. So let's bring that 7x onto the left. And we can do that by subtracting 7x from both sides. Minus 7x, plus, it's a minus 7x. The right-hand side, these two 7x's will cancel out. And on the left-hand side we have minus 7x plus x. Well, that's minus 6x plus 2 is equal to, and on the right all we have left is 7. Now we just have to get rid of this 2. And we can just do that by subtracting 2 from both sides. And we're left with minus 6x packs is equal to 6. Now it's a level one problem. We just have to multiply both sides times the reciprocal of the coefficient on the left-hand side. And the coefficient's negative 6. So we multiply both sides of the equation by negative 1/6." + }, + { + "Q": "i dont understand anything, in 3:29 how did -7x+x+2 = 7x+7-7x turn into -6x+2=7?\nhow did -7x turn into -6x? it seems impossible to me", + "A": "-7x+x=-6x", + "video_name": "9IUEk9fn2Vs", + "timestamps": [ + 209 + ], + "3min_transcript": "We either just multiply both sides by 1/5, or you could just do that as dividing by 5. If you multiply both sides by 1/5. The left-hand side becomes x. And the right-hand side, 3 times 1/5, is equal to 3/5. So what did we do here? This is just like, this actually turned into a level two problem, or actually a level one problem, very quickly. All we had to do is multiply both sides of this equation by x. And we got the x's out of the denominator. Let's do another problem. Let's have -- let me say, x plus 2 over x plus 1 is equal to, let's say, 7. So, here, instead of having just an x in the denominator, we have a whole x plus 1 in the denominator. To get that x plus 1 out of the denominator, we multiply both sides of this equation times x plus 1 over 1 times this side. Since we did it on the left-hand side we also have to do it on the right-hand side, and this is just 7/1, times x plus 1 over 1. On the left-hand side, the x plus 1's cancel out. And you're just left with x plus 2. It's over 1, but we can just ignore the 1. And that equals 7 times x plus 1. And that's the same thing as x plus 2. And, remember, it's 7 times the whole thing, x plus 1. So we actually have to use the distributive property. And that equals 7x plus 7. So now it's turned into a, I think this is a level three linear equation. And now all we do is, we say well let's get all the x's on the other side of the equation. So I'm going to choose to get the x's on the left. So let's bring that 7x onto the left. And we can do that by subtracting 7x from both sides. Minus 7x, plus, it's a minus 7x. The right-hand side, these two 7x's will cancel out. And on the left-hand side we have minus 7x plus x. Well, that's minus 6x plus 2 is equal to, and on the right all we have left is 7. Now we just have to get rid of this 2. And we can just do that by subtracting 2 from both sides. And we're left with minus 6x packs is equal to 6. Now it's a level one problem. We just have to multiply both sides times the reciprocal of the coefficient on the left-hand side. And the coefficient's negative 6. So we multiply both sides of the equation by negative 1/6." + }, + { + "Q": "In 1:16 where where doe's the speaker get 1/5 from ? And do I have to watch all of the pre-algebra videos before I understand where 1/5 comes from ?", + "A": "Thanx, then I ll have to watch all of the pre-al videos.", + "video_name": "9IUEk9fn2Vs", + "timestamps": [ + 76 + ], + "3min_transcript": "Welcome to the presentation on level four linear equations. So, let's start doing some problems. Let's say I had the situation-- let me give me a couple of problems-- if I said 3 over x is equal to, let's just say 5. So, what we want to do -- this problem's a little unusual from everything we've ever seen. Because here, instead of having x in the numerator, we actually have x in the denominator. So, I personally don't like having x's in my denominators, so we want to get it outside of the denominator into a numerator or at least not in the denominator as So, one way to get a number out of the denominator is, if we were to multiply both sides of this equation by x, you see that on the left-hand side of the equation these two x's will cancel out. And in the right side, you'll just get 5 times x. So this equals -- the two x's cancel out. And you get 3 is equal to 5x. Now, we could also write that as 5x is equal to 3. We either just multiply both sides by 1/5, or you could just do that as dividing by 5. If you multiply both sides by 1/5. The left-hand side becomes x. And the right-hand side, 3 times 1/5, is equal to 3/5. So what did we do here? This is just like, this actually turned into a level two problem, or actually a level one problem, very quickly. All we had to do is multiply both sides of this equation by x. And we got the x's out of the denominator. Let's do another problem. Let's have -- let me say, x plus 2 over x plus 1 is equal to, let's say, 7. So, here, instead of having just an x in the denominator, we have a whole x plus 1 in the denominator. To get that x plus 1 out of the denominator, we multiply both sides of this equation times x plus 1 over 1 times this side. Since we did it on the left-hand side we also have to do it on the right-hand side, and this is just 7/1, times x plus 1 over 1. On the left-hand side, the x plus 1's cancel out. And you're just left with x plus 2. It's over 1, but we can just ignore the 1. And that equals 7 times x plus 1. And that's the same thing as x plus 2. And, remember, it's 7 times the whole thing, x plus 1. So we actually have to use the distributive property. And that equals 7x plus 7. So now it's turned into a, I think this is a level three linear equation. And now all we do is, we say well let's get all the x's on" + }, + { + "Q": "At 2:53, how was Sal able to tell whether it was sin or cosine?", + "A": "When x is 0, the value of the cosine equation would be 1, and (0, 1) is not a point on the graph. When x is 0, the value of the sine equation would be -2, and (0, -2) is a valid point on the graph. Thus, the cosine equation can be eliminated.", + "video_name": "yHo0CcDVHsk", + "timestamps": [ + 173 + ], + "3min_transcript": "Going from negative 2 to 1, it went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So immediately, we can say, well, look. This is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write \"cosine\" first. Cosine maybe some coefficient times x plus the midline. The midline-- we already figured out-- was minus 2 or negative 2. So it could take that form or it could take f of x is equal to 3 times-- it could be sine of x or sine of some coefficient times x. So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1. Whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0-- so if x is 0, k times 0 is going to be 0-- sine of 0 is 0. So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since, when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. And we just really need to figure out-- what could this constant actually be? at the period of this function. Let's see. If we went from this point-- where we intersect the midline-- and we have a positive slope, the next point that we do that is right over here. So our period is 8. So what coefficient could we have here to make the period of this thing be equal to 8? Well, let us just remind ourselves what the period of sine of x is. So the period of sine of x-- so I'll write \"period\" right over here-- is 2pi. You increase your angle by 2 pi radians or decrease it. you're back at the same point on the unit circle. So what would be the period of sine of kx? Well, now, your x, your input is increasing k times faster." + }, + { + "Q": "At 0:24 why did the problem say the time when you don't need it? And why did it say how many wheels?", + "A": "Because the skill this video is trying to teach is how to look at a word problem and find the information that you need. So Sal said the time and how many wheels so you could find out what was the important part of the problem.", + "video_name": "6QZCj4O9sk0", + "timestamps": [ + 24 + ], + "3min_transcript": "The local grocery store opens at nine. Its parking lot has six rows. Each row can fit seven cars. Each car has four wheels. How many cars can the parking lot fit? And I encourage you to pause the video and think about this yourself. Try to figure it out on your own. So, let's re-read this. The local grocery store opens up at nine. Well, that doesn't really matter. If we're thinking about how many cars can the parking lot fit. So we don't really have to care about that. We also dont have to care about how many wheels each car has. They're not asking us how many wheels can fit in the parking lot. So we can ignore that. What we really care about is how many rows we have. And how many cars can fit in each row. What we have is -- We have six rows and each row can fit seven cars. We're going to six groups of seven. Or, another way of thinking about it. We're going to have six times seven cars can fit in the parking lot. What is this going to be equal to? This is literally six sevens added up. This is the same thing as one, two, three, four, five, six. Now we're going to add these up. Seven plus seven is fourteen. Twenty-one, twenty-eight, thirty-five, forty-two. Six times seven is equal to forty-two. So forty-two cars can fit in the parking lot. Don't believe me? I made a little diagram here. We have six rows. This is the first row. Third.. Fourth.. Fifth.. Sixth. Each row can fit seven cars. You see it here. One; Let me make that a little brighter. One.. Three.. Four, five, six, seven. How many cars are there? You have seven. Fourteen.. Twenty-one.. Twenty-eight.. Thirty-five.. Forty-two total cars. Six rows of seven." + }, + { + "Q": "At 2:13, why is 3^2 not the same as 2^3?", + "A": "It doesn t work that way. 3*2 may be the same as 2*3, but exponentiation does NOT have the same property. Think about it. 2^3 implies 2*2*2, which we can calculate to be 8. 3^2 implies 3*3 which we can calculate to be 9. Obviously, 9 does not equal 8. This hold true for all exponents. In general, a^b does not equal b^a. (There are exceptions. One that I know is 2^4 does equal 4^2. Both are 16. This is only one instance. Otherwise, they usually are not the same.)", + "video_name": "XZRQhkii0h0", + "timestamps": [ + 133 + ], + "3min_transcript": "You already know that we can view multiplication as repeated addition. So, if we had 2 times 3 (2 \u00d7 3), we could literally view this as 3 2's being added together. So it could be 2 + 2 + 2. Notice this is [COUNTING: 1, 2] 3 2's. And when you add those 2's together, you get 6. What we're going to introduce you to in this video is the idea of repeated multiplication \u2013 a new operation that really can be viewed as repeated multiplication. And that's the operation of taking an 'exponent.' And it sounds very fancy. But we'll see with a few examples that it's not too bad. So now, let's take the idea of 2 to the 3rd power (2^3) \u2013 which is how we would say this. (So let me write this down in the appropriate colors.) So 2 to the 3rd power. (2^3.) So you might be tempted to say, \"Hey, maybe this is 2 \u00d7 3, which would be 6.\" this is repeated multiplication. So if I have 2 to the 3rd power, (2^3), this literally means multiplying 3 2's together. So this would be equal to, not 2 + 2 + 2, but 2 \u00d7 ... (And I\u2019ll use a little dot to signify multiplication.) ... 2 \u00d7 2 \u00d7 2. Well, what's 2 \u00d7 2 \u00d7 2? Well that is equal to 8. (2 \u00d7 2 \u00d7 2 = 8.) So 2 to the 3rd power is equal to 8. (2^3 = 8.) Let's try a few more examples here. What is 3 to the 2nd power (3^2) going to be equal to? And I'll let you think about that for a second. I encourage you to pause the video. So let's think it through. This literally means multiplying 2 3's. So let's multiply 3 \u2013 (Let me do that in yellow.) So this is going to be equal to 9. Let\u2019s do a few more examples. What is, say, 5 to the \u2013 let's say \u2013 5 to the 4th power (5^4)? And what you'll see here is this number is going to get large very, very, very fast. So 5 to the 4th power (5^4) is going to be equal to multiplying 4 5's together. So 5^4 = 5 \u00d7 5 \u00d7 5 \u00d7 5. Notice, we have [COUNTING: 1, 2, 3] 4 5's. And we are multiplying them. This is not 5 \u00d7 4. This is not 20. This is 5 \u00d7 5 \u00d7 5 \u00d7 5. So what is this going to be? Well 5 \u00d7 5 is 25. (5 \u00d7 5 = 25.) 25 \u00d7 5 is 125. (25 \u00d7 5 = 125.) 125 \u00d7 5 is 625. (125 \u00d7 5 = 625.)" + }, + { + "Q": "I really do not get this one. At 4:00 he points at a and says this is the heighth of the parallelogram. The parallelogram first starts with two sides of A and two sides of C. After comparing them he says the area is A^2 , this insinuates that Side A and Side C are the same length and that is just not true?", + "A": "Actually it is true. As he notes at 3:52 and repeats twice is that the height of the paralllelogram is a also which is shown in the triangle on the left. area of a parallelogram is base times height, so it is a \u00e2\u0080\u00a2 a or a^2. It does not have anything to do with the sides being the same length, and with the drawing, they cannot be the same length because c forms the hypotenuse of the right triangle which by definition must be longer than either leg.", + "video_name": "rcBaqkGp7CA", + "timestamps": [ + 240 + ], + "3min_transcript": "And this and the right angle is now here. So all I did is I rotated this by 90 degrees counterclockwise. Now, what I want to do is construct a parallelogram. I'm going to construct a parallelogram by essentially-- and let me label. So this is height c right over here. Let me do that white color. This is height c. Now, what I want to do is go from this point and go up c as well. Now, so this is height c as well. And what is this length? What is the length over here from this point to this point going to be? Well, a little clue is this is a parallelogram. This line right over here is going to be parallel to this line. And since it's traveling the same distance in the x direction or in the horizontal direction and the vertical direction, this is going to be the same length. So this is going to be of length a. Now, the next question I have for you is, what is the area of this parallelogram that I have just constructed? Well, to think about that, let's redraw this part of the diagram so that the parallelogram is sitting on the ground. So this is length a. This is length c. This is length c. And if you look at this part right over here, it gives you a clue. I'll use this green color. The height of the parallelogram is given right over here. This side is perpendicular to the base. So the height of the parallelogram is a as well. Well, the area of a parallelogram is just the base times the height. So the area of this parallelogram right over here is going to be a squared. Now, let's do the same thing. But let's rotate our original right triangle. Let's rotate it the other way. So let's rotate it 90 degrees clockwise. And this time, instead of pivoting on this point, we're going to pivot on that point right over there. So what are we going to get? So the side of length c if we rotate it like that, it's going to end up right over here. I'll try to draw it as close to scale as possible. So that side has length c. Now, the side of length of b is going to pop out and look something like this. It's going to be parallel to that. This is going to be a right angle." + }, + { + "Q": "At 0:15, what does F(x) mean?", + "A": "F(x) is just a notation to express a function in terms of x. It is the same as y = function, but that tends to be used in lower-level mathematics.", + "video_name": "1LxhXqD3_CE", + "timestamps": [ + 15 + ], + "3min_transcript": "" + }, + { + "Q": "The only thing I don't understand is the (x-c) part at 3:15, why put the c and not simply use x, x^2, x^3 and so on, like on the Mclaurin series?", + "A": "It s a shift. It s like shifting the parabola function, y = x^2, three places to the left. You d write it as y = (x+3)^2. To shift it c to the left, you d use (x-c)^2. Or,in the case of the Taylor expansion, multiply the derivative(s) by (x-c).", + "video_name": "1LxhXqD3_CE", + "timestamps": [ + 195 + ], + "3min_transcript": "" + }, + { + "Q": "At 7:42 can someone explain how he got x+sqrt of 2=0 and the same for x-sqrt2= 0?", + "A": "Sal recognised that the binomial x^2 - 2 could be viewed as the difference of two squares, so he factored it into sum and difference of the two numbers being squared, ( x and \u00e2\u0088\u009a2 ).", + "video_name": "x9lb_frpkH0", + "timestamps": [ + 462 + ], + "3min_transcript": "times x minus the square root of two. I'm just recognizing this as a difference of squares. And, once again, we just want to solve this whole, all of this business, equaling zero. All of this equaling zero. So how can this equal to zero? Well any one of these expressions, if I take the product, and if any one of them equals zero then I'm gonna get zero. So, x could be equal to zero. X could be equal to zero, and that actually gives us a root. When x is equal to zero, this polynomial is equal to zero, and that's pretty easy to verify. Let's see, can x-squared plus nine equal zero? X-squared plus nine equal zero. Well, if you subtract nine from both sides, you get x-squared is equal to negative nine. And that's why I said, there's no real solution to this. So, no real, let me write that, no real solution. There are some imaginary solutions, but no real solutions. Now, can x plus the square root of two equal zero? Sure, if we subtract square root of two from both sides, you get x is equal to the negative square root of two. And can x minus the square root of two equal zero? Sure, you add square root of two to both sides, you get x is equal to the square root of two. So, there we have it. We have figured out our zeros. X could be equal to zero. P of zero is zero. P of negative square root of two is zero, and p of square root of two is equal to zero. So, those are our zeros. Their zeros are at zero, negative squares of two, and positive squares of two. And so those are going to be the three times that we intercept the x-axis. And what is the smallest of those intercepts? Well, the smallest number here is negative square root, negative square root of two. And you could tackle it the other way. You could take this part right over... Yeah, this part right over here and you could add those two middle terms, and I encourage you to do that. But just to see that this makes sense that zeros really are the x-intercepts. I went to Wolfram|Alpha and I graphed this polynomial and this is what I got. So, this is what I got, right over here. If you see a fifth-degree polynomial, say, it'll have as many as five real zeros. But, if it has some imaginary zeros, it won't have five real zeros. Instead, this one has three. And that's because the imaginary zeros, which we'll talk more about in the future, they come in these conjugate pairs. So, if you don't have five real roots, the next possibility is that you're going to have three real roots. And, if you don't have three real roots, the next possibility is you're gonna have one real root. So, that's an interesting thing to think about. And so, here you see, your three real roots. You see your three real roots which correspond to the x-values at which the function is equal to zero," + }, + { + "Q": "at approximately 14:20, how would you write\n-1 (x^2 + 5x - 24) as a two parentheses group? like this? ---------> -1 (x-3)(x+8)\nor, do you leave it as is?", + "A": "Yes, you can write it the way you suggested: -1 (x-3)(x+8) Exactly how you would write it depends on what you need to do to reach the final answer. You can also use the distributive property to distribute the -1 into ONE (not both) of the factors: -1 (x-3)(x+8) = (3-x)(x+8) however, doing this is rather unusual.", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 860 + ], + "3min_transcript": "And then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8, times x plus 7. This is often one of the hardest concepts people learn in algebra, because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors when one is positive, one is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more. Let's say we had negative x squared-- everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x, plus 24. How do we do this? Well, the easiest way I can think of doing it is factor we've been doing before. So this is the same thing as negative 1 times positive x squared, plus 5x, minus 24. Right? I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1. And you get that right there. Now, same game as before. I need two numbers, that when I take their product I get negative 24. So one will be positive, one will be negative. When I take their sum, it's going to be 5. So let's think about 24 is 1 and 24. Let's see, if this is negative 1 and 24, it'd be positive 23, if it was the other way around, it'd be negative 23. Doesn't work. What about 2 and 12? be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works! So if we pick negative 3 and 8, negative 3 and 8 work. Because negative 3 plus 8 is 5. Negative 3 times 8 is negative 24. So this is going to be equal to-- can't forget that negative 1 out front, and then we factor the inside. Negative 1 times x minus 3, times x plus 8. And if you really wanted to, you could multiply the negative 1 times this, you would get 3 minus x if you did. Or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared" + }, + { + "Q": "At 0:57, does he use the FOIL method?", + "A": "Yep! First Outside Inside Last", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 57 + ], + "3min_transcript": "In this video I want to do a bunch of examples of factoring a second degree polynomial, which is often called a quadratic. Sometimes a quadratic polynomial, or just a quadratic itself, or quadratic expression, but all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, in all of the examples we'll do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x, plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a, and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, Or if we want to add these two in the middle right here, because they're both coefficients of x. We could right this as x squared plus-- I can write it as b plus a, or a plus b, x, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And, of course, this is the same thing as this. Is there some a and b where a plus b is equal to 10? And a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer. And normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3-- that doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9." + }, + { + "Q": "at 14:46 we do not have to multiply negative 1 to both (x-3)(x+8) ?", + "A": "No, because -x*-x would equal positive x\u00c2\u00b2. The goal is to get -x\u00c2\u00b2. So, -x*x = -x\u00c2\u00b2 Hope this helps!", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 886 + ], + "3min_transcript": "we've been doing before. So this is the same thing as negative 1 times positive x squared, plus 5x, minus 24. Right? I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1. And you get that right there. Now, same game as before. I need two numbers, that when I take their product I get negative 24. So one will be positive, one will be negative. When I take their sum, it's going to be 5. So let's think about 24 is 1 and 24. Let's see, if this is negative 1 and 24, it'd be positive 23, if it was the other way around, it'd be negative 23. Doesn't work. What about 2 and 12? be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works! So if we pick negative 3 and 8, negative 3 and 8 work. Because negative 3 plus 8 is 5. Negative 3 times 8 is negative 24. So this is going to be equal to-- can't forget that negative 1 out front, and then we factor the inside. Negative 1 times x minus 3, times x plus 8. And if you really wanted to, you could multiply the negative 1 times this, you would get 3 minus x if you did. Or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared, minus 18x, plus 72. Now we just have to think of two numbers, that when I multiply them I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18. So they're the same sign, and their sum is a negative number, they both must be negative. And we could go through all of the factors of 72. But the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9, doesn't work. That turns into 17. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17." + }, + { + "Q": "At 13:55 it says that we turn 24 into 1 and 24,and then that if it is negative 1 and 24 it would be positive 23. Can you explain this?", + "A": "(a+b)^2 = a^2 + 2ab + b^2 Sal meant that 2ab would = positive 23 if a and b were -1 and +24", + "video_name": "eF6zYNzlZKQ", + "timestamps": [ + 835 + ], + "3min_transcript": "And then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8, times x plus 7. This is often one of the hardest concepts people learn in algebra, because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors when one is positive, one is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more. Let's say we had negative x squared-- everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x, plus 24. How do we do this? Well, the easiest way I can think of doing it is factor we've been doing before. So this is the same thing as negative 1 times positive x squared, plus 5x, minus 24. Right? I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1. And you get that right there. Now, same game as before. I need two numbers, that when I take their product I get negative 24. So one will be positive, one will be negative. When I take their sum, it's going to be 5. So let's think about 24 is 1 and 24. Let's see, if this is negative 1 and 24, it'd be positive 23, if it was the other way around, it'd be negative 23. Doesn't work. What about 2 and 12? be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works! So if we pick negative 3 and 8, negative 3 and 8 work. Because negative 3 plus 8 is 5. Negative 3 times 8 is negative 24. So this is going to be equal to-- can't forget that negative 1 out front, and then we factor the inside. Negative 1 times x minus 3, times x plus 8. And if you really wanted to, you could multiply the negative 1 times this, you would get 3 minus x if you did. Or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared" + }, + { + "Q": "I instantly went from getting it to COMPLETELY LOST.\n\nAt 2:14 he says, \"But we have i times i, or i squared, which is negative one.\" He lost me right there. I have no idea how he can say i squared is negative one. Can anyone explain this to me?", + "A": "The definition of i is the number whose square is -1. What he said was completely valid. You should watch the videos on i.", + "video_name": "Z8j5RDOibV4", + "timestamps": [ + 134 + ], + "3min_transcript": "We're asked to divide. And we're dividing six plus three i by seven minus 5i. And in particular, when I divide this, I want to get another complex number. So I want to get some real number plus some imaginary number, so some multiple of i's. So let's think about how we can do this. Well, division is the same thing -- and we rewrite this as six plus three i over seven minus five i. These are clearly equivalent; dividing by something is the same thing as a rational expression where that something is in the denominator, right over here. And so how do we simplify this? Well, we have a tool in our toolkit that can make sure that we don't have an imaginary or complex number in the denominator. And that's the complex conjugate. If we multiply both the numerator and the denominator of this expression by the complex conjugate of the denominator, then we will have a real number in the denominator. So let's do that. Let's multiply the numerator and the denominator by the conjugate of this. So seven PLUS five i. Seven plus five i is the complex conjegate of seven minus five i. And anything divided by itself is going to be one (assuming you're not dealing with zero; zero over zero is undefined). But seven plus five i over seven plus five i is one. So we're not changing the value of this. But what this does is it allows us to get rid of the imaginary part in the denominator. So let's multiply this out. Our numerator -- we just have to multiply every part of this complex number times every part of this complex number. You can think of it as FOIL if you like; we're really just doing the distributive property twice. We have six times seven, which is forty two. And then we have six times five i, which is thirty i. So plus thirty i. And then we have three i times seven, so that's plus twenty-one i. Three times five is fifteen. But we have i times i, or i squared, which is negative one. So it would be fifteen times negative one, or minus 15. So that's our numerator. And then our denomenator is going to be -- Well, we have a plus b times a minus b. (You could think of it that way. Or we could just do what we did up here. Actually, let's just do what we did up here so you don't have to remember that difference of squares pattern and all that.) Seven times seven is forty-nine. Let's think of it in the FOIL way, if that is helpful for you. So first we did the 7X7. And we can do the outer terms. 7 X 5i is +35i. Then we can do the inner terms. -5i X 7 is -35i. These two are going to cancel out. And then -5i X 5i is -25i^2 (\"negative twenty five i squared\")." + }, + { + "Q": "At 1:18 sal says 0/0 is undefined why?", + "A": "Division by zero is an operation for which you cannot find an answer, so it is disallowed. You can understand why if you think about how division and multiplication are related. 12 divided by 6 is 2 because: 6 times 2 is 12 Now image 12/0: 12 divided by 0 is x would mean that: 0 times x = 12 But no value would work for x because 0 times any number is 0. So division by zero doesn t work.", + "video_name": "Z8j5RDOibV4", + "timestamps": [ + 78 + ], + "3min_transcript": "We're asked to divide. And we're dividing six plus three i by seven minus 5i. And in particular, when I divide this, I want to get another complex number. So I want to get some real number plus some imaginary number, so some multiple of i's. So let's think about how we can do this. Well, division is the same thing -- and we rewrite this as six plus three i over seven minus five i. These are clearly equivalent; dividing by something is the same thing as a rational expression where that something is in the denominator, right over here. And so how do we simplify this? Well, we have a tool in our toolkit that can make sure that we don't have an imaginary or complex number in the denominator. And that's the complex conjugate. If we multiply both the numerator and the denominator of this expression by the complex conjugate of the denominator, then we will have a real number in the denominator. So let's do that. Let's multiply the numerator and the denominator by the conjugate of this. So seven PLUS five i. Seven plus five i is the complex conjegate of seven minus five i. And anything divided by itself is going to be one (assuming you're not dealing with zero; zero over zero is undefined). But seven plus five i over seven plus five i is one. So we're not changing the value of this. But what this does is it allows us to get rid of the imaginary part in the denominator. So let's multiply this out. Our numerator -- we just have to multiply every part of this complex number times every part of this complex number. You can think of it as FOIL if you like; we're really just doing the distributive property twice. We have six times seven, which is forty two. And then we have six times five i, which is thirty i. So plus thirty i. And then we have three i times seven, so that's plus twenty-one i. Three times five is fifteen. But we have i times i, or i squared, which is negative one. So it would be fifteen times negative one, or minus 15. So that's our numerator. And then our denomenator is going to be -- Well, we have a plus b times a minus b. (You could think of it that way. Or we could just do what we did up here. Actually, let's just do what we did up here so you don't have to remember that difference of squares pattern and all that.) Seven times seven is forty-nine. Let's think of it in the FOIL way, if that is helpful for you. So first we did the 7X7. And we can do the outer terms. 7 X 5i is +35i. Then we can do the inner terms. -5i X 7 is -35i. These two are going to cancel out. And then -5i X 5i is -25i^2 (\"negative twenty five i squared\")." + }, + { + "Q": "1:50 i still dont understand what the difference is between midrange and range? help?", + "A": "The midrange is an attempt to show in a single number how well or bad a class is doing, so it s pretty similar to the average and the median. The range is only used to show how big the difference is between the best and the worst student. It can t be used to judge how well or bad a class is doing. Only if all of the students are on a similar level or if there are huge differences between the students.", + "video_name": "DGZNaKnbQo0", + "timestamps": [ + 110 + ], + "3min_transcript": "In this chart right here we're given scores on midterm and final exams, where the vertical axis is the score in points. And then each of these pairs of bar charts give us for an individual, where the blue bar is, for example, how Ishaan did on the midterm, the yellow is how he did on the final. For Emily, the blue is how she did on the midterm, yellow is how she did on the final. And we have a bunch of interesting questions here. The first question, what was the median score for the final exam? So just as a review, median literally means what was the middle score? So really we should list all the scores for the final exam and sort them in order and then figure out what the middle score actually was. So let's look at all the scores on final exam. Sp you have 100 here. Ishaan got 100 on the final exam. Remember, this yellow bar is the final exam. So there's 100. Emily also got 100 on the final exam. It looks like it was an easy final exam. Daniel also got 100 on the final exam. And then, let's see, Jessica, it looks like she got a 75. So if we were to sort these in order, and let's say we did it in increasing order, you could write-- well, the lowest score was a 75, then you have an 80, and then you have three 100's-- 100, another 100, and another 100. So there's five scores right over here, so you will have a middle. If you had an even number, then you would take the mean of the two center values. But here you have one center value, and when you order it like this, it's pretty clear that your center value, your middle value, is 100. So the median score for the final exam is 100. And that's because you had so many hundreds here that the median, the middle score, was still 100. What is the midrange of the midterm scores? I'll do it in blue in honor of the color of the bars for the midterm. So the midrange is the mean of your highest and lowest scores. So let's calculate this. So let's go to the midrange. or the average of your highest and lowest scores, so the midrange of midterm. So let's see, the highest midterm score, looking at the blue, the highest one is right here. So Jessica got 100 on the midterm, so that's your highest score. Your lowest score on the midterm looks like this one right over here. Daniel got a 60. And so the midrange is going to be the mean, the arithmetic mean of these two numbers. So you add 100 plus 60, divide by 2, you get 160 over 2, or 80. So this right over here is going to be 80. What was the average student score for the final exam? Well, for, that we just have to add up the scores on the final exams and then divide by the number of scores we have. So we might be able to do that in our heads. Well, we could-- let me just write it over here. So we have 100, plus 100, plus 100, plus 75, plus 80." + }, + { + "Q": "At 3:49, how is the abs(-x^2/3) the same as abs(x^2/3) ? Wouldn't the negative sign have to be like this: abs{(-x^2)/3}, for the negative sign to disappear since -x times -x is positive x^2? I know this is very basic, but it's confusing me...", + "A": "Let s solve it in the normal way: |-x\u00c2\u00b2/3| < 1 \u00e2\u0086\u0094 -1 < -x\u00c2\u00b2/3 < 1 \u00e2\u0086\u0094 -3 < -x\u00c2\u00b2 < 3 \u00e2\u0086\u0094 3 > x\u00c2\u00b2 > -3 Because x\u00c2\u00b2 >= 0 > -3 for every x so we just need to take care x\u00c2\u00b2 < 3 \u00e2\u0086\u0094 -\u00e2\u0088\u009a3 < x < \u00e2\u0088\u009a3.", + "video_name": "aiwy2fNF_ZQ", + "timestamps": [ + 229 + ], + "3min_transcript": "this would be equal to-- so the first term is 1/3 times all of this to the 0-th power. So it's just going to be 1/3. And so each successive term is just going to be the previous term times our common ratio. So 1/3 times negative x squared over 3 is going to be negative 1/9 x squared. To go from that to that, you have to multiply by-- let's see, 1/3 to negative 1/3, you have to multiply it by negative 1/3. And we multiplied by x squared as well. Now in our next term, we're going to multiply by negative x squared over 3 again. So it's going to be plus-- a negative times a negative is a positive-- plus 1/27 x to the fourth. x squared times x squared, x to the fourth power. And we just keep going on and on and on. And when this converges, so over the interval of convergence, Now, what is the interval of convergence here? And I encourage you to pause the video and think about it. Well, the interval of convergence is the interval over which your common ratio, the absolute value of your common ratio, is less than 1. So let me write this right over here. So our absolute value of negative x squared over 3 has to be less than 1. Well, the absolute value, this is going to be a negative number. This is the same thing as saying-- let me scroll down a little bit. This is the same thing as saying that the absolute value of x squared over 3 has to be less than 1. And this is another way of saying-- this is going to be positive no matter what. Or I guess I should say, this is going to be non-negative no matter what. So this is another way of saying that x squared over 3 has to be less than 1. Right? I don't want to confuse you in this step right over here. But the absolute value of x squared over 3 is just going to be x squared over 3, because this is never going to take on a negative value. And so we can multiply both sides by 3. I'll go up here now to do it. Multiply both sides by 3 to say that x squared needs to be less than 3. And so that means that the absolute value of x needs to be less than the square root of 3. Or we could say that x is greater than the negative square root of 3, and it is less than the square root of 3. So this is the interval of convergence. This is the interval of convergence for this series," + }, + { + "Q": "How come 13/10. At 2:40?", + "A": "A one above a zero is is 10/10 s where are the 0.3 is 3/10 s. If you add 3/10 s to 10/10 s you will get 13/10 s.", + "video_name": "3szFVS5p_7A", + "timestamps": [ + 160 + ], + "3min_transcript": "So we still have 2 tens. So this is still going to be 2 tens. Now we have plus 0 ones. And we essentially wanted to write that 1 that we took away from the ones place in terms of tenths. So if we were to write this in terms of tenths, it would be 10/10 plus the 3/10 that were already there. And so this is going to be equal to 13/10. Let me write that down. So this is equal to 20. That's the color you can't see. This is equal to 20 plus 0 ones, so 2 tens plus 0 ones plus 13 tens. Let's do another example with this exact same number. So once again, 21.3. And I'll write it out again. This is equal to 20 plus 1. Plus 1 plus 3/10, plus 3 over 10. Now, I could take 1 from the tens place so that this becomes just 1. Now what do I do with that 10? Well, let's say with that 10 I give 9 of it to the ones place. So I give 9 of it to the ones place so that this becomes 10. And I still have 1 left over, and I give it to the tenths place, so that's going to become 13/10. So what did I just do? Well, I could rewrite this. Let me be clear what I did. This is the same thing as 1 plus 9. Actually, let me write it this way. 1 plus 9 plus 1. That's obviously the same thing-- 10 And of course, we have what we have in our ones place, plus 1 plus 3/10. And what I want to do is I want to take this 9, the 9 that I took from the tens place and give to the ones place. And I'm going to take this 1 that I took from the tens place and give it to the tenths place. So 1 is the same thing as 10/10. And so when you regroup this value, you get this as being equal to 10 plus-- 9 plus 1 is 10, and then 10/10 plus 3/10 is 13/10. So that's all that happened here. I changed the value in the places. I took 1 ten away. I had 2 tens. Now I'm only left with 1 ten. And that extra 10 of value, I regrouped it. I gave 9 to the ones place." + }, + { + "Q": "If you did not simplified 2xh+x*/h and you are finding the limit of h as it get closer to 0, doesn't the question become undefined number since it is dividing by zero. I tried it with a function of x+2 and it gave me h/h after all the simplifying. Did I do something wrong? Video(5:22 - 8:28)", + "A": "The limit takes care of that issue, we are not dividing by 0, but by a number that approaches 0. The slope of the function x+2 is 1, so it makes sense you ended with the answer of h/h since the Limit as h approaches 0 of the function h/h = 1", + "video_name": "IePCHjMeFkE", + "timestamps": [ + 322, + 508 + ], + "3min_transcript": "" + }, + { + "Q": "At 7:46 how does he get 6 from 6+delta(x)??", + "A": "Basically we re asking what happens as the \u00ce\u0094x approaches zero. So pretend that \u00ce\u0094x is actually zero and then you have 6 + 0 and 6 + 0 = 6.", + "video_name": "IePCHjMeFkE", + "timestamps": [ + 466 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:20, why do you have to move left, why can't you move right. Is it because the place values get lower when you go to the eftl", + "A": "Exactly. If you go to the left, the place value gets 20 times greater. For example, there is the ones place, the tens place, hundreds place, thousands place, ten thousands place, hundred thousands place, millions place, ten millions place, and you get the idea. But, if you go to the right side, the place value gets 1/10 smaller.", + "video_name": "iK0y39rjBgQ", + "timestamps": [ + 80 + ], + "3min_transcript": "Write 14,897 in expanded form. Let me just rewrite the number, and I'll color code it, and that way, we can keep track of our digits. So we have 14,000. I don't have to write it-- well, let me write it that big. 14,000, 800, and 97-- I already used the blue; maybe I should use yellow-- in expanded form. So let's think about what place each of these digits are in. This right here, the 7, is in the ones place. The 9 is in the tens place. This literally represents 9 tens, and we're going to see this in a second. This literally represents 7 ones. The 8 is in the hundreds place. It literally represents 4,000. And then the 1 is in the ten-thousands place. And you see, every time you move to the left, you move one place to the left, you're multiplying by 10. Ones place, tens place, hundreds place, thousands place, ten-thousands place. Now let's think about what that really means. If this 1 is in the ten-thousands place, that means that it literally represents-- I want to do this in a way that my arrows don't get mixed up. Actually, let me start at the other end. Let me start with what the 7 represents. The 7 literally represents 7 ones. Or another way to think about it, you could say it represents 7 times 1. All of these are equivalent. They represent 7 ones. That's why I'm doing it from the right, so that the arrows don't have to cross each other. So what does the 9 represent? It represents 9 tens. You could literally imagine you have 9 actual tens. You could have a 10, plus a 10, plus a 10. Do that nine times. That's literally what it represents: 9 actual tens. 9 tens, or you could say it's the same thing as 9 times 10, or 90, either way you want to think about it. So let me write all the different ways to think about it. It represents all of these things: 9 tens, or 9 times 10, or 90. So then we have our 8. Our 8 represents-- we see it's in the hundreds place. It represents 8 hundreds. Or you could view that as being equivalent to 8 times 100-- a hundred, not a thousand-- 8 times 100, or 800. That 8 literally represents 8 hundreds, 800." + }, + { + "Q": "Okay... he lost me at 0:14. I have no idea what's going on, can someone help me?", + "A": "a difference of square is a binomial in which both the terms are perfect squares and they are subtracted a2-b2 if you have a difference of squares expression here is how you would factor it a2-b2=(a+b)(a-b) in this case it is x2-49y2 a=x b=7y x2-49y2=(x+7y)(x-7y)", + "video_name": "tvnOWIoeeaU", + "timestamps": [ + 14 + ], + "3min_transcript": "Factor x squared minus 49y squared. So what's interesting here is that well x squared is clearly a perfect square. It's the square of x. And 49y squared is also a perfect square. It's the square of 7y. So it looks like we might have a special form here. And to remind ourselves, let's think about what happens if we take a plus b times a minus b. I'm just doing it in the general case so we can see a pattern here. So over here, this would be a times a, which would be a squared plus a times negative b, which would be negative ab plus b times a or a times b again, which would be ab. And then you have b times negative b, so it would b minus b squared. Now these middle two terms cancel out. Negative ab plus ab, they cancel out and you're left with just a squared minus b squared. And that's the exact pattern we have here. We have an a squared minus a b squared. So in this case, a is equal to x and b is equal to 7y. So we can expand this as the difference of squares, or actually this thing right over here is the difference of squares. So we expand this like this. So this will be equal to x plus 7y times x minus 7y. And once again, we're just pattern matching based on this realization right here. If I take a plus b times a minus b, I get a difference of squares. This is a difference of squares. So when I factor it, it must come out to the result of something that looks like a plus b times a minus b or x plus 7y times x minus 7y." + }, + { + "Q": "When, at 1:09 he shows what happens as a perfect square for A and B (that b is 7y, because 49 is a perfect square) would that work for say, 5, where for example it might be b=sqrt(5)? Or am I jumping to conclusions?", + "A": "it wouldn t be a perfect square, so no.", + "video_name": "tvnOWIoeeaU", + "timestamps": [ + 69 + ], + "3min_transcript": "Factor x squared minus 49y squared. So what's interesting here is that well x squared is clearly a perfect square. It's the square of x. And 49y squared is also a perfect square. It's the square of 7y. So it looks like we might have a special form here. And to remind ourselves, let's think about what happens if we take a plus b times a minus b. I'm just doing it in the general case so we can see a pattern here. So over here, this would be a times a, which would be a squared plus a times negative b, which would be negative ab plus b times a or a times b again, which would be ab. And then you have b times negative b, so it would b minus b squared. Now these middle two terms cancel out. Negative ab plus ab, they cancel out and you're left with just a squared minus b squared. And that's the exact pattern we have here. We have an a squared minus a b squared. So in this case, a is equal to x and b is equal to 7y. So we can expand this as the difference of squares, or actually this thing right over here is the difference of squares. So we expand this like this. So this will be equal to x plus 7y times x minus 7y. And once again, we're just pattern matching based on this realization right here. If I take a plus b times a minus b, I get a difference of squares. This is a difference of squares. So when I factor it, it must come out to the result of something that looks like a plus b times a minus b or x plus 7y times x minus 7y." + }, + { + "Q": "At 1:29, Sal says that 0.1 is bigger than 0.070. How is that possible when 0.070 has more digits than 0.1", + "A": "One of the easiest ways to compare decimals, it to give them the same number of decimal digits. Adding zeros on the right of the decimal does not change the original value of the number. Change 0.1 into 0.100 You are now comparing 0.100 to 0.070 100 is bigger than 70. Thus, 0.1 is larger then 0.070 Another way to look at this is the the further to the right the number is following the decimal point, the smaller the number. The 1 in 0.1 = 1/10. The 7 in 0.070 = 7/100 7/100 is much smaller than 1/10 Hope this helps.", + "video_name": "gAV9kwvoD6s", + "timestamps": [ + 89 + ], + "3min_transcript": "Let's compare 0.1 to 0.070. So this 1 right over here, it is in the tenths place. So it literally represents 1 times 1/10, which is obviously the same thing as 1/10. Now, when we look at this number right over here, it has nothing in the tenths place. It has 7 in the hundredths place. So this is the hundredths place right over here. And then it also has nothing in the thousandths place. So this number can be rewritten as 7 times 1/100, or 7/100. And now we could compare these two numbers. And there's two ways you could think about it. You could try to turn 1/10 into hundredths. And the best way to do that, if you want the denominator to be increased by a factor of 10, you need to do the same thing to the numerator. So all I did is I multiplied the numerator and denominator by 10. Ten 100's is the exact same thing as 1/10. And here it becomes very clear, 10/100 Another way you could think about this is, look, if you were to increment by hundredths here, you would start at 7/100, 8/100, 9/100, and then you would get to 10/100. So then you would get to that number. So this number, multiple ways you could think about it, is definitely larger. So let me write this down. This is definitely larger, greater than. This is greater than that. The greater than symbol opens to the larger value. So here we have 0.093 and here we have 0.01. So let's just think about this a little bit. So this 9-- get a new color here. This 9 is not in the tenths, the hundredths. It's in the thousandths place. It's in the thousands place. And this 3 is in the-- I'm running out of colors again. This 3 is in the ten thousandths place. So the 3 is in the ten thousandths place. And if you just wanted to write it in terms of ten thousandths, you can multiply the 9 and 1,000 by 0. And so it becomes 90/10,000. And if you want to add them together, you could, of course, write this as 93/10,000. Ten thousandths. I always have trouble with that \"-ths\" at the end. Now, let's think about this number right over here, 0.01. Well, this 1 right over here is in the hundredths place. It's in the hundredths place. So it literally represents 1/100. So how can we compare 1/100 to 93/10,000? So the best way to think about it is, well, what's 1/100 in terms of ten thousandths? Well, let's just multiply both the numerator and the denominator here by 10 twice. Or you could say, let's multiply them both by 100." + }, + { + "Q": "at 4:52 he says over 2 does that apply all the time or just for this instance?", + "A": "The midpoint formula is ((x1+x2)/2,(y1+y2)/2). This applies all the time.", + "video_name": "Efoeqb6tC88", + "timestamps": [ + 292 + ], + "3min_transcript": "Well along the imaginary axis we're going from negative one to three so the distance there is four. So now we can apply the Pythagorean theorem. This is a right triangle, so the distance is going to be equal to the distance. Let's just say that this is x right over here. x squared is going to be equal to seven squared, this is just the Pythagorean theorem, plus four squared. Plus four squared or we can say that x is equal to the square root of 49 plus 16. I'll just write it out so I don't skip any steps. 49 plus 16, now what is that going to be equal to? That is 65 so x, that's right, 59 plus another 6 is 65. x is equal to the square root of 65. There's no factors that are perfect squares here, this is just 13 times five so we can just leave it like that. x is equal to the square root of 65 so the distance in the complex plane between these two complex numbers, square root of 65 which is I guess a little bit over eight. Now what about the complex number that is exactly halfway between these two? Well to figure that out, we just have to figure out what number has a real part that is halfway between these two real parts and what number has an imaginary part that's halfway between these two imaginary parts. So if we had some, let's say that some complex number, let's just call it a, is the midpoint, it's real part is going to be the mean of these two numbers. So it's going to be two plus negative five. Two plus negative five over two, over two, and it's imaginary part is going to be the mean of these two numbers so plus, plus three minus one. and this is equal to, let's see, two plus negative five is negative three so this is negative 3/2 plus this is three minus 1 is negative, is negative two over two is let's see three, make sure I'm doing this right. Three, something in the mean, three minus one is two divided by two is one, so three plus three. Negative 3/2 plus i is the midpoint between those two and if we plot it we can verify that actually makes sense. So real part negative 3/2, so that's negative one, negative one and a half so it'll be right over there and then plus i so it's going to be right over there." + }, + { + "Q": "At 4:06, What do you mean by rotating the plane in infinite directions?", + "A": "the plane could be facing any direction but you don t know by the representation.", + "video_name": "J2Qz-7ZWDAE", + "timestamps": [ + 246 + ], + "3min_transcript": "So it doesn't seem like just a random third point is sufficient to define, to pick out any one of these planes. But what if we make the constraint that the three points are not all on the same line. Obviously, two points will always define a line. But what if the three points are not collinear. So instead of picking C as a point, what if we pick-- Is there any way to pick a point, D, that is not on this line, that is on more than one of these planes? We'll, no. If I say, well, let's see, the point D-- Let's say point D is right over here. So it sits on this plane right over here, one of the first ones that I drew. So point D sits on that plane. Between point D, A, and B, there's only one plane that all three of those points sit on. So a plane is defined by three non-colinear points. So D, A, and B, you see, do not sit on the same line. A and B can sit on the same line. D and A can sit on the same line. But A, B, and D does not sit on-- They are non-colinear. So for example, right over here in this diagram, we have a plane. This plane is labeled, S. But another way that we can specify plane S is we could say, plane-- And we just have to find three non-collinear points on that plane. So we could call this plane AJB. We could call it plane JBW. We could call it plane-- and I could keep going-- plane WJA. But I could not specify this plane, uniquely, by saying plane ABW. And the reason why I can't do this is because ABW are all on the same line. I could keep rotating around the line, just as we did over here. It does not specify only one plane." + }, + { + "Q": "at 3:26 Sal said that 49+9 is 57 but it is 58 right?", + "A": "Yes. This is a known error in the video. And, a box does pop up and tell you the error and correct info.", + "video_name": "-U53eHKCLcg", + "timestamps": [ + 206 + ], + "3min_transcript": "Greek letter delta is just shorthand for change in y and this once again comes straight out of the distance formula which really comes out of the Pythagorean theorem. It'll become a little bit more obvious when I draw the change in y and I draw the change in x on this diagram. So what's our change in y? Well we're starting at our initial point. We're starting at y is equal to nine and we are, to get to the y value of our terminal point, we're going down to y is equal to two. So we have a change in y our change in y is equal to, going from nine to two, our change in y is negative seven. Similarly, our change in x we're going from x is equal to two, to x is equal to five. So our change in the horizontal direction is plus three. So our change in x where we can either think of this as the horizontal component of the vector. This is equal to positive three we have drawn a right triangle and so we can use a Pythagorean theorem to figure out, to figure out the length of the hypotenuse and you might say wait, wait, a length of a side of a triangle can't have a negative value and that's why these squareds are valuable because it doesn't matter if you're taking a negative seven squared or a positive seven squared, you're going to get a positive value here and if you really just view this as a triangle, all you care about is the length of this side right over here or it's the magnitude of this side or the absolute value of it which is just going to be positive seven and so we can say this is going to be equal to the magnitude of our vector is going to be equal to so three squared is nine, nine, and then negative seven squared is positive 49. So plus 49 or once again you could view this as our change in y squared which is negative seven squared or you could say, well just look, we don't wanna think of a side as having a negative value, the negative really just says, hey we're going from the top to the bottom it gives us our direction, but if we just say the length of it it's seven, well you're there if you just use a Pythagorean theorem. Seven squared would also be 49 and so either way you get the magnitude of our vector is equal to the square root of nine plus 49 is going to be 57, I don't think I can simplify this radical too much, no that's it. So the magnitude of this vector is the square root of 57." + }, + { + "Q": "At 3:33 the value would be square root of 58.", + "A": "Hello Soham, Correct, this is a know problem based on the number of comment it has received... Regards, APD", + "video_name": "-U53eHKCLcg", + "timestamps": [ + 213 + ], + "3min_transcript": "Greek letter delta is just shorthand for change in y and this once again comes straight out of the distance formula which really comes out of the Pythagorean theorem. It'll become a little bit more obvious when I draw the change in y and I draw the change in x on this diagram. So what's our change in y? Well we're starting at our initial point. We're starting at y is equal to nine and we are, to get to the y value of our terminal point, we're going down to y is equal to two. So we have a change in y our change in y is equal to, going from nine to two, our change in y is negative seven. Similarly, our change in x we're going from x is equal to two, to x is equal to five. So our change in the horizontal direction is plus three. So our change in x where we can either think of this as the horizontal component of the vector. This is equal to positive three we have drawn a right triangle and so we can use a Pythagorean theorem to figure out, to figure out the length of the hypotenuse and you might say wait, wait, a length of a side of a triangle can't have a negative value and that's why these squareds are valuable because it doesn't matter if you're taking a negative seven squared or a positive seven squared, you're going to get a positive value here and if you really just view this as a triangle, all you care about is the length of this side right over here or it's the magnitude of this side or the absolute value of it which is just going to be positive seven and so we can say this is going to be equal to the magnitude of our vector is going to be equal to so three squared is nine, nine, and then negative seven squared is positive 49. So plus 49 or once again you could view this as our change in y squared which is negative seven squared or you could say, well just look, we don't wanna think of a side as having a negative value, the negative really just says, hey we're going from the top to the bottom it gives us our direction, but if we just say the length of it it's seven, well you're there if you just use a Pythagorean theorem. Seven squared would also be 49 and so either way you get the magnitude of our vector is equal to the square root of nine plus 49 is going to be 57, I don't think I can simplify this radical too much, no that's it. So the magnitude of this vector is the square root of 57." + }, + { + "Q": "at about 1:50, how do you get 2y=y+3? also, when you subtract y from both sides, you still need to divide by the 2, so wouldn't y be equal to 1 and 1/2? sorry if this is a stupid question but I could really use the help!:)", + "A": "To answer the first part of your question, we just get the +3 by combining -4 and +7. As for the second part, when you subtract y from 2y, you just end up with y, so you don t have any number you need to divide the 3 by. Hope that helped!", + "video_name": "uzyd_mIJaoc", + "timestamps": [ + 110 + ], + "3min_transcript": "Use substitution to solve for x and y. And we have a system of equations here. The first equation is 2y is equal to x plus 7. And the second equation here is x is equal to y minus 4. So what we want to do, when they say substitution, what we want to do is substitute one of the variables with an expression so that we have an equation and only one variable. And then we can solve for it. Let me show you what I'm talking about. So let me rewrite this first equation. 2y is equal to x plus 7. And we have the second equation over here, that x is equal to y minus 4. So if we're looking for an x and a y that satisfies both constraints, well we could say, well look, at the x and y have to satisfy both constraints, both of these constraints have to be true. So x must be equal to y minus 4. So anywhere in this top equation where we see an x, anywhere we see an x, we say well look, that x by the second constraint has to be equal to y minus 4. So everywhere we see an x, we can substitute it So let's do that. So if we substitute y minus 4 for x in this top equation, the top equation becomes 2y is equal to instead of an x, the second constraint tells us that x needs to be equal to y minus 4. So instead of an x, we'll write a y minus 4, and then we have a plus 7. All I did here is I substituted y minus 4 for x. The second constraint tells us that we need to do it. y minus 4 needs to be equal to x or x needs to be equal to y minus 4. The value here is now we have an equation, one equation with one variable. We can just solve for y. So we get 2y is equal to y, and then we have minus 4 plus 7. So y plus 3. We can subtract y from both sides of this equation. The left hand side, 2y minus y is just y. And then we could go back and substitute into either of these equations to solve for x. This is easier right over here, so let's substitute right over here. x needs to be equal to y minus 4. So we could say that x is equal to 3 minus 4 which is equal to negative 1. So the solution to this system is x is equal to negative 1 and y is equal to 3. And you can verify that it works in this top equation 2 times 3 is 6 which is indeed equal to negative 1 plus 7. Now I want to show you that over here we substituted-- we had an expression that, or we had an equation, that explicitly solved for x. So we were able to substitute the x's. What I want to show you is we could have done it the other way around. We could have solved for y and then substituted for the y's." + }, + { + "Q": "at 1:10 couldn't \"a\" be a different value", + "A": "Yes, it can. A variable can be any number you choose it to be. For example, X is the most used variable but, in the video Sal used a as the variable. This has no effect in the equation.", + "video_name": "P6_sK8hRWCA", + "timestamps": [ + 70 + ], + "3min_transcript": "Let's say that you started off with 3 apples. And then I were to give you another 7, another 7 apples. So my question to you-- and this might be very obvious-- is how many apples do you now have? And I'll give you a second to think about that. Well, this is fairly basic. You had 3 apples. Now, I'm going to give you 7 more. You now have 3 plus 7. You now have 10 apples. But let's say I want to do the same type of thinking, but I'm too lazy to write the word \"apples.\" Let's say instead of writing the word \"apples,\" I just use the letter a. And let's say this is, say, a different scenario. You start off with 4 apples. And to that, I add another 2 apples. How many apples do you now have? Instead of writing apples, I'm just going to write a's here. So how many of these a's do you now have? And once again, I'll give you a few seconds This also might be a little bit of common sense for you. If you had 4 of these apples or whatever these a's represented, if you had 4 of them and then you add 2 more of them, you're now going to have 6 of these apples. But once again, we started off assuming that a's represent apples. But they could have represented anything. If you have 4 of whatever a represents, and then you have another 2 of whatever a represents, you'll now have 6 of whatever a represents. Or if you just think of it if I have 4 a's, and then I add another 2 a's, I'm going to have 6 a's. You can literally think of 4 a's as a plus a plus a plus a. And if to that, I add another 2 a's-- so plus a plus a, that's 2 a's right over there-- how many a's do I now have? Well, that's 1, 2, 3, 4, 5, 6. I now have 6 a's. So thinking of it that way, let's get a little bit more abstract. Let's say that I have 5 x's, whatever x represents. So I have 5 of whatever that number is. And from that, I subtract 2 of whatever that number is. What would this evaluate to? How many of these x's would I now have? So it's essentially 5x minus 2x is going to be what times x? Once again, I'll give you a few seconds to think about it. Well, if I have 5 of something and I subtract 2 of those away, I'm going to have 3 of that something left. So this is going to be equal to 3x. 5x minus 2x is equal to 3x. And if you really think about what that means, five x's are just x plus x plus x plus x plus x. And then we're going to take away two of those x's. So take away one x, take away two x's. You are going to be left with three x's." + }, + { + "Q": "This might seem like a pretty arbitrary question, but at 1:37 Sal uses the \"/\" symbol to signify x(a+3-b) over (a+3-b). I've been seeing that \"over than\" symbol a lot in algebra and I was wondering if there was any difference between the over than symbol, \"/\", and the division symbol I'm used to seeing, \"\u00c3\u00b7\"?", + "A": "They mean the same thing. I m honestly not sure why, but once you get past pre-algebra, you stop seeing the \u00c3\u00b7 symbol.", + "video_name": "adPgapI-h3g", + "timestamps": [ + 97 + ], + "3min_transcript": "- [Voiceover] So we have an equation. It says, a-x plus three-x is equal to b-x plus five. And what I want to do together is to solve for x, and if we solve for x it's going to be in terms of a, b, and other numbers. So pause the video and see if you can do that. All right now, let's do this together, and what I'm going to do, is I'm gonna try to group all of the x-terms, let's group all the x-terms on the left-hand side. So, I already have a-x and three-x on the left-hand side. Let's get b-x onto the left-hand side as well, and I can do that by subtracting b-x from both sides. And if I subtract b-x from both sides, I'm going to get on the right-hand side, I'm going to have a, or on the left-hand side, a-x plus three-x minus b-x, so I can do that in that color for fun, minus b-x, and that's going to be equal to... Well, b-x minus b-x is just zero, and I have five. It is equal to five. I can factor an x out of this left-hand side of this equation, out of all of the terms. So, I can rewrite this as x times... Well, a-x divided by x is a. Three-x divided by x is three, and then negative b-x divided by x is just going to be negative b. I could keep writing it in that pink color. And that's all going to be equal to five. And now, to solve for x I can just divide both sides by, the thing that x is being multiplied by, by a plus three minus b. So, I can divide both sides by a plus three minus b. A plus three minus b. On this side, they cancel out. And, I have x is equal to five over a plus three minus b, and we are done. Let's do one more of these. We have a... Here we have a times the quantity, five minus x, is equal to b-x minus eight. So, once again, pause the video and see if you can solve for x. Well, the way I like to approach these is let's just expand everything all out. So, let me just distribute this a, and then I'm gonna collect all the x-terms on one side, and all of the non-x-terms on the other side, and essentially do what I just did in the last example. So, let's first distribute this a. So the left-hand side becomes five-a, I could say a times five or five-a, minus a-x, a-x, that is going to be equal to b-x minus eight. Now we can subtract b-x from both sides. So, we're gonna subtract b-x from the left-hand side, b-x from the right-hand side. And, once again, the whole reason I'm doing that, I want all the x-terms on the left, and all the non-x-terms on the right. And, actually, since I want all the non-x-terms on the right," + }, + { + "Q": "At 4:06, Sal multiplies the numbers by -1. Do you have to do this?", + "A": "If you didn t do it, you would get: x = (-8-5a)/((-a-b) This fraction is not fully reduced. You would need to factor out a -1 from both the numerator and denominator to reduce the fraction.", + "video_name": "adPgapI-h3g", + "timestamps": [ + 246 + ], + "3min_transcript": "We have a... Here we have a times the quantity, five minus x, is equal to b-x minus eight. So, once again, pause the video and see if you can solve for x. Well, the way I like to approach these is let's just expand everything all out. So, let me just distribute this a, and then I'm gonna collect all the x-terms on one side, and all of the non-x-terms on the other side, and essentially do what I just did in the last example. So, let's first distribute this a. So the left-hand side becomes five-a, I could say a times five or five-a, minus a-x, a-x, that is going to be equal to b-x minus eight. Now we can subtract b-x from both sides. So, we're gonna subtract b-x from the left-hand side, b-x from the right-hand side. And, once again, the whole reason I'm doing that, I want all the x-terms on the left, and all the non-x-terms on the right. And, actually, since I want all the non-x-terms on the right, So I'm kind of doing two steps at once, here, but hopefully it makes sense. I'm trying to get rid of the b-x here, and I'm trying to get rid of the five-a here. So, I subtract five-a there, and I'll subtract five-a there, and then let's see what this give us. So, the five-a's cancel out. And, on the left-hand side, I have negative a-x, negative a-x, minus b-x, minus, you know, in that same green color, minus b-x. And on the right-hand side, This is going to be equal to, the b-x's cancel out, and I have negative eight minus five-a. Negative eight minus, in that same magenta color, minus five-a. And let's see, I have all my x's on one side, all my non-x's on the other side. And here I can factor out an x, And, actually, one thing that might be nice. Let me just multiply both sides by negative one. If I multiple both sides by negative one, I get a-x plus b-x, plus b-x is equal to eight plus five-a. That just gets rid of all of those negative signs. And now I can factor out an x here. So let me factor out an x, and I get x times a plus b. A plus b is going to be equal to eight plus five-a. Eight plus five-a. And we're in the home stretch now. We can just divide both sides by a plus b. So we could divide both sides by a plus b. A plus b. And we're going to be left with, x is equal to" + }, + { + "Q": "At 1:21 , why does it matter where p is?", + "A": "because the P is the variable for the pizzas. It is with the 8.5 because each pizza is 8.50. If it were with the 42.5, that would mean we are saying the cost of the pizza is 42.50 apiece.", + "video_name": "2REbsY4-S70", + "timestamps": [ + 81 + ], + "3min_transcript": "- [Voiceover] Anna wants to celebrate her birthday by eating pizza with her friends. For $42.50 total, they can buy p boxes of pizza. Each box of pizza costs $8.50. Select the equation that matches this situation. So before I even look at these, let's see if I can make sense of the sentence here. So for $42.50 total, and I'll just write 42.5, especially because in all these choices they didn't write 42.50, they just wrote 42.5 which is equivalent. So 42.50 that's the total amount they spent on pizza and if I wanted to figure out how many boxes of pizza they could buy, I could divide the total amount they spend, I could divide that by the price per box. That would give me the number of boxes. So this is the total, total dollars. This right over here is the dollar per box and then this would give me the number of boxes. Now other ways that I could think about it. I could say, well what's the total that they spend? So 42.50, but what's another way of thinking about the total they spend? Well you could have the amount they spend per box, times the number of boxes. So this is the total they spend and this another way of thinking about the total they spend, so these two things must be equal. So let's see, if I can see anything here that looks like this, well actually this first choice, this, is exactly, is exactly what I wrote over here. Let's see this choice right over here. P is equal to 8.5 x 42.5. Well we've already been able to write an equation that has explicitly, that has just a p on one side and so when you solve for just a p on one side, you get this thing over here, not this thing, so we could rule that out. Over here it looks kind of like this, except the p is on the wrong side. This has 8.5p is equal to 42.5, If we try to get the p on the other side here, you could divide both sides by p, but then you would get p divided by p is one. You would get 42.5 is equal to 8.5/p which is not true. We have 8.5 times p is equal to 42.5, so this is, this is not going to be the case. One thing to realize, no matter what you come up with, if you came up with this first, or if you came up with this first, you can go between these two with some algebraic manipulations. So for example, to go from this blue one to what I wrote in red up here, you just divide both sides by 8.5. So you divide by 8.5 on the left, you divide 8.5 on the right. Obviously to keep the equal sign you have to do the same thing to the left and right, but now you would have 42.5/8.5 is equal to, is equal to p. Which is exactly what we have over there. Let's do one more of these. Good practice." + }, + { + "Q": "Around 1:30, he explains that we need to use the pythagorean theory to find the radius r. But can't we just estimate the no. of units from the centre (point -1, 1) to the point (7.5, 1), which also lies on the circumference?", + "A": "sometimes you will need precision in your answers. Pythagorean Theorem will give you as much precision as you need", + "video_name": "iX5UgArMyiI", + "timestamps": [ + 90 + ], + "3min_transcript": "- [Voiceover] So we have a circle here and they specified some points for us. This little orangeish, or, I guess, maroonish-red point right over here is the center of the circle, and then this blue point is a point that happens to sit on the circle. And so with that information, I want you to pause the video and see if you can figure out the equation for this circle. Alright, let's work through this together. So let's first think about the center of the circle. And the center of the circle is just going to be the coordinates of that point. So, the x-coordinate is negative one and then the y-coordinate is one. So center is negative one comma one. And now, let's think about what the radius of the circle is. Well, the radius is going to be the distance between the center and any point on the circle. So, for example, for example, this distance. The distance of that line. Let's see I can do it thicker. A thicker version of that. This line, right over there. Something strange about my pen tool. It's making that very thin. Let me do it one more time. Okay, that's better. (laughs) The distance of that line right over there, that is going to be the radius. So how can we figure that out? Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean Theorem. To figure out the length of that line, so this is the radius, we could figure out a change in x. So, if we look at our change in x right over here. Our change in x as we go from the center to this point. So this is our change in x. And then we could say that this is our change in y. That right over there is our change in y. And so our change in x-squared plus our change in y-squared is going to be our radius squared. That comes straight out of the Pythagorean Theorem. This is a right triangle. is going to be equal to our change in x-squared plus our change in y-squared. Plus our change in y-squared. Now, what is our change in x-squared? Or, what is our change in x going to be? Our change in x is going to be equal to, well, when we go from the radius to this point over here, our x goes from negative one to six. So you can view it as our ending x minus our starting x. So negative one minus negative, sorry, six minus negative one is equal to seven. So, let me... So, we have our change in x, this right over here, is equal to seven. If we viewed this as the start point and this as the end point, it would be negative seven, but we really care about the absolute value in the change of x, and once you square it" + }, + { + "Q": "Is obtuse larger than acute?\n\nI LIKE KHAN ACADAMEY!\nAT 5:04", + "A": "Yes, an obtuse angle has one side of the triangle larger than 90*. An acute angle is an angle that is 89* or less. A right angle is exactly 90*.", + "video_name": "ALhv3Rlydig", + "timestamps": [ + 304 + ], + "3min_transcript": "" + }, + { + "Q": "did sal make a mistake by telling indefinite integrals at 0:31 or have I misunderstood ?\n\nAnd one more, how can you just get what dx is ? I mean , I know how to take d/dx but dx ? Please explain succinctly.", + "A": "If either of the bounds of an integral are + or - infinity, then the integral is improper. The video show how to use limits to solve such improper integrals. dx represents an infinitesimal change in x. If you can measure the change in x, the change is not infinitesimal and the change is called \u00ce\u0094x. If you CANNOT measure the change in x, the change IS infinitesimal, and is called dx.", + "video_name": "9JX2s90_RNQ", + "timestamps": [ + 31 + ], + "3min_transcript": "Right here we have the graph of y is equal to 250 over 25 plus x squared. And what I'm curious about in this video is the total area under this curve and above the x-axis. So I'm talking about everything that I'm shading in white here, including what we can't see, as we keep moving to the right and we keep moving to the left. So I'm talking about from x at negative infinity all the way to x at infinity. So first, how would we actually denote this? Well, it would be an improper integral. We would denote this area as the indefinite integral from x is equal to negative infinity to x is equal to infinity of our function, 250 over 25 plus x squared, dx. Now, we've already seen improper integrals where one of our boundaries was infinity. But how do you do it when you have one boundary at positive infinity and one boundary at negative infinity? You can't take a limit to two different things. And so the way that we're going to tackle this into two different improper integrals, one improper integral that describes this area right over here in blue from negative infinity to 0. So we'll say that this is equal to the improper integral that goes from negative infinity to 0 of 250 over 25 plus x squared dx, plus the improper integral that goes from 0 to positive infinity. So plus the improper, or the definite, integral from 0 to positive infinity of 250 over 25 plus x squared dx. And now we can start to make sense of this. So what we have in blue can be rewritten. This is equal to the limit as n approaches negative infinity of the definite integral from n to 0 of 250 over 25 Plus-- and I'm running out of real estate here-- the limit as-- since I already used n, let me use m now-- the limit as m approaches positive infinity of the definite integral from 0 to m of 250 over 25 plus x squared dx. So now all we have to do is evaluate these definite integrals. And to do that, we just have to figure out an antiderivative of 250 over 25 plus x squared. So let's try to figure out what that is. I'll do it over here on the left. So we need to figure out the antiderivative of 250 over 25 plus x squared. And it might already jump out at you that trig substitution might be a good thing to do. You see this pattern of a squared plus x squared, where in this case, a would be 5." + }, + { + "Q": "At 5:03, Sal said \"6.5 divided by 450\". Did he make a mistake? Can someone explain this?", + "A": "Yes he made a mistake, and I did not see a correction box, It appears that he should have said 675 divided by 450", + "video_name": "EtefJ85R1OQ", + "timestamps": [ + 303 + ], + "3min_transcript": "So when we figured out that the common ratio is 1.5, that tells us that our function is going to be of the form F of t is equal to a times-- instead of writing an r there, we now know that r is 1.5 to the t power. 1.5 to the t power. Write a formula for this function. Well we've almost done that, but we haven't figured out what a is. And to figure out what a is, we could just substitute-- we know what F of 1 is. When t is equal to 1, F is equal to 300. And so we should be able to use that information to solve for a. We could have used any of these data points to solve for a. So let's do that. F of 1 is equal to a times 1.5 to the first power, or a times 1.5. And that is going to be equal to-- they tell us that F of 1 is equal to 300. And so another way of writing this is we could say 1.5 times a is equal to 300. And we get a is equal to 200. And so our function, our formula for our function is-- let me write it in black so we can see it-- is going to be 200-- that's our a-- times 1.5 to the t power. Now another way-- well actually let's just think about the next question. What is the fine in euros for Sarah's speeding ticket if she pays it on time? So paying it on time, that implies that t is equal to 0. Or another way of thinking about it, we need to figure out her fine for t equals 0. So we need to figure out F of 0. So what's F of 0? It's 200 times 1.5 to the 0 power. 1.5 to the 0 power is 1, so that's just going to be equal to 200 euro. Now another way of thinking about it is, well look, let's look at the common ratio. To go from 6.5 to 450, you're essentially To go from 450 to 300, you're dividing by the common ratio. So then to go from t equals 1 to t equals 0, you would divide by the common ratio again. And you would get to 200. Or another way of thinking about it is to go to successive months every time we are multiplying by the common ratio. Every time we are multiplying by the common ratio." + }, + { + "Q": "0:44 Sal said exponential function f(t) = a . r ^t, is it always necessary, i mean can't exponential function f(t) = a - r^t or\nf(t) a + r^t ?\nWhat is exact definition of exponential function ?", + "A": "The a term doesn t vary exponentially (or at all).", + "video_name": "EtefJ85R1OQ", + "timestamps": [ + 44 + ], + "3min_transcript": "Sarah Swift got a speeding ticket on her way home from work. If she pays her fine now, there will be no added penalty. If she delays her payment then a penalty will be assessed for the number of months t that she delays paying her fine. Her total fine F in euros is indicated in the table below. These numbers represent an exponential function. So they give us the number of months that the payment is delayed, and then the amount of fine. And this is essentially data points from an exponential function. And just to remind ourselves what an exponential function would look like, this tells us that are fine as our function of the months delayed is going to be equal to some number times some common ratio to the t power. This exponential function is essentially telling us that our function is going to have this form right over here. So let's see if we can answer their questions. So the first question is, what is the common ratio So the reason why r right over here is called the common ratio is it's the ratio that if you look at any two-- say if you were to increment t by 1, the ratio of that to F of t-- that ratio should be consistent for any t. So let me give you an example here. The ratio of F of 2 to F of 1 should be equal to the ratio of F of 3 to f of 2, which would be the same as the ratio of F of 4 to F of 3. Or in general terms, the ratio of F of t plus 1 to the ratio of F of t should be equal to all of these things. That would be the common ratio. So let's see what that is. If we just look at the form. If we just look at this right over here. So what's the ratio of F of 2 to F of 1? Well that's 1.5. 675 divided by 450? That's 1.5. 1012.50 divided by 675? That's 1.5. So the common ratio in all of these situations is 1.5. So the common ratio over here is 1.5. And another way-- and just to make it clear why this r right over here is called the common ratio-- is let's just do this general form. So f of t plus 1? Well that's just going to be a times r to the t plus 1 power. And F of t is a times r to the t power. So what is this going to be? This is going to be-- let's see-- this is going to be r to the t plus 1 minus t, which is just going to be equal to r to the first power, which is just equal to r." + }, + { + "Q": "At 1:39, why can't we use product rule or the quotient rule to find the derivative of (1/lnb)(lnx) or lnx/lnb?", + "A": "The KEY thing to notice in this problem is that 1 / ln (b) is just a constant value. And so this problem is not much different than say 5 * ln x. You can pull the constant value out of the expression and just differentiate ln x just like Sal does. This will make the problem simpler.", + "video_name": "ssz6TElXEOM", + "timestamps": [ + 99 + ], + "3min_transcript": "We already know that the derivative with respect to x of the natural log of x is equal to 1/x. But what about the derivative, not of the natural log of x, but some logarithm with a different base? So maybe you could write log base b of x where b is an arbitrary base. How do we evaluate this right over here? And the trick is to write this using the change of base formula. So we could write it in terms of logarithms. We know that log-- I'm just going to restate the change of base formula. And I'm going to change from log base b to log base e, which is essentially the natural log. So the change of base formula, we prove it elsewhere on the site. Feel free to search for it on the Khan Academy. The change of base formula tells us that log base b of x is equal to the natural log, if we want to go to log base e. The natural log of x over the natural-- so it makes it clear what I'm doing. Log base e of x over log base e of b, which is the exact same thing as the natural log of x over the natural log of b. So all we have to do is rewrite this thing. This is equal to the derivative with respect to x of the natural log of x over the natural log of b. Or we could even write it as 1 over the natural log of b times the natural log of x. And now this becomes pretty straightforward. Because what we have right here, 1 over the natural log of b, this is just a constant that's multiplying the natural log of x. So we could take it out of the derivative. So this is the same thing as 1 over the natural log of b times the derivative with respect to x of the natural log of x. This thing right over here is just going to be equal to 1/x. So we end up with 1 over the natural log of b times 1/x. So we end up with 1 over the natural log of b times 1/x, or 1 over the natural log of b, which is just a number times x. So if someone asks you what is the derivative with respect to x of log base 5 of x, well, now you know. It's going to be 1 over the natural log of 5 times x, just like that." + }, + { + "Q": "how does he evaluate sin(at) at infinity? at 4:51 he doesn't take that into account", + "A": "sin(at) is a periodic function and it oscillates between -1 and 1. The same thing is for cos(at). So the expression in parentheses is always confined between some two numbers no matter how big t is. This expression is multiplied by e^(-st). When t goes to infinity, e^(-st) goes to zero. Zero times some finite number is still zero. Hope it helps!", + "video_name": "-cApVwKR1Ps", + "timestamps": [ + 291 + ], + "3min_transcript": "Because the t's are involved in evaluating the boundaries, since we're doing our definite integral or improper integral. So let's evaluate the boundaries now. And we could've kept them along with us the whole time, right? And just factored out this term right here. So let's evaluate this from 0 to infinity. And this should simplify things. So the right-hand side of this equation, when I evaluate it at infinity, what is e to the minus infinity? Well, that is 0. We've established that multiple times. And now it approaches 0 from the negative side, but it's still going to be 0, or it approaches 0. What's sine of infinity? Well, sine just keeps oscillating, between negative 1 and plus 1, and so does cosine. Right? So this is bounded. So this thing is going to overpower these. And if you're curious, you can graph it. This kind of forms an envelope around these oscillations. So the limit, as this approaches infinity, is going And that makes sense, right? These are bounded between 0 and negative 1. And this approaches 0 very quickly. So it's 0 times something bounded between 1 and negative 1. Another way to view it is the largest value this could equal is 1 times whatever coefficient's on it, and then this is going to 0. So it's like 0 times 1. Anyway, I don't want to focus too much on that. You can play around with that if you like. Minus this whole thing evaluated at 0. So what's e to the minus 0? Well, e to the minus 0 is 1. Right? That's e to the 0. We have a minus 1, so it becomes plus 1 times-- now, sine of 0 is 0. Minus 1 over s squared, cosine of 0. Let's see. 1 over s squared, times 1. So that is equal to minus 1 over s squared. And I think I made a mistake, because I shouldn't be having a negative number here. So let's backtrack. Maybe this isn't a negative number? Let's see, infinity, right? This whole thing is 0. When when you put 0 here, this becomes a minus 1. Yeah. So either this is a plus or this is a plus. Let's see where I made my mistake. e to the minus st-- oh, I see where my mistake is. Right up here. Where I factored out a minus e to the minus st, right?" + }, + { + "Q": "At 1:40 why does Sal factor out -e^(-st) and not (-e^(-st))/s? Isn't it possible to factor that out as well?", + "A": "Yes, but it helps better with the evaluation later on since you d end up with something weird like 1/infinity/infinity or something like that.", + "video_name": "-cApVwKR1Ps", + "timestamps": [ + 100 + ], + "3min_transcript": "Welcome back. We were in the midst of figuring out the Laplace transform of sine of at when I was running out of time. This was the definition of the Laplace transform of sine of at. I said that also equals y. This is going to be useful for us, since we're going to be doing integration by parts twice. So I did integration by parts once, then I did integration by parts twice. I said, you know, don't worry about the boundaries of the integral right now. Let's just worry about the indefinite integral. And then after we solve for y-- let's just say y is the indefinite version of this-- then we can evaluate the boundaries. And we got to this point, and we made the realization, after doing two integration by parts and being very careful not to hopefully make any careless mistakes, we realized, wow, this is our original y. If I put the boundaries here, that's the same thing as the Laplace transform of sine of at, right? That's our original y. So now-- and I'll switch colors just avoid monotony-- this is equal to, actually, let me just-- this is y. So let's add a squared over sine squared y to both sides of this. So this is equal to y plus-- I'm just adding this whole term to both sides of this equation-- plus a squared over s squared y is equal to-- so this term is now gone, so it's equal to this stuff. And let's see if we can simplify this. So let's factor out an e to the minus st. Actually, let's factor out a negative e to the minus st. So it's minus e to the minus st, times sine of-- well, let me just write 1 over s, sine of at, minus 1 over s squared, cosine of at. And so this, we can add the coefficient. So we get 1 plus a squared, over s squared, times y. But that's the same thing as s squared over s squared, plus a squared over s squared. So it's s squared plus a squared, over s squared, y is equal to minus e to the minus st, times this whole thing, sine of at, minus 1 over s squared, cosine of at. And now, this right here, since we're doing everything with respect to dt, this is just a constant, right? So we can say a constant times the antiderivative is equal to this. This is as good a time as any to evaluate the boundaries. If this had a t here, I would have to somehow get them back" + }, + { + "Q": "At 3:59, shouldn't Sal get 62__ something?\n\nA confuzzled child.", + "A": "No, he got the right answer but he made his 0 look like a 6. I also got confused but saw his mistake.", + "video_name": "TvSKeTFsaj4", + "timestamps": [ + 239 + ], + "3min_transcript": "We can divide both sides of this equation by 0.25, or if you recognize that four quarters make a dollar, you could say, let's multiply both sides of this equation by 4. You could do either one. I'll do the first, because that's how we normally do algebra problems like this. So let's just multiply both by 0.25. That will just be an x. And then the right-hand side will be 150 divided by 0.25. And the reason why I wanted to is really it's just good practice dividing by a decimal. So let's do that. So we want to figure out what 150 divided by 0.25 is. And we've done this before. When you divide by a decimal, what you can do is you can make the number that you're dividing into the other number, you can turn this into a whole number by essentially shifting the decimal two to the right. But if you do that for the number in the denominator, you also have to do that to the numerator. So right now you can view this as 150.00. decimal two to the right. Then you'd also have to do that with 150, so then it becomes 15,000. Shift it two to the right. So our decimal place becomes like this. So 150 divided by 0.25 is the same thing as 15,000 divided by 25. And let's just work it out really fast. So 25 doesn't go into 1, doesn't go into 15, it goes into 150, what is that? Six times, right? If it goes into 100 four times, then it goes into 150 six times. 6 times 0.25 is-- or actually, this is now a 25. We've shifted the decimal. This decimal is sitting right over there. So 6 times 25 is 150. You subtract. You get no remainder. Bring down this 0 right here. 25 goes into 0 zero times. 0 times 25 is 0. Subtract. No remainder. 25 goes into 0 zero times. 0 times 25 is 0. Subtract. No remainder. So 150 divided by 0.25 is equal to 600. And you might have been able to do that in your head, because when we were at this point in our equation, 0.25x is equal to 150, you could have just multiplied both sides of this equation times 4. 4 times 0.25 is the same thing as 4 times 1/4, which is a whole. And 4 times 150 is 600. So you would have gotten it either way. And this makes total sense. If 150 is 25% of some number, that means 150 should be 1/4 of that number. It should be a lot smaller than that number, and it is. 150 is 1/4 of 600. Now let's answer their actual question. Identify the percent. Well, that looks like 25%, that's the percent. The amount and the base in this problem." + }, + { + "Q": "at 1:52 ,why did (x+5)^2 became (x- -5)^2 ? I'm doomedddd", + "A": "Sal wrote it that way to get it into the form of the equation for a circle, which is (x - h)^2 + (y - k)^2 = r^2. In other words, the center of the circle is at (h,k), where h and k are the numbers being SUBTRACTED from x and y. Hope this saves you from being doomedddd!", + "video_name": "thDrJvWNI8M", + "timestamps": [ + 112 + ], + "3min_transcript": "- [Voiceover] Whereas to graph the circle x plus five squared plus y minus 5 squared equals four. I know what you're thinking. What's all of this silliness on the right hand side? This is actually just the view we use when we're trying to debug things on Khan Academy. But we can still do the exercise. So it says drag the center point and perimeter of the circle to graph the equation. So the first thing we want to think about is well what's the center of this equation? Well the standard form of a circle is x minus the x coordinate of the center squared, plus y minus the y coordinate of the center squared is equal to the radius squared. So x minus the x coordinate of the center. So the x coordinate of the center must be negative five. Cause the way we can get a positive five here's by subtracting a negative five. So the x coordinate must be negative five and the y coordinate must be positive five. Cause y minus the y coordinate of the center. So y coordinate is positive five and then the radius squared is going to be equal to four. is equal to two. And the way it's drawn right now, we could drag this out like this, but this the way it's drawn, the radius is indeed equal to two. And so we're done. And I really want to hit the point home of what I just did. So let me get my little scratch pad out. Sorry for knocking the microphone just now. That equation was x plus five squared plus y minus five squared is equal to four squared. So I want to rewrite this as, this is x minus negative five, x minus negative five squared, plus y minus positive five, positive five squared is equal to, instead of writing it as four I'll write it as two squared. So this right over here tells us x equals negative five y equals five and the radius is going to be equal to two. And once again, this is no magic here. This is not just, I don't want you to just memorize this formula. I want you to appreciate that this formula comes straight out of the Pythagorean Theorem, straight out of the distance formula, which comes out of the Pythagorean Theorem. Remember, if you have some center, in this case is the point negative five comma five, so negative five comma five, and you want to find all of the x's and y's that are two away from it. So you want to find all the x's and y's that are two away from it. So that would be one of them, x comma y. This distance is two. And there's going to be a bunch of them. And when you plot all of them together, you're going to get a circle with radius two around that center. Plus think about how we got that actual formula. Well the distance between that coordinate, between any of these x's and y's, it could be an x and y here, it could be an x and y here," + }, + { + "Q": "At around 3:36 Sal said that at 0 hours there were 550 pages left to read. Couldn't of Naoya also read in minutes? So there aren't really exactly 550 pages in the book right?", + "A": "Merely changing to minutes would not change the y-intercept because the rate would be equivalent (55 pages/hour = 55/60 or 11/12 pages/minute). So after 240 minutes (4*60), Naoya still had 330 pages to read. so 240 * 11/12 = 220 pages. Think about what is happening, you are changing pages/hour by dividing by 60 then you are multiplying by 60 to convert to minutes, the two 60s cancel each other out so 55/60 * 4 * 60 is the same as 55 * 4.", + "video_name": "W3flX500w5g", + "timestamps": [ + 216 + ], + "3min_transcript": "would be 385. Now let's see whether this makes sense. So when our change in time, this triangle is just the Greek letter Delta, means \"change in.\" When our change in time is plus one, plus one hour, our change in pages left to read is going to be equal to negative 55 pages. And that makes sense, the pages left to read goes down every hour. We're measuring not how much he's read, we're measuring how much he has left to read. So that should go down by 55 pages every hour. Or if we were to go backwards through time, it should go up. So at two hours he should have 55 more pages to read. So what's 385 plus 55? We'll let's see, 385 plus 5 is 390, plus 50 is 440. So he'd have 440 pages, and all I did is I added 55. he would have 55 more pages than after reading for two hours. So 440 plus 55 is 495. And then before he started reading, or right when he started reading, he would have had to read even 55 more pages, 'cause after one hour, he would have read those 55 pages. So 495 plus 55 is going to be, let's see, it's gonna be, add 5, you get to 500 plus another 50 is 550 pages. So at time equals zero had had 550 pages to read. So that's how long the book is. But how long does it take Naoya to read the entire book? Well we could keep going. We could say, \"Okay, at the fifth hour, \"this thing's gonna go down by 55.\" So let's see, if this goes down by 50, if this goes down by 50, we're going to get to 280, but then you go down five more, it's gonna go to 275, and we could keep going on and on and on. \"He's got 330 pages left to read, and he's gonna,\" Let's see, let me write this, let me write the units down. \"Pages, and he's reading at a rate of 55 pages per hour, \"pages per hour, this is the same thing, \"this is going to be equal to 330 pages \"times one over 55 hours per page.\" I'm dividing by something, the same thing as multiplying by its reciprocal, so 55 pages per hours, if you divide by that, that's the same thing as multiplying by 1/55th of an hour per page, is one way to think about it. And so what do you get? The pages cancel out, pages divided by pages, and you have 330 divided by 55 hours. 330 divided by 55 hours. And what's that going to be? Let's see, 30 divided by 5 is 6, 300 divided by 50 is 6, so this is going to be equal to 6 hours." + }, + { + "Q": "At 2:23, Sal talks about using ratios of the sides to find the area of the equilateral triangle. Could you use the pythagorean theorem to find the height of the equilateral triangle and then calculate the area that way?", + "A": "yes, you could do that.", + "video_name": "QVxqgxVtKbs", + "timestamps": [ + 143 + ], + "3min_transcript": "Let's say I have an equilateral triangle where the length of each side is 14. So this is an equilateral triangle. All of the sides have length 14. And inside that I have another equilateral triangle-- right over here-- where the length of each of the sides is 4. Now what I'm curious about, is the area of the region-- let me color this in a different color-- is the area of the region that I'm shading in right here. So it's the area inside the larger equilateral triangle, but outside of the smaller equilateral triangle. So let's think about how we would do this. And I encourage you to pause this and try this on your own. Well the shaded area is going to be equal to the large equilateral triangle's area minus the area of the small equilateral triangle. the area of each of these equilateral triangles are. And so to do it, we remember that the area of a triangle is equal to 1/2 base times height. But how do we figure out the height of an equilateral triangle? So for example, if I have an equilateral triangle like this-- let me draw it big so I can dissect it little bit-- so I have an equilateral triangle like this. The length of each of the sides are s. And I always have to re-prove it for myself. Just because I always forget the formula. We remember that the angles are 60 degrees, 60 degrees, and 60 degrees. They're all equal. And what I like to do to find out the area of this, in order to figure out the height, is I drop an altitude. So I drop an altitude just like here, and it would split the side in two. I know it doesn't look like it perfectly because I didn't draw it to scale. But it would split it in two. It would form these right angles. split my equilateral triangle into two 30-60-90 triangles. And that's useful because I know the ratio of the sides of a 30-60-90 triangle. If this is s and I've just split this in two, this orange section right over here is going to be s/2. This is also going to be s/2 right over here. They obviously add up to s. And then we know from 30-60-90 triangles, that the side opposite the 60-degree side is square root of 3 times the shortest side. So this altitude right over here, is going to be square root of 3s/2. And now we can figure out a generalized formula for the area of an equilateral triangle. It's going to be equal to 1/2 times the base. Well the base is going to be s. So the base is s. And the height is square root of 3s over 2." + }, + { + "Q": "At 3:20, how is 5 tenths equal to 500 thousandths?", + "A": "Because the zeros after the decimal point don t carry any information if no other number follows them. For instance, 5 is equal to 5.0 or 5.00 or 5.000 and so on. But if there is a number after the zeros, then they are significant and can t be removed. Like this: 5.00002 is NOT equal to 5.2. That s why, in your example, 0.5 = 0.500. Since no other number comes after the final zeros, they are not important and can as well be removed.", + "video_name": "G7QiIkYfeME", + "timestamps": [ + 200 + ], + "3min_transcript": "And then finally, we have 7/1000, that's the 1000th place. So we could write that, plus 7/1000. So if we would write down everything that I just spoke out loud, we would say that this is 20-- let me write that a little bit neater. This is 20,000. 20,000 and 5/10. And 5-- let me write out the word-- and 5/10 and 7/1000. Now, this isn't the only way to say this. Another way of thinking about it is to try to merge the 5/10 and the 7/1000 in terms of thousandths. So let's think about this. So we could write this as-- so once again, we would have our 20,000. write our 5/10 in terms of thousandths. And the easiest way to do it is to multiply the numerator and denominator, both here, by 100. So then we will have-- so this 5/10 is the same thing as 500 over 1,000. And the 7/1000 is still 7/1000. And these two combined are 507/1000. So we could just call this 20,000 and 507/1000. so let's write that down. So we could just say this is 20,000 and 507/1000. actually represents 1,000. So we got 20 thousands, that's that right over there, and 507/1000." + }, + { + "Q": "at 2:20 Sal says that a 1cm x 1cm x 1cm cube is 1 milliliter...why wouldn't this be 1 centimeter?", + "A": "The cube is equal to 1 cubic centimetre. A cubic centimetre is equal to 1 milliliter .", + "video_name": "LhMEqsL_M5o", + "timestamps": [ + 140 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:10, where did you get -66?", + "A": "He s factoring by grouping . He starts by taking a (the coefficient of f^2) and multiplies it by b (the coefficient of f). Hope this helps :)", + "video_name": "d-2Lcp0QKfI", + "timestamps": [ + 70 + ], + "3min_transcript": "We need to factor negative 12f squared minus 38f, plus 22. So a good place to start is just to see if, is there any common factor for all three of these terms? When we look at them, they're all even. And we don't like a negative number out here. So let's divide everything, or let's factor out a negative 2. So this expression right here is the same thing as negative 2 times-- what's negative 12f squared divided by negative 2? It's positive 6f squared. Negative 38 divided by negative 2 is positive 19, so it'll be positive 19f. And then 22 divided by negative 22-- oh, sorry, 22 divided by negative 2 is negative 11. So we've simplified it a bit. We have the 6f squared plus 19f, minus 11. We'll just focus on that part right now. And the best way to factor this thing, since we don't have a 1 here as the coefficient on the f squared, is to factor it by grouping. times negative 11. So two numbers, so a times b, needs to be equal to 6 times negative 11, or negative 66. And a plus b needs to be equal to 19. So let's try a few numbers here. So let's see, 22, I'm just thinking of numbers that are roughly 19 apart, because they're going to be of different signs. So 22 and 3, I think will work. Right. If we take 22 times negative 3, that is negative 66, and 22 plus negative 3 is equal to 19. And the way I kind of got pretty close to this number is, well, you know, they're going to be of different signs, so the positive versions of them have to be about 19 apart, and that worked out. 22 and negative 3. So now we can rewrite this 19f right here as the sum of That's the same thing as 19f. I just kind of broke it apart. And, of course, we have the 6f squared and we have the minus 11 here. Now, you're probably saying, hey Sal, why did you put the 22 here and the negative 3 there? Why didn't you do it the other way around? Why didn't you put the 22 and then the negative 3 there? And my main motivation for doing it, I like to put the negative 3 on the same side with the 6 because they have the common factor of the 3. I like to put the 22 with the negative 11, they have the same common factor of 11. So that's why I decided to do it that way. So now let's do the grouping. And, of course, you can't forget this negative 2 that we have sitting out here the whole time. So let me put that negative 2 out there, but that'll just kind of hang out for awhile. But let's do some grouping. So let's group these first two. And then we're going to group this-- let me get a nice color here-- and then we're going to group this second two." + }, + { + "Q": "At 1:27, How did he get 9?", + "A": "Ah, 45 - 36 = 9", + "video_name": "EFVrAk61xjE", + "timestamps": [ + 87 + ], + "3min_transcript": "We are asked to approximate the principal root, or the positive square root of 45, to the hundredths place. And I'm assuming they don't want us to use a calculator. Because that would be too easy. So, let's see if we can approximate this just with our pen and paper right over here. So the square root of 45, or the principal root of 45. 45 is not a perfect square. It's definitely not a perfect square. Let's see, what are the perfect squares around it? We know that it is going to be less than-- the next perfect square above 45 is going to be 49 because that is 7 times 7-- so it's less than the square root of 49 and it's greater than the square root of 36. And so, the square root of 36, the principal root of 36 I should say, is 6. And the principal root of 49 is 7. So, this value right over here is going to be between 6 and 7. And it's nine away from 36. So, the different between 36 and 49 is 13. So, it's a total 13 gap between the 6 squared and 7 squared. And this is nine of the way through it. So, just as a kind of approximation maybe-- and it's not going to work out perfectly because we're squaring it, this isn't a linear relationship-- but it's going to be closer to 7 than it's going to be to 6. At least the 45 is 9/13 of the way. It looks like that's about 2/3 of the way. So, let's try 6.7 as a guess just based on 0.7 is about 2/3. It looks like about the same. Actually, we could calculate this right here if we want. So 9/13 as a decimal is going to be what? It's going to be 13 into 9. We're going to put some decimal places right over here. 13 doesn't go into 9 but 13 does go into 90. And it goes into 90-- let's see, does it go into it seven times-- it goes into it six times. So, 6 times 3 is 18. 6 times 1 is 6, plus 1 is 7. And then you subtract, you get 12. So, went into it almost exactly seven times. So, this value right here is almost a 0.7. And so if you say, how many times does 13 go into 120? It looks like it's like nine times? Yeah, it would go into it nine times. 9 times 3. Get rid of this. 9 times 3 is 27. 9 times 1 is 9, plus 2 is 11. You have a remainder of 3. It's about 0.69." + }, + { + "Q": "At 1:24 sal writes the fraction 9/13. Why did he put a nine instead of four? Can someone please help me??", + "A": "because it is 9 away from the first root over the total amount of numbers in between", + "video_name": "EFVrAk61xjE", + "timestamps": [ + 84 + ], + "3min_transcript": "We are asked to approximate the principal root, or the positive square root of 45, to the hundredths place. And I'm assuming they don't want us to use a calculator. Because that would be too easy. So, let's see if we can approximate this just with our pen and paper right over here. So the square root of 45, or the principal root of 45. 45 is not a perfect square. It's definitely not a perfect square. Let's see, what are the perfect squares around it? We know that it is going to be less than-- the next perfect square above 45 is going to be 49 because that is 7 times 7-- so it's less than the square root of 49 and it's greater than the square root of 36. And so, the square root of 36, the principal root of 36 I should say, is 6. And the principal root of 49 is 7. So, this value right over here is going to be between 6 and 7. And it's nine away from 36. So, the different between 36 and 49 is 13. So, it's a total 13 gap between the 6 squared and 7 squared. And this is nine of the way through it. So, just as a kind of approximation maybe-- and it's not going to work out perfectly because we're squaring it, this isn't a linear relationship-- but it's going to be closer to 7 than it's going to be to 6. At least the 45 is 9/13 of the way. It looks like that's about 2/3 of the way. So, let's try 6.7 as a guess just based on 0.7 is about 2/3. It looks like about the same. Actually, we could calculate this right here if we want. So 9/13 as a decimal is going to be what? It's going to be 13 into 9. We're going to put some decimal places right over here. 13 doesn't go into 9 but 13 does go into 90. And it goes into 90-- let's see, does it go into it seven times-- it goes into it six times. So, 6 times 3 is 18. 6 times 1 is 6, plus 1 is 7. And then you subtract, you get 12. So, went into it almost exactly seven times. So, this value right here is almost a 0.7. And so if you say, how many times does 13 go into 120? It looks like it's like nine times? Yeah, it would go into it nine times. 9 times 3. Get rid of this. 9 times 3 is 27. 9 times 1 is 9, plus 2 is 11. You have a remainder of 3. It's about 0.69." + }, + { + "Q": "How do you know which numbers are a and b in the equation? At 13:21 he says that a=3 and b=4 (because he took the square roots) but how do you know that a isn't 4 and b isn't 3? For ellipses you know that a is the bigger number, but what do you know for hyperbolas? Thanks!", + "A": "a is 3 so 3 squared gives u nine and b is 4 so 4 squared gives u 16.", + "video_name": "S0Fd2Tg2v7M", + "timestamps": [ + 801 + ], + "3min_transcript": "the focal length is the same on either side of the center of the hyperbola depending on how you may view it, but I think that's not too much of a stretch of a statement for you to for you to accept. So if this distance is the same as this distance, then the magenta distance minus this blue distance is going to be equal to this green distance. And this green distance is what? That's 2a. We saw that at the beginning of this video. So this, once again, is also equal to 2a. Anyway, I'll leave you there right now. Actually, let's actually just do one problem, just because I like to make one concrete. Because I told you at the beginning that if you wanted to find the-- so if you have an ellipse-- so if you have-- this is an ellipse, x squared over a squared plus y squared over b squared is equal to 1, we learned that the-- that's over b squared-- this is an ellipse. square root of a squared minus b squared. Now for a hyperbola, you kind of see that there's a very close relation between the ellipse and the hyperbola, but it is kind of a fun thing to ponder about. And a hyperbola's equation looks like this. x squared over a squared minus y squared over b squared, or it could be y squared over b squared minus x squared over a square is equal to 1. It turns out, and I'll prove this to you in the next video, it's a little bit of a hairy math problem, that the focal length of a hyperbola is equal to the square root of the sum of these two numbers, is equal to the sum of a squared plus b squared. So if I were to give you-- so notice the difference. It's just a difference in sign. You're taking the difference of those two denominators, and now you're taking the sum of the two denominators. So if I were to give you the following hyperbola. x squared over 9 plus y squared over 16 is equal to 1. we could just figure out the focal length just by plugging into the formula. The focal length is equal to the square root of a squared plus b squared. This is squared, right? a is three. b is 4. So 9 plus 16 is 25, which is equal to 5. And so if we were to graph this-- that's my y-axis, that's my x-axis-- and the focal length is the distance to, in this case, to the left and the right of the origin. If it was kind of an up and down opening hyperbola, it would be above and below the origin, so this is a-- oh sorry, this should be a plus. We're doing with a hyperbola, that should be a minus. Don't want to confuse you. What I had written before, with a plus, that would have been an ellipse. A minus is the hyperbola. So the two asymptotes-- this is centered at the origin, it hasn't been shifted-- are going to be 16 over 9, so it's going" + }, + { + "Q": "at 2:43 sal said standard but wrote slandard why", + "A": "He probably made a mistake...you should report it is the tag named report a mistake ....", + "video_name": "ZgFXL6SEUiI", + "timestamps": [ + 163 + ], + "3min_transcript": "That doesn't mean that you won't ever see it while you're doing algebra or mathematics. But we just wouldn't call this a polynomial because it has a negative and a fractional exponent in it. Or if I were to give you the expression y times the square root of y minus y squared. Once again, this is not a polynomial, because it has a square root in it, which is essentially raising something to the 1/2 power. So all of the exponents on our variables are going to have to be non-negatives. Once again, neither of these are polynomials. Now, when we're dealing with polynomials, we're going to have some terminology. And you may or may not already be familiar with it, so I'll expose it to you right now. The first terminology is the degree of the polynomial. in the polynomial. So for example, that polynomial right there is a third degree polynomial. Now why is that? No need to keep writing it. Why is that a third degree polynomial? Because the highest exponent that we have in there is the x to the third term. So that's where we get it's a third degree polynomial. This right here is a second degree polynomial. And this is the second degree term. Now a couple of other terminologies, or words, that we need to know regarding polynomials, are the constant versus the variable terms. And I think you already know, these are variable terms right here. This is a constant term. That right there is a constant term. properly is to understand the coefficients of a polynomial. So let me write a fifth degree polynomial here. And I'm going to write it in maybe a non-conventional form right here. I'm going to not do it in order. So let's just say it's x squared minus 5x plus 7x to the fifth minus 5. So, once again, this is a fifth degree polynomial. Why is that? Because the highest exponent on a variable here is the 5 So this tells us this is a fifth degree polynomial. And you might say, well why do we even care about that? And at least, in my mind, the reason why I care about the degree of a polynomial is because when the numbers get large, the highest degree term is what really dominates all of the other terms. It will grow the fastest, or go negative the fastest, depending on whether there's a" + }, + { + "Q": "At 2:50 why did Sal call the binomial a second degree polynomial? Didn't he say that it was not a polynomial", + "A": "We reserve the word polynomial for expressions in which all the terms have positive, non-fractional exponents. When Sal earlier said some expressions did not qualify as polynomials, he was excluding one that included a square root (which is a fractional exponent of 1/2) and also excluding one that had a negative exponent. He wasn t excluding ordinary binomials, which are included in the definition of polynomial.", + "video_name": "ZgFXL6SEUiI", + "timestamps": [ + 170 + ], + "3min_transcript": "That doesn't mean that you won't ever see it while you're doing algebra or mathematics. But we just wouldn't call this a polynomial because it has a negative and a fractional exponent in it. Or if I were to give you the expression y times the square root of y minus y squared. Once again, this is not a polynomial, because it has a square root in it, which is essentially raising something to the 1/2 power. So all of the exponents on our variables are going to have to be non-negatives. Once again, neither of these are polynomials. Now, when we're dealing with polynomials, we're going to have some terminology. And you may or may not already be familiar with it, so I'll expose it to you right now. The first terminology is the degree of the polynomial. in the polynomial. So for example, that polynomial right there is a third degree polynomial. Now why is that? No need to keep writing it. Why is that a third degree polynomial? Because the highest exponent that we have in there is the x to the third term. So that's where we get it's a third degree polynomial. This right here is a second degree polynomial. And this is the second degree term. Now a couple of other terminologies, or words, that we need to know regarding polynomials, are the constant versus the variable terms. And I think you already know, these are variable terms right here. This is a constant term. That right there is a constant term. properly is to understand the coefficients of a polynomial. So let me write a fifth degree polynomial here. And I'm going to write it in maybe a non-conventional form right here. I'm going to not do it in order. So let's just say it's x squared minus 5x plus 7x to the fifth minus 5. So, once again, this is a fifth degree polynomial. Why is that? Because the highest exponent on a variable here is the 5 So this tells us this is a fifth degree polynomial. And you might say, well why do we even care about that? And at least, in my mind, the reason why I care about the degree of a polynomial is because when the numbers get large, the highest degree term is what really dominates all of the other terms. It will grow the fastest, or go negative the fastest, depending on whether there's a" + }, + { + "Q": "at around 14:40, wouldn't the width of the green square be 1/2b ?", + "A": "You can t assume that it is half of b from the picture. Never trust the picture. This is why Sal denoted it as c.", + "video_name": "ZgFXL6SEUiI", + "timestamps": [ + 880 + ], + "3min_transcript": "Well, this blue part right here, the area there is x times y. And then what's the area here? It's going to be x times z. So plus x times z. We have one x times z, and then we have another x times z. So I could just add an x times z here. Or I could just write, say, plus 2 times x times z. And here we have a polynomial that represents the area of this figure right there. Now let's do this next one. What's the area here? Well I have an a times a b. ab. This looks like an a times a b again, plus ab. That looks like an ab again, plus ab. I think they've drawn it actually, Well, I'm going to ignore this c right there. Maybe they're telling us that this right here is c. Because that's the information we would need. Maybe they're telling us that this base right there, that this right here, is c. Because that would help us. But if we assume that this is another ab here, which I'll assume for this purpose of this video. And then we have that last ab. And then we have this one a times c. This is the area of this figure. And obviously we can add these four terms. This is 4ab and then we have plus ac. And I made the assumption that this was a bit of a typo, that that c where they were actually telling us the width of this little square over here. We don't know if it's a square, that's only if a and c are the same. Now let's do this one. So how do we figure out the area of the pink area? would be 2xy, and then we could subtract out the area of these squares. So each square has an area of x times x, or x squared. And we have two of these squares, so it's minus 2x squared. And then finally let's do this one over here. So that looks like a dividing line right there. So the area of this point, of this area right there, is a times b, so it's ab. And then the area over here looks like it will also be ab. So plus ab. And the area over here is also ab. So the area here is 3ab. Anyway, hopefully that gets us pretty warmed up with polynomials." + }, + { + "Q": "At about 6:08 you talked about descending order, what would the degree of a term without an exponent be?", + "A": "If the term is something like 2x, then there is an exponent on that variable. If one is not written, it s implied that it is to the 1st power. If there is no variable at all, like in 4, you would say that it is degree 0. The only other weird case is that a term of 0 is said to have no degree.", + "video_name": "ZgFXL6SEUiI", + "timestamps": [ + 368 + ], + "3min_transcript": "But it's going to dominate everything else. It really gives you a sense for how quickly, or how fast the whole expression would grow or decrease in the case if it has a negative coefficient. Now I just used the word coefficient. What does that mean? Coefficient. And I've used it before, when we were just doing linear equations. And coefficients are just the constant terms that are multiplying the variable terms. So for example, the coefficient on this term right here is negative 5. You have to remember we have a minus 5, so we consider negative 5 to be the whole coefficient. The coefficient on this term is a 7. There's no coefficient here; it's just a constant term of negative 5. And then the coefficient on the x squared term is 1. The coefficient is 1. It's implicit. You're assuming it's 1 times x squared. idea of the standard form of a polynomial. Now none of this is going to help you solve a polynomial just yet, but when we talk about solving polynomials, I might use some of this terminology, or your teacher might use some of this terminology. So it's good to know what we're talking about. The standard form of a polynomial, essentially just list the terms in order of degree. So this is in a non-standard form. If I were to list this polynomial in standard form, I would put this term first. So I would write 7x to the fifth, then what's the next smallest degree? Well, they have this x squared term. I don't have an x to the fourth or an x to the third here. So that'll be plus 1-- well I don't have to write 1-- plus x squared. And then I have this term, minus 5x. And then I have this last term right here, minus 5. it in descending order of degree. Now let's do a couple of operations with polynomials. And this is going to be a super useful toolkit later on in your algebraic, or really in your mathematical careers. So let's just simplify a bunch of polynomials. And we've kind of touched on this in previous videos. But I think this will give you a better sense, especially when we have these higher degree terms over here. So let's say I wanted to add negative 2x squared plus 4x minus 12. And I'm going to add that to 7x plus x squared. Now the important thing to remember when you simplify these polynomials is that you're going to add the terms of the same variable of like degree. I'll do another example in a second where I have multiple variables getting involved in the situation." + }, + { + "Q": "At about 3:40, Sal says that 5 is the highest exponent on a variable. Does this mean that a degree can only come from the highest exponent on a variable, or can it be on a normal number as well?", + "A": "The degree of a polynomial is by definition the largest exponent of the variable. So, yes we only consider the exponent of the the variable. So, x\u00c2\u00b2 + 50\u00c2\u00b3\u00c2\u00b9\u00e2\u0081\u00b7 would still be a second degree polynomial.", + "video_name": "ZgFXL6SEUiI", + "timestamps": [ + 220 + ], + "3min_transcript": "in the polynomial. So for example, that polynomial right there is a third degree polynomial. Now why is that? No need to keep writing it. Why is that a third degree polynomial? Because the highest exponent that we have in there is the x to the third term. So that's where we get it's a third degree polynomial. This right here is a second degree polynomial. And this is the second degree term. Now a couple of other terminologies, or words, that we need to know regarding polynomials, are the constant versus the variable terms. And I think you already know, these are variable terms right here. This is a constant term. That right there is a constant term. properly is to understand the coefficients of a polynomial. So let me write a fifth degree polynomial here. And I'm going to write it in maybe a non-conventional form right here. I'm going to not do it in order. So let's just say it's x squared minus 5x plus 7x to the fifth minus 5. So, once again, this is a fifth degree polynomial. Why is that? Because the highest exponent on a variable here is the 5 So this tells us this is a fifth degree polynomial. And you might say, well why do we even care about that? And at least, in my mind, the reason why I care about the degree of a polynomial is because when the numbers get large, the highest degree term is what really dominates all of the other terms. It will grow the fastest, or go negative the fastest, depending on whether there's a But it's going to dominate everything else. It really gives you a sense for how quickly, or how fast the whole expression would grow or decrease in the case if it has a negative coefficient. Now I just used the word coefficient. What does that mean? Coefficient. And I've used it before, when we were just doing linear equations. And coefficients are just the constant terms that are multiplying the variable terms. So for example, the coefficient on this term right here is negative 5. You have to remember we have a minus 5, so we consider negative 5 to be the whole coefficient. The coefficient on this term is a 7. There's no coefficient here; it's just a constant term of negative 5. And then the coefficient on the x squared term is 1. The coefficient is 1. It's implicit. You're assuming it's 1 times x squared." + }, + { + "Q": "At around 2:07 Sal said that the degree of the polynomial is the value of the largest exponent, whereas my algebra teacher says it is the sum of the exponent values. Who is wrong? Thanks.", + "A": "The both are! When working with polynomials of 1 variable, what Sal said is correct, assuming the polynomial is in canonical form, that is, you don t have terms like (x^3)(x^4), which in canonical form would be written x^7. When you have polynomials of more than one variable, you need to sum the exponents of each term, for example, (x^4)(y) has degree 5 and (x^2)(y^2)(z^2) has degree 6.", + "video_name": "ZgFXL6SEUiI", + "timestamps": [ + 127 + ], + "3min_transcript": "In this video I want to introduce you to the idea of a polynomial. It might sound like a really fancy word, but really all it is is an expression that has a bunch of variable or constant terms in them that are raised to non-zero exponents. So that also probably sounds complicated. So let me show you an example. If I were to give you x squared plus 1, this is a polynomial. This is, in fact, a binomial because it has two terms. The term polynomial is more general. It's essentially saying you have many terms. Poly tends to mean many. This is a binomial. If I were to say 4x to the third minus 2 squared plus 7. This is a trinomial. I have three terms here. Let me give you just a more concrete sense of what is and is not a polynomial. For example, if I were to have x to the negative 1/2 plus 1, That doesn't mean that you won't ever see it while you're doing algebra or mathematics. But we just wouldn't call this a polynomial because it has a negative and a fractional exponent in it. Or if I were to give you the expression y times the square root of y minus y squared. Once again, this is not a polynomial, because it has a square root in it, which is essentially raising something to the 1/2 power. So all of the exponents on our variables are going to have to be non-negatives. Once again, neither of these are polynomials. Now, when we're dealing with polynomials, we're going to have some terminology. And you may or may not already be familiar with it, so I'll expose it to you right now. The first terminology is the degree of the polynomial. in the polynomial. So for example, that polynomial right there is a third degree polynomial. Now why is that? No need to keep writing it. Why is that a third degree polynomial? Because the highest exponent that we have in there is the x to the third term. So that's where we get it's a third degree polynomial. This right here is a second degree polynomial. And this is the second degree term. Now a couple of other terminologies, or words, that we need to know regarding polynomials, are the constant versus the variable terms. And I think you already know, these are variable terms right here. This is a constant term. That right there is a constant term." + }, + { + "Q": "9:43 Why is is a^2(x^2-2xf+f^2+y^2)? Why is the \"a\" outside of the parentheses? In other words, why isn't it just a^2x^2-2xf+f^2+y^2? Why are there parentheses?", + "A": "Because without the parentheses, the a^2 will not be distributed to the equation that was inside the square root. Without it you just get a^2 times x^2. With them you have to multiply each factor by a^2.", + "video_name": "HPRFmu7JsKU", + "timestamps": [ + 583 + ], + "3min_transcript": "of the distances between these two points, and then see how it relates to the equation of the hyperbola itself. The a's and the b's. Let's take this 4a put it on this side, so you get 4xf minus 4a squared is equal to 4a times the square root of-- well let's just multiply this out 'cause we'll probably have to eventually --x squared minus 2xf plus f squared plus y squared. That's this just multiplied out. That's the y squared right there. We could divide both sides of this by 4. All I'm trying to do is just simplify this as much as possible, so then this becomes xf minus a squared is equal to a times the square root of this whole thing. x squared minus 2xf plus f squared plus why squared. equation right here. And then if you square both sides, this side becomes x squared f squared minus 2a squared xf plus a to the fourth. That's this side squared. And that's equal to, if you square the right hand side, a squared times the square of a square root is just that expression, x squared minus 2xf plus f squared plus y squared. This really is quite hairy. And let's see what we can do now. Let's divide both sides of this equation by a squared, and then you get x squared-- I'm really just trying to simplify this as much as possible --over a squared minus-- so the a square well that's just a squared. So a squared is equal to x squared minus 2xf plus f squared plus y squared. Well good. There's something to cancel out. There's a mine 2xf on both sides of this equation so let's cancel that out. Simplify our situation a little bit. And let's see, we have. so we could do is subtract this x squared from this. So you get-- let me rewrite it --so you get x squared f squared over a squared minus x squared. And let's bring this y to this side of the equation too. So minus y squared. That's all I did, I just brought that to that side. And then let's bring-- and I'm kind of skipping a couple of steps, but I don't want to take too long --let's take this a" + }, + { + "Q": "At 2:00 how did he get 36x?", + "A": "When Sal was doing the distributive property 9x(x+4) 9x times x is 9x squared and 9x times 4 is 36x.", + "video_name": "vl9o9XEfXtw", + "timestamps": [ + 120 + ], + "3min_transcript": "The volume of a box is 405 cube units, or I guess cubic units. So they just want to keep it general. It could've been in cubic feet, or cubic meters, or cubic centimeters, or cubic miles. They just want to keep it as units, keep it as general as possible. The length is x units, the width is x plus 4 units, and the height is 9 units. So let me draw this box here. Let me draw a little box here, so we have a nice little visualization. So they tell us, that the length is x. Maybe we could call this the length right there. They say the width is x plus 4, and the height is 9 of this box. In units, what are the dimensions of the box? Well, they also tell us that the volume is 405. So the volume, 405-- let me do it this way. So if we wanted to calculate the volume, what would it be? Well it would be the width-- it would be x plus 4 times the That's, literally, the volume of the box. Now they also tell us that the volume of the box is 405 cubic units, is equal to 405. So now we just solve for x. So what do we get here? If we distribute this x into this x plus 4. Actually, if we distribute a 9x. Let me just rewrite it. This is the same thing as 9x times x plus 4 is equal to 405. 9x times x is equal to 9x squared. 9x times 4 is equal to 36x, is equal to 405. Now we want our quadratic expression to be equal to 0. So let's subtract 405 from both sides of this equation. So when you do that, your right-hand side equals 0, and Now, is there any common factor to these numbers right here? Well 405, 4 plus 0 plus 5 is 9, so that is divisible by 9. So all of these are divisible by 9. Let's just figure out what 405 divided by 9 is. So 9 goes into 405-- 9 goes into 40 4 times. 4 times 9 is 36. Subtract you get 45. 9 goes into 45 5 times. 5 times 9 is 45. Subtract, you get 0. So it goes 45 times. So if we factor out a 9 here, we get 9 times x squared-- actually even better, you don't even have to factor out of 9. If you think about it, you can divide both sides of this equation by 9. So if you can divide all of the terms by 9, it won't" + }, + { + "Q": "at 4:05 how did he get -16?", + "A": "exponents with negitive bases", + "video_name": "vEZea0EThus", + "timestamps": [ + 245 + ], + "3min_transcript": "times a negative, you're going to get a positive. And so when you do it an even number of times, doing it a multiple-of-two number of times. So the negatives and the negatives all cancel out, I guess you could say. Or when you take the product of the two negatives, you keep getting positives. So this right over here is going to give you a positive value. So there's really nothing new about taking powers of negative numbers. It's really the same idea. And you just really have to remember that a negative times a negative is a positive. And a negative times a positive is a negative, which we already learned from multiplying negative numbers. Now there's one other thing that I want to clarify \u2013 because sometimes there might be ambiguity if someone writes this. Let's say someone writes that. And I encourage you to actually pause the video and think about with this right over here And, if you given a go at that, think about whether this should mean something different then that. Well this one can be a little bit and big ambiguous and if people are strict about order of operations, you should really be thinking about the exponent before you multiply by this -1. You could this is implicitly saying -1 \u00d7 2^3. So many times, this will usually be interpreted as negative 2 to the third power, which is equal to -8, while this is going to be interpreted as -2 to the third power. Now that also is equal to -8. You might say well what's what's the big deal here? Well what if this was what if these were even exponents. So what if someone had give myself some more space here. What if someone had these to express its -4 or a -4 squared or -4 squared. This one clearly evaluates to 16 \u2013 positive 16. This one could be interpreted as is. Especially if you look at order of operations, and you do your exponent first, this would be interpreted as -4 times 4, which would be -16. So it's really important to think about this properly. And if you want to write the number negative if you want the base to be negative 4, put parentheses around it and then write the exponent." + }, + { + "Q": "Around 1:00, Sal mentioned \"functions that are not equations.\" Could anyone help clarify how there are some functions that are not equations?", + "A": "Functions aren t equations. They re mappings between sets. Here s a function: f: \u00e2\u0084\u009d \u00e2\u0086\u0092 \u00e2\u0084\u009d, x \u00e2\u0086\u00a6 2x + 1 It s a function from the set of real numbers to the set of real numbers, and a real number x is sent to the real number 2x + 1. A function may be represented with an equation. For example: f(x) = 2x + 1 defines (mostly) the same function as above. (It doesn t explicitly define the domain and range of the function.)", + "video_name": "l3iXON1xEC4", + "timestamps": [ + 60 + ], + "3min_transcript": "SALMAN KHAN: I'm here with Jesse Roe of Summit Prep. What classes do you teach? JESSE ROE: I teach algebra, geometry, and algebra II. SALMAN KHAN: And now you're with us, luckily, for the summer, doing a whole bunch of stuff as a teaching fellow. JESSE ROE: Yeah, as a teaching fellow I've been helping with organizing and developing new content, mostly on the exercise side of the site. SALMAN KHAN: And the reason why we're doing this right now is you had some very interesting ideas or questions. JESSE ROE: Yeah, so as an algebra teacher, when I introduce that concept of algebra to students, I get a lot of questions. One of those questions is, what's the difference between an equation and a function? SALMAN KHAN: The difference between an equation verses a function, that's an interesting question. Let's pause it and let the viewers try to think about it a little bit. And then maybe we'll give a stab at it. JESSE ROE: Sounds great. So Sal, how would you answer this question? What's the difference between an equation and a function? SALMAN KHAN: Let me think about it a little bit. So let me think. I think there's probably equations that are not functions and functions that are not equations. So let me think of it that way. So I'm going to draw-- if this is the world of equations right over here, so this is equations. And then over here is the world of functions. That's the world of functions. I do think there is some overlap. We'll think it through where the overlap is, the world of functions. So an equation that is not a function that's sitting out here, a simple one would be something like x plus 3 is equal to 10. I'm not explicitly talking about inputs and outputs or relationship between variables. I'm just stating an equivalence. The expression x plus 3 is equal to 10. So this, I think, traditionally would just be an equation, would not be a function. Functions essentially talk about relationships between variables. You get one or more input variables, and we'll give you only one output variable. And you can define a function. And I'll do that in a second. You could define a function as an equation, but you can define a function a whole bunch of ways. You can visually define a function, maybe as a graph-- so something like this. And maybe I actually mark off the values. So that's 1, 2, 3. Those are the potential x values. And then on the vertical axis, I show what the value of my function is going to be, literally my function of x. And maybe that is 1, 2, 3. And maybe this function is defined for all non-negative values. So this is 0 of x. And so let me just draw-- so this right over here, at least for what I've drawn so far, defines that function. I didn't even have to use an equal sign. If x is 2, at least the way I drew it, y is equal to 3. You give me that input. I gave you the value of only one output. So that would be a legitimate function definition. Another function definition would be very similar to what you do in a computer program," + }, + { + "Q": "hey at 1:40 doesn't 3 ^1 equal zero?", + "A": "Anything to the first power is equal to that anything. Take 3^2, that s 3 * 3, right? So 3^1 = 3. And note that 3^0 = 1, not 0.", + "video_name": "6WMZ7J0wwMI", + "timestamps": [ + 100 + ], + "3min_transcript": "In this video, I want to introduce you to the idea of an exponential function and really just show you how fast these things can grow. So let's just write an example exponential function here. So let's say we have y is equal to 3 to the x power. Notice, this isn't x to the third power, this is 3 to the x power. Our independent variable x is the actual exponent. So let's make a table here to see how quickly this thing grows, and maybe we'll graph it as well. So let's take some x values here. Let's start with x is equal to negative 4. Then we'll go to negative 3, negative 2, 0, 1, 2, 3, and 4. And let's figure out what our y-values are going to be for each of these x-values. Now, here, y is going to be 3 to the negative 4 power, which is equal to 1 over 3 to the fourth power. So this is equal to 1/81. When x is equal to negative 3, y is 3. We'll do this in a different color. This color is hard to read. y is 3 to the negative 3 power. Well, that's 1 over 3 to the third power, which is equal to 1/27. So we're going from a super-small number to a less super-small number. And then 3 to the negative 2 power is going to be 1/9, right? 1 over 3 squared, and then we have 3 to the 0 power, which is just equal to 1. So we're getting a little bit larger, a little bit larger, but you'll see that we are about to explode. Now, we have 3 to the first power. That's equal to 3. So we have 3 to the second power, right? y is equal to 3 to the second power. That's 9. 3 to the third power, 27. 3 to the fourth power, 81. If we were to put the fifth power, 243. quickly we're exploding. Let me draw my axes here. So that's my x-axis and that is my y-axis. And let me just do it in increments of 5, because I really want to get the general shape of the graph here. So let me just draw as straight a line as I can. Let's say this is 5, 10, 15. Actually, I won't get to 81 that way. I want to get to 81. Well, that's good enough. Let me draw it a little bit differently than I've drawn it. So let me draw it down here because all of these values, you might notice, are positive values because I have a positive base. So let me draw it like this. Good enough. And then let's say I have 10, 20, 30, 40, 50, 60, 70, 80." + }, + { + "Q": "At 6:32, Sal says function is undefined at x2. He meant is non-differentiable, because it is defined right?", + "A": "I definitely think that Sal is trying to say where the DERIVATIVE is undefined. Because when the function is undefined at a point, we will not have a critical point because this point does not exist: ie, it is not defined", + "video_name": "lDY9JcFaRd4", + "timestamps": [ + 392 + ], + "3min_transcript": "it's going to be a minimum or maximum. And we see the intuition here. If you have-- so non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or a maximum point, at some point x is equal to a, and x isn't the endpoint of some interval, this tells you something interesting. Or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0. Or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0, derivative is 0, derivative is undefined. And we have a word for these points where the derivative is We called them critical points. So for the sake of this function, the critical points are, we could include x sub 0, we could include x sub 1. At x sub 0 and x sub 1, the derivative is 0. And x sub 2, where the function is undefined. Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. But can we say it the other way around? If we find a critical point, where the derivative is 0, or the derivative is undefined, is that going to be a maximum or minimum point? And to think about that, let's imagine this point right over here. So let's call this x sub 3. if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0. So based on our definition of critical point, x sub 3 would also be a critical point. But it does not appear to be a minimum or a maximum point. So a minimum or maximum point that's not an endpoint, it's definitely going to be a critical point. But being a critical point by itself does not mean you're at a minimum or maximum point. So just to be clear that all of these points were at a minimum or maximum point. This were at a critical point, all of these are critical points. But this is not a minimum or maximum point. In the next video, we'll start to think about how you can differentiate, or how you can tell, whether you have a minimum or maximum at a critical point." + }, + { + "Q": "7:12 We can create tangent lines at a point that crosses the function?", + "A": "Yes, a line can be tangent at one point on a curve but then cross it later when the curve takes a U-turn later on down the road.", + "video_name": "lDY9JcFaRd4", + "timestamps": [ + 432 + ], + "3min_transcript": "it's going to be a minimum or maximum. And we see the intuition here. If you have-- so non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or a maximum point, at some point x is equal to a, and x isn't the endpoint of some interval, this tells you something interesting. Or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0. Or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0, derivative is 0, derivative is undefined. And we have a word for these points where the derivative is We called them critical points. So for the sake of this function, the critical points are, we could include x sub 0, we could include x sub 1. At x sub 0 and x sub 1, the derivative is 0. And x sub 2, where the function is undefined. Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. But can we say it the other way around? If we find a critical point, where the derivative is 0, or the derivative is undefined, is that going to be a maximum or minimum point? And to think about that, let's imagine this point right over here. So let's call this x sub 3. if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0. So based on our definition of critical point, x sub 3 would also be a critical point. But it does not appear to be a minimum or a maximum point. So a minimum or maximum point that's not an endpoint, it's definitely going to be a critical point. But being a critical point by itself does not mean you're at a minimum or maximum point. So just to be clear that all of these points were at a minimum or maximum point. This were at a critical point, all of these are critical points. But this is not a minimum or maximum point. In the next video, we'll start to think about how you can differentiate, or how you can tell, whether you have a minimum or maximum at a critical point." + }, + { + "Q": "At 4:50, why don't the endpoints count as maxima/minima?", + "A": "They can be maxima/minima on a closed interval (but not always), but in the video, we are not interested in them because we only want to show point (non-endpoint) that is min/max will have f (a)= 0 or undefined. Endpoints are max/min but they don t necessarily have f (a)=0 or undefined. Hope that helps.", + "video_name": "lDY9JcFaRd4", + "timestamps": [ + 290 + ], + "3min_transcript": "I'm not being very rigorous. But you can see it just by looking at it. So that's fair enough. We've identified all of the maxima and minima, often called the extrema, for this function. Now how can we identify those, if we knew something about the derivative of the function? Well, let's look at the derivative at each of these points. So at this first point, right over here, if I were to try to visualize the tangent line-- let me do that in a better color than brown. If I were to try to visualize the tangent line, it would look something like that. So the slope here is 0. So we would say that f prime of x0 is equal to 0. The slope of the tangent line at this point is 0. What about over here? Well, once again, the tangent line would look something like that. So once again, we would say f prime at x1 is equal to 0. What about over here? We have a positive slope going into it, and then it immediately jumps to being a negative slope. So over here, f prime of x2 is not defined. Let me just write undefined. So we have an interesting-- and once again, I'm not rigorously proving it to you, I just want you to get the intuition here. We see that if we have some type of an extrema-- and we're not talking about when x is at an endpoint of an interval, just to be clear what I'm talking about when I'm talking about x as an endpoint of an interval. We're saying, let's say that the function is where you have an interval from there. So let's say a function starts right over there, and then This would be a maximum point, but it would be an end point. We're not talking about endpoints right now. We're talking about when we have points in between, or when our interval is infinite. So we're not talking about points like that, or points like this. We're talking about the points in between. it's going to be a minimum or maximum. And we see the intuition here. If you have-- so non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or a maximum point, at some point x is equal to a, and x isn't the endpoint of some interval, this tells you something interesting. Or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0. Or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0, derivative is 0, derivative is undefined. And we have a word for these points where the derivative is" + }, + { + "Q": "This is a bit confusing. At 2:24 he said that we don't have more than one local minimum but at the far right of the graph of the function shouldn't the part of the graph that fall under the x-axis count as a local minima, or am I just not paying attention", + "A": "A local minimum has higher points at both sides. This point only has higher one one of the sides so it is not a minimum.", + "video_name": "lDY9JcFaRd4", + "timestamps": [ + 144 + ], + "3min_transcript": "I've drawn a crazy looking function here in yellow. And what I want to think about is when this function takes on the maximum values and minimum values. And for the sake of this video, we can assume that the graph of this function just keeps getting lower and lower and lower as x becomes more and more negative, and lower and lower and lower as x goes beyond the interval that I've depicted right over here. So what is the maximum value that this function takes on? Well we can eyeball that. It looks like it's at that point right over there. So we would call this a global maximum. Function never takes on a value larger than this. So we could say that we have a global maximum at the point x0. Because f of of x0 is greater than, or equal to, f of x, for any other x in the domain. And that's pretty obvious, when you look at it like this. Now do we have a global minimum point, Well, no. This function can take an arbitrarily negative values. It approaches negative infinity as x approaches negative infinity. It approaches negative infinity as x approaches positive infinity. So we have-- let me write this down-- we have no global minimum. Now let me ask you a question. Do we have local minima or local maxima? When I say minima, it's just the plural of minimum. And maxima is just the plural of maximum. So do we have a local minima here, or local minimum here? Well, a local minimum, you could imagine means that that value of the function at that point is lower than the points around it. So right over here, it looks like we have a local minimum. And I'm not giving a very rigorous definition here. But one way to think about it is, we can say that we have a local minimum point at x1, is less than an f of x for any x in this region right over here. And it's pretty easy to eyeball, too. This is a low point for any of the values of f around it, right over there. Now do we have any other local minima? Well it doesn't look like we do. Now what about local maxima? Well this one right over here-- let me do it in purple, I don't want to get people confused, actually let me do it in this color-- this point right over here looks like a local maximum. Not lox, that would have to deal with salmon. Local maximum, right over there. So we could say at the point x1, or sorry, at the point x2, we have a local maximum point at x2. Because f of x2 is larger than f of x for any" + }, + { + "Q": "I don't understand how the three becomes negative. I thought it was just a subtraction sign? 2:11", + "A": "it s because it s part of the number. If it has a negative sign in front of it it s because the number is negative ex. 4 - 3 = 1 the 3 is a negative.", + "video_name": "xMsG9hvqzbY", + "timestamps": [ + 131 + ], + "3min_transcript": "Like the last video, I want to start with two warm up problems. And then we'll do an actual word. And you're going to see these are going to be a little bit more involved than the equations in the last video. But we're still going to be doing the exact same operations, or what we could consider legitimate operations, to get our answer. So here we have 3 times x minus 1 is equal to 2 times x plus 3. So let's see what we can do here. The first thing I like to do-- and there's no definite right way to do it-- there's several ways you could do these problems-- but I like to distribute out the 3 and the 2. So 3 times x minus 1, that's the same thing as 3x minus 3-- I just distributed the 3-- is equal to-- distribute out the 2. 2 times x, plus 2 times 3, which is 6. Now what I like to do is get all of my constant terms on the same side of the equation and all my variable terms on the same side of the equation. So let's see if we can get rid of this 2x term on So let's subtract 2x. I'm going to do a slightly different notation this time because you might see it done this way. Or you might find it easier to visualize it this way. It doesn't matter. It's the same thing we did in the last video. But I want to subtract 2x from this side of the equation. But if I subtract 2x from the side of the equation, I also have to subtract 2x from that side of the equation. So then when we subtract 2x from both sides of the equation, what do we get? Here we get 3x minus 2x. That's just 1x, or x minus 3. 2x minus 2x is no x's, or 0. Then you just have the 6. So we get x minus 3 is equal to 6. That was by getting rid of the 2x from the right hand side; subtracting it from both sides of this equation. Now we have this negative 3 on the left hand side. I just want an x there. So to get rid of that we can add 3 to both sides of this equation. the equation 3 is equal to 3. 3 is, obviously, equal to 3. Negative 2x is, obviously, equal to negative 2x. You could do it either way. But if you add 3 to both sides of this equation the left hand side of the equation becomes just an x, because these two guys cancel out. x equal, and 6 plus 3 is 9. And we are done. And we can even check our answer. 3 times 9 minus 1 is what? This is 3 times 8. This is 24. So that's what the left hand side equals. What does the right hand side equal? That is 2 times 9 plus 3. That's the 2 times 12, which is also equals 24. So it all works out. x is equals to 9. Next problem. z over 16 is equal to 2 times 3z plus 1, all of that over 9." + }, + { + "Q": "In the second problem at 3:37, why does he write 6z + 2, instead of 6z + 1, when it's written 1 in the equation?", + "A": "Well, that s BECAUSE IT s 2(3z+1)=2(3z)+2(1) IT s SO SIMPLE!", + "video_name": "xMsG9hvqzbY", + "timestamps": [ + 217 + ], + "3min_transcript": "the equation 3 is equal to 3. 3 is, obviously, equal to 3. Negative 2x is, obviously, equal to negative 2x. You could do it either way. But if you add 3 to both sides of this equation the left hand side of the equation becomes just an x, because these two guys cancel out. x equal, and 6 plus 3 is 9. And we are done. And we can even check our answer. 3 times 9 minus 1 is what? This is 3 times 8. This is 24. So that's what the left hand side equals. What does the right hand side equal? That is 2 times 9 plus 3. That's the 2 times 12, which is also equals 24. So it all works out. x is equals to 9. Next problem. z over 16 is equal to 2 times 3z plus 1, all of that over 9. Let's multiply both sides of this equation by 9. So if you multiply both sides of this equation by 9, what do we get? We get 9 over 16z is equal to-- this 9 and that 9 will cancel out-- 2 times 3z plus 1. Now let's distribute this 2. So we get 9 over 16z is equal to 2 times 3z is 6z plus 2. Now let's get all of z's on the same side of the equation. So let's subtract 6z from both sides of the equation. So let's subtract minus 6z there, and that, of course, equals minus 6z there. And what do we get? On the left hand side, we get 9/16 minus 6. 6 is equal to what over 16? 60 plus 36. It's equal to 96/16. So it's 9 minus 96 over 16z. This is just 6-- I just rewrote minus 6 here-- is equal to-- these two cancel out. That's why I subtracted 6z in the first place. So it's going to equal 2. So what does this equal right over here? Let me do it in orange. 9 minus 96. The difference between 9 and 96 is going to be 87. And, of course, we're subtracting the larger from the smaller. So it's going to be negative 87 over 16z is equal to 2. And we're almost there. We just have to multiply both sides of this equation by the inverse of this coefficient. So multiply both sides of this equation by negative 16/87." + }, + { + "Q": "At 5:45 shouldn't \"a sub n\" equal a*n!", + "A": "Yes, but a equals 1. a sub n is not a. I hope this clarifies the video!", + "video_name": "dIGLhLMsy2U", + "timestamps": [ + 345 + ], + "3min_transcript": "Well let's think about what's going on. To go from 1 to 2, I multiplied by 2. To go from 2 to 6, I multiplied by 3. To go from 6 to 24, I multiplied by 4. So I'm always multiplying not by the same amount. You have to multiply by the same amount in order for it to be a geometric sequence. Here I'm multiplying it by a different amount. So this sequence that I just constructed has the form, I have my first term, and then my second term is going to be 2 times my first term, and then my third one is going to be 3 times my second term, so 3 times 2 times a. My fourth one is 4 times the third term, so 4 times 3 times 2 times a. So this sequence, which is not a geometric sequence, we can still define it explicitly. We could say that its set or it's the sequence a sub n from n equals 1 to infinity with a sub n being equal to, let's see the fourth one is essentially 4 factorial times a. Well, actually, if we look at this particular, these particular numbers our a is 1. So this is actually, let me write this, this is 1, this is 2 times 1, this is 3 times 2 times 1, this is 4 times 3 times 2 times 1. And so a sub n is just equal to n factorial. This right over here, which is not a geometric sequence, describes exactly this sequence right over here. Just to get some practice with-- Here we've defined it explicitly, but we can also We could also say-- do it in white-- we could also say that a sub n takes us from n equals 1 to infinity, with a sub 1, or maybe at a sub 1 is equal to 1. That's our first term. And then each successive term is going to be equal to the previous term times n. So the second term is equal to the previous term times 2. The nth term is going to be the previous turn times n So this is another valid way of defining it." + }, + { + "Q": "There may be a mistake at 5:47. You say it is equal to n!, when it is equal to n!a.", + "A": "No mistake. Here a = 1. And just like how we don t write 1x\u00c2\u00b2 + 1x - 2, but rather x\u00c2\u00b2 + x - 2 we don t write n! as 1n! Does that make sense? (If a was not equal to 1, then we would need to include it)", + "video_name": "dIGLhLMsy2U", + "timestamps": [ + 347 + ], + "3min_transcript": "Well let's think about what's going on. To go from 1 to 2, I multiplied by 2. To go from 2 to 6, I multiplied by 3. To go from 6 to 24, I multiplied by 4. So I'm always multiplying not by the same amount. You have to multiply by the same amount in order for it to be a geometric sequence. Here I'm multiplying it by a different amount. So this sequence that I just constructed has the form, I have my first term, and then my second term is going to be 2 times my first term, and then my third one is going to be 3 times my second term, so 3 times 2 times a. My fourth one is 4 times the third term, so 4 times 3 times 2 times a. So this sequence, which is not a geometric sequence, we can still define it explicitly. We could say that its set or it's the sequence a sub n from n equals 1 to infinity with a sub n being equal to, let's see the fourth one is essentially 4 factorial times a. Well, actually, if we look at this particular, these particular numbers our a is 1. So this is actually, let me write this, this is 1, this is 2 times 1, this is 3 times 2 times 1, this is 4 times 3 times 2 times 1. And so a sub n is just equal to n factorial. This right over here, which is not a geometric sequence, describes exactly this sequence right over here. Just to get some practice with-- Here we've defined it explicitly, but we can also We could also say-- do it in white-- we could also say that a sub n takes us from n equals 1 to infinity, with a sub 1, or maybe at a sub 1 is equal to 1. That's our first term. And then each successive term is going to be equal to the previous term times n. So the second term is equal to the previous term times 2. The nth term is going to be the previous turn times n So this is another valid way of defining it." + }, + { + "Q": "On problem 51. or (2:55)\n\nCan you prove to me why all the four right triangles are congruent? I know they are... but why? D:", + "A": "The shape in the middle is a square, so all four sides are equal. When it is placed inside another square, the triangles that it creates (4), are all congruent, because the four side lengths of a square are equivalent. If the shape in the middle were a rectangle that is not a square, there would be two congruent triangles, and another two congruent triangles, which are different from the first two. So, because they are both squares, the triangles are congruent.", + "video_name": "6EY0E3z-hsU", + "timestamps": [ + 175 + ], + "3min_transcript": "Fair enough. Now we can also say that the area of this larger square, and it's a bit of an optical illusion, it looks like it's tilted to the left because of the way it's drawn. But anyway, that the area of this larger square is also the area of these four triangles plus the area of this smaller square. So this, the area of the larger square, which we figured out just by taking one side of it and squaring it, that should be equal to the area of the four smaller triangles. So there's four of them. And what's the area of each of them. Let's see, let's just pick this one. 1/2 base times height. So it's 1/2 times a times b. So 1/2 ab is one of these and I multiply by 4 to get all four of these triangles. And then we want to add the area of this inside square. And that's just going to be c squared. Let's see if we can simplify this. So you get a squared plus 2ab plus b squared is equal to 4 times 1/2 is 2ab plus c squared. Well, we could subtract 2ab from both sides of this equation. The top and the bottom of this equation the way I've written it. But if we do that, subtract 2ab from there, subtract 2ab from there, and you're left with a squared plus b squared is equal to c squared, which is the Pythagorean theorem. And we've proved it. So let's see which of their choices matches what we did. OK, which statement would not be used in the proof of the Pythagorean theorem. The area of a triangle equals 1/2 ab. We used that. The four right triangles are congruent. The area of the inner square is equal to half of the area of the larger square. We didn't use that. I think this is the one that would not be used in the proof. Choice D, the area of the larger square is equal to the sum of the squares of the smaller square and the four congruent triangles. No, that that was the crux of the proof. So we definitely used that. So C is our answer. That's the statement that would not be used in the proof. I'm learning to copy and paste ahead of time. So I don't waste your time. All right, a right triangle's hypotenuse has length 5. If one leg has length 2, what is the length of the other leg? Pythagorean theorem, x squared plus 2 squared is equal to 5 squared, because 5 is the hypotenuse. x squared plus 4 is equal to 25." + }, + { + "Q": "At 1:16, how is a a^2 + b^2 equal to a^2 + 2ab +b^2?", + "A": "You misunderstood. It says that (a+b)^2=a^2+2ab+b^2. => (a+b)^2=(a+b)(a+b) Distribute => a*a+a*b+b*a+b*b => a^2+ab+ab+b^2 => a^2+2ab+b^2.", + "video_name": "6EY0E3z-hsU", + "timestamps": [ + 76 + ], + "3min_transcript": "We're on problem 51. And they say, a diagram from a proof of the Pythagorean theorem is pictured below. And they they say, which statement would not be used in the proof of the Pythagorean theorem? So since they have drawn this diagram out, I think we might as well just kind of do the proof and then we can look at their choices and see which ones kind of match up to what we did. Hopefully, they do it the same way. And this is a pretty neat proof of the Pythagorean theorem. I don't think I've done it yet. So I might as well do it now. Well, let's figure out what the area of this large square is right. Well there's two ways to think about it you could just say, OK, this is a square. That's a, that's b. Well this is going to be b as well. This is going to be a as well. So the area of the square is going to be the length of one of its sides squared. So we could say the whole square's area is a plus b squared. Fair enough. Now we can also say that the area of this larger square, and it's a bit of an optical illusion, it looks like it's tilted to the left because of the way it's drawn. But anyway, that the area of this larger square is also the area of these four triangles plus the area of this smaller square. So this, the area of the larger square, which we figured out just by taking one side of it and squaring it, that should be equal to the area of the four smaller triangles. So there's four of them. And what's the area of each of them. Let's see, let's just pick this one. 1/2 base times height. So it's 1/2 times a times b. So 1/2 ab is one of these and I multiply by 4 to get all four of these triangles. And then we want to add the area of this inside square. And that's just going to be c squared. Let's see if we can simplify this. So you get a squared plus 2ab plus b squared is equal to 4 times 1/2 is 2ab plus c squared. Well, we could subtract 2ab from both sides of this equation. The top and the bottom of this equation the way I've written it. But if we do that, subtract 2ab from there, subtract 2ab from there, and you're left with a squared plus b squared is equal to c squared, which is the Pythagorean theorem. And we've proved it. So let's see which of their choices matches what we did. OK, which statement would not be used in the proof of the Pythagorean theorem. The area of a triangle equals 1/2 ab. We used that. The four right triangles are congruent." + }, + { + "Q": "What does he mean at 5:39. \"But if I just cancelled these two things out, the new function would be defined when x = -8\". Why -8?", + "A": "Because when x= -8, the denominator is 0, so the function is undefined. But if we cancel the factors (x+8), then we could plug in x= -8 without dividing by 0, so the function would defined. So since one is defined at x= -8, and the other is not, cancelling the (x+8) changes the function, which we don t want.", + "video_name": "u9v_bakOIcU", + "timestamps": [ + 339 + ], + "3min_transcript": "Both of them have an x plus 8 in them. So we can factor out an x plus 8. So if we factor out an x plus 8, we're left with 2x minus 1, put parentheses around it, times the factored out x plus 8. So we've simplified the numerator. The numerator can be rewritten. And you could have gotten here using the quadratic formula as well. The numerator is 2x minus 1 times x plus 8. And now see if you can factor the denominator. And this one's more straightforward. The coefficient here is 1. So we just have to think of two numbers that when I multiply them, I get 16. And when I add them, I get 10. And the obvious one is 8 and 2, positive 8 and positive 2. So we can write this as x plus 2 times x plus 8. We can divide the numerator and denominator by x plus 8, assuming that x does not equal negative 8. Because this function right over here that's defined by f divided by g, it is not defined when g of x is equal to 0, because then you have something divided by 0. And the only times that g of x is equal to 0 is when x is equal to negative 2 or x is equal to negative 8. So if we divide the numerator and the denominator by x plus 8 to simplify it, in order to not change the function definition, we have to still put the constraint that x cannot be equal to negative 8. That the original function, in order to not change it-- because if I just cancelled these two things out, the new function with these canceled would be defined when x is equal to negative 8. But we want this simplified thing to be the same exact function. And this exact function is not defined when x is equal to negative 8. So now we can write f/g of x, which is really just f of x divided by g of x, is equal to 2x minus 1 You have to put the condition there that x cannot be equal to negative 8. If you lost this condition, then it won't be the exact same function as this, because this is not defined when x is equal to negative 8." + }, + { + "Q": "@ 2:16 where does -16 come from?", + "A": "Ok, so he s trying to factor the trinomial (ax^2+ bx + c). This method is what I learned as the long method. He is just finding the LCM of a and c so he can split bx so it is easier to factor it out. It is just finding or converting numbers into more factor able numbers and group them. This is what Sal calls factoring by grouping. I hope this helps.", + "video_name": "u9v_bakOIcU", + "timestamps": [ + 136 + ], + "3min_transcript": "f of x is equal to 2x squared plus 15x minus 8. g of x is equal to x squared plus 10x plus 16. Find f/g of x. Or you could interpret this is as f divided by g of x. And so based on the way I just said it, you have a sense of what this means. f/g, or f divided by g, of x, by definition, this is just another way to write f of x divided by g of x. You could view this as a function, a function of x that's defined by dividing f of x by g of x, by creating a rational expression where f of x is in the numerator and g of x is in the denominator. And so this is going to be equal to f of x-- we have right up here-- is 2x squared 15x minus 8. And g of x-- I will do in blue-- is right over here, g of x. So this is all going to be over g And you could leave it this way, or you could actually try to simplify this a little bit. And the easiest way to simplify this would see if we could factor the numerator and the denominator expressions into maybe simpler expressions. And maybe some of them might be on-- maybe both the numerator and denominator is divisible by the same expression. So let's try to factor each of them. So first, let's try the numerator. And I'll actually do it up here. So let's do it. Actually, I'll do it down here. So if I'm looking at 2x squared plus 15x minus 8, we have a quadratic expression where the coefficient is not 1. And so one technique to factor this is to factor by grouping. You could also use the quadratic formula. And when you factor by grouping, you're going to split up this term, this 15x. And you're going to split up into two terms where the coefficients are, if I were to take the product of those coefficients, of the first and the last terms. And we proved that in other videos. So essentially, we want to think of two numbers that add up to 15, but whose product is equal to negative 16. And this is just the technique of factoring by grouping. It's really just an attempt to simplify this right over here. So what two numbers that, if I take their product, I get negative 16. But if I add them, I get 15? Well, if I take the product and get a negative number, that means they have to have a different sign. And so that means one of them is going to be positive, one of them is going to be negative, which means one of them is going to be larger than 15 and one of them is going to be smaller than 15. And the most obvious one there might be 16, positive 16, and negative 1. If I multiply these two things, I definitely get negative 16. If I add these two things, I definitely get 15. So what we can do is we can split this. We can rewrite this expression as 2x squared plus 2x squared" + }, + { + "Q": "At 6:06 Khan is explaining why x can't be -8, but he didn't say that it can't be -2. So, does this mean that x can be -2? This will also give you a 0 in the denominator.", + "A": "No, x cannot be -2 either, but it is not necessary to specify it because x - 2 is still in the simplified expression. You have to put the x does not equal -8 because the term that would have made that clear has vanished.", + "video_name": "u9v_bakOIcU", + "timestamps": [ + 366 + ], + "3min_transcript": "Both of them have an x plus 8 in them. So we can factor out an x plus 8. So if we factor out an x plus 8, we're left with 2x minus 1, put parentheses around it, times the factored out x plus 8. So we've simplified the numerator. The numerator can be rewritten. And you could have gotten here using the quadratic formula as well. The numerator is 2x minus 1 times x plus 8. And now see if you can factor the denominator. And this one's more straightforward. The coefficient here is 1. So we just have to think of two numbers that when I multiply them, I get 16. And when I add them, I get 10. And the obvious one is 8 and 2, positive 8 and positive 2. So we can write this as x plus 2 times x plus 8. We can divide the numerator and denominator by x plus 8, assuming that x does not equal negative 8. Because this function right over here that's defined by f divided by g, it is not defined when g of x is equal to 0, because then you have something divided by 0. And the only times that g of x is equal to 0 is when x is equal to negative 2 or x is equal to negative 8. So if we divide the numerator and the denominator by x plus 8 to simplify it, in order to not change the function definition, we have to still put the constraint that x cannot be equal to negative 8. That the original function, in order to not change it-- because if I just cancelled these two things out, the new function with these canceled would be defined when x is equal to negative 8. But we want this simplified thing to be the same exact function. And this exact function is not defined when x is equal to negative 8. So now we can write f/g of x, which is really just f of x divided by g of x, is equal to 2x minus 1 You have to put the condition there that x cannot be equal to negative 8. If you lost this condition, then it won't be the exact same function as this, because this is not defined when x is equal to negative 8." + }, + { + "Q": "at 2:19 why has he divided 1* 10^4 upon 7*10^5 when it is written in the question 7*10^5 than 1*10^4?", + "A": "To find the quotient of exponents you subtract the exponents from the other. If you need more help, you can look at some other of Sal s Exponents videos.", + "video_name": "DaoJmvqU3FI", + "timestamps": [ + 139 + ], + "3min_transcript": "Let's do a few more examples from the orders of magnitude exercise. Earth is approximately 1 times 10 to the seventh meters in diameter. Which of the following could be Earth's diameter? So this is just an approximation. It's an estimate. And they're saying, which of these, if I wanted to estimate it, would be close or would be 1 times 10 to the seventh? And the key here is to realize that 1 times 10 to the seventh is the same thing as one followed by seven zeroes. One, two, three, four, five, six, seven. Let me put some commas here so we make it a little bit more Or another way of talking about it is that it is, 1 times 10 to the seventh, is the same thing as 10 million. So which of these, if we were to really roughly estimate, we would go to 10 million. Well, this right over here is 1.271 million, or 1,271,543. If I were to really roughly estimate it, I might go to one million, but I'm not going to go to 10 million. This is 12,715,430. If I were to roughly estimate this, well, yeah. I would go to 10 million. 10 million is if I wanted really just one digit to represent it, if I were write this in scientific notation. This right over here is 1.271543 times 10 to the seventh. Let me write that down. 12,715,430. If I were to write this in scientific notation as 1.271543 times 10 to the seventh. And when you write it this way, you say, hey, well, yeah, if I was to really estimate this and get pretty rough with it, and I just rounded this down, I would make this 1 times 10 to the seventh. So this really does look like our best choice. Now let me just verify. Well, this right over here, if I were to write it, I would go to 100 million, or 1 times 10 to the eighth. That's way too big. And this, if I were to write it, I would go to a billion, or 1 times 10 to the ninth. So that's also too big. So once again, this feels like the best answer. So here we're asked, how many times larger is 7 times 10 to the fifth than 1 times 10 to the fourth? Well, we could just divide to think about that. So 7 times 10 to the fifth divided by 1 times 10 to the fourth. Well, this is the same thing as 7 over 1 times 10 to the fifth over 10 to the fourth, which is just going to be equal to-- well, 7 divided by 1 is 7. And 10 to the fifth, that's multiplying five 10's. And then you're dividing by four 10's. You're going to have one 10 left over. Or, if you remember your exponent properties, this would be the same thing as 10 to the 5 minus 4 power, or 10 to the first power. So this right over here, all of this business, is going to simplify to 10 to the first, or I could actually write it this way. This would be the same thing as 10 to the 5 minus 4," + }, + { + "Q": "Quick Question at 5:55 he writes cos^2 and theta and sin^2 theta now based on how he wrote it I am wondering if it is cos^(2 theta) or cos^(2) Theta?", + "A": "cos^2(\u00ce\u00b8) is the trig representation for (cos(\u00ce\u00b8))^2.", + "video_name": "n0DLSIOYBsQ", + "timestamps": [ + 355 + ], + "3min_transcript": "Say you know cosine theta then you use this to figure out sine of theta, then you can figure out tangent of theta because tangent of theta is just sine over cosine. If you're a little bit confused as to why this is called the Pythagorean identity, well it really just falls out of where the equation of a circle even came from. If we look at this point right over here, we look at this point right over here, which we're saying is the x coordinate is cosine theta and the y coordinate is sine of theta, what is the distance between that point and the origin? Well to think about that we can construct a right triangle. This distance right over here. So that we could deal with any quadrant I'll make it the absolute value of the cosine theta is this distance right over here. And this distance right over here is the absolute value of the sine of theta. for this first quadrant here but if I went into the other quadrants and I were to setup a similar right triangle then the absolute value is at play. What do we know from the Pythagorean theorem? This is a right triangle here, the hypotenuse has length one, so we know that this expression squared, the absolute value of cosine of theta squared, plus this expression squared, which is this length, plus the absolute value of the sine of theta squared needs to be equal to the length of the hypotenuse squared, which is the same thing which is going to be equal to one squared. Or we could say, this is the same thing. If you're going to square something the sign, if negative it's going to be negative times a negative so it's just going to be positive so this is going to be the same thing as saying that the cosine squared theta plus sine squared theta is equal to one. This is why it's called the Pythagorean identity. a circle comes from, it comes straight out of the Pythagorean theorem where your hypotenuse has length one." + }, + { + "Q": "at 6:30 Sal mentions break even\nwhat does that mean?", + "A": "Break even is the point when revenue equals expenses and so there is no profit or loss incurred. If the expense amounts to $25 then the break even point is when revenue is $25 also. At this point there will be no loss or profit as 25-25 = 0", + "video_name": "5EdbPz1ZVn0", + "timestamps": [ + 390 + ], + "3min_transcript": "So then our graph is going to be out here someplace. But we could just figure it out algebraically. At the end of the year, m will be equal to 12. When m is equal to 12, how much is our membership? The price of our membership is going to be our $200 membership fee plus 39 times the number of months, times 12. What's 39 times 12? 2 times 9 is 18. 2 times 3 is 6 plus 1 is 70. I have a 0. 1 times 9 is 9. 1 times 3-- we want to ignore this. 1 times 3 is 3. So we have 8. 7 plus 9 is 16. 1 plus 3 is 4. So the price of our membership is $200 plus 39 times 12, which is $468. So it's equal to $668 at the end of our year. So if you went all the way out to 12, you would have to plot 668 someplace here on our line, if we just Let's do one more of these. Bobby and Petra are running a lemonade stand and they charge $0.45 for each glass of lemonade. In order to break even, they must make $25. How many glasses of lemonade must they sell to break even? So let me just do it with y and x. y is equal to the amount they make. Not max-- the amount they make. Let x is equal to the number of glasses they sell. What is y as a function of x? So y is equal to-- well for every glass they sell, they the number of glasses. There's not any kind of minimum fee that they need to charge or they don't say any kind of minimum cost that they have to spend to run this place. How much in order to break even for each a glass of lemonade? They need to make $25. So in order to break even, they must make $25. So how many glasses of lemonade do they need to sell? y needs to be equal to $25. How many glasses do they need to sell? Well you just set this equation. You say 0.45x has to be equal to 25. We can divide both sides by 0.45." + }, + { + "Q": "like sal said in 5:36 are all sums of 3 angles 180", + "A": "No, not all angles will sum to 180. Ex: three angles could be 10, 20 and 30. They don t sum to 180. Only three angles of a triangle will always sum to 180.", + "video_name": "BTnAlNSgNsY", + "timestamps": [ + 336 + ], + "3min_transcript": "And these come out of the fact that the angle formed between DB and BC, that is a 90-degree angle. Now, we have other words when our two angles add up to other things. So let's say, for example, I have one angle over here. Let me put some letters here so we can specify it. So let's say this is X, Y, and Z. And let's say that the measure of angle XYZ is equal to 60 degrees. And let's say that you have another angle that looks like this. And I'll call this, let's say, maybe MNO. So if you were to add the two measures of these-- so let me write this down. The measure of angle MNO plus the measure of angle XYZ, this is going to be equal to 120 degrees plus 60 degrees, which is equal to 180 degrees. So if you add these two things up, you essentially are able to go all halfway around the circle. Or you could go throughout the entire half circle or semicircle for a protractor. And when you have two angles that add up to 180 degrees, we call them supplementary. I know it's a little hard to remember sometimes. 90 degrees is complementary. They're just complementing each other. And then if you add up to 180 degrees, you have supplementary. You have supplementary angles. are adjacent so that they share a common side-- so let me draw that over here. So let's say you have one angle that looks like this. And that you have another angle. So let me put some letters here again. And I'll start reusing letters. So let's say that this is ABC. And you have another angle that looks like this. I already used C. Once again, let's say that this is 50 degrees. And let's say that this right over here is 130 degrees. Clearly, angle DBA plus angle ABC, if you add them together, you get 130 degrees plus 50 degrees, which is 180 degrees. So they are supplementary. So let me write that down. Angle DBA and angle ABC are supplementary." + }, + { + "Q": "At 2:30 Sal describes the term 'Complementary Angle\" and states that if two angles add up to 90 degrees then they are complementary. Do the two angles that add up to 90 degrees have to be adjacent?", + "A": "No. (Complementary simply means that, if put next to each other, they would create an angle of 90 degrees; they don t actually have to be next to each other.) The same applies to supplementary angles.", + "video_name": "BTnAlNSgNsY", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's say I have an angle ABC, and it looks something like this. So its vertex is going to be at B. Maybe A sits right over here, and C sits right over there. And then also let's say that we have another angle called DBA. I want to have the vertex once again at B. So let's say it looks like this. So this right over here is our point D. That is our point D. And let's say that we know that the measure of angle DBA is equal to 40 degrees. So this angle right over here, its measure is equal to 40 degrees. And let's say that we know that the measure of angle ABC is equal to 50 degrees. So there's a bunch of interesting things happening here. The first interesting thing that you might realize is that both of these angles share a side. or rays-- but if you view them as rays, they both share this ray BA. And when you have two angles like this that share the same side, these are called adjacent angles. Because the word \"adjacent\" literally means next to. These are adjacent. They are adjacent angles. Now there's something else that you might notice that's interesting here. We know that the measure of angle DBA is 40 degrees and the measure of angle ABC is 50 degrees. And you might be able to guess what the measure of angle DBC is. If we drew a protractor over here-- I'm not going to draw it. It will make my drawing all messy. Well, maybe I'll draw it really fast. So if you had a protractor right over here, clearly this is opening up to 50 degrees. And this is going another 40 degrees. So if you wanted to say what the measure of angle DBC is, And let me delete all of this stuff right here to keep things clean. So the measure of angle DBC would be equal to 90 degrees. And we already know that 90 degrees is a special angle. This is a right angle. There's also a word for two angles whose sum add up to 90 degrees, and that is complementary. So we can also say that angles DBA and angles ABC are complementary. And that is because their measures add up to 90 degrees. So the measure of angle DBA plus the measure of angle ABC is equal to 90 degrees. They form a right angle when you add them up." + }, + { + "Q": "I thought you can't divide by variable 7:00\nor am I missing something here?", + "A": "You can divide by a variable, so long as the variable is not 0 (ie b does not equal k).", + "video_name": "okXVhDMuGFg", + "timestamps": [ + 420 + ], + "3min_transcript": "so we have all the y's on the left-hand side, so, plus 2yb, that's gonna give us a 2yb on the left-hand side, plus 2yb. So what is this going to be equal to? And I'm starting to run into my graph, so let me give myself a little bit more real estate over here. So on the left-hand side, what am I going to have? This is the same thing as 2yb minus 2yk, which is the same thing, actually let me just write that down. That's going to be 2y-- Do it in green, actually, well, yeah, why not green? That's going to be-- Actually, let me start a new color. (chuckles) That's going to be 2yb minus 2yk. You can factor out a 2y, and it's gonna be 2y times, b minus k. So let's do that. So we could write this as 2 times, b minus k, y if you factor out a 2 and a y, so that's that piece right over there. These things cancel out. Now, on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise. On the right-hand side, you have x minus a, squared, and then, let's see, these characters cancel out, and you're left with b squared minus k squared, so these two are gonna be b squared minus k squared, plus b squared minus k squared. Now, I said all I want is a y on the left-hand side, so let's divide everything by two times, b minus k. So, let's divide everything, two times, b minus k, so, two times, b minus k. And I'm actually gonna divide this whole thing by two times, b minus k. Now, obviously on the left-hand side, this all cancels out, you're left with just a y, and then it's going to be y equals, y is equal to one over, and notice, b minus k is the difference between the y-coordinate of the focus, and the y-coordinate, I guess you could say, of the line, y equals k, so it's one over, two times that, times x minus a, squared. So if you knew what b minus k was, this would just simplify to some number, some number that's being multiplied times x minus a, squared, so hopefully this is starting to look like the parabolas that you remember from your childhood, (chuckles) if you do remember parabolas from your childhood. Alright, so then let's see if we could simplify this thing on the right, and you might recognize, b squared minus k squared, that's a difference of squares, that's the same thing as b plus k, times b minus k, so the b minus k's cancel out, and we are just left with," + }, + { + "Q": "At 2:30, when he says that the absolute value could also work in ensuring that the value is positive, can someone explain to me why the absolute value isn't used in the equation?", + "A": "In the video, Sal does some algebraic manipulation to achieve the formula of the parabola. It is much easier to derive the parabola if he were to square the expression and take the square root than to take the absolute value of the expression. Both methods yield the same value of the expression; however, the latter method (squaring then taking the square root) allows for more easier manipulation. Hope this helps!", + "video_name": "okXVhDMuGFg", + "timestamps": [ + 150 + ], + "3min_transcript": "- [Voiceover] What I have attempted to draw here in yellow is a parabola, and as we've already seen in previous videos, a parabola can be defined as the set of all points that are equidistant to a point and a line, and the point is called the focus of the parabola, and the line is called the directrix of the parabola. What I want to do in this video, it's gonna get a little bit of hairy algebra, but given that definition, I want to see, and given that definition, and given a focus at the point x equals a, y equals b, and a line, a directrix, at y equals k, to figure out what is the equation of that parabola actually going to be, and it's going to be based on a's, b's, and k's, so let's do that. So let's take a arbitrary point on the parabola. Let's say we take this point right over here, and its x-coordinate is x, and its y-coordinate is y, and by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix, That means that the distance to the directrix, which I'm drawing here in blue, has to be the same as the distance to the focus, which I am drawing in magenta, and when we take the distance to the directrix, we literally just drop a perpendicular, that is, that's going to be the shortest distance to that line, but the distance to the focus, well we see that's at a bit of an angle, and we might have to use the distance formula, which is really just the Pythagorean Theorem. So let's do that. This distance has to be the same as that distance. So, what's this blue distance? Well, that's just gonna be our change in y. It's going to be this y, minus k. It's just this distance. So it's going to be y minus k. Now we have to be careful. The way I've just drawn it, yes, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances, but you can definitely have a parabola where lower than the y-coordinate of the directrix, in which case this would be negative. So what we really want is the absolute value of this, or, we could square it, and then we could take the square root, the principle root, which would be equivalent to taking the absolute value of y minus k. So that's this distance right over here, and by the definition of a parabola, in order for (x,y) to be sitting on the parabola, that distance needs to be the same as the distance from (x,y) to (a,b), to the focus. So what's that going to be? Well, we just apply the distance formula, or really, just the Pythagorean Theorem. It's gonna be our change in x, so, x minus a, squared, plus the change in y, y minus b, squared, and the square root of that whole thing, the square root of all of that business. Now, this right over here is an equation of a parabola. It doesn't look like it, it looks really hairy," + }, + { + "Q": "In practice sessions video (using hint), of unit vectors at around 2:28 ,he said that we should divide 3 by 5 and also 4 by 5...why should we divide 3 by 5 and 4 by 5?", + "A": "(Note that this question actually refers to the next video.) The problem asked us to find the UNIT vector. In other words, a vector of length 1 unit. We re given a vector of length 5 units (which is 5 times what we want), so we have to scale it down by dividing each of its components by 5.", + "video_name": "9ylUcCOTH8Y", + "timestamps": [ + 148 + ], + "3min_transcript": "We've already seen that you can visually represent a vector as an arrow, where the length of the arrow is the magnitude of the vector and the direction of the arrow is the direction of the vector. And if we want to represent this mathematically, we could just think about, well, starting from the tail of the vector, how far away is the head of the vector in the horizontal direction? And how far away is it in the vertical direction? So for example, in the horizontal direction, you would have to go this distance. And then in the vertical direction, you would have to go this distance. Let me do that in a different color. You would have to go this distance right over here. And so let's just say that this distance is 2 and that this distance is 3. We could represent this vector-- and let's call this vector v. We could represent vector v as an ordered list or a 2-tuple of-- so we could say we and 3 in the vertical direction. So you could represent it like that. You could represent vector v like this, where it is 2 comma 3, like that. And what I now want to introduce you to-- and we could come up with other ways of representing this 2-tuple-- is another notation. And this really comes out of the idea of what it means to add and scale vectors. And to do that, we're going to define what we call unit vectors. And if we're in two dimensions, we define a unit vector for each of the dimensions we're operating in. If we're in three dimensions, we would define a unit vector for each of the three dimensions that we're operating in. And so let's do that. So let's define a unit vector i. And the way that we denote that is the unit vector we put this hat on top of it. So the unit vector i, if we wanted to write it in this notation right over here, we would say it only goes 1 unit in the horizontal direction, and it doesn't go at all in the vertical direction. So it would look something like this. That is the unit vector i. And then we can define another unit vector. And let's call that unit vector-- or it's typically called j, which would go only in the vertical direction and not in the horizontal direction. And not in the horizontal direction, and it goes 1 unit in the vertical direction. So this went 1 unit in the horizontal. And now j is going to go 1 unit in the vertical. So j-- just like that. Now any vector, any two dimensional vector, we can now represent as a sum of scaled up versions of i and j." + }, + { + "Q": "at 2:58 Sal says that any real number is greater or equal to 0 but isn't that a whole number?\nisn't a real number greater than 0?", + "A": "Natural numbers: 1, 2, 3... Whole numbers: 0, 1, 2, 3... Integers: ...-3, -2, -1, 0, 1, 2, 3... Real: Includes all positive and negative numbers including the decimals and fractional values.", + "video_name": "qFFhdLlX220", + "timestamps": [ + 178 + ], + "3min_transcript": "Now, the one thing I'm going to do here-- actually, I won't talk about it just yet, of how we're going to do it differently than we did it in the last video. This radical right here can be rewritten as-- so this is going to be 3 times the square root, or the principal root, I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and then taking the product. And so then this over here is going to be times the square root of, or the principal root of, x squared times the principal root of x. And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30-- and I'm just going to switch the order here-- times the absolute value of x. And then you have the square root of 5, And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is, if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers-- then x has to be greater than or equal to 0. So maybe I could write it this way. The domain here is that x is any real number greater than or equal to 0. And the reason why I say that is, if you put a negative number in here and you cube it, And then at least in the real numbers, you won't get an actual value. You'll get a square root of a negative number here. So if you make this-- if you assume this right here, we're dealing with the real numbers. We're not dealing with any complex numbers. When you open it up to complex numbers, then you can expand the domain more broadly. But if you're dealing with real numbers, you can say that x is going to be greater than or equal to 0. And then the absolute value of x is just going to be x, because it's not going to be a negative number. And so if we're assuming that the domain of x is-- or if this expression is going to be evaluatable, or it's going to have a positive number, then this can be written as 30x times the square root of 5x. If you had the situation where we were dealing with complex numbers-- and if you don't know what a complex number is, or an imaginary number," + }, + { + "Q": "Why does Sal keep saying \"the principal root of...\" as opposed to \"the square root of...\"? At 1:30 I heard him say \"the square root of, or the principal root of...\" so does that mean they're the same thing? Because it appears as if he sort of corrected himself.", + "A": "Each square root actually has 2 roots. Consider: 5^2 = 25, but if you square (-5), it is also 25. So, 5^2 = 25 and (-5)^2 = 25. Now, consider square root of 25. Is it 5 or is it -5? We need to know what value to use. So, when you see sqrt(25) , it is understood that the answer should be the principal root or the positive root = 5. If you see - sqrt(25) , the minus sign in front of the radical tells you that your answer should be the negative root = -5. Hope this helps.", + "video_name": "qFFhdLlX220", + "timestamps": [ + 90 + ], + "3min_transcript": "What I want to do in this video is resimplify this expression, 3 times the principal root of 500 times x to the third, and take into consideration some of the comments that we got out on YouTube that actually give some interesting perspective on how you could simplify this. So just as a quick review of what we did in the last video, we said that this is the same thing as 3 times the principal root of 500. And I'm going to do it a little bit different than I did in the last video, just to make it interesting. This is 3 times the principal root of 500 times the principal root of x to the third. And 500-- we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5. Or even better, we could rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. Now, the one thing I'm going to do here-- actually, I won't talk about it just yet, of how we're going to do it differently than we did it in the last video. This radical right here can be rewritten as-- so this is going to be 3 times the square root, or the principal root, I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and then taking the product. And so then this over here is going to be times the square root of, or the principal root of, x squared times the principal root of x. And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30-- and I'm just going to switch the order here-- times the absolute value of x. And then you have the square root of 5, And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is, if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers-- then x has to be greater than or equal to 0. So maybe I could write it this way. The domain here is that x is any real number greater than or equal to 0. And the reason why I say that is, if you put a negative number in here and you cube it," + }, + { + "Q": "Wait- at 2:37, is tau 2 pi?", + "A": "Yes, tau is equivalent to twice pi.", + "video_name": "FtxmFlMLYRI", + "timestamps": [ + 157 + ], + "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211." + }, + { + "Q": "Sorry, I don't get 0:48-0:52.\nWhy do you need a distance and to pick a point?", + "A": "the distance is the radius and the point is the center", + "video_name": "FtxmFlMLYRI", + "timestamps": [ + 48, + 52 + ], + "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211." + }, + { + "Q": "at 1:19 what is a 4 dimensional object", + "A": "Actually, Steven Hawking has speculated that all objects have 4-dimensions, and the fourth dimension is time. Like, as an object grows older, the volume (or whatever you d call it) of it s fourth dimension would increase.", + "video_name": "FtxmFlMLYRI", + "timestamps": [ + 79 + ], + "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211." + }, + { + "Q": "At 00:50,...is multiplying out the Binomial the only way to get rid of the bracket. If we can distribute the exponent in for example (2r)^2 to be 4r^2.....why can't we just distribute the exponent through (x+9)^2 ?", + "A": "Because (a + b)^2 isn t the same thing as a^2 + b^2. Example: (2 + 2)^2 = 4^2 = 16 =/= 2^2 + 2^2 = 4 + 4 = 8 Multiplying the binomial is the correct way.", + "video_name": "bFtjG45-Udk", + "timestamps": [ + 50 + ], + "3min_transcript": "And now I want to do a bunch of examples dealing with probably the two most typical types of polynomial multiplication that you'll see, definitely, in algebra. And the first is just squaring a binomial. So if I have x plus 9 squared, I know that your temptation is going to say, oh, isn't that x squared plus 9 squared? And I'll say, no, it isn't. You have to resist every temptation on the planet to do this. It is not x squared plus 9 squared. Remember, x plus 9 squared, this is equal to x plus 9, times x plus 9. This is a multiplication of this binomial times itself. You always need to remember that. It's very tempting to think that it's just x squared plus 9 squared, but no, you have to expand it out. And now that we've expanded it out, we can use some of the skills we learned in the last video to actually multiply it. multiplied the trinomial last time, let's multiply x plus 9, times x plus a magenta 9. And I'm doing it this way just to show you when I'm multiplying by this 9 versus this x. But let's just do it. So we go 9 times 9 is 81. Put it in the constants' place. 9 times x is 9x. Then we have-- go switch to this x term-- we have a yellow x. x times 9x is 9x. Put it in the first degree space. x times x is x squared. And then we add everything up. And we get x squared plus 18x plus 81. So this is equal to x squared plus 18x plus 81. Now you might see a little bit of a pattern here, and I'll actually make the pattern explicit in a second. You have x squared. You have this x times this x, gives you x squared. You have the 9 times the 9, which is 81. And then you have this term here which is 18x. How did we get that 18x? Well, we multiplied this x times 9 to get 9x, and then we multiplied this 9 times x to get another 9x. And then we added the two right here to get 18x. So in general, whenever you have a squared binomial-- let I'll do it in very general terms. Let's say we have a plus b squared. Let me multiply it this way again, just to give you the hang of it. This is equal to a plus b, times a plus-- I'll do a green b right there. So we have to b times b is b squared. Let's just assume that this is a constant term. I'll put it in the b squared right there." + }, + { + "Q": "At 4:57, why does Khan use a^2+2ab+b^2 for (a-b)^2 instead of a^2-2ab+b^2? Am I wrong? Why? Thank you!", + "A": "no, he doesn t. he s got it all right", + "video_name": "bFtjG45-Udk", + "timestamps": [ + 297 + ], + "3min_transcript": "So this would be a constant, this would be analogous to our 81. a is a variable that we-- actually let me change that up even better. Let me make this into x plus b squared, and we're assuming b is a constant. So it would be x plus b, times x plus a green b, right there. So assuming b's a constant, b times b is b squared. b times x is bx. And then we'll do the magenta x. x times b is bx. And then x times x is x squared. So when you add everything, you're left with x squared plus 2bx, plus b squared. So what you see is, the end product, what you have when product of x and b, plus b squared. So given that pattern, let's do a bunch more of these. And I'm going to do it the fast way. So 3x minus 7 squared. Let's just remember what I told you. Just don't remember it, in the back of your mind, you should know why it makes sense. If I were to multiply this out, do the distributive property twice, you know you'll get the same answer. So this is going to be equal to 3x squared, plus 2 times 3x, times negative 7. Right? We know that it's 2 times each the product of these terms, plus negative 7 squared. And if we use our product rules here, 3x squared is the This right here, you're going to have a 2 times a 3, which is 6, times a negative 7, which is negative 42x. And then a negative 7 squared is plus 49. That was the fast way. And just to make sure that I'm not doing something bizarre, let me do it the slow way for you. 3x minus 7, times 3x minus 7. Negative 7 times negative 7 is positive 49. Negative 7 times 3x is negative 21x. 3x times negative 7 is negative 21x. 3x times 3x is 9 x squared. Scroll to the left a little bit. Add everything. You're left with 9x squared, minus 42x, plus 49." + }, + { + "Q": "Why did Sal distribute the 4x^2 at 6:25 instead of just writing (4x^2+y^2)(4x^2+y^2) like he did for the previous questions?", + "A": "It s more detailed that way.", + "video_name": "bFtjG45-Udk", + "timestamps": [ + 385 + ], + "3min_transcript": "product of x and b, plus b squared. So given that pattern, let's do a bunch more of these. And I'm going to do it the fast way. So 3x minus 7 squared. Let's just remember what I told you. Just don't remember it, in the back of your mind, you should know why it makes sense. If I were to multiply this out, do the distributive property twice, you know you'll get the same answer. So this is going to be equal to 3x squared, plus 2 times 3x, times negative 7. Right? We know that it's 2 times each the product of these terms, plus negative 7 squared. And if we use our product rules here, 3x squared is the This right here, you're going to have a 2 times a 3, which is 6, times a negative 7, which is negative 42x. And then a negative 7 squared is plus 49. That was the fast way. And just to make sure that I'm not doing something bizarre, let me do it the slow way for you. 3x minus 7, times 3x minus 7. Negative 7 times negative 7 is positive 49. Negative 7 times 3x is negative 21x. 3x times negative 7 is negative 21x. 3x times 3x is 9 x squared. Scroll to the left a little bit. Add everything. You're left with 9x squared, minus 42x, plus 49. Let's do one more, and we'll do it the fast way. So if we have 8x minus 3-- actually, let me do one which has more variables in it. Let's say we had 4x squared plus y squared, and we wanted to square that. Well, same idea. This is going to be equal to this term squared, 4x squared, squared, plus 2 times the product of both terms, 2 times 4x squared times y squared, plus y squared, this term, squared. And what's this going to be equal to? This is going to be equal to 16-- right, 4 squared is 16-- x squared, squared, that's 2 times 2, so it's x to the fourth power. And then plus, 2 times 4 times 1, that's 8x squared y squared." + }, + { + "Q": "At 1:46 why can Sal simplify any further. Cant he take the square root", + "A": "We don t know the value of x and you can t apply the square root individually like sqrt(4x^2/9+4) =2x/3+2 , it is wrong, so you can t simplify any further from there.", + "video_name": "hl58vTCqVIY", + "timestamps": [ + 106 + ], + "3min_transcript": "In the last hyperbola video I didn't get a chance to do some concrete examples. So I'll do that right now. So, let's say I had the hyperbola y squared over 4 minus x squared over, I don't know, let me think of a good number. Let's say, x squared over 9 is equal to 1. So the first thing to figure out about this hyperbola is, what are its asymptotes? And, once again, I always forget the formulas. And I just try to solve for y and see what happens when x approaches positive or negative infinity. So if you solve for y, you can add x squared over 9 to both sides. And you get y squared over 4 is equal to x squared over 9 plus 1. Now, I can multiply 4 times both sides. And you get y squared is equal to 4 over 9 times I distribute the 4, take the positive and negative square root both sides. y is equal to the plus or minus square root of 4 over 9x squared plus 4. And you can't really simplify this anymore. But we can think about, what does this approach as x approaches positive or negative infinity. So, as x approaches plus or minus infinity, what does this roughly equal? What does this approximate? What does the graph get a lot closer to? Well, then, y is approximately equal to just the square Because this becomes super huge and relative to this term, this starts to matter less and less and less. And that's why we get closer and closer to the asymptotes. Because when this number is, like, a trillion, or a google, then this number is almost insignificant. square root of this, and you'll just be a little bit above the graph. Because you have this extra plus-4 there. So as you approach positive or negative infinity, this equation is approximately equal to the plus or minus square root of 4 over 9x squared. And so, that is -- so y would be approximately equal We can take the square root of this. Plus or minus the square root of 4/9 is 2/3, right? Square root of 4 over square root of 9, times x. So, these are the asymptotes. There's two lines here. There's y is equal to 2/3 x. And then there's y is equal to minus 2/3 x. So let's draw those two lines. Let me draw my axes. Let's make that my y axis. Make that the x axis. Let me switch some colors, just to make things interesting. So let me draw the first one." + }, + { + "Q": "So at 6:34, are (0, 2) and (0, -2) the foci of the hyperbola?", + "A": "Those are the vertices of the hyperbola.", + "video_name": "hl58vTCqVIY", + "timestamps": [ + 394 + ], + "3min_transcript": "Because it will never, a hyperbola will never cross the asymptotes. It's not like it can go out here and across this asymptote. We already know that the graph of this parabola -- and you can try other points, if you want, just to verify. It's going to look something like this. It's going to go and then -- nope, I want to make it so it never touches. It's going to get really close, but no, I touched it. It's going to get really close but never touch. And then on this side it's going to get really close, but never touch. And I don't want to touch it. And then on the top side it's going to do the same thing, it's going to get really close, and as you approach infinity it's never going to touch it. And as you get reall close, it'll get infiniitely close but never touch it. So that's what this parabola -- this hyperbola -- is going to look like. And I did it by just trying to see if x could be equal to 0. And I encourage you to try what happens when y equals 0. And you'll get no solution. And that makes sense because this hyperbola never crosses y equals 0, right? It never crosses the x axis. And this should also be intuitive, because if we positive or negative infinity, we saw that we always did have this plus 4 sitting here. We said, oh, well, as x gets super large or super negative, this starts to matter less and less. But we will always be slightly larger than this number. Especially in the positive quadrant, right? We're always going to be -- so the positive quadrant is always going to be slightly larger than the asymptote. And even when we take the positive square root, I guess, When we take the positive square root, we'll always be larger than either of the asymptotes. And, likewise, when you take the negative square root, you're always going to be a little bit smaller than Because this number is going to be a little bit bigger than this number. Then we take the negative square root, you're going to be a little bit smaller, and that's why we're a little bit below. I don't know which one's more intutive for you. Maybe just the -- trying it when x is equal to 0 and when y equals 0, and see what points you get and say, oh, then I'm in kind of a vertical hyperbola as opposed to a horizontal one. video right there. And then I'll do another video where I actually shift the hyperbola. And shifting it is actually no different than shifting an ellipse a circle. You just have, you know, y minus something squared, and x plus something, or x minus something squared and that just tells you where you shift the origin. This hyperbola, of course, is just centered at the origin. Anyway, see you in the next video." + }, + { + "Q": "I don't get it at 3:30. If I type in (40)sin40/30 into my calculator I get approx 0.93. If I type 4/3sin40 I get approx 0.86.", + "A": "It should be 40 sin (40) /30 = .86. Be careful with where parentheses go. What you actually did was 40 sin (40/30) to get .93.", + "video_name": "IJySBMtFlnQ", + "timestamps": [ + 210 + ], + "3min_transcript": "3 sides of a triangle. So a, b, c to an angle. So, for example, if I do 2 sides and the angle in between them, I can figure out the third side. Or if I know all 3 sides, then I can figure out this angle. But that's not the situation that we have over here. We're trying to figure out this question mark and we don't know 3 of the sides. We're trying to figure out an angle but we don't know 3 of the sides. The Law of Cosine just doesn't seem, at least in an obvious way, that it's going to help me. I could also try to find this angle. Once again, we don't know all 3 sides to be able to solve for the angle. So maybe Law of Sines could be useful. So the Law of Sines, the Law of Sines. Let's say that this is, the measure of this angle is a, the measure of this angle is lower case b, the measure of this angle is lower case c, length of this side is capital C, length of this side is capital A, The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information" + }, + { + "Q": "At 3:07, is that \"theta\" like a variable?", + "A": "Yes. Theta can be used like x. But it is usually used for angles so you will see it quite a bit in trigonometry.", + "video_name": "IJySBMtFlnQ", + "timestamps": [ + 187 + ], + "3min_transcript": "3 sides of a triangle. So a, b, c to an angle. So, for example, if I do 2 sides and the angle in between them, I can figure out the third side. Or if I know all 3 sides, then I can figure out this angle. But that's not the situation that we have over here. We're trying to figure out this question mark and we don't know 3 of the sides. We're trying to figure out an angle but we don't know 3 of the sides. The Law of Cosine just doesn't seem, at least in an obvious way, that it's going to help me. I could also try to find this angle. Once again, we don't know all 3 sides to be able to solve for the angle. So maybe Law of Sines could be useful. So the Law of Sines, the Law of Sines. Let's say that this is, the measure of this angle is a, the measure of this angle is lower case b, the measure of this angle is lower case c, length of this side is capital C, length of this side is capital A, The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information" + }, + { + "Q": "I'm lost. 3:42 shows Sal making sin go to the other side by using sin-1. I know it's a cofunction, but is it like a reciprocal? Can someone explain how sin becomes sine-1 when moved to the other side? (Ex: 3x=2, you divide 3 on both sides to leave x, to make x= 2/3)", + "A": "1. Sin(a) \u00c3\u00b7 A = sin(b) \u00c3\u00b7 B 2. Sin(a) = sin(b) \u00c3\u00b7 B * A 3. a = arcsin( sin(b) \u00c3\u00b7 B * A) Since the problem is asking for a specific angle, you have to pull the arcsin to get the specfic angle in this equation. This works for cos, tan, and sin: Sin(x) = y and arcsin(y) = x", + "video_name": "IJySBMtFlnQ", + "timestamps": [ + 222 + ], + "3min_transcript": "The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information So, let's get a calculator out and see if we can calculate it. Let me just verify, I am in degree mode. Very important. All right, now I'm going to take the inverse sine of 4/3 times sine of 40 degrees, and that gets me, and I deserve a little bit of a drum roll, 58, well if we round to the nearest, let's just maintain our precision here. So 58.99 degrees roughly. This is approximately equal to 58.99 degrees. So, if that is 58.99 degrees, what is this one? It's going to 180 minus this angle's measure minus that angle's measure. Let's calculate that. It's going to 180 degrees minus this angle, so minus 40," + }, + { + "Q": "At 10:00 the right hand rule , what happens when i change the letters from\na to b and b to a , where is the explanation that\na cross b = b cross a , is true or not ?", + "A": "Compute (a1, a2, a3) x (b1, b2, b3) = a x b and (b1, b2, b3) x (a1, a2, a3) = b x a. Are they the same? How do they compare?", + "video_name": "pJzmiywagfY", + "timestamps": [ + 600 + ], + "3min_transcript": "That forms a plane. If you take a cross b, you get a third vector that's orthogonal to those two. And so a cross b will pop out like this. It'll be orthogonal to both of them and look like that. And so this vector right there is a cross b. And you might say, Sal, how did you know-- I mean, there's multiple vectors that are orthogonal. Obviously, the length of the vector, and I didn't specify that there, but it could pop straight up like that or why didn't it-- you know, you just as easily could have popped straight down like that. That also would be orthogonal to a and b. And the way that a cross b is defined, you can essentially figure out the direction visually by using what's called the right hand rule. And the way I think about it is you take your right hand and let me see if I can draw a suitable right hand. Point your index finger in the direction of a. So if your index finger is in the direction of a and then I So my middle finger, in this case, is going to go something like that. My middle finger is going to do something like that. And then my other fingers do nothing. Then my thumb will go in the direction of a cross b. You could see that there. My thumb is in the direction of a cross b. And assuming that you are anatomically similar to me, then you still get the same result. Let me draw it all. So this is vector a. Vector b goes in that direction. Hopefully you don't have a thumb hanging down here. You know that a cross b in this example will point up and it's orthogonal to both. To kind of satisfy you a little bit, that the vector's definitely orthogonal or that this thing is definitely orthogonal to both of these, let's just play with it and see that that definitely is the case. And what is orthogonal? What is in our context, the definition of orthogonal? If a and b are orthogonal, that means that a dot b is Remember, the difference between orthogonal and perpendicular is that orthogonal also applies to 0 vectors. So these could also be 0 vectors. Notice that I didn't say that any of these guys up here had to be nonzero. Well, in a little bit, we'll talk about the angle between vectors and then you have to assume nonzero. But if you're just taking a cross product, nothing to stop you from taking-- no reason why any of these numbers can't be 0. But let me show you that a cross b is definitely orthogonal to both a and b. I think that might be somewhat satisfying to you. So let me copy a cross b here. I don't feel like rewriting it. OK. Let me paste it. OK, bundle up little other stuff with it." + }, + { + "Q": "in the video at 3:25 I want 2 know how to make a hexaflexagon, but it goes to fast. Can you make a slow video on how to make a hexaflexagon?", + "A": "Hleyendecker 2020, the major predicament of alacritous motion in the video can be eradicated! Because Vi shows numerous ways of construction on the hexaflexagon, at 3:22-3:27, a reasonably stagnant-paced clip is displayed to contrive a hexaflexagon in a rare moment of slowness.", + "video_name": "VIVIegSt81k", + "timestamps": [ + 205 + ], + "3min_transcript": "you decide to try this three-way fold the other way, with flappy parts up, and are collapsing it down when suddenly the inside of your hexagon decides to open right up What, you close it back up and undo it. Everything seems the same as before, the center is not open-uppable. But when you fold it that way again, it, like, flips inside-out. Weird. This time, instead of going backwards, you try doing it again and again and again and again. And you want to make one that's a little less messy, so you try with another strip and tape it nicely into a twisty-foldy loop. You decide that it would be cool to colour the sides, so you get out a highlighter and make one yellow. Now you can flip from yellow side to white side. Yellow side, white side, yellow side, white side Hmm. White side? What? Where did the yellow side go? So you go back and this time you colour the white side green, and find that your piece of paper has three sides. Yellow, white and green. Now this thing is definitely cool. Therefore, you need to name it. And since it's shaped like a hexagon and you flex it and flex rhymes with hex, hexaflexagon it is. That night, you can't sleep because you keep thinking And the next day, as soon as you get to your math class you pull out your paper strips. You had made this sort of spirally folded paper that folds into again, the shape of a piece of paper, and you decide to take that and use it like a strip of paper to make a hexaflexagon. Which would totally work, but it feels sturdier with the extra paper. And you color the three sides and are like, orange, yellow, pink. And you're sort of trying to pay attention to class. Math, yeah. Orange, yellow, pink. Orange, yellow, white? Wait a second. Okay, so you colour that one green. And now it;s orange, yellow, green, Orange, yellow, green. Who knows where the pink side went? Oh, there it is. Now it's back to orange, yellow, pink. Orange, yellow, pink. Hmm. Blue. Yellow, pink, blue. Yellow, pink, blue. Yellow, pink, huh. With the old flexagon, you could only flex it one way, flappy way up. But now there's more flaps. So maybe you can fold it both ways. Yes, one goes from pink to blue, but the other, from pink to orange. And now, one way goes from orange to yellow, but the other way goes from orange to neon yellow. to one of your new friends, Bryant Tuckerman. You start with the original, simple, three-faced hexaflexagon, which you call the trihexaflexagon. and he's like, whoa! and wants to learn how to make one. and you are like, it's easy! Just start with a paper strip, fold it into equilateral traingles, and you'll need nine of them, and you fold them around into this cycle and make sure it's all symmetric. The flat parts are diamonds, and if they're not, then you're doing it wrong. And then you just tape the first triangle to the last along the edge, and you're good. But Tuckerman doesn't have tape. After all, it was invented only 10 years ago. So he cuts out ten triangles instead of nine, and then glues the first to the last. Then you show him how to flex it by pinching around a flappy part and pushing in on the opposite side to make it flat and traingly, and then opening from the centre. You decide to start a flexagon committee together to explore the mysteries of flexagotion, But that will have to wait until next time." + }, + { + "Q": "How do we actually fold a hexaflexagon at 0:34? She moves and talks too fast.", + "A": "If you go to the bottom right corner of the video and click on the cog icon (settings), and click on the speed option, you can slow the video down with the options that come up. Hope that helps!", + "video_name": "VIVIegSt81k", + "timestamps": [ + 34 + ], + "3min_transcript": "So say you just moved from England to the US and you've got your old school supplies from England and your new school supplies from the US and it's your first day of school and you get to class and find that your new American paper doesn't fit in your old English binder. The paper is too wide, and hangs out. So you cut off the extra and end up with all these strips of paper. And to keep yourself amused during your math class you start playing with them. And by you, I mean Arthur H. Stone in 1939. Anyway, there's lots of cool things you do with a strip of paper. You can fold it into Shapes and more shapes. Maybe spiral it around snugly like this. Maybe make it into a square. Maybe wrap it into a hexagon with a nice symmetric sort of cycle to the flappy parts. In fact, there's enough space here to keep wrapping the strip, and the your hexagon is pretty stable. and you're like. \"I don't know, hexagons aren't too exciting, but I guess it has symmetry or something.\" Maybe you could kinda fold it so the flappy parts are down and the unflappy parts are up. That's symmetric, and it collapses down into these three triangles, which collapse down into one triangle, and collapsible hexagons are, you suppose, cool enough to at least amuse you a little but during your class. you decide to try this three-way fold the other way, with flappy parts up, and are collapsing it down when suddenly the inside of your hexagon decides to open right up What, you close it back up and undo it. Everything seems the same as before, the center is not open-uppable. But when you fold it that way again, it, like, flips inside-out. Weird. This time, instead of going backwards, you try doing it again and again and again and again. And you want to make one that's a little less messy, so you try with another strip and tape it nicely into a twisty-foldy loop. You decide that it would be cool to colour the sides, so you get out a highlighter and make one yellow. Now you can flip from yellow side to white side. Yellow side, white side, yellow side, white side Hmm. White side? What? Where did the yellow side go? So you go back and this time you colour the white side green, and find that your piece of paper has three sides. Yellow, white and green. Now this thing is definitely cool. Therefore, you need to name it. And since it's shaped like a hexagon and you flex it and flex rhymes with hex, hexaflexagon it is. That night, you can't sleep because you keep thinking And the next day, as soon as you get to your math class you pull out your paper strips. You had made this sort of spirally folded paper that folds into again, the shape of a piece of paper, and you decide to take that and use it like a strip of paper to make a hexaflexagon. Which would totally work, but it feels sturdier with the extra paper. And you color the three sides and are like, orange, yellow, pink. And you're sort of trying to pay attention to class. Math, yeah. Orange, yellow, pink. Orange, yellow, white? Wait a second. Okay, so you colour that one green. And now it;s orange, yellow, green, Orange, yellow, green. Who knows where the pink side went? Oh, there it is. Now it's back to orange, yellow, pink. Orange, yellow, pink. Hmm. Blue. Yellow, pink, blue. Yellow, pink, blue. Yellow, pink, huh. With the old flexagon, you could only flex it one way, flappy way up. But now there's more flaps. So maybe you can fold it both ways. Yes, one goes from pink to blue, but the other, from pink to orange. And now, one way goes from orange to yellow, but the other way goes from orange to neon yellow." + }, + { + "Q": "(6:27) what sal says is wrong. it is less likely for the next toss to be heads if you look at is as a whole. lets say you toss the coin 100 times and you get 90 heads and 10 tails. it is more likely to get that then 90 heads and then 10 tails because there is only combination so theoretically, it is less likely to get a head is less likely in the next toss.", + "A": "True, but nonetheless the more times you flip the more you approach 50/50 results. This person just assumed that there was some underlying force that pulled everything together into theoretical accuracy. But the only way you approach this truth is through sheer quantity of data, there is no invisible force that makes anything more likely.", + "video_name": "VpuN8vCQ--M", + "timestamps": [ + 387 + ], + "3min_transcript": "Let me differentiate. And I'll use this example. So let's say-- let me make a graph. And I'll switch colors. This is n, my x-axis is n. This is the number of trials I take. And my y-axis, let me make that the sample mean. And we know what the expected value is, we know the expected value of this random variable is 50. Let me draw that here. This is 50. So just going to the example I did. So when n is equal to-- let me just [INAUDIBLE] here. So my first trial I got 55 and so that was my average. I only had one data point. Then after two trials, let's see, then I have 65. which is 60. So then my average went up a little bit. Then I had a 45, which will bring my average down a little bit. I won't plot a 45 here. Now I have to average all of these out. What's 45 plus 65? Let me actually just get the number just so you get the point. So it's 55 plus 65. It's 120 plus 45 is 165. Divided by 3. 3 goes into 165 5-- 5 times 3 is 15. It's 53. No, no, no. 55. So the average goes down back down to 55. And we could keep doing these trials. So you might say that the law of large numbers tell this, OK, after we've done 3 trials and our average is there. So a lot of people think that somehow the gods of probability are going to make it more likely that we get fewer That somehow the next couple of trials are going to have to be down here in order to bring our average down. And that's not necessarily the case. Going forward the probabilities are always the same. The probabilities are always 50% that I'm going to get heads. It's not like if I had a bunch of heads to start off with or more than I would have expected to start off with, that all of a sudden things would be made up and I would get more tails. That would the gambler's fallacy. That if you have a long streak of heads or you have a disproportionate number of heads, that at some point you're going to have-- you have a higher likelihood of having a disproportionate number of tails. And that's not quite true. What the law of large numbers tells us is that it doesn't care-- let's say after some finite number of trials your average actually-- it's a low probability of this happening, but let's say your average is actually up here. Is actually at 70. You're like, wow, we really diverged a good bit from the expected value. But what the law of large numbers says, well, I don't care how many trials this is. We have an infinite number of trials left." + }, + { + "Q": "Why at 4:53 in the video Sal puts down four squared raised to the seventh power equal to four to the seventh power", + "A": "its an error. (4^2)^7 is actually 4^14 he corrected himself at 5:23", + "video_name": "dC1ojsMi1yU", + "timestamps": [ + 293 + ], + "3min_transcript": "times this thing to the second power. Eight to the seventh to the second power, and then here, negative two times two is negative four, so that's A to the negative four times, eight to the seven times two is 14, eight to the 14th power. In other videos, we go into more depth about why this should hopefully make intuitive sense. Here you have eight to the seventh times eight to the seventh. Well, you would then add the two exponents, and you would get to eight to the 14th, so however many times you have eight to the seventh, you would just keep adding the exponents, or you would multiply by seven that many times. Hopefully that didn't sound too confusing, but the general idea is if you raise something to exponent and then another exponent, you can multiply those exponents. Let's do one more example where we are dealing with quotients, which that first example could have even been perceived as. So let's say we have divided by four squared, and we're gonna raise all of that to the seventh power. Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power, so if you have the difference of two things and you're raising it to some power, that's the same thing as a numerator raised to that power divided by the denominator raised to that power. Well, what's our numerator going to be? Well, we've done this drill before. It'd be two to the negative 10 times seventh power, so this would be equal to two to the negative 70th power, and then in the denominator, four to the second power, then that raised to the seventh power. Well, two times seven is 14, so that's going to be four to the 17th power. Now, we actually could think There's multiple ways that you could rewrite this, but one thing you could do is say, \"Hey, look, \"four is a power of two.\" So you could rewrite this as this is equal to two to the negative 70th power over, instead of writing four to the 17th power, why did I write the 17th power? It should be four to the 14th power. Let me correct that. Instead of writing four to the 14th power, I instead could write, so this is two, get the colors right. This is two to the negative 70th over, instead of writing four, I could write two squared to the 14th power. Four is the same thing as two squared, and so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second," + }, + { + "Q": "At 3:25, Sal says \"...the magnitude or absolute value of Z1...\". He denotes it like |Z1|. Why doesn't he denote it as ||Z1||?", + "A": "Notations are subject to a bit of flexibility. I had a physics teacher in college that consistently referred to the absolute values of vectors. In fact, absolute value is a type of magnitude. So you could legitimately denote it |*| or ||*||. But here, I think Sal opted for the simpler and more familiar notation.", + "video_name": "FwuPXchH2rA", + "timestamps": [ + 205 + ], + "3min_transcript": "it would be b. This is a real number, but this tells us how much the i is scaled up in the complex number z right over there. Now, one way to visualize complex numbers, and this is actually a very helpful way of visualizing it when we start thinking about the roots of numbers, especially the complex roots, is using something called an Argand diagram. So this is this. And so it looks a lot like the coordinate axes and it is a coordinate axes. But instead of having an x and y-axis it has a real and an imaginary axis. So in the example of z being a plus bi, we would plot it really as a position vector, where you have the real part on the horizontal axis. So let's say this is a and then the imaginary part along the vertical axis, or the imaginary axis. And so we would represent, in an Argand diagram, the vector z as a position vector that starts at 0 and that has a tip at the coordinate a comma b. So this right here is our complex number. This right here is a representation in our Argand diagram of the complex number a plus bi, or of z. Now when you draw it this way, when you draw it as a position vector, and if you're familiar with polar coordinates, you're probably thinking, hey, I don't have to represent this complex number just as coordinates, just as an a plus bi. Maybe I could represent this as some angle here, let's call that angle phi, and some the distance here, let's call that r, which is kind of the magnitude And you could. If you gave some angle and some distance, that would also specify this point in the complex plane. and this right here is called the magnitude, or sometimes the modulus, or the absolute value of the complex number. So let's think about it a little bit. Let's think about how we would actually calculate these values. So r, which is the modulus, or the magnitude. It's denoted by the magnitude or the absolute value of z1. What's this going to be. Well, we have a right triangle here. This side is b, length b. The base right here has length a. So to calculate r, we can just use the Pythagorean Theorem. r squared is going to be equal to a squared plus b squared. Or r is going to be equal to the square root of a squared plus b squared. If we want to figure out the argument, this is going to be equal to what?" + }, + { + "Q": "13:04 I've tried to make a non-congruent triangle that complies with SSA in CAD and I haven\u00c2\u00b4t being able to do so. Was Sal wrong or there are so few possible triangles that are non-congruent and SSA that I can't find them by trial and error? In any case, many triangles that are SSA are also congruent.", + "A": "Well, the answer was in the More on why SSA is not a postulate video... :S. I should be more careful when asking questions from now on. Equally thanks for your reply.", + "video_name": "8Ld8Csu4sEs", + "timestamps": [ + 784 + ], + "3min_transcript": "is going to touch this one right over there. And the only way it's going to touch that one right over there is if it starts right over here, because we're constraining this angle right over here. We're constraining that angle. And so it looks like angle, angle, side does indeed imply congruency. So that does imply congruency. So let's just do one more just to kind of try out all of the different situations. What if we have-- and I'm running out of a little bit of real estate right over here at the bottom-- what if we tried out side, side, angle? So once again, draw a triangle. So it has one side there. It has another side there. And then-- I don't have to do those hash marks just yet. So one side, then another side, and then another side. And what happens if we know that there's another triangle that has two of the sides the same and then the angle after it? So for example, we would have that side just like that, But we're not constraining the angle. We aren't constraining this angle right over here, but we're constraining the length of that side. So let me color code it. So that blue side is that first side. Then we have this magenta side right over there. So this is going to be the same length as this right over here. But let me make it at a different angle to see if I can disprove it. So let's say it looks like that. Or actually let me make it even more interesting. Let me try to make it like that. So it's a very different angle. But now, it has to have the same angle out here. It has to have that same angle out here. So it has to be roughly that angle. So it actually looks like we can draw a triangle that is not congruent that has two sides being the same length and then an angle is different. For example, this is pretty much that. I made this angle smaller than this angle. These two sides are the same. is that this green side is going to be shorter on this triangle right over here. So you don't necessarily have congruent triangles with side, side, angle. So this is not necessarily congruent, not necessarily, or similar. It gives us neither congruency nor similarity." + }, + { + "Q": "When Sal talks the definition of similar (at 2:36) , he says that in geometry similar things have same shape. However, I still wonder if he meant just having same type of form (like triangle, pentagon or rectangle) or if the angles of the figure also have to be the same so two figure can be similar geometrically?", + "A": "Two shapes are similar if:the corresponding sides are proportional and the angles are congruent. The proportionality constant can be one, in which case the sides are the same length, but the definition of similar is the first statement.", + "video_name": "8Ld8Csu4sEs", + "timestamps": [ + 156 + ], + "3min_transcript": "But when you think about it, you can have the exact same corresponding angles, having the same measure or being congruent, but you could actually scale one of these triangles up and down and still have that property. For example, if I had this triangle right over here, it looks similar-- and I'm using that in just the everyday language sense-- it has the same shape as these triangles right over here. And it has the same angles. That angle is congruent to that angle, this angle down here is congruent to this angle over here, and this angle over here is congruent to this angle over here. So all of the angles in all three of these triangles The corresponding angles have the same measure. But clearly, clearly this triangle right over here is not the same. It is not congruent to the other two. The sides have a very different length. This side is much shorter than this side right over here. This side is much shorter than that side over there. So with just angle, angle, angle, you cannot say that a triangle has the same size and shape. It does have the same shape but not the same size. So this does not imply congruency. So angle, angle, angle does not imply congruency. What it does imply, and we haven't talked about this yet, is that these are similar triangles. So angle, angle, angle implies similar. So let me write it over here. It implies similar triangles. And similar-- you probably are use to the word in just everyday language-- but similar has a very specific meaning in geometry. And similar things have the same shape but not necessarily the same size. So anything that is congruent, because it has the same size and shape, is also similar. But not everything that is similar is also congruent. So for example, this triangle is similar-- all of these triangles are similar to each other, but they aren't all congruent. These two are congruent if their sides But if we know that their sides are the same, then we can say that they're congruent. But neither of these are congruent to this one right over here, because this is clearly much larger. It has the same shape but a different size. So we can't have an AAA postulate or an AAA axiom to get to congruency. What about side, angle, side? So let's try this out, side, angle, side. So let's start off with one triangle right over here. So let's start off with a triangle that looks like this. I have my blue side, I have my pink side, and I have my magenta side. And let's say that I have another triangle that has this blue side. It has the same side, same length as that blue side. So let me draw it like that. It has the same length as that blue side. So that length and that length are going to be the same. It has a congruent angle right after that. So this angle and the next angle for this triangle are going to have the same measure," + }, + { + "Q": "at about 3:15 in the video. Can someone explain the difference between direct and joint? thanks", + "A": "i believe that the difference is that in the direct variation you are dealing with only TWO variables (x and y) and one constant number (k); in the joint variation you are dealing with THREE variables, like Sal said, the example would be the area (A) of a rectangle, which is the width (w) times the length (l), so if you had to insert these into a table, you would have to have THREE columns.", + "video_name": "v-k5L0BPOmc", + "timestamps": [ + 195 + ], + "3min_transcript": "think about the telltale signs of direct variation. If x increases, y should increase. So if x increases. Let me do that in the same yellow. So the telltale signs of direct variation, if x increases then y will increase and vice versa. The other telltale sign is. Is if you increase x by some, by some factor. So, if you have x going to 3x then y should also increase by that same factor. And we could see that with some examples. So, I mean, you could pick a K, let's say that, let's say that K was one. So if y is equal to x, if you take, if x goes from one to three, then y is also going to go from one to three. So that's all we're talking about here. Let me actually, y should actually to three times y, that's what I'm talking about. If you triple x, you're also gonna end up tripling y. Inverse variation. You have y being equal to some constant times one over x. multiply both sides by x you get x times y is equal to some constant. And you could switch the x's and the y's around as well for inverse variation. Now what are the tale tale signs? Well if you increase x, if x goes up, then what happens to y? If x goes up then this becomes a smaller value cuz it's one over x so then y will go down. Then y will go down. And if you take X and if you're to say increase it by a factor of three then what's going to happen to Y? Well if you increase this by a factor of three, you're actually going to decrease this whole value by a factor of one-third, so Y is going to go, so then you're going to have one-third of y. So that's, these are the tell-tale signs for inverse variation. Now finally they talk about something called joint variation, and this one you won't necessarily see in introductory algebra course. So if I told you, if I told you that area of a rectangle is equal to the width of a rectangle times the length of rectangle, this is an example of joint variation. Area is proportionally to two. Is the proportional to two different quantities? So the main tell tale sign here for joint variation frankly is you're gonna be dealing with more than two variables. Joint, Joint, Joint variation. So when you look at this example, there only gi, giving us two variables. So you can rule out joint variation just right from the get go. Now let's look at the tell tale signs. So as x is increasing, as x goes from one to two, what is happening to y? Y went from 12 to six. So as x is going up by a factor of two, y is going, is, is going by a factor of one half. Or y is being multiplied by one half. So as x goes from one to three, it's being multiplied by three." + }, + { + "Q": "at 3:46 , Can you really just square the 1/3 and then plug it into the radical? I have never seen that before.", + "A": "Yes as you re not changing the equation in any way. You re basically squaring a positive multiple then when you place it inside the radical you re square rooting it again so it s exactly the same multiple. (1/3)^2 = 1/9 sqrt(1/9) = 1/3", + "video_name": "WAoaBTWKLoI", + "timestamps": [ + 226 + ], + "3min_transcript": "So this is u in terms of x. So everywhere we see a u up here we can replace it with this expression. And we are essentially done. We would have written this in terms of x. Now, there's another technique you might sometimes see in a calculus class where someone says, OK, we know that u is equal to cosine theta. We know this relationship. How can we express u in terms of x? And we'll say, let's draw a right triangle. They'll draw a right triangle like this. They'll draw a right triangle, and they'll say, OK, look, sine of theta is x over 3. So if we say that this is theta right over here, sine of theta is the same thing as opposite over hypotenuse. Opposite over hypotenuse is equal to x over 3. So let's say that this is x and then this right over here is 3. Then the sine of theta will be x over 3. So we look at that first substitution right over here. But in order to figure out what u is in terms of x, we need to figure out what cosine of theta is. So we have to figure out what this adjacent side is. Well, we can just use the Pythagorean theorem for that. Pythagorean theorem would tell us that this is going to be the square root of the hypotenuse squared, which is 9, minus the other side squared, minus x squared. So from this, we fully solved the right triangle in terms of x. We can realize that cosine of theta is going to be equal to the adjacent side, square root of 9 minus x squared, over the hypotenuse, over 3, which is the same thing as 1/3 times the square root of 9 minus x squared, which is the same thing if we square 1/3 and put it into the radical. So we're essentially going to take the square-- 1/3 is the same thing as the square root of 1/9. So can rewrite this as the square root of 1/9 times 9 minus x squared. Essentially, we just brought the 1/3 third into the radical. Now it's 1/9. And so now this is going to be the same thing which is exactly this thing right over here. x squared over 9 is the same thing as x over 3 squared. So either way, you get the same result. I find using the trig identity right over here to express cosine of theta in terms of sine theta and then just do the substitution to be a little bit more straightforward. But now we can just substitute into the original thing. So either of these-- I can write it as either way-- this thing right over here, this is the same thing as 1 minus x squared over 9 to the 1/2 power. That's what u is equal to. And everywhere we see u, we just substitute it with this thing. So our final answer in terms of x is going to be equal to 243 times u to the fifth, this to the fifth power over 5. This to the fifth power is 1 minus x squared over 9. It was to the 1/2, but if we raise that to the fifth power," + }, + { + "Q": "is it me or, It appears that at 5:20 sal gave the final answer as if it was (-u^5/5+u^3/3) instead of ( u^5/5-u^3/3) after unwinding the trig and u substitution. please correct me if I'm wrong.\nRegards", + "A": "I don t see any problems with the final answer.", + "video_name": "WAoaBTWKLoI", + "timestamps": [ + 320 + ], + "3min_transcript": "So we have to figure out what this adjacent side is. Well, we can just use the Pythagorean theorem for that. Pythagorean theorem would tell us that this is going to be the square root of the hypotenuse squared, which is 9, minus the other side squared, minus x squared. So from this, we fully solved the right triangle in terms of x. We can realize that cosine of theta is going to be equal to the adjacent side, square root of 9 minus x squared, over the hypotenuse, over 3, which is the same thing as 1/3 times the square root of 9 minus x squared, which is the same thing if we square 1/3 and put it into the radical. So we're essentially going to take the square-- 1/3 is the same thing as the square root of 1/9. So can rewrite this as the square root of 1/9 times 9 minus x squared. Essentially, we just brought the 1/3 third into the radical. Now it's 1/9. And so now this is going to be the same thing which is exactly this thing right over here. x squared over 9 is the same thing as x over 3 squared. So either way, you get the same result. I find using the trig identity right over here to express cosine of theta in terms of sine theta and then just do the substitution to be a little bit more straightforward. But now we can just substitute into the original thing. So either of these-- I can write it as either way-- this thing right over here, this is the same thing as 1 minus x squared over 9 to the 1/2 power. That's what u is equal to. And everywhere we see u, we just substitute it with this thing. So our final answer in terms of x is going to be equal to 243 times u to the fifth, this to the fifth power over 5. This to the fifth power is 1 minus x squared over 9. It was to the 1/2, but if we raise that to the fifth power, over 5 minus this to the third power, 1 minus x squared over 9 to the 3/2 raising this to the third power-- that's this right over here-- over 3, and then all of that plus c. And we're done. It's messy, but using first trig substitution then u substitution, or trig substitution then rearranging using a couple of our techniques for manipulating these powers of trig functions, we're able to get into a form where we could use u substitution, and then we were able to unwind all the substitutions and actually evaluate the indefinite integral." + }, + { + "Q": "At around 7:30, when he is discussing the trig. ratios, is there any particular reason why, for example, we use adj/hyp instead of hyp/adj? If we used the reciprocal of any of these ratios would it matter?", + "A": "For each of the main trig functions, sine, cosine and tangent, there is another trig function that is its reciprocal: secant is 1/cosine, cosecant is 1/sine, and cotangent is 1/tangent. For an acute angle in a right triangle, adj/hyp is cosine. You can t use hyp/adj for cosine because that s something entirely different (secant) with a graph that looks nothing like the graph of cosine.", + "video_name": "QuZMXVJNLCo", + "timestamps": [ + 450 + ], + "3min_transcript": "And I keep stating from theta's point of view because that wouldn't be the case for this other angle, for angle B. From angle B's point of view, this is the adjacent side over the hypotenuse. And we'll think about that relationship later on. But let's just all think of it from theta's point of view right over here. So from theta's point of view, what is this? Well theta's right over here. Clearly AB and DE are still the hypotenuses-- hypoteni. I don't know how to say that in plural again. And what is AC, and what are DF? Well, these are adjacent to it. They're one of the two sides that make up this angle that is not the hypotenuse. So this we can view as the ratio, in either of these triangles, between the adjacent side-- so this is relative. Once again, this is opposite angle B, but we're only thinking about angle A right here, or the angle that measures theta, or angle D right over here-- relative to angle A, Relative to angle D, DF is adjacent. So this ratio right over here is the adjacent over the hypotenuse. And it's going to be the same for any right triangle that has an angle theta in it. And then finally, this over here, this is going to be the opposite side. Once again, this was the opposite side over here. This ratio for either right triangle is going to be the opposite side over the adjacent side. And I really want to stress the importance-- and we're going to do many, many more examples of this to make this very concrete-- but for any right triangle that has an angle theta, the ratio between its opposite side and its hypotenuse is going to be the same. That comes out of similar triangles. We've just explored that. The ratio between the adjacent side to that angle that is theta and the hypotenuse is going to be the same, for any of these triangles, as long as it has that angle theta in it. theta, between the opposite side and the adjacent side, between the blue side and the green side, is always going to be the same. These are similar triangles. So given that, mathematicians decided to give these things names. Relative to the angle theta, this ratio is always going to be the same, so the opposite over hypotenuse, they call this the sine of the angle theta. Let me do this in a new color-- by definition-- and we're going to extend this definition in the future-- this is sine of theta. This right over here, by definition, is the cosine of theta. And this right over here, by definition, is the tangent of theta. And a mnemonic that will help you remember this-- and these really are just definitions. People realized, wow, by similar triangles, for any angle theta, this ratio is always going to be the same." + }, + { + "Q": "where did Sal get the number 7 1/2 from at 1:44?", + "A": "7 1/2 is half of 15", + "video_name": "h0FFEBHBufo", + "timestamps": [ + 104 + ], + "3min_transcript": "We're on problem 21. In the figure below, n is a whole number. What is the smallest possible value for n? OK, both of these sides are n. And so this is something that's actually really good to get this intuition. Because this shows up on all sorts of standardized tests. And so let's think about how small can we make n. So the lower the top of this pyramid becomes, the smaller n becomes. If we pushed this top of the pyramid really high then n would have to be really big. For example, if we made the triangle into that. Then clearly, this length is shorter than that length. We want to keep lowering it to get as small a possible n. But what happens if we lowered it all the way? If we flattened this triangle. If we just flattened it all the way down. So essentially this would be the top of it and this would be n and this would be n. I hope you're visualizing that properly. I've flattened the triangle. So these two sides would just go flat with the base. And so this top if I were to do it in So this is as small as n could get. One could argue whether this is a triangle at all anymore. It's really a line now, because I've squished out all of the area in there. But even in this case, n would have to be, in the smallest case, it would be 7.5. Each n would be half of this 15. So, as we push this base down, that's kind of the limit that n approaches. n cannot be any smaller than 7.5. And they tell us that n is a whole number. So n has to be greater than 7.5 in order for this to be a triangle. And n is a whole number, so n is equal to 8. That's choice C. And that's an important intuition to have in general. That the third side of a triangle can never be bigger than two of the other sides combined. Then you're dealing with something else. You're not dealing with a triangle. then you're actually dealing with a line. Because you'd have to squish out all of the area of the triangle in order to get there. Anyway, I like that problem. Next question. I think, just eyeballing it, they want us to do the same thing. Same type of intuition. And I had my rant in the last video about how they weren't doing problems that give you intuition or that test your intuition. But I'll take that back, because I think that's what they are testing now. Which of the following sets of numbers could represent the lengths of the sides of a triangle. 2, and 2, and 5. So this is the same thing again. How can I have two sides of a triangle combined being shorter than the third side. If I had a side of length 2, and then I had another side of length 2, there's no way that this last side could be 5. Even if I completely flattened this triangle, 2 and 2, the longest that this last side, this third" + }, + { + "Q": "At 05:40: \"So a note and the note with twice the frequency, totally are like the same note, am I right?\" Can anybody explain this statement to me, please?", + "A": "The first note is a note on a keyboard (or whatever she was playing at the time) then the second note is twice the frequency, or 1 octave up. so technically, they are the same note, even though they are not pitch. one 8(how many notes in a octave) x 2 (doubling the frequency) = 16(twice the frequency, or 1 octave up from the first note.", + "video_name": "i_0DXxNeaQ0", + "timestamps": [ + 340 + ], + "3min_transcript": "So he played 1/2 the length and found the note was an octave higher. He thought that was pretty neat. So then he tried the next simplest ratio and played 1/3 of the string. If the full length was C, then 1/3 the length would give the note G, an octave and a fifth above. The next ratio to try was 1/4 of the string, but we can already figure out what note that would be. In 1/2 the string was C an octave up, then 1/2 of that would be C another octave up. And 1/2 of that would be another octave higher, and so on and so forth. And then 1/5 of the string would make the note E. But wait. Let's play that again. It's a C Major chord. So what about 1/6? We can figure that one out, too, using ratios we already know. 1/6 is the same as 1/2 of 1/3. And 1/3 third was this G. So 1/6 is the G an octave up. Check it out. 1/7 will be a new note, because 7 is prime. And Pythagoras found that it was this B-flat. Then 8 is 2 times 2 times 2. And 1/9 is 1/3 of 1/3. So we go an octave and a fifth above this octave and a fifth. And the notes get closer and closer until we have all the notes in the chromatic scale. And then they go into semi-tones, et cetera. But let's make one thing clear. This is not some magic relationship between mathematical ratios and consonant intervals. It's that these notes sound good to our ear because our ears hear them together in every vibration that reaches the cochlea. Every single note has the major chord secretly contained within it. So that's why certain intervals sound consonant and others dissonant and why tonality is like it is and why cultures that developed music independently of each other still created similar scales, chords, and tonality. This is called the overtone series, by the way. And, because of physics, but I don't really know why, a string 1/2 the length vibrates twice as fast, which, hey, makes this series the same as that series. If this were A440, meaning that this is a swing that likes to swing 440 times a second, And here's E at three times the original frequency, 1320. The thing about this series, what with making the string vibrate with different lengths at different frequencies, is that the string is actually vibrating in all of these different ways even when you don't hold it down and producing all of these frequencies. You don't notice the higher ones, usually, because the lowest pitch is loudest and subsumes them. But say I were to put my finger right in the middle of the string so that it can't vibrate there, but didn't actually hold the string down there. Then the string would be free to vibrate in any way that doesn't move at that point, while those other frequencies couldn't vibrate. And if I were to touch it at the 1/3 point, you'd expect all the overtones not divisible by 3 to get dampened. And so we'd hear this and all of its overtones. The cool part is that the string is pushing it around the air at all these different frequencies. And so the air is pushing around your ear at all these different frequencies. And then the basilar membrane is vibrating in sympathy with all these frequencies. And your ear puts it together and understands it as one sound." + }, + { + "Q": "at 1:56 Vi says cochlea, what does the cochlea do?", + "A": "Sound goes into your ear, past your ear drum, past the bones in your ear , into your cochlea and through the nerve to the brain. The cochlea basically processes the sound so you can understand it.", + "video_name": "i_0DXxNeaQ0", + "timestamps": [ + 116 + ], + "3min_transcript": "[PIANO ARPEGGIOS] When things move, they tend to hit other things. And then those things move, too. When I pluck this string, it's shoving back and forth against the air molecules around it and they push against other air molecules that they're not literally hitting so much as getting too close for comfort until they get to the air molecules in our ears, which push against some stuff in our ear. And then that sends signals to our brain to say, Hey, I am getting pushed around here. Let's experience this as sound. This string is pretty special, because it likes to vibrate in a certain way and at a certain speed. When you're putting your little sister on a swing, you have to get your timing right. It takes her a certain amount of time to complete a swing and it's the same every time, basically. If you time your pushes to be the same length of time, then even general pushes make your swing higher and higher. That's amplification. If you try to push more frequently, you'll just end up pushing her when she's swinging backwards and instead of going higher, you'll dampen the vibration. It wants to swing at a certain speed, frequency. If I were to sing that same pitch, the sound waves I'm singing will push against the string at the right speed to amplify the vibrations so that that string vibrates while the other strings don't. It's called a sympathy vibration. Here's how our ears work. Firstly, we've got this ear drum that gets pushed around by the sound waves. And then that pushes against some ear bones that push against the cochlea, which has fluid in it. And now it's sending waves of fluid instead of waves of air. But what follows is the same concept as the swing thing. The fluid goes down this long tunnel, which has a membrane called the basilar membrane. Now, when we have a viola string, the tighter and stiffer it is, the higher the pitch, which means a faster frequency. The basilar membrane is stiffer at the beginning of the tunnel and gradually gets looser so that it vibrates at high frequencies at the beginning of the cochlea and goes through the whole spectrum down to low notes at the other end. So when this fluid starts getting pushed around there's a certain part of the ear that vibrates in sympathy. The part that's vibrating a lot is going to push against another kind of fluid in the other half of the cochlea. And this fluid has hairs in it which get pushed around by the fluid, and then they're like, Hey, I'm middle C and I'm getting pushed around quite a bit! Also in humans, at least, it's not a straight tube. The cochlea is awesomely spiraled up. OK, that's cool. But here are some questions. You can make the note C on any instrument. And the ear will be like, Hey, a C. But that C sounds very different depending on whether I sing it or play it on viola. Why? And then there's some technicalities in the mathematics of swing pushing. It's not exactly true that pushing with the same frequency that the swing is swinging is the only way to get this swing to swing. You could push on just every other swing. And though the swing wouldn't go quite as high as if you pushed every time, it would still swing pretty well. In fact, instead of pushing every time or half the time, you could push once every three swings or four, and so on. There's a whole series of timings that work," + }, + { + "Q": "Hey guys, did you notice that at 1:05 Sal said tenths when he ment thousandths!", + "A": "No! I didn t thanks for letting me know! ;P ;) :)", + "video_name": "BINElq3DFkg", + "timestamps": [ + 65 + ], + "3min_transcript": "Let's once again see if we can order now a different set of decimals from least to greatest, and once again I encourage you to pause this video and try to do this on your own. So let's go to the most significant place, the ones place here. None of these have any ones. So then we can go to the next most significant place, which is the tenths place. This has five tenths. This has six tenths. This has one tenth. This has five tenths. This has one tenth. So if we just look at the tenths place, the ones that have the fewest tenths-- this has only one tenth, this one only has one tenth, this one has five tenths, this one has five tenths, and then this one has six tenths. So I've ordered it by what's going on in the tenths place. Now, both of these have the same number of tenths. Let's move to the hundredths place to figure out which of these is larger. This one has six hundredths. This has five hundredths, so this one is larger. It has more hundredths. Same number of tenths, more hundredths. And hundredths are obviously more significant than thousandths, so it It matters that this one has more tenths, and actually this one has more thousandths as well. But now let's go look at these two. These have the same number of tenths. They both have five tenths. But this one has six hundredths, while this one only has two hundredths, so this one is larger. And then finally, this one of course, had six tenths, so this one had the most tenths. So we don't even have to look at the other places here. And we're done." + }, + { + "Q": "How do you know when to leave the circle open or closed on the number line at 3:59", + "A": "You leave the circle open when you need to exclude it from your inequality (less than or greater than inequalities). You shade it in when it is included in the inequality (commonly known as less than or equal to OR greater than or equal).", + "video_name": "FZ2APP6-grU", + "timestamps": [ + 239 + ], + "3min_transcript": "sign right there. So if we want to solve for x, how many tiles can he buy? We can divide both sides of this inequality by 3. And because we're dividing or multiplying-- you could imagine we're multiplying by 1/3 or dividing by 3 -- because this is a positive number, we do not have to swap the inequality sign. So we are left with x is less than 1,000 over three, which is 333 and 1/3. So he has to buy less than 333 and 1/3 tiles, that's how many tiles, and each tile is one square foot. So if he can buy less than 333 and 1/3 tiles, then the patio also has to be less than 333 and 1/3 square feet. And we're done." + }, + { + "Q": "How did Sal just throw in the factor of dx at 3:45? The volume of a real shell is not the area of its outer surface times the depth of the shell, so why should that be the volume when the shell is infinitesimally thin?", + "A": "The volume of the shell, as stated in previous videos, is the circumference times the height of the shell times the width. Here, the width is dx. However, it does not really matter in the end, because you are just taking the definite integral to find the area. Hope that helps!", + "video_name": "SfWrVNyP9E8", + "timestamps": [ + 225 + ], + "3min_transcript": "What is that distance going to be? Well, it's the horizontal distance between x equals 2 and whatever the x value is right over here. So it's going to be 2 minus our x value. So this radius, this distance right over here, is going to be 2 minus x. And so the circumference is going to be that times 2 pi. 2 pi r gives us the circumference of that circle. So 2 pi times 2 minus x. And then if we want the surface area of the outside of our shell, so the area is going to be the circumference 2 pi times 2 minus x times the height of each shell. Now, what is the height of each shell? It's going to be the vertical distance expressed as functions of y. So it's going to be the top boundary is y is equal to square root of x, the bottom boundary is So it's going to be square root of x minus x squared. Let me do this in the yellow. So it's going to be square root of x minus x squared. And so if you want the volume of a given shell-- I'll write all this in white-- it's going to be 2 pi times 2 minus x times square root of x minus x squared. So this whole expression, I just rewrote it, is the area, the outside surface area, of one of these shells. If we want the volume, we have to get a little bit of depth, multiply by how deep the shell is, so times dx. And if we want the volume of this whole thing, we just have to solve all the shells for all of the x's in this interval and take the limit as the dx's get smaller and smaller and we have more and more shells. And so, what's our interval? Well our x's are going to go between 0 and 1." + }, + { + "Q": "3:10 Why would you subtract 90 degrees from that equation? I haven't exactly figured that out.", + "A": "Sal subtracted 90 degrees from both sides of the equation to simplify the equation. What I would do instead (personally) is add 90 and 32 to get 122, and then subtract 122 from both sides. 180-122= 58, so either way, you get the same answer. Hope that helps you!", + "video_name": "iqeGTtyzQ1I", + "timestamps": [ + 190 + ], + "3min_transcript": "And now we have three angles in the triangle, and we just have to solve for theta. Because we know this angle plus this angle plus this angle are going to be equal to 180 degrees. So you have 90 minus theta plus 90 degrees plus 32 degrees-- so I'm going to do that in a different color-- is going to be equal to 180 degrees. The sum of the measures of the angle inside of a triangle add up to 180 degrees. That's all we're doing over here. And so let's see if we can simplify this a little bit. So these two guys-- 90 plus 90's going to be 180, so you get 180 minus theta plus 32 is equal to 180 degrees. And then what else do we have? We have 180 on both sides. We can subtract that from both sides. So that cancels out. That goes to 0. You can add theta to both sides. And you get 32 degrees is equal to theta, or theta is equal to 32 degrees. So it's going to actually be the same measure as this angle right over here. That's one way to do the problem. There's other ways that we could have done the problem. Actually, there's a ton of ways we could have done this. We could have looked at this big triangle over here. And we could've said, look. If this is 90 degrees over here, this is 32 degrees over here, this angle up here is going to be 180 minus 90 degrees minus 32 degrees. Because they all have to add up to 180 degrees. And I just kind of skipped a step there. Actually, let me not skip a step. Let me call this x. If we call the measure of that angle x, we would have x plus 90. I'm looking at the biggest triangle in this diagram right here. x plus 90 plus 32 is going to be equal to 180 degrees. So if you subtract 90 from both sides, you get x plus 32 is equal to 90. And then if you subtract 32 from both sides, you get x is equal to-- what is this-- 58 degrees. Fair enough. Now, what else can we figure out? Well, if this angle over here is a right angle-- and I'm just redoing the problem over again just to show you that there's multiple ways to get the answer. We were given that this is a right angle. If that is 90 degrees, then this angle over here is supplementary to it, and it also has to be 90 degrees. So then we have this angle plus 90 degrees plus this angle have to equal 180. Maybe we could call that y. So y plus 58 plus 90 is equal to 180." + }, + { + "Q": "@2:56 how does it simplify to 1, not the best at algebra", + "A": "1/x * x = 1/x * x/1 = (1*x)/(x*1) = x/x = 1 We multiply 1/x with x, so there s one x at the numerator and denominator.", + "video_name": "iw5eLJV0Sj4", + "timestamps": [ + 176 + ], + "3min_transcript": "f of x times g of x minus the antiderivative of, instead of having f and g prime, you're going to have f prime and g. So f prime of x times g of x dx. And we've seen this multiple times. So when you figure out what should be f and what should be g, for f you want to figure out something that it's easy to take the derivative of and it simplifies things, possibly if you're taking the derivative of it. And for g prime of x, you want to find something where it's easy to take the antiderivative of it. So good candidate for f of x is natural log of x. If you were to take the derivative of it, it's 1 over x. Let me write this down. So let's say that f of x is equal to the natural log of x. Then f prime of x is equal to 1 over x. And let's set g prime of x is equal to 1. That means that g of x could be equal to x. And so let's go back right over here. So this is going to be equal to f of x times g of x. Well, f of x times g of x is x natural log of x. So g of x is x, and f of x is the natural log of x, I just like writing the x in front of the natural log of x to avoid ambiguity. So this is x natural log of x minus the antiderivative of f prime of x, which is 1 over x times g of x, which is x, which is xdx. Well, what's this going to be equal to? Well, what we have inside the integrand, this is just 1 over x times x, which is just equal to 1. So this simplifies quite nicely. This is going to end up equaling x natural log of x or the antiderivative of 1dx, or the integral of 1dx, or the antiderivative of 1 is just minus x. And this is just an antiderivative of this. If we want to write the entire class of antiderivatives we just have to add a plus c here, and we are done. We figured out the antiderivative of the natural log of x. I encourage you to take the derivative of this. For this part, you're going to use the product rule and verify that you do indeed get natural log of x when you take the derivative of this." + }, + { + "Q": "at 5:21 i understand that A to the B+C power equals A to the B power times A to the C power but what about if it was negative, as in A to the B-C power. would that make it A to the B power divided by A to the C power or would that just make C negative?", + "A": "Yes, you are correct. A to the B-C power would give you A to the B power divided by A to the C power. Sal actually explains this property at 7:19. Another way of doing this is: A^(B-C) = A^B x A^-C and since A^-C = 1/A^C we can rewrite this as A^B/A^C (A to the B power divided by A to the C power).", + "video_name": "vSijVSL3ChU", + "timestamps": [ + 321 + ], + "3min_transcript": "isn't a huge stretch here. I just literally multiplied and divide by, divided by 10, times this t over 10. But when I write it this way, an exponent property might jump out at you. If I have, if I have a to the b, and then I raise that to the c, that's going to be a to the bc. Or, another way around, a to the bc is going to be a to the b to the c. And so, this piece right over here, I can rewrite it as two to the 10th, and then raise that to the t over 10 power. To the t over 10 power. and then raise that to the t over 10, that's gonna be the same thing as two to the 10 times t over 10. And of course, we still have the one over 32 over here. One over, one over 32. I'm tempted to write that as two to the negative fifth power, but I won't do that just yet. Actually, let's just, let's just keep it, let's just keep it as two to the 10th power, just for simplicity right now. Later we can, you might know that that's gonna be 1,024. But let's just, let's see what else we can do. So we know this is going to be some num- ... Actually, let me just write out as 1,024. So we have one over 32 times 1,024 to the t over 10, to the t over 10 power. So it seems like we're getting close. If there was no minus one here, we're essentially done. But now there's this minus one. So how do we deal with that? Well, we can do a similar type of strategy. We can subtract one, and then we could add, and then we could add one. Then we're not actually changing the value. Just as we multiplied by 10 and divide by 10, we're not changing the value up here. If you subtract one and add one to the exponent, you're not changing its value. And so, what is this going to be? We wanna leave this minus one here. But we wanna get rid of, And here, we just have to remind ourselves that, if we have a to the b times a to the c, that's going to be equal to a to the b plus c. If you have the same base, multiplied, same base raised to different exponents and you multiply them, you could just add the exponents. And so you could also go the other way around. If you have a to the b plus c, you could break it up into a to the b times a to the c. So, this business right over here, this business right over here, this is 1,024 to the t over 10 minus one, plus one. So we can break this up as, we can break this up as, 1,024, 1,024 to the t over 10 minus one, that's this part here, and then times 1,024 to the one. Times ... Let me make this in a different color. So, let's see green." + }, + { + "Q": "I dont understand the way Sal's doing these. I paused at 0:06 and factored it by grouping, because it seemed like the obvious way to go. Like this:\n30x^2 + 11xy + y^2\n30x^2 + 5xy + 6xy + y^2\n5x(6x+y) + y(6x+y)\n=(5x+y)(6x+y)\nMultiplying it out, i get back to the beginning.\nI did the problem in the previous video the same way. Is this correct? Or am i doing it wrong and getting the right answers just by accident?", + "A": "I did it the same way you did.", + "video_name": "0xrvRKHoO2g", + "timestamps": [ + 6 + ], + "3min_transcript": "Let's see if we can use our existing factoring skills to factor 30x squared plus 11xy plus y squared. And I encourage you to pause the video and see if you can handle it yourself. Now, the first hint I will give you-- and this might open up what's going on here-- is to maybe rearrange this a little bit. We could rewrite this as y squared plus 11xy plus 30x squared. And my whole motivation for doing that-- there are ways to factor a quadratic where your first coefficient, your coefficient on this first term, is something other than 1. But we haven't seen that yet. And so rearranging it this way, this got us a little bit more into our comfort zone. Now our coefficient is a 1 on the y squared term. So now we can start to think of this in the same form that we've looked at some of the other factoring problems. Can we think of two numbers whose product is 30x squared and whose sum is 11x? We have y squared, some coefficient on y. And then in terms of y, this isn't in any way dependent on y. So one way to think about this, if you knew what x was, then this would be a quadratic in terms of y. And that's how we're really thinking about it here. So can we find two numbers whose product is 30x squared and two numbers whose sum is the coefficient on this y term right here, whose sum is 11x? So let's just think about all of the different possibilities. If we were just thinking about two numbers whose product was 30 and whose sum was 11, we would be thinking of 5 and 6. 5 times 6 is 30. 5 plus 6 is 11. It's some trial and error. You could have tried 3 and 10. Well, that would have been-- 13 would be their sum. You could have tried 2 and 15. That wouldn't have worked. But 5 and 6 does work here, so we've already seen that multiple times. So 5 and 6 would work for 30, but we have 30x squared. Well, 5x times 6x is 30x squared, and 5x plus 6x is 11x. So this actually works. So then our factoring or our factorization of this expression is just going to be y plus 5x times y plus 6x. And I'll leave it up to you to verify that this does indeed, when you multiply it out, equal this up here." + }, + { + "Q": "At 2:55, what is meant by derivative of something \"with respect to\" something else?", + "A": "With respect to is generally used to describe the term you are talking about. For example, say you have: f(x) = (x^2)/3 with respect to x, the function is the same, (x^2)/3 BUT with respect to x^2, the function is x/3 Taking the derivative of a function with respect to something basically means you are determining what the derivative function is doing to the term that you re talking about. Hope that helps!", + "video_name": "Mci8Cuik_Gw", + "timestamps": [ + 175 + ], + "3min_transcript": "So this is the x power in yellow. And so let's do that right over here. So instead of taking the derivative with respect to x of 2 to the x, let's say, let's just take the derivative with respect to x of the exact same expression rewritten, of e to the natural log of 2 raised to the x power. Let me put this x in that same color, dx. Now we know from our exponent properties if we raise something to some power, and then raise that to another power, we can take the product of the two powers. Let me rewrite this just to remember. If I have a to b, and then I raise that to the c power, this is the exact same thing as a to the b times c power. So we can utilize that exponent property right here to rewrite this as being equal to the derivative with respect And what's neat about this is now we've got this into a form of e to the something. So we can essentially use the chain rule to evaluate this. So this derivative is going to be equal to the derivative of e to the something with respect to that something. Well, the derivative e to the something with respect to that something is just e to that something. So it's going to be equal to e to the natural log of 2 times x. So let me make it clear what I just did here. This right over here is the derivative of e to the natural log of 2 times x with respect to the natural log of 2-- let me make it a little bit clearer-- with respect So we took the derivative of e to the something with respect to that something-- that's this right here, it's just e to that something. And then we're going to multiply that by, this is just an application of the chain rule, of the derivative of that something with respect to x. So the derivative of natural log of 2 times x with respect to x is just going to be natural log of 2. This is just going to be natural log of 2. The derivative of a times x is just going to be equal to a. This is just the coefficient on the x. And just to be clear, this is the derivative of natural log of 2 times x with respect to x. So we're essentially done. But we can simplify this even further. This thing right over here can be rewritten. And let me draw a line here just to make it clear that this equals sign is a continuation from what we did up there. But this e to the natural log of 2x, we can rewrite that, using this exact same exponent property, as e to the natural log of 2, and then" + }, + { + "Q": "at 2:17, we could just treat e^(ln 2)^x as a function and use the chain rule to differentiate it, couldn't we?\nbut the ans is different", + "A": "Sal has been solving e^(ln2*x) not e^(ln2)^x. I have hope it is a typo.", + "video_name": "Mci8Cuik_Gw", + "timestamps": [ + 137 + ], + "3min_transcript": "Let's see if we can take the derivative with respect to x of 2 to the x power. And you might say, hold on a second. We know how to take the derivative of e to the x. But what about a base like 2? We don't know what to do with 2. And the key here is to rewrite 2 to the x so that we essentially have it as e to some power. And the key there is to rewrite 2. So how can we rewrite 2 so it is e to some power? Well, let's think about what e to the natural log of 2 power is. The natural log of 2 is the power that I would have to raise e to to get to 2. So if we actually raise e to that power, we are going to get to 2. So what we could do, instead of writing 2 to the x, we could rewrite this as e. We could rewrite 2 as e to the natural log of 2, So this is the x power in yellow. And so let's do that right over here. So instead of taking the derivative with respect to x of 2 to the x, let's say, let's just take the derivative with respect to x of the exact same expression rewritten, of e to the natural log of 2 raised to the x power. Let me put this x in that same color, dx. Now we know from our exponent properties if we raise something to some power, and then raise that to another power, we can take the product of the two powers. Let me rewrite this just to remember. If I have a to b, and then I raise that to the c power, this is the exact same thing as a to the b times c power. So we can utilize that exponent property right here to rewrite this as being equal to the derivative with respect And what's neat about this is now we've got this into a form of e to the something. So we can essentially use the chain rule to evaluate this. So this derivative is going to be equal to the derivative of e to the something with respect to that something. Well, the derivative e to the something with respect to that something is just e to that something. So it's going to be equal to e to the natural log of 2 times x. So let me make it clear what I just did here. This right over here is the derivative of e to the natural log of 2 times x with respect to the natural log of 2-- let me make it a little bit clearer-- with respect" + }, + { + "Q": "At 3:30 and onwards, how come the 1 in \"anti-deriv: 1+sinx\" isnt accounted for? shouldn't its anti-derivative be x?", + "A": "It s not 1+sin(x), it s 1*sin(x). Which equals sin(x). Or 1*1*1*1*1*1*1*1*sin(x), depending on weather.", + "video_name": "bZ8YAHDTFJ8", + "timestamps": [ + 210 + ], + "3min_transcript": "If we assign g prime of x to be cosine of x, once again, if we take its antiderivative, that sine of x, it's not any more complicated. If we did it the other way around, if we set f of x to be cosine of x, then we're taking its derivative here. That's not that much more complicated. But if we set g prime of x equaling to x and then we had to take its antiderivative, we get x squared over 2, that is more complicated. So let me make it clear over here. We are assigning f of x to be equal to x. And that means that the derivative of f is going to be equal to 1. We are assigning-- I'll write it right here-- g prime of x to be equal to cosine of x, which means g of x is equal to sine of x, the antiderivative of cosine of x. Now let's see, given these assumptions, let's see if we can apply this formula. Let's see, the right-hand side says f of x times g of x. So f of x is x. g of x is sine of x. And then from that, we are going to subtract the antiderivative of f prime of x-- well, that's just 1-- times g of x, times sine of x dx. Now this was a huge simplification. Now I went from trying to solve the antiderivative of x cosine of x to now I just have to find the antiderivative of sine of x. And we know the antiderivative of sine of x dx is just equal to negative cosine of x. And of course, we can throw the plus c in now, now that we're pretty done with taking all of our antiderivatives. So all of this is going to be equal to x of this, which is just negative cosine of x. And then we could throw in a plus c right at the end of it. And doesn't matter if we subtract a c or add the c. We're saying this is some arbitrary constant which could even be negative. And so this is all going to be equal to-- we get our drum roll now-- it's going to be x times sine of x, subtract a negative, that becomes a positive, plus cosine of x plus c. And we are done. We were able to take the antiderivative of something that we didn't know how to take the antiderivative of before. That was pretty interesting." + }, + { + "Q": "At 1:52 Sal writes g'(x) = cosx and g(x) = sinx and not {sinx+C}. Can someone explain?", + "A": "Sal chose a convenient antiderivative. Any choice for an antiderivative would yield the same final result. Try the same process choosing a nonzero number for the constant. See what result you get.", + "video_name": "bZ8YAHDTFJ8", + "timestamps": [ + 112 + ], + "3min_transcript": "In the last video, I claimed that this formula would come handy for solving or for figuring out the antiderivative of a class of functions. Let's see if that really is the case. So let's say I want to take the antiderivative of x times cosine of x dx. Now if you look at this formula right over here, you want to assign part of this to f of x and some part of it to g prime of x. And the question is, well do I assign f of x to x and g prime of x to cosine of x or the other way around? Do I make f of x cosine of x and g prime of x, x? And that thing to realize is to look at the other part of the formula and realize that you're essentially going to have to solve this right over here. And here where we have the derivative of f of x times g of x. So what you want to do is assign f of x so that the derivative of f of x is actually simpler than f of x. And assign g prime of x that, if you were to take its antiderivative, it doesn't really become any more complicated. So in this case, if we assign f of x to be equal to x, f prime of x is definitely simpler, If we assign g prime of x to be cosine of x, once again, if we take its antiderivative, that sine of x, it's not any more complicated. If we did it the other way around, if we set f of x to be cosine of x, then we're taking its derivative here. That's not that much more complicated. But if we set g prime of x equaling to x and then we had to take its antiderivative, we get x squared over 2, that is more complicated. So let me make it clear over here. We are assigning f of x to be equal to x. And that means that the derivative of f is going to be equal to 1. We are assigning-- I'll write it right here-- g prime of x to be equal to cosine of x, which means g of x is equal to sine of x, the antiderivative of cosine of x. Now let's see, given these assumptions, let's see if we can apply this formula. Let's see, the right-hand side says f of x times g of x. So f of x is x. g of x is sine of x. And then from that, we are going to subtract the antiderivative of f prime of x-- well, that's just 1-- times g of x, times sine of x dx. Now this was a huge simplification. Now I went from trying to solve the antiderivative of x cosine of x to now I just have to find the antiderivative of sine of x. And we know the antiderivative of sine of x dx is just equal to negative cosine of x. And of course, we can throw the plus c in now, now that we're pretty done with taking all of our antiderivatives. So all of this is going to be equal to x" + }, + { + "Q": "At 3:40, why isn't the anti derivative of 1 added to -cos x..won't it be x times -cos x?", + "A": "Hmm, are you trying to apply the product rule ? That is only for derivatives. In this case, we are doing the anti-derivative (integral). So when he wants to do the anti-derivative: \u00e2\u0088\u00ab1\u00e2\u008b\u0085sin(x)dx = \u00e2\u0088\u00absin(x)dx = -cos(x) + C the 1 times inside the integral has no effect, since 1\u00e2\u008b\u0085sin(x) = sin(x). Or maybe you had some other reason in mind?", + "video_name": "bZ8YAHDTFJ8", + "timestamps": [ + 220 + ], + "3min_transcript": "If we assign g prime of x to be cosine of x, once again, if we take its antiderivative, that sine of x, it's not any more complicated. If we did it the other way around, if we set f of x to be cosine of x, then we're taking its derivative here. That's not that much more complicated. But if we set g prime of x equaling to x and then we had to take its antiderivative, we get x squared over 2, that is more complicated. So let me make it clear over here. We are assigning f of x to be equal to x. And that means that the derivative of f is going to be equal to 1. We are assigning-- I'll write it right here-- g prime of x to be equal to cosine of x, which means g of x is equal to sine of x, the antiderivative of cosine of x. Now let's see, given these assumptions, let's see if we can apply this formula. Let's see, the right-hand side says f of x times g of x. So f of x is x. g of x is sine of x. And then from that, we are going to subtract the antiderivative of f prime of x-- well, that's just 1-- times g of x, times sine of x dx. Now this was a huge simplification. Now I went from trying to solve the antiderivative of x cosine of x to now I just have to find the antiderivative of sine of x. And we know the antiderivative of sine of x dx is just equal to negative cosine of x. And of course, we can throw the plus c in now, now that we're pretty done with taking all of our antiderivatives. So all of this is going to be equal to x of this, which is just negative cosine of x. And then we could throw in a plus c right at the end of it. And doesn't matter if we subtract a c or add the c. We're saying this is some arbitrary constant which could even be negative. And so this is all going to be equal to-- we get our drum roll now-- it's going to be x times sine of x, subtract a negative, that becomes a positive, plus cosine of x plus c. And we are done. We were able to take the antiderivative of something that we didn't know how to take the antiderivative of before. That was pretty interesting." + }, + { + "Q": "At 6:26 Sal Khan says \"... our variance is essentially the probability of success times the probability of failure.\" Mathematically I understand this (Khan walks us through the derivation) but conceptually I don't. If variance is some measure of the spread of values around the mean, how does the product of the probably of success and failure describe the variance?", + "A": "To me, p(1-p) seems like an algebraic simplification without a conceptual component to it. Sometimes algebraic simplifications lead to more conceptual insight, but this one really doesn t.", + "video_name": "ry81_iSHt6E", + "timestamps": [ + 386 + ], + "3min_transcript": "This right here is going to be the variance. Now let's actually work this out. So this is going to be equal to 1 minus p. Now 0 minus p is going to be negative p. If you square it you're just going to get p squared. So it's going to be p squared. Then plus p times-- what's 1 minus p squared? 1 minus p squared is going to be 1 squared, which is just 1, minus 2 times the product of this. So this is going to be minus 2p right over here. And then plus negative p squared. So plus p squared just like that. And now let's multiply everything out. This is going to be, this term right over here is going to be p squared minus p to the third. going to be plus p times 1 is p. p times negative 2p is negative 2p squared. And then p times p squared is p to the third. Now we can simplify these. p to the third cancels out with p to the third. And then we have p squared minus 2p squared. So this right here becomes, you have this p right over here, so this is equal to p. And then when you add p squared to negative 2p squared you're left with negative p squared minus p squared. And if you want to factor a p out of this, this is going to be equal to p times, if you take p divided p you get a 1, p square divided by p is p. So p times 1 minus p, which is a pretty neat, clean formula. So our variance is p times 1 minus p. And if we want to take it to the next level and figure out square root of the variance, which is equal to the square root of p times 1 minus p. And we could even verify that this actually works for the example that we did up here. Our mean is p, the probability of success. We see that indeed it was, it was 0.6. And we know that our variance is essentially the probability of success times the probability of failure. That's our variance right over there. The probability of success in this example was 0.6, probability of failure was 0.4. You multiply the two, you get 0.24, which is exactly what we got in the last example. And if you take its square root for the standard deviation, which is what we do right here, it's 0.49. So hopefully you found that helpful, and we're going to build on this later on in some of our inferential statistics." + }, + { + "Q": "at around 5:37, Sal divided the 2 and the 12, but, don't you have to divide the 27 too or am I just forgetting a rule?", + "A": "(27*12)/2 = (3*3*3*2*2*3)/2 now you can cancel out factors", + "video_name": "8C5kAIKLcZo", + "timestamps": [ + 337 + ], + "3min_transcript": "And that makes sense because we should have more feet than yards. And actually, this should be three times more, so everything makes sense. 27/2 is 3 times 9/2. So now we have 27/2 feet, and now we want to convert this to inches. And we just have to remember there are 12 inches per yard. And we're going to want to multiply by 12, because however many feet we have, we're going to have 12 times as many inches. If we have 1 foot, we're going to have 12 inches, 2 feet, 24 inches. 27/2 feet, we're going to multiply it by 12 to get the number of inches. Since this is going to be times 12, and we'll make sure the dimensions work out: 12 inches per foot. And the feet and the foot, this is just the plural and It's the same dimension. This will cancel out. So this will be-- if we just rearrange the multiplication, view it as everything is getting multiplied, and when you just multiply a bunch of things, order doesn't matter. So this is equal to 27/2 times 12 feet. I'm just swapping the order. Feet times inches divided by feet, or foot, just the singular of the same word. The feet and the foot cancel out, they're the same unit. And you have 27 times 12 divided by 2 inches. And what we could do here is that our final answer is going to be 27 times 12/2 inches. And before we multiply the 27 times 12 and then divide by 2, you immediately see, well, I can just divide 12 by 2, and 2 by 2, and it makes our computation simpler. It becomes 27 times 6 inches, and let's figure out what that is. 27 times 6. 7 times 6 is 42. 2 times 6 is 12, plus 4 is 16. This is equal to 162 inches, which makes sense. 4 and 1/2 yards, that gets us to this number right here: 27 divided by 2 is 13 and 1/2 feet. You multiply that by 12, it makes sense. You're going to have a bunch of inches. 162 inches." + }, + { + "Q": "In 0:26 what does compute mean", + "A": "Compute means to calculate, find, or figure out. He will show how to find the answer.", + "video_name": "twMdew4Zs8Q", + "timestamps": [ + 26 + ], + "3min_transcript": "Let's multiply 9 times 8,085. That should be a pretty fun little calculation to do. So like always, let's just rewrite this. So I'm going to write the 8,085. I'm going to write the 9 right below it and write our little multiplication symbol. And now, we're ready to compute. So first we can tackle 9 times 5. Well, we know that 9 times 5 is 45. We can write the 5 in the ones place and carry the 5 to the tens place. So 9 times 5 is 45. Now we're ready to move on to 9 times 8. And we're going to calculate 9 times 8 and then add the 4 that we just carried. So 9 times 8 is 72, plus the 4 is 76. So we'll write the 6 right here the tens place and carry the 7. looking for a suitable color. 9 times 0 100's plus-- and this is a 7 in the hundreds place, so that's actually 700. Or if we're just kind of going with the computation, 9 times 0 plus 7. Well, 9 times 0 is 0, plus 7 is 7. And then, finally, we have-- and once again, I'm looking for a suitable color-- 9 times 8. This is the last thing we have to compute. We already know that 9 times 8 is 72. And we just write the 72 right down here, and we're done. 8,085 times 9 is 72,765. Let's do one more example just to make sure that this is really clear in your brain, at least the process for doing this. And I also want you to think about why this works. So let's try 7 times 5,396. I'm going to rewrite it-- 5,396 times 7. First, we'll think about what 7 times 6 is. We know that's 42. We'll put the 2 in the ones place. 4 we will carry. Then we need to concern ourselves with 7 times 9. But then, we have to calculate that and then add the 4. 7 times 9 is 63, plus 4 is 67. So we put the 7 down here and carry the 6. Then we have to worry about 7 times 3 plus this 6 that we had just finished carrying. 7 times 3 is 21, plus 6 is 27." + }, + { + "Q": "why at around 3:00 is the x2 put inside of the square root symbol would it not be in the front like the 8", + "A": "You are not seeing it correctly. The x2 is not inside the radical.", + "video_name": "Z3db5itCIiQ", + "timestamps": [ + 180 + ], + "3min_transcript": "And I've simplified a little bit, I've done no rationalizing just yet, and it looks like there is a little more simplification I can do first. Because everything in the numerator and everything in the denominator is divisible by 2. So lets divide the numerator by 2. So if you divide the numerator by 2, 16 divided by 2, or you could view it as multipying the numerator and denominator by one half. So 16 times one half is 8. 2X squared times one half is just X squared. And then 2 times the principle square root of 2 times one half is just the square root of 2. It is 1 square roots of 2. So this whole thing has simplified to 8 plus X squared, all of that over the square root of 2. And now lets rationalize this. So lets do that. So times the principle square root of 2 over the principle square root of 2. Now just to show that it works on the denominator what is the principle square root of 2 times the principle square root of 2? Well its going to be 2. And in our numerator, we are going to distribute this term onto both terms in this expression, so you have 8 times the principle square root of 2 plus the square root of 2 times X squared. And we could consider this done, we have simplified the expression, or if you want you could break it up. You could say this is the same thing as 8 square roots of 2 over 2, which is 4 square roots of 2, plus the square root of 2 times X squared over 2. So depending on your tastes, you might view this as more simple or this as more simple but both are equally valid. We could have rationalized right from the get go. Let me start with our original problem. So our original problem was 16 + 2X squared, all of that over the principle square root of 8. We could have rationalized from the get go by multiplying the numerator and the denominator by the principle square root of 8. And so in our denominator we'll just get 8. And then in our numerator we would get 16 times the principle square root of 8, plus 2 times the principle square root of 8X squared. And now we could try and simplify this a little bit more. You can say, well, everything in the numerator and denominator is divisible by 2, so the 16 could become an 8 if you divide by 2. The 2 becomes a 1. And this 8 becomes a 4. And then you get 8 square roots of 8 plus the square root of 8X squared. All of this over 4." + }, + { + "Q": "At 2:05 Sal puts in the -9y^2x at the end of the simplified polynomial equation. Could he have put -9y^2x at the beginning? Why did he put the -9y^2x where he did?", + "A": "The terms in the polynomial can be listed in any order. For example - these are all the same: 4x^2y - 10xy + 45 - 9y^2x (this is Sal s version) 4x^2y - 10xy - 9y^2x + 45 45 - 9y^2x - 10xy + 4x^2y - 9y^2x + 45 + + 4x^2y - 10xy etc.", + "video_name": "AqMT_zB9rP8", + "timestamps": [ + 125 + ], + "3min_transcript": "We've got 4x squared y minus 3xy plus 25 minus the entire expression 9y squared x plus 7xy minus 20. So when we're subtracting this entire expression, that's equivalent to subtracting each of these terms individually if we didn't have the parentheses. Or another way of thinking about it-- we could distribute this negative sign. Or you could view this as a negative 1 times this entire expression. And we can distribute it. So let's do that. So let me write this first expression here. I'm going to write it unchanged. So it is 4x squared y minus 3xy plus 25. And now let me distribute the negative 1, or the negative sign times all of this stuff. So negative 1 times 9y squared x is negative 9y squared x. Negative 1 times 7xy is negative 7xy. And then negative 1 times 20 is positive 20. And we just want to group like terms. So let's see, is there another x squared y term anywhere? No, I don't see one. So I'll just rewrite this. So we have 4x squared y. Now, is there another xy term? Yeah, there is. So we can group negative 3xy and negative 7xy. Negative 3 of something minus another 7 of that something is going to be negative 10 of that something. So it's negative 10xy. And then we have a 25, which is just a constant term. Or an x to the 0 term. It's 25x to the 0. You could view it that way. And there's another constant term right over here. We can always add 25 to 20. That gives us 45. And then we have this term right over here, which clearly can't be merged with anything else. So minus 9y squared. Let me do that in that original color. Minus 9-- I'm having trouble shifting And we are done." + }, + { + "Q": "At 2:50, why not just keep it (a+2)(a-2) and then cancel out (a+2) from the top and bottom? Why does this not work to simplify?", + "A": "I see. So (sorry if this is worded wrong) you have to combine the -(a-3) before simplifying for the same reason you can t cancel out the two a s in (a + b)/a ?", + "video_name": "IKsi-DQU2zo", + "timestamps": [ + 170 + ], + "3min_transcript": "is the least common multiple of this expression, and that expression, and it could be a good common denominator. Let's set that up. This will be the same thing as being equal to this first term right here, a minus 2 over a plus 2, but we want the denominator now to be a plus 2 times a plus 2-- we wanted it to be a plus 2 squared. So, let's multiply this numerator and denominator by a plus 2, so its denominator is the same thing as this. Let's multiply both the numerator and the denominator by a plus 2. We're going to assume that a is not equal to negative 2, that would have made this undefined, and it would have also made this undefined. Throughout this whole thing, we're going to assume that a cannot be equal to negative 2. The domain is all real numbers, a can be any real number except for negative 2. So, the first term is that-- extend the line a little bit-- denominator is already the common denominator. Minus a minus 3 over-- and we could write it either as a plus 2 times a plus 2, or as this thing over here. Let's write it in the factored form, because it'll make it easier to simplify later on: a plus 2 times a plus 2. And now, before we-- let's set this up like this-- now, before we add the numerators, it'll probably be a good idea to multiply this out right there, but let me write the denominator, we know what that is: it is a plus 2 times a plus 2. Now this numerator: if we have a minus 2 times a plus 2, we've seen that pattern before. We can multiply it out if you like, but we've seen it enough hopefully to recognize that this is going to be a squared minus 2 squared. This is going to be a squared minus 4. You can multiply it out, and the middle terms cancel out-- the negative 2 times a cancels out the a times 2, and you're just left with a squared minus 4-- that's that over there. be very careful here-- you're subtracting a minus 3, so you want to distribute the negative sign, or multiply both of these terms times negative 1. So you could put a minus a here, and then negative 3 is plus 3, so what does this simplify to? You have a squared minus a plus-- let's see, negative 4 plus 3 is negative 1, all of that over a plus 2 times a plus 2. We could write that as a plus 2 squared. Now, we might want to factor this numerator out more, to just make sure it doesn't contain a common factor with the denominator. The denominator is just 2a plus 2 is multiplied by themselves. And you can see from inspection a plus 2 will not" + }, + { + "Q": "at 4:37, a^2-a is just a. I don't understand why the answer isn't just a-1. I guess that is another one of those \"algebra\" rules that people are just supposed to magically know?", + "A": "You cannot subtract a^2 and a. Because they are not like terms. Powers are related to mutliplication, and will not be affected using addition and subtraction. And no, you are not supposed to magically know this - lol - but you should have learned about combining like terms before you started working on rational expressions. If you missed that in one of your previous courses, you can certainly go back and review this skill. It should be listed under the Algebra I content.", + "video_name": "IKsi-DQU2zo", + "timestamps": [ + 277 + ], + "3min_transcript": "denominator is already the common denominator. Minus a minus 3 over-- and we could write it either as a plus 2 times a plus 2, or as this thing over here. Let's write it in the factored form, because it'll make it easier to simplify later on: a plus 2 times a plus 2. And now, before we-- let's set this up like this-- now, before we add the numerators, it'll probably be a good idea to multiply this out right there, but let me write the denominator, we know what that is: it is a plus 2 times a plus 2. Now this numerator: if we have a minus 2 times a plus 2, we've seen that pattern before. We can multiply it out if you like, but we've seen it enough hopefully to recognize that this is going to be a squared minus 2 squared. This is going to be a squared minus 4. You can multiply it out, and the middle terms cancel out-- the negative 2 times a cancels out the a times 2, and you're just left with a squared minus 4-- that's that over there. be very careful here-- you're subtracting a minus 3, so you want to distribute the negative sign, or multiply both of these terms times negative 1. So you could put a minus a here, and then negative 3 is plus 3, so what does this simplify to? You have a squared minus a plus-- let's see, negative 4 plus 3 is negative 1, all of that over a plus 2 times a plus 2. We could write that as a plus 2 squared. Now, we might want to factor this numerator out more, to just make sure it doesn't contain a common factor with the denominator. The denominator is just 2a plus 2 is multiplied by themselves. And you can see from inspection a plus 2 will not number right here would be divisible by 2, it's not divisible by 2. So, a plus 2 is not one of the factors here, so there's not going to be any more simplification, even if we were able to factor this thing, and the numerator out. we're done. We have simplified the rational expression, and the domain is for all a's, except for a cannot, or, all a's given that a does not equal negative 2-- all a's except for negative 2. And we are done." + }, + { + "Q": "what is a real number 1:54\nis there such a thing as a fake number", + "A": "Imaginary numbers exist, and they exist because sometimes one needs to express the square root of a negative number, which doesn t exist. i is the central imaginary number, and it stands for the square root of negative one.", + "video_name": "IKsi-DQU2zo", + "timestamps": [ + 114 + ], + "3min_transcript": "Find the difference. Express the answer as a simplified rational expression, and state the domain. We have two rational expressions, and we're subtracting one from the other. Just like when we first learned to subtract fractions, or add fractions, we have to find a common denominator. The best way to find a common denominator, if were just dealing with regular numbers, or with algebraic expressions, is to factor them out, and make sure that our common denominator has all of the factors in it-- that'll ensure that it's divisible by the two denominators here. This guy right here is completely factored-- he's just a plus 2. This one over here, let's see if we can factor it: a squared plus 4a plus 4. Well, you see the pattern that 4 is 2 squared, 4 is 2 times 2, so a squared plus 4a plus 4 is a plus 2 times a plus 2, or a plus 2 squared. We could say it's a plus 2 times a plus 2-- that's what a squared plus 4a plus 4 is. This is obviously divisible by itself-- everything is divisible by itself, except, I guess, for 0, is divisible by is the least common multiple of this expression, and that expression, and it could be a good common denominator. Let's set that up. This will be the same thing as being equal to this first term right here, a minus 2 over a plus 2, but we want the denominator now to be a plus 2 times a plus 2-- we wanted it to be a plus 2 squared. So, let's multiply this numerator and denominator by a plus 2, so its denominator is the same thing as this. Let's multiply both the numerator and the denominator by a plus 2. We're going to assume that a is not equal to negative 2, that would have made this undefined, and it would have also made this undefined. Throughout this whole thing, we're going to assume that a cannot be equal to negative 2. The domain is all real numbers, a can be any real number except for negative 2. So, the first term is that-- extend the line a little bit-- denominator is already the common denominator. Minus a minus 3 over-- and we could write it either as a plus 2 times a plus 2, or as this thing over here. Let's write it in the factored form, because it'll make it easier to simplify later on: a plus 2 times a plus 2. And now, before we-- let's set this up like this-- now, before we add the numerators, it'll probably be a good idea to multiply this out right there, but let me write the denominator, we know what that is: it is a plus 2 times a plus 2. Now this numerator: if we have a minus 2 times a plus 2, we've seen that pattern before. We can multiply it out if you like, but we've seen it enough hopefully to recognize that this is going to be a squared minus 2 squared. This is going to be a squared minus 4. You can multiply it out, and the middle terms cancel out-- the negative 2 times a cancels out the a times 2, and you're just left with a squared minus 4-- that's that over there." + }, + { + "Q": "6:09 .... did Vi just say \"damn\"?", + "A": "No she said Bam!", + "video_name": "EdyociU35u8", + "timestamps": [ + 369 + ], + "3min_transcript": "Which makes you really want to know what you get if you do the first thing. But instead of always starting with zigging out, you alternate starting zig out and zig in. And it's kind of bumping into itself. And this would definitely be more perfect with graph paper OK, now it's starting to look like a triangle? On the one hand, not nearly as cool as a spirally thing you get when your zigs always go the same way. On the other hand, why would you get a triangle? And it's like a solid triangle too. If you kept doing this forever, would the triangle just fill up completely? These didn't do that. Although with this thing you have sections that are starting to fill up. Maybe here at some point that will happen, though it seems like it's just full of holes forever. You kind of wish you had a way to take some graph paper and skip all the way to what happens later in the sequence. Maybe if you had some sort of diagram, like-- this has a line with one right turn. The next one goes right, right, left. And then the next goes right, right, left, right, right, left, left. Is there a rule? Maybe it's just like the zig zag zag zigs. OK, but there's probably some rule. Suddenly, a note lands on your desk from your friend Sam. Who writes, looks like you're concentrating pretty hard. Don't tell me you're actually doing math. As, if. You write back, no way. I'm just doodling this. And just to make extra clear it's not math, you turn it into a dragon, and name it the dragon curve. Yes. You don't want to crumple up your awesome dragon doodle, but you do have to throw it two rows over. So you neatly fold it into a note spear. Which just means you're folding it in half again and again, until it's easy to javelin across the room the moment the teachers back is turned-- Bam! Yes! Perfect landing. You watch as Sam unfolds it. And suddenly you feel like you see something familiar. Some sort of similarity between the paper and-- is that possible? You take your diagram, and fold it in half. And half again. And again. And wow, not only does it look like it's doing the same thing, probably, but it's also showing a new way to do it. you can just copy the old one, and add it 90 degrees from the other, which is totally traceable. Bam. Well, as long as you can keep track of what end to start from. And you don't even have to keep track of what order to draw the lines in. You just need to keep things roughly on a square grid so things line up. Until it gets too big for your paper and you have to dragon-ize it. It's funny, because one way it gets bigger and bigger. If you go on forever, it'll be infinitely big. But with the first way, it stays basically the same size. You just draw more details. Which means if you do it forever, the line will still get infinitely long, but the total size will stay the same. Will that even work? An infinitely long line all squiggled up into a finite area? And then with folding paper, the whole thing gets smaller and smaller until maybe it disappears entirely. Which you suppose makes sense because the edge of the paper stays the same length, no matter how you fold it. You can't make it longer and longer like copying it, where the length doubles each time." + }, + { + "Q": "What does Vi mean at 0:19?", + "A": "Pi is not infinite. A number is infinite, by definition, if it is greater than all positive integers. Pi is less than 4, so pi is not infinite.", + "video_name": "5iUh_CSjaSw", + "timestamps": [ + 19 + ], + "3min_transcript": "Voiceover:Hello and welcome to that one day of the year when, well, everyone else is building up how great Pi is. I'm here to tear it down, because you deserve the truth. Forget about the part where Pi isn't the correct circle concept. This Pi day, I'm not about how people worship Pi for being infinite for going on forever. First of all, Pi is not infinite. It is more three, but you know, less than four. There are cultures where three is the biggest number, so I don't want to be insensitive, but trust me on this four is not infinite and neither is Pi. I know it's not about it's magnitude, it's about all those digits, infinite digits going on forever, but first of all it doesn't go anywhere. It just is. There's no time element. If you had a number line, Pi would be exactly one point on that number line sitting perfect still right now. It's not going to start wondering off on an infinite journey that takes forever, or even on a finite journey that takes forever, or an infinite journey that takes finite time. Secondly, yeah, so it's got infinite digits. So what, one-third has infinite digits. There's exactly as exactly as many digits in one-third and in Pi as in 99.9999 repeating. Oh, and there's also as many digits as in numbers like, say, five, I know, big number. It's even more than four, so, it's piratically like double infinity. Which, it actuality kind of is, because in decimal innovation there's secretly infinite zeros in all of these places. Zero's going out to forever. Ooh! So mysterious, and then zero's going the other way too. Which is actually not any more zeros than if they only went one way. No. Pi is not especially infinite in any way, it's more like in-between-finite. There's an infinite number of rational numbers. for any two factions you can find another fraction that's between them again and again and again. There's never any fractions that are right next to each other on the number line. But, despite there's a infinite amount of rational numbers, Pi isn't one of them. and you can find an infinite number of rational numbers that are closer to Pi on either side. Pi is between all of them in one of the gaps. It isn't infinite. It's in-between-finite. So what, you think that's special, as if there's just one hole in the rational number line exactly where Pi is, and once you plug that in with a super special number, you're good to go? Maybe, a few more for E and [towel] and square root too. No! Super nope! The in-between-[finiteness] of Pi, its irrationality is an incredibility un-special property. Turns out, most real numbers are irrational. It's the nicely packaged rational numbers that are weird. In fact, if you threw a dart and picked a random number off the number line, the chance of getting a rational number is exactly zero. I'll get into kinds of infinities some other time, but [unintelligible] to say the number of rational numbers, like the number of digits in Pi, is the small and unimpressive countable infinity. While the number of irrational numbers is so much bigger than countable infinity," + }, + { + "Q": "Isn't the triangle person from 4:00 Wind from wind and mr. ug?", + "A": "No, the triangle is actually Vi s symbol for herself. And, as you probably know, Vi loves triangles.", + "video_name": "5iUh_CSjaSw", + "timestamps": [ + 240 + ], + "3min_transcript": "and you can find an infinite number of rational numbers that are closer to Pi on either side. Pi is between all of them in one of the gaps. It isn't infinite. It's in-between-finite. So what, you think that's special, as if there's just one hole in the rational number line exactly where Pi is, and once you plug that in with a super special number, you're good to go? Maybe, a few more for E and [towel] and square root too. No! Super nope! The in-between-[finiteness] of Pi, its irrationality is an incredibility un-special property. Turns out, most real numbers are irrational. It's the nicely packaged rational numbers that are weird. In fact, if you threw a dart and picked a random number off the number line, the chance of getting a rational number is exactly zero. I'll get into kinds of infinities some other time, but [unintelligible] to say the number of rational numbers, like the number of digits in Pi, is the small and unimpressive countable infinity. While the number of irrational numbers is so much bigger than countable infinity, cantabile infinity looks like zero. So, I don't know why anyone would make a fus about the grand infinities and forever, as about the boring little number like Pi. And of course, those are just the first couple of kinds of infinities in an infinite number of infinities in their correspondingly more in-between-[finiter] numbers like the [infinitesimals]. So, don't let Pi impress you by being a member of an unaccountably infinite set of in-between-infinite number either. The only thing even a little weird about Pi is that you do get an irrational number by taking such a simple ratio of such a simple geometric object. Surely that never happens with other simple ratios of other simple geometric objects. Oh wait! Their in everything! What are the chances? No! Let's pretend math equals arithmetic, and then get all surprised and amazed when the moment you leave arithmetic that you get a non-arithmetic number as if it were some odd unpredictable phenomenon. That way, by the time you get to calculus you won't have any idea what's going on to pass your class without ever realizing that you were dealing with infinities two levels deeper than the infinity you think is so cool when Pi does it. Pi is not special. Yeah, Pi can be fun, and I'd never deny you your deserts, but maybe try some real food once in a while." + }, + { + "Q": "At 4:54, how does this equation with all the ones = (x=(x+x)/1)?", + "A": "Since the fractal fraction is infinitely big, you can say that since infinity minus 1 is still infinity,", + "video_name": "a5z-OEIfw3s", + "timestamps": [ + 294 + ], + "3min_transcript": "Or 8 square root 13. Or you could even make each layer different. 7, 8, 9, 10, 11. Now look how confusing this is. Awesome. Say you wanted to actually solve one of these things. Say you started with this puzzle. What is 1/1 plus 1, but each 1 is over 1 plus 1? And so on, all the way to infinity. You could try doing it by hand, thinking maybe it'll converge on something. 1/1 plus 1 is 1/2. So the next layer, these are 1/2, add up to 1. So this is 1/1, 1. Three layers, back to 1/2. Uh oh. Any whole number of layers is going to give either 1 or 1/2. So what could this possibly be? Well, you could try doing algebra to it. Say all this equals x. Look, you've isolated x on one side, and everything else on the other, and it doesn't help one bit. Take that math teacher. OK, but if all this is x, then all this-- which is the same as all this-- is x. You can write this as 1/x plus x, which completely works. You could generate it all again by replacing x with 1/x plus x. x's on the wrong side of the equation, you can solve it and get the boring way to write this number if you wanted. One last fraction, this one with a caution sign. Say you want something to equal 1. Split 1 into 1/2 plus 1/2. Now these 1's could be replaced with 1/2 plus 1/2. Each time you do this, it works. What happens if you go to infinity? It's weird because if you look at any number of layers of 2's, to see if it converges to something, the result is always 2 for each fraction. Which might make you think that at infinity it's also 2 for each fraction, and therefore 1 equals 4? And just looking at this and trying to take it backwards, you might say, all this equals x, and all this equals x. So it's x plus x/2. Just try and solve that equation. The problem is, half of something plus half of something always equals that something, no matter what the x. So this could be anything, it's undefined. Or say you want to make something with all 1's Now x equals x plus x/1, or x equals x plus x. You can algebra your way to a contradiction, and as far as algebra is concerned, this is undefined. two numbers I know that fit this description, infinity and 0. This I suppose could be either, or both at once, or nothing I don't know. Why does it do that? Maybe because the numerator got lost up there, and could have been anything. Interesting though that even here, when the denominator got lost in infinity, you can still solve this back to 5. That's, to me, the cool part about algebra. Unlike the neat little problems they put in grade school textbooks, not all problems can be solved, and it's not always obvious when there's an answer and when there's not. Weird stuff happens all the time. And most importantly, algebra isn't a dead ancient thing. There are things no one's ever done before, that you can do with the simplest concepts. As simple as that x is what x is." + }, + { + "Q": "How come at 5:00 Vi says 2 = 1?", + "A": "no, the equation is x=2x. so you bring x on the right side to the left side to get x/x=2. Then, any number divided by that number is always 1. So, 1=2", + "video_name": "a5z-OEIfw3s", + "timestamps": [ + 300 + ], + "3min_transcript": "Or 8 square root 13. Or you could even make each layer different. 7, 8, 9, 10, 11. Now look how confusing this is. Awesome. Say you wanted to actually solve one of these things. Say you started with this puzzle. What is 1/1 plus 1, but each 1 is over 1 plus 1? And so on, all the way to infinity. You could try doing it by hand, thinking maybe it'll converge on something. 1/1 plus 1 is 1/2. So the next layer, these are 1/2, add up to 1. So this is 1/1, 1. Three layers, back to 1/2. Uh oh. Any whole number of layers is going to give either 1 or 1/2. So what could this possibly be? Well, you could try doing algebra to it. Say all this equals x. Look, you've isolated x on one side, and everything else on the other, and it doesn't help one bit. Take that math teacher. OK, but if all this is x, then all this-- which is the same as all this-- is x. You can write this as 1/x plus x, which completely works. You could generate it all again by replacing x with 1/x plus x. x's on the wrong side of the equation, you can solve it and get the boring way to write this number if you wanted. One last fraction, this one with a caution sign. Say you want something to equal 1. Split 1 into 1/2 plus 1/2. Now these 1's could be replaced with 1/2 plus 1/2. Each time you do this, it works. What happens if you go to infinity? It's weird because if you look at any number of layers of 2's, to see if it converges to something, the result is always 2 for each fraction. Which might make you think that at infinity it's also 2 for each fraction, and therefore 1 equals 4? And just looking at this and trying to take it backwards, you might say, all this equals x, and all this equals x. So it's x plus x/2. Just try and solve that equation. The problem is, half of something plus half of something always equals that something, no matter what the x. So this could be anything, it's undefined. Or say you want to make something with all 1's Now x equals x plus x/1, or x equals x plus x. You can algebra your way to a contradiction, and as far as algebra is concerned, this is undefined. two numbers I know that fit this description, infinity and 0. This I suppose could be either, or both at once, or nothing I don't know. Why does it do that? Maybe because the numerator got lost up there, and could have been anything. Interesting though that even here, when the denominator got lost in infinity, you can still solve this back to 5. That's, to me, the cool part about algebra. Unlike the neat little problems they put in grade school textbooks, not all problems can be solved, and it's not always obvious when there's an answer and when there's not. Weird stuff happens all the time. And most importantly, algebra isn't a dead ancient thing. There are things no one's ever done before, that you can do with the simplest concepts. As simple as that x is what x is." + }, + { + "Q": "At 4:47, I'm a little confused why he multiplied each side by 2... I keep rewatching it and trying to understand but it's just not clicking with me", + "A": "He s trying to relate the inequality to the definition of f(x). Since f(x)=2x in the region of interest, and the inequality only has a x, he is multiplying the inequality by 2 in order to get the definition of f(x) in the inequality.", + "video_name": "0sCttufU-jQ", + "timestamps": [ + 287 + ], + "3min_transcript": "this is 5 minus delta. So that's our range we're going to think about. We're going to think about it in the abstract at first. And then we're going to try to come up with a formula for delta in terms of epsilon. So how could we describe all of the x's that are in this range but not equal to 5 itself? Because we really care about the things that are within delta of 5, but not necessarily equal to 5. This is just a strictly less than. They're within a range of C, but not equal to C. Well, that's going to be all of the x's that satisfy x minus 5 is less than delta. That describes all of these x's right over here. And now, what we're going to do, and the way these proofs typically go, is we're going to try to manipulate the left-hand side of the inequality, so it starts to look something like this, or it starts to look exactly like that. of the inequality is going to be expressed in terms of delta. And then we can essentially say well, look. If the right-hand side is in terms of delta and the left-hand side looks just like that, that really defines how we can express delta in terms of epsilon. If that doesn't make sense, bear with me. I'm about to do it. So, if we want x minus 5 to look a lot more like this, when x is not equal to 5-- in all of this, this whole interval, x is not equal to 5-- f of x is equal to 2x, our proposed limit is equal to 10. So if we could somehow get this to be 2x minus 10, then we're in good shape. And the easiest way to do that is to multiply both sides of this inequality by 2. And 2 times the absolute value of something, that's the same thing as the absolute value of 2 times that thing. If I were to say 2 times the absolute value of a, that's the same thing as the absolute value of 2a. this is just going to be the absolute value of 2x minus 10. And it's going to be less than on the right-hand side, you just end up with a 2 delta. Now, what do we have here on the left-hand side? Well, this is f of x as long as x does not equal 5. And this is our limit. So we can rewrite this as f of x minus L is less than 2 delta. And this is for x does not equal 5. This is f of x, this literally is our limit. Now this is interesting. This statement right over here is almost exactly what we want right over here, except the right sides are just different. In terms of epsilon, this has it in terms of delta. So, how can we define delta so that 2 delta is essentially going to be epsilon? Well, this is our chance." + }, + { + "Q": "At 2:50, why does he go from 9 down to 1 to solve for the denominator. In other problems I thought you would continue from (in this case) 28, 27 ........3,2,1. Why is this different?\nSorry if this is confusing!", + "A": "The problem is different since there are 9 cards in a hand and to fill each slot it will be 9! which is 9 down to 1", + "video_name": "SbpoyXTpC84", + "timestamps": [ + 170 + ], + "3min_transcript": "there's 36. But then that's now part of my hand. Now for the second slot, how many will there be left to pick from? Well, I've already picked one, so there will only be 35 to pick from. And then for the third slot, 34, and then Then 33 to pick from, 32, 31, 30, 29, and 28. So you might want to say that there are 36 times 35, times 34, times 33, times 32, times 31, times 30, times 29, times 28 possible hands. Now, this would be true if order mattered. This would be true if I have card 15 here. Maybe I have a-- let me put it here-- maybe I have a 9 of spades here, and then I have a bunch of cards. And maybe I have-- and that's one hand. And then I have another. So then I have cards one, two, three, four, five, six, seven, eight. Or maybe another hand is I have the eight cards, 1, 2, 3, 4, 5, 6, 7, 8, and then I have the 9 of spades. If we were thinking of these as two different hands, because we have the exact same cards, but they're in different order, then what I just calculated would make a lot of sense, because we did it based on order. But they're telling us that the cards can be sorted however the player chooses, so order doesn't matter. So we're overcounting. We're counting all of the different ways that the same number of cards can be arranged. So in order to not overcount, we have to divide this by the ways in which nine cards can be rearranged. So we have to divide this by the way nine cards can be rearranged. So how many ways can nine cards be rearranged? If I have nine cards and I'm going to pick one of nine to be in the first slot, well, that means I have 9 ways to put something in the first slot. Then in the second slot, I have 8 ways of putting a card first, so I have 8 left. Then 7, then 6, then 5, then 4, then 3, then 2, then 1. That last slot, there's only going to be 1 card left to put in it. So this number right here, where you take 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1, or 9-- you start with 9 and then you multiply it by every number less than 9. Every, I guess we could say, natural number less than 9. This is called 9 factorial, and you express it as an exclamation mark. So if we want to think about all of the different ways that we can have all of the different combinations for hands, this is the number of hands if we cared about the order, but then we want to divide by the number of ways we can order things so that we don't overcount. And this will be an answer and this will be the correct answer. Now this is a super, super duper large number. Let's figure out how large of a number this is." + }, + { + "Q": "At 6:16 why did he subtract (36-9)!?", + "A": "36! means 36*35*34*33*32*31*30*29*28*27*26*25*24*23*22....*1 But since we want it to stop at 28, we have to cancel the rest out by dividing with 27!(27*26*25*24*23*22*21*20*19*18*17....*1)", + "video_name": "SbpoyXTpC84", + "timestamps": [ + 376 + ], + "3min_transcript": "times 35, times 34, times 33, times 32, times 31, times 30, times 29, times 28, divided by 9. Well, I can do it this way. I can put a parentheses-- divided by parentheses, 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1. Now, hopefully the calculator can handle this. And it gave us this number, 94,143,280. Let me put this on the side, so I can read it. So this number right here gives us 94,143,280. That there are 94,143,280 possible 9 card hands in this situation. Now, we kind of just worked through it. We reasoned our way through it. There is a formula for this that does essentially the exact same thing. And the way that people denote this formula is to say, look, we have 36 things and we are going to choose 9 of them. And we don't care about order, so sometimes it'll be written as n choose k. Let me write it this way. So what did we do here? We have 36 things. We chose 9. So this numerator over here, this was 36 factorial. But 36 factorial would go all the way down to 27, 26, 25. It would just keep going. But we stopped only nine away from 36. So this is 36 factorial, so this part right here, that part right there, is not just 36 factorial. What is 36 minus 9? It's 27. So 27 factorial-- so let's think about this-- 36 factorial, it'd be 36 times 35, you keep going all the way, times 28 times 27, going all the way down to 1. That is 36 factorial. Now what is 36 minus 9 factorial, that's 27 factorial. So if you divide by 27 factorial, 27 factorial is 27 times 26, all the way down to 1. Well, this and this are the exact same thing. This is 27 times 26, so that and that would cancel out. So if you do 36 divided by 36, minus 9 factorial, you just get the first, the largest nine terms of 36 factorial, which is exactly what we have over there. And then we divided it by 9 factorial." + }, + { + "Q": "At time5:14 min, shouldn't the answer be 362880...i think u shud verify...u mistakenly pressed smth on ur calculator..i calculated it to be 36880", + "A": "9! = 362880. That is how many ways there are to arrange 9 objects. But Sal is calculating the number of 9-card hands, so he needs to start with 36 objects. He is dividing the number of permutations of 9 cards (from a set of 36), by the number of ways to arrange those 9 cards.", + "video_name": "SbpoyXTpC84", + "timestamps": [ + 314 + ], + "3min_transcript": "first, so I have 8 left. Then 7, then 6, then 5, then 4, then 3, then 2, then 1. That last slot, there's only going to be 1 card left to put in it. So this number right here, where you take 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1, or 9-- you start with 9 and then you multiply it by every number less than 9. Every, I guess we could say, natural number less than 9. This is called 9 factorial, and you express it as an exclamation mark. So if we want to think about all of the different ways that we can have all of the different combinations for hands, this is the number of hands if we cared about the order, but then we want to divide by the number of ways we can order things so that we don't overcount. And this will be an answer and this will be the correct answer. Now this is a super, super duper large number. Let's figure out how large of a number this is. times 35, times 34, times 33, times 32, times 31, times 30, times 29, times 28, divided by 9. Well, I can do it this way. I can put a parentheses-- divided by parentheses, 9 times 8, times 7, times 6, times 5, times 4, times 3, times 2, times 1. Now, hopefully the calculator can handle this. And it gave us this number, 94,143,280. Let me put this on the side, so I can read it. So this number right here gives us 94,143,280. That there are 94,143,280 possible 9 card hands in this situation. Now, we kind of just worked through it. We reasoned our way through it. There is a formula for this that does essentially the exact same thing. And the way that people denote this formula is to say, look, we have 36 things and we are going to choose 9 of them. And we don't care about order, so sometimes it'll be written as n choose k. Let me write it this way. So what did we do here? We have 36 things. We chose 9. So this numerator over here, this was 36 factorial. But 36 factorial would go all the way down to 27, 26, 25. It would just keep going. But we stopped only nine away from 36. So this is 36 factorial, so this part right here, that part right there, is not just 36 factorial." + }, + { + "Q": "At 10:09 How do you find the surface of a rectangle", + "A": "Surface area of a rectangle= its length multiply wide.", + "video_name": "I9eLKDbc8og", + "timestamps": [ + 609 + ], + "3min_transcript": "" + }, + { + "Q": "At about 2:23, to figure out the 1.5 portion Sal says \"15 times 15 is 225\" and gets 2.25 that way. What is this process called?", + "A": "It s actually 1.50 times 1.50 and the answer is 2.25.", + "video_name": "I9eLKDbc8og", + "timestamps": [ + 143 + ], + "3min_transcript": "The surface area of a cube is equal to the sum of the areas of its six sides. Let's just visualize that. I like to visualize things. So if that's the cube, we can see three sides. Three sides are facing us. But then if it was transparent, we see that there are actually six sides of a cube. So there's this one-- one, two, three in front-- and then one-- this is the bottom. This is in the back, and this is also in the back. So you have three sides of the cube. So I believe what they're saying. The surface area of a cube with side length x-- so if this is x, if this is x, if this is x-- is given by the expression 6x squared. That also makes sense. The area of each side is going to be x times x is x squared, and there's six of them. So it's going to be 6x squared. Jolene has two cube-shaped containers that she wants to paint. One cube has side length 2. So this is one cube right over here. I'll do my best to draw it. So this right over here has side length 2, The other cube has side length 1.5. So the other cube is going to be a little bit smaller. It has side length 1.5. So it's 1.5 by 1.5 by 1.5. What is the total surface area that she has to paint? Well, we know that the surface area of each cube is going to be 6x squared, where x is the dimensions of that cube. So the surface area of this cube right over here is going to be 6. And now-- let me do it in that color of that cube-- it's going to be 6 times x, where x is the dimension of the cube. And then the cube all has the same dimensions, so its length, width, and depth is all the same. So for this cube, the surface area is going to be 6 times 2 squared. And then the surface area of this cube is going to be 6 times 1.5 squared. it's going to be the sum of the two cubes. So we're just going to add these two things. And so if we were to compute this first one right over here, this is going to be 6 times 4. This is 24. And this one right over here, this is going to be a little bit hairier. Let's see. 15 times 15 is 225. So 1.5 times 1.5 is 2.25. So 1.5 squared is 2.25. And 2.25 times 6-- so let me just multiply that out. 2.25 times 6. Let's see. We're going to have 6 times 5 is 30. 6 times 2 is 12, plus 3 is 15. 6 times 2 is 12, plus 1 is 13. I have two numbers behind the decimal-- 13.5. So it's going to be 13.5." + }, + { + "Q": "At 2:01 what is that line over repeating decimals called ?", + "A": "The line over the repeating decimal can be called a vinculum . In a repeating decimal, the vinculum is used to indicate the group of repeating digits.", + "video_name": "d9pO2z2qvXU", + "timestamps": [ + 121 + ], + "3min_transcript": "Which of the following real numbers are irrational? Well, irrational just means it's not rational. It means that you cannot express it as the ratio of two integers. So let's see what we have here. So we have the square root of 8 over 2. If you take the square root of a number that is not a perfect square, it is going to be irrational. And then if you just take that irrational number and you multiply it, and you divide it by any other numbers, you're still going to get an irrational number. So square root of 8 is irrational. You divide that by 2, it is still irrational. So this is not rational. Or in other words, I'm saying it is irrational. Now, you have pi, 3.14159-- it just keeps going on and on and on forever without ever repeating. So this is irrational, probably the most famous of all of the irrational numbers. 5.0-- well, I can represent 5.0 as 5/1. So 5.0 is rational. It is not irrational. 0.325-- well, this is the same thing as 325/1000. So this is rational. Just as I could represent 5.0 as 5/1, both of these are rational. They are not irrational. Here I have 7.777777, and it just keeps going on and on and on forever. And the way we denote that, you could just say these dots that say that the 7's keep going. Or you could say 7.7. And this line shows that the 7 part, the second 7, just keeps repeating on forever. Now, if you have a repeating decimal-- in other videos, we'll actually convert them into fractions-- but a repeating decimal can be represented as a ratio of two integers. Just as 1/3 is equal to 0.333 on and on and on. Or I could say it like this. I could say 3 repeating. We can also do the same thing for that. I won't do it here, but this is rational. 8 and 1/2? Well, that's the same thing. 8 and 1/2 is the same thing as 17/2. So it's clearly rational. So the only two irrational numbers are the first two right over here." + }, + { + "Q": "At 0:19, when Sal said 8/2 is not a perfect square, what's the difference between a perfect square and a non-perfect square?", + "A": "A perfect square is when a whole number is the square root of the number. like \u00e2\u0088\u009a49=7 But \u00e2\u0088\u009a74 is a non-perfect square.", + "video_name": "d9pO2z2qvXU", + "timestamps": [ + 19 + ], + "3min_transcript": "Which of the following real numbers are irrational? Well, irrational just means it's not rational. It means that you cannot express it as the ratio of two integers. So let's see what we have here. So we have the square root of 8 over 2. If you take the square root of a number that is not a perfect square, it is going to be irrational. And then if you just take that irrational number and you multiply it, and you divide it by any other numbers, you're still going to get an irrational number. So square root of 8 is irrational. You divide that by 2, it is still irrational. So this is not rational. Or in other words, I'm saying it is irrational. Now, you have pi, 3.14159-- it just keeps going on and on and on forever without ever repeating. So this is irrational, probably the most famous of all of the irrational numbers. 5.0-- well, I can represent 5.0 as 5/1. So 5.0 is rational. It is not irrational. 0.325-- well, this is the same thing as 325/1000. So this is rational. Just as I could represent 5.0 as 5/1, both of these are rational. They are not irrational. Here I have 7.777777, and it just keeps going on and on and on forever. And the way we denote that, you could just say these dots that say that the 7's keep going. Or you could say 7.7. And this line shows that the 7 part, the second 7, just keeps repeating on forever. Now, if you have a repeating decimal-- in other videos, we'll actually convert them into fractions-- but a repeating decimal can be represented as a ratio of two integers. Just as 1/3 is equal to 0.333 on and on and on. Or I could say it like this. I could say 3 repeating. We can also do the same thing for that. I won't do it here, but this is rational. 8 and 1/2? Well, that's the same thing. 8 and 1/2 is the same thing as 17/2. So it's clearly rational. So the only two irrational numbers are the first two right over here." + }, + { + "Q": "At 0:15, Sal says:\nIf you take the square root of a number that is not a perfect square, it is going to be irrational.\n\nWhat about sqrt 2.25? Isn't that rational? Does the rule he stated only apply to integers?", + "A": "Sal is correct. The definition of perfect square is a number times itself is the square. Since 1.5*1.5=2.25, 1.5^2=2.25, therefore 1.5 is the number multiplied by itself, so 2.25 is the perfect square 1.5. So in this case it works. The hope is that the perfect square will be the nice and easy whole numbers.", + "video_name": "d9pO2z2qvXU", + "timestamps": [ + 15 + ], + "3min_transcript": "Which of the following real numbers are irrational? Well, irrational just means it's not rational. It means that you cannot express it as the ratio of two integers. So let's see what we have here. So we have the square root of 8 over 2. If you take the square root of a number that is not a perfect square, it is going to be irrational. And then if you just take that irrational number and you multiply it, and you divide it by any other numbers, you're still going to get an irrational number. So square root of 8 is irrational. You divide that by 2, it is still irrational. So this is not rational. Or in other words, I'm saying it is irrational. Now, you have pi, 3.14159-- it just keeps going on and on and on forever without ever repeating. So this is irrational, probably the most famous of all of the irrational numbers. 5.0-- well, I can represent 5.0 as 5/1. So 5.0 is rational. It is not irrational. 0.325-- well, this is the same thing as 325/1000. So this is rational. Just as I could represent 5.0 as 5/1, both of these are rational. They are not irrational. Here I have 7.777777, and it just keeps going on and on and on forever. And the way we denote that, you could just say these dots that say that the 7's keep going. Or you could say 7.7. And this line shows that the 7 part, the second 7, just keeps repeating on forever. Now, if you have a repeating decimal-- in other videos, we'll actually convert them into fractions-- but a repeating decimal can be represented as a ratio of two integers. Just as 1/3 is equal to 0.333 on and on and on. Or I could say it like this. I could say 3 repeating. We can also do the same thing for that. I won't do it here, but this is rational. 8 and 1/2? Well, that's the same thing. 8 and 1/2 is the same thing as 17/2. So it's clearly rational. So the only two irrational numbers are the first two right over here." + }, + { + "Q": "At 5:17 why do you have to times everything with -2?", + "A": "Because by doing so, we can get 200m in one equation and -200m in the other, which allows us to solve using elimination. We can add the two equations and be left with only one variable, w.", + "video_name": "VuJEidLhY1E", + "timestamps": [ + 317 + ], + "3min_transcript": "\" In green. Well, let's think about the total number of bags that the men ate. You had 200 men, [Let me scroll over a little bit] and they each ate m bags per man. \" \"So the man at this first party collectively ate 200 times m bags. If m is 10 bags per man, then this would be 2000. If m was 5 bags per man, then this would be 5000. We don't know what m is, but 200 times m is the total eaten by the man.\"" + }, + { + "Q": "During the video, specifically 1:00, Sal does not explain WHAT a line of best fit/line of fit is. My question: How does one find a line of best fit, and how do you know that it is true?", + "A": "Well, a line of best fit is sort of like an estimate. Nobody can truly find the line of best fit (unless all the points are on the same line). Usually, the line of best fit, assuming the points follow a pattern, is the equation of the line connecting the first point and the last point.", + "video_name": "ioieTr41L24", + "timestamps": [ + 60 + ], + "3min_transcript": "Find the line of best fit, or mark that there is no linear correlation. So let's see, we have a bunch of data points, and we want to find a line that at least shows the trend in the data. And this one seems a little difficult because if we ignore these three points down here, maybe we could do a line that looks something like this. It seems like it kind of approximates this trend, although this doesn't seem like a great trend. And if we ignored these two points right over, we could do something like maybe something like that. But we can't just ignore points like that. So I would say that there's actually no good line of best fit here. So let me check my answer. Let's try a couple more of these. Find the line of best-- well, this feels very similar. It really feels like there's no-- I mean, I could do that, but I'm ignoring these two points. I could do something like that, then I'd be ignoring these points. So I'd also say no good best fit line exists. So let's try one more. So here it looks like there's very clearly this trend. And I could try to fit it a little bit better than it's fit right now. quite well. I could maybe drop this down a little bit, something like that. Let's check my answer. A good best fit line exists. Let me check my answer. We got it right." + }, + { + "Q": "At 1:16 i lost you... why do we regroup? I need another example to kinda clarify it.", + "A": "Because you don t have enough in the tens place, so you need 1 from the hundred place to add to the tens place.", + "video_name": "X3JqIZR1XcY", + "timestamps": [ + 76 + ], + "3min_transcript": "Let's think about different ways that we can represent the number 675. So the most obvious way is to just look at the different place values. So the 6 is in the hundreds place. It literally represents 600. So that's 600. I'm going to do that in the red color-- 600. The 7 is in the tens place. It represents 7 tens, or 70. And then the 5 in the ones place. It represents 5. So let me copy and paste this and then think about how we can regroup the value in the different places to represent this in different ways. So let me copy and let me paste it, and maybe I'll do it three times. So let me do it once, and let me do it one more time. So one thing that we could do is we could regroup from one place to the next. So, for example, we could take if we wanted to-- we could take 1 from the hundreds place. That's essentially taking 100 away. So this is really making this a 500. And we could give that 100 to-- well, we could actually give it to either place, but let's give it to the tens place. So we're going to give 100 to the tens place. Now, if you give 100 to the tens place and you already had 70 there, what's it going to be equal to? Well, it's going to be equal to 170. Well, how would we represent that is tens? Well, 170 is 17 tens. So we could just say that 7 becomes 17. Now, we could keep doing that. We could regroup some of this value in the tens place to the ones place. So, for example, we could give 10 from the tens place and give it to the ones place. So let's take 10 away from here. So that becomes 160. This becomes 16. And let's give that 10 to the ones place. Well, 10 plus 5 is 15. So this 5 is now a 15. Let's do another scenario. Let's do something nutty. Let's take 200 from the hundreds place. So this is going to now 4, and this is going to become 400. That's what this 4 now represents. And let's give 100 to the tens place. And let's give another 100 to the ones place. So in other words, I'm just regrouping that 200. Those 200's, I've taken from the hundreds place, and I'm going to give it to these other places. So now the tens place is going to be 170. We're going to have 170 here, which is 17 tens. So you could say that the tens place is now 17." + }, + { + "Q": "At 1:50, why does Sal square the denominator?", + "A": "because it was originally 2c in the denominator INSIDE the big parentheses with the squared exponent. In order to move the 2c outside of the parentheses the operation around those parentheses must operate on the 2c. Since the exponent says square everything inside these parentheses you must square the 2c to eliminate those parentheses around the denominator. Notice after the 2c is squared, the parentheses with the squared exponent still surrounds the numerator, but no longer surrounds the denominator.", + "video_name": "nZu7IZLhJRI", + "timestamps": [ + 110 + ], + "3min_transcript": "the last video, I claimed that this result we got for the area of a triangle that had sides of length a, b, and c is equivalent to Heron's formula. And what I want to do in this video is show you that this is equivalent to Heron's formula by essentially just doing a bunch of algebraic manipulation. So the first thing we want to do-- let's just spring this 1/2 c under the radical sign. So 1/2 c, that's the same thing as the square root of c squared over 4. You take the square root of that you get 1/2 c. So this whole expression is equal to-- instead of drawing the radical, I'll just write the square root of this, of c squared over 4 times all of this. I'll just copy and paste it. Copy and paste. So times all of that. And of course, it has to be distributed. And then we have to close the square root. Let me just distribute the c squared over 4. This is going to be equal to the square root. This is going to be hairy, but I think you'll find it satisfying to see how this could turn into something as simple as Heron's formula. The square root of c squared over 4 times a squared is c squared a squared over 4, minus c squared over 4. I'm just distributing this. And I'm going to write it as the numerator squared over the denominator squared. So times c squared plus a squared minus b squared, squared. Over-- if I square the denominator that's 4c squared. And we immediately see that c squared and that c squared are going to cancel out. Let me close all of the parentheses just like that. And, of course, this 4 times that 4, that's going to That's the same thing as 4 squared. And I'm instead of writing 16, you'll see why I'm writing that. Now this I can rewrite. This is going to be equal to the square root-- I'm arbitrarily switching colors-- of ca over 2 squared. This is the same thing as that. I'm just writing it as the whole thing squared. If I square that, that's the c squared a squared over 2 squared over 4, minus-- and I'm going to write this whole thing as an expression squared. So that's c squared plus a squared minus b squared, over 4. And we are squaring both the numerator and the denominator. Now this might look a little bit interesting to you. Let me make the parentheses in a slightly different color. You might remember from factoring polynomials that if I have something of the form x squared minus y squared, that" + }, + { + "Q": "at 2:50 suppose the question was log10(16)+log10(2) would you than multiply 16*2 or would you still divided it?", + "A": "log (a) + log (b) = log (ab), provided that both a and b are positive. log (a) - log (b) = log (a/b), provided that both a and b are positive. Thus, log (16) + log (2) = log(16*2) = log (32)", + "video_name": "Kv2iHde7Xgw", + "timestamps": [ + 170 + ], + "3min_transcript": "properties that we know over here. We also know that if we have a logarithm-- let me write it this way, actually-- if I have b times the log base a of c, this is equal to log base a of c to the bth power. And we also know, and this is derived really straight from both of these, is that if I have log base a of b minus log base a of c, that this is equal to the log base a of b over c. And this is really straight derived from these two right over here. Now with that out of the way, let's see what we can apply. So right over here, we have all the logs are the same base. And we have logarithm of x plus logarithm of 3. So by this property right over here, the sum of logarithms with the same base, this is going to be equal to log base 3-- sorry, log base 10 of 3 times x, of 3x. Then, based on this property right over here, this thing could be rewritten-- so this is going to be equal to-- this thing can be written as log base 10 of 4 to the second power, which is really just 16. So this is just going to be 16. And then we still have minus logarithm base 10 of 2. And now, using this last property, we know we have one logarithm minus another logarithm. This is going to be equal to log base 10 of 16 over 2, 16 divided by 2, which is the same thing as 8. So the right-hand side simplifies to log base 10 of 8. The left-hand side is log base 10 of 3x. And 10 to the same power is going to be equal to 8. So 3x must be equal to 8. 3x is equal to 8, and then we can divide both sides by 3. Divide both sides by 3, you get x is equal to 8 over 3. One way, this little step here, I said, look, 10 to the-- this is an exponent. If I raise 10 to this exponent, I get 3x, 10 to this exponent, I get 8. So 8 and 3x must be the same thing. One other way you could have thought about this is, let's take 10 to this power, on both sides. So you could say 10 to this power, and then 10 to this power over here. If I raise 10 to the power that I need to raise 10 to to get to 3x, well, I'm just going to get 3x. If I raise 10 to the power that I need to raise 10 to to get 8," + }, + { + "Q": "at 2:41, could i cut the log's base and get 3x= 16-2, simplifying more early than the exposed in video or is it mandatory to use the properties explained?\nMost things in mathematics can be simplified before to reach resolution.\nThis will be the first out rule.", + "A": "No, you cannot the simplify the log base at log 3x = log 16 - log 2 Simplifying the log base should be done after the two logs on the right are combined. If you finish the work on your question you will find that you get 14/3 and not the correct answer of 8/3.", + "video_name": "Kv2iHde7Xgw", + "timestamps": [ + 161 + ], + "3min_transcript": "properties that we know over here. We also know that if we have a logarithm-- let me write it this way, actually-- if I have b times the log base a of c, this is equal to log base a of c to the bth power. And we also know, and this is derived really straight from both of these, is that if I have log base a of b minus log base a of c, that this is equal to the log base a of b over c. And this is really straight derived from these two right over here. Now with that out of the way, let's see what we can apply. So right over here, we have all the logs are the same base. And we have logarithm of x plus logarithm of 3. So by this property right over here, the sum of logarithms with the same base, this is going to be equal to log base 3-- sorry, log base 10 of 3 times x, of 3x. Then, based on this property right over here, this thing could be rewritten-- so this is going to be equal to-- this thing can be written as log base 10 of 4 to the second power, which is really just 16. So this is just going to be 16. And then we still have minus logarithm base 10 of 2. And now, using this last property, we know we have one logarithm minus another logarithm. This is going to be equal to log base 10 of 16 over 2, 16 divided by 2, which is the same thing as 8. So the right-hand side simplifies to log base 10 of 8. The left-hand side is log base 10 of 3x. And 10 to the same power is going to be equal to 8. So 3x must be equal to 8. 3x is equal to 8, and then we can divide both sides by 3. Divide both sides by 3, you get x is equal to 8 over 3. One way, this little step here, I said, look, 10 to the-- this is an exponent. If I raise 10 to this exponent, I get 3x, 10 to this exponent, I get 8. So 8 and 3x must be the same thing. One other way you could have thought about this is, let's take 10 to this power, on both sides. So you could say 10 to this power, and then 10 to this power over here. If I raise 10 to the power that I need to raise 10 to to get to 3x, well, I'm just going to get 3x. If I raise 10 to the power that I need to raise 10 to to get 8," + }, + { + "Q": "why did you subtracted the -3 with the 4x in 6:10? pls help.", + "A": "(x + 7)(4x - 3) is equal to 4x(x + 7) - 3(x + 7). You just distributed it. They re still the same and it s appropriate to write it in that way.", + "video_name": "X7B_tH4O-_s", + "timestamps": [ + 370 + ], + "3min_transcript": "So I grouped the 28 on the side of the 4. And you're going to see what I mean in a second. If we, literally, group these so that term becomes 4x squared plus 28x. And then, this side, over here in pink, it's plus negative 3x minus 21. Once again, I picked these. I grouped the negative 3 with the 21, or the negative 21, because they're both divisible by 3. And I grouped the 28 with the 4, because they're both divisible by 4. And now, in each of these groups, we factor as much out as we can. So both of these terms are divisible by 4x. So this orange term is equal to 4x times x-- 4x squared divided by 4x is just x-- plus 28x divided by 4x is just 7. Remember, you factor out everything that you can factor out. Well, both of these terms are divisible by 3 or negative 3. So let's factor out a negative 3. And this becomes x plus 7. And now, something might pop out at you. We have x plus 7 times 4x plus, x plus 7 times negative 3. So we can factor out an x plus 7. This might not be completely obvious. You're probably not used to factoring out an entire binomial. But you could view this could be like a. Or if you have 4xa minus 3a, you would be able to factor out an a. And I can just leave this as a minus sign. Let me delete this plus right here. Because it's just minus 3, right? Plus negative 3, same thing as minus 3. So what can we do here? We have an x plus 7, times 4x. Let's factor out the x plus 7. We get x plus 7, times 4x minus 3. Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping. And we factored it into two binomials. Let's do another example of that, because it's a little But once you get the hang of it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill. We want to find a times b that is equal to 1 times 6, which is equal to 6. And we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. What are the-- well, the obvious one is 1 and 6, right?" + }, + { + "Q": "At 8:40, why is it (x+1)(6x+1) instead of 7(x+1)?", + "A": "Multiply the factors... only the correct factors will create the original polynomial of 6x^2 + 7x + 1 7(x+1) = 7x + 7. This is not the original polynomial. So, these factors can not be correct. (x+1)(6x+1) = 6x^2 + x + 6x + 1 = 6x^2 + 7x + 1. This matches the original polynomial. so, these are the correct factor. Hope this helps.", + "video_name": "X7B_tH4O-_s", + "timestamps": [ + 520 + ], + "3min_transcript": "1 plus 6 is 7. So we have a is equal to 1. Or let me not even assign them. The numbers here are 1 and 6. Now, we want to split this into a 1x and a 6x. But we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squar ed here, plus-- and so I'm going to put the 6x first because 6 and 6 share a factor. And then, we're going to have plus 1x, right? 6x plus 1x equals 7x . That was the whole point. They had to add up to 7 . And then we have the final plus 1 there. Now, in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times-- 6x squar ed divided by 6x is just an x. 6x divided by 6x is just a 1. to have a plus here. But this second group, we just literally have a x plus 1. Or we could even write a 1 times an x plus 1. You could imagine I just factored out of 1 so to speak. Now, I have 6x times x plus 1, plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. And now, I'm going to actually explain why this little magical system actually works. Let me take an example. I'll do it in very general terms. Let's say I had ax plus b, times cx-- actually, I'm I think that'll confuse you, because I use a's and b's here. They won't be the same thing. So let me use completely different letters. Let's say I have fx plus g, times hx plus, I'll use j instead of i. You'll learn in the future why don't like using i as a variable. So what is this going to be equal to? Well, it's going to be fx times hx which is fhx. And then, fx times j. So plus fjx. And then, we're going to have g times hx. So plus ghx. And then g times j. Plus gj. Or, if we add these two middle terms, you have fh times x," + }, + { + "Q": "At 0:40, Sal mentions that the technique of factoring by grouping becomes obsolete once you learn the quadratic formula. So, if I have a quadratic and I need to factor it (not just find the zeros, but actually know what it looks like factored, like if I'm trying to simplify a larger rational expression), is there any way to factor it with just the quadratic formula? Or would you have to use one of the factoring techniques?", + "A": "at 0:40 he is just saying the quadratic equation is easier than rooting.", + "video_name": "X7B_tH4O-_s", + "timestamps": [ + 40 + ], + "3min_transcript": "In this video, I want to focus on a few more techniques for factoring polynomials. And in particular, I want to focus on quadratics that don't have a 1 as the leading coefficient. For example, if I wanted to factor 4x squared plus 25x minus 21. Everything we've factored so far, or all of the quadratics we've factored so far, had either a 1 or negative 1 where this 4 is sitting. All of a sudden now, we have this 4 here. So what I'm going to teach you is a technique called, factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick. To some degree, it'll become obsolete once you learn the quadratic formula, because, frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique. And then at the end of this video, I'll actually show you why it works. So what we need to do here, is we need to think of two numbers, a and b, where a times b is equal 4 times So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a plus b, need to be equal to 25. Let me be very clear. This is the 25, so they need to be equal to 25. This is where the 4 is. So we go, 4 times negative 21. That's a negative 21. So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17. Doesn't work. Let's try some other combinations. 1 and 84, too far apart when you take their difference. Because that's essentially what you're going to do, if one is negative and one is positive. Too far apart. Let's see you could do 3-- I'm jumping the gun. 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40. 2 plus negative 42 is negative 40-- too far apart. 3 and-- Let's see, 3 goes into 84-- 3 goes into 8 2 times. 2 times 3 is 6. 8 minus 6 is 2. Bring down the 4. Goes exactly 8 times. So 3 and 28. This seems interesting. And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25." + }, + { + "Q": "At 1:20, wouldn't it be a*c instead of a*b? you said a*b but did 4*-21. It just confused me when I did a*b on my homework and it was incorrect", + "A": "Yes... You need to multiply A*C from the quadratic Ax^2 + Bx + C. Unfortunately, Sal chose the variables a and b to represent the numbers you are seeking to find after multiplying AC. His choice does make the video confusing because we refer to A, B and C as the coefficients in the trinomial. Call the 2 number m and n instead of a and b . You still need to find two numbers that multiply to = AC and also add to = the middle term B. Hope this helps.", + "video_name": "X7B_tH4O-_s", + "timestamps": [ + 80 + ], + "3min_transcript": "In this video, I want to focus on a few more techniques for factoring polynomials. And in particular, I want to focus on quadratics that don't have a 1 as the leading coefficient. For example, if I wanted to factor 4x squared plus 25x minus 21. Everything we've factored so far, or all of the quadratics we've factored so far, had either a 1 or negative 1 where this 4 is sitting. All of a sudden now, we have this 4 here. So what I'm going to teach you is a technique called, factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick. To some degree, it'll become obsolete once you learn the quadratic formula, because, frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique. And then at the end of this video, I'll actually show you why it works. So what we need to do here, is we need to think of two numbers, a and b, where a times b is equal 4 times So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a plus b, need to be equal to 25. Let me be very clear. This is the 25, so they need to be equal to 25. This is where the 4 is. So we go, 4 times negative 21. That's a negative 21. So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17. Doesn't work. Let's try some other combinations. 1 and 84, too far apart when you take their difference. Because that's essentially what you're going to do, if one is negative and one is positive. Too far apart. Let's see you could do 3-- I'm jumping the gun. 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40. 2 plus negative 42 is negative 40-- too far apart. 3 and-- Let's see, 3 goes into 84-- 3 goes into 8 2 times. 2 times 3 is 6. 8 minus 6 is 2. Bring down the 4. Goes exactly 8 times. So 3 and 28. This seems interesting. And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25." + }, + { + "Q": "At 4:00 Sal states that y=1/k and that this relationship is true for y' and makes a substitution. Would this relationship extend to second, third (etc.) derivatives? Could relationships like this be established for other equations and their derivatives? Feel free to give me a problem!", + "A": "y=f(x)=1/x. y=1/k for x=k only. You can use it for substitution for y in 2nd, 3rd or nth derivative as long as x=k then y=1/k. Let say you want to use x=2, then y is not 1/k anymore, but y=1/2.", + "video_name": "FJ7AMaR9miI", + "timestamps": [ + 240 + ], + "3min_transcript": "the slope of the tangent line? Well, to figure out the slope of the tangent line, let's take the derivative. So if we write f of x, instead writing it as 1/x, I'll write it as x to the negative 1 power. That makes it a little bit more obvious that we're about to use the power rule here. So the derivative of f at any point x is going to be equal to-- well, it's going to be the exponent here is negative 1. So negative 1 times x to the-- now we decrement the exponent to the negative 2 power. Or I could say it's negative x to the negative 2. Now, what we care about is the slope when x equals k. So f prime of k is going to be equal to negative k to the negative 2 power. Or another way of thinking about it, this is equal to negative 1 So this right over here is the slope of the tangent line at that point. Now, let's just think about what the equation of the tangent line is. And we could think about it in slope-intercept form. So we know the equation of a line in slope-intercept form is y is equal to mx plus b, where m is the slope and b is the y-intercept. So if we can get it in this form, then we know our answer. We know what the y-intercept is going to be. It's going to be b. So let's think about it a little bit. This equation, so we could say y is equal to our m, our slope of the tangent line, when x is equal to k, we just figure out to be this business. It equals this thing right over here. So let me write that in blue. Negative 1 over k squared times x plus b. Well, we know what y is when x is equal to k. And so we can use that to solve for b. We know that y is equal to 1/k when x is equal to k. So this is going to be equal to negative 1-- that's not the same color. Negative 1 over k squared times k plus b. Now, what does this simplify to? See, k over k squared is the same thing as 1/k, so this is going to be negative 1/k. So this part, all of this simplifies to negative 1/k. So how do we solve for b? Well, we could just add 1/k to both sides" + }, + { + "Q": "At 2:13 he states that 90 x -1/3 = -30 . Can anyone please explain how he got that answer? I understand the finding the common ratio step, but what I don't understand is how whenever I find the common ratio my math doesn't add up. I think I'm multiplying my fractions wrong.", + "A": "a * (-(b/c)) = a*(-b)*(1/c) = -(ab)/c. 90*(-(1/3)) = 90*(-1)*(1/3) = -90/3", + "video_name": "pXo0bG4iAyg", + "timestamps": [ + 133 + ], + "3min_transcript": "In this video I want to introduce you to the idea of a geometric sequence. And I have a ton of more advanced videos on the topic, but it's really a good place to start, just to understand what we're talking about when someone tells you a geometric sequence. Now a good starting point is just, what is a sequence? And a sequence is, you can imagine, just a progression of numbers. So for example, and this isn't even a geometric series, if I just said 1, 2, 3, 4, 5. This is a sequence of numbers. It's not a geometric sequence, but it is a sequence. A geometric sequence is a special progression, or a special sequence, of numbers, where each successive number is a fixed multiple of the number before it. Let me explain what I'm saying. So let's say my first number is 2 and then I multiply 2 by So I multiply it by 3, I get 6. And then I multiply 6 times the number 3, and I get 18. Then I multiply 18 times the number 3, and I get 54. And I just keep going that way. So I just keep multiplying by the number 3. So I started, if we want to get some notation here, this is my first term. We'll call it a1 for my sequence. And each time I'm multiplying it by a common number, and that number is often called the common ratio. So in this case, a1 is equal to 2, and my common ratio is equal to 3. So if someone were to tell you, hey, you've got a geometric sequence. a1 is equal to 90 and your common ratio is equal to negative 1/3. The second term is negative 1/3 times 90. Which is what? That's negative 30, right? 1/3 times 90 is 30, and then you put the negative number. Then the next number is going to be 1/3 times this. So negative 1/3 times this. 1/3 times 30 is 10. The negatives cancel out, so you get positive 10. Then the next number is going to be 10 times negative 1/3, or negative 10/3. And then the next number is going to be negative 10/3 times negative 1/3 so it's going to be positive 10/3. And you could just keep going on with this sequence. So that's what people talk about when they mean a geometric sequence. I want to make one little distinction here. This always used to confuse me because the terms are used very often in the same context. These are sequences. These are kind of a progression of numbers." + }, + { + "Q": "Sal, at 3:33 you said a series is a sum of a sequence, but could it also be a group of the sums from a sequence?", + "A": "Yeah, I guess those can be called sub-series of a sequence.", + "video_name": "pXo0bG4iAyg", + "timestamps": [ + 213 + ], + "3min_transcript": "The second term is negative 1/3 times 90. Which is what? That's negative 30, right? 1/3 times 90 is 30, and then you put the negative number. Then the next number is going to be 1/3 times this. So negative 1/3 times this. 1/3 times 30 is 10. The negatives cancel out, so you get positive 10. Then the next number is going to be 10 times negative 1/3, or negative 10/3. And then the next number is going to be negative 10/3 times negative 1/3 so it's going to be positive 10/3. And you could just keep going on with this sequence. So that's what people talk about when they mean a geometric sequence. I want to make one little distinction here. This always used to confuse me because the terms are used very often in the same context. These are sequences. These are kind of a progression of numbers. negative 10/3. Then, I'm sorry, this is positive 10/9, right? Negative 1/3 times negative 10/3, negatives cancel out. 10/9. Don't want to make a mistake here. These are sequences. You might also see the word a series. And you might even see a geometric series. A series, the most conventional use of the word series, means a sum of a sequence. So for example, this is a geometric sequence. A geometric series would be 90 plus negative 30, plus 10, plus negative 10/3, plus 10/9. So a general way to view it is that a series is the sum of a sequence. I just want to make that clear because that used to confuse But anyway, let's go back to the notion of a geometric sequence, and actually do a word problem that deals with one of these. So they're telling us that Anne goes bungee jumping off of a bridge above water. On the initial jump, the cord stretches by 120 feet. So on a1, our initial jump, the cord stretches by 120 feet. We could write it this way. We could write, jump, and then how much the cord stretches. So on the initial jump, on jump one, the cord stretches 120 feet." + }, + { + "Q": "At 8:56, he says the formula to find the 12th bounce is (120)(0.6)^n. I thought it was (120)(0.6)^(n - 1). I am kind of confused about that...", + "A": "In the video, he counts the zero bounce so you can subtract 1 from both sides and then they cancel out. ex. Jumps: a^n=120(0.6)^n-1 Bounces: a^n-1=120(0.6)^n-1 the -1 s cancel so: a^n=120(0.6)^n", + "video_name": "pXo0bG4iAyg", + "timestamps": [ + 536 + ], + "3min_transcript": "So you have 0.6 to the 0th power, and you've just got a 1 here. And that's exactly what happened on the first jump. Then on the second jump, you put a 2 minus 1, and notice 2 minus 1 is the first power, and we have exactly one 0.6 here. So I figured it was n minus 1 because when n is 2, we have one 0.6, when n is 3, we have two 0.6's multiplied by themselves. When n is 4, we have 0.6 to the third power. So whatever n is, we're taking 0.6 to the n minus 1 power, and of course we're multiplying that times 120. Now and the question they also ask us, what will be the rope stretch on the 12th bounce? And over here I'm going to use the calculator. and actually let me correct this a little bit. bounce, and we could call the jump the zeroth bounce. Let me change that. This isn't wrong, but I think this is where they're going with the problem. So you can view the initial stretch as the zeroth bounce. So instead of labeling it jump, let me label it bounce. So the initial stretch is the zeroth bounce, then this would be the first bounce, the second bounce, the third bounce. And then our formula becomes a lot simpler. Because if you said the stretch on nth bounce, then the formula just becomes 0.6 to the n times 120, right? On the zeroth bounce, that was our original stretch, you get 0.6 to the 0, that's 1 times 120. 0.6 times the previous stretch, or the previous bounce. So this has it in terms of bounces, which I think is what the questioner wants us to do. So what about the 12th bounce? Using this convention right there. So if we do the 12th bounce, let's just get our calculator out. We're going to have 120 times 0.6 to the 12th power. And hopefully we'll get order of operations right, because exponents take precedence over multiplication, so it'll just take the 0.6 to the 12th power only. And so this is equal to 0.26 feet. So after your 12th bounce, she's going to be barely moving. She's going to be moving about 3 inches on that 12th bounce." + }, + { + "Q": "at 4:32 why do we have to multiply 50 by 1 hour? I thought all we needed to do was divide 3600 by 50 . Also i did not get whether it was 72 km per second or 1 km per 72 seconds. Although i had some questions this was a fantastic video that triggered a much needed Eureka! moment!! Would i need to multiply 50 by 2 hr if a question said 20/km per 2 hrs or is that not mathematically correct to use 2 hours as a unit", + "A": "its 1km per 72 seconds. 1/72", + "video_name": "d5lcGCbV5cM", + "timestamps": [ + 272 + ], + "3min_transcript": "of kilometers per second. So how could we write 50 kilometers per hour, in terms of kilometers per second? Well it's always good, actually, as a first approximation, to just think about it. If you went this far in an hour, then the number of kilometers you go in a second, is that going to be less, or more? Well a second's a much, much shorter period of time. There's 3,600 seconds in an hour. So you're going to go 1/3,600 of this distance. But let's think about how we would actually work out with the units. Well, we want to get rid of this hours in the denominator. And the plural, obviously the grammar doesn't hold up with the algebra, but this could be hour or hours. So we could think about well, 1 hour-- I'll write an hour in the numerator that's going to cancel with this hour in the denominator. But we want it in terms of seconds. So 1 hour is equal to how many seconds? This is what I meant by saying that using dimensional analysis, which is what I'm doing right now, we can essentially manipulate these units, as we would traditionally do with a variable. So we have hours divided by hours. And so when we do the multiplication, we can multiply the numeric parts. So we have 50 times 1, divided by 3,600. Let me write that. 50 times 1 over 3,600. And then our units left are kilometers per second. Or I could say seconds. So we can play around with the plural and singular parts of it, but I'll just write it as kilometers per second. And so this is 50/3,600. And this fits our intuition. In a second, you're going to go 1/3,600 as far as you would go But let's actually think about what this is equal to. 50/3,600-- so this is going to be the same thing, as-- Let me just simplify it over here. So 50/3,600 is the same thing as 5/360, which is the same thing as-- let me write it this way-- 10/720. And I did that way because that makes it clear that that's the same thing as 1/72. So you could write this as, you're going, this is equal to 1/72 of a kilometer per second. Now I would claim that this is not so reasonable of units for this example right over here." + }, + { + "Q": "at 4:10 , just have a silly question: are addition and subtraction between a matrix and a scalar undefined?", + "A": "Yes, addition and subtraction between a scalar and a matrix (or even between matrices of different dimensions) is undefined. That is why previous to adding, the scalar is multiplied by the Identity Matrix, so that at the time of the addition, you are adding two matrices of the same size.", + "video_name": "rfm0wQObxjk", + "timestamps": [ + 250 + ], + "3min_transcript": "associated with it. Because if v is equal to 0, any eigenvalue will work for that. So normally when we're looking for eigenvectors, we start with the assumption that we're looking for non-zero vectors. So we're looking for vectors that are not equal to the 0 vector. So given that, let's see if we can play around with this equation a little bit and see if we can at least come up with eigenvalues maybe in this video. So we subtract Av from both sides, we get the 0 vector is equal to lambda v minus A times v. Now, we can rewrite v as-- v is just the same thing as the identity matrix times v, right? v is a member of Rn. The identity matrix n by n. You just multiply and we're just going to get v again. So if I rewrite v this way, at least on this part of the expression-- and let me swap sides-- so then I'll get lambda times-- instead of v I'll write the identity is equal to the 0 vector. Now I have one matrix times v minus another matrix times v. Matrix vector products, they have the distributive property. So this is equivalent to the matrix lambda times the identity matrix minus A times the vector v. And that's going to be equal to 0, right? This is just some matrix right here. And the whole reason why I made this substitution is so I can write this as a matrix vector product instead of just a scalar vector product. And that way I was able to essentially factor out the v and just write this whole equation as essentially, some matrix vector product is equal to 0. Now, in order-- if we assume that this is the case, and we're assuming-- remember, we're assuming that v So what does this mean? So we know that v is a member of the null space of this matrix right here. Let me write this down. v is a member of the null space of lambda I sub n minus A. I know that might look a little convoluted to you right now, but just imagine this is just some matrix B. It might make it simpler. This is just some matrix here, right? That's B. Let's make that substitution. Then this equation just becomes Bv is equal to 0. Now, if we want to look at the null space of this, the null space of B is all of the vectors x that are a member of Rn such that B times x is equal to 0. Well, v is clearly one of those guys, right? Because B times v is equal to 0." + }, + { + "Q": "it is 5:00 pm ET, why is the exponent 3/5 after you multiply 3 times 1/5 wouldn't you multiply the numerator and denominator of the exponent against the power of 3 and get 3/15 which can be reduced to 1/5, thouroughly confused over here!", + "A": "3 does not equal 3/3. 3 as a fraction = 3/1 Thus, 3 times 1/5 = 3/1 * 1/5 = 3/5 Hope this helps.", + "video_name": "Ht-YXje4R2g", + "timestamps": [ + 300 + ], + "3min_transcript": "Well, we could find a common denominator. It would be 10, so that's the same thing as-- actually let me just write it this way-- this is the same thing as 6 to the-- instead of 1/2, we can write it as 5/10. Plus 3/5 is the same thing as 6/10 power, which is the same thing-- and we deserve a little bit of a drum roll here, this wasn't that long of a problem-- 6 to the 11/10 power. I'll just write it all, 11/10 power. And so, that looks pretty simplified to me. I guess we're done." + }, + { + "Q": "At 2:45: Can 6^11/10 be re-written as 6^1/10?\n10/10 = 1 Leaving 1/10\n6^1 = 6 so the remaining 1/10 is left over.", + "A": "Actually, we can t change the exponent from 11/10 to 1/10 by subtracting one from the exponent. We could do this however: 6^(11/10) 11/10=10/10+1/10=1+1/10. 6^1=6. 6^(11/10) 6^(10/10+1/10) 6^(1+1/10) 6^1*6^(1/10) 6*6^(1/10) I hope this helps!", + "video_name": "Ht-YXje4R2g", + "timestamps": [ + 165 + ], + "3min_transcript": "Let's see if we can simplify 6 to the 1/2 power times the fifth root of 6 and all of that to the third power. And I encourage you to pause this video and try it on your own. So let me actually color code these exponents, just so we can keep track of them a little better. So that's the 1/2 power in blue. This is the fifth root here in magenta. And let's see. In green, let's think about this third power. So one way to think about this fifth root is that this is the exact same thing as raising this 6 to the 1/5 power, so let's write it like that. So this part right over here, we could rewrite as 6 to the 1/5 power, and then that whole thing gets raised to the third power. And of course, we have this 6 to the 1/2 power out here, 6 to the 1/2 power times all of this business right over here. and then raise that whole thing to another exponent? Well we've already seen in our exponent properties, that's the equivalent of raising this to the product of these two exponents. So this part right over here could be rewritten as 6 to the-- 3 times 1/5 is 3/5-- 6 to the 3/5 power. And of course, we're multiplying that times 6 to the 1/2 power. 6 to the 1/2 power times 6 to the 3/5 power. And now, if you're multiplying some base to this exponent and then the same base again to another exponent, we know that this is going to be the same thing. And actually we could put these equal signs the whole way, because these all equal each other. This is the same thing as 6 being raised to the 1/2 plus 3/5 power, 1/2 plus 3 over 5. Well, we could find a common denominator. It would be 10, so that's the same thing as-- actually let me just write it this way-- this is the same thing as 6 to the-- instead of 1/2, we can write it as 5/10. Plus 3/5 is the same thing as 6/10 power, which is the same thing-- and we deserve a little bit of a drum roll here, this wasn't that long of a problem-- 6 to the 11/10 power. I'll just write it all, 11/10 power. And so, that looks pretty simplified to me. I guess we're done." + }, + { + "Q": "ok so at 1:10 say got 4 but how did he get that out of 6?", + "A": "6 divided by 3 is 2, so 2 is 1/3 of 6. 4 is 2/3 of 6.", + "video_name": "6dyWKD_JPhI", + "timestamps": [ + 70 + ], + "3min_transcript": "The graph below contains the rectangle ABCP. Draw the image of ABCP under a dilation whose center is at P and a scale factor is 1 and 2/3. What are the lengths of the side AB and its image? So we're going to do a dilation centered at P. So if we're centering a dilation at P and its scale factor is 1 and 2/3, that means once we perform the dilation, every point is going to be 1 and 2/3 times as far away from P. Well P is 0 away from P, so its image is still going to be at P. So let's put that point right over there. Now point C is going to be 1 and 2/3 times as far as it is right now. So let's see, right now it is 6 away. It's at negative 3. And P, its x-coordinate is the same, but in the y direction, C is at negative 3. So it's 6 less. We want to be 1 and 2/3 times as far away. So what's 1 and 2/3 of 6? Well, 2/3 of 6 is 4, so it's going to be 6 plus 4. You're going to be 10 away. So 3 minus 10, that gets us to negative 7. So that gets us right over there. Now point A, right now it is 3 more in the horizontal direction than point P's x-coordinate. So we want to go 1 and 2/3 as far. So what is 1 and 2/3 times 3? Well that's going to be 3 plus 2/3 of 3, which is another 2. So that's going to be 5. So we're going to get right over there. Then we could complete the rectangle. And notice point B is now 1 and 2/3 times as far in the horizontal direction. It was 3 away in the horizontal direction, now it is 5 away from P's x-coordinate. And in the vertical direction, in the y direction, Now it is 1 and 2/3 times as far. It is 10 below P's y-coordinate. So then let's answer these questions. The length of segment AB-- well, we already saw that. That is, we're going from 3 to negative 3. That is 6 units long. And its image, well it's 1 and 2/3 as long. We see it over here. We're going from 3 to negative 7. 3 minus negative 7 is 10. It is 10 units long. We got it right." + }, + { + "Q": "I entered this limit (as an equation) onto the Desmos graphing calculator, and it came out similar to the graph in the video at 3:31, with one exception: At x=0, there was a straight line going from 0,2 to 0,1. Is this due to something in this limit, or is it just a glitch in the calculator?", + "A": "I d guess that the calculator defines 0/0 as 1. As it evaluated the function at 0, it got 1 instead of 2. The calculator basically calculates the coordinates of a lot of points and then connects the dots, thus causing the downward spike. That s my guess of what s happening, anyways.", + "video_name": "t7NvlTgMsO8", + "timestamps": [ + 211 + ], + "3min_transcript": "Get my calculator out. So I want to evaluate x squared over 1 minus cosine of x when x is equal to 0.1. Let me actually verify that I'm in radian mode, because otherwise, I might get a strange answer. So I am in radian mode. Let me evaluate it. So I'm going to have 0.1 squared divided by 1 minus cosine of 0.1, and this gets me 2.0016. And let's see, they want us to round to the nearest thousandth. So that would be 2.002. Type that in, 2.002. And so it looks like the limit is approaching 2. crossed 2.005 from 2.007 to 2.002. So let's check our answer, and we got it right. I always find it fun to visualize these things. And that's what a graphing calculator is good for. It can actually graph things. So let's graph this right over here. So go in to graph mode. Let me redefine my function here. So let's see, it's going to be x squared divided by 1 minus cosine of x. And then let me make sure that the range of my graph is right. So I'm zoomed in at the right the part that I care about. So let me go to the range. And let's see, I care about approaching x from the-- or approaching 0 from the positive direction, but as long as I see values around 0, I should be fine. So I could make my minimum x-value negative 1. Let me make my maximum x-value-- the maximum x-value here is 1, but just to get some space here I'll make this 1.5. So the x-scale is 1. y minimum, it seems like we're approaching 2. So the y max can be much smaller. Let's see, let me make y max 3. And now let's graph this thing. So let's see what it's doing. And actually it looks -- whether you're approaching from the positive direction or from the negative direction-- it looks like the value of the function approaches 2. But this problem, we're only caring about-- as we have x-values that are approaching 0 from values larger than 0. So this is the one-sided limit that we care about. But the 2 shows up right over here, as well." + }, + { + "Q": "In 2:07 can the hypotenuse be the butom side if the triangle is fliped?", + "A": "Yes there can be a hypotenuse no matter which way the triangle is flipped. The only property is that there should be a 90 degree angle in the triangle. The side opposite to that will be the hypotenuse. :)", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 127 + ], + "3min_transcript": "In this video we're going to get introduced to the Pythagorean theorem, which is fun on its own. But you'll see as you learn more and more mathematics it's one of those cornerstone theorems of really all of math. It's useful in geometry, it's kind of the backbone of trigonometry. You're also going to use it to calculate distances between points. So it's a good thing to really make sure we know well. So enough talk on my end. Let me tell you what the Pythagorean theorem is. So if we have a triangle, and the triangle has to be a right triangle, which means that one of the three angles in the triangle have to be 90 degrees. And you specify that it's 90 degrees by drawing that little box right there. So that right there is-- let me do this in a different color-- a 90 degree angle. And a triangle that has a right angle in it is called a right triangle. So this is called a right triangle. Now, with the Pythagorean theorem, if we know two sides of a right triangle we can always figure out the third side. And before I show you how to do that, let me give you one more piece of terminology. The longest side of a right triangle is the side opposite the 90 degree angle-- or opposite the right angle. So in this case it is this side right here. This is the longest side. And the way to figure out where that right triangle is, and kind of it opens into that longest side. That longest side is called the hypotenuse. And just so we always are good at identifying the hypotenuse, let me draw a couple of more right triangles. So let's say I have a triangle that looks like that. Let me draw it a little bit nicer. So let's say I have a triangle that looks like that. And I were to tell you that this angle right here is 90 degrees. In this situation this is the hypotenuse, because it is opposite the 90 degree angle. It is the longest side. Let me do one more, just so that we're good at recognizing the hypotenuse. So let's say that that is my triangle, and this is the 90 degree angle right there. And I think you know how to do this already. You go right what it opens into. That is the hypotenuse. That is the longest side. So once you have identified the hypotenuse-- and let's say that that has length C." + }, + { + "Q": "This doesn't have much to do with the video, but at 5:28, Sal says we take the positive square root of both sides. Is there a negative square root?", + "A": "Yes, for example, the positive square root of 25 is 5 and the negative square root is -5. When you square negative numbers, you get a positive answer, therefore the square root of a positive number will have both a positive and a negative.", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 328 + ], + "3min_transcript": "theorem tells us. So let's say that C is equal to the length of the hypotenuse. So let's call this C-- that side is C. Let's call this side right over here A. And let's call this side over here B. So the Pythagorean theorem tells us that A squared-- so the length of one of the shorter sides squared-- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared. Now let's do that with an actual problem, and you'll see that it's actually not so bad. So let's say that I have a triangle that looks like this. Let me draw it. Let's say this is my triangle. It looks something like this. And let's say that they tell us that this is the right angle. That this length right here-- let me do this in different length right here is 4. And they want us to figure out that length right there. Now the first thing you want to do, before you even apply the Pythagorean theorem, is to make sure you have your hypotenuse straight. You make sure you know what you're solving for. And in this circumstance we're solving for the hypotenuse. And we know that because this side over here, it is the side opposite the right angle. If we look at the Pythagorean theorem, this is C. This is the longest side. So now we're ready to apply the Pythagorean theorem. It tells us that 4 squared-- one of the shorter sides-- plus 3 squared-- the square of another of the shorter sides-- is going to be equal to this longer side squared-- the hypotenuse squared-- is going to be equal to C squared. And then you just solve for C. That is 16. And 3 squared is the same thing as 3 times 3. So that is 9. And that is going to be equal to C squared. Now what is 16 plus 9? It's 25. So 25 is equal to C squared. And we could take the positive square root of both sides. I guess, just if you look at it mathematically, it could be negative 5 as well. But we're dealing with distances, so we only care about the positive roots. So you take the principal root of both sides and you get 5 is equal to C. Or, the length of the longest side is equal to 5. Now, you can use the Pythagorean theorem, if we give you two of the sides, to figure out the third side no matter what the third side is. So let's do another one right over here. Let's say that our triangle looks like this." + }, + { + "Q": "10:09 instead of it being multiple perfect sqauares wouldnt yoiu just leave it alone and put n/a when finished because my teacher said that its no pyth. anymore unless it in a perfect square", + "A": "No you wouldn t. If it wasn t a perfect square, you would put it under the radical sign, and that would be your answer. Hope I helped! :)", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 609 + ], + "3min_transcript": "Those cancel out. On the left-hand side we're left with just a B squared is equal to-- now 144 minus 36 is what? 144 minus 30 is 114. And then you subtract 6, is 108. So this is going to be 108. So that's what B squared is, and now we want to take the principal root, or the positive root, of both sides. And you get B is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same So we have the square root of 108 is the same thing as the square root of 2 times 2 times-- well actually, I'm not done. 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so, we have a couple of perfect squares in here. Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there. And, you know, you wouldn't have to do all of this on paper. You could do it in your head. What is this? 2 times 2 is 4. 4 times 9, this is 36. So this is the square root of 36 times the square root of 3. The principal root of 36 is 6. So this simplifies to 6 square roots of 3. So the length of B, you could write it as the square root of 108, or you could say it's equal to 6 times the square root of 3. This is 12, this is 6. And the square root of 3, well this is going to be a 1 point something something. So it's going to be a little bit larger than 6." + }, + { + "Q": "At 10:35 did anyone else notice that that triangle wasn't a right triangle?", + "A": "No those are all right triangles, but are set at different angles.", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 635 + ], + "3min_transcript": "Those cancel out. On the left-hand side we're left with just a B squared is equal to-- now 144 minus 36 is what? 144 minus 30 is 114. And then you subtract 6, is 108. So this is going to be 108. So that's what B squared is, and now we want to take the principal root, or the positive root, of both sides. And you get B is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same So we have the square root of 108 is the same thing as the square root of 2 times 2 times-- well actually, I'm not done. 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so, we have a couple of perfect squares in here. Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there. And, you know, you wouldn't have to do all of this on paper. You could do it in your head. What is this? 2 times 2 is 4. 4 times 9, this is 36. So this is the square root of 36 times the square root of 3. The principal root of 36 is 6. So this simplifies to 6 square roots of 3. So the length of B, you could write it as the square root of 108, or you could say it's equal to 6 times the square root of 3. This is 12, this is 6. And the square root of 3, well this is going to be a 1 point something something. So it's going to be a little bit larger than 6." + }, + { + "Q": "At 5:25 Sal refers to the square root of 25 that is positive and calls it the principle root. Is that the name of the positive answer of a square root? Is there a name for the negative answer of the square root?", + "A": "The principle square root of a number is the positive square root. If you want to specify the negative answer of a square root its simply the negative square root You could also say the positive square root to identify the principle square root. The positive square root is used more often that the negative so it gets a special name", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 325 + ], + "3min_transcript": "theorem tells us. So let's say that C is equal to the length of the hypotenuse. So let's call this C-- that side is C. Let's call this side right over here A. And let's call this side over here B. So the Pythagorean theorem tells us that A squared-- so the length of one of the shorter sides squared-- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared. Now let's do that with an actual problem, and you'll see that it's actually not so bad. So let's say that I have a triangle that looks like this. Let me draw it. Let's say this is my triangle. It looks something like this. And let's say that they tell us that this is the right angle. That this length right here-- let me do this in different length right here is 4. And they want us to figure out that length right there. Now the first thing you want to do, before you even apply the Pythagorean theorem, is to make sure you have your hypotenuse straight. You make sure you know what you're solving for. And in this circumstance we're solving for the hypotenuse. And we know that because this side over here, it is the side opposite the right angle. If we look at the Pythagorean theorem, this is C. This is the longest side. So now we're ready to apply the Pythagorean theorem. It tells us that 4 squared-- one of the shorter sides-- plus 3 squared-- the square of another of the shorter sides-- is going to be equal to this longer side squared-- the hypotenuse squared-- is going to be equal to C squared. And then you just solve for C. That is 16. And 3 squared is the same thing as 3 times 3. So that is 9. And that is going to be equal to C squared. Now what is 16 plus 9? It's 25. So 25 is equal to C squared. And we could take the positive square root of both sides. I guess, just if you look at it mathematically, it could be negative 5 as well. But we're dealing with distances, so we only care about the positive roots. So you take the principal root of both sides and you get 5 is equal to C. Or, the length of the longest side is equal to 5. Now, you can use the Pythagorean theorem, if we give you two of the sides, to figure out the third side no matter what the third side is. So let's do another one right over here. Let's say that our triangle looks like this." + }, + { + "Q": "At the time of 5:40 he mentions the squre root. Can anyone explain what that is or show me a video of how to find the sqaure root", + "A": "A square root a number to the 1/2 power. like 25^1/2=5 because 5^2=25. (E.G. square root of 49 is 7 because 7^2 (7x7) is 49", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 340 + ], + "3min_transcript": "length right here is 4. And they want us to figure out that length right there. Now the first thing you want to do, before you even apply the Pythagorean theorem, is to make sure you have your hypotenuse straight. You make sure you know what you're solving for. And in this circumstance we're solving for the hypotenuse. And we know that because this side over here, it is the side opposite the right angle. If we look at the Pythagorean theorem, this is C. This is the longest side. So now we're ready to apply the Pythagorean theorem. It tells us that 4 squared-- one of the shorter sides-- plus 3 squared-- the square of another of the shorter sides-- is going to be equal to this longer side squared-- the hypotenuse squared-- is going to be equal to C squared. And then you just solve for C. That is 16. And 3 squared is the same thing as 3 times 3. So that is 9. And that is going to be equal to C squared. Now what is 16 plus 9? It's 25. So 25 is equal to C squared. And we could take the positive square root of both sides. I guess, just if you look at it mathematically, it could be negative 5 as well. But we're dealing with distances, so we only care about the positive roots. So you take the principal root of both sides and you get 5 is equal to C. Or, the length of the longest side is equal to 5. Now, you can use the Pythagorean theorem, if we give you two of the sides, to figure out the third side no matter what the third side is. So let's do another one right over here. Let's say that our triangle looks like this. Let's say this side over here has length 12, and let's say that this side over here has length 6. And we want to figure out this length right over there. Now, like I said, the first thing you want to do is identify the hypotenuse. And that's going to be the side opposite the right angle. We have the right angle here. You go opposite the right angle. The longest side, the hypotenuse, is right there. So if we think about the Pythagorean theorem-- that A squared plus B squared is equal to C squared-- 12 you could view as C. This is the hypotenuse. The C squared is the hypotenuse squared. So you could say 12 is equal to C. And then we could say that these sides, it doesn't matter whether you call one of them A or one of them B." + }, + { + "Q": "so i stopped at 5:48 and i was wondering why each equation has to be like squared such as his example 4 squared + 3 squared = c squared, how come? please no long answers im in sixth grade", + "A": "If you put squares on each side of the right triangle, the sum of the smaller squares would equal the largest square. Because the area of a square is (side)^2, and we are looking for the side, we square the sides.", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 348 + ], + "3min_transcript": "length right here is 4. And they want us to figure out that length right there. Now the first thing you want to do, before you even apply the Pythagorean theorem, is to make sure you have your hypotenuse straight. You make sure you know what you're solving for. And in this circumstance we're solving for the hypotenuse. And we know that because this side over here, it is the side opposite the right angle. If we look at the Pythagorean theorem, this is C. This is the longest side. So now we're ready to apply the Pythagorean theorem. It tells us that 4 squared-- one of the shorter sides-- plus 3 squared-- the square of another of the shorter sides-- is going to be equal to this longer side squared-- the hypotenuse squared-- is going to be equal to C squared. And then you just solve for C. That is 16. And 3 squared is the same thing as 3 times 3. So that is 9. And that is going to be equal to C squared. Now what is 16 plus 9? It's 25. So 25 is equal to C squared. And we could take the positive square root of both sides. I guess, just if you look at it mathematically, it could be negative 5 as well. But we're dealing with distances, so we only care about the positive roots. So you take the principal root of both sides and you get 5 is equal to C. Or, the length of the longest side is equal to 5. Now, you can use the Pythagorean theorem, if we give you two of the sides, to figure out the third side no matter what the third side is. So let's do another one right over here. Let's say that our triangle looks like this. Let's say this side over here has length 12, and let's say that this side over here has length 6. And we want to figure out this length right over there. Now, like I said, the first thing you want to do is identify the hypotenuse. And that's going to be the side opposite the right angle. We have the right angle here. You go opposite the right angle. The longest side, the hypotenuse, is right there. So if we think about the Pythagorean theorem-- that A squared plus B squared is equal to C squared-- 12 you could view as C. This is the hypotenuse. The C squared is the hypotenuse squared. So you could say 12 is equal to C. And then we could say that these sides, it doesn't matter whether you call one of them A or one of them B." + }, + { + "Q": "At 5:32, Sal says it could be negative five as well. But isn't it impossible for a length to equal a nonpositive number? Even though he said, that in this case, we only need the positive root, wouldn't that imply that it could be negative in other cases? Could he be speaking only of geometrical properties? Or is he just absent-mindedly pointing out the mathematical property of equality that a square root of a number can be both negative and positive?", + "A": "There are two numbers you can square to get 25. You can square 5: (5)^2=25 or -5: (-5)^2=25. So there are actually two answersfor the square root of 25: 5 AND -5. But as he pointed out, we re dealing with distances, so we know that the answer can t be -5. That s why he said we re taking the positive square root.", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 332 + ], + "3min_transcript": "length right here is 4. And they want us to figure out that length right there. Now the first thing you want to do, before you even apply the Pythagorean theorem, is to make sure you have your hypotenuse straight. You make sure you know what you're solving for. And in this circumstance we're solving for the hypotenuse. And we know that because this side over here, it is the side opposite the right angle. If we look at the Pythagorean theorem, this is C. This is the longest side. So now we're ready to apply the Pythagorean theorem. It tells us that 4 squared-- one of the shorter sides-- plus 3 squared-- the square of another of the shorter sides-- is going to be equal to this longer side squared-- the hypotenuse squared-- is going to be equal to C squared. And then you just solve for C. That is 16. And 3 squared is the same thing as 3 times 3. So that is 9. And that is going to be equal to C squared. Now what is 16 plus 9? It's 25. So 25 is equal to C squared. And we could take the positive square root of both sides. I guess, just if you look at it mathematically, it could be negative 5 as well. But we're dealing with distances, so we only care about the positive roots. So you take the principal root of both sides and you get 5 is equal to C. Or, the length of the longest side is equal to 5. Now, you can use the Pythagorean theorem, if we give you two of the sides, to figure out the third side no matter what the third side is. So let's do another one right over here. Let's say that our triangle looks like this. Let's say this side over here has length 12, and let's say that this side over here has length 6. And we want to figure out this length right over there. Now, like I said, the first thing you want to do is identify the hypotenuse. And that's going to be the side opposite the right angle. We have the right angle here. You go opposite the right angle. The longest side, the hypotenuse, is right there. So if we think about the Pythagorean theorem-- that A squared plus B squared is equal to C squared-- 12 you could view as C. This is the hypotenuse. The C squared is the hypotenuse squared. So you could say 12 is equal to C. And then we could say that these sides, it doesn't matter whether you call one of them A or one of them B." + }, + { + "Q": "How is it possible that the hypotenuse is always opposite of the right angle? Also at 0:40 he says has to be right triangle, is it possible to do it with any other kinds of triangles?", + "A": "because hypotenuse is defined as the side which is opposite to the perpendicular! and it is a universal fact that it would be the largest side of a right angled triangle. No,you can t Pythagoras theorem is based on right triangles and is the base for trigonometry which is studied in higher classes", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 40 + ], + "3min_transcript": "In this video we're going to get introduced to the Pythagorean theorem, which is fun on its own. But you'll see as you learn more and more mathematics it's one of those cornerstone theorems of really all of math. It's useful in geometry, it's kind of the backbone of trigonometry. You're also going to use it to calculate distances between points. So it's a good thing to really make sure we know well. So enough talk on my end. Let me tell you what the Pythagorean theorem is. So if we have a triangle, and the triangle has to be a right triangle, which means that one of the three angles in the triangle have to be 90 degrees. And you specify that it's 90 degrees by drawing that little box right there. So that right there is-- let me do this in a different color-- a 90 degree angle. And a triangle that has a right angle in it is called a right triangle. So this is called a right triangle. Now, with the Pythagorean theorem, if we know two sides of a right triangle we can always figure out the third side. And before I show you how to do that, let me give you one more piece of terminology. The longest side of a right triangle is the side opposite the 90 degree angle-- or opposite the right angle. So in this case it is this side right here. This is the longest side. And the way to figure out where that right triangle is, and kind of it opens into that longest side. That longest side is called the hypotenuse. And just so we always are good at identifying the hypotenuse, let me draw a couple of more right triangles. So let's say I have a triangle that looks like that. Let me draw it a little bit nicer. So let's say I have a triangle that looks like that. And I were to tell you that this angle right here is 90 degrees. In this situation this is the hypotenuse, because it is opposite the 90 degree angle. It is the longest side. Let me do one more, just so that we're good at recognizing the hypotenuse. So let's say that that is my triangle, and this is the 90 degree angle right there. And I think you know how to do this already. You go right what it opens into. That is the hypotenuse. That is the longest side. So once you have identified the hypotenuse-- and let's say that that has length C." + }, + { + "Q": "At 9:35 it says the term perfect square. What does that mean?", + "A": "By perfect square, you mean to say that if the square root of a number is taken the result would be a whole number. For example, 4,9,16 and 25 are perfect squares, since if you take the square root of those numbers, you would get 2,3,4 and 5, which are whole numbers. Hoped it helped! :)", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 575 + ], + "3min_transcript": "Those cancel out. On the left-hand side we're left with just a B squared is equal to-- now 144 minus 36 is what? 144 minus 30 is 114. And then you subtract 6, is 108. So this is going to be 108. So that's what B squared is, and now we want to take the principal root, or the positive root, of both sides. And you get B is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same So we have the square root of 108 is the same thing as the square root of 2 times 2 times-- well actually, I'm not done. 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so, we have a couple of perfect squares in here. Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there. And, you know, you wouldn't have to do all of this on paper. You could do it in your head. What is this? 2 times 2 is 4. 4 times 9, this is 36. So this is the square root of 36 times the square root of 3. The principal root of 36 is 6. So this simplifies to 6 square roots of 3. So the length of B, you could write it as the square root of 108, or you could say it's equal to 6 times the square root of 3. This is 12, this is 6. And the square root of 3, well this is going to be a 1 point something something. So it's going to be a little bit larger than 6." + }, + { + "Q": "At around the 10:00 minute mark he starts talking about principal roots and stuff. Can someone help explain that to me?", + "A": "Principle root simply means that the root is positive, not negative.", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 600 + ], + "3min_transcript": "Those cancel out. On the left-hand side we're left with just a B squared is equal to-- now 144 minus 36 is what? 144 minus 30 is 114. And then you subtract 6, is 108. So this is going to be 108. So that's what B squared is, and now we want to take the principal root, or the positive root, of both sides. And you get B is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same So we have the square root of 108 is the same thing as the square root of 2 times 2 times-- well actually, I'm not done. 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so, we have a couple of perfect squares in here. Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there. And, you know, you wouldn't have to do all of this on paper. You could do it in your head. What is this? 2 times 2 is 4. 4 times 9, this is 36. So this is the square root of 36 times the square root of 3. The principal root of 36 is 6. So this simplifies to 6 square roots of 3. So the length of B, you could write it as the square root of 108, or you could say it's equal to 6 times the square root of 3. This is 12, this is 6. And the square root of 3, well this is going to be a 1 point something something. So it's going to be a little bit larger than 6." + }, + { + "Q": "at 7:20 couldn't you just do a+b=c instead of A2+B2=C2.", + "A": "You can t just use a + b = c because you are trying to take the square root of both sides and just eliminate the squares. But the square root of both sides of a^2 + b^2 = c^2 gives you sqrt(a^2 + b^2) on the left and you have to do what s inside the parentheses first.", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 440 + ], + "3min_transcript": "That is 16. And 3 squared is the same thing as 3 times 3. So that is 9. And that is going to be equal to C squared. Now what is 16 plus 9? It's 25. So 25 is equal to C squared. And we could take the positive square root of both sides. I guess, just if you look at it mathematically, it could be negative 5 as well. But we're dealing with distances, so we only care about the positive roots. So you take the principal root of both sides and you get 5 is equal to C. Or, the length of the longest side is equal to 5. Now, you can use the Pythagorean theorem, if we give you two of the sides, to figure out the third side no matter what the third side is. So let's do another one right over here. Let's say that our triangle looks like this. Let's say this side over here has length 12, and let's say that this side over here has length 6. And we want to figure out this length right over there. Now, like I said, the first thing you want to do is identify the hypotenuse. And that's going to be the side opposite the right angle. We have the right angle here. You go opposite the right angle. The longest side, the hypotenuse, is right there. So if we think about the Pythagorean theorem-- that A squared plus B squared is equal to C squared-- 12 you could view as C. This is the hypotenuse. The C squared is the hypotenuse squared. So you could say 12 is equal to C. And then we could say that these sides, it doesn't matter whether you call one of them A or one of them B. Let's say A is equal to 6. And then we say B-- this colored B-- is equal to question mark. And now we can apply the Pythagorean theorem. A squared, which is 6 squared, plus the unknown B squared is equal to the hypotenuse squared-- is equal to C squared. Is equal to 12 squared. And now we can solve for B. And notice the difference here. Now we're not solving for the hypotenuse. We're solving for one of the shorter sides. In the last example we solved for the hypotenuse. We solved for C. So that's why it's always important to recognize that A squared plus B squared plus C squared, C is the length So let's just solve for B here. So we get 6 squared is 36, plus B squared, is equal to 12 squared-- this 12 times 12-- is 144." + }, + { + "Q": "at 4:22, why does he cross out the 2s?", + "A": "Do u know why in the video he subtracted 2x and not 5x", + "video_name": "f15zA0PhSek", + "timestamps": [ + 262 + ], + "3min_transcript": "" + }, + { + "Q": "At 1:16 what exactly does he mean by 7 away from 0, what if it was negative?", + "A": "The simplest way to explain absolute value is forget the sign . Thus, |-7| = |7| = 7", + "video_name": "hKkBlcnU9pw", + "timestamps": [ + 76 + ], + "3min_transcript": "Let's do some examples comparing absolute values. So let's say we were to ask ourselves how the absolute value of negative 9, I should say, how that compares to the absolute value of-- let me think of a good number-- let's say the absolute value of negative 7. So let's think about this a little bit, and let's think about what negative 9 looks like, or where it is on the number line, where negative 7 is on the number line. Let's look at what the absolute values mean, and then we should probably be able to do this comparison. So there's a couple of ways to think about it. One is you could draw them on the number line. So if this is 0, if this is negative 7, and then this is negative 9 right over here. Now, when you take the absolute value of a number, you're really saying how far is that number from 0, whether it's to the left or to the right of 0. So, for example, negative 9 is 9 to the left of 0. This evaluates to 9, Negative 7 is exactly 7 to the left of 0. So the absolute value of negative 7 is positive 7. And so if you were to compare 9 and 7, this is a little bit more straightforward. 9 is clearly greater than 7. And if you ever get confused with the greater than or less than symbols, just remember that the symbol is larger on the left-hand side. So that's the greater than side. If I were to write this-- and this is actually also a true statement. If you took these without the absolute value signs, it is also true that negative 9 is less than negative 7. Notice the smaller side is on the smaller number. And so that's the interesting thing. Negative 9 is less than negative 7, but their absolute value, since negative 9 is further to the left of 0, it is-- the absolute value than the absolute value of negative 7. Another way to think about it is if you take the absolute value of a number, it's really just going to be the positive version of that number. So if you took the absolute value of 9, that equals 9. Or the absolute value of negative 9, that is also equal to 9. Well, when you think of it visually, that's because both of these numbers are exactly 9 away from 0. This is 9 to the right of 0, and this is 9 to the left of 0. Let's do a few more of these. So let's say that we wanted to compare the absolute value of 2 to the absolute value of 3. Well, the absolute value of a positive number is just going to be that same value. 2 is two to the right of 0, so this is just going to evaluate to 2. And then the absolute value of 3, that's just going to evaluate to 3. It's actually pretty straightforward. So 2 is clearly the smaller number here. And so we clearly get 2 is less than 3," + }, + { + "Q": "At ~8:25 Sal says that sin(x) reflected over the y-axis is equal to sin(-x). It looks to me that you could just as correctly said that sin(x) reflected over the x-axis is equal to sin(-x). Is this right?", + "A": "The sine function is a member of a special family of functions we call odd functions. If f(x) is odd, f(-x)=-f(x). You are right about your statement, because it is another property of odd functions.", + "video_name": "0zCcFSO8ouE", + "timestamps": [ + 505 + ], + "3min_transcript": "well it's common sense the amplitude here was 1 but now you're swaying from that middle position twice as far because you're multiplying by 2 Now let's go back to sin(x) and let's change it in a different way Let's graph sin(-x) so now let me once again put some graph paper here And now my goal is to graph sin(-x) y=sin(-x) so at least for the time being I've got rid of that 2 there and I'm just going straight from sin(x) to sin(-x) So let's think about how the values are going to work out So when x is 0 this is still going to be sin(0) which is 0 But then what as x increases, what happens when x is \u03c0/2 we're going to have to multiply by this negative so when x is \u03c0/2 we're really taking sin(-\u03c0/2) but what's sin(-\u03c0/2) but we can see over here here it's -1 It's - 1 and then when x = \u03c0 well sin(-\u03c0) we see this is 0 When x is 3\u03c0/2 well it's going to be sin(-3\u03c0/2) which is 1 Once again when x is 2\u03c0 it's going to be sin(-2\u03c0) is 0 So notice what was happening as I was trying to graph between 0 and 2\u03c0 I kept referring to the points in the negative direction so you can imagine taking this negative side right over here between 0 and -2\u03c0 and then flipping it over to get this one right over here that's what that -x seems to do you say when x = -\u03c0/2 where you have the negative in front of it so it's going to be sin(\u03c0/2) so it's going to be equal to 1 and you can flip this over the y-axis so essentially what we have done is we have flipped it we have reflected the graph of sin(x) over the y-axis So we have reflected it over the y-axis This is the y-axis so hopefully you see that reflection that's what that -x has done So now let's think about kind of the combo Having the 2 out the front and the -x right over there so let me put the graph on the axis there one more time And now let's try to do what was asked of us" + }, + { + "Q": "At 1:27, Sal says that if you take away 2 x's from one side, that side will go up; it has less weight. But wouldn't that side go down if x was a negative number?", + "A": "Yeah, but you can t have negative mass.", + "video_name": "Ye13MIPv6n0", + "timestamps": [ + 87 + ], + "3min_transcript": "So we have our scale again. And we've got some masses on the left hand side and some masses on the right hand side. And we see that our scale is balanced. We have the same total mass on the left hand side that we have on the right hand side. Instead of labeling the mystery masses as question mark, I've labeled them all x. And since they all have an x on it, we know that each of these have the same mass. But what I'm curious about is, what is that mass? What is the mass of each of these mystery masses, I guess we could say? And so I'll let think about that for a second. How would you figure out what this x value actually is? How many kilograms is the mass of each of these things? What could you do to either one or both sides of this scale? I'll give you a few seconds to think about that. So you might be tempted to say, well if I could end up with just one mystery mass on the left hand side, and if I keep my scale balanced, then that thing's going to be equal to whatever I have on the right hand side. And that part would actually be a true statement. on the left hand side, you might say, well why don't I just remove two of them? You might just say, well why don't I just remove-- let me do it a good color for removing-- why don't I just remove that one and that one? And then I'll just be left with that right over there. But if you just removed these two, then the left hand side is going to become lighter or it's going to have a lower mass than the right hand side. So it's going to move up and the right hand side is going to move down. And then you might say, OK, I understand. Whatever I have to do to the left hand side, I have to do to the right hand side in order to keep my scale balanced. So you might say, well why don't I remove two of these mystery masses from the right hand side? But that's a problem too because you don't know what this mystery mass is. You could try to remove two from this, but how many of these blocks represent a mystery mass? We actually don't know. But you might then say, well let's see, I've got three of these things here. If I essentially multiply what I have here by 1/3 and if I only leave a 1/3 of the stuff here, then the scale should be balanced. If this has the total mass as this, then 1/3 of this total mass is going to be the same thing as 1/3 of that total mass. So let's just keep only 1/3 of this here. So that's the equivalent to multiplying by 1/3. So if we're only going to keep 1/3 there, we're going to be left with only one of the masses. And if we only keep 1/3 here, let's see, we have one, two, three, four, five, six, seven, eight, nine masses. If we multiply this by 1/3, or if we only keep 1/3 of it there, 1/3 times 9 is 3. So we're going to remove these . And so we have 1/3 of what we originally had on the right hand side and 1/3 of what we originally had on the left hand side. And they will be balanced because we took 1/3 of the same total masses. And so what you're left with is just one of these mystery masses, this x thing right over here, whatever x might be." + }, + { + "Q": "At 6:25, why did he plug in 2 and 4 into the original equation to find the minimum?? Shouldn't he have just picked one number??", + "A": "The derivative is a second degree polynomial thus it is a parabola .......... when sal found its two roots at 1 and 3 ........ it is understood that the vertex of parabola will exactly be between them because the symmetry of parabola.... i.e. x=2 ....and for y value he plugged x=2 .....3(2)^2-12(2)+9.......3(4)-12(2)+9....", + "video_name": "SE1ltVuE5yM", + "timestamps": [ + 385 + ], + "3min_transcript": "Let's see. What two numbers, when you take a product, get 3, and when you add them, you get negative 4? Well, that's going to be t minus 3 times t minus 1 is equal to 0. How can this expression be equal to 0? Well if either of these are equal to 0, if either t minus 3 is 0 or t minus 1 is 0, it's going to be equal to 0. So t could be equal to 3, or t could be equal to 1. If t is 3 or t is 1, either of these are equal to 0, or this entire expression up here is going to be equal to 0. And since our coefficient on the t squared term is positive, we know this is going to be an upward opening parabola. So let's see if we can plot velocity as a function of time. So that is my velocity axis. This right over here is my time axis. And let's say this is 1 times 1 second, or I'm assuming this is in seconds-- 2, 3, 4. just because 1 and 3 are significant-- 1, 2, and 3. And they're not going to be-- I'm going to squash to the vertical scale a little bit. But this right over here, let's say that is 9, a velocity of 9. And so when t equals 0, our velocity is 9. When t equals 1, then our velocity is going to be 0. We get that right over here. 3 minus 12 plus 9, that's 0. And when t is equal to 3 our velocity is 0 again. Our vertex is going to be right in between those, when t is equal to 2-- right in between these two 0's. And we could figure out what that velocity is if we like. It's going to be 3 times 4 minus 12 times 2 plus 9. So what is that? That's 12 minus 24 plus 9. So that is negative 12 plus 9. Did I do that-- 12, yep, negative 3. So you're going to be-- negative 3 might be-- that's 9, so that's positive. So it might be something like this. So the graph of our velocity as a function of time is going to look something like this. And we only care about positive time. It's going to look something like this. So let's think. Remember, this is velocity. This is our velocity as a function of time. Now let's think about when is the velocity less than 0 and the acceleration is less than 0? So let's think about this case right over here? When is this the case? Both of them are going to be less than 0. Well, velocity is the less than 0 over this entire interval, this entire magenta interval. But the acceleration isn't less than 0 that entire time. Remember, the acceleration is the rate of change of velocity." + }, + { + "Q": "At 2:31, didn't Sal do the dot inaccurately?", + "A": "It is off by just a little, but this is just due to human error and the fact that he didn t have a grid to draw on. It is implied that the dot is at x = 9.585 s and y = 100 m.", + "video_name": "EKvHQc3QEow", + "timestamps": [ + 151 + ], + "3min_transcript": "Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about. Instantaneous rates of change. Differential calculus. Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis I'll have distance. I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance. And in this axis, we'll say time. but I'll just say x is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way, his average speed is just going to be his change in distance over his change in time. And using the variables that are over here, we're saying y is distance. So this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you This is the slope between these two points. If I have a line that connects these two points, this is the slope of that line. The change in distance is this right over here. Change in y is equal to 100 meters. And our change in time is this right over here. So our change in time is equal to 9.58 seconds. We started at 0, we go to 9.58 seconds. Another way to think about it, the rise over the run you might have heard in your algebra class. It's going to be 100 meters over 9.58 seconds. So this is 100 meters over 9.58 seconds. And the slope is essentially just rate of change, or you could view it as the average rate of change between these two points. And you'll see, if you even just follow the units, it gives you units of speed here. It would be velocity if we also specified the direction. And we can figure out what that is, let me get the calculator out." + }, + { + "Q": "People say that we see math in our everyday lives -- and while I understand how this concept applies to beginning math, pre-algebra, algebra, and trig, how does this apply to calculus? At 1:01, Sal says that differential calculus is all about finding instantaneous rate of change, but is that the only \"everyday use\"? Or is calculus simply a concept that is used in other subjects, or even professions, like engineering?\n\nThanks!", + "A": "I think the best way to find a good answer to this question is to just keep watching the videos! If you attend college for any engineering discipline, you have to learn calculus before you even begin learning the specifics of your discipline (whether it be mechanical, electrical, civil, computer, computer science, etc...). The best way to understand what every day things calculus will enable you to do is to learn calculus and start doing incredible things every day :-)", + "video_name": "EKvHQc3QEow", + "timestamps": [ + 61 + ], + "3min_transcript": "This is a picture of Isaac Newton, super famous British mathematician and physicist. This is a picture of a Gottfried Leibnitz, super famous, or maybe not as famous, but maybe should be, famous German philosopher and mathematician, and he was a contemporary of Isaac Newton. These two gentlemen together were really the founding fathers of calculus. And they did some of their-- most of their major work in the late 1600s. And this right over here is Usain Bolt, Jamaican sprinter, whose continuing to do some of his best work in 2012. And as of early 2012, he's the fastest human alive, and he's probably the fastest human that has ever lived. And you might have not made the association with these three You might not think that they have a lot in common. But they were all obsessed with the same fundamental question. And this is the same fundamental question that differential calculus addresses. And the question is, what is the instantaneous rate of change of something? Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about. Instantaneous rates of change. Differential calculus. Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis I'll have distance. I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance. And in this axis, we'll say time. but I'll just say x is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way, his average speed is just going to be his change in distance over his change in time. And using the variables that are over here, we're saying y is distance. So this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you" + }, + { + "Q": "At 6:00 I am wondering what happened to miles/Meters/ they got confused in my mind. Which one is using?", + "A": "If you re asking about how he got from 10.4 m/s to 23.5 mi/hr, ... 10.4 meters / second, when converted to miles/hour, is roughly 23.5 miles/hour. Since most of us (in the USA) drive in cars that have speedometers reading out in miles/hours, it s easier for some to be able to relate to 23 miles per hour, versus 10.4 m/sec. However, they are equivalent speeds (velocities), just one uses metric (SI) units, and the other uses US units.", + "video_name": "EKvHQc3QEow", + "timestamps": [ + 360 + ], + "3min_transcript": "So we're going 100 meters in the 9.58 seconds. So it's 10.4, I'll just write 10.4, I'll round to 10.4. So it's approximately 10.4, and then the units are meters per second. And that is his average speed. And what we're going to see in a second is how average speed is different than instantaneous speed. How it's different than what the speed he might be going at any given moment. And just to have a concept of how fast this is, let me get the calculator back. This is in meters per second. If you wanted to know how many meters he's going in an hour, well there's 3,600 seconds in an hour. So he'll be able to go this many meters 3,600 times. So that's how many meters he can, if he were able to somehow keep up that speed in an hour. This is how fast he's going meters per hour. And then, if you were to say how many miles per hour, but roughly 1600 meters per mile. So let's divide it by 1600. And so you see that this is roughly a little over 23, about 23 and 1/2 miles per hour. So this is approximately, and I'll write it this way-- this is approximately 23.5 miles per hour. And relative to a car, not so fast. But relative to me, extremely fast. Now to see how this is different than instantaneous velocity, let's think about a potential plot of his distance relative to time. He's not going to just go this speed immediately. He's not just going to go as soon as the gun fires, he's not just going to go 23 and 1/2 miles per hour all the way. He's going to accelerate. So at first he's going to start off going a little bit slower. So the slope is going to be a little bit lot lower than the average slope. He's going to go a little bit slower, then he's going to start accelerating. And so his speed, and you'll see the slope here And then maybe near the end he starts tiring off a little bit. And so his distance plotted against time might be a curve that looks something like this. And what we calculated here is just the average slope across this change in time. What we could see at any given moment the slope is actually different. In the beginning, he has a slower rate of change of distance. Then over here, then he accelerates over here, it seems like his rate of change of distance, which would be roughly-- or you could view it as the slope of the tangent line at that point, it looks higher than his average. And then he starts to slow down again. When you average it out, it gets to 23 and 1/2 miles per hour. And I looked it up, Usain Bolt's instantaneous velocity, his peak instantaneous velocity, is actually closer to 30 miles per hour. So the slope over here might be 23 whatever miles per hour. But the instantaneous, his fastest point in this 9.58 seconds is closer to 30 miles per hour." + }, + { + "Q": "At 2:04 shouldn't it be -1 x -4 = 4 and not 5?? so the real answer should be x^2+3 right?", + "A": "Correct. This is a known problem and as such has a pop-up box in the lower right-hand corner to say so. It appears on screen shortly after Sal s mistake.", + "video_name": "KvMyZY9upuA", + "timestamps": [ + 124 + ], + "3min_transcript": "- [Instructor] We're told that f of x is equal to two x times the square root of five minus four. And we're also told that g of x is equal to x squared plus two x times the square root of five minus one. And they want us to find g minus f of x. So pause this video, and see if you can work through that on your own. So the key here is to just realize what this notation means. G minus f of x is the same thing as g of x minus f of x. And so, again, if this was helpful to you, once again I encourage you to pause the video. All right, now let's work through this again. So this is going, or, I guess the first time, but now that we know that this is equal to g of x minus f of x. So what is g of x? Well, that's the same thing as x squared plus two x times the square root of five minus one. And what is f of x? Well, it's going to be two x And we are subtracting f of x from g of x. So let's subtract, this is f of x, from g of x. And so now it's just going to be a little bit of algebraic simplification. So this is going to be equal to, this is equal to x squared plus two x times the square root of five minus one. And now we just have to distribute this negative sign. So negative one times two x times the square root of five is, we're gonna have minus two x times the square root of five. And then the negative of negative four is positive four. Now let's see if we can simplify this some. So this is going to be equal to, we only have one x squared term, so that's that one there. So we have x squared. Now let's see, we have two x times the square root of five. And then we have another, oh, and then we subtract two x times the square root of five. So these two cancel out with each other. So those cancel out. And then we have minus one plus four. So if we have negative one and then we add four to it, So if we just fact, if we take this and this into consideration, four minus one is going to be equal to three, and we're done. That's what g minus f of x is equal to, x squared plus three." + }, + { + "Q": "At 0:23 why did you put diffret divison structures?", + "A": "because if you don t know hat problem by heart before you do it its easier to put it into that structure", + "video_name": "AjYil74WrVo", + "timestamps": [ + 23 + ], + "3min_transcript": "In the United States, 13 out of every 20 cans are recycled. What percent of cans are recycled? So 13 out of every 20 are recycled. So 13/20, or 13 over 20, could also be viewed as 13 divided 20, or 13 divided by 20. And if we do this, we'll get a decimal, and it's fairly straightforward to convert that decimal into a percentage. So 13 divided 20. We have the smaller number in this case being divided by the larger number. So we're going to get a value less than 1. Since we're going to get a value less than 1, let's put a decimal right over here. And let's add a couple of zeroes, as many zeroes as we would need. And we could say, hey, look, 20 goes into 13 zero times. 0 times 20 is 0. And then 13 minus 0 is 13. Now you bring down a 0. 20 goes into 130. So 6 times 20 is 120. So it's going to six times. 6 times 20 is 120. You subtract. You get a 10. Let's bring down another 0. 20 goes into 100 five times. 5 times 20 is 100. And we are done. So this, written as a decimal, is 0.65. So as a decimal, it's 0.65. And if you want to write it as a percentage you essentially multiply this number by 100. Or another way you could say is you shift the decimal point over two spots to the right. So this is going to be equal to 65%. Now, there's another way you could have done it. You could have said, look, percent literally means per hundred. So 13 out of 20 is going to be equal to what over 100? To go from 20 to 100, you would multiply by 5. So let's multiply the numerator by 5 as well. And 13 times 5, let's see, that's 15 plus 50, which is 65. So this would have been a faster way to do it, especially if you recognize it's pretty easy to go from 20 to 100. You multiply it by 5. So we would do the same thing with the 13. And so you would get 65/100, which is the same thing as 65 per-- let me write this percent symbol-- 65%. And just a reminder, percent literally means per hundred, 65 per hundred, 65%." + }, + { + "Q": "3:09 why is y zero?", + "A": "To find the two intercepts, you have to set x = 0 (to find the y intercept) and y = 0 (to find the x intercept). He is just finding the x intercept at this point.", + "video_name": "6CFE60iP2Ug", + "timestamps": [ + 189 + ], + "3min_transcript": "And in point-slope form, if you know that some, if you know that there's an equation where the line that represents the solutions of that equation has a slope M. Slope is equal to M. And if you know that X equals, X equals A, Y equals B, satisfies that equation, then in point-slope form you can express the equation as Y minus B is equal to M times X minus A. This is point-slope form and we do videos on that. But what I really want to get into in this video is another form. And it's a form that you might have already seen. And that is standard form. Standard. Standard form. And standard form takes the shape of AX plus BY is equal to C, And what I want to do in this video, like we've done in the ones on point-slope and slope-intercept is get an appreciation for what is standard form good at and what is standard form less good at? So let's give a tangible example here. So let's say I have the linear equation, it's in standard form, 9X plus 16Y is equal to 72. And we wanted to graph this. So the thing that standard form is really good for is figuring out, not just the y-intercept, y-intercept is pretty good if you're using slope-intercept form, but we can find out the y-intercept pretty clearly from standard form and the x-intercept. The x-intercept isn't so easy to figure out from these other forms right over here. So how do we do that? Well to figure out the x and y-intercepts, let's just set up a little table here, X comma Y, and so the x-intercept is going to happen when Y is equal to zero. when X is equal to zero. So when Y is zero, what is X? So when Y is zero, 16 times zero is zero, that term disappears, and you're left with 9X is equal to 72. So if nine times X is 72, 72 divided by nine is eight. So X would be equal to eight. So once again, that was pretty easy to figure out. This term goes away and you just have to say hey, nine times X is 72, X would be eight. When Y is equal to zero, X is eight. So the point, let's see, Y is zero, X is one, two, three, four, five, six, seven, eight. That's this point, that right over here. This point right over here is the x-intercept. When we talk about x-intercepts we're referring to the point where the line actually intersects the x-axis. Now what about the y-intercept? Well, we said X equals zero, this disappears. And we're left with 16Y is equal to 72. And so we could solve," + }, + { + "Q": "The method at 7:29 is nice but I can't make it work for something like:\n(4x^2 + y^2)^2\n\nI end up with 16x^2 + 8x^2 * 2y^2 + y ^4 but it should be 16x^2 + 8x^2 * y^2 + y ^4", + "A": "_4x^2 + _______ + y^2 x 4x^2 + _______ + y^2 --------------------------------- _______4(x^2)(y^2) + y^4 + 16x^4 + 4(x^2)(y^2) ---------------------------------- 16x^4 + 8(x^2)(y^2) + y^4", + "video_name": "fGThIRpWEE4", + "timestamps": [ + 449 + ], + "3min_transcript": "And soon you're going to get used to this. We can do it in one step. You're actually multiplying every term in this one by every term in that one, and we'll figure out faster ways But I really want to show you the idea here. So what's this going to equal? This is going to equal x squared. This right here is going to be minus 3x. This is going to be plus 2x. And then this right here is going to be minus 6. And so this is going to be x squared minus 3 of something, plus 2 of something, that's minus 1 of that something. Minus x, minus 6. We've multiplied those two. Now before we move on and do another problem, I want to show you that you can kind of do this in your head as well. You don't have to go through all of these steps. I just want to show you really that this is just the distributive property. The fast way of doing it, if you had x minus 3, times x plus 2, you literally just want to multiply every term So you'd say, this x times that x, so you'd have x squared. Then you'd have this x times that 2, so plus 2x. Then you'd have this minus 3 times that x, minus 3x. And then you have the minus 3, or the negative 3, times 2, which is negative 6. And so when you simplify, once again you get x squared minus x minus 6. And it takes a little bit of practice to really get used to it. Now the next thing I want to do-- and the principal is really the exact same way-- but I'm going to multiply a binomial times a trinomial, which many people find daunting. But we're going to see, if you just stay calm, it's not too bad. 3x plus 2, times 9x squared, minus 6x plus 4. Now you could do it the exact same way that we did the previous video. We could literally take this 3x plus 2, distribute it onto of these terms, and then you're going to distribute each of those terms into 3x plus 2. It would take a long time and in reality, you'll never do it quite that way. But you will get the same answer we're going to get. When you have larger polynomials, the easiest way I can think of to multiply, is kind of how you multiply long numbers. So we'll write it like this. 9x squared, minus 6, plus 4. And we're going to multiply that times 3x plus 2. And what I imagine is, when you multiply regular numbers, you have your ones' place, your tens' place, your hundreds' place. Here, you're going to have your constants' place, your first degree place, your second degree place, your third degree place, if there is one. And actually there will be in this video. So you just have to put things in their proper place. So let's do that. So you start here, you multiply almost exactly like you would do traditional multiplication. 2 times 4 is 8." + }, + { + "Q": "At 9:40 is it ok if you use the same technique for multiplying to binomials?", + "A": "It depends on what you are doing.", + "video_name": "fGThIRpWEE4", + "timestamps": [ + 580 + ], + "3min_transcript": "of these terms, and then you're going to distribute each of those terms into 3x plus 2. It would take a long time and in reality, you'll never do it quite that way. But you will get the same answer we're going to get. When you have larger polynomials, the easiest way I can think of to multiply, is kind of how you multiply long numbers. So we'll write it like this. 9x squared, minus 6, plus 4. And we're going to multiply that times 3x plus 2. And what I imagine is, when you multiply regular numbers, you have your ones' place, your tens' place, your hundreds' place. Here, you're going to have your constants' place, your first degree place, your second degree place, your third degree place, if there is one. And actually there will be in this video. So you just have to put things in their proper place. So let's do that. So you start here, you multiply almost exactly like you would do traditional multiplication. 2 times 4 is 8. 2 times negative 6x is negative 12x. And we'll put a plus there. That was a plus 8. 2 times 9x squared is 18x squared, so we'll put that in the x squared place. Now let's do the 3x part. I'll do that in magenta, so you see how it's different. 3x times 4 is 12x, positive 12x. 3x times negative 6x, what is that? The x times the x is x squared, so it's going to go over here. And 3 times negative 6 is negative 18. And then finally 3x times 9x squared, the x times the x 3 times 9 is 27. I wrote it in the x third place. And once again, you just want to add the like terms. So you get 8. There's no other constant terms, so it's just 8. Negative 12x plus 12x, these cancel out. 18x squared minus 18x squared cancel out, so we're just left over here with 27x to the third. So this is equal to 27x to the third plus 8. And we are done. And you can use this technique to multiply a trinomial times a binomial, a trinomial times a trinomial, or really, you know, you could have five terms up here. A fifth degree times a fifth degree. This will always work as long as you keep things in their proper degree place." + }, + { + "Q": "At 2:50 Sal says this is a fifth degree polynomial. What is that?", + "A": "A 5th degree polynomial is a polynomial that has the 5th power as the highest exponent in one of its term. Ex: 2x^5 _+ 2 x^5 + x^4 + 3", + "video_name": "fGThIRpWEE4", + "timestamps": [ + 170 + ], + "3min_transcript": "Remember, x to the 1, times x to the 1, add the exponents. I mean, you know x times x is x squared. So this first term is going to be 8x squared. And the second term, negative 5 times 2 is negative 10x. Not too bad. Let's do a slightly more involved one. Let's say we had 9x to the third power, times 3x squared, minus 2x, plus 7. So once again, we're just going to do the distributive property here. So we're going to multiply the 9x to the third times each of these terms. So 9x to the third times 3x squared. I'll write it out this time. In the next few, we'll start doing it a little bit in our heads. So this is going to be 9x to the third times 3x squared. way-- minus 2x times 9x to the third, and then plus 7 times 9x to the third. So sometimes I wrote the 9x to the third first, sometimes we wrote it later because I wanted this negative sign here. But it doesn't make a difference on the order that you're multiplying. So this first term here is going to be what? 9 times 3 is 27 times x to the-- we can add the exponents, we learned that in our exponent properties. This is x to the fifth power, minus 2 times 9 is 18x to the-- we have x to the 1, x to the third-- x to the fourth power. Plus 7 times 9 is 63x to the third. So we end up with this nice little fifth degree polynomial. Now let's do one where we are multiplying two binomials. This you're going to see very, very, very frequently in algebra. So let's say you have x minus 3, times x plus 2. And I actually want to show you that all we're doing here is the distributive property. So let me write it like this: times x plus 2. So let's just pretend that this is one big number here. And it is. You know, if you had x's, this would be some number here. So let's just distribute this onto each of these variables. So this is going to be x minus 3, times that green x, plus x minus 3, times that green 2. All we did is distribute the x minus 3. This is just the distributive property. Remember, if I had a times x plus 2, what would" + }, + { + "Q": "At around 4:48: Why do you subtract - 10x^2y dy/dx from both sides?", + "A": "When you solve an equation for a variable, you have to move all the terms with that variable to one side of the equation and all of the other terms to the opposite side. Then you can factor out the variable you re solving for and divide by the term in parenthesis: a simplified example would be: -2xy - 3 = 5y (to solve for y) -3 = 5y + 2xy -3 = y(5+2x) -3 / (5+2x) = y", + "video_name": "1DcsREjyoiM", + "timestamps": [ + 288 + ], + "3min_transcript": "of these terms. So if you distribute this purple thing onto this term right over here, you get 3 times 2x, which is 6x, times x squared plus y squared, squared. And then if you distribute this purple stuff on to this one right over here you get plus, let's see, 2y times 3 is 6y times x squared plus y squared. Let me make sure. 2y times 3 is 6y times x squared plus y squared, squared, and then I'll keep the dy dx in that green color. dy dx is equal to, well, we can multiply the 5 times this business right over here. And so everything that's not a dy dx term maybe I will do in purple now. So you do 5 times this stuff right over here, which gives you 10xy squared. is going to be plus 10x squared y dy dx. Did I do that right? Yep, that looks just about right. And now we have to solve for dy dx. What I'm going to do is I'm going to subtract this 10x squared. I'm going to subtract the 10x squared y dy dx from both sides. 10x squared y dy dx. Derivative. That's not green. Derivative of y with respect to x. Going to subtract that from both sides so that I can get it on the left hand side. dy dx. And I'm going to subtract this business, this 6x times all this craziness from both sides. So minus 6x times x squared plus y squared, squared. Let me subtract it from here as well. Minus 6x x squared plus y squared squared. Well, these guys cancel out. On the left hand side right over here, we are left with 6y times x squared plus y squared, squared minus 10x squared y times dy dx, the derivative of y with respect to x. The derivative of y with respect to x is equal to-- these characters cancel out-- and we are left with 10xy squared minus 6x times x squared plus y squared, squared. And now if we want to solve for dy dx, we just divide both sides of this equation by this business right over here. And you get the derivative of y with respect to x. And we deserve a drum roll now. The derivative of y with respect to x" + }, + { + "Q": "At 1:20, I just can't understand why the derivative,at the end of the application of the chain rule, is equal to ...2y*dy/dx. Why it's not just 2dy/dx??", + "A": "When you take a derivative, you bring down the power, and subtract one from the exponent. So derivative of y^3 would be 3y^2(dy/dx) and derivative of y^5 would be 5y^4(dy/dx). In your case, it was y^2, thus its derivative is 2y*(dy/dx)", + "video_name": "1DcsREjyoiM", + "timestamps": [ + 80 + ], + "3min_transcript": "Once again, I have some crazy relationship between x and y. And just to get a sense of what this might look like, if you plot all the x's and y's that satisfy this relationship, you get this nice little clover pattern. And I plotted this off of Wolfram Alpha. But what I'm curious about in this video, as you might imagine from the title, is to figure out the rate at which y is changing with respect to x. And we're going to have to do it implicitly. We're going to have to find the implicit derivative of this. We're going to have to derive this implicitly or take the derivative of it implicitly. So let's apply our derivative operator to both sides of the derivative with respect to x on the left and the derivative with respect to x on the right. So once again, we apply our chain rule. The derivative of something to the third power with respect to that something is going to be three times that something squared. And then we have to multiply that times the derivative of the something with respect to x. is going to be 2x, that's the derivative of x squared with respect to x, plus the derivative of y squared with respect to y is going to be 2y times the derivative of y with respect to x. Once again, we're applying the chain rule right over here. The derivative of something squared with respect to the something, which is 2y, times the derivative of the something with respect to x, which is dy dx. Now, that is going to be equal to what we have on the right hand side. So we have a 5 times x squared times y squared. We can take the 5 out of the picture for now. Take the 5 onto the-- take it out of the derivative. The derivative of 5 times something is the same thing as 5 times the derivative. And now we can apply the product rule. So it's going to be 5 times the derivative of x squared is just going to be 2x times y squared. times the second function. Plus the first function, not taking its derivative, x squared times the derivative of the second function. Well, what's the derivative of y squared with respect to x? Well, we already figured it out. It's the derivative of y squared with respect to y, which is 2y times the derivative of y with respect Let me make it clear what I just did. This is this, and then this is when I took its derivative. So that is when I applied the derivative operator. Similarly, that is that. And when I applied the derivative operator, I got that. The derivative with respect to x right over there. So let's see if we can somehow solve for dy dx. So what I'm going to do on the left hand side" + }, + { + "Q": "@2:40 is Sal making an assumption that AG is the longest part ?", + "A": "He may be eyeballing it, but you can also tell by the fact that line AG goes all the way past point F on its side, which is also the point that bisects line AE. That shows that AG is longer by proportion than line GD, or the longer part of the median if that made sense :)", + "video_name": "k45QTFCHSVs", + "timestamps": [ + 160 + ], + "3min_transcript": "Let me make sure I have enough space. This entire distance right over here is 18. They tell us that. So the area of AEC is going to be equal to 1/2 times the base-- which is 18-- times the height-- which is 12-- which is equal to 9 times 12, which is 108. That's the area of this entire right triangle, triangle AEC. If we want the area of BGC or any of these smaller of the six triangles-- if we ignore this little altitude right over here, the ones that are bounded by the medians-- then we just have to divide this by 6. Because they all have equal area. We've proven that in a previous video. So the area of BGC is equal to the area of AEC, the entire triangle, divided by 6, which is 108 divided by 6. You get 10 and then 48. Looks like it would be 18. And that's right because it would be-- 108 is the same thing as 18 times 6. So we did our first part. The area of that right over there is 18. And if we wanted, we could say, hey, the area of any of these triangles-- the ones that are bounded by the medians-- this is going to be 18. This is going to be 18. This entire FGE triangle is going to be 18, but we did this first part right over there. Now they ask us, what is the length of AG? So AG is the distance. It's the longer part of this median right over here. And to figure out what AG is, we just have to remind ourselves that the centroid is always 2/3 along the way of the medians, or it divides the median into two segments that have a ratio of 2 to 1. So if we know the entire length of this median, we could just take 2/3 of that. And that'll give us the length of AG. And we know that F and D are the midpoints. So for example, we know this AE is 12. That was given. We know that ED is half of this 18. So ED right over here-- I'll do this in a new color. ED is going to be 9. So then we could just use the Pythagorean theorem to figure out what AD is. AD is the hypotenuse of this right triangle. So we're looking at triangle AED right now. Let me write this down. We know that 12 squared plus 9 squared is going to be equal to AD squared. 12 squared is 144. 144 plus 81." + }, + { + "Q": "When you are multiplying a fraction with the matrix, is it necessary to simplify if allowed?\nRefer to 9:47...", + "A": "No, it is not necessary.", + "video_name": "iUQR0enP7RQ", + "timestamps": [ + 587 + ], + "3min_transcript": "But let's apply this to a real problem, and you'll see that it's actually not so bad. So let's change letters, just so you know it doesn't always have to be an A. Let's say I have a matrix B. And the matrix B is 3-- I'm just going to pick random numbers-- minus 4, 2 minus 5. Let's calculate B inverse. So B inverse is going to be equal to 1 over the determinant of B. What's the determinant? It's 3 times minus 5 minus 2 times minus 4. So 3 times minus 5 is minus 15, minus 2 times minus 4. 2 times minus 4 is minus 8. We're going to subtract that. So it's plus 8. And we're going to multiply that times what? And we just make these two terms negative. Minus 2 and 4. 4 was minus 4, so now it becomes 4. And let's see if we can simplify this a little bit. So B inverse is equal to minus 15 plus 8. That's minus 7. So this is minus 1/7. So the determinant of B-- we could write B's determinant-- is equal to minus 7. So that's minus 1/7 times minus 5, 4, minus 2, 3. Which is equal to-- this is just a scalar, this is just a number, so we multiply it times each of the elements-- so that is equal to minus, minus, plus. That's 5/7. 5/7 minus 4/7. Let's see. And then minus 3/7. It's a little hairy. We ended up with fractions here and things. But let's confirm that this really is the inverse of the matrix B. Let's multiply them out. So before I do that I have to create some space. I don't even need this anymore. OK. So let's confirm that that times this, or this times that, is really equal to the identity matrix. So let's do that. So let me switch colors. So B inverse is 5/7, if I haven't made any careless mistakes. Minus 4/7. 2/7. And minus 3/7." + }, + { + "Q": "when he fins the determinant of B he says that it is 1/-7 and that is what he uses to multiply with. So why does he change it to -7 when he puts the B in absolute value signs over on the side (9:34) ?", + "A": "When dealing with matrices the absolute value sign actually means the determinant and does not mean to take the absolute of the number.", + "video_name": "iUQR0enP7RQ", + "timestamps": [ + 574 + ], + "3min_transcript": "But let's apply this to a real problem, and you'll see that it's actually not so bad. So let's change letters, just so you know it doesn't always have to be an A. Let's say I have a matrix B. And the matrix B is 3-- I'm just going to pick random numbers-- minus 4, 2 minus 5. Let's calculate B inverse. So B inverse is going to be equal to 1 over the determinant of B. What's the determinant? It's 3 times minus 5 minus 2 times minus 4. So 3 times minus 5 is minus 15, minus 2 times minus 4. 2 times minus 4 is minus 8. We're going to subtract that. So it's plus 8. And we're going to multiply that times what? And we just make these two terms negative. Minus 2 and 4. 4 was minus 4, so now it becomes 4. And let's see if we can simplify this a little bit. So B inverse is equal to minus 15 plus 8. That's minus 7. So this is minus 1/7. So the determinant of B-- we could write B's determinant-- is equal to minus 7. So that's minus 1/7 times minus 5, 4, minus 2, 3. Which is equal to-- this is just a scalar, this is just a number, so we multiply it times each of the elements-- so that is equal to minus, minus, plus. That's 5/7. 5/7 minus 4/7. Let's see. And then minus 3/7. It's a little hairy. We ended up with fractions here and things. But let's confirm that this really is the inverse of the matrix B. Let's multiply them out. So before I do that I have to create some space. I don't even need this anymore. OK. So let's confirm that that times this, or this times that, is really equal to the identity matrix. So let's do that. So let me switch colors. So B inverse is 5/7, if I haven't made any careless mistakes. Minus 4/7. 2/7. And minus 3/7." + }, + { + "Q": "at 6:25, why do you subract the two equations and not add them?", + "A": "you can do either or as long as you are left one variable eliminated", + "video_name": "xCIHAjsZCE0", + "timestamps": [ + 385 + ], + "3min_transcript": "" + }, + { + "Q": "at 5:00, why do you subtract 300 instead of 200?", + "A": "Because he is subtracting the blue/teal equation from the red/pink equation. Subtracting the red equation from itself would just get you 0=0, which is true, but not very useful.", + "video_name": "xCIHAjsZCE0", + "timestamps": [ + 300 + ], + "3min_transcript": "" + }, + { + "Q": "At 7:50, Khan says \"one equation for one unknown.\"\nDoes this mean that if two equations for 2 unkowns and 1 equation for 1 unknown is possible to be solved, then for 3 unknowns you have to get 3 equations? And so on and so forth?", + "A": "That is exactly correct. Good job :-)", + "video_name": "xCIHAjsZCE0", + "timestamps": [ + 470 + ], + "3min_transcript": "" + }, + { + "Q": "@ 5:09 why Sal is subtracting 500a and 300c from 500a + 200c?", + "A": "The whole point of solving systems of equations by elimination is to get rid of one variable by making it subtract to be 0, so if you have a positive 500 and a - 500, then the two add to be zero which gets rid of a, so when we add the coefficients of the c variable, we can find a value for c, then substitute it in to find a.", + "video_name": "xCIHAjsZCE0", + "timestamps": [ + 309 + ], + "3min_transcript": "" + }, + { + "Q": "What is the angle between two hands of a clock at 4:30", + "A": "If the total clock face represents 360\u00c2\u00b0, then each number on the clock represents an angle of 30\u00c2\u00b0 and each minute, 6\u00c2\u00b0. So at 4:30 the minute hand is at 30*6\u00c2\u00b0 = 180\u00c2\u00b0 and the hour hand is at 4.5*30\u00c2\u00b0 = 135\u00c2\u00b0. Thus the angle between them is 180\u00c2\u00b0-135\u00c2\u00b0 = 45\u00c2\u00b0", + "video_name": "2mzuFKCuDg4", + "timestamps": [ + 270 + ], + "3min_transcript": "So let's check our answer and make sure we actually got this one right. Let's do one more, just for fun. Which of these angles has a measure of 60 degrees? So there's a couple that we can immediately rule out. This angle here is clearly more than 90 degrees. It's an obtuse angle. This angle here is 180 degrees, so we're going to have to pick between these two. And we can assume that these are all at scale. So if we know that a right angle would have this line straight up and down, this looks like it's 2/3 of the way there. So I would go with this one right over there. Now, you might say, well, what about this one? Well, this one looks like it's about 1/3 of a right angle. If we were to do 2/3, then 3/3, it looks like we would get to a right angle. So this one looks much closer to 30 degrees. So which of these angles has a measure of 60? That one right over there. You could immediately rule out that one and that one, and then these take a little bit closer inspection." + }, + { + "Q": "At 3:51 where did he get 10x from? He seems to have pulled it out of thin air.", + "A": "The 10x is created from multiplying the 2 binomials (x+3)(x+7) Let s do the multiplication (use FOIL to multiply the binomials). (x+3)(x+7) = x^2 + 7x + 3x + 21 Combine the middle 2 terms... they create the 10x. x^2 + 7x + 3x + 21 = x^2 + 10x + 21 Hope this helps.", + "video_name": "SjN3_xCJamA", + "timestamps": [ + 231 + ], + "3min_transcript": "is Y to the fourth power. And so what we could say is, if we wanted to say factors of negative six X to the third, Y to the fourth, we could say that 3xy is a factor of this just as an example; so let me write that down. We could write that 3xy is a factor of, is a factor of... of negative six X to the third power, Y to the fourth, or we could phrase that the other way around. We could say that negative six X to the third, Y to the fourth, is divisible by, is divisible by 3xy. So hopefully you're seeing the parallels. If I'm taking these two monomials with integer coefficients and I multiply 'em and I get this other, in this case, and there's actually other factors of this, but I could say either one of these is a factor of this monomial, or we could say that negative six X to the third, Y to the four is divisible by one of its factors. And we could even extend this to binomials or polynomials. For example, if I were to take, if I were to take, let me scroll down a little bit, whoops, if I were to take, let me say X plus three and I wanted to multiply it times X plus seven, we know that this is going to be equal to, if I were to write it as a trinomial, it's gonna be X times X, so X squared, and then it's gonna be three X plus seven X, so plus 10x; and if any of this looks familiar, we have a lot of videos where we go in detail of multiplying binomials like this. And then three times seven is 21. So because I multiplied these two, in this case binomials, or we could consider themselves to be polynomials, polynomials or binomials with integer coefficients. Notice the coefficients here, they're one, one. The constants here, they're all integers. Because I'm dealing with all integers here, we could say that either one of these binomials is a factor of this trinomial, or we could say this trinomial is divisible by either one of these. So let me write that down. So I could say, I'll just pick on X plus seven. We could say that X plus seven is a factor, is a factor of X squared plus 10x plus 21; or we could say that X squared plus 10x plus 21 is divisible by, is divisible by" + }, + { + "Q": "At 3:59ish, Sal switches the subtraction to addition and adds -1. Do we always use 1? And why do we change the sign? What happens if you do not use a scaler?", + "A": "the jest of it is this ... a - b = a + (-1)b = a + -b", + "video_name": "WR9qCSXJlyY", + "timestamps": [ + 239 + ], + "3min_transcript": "in which I'm adding the matrices does not matter So this is just like adding numbers. A plus B is just the same thing as B plus A. What we'll see is this won't be true for every matrix operation that we study and in particular this will not be true for matrix multiplication. But if you add these two things, using the definition we just came up with, adding corresponding terms, you'll get the exact same result. Up here we added one plus five and we got six Her we'll add five plus one and we'll get six. We get the same result because one plus five is the same thing as five plus one. Here we have zero plus negative seven you get negative seven. So you're going to get the exact same thing as we got up here. So when you're adding matrices, if you were to call -if you were to call this matrix right over here matrix A which we normally denote with a capital, bolder letter, and you call this matrix right over here Matrix B Then when we take the sum of A plus B which is this thing right over here, and we see it's the exact same thing as B, as Matrix B plus Matrix A. What if I wanted to subtract matrices? So let's once again think about matrices that have the same dimensions. So let's say I'm gonna do then two two-by-two matrices. So let's say it's zero, one, three, two, and from that I want to subtract negative one, three, zero, and five. So you might say well maybe we just subtract corresponding entries. And that indeed is how you can define matrix subtraction. In fact you don't even have to define matrix subtraction, you can let this fall out of what we did with scalar multiplication and matrix addition. We can view as the exact same thing -this as the exact same thing- as taking zero, one, three, two and to that we add negative one, negative one times negative one, three, zero, five. And if you work out the math you're going to get subtracting the corresponding terms. So this is going to be -what is this going to be? Zero minus negative one is positive one, one minus three is negative two, three minus zero is three, two minus five is negative three. And you'll see that you get the exact same thing here. When you multiply negative one times negative one you get positive one, positive one plus zero is one. Negative one times three plus one is negative two. Fair enough. There might be a question that is lingering in your brain right now. \"Okay Sal, I understand when I'm adding or subtracting matrices with the same dimensions I just add or subtract the corresponding terms. But what happens when I have matrices with different dimensions?\" So, for example, what about the scenario where I want to add the matrix one, zero, three, five, zero, one to the matrix -so this a three-by-two matrix-" + }, + { + "Q": "At 5:08 he says '0', but he writes '6' . Which one is he saying??", + "A": "It was a sloppy 0. The actual value is irrelevant since the two matrices have different dimensions.", + "video_name": "WR9qCSXJlyY", + "timestamps": [ + 308 + ], + "3min_transcript": "What if I wanted to subtract matrices? So let's once again think about matrices that have the same dimensions. So let's say I'm gonna do then two two-by-two matrices. So let's say it's zero, one, three, two, and from that I want to subtract negative one, three, zero, and five. So you might say well maybe we just subtract corresponding entries. And that indeed is how you can define matrix subtraction. In fact you don't even have to define matrix subtraction, you can let this fall out of what we did with scalar multiplication and matrix addition. We can view as the exact same thing -this as the exact same thing- as taking zero, one, three, two and to that we add negative one, negative one times negative one, three, zero, five. And if you work out the math you're going to get subtracting the corresponding terms. So this is going to be -what is this going to be? Zero minus negative one is positive one, one minus three is negative two, three minus zero is three, two minus five is negative three. And you'll see that you get the exact same thing here. When you multiply negative one times negative one you get positive one, positive one plus zero is one. Negative one times three plus one is negative two. Fair enough. There might be a question that is lingering in your brain right now. \"Okay Sal, I understand when I'm adding or subtracting matrices with the same dimensions I just add or subtract the corresponding terms. But what happens when I have matrices with different dimensions?\" So, for example, what about the scenario where I want to add the matrix one, zero, three, five, zero, one to the matrix -so this a three-by-two matrix- Five, seven, negative one, zero. What would we define this as? Well it turns out that the mathematical mainstream does not define this. This is undefined. This is undefined. So we do not define matrix addition, or matrix subtraction, when the matrices have different dimensions. There didn't seem to be any reasonable way to do this, that would actually be useful and logically consistent in some nice way." + }, + { + "Q": "Greetings from Venezuela...Very helpful explanation, thanks Sal! Just a question: at 3:28, Sal is taking the derivative of x with respect to theta. The derivative of sine of theta is the cosine of theta, agreed. But why is the sqrt(3)/sqrt(2) not derived?", + "A": "because sqrt(3)/sqrt(2) is a constant. Sorta like why you wouldnt derive pi, or any other number that isnt being multiplied by a variable such as X.", + "video_name": "n4EK92CSuBE", + "timestamps": [ + 208 + ], + "3min_transcript": "We could do either way. But this, all of a sudden, this thing right here, starts to look a little bit like this. Maybe I can do a little bit of algebraic manipulation to make this look a lot like that. So the first thing, I would like to have a 1 here-- at least, that's how my brain works-- so let's factor out a 3 out of this denominator. So this is the same thing as the integral of 1 over the square root of-- let me factor out a 3 out of this expression. 3 times 1 minus 2/3x squared. I did nothing fancy here. I just factored the 3 out of this expression, that's all I did. But the neat thing now is, this expression looks a lot like that expression. In fact, if I substitute, if I say that this thing right here, this 2/3x squared, if I set it equal to sine squared theta, I So let's do that. Let's set 2/3x squared, let's set that equal to sine squared of theta. So if we take the square root of both sides of this equation, I get the square root of 2 over the square root of 3 times x is equal to the sine of theta. If I want to solve for x, what do I get? And, well, we're going to have to solve for both x and for theta, so let's do it both ways. First, let's solve for theta. If we solve for theta, you get that theta is equal to the arcsine, or the inverse sine, of square root of 2 over square root of 3x. That's if you solve for theta. Now, if you solve for x, you just multiply both sides of this equation times the inverse of this and you get x is equal to-- divide both sides of the equation by this or multiply it square root of 2 times the sine of theta. And we were going to substitute this with sine squared of theta, but we can't leave this dx out there. We have to take the integral with respect to d theta. So what's dx with respect to d theta? So the derivative of x with respect to theta is equal to square root of 3 over square root of 2. Derivative of this with respect to theta is just cosine of theta, and if we want to write this in terms of dx, we could just write that dx is equal to square root of 3 over the square root of 2 cosine of theta d theta. Now we're ready to substitute. So we can rewrite this expression up here-- I'll do it in this reddish color-- I was using that, let me do it in the blue color. We can rewrite this expression up here now. It's an indefinite integral of-- dx is on the" + }, + { + "Q": "At 0:58 he says you can use either sin^2 or cos^2, but won't that result in arccos at 7:05 instead of arcsin, which is a different answer? Or am I making a mistake?", + "A": "I didn t watch the video, but arccos and arcsin differ by a constant and a negative sign only. So you ll end up with, say, -arccos instead of arcsin, and the constant difference will be absorbed by your arbitrary constant. Both are equally fine antiderivatives.", + "video_name": "n4EK92CSuBE", + "timestamps": [ + 58, + 425 + ], + "3min_transcript": "Let's say I have the indefinite integral 1 over the square root of 3 minus 2x squared. Of course I have a dx there. So right when I look at that, there's no obvious traditional method of taking this antiderivative. I don't have the derivative of this sitting someplace else in the integral, so I can't do traditional u-substitution. But what I can do is I could say, well, this almost looks like some trig identities that I'm familiar with, so maybe I can substitute with trig functions. So let's see if I can find a trig identity that looks similar to this. Well, our most basic trigonometric identity-- this comes from the unit circle definition-- is that the sine squared of theta plus the cosine squared of theta is equal to 1. And then if we subtract cosine squared of theta from both sides, we get-- or if we subtract sine squared of theta from both sides, we could do either-- we could get cosine We could do either way. But this, all of a sudden, this thing right here, starts to look a little bit like this. Maybe I can do a little bit of algebraic manipulation to make this look a lot like that. So the first thing, I would like to have a 1 here-- at least, that's how my brain works-- so let's factor out a 3 out of this denominator. So this is the same thing as the integral of 1 over the square root of-- let me factor out a 3 out of this expression. 3 times 1 minus 2/3x squared. I did nothing fancy here. I just factored the 3 out of this expression, that's all I did. But the neat thing now is, this expression looks a lot like that expression. In fact, if I substitute, if I say that this thing right here, this 2/3x squared, if I set it equal to sine squared theta, I So let's do that. Let's set 2/3x squared, let's set that equal to sine squared of theta. So if we take the square root of both sides of this equation, I get the square root of 2 over the square root of 3 times x is equal to the sine of theta. If I want to solve for x, what do I get? And, well, we're going to have to solve for both x and for theta, so let's do it both ways. First, let's solve for theta. If we solve for theta, you get that theta is equal to the arcsine, or the inverse sine, of square root of 2 over square root of 3x. That's if you solve for theta. Now, if you solve for x, you just multiply both sides of this equation times the inverse of this and you get x is equal to-- divide both sides of the equation by this or multiply it" + }, + { + "Q": "at 1:43 when it is like (5)(4) is that multiplication because that confused me because cant u just writ it like (5 times 4) or does it have to be like (5)(4)", + "A": "Yes, (5)(4) is just the same thing as 5 times 4 . This is just another way to write it, which you will see quite often in math. You can write it however you prefer, but you should get used to seeing it written this way as well, because it will show up often. I hope this helps.", + "video_name": "GiSpzFKI5_w", + "timestamps": [ + 103 + ], + "3min_transcript": "We're asked to simplify 8 plus 5 times 4 minus, and then in parentheses, 6 plus 10 divided by 2 plus 44. Whenever you see some type of crazy expression like this where you have parentheses and addition and subtraction and division, you always want to keep the order of operations in mind. Let me write them down over here. So when you're doing order of operations, or really when you're evaluating any expression, you should have this in the front of your brain that the top priority goes to parentheses. And those are these little brackets over here, or however Those are the parentheses right there. That gets top priority. Then after that, you want to worry about exponents. There are no exponents in this expression, but I'll just write it down just for future reference: exponents. One way I like to think about it is parentheses always takes top priority, but then after that, we go in descending order, or I guess we should say in-- well, yeah, in When I say fast, how fast it grows. When I take something to an exponent, when I'm taking something to a power, it grows really fast. Then it grows a little bit slower or shrinks a little bit slower if I multiply or divide, so that comes next: multiply or divide. Multiplication and division comes next, and then last of all comes addition and subtraction. So these are kind of the slowest operations. This is a little bit faster. This is the fastest operation. And then the parentheses, just no matter what, always take priority. So let's apply it over here. Let me rewrite this whole expression. So it's 8 plus 5 times 4 minus, in parentheses, 6 plus 10 divided by 2 plus 44. So we're going to want to do the parentheses first. We have parentheses there and there. Now this parentheses is pretty straightforward. could really just view this as 5 times 4. So let's just evaluate that right from the get go. So this is going to result in 8 plus-- and really, when you're evaluating the parentheses, if your evaluate this parentheses, you literally just get 5, and you evaluate that parentheses, you literally just get 4, and then they're next to each other, so you multiply them. So 5 times 4 is 20 minus-- let me stay consistent with the colors. Now let me write the next parenthesis right there, and then inside of it, we'd evaluate this first. Let me close the parenthesis right there. And then we have plus 44. So what is this thing right here evaluate to, this thing inside the parentheses? Well, you might be tempted to say, well, let me just go left to right. 6 plus 10 is 16 and then divide by 2 and you would get 8. But remember: order of operations. Division takes priority over addition, so you actually want" + }, + { + "Q": "At 3:00 what did he mean?", + "A": "First solve the problem of first bracket. Then solve the other problems.I think you got it,", + "video_name": "GiSpzFKI5_w", + "timestamps": [ + 180 + ], + "3min_transcript": "When I say fast, how fast it grows. When I take something to an exponent, when I'm taking something to a power, it grows really fast. Then it grows a little bit slower or shrinks a little bit slower if I multiply or divide, so that comes next: multiply or divide. Multiplication and division comes next, and then last of all comes addition and subtraction. So these are kind of the slowest operations. This is a little bit faster. This is the fastest operation. And then the parentheses, just no matter what, always take priority. So let's apply it over here. Let me rewrite this whole expression. So it's 8 plus 5 times 4 minus, in parentheses, 6 plus 10 divided by 2 plus 44. So we're going to want to do the parentheses first. We have parentheses there and there. Now this parentheses is pretty straightforward. could really just view this as 5 times 4. So let's just evaluate that right from the get go. So this is going to result in 8 plus-- and really, when you're evaluating the parentheses, if your evaluate this parentheses, you literally just get 5, and you evaluate that parentheses, you literally just get 4, and then they're next to each other, so you multiply them. So 5 times 4 is 20 minus-- let me stay consistent with the colors. Now let me write the next parenthesis right there, and then inside of it, we'd evaluate this first. Let me close the parenthesis right there. And then we have plus 44. So what is this thing right here evaluate to, this thing inside the parentheses? Well, you might be tempted to say, well, let me just go left to right. 6 plus 10 is 16 and then divide by 2 and you would get 8. But remember: order of operations. Division takes priority over addition, so you actually want here like this. You could imagine putting some more parentheses. Let me do it in that same purple. You could imagine putting some more parentheses right here to really emphasize the fact that you're going to do the division first. So 10 divided by 2 is 5, so this will result in 6, plus 10 divided by 2, is 5. 6 plus 5. Well, we still have to evaluate this parentheses, so this results-- what's 6 plus 5? Well, that's 11. So we're left with the 20-- let me write it all down again. We're left with 8 plus 20 minus 6 plus 5, which is 11, plus 44. And now that we have everything at this level of operations, we can just go left to right. So 8 plus 20 is 28, so you can view this as 28 minus 11 plus 44." + }, + { + "Q": "At 04:09, if the hypotenuse is irrational does this mean that the hypotenuse is a multipal of \"root 2\" or can it also be another irrational number?", + "A": "root two", + "video_name": "X1E7I7_r3Cw", + "timestamps": [ + 249 + ], + "3min_transcript": "I want the context, because in school today if you bring out the ruler and compass and are like, \"Let's do some geometry! Let's draw two lines at 90 degree angles using a straight-edge and compass! Here's a happy square!\" Then you've probably had years of math class already and think of geometry as being harder than adding big numbers together. You probably think that zero is a simple, easy concept and have heard of decimals too. Well, here's now, 2012. Here's Einstein, Euler, Newton and Da Vinci - - that sure was a while ago! Now let's go all the way back to when Arabic numerals were invented and brought to the West by Fibbonacci. Before that, arithmetic was nightmarishly hard, so if you can multiply multi-digit numbers together you can go back in time and impress the beans out of Pythagoras. And before that there was no concept of zero, except in India where zero was discovered around here. And if you keep going back you get to the year one, (there's no year zero, of course, because zero hadn't been invented) and back a bit more you get to folk like Aristotle, Euclid, Archimedes and then finally Pythagoras, all the way back in 6th century BCE. Point is, you can do some pretty cool mathematics without having a good handle on arithmetic and people did for a long time. you need to memorize your multiplication table and graph a parabola before you can learn real mathematics, they are lying to you. In Pythagoras's time there were no variables, no equations or formulas like we see today, Pythagoras's theorem wasn't 'a squared plus b squared equals c squared,' it was 'The squares of the legs of a right triangle have the same area as the square of the hypotenuse,' all written out. And when he said 'square' he meant 'square.' One leg's square plus the other leg's square equals hypotenuse's square. Three literally squared plus four made into a square. Those two squares have the same area as a five by five square. You can cut out the nine squares here and the sixteen here and fit them together where these 25 squares are, and in the same way, you can cut out the 25 hypotenuse squares and fit them into the two leg squares. Pythagoras thought you could do this trick with any right triangle, that it was just a matter of figuring out how many pieces to cut each side into. There was a relationship between the length of one side and the length of another and he wanted to find it on this map. But the trouble began with the simplest right triangle one where both the legs are the same length, one where both the legs' squares are equal. hypotenuse is something that, when squared, gives two. So what's the square root of two and how do we make it into a whole number ratio? Square root two is very close to 1.4 which would be a whole number ratio of 10:14 but 10 squared plus 10 squared is definitely not 14 squared, and a ratio of 1,000 to 1,414 is even closer, and a ratio of 100,000,000 to 141,421,356 is very close indeed but still not exact, so what is it? Pythagoras wanted to find the perfect ratio he knew it must exist, but meanwhile someone from his very own Pythagorean brotherhood proved there wasn't a ratio, the square root of two is irrational, that in decimal notation (once decimal notation was invented) the digits go on forever. Usually this proof is given algebraically, something like this, which is pretty simple and beautiful if you know algebra, but the Pythagoreans didn't. So I like to imagine how they thought of this proof, no algebra required. Okay, so Pythagoras is all like, \"There's totally a ratio, you can make this with whole numbers.\" And this guy's like,\"Is not!\" \"Is too!\" \"Is not!\" \"Is too!\" \"Fine have it your way." + }, + { + "Q": "refering to 2:00\n\nwas he actually that crazy, he meaning pathagros", + "A": "Mostly, but you have to remember that he was a mathematician, and mathematicians love elegance and simplicity. Adding irrationals to mathematics made it far more ugly. And nobody likes to find that what they had believed to be true to be wrong.", + "video_name": "X1E7I7_r3Cw", + "timestamps": [ + 120 + ], + "3min_transcript": "Ok, so I've been learning about Pythagoras and the dirt on him is just too good. You've probably heard of the Pythagorean Theorem but not the part where Pythagoras was a crazy cult leader who thought he'd made a deal with a god thousands of years ago and could remember all of his past lives. Oh, and he killed a guy. I mean maybe it was a long time ago and he was afraid of beans. As in beans, they just like freaked him out or something, I don't know. But mostly I want to talk about the murdery part. See, Pythagoras and his cult of Pythagoreans had this cool-kids club where they'd talk about proportions all day. They'd be like, \"Hey, I drew a two by three rectangle using a straight-edge and compass. Isn't that awesome?\" And then someone would be like, \"Hey guys, I have a box that's two by three and a half?\" And the cool kids would be like, \"Three and a half? That's not a number! Get out of our club!\" And then they'd make the units half the length and call it four by seven, and everything was okay. Even if your box is 2.718 by 6.28 you can just divide your units into thousandths and you'd get a box that's a nice, even 2,718 by 6,280. It's not a simple proportion but hey, the box still has a whole number proportion, so Pythagoras is happy. Unless it's a box of beans, then he freaks out. I'd like to imagine what it would be like to Maybe you think of numbers as being on a line. Numbers one way, zero, and negative numbers the other, and there are numbers between them: fractions, rationals, filling in the gaps. But Pythagoras didn't think about numbers like this at all. They weren't points in a continuum they were each their own, separate being, which was still pretty modern because before that people only thought of numbers as adjectives, numbers of. In Pythagoras's world, there is no number between seven and eight, and there is no number three over two so much as a relationship between three and two, a proportion. Six to four has the same relationship because the numbers share this evenness, which when accounted for makes it three to two. The universe to Pythagoras was made up of these relationships. Mathematics wasn't numbers, mathematics was between the numbers. Though while people admire how much Pythagoras loved proportions, there's a dark flip side to that obsession. How far was he willing to go to protect the proportions he loved? Would he kill for them? Would he die for them? And the answer was he'd go pretty far until beans got involved. It's time for Time Line Time. I want the context, because in school today if you bring out the ruler and compass and are like, \"Let's do some geometry! Let's draw two lines at 90 degree angles using a straight-edge and compass! Here's a happy square!\" Then you've probably had years of math class already and think of geometry as being harder than adding big numbers together. You probably think that zero is a simple, easy concept and have heard of decimals too. Well, here's now, 2012. Here's Einstein, Euler, Newton and Da Vinci - - that sure was a while ago! Now let's go all the way back to when Arabic numerals were invented and brought to the West by Fibbonacci. Before that, arithmetic was nightmarishly hard, so if you can multiply multi-digit numbers together you can go back in time and impress the beans out of Pythagoras. And before that there was no concept of zero, except in India where zero was discovered around here. And if you keep going back you get to the year one, (there's no year zero, of course, because zero hadn't been invented) and back a bit more you get to folk like Aristotle, Euclid, Archimedes and then finally Pythagoras, all the way back in 6th century BCE. Point is, you can do some pretty cool mathematics without having a good handle on arithmetic and people did for a long time." + }, + { + "Q": "At 8:00, was Pythagoras really scared of beans?", + "A": "There are several theories on why Pythagoras did not allow his followers to eat beans. One more likely theory states that he believed that beans contained the souls of humans, but we aren t sure.", + "video_name": "X1E7I7_r3Cw", + "timestamps": [ + 480 + ], + "3min_transcript": "\"We are. If there's a ratio in simlest form at least one of the numbers is odd and since the hypotenuse has to literally be divisible by two, then the leg must be the odd one. So what if I proved the leg had to be even?\" \"You just proved it's not. It can't be both.\" \"Unless it doesn't exist! What you forget Pythagoras is that if this is a square then the two sides are the same. Just as this is divsible right down the center so too is it divisible the other way! And the number squares on this side, which are the number of squares in just one leg is an even number. And for a number of squares to be even what does the number have to be, Pythagoras, oh my brother?\" \"If leg squared is even then the leg is even. But it can't be even, because it's already odd.\" \"Unless it doesn't exist.\" \"But if they're both even you can divide both by two and start again, but this still has to be even which means this still has to be even, which means you can divide by two again, but then it has to be even so everything is even forever and you never find the perfect ratio. Aww, beans\" He had a vision, a beautiful vision of a world made up of relationships between numbers. The Pythagoreans still believed, wanted to believe that irrationality was somehow false and the world was as they wanted it. So this proof stayed secret. Until someone spilled the beans. According to some, it was all a guy named Hippasus and Pythagoras threw him of a boat to drown him as punishment for ruining what had been perfect. Or maybe it was someone else who discovered it or Hippasus or someone else who was killed by the Pythagoreans long after Pythagoras was dead or maybe they just got exiled, who knows? And how did Pythagoras die? Well, according to one guy some guys got mad because they didn't get into the cool kids' club. So they set Pythagoras' house on fire. And Pythagoras was running away and they were chasing him, but then they came upon a field and not just any field, but a field of beans. And Pythagoras turned around to face his pursuers and proclaimed: \"Better to be slaughetered by enemies than to trample on beans!\" And he was. Others say he ran off and starved himself to death. Or just got caught by his enemies because he ran around the bean field instead of through it or who knows what happened. People claim Pythagoras didn't like beans because he thought they were bad for digestion, or gave you bad dreams because he didn't want a clubhouse full of flatulating mathematicians or he just didn't like them metaphorically. He and his followers were or weren't vegetarian did or didn't sacrifice animals possibly were only allowed to eat certain colors of birds I mean he definitely had a lot of rules to follow but just what they were and what they meant is lost to history. I'd like to give you a colorful story about exactly what happened with Pythagoras, but somehow that kind of truth doesn't last. What I do know is that the square root of two is irrational, that there's no way to have the length of a side of a square and of the square's diagonal both be whole numbers. Mathematical truth is truth that indures. This proof is just as good now as it was 2500 years ago, I mean it's awesome and it shows that there's more to the world than whole numbers and shame on the Pythagoreans who didn't have the beans to admit it." + }, + { + "Q": "At 8:30, why does Sal keep expanding everything out? I do not understand it.", + "A": "he is using this as a complete example to show how it works. He is also using the sigma, which is a sum of all integers from the number on the bottom to n.", + "video_name": "iPwrDWQ7hPc", + "timestamps": [ + 510 + ], + "3min_transcript": "to keep switching colors, but hopefully it's worth it, a plus b. Let's take that to the 4th power. The binomial theorem tells us this is going to be equal to, and I'm just going to use this exact notation, this is going to be the sum from k equals 0, k equals 0 to 4, to 4 of 4 choose k, 4 choose k, 4 choose ... let me do that k in that purple color, 4 choose k of a to the 4 minus k power, 4 minus k power times b to the k power, b to the k power. Now what is that going to be equal to? Well, let's just actually just do the sum. This is going to be equal to, so we're going to start at k equals 0, so when k equals 0, it's going to be 4 choose 0, times a to the 4 minus 0 power, well, that's just going to be a to the 4th power, times b to the 0 power. b to the 0 power is just going to be equal to 1, so we could just put a 1 here if we want to, or we could just leave it like that. This is what we get when k equals 0. Then to that, we're going to add when k equals 1. k equals 1 is going to be, the coefficient is going to be 4 choose 1, and it's going to be times a to the 4 minus 1 power, so a to the 3rd power, and I'll just stick with that color, times b to the k power. Well, now, k is 1b to the 1st power. Then to that, we're going to add, we're going to add 4 choose 2, 4 choose 2 times a to the ... 4 minus 2 is 2. I think you see a pattern here. a to the 4th, a to the 3rd, a squared, and then times b to the k. Well, k is 2 now, so b squared, and you see a pattern again. You could say b to the 0, b to the 1, b squared, and we only have two more terms to add here, plus 4 choose 3, 4 choose 3 times 4 minus 3 is 1, times a, or a to the 1st, I guess we could say, and then b to the 3rd power, times a to the 1st b to the third, and then only one more term, plus 4 choose, 4 choose 4. k is now 4. This is going to be our last term right now. We're getting k goes from 0 all the way to 4, 4 choose 4. a to the 4 minus 4, that's just going to be 1, a to the 0, that's just 1, so we're going to be left with just b to the k power," + }, + { + "Q": "At 4:39 point, when writing \"n choose k\" for the first time, you say, \"We'll review that in a second. This comes straight of out ?\" I didn't hear that part. It comes straight out of WHAT?", + "A": "There is a term called Combination,which states that each of the different groups or selection which can be made out by taking some or all of a number of things at a time.or simply Selection of r terms out of n terms....that part is derived from this very term.....selection of r terms out of n terms..... nCr = n!/r!(n-r)! ....", + "video_name": "iPwrDWQ7hPc", + "timestamps": [ + 279 + ], + "3min_transcript": "plus b to the 3rd power. Just taking some of the 3rd power, this already took us a little reasonable amount of time, and so you can imagine how painful it might get to do something like a plus b to the 4th power, or even worse, if you're trying to find a plus b to the 10th power, or to the 20th power. This would take you all day or maybe even longer than that. It would be incredibly, incredibly painful. That's where the binomial theorem becomes useful. What is the binomial theorem? The binomial theorem tells us, let me write this down, binomial theorem. Binomial theorem, it tells us that if we have a binomial, and I'll just stick with the a plus b for now, if I have, and I'm going to try to color code this a little bit, if I have the binomial a plus b, and I'm going to raise it the nth power, I'm going to raise this to the nth power, the binomial theorem tells us that this is going to be equal to, and the notation is going to look a little bit complicated at first, but then we'll work through an actual example, is going to be equal to the sum from k equals 0, k equals 0 to n, this n and this n are the same number, of ... I don't want to ... that's kind of a garish color ... of n choose k, n choose k, and we'll review that in a second; this comes straight out of combinatorics; n choose k times a to the n minus k, n minus k, times b, times b to the k, Now this seems a little bit unwieldy. Let's just review, remind ourselves what n choose k actually means. If we say n choose k, I'll do the same colors, n choose k, we remember from combinatorics this would be equal to n factorial, n factorial over k factorial, over k factorial times n minus k factorial, n minus k factorial, so n minus k minus k factorial, let me color code this, n minus k factorial. Let's try to apply this. Let's just start applying it to the thing that started to intimidate us, say, a plus b to the 4th power. Let's figure out what that's going to be. Let's try this. So a, and I'm going to try to keep it color-coded so you know what's going on, a plus b," + }, + { + "Q": "Why 4! / 0!4! = 1? it's just ( 4 * 3 * 2* 1 ) / ( 0 * 4 * 3 * 2 * 1 ) = 24 / 0\nwhich is undefined. Why sal says it's equal to 1? at 9:37", + "A": "0 factorial does not equal zero. By definition it equals 1.", + "video_name": "iPwrDWQ7hPc", + "timestamps": [ + 577 + ], + "3min_transcript": "4 minus 2 is 2. I think you see a pattern here. a to the 4th, a to the 3rd, a squared, and then times b to the k. Well, k is 2 now, so b squared, and you see a pattern again. You could say b to the 0, b to the 1, b squared, and we only have two more terms to add here, plus 4 choose 3, 4 choose 3 times 4 minus 3 is 1, times a, or a to the 1st, I guess we could say, and then b to the 3rd power, times a to the 1st b to the third, and then only one more term, plus 4 choose, 4 choose 4. k is now 4. This is going to be our last term right now. We're getting k goes from 0 all the way to 4, 4 choose 4. a to the 4 minus 4, that's just going to be 1, a to the 0, that's just 1, so we're going to be left with just b to the k power, We're almost done. We've expanded it out. We just need it figure out what 4 choose 0, 4 choose 1, 4 choose 2, et cetera, et cetera are, so let's figure that out. We could just apply this over and over again. So 4 choose 0, 4 choose 0 is equal to 4 factorial over 0 factorial times 4 minus 0 factorial. That's just going to be 4 factorial again. 0 factorial, at least for these purposes, we are defining to be equal to 1, so this whole thing is going to be equal to 1, so this coefficient is 1. Let's keep going here. So 4 choose 1 is going to be 4 factorial over 1 factorial times 4 minus 1 factorial, 4 minus 1 factorial, so 3 factorial. What's this going to be? 1 factorial is just going to be 1. 3 factorial is 3 times 2 times 1. 4 times 3 times 2 times 1 over 3 times 2 times 1 is just going to leave us with 4. This right over here is just going to be 4. Then we need to figure out what 4 choose 2 is. 4 choose 2 is going to be 4 factorial over 2 factorial times what's 4 minus ... this is going to be n minus k, 4 minus 2 over 2 factorial. So what is this going to be? Let me scroll over to the right a little bit. This is going to be 4 times 3 times 2 times 1 over 2 factorial is 2, over 2 times 2. This is 2, this is 2, so 2 times 2 is same thing as 4. We're left with 3 times 2 times 1, which is equal to 6. That's equal to 6. Then what is 4 choose 3? I'll use some space down here. So 4 choose 3," + }, + { + "Q": "At 4:21, what is the sideways W symbol, and what does it mean?", + "A": "The sideways W is a Greek letter known as a Sigma. It indicates a pattern.", + "video_name": "iPwrDWQ7hPc", + "timestamps": [ + 261 + ], + "3min_transcript": "so 2ab squared, and then b times a squared is ba squared, or a squared b, a squared b. I'll multiply b times all of this stuff. Now let's multiply a times all this stuff. a times b squared is ab squared, ab squared. a times 2ab is 2a squared b, 2a squared b, and then a times a squared is a to the 3rd power. Now when we add all of these things together, we get, we get a to the 3rd power plus, let's see, we have 1 a squared b plus another, plus 2 more a squared b's. That's going to be 3a squared b plus 3ab squared. 2ab squared plus another ab squared plus b to the 3rd power. Just taking some of the 3rd power, this already took us a little reasonable amount of time, and so you can imagine how painful it might get to do something like a plus b to the 4th power, or even worse, if you're trying to find a plus b to the 10th power, or to the 20th power. This would take you all day or maybe even longer than that. It would be incredibly, incredibly painful. That's where the binomial theorem becomes useful. What is the binomial theorem? The binomial theorem tells us, let me write this down, binomial theorem. Binomial theorem, it tells us that if we have a binomial, and I'll just stick with the a plus b for now, if I have, and I'm going to try to color code this a little bit, if I have the binomial a plus b, and I'm going to raise it the nth power, I'm going to raise this to the nth power, the binomial theorem tells us that this is going to be equal to, and the notation is going to look a little bit complicated at first, but then we'll work through an actual example, is going to be equal to the sum from k equals 0, k equals 0 to n, this n and this n are the same number, of ... I don't want to ... that's kind of a garish color ... of n choose k, n choose k, and we'll review that in a second; this comes straight out of combinatorics; n choose k times a to the n minus k, n minus k, times b, times b to the k," + }, + { + "Q": "at 2:35, when looking to draw the vector [1,2], I don't understand why the x component should be 1 and the Y component should be 2. Isn't the desired output, based on y = 1 and x = 2?", + "A": "There s no mistake in there. The input coordinates are (2,1). And according to the partial derivative of the given function, the output is a vector field with x-component equal to the ordinate and y- component equal to the abscissa. So, the output vectors are given by = yi + xj , where i and j are the basis vectors for x and y axes. So the output for (2,1) will be 1i+2j", + "video_name": "ZTbTYEMvo10", + "timestamps": [ + 155 + ], + "3min_transcript": "y equals two over x. And that's where you would see something like this. So all of these lines, they're representing constant values for the function. And now I want to take a look at the gradient field. And the gradient, if you'll remember, is just a vector full of the partial derivatives of f. And let's just actually write it out. The gradient of f, with our little del symbol, is a function of x and y. And it's a vector-valued function whose first coordinate is the partial derivative of f with respect to x. And the second component is the partial derivative with respect to y. So when we actually do this for our function, we take the partial derivative with respect to x. X looks like a variable. Y looks like a constant. The derivative of this whole thing is just equal to that constant, y. And then kind of the reverse for when you take the partial derivative with respect to y. X looks like a constant. And the derivative is just that constant, x. And this can be visualized as a vector field in the xy plane as well. You know, at every given point, xy, so you kind of go like x equals two, y equals one, let's say. So that would be x equals two, y equals one. You would plug in the vector and see what should be output. And at this point, the point is two, one. The desired output kind of swaps those. So we're looking somehow to draw the vector one, two. So you would expect to see the vector that has an x component of one and a y component of two. Something like that. But it's probably gonna be scaled down because of the way we usually draw vector fields. And the entire field looks like this. So I'll go ahead and erase what I had going on. Since this is a little bit clearer. And remember, we scaled down all the vectors. The color represents length. So red here is super-long. Blue is gonna be kind of short. And one thing worth noticing. if the vector is crossing a contour line, it's perpendicular to that contour line. Wherever you go. this vector's perpendicular to the contour line. Over here, perpendicular to the contour line. And this happens everywhere. And it's for a very good reason. And it's also super-useful. So let's just think about what that reason should be. Let's zoom in on a point. So I'm gonna clear up our function here. Clear up all of the information about it. And just zoom in on one of those points. So let's say like right here. We'll take that guy and kind of imagine zooming in and saying what's going on in that region? So you've got some kind of contour line. And it's swooping down like this. And that represents some kind of value. Let's say that represents the value f equals two. And, you know, it might not be a perfect straight line. But the more you zoom in, the more it looks like a straight line. And when you want to interpret the gradient vector." + }, + { + "Q": "at 1:21 didnt understand the formulae", + "A": "The formula is showing there is a correlation between the angle of the sector and the area of the sector. If we know the angle and the area of the whole circle we can find the area of the sector, since they are similar. For more detail: 0:28", + "video_name": "u8JFdwmBvvQ", + "timestamps": [ + 81 + ], + "3min_transcript": "A circle with area 81 pi has a sector with a 350-degree central angle. So this whole sector right over here that's shaded in, this pale orange-yellowish color, that has a 350-degree central angle. So you see the central angle, it's a very large angle. It's going all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle-- over 360. So the area of the sector over the total area over the total degrees in a circle. And then we just can solve for area of a sector by multiplying both sides by 81 pi. 81 pi, 81 pi-- so these cancel out. 350 divided by 360 is 35/36. And so our area, our sector area, is equal to-- let's see, in the numerator, we have 35 times-- instead of 81, let's see, that's going to be 9 times 9 pi. And in the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9, and so we are left with 35 times 9. And neither of these are divisible by 4, so that's about as simplified as we can get it. 35 times 9, it's going to be 350 minus 35, which would be 315, I guess. Did I do that right? Yeah, it's going to be 270 plus 45, which is 315 pi over 4. 315 pi over 4 is the area of the sector." + }, + { + "Q": "At 1:40 where did he get nine times nine from? Also he divided the 350/360 by ten should he divide the other side by ten also?", + "A": "Hey Janet, 9*9 is the same thing as 81. With fractions, if you divide the numerator by 10, you divide the denominator by 10 as well. You don t need to divide the other side, it s just simplifying a fraction and still the same number after all.", + "video_name": "u8JFdwmBvvQ", + "timestamps": [ + 100 + ], + "3min_transcript": "A circle with area 81 pi has a sector with a 350-degree central angle. So this whole sector right over here that's shaded in, this pale orange-yellowish color, that has a 350-degree central angle. So you see the central angle, it's a very large angle. It's going all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle-- over 360. So the area of the sector over the total area over the total degrees in a circle. And then we just can solve for area of a sector by multiplying both sides by 81 pi. 81 pi, 81 pi-- so these cancel out. 350 divided by 360 is 35/36. And so our area, our sector area, is equal to-- let's see, in the numerator, we have 35 times-- instead of 81, let's see, that's going to be 9 times 9 pi. And in the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9, and so we are left with 35 times 9. And neither of these are divisible by 4, so that's about as simplified as we can get it. 35 times 9, it's going to be 350 minus 35, which would be 315, I guess. Did I do that right? Yeah, it's going to be 270 plus 45, which is 315 pi over 4. 315 pi over 4 is the area of the sector." + }, + { + "Q": "At 0:28, Mr. Khan mentions a ratio. What is that?", + "A": "He basically created a proportion using the values given in the circle.", + "video_name": "u8JFdwmBvvQ", + "timestamps": [ + 28 + ], + "3min_transcript": "A circle with area 81 pi has a sector with a 350-degree central angle. So this whole sector right over here that's shaded in, this pale orange-yellowish color, that has a 350-degree central angle. So you see the central angle, it's a very large angle. It's going all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle-- over 360. So the area of the sector over the total area over the total degrees in a circle. And then we just can solve for area of a sector by multiplying both sides by 81 pi. 81 pi, 81 pi-- so these cancel out. 350 divided by 360 is 35/36. And so our area, our sector area, is equal to-- let's see, in the numerator, we have 35 times-- instead of 81, let's see, that's going to be 9 times 9 pi. And in the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9, and so we are left with 35 times 9. And neither of these are divisible by 4, so that's about as simplified as we can get it. 35 times 9, it's going to be 350 minus 35, which would be 315, I guess. Did I do that right? Yeah, it's going to be 270 plus 45, which is 315 pi over 4. 315 pi over 4 is the area of the sector." + }, + { + "Q": "The result of the composition of Ix and g has to be the same as the result of the composition of h and f and g when one inputs a 'y' value and gets an 'x' value (at time marker 16:40). But how is showing that if the results are the same, then the functions are the same as well? (abstractly speaking)", + "A": "Speaking any kind of way (abstractly or otherwise), x = g(y) = I_X(g(y)) = h(f(g(y))) = h((I_Y)(y)) = h(y). So g(y) = h(y), and g = h. (I don t really understand this either yet.) : |", + "video_name": "-eAzhBZgq28", + "timestamps": [ + 1000 + ], + "3min_transcript": "I could do the same thing here with h. I just take a point here, apply h, then apply f back. I should just go back to that point. That's all of what this is saying. So let's go back to the question of whether g is unique. Can we have two different inverse functions g and h? So let's start with g. Remember g is just a mapping from Y to X. So this is going to be equal to, this is the same thing as the composition of the identity function over x with g. To show you why that's the case, remember g just goes from-- these diagrams get me confused very quickly-- so let's say this is x and this is y. Remember g is a mapping from y to x. So g will take us there. There's a mapping from y to x. mapping, or the identity function in composition with this. Because all this is saying is you apply g, and then you apply the identity mapping on x. So obviously you're going to get to the exact same mapping or the exact same point. So these are equivalent. But what is another way of writing the identity mapping on x? What's another way of writing that? Well by definition, if h is another inverse of f, this is true. So I can replace this in this expression with a composition of h with f. So this is going to be equal to the composition of h with f, and the composition of that with g. You might want to put parentheses here. I'll do it very lightly. You might want to put parentheses there. But I showed you a couple of videos ago that the composition of functions, or of transformations, is associative. It doesn't matter if you put the parentheses there or if you put the parentheses there. Actually I'll do that. I'll put the parentheses there at first just so you can as that right there. But we know that composition is associative. So this is equal to the composition of h with the composition of f and g. Now what is this equal to, the composition of f and g? Well it's equal to, by definition, it's equal to the identity transformation over y. So this is equal to h composed with, or the composition of h with, the identity function over y with this right here. Now what is this going to be? Remember h is a mapping from y to x. Let me redraw it. So that's my x and that is my y. h could take some element in y and gives me some element in x. If I take the composition of the identity in y-- so that's essentially I take some element, let me do it this" + }, + { + "Q": "3:41 OK really confused. why do the repeating numbers start at 4 and not 1. Sal writes 414141... but shouldnt it be 14141414...\ncan anyone explain this?", + "A": "x=.714141414... If you multiplied by x by 10, then that would move it one place to the right, or 7.14141414... and the 141414... would start immediately after the decimal point. Since you have to multiply by 100, however, you move it two places to the right and get: 7 1. 4 1 4 1 4 1 4. The numbers after the decimal must start with the 4 first since you moved it two places to the right. The 141414... pattern is still there. You re just starting at a different place.", + "video_name": "Ihws0d-WLzU", + "timestamps": [ + 221 + ], + "3min_transcript": "are left with x is equal to 36 over 99. And both the numerator and the denominator is divisible by 9, so we can reduce this. If we divide the numerator by 9, we get 4. The denominator by 9-- we get 11. So 0.363636 forever and forever repeating is 4/11. Now let's do another interesting one. Let's say we have the number 0.714, and the 14 is repeating. And so this is the same thing. So notice, the 714 isn't going to repeat. Just the 14 is going to repeat. So this is 0.7141414, on and on and on and on. So let's set this equal to x. the decimal all the way clear of 714. But you actually don't want to do that. You want to shift it just enough so that the repeating pattern can be right under itself when you do the subtraction. So again in this situation, even though we have three numbers behind the decimal point, because only two of them are repeating, we only want to multiply by 10 to the second power. So once again, you want to multiply by 100. So you get 100x is equal to-- we're moving the decimal two to the right, one, two-- so it's going to be 71.4141, on and on and on and on. So it's going to be 71.4141414 and on and on and on. And then let me rewrite x right below this. We have x is equal to 0.7141414. And notice, now the 141414's, they're So it's going to work out when we subtract. So let's subtract these things. 100x minus x is 99x. And this is going to be equal to-- these 1414's are going to cancel with those 1414's. And we have 71.4 minus 0.7. And we can do this in our head, or we can borrow if you like. This could be a 14. This is a 0. So you have 14 minus 7 is 7 and then 70 minus 0. So you have 99x is equal to 70.7. And then we can divide both sides by 99. And you could see all of the sudden something strange is happening because we still have a decimal. But we can fix that up at the end. So let's divide both sides by 99. You get x is equal to 70.7 over 99. Now obviously, we haven't converted this into a pure fraction yet. We still have a decimal in the numerator. But that's pretty easy to fix." + }, + { + "Q": "This video was really helpful, I just have one question. How do you whether to use 10x or 100x or 1000x at 1:01 in the video?", + "A": "How many different numbers are repeating? If it is .333... , one repeating times 10, if it is ,232323... two repeating, so times 100, three repeating .234234234... times 1000. A zero for each repeating number.", + "video_name": "Ihws0d-WLzU", + "timestamps": [ + 61 + ], + "3min_transcript": "In the last video, we did some examples where we had one digit repeating on and on forever, and we were able to convert those into fractions. In this video, we want to tackle something a little bit more interesting, which is multiple digits repeating on and on forever. So let's say I had 0.36 repeating, which is the same thing as 0 point-- since the bar's over the 3 and the 6, both of those repeat-- 363636. And it just keeps going on and on and on like that forever. Now the key to doing this type of problem is, so like we did in the last video, we set this as equal to x. And instead of just multiplying it by 10-- 10 would only shift it one over-- we want to shift it over enough so that when we line them up, the decimal parts will still line up with each other. And to do that we, want to actually shift the decimal space two to the right. And to shift it two to the right, we have to have multiplied by 100 or 10 to the second power. So 100x is going to be equal to what? So 100x is going to be equal to-- the decimal is going to be there now, so it's going to be 36.363636 on and on and on forever. And then let me rewrite x over here. We're going to subtract that from the 100x. x is equal to 0.363636 repeating on and on forever. And notice when we multiplied by 100x, the 3's and the 6's still line up with each other when we align the decimals. And you want to make sure you get the decimals lined up appropriately. And the reason why this is valuable is now that when we subtract x from 100x, the repeating parts will cancel out. So let's subtract. Let us subtract these two things. So on the left-hand side, we have 100x minus x. So that gives us 99x. And then we get, on the right-hand side, this part cancels out with that part. And we're just left with 36. are left with x is equal to 36 over 99. And both the numerator and the denominator is divisible by 9, so we can reduce this. If we divide the numerator by 9, we get 4. The denominator by 9-- we get 11. So 0.363636 forever and forever repeating is 4/11. Now let's do another interesting one. Let's say we have the number 0.714, and the 14 is repeating. And so this is the same thing. So notice, the 714 isn't going to repeat. Just the 14 is going to repeat. So this is 0.7141414, on and on and on and on. So let's set this equal to x." + }, + { + "Q": "At 4:04 can we multiply by 10?", + "A": "If we multiply by 10, the decimal point would only shift to the right one point. we need the point shifted over twice, so we multiply by 100.", + "video_name": "Ihws0d-WLzU", + "timestamps": [ + 244 + ], + "3min_transcript": "are left with x is equal to 36 over 99. And both the numerator and the denominator is divisible by 9, so we can reduce this. If we divide the numerator by 9, we get 4. The denominator by 9-- we get 11. So 0.363636 forever and forever repeating is 4/11. Now let's do another interesting one. Let's say we have the number 0.714, and the 14 is repeating. And so this is the same thing. So notice, the 714 isn't going to repeat. Just the 14 is going to repeat. So this is 0.7141414, on and on and on and on. So let's set this equal to x. the decimal all the way clear of 714. But you actually don't want to do that. You want to shift it just enough so that the repeating pattern can be right under itself when you do the subtraction. So again in this situation, even though we have three numbers behind the decimal point, because only two of them are repeating, we only want to multiply by 10 to the second power. So once again, you want to multiply by 100. So you get 100x is equal to-- we're moving the decimal two to the right, one, two-- so it's going to be 71.4141, on and on and on and on. So it's going to be 71.4141414 and on and on and on. And then let me rewrite x right below this. We have x is equal to 0.7141414. And notice, now the 141414's, they're So it's going to work out when we subtract. So let's subtract these things. 100x minus x is 99x. And this is going to be equal to-- these 1414's are going to cancel with those 1414's. And we have 71.4 minus 0.7. And we can do this in our head, or we can borrow if you like. This could be a 14. This is a 0. So you have 14 minus 7 is 7 and then 70 minus 0. So you have 99x is equal to 70.7. And then we can divide both sides by 99. And you could see all of the sudden something strange is happening because we still have a decimal. But we can fix that up at the end. So let's divide both sides by 99. You get x is equal to 70.7 over 99. Now obviously, we haven't converted this into a pure fraction yet. We still have a decimal in the numerator. But that's pretty easy to fix." + }, + { + "Q": "At 9:22, how do you turn 0.00123 with the 3 repeating into a decimal?", + "A": "x = 0.001233... 100000x = 123.33... 10000x = 12.33... 90000x = 111 x = 111/90000 = 37/30000", + "video_name": "Ihws0d-WLzU", + "timestamps": [ + 562 + ], + "3min_transcript": "And then you can divide both sides of this by 999. And you are left with x is equal to 3,254/999. And so obviously, this is an improper fraction. The numerator is larger than the denominator. You could convert this to a proper fraction if you like. One way, you could have just tried to figure out what to the 0.257 repeating forever is equal to and just had the 3 being the whole number part of a mixed fraction. Or you could just divide 999 into 3,254. Actually, we could do that pretty straightforwardly. It goes into it three times, and the remainder-- well, let me just do it, just to go through the motions. So 999 goes into 3,254. It'll go into it three times. And we know that because this is originally 3.257, So 3 times 9 is 27. But we have to add the 2, so it's 29. 3 times 9 is 27. We have a 2, so it's 29. And so we are left with, if we subtract, if we regroup or borrow or however we want to call it, this could be a 14. And then this could be a 4. Let me do this in a new color. And then the 4 is still smaller than this 9, so we need to regroup again. So then this could be a 14, and then this could be a 1. But this is smaller than this 9 right over here, so we regroup again. This would be an 11, and then this is a 2. 14 minus 7 is 7. 14 minus 9 is 5. 11 minus 9 is 2. So we are left with-- did I do that right? Yep-- so this is going to be equal to 3 and 257/999." + }, + { + "Q": "at 4:00, he could have just multiplied it by ten. that would have been enough to get the seven on the other side of the decimal point. why'd he multiply it by 100?", + "A": "That s because you still want the decimal points and the repeating part (which was 141414...in this case) to line up.", + "video_name": "Ihws0d-WLzU", + "timestamps": [ + 240 + ], + "3min_transcript": "are left with x is equal to 36 over 99. And both the numerator and the denominator is divisible by 9, so we can reduce this. If we divide the numerator by 9, we get 4. The denominator by 9-- we get 11. So 0.363636 forever and forever repeating is 4/11. Now let's do another interesting one. Let's say we have the number 0.714, and the 14 is repeating. And so this is the same thing. So notice, the 714 isn't going to repeat. Just the 14 is going to repeat. So this is 0.7141414, on and on and on and on. So let's set this equal to x. the decimal all the way clear of 714. But you actually don't want to do that. You want to shift it just enough so that the repeating pattern can be right under itself when you do the subtraction. So again in this situation, even though we have three numbers behind the decimal point, because only two of them are repeating, we only want to multiply by 10 to the second power. So once again, you want to multiply by 100. So you get 100x is equal to-- we're moving the decimal two to the right, one, two-- so it's going to be 71.4141, on and on and on and on. So it's going to be 71.4141414 and on and on and on. And then let me rewrite x right below this. We have x is equal to 0.7141414. And notice, now the 141414's, they're So it's going to work out when we subtract. So let's subtract these things. 100x minus x is 99x. And this is going to be equal to-- these 1414's are going to cancel with those 1414's. And we have 71.4 minus 0.7. And we can do this in our head, or we can borrow if you like. This could be a 14. This is a 0. So you have 14 minus 7 is 7 and then 70 minus 0. So you have 99x is equal to 70.7. And then we can divide both sides by 99. And you could see all of the sudden something strange is happening because we still have a decimal. But we can fix that up at the end. So let's divide both sides by 99. You get x is equal to 70.7 over 99. Now obviously, we haven't converted this into a pure fraction yet. We still have a decimal in the numerator. But that's pretty easy to fix." + }, + { + "Q": "at around 5:02, why didnt he multiply 4 into the number to the right of the greater than symbol? If you are making a change to one side wouldn't you have to do it to the other, because it would affect the end result?", + "A": "He did. He said, let s multiply both sides by 4 and he pointed to both sides of the inequality. 4 ( p\u00c2\u00b2 - 17/4 p+ 1 > 0 ) becomes 4 \u00e2\u0088\u0099 p\u00c2\u00b2 - 4 \u00e2\u0088\u0099 17/4 p + 4 \u00e2\u0088\u0099 1 > 4 \u00e2\u0088\u0099 0 which is 4p\u00c2\u00b2 - 17p + 4 > 0 because zero times anything is still zero He probably didn t mention zero times 4 is zero because he guessed we knew that by the time we are doing quadratic inequalities because we run into it so much when solving quadratics and other equations.", + "video_name": "GDppV18XDCs", + "timestamps": [ + 302 + ], + "3min_transcript": "So we have just set up the first part. We have written an inequality that models the situation. Now let's actually solve this inequality. And so to do that, I will just expand 1 minus p squared out. 1 minus p squared is the same thing as-- well, I'll just multiply it out. So this is going to be 1 squared minus 2p plus p squared. And that's going to be greater than 2 and 1/4 p. Now let's see. If we subtract 2 and 1/4 p from both sides, we're going to be left with-- and I'm going to reorder this. We're going to get p squared. So you have minus 2p minus 2 and 1/4 p, so that's going to get us minus 4 and 1/4 p. greater than 0. And so let's think about solving this quadratic right over here. And under which circumstances is this greater than 0? To think about it, let's factor it. And actually, before we factor it, let's simplify it a little bit. I don't like having this 17/4 right over here, so let's multiply both sides times 4. And since 4 is a positive number, it's not going to change the sign, the direction of this inequality. So we could rewrite this as 4p squared minus 17p plus 4 is greater than 0. What are the roots of this? And we could use the quadratic formula if we wanted to do it really quick. We could probably do it other ways. the square root of negative 17 squared-- b squared-- so that's 289 minus 4 times a times c. Well a times c is 16 times 4, so minus 64. All of that over 2 times a-- all of that over 8. So that's 17 plus or minus-- let's see, this is the square root of 225 over 8, which is equal to 17 plus or minus 15 over 8, which is equal to-- let's see, 17 minus 15 over 8 is 2/8-- which is equal to 2/8 or 1/4. So that's one of them. That's when we take the minus. And if we add 17 plus 15, that gets us to 32 divided by 8" + }, + { + "Q": "in 1:16, why did the voice changed?", + "A": "Yes, Vi did get farther from her mic, thats why her voice got deeper and quieter", + "video_name": "4tsjCND2ZfM", + "timestamps": [ + 76 + ], + "3min_transcript": "So say your vector field green bean casserole is in the oven, and now it's time to think about a nice, crispy onion topping. Normal people might just use, for instance, French's French fried onions in a can, put super awesome people use a real French person, and real fresh onions, to make their own fresh onion toroids. And they fry free linked with the Brunnian property to get Borromean onion rings. The Borromean rings show up in many forms, they come flat and in 3D, round, rectangly, triangly. But, the important thing is not the way the rings appear, but the way they are connected to each other. The thing about the Borromean rings is that no two of the rings are actually linked together. Ignore the pink and look at just the green and brown. They're sitting on top of each other, not linked. And if you just look at the green and pink, or pink and brown, it's the same thing. And yet, all three together are linked inseparably. So to make your Borromean rings out of onion rings, you will have to cut one of your rings and then fasten it back together with a toothpick or something, which can be removed after frying. Or you can use the fourth dimension. And luckily I have a four-dimensional guest to help me out. If you're stuck in three dimensions, you can think of it like this. Now, the third ring which I have cut, is going to go outside of the outside ring, but inside of the inside ring. Each ring is wholly out of, and wholly inside of the other two rings so that no two are linked, but all three are. You can also think of laying two on each other flat, one on top of the other. And then having the third weave through them, so that it goes over the one on top, and under the one on bottom. The result can be made to be flatter or more spherical, in some you can see the relationship that Borromean rings have with braids. Sure the orange, yellow, and red ribbons are all twisted together, but no two strands are twisted together. If I pull out the orange one, the other two fall apart. Some people and cultures and stuff think of this togetherness property as a metaphor for unity. So when you eat Borromean onion rings, you get to feel all deep and symbolic. But don't forget to save enough to put on top of your green bean matherole. And there we go. At this point I've got a gelatinous cranberry cylinder, bread spheres with butter prism, masked potatoes, a vector field green bean matherole with Borromean onion rings, All I need is a double helix cut ham, and of course, the crowning glory of this feast which I will tell you about next time." + }, + { + "Q": "At 2:37 What does beat a dead horse mean?", + "A": "It is an idiom, meaning it is not to be taken literally, and it means to waste time and/or energy repeating something in excess or doing something otherwise that will be non-helpful or productive.", + "video_name": "AuD2TX-90Cc", + "timestamps": [ + 157 + ], + "3min_transcript": "We could say, and one tenth and five hundredths, or we could just say, look, this is fifteen hundredths. One tenth is ten hundredths. So one tenth and five hundredths is fifteen hundredths. So maybe I can write it like this: sixty-three and fifteen hundredths. Just like that. Now, it might have been a little bit more natural to say, how come I don't say one tenth and then five And you could, but that would just make it a little bit harder for someone's brain to process it when you say it. So it could have been sixty-three-- so let me copy and paste that. It could be sixty-three and, and then you would write, one Sixty-three and one tenth and five hundredths is hard for most people's brains to process. But if you say, fifteen hundredths, people get what you're saying. Not to beat a dead horse, but this right here, this is 1/10 right here and then this is 5/100, 5 over 100. But if you were to add these two, If you were to add 1/10 plus 5/100 -- so let's do that. If you were to add 1/10 plus 5/100, how would you do it? You need a common denominator. numerator and denominator of this character by 10. You get 10 on the top and 100 on the bottom. 1/10 is the same thing as 10 over 100. 10/100 plus 5/100 is equal to 15 over 100, so this piece right here is equal to 15/100. And that's why we say sixty-three and fifteen hundredths." + }, + { + "Q": "At 4:14, how come he didn't turn the fraction into a decimal?", + "A": "Sal solves two problems where a missing side length is found by proving that triangles are similar and using this to find the measure.", + "video_name": "R-6CAr_zEEk", + "timestamps": [ + 254 + ], + "3min_transcript": "Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it in just the way that we've written down the similarity. If this is true, then BC is the corresponding side to DC. So we know that the length of BC over DC right over here is going to be equal to the length of-- well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to-- what's the corresponding side to CE? The corresponding side over here is CA. This is last and the first. Last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. And now, we can just solve for CE. Well, there's multiple ways that you could think about this. You could cross-multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE. 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2/5, or 2.4. And we're done. We were able to use similarity to figure out this side just knowing that the ratio between the corresponding sides are going to be the same. Now, let's do this problem right over here. Let's do this one. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also, in both triangles-- so I'm looking at triangle CBD and triangle CAE-- they both share this angle up here." + }, + { + "Q": "At 2:51, how do you figure out which line segment to put together when you are trying to figure out the missing length?", + "A": "Since you are looking for the side CE, notice that it is the third letter to first letter of second triangle. With the congruency statement, the same two letters are CA, so they are one of three pair of congruent sides. Does this answer your question?", + "video_name": "R-6CAr_zEEk", + "timestamps": [ + 171 + ], + "3min_transcript": "but we don't have to. So we already know that they are similar. And actually, we could just say it. Just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar, even before doing that. So we already know that triangle-- I'll color-code it so that we have the same corresponding vertices. And that's really important-- to know what angles and what sides correspond to what side so that you don't mess up your, I guess, your ratios or so that you do know what's corresponding to what. So we know triangle ABC is similar to triangle-- so this vertex A corresponds to vertex E over here. It's similar to vertex E. And then, vertex B right over here corresponds to vertex D. EDC. Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it in just the way that we've written down the similarity. If this is true, then BC is the corresponding side to DC. So we know that the length of BC over DC right over here is going to be equal to the length of-- well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to-- what's the corresponding side to CE? The corresponding side over here is CA. This is last and the first. Last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. And now, we can just solve for CE. Well, there's multiple ways that you could think about this. You could cross-multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE. 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2/5, or 2.4." + }, + { + "Q": "at 6:16, would it work if you wrote CA/CB = CE/CD instead of CB/CA=CD/CE? because I got pretty confused.", + "A": "yes it would work", + "video_name": "R-6CAr_zEEk", + "timestamps": [ + 376 + ], + "3min_transcript": "And we're done. We were able to use similarity to figure out this side just knowing that the ratio between the corresponding sides are going to be the same. Now, let's do this problem right over here. Let's do this one. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also, in both triangles-- so I'm looking at triangle CBD and triangle CAE-- they both share this angle up here. but we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other. And once again, this is an important thing to do, is to make sure that you write it in the right order when you write your similarity. We now know that triangle CBD is similar-- not congruent-- it is similar to triangle CAE, which means that the ratio of corresponding sides are going to be constant. So we know, for example, that the ratio between CB to CA-- so let's write this down. is going to be equal to the ratio of CD over CE. And we know what CB is. CB over here is 5. We know what CA is. And we have to be careful here. It's not 3. CA, this entire side is going to be 5 plus 3. So this is going to be 8. And we know what CD is. CD is going to be 4. And so once again, we can cross-multiply. We have 5CE. 5 times CE is equal to 8 times 4. 8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to think about that, 6 and 2/5." + }, + { + "Q": "At 2:20 Did When Sal said Sin 32, Does that also mean Sin A?", + "A": "at 2:20 when sal says sin32, yes that is equivalent to sinA", + "video_name": "yiH6GoscimY", + "timestamps": [ + 140 + ], + "3min_transcript": "We are told that the cosine of 58 degrees is roughly equal to 0.53. And that's roughly equal to, because it just keeps going on and on. I just rounded it to the nearest hundredth. And then we're asked, what is the sine of 32 degrees? And I encourage you to pause this video and try it on your own. And a hint is to look at this right triangle. One of the angles is already labeled 32 degrees. Figure out what all of the angles are, and then use the fundamental definitions, your sohcahtoa definitions, to see if you can figure out what sine of 32 degrees is. So I'm assuming you've given a go at it. Let's work it through now. So we know that the sum of the angles of a triangle add up to 180. Now in a right angle, one of the angles is 90 degrees. So that means that the other two must add up to 90. These two add up to 90 plus another 90 is going to be 180 degrees. Or another way to think about is that the other two non-right angles are going to be complementary. So what plus 32 is equal to 90? So this right over here is going to be 58 degrees. Well, why is that interesting? Well, we already know what the cosine of 58 degrees is equal to. But let's think about it in terms of ratios of the lengths of sides of this right triangle. Let's just write down sohcahtoa. Soh, sine, is opposite over hypotenuse. Cah, cosine, is adjacent over hypotenuse. Toa, tangent, is opposite over adjacent. So we could write down the cosine of 58 degrees, which we already know. If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse. This is a 58 degree angle. The side that is adjacent to it is-- let me do it in this color-- is side BC right over here. It's one of the sides of the angle, the side of the angle that is not the hypotenuse. The other side, this over here, is a hypotenuse. So this is going to be the adjacent, the length of the adjacent side, BC, over the length of the hypotenuse. Now let's think about what the sine of 32 degrees would be. Well, sine is opposite over hypotenuse. So now we're looking at this 32 degree angle. What side is opposite it? Well, it opens up onto BC. And what's the length of the hypotenuse? It's AB. Notice, the sine of 32 degrees is BC over AB. The cosine of 58 degrees is BC over AB. Or another way of thinking about it, the sine of this angle is the same thing as the cosine of this angle. So we could literally write the sine-- I want to do that in that pink color-- the sine of 32 degrees" + }, + { + "Q": "At 4:29 What is the exact difference between obtuse and acute triangles", + "A": "in the obtuse triangle, it has one obtuse angle (bigger than a right angle) and in an acute angle, all angles are smaller that right angle. Ex: acute:all angles 60 ( 60+60+60=180) Ex: obtuse: one angle 120 another 35 and another 25 Hope this helped! Please vote this", + "video_name": "D5lZ3thuEeA", + "timestamps": [ + 269 + ], + "3min_transcript": "so it meets the constraints for an isosceles. So by that definition, all equilateral triangles are also isosceles triangles. But not all isosceles triangles are equilateral. So for example, this one right over here, this isosceles triangle, clearly not equilateral. All three sides are not the same. Only two are. But both of these equilateral triangles meet the constraint that at least two of the sides are equal. Now down here, we're going to classify based on angles. An acute triangle is a triangle where all of the angles are less than 90 degrees. So for example, a triangle like this-- maybe this is 60, let me draw a little bit bigger so I can draw the angle measures. I want to make it a little bit more obvious. So let's say a triangle like this. If this angle is 60 degrees, maybe this one right over here is 59 degrees. And then this angle right over here is 61 degrees. Notice they all add up to 180 degrees. This would be an acute triangle. Notice all of the angles are less than 90 degrees. A right triangle is a triangle that has one angle that is exactly 90 degrees. So for example, this right over here would be a right triangle. Maybe this angle or this angle is one that's 90 degrees. And the normal way that this is specified, people wouldn't just do the traditional angle measure and write 90 degrees here. They would draw the angle like this. And that tells you that this angle right over here is 90 degrees. And because this triangle has a 90 degree angle, and it could only have one 90 degree angle, this is a right triangle. So that is equal to 90 degrees. Now you could imagine an obtuse triangle, based on the idea that an obtuse angle is larger than 90 degrees, an obtuse triangle is a triangle that has one angle that is larger than 90 degrees. So let's say that you have a triangle that looks like this. Maybe this is 120 degrees. And then let's see, let me make sure that this would make sense. Maybe this is 25 degrees. Or maybe that is 35 degrees. And this is 25 degrees. Notice, they still add up to 180, or at least they should." + }, + { + "Q": "At 8:00, wouldn't you have to times the number of years and the rate the interest is compounded for raising everything?", + "A": "Yes, I suppose. But in this video the loan is only for one year.", + "video_name": "BKGx8GMVu88", + "timestamps": [ + 480 + ], + "3min_transcript": "your going have to multiply by this again. Times 1.083 repeating, and so that would get you 1.083 repeating squared. If you went all the way down 12 months ... let me get myself some space here. If you went all the way down 12 months ... let me just. I should way from the beginning 12 months, so another 10 months. What's the total interest you would have to pay over a year if you weren't able to keep coming up with the money? If you had to keep re-borrowing it. I kept compounding that interest. Well, you're going have to pay 1.083 to the ... this is for 1 month. You could view this as to the first power. This is for 2 months, so you're going have to pay this to the 12th power. We have compounded over 12 periods, 8 1/3% over 12 periods. If you wanted to write it in this form right over here, this would be the same Our original principal times 1 plus 100% divided by 12. Now we've divided our 100% into 12 periods, and we're going to compound that 12 times. We're going to take that to the 12th power. What is this going to equal to? This buisness over here. We can get a calculator out for that. I'll get my TI-85 out. What is this going to be equal to? We could do it a couple of ways. This is 1.083 repeating. Let's get our calculator out. We could do it a couple of ways. Let me write it this way. Your going to get the same value. I don't have to rewrite this one. I just did that there to kind of hopefully you'd see the kind of structure in this expression. 1 plus ... 100% is the same thing as 1. 1 divided by 12 to the 12th power. 2.613, I'll just round. interesting game you all most forgot about your financial troubles, and you're just intrigued by what happens if we keep going this. Here we compounded just ... we have 100% over here. Here we do 50% every 6 months. Here we do a 12th of 100%, 8 1/3% every 12 months until we get to this number. What happens if we did every day? Every day. If I borrowed a one dollar, and I'd say well gee I'm just going to ... each day I'm going to charge you charge you one three hundred sixty-fifth of a 100%. So, 100% divided 365, and I'm going to compound that 365 times. You're curious mathematically. You say well, what do we get then? What do we get after a year?" + }, + { + "Q": "At 0:35 couldn't you just make the two triangles into one rectangle with a height of 3\" and a width of 5\"?", + "A": "No, but you could make the two triangles into a rectangle with a height of 3 and a width of 4 (the entire base of the pentagon is 8, so half of that would be 4)", + "video_name": "7S1MLJOG-5A", + "timestamps": [ + 35 + ], + "3min_transcript": "Find the area and perimeter of the polygon. So let's start with the area first. So the area of this polygon-- there's kind of two parts of this. First, you have this part that's kind of rectangular, or it is rectangular, this part right over here. And that area is pretty straightforward. It's just going to be base times height. So area's going to be 8 times 4 for the rectangular part. And then we have this triangular part up here. So we have this area up here. And for a triangle, the area is base times height times 1/2. And that actually makes a lot of sense. Because if you just multiplied base times height, you would get this entire area. You would get the area of that entire rectangle. And you see that the triangle is exactly 1/2 of it. If you took this part of the triangle and you flipped it over, you'd fill up that space. If you took this part of the triangle and you flipped it over, you'd fill up that space. So the triangle's area is 1/2 of the triangle's base times the triangle's height. So plus 1/2 times the triangle's base, is 4 inches. And so let's just calculate it. This gives us 32 plus-- oh, sorry. That's not 8 times 4. I don't want to confuse you. The triangle's height is 3. 8 times 3, right there. That's the triangle's height. So once again, let's go back and calculate it. So this is going to be 32 plus-- 1/2 times 8 is 4. 4 times 3 is 12. And so our area for our shape is going to be 44. Now let's do the perimeter. The perimeter-- we just have to figure out what's the sum of the sides. How long of a fence would we have to build if we wanted to make it around this shape, right along the sides of this shape? So the perimeter-- I'll just write P for perimeter. It's going to be equal to 8 plus 4 plus 5 plus this 5, this edge So I have two 5's plus this 4 right over here. So you have 8 plus 4 is 12. 12 plus 10-- well, I'll just go one step at a time. 12 plus 5 is 17. 17 plus 5 is 22. 22 plus 4 is 26. So the perimeter is 26 inches. And let me get the units right, too. Because over here, I'm multiplying 8 inches by 4 inches. So you get square inches. 8 inches by 3 inches, so you get square inches again. So this is going to be square inches. So area is 44 square inches. Perimeter is 26 inches. And that makes sense because this is a two-dimensional measurement. It's measuring something in two-dimensional space, so you get a two-dimensional unit. This is a one-dimensional measurement. It's only asking you, essentially, how long would a string have to be to go around this thing. And so that's why you get one-dimensional units." + }, + { + "Q": "At at 4:12 - 4:14, why did Sal write x^4 '+'... and not '-' ...?", + "A": "Addition does not change signs of the original values. If he had used a minus sign, it would need to be distributed across the term terms in the parentheses, which would result in those values having the wrong signs. Hope this helps.", + "video_name": "yAH3722GrP8", + "timestamps": [ + 252, + 254 + ], + "3min_transcript": "as this up here, we're just switching the order of the multiplication. You can kind of view it as we're distributing from the right. And so if you do this, you get x squared times x squared, which is x to the fourth, that's that times that, and then minus x squared times the principal square root of 6. And then over here you have square root of 2 times x squared, so plus x squared times the square root of 2. And then you have square root of 2 times the square root of 6. And we have a negative sign out here. Now if you take the square root of 2-- let me do this on the side-- square root of 2 times the square root of 6, we know from simplifying radicals that this is the exact same thing as the square root of 2 times 6, or the principal square root of 12. So the square root of 2 times square root of 6, we have a negative sign out here, it becomes minus the square root of 12. You have an x to the fourth term. And then here you have-- well depending on how you want to view it, you could say, look, we have to second degree terms. We have something times x squared, and we have something else times x squared. So if you want, you could simplify these two terms over here. So I have square root of 2 x squareds and then I'm going to subtract from that square root of 6 x squareds. So you could view this as square root of 2 minus the square root of 6, or the principal square root of 2 minus the principal square root of 6, x squared. And then, if you want, square root of 12, you might be able to simplify that. 12 is the same thing as 3 times 4. So the square root of 12 is equal to square root of 3 times square root of 4. And the square root of 4, or the principal square root of 4 I should say, is 2. So the square root of 12 is the same thing as 2 square roots of 3. we could write minus 2 times the principal square root of 3. And then out here you have an x to the fourth plus this. And you see, if you distributed this out, if you distribute this x squared, you get this term, negative x squared, square root of 6, and if you distribute it onto this, you'd get that term. So you could debate which of these two is more simple. Now I mentioned that this way I just did the distributive property twice. Nothing new, nothing fancy. But in some classes, you will see something called FOIL. And I think we've done this in previous videos. FOIL. I'm not a big fan of it because it's really a way to memorize a process as opposed to understanding that this is really just from the common-sense distributive property. But all this is is a way to make sure that you're multiplying everything times everything when you're multiplying two binomials times each other like this. And FOIL just says, look, first multiply the first term. So x squared times x squared is x to the fourth." + }, + { + "Q": "Could someone please tell me why at 3:40 -x^2*sqrt6+x^2*sqrt2=(sqrt2-sqrt6)x^2?\n\nShouldn't it be -x^2*sqrt6+x^2*sqrt2=sqrt2-sqrt6 since one of the x^2 is negative while the other is positive? Shouldn't the x^2s cancel out then.\n\nI'm not sure if I've just made a careless error or am just missing something here, but I don't know why you get a negative x^2 or an x^2 at all. Please help explain this.", + "A": "So lets pretend for a minute that instead the expression was -6z + 2z We can rearrange them using the commutative property: = 2z - 6z And then factor out the z, using the distributive property: = z (2 - 6) Now we can replace z with x^2, and the numbers with their square roots and we can still do the same thing: -Sqrt(6)x^2 + Sqrt(2)x^2 = Sqrt(2)x^2 - Sqrt(6)x^2 = (Sqrt(2)-Sqrt(6))x^2", + "video_name": "yAH3722GrP8", + "timestamps": [ + 220 + ], + "3min_transcript": "as this up here, we're just switching the order of the multiplication. You can kind of view it as we're distributing from the right. And so if you do this, you get x squared times x squared, which is x to the fourth, that's that times that, and then minus x squared times the principal square root of 6. And then over here you have square root of 2 times x squared, so plus x squared times the square root of 2. And then you have square root of 2 times the square root of 6. And we have a negative sign out here. Now if you take the square root of 2-- let me do this on the side-- square root of 2 times the square root of 6, we know from simplifying radicals that this is the exact same thing as the square root of 2 times 6, or the principal square root of 12. So the square root of 2 times square root of 6, we have a negative sign out here, it becomes minus the square root of 12. You have an x to the fourth term. And then here you have-- well depending on how you want to view it, you could say, look, we have to second degree terms. We have something times x squared, and we have something else times x squared. So if you want, you could simplify these two terms over here. So I have square root of 2 x squareds and then I'm going to subtract from that square root of 6 x squareds. So you could view this as square root of 2 minus the square root of 6, or the principal square root of 2 minus the principal square root of 6, x squared. And then, if you want, square root of 12, you might be able to simplify that. 12 is the same thing as 3 times 4. So the square root of 12 is equal to square root of 3 times square root of 4. And the square root of 4, or the principal square root of 4 I should say, is 2. So the square root of 12 is the same thing as 2 square roots of 3. we could write minus 2 times the principal square root of 3. And then out here you have an x to the fourth plus this. And you see, if you distributed this out, if you distribute this x squared, you get this term, negative x squared, square root of 6, and if you distribute it onto this, you'd get that term. So you could debate which of these two is more simple. Now I mentioned that this way I just did the distributive property twice. Nothing new, nothing fancy. But in some classes, you will see something called FOIL. And I think we've done this in previous videos. FOIL. I'm not a big fan of it because it's really a way to memorize a process as opposed to understanding that this is really just from the common-sense distributive property. But all this is is a way to make sure that you're multiplying everything times everything when you're multiplying two binomials times each other like this. And FOIL just says, look, first multiply the first term. So x squared times x squared is x to the fourth." + }, + { + "Q": "At 3:21 why is it sqrt2 - sqrt6 and not the other way around? Or would it still be the same either way?", + "A": "It s equivalent, sal choose to do it like that so you only have to use one operation symbol i.e. rather than: -sqrt6 + sqrt2 He choose: sqrt2 - sqrt6 But both are equivalent", + "video_name": "yAH3722GrP8", + "timestamps": [ + 201 + ], + "3min_transcript": "So let's do that. So we get x squared minus the principal square root of 6 times this term-- I'll do it in yellow-- times x squared. And then we have plus this thing again. We're just distributing it. It's just like they say. It's sometimes not that intuitive because this is a big expression, but you can treat it just like you would treat a variable over You're distributing it over this expression over here. And so then we have x squared minus the principal square root of 6 times the principal square root of 2. And now we can do the distributive property again, but what we'll do is we'll distribute this x squared onto each of these terms and distribute the square root of 2 onto each of these terms. It's the exact same thing as here, it's just you could imagine writing it like this. x plus y times a is still going to be ax plus ay. as this up here, we're just switching the order of the multiplication. You can kind of view it as we're distributing from the right. And so if you do this, you get x squared times x squared, which is x to the fourth, that's that times that, and then minus x squared times the principal square root of 6. And then over here you have square root of 2 times x squared, so plus x squared times the square root of 2. And then you have square root of 2 times the square root of 6. And we have a negative sign out here. Now if you take the square root of 2-- let me do this on the side-- square root of 2 times the square root of 6, we know from simplifying radicals that this is the exact same thing as the square root of 2 times 6, or the principal square root of 12. So the square root of 2 times square root of 6, we have a negative sign out here, it becomes minus the square root of 12. You have an x to the fourth term. And then here you have-- well depending on how you want to view it, you could say, look, we have to second degree terms. We have something times x squared, and we have something else times x squared. So if you want, you could simplify these two terms over here. So I have square root of 2 x squareds and then I'm going to subtract from that square root of 6 x squareds. So you could view this as square root of 2 minus the square root of 6, or the principal square root of 2 minus the principal square root of 6, x squared. And then, if you want, square root of 12, you might be able to simplify that. 12 is the same thing as 3 times 4. So the square root of 12 is equal to square root of 3 times square root of 4. And the square root of 4, or the principal square root of 4 I should say, is 2. So the square root of 12 is the same thing as 2 square roots of 3." + }, + { + "Q": "At 2:10 he says that if r=1 denominator is 0, and we can't divide by zero. But, in that case numerator would also be 0, since a-a*(1)^n=0. Isn't lim 0/0=1?", + "A": "No, the limit of 0/0 is undefined, and since the limit is for the variable n and not for r, you cannot use any of the limit techniques to get rid of the 0/0.", + "video_name": "b-7kCymoUpg", + "timestamps": [ + 130 + ], + "3min_transcript": "In a previous video, we derived the formula for the sum of a finite geometric series where a is the first term and r is our common ratio. What I want to do in this video is now think about the sum of an infinite geometric series. And I've always found this mildly mind blowing because, or actually more than mildly mind blowing, because you're taking the sum of an infinite things but as we see, you can actually get a finite value depending on what your common ratio is. So there's a couple of ways to think about it. One is, you could say that the sum of an infinite geometric series is just a limit of this as n approaches infinity. So we could say, what is the limit as n approaches infinity of this business, of the sum from k equals zero to n of a times r to the k. as n approaches infinity right over here. So that would be the same thing as the limit as n approaches infinity of all of this business. Let me just copy and paste that so I don't have to keep switching colors. So copy and then paste. So what's the limit as n approaches infinity here? Let's think about that for a second. I encourage you to pause the video, and I'll give you one hint. Think about it for r is greater than one, for r is equal to one, and actually let me make it clear-- let's think about it for the absolute values of r is greater than one, the absolute values of r equal to one, and then the absolute value of r less than one. Well, I'm assuming you've given a go at it. So if the absolute value of r is greater than one, as this exponent explodes, as it approaches infinity, this number is just going to become massively, And so the whole thing is just going to become, or at least you could think of the absolute value of the whole thing, is just going to become a very, very, very large number. If r was equal to one, then the denominator is going to become zero. And we're going to be dividing by that denominator, and this formula just breaks down. But where this formula can be helpful, and where we can get this to actually give us a sensical result, is when the absolute value of r is between zero and one. We've already talked about, we're not even dealing with the geometric, we're not even talking about a geometric series if r is equal to zero. So let's think about the case where the absolute value of r is greater than zero, and it is less than one. What's going to happen in that case? Well, the denominator is going to make sense, right over here. And then up here, what's going to happen? Well, if you take something with an absolute value less than one, and you take it to higher and higher and higher" + }, + { + "Q": "@9:52 I don't understand how a smaller denomater is bigger than a larger denmotor ?", + "A": "the larger the denominator the smaller the piece The smaller the denominator the larger the piece.", + "video_name": "wbAxarp_Ug4", + "timestamps": [ + 592 + ], + "3min_transcript": "" + }, + { + "Q": "at 5:30 he Sal says 3 forth of the pizza has cheese, why does he put the number like that, one on top of the other, what does it mean and why not put the 4 on top instead of the 3?", + "A": "first i think you meant at 2:30. Sal s pizza had 1/4 with olives & 3/4 with cheese, right? now it makes sense 3 of the 4 slices of pizza are cheese while 1 slice is olives. now if it was 4 out of the 3 pieces have cheese that would mean you would have 3 slices of pizza with 4 of those 3 slices being cheese, that would make things a bit confusing.", + "video_name": "kZzoVCmUyKg", + "timestamps": [ + 330 + ], + "3min_transcript": "" + }, + { + "Q": "I still don't understand, at 0:33, why did he make reference of the pizza as an example of fractions?", + "A": "the circle is the most simple to esplain fractions.", + "video_name": "kZzoVCmUyKg", + "timestamps": [ + 33 + ], + "3min_transcript": "" + }, + { + "Q": "At 4:00 what if you were given only two points where the function intersects the midline twice? How do I find period with those two points?", + "A": "It depends on which two points you re given, but if you re referring to two consecutive points where the function intersects the midline, then the horizontal distance between them would be 1/2 of the period. Can you see it?", + "video_name": "s4cLM0l1gd4", + "timestamps": [ + 240 + ], + "3min_transcript": "And the midline is in the middle, so it's going to be the same amount whether you go above or below. One way to say it is, well, at this maximum point, right over here, how far above the midline is this? Well, to get from 1 to 4 you have to go-- you're 3 above the midline. Another way of thinking about this maximum point is y equals 4 minus y equals 1. Well, your y can go as much as 3 above the midline. Or you could say your y-value could be as much as 3 below the midline. That's this point right over here, 1 minus 3 is negative 1. So your amplitude right over here is equal to 3. You could vary as much as 3, either above the midline or below the midline. Finally, the period. And when I think about the period I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's a nice-- when x is at negative 2, y is it And so what I want to do is keep traveling along this curve until I get to the same y-value but not just the same y-value but I get the same y-value that I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here because, once again, y is equal to 1? You haven't completed a cycle here because notice over here where our y is increasing as x increases. Well here our y is decreasing as x increases. Our slope is positive here. Our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals 1. Or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. So that gets us to right over there. So the change in x needed to complete one cycle. That is your period. So to go from negative 2 to 0, your period is 2. So your period here is 2. And you could do it again. So we're at that point. Let's see, we want to get back to a point where we're at the midline-- and I just happen to start right over here at the midline. I could have started really at any point. You want to get to the same point but also where the slope is the same. We're at the same point in the cycle once again. So I could go-- so if I travel 1 I'm at the midline again but I'm now going down. So I have to go further. Now I am back at that same point in the cycle. I'm at y equals 1 and the slope is positive. And notice, I traveled. My change in x was the length of the period. It was 2." + }, + { + "Q": "Why go through all the trouble described by Sal around 04:00 to find the period of a function if one can simply measure the distance between two consecutive maximums or two consecutive minimums?", + "A": "How would you go about measuring it precisely?", + "video_name": "s4cLM0l1gd4", + "timestamps": [ + 240 + ], + "3min_transcript": "And the midline is in the middle, so it's going to be the same amount whether you go above or below. One way to say it is, well, at this maximum point, right over here, how far above the midline is this? Well, to get from 1 to 4 you have to go-- you're 3 above the midline. Another way of thinking about this maximum point is y equals 4 minus y equals 1. Well, your y can go as much as 3 above the midline. Or you could say your y-value could be as much as 3 below the midline. That's this point right over here, 1 minus 3 is negative 1. So your amplitude right over here is equal to 3. You could vary as much as 3, either above the midline or below the midline. Finally, the period. And when I think about the period I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's a nice-- when x is at negative 2, y is it And so what I want to do is keep traveling along this curve until I get to the same y-value but not just the same y-value but I get the same y-value that I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here because, once again, y is equal to 1? You haven't completed a cycle here because notice over here where our y is increasing as x increases. Well here our y is decreasing as x increases. Our slope is positive here. Our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals 1. Or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. So that gets us to right over there. So the change in x needed to complete one cycle. That is your period. So to go from negative 2 to 0, your period is 2. So your period here is 2. And you could do it again. So we're at that point. Let's see, we want to get back to a point where we're at the midline-- and I just happen to start right over here at the midline. I could have started really at any point. You want to get to the same point but also where the slope is the same. We're at the same point in the cycle once again. So I could go-- so if I travel 1 I'm at the midline again but I'm now going down. So I have to go further. Now I am back at that same point in the cycle. I'm at y equals 1 and the slope is positive. And notice, I traveled. My change in x was the length of the period. It was 2." + }, + { + "Q": "At 6:27, how did you get 110 degrees?", + "A": "180 degrees is a straight line and 70 degrees is an angle. Since the 70 degree angle and the other angle he is trying to find are supplementary it means that the angles must add up to 180 degrees. So 180 degrees minus 70 degrees is equal to 110 degrees.", + "video_name": "gRKZaojKeP0", + "timestamps": [ + 387 + ], + "3min_transcript": "From this perspective it's kind of the top right angle and each intersection is the same. Now the same is true of the other corresponding angles. This angle right here in this example, it's the top left angle will be the same as the top left angle right over here. This bottom left angle will be the same down here. If this right here is 70 degrees, then this down here will also be 70 degrees. And then finally, of course, this angle and this angle will also be the same. So corresponding angles -- let me write these -- these are corresponding angles are congruent. Corresponding angles are equal. And that and that are corresponding, that and that, that and that, and that and that. Now, the next set of equal angles to realize are sometimes opposite angles. But if you take this angle right here, the angle that is vertical to it or is opposite as you go right across the point of intersection is this angle right here, and that is going to be the same thing. So we could say opposite -- I like opposite because it's not always in the vertical direction, sometimes it's in the horizontal direction, but sometimes they're referred to as vertical angles. Opposite or vertical angles are also equal. So if that's 70 degrees, then this is also 70 degrees. And if this is 70 degrees, then this right here is also 70 degrees. So it's interesting, if that's 70 degrees and that's 70 degrees, and if this is 70 degrees and that is also 70 degrees, so no matter what this is, this will also be the same thing because this is the same as that, that is the same as that. Now, the last one that you need to I guess kind of realize are green angle right there. You can see that when you add up the angles, you go halfway around a circle, right? If you start here you do the green angle, then you do the orange angle. You go halfway around the circle, and that'll give you, it'll get you to 180 degrees. So this green and orange angle have to add up to 180 degrees or they are supplementary. And we've done other videos on supplementary, but you just have to realize they form the same line or a half circle. So if this right here is 70 degrees, then this orange angle right here is 110 degrees, because they add up to 180. Now, if this character right here is 110 degrees, what do we know about this character right here? Well, this character is opposite or vertical to the 110 degrees so it's also 110 degrees." + }, + { + "Q": "1:47 What is a nonlinear correlation?", + "A": "nonlinear correlation = non linear correlation linear correlation --> correlation assuming there is a line (straight). non --> not doing non-linear correlation --> correlation that assumes the line is not a straight line.", + "video_name": "Jpbm5YgciqI", + "timestamps": [ + 107 + ], + "3min_transcript": "The graphs below show the test grades of the students in Dexter's class. The first graph shows the relationship between test grades and the amount of time the students spent studying. So this is study time on this axis and this is the test grade on this axis. And the second graph shows the relationship between test grades and shoes size. So shoe size on this axis and then test grade. Choose the best description of the relationship between the graphs. So first, before looking at the explanations, let's look at the actual graphs. So this one on the left right over here, it looks like there is a positive linear relationship right over here. I could almost fit a line that would go just like that. And it makes sense that there would be, that the more time that you spend studying, the better score that you would get. Now for a certain amount of time studying, some people might do better than others, but it does seem like there's this relationship. really much of a relationship. You see the shoe sizes, for a given shoe size, some people do not so well and some people do very well. Someone with a size 10 and 1/2, it looks like, someone it looks like they flunked the exam. Someone else, looks like they got A minus or a B plus And it really would be hard to somehow fit a line here. No matter how you draw a line, these dots don't seem to form a trend. So let's see which of these choices apply. There's a negative linear relationship between study time and score. No, that's not true. It looks like there's a positive linear relationship. The more you study, the better your score would be. A negative linear relationship would trend downwards like that. There is a non-linear relationship between study time and score and a negative linear relationship between shoe size and score. Well that doesn't seem right either. A non-linear relationship, it would not be easy to fit a line to it. And this one seems like a line would be very reasonable. between shoe size and score. So I wouldn't pick this one either. There's a positive linear relationship between study time and score. That's right. And no relationship between shoe size and score. Well, I'm going to go with that one. Both graphs show positive linear trends of approximately equal strength. No, not at all. This one doesn't show a linear relationship of really any strength." + }, + { + "Q": "at 2:05 dose he mean negitive", + "A": "Sal said 4+7 to see how many blocks were in between the points. At 2:05 if he had done 4+(-7) he would have gotten -3 which is impossible. there can t be -3 blocks between her house and the mall. Hope this helped.", + "video_name": "PC_FoyewoIs", + "timestamps": [ + 125 + ], + "3min_transcript": "Milena's town is built on a grid similar to the coordinate plane. She is riding her bicycle from her home at point negative 3, 4 to the mall at point negative 3, negative 7. Each unit on the graph denotes one city block. Plot the two points, and find the distance between Milena's home and the mall. So let's see, she's riding her bicycle from her home at the point negative 3, 4. So let's plot negative 3, 4. So I'll use this point right over here. So negative 3 is our x-coordinate. So we're going to go 3 to the left of the origin 1, 2, 3. That gets us a negative 3. And positive 4 is our y-coordinate. So we're going to go 4 above the origin. Or I should say, we're going to go 4 up. So we went negative 3, or we went 3 to the left. That's negative 3, positive 4. Or you could say we went positive 4, negative 3. This tells us what we do in the horizontal direction. This tells us what we do in the vertical direction. Now let's figure out where the mall is. It's at the point negative 3, negative 7. So negative 3, we went negative 3 along the horizontal direction and then negative 7 along the vertical direction. So we get to negative 3, negative 7 right over there. And now we need to figure out the distance between her home and the mall. Now, we could actually count it out, or we could just compute it. If we wanted to count it out, it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 blocks. So we could type that in. And another way to think about it is they have the exact same x-coordinate. They're both at the x-coordinate negative 3. The only difference between these two is what is happening in the y-coordinate. This is at a positive 4. This is at a negative 7. Positive 4, negative 7. So we're really trying to find the distance between 4 and negative 7. So if I were to say 4 minus negative 7, So we have 4 minus negative 7, which is the same thing as 4 plus 7, which is 11. Let's do a couple more. Carlos is hanging a poster in the area shown by the red rectangle. He is placing a nail in the center of the blue line. In the second graph, plot the point where he places the nail. So he wants to place a nail in the center of the blue line. The blue line is 6 units long. The center is right over here. That's 3 to the right, 3 to the left. So he wants to put the nail at the point x equals 0, y is equal to 4. So he wants to put it at x is equal to 0, y is equal to 4. That's this point right over here. So let's check our answer. Let's do one more." + }, + { + "Q": "@4:05 why didn't Sal write 20u^2v/10uv^2? why did we drop the exponent?", + "A": "That did confuse me a bit, but the reason why is because we have to put the GCF in the bottom for it to be the same as 10uv(2u-v). If we did 20u^2v/10uv^2 along with the other part, we would be left with our answer being 10uv(2u/v-1), which is not the same as 10uv(2u-v).", + "video_name": "499MvHFrqUU", + "timestamps": [ + 245 + ], + "3min_transcript": "If we wrote 2 times 5 times u times v, and we say that's going to be-- this expression is equal to this times what? Well if you factor the 2 times 5 times u times v out, all you're going to be left with in this first term is the 2 times u, so 2u here. And in the second term, all you're going to be left with is a v. All this other stuff gets factored out. All you're going to be left with is a v. Hopefully you see, if I multiply 2 times 5 times u times v times 2u, I'm going to get this first term here. So if I were to distribute it, I would get this first term. And if I multiply 2 times 5 times u times v times this v over here, I'm going to get this second term. So this expression, and that expression is the exact same thing. We have factored it out, now we can simplify it a little bit. 2 times 5 times u times v we rewrite as 10uv. a 2u and then a minus v. And we're done! We have factored the expression. Now you won't be doing it to this granular level, but this is the best way to think about it. Eventually you're going to say, hey, wait, look, the largest number that divides both of these is a 10. Because you could see 10 goes into the 20, 10 goes into 10. And, let's see, a u goes into both of these, and a v goes So let me factor out a 10uv, and then if I divide this thing by 10uv, I'm going to be left with 2u. And if I divide this by 10uv, I'm going to just be left with a v. So that's another way to think about it. Let me do that right now, so we could say that this is the same thing. Another way of approaching it, you could have said that this is the same thing as-- Well, the largest number that divides both of these is 10uv, and that's going to be times 20u squared v over 10uv minus this thing. This expression and this are obviously the same thing. If I were to distribute the 10uv it would cancel out with each of these in the denominator right there. So they're the same thing, but we can do is we can simplify this. We could say that 20 divided by 10 is just 2, u squared divided by u is just a u, v divided by v is just 1, 10 divided by 10 is 1, u divide by u is u, v squared divided by v is just a v to the first power. So you're left with 10uv times the quantity 2u minus v. Either way you get the same answer." + }, + { + "Q": "At 5:19, Sal draws a \"cone\", however it appears to be two cones on top of each other (tip to tip). Why is this?", + "A": "The correct term for the solid is double-napped cone . Essentially two congruent cones with the same axis and a common vertex. Sal just didn t use the formal name for the solid.", + "video_name": "0A7RR0oy2ho", + "timestamps": [ + 319 + ], + "3min_transcript": "I want to go right through the-- that's pretty good. These are asymptotes. Those aren't the actual hyperbola. But a hyperbola would look something like this. They get to be right here and they get really close to the asymptote. They get closer and closer to those blue lines like that and it happened on this side too. The graphs show up here and then they pop over and they show up there. This magenta could be one hyperbola; I haven't done true justice to it. Or another hyperbola could be on, you could kind of call it a vertical hyperbola. That's not the exact word, but it would look something like that where it's below the asymptote here. It's above the asymptote there. So this blue one would be one hyperbola and then the magenta one would be a different hyperbola. So those are the different graphs. So the one thing that I'm sure you're asking is why are Why are they not called bolas or variations of circles or whatever? And in fact, wasn't even the relationship. It's pretty clear that circles and ellipses are somehow related. That an ellipse is just a squished circle. And maybe it even seems that parabolas and hyperbolas are somewhat related. This is a P once again. They both have bola in their name and they both kind of look like open U's. Although a hyperbola has two of these going and kind of opening in different directions, but they look related. But what is the connection behind all these? And that's frankly where the word conic comes from. So let me see if I can draw a three-dimensional cone. So this is a cone. That's the top. I could've used an ellipse for the top. Looks like that. Actually, it has no top. It would actually keep going on forever in that direction. This could be the bottom part of it. So let's take different intersections of a plane with this cone and see if we can at least generate the different shapes that we talked about just now. So if we have a plane that goes directly-- I guess if you call this the axis of this three-dimensional cone, so this is the axis. So if we have a plane that's exactly perpendicular to that axis-- let's see if I can draw it in three dimensions. The plane would look something like this. So it would have a line. This is the front line that's closer to you and then they would have another line back here. That's close enough. And of course, you know these are infinite planes, so it goes off in every direction. If this plane is directly perpendicular to the axis of these and this is where the plane goes behind it. The intersection of this plane and this cone is going to look like this." + }, + { + "Q": "At 9:00, as in trigonometry equation, why don't we put negative sign in front of the coefficient instead of the..mm....unknown? Ex. x= -3 cos (t), instead of x= 3 cos (-t)", + "A": "Because trignometric functions are not commutative. a*cos(t) is not always the same as cos(a*t), even though sometimes it is.", + "video_name": "IReD6c_njOY", + "timestamps": [ + 540 + ], + "3min_transcript": "We get all the way-- Oh. That's the same color I used before. Let me see this. Let me do this color. We're here. Now notice: in the first one, when we went from t equals 0 to t equals pi over 2, we went from here to there. We went kind of a quarter of the way around the ellipse. But now when we went from t equals zero to pi over 2, where did we go? Went halfway around the ellipse. We went all the way from there, all the way over there. And likewise, when we went from t equals pi over 2 to t equals pi with this set of parametric equations, we went another quarter of the ellipse. We went from there to there. But here, when we go from t equals pi ever 2 to t equals pi, we go all of this way. We go back to the beginning part of our ellipse. has the exact same shape of its path as this set of parametric equations. Except it's going around it at twice as fast of a rate. For every time when t increases by pi over 2 here, we go by-- we kind of go a quarter way around the ellipse. But when t increases by pi over 2 here, we go halfway around the ellipse. So the thing to realize-- and I know I've touched on this before --is that even though both of these sets of parametric equations, when you do the algebra, they can kind of be converted into this shape. You lose the information about where our particle is as it's rotating around the ellipse or how fast it's rotating And that's why you need these parametric equations. We can even set up a parametric equation that goes in the other direction. Instead of having these-- and I encourage you to play with that --but if you instead of this, if you just put a minus sign right here. Instead of going in that direction, it would go in this direction. It would go in a clockwise direction. So one thing that you've probably been thinking from the beginning is OK, I was able to go from my parametric equations to this equation of ellipse in terms of just x and y. Can you go back the other way? Could you go from this to this? And, I think you might realize now, that the answer is no. Because there's no way, just with the information that you're given here, to know that you should go to this parametric equation or this parametric equation or any of an infinite number of parametric equations. I mean anything of the form x is equal to 3 cosine of really anything times t and y is equal to 3 times cosine of-- As long as it's the same anything-- I drew the two squiggly marks the same. --as long as these two things are the same, then you" + }, + { + "Q": "Why at 2:44 does he write 2 dot 2 dot?", + "A": "he is trying to say that 2 multiply by 2 multiply by 2 multiply by 2", + "video_name": "lxjmR4pYIVU", + "timestamps": [ + 164 + ], + "3min_transcript": "So let's do it that way first. So multiples of six are 6, 12, 18, 24 30. And I could keep going if we don't find any common multiples out of this group here with any of the multiples in eight. And the multiples of eight are 8, 16, 24, and it looks like we're done. And we could keep going obviously-- 32, so on and so forth. But I found a common multiple and this is their smallest common multiple. They have other common multiples-- 48 and 72, and we could keep adding more and more multiple. But this is their smallest common multiple, their least common multiple. So it is 24. Another way that you could have found at least common multiple is you could have taken the prime factorization of six and you say, hey, that's 2, and 3. So the least common multiple has to have at least 1, 2, and 1, 3 in its prime factorization in order for it And you could have said, what's the prime factorization of 8? It is 2 times 4 and 4 is 2 times 2. So in order to be divisible by 8, you have to have at least three 2's in the prime factorization. So to be divisible by 6, you have to have a 2 times a 3. And then to be divisible by 8, you have to have at least three 2's. You have to have two times itself three times I should say. Well, we have one 2 and let's throw in a couple more. So then you have another 2 and then another 2. So this part right over here makes it divisible by 8. And this part right over here makes it divisible by 6. If I take 2 times 2 times 2 times 3, that does give me 24. So our least common multiple of 8 and 6, which is also the least common denominator of these two fractions is going to be 24. So what we want to do is rewrite each of these fractions with 24 as the denominator. So I'll start with 2 over 8. Well, to get the denominator be 24, we have to multiply it by 3. 8 times 3 is 24. And so if we don't want to change the value of the fraction, we have to multiply the numerator and denominator by the same thing. So let's multiply the numerator by 3 as well. 2 times 3 is 6. So 2/8 is the exact same thing as 6/24. To see that a little bit clearer, you say, look, if I have 2/8, and if I multiply this times 3 over 3, that gives me 6/24. And this are the same fraction because 3 over 3 is really just 1. It's one whole. So 2/8 is 6/24 let's do the same thing with 5/6. So 5 over 6 is equal to something over 24." + }, + { + "Q": "from 3:30 to 3:34 what was he saying can you please explain", + "A": "He was just saying... when he did 2/8 times 3 on the top and 3 on the bottom is the exact same as multiplying it by 3/3 :D Hope that helps =)", + "video_name": "lxjmR4pYIVU", + "timestamps": [ + 210, + 214 + ], + "3min_transcript": "So let's do it that way first. So multiples of six are 6, 12, 18, 24 30. And I could keep going if we don't find any common multiples out of this group here with any of the multiples in eight. And the multiples of eight are 8, 16, 24, and it looks like we're done. And we could keep going obviously-- 32, so on and so forth. But I found a common multiple and this is their smallest common multiple. They have other common multiples-- 48 and 72, and we could keep adding more and more multiple. But this is their smallest common multiple, their least common multiple. So it is 24. Another way that you could have found at least common multiple is you could have taken the prime factorization of six and you say, hey, that's 2, and 3. So the least common multiple has to have at least 1, 2, and 1, 3 in its prime factorization in order for it And you could have said, what's the prime factorization of 8? It is 2 times 4 and 4 is 2 times 2. So in order to be divisible by 8, you have to have at least three 2's in the prime factorization. So to be divisible by 6, you have to have a 2 times a 3. And then to be divisible by 8, you have to have at least three 2's. You have to have two times itself three times I should say. Well, we have one 2 and let's throw in a couple more. So then you have another 2 and then another 2. So this part right over here makes it divisible by 8. And this part right over here makes it divisible by 6. If I take 2 times 2 times 2 times 3, that does give me 24. So our least common multiple of 8 and 6, which is also the least common denominator of these two fractions is going to be 24. So what we want to do is rewrite each of these fractions with 24 as the denominator. So I'll start with 2 over 8. Well, to get the denominator be 24, we have to multiply it by 3. 8 times 3 is 24. And so if we don't want to change the value of the fraction, we have to multiply the numerator and denominator by the same thing. So let's multiply the numerator by 3 as well. 2 times 3 is 6. So 2/8 is the exact same thing as 6/24. To see that a little bit clearer, you say, look, if I have 2/8, and if I multiply this times 3 over 3, that gives me 6/24. And this are the same fraction because 3 over 3 is really just 1. It's one whole. So 2/8 is 6/24 let's do the same thing with 5/6. So 5 over 6 is equal to something over 24." + }, + { + "Q": "During 1:25-1:50, the LCD should be 0, shouldn't it?", + "A": "Not exactly.. the way you get LCD ( Lowest common Denominator ) is by factorizing the denominators of fractions until you get something common. Sal was just looking for something common. and if you noticed he stopped at 24. I hope this helps.", + "video_name": "lxjmR4pYIVU", + "timestamps": [ + 85, + 110 + ], + "3min_transcript": "We're asked to rewrite the following two fractions as fractions with a least common denominator. So a least common denominator for two fractions is really just going to be the least common multiple of both of these denominators over here. And the value of doing that is then if you can make these a common denominator, then you can add the two fractions. And we'll see that in other videos. But first of all, let's just find the least common multiple. Let me write it out because sometimes LCD could meet other things. So least common denominator of these two things is going to be the same thing as the least common multiple of the two denominators over here. The least common multiple of 8 and 6. And a couple of ways to think about least common multiple-- you literally could just take the multiples of 8 and 6 So let's do it that way first. So multiples of six are 6, 12, 18, 24 30. And I could keep going if we don't find any common multiples out of this group here with any of the multiples in eight. And the multiples of eight are 8, 16, 24, and it looks like we're done. And we could keep going obviously-- 32, so on and so forth. But I found a common multiple and this is their smallest common multiple. They have other common multiples-- 48 and 72, and we could keep adding more and more multiple. But this is their smallest common multiple, their least common multiple. So it is 24. Another way that you could have found at least common multiple is you could have taken the prime factorization of six and you say, hey, that's 2, and 3. So the least common multiple has to have at least 1, 2, and 1, 3 in its prime factorization in order for it And you could have said, what's the prime factorization of 8? It is 2 times 4 and 4 is 2 times 2. So in order to be divisible by 8, you have to have at least three 2's in the prime factorization. So to be divisible by 6, you have to have a 2 times a 3. And then to be divisible by 8, you have to have at least three 2's. You have to have two times itself three times I should say. Well, we have one 2 and let's throw in a couple more. So then you have another 2 and then another 2. So this part right over here makes it divisible by 8. And this part right over here makes it divisible by 6. If I take 2 times 2 times 2 times 3, that does give me 24. So our least common multiple of 8 and 6, which is also the least common denominator of these two fractions is going to be 24. So what we want to do is rewrite each of these fractions with 24 as the denominator. So I'll start with 2 over 8." + }, + { + "Q": "why does x have to be greater than zero in the domain?\n\nTime on the video: 6:39", + "A": "The log of 0 is undefined. The log of negative numbers involves rather difficult complex numbers, so it is usually treated as undefined at this stage of studying math.", + "video_name": "DuYgVVU_BwY", + "timestamps": [ + 399 + ], + "3min_transcript": "is so I get nice clean results that I can plot by hand. So let's actually graph it. Let's actually graph this thing over here. So the y's go between negative 2 and 2. The x's go from 1/25th all the way to 25. So let's graph it. So that is my y-axis, and this is my x-axis. Draw it like that. That is my x-axis. And then the y's start at 0. Then, you get to positive 1, positive 2. And then you have negative 1. And you have negative 2. And then on the x-axis, it's all positive. And I'll let you think about whether the domain here is-- well, when you think about it-- is a logarithmic function defined for an x that is not positive? No. You could raise five to an infinitely negative power to get a very, very, very, very small number that approaches zero, but you can never get-- there's no power that you can raise 5 to to get 0. So x cannot be 0. And there's no power then you could raise 5 to get another negative number. So x can also not be a negative number. So the domain of this function right over here-- and this is relevant, because we want to think about what we're graphing-- the domain here is x has to be greater than zero. Let me write that down. The domain here is that x has to be greater than 0. So we're only going to be able to graph this function in the positive x-axis. So with that out of the way, x gets as large as 25. So let me graph-- we put those points here. So that is 5, 10, 15, 20, and 25. And then let's plot these. So the first one is in blue. When x is 1/25 and y is negative 2-- is going to be really close to there-- Then y is negative 2. So it's going to be like right over there, not quite at the y-axis. We're at 1/25 to the right of the y-axis. So that's right over there. That is 1 over 25, comma negative 2 right over there. Then, when x is one fifth, which is slightly further to the right, one fifth y is negative 1. So right over there. So this is one fifth, negative 1. Then when x is 1, y is 0. So 1 might be right there. So this is the point 1,0. And then when x is 5, y is 1. When x is 5, I covered it over here, when this is five, y is 1. So that's the point 5,1. And then finally, when x is 25, y is 2." + }, + { + "Q": "At 11:30 or so, why do we ignore the normalization scalar that's being multiplied with the vectors? We can keep it aside until the end like that? Looks cool.", + "A": "The Gram-Schmidt method is a way to find an orthonormal basis. To do this it is useful to think of doing two things. Given a partially complete basis we first find any vector that is orthogonal to these. Second we normalize. Then we repeat these two steps until we have filled out our basis. There are formulas that you could write down where both steps are taken care of in a long computation but it is useful to understand the process as a multiple step procedure which is what Sal is doing.", + "video_name": "ZRRG386v6DI", + "timestamps": [ + 690 + ], + "3min_transcript": "Let me switch colors . Minus v3 , which is 1, 1 0, 0 dotted with u2, dotted with the square root of 2/3 times 0, 1, 1/2, minus 1/2 times u2, times the vector u2, times the square root of 2/3, times the vector 0, 1, 1/2, minus 1/2. And what do we get? Let's calculate this. So we could take the-- so this is going to be equal to the vector 1, 1, 0, 0, minus-- so the 1 over the square root of 2 and the 1 over the square root of 2, multiply them. You're going to get a 1/2. And then when you take the dot product of these two, 1 times 0-- let's see, this is actually all going to be, if gets 0, right? So this guy, v3, was actually already orthogonal to u1. This will just go straight to 0, which is nice. We don't have to have a term right there. I took the dot product 1 times 0 plus 1 times 0 plus 0 times 1 plus 0 times 1, all gets zeroed. So this whole term drops out. We can ignore it, which makes our computation simpler. And then over here we have minus the square root of 2/3 times the square root of 2/3 is just 2/3 times the dot product of these two guys. So that's 1 times 0, which is 0, plus 1 times 1, which is 1, plus 0 times 1/2, which is 0, plus 0 times minus 1/2, which is 0, so we just get a 1 there, times the vector 0, 1, 1/2, minus 1/2. And then what do we get? We get-- this is the home stretch-- 1, 1, 0, 0 minus 2/3 So 2/3 time 0 is 0. 2/3 times 1 is 2/3. 2/3 times 1/2 is 1/3. And then 2/3 times minus 1/2 is minus 1/3. So then this is going to be equal to 1 minus 0 is 1, 1 minus 2/3 is 1/3, 0 minus 1/3 is minus 1/3, and then 0 minus minus 1/3 is positive 1/3. So this vector y3 is orthogonal to these two other vectors, which is nice, but it still hasn't been normalized. So we finally have to normalize this guy, and then we're done. Then we have an orthonormal basis. We'll have u1, u2, and now we'll find u3. So the length of my vector y-- actually, let's do something" + }, + { + "Q": "At 03:51, the last vector sal just wrote (0,0,1,1) is the vector v1, shouldn't it be the vector v2 instead (0,1,1,0)? Or is it the formula above that should be Y2=V2-Proj(V1)*u1 instead of v2 in the end?", + "A": "The projection of v2 on v1 is in the direction of v1, so it s magnitude is multiplied by u1 = v1/||v1||.", + "video_name": "ZRRG386v6DI", + "timestamps": [ + 231 + ], + "3min_transcript": "So I can say that V is now equal to the span of the vectors u1, v2, and v3. Because I can replace v1 with this guy, because this guy is just a scaled-up version of this guy. So I can definitely represent him with him, so I can represent any linear combination of these guys with any linear combination of those guys right there. Now, we just did our first vector. We just normalized this one. But we need to replace these other vectors with vectors that are orthogonal to this guy right here. So let's do v2 first. So let's replace-- let's call it y2 is equal to v2 minus the projection of v2 onto the space spanned by u1 or onto-- you know, I could call it c times u1, or in the past videos, we called that subspace V1, but the space spanned by u1. which is 0, 1, 1, 0, minus-- v2 projected onto that space is just a dot product of v2, 0, 1, 1, 0, with the spanning vector of that space. And there's only one of them, so we're only going to have one term like this with u1, so dotted with 1 over the square root of 2 times 0, 0, 1, 1, and then all of that times u1. So 1 over the square root of 2 times the vector 0, 0, 1, 1. And so this is going to be equal to v2, which is 0, 1, 1, 0. The square root of 2, let's factor them out. So then you just get-- or kind of reassociate them out. over the square root of 2 is minus 1/2. You times-- what's the dot product of these two guys? You get 0 times 0 plus 1 times 0, which is still 0, plus 1 times 1 plus 0 times 0. So you're just going to have times 1 times this out here: 0, 0, 1, 1. I'll write that a little bit neater. I'm getting careless. 1, 1. So this is just going to be equal to 0, 1, 1, 0 minus-- 1/2 times 0 is 0. 1/2 times 0 is 0. Then I have two halves here. So y2 is equal to-- let's see, 0 minus 0 is 0, 1 minus 0 is 1, 1 minus 1/2 is 1/2, and then 0 minus 1/2 is minus 1/2." + }, + { + "Q": "I don't get it at 0:34 what does he mean?", + "A": "Compute means to calculate, find, or figure out. He will show how to find the answer.", + "video_name": "k68CPfcehTE", + "timestamps": [ + 34 + ], + "3min_transcript": "Let's try to calculate 3 times 32. And I like to rewrite it, and this is one way of doing it. I like to rewrite it where I have a larger number on top. So in this case it's 32. And I write the smaller number right below it. And since the smaller number is only one digit, it's only a ones digit, I put that below the ones place on the larger number. So I'll put the 3 right over here. And of course, we can't forget our multiplication symbol. And this is essentially a way of saying the same thing. You could read this as 32 times 3. But 32 times 3 is the exact same value as 3 times 32. It doesn't matter what order you multiply in. Now let's try to compute it. And once again, this is only one way of doing it. There's many ways of doing it. And I want you to think about why this works. We'll start with this 3 down here, and we're going to multiply it times each of the digits in 32. So we'll start with 3 times 2. Well, 3 times 2 from our multiplication tables, and you can figure it out even if you didn't know your multiplication tables, is 6. So 3 times 2, I'll write 6 right over here in the ones place. Well, once again, we know that 3 times 3 is 9. And since I'm multiplying times the tens place right over here, I'm going to put it in the tens place right like this. We got 32 times 3 is 96. And I really encourage you to think about why this worked. And I'll give you a little bit of a hint here. I'll give you a little bit of a hint about why this worked. Remember, 3 times 32 is the same thing as 3 times 30 plus 3 times 2. And if you look at it that way, that's essentially what this process did. We did 3 times 2 is 6. 3 times 30 is 90. You add them together, you get 90 plus 6 is 96." + }, + { + "Q": "why does he say minus is it not negative at around 5:20", + "A": "Any number minus another number is the same as saying that number plus the negative form of the other number. example: 7 - 5 is really the same as 7 + -5 Minus and negative are really just the same thing. I hope this helps you1 Unikitty <[:)]", + "video_name": "d8lP5tR2R3Q", + "timestamps": [ + 320 + ], + "3min_transcript": "or if I said negative 1 times negative 1 is equal to positive 1 as well. Or if I said 1 times negative 1 is equal to negative 1, or negative 1 times 1 is equal to negative 1. You see how on the bottom two problems I had two different signs, positive 1 and negative 1? And the top two problems, this one right here both 1s are positive. And this one right here both 1s are negative. So let's do a bunch of problems now, and hopefully it'll hit the point home, and you also could try to do along the practice problems and also give the hints and give you what rules to use, so that should help you as well. So if I said negative 4 times positive 3, well 4 times So different signs mean negative. So negative 4 times 3 is a negative 12. That makes sense because we're essentially saying what's negative 4 times itself three times, so it's like negative 4 plus negative 4 plus negative 4, which is negative 12. If you've seen the video on adding and subtracting negative numbers, you probably should watch first. Let's do another one. What if I said minus 2 times minus 7. And you might want to pause the video at any time to see if you know how to do it and then restart it to see what the answer is. Well, 2 times 7 is 14, and we have the same sign here, so it's a positive 14 -- normally you wouldn't have to write the positive but that makes it a little bit more explicit. And what if I had -- let me think -- 9 times negative 5. And once again, the signs are different so it's a negative. And then finally what if it I had -- let me think of some good numbers -- minus 6 times minus 11. Well, 6 times 11 is 66 and then it's a negative and negative, it's a positive. Let me give you a trick problem. What is 0 times negative 12? Well, you might say that the signs are different, but 0 is actually neither positive nor negative. And 0 times anything is still 0. It doesn't matter if the thing you multiply it by is a negative number or a positive number. 0 times anything is still 0. So let's see if we can apply these same rules to division. It actually turns out that the same rules apply." + }, + { + "Q": "at 5:52, I'm a little confused about the way Sal explains how there are 7 cubes that are not yellow or something. What does that mean?", + "A": "It s easier to see from the table over on the left. When you want to count all the shapes that are either yellow or cubes, you add the seven yellow spheres, the five yellow cubes, and the eight green cubes.", + "video_name": "QE2uR6Z-NcU", + "timestamps": [ + 352 + ], + "3min_transcript": "that I've drawn? This Venn diagram is just a way to visualize the different probabilities. And they become interesting when you start thinking about where sets overlap, or even where they don't overlap. So here we are thinking about things that are members of the set yellow. So they're in this set, and they are cubes. So this area right over here-- that's the overlap of these two sets. So this area right over here-- this represents things that are both yellow and cubes, because they are inside both circles. So this right over here-- let me rewrite it right over here. So there's five objects that are both yellow and cubes. Now let's ask-- and this is probably the most interesting thing to ask-- what is the probability of getting something that is yellow or or a cube, a cube of any color? or a cube of any color-- well, we still know that the denominator here is going to be 29. These are all of the equally likely possibilities that might jump out of the bag. But what are the possibilities that meet our conditions? Well, one way to think about it is, well, the probability-- there's 12 things that would meet the yellow condition. So that would be this entire circle right over here-- 12 things that meet the yellow condition. So this right over here is 12. This is the number of yellow. That is 12. And then to that, we can't just add the number of cubes, because if we add the number of cubes, we've already counted these 5. These 5 are counted as part of this 12. One way to think about it is there are 7 yellow objects that are not cubes. Those are the spheres. There are 5 yellow objects that are cubes. And then there are 8 cubes that are not yellow. That's one way to think about. we counted all of this. So we can't just add the number of cubes to it, because then we would count this middle part again. So then we have to essentially count cubes, the number of cubes, which is 13. So the number of cubes, and we'll have to subtract out this middle section right over here. Let me do this. So subtract out the middle section right over here. So minus 5. So this is the number of yellow cubes. It feels weird to write the word yellow in green. The number of yellow cubes-- or another way to think about it-- and you could just do this math right here. 12 plus 13 minus 5 is 20. Did I do that right? 12 minus, yup, it's 20. So that's one way. You just get this is equal to 20 over 29. But the more interesting thing than even the answer of the probability of getting that," + }, + { + "Q": "Why he needs to minus 5/29 at 7.25 at 7:24?", + "A": "He subtracted 5/29 so he wouldn t count the overlapping area twice.", + "video_name": "QE2uR6Z-NcU", + "timestamps": [ + 444 + ], + "3min_transcript": "or a cube of any color-- well, we still know that the denominator here is going to be 29. These are all of the equally likely possibilities that might jump out of the bag. But what are the possibilities that meet our conditions? Well, one way to think about it is, well, the probability-- there's 12 things that would meet the yellow condition. So that would be this entire circle right over here-- 12 things that meet the yellow condition. So this right over here is 12. This is the number of yellow. That is 12. And then to that, we can't just add the number of cubes, because if we add the number of cubes, we've already counted these 5. These 5 are counted as part of this 12. One way to think about it is there are 7 yellow objects that are not cubes. Those are the spheres. There are 5 yellow objects that are cubes. And then there are 8 cubes that are not yellow. That's one way to think about. we counted all of this. So we can't just add the number of cubes to it, because then we would count this middle part again. So then we have to essentially count cubes, the number of cubes, which is 13. So the number of cubes, and we'll have to subtract out this middle section right over here. Let me do this. So subtract out the middle section right over here. So minus 5. So this is the number of yellow cubes. It feels weird to write the word yellow in green. The number of yellow cubes-- or another way to think about it-- and you could just do this math right here. 12 plus 13 minus 5 is 20. Did I do that right? 12 minus, yup, it's 20. So that's one way. You just get this is equal to 20 over 29. But the more interesting thing than even the answer of the probability of getting that, that we figured out earlier in the video. So let's think about this a little bit. We can rewrite this fraction right over here. We can rewrite this as 12 over 29 plus 13 over 29 minus 5 over 29. And this was the number of yellow over the total possibilities. So this right over here was the probability of getting a yellow. This right over here was the number of cubes over the total possibilities. So this is plus the probability of getting a cube. And this right over here is the number of yellow cubes over the total possibilities. So this right over here was minus the probability of yellow, and a cube. I'm not going to write it that way. Minus the probability of yellow-- I'll" + }, + { + "Q": "at 0:53 how is it 8:20,should it be 8:12", + "A": "In this video, Sal is asking for the ratio of Apples to Fruit. In other words, how many apples compared to how many pieces of fruit total. Since 8 out of the 20 pieces of fruit are apples, the answer for this question is 8:20. However, if the question was What is the ratio of Apples to Oranges? your answer of 8:12 would be correct since there are 8 apples to 12 oranges.", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 53, + 500, + 492 + ], + "3min_transcript": "Voiceover:Let's think about another scenario involving ratios. In this case, let's think about the ratio of the number of apples. Number of apples to ... Instead of taking the ratio of the number of apples to the number of oranges, let's take the number of apples to the number of fruit. The number of fruit that we have over here. And I encourage you to pause the video and think about that on your own. Well, how many total apples do we have? We have 2, 4, 6, 8 apples. So we're going to have 8 apples. And then how much total fruit do we have? Well we have 8 apples and we have 3, 6, 9, 12 oranges. So our total fruit is 8 plus 12. We have 20 pieces of fruit. So this ratio is going to be 8 to ... 8 to 20. Or, if we want to write this in a more reduced form, we can divide both of these by 4. 4 is their greatest common divisor. 8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to." + }, + { + "Q": "could you do the same with 5:2 as 5/2?", + "A": "Yes, they are equivalent ways of writing the same ratio.", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 302 + ], + "3min_transcript": "As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to. Or we could say it's 2/5, the fraction 2/5, which would sometimes be read as 2 to 5. This is also, when it's written this way, you could also read that as a ratio, depending on the context. In a sentence like this I would read this as 2/5 of the fruit are apples." + }, + { + "Q": "there are 18 monkeys, 6 gorillas, and 15 apes what ratio is same to 5:13? Please someone help me i am having trouble", + "A": "The total number of animals is 18 + 6 + 15 = 39, and there are 15 apes. Note that 5:13 is equivalent to 15:39, from multiplying each number in this ratio by 3. So 5:13 is ratio of the number of apes to the total number of animals Have a blessed, wonderful day!", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 313 + ], + "3min_transcript": "As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to. Or we could say it's 2/5, the fraction 2/5, which would sometimes be read as 2 to 5. This is also, when it's written this way, you could also read that as a ratio, depending on the context. In a sentence like this I would read this as 2/5 of the fruit are apples." + }, + { + "Q": "how did sal reduce the original number/ 8:20 to 2:5", + "A": "common factor of 8 and 20 is 4 and 8/4 = 2 and 20/4 = 5 so 2:5. does this help you?", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 500, + 125 + ], + "3min_transcript": "" + }, + { + "Q": "he said for the first question that the fruit was 8:20 when it was 8:12. why did he say that? or was it a mistake?", + "A": "You are confusing fruit vs oranges . By fruit, Sal is referring to apples + oranges = 8+12 = 20 The ratio of apples to oranges = 8 : 12 The ratio of apples to fruit = 8 : 20 Hope this helps.", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 500, + 492 + ], + "3min_transcript": "" + }, + { + "Q": "is 2:5 the same as 2/5?", + "A": "Yes, those are the same.", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 125 + ], + "3min_transcript": "Voiceover:Let's think about another scenario involving ratios. In this case, let's think about the ratio of the number of apples. Number of apples to ... Instead of taking the ratio of the number of apples to the number of oranges, let's take the number of apples to the number of fruit. The number of fruit that we have over here. And I encourage you to pause the video and think about that on your own. Well, how many total apples do we have? We have 2, 4, 6, 8 apples. So we're going to have 8 apples. And then how much total fruit do we have? Well we have 8 apples and we have 3, 6, 9, 12 oranges. So our total fruit is 8 plus 12. We have 20 pieces of fruit. So this ratio is going to be 8 to ... 8 to 20. Or, if we want to write this in a more reduced form, we can divide both of these by 4. 4 is their greatest common divisor. 8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to." + }, + { + "Q": "If ratios and fractions are technically the same then why do you use ratios or if you are thinking the other way around why do you use fractions? The only thing different I see if that fractions are used more often and they are written differently. ratios: 2:3 fractions: 2/3.", + "A": "You should think of ratios as a special usage of fractions. They are used to compare similar quantities. Fractions can be used for many more purposes.", + "video_name": "UK-_qEDtvYo", + "timestamps": [ + 123 + ], + "3min_transcript": "Voiceover:Let's think about another scenario involving ratios. In this case, let's think about the ratio of the number of apples. Number of apples to ... Instead of taking the ratio of the number of apples to the number of oranges, let's take the number of apples to the number of fruit. The number of fruit that we have over here. And I encourage you to pause the video and think about that on your own. Well, how many total apples do we have? We have 2, 4, 6, 8 apples. So we're going to have 8 apples. And then how much total fruit do we have? Well we have 8 apples and we have 3, 6, 9, 12 oranges. So our total fruit is 8 plus 12. We have 20 pieces of fruit. So this ratio is going to be 8 to ... 8 to 20. Or, if we want to write this in a more reduced form, we can divide both of these by 4. 4 is their greatest common divisor. 8 divided by 4 is 2 and 20 divided by 4 is 5. So 2 to 5. Now, does this make sense? Well, if we divide ... If we divide everything into groups of 4. So ... Or if we divide into 4 groups, I should say. So 1 group, 2 groups, 3 groups, and 4 groups. That's the largest number of groups that we can divide these into so that we don't have to cut up the apples or the oranges. We see that in each group, for every 2 apples we have 1, 2, 3, 4, 5 pieces of fruit. For every 2 apples we have 5 pieces of fruit. This is actually a good opportunity for us to introduce another way of representing ... Another way of representing ratios, and that's using fraction notation. So we could also represent this ratio as As the fraction 2 over 5. Whenever we put it in the fraction it's very important to recognize what this represents. This is telling us the fraction of fruit that are apples. So we could say 2/5 of the fruit ... Of the fruit ... Of the number of fruit, I guess I could say. Of number of fruit ... Of fruit is equal to the number of apples. Right, I'm just going to say 2/5 of fruit if we're just speaking in more typical terms. 2/5 of fruit are apples. Are, are apples. So, once again, this is introducing another way of representing ratios. We could say that the ratio of apples to fruit, once again, it could be 2 to 5 like that. It could be 2, instead of putting this little colon there we could literally write out the word to." + }, + { + "Q": "At 1:21 why can he only measure length?", + "A": "He can only measure length because it is a line. A line is a one-dimensional object. A two dimensional objects allows you to measure width and length, and a three-dimensional object allows you to measure its height, width, and length", + "video_name": "xMz9WFvox9g", + "timestamps": [ + 81 + ], + "3min_transcript": "Human beings have always realized that certain things are longer than other things. For example, this line segment looks longer than this line segment. But that's not so satisfying just to make that comparison. You want to be able to measure it. You want to be able to quantify how much longer the second one is than the first one. And how do we go about doing that? Well, we define a unit length. So if we make this our unit length, we say this is one unit, then we could say how many of those the lengths are each of these lines? So this first line looks like it is-- we could do one of those units and then we could do it again, so it looks like this is two units. While this third one looks like we can get-- let's see that's 1, 2, 3 of the units. So this is three of the units. And right here, I'm just saying units. Sometimes we've made conventions to define a centimeter, where the unit might look something like this. And it's going to look different depending on your screen. Or we might have an inch that looks something like this. be able to fit on this screen based on how big I've just drawn the inch or a meter. So there's different units that you could use to measure in terms of. But now let's think about more dimensions. This is literally a one-dimensional case. This is 1D. Why is it one dimension? Well, I can only measure length. But now let's go to a 2D case. Let's go to two dimensions where objects could have a length and a width or a width and a height. So let's imagine two figures here that look like this. So let's say this is one of them. This is one of them. And notice, it has a width and it has a height. Or you could view it as a width and the length, depending on how you want to view it. So let's say this is one figure right over here. And let's say this is the other one. So this is the other one right over here. Try to draw them reasonably well. And we want to say, well, how much in two dimensions space is this taking up? Or how much area are each of these two taking up? Well, once again, we could just make a comparison. This second, if you viewed them as carpets or rectangles, the second rectangle is taking up more of my screen than this first one, but I want to be able to measure it. So how would we measure it? Well, once again, we would define a unit square. Instead of just a unit length, we now have two dimensions. We have to define a unit square. And so we might make our unit square. And the unit square we will define as being a square, where its width and its height are both equal to the unit length. So this is its width is one unit and its height is one unit. And so we will often call this 1 square unit. Oftentimes, you'll say this is 1 unit. And you put this 2 up here, this literally means 1 unit squared." + }, + { + "Q": "At 2:36, what is a \"unit Square\"?", + "A": "Its the unit of the shape ( e.g. feet, inches ) squared. The square is the little 2 at the top. For example 56 ft squared.", + "video_name": "xMz9WFvox9g", + "timestamps": [ + 156 + ], + "3min_transcript": "be able to fit on this screen based on how big I've just drawn the inch or a meter. So there's different units that you could use to measure in terms of. But now let's think about more dimensions. This is literally a one-dimensional case. This is 1D. Why is it one dimension? Well, I can only measure length. But now let's go to a 2D case. Let's go to two dimensions where objects could have a length and a width or a width and a height. So let's imagine two figures here that look like this. So let's say this is one of them. This is one of them. And notice, it has a width and it has a height. Or you could view it as a width and the length, depending on how you want to view it. So let's say this is one figure right over here. And let's say this is the other one. So this is the other one right over here. Try to draw them reasonably well. And we want to say, well, how much in two dimensions space is this taking up? Or how much area are each of these two taking up? Well, once again, we could just make a comparison. This second, if you viewed them as carpets or rectangles, the second rectangle is taking up more of my screen than this first one, but I want to be able to measure it. So how would we measure it? Well, once again, we would define a unit square. Instead of just a unit length, we now have two dimensions. We have to define a unit square. And so we might make our unit square. And the unit square we will define as being a square, where its width and its height are both equal to the unit length. So this is its width is one unit and its height is one unit. And so we will often call this 1 square unit. Oftentimes, you'll say this is 1 unit. And you put this 2 up here, this literally means 1 unit squared. could've been a centimeter. So this would be 1 square centimeter. But now we can use this to measure these areas. And just as we said how many of this unit length could fit on these lines, we could say, how many of these unit squares can fit in here? And so here, we might take one of our unit squares and say, OK, it fills up that much space. Well, we need more to cover all of it. Well, there, we'll put another unit square there. We'll put another unit square right over there. We'll put another unit square right over there. Wow, 4 units squares exactly cover this. So we would say that this has an area of 4 square units or 4 units squared. Now what about this one right over here? Well, here, let's seem I could fit 1, 2, 3, 4, 5, 6, 7, 8, and 9. So here I could fit 9 units, 9 units squared." + }, + { + "Q": "at 6:11 why did he only count the first cubes showing e forgot the back that you cannot see", + "A": "He didn t forget. He showed that there are two layers of 2 x 2 blocks. Each block is 1 unit^3 and there are 8 blocks so the volume is 8 units^3", + "video_name": "xMz9WFvox9g", + "timestamps": [ + 371 + ], + "3min_transcript": "We live in a three-dimensional world. Why restrict ourselves to only one or two? So let's go to the 3D case. And once again, when people say 3D, they're talking about 3 dimensions. They're talking about the different directions that you can measure things in. Here there's only length. Here there is length and width or width and height. And here, there'll be width and height and depth. So once again, if you have, let's say, an object, and now we're in three dimensions, we're in the world we live in that looks like this, and then you have another object that looks like this, it looks like this second object takes up more space, more physical space than this first object does. But how do we actually measure that? And remember, volume is just how much space something takes up in three dimensions. Area is how much space something takes up in two dimensions. Length is how much space something takes up in one dimension. But when we think about space, we're normally thinking about three dimensions. So how much space would you take up in the world that we live in? So just like we did before, we can define, instead of a unit length or unit area, we can define a unit volume or unit cube. So let's do that. Let's define our unit cube. And here, it's a cube so its length, width, and height are going to be the same value. So my best attempt at drawing a cube. And they're all going to be one unit. So it's going to be one unit high, one unit deep, and one unit wide. And so to measure volume, we could say, well, how many of these unit cubes can fit into these different shapes? won't be able to actually see all of them. I could essentially break it down into-- so let me see how well I can do this so that we can count them all. It's a little bit harder to see them all because there's some cubes that are behind us. But if you think of it as two layers, so one layer would look like this. One layer is going to look like this. So imagine two things like this stacked on top of each other. So this one's going to have 1, 2, 3, 4 cubes. Now, this is going to have two of these stacked on top of each other. So here you have 8 unit cubes. Or you could have 8 units cubed volume. What about here? If we try to fit it all in-- let me see how well I could draw this. It's going to look something like this. And obviously, this is kind of a rough drawing. And so if we were to try to take this apart, you would essentially have a stack of three sections that" + }, + { + "Q": "Let's say I was faced with a similar question to the one at 0:00. I used the vertical motion model and the quadratic formula to solve for the ball's time in motion. What would I do if both solutions of the equation are positive?", + "A": "The only way for your scenario could be true is if the y intercept were negative, so he would be shooting the ball from below ground level. The smaller number would be here the ball initially reaches ground level, and the second would be where the ball hits the ground. So you would have to have an equation such as f(x) = -16x^2 + 20t - 40. Even if you shot it off at ground level, one solution would be zero (from the origin). In either case, the time would be the highest x value.", + "video_name": "OZtqz_xw0SQ", + "timestamps": [ + 0 + ], + "3min_transcript": "A ball is shot into the air from the edge of a building, 50 feet above the ground. Its initial velocity is 20 feet per second. The equation h-- and I'm guessing h is for height-- is equal to negative 16t squared plus 20t plus 50 can be used to model the height of the ball after t seconds. And I think in this problem they just want us to accept this formula, although we do derive formulas like this and show why it works for this type of problem in the Khan Academy physics playlist. But for here, we'll just go with the flow on this example. So they give us the equation that can be used to model the height of the ball after t seconds, and then say about how long does it take for the ball to hit the ground. So if this is the height, the ground is when the height is equal to 0. So hitting the ground means-- this literally means that h is equal to 0. So we need to figure out at which times does h equal 0. So we're really solving the equation 0 is equal to negative 16t squared plus 20t plus 50. everything here is divisible at least by 2. And let's divide everything by negative 2, just so that we can get rid of this negative leading coefficient. So you divide the left hand side by negative 2, you still get a 0. Negative 16 divided by negative 2 is 8. So 8t squared. 20 divided by negative 2 is negative 10. Minus 10t. 50 divided by negative 2 is minus 25. And so we have 8t squared minus 10t minus 25 is equal to 0. Or if you're comfortable with this on the left hand side, we can put on the left hand side. We could just say this is equal to 0. And now we solve. And we could complete this square here, or we can just apply the quadratic formula, which is derived from completing the square. And we have this in standard form. We know that this is our a. This right over here is our b. And the quadratic formula tells us that the roots-- and in this case, it's in terms of the variable t-- are going to be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. So if we apply it, we get t is equal to negative b. b is negative 10. So negative negative 10 is going to be positive 10. Plus or minus the square root of negative 10 squared. Well, that's just positive 100, minus 4 times a, which is 8, times c, which is negative 25. And all of that over 2a. a Is 8. So 2 times 8 is 16. And this over here, we have a-- let's see if we can simplify this a little bit. The negative sign, negative times a negative, these are going to be positive. 4 times 25 is 100, times 8 is 800." + }, + { + "Q": "At 3:46, Sal gives a formula.... is that formula a generic one or is it restricted to n=7?", + "A": "It is generic . Sal chose 7 just for example purposes.", + "video_name": "LwhJVURumAA", + "timestamps": [ + 226 + ], + "3min_transcript": "when this is 7, all the way to 14. You could factor out a 2. And so this is going to become 2 times 1 plus 2 plus 3 all the way to 7. And so you can rewrite this piece right over here as 2 times the sum-- so we're essentially just factoring out the 2-- 2 times the sum, which is the sum from n equals 1 to 7 of n. So this is this piece. We still have this 28 that we have to add. So we have this 28. And we draw the parentheses so you don't think that the 28 is part of this right over here. And now we can do the same thing with this. 3 times n-- we're taking from n equals 1 to 7 of 3 n squared. Doing the same exact thing as we just did in magenta, this is going to be equal to 3 times the sum from n We're essentially factoring out the 3. We're factoring out the 2. n squared. And once again, we can put parentheses just to clarify things. Now, at this point, there are formulas to evaluate each of these things. There's a formula to evaluate this thing right over here. There's a formula to evaluate this thing over here. And you can look them up. And actually, I'll give you the formulas, in case you're curious. This formula, one expression of this formula is that this is going to be n to the third over 3 plus n squared over 2 plus n over 6. That's one formula for that. And one formula for this piece right over here, going from n equals 1 to 7-- sorry. Let me make it clear. This n is actually what your terminal value should be. So this should be 7 to the third power over 3-- I was just mindlessly using the formula-- 7 to the third over 3 plus 7 squared over 2 plus 7/6. So that's this sum. And this sum, you could view it as the average of the first and the last terms. So the first term is 1. The last term is 7. So take their average and then multiply it times the number of terms you have. So times-- you have 7 terms. So what is this middle one going to evaluate to? Well, 1 times-- and of course, we have this 2 out front. This green is just this part right over here. So you have 2 times this. And over here, you have 3 times this business right over here. So if we evaluate this one, 2 times-- let's see. 1 plus 7 is 8, divided by 2 is 4. 4 times 2 is 8. Times 7, it's 56. So that becomes 56. Now, this-- let's see." + }, + { + "Q": "At about 1:11 Sal(?) says 1/2 times 1/2 = 1/4. Then at about 1:26 1/2 is added to 1/4 to get 3/4 chance of winning. Why is one added and the other multiplied?", + "A": "The first case requires Brit to win both of the two times, whereas the second case requires Sal to win either time.", + "video_name": "tDdtAF3WtIY", + "timestamps": [ + 71, + 86 + ], + "3min_transcript": "Brit:How are we ever going to figure that out exactly? Let's just split the pot and call it even. Sal:Let's think about it this way, the next flip I have a 50% chance. Brit:I'm going to need to draw this out. Sal:Draw it out, draw it with trees. Brit:Okay so the next flip we'll go trees, so. Sal:It's either be heads or tails. Brit:Let's say it's heads or tails. Sal:Right, if it's heads Sal wins. Brit:If it's heads, Sal wins. Sal:And I get eight. Brit:And you get the whole pot, but if it's tails... Sal:If it's tails, then we keep playing. Brit:Oh so I need to do- Sal:Do another branch. Brit: Another branch. Brit:Oh I like this. Sal:So see they're going to be heads or tails. Now, at this point if it's heads I win. Brit:Sal wins. Sal:Sal wins, and if it's tails you win. Brit:Tails Brit wins. Sal:Yes. Brit:See, there's a chance I'm going to win, Sal:There is a chance, but your chance is not half. Your chance is substantially less than half. There's a one half chance of this happening. So this one has a half, and this one has a half. Sal:actually you could write, that's our first flip. then each of these outcomes are one half, one over two, one over two. So you have, to get two tails in a row, there's a one half time one half probability or there's only one fourth chance of this happening. And there's a one fourth chance of us getting a tails and another heads. So I have a one half plus one fourth chance of winning. One half plus one fourth, that's two fourths plus one fourth, I have a three fourths chance of winning. So I say, give me three fourths. I have a half chance of winning the next one. Brit:But there's eight chips here. Sal:Yeah, so this is how we think about it. I have a half chance of winning the next one, four chips for that. And then if I don't win the next one, I still have a half chance of winning that one. So I should get six chips, which is three fourths of eight. Brit:And I just get two. Sal:Unfortunately for you, that would be the case. Brit:You know I wanted half, but I can't argue with this." + }, + { + "Q": "At 0:32, they talk about adding more branches for each game. I wonder what would happen if they played 6 games instead of three?", + "A": "Exactly the same would happen, only the difference between chance of winnings would not be that much.", + "video_name": "tDdtAF3WtIY", + "timestamps": [ + 32 + ], + "3min_transcript": "Brit:How are we ever going to figure that out exactly? Let's just split the pot and call it even. Sal:Let's think about it this way, the next flip I have a 50% chance. Brit:I'm going to need to draw this out. Sal:Draw it out, draw it with trees. Brit:Okay so the next flip we'll go trees, so. Sal:It's either be heads or tails. Brit:Let's say it's heads or tails. Sal:Right, if it's heads Sal wins. Brit:If it's heads, Sal wins. Sal:And I get eight. Brit:And you get the whole pot, but if it's tails... Sal:If it's tails, then we keep playing. Brit:Oh so I need to do- Sal:Do another branch. Brit: Another branch. Brit:Oh I like this. Sal:So see they're going to be heads or tails. Now, at this point if it's heads I win. Brit:Sal wins. Sal:Sal wins, and if it's tails you win. Brit:Tails Brit wins. Sal:Yes. Brit:See, there's a chance I'm going to win, Sal:There is a chance, but your chance is not half. Your chance is substantially less than half. There's a one half chance of this happening. So this one has a half, and this one has a half. Sal:actually you could write, that's our first flip. then each of these outcomes are one half, one over two, one over two. So you have, to get two tails in a row, there's a one half time one half probability or there's only one fourth chance of this happening. And there's a one fourth chance of us getting a tails and another heads. So I have a one half plus one fourth chance of winning. One half plus one fourth, that's two fourths plus one fourth, I have a three fourths chance of winning. So I say, give me three fourths. I have a half chance of winning the next one. Brit:But there's eight chips here. Sal:Yeah, so this is how we think about it. I have a half chance of winning the next one, four chips for that. And then if I don't win the next one, I still have a half chance of winning that one. So I should get six chips, which is three fourths of eight. Brit:And I just get two. Sal:Unfortunately for you, that would be the case. Brit:You know I wanted half, but I can't argue with this." + }, + { + "Q": "At about 1:20 he says, \"P = 0.25p equals to 1.25p\". How did it become \"1.25p\" from \"P = 0.25\"? Thanks!", + "A": "Sal is simplifying the expression: P + 0.25P. Remember P has a coefficient of 1. P and 1P are the same. Sal is just adding like terms: P + 0.25P = 1P + 0.25P. Add 1+0.25 and you get 1.25. Thus, 1P + 0.25P = 1.25P. Hope this helps.", + "video_name": "ao9cx8JlJIU", + "timestamps": [ + 80 + ], + "3min_transcript": "Handsome Jack is buying a pony made of diamonds. The price of the pony is P dollars. And Jack also has to pay a 25% diamond pony tax. Match the expressions to their meaning for Handsome Jack. And they say multiple expressions may fit the same description. So in this bucket, we have the price of the diamond pony before tax. Well, they already tell us that the price of the diamond pony is P dollars. So that's P right over here. Now over here, they say the amount of tax Handsome Jack pays. Well, he pays a 25% diamond pony tax. So whatever the price is, he's going to pay 25% of that. Or another way of thinking about it, he's going to pay 25% is the same thing as 0.25. So 0.25 times P is the amount of tax he's going to pay on this diamond pony. Now, they say Handsome Jack's total bill for the diamond pony. Well, he's going to pay P for the pony plus 0.25P in taxes. P for the pony plus 0.25P for the taxes. So that's that one over there. But if we look at this, you could view this literally as 1P plus 0.25P's. Well, that's the same thing as 1.25P. So that's the same thing as this right over here. So it's 1.25P. And these other three don't seem to fit in any of these categories. So I'm going to put it into the not used. We're required to categorize everything. So let me put this in the not used. It's falling off the screen, I realize. Let me put this in the not used. And then let me put this in the not used. Let's check our answer. We got it right." + }, + { + "Q": "Since we are dividing by 4 at 1:16 wouldn't we write 4 at the beginning of the equation like this 4(x^2+10x-75=0?", + "A": "If you use factoring, you would create your format: 4(x^2+10x-75) = 0 This is done sometimes, but it actually easier to complete the square if the 4 is gone completely. This can be done by dividing the entire equation by 4, which is the technique that Sal used. Hope this helps.", + "video_name": "TV5kDqiJ1Os", + "timestamps": [ + 76 + ], + "3min_transcript": "We're asked to complete the square to solve 4x squared plus 40x minus 300 is equal to 0. So let me just rewrite it. So 4x squared plus 40x minus 300 is equal to 0. So just as a first step here, I don't like having this 4 out front as a coefficient on the x squared term. I'd prefer if that was a 1. So let's just divide both sides of this equation by 4. So let's just divide everything by 4. So this divided by 4, this divided by 4, that divided by 4, and the 0 divided by 4. Just dividing both sides by 4. So this will simplify to x squared plus 10x. And I can obviously do that, because as long as whatever I do to the left hand side, I also do the right hand side, that will make the equality continue to be valid. So that's why I can do that. So 40 divided by 4 is 10x. And then 300 divided by 4 is what? That is 75. Let me verify that. 7 times 4 is 28. You subtract, you get a remainder of 2. Bring down the 0. 4 goes into 20 five times. 5 times 4 is 20. Subtract zero. So it goes 75 times. This is minus 75 is equal to 0. And right when you look at this, just the way it's written, you might try to factor this in some way. But it's pretty clear this is not a complete square, or this is not a perfect square trinomial. Because if you look at this term right here, this 10, half of this 10 is 5. And 5 squared is not 75. So this is not a perfect square. So what we want to do is somehow turn whatever we have on the left hand side into a perfect square. And I'm going to start out by kind of getting this 75 out You'll sometimes see it where people leave the 75 on the left hand side. I'm going to put on the right hand side just so it kind of clears things up a little bit. So let's add 75 to both sides to get rid of the 75 from the left hand side of the equation. plus 75. Those guys cancel out. And I'm going to leave some space here, because we're going to add something here to complete the square that is equal to 75. So all I did is add 75 to both sides of this equation. Now, in this step, this is really the meat of completing the square. I want to add something to both sides of this equation. I can't add to only one side of the equation. So I want to add something to both sides of this equation so that this left hand side becomes a perfect square. And the way we can do that, and saw this in the last video where we constructed a perfect square trinomial, is that this last term-- or I should say, what we see on the left hand side, not the last term, this expression on the left hand side, it will be a perfect square if we have a constant term that is the square of half of the coefficient on the first degree So the coefficient here is 10. Half of 10 is 5." + }, + { + "Q": "At 5:35, Sal says that the radius will be smaller than it was before, but I'm confused because I thought that the radius of a unit sphere is always constant. Does he mean the z-coordinate or something like that instead of the radius?", + "A": "think in 3D... take another plane // to xy plane... and it intersects the sphere above the origin... it is going to form a smaller circle, and obviously its radius is going to be smaller than unity.", + "video_name": "E_Hwhp74Rhc", + "timestamps": [ + 335 + ], + "3min_transcript": "And the radius here is always 1. It's a unit sphere. So given this parameter s, what would be your x- and y-coordinates? And now we're thinking about it right if we're sitting in the xy-plane. Well, the x-coordinate-- this goes back to the unit circle definition of our trig functions. The x-coordinate is going to be cosine of s. It would be the radius, which is 1, times the cosine of s. And the y-coordinate would be 1 times the sine of s. That's actually where we get our definitions for cosine and sine from. So that's pretty straightforward. And in this case, z is obviously equal to 0. So if we wanted to add our z-coordinate here, z is 0. We are sitting in the xy-plane. But now, let's think about what happens if we go above and below the xy-plane. Remember, this is in any plane that is parallel to the xy-plane. This is saying how we are rotated around the z-axis. Now, let's think about if we go above and below it. And to figure out how far above or below it, And this new parameter I'm going to introduce is t. t is how much we've rotated above and below the xy-plane. Now, what's interesting about that is if we take any other cross section that is parallel to the xy-plane now, we are going to have a smaller radius. Let me make that clear. So if we're right over there, now where this plane intersects our unit sphere, the radius is smaller. The radius is smaller than it was before. Well, what would be this new radius? Well, a little bit of trigonometry. It's the same as this length right over here, which is going to be cosine of t. So the radius is going to be cosine of t. And it still works over here because if t goes all the way to 0, cosine of 0 is 1. when we're in the xy-plane. So the radius over here is going to be-- so that right over there is cosine of 0. So this is when t is equal to 0. And we haven't rotated above or below the xy-plane. But if we have rotated above the xy-plane, the radius has changed. It is now cosine of t. And now we can use that to truly parameterize x and y anywhere. So now, let's look at this cross section. So we're not necessarily in the xy-plane, we're in something that's parallel to the xy-plane. And so if we're up here, now all of a sudden, the cross section-- if we view it from above, might look something like this. It might look something like this. We're viewing it from above, this cross section right over here. Our radius right over here is cosine of t. And so given that-- I guess altitude" + }, + { + "Q": "At 1:00, why is the Celsius scale called the Celsius scale and why is the Fahrenheit scale called the Fahrenheit scale?", + "A": "The Celcius (or centigrade) scale is named for Anders Celsius (1701 - 1744) who created and defined a similar but upside down (0 was boiling water, 100 freezing water. The Fahrenheit Scale is named for Daniel Fahrenheit (1686-1736), based on one he first proposed in 1724.", + "video_name": "aASUZqJCHHA", + "timestamps": [ + 60 + ], + "3min_transcript": "Look at the two thermometers below. Identify which is Celsius and which is Fahrenheit, and then label the boiling and freezing points of water on each. Now, the Celsius scale is what's used in the most of the world. And the easy way to tell that you're dealing with the Celsius scale is on the Celsius scale, 0 degrees is freezing of water at standard temperature and pressure, and 100 degrees is the boiling point of water at standard temperature and pressure. Now, on the Fahrenheit scale, which is used mainly in the United States, the freezing point of water is 32 degrees, As you could tell, Celsius, the whole scale came from using freezing as 0 of regular water at standard temperature and pressure and setting 100 to be boiling. On some level, it makes a little bit more logical sense, but at least here in the U.S., we still use Fahrenheit. Now let's figure out which of these are Fahrenheit and which Now remember, regardless of which thermometer you're using, water will always actually boil at the exact So Fahrenheit, 32 degrees, this has to be the same thing as Celsius 0 degrees. So let's see what happens. So when this temperature right here is 0, this one over here, it looks like it's negative something. So this one right here doesn't look like Celsius. Here, if we say this is Celsius, this looks pretty close to 32 on this one. Let me do that in a darker color. So this one right here looks like Celsius, and this one right here looks like Fahrenheit. needs to be the same thing as 32 degrees Fahrenheit. In both cases, this is where water freezes, the freezing point. That is water freezing. So if this is the Celsius scale, this is where water will boil, 100 degrees Celsius, and that looks like it is right about 212 on the other scale. So right there is where water is boiling at standard temperature and pressure. So this thing on the right, right here, I guess I'll circle it in orange, that is Celsius. And then the one on the left, I'll do it in magenta, the one on the left is Fahrenheit." + }, + { + "Q": "at 2:02, I tried to do that prob on my own, but i failed.", + "A": "It s trying to show you WHY multiplying fractions works the way it does. Let s move on to how to actually multiply with fractions. First, remember that every whole number can be rewritten as a fraction. 5 is the same thing as 5/1. So what is 10 x 1/2? Think of it as 10/1 x 1/2. Now, multiply the two numerators together. 10 x 1 = 10. Now multiply the two denominators: 1 x 2 = 2. Put them together... 10/2 and simplify if possible: 10/2 is the same as 5.", + "video_name": "hr_mTd-oJ-M", + "timestamps": [ + 122 + ], + "3min_transcript": "Let's think a little bit about what it means to multiply fractions. Say I want to multiply 1/2 times 1/4. Well, one way to think about this is we could view this as 1/2 of a 1/4. And what do I mean there? Well let me take a whole, let me take a whole here, and let me divide it into fourths. So let me divide it into fourths, so I'll divided into 4 equal sections. And so 1/4 would be 1 of these 4 equal sections. But we want to take 1/2 of that. So how do we take half of that? Well, we could divide this into 2 equal sections, and then just take 1 of them. So divide it into 2 equal sections, and then take 1 of them. So we're taking this pink area, this whole pink area is 1/4, and now we're going to take 1/2 of it. We're now going to take 1/2 of it. So that's this yellow square right over here. Well, it now represents 1 out of 1, 2, 3, 4, 5, 6, 7, 8 equal sections. So this right over here, this represents 1/8 of the whole. And so we see conceptually that 1/2 times 1/4, it completely makes sense, that 1/2 of 1/4 should be 1/8. And it hopefully makes sense that you get this 8 by multiplying the 2 times the 4. You started with 4 equal sections, but then you divided each of those 4 equal sections into 2 equal sections. So then you have 8 total equal sections that you split your whole into. Let's do another example, but now let's multiply two fractions that don't have 1's in the numerator. So let's multiply, let's multiply 2/3 times 4/5. And I encourage you now to pause the video and do something very similar to what I just did. to represent 2/3 of that 4/5 and see what fraction of the whole you actually have. So pause now. So let's think about this. Let's represent 4/5. So if I have a whole like this, let me try to divide it into 5 equal sections. 5 equal sections, so let's say that is 1 equal section, that is 2 equal sections, that is 3, 4, and 5-- I can do a better This is always the hard part. I'm trying my best to make them look, at least, like equal sections-- 2, 3, 4, and 5. I think you get the point here. I'm trying to make them equal sections. And we want 4/5. So we want 4 of these 5 equal sections. So this would be 1 of the 5 equal sections, 2 of them, 3 of them, and then 4 of them. So that right over there is 4/5." + }, + { + "Q": "(0:48) why in multiplying fractions it is taking away, not increasing?", + "A": "The reason is when you multiply fractions the number being multiplied gets smaller because a fraction is a part of a whole. When you multiply a number by one: 5x1 You get 5 the same number you multiplied. So, if you multiply a number by a fraction ,the product should be less then the number you multiplied since a fraction smaller than 1. Same with decimals (if less than 1) since fractions are decimals.", + "video_name": "hr_mTd-oJ-M", + "timestamps": [ + 48 + ], + "3min_transcript": "Let's think a little bit about what it means to multiply fractions. Say I want to multiply 1/2 times 1/4. Well, one way to think about this is we could view this as 1/2 of a 1/4. And what do I mean there? Well let me take a whole, let me take a whole here, and let me divide it into fourths. So let me divide it into fourths, so I'll divided into 4 equal sections. And so 1/4 would be 1 of these 4 equal sections. But we want to take 1/2 of that. So how do we take half of that? Well, we could divide this into 2 equal sections, and then just take 1 of them. So divide it into 2 equal sections, and then take 1 of them. So we're taking this pink area, this whole pink area is 1/4, and now we're going to take 1/2 of it. We're now going to take 1/2 of it. So that's this yellow square right over here. Well, it now represents 1 out of 1, 2, 3, 4, 5, 6, 7, 8 equal sections. So this right over here, this represents 1/8 of the whole. And so we see conceptually that 1/2 times 1/4, it completely makes sense, that 1/2 of 1/4 should be 1/8. And it hopefully makes sense that you get this 8 by multiplying the 2 times the 4. You started with 4 equal sections, but then you divided each of those 4 equal sections into 2 equal sections. So then you have 8 total equal sections that you split your whole into. Let's do another example, but now let's multiply two fractions that don't have 1's in the numerator. So let's multiply, let's multiply 2/3 times 4/5. And I encourage you now to pause the video and do something very similar to what I just did. to represent 2/3 of that 4/5 and see what fraction of the whole you actually have. So pause now. So let's think about this. Let's represent 4/5. So if I have a whole like this, let me try to divide it into 5 equal sections. 5 equal sections, so let's say that is 1 equal section, that is 2 equal sections, that is 3, 4, and 5-- I can do a better This is always the hard part. I'm trying my best to make them look, at least, like equal sections-- 2, 3, 4, and 5. I think you get the point here. I'm trying to make them equal sections. And we want 4/5. So we want 4 of these 5 equal sections. So this would be 1 of the 5 equal sections, 2 of them, 3 of them, and then 4 of them. So that right over there is 4/5." + }, + { + "Q": "from 0:19 to 2:14, isn't there another way ?", + "A": "To be honest, drawing fraction models are slower. You can just multiply the the two fractions (should be improper for both) and get the right answer. Ex) 1/2*1/2= 1*1/2*2 = 1/4", + "video_name": "hr_mTd-oJ-M", + "timestamps": [ + 19, + 134 + ], + "3min_transcript": "Let's think a little bit about what it means to multiply fractions. Say I want to multiply 1/2 times 1/4. Well, one way to think about this is we could view this as 1/2 of a 1/4. And what do I mean there? Well let me take a whole, let me take a whole here, and let me divide it into fourths. So let me divide it into fourths, so I'll divided into 4 equal sections. And so 1/4 would be 1 of these 4 equal sections. But we want to take 1/2 of that. So how do we take half of that? Well, we could divide this into 2 equal sections, and then just take 1 of them. So divide it into 2 equal sections, and then take 1 of them. So we're taking this pink area, this whole pink area is 1/4, and now we're going to take 1/2 of it. We're now going to take 1/2 of it. So that's this yellow square right over here. Well, it now represents 1 out of 1, 2, 3, 4, 5, 6, 7, 8 equal sections. So this right over here, this represents 1/8 of the whole. And so we see conceptually that 1/2 times 1/4, it completely makes sense, that 1/2 of 1/4 should be 1/8. And it hopefully makes sense that you get this 8 by multiplying the 2 times the 4. You started with 4 equal sections, but then you divided each of those 4 equal sections into 2 equal sections. So then you have 8 total equal sections that you split your whole into. Let's do another example, but now let's multiply two fractions that don't have 1's in the numerator. So let's multiply, let's multiply 2/3 times 4/5. And I encourage you now to pause the video and do something very similar to what I just did. to represent 2/3 of that 4/5 and see what fraction of the whole you actually have. So pause now. So let's think about this. Let's represent 4/5. So if I have a whole like this, let me try to divide it into 5 equal sections. 5 equal sections, so let's say that is 1 equal section, that is 2 equal sections, that is 3, 4, and 5-- I can do a better This is always the hard part. I'm trying my best to make them look, at least, like equal sections-- 2, 3, 4, and 5. I think you get the point here. I'm trying to make them equal sections. And we want 4/5. So we want 4 of these 5 equal sections. So this would be 1 of the 5 equal sections, 2 of them, 3 of them, and then 4 of them. So that right over there is 4/5." + }, + { + "Q": "Why didn't he just make the -6 a + 6 at 0:25? The way I am being taught right now is you take the first set of complex numbers leave them the same (2-3i stays the same) and then do the opposite of -(6-18) and this would be +6+18 when you remove the parentheses. The full equation should look something like this when the parentheses are removed: 2-3i+6+18. Why didn't he do it this way?", + "A": "That would be true if it were (-6-18i) but look where the parentheses fall: -(6-18i). The basic rules of algebra apply and you have to distribute: -6+18i, as Sal had.", + "video_name": "tvXRaZbIjO8", + "timestamps": [ + 25 + ], + "3min_transcript": "We're asked to subtract. And we have the complex number 2 minus 3i. And from that, we are subtracting 6 minus 18i. So the first thing I'd like to do here is to just get rid of these parentheses. So we just have a bunch of real parts and imaginary parts that we can then add up together. So we have 2 minus 3i. And then we're subtracting this entire quantity. And to get rid of the parentheses, we can just distribute the negative sign. Or another way to think about it, we can say that this is negative 1 times all of this. So we can just distribute the negative sign. And negative 1 times 6 is negative 6. Let me do these in magenta. So this is negative 6. And then negative 1 times negative 18i-- well, that's just going to be positive 18i. Negative times a negative is a positive. And now we want to add the real parts, and we want to add the imaginary parts. So here's a real part here, 2. And then we have a minus 6. So we have 2 minus 6. And we want to add the imaginary parts. Let me do that in a different color. We have a negative 3i right over here. So negative 3i, or minus 3i right over there. And then we have a plus 18i or positive 18i. If you add the real parts, 2 minus 6 is negative 4. And you add the imaginary parts. If I have negative 3 of something and to that I add 18 of something, well, that's just going to leave me with 15 of that something. Or another way you could think about it, if I have 18 of something and I subtract 3 of that something, I'll have 15 of that something. And in this case, the something is i, is the imaginary unit. So this is going to be plus 15i. And we are done." + }, + { + "Q": "I love this song but what does Tau mean 0:50", + "A": "Tau in this case means the ratio between the circumference of a circle and its radius.", + "video_name": "FtxmFlMLYRI", + "timestamps": [ + 50 + ], + "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211." + }, + { + "Q": "1:14 - 1:20:\n\"Well, depending on the dimension, Because it's 1, 2, 3, There's 4 and even more.\"\nWhat is the fourth dimension?", + "A": "The 4th dimension is time. As far as we know, it is the biggest dimension perceivable by humans. You can t see time, but you can measure it, and tell the difference between long and short periods of time.", + "video_name": "FtxmFlMLYRI", + "timestamps": [ + 74, + 80 + ], + "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211." + }, + { + "Q": "AT 0:07 sal say to \"massage the equation\", but what does that mean?", + "A": "It means manipulate the equation, converting it into an equivalent form that makes it easier to solve.", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 7 + ], + "3min_transcript": "Let's solve a few more systems of equations using elimination, but in these it won't be kind of a one-step elimination. We're going to have to massage the equations a little bit in order to prepare them for elimination. So let's say that we have an equation, 5x minus 10y is equal to 15. And we have another equation, 3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution, we could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, you know, with elimination, you were subtracting the left-hand side of one equation from another, or adding the two, and then adding the two right-hand sides. And I could do that, because it was essentially adding the same thing to both sides of the equation. But here, it's not obvious that that If we added these two left-hand sides, you would get 8x minus 12y. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is, we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. And you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y. Because if this is a positive 10y, it'll cancel out when I add the left-hand sides of this equation. So what can I multiply this equation by? negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times negative 5. So you multiply the left-hand side by negative 5, and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And then negative 5 times negative 2y is plus 10y, is equal to 3 times negative 5 is negative 15. And now, we're ready to do our elimination. If we add this to the left-hand side of the yellow equation, and we add the negative 15 to the right-hand" + }, + { + "Q": "@ 10:26 in the video he says 5 is the same thing as 20/4 + 15/4. How is 5 the same as 20/4? Thanks.", + "A": "Think of a fraction as literally the numerator divided by the denominator. When 20 is in the numerator, it s the same as saying 20 divided by 4. 20/4= 5.", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 626 + ], + "3min_transcript": "25, which is also equal to this expression. So let's add the left-hand sides and the right-hand sides. Because we're really adding the same thing to both sides of the equation. So the left-hand side, the x's cancel out. 35x minus 35x. That was the whole point. They cancel out, and on the y's, you get 49y plus 15y, that is 64y. 64y is equal to 105 minus 25 is equal to 80. Divide both sides by 64, and you get y is equal to 80/64. And let's see, if you divide the numerator and the denominator by 8-- actually you could probably do 16. 16 would be better. But let's do 8 first, just because we know our 8 times tables. So that becomes 10/8, and then you can divide this by 2, and If you divided just straight up by 16, you would've gone straight to 5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these equations, or into one of the original equations. Let's substitute into the second of the original equations, where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? Is equal to 5. Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1." + }, + { + "Q": "At 6:30 Sal makes the bottom equation negative and the top one positive. Would it make a difference if I made the top one negative and the bottom one positive.", + "A": "It doesn t matter. You want one equation to be negative and one equation to be positive so when you add them, one of the variables become 0 (eliminated). You could have both positives or both negatives and use subtraction. Subtraction is easier to mess up when subtracting negatives. So I recommend making one positive and one negative then use addition.", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 390 + ], + "3min_transcript": "Let me write that. Negative 10y is equal to 15. Divide both sides by negative 10. And we are left with y is equal to 15/10, is negative 3/2. So if you were to graph it, the point of intersection would be the point 0, negative 3/2. And you can verify that it also satisfies this equation. The original equation over here was 3x minus 2y is equal to 3. 3 times 0, which is 0, minus 2 times negative 3/2 is, this is 0, this is positive 3. Right? These cancel out, these become positive. Plus positive 3 is equal to 3. So this does indeed satisfy both equations. Let's do another one of these where we have to multiply, and the variables. Let's do another one. Let's say we have 5x plus 7y is equal to 15. And we have 7-- let me do another color-- 7x minus 3y is equal to 5. Now once again, if you just added or subtracted both the left-hand sides, you're not going to eliminate any variables. These aren't in any way kind of have the same coefficient or the negative of their coefficient. So let's pick a variable to eliminate. Let's say we want to eliminate the x's this time. And you could literally pick on one of the It doesn't matter. You can say let's eliminate the y's first. But I'm going to choose to eliminate the x's first. And so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients, or their add the left-hand sides, they're going to eliminate each other. Now, there's nothing obvious-- I can multiply this by a fraction to make it equal to negative 5. Or I can multiply this by a fraction to make it equal to negative 7. But even a more fun thing to do is I can try to get both of them to be their least common multiple. I could get both of these to 35. And the way I can do it is by multiplying by each other. So I can multiply this top equation by 7. And I'm picking 7 so that this becomes a 35. And I can multiply this bottom equation by negative 5. And the reason why I'm doing that is so this becomes a negative 35. Remember, my point is I want to eliminate the x's. So if I make this a 35, and if I make this a negative 35, then I'm going to be all set." + }, + { + "Q": "At \"10:32\" how does he get 7x= 35/4 ? Wouldn't the it be 7x=20/4 because 5+15/4=20/4.", + "A": "No 5 = 20/4. 15/4 + 20/4 = 35/4...", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 632 + ], + "3min_transcript": "If you divided just straight up by 16, you would've gone straight to 5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these equations, or into one of the original equations. Let's substitute into the second of the original equations, where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? Is equal to 5. Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So the point of intersection of this right here is both x and y are going to be equal to 5/4. So if you looked at it as a graph, it'd be 5/4 comma 5/4. And let's verify that this satisfies the top equation. And if you take 5 times 5/4, plus 7 times 5/4, what do you get? It should be equal to 15. So this is equal to 25/4, plus-- what is this? This is plus 35/4. Which is equal to 60/4, which is indeed equal to 15. So it does definitely satisfy that top equation. And you could check out this bottom equation for yourself, but it should, because we actually used this bottom equation to figure out that x is equal to 5/4." + } +] \ No newline at end of file diff --git a/MathSc-Timestamp/test500.json b/MathSc-Timestamp/test500.json new file mode 100644 index 0000000000000000000000000000000000000000..a0654b3f575113180edea96c6b345692c713152d --- /dev/null +++ b/MathSc-Timestamp/test500.json @@ -0,0 +1,5027 @@ +[ + { + "Q": "at 7:00 why is Newtons the number of particles?? isnt newton kg*m/s^2\nat 6:15 Sal say its number of molecules. how is that possible if newton is kg*m/s^2?", + "A": "There are no Newtons in this video. N stands for Number, as in Number of molecules. It s a variable, not a unit.", + "video_name": "x34OTtDE5q8", + "timestamps": [ + 420, + 375 + ], + "3min_transcript": "I think you also have the sense that-- what would have more energy? A 100 degree cup of tea, or a 100 degree barrel of tea. I want to make them equivalent in terms of what they're holding. I think you have a sense. Even though they're the same temperature, they're both pretty warm-- let's say this is 100 degrees Celsius, so they're both boiling-- that the barrel, because there's more of it, is going to have more energy. It's equally hot, and there's just more molecules there. That's what temperature is. Temperature, in general, is a measure roughly equal to some energy-- per molecule. So the average kinetic energy of the system divided by the total number of molecules we have. Another way we could talk about is, temperature is essentially energy per molecule. So something that has a lot of molecules, where N is the number of molecules. Another way we could view this is that the kinetic energy of the system is going to be equal to the number of molecules times the temperature. This is just a constant-- times 1 over K, but we don't even know what this is, so we could say that's still a constant-- so the kinetic energy of the system is going to be equal to some constant times the number of particles We don't know what this is, and we're going to figure this out later. This is another interesting concept. We said that pressure times volume is proportional to the kinetic energy of the system-- the aggregate, if you take all of the molecules and combine their kinetic energies. These aren't the same K's-- I could put another constant here and call that K1. And we also know that the kinetic energy of the system is equal to some other constant times the number of molecules I have times the temperature. If you think about it, you could also say that this is proportional to this, and this is proportional to this. You could say that pressure times volume is proportional", + "qid": "x34OTtDE5q8_420_375" + }, + { + "Q": "so from 00:01 to 23:17 he talking about the common cold and flu (influenza)? :|", + "A": "He s talking about Viruses in general and not about a specific one. At the beginning he says that because he has a cold that he s going to talk about Viruses. In between those times he covers how a most viruses interact with living cells and also how a retrovirus might.", + "video_name": "0h5Jd7sgQWY", + "timestamps": [ + 1, + 1397 + ], + "3min_transcript": "Considering that I have a cold right now, I can't imagine a more appropriate topic to make a video on than a virus. And I didn't want to make it that thick. A virus, or viruses. And in my opinion, viruses are, on some level, the most fascinating thing in all of biology. Because they really blur the boundary between what is an inanimate object and what is life? I mean if we look at ourselves, or life as one of those things that you know it when you see it. If you see something that, it's born, it grows, it's constantly changing. Maybe it moves around. Maybe it doesn't. But it's metabolizing things around itself. It reproduces and then it dies. You say, hey, that's probably life. And in this, we throw most things that we see-- or we throw in, us. We throw in bacteria. We throw in plants. I mean, I could-- I'm kind of butchering the taxonomy system here, but we tend to know life when we see it. information inside of a protein. Inside of a protein capsule. So let me draw. And the genetic information can come in any form. So it can be an RNA, it could be DNA, it could be single-stranded RNA, double-stranded RNA. Sometimes for single stranded they'll write these two little S's in front of it. Let's say they are talking about double stranded DNA, they'll put a ds in front of it. But the general idea-- and viruses can come in all of these forms-- is that they have some genetic information, some chain of nucleic acids. Either as single or double stranded RNA or single or double stranded DNA. And it's just contained inside some type of protein structure, which is called the capsid. And kind of the classic drawing is kind of an icosahedron type looking thing. Let me see if I can do justice to it. It looks something like this. And not all viruses have to look exactly like this. And we're really just scratching the surface and understanding even what viruses are out there and all of the different ways that they can essentially replicate themselves. We'll talk more about that in the future. And I would suspect that pretty much any possible way of replication probably does somehow exist in the virus world. But they really are just these proteins, these protein capsids, are just made up of a bunch of little proteins put together. And inside they have some genetic material, which might be DNA or it might be RNA. So let me draw their genetic material. The protein is not necessarily transparent, but if it was, you would see some genetic material inside of there. So the question is, is this thing life? It seems pretty inanimate. It doesn't grow. It doesn't change. It doesn't metabolize things. This thing, left to its own devices, is just It's just going to sit there the way a book on a table just sits there. It won't change anything.", + "qid": "0h5Jd7sgQWY_1_1397" + }, + { + "Q": "At 3:10, the ice has 2 equal forces acting upon it at opposite sides, but wouldn't the ice just go upwards instead of remaining stationary because of friction?", + "A": "No. Friction, in this case, points sideways, so it can t make it go up. The reaction force does point upwards, but it s as big as gravity, so it also can t do that.", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 190 + ], + "3min_transcript": "I can keep observing that rock. And it is unlikely to move, assuming that nothing happens to it. If there's no force applied to that rock, that rock will just stay there. So the first part is pretty obvious. So, \"Every body persists in a state of being at rest\"-- I'm not going to do the second part-- \"except insofar as there's some force being applied to it.\" So clearly a rock will be at rest, unless there's some force applied to it, unless someone here tries to push it or roll it or do something to it. What's less intuitive about the first law is the second part. \"Every body persists in,\" either, \"being in a state of rest or moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.\" So this Newton's first law-- and I think I should do a little aside here, because, this right here is Newton. And if this is Newton's first law, Well, the reason is is because Newton's first law is really just a restatement of this guy's law of inertia. And this guy, another titan of civilization really, this is Galileo Galilei. And he is the first person to formulate the law of inertia. And Newton just rephrased it a little bit and packaged it with his other laws. But he did many, many, many other things. So you really have to give Galileo credit for Newton's first law. So that's why I made him bigger than here. But I was in the midst of a thought. So we understand if something is at rest, it's going to stay at rest, unless there's some force that acts on it. And in some definitions, you'll see unless there's some unbalanced force. And the reason why they say unbalanced is, because you could have two forces that act on something and they might balance out. For example, I could push on this side of the rock with a certain amount of force. And if you push on this side of the rock with the exact same amount of force, the rock won't move. And the only way that it would move if there's a lot more so if you have an unbalanced force. So if you have a ton of-- and maybe the rock is a bad analogy. Let's take ice, because ice is easier to move, or ice on ice. So there's ice right here. And then, I have another block of ice sitting on top of that ice. So once again, we're familiar with the idea, if there's no force acting on it that ice won't move. But what happens if I'm pushing on the ice with a certain amount of force on that side, and you're pushing on the ice on that side with the same amount of force? The ice will still not move. So this right here, this would be a balanced force. So the only way for the ice to change its condition, to change its restful condition is if the force is unbalanced. So if we add a little bit of force on this side, so it more than compensates the force pushing it this way, then you're going to see the ice block start to move, start to really accelerate in that direction.", + "qid": "CQYELiTtUs8_190" + }, + { + "Q": "At 8:12 Sal says \"if gravity disappeared, and you had no air...\". What's the relation between gravity and air?", + "A": "He uses gravity and air as two examples of forces that occur naturally all around us.", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 492 + ], + "3min_transcript": "so if you have the actual water molecules in a lattice structure in the ice cube, and then here are the water molecules in a lattice structure on the ice, on the actual kind of sea of ice that it's traveling on-- they do kind of bump and grind into each other. Although they're both smooth, there are imperfections here. They bump and grind. They generate a little bit of heat. And they'll, essentially, be working against the movement. So there's a force of friction that's being applied to here. And that's why it's stopping. Not only a force of friction, you also have some air resistance. The ice block is going to be bumping into all sorts of air particles. It might not be noticeable at first, but it's definitely going to keep it from going on forever. Same thing with the ball being tossed to the air. Obviously, at some point, it hits the ground because of gravity. So that's one force acting on it. But even once it hits the ground, it doesn't keep rolling forever, once again, because of the friction, especially if there's grass The grass is going to stop it from going. And even while it's in the air, it's going to slow down. It's not going to have a constant velocity. Because you have all of these air particles that are going to bump into it and exert force to slow it down. So what was really brilliant about these guys is that they could imagine a reality where you didn't have gravity, where you did not have air slowing things down. And they could imagine that in that reality, something would just keep persisting in its motion. And the reason why Galileo, frankly, was probably good at thinking about that is that he studied the orbits of planets. And he could, or at least he's probably theorized that, hey, maybe there's no air out there. And that maybe that's why these planets can just keep going round and round in orbit. And I should say their speed, because their direction is changing, but their speed never slows down, because there's nothing in the space to actually slow down those planets. Because on some level, it's super-duper obvious. But on a whole other level, it's completely not obvious, especially this moving uniformly straightforward. And just to make the point clear, if gravity disappeared, and you had no air, and you threw a ball, that ball literally would keep going in that direction forever, unless some other unbalanced force acted to stop it. And another way to think about it-- and this is an example that you might see in everyday life-- is, if I'm in an airplane that's going at a completely constant velocity and there's absolutely no turbulence in the airplane. So if I'm sitting in the airplane right over here. And it's going at a constant velocity, completely smooth, no turbulence. There's really no way for me to tell whether that airplane is moving without looking out the window. Let's assume that there's no windows in that airplane. It's going at a constant velocity.", + "qid": "CQYELiTtUs8_492" + }, + { + "Q": "At around 5:05 Sal says everything will eventually stop, if so, how is the earth constantly orbiting the sun? What is causing the earth to move?", + "A": "The Earth continues to spin upon its axis because there are no outside forces acting to stop its rotation.", + "video_name": "CQYELiTtUs8", + "timestamps": [ + 305 + ], + "3min_transcript": "so if you have an unbalanced force. So if you have a ton of-- and maybe the rock is a bad analogy. Let's take ice, because ice is easier to move, or ice on ice. So there's ice right here. And then, I have another block of ice sitting on top of that ice. So once again, we're familiar with the idea, if there's no force acting on it that ice won't move. But what happens if I'm pushing on the ice with a certain amount of force on that side, and you're pushing on the ice on that side with the same amount of force? The ice will still not move. So this right here, this would be a balanced force. So the only way for the ice to change its condition, to change its restful condition is if the force is unbalanced. So if we add a little bit of force on this side, so it more than compensates the force pushing it this way, then you're going to see the ice block start to move, start to really accelerate in that direction. This, you know, something that's at rest will stay at rest, unless it's being acted on by an unbalanced force. What's less obvious is the idea that something moving uniformly straightforward, which is another way of saying something having a constant velocity. What he's saying is, is that something that has a constant velocity will continue to have that constant velocity indefinitely, unless it is acted on by an unbalanced force. And that's less intuitive. Because everything in our human experience-- even if I were to push this block of ice, eventually it'll stop. It won't just keep going forever, even assuming that this ice field is infinitely long, that ice will eventually stop. Or if I throw a tennis ball. That tennis ball will eventually stop. It'll eventually grind to a halt. We've never seen, at least in our human experience, it looks like everything will eventually stop. So this is a very unintuitive thing to say, that something in motion will just keep going in motion indefinitely. Everything in human intuition says if you want something to keep going in motion, you have to keep putting more force, keep putting more energy into it for it to keep going. Your car won't go forever, unless you keep, unless the engine keeps burning fuel to drive and consuming energy. So what are they talking about? Well, in all of these examples-- and I think this is actually a pretty brilliant insight from all of these fellows is that-- all of these things would have gone on forever. The ball would keep going forever. This ice block would be going on forever, except for the fact that there are unbalanced forces acting on them to stop them. So in the case of ice, even though ice on ice doesn't have a lot of friction, there is some friction between these two. And so you have, in this situation, the force of friction is going to be acting against the direction of the movement of the ice.", + "qid": "CQYELiTtUs8_305" + }, + { + "Q": "at 6:15 he used H2O as a base can we use the HSO4- from the process where we generate electrophile?", + "A": "HSO\u00e2\u0082\u0084\u00e2\u0081\u00bb can act as a base, but H\u00e2\u0082\u0082O is stronger and it is present in much larger amounts.", + "video_name": "rC165FcI4Yg", + "timestamps": [ + 375 + ], + "3min_transcript": "And then over here, we would have an oxygen with three lone pairs of electrons, giving that a negative 1 formal charge. And the nitrogen, of course, is still going to have a plus 1 formal charge like that. All right, let me go ahead and highlight those electrons. So once again, these pi electrons are going to be attracted to the positive charge, nucleophile-electrophile. And those pi electrons are going to form this bond right here to our nitro group. Well, once again, as we've seen several times before, we took away a bond from this carbon. So that's where our plus 1 the formal charge is going to go like that. And so we can draw some resonance structures. So let's go ahead and show a resonance structure for this. We could move these pi electrons in over here. So let's go ahead and draw that. So we had a hydrogen up here. And you could just show a nitro group as NO2. So I'm just going to go ahead and do that to save some time. And I'm saying that those pi electrons moved over to here. So let me go ahead and highlight those. So these pi electrons in blue move over to here, took a bond away from that carbon. So now we can put a plus 1 formal charge at that carbon like that. We can draw yet another resonance structure. So I could show these electrons over here moving to here. So let me go ahead and draw that. So we have our ring. We have our nitro group already on our ring. We have some pi electrons right here. And we have some more pi electrons moving from here to here, which, of course, takes a bond away from this top carbon. So that's where our positive 1 formal charge is now. So now we have our three resonance structures. And remember, once again, that the sigma complex is a hybrid of these three. And we're now ready for our last step, So if we go back up to here, we think, what could function as a base? Well, the water molecule here could function as a base. So a lone pair of electrons on our water molecule are going to take that proton, which would cause these electrons to move in here to reform your aromatic ring. So let's go ahead and show that. So we're going to reform our benzene ring here. And we took off the proton. So deprotonation of the sigma complex yields our product with a nitro group substituted in. So let me go ahead and highlight those electrons again. So this time I'll use green. So these electrons right in here, when that sigma complex is deprotonated, those electrons are going to move in here to restore the aromatic ring, and we have created our product. We have added in our nitro group.", + "qid": "rC165FcI4Yg_375" + }, + { + "Q": "At 10:30 he says that the Egg gets all the organelles. Does that mean that we get all of the organelles from our mother? (I know about the maternal inheritance of mitochondria.) If so, what happens with the organelles in the sperm cells?", + "A": "No such thing. Organelles are made as instructed by DNA. The part of mitochondrial genome that is still within it is maternally inherited. All others are synthesised using simple molecules, as instructed by maternal and paternal DNA in the nucleus.", + "video_name": "TX7-Kdn6lJQ", + "timestamps": [ + 630 + ], + "3min_transcript": "and were then pulled in half, but not here. In meiosis, each chromosome lines up next to it's homologous pair partner that it's already swapped a few genes with. Now, the homologous pairs get pulled apart and migrate to either end of the cell and that's anaphase I. The final phase of the first round, telophase I rolls out in pretty much the same way as mitosis. The nuclear membrane reforms, the nucleoli form within them, the chromosomes fray out back into chromatin, a crease forms between the two new cells called cleavage and then the two new nuclei move apart from each other, the cells separate in a process called cytokinesis, literally again, cell movement and that is the end of round one. We now have two haploid cells, each with 23 double chromosomes that are new, unique combinations of the original chromosome pairs. In these new cells, the chromosomes are still duplicated and still connected at the centromeres. They still look like X's, but remember, the aim is to end up with four cells. Here, the process is exactly the same as mitosis, except that the aim here isn't to duplicate the double chromosomes, but instead to pull them apart into separate single strand chromosomes. Because of this, there's no DNA replication involved in prophase II. Instead, the DNA just clumps up again into chromosomes and the infrastructure for moving them, the microtubules are put back in place. In metaphase II, the chromosomes are moved into alignment into the middle of the cell and in anaphase II, the chromotids are pulled apart into separate single chromosomes. The chromosomes uncoil into chromatin, the crease form in cleavage and the final separation of cytokinesis then mark the end of telophase II. From one original cell with 46 chromosomes, we now have four new cells with 23 single chromosomes each. If these are sperm, all four of the resulting cells are the same size, but they each have slightly different genetic information and half will be for making girls and half will be for making boys, but if this is the egg making process, and the result is only one egg. To rewind a little, during telophase I, more of the inner goodness of a cell, the cytoplasm, the organelles heads into one of the cells that gets split off then to the other one. In telophase II, when it's time to split again, the same thing happens with more stuff going into one of the cells than the other. This big ol' fat remaining cell becomes the egg with more of the nutrients and cytoplasm and organelles that it will take to make a new embryo. The other three cells that were produced, the little ones, are called polar bodies and they're totally useless in people, though they are useful in plants. In plants, those polar bodies actually, also get fertilized too and they become the endosperm. That's the starchy, protein-ey stuff that we grind into wheat, or pop into popcorn and it's basically the nutrients that feed the plant embryo, the seed. And that's all there is to it. I know you were probably were really excited when I started talking about reproduction, but then I rambled on for a long time", + "qid": "TX7-Kdn6lJQ_630" + }, + { + "Q": "At 3:21 he mentioned frequency as how close the peaks are, instead that distance between two peaks is called wavelength. Frequency is just the number of waves in a specific time period. Feel free to comment, I am not 100% sure!", + "A": "Yes ,you are correct.", + "video_name": "6GB_kcdVMQo", + "timestamps": [ + 201 + ], + "3min_transcript": "and it makes a very distinct sound. Let's imagine that these two lines right here are your hands. When you clap your hands, the lines move towards each other, so your hands are moving towards each other. They're moving towards each other fairly quickly. In between your hands are a whole bunch of little air molecules, which I'm drawing. These air molecules, which I'm drawing right now, let's imagine that they're these little purple dots. So, in between your hands are a whole bunch of these air molecules. They're just floating around doing their thing. Then all of a sudden, the hands are moving towards each other, and all of a sudden, this space that these air molecules occupy gets a lot smaller. A little bit later in time, as the hands are moving towards each other - so here we are just drawing the hands almost about to touch. What happened was all these air molecules that are just floating around, they had all this space.. Now all of a sudden, they're really compressed, so they're really, really close together, They're very compacted now. You can imagine that as your hands are even closer together, that the air molecules get even more compacted. Basically, what is effectively going on is the air molecules here are getting pressurized. As you bring your hands together, you're actually adding all the molecules up, and it creates this pressure. This area of pressure actually tries to escape. It tries to escape and it kinda goes this way. It tries to escape out wherever it can. As it's escaping, it creates these areas of high and low pressure. That's what I'm representing here by these lines. These areas of high and low pressure are known as sound waves. We can have different types of sound waves. We can have sound waves that are really, really close together, or really far away from each other. If we draw this graphically, Basically, what I'm drawing here is, up here would be an area of high pressure. Over here would be an area of low pressure. Basically, there are just areas of high and low pressure. How close these peaks are together is the frequency. If I clap my hands even faster together, or if there's something else that's a higher frequency, a higher-pitched sound, the sound waves would be closer to one another, and it would look something like this. Depending on the frequency of the sound wave, it's perceived to be a different noise. Let's imagine that this sound wave right here is F1, and that this one over here is F2. Sound waves of lower frequency actually travel further. This actually happens in the ear, so these lower frequency sound waves actually travel further, and they actually penetrate deeper into the cochlea,", + "qid": "6GB_kcdVMQo_201" + }, + { + "Q": "Does the equation at 4:56 imply that there is no magnetic force when a charge isn't moving? If so, how does a paperclip feel a magnetic force towards a magnet when both objects are held stationary?", + "A": "Go to youtube and search for veritasium how do magnets work and watch the pair of videos.", + "video_name": "NnlAI4ZiUrQ", + "timestamps": [ + 296 + ], + "3min_transcript": "So that's fine, you say, Sal, that's nice. You drew these field lines. And you've probably seen it before if you've ever dropped metal filings on top of a magnet. They kind of arrange themselves But you might say, well, that's kind of useful. But how do we determine the magnitude of a magnetic field at any point? And this is where it gets interesting. The magnitude of a magnetic field is really determined, or it's really defined, in terms of the effect that it has on a moving charge. So this is interesting. I've kind of been telling you that we have this different force called magnetism that is different than the electrostatic force. But we're defining magnetism in terms of the effect that it has on a moving charge. And that's a bit of a clue. And we'll learn later, or hopefully you'll learn later as you advance in physics, that magnetic force or a magnetic field is nothing but an electrostatic field moving at a very high speed. Or you could almost view it as they are the same thing, just from different frames of reference. I don't want to confuse you right now. But anyway, back to what I'll call the basic physics. So if I had to find a magnetic field as B-- so B is a vector and it's a magnetic field-- we know that the force on a moving charge could be an electron, a proton, or some other type of moving charged particle. And actually, this is the basis of how they-- you know, when you have supercolliders-- how they get the particles to go in circles, and how they studied them by based on how they get deflected by the magnetic field. But anyway, the force on a charge is equal to the magnitude of the charge-- of course, this could be positive or negative-- times, and this is where it gets interesting, the velocity of the charge cross the magnetic field. So you take the velocity of the charge, you could either multiply it by the scalar first, or you could take the cross product then multiply it by the scalar. this isn't a vector. But you essentially take the cross product of the velocity and the magnetic field, multiply that times the charge, and then you get the force vector on that particle. Now there's something that should immediately-- if you hopefully got a little bit of intuition about what the cross product was-- there's something interesting going on here. The cross product cares about the vectors that are perpendicular to each other. So for example, if the velocity is exactly perpendicular to the magnetic field, then we'll actually get a number. If they're parallel, then the magnetic field has no impact on the charge. That's one interesting thing. And then the other interesting thing is when you take the cross product of two vectors, the result is perpendicular to both of these vectors. So that's interesting. A magnetic field, in order to have an effect on a charge, has to be perpendicular to its you velocity. And then the force on it is going to be perpendicular to", + "qid": "NnlAI4ZiUrQ_296" + }, + { + "Q": "i dont get what a magnetic mono-pole is ( 0:42 - 0:46 ) ?", + "A": "Thanx Matt :)", + "video_name": "NnlAI4ZiUrQ", + "timestamps": [ + 42, + 46 + ], + "3min_transcript": "We know a little bit about magnets now. Let's see if we can study it further and learn a little bit about magnetic field and actually the effects that they have on moving charges. And that's actually really how we define magnetic field. So first of all, with any field it's good to have a way to visualize it. With the electrostatic fields we drew field lines. So let's try to do the same thing with magnetic fields. Let's say this is my bar magnet. This is the north pole and this is the south pole. Now the convention, when we're drawing magnetic field lines, is to always start at the north pole and go towards the south pole. And you can almost view it as the path that a magnetic north monopole would take. So if it starts here-- if a magnetic north monopole, even though as far as we know they don't exist in nature, although they theoretically could, but let's just say for the sake of argument that we do have a magnetic north monopole. If it started out here, it would want to run away from this north pole and would try to get to the south pole. something like this. If it started here, maybe its path would look something like this. Or if it started here, maybe its path would look something like this. I think you get the point. Another way to visualize it is instead of thinking about a magnetic north monopole and the path it would take, you could think of, well, what if I had a little compass here? Let me draw it in a different color. Let's say I put the compass here. That's not where I want to do it. Let's say I do it here. The compass pointer will actually be tangent to the field line. So the pointer could look something like this at this point. It would look something like this. And this would be the north pole of the pointer and this would be the south pole of the pointer. Or you could-- that's how north and south were defined. People had compasses, they said, oh, this is the north seeking pole, and it points in that direction. of the larger magnet. And that's where we got into that big confusing discussion of that the magnetic geographic north pole that we're used to is actually the south pole of the magnet that we call Earth. And you could view the last video on Introduction to Magnetism to get confused about that. But I think you see what I'm saying. North always seeks south the same way that positive seeks negative, and vice versa. And north runs away from north. And really the main conceptual difference-- although they are kind of very different properties-- although we will see later they actually end up being the same thing, that we have something called an electromagnetic force, once we start learning about Maxwell's equations and relativity and all that. But we don't have to worry about that right now. But in classical electricity and magnetism, they're kind of a different force. And the main difference-- although you know, these field lines, you can kind of view them as being similar-- is that magnetic forces always come in dipoles, soon. while you could have electrostatic forces that are monopoles.", + "qid": "NnlAI4ZiUrQ_42_46" + }, + { + "Q": "At 0:43 where was the particle before expansion? What were its surroundings?", + "A": "there re no surroundings. The universe is only expanding on the inside because there is no outside. If there were an outside of the universe, it s the same size it s always been.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 43 + ], + "3min_transcript": "Right now, the prevailing theory of how the universe came about is commonly called the Big Bang theory. And really is just this idea that the universe started as kind of this infinitely small point, this infinitely small singularity. And then it just had a big bang or it just expanded from that state to the universe that we know right now. And when I first imagined this-- and I think if it's also a byproduct of how it's named-- Big Bang, you kind of imagine this type of explosion, that everything was infinitely packed in together and then it exploded. And then it exploded outward. And then as all of the matter exploded outward, it started to condense. And then you have these little galaxies and super clusters of galaxies. And they started to condense. And then within them, planets condensed and stars condensed. And then we have the type of universe that we have right now. has a couple of problems. One is when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding. But you're like, whoa, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, And two, there is no outside space. We are not expanding into another space. And I'm going to explain that. Hopefully, we'll see why that is the case right now. So the best way to view it-- and we're going to view it by analogy. If I were tell you that I have a two-dimensional space that has a finite area, so it has a finite area-- so it's not infinite. And it also has no edge. This once again, when you first look at it, seems difficult. How do I just construct something that has a finite area, but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved, what happens? And I think the easiest example of that", + "qid": "eUF59jCFcyQ_43" + }, + { + "Q": "At 10:30 , if three points are in the universe , and after some significant amount of time , the universe expands and the three points get separated farther apart. So , is the EARTH getting any farther apart from the SUN and the MOON getting farther apart from EARTH??", + "A": "No, gravitational forces overpower the force of expansion at closer ranges.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 630 + ], + "3min_transcript": "means that if you go up and you just keep going up, you'll eventually come back from the bottom. So if you keep going all the way up, you'll eventually come back to the point that you were. It might be an unbelievably large distance, but you'll eventually get back where you were. If you go to the right, you'll eventually come back all the way around to the point where you were. And if you were to go into the page-- so if you were to go into the page-- let me draw it that way-- if you go into the page, you would eventually come back from above the page and come back to the point that you are. So that's what this implication would be. That you would eventually get back to where you are. So let's go back to the question of an expanding universe, a expanding universe that's not expanding into any other space. That is all of the space, but it's still expanding. Well, this is the model. So you could imagine shortly after the Big Bang, our four-dimensional sphere looked like this. Maybe right at the Big Bang, it was like this little unbelievably small sphere. Then a little bit later, it's this larger sphere. Let me just shade it in to show you that it's kind of popping out of the page, that's it's a sphere. And then at a later time, the sphere might look like this. The sphere might look like this. Now, your temptation might be to say, wait, Sal, isn't this stuff outside of this sphere, isn't that some type of a space that it's expanding into? Isn't that somehow part of the universe? And I would say if you're talking in three dimensions, no, it's not. The entire universe is this surface. It is this surface of this four-dimensional sphere. If you start talking about more dimensions, then, yes, you could talk about maybe things outside of our three-dimensional universe. So as this expands in space/time-- so one way to view the fourth dimension getting further and further apart. And I'll talk about more evidence in future videos for why the Big Bang is the best theory we have out there right now. But as you could imagine, if you have two points on this sphere that are that far apart, as this sphere expands, this four-dimensional sphere, as this bubble blows up or this balloon blows up, those two points are just-- let me draw three points. Let's say those are three points. Those three points are just going to get further and further apart. And that's actually one of the main points that-- or one of the first reasons why it made sense to believe the Big Bang-- is that everything is expanding, not from some central point. But everything is expanding from everything. That if you go in any direction from any point in the universe, everything else is expanding away. And the further away you go, it looks like the faster it's expanding away from you. So I'll leave you there, something for you to kind of think about a little bit. And then we'll build on some of this to think about what it means to kind of observe the observable universe.", + "qid": "eUF59jCFcyQ_630" + }, + { + "Q": "At 10:14 Sal talks about the sphere getting bigger and the points are getting further away from each other. So, if the universe is a sphere and it is continually expanding, Will we (earth) ever possibly get further away from the sun? Will the sun engulf us all before that happens?", + "A": "Earth is held near the sun by gravity, which keeps them together even as the space they are in stretches.", + "video_name": "eUF59jCFcyQ", + "timestamps": [ + 614 + ], + "3min_transcript": "means that if you go up and you just keep going up, you'll eventually come back from the bottom. So if you keep going all the way up, you'll eventually come back to the point that you were. It might be an unbelievably large distance, but you'll eventually get back where you were. If you go to the right, you'll eventually come back all the way around to the point where you were. And if you were to go into the page-- so if you were to go into the page-- let me draw it that way-- if you go into the page, you would eventually come back from above the page and come back to the point that you are. So that's what this implication would be. That you would eventually get back to where you are. So let's go back to the question of an expanding universe, a expanding universe that's not expanding into any other space. That is all of the space, but it's still expanding. Well, this is the model. So you could imagine shortly after the Big Bang, our four-dimensional sphere looked like this. Maybe right at the Big Bang, it was like this little unbelievably small sphere. Then a little bit later, it's this larger sphere. Let me just shade it in to show you that it's kind of popping out of the page, that's it's a sphere. And then at a later time, the sphere might look like this. The sphere might look like this. Now, your temptation might be to say, wait, Sal, isn't this stuff outside of this sphere, isn't that some type of a space that it's expanding into? Isn't that somehow part of the universe? And I would say if you're talking in three dimensions, no, it's not. The entire universe is this surface. It is this surface of this four-dimensional sphere. If you start talking about more dimensions, then, yes, you could talk about maybe things outside of our three-dimensional universe. So as this expands in space/time-- so one way to view the fourth dimension getting further and further apart. And I'll talk about more evidence in future videos for why the Big Bang is the best theory we have out there right now. But as you could imagine, if you have two points on this sphere that are that far apart, as this sphere expands, this four-dimensional sphere, as this bubble blows up or this balloon blows up, those two points are just-- let me draw three points. Let's say those are three points. Those three points are just going to get further and further apart. And that's actually one of the main points that-- or one of the first reasons why it made sense to believe the Big Bang-- is that everything is expanding, not from some central point. But everything is expanding from everything. That if you go in any direction from any point in the universe, everything else is expanding away. And the further away you go, it looks like the faster it's expanding away from you. So I'll leave you there, something for you to kind of think about a little bit. And then we'll build on some of this to think about what it means to kind of observe the observable universe.", + "qid": "eUF59jCFcyQ_614" + }, + { + "Q": "at 3:57, I thought that each element had a set number of electrons. So is that a hypothetical question orrr...", + "A": "In a NEUTRAL atom the number of protons is equal to the number of electrons. But this does not always have to be the case. Atoms can and do gain or lose electrons. This is the whole point of the video. An ion is an atom that does not have the same number of electrons and protons, so it has a charge.", + "video_name": "zTUnjPALX_U", + "timestamps": [ + 237 + ], + "3min_transcript": "what element you're dealing with, so now if you look at what element has five protons we're dealing with boron. So this is going to be boron. Neutral boron would have five protons and five electrons. But this one has one extra electron, so it has one extra negative charge. So you can write it like this, one minus. Or you could just say it has a negative charge. So this is a boron ion right over here. As soon as you have an imbalance between protons and electrons you no longer would call it an atom, you would call it an actual ion. Now let's do an example question dealing with this. So our question tells us... Our question ... our question tells us ... So let's just look up platinum on our periodic table. Platinum is sitting right over here if you can see it. So an atom of platinum has a mass number of 195. And 195 looks pretty close to that atomic mass we have there. And it contains 74 electrons. 74 electrons. How many protons and neutrons does it contain and what is its charge? Alright, so let's think about this a little bit. So we're dealing with platinum. So by definition platinum has 78 protons, so we know that. It has 78 protons. They're telling us it has 74 electrons. 74 electrons. protons than electrons. So you're going to have a positive four charge. Four more of the positive thing than you have of the negative things. So you could write this as platinum with a plus four charge. This is a platinum ion, a positive platinum ion. The general term when we're talking about a positive ion, we're talking about a cation. That is a positive ion. Up there when we talked about boron being negative, a negative ion, that is an anion. This is just to get ourselves used to some of the terminology. But we're not done answering the question. They say an atom of platinum has a mass number of 195", + "qid": "zTUnjPALX_U_237" + }, + { + "Q": "Hey Sal! at 5:03, you mentioned that there's a 'propyl' functional group on Carbon-3...just to clarify, it's an 'ethyl' group, not propyl (C2H5) :)", + "A": "people with exceptional talent are prone to commit minor mistakes more often - my maths tution teacher", + "video_name": "GFiizJ-jGVw", + "timestamps": [ + 303 + ], + "3min_transcript": "So this molecule is zusammen. Which on some levels, you can think of as the same thing as cis, but cis and trans stops applying when you start having more than two functional groups. In this case, we have three. So we would call this Z-4-methylhept-3-ene. And that's because the higher priority functional groups are on the same side of the double bond. Now let's do this one over here. And someone pointed out, rightly, that I had misspelled zusammen in the last video. It's actually spelled like this, zusammen. I had spelled it with two s's and one m. I guess you can forgive me. I don't speak German. But anyway, I thought I would point that out. Now let's try to label this thing right over here. So the first thing, this once again is an alkene. Let's identify the longest carbon chain here. So it looks like one, two, three, four, five, six, seven, eight carbons. Double bonds are closer to the left hand side. One, two, three, four, five, six, seven, eight. So just the main chain is oct-- let me make sure I have some space here-- it is oct-3-ene. And then we have, well we have one functional group sitting off of the main chain. We have this bromine sitting right over there on the third carbon. So we would call this 3-bromooct-3-ene. And now we have to figure out is it entgegen or zusammen. So if we look on the carbon on the right hand side, it's pretty obvious that this is the only functional group. We just have a hydrogen there. So let me circle it in the magenta. And then on the left hand side we have two functional groups. We have this [UNINTELLIGIBLE] bromo or we have a bromine sitting right there. And then we have this propyl group. tempted to say it takes higher priority. But remember, in the Cahn-Ingold-Prelog system, you give higher priority to the atom that has a higher atomic number. Bromine has an atomic number of 35. Carbon has an atomic number of only 6. So Bromine is actually higher priority. So this is the higher priority functional group right over here. So now for deciding whether it's entgegen or zusammen, we see that our higher priority groups are apart. They're on opposite sides of the double bond. This one is on top. This one is below. We are apart. So this is entgegen. Or we would write this is E-3-bromooct-3-ene. And E is for-- just as a bit of a refresher-- it's for entgegen, a word that I enjoy saying, entgegen.", + "qid": "GFiizJ-jGVw_303" + }, + { + "Q": "At 6:17, wouldn't the lack of Hydrogen in the middle Carbon (Carbocation) make it negative instead of positive?", + "A": "No. The middle carbon could be positive, negative, or a radical. It all depends on whether the carbon has lost an H\u00e2\u0081\u00ba, H\u00e2\u0081\u00bb, or H.", + "video_name": "Z4F88tTx9-8", + "timestamps": [ + 377 + ], + "3min_transcript": "Next let's look at acetone. So oxygen is more electronegative than carbon so oxygen is going to withdraw some electron density away from this carbon here and this carbon would be partially positive, so this carbon is the electrophilic portion of this compound. Next let's look at a carbocation where there's a full positive charge on this carbon so this carbon has only three bonds to it which gives it a full positive charge. Obviously a full positive charge is going to love electrons. Opposite charges attract, so this carbon is the electrophilic portion of this ion. And finally let's look at this compound, right. We know that oxygen is more electronegative than carbon so oxygen withdraws some electron density away from structure here, so let me take these pi electrons and move them out onto the oxygen, so let's draw a resonance structure so I put in my double bond. Now if I'm showing those pi electrons moving off onto the oxygen I would need three lone pairs of electrons on that top oxygen giving it a negative one formal charge. I took a bond away from this carbon in magenta which is this carbon which gives it a plus one formal charge, so that's one of the possible resonance structures that you can draw and of course we know the carbon in magenta is an electrophilic center, but I could draw another resonance structure so let me go ahead and do that, put in my brackets over here. I could take these pi electrons, I'll show it on this one actually, these pi electrons and move them over to here, so let's draw the resulting resonance structure. So I'd have a double bond here now so let me put that in here, draw in the hydrogen, put in my brackets, and I removed a bond, we took a bond away from, let me use blue for this, from this carbon, so this carbon now has a plus one formal charge, so the carbon in blue is this carbon over here, so let me draw in a plus one formal charge, so that is also electrophilic, right, a full positive charge is going to be attracted to a negative charge, so this compound actually has, this compound actually has two electrophilic centers, so this carbon here and also this carbon.", + "qid": "Z4F88tTx9-8_377" + }, + { + "Q": "Where does the term \"spectrophotometry\"come from? 0:04", + "A": "Spectro- meaning to look photo- meaning light -metry- meaning measurement", + "video_name": "qbCZbP6_j48", + "timestamps": [ + 4 + ], + "3min_transcript": "What I want to do in this video is to talk a little bit about spectrophotometry. Spectrophotometry sounds fairly sophisticated, but it's really based on a fairly simple principle. So let's say we have two solutions that contain some type of solute. So that is solution one, and then this is solution two. And let's just assume that our beakers have the same width. Now let's say solution 1-- let me put it right here, number 1, and number 2. Now let's say that solution 1 has less of the solute in it. So that's the water line right there. So this guy has less of it. And let's say it's yellow or to our eyes it looks yellow. So this has less of it. Actually, let me do it this way. Let me shade it in like this. So it has less of it. And let's say solution number 2 has more of the solute. So it's more. So I'll just kind of represent that as more So the concentration of the solute is higher here. So let me write higher concentration. And let's say this is a lower concentration. Now let's think about what will happen if we shine some light through each of these beakers. And let's just assume that we are shining at a wavelength of light that is specifically sensitive to the solute that we have dissolved in here. I'll just leave that pretty general right now. So let's say I have some light here of some intensity. So let's just call that the incident intensity. I'll say that's I0. So it's some intensity. What's going to happen as the light exits the other side of this beaker right here? Well, some of it is going to be at absorbed. Some of this light, at certain frequencies, is going to be And so you're actually going to have less light coming out from the other side. Especially less of those specific frequencies that these molecules in here like to absorb. So your're going to have less light come out the other side. I'll call this I1. Now in this situation, if we shine the same amount of light-- so I0-- that's supposed to be an arrow there, but my arrow is kind of degrading. If we shined the same amount of light into this beaker-- so it's the same number, that and that is the same-- the same intensity of light, what's going to happen? Well more of those specific frequencies of light are going to be absorbed as the light travels through this beaker. It's just going to bump into more molecules because it's a higher concentration here.", + "qid": "qbCZbP6_j48_4" + }, + { + "Q": "At 3:12, couldn't you just round the numbers like 1.0079 to just 1.01?", + "A": "No, that is not being done correctly. Unlike what Sal does in many of his videos, you cannot round to whatever amount you find convenient. You MUST respect the number of significant digits you have. Almost any chemistry will mark a problem wrong if you do not round in such a way that your number of significant digits requires. You cannot round less than that, you cannot round more than that.", + "video_name": "UPoXG1Z3sI8", + "timestamps": [ + 192 + ], + "3min_transcript": "this is going to be 12.011 atomic mass units, and the contribution from hydrogen, the contribution from hydrogen, let me do that in yellow... The contribution from hydrogen, from the six hydrogens, is going to be six times the atomic weight of hydrogen, which is 1.0079, by this periodic table that I have right here, once again, the weighted average of all the isotopes, et cetera, et cetera, 1.0079. 1.0079, and so, the molecular weight of this whole thing, a typical benzene molecule, if you were to take the weighted average of all of its molecular masses, based on the prevalence of the different isotopes on Earth, well, you would just say it's just going to be the sum of these two things. It is going to be six times 12.011 is that going to give us? Well, let's see, let's get a calculator out, so let me clear that, so 12.011 times six gives us 72.066, so this right over here. is 72.066, and actually, I probably could have done that in my head, well anyway, let's look at what we have from the hydrogen, so the hydrogen are going to be six times 1.0079 gets us to 6.0474, plus 6.0474, but since I only go to the thousandth place in terms of precision here, if I care about significant figures, significant digits, then I'm only going to go three spaces to the right of the decimal, but let's add these two together, so I'm going to get 6.0474, plus 72.066 equals, and I'm just going to go three to the right, so 78.113, so this is equal to, 78.113 atomic mass units, that's the molecular mass of a molecule of benzene, now what percentage is from the carbon? Well, it's going to be equal to the 72.066 over the 78.113 which is equal to, all right, so let me just clear this.", + "qid": "UPoXG1Z3sI8_192" + }, + { + "Q": "Based on the pneumonia videos, pneumonia is the infection of the flu virus in the lungs in the situation described at 10:04, and pneumonitis (inflammation of alveolar walls) is secondary to this infection. At 10:04 when he describes inflammation of the alveolar walls, is that an instance where pneumonitis and pneumonia are interchangeable? Or, should inflammation of alveolar walls always be called pneumonia if it is caused by pneumonia?", + "A": "The disease entity is called pneumonia clinically and usually there is further description such as bacterial pneumonia, viral pneumonia, fungal pneumonia, etc. At the microscopic, pathologic level, the changes seen are called pneumonitis. When there is no infection, sometimes the disease is described based upon the pathologic changes hence things like aspiration pneumonitis.", + "video_name": "6vy5CX6vK0I", + "timestamps": [ + 604, + 604 + ], + "3min_transcript": "These are high risk individuals. So why do we care so much about these high risk individuals? Well, it's because they develop complications of flu. And this is what it really boils down to. You remember we initially talked about all the hundreds of thousands of people in the US and around the world that get hospitalized for the flu. And then the numbers of people that die from the flu. Well, overwhelmingly it's people in this group. This high risk group. And the things that they get, the kind of complications they get, are many. Actually, flu leads to many different types of complications. And I'm going to draw out just a few of them for you. I don't want to give you an exhaustive list. But I want you to at least get an appreciation for the kinds of things we're talking about. So, for example, let's say these are your lungs. I'm drawing two branches of your lungs. And this is going to your left lung So this is your trachea splitting up. And you know that the flu, the influenza, is going to affect the cells in your respiratory tree. So it's going to affect these cells and it's going to cause inflammation. You're going to get a big immune response. And if that response is really big, let's say you have a big response, and if it's around these airways here, these bronchioles-- let me actually extend this out a little bit, so you can at least appreciate where the arrow's going. If the response is really strong in the bronchioles, we call that bronchitis. So someone might actually develop bronchitis as a result of getting the flu. Now someone else might actually have a big inflammatory reaction in these little air sacs. Your lungs end in thousands and thousands of little air sacs. And if that happens, then you might call that pneumonia. You might say, well, this person has pneumonia. Actually, lots and lots of it. Smooth muscle that wraps around the bronchioles. And sometimes with the flu you actually can trigger twitichiness of that smooth muscle. It starts spasming. And when that happens we know that sometimes as an asthma attack. So you can actually get an asthma attack related to the flu. So all sorts of things like this can happen. And it's awful. These are things that can actually land you in the hospital. Or can cause death, as well. So these are the kinds of complications. And there are other ones. Things like ear infections and sinus infections and many, many other things as well. But here I just wanted to show you a few of the complications that people get. And show you and remind you that it's usually the high risk people that you have to worry about.", + "qid": "6vy5CX6vK0I_604_604" + }, + { + "Q": "At 5:15, why would you not want to take one of these devices apart? What could be in items like this that you would need to be afraid of?", + "A": "Underneath the bottom cover you find the electric heater and water hoses. This is a dangerous combination (water and electricity). The manufacturer wants to keep you away from the chance of messing up the connections and putting a defective coffeemaker back in service. In this demo video, the coffeemaker takes a one-way trip to full disassembly. It will never make another cup.", + "video_name": "XQTIKNXDAao", + "timestamps": [ + 315 + ], + "3min_transcript": "It would make it easy to pull the mold out this way. They probably also had, it was probably a three part mold and there was a section that also came out in this direction. Then this is just another injection molded part that snaps onto this one, as we've seen. This is the part that holds the handle on. Very important part. I think they definitely paid the extra money for a stainless piece there because it's really important that that doesn't come loose and it probably gets fairly wet, so if it was made out of regular steel or another material it might rust and could potentially come apart. We wouldn't want hot coffee on us, now would we. Alright, so that's the coffee kettle. So, inside, here's our coffee maker. We know that hot water... We've got a container here and in this container, is a space where we put our coffee filter and then we put our coffee grounds and we fill this with water and then we close the top and we turn it on and we wait. What happens is that water that we pour in drains down a little hole on the inside there, you can see it right there. Let me point to it with the screw driver. It drains down that hole and it goes down into this underside, so we'll take a look at the underside and see what happens down there. Okay, so, I've modified a screw driver. This was a low-cost screw driver. It was a 99 cent one, so I modified the end of it so I could take out these safety screws. Don't do this at home unless you have a professional with you because this is not meant to be taken apart. That's why they use these special screw heads, so you won't take it apart. There we go. Again, this is an injection molded part. This is a co-molded part, it looks like. Which means that there were two different materials molded together. Let's see if I can knock that screw out. Okay, it wants to stay, that's fine. This material here is, these feet are made out of a softer material and this is a polypropylene material. So it's a plastic, a low-cost plastic. So the mold comes together and they injection mold this material and then once this material has begun to harden, they injection mold the softer material, so it's co-molded or it's a dual molded part. You can see other parts are done like this, like sometimes you'll see toothbrushes that have soft saniprene and then the hard toothbrush and they're molded in one mold. It's a dual shot mold. In any case, so that's the bottom.", + "qid": "XQTIKNXDAao_315" + }, + { + "Q": "At about 3:00 Sal talks about plaque and how it can damage your heart. So, if you do get plaque in your blood vessel, how do you get rid of it?", + "A": "Presumably exercise, a low fat, healthy diet and stopping smoking if one is doing so already.", + "video_name": "vYnreB1duro", + "timestamps": [ + 180 + ], + "3min_transcript": "so this was my ....this is what I thought people were talking about when they were saying clogging of the arteries and maybe when they got clogged enough, the stopped blood flow to the rest of the body somehow and that would actually kill the person. I want to make it very clear right now. Those are not the arteries that people are talking about getting clogged, when people talk about heart disease or heart attacks. The arteries that they are talking about are the arteries that actually provide blood to the heart. Remember the heart itself is a muscle. It itself needs oxygen. So you have these arteries right over here, the red tubes. Those are arteries. and then the blue ones are veins. They're taking the de-oxygenated blood away from the tissue of the heart. And these are called coronary arteries. And this one over here at least from the point of view of me or you looks like it's on the right. This right over here is called the left coronary artery or LCA. And this right over here in red is called the right coronary arteries or the RCA. And so when people talk about arteries getting blocked or getting clogged, they're talking about the coronary arteries. They're talking about the things that supply blood to the heart. So let's zoom in on one of them....Maybe we can zoom in right over here, that part of the artery. That's the tube....clear where I am zooming in. I am zooming in right over here. So over time, I am not going into the details how this happened. It is subject for another video. You can have these plaques build up along the walls of the artery. So over time if a person doesn't have the right diet, or maybe they just have a predisposition to it, And the plaques, the material inside of them are lipids, so things like fat, cholesterol and also dead white blood cells, which is this kind of messy substance right over here. This is what we call a plaque. And the formation of these plaques that obstruct the actual blood vessel, that actually obstruct the artery. We call it.....make it clear you see that. This is kind of tube over here. Let me draw the blood So this formation of these plaques we call atherosclerosis. So you can imagine if you have these things build up,", + "qid": "vYnreB1duro_180" + }, + { + "Q": "Around 2:00, for the equation for the very first question, why is molarity used instead of the number of moles present?", + "A": "HCl will dissociate completely and form 0.500 moles of H3O+. Molarity is the number of moles present, i.e. the concentration.", + "video_name": "JoGQYSTlOKo", + "timestamps": [ + 120 + ], + "3min_transcript": "- [Voiceover] Let's say we're doing a titration and we start with 20 mL of .500 molar HCl. So we're starting with a strong acid, and to the strong acid, we're going to add a solution of a strong base. We're going to add a .500 molar solution of NaOH, and as we add the base, the pH is going to increase, and we can show this on our titration curve. So we put the pH on the y-axis, and on the x-axis we put the volume of base that we are adding. So in part A, our goal is to find the pH before we've added any of our sodium hydroxide, so the addition of 0.0 mL of our base. So if we haven't added any base, the only thing we have present is acid. We know that HCl is a strong acid, so it is going to ionize 100%. So if HCl donates a proton to H2O, then we get H3O plus, and we would get the conjugate base to HCl, which is Cl minus. let's go ahead and write that, .500 molar HCl, and since we know this is 100% ionization because we have a strong acid, that's the same concentration of hydronium ions that we'll have in solution, so we have .500 molar for the concentration of hydronium ions. Now it's easy to find the pH because we know that the pH is equal to the negative log of the concentration of hydronium ions. So let's go ahead and do the calculation. We take the negative log of the concentration of hydronium which is .500, so that gives us .301. So we plug this into here, and we get a pH equal to 0.301. So we can find that point on our titration curve. We've added 0.0 mL of base, and our pH looks like it's just above zero here on our titration curve, and we calculated it to be .301. Let's find another point on our titration curve, this time after we add 10 mL of our base. So we find 10 mL of our base right here, and we're trying to find this point. So what is the pH after we add 10 mL? Well it looks like it's pretty close to one. Let's see if we can calculate what the pH is. So if we're adding base, we know that the base that we're adding, the hydroxide ions that we're adding, are going to neutralize the hydronium ions that are already present. So first, let's calculate how many moles of hydronium ions that we had present here. So the concentration of hydronium ions is .500.", + "qid": "JoGQYSTlOKo_120" + }, + { + "Q": "At 1:20, shouldn't it be hydroxonium not hydronium?", + "A": "Hydroxonium and hydronium mean the same thing and both terms are in use.", + "video_name": "JoGQYSTlOKo", + "timestamps": [ + 80 + ], + "3min_transcript": "- [Voiceover] Let's say we're doing a titration and we start with 20 mL of .500 molar HCl. So we're starting with a strong acid, and to the strong acid, we're going to add a solution of a strong base. We're going to add a .500 molar solution of NaOH, and as we add the base, the pH is going to increase, and we can show this on our titration curve. So we put the pH on the y-axis, and on the x-axis we put the volume of base that we are adding. So in part A, our goal is to find the pH before we've added any of our sodium hydroxide, so the addition of 0.0 mL of our base. So if we haven't added any base, the only thing we have present is acid. We know that HCl is a strong acid, so it is going to ionize 100%. So if HCl donates a proton to H2O, then we get H3O plus, and we would get the conjugate base to HCl, which is Cl minus. let's go ahead and write that, .500 molar HCl, and since we know this is 100% ionization because we have a strong acid, that's the same concentration of hydronium ions that we'll have in solution, so we have .500 molar for the concentration of hydronium ions. Now it's easy to find the pH because we know that the pH is equal to the negative log of the concentration of hydronium ions. So let's go ahead and do the calculation. We take the negative log of the concentration of hydronium which is .500, so that gives us .301. So we plug this into here, and we get a pH equal to 0.301. So we can find that point on our titration curve. We've added 0.0 mL of base, and our pH looks like it's just above zero here on our titration curve, and we calculated it to be .301. Let's find another point on our titration curve, this time after we add 10 mL of our base. So we find 10 mL of our base right here, and we're trying to find this point. So what is the pH after we add 10 mL? Well it looks like it's pretty close to one. Let's see if we can calculate what the pH is. So if we're adding base, we know that the base that we're adding, the hydroxide ions that we're adding, are going to neutralize the hydronium ions that are already present. So first, let's calculate how many moles of hydronium ions that we had present here. So the concentration of hydronium ions is .500.", + "qid": "JoGQYSTlOKo_80" + }, + { + "Q": "Why do seismic waves travel faster through denser material (4:45) ?", + "A": "You can think that in a dense material the molecules are very closer together and when a wave hits one molecule it takes less time for it to reach the next one and the wave then travles faster. In air the molecules are far apart and the wave than travles slower.", + "video_name": "yAQSucmHrAk", + "timestamps": [ + 285 + ], + "3min_transcript": "All of a sudden, the waves were reaching there faster. The slope of this line changed. It took less time for each incremental distance. So for some reason, the waves that we're going at these farther stations, the stations that were more than 200 kilometers away, somehow they were accelerated. Somehow they were able to move faster. And he's the one that realized that this was because the waves that were getting to these further stations must have traveled through a more dense layer of the earth. So let's just think about it. So if we have a more dense layer, it will fit this information right over here. So if we have a layer like this, which we now know to be the crust, and then you have a denser layer, which we now know to be the mantle, then what you would have is-- so you have your earthquake right over here, closer by, while you're still within the crust, it would be proportional. And then let's say that this is exactly, this right here is 200 kilometers away. But then if you go any further, the waves would have to travel. They would travel, so they would go like this. And then they would get refracted even harder. So they would get refracted. So they would be a little bit curved ahead of time. But then they're going to a much denser material. Or it's not gradually dense, it's actually kind of a all of a sudden a considerably more dense material, so it will get refracted even more. And then it'll go over here. And since it was able to travel all of this distance in a denser material, it would have traveled faster along this path. And so it would get to this distance on the surface that's more than 200 kilometers away, it would get there faster. And so he said that there must be a denser layer that those waves are traveling through, which we now know to be the mantle. And the boundary between what we now know to be the crust and this denser layer, It's called the Mohorovicic discontinuity. And sometimes this is called the Moho for short. So that boundary between the crust and the mantle is now named for him. But this was a huge discovery, because not only was he able to tell us, based on the data-- based on, kind of, indirect data, just based on earthquakes happening, and measuring when the earthquakes reach different points of the earth-- that there probably is a denser layer. And if you do the math, under continental crust that denser layer is about 35 kilometers down. He was able to tell us that there is that layer. But even more importantly, he was able to give the clue that just using information from earthquakes, we could essentially figure out the actual composition of the earth. Because no one has ever dug that deep. No one has ever dug into the mantle, much less the outer core or the inner core. In the next few videos, we're going to kind of take this insight, that we can use information from earthquakes, to actually think about how we know that there is an outer liquid core", + "qid": "yAQSucmHrAk_285" + }, + { + "Q": "At 4:27, Jay numbers one of the carbons as Beta Carbon 3. I don't understand why Carbon 3 is a Beta-carbon. It's not connected to the alpha carbon.", + "A": "The \u00ce\u00b1-carbon is the carbon bearing the leaving group (C-2). So the \u00ce\u00b2-carbons are the ones next to it (C-1 and C-3).", + "video_name": "uCW6154hPkc", + "timestamps": [ + 267 + ], + "3min_transcript": "The chlorine has to be up axial, and so if I go around to carbon six, so this would be carbon one here, two, three, four, five, and six, I have a methyl group going away from me in space, so this would be going down, so I'm gonna draw in a Me for a methyl group right here, so it's down axial. So now let's draw in some hydrogens on our beta carbons, so let me highlight our beta carbons here. I'll use red, so this would be, what I've marked is being beta one, so I have two hydrogens on that carbon, so I'll draw those in here, and then my other beta carbon which I called beta two up here, so I only have, so this is beta two, I have only one hydrogen, and it is equatorial. So let's go to a video, so we can analyze which one of these protons will participate in our E two mechanism. Here's our chair conformation, we have the yellow chlorine up and axial. When we go to the beta one carbon, the hydrogen in green is the only one that's anti-periplanar to the halogen. If I turn to the side here, it's easier to see that we have all four of those atoms in the same plane, so the green proton is anti-periplanar to the chlorine. The other hydrogen, the one in white, is not anti-periplanar, so it will not participate in our E two mechanism. We go to the beta two carbon, and this hydrogen in white is not anti-periplanar, and when we look at the down axial position, it's occupied by a methyl group, so that is where a hydrogen would need to be if it were to participate in an E two mechanism. For E two elimination in cyclohexanes, the halogen must be axial, so here is our halogen that's axial, and when the halogen is axial in this chair conformation, is this one in green as we saw in the video, so if a strong base comes along, and takes the proton in green, the electrons in here would move in to form our double bond, and these electrons come off onto the chlorine, so a double bond forms between the alpha and the beta one carbons, which would give us this as our only product, so we don't get a double bond forming between our alpha and our beta two carbon because we would need to have a hydrogen where our methyl group is, so this, if we did have a hydrogen here, this hydrogen would be anti-periplanar to our halogen, but in this case, we get only one product, so this is the only product observed, and we figured that out because we drew our chair conformation. Let's do another E two mechanism for a substituted cyclohexane, and I'll start by numbering my cyclohexane, so that's carbon one, and this is carbon two, this is carbon three, and this is carbon four.", + "qid": "uCW6154hPkc_267" + }, + { + "Q": "at 9:52, why does he say that because the iso-propyl group is axial can't participate in the mechanism?", + "A": "In order to get elimination of HCl, the Cl onC2 and the \u00ce\u00b2 H must be in a trans diaxial conformation. If the Cl is axial, the isopropyl group on C1 is also axial and the \u00ce\u00b2 H on C1 is equatorial. There is no axial H on C1, because the isopropyl group has replaced it. The axial \u00ce\u00b2 H must therefore come from C3.", + "video_name": "uCW6154hPkc", + "timestamps": [ + 592 + ], + "3min_transcript": "and next to that would be a beta carbon, and this beta hydrogen is anti-periplanar to this halogen. We have another beta carbon over here with another hydrogen that's anti-periplanar to this halogen. Finally, let's draw our two products, so let's take a proton from the beta one position first, so our base, let me draw it in here, so our strong base is going to take this proton, and so these electrons would move into here to form our double bond, and these electrons come onto the chlorine to form the chloride anion as our leaving group. So this would form a double bond between what I called carbons two and three 'cause this is carbon one, this is carbon two, this is carbon three, and this is carbon four, so we have a methyl group that's up at carbon one, so let me draw in our methyl group up at carbon one, so we put that on a wedge, so if that's carbon one, then this is carbon two, and this is carbon three, and that's where our double bond forms, so the double bond forms between carbons two and three, so let me go ahead and put that in, going away from us, we put that on a dash, so that's one product. If our base took this proton, then the electrons would move into here, and these electrons would come off onto the chlorine, so if we took a proton from the beta two carbon, we would form a double bond between carbons three and four, so here's carbons three and four, so I put a double bond in there. I still have a methyl group that's going up at what I called carbon one, and my isopropyl group is at carbon four, but since now, this carbon is sp two hybridized, I need to draw in this isopropyl group on a straight line, so sometimes, students would put this isopropyl group in on a wedge or a dash, but you're trying to show the planar geometry around this carbon, so a straight line is what you need, and so those are the two products. Let's do one more, and you can see this substrate in the previous example. The only difference is this time, the chlorine is on a wedge instead of a dash, so if I number my ring one, two, three, four, I've already put in both chair conformations to save time, so that's carbon one, this is carbon two, this is carbon three, and this is carbon four. On the other chair conformation, this is carbon one, two, three, and four, and notice for the chair conformation on the left, we have the chlorine in the axial position, so this would be the alpha carbon, and the carbons next to the alpha carbon would be the beta carbons, so this one on the right is a beta carbon, and the one on the left is a beta carbon. We need a proton that's anti-periplanar, and the only one that fits would be this hydrogen right here, so if a base takes that proton, so let me draw in our strong base, taking this proton, these electrons would move into here to form our double bond, the electrons come off onto our chlorine, and a double bond forms between carbons two and three,", + "qid": "uCW6154hPkc_592" + }, + { + "Q": "At 1:52 someone mentioned protozoans. What are protozoans", + "A": "In some systems of biological classification, the Protozoa are a diverse group of unicellular eukaryotic organisms. Historically, protozoa were defined as single-celled organisms with animal-like behaviours, such as motility and predation.", + "video_name": "1aJBToJrlvA", + "timestamps": [ + 112 + ], + "3min_transcript": "Man: This is an animal. This is also an animal. Animal. Animal carcass. Animal. Animal carcass again. Animal. The thing that all of these other things have in common is that they're made out of the same basic building block, the animal cell. (music) Animals are made up of your run-of-the-mill eukaryotic cells. These are called eukaryotic because they have a true kernel in the Greek, a good nucleus. That contains the DNA and calls the shots for the rest of the cell, also containing a bunch of organelles. There's a bunch of different kinds of organelles and they all have very specific functions. All of this is surrounded by the cell membrane. Of course, plants are eukaryotic cells too, but theirs are set up a little bit differently. Of course, they have oranelles that allow them to make their own food, which is super nice. We don't have those. Also, their cell membrane is actually a cell wall that's made of cellulose. It's rigid which is why plants can't dance. we did a whole video on it and you can click on it hit here, if it's online yet; it might not be. A lot of the stuff in this video is going to apply to all eukaryotic cells, which includes plants and fungi and protists. Rigid cell walls, that's cool and all, but one of the reasons that animals have been so successful is that their flexible membrane, in addition to allowing them the ability to dance, gives animals the flexibility to create a bunch of different cell types and organ types and tissue types that could never be possible in a plant. Cell walls that protect plants and give them structure prevent them from evolving complicated nerve structures and muscle cells that allow animals to be such a powerful force for eating plants. Animals can move around, find shelter and food, find things to mate with, all that good stuff. In fact, the ability to move oneself around using specialized muscle tissue has been 100% trademarked by kindgom animalia. Voiceover: What about protozoans? Man: Excellent point. What about protozoans? They don't have specialized muscle tissue. They move around with cilia and flagella and that kind of thing. Robert Hooke discovered cells with his kind of crude beta version microscope. He called them cells because they looked like bare, spartan Monk's bedrooms with not much going on inside. Hooke was a smart guy and everything, but he could not have been more wrong about what was going on inside of a cell. There is a whole lot going on inside of a eukaryotic cell. It's more like a city than a Monk's cell. In fact, let's go with that. A cell is like a city. It has defined geographical limits, a ruling government, power plants, roads, waste treatment plants, a police force, industry, all the things a booming metropolis needs to run smoothly. But, this city does not have one of those hippy governments where everybody votes on stuff and talks things out at town hall meetings Nope. Think fascist Italy circa 1938. Think Kim Jong-Il's, I mean Kim Jong-Un's North Korea and you might be getting a closer idea of how eukaryotic cells do their business. Let's start out with city limits. As you approach the city of Eukaryopolis, there's a chance that you will notice something that a traditional city never has, which is either cilia or flagella.", + "qid": "1aJBToJrlvA_112" + }, + { + "Q": "At 6:17, why is the pz, px, py etc. used and what do the subscripts stand for?", + "A": "In the p subshell there are three p orbitals: the px, py, and pz orbitals. These three orbitals are identical, except that they point in different directions (they are orthogonal to each other). The subscripts distinguish the p orbitals based on their orientation; if you draw an imaginary x-y-z axis with the origin at your atom of interest, then the px orbital points along the x-axis, the py orbital points along the y axis, and the pz orbital points along the z axis.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 377 + ], + "3min_transcript": "it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves. that you can configure these dumbbells. One in the z direction, up and down. One in the x direction, left or right. And then one in the y direction, this way, forward and backwards, right? And so if you were to draw-- let's say you wanted to draw the p-orbitals. So this is what you fill next. And actually, you fill one electron here, another electron here, then another electron there. Then you fill another electron, and we'll talk about spin and things like that in the future. But, there, there, and there. And that's actually called Hund's rule. Maybe I'll do a whole video on Hund's rule, but that's not relevant to a first-year chemistry lecture. But it fills in that order, and once again, I want you to have the intuition of what this would look like. Look. I should put look in quotation marks, because it's very abstract. But if you wanted to visualize the p orbitals-- let's say we're looking at the electron configuration for, let's say, carbon.", + "qid": "FmQoSenbtnU_377" + }, + { + "Q": "In 4:50, does the red and blue represent + 1/2 and -1/2 spin?", + "A": "No, the different colours do not represent the spin as if you look at the 4 lobes of the d orbital (for example the dxz sub-shell), there are two red and two blue sections, but at most it will only contain 2 electrons, which can be found in any of the 4 lobes and not one in each section or even pair of sections.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 290 + ], + "3min_transcript": "well I don't know if Tetris is the thing-- but if I'm stacking cubes, I lay out cubes from low energy, if this is the floor, I put the first cube at the lowest energy state. And let's say I could put the second cube But I only have this much space to work with. So I have to put the third cube at the next highest energy state. In this case our energy would be described as potential energy, right? This is just a classical, Newtonian physics example. But that's the same idea with electrons. Once I have two electrons in this 1s orbital -- so let's say the electron configuration of helium is 1s2 -- the third electron I can't put there anymore, because there's only room for two electrons. The way I think about it is these two electrons are now going to repel the third one I want to add. So then I have to go to the 2s orbital. it would look something like this, where I have a high probability of finding the electrons in this shell that's essentially around the 1s orbital, right? So right now, if maybe I'm dealing with lithium right now. So I only have one extra electron. So this one extra electron, that might be where I observed that extra electron. But every now and then it could show up there, it could show up there, it could show up there, but the high probability is there. So when you say where is it going to be 90% of the time? It'll be like this shell that's around the center. Remember, when it's three-dimensional you would kind of cover it up. So that's what they drew here. They do the 1s. It's just a red shell. And then the 2s. The second energy shell is just this blue shell over it. And you can see it a little bit better in, actually, the higher energy orbits, the higher energy shells, where the seventh s energy shell is this red area. And so I think you get the idea that each of those are energy shells. So you kind of keep overlaying the s energy orbitals around each other. But you probably see this other stuff here. And the general principle, remember, is that the electrons fill up the orbital from lowest energy orbital to higher energy orbital. So the first one that's filled up is the 1s. This is the 1. This is the s. So this is the 1s. It can fit two electrons. Then the next one that's filled up is 2s. It can fill two more electrons. And then the next one, and this is where it gets interesting, you fill up the 2p orbital. 2p orbital. That's this, right here. 2p orbitals. And notice the p orbitals have something, p sub z, p sub x, p sub y. What does that mean? Well, if you look at the p-orbitals, they have these dumbbell shapes. They look a little unnatural, but I think in future videos we'll show you how they're analogous to standing waves.", + "qid": "FmQoSenbtnU_290" + }, + { + "Q": "how can we place helium[inert gas]in the second group \"10:20\"?", + "A": "becouse it has two electrons in its outer shell just like the other elements of the second group. the rest of the noblegasses have 8 electrons in their outer shell", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 620 + ], + "3min_transcript": "if we just wanted to visualize this orbital right here, the p orbitals? So we have two electrons. So one electron is going to be in a-- Let's say if this is, I'll try to draw some axes. That's too thin. So if I draw a three-dimensional volume kind of axes. If I were to make a bunch of observations of, say, one of the electrons in the p orbitals, let's say in the pz dimension, sometimes it might be here, sometimes it might be there, sometimes it might be there. And then if you keep taking a bunch of observations, you're going to have something that looks like this bell shape, this barbell shape right there. And then for the other electron that's maybe in the x direction, you make a bunch of observations. Let me do it in a different, It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2.", + "qid": "FmQoSenbtnU_620" + }, + { + "Q": "configuration of berilium is 2s1 or 2s2 at 11:30", + "A": "Beryllium has four electrons, so its configuration is 1s\u00c2\u00b22s\u00c2\u00b2.", + "video_name": "FmQoSenbtnU", + "timestamps": [ + 690 + ], + "3min_transcript": "It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. Let me just do that. Let me take helium out of the picture. These take the p orbitals. And actually, for the sake of figuring out these, you should take helium and throw it right over there. Right? The periodic table is just a way to organize things so it makes sense, but in terms of trying to figure out orbitals, you could take helium. Let me do that. The magic of computers. Cut it out, and then let me paste it right over there. Right? And now you see that helium, you get 1s and then you get 2s, so helium's configuration is -- Sorry, you get 1s1, then 1s2. We're in the first energy shell. Right? So the configuration of hydrogen is 1s1. You only have one electron in the s subshell of the first energy shell. The configuration of helium is 1s2. The configuration of lithium is 1s2. That's where the first two electrons go. And then the third one goes into 2s1, right? And then I think you start to see the pattern. And then when you go to nitrogen you say, OK, it has three in the p sub-orbital. So you can almost start backwards, right? So we're in period two, right? So this is 2p3. Let me write that down. So I could write that down first. 2p3. So that's where the last three electrons go into the p orbital. Then it'll have these two that go into the 2s2 orbital. And then the first two, or the electrons", + "qid": "FmQoSenbtnU_690" + }, + { + "Q": "I have a problem with how we include friction in this problem. If the entire premise of the easy style of doing these problems is to treat the system as a single object, shouldn't we use 20 kg for the friction component, and not 12?\n\nIt makes sense to me that the box is being pressed down on the table by the additional force, and the normal force pressing back upwards would also be 20 kg * 9.8, rather than 12 kg *9.8.\n\nThis occurs at 4:40", + "A": "How is the box pressing down with 20 * 9.8 N of force? The force from the 3 and 5 kg blocks on the 12 kg block comes from the ropes that are going over the pulleys so it is horizontal and not not down so it doesn t add to the normal force.", + "video_name": "ibdidr-bEvI", + "timestamps": [ + 280 + ], + "3min_transcript": "to reduce the acceleration? Yeah, there's this force of gravity over here. This force of gravity on the three kilogram mass is trying to prevent the acceleration because it's pointing opposite the direction of motion. The motion of this system is upright and down across this direction but this force is pointing opposite that direction. This force of gravity right here. So, I'm gonna have to subtract three kilograms times 9.8 meters per second squared. Am I gonna have any other forces that try to prevent the system from moving? You might think the force of gravity on this 12 kilogram box, but look, that doesn't really, in and of itself, prevent the system from moving or not moving. That's perpendicular to this direction. I've called the direction of motion, this positive direction. If it were a force this way, if it were a force this way or a force that way it'd try to cause acceleration of the system. by the normal force, so I don't even have to worry about that force. So, are there any forces associated with the 12 kilogram box that try to prevent motion? It turns out there is. There is going to be a force of friction between the table because there's this coefficient of kinetic friction. So, I've got a force this way, this kinetic frictional force, that's gonna be, have a size of Mu K times f n. That's how you find the normal force and so this is gonna be minus, the Mu K is 0.1 and the normal force will be the normal force for this 12 kilogram mass. So, I'll use 12 kilograms times 9.8 meters per second squared. You might object, you might say, \"Hey, hold on, 12 times 9.8, that's the force of gravity. \"Why are you using this force? \"I thought you said we didn't use it?\" Well, we don't use this force by itself, but it turns out this force of friction depends on this force. So, we're really using a horizontal force, which is why we've got this negative sign here, but it's a horizontal force. It just so happens that this horizontal force depends on a vertical force, which is the normal force. And so that's why we're multiplying by this .1 that turns this vertical force, which is not propelling the system, or trying to stop it, into a horizontal force which is trying to reduce the acceleration of the system. That's why I subtracted and then I divide by the total mass and my total mass is gonna be three plus 12 plus five is gonna be 20 kilograms. Now, I can just solve. If I solve this, I'll get that the acceleration of this system is gonna be 0.392 meters per second squared. So, this is a very fast way. Look it, this is basically a one-liner. If you could put this together right, it's a one-liner. There's much less chance for error than when you're trying to solve three equations with three unknowns. This is beautiful.", + "qid": "ibdidr-bEvI_280" + }, + { + "Q": "at 2:59, what is a photon?", + "A": "Its the basic unit of light. Its the smallest amount of light that you can play around(create, reflect, refract) with.", + "video_name": "y55tzg_jW9I", + "timestamps": [ + 179 + ], + "3min_transcript": "In a vacuum. There's nothing there, no air, no water, no nothing, that's where the light travels the fastest. And let's say that this medium down here, I don't know, let's say it's water. Let's say that this is water. All of this. This was all water over here. This was all vacuum right up here. So what will happen, and actually, that's kind of an unrealistic-- well, just for the sake of argument, let's say we have water going right up against a vacuum. This isn't something you would normally just see in nature but let's just think about it a little bit. Normally, the water, since there's no pressure, it would evaporate and all the rest. But for the sake of argument, let's just say that this is a medium where light will travel slower. What you're going to have is is this ray is actually going to switch direction, it's actually going to bend. Instead of continuing to go in that same direction, it's going to bend a little bit. It's going to go down, in that direction is the refraction. That's the refraction angle. Refraction angle. Or angle of refraction. This is the incident angle, or angle of incidence, and this is the refraction angle. Once again, against that perpendicular. And before I give you the actual equation of how these two things relate and how they're related to the speed of light in these two media-- and just remember, once again, you're never going to have vacuum against water, the water would evaporate because there's no pressure on it and all of that type of thing. But just to--before I go into the math of actually how to figure out these angles relative to the velocities of light in the different media I want to give you an intuitive understanding of not why it bends, 'cause I'm not telling you actually how light works this is really more of an observed property and light, as we'll learn, as we do more and more videos about it, can get pretty confusing. Sometimes you want to treat it as a ray, sometimes you want to treat it as a wave, sometimes you want to treat it as a photon. I actually like to think of it as kind of a, as a bit of a vehicle, and to imagine that, let's imagine that I had a car. So let me draw a car. So we're looking at the top of a car. So this is the passenger compartment, and it has four wheels on the car. We're looking at it from above. And let's say it's traveling on a road. It's traveling on a road. On a road, the tires can get good traction. The car can move pretty efficiently, and it's about to reach an interface it's about to reach an interface where the road ends and it will have to travel on mud. It will have to travel on mud. Now on mud, obviously, the tires' traction will not be as good. The car will not be able to travel as fast. So what's going to happen? Assuming that the car, the steering wheel isn't telling it to turn or anything, the car would just go straight in this direction. But what happens right when--which wheels are going to reach the mud first? Well, this wheel. This wheel is going to reach the mud first.", + "qid": "y55tzg_jW9I_179" + }, + { + "Q": "At 6:30, don't you need to take the inverse of that expression to get the equivalent resistance of the 2 resistors in parallel?", + "A": "There are two formulas for computing 2 parallel resistors. The one I used is Rp = (R1 R2)/(R1+R2). The other formula is the one with all the reciprocals. That s the one you are thinking of: 1/Rp = 1/R1 + 1/R2. Both give the same answer.", + "video_name": "j-iR7puLj6M", + "timestamps": [ + 390 + ], + "3min_transcript": "If you look here, I have two batteries that are hooked up, their inputs, their positive side is hooked up together and their negative side is hooked up together, so they're actually just acting like one, big battery, so let me draw that. I'm going to draw the circuit again so it looks like this. Here's my combined big battery. And it goes to... Relabel these again so we don't get mixed up. Okay. This is R1. This is R2. So this circuit looks a little simpler, and I'm gonna look at it again, see if I can do any more simplification, so what I recognize right here, right in this area right here, R1 and R2 are in parallel. They have the same voltage on their terminals. That means they're in parallel. I know how to simplify parallel resistors. We'll just use the parallel resistor equation that I have in my head, and that looks like this, let's go to this color here. Okay, so parallel resistors, R1 in parallel with R2. I made up this symbol, two vertical lines, that means they're in parallel, and the formula for two parallel resistors is R1 times R2, over R1 plus R2. Now I'll plug in the values. over here at our schematic, they're the same value, and that has a special thing when in parallel resistors so it's actually R R over 2R. Because those resistors are the same. And you can see I can cancel that and I can cancel that and two parallel resistors, if the resistors are equal, is equal to half the resistance. And let's plug in the real values, 1.4 ohms over 2 equals 0.7 ohms. That's the equivalent resistance of these two resistors in parallel. So this is a good time to redraw this circuit again. Let's do it again. Here's our battery. This time I'm going to draw the equivalent resistance. Then we have R3,", + "qid": "j-iR7puLj6M_390" + }, + { + "Q": "At 9:07 does mol mean molecule or is that what it is called.", + "A": "mol means mole (which is Avagadro s number of an atom or molecule).", + "video_name": "-QpkmwIoMaY", + "timestamps": [ + 547 + ], + "3min_transcript": "atmospheres-- let me get my calculator out, get the calculator out, put it right over there-- so it's going to be 23.8 times 1 over 760, or just divided by 760. And we have three significant digits, so it looks like 0.0313. So this is equal to 0.0313 atmospheres. That is our vapor pressure. So let's just deal with this right here. So the number of molecules of water that are going to be in the air in the gaseous state, in the vapor state, is going to be equal to our vapor pressure. That's our equilibrium pressure. If more water molecules evaporate after that point, then we're going to have a higher pressure, which will state, so we'll go kind of past the equilibrium, which is not likely. Or another way to think about it-- more water molecules are not going to of evaporate at a faster rate than they are going to condense beyond that pressure. Anyway, the pressure here is 0.0313 atmospheres. The volume here-- they told us right over here-- so that's the volume-- 4.25. 4.25 times 10 to the fourth liters. And then we want to divide that by-- and you want to make sure that your universal gas constant has the right units, I just looked mine up on Wikipedia-- 0.08-- see everything has three significant digits. So let me just allow that more significant digits and we'll just round at the end. 0.082057, and the units here are liters atmospheres per And this makes sense. This liter will cancel out with that liter. That atmospheres cancels out with that atmospheres. I'm about to multiply it by temperature right here in kelvin. We'll cancel out there. And then we'll have a 1 over moles in the denominator. A 1 over moles in the denominator will just be a moles because you're going to invert it again. So that gives us our answer in moles. And so finally our temperature-- and you've got to remember you've got to do it in kelvin. So 25 degrees Celsius-- let me right it over here-- 25 degrees Celsius is equal to, you just add 273 to it, so this is equal to 298 kelvin. So times 298 kelvin. And now we just have to calculate this. So let's do that.", + "qid": "-QpkmwIoMaY_547" + }, + { + "Q": "At 8:30, why is the volume of the room used instead of the volume of water?", + "A": "As the liquid water sits in the container, it releases water vapour which spreads throughout the room until it is at an equilibrium with the liquid water. There will be equal amounts of water vapour in every spot of the room and Sal wants to know how many total molecules of vapour there are. So, he takes the entire volume of the room and not just that of the liquid water.", + "video_name": "-QpkmwIoMaY", + "timestamps": [ + 510 + ], + "3min_transcript": "Now, the hardest thing about this is just making sure you have your units right and you're using the right ideal gas constant for the right units, and we'll do that right here. So what I want to do, because the universal gas constant that I have is in terms of atmospheres, we need to figure out this vapor pressuree- this equilibrium pressure between vapor and liquid-- we need to write this down in terms of atmospheres. So let me write this down. So the vapor pressure is equal to 23.8 millimeters of mercury. And you can look it up at a table if you don't have this One atmosphere is equivalent to 760 millimeters of mercury. atmospheres-- let me get my calculator out, get the calculator out, put it right over there-- so it's going to be 23.8 times 1 over 760, or just divided by 760. And we have three significant digits, so it looks like 0.0313. So this is equal to 0.0313 atmospheres. That is our vapor pressure. So let's just deal with this right here. So the number of molecules of water that are going to be in the air in the gaseous state, in the vapor state, is going to be equal to our vapor pressure. That's our equilibrium pressure. If more water molecules evaporate after that point, then we're going to have a higher pressure, which will state, so we'll go kind of past the equilibrium, which is not likely. Or another way to think about it-- more water molecules are not going to of evaporate at a faster rate than they are going to condense beyond that pressure. Anyway, the pressure here is 0.0313 atmospheres. The volume here-- they told us right over here-- so that's the volume-- 4.25. 4.25 times 10 to the fourth liters. And then we want to divide that by-- and you want to make sure that your universal gas constant has the right units, I just looked mine up on Wikipedia-- 0.08-- see everything has three significant digits. So let me just allow that more significant digits and we'll just round at the end. 0.082057, and the units here are liters atmospheres per", + "qid": "-QpkmwIoMaY_510" + }, + { + "Q": "As at 1:38 minutes it is said that this emission spectrum is unique to hydrogen atom , which means we have different emission spectrum's for different atoms , so does that in turn mean that we have different energies for same energy levels in different atoms ?", + "A": "Yes. Some even swap places.", + "video_name": "Kv-hRvEOjuA", + "timestamps": [ + 98 + ], + "3min_transcript": "- [Voiceover] I'm sure that most of you know the famous story of Isaac Newton where he took a narrow beam of light and he put that narrow beam of light through a prism and the prism separated the white light into all the different colors of the rainbow. And so if you did this experiment, you might see something like this rectangle up here so all of these different colors of the rainbow and I'm gonna call this a continuous spectrum. It's continuous because you see all these colors right next to each other. So they kind of blend together. So that's a continuous spectrum If you did this similar thing with hydrogen, you don't see a continuous spectrum. So, if you passed a current through a tube containing hydrogen gas, the electrons in the hydrogen atoms are going to absorb energy and jump up to a higher energy level. When those electrons fall down to a lower energy level they emit light and so we talked about this in the last video. This is the concept of emission. If you use something like a prism or a defraction grading to separate out the light, for hydrogen, you don't get a continuous spectrum. So, since you see lines, we call this a line spectrum. So this is the line spectrum for hydrogen. So you see one red line and it turns out that that red line has a wave length. That red light has a wave length of 656 nanometers. You'll also see a blue green line and so this has a wave length of 486 nanometers. A blue line, 434 nanometers, and a violet line at 410 nanometers. And so this emission spectrum is unique to hydrogen and so this is one way to identify elements. And so this is a pretty important thing. And since line spectrum are unique, this is pretty important to explain where those wavelengths come from. And we can do that by using the equation we derived in the previous video. So I call this equation the Balmer Rydberg equation. And you can see that one over lamda, is equal to R, which is the Rydberg constant, times one over I squared, where I is talking about the lower energy level, minus one over J squared, where J is referring to the higher energy level. For example, let's say we were considering an excited electron that's falling from a higher energy level n is equal to three. So let me write this here. So we have an electron that's falling from n is equal to three down to a lower energy level, n is equal to two. All right, so it's going to emit light when it undergoes that transition. So let's look at a visual representation of this. Now let's see if we can calculate the wavelength of light that's emitted. All right, so if an electron is falling from n is equal to three to n is equal to two, I'm gonna go ahead and draw an electron here. So an electron is falling from n is equal to three energy level down to n is equal to two, and the difference in those two energy levels", + "qid": "Kv-hRvEOjuA_98" + }, + { + "Q": "At 15:30, Sal explains that water is hydrolyzed to resupply photosystem II with electrons it transferred to photosystem I. I always learned this reaction was carried out by a photoactive enzyme on the phospholipid bilayer but Sal seems to indicate its actually carried out by the photosystem. Any ideas?", + "A": "The reaction is carried out by a poorly understood Oxygen Evolving Complex (OEC) which is very integrally attached to the photosystem II and its working is strongly coupled with the working of PSII and therefore it is actually considered to be single complex.", + "video_name": "GR2GA7chA_c", + "timestamps": [ + 930 + ], + "3min_transcript": "the stroma to the lumen. Then the hydrogen protons want to go back. They want to-- I guess you could call it-- chemiosmosis. They want to go back into the stroma and then that drives ATP synthase. Right here, this is ATP synthase. ATP synthase to essentially jam together ADPs and phosphate groups to produce ATP. Now, when I originally talked about the light reactions and dark reactions I said, well the light reactions have two byproducts. It has ATP and it also has-- actually it has three. It has ATP, and it also has NADPH. NADP is reduced. It gains these electrons and these hydrogens. So where does that show up? Well, if we're talking about non-cyclic oxidative photophosphorylation, or non-cyclic light reactions, the final electron acceptor. energy states, the final electron acceptor is NADP plus. So once it accepts the electrons and a hydrogen proton with it, it becomes NADPH. Now, I also said that part of this process, water-- and this is actually a very interesting thing-- water gets oxidized to molecular oxygen. So where does that happen? So when I said, up here in photosystem I, that we have a chlorophyll molecule that has an electron excited, and it goes into a higher energy state. And then that electron essentially gets passed from one guy to the next, that begs the question, what can we use to replace that electron? And it turns out that we use, we literally use, the electrons in water. So over here you literally have H2O. So you can kind of imagine it donates two hydrogen protons and two electrons to replace the electron that got excited by the photons. Because that electron got passed all the way over to photosystem I and eventually ends up in NADPH. So, you're literally stripping electrons off of water. And when you strip off the electrons and the hydrogens, you're just left with molecular oxygen. Now, the reason why I want to really focus on this is that there's something profound happening here. Or at least on a chemistry level, something profound is happening. You're oxidizing water. And in the entire biological kingdom, the only place where we know something that is strong enough of an oxidizing agent to oxidize water, to literally take away electrons from water. Which means you're really taking electrons away from oxygen. So you're oxidizing oxygen. The only place that we know that an oxidation agent is", + "qid": "GR2GA7chA_c_930" + }, + { + "Q": "At 9:28 why exactly does he write 3d6 when he should be writing 4d6?", + "A": "He writes 3d\u00e2\u0081\u00b6 because the 3d subshell id filled after the 4s subshell.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 568 + ], + "3min_transcript": "This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy.", + "qid": "NYtPw0WiUCo_568" + }, + { + "Q": "At 5:34, why does carbon have 4 valence electrons instead of 2 when the 2s2 shell is filled already?", + "A": "To expand on Just Keith s answer and clarify a bit more, only the level 1 shell is filled in carbon. The 2s sublevel is filled, but the electrons in it are still much closer in energy to the 2p sublevel than they are to those in the filled level 1.", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 334 + ], + "3min_transcript": "It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange. Hydrogen, hydrogen, hydrogen and actually let me do it the way I was doing it first. You could imagine something like this where carbon could bond just based on what we've learned about Valence electrons and Lewis dot structures. You could say, well, I would predict that maybe a molecule like this could form where a carbon shares it's four Valence electrons with four different hydrogens and in exchange it shares an electron from each of those four hydrogens and so the carbon can feel like it has eight Valence electrons. Each of the hydrogens can feel like it has two Valence electrons. If you did this, if you say, \"Well there should be some molecule out there\" \"in nature that seems pretty stable like this.\" You would be absolutely correct. This is methane and the way that this would be depicted with the Lewis dot structure is this right what I did over here is less conventional notation. Each of these electron pair so that electron pair", + "qid": "NYtPw0WiUCo_334" + }, + { + "Q": "At 5:00, why am I not supposed to write the electron confu-thingy as Be 2p2?\nIts confusing.", + "A": "Be is not a noble gas, the square bracket notation is only used with noble gasses", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 300 + ], + "3min_transcript": "but the important thing to realize and actually for the example of hydrogen sodium is that all of these group one elements are going to have one Valence electron. They're going to have one electron that they tend to use when they are either getting lost to form an ion or that they might be able to use to form a covalent bond. Now let's think about helium and helium's an interesting character because all of the rest of the noble gases have eight Valence electrons which makes them very stable but helium only has two Valence electrons. The reason why it's included here is because helium is also very stable because for that first shell, you only need two electrons to fill full, to fill stable. Helium has two Valence electrons, its electron configuration is one S two. Once again the reason why it's out here with the noble gases is because it's very stable and very inert like the noble gases that's why we now use those helium for balloons It's not going to blow up like the Hindenburg but you might say, well, if it has two Valence electrons maybe it should be in group two because wouldn't all of the group two elements have two Valence electrons? That actually would be a very reasonable argument and we've seen that already. One can make a very reasonable argument to put helium in group two for that reason. All of the elements in group two are going to have two Valence electrons. Now, let's jump to one of the most interesting and versatile elements in the periodic table, the one that really forms the basis of life as we know it, and that's carbon. I encourage you to pause this video and based on what we've just talked about, think about how many Valence electrons carbon has and what its Lewis dot structure could look like. Well carbon's electron configuration is going to be the same as helium plus How many electrons does it have in its outer most shell that has not been completed yet? Well it has these four, two plus two. We could depict them as one, two, three, four Valence electrons. Why is this interesting? Well we can now think about especially if we see carbon's Valence electrons and we see hydrogen is this Lewis dot structure, we can begin to predict what types of molecules carbon and hydrogen could form together. For example, carbon would like to get to A, it would like to pretend like it has electrons so it feels more stable like the noble gas neon and hydrogen would like to at least feel like it has two electrons in its outer most shell so it can feel more stable like helium. If these are carbon atoms and if these are hydrogen atoms, I'll do the hydrogen orange.", + "qid": "NYtPw0WiUCo_300" + }, + { + "Q": "9:28 Why is it 3d^2 and not 4d^2? Why is it written [Ne] when he's talking about iron (Fe)?", + "A": "Because it is? The 4s and 3d orbitals are similar in energy so they are filled around the same time, but the 4d orbitals are quite a bit higher than those two. Why is what written [Ne]? That element in square brackets notation represents the electron configuration of the noble gas from the row above, it saves time when you get to much heavier elements by removing redundant information. Argon is the noble gas in the row above iron so you use [Ar] to represent the following: 1s^2 2s^2 2p^6 3s^2 3p^6", + "video_name": "NYtPw0WiUCo", + "timestamps": [ + 568 + ], + "3min_transcript": "This would be represented as a covalent bond. That would be represented as covalent bond. That would be represented as a covalent bond. Each of these bonds or the sharing of essentially two electrons, the two electrons. Carbon can feel like it has two, four, six, eight electrons even it's sharing. Each of the hydrogens can feel like they have two electrons which gets it into a more stable state. In any of the elements, in carbon's group, they are all going to have four Valence electrons. For example tin, even though neutral tin is going to have 50 electrons, the Valence electrons, the ones that are going to react are going to be the one, two, three, four in its outer shell. One, two, three, four and so you might predict well maybe it could form bonds not too different or it might react in a similar way in some ways could react something similar to say carbon. People even think, there could be life forms in other planets that aren't based on carbon but actually are based on silicon because silicon would have similar types of bond that it can form similar types of structures to carbon for this exact reason. Now what about, I kind of said that you have your transition metals, you d-block right over here and actually your f-blocks are going to be thrown in here as well and these are special cases. These get a little bit more involved because as we already learned that once you're in the fourth period, Let's say we want to do the electron configuration. of say iron. Iron's electron configuration, we could start with argon as a base and then we're now in the d-block but we're not going to fill the four D suborbital. We're not going to back up and back fill into the three d-suborbital. It's one, two, three, four, five, six. It's three D six. This is where this gets a little bit more vigorous. What are the highest energy electrons? Well those are these D electrons right over here. What are the furthest out? Well they're the ones in the fourth sub shell, these four S two. That's why iron's reactivity is a little bit at least based on just this superficial electron configuration. It's a little bit harder to predict. Iron is known to lose one electron, known to lose two electrons, known to lose three electrons and so those could be some combination of these highest energy electrons, both the ones that are furthest out and the ones that are highest energy.", + "qid": "NYtPw0WiUCo_568" + }, + { + "Q": "At 5:20, is the esophagus a part of the lower or upper respiratory tract, or is it not included because it is meant for food and water?", + "A": "It is not included because it is meant for food and water, it is part of the digestive tract.", + "video_name": "Z-yv3Yq4Aw4", + "timestamps": [ + 320 + ], + "3min_transcript": "is kind of the more medical word, I guess. And sitting over the larynx is the epiglottis. And the epiglottis is basically like a lid kind of protecting the larynx from making sure that food and water don't go into it. Now, there's another tube I just alluded to, and it's sitting right here, and this purple tube, is our esophagus So the esophagus is basically, it's fantastic for things like food and water. You want food and water to go down the esophagus because it's going to lead to the stomach. So you want food and water to go that way, but you don't want food and water to go into the larynx. And so you want to make sure that the epiglottis, that lid, is working really well. And if you're swallowing food and water, this epiglottis will literally just kind of close up and protect your larynx. But in this case, that's not happening. We're not actually food and water, we're a little molecule of oxygen, Let's see what happens to it. I'm going to drag up the canvas a little bit. Let's make a little bit of space, and I want to just stop it right there because I want to show you that the air molecule, the oxygen molecule has already kind of made an interesting crossroads. It's actually kind of broken an important boundary, and that's this boundary right here. And on the top of this boundary, I've included the larynx and of course, all the other stuff we just talked about-- the mouth and the nose-- and this is considered our upper respiratory tract. So anything above this dashed line is our upper respiratory tract, and then, of course, you can then guess that anything below the line must then be our lower respiratory tract. So this is an important boundary because people will talk about the upper and lower tract, and I want to make sure you know what is on which side. So on the top of it, is the larynx and everything Let me label that here. The trachea is right here, the wind pipe or the trachea, and everything below that, which, of course, mainly includes things like the lungs, but as we'll see a few other structures that we're going to name. So I'm going to keep moving down, but now you know that important boundary exists. So now let me just make a little bit more space you can see the entire lungs. You can see the molecule is going to go through the trachea, and actually, I have my left lung incompletely drawn. Let me just finish it off right there. So we have our right and left lung, right? These are the two lungs, and our air is going to just kind of slowly pass down-- our molecule of oxygen is going to pass down, and it's going to go either into the right lung or the left lung. Now here, I want to make sure I just take a quick pause and show you the naming structure. And the important word of the day", + "qid": "Z-yv3Yq4Aw4_320" + }, + { + "Q": "I'm pretty sure at 2:17 the C has only 2 methyl groups on it; the third one on the right is supposed to be the bond that was connected to X, right?", + "A": "Don t they have to be hydrogens? SN2 can t occur with a tertiary carbon, or occurs so little to be negligible.", + "video_name": "X9ypryY7hrQ", + "timestamps": [ + 137 + ], + "3min_transcript": "One way to make ethers is to use the Williamson ether synthesis, which is where you start with an alcohol, and you add a strong base to deprotonate the alcohol. Once you deprotonate the alcohol, you add an alkyl halide, and primary alkyl halides work the best. We'll talk about why in a minute. And what happens is you end up putting the R prime group from your alkyl halide on to what used to be your alcohol to form your ether like that. So let's look at the mechanism for the Williamson ether synthesis, where you start with your alcohol. We know that alcohols can function as weak acids. So if you react an alcohol with a strong base, something like sodium hydride, we know that the hydride portion of the molecule is going to function as a strong base. This lone pair of electrons is going to take that proton, which is going to kick these electrons off onto the oxygen. So if we're drawing the product of that acid-based reaction, we now have an oxygen with three lone pairs of electrons around it, giving it a negative 1 formal charge. And we call that an alkoxide anion, which charged sodium ion floating around. So there's some electrostatic or ionic interaction between those opposite charges. And here's where you introduce your alkyl halide. So if we draw our alkyl halide, it would look like this. And we know that there's an electronegativity difference between our halogen and our carbon, where our halogen is going to be partially negative, and our carbon is going to be partially positive. Partially positive carbon means that that carbon wants electrons. It's going to function as an electrophile in the next step of the mechanism. And a lone pair of electrons in the oxygen is going to function as a nucleophile. So opposite charges attract. A lone pair of electrons on our nucleophile are going to attack our electrophile, our carbon. At the same time, the electrons in the bond between the carbon the halogen are going to kick off onto the halogen like that. So this is an SN2-type mechanism, because that has the decreased steric hindrance compared to other alkyl halides. So what will happen is, after nucleophilic attack, we're going to attach our oxygen to our carbon like that, and we form our ether. So if we wanted to, we could just rewrite our ether like this to show it as we added on an R prime group like that. Let's look at an example of the Williamson ether synthesis. So if I start with a molecule over here on the left, and it's kind of an interesting-looking molecule. It's called beta-naphthol. And so beta-naphthol has two rings together like this, and then there's an OH coming off of one of the rings, So that's beta-naphthol. And in the first part, we're going to add potassium hydroxide as our base.", + "qid": "X9ypryY7hrQ_137" + }, + { + "Q": "9:13, when did the first oceans come from , how did they form", + "A": "We re not sure, but water is fairly common in the solar system. The water on earth might have come mostly from comets striking the earth.", + "video_name": "nYFuxTXDj90", + "timestamps": [ + 553 + ], + "3min_transcript": "And so we have rocks from, that are roughly 3.8 billion years So we kind of put that as the beginning of the Archean Eon. And so there's two things there. One, rocks have survived from the beginning of the Archean And also, that's roughly when we think that the first life existed. And so we're now in the Archean Eon. And you might say, oh, maybe Earth is a more pleasant place now. But it would not be. It still has no to little oxygen in the environment. If you were to go to Earth at that time, it might have looked something like this. It would have a been a reddish sky. You would have had nitrogen and methane and carbon dioxide in the atmosphere. There would have been nothing for you to breathe. There still would have been a lot of volcanic activity. This right here, these are pictures of stromatolites. And these are formed from bacteria And over time, these things get built up. But the most significant event in the Archean period, at least in my humble opinion, was what we believe started to happen about 3.5 billion years ago. And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun, to actually do photosynthesis. And the real fascinating byproduct of that, other than the fact that they can now use energy directly from the sun, is that it started to produce oxygen, so starts to produce oxygen. And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere. Because you had all of this iron that was dissolved in the oceans. And let me be clear. the next several billion years, it all occurred in the ocean. We had no ozone layer now. The land was being irradiated. The land was just a completely inhospitable environment for life. So all of this was occurring in the ocean. And so the first oxygen that actually got produced, it actually, instead of just being released into the atmosphere, it ended up bonding with the iron that was dissolved in the ocean at that time. So it actually didn't have a chance to accumulate in the atmosphere. And when we fast forward past the Archean period, we're going to see, that once a lot of that iron was oxidized and the oxygen really did start to get released in the atmosphere, it actually had-- it's funny to say-- a cataclysmic effect or a catastrophic effect on the other anaerobic life on the planet at the time. And it's funny to say that because it was a catastrophe for them. But it was kind of a necessary thing that had to happen for us to happen. So for us, it was a blessing that this cyanobacteria", + "qid": "nYFuxTXDj90_553" + }, + { + "Q": "at 6:31, can I write H2O4S instead of H2SO4 like Sal?", + "A": "yes most people go alphabetically when arranging molecular formulas.", + "video_name": "sXOIIEZh6qg", + "timestamps": [ + 391 + ], + "3min_transcript": "There's a city in Louisiana which we used to drive to all the time -- I think we had some family friends there --called Port Sulfur. And they did a lot of sulfur processing there. And lucky for the residents, at least at the time -- I apologize if they fixed the issue -- it smelled like sulfur, which smells like rotten eggs. But anyway, one mole of sulfur. One mole of sulfur. So sulfur's atomic mass is 32 atomic mass units per sulfur atom. So a whole mole of it is going to have a mass of 32 grams. So a mole of sulfur-- not a mule. Maybe I should invent a new unit called the mule. So a mole of sulfur is 32 grams. So how many moles do we have? We have a little bit more than one, but let's be precise here, because everything else is a little bit of a decimal. So if we have 32.65 grams of sulfur -- divided by 32 grams per mole we have 1.02 moles of sulfur. This was hydrogen up here. So here, you should hopefully see a pretty good ratio, here. For every one sulfur atom-- I mean, the ratio worked exactly out and that's because I did this problem before. I actually made up this problem before we worked, so I made it so the numbers worked out. But one mole of sulfur, for every mole of sulfur, so for every 6.02 times 10 to the 23 sulphur atoms, you have two moles of hydrogen, right? This ratio is 1:2. Two times 1.02 is 2.04. And then, for every one mole of sulfur, you have four moles of oxygen. Right? Literally, if you multiply this times four you get 4.08. So the ratio of hydrogen to sulfur to oxygen we have two hydrogens and we have four oxygens. So the empirical formula of this is H2. And then we have one sulfur. And then we have four oxygens. And this is sulfuric acid, one of the things you would least like poured on you most of the time. Anyway, hope you found that useful.", + "qid": "sXOIIEZh6qg_391" + }, + { + "Q": "Shouldn't a mole of different gases have different volumes not all the same of 22.4 at 11:50? Is it because in ideal gases we disregard the gas's volume?", + "A": "Most gases approximate ideal gas behavior as long as the pressure is not to high and the temperature is not too low.", + "video_name": "GwoX_BemwHs", + "timestamps": [ + 710 + ], + "3min_transcript": "Pressure is 1 atmosphere, but remember we're dealing with atmospheres. 1 atmosphere times volume-- that's what we're solving for. I'll do that and purple-- is equal to 1 mole times R times temperature, times 273. Now this is in Kelvin; this is in moles. We want our volume in liters. So which version of R should we use? Well, we're dealing with atmospheres. We want our volume in liters, and of course, we have moles and Kelvin, so we'll use this version, 0.082. So this is 1, so we can ignore the 1 there, the 1 there. So the volume is equal to 0.082 times 273 degrees Kelvin, So if I have any ideal gas, and all gases don't behave ideally ideal, but if I have an ideal gas and it's at standard temperature, which is at 0 degrees Celsius, or the freezing point of water, which is also 273 degrees Kelvin, and I have a mole of it, and it's at standard pressure, 1 atmosphere, that gas should take up exactly 22.4 liters. And if you wanted to know how many meters cubed it's going to take up, well, you could just say 22.4 liters times-- now, how many meters cubed are there -- so for every 1 meter cubed, you have 1,000 liters. I know that seems like a lot, but it's true. Just think about how big a meter cubed is. If you have something at 1 atmosphere, a mole of it, and at 0 degrees Celsius. Anyway, this is actually a useful number to know sometimes. They'll often say, you have 2 moles at standard temperature and pressure. How many liters is it going to take up? Well, 1 mole will take up this many, and so 2 moles at standard temperature and pressure will take up twice as much, because you're just taking PV equals nRT and just doubling. Everything else is being held constant. The pressure, everything else is being held constant, so if you double the number of moles, you're going to double the volume it takes up. Or if you half the number of moles, you're going to half the volume it takes up. So it's a useful thing to know that in liters at standard temperature and pressure, where standard temperature and pressure is defined as 1 atmosphere and 273 degrees Kelvin, an ideal gas will take up 22.4 liters of volume.", + "qid": "GwoX_BemwHs_710" + }, + { + "Q": "At 1:56 how come we don't put 2 more lone pairs of electrons on Be?", + "A": "Total number of electrons in BeCL2 = 16 Which are all used up. ( 6 each to Cl + 4 electrons as bonds)", + "video_name": "97POZGcfoY8", + "timestamps": [ + 116 + ], + "3min_transcript": "This next set of videos, we're going to predict the shapes of molecules and ions by using VSEPR, which is an acronym for valence shell electron pair repulsion. And really all this means is that electrons, being negatively charged, will repel each other. Like charges repel, and so when those electrons around a central atom repel each other, they're going to force the molecule or ion into a particular shape. And so the first step for predicting the shape of a molecule or ion is to draw the dot structure to show your valence electrons. And so let's go ahead and draw a dot structure for BeCl2. So you find beryllium on the periodic table. It's in group 2, so two valence electrons. Chlorine is in group 7, and we have two of them. So 2 times 7 is 14. And 14 plus 2 gives us a total of 16 valence electrons that we need to account for in our dot structure. So you put the less electronegative atom So beryllium goes in the center. We know it is surrounded by two chlorines, And we just represented four valence electrons. So here's two valence electrons. And here's another two for a total of four. So, instead of 16, we just showed four. So now we're down to 12 valence electrons that we need to account for. So 16 minus 4 is 12. We're going to put those left over electrons on our terminal atoms, which are our chlorines. And chlorine is going to follow the octet rule. Each chlorine is already surrounded by two valence electrons, so each chlorine needs six more. So go ahead and put six more valence electrons on each chlorine. And, since I just represented 12 more electrons there, now we're down to 0 valence electron. So this dot structure has all of our electrons in it. And some of you might think, well, why don't you keep going? Why don't you show some of those lone pairs of electrons in chlorine moving in to share them with the beryllium to give it an octet of electrons? And the reason you don't is because of formal charge. So let's go ahead and assign a formal charge So remember each of our covalent bonds consists of two electrons. So I go ahead and put that in. And if I want to find formal charge, I first think about the number of the valence electrons in the free atom. And that would be two, four-- four berylliums. So we have two electrons in the free atom. And then we think about the bonded atom here, so when I look at the covalent bond, I give one of those electrons to chlorine and one of those electrons to beryllium. And I did the same thing for this bond over here, and so you can see that it is surrounded by two valence electrons. 2 minus 2 gives us a formal charge of 0. And so that's one way to think about why you would stop here for the dot structure. So it has only two valence electrons, so even though it's in period 2, it doesn't necessarily have to follow the octet rule. It just has to have less than eight electrons. And so, again, formal charge helps you understand why you can stop here for your dot structure. Let me go ahead and redraw our molecule", + "qid": "97POZGcfoY8_116" + }, + { + "Q": "At 6:45 Sal says that the base must have the same number of moles as the weak acid. Doesn't that statement assume the acid is monoprotic (donates 1 H+)? How would the calculation change for a diprotic acid (donates 2 H+)? The moles for the acid would just be doubled, right?", + "A": "You re right, so the normality of the base must equal the normality of the acid. Therefore, for diprotic molecules you would need twice as much base.", + "video_name": "BBIGR0RAMtY", + "timestamps": [ + 405 + ], + "3min_transcript": "And when we're adding more hydrogens, we're getting really acidic really fast. But we have a lot of the conjugate acid there in the solution already. So we're going to have an acidic equivalence point. Now, let me give you an actual problem, just to hit all the points home. Because everything I've done now has been very hand-wavey, and no numbers. So let me draw one. Let me draw a weak acid. And you'll recognize it because you're good at this now. But I'll deal with some real numbers here. So let's say that's a pH of 7. We're going to titrate it. It starts off at a low pH because it's a weak acid. And as we titrate it, it's pH goes up. And then it hits the equivalence point and it goes like that. The equivalence point is right over here. And let's say our reagent that we were adding is sodium hydroxide. And let's say it's a 0.2 molar solution. I'll use 700 milliliters of sodium hydroxide is our equivalence point. Right there. So the first question is how much of our weak acid did we have? So what was our original concentration of our weak acid? This is just a general placeholder for the acid. So original concentration of our weak acid. Well, we must have added enough moles of OH at the equivalent point to cancel out all of the moles of the weak acid in whatever hydrogen was out there. But the main concentration was from the weak acid. This 700 milliliters of our reagent must have the same number of moles as the number of moles of weak acid we started off with. And let's say our solution at the beginning was 3 liters. 3 liters to begin with, before we started titrating. the solution. But let's just say that in the beginning, we started with 3 liters. So how many moles have we sopped up? Well, how many moles of OH are there in 700 milliliters of our solution? Well, we know that we have 0.2 moles per liter of OH. And then we know that we don't have-- times 0.7 liters, right? 700 milliliters is 0.7 liters. So how many moles have we added to the situation? 2 times 7 is 14. And we have 2 numbers behind the decimal. So it's 0.14. So 700 milliliters of 0.2 molar sodium hydroxide, and we have 700 milliliters of it, or 0.7 liters. We're going to have 0.14 moles of, essentially, OH that we", + "qid": "BBIGR0RAMtY_405" + }, + { + "Q": "how does bacteria get its energy 1:40", + "A": "Bacteria obtain energy by either ingesting other organisms and organic compounds or by producing their own food. The bacteria that produce their own food are called autotrophs. Bacteria that must consume other organic molecules for energy are called heterotrophs.", + "video_name": "dQCsA2cCdvA", + "timestamps": [ + 100 + ], + "3min_transcript": "- [Voiceover] I would like to welcome you to Biology at Khan Academy. And biology, as you might now, is the study of life. And I can't really imagine anything more interesting than the study of life. And when I say \"life,\" I'm not just talking about us, human beings. I'm talking about all animals. I'm talking about plants. I'm talking about bacteria. And it really is fascinating. How do we start off with inanimate molecules and atoms? You know, this right here is a molecule of DNA. How do we start with things like that, and we get the complexity of living things? And you might be saying, well, what makes something living? Well, living things convert energy from one form to another. They use that energy to grow. They use that energy to change. And I guess growth is a form of change. They use that energy to reproduce. And these are all, in and of themselves, How do they do this? You know, we look around us. How do we, you know, eat a muffin? And how does that allow us to move around and think and do all the things we do? Where did the energy from that muffin come from? How are we similar to a plant or an insect? And we are eerily or strangely similar to these things. We actually have a lot more in common with, you know, that tree outside your window, or that insect, that bee, that might be buzzing around, than you realize. Even with the bacteria that you can only even see at a microscopic level. In fact, we have so much bacteria as part of what makes us, us. So these are fascinating questions. How did life even emerge? And so over the course of what you see in Biology on Khan Academy, we're going to answer these fundamental, fascinating questions. We're going to think about things like energy and the role of energy in life. We're going to thing about important molecules in biology. DNA and its role in reproduction and containing information. And we're going to study cells, which are the basic building block of life. And as we'll see, even though we view cells as these super, super small, small things, cells in and of themselves are incredibly complex. And if you compare them to an atomic scale, they're quite large. In fact, this entire blue background that I have there, that's the surface of an immune cell. And what you see here emerging from it, these little yellow things. These are HIV viruses, emerging from an immune cell. So even though you imagine cells as these very, very small microscopic things, this incredible complexity. Even viruses. Viruses are one of these fascinating things that kind of are right on the edge between life and nonlife. They definitely reproduce, and they definitely evolve. But they don't necessarily have a metabolism.", + "qid": "dQCsA2cCdvA_100" + }, + { + "Q": "At 11:20:\n\nWhat did 12 ever do?\nHow was it activated?", + "A": "Factor 12 is the first factor that is activated in the intrinsic pathway. It is activated by a called Kallikrein. Factor 12 then simply becomes a catalyst to convert 11 from its inactive form to its active form.", + "video_name": "FNVvQ788wzk", + "timestamps": [ + 680 + ], + "3min_transcript": "which is actually one of these little yellow guys. And that tissue factor activates VII, which activates X, so you get a shot - a spark that shoots down this way and activates a little bit of X. And then X will activate a little bit of thrombin, and then thrombin will get the intrisic workhorse going. And how will thrombin do that? Well thrombin actually activates a whole bunch of these guys, and to remember the ones that it activates, you just need to take the five odd numbers starting at five. So what is that? That's V, VII, IX, XI, and XIII. Actually, this is just almost right, but it actually turns out that it's not IX, it's VIII because it couldn't be quite that easy. So those are the five that it activates. So let's draw that in here in our drawing. So let's draw that in the form of blue arrows because thrombin is blue. We said it's going to activate VII. We said it's going to activate not IX, but VIII, so this will be an awkward arrow to draw. We said it's going to activate XI, and we said it's going to activate XIII. Where's our XIII? Well, we haven't actually drawn it in yet, so let's quickly chat about that. The end goal of this whole cascade is to get these fibrin molecules, and these fibrin molecules together will form some strands. It actually turns out that there's one more step, which is to connect these strands together. So we're going to want to connect these strands together with some cross links. These cross links will just hold them together so that they actually form a tight mesh. It turns out that it's this step right here, which is enabled by factor XIII. So let's draw the final thrombin activity, which is to activate XIII. it's going to activate all the necessary things in this intrinsic pathway to get it going. You might actually be wondering about XII up there because thrombin is not hitting him, and actually it turns out that if you remove a person's factor XII, they can still clot pretty well. So it's clear that XII is not a totally necessary part of this intrinsic pathway. And to be clear again, with our use of arrows, this green arrow here is different from these white arrows in the sense that here, we are saying that fibrin is going to become fibrin strands, which is going to become interlaced fibrin strands. So if this was all there was to the story, then every time you had a little bit of damage to your endothelium, you would cause the extrinsic pathway to fire. So you'd create a little activated VII. You would activate some X, which would activate some II, which is thrombin, which would start to create fibrin from fibrinogen. And moreover, the thrombin would have", + "qid": "FNVvQ788wzk_680" + }, + { + "Q": "At 4:08 why would it be an SN2 if that is a tertiary carbon and SN2 rxn only happens in primary and secondary carbons?", + "A": "its not a tertiary carbon, its given at the start that its a secondary or primary carbon", + "video_name": "LccmkSz-Y-w", + "timestamps": [ + 248 + ], + "3min_transcript": "And these electrons would kick off on to that chlorine. So when we draw the next intermediate here, we would now have our oxygen, still bonded to a hydrogen, still with a plus 1 formal charge like that. And now our sulfur is double bonded to our oxygen again with two lone pairs of electrons on the oxygen. The sulfur is also bonded to one chlorine now, so one of the chlorines left. And we can go ahead and draw in that chlorine. So one of the chlorines left here. It's a negatively charged chloride anion. And then still there's a lone pair of electrons on our sulfur like that. So at this part of the mechanism, the pyridine comes along. So if we go ahead and draw the dot structure for pyridine. It's a base, and so it looks like a benzene ring, except we have a nitrogen here instead. And there'd be a lone pair of electrons on this nitrogen. And so that lone pair of electrons and take this proton here on the oxygen. And that would kick these electrons back off onto this oxygen. So when we go ahead and draw that-- let's go ahead and get some more room here-- so what would we get? We would now have our carbon bonded to our oxygen. Our oxygen now has two lone pairs of electrons around it. And we have our sulfur, and our chlorine, and our lone pair of electrons on the sulfur. And now we've made a better leaving group. So this is a better leaving group than the OH was in the beginning. And if we think about an SN2 type mechanism now, we know that the bond between carbon and oxygen is polarized, right? Oxygen being more electronegative, it will be partially negative. And this carbon here be partially positive. And so now we can think about our SN2 type mechanism. Our nucleophile will be this chloride anion up here that we formed in the mechanism. and it's going to attack our partially positive carbon. An SN2 type mechanism. So as the chloride attacks, this stuff on the right is going to leave. So the electrons in magenta are actually going to move in here, and then these electrons are going to kick off onto that chlorine. So when we draw the product, we can go ahead and show the chlorine has now added on to our carbon on the left. And on the right, if you follow the movement of those electrons, they're going to form sulfur dioxides. So SO2. And also the chloride anions, so the Cl minus, like that. And so we've done it. We've substituted our chlorine atom for the OH and formed an alkyl halide. So this is just a better way of forming an alkyl chloride from an alcohol. So if we look at an example, we'll just take something like ethanol here.", + "qid": "LccmkSz-Y-w_248" + }, + { + "Q": "at 1:35 you said that a bullet goes as fast as a jet. how fast does a jet go in miles?", + "A": "Legally around 300MPH for the big commercial jets, Regular Jets can go around 600MPH legally, but when we talking Illegally were talking 1000MPH+", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 95 + ], + "3min_transcript": "My goal in this video and the next video is to start giving a sense of the scale of the earth and the solar system. And as we see, as we start getting into to the galaxy and the universe, it just becomes almost impossible to imagine. But we'll at least give our best shot. So I think most of us watching this video know that this right here is earth. And just to get a sense of scale here, I think probably the largest distance that we can somehow relate to is about 100 miles. You can get into a car for an hour, hour and a half, and go about 100 miles. And on the earth that would be about this far. It would be a speck that would look something like that. That is 100 miles. And also to get us a bit of scale, let's think about a speed that at least we can kind of comprehend. And that would be, maybe, the speed of a bullet. Maybe we can't comprehend it, but I'll say this is the fastest thing that we could maybe comprehend. depending on the type of gun and all of that-- about 280 meters per second, which is about 1,000 kilometers per hour. And this is also roughly the speed of a jet. So just to give a sense of scale here, the earth's circumference-- so if you were to go around the planet-- is about 40,000 kilometers. So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the earth. You might have taken a 12- or 15-hour flight that gets you-- not all the way around the earth-- but gets you pretty far. San Francisco to Australia, or something like that. So right now these aren't scales that are too crazy. Although, even for me, the earth itself is a pretty mind-blowingly large object. Now, with that out of the way let's think about the sun. Because the sun starts to approach something far huger. So this obviously here is the sun. And I think most people appreciate that the sun is much larger than the earth, and that it's pretty far away from the earth. But I don't think most people, including myself, fully appreciate how large the sun is or how far it is away from the earth. So just to give you a sense, the sun is 109 times the circumference of the earth.", + "qid": "GZx3U0dbASg_95" + }, + { + "Q": "At 6:15, when you begin to talk about AUs, the measurement is about equal to the distance between the earth and the sun. Was this done on purpose?", + "A": "yes, that s the definition of the AU", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 375 + ], + "3min_transcript": "So it's 4,360 hours to circumnavigate the sun, going at the speed of a bullet or a jetliner. And so that is-- 24 hours in the day-- that is 181 days. It would take you roughly half a year to go around the sun at the speed of a jetliner. Let me write this down. Half a year. The sun is huge. Now, that by itself may or may not be surprising--and actually let me give you a sense of scale here, because I have this other diagram of a sun. And we'll talk more about the rest of the solar system in the next video. But over here, at this scale, the sun, at least on my screen-- if I were to complete it, it would probably be about 20 inches in diameter. than a raindrop. If I were to draw it on this scale, where the sun is even smaller, the earth would be about that big. Now, what isn't obvious, because we've all done our science projects in third and fourth grade--or we always see these diagrams of the solar system that look something like this-- is that these planets are way further away. Even though these are depicted to scale, they're way further away from the sun than this makes it look. So the earth is 150 million kilometers from the sun. So if this is the sun right here, at this scale you wouldn't even be able to see the earth. It wouldn't even be a pixel. But it would be 150 million kilometers from the earth. and we'll be using that term in the next few videos just because it's an easier way to think about distance-- sometimes abbreviated AU, astronomical unit. And just to give a sense of how far this is, light, which is something that we think is almost infinitely fast and that is something that looks instantaneous, that takes eight minutes to travel from the sun to the earth. If the sun were to disappear, it would take eight minutes for us to know that it disappeared on earth. Or another way, just to put it in the sense of this jet airplane-- let's get the calculator back out. So we're talking about 150 million kilometers.", + "qid": "GZx3U0dbASg_375" + }, + { + "Q": "At 3:32, how do you know the speed of a bullet.", + "A": "The muzzle velocity of standard rounds for guns are well known. The manufacturing is well controlled so that they will have predictable trajectories when fired.", + "video_name": "GZx3U0dbASg", + "timestamps": [ + 212 + ], + "3min_transcript": "You might have taken a 12- or 15-hour flight that gets you-- not all the way around the earth-- but gets you pretty far. San Francisco to Australia, or something like that. So right now these aren't scales that are too crazy. Although, even for me, the earth itself is a pretty mind-blowingly large object. Now, with that out of the way let's think about the sun. Because the sun starts to approach something far huger. So this obviously here is the sun. And I think most people appreciate that the sun is much larger than the earth, and that it's pretty far away from the earth. But I don't think most people, including myself, fully appreciate how large the sun is or how far it is away from the earth. So just to give you a sense, the sun is 109 times the circumference of the earth. if we said, OK, if I'm traveling at the speed of a bullet or the speed of a jetliner, it would take me 40 hours to go around the earth. Well, how long would it take to go around the sun? So if you were to get on a jet plane and try to go around the sun, or if you were to somehow ride a bullet and try to go around the sun-- do a complete circumnavigation of the sun-- it's going to take you 109 times as long as it would have taken you to do the earth. So it would be 100 times-- I could do 109, but just for approximate-- it's roughly 100 times the circumference of the earth. So 109 times 40 is equal to 4,000 hours. And just to get a sense of what 4,000 is-- actually, since I have the calculator out, let's do the exact calculation. It's 109 times the circumference of the earth times 40 hours. So it's 4,360 hours to circumnavigate the sun, going at the speed of a bullet or a jetliner. And so that is-- 24 hours in the day-- that is 181 days. It would take you roughly half a year to go around the sun at the speed of a jetliner. Let me write this down. Half a year. The sun is huge. Now, that by itself may or may not be surprising--and actually let me give you a sense of scale here, because I have this other diagram of a sun. And we'll talk more about the rest of the solar system in the next video. But over here, at this scale, the sun, at least on my screen-- if I were to complete it, it would probably be about 20 inches in diameter.", + "qid": "GZx3U0dbASg_212" + }, + { + "Q": "At 9:13 when both sides are square rooted, how come the 19.6 doesnt become 4.42?", + "A": "It is true that you could write 4.42 instead of \u00e2\u0088\u009a19.6, but you would lose the accuracy of the number. 19.6 is not equal to 4.42, rather 4.4271887242357310647984509622058. And yet, though it is more accurate, it is still not. If you get an irrational number, just keep it in the \u00e2\u0088\u009a form.", + "video_name": "2ZgBJxT9pbU", + "timestamps": [ + 553 + ], + "3min_transcript": "So let me write this over here. So this is negative 9.8. So we have 2 times negative 9.8-- let me just multiply that out. So that's negative 19.6 meters per second squared. And then what's our displacement going to be? What's the displacement over the course of dropping this rock off of this ledge or off of this roof? So you might be tempted to say that our displacement is h. But remember, these are vector quantities, so you want to make sure you get the direction right. From where the rock started to where it ends, what's it doing? It's going to go a distance of h, but it's going to go a distance of h downwards. And our convention is down is negative. So in this example, our displacement from when it leaves your hand to when it hits the ground, the displacement is going to be equal to negative h. It's going to travel a distance of h, but it's going to travel that distance downwards. Our convention is very important here. So our displacement over here is going to be negative h meters. So this is the variable, and this is the shorthand for meters. So when you multiply these two things out, lucky for us these negatives cancel out, and you get 19.6h meters squared per second squared is equal to our final velocity squared. And notice, when you square something you lose the sign information. If our final velocity was positive, you square it, you still get a positive value. If it was negative and you square it, you still get a positive value. But remember, in this example, we're going to be moving downward. So we want the negative version of this. So to really figure out our final velocity, of both sides of this equation. So if we were to take the square root of both sides of this, you take the square root of that side, you take the square root of that side, you will get-- and I'll flip them around-- your final velocity, we could say, is equal to the square root of 19.6h. And you can even take the square root of the meter squared per second squared, treat them almost like variables, even though they're units. And then outside of the radical sign, you will get a meters per second. And the thing I want to be careful here is if we just take the principal root here, the principal root here is the positive square root. But we know that our velocity is going to be downwards here, because that is our convention. So we want to make sure we get the negative square root. So let's try it out with some numbers. We've essentially solved what we set out to solve at the beginning of this video, how fast would we be falling, as a function of the height. Well, let's try it out with something.", + "qid": "2ZgBJxT9pbU_553" + }, + { + "Q": "at 3:34 how come it is called the swimmers view?", + "A": "Because one arm is raised up by the patient s head as if they were swimming the freestyle stroke.", + "video_name": "cbkTTluHaTw", + "timestamps": [ + 214 + ], + "3min_transcript": "is blocking it. DR. MAHADEVAN: Exactly. You can see that that big white thing there is the shoulder that's gotten in the way. And it's making it hard to see. SAL KHAN: They shouldn't have worn those lead shoulder pads. DR. MAHADEVAN: [LAUGHING] And it's making it hard to see whether there's something going on down there right now. So it's really a mystery, as you've shown. SAL KHAN: So how do you solve this problem? DR. MAHADEVAN: If you look over at the other film, it's what we call a swimmer's view. And what we've asked the patient to do is raise one arm up and lower the other. And in doing so, you kind of clear that lower cervical spine and allow better visualization of the entire spine. SAL KHAN: I see. And you're taking it from the direction of the raised arm, on the side of raised arm. DR. MAHADEVAN: You take it from the side. And you can see. SAL KHAN: This is the raised arm right over here. DR. MAHADEVAN: Exactly, that's the raised arm. SAL KHAN: I see. And the other arm on the further side of the patient is down. And that's what allows us to get to the shoulder in a position, so it doesn't block like it does in this left view. DR. MAHADEVAN: Exactly, exactly. SAL KHAN: I see. And over here, it is much clearer And OK. So let me see. So we can count. This is number one right up here. DR. MAHADEVAN: That's one. SAL KHAN: One, two, three, four, five, six-- yeah, we already got to six. We didn't see six over here. And then we got seven. DR. MAHADEVAN: Exactly. SAL KHAN: OK. And so you would call this is an adequate view for what we're trying-- of the neck, because now we can look at all the way DR. MAHADEVAN: Absolutely. We can get all the way down to seven. And ideally, you want to see the top of one, which comes-- actually, in this counting system, we go one through seven. And then we start back at one again, because we're starting with the thoracic vertebrae. SAL KHAN: Oh, look at that. It's like with those streets, where they restart numbering. And you can't find it. So it becomes one again. DR. MAHADEVAN: Exactly. SAL KHAN: Did I number that right? And again, we're looking more to the front. You've got your numbers perfectly on every spinous process, the little bump that you can feel, if you press on the back of your neck. But what we're really interested is the alignment of the front of the vertebral bodies. SAL KHAN: So this is one. This is two, three, four, five, six, seven. Where's the top of one? DR. MAHADEVAN: If you just continue down right there. And it sometimes is difficult to see. But exactly, you want to see that there's alignment right in front of-- I'll assume that there's something here that I can't really see. But you're an expert. So maybe you see things that I don't. OK, so now what do we do with this? DR. MAHADEVAN: Now, we've shown you that you can get a swimmer's view. And it can show you all the way down to C7, T1. But on the original view, as you've shown, we can't see that. So what we did for this patient was get a swimmer's view. SAL KHAN: I see. So this is adequate. And we have this other slide right over here. We have this other one right over here. And why is this one interesting? DR. MAHADEVAN: This is the same patient. And now we've taken that same view that we talked about before, the swimmer's view, where you've got-- SAL KHAN: This is the same patient as this patient right over here, not this patient over here SAL KHAN: Because that one looked overall pretty healthy. DR. MAHADEVAN: That was a normal swimmer's view. But here is an abnormal swimmer's view.", + "qid": "cbkTTluHaTw_214" + }, + { + "Q": "At 3:58,Sal said that the satellite changes its direction but not its speed.So if that same planet is moving around the sun, shouldn't the speed of the satellite vary if it has to go around the planet?", + "A": "I think when the satellite is in the geo stationary orbit its velocity is constant other wise the satellite velocity changes like a planet around the sun", + "video_name": "D1NubiWCpQg", + "timestamps": [ + 238 + ], + "3min_transcript": "impact the object's velocity. That would be true. But we wrote \"speed\" here. Speed is the magnitude of velocity. It does not take into account the direction. And to see why this second statement is false, you could think about a couple of things. And we'll do more videos on the intuition of centripetal acceleration and centripetal forces, inward forces, if this does not make complete intuitive sense to you just at this moment. But imagine we're looking at an ice skating rink from above. And you have an ice skater. This is the ice skater's head. And they are traveling in that direction. Now imagine right at that moment, they grab a rope that is nailed to a stake in the ice skating rink right over there. We're viewing all of this from above, and this right over here Now what is going to happen? Well, the skater is going to travel. Their direction is actually going to change. And they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. And when they let go of the rope, they were traveling in when they let go. They'll keep going on in that direction. And if we assume very, very, very small frictions from the ice skating rink, they'll actually have the same speed. So the force, the inward force, the tension from the rope pulling on the skater in this situation, would have only changed the skater's direction. So and unbalanced force doesn't necessarily have to impact the object's speed. It often does. But in that situation, it would have only impacted the skater's direction. Another situation like this-- and once again, this involves centripetal acceleration, inward forces, inward acceleration-- is a satellite in orbit, or any type of thing in orbit. So if that is some type of planet, and this is one of the planet's moons right over here, the reason why it stays in orbit is because the pull of gravity keeps making the object change its direction, but not its speed. So this was its speed right here. If the planet wasn't there, it would just keep going on in that direction forever and forever. But the planet right over here, there's an inward force of gravity. And we'll talk more about the force of gravity in the future. But this inward force of gravity is going to accelerate this object inwards while it travels. And so after some period of time, this object's velocity vector-- if you add the previous velocity with how much it's changed its new velocity vector. Now this is after its traveled a little bit-- its new velocity vector might look something like this. And it's traveling at the exact right speed so that the force of gravity is always at a right angle to its actual trajectory. It's the exact right speed so it doesn't go off into deep space and so it doesn't plummet into the earth. And we'll cover that in much more detail. But the simple answer is, unbalanced force on a body", + "qid": "D1NubiWCpQg_238" + }, + { + "Q": "At 6:00 what is the difference between MORULA and BLASTOCYST??", + "A": "A morula is a special kind of blastocyst, it has 16 cells assembled in a solid block of cells and looks like a mulberry/ Latin: morula. A proper blastocyst is a few days older, has more cell and forms a hollow sphere.", + "video_name": "-yCIMk1x0Pk", + "timestamps": [ + 360 + ], + "3min_transcript": "to start filling in some of this gap between the embryoblast and the trophoblast, so you're going to start having some fluid that comes in there, and so the morula will eventually look like this, where the trophoblast, or the outer membrane, is kind of this huge sphere of cells. And this is all happening as they keep replicating. Mitosis is the mechanism, so now my trophoblast is going to look like that, and then my embryoblast is going to look like this. Sometimes the embryoblast-- so this is the embryoblast. Sometimes it's also called the inner cell mass, so let me write that. And this is what's going to turn into the organism. And so, just so you know a couple of the labels that are organism, and we are mammals, we call this thing that the morula turned into is a zygote, then a morula, then the cells of the morula started to differentiate into the trophoblast, or kind of the outside cells, and then the embryoblast. And then you have this space that forms here, and this is just fluid, and it's called the blastocoel. A very non-intuitive spelling of the coel part of blastocoel. But once this is formed, this is called a blastocyst. That's the entire thing right here. Let me scroll down a little bit. This whole thing is called the blastocyst, and this is the case in humans. Now, it can be a very confusing topic, because a lot of times in a lot of books on biology, you'll say, hey, you go from the morula to the blastula or the Let me write those words down. So sometimes you'll say morula, and you go to blastula. Sometimes it's called the blastosphere. And I want to make it very clear that these are essentially the same stages in development. These are just for-- you know, in a lot of books, they'll start talking about frogs or tadpoles or things like that, and this applies to them. While we're talking about mammals, especially the ones that are closely related to us, the stage is the blastocyst stage, and the real differentiator is when people talk about just blastula and blastospheres. There isn't necessarily this differentiation between these outermost cells and these embryonic, or this embryoblast, or this inner cell mass here. But since the focus of this video is humans, and really that's where I wanted to start from, because that's what we are and that's what's interesting, we're going to", + "qid": "-yCIMk1x0Pk_360" + }, + { + "Q": "At 13:20, I am a little confused on how you went from 0 ice to 0 degrees water ?", + "A": "For each kg of ice, you need a certain amount of energy to melt it to water. During the melting, the temperature stays the same. The amount of energy you need is given by L, the latent heat of fusion, which is in units joules per kilogram.", + "video_name": "zz4KbvF_X-0", + "timestamps": [ + 800 + ], + "3min_transcript": "Which is this right here, that distance right here. I forgot to figure out how much energy to turn that 100 degree water into 100 degree vapor. So that's key. So I really should have done that up here before I calculated the vapor. But I'll do it down here. So to do 100 degree water to 100 degree vapor. That's this step right here, this is the phase change. I multiply the heat of vaporization which is 2,257 joules per gram times 200 grams. And this is equal to 451,400. I'll do it in that blue color. for our sample of 200 grams. This piece right here was 83,000 joules. This piece right here was 3,780 joules. So to know the total amount of energy the total amount of heat that we had to put in the system to go from minus 10 degree ice all the way to 110 degree vapor, we just add up all of the energies which we had to do in all of these steps. Let's see. And I'll do them in order this time. So to go from minus 10 degree ice to zero degree ice. Of course we have 200 grams of it. It was 4,100. Plus the 67,000. So plus 67,110. Plus 83,000. That's to go from 0 degree water to 100 degree water. Plus 83,560. So we're at 154,000 right now, and just to get to 100 degree water. And then we need to turn that 100 degree water into 100 degree vapor. So you add the 451,000. So, plus 451,400 is equal to 606. And then finally, we're at 100 degree vapor, and we want to convert that to 110 degree vapor. So it's another 3,700 joules. So plus 3,780 is equal to 609,950 joules. So this whole thing when we're dealing with 200 grams", + "qid": "zz4KbvF_X-0_800" + }, + { + "Q": "At 9:55 Sal only crossed out the I's to make the equation simpler. why is that and why didn't he cross out the R's?", + "A": "This is because I(current) is the same throughout the circuit. so Sal cancelled the I s. But the resistance keeps on changing after going through each of the resistor. so he simply cannot cancel out the R s.", + "video_name": "ZrMw7P6P2Gw", + "timestamps": [ + 595 + ], + "3min_transcript": "drew in the previous diagram, although now I will assign numbers to it. Let's say that this resistance is 20 ohms and let's say that this resistance is 5 ohms. What I want to know is, what is the current through the system? First, we'll have to figure out what the equivalent resistance is, and then we could just use Ohm's law to figure out the current in the system. So we want to know what the current is, and we know that the convention is that current flows from the positive terminal to the negative terminal. So how do we figure out the equivalent resistance? Well, we know that we just hopefully proved to you that the total resistance is equal to 1 over this resistor plus 1 over this resistor. So 1 over-- I won't keep writing it. What's 1 over 20? Well, actually, let's just make it a fraction. That's 4/20, right? So 1 over our total resistance is equal to 5/20, which is 1/4. So if 1/R is equal to 1/4, R must be equal to 4. R is equal to 4 ohms. So we could redraw this crazy circuit as this. I'll try to draw it small down here. We could redraw this where this resistance is 4 ohms and this is 16 volts. We could say that this whole thing combined is really just a resistor that is 4 ohms. Well, if we have a 16-volt potential difference, current is flowing that way, even though that's not what the electrons are doing. And that's what our resistance is, 4 ohms. What is the current? V equals IR, Ohms law. It equals the current times 4 ohms. So current is equal to 16 divided by 4, is equal to 4 amps. So let's do something interesting. Let's figure out what the current is flowing through. What's this? What's the current I1 and what's this current I2? Well, we know that the potential difference from here to here is also 16 volts, right? Because this whole thing is essentially at the same potential and this whole thing is essentially at the same potential, so you have 16 volts across there. 16 volts divided by 20 ohms, so let's call this I1. So I1 is equal to 16 volts divided by 20 ohms, which is equal to what? 4/5. So it equals 4/5 of an ampere, or 0.8 amperes.", + "qid": "ZrMw7P6P2Gw_595" + }, + { + "Q": "At 4:07 when the ring breaks do the electrons in magenta need to rotate in order to form a bond with the other halogen atom(anti addition)?", + "A": "Nothing needs to rotate, the magenta electrons are not forming a new bond they are going to the Br that was in the ring. The electrons in the new C-Br bond comes from the bromide anion. It would help if Jay coloured these in.", + "video_name": "Yiy84xYQ3es", + "timestamps": [ + 247 + ], + "3min_transcript": "on the right and my halogen, like that. And these electrons over here, I'm going to mark in red. So this lone pair of electrons on my halogen are going to form a bond with the carbon on the left, And that halogen still has two lone pairs of electrons on it. So I'm going to put in those two lone pairs of electrons. And that halogen has a plus 1 formal charge. This is called a cyclic halonium ion, and it's been proven to occur in this mechanism. If I think about that positively charged halogen, halogens are very electronegative, and they want electrons. So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here losing a little bit of electron density, giving it a partial positive charge. And so that's going to function as our electrophile in the next step. in the previous step. So we had a halogen that had 3 lone pairs of electrons around it. It picked up the electrons in blue. Right? So now, it has 4 lone pairs of electrons-- 8 total electrons-- giving it a negative 1 formal charge, meaning it can now function as a nucleophile. So if I think about this cyclic halonium ion here, the halogen on top is going to prevent the nucleophile from attacking from the top. It's going to have to attack from below here. So this negatively charged halide anion is going to nucleophilic attack this electrophile here-- this carbon. And that's going to kick these electrons in magenta off onto this halogen here. So let's go ahead and draw the results of that nucleophilic attack. All right. So now, I'm going to have my 2 carbons still bonded to each other like that. And the top halogen has swung over here to the carbon It picked up the electrons in magenta. So that's what the carbon on the left will look like. The carbon on the right is still bonded to 2 other things. And the halide anion had to add from below. So now we're going to have this halogen down here. Like that. And so now we understand why it's an anti addition of my 2 halogen atoms. Let's go ahead and do a reaction. So we're going to start with cyclohexane as our reactant And we're going to react cyclohexane with bromine-- so Br2. Now, if I think about the first step of the mechanism, I know I'm going to form a cyclic halonium ion. So I'm going to draw that ring. And I'm going to show the formation of my cyclic halonium ion. It's called a bromonium ion. So I'm going to form a ring like this.", + "qid": "Yiy84xYQ3es_247" + }, + { + "Q": "At 5:19, how doe he say that the lone pair will move up angularly away from the Central Atom O ?", + "A": "The electron pairs in water point towards the corners of a tetrahedron. The bonding pairs are in the plane of the paper. One lone pair is coming angularly out of the paper, and the other lone pair is pointing angularly behind the paper.", + "video_name": "q3g3jsmCOEQ", + "timestamps": [ + 319 + ], + "3min_transcript": "", + "qid": "q3g3jsmCOEQ_319" + }, + { + "Q": "At 8:07 i did not quite understand why the tetrachloromethane molecule will have 0 D??", + "A": "Same case as in CO2. Since the power of Cl in C(CL)4 attracting the electron to themselves have the same magnitude individually, they cancel each other. Hope this helps.", + "video_name": "q3g3jsmCOEQ", + "timestamps": [ + 487 + ], + "3min_transcript": "", + "qid": "q3g3jsmCOEQ_487" + }, + { + "Q": "I know this might be a simple math question but dont you have to divide the entire side of the equation by 4 instead of just the 100 @ 8:52 ?", + "A": "this might help: since there are no variables or anything complicated, the entire side is just multiplying numbers. because of the order of operations (BEDMAS, or BEMDAS, or PEMDAS, or however you learned it), it doesn t matter whether you divide the whole thing by at the end, or divide one of the numbers by 4 in the middle. you re just dividing by 4! try it on a calculator, you get the same result :)", + "video_name": "d4bqNf37mBY", + "timestamps": [ + 532 + ], + "3min_transcript": "There you go. And I can erase some stuff that you're not seeing off the screen. And now I'm ready. So we have 100 moles. So we just do our PV is equal to nRT. We're trying to solve for P. P times 4 meters cubed is equal to n. n is the number of moles. We have 100 moles. Is equal to 100 moles times R. I'll put a blank there for R, because we have to figure out which R we want to use. Times temperature, remember we have to do it in Kelvin. So 0 degrees Celsius is 273 Kelvin. And then which R do we use? I always like to write my R's down here. So we're dealing with meters cubed, we're not dealing with liters, so let's use this one. 8.3145 meters cubed pascals per mole Kelvin. these are in meters cubed pascals divided by moles Kelvin. And then our temperature was 273 Kelvin. Now let's do a little dimensional analysis to make sure that we're doing things right. These meters cancel out with those meters. We divide both sides of the equation by meters. These moles cancel with these moles. The moles of the numerator, the moles of the denominator. Kelvin in the numerator, Kelvin in the denominator. And all we're left with is pascals. Which is good because that is a unit of pressure. So if we divide both sides of this equation by 4. I'll just divide the 100 by 4. 25 times 8.3145 times 273. Which is nice, because that's a unit of pressure. So, let's do the math, 25 times 8.3145 times 273 is equal to 56,746 pascals. And that might seem like a crazy number. But the pascal is actually a very small amount of pressure. It actually turns out that 101,325 pascals is equal to one atmosphere. So if we want to figure out how many atmospheres this is, we could just divide that. Let me look it up on this table. Yes, 101,325.", + "qid": "d4bqNf37mBY_532" + }, + { + "Q": "At 4:28, why did he change oxygen from 16 to 32?", + "A": "The relative atomic mass of one oxygen atom (O) is 16, while the relative atomic mass of one oxygen molecule (O2) is twice that at 32. It depends on if you are talking about the atom or the molecule.", + "video_name": "d4bqNf37mBY", + "timestamps": [ + 268 + ], + "3min_transcript": "So let's do it in grams. Because when we talk about molecular mass it's always in grams. It doesn't have to be. But it makes it a lot simpler to convert between atomic mass units and mass in our world. So this is 2/3 of 2100, that's 1400 grams of N2. Now what's the molar mass of this nitrogen molecule? Well we know that the atomic mass of nitrogen is 14. So this molecule has two nitrogens. So its atomic mass is 28. So one of these molecules will have a mass of 28 atomic mass units. Or one mole of N2 would have a mass of 28 grams. So one mole is 28 grams. We have 1400 grams -- or we say grams per mole, if we want to keep our units right. by 28 grams per mole we should get the number of moles. So 1400 divided by 28 is equal to 50. That worked out nicely. So we have 50 moles of N2. We could write that right there. All right. Let's do oxygen next. So we do the same process over again. 30% is oxygen. So let's do oxygen down here, O2. So we take 30%. Remember, these percentages I gave you, these are the percentages of the total mass, not the percentage of the moles. So we have to figure out what the moles are. So 30.48% of 2100 grams is equal to about 640. So this is equal to 640 grams. And then what is the mass of one mole of the oxygen gas molecule? The atomic mass of one oxygen atom is 16. You can look it up on the periodic table, although you should probably be pretty familiar with it by now. So the atomic mass of this molecule is 32 atomic mass units. So one mole of O2 is going to be 32 grams. We have 640 grams. So how many moles do we have? 640 divided by 32 is equal to 20. We have 20 moles of oxygen. Let me write that down. We have 20 moles. Now we just have to figure out the hydrogen.", + "qid": "d4bqNf37mBY_268" + }, + { + "Q": "At 7:52 how did you get 3 different R equations? And how can you be sure that you are choosing the right one, is it that you choose which ever one matches your units ?", + "A": "You re right - you always want to choose the R equation that matches the units you re using (or, you have to convert your units to match your R equation!).", + "video_name": "d4bqNf37mBY", + "timestamps": [ + 472 + ], + "3min_transcript": "was a super small fraction of the total mass of the gas that we have inside of the container, we actually have more actual particles, more actual molecules of hydrogen than we do of oxygen. That's because each molecule of hydrogen only has an atomic mass of 2 atomic mass units, while each molecule of oxygen has 32 because there's two oxygen atoms. So already we're seeing we actually have more particles do the hydrogen than do the oxygen. And the particles are what matter, not the mass, when we talk about part pressure and partial pressure. So the first thing we can think about is how many total moles of gas, how many total particles do we have bouncing around? 20 moles of oxygen, 30 moles of hydrogen, 50 moles of nitrogen gas. Add them up. We have 100 moles of gas. So if we want to figure out the total pressure first, we can just apply this 100 moles. Let me erase this. There you go. And I can erase some stuff that you're not seeing off the screen. And now I'm ready. So we have 100 moles. So we just do our PV is equal to nRT. We're trying to solve for P. P times 4 meters cubed is equal to n. n is the number of moles. We have 100 moles. Is equal to 100 moles times R. I'll put a blank there for R, because we have to figure out which R we want to use. Times temperature, remember we have to do it in Kelvin. So 0 degrees Celsius is 273 Kelvin. And then which R do we use? I always like to write my R's down here. So we're dealing with meters cubed, we're not dealing with liters, so let's use this one. 8.3145 meters cubed pascals per mole Kelvin. these are in meters cubed pascals divided by moles Kelvin. And then our temperature was 273 Kelvin. Now let's do a little dimensional analysis to make sure that we're doing things right. These meters cancel out with those meters. We divide both sides of the equation by meters. These moles cancel with these moles. The moles of the numerator, the moles of the denominator. Kelvin in the numerator, Kelvin in the denominator. And all we're left with is pascals. Which is good because that is a unit of pressure. So if we divide both sides of this equation by 4. I'll just divide the 100 by 4. 25 times 8.3145 times 273.", + "qid": "d4bqNf37mBY_472" + }, + { + "Q": "At 3:43 you say that the star will enlarge its radius like a red giant, but not get to the same size as a sun-sized star becoming a red giant. Does it also a experience a color change toward red, and if so what color was it in the first place? Does the creation of heavier elements in the core affect the color as well?", + "A": "No, as the stars surface gets farther from the core, it gets cooler. Color is entirely related to temperature the cooler surface glows red instead of yellow.", + "video_name": "UhIwMAhZpCo", + "timestamps": [ + 223 + ], + "3min_transcript": "But anyway, let's think about what happens. And so far, just the pattern of what happens, it's going to happen faster because we have more pressure, more gravity, more temperature. But it's going to happen in pretty much the same way as what we saw with a star the mass of the sun. Eventually that helium-- sorry, that hydrogen is going to fuse into a helium core that's going to have a hydrogen shell around it. It's going to have a hydrogen shell around it, hydrogen fusion shell around it. And then you have the rest of the star around that. So let me label it. This right here is our helium core. And more and more helium is going to be built up as this hydrogen in this shell fuses. And in a star the size of our sun or the mass of our sun, this is when it starts to become a red giant. Because this core is getting denser and denser and denser as more and more helium is produced. And as it gets denser and denser and denser, being put on the hydrogen, on this hydrogen shell out here, where we have fusion still happening. And so that's going to release more outward energy to push out the radius of the actual star. So the general process, and we're going to see this as the star gets more and more massive, is we're going to have heavier and heavier elements forming in the core. Those heavier and heavier elements, as the star gets denser and denser, will eventually ignite, kind of supporting the core. But because the core itself is getting denser and denser and denser, material is getting pushed further and further out with more and more energy. Although if the star is massive enough, it's not going to be able to be pushed out as far as you will have in kind of a red giant, with kind of a sun-like star. But let's just think about how this pattern is going to continue. So eventually, that helium, once it gets dense enough, it's going to ignite and it's going to fuse into carbon. And you're going to have a carbon core forming. So that is carbon core. That's a carbon core. Around that, you have a helium core. you have a shell of helium fusion-- that's helium, not hydrogen-- turning into carbon, making that carbon core denser and hotter. And then around that, you have hydrogen fusion. Have to be very careful. You have hydrogen fusion. And then around that, you have the rest of the star. And so this process is just going to keep continuing. Eventually that carbon is going to start fusing. And you're going to have heavier and heavier elements form the core. And so this is a depiction off of Wikipedia of a fairly mature massive star. And you keep forming these shells of heavier and heavier elements, and cores of heavier and heavier elements until eventually, you get to iron. And in particular, we're talking about iron 56. Iron with an atomic mass of 56. Here on this periodic table that 26 is its atomic number. It's how many protons it has. 56, you kind of view it as a count of the protons", + "qid": "UhIwMAhZpCo_223" + }, + { + "Q": "4:00-4:30 are there others stars that have different elements or have we discovered all of the elements ?", + "A": "we have discovered all the elements that naturally exist, and even if we hadn t there would have to be stars as big as the milky way to fuse elements beyond iron.", + "video_name": "UhIwMAhZpCo", + "timestamps": [ + 240, + 270 + ], + "3min_transcript": "But anyway, let's think about what happens. And so far, just the pattern of what happens, it's going to happen faster because we have more pressure, more gravity, more temperature. But it's going to happen in pretty much the same way as what we saw with a star the mass of the sun. Eventually that helium-- sorry, that hydrogen is going to fuse into a helium core that's going to have a hydrogen shell around it. It's going to have a hydrogen shell around it, hydrogen fusion shell around it. And then you have the rest of the star around that. So let me label it. This right here is our helium core. And more and more helium is going to be built up as this hydrogen in this shell fuses. And in a star the size of our sun or the mass of our sun, this is when it starts to become a red giant. Because this core is getting denser and denser and denser as more and more helium is produced. And as it gets denser and denser and denser, being put on the hydrogen, on this hydrogen shell out here, where we have fusion still happening. And so that's going to release more outward energy to push out the radius of the actual star. So the general process, and we're going to see this as the star gets more and more massive, is we're going to have heavier and heavier elements forming in the core. Those heavier and heavier elements, as the star gets denser and denser, will eventually ignite, kind of supporting the core. But because the core itself is getting denser and denser and denser, material is getting pushed further and further out with more and more energy. Although if the star is massive enough, it's not going to be able to be pushed out as far as you will have in kind of a red giant, with kind of a sun-like star. But let's just think about how this pattern is going to continue. So eventually, that helium, once it gets dense enough, it's going to ignite and it's going to fuse into carbon. And you're going to have a carbon core forming. So that is carbon core. That's a carbon core. Around that, you have a helium core. you have a shell of helium fusion-- that's helium, not hydrogen-- turning into carbon, making that carbon core denser and hotter. And then around that, you have hydrogen fusion. Have to be very careful. You have hydrogen fusion. And then around that, you have the rest of the star. And so this process is just going to keep continuing. Eventually that carbon is going to start fusing. And you're going to have heavier and heavier elements form the core. And so this is a depiction off of Wikipedia of a fairly mature massive star. And you keep forming these shells of heavier and heavier elements, and cores of heavier and heavier elements until eventually, you get to iron. And in particular, we're talking about iron 56. Iron with an atomic mass of 56. Here on this periodic table that 26 is its atomic number. It's how many protons it has. 56, you kind of view it as a count of the protons", + "qid": "UhIwMAhZpCo_240_270" + }, + { + "Q": "At 7:00, Sal says that astronauts can't tell whether they're in free fall near an object with a gravitational pull, or in deep space without any noticeable gravitational forces. How is this possible? Won't they feel the difference in acceleration?", + "A": "How can they feel the acceleration of free fall? The astronauts on the ISS are in free fall. What do you think they feel?", + "video_name": "oIZV-ixRTcY", + "timestamps": [ + 420 + ], + "3min_transcript": "If they were to just slow themselves down, if they were to just brake relative to the Earth, and if they were to just put their brakes on right over there, they would all just plummet to the Earth. So there's nothing special about going 300 or 400 miles up into space, that all of a sudden gravity disappears. The influence of gravity, actually on some level, it just keeps going. You can't, it might become unnoticeably small at some point, but definitely for only a couple of hundred miles up in the air, there is definitely gravity there. It's just they're in orbit, they're going fast enough. So if they just keep falling, they're never going to hit the Earth. And if you want to simulate gravity, and this is actually how NASA does simulate gravity, is that they will put people in a plane, and they call it the vomit rocket because it's known to make people sick, and they'll make them go in a projectile motion. So if this is the ground, in a projectile path or in a parabolic path I should say, so the plane will take off, and it or in a parabolic path. And so anyone who's sitting in that plane will experience free fall. So if you've ever been in, if you've ever right when you jump off of a or if you've ever bungee jumped or skydived or even the feeling when a roller coaster is going right over the top, and it's pulling you down, and your stomach feels a little ill, that feeling of free fall, that's the exact same feeling that these astronauts feel because they're in a constant state of free fall. But that is an indistinguishable feeling from, if you were just in deep space and you weren't anywhere close any noticeable mass, that is an identical feeling that you would feel to having no gravitational force around you. So hopefully that clarifies things a little bit. To someone who's just sitting in the space shuttle, and if they had no windows, there's no way of them knowing whether they are close to a massive object and they're just in free fall around it, they're in orbit, or whether they're and they really are in a state of or in a place where there's very little gravity.", + "qid": "oIZV-ixRTcY_420" + }, + { + "Q": "at 1:03 where did he get a and b?", + "A": "For Ax^2 +Bx +C a*b = AC a + b = B He set the equality, then factored the resultants to obtain candidates for a and b. This takes experience and trial and error.", + "video_name": "dstNU7It-Ro", + "timestamps": [ + 63 + ], + "3min_transcript": "Simplify the rational expression and state the domain. Once again, we have a trinomial over a trinomial. To see if we can simplify them, we need to factor both of them. That's also going to help us figure out the domain. The domain is essentially figuring out all of the valid x's that we can put into this expression and not get something that's undefined. Let's factor the numerator and the denominator. So let's start with the numerator there, and since we have a 2 out front, factoring by grouping will probably be the best way to go, so let's just rewrite it here. I'm just working on the numerator right now. 2x squared plus 13x plus 20. We need to find two numbers, a and b, that if I multiply them, a times b, needs to be equal to-- let me write it over here on the right. a times b needs to be equal to 2 times 20, so it has to be equal to positive 40. And then a plus b has to be equal to 13. are 5 and 8, right? 5 times 8 is 40. 5 plus 8 is 13. We can break this 13x into a 5x and an 8x, and so we can rewrite this as 2x squared. It'll break up the 13x into-- I'm going to write the 8x first. I'm going to write 8x plus 5x. The reason why I wrote the 8x first is because the 8 shares common factors with the 2, so maybe we can factor out a 2x here. It'll simplify it a little bit. 5 shares factors with the 20, so let's see where this goes. We finally have a plus 20 here, and now we can group them. That's the whole point of factoring by grouping. You group these first two characters right here. Let's factor out a 2x, so this would become 2x times-- well, 2x squared divided by 2x is just going to be x. 8x divided by 2x is going to be plus 4. And if we factor out a 5, what do we get? We get plus 5 times x plus 4. 5x divided by 5 is x, 20 divided by 5 is 4. We have an x plus 4 in both cases, so we can factor that out. We have x plus 4 times two terms. We can undistribute it. This thing over here will be x plus four times-- let me do it in that same color-- 2x plus 5. And we've factored this numerator expression right there. Now, let's do the same thing with the denominator I'll do that in a different-- I don't want to run out of colors. So the denominator is right over here, let's do the same exercise with it. We have 2x squared plus 17x plus 30.", + "qid": "dstNU7It-Ro_63" + }, + { + "Q": "At 2:30 he takes the derivative of e^(-2x^2) with respect to (-2x^2). Why doesn't the power rule apply here?", + "A": "The power rule applies when the base is a variable and the power is a number. This is an exponential, the base is a number and the variable is in the power. They are very different functions, and therefore have different derivatives.", + "video_name": "MUQfl385Yug", + "timestamps": [ + 150 + ], + "3min_transcript": "Let's say that f of x is equal to x times e to the negative two x squared, and we want to find any critical numbers for f. I encourage you to pause this video and think about, can you find any critical numbers of f. I'm assuming you've given a go at it. Let's just remind ourselves what a critical number is. We would say c is a critical number of f, if and only if. I'll write if with two f's, short for if and only if, f prime of c is equal to zero or f prime of c is undefined. If we look for the critical numbers for f we want to figure out all the places where the derivative of this with Let's think about how we can find the derivative of this. f prime of x is going to be, well let's see. We're going to have to apply some combination of the product rule and the chain rule. It's going to be the derivative with respect to x of x, so it's going to be that, times e to the negative two x squared plus the derivative with respect to x of e to the negative two x squared times x. This is just the product rule right over here. Derivative of the x times e to the negative of two x squared plus the derivative of e to the What is this going to be? Well all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to one. This first part is going to be equal to e to the negative two x squared. Now the derivative of e to the negative two x squared over here. I'll do this in this pink color. This part right over here, that is going to be equal to- We'll just apply the chain rule. Derivative of e to the negative two x squared with respect to negative two x squared, well that's just going to be e to the negative two x squared. We're going to multiply that times the derivative of negative two x squared with respect to x. That's going to be what, negative four x. Times negative four x, and of course we have this x over here. We have that x over there and let's see,", + "qid": "MUQfl385Yug_150" + }, + { + "Q": "At about 2:29, Sal is saying that 3a(a^5)(a^2) is 3(a^1*a^7). How? I am probably just confused on this topic, though.", + "A": "a = a^1 3\u00e2\u0080\u00a2a^1\u00e2\u0080\u00a2a^5\u00e2\u0080\u00a2a^2 = 3\u00e2\u0080\u00a2a^(1 + 5 + 2) = 3\u00e2\u0080\u00a2a^8", + "video_name": "-TpiL4J_yUA", + "timestamps": [ + 149 + ], + "3min_transcript": "Simplify 3a times a to the fifth times a squared. So the exponent property we can use here is if we have the same base, in this case, it's a. If we have it raised to the x power, we're multiplying it by a to the y power, then this is just going to be equal to a to the x plus y power. And we'll think about why that works in a second. So let's just apply it here. Let's start with the a to the fifth times a squared. So if we just apply this property over here, this will result in a to the fifth plus two-th power. So that's what those guys reduce to, or simplify to. And of course, we still have the 3a out front. Now what I want to do is take a little bit of an aside and realize why this worked. Let's think about what a to the fifth times a squared means. A to the fifth literally means a times a times a times a times a. Now, a squared literally means a times a. So we're multiplying these five a's times these two a's. And what have we just done? We're multiplying a times itself five times, and then another two times. We are multiplying a times itself. So let me make it clear. This over here is a to the fifth. This over here is a squared. When you multiply the two, you're multiplying a by itself itself seven times. 5 plus 2. So this is a to the seventh power. a to the 5 plus 2 power. So this simplifies to 3a times a to the seventh power. Now you might say, how do I apply the property over here? What is the exponent on the a? And remember, if I just have an a over here, this is equivalent to a to the first power. So I can rewrite 3a is 3 times a to the first power. A to the first power-- and the association property of multiplication, I can do the multiplication of the a's before I worry about the 3's. So I can multiply these two guys first. So a to the first times a to the seventh-- I just have to add the exponents because I have the same base and I'm taking the product-- that's going to be a to the eighth power. And I still have this 3 out front. So 3a times a to the fifth times a squared simplifies to 3a to the eighth power.", + "qid": "-TpiL4J_yUA_149" + }, + { + "Q": "is there other basic rigid motions other than reflect,translate, and rotate?\n\nas said in 1:07", + "A": "Those three translations are the three basic geometric translations besides dilation.", + "video_name": "EDlZAyhWxhk", + "timestamps": [ + 67 + ], + "3min_transcript": "So we have another situation where we want to see whether these two figures are congruent. And the way we're going to test that is by trying to transform this figure by translating it, rotating it, and reflecting it. So the first thing that might-- let me translate it. So if these two things are congruent, looks like point E and the point that I'm touching right now, those would correspond to each other. Let me try to rotate it a little bit. So let me rotate-- whoops, I don't want to rotate there, I want to rotate around point E since I already have those on top of each other. So just like that. And it looks like now, if I reflect it across that line, I'm going to be-- oh no, I'm not there. You see, this is tricky. See, when I reflect it, this point, this point, this point, and this point seem to be in the exact same place, but point C does not correspond with that point right over there. So these two polygons are not congruent. And this is why it's important to do this, to make sure the rotations work out. So these two are not congruent. through translations, rotations, and reflections. So are these polygons congruent? No.", + "qid": "EDlZAyhWxhk_67" + }, + { + "Q": "At 1:09 Sal says that he doesn't like using FOIL. FOIL is really easy and is much less confusing then the way Sal did the distributive property twice even though you get the same answer. Why does Sal not like FOIL?", + "A": "FOIL won t help you if you have to expand a product that isn t two binomials multiplied together; for example, two trinomials multiplied together. It s usually better to understand what you re doing instead of relying on mnemonics. For example: (a + b + c) * (d + e + f) = ad + ae + af + bd + be + bf + cd + ce + cf", + "video_name": "JKvmAexeMgY", + "timestamps": [ + 69 + ], + "3min_transcript": "f of x is equal to 7x minus 5. g of x is equal to x to the third power plus 4x. And then they ask us to find f times g of x So the first thing to realize is that this notation f times g of x is just referring to a function that is a product of f of x and g of x. So by definition, this notation just means f of x times g of x. And then we just have to substitute f of x with this definition, g of x with this definition, and then multiply out these algebraic expressions. f of x is right over there. And g of x, is right over there. So let's do it. So this is going to be equal to-- switch back to the orange color. It's going to be equal to f of x, which is 7x minus 5 times g of x, and g of x is x to the third power plus 4x. that each have two terms. You could use FOIL if you like. I don't like using FOIL because you might forget what it's even about. Foil is really just using the distributive property twice. So for example, you take this expression. Whatever you have out here, if you had a 9 out here, or an a, or an x, or anything. Now you have 7x minus 5. If you're multiplying it times this expression, you would multiply this times each term over here. So when you multiply 7x minus 5 times x to the third, you get-- I'll write it this way. You get x to the third times-- actually, let me write it the other way. You get 7x minus 5 times x to the third. And then you have plus 7x minus 5 times 4x. And now we can do the distributive property again. the things we distribute on the right hand side. It's the same exact idea. We could put the x to the third here as well. And when we distribute, you multiply x to the third times 7x and times negative 5. x to the third times 7x is 7x to the fourth power. X to the third times negative 5 is minus 5x to the third. And then you do it over here. You distribute the 4x over the 7x. 4x times 7x is plus 28x squared. 4x times negative 5 is minus 20x. And let's see if we can simplify this. We only have one fourth degree term, one third degree term, one second degree term, and one first degree term. Actually, we can't simplify this anymore. And we're done. This is the product of those two function definitions. This is f times g of x. It is a new function created by multiplying the other two functions.", + "qid": "JKvmAexeMgY_69" + }, + { + "Q": "At 5:00, doesn't the \"function\" Sal draws fail the vertical line test? Why does he still call it a function?", + "A": "That s called a piecewise function - they re defined like this: f(x) = { x + 1 for x > 1 x - 1 for x < 1 x for x = 1 }", + "video_name": "8VgmBe3ulb8", + "timestamps": [ + 300 + ], + "3min_transcript": "from the beginning-- because this is really the definition of an even function-- is when you look at this, you're like hey, what does this mean? f of x is equal to f of negative x. And all it does mean is this. It means that if I were to take f of 2, f of 2 is 4. So let me show you with a particular case. f of 2 is equal to f of negative 2. And this particular case for f of x is equal to x squared, they are both equal to 4. So really, this is just another way of saying that the function can be reflected, or the left side of the function is the reflection of the right side of the function across the vertical axis, across the y-axis. Now just to make sure we have a decent understanding here, let me draw a few more even functions. And I'm going to draw some fairly wacky things just so you really kind of learn to visually recognize them. So let's say a function like this, it does something like that. And then on this side, it does the same thing. It's the reflection, so it jumps up here, then it goes like this, then it goes like this. I'm trying to draw it so it's the mirror image of each other. This is an even function. You take what's going on on the right hand side of this function and you literally just reflect it over the y-axis, and you get the left hand side of the function. And you could see that even this holds. If I take some value-- let's say that this value right here is, I don't know, 3. And let's say that f of 3 over here is equal to, let's say, that that is 5. So this is 5. We see that f of negative 3 is also going to be equal to 5. And I can draw, let me just draw one more to really make sure. I'll do the axis in that same green color. Let me do one more like this. And you could have maybe some type of trigonometric looking function that looks like this, that looks like that. And it keeps going in either direction. So something like this would also be even. So all of these are even functions. Now, you are probably thinking, well, what is an odd function? And let me draw an odd function for you. So let me draw the axis once again. x-axis, y-axis, or the f of x-axis. And to show you an odd function, I'll give you a particular odd function, maybe the most famous of the odd functions. This is probably the most famous of the even functions. And it is f of x-- although there are probably", + "qid": "8VgmBe3ulb8_300" + }, + { + "Q": "AT 9:39, why is there a gap between the reflection about the y-axis and the reflection about the x-axis?\nAt 6:58, there was no gap.", + "A": "It s just a different equation he s graphing (I m not sure what the equation is; if anyone knows I d love to hear). It does look rather strange as a function, but it s still classified as an odd function.", + "video_name": "8VgmBe3ulb8", + "timestamps": [ + 579, + 418 + ], + "3min_transcript": "we figured out f of 2 is 8. 2 to the third power is 8. We know that f of negative 2 is negative 8. Negative 2 to the third power is negative 8. So you have the negative of negative 8, negatives cancel out, and it works out. So in general, you have an odd function. So here's the definition. You are dealing with an odd function if and only if f of x for all the x's that are defined on that function, or for which that function is defined, if f of x is equal to the negative of f of negative x. Or you'll sometimes see it the other way if you multiply both sides of this equation by negative 1, you would get negative f of x is equal to f of negative x. And sometimes you'll see it where it's swapped around where they'll say f of negative x is equal to-- let me write that careful-- I just swapped these two sides. So let me just draw you some more odd functions. So I'll do these visually. So let me draw that a little bit cleaner. So if you have maybe the function does something wacky like this on the right hand side. If it was even, you would reflect it there. But we want to have an odd function, so we're going to reflect it again. So the rest of the function is going to look like this. So what I've drawn in the non-dotted lines, this right here is an odd function. And you could even look at the definition. If you take some value, a, and then you take f of a, which would put you up here. This right here would be f of a. If you take the negative value of that, if you took negative a here, f of negative a is going to be down here. the same distance from the horizontal axis. It's not completely clear the way I drew it just now. So it's maybe going to be like right over here. So this right over here is going to be f of negative a, which is the same distance from the origin as f of a, it's just the negative. I didn't completely draw it to scale. Let me draw one more of these odd functions. I think you might get the point. Actually, I'll draw a very simple odd function, just to show you that it doesn't always have to be something crazy. So a very simple odd function would be y is equal to x, something like this. Whoops. y is equal going through the origin. You reflect what's on the right onto to the left. You get that. And then you reflect it down, you get all of this stuff in the third quadrant. So this is also an odd function. Now, I want to leave you with a few things that are not odd functions and that sometimes might", + "qid": "8VgmBe3ulb8_579_418" + }, + { + "Q": "Would 3 quadrupled by shown as a radical with a little four in the \"notch\", as a cube root is described at 4:32?", + "A": "4th roots would have a little 4 in the notch of the radical symbol. 5th roots would have a 5 in that position. 6th roots would have a 6 in that position. See the pattern?", + "video_name": "87_qIofPwhg", + "timestamps": [ + 272 + ], + "3min_transcript": "Well the volume is going to be two, times two, times two, which is two to the third power or two cubed. This is two cubed. That's why they use the word cubed because this would be the volume of a cube where each of its sides have length two and this of course is going to be equal to eight. But what if we went the other way around? What if we started with the cube? What if we started with this volume? What if we started with a cube's volume and let's say the volume here is eight cubic units, so volume is equal to eight and we wanted to find the lengths of the sides. So we wanted to figure out what X is cause that's X, that's X, and that's X. It's a cube so all the dimensions have the same length. Well there's two ways that we could express this. We could say that X times X times X or we could use the cube root symbol, which is a radical with a little three in the right place. Or we could write that X is equal to, it's going to look very similar to the square root. This would be the square root of eight, but to make it clear, they were talking about the cube root of eight, we would write a little three over there. In theory for square root, you could put a little two over here, but that'd be redundant. If there's no number here, people just assume that it's the square root. But if you're figuring out the cube root or sometimes you say the third root, well then you have to say, well you have to put this little three right over here in this little notch in the radical symbol right over here. And so this is saying X is going to be some number that if I cube it, I get eight. So with that out of the way, let's do some examples. Let's say that I have... Let's say that I want to calculate the cube root of 27. Well if say that this is going to be equal to X, this is equivalent to saying that X to the third or that 27 is equal to X to the third power. So what is X going to be? Well X times X times X is equal to 27, well the number I can think of is three, so we would say that X, let me scroll down a little bit, X is equal to three. Now let me ask you a question. Can we write something like... Can we pick a new color? The cube root of, let me write negative 64. I already talked about that if we're talking the square root, it's fairly typical that hey you put a negative number in there at least until we learn about imaginary numbers, we don't know what to do with it. But can we do something with this? Well if I cube something, can I get a negative number? Sure. So if I say this is equal to X,", + "qid": "87_qIofPwhg_272" + }, + { + "Q": "On the third example, at about 7:25, Sal says that the length marked 2 and the segment parallel to it are equal. He says both are length 2. He says \"We know that these are both length 2 [because] these are all 90 degree angles..\" -- I don't understand how that lets us assume they are equal.", + "A": "Actually, you are right. He can t assume they are equal. BUT, the reason they wouldn t be equal is because the length of either the vertical purple side or the white side (or both) would have been changed. In that case, though, the green side would have changed exactly in the same amount that the other two (white an vertical purple) would have changed, except that if they got shorter, it would have gotten longer, and vice versa.", + "video_name": "vWXMDIazHjA", + "timestamps": [ + 445 + ], + "3min_transcript": "Lets do one more. So here I have a bizarre looking, a bizarre looking shape, and we need to figure out its perimeter. And it it first seems very daunting because they have only given us this side and this side and they have only given us this side right over here. And one thing that we are allowed to assume in this and you don't always have to make you can't always make that assumption and I just didn't draw it here I had time because it would had really crowded out this this diagram. Is it all of the angles in this diagrams are right angles,so i could have drawn a right angle here a right angle here, a right angle there, right angle there, but as you can see it kind of makes things a little bit, it makes things a little bit messy. But how do we figure out the perimeter if we don't know these little distances, if we don't know these little distances here. And the secret here is to kind of shift the sides because all we want to care about is the sum of the sides of the sides. So what I will do is a little exercise in shifting the sides. So this side over here I am going to shift I am going to shift and put it right over there. Then let me keep using different colors, and then this side right over here I am going to shift it and put it right up here. Then finally Iam going to have this side right over here, I can shift it and put it right over there and I think you see what is going on right now. Now all of these sides combined are going to be the same as this side kind of building, even you know this thing was not a rectangle,its its perimeter is going to be a little bit interesting. All we have to think about is this 2 right over here, now lets think about all of these sides that is going up and down. So this side i can shift it all the way to the right and go right over here. Let me make it clear all inside goes all the way to the end, right that it is the exact same all insde. Now this white side I can shift all the way to the right over there, then this green side I can shift right over there and then I have, and then I can shift, and then i can shift this. so I have not, I have not done anything yet, let me be clear I have not done anything yet with that and that I have not shift them over and let me take this side right over here and shift it over. So let me take this entire thing and shift over there and shift it over there. So before I count these two pieces right over here and we know that each have length 2 this 90 degrees angle, so this has link to and this has link to. Before I count these two pieces, I shifted everything else so I was able to form a rectangle. So at least counting everything else I have 7 plus 6, so lets see 7 plus 6 all of these combined are also going to be 7, plus 7, and all of these characters combined are all also going to be 6, plus 6, and then finally I have this 2, right here that I have not counted before, this 2, plus this 2, plus this 2. And then we have our perimeter, so what is this giving us,", + "qid": "vWXMDIazHjA_445" + }, + { + "Q": "At 3:02 when Sal mentions altitude, what does he mean exactly?", + "A": "Altitude is a geographic term used to describe the height of land.", + "video_name": "vWXMDIazHjA", + "timestamps": [ + 182 + ], + "3min_transcript": "it is going to be equal to the perimeter of the 5 triangles is equal to perimeter of 5 outer triangles. Just call them 5 triangles like this minus their basis, right, if i take the perimeter of all of these sides If i added up the part that should not be part of the perimeter of the star would be this part,that part, that part,that part, that part and that part. those are not the part, those are not the part of the perimeter of the star so should be the perimeter of the 5 triangles minus the links of their bases links of their 5 bases. So what is the perimeter of the 5 triangles? well, the perimeter of each of them is 30, perimeter of 5 of them is going to be 5 times 30 which is 150, now we want to subtract out the links of their 5 bases right over here. So this inner pentagon has a perimeter 50, that is the sum of the 5 bases. So that right over here is 50, so the perimeter of the star is going to be 150 minus 50, or or 100. All we need is to get the perimeter of all triangles, subtracted out these bases which was the perimeter of the inner pentagon and we are done. Now lets do the next problem. What is the area of this this quadrilateral, something that has 4 sides of ABCD? And this is a little bit we have not seen a figure quite like this just yet, it on the right hand side looks like a rectangle, and on the left hand side looks like a triangle and this is actually trapezoid, but we can actually as you could imagine the way we figure out the area of several triangles splitting it up into pieces we can recognise. And the most obvious thing to do here is started A and just drop a rock at 90 degrees and we could call this point E. And what is interesting here is we can split this up into something we recognize a rectangle and a right triangle. But you might say how do, how do we figure out what these you know we have this side and that side, so we can figure out the area of this rectangle pretty straight forwardly. But how would we, how would we figure out the area of this triangle? Well if this side is 6 then that means that this that EC is also going to be 6. If AB is 6, notice we have a rectangle right over her, opposite side of a rectangle are equal. So if AB equals 6, implies that EC is equal to 6, EC is equal to 6, so EC is equal to 6 and if EC is equal to 6 then that tells us that DE is going to be 3. DE is going to be 3, this distance right over here is going to be 3.", + "qid": "vWXMDIazHjA_182" + }, + { + "Q": "Sal @ 18:00 c2= 1/3 (x2 - 2x1) you forgot the 2 from the equation above. love you Sal. you are the greatest.", + "A": "Sal corrected this error at the end of the video.", + "video_name": "Qm_OS-8COwU", + "timestamps": [ + 1080 + ], + "3min_transcript": "So we have c1 times this vector plus c2 times the b vector 0, 3 should be able to be equal to my x vector, should be able to be equal to my x1 and x2, where these are just arbitrary. So let's see if I can set that to be true. So if this is true, then the following must be true. c1 times 1 plus 0 times c2 must be equal to x1. We just get that from our definition of multiplying vectors times scalars and adding vectors. And then we also know that 2 times c2-- sorry. c1 times 2 plus c2 times 3, 3c2, should be equal to x2. Now, if I can show you that I can always find c1's and c2's any point in R2 using just these two vectors. So let me see if I can do that. So this is just a system of two unknowns. This is just 0. We can ignore it. So let's multiply this equation up here by minus 2 and put it here. So we get minus 2, c1-- I'm just multiplying this times minus 2. We get a 0 here, plus 0 is equal to minus 2x1. And then you add these two. You get 3c2, right? These cancel out. You get 3-- let me write it in a different color. You get 3c2 is equal to x2 minus 2x1. Or divide both sides by 3, you get c2 is equal to 1/3 x2 minus x1. Now we'd have to go substitute back in for c1. this first equation that says c1 plus 0 is equal to x1, so c1 is equal to x1. So that one just gets us there. So c1 is equal to x1. So you give me any point in R2-- these are just two real numbers-- and I can just perform this operation, and I'll tell you what weights to apply to a and b to get to If you say, OK, what combination of a and b can get me to the point-- let's say I want to get to the point-- let me go back up here. Oh, it's way up there. Let's say I'm looking to get to the point 2, 2. So x1 is 2. Let me write it down here. Say I'm trying to get to the point the vector 2, 2. What combinations of a and b can be there? Well, I know that c1 is equal to x1, so that's equal to 2, and c2 is equal to 1/3 times 2 minus 2.", + "qid": "Qm_OS-8COwU_1080" + }, + { + "Q": "At 2:47, he starts using numbers like two and eight that didn't fit with what he was doing a minute before. Are those numbers suppose to be there to help solve the equation or just random numbers he pulled out?", + "A": "They are just numbers he chose to plug in to the equation so we could see that the property was true.", + "video_name": "PupNgv49_WY", + "timestamps": [ + 167 + ], + "3min_transcript": "", + "qid": "PupNgv49_WY_167" + }, + { + "Q": "At 4:41, in the third line, Sal writes f(x+h) without the denominator of h that it had in line 2. Why isn't this a mistake?\nIf I did the same thing with numbers, it would be as if I rewrote (3 + 5)/2 as 3 + 5/2. That IS a mistake. What's different here?", + "A": "He factored it out, which is not a mistake. This would be like doing the following: (6+10)/5 = 2 \u00e2\u0088\u0099 [(3+5)/5] This is valid because: (ab)/c = a \u00e2\u0088\u0099 (b/c)", + "video_name": "L5ErlC0COxI", + "timestamps": [ + 281 + ], + "3min_transcript": "I just added and subtracted the same thing, but now this thing can be manipulated in interesting algebraic ways to get us to what we all love about the product rule. And at any point you get inspired, I encourage you to pause this video. Well to keep going, let's just keep exploring this expression. So all of this is going to be equal to, it's all going to be equal to the limit as H approaches zero. So the first thing I'm gonna do is I'm gonna look at, I'm gonna look at this part, this part of the expression. And in particular, let's see, I am going to factor out an F of X plus H. So if you factor out an F of X plus H, this part right over here is going to be F of X plus H, F of X plus H, times you're going to be left with G of X plus H. G of X plus H, that's that there, minus G of X, minus G of X, oops, I forgot the parentheses. Oops, it's a different color. I got a new software program and it's making it hard for me to change colors. My apologies, this is not a straightforward proof and the least I could do is change colors more smoothly. Alright, (laughing) G of X plus H minus G of X, that's that one right over there, and then all of that over this H. All of that over H. So that's this part here and then this part over here this part over here, and actually it's still over H, so let me actually circle it like this. So this part over here I can write as. actually here let me, let me factor out a G of X here. So plus G of X plus G of X times this F of X plus H. Times F of X plus H minus this F of X. Minus that F of X. All of that over H. All of that over H. Now we know from our limit properties, the limit of all of this business, well that's just going to be the same thing as the limit of this as H approaches zero plus the limit of this as H approaches zero. And then the limit of the product is going to be the same thing as the product of the limits. So if I used both of those limit properties, I can rewrite this whole thing as the limit, let me give myself some real estate, the limit as H approaches zero of F of X plus H,", + "qid": "L5ErlC0COxI_281" + }, + { + "Q": "So if we can claim that the the limit as h\u00e2\u0086\u00920 of f(x+h) is f(x), as was stated in the video at 7:30, Why can't you evaluate it as f(x+h)g(x+h)-f(x)g(x)/h = f(x)g(x+h)-f(x)g(x)/h = f(x)((g(x+h)-g(x))/h), which would be equal to f(x)g'(x). This result is clearly wrong, but I can't see where exactly I've made a mistake.", + "A": "You need to put the entirety of the first expression in parentheses as it all must be divided by h", + "video_name": "L5ErlC0COxI", + "timestamps": [ + 450 + ], + "3min_transcript": "actually here let me, let me factor out a G of X here. So plus G of X plus G of X times this F of X plus H. Times F of X plus H minus this F of X. Minus that F of X. All of that over H. All of that over H. Now we know from our limit properties, the limit of all of this business, well that's just going to be the same thing as the limit of this as H approaches zero plus the limit of this as H approaches zero. And then the limit of the product is going to be the same thing as the product of the limits. So if I used both of those limit properties, I can rewrite this whole thing as the limit, let me give myself some real estate, the limit as H approaches zero of F of X plus H, times the limit as H approaches zero, of all of this business, G of X plus H minus G of X, minus G of X, all of that over H, I think you might see where this is going. Very exciting. Plus, plus the limit, let me write that a little bit more clearly. Plus the limit as H approaches zero of G of X, our nice brown colored G of X, times, now that we have our product here, the limit, the limit as H approaches zero of F of X plus H. Of F of X plus H minus F of X, all of that, all of that over H. And let me put the parentheses where they're appropriate. So that, that, that, that. And all I did here, the limit, the limit of this sum, that's gonna be the sum of the limits, that's gonna be the limit of this plus the limit of that, and then the limit of the products is gonna be the same thing as the product of the limits. So I just used those limit properties here. But now let's evaluate them. What's the limit, and I'll do them in different colors, what's this thing right over here? The limit is H approaches zero of F of X plus H. Well that's just going to be F of X. Now, this is the exciting part, what is this? The limit is H approaches zero of G of X plus H minus G of X over H. that's the definition of our derivative. That's the derivative of G. So this is going to be, this is going to be the derivative of G of X, which is going to be G prime of X.", + "qid": "L5ErlC0COxI_450" + }, + { + "Q": "3:50 wouldn't length 'a' equal length 'b' then?", + "A": "You can t make that assumption. As you shift to a different point on the circle, the length of a and b will vary. If the point was very close to the x-axis, then b would be very short compared to a . If the point was close to the y-axis, then b would be very long compared to a . Hope this makes sense.", + "video_name": "1m9p9iubMLU", + "timestamps": [ + 230 + ], + "3min_transcript": "And then this is the terminal side. So this is a positive angle theta. And what I want to do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates a comma b. The x value where it intersects is a. The y value where it intersects is b. And the whole point of what I'm doing here is I'm going to see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I want to do is I want to make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here. And let me make it clear that this is a 90-degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. you is, what is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of 1. So the hypotenuse has length 1. Now, what is the length of this blue side right over here? You could view this as the opposite side to the angle. Well, this height is the exact same thing as the y-coordinate of this point of intersection. So this height right over here is going to be equal to b. The y-coordinate right over here is b. This height is equal to b. Now, exact same logic-- what is the length of this base going to be? The base just of the right triangle? Well, this is going to be the x-coordinate If you were to drop this down, this is the point x is equal to a. Or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine-- let me use the same green-- what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our soh cah toa definition. That's the only one we have now. We are actually in the process of extending it-- soh cah toa definition of trig functions. And the cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side-- for this angle, the adjacent side has length a. So it's going to be equal to a over-- what's", + "qid": "1m9p9iubMLU_230" + }, + { + "Q": "At 2:17, Sal distributes the (2x-2y) to the -(dy/dx). I understand that the answer originally is (-2x+2y)(dy/dx), and that he's just rearranging things when he writes it as (2y-2x)(dy/dx). But I don't understand how he gets (-2x+2y)(dy/dx) in the first place! I would have written (2x-2y)(-(dy/dx)). Is he simply applying the negative sign of the (dy/dx) to the (2x-2y)? That's what it looks like, but I don't understand why that is being done. Why can't the negative stay with the (dy/dx)?", + "A": "He is multiplying (2x - 2y)(-(dy/dx)) by one, but by a special form of one which is (-1)(-1). Now he has (-1)(-1)(2x - 2y)(-(dy/dx)). Since we can multiply in any order let s shift things to: (-1)(2x -2y)(-1)(-(dy/dx)). If you multiply the first two terms together, and the 3rd and 4th term together you get: (2y - 2x)(dy/dx).", + "video_name": "9uxvm-USEYE", + "timestamps": [ + 137 + ], + "3min_transcript": "Let's get some more practice doing implicit differentiation. So let's find the derivative of y with respect to x. We're going to assume that y is a function of x. So let's apply our derivative operator to both sides of this equation. So let's apply our derivative operator. And so first, on the left hand side, we essentially are just going to apply the chain rule. First we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1, and the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x. We're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do. Let's make it clear. This right over here, I can rewrite as 2x minus 2y. So let me do that so I can save some space. This is 2x minus 2y If I just distribute the 2. onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides.", + "qid": "9uxvm-USEYE_137" + }, + { + "Q": "5:34 So is codomain the same as range?", + "A": "No, range is a subspace (or subset) of codomain. Range is the specific mapping from the elements in set X to elements in set Y, where codomain is every element in set Y. To concretize: Let f map X --> Y Where X = { 0, 1} Where Y = { 2, 3, 4, 5, 6...}. Let f(0) = 2 & f(1) = 3. Clearly, the range of X = [ 1, 2} where as the codomain of X = { 1, 2 ,3, 4, 5, 6..}.", + "video_name": "BQMyeQOLvpg", + "timestamps": [ + 334 + ], + "3min_transcript": "This statement you've probably never seen before, but I like it because it shows the mapping or the association more, while this association I think that you're putting an x into a little meat grinder or some machine that's going to ground up the x or square the x, or do whatever it needs to do to the x. This notation to me implies the actual mapping. You give me an x, and I'm going to associate another number in real numbers called x squared. So it's going to be just another point. And just as a little bit of terminology, and I think you've seen this terminology before, the set that you are mapping from is called the domain and it's part of the function definition. I, as the function creator, have to tell you that every valid input here has to be a set of real numbers. Now the set that I'm mapping to is called the codomain. Sal, when I learned all of this function stuff in algebra II or whenever you first learned it, we never used this codomain word. We have this idea of range, I learned the word range when I was in 9th or 10th grade. How does this codomain relate to range? And it's a very subtle notation. So the codomain is a set that you're mapping to, and in this example this is the codomain. In this example, the real numbers are the domain and the codomain. So the question is how does the range relate to this? So the codomain is the set that can be possibly mapped to. You're not necessarily mapping to every point in the codomain. I'm just saying that this function is generally mapping from members of this set to that set. It could be equal to the codomain. It's some subset. A set is a subset of itself, every member of a set is also a member of itself, so it's a subset of itself. So range is a subset of the codomain which the function actually maps to. So let me give you an example. Let's say I define the function g, and it is a mapping from the set of real numbers. Let me say it's a mapping from R2 to R.", + "qid": "BQMyeQOLvpg_334" + }, + { + "Q": "why is 0 not in the range? it is a real number and the product of any number and zero would be zero, so I don\u00c2\u00b4t get why Sal says in 13:36 that it is not a member of the range?!", + "A": "Elements of the range are not numbers, but triples of numbers. So asking whether or not 0 is in the range makes no sense. Yes it is true that we can make the third coordinate 0, but can we make the first coordinate 5, the second coordinate 1, and the third coordinate 0 all at the same time? The answer is no.", + "video_name": "BQMyeQOLvpg", + "timestamps": [ + 816 + ], + "3min_transcript": "Let's take our h of-- let me use my other notation-- let's say that I said h, and I wanted to find the mapping from the point in R2, let's say the point 2 comma 3. And then my function tells me that this will map to the point in R3. I add the two terms, the 2 plus 3, so it's 5. I'd find the difference between x2 and x1-- so 3 minus 2 is 1-- and then I multiply the two, 6. So clearly this will be in the range, this is a member of the range. So for example the point 2, 3, which might be right there, will be mapped to the three dimensional point, it's kind but I think you get the idea, would be mapped to the three dimensional point 5, 1, 6. So this is definitely a member of the range. Now my question to you, if I have some point in R3, let's say that this is the point 5, 1, 0. Is this point a member of the range? It's definitely a member of the codomain, it's in R3. It's definitely in here, and this by definition is the codomain. But is this in our range? 5 has to be the sum of two numbers, the 1 has to be the difference of two numbers, and then the 0 would have to be the product of two numbers. And clearly we know 5 is the sum, and 1 is the difference, we're dealing with 2 and 3, and there's no way you can get the product of those numbers to be equal to 0. So the range would be the subset of all of these points in R3, so there'd be a ton of points that aren't in the range, and there'll be a smaller subset of R3 that is in the range. Now I want to introduce you to one more piece of terminology when it comes to functions. These functions up here, this function that mapped from points in R2 to R, so its codomain was R. This function up here is probably the most common function you see in mathematics, this is also mapping to R. These functions that map to R are called scalar value or real value, depending on how you want to think about it. But if they map to a one dimensional space, we call them a scalar valued function, or a real valued function.", + "qid": "BQMyeQOLvpg_816" + }, + { + "Q": "What does he mean at 12:42?", + "A": "The function h maps an ordered pair onto an ordered triple. He is observing that the result of the function is an ordered triple of real numbers, an element of the range R^3.", + "video_name": "BQMyeQOLvpg", + "timestamps": [ + 762 + ], + "3min_transcript": "And notice I'm going from a space that has two dimensions to a space it has three dimensions, or three But I can always associate some point with x1, x2 with some point in my R3 there. A slightly trickier question here is, what is the range? Can I always associate every point-- maybe this wasn't the best example because it's not simple enough -- but can I associate every point in R3-- so this is my codomain, my domain was R2, and my function goes from R2 to R3, so that's h. And so my range, as you could see, it's not like every coordinate you can express in this way in some way. Let me give you an example. Let's take our h of-- let me use my other notation-- let's say that I said h, and I wanted to find the mapping from the point in R2, let's say the point 2 comma 3. And then my function tells me that this will map to the point in R3. I add the two terms, the 2 plus 3, so it's 5. I'd find the difference between x2 and x1-- so 3 minus 2 is 1-- and then I multiply the two, 6. So clearly this will be in the range, this is a member of the range. So for example the point 2, 3, which might be right there, will be mapped to the three dimensional point, it's kind but I think you get the idea, would be mapped to the three dimensional point 5, 1, 6. So this is definitely a member of the range. Now my question to you, if I have some point in R3, let's say that this is the point 5, 1, 0. Is this point a member of the range? It's definitely a member of the codomain, it's in R3. It's definitely in here, and this by definition is the codomain. But is this in our range? 5 has to be the sum of two numbers, the 1 has to be the difference of two numbers, and then the 0 would have to be the product of two numbers. And clearly we know 5 is the sum, and 1 is the difference, we're dealing with 2 and 3, and there's no way you can get the product of those numbers to be equal to 0.", + "qid": "BQMyeQOLvpg_762" + }, + { + "Q": "At 2:30, why does a number to the negative power a decimal?", + "A": "When you have a negative exponent, you re essentially dividing the current number, like 5 squared, by the base, or 5. If you keep doing that, you get 1, 0.2, 0.04, and so on. I hope this helped you!", + "video_name": "6phoVfGKKec", + "timestamps": [ + 150 + ], + "3min_transcript": "Express 0.0000000003457 in scientific notation. So let's just remind ourselves what it means to be in scientific notation. Scientific notation will be some number times some power of 10 where this number right here-- let me write it this way. It's going to be greater than or equal to 1, and it's going to be less than 10. So over here, what we want to put here is what that leading number is going to be. And in general, you're going to look for the first non-zero digit. And this is the number that you're going to want to start off with. This is the only number you're going to want to put ahead of or I guess to the left of the decimal point. So we could write 3.457, and it's going to be multiplied by 10 to something. Now let's think about what we're going to have to multiply it by. To go from 3.457 to this very, very small number, to move the decimal to the left a bunch. You have to add a bunch of zeroes to the left of the 3. You have to keep moving the decimal over to the left. To do that, we're essentially making the number much much, much smaller. So we're not going to multiply it by a positive exponent of 10. We're going to multiply it times a negative exponent of 10. The equivalent is you're dividing by a positive exponent of 10. And so the best way to think about it, when you move an exponent one to the left, you're dividing by 10, which is equivalent to multiplying by 10 to the negative 1 power. Let me give you example here. So if I have 1 times 10 is clearly just equal to 10. 1 times 10 to the negative 1, that's equal to 1 times 1/10, which is equal to 1/10. to 0-- let me actually-- I skipped a step right there. Let me add 1 times 10 to the 0, so we have something natural. So this is one times 10 to the first. One times 10 to the 0 is equal to 1 times 1, which is equal to 1. 1 times 10 to the negative 1 is equal to 1/10, which is equal to 0.1. If I do 1 times 10 to the negative 2, 10 to the negative 2 is 1 over 10 squared or 1/100. So this is going to be 1/100, which is 0.01. What's happening here? When I raise it to a negative 1 power, I've essentially moved the decimal from to the right of the 1 to the left of the 1. I've moved it from there to there. When I raise it to the negative 2, I moved it two over to the left. So how many times are we going to have to move it over to the left to get this number right over here?", + "qid": "6phoVfGKKec_150" + }, + { + "Q": "In 9:15 - he simplified 0.1 into 1/10, but it could have been into 10/100 as well... If you do it like that, you have root10 / root100 = root10 / 10\n\nSeems correct to me but has different answer.... why not do it this way?", + "A": "You can do it either way, but depending on what the problem is, it might be easier to do it your way or to do it Sal s way. For instance, if you had 0.2 , then that would simplify to 1/5 Sal s way; but your way, it would simplify to 20/200 , which would turn into root20 / root200 , and that s a little more messy. So, sometimes your way is easier, and sometimes Sal s is. But you can do it whichever way you prefer; they both give correct answers.", + "video_name": "BpBh8gvMifs", + "timestamps": [ + 555 + ], + "3min_transcript": "which is equal to 1/2. Which is clearly rational. It can be expressed as a fraction. So that's clearly rational. Part G is the square root of 9/4. Same logic. This is equal to the square root of 9 over the square root of 4, which is equal to 3/2. Let's do part H. The square root of 0.16. recognize that, gee, if I multiply 0.4 times 0.4, I'll get this. But I'll show you a more systematic way of doing it, if that wasn't obvious to you. So this is the same thing as the square root of 16/100, right? That's what 0.16 is. So this is equal to the square root of 16 over the square root of 100, which is equal to 4/10, which is equal to 0.4. Let's do a couple more like that. Part I was the square root of 0.1, which is equal to the square root of 1/10, which is equal to the square root of 1 over the square root of 10, which is equal to 1 over-- now, the square root of 10-- 10 is just 2 times 5. So that doesn't really help us much. A lot of math teachers don't like you leaving that radical But I can already tell you that this is irrational. You'll just keep getting numbers. You can try it on your calculator, and it will never repeat. Your calculator will just give you an approximation. Because in order to give the exact value, you'd have to have an infinite number of digits. But if you wanted to rationalize this, just to show you. If you want to get rid of the radical in the denominator, you can multiply this times the square root of 10 over the square root of 10, right? This is just 1. So you get the square root of 10/10. These are equivalent statements, but both of them are irrational. You take an irrational number, divide it by 10, you still have an irrational number. Let's do J. We have the square root of 0.01. This is the same thing as the square root of 1/100. Which is equal to the square root of 1 over the square root of 100, which is equal to 1/10, or 0.1.", + "qid": "BpBh8gvMifs_555" + }, + { + "Q": "At 6:31 Sal said that Pi never repeats. however, doesn't Pi always repeat?", + "A": "What he meant was pi doesn t have a pattern. You never see the same set of numbers twice. Unlike a repeating decimal such as 0.3838383838.... and so on.", + "video_name": "qfQv8GzyjB4", + "timestamps": [ + 391 + ], + "3min_transcript": "will cancel out. And we just have to figure out what 34,028 minus 340 is. So let's just figure this out. 8 is larger than 0, so we won't have to do any 2 is less than 4. So we will have to do some regrouping, but we can't borrow yet because we have a 0 over there. And 0 is less than 3, so we have to do some regrouping there or some borrowing. So let's borrow from the 4 first. So if we borrow from the 4, this becomes a 3 and then this becomes a 10. And then the 2 can now borrow from the 10. This becomes a 9 and this becomes a 12. And now we can do the subtraction. 8 minus 0 is 8. 12 minus 4 is 8. 9 minus 3 is 6. 3 minus nothing is 3. 3 minus nothing is 3. So 9,900x is equal to 33,688. So we get 33,688. Now, if we want to solve for x, we just divide both sides by 9,900. Divide the left by 9,900. Divide the right by 9,900. And then, what are we left with? We're left with x is equal to 33,688 over 9,900. Now what's the big deal about this? Well, x was this number. x was this number that we started off with, this number that just kept on repeating. And by doing a little bit of algebraic manipulation and subtracting one multiple of it from another, we're able to express that same exact x as a fraction. Now this isn't in simplest terms. I mean they're both definitely divisible by 2 and it looks like by 4. So you could put this in lowest common form, but we All we care about is the fact that we were able to represent x, we were able to represent this number, as a fraction. As the ratio of two integers. So the number is also rational. It is also rational. And this technique we did, it doesn't only apply to this number. Any time you have a number that has repeating digits, you could do this. So in general, repeating digits are rational. The ones that are irrational are the ones that never, ever, ever repeat, like pi. And so the other things, I think it's pretty obvious, this isn't an integer. The integers are the whole numbers that we're dealing with. So this is someplace in between the integers. It's not a natural number or a whole number, which depending on the context are viewed as subsets of integers. So it's definitely none of those. So it is real and it is rational. That's all we can say about it.", + "qid": "qfQv8GzyjB4_391" + }, + { + "Q": "How did sal in 3:55 get 12.5 from 128,000? -dazed and confused", + "A": "Well, at 3:50 Sal got 12.8 square meters (m^2), in your question you said 12.5 (maybe a typo?), but the conversion is a simple conversion from square centimeters (cm^2) to square meters (m^2).", + "video_name": "byjmR7JBXKc", + "timestamps": [ + 235 + ], + "3min_transcript": "Let me write this down. 400 times 320. Let's think about it. 4 times 32 is going to be 120, plus 8, 128. And I have 1, 2, 3 zeroes. 1, 2, 3. So it's going to be 128,000 centimeters squared. Now that's a lot of square centimeters. What would we do if we wanted to convert it into meters? Well, we just have to figure out how many square centimeters are there in a square meter. So let's think about it this way. A meter is equal to-- 1 meter is equal to 100 centimeters. So a square meter, so that's right over there. is the same thing as 100 centimeters by 100 centimeters. And so if you were to calculate this area in centimeters, 100 times 100 is 10,000, is equal to 10,000 centimeters squared. So you have 10,000 square centimeters for every square meter. And so, if you want to convert 128,000 centimeters squared to meters squared, you would divide by 10,000. So dividing that by 10,000 would give us 12.8 square meters. Now, another way you could've done it, and maybe this would have been easier, is to convert it up here. Instead of saying 400 centimeters times 320 centimeters, you would say, well, And 320 centimeters, well, that's 3.2 meters. And you would say, OK, 4 times 3.2, that is 12.8 square meters. But either way, the area of the living room in the real world in meters squared, or square meters, is 12.8.", + "qid": "byjmR7JBXKc_235" + }, + { + "Q": "How many more proportion can we make from 2:4::3:6\nPlz answer. My answer is total 8 including ratio given above.", + "A": "Since a ratio is a fraction, we can create a infinite number of proportions from one ratio. Take any ratio, multiply its 2 parts by the same number and you will get an equivalent ratio. Since the 2 ratios are equal, you have a proportion. The number you select to multiply with can be any number. Since there are an infinite set of numbers, you can create an infinite set of ratios.", + "video_name": "qYjiVWwefto", + "timestamps": [ + 124, + 186 + ], + "3min_transcript": "What I want to introduce you to in this video is the notion of a proportional relationship. And a proportional relationship between two variables is just a relationship where the ratio between the two variables is always going to be the same thing. So let's look at an example of that. So let's just say that we want to think about the relationship between x and y. And let's say that when x is one, y is three, and then when x is two, y is six. And when x is nine, y is 27. Now this is a proportional relationship. Why is that? Because the ratio between y and x is always the same thing. And actually the ratio between y and x or, you could say the ratio between x and y, is always the same thing. So, for example-- if we say the ratio y over x-- this is always equal to-- it could be three over one, which is just three. It could be six over two, It could be 27 over nine, which is also just three. So you see that y over x is always going to be equal to three, or at least in this table right over here. And so, or at least based on the data points we have just seen. So based on this, it looks like that we have a proportional relationship between y and x. So this one right over here is proportional. So given that, what's an example of relationships that are not proportional. Well those are fairly easy to construct. So let's say we had-- I'll do it with two different variables. So let's say we have a and b. And let's say when a is one, b is three. And when a is two, b is six. And when a is 10, b is 35. when a is one, b is three so the ratio b to a-- you could say b to a-- you could say well when b is three, a is one. Or when a is one, b is three. So three to one. And that's also the case when b is six, a is two. Or when a is two, b is six. So it's six to two. So these ratios seem to be the same. They're both three. But then all of sudden the ratio is different right over here. This is not equal to 35 over 10. So this is not a proportional relationship. In order to be proportional the ratio between the two variables always has to be the same. So this right over here-- This is not proportional. Not proportional. So the key in identifying a proportional relationship is look at the different values that the variables take on when one variable is one value,", + "qid": "qYjiVWwefto_124_186" + }, + { + "Q": "At 3:49, why did he divide 2xvv' by the entire right hand side as opposed to subtracting v^2 from both sides?", + "A": "When solving a separable equation you don t want to have a term you are adding to the dy/dx (or in this case dv/dx) term. The problem is that it will make it harder to separate the variables but lets try: 2x dv/dx - v^2 = 1 2 dv/dx - v^2/x = 1/x 2 dv - ((v^2/x) dx) = 1/x dx So by creating an additional term on the left side of the equation you have a mix of terms and so you can t integrate.", + "video_name": "6YRGEsQWZzY", + "timestamps": [ + 229 + ], + "3min_transcript": "And now, we can substitute the derivative of y with respect to x is just this. And then the right hand side of the equation is this. But we can substitute v for y over x. So let's do that. And so we get v plus x. Instead of dv [? dv ?] x, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared, we're making the substitution v is equal to y over x. All of that over 2v. Now, let's see what we can do. This is where we just get our algebra hat on, and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. So we'll get 2v squared plus 2xv v prime-- 2v times x, yep, Now let's see, let's subtract 2v squared from both sides of this. And we will be left with 2xv v prime is equal to 1 plus-- let's see, we're subtracting 2v squared from both sides. So we're just left with a 1 plus v squared here, right? 3v squared minus 2v squared is just v squared. And let's see, we want it to be separable, so let's put all the v's on the left hand side. So we get 2xv v prime divided by 1 plus v squared is equal to 1. And let's divide both sides by x. So we get the x's on the other side. So then we get 2v-- and I'll now switch back Instead of v prime, I'll write dv dx. 2v times the derivative of v with respect to x divided by 1 plus v squared is equal to-- I'm dividing both sides by x, notice I didn't write the x on this side-- so that is equal to 1 over x. And then, if we just multiply both sides of this times dx, we've separated the two variables and we can integrate So let's do that. Let's go up here. I'll switch to a different color, so you know I'm working on a different column now. So multiply both sides by dx. I get 2v over 1 plus v squared dv is equal to 1 over x dx. And now we can just integrate both sides of this equation. This is a separable equation in terms of v and x. And what's the integral of this? At first, you might think, oh boy, this is complicated. This is difficult, maybe some type of trig function.", + "qid": "6YRGEsQWZzY_229" + }, + { + "Q": "At 1:46 Sal says that this property extends to negative exponents as well. But 0^-1 power would be 1/0^1 = 1/0 which is undefined right?", + "A": "Yes you are correct. By definition, a negative exponent implies a reciprocal. That is, any number, x^(-1) is equivalent to 1/x.", + "video_name": "PwDnpb_ZJvc", + "timestamps": [ + 106 + ], + "3min_transcript": "So let's think a little bit about powers of 0. So what do you think 0 to the first power is going to be? And I encourage you to pause this video. Well, let's just think about it. One definition of exponentiation is that you start with a 1, and then, you multiply this number times a 1 one time. So this is literally going to be 1 times-- let me do it in the right color-- it's 1 times 0. You're multiplying the 1 by 0 one time. 1 times 0, well, that's just going to be equal to 0. Now, what do you think 0 squared or 0 to the second power is going to be equal to? Well, once again, one way of thinking about this is that you start with a 1, and we're going to multiply it by 0 two times. So times 0 times 0. Well, what's that going to be? once again, you are going to get 0. And I think you see a pattern here. If I take 0 to any non-zero number-- so to the power of any non-zero, so this is some non-zero number, then this is going to be equal to 0. Now, this raises a very interesting question. What happens at 0 to the 0-th power? So here, 0 to the millionth power is going to be 0. 0 to the trillionth power is going to be 0. Even negative or fractional exponents, which we haven't talked about yet, as long as they're non-zero, this is just going to be equal to 0, kind of makes sense. But now, let's think about what 0 to the 0-th power is, because this is actually a fairly deep question. Well, actually, why don't you pause the video and think a little bit about what 0 to the 0-th power should be. Well, there's two trains of thought here. You could say, look, 0 to some non-zero number is 0. So why don't we just extend this to all numbers and say 0 to any number should be 0. And so maybe you should say that 0 to the 0-th power is 0. But then, there was another train of logic that we've already learned, that any non-zero number, if you take any non-zero number, and you raise it to the 0-th power. We've already established that you start with a 1, and you multiply it times that non-zero number 0 times. So this is always going to be equal to 1 for non-zero numbers. So maybe say, hey, maybe we should extend this to all numbers, including 0. So maybe 0 to the 0-th power should be 1.", + "qid": "PwDnpb_ZJvc_106" + }, + { + "Q": "at 2:54, why did sal write 8? wasn't it 7?", + "A": "No it is f(7) = 8, so at x = 7, y reaches it maximum height of 8. Range has to do with possible ys.", + "video_name": "sXP7VhU1gYE", + "timestamps": [ + 174 + ], + "3min_transcript": "is greater than or equal to negative 6. Or we could say negative 6 is less than or equal to x, which is less than or equal to 7. If x satisfies this condition right over here, the function is defined. So that's its domain. So let's check our answer. Let's do a few more of these. The function f of x is graphed. What is its domain? Well, exact similar argument. This function is not defined for x is negative 9, negative 8, all the way down or all the way up I should say to negative 1. At negative 1, it starts getting defined. f of negative 1 is negative 5. So it's defined for negative 1 is less than or equal to x. And it's defined all the way up to x equals 7, including x equals 7. So this right over here, negative 1 is less than or equal to x is less than or equal to 7, the function is defined for any x that satisfies this double inequality right over here. Let's do a few more. What is its range? So now, we're not thinking about the x's for which this function is defined. We're thinking about the set of y values. Where do all of the y values fall into? Well, let's see. The lowest possible y value or the lowest possible value of f of x that we get here looks like it's 0. The function never goes below 0. So f of x-- so 0 is less than or equal to f of x. It does equal 0 right over here. f of negative 4 is 0. And then the highest y value or the highest value that f of x obtains in this function definition is 8. f of 7 is 8. It never gets above 8, but it does equal 8 right over here when x is equal to 7. So 0 is less than f of x, which is less than or equal to 8. So that's its range. Let's do a few more. This is kind of fun. The function f of x is graphed. So once again, this function is defined for negative 2. Negative 2 is less than or equal to x, which is less than or equal to 5. If you give me an x anywhere in between negative 2 and 5, I can look at this graph to see where the function is defined. f of negative 2 is negative 4. f of negative 1 is negative 3. So on and so forth, and I can even pick the values in between these integers. So negative 2 is less than or equal to x, which is less than or equal to 5.", + "qid": "sXP7VhU1gYE_174" + }, + { + "Q": "At 2:05 Sal says that the line he just drew was called a transversal, because it transversed across the other two lines. But what does it mean to transverse?", + "A": "To transverse means to intercept.", + "video_name": "H-E5rlpCVu4", + "timestamps": [ + 125 + ], + "3min_transcript": "Let's say we have two lines over here. Let's call this line right over here line AB. So A and B both sit on this line. And let's say we have this other line over here. We'll call this line CD. So it goes through point C and it goes through point D. And it just keeps on going forever. And let's say that these lines both sit on the same plane. And in this case, the plane is our screen, or this little piece of paper that we're looking at right over here. And they never intersect. So they're on the same plane, but they never intersect each other. If those two things are true, and when they're not the same line, they never intersect and they can be on the same plane, then we say that these lines are parallel. They're moving in the same general direction, in fact, the exact same general direction. If we were looking at it from an algebraic point of view, we would say that they have the same slope, but they have different y-intercepts. They involve different points. they would intersect that at a different point, but they would have the same exact slope. And what I want to do is think about how angles relate to parallel lines. So right over here, we have these two parallel lines. We can say that line AB is parallel to line CD. Sometimes you'll see it specified on geometric drawings like this. They'll put a little arrow here to show that these two lines are parallel. And if you've already used the single arrow, they might put a double arrow to show that this line is parallel to that line right over there. Now with that out of the way, what I want to do is draw a line that intersects both of these parallel lines. So here's a line that intersects both of them. Let me draw a little bit neater than that. So let me draw that line right over there. Well, actually, I'll do some points over here. Well, I'll just call that line l. And this line that intersects both of these parallel lines, This is a transversal line. It is transversing both of these parallel lines. This is a transversal. And what I want to think about is the angles that are formed, and how they relate to each other. The angles that are formed at the intersection between this transversal line and the two parallel lines. So we could, first of all, start off with this angle right over here. And we could call that angle-- well, if we made some labels here, that would be D, this point, and then something else. But I'll just call it this angle right over here. We know that that's going to be equal to its vertical angles. So this angle is vertical with that one. So it's going to be equal to that angle right over there. We also know that this angle, right over here, is going to be equal to its vertical angle, or the angle that is opposite the intersection. So it's going to be equal to that. And sometimes you'll see it specified like this, where you'll see a double angle mark like that. Or sometimes you'll see someone write", + "qid": "H-E5rlpCVu4_125" + }, + { + "Q": "At 5:00, if we were given the measure of the yellow angle in the second diagram, could we find out what the measure of the green angle was? If so, how?", + "A": "No, you would not. Since the two angles exist on different lines, and the lines are not parallel, then there is no way to know for certain if the angles would compliment each other or not.", + "video_name": "H-E5rlpCVu4", + "timestamps": [ + 300 + ], + "3min_transcript": "and these two are equal right over here. Now the other thing we know is we could do the exact same exercise up here, that these two are going to be equal to each other and these two are going to be equal to each other. They're all vertical angles. What's interesting here is thinking about the relationship between that angle right over there, and this angle right up over here. And if you just look at it, it is actually obvious what that relationship is-- that they are going to be the same exact angle, that if you put a protractor here and measured it, you would get the exact same measure up here. And if I drew parallel lines-- maybe I'll draw it straight left and right, it might be a little bit more obvious. So if I assume that these two lines are parallel, and I have a transversal here, what I'm saying is that this angle is going to be the exact same measure as that angle there. And to visualize that, just imagine tilting this line. And as you take different-- so it If you take the line like this and you look at it over here, it's clear that this is equal to this. And there's actually no proof for this. This is one of those things that a mathematician would say is intuitively obvious, that if you look at it, as you tilt this line, you would say that these angles are the same. Or think about putting a protractor here to actually measure these angles. If you put a protractor here, you'd have one side of the angle at the zero degree, and the other side would specify that point. And if you put the protractor over here, the exact same thing would happen. One side would be on this parallel line, and the other side would point at the exact same point. So given that, we know that not only is this side equivalent to this side, it is also equivalent to this side over here. And that tells us that that's also equivalent to that side over there. So all of these things in green are equivalent. And by the same exact argument, this angle And that's going to be the same as this angle, because they are opposite, or they're vertical angles. Now the important thing to realize is just what we've deduced here. The vertical angles are equal and the corresponding angles at the same points of intersection are also equal. And so that's a new word that I'm introducing right over here. This angle and this angle are corresponding. They represent kind of the top right corner, in this example, of where we intersected. Here they represent still, I guess, the top or the top right corner of the intersection. This would be the top left corner. They're always going to be equal, corresponding angles. And once again, really, it's, I guess, for lack of a better word, it is a bit obvious. Now on top of that, there are other words that people will see. We've essentially just proven that not only is this angle equivalent to this angle, but it's also equivalent to this angle right over here. And these two angles-- let me label them so that we can make some headway here.", + "qid": "H-E5rlpCVu4_300" + }, + { + "Q": "Hi friends :-) I have a question for you-\n1> at 6:27 \"Sal\" proved us that alternate interior angles are equal , right? so, here is my question ?\nwe know that :- b=c,f=g\nnow b and c are vertical angles and we also know that b = f (corresponding angle) so why can't we say that c = f ,right? and why not alternate exterior angles?", + "A": "Ok, so you said b=c,f=g. Which says b=c and f=g. If you look back at Sals blackboard on the video there is no comma, it simply says b=c=f=g which means they are ALL equal, so yes , c does equal f and c equals g, and b equals g etc. etc.. As for alternate exterior angles, the same rule applies. a=d=e=h. They are all equal.", + "video_name": "H-E5rlpCVu4", + "timestamps": [ + 387 + ], + "3min_transcript": "If you take the line like this and you look at it over here, it's clear that this is equal to this. And there's actually no proof for this. This is one of those things that a mathematician would say is intuitively obvious, that if you look at it, as you tilt this line, you would say that these angles are the same. Or think about putting a protractor here to actually measure these angles. If you put a protractor here, you'd have one side of the angle at the zero degree, and the other side would specify that point. And if you put the protractor over here, the exact same thing would happen. One side would be on this parallel line, and the other side would point at the exact same point. So given that, we know that not only is this side equivalent to this side, it is also equivalent to this side over here. And that tells us that that's also equivalent to that side over there. So all of these things in green are equivalent. And by the same exact argument, this angle And that's going to be the same as this angle, because they are opposite, or they're vertical angles. Now the important thing to realize is just what we've deduced here. The vertical angles are equal and the corresponding angles at the same points of intersection are also equal. And so that's a new word that I'm introducing right over here. This angle and this angle are corresponding. They represent kind of the top right corner, in this example, of where we intersected. Here they represent still, I guess, the top or the top right corner of the intersection. This would be the top left corner. They're always going to be equal, corresponding angles. And once again, really, it's, I guess, for lack of a better word, it is a bit obvious. Now on top of that, there are other words that people will see. We've essentially just proven that not only is this angle equivalent to this angle, but it's also equivalent to this angle right over here. And these two angles-- let me label them so that we can make some headway here. for the angles themselves. So let's call this lowercase a, lowercase b, lowercase c. So lowercase c for the angle, lowercase d, and then let me call this e, f, g, h. So we know from vertical angles that b is equal to c. But we also know that b is equal to f because they are corresponding angles. And that f is equal to g. So vertical angles are equivalent, corresponding angles are equivalent, and so we also know, obviously, that b is equal to g. And so we say that alternate interior angles are equivalent. So you see that they're kind of on the interior of the intersection. They're between the two lines, but they're on all opposite sides of the transversal. Now you don't have to know that fancy word, alternate interior angles, you really just have to deduce what we just saw over here. Know that vertical angles are going to be equal and corresponding angles are going to be equal. And you see it with the other ones, too. We know that a is going to be equal to d, which is going to be equal to h, which is going to be equal to e.", + "qid": "H-E5rlpCVu4_387" + }, + { + "Q": "1:47 why did you add -4x on that side? Not as in addition but you rewrote it as a subtraction sentence using -4x . -8 doesn't have a coefficient so why did Sal do that? I thought it was X minus X or X plus X. Really Confused Here !", + "A": "In order to graph an equation, it is easiest if it is in y=mx+b form. Therefore you want y alone on the left. Since the 4x is hanging out with the 2y, by subtracting it to both sides it helps get y all by itself.", + "video_name": "V6Xynlqc_tc", + "timestamps": [ + 107 + ], + "3min_transcript": "We're asked to convert these linear equations into slope-intercept form and then graph them on a single coordinate plane. We have our coordinate plane over here. And just as a bit of a review, slope-intercept form is a form y is equal to mx plus b, where m is the slope and b is the intercept. That's why it's called slope-intercept form. So we just have to algebraically manipulate these equations into this form. So let's start with line A, so start with a line A. So line A, it's in standard form right now, it's 4x plus 2y is equal to negative 8. The first thing I'd like to do is get rid of this 4x from the left-hand side, and the best way to do that is to subtract 4x from both sides of this equation. So let me subtract 4x from both sides. The left hand side of the equation, these two 4x's cancel out, and I'm just left with 2y is equal to. or negative 8 minus 4, however you want to do it. Now we're almost at slope-intercept form. We just have to get rid of this 2, and the best way to do that that I can think of is divide both sides of this equation by 2. So let's divide both sides by 2. So we divide the left-hand side by 2 and then divide the right-hand side by 2. You have to divide every term by 2. And then we are left with y is equal to negative 4 divided by 2 is negative 2x. Negative 8 divided by 2 is negative 4, negative 2x minus 4. So this is line A, let me graph it right now. So line A, its y-intercept is negative 4. So the point 0, negative 4 on this graph. If x is equal to 0, y is going to be equal to negative 4, you So 0, 1, 2, 3, 4. That's the point 0, negative 4. That's the y-intercept for line A. And then the slope is negative 2x. So that means that if I change x by positive 1 that y goes down by negative 2. So let's do that. So if I go over one in the positive direction, I have to go down 2, that's what a negative slope's going to do, negative 2 slope. If I go over 2, I'm going to have to go down 4. If I go back negative 1, so if I go in the x direction negative 1, that means in the y direction I go positive two, because two divided by negative one is still negative two, so I go over here. If I go back 2, I'm going to go up 4. Let me just do that. Back 2 and then up 4. So this line is going to look like this.", + "qid": "V6Xynlqc_tc_107" + }, + { + "Q": "At 6:01 we have,\n\n= exp [u*ln(2)] / ln(2) + C\n\nWhy can't we substitute u=ln(x) now and use ln(x)ln(2)=ln(x+2) ? i.e.,\n\n= exp [ln(x+2)] / ln(2) + C\n= (x+2) / ln(2) + C\nCan anyone see what I'm doing wrong?", + "A": "You ve confused the property a little bit; ln(x)*ln(2) is not equal to ln(x+2). Rather, ln(x) + ln(2) = ln(2*x), and ln(x)*ln(2) = ln(2^(ln(x))). Hope that helps.", + "video_name": "C5Lbjbyr1t4", + "timestamps": [ + 361 + ], + "3min_transcript": "product of those exponents. So this is equal to, let me switch colors, this is equal to the integral of e, to the u, e to the, let me write it this way. e to the natural log of 2 times u. I'm just multiplying these two exponents. I raise something to something, then raise it again, we know from our exponent rules, it's just a product of those two exponents. du. Now, this is just a constant factor, right here. This could be, you know, this could just be some number. We could use a calculator to figure out what this is. We could set this equal to a. But we know in general that the integral, this is pretty straightforward, we've now put it in this form. The antiderivative of e to the au, du, is just 1 over a e to the au. This comes from this definition up here, and of course plus If we take the derivative of this, we take the derivative of the inside, which is just going to be a. We multiply that times the one over a, it cancels out, and we're just left with e to the au. So this definitely works out. So the antiderivative of this thing right here is going to be equal to 1 over our a, it's going to be 1 over our constant term, 1 over the natural log of 2 times our whole expression, e e. And I'm going to do something. This is just some number times u, so I can write it as u times some number. And I'm just doing that to put in a form that might help us simplify it a little bit. So it's e to the u times the natural log of 2, right? All I did, is I swapped this order. I could have written this as e to the natural log of 2 times u. If this an a, a times u is the same thing as u times a. Plus c. So this is our answer, but we have to kind of reverse the antiderivative with respect to x. But before I do that, let's see if I can simplify this a little bit. What is, if I have, just from our natural log properties, or logarithms, a times the natural log of b. We know this is the same thing as the natural log of b to the a. Let me draw a line here. Right? That this becomes the exponent on whatever we're taking the natural log of. So u, let me write this here, u times the natural log of 2, is the same thing as the natural log of 2 to the u. So we can rewrite our antiderivative as being equal to 1 over the natural log of 2, that's just that part here, times e to the, this can be rewritten based on this logarithm property, as the natural log of 2 to the u, and of course we still have our plus c there. Now, what is e raised to the natural log of 2 to the u? The natural log of 2 to the u is the power that you have to", + "qid": "C5Lbjbyr1t4_361" + }, + { + "Q": "At 5:00, why does the integral of e^(au) become 1/a * e^(au)? Where did the 1/a come from? Thanks!", + "A": "The chain rule, I presume would be the answer, since the expression uses a function of a function ( the thing you wrote up). Hope that helps :)", + "video_name": "C5Lbjbyr1t4", + "timestamps": [ + 300 + ], + "3min_transcript": "So let's see. How can we redefine this right here? Well, 2, 2 is equal to what? 2 is the same thing as e to the natural log of 2, right? The natural log of 2 is the power you have to raise e to to get 2. So if you raise e to that power, you're, of course going to get 2. This is actually the definition of really, the natural log. You raise e to the natural log of 2, you're going to get 2. So let's rewrite this, using this-- I guess we could call this this rewrite or-- I don't want to call it quite a substitution. It's just a different way of writing the number 2. So this will be equal to, instead of writing the number 2, I could write e to the natural log of 2. And all of that to the u du. And now what is this equal to? Well, if I take something to an exponent, and then to another product of those exponents. So this is equal to, let me switch colors, this is equal to the integral of e, to the u, e to the, let me write it this way. e to the natural log of 2 times u. I'm just multiplying these two exponents. I raise something to something, then raise it again, we know from our exponent rules, it's just a product of those two exponents. du. Now, this is just a constant factor, right here. This could be, you know, this could just be some number. We could use a calculator to figure out what this is. We could set this equal to a. But we know in general that the integral, this is pretty straightforward, we've now put it in this form. The antiderivative of e to the au, du, is just 1 over a e to the au. This comes from this definition up here, and of course plus If we take the derivative of this, we take the derivative of the inside, which is just going to be a. We multiply that times the one over a, it cancels out, and we're just left with e to the au. So this definitely works out. So the antiderivative of this thing right here is going to be equal to 1 over our a, it's going to be 1 over our constant term, 1 over the natural log of 2 times our whole expression, e e. And I'm going to do something. This is just some number times u, so I can write it as u times some number. And I'm just doing that to put in a form that might help us simplify it a little bit. So it's e to the u times the natural log of 2, right? All I did, is I swapped this order. I could have written this as e to the natural log of 2 times u. If this an a, a times u is the same thing as u times a. Plus c. So this is our answer, but we have to kind of reverse", + "qid": "C5Lbjbyr1t4_300" + }, + { + "Q": "at 0:42,what does reciprical mean?", + "A": "The reciprocal of a fraction is simply the fraction flipped over. For example, the reciprocal of 6/9 is 9/6.", + "video_name": "yb7lVnY_VCY", + "timestamps": [ + 42 + ], + "3min_transcript": "Tommy is studying for final exams this weekend. He will spend 1/5 of the weekend studying. What fraction of the weekend will he spend studying for each of his 4 subjects if he spends the same amount of time studying for each subject? So the total amount of time he's going to spend studying this weekend is 1/5 of the weekend. And he has to divide that into 4 equal sections. And he's going to spend that much time on each subject. So he's going to divide this by 4. Now, we've already seen that dividing by a number is the same thing as multiplying by its reciprocal. You might say, hey, well, what's the reciprocal of 4? You just have to remind yourself that 4 is the same thing as 4/1. So 1/5 divided by 4/1 is the same thing as 1/5 times 1/4. And you could also view this as 1/4 of 1/5 or 1/5 But here we multiply our numerators to get 1. And then we multiply our denominators, 4 times 5 is 20. So you get 1/20 of the weekend will be spent studying for each subject. Now, let's also try to think about this visually. Let's imagine that this is his entire weekend. And I've divided it into 5 equal sections. And so we already know that the total amount of his weekend spent studying is 1/5. So that's the total amount studying for the weekend is 1/5. Now, he has to divide this into 4 equals section. So let's do that. He's got four subjects, and he's going to spend the same amount of time on each of the 4 subjects. So he's going to divide this into 4 equal sections. So how much time does he spend on one subject? Well, in each subject, that would be this little area that I'm doing in yellow right over here. And what is that? Well, that's 1 over-- how many equal sections are there Well, I've just drawn out the grid. You had 5 rows, and now you have 4 columns. So 5 rows times 4 columns, you have 20 equal sections. So once again, looking at it visually, he's spending 1/20 of his weekend on each of the 4 subjects. And then if you do this for 4 subjects, that means that in this whole weekend, 1/5 will be spent studying. But the question that they're asking, he's spending 1/20 of the weekend on each subject.", + "qid": "yb7lVnY_VCY_42" + }, + { + "Q": "1:15 what does he mean", + "A": "He means that if the weekend was a square split into 20 parts, one of those parts would be spent studying for a single subject (math, science, english, history, etc)", + "video_name": "yb7lVnY_VCY", + "timestamps": [ + 75 + ], + "3min_transcript": "Tommy is studying for final exams this weekend. He will spend 1/5 of the weekend studying. What fraction of the weekend will he spend studying for each of his 4 subjects if he spends the same amount of time studying for each subject? So the total amount of time he's going to spend studying this weekend is 1/5 of the weekend. And he has to divide that into 4 equal sections. And he's going to spend that much time on each subject. So he's going to divide this by 4. Now, we've already seen that dividing by a number is the same thing as multiplying by its reciprocal. You might say, hey, well, what's the reciprocal of 4? You just have to remind yourself that 4 is the same thing as 4/1. So 1/5 divided by 4/1 is the same thing as 1/5 times 1/4. And you could also view this as 1/4 of 1/5 or 1/5 But here we multiply our numerators to get 1. And then we multiply our denominators, 4 times 5 is 20. So you get 1/20 of the weekend will be spent studying for each subject. Now, let's also try to think about this visually. Let's imagine that this is his entire weekend. And I've divided it into 5 equal sections. And so we already know that the total amount of his weekend spent studying is 1/5. So that's the total amount studying for the weekend is 1/5. Now, he has to divide this into 4 equals section. So let's do that. He's got four subjects, and he's going to spend the same amount of time on each of the 4 subjects. So he's going to divide this into 4 equal sections. So how much time does he spend on one subject? Well, in each subject, that would be this little area that I'm doing in yellow right over here. And what is that? Well, that's 1 over-- how many equal sections are there Well, I've just drawn out the grid. You had 5 rows, and now you have 4 columns. So 5 rows times 4 columns, you have 20 equal sections. So once again, looking at it visually, he's spending 1/20 of his weekend on each of the 4 subjects. And then if you do this for 4 subjects, that means that in this whole weekend, 1/5 will be spent studying. But the question that they're asking, he's spending 1/20 of the weekend on each subject.", + "qid": "yb7lVnY_VCY_75" + }, + { + "Q": "In 4:29, he puts (-1/3) -3x= 11 (-1/3) . What I'm confused about, is he taught us in I believe Linear Equations level 2, that you should do it the other way around. Example: (-3/1) -3x=11 (-3/1) Am I making sense lol? I mean I understand that if you do it out, it won't work but still its a bit confusing since it contradicts what I learned earlier.", + "A": "You make sense, except you probably confused yourself Sal doesn t take the recipricol of 1/2 he multiples 1/2 Hope this helps!", + "video_name": "Zn-GbH2S0Dk", + "timestamps": [ + 269 + ], + "3min_transcript": "the reciprocal of 10, which is the coefficient on x, times 1 over 10. You could also, some people would say, well, we're just dividing both side by 10 which is essential what we're doing. If you divide by 10, that's the same thing as multiplying by 1 over 10. Well, anyway, the left-hand side, 1 over 10 times 10. Well, that equals 1, so we're just left with x equals negative 8 over 10. And that can be reduced further. They both share the common factor 2. So you divide by 2. So it's minus 4 over 5. I think that's right, assuming that I didn't make any careless mistakes. Let's do another problem. Let's say I had 5, that's a 5x minus 3 minus And in general if you wanna work this out before I give you how I do it that now's a good time to actually pause the video. And you could, you could try to work it out and then, play it again and, and see what I have to say about it. But assuming you wanna hear it, let me go and do it. So let's do the same thing. We, first of all, we can merge these two Xs on the left-hand side. Remember, you can't add the 5 and the 3 because the 3 is just a constant term while the 5 is 5 times x. But the 5 times x and the negative 7 times actually can merge. So 5, you just add the coefficient. So, it's 5 and negative 7. So, that becomes negative 2x minus 3 is equal to x plus 8. Now, if we wanna take this x that's on the right-hand side and put it over the left-hand side, we can just subtract x from both sides. to, these two Xs cancel out, is equal to 8. Now, we can just add 3 to both sides to get rid of that constant term 3 on left hand-side. These two 3's will cancel out. And you get minus 3x is equal to 11. Now, you just multiply both sides by negative one-third. And once again, this is just the same thing as dividing both sides by negative 3. And you get x equals negative 11 over 3. Actually let's, let's, just for fun, let's check this just to see. And the cool thing about algebra is if you have enough time, you can always make sure you got the right answer. So we have 5x, so we have 5 times negative 11 over 3. So that's, I'm just, I'm just gonna take this and substitute it back into the original equation.", + "qid": "Zn-GbH2S0Dk_269" + }, + { + "Q": "Why do you have to change the sign > to ?(at 5:13)", + "A": "I use a simple example whenever I can. Say 10>0. If you change the sign on 10, then -10<0. The same exact change happens if there are multiple numbers, as long as you change all of the signs. BTW, if you have numbers on both sides of inequality, say, 10>9, you just change signs on both sides and flip the arrow. -10<-9. Hope this helps.", + "video_name": "xdiBjypYFRQ", + "timestamps": [ + 313 + ], + "3min_transcript": "So just visually looking at it, what x values make this true? Well, this is true whenever x is less than minus 3, right, or whenever x is greater than 2. Because when x is greater than 2, f of x is greater than 0, and when x is less than negative 3, f of x is greater than 0. So we would say the solution to this quadratic inequality, and we pretty much solved this visually, is x is less than minus 3, or x is greater than 2. And you could test it out. You could try out the number minus 4, and you should get f of x being greater than 0. You could try it out here. Or you could try the number 3 and make sure that this works. try out the number 0 and make sure that 0 doesn't work, right, because 0 is between the two roots. It actually turns out that when x is equal to 0, f of x is minus 6, which is definitely less than 0. So I think this will give you a visual intuition of what this quadratic inequality means. Now with that visual intuition in the back of your mind, let's do some more problems and maybe we won't have to go through the exercise of drawing it, but maybe I will draw it just to make sure that the point hits home. Let me give you a slightly trickier problem. Let's say I had minus x squared minus 3x plus 28, let me say, is greater than 0. Well I want to get rid of this negative sign in front of the x squared. I just don't like it there because it makes it look more confusing to factor. I'm going to multiply everything by negative 1. I get x squared plus 3x minus 28, and when you multiply or to swap the sign. So this is now going to be less than 0. And if we were to factor this, we get x plus 7 times x minus 4 is less than 0. So if this was equal to 0, we would know that the two roots of this function -- let's define the function f of x -- let's define the function as f of x is equal to -- well we can define it as this or this because they're the same thing. But for simplicity let's define it as x plus 7 times x minus 4. That's f of x, right? Well, after factoring it, we know that the roots of this, the roots are x is equal to minus 7, and x is equal to 4.", + "qid": "xdiBjypYFRQ_313" + }, + { + "Q": "At about 4:10. Why do the negatives cancel out. Because wouldn't it be -4/6??", + "A": "Review your sign rules. A negative divided by a negative = a positive. So, -4 / (-6) = +4/6", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 250 + ], + "3min_transcript": "Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is I went from 1 to negative 3, that's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well I'm going from this point, or from this x value, all the way-- let me do that in a different color-- all the way back like this. So I'm going to the left, so it's going to be a negative change in x, and I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. And you can even see I started it at x is equal to 3, and I went all the way to x is equal to negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out and what's 4 over 6? So it's the same value, you just have to be consistent. If this is my start point, I went down 4, and then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4, so it would be a change in y would be 4, and then my change in x would be 6. And either way, once again, change in y over change in x is going to be 4 over 6, 2/3. So no matter which point you choose, as long as you kind of think about it in a consistent way, you're going to get the same value for slope.", + "qid": "R948Tsyq4vA_250" + }, + { + "Q": "at 0:55 when you are talking about the starting point, what if a certain graph/problem does not give you a certain starting point. Would you start at (0,0)", + "A": "A problem would not be worded like that, it would be more clear.", + "video_name": "R948Tsyq4vA", + "timestamps": [ + 55 + ], + "3min_transcript": "Find the slope of the line in the graph. And just as a bit of a review, slope is just telling us how steep a line is. And the best way to view it, slope is equal to change in y over change in x. And for a line, this will always be constant. And sometimes you might see it written like this: you might see this triangle, that's a capital delta, that means change in, change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here-- let me do it in a more vibrant color-- so let's say we start at that point right there. And we want to go to another point that's pretty that point right there. We could literally pick any two points on this line. I'm just picking ones that are nice integer coordinates, so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I can just count it out. I went 1 steps, 2 steps, 3 steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this, change in x, delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1, or you could just say 1, 2. So my change in y, is equal to positive 2. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick-- let me clear this out-- let's say I didn't pick those two points, let me pick some other points, and I'll even go in a different direction. I want to show you that you're going to get the same answer. Let's say I've used this as my starting point, and I want to go all the way over there. Well, let's think about the change in y first. So the change in y, I'm going down by how many units? 1, 2, 3, 4 units, so my change in y, in this example, is", + "qid": "R948Tsyq4vA_55" + }, + { + "Q": "After looking at 5:08, does that mean the slope of a graph basically is a rate of change in a table and graph? In a nutshell, the rate of change and the slope of a graph is the same, right?", + "A": "Yes you are correct. Adding on to that, it can really be applied to Physics. For example, if you found the slope for a velocity over time graph, you get the velocity. See, isnt that cool? like how everything is connected in this universe.", + "video_name": "MeU-KzdCBps", + "timestamps": [ + 308 + ], + "3min_transcript": "divided by increase in horizontal, this is what mathematicians use to describe the steepness of lines. And this is called the slope. So this is called the slope of a line. And you're probably familiar with the notion of the word slope being used for a ski slope, and that's because a ski slope has a certain inclination. It could have a steep slope or a shallow slope. So slope is a measure for how steep something is. And the convention is, is we measure the increase in vertical for a given in increase in horizontal. So six two over one is equal to six over three is equal to two, this is equal to the slope of this magenta line. So let me write this down. So this slope right over here, the slope of that line, is going to be equal to two. And one way to interpret that, for whatever amount you increase in the horizontal direction, you're going to increase twice as much in the vertical direction. What would be the slope of the blue line? Well, let me rewrite another way that you'll typically see the definition of slope. And this is just the convention that mathematicians have defined for slope but it's a valuable one. What is are is our change in vertical for a given change in horizontal? And I'll introduce a new notation for you. So, change in vertical, and in this coordinate, the vertical is our Y coordinate. divided by our change in horizontal. And X is our horizontal coordinate in this coordinate plane right over here. So wait, you said change in but then you drew this triangle. Well this is the Greek letter delta. This is the Greek letter delta. And it's a math symbol used to represent change in. So that's delta, delta. And it literally means, change in Y, change in Y, change in X. So if we want to find the slope of the blue line, we just have to say, well how much does Y change for a given change in X? So, the slope of the blue line. So let's see, let me do it this way. Let's just start at some point here. And let's say my X changes by two so my delta X is equal to positive two. What's my delta Y going to be? What's going to be my change in Y? Well, if I go by the right by two, to get back on the line, I'll have to increase my Y by two. So my change in Y is also going to be plus two. So the slope of this blue line, the slope of the blue line, which is change in Y over change in X. We just saw that when our change in X is positive two, our change in Y is also positive two. So our slope is two divided by two, which is equal to one.", + "qid": "MeU-KzdCBps_308" + }, + { + "Q": "At 1:44 how is it a ray without a arrow on the other end?", + "A": "Think of it with everyday things. Take for example sun rays. they go in one direction until they hit an object (us, the atmosphere or a mirror ect.). Do they have arrows? As long as one line looks like it is going to continue on and on, they it should be considered a ray.", + "video_name": "DkZnevdbf0A", + "timestamps": [ + 104 + ], + "3min_transcript": "Any pair of points can be connected by a line segment. That's right. Connect two pairs of black points in a way that creates two parallel line segments. So let's see if we can do that. So I could create one segment that connects this point to this point and then another one that connects this point to this point. And they look pretty parallel. In fact, I think this is the right answer. If we did it another way, if we had connected that point to that point and this point to this point, then it wouldn't look so parallel. These clearly, if they were to keep going, they would intersect at some point. So let me set it back up the way I did it the first time. Let me make these two points parallel. And these are line segments because they have two end points. They each have two end points. And they continue forever. Well they don't continue forever. They continue forever in no directions, in zero directions. If it was a ray, it would continue forever If it's a line, it continues forever In fact, it wouldn't have end points because it would just continue forever in both of these directions. Let's do one more. Drag the ray so it has an endpoint at A, so we want to make its endpoint at A where the ray terminates and goes through one of the other black points. The ray should also be parallel to the pink line. So I have two options. I could make it go through this black point, but it's clearly not parallel. In fact, it looks perpendicular here. So let's try to make it go through this point. Well, yes, when I do that, it does indeed look like my ray is parallel to the pink line. And this is a ray because it has one endpoint. This is where the ray terminates. It's an endpoint. It literally ends there. And it continues forever in one direction. In this case, the direction is to the right. It continues forever to the right. So it continues forever in one-- continues forever in one direction.", + "qid": "DkZnevdbf0A_104" + }, + { + "Q": "at 5:28 , I dont understand why Sal put +(-14). it is so confusing.", + "A": "He meant 6 plus negative 14, which is 6 minus 14. Hopefully this helped you.", + "video_name": "-4bTgmmWI9k", + "timestamps": [ + 328 + ], + "3min_transcript": "Our goal is four right over here. So this is one, two, three, four, get to positive four. That's positive four right there. We're starting at negative two... Let me do this in a different color. We're starting at negative two. We're saying four is the same thing as negative two, negative two is right over here, negative two plus some amount. And it's clear we're gonna be moving to the right by, let's see, we're gonna move to the right by one, two, three, four, five, six. So we moved to the right by six. So we added six. So negative two plus six is equal to four. This is fascinating. Actually, let's just do several more of these, I can't stop. (laughs) Alright. So let's say we wanted to figure out, Six plus blank is equal to negative eight. Like always, try to pause the video and figure out what this blank is going to be. Let me throw my number line back here. So my number line. And one way to think about it is I am starting at six. So it's five, this is six right over here. And I'm gonna add something to get to negative eight. To get to negative eight this is negative five, negative six, negative seven, negative eight. I want to get right over here, I want to get to negative eight. So what do I have to do to get from six to negative eight? To go from six to negative eight. We're clear I have to go to the left on the number line. And how much do I have to go to the left? Well, let's count it. I have to go one, two, three, four, five, six, seven, eight, nine, So going 14 to the left, you could say that I just subtracted 14. And if we phrase it as six plus what is equal to negative eight? Well, six plus negative 14. If this said six minus something I could have just said six minus 14, but if it's saying six plus what it's going to be six plus negative 14.", + "qid": "-4bTgmmWI9k_328" + }, + { + "Q": "what is a checker board pattern? Sal mentions it at 0:20", + "A": "A checkerboard pattern is when every other square is black or white both in the horizontal and vertical directions. With numbers instead of colors, it could look like this: 1 0 1 0 1 0 1 0 1", + "video_name": "u00I3MCrspU", + "timestamps": [ + 20 + ], + "3min_transcript": "As a hint, I will take the determinant of another 3 by 3 matrix. But it's the exact same process for the 3 by 3 matrix that you're trying to find the determinant of. So here is matrix A. Here, it's these digits. This is a 3 by 3 matrix. And now let's evaluate its determinant. So what we have to remember is a checkerboard pattern when we think of 3 by 3 matrices: positive, negative, positive. So first we're going to take positive 1 times 4. So we could just write plus 4 times 4, the determinant of 4 submatrix. And when you say, what's the submatrix? Well, get rid of the column for that digit, and the row, and then the submatrix is what's left over. So we'll take the determinant of its submatrix. So it's 5, 3, 0, 0. Then we move on to the second item in this row, in this top row. But the checkerboard pattern says we're going to take the negative of it. So it's going to be negative of negative 1-- times the determinant of its submatrix. You get rid of this row, and this column. You're left with 4, 3, negative 2, 0. And then finally, you have positive again. Positive times 1. This 1 right over here. Let me put the positive in that same blue color. So positive 1, or plus 1 or positive 1 times 1. Really the negative is where it got a little confusing on this middle term. But positive 1 times 1 times the determinant of its submatrix. So it's submatrix is this right over here. You get rid of the row, get rid of the column 4, 5, negative 2, 0. So the determinant right over here is going to be 5 times 0 minus 3 times 0. And all of that is going to be multiplied times 4. Well this is going to be 0 minus 0. So this is all just a 0. So 4 times 0 is just a 0. So this all simplifies to 0. Now let's do this term. We get negative negative 1. So that's positive 1. So let me just make these positive. Positive 1, or we could just write plus. Let me just write it here. So positive 1 times 4 times 0 is 0. So 4 times 0 minus 3 times negative 2. 3 times negative 2 is negative 6. So you have 4-- oh, sorry, you have 0 minus negative 6, which is positive 6. Positive 6 times 1 is just 6. So you have plus 6. And then finally you have this last determinant.", + "qid": "u00I3MCrspU_20" + }, + { + "Q": "Wouldn't it be 1/2 +pi^2/8? if not, what happened to the +1 at 10:23 ?could someone please help, because I don't understand where that 1 went.", + "A": "The 1 was multiplied by the sin(t)*cos(t) function inside the integral. The square root of negative sine squared plus cosine squared is one.", + "video_name": "uXjQ8yc9Pdg", + "timestamps": [ + 623 + ], + "3min_transcript": "curtain that has our curve here as kind of its base, and has this function, this surface as it's ceiling. So we go back down here, and let me rewrite this whole thing. So this becomes the integral from t is equal to o to t is equal to pi over 2-- I don't like this color --of cosine of t, sine of t, cosine times sine-- that's just the xy --times ds, which is this expression right here. And now we can write this as-- I'll go switch back to that color I don't like --the derivative of x with respect to t is minus sine of t, and we're going to square it, plus the derivative of y with respect to t, that's cosine of t, and we're going to square it-- let me make my radical a little bit bigger --and then all of that times dt. you realize that this right here, and when you take a negative number and you squared it, this is the same thing. Let me rewrite, do this in the side right here. Minus sine of t squared plus the cosine of t squared, this is equivalent to sine of t squared plus cosine of t squared. You lose the sign information when you square something; it just becomes a positive. So these two things are equivalent. And this is the most basic trig identity. This comes straight out of the unit circle definition: sine squared plus cosine squared, this is just equal to 1. So all this stuff under the radical sign is just equal to 1. And we're taking the square root of 1 which is just 1. So all of this stuff right here will just become 1. And so this whole crazy integral simplifies a good bit pi over 2 of-- and I'm going to switch these around just because it will make it a little easier in the next step --of sine of t times cosine of t, dt. All I did, this whole thing equals 1, got rid of it, and I just switched the order of that. It'll make the next up a little bit easier to explain. Now this integral-- You say sine times cosine, what's the antiderivative of that? And the first thing you should recognize is, hey, I have a function or an expression here, and I have its derivative. The derivative of sine is cosine of t. So you might be able to a u substitution in your head; it's a good skill to be able to do in your head. But I'll do it very explicitly here. So if you have something that's derivative, you define that something as u. So you say u is equal to sine of t and then du, dt, the derivative of u with respect to t is equal to cosine of t.", + "qid": "uXjQ8yc9Pdg_623" + }, + { + "Q": "At 19:40, it is said that \"2x -y -z +3x\" must be equal to 0 in order for b to be valid.\nIsn't it that we have then:\nif b is valid => 5x -y -z = 0\n(i.e. not yet \"<=>\". Necessary condition only).\nThen b is in the plan of equation 5x -y -z = 0, in other words:\n\"the plan of equation 5x -y -z = 0\" is included in C(A). Finally, as C(A) is a plan (dim C(A) = 2), then C(A) = \"the plan of equation 5x -y -z = 0\". (now we have \"<=>\" . b is valid <=> 5x -y -z = 0 )", + "A": "He states Ax = b at 14:00 ish, and asks what are all the possible b s, which form C(A). So if you find out that Ax forms a plane, that equality gives you equivalence .. Ax forms a plane <=> All the b s form a plane, because LHS = RHS (or in other words the validity of b is given). At least that s how I d interpret it.", + "video_name": "EGNlXtjYABw", + "timestamps": [ + 1180 + ], + "3min_transcript": "So let me from the get go try to zero out this third row. And the best way to zero out this third row is to just replace the third row. So the first row-- well, I won't even write the first row. The second row is 0, 1, minus 2, minus 1, and 2x minus y. I'm not even going to worry about the first row right now. But let's replace the third row, just in our attempt to go into reduced row echelon form. Let's replace it with the second row minus the third row. So you get 2x minus y minus z plus 3x. I just took this minus this. So minus z plus 3x. So 0 minus 0 is 0. 1 minus 1 is 0. Minus 2 minus minus 2 is 0, and that's also 0. So we're only going to have a valid solution to Ax equals b What happens if he's not equal to zero? Then we're going to have a bunch of zeroes equaling some number, which tells us that there's no solution. So if I pick a b where this guy does not equal zero, then I'll have no solution. If this guy equals 5, if I pick x, y, and z's such as that this expression is equal to 5, then Ax equal to b will have no solution, because it will have 0 is equal to 5. So this has to equal 0. So 2x minus y minus z plus 3x must be equal to 0 in order for b to be valid, in order for b to be a member of the column space of A, in order for it to be a valid vector that Ax can become, or the product A times x can become for some x. So what does this equal to? If we add the 2x plus the 3x, I get 5x minus y minus z is figured out the basis vectors. We said oh, you know what? The basis vectors, they have to be in the column space themselves by definition. So let me find a normal vector to them both by taking the cross product. I did that, and I said the cross product times any valid vector in our space minus one of the basis vectors has to be equal to zero, and then I got this equation. Or we could have done it the other way. We could've actually literally solved this equation setting our b equal to this. We said what b's will give us a valid solution? And our only valid solution will be obtained when this guy has to be equal to zero, because the rest of his row became zero. And when we set that equal to zero, we got the exact same equation. So, hopefully, you found this to be mildly satisfying, because we were able to tackle the same problem from two", + "qid": "EGNlXtjYABw_1180" + }, + { + "Q": "I'm so confused.. at about 11:00 Sal decided to factor out 27\u00c2\u00b712x^2 - 4x^6 = 0 to 4x^2(27\u00c2\u00b73-x^4) = 0 .. When I was doing it on my own I multiplied the constants and got 324x^2-4x^6 = 0, factoring out to 4x^2(324-x^4) = 0... Is this an example of where BEDMAS is really important?", + "A": "You forgot to divide the 324 in your final equation by 4. :-)", + "video_name": "zC_dTaEY2AY", + "timestamps": [ + 660 + ], + "3min_transcript": "But if the derivative is equal to 0, the second derivative is equal to 0, you cannot assume that is an inflection point. So what we're going to do is, we're going to find all of the point at which this is true, and then see if we actually do have a sign change in the second derivative of that point, and only if you have a sign change, then you can say it's an inflection point. So let's see if we can do that. So just because a second derivative is 0, that by itself does not tell you it's an inflection point. It has to have a second derivative of 0, and when you go above or below that x, the second derivative has to actually change signs. Only then. So we can say, if f prime changes signs around x, then we can say that x is an inflection. And if it's changing signs around x, then it's definitely if it's negative before x, has to be positive after x,or if it's positive before x, has to be negative after x. So let's test that out. So the first thing we need to do is find these candidate points. Remember, the candidate points are where the second derivative is equal to 0. We're going to find those points, and then see if this is true, that the sign actually changes. We want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where this our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12x squared minus 4x to the sixth is equal to 0. Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So 4x squared. Now we'll have 27 times, if we factor 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the fourth is equal to 0. So the x's that will make this equal to 0 will satisfy either, I'll switch colors, either 4x squared is equal to 0, or, now 27 times 3, I can do that in my head. That's 81. 20 times 3 is 60, 7 times 3 is 21, 60 plus 21is 81. Or 81 minus x to the fourth is equal to 0. Any x that satisfies either of these will make this entire expression equal 0. Because if this thing is 0, the whole thing is If this thing is 0, the whole thing is going to be equal to 0. Let me be clear, this is 81 right there. So let's solve this. This is going to be 0 when x is equal to 0, itself.", + "qid": "zC_dTaEY2AY_660" + }, + { + "Q": "At 3:41, I thought that you put dots over the top of repeating decimals. Can you do both?", + "A": "Some students learn to write repeating decimals by putting dots over the repeating numbers, but in this case Sal puts a line over the repeating numbers. However If you want to use Sal s method you can but if you were taught to put dots, its better you do that and yes, you CAN use both lines or dots for your answer. Hope that helped !!", + "video_name": "Y2-tz27nKoQ", + "timestamps": [ + 221 + ], + "3min_transcript": "Well, we could take 100 from the 100's place, and make it 10 10's. And then we could take 1 of those 10's from the 10's place and turn it into 10 1's. And so 9 10's minus 8 10's is equal to 1 10. And then 10 -1 is 9. So it's equal to 19. You probably \u2013 You might have been able to do that in your head. And then we have \u2013 And I see something interesting here \u2013 because when we bring down our next 0, we see 190 again. We saw 190 up here. But let's just keep going. So 27 goes into 190 \u2013 And we already played this game. It goes into it 7 times. 7 \u00d7 27 \u2013 we already figured out \u2013 was 189. We subtracted. We had a remainder of 1. Then we brought down another 0. We said 27 goes into 10 0 times. 0 \u00d7 27 is 0. Subtract. Then you have \u2013 but we've got to bring down another 0. So you have 27, which goes into 100 \u2013 (We've already done this.) \u20133 times. So you see something happening here. It's 0.703703. And we're just going to keep repeating 703. This is going to be equal to 0.703703703703 \u2013 on and on and on forever. So the notation for representing a repeating decimal like this is to write the numbers that repeat \u2013 in this case 7, 0, and 3 \u2013 and then you put a line over all of to indicate that they repeat. So you put a line over the 7, the 0, and the 3, which means that the 703 will keep repeating on and on and on. So let's actually input it into the answer box now. So it's 0.703703. the first six digits of the decimal in your answer. And they don't tell us to round or approximate \u2013 because, obviously, if they said to round to that smallest, sixth decimal place, then you would round up because the next digit is a 7. But they don't ask us to round. They just say, \"Include only the first six digits of the decimal in your answer.\" So that should do the trick. And it did.", + "qid": "Y2-tz27nKoQ_221" + }, + { + "Q": "0:42, 3 doesn't go into 11 3 times. or am i missing something?", + "A": "You are missing something. Since 117 (if I add digits, they =9, so I know that it is divisible by three and 9). When he says 3 goes into 11 3 times, he then multiplies and subtracts (11-9) and drops down the remainder of 2 and divides 27 by 3", + "video_name": "cw3mp8oNASk", + "timestamps": [ + 42 + ], + "3min_transcript": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can.", + "qid": "cw3mp8oNASk_42" + }, + { + "Q": "At 6:13, Sal got 2 differant answers for 2 sides of a square. How did he get the 2 answers?", + "A": "To get the dimension on the left side of the large rectangle, he added the lengths of a side of each square on the left. To get the dimension on the right side of the large rectangle, he added the lengths of a side of each square on the right. Since we know opposite sides of a rectangle have equal length, we can set up an equation (13x+7y=8x+9y) to solve for the ratio of x to y (x=2/5*y).", + "video_name": "1uWZNW5PF-s", + "timestamps": [ + 373 + ], + "3min_transcript": "5x plus 3y is going to be that entire length right over there. And we can also go to this side right over here where we have this length-- let me do that same color. This length is 3x plus 2y. This is x plus y. And this is y. So if you add 3x plus 2y plus x plus y plus y, you get 4x plus-- what is that-- 4y, right? 2y, 3y, 4y. And then we can express this character's dimensions in terms of x and y because this is going to be 5x plus 3y. Then you're going to have 2x plus y. And then you're going to have x. So you add the x's together. 5x plus 2x is 7x, plus x is 8x. And then you add the y's together, 3y plus y, and then you don't have a y there. So that's going to be plus 4y. And then finally, we have this square right over here. Its dimensions are going to be the y plus the 4x plus 4y. So that's 4x plus 5y. And then if we think about the dimensions of this actual rectangle over here, if we think about its height right over there, that's going to be 5x plus 3y plus 8x plus 4y. So 5 plus 8 is 13. So it's 13x plus 3 plus 4 is 7y. So that's its height. But we can also think about its height by going on the other side of it. And maybe this will give us some useful constraints because this is going to have to be the same length as this over here. And so if we add 4x plus 4x, we get 8x. So these are going to have to be equal to each other, so that's an interesting constraint. So we have 13x plus 7y is going to have to equal 8x plus 9y. And we can simplify this. If you subtract 8x from both sides, you get 5x. And if you subtract 7y from both sides, you get 5x is equal to 2y. Or you could say x is equal to 2/5 y. In order for these to show up as integers, we have to pick integers here. But let's see if we have any other interesting constraints if we look at the bottom and the top of this, if this gives us any more information. So if we add 5x plus 3y plus 3x plus 2y plus 4x plus 4y,", + "qid": "1uWZNW5PF-s_373" + }, + { + "Q": "At 1:10, I don't get why do you start multiplying (-3x -2). It's getting me confused.", + "A": "If you have a variable, like X in a fraction, you usually want to get that variable isolated. So in order to get X by itself you have to multiply both sides by the denominator. When that happens, the side with the fraction has its denominator cancelled out, but what you do to one side, you must do to the other. So that is why he had to multiply both sides by (-3x-2). Hope this helped!!", + "video_name": "PPvd4X3Wv5I", + "timestamps": [ + 70 + ], + "3min_transcript": "So we have 14x plus 4 over negative 3x minus 2 is equal to 8. And I'll give you a few moments to see if you can tackle it on your own. So this equation right here, at first it doesn't look like a straightforward linear equation. We have one expression on top of another expression. But as we'll see, we can simplify this to turn it into a linear equation. So the first thing that I want to do is, I don't like this negative 3x plus 2 sitting here in the denominator, it makes me stressed. So I want to multiply both sides of this equation times negative 3x minus 2. What does that do for us? Well, on the left-hand side, you have this negative 3x minus 2, it's going to be over negative 3x minus 2, they will cancel out. And so you're left with, on the left-hand side, your 14x plus 4. have to multiply 8 times negative 3x minus 2. So you are left with-- well, 8 times negative 3x is negative 24x, and then 8 times negative 2 is negative 16. And there you have it. We have simplified this to just a traditional linear equation. We've got variables on both sides, so we can just keep simplifying. So the first thing I want to do, let's just say we want to put all of our x terms on the left-hand side. So I want to get rid of this negative 24x right over here. So the best way to do that, I'm going to add 24x to the right-hand side. I can't just do it to the right-hand side, I have to also do it to the left-hand side. And so I am left with, on the left-hand side, 14x plus 24x is 38x. And then I have the plus 4-- Is equal to, well, negative 24x plus 24x. are left with just the negative 16. Now we just have to get rid of this 4 here. Let's subtract that 4 from both sides. And we are left with-- and this is the home stretch now-- we are left with 38x is equal to negative 16 minus 4 is negative 20. And so we can divide both sides of this equation by 38. And we are left with x is equal to negative 20 over 38, which can be simplified further. Both the numerator the denominator is divisible by 2. So let's divide the numerator and denominator by 2, and we get negative 10 over 19. x is equal to negative 10 over 19, and we are done. I encourage you to validate this for yourself. It's a little bit of a hairy number right over here,", + "qid": "PPvd4X3Wv5I_70" + }, + { + "Q": "At 4:04: Why can you rewrite d/Dt [pi*r^2] as pi*d/dt [r(t)?", + "A": "Because pi is a constant, and you can do that with constants when you are taking derivatives.", + "video_name": "kQF9pOqmS0U", + "timestamps": [ + 244 + ], + "3min_transcript": "Well, they say at what rate is the area of the circle growing? So we need to figure out at what rate is the area of the circle-- where a is the area of the circle-- at what rate is this growing? This is what we need to figure out. So what might be useful here is if we can come up with a relationship between the area of the circle and the radius of the circle and maybe take the derivative with respect to time. And we'll have to use a little bit of the chain rule So what is the relationship at any given point in time between the area of the circle and the radius of the circle? Well, this is elementary geometry. The area of a circle is going to be equal to pi times the radius of the circle squared. Now what we want to do is figure out the rate at which the area is changing with respect to time. So why don't we take the derivative of both sides of this with respect to time? And let me give myself a little more real estate. Actually, let me just rewrite what I just had. So pi r squared. I'm going to take the derivative of both sides of this with respect to time. So the derivative with respect to time. I'm not taking the derivative with respect to r, I'm taking the derivative with respect to time. So on the left-hand side right over here, I'm going to have the derivative of our area. Actually, let me just write it in that green color. I'm going to have the derivative of our area with respect to time on the left-hand side. And on the right-hand side, what do I have? Well, if I'm taking the derivative of a constant times something, I can take the constant out. So let me just do that. Pi times the derivative with respect to time of r squared. And to make it a little bit clearer what I'm about to do, why I'm using the chain rule, we're assuming that r is a function of time. If r wasn't a function of time then area wouldn't be a function of time. So instead of just writing r, let me make it explicit I'll write r of t. So it's r of t, which we're squaring. And we want to find the derivative of this with respect to time. And here we just have to apply the chain rule. We're taking the derivative of something squared with respect to that something. So the derivative of that something squared with respect to the something is going to be 2 times that something to the first power. Let me make it clear. This is the derivative of r of t squared with respect to r of t. The derivative of something squared with respect to that something. If it was a derivative of x squared with respect to x, we'd have 2x. If it was the derivative of r of t squared with respect to r of t, it's 2r of t. But this doesn't get us just the derivative with respect to time. This is just the derivative with respect to r of t. The derivative at which this changes with respect to time, we have to multiply this times the rate at which r of t", + "qid": "kQF9pOqmS0U_244" + }, + { + "Q": "at 1:36 why does sal multiply a/b with n/n and m/n with b/b", + "A": "I think he did it so he would have the same denominators for both fractions.", + "video_name": "HKUJkMQsGkM", + "timestamps": [ + 96 + ], + "3min_transcript": "What I want to do in this video is think about whether the product or sums of rational numbers are definitely going to be rational. So let's just first think about the product of rational numbers. So if I have one rational number and-- actually, let me instead of writing out the word rational, let me just represent it as a ratio of two integers. So I have one rational number right over there. I can represent it as a/b. And I'm going to multiply it times another rational number, and I can represent that as a ratio of two integers, m and n. And so what is this product going to be? Well, the numerator, I'm going to have am. I'm going to have a times m. And in the denominator, I'm going to have b times n. Well a is an integer, m is an integer. So you have an integer in the numerator. And b is an integer and n is an integer. So you have an integer in the denominator. So now the product is a ratio of two integers right over here, so the product is also rational. So this thing is also rational. you're going to end up with a rational number. Let's see if the same thing is true for the sum of two rational numbers. So let's say my first rational number is a/b, or can be represented as a/b, and my second rational number can be represented as m/n. Well, how would I add these two? Well, I can find a common denominator, and the easiest one is b times n. So let me multiply this fraction. We multiply this one times n in the numerator and n in the denominator. And let me multiply this one times b in the numerator and b in the denominator. Now we've written them so they have a common denominator of bn. And so this is going to be equal to an plus bm, So b times n, we've just talked about. This is definitely going to be an integer right over here. And then what do we have up here? Well, we have a times n, which is an integer. b times m is another integer. The sum of two integers is going to be an integer. So you have an integer over in an integer. You have the ratio of two integers. So the sum of two rational numbers is going to give you another. So this one right over here was rational, and this one is right over here is rational. So you take the product of two rational numbers, You take the sum of two rational numbers, you get a rational number.", + "qid": "HKUJkMQsGkM_96" + }, + { + "Q": "At 0:55 could b^1 also be b^0?", + "A": "Only if b = 1.", + "video_name": "X6zD3SoN3iY", + "timestamps": [ + 55 + ], + "3min_transcript": "Simplify 25 a to the third and a to the third is being raised to the third power, times b squared and all of that over 5 a squared, b times b squared So we can do this in multiple ways, simplify different parts. What I want to do is simplify this part right over here. a to the third power, and we're raising that to the third power. So this is going to be from the power property of exponents, or the power rule this is going to be the same thing as a to the 3 times 3 power So this over here (let me scroll up a little bit) is going to be equal to a the 3 times 3 power, or a to the ninth power. We could also simplify this b times b squared over here. This b times b squared, that is the same thing as b to the first power remember, b is just b to the first power. So it's b to the first power times b to the second power. So b to the first times b to the second power is just equal to b to the one plus two power, which is equal to b to the third power and then last thing we could simplify, just right off the bat just looking at this: or we could say it's going to give us 5 over 1 if you view it as dividing the numerator and the denominator both by 5. So what does our expression simiply to? We have 5a to the ninth, and then we still have this b squared here, b squared. All of that over a squared times b to the third power... times b to the third power. Now, we can use the quotient property of exponents. You have an a to the ninth Let me use a slightly different color. We have an a to the ninth. over a squared. What's that going to simplify to? Well, that's going to simplify to be the same thing, a to the ninth over a squared, the same thing as a to the nine minus two, which is equal to a to the seventh power. and this will get a little bit interesting here So that simplifies too. So b squared over b to the third is equal to b to the two minus three power, which is equal to b to the negative one power. And we'll leave it alone like that right now. So this whole expression simplifies to It simplifies to: 5 times a to the seventh power (because this simplifies to a to the seventh) a to the seventh, times (the bs right here simplify too) b to the negative one. We could leave it like that, you know, that's pretty simple but we may not want a negative exponent there we just have to remeber that b to the negative one power is the same thing as one over b. Now if we remember that, then we can rewrite this entire expression as, the numerator will have a five and will have a to the seventh 5 a to the seventh. And then the denominator will have the b. So we're multiplying this times one over b.", + "qid": "X6zD3SoN3iY_55" + }, + { + "Q": "when he means the length of vector.. does he mean the magnitude of the vector. is it another word for magnitude or is it completely different things. please explain... thanks (4:44 min)", + "A": "i think it s exactly the same thing", + "video_name": "WNuIhXo39_k", + "timestamps": [ + 284 + ], + "3min_transcript": "", + "qid": "WNuIhXo39_k_284" + }, + { + "Q": "1:58 Why is 3+5i a complex \"number\"?\nIt consists of 2 numbers... Real and Imaginary.\nCouldn't it be called \"complex numbers\"?", + "A": "Z is the complex number, comprised of a Real Part (5) and an Imaginary Part (3i).", + "video_name": "SP-YJe7Vldo", + "timestamps": [ + 118 + ], + "3min_transcript": "Voiceover:Most of your mathematical lives you've been studying real numbers. Real numbers include things like zero, and one, and zero point three repeating, and pi, and e, and I could keep listing real numbers. These are the numbers that you're kind of familiar with. Then we explored something interesting. We explored the notion of what if there was a number that if I squared it I would get negative one. We defined that thing that if we squared it we got negative one, we defined that thing as i. So we defined a whole new class of numbers which you could really view as multiples of the imaginary unit. So imaginary numbers would be i and negative i, and pi times i, and e times i. This might raise another interesting question. What if I combined imaginary and real numbers? What if I had numbers that were essentially sums or differences of real or imaginary numbers? Let's say I call it z, and z tends to be the most used variable when we're talking about what I'm about to talk about, complex numbers. Let's say that z is equal to, is equal to the real number five plus the imaginary number three times i. So this thing right over here we have a real number plus an imaginary number. You might be tempted to add these two things, but you can't. They won't make any sense. These are kind of going in different, we'll think about it visually in a second, but you can't simplify this anymore. You can't add this real number to this imaginary number. A number like this, let me make it clear, that's real and this is imaginary, imaginary. A number like this we call a complex number, a complex number. It has a real part and an imaginary part. or someone will say what's the real part? What's the real part of our complex number, z? Well, that would be the five right over there. Then they might say, \"Well, what's the imaginary part? \"What's the imaginary part of our complex number, z? And then typically the way that this function is defined they really want to know what multiple of i is this imaginary part right over here. In this case it is going to be, it is going to be three. We can visualize this. We can visualize this in two dimensions. Instead of having the traditional two-dimensional Cartesian plane with real numbers on the horizontal and the vertical axis, what we do to plot complex numbers is we on the vertical axis we plot the imaginary part, so that's the imaginary part. On the horizontal axis we plot the real part. We plot the real part just like that.", + "qid": "SP-YJe7Vldo_118" + }, + { + "Q": "I don't understand the end behaviors @4:25", + "A": "End behavior is what the function looks like when it reaches really high or really low values of x.", + "video_name": "tZKzaF28sOk", + "timestamps": [ + 265 + ], + "3min_transcript": "So when we talk about end behavior, we're talking about the idea of what is this function? What does this polynomial do as x becomes really, really, really, really positive and as x becomes really, really, really, really negative? And kind of fully recognizing that some weird things might be happening in the middle. But we just want to think about what happens at extreme values of x. Now obviously for the second degree polynomial nothing really weird happens in the middle. But for a third degree polynomial, we sort of see that some interesting things can start happening in the middle. But the end behavior for third degree polynomial is that if a is greater than 0-- we're starting really small, really low values-- and as a becomes positive, we get to really high values. If a is less than 0 we have the opposite. And these are kind of the two prototypes for polynomials. Because from there we can start thinking about any degree polynomial. So let's just think about the situation of a fourth degree polynomial. plus bx to the third plus cx squared plus dx plus-- I don't want to write e because e has other meanings in mathematics. I'll say plus-- I'm really running out of letters here. I'll just use f, although this isn't the function f. This is just a constant f right over here. So let's just think about what this might look like. Let's think about its end behavior, and we could think about it relative to a second degree polynomial. So its end behavior, if x is really, really, really, really negative, x to the fourth is still going to be positive. And if a is greater than 0 when x is really, really, really negative, we're going to have really, really positive values, just like a second degree. And when x is really positive, same thing. x to the fourth is going to be positive, times a is still going to be positive. So its end behavior is going to look Now, it might do-- in fact it probably will do some funky stuff in between. It might do something that looks kind of like that in between. But we care about the end behavior. I guess you could call the stuff that I've dotted lined in the middle, this is called the non-end behavior, the middle behavior. This will obviously be different than a second degree polynomial. But what happens at the ends will be the same. And the reason why, when you square something, or you raise something to the fourth power, you raise anything to any even power for a very large-- as long as a is greater than 0, for very large positive values, you're going to get positive values. And for very large negative values, you're going to get very large positive values. You take a negative number, raise it to the fourth power, or the second power, you're going to get a positive value. Likewise, if a is less than 0, you're going to have very similar end behavior to this case. For a polynomial where the highest degree", + "qid": "tZKzaF28sOk_265" + }, + { + "Q": "At 3:05, why is it just answer 17 but at 3:31, 5 is 25?", + "A": "He is squaring each number. So at 3:05 he squares the squared root of 17, the square root of 17x the square root of 17 equals 17. The square root of 17 is a number slightly bigger than 4, because 4x4 equals 16, so this is just a little bit more than that. At 3:31 he square 5. 5x5=25 The concept is that if you square each number you can compare the numbers without the radical signs........", + "video_name": "KibTbfkoPTs", + "timestamps": [ + 185, + 211 + ], + "3min_transcript": "Because their squares are not going to be irrational numbers. It's going to be much easier to compare, and then we can order them. Because if we order the squares, then they'll tell us what happens if we order their square roots. What am I talking about? Well, I'm just gonna square each of these. So if I take this to the second power, this is going to be four square roots of two, times four square roots of two. You can change the order of multiplication. That's four times four times the square root of two times the square root of two. Now, four times four is 16. Square root of two times square root of two, well, that's just going to be two. So it's gonna be 16 times two which is equal to 32. Now what about two square roots of three? Well, same idea. Let's square it, let's square it. And i'll do this one a little bit faster. So if we square two square roots of three, times square root of three squared. So it's going to be two squared times the square root of three squared. Well, two squared is going to be four. Square root of three squared is going to be three. So this is going to be equal to 12. That's this thing squared. If this step seems a little bit confusing, if you have the product of two things raised to a power, that's the same thing as raising each of them to that power, and then taking the product. And you can actually see, I worked it out here, why that actually makes sense. Notice when I just changed the order of multiplication you had four times four, or four squared, times square root of two squared, which is going to be two. So let's keep doing that. So what is this value squared? It's gonna be three squared, which is nine, times square root of two squared, which is two. What's the square root of 17 squared? That's just going to be seventeen. Do that in blue. This is just going to be 17. What is three square roots of three squared? It's gonna be three squared, which is nine, times square root of three squared. The square root of three times the square root of three is three. So it's gonna be nine times three, or 27. And what is five squared? This is pretty straightforward. That's going to be 25. So let's order them from least to greatest. Which of them, when I square it, gives me the smallest value? Compare 32 to 12 to 18 to 17 to 27 to 25. 12 is the smallest value. So if their square is the smallest, and these are all positive numbers, then this is going to be the smallest value out of all of them. Let me write that first. Two square roots of three. So I've covered that one.", + "qid": "KibTbfkoPTs_185_211" + }, + { + "Q": "At 3:28, Sal says that he can do multiplication in any order, but thought you had to go from left to right.", + "A": "You can do multiplication in any order because of the commutative property of multiplication, which means that you ll get the same answer no matter what order you put it in.", + "video_name": "XHHYA2Ug9lk", + "timestamps": [ + 208 + ], + "3min_transcript": "and now I'm going to write the dot for times. A times b times c, a times b times c. Now if I told you that these were all positive numbers, then you say, \"OK a times b times c is going \"to be positive.\" \"A times b would be positive, and that times c is positive.\" Now what would happen if I were to tell you that they were all negative numbers. What if a, b, and c were all negative? Well if there were all negative, let me write it that way. Let me actually write, let me write, a, b, and c, a, b, and c, they're all going to be negative. So if that's the case, what is this product going to be equal to? Well you're going to have a negative here, times a negative. So a negative times a negative, a times b, if you do that first, and we can when we multiply these numbers. That's going to give you a positive. But then you're going to multiply that times c. You're going to multiply that times c, which is a negative. So you're going to have a positive times a negative, which is going to be a negative. So this one, if a, b, and c are all less than zero, then the product, a, b, c is going to be less than zero as well. This whole thing is going to be negative. Now, if I did something else. If I said, \"There's other ways \"that I can make the product negative.\" If a is, let's say that a, actually let me just write it this way. Let's say that a is positive, b is negative. B is negative, and c is positive. And c is positive. Well here, positive times a negative, if you do this first. Positive times a negative is going to give you a negative. And then a negative times a positive, different signs, is going to give you a negative. this whole thing is going to be a negative. But let's keep on doing, we said, look if all of them are negative, then this thing would be negative, but that's because I had three numbers here. What if I had four numbers here? What if I had times, What if I had times d here? And if I told you all of these numbers were negative? Let's think about it, if negative, negative, negative, negative. And I can do the multiplication in any order, but I'll just go left to right. A times b, negative times a negative. That would yield a positive. Now if you multiply that product times c, positive times a negative, positive times a negative, positive times a negative that would give you a negative. And then you multiply this negative times this negative, so this whole product, a, b, c, is going to be negative. But then we multiply it times a negative. Well a negative times a negative is going to be a positive. So this whole thing is going to be a positive. And so you're probably seeing a pattern here.", + "qid": "XHHYA2Ug9lk_208" + }, + { + "Q": "i kind of get but killo means 1,000 or 1,000,000 on 00:14", + "A": "Kilo means 1,000.", + "video_name": "9iulv2QvKwo", + "timestamps": [ + 14 + ], + "3min_transcript": "What I want to do in this video is convert this amount of kilometers into meters and centimeters. So we'll first start with the 2 kilometers. And I encourage you to now pause this video and try to do this on your own. Well, the one thing that we know is that a kilometer literally means 1,000 meters. So you could literally view this is as 2 times 1,000 meters. Let me write that down. So this is going to be equal to 2 times 1,000 meters, which is equal to 2,000 meters. If we wanted to convert the 11 kilometers into meters, it's the same thing. 11 kilometers-- this right over here-- means 1,000 meters. So you could think of it as 11 times 1,000 meters. So 11 times 1,000 is going to be 11,000 meters. Now let's convert these distances into centimeters. And here we just have to remember that 1 meter is equal to 100 centimeters. Let me write that down. And that's because the prefix centi means one hundredth. Another way you can write it is that one centimeter is equal to one hundredth of a meter. But here we have a certain number of meters, and each of those meters are going to be equivalent to 100 centimeters. So if we wanted to write 2,000 meters as centimeters, we could say, well, we have 2,000 meters. Each of those are going to be equivalent to 100 centimeters. And so this is going to be equal to 2,000 times 100. Well, that's going to be 2, and since where we have the three zeroes from the 2,000. And then every time you multiply by 10, you're going to add another zero. Or you're going to have another zero at the end of it. And we're going to be at multiplying by 100 is equivalent to multiplying by 10 twice. So we're going to have two more zeroes. So this is going to be 200,000 centimeters. with the kilometers, the 11 kilometers, which are 11,000 meters. And once again, I encourage you to pause the video and try to convert it into centimeters. Well, same idea. You have 11,000 meters, and each of them are equivalent to 100 centimeters. So this is going to be 11, and let's see, we have one, two, three, four, five zeroes. So this gets us to 1,100,000 centimeters.", + "qid": "9iulv2QvKwo_14" + }, + { + "Q": "2:32 how is the opposite and adjacent line of theta can become sin theta and cos theta? i thought sin is like opposite over adjacent something like that..", + "A": "In the unit circle, the hypotenuse always equals 1 (it s the radius of the unit circle). Since sin(\u00ce\u00b8)=opposite/hypotenuse, and the hypotenuse equals 1, you can say that sin(\u00ce\u00b8)=opposite/1, or sin(\u00ce\u00b8)=opposite. It s the same idea for cosine. This only works for the unit circle, though. You can watch the videos on the unit circle if you haven t already.", + "video_name": "Idxeo49szW0", + "timestamps": [ + 152 + ], + "3min_transcript": "me-- He's just saying the tangent of some angle is equal to x. And I just need to figure out what that angle is. So let's do an example. So let's say I were walk up to you on the street. There's a lot of a walking up on a lot of streets. I would write -- And I were to say you what is the arctangent of minus 1? Or I could have equivalently asked you, what is the inverse tangent of minus 1? These are equivalent questions. And what you should do is you should, in your head-- If you don't have this memorized, you should draw the unit circle. Actually let me just do a refresher of what tangent is even asking us. The tangent of theta-- this is just the straight-up, vanilla, non-inverse function tangent --that's equal to the sine of theta over the cosine of theta. And the sine of theta is the y-value on the unit function-- on the unit circle. And so if you draw a line-- Let me draw a little unit circle here. So if I have a unit circle like that. And let's say I'm at some angle. Let's say that's my angle theta. And this is my y-- my coordinates x, y. We know already that the y-value, this is the sine of theta. Let me scroll over here. Sine of theta. And we already know that this x-value is the cosine of theta. So what's the tangent going to be? It's going to be this distance divided by this distance. Or from your algebra I, this might ring a bell, because we're starting at the origin from the point 0, 0. This is our change in y over our change in x. Or it's our rise over run. Or you can kind of view the tangent of theta, or it really The slope. So you could write slope is equal to the tangent of theta. So let's just bear that in mind when we go to our example. If I'm asking you-- and I'll rewrite it here --what is the inverse tangent of minus 1? And I'll keep rewriting it. Or the arctangent of minus 1? I'm saying what angle gives me a slope of minus 1 on the unit circle? So let's draw the unit circle. Let's draw the unit circle like that. Then I have my axes like that. And I want a slope of minus 1. A slope of minus 1 looks like this. If it was like that, it would be slope of plus 1. So what angle is this? So in order to have a slope of minus 1, this distance is the same as this distance. And you might already recognize that this is a right angle.", + "qid": "Idxeo49szW0_152" + }, + { + "Q": "1:16-1:46 he does the translation visually. Is there any way to do it with an equation.", + "A": "Yes, to translate a figure 2 spots to the right you just add 2 to its x value. So E (3+2,3) -> E (5,3).", + "video_name": "XiAoUDfrar0", + "timestamps": [ + 76, + 106 + ], + "3min_transcript": "- [Voiceover] What I hope to introduce you to in this video is the notion of a transformation in mathematics, and you're probably used to the word in everyday language. Transformation means something is changing, it's transforming from one thing to another. What would transformation mean in a mathematical context? Well, it could mean that you're taking something mathematical and you're changing it into something else mathematical, that's exactly what it is. It's talking about taking a set of coordinates or a set of points, and then changing them into a different set of coordinates or a different set of points. For example, this right over here, this is a quadrilateral we've plotted it on the coordinate plane. This is a set of points, not just the four points that represent the vertices of the quadrilateral, but all the points along the sides too. There's a bunch of points along this. You could argue there's an infinite, or there are an infinite number of points along this quadrilateral. This right over here, the point X equals 0I equals negative four, this is a point on the quadrilateral. Now, we can apply a transformation to this, which just means moving all the points in the same direction, and the same amount in that same direction, and I'm using the Khan Academy translation widget to do it. Let's translate, let's translate this, and I can do it by grabbing onto one of the vertices, and notice I've now shifted it to the right by two. Every point here, not just the orange points has shifted to the right by two. This one has shifted to the right by two, this point right over here has shifted to the right by two, every point has shifted in the same direction by the same amount, that's what a translation is. Now, I've shifted, let's see if I put it here every point has shifted to the right one and up one, they've all shifted by the same amount in the same directions. That is a translation, but you could imagine a translation is not the only kind of transformation. In fact, there is an unlimited variation, there's an unlimited number different transformations. I have another set of points here that's represented by quadrilateral, I guess we could call it CD or BCDE, and I could rotate it, and I rotate it I would rotate it around the point. So for example, I could rotate it around the point D, so this is what I started with, if I, let me see if I can do this, I could rotate it like, actually let me see. So if I start like this I could rotate it 90 degrees, I could rotate 90 degrees, so I could rotate it, I could rotate it like, that looks pretty close to a 90-degree rotation. So, every point that was on the original or in the original set of points I've now shifted it relative to that point that I'm rotating around. I've now rotated it 90 degrees, so this point has now mapped to this point over here. This point has now mapped to this point over here, and I'm just picking the vertices because those are a little bit easier to think about.", + "qid": "XiAoUDfrar0_76_106" + }, + { + "Q": "at 5:09, i dont understand how it is the same distance", + "A": "Most transformations, translations, rotations, and reflections all end up with an image that is congruent to the pre-image, so same parts of congruent figures are congruent. The video has the rotation slightly off.", + "video_name": "XiAoUDfrar0", + "timestamps": [ + 309 + ], + "3min_transcript": "The point of rotation, actually, since D is actually the point of rotation that one actually has not shifted, and just 'til you get some terminology, the set of points after you apply the transformation this is called the image of the transformation. So, I had quadrilateral BCDE, I applied a 90-degree counterclockwise rotation around the point D, and so this new set of points this is the image of our original quadrilateral after the transformation. I don't have to just, let me undo this, I don't have to rotate around just one of the points that are on the original set that are on our quadrilateral, I could rotate around, I could rotate around the origin. I could do something like that. Notice it's a different rotation now. It's a different rotation. I could rotate around any point. Now let's look at another transformation, and that would be the notion of a reflection, You imagine the reflection of an image in a mirror or on the water, and that's exactly what we're going to do over here. If we reflect, we reflect across a line, so let me do that. This, what is this one, two, three, four, five, this not-irregular pentagon, let's reflect it. To reflect it, let me actually, let me actually make a line like this. I could reflect it across a whole series of lines. Woops, let me see if I can, so let's reflect it across this. Now, what does it mean to reflect across something? One way I imagine is if this was, we're going to get its mirror image, and you imagine this as the line of symmetry that the image and the original shape they should be mirror images across this line we could see that. Let's do the reflection. There you go, and you see we have a mirror image. This is this far away from the line. This, its corresponding point in the image This point over here is this distance from the line, and this point over here is the same distance but on the other side. Now, all of the transformations that I've just showed you, the translation, the reflection, the rotation, these are called rigid transformations. Once again you could just think about what does rigid mean in everyday life? It means something that's not flexible. It means something that you can't stretch or scale up or scale down it kind of maintains its shape, and that's what rigid transformations are fundamentally about. If you want to think a little bit more mathematically, a rigid transformation is one in which lengths and angles are preserved. You can see in this transformation right over here the distance between this point and this point, between points T and R, and the difference between their corresponding image points, that distance is the same. The angle here, angle R, T, Y, the measure of this angle over here,", + "qid": "XiAoUDfrar0_309" + }, + { + "Q": "At around 4:25, Sal said the square root couldn't be a negative number, so x would have had to be 3 or more. Wouldn't it have to be 4 or more because 3 - 3 = 0?", + "A": "As Sal says: there is nothing wrong with the expression sqrt(0). Its just zero.", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 265 + ], + "3min_transcript": "And what's 1 over 0? I don't know what it is, so this is undefined. No one ever took the trouble to define what 1 over 0 should be. And they probably didn't do, so some people probably thought about what should be, but they probably couldn't find out with a good definition for 1 over 0 that's consistent with the rest of mathematics. So 1 over 0 stays undefined. So f of 0 is undefined. So we can't put 0 in and get a valid answer for f of 0. So here we say the domain is equal to -- do little brackets, that shows kind of the set of what x's apply. That's those little curly brackets, I didn't draw it that well. x is a member of the real numbers still, such that So here I just made a slight variation on what I had before. Before we said when f of x is equal to x squared that x is just any real number. Now we're saying that x is any real number except for 0. This is just a fancy way of saying it, and then these curly brackets just mean a set. Let's do a couple more ones. Let's say f of x is equal to the square root of x minus 3. So up here we said, well this function isn't defined when we get a 0 in the denominator. But what's interesting about this function? Can we take a square root of a negative number? Well until we learn about imaginary and complex numbers we can't. So here we say well, any x is valid here except for the x's So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the", + "qid": "U-k5N1WPk4g_265" + }, + { + "Q": "Mr Sal at 3:43 you said\nF(x)=square root of x-3greater or equal to 0\nHow can it be equal to 0, if I plug 0 in place of x i would get -3 which is wrong??", + "A": "Notice the subtle distinction. Sal did not say that x was greater than or equal to 0, but that x-3 was. So x must be greater than or equal to 3.", + "video_name": "U-k5N1WPk4g", + "timestamps": [ + 223 + ], + "3min_transcript": "And what's 1 over 0? I don't know what it is, so this is undefined. No one ever took the trouble to define what 1 over 0 should be. And they probably didn't do, so some people probably thought about what should be, but they probably couldn't find out with a good definition for 1 over 0 that's consistent with the rest of mathematics. So 1 over 0 stays undefined. So f of 0 is undefined. So we can't put 0 in and get a valid answer for f of 0. So here we say the domain is equal to -- do little brackets, that shows kind of the set of what x's apply. That's those little curly brackets, I didn't draw it that well. x is a member of the real numbers still, such that So here I just made a slight variation on what I had before. Before we said when f of x is equal to x squared that x is just any real number. Now we're saying that x is any real number except for 0. This is just a fancy way of saying it, and then these curly brackets just mean a set. Let's do a couple more ones. Let's say f of x is equal to the square root of x minus 3. So up here we said, well this function isn't defined when we get a 0 in the denominator. But what's interesting about this function? Can we take a square root of a negative number? Well until we learn about imaginary and complex numbers we can't. So here we say well, any x is valid here except for the x's So we have to say that x minus 3 has to be greater than or equal to 0, right, because you could have the square to 0, that's fine, it's just 0. So x minus 3 has to be greater than or equal to 0, so x has to be greater than or equal to 3. So here our domain is x is a member of the real numbers, such that x is greater than or equal to 3. Let's do a slightly more difficult one. What if I said f of x is equal to the square root of the", + "qid": "U-k5N1WPk4g_223" + }, + { + "Q": "From @7:45 to @10:36 , what are you trying to achieve when you convert the matrix to reduced row-echelon form?\n\nI thought you were trying to turn row 2 and 3 to zeros, like you did with the 2x2 matrix.\n\nWhy do you have to convert it to row-echelon form?\n\nThank for the videos btw, they've been really helpful.", + "A": "rref is the solution for the system of equations represented by the augmented matrix. You re finding the vectors v such that Av = lambda*v - i.e., the solution for (lambda*I - A)v = 0; i.e., rref ( [ (lambda*I - A) | 0 ] ).", + "video_name": "3Md5KCCQX-0", + "timestamps": [ + 465, + 636 + ], + "3min_transcript": "And they're the eigenvectors that correspond to eigenvalue lambda is equal to 3. So if you apply the matrix transformation to any of these vectors, you're just going to scale them up by 3. Let me write this way. The eigenspace for lambda is equal to 3, is equal to the span, all of the potential linear combinations of this guy and that guy. So 1/2, 1, 0. And 1/2, 0, 1. So that's only one of the eigenspaces. That's the one that corresponds to lambda is equal to 3. Let's do the one that corresponds to lambda is equal to minus 3. So if lambda is equal to minus 3-- I'll do it up here, I think I have enough space-- lambda is equal to minus 3. This matrix becomes-- I'll do the diagonals-- minus 3 plus 1 is minus 2. Minus 3 minus 2 is minus 5. And all the other things don't change. Minus 2, minus 2, 1. Minus 2, minus 2 and 1. And then that times vectors in the eigenspace that corresponds to lambda is equal to minus 3, is going to be equal to 0. I'm just applying this equation right here which we just derived from that one over there. So, the eigenspace that corresponds to lambda is equal to minus 3, is the null space, this matrix right here, are all the vectors that satisfy this equation. So what is-- the null space of this is the same thing as the null space of this in reduced row echelon form So let's put it in reduced row echelon form. So the first thing I want to do, I'm going to keep my first row the same. I'm going to write a little bit smaller than I normally do because I think I'm going to run out of space. So minus 2, minus 2, minus 2. I will skip some steps. Let's just divide the first row by minus 2. So we get 1, 1, 1. And then let's replace this second row with the second row plus this version of the first row. So this guy plus that guy is 0 minus 5 plus minus-- or let me Let me replace it with the first row minus the second row. So minus 2 minus minus 2 is 0. Minus 2 minus minus 5 is plus 3. And then minus 2 minus 1 is minus 3. And then let me do the last row in a different color for fun. And I'll do the same thing. I'll do this row minus this row. So minus 2 minus minus 2 is a 0. Minus 2 plus 2. Minus 2 minus 1 is minus 3. And then we have minus 2 minus minus 5.", + "qid": "3Md5KCCQX-0_465_636" + }, + { + "Q": "At 3:35, why is it i*\u00e2\u0088\u009a4*13 and not i^2*\u00e2\u0088\u009a4*13? I thought that by definition i^2= -1", + "A": "i^2 = -1 i = \u00e2\u0088\u009a(-1)", + "video_name": "s03qez-6JMA", + "timestamps": [ + 215 + ], + "3min_transcript": "things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i. And then we're going to multiply that times the square root of 4 times 13. And this is going to be equal to i times the square root of 4. i times the square root of 4, or the principal square root of 4 times the principal square root of 13. The principal square root of 4 is 2. So this all simplifies, and we can switch the order, over here. This is equal to 2 times the square root of 13. 2 times the principal square root of 13, I should say, times i. And I just switched around the order. It makes it a little bit easier to read if I put the i after the numbers over here. But I'm just multiplying i times 2 times the square root of 13. That's the same thing as multiplying 2 times the principal square root of 13 times i. And I think this is about as simplified as we can get here.", + "qid": "s03qez-6JMA_215" + }, + { + "Q": "At 2:04, Sal says that I can not split sqrt(-1 x -52) into sqrt(-1) x sqrt(-52). Can I go the opposite way? Would it be mathematically correct to simplify: sqrt(-1) x sqrt(-52) into sqrt(-1 x -52)?", + "A": "I believe it would be mathematically correct, because sqrt(-1 x -52) is sqrt(52). However, sqrt(-1) x sqrt(-52) is not the same as sqrt(-1 x -52) because sqrt(-1) x sqrt(-52) is equal to sqrt(-1) x sqrt(52) x sqrt(-1) = -1 x sqrt(52) which is not the same as sqrt(-52). Sorry you got an answer to your question 2 years after you asked it, hopefully this helps! ;-}", + "video_name": "s03qez-6JMA", + "timestamps": [ + 124 + ], + "3min_transcript": "We're asked to simplify the principal square root of negative 52. And we're going to assume, because we have a negative 52 here inside of the radical, that this is the principal branch of the complex square root function. That we can actually put, input, negative numbers in the domain of this function. That we can actually get imaginary, or complex, results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times-- or I should say, the principal square root of negative 1 times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this, or I should say, you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. So far, I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not OK. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now with that said, we can do it if only one of them are negative or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to-- Well, we have our i, now. The principal square root of negative 1 is i. The other square root of negative 1 is negative i.", + "qid": "s03qez-6JMA_124" + }, + { + "Q": "on 0:59 he is a horrible stock investor\nVote up if u agree", + "A": "There isn t enough info to know. -- The problem didn t tell you the time frame. If the portfolio went up in 1 month by 25%, that would be fantastic! -- If the overall market is down for the year by 20% and this person s portfolio went up 25%, that would also be fantastic! -- If it took 20 years for his portfolio to grow 25%, that would not be good, unless his goal was to project his portfolio (low risk). Then, it would be good.", + "video_name": "X2jVap1YgwI", + "timestamps": [ + 59 + ], + "3min_transcript": "Let's do some more percentage problems. Let's say that I start this year in my stock portfolio with $95.00. And I say that my portfolio grows by, let's say, 15%. How much do I have now? I think you might be able to figure this out on your own, but of course we'll do some example problems, just in case it's a little confusing. So I'm starting with $95.00, and I'll get rid of the dollar sign. We know we're working with dollars. 95 dollars, right? And I'm going to earn, or I'm going to grow just because I was an excellent stock investor, that 95 dollars So to that 95 dollars, I'm going to add another 15% of 95. So we know we write 15% as a decimal, as 0.15, so 95 plus 0.15 of 95, so this is times 95-- that dot is just a times sign. It's not a decimal, it's a times, it's a little higher than a decimal-- So 95 plus 0.15 times 95 is what we have now, right? Because we started with 95 dollars, and then we made another 15% times what we started with. Hopefully that make sense. Another way to say it, the 95 dollars has grown by 15%. So let's just work this out. This is the same thing as 95 plus-- what's 0.15 times 95? Let's see. So let me do this, hopefully I'll have enough space here. 95 times 0.15-- I don't want to run out of space. of space-- 95 times 0.15. 5 times 5 is 25, 9 times 5 is 45 plus 2 is 47, 1 times 95 is 95, bring down the 5, 12, carry the 1, 15. And how many decimals do we have? 1, 2. 15.25. Actually, is that right? I think I made a mistake here. See 5 times 5 is 25. 5 times 9 is 45, plus 2 is 47. And we bring the 0 here, it's 95, 1 times 5, 1 times 9, then we add 5 plus 0 is 5, 7 plus 5 is 12-- oh. I made a mistake.", + "qid": "X2jVap1YgwI_59" + }, + { + "Q": "4:48 Hey, does anyone know why Sal puts the = sign like a smiley face? =D", + "A": "It s not an equal sign, but rather an arrow. He just draws it in a way that it doesn t look quite connected. This is actually a common way to note progression of steps in mathematics.", + "video_name": "X2jVap1YgwI", + "timestamps": [ + 288 + ], + "3min_transcript": "So I'll ask you an interesting question? How did I know that 15.25 was a mistake? Well, I did a reality check. I said, well, I know in my head that 15% of 100 is 15, so if 15% of 100 is 15, how can 15% of 95 be more than 15? I think that might have made sense. The bottom line is 95 is less than 100. So 15% of 95 had to be less than 15, so I knew my answer of 15.25 was wrong. And so it turns out that I actually made an addition error, and the answer is 14.25. So the answer is going to be 95 plus 15% of 95, which is the same thing as 95 plus 14.25, well, that equals what? 109.25. especially this 2 here. 109.25. So if I start off with $95.00 and my portfolio grows-- or the amount of money I have-- grows by 15%, I'll end up with $109.25. Let's do another problem. Let's say I start off with some amount of money, and after a year, let's says my portfolio grows 25%, and after growing 25%, I now have $100. How much did I originally have? Notice I'm not saying that the $100 is growing by 25%. by 25%, and I end up with $100 after it grew by 25%. To solve this one, we might have to break out a little bit of algebra. So let x equal what I start with. So just like the last problem, I start with x and it grows by 25%, so x plus 25% of x is equal to 100, and we know this 25% of x we can just rewrite as x plus 0.25 of x is equal to 100, and now actually we have a level-- actually this might be level 3 system, level 3 linear equation-- but the bottom", + "qid": "X2jVap1YgwI_288" + }, + { + "Q": "At 4:21, Sal says that that 0i is the same thing as 0 + i. Wouldn't be 0 times i? I am confused.", + "A": "No, he says that i is the same thing as 0 + i , which is true. You are right that 0i is just 0 times i , not 0 + i . So 0i is the same thing as just 0 .", + "video_name": "A_ESfuN1Pkg", + "timestamps": [ + 261 + ], + "3min_transcript": "And 9 is a real number. So we could just add those up. So 2 plus negative 7 would be negative 5. Negative 5 plus 9 would be 4. So the real numbers add up to 4. And now we have these imaginary numbers. So 3 times i minus 5 times i. So if you have 3 of something and then I were to subtract 5 of that same something from it, now you're going to have negative 2 of that something. Or another way of thinking about it is the coefficients. 3 minus 5 is negative 2. So three i's minus five i's, that's going to give you negative 2i. Now you might say, well, can we simplify this any further? Well no, you really can't. This right over here is a real number. 4 is a number that we've been dealing with throughout our mathematical careers. And so what we really consider this is this 4 minus 2i, we can now consider this entire expression to really be a number. So this is a number that has a real part and an imaginary part. And numbers like this we call complex numbers. It is a complex number. Why is it complex? Well, it has a real part and an imaginary part. And you might say, well, gee, can't any real number be considered a complex number? For example, if I have the real number 3, can't I just write the real number 3 as 3 plus 0i? And you would be correct. Any real number is a complex number. You could view this right over here as a complex number. a subset of the complex numbers. Likewise, imaginary numbers are a subset of the complex numbers. For example, you could rewrite i as a real part-- 0 is a real number-- 0 plus i. So the imaginaries are a subset of complex numbers. Real numbers are a subset of complex numbers. And then complex numbers also have all of the sums and differences, or all of the numbers that have both real and imaginary parts.", + "qid": "A_ESfuN1Pkg_261" + }, + { + "Q": "At 7:40, why did he set a maximum y value? Isn't that already determined since x cannot be greater than 10?", + "A": "Right. He knows that the maximum x-value is 10, but he doesn t know what the maximum y-value is. The y-value represents the volume of the box, which can get pretty big depending on the dimensions. (width, depth, height)", + "video_name": "MC0tq6fNRwU", + "timestamps": [ + 460 + ], + "3min_transcript": "You're not going to have any height here. So you're not going to have any volume, so our volume would be 0. What is our volume when x is equal to 10? Well, if x equaled 10, then the width here that I've drawn in pink would be 0. So once again, we would have no volume. And this term right over here, if we just look at it algebraically would also be, equal to 0, so this whole thing would be equal to 0. So someplace in between x equals 0 and x equals 10 we should achieve our maximum volume. And before we do it analytically with a bit of calculus, let's do it graphically. So I'll get my handy TI-85 out. So let me get my TI-85 out. And so first let me set my range appropriately before I attempt to graph it. So I'll put my graph function. Let me set my range. So my minimum x-value, let me make that 0. We know that x cannot be less than 0. My maximum x-value, well, 10 seems pretty good. My minimum y-value, this is essentially I'm not going to have negative volume, so let me set that equal 0. And my maximum y-value, let's see what would be reasonable here. I'm just going to pick some a random x and see what type of a volume I get. So if x were to be 5, it would be 5 times 20 minus 10, which is 10. So that would be-- did I do that right? Yeah, 20 minus 2 times 5, so that would be 10, and then times 30 minus 2 times 5, which would be 20. So it would be 5 times 10 times 20. So you'd get a volume of 1,000 cubic inches. And I just randomly picked the number 5. So let me give my maximum y-value a little higher than that just in case to that isn't the maximum value. I just randomly picked that. So let's say yMax is 1,500, and if for whatever reason our graph doesn't fit in there, then maybe we can make our yMax even larger. So I think this is going to be a decent range. Now let's actually input the function itself. So our volume is equal to x times 20 And that looks about right. So now I think we can graph it. So 2nd, and I want to select that up there, so I'll do Graph. And it looks like we did get the right range. So this tells us volume is a function of x between x is 0 and x is 10, and it does look like we hit a maximum point right around there. to use the Trace function to figure out roughly what that maximum point is. So let me trace this function. So I can still go higher, higher. So over there my volume is 1,055.5. Then I can get to 1,056. So let's see, this was 1,056.20, this is 1,056.24, then I go back to 1,055. So at least based on the level of zoom that I have my calculator right now,", + "qid": "MC0tq6fNRwU_460" + }, + { + "Q": "There is one point at \"2:25\" that I really want it to be clarified. I think that \"integral f(t) d(t) from a to x\" should be \" F(x) - F(a)\" instead of \"F(x)\". Please help me Im so confused!", + "A": "The integral f(t) d(t) from a to x would equal F(x)-F(a). I believe Sal is simply using F to put the integral as a function of x. Any letter or symbol would do, some calculus books write it as A(x) instead of F(x). They tend to use A because the formula seen in the video is an area accumulation formula (A for area). Whatever x you put into F(x) would give you the amount of area you have accumulated from a to x.", + "video_name": "C7ducZoLKgw", + "timestamps": [ + 145 + ], + "3min_transcript": "", + "qid": "C7ducZoLKgw_145" + }, + { + "Q": "At 0:50 can anyone tell me why the yellow line's slope is going to be the negative inverse?", + "A": "Because they are both perpendicular lines!!", + "video_name": "0671cRNjeKI", + "timestamps": [ + 50 + ], + "3min_transcript": "We are asked which of these lines are perpendicular. And it has to be perpendicular to one of the other lines, you can't be just perpendicular by yourself. And perpendicular line, just so you have a visualization for what for perpendicular lines look like, two lines are perpendicular if they intersect at right angles. So if this is one line right there, a perpendicular line will look like this. A perpendicular line will intersect it, but it won't just be any intersection, it will intersect at right angles. So these two lines are perpendicular. Now, if two lines are perpendicular, if the slope of this orange line is m-- so let's say its equation is y is equal to mx plus, let's say it's b 1, so it's some y-intercept-- then the equation of this yellow line, its slope is going to be the negative inverse of this guy. This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept. perpendicular, the product of their slopes is going to be negative 1. And so you could write that there. m times negative 1 over m, that's going to be-- these two guys are going to cancel out-- that's going to be equal to negative 1. So let's figure out the slopes of each of these lines and figure out if any of them are the negative inverse of any of the other ones. So line A, the slope is pretty easy to figure out, it's already in slope-intercept form, its slope is 3. So line A has a slope of 3. Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it. So let's do line B over here. Line B, we have x plus 3y is equal to negative 21. Let's subtract x from both sides so that it ends up on the right-hand side. So we end up with 3y is equal to negative x minus 21. and we get y is equal to negative 1/3 x minus 7. So this character's slope is negative 1/3. So here m is equal to negative 1/3. So we already see they are the negative You take the inverse of 3, it's 1/3, and then it's the negative of that. Or you take the inverse of negative 1/3, it's negative 3, and then this is the negative of that. So these two lines are definitely perpendicular. Let's see the third line over here. So line C is 3x plus y is equal to 10. If we subtract 3x from both sides, we get y is equal to", + "qid": "0671cRNjeKI_50" + }, + { + "Q": "at 5:08\nisn't this the formula of Explicit geometric sequence ?", + "A": "In a geometric sequence, the input value is multiplied. Here it is the power.", + "video_name": "G2WybA4Hf7Y", + "timestamps": [ + 308 + ], + "3min_transcript": "two to the first power. So it's going to be one-fourth two. What is h of two going to be equal to? Well, it's going to be one-fourth times two squared, so it's going to be times two times two. Or, we could just view this as this is going to be two times h of one. And actually I should have done this when I wrote this one out, but this we can write as two times h of zero. So notice, if we were to take the ratio between h of two and h of one, it would be two. If we were to take the ratio between h of one it would be two. That is the common ratio between successive whole number inputs into our function. So, h of I could say plus one over h of n is going to be equal to is going to be equal to actually I can work it out mathematically. One-fourth times two to the n plus one one-fourth two to the n. Two to the n plus one, divided by two to the n is just going to be equal to two. That is your common ratio. So for the function h. For the function f, our common ratio is three. If we were to go the other way around, if someone said, hey, I have some function whose initial value, so let's say, I have some function, I'll do this in a new color, I have some function, g, and we know that its initial initial value is five. And someone were to say its common ratio its common ratio is six, what would this exponential function look like? And they're telling you this is an exponential function. Well, g of let's say x is the input, is going to be equal to our initial value, which is five. That's not a negative sign there, Our initial value is five. And then times our common ratio to the x power. So once again, initial value, right over there, that's the five. And then our common ratio is the six, right over there. So hopefully that gets you a little bit familiar with some of the parts of an exponential function, why they are called what they are called.", + "qid": "G2WybA4Hf7Y_308" + }, + { + "Q": "At 3:55 What does he mean by principle root?", + "A": "The principal root is just the positive square root. For instance: \u00e2\u0088\u009a9 = \u00c2\u00b13 But the principal square root of 9 is just 3.", + "video_name": "McINBOFCGH8", + "timestamps": [ + 235 + ], + "3min_transcript": "are equal. So these are the two sides that are equal. And then if the two sides are equal, we have proved to ourselves that the base angles are equal. And if we called the measure of these base angles x, then we know that x plus x plus 90 have to be equal to 180. Or if we subtract 90 from both sides, you get x plus x is equal to 90 or 2x is equal to 90. Or if you divide both sides by 2, you get x is equal to 45 degrees. So a right isosceles triangle can also be called-- and this is the more typical name for it-- it can also be called a 45-45-90 triangle. And what I want to do this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle. Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x. And then we can use the Pythagorean Theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the square of length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. We can take the principal root of both sides of that. I wanted to just change it to yellow. Last, take the principal root of both sides of that. The left-hand side you get, principal root of 2 is just square root of 2, and then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2 So if you have a right isosceles triangle, whatever the two legs are, they're going to have the same length. That's why it's isosceles. The hypotenuse is going to be square root of 2 times that. So c is equal to x times the square root of 2. So for example, if you have a triangle that looks like this. Let me draw it a slightly different way. It's good to have to orient ourselves in different ways every time. So if we see a triangle that's 90 degrees, 45 and 45 like this, and you really just have to know two of these angles to know what the other one is going to be, and if I tell you that this side right over here is 3-- I actually don't even have to tell you that this other side's going to be 3. This is an isosceles triangle, so those two legs are going to be the same. And you won't even have to apply the Pythagorean theorem if you know this-- and this is a good one to know-- that the hypotenuse here, the side opposite the 90 degree side, is just going to be square root of 2 times the length of either of the legs. So it's going to be 3 times the square root of 2.", + "qid": "McINBOFCGH8_235" + }, + { + "Q": "At 4:59, why is it bn and vn, instead of bk and vk?", + "A": "There s no error here. b belongs to Rn, otherwise original equation Ax=b, where A is nxk matrix would not make any sense. v is a projection of b to the column space, but it still is a member of Rn, hence it also has n components.", + "video_name": "MC7l96tW8V8", + "timestamps": [ + 299 + ], + "3min_transcript": "Or another way to view it, when I say close, I'm talking about length, so I want to minimize the length of-- let me write this down. I want to minimize the length of b minus A times x-star. Now, some of you all might already know where this is going. But when you take the difference between 2 and then take its length, what does that look like? Let me just call Ax. Ax is going to be a member of my column space. Let me just call that v. Ax is equal to v. You multiply any vector in Rk times your matrix A, you're So any Ax is going to be in your column space. And maybe that is the vector v is equal to A times x-star. And we want this vector to get as close as possible to this as long as it stays-- I mean, it has to be in my column space. But we want the distance between this vector and this vector to be minimized. Now, I just want to show you where the terminology for this will come from. I haven't given it its proper title yet. If you were to take this vector-- let just call this vector v for simplicity-- that this is equivalent to the length of the vector. You take the difference between each of the elements. So b1 minus v1, b2 minus v2, all the way to bn minus vn. thing as this. This is going to be equal to the square root. Let me take the length squared, actually. The length squared of this is just going to be b1 minus v1 squared plus b2 minus v2 squared plus all the way to bn minus vn squared. And I want to minimize this. So I want to make this value the least value that it can be possible, or I want to get the least squares estimate here. And that's why, this last minute or two when I was just explaining this, that was just to give you the motivation for why this right here is called the least squares estimate, or the least squares solution, or the least squares approximation for the equation Ax equals b. There is no solution to this, but maybe we can find some", + "qid": "MC7l96tW8V8_299" + }, + { + "Q": "@ 4:05 he says it goes down 2/3's then he says its 1 and 1/3. How does he go from 2/3's to 1/3rd? I get the whole thing that 1/ 1/3rd makes 4/3rds, but he went down 2/3rds??", + "A": "When he went down 2/3 it was from 2! The line was 2/3 below 2. So if you look at it in terms of the line being above 1 the line is actually 1/3 above 1. So therefore the line is at 1 1/3.", + "video_name": "9wOalujeZf4", + "timestamps": [ + 245 + ], + "3min_transcript": "That's our end point. That's our starting point. So if you simplify this, b minus b is 0. 1 minus 0 is 1. So you get m/1, or you get it's equal to m. So hopefully you're satisfied and hopefully I didn't confuse you by stating it in the abstract with all of these variables here. But this is definitely going to be the slope and this is definitely going to be the y-intercept. Now given that, what I want to do in this exercise is look at these graphs and then use the already drawn graphs to figure out the equation. So we're going to look at these, figure out the slopes, figure out the y-intercepts and then know the equation. So let's do this line A first. So what is A's slope? Let's start at some arbitrary point. Let's start right over there. We want to get even numbers. If we run one, two, three. One, two, three. Our delta y-- and I'm just doing it because I want to hit an even number here-- our delta y is equal to-- we go down by 2-- it's equal to negative 2. So for A, change in y for change in x. When our change in x is 3, our change in y is negative 2. So our slope is negative 2/3. When we go over by 3, we're going to go down by 2. Or if we go over by 1, we're going to go down by 2/3. You can't exactly see it there, but you definitely see it when you go over by 3. So that's our slope. We've essentially done half of that problem. Now we have to figure out the y-intercept. So that right there is our m. Now what is our b? Our y-intercept. Well where does this intersect the y-axis? Well we already said the slope is 2/3. So this is the point y is equal to 2. gone down by 2/3. So this right here must be the point 1 1/3. Or another way to say it, we could say it's 4/3. That's the point y is equal to 4/3. Right there. A little bit more than 1. About 1 1/3. So we could say b is equal to 4/3. So we'll know that the equation is y is equal to m, negative 2/3, x plus b, plus 4/3. That's equation A. Let's do equation B. Hopefully we won't have to deal with as many fractions here. Equation B. Let's figure out its slope first. Let's start at some reasonable point. We could start at that point. Let me do it right here. B. Equation B. When our delta x is equal to-- let me write it this way, delta x. So our delta x could be 1. When we move over 1 to the right, what happens", + "qid": "9wOalujeZf4_245" + }, + { + "Q": "Hold on.. how did he get 4/3 as the y-intercept in 4:05..?", + "A": "The slope = -2/3. Sal has a point at (-1, 2). He needs to move one uni to the right to get to the y-axis. The slope tells him that as he moves 1 unit to the right, the y-coordinate will decrease by 2/3. Sal finds the y-intercept by doing: 2 - 2/3 = 6/3 - 2/3 = 4/3 Hope this helps.", + "video_name": "9wOalujeZf4", + "timestamps": [ + 245 + ], + "3min_transcript": "That's our end point. That's our starting point. So if you simplify this, b minus b is 0. 1 minus 0 is 1. So you get m/1, or you get it's equal to m. So hopefully you're satisfied and hopefully I didn't confuse you by stating it in the abstract with all of these variables here. But this is definitely going to be the slope and this is definitely going to be the y-intercept. Now given that, what I want to do in this exercise is look at these graphs and then use the already drawn graphs to figure out the equation. So we're going to look at these, figure out the slopes, figure out the y-intercepts and then know the equation. So let's do this line A first. So what is A's slope? Let's start at some arbitrary point. Let's start right over there. We want to get even numbers. If we run one, two, three. One, two, three. Our delta y-- and I'm just doing it because I want to hit an even number here-- our delta y is equal to-- we go down by 2-- it's equal to negative 2. So for A, change in y for change in x. When our change in x is 3, our change in y is negative 2. So our slope is negative 2/3. When we go over by 3, we're going to go down by 2. Or if we go over by 1, we're going to go down by 2/3. You can't exactly see it there, but you definitely see it when you go over by 3. So that's our slope. We've essentially done half of that problem. Now we have to figure out the y-intercept. So that right there is our m. Now what is our b? Our y-intercept. Well where does this intersect the y-axis? Well we already said the slope is 2/3. So this is the point y is equal to 2. gone down by 2/3. So this right here must be the point 1 1/3. Or another way to say it, we could say it's 4/3. That's the point y is equal to 4/3. Right there. A little bit more than 1. About 1 1/3. So we could say b is equal to 4/3. So we'll know that the equation is y is equal to m, negative 2/3, x plus b, plus 4/3. That's equation A. Let's do equation B. Hopefully we won't have to deal with as many fractions here. Equation B. Let's figure out its slope first. Let's start at some reasonable point. We could start at that point. Let me do it right here. B. Equation B. When our delta x is equal to-- let me write it this way, delta x. So our delta x could be 1. When we move over 1 to the right, what happens", + "qid": "9wOalujeZf4_245" + }, + { + "Q": "At 4:00 and 10:00, why is is cos(2a) a minus but sin(2a) is a plus?\n\nAlso which video did I miss? I am so confused here. I don't ever remember learning this in my high school precalc class.", + "A": "You have to review the formula of cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b) From there, you can write cos(2a) as cos(a+a), then go on. You see why the minus in here. That is trigonometry. Go back to the beginning and learn the formula come from.", + "video_name": "a70-dYvDJZY", + "timestamps": [ + 240, + 600 + ], + "3min_transcript": "Because cosine of minus c is the same thing as the cosine of c. Times the cosine of c. And then, minus the sine of c. Instead of writing this, I could write this. Minus the sine of c times the cosine of a. So that we kind of pseudo proved this by knowing this and this ahead of time. Fair enough. And I'm going to use all of these to kind of prove a bunch of more trig identities that I'm going to need. So the other trig identity is that the cosine of a plus b is equal to the cosine of a-- you don't mix up the cosines and the sines in this situation. Cosine of a times the sine of b. And this is minus-- well, sorry. I just said you don't mix it up and then I mixed them up. Times the cosine of b minus sine of a times the sine of b. well, you use these same properties. Cosine of minus b, that's still going to be cosine on b. So that's going to be the cosine of a times the cosine-- cosine of minus b is the same thing as cosine of b. But here you're going to have sine of minus b, which is the same thing as the minus sine of b. And that minus will cancel that out, so it'll be plus sine of a times the sine of b. When you have a plus sign here you get a minus there. When you don't minus sign there, you get a plus sign there. But fair enough. I don't want to dwell on that too much because we have many more identities to show. So what if I wanted an identity for let's say, the cosine of 2a? So the cosine of 2a. Well that's just the same thing as the cosine of a plus a. If my second a is just my b, then this is just equal to cosine of a times the cosine of a minus the sine of a times the sine of a. My b is also an a in this situation, which I could rewrite as, this is equal to the cosine squared of a. I just wrote cosine of a times itself twice or times itself. Minus sine squared of a. This is one I guess identity already. Cosine of 2a is equal to the cosine squared of a minus the sine squared of a. Let me box off my identities that we're showing in this video. So I just showed you that one. What if I'm not satisfied? What if I just want it in terms of cosines? Well, we could break out the unit circle definition of our trig functions. This is kind of the most fundamental identity. The sine squared of a plus the cosine squared", + "qid": "a70-dYvDJZY_240_600" + }, + { + "Q": "At 1:25, why is f(x) = D?", + "A": "i think you are getting confused by his arrows...", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 85 + ], + "3min_transcript": "In this video, I want to familiarize you with the idea of a limit, which is a super important idea. It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really, really, really simple idea. So let me draw a function here, actually, let me define a function here, a kind of a simple function. So let's define f of x, let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal look, I have the same thing in the numerator and denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true, the difference between f of x equals 1 and this thing right over here, is that this thing can never equal-- this thing is undefined when x is equal to 1. Because if you set, let me define it. Let me write it over here, if you have f of, sorry not f of 0, if you have f of 1, what happens. which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here.", + "qid": "riXcZT2ICjA_85" + }, + { + "Q": "Wait a minute, at 9:20 Sal says that when x approaches to 2, the value of g(x) would be getting really close of 4, but shouldn't it be 1? Since when x = 2 <=> g(x) = 1?", + "A": "g(2) is in fact 1, given by the dot. However, the limit is different. Remeber g(x) = y. To understand limit, try putting your pencil on the graph and trace it. As you trace it toward x=2, you see that y=g(x) is getting toward 4 and not 1.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 560 + ], + "3min_transcript": "this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1, as we get closer and closer to it. And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1. The limit of g of x as x approaches 2 is equal to 4. And you could even do this numerically using a calculator, and let me do that, because I think that will be interesting. So let me get the calculator out, let me get my trusty TI-85 out. So here is my calculator, and you could numerically say, OK, what's it going to approach as you approach x equals 2. So let's try 1.94, for x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared.", + "qid": "riXcZT2ICjA_560" + }, + { + "Q": "in the video at 4:24 Sal says that we can get infinitely closer to one. If we can get infinitely closer to one doesn't that mean that we can never approach one?", + "A": "True. Yes, we can always get closer and closer to one but the function actually never reaches one.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 264 + ], + "3min_transcript": "And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here. you go, and let's say that even though this was a function definition, you'd go, OK x is equal to 1, oh wait there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1. And now this is starting to touch on the idea of a limit. So as x gets closer and closer to 1. So as we get closer and closer x is to 1, what is the function approaching. Well, this entire time, the function, what's a getting closer and closer to. On the left hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here from the right hand side, you get the same thing. familiar with this idea as we do more examples, that the limit as x and L-I-M, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1. And our function is going to be equal to 1, it's getting closer and closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look what is a function approaching as x gets closer and closer to 1. Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, let me just for the sake of variety,", + "qid": "riXcZT2ICjA_264" + }, + { + "Q": "At 7:30 Sal has drawn a parabola with a gap at the point (2,4).\n\nA point takes up zero space right? It has no size? So how can something have a gap in it if the gap doesn't cover any actual space? How big is the gap? Surely a gap is between two points.\n\nCan someone please help me understand this?", + "A": "At this point it has size 0. The gap is just for understanding that there is no defined point at that x coordinate.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 450 + ], + "3min_transcript": "Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared, when x does not equal, I don't know when x does not equal 2. And let's say that when x equals 2 it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous, it has a discontinuity. Let me graph it. So this is my y equals f of x axis, this is my x-axis right over here. Let me draw x equals 2, x, let's say this is x equals 1, this is x equals 2, this is negative 1, this is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. let me draw a better version of the parabola. So it'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric, let me redraw it because that's kind of ugly. And that's looking better. OK, all right, there you go. All right, now, this would be the graph of just x squared. But this can't be. It's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2 the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that. this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of f of x is equal to x squared, except when you get to 2, it has this gap, because you don't use the f of x is equal to x squared when x is equal to 2. You use f of x-- or I should say g of x-- you use g of x is equal to 1. Have I been saying f of x? I apologize for that. You use g of x is equal to 1. So then then at 2, just at 2, just exactly at 2, it drops down to 1. And then it keeps going along the function g of x is equal to, or I should say, along the function", + "qid": "riXcZT2ICjA_450" + }, + { + "Q": "At 1:17; what's the difference between infinity and 1/0 ?", + "A": "Infinity (symbol: \u00e2\u0088\u009e) is an abstract concept describing something without any bound or larger than any number. In mathematics, infinity is often treated as a number While the expression a/0 has no meaning, as there is no number which, multiplied by 0, gives a (assuming a\u00e2\u0089\u00a00), and so division by zero is undefined. Since any number multiplied by zero is zero, the expression 0/0 also has no defined value; You can also prove it by the concept of limits. Try it out yourself.", + "video_name": "riXcZT2ICjA", + "timestamps": [ + 77 + ], + "3min_transcript": "In this video, I want to familiarize you with the idea of a limit, which is a super important idea. It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really, really, really simple idea. So let me draw a function here, actually, let me define a function here, a kind of a simple function. So let's define f of x, let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal look, I have the same thing in the numerator and denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true, the difference between f of x equals 1 and this thing right over here, is that this thing can never equal-- this thing is undefined when x is equal to 1. Because if you set, let me define it. Let me write it over here, if you have f of, sorry not f of 0, if you have f of 1, what happens. which is, let me just write it down, in the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is you the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent, both of these are going to be equal to 1 for all other X's other than one, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function. So let me graph it. So that, is my y is equal to f of x axis, y is equal to f of x axis, and then this over here And then let's say this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1, right up there I could do negative 1. but that matter much relative to this function right over here. And let me graph it. So it's essentially for any x other than 1 f of x is going to be equal to 1. So it's going to be, look like this. It's going to look like this, except at 1. At 1 f of x is undefined. So I'm going to put a little bit of a gap right over here, the circle to signify that this function is not We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us It's literally undefined, literally undefined when x is equal to 1. So this is the function right over here.", + "qid": "riXcZT2ICjA_77" + }, + { + "Q": "At 1:33, isn't that the Riemann's zeta function?", + "A": "Good question! That would be Riemann s zeta function, evaluated at z = 2. In general, this function is sum n=1 to infinity of 1/(n^z), where z is a complex number (that is, a number of the form a+bi where a and b are real numbers and i is the square root of -1). This function is defined when the real part of z (that is, a) is greater than 1. Have a blessed, wonderful day!", + "video_name": "k9MEOgcc5KY", + "timestamps": [ + 93 + ], + "3min_transcript": "- [Voiceover] Let's say that you have an infinite series, S, which is equal to the sum from n equals one, let me write that a little bit neater. n equals one to infinity of a sub n. This is all a little bit of review. We would say, well this is the same thing as a sub one, plus a sub two, plus a sub three, and we would just keep going on and on and on forever. Now what I want to introduce to you is the idea of a partial sum. This right over here is an infinite series. But we can define a partial sum, so if we say S sub six, this notation says, okay, if S is an infinite series, S sub six is the partial sum of the first six terms. So in this case, this is going to be we're not going to just keep going on forever, this is going to be a sub one, plus a sub two, plus a sub three, plus a sub four, plus a sub five, And I can make this a little bit more tangible if you like. So let's say that S, the infinite series S, is equal to the sum from n equals one, to infinity of one over n squared. In this case it would be one over one squared, plus one over two squared, plus one over three squared, and we would just keep going on and on and on forever. But what would S sub -- I should do that in that same color. What would S -- I said I would change color, and I didn't. What would S sub three be equal to? The partial sum of the first three terms, and I encourage you to pause the video and try to work through it on your own. Well, it's just going to be the first term one, plus the second term, is going to be the sum of the first three terms, and we can figure that out, that's to see if you have a common denominator here, it's going to be 36. It's going to be 36/36, plus 9/36, plus 4/36, so this is going to be 49/36. 49/36. So the whole point of this video, is just to appreciate this idea of a partial sum. And what we'll see is, that you can actually express what a partial sum might be algebraically. So for example, for example, let's give ourselves a little bit more real estate here. Let's say, let's go back to just saying we have an infinite series, S, that is equal to the sum from n equals one to infinity of a sub n. And let's say we know the partial sum, S sub n, so the sum of the first n terms", + "qid": "k9MEOgcc5KY_93" + }, + { + "Q": "At 0:28 Sal Said that an Odd Function Implies j(a) = - j(-a). Is this equivalent to -j(a) = j(-a) the more well known definition of an odd function? Or did Sal make a mistake?", + "A": "Multiply both sides by -1. They re the same.", + "video_name": "zltgXTlUVLw", + "timestamps": [ + 28 + ], + "3min_transcript": "Which of these functions is odd? And so let's remind ourselves what it means for a function to be odd. So I have a function-- well, they've already used f, g, and h, so I'll use j. So function j is odd. If you evaluate j at some value-- so let's say j of a. And if you evaluate that j at the negative of that value, and if these two things are the negative of each other, then my function is odd. If these two things were the same-- if they didn't have this negative here-- then it would be an even function. So let's see which of these meet the criteria of being odd. So let's look at f of x. So we could pick a particular point. So let's say when x is equal to 2. So we get f of 2 is equal to 2. Now, what is f of negative 2? f of negative 2 looks like it is 6. f of negative 2 is equal to 6. So these aren't the negative of each other. In order for this to be odd, f of negative 2 have had to be equal to negative 2. So f of x is definitely not odd. So all I have to do is find even one case that violated this constraint to be odd. And so I can say it's definitely not odd. Now let's look at g of x. So I could use the same-- let's see, when x is equal to 2, we get g of 2 is equal to negative 7. Now let's look at when g is negative 2. So we get g of negative 2 is also equal to negative 7. So here we have a situation-- and it looks like that's the case for any x we pick-- that g of x is going to be equal to g of negative x. So g of x is equal to g of negative x. It's symmetric around the y-- or I should say the vertical axis-- right over here. So which of these functions is odd? Definitely not g of x. So our last hope is h of x. Let's see if h of x seems to meet the criteria. I'll do it in this green color. So if we take h of 1-- and we can look at it even visually. So h of 1 gets us right over here. h of negative 1 seems to get us an equal amount, an equal distance, negative. So it seems to fit for 1. For 2-- well, 2 is at the x-axis. But that's definitely h of 2 is 0. h of negative 2 is 0. But those are the negatives of each other. 0 is equal to negative 0. If we go to, say, h of 4, h of 4 is this negative number. And h of negative 4 seems to be a positive number of the same magnitude. So once again, this is the negative of this. So it looks like this is indeed an odd function.", + "qid": "zltgXTlUVLw_28" + }, + { + "Q": "At 1:02,can a kite also be a diamond?", + "A": "yes, as it is of the same shape", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 62 + ], + "3min_transcript": "", + "qid": "inlMrf2d-k4_62" + }, + { + "Q": "at around 2:00 to 2:11 what is adjacent", + "A": "Next to or adjoining something else.", + "video_name": "inlMrf2d-k4", + "timestamps": [ + 120, + 131 + ], + "3min_transcript": "", + "qid": "inlMrf2d-k4_120_131" + }, + { + "Q": "In 0:39, I don't understand the part where it says that there are a bunch of different Xs. How is that possible?", + "A": "Multiple variables (Same ones) can be in an inequality. Ex: x + x + 1 =23", + "video_name": "UTs4uZhu5t8", + "timestamps": [ + 39 + ], + "3min_transcript": "what i want to do in this video is a handful of fairly simple inequality videos. But the real value of it, I think, will be just to get you warmed up in the notation of inequality. So, let's just start with one. we have x minus 5 is less than 35. So let's see if we can find all of the x's that will satisfy this equation. And that's one of the distinctions of an inequality. In an equation, you typically have one solution, or at least the ones we've solved so far. In the future, we'll see equations where they have more than one solution. But in the ones we've solved so far, you solved for a particular x. In the inequalities, there's a whole set of x's that will satisfy this inequality. So they're saying, what are all the x's, that when you subtract 5 from them, it's going to be less than 35? And we can already think about it. I mean 0 minus 5. That's less than 35. Minus 100 minus 5. That's less than 35. 5 minus 5. That's less than 35. So there's clearly a lot of x's that will satisfy that. essentially encompasses all of the x's. So the way we do that is essentially the same way that we solved any equations. We want to get just the x terms, in this case, on the left-hand side. So I want to get rid of this negative 5, and I can do that by adding 5 to both sides of this equation. So I can add 5 to both sides of this equation. That won't change the inequality. It won't change the less than sign. If something is less than something else, something plus 5 is still going to be less than the other thing plus 5. So on the left-hand side, we just have an x. This negative 5 and this positive 5 cancel out. x is less than 35 plus 5, which is 40. And that's our solution. And to just visualize the set of all numbers that represents, let me draw a number line here. And I'll do it around-- let's say that's 40, And then we could go below 40. 39, 38. You can just keep going below 40. It just keeps going on in both directions. And any x that is less than 40 will satisfy this. So it can't be equal to 40, because if x is equal to 40, 40 minus 5 is 35. It's not less than 35. So x has to be less than 40. And to show this on the number line, we do a circle around 40 to show that we're not including 40. But then we can shade in everything below 40. So everything that's just exactly below 40 is included So everything I've shaded in yellow is included in our solution set. So 39, 39.999999, repeating, which is about as close as you can get to 40 as possible, that's in our solution set. But 40 is not. That's why we put that open circle around it. Let's do another one. Let me do it in another color as well.", + "qid": "UTs4uZhu5t8_39" + }, + { + "Q": "At 4:01, isn't it supposed to be 6x-2 and not 6x+2?", + "A": "I thought so too. Probably we can t see it clearly, when I rewatched I thought I saw a hint of the vertical line in the +.", + "video_name": "_HJljJuVHLw", + "timestamps": [ + 241 + ], + "3min_transcript": "OK, so first of all, we have to remember what is the sum of the interior angles of a pentagon? And that's where I always draw an arbitrary pentagon. Let me see if I can do that. Actually, there's a polygon tool here. How does it work? I'm just trying to draw a pentagon. I don't know if that's any different than the line tool, So how many triangles can I draw in a pentagon? And that tells me what my total interior angles are. And there is a formula for that, but I like relying on your reasoning more than the formula, because you might forget the formula, or even worse, you might remember it, but not have the confidence to use it, or you might remember it wrong ten years in the future. So the best thing to do, if you have a polygon, is to count the triangles in it. Straightforward enough. So a pentagon has three triangles in it. So the sum of its interior angles are going to be 3 times 180, because it has 3 triangles in it. Each triangle has 180 degrees. And I know you can't see what I just wrote. So the sum of all of these angles are going to be the sum of all of the angles in all three triangles. So it's 3 times 180 is equal to 540 degrees. So that's the sum of all of these interior angles. Now, they say that each of them are 2x, 6x, et cetera. So the sum of all of these terms have to be equal 540. I'm going to write them vertically. It makes them easier to add. So if we write there 2x, 6x, 4x minus 6, 2x minus 16, and 6x plus 2. This is going to be the largest, right? So let's add this up. Minus 6, minus 16, that's minus 22. Plus 2 is minus 20. That's right. And 2x plus 6x that is 8x plus 4 is 12, 12 plus 2 is 14, 14 plus 6 is 20. So we have 20x minus 20 is equal to 540 degrees. 20x minus 20 is equal to 540. Let's divide both sides of this equation by 20. So you get x 1 minus 1 is equal to-- it would be 54 divided by 2, which is equal to 27.", + "qid": "_HJljJuVHLw_241" + }, + { + "Q": "at 4:10 Sal mentions that it is ar^k power. Why do the k power instead of nth power?", + "A": "k is the index, where k=0,1,2,3,...n So it would mean ar\u00e2\u0081\u00b0+ar\u00c2\u00b9+ar\u00c2\u00b2+ar\u00c2\u00b3+...+ar^n", + "video_name": "CecgFWTg9pQ", + "timestamps": [ + 250 + ], + "3min_transcript": "Well, we'll start with whatever our first term is. And over here if we want to speak in general terms we could call that a, our first term. So we'll start with our first term, a, and then each successive term that we're going to add is going to be a times our common ratio. And we'll call that common ratio r. So the second term is a times r. Then the third term, we're just going to multiply this one times r. So it's going to be a times r squared. And then we can keep going, plus a times r to the third power. And let's say we're going to do a finite geometric series. So we're not going to just keep on going forever. Let's say we keep going all the way until we get to some a times r to the n. a times r to the n-th power. And I encourage you to pause the video and try it on your own. Well, we could think about it this way. And I'll give you a little hint. You could view this term right over here as a times r to the 0. And let me write it down. This is a times r to the 0. This is a times r to the first, r squared, r third, and now the pattern might be emerging for you. So we can write this as the sum, so capital sigma right over here. We can start our index at 0. So we could say from k equals 0 all the way to k equals n of a times r to the k-th power. a general way to represent a geometric series where r is some non-zero common ratio. It can even be a negative value.", + "qid": "CecgFWTg9pQ_250" + }, + { + "Q": "I've been thinking about this for a while and I can't figure out how to find the limit of an asymptote. At 6:20 as x=>3- is it negative infinity or undefined?", + "A": "Recall that an asymptote is just a line that a given function or curve tends to (gets closer and closer to). At 6:20, the asymptote is x = 3. However, if you actually meant to find the limit of f(x) as x -> 3\u00e2\u0081\u00bb, it is -\u00e2\u0088\u009e.", + "video_name": "nOnd3SiYZqM", + "timestamps": [ + 380 + ], + "3min_transcript": "As x equals 4 from below-- So when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999-- we're getting closer and closer to our function equaling negative 5. So the limit as we approach 4 from below-- this one-sided limit from the left, we could say-- this is going to be equal to negative 5. And if we were to ask ourselves the limit of f of x, as x approaches 4 from the right, from values larger than 4, well, same exercise. f of 5 gets us here. f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of 4 is actually defined, but we're getting closer and closer to it. Even if f of 4 was not defined on either side, we would be approaching negative 5. So this is also approaching negative 5. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say-- so these two things are equal. And because these two things are equal, we know that the limit of f of x, as x approaches 4, is equal to 5. Let's look at a few more examples. So let's ask ourselves the limit of f of x-- now, this is our new f of x depicted here-- as x approaches 8. And let's approach 8 from the left. As x approaches 8 from values less than 8. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to 8. If x is 7.5, f of 7.5 is here. So it looks like our value of f of x is getting closer and closer and closer to 3. So it looks like the limit of f of x, as x approaches 8 from the negative side, is equal to 3. What about from the positive side? What about the limit of f of x as x approaches 8 from the positive direction or from the right side? Well, here we see as x is 9, this is our f of x. As x is 8.5, this is our f of 8.5. It seems like we're approaching f of x equaling 1. So notice, these two limits are different. So the non-one-sided limit, or the two-sided limit, does not exist at f of x or as we approach 8. So let me write that down. The limit of f of x, as x approaches 8--", + "qid": "nOnd3SiYZqM_380" + }, + { + "Q": "Why is around 2:30 mins in... (1/3) dividing 3x but on the other side it is multiplying!", + "A": "He s saying that multiplying both sides of the equation by the fraction (1/3) is the same as dividing both sides of the equation by 3.", + "video_name": "kbqO0YTUyAY", + "timestamps": [ + 150 + ], + "3min_transcript": "So once again, we have three equal, or we say three identical objects. They all have the same mass, but we don't know what the mass is of each of them. But what we do know is that if you total up their mass, it's the same exact mass as these nine objects And each of these nine objects have a mass of 1 kilograms. So in total, you have 9 kilograms on this side. And over here, you have three objects. They all have the same mass. And we don't know what it is. We're just calling that mass x. And what I want to do here is try to tackle this a little bit more symbolically. In the last video, we said, hey, why don't we just multiply 1/3 of this and multiply 1/3 of this? And then, essentially, we're going to keep things balanced, because we're taking 1/3 of the same mass. This total is the same as this total. That's why the scale is balanced. Now, let's think about how we can represent this symbolically. So the first thing I want you to think about is, can we set up an equation that expresses that we have these three things of mass x, Can we express that as an equation? And I'll give you a few seconds to do it. Well, let's think about it. Over here, we have three things with mass x. So their total mass, we could write as-- we could write their total mass as x plus x plus x. And over here, we have nine things with mass of 1 kilogram. I guess we could write 1 plus 1 plus 1. That's 3. Plus 1 plus 1 plus 1 plus 1. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9. And actually, this is a mathematical representation. If we set it up as an equation, it's an algebraic representation. It's not the simplest possible way we can do it, but it is a reasonable way to do it. If we want, we can say, well, if I have an x plus another x plus another x, I have three x's. So I could rewrite this as 3x. And 3x will be equal to? Well, if I sum up all of these 1's right over here-- 1 We have 9 of them, so we get 3x is equal to 9. And let me make sure I did that. 1, 2, 3, 4, 5, 6, 7, 8, 9. So that's how we would set it up. And so the next question is, what would we do? What can we do mathematically? Actually, to either one of these equations, but we'll focus on this one right now. What can we do mathematically in order to essentially solve for the x? In order to figure out what that mystery mass actually is? And I'll give you another second or two to think about it. Well, when we did it the last time with just the scales we said, OK, we've got three of these x's here. We want to have just one x here. So we can say, whatever this x is, if the scale stays balanced, it's going to be the same as whatever we have there. There might be a temptation to subtract two of the x's maybe from this side, but that won't help us. And we can even see it mathematically over here. If we subtract two x's from both sides, on the left-hand side you're going to have 3x minus 2x. And on the right-hand side, you're going to have 9 minus 2x.", + "qid": "kbqO0YTUyAY_150" + }, + { + "Q": "I got lost at 3:00 when Sal wrote 9 - 2x", + "A": "pretend 5=x. 9-2x=-1", + "video_name": "kbqO0YTUyAY", + "timestamps": [ + 180 + ], + "3min_transcript": "Can we express that as an equation? And I'll give you a few seconds to do it. Well, let's think about it. Over here, we have three things with mass x. So their total mass, we could write as-- we could write their total mass as x plus x plus x. And over here, we have nine things with mass of 1 kilogram. I guess we could write 1 plus 1 plus 1. That's 3. Plus 1 plus 1 plus 1 plus 1. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9. And actually, this is a mathematical representation. If we set it up as an equation, it's an algebraic representation. It's not the simplest possible way we can do it, but it is a reasonable way to do it. If we want, we can say, well, if I have an x plus another x plus another x, I have three x's. So I could rewrite this as 3x. And 3x will be equal to? Well, if I sum up all of these 1's right over here-- 1 We have 9 of them, so we get 3x is equal to 9. And let me make sure I did that. 1, 2, 3, 4, 5, 6, 7, 8, 9. So that's how we would set it up. And so the next question is, what would we do? What can we do mathematically? Actually, to either one of these equations, but we'll focus on this one right now. What can we do mathematically in order to essentially solve for the x? In order to figure out what that mystery mass actually is? And I'll give you another second or two to think about it. Well, when we did it the last time with just the scales we said, OK, we've got three of these x's here. We want to have just one x here. So we can say, whatever this x is, if the scale stays balanced, it's going to be the same as whatever we have there. There might be a temptation to subtract two of the x's maybe from this side, but that won't help us. And we can even see it mathematically over here. If we subtract two x's from both sides, on the left-hand side you're going to have 3x minus 2x. And on the right-hand side, you're going to have 9 minus 2x. minus 2 something is just 1 of something. So you will just have an x there if you get rid of two of them. But on the right-hand side, you're going to get 9 minus 2 So the x's still didn't help you out. You still have a mystery mass on the right-hand side. So that doesn't help. So instead, what we say is-- and we did this the last time. We said, well, what if we took 1/3 of these things? If we take 1/3 of these things and take 1/3 of these things, we should still get the same mass on both sides because the original things had the same mass. And the equivalent of doing that mathematically is to say, why don't we multiply both sides by 1/3? Or another way to say it is we could divide both sides by 3. Multiplying by 1/3 is the same thing as dividing by 3. So we're going to multiply both sides by 1/3. When you multiply both sides by 1/3-- visually over here, if you had three x's, you multiply it by 1/3, you're only going to have one x left. If you have nine of these one-kilogram boxes, you multiply it by 1/3, you're only going to have three left. And over here, you can even visually-- if you divide by 3, which is the same thing as multiplying", + "qid": "kbqO0YTUyAY_180" + }, + { + "Q": "Sal uses a calculator throughout the video however, calculators are not allowed in my school so can someone explain how he got 10.7 around the 2:40 mark?", + "A": "Honestly, if you re not allowed calculators, you should probably just leave the answer in terms of tangent, sine, or cosine unless it s an easy value to find. 65\u00c2\u00b0 isn t an easy value to find, so this should be an acceptable answer: a = 5*tan(65\u00c2\u00b0) (this is actually a more exact answer than 10.7)", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 160 + ], + "3min_transcript": "So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse.", + "qid": "l5VbdqRjTXc_160" + }, + { + "Q": "At 5:54 he says \" you could of solved this using the Pythagorean theorem... But there is an issue if im not mistaken:\n\n10.7*10.7 + 5*5 does not equal to 11.8?", + "A": "I don t think there is an issue he just wanted to solve using the trigonometric functions because that is what we had just been working on. Just to show, 10.7*10.7 = 114.49 5*5 = 25 114.49+25=139.49 And the square root of 139.49 = 11.8 a^2+b^2=c^2 so don t forget to square root everything.", + "video_name": "l5VbdqRjTXc", + "timestamps": [ + 354 + ], + "3min_transcript": "And so what trigonometric ratios-- or we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that XY is exactly 5 and we don't have to deal with this approximation, let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well we see from SohCahToa cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees is equal to the length of the adjacent side, which is 5 over the length of the hypotenuse, which has a length of b. And then we can try to solve for b. You multiply both sides times b, you're left with b times cosine of 65 degrees is equal to 5. And then to solve for b, you could divide both sides by cosine of 65 degrees. So we're just dividing-- we have to figure it out what our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees. And we're left with b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is 5 divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that 5 squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out So I'll give you a few seconds to think about what the measure of angle W is. Well here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is-- well we can simplify the left-hand side right over here. 65 plus 90 is 155. So angle W plus 155 degrees is equal to 180 degrees. And then we get angle W-- if we subtract 155 from both sides-- angle W is equal to 25 degrees. And we are done solving the right triangle shown below.", + "qid": "l5VbdqRjTXc_354" + }, + { + "Q": "At 2:28, how come he used multiplications to solve pentagon when there was more than 1 of the same numbers, when he didn't use multiplication when he was working out the rectangle", + "A": "For the rectangle, he could ve done 2*5 + 2*3 to get the perimeter of the rectangle. I think he used multiplication for the pentagon because he would have to write 2 five times, which would take too much space. If we understand that we re adding 2 five times, that just 2 multiplied by five.", + "video_name": "9uwLgf84p5w", + "timestamps": [ + 148 + ], + "3min_transcript": "When people use the word \"perimeter\" in everyday language, they're talking about the boundary of some area. And when we talk about perimeter in math, we're talking about a related idea. But now we're not just talking about the boundary. We're actually talking about the length of the boundary. How far do you have to go around the boundary to essentially go completely around the figure, completely go around the area? So let's look at this first triangle right over here. It has three sides. That's why it's a triangle. So what's its perimeter? Well, here, all the sides are the same, so the perimeter for this triangle is going to be 4 plus 4 plus 4, and whatever units this is. If this is 4 feet, 4 feet and 4 feet, then it would be 4 feet plus 4 feet plus 4 feet is equal to 12 feet. Now, I encourage you to now pause the video and figure out the parameters of these three figures. Well, it's the exact same idea. We would just add the lengths of the sides. So let's say that these distances, let's So let's say this is 3 meters, and this is also 3 meters. This is a rectangle here, so this is 5 meters. This is also 5 meters. So what is the perimeter of this rectangle going to be? What is the distance around the rectangle that bounds this area? Well, it's going to be 3 plus 5 plus 3 plus 5, which is equal to-- let's see, that's 3 plus 3 is 6, plus 5 plus 5 is 10. So that is equal to 16. And if we're saying these are all in meters, these are all in meters, then it's going to be 16 meters. Now, what about this pentagon? Let's say that each of these sides are 2-- and I'll make up a unit here. Let's say they're 2 gnus. That's a new unit of distance that I've just invented-- 2 gnus. Well, it's 2 plus 2 plus 2 plus 2 plus 2 gnus. Or we're essentially taking 1, 2, 3, 4, 5 sides. Each have a length of 2 gnus. So the perimeter here, we could add 2 repeatedly five times. Or you could just say this is 5 times 2 gnus, which is equal to 10 gnus, where gnu is a completely made-up unit of length that I just made up. Here we have a more irregular polygon, but same exact idea. How would you figure out its perimeter? Well, you just add up the lengths of its sides. And here I'll just do it unitless. I'll just assume that this is some generic units. And here the perimeter will be 1 plus 4 plus 2 plus 2-- let me scroll over to the right a little bit-- plus 4 plus 6.", + "qid": "9uwLgf84p5w_148" + }, + { + "Q": "At 2:31, Sal writes x^(n+1). Does the denominator \"n+1\" apply to the whole thing or just x? (i.e. does it read ((x^(n+1))/(n+1) or (x/(n+1))^(n+1)?) Does it matter either way? Thanks!", + "A": "The whole thing. It is [(x^(n+1)]/(n+1). And yes it matters!", + "video_name": "QxbJsg-Vdms", + "timestamps": [ + 151 + ], + "3min_transcript": "And then we have plus c. The derivative of a constant with respect to x-- a constant does not change as x changes, so it is just going to be 0, so plus 0. And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing-- and this is a very general terms-- is equal to x to the n. So given that, what is the antiderivative-- let me switch colors here. What is the antiderivative of x to the n? And remember, this is just the kind of strange-looking notation we use. It'll make more sense when we start doing definite integrals. But what is the antiderivative of x to the n? And we could say the antiderivative with respect to x, if we want to. And another way of calling this is the indefinite integral. is the derivative of what? Well, we just figured it out. It's the derivative of this thing, and we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0, plus 1, plus 2, plus pi, plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this-- you could call it the reverse power So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well, all we have to say is, well, look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was when you increment it by 1, we divide by that same value, divided by 5 plus 1. And of course, we want to encapsulate all of the possible antiderivatives, so you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now we'll do it in blue. Let's try the antiderivative of-- let's make it interesting. Let's make it 5 times x to the negative 2 power dx.", + "qid": "QxbJsg-Vdms_151" + }, + { + "Q": "at 0:35 , how does he split 60 into 6 * 10 ?", + "A": "because 6 * 10 is 60, and the order of multiplication doesn t matter.", + "video_name": "jb8mFpA1YI8", + "timestamps": [ + 35 + ], + "3min_transcript": "Let's see if we can figure out 3 times 60. Well, there's a couple of ways you could think about it. You could literally view this as 60 three times. So you could view this as 60 plus 60 plus 60. And you might be able to compute this in your head. 60 plus 60 is 120, plus another 60 is 180. And you'd be done. But another way to think about this is that 3 times 60 is the same thing as 3 times-- instead of thinking of it as 60, you could think of 60 as 6 times 10. 3 times 6 times 10. Now, when you're multiplying three numbers like this, it doesn't matter what order you multiply them in. So we could multiply the 3 times 6 first and get 18 and then multiply that times 10. And 18 times 10 is just going to be 180. It's going to be 18 with another zero. So this is going to be 180. Now, the more practice you get here, you'll realize, but I have to worry about this 0 right over here. So I'm going to put one more zero at the end. It's going to be 180. Same answer that we got right over there. Let's do another one of these. So let's say we want to multiply 50 times 7. And I encourage you to pause the video and think about it yourself, and then unpause it and see what I do. So 50, well, there's a couple of ways you could think about it. One, you could literally try to add 50 seven times. Adding 7 fifty times would take forever, but you could literally say 50 plus 50 plus 50 plus 50-- let's see, that's four-- plus 50 plus 50. Let's see, that is six. I'll do one more right over here. 50 right over here. So this is 50 seven times. If you add together 50 plus 50 is 100, 150, 200, 250, 300, 350. So you could do it that way. You just need to realize that 50 is the same thing as 10 times 5. So we could write this as 10 times 5, and then we're multiplying that times 7. Once again, the order that we multiply does not matter. So we can multiply the 5 times the 7 first. We know that that is 35, and we're going to multiply that times 10. 10 times 35, well, we're just going to stick a zero at the end of the 35. It's going to be equal to 350. Now I want to do that zero in that same color. It's going to be 350. Now, you might realize, hey, look, I could have just looked at this 5 right over here, multiplied the 5 times the 7, and have gotten the 35. And then, not forgetting that it's actually not a 5, it's a 50. So I have to multiply by 10 again, or I have to throw that 0 at the end of it to get that 0 right over here. So 50 times 7 is 350.", + "qid": "jb8mFpA1YI8_35" + }, + { + "Q": "at 6:47\nDose Sal notice this too?: b, h, c, and i are all right angles.", + "A": "If they were, yes. However, it is not given that those angles are right angles. For all we know, they could be 89 or 91 degrees or anything else close to right. Great observation though.", + "video_name": "95logvV8nXY", + "timestamps": [ + 407 + ], + "3min_transcript": "So let's say j is equal to m plus n. And then finally, we can split up h. Remember, it's this whole thing. Let's say that h is the same thing as o plus p plus q. This is o, this is p, this is q. And once again, I wanted to split up these interior angles if they're not already an angle of a triangle. I want to split them up into angles that are parts of these triangles. So we have h is equal to o, plus p, plus q. And the reason why that's interesting is now we can write the sum of these interior angles as the sum of a bunch of angles that are part of these triangles. And then we can use the fact that, for any one triangle, they add up to 180 degrees. So this expression right over her is going to be g. g is that angle right over here. We didn't make any substitutions. let me write the whole thing. So we have 900 minus, and instead of a g, well, actually I'm not making a substitution, so I can write g plus, and instead of an h I can write that h is o plus p plus q. And then plus i. i is sitting right over there. Plus i. And then plus j. And I kind of messed up the colors. The magenta will go with i. And then j is this expression right over here. So j is equal to m plus n instead of writing a j right there. And then finally, we have our f. And f, we've already seen, is equal to k plus l. So plus k plus l. So once again, I just rewrote this part right over here, in terms of these component angles. And now something very interesting is going to happen, because we know what these sums are going to be. They are the measures of the angle for this first triangle over here, for this triangle right over here. So g plus o plus k is 180 degrees. So g-- let me do this in a new color. So for this triangle right over here, we know that g plus o plus k are going to be equal to 180 degrees. So if we cross those out, we can write 180 instead. And then we also know-- let me see, I'm definitely running out of colors-- we know that p, for this middle triangle right over here, we know that p plus l plus m is 180 degrees. So you take those out and you know that sum is going to be equal to 180 degrees. And then finally, this is the home stretch here. We know that q plus n plus i is 180 degrees in this last triangle.", + "qid": "95logvV8nXY_407" + }, + { + "Q": "At 02:35, Sal says that pi/2 there is equal to 3.5 pi over seven, how does that work out? Really trying to wrap my head around this. Why did he choose the number 3.5?", + "A": "Sal noted that quadrant I contains angles from 0 (X-axis) to pi/2 radians (Y-axis). Because he was discussing the angle of 2pi / 7 radians, he converted pi / 2 to sevenths. Half of 7 is 3.5. So he checked to see if the given angle was between zero radians and 3.5pi / 7. Because the denominators of the angles were the same, it was then easy to compare the numerators and see that 2 pi / 7 radians is less than 3.5pi / 7 radians . Ttherefore the angle of 2 pi / 7 lies in quadrant 1.", + "video_name": "fYQ3GRSu4JU", + "timestamps": [ + 155 + ], + "3min_transcript": "If we go straight up, if we rotate it, essentially, if you want to think in degrees, if you rotate it counterclockwise 90 degrees, that is going to get us to pi over two. That would have been a counterclockwise rotation of pi over two radians. Now is three pi over five greater or less than that? Well, three pi over five, three pi over five is greater than, or I guess another way I can say it is, three pi over six is less than three pi over five. You make the denominator smaller, making the fraction larger. Three pi over six is the same thing as pi over two. So, let me write it this way. Pi over two is less than three pi over five. It's definitely past this. We're gonna go past this. Does that get us all the way over here? If we were to go, essentially, be pointed in the opposite direction. Instead of being pointed to the right, making a full, that would be pi radians. That would be pi radians. But this thing is less than pi. Pi would be five pi over five. This is less than pi radians. We are going to sit, we are going to sit someplace, someplace, and I'm just estimating it. We are gonna sit someplace like that. And so we are going to sit in the second quadrant. Let's think about two pi seven. Two pi over seven, do we even get past pi over two? Pi over two here would be 3.5 pi over seven. We don't even get to pi over two. We're gonna end up, we're gonna end up someplace, someplace over here. This thing is, it's greater than zero, so we're gonna definitely start moving counterclockwise, but we're not even gonna get to... This thing is less than pi over two. This is gonna throw us in the first quadrant. What about three radians? three is a little bit less than pi. Right? Three is less than pi but it's greater than pi over two. How do we know that? Well, pi is approximately 3.14159 and it just keeps going on and on forever. So, three is definitely closer to that than it is to half of that. It's going to be between pi over two, and pi. It's gonna be, if we start with this magenta ray, we rotate counterclockwise by three radians, we are gonna get... Actually, it's probably gonna be, it's gonna look something, it's gonna be something like this. But for the sake of this exercise, we have gotten ourselves, once again, into the second quadrant.", + "qid": "fYQ3GRSu4JU_155" + }, + { + "Q": "Infinagons?? 3:05", + "A": "infinigons are polygons that have an infinite number of sides.", + "video_name": "D2xYjiL8yyE", + "timestamps": [ + 185 + ], + "3min_transcript": "is something that looks like a circle, but is not a circle. I mean, what does circle mean anyway? There's this loopy line thing, And then there's this solid disk shape. And while they're related and all, they're not actually the same mathematical object. This troll proof is so much fun because, as you repeat this zigzag move, all the way to zig-finity, it approaches the shape of a circle in the disk sense, and it approaches the area of a circle, but it does not approach a circle. It's all crinkled up. And you imagine that if you stuck a straw in this thing and inflated it-- that is, added as much area as you can while keeping the perimeter as 4-- all the infinite wrinkles would smooth out, and you'd get a circle with perimeter 4 and diameter 4 over pi. In fact, you could inflate the square to get the same thing. Probably has something to do with how circles are the shape with the most area possible given their perimeter, and why soap bubbles like to be spheres, and raindrops are actually pretty sphere-ish too. But anyway, you decide the best way to respond to your friend is to try applying the same fake proof Maybe you can choose another irrational number, like square root 2. In fact, square root 2 would work great, because it's also a common geometry number. The ratio of the diagonal of a square to its perimeter. I mean, like, if this square has side length 1, then you've got this right triangle. And a squared plus b squared equals c squared. In fact, that's how you decide to start your proof for your friend. Take a right isosceles triangle. Each leg has length 1. So the hypotenuse is square root 2. Put this all in the square of perimeter 4. Now we can approach a triangle in the same way. And each time, the perimeter of the whole shape is still 4. And thus, by the time we get to a triangle, the length of the hypotenuse must be 2. Thus, the square root of 2 is 2. So you pass that over to you friend. Will he notice that you're approaching the area of a triangle without approaching a triangle? That a triangle has three sides, but the shape you end up with is a polygon with infinite sides? An infini-gon? But not a boring one, like how a regular polygon approaches because infini-gons are more fun when there's actual angles between sides. Zigzagging along to make what you've decided to call a zig-fini-gon. You begin to wonder what other zig-fini-gons you can make. Maybe if you started with a star, and zigged all the points in. And did that again and again to infinity. You get something that looks like a pentagon, but it's actually a zig-finite star with the same perimeter as the original star. Or maybe you could have a rule where at each step you zig down only part way. And then your zig-fini-gons will have more texture to it. Maybe you could throw some zags in there too. There's something fractal-ly about it, except the perimeter never changes. You could do that to a triangle, or maybe make a square that turns into a zig-fini-square. Uh oh, teacher's walking around. Better draw some axes, and pretend to be doing math. So you turn the idea sideways, and start at zero. Go to y equals 1 at x equals 1, then back to 0 at x equals 2. The next iteration is like folding this point down to 0. So the function zigzags from 0 to 1/2, to 0 to 1/2, to 0.", + "qid": "D2xYjiL8yyE_185" + }, + { + "Q": "At 3:48, why can't x be negative? .", + "A": "The domain (values of x) is any real number. It s the range (values of y) that cannot be negative. That s because y = x^2 , and we know that squaring anything (whether x is positive or negative or zero to begin with) cannot produce a negative result.", + "video_name": "96uHMcHWD2E", + "timestamps": [ + 228 + ], + "3min_transcript": "and I have my function f, and I'm gonna output f(x). And let's say this def... The function definition here, the thing that tries to figure out, \"okay, given an x, what f(x) do I produce?\", the definition says \"f(x) is going to be equal\" \"to whatever my input is, squared.\" Well, just as a little bit of review, we know what the domain here is going to be. The domain is the set of all valid inputs. So what are the valid inputs here? Well, I could take any real number and input into this, and I could take any real number and I can square it, there's nothing wrong with that, and so the domain is all real numbers. All, all real, all real numbers. But what's the range? Maybe I'll do that in a different color just to highlight it. What is going to be the range here, what is the set of all possible outputs? Well if you think about, actually, to help us think about, let me actually draw a graph here. What this looks like. So the graph of \"f(x) is equal to x squared\" is going to look something like this. So, it's gonna look, it's going to look something like this. I'm obviously hand-drawing it, so it's not perfect. It's gonna be a parabola with a, with a vertex right here at the origin. So this is the graph, this is the graph, \"y is equal to f(x),\" this of course is the x-axis, this of course is the y-axis. So let's think about it, what is the set of all possible outputs? Well in this case, the set of all possible outputs is the set of all possible y's here. Well, we see, y can take on any non-negative value. y could be zero, y could one, y could be pi, y could be e, but y cannot be negative. So the range here is, the range... We could, well we could say it a couple of ways, we could say, \"f(x)\", let me write it this way. \"f(x) is a member of the real numbers\" \"or equal to zero.\" We could write it that way, if we wanted to write it in a less mathy notation, we could say that \"f(x) is going to be\" \"greater than or equal to zero.\" f(x) is not going to be negative, so any non-negative number, the set of all non-negative numbers, that is our range. Let's do another example of this, just to make it a little bit, just to make it a little bit, a little bit clearer. Let's say that I had, let's say that I had g(x), let's say I have g(x), I'll do this in white, let's say it's equal to \"x squared over x.\" So we could try to simplify g(x) a little bit, we could say, \"look, if I have x squared\" \"and I divide it by x, that's gonna,\" \"that's the same thing as g(x) being equal to x.\" \"x squared over x\" is x, but we have to be careful.", + "qid": "96uHMcHWD2E_228" + }, + { + "Q": "at 7:20 why do you ignore the denominators when solving the equation?", + "A": "Both sides of the equation have the same denominator, so multiplying the numerators on both sides by the denominator cancels them out, which has the same effect as ignoring them. Example: x/3 = y/3 Multiply both sides by the denominator 3: 3 * (x/3) = (y/3) * 3 3x/3 = 3y/3 Simplify: x = y It s the same result as ignoring the denominators.", + "video_name": "S-XKGBesRzk", + "timestamps": [ + 440 + ], + "3min_transcript": "it actually does add up to this, then I'm done and I will have fully decomposed this fraction. I guess is the way-- I don't know if that's the correct terminology. So let's try to do that. So if I were to add these two terms, what do I get? When you add anything, you find the common denominator, and the common denominator, the easiest common denominator, is to multiply the two denominators, so let me write this here. So a over x plus 5 plus b over x minus 8 is equal to-- well, let's get the common denominator-- it's equal to x plus 5 times x minus 8. And then the a term, we would-- a over x plus 5 is the same thing as a times x minus 8 over this whole thing. I mean, if I just wrote this right here, you would just And then you could add that to the common denominator, x plus 5 times x minus 8, and it would be b times x plus 5. Important to realize, that, look. This term is the exact same thing as this term if you just cancel the x minus 8 out, and this term is the exact same thing as this term if you just cancel the x plus 5 out. But now that we have an actual common denominator, we can add them together, so we get-- let me just write the left side here over-- a over x plus 5-- I'm sorry. I want to write this over here. I want to write x plus 3 over plus 5 times x minus 8 is equal to is equal to the sum of these two things on top. a times x minus 8 plus b times x plus 5, all of that over So the denominators are the same, so we know that this, when you add this together, you have to get this. So if we want to solve for a and b, let's just set that equality. We can ignore the denominators. So we can say that x plus 3 is equal to a times x minus 8 plus b times x plus 5. Now, there's two ways to solve for a and b from this point going forward. One is the way that I was actually taught in the seventh or eighth grade, which tends to take a little longer, then there's a fast way to do it and it never hurts to do the fast way first. If you want to solve for a, let's pick an x that'll make this term disappear. So what x would make this term disappear? Well, if I say x is minus 5, then this becomes 0, and", + "qid": "S-XKGBesRzk_440" + }, + { + "Q": "why does he put hash marks on the angle markers ? like at 4:40", + "A": "He puts hash marks on the angle markers to show that the marked angles are congruent. If he didn t put hash marks on the angle markers, the (previously marked) angles would be considered congruent to the other two angles.", + "video_name": "wRBMmiNHQaE", + "timestamps": [ + 280 + ], + "3min_transcript": "we also know that angle DBA --we know that this is DBA right over here-- we also know that angle DBA and angle DBC are supplementary this angle and this angle are supplementary, their outer sides form a straight angle, they are adjacent so they are supplementary which tells us that angle DBA, this angle right over here, plus angle DBC, this angle over here, is going to be equal to 180 degrees. Now, from this top one, this top statement over here, we can subtract angle DBC from both sides and we get angle CBE is equal to 180 degrees minus angle DBC that's this information right over here, I just put or subtracted it from both sides of the equation and this right over here, if I do the exact same thing, subtract angle DBC from both sides of the equation, I get angle DBA is equal to 180 degrees --let me scroll over to the right a little bit-- is equal to 180 degrees minus angle DBC. So clearly, angle CBE is equal to 180 degrees minus angle DBC angle DBA is equal to 180 degrees minus angle DBC so they are equal to each other! They are both equal to the same thing so we get, which is what we wanted to get, angle CBE is equal to angle DBA. Angle CBE, which is this angle right over here, is equal to angle DBA and sometimes you might see that shown like this; so angle CBE, that's its measure, and you would say that And we have other vertical angles whatever this measure is, and sometimes you will see it with a double line like that, that you can say that THAT is going to be the same as whatever this angle right over here is. You will see it written like that sometimes, I like to use colors but not all books have the luxury of colors, or sometimes you will even see it written like this to show that they are the same angle; this angle and this angle --to show that these are different-- sometimes they will say that they are the same in this way. This angle is equal to this vertical angle, is equal to its vertical angle right over here and that this angle is equal to this angle that is opposite the intersection right over here. What we have proved is the general case because all I did here is I just did two general intersecting lines I picked a random angle, and then I proved that it is equal to the angle that is vertical to it.", + "qid": "wRBMmiNHQaE_280" + }, + { + "Q": "Is RSH at 2:39 a real theorem? Or just another name for the HL postulate?", + "A": "RSH is actually the HL congruence Theorem", + "video_name": "q7eF5Ci944U", + "timestamps": [ + 159 + ], + "3min_transcript": "we'll set up some triangles here since we know a lot about triangles now. And we'll set up the triangles by drawing two more radii, radius OC and radius OA. And that's useful for us because we know that they're both radii for the same circle. So they have to be the same length. The radius doesn't change on a circle. So those two things are the same length. And you might already see where this is going. Let me label this point here. Let me call this M because we're hoping that ends up being the midpoint of AC. Triangle AMO is a right triangle. This is its hypotenuse. AO is its hypotenuse. Triangle OMC is a right triangle, and this is its hypotenuse right over there. And so already showed that the hypotenuses have the same length, and both of these right triangles share segment or side OM. So OM is clearly equal to itself. And in a previous video, not the same video where we explained this thing. I think the video is called \"More on why SSA is not a postulate.\" In that video, we say that SSA is not a postulate. So SSA, not a congruency postulate. But we did establish in that video that RSH is a congruency postulate. And RSH tells us that if we have a right triangle-- that's where the R comes from-- if we have a right triangle, and we have one of the sides are congruent, and the hypotenuse is congruent, then we have two congruent triangles. And if you look at this right over here, we have two right triangles. AMO is a right triangle. CMO is a right triangle. They have one leg that's congruent, right over here, MO, and then both of their hypotenuses are congruent to each other. So, by RSH, we know that triangle AMO And so if we know that they're congruent, then their corresponding sides have to be congruent. So based on that, we then know that AM is a corresponding side. Let me do that in a different color. AM is a corresponding side to MC. So we know that AM must be equal to MC because they're corresponding sides. These are corresponding sides. So congruency implies that these are equal. And if those are equal, then we know that OD is bisecting AC. So we've established what we need to do. Another way that we could have proven it without RSH, is just straight up with the Pythagorean theorem. We already know, just by setting up these two radii right over here, we know that OA-- so we draw a little line here.", + "qid": "q7eF5Ci944U_159" + }, + { + "Q": "You lost me from 0:36", + "A": "It is like that because, let s say if x = 0 and we know that y = x+2 so: y = x+2 y=0+2 y=2 So, there the coordinates will be (0,2) Hope it made sense.", + "video_name": "RLyXTj2j_c4", + "timestamps": [ + 36 + ], + "3min_transcript": "- We're asked to use the reflect tool to find the image of quadrilateral PQRS, that's this quadrilateral right over here, for a reflection over the line y is equal to x plus two. All right, so let's use the reflect tool. So let me scroll down. So let me click on Reflect, it brings up this tool, and I want to reflect across the line y is equal to x plus two. So let me move this so it is the line y equals x plus two and to think about it, let's see, it's going to have a slope of one, the coefficient on the x term is one, so it's going to have a slope of one, so let me see if I can give this a slope of one. Is this a slope of one? Let me put it a little bit -- yep, it looks like a slope of one as the line moves one to the right. We go from one point of the line to the other, you have to go one to the right, and one up or two to the right, and two up, however much you move to the right in the x direction, you have to move that same amount in the vertical direction. So now it has a slope of one, and the y intercept is going to be the point x equals zero, y equals two, When x is equal to zero, y is going to be equal to two, so let me move this. So we see that we now go through that point. When x is equal to zero, y is equal to two. And now, we just need to reflect PQRS, this quadrilateral, over this line, so let's do that. There you go, we did it. The things, the point, like point S right over here that was to the top and left of the line, its reflection, the corresponding point in the image is now to the right and the bottom of the line. The things that were to the right and the bottom of our line, like point P, it's the corresponding point in the reflection is now on the other side of the line. So there you go, I think we're done, and we got it right.", + "qid": "RLyXTj2j_c4_36" + }, + { + "Q": "8:35 does that also mean that g(x)-h(x) and h(x)-g(x) are also solutions?", + "A": "Yes, it does. 0 - 0 does equal 0. In fact any linear combination of g and h will be solutions. You can do a*g(x) + b*h(x), where a and b are any constants, and that will be a solution.", + "video_name": "UFWAu8Ptth0", + "timestamps": [ + 515 + ], + "3min_transcript": "That's just g prime prime, plus h prime prime, plus B times-- the first derivative of this thing-- g prime plus h prime, plus C times-- this function-- g plus h. And now what can we do? Let's distribute all of these constants. We get A times g prime prime, plus A times h prime prime, plus B times the first derivative of g, plus B times the first derivative of h, plus C times g, plus C times h. And now we can rearrange them. And we get A-- let's take this one; let's take all the g terms-- A times the second derivative of g, plus B times the first derivative, plus C times g-- that's these three terms-- plus A times the second derivative of h, plus B And now we know that both g and h are solutions of the original differential equation. So by definition, if g is a solution of the original differential equation, and this was the left-hand side of that differential equation, this is going to be equal to 0, and so is this going to be equal to 0. So we've shown that this whole expression is equal to 0. So if g is a solution of the differential equation-- of this second order linear homogeneous differential equation-- and h is also a solution, then if you were to add them together, the sum of them is also a solution. So in general, if we show that g is a solution and h is a solution, you can add them. And we showed before that any constant times them is also a solution. So you could also say that some constant times g of x And maybe the constant in one of the cases is 0 or something. I don't know. But anyway, these are useful properties to maybe internalize for second order homogeneous linear differential equations. And in the next video, we're actually going to apply these properties to figure out the solutions for these. And you'll see that they're actually straightforward. I would say a lot easier than what we did in the previous first order homogeneous difference equations, or the exact equations. This is much, much easier. I'll see you in the next video.", + "qid": "UFWAu8Ptth0_515" + }, + { + "Q": "At 0:15, how does the second prize relate to the first prize? Doesn't the first prize have a predetermined ticket, thus making it an independent event?", + "A": "The first prize is a independent event, yes. But the second prize is dependent on the first prize because the ticket drawn for the first prize is not but back in.", + "video_name": "Za7G_eWKiF4", + "timestamps": [ + 15 + ], + "3min_transcript": "The marching band is holding a raffle at a football game with two prizes. After the first ticket is pulled out and the winner determined, the ticket is taped to the prize. The next ticket is pulled out to determine the winner of the second prize. Are the two events independent? Explain. Now before we even think about this exact case, let's think about what it means for events to be independent. It means that the outcome of one event doesn't affect the outcome of the other event. Now in this situation, the first event-- after the first ticket is pulled out and the winner determined-- the ticket is taped to the prize. Then the next ticket is pulled out to determine the winner of the second prize. Now, the winner of the second prize-- the possible winners, the possible outcomes for the second prize, is dependent on who was pulled out for the first prize. You can imagine if there's three tickets, let's say there's tickets A, B, and C in the bag. That's for the first prize. Now, when we think about who could be pulled out for the second prize, it's only going to be tickets B or C. Now the first prize could have gone the other way. It could have been A, B, and C. The first prize could have gone to ticket B. And then the possible outcomes for the second prize would be A or C. So the possible outcomes for the second event, for the second prize, are completely dependent on what happened or what ticket was pulled out for the first prize. So these are not independent events. The second event-- the outcomes for it, are dependent on what happened in the first event. So they are not independent. after the first ticket was pulled out, if they just wrote down the name or something, and then put that ticket back in. Instead they taped it to the prize. But if they put that ticket back in, then the second prize, it would have still had all the tickets there. It wouldn't have mattered who was picked out in the first time because their name was just written down, but their ticket was put back in. And then you would have been independent. So if you had replaced the ticket, you would have been independent. But since they didn't replace the ticket, they taped it to the prize, these are not independent events.", + "qid": "Za7G_eWKiF4_15" + }, + { + "Q": "In 1:37 Sal wrote 59+29+x=180, and he wrote 180-59-29=x, why he is subtracting if the original operation is addition? Can you add those two numbers, and subtract the total by 180 and you will get the missing angle?", + "A": "Both ways end up giving you the same answer for x. You can just use the way that is the easiest for you to use.", + "video_name": "eTwnt4G5xE4", + "timestamps": [ + 97 + ], + "3min_transcript": "We're given a bunch of lines here that intersect in all different ways and form triangles. And what I want to do in this video, we've been given the measures of some of the angles, this angle, that angle, and that angle. And what we want to do in this video is figure out what the measure of this angle is. And we're going to call that measure x. And so I encourage you to pause the video right now and try it yourself. And then I'm going to give you the solution. So I'm assuming you've unpaused it. And you've solved it or you've given it at least a good shot of it. And what's fun about these is there's multiple ways to solve these. And you kind of just have to keep figuring out what you can figure out. So let's say you start on the left-hand side right over here. If this is 121 degrees, then you'd say, well look, this angle right over here is supplementary to this angle right over there. So this is 121 degrees plus this green angle, that has to be equal to 180 degrees. So this is going to be 180 minus 121. 80 minus 20 would be 60. So that's going to be 59 degrees. So let me write that down. That's going to be 59 degrees. Now we see that we have two angles of a triangle. If you have two angles of a triangle, you can figure out the third angle, because they need to add up to 180. Or you could say that this angle right over here-- so we'll call that question mark-- we know that 59 plus 29 plus question mark needs to be equal to 180 degrees. And if we subtract the 15 out of the 29 from both sides, we get question mark is equal to 180 minus 59 minus 29 degrees. So that is going to be 180 minus 59 minus 29, let's see, 180 minus 59, we already know, is 121. And then 121 minus 29. So if you subtract just 20, you get 101. You subtract another 9, you get 92. This is equal to 92 degrees. Well, this right here is equal to 92 degrees. This angle right here is vertical with that angle. So it is also going to be equal to 92 degrees. And now we're getting pretty close. We can zoom in on this triangle down here. And let me save some space here. So let me just say that that over there is also going to be 92 degrees. And at this triangle down here, we have two of the sides of the triangle. We just have to figure out the third. And actually, we don't even have to do much math here, because we have two of the angles of this triangle. We have to figure out the third angle. So over here, we have one angle that's 92, one angle that's 29. The other one will be 180 minus 92 minus 29. And we don't even have to do any math here, because essentially, this is the exact same angles that we have in this triangle right over here. We have a 92 degree angle, we have a 29 degree angle, and the other one is 59 degrees. So in this case, it has to be also", + "qid": "eTwnt4G5xE4_97" + }, + { + "Q": "At 1:08 - 1:13, Sal mentions radians. What are they?", + "A": "Radians are basically units used for measuring angles. In a circle, if finding the radians of a circle, it would be that particular sectors arc length.", + "video_name": "D-EIh7NJvtQ", + "timestamps": [ + 68, + 73 + ], + "3min_transcript": "We already know that an angle is formed when two rays share a common endpoint. So, for example, let's say that this is one ray right over here, and then this is one another ray right over here, and then they would form an angle. And at this point right over here, their common endpoint is called the vertex of that angle. Now, we also know that not all angles seem the same. For example, this is one angle here, and then we could have another angle that looks something like this. And viewed this way, it looks like this one is much more open. So I'll say more open. And this one right over here seems less open. So to avoid having to just say, oh, more open and less open and actually becoming a little bit more exact about it, we'd actually want to measure how open an angle is, or we'd want to have a measure of the angle. Now, the most typical way that angles are measured, The most typical unit is in degrees, but later on in high school, you'll also see the unit of radians being used, especially when you learn trigonometry. But the degrees convention really comes from a circle. So let's draw ourselves a circle right over here, so that's a circle. And the convention is that-- when I say convention, it's just kind of what everyone has been doing. The convention is that you have 360 degrees in a circle. So let me explain that. So if that's the center of the circle, and if we make this ray our starting point or one side of our angle, if you go all the way around the circle, that represents 360 degrees. And the notation is 360, and then this little superscript circle represents degrees. This could be read as 360 degrees. And no one knows for sure, but there's hints in history, and there's hints in just the way that the universe works, or at least the Earth's rotation around the sun. You might recognize or you might already realize that there are 365 days in a non-leap year, 366 in a leap year. And so you can imagine ancient astronomers might have said, well, you know, that's pretty close to 360. And in fact, several ancient calendars, including the Persians and the Mayans, had 360 days in their year. And 360 is also a much neater number than 365. It has many, many more factors. It's another way of saying it's divisible by a bunch of things. But anyway, this has just been the convention, once again, what history has handed us, that a circle is viewed to have 360 degrees.", + "qid": "D-EIh7NJvtQ_68_73" + }, + { + "Q": "at the 9:23 shouldn't it be -1 = f\"(y)", + "A": "He corrects himself at 9:41", + "video_name": "Pb04ntcDJcQ", + "timestamps": [ + 563 + ], + "3min_transcript": "respect to y, is just sine of x. Plus the derivative of e to the y is e to the y. x squared So it's just x squared e to the y, plus-- what's the partial of f of y, with respect to y? It's going to be f prime of y. Well, what did we do? We took M, we integrated with respect to x, and we said, well, we might have lost some function of y, so we added that to it. And then we took the partial of that side that we've almost constructed, and we took the partial of that, with respect to y. Now, we know, since this is exact, that that is going to equal our N. So our N is up there. Cosine of x plus-- So that's going to be equal to-- I want to make sure I can read it up there-- to our N, right? Oh no, sorry. N is up here. Our N is up here. Sine of x-- let me write that-- sine of x plus x So sine of x plus x squared, e to the y, minus 1. That was just our N, from our original differential equation. And now we can solve for f prime of y. So let's see, we get sine of x plus x squared, e to the y, plus f prime of y, is equal to sine of x plus x squared, e to the y, minus 1. So let's see, we can delete sine of x from both sides. We can delete x squared e to the y from both sides. And then what are we left with? We're left with f prime of y is equal to 1. And then we're left with f of y is equal to-- well, it So what is our psi now? We wrote our psi up here, and we had this f of y here, so we So psi is a function of x and y-- we're actually pretty much almost done solving it-- psi is a function of x and y is equal to y sine of x, plus x squared, e to the y, plus y-- oh, sorry, this is f prime of y, minus 1. So this is a minus 1. So this is a minus y plus c. So this is going to be a minus y plus c. So we've solved for psi. And so what does that tell us? Well, we said that original differential equation, up here, using the partial derivative chain rule, that original differential equation, can be rewritten now", + "qid": "Pb04ntcDJcQ_563" + }, + { + "Q": "at 3:10 he breaks 10 down into it's prime numbers. if there is a number all alone do you always have to break it down?", + "A": "I quess you mean break down 10 in its primenumbers like 10=2times5. this is not necessary. remember that the goal was to make the fraction as simple as possible. In that case you make the denomenator as simple as possible. not the numerator. for instance: 2/4 you make it 1/2 3/9 you make it 1/3 6/3 you make it 3/1 so remember 10 is the same as 10/1 which you would simplify to 10. and besides 10 is easier to write than 2times5.", + "video_name": "gcnk8TnzsLc", + "timestamps": [ + 190 + ], + "3min_transcript": "", + "qid": "gcnk8TnzsLc_190" + }, + { + "Q": "At 1:11 how do you know that b is on the y intercept and how do you know y will be on zero?", + "A": "1) What is x where the y axis crosses the x axis? What is the x coordinate of the y axis? 2) What is x where any line intersects the y axis? Now, all the points (x, y) on the line satisfy y = mx + b . What is the x coordinate of the point on the line where the line intersects the y axis? What is the equation for the point on the line at x=0? (y = m*0 + b; or y = b).", + "video_name": "uk7gS3cZVp4", + "timestamps": [ + 71 + ], + "3min_transcript": "We are asked to graph y is equal to 1/3x minus 2. Now, whenever you see an equation in this form, this is called slope-intercept form. And the general way of writing it is y is equal to mx plus b, where m is the slope. And here in this case, m is equal to 1/3-- so let me write that down-- m is equal to 1/3, and b is the y-intercept. So in this case, b is equal to negative 2. And you know that b is the y-intercept, because we know that the y-intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y-intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y-intercept. In this case it is negative 2, so that means that this line it's this point right here. Negative 1, negative 2, this is the point 0, negative 2. If you don't believe me, there's nothing magical about this, try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y-intercept right there. Now, this 1/3 tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1/3, so that right there, is the slope. So it tells us that 1/3 is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y So let me graph that. So we know that this point is on the graph, that's the y-intercept. The slope tells us that if x changes by 3-- so let me go 3 three to the right, 1, 2, 3-- that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio, so 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line, and the line is the graph of this equation up here. So let me graph it. So it'll look something like that. And you're done.", + "qid": "uk7gS3cZVp4_71" + }, + { + "Q": "at 8:00 Sal shows that lim u->0 of ln (Z)= ln (lim u->0 of Z). Is this a property of limits that I don't remember?", + "A": "It s the property of limits having to do with continuous functions. If f is continuous, then lim_{x->c}f(g(x)) = f(lim_{x->c}g(x)). basically you can move the limit inside a continuous function.", + "video_name": "yUpDRpkUhf4", + "timestamps": [ + 480 + ], + "3min_transcript": "And we know this is an exponent property, which I'll now do in a different color. We know that a to the bc is equal to a to the b to the c power. So that tells us that this me is equal to the limit as u approaches 0 of the natural log of 1 plus u to the 1/u, because this is one over xu, right? 1/u, and then all of that to the 1/x. And how did I do that? Just from this exponent property, right? If I were to simplify this, I would have 1/x times 1/u, and that's where I get this 1 over xu. If I have b to the a I can put that a out front. So I could take this 1/x and put it in front of the natural log. So now what do I have? We're almost there. We have the limit as u approaches 0. Take that 1/x, put it in front of the natural log sign. 1/x times the natural log of 1 plus u to the 1/u. Fair enough. When we're taking the limit as u approaches 0, x, this term doesn't involve it at all. So we could take this out in front, because the limit doesn't affect this term. And then we're essentially saying what happens to this expression as the limit as u approaches 0. So this thing is equivalent to 1/x times the natural log of And by now hopefully you would recognize that this is the definition. This limit comes to e, if you remember anything from compound interest. You might remember it as the limit-- as n approaches infinity of 1 plus 1 over n to the n. But these things are equivalent. If you just took the substitution u is equal to 1/n, you would get this. You would just get this. So this expression right here is e That expression is e. So we're getting close. So this whole thing is equivalent to 1/x times", + "qid": "yUpDRpkUhf4_480" + }, + { + "Q": "At 3:21,how is S the same thing as 9/9?", + "A": "s is not 9/9. if it was, Sal would have gotten rid of the s. any number times 1 is itself, and 9/9 = 1. so 9/9 * s = s * 1", + "video_name": "vBlR2xNAGmo", + "timestamps": [ + 201 + ], + "3min_transcript": "But I'll just do it by hand this time, just so that we don't have to resort to some magical formulas So let's say that we define some sum, this one over here (let's call it S) Let's say that S is equal to what we have in parentheses over here It's going to be equal to four ninths, plus four ninths squared, plus four ninths to the third all the way to infinity Now let's also say that we multiply S by four ninths What's four ninths S going to look like? So then, I'm just essentially multiplying every term here by four ninths So if I take this first term and multiply it by four ninths, what am I going to get? If I take the second term and multiply it by four ninths, I'm going to get four ninths to the third power And we are going to go all the way to infinity So this is interesting When I multiply four ninths times this I get all of the terms here except for this first four ninths Now, this is kind of the magic of how we can actually find the sum of an infinite geometric series We can subtract this term right over here (this pink line) from this green line If we do that, clearly this is equal to that and this is equal to that So if we subtract this from that its equivalent to subtracting the pink from the green So we get S minus four ninths S is equal to... Well, every other term, this guy minus this guy is going to cancel out and on the right hand side you're only going to be left with this four ninths over here Then this four ninths, we can (S is the same thing as nine over nine) write this as nine over nine S minus four ninths S is equal to four ninths So nine over nine minus four over nine of something gives us five over nine So this becomes five ninths S is equal to four ninths Then to solve for S (and this is kind of magical but it's actually quite logical) Multiply both sides times the inverse of this, so times nine fifths on both sides These guys cancel out, and we get S is equal to four fifths That's really neat! We've just shown that this whole thing over here is equal to four fifths", + "qid": "vBlR2xNAGmo_201" + }, + { + "Q": "Why did Sal draw two lines over the angles at 1:18?", + "A": "To show that both angles became one angle", + "video_name": "jRrRqMJbHKc", + "timestamps": [ + 78 + ], + "3min_transcript": "All right. We're on problem 26. For the quadrilateral shown below, a quadrilateral has four sides, measure of angle A plus the measure of angle C is equal to what? And here, you should know that the sum of all the angles in a quadrilateral are equal to 360 degrees. And you might say, OK, I'll add that to my memory bank of things to memorize. Like the angles in a triangle are equal to 180. And I'll show you no, you don't have to memorize that. Because if you imagine any quadrilateral, let me draw a quadrilateral for you. And this is true of any polygon. So let's say this is some quadrilateral. You don't have to memorize that the sum of the angles is equal to 360. Although it might be useful for a quadrilateral. But I'll show you how to always prove it for any polygon. You just break it up into triangles. Then you only have to memorize one thing. If you break it up into triangles, this angle plus that angle plus that angle has to be equal to 180. equal to 180. So the angles in the quadrilateral itself are this angle and this angle. And then this angle and this angle. Well this one is just the sum of those two, and this one's just the sum of those two. So if these three added up to 180. And these three added up to 180. This plus this plus this, plus this will add up to 360. And you can do that with an arbitrarily shaped polygon. Let's do five sides, let's do a pentagon. So one, two, three, four, five sides. Wow, how many angles are there in a pentagon. Just break it up into triangles. How many triangles can you fit in it? Let's see. One, two. Each of these triangles, their angles, they add up to 180. So if you want to know that, that, that, plus that, that, that, plus that, that, and that. And that also would be the angle measures of the polygon. Because these three angles add up to that angle. That's that. Those angles add up to that one. Those angles add up to that one, and those angles add up So now hopefully, if I gave you a 20 sided polygon, you can figure out how many times can I fit triangles into it. And you'll know how many angles there are. And the sum of all of them. But anyway, back to the quadrilateral. A quadrilateral, the sum of the angles are going to be 360 degrees. So, if we say, measure of angle A, plus measure of angle C, plus these two angles. Let me write it down. Plus 95 plus 32 is going to be equal to 360. So I'll just write A plus C, just a quick notation. Let's see, 95 plus 32 is 127. Plus 127 is equal to 360. A plus C is equal to 360 minus 127.", + "qid": "jRrRqMJbHKc_78" + }, + { + "Q": "At 4:36 shouldn't the integral be equal to e^ (1 - x^2)/ 2*(1 - x^2) ?", + "A": "No, I m not sure where you re getting (1 - x\u00c2\u00b2). This is an integral best done with u-substition: u = - x\u00c2\u00b2 du = - 2x dx so 1/2\u00e2\u0088\u00ab -2x e^-x\u00c2\u00b2 dx = 1/2 \u00e2\u0088\u00ab e^u du = 1/2 e^u + C = 1/2 e^-x\u00c2\u00b2 + C", + "video_name": "DL-ozRGDlkY", + "timestamps": [ + 276 + ], + "3min_transcript": "a separable differential equation. Differential equation. And it's usually the first technique that you should try. Hey, can I separate the Ys and the Xs and as I said, this is not going to be true of many, if not most differential equations. But now that we did this we can integrate both sides. So let's do that. So, I'll find a nice color to integrate with. So, I'm going to integrate both sides. Now if you integrate the left hand side what do you get? You get and remember, we're integrating with respect to Y here. So this is going to be Y squared over two and we could put some constant there. I could call that plus C one. And if you're integrating that that's going to be equal to. Now the right hand side we're integrating with respect to X. And let's see, you could do U substitution or you could recognize that look, the derivative of negative X squared So if that was a two there and if you don't want to change the value of the integral you put the 1/2 right over there. And so now you could either do U substitution explicitly or you could do it in your head where you said U is equal to negative X squared and then DU will be negative to X, DX or you can kind of do this in your head at this point. So I have something and it's derivative so I really could just integrate with respect to that something too with respect to that U. So this is going to be 1/2. This 1/2 right over here. The anti-derivative. This is E to the negative X squared and then of course, I might have some other constant. I'll just call that C two. And once again, if this part over here what I just did seemed strange, the U substitution, you might want to review that piece. Now, what can I do here? We'll have a constant on the left hand side. It's an arbitrary constant. We don't know what it is. we could call it. So, let me just subtract C one from both sides. So if I just subtract C one from both sides I have an arbitrary so this is gonna cancel, and I have C two, sorry. Let me. So, this is C one. So these are going to cancel and C two minus C one. These are both constants, arbitrary constants and we don't know what they are yet. And so, we could just rewrite this as on the left hand side we have Y squared over two is equal to on the right hand side. I'll write 1/2 E. Let me write that in blue just because I wrote it in blue before. 1/2 E to the negative X squared and I'll just say C two minus C one. Let's just call that C. So if you take the sum of those two things let's just call that C. And so now, this is kind of a general solution. We don't know what this constant is", + "qid": "DL-ozRGDlkY_276" + }, + { + "Q": "At 07:42: You will never reach _____ steps or 1/3: What is the blank?", + "A": "The blank is infinity", + "video_name": "TINfzxSnnIE", + "timestamps": [ + 462 + ], + "3min_transcript": "Take 0.3 repeating, a repeating decimal equal to 1/3. Multiply it by 3. Obviously, by definition, 3/3 is 1, and 0.3 repeating times 3 is 0.9 repeating, which you might have noticed is also 1. The only assumption here is that 0.3 repeating equals 1/3. Maybe you don't like decimal notation in general, which brings us to reason number 9, this sum of an infinite series thing. 9/10 plus 9/100 plus 9/1000. And we can sum this series and get 1. But I can see why you might be unhappy with this. It recalls Zeno's paradoxes. How can you get across a room, when first you have to walk halfway, and then half of that, and so on. Or, how can you shoot an arrow into a target, when first it needs to go halfway, but before it can get halfway, it needs to go half of halfway, and before that, half of half of halfway, and half of half of half of halfway, and so on. Anyway, it's 1/2 plus 1/4 plus 1/8, dot, dot, dot, dot, dot, to get 1. Each time, you fall short of 1. So how can you ever do anything? Luckily, infinity has got our backs. I mean, that's like the definition of infinity, a numbers so large, you can never get there, no matter how many steps you do, no matter how high you count. This way of writing numbers with this dot, dot, dot business, or with a bar over the repeating part, is a shorthand for an infinite series, whether it be 9/10 plus 9/100, and so on to get 1. Or 3/10 plus 3/100, and so on, to get 1/3. No matter how many 3s you write down, it will always be less than 1/3, but it will also always be less than infinity 3s. Infinity is what gets us there when no real number can. The binary equivalent of 0.9 repeating is 0.1 repeating. That's exactly 1/2, plus 1/4, plus 1/8, and so on. That's how we know a dotted, dotted, dot, dot, dot The ultimate reason that 0.9 repeating equals 1 is because it works. It's consistent, just like 1 plus 1 equals 2 is consistent, and just like 1 divided by 0 equals infinity isn't. Mathematics is about making up rules and seeing what happens. And it takes great creativity to come up with good rules. The only difference between mathematics and art is that if you don't follow your invented rules precisely in mathematics, people have a tendency to tell you you're wrong. Some rules give you elementary algebra and real numbers, and these rules can't tell the difference between 0.9 repeating and 1, just like they can't tell the difference between 0.5 and 1/2, or between 0 and negative 0. I hope you see now that the view that 9.9 repeating does not equal 10 is simply un-- 9.9 repeating --able. If you started this video thinking, I h-- 7.9 repeating that 7.9 repeating is 8, I hope now, you're thinking, oh, sweet, 4.9 repeating is 5? High 4.9 repeating!", + "qid": "TINfzxSnnIE_462" + }, + { + "Q": "At 2:05, how did you get pi. When he said m 10 instead of 8r - 13 < 10. Hope this helped.", + "video_name": "x5EJG_rAtkY", + "timestamps": [ + 213 + ], + "3min_transcript": "This quantity right here has to be between negative 2 and 1/2. It has to be greater than negative 2 and 1/2 right there. And it has to be less than 2 and 1/2, so that's all I wrote there. So let's solve each of these independently. Well, this first went over here, you've learned before that I don't like improper fractions, and I don't like fractions in general. So let's make all of these fractions. Sorry, I don't like mixed numbers. I want them to be improper fractions. So let's turn all of these into improper fractions. So if I were to rewrite it, we get 2r minus 3 and 1/4 is the same thing as 3 times 4 is 12, plus 1 is 13. 2r minus 13/4 is less than-- 2 times 2 is 4, plus 1 is five-- is less than 5/2. So that's the first equation. And then the second question-- and do the same thing here-- we have 2r minus 13 over 4 has to be a greater All right, now let's solve each of these independently. To get rid of the fractions, the easiest thing to do is to multiply both sides of this equation by 4. That'll eliminate all of the fractions, so let's do that. Let's multiply-- let me scroll to the left a little bit-- let's multiply both sides of this equation by 4. 4 times 2r is 8r, 4 times negative 13 over 4 is negative 13, is less than-- and I multiplied by a positive number so I didn't have to worry about swapping the inequality-- is less than 5/2 times 4 is 10, right? You get a 2 and a 1, it's 10. So you get 8r minus 13 is less than 10. Now we can add 13 to both sides of this equation so that we get rid of it on the left-hand side. Add 13 to both sides and we get 8r-- these guys cancel out-- is less than 23, and then we divide And once again, we didn't have to worry about the inequality because we're dividing by a positive number. And we get r is less than 23 over 8. Or, if you want to write that as a mixed number, r is less than-- what is that-- 2 and 7/8. So that's one condition, but we still have to worry about this other condition. There was an and right here. Let's worry about it. So our other condition tells us 2r minus 13 over 4 has to be greater than negative 5/2. Let's multiply both sides of this equation by 4. So 4 times 2r is 8r. 4 times negative 13 over 4 is negative 13, is greater than negative 5/2 times 4 is negative 10. Now we add 13 to both sides of this equation. The left-hand side-- these guys cancel out, you're just", + "qid": "x5EJG_rAtkY_213" + }, + { + "Q": "At 3:48 I don't understand how the 2 gets under c^2 and disappears from the other side. Is it being added, subtracted, multiplied, or divided?", + "A": "The 2 dissapears on the left side because the expression is divided by 2 on both sides.", + "video_name": "tSHitjFIjd8", + "timestamps": [ + 228 + ], + "3min_transcript": "So the only right triangle in which the other two angles are equal is a 45-45-90 triangle. So what's interesting about a 45-45-90 triangle? Well other than what I just told you-- let me redraw it. I'll redraw it like this. So we already know this is 90 degrees, this is 45 degrees, this is 45 degrees. And based on what I just told you, we also know that the sides that the 45 degree angles don't share are equal. So this side is equal to this side. And if we're viewing it from a Pythagorean theorem point of view, this tells us that the two sides that are not the hypotenuse are equal. So this is a hypotenuse. We know from the Pythagorean theorem-- let's say the hypotenuse is equal to C-- the Pythagorean theorem tells us that A squared plus B squared is equal to C squared. Right? Well we know that A equals B, because this is a 45-45-90 triangle. So we could substitute A for B or B for A. But let's just substitute B for A. So we could say B squared plus B squared is equal to C squared. Or 2B squared is equal to C squared. Or B squared is equal to C squared over 2. Or B is equal to the square root of C squared over 2. the numerator and the square root of the denominator-- C over the square root of 2. And actually, even though this is a presentation on triangles, I'm going to give you a little bit of actually information on something called rationalizing denominators. So this is perfectly correct. We just arrived at B-- and we also know that A equals B-- but that B is equal to C divided by the square root of 2. It turns out that in most of mathematics, and I never understood quite exactly why this was the case, people don't like square root of 2s in the denominator. Or in general they don't like irrational numbers in the denominator. Irrational numbers are numbers that have decimal places that never repeat and never end. So the way that they get rid of irrational numbers in the denominator is that you do something called rationalizing the denominator. And the way you rationalize a denominator-- let's take our example right now. If we had C over the square root of 2, we just multiply both the numerator and the denominator by the", + "qid": "tSHitjFIjd8_228" + }, + { + "Q": "5:01 what are assymptotes? (I don't think I spelled that right)\nSal keeps mentioning them but I think if I don't know what they are I'm not going anywhere with hyperbolas....", + "A": "an asymptote is a line that the function gets infinitely close to but never touches. You draw them on your graph before you draw your hyperbola (using dotted lines).", + "video_name": "pzSyOTkAsY4", + "timestamps": [ + 301 + ], + "3min_transcript": "you'll probably get confused. Because sometimes they always use the a under the x and the b under the y, or sometimes they always use the a under the positive term and to b under the negative term. So if you just memorize, oh, a divided by b, that's the slope of the asymptote and all of that, you might be using the wrong a and b. So I encourage you to always re-prove it to yourself. And that's what we're going to do right here. It actually doesn't take too long. So these are both hyperbolas. And what I like to do whenever I have a hyperbola is solve for y. So in this case, if I subtract x squared over a squared from both sides, I get-- let me change the color-- I get minus y squared over b squared. That stays there. Is equal to 1 minus x squared over a squared. And then, let's see, I want to get rid of this minus, and I want to get rid of this b squared. So let's multiply both sides of this equation times minus b squared. minus and the b squared go away, and you're just left with y squared is equal to minus b squared. And then minus b squared times a plus, it becomes a plus b squared over a squared x squared. We're almost there. And then you get y is equal to-- and I'm doing this on purpose-- the plus or minus square root, because it can be the plus or minus square root. Of-- and let's switch these around, just so I have the positive term first. b squared over a squared x squared minus b squared. Now you said, Sal, you said this was simple. This looks like a really complicated thing. But remember, we're doing this to figure out asymptotes of the hyperbola, just to kind of give you a sense of where we're going. that other hyperbola. So a hyperbola, if that's the x, that's the y-axis, it has two asymptotes. And the asymptotes, they're these lines that the hyperbola will approach. So if those are the two asymptotes-- and they're always the negative slope of each other-- we know that this hyperbola's is either, and we'll show in a second which one it is, it's either going to look something like this, where as we approach infinity we get closer and closer this line and closer and closer to that line. And here it's either going to look like that-- I didn't draw it perfectly; it never touches the asymptote. It just gets closer and closer and closer, arbitrarily It's either going to look like that, where it opens up to the right and left. Or our hyperbola's going to open up and down. And once again, as you go further and further, and asymptote means it's just going to get closer and closer to one of these lines without ever touching it. It will get infinitely close as you get infinitely far away,", + "qid": "pzSyOTkAsY4_301" + }, + { + "Q": "At 4:25 when multiplying b^2, why does the x^2 get moved from the numerator?", + "A": "x^2 is still part of the numerator - just think of it as x^2/1, multiplied by b^2/a^2. you could also write it as a^2*x^2/b^2, all as one fraction... it means the same thing (multiply x^2 and a^2 and divide by b^2 ->> since multiplication and division occur at the same level of the order of operations, both ways of writing it out are totally equivalent!). hope that helps", + "video_name": "pzSyOTkAsY4", + "timestamps": [ + 265 + ], + "3min_transcript": "minus y squared over b squared is equal to 1. And notice the only difference between this equation and this one is that instead of a plus y squared, we have a minus y squared here. So that would be one hyperbola. The other one would be if the minus sign was the other way around. If it was y squared over b squared minus x squared over a squared is equal to 1. So now the minus is in front of the x squared term instead of the y squared term. And what I want to do now is try to figure out, how do we graph either of these parabolas? Maybe we'll do both cases. And in a lot of text books, or even if you look it up over the web, they'll give you formulas. But I don't like those formulas. One, because I'll always forget it. And you'll forget it immediately after taking the test. You might want to memorize it if you just want to be able to do the test a little bit faster. But you'll forget it. you'll probably get confused. Because sometimes they always use the a under the x and the b under the y, or sometimes they always use the a under the positive term and to b under the negative term. So if you just memorize, oh, a divided by b, that's the slope of the asymptote and all of that, you might be using the wrong a and b. So I encourage you to always re-prove it to yourself. And that's what we're going to do right here. It actually doesn't take too long. So these are both hyperbolas. And what I like to do whenever I have a hyperbola is solve for y. So in this case, if I subtract x squared over a squared from both sides, I get-- let me change the color-- I get minus y squared over b squared. That stays there. Is equal to 1 minus x squared over a squared. And then, let's see, I want to get rid of this minus, and I want to get rid of this b squared. So let's multiply both sides of this equation times minus b squared. minus and the b squared go away, and you're just left with y squared is equal to minus b squared. And then minus b squared times a plus, it becomes a plus b squared over a squared x squared. We're almost there. And then you get y is equal to-- and I'm doing this on purpose-- the plus or minus square root, because it can be the plus or minus square root. Of-- and let's switch these around, just so I have the positive term first. b squared over a squared x squared minus b squared. Now you said, Sal, you said this was simple. This looks like a really complicated thing. But remember, we're doing this to figure out asymptotes of the hyperbola, just to kind of give you a sense of where we're going.", + "qid": "pzSyOTkAsY4_265" + }, + { + "Q": "3:20 Commutative property of addition 704 = 700 + 4 or 4 + 700\n3:30 Associative property of addition 18 + (4 + 700) = (18 + 4) + 700", + "A": "that is 3th grade math for asossitive property and 6th grade math for comunitive property", + "video_name": "jAfJcgPGqgI", + "timestamps": [ + 200, + 210 + ], + "3min_transcript": "to get to 500? And I could have done that in my head. Okay, I need to add 20. 355 minus 20 is 335. But now this problem is much, much simpler to compute. 335 plus 500, well, it's going to be three hundreds plus five hundreds, it's going to be equal to 800 and then we have 35. 835. Now, to make it a little bit clearer what we did, remember, we wanted to take 20 from here and put it over here, so we could break up 355, we could say that it's going to be equal to, it's going to be equal to, let's break it up into 335 and 20, and remember, the whole reason why I picked 20 is because I'm going to want to add that to 480, but I'm just doing it step-by-step here, so plus 480, and now I can just change the order with which I add, 335 plus 20 plus 480, and instead of adding the 335 and 20 first, I could add the 20 and 480 first, so I could add these two characters first, and so then I'm going to be left with, this is going to be equal to 335 plus, what's 20 plus 480? Well, that was the whole point. I picked 20 so that I can get to 500. 20 plus 480 is going to be 500 and then, now, you can add them. This is going to be 835. So this is just a longer way of saying what I did here. I took 20 from 355, so I can make the 480 into a 500. Let's do a couple more examples of this, and remember, the key is just thinking about how could I add or take away from one of the numbers to make them simpler? So there's a couple ways we could tackle this. One way, we could try to get the 704 to be equal to 700. So, we could say that this is the same thing as 18 plus, I could write 700 plus 4, or I could write it as 4 plus 700, like that, and then I could put the parentheses around this first, and then I could just switch the order in which I add, so this is the same thing as 18 plus 4 plus 700, and now I could add the 18 and the 4 first. Now what's 18 plus 4? It's 22, and then I have plus 700, and now this is pretty easy to compute, and all of these are going to be equal to each other, so let me just write it like that. 22 plus 700, I could do that in my head.", + "qid": "jAfJcgPGqgI_200_210" + }, + { + "Q": "At 2:18 talking about the passage from the purple curve to the yellow segment of the function, he said a slope is not defined there because we could draw a lot.\nYet a unique limit for that point exists, so it should also exist a derivative right?\n(The same happens between the blue and the orange segments at the end.)", + "A": "just because a limit exists does not mean that a function is differentiable, although it is one of the conditions of that. For a function to be differentiable, the derivative from the left side of the point must be equal to the derivative from the right using one sided limits.", + "video_name": "eVme7kuGyuo", + "timestamps": [ + 138 + ], + "3min_transcript": "So I've got this crazy discontinuous function here, which we'll call f of x. And my goal is to try to draw its derivative right over here. So what I'm going to need to think about is the slope of the tangent line, or the slope at each point in this curve, and then try my best to draw that slope. So let's try to tackle it. So right over here at this point, the slope is positive. And actually, it's a good bit positive. And then as we get larger and larger x's, the slope is still positive, but it's less positive-- and all the way up to this point right over here, where it becomes 0. So let's see how I could draw that over here. So over here we know that the slope must be equal to 0-- right over here. Remember over here, I'm going to try to draw y is equal to f prime of x. And I'm going to assume that this is some type of a parabola. But let's say that, so let's see, here the slope is quite positive. So let's say the slope is right over here. And then it gets less and less and less positive. And I'll assume it does it in a linear fashion. That's why I had to assume that it's some type of a parabola. So it gets less and less and less positive. Notice here, for example, the slope is still positive. And so when you look at the derivative, the slope is still a positive value. But as we get larger and larger x's up to this point, the slope is getting less and less positive, all the way to 0. And then the slope is getting more and more negative. And at this point, it seems like the slope is just as negative as it was positive there. So at this point right over here, the slope is just as negative as it was positive right over there. So it seems like this would be a reasonable view of the slope of the tangent line over this interval. Here the slope seems constant. Our slope is a constant positive value. So once again, our slope here is a constant positive line. Let me be careful here because at this point, our slope won't really be defined, because our slope, you could draw multiple tangent lines at this little pointy point. So let me just draw a circle right over there. But then as we get right over here, the slope seems to be positive. So let's draw that. The slope seems to be positive, although it's not as positive as it was there. So the slope looks like it is-- I'm just trying to eyeball it-- so the slope is a constant positive this entire time. We have a line with a constant positive slope. So it might look something like this.", + "qid": "eVme7kuGyuo_138" + }, + { + "Q": "Again 11:53:\nHow do I get from [x]B to [Ax]B ?? that is [x]B = [Ax]B algebraically?", + "A": "Sal is not saying that [Ax]_B = [x]_B. He wrote that D[x]_B = [Ax]_B. We have the rule that some vector v can be expressed in alternative coordinate systems by: C [v]_B = v, and [v]_B = C^-1 v. Ax is some vector. Therefore, we can apply the rule to it. x is also some vector. Therefore, we can apple the rule to it.", + "video_name": "PiuhTj0zCf4", + "timestamps": [ + 713 + ], + "3min_transcript": "Let's see what D times xB is equal to. So let's say if we take D times xB, so this thing right here should be equal to D times the representation or the coordinates of x with respect to the basis B. That's what we're claiming. We're saying that this guy is equal to D times the representation of x with respect to the coordinates with respect to the basis B. Let me write all of this down. I'll do it right here, because I think it's nice to have this graphic up here. So we can say that D times xB is equal to this thing right here. It's the same thing as the transformation of x represented in coordinates with respect to B, or in these nonstandard coordinates. represented in this coordinate system, represented in coordinates with respect to B. We see that right there. But what is the transformation of x? That's the same thing as A times x. That's kind of the standard transformation if x was represented in standard coordinates. So this is equal to x in standard coordinates times the matrix A. Then that will get us to this dot in standard coordinates, but then we want to convert it to these nonstandard coordinates just like that. Now, if we have this, how can we just figure out what the vector Ax should look like? What this vector should look like? Well, we can look at this equation right here. We have this. This is the same thing as this. we want to go the other way. We have this. We have that right there. That's this right there. We want to get just this dot represented in regular standard coordinates. So what do we do? We multiply it by C. Let me write it this way. If we multiply both sides of this equation times C, what do we get? We get this right here. I was looking at the right equation the first time. We have this right here, which is the same-- first intuition is always right. We have this, which is the same thing as this right here. So this can be rewritten. This thing can be rewritten as C inverse-- we don't have an x here. We have an Ax here, so C inverse times Ax. The vector Ax represented in these nonstandard coordinates", + "qid": "PiuhTj0zCf4_713" + }, + { + "Q": "At 14:00 why did Sal do b-a instead of b+a ? I thought you had to add vectors together to get the resultant vector. Can someone clarify this for me ?", + "A": "I have the same question. I don t have a good intuitive answer. But if you plot the vectors mentioned in the video we can see that a-b or b-a is the only vector that passes through the tip of the 2 vectors.", + "video_name": "hWhs2cIj7Cw", + "timestamps": [ + 840 + ], + "3min_transcript": "all and I go up. So my vector b will look like that. Now I'm going to say that these are position vectors, that we draw them in standard form. When you draw them in standard form, their endpoints represent some position. So you can almost view these as coordinate points in R2. This is R2. All of these coordinate axes I draw are going be R2. Now what if I asked you, give me a parametrization of the line that goes through these two points. So essentially, I want the equation-- if you're thinking in Algebra 1 terms-- I want the equation for the line that goes through these two points. So the classic way, you would have figured out the slope and all of that, and then you would have substituted back in. But instead, what we can do is, we can say, hey look, this line that goes through both of those points-- you could that's a better-- Both of these vectors lie on this line. Now, what vector can be represented by that line? Or even better, what vector, if I take any arbitrary scalar-- can represent any other vector on that line? Now let me do it this way. What if I were to take-- so this is vector b here-- what if I were to take b minus a? We learned in, I think it was the previous video, that b minus a, you'll get this vector right here. You'll get the difference in the two vectors. This is the vector b minus the vector a. And you just think about it. What do I have to add to a to get to b? I have to add b minus a. So if I can get the vector b minus a-- right, we know how We just subtract the vectors, and then multiply it by any scalar, then we're going to get any point along that line. So what happens if we take t, so some scalar, times our vector, times the vectors b minus a? What will we get then? So b minus a looks like that. But if we were to draw it in standard form-- remember, in standard form b minus a would look something like this. It would start at 0, it would be parallel to this, and then from 0 we would draw its endpoint. So if we just multiplied some scalar times b minus a, we would actually just get points or vectors that lie on this line. Vectors that lie on that line right there. Now, that's not what we set out to do. We wanted to figure out an equation, or parametrization, if you will, of this line, or this set. Let's call this set l. So we want to know what that set is equal to. So in order to get there, we have to start with this, which", + "qid": "hWhs2cIj7Cw_840" + }, + { + "Q": "At 0:42why did Sal write 4000+500=3000", + "A": "Look at the original problem at the top of the screen. Sal is not saying 4000+500=3000. He is writing the problem out using numbers. The original problem as + ? that Sal has not yet written in. But, by the end of the video, he does.", + "video_name": "a_mzIWvHx_Y", + "timestamps": [ + 42 + ], + "3min_transcript": "We have 4,5000 equals 3 thousands plus how many hundreds, question mark hundreds? So let's write this left-hand side, but I'm going to write it out in terms of thousands and hundreds. So I'll write the thousands in orange. So this is equal to 4 thousands, which is the same thing as just 4,000, plus 500, which you could also view as 5 hundreds. So this is the left-hand side. Now let's look at the right-hand side. We have 3 thousands. So it's 3 thousands. Now let's not even look at this right now. Let's just think about what do we have to add to this right-hand side in order to get the same thing that we have on the left-hand side? Well, if you compare the 3,000 and the 4,000, you see you have an extra 1,000 over here. So let's add an extra 1,000 on the right. So we're going to add one extra 1,000. And now we just have 3,000 plus 1,000. This makes it the 4,000. But then, of course, we also need another 500. So we're going to need plus a 500 right over here. we need to say 4,000 plus 500 is equal to 3,000 plus 1,500. Now, the way they've set this up, we need to express-- so it almost looks the same. On the left-hand side, this is 4,500. So this right over here, this is the same thing as 4,500. This is this right over here. And on the right-hand side, we have 3 thousands. So that's this right over here. That's the 3 thousands. And then we just need to express this as hundreds. So 1,500, this is the same thing as 15 hundreds. So let's rewrite everything. We can rewrite this as saying 4,500, just to get the exact same form that they wrote it over there. So we could write 4,500 is equal to 3 thousands plus-- now This is 15 hundreds. Literally, if you took 15 times 100, it's going to be equal to 1,500. So this could be viewed as 15 hundreds, so plus 15 hundreds. So in this situation, the question mark is equal to 15.", + "qid": "a_mzIWvHx_Y_42" + }, + { + "Q": "at 6:30 what does he mean?", + "A": "He is doing what is called FOIL. in the problem (x+5)(x+1), you first multiply the x in both parentheses, then the x in the first and the 1 in the second, then the 5 and the x and then the 5 and 1. the FOIL means: F-first (the two x s) O-outside (the x and the 1) I-inside (5 and x) L-last (5 and 1) Hope this helps clear things a little!", + "video_name": "7Uos1ED3KHI", + "timestamps": [ + 390 + ], + "3min_transcript": "So to make them the same, I also have to add the extra condition that x cannot equal negative 3. So likewise, over here, if this was a function, let's say we wrote y is equal to 9x plus 3 over 12x plus 4 and we wanted to graph it, when we simplify it, the temptation is oh, well, we factored out a 3x plus 1 in the numerator and They cancel out. The temptation is to say, well, this is the same graph as y is equal to the constant 3/4, which is just a horizontal line at y is equal to 3/4. But we have to add one condition. We have to eliminate-- we have to exclude the x-values that would have made this thing right here equal to zero, and that would have been zero if x is equal to negative 1/3. If x is equal to negative 1/3, this or this denominator would be equal to zero. So even over here, we'd have to say x cannot be equal to That condition is what really makes that equal to that, that x cannot be equal to negative 1/3. Let's do a couple more of these. And I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Or actually, even better, let me do this a little bit. x squared plus 6x plus 5 over x squared minus x minus 2. So once again, we want to factor the numerator and the denonminator, just like we did with traditional numbers when we first learned about fractions and lowest terms. So multiply them equal 5 and I add them equal 6? Well, the numbers that pop in my head are 5 and 1. So the numerator is x plus 5 times x minus 1. And then our denonminator, two numbers. Multiply negative 2, add a negative 1. Negative 2 and positive 1 pop out of my head. So this is a positive 1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive 1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor in the numerator and the denonminator. These cancel out. So you could say that this is equal to x plus 5 over x minus 2. But for them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to", + "qid": "7Uos1ED3KHI_390" + }, + { + "Q": "How do we do it from slope-intersect form? 00:59", + "A": "Starting from 5x + 3y = 7, you subtract 5x and divide by 3, so 3y = - 5x + 7 or y = -5/3x + 7/3. He graphed 7/3 as the y intercept, then go down 5 over 3 to get second point. These are not very neat numbers to work with.", + "video_name": "MRAIgJmRmag", + "timestamps": [ + 59 + ], + "3min_transcript": "Solve the system of linear equations by graphing, and they give us two equations here. 5x plus 3y is equal to 7, and 3x minus 2y is equal to 8. When they say, \"Solve the system of linear equations,\" they're really just saying find an x and a y that satisfies both of these equations. And when they say to do it by graphing, we're essentially going to graph this first equation. Remember, the graph is really just depicting all of the x's and y's that satisfy this first equation, and then we graph the second equation that's depicting all of the x's and y's that satisfy that one. So if we were looking for an x and a y that satisfies both, that point needs to be on both equations or it has to be on both graphs. So it'll be the intersection of the two graphs. So let's try to see if we can do that. So let's focus on this first equation, and I want to graph it. So I have 5x plus 3y is equal to 7. There's a couple of ways we could graph this. We could put this in slope-intercept form, You just really need two points to graph a line. So let me just set some points over here. Let's say x and y. When x is equal to 0, what does y need to be equal to? So when x is equal to 0, we have-- let me do it over here-- we have 5 times 0, plus 3 times y, is equal to 7. That's just 0 over there. So you have 3y is equal to 7. Divide both sides by 3, you get y is equal to 7/3, which is the same thing as 2 and 1/3 if we want to write it as a mixed number. Now let's set y equal to 0. So if we set y as equal to 0, we get 5x, plus 3 times 0, is equal to 7. Or in this part right over here, it just becomes 0. So we have 5x is equal to 7. Divide both sides by 5, and we get So let's graph both of these points, and then we should be able to graph this line, or at least a pretty good approximation of that line. So we have the point, 0, 2 and 1/3. So that's that point right over there. So I'll call it 0, 7/3 right over there, and then we have the point, 7/5, 0, or 1 and 2/5, 0. So 1 and 2/5. 2/5 is a little less than a half. So 1 and 2/5, 0. So our line is going to look something like this. I just have to connect the dots. It's always hard to draw the straight line. I'll draw it as a dotted line. So it would look something like this. Normally, when you have to solve a system of equations", + "qid": "MRAIgJmRmag_59" + }, + { + "Q": "At 13:23, why doesn't he divide the number inside the radical by 2?", + "A": "Terms are things we add or subtract. They are held together by multiplication and division. The numerator only has 2 terms : -12 and 2*sqrt39. Both of those terms were divided by 2 to get -6 and sqrt39. But, if the numerator had been -12 + 2 + sqrt39, (in other words, 3 terms) and then we divided by 2, we would get -6 + 1 + (sqrt39)/2 ( Of course, that would simplify to -5 + (sqrt39)/2. )", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 803 + ], + "3min_transcript": "", + "qid": "i7idZfS8t8w_803" + }, + { + "Q": "Isn't there a small mistake at 16:10 ? Sal says \"a little less than one\", where the graph shows a little less than 0 ...", + "A": "he says maybe close to zero but i little bit less than that", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 970 + ], + "3min_transcript": "", + "qid": "i7idZfS8t8w_970" + }, + { + "Q": "at 4:16 why did it change to 10 why didn't you put 100?i,m sorry i am so very new to this", + "A": "the 100 was under a square root and then he took the square root of 100 to get 10", + "video_name": "i7idZfS8t8w", + "timestamps": [ + 256 + ], + "3min_transcript": "", + "qid": "i7idZfS8t8w_256" + }, + { + "Q": "At 7:11, Sal describes the system as existing in R4, but isn't it also safe to describe this as 4 column vectors in R3?", + "A": "For this system of equations Ax = b , A , not the system, is 4 column vectors. x is an R^4 variable vector, each row is a plane in R^4 , and Ax is constrained to b , an R^3 column vector of constants.", + "video_name": "JVDrlTdzxiI", + "timestamps": [ + 431 + ], + "3min_transcript": "entry is in a lower row than that one. So it's in a column to the right of this one right there. And I just inspected, this looks like a-- this column two looks kind of like a free variable-- there's no pivot entry there, no pivot entry there. But let's see, let's map this back to our system of equations. These are just numbers to me and I just kind of mechanically, almost like a computer, put this in reduced row echelon form. Actually, almost exactly like a computer. But let me put it back to my system of linear equations, to see what our result is. So we get 1 times x1, let me write it in yellow. So I get 1 times x1, plus 2 times x2, plus 0 times x3, plus 3 times x4 is equal to 4. Obviously I could ignore this term right there, I didn't even have to write it. Actually. Then I get 0 times x1, plus 0 times x2, plus 1 times x3, so 1 times x3, minus 2 times x4, is equal to 4. And then this last term, what do I get? I get 0 x1, plus 0 x2 plus 0 x3 plus 0 x4, well, all of that's equal to 0, and I've got to write something on the left-hand side. So let me just write a 0, and that's got to be equal to minus 4. Well this doesn't make any sense whatsoever. 0 equals minus 4. This is this is a nonsensical constraint, this is impossible. 0 can never equal minus 4. This is impossible. Which means that it is essentially impossible to find an intersection of these three systems of equations, or a solution set that satisfies all of them. When we looked at this initially, at the beginning of the of the video, we said there are only three equations, we have four unknowns, maybe there's going But turns out that these three-- I guess you can call them these three surfaces-- don't intersect in r4. These are all four dimensional, we're dealing in r4 right here, because we have-- I guess each vector has four components, or we have four variables, is the way you could think about it. And it's always hard to visualize things in r4. But if we were doing things in r3, we can imagine the situation where, let's say we had two planes in r3. So that's one plane right there, and then I had another completely parallel plane to that one. So I had another completely parallel plane to that first one. Even though these would be two planes in r3, so let me give So let's say that this first plane was represented by the equation 3x plus 6y plus 9z is equal to 5, and the second", + "qid": "JVDrlTdzxiI_431" + }, + { + "Q": "11:30. To have an infinte number of solutions does one needs to have free variables AND a row of all Zeroes? The video was unclear on this point but alluded to it.", + "A": "row of 0 s is not a necessary condition, e.g. x1 - x2 = 5 x1 - x2 + x3 = 3 reduces to 1 -1 0 | 5 0 0 1 | -2 ` with x2 being free. The solution is (x1, x2, x3) = (5, 0, -2) + x2(1, 1, 0) If you think of a row being a constraint to the solution(s), a row of zero s seems to indicate one that is a linear combination of the remaining one s i.e. it is a superfluous constraint.", + "video_name": "JVDrlTdzxiI", + "timestamps": [ + 690 + ], + "3min_transcript": "of parallel equations, they won't intersect. And you're going to get, when you put it in reduced row echelon form, or you just do basic elimination, or you solve the systems, you're going to get a statement that zero is equal to something, and that means that there are no solutions. So the general take-away, if you have zero equals something, no solutions. If you have the same number of pivot variables, the same number of pivot entries as you do columns, so if you get the situations-- let me write this down, this is good to know. if you have zero is equal to anything, then that means no solution. If you're dealing with r3, then you probably have parallel planes, in r2 you're dealing with parallel lines. If you have the situation where you have the same number of pivot entries as columns, so it's just 1, 1, 1, 1, this I think you get the idea. That equals a, b, c, d. Then you have a unique solution. Now if, you have any free variables-- so free variables look like this, so let's say we have 1, 0, 1, 0, and then I have the entry 1, 1, let me be careful. 0, let me do it like this. 1, 0, 0, and then I have the entry 1, 2, and then I have a bunch of zeroes over here. And then this has to equal zero-- remember, if this was a bunch of zeroes equaling some variable, then I would have no solution, or equalling some constant, let's say this is equal to 5, this is equal to 2. If this is our reduced row echelon form that we eventually get to, then we have a few free variables. This is a free, or I guess we could call this column a free Because it has no pivot entries. These are the pivot entries. So this is variable x2 and that's variable x4. Then these would be free, we can set them equal to anything. So then here we have unlimited solutions, or no unique solutions. And that was actually the first example we saw. And these are really the three cases that you're going to see every time, and it's good to get familiar with them so you're never going to get stumped up when you have something like 0 equals minus 4, or 0 equals 3. Or if you have just a bunch of zeros and a bunch of rows. I want to make that very clear. Sometimes, you see a bunch of zeroes here, on the left-hand side of the augmented divide, and you might say, oh maybe I have no unique solutions, I have an infinite number of solutions. But you have to look at this entry right here. Only if this whole thing is zero and you have free variables, then you have an infinite number of solutions. If you have a statement like, 0 is equal to a, if this is equal to 7 right here, then all of the sudden you would", + "qid": "JVDrlTdzxiI_690" + }, + { + "Q": "why at 1:18 does neg. numerator over pos. denominator (-3/+7) become the whole fraction neg. ?", + "A": "You may be aware of the mathematical rule which dictates that if you divide a negative by a positive, you get a negative. A fraction basically means to divide the top number, or the numerator, by three bottom number, the denominator. EXAMPLE: -2/4 We know that +2/+4 is 1/2, so we can simplify -2/4 to -1/2, or -0.5 I hope this this helps!", + "video_name": "pi3WWQ0q6Lc", + "timestamps": [ + 78 + ], + "3min_transcript": "Let's do a few examples multiplying fractions. So let's multiply negative 7 times 3/49. So you might say, I don't see a fraction here. This looks like an integer. But you just to remind yourself that the negative 7 can be rewritten as negative 7/1 times 3/49. Now we can multiply the numerators. So the numerator is going to be negative 7 times 3. And the denominator is going to be 1 times 49. 1 times 49. And this is going to be equal to-- 7 times 3 is 21. And one of their signs is negative, so a negative times a positive is going to be a negative. So this is going to be negative 21. You could view this as negative 7 plus negative 7 plus negative 7. And that's going to be over 49. And this is the correct value, but we can simplify it more because 21 and 49 both share 7 as a factor. So let's divide both the numerator and the denominator by 7. Divide the numerator and the denominator by 7. And so this gets us negative 3 in the numerator. And in the denominator, we have 7. So we could view it as negative 3 over 7. Or, you could even do it as negative 3/7. Let's do another one. Let's take 5/9 times-- I'll switch colors more in this one. That one's a little monotonous going all red there. 5/9 times 3/15. So this is going to be equal to-- we multiply the numerators. So it's going to be 5 times 3. 5 times 3 in the numerator. And the denominator is going to be 9 times 15. 9 times 15. you see that there is already common factors in the numerator and the denominator. Both the numerator and the denominator, they're both divisible by 5 and they're both divisible by 3, which essentially tells us that they're divisible by 15. So we can divide the numerator and denominator by 15. So divide the numerator by 15, which is just like dividing by 5 and then dividing by 3. So we'll just divide by 15. Divide by 15. And this is going to be equal to-- well, 5 times 3 is 15. Divided by 15 you get 1 in the numerator. And in the denominator, 9 times 15 divided by 15. Well, that's just going to be 9. So it's equal to 1/9. Let's do another one. What would negative 5/9 times negative 3/15 be? Well, we've already figured out what positive 5/9 times positive 3/15 would be. So now we just have to care about the sign.", + "qid": "pi3WWQ0q6Lc_78" + }, + { + "Q": "at 5:05 is he writing out what he did in the model?", + "A": "Yep, that is correct. Sal (the speaker) just replaced the equation with absolute values, which are the lines on both sides of the subtraction.", + "video_name": "DPuK6ZgBGmE", + "timestamps": [ + 305 + ], + "3min_transcript": "Once again, this is two, this is three. She deviates. Her absolute deviation is three. And then we wanna take the mean of the absolute deviation. That's the M in MAD, in Mean Absolute Deviation. This is Manueala's absolute deviation, Sophia's absolute deviation, Jada's absolute deviation, Tara's absolute deviation. We want the mean of those, so we divide by the number of datapoints, and we get zero plus one, plus two, plus three, is six over four. Six over four, which is the same thing as 1 1/2. Or, lemme just write it in all the different ways. We could write it as three halves, or 1 1/2, or 1.5. Which gives us a measure of how much do these datapoints vary from the mean of four. I know what some of you are thinking. \"Wait, I thought there was a formula \"associated with the mean absolute deviation. \"It seems really complex. \"It has all of these absolute-value signs That's all we did. When we write all those absolute-value signs, that's just a fancy way of looking at each datapoint, and thinking about how much does it deviate from the mean, whether it's above or below. That's what the absolute value does. It doesn't matter, if it's three below, we just say three. If it's two above, we just say two. We don't put a positive or negative on. Just so you're comfortable seeing how this is the exact same thing you would've done with the formula, let's do it that way, as well. So the mean absolute deviation is going to be equal to. Well, we'll start with Manueala. How many bubbles did she blow? She blew four. From that you subtract the mean of four, take the absolute value. That's her absolute deviation. Of course, this does evaluate to this zero, to zero here. Then you take the absolute value. Sophia blew five bubbles, and the mean is four. Then you do that for Jada. Jada blew six bubbles; the mean is four. And then you do it for Tara. Then you divide it by the number of datapoints you have. Lemme make it very clear. This right over here, this four, is the mean. This four is the mean. You're taking each of the datapoints, and you're seeing how far it is away from the mean. You're taking the absolute value 'cause you just wanna figure out the absolute distance. Now you see, or maybe you see. Four minus four, this is. Four minus four, that is a zero. That is that zero right over there. Five minus four, absolute value of that? That's going to be. Lemme do this in a new color. This is just going to be one. This thing is the same thing as that over there. We were able to see that just by inspecting this graph, or this chart. And then, six minus four, absolute value of that, that's just going to be two. That two is that two right over here, which is the same thing as this two right over there. And then, finally, our one minus four, this negative three,", + "qid": "DPuK6ZgBGmE_305" + }, + { + "Q": "At the 1:53 mark he says something like \"you could probably draw a better freehand diagram\". My opinion: I definitely cannot. This guy seems to be good at explaining math in these videos-who is he?", + "A": "He s Richard Rusczyk (yeah, I probably butchered his name), but if you look him up on google you can find more info about him. He s pretty famous.", + "video_name": "rcLw4BlxaRs", + "timestamps": [ + 113 + ], + "3min_transcript": "We've got some 3D geometry here so we're going to have read carefully, visualize what's going on because it's kind of hard to draw in 3D. We got six spheres with a radius 1. Their centers are at the vertices of a regular hexagon that has side length 2. So we're starting with a regular hexagon, and we're going to put spheres centered at each of the vertices. And since the radius of each sphere is 1, side length is 2, that means each of these spheres is going to be tangent to its two neighbors. So we start off with a hexagon, six spheres, each one tangent to each of its two neighbors. And then we're going to have a larger sphere centered at the center of the hexagon such that it's tangent to each of the little spheres. Now, each of the little spheres will touch the inside of this giant sphere. And then we bring out an eighth sphere that's externally tangent to the six little ones. So we got out six little ones down here around the hexagon, and we're going to take this new sphere and just set it right on top of those six. And it's going to touch-- right at the top of it, So we have at least somewhat of a picture of what's going on here, and we want the radius of this last sphere that we dropped in at the top there. And now one thing I like to do with these 3D problems is I like to take 2D cross sections, turn 3D problems into 2D problems. So when I have a problem with a whole bunch of spheres, I like to throw my cross sections through centers of those spheres and through points of tangency whenever I have tangent spheres. Now, a natural place to start here, of course, is the hexagon. We take the cross section with the hexagon, because that's going to go through the centers of seven of these spheres and all kinds of points tangency. So to start off, we'll draw a regular hexagon. And you're going to have to bear with me. On the test, of course, you've got your ruler, you got your protractor, you got your compass so you can draw a perfect diagram. You could probably draw a freehand better diagram better than I can, too. But when we take cross sections of our spheres, we make circles. And we include the points of tangency in this cross section. A cross section of that is a circle that touches each of these little circles. All right, and there we go. This is the cross section through the hexagon. Now we can label some lengths. We know that the radii of the little spheres is 1, and one thing that's really nice about regular hexagons is you can break them up into equilateral triangles. So this is an equilateral triangle. Here's the center of the hexagon center and the big circle, and I can extend this out to the point of tangency of small sphere and the big one. So we know this is 1 because it's a radius of the small sphere. This is 1. This is an equilateral triangle so this side is the same as this side. So it tells us that this is 1, and now we know that the radius of the giant sphere is 3. So we've got the radius of the giant sphere. We got the radii of all these little spheres. All we have left is that eighth sphere we sat on top. And, of course, that sphere's not in this diagram. It's sitting right up here.", + "qid": "rcLw4BlxaRs_113" + }, + { + "Q": "At 8:16, how did he get A^2=3/4 h^2?", + "A": "Simple algebraic solving techniques. if you take h to be c like it is normally represented, and assume a^2 + b^2 = c^2... and take the 30-60-90 side ratio definition where the side opposite the 30 degree angle is c/2, or hypotenuse/2, then (c/2)^2 + a^2 = c^2. Subtract (c/2)^2 from both sides... you get a^2 = c^2 -(c/2)^2. Fraction stuff... a^2 = (4c/4)^2 - (c/2)^2 so a^2 = (3/4c)^2.", + "video_name": "Qwet4cIpnCM", + "timestamps": [ + 496 + ], + "3min_transcript": "Because that's h over 2, and this is also h over 2. Right over here. So if we go back to our original triangle, and we said that this is 30 degrees and that this is the hypotenuse, because it's opposite the right angle, we know that the side opposite the 30 degree side is 1/2 of the hypotenuse. And just a reminder, how did we do that? Well we doubled the triangle. Turned it into an equilateral triangle. Figured out this whole side has to be the same as the hypotenuse. And this is 1/2 of that whole side. So it's 1/2 of the hypotenuse. So let's remember that. The side opposite the 30 degree side is 1/2 of the hypotenuse. Let me redraw that on another page, because I think this is getting messy. So going back to what I had originally. This is a right angle. This is the hypotenuse-- this side right here. If this is 30 degrees, we just derived that the side opposite that this is equal to 1/2 the hypotenuse. If this is equal to 1/2 the hypotenuse then what is this side equal to? Well, here we can use the Pythagorean theorem again. We know that this side squared plus this side squared-- let's call this side A-- is equal to h squared. So we have 1/2 h squared plus A squared is equal to h squared. This is equal to h squared over 4 plus A squared, is equal to h squared. Well, we subtract h squared from both sides. We get A squared is equal to h squared minus h squared over 4. This is equal to 3/4 h squared. And once going that's equal to A squared. I'm running out of space, so I'm going to go all the way over here. So take the square root of both sides, and we get A is equal to-- the square root of 3/4 is the same thing as the square root of 3 over 2. And then the square root of h squared is just h. And this A-- remember, this is an area. This is what decides the length of the side. I probably shouldn't have used A. But this is equal to the square root of 3 over 2, times h. We've derived what all the sides relative to the hypotenuse are of a 30-60-90 triangle. So if this is a 60 degree side.", + "qid": "Qwet4cIpnCM_496" + }, + { + "Q": "at 0:23, where did you get the radical 2 over 2 from?", + "A": "That was explained in the earlier video about 45-45-90 triangles.", + "video_name": "Qwet4cIpnCM", + "timestamps": [ + 23 + ], + "3min_transcript": "Sorry for starting the presentation with a cough. I think I still have a little bit of a bug going around. But now I want to continue with the 45-45-90 triangles. So in the last presentation we learned that either side of a 45-45-90 triangle that isn't the hypotenuse is equal to the square route of 2 over 2 times the hypotenuse. Let's do a couple of more problems. So if I were to tell you that the hypotenuse of this triangle-- once again, this only works for 45-45-90 triangles. And if I just draw one 45 you know the other angle's got to be 45 as well. If I told you that the hypotenuse here is, let me say, 10. We know this is a hypotenuse because it's opposite the right angle. And then I would ask you what this side is, x. Well we know that x is equal to the square root of 2 over 2 times the hypotenuse. So that's square root of 2 over 2 times 10. 10 divided by 2. So x is equal to 5 square roots of 2. And we know that this side and this side are equal. I guess we know this is an isosceles triangle because these two angles are the same. So we also that this side is 5 over 2. And if you're not sure, try it out. Let's try the Pythagorean theorem. We know from the Pythagorean theorem that 5 root 2 squared, plus 5 root 2 squared is equal to the hypotenuse squared, where the hypotenuse is 10. Is equal to 100. Or this is just 25 times 2. So that's 50. But this is 100 up here. Is equal to 100. And we know, of course, that this is true. So it worked. We proved it using the Pythagorean theorem, and that's actually how we came up with this formula in the first place. Maybe you want to go back to one of those presentations if you forget how we came up with this. type of triangle. And I'm going to do it the same way, by just posing a problem to you and then using the Pythagorean theorem to figure it out. This is another type of triangle called a 30-60-90 triangle. And if I don't have time for this I will do another presentation. Let's say I have a right triangle. That's not a pretty one, but we use what we have. That's a right angle. And if I were to tell you that this is a 30 degree angle. Well we know that the angles in a triangle have to add up to 180. So if this is 30, this is 90, and let's say that this is x. x plus 30 plus 90 is equal to 180, because the angles in", + "qid": "Qwet4cIpnCM_23" + }, + { + "Q": "At 2:25, Sal uses the chain rule for the derivative of (2+x^3)^-1. Would it not work if he just used the power rule and left it at that?", + "A": "No, the power rule applies only when you have x to the n, not when you have some function of x raised to the n. You can see this with an example like (x^2)^3. If you just apply the power rule, you get 3(x^2)^2, but we know that s wrong because (x^2)^3 is x^6, so the answer has to be 6x^5 or something equivalent. You get the right result when you apply the chain rule.", + "video_name": "GH8-URjRQpQ", + "timestamps": [ + 145 + ], + "3min_transcript": "We have the curve y is equal to e to the x over 2 plus x to the third power. And what we want to do is find the equation of the tangent line to this curve at the point x equals 1. And when x is equal to 1, y is going to be equal to e over 3. It's going to be e over 3. So let's try to figure out the equation of the tangent line to this curve at this point. And I encourage you to pause this video and try this on your own first. Well, the slope of the tangent line at this point is the same thing as the derivative at this point. So let's try to find the derivative of this or evaluate the derivative of this function right over here at this point. So to do that, first I'm going to rewrite it. You could use the quotient rule if you like, but I always forget the quotient rule. The product rule is much easier for me to remember. So I can rewrite this as y is equal to-- and I might as well color code it-- is equal to e to the x times 2 plus x And so the derivative of this, so let me write it here. So y prime is going to be equal to the derivative of this part of it, e to the x. So the derivative of e to the x is just e to the x. Just let me write that. So we're going to take the derivative of it. And that's what's amazing about e to the x, is that the derivative of e to the x is just e to the x times this thing. So times 2 plus x to the third to the negative 1. And then to that we're going to add this thing. So not its derivative anymore. We're just going to add e to the x times the derivative of this thing right over here. So we're going to take the derivative. So we can do the chain rule. It's going to be the derivative of 2 power with respect to 2 plus x to the third times the derivative of 2 plus x to the third with respect to x. So this is going to be equal to negative-- I'll write it this way-- negative 2 plus x to the third to the negative 2 power. And then we're going to multiply that times the derivative of 2 plus x to the third with respect to x. Well, derivative of this with respect to x is just 3x squared. And of course, we could simplify this a little bit if we like. But the whole point of this is to actually find the value of the derivative at this point. So let's evaluate. Let's evaluate y prime when x is equal to 1. Y prime of 1 when x is equal to 1.", + "qid": "GH8-URjRQpQ_145" + }, + { + "Q": "At 1:10. why does he make 4.1 to 41?", + "A": "Moving the decimal around is like multiplying or dividing by 10, so he notes that 4.1 hundredths is the same as 41 thousandths (4.1 x 10^-2 = 41 x 10^-3).", + "video_name": "ios3QL9t9LQ", + "timestamps": [ + 70 + ], + "3min_transcript": "- [Voiceover] What I want to do in this video is get a little bit of practice subtracting in scientific notation. So let's say that I have 4.1 x 10 to the -2 power. 4.1 x 10 to the -2 power and from that I want to subtract, I want to subtract 2.6, 2.6 x 10 to the -3 power. Like always, I encourage you to pause this video and see if you can solve this on your own and then we could work through it together. All right, I'm assuming you've had a go it. So the easiest thing that I can think of doing is try to convert one of these numbers so that it has the same, it's being multiplied by the same power of ten as the other one. What I could think about doing, well can we express 4.1 times 10 to the -2? Can we express it as something times 10 to the -3? So we have 4.1 times 10 to the -2. we would divide by 10, but we can't just divide by 10. That would literally change the value of the number. In order to not change it, we want to multiply by 10 as well. So we're multiplying by 10 and dividing by 10. I could have written it like this. I could have written 10/10 times, let me write this a little bit neater. I could have written 10/10 x this and then you take 10 x 4.1, you get 41, and then 10 to the -2 divided by 10 is going to be 10 to the -3. So this right over here, this is equal to 10 x 4.1 is 41 times 10 to the -3. And that makes sense. 41 thousandths is the same thing as 4.1 hundredths and all we did is we multiplied this times 10 and we divided this times 10. So let's rewrite this. We can rewrite it now as 41 X 10 to the -3 So now we have two things. We have 41 X 10 to the -3 - 2.6 x 10 to the -3. Well this is going to be the same thing as 41 - 2.6. - 2.6, let me do it in that same color. That was purple. - 2.6, 10 to the -3. 10, whoops, 10 to the -3. There's 10 to the -3 there, 10 to the -3 there. One way to think about it, I have just factored out a 10 to the -3. Now what's 41 - 2.6? Well 41 - 2 is 39, and then -.6 is going to be 38.4.", + "qid": "ios3QL9t9LQ_70" + }, + { + "Q": "I was just wondering, at 1:05, why is point B said to be 2/5 of the way from A? Shouldn't it be that the total distance from A to C is 7 (add both sides of the ratio to find the total number of parts) and then B should be 2 parts out of this total distance? Thus, B is 2/7 of the way? I'm not entirely sure of whether or not I am correct, but I'm assuming I've made some error.", + "A": "Your error was in saying the total distance from A to C is 7. The problem states that AC is 5 when it says the ratio of AB to AC is 2 : 5. In other words, the 2 : 5 ratio is a ratio of a part to the whole, not a ratio of the lengths of 2 portions of the whole.", + "video_name": "lEGS5ECgFxE", + "timestamps": [ + 65 + ], + "3min_transcript": "A, B and C are collinear, and B is between A and C. The ratio of AB to AC is 2 to 5. If is A is at negative 6 comma 9, and B is at negative 2 comma 3, what are the coordinates of point C? And I encourage you to now pause this video and try this on your own. So let's try to visualize this. So A is at negative 6 comma 9. B, let's see, it's less negative in the horizontal direction, it's lower in the vertical direction. So we could put B right over here. B is at the point negative 2 comma 3. And that C, it's going to be collinear, so we're going to go along the same line, let me draw that line right now. So it's going to be collinear and they tell us that the ratio of AB to AC is 2 to 5. So let's say C is-- I'm just trying to eyeball it right now-- let's put C right over here. And we don't know C's coordinates. Well the way we could think about it is to break it up into horizontal change in coordinate and vertical change in coordinate, and apply the same ratio. So for example, what is the horizontal change in coordinates going from A to B? Well let's draw that. Going from A to B, so this is A's x-coordinate, it's at negative 6. B's x-coordinate is at negative 2. So it's this change right over here. This is the horizontal change that we care about. Now what is that? Well if you start at negative 6 and you go to negative 2, you have increased by 4. Another way of thinking about it is negative 2 minus negative 6 Now the ratio between this change and the change of the x-coordinate between A and C is going to be 2 to 5. So let's call that change, this entire change, let's call this x. So we could say that the ratio between 4 and x is equal to-- and notice this is the horizontal change from A to B just if you look on the horizontal axis-- so the ratio of that, which is 4, to the horizontal change between A and C, well that's going to have to be the same ratio. So it's going to be 2/5. Now to solve for x, a fun thing might be to just take the reciprocal of both sides. So x/4 is equal to 5/2. We can multiply both sides times 4 and we are left with x is equal to 5 times 4 divided by 2,", + "qid": "lEGS5ECgFxE_65" + }, + { + "Q": "At the 0:53 problem how does Sal get to an Answer of 1/3 ? What is the work not shown here ?", + "A": "Let s start with 8^x =2. When the variable is in the exponent, it is useful (where possible) to express both sides of the equation using the same base. Since on the righthand side there is a 2 to the first power, ask yourself whether 8 can be expressed as a power of 2? So we end up with (2^3)^x which is the same as 2^3x. And we then have 2^3x = 2^1 as the equation. The bases are the same, so the exponents must be equal. Therefore 3x = 1 so x = 1/3 Hope you find this useful!", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 53 + ], + "3min_transcript": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3.", + "qid": "eTWCARmrzJ0_53" + }, + { + "Q": "on 2:37 how does negitive turn into a fraction", + "A": "That is a basic property of exponents. The rule is: a\u00e2\u0081\u00bb\u00e1\u00b5\u0087 = 1/a\u00e1\u00b5\u0087", + "video_name": "eTWCARmrzJ0", + "timestamps": [ + 157 + ], + "3min_transcript": "So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3. What would be the log base 8 of 1/2? What does this evaluate to? Let me clean this up so that we have some space to work with. So as always, we're saying, what power do I have to raise 8 to to get to 1/2? So let's think about that a little bit. We already know that 8 to the one-third power is equal to 2. If we want the reciprocal of 2 right over here, we have to just raise 8 to the negative one-third. So let me write that down. 8 to the negative one-third power is going to be equal to 1 over 8 to the one-third power. And we already know the cube root of 8, or 8 to the one-third power, is equal to 2. This is equal to 1/2.", + "qid": "eTWCARmrzJ0_157" + }, + { + "Q": "The equation is -4x+7. Shortly after the 4:00 mark, Sal replaces the x with -1 and then says, \"4 times -1 = -4\". Shouldn't it be -4 * -1?", + "A": "He misspoke and says 4*-1=4, but what he really meant is -4*-1=4 and he completes the equation as if he had said that correctly. It does not change the problem because he just misspoke and didn t write the incorrect statement down", + "video_name": "nGCW5teACC0", + "timestamps": [ + 240 + ], + "3min_transcript": "all of that over your change in the horizontal axis. Well, your change in the horizontal axis is this distance, which is the same thing as this distance. Notice your change in vertical over change in horizontal, change in vertical over change in horizontal, reviewing the green point as the endpoint. So it's going to be x minus negative 1. And this is the exact same expression. These are the exact same expression. You can simplify them, minus negative 1, and this becomes plus. This could become a plus 1. But these are the exact same expression. So this is the expression, really, for the slope between negative 1 and g of negative 1 and an arbitrary x. Well, we already know that no matter what x you pick, the slope between x, g of , and this point right over here is It's going to be the slope of the line. It's going to be equal to negative 4. This thing is going to be equal to negative 4. It's going to be equal to negative 4. Doesn't matter how close x gets, and weather x comes from the right or whether x comes from the left. So this thing, taking the limit of this, this just gets you to negative 4. It's really just the slope of the line. So even if you were to take the limit as x approaches negative 1, as x gets closer and closer and closer to negative 1, well then, these points are just going to get closer and closer and closer. But every time you calculate the slope, it's just going to be the slope of the line, which Now, you could also do this algebraically. And let's try to do it algebraically. So let's actually just take the limit as x approaches negative 1 of g of x. Well, they already told us what g of x is. It is negative 4x plus 7, minus g of negative 1. Negative 1 times 4 is positive 4. Positive 4 plus 7 is 11. All of that over x plus 1, all of that over x plus 1. And that's really x minus negative 1, is you want to think of it that way. But I'll just write x plus 1 this way here. So this is going to be equal to the limit as x approaches negative 1 of, in our numerator-- let's see. 7 minus 11 is negative 4. We can factor out a negative 4. It's a negative 4 times x plus 1, all of that over x plus 1. And then since we're just trying to find the limit as x approaches negative 1, so we can cancel those out. And this is going to be non-zero for any x value other than negative 1. And so this is going to be equal to negative 4.", + "qid": "nGCW5teACC0_240" + }, + { + "Q": "Wait is this stuff a joke? Especially 3:42 are there really such diseases?", + "A": "Don t worry, all of them are made up, except maybe the mind-blown syndrom. If you show hexaflexagons to your friends, they could very well be disbelieving at the amazing-ness.", + "video_name": "AmN0YyaTD60", + "timestamps": [ + 222 + ], + "3min_transcript": "clockwise down under the flaps, even if you flip it over or flex it. Well, in this hexaflexagon, it flows counter-clockwise. They're mirror images. The chirality is decided when you fold and tape your triangles into a twisty loop, and once taped, it is impossible to change from one to the other without cutting it apart, at least in three-dimensional euclidean space. A change in chirality could be a sign that your flexagon has been flipped through four-dimensional space and is possibly a highly dangerous multi-dimensional portal. With experience, a hexaflexagon master can construct a hexaflexagon in mere seconds. Some forgo tape and scissors entirely by folding a double strip that's too long and tucking the extra in. This is an advanced technique that should not be attempted without prior training. Beware topological changes. This family seems safe from this philosoraptor, because they live on separate planets with a cold, empty vacuum of space between them. But after a single flex, the unfortunate victims are now doomed, protected only by the inconsequential barrier of their domicile. Your stars might explode, your frowns may become smiles, become the roundest of circles. Perfectly healthy snakes may turn into snake loops, or worse, become decapitated. Either state is fatal for the snake, as having no head can lead to starvation. This can be avoided by simply marking where connections will be across neighboring triangles first. Afterwards, the lines can be filled in however you like. Be aware that with the trihexaflexagon, there are two variations to each face. So you can simply draw one side where triangles connect, and flip and draw the other. But in the hexa-hexaflexagon, the main three faces each appear four different ways. If you use hexaflexagons, keep an eye out for signs of dependency. Overuse can lead to addiction and possibly an overdose. Some users of hexaflexagons report confusion, mind-blown syndrome, hexaflexaperplexia, hexaflexadyslexia, hexaflexaperfectionism, and hexaflexa-Mexican-food-cravings. If you find yourself experiencing any of these symptoms, stop flexagon use immediately, and see the head of your math department. With proper precautions, flexagating can be a great part of your life. Follow these simple safety guidelines, experience.", + "qid": "AmN0YyaTD60_222" + }, + { + "Q": "At 1:43 what is the difference in tangent line and secant line?", + "A": "A tangent line touches the curve at one point whereas the secant line intersects at two points. The secant slope can be found by simple slope eq: y2 - y1/x2-x1 The tangent slope is found by f(x) - f(a)/ x - a. Hope this helps", + "video_name": "BYTfCnR9Sl0", + "timestamps": [ + 103 + ], + "3min_transcript": "What I want to do in this video is to see whether the power rule is giving us results that at least seem reasonable. This is by no means a proof of the power rule, but at least we'll feel a little bit more comfortable using it. So let's say that f of x is equal to x. The power rule tells us that f prime of x is going to be equal to what? Well, x is the same thing as x to the first power. So n is implicitly 1 right over here. So we bring the 1 out front. It'll be 1 times x to the 1 minus 1 power. So it's going to be 1 times x to the 0 power. x to the 0 So it's just going to be equal to 1. Now, does that makes conceptual sense if we actually try to visualize these functions? So let me actually try to graph these functions. So that's my y-axis. This is my x-axis. And let me graph y equals x. So y is equal to f of x here. So y is equal to x. So it looks something like that. Or this is f of x is equal to x, or y is equal to this f of x right over there. Now, actually, let me just call that f of x just to not confuse you. So this right over here is f of x is equal to x that I graphed right over here. y is equal to f of x, which is equal to x. And now, let me graph the derivative. Let me graph f prime of x. That's saying it's 1. That's saying it's 1 for all x. Regardless of what x is, it's going to be equal to 1. Is this consistent with what we know about derivatives and slopes and all the rest? Well, let's look at our function. What is the slope of the tangent line right at this point? Well, right over here, this has slope 1 continuously. Or it has a constant slope of 1. Slope is equal to 1 no matter what x is. It's a line. And for a line, the slope is constant. So over here, the slope is indeed 1. If you go to this point over here, the slope is indeed 1. If you go over here, the slope is indeed 1. So we've got a pretty valid response there. So let's say I have g of x is equal to x squared. The power rule tells us that g prime of x would be equal to what? Well, n is equal to 2. So it's going to be 2 times x to the 2 minus 1. Or it's going to be equal to 2 x to the first power. It's going to be equal to 2x. So let's see if this makes a reasonable sense. And I'm going to try to graph this one a little bit more precisely. Let's see how precisely I can graph it. So this is the x-axis, y-axis. Let me mark some stuff off here. So this is 1, 2, 3, 4, 5. This is 1, 2, 3, 4. 1, 2, 3, 4. So g of x.", + "qid": "BYTfCnR9Sl0_103" + }, + { + "Q": "At 2:16, it could have been (Star) + 1 = ... I don't know... Smiley Face). We didn't have to say \"y = x + 1\".", + "A": "It could have been. But people thought letters were more simpler.", + "video_name": "Tm98lnrlbMA", + "timestamps": [ + 136 + ], + "3min_transcript": "I'm here with Jesse Ro, whose a math teacher at Summit San Jose and a Khan Academy teaching fellow and you had some interesting ideas or questions. Yeah, one question that students ask a lot when they start Algebra is why do we need letters, why can't we just use numbers for everything? Why letters? So why do we have all these Xs and Ys and Zs and ABCs when we start dealing with Algebra? Yeah, exactly. That's interesting, well why don't we let people think about that for a second. So Sal, how would you answer this question? Why do we need letters in Algebra? So why letters. So there are a couple of ways I'd think about it. One is if you have an unknown. So if I were to write X plus three is equal to ten the reason why we're doing this is that we don't know what X is It's literally an unknown. And so we're going to solve for it in some way. But it did not have to be the letter X. We could have literally written blank plus three is equal to ten. Or we could have written Question Mark plus three is equal to ten. So it didn't have to be letters, but we needed some type of symbol. But until you know it, you need some type of a symbol to represent whatever that number is. Now we can go and solve this equation and then know what that symbol represents. But if we knew it ahead of time, it wouldn't be an unknown. It wouldn't be something that we didn't know. So that's one reason why I would use letters and where just numbers by itself wouldn't be helpful. The other is when you're describing relationships between numbers. So I could do something like - I could say - that whenever you give me a three, I'm going to give you a four. And I could say, if you give me a five, I'm going to give you a six. And i could keep going on and on forever. If you give me a 7.1, I'm going to give you an 8.1. And I could keep listing this on and on forever. Maybe you could give me any number, and I could tell you what I'm going to give you. But I would obviously run out of space and time if I were to list all of them. And we could do that much more elegantly if we used letters to describe the relationship. And so I say, look, whatever you give me, I'm going to add one to it. And that's what I'm going to give back to you. And so now, this very simple equation here can describe an infinite number of relationships between X or an infinite number of corresponding Ys and Xs. So now someone knows whatever X you give me you give me three, I add one to it, and I'm going to give you four. You give me 7.1, I'm going to add one to it and give you 8.1. So there is no more elegant way that you could've done it than by using symbols. With that said, I didn't have to use Xs and Ys. This is just a convention that kind of comes to use from history. I could've defined what you give me as Star and what I give you as Smiley Face and this also would've been a valid way to express this. So the letters are really just symbols. Nothing more.", + "qid": "Tm98lnrlbMA_136" + }, + { + "Q": "How does Sal know at 6:34, 6:38, and 6:46 that y=x^2, xy=12, and 5/x+y=10 are not linear equations without graphing them first?", + "A": "Linear equations have specific formats. For example, here are some of their formats: 1) Ax + By = C where A, B and C are integers 2) y = mx + b where m and b are numbers None of Sal s equations look like the examples above. y = x^2: this has an exponent on x which makes it non-linear xy = 12: this has x and y being multiplied which doesn t occur in a linear equation 5/x + y = 10: this has x in a denominator which doesn t occur in a linear equation Hope this helps.", + "video_name": "AOxMJRtoR2A", + "timestamps": [ + 394, + 398, + 406 + ], + "3min_transcript": "You can verify that. Four times zero minus three times negative four well that's gonna be equal to positive twelve. And let's see, if y were to equal zero, if y were to equal zero then this is gonna be four times x is equal to twelve, well then x is equal to three. And so you have the point zero comma negative four, zero comma negative four on this line, and you have the point three comma zero on this line. Three comma zero. Did I do that right? So zero comma negative four and then three comma zero. These are going to be on this line. Three comma zero is also on this line. So this is, this line is going to look something like-- something like, I'll just try to hand draw it. Something like that. So once again, all of the xy-- all of the xy pairs that satisfy this, Now what are some examples, maybe you're saying \"Wait, wait, wait, isn't any equation a linear equation?\" And the simple answer is \"No, not any equation is a linear equation.\" I'll give you some examples of non-linear equations. So a non-- non-linear, whoops let me write a little bit neater than that. Non-linear equations. Well, those could include something like y is equal to x-squared. you will see that this is going to be a curve. it could be something like x times y is equal to twelve. This is also not going to be a line. Or it could be something like five over x plus y is equal to ten. This also is not going to be a line. So now, and at some point you could-- I encourage you to try to graph these things, they're actually quite interesting. But given that we've now seen examples of linear equations and non-linear equations, linear equations. One way to think about is it's an equation that if you were to graph all of the x and y pairs that satisfy this equation, you'll get a line. And that's actually literally where the word linear equation comes from. But another way to think about it is it's going to be an equation where every term is either going to be a constant, so for example, twelve is a constant. It's not going to change based on the value of some variable, twelve is twelve. Or negative three is negative three. So every term is either going to be a constant or it's going to be a constant times a variable raised to the first power. So this is the constant two times x to the first power. This is the variable y raised to the first power. You could say that bceause this is just one y. We're not dividing by x or y, we're not multiplying, we don't have a term that has x to the second power, or x to the third power, or y to the fifth power. We just have y to the first power, we have x to the first power. We're not multiplying x and y together like we did over here.", + "qid": "AOxMJRtoR2A_394_398_406" + }, + { + "Q": "At 1:58, why does he divide by ten?", + "A": "he divided by ten because if you look after you solve the 9x6 which was what was in the parentheses you then solve the 54/10 because in order of operation division is after parentheses, exponents, and multiplication and since there is none of that you divide which comes after multiplication in order of operations.", + "video_name": "STyoP3rCmb0", + "timestamps": [ + 118 + ], + "3min_transcript": "Let's see if we can multiply 9 times 0.6. Or another way to write it, we want to calculate 9 times 0.6. I'll write it like this-- 0.6. We want to figure out what this is equal to. And I encourage you to pause the video and try to figure it out on your own. And I'll give you a little bit of a hint. 0.6 is the same thing as 6 divided by 10. We know that if we start with 6, which we could write as 6.0, and if you were to divide it by 10, dividing by 10 is equivalent to moving the decimal place one place to the left. So 6 divided by 10 is 0.6. We are moving the decimal one place to the left. So I'm assuming you given a go at it. But what I'm going to do is use this that we already know to rewrite what we're trying to multiply. 0.6 is 6 divided by 10. And this expression right over here, we could either do the 6 divided by 10 first, in which case we would get 0.6, and this would turn into this problem. Or we could do the 9 times 6 first. And so let's do 9 times 6, which we know how to calculate, and then divide by 10, which we also know how to do. That. all about just moving the decimal place. So we could write 9 times 6. 9 times 6, we already know, is 54. I'll do that in orange-- is going to be 54. So this right over here is 54. And now to get to this expression, we have to divide by 10. We have to divide by 10. And what happens when we divide something by 10? And we've seen this in previous videos, why this is the case. Each place represents 10 times as much as the place to its right, or each place represents 1/10 of the place So 54 divided by 10, this is going to be-- you could start with 54. And I'll put a 0 here after the decimal. And when you divide by 10, that's equivalent of shifting the decimal one to the left. This is going to be equal to 5.4. And that should make sense to you. 5 times 10 is 50. 0.4 times 10 is 4. So it makes sense that 54 divided by 10-- I shouldn't say equal. I'd write 54 divided by 10 is equal to 5.4. So this right over here is equal to 5.4, and that's what this is. This is equal to 5.4. Notice, 9 times 6 is 54.", + "qid": "STyoP3rCmb0_118" + }, + { + "Q": "At 3:00 how is how is 9 * 0.6=5.4 . isn't the product of a multiplication problem always bigger then the numbers you are multiplying?", + "A": "not always because when you multiply by a number less than 1 in a multiplication the biggest number gets smaller. 0.6 is 6/10 * 9 is 54/10 and that is 5.4", + "video_name": "STyoP3rCmb0", + "timestamps": [ + 180 + ], + "3min_transcript": "0.6 is 6 divided by 10. And this expression right over here, we could either do the 6 divided by 10 first, in which case we would get 0.6, and this would turn into this problem. Or we could do the 9 times 6 first. And so let's do 9 times 6, which we know how to calculate, and then divide by 10, which we also know how to do. That. all about just moving the decimal place. So we could write 9 times 6. 9 times 6, we already know, is 54. I'll do that in orange-- is going to be 54. So this right over here is 54. And now to get to this expression, we have to divide by 10. We have to divide by 10. And what happens when we divide something by 10? And we've seen this in previous videos, why this is the case. Each place represents 10 times as much as the place to its right, or each place represents 1/10 of the place So 54 divided by 10, this is going to be-- you could start with 54. And I'll put a 0 here after the decimal. And when you divide by 10, that's equivalent of shifting the decimal one to the left. This is going to be equal to 5.4. And that should make sense to you. 5 times 10 is 50. 0.4 times 10 is 4. So it makes sense that 54 divided by 10-- I shouldn't say equal. I'd write 54 divided by 10 is equal to 5.4. So this right over here is equal to 5.4, and that's what this is. This is equal to 5.4. Notice, 9 times 6 is 54. Now you might see a little pattern here. Between these two numbers, I had exactly one number to the right of the decimal. When I take its product, let's say I ignored the decimal. I just said 9 times 6, I would've gotten 54. But then I have to divide by 10 in order to take account of the decimal, take account of the fact this wasn't a 6. This was a 6/10. And so I have one number to the right of the decimal here. And I want to you to think about that whether that's a general principle. Can we just count the total numbers of digits to the right of the decimals and then our product is going to have the same number of digits to right of the decimal? I'll let you to think about that.", + "qid": "STyoP3rCmb0_180" + }, + { + "Q": "in the video @ 4:54 i do not understand the part where the instructor of the video multiplied 4*84 to get the sum of the previous four tests.\n\n80 + 81 + 87 + 88 = 4 * 84 ? <--- how is this possible, i think its cool but do not understand how that works. thanks!!", + "A": "80+81+87+88=80+80+1+80+7+80+8=(80+80+80+80)+(0+1+7+8)=4*80+16=4*(80+4)=4*84", + "video_name": "9VZsMY15xeU", + "timestamps": [ + 294 + ], + "3min_transcript": "8 plus 8 is 16. I just ran eight miles, so I'm a bit tired. And, 4/8, so that's 32. Plus 1 is 33. And now we divide this number by 4. 4 goes into 336. Goes into 33, 8 times. 8 times 4 is 32. 33 minus 32 is 1, 16. So the average is equal to 84. So depending on what school you go to that's either a B or a C. So, so far my average after the first four exams is an 84. Now let's make this a little bit more difficult. We know that the average after four exams, at four exams, is equal to 84. average an 88, to average an 88 in the class. So let's say that x is what I get on the next test. So now what we can say is, is that the first four exams, I could either list out the first four exams that I took. Or I already know what the average is. So I know the sum of the first four exams is going to 4 times 84. And now I want to add the, what I get on the 5th exam, x. And I'm going to divide that by all five exams. So in other words, this number is the average of my first five exams. We just figured out the average of the first four exams. We add what I got on the fifth exam, and then we divide it by 5, because now we're averaging five exams. And I said that I need to get in an 88 in the class. And now we solve for x. Let me make some space here. So, 5 times 88 is, let's see. 5 times 80 is 400, so it's 440. 440 equals 4 times 84, we just saw that, is 320 plus 16 is 336. 336 plus x is equal to 440. Well, it turns out if you subtract 336 from both sides, you get x is equal to 104. So unless you have a exam that has some bonus problems on it, it's probably impossible for you to get ah an 88 average in the class after just the next exam.", + "qid": "9VZsMY15xeU_294" + }, + { + "Q": "at time 10:45, can anyone explain to me how we get the +/- square root of 61/20. specifically, the reason why we have +/- the square root. what rule is that?", + "A": "I don t know if this will help or not but I ll try to explain. So basically you re wondering why should there be a positive + and negative - square root, right? Think about any squared number really does have two possible square roots, the positive and the negative ones. For instance \u00c2\u00b1\u00e2\u0088\u009a9= -3 or 3 Because if we square (-3)^2=9 (\u00e2\u0086\u0092Notice that this different from -3^2 which is equal to =-9) and 3^3=9", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 645 + ], + "3min_transcript": "got to do to the other side. So we added a 9/4 here, let's add a 9/4 over there. And what does our equation become? We get x squared minus 3x plus 9/4 is equal to-- let's see if we can get a common denominator. So, 4/5 is the same thing as 16/20. Just multiply the numerator and denominator by 4. Plus over 20. 9/4 is the same thing if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it. Crazy number. Now this, at least on the left hand side, is a perfect square. This is the same thing as x minus 3/2 squared. And it was by design. Negative 3/2 times negative 3/2 is positive 9/4. Negative 3/2 plus negative 3/2 is equal to negative 3. So this squared is equal to 61/20. We can take the square root of both sides and we get x minus 3/2 is equal to the positive or the negative square root of 61/20. And now, we can add 3/2 to both sides of this equation and you get x is equal to positive 3/2 plus or minus the And this is a crazy number and it's hopefully obvious you would not have been able to-- at least I would not have been able to-- get to this number just by factoring. And if you want their actual values, you can get your calculator out. And then let me clear all of this. And 3/2-- let's do the plus version first. So we want to do 3 divided by 2 plus the second square root. We want to pick that little yellow square root. So the square root of 61 divided by 20, which is 3.24. This crazy 3.2464, I'll just write 3.246. So this is approximately equal to 3.246, and that was just", + "qid": "bNQY0z76M5A_645" + }, + { + "Q": "At 4:10, why does it become +/- 3, instead of just 3?", + "A": "the sqare root of 9 = +/-3 because in algebra, whenever you square a positive OR negative number, the answer is always positive.", + "video_name": "bNQY0z76M5A", + "timestamps": [ + 250 + ], + "3min_transcript": "to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2, and if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2-- same idea there-- and then you square it. You square a, you get positive 4. So we add positive 4 here. Add a 4. Now, from the very first equation we ever did, you should know that you can never do something to just one side You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4 It's going to be equal to 5 plus 4. We added 4 on the left-hand side because we wanted this to be a perfect square. But if you add something to the left-hand side, you've got to add it to the right-hand side. And now, we've gotten ourselves to a problem that's just like the problems we did in the last video. What is this left-hand side? Let me rewrite the whole thing. We have x squared minus 4x plus 4 is equal to 9 now. All we did is add 4 to both sides of the equation. But we added 4 on purpose so that this left-hand side becomes a perfect square. What number when I multiply it by itself is equal to 4 and when I add it to itself I'm equal to negative 2? Well, we already answered that question. It's negative 2. So we get x minus 2 times x minus 2 is equal to 9. Or we could have skipped this step and written x minus 2 squared is equal to 9. minus 2 is equal to plus or minus 3. Add 2 to both sides, you get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without completing the square. We could've started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, then their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0.", + "qid": "bNQY0z76M5A_250" + }, + { + "Q": "i do not understand from 0:22 to 0:56", + "A": "Sal realised that he ll be unable to find the square root of the quantity within the radical if it s negative. So he sets up the inequality to show that that quantity (2x-8) has to be either zero or positive.", + "video_name": "4h54s7BBPpA", + "timestamps": [ + 22, + 56 + ], + "3min_transcript": "Find the domain of f of x is equal to the principal square root of 2x minus 8. So the domain of a function is just the set of all of the possible valid inputs into the function, or all of the possible values for which the function is defined. And when we look at how the function is defined, right over here, as the square root, the principal square root of 2x minus 8, it's only going to be defined when it's taking the principal square root of a non-negative number. And so 2x minus 8, it's only going to be defined when 2x minus 8 is greater than or equal to 0. It can be 0, because then you just take the square root of 0 is 0. It can be positive. But if this was negative, then all of a sudden, this principle square root function, which we're assuming is just the plain vanilla one for real numbers, it would not be defined. So this function definition is only defined when 2x minus 8 is greater than or equal to 0. And then we could say if 2x minus 8 has to be greater than or equal to 0, what it's saying about what x has to be. So if we add 8 to both sides of this inequality, you get-- so let me just add 8 to both sides. These 8's cancel out. You get 2x is greater than or equal to 8. 0 plus 8 is 8. And then you divide both sides by 2. Since 2 is a positive number, you don't have to swap the inequality. So you divide both sides by 2. And you get x needs to be greater than or equal to 4. So the domain here is the set of all real numbers that are greater than or equal to 4. x has to be greater than or equal to 4. Or another way of saying it is this function is defined when x is greater than or equal to 4. And we're done.", + "qid": "4h54s7BBPpA_22_56" + }, + { + "Q": "at 1:45 i still do not really understand why after adding 64 you must also subtract 64 from 9. Could someone explain I'm a bit confused.", + "A": "When we make changes to an expression, we need to make sure that the new version is equivalent to the prior version. If you just add 64, you have changed the value of the original expression. The new version would not longer be equal to the original. The identify property of addition says we can add zero to any expression and it doesn t change the value of the expression. Thus, by using: +64 and -64, Sal is adding zero to the expression (+64 - 64 =0). Hope this helps.", + "video_name": "sh-MP-dVhD4", + "timestamps": [ + 105 + ], + "3min_transcript": "- [Voiceover] Let's see if we can take this quadratic expression here, X squared plus 16 X plus nine and write it in this form. You might be saying, hey Sal, why do I even need to worry about this? And one, it is just good algebraic practice to be able to manipulate things, but as we'll see in the future, what we're about to do is called completing the square. It's a really valuable technique for solving quadratics and it's actually the basis for the proof of the quadratic formula which you'll learn in the future. So it's actually a pretty interesting technique. So how do we write this in this form? Well one way to think about is if we expanded this X plus A squared, we know if we square X plus A it would be X squared plus two A X plus A squared, and then you still have that plus B, right over there. So one way to think about it is, let's take this expression, this X squared plus 16 X plus nine, and I'm just gonna write it with a few spaces in it. X squared plus 16 X And so, if we say alright, we have an X squared here. We have an X squared here. If we say that two A X is the same thing as that, then what's A going to be? So this is two A times X. Well, that means two A is 16 or that A is equal to 8. And so if I want to have an A squared over here, well if A is eight, I would add an eight squared which would be a 64. Well I can't just add numbers willy nilly to an expression without changing the value of an expression so if don't want to change the value of the expression, I still need to subtract 64. So notice, all that I have done now, is I just took our original expression and I added 64 and I subtracted 64, so I have not changed the value of that expression. But what was valuable about me doing that, is now this first part of the expression, it fits the pattern of a perfect square quadratic right over here. We have X squared plus two A X, where A is 8, plus A squared, 64. Once again, how did I get 64? I took half of the 16 and I squared it to get to the 64. And so the stuff that I just squared off, this is going to be X plus eight squared. X plus eight, squared. Once again I know that because A is eight, A is eight, so this is X plus eight squared, and then all of this business on the right hand side. What is nine minus 64? Well 64 minus nine is 55, so this is going to be negative 55. So minus 55, and we're done. We've written this expression in this form, and what's also called completing the square.", + "qid": "sh-MP-dVhD4_105" + }, + { + "Q": "In 0:44 , how did Khan get 2ax from (x+a)^2? Can someone explain how and why it works?", + "A": "(x+a)\u00c2\u00b2=(x+a)(x+a). When you multiply it out, you get x\u00c2\u00b2+ax+ax+a\u00c2\u00b2. Simplify it you get x\u00c2\u00b2+2ax+a\u00c2\u00b2", + "video_name": "sh-MP-dVhD4", + "timestamps": [ + 44 + ], + "3min_transcript": "- [Voiceover] Let's see if we can take this quadratic expression here, X squared plus 16 X plus nine and write it in this form. You might be saying, hey Sal, why do I even need to worry about this? And one, it is just good algebraic practice to be able to manipulate things, but as we'll see in the future, what we're about to do is called completing the square. It's a really valuable technique for solving quadratics and it's actually the basis for the proof of the quadratic formula which you'll learn in the future. So it's actually a pretty interesting technique. So how do we write this in this form? Well one way to think about is if we expanded this X plus A squared, we know if we square X plus A it would be X squared plus two A X plus A squared, and then you still have that plus B, right over there. So one way to think about it is, let's take this expression, this X squared plus 16 X plus nine, and I'm just gonna write it with a few spaces in it. X squared plus 16 X And so, if we say alright, we have an X squared here. We have an X squared here. If we say that two A X is the same thing as that, then what's A going to be? So this is two A times X. Well, that means two A is 16 or that A is equal to 8. And so if I want to have an A squared over here, well if A is eight, I would add an eight squared which would be a 64. Well I can't just add numbers willy nilly to an expression without changing the value of an expression so if don't want to change the value of the expression, I still need to subtract 64. So notice, all that I have done now, is I just took our original expression and I added 64 and I subtracted 64, so I have not changed the value of that expression. But what was valuable about me doing that, is now this first part of the expression, it fits the pattern of a perfect square quadratic right over here. We have X squared plus two A X, where A is 8, plus A squared, 64. Once again, how did I get 64? I took half of the 16 and I squared it to get to the 64. And so the stuff that I just squared off, this is going to be X plus eight squared. X plus eight, squared. Once again I know that because A is eight, A is eight, so this is X plus eight squared, and then all of this business on the right hand side. What is nine minus 64? Well 64 minus nine is 55, so this is going to be negative 55. So minus 55, and we're done. We've written this expression in this form, and what's also called completing the square.", + "qid": "sh-MP-dVhD4_44" + }, + { + "Q": "at 3:19 he said 2x5. = 12 and I am pretty sure it = 10 so I'm just confused", + "A": "Yeah, your right, he made a mistake, but if you watch for about two seconds longer, he actually corrects himself. \u00f0\u009f\u0091\u008dgood job on catching his mistake though.", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 199 + ], + "3min_transcript": "Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5. 6 minus 5 is 1. Bring down the 3. 5 goes into 13 two times. And you could have immediately said 5 goes into 63 twelve times, but this way, at least to me, it's a little bit more obvious. And then 2 times 5 is 12, and then we have sorry! 2 times 5 is 10. That tells you not to switch gears in the middle of a math problem. 2 times 5 is 10, and then you subtract, and you have a remainder of 3. So 63/5 is the same thing as 12 wholes and 3 left over, or 3/5 left over. And if you wanted to go back from this to that, just think: 12 is the same thing as 60 fifths, or 60/5.", + "qid": "RPhaidW0dmY_199" + }, + { + "Q": "At 1:51 you simplify the numerator of one fraction, but then simplify the denominator of the other fraction. How is it that you can do this? I thought you had to simplify the denominator of the same fraction.", + "A": "You can do this if you are multiplying fractions. If you multiply first, you get (7/4)*(36/5)=(7*36)/(4*5). Now you have all the numbers from the original two fractions in the same fraction. So it really doesn t matter at what stage you simplify. I hope this helped you. And remember that this is allowed only in multiplication, not in any other operation.", + "video_name": "RPhaidW0dmY", + "timestamps": [ + 111 + ], + "3min_transcript": "Multiply 1 and 3/4 times 7 and 1/5. Simplify your answer and write it as a mixed fraction. So the first thing we want to do is rewrite each of these mixed numbers as improper fractions. It's very difficult, or at least it's not easy for me, to directly multiply mixed numbers. One can do it, but it's much easier if you just make them improper fractions. So let's convert each of them. So 1 and 3/4 is equal to-- it's still going to be over 4, so you're still going to have the same denominator, but your numerator as an improper fraction is going to be 4 times 1 plus 3. And the reason why this makes sense is 1 is 4/4, or 1 is 4 times 1 fourths, right? 1 is the same thing as 4/4, and then you have three more fourths, so 4/4 plus 3/4 will give you 7/4. So that's the same thing as 1 and 3/4. Same exact process. We're going to still be talking in terms of fifths. That's going to be the denominator. You take 5 times 7, because think about it. 7 is the same thing as 35/5. So you take 5 times 7 plus this numerator right here. So 7 is 35/5, then you have one more fifth, so you're going to have 35 plus 1, which is equal to 36/5. So this product is the exact same thing as taking the product of 7/4 times 36/5. And we could multiply it out right now. Take the 7 times 36 as our new numerator, 4 times 5 as our new denominator, but that'll give us large numbers. I can't multiply 7 and 36 in my head, or I So let's see if we can simplify this first. Both our numerator and our denominator have numbers that are divisible by 4, so let's divide both the numerator and the denominator by 4. So in the numerator, we can divide the 36 by 4 and get 9. If you divide something in the numerator by 4, you need to divide something in the denominator by 4, and the 4 is the obvious guy, so 4 divided by 4 is 1. So now this becomes 7 times 9, and what's the 7 times 9? It's 63, over 1 times 5. So now we have our answer as an improper fraction, but they want it as a mixed number or as a mixed fraction. So what are 63/5? So to figure that out-- let me pick a nice color here-- we take 5 into 63. 5 goes into 6 one time. 1 times 5 is 5.", + "qid": "RPhaidW0dmY_111" + }, + { + "Q": "At 1:23 , how many ounces is a ton?", + "A": "There is 32,000 ounces in a ton.", + "video_name": "Dj1rbIP8PHM", + "timestamps": [ + 83 + ], + "3min_transcript": "Let's talk a little bit about the US customary units of weight. So the one that's most typically used is the pound, especially for things of kind of a human scale. And to understand what a pound is, most playing balls are roughly about a pound. So, for example, a soccer ball-- my best attempt to draw a soccer ball. So let's say that this is a soccer ball right over here. And then of course it has some type of pattern on it. So you could imagine a soccer ball is about a pound. So it's roughly one pound. And a pound will often be shorthanded with this \"lb.\" right over here. So it's about a pound. A football, an American football, is also a little under a pound. But we could say it is about a pound, just so we get a sense of what a pound actually represents. Now, if you want to go to scales smaller than a pound, you would think about using the ounce. thinking about weight is that one pound-- let me write this-- is equal to 16 ounces. Or another way of thinking about it is that 1 ounce is equal to 1/16 of a pound. And if you want to visualize things that weigh about an ounce, you could imagine a small box of matches weigh about an ounce. So a small box of matches might weigh about an ounce. Maybe a small AA battery would weigh about an ounce. But that gives you a sense of it. So if you were to take 16 of these together, they would be about the weight of a soccer ball. 16 of these things together, they would be about-- they would be about the weight of a soccer ball or a football. that are larger than a pound, then we would go to the ton. And a ton is equal to-- 1 ton is equal to 2,000 pounds. And you have to be a little bit careful with the ton. We're talking about the US customary units, and this is where we're talking about 2,000 pounds. But when we're talking on a more international level, this is sometimes called the short ton. There's also a long ton. There's also the metric ton. But here we're talking about US customary units, which is the short ton. So one ton is 2,000 pounds. And to get a sense of something that weighs 2,000 pounds, or to get a sense of what 2,000 pounds is like, or what might be measured in tons, a car is a good example. Your average midsize sedan would weigh about a little under to a little over 2 tons, so a little", + "qid": "Dj1rbIP8PHM_83" + }, + { + "Q": "What does he mean at 0:31 when he says fair coin?", + "A": "He means a coin with one head and one tail that has an equal chance of flipping one or the other.", + "video_name": "cqK3uRoPtk0", + "timestamps": [ + 31 + ], + "3min_transcript": "Voiceover:Let's say we define the random variable capital X as the number of heads we get after three flips of a fair coin. So given that definition of a random variable, what we're going to try and do in this video is think about the probability distributions. So what is the probability of the different possible outcomes or the different possible values for this random variable. We'll plot them to see how that distribution is spread out amongst those possible outcomes. So let's think about all of the different values that you could get when you flip a fair coin three times. So you could get all heads, heads, heads, heads. You could get heads, heads, tails. You could get heads, tails, heads. You could get heads, tails, tails. You could have tails, heads, heads. You could have tails, head, tails. You could have tails, tails, heads. And then you could have all tails. when you do the actual experiment there's eight equally likely outcomes here. But which of them, how would these relate to the value of this random variable? So let's think about, what's the probability, there is a situation where you have zero heads. So what's the probably that our random variable X is equal to zero? Well, that's this situation right over here where you have zero heads. It's one out of the eight equally likely outcomes. So that is going to be 1/8. What's the probability that our random variable capital X is equal to one? Well, let's see. Which of these outcomes gets us exactly one head? We have this one right over here. We have that one right over there. We have this one right over there. So three out of the eight equally likely outcomes provide us, get us to one head, which is the same thing as saying that our random variable equals one. So this has a 3/8 probability. So what's the probability, I think you're getting, maybe getting the hang of it at this point. What's the probability that the random variable X is going to be equal to two? Well, for X to be equal to two, we must, that means we have two heads when we flip the coins three times. So that's this outcome meets this constraint. This outcome would get our random variable to be equal to two. And this outcome would make our random variable equal to two. And this is three out of the eight equally likely outcomes. So this has a 3/8 probability. And then finally we could say what is the probability that our random variable X is equal to three? Well, how does our random variable X equal three? Well we have to get three heads when we flip the coin. So there's only one out of the eight", + "qid": "cqK3uRoPtk0_31" + }, + { + "Q": "At 3:28 why is the probability range between 0 and 1? I understand that beyond 1 we have a certainty of something happening, but why 1?", + "A": "In statistics, 1=100%. One hundred percent is is absolute certainty. You can t be more certain than that.", + "video_name": "cqK3uRoPtk0", + "timestamps": [ + 208 + ], + "3min_transcript": "when you do the actual experiment there's eight equally likely outcomes here. But which of them, how would these relate to the value of this random variable? So let's think about, what's the probability, there is a situation where you have zero heads. So what's the probably that our random variable X is equal to zero? Well, that's this situation right over here where you have zero heads. It's one out of the eight equally likely outcomes. So that is going to be 1/8. What's the probability that our random variable capital X is equal to one? Well, let's see. Which of these outcomes gets us exactly one head? We have this one right over here. We have that one right over there. We have this one right over there. So three out of the eight equally likely outcomes provide us, get us to one head, which is the same thing as saying that our random variable equals one. So this has a 3/8 probability. So what's the probability, I think you're getting, maybe getting the hang of it at this point. What's the probability that the random variable X is going to be equal to two? Well, for X to be equal to two, we must, that means we have two heads when we flip the coins three times. So that's this outcome meets this constraint. This outcome would get our random variable to be equal to two. And this outcome would make our random variable equal to two. And this is three out of the eight equally likely outcomes. So this has a 3/8 probability. And then finally we could say what is the probability that our random variable X is equal to three? Well, how does our random variable X equal three? Well we have to get three heads when we flip the coin. So there's only one out of the eight So it's a 1/8 probability. So now we just have to think about how we plot this, to see how this is distributed. So let me draw... So over here on the vertical axis this will be the probability. Probability. And it's going to be between zero and one. You can't have a probability larger than one. So just like this. So let's see, if this is one right over here, and let's see everything here looks like it's in eighths so let's put everything in terms of eighths. So that's half. This is a fourth. That's a fourth. That's not quite a fourth. This is a fourth right over here. And then we can do it in terms of eighths. So that's a pretty good approximation. And then over here we can have the outcomes.", + "qid": "cqK3uRoPtk0_208" + }, + { + "Q": "In 1:12, How did Sal get 1 from i^25?", + "A": "Sal got one from i^100, not i^25. he said that since 4x25=100, then i^100 is the same thing as (i^4)^25. i^4 is equal to 1, so 1^25=1. Finally, Sal deduced that i^100=1", + "video_name": "QiwfF83NWNA", + "timestamps": [ + 72 + ], + "3min_transcript": "Now that we've seen that as we take i to higher and higher powers, it cycles between 1, i, negative 1, negative i, then back to 1, i, negative 1, and negative i. I want to see if we can tackle some, I guess you could call them, trickier problems. And you might see these surface. And they're also kind of fun to do to realize that you can use the fact that the powers of i cycle through these values. You can use this to really, on a back of an envelope, take arbitrarily high powers of i. So let's try, just for fun, let's see what i to the 100th power is. And the realization here is that 100 is a multiple of 4. So you could say that this is the same thing as i to the 4 times 25th power. And this is the same thing, just from our exponent properties, as i to the fourth power raised to the 25th power. If you have something raised to an exponent, and then that is raised to an exponent, that's the same thing as multiplying the two exponents. that's pretty straightforward. i to the fourth is just 1. i to the fourth is 1, so this is 1. So this is equal to 1 to the 25th power, which is just equal to 1. So once again, we use this kind of cycling ability of i when you take its powers to figure out a very high exponent of i. Now let's say we try something a little bit stranger. Let's try i to the 501st power. Now in this situation, 501, it's not a multiple of 4. So you can't just do that that simply. But what you could do, is you could write this as a product of two numbers, one that is i to a multiple of fourth power. And then one that isn't. And so you could rewrite this. 500 is a multiple of 4. So you could write this as i to the 500th power times i Right? You have the same base. When you multiply, you can add exponents. So this would be i to the 501st power. as-- i to the 500th power is the same thing as i to the fourth power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the fourth to the 125th power. And then that times i to the first power. Well, i to the fourth is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the first. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day, but you can use this cycling to realize look, i to the 500th is just going to be 1. And so i to the 501th is just going to be i times that. So i to any multiple of 4-- let me write this generally. So if you have i to any multiple of 4, so this right over here is-- well, we'll just restrict k to be non-negative right now. k", + "qid": "QiwfF83NWNA_72" + }, + { + "Q": "At 5:20 Sal says regarding i^96 that \"This is i^4, and then that to the 16th power\". Shouldn't he have said that \"This is i^4, and then that to the 24th power\" instead?", + "A": "Yeah... But the result wasn t wrong.", + "video_name": "QiwfF83NWNA", + "timestamps": [ + 320 + ], + "3min_transcript": "So if we have i to any multiple of 4, right over here, we are going to get 1, because this is the same thing as i to the fourth power to the k-th power. And that is the same thing as 1 to the k-th power, which is clearly equal to 1. And if we have anything else-- if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight, that 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over. i to the first power. This is a multiple of 4-- this right here is a multiple of 4-- and I know that because any 1,000 is multiple of 4, any 100 is a multiple of 4, and then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the first power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1, and we're just going to be left with i to the first power, or just i. Let's do another one. i to the 90-- let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99? So this is the same thing as i to the 96th power times i to the third power, right? If you multiply these, same base, add the exponent, you would get i to the 99th power. i to the 96th power, since this is a multiple of 4, this is i to the fourth, and then that to the 16th power. So that's just 1 to the 16th, so this is just 1. And then you're just left with i to the third power. And you could either remember that i to the third power is equal to-- you can just remember that it's equal to negative i. Or if you forget that, you could just say, look, this is the same thing as i squared times i. This is equal to i squared times i. i squared, by definition, is equal to negative 1. So you have negative 1 times i is equal to negative i. Let me do one more just for the fun of it. Let's take i to the 38th power.", + "qid": "QiwfF83NWNA_320" + }, + { + "Q": "At 7:46, I would like to make sure that I have the correct answer for the \"cliffhanger\". Is it 20x^9\nThanks", + "A": "Yes, you have the correct answer. Good job. Looks like they cut the video off too quickly.", + "video_name": "iHnzLETGz2I", + "timestamps": [ + 466 + ], + "3min_transcript": "Negative 9x to the fifth power times negative three, use parentheses there, when you have a negative in front, you always wanna use parentheses. Let's do x to the 107th power. If I would have showed you this before this video, you would have said oh my goodness, there's nothing I can do, I'm boxed, there's no way out. But now you know that it's as simple as follow the rules. We're going to multiply the coefficients, negative nine times negative three is 27. Two negatives is a positive and nine times three is 27. I'm gonna add my powers. Five plus 107 is a hundred, ooh, not two, that was almost a mistake I made there. Let's get rid of that, give me a second chance here. Life's all about second chances, And so, this crazy expression, which is two monomials, here's the first, here's the second, when we multiply and simplify we get another monomial, which is 27x to the 112th. I'm gonna leave you on a cliffhanger here. Which, I'm gonna show you a problem. What variable should we use? You notice I've been trying to vary the variables up to show you that it just doesn't matter. That's an ugly five, let's get rid of that. Give me a second chance with that one too. So let's look at 5x to the third power, times 4x to the sixth power. And I'm gonna show you a wrong answer. I had a student that asked to do this, and here's the wrong answer that they gave me. They told me 9x to the 18th power. What did they do wrong? What did they do wrong? I want you to think to yourself, what have we been talking about? What did they do with the five and the four to get the nine? What should they have done? What did they do with the three and the six to get the 18, and what should they have done? That's multiplying monomials by monomials.", + "qid": "iHnzLETGz2I_466" + }, + { + "Q": "at 2:04 when p^2 = 2p, why wouldn't you solve it as sqrt(p^2) = sqrt(2p)", + "A": "You could do it that way, you d get p = \u00e2\u0088\u009a(2p). The solution would be the same (0 = \u00e2\u0088\u009a0, 2 = \u00e2\u0088\u009a4). It s just easier the way Sal does it, p(p-2) = 0, where you can clearly see the solutions are 0/2.", + "video_name": "ZIqW_sXymrM", + "timestamps": [ + 124 + ], + "3min_transcript": "- [Voiceover] So let's try to find the solutions to this equation right over here. We have the quantity two X minus three squared, and that is equal to four X minus six, and I encourage you to pause the video and give it a shot. And I'll give you a little bit of a hint, you could do this in the traditional way of expanding this out, and then turning it into kind of a classic quadratic form, but there might be a faster or a simpler way to do this if you really pay attention to the structure of both sides of this equation. Well let's look at this, we have two X minus three squared on the left-hand side, on the right-hand side we have four X minus six. Well four X minus six, that's just two times two X minus three, let me be clear there, so this is the same thing as two X minus three squared is equal to, four X minus six, if I factor out a two, that's two times two X minus three. And so this is really interesting, we have something squared is equal to two times that something. let me be very clear here, so the stuff in blue squared is equal to two times the stuff in blue. So if we can solve for what the stuff in blue could be equal to, then we could solve for X, and I'll show you that right now. So let's say, let's just replace two X minus three, we'll do a little bit of a substitution, let's replace that with P. So let's say that P is equal to two X minus three. Well then this equation simplifies quite nicely, the left-hand side becomes P squared, P squared is equal to two times P, 'cause once again two X minus three is P, two times P. And now we just have to solve for P. And I'll switch to just one color now. So we can write this as, if we subtract two P from both sides, we can get P squared minus two P and we can factor out a P, so we get P times P minus two is equal to zero. And we've seen this shown multiple times, if I have the product of two things and they equal to zero, at least one of them needs to be equal to zero, so either P is equal to zero, or P minus two is equal to zero. Well if P minus two is equal to zero, then that means P is equal to two. So either P equals zero, or P equals two. Well we're not quite done yet, because we wanted to solve for X, and not for P. But luckily we know that two X minus three is equal to P. So now we could say either two X minus three is going to be equal to this P value, is going to be equal to zero, or two X minus three is going to be equal to this P value, is going to be equal to two. And so this is pretty straightforward to solve,", + "qid": "ZIqW_sXymrM_124" + }, + { + "Q": "At 4:06, why did Sal multiply 3x and 12 by 1/3?\n\nCouldn't you divide each side by 3 to isolate the variable?\n\nI was taught to use inverse operations. The inverse operation of multiplication is division.\n\nSo why did Sal not use inverse operations to solve 3x=12?", + "A": "Multiplying by 1/3 is the exact same thing as dividing by 3. Remember how to divide fractions? If you have 2/3 divided by 5/6 it is the same as 2/3 times 6/5, right. Remember you can write whole numbers, such as 12 and 3 as 12/1 and 3/1, So 12/3 = 12/1 / 3/1 = 12/1 * 1/3 = 12 * 1/3. So you see, multiplying by 1/3 is the same a dividing by 3. Great Question! Keep Studying! 12/3 = 12/1 * 1/3", + "video_name": "_y_Q3_B2Vh8", + "timestamps": [ + 246 + ], + "3min_transcript": "but remember, if you just took these blocks off of the left-hand side, and it was balanced before, now the left-hand side will be lighter and it will move up. But we want to keep it balanced so we can keep saying 'equal'. That this mass is equal to that mass. So, if we remove 2 block from the left-hand side, we need to remove 2 from the right-hand side. So, we can remove two there, and then we can remove two over there. Mathematically, what we are doing is: we are subtracting 2 kilograms from each side. We are subtracting 2 from this side, So on the left-hand side we now have 3x + 2, minus 2 we are left with just 3x, and on the right-hand side we had 14 and we took away 2 (let me write this:) we took away 2, so we are going to be left with 12 blocks. And you see that there, the ones that I have not crossed out, there are 12 left, and here you have 3 of those X-blocks. Since we removed the same amount from both sides, Now, this turns into a problem very similar to what we saw in the last video, so now I ask you: what can we do to isolate one x, to only have one 'X' on the left-hand side of the scale, while keeping the scale balanced? The easiest way to think about it is: If I want one X on this left-hand side, that is a third of the total X's here. So what if I were to multiply the left-hand side by one-third -- -- but if I want to keep the scale balanced, I have to multiply the right-hand side by one-third. If we can do that mathematically, Over here I can multiply the left-hand side by 1/3, and if I want to keep my scale balanced I also have to multiply the right-hand side by 1/3. Multiplying it physically literally means: just keeping a third of what we had originally We would get rid of two of these. -- there are 12 blocks left over after removing those first two -- so, 1/3 of 12: we are only going to have four of these little 1kg boxes left. Let me remove all but four. (so, remove those, and those...) And I have left [counts them] 4 here. And so, what you are left with, the only thing you have left, is this 'X' - I will shade it in to show this is the one we actually have left - and then we have these 1 kilogram boxes. You see it mathematically over here: 1/3 * 3x -- or you could have said 3x divided by 3 -- either way, that gives us -- these threes cancel out, so that would give you an 'X' and on the right-hand side: 12 * 1/3 - which is the same as 12/3, is equal to 4. And, since we did the same thing to both sides, the scale is still balanced.", + "qid": "_y_Q3_B2Vh8_246" + }, + { + "Q": "At 3:30 just when I learned that parenthesis most be solved first. But, I also learned about the Distributed Property. SMH", + "A": "Whenever you have parenthesis, do the math inside the parenthesis first, then use the distributive property. 4(11-3)= 4(8)=32 If you want something to help you remember the order of operations, try PEMDAS. Parentheses Exponents (if necessary) Multiplication (this includes the distributive property) Division Addition Subtraction", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 210 + ], + "3min_transcript": "So is equal to 12. You could imagine that I did the reverse distributive property out here. I factored out an x. But the way my head thinks about it is, I have 1.3 of something minus 0.7 of something, that's going to be equal to 1.3 minus 0.7 of those somethings, that x. And of course 1.3 minus 0.7 is 0.6 times x of my somethings is equal to 12. And now, this looks just like one of the problems we did in the last video. We have a coefficient times x is equal to some other number. Well, let's divide both sides of this equation by that coefficient. Divide both sides by 0.6. So the left-hand side will just become an x. X is equal to-- and what is 12 divided by 0.6? 0.6 goes into 12-- let's add some decimal points here-- 6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12. We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2.", + "qid": "tuVd355R-OQ_210" + }, + { + "Q": "@4:42 I didn't understand how you got 3/2. The way i did it was i divided 3 by 2 and got 1.5 .", + "A": "they re really the same thing, 1.5=3/2", + "video_name": "tuVd355R-OQ", + "timestamps": [ + 282 + ], + "3min_transcript": "We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something.", + "qid": "tuVd355R-OQ_282" + }, + { + "Q": "At 1:40 shouldn't it be (y^3)/2 (instead of (y^3)/3)?", + "A": "Okay, this is corrected later at 2:46", + "video_name": "0pv0QtOi5l8", + "timestamps": [ + 100 + ], + "3min_transcript": "Welcome back. In the last video we were just figuring out the volume under the surface, and we had set up these integral bounds. So let's see how to evaluate it now. And look at this. I actually realized that I can scroll things, which is quite useful because now I have a lot more board space. So how do we evaluate this integral? Well, the first integral I'm integrating with respect to x. I'm adding up the little x sums. So I'm forming this rectangle right here. Or you could kind of view it I'm holding y constant and integrating along the x-axis. I should switch colors. So what's the antiderivative of x y squared with respect to x? Well it's just x squared over 2. And then I have the y squared-- that's just a constant-- all over 2. And I'm going to evaluate that from x is equal to 1 to x is equal to the square root of y, which you might be daunted by. But you'll see that it's actually not that bad once you evaluate them. This is y is equal to 0 to y is equal 1. dy. Now, if x is equal to 1 this expression becomes y squared over 2. Right? y squared over 2, minus-- now if x is equal to square root of y, what does this expression become? If x is equal to the square root of y, then x squared is just y. And then y times y squared is y to the third. Right? So it's y to the third over 3. And now I take the integral with respect to y. So now I sum up all of these rectangles in the y direction. 0, 1. This is with respect to y. And that's cool, right? Because when you take the first integral with respect to x you end up with a function of y anyway, so you might as well have your bounds as functions of y's. It really doesn't make it any more difficult. But anyway, back to the problem. What is the antiderivative of y squared over 2 minus Well the antiderivative of y squared-- and you have to divide by 3, so it's y cubed over 6. Minus y to the fourth-- you have to divide by 4. Minus y to the fourth over-- did I mess up some place? No, I think this is correct. y to the fourth over 12. How did I get a 3 here? That's where I messed up. This is a 2, right? Let's see. x is square root of y. Yeah, this is a 2. I don't know how I ended up. Square root of y squared is y, times y squared y to the third over 2. Right. And then when I take the integral of this it's 4 times 2.", + "qid": "0pv0QtOi5l8_100" + }, + { + "Q": "At 8:34, she sang something about unary. i dont that that makes sense. help me!", + "A": "Pick a number. 13 for example. In the decimal system (base ten) it is written 13, 1 ten and three 1s. In the binary system (base two) it is written 1101, one 8, one 4, no 2s, one 1. In the unary system (base one) it is written 0000000000000, one 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, and another 1.", + "video_name": "sxnX5_LbBDU", + "timestamps": [ + 514 + ], + "3min_transcript": "backwards E, half a heart on a plate, and a simple, boring, short and straight line. So yeah. If you want to add up all the numbers between 1 and anything, this trick works. For example, all the numbers between 1 and 12. 1 plus 12 is 13. 2 plus 11 is 13. All the way in to 6 plus 7. That's 6 times 13, which I do my head as 60 plus 18 equals 78. So 78 is the 12th triangular number, while 5,050 is the 100th triangular number. At least it's not \"100 Days of Christmas.\" I'll stick with \"Bottles of Beer.\" [SINGING] On the tenth day of Christmas, my true love gave to me, the base of our Arabic numeral system, the base of a nonary numeral and the base of senary, and a number five, quaternary's base, ternary's base also, and binary two, and the base of unary. Here's my favorite way of visualizing the triangular numbers. Say you've got them in this configuration where they make a nice right equilateral triangle, or half a square. Finding the area of a square is easy, because you just square the length of it. In this case, 12 times 12. And the triangle is half of that. Only not really, because half the square means you only get half of this diagonal, so you've got to add back in the other half. But that's easy because there's 12 things in the diagonal, So to get the n-th triangular number, just take n squared over 2 plus n over 2, or n squared plus n over 2. [SINGING] On the eleventh day of Christmas, my true love gave to me the number my amp goes up to, the number of fingers on my hands, the German word for no, what I did after I eat, the number of heads on a hydra, at least until you start cutting them off, the number of strings on a guitar, the number I like to do high, the amount of horsemen of the Apocalypse, the number of notes in a triad, the number of pears in a pair of pears, and the number of partridges in a pear tree. [SINGING] On the twelfth day of Christmas,", + "qid": "sxnX5_LbBDU_514" + }, + { + "Q": "7:43. Isn't 1 -3 -2, not 2, or are you saying the absolute value?", + "A": "Sal is using absolute value in this video, and that is why your a little confused. \u00f0\u009f\u0098\u0089", + "video_name": "GdIkEngwGNU", + "timestamps": [ + 463 + ], + "3min_transcript": "It's just gonna be one. And you see that here visually. This point is just one away. It's just one away from three. This point is just one away from three. Four minus three is one. Absolute value of that is one. This point is just one away from three. Four minus three, absolute value. That's another one. every data point was exactly one away from the mean. And we took the absolute value so that we don't have negative ones here. We just care how far it is in absolute terms. So you have four data points. Each of their absolute deviations is four away. So the mean of the absolute deviations are one plus one plus one plus one, which is four, over four. So it's equal to one. One way to think about it is saying, on average, the mean of the distances of these points away from the actual mean is one. And that makes sense because all of these are exactly one away from the mean. Now, let's see how, what results we get for this data set right over here. Let me actually get some space over here. At any point, if you get inspired, I encourage you to calculate the Mean Absolute Deviation on your own. So let's calculate it. The Mean Absolute Deviation here, I'll write MAD, is going to be equal to ... Well, let's figure out the absolute deviation of each of these points from the mean. It's the absolute value of one minus three, that's this first one, plus the absolute deviation, so one minus three, that's the second one, then plus the absolute value of six minus three, that's the six, then we have the four, plus the absolute value of four minus three. Then we have four points. So one minus three is negative two. Absolute value is two. And we see that here. This is two away from three. We just care about absolute deviation. We don't care if it's to the left or to the right. Then we have another one minus three is negative two. It's absolute value, so this is two. That's this. This is two away from the mean. Then we have six minus three. Absolute value of that is going to be three. We see this six is three to the right of the mean. We don't care whether it's to the right or the left. And then four minus three. Four minus three is one, absolute value is one. And we see that. It is one to the right of three. And so what do we have? We have two plus two is four, plus three is seven, plus one is eight, over four, which is equal to two. So the Mean Absolute Deviation ... It fell off over here. Here, for this data set, the Mean Absolute Deviation is equal to two, while for this data set, the Mean Absolute Deviation is equal to one. And that makes sense. They have the exact same means. They both have a mean of three. But this one is more spread out. The one on the right is more spread out because, on average, each of these points are two away from three, while on average, each of these points are one away from three. The means of the absolute deviations on this one is one.", + "qid": "GdIkEngwGNU_463" + }, + { + "Q": "At 3:10 why does Sal use the \"ln\"? i know ln means natural log but why is this used?", + "A": "All the formulas in calculus involving log use the log with base e or ln. The formula used here is \u00e2\u0088\u00ab (1/x) = ln|x|", + "video_name": "Zp5z0wa0kgo", + "timestamps": [ + 190 + ], + "3min_transcript": "so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du? and we use the absolute value so it'll be defined even for negative use -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!", + "qid": "Zp5z0wa0kgo_190" + }, + { + "Q": "At 2:56, I'm really confused about where the 1 came from. Can someone please explain? Thank you", + "A": "Example: 5/3 = (1/3)*5, right? You had du/u and did the same thing as above, so du/u = (1/u)*du.", + "video_name": "Zp5z0wa0kgo", + "timestamps": [ + 176 + ], + "3min_transcript": "so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du? and we use the absolute value so it'll be defined even for negative use -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!", + "qid": "Zp5z0wa0kgo_176" + }, + { + "Q": "At 3:24 where does the du go? I do understand that 1/u = ln |u| then we put the constant +C . Bu I can not seem to understand what happens to du and where it goes...? Thank you!", + "A": "It disappears the same way dx does when you do a regular integration without a u-substitution. For example, the integral of x dx is (x^2)/2 + C, and the integral of (1/x)dx is ln|x| + C. We re using the same process when we integrate after a u-substitution, but now we re integrating with respect to u, so the integration needs du to work instead of dx.", + "video_name": "Zp5z0wa0kgo", + "timestamps": [ + 204 + ], + "3min_transcript": "so 4x^3 plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x^3 power. When someone writes du over dx, like this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but often times, you can kind of pseudo-manipulate them like fractions. So if you want to go from here to there, you can kind of pretend that you're multiplying both sides by dx. But these are equivalent statements and we want to get it in differential form in order to do proper use of u-substitution. And the reason why this is useful -- our original integral we can rewrite as 4x^3 dx over x^4 plus 7. And then it's pretty clear what's du and what's u. U, which we set to be equal to x^4 plus 7. And then du is equal to this. It's equal to 4x^3 dx. We saw it right over here. So we could rewrite this integral -- I'll try to stay consistent with the colors -- as the indefinite integral, well we have in magenta right over here, that's du over -- try to stay true to the colors -- over x^4 plus 7, which is just u. Or, we could rewrite this entire thing as the integral of 1 over u du. Well, what is the indefinite integral of 1 over u du? and we use the absolute value so it'll be defined even for negative use -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!", + "qid": "Zp5z0wa0kgo_204" + }, + { + "Q": "At 10:45 he takes the \u00e2\u0088\u009a200 and converts it to 10\u00e2\u0088\u009a2. That lost me completely. What math should I learn so that makes sense? Thank you in advance!", + "A": "The key to simplifying square roots is to take out any perfect squares and leave any non perfect squares inside. So if we notice that 200 = 100 \u00e2\u0080\u00a2 2, we see that 10^2 can be taken out as a 10 and 2 has to stay in. If we prime factor it, we have 2 \u00e2\u0080\u00a2 2 \u00e2\u0080\u00a2 2 \u00e2\u0080\u00a2 5 \u00e2\u0080\u00a2 5, and again \u00e2\u0088\u009a200 = \u00e2\u0088\u009a4 \u00e2\u0080\u00a2 \u00e2\u0088\u009a2 \u00e2\u0080\u00a2 \u00e2\u0088\u009a25 or 2 \u00e2\u0080\u00a2 5 \u00e2\u0088\u009a2", + "video_name": "E4HAYd0QnRc", + "timestamps": [ + 645 + ], + "3min_transcript": "So this is going to be--all right, this is 10/5, which is equal to 2. So the variance here-- let me make sure I got that right. Yes, we have 10/5. So the variance of this less-dispersed data set is a lot smaller. The variance of this data set right here is only 2. So that gave you a sense. That tells you, look, this is definitely a less-dispersed data set then that there. Now, the problem with the variance is you're taking these numbers, you're taking the difference between them and the mean, then you're squaring it. It kind of gives you a bit of an arbitrary number, and if you're dealing with units, let's say if these are distances. So this is negative 10 meters, 0 meters, 10 meters, this is 8 meters, so on and so forth, then when you square it, you get your variance in terms of meters squared. It's kind of an odd set of units. So what people like to do is talk in terms of standard or the square root of sigma squared. And the symbol for the standard deviation is just sigma. So now that we've figured out the variance, it's very easy to figure out the standard deviation of both of these The standard deviation of this first one up here, of this first data set, is going to be the square root of 200. The square root of 200 is what? The square root of 2 times 100. This is equal to 10 square roots of 2. That's that first data set. Now the standard deviation of the second data set is just going to be the square root of its variance, which is just 2. this first data set. This is 10 roots of 2, this is just the root of 2. So this is 10 times the standard deviation. And this, hopefully, will make a little bit more sense. This has 10 times more the standard deviation than this. And let's remember how we calculated it. Variance, we just took each data point, how far it was away from the mean, squared that, took the average of those. Then we took the square root, really just to make the units look nice, but the end result is we said that that first data set has 10 times the standard deviation as the second data set. So let's look at the two data sets. This has 10 times the standard deviation, which makes sense", + "qid": "E4HAYd0QnRc_645" + }, + { + "Q": "At 5:41 why do we multiply P(A) by P(B)", + "A": "As the events A and B are independent, meaning the outcome of event A does not affect the outcome of event B, then we can calculate the probability of both A and B taking place by multiplying the P(A) with the probability of B. This holds for independent events. But we must be careful about what we are calculating according to the problem we are trying to solve.", + "video_name": "RI874OSJp1U", + "timestamps": [ + 341 + ], + "3min_transcript": "", + "qid": "RI874OSJp1U_341" + }, + { + "Q": "At 7:35, Sal enters some values into the calculator which are close to -4. What would happen on the calculator if we tried to enter -4 into the function replacing x??", + "A": "When we substitute -4 for x, the equation becomes (-4)^2/(-4)^2-16. We can simplify this expression to 16/16-16 which then becomes 16/0. It s impossible to divide by 0 because nothing times 0 is equal to 16. There is no way to get 16 (or any other number) by multiplying by 0. The expression x^2/x^2-16 becomes undefined when x=-4 and when x=4 because there are no possible outputs from those inputs.", + "video_name": "2N62v_63SBo", + "timestamps": [ + 455 + ], + "3min_transcript": "", + "qid": "2N62v_63SBo_455" + }, + { + "Q": "At around 7:10, why did Sal do 16-2=14 instead of 2+16=18?", + "A": "At 3:00 and at 4:12 or so, he explanes that the change in y is 14 because one point is at +2 and the other is at +16. and if you count that on the graph they are 14 spaces apart.", + "video_name": "WkspBxrzuZo", + "timestamps": [ + 430 + ], + "3min_transcript": "So let me save some space here. So we have 1, 2, 3, 4. It's 4 comma-- 1, 2. So 4 comma 2 is right over here. 4 comma 2. Then we have the point negative 3 comma 16. So let me draw that over here. So we have negative 1, 2, 3. And we have to go up 16. So this is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. So it goes right over here. So this is negative 3 comma 16. Negative 3 comma 16. So the line that goes between them is going to look something like this. Try my best to draw a relatively straight line. That line will keep going. So the line will keep going. So that's my best attempt. And now notice, it's downward sloping. As you increase an x-value, the line goes down. It's going from the top left to the bottom right. As x gets bigger, y gets smaller. And just to visualize our change in x's and our change in y's that we dealt with here, when we started at 4 and we ended at-- or when we started at 4 comma 2 and ended at negative 3 comma 16, that was analogous to starting here and ending over there. And we said our change in x was negative 7. We had to move back. Our run we had to move in the left direction by 7. That's why it was a negative 7. And then we had to move in the y-direction. We had to move in the y-direction positive 14. So that's why our rise was positive. So it's 14 over negative 7, or negative 2. When we did it the other way, we started at this point. We started at this point, and then ended at this point. Started at negative 3, 16 and ended at that point. So in that situation, our run was positive 7. And now we have to go down in the y-direction since we switched the starting and the endpoint. And now we had to go down negative 14. Either way, we got the same slope.", + "qid": "WkspBxrzuZo_430" + }, + { + "Q": "At 1:24, Sal mentions that 2 is a factor of (x^3 - 8). How did he come up with that?", + "A": "If 2 is a root of a polynomial, then (x-2) is a factor. The polynomial is x^3-8. Set that equal to 0 and solve: x^3-8=0 x^3=8 (x^3)^(1/3)=8^(1/3) x=2. So 2 must be a root, and therefore (x-2) is a factor. You can divide x^3-8 by (x-2) and get the quotient of x^2+2x+4. This is how you derive the difference of cubes formula.", + "video_name": "6FrPLJY0rqM", + "timestamps": [ + 84 + ], + "3min_transcript": "Let's see if we can tackle a more complicated partial fraction decomposition problem. I have 10x squared plus 12x plus 20, all of that over x to the third minus 8. The first thing to do with any of these rational expressions that you want to decompose is to just make sure that the numerator is of a lower degree than the denominator, and if it's not, then you just do the algebraic long division like we did in the first video. But here, you can do from [UNINTELLIGIBLE] the highest degree term here is a second-degree term, here it's a third-degree term, so we're cool. This is a lower degree than that one. If it was the same or higher, we would do a little long division. The next thing to do, if we're going to decompose this into its components, we have to figure out the factors of the denominator right here, so that we can use those factors as the denominators in each of the components, and a third-degree polynomial is much, much, much harder to factor than a second-degree polynomial, normally. pop out at you-- if it doesn't immediately, hopefully what I'm about to say will make it pop out at you in the future-- is you should always think about what number, when you substitute into a polynomial, will make it equal to 0. And in this case, what to the third power minus 8 equals 0? And hopefully 2 pops out at you. And this is something you can only do really through inspection or through experience. And you'll immediately see 2 to the third minus 8 is 0, so 2 is a 0 of this, or 2 makes this expression 0, and that tells us that x minus 2 is a factor. So we can rewrite this right here as 10x squared plus 12x plus 20 over x minus 2 times something something We don't know what that something is yet. And I just want to hit the point home of why this is true, or what the intuition vibe has true. expression 0. And we know that 2 would make this factored expression 0, because when you put a 2 right here, this factor become 0, so it'll make the whole thing 0. And so that's why, that's the intuition where, if you substitute a number here and it makes this 0, you do x minus that number here, and we know that that will be a factor of the thing. Well, anyway, the next step if we really want to decompose this rational expression is to figure out what this part of it is, and the way to do that is with algebraic long division. We essentially just divide x minus 2 into x to the third minus 8 to get this, so let's do that. So you get x minus 2 goes into x to the third-- and actually, what I'm going to do is, I'm going to write-- I leave space for the second-degree term, which is 0, the first-degree term, and then minus 8 is the constant term, so minus 8-- I", + "qid": "6FrPLJY0rqM_84" + }, + { + "Q": "In 2:45, who is blaise pascal?", + "A": "Blaise Pascal (1623-1662) was a French mathematician, physicist, writer, philosopher, and inventor. Pascal s earliest work was in the field of fluids, where the Pascal (Pa), a unit of pressure, was named after him. He then worked to invent the calculator. He became the second inventor of a calculator. Additionally, he basically created the field of projective geometry and worked on probabilities.", + "video_name": "Yhlv5Aeuo_k", + "timestamps": [ + 165 + ], + "3min_transcript": "\"Hmmm what about 29...? pretty sure it's prime.\" But as a number theorist, you'll be shocked to know it takes me a moment to figure these out. But, even though you have your primes memorised up to at least 1000 that doesn't change that primes, in general, are difficult to find. I mean if I ask you to find the highest even number, you'd say, \"that's silly, just give me the number you think is the highest and i'll just add 2.... BAM!!\" But guess what the highest prime number we know is? 2 to the power of 43,112,609 - 1. Just to give you an idea about how big a deal primes are, the guy that found this one won a $100,000 prize for it! We even sent our largest known prime number into space because scientists think aliens will recognise it as something important and not just some arbitrary number. So they will be able to figure out our alien space message... So if you ever think you don't care about prime numbers because they're 'not useful', remember that we use prime numbers to talk to aliens, I'm not even making this up! Anyway, the point is you started doodling because you were bored but ended up discovering some neat patterns. See how the primes tend to line up on the diagonals? Why do they do that?... also this sort of skeletal structure reminds me of bones so lets call these diagonal runs of primes: Prime Ribs! But how do you predict when a Prime Rib will end? I mean, maybe this next number is prime... (but my head is too fuzzy for now this right now so you tell me.) Anyway...Congratulations, You've discovered the Ulam Spiral! So that's a little mathematical doodling history for you. Yyou can stop being Ulam now... or you can continue. Maybe you like being Ulam. (thats fine) However you could also be Blaise Pascal. Here's another number game you can do using Pascal's triangle.(I don't know why I'm so into numbers today but I have a cold so if you'll just indulge my sick predelections maybe I'll manage to infect you with my enthusiasm :D Pascal's Triangle is the one where you get the next row in the triangle by adding two adjacent numbers. Constructing Pascal's Triangle is, in itself a sort of number game because it's not just don't have to do all the adding. I don't know if this was discovered through doodling but it was discovered independantly in: France, Italy, Persia, China and probably other places too so it's possible someone did. Right... so I don't actually care about the individual numbers right now. So, if you still Ulam, you pick a property and highlight it(e.g. if it's even or odd) If you circle all the odd numbers you'll get a form which might be starting to look familiar. And it makes sense you'd get Sierpinski's Triangle because when you add an odd number and an even number, you get an odd number. (odd + odd) = even and (even + even) = even... So it's just like the crash and burn binary tree game. The best part about it is that, if you know these properties, you can forget about the details of the numbers You don't have to know that a space contains a 9 to know that it's going to be odd. Now, instead of two colours, let's try three. we'll colour them depending on what the remainder is when you divide them by three(instead of by two). Here's a chart! :) So, all the multiples of three are coloured red, remainder of one will be coloured black and", + "qid": "Yhlv5Aeuo_k_165" + }, + { + "Q": "I'm confused. Khan says there is only one x term at 1:55. Can someone help me with this?", + "A": "Look at the second line. There is a 3x, a -3x, and there is a x by itself near the middle. Since the -3x cancels out the 3x, the x by itself is left. hope this helps :)", + "video_name": "DMyhUb1pZT0", + "timestamps": [ + 115 + ], + "3min_transcript": "We're asked to simplify this huge, long expression here. x to the third plus 3x minus 6-- that's in parentheses-- plus negative 2x squared plus x minus 2. And then minus the quantity 3x minus 4. So a good place to start, we'll just rewrite this and see if we can eliminate the parentheses in this step. So let's just start at the beginning. We have the x to the third right over there. So x to the third and then plus 3x-- I'll do that in pink-- plus 3x. And then we have a minus 6. And we don't have to put the parentheses around there, those don't really change anything. And we don't have to even write these-- do anything with these parentheses. We can eliminate them. Just because there's a positive sign out here we don't have to distribute anything. Distributing a positive sign doesn't do anything to these numbers. So then plus, we have a negative 2x squared. So this term right here is negative 2x squared, or minus x squared. And then we have a plus x. We have a plus x. Then we have a negative sign times this whole expression. So we're going to have to distribute the negative sign. So it's a positive 3x, but it's being multiplied by negative 1. So it's really a negative 3x. So minus 3x, then you have a negative-- you can imagine this is a negative 1 implicitly out here-- negative 1 times negative 4. That's a positive 4. So plus 4. Now, we could combine terms of similar degree, of the same degree. Now, first we have an x to the third term and I think it's the only third degree term here, because we have x being raised to the third power. So let me just rewrite it here. We have x to the third. And now let's look at our x squared terms. Looks like we only have one. We only have this term right here. So we have minus 2x squared. And then what about our x terms? We have a 3x plus an x minus a 3x again. So that 3x minus the 3x would cancel out, and you're just So plus x. And then finally our constant terms. Negative 6 minus 2 plus 4. Negative 6 minus 2 gets us to negative 8. Plus 4 is negative 4. We have simplified the expression. Now we just have a four term polynomial.", + "qid": "DMyhUb1pZT0_115" + }, + { + "Q": "at 7:36 how do you get your limits of integration? I understand that you did by inspection, but how would you do it by making the two equations equal to each other ?", + "A": "In that example, Sal is integrating with respect to y. To solve for zero and one analytically, we could make the equations equivalent to each other and solve. Here s the example: sqrt(x) = x^2 x = x^4 This last equation is true only when x = 0 or x = 1.", + "video_name": "WAPZihVUmzE", + "timestamps": [ + 456 + ], + "3min_transcript": "It's going to be 2 minus whatever the x value is right over here. So the x value right over here is going to be y squared. Remember, we want this is a function of y. So our outer radius, this whole distance, is going to be 2 minus the x value here as a function of y. That x value is y squared. So the outer radius is 2 minus x-- sorry. 2 minus y squared. We want it as a function of y. And then our inner radius is going to be equal to what? Well, that's going to be the difference, the horizontal distance, between this vertical line and our inner function, our inner boundary. So it's going to be the horizontal distance between two and whatever x value this is. But this x value as a function of y is just square root of y. So it's going to be 2 minus square root of y. And so now, we can come up with an expression for area. I'll leave the pi there. So it's going to be pi-- right over here-- it's going to be pi times outer radius squared. Well, the outer radius is 2 minus y squared-- and let me just-- well, I'll just write it. 2 minus y squared-- and we're going to square that-- squared, minus pi times the inner radius squared. Well, we already figured that out. The inner radius is 2 minus square root of y. And we're going to square that one, too. So this gives us the area of one of our rings as a function of y, the top of the ring, where I shaded in orange. And now, if we want the volume of one of those rings, we have to multiply it by its depth or its height the way we've drawn it right over here. And its height-- we've done this multiple times already right over here-- is an infinitesimal change in y. So we're going to multiply all that business times dy. This is the volume of one of our rings. So we're going to sum up all of the rings over the interval. And when you take the integral sign, it's a sum where you're taking the limit as you have an infinite number of rings that become infinitesimally small in height or depth, depending on how you view it. And what's our interval? So we've looked at this multiple times. These two graphs-- you could do it by inspection. You could try to solve it in some way, but it's pretty obvious that they intersect at-- remember we care about our y interval. They intersect at y is equal to 0 and y is equal to 1. And there you have it. We've set up our integral for the volume of this shape right over here. I'll leave you there, and in this next video, we will just evaluate this integral.", + "qid": "WAPZihVUmzE_456" + }, + { + "Q": "Are du and dx able to just be treated like normal variables?\n@5:30 he treats the dx as another variable, to be multiplied with the 7", + "A": "Yes, when using Leibnitz notation, you can think of them as quantifiable numbers. What they really mean is this - an infinitely small change in the given variable. For example, du is an infinitely small change in u. dx is an infinitely small change in x. Then, if we think of it this way, du/dx as giving the slope makes sense! The change in u over the change in x gives slope! Long story short, you can treat them as actual numbers - infinitely small changes in a variable.", + "video_name": "oqCfqIcbE10", + "timestamps": [ + 330 + ], + "3min_transcript": "It makes it a little bit easier for us to kind of do the reverse power rule here. So we can rewrite this as equal to 1/7 times the integral of u to the 1/2 power du. And let me just make it clear. This u I could have written in white if I want it the same color. And this du is the same du right over here. So what is the antiderivative of u to the 1/2 power? Well, we increment u's power by 1. So this is going to be equal to-- let me not forget this 1/7 out front. So it's going to be 1/7 times-- if we increment the power here, it's going to be u to the 3/2, 1/2 plus 1 is 1 and 1/2 or 3/2. So it's going to be u to the 3/2. times the reciprocal of 3/2, which is 2/3. And I encourage you to verify the derivative of 2/3 u to the 3/2 is indeed u to the 1/2. And so we have that. And since we're multiplying 1/7 times this entire indefinite integral, we could also throw in a plus c right over here. There might have been a constant. And if we want, we can distribute the 1/7. So it would get 1/7 times 2/3 is 2/21 u to the 3/2. And 1/7 times some constant, well, that's just going to be some constant. And so I could write a constant like that. I could call that c1 and then I could call this c2, but it's really just some arbitrary constant. Oh, actually, no we aren't done. We still just have our entire thing in terms of u. So now let's unsubstitute it. So this is going to be equal to 2/21 times u to the 3/2. u is equal to 7x plus 9. Let me put a new color here just to ease the monotony. So it's going to be 2/21 times 7x plus 9 to the 3/2 power plus c. And we are done. We were able to take a kind of hairy looking integral and realize that even though it wasn't completely obvious at first, that u-substitution is applicable.", + "qid": "oqCfqIcbE10_330" + }, + { + "Q": "At 1:10, Sal says that the perimeter is length+width+length+width, can't you do length*2 and width*2 ?", + "A": "Yup... depends on weather you are faster at addition or multiplication:)", + "video_name": "CDvPPsB3nEM", + "timestamps": [ + 70 + ], + "3min_transcript": "The perimeter of a rectangular fence measures 0.72 kilometers. The length of the fenced area is 160 meters. What is its width? So we have a rectangular fence. Let me draw it. So that is a rectangle. You can imagine we're looking from above. This line is the top of the fence. And if you take its perimeter, the perimeter is the distance around the fence. If you take this distance plus this distance plus that distance plus that distance, it's going to be 0.72 kilometers. That's the total distance of all of the sides. Now, the length of the fenced area is 160 meters. Let's call this the length. So this distance right here is 160 meters. Since it's a rectangle, that distance over there is also going to be 160 meters, and we need to figure out its width. They want us to figure out the width. The width is this distance right over here, which is also Now, what is the perimeter of the fence? The perimeter is the sum of this length plus this length plus that length plus that length. So the perimeter-- let me write it in orange-- is going to be equal to w, the width, plus 160 meters, plus the width, plus 160 meters. And let's assume w is in meters. So we could add it all up. So if you were to add all of these up, you'd get a certain If you were to add all of these up, you have a w plus a w, so you'd be 2w, plus-- 160 plus 160 is 320-- so plus 320. We're assuming that w is in meters. Now, they also told us that the perimeter of the fence is 0.72 kilometers. So the perimeter-- let me do it in that same orange color. The perimeter is also equal to 0.72 kilometers, and we can abbreviate that with km. Now, if we wanted to solve an equation, if we wanted to set this thing equal to the perimeter they gave us, we have to make sure that the units are the same. Here it's in meters. Here it's in kilometers. So let's convert this 0.72 kilometers to meters. And the way to do that, we want the kilometers in the denominator so it cancels out with the kilometers, and you want meters in the numerator. Now, how many meters equal a kilometer? Well, it's 1,000 meters for every 1 kilometer.", + "qid": "CDvPPsB3nEM_70" + }, + { + "Q": "At 8:24 Sal combines X(-b-2a) and X(-a-2b) and is left with only a single X coefficient, I was sure this would be 2X. Can someone please explain why it remains a single X coefficient? Thanks.", + "A": "Because the both 2s in -b-2a and -a-2b would be simplified and that would make it a single X coefficient.", + "video_name": "Vc09LURoMQ8", + "timestamps": [ + 504 + ], + "3min_transcript": "as being equal to that, so let's do that. We get x plus b times c, over a plus 2b is equal to x minus a times c over negative b minus 2a. Well, both sides are divisible by c, we can assume that c is non-zero, so this triangle actually exists, it actually exists in two dimensions, so we can divide both sides by c and we get that. Let's see, now we can cross-multiply. We can multiply x plus b times this quantity right here, and that's going to be equal to a plus 2b times this quantity over here. We can just distribute it, so it's going to be x times all of this business, it's going to be x times negative b minus 2a, plus b times all of this, so I just multiply that times that, is going to equal to this times that, so it's going to be x times all of this business. X times a plus 2b, minus a times that, so distribute the a, minus a squared minus 2ab. Let's see if we can simplify this. We have a negative 2ab on both sides, so let's just subtract that out, so we cancelled things out, let's subtract this from both sides of the equation, so we'll have a minus x times a plus 2b, and I'll subtract that from here, but I'll write it a little different, it'll simplify things, so this will become x times, I'll distribute the negative inside the a plus 2b, so negative a minus 2b, and let's add a b squared to both sides, so plus b squared, plus b squared. and all of the constants on the other. This becomes on the left-hand side, the coefficient on x, I have negative b minus 2b is negative 3b negative 2a minus a is negative 3a, times x is equal to, these guys cancel out, and these guys cancel out, is equal to b squared minus a squared. Let's see if we can factor, this already looks a little suspicious in a good way, something that we should be able to solve. This can be factored. We can factor out a negative three. We can actually factor out a three. We'll get three times b minus a, actually, let's factor out a negative three, so negative three times b plus a times x is equal to, now this is the same thing as b plus a", + "qid": "Vc09LURoMQ8_504" + }, + { + "Q": "at 1:24, how can there be 4 possibilities, one person only have 3 options and not his own self. right?", + "A": "The 4 means that the first person involved in shaking hands can be any of the 4 people. The 3 means that the second person involved in shaking hands can be any of the remaining 3 people not counting the person identified as the first person.", + "video_name": "boH4l1SgJbM", + "timestamps": [ + 84 + ], + "3min_transcript": "- Let's say that there are four people in a room. And you're probably tired of me naming the people with letters, but I'm going to continue doing that. So the four people in the room are people A, B, C, and D. And they are all told, \"You don't know each other. \"So I want you to all meet each other. \"You need to shake the hand, exactly once, \"of every other person in the room so that you all meet.\" So my question to you is, if each of these people need to shake the hand of every other person exactly once, how many handshakes are going to occur? The number of handshakes that are going to occur. So, like always, pause the video and see if you can make sense of this. Alright, I'm assuming you've had a go at it. So one way to think about it is, if you say there's a handshake, two people are party to a handshake. We're not talking about some new three-person handshake or four-person handshake, we're just talking about the traditional, two people shake their right hands. another person in this party. There's four possibilities of one party. And if we assume people aren't shaking their own hands, which we are assuming, they're always going to shake someone else's hand. For each of these four possibilities who's this party, there's three possibilities who's the other party. And so you might say that there's four times three handshakes. Since there's four times three possible handshakes. And what I'd like you to do is think a little bit about whether this is right, whether there would actually be 12 handshakes. You might have thought about it, and you might say, there's four times three, this is actually counting the permutations. This is counting how many ways can you permute four people into two buckets, the two buckets of handshakers, where you care about which bucket they are in. Whether they're handshaker number one, or handshaker number two. and B being the number two handshaker as being different than B being the number one handshaker and A being the number two handshaker. But we don't want both of these things to occur. We don't want A to shake B's hand, where A is facing north and B is facing south. And then another time, they shake hands again where now B is facing north and A is facing south. We only have to do it once. These are actually the same thing, so there's no reason for both of these to occur. So we are going to be double counting. So what we really want to do is think about combinations. One way to think about it is, you have four people. In a world of four people or a pool of four people, how many ways can you choose two?", + "qid": "boH4l1SgJbM_84" + }, + { + "Q": "AT 9:20 would 5/4 x^-1 y also be correct", + "A": "It generally would be considered incomplete. Final answers should have positive exponents. And, we would multiply the items together to show them as one fraction. This is why Sal is showing the answer as 5y / (4x)", + "video_name": "AR1uqNbjM5s", + "timestamps": [ + 560 + ], + "3min_transcript": "And we would have simplified this about as far as you can go. Let's do one more of these. I think they're good practice and super-valuable experience later on. Let's say I have 25xy to the sixth over 20y to the fifth x squared. So once again, we can rearrange the numerators and the denominators. So this you could rewrite as 25 over 20 times x over x squared, right? We could have made this bottom 20x squared y to the fifth-- it doesn't matter the order we do it in-- times y to the sixth over y to the fifth. actually just simplify fractions. 25 over 20, if you divide them both by 5, this is equal to 5 over 4. x divided by x squared-- well, there's two ways you could think about it. That you could view as x to the negative 1. You have a first power here. 1 minus 2 is negative 1. So this right here is equal to x to the negative 1 power. Or it could also be equal to 1 over x. These are equivalent. So let's say that this is equal into 1 over x, just like that. And it would be. x over x times x. One of those sets of x's would cancel out and you're just left with 1 over x. And then finally, y to the sixth over y to the fifth, that's y to the 6 minus 5 power, which is just y to the first power, or just y, so times y. rational expression, you have 5 times 1 times y, which would be 5y, all of that over 4 times x, right? This is y over 1, so 4 times x times 1, all of that over 4x, and we have successfully simplified it.", + "qid": "AR1uqNbjM5s_560" + }, + { + "Q": "At 0:20 can we move the x out in front first, instead of applying the quotient log property first? And then apply the quotient log property?\n\nIf not...is there a certain order in which log properties should be applied (like BEDMAS)??? Thanks in advance :)", + "A": "You can t move the x in front first, because not the whole factor is to the xth power. if we had log(25/y)^x we could. otherwise the whole term, so both log25 and log5 would be times 5, which is wrong. I hope this helps you, English is not my first language, so I don t know if I explained it well enough!", + "video_name": "RhzXX5PbsuQ", + "timestamps": [ + 20 + ], + "3min_transcript": "We're asked to simplify log base 5 of 25 to the x power over y. So we can use some logarithm properties. And I do agree that this does require some simplification over here, that having this right over here inside of the logarithm is not a pleasant thing to look at. So the first thing that we realize-- and this is one of our logarithm properties-- is logarithm for a given base-- so let's say that the base is x-- of a/b, that is equal to log base x of a minus log base x of b. And here we have 25 to the x over y. So we can simplify. So let me write this down. I'll do this in blue. Log base 5 of 25 to the x over y using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying. is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did it right over there. So this part right over here can be rewritten as x times the logarithm base 5 of 25. And then, of course, we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about. This part right here is asking us, what power do I have to raise 5 to to get to 25? So we have to raise 5 to the second power to get to 25. So this simplifies to 2. So then we are left with, this is equal to-- and I'll write it in front of the x now-- 2 times x minus log base 5 of y.", + "qid": "RhzXX5PbsuQ_20" + }, + { + "Q": "why do you find the square root of (4x+1 )squared and 8 around 1:30?", + "A": "Since we are trying to solve for x, we want to isolate x on one side of the equation. in order to do that, we have to cancel out as much stuff on the left hand side of the equation as possible. In this example, if we take (4x+1)^2 = 8 and we take the square root of both sides, we are left with 4x+1=sqrt(8), and we are one step closer to isolating x.", + "video_name": "55G8037gsKY", + "timestamps": [ + 90 + ], + "3min_transcript": "In this video, I'm going to do several examples of quadratic equations that are really of a special form, and it's really a bit of warm-up for the next video that we're going to do on completing the square. So let me show you what I'm talking about. So let's say I have 4x plus 1 squared, minus 8 is equal to 0. Now, based on everything we've done so far, you might be tempted to multiply this out, then subtract 8 from the constant you get out here, and then try to factor it. And then you're going to have x minus something, times x minus something else is equal to 0. And you're going to say, oh, one of these must be equal to 0, so x could be that or that. We're not going to do that this time, because you might see something interesting here. We can solve this without factoring it. And how do we do that? Well, what happens if we add 8 to both sides of this equation? squared, and these 8's cancel out. The right-hand becomes just a positive 8. Now, what can we do to both sides of this equation? And this is just kind of straight, vanilla equation-solving. This isn't any kind of fancy factoring. We can take the square root of both sides of this equation. We could take the square root. So 4x plus 1-- I'm just taking the square root of both sides. You take the square root of both sides, and, of course, you want to take the positive and the negative square root, because 4x plus 1 could be the positive square root of 8, or it could be the negative square root of 8. So 4x plus 1 is equal to the positive or negative square root of 8. Instead of 8, let me write 8 as 4 times 2. We all know that's what 8 is, and obviously the square root of 4x plus 1 squared is 4x plus 1. or the square root of 4, which is 2-- is equal to the plus or minus times 2 times the square root of 2, right? Square root of 4 times square root of 2 is the same thing as square root of 4 times the square root of 2, plus or minus the square root of 4 is that 2 right there. Now, it might look like a really bizarro equation, with this plus or minus 2 times the square of 2, but it really isn't. These are actually two numbers here, and we're actually simultaneously solving two equations. We could write this as 4x plus 1 is equal to the positive 2, square root of 2, or 4x plus 1 is equal to negative 2 times the square root of 2. This one statement is equivalent to this right here, because we have this plus or minus here, this or statement. Let me solve all of these simultaneously.", + "qid": "55G8037gsKY_90" + }, + { + "Q": "What does Sal mean when he says, \"vanilla like\"? I am guessing he means not-so-complex-looking, like at 4:48.", + "A": "In the U.S, an object can be described to be as plain as vanilla , which means that something is plain and simple like vanilla ice cream. You are correct in that he means not-so-complex-looking.", + "video_name": "55G8037gsKY", + "timestamps": [ + 288 + ], + "3min_transcript": "equation, what do I have? On the left-hand side, I'm just left with 4x. On the right-hand side, I have-- you can't really mathematically, I mean, you could do them if you had a calculator, but I'll just leave it as negative 1 plus or minus the square root, or 2 times the square root of 2. That's what 4x is equal to. If we did it here, as two separate equations, same idea. Subtract 1 from both sides of this equation, you get 4x is equal to negative 1 plus 2, times the square root of 2. This equation, subtract 1 from both sides. 4x is equal to negative 1 minus 2, times the square root of 2. This statement right here is completely equivalent to these two statements. Now, last step, we just have to divide both sides by 4, so you divide both sides by 4, and you get x is equal to root of 2, over 4. Now, this statement is completely equivalent to dividing each of these by 4, and you get x is equal to negative 1 plus 2, times the square root 2, over 4. This is one solution. And then the other solution is x is equal to negative 1 minus 2 roots of 2, all of that over 4. That statement and these two statements are equivalent. And if you want, I encourage you to-- let's substitute one of these back in, just so you feel confident that something as bizarro as one of these expressions can be a solution to a nice, vanilla-looking equation like this. So let's substitute it back in. 4 times x, or 4 times negative 1, plus 2 root 2, over 4, plus 1 squared, minus 8 is equal to 0. plus 2 roots 2, plus 1, squared, minus 8 is equal to 0. This negative 1 and this positive 1 cancel out, so you're left with 2 roots of 2 squared, minus 8 is equal to 0. And then what are you going to have here? So when you square this, you get 4 times 2, minus 8 is equal to 0, which is true. 8 minus 8 is equal to 0. And if you try this one out, you're going to get the exact same answer. Let's do another one like this. And remember, these are special forms where we have squares of binomials in our expression. And we're going to see that the entire quadratic formula is actually derived from a notion like this, because you can actually turn any, you can turn any, quadratic equation into a perfect square equalling something else.", + "qid": "55G8037gsKY_288" + }, + { + "Q": "Around 3:10 why not take a proportion like 37% instead of 30%?", + "A": "Look at the null hypothesis. A value of 37% does not satisfy the null hypothesis of p<=0.30. When choosing a value of p, we need to satisfy the null hypothesis, and to choose a conservative number (one which will be less likely to reject the null hypothesis. In this case the largest value allowed by the null, which is 0.30).", + "video_name": "1JT9oODsClE", + "timestamps": [ + 190 + ], + "3min_transcript": "hypothesis for the population. And the given that assumption, what is the probability that 57 out of 150 of our samples actually have internet access. And if that probability is less than 5%, if it's less than our significance level, then we're going to reject the null hypothesis in favor of the alternative one. So let's think about this a little bit. So we're going to start off assuming-- we're going to assume the null hypothesis is true. And in that assumption we're going to have to pick a population proportion or a population mean-- we know that for Bernoulli distributions do the same thing. And what I'm going to do is I'm going to pick a proportion so high so that it maximizes the probability of getting this over here. And we actually don't even know what that number is. And actually so that we can think about a little more proportion even is. We had 57 people out of 150 having internet access. So 57 households out of 150. So our sample proportion is 0.38, so let me write that over here. Our sample proportion is equal to 0.38. So when we assume our null hypothesis to be true, we're going to assume a population proportion that maximizes the probability that we get this over here. So the highest population proportion that's within our null hypothesis that will maximize the probability of getting this is actually if we are right at 30%. So if we say our population proportion, we're going to assume this is true. This is our null hypothesis. We're going to assume that it is 0.3 or 30%. And I want you understand that-- 29% would have been a 28% that would have been a null hypothesis. But for 29% or 28%, the probability of getting this would have been even lower. So it wouldn't have been as strong of a test. If we take the maximum proportion that still satisfies our null hypothesis, we're maximizing the probability that we get this. So if that number is still low, if it's still less than 5%, we can feel pretty good about the alternative hypothesis. So just to refresh ourselves we're going to assume a population proportion of 0.3, and if we just think about the distribution-- sometimes it's helpful to draw these things, so I will draw it. So this is what the population distribution looks like based on our assumption, based on this assumption right over here. Our population distribution has-- or maybe I should write 30% have internet access. And I'll signify that with a 1. And then the rest don't have internet access. 70% do not have internet access.", + "qid": "1JT9oODsClE_190" + }, + { + "Q": "at 1:50 how do you divide?", + "A": "You don t, you multiply the improper fraction by the recipricole of 1/5, which is 5/1", + "video_name": "xoXYirs2Mzw", + "timestamps": [ + 110 + ], + "3min_transcript": "Tracy is putting out decorative bowls of potpourri in each room of the hotel where she works. She wants to fill each bowl with 1/5 of a can of potpourri. If Tracy has 4 cans of potpourri, in how many rooms can she place a bowl of potpourri? So she has 4 cans, and she wants to divide this 4 cans into groups of 1/5 of a can. So if you have 4 of something and you're trying to divide it into groups of a certain amount, you would divide by that amount per group. So you want to divide 4 by 1/5. You want to divide 4 cans of potpourri into groups of 1/5. So let's visualize this. Let me draw one can of potpourri right over here. So one can of potpourri can clearly be cut up into 5/5. We have it right over here. 1, 2, 3, 4, 5. So 1 can of potpourri can fill 5 bowls Now, we have 4 cans. So let me paste these. So 2, 3, and 4. So how many total bowls of potpourri can Tracy fill? Well, she's got 4 cans. So this is going to be equal to-- let me do this is the right color-- this is going to be equal to, once again, she has 4 cans. And then for each of those cans, she can fill 5 bowls of potpourri because each bowl only requires a 1/5 of those cans. So this is going to be the same thing as 4 times 5. Or we can even write this as 4 times 5 over 1. 5 is the same thing as 5/1, which is the same thing as 4 times 5, which, of course, is equal to 20. she can fill 20 bowls of potpourri. Now, just as a review here, we've already seen that dividing by a number is equal to multiplying by its reciprocal. And we see that right over here. Dividing by 1/5 is the same thing as multiplying by the reciprocal of 1/5, which is 5/1. So she could fill up 20 bowls of potpourri.", + "qid": "xoXYirs2Mzw_110" + }, + { + "Q": "At 8:15, Isn't PI transcendental number, so it is not a real number.\nSo is the graph wrong ?", + "A": "\u00cf\u0080 is both transcendental and real. Transcendental just means that the number isn t the root of a polynomial with rational coefficients. \u00e2\u0088\u009a2 is irrational, but not transcendental, since it s the root of x\u00c2\u00b2-2=0.", + "video_name": "sjUhr0HkLUg", + "timestamps": [ + 495 + ], + "3min_transcript": "", + "qid": "sjUhr0HkLUg_495" + }, + { + "Q": "At 6:02 he writes the range for a general arcsin function, but just to confirm, that's the range for all arcsin functions?? Or am I missing something. Because for the example problem he did at the end, he didn't state the domain and range, but I thought restricting the range was necessary to make the sine function valid. Could someone please clarify?", + "A": "i think you mean all inverse trig functions rather than all arcsin functions. and to answer your question, the range of all inverse trig functions are not the same due to the shape of the graph.", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 362 + ], + "3min_transcript": "Let me go over here. You're like, look pi over 2 worked. 45 degrees worked. But I could just keep adding 360 degrees or I could keep just adding 2 pi. And all of those would work because those would all get me to that same point of the unit circle, right? And you'd be correct. And so all of those values, you would think, would be valid answers for this, right? Because if you take the sine of any of those angles-- You could just keep adding 360 degrees. If you take the sine of any of them, you would get square root of 2 over 2. And that's a problem. You can't have a function where if I take the function-- I can't have a function, f of x, where it maps to multiple values, right? Where it maps to pi over 4, or it maps to pi over 4 plus 2 pi or pi over 4 plus 4 pi. So in order for this to be a valid function-- In order for the inverse sine function to be valid, I have to restrict its range. And the way that-- We'll just restrict its range to So let's restrict its range. Actually, just as a side note, what's its domain restricted to? So if I'm taking the arcsine of something. So if I'm taking the arcsine of x, and I'm saying that that is equal to theta, what's the domain restricted to? What are the valid values of x? x could be equal to what? Well if I take the sine of any angle, I can only get values between 1 and negative 1, right? So x is going to be greater than or equal to negative 1 and then less than or equal to 1. That's the domain. Now, in order to make this a valid function, I have to restrict the range. The possible values. I have to restrict the range. Now for arcsine, the convention is to restrict it to the first and fourth quadrants. To restrict the possible angles to this area right here along the unit circle. So theta is restricted to being less than or equal to pi over So given that, we now understand what arcsine is. Let's do another problem. Clear out some space here. Let me do another arcsine. So let's say I were to ask you what the arcsine of minus the square root of 3 over 2 is. Now you might have that memorized. And say, I immediately know that sine of x, or sine of theta is square root of 3 over 2. And you'd be done. But I don't have that memorized. So let me just draw my unit circle. And when I'm dealing with arcsine, I just have to draw the first and fourth quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I?", + "qid": "JGU74wbZMLg_362" + }, + { + "Q": "At 1:50, Sal says that the sin is the height, but I'm pretty sure that is not usually sin. What special circumstances makes sin the height?", + "A": "Hello Kevin, Height, including negative heights, is an appropriate description for SIN(X). You see, SIN(X) gives the vertical component and COS(X) gives the horizontal component. If this is a fuzzy answer it will make more sense when you get to the unit circle. Regards, APD", + "video_name": "JGU74wbZMLg", + "timestamps": [ + 110 + ], + "3min_transcript": "If I were to walk up to you on the street and say you, please tell me what-- so I didn't want to write that thick --please tell me what sine of pi over 4 is. And, obviously, we're assuming we're dealing in radians. You either have that memorized or you would draw the unit circle right there. That's not the best looking unit circle, but you get the idea. You'd go to pi over 4 radians, which is the same thing as 45 degrees. You would draw that unit radius out. And the sine is defined as a y-coordinate on the unit circle. So you would just want to know this value right here. And you would immediately say OK. This is a 45 degrees. Let me draw the triangle a little bit larger. The triangle looks like this. This is 45. That's 45. This is 90. And you can solve a 45 45 90 triangle. The hypotenuse is 1. This is x. They're going to be the same values. Their base angles are the same. So you say, look. x squared plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1/2. x is equal to the square root of 1/2, which is one over the square root of 2. I can put that in rational form by multiplying that by the square root of 2 over 2. And I get x is equal to the square root of 2 over 2. So the height here is square root of 2 over 2. And if you wanted to know this distance too, it would also be the same thing. But we just cared about the height. Because the sine value, the sine of this, is just this height right here. The y-coordinate. And we got that as the square root of 2 over 2. This is all review. We learned this in the unit circle video. But what if someone else-- Let's say on another day, I arcsine of the square root of 2 over 2 is. What is the arcsine? You're like I know what the sine of an angle is, but this is some new trigonometric function that Sal has devised. And all you have to realize, when they have this word arc in front of it-- This is also sometimes referred to as the inverse sine. This could have just as easily been written as: what is the inverse sine of the square root of 2 over 2? All this is asking is what angle would I have to take the sine of in order to get the value square root of 2 over 2. This is also asking what angle would I have to take the sine of in order to get square root of 2 over 2. I could rewrite either of these statements as saying square-- Let me do it.", + "qid": "JGU74wbZMLg_110" + }, + { + "Q": "4:12-4:20 i got confused who can help me ???", + "A": "oh ok thanks for the help ..i really appreciated it", + "video_name": "-YI6UC4qVEY", + "timestamps": [ + 252, + 260 + ], + "3min_transcript": "This combination. Let me square it off for you. So that right there is Heron's Formula. And if that looks a little bit daunting-- it is a little bit more daunting, clearly, than just 1/2 times base times height. Let's do it with an actual example or two, and actually see this is actually not so bad. So let's say I have a triangle. I'll leave the formula up there. So let's say I have a triangle that has sides of length 9, 11, and 16. So let's apply Heron's Formula. S in this situation is going to be the perimeter divided by 2. So 9 plus 11 plus 16, divided by 2. Which is equal to 9 plus 11-- is 20-- plus 16 is 36, divided by 2 is 18. the square root of S-- 18-- times S minus a-- S minus 9. 18 minus 9, times 18 minus 11, times 18 minus 16. And then this is equal to the square root of 18 times 9 times 7 times 2. Which is equal to-- let's see, 2 times 18 is 36. So I'll just rearrange it a bit. This is equal to the square root of 36 times 9 times 7, which is equal to the square root of 36 times the square The square root of 36 is just 6. This is just 3. And we don't deal with the negative square roots, because you can't have negative side lengths. And so this is going to be equal to 18 times the square root of 7. So just like that, you saw it, it only took a couple of minutes to apply Heron's Formula, or even less than that, to figure out that the area of this triangle right here is equal to 18 square root of seven. Anyway, hopefully you found that pretty neat.", + "qid": "-YI6UC4qVEY_252_260" + }, + { + "Q": "Principal root of a aquare root fnction? What does he mean by this? It is approximately 5:45...", + "A": "Principal root is the positive one of the two roots", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 345 + ], + "3min_transcript": "and complex numbers and all the rest-- you have to expand the definition of what this radical means. So when you are taking the square root of really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function, or this is now defined for complex inputs or the domain, it can also generate imaginary or complex outputs, or I guess you could call that the range. And if you assume that, then really straight from this you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only-- and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true-- So we could apply this when x is greater than or equal to 0. then negative x is clearly a negative number, or I guess it could also be 0. It's a negative number. And then we can apply this right over here. If x was less than 0, then we would be doing all of this nonsense up here. And we start to get nonsensical answers. And if you look at it this way and you say hey, look, i can be the square root of negative one, if it's the principal branch of the complex square root function. Then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really the real fault in this logic when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the complex to include negative numbers in the domain and to include imaginary numbers, then you can do this. You can say the square root of negative x is the square root of negative 1 times-- Or you should say the principal square root of negative x-- I should be particular my words-- is the same thing as the principal square root of negative 1 times the principal square root of x when x is greater than or equal to 0. And I don't want to confuse you, if x is greater than or equal to 0, this negative x, that is clearly a negative, or I guess you should say a non positive number.", + "qid": "rYG1D5lUE4I_345" + }, + { + "Q": "I think Sal make a mistake on (vid @ 5:11) when he write the greater than sign! it should be Less than", + "A": "No, Sal is correct. If he had: i sqrt(x) where X<0, then X is negative. Backup thru Sal steps. If X is negative * (-1) = +X. And he would have started with sqrt(x), not sqrt(-x). He is also trying to highlight that if you had something like: sqrt(12), you would not make this into i sqrt(-12). The imaginary number is not needed if the radical contains a positive number to start with.", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 311 + ], + "3min_transcript": "saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. When we take traditional principal square roots. So when you take the principle square root of 4. We know that this is positive 2, that 4 actually has two square roots. There's negative 2 also is a square root of 4. If you have negative 2 times negative 2 it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non imaginary, non complex numbers, you could really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers-- and complex numbers and all the rest-- you have to expand the definition of what this radical means. So when you are taking the square root of really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function, or this is now defined for complex inputs or the domain, it can also generate imaginary or complex outputs, or I guess you could call that the range. And if you assume that, then really straight from this you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only-- and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true-- So we could apply this when x is greater than or equal to 0. then negative x is clearly a negative number, or I guess it could also be 0. It's a negative number. And then we can apply this right over here. If x was less than 0, then we would be doing all of this nonsense up here. And we start to get nonsensical answers. And if you look at it this way and you say hey, look, i can be the square root of negative one, if it's the principal branch of the complex square root function. Then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really the real fault in this logic when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the complex", + "qid": "rYG1D5lUE4I_311" + }, + { + "Q": "At 6:41 how can it only apply when x is greater or equal to zero, given that it is a negative number?", + "A": "When the number is positive, it applies as a normal (square) root. For example, the square root of 4 is 2, but the square root of -4 could be seen as 2i, because i=-1. Therefore, x must be greater than or equal to zero in order to have that negative number, and in turn, contain i. Like he says at 5:58, the two negative numbers are where it goes wrong. x cannot be negative.", + "video_name": "rYG1D5lUE4I", + "timestamps": [ + 401 + ], + "3min_transcript": "and complex numbers and all the rest-- you have to expand the definition of what this radical means. So when you are taking the square root of really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function, or this is now defined for complex inputs or the domain, it can also generate imaginary or complex outputs, or I guess you could call that the range. And if you assume that, then really straight from this you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only-- and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true-- So we could apply this when x is greater than or equal to 0. then negative x is clearly a negative number, or I guess it could also be 0. It's a negative number. And then we can apply this right over here. If x was less than 0, then we would be doing all of this nonsense up here. And we start to get nonsensical answers. And if you look at it this way and you say hey, look, i can be the square root of negative one, if it's the principal branch of the complex square root function. Then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really the real fault in this logic when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the complex to include negative numbers in the domain and to include imaginary numbers, then you can do this. You can say the square root of negative x is the square root of negative 1 times-- Or you should say the principal square root of negative x-- I should be particular my words-- is the same thing as the principal square root of negative 1 times the principal square root of x when x is greater than or equal to 0. And I don't want to confuse you, if x is greater than or equal to 0, this negative x, that is clearly a negative, or I guess you should say a non positive number.", + "qid": "rYG1D5lUE4I_401" + }, + { + "Q": "At 2:41, what does Sal mean by \" And notice, both of these numbers are exactly 10 away from the number 5?\" Why are the numbers 10 units away from 5? Why isn't it 10 away from 0?", + "A": "To be 10 away from zero, the problem would need to be: |x| = 10. Notice the original equation: |x - 5| = 10. You need to take into account the -5 inside the absolute value. That s where the 5 comes from. Hope this helps.", + "video_name": "u6zDpUL5RkU", + "timestamps": [ + 161 + ], + "3min_transcript": "So let's say I have the equation the absolute value of x minus 5 is equal to 10. And one way you can interpret this, and I want you to think about this, this is actually saying that the distance between x and 5 is equal to 10. So how many numbers that are exactly 10 away from 5? And you can already think of the solution to this equation, but I'll show you how to solve it systematically. Now this is going to be true in two situations. Either x minus 5 is equal to positive 10. If this evaluates out to positive 10, then when you take the absolute value of it, you're going to get positive 10. Or x minus 5 might evaluate to negative 10. If x minus 5 evaluated to negative 10, when you take the absolute value of it, you would get 10 again. So x minus 5 could also be equal to negative 10. Now, to solve this one, add 5 to both sides of this equation. You get x is equal to 15. To solve this one, add 5 to both sides of this equation. x is equal to negative 5. So our solution, there's two x's that satisfy this equation. x could be 15. 15 minus 5 is 10, take the absolute value, you're going to get 10, or x could be negative 5. Negative 5 minus 5 is negative 10. Take the absolute value, you get 10. And notice, both of these numbers are exactly 10 away from the number 5. Let's do another one of these. Let's do another one. Let's say we have the absolute value of x plus 2 is equal to 6. So what does that tell us? That tells us that either x plus 2, that the thing inside Or the thing inside of the absolute value sign, the x plus 2, could also be negative 6. If this whole thing evaluated to negative 6, you take the absolute value, you'd get 6. So, or x plus 2 could equal negative 6. And then if you subtract 2 from both sides of this equation, you get x could be equal to 4. If you subtract 2 from both sides of this equation, you get x could be equal to negative 8. So these are the two solutions to the equation. And just to kind of have it gel in your mind, that absolute value, you can kind of view it as a distance, you could rewrite this problem as the absolute value of x minus negative 2 is equal to 6. And so this is asking me, what are the x's that are exactly 6 away from negative 2? Remember, up here we said, what are the x's that are exactly 10 away from positive 5?", + "qid": "u6zDpUL5RkU_161" + }, + { + "Q": "At 1:39, Sal said that 0 to the zeroth power is undefined. Why?", + "A": "Well, any number to the zeroth power is 1 right? So 0 to the zeroth power would be 1. But zero to any power is 0. So 0 to the zeroth power would be 0. Which is it? It can t be both, so it is undefined.", + "video_name": "NEaLgGi4Vh4", + "timestamps": [ + 99 + ], + "3min_transcript": "Evaluate the expression 4n to the first power minus 2n to the zeroth power for n equals 1 and n equals 5. So, let's do n equals 1 first. So, every place we see an n, let's put a 1 in there. So this is equal to 4 times 1 to the first power minus 2 times 1 to the zeroth power. Every place we saw an n, we have an n there. We have an n to the first, one to the first. So remember, we want to do our order of operations. And I'm going to write it this way just to make it clear that we do parentheses then exponentiation. Then multiplication and division are at the same level, they get the same priority. And then addition and subtraction get the same priority. So, we want to do our exponents before we do anything else. There are no parentheses here. So, we first want to evaluate one to the first power. And anything to the first power is just that anything. If I just have any number to the first power, that is just going to be equal to that number. So, 1 to the first power is just equal to 1. Let me write it. 4 times 1 minus 2 times 1 to the zeroth power. Now, any number other than 0 to the zeroth power is going to be 1. So, if I tell you x to the zeroth power is equal to what? And I tell you that x is not equal to 0. What is x to the zeroth power? You should immediately know that it is equal to 1. I exclude 0 because 0 to the zeroth power is usually undefined. Or for our purposes, you can accept it to be undefined. And maybe in the future, I'll do a video on explaining why it's undefined. There's some good arguments for why this should be 0. Because 0 to pretty much any other power is a 0. But then there's a really good argument for why it should be 1. Because any other number to the zeroth power is 1. So that's why, for the most part, we like to keep this undefined. So with that said, 1 to the zeroth power Anything other than 0 to the zeroth power is 1. So, we have 4 times 1, which is 4, minus 2 times 1, which is 2. So, 4 minus 2 is just equal to 2. Now, let's do the same thing with n equals 5. So it becomes 4 times 5 to the first power minus 2 times 5 to the zeroth power. Same thing. 5 to the first power is just going to be equal to 5. So, I want to do that in a different color. 5 to the first power is just going to be equal to 5. 5 to the zeroth power is just going to be equal to 1. And so, this expression becomes 4 times 5 minus 2 times 1. Or 4 times 5 is 20, minus 2 times 1, that is 2. And this is equal to 18. And we're done.", + "qid": "NEaLgGi4Vh4_99" + }, + { + "Q": "At 9:00, I am not sure but why didn't Sal solve those 2 equations using matrices? I mean before these vector videos, he was using matrices to solve these kinds of equations. Or is it that these equations can not be solved through matrices? Thanks In Advance.", + "A": "If he used a matrix, it would remove the C constants and sort of defeat the purpose for doing what he was doing. He just wanted to keep the constants intact so you could see what was happening.", + "video_name": "Alhcv5d_XOs", + "timestamps": [ + 540 + ], + "3min_transcript": "In order for them to be linearly dependent, that means that if some constant times 2,1 plus some other constant times this second vector, 3,2 where this should be equal to 0. Where these both aren't necessarily 0. Before I go up for this problem, let's remember what we're going to find out. If either of these are non-zero, if c1 or c2 are non-zero, then this implies that we are dealing with a dependent, linearly dependent set. If c1 and c2 are both 0, if the only way to satisfy this equation -- I mean you can always satisfy it by sitting guys 0, then we're dealing with a linearly independent set. Let's try to do some math. And this'll just take us back to our Algebra 1 days. In order for this to be true, that means 2 times c1 plus 3 times c2 is equal to -- when I say this is equal to 0, it's really the 0 vector. I can rewrite this as 0,0. So 2 times c1 plus 3 times c2 would be equal to that 0 there. And then we'd have 1 times c1 plus 2 times c2 is equal to that 0. And now this is just a system, two equations, two unknowns. A couple of things we could do. Let's just multiply this top equation by 1/2. equal to 0. And then if we subtract the green equation from the red equation this becomes 0. 2 minus 1 and 1/2-- 3/2 is 1 and 1/2 --of this is just 1/2 c2 is equal to 0. And this is easy to solve. c2 is equal to 0. So what's c1? Well, just substitute this back in. c2 is equal to 0. So this is equal to 0. So c1 plus 0 is equal to 0. So c1 is also equal to 0. We could have substituted it back into that top equation as well. So the only solution to this equation involves both c1 and c2 being equal to 0. So they both have to be 0. So this is a linearly independent set of vectors.", + "qid": "Alhcv5d_XOs_540" + }, + { + "Q": "At 20:00, why does a derivative of 0 mean a maximum/minimum? I get why, but I thought y=x^2 has only one minimum at x=0.", + "A": "Yes and if you let d/dx x^2 = 2x =0, you get x = 0 (and y = 0).", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 1200 + ], + "3min_transcript": "Because the problem wants us to find that point, the maximum point of intersection. They call this the extreme normal line. The extreme normal line is when our second quadrant intersection essentially achieves a maximum point. I know they call it the smallest point, but it's the smallest negative value, so it's really a maximum point. So how do we figure out that maximum point? Well, we have our second quadrant intersection as a function of our first quadrant x. I could rewrite this as, my second quadrant intersection as a function of x0 is equal to minus x minus 1 over 2 x0. So this is going to reach a minimum or a maximum point when its derivative is equal to 0. This is a very unconventional notation, and that's probably the hardest thing about this problem. But let's take this derivative with respect to x0. So my second quadrant intersection, the derivative of straightforward. It's equal to minus 1, and then I have a minus 1/2 times, this is the same thing as x to the minus 1. So it's minus 1 times x0 to the minus 2, right? I could have rewritten this as minus 1/2 times x0 to the minus 1. So you just put its exponent out front and decrement it by 1. And so this is the derivative with respect to my first quadrant intersection. So let me simplify this. So x, my second quadrant intersection, the derivative of it with respect to my first quadrant intersection, is equal to minus 1, the minus 1/2 and the minus 1 become a positive when you multiply them, and so plus 1/2 over x0 squared. Now, this'll reach a maximum or minimum when it equals 0. problem right there. Well, we add one to both sides. We get 1 over 2 x0 squared is equal to 1, or you could just say that that means that 2 x0 squared must be equal to 1, if we just invert both sides of this equation. Or we could say that x0 squared is equal to 1/2, or if we take the square roots of both sides of that equation, we get x0 is equal to 1 over the square root of 2. So we're really, really, really close now. We've just figured out the x0 value that gives us our extreme normal line. This value right here. Let me do it in a nice deeper color. This value right here, that gives us the extreme normal line, that over there is x0 is equal to 1 over the square root of 2. Now, they want us to figure out the equation of the", + "qid": "viaPc8zDcRI_1200" + }, + { + "Q": "At 18:45 How does khan figure out that when derivative of -x_0-1/2x_0 is equal to 0 is actually minimum or maximum of the function?", + "A": "derivative = 0 ==> (implies) a maximum or minimum", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 1125 + ], + "3min_transcript": "So minus x0 minus 1 over 4 mine x0. Now what do we have? So let's see. We have a minus 1 over 4 x0, minus 1 over 4 x0. So this is equal to minus x0, minus x0, minus 1 over 2 x0. So if I take minus 1/4 minus 1/4, I get minus 1/2. And so my second quadrant intersection, all this work I did got me this result. My second quadrant intersection, I hope I don't run out of space. My second quadrant intersection, of the normal line and the parabola, is minus x0 minus 1 over 2 x0. Now this by itself is a pretty neat result we just got, but Because the problem wants us to find that point, the maximum point of intersection. They call this the extreme normal line. The extreme normal line is when our second quadrant intersection essentially achieves a maximum point. I know they call it the smallest point, but it's the smallest negative value, so it's really a maximum point. So how do we figure out that maximum point? Well, we have our second quadrant intersection as a function of our first quadrant x. I could rewrite this as, my second quadrant intersection as a function of x0 is equal to minus x minus 1 over 2 x0. So this is going to reach a minimum or a maximum point when its derivative is equal to 0. This is a very unconventional notation, and that's probably the hardest thing about this problem. But let's take this derivative with respect to x0. So my second quadrant intersection, the derivative of straightforward. It's equal to minus 1, and then I have a minus 1/2 times, this is the same thing as x to the minus 1. So it's minus 1 times x0 to the minus 2, right? I could have rewritten this as minus 1/2 times x0 to the minus 1. So you just put its exponent out front and decrement it by 1. And so this is the derivative with respect to my first quadrant intersection. So let me simplify this. So x, my second quadrant intersection, the derivative of it with respect to my first quadrant intersection, is equal to minus 1, the minus 1/2 and the minus 1 become a positive when you multiply them, and so plus 1/2 over x0 squared. Now, this'll reach a maximum or minimum when it equals 0.", + "qid": "viaPc8zDcRI_1125" + }, + { + "Q": "At 14:03, I don't understand what Sal does to get the expression 4/x0(x0^4+1/2(x0^2)+1/16) under the squareroot. It seems to me that he divides the whole expression by 4, but I still can't make sense out of it...", + "A": "He s factoring out 4/(x0)^2 from the three terms. Unfortunately, he turns around the expression so that in the factored form the first and third terms are exchanged, which might be contributing to your confusion (I did a double take as well!). If you try factoring out 4/(x0)^2 from the three terms but keep them in the same order, you end up with: 4/(x0)^2 [1/16 + (1/2)(x0)^2 + (x0)^4]. You can see it s the same as what he has. Hope that helps!", + "video_name": "viaPc8zDcRI", + "timestamps": [ + 843 + ], + "3min_transcript": "So let's see if I can simplify this. We're just figuring out where the normal line and the parabola intersect. Now, what do we get here. This looks like a little hairy beast here. Let me see if I can simplify this a little bit. So let us factor out-- let me rewrite this. I can just divide everything by 1/2, so this is minus 1 over 4 x0, I just divided this by 2, plus or minus 1/2, that's just this 1/2 right there, times the square root, let me see what I can simplify out of here. So if I factor out a 4 over x0 squared, then what does This term right here will become an x to the fourth, x0 to the fourth, plus, now, what does this term become? This term becomes a 1/2 x0 squared. And just to verify this, multiply 4 times 1/2, you get 2, and then the x0 squares cancel out. So write this term times that, will equal 2, and then you have plus-- now we factored a four out of this and the x0 squared, so plus 1/16. Let me scroll over a little bit. And you can verify that this works out. If you were to multiply this out, you should get this business right here. I see the home stretch here, because this should actually factor out quite neatly. So what does this equal? parabola is equal to this. Minus 1 over 4 x0 plus or minus 1/2 times the square root of this business. And the square root, this thing right here is 4 over x0 squared. This is actually, lucky for us, a perfect square. And I won't go into details, because then the video will get too long, but I think you can recognize that this is x0 squared, plus 1/4. If you don't believe me, square this thing right here. You'll get this expression right there. And luckily enough, this is a perfect square, so we can actually take the square root of it. And so we get, the point at which they intersect, our normal line and our parabola, and this is quite a hairy problem. The points where they intersect is minus 1 over 4 x0, plus or minus 1/2 times the square root of this. The square root of this is the square root of this, which is just 2 over x0 times the square root of this, which is", + "qid": "viaPc8zDcRI_843" + }, + { + "Q": "At 0:21, you mention \"Fibonacci numbers\"; 1,1,2,5,8,13,21,34,...; what are those? And why are they called that?", + "A": "In 1175 A.D., the man who invented the Fibonacci Numbers, Fibonacci, was born. Throughout his life he encountered many circumstances where these numbers appear (like in the spirals of a pineapple or pinecone). This pattern frequently shows up in nature and is extremely fun to find out where it exactly fits in.", + "video_name": "gBxeju8dMho", + "timestamps": [ + 21 + ], + "3min_transcript": "Dear Nickelodeon, I've gotten over how SpongeBob's pants are not actually square. I can ignore most of the time that Gary's shell is not a logarithmic spiral. But what I cannot forgive is that SpongeBob's pineapple house is a mathematical impossibility. There's three easy ways to find spirals on a pineapple. There's the ones that wind up it going right, the ones that spiral up to the left, and the ones that go almost straight up-- keyword almost. If you count the number of spirals going left and the number of spirals going right, they'll be adjacent Fibonacci numbers-- 3 and 5, or 5 and 8, 8 and 13, or 13 and 21. You claim that SpongeBob Squarepants lives in a pineapple under the sea, but does he really? A true pineapple would have Fibonacci spiral, so let's take a look. Because these images of his house don't let us pick it up and turn it around to count the number of spirals going around it, it might be hard to figure out whether it's mathematically a pineapple or not. But there's a huge clue in the third spiral, the one going upwards. In this pineapple there's 8 to the right, 13 to the left. You can add those numbers together to get how many spirals are in the set spiraling In this case, 21. The three sets of spirals in any pineapple are pretty much always adjacent Fibonacci numbers. The rare mutant cases might show Lucas numbers or something, but it will always be three adjacent numbers in a series. What you'll never have is the same number of spirals both ways. Pineapples, unlike people, don't have bilateral symmetry. You'll never have that third spiral be not a spiral, but just a straight line going up a pineapple. Yet, when we look at SpongeBob's supposed pineapple under the sea, it clearly has lines of pineapple things going straight up. It clearly has bilateral symmetry. It clearly is not actually a pineapple at all, because no pineapple could possibly grow that way. Nickelodeon, you need to take a long, hard look in the mirror and think about the way you're misrepresenting the universe to your viewers. This kind of mathematical oversight is simply irresponsible. Sincerely, Vi Hart.", + "qid": "gBxeju8dMho_21" + }, + { + "Q": "At 2:04, could you multiply with a decimal? I mean, fractions and decimals are the same things, right?", + "A": "I don t understand your question properly but let say (5.5)^-1 = 1 / 5.5 let say (1/1.3)^-1 = 1.3", + "video_name": "JnpqlXN9Whw", + "timestamps": [ + 124 + ], + "3min_transcript": "", + "qid": "JnpqlXN9Whw_124" + }, + { + "Q": "Why are vectors represented as column form and not row form ? For eg. at 4:00 of this video why wasn't vector V represented as [5,0] in row form.", + "A": "It is simply a convention that most people agree upon, many authors of books write row vectors instead of column vectors. The advantage of the column vector is that it is easy to see how to do matrix multiplication. Further on in linear algebra, one learns about the transpose map and it s relation to functions that operate on vectors to give scalars (linear functionals). Then it becomes a bigger deal whether or not your vector is column or row. For now try not to worry about it too much.", + "video_name": "br7tS1t2SFE", + "timestamps": [ + 240 + ], + "3min_transcript": "And then the direction that the arrow is pointed in specifies it's direction. So this right over here could represent visually this vector. If we say that the horizontal axis is say east, or the positive horizontal direction is moving in the east, this would be west, that would be north, and then that would be south. Now, what's interesting about vectors is that we only care about the magnitude in the direction. We don't necessarily not care where we start, where we place it when we think about it visually like this. So for example, this would be the exact same vector, or be equivalent vector to this. This vector has the same length. So it has the same magnitude. It has a length of 5. And its direction is also due east. So these two vectors are equivalent. Now one thing that you might say is, well, that's fair enough. But how do we represent it with a little bit more mathematical notation? So we don't have to draw it every time. And we could start performing operations on it. want a variable to represent a vector, is usually a lowercase letter. If you're publishing a book, you can bold it. But when you're doing it in your notebook, you would typically put a little arrow on top of it. And there are several ways that you could do it. You could literally say, hey 5 miles per hour east. But that doesn't feel like you can really operate on that easily. The typical way is to specify, if you're in two dimensions, to specify two numbers that tell you how much is this vector moving in each of these dimensions? So for example, this one only moves in the horizontal dimension. And so we'll put our horizontal dimension first. So you might call this vector 5, 0. It's moving 5, positive 5 in the horizontal direction. And it's not moving at all in the vertical direction. And the notation might change. You might also see notation, and actually in the linear algebra context, it's more typical to write it as a column vector like this-- 5, 0. represents how much we're moving in the horizontal direction. And the second coordinate represents how much are we moving in the vertical direction. Now, this one isn't that interesting. You could have other vectors. You could have a vector that looks like this. Let's say it's moving 3 in the horizontal direction. And positive 4. So 1, 2, 3, 4 in the vertical direction. So it might look something like this. So this could be another vector right over here. Maybe we call this vector, vector a. And once again, I want to specify that is a vector. And you see here that if you were to break it down, in the horizontal direction, it's shifting three in the horizontal direction, and it's shifting positive four in the vertical direction.", + "qid": "br7tS1t2SFE_240" + }, + { + "Q": "At 5:20, you take the derivative of h/500. Aren't you supposed to use the quotient rule for that?", + "A": "You don t have a variable in the denominator. That 500 m is a constant. So you would treat it as ( \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0085\u00e2\u0082\u0080\u00e2\u0082\u0080 ) h", + "video_name": "_kbd6troMgA", + "timestamps": [ + 320 + ], + "3min_transcript": "is equal to the opposite side-- the opposite side is equal to h-- over the adjacent side, which we know is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between d theta dt, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of both sides of this with respect to t implicitly. So let's do that. And actually, let me move over this h over 500 a little bit. So let me move it over a little bit so I have space to show the derivative operator. So let's write it like that. And now let's take the derivative with respect to t. So d dt. I'm going to take the derivative with respect to t on the left. We're going to take the derivative with respect So what's the derivative with respect to t of tangent of theta? Well, we're just going to apply the chain rule here. It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to t, times d theta dt. Once again, this is just the derivative of the tangent, the tangent of something with respect to that something times the derivative of the something with respect to t. Derivative of tangent theta with respect to theta times the derivative theta with respect to t gives us the derivative of tangent of theta with respect to t, which is what we want when we use this type of a derivative operator. We're taking the derivative with respect to t, not just the derivative with respect to theta. So this is the left hand side. it's just going to be 1 over 500 dh dt. So 1 over 500 dh dt. We're literally saying it's just 1 over 500 times the derivative of h with respect to t. But now we have our relationship. We have the relationship that we actually care about. We have a relationship between the rate at which the height is changing with respect to time and the rate at which the angle is changing with respect to time and our angle at any moment. So we can just take these values up here, throw it in here, and then solve for the unknown. So let's do that. Let's do that right over here. So we get secant squared of theta. So we get secant squared. Right now our theta is pi over 4. Secant squared of pi over 4. Let me write those colors in to show you that I'm putting these values in. Secant squared of pi over 4. Secant times d theta dt.", + "qid": "_kbd6troMgA_320" + }, + { + "Q": "I've heard slope-intercept form is y=mx+b. At 6:05, how would that fit in?", + "A": "If you distribute across the parenthesis, and then add/subtract the term on the y side (from both), you should get an answer in slope-intercept form.", + "video_name": "K_OI9LA54AA", + "timestamps": [ + 365 + ], + "3min_transcript": "Let's say that someone tells you that I'm dealing with some line where the slope is equal to 2, and let's say it goes through the point negative 7, 5. So very quickly, you could use this information and your knowledge of point-slope form to write this in this form. You would just say, well, an equation that contains this point and has this slope would be y minus b, which is 5-- y minus the y-coordinate of the point that this line contains-- is equal to my slope times x minus the x-coordinate that this line contains. So x minus negative 7. And just like that, we have written an equation that this point right over here. And if we don't like the x minus negative 7 right over here, we could obviously rewrite that as x plus 7. But this is kind of the purest point-slope form. If you want to simplify it a little bit, you could write it as y minus 5 is equal to 2 times x plus 7. And if you want to see that this is just one way of expressing the equation of this line-- there are many others, and the one that we're most familiar with is y-intercept form-- this can easily be converted to y-intercept form. To do that, we just have to distribute this 2. So we get y minus 5 is equal to 2 times x plus 2 times 7, so that's equal to 14. And then we can get rid of this negative 5 on the left by adding 5 to both sides of this equation. And then we are left with, on the left-hand side, y and, on the right-hand side, 2x plus 19. So this right over here is slope-intercept form. You have your slope and your y-intercept. So this is slope-intercept form.", + "qid": "K_OI9LA54AA_365" + }, + { + "Q": "At 1:00 in the video, what do the triangles mean when he is talking about what the slope equals?", + "A": "The triangles are supposed to mean The Change in , so it is The Change in Y/ The Change in X , also known as slope, or known as in Slope-intercept form, m.", + "video_name": "K_OI9LA54AA", + "timestamps": [ + 60 + ], + "3min_transcript": "So what I've drawn here in yellow is a line. And let's say we know two things about this line. We know that it has a slope of m, and we know that the point a, b is on this line. And so the question that we're going to try to answer is, can we easily come up with an equation for this line using this information? Well, let's try it out. So any point on this line, or any x, y on this line, would have to satisfy the condition that the slope between that point-- so let's say that this is some point x, y. It's an arbitrary point on the line-- the fact that it's on the line tells us that the slope between a, b and x, y must be equal to m. So let's use that knowledge to actually construct an equation. So what is the slope between a, b and x, y? Well, our change in y-- remember slope is just change in y over change in x. Let me write that. Slope is equal to change in y over change in x. This little triangle character, that's the Greek letter Delta, Our change in y-- well let's see. If we start at y is equal to b, and if we end up at y equals this arbitrary y right over here, this change in y right over here is going to be y minus b. Let me write it in those same colors. So this is going to be y minus my little orange b. And that's going to be over our change in x. And the exact same logic-- we start at x equals a. We finish at x equals this arbitrary x, whatever x we happen to be at. So that change in x is going to be that ending point minus our starting point-- minus a. And we know this is the slope between these two points. That's the slope between any two points on this line. And that's going to be equal to m. So this is going to be equal to m. And so what we've already done here is actually create an equation that describes this line. It might not be in any form that you're used to seeing, any x, y that satisfies this equation right over here will be on the line because any x, y that satisfies this, the slope between that x, y and this point right over here, between the point a, b, is going to be equal to m. So let's actually now convert this into forms that we might recognize more easily. So let me paste that. So to simplify this expression a little bit, or at least to get rid of the x minus a in the denominator, let's multiply both sides by x minus a. So if we multiply both sides by x minus a-- so x minus a on the left-hand side and x minus a on the right. Let me put some parentheses around it. So we're going to multiply both sides by x minus a. The whole point of that is you have x minus a divided by x", + "qid": "K_OI9LA54AA_60" + }, + { + "Q": "At 1:55, Sal says \"their greatest common factor is 3\" what does greatest common factor even mean?", + "A": "The Greatest Common Factor means when you take two numbers and find a factor that both these numbers have in common. For example: What is the greatest common factor of 6 and 9? Well, the first step is to list their factors: The factors for 6 are: 1, 2, 3, and 6. The factors for 9 are: 1, 3, 3, 9. Now what number is the same in both these factor lists? 3! So the Greatest Common Factor will be 3.", + "video_name": "Bt60JVZRVCI", + "timestamps": [ + 115 + ], + "3min_transcript": "Let's think about what fraction of this grid is actually shaded in pink. So the first thing we want to think about is how many equal sections do we have here? Well, this is a 1, 2, 3, 4, 5 by 1, 2, 3 grid. So there's 15 sections here. You could also count it-- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. So there are 15 equal sections here. And how many of those equal sections are actually shaded in this kind of pinkish color? Well, We have 1, 2, 3, 4, 5, 6. So it's 6/15 is shaded in. But I want to simplify this more. I have a feeling that there's some equivalent fractions that represent the exact same thing as 6/15. And to get a sense of that, let me redraw this a little bit, where I still shade in six of these rectangles, but I'll shade them a little bit in one chunk. So let me throw in another grid right over here, and let me attempt to shade in the rectangles So that is 1-- 1 rectangle. I'll even make my thing even bigger. All right, 1 rectangle, 2 rectangles, 3 rectangles-- halfway there-- 4 rectangles, 5 rectangles shaded in and now 6 rectangles shaded in. So this right over here, what I just did, this is still 6 rectangles of the 15 rectangles So this is still 6/15. These are representing the same thing. But how can I simplify this even more? Well, when you look at it numerically, you see that both 6 and 15 are divisible by 3. In fact, their greatest common factor is 3. So what happens if we divide the numerator and denominator by 3? we're not going to be changing the value of the fraction. So let's divide the numerator by 3 and divide the denominator by 3. And what do we get? We get 2 over 5. Now how does this make sense in the context of this diagram right here? Well, we started off with 6 shaded in. You divide by 3, you have 2 shaded in. So you're essentially saying, hey, let's group these into sections of 3. So let's say that this right over here is one section of 3. This is one section of 3 right over here. So that's one section of 3. And then this is another section of 3 right over here. And so you have two sections of 3. And actually let me color it in a little bit better. So you have two sections of 3. And if you were to combine them, it looks just like this.", + "qid": "Bt60JVZRVCI_115" + }, + { + "Q": "I'm confused at 2:10. Sal says that the sequence converges but I learned that when we calculate limits, if you have infinity over infinity, its called a \"indetermination\". Though, what Sal said makes perfectly sense...", + "A": "Recall when we want to find the horizontal asymptote, we take the limit as x approaches infinity. The method was to multiply the numerator and denominator by 1 over the highest term of x on the denominator. Similarly, we find the limit as n approaches infinity to find out if it diverges or converges. So if we multiply by 1/n\u00c2\u00b2 for both the top and bottom, it will converge to 1.", + "video_name": "muqyereWEh4", + "timestamps": [ + 130 + ], + "3min_transcript": "So we've explicitly defined four different sequences here. And what I want you to think about is whether these sequences converge or diverge. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. And diverge means that it's not approaching some value. So let's look at this. And I encourage you to pause this video and try this on your own before I'm about to explain it. So let's look at this first sequence right over here. So the numerator n plus 8 times n plus 1, the denominator n times n minus 10. So one way to think about what's happening as n gets larger and larger is look at the degree of the numerator and the degree of the denominator. And we care about the degree because we want to see, look, is the numerator growing faster than the denominator? In which case this thing is going to go to infinity and this thing's going to diverge. Or is maybe the denominator growing faster, in which case this might converge to 0? Or maybe they're growing at the same level, and maybe it'll converge to a different number. So let's multiply out the numerator and the denominator So n times n is n squared. n times 1 is 1n, plus 8n is 9n. And then 8 times 1 is 8. So the numerator is n squared plus 9n plus 8. The denominator is n squared minus 10n. And one way to think about it is n gets really, really, really, really, really large, what dominates in the numerator-- this term is going to represent most of the value. And this term is going to represent most of the value, as well. These other terms aren't going to grow. Obviously, this 8 doesn't grow at all. But the n terms aren't going to grow anywhere near as fast as the n squared terms, especially for large n's. So for very, very large n's, this is really going to be approaching n squared over n squared, or 1. So it's reasonable to say that this converges. So this one converges. And once again, I'm not vigorously proving it here. Or I should say I'm not rigorously proving it over here. in the numerator and the denominator. So now let's look at this one right over here. So here in the numerator I have e to the n power. And here I have e times n. So this grows much faster. I mean, this is e to the n power. Imagine if when you have this as 100, e to the 100th power is a ginormous number. e times 100-- that's just 100e. Grows much faster than this right over here. So this thing is just going to balloon. This is going to go to infinity. So we could say this diverges. Now let's look at this one right over here. Well, we have a higher degree term. We have a higher degree in the numerator than we have in the denominator. n squared, obviously, is going to grow much faster than n. So for the same reason as the b sub n sequence, this thing is going to diverge. The numerator is going to grow much faster", + "qid": "muqyereWEh4_130" + }, + { + "Q": "At 00:34, couldn't I just do 64+31+50+x=180?", + "A": "You sure can! There are many ways to figure out each of the angles.", + "video_name": "hmj3_zbz2eg", + "timestamps": [ + 34 + ], + "3min_transcript": "What I want to do now is just a series of problems that really make sure that we know what we're doing with parallel lines and triangles and all the rest. And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here-- so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it just knowing what you know about the sums of the measures of the angles inside of a triangle, and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself because I'm about to give you the solution. So the first thing you might say-- and this is a general way to think about a lot of these problems where they give you some angles and you have to figure out some other angles based on the sum of angles and a triangle equaling 180, or this one doesn't have parallel lines on it. and supplementary lines and complementary lines-- is to just fill in everything that you can figure out, and one way or another, you probably would be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given 2 of the angles. And if you have 2 of the angles in a triangle, you can always figure out the third angle because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say, x plus, what is this, 114. X plus 114 is equal to 180 degrees. We could subtract 114 from both sides of this equation, and we get x is equal to 180 minus 114. So 80 minus 14. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. Let me write x is equal to 66 degrees. Well if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here, y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides, and we get y is equal to-- these cancel out-- 180 minus 66 is 114.", + "qid": "hmj3_zbz2eg_34" + }, + { + "Q": "At 3:00 they wrote two to the sixty third power. Couldn't you also write it as 2^63?", + "A": "Yes, the two symbols mean the same thing", + "video_name": "UCCNoXqCGZQ", + "timestamps": [ + 180 + ], + "3min_transcript": "8, 2's multiplied together. 9, 2's. 10, 11, 12, 13. So all of this stuff multiplied together. 8,192 grains of rice is what we should see right over here. Voiceover:And you know, I had fun last night and I was up late, but there you go. Voiceover:Did you really count out 8,192 grains of rice? Voiceover:More or less. Voiceover:Okay. Let's just say you did. Voiceover:What if we just went, you know, 4 steps ahead. How much rice would be here? Voiceover:4 steps ahead, so we're going to multiple by 2, then multiple by 2 again, then multiply by 2 again, the multiply by 2 again. So it's this number times ... Let's see, 2 times 2 is 4. Times 2 is 8, times 2 is 16. So it's going to get us like 120, like 130,000 or around there. Voiceover:131,672. Voiceover:You had a lot of time last night. We're not even halfway across the board yet. Voiceover:We're not. Voiceover:This is a lot of ... You could throw a party. Voiceover:What about the last square? This is 63 steps. Voiceover:We're going to take 2 times 2 and we're going to do 63 of those. So this is going to be a huge number. And actually, it would be neat if there was a notation for that. Voiceover:I didn't count this one out but it is the size of Mount Everest, the pile of rice. And it would feed 485 trillion people. I mean, you know, this was a little bit of a pain for me to write all of these 2's. Voiceover:So was this. Voiceover:If I were the mathematical community I would want some type of notation. Voiceover:You kind of got on it here. I like this dot, dot, dot and the 63. This I understand this. Voiceover:Yeah, you could understand this but this is still a little bit ... This is a little bit too much. What if, instead, we just wrote ... Voiceover:Mathematicians love being efficient, right? They're lazy. Voiceover:Yeah, they have things to do. They have to go home and count grains of rice. (laughter) Voiceover:Yeah. So that is, take 63, 2's and multiply them Voiceover:This is the first square on our board. We have 1 grain of rice. And when we double it we have 2 grains of rice. Voiceover:Yup. Voiceover:And we double it again we have 4. I'm thinking this is similar to what we were doing, it's just represented differently. Voiceover:Yeah, well, I mean, this one, the one you were making, right, every time you were kind of adding these popsicle sticks, you're kind of branching out. 1 popsicle stick now becomes 2 popsicles sticks. Then you keep doing that. 1 popsicle stick becomes 2 but now you have 2 of them. So here you have 1, now you have 1 times 2. Now each of these 2 branch into 2, so now you have 2 times 2, or you have 4 popsicle sticks. Every stage, every branch, you're multiplying by 2 again. Voiceover:I basically just continue splitting just like a tree does. Voiceover:Yup. Voiceover:Now I can really see what 2 to the power of 3 looks like. Voiceover:And that's what we have here. 1 times 2 times 2 times 2, which is 8. This is 2 to the third power.", + "qid": "UCCNoXqCGZQ_180" + }, + { + "Q": "5:61 Is there something wrong here?", + "A": "if you are writing in decimals it is a decimal dot (looks like a period) not a colon. If it is time, the time is impossible because there can be only 60 minutes per hour actually 59 before the next hour. Good Question! Keep it Up!", + "video_name": "AGFO-ROxH_I", + "timestamps": [ + 361 + ], + "3min_transcript": "So 3 goes into 10,560. It doesn't go into 1. It goes into 10 three times. 3 times 3 is 9. And we subtract. We get 1. Bring down this 5. It becomes a 15. 3 goes into 15 five times. 5 times 3 is 15. We have no remainder, or 0. You bring down the 6. 3 goes into 6 two times. Let me scroll down a little bit. 2 times 3 is 6. Subtract. No remainder. Bring down this last 0. 3 goes into 0 zero times. 0 times 3 is 0. And we have no remainder. So 2 miles is the equivalent to 3,520 yards. That's the total distance he has to travel. Now we want to figure out how many laps there are. We want this in terms of laps, not in terms of yards. So we want the yards to cancel out. And we want laps in the numerator, right? Because when you multiply, the yards will cancel out, and we'll just be left with laps. Now, how many laps are there per yard or yards per lap? Well, they say the distance around the field is 300 yards. So we have 300 yards for every 1 lap. So now, multiply this right here. The yards will cancel out, and we will get 3,520. Let me do that in a different color. We will get 3,520, that right there, times 1/300. When you multiply it times 1, it just becomes 3,520 divided by 300. And in terms of the units, the yards canceled out. We're just left with the laps. So 3,520 divided by 300. Well, we can eyeball this right here. What is 11 times 300? Let's just approximate this right here. So if we did 11 times 300, what is that going to be equal to? Well, 11 times 3 is 33, and then we have two zeroes here. So this will be 3,300. So it's a little bit smaller than that. If we have 12 times 300, what is that going to be? 12 times 3 is 36, and then we have these two zeroes, so it's equal to 3,600. So this is going to be 11 point something. It's larger than 11, right? 3,520 is larger than 3,300. So when you divide by 300 you're going to get something larger than 11. But this number right here is smaller than 3,600 so when you divide it by 300, you're going to get something a little bit smaller than 12.", + "qid": "AGFO-ROxH_I_361" + }, + { + "Q": "I watched the videos on the binomial theorem but they didn't show what he is doing here... I can't follow his thinking process at around 2:00 when he keeps expanding the binomial. Any help for me? I wouldn't be able to write out the whole thing by myself is what I mean.", + "A": "The expansion for (a + b)^n always starts with a^n and always ends with b^n. The stuff in the middle comes from the binomial expansion. Check it out in the precalc section! It is most excellent. :)", + "video_name": "dZnc3PtNaN4", + "timestamps": [ + 120 + ], + "3min_transcript": "", + "qid": "dZnc3PtNaN4_120" + }, + { + "Q": "Since C can be negative (when pi/2 < theta < pi) how do we know the value in the radical at 4:45 will remain positive?\n\nEdit: Nevermind, I see my error. C can never be less than -1 on the unit circle, so that radical will never be negative.", + "A": "you made a small mistake in your edit: if C<-1, the radical becomes more positive, as the formula has 1-C, not 1+C in it, so C would have to become larger than 1 for the radical to become negative.", + "video_name": "yV4Xa8Xtmrc", + "timestamps": [ + 285 + ], + "3min_transcript": "this yellow sine squared theta, and all of this is equal to C or we could get that C is equal to one minus two sine squared theta. And what's useful about this is we just have to solve for sine of theta. So let's see, I could multiply both sides by a negative just so I can switch the order over here. So I could write this as negative C is equal to two sine squared theta minus one and just multiply both sides by a negative and then let's see I could add one to both sides, if I add one to both sides, and I'll go over here, if I add one to both sides, I could divide both sides by two, and then so I get sine squared theta is equal to one minus C over two, or I could write that sine of theta is equal to the plus or minus square root of one minus C over two. So that leads to a question, Is it both? Is it the plus and minus square root? Or is it just one of those? And I encourage you to pause the video again in case you haven't already figured it out, and look at the information here, and think about whether they give us the information of whether we should be looking at the positive or negative sine. Well they tell us that theta is between zero and pi. So if I were to draw a unit circle here, between zero and pi radians. and pi is going all the way over here. So this angle places its terminal ray either in the first or second quadrants. So, it could be an angle like this, it could be an angle like this, it cannot be an angle like this, and we know that the sine of an angle is the Y coordinate, and so we know that for the first of second quadrant the Y coordinate is going to be non-negative. So we would want to take the positive square root, so we would get sine of theta, is equal to the principal root, or we could even think of it as the positive square root of one minus C over two. So let's go back to our... Make sure we can check our answers. So sine of theta is equal to the square root of one minus capital C, all of that over two,", + "qid": "yV4Xa8Xtmrc_285" + }, + { + "Q": "At around 7:53, I noticed Sal wrote XY/AB = k = BC/YZ. Shouldn't the \"BC/YZ\" be \"YZ/BC\" instead? As if XY/AB = k, then BC/YZ must equal to 1/k.", + "A": "Yes, nice catch! Sal only messes up once in a while. Just submit that in the Report a mistake In the video , and just say, at 7:53, Sal says XY/AB = k = BC/YZ, but should say BC/YZ be YZ/BC, if you are really worried about it that much!", + "video_name": "7bO0TmJ6Ba4", + "timestamps": [ + 473 + ], + "3min_transcript": "Let me draw it like this. Actually, I want to leave this here so we can have our list. So let's draw another triangle ABC. So this is A, B, and C. And let's say that we know that this side, when we go to another triangle, we know that XY is AB multiplied by some constant. So I can write it over here. XY is equal to some constant times AB. Actually, let me make XY bigger, so actually, it That constant could be less than 1 in which case it would be a smaller value. But let me just do it that way. So let me just make XY look a little bit bigger. So let's say that this is X and that is Y. So let's say that we know that XY over AB Or if you multiply both sides by AB, you would get XY is some scaled up version of AB. So maybe AB is 5, XY is 10, then our constant would be 2. We scaled it up by a factor of 2. And let's say we also know that angle ABC is congruent to angle XYZ. I'll add another point over here. So let me draw another side right over here. So this is Z. So let's say we also know that angle ABC is congruent to XYZ, and let's say we know that the ratio between BC and YZ is also this constant. The ratio between BC and YZ is also equal to the same constant. So an example where this 5 and 10, maybe this is 3 and 6. The constant we're kind of doubling So is this triangle XYZ going to be similar? Well, if you think about it, if XY is the same multiple of AB as YZ is a multiple of BC, and the angle in between is congruent, there's only one triangle we can set up over here. We're only constrained to one triangle right over here, and so we're completely constraining the length of this side, and the length of this side is going to have to be that same scale as that over there. And so we call that side-angle-side similarity. So once again, we saw SSS and SAS in our congruence postulates, but we're saying something very different here. We're saying that in SAS, if the ratio between corresponding sides of the true triangle are the same, so AB and XY of one corresponding side and then", + "qid": "7bO0TmJ6Ba4_473" + }, + { + "Q": "So at 9:00, if you have log base 2 (32/sqrt 8), you're NOT supposed to simplify (32/sqrt 8) into (4*sqrt 8)?", + "A": "Yes, you could do that and still get to the same answer, but sal just skipped it", + "video_name": "TMmxKZaCqe0", + "timestamps": [ + 540 + ], + "3min_transcript": "Just so you know logarithms are useful. Let's do some examples. Let's just let's just rewrite a bunch of things in simpler forms. So if I wanted to rewrite the log base 2 of the square root of-- let me think of something. Of 32 divided by the cube-- no, I'll just take the square root. Divided by the square root of 8. How can I rewrite this so it's reasonably not messy? Well let's think about this. This is the same thing, this is equal to-- I don't know I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right? Minus the logarithm base 2 of the square root of 8. Right? Let's see. Well here once again we have a square root here, so we could say this is equal to 1/2 times log base 2 of 32. Minus this 8 to the 1/2, which is the same thing is 1/2 log base 2 of 8. We learned that property in the beginning of this presentation. And then if we want, we can distribute this original 1/2. This equals 1/2 log base 2 of 32 minus 1/4-- because we have to distribute that 1/2-- minus 1/4 log base 2 of 8. This is 5/2 minus, this is 3. 3 times 1/4 minus 3/4. Or 10/4 minus 3/4 is equal to 7/4. I probably made some arithmetic errors, but you get the point.", + "qid": "TMmxKZaCqe0_540" + }, + { + "Q": "I don't get why he corrected from -1/2 to -1/4 in the very end (9:48)? Can someone please explain?", + "A": "He did not distribute the 1/2 to the -1/2 log\u00e2\u0082\u00828 term (he wrote -1/2 log\u00e2\u0082\u00828, which was unchanged from the original term), so he corrected it to -1/4 log\u00e2\u0082\u00828. Hope this helps!", + "video_name": "TMmxKZaCqe0", + "timestamps": [ + 588 + ], + "3min_transcript": "I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right? Minus the logarithm base 2 of the square root of 8. Right? Let's see. Well here once again we have a square root here, so we could say this is equal to 1/2 times log base 2 of 32. Minus this 8 to the 1/2, which is the same thing is 1/2 log base 2 of 8. We learned that property in the beginning of this presentation. And then if we want, we can distribute this original 1/2. This equals 1/2 log base 2 of 32 minus 1/4-- because we have to distribute that 1/2-- minus 1/4 log base 2 of 8. This is 5/2 minus, this is 3. 3 times 1/4 minus 3/4. Or 10/4 minus 3/4 is equal to 7/4. I probably made some arithmetic errors, but you get the point.", + "qid": "TMmxKZaCqe0_588" + }, + { + "Q": "At 8:25-8:38 why did he remove the exponent 1/2 and put it on the left so that it turns into 1/2_log(32/sqrt(8))? He mentioned a property but which one is it?", + "A": "Because, one of the property of logarithm is: log of a^b = b log a (log of a power b equals b log of a) in this video, log of some thing power to 1/2, then equals to 1/2 log of some thing. Hope it helps.", + "video_name": "TMmxKZaCqe0", + "timestamps": [ + 505, + 518 + ], + "3min_transcript": "We could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just type in 357 in your calculator and press the log button and you're going to get bada bada bam. Then you can clear it, or if you know how to use the parentheses on your calculator, you could do that. But then you type 17 on your calculator, press the log button, go to bada bada bam. And then you just divide them, and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth. This one, to me it's the most useful, but it doesn't completely-- it does fall out of, obviously, the exponent properties. But it's hard for me to describe the intuition simply, so you probably want to watch the proof on it, if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. Just so you know logarithms are useful. Let's do some examples. Let's just let's just rewrite a bunch of things in simpler forms. So if I wanted to rewrite the log base 2 of the square root of-- let me think of something. Of 32 divided by the cube-- no, I'll just take the square root. Divided by the square root of 8. How can I rewrite this so it's reasonably not messy? Well let's think about this. This is the same thing, this is equal to-- I don't know I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1/2 power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1/2 times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing. And we learned that in the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm-- let's keep the 1/2 out. That's going to equal, parentheses, logarithm-- oh, I forgot my base. Logarithm base 2 of 32 minus, right?", + "qid": "TMmxKZaCqe0_505_518" + }, + { + "Q": "10:20 What is the purpose of functions in a practical real life scenario?", + "A": "Functions are used all the time. -- Computer programs are functions -- the cash register at a store uses functions to determine what you owe -- when you calculate the tip on a bill in a restaurant, you are using a function. Those a just a few. There are many more.", + "video_name": "5fkh01mClLU", + "timestamps": [ + 620 + ], + "3min_transcript": "plus 5/2, plus 5/2. I like to change my notation just so you get familiar with both. So the equation becomes 5/2 is equal to-- that's a 0-- is equal to b. b is 5/2. So the equation of our line is y is equal to 5/6 x plus b, which we just figured out is 5/2, plus 5/2. We are done. Let's do another one. We have a graph here. Let's figure out the equation of this graph. This is actually, on some level, a little bit easier. What's the slope? Slope is change in y over change it x. So let's see what happens. When we move in x, when our change in x is 1, so that is our change in x. So change in x is 1. I'm just deciding to change my x by 1, increment by 1. It looks like y changes exactly by 4. It looks like my delta y, my change in y, is equal to 4 when my delta x is equal to 1. So change in y over change in x, change in y is 4 when change in x is 1. So the slope is equal to 4. Now what's its y-intercept? Well here we can just look at the graph. It looks like it intersects y-axis at y is equal to negative 6, or at the point 0, negative 6. So we know that b is equal to negative 6. So we know the equation of the line. The equation of the line is y is equal to the slope times x plus the y-intercept. I should write that. equation of our line. Let's do one more of these. So they tell us that f of 1.5 is negative 3, f of negative 1 is 2. What is that? Well, all this is just a fancy way of telling you that the point when x is 1.5, when you put 1.5 into the function, the function evaluates as negative 3. So this tells us that the coordinate 1.5, negative 3 is on the line. Then this tells us that the point when x is negative 1, f of x is equal to 2. This is just a fancy way of saying that both of these two points are on the line, nothing unusual. I think the point of this problem is to get you familiar with function notation, for you to not get intimidated if you see something like this. If you evaluate the function at 1.5, you get negative 3.", + "qid": "5fkh01mClLU_620" + }, + { + "Q": "At 3:00, why did Sal put the answer to x^2 divided by x above the 3x? Why not above the x^2?", + "A": "I agree with you at 3:00. It would be better and less confused, if Sal put the x right on top of x^2 , even though it is not a require. In fact, x can be put any where within the line.", + "video_name": "FXgV9ySNusc", + "timestamps": [ + 180 + ], + "3min_transcript": "for more complicated problems. So you could have also written this as 2 goes into 2x plus 4 how many times? And you would perform this the same way you would do traditional long division. You'd say 2-- you always start with the highest degree term. 2 goes into the highest degree term. You would ignore the 4. 2 goes into 2x how many times? Well, it goes into 2x x times and you put the x in the x place. x times 2 is 2x. And just like traditional long division, you now subtract. So 2x plus 4 minus 2x is what? It's 4, right? And then 2 goes into 4 how many times? It goes into it two times, a positive two times. Put that in the constants place. 2 times 2 is 4. You subtract, remainder 0. So this might seem overkill for what was probably a do it in a few steps. We're now going to see that this is a very generalizable process. You can do this really for any degree polynomial dividing into any other degree polynomial. Let me show you what I'm talking about. So let's say we wanted to divide x plus 1 into x squared plus 3x plus 6. So what do we do here? So you look at the highest degree term here, which is an x, and you look at the highest degree term here, which is an x squared. So you can ignore everything else. And that really simplifies the process. You say x goes into x squared how many times? Well, x squared divided by x is just x, right? x goes into x squared x times. You put it in the x place. This is the x place right here or the x to So x times x plus 1 is what? x times x is x squared. x times 1 is x, so it's x squared plus x. And just like we did over here, we now subtract. And what do we get? x squared plus 3x plus 6 minus x squared-- let me be very careful-- this is minus x squared plus x. I want to make sure that negative sign only-- it applies to this whole thing. So x squared minus x squared, those cancel out. 3x, this is going to be a minus x. Let me put that sign there. So this is minus x squared minus x, just to be clear. We're subtracting the whole thing. 3x minus x is 2x. And then you bring down the 6, or 6 minus 0 is nothing. So 2x plus 6. Now, you look at the highest degree term, an x and a 2x. How many times does x go into 2x?", + "qid": "FXgV9ySNusc_180" + }, + { + "Q": "At around 8:00, you need to restrict x=-4 to be equal to the first expression. So if your long division works out with no remainder, you have to make restrictions for your original denominator, or am I missing something?", + "A": "You are correct. The original expression is undefined at x = -4, so the correct answer would be: (x\u00c2\u00b2 +5x + 4) / (x + 4) = x + 1; x \u00e2\u0089\u00a0 -4", + "video_name": "FXgV9ySNusc", + "timestamps": [ + 480 + ], + "3min_transcript": "Let's do another one of these. They're kind of fun. So let's say that we have-- we want to simplify x squared plus 5x plus 4 over x plus 4. So once again, we can do our algebraic long division. We can divide x plus 4 into x squared plus 5x plus 4. And once again, same exact process. Look at the highest degree terms in both of them. x goes into x squared how many times? It goes into it x times. Put it in the x place. This is our x place right here. X times x is x squared. x times 4 is 4x. So let me just put a negative sign there. And then these cancel out. 5x minus 4x is x. 4 minus 0 is plus 4. x plus 4, and then you could even see this coming. You could say x plus 4 goes into x plus 4 obviously one time, or if you were not looking at the constant terms, you would completely just say, well, x goes into x how many times? Well, one time. Plus 1. 1 times x is x. 1 times 4 is 4. We're going to subtract them from up here, so it cancels out, so we have no remainder. So this right here simplifies to-- this is equal to x plus 1. And there's other ways you could have done this. We could have tried to factor this numerator. x squared plus 5x plus 4 over x plus 4. This is the same thing as what? We could have factored this numerator as x plus 4 4 times 1 is 4. 4 plus 1 is 5, all of that over x plus 4. That cancels out and you're left just with x plus 1. Either way would have worked, but the algebraic long division will always work, even if you can't cancel out factors like that, even if you did have a remainder. In this situation, you didn't. So this was equal to x plus 1. Let's do another one of these just to make sure that you really-- because this is actually a very, very useful skill to have in your toolkit. So let's say we have x squared-- let me Let's say we had 2x squared-- I could really make these numbers up on the fly. 2x squared minus 20x plus 12 divided by-- actually, let's", + "qid": "FXgV9ySNusc_480" + }, + { + "Q": "about 04:00, does this mean that every sum of normal distributed, independent, and squared variables will have the chi-square distribuition?", + "A": "Indeed it does. This is mainly because the sum of normally-distributed, independent, and squared random variables is the very definition of the chi-square distribution.", + "video_name": "dXB3cUGnaxQ", + "timestamps": [ + 240 + ], + "3min_transcript": "here is going to be an example of the chi-square distribution. Actually what we're going to see in this video is that the chi-square, or the chi-squared distribution is actually a set of distributions depending on how many sums you have. Right now, we only have one random variable that we're squaring. So this is just one of the examples. And we'll talk more about them in a second. So this right here, this we could write that Q is a chi-squared distributed random variable. Or that we could use this notation right here. Q is-- we could write it like this. So this isn't an X anymore. This is the Greek letter chi, although it looks a lot like a curvy X. So it's a member of chi-squared. And since we're only taking one sum over here-- we're only taking the sum of one independent, normally distributed, standard or normally distributed variable, we say that this only has 1 degree of freedom. So this right here is our degree of freedom. We have 1 degree of freedom right over there. So let's call this Q1. Let's say I have another random variable. Let's call this Q-- let me do it in a different color. Let me do Q2 in blue. Let's say I have another random variable, Q2, that is defined as-- let's say I have one independent, standard, normally distributed variable. I'll call that X1. And I square it. And then I have another independent, standard, normally distributed variable, X2. And I square it. So you could imagine both of these guys have distributions like this. And they're independent. So get to sample Q2, you essentially sample X1 from this distribution, square that value, sample X2 and then add the two. And you're going to get Q2. This over here-- here we would write-- so this is Q1. Q2 here, Q2 we would write is a chi-squared, distributed random variable with 2 degrees of freedom. Right here. 2 degrees of freedom. And just to visualize kind of the set of chi-squared distributions, let's look at this over here. So this, I got this off of Wikipedia. This shows us some of the probability density functions for some of the chi-square distributions. This first one over here, for k of equal to 1, that's the degrees of freedom. So this is essentially our Q1. This is our probability density function for Q1. And notice it really spikes close to 0. And that makes sense. Because if you are sampling just once from this standard normal distribution,", + "qid": "dXB3cUGnaxQ_240" + }, + { + "Q": "So at 2:05, Sal put -2y after the -6x^2. But when I was doing the problem myself, I put 8xy first and then -2y. Is the way I ordered the expression wrong and if it is...why? This was my answer:\n( 6x^2y - 6x^2 + 8xy - 2y +4 ) This was Sal's answer:\n( 6x^2y - 6x^2 - 2y + 8xy +4 )", + "A": "Hey Tushar Gaddi!, it does not matter in what order you put it in as long as you dont change any of the negative( - ) or posiive( + ) signs. the way you wrote it was just fine. for problems like this this is why the Order Of Operations comes into play. no matter what order you put it in you will get the same answer. Hope This Helps! Good Luck and Have Fun! :)", + "video_name": "jroamh6SIo0", + "timestamps": [ + 125 + ], + "3min_transcript": "So let's get some practice simplifying polynomials, especially in the case where we have more than one variable over here. So I have 4x squared y minus 3x squared minus 2y. So that entire expression plus the entire expression 8xy minus 3x squared plus 2x squared y plus 4. So the first thing that jumps out at me is that I'm just adding this expression to this expression. So to a large degree, these parentheses don't matter. So I can just rewrite it as 4x squared y minus 3x squared minus 2y plus 8xy minus 3x squared plus 2x squared y plus 4. Now we can try to group similar terms or like terms. So let's think about what we have over here. So this first term right here is a 4x squared y. So can I add this to any of the other terms here? Do we have any other x squared y terms? is another x squared y term. If I have 4 of something-- in this case, I have 4x squared y's and I add 2x squared y's to it, how many x squared y's do I now have? Well, 4 plus 2-- I now have 6x squared y's. Now let's move on to this term. So I have negative 3x squareds. Do I have any more x squareds in this expression right over here? Well, sure, I have another negative 3x squared. So if I have negative 3 of something and then I have another negative 3, I end up with negative 6 So it's negative 6x squared. Now let's think about this negative 2y term. Are there any other y's over here? Well, it doesn't look like there are. This is an 8xy. This is a 4. There's no just y's. So I'll just rewrite it, negative 2y. And then 8xy-- well, once again, it doesn't seem like that can be added to anything else. So let's just write that over again. And then finally, we just have the constant term plus 4. And it pretty much looks like we're done. We have simplified this as much as we can.", + "qid": "jroamh6SIo0_125" + }, + { + "Q": "At 4:15 of the movie above suddenly you drop the unit vectors(i,j,k) from the result of bxc. Is it ok to do that? If it is ok, then I want to know why is it possible to do that.", + "A": "i, j, and k are (1, 0, 0), (0. 1, 0), and (0, 0, 1). a = (a1, a2, a3) = a1i + a2j + a3k (the resultant of the components of a).", + "video_name": "b7JTVLc_aMk", + "timestamps": [ + 255 + ], + "3min_transcript": "of three-by-threes. And we'll put b's x term, b's y coefficient, and b's z component. And then you do the same thing for the c, cx, cy, cz. And then this is going to be equal to-- so first you have the i component. So it's going to be the i component times b. So you ignore this column and this row. So bycz minus bzcy. So I'm just ignoring all of this. And I'm looking at this two-by-two over here, minus bzcy. And then we want to subtract the j component. Remember, we alternate signs when we take our determinant. Subtract that. And then we take out that column and that row, so it's going to be bxcz-- this is a little monotonous, bxcz minus bzcx. And then finally, plus the k component. OK, we're going to have bx times cy minus bycx. We just did the dot product, and now we want to take the-- oh, sorry, we just did the cross product. I don't want to get you confused. We just took the cross product of b and c. And now we need take the cross product of that with a, or the cross product of a with this thing right over here. Instead of rewriting the vector, let me just set up another matrix here. So let me write my i j k up here. And then let me write a's components. And then let's clean this up a little bit. We're just looking at-- no, I want to do that in black. Let's do this in black, so that we can kind of erase that. Now this is a minus j times that. So what I'm going to do is I'm going to get rid of the minus and the j, but I am going to rewrite this with the signs swapped. So if you swap the signs, it's actually bzcx minus bxcz. So let me delete everything else. So I just took the negative and I multiplied it by this. I hope I'm not making any careless mistakes here, so let me just check and make my brush size little bit bigger, so I can erase that a little more efficiently. And then we also want to get rid of that right over there. Now let me get my brush size back down to normal size.", + "qid": "b7JTVLc_aMk_255" + }, + { + "Q": "Doesn't he mean cross product at 2:00?", + "A": "Yes, he meant cross product.", + "video_name": "b7JTVLc_aMk", + "timestamps": [ + 120 + ], + "3min_transcript": "What I want to do with this video is cover something called the triple product expansion-- or Lagrange's formula, sometimes. And it's really just a simplification of the cross product of three vectors, so if I take the cross product of a, and then b cross c. And what we're going to do is, we can express this really as sum and differences of dot products. Well, not just dot products-- dot products scaling different vectors. You're going to see what I mean. But it simplifies this expression a good bit, because cross products are hard to take. They're computationally intensive and, at least in my mind, they're confusing. Now this isn't something you have to know if you're going to be dealing with vectors, but it's useful to know. My motivation for actually doing this video is I saw some problems for the Indian Institute of Technology entrance exam that seems to expect that you know Lagrange's formula, or the triple product expansion. So let's see how we can simplify this. So to do that, let's start taking the cross product of b and c. just going to assume-- let's say I have vector a. That's going to be a, the x component of vector a times the unit of vector i plus the y component of vector a times the unit vector j plus the z component of vector a times unit vector k. And I could do the same things for b and c. So if I say b sub y, I'm talking about what's scaling the j component in the b vector. So let's first take this cross product over here. And if you've seen me take cross products, you know that I like to do these little determinants. Let me just take it over here. So b cross c is going to be equal to the determinant. And I put an i, j, k up here. This is actually the definition of the cross product, so no proof necessary to show you why this is true. This is just one way to remember the dot product, of three-by-threes. And we'll put b's x term, b's y coefficient, and b's z component. And then you do the same thing for the c, cx, cy, cz. And then this is going to be equal to-- so first you have the i component. So it's going to be the i component times b. So you ignore this column and this row. So bycz minus bzcy. So I'm just ignoring all of this. And I'm looking at this two-by-two over here, minus bzcy. And then we want to subtract the j component. Remember, we alternate signs when we take our determinant. Subtract that. And then we take out that column and that row, so it's going to be bxcz-- this is a little monotonous,", + "qid": "b7JTVLc_aMk_120" + }, + { + "Q": "At 20:00 couldn't the graph left of zero be the right half of upward concavity ( / )?", + "A": "No, because slope is zero at 0. Isn t it?", + "video_name": "hIgnece9ins", + "timestamps": [ + 1200 + ], + "3min_transcript": "", + "qid": "hIgnece9ins_1200" + }, + { + "Q": "At 5:20, what is he talking about when he says \"This business\"?", + "A": "Occasionally in these videos Sal saves a little time by using the words this business as a way to refer to some expression he s manipulating on the screen. In this video, as he says that, Sal is putting brackets around the expression \u00cf\u0080h^3/12, so that s what he means here by this business.", + "video_name": "Xe6YlrCgkIo", + "timestamps": [ + 320 + ], + "3min_transcript": "So times h. So how can we figure out the area of the water surface, preferably in terms of h? Well we see right over here, the diameter across the top of the cone is 4 centimeters. And the height of the whole cup is 4 centimeters. And so that ratio is going to be true of any-- at any depth of water. It's always going to have the same ratio between the diameter across the top and the height. Because these are lines right over here. So at any given point, the ratio between this and this is going to be the same. So at any given point, the diameter across the surface of the water-- if the depth is h, the diameter across the surface of the water is also going to be h. And so from that we can figure out what are the radius is going to be. The radius is going to be h over 2. is going to be pi r squared, pi times the radius squared. h over 2 squared. That's the area of the surface of the water. And of course we still have the 1/3 out here. And we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1/3 times pi h squared over 4 times another h, which is equal to-- we have pi, h to the third power over 12. So that is our volume. Now what we want to do is relate the volume, how fast the volume is changing with respect to time, and how fast the height is changing with respect to time. So we care with respect to time, since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time. let me move this over. Let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well the derivative with respect to time of our volume, we could just rewrite that as dV dt, this thing right over here. This is dV dt, and this is going to be equal to-- well we could take the constants out of this-- this is going to be equal to pi over 12 times the derivative with respect to t of h, of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time.", + "qid": "Xe6YlrCgkIo_320" + }, + { + "Q": "At 3:20, why is it 1/15", + "A": "To change 16/15 into a mixed number, you divide. 15 goes into 16 once (that s the whole number) Subtract 16 - 15 = 1. We have a remainder of 1 that becomes the new numerator Thus, you get 1 1/15 Hope this helps.", + "video_name": "bcCLKACsYJ0", + "timestamps": [ + 200 + ], + "3min_transcript": "as something over 30. So nine over 10. How would I write that as something over 30? Well I multiply the denominator, I'm multiplying the denominator by three. So I've just multiplied the denominator by three. So if I don't want to change the value of the fraction, I have to do the same thing to the numerator. I have to multiply that by three as well because now I'm just multiplying the numerator by three and the denominator by three, and that doesn't change the value of the fraction. So nine times three is 27. So once again, 9/10 and 27/30 represent the same number. I've just written it now with a denominator of 30, and that's useful because I can also write 1/6 with a denominator of 30. Let's do that. So 1/6 is what over 30? I encourage you to pause the video and try to think about it. So what did we do go from six to 30? We had to multiply by five. So if we multiply the denominator by five, so one times five, one times five is five. So 9/10 is the same thing as 27/30, and 1/6 is the same thing as 5/30. And now we can add, now we can add and it's fairly straightforward. We have a certain number of 30ths, added to another number of 30ths, so 27/30 + 5/30, well that's going to be 27, that's going to be 27 plus five, plus five, plus 5/30, plus 5/30, which of course going to be equal to 32/30. 32 over 30, and if we want, we could try to reduce this fraction. We have a common factor of 32 and 30, they're both divisible by two. So if we divide the numerator and the denominator by two, denominator divided by two is 15. So, this is the same thing as 16/15, and if I wanted to write this as a mixed number, 15 goes into 16 one time with a remainder one. So this is the same thing as 1 1/15. Let's do another example. Let's say that we wanted to add, we wanted to add 1/2 to to 11/12, to 11 over 12. And I encourage you to pause the video and see if you could work this out. Well like we saw before, we wanna find a common denominator. If these had the same denominator, we could just add them immediately, but we wanna find a common denominator because right now they're not the same. Well what we wanna find is a multiple, a common multiple of two and 12, and ideally we'll find the lowest common multiple of two and 12, and just like we did before, let's start with the larger of the two numbers, 12.", + "qid": "bcCLKACsYJ0_200" + }, + { + "Q": "At 9:22 How do you know to put the number before the decimal, or at 5:06 the zero?", + "A": "It depends if the denominator divides before the decimal point or after. For instance, if you had 31/15, it would come out to 2.1. I hope that this was helpful! :)", + "video_name": "Gn2pdkvdbGQ", + "timestamps": [ + 562, + 306 + ], + "3min_transcript": "That's the same thing as 35/1,000. And you're probably saying, Sal, how did you know it's 35/1000? Well because we went to 3-- this is the 10's place. Tenths not 10's. This is hundreths. This is the thousandths place. So we went to 3 decimals of significance. So this is 35 thousandths. If the decimal was let's say, if it was 0.030. There's a couple of ways we could say this. Well, we could say, oh well we got to 3-- we went to the thousandths Place. So this is the same thing as 30/1,000. We could have also said, well, 0.030 is the same thing as 0.03 because this 0 really doesn't add any value. So this is the same thing as 3/100. So let me ask you, are these two the same? Sure they are. If we divide both the numerator and the denominator of both of these expressions by 10 we get 3/100. Let's go back to this case. Are we done with this? Is 35/1,000-- I mean, it's right. That is a fraction. 35/1,000. But if we wanted to simplify it even more looks like we could divide both the numerator and the denominator by 5. And then, just to get it into simplest form, that equals 7/200. And if we wanted to convert 7/200 into a decimal using the technique we just did, so we would do 200 goes into 7 and figure it out. We should get 0.035. I'll leave that up to you as an exercise. to convert a fraction into a decimal and maybe vice versa. And if you don't, just do some of the practices. And I will also try to record another module on this or another presentation. Have fun with the exercises.", + "qid": "Gn2pdkvdbGQ_562_306" + }, + { + "Q": "In 1:40 how come we don't multiply by -1 to make -3p a positive and flip the inequality? I've seen it in other problems before.", + "A": "You can either multiply with -1, divide by -3, or just swap the numbers (the left goes to the right and vice versa), do as you like.", + "video_name": "0YErxSShF0A", + "timestamps": [ + 100 + ], + "3min_transcript": "Solve for z. 5z plus 7 is less than 27 or negative 3z is less than or equal to 18. So this is a compound inequality. We have two conditions here. So z can satisfy this or z can satisfy this over here. So let's just solve each of these inequalities. And just know that z can satisfy either of them. So let's just look at this. So if we look at just this one over here, we have 5z plus 7 is less than 27. Let's isolate the z's on the left-hand side. So let's subtract 7 from both sides to get rid of this 7 on the left-hand side. And so our left-hand side is just going to be 5z. Plus 7, minus 7-- those cancel out. 5z is less than 27 minus 7, is 20. So we have 5z is less than 20. Now we can divide both sides of this inequality by 5. And we don't have to swap the inequality because we're And so we get z is less than 20/5. z is less than 4. Now, this was only one of the conditions. Let's [? look at ?] the other one over here. We have negative 3z is less than or equal to 18. Now, to isolate the z, we could just divide both sides of this inequality by negative 3. But remember, when you divide or multiply both sides of an inequality by a negative number, you have to swap the inequality. So we could write negative 3z. We're going to divide it by negative 3. And then you have 18. We're going to divide it by negative 3. But we're going to swap the inequality. So the less than or equal will become greater than or equal to. And so these guys cancel out. Negative 3 divided by negative 3 is 1. So we have z is greater than or equal to 18 over negative 3 is negative 6. And remember, it's this constraint or this constraint. And this constraint right over here boils down to this. So our solution set-- z is less than 4 or z is greater than or equal to negative 6. So let me make this clear. Let me rewrite it. So z could be less than 4 or z is greater than or equal to negative 6. It can satisfy either one of these. And this is kind of interesting here. Let's plot these. So there's a number line right over there. Let's say that 0 is over here. We have 1, 2, 3, 4 is right over there. And then negative 6. We have 1, 2, 3, 4, 5, 6. That's negative 6 over there. Now, let's think about z being less than 4. We would put a circle around 4, since we're not including 4. And it'd be everything less than 4.", + "qid": "0YErxSShF0A_100" + }, + { + "Q": "At 7:24, I still do not quite understand why the integral of f(x) between 0 and 1 is the same as the integral of f(x) between 1 and 2? I mean I can kind of visualise it, but how can I prove this is true? Thanks a lot:)", + "A": "That is the case because of symmetry. The area between 0 and 1 is the mirror image (along y = 1) of the area between 1 and 2. The easiest way to prove it is to calculate both integrals and see that the result is the same. Int from 0 to 1 (1-x)*cos(pi*x) = Int from 1 to 2 (x-1)*cos(pi*x) = 2/pi^2", + "video_name": "CZdziIlYIfI", + "timestamps": [ + 444 + ], + "3min_transcript": "Don't want you to get that wrong. Cosine pi 0 is cosine of 0. So that's 1. Cosine of pi is negative 1. So when x is equal to 1, this becomes cosine of pi. So then the value of the function is negative 1. It'll be over here. And then cosine of 2 pi, 2 times pi, is then 1 again. So it'll look like this. This is at 1/2. When you put it over here, it'll become pi over 2. Cosine of pi over 2 is 0. So it'll look like this. Let me draw it as neatly as possible. So it will look like this. Cosine, and then it'll keep doing that, and then it'll go like this. So it is also periodic. So if we wanted to figure out the integral of the product from negative 10 to 10, can we simplify that? And it looks like it would just be, because we have this interval, let's look at this interval over here. Let's look at just from 0 to 1. So just from 0 to 1, we're going to take this function and take the product of this cosine times essentially 1 minus x, and then find the area under that curve, whatever it might be. Then when we go from 1 to 2, when we take the product of this and x minus 1, it's actually going to be the same area, because these two, going from 0 to 1 and going from 1 to 2, it's completely symmetric. You can flip it over this line of symmetry, and both functions are completely symmetric. So you're going to have the same area when you take their product. So what we see is, over every interval, from 2 to 3 is clearly the same thing as the integral from 0 to 1. Both functions look identical over that interval. But it will also be the same as going from 1 to 2, because it's completely symmetric. When you take the products of the function, that function will be completely symmetric around this axis. So the integral from here to here will be the same as the integral from there to there. So with that said, we can rewrite this thing over here. So what we want to evaluate, pi squared over 10 times the integral from negative 10 to 10 of f of x cosine of pi x, using the logic we just talked about. This is going to be the same thing as being equal to pi squared over 10 times the integral from 0 to 1, but 20 times that, because we have 20 integers between negative 10 and 10.", + "qid": "CZdziIlYIfI_444" + }, + { + "Q": "At 0:14 how does he turn the seven-sixths to negative.", + "A": "Since there is a subtraction sign after the - 3/4, this could be viewed as adding negatives. Since 7/6 ad 3/6 have common denominators already there, we could block them up to solve the problem easier. -7/6 + -3/6 = -10/6 Then you can find the solution by scaling the numerator and denominators of -10/6 and -3/4 and then add the two together.", + "video_name": "9tmtDBpqq9s", + "timestamps": [ + 14 + ], + "3min_transcript": "We have negative 3/4 minus 7/6 minus 3/6. And there's many ways to do this. But it immediately jumps out at me that these last two numbers have a 6 in the denominator. So I'm going to worry about these first. I'm going to view this as negative 7/6 minus 3/6. So if we have negative 7/6 minus 3/6, that's going to be the same thing as negative 7 minus 3 over 6. And of course, we have this negative 3/4 out front that we're going to add to whatever we get over here. So this is these two terms that I'm adding together. Negative 7 minus 3 is negative 10. So it's negative 10 over 6. And then I'm going to have to add that to negative 3/4. And now I have to worry about finding a common denominator. Let me write that so they have a similar size. What is the smallest number that is a multiple of both 4 and 6? Well, it might jump out at you that it's 12. You can literally just go through the multiples of 4. Or you could look at the prime factorization of both of these numbers. And what's the smallest number that has all of the prime factors of both of these? So you need two 2s, and you need a 2 and a 3. So if you have two 2s and a 3, that's 4 times 3 is 12. So let's rewrite this as something over 12 plus something over 12. Well, to get your denominator from 4 to 12, you have to multiply by 3. So let's multiply our numerator by 3 as well. So if we multiply negative 3 times 3, you're going to have negative 9. And to get your denominator from 6 to 12, you have to multiply by 2. So let's multiply our numerator by 2 as well so that we don't change the value of the fraction. So that's going to be negative 20. Our common denominator is 12. And so this is going to be negative 9 plus negative 20, or we could even write that as minus 20, over 12, which is equal to-- and we deserve a drum roll now. This is negative 29 over 12. And 29 is a prime number, so it's not going to share any common factors other than 1 with 12. So we also have this in the most simplified form.", + "qid": "9tmtDBpqq9s_14" + }, + { + "Q": "At 1:30, why does he say \"the negative direction\"?", + "A": "lim : Means you are inputting smaller negative values. x\u00e2\u0086\u00920- Relating the above limt to the example, ln(n \u00e2\u0089\u00a4 0) is undifined. Hence the given limit condition: lim x\u00e2\u0086\u00920+", + "video_name": "CDf_aE5yg3A", + "timestamps": [ + 90 + ], + "3min_transcript": "- [Voiceover] What I would like to tackle in this video is what I consider to be a particularly interesting limits problem. Let's say we want to figure out the limit as X approaches zero from the positive direction of sine of X. This is where it's about to get interesting. Sine of X to the one over the natural log of X power and I encourage you to pause this video and see if you can have a go at it fully knowing that this is a little bit of a tricky exercise. I'm assuming you have attempted. Some of you might have been able to figure out on the first pass. I will tell you that the first time that I encountered something like this, I did not figure it out at the first pass so definitely do not feel bad if you fall into that second category. What many of you all probably did is you said okay, let me think about it. Let me just think about the components here. If I were to think about the limit, if I were to think about the limit as X approaches zero from the positive direction of sine of X, well that's pretty straightforward. That's going to be zero, is going to approach zero but then if you say, and you could say, I guess I should say. The limit as X approaches zero from the positive direction of one over natural log of X and this is why we have to think about it from the positive direction. It doesn't make sense to approach it from the negative direction. You can't take the natural log of a negative number. That's not in the domain for the natural log but as you get closer and closer to zero from the negative direction, the natural log of those values, you have to raise E to more larger and larger negative values. This part over here is going to approach negative infinity. It's going to go to negative infinity. One over negative infinity, one divided by super large or large magnitude negative numbers, well, that's just going to approach zero. You could say that this right over here is also going to be, is also going to be equal to zero. That doesn't seem to help us much because if this thing is going to zero and that thing is going to zero, it's kind of an implication that well to the zero power but we don't really know what zero, let me do the some, those color. Zero to the zero power but this is one of those great fun things to think about in mathematics. There's justifications why this could be zero, justifications why this could be one. We don't really know what to make of this. This isn't really a satisfying answer. Something at this point might be going into your brain. We have this thing that we've been exposed to called L'Hopital's rule. If you have not been exposed to it, I encourage you to watch the video, the introductory video on L'Hopital's rule. In L'Hopital's rule, let me just write it down. L'Hopital's rule helps us out with situations where when we try to superficially evaluate the limit, we get indeterminate forms things like zero over zero. We get infinity over infinity. We get negative infinity over negative infinity and we go into much more detail into that video.", + "qid": "CDf_aE5yg3A_90" + }, + { + "Q": "I'm a bit confused, I thought DC and EF were the same because both triangles are similar 3:30. Can someone explain?", + "A": "If the triangles are congruent, then corresponding parts of corresponding triangles are congruent, however, if triangles are similar (AA is one method to show this), then the corresponding sides are proportional ( a common scale factor to get from each side to its corresponding side on the other triangle).", + "video_name": "TugWqiUjOU4", + "timestamps": [ + 210 + ], + "3min_transcript": "we're dealing with this right triangle right over here. That's the only right triangle that angle ADC is part of. And so what side is opposite angle ADC? Well, it's side CA, or I guess I say AC, side AC. So that is opposite. And what side is adjacent? Well, this side, CD. CD, or I guess I could call it DC, whatever I want to call it. DC, or CD, is adjacent. Now how did I know that this side is adjacent and not side DA? Because DA is the hypotenuse. They both, together, make up the two sides of this angle. But the adjacent side is one of the sides of the angle that is not the hypotenuse. AD or DA in the sohcahtoa context we would consider to be the hypotenuse. For this angle, this is opposite, this is adjacent, this is hypotenuse. Tangent of this angle is opposite over adjacent-- AC Now is that what they wrote here? No. They wrote AC over EF. Well, where's EF? EF is nowhere to be seen either in this triangle, or even in this figure. EF is this thing right over here. EF is this business right over here. That's EF. It's in a completely different triangle in a completely different figure. We don't even know what scale this is drawn at. There's no way the tangent of this angle is related to this somewhat arbitrary number that's over here. They haven't labelled it. This thing might be a million miles long for all we know. This thing really could be any number. So this isn't the case. We would have to relate it to something within this triangle, or something that's the same length. So if somehow we could prove that EF is the same length as DC, then we could go with that. But there's no way. This is a completely different figure, a completely different diagram. This is a similar triangle to this, but we don't know anything about the lengths. A similar triangle just lets us know that the angles are all might be the same, but it doesn't tell us what this number right over here, doesn't tell us that this side is somehow congruent to DC. So we can't go with this one. Now let's think about the sine of CBA. So the sine-- let me do this in a different color. So the sine of angle CBA. So that's this angle right over here, CBA. Well, sine is opposite over hypotenuse. I guess let me make it clear which I'll do this in yellow. We're now looking at this triangle right over here. The opposite side is AC. That's what the angle opens up into. So it's going to be equal to AC. And what is the hypotenuse? What is the hypotenuse here? Well, the hypotenuse-- so let me see,", + "qid": "TugWqiUjOU4_210" + }, + { + "Q": "What does R^2 (at 0:59) mean?", + "A": "Basically, it s a coordinate space analogous to the xy plane that we all know and love from graphing functions in algebra 2.", + "video_name": "8QihetGj3pg", + "timestamps": [ + 59 + ], + "3min_transcript": "So I have two 2-dimensional vectors right over here, vector a and vector b. And what I want to think about is how can we define or what would be a reasonable way to define the sum of vector a plus vector b? Well, one thing that might jump at your mind is, look, well, each of these are two dimensional. They both have two components. Why don't we just add the corresponding components? So for the sum, why don't we make the first component of the sum just a sum of the first two components of these two vectors. So why don't we just make it 6 plus negative 4? Well, 6 plus negative 4 is equal to 2. And why don't we just make the second component the sum of the two second components? So negative 2 plus 4 is also equal to 2. So we start with two 2-dimensional vectors. You add them together, you get another two 2-dimensional vectors. If you think about it in terms of real coordinates bases, both of these are members of R2-- I'll write this down here just so we get used to the notation. which is just another way of saying that these are both two tuples. They are both two-dimensional vectors right over here. Now, this might make sense just looking at how we represented it, but how does this actually make visual or conceptual sense? And to do that, let's actually plot these vectors. Let's try to represent these vectors in some way. Let's try to visualize them. So vector a, we could visualize, this tells us how far this vector moves in each of these directions-- horizontal direction and vertical direction. So if we put the, I guess you could say the tail of the vector at the origin-- remember, we don't have to put the tail at the origin, but that might make it a little bit easier for us to draw it. We'll go 6 in the horizontal direction. 1, 2, 3, 4, 5, 6. And then negative 2 in the vertical. So negative 2. So vector a could look like this. And once again, the important thing is the magnitude and the direction. The magnitude is represented by the length of this vector. And the direction is the direction that it is pointed in. And also just to emphasize, I could have drawn vector a like that or I could have put it over here. These are all equivalent vectors. These are all equal to vector a. All I really care about is the magnitude and the direction. So with that in mind, let's also draw vector b. Vector b in the horizontal direction goes negative 4-- 1, 2, 3, 4, and in the vertical direction goes 4-- 1, 2, 3, 4. So its tail if we start at the origin, if its tail is at the origin, its head would be at negative 4, 4. So let me draw that just like that.", + "qid": "8QihetGj3pg_59" + }, + { + "Q": "Isn't it supposed to be i=\u00c2\u00b1sqrt(-1) at \"0:56\"???", + "A": "But isn t (+sqrt(-1))^2=(-sqrt(-1))^2 or is it just defined and can t be changed???", + "video_name": "ysVcAYo7UPI", + "timestamps": [ + 56 + ], + "3min_transcript": "In this video, I want to introduce you to the number i, which is sometimes called the imaginary, imaginary unit What you're gonna see here, and it might be a little bit difficult, to fully appreciate, is that its a more bizzare number than some of the other wacky numbers we learn in mathematics, like pi, or e. And its more bizzare because it doesnt have a tangible value in the sense that we normally, or are used to defining numbers. \"i\" is defined as the number whose square is equal to negative 1. This is the definition of \"i\", and it leads to all sorts of interesting things. Now some places you will see \"i\" defined this way; \"i\" as being equal to the principle square root of negative one. I want to just point out to you that this is not wrong, it might make sense to you, you know something squared is negative one, then maybe its the principle square root of negative one. And so these seem to be almost the same statement, some people will even go so far as to say this is wrong, and it actually turns out that they are wrong to say that this is wrong. But, when you do this you have to be a little bit careful about what it means to take a principle square root of a negative number, and it being defined for imaginary, and we'll learn in the future, complex numbers. But for your understanding right now, you dont have to differentiate them, you don't have to split hairs between any of these definitions. Now with this definition, let us think about what these different powers of \"i\" are. because you can imagine, if something squared is negative one, if I take it to all sorts of powers, maybe that will give us weird things. And what we'll see is that the powers of \"i\" are kind of neat, because they kind of cycle, where they do cycle, through a whole set of values. So I could start with, lets start with \"i\" to the zeroth power. And so you might say, anything to the zeroth power is one, so \"i\" to the zeroth power is one, and that is true. And you could actually derive that even from this definition, but this is pretty straight forward; anything to the zeroth power, including \"i\" is one. well anything to the first power is just that number times itself once. So that's justgoing to be \"i\". Really by the definition of what it means to take an exponent, so that completely makes sense. And then you have \"i\" to the second power. \"i\" to the second power, well by definition, \"i\" to the second power is equal to negative one. Lets try \"i\" to the third power ill do this in a color i haven't used. \"i\" to the third power, well that's going to be \"i\" to the second power times \"i\" And we know that \"i\" to the second power is negative one, so its negative one times \"i\" let me make that clear. This is the same thing as this, which is the same thing as that, \"i\" squared is negative one. So you multiply it out, negative one times \"i\" equals negative \"i\". Now what happens when you take \"i\" to the fourth power,", + "qid": "ysVcAYo7UPI_56" + }, + { + "Q": "can you write that congruency sign (at 1:39) on a keyboard?", + "A": "In Windows, like in Microsoft Word or Microsoft Excel, you can your font to Symbol. Then, that symbol is SHIFT+2 (or the @ key on your keyboard.). As a side note, with the Symbol font you get most of the Greek letters..alpha, beta, delta, pi, etc.", + "video_name": "CJrVOf_3dN0", + "timestamps": [ + 99 + ], + "3min_transcript": "Let's talk a little bit about congruence, congruence And one to think about congruence, it's really kind of equivalence for shapes So, when in algebra when something is equal to another thing it means that their quantities are the same But when we're all of the sudden talking about shapes and we say that those shapes are the same, the shapes are the same size and shape then we say that they're congruent And just to see a simple example here: I have this triangle, right over there and let's say I have this triangle right over here And if you are able to shift, you are able to shift this triangle and flip this triangle, you can make it look exactly like this triangle As long as you're not changing the lengths of any of the sides or the angles here But you can flip it, you can shift it, you can rotate it So you can shift, let me write this, you can shift it, you can flip it and you can rotate If you can do those three procedures to make these And if you say that a triangle is congruent, let me label this So, let's call this triangle ABC Now let's call this D, let me call it XYZ XY and Z So, if we were to say, if we make the claim that both of these triangles are congruent So, if we say triangle ABC is congruent And the way you specify it, it almost look like an equal sign But it's equal sign with a curly thing on top Let me write it a little bit either So, we would write it like this If we know that triangle ABC is congruent to triangle XYZ That means their corresponding sides have the same length And their corresponding angles have the same measure So, if we make this assumption or someone tells us that this is true then we know, for example, that AB is going to equal to XY And we could do this like this, and I'm assuming this are the corresponding sides And you can see that actually we've defined these triangles A corresponds to X, B corresponds to Y and C corresponds to Z right over there So, side AB is gonna have the same length as XY Then you can sometimes if you don't have the colors you can denote it just like that These two length are- or this two lines segments have the same length And you can actually say this, you don't always see this written this way You could also make the statement that line segment AB is congruent to line segment XY But congruence of line segments really just means that their lengths are equivalent So, these two things mean the same thing", + "qid": "CJrVOf_3dN0_99" + }, + { + "Q": "At 5:10, how did Sal come to know that (23/4) is (5 3/4) ?", + "A": "Divide it by 4; you have 20 as the nearest multiple so 3 remains giving 5 3/4 With division practice you become better", + "video_name": "w56Vuf9tHfA", + "timestamps": [ + 310 + ], + "3min_transcript": "that really uses our knowledge of the vertex of a parabola to be able to figure out where the focus and the directrix is going to be. So let's think about the vertex of this parabola right over here. Remember, the vertex, if the parabola is upward opening like this, the vertex is this minimum point. If it is downward opening, it's going to be this maximum point. And so when you look over here, you see that you have a negative one-third in front of the x minus one squared. So this quantity over here is either going to be zero or negative. It's not going to add to 23 over four, it's either gonna add nothing or take away from it. So this thing's going to hit a maximum point, when this thing is zero, when this thing is zero, and that's just gonna go down from there and when this thing is zero, y is going to be equal to 23 over four. So our vertex is going to be that maximum point. Well, when x equals one. When x equals one, you get one minus one squared. So zero squared times negative one-third, this is zero. So when x is equal to one, we're at our maximum y value of 23 over four which five and three-fourths. Actually, let me write that as a . Actually, I'll leave just that's our vertex. and it is a downward opening parabola. So actually, let me start to draw this. So we'd get some axis here. So we have to go all the way up to five and three-fourths. So. Let's make this our y, this is our y axis. This is the x axis. That's the x axis. We're gonna see, we're gonna go to one. Let's call that one. Let's call that two. And then I wanna get, let's see, if I go to five and three-fourths, let's go up to, let's see one, two, three, four We can label 'em. One, two, three, four five, six and seven and so our vertex is right over here. One comma 23 over four, so that's five and three-fourths. So it's gonna be right around right around there and as we said, since we have a negative value in front of this x minus one squared term, I guess we could call it, this is going to be a downward opening parabola. This is going to be a maximum point. So our actual parabola is going to look is going to look something it's gonna look something like this. It's gonna look something like this and we could, obviously, I'm hand drawing it, so it's not going to be exactly perfect, but hopefully you get the general idea of what the parabola is going look like and actually, let me just do part of it,", + "qid": "w56Vuf9tHfA_310" + }, + { + "Q": "At 11:01 you just end saying that it is .5 + .495 to get .995 to look up on the z table. Why .995? Just a few seconds before that you said the interval is symmetric about the mu and the right half was .495. Since that is .495 and .495 + .495 = .99 which is the confidence level we want, why do .5 + .495? That lost me.", + "A": "For z = 2.58, probability (area) is .9951. But this is the area from minus infinity to +2.58 SDs. We want the area from the 0 to 2.58 SDs (so we can double it), so we subtract .5. Then we have .4951*2 = .9902, approximately .99 = 99%", + "video_name": "SeQeYVJZ2gE", + "timestamps": [ + 661 + ], + "3min_transcript": "chance, or how many-- let me think of it this way. How many standard deviations away from the mean do we have to be that we can be 99% confident that any sample from the sampling distribution will be in that interval? So another way to think about think it, think about how many standard deviations we need to be away from the mean, so we're going to be a certain number of standard deviations away from the mean such that any sample, any mean that we sample from here, any sample from this distribution has a 99% chance of being plus or minus that many standard So it might be from there to there. So that's what we want. We want a 99% chance that if we pick a sample from the sampling distribution of the sample mean, it will be within this many standard deviations of the actual mean. And to figure that out let's look at an actual Z-table. So another way to think about it if we want 99% confidence, if we just look at the upper half right over here, that orange area should be 0.475, because if this is 0.475 then this other part's going to be 0.475, and we will get to our-- oh sorry, we want to get to 99%, so it's not going to be 0.475. We're going to have to go to 0.495 if we want 99% So this area has to be 0.495 over here, because if that is, that over here will also be. So that their sum will be 99% of the area. Now if this is 0.495, this value on the z table right here will have to be 0.5, because all of this area, if you include all of this is going to be 0.5. So it's going to be 0.5 plus 0.495. Let me make sure I got that right. 0.995. So let's look at our Z-table. So where do we get 0.995. on our z table? 0.995. is pretty close, just to have a little error, it will be right over here-- this is 0.9951. So another way to think about it is 99-- so this value right here gives us the whole cumulative area up to that, up to our mean. So if you look at the entire distribution like this, this is the mean right over here. This tells us that at 2.5 standard deviations above the mean, so this is 2.5 standard deviations above the mean. So this is 2.5 times the standard deviation of the sampling distribution.", + "qid": "SeQeYVJZ2gE_661" + }, + { + "Q": "At 6:53 Why did Sal say that second deravative of why is also same as the first one ?", + "A": "If y = e^x, then the first derivative y is also equal to e^x (e^x is its own derivative). The derivative of the first derivative, known as the second derivative y , is therefore also equal to e^x. Thus, the first derivative of y is equal to the second derivative of y. Also why is how we pronounce the letter y.", + "video_name": "6o7b9yyhH7k", + "timestamps": [ + 413 + ], + "3min_transcript": "So let's verify that. So we first have the second derivative of y. So that's that term right over there. So we have nine e to the negative three x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative three x. Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative three x. Three e to the negative three x. So these two terms right over here, nine e to the negative three x, essentially minus six e to the negative three x, that's gonna be three e to the negative three x. Which is indeed equal to three e to the negative three x. So y one is indeed a solution to this differential equation. But as we'll see, it is not the only solution to this differential equation. For example, let's say y two is equal to e to the x is also a solution to this differential equation. And I encourage you to pause the video again and verify that it's a solution. So assuming you've had a go at it. So the first derivative of this is pretty straight-forward, is e to the x. Second derivative, one of the profound things of the exponential function, the second derivative here is also e to the x. in those same colors. So the second derivative is going to be e to the x plus two times e to the x is indeed going to be equal to three times e to the x. This is absolutely going to be true. E to the x plus two e to the x is three e to the x. So y two is also a solution to this differential equation. So that's a start. In the next few videos, we'll explore this more. We'll start to see what the solutions look like, what classes of solutions are, techniques for solving them, visualizing solutions to differential equations, and a whole toolkit for kind of digging in deeper.", + "qid": "6o7b9yyhH7k_413" + }, + { + "Q": "Is there a test anywhere with questions similar to 10:08? I enjoyed simplifying the expression but I want more challenges like that to see if I can do it again.\n\nThe test \"Practice Multiply Powers\" only has simple questions", + "A": "Keep working thru this section. There are later exercise sets that combine 2 or more of the exponent properties which makes them more challenging.", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 608 + ], + "3min_transcript": "So just to review the properties we've learned so far in this video, besides just a review of what an exponent is, if I have x to the a power times x to the b power, this is going to be equal to x to the a plus b power. We saw that right here. x squared times x to the fourth is equal to x to the sixth, 2 plus 4. We also saw that if I have x times y to the a power, this is the same thing is x to the a power times y to the a power. We saw that early on in this video. We saw that over here. 3x to the third is the same thing as 3 to the third times x to the third. That's what this is saying right here. 3x to the third is the same thing is 3 to the third times x to the third. is, if you have x to the a and then you raise that to the bth power, that's equal to x to the a times b. And we saw that right there. a to the third and then raise that to the fourth power is the same thing is a to the 3 times 4 or a to the 12th power. So let's use these properties to do a handful of more complex problems. Let's say we have 2xy squared times negative x squared y squared times three x squared y squared. And we wanted to simplify this. squared times y squared. So if we take this whole thing to the squared power, this is like raising each of these to the second power. So this part right here could be simplified as negative 1 squared times x squared squared, times y squared. And then if we were to simplify that, negative 1 squared is just 1, x squared squared-- remember you can just multiply the exponents-- so that's going to be x to the fourth y squared. That's what this middle part simplifies to. And let's see if we can merge it with the other parts. The other parts, just to remember, were 2 xy squared, and then 3x squared y squared. Well now we're just going ahead and just straight up multiplying everything. And we learned in multiplication that it doesn't matter which order you multiply things in. So I can just rearrange. We're just going and multiplying 2 times x times y", + "qid": "zM_p7tfWvLU_608" + }, + { + "Q": "at 11:12, Sal is rearranging the problem. He begins with 2x3. Where is the 3 coming from? Please enlighten me on this.", + "A": "The 3 is coming from that last term in parentheses (3x^2y^2). First he multiplied all the numbers together (the middle term didn t have a number), then the letters.........", + "video_name": "zM_p7tfWvLU", + "timestamps": [ + 672 + ], + "3min_transcript": "is, if you have x to the a and then you raise that to the bth power, that's equal to x to the a times b. And we saw that right there. a to the third and then raise that to the fourth power is the same thing is a to the 3 times 4 or a to the 12th power. So let's use these properties to do a handful of more complex problems. Let's say we have 2xy squared times negative x squared y squared times three x squared y squared. And we wanted to simplify this. squared times y squared. So if we take this whole thing to the squared power, this is like raising each of these to the second power. So this part right here could be simplified as negative 1 squared times x squared squared, times y squared. And then if we were to simplify that, negative 1 squared is just 1, x squared squared-- remember you can just multiply the exponents-- so that's going to be x to the fourth y squared. That's what this middle part simplifies to. And let's see if we can merge it with the other parts. The other parts, just to remember, were 2 xy squared, and then 3x squared y squared. Well now we're just going ahead and just straight up multiplying everything. And we learned in multiplication that it doesn't matter which order you multiply things in. So I can just rearrange. We're just going and multiplying 2 times x times y x squared times y squared. So I can rearrange this, and I will rearrange it so that it's in a way that's easy to simplify. So I can multiply 2 times 3, and then I can worry about the x terms. Let me do it in this color. Then I have times x times x to the fourth times x squared. And then I have to worry about the y terms, times y squared times another y squared times another y squared. And now what are these equal to? Well, 2 times 3. You knew how to do that. That's equal to 6. And what is x times x to the fourth times x squared. Well, one thing to remember is x is the same thing as x to the first power. Anything to the first power is just that number. So you know, 2 to the first power is just 2.", + "qid": "zM_p7tfWvLU_672" + }, + { + "Q": "At 1:18, he says that in order to be divisible by both a number has to have 2 2s, a 3 etc... but why 2 2s and just 1 3?", + "A": "12 needed 2 2s and one 3 and 20 needed a 5 and 2 2s but the 2s were thare and he use prime numbers and 2, 3, and 5s are prime", + "video_name": "zWcfVC-oCNw", + "timestamps": [ + 78 + ], + "3min_transcript": "- [Voiceover] In this video I wanna do a bunch of example problems that show up on standardized exams and definitely will help you with our divisibility module because it's asking you questions like this. And this is just one of the examples. All numbers divisible by both 12 and 20 are also divisible by: And the trick here is to realize that if a number is divisible by both 12 and 20, it has to be divisible by each of these guy's prime factors. So let's take the prime factorization, the prime factorization of 12, let's see, 12 is 2 times 6. 6 isn't prime yet so 6 is 2 times 3. So that is prime. So any number divisible by 12 needs to be divisible by 2 times 2 times 3. So its prime factorization needs to have a 2 times a 2 times a 3 in it, any number that's divisible by 12. Now any number that's divisible by 20 needs to be divisible by, let's take it's prime factorization. 2 times 10 10 is 2 times 5. So any number divisible by 20 2 times 2 times 5. Or another way of thinking about it, it needs to have two 2's and a 5 in its prime factorization. If you're divisible by both, you have to have two 2's, a 3, and a 5, two 2's and a 3 for 12, and then two 2's and a 5 for 20, and you can verify this for yourself that this is divisible by both. Obviously if you divide it by 20, let me do it this way. Dividing it by 20 is the same thing as dividing by 2 times 2 times 5, so you're going to have the 2's are going to cancel out, the 5's are going to cancel out. You're just going to have a 3 left over. So it's clearly divisible by 20, and if you were to divide it by 12, you'd divide it by 2 times 2 times 3. This is the same thing as 12. And so these guys would cancel out and you would just have a 5 left over so it's clearly divisible by both, and this number right here is 60. It's 4 times 3, which is 12, times 5, it's 60. of 12 and 20. Now this isn't the only number that's divisible by both 12 and 20. You could multiply this number right here by a whole bunch of other factors. I could call them a, b, and c, but this is kind of the smallest number that's divisible by 12 and 20. Any larger number will also be divisible by the same things as this smaller number. Now with that said, let's answer the questions. All numbers divisible by both 12 and 20 are also divisible by: Well we don't know what these numbers are so we can't really address it. They might just be 1's or they might not exist because the number might be 60. It might be 120. Who knows what this number is? So the only numbers that we know can be divided into this number, well we know 2 can be, we know that 2 is a legitimate answer. 2 is obviously divisible into 2 times 2 times 3 times 5. We know that 2 times 2 is divisible into it, cuz we have the 2 times 2 over there.", + "qid": "zWcfVC-oCNw_78" + }, + { + "Q": "at 2:42, to isolate the -2, why did you have to divide and not subtract?", + "A": "be careful with leading negative numbers. Try to think of this as a negative number, not subtraction. The negative two is being multiplied by the absolute value. In order to cancel this number we do the opposite of multiplying by -2 which is dividing by -2.", + "video_name": "15s6B7K9paA", + "timestamps": [ + 162 + ], + "3min_transcript": "If these two things are equal, and if I want to keep them equal, if I subtract 6 from the right-hand side, I've got to subtract-- or if I subtract 6 times the absolute value of x plus 10 from the right-hand side, I have to subtract the same thing from the left-hand side. So we're going to have minus 6 times the absolute value of x plus 10. And likewise, I want to get all my constant terms, I want to get this 4 out of the left-hand side. So let me subtract 4 from the left, and then I have to also do it on the right, otherwise my equality wouldn't hold. And now let's see what we end up with. So on the left-hand side, the 4 minus 4, that's 0. You have 4 of something minus 6 of something, that means you're going to end up with negative 2 of that something. Negative 2 of the absolute value of x plus 10. Remember, this might seem a little confusing, but remember, if you had 4 apples and you subtract 6 apples, you now have negative 2 apples, Same way, you have 4 of this expression, you take away 6 of this expression, you now have negative 2 of this expression. Let me write it a little bit neater. So it's negative 2 times the absolute value of x plus 10 is equal to, well the whole point of this, of the 6 times the absolute value of x plus 10 minus 6 times the absolute value of x plus 10 is to make those cancel out, and then you have 10 minus 4, which is equal to 6. Now, we want to solve for the absolute value of x plus 10. So let's get rid of this negative 2, and we can do that by dividing both sides by negative 2. You might realize, everything we've done so far is just treating this red expression as almost just like a variable, and we're going to solve for that red expression and then take it from there. So negative 2 divided by negative 2 is 1. 6 divided by negative 2 is negative 3. So we get the absolute value of x plus 10 is equal to negative 3. You might say maybe this could be the positive version or the negative, but remember, absolute value is always non-negative. If you took the absolute value of 0, you would get 0. But the absolute value of anything else is going to be positive. So this thing right over here is definitely going to be greater than or equal to 0. Doesn't matter what x you put in there, when you take its absolute value, you're going to get a value that's greater than or equal to 0. So there's no x that you could find that's somehow-- you put it there, you add 10, you take the absolute value of it, you're actually getting a negative value. So this right over here has absolutely no solution. And I'll put some exclamation marks there for emphasis.", + "qid": "15s6B7K9paA_162" + }, + { + "Q": "At 1:42 Sal says to move the 0 degrees on the protractor to one side of the angle. How do you know which side to use?", + "A": "One side is in the angle, if the angle is right or acute, than use the side that flows into the actual angle, obtuse angles can vary.", + "video_name": "wJ37GJyViU8", + "timestamps": [ + 102 + ], + "3min_transcript": "This is the video for the measuring angles module because, clearly, at the time that I'm doing this video, there is no video for the measuring angles module. And this is a pretty neat module. This was made by Omar Rizwan, one of our amazing high school interns that we had this past summer. This is the summer of 2011. And what it really is, is it makes you measure angles. And he made this really cool protractor tool here so that you actually use this protractor to measure the angles there. And so the trick here is you would actually measure it the way you would measure any angle using an actual physical protractor. You'd want to put the center of the protractor right at the vertex of where the two lines intersect. You can view it as the vertex of the angle. And then you'd want to rotate it so that, preferably, this edge, this edge at 0 degrees, is at one of these sides. So let's do it so that this edge right over here is right along this line. So let me rotate it. So then-- I've got to rotate it a little bit further, maybe So that looks about right. And then if you look at it this way, you can see that the angle-- and I don't have my Pen tool here. I'm just using my regular web browser-- if you look at the angle here, you see that the other line goes to 130 degrees. So this angle that we need to measure here is 130 degrees, assuming you can read sideways. So that is 130 degrees. Let me check my answer. Very good, I got it right. It would have been embarrassing if I didn't. Let's do the next question. I'll do a couple of examples like this. So once again, let us put the center of the protractor right at the vertex right over there. And let's get this 0 degrees side to be on one of these sides so that this angle will be within the protractor. So let me rotate it this way. And this really is pretty cool what Omar did with this module. So let's see. Let's do it one more time. That's too far. And then you can see that the angle right over here, if we look at where the other line points to, it is 40 degrees. Check answer-- very good. Let's do another one. This is fun. So let's get our protractor right over there. And you don't always have to do it in that same order. You could rotate it first so that the 0 degrees is-- and what you want to do is you want to rotate the 0 degrees to one of the sides so that the angle is still within the protractor. So let's rotate it around. So if you did it like that-- so you don't always have to do it in that same order. Although I think it's easier to rotate it when you have the center of the protractor at the vertex of the angle. So we have to rotate it a little bit more. So 0 degrees is this line. And then as we go further and further up, I guess, since this is on its side, it looks like this other line gets us to 150 degrees. And hopefully you're noticing that the higher the degrees, the more open this angle is.", + "qid": "wJ37GJyViU8_102" + }, + { + "Q": "At 1:22 in the video do you do four times eleven?", + "A": "Yep! 4(11) = 44", + "video_name": "gl_-E6iVAg4", + "timestamps": [ + 82 + ], + "3min_transcript": "Rewrite the expression 4 times, and then in parentheses we have 8 plus 3, using the distributive law of multiplication over addition. Then simplify the expression. So let's just try to solve this or evaluate this expression, then we'll talk a little bit about the distributive law of multiplication over addition, usually just called the distributive law. So we have 4 times 8 plus 8 plus 3. Now there's two ways to do it. Normally, when you have parentheses, your inclination is, well, let me just evaluate what's in the parentheses first and then worry about what's outside of the parentheses, and we can do that fairly easily here. We can evaluate what 8 plus 3 is. 8 plus 3 is 11. So if we do that-- let me do that in this direction. So if we do that, we get 4 times, and in parentheses we have an 11. 8 plus 3 is 11, and then this is going to be equal to-- evaluate it that way. But they want us to use the distributive law of multiplication. We did not use the distributive law just now. We just evaluated the expression. We used the parentheses first, then multiplied by 4. In the distributive law, we multiply by 4 first. And it's called the distributive law because you distribute the 4, and we're going to think about what that means. So in the distributive law, what this will become, it'll become 4 times 8 plus 4 times 3, and we're going to think about why that is in a second. So this is going to be equal to 4 times 8 plus 4 times 3. A lot of people's first instinct is just to multiply the 4 times the 8, but no! You have to distribute the 4. You have to multiply it times the 8 and times the 3. This is the distributive property in action right here. Distributive property in action. And then when you evaluate it-- and I'm going to show you in kind of a visual way why this works. But then when you evaluate it, 4 times 8-- I'll do this in a different color-- 4 times 8 is 32, and then so we have 32 plus 4 times 3. 4 times 3 is 12 and 32 plus 12 is equal to 44. That is also equal to 44, so you can get it either way. But when they want us to use the distributive law, you'd distribute the 4 first. Now let's think about why that happens. Let's visualize just what 8 plus 3 is. Let me draw eight of something. So one, two, three, four, five, six, seven, eight, right? And then we're going to add to that three of something, of", + "qid": "gl_-E6iVAg4_82" + }, + { + "Q": "At 8:16, when you have the area of 324 pi. Why can't you substitute pi with 3 because 3.14 rounded is 3 and multiply 3 by 324?", + "A": "The answer will not be accurate", + "video_name": "mLE-SlOZToc", + "timestamps": [ + 496 + ], + "3min_transcript": "There's actually an infinite number of points you could pick here. And so, when we talk about the probability that the point also lies in the smaller circle, we're really thinking about the percentage of the points in the larger circle that are also in the smaller circle. Or another way to think about it is the probability that if we pick a point from this larger circle, the probably that it's also in the smaller circle is really just going to be the percentage of the larger circle that is the smaller circle. I know that might sound confusing, but we're really just have to figure out the areas for both of them, and it's really just going to be the ratios so let's think about that. So there's a temptation to just use this 36 pi up here, but we have to remember, this was the circumference, and we need to figure out the area of both of these circles. And so for area, we need to know the radius, because area is pi r squared. So we can figure out the radius from the circumference by saying, well, circumference is equal to 2 times pi times the radius of the circle. is equal to 2 times pi times the radius, we can divide both sides by 2 pi, and on the left hand side, 36 divided by 2 is 18 the pi's cancel out, we get our radius as being equal to 18 for this larger circle. So if we want to know its area, its area is going to be pi r squared, which is equal to pi times 18 squared. And let's figure out what 18 squared is. 18 times 18, 8 times 8 is 64, eight times 1 is 8 plus 6 is 14, and then we put that 0 there because we're now in the tens place, 1 times 8 is 8, 1 times 1 is 1. And really, this is a 10 times the 10, and that's why it gives us 100. Anyway, 4 plus 0 is a 4, 4 plus 8 is a 12, So the area here is equal to pi times 324, or we could say 324 pi. So the area of the entire larger circle, the part that I shaded in yellow, including what's kind of under this orange circle, if you want to view it that way, this area right over here is equal to 324 pi. So the probability that a point that we select from this larger circle is also in the smaller circle is really just a percentage of the larger circle that is the smaller circle. So our probability-- I'll just write it like this-- the probability that the point also lies in the smaller circle-- so all of that stuff The probability of that is going to be equal to the percentage of this larger circle that is this smaller one, and that's going to be--", + "qid": "mLE-SlOZToc_496" + }, + { + "Q": "At 1:28 Sal says that we must divide both the sides with 5 but instead of that we can transpose 5.It is much easier that way!", + "A": "fair enough, but I think that dividing both the sides by 5 is good for beginners because the fundamental belief of algebra, that if you do the same thing to both the sides of an equation,you won t change anything seems much more prominent in dividing both sides by 5 than in transposing.", + "video_name": "c6-FJRda_Vc", + "timestamps": [ + 88 + ], + "3min_transcript": "y is directly proportional to x. If y equals 30 when x is equal to 6, find the value of x when y is 45. So let's just take this each statement at a time. y is directly proportional to x. That's literally just saying that y is equal to some constant times x. This statement can literally be translated to y is equal to some constant times x. y is directly proportional to x. Now, they tell us, if y is 30 when x is 6-- and we have this constant of proportionality-- this second statement right over here allows us to solve for this constant. When x is 6, they tell us y is 30 so we can figure out what this constant is. We can divide both sides by 6 and we get this left-hand side is 5-- 30 divided by 6 is 5. 5 is equal to k or k is equal to 5. So the second sentence tells us, this gives us the information that y is equal to 5 times x. y is 30 when x is 6. And then finally, they say, find the value of x when y is 45. So when y is 45 is equal to-- so we're just putting in 45 for y-- 45 is equal to 5x. Divide both sides by 5 to solve for x. We get 45 over 5 is 9, and 5x divided by 5 is just x. So x is equal to 9 when y is 45.", + "qid": "c6-FJRda_Vc_88" + }, + { + "Q": "Can someone explain to me how he got the 6i in (9+6i-1) at 5:50? Thanks!", + "A": "Shivanie, He was multiplying (3+i)(3+i) Using FOIL you get First 3*3 = 9 Outside 3*i = 3i Inside i*3 = 3i Last i*i = -1 So you get 9+3i+3i-1 And the 3i+3i = 6i so you get 9+6i-1 I hope that is of help to you.", + "video_name": "dnjK4DPqh0k", + "timestamps": [ + 350 + ], + "3min_transcript": "Those are the two roots. Now what I want to do is a verify that these work. Verify these two roots. So this one I can rewrite as 3 plus i over 2. These are equivalent. All I did-- you can see that this is just dividing both of these by 2. Or if you were to essentially factor out the 1/2, you could go either way on this expression. And this one over here is going to be 3 minus i over 2. Or you could go directly from this. This is 3 plus or minus i over 2. So 3 plus i over 2. Or 3 minus i over 2. This and this or this and this, or this. These are all equal representations of both of the roots. But let's see if they work. So I'm first going to try this character right over here. It's going to get a little bit hairy, because we're going to have to square it and all the rest. But let's see if we can do it. 2 times this quantity squared. So 2 times 3 plus i over 2 squared plus 5. And we want to verify that that's the same thing as 6 times this quantity, as 6 times 3 plus i over 2. So what is 3 plus i squared? So this is 2 times-- let me just square this. So 3 plus i, that's going to be 3 squared, which is 9, plus 2 times the product of three and i. So 3 times i is 3i, times 2 is 6i. So plus 6i. And if that doesn't make sense to you, I encourage you to kind of multiply it out either with the distributive property or FOIL it out, and you'll get the middle term. You'll get 3i twice. When you add them, you get 6i. I And then plus i squared, and i squared is negative 1. Minus 1. All of that over 4, plus 5, is equal to-- well, you get a 3 here and you get a 1 here. And 3 distributed on 3 plus i is equal to 9 plus 3i. And what we have over here, we can simplify it just to save some screen real estate. 9 minus 1 is 8. So if I get rid of this, this is just 8 plus 6i. We can divide the numerator and the denominator right here by 2. So the numerator would become 4 plus 3i, if we divided it by 2, and the denominator here is just going to be 2. This 2 and this 2 are going to cancel out. So on the left hand side, we're left with 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works.", + "qid": "dnjK4DPqh0k_350" + }, + { + "Q": "Can someone explain to me what Sal is doing at 5:29 onwards?", + "A": "Sal is using the same FOIL technique except now there are complex numbers. ((3 + i) / 2)^2 can also be written as ((3 + i)*(3 + i)) / (2*2). By using FOIL the numerator will become... F: 3*3 = 9 O: 3*i = 3i I: 3*i = 3i L: i*i = -1 (3 + i)(3 + i) ----> 9 + 3i + 3i - 1 ----> 8 + 6i 2(8 + 6i) / 4 ----> 4(4 + 3i) / 4 ----> 4 + 3i I hope this helps", + "video_name": "dnjK4DPqh0k", + "timestamps": [ + 329 + ], + "3min_transcript": "I could even do it one step-- that's the same thing as negative 1 times 4 under the radical, which is the same thing as the square root of negative 1 times the square root of 4. And the principal square root of negative 1 is i times the principal square root of 4 is 2. So this is 2i, or i times 2. So this right over here is going to be 2i. So we are left with x is equal to 6 plus or minus 2i over 4. And if we were to simplify it, we could divide the numerator and the denominator by 2. And so that would be the same thing as 3 plus or minus i over 2. Or if you want to write them as two distinct complex numbers, you could write this as 3 plus i over 2, or 3/2 plus 1/2i. That's if I take the positive version of the i there. Or we could view this as 3/2 minus 1/2i. Those are the two roots. Now what I want to do is a verify that these work. Verify these two roots. So this one I can rewrite as 3 plus i over 2. These are equivalent. All I did-- you can see that this is just dividing both of these by 2. Or if you were to essentially factor out the 1/2, you could go either way on this expression. And this one over here is going to be 3 minus i over 2. Or you could go directly from this. This is 3 plus or minus i over 2. So 3 plus i over 2. Or 3 minus i over 2. This and this or this and this, or this. These are all equal representations of both of the roots. But let's see if they work. So I'm first going to try this character right over here. It's going to get a little bit hairy, because we're going to have to square it and all the rest. But let's see if we can do it. 2 times this quantity squared. So 2 times 3 plus i over 2 squared plus 5. And we want to verify that that's the same thing as 6 times this quantity, as 6 times 3 plus i over 2. So what is 3 plus i squared? So this is 2 times-- let me just square this. So 3 plus i, that's going to be 3 squared, which is 9, plus 2 times the product of three and i. So 3 times i is 3i, times 2 is 6i. So plus 6i. And if that doesn't make sense to you, I encourage you to kind of multiply it out either with the distributive property or FOIL it out, and you'll get the middle term. You'll get 3i twice. When you add them, you get 6i. I And then plus i squared, and i squared is negative 1. Minus 1. All of that over 4, plus 5, is equal to-- well,", + "qid": "dnjK4DPqh0k_329" + }, + { + "Q": "he lost me at 2:32, what is he saying. Please help", + "A": "(anything)%=(anything)/100 I hope you understand now!", + "video_name": "-gB1y-PMWfs", + "timestamps": [ + 152 + ], + "3min_transcript": "It's 0.18. You could view this as 1 tenth and 8 hundredths, which is the same thing, or 10 hundredths and 8 hundredths, which is 18 hundredths. So this is written in decimal form. And if we write it as a simplified fraction, we need to see if there is a common factor for 18 and 100. And they're both even numbers, so we know they're both divisible by 2, so let's divide both the numerator and the denominator by 2. So we have 18 divided by 2 over 100 divided by 2. And we're going to get 18 divided by 2 is 9. 100 divided by 2 is 50. And I don't think these guys share any common factors. 50 is not divisible by 3. 9 is only divisible by 3 and 1 and 9. So this is the fraction in simplest form. So we have 18% is the same thing as 0.18, which is the Now, I went through a lot of pain here to show you that this really just comes from the word, from percent, from per 100. But if you ever were to see this in a problem, the fast way to do this is to immediately say, OK, if I have 18%, you should immediately say, anything in front of the percent-- that's that anything, whatever this anything is-- it should be equal to that anything. In this case it's 18/100. And another way to think about it, you could view this as 18.0%. I just added a trailing zero there, just so that you see the decimal, really. But if you want to express this as a decimal without the percent, you just move the decimal to the left two spaces. this becomes 0.18. Or you could immediately say that 18% as a fraction is 18/100. When you put it in simplified form, it's 9/50. But you should also see that 18/100, and we have seen this, is the exact same thing as 18 hundredths, or 0.18. Hopefully, this made some connections for you and didn't confuse you.", + "qid": "-gB1y-PMWfs_152" + }, + { + "Q": "At about 1:17, he showed that he was converting the decimals to whole numbers. That method really confuses me. Can somebody please show me a way to solve the problem while keeping the decimals and solving it that way? would it be possible to instead of changing the decimals, keep the decimals and work out the problem with the decimals?", + "A": "You can do 0.6 / 1.2 But, the 1st step in decimal division, is to change the 1.2 into a whole number. Shift the decimal places one place to right 6 / 12. Then, do the long division. You will get c = 0.5", + "video_name": "a3acutLstF8", + "timestamps": [ + 77 + ], + "3min_transcript": "Let's get some practice solving some equations, and we're gonna set up some equations that are a little bit hairier than normal, they're gonna have some decimals and fractions in them. So let's say I had the equation 1.2 times c is equal to 0.6. So what do I have to multiply times 1.2 to get 0.6? And it might not jump out immediately in your brain but lucky for us we can think about this a little bit methodically. So one thing I like to do is say okay, I have the c on the left hand side, and I'm just multiplying it by 1.2, it would be great if this just said c. If this just said c instead of 1.2c. So what can I do there? Well I could just divide by 1.2 but as we've seen multiple times, you can't just do that to the left hand side, that would change, you no longer could say that this is equal to that if you only operate on one side. So you have to divide by 1.2 on both sides. So on your left hand side, 1.2c divided by 1.2, well that's just going to be c. You're just going to be left with c, Now what is that equal to? There's a bunch of ways you could approach it. The way I like to do it is, well let's just, let's just get rid of the decimals. Let's just multiply the numerator and denominator by a large enough number so that the decimals go away. So what happens if we multiply the numerator and the denominator by... Let's see if we multiply them by 10, you're gonna have a 6 in the numerator and 12 in the denominator, actually let's do that. Let's multiply the numerator and denominator by 10. So once again, this is the same thing as multiplying by 10 over 10, it's not changing the value of the fraction. So 0.6 times 10 is 6, and 1.2 times 10 is 12. So it's equal to six twelfths, and if we want we can write that in a little bit of a simpler way. We could rewrite that as, divide the numerator and denominator by 6, you get 1 over 2, And if you look back at the original equation, 1.2 times one half, you could view this as twelve tenths. Twelve tenths times one half is going to be equal to six tenths, so we can feel pretty good that c is equal to one half. Let's do another one. Let's say that we have 1 over 4 is equal to y over 12. So how do we solve for y here? So we have a y on the right hand side, and it's being divided by 12. Well the best way I can think of of getting rid of this 12 and just having a y on the right hand side is multiplying both sides by 12. We do that in yellow. So if I multiply the right hand side by 12, I have to multiply the left hand side by 12. And once again, why did I pick 12? Well I wanted to multiply by some number, that when I multiply it by y over 12", + "qid": "a3acutLstF8_77" + }, + { + "Q": "I still don't understand how does the computer program calculate the \"pseudo-sample variance\" @4:10 if we don't know mu's value. Can someone please explain?", + "A": "In real life we generally don t know the value of \u00ce\u00bc. However, in a simulation, we are making up the data, and we do in fact know \u00ce\u00bc. What were doing is: 1. Set \u00ce\u00bc and create some data from a distribution with that mean. 2. Pretend that we don t know \u00ce\u00bc, and calculate the mean and standard deviation. 3. Remember that we know \u00ce\u00bc, and perform the calculations shown in the video.", + "video_name": "F2mfEldxsPI", + "timestamps": [ + 250 + ], + "3min_transcript": "In the vertical axis, using this denominator, dividing by n, we calculate two different variances. One variance, we use the sample mean. The other variance, we use the population mean. And this, in the vertical axis, we compare the difference between the mean calculated with the sample mean versus the mean calculated with the population mean. So for example, this point right over here, when we calculate our mean with our sample mean, which is the normal way we do it, it significantly underestimates what the mean would have been if somehow we knew what the population mean was and we could calculate it that way. And you get this really interesting shape. And it's something to think about. And he recommends some thinking about why or what kind of a shape this actually is. The other interesting thing is when you look at it this way, it's pretty clear this entire graph is sitting below the horizontal axis. So we're always, when we calculate our sample variance which we typically do, we're always getting a lower variance than when we use the population mean. Now this over here, when we divide by n minus 1, we're not always underestimating. Sometimes we are overestimating it. And when you take the mean of all of these variances, And here we're overestimating it a little bit more. And just to be clear what we're talking about in these three graphs, let me take a screen shot of it and explain it in a little bit more depth. So just to be clear, in this red graph right over here, let me do this. A color close to at least. So this orange, what this distance is for each of these samples, we're calculating the sample variance using, so let me, using the sample mean. And in this case, we are using n as our denominator. In this case right over here. or I guess you could call this some kind of pseudo sample variance, if we somehow knew the population mean. This isn't something that you see a lot in statistics. But it's a gauge of how much we are underestimating our sample variance given that we don't have the true population mean at our disposal. And so this is the distance. This is the distance we're calculating. And you see we're always underestimating. Here we overestimate a little bit. And we also underestimate. But when you take the mean, when you average them all out, it converges to the actual value. So here we're dividing by n minus 1, here we're dividing by n minus 2.", + "qid": "F2mfEldxsPI_250" + }, + { + "Q": "At 2:46, how can one list the factors of \"a\" if it has already been declared prime?", + "A": "It hasn t been declared prime. a/b is reduced to lowest terms. 27/32 is reduced to lowest terms. It s (3*3*3)/(2*2*2*2*2).", + "video_name": "W-Nio466Ek4", + "timestamps": [ + 166 + ], + "3min_transcript": "Well, this being rational says I can represent the square root of p as some fraction, as some ratio of two integers. And if I can represent anything as a ratio of two integers, I can keep dividing both the numerator and the denominator by the common factors until I eventually get to an irreducible fraction. So I'm assuming that's where we are right here. So this cannot be reduced. And this is important for our proof-- cannot be reduced, which is another way of saying that a and b are co-prime, which is another way of saying that a and b share no common factors other than 1. So let's see if we can manipulate this a little bit. Let's take the square of both sides. We get p is equal to-- well, a/b, the whole thing squared, that's the same thing as a squared over b squared. We can multiply both sides by b squared, and we get b squared times p is equal to a squared. Well, b is an integer, so b squared must be an integer. So an integer times p is equal to a squared. Well, that means that p must be a factor of a squared. Let me write this down. So a squared is a multiple of p. Now, what does that tell us about a? Does that tell us that a must also be a multiple of p? Well, to think about that, let's think about the prime factorization of a. Let's say that a can be-- and any number-- can be rewritten as a product of primes. Or any integer, I should say. So let's write this out as a product of primes right over here. So let's say that I have my first prime factor times my second prime factor, all the way to my nth prime factor. I'm just saying that a is some integer right over here. So that's the prime factorization of a. What is the prime factorization of a squared going to be? Well, a squared is just a times a. Its prime factorization is going to be f1 times f2, all the way to fn. And then that times f1 times f2 times, all the way to fn. Or I could rearrange them if I want. f1 times f1 times f2 times f2, all the way to fn times fn. Now, we know that a squared is a multiple of p. p is a prime number, so p must be one of these numbers in the prime factorization. p could be f2, or p could be f1, but p needs to be one of these numbers in the prime factorization. So p needs to be one of these factors.", + "qid": "W-Nio466Ek4_166" + }, + { + "Q": "At 1:08, when sal divides both sides of 24x/24x wouldn't you be left with 1x? so then you move the variables to one side making 1x-1x making it 0x. ahh i think i just answered my own question. 0xanything is 0. right?", + "A": "he s not dividing though, he s subtracting", + "video_name": "zKotuhQWIRg", + "timestamps": [ + 68 + ], + "3min_transcript": "Solve for x. We have 8 times the quantity 3x plus 10 is equal to 28x minus 14 minus 4x. So like every equation we've done so far, we just want to isolate all of the x's on one side of this equation. But before we do that, we can actually simplify each of these sides. On the left-hand side, we can multiply the quantity 3x plus 10 times 8. So we're essentially just distributing the 8, the distributive property right here. So this is the same thing as 8 times 3x, which is 24x, plus 8 times 10, which is 80, is equal to-- and over here, we have 28x minus 14 minus 4x. So we can combine the 28x and the minus 4x. If we have 28x minus 4x, that is 24x And then you have the minus 14 right over here. Now, the next thing we could-- and it's already looking a little bit suspicious, but just to confirm that it's as suspicious as it looks, let's try to subtract 24x from both sides of this equation. And if we do that, we see that we actually remove the x's and we have a 24x there. You might say, hey, let's put all the x's on the left-hand side. So let's get rid of this 24x. So you subtract 24x right over there, but you have to do it to the left-hand side as well. On the left-hand side, these guys cancel out, and you're left with just 80-- these guys cancel out as well-- is equal to a negative 14. Now, this looks very bizarre. It's making a statement that 80 is equal to negative 14, which we know is not true. This does not happen. 80 is never equal to negative 14. They're just inherently inequal. So this equation right here actually has no solution. This has no solution. There is a no x-value that will make 80 equal to negative 14.", + "qid": "zKotuhQWIRg_68" + }, + { + "Q": "At 4:06 he says that the denominator is zero but isn't the square root of 0 + 1 just equal one. would the graph be different?", + "A": "Yes, but he was finding the limit of 0/(sqrt(0+1)), which is 0.", + "video_name": "xks4cETlN58", + "timestamps": [ + 246 + ], + "3min_transcript": "Now, for positive x'es the absolute value of x is just going to be x. This is going to be x divided by x, so this is just going to be 1. Similarly, right over here, we take the limit as we go to negative infinity, this is going to be the limit of x over the absolute value of x as x approaches negative infinity. Remember, the only reason I was able to make this statement is that f(x) and this thing right over here become very very similar, you can kind of say converge to each other, as x gets very very very large or x gets very very very very negative. Now, for negative values of x the absolute value of x is going to be positive, x is obviously going to be negative and we're just going to get negative 1. And so using this, we can actually try to graph our function. So let's say, that is my y axis, this is my x axis, and we see that we have 2 horizontal asymptotes. We have 1 horizontal asymptote at y=1, so let's say this right over here is y=1, let me draw that line as dotted line, we're going to approach this thing, and then we have another horizontal asymptote at y=-1. So that might be right over there, y=-1. And if we want to plot at least 1 point we can think about what does f(0) equal. So, f(0) is going to be equal to 0 over the square root of 0+1, or 0 squared plus 1. Well that's all just going to be equal to zero. So we have this point, right over here, and we know that as x approaches infinity, we're approaching this blue, horizontal asymptote, Let me do it a little bit differently. There you go. I'll clean this up. So it might look something like this. That's not the color I wanted to use. So it might look something like that. We get closer and closer to that asymptote as x gets larger and larger and then like this -- we get closer and closer to this asymptote as x approaches negative infinity. I'm not drawing it so well. So that right over there is y=f(x). And you can verify this by taking a calculator, trying to plot more points or using some type of graphing calculator or something. But anyway, I just wanted to tackle another situation we're approaching infinity and or negative infinity and we're trying to determine the horizontal asymptotes. And remember, the key is just to say what terms dominate", + "qid": "xks4cETlN58_246" + }, + { + "Q": "Wait, at 0:37 Sal says that for the first flip there's 2 possibilities, same on the second and the third. So 2 and 2 and 2 should be 6, right?", + "A": "The 2 s should be multiplied instead of added together. This should be seen in layers like this: There are 2 possibility for the first flip. For every possibility of the first flip, there are 2 possibility for the second flip. So there are a total of 2 * 2 possibilities for the first two flips. For every possibility of the first two flips, there are 2 possibility for the third flip. So there are a total of (2 * 2) * 2 possibilities for the three flips.", + "video_name": "mkyZ45KQYi4", + "timestamps": [ + 37 + ], + "3min_transcript": "", + "qid": "mkyZ45KQYi4_37" + }, + { + "Q": "At 2:03, how do you add 7 + 1/100?", + "A": "He isn t. He s multiplying them. But if you want to add them, you d get 7 1/100 (seven and one hundredth).", + "video_name": "he4kcTujy30", + "timestamps": [ + 123 + ], + "3min_transcript": "Let's say I have the number 905.074. So how could I expand this out? And what does this actually represent? So let's just think about each of the place values here. The 9 right over here, this is in the hundreds place. This literally represents nine hundreds. So we could rewrite that 9 as nine hundreds. Let me write it two ways. We could write it as 900, which is the same thing as 9 times 100. Now, there's a 0. That's just going to represent zero tens. But zero tens is still just 0. So we don't have to really worry about that. It's not adding any value to our expression or to our number. Now we have this 5. This 5 is in the ones place. It literally represents five ones, or you could just say it represents 5. Now, if we wanted to write it as five ones, we could say well, that's going to be 5 times 1. 100 plus 5 times 1. And you might say hey, how do I know whether I should multiply or add first? Should I do this addition before I do this multiplication? And I'll always remind you, order of operations. In this scenario, you would do your multiplication before you do your addition. So you would multiply your 5 times 1 and your 9 times 100 before adding these two things together. But let's move on. You have another 0. This 0 is in the tenths place. This is telling us the number of tenths we're going to have. This is zero tenths, so it's really not adding much, or it's not adding anything. Now we go to the hundredths place. So this literally represents seven hundredths. So we could write this as 7/100, or 7 times 1/100. So we go to the thousandths place. And we have four thousandths. So that literally represents 4 over 1,000, or 4 times 1/1000. Notice this is coming from the hundreds place. You have zero tens, but I'll write the tens place there just so you see it. So it's zero tens, so I didn't even bother to write that down. Then you have your ones place. You have five ones. Then you have zero tenths. So I didn't write that down. Then you have seven hundredths and then you have four thousandths. We've written this out, really just understanding what this number represents.", + "qid": "he4kcTujy30_123" + }, + { + "Q": "at 0:49 isnt it the other way because it would be 1.5 in real life problem", + "A": "Do you mean 1*5? 1.5 is a decimal number for 1 1/2. Any dot used for multiplication must be raised. Anyway... I think Sal is using the bar under each number to separate the digit from its place value. I don t believe he is using it for division if that is what you were thinking.", + "video_name": "BItpeFXC4vA", + "timestamps": [ + 49 + ], + "3min_transcript": "So I have a number written here. It's a 2, a 3, and a 5. And we already have some experience with numbers like this. We can think about 'what does it represent'. And to think about that we just have to look at the actual place values. So this right-most place right over here. This is the ones place. So this 5 represents five ones, or I guess you could say that's just going to be 5. This 3, this is in the tens place. This is the tens place, so we have three tens. So that's just going to be 30. And the 2 is in the hundreds place. So putting a 2 there means that we have two hundreds. So this number we can view as two hundred, thirty, five. Or you could view it as two hundred plus thirty plus five. Now what I want to do in this video is think about place values to the right of the ones place. And you might say 'wait, wait, I always thought that the ones place was the place furthest to the right.' Well everything that we've done so far, it has been. But to show that you can go even further to the right We call that a 'decimal point'. And that dot means that anything to the right of this is going to be place values that are smaller, I guess you could say, than the ones place. So right to the left you have the ones place and the tens place and the hundreds place, and if you were to keep going you'd go to the thousands place and the ten thousands place. But then if you go to the right of the decimal point now you're going to divide by 10. So what am I talking about? Well, right to the right of the decimal point you are going to have-- find a new color-- this is going to be the tenths place. Well what does that mean? Well whatever number I write here that tells us how many tenths we're dealing with. So if I were to write the number 4 right over here, now my number is 2 hundreds plus 3 tens plus 5 ones plus 4 tenths. Or you could write this as 4 tenths. Not tens, 4 tenths. Or 4 tenths is the same thing as this right over here. So this is a super important idea in mathematics. I can now use our place values to represent fractions. So this right over here, this 'point 4', this is 4/10. So another way to write this number-- I could write it this way, I could write it as two hundred, thirty-- let me do the thirty in blue-- two hundred and thirty five and four tenths. So I could write it like this, as a mixed number. So this up here would be a decimal representation: 235.4 And this right over here would be a mixed number representation: 235 and 4/10 but they all represent 200 plus 30 plus 5 plus 4/10.", + "qid": "BItpeFXC4vA_49" + }, + { + "Q": "Instead of using his \"aside\" with the triangle beginning at 1:49, I did something similar but using x/3 instead of 'S'. In other words, I said:\nUsing the Pythagorean Theorem: ((x/3)/2)^2+h^2 = (x/3)^2\nh^2 = (x/3)^2-(x/6)^2\nh = x/3-x/6\nh=x/6\n\nObviously this differs from Sal's answer but I can't figure out why?", + "A": "Your mistake was when you square rooted, \u00e2\u0088\u009a(a\u00c2\u00b2+b\u00c2\u00b2) is NOT \u00e2\u0088\u009aa\u00c2\u00b2 + \u00e2\u0088\u009ab\u00c2\u00b2 and is not a+b Here is the correct way to do it: h\u00c2\u00b2 = (\u00e2\u0085\u0093x)\u00c2\u00b2- (\u00e2\u0085\u0099x)\u00c2\u00b2 h\u00c2\u00b2 = \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0089 x\u00c2\u00b2 - \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0083\u00e2\u0082\u0086 x\u00c2\u00b2 h\u00c2\u00b2 = \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0081\u00e2\u0082\u0082 x\u00c2\u00b2 h = x / \u00e2\u0088\u009a12 h = x / (2\u00e2\u0088\u009a3) h = \u00e2\u0085\u0099 x\u00e2\u0088\u009a3", + "video_name": "IFU7Go6Qg6E", + "timestamps": [ + 109 + ], + "3min_transcript": "Let's say that I have a 100 meter long wire. So that is my wire right over there. And it is 100 meters. And I'm going to make a cut someplace on this wire. And so let's say I make the cut right over there. With the left section of wire-- I'm going to obviously cut it in two-- with the left section, I'm going to construct an equilateral triangle. And with the right section, I'm going to construct a square. And my question for you and for me is, where do we make this cut in order to minimize the combined areas of this triangle and this square? Well, let's figure out. Let's define a variable that we're trying to minimize, or that we're trying to optimize with respect to. So let's say that the variable x is the number of meters that we decide to cut from the left. So if we did that, then this length for the triangle would be, well, if we use x up for the left hand side, we're going to have 100 minus x for the right hand side. And so what would the dimensions of the triangle and the square Well, the triangle sides are going to be x over 3, x over 3, and x over 3 as an equilateral triangle. And the square is going to be 100 minus x over 4 by 100 minus x over 4. Now it's easy to figure out an expression for the area of the square in terms of x. But let's think about what the area of an equilateral triangle might be as a function of the length of its sides. So let me do a little bit of an aside right over here. So let's say we have an equilateral triangle. Just like that. And its sides are length s, s, and s. is 1/2 times the base times the height. So in this case, the height we could consider to be altitude, if we were to drop an altitude just like this. This length right over here, this is the height. And this would be perpendicular, just like that. So our area is going to be equal to one half times our base is s. 1/2 times s times whatever our height is, times our height. Now how can we express h as a function of s? Well, to do that we just have to remind ourselves that what we've drawn over here is a right triangle. It's the left half of this equilateral triangle. And we know what this bottom side of this right triangle is. This altitude splits this side exactly into two.", + "qid": "IFU7Go6Qg6E_109" + }, + { + "Q": "At 4:28, what is a basis?", + "A": "At 4:28, A basis is a System which consist of the MINIMAL amount of VECTORS which are needed to define a room (space, coordinate system). All VECTORS in from a BASIS must be liear dependant from eachother.", + "video_name": "C2PC9185gIw", + "timestamps": [ + 268 + ], + "3min_transcript": "We're just assuming that A has at least n linearly independent eigenvectors. In general, you could take scaled up versions of these and they'll also be eigenvectors. Let's see, so the transformation of vn is going to be equal to A times vn. And because these are all eigenvectors, A times vn is just going to be lambda n, some eigenvalue times the vector, vn. Now, what are these also equal to? Well, this is equal to, and this is probably going to be unbelievably obvious to you, but this is the same thing as lambda 1 times vn plus 0 times v2 plus all the way to 0 times vn. And this right here is going to be 0 times v1 plus lambda 2 times v2 plus all the way, 0 times all of the other vectors vn. And then this guy down here, this is going to be 0 times v1 these eigenvectors, but lambda n times vn. This is almost stunningly obvious, right? I just rewrote this as this plus a bunch of zero vectors. But the reason why I wrote that is, because in a second, we're going to take this as a basis and we're going to find coordinates with respect to that basis, and so this guy's coordinates will be lambda 1, 0, 0, because that's the coefficients on our basis vectors. So let's do that. So let's say that we define this as some basis. So B is equal to the set of-- actually, I don't even have to write it that way. Let's say I say that B, I have some basis B, that's equal to that. What I want to show you is that when I do a change of basis-- we've seen this before-- in my standard coordinates or in coordinates with respect to the standard basis, you give me some vector in Rn, I'm going to multiply it times A, and you're going to have the It's also going to be in Rn. Now, we know we can do a change of basis. And in a change of basis, if you want to go that way, you multiply by C inverse, which is-- remember, the change of basis matrix C, if you want to go in this direction, you multiply by C. The change of basis matrix is just a matrix with all of these vectors as columns. It's very easy to construct. But if you change your basis from x to our new basis, you multiply it by the inverse of that. We've seen that multiple times. If they're all orthonormal, then this is the same thing as We can't assume that, though. And so this is going to be x in our new basis. And if we want to find some transformation, if we want to find the transformation matrix for T with respect to our new basis, it's going to be some matrix D. And if you multiply D times x, you're going to get this guy, but you're going to get the B representation of that guy. The transformation of the vector x is B representation.", + "qid": "C2PC9185gIw_268" + }, + { + "Q": "Isn't a line segment that has no length a point? Refering to 4:50", + "A": "Yes. A point really has no size, and since lines have no width, if a line also had no length, it would not be a line, it would be a point.", + "video_name": "Oc8sWN_jNF4", + "timestamps": [ + 290 + ], + "3min_transcript": "If it were a regular line, not an infinitely spiked one, scaling up by three would make it three times as much drawing, as expected. But if that spiky line were supposed to represent an infinitely spiked magical fortress city of dragon dungeon doom, by scaling it up in this way, you'd be losing details, making these long lines that should have had spiky bumps in them. Theoretically, no matter how much you scale up the city or no matter how finely you look at it, you'll never get any flat sections. This whole thing scaled down is the same as this section, which is the same as this section, which is the same as this. Three times as big is four times as much stuff. Not three, like if it were a normal 1D line, and certainly not nine, like that 2D area on the inside. Somehow, the infinity fractal-ness of the thing makes it behave differently from all 1D things and all 2D things. You convince yourself that all 1D things got twice as big when you make them twice as big, because you could think And you know how line segments behave. And you convince yourself all 2D things scaled up by two get four times as much stuff, because 2D things can be thought of as being mad of squares, and you know how squares behave. But then there is this which has no straight lines in it. And there's no square areas in it, either. More than three to the one, less than three to the two. It behaves as if it's between one and two dimensions. You think back to Sierpinski's triangle. Maybe it can be thought of as being made out of straight line segments, though there's an infinite amount of them and they get infinitely small. When you make it twice as tall, if you just make all the lines of this drawing twice as long, you're missing detail again. But the tiny lines too small to draw are also twice as long and now visible. And so on, all the way down to the infinitely small line segments. You wonder if your similar line thing works on lines that don't actually have length. Wait. Lines that don't have length? Is that a thing? First, though, you figure out that when you make it twice as Not two, like a 1D triangle outline. Not four, like a solid 2D triangle. But somewhere in between. And the in between-ness seems to be true, no matter which way you make it-- out of lines, or by subtracting 2D triangles, or with squiggles. They all end up the same. An object in fractional dimension. No longer 1D because of infinity infinitely small lines. Or no longer 2D because of subtracting out all the area. Or being an infinitely squiggled up line that's too infinate and squiggled to be a line anymore, but doesn't snuggle into itself enough to have any 2D area, either. Though in the dragon curve, it does seem to snuggle up into itself. Hm. If you pretend this is the complete dragon curve and iterate this way, there's twice as much stuff. That's what you'd expect from a 1D line if it were scaling it by two. But let's see, this is scaling up by, well, not quite two. Let's see. I suppose if you did it perfectly, it's supposed to be an equilateral right triangle. So square root 2. If it were two dimensional, you'd", + "qid": "Oc8sWN_jNF4_290" + }, + { + "Q": "At 5:00, Vi mentions a line that has no length. Wouldn't that be resembling a point? A point has no length whatsoever, or width, hence only trapped in 1st dimension. So does a potential line with no length. So... that means it's a case of a=b and b=c so a=c, right?", + "A": "A point is zero-D it has no length", + "video_name": "Oc8sWN_jNF4", + "timestamps": [ + 300 + ], + "3min_transcript": "If it were a regular line, not an infinitely spiked one, scaling up by three would make it three times as much drawing, as expected. But if that spiky line were supposed to represent an infinitely spiked magical fortress city of dragon dungeon doom, by scaling it up in this way, you'd be losing details, making these long lines that should have had spiky bumps in them. Theoretically, no matter how much you scale up the city or no matter how finely you look at it, you'll never get any flat sections. This whole thing scaled down is the same as this section, which is the same as this section, which is the same as this. Three times as big is four times as much stuff. Not three, like if it were a normal 1D line, and certainly not nine, like that 2D area on the inside. Somehow, the infinity fractal-ness of the thing makes it behave differently from all 1D things and all 2D things. You convince yourself that all 1D things got twice as big when you make them twice as big, because you could think And you know how line segments behave. And you convince yourself all 2D things scaled up by two get four times as much stuff, because 2D things can be thought of as being mad of squares, and you know how squares behave. But then there is this which has no straight lines in it. And there's no square areas in it, either. More than three to the one, less than three to the two. It behaves as if it's between one and two dimensions. You think back to Sierpinski's triangle. Maybe it can be thought of as being made out of straight line segments, though there's an infinite amount of them and they get infinitely small. When you make it twice as tall, if you just make all the lines of this drawing twice as long, you're missing detail again. But the tiny lines too small to draw are also twice as long and now visible. And so on, all the way down to the infinitely small line segments. You wonder if your similar line thing works on lines that don't actually have length. Wait. Lines that don't have length? Is that a thing? First, though, you figure out that when you make it twice as Not two, like a 1D triangle outline. Not four, like a solid 2D triangle. But somewhere in between. And the in between-ness seems to be true, no matter which way you make it-- out of lines, or by subtracting 2D triangles, or with squiggles. They all end up the same. An object in fractional dimension. No longer 1D because of infinity infinitely small lines. Or no longer 2D because of subtracting out all the area. Or being an infinitely squiggled up line that's too infinate and squiggled to be a line anymore, but doesn't snuggle into itself enough to have any 2D area, either. Though in the dragon curve, it does seem to snuggle up into itself. Hm. If you pretend this is the complete dragon curve and iterate this way, there's twice as much stuff. That's what you'd expect from a 1D line if it were scaling it by two. But let's see, this is scaling up by, well, not quite two. Let's see. I suppose if you did it perfectly, it's supposed to be an equilateral right triangle. So square root 2. If it were two dimensional, you'd", + "qid": "Oc8sWN_jNF4_300" + }, + { + "Q": "At 5:17, how did you get those numbers if young are adding 2, then, then 6.", + "A": "Each number is 2 more than the previous number. If x is the first number, then the 2nd number is x+2. The third number is 2 more than the previous number, which is the 2nd number, which also is x+2. So the 3rd number is (x+2)+2 = x+3. The 4th number is the 3rd + 2, which is (x+4)+2 = x+6", + "video_name": "8CJ6Qdcoxsc", + "timestamps": [ + 317 + ], + "3min_transcript": "So we can rewrite those literally as 4x. And then we have 2 plus 4, which is 6, plus another 6 is 12. 4x plus 12 is equal to 136. So to solve for x, a good starting point would be to just to isolate the x terms on one side of the equation or try to get rid of this 12. Well to get rid of that 12, we'd want to subtract 12 from the left-hand side. But we can't just do it from the left-hand side. Then this equality wouldn't hold anymore. If these two things were equal before subtracting the 12, well then if we want to keep them equal, if we want the left and the right to stay equal, we've got to subtract 12 from both sides. So subtracting 12 from both sides gives us, well on the left-hand side, we're just left with 4x. And on the right-hand side, we are left with 136 minus 12 is 124. Yeah, 124. Well, we just divide both sides by 4 to solve for x. And we get-- do that in the same, original color-- x is equal to 124 divided by 4. 100 divided by 4 is 25. 24 divided by 4 is 6. 25 plus 6 is 31. And if you don't feel like doing that in your head, you could also, of course, do traditional long division. Goes into 124-- 4 doesn't go into 1. 4 goes into 12 three times. 3 times 4 is 12. You subtract, bring down the next 4. 4 goes into 4 one time. You get no remainder. So x is equal to 31. So x is the smallest of the four integers. So this right over here, x is 31. x plus 2 is going to be 33. And x plus 6 is going to be 37. So our four consecutive odd integers are 31, 33, 35, and 37.", + "qid": "8CJ6Qdcoxsc_317" + }, + { + "Q": "I don't understand what he says at 6:09, please explain!", + "A": "The length of one cycle of the standard cosine function is 2pi. So we need to ask ourselves: How does 2pi compare to the length of one cycle in the problem? That s why we set up the ratio of 2pi to 365, or written in fractional terms, 2pi/365.", + "video_name": "mVlCXkht6hg", + "timestamps": [ + 369 + ], + "3min_transcript": "either actual degrees or radians, which trigonometric function starts at your maximum point? Well cosine of zero is one. The cosine starts at your maximum point. Sine of zero is zero, so I'm going to use cosine here. I'm going to use a cosine function. So, temperature as a function of days. There's going to be some amplitude times our cosine function and we're going to have some argument to our cosine function and then I'm probably going to have to shift it. So let's think about how we would do that. Well, what's the mid line here? The mid line is the halfway point between our high and our low. So our midpoint, if we were to visualize it, looks just like so. That is our mid line right over there. And what value is this? Well what's the average of 29 and 14? 29 plus 14 is 43 divided by two is 21.5 degrees Celsius. shifted up our function by that amount. If we just had a regular cosine function our mid line would be at zero, but now we're at 21.5 degrees Celsius. I'll just write plus 21.5, that's how much we've shifted it up. Now, what's the amplitude? Well our amplitude is how much we diverge from the mid line. Over here we're 7.5 above the mid line so that's plus 7.5. Here we're 7.5 below the mid line, so minus 7.5. So our amplitude is 7.5, the maximum amount we go away from the mid line is 7.5. So that's our amplitude. And now let's think about our argument to the cosine function right over here. It's going to be a function of the days. And what do we want? When 365 days have gone by, we want this entire argument to be two pi. whole thing to evaluate to two pi. We could put two pi over 365 in here. You might remember your formulas, I always forget them that's why I always try to reason through them again. The formulas, you want two pi divided by your period and all the rest, but I just like to think, \"Okay, look. \"After one period, which is 365 days, I want the whole \"argument over here to be two pi. \"I want to go around the unit circle once and so if this \"is two pi over 365, when you multiply it by 365 \"your argument here is going to be two pi.\" Just like that we've done the first part of this question. We have modeled the average high temperature in Santiago as a function of days after January seventh. In the next video we'll answer this second question. I encourage you to do it ahead of time before watching that next video and I'll give you one clue.", + "qid": "mVlCXkht6hg_369" + }, + { + "Q": "At 2:28 why does -24x become -25x when you subtract x", + "A": "Because a negative minus a positive is basically adding a negative to a negative.", + "video_name": "711pdW8TbbY", + "timestamps": [ + 148 + ], + "3min_transcript": "In this video, we're going to start having some experience solving radical equations or equations that involve square roots or maybe even higher-power roots, but we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's say I have the equation the square root of x is equal to 2 times x minus 6. Now, one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that, I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8.", + "qid": "711pdW8TbbY_148" + }, + { + "Q": "At 1:34, you did some factoring in your head...hard to follow to get the 24x. Could you expound upon that? I know - watch the factoring videos - but that one lost me. It wasn't quite clear WHY you multiply the 2x and the 6 to get the 24.", + "A": "Because the square factor form is as following: (ax-b)^2 ax-b x ax-b _________ ax^2 -abx -bax +b^2 Focus in the part (-abx -bax), since multiplication scalar is commutative, they are twice themselves. Rewrite them in order, (-abx -abx), which is equal to 2(-abx). If a = 2 and b = 6, representing them in the expression 2(-2*6x), 2(-12x) that is equal to -24x.", + "video_name": "711pdW8TbbY", + "timestamps": [ + 94 + ], + "3min_transcript": "In this video, we're going to start having some experience solving radical equations or equations that involve square roots or maybe even higher-power roots, but we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's say I have the equation the square root of x is equal to 2 times x minus 6. Now, one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that, I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8.", + "qid": "711pdW8TbbY_94" + }, + { + "Q": "Around 4:50 when Sal is subtracting \"all of this business\", i.e. -e^xcosx+Se^x cosx dx, why does the plus sign become a minus sign? Its e^x sinx - -e^xcosx... so that becomes plus. But why does the plus sign before the antiderivative become a minus sign??", + "A": "When he was taking e^x(sin(x)) - \u00e2\u0088\u00ab e^x(sin(x))dx, the minus sign acts as a negative sign so it would end up something like this in his thinking: e^x(sin(x)) + [-( \u00e2\u0088\u00ab e^x(sin(x))dx)]. You would take the integral of e^x(sin(x))dx to get -e^xcos(x) + \u00e2\u0088\u00ab (e^xcos(x))dx and factor the negative into the whole equation to get: e^x(sin(x)) + e^xcos(x) - \u00e2\u0088\u00ab (e^xcos(x))dx", + "video_name": "LJqNdG6Y2cM", + "timestamps": [ + 290 + ], + "3min_transcript": "prime of xg of x. F prime of x is e to x. And then g of x is negative cosine of x. So I'll put the cosine of x right over here, and then the negative, we can take it out of the integral sign. And so we're subtracting a negative. That becomes a positive. And of course, we have our dx right over there. And you might say, Sal, we're not making any progress. This thing right over here, we now expressed in terms of an integral that was our original integral. We've come back full circle. But let's try to do something interesting. Let's substitute back this-- all right, let me write it this way. Let's substitute back this thing up here. Let's substitute this for this in our original equation. And let's see if we got anything interesting. So what we'll get is our original integral, on the left hand side here. The indefinite integral or the antiderivative of e to the x cosine of x dx is equal to e to the x sine of x, minus all of this business. So let's just subtract all of this business. We're subtracting all of this. So if you subtract negative e to the x cosine of x, it's going to be positive. It's going to be positive e to the x, cosine of x. And then remember, we're subtracting all of this. So then we're going to subtract. So then we have minus the antiderivative of e to the x, Now this is interesting. Just remember all we did is, we took this part right over here. We said, we used integration by parts to figure out that it's the same thing as this. So we substituted this back in. When you subtracted it. When you subtracted this from this, we got this business right over here. Now what's interesting here is we have essentially an equation where we have our expression, our original expression, twice. We could even assign this to a variable and essentially solve for that variable. So why don't we just add this thing to both sides of the equation? Let's just add the integral of e to the x cosine of x dx to both sides. e to the x, cosine of x, dx. And what do you get? Well, on the left hand side, you have two times our original integral. e to the x, cosine of x, dx is equal to all of this business.", + "qid": "LJqNdG6Y2cM_290" + }, + { + "Q": "Why didn't Sal explain why -8/7 is the same that -(8/7) at 8:30. It isn't so simple to understand.", + "A": "just think about it. -8/7 is equal to 8/-7 and also equal to the negative value of 8/7. so -8/7 = 8/-7 = -(8/7)", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 510 + ], + "3min_transcript": "", + "qid": "XoEn1LfVoTo_510" + }, + { + "Q": "At 4:14, EX1. y/2-8=10;", + "A": "y/2 - 8 = 10 Add 8 to both sides y/2 = 18 Multiply both sides by 2 y = 36", + "video_name": "XoEn1LfVoTo", + "timestamps": [ + 254 + ], + "3min_transcript": "", + "qid": "XoEn1LfVoTo_254" + }, + { + "Q": "where does he get 104 at 1:26? 80-24? where does he even get that.", + "A": "to add or subtract the fractions like 85/13 and 8/1 are we take the common denominators of both of the fractions and to maintain the equality the changes which we make in the denominator like to make the denominator of fraction 8/1 we multiplied the denominator by 13 so we do the same thing in the numerator thats how sal got 104", + "video_name": "KV_XLL4K2Fw", + "timestamps": [ + 86 + ], + "3min_transcript": "The following line passes through the point 5 comma 8, and the equation of the line is y is equal to 17/13x plus b. What is the value of the y-intercept b? So we know that this point, this x and y value must satisfy this equation, so we know that when x is equal to 5, y is equal to 8. So we can say-- so when x is equal to 5, y is equal to 8. So we can say 8 must be equal to 17/13 times x times 5 plus b, and then we can solve for b. So if we simplify this a little bit, we get 8 is equal to-- let's see, 5 times 17 is 50, plus 35 is 85-- is 85/13 plus b. Then to solve for b, we just subtract 85/13 from both sides. 85/13 Is equal to b. And now we just have to subtract these two numbers. So 8 is the same thing as-- let's see. 80 plus 24 is 104, so it is 104/13-- this is the same thing as 8-- minus 85/13, which is going to be-- let's see. This is 19/13. Is that right? Yes, if this was 105, then it would be 20/13. So this is 19/13, so it's equal to 19/13, which is equal to b. So the equation of this line is going to be y is equal to 17/13x plus 19/13.", + "qid": "KV_XLL4K2Fw_86" + }, + { + "Q": "Around 9:50 he mentions that v3 is a linear combination because it is v1 and v2 added together which gives a vector in between the angle of the two. If I took v1 - v2 is it still a linear combination because it is outside the angle? Or is it still a linear combination as it is in R^2", + "A": "It is a linear combination: the constant you multiplied v2 by was -1.", + "video_name": "CrV1xCWdY-g", + "timestamps": [ + 590 + ], + "3min_transcript": "call this vector 2, is equal to vector 3. So vector 3 is a linear combination of these other two vectors. So this is a linearly dependent set. And if we were to show it, draw it in kind of two space, and it's just a general idea that-- well, let me see. Let me draw it in R2. There's a general idea that if you have three two-dimensional vectors, one of them is going to be redundant. Well, one of them definitely will be redundant. For example, if we do 2, 3, if we do the vector 2, 3, that's the first one right there. I draw it in the standard position. And I draw the vector 7, 2 right there, I could show you combination of these two vectors. We can even do a kind of a graphical representation. I've done that in the previous video, so I could write that the span of v1 and v2 is equal to R2. That means that every vector, every position here can be represented by some linear combination of these two guys. Now, the vector 9, 5, it is in R2. It is in R2, right? Clearly. I just graphed it on this plane. It's in our two-dimensional, real number space. Or I guess we could call it a space or in our set R2. It's right there. So we just said that anything in R2 can be represented by a linear combination of those two guys. So clearly, this is in R2, so it can be represented as a linear combination. So hopefully, you're starting to see the relationship between span and linear independence or linear dependence. Let me do another example. Let's say I have the vectors-- let me do a new color. Let's say I have the vector-- and this one will be a little bit obvious-- 7, 0, so that's my v1, and then I have my second vector, which is 0, minus 1. That's v2. Now, is this set linearly independent? Is it linearly independent? Well, can I represent either of these as a combination of the other? And really when I say as a combination, you'd have to scale up one to get the other, because there's only two vectors here. If I am trying to add up to this vector, the only thing I", + "qid": "CrV1xCWdY-g_590" + }, + { + "Q": "At 7:45 when Sal asks if the vectors are dependent or independent, don't we know that they can't be independent solely based on the fact that they are 3 vectors that are only written in two dimensions (a 2x1 matrix)?", + "A": "Yes, you can say that. But Sal hasn t proved that that is the case yet, and he is just trying to introduce linear dependence right now.", + "video_name": "CrV1xCWdY-g", + "timestamps": [ + 465 + ], + "3min_transcript": "Anything in this plane going in any direction can be-- any vector in this plane, when we say span it, that means that any vector can be represented by a linear combination of this vector and this vector, which means that if this vector is on that plane, it can be represented as a linear combination of that vector and that vector. So this green vector I added isn't going to add anything to the span of our set of vectors and that's because this is a linearly dependent set. This one can be represented by a sum of that one and that one because this one and this one span this plane. In order for the span of these three vectors to kind of get more dimensionality or start representing R3, the third vector will have to break out of that plane. It would have to break out of that plane. And if a vector is breaking out of that plane, that means it's a vector that can't be represented anywhere on that Where it's outside, it can't be represented by a linear combination of this one and this one. So if you had a vector of this one, this one, and this one, and just those three, none of these other things that I drew, that would be linearly independent. Let me draw a couple more examples for you. That one might have been a little too abstract. So, for example, if I have the vectors 2, 3 and I have the vector 7, 2, and I have the vector 9, 5, and I were to ask you, are these linearly dependent or independent? So at first you say, well, you know, it's not trivial. Let's see, this isn't a scalar multiple of that. That doesn't look like a scalar multiple of either of Maybe they're linearly independent. But then, if you kind of inspect them, you kind of see call this vector 2, is equal to vector 3. So vector 3 is a linear combination of these other two vectors. So this is a linearly dependent set. And if we were to show it, draw it in kind of two space, and it's just a general idea that-- well, let me see. Let me draw it in R2. There's a general idea that if you have three two-dimensional vectors, one of them is going to be redundant. Well, one of them definitely will be redundant. For example, if we do 2, 3, if we do the vector 2, 3, that's the first one right there. I draw it in the standard position. And I draw the vector 7, 2 right there, I could show you", + "qid": "CrV1xCWdY-g_465" + }, + { + "Q": "I'm confused with these factorials (!) In the problem around 6:12 how does 5!/4!=5?", + "A": "5! = 5\u00e2\u0080\u00a24\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 4! = 4\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 5! 5\u00e2\u0080\u00a24\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 \u00e2\u0094\u0080\u00e2\u0094\u0080 = \u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080\u00e2\u0094\u0080 4! 4\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 5! \u00e2\u0094\u0080\u00e2\u0094\u0080 = 5 4!", + "video_name": "WWv0RUxDfbs", + "timestamps": [ + 372 + ], + "3min_transcript": "Well actually over zero factorial times five minus zero factorial. Well zero factorial is one, by definition, so this is going to be five factorial, over five factorial, which is going to be equal to one. Once again I like reasoning through it instead of blindly applying a formula, but I just wanted to show you that these two ideas are consistent. Let's keep going. I'm going to do x equals one all the way up to x equals five. If you are inspired, and I encourage you to be inspired, try to fill out the whole thing, what's the probability that x equals one, two, three, four or five. So let's go to the probability that x equals two. Or sorry, that x equals one. The probability that x equals one is going to be equal to... Well how do you get one head? It could be, the first one could be head The second one could be head and then the rest of them are gonna be tails. I could write them all out but you can see that there's five different places to have that one head. So five out of the 32 equally likely outcomes involve one head. Let me write that down. This is going to be equal to five out of 32 equally likely outcomes. Which of course is the same thing, this is going to be the same thing as saying I got five flips, and I'm choosing one of them to be heads. So that over 32. You could verify that five factorial over one factorial times five minus-- Actually let me just do it just so that you don't have to take my word for it. So five choose one is equal to five factorial over one factorial, which is just one, times five minus four-- Sorry, five minus one factorial. which is just going to be equal to five. All right, we're making good progress. Now in purple let's think about the probability that our random variable x is equal to two. Well this is going to be equal to, and now I'll actually resort to the combinatorics. You have five flips and you're choosing two of them to be heads. Over 32 equally likely possibilities. This is the number of possibilities that result in two heads. Two of the five flips have chosen to be heads, I guess you can think of it that way, by the random gods, or whatever you want to say. This is the fraction of the 32 equally likely possibilities, so this is the probability that x equals two. What's this going to be? I'll do it right over here. And actually no reason for me to have to keep switching colors.", + "qid": "WWv0RUxDfbs_372" + }, + { + "Q": "@0:34\n\nI understand that it doesn't produce the correct answer, in fact it seems to produce the reciprocal of the correct answer, but why doesn't it work to multiply the left and right sides by the fraction x/x (which would be equal to 1) which would leave us with 10x on the left and 15x on the right?\n\nI'm sure that I'm missing something simple, as usual.", + "A": "If you multiply both sides by x/x, here s what you would get: 7x/x - 10x/x^2 = 2x/x + 15x/x^2 This just gives you a much more complicated equation to try and solve. You need to multiply by x to eliminate the fractions and get the variable into the numerator.", + "video_name": "Z7C69xP08d8", + "timestamps": [ + 34 + ], + "3min_transcript": "So I have the equation 7 minus 10/x is equal to 2 plus 15/x. And so this isn't the type of equation that you might think that you're used to solving. But I'll give you a few moments to see if you can solve it on your own. Well, what we'll see is we can do a quick multiplication of both sides to actually simplify this to a form that we are more used to looking at. So what's probably bothering you, because it's bothering me, is these x's that we have in the denominators right over here. We're like, well, how do we deal with that? Well, whenever we see an x in the denominator, the temptation is to multiply it by x. But we can't just multiply one of the terms by x. We have to multiply the entire side by x. So we could multiply this entire side by x. But we can't just multiply the left-hand side by x. We'd also want to multiply the right-hand side by x. And so what will that give us? Well, we distribute the x. We get x times 7 is 7x. And then x times negative 10/x, well, So you get negative 10 right over there. So the left-hand side simplifies to 7x minus 10. And then your right-hand side, once again, distribute the x. x times 2 is 2x. x times 15/x, well, x times something over x is just going to be the something. x times 15/x is just going to be 15-- plus 15. So now we've simplified this to a linear equation. We have the variable on both sides. So we just have to do some of the techniques that we already know. So the first thing that I like to do is maybe get all my x's on the left-hand side. So I want to get rid of this 2x right over here. So I subtract 2x from the right-hand side. Now, and I always remind you, I can't do that just to the right-hand side. If I did it just to the right-hand side, it wouldn't be an equality anymore. You have to do that to the left-hand side as well. And so we are left with-- let me get that pink color again. well, you're going to have 5 of that something, minus 10. These two x's negate each other. And you're left with equals 15. Now we can get rid of this negative 10 by adding 10 to both sides. You know, I like that green color when I do stuff to both sides. So I can add 10 to both sides. And I'm left with 5x-- these negate each other-- is equal to 25. And this is the home stretch. You see where this is going. We can divide both sides by 5. And we are left with x is equal to 5. Now let's verify that this actually worked. So let's go back to the original equation. We have 7 minus 10/5. This needs to be equal to-- I'm just taking our 5 and substituting it back here. This needs to be equal to 2 plus 15/5.", + "qid": "Z7C69xP08d8_34" + }, + { + "Q": "At 2:30, Sal crossed out -10 & +10. He said \"these negate each other\". What does negate mean?", + "A": "It means they cancel each other out to make 0.", + "video_name": "Z7C69xP08d8", + "timestamps": [ + 150 + ], + "3min_transcript": "So I have the equation 7 minus 10/x is equal to 2 plus 15/x. And so this isn't the type of equation that you might think that you're used to solving. But I'll give you a few moments to see if you can solve it on your own. Well, what we'll see is we can do a quick multiplication of both sides to actually simplify this to a form that we are more used to looking at. So what's probably bothering you, because it's bothering me, is these x's that we have in the denominators right over here. We're like, well, how do we deal with that? Well, whenever we see an x in the denominator, the temptation is to multiply it by x. But we can't just multiply one of the terms by x. We have to multiply the entire side by x. So we could multiply this entire side by x. But we can't just multiply the left-hand side by x. We'd also want to multiply the right-hand side by x. And so what will that give us? Well, we distribute the x. We get x times 7 is 7x. And then x times negative 10/x, well, So you get negative 10 right over there. So the left-hand side simplifies to 7x minus 10. And then your right-hand side, once again, distribute the x. x times 2 is 2x. x times 15/x, well, x times something over x is just going to be the something. x times 15/x is just going to be 15-- plus 15. So now we've simplified this to a linear equation. We have the variable on both sides. So we just have to do some of the techniques that we already know. So the first thing that I like to do is maybe get all my x's on the left-hand side. So I want to get rid of this 2x right over here. So I subtract 2x from the right-hand side. Now, and I always remind you, I can't do that just to the right-hand side. If I did it just to the right-hand side, it wouldn't be an equality anymore. You have to do that to the left-hand side as well. And so we are left with-- let me get that pink color again. well, you're going to have 5 of that something, minus 10. These two x's negate each other. And you're left with equals 15. Now we can get rid of this negative 10 by adding 10 to both sides. You know, I like that green color when I do stuff to both sides. So I can add 10 to both sides. And I'm left with 5x-- these negate each other-- is equal to 25. And this is the home stretch. You see where this is going. We can divide both sides by 5. And we are left with x is equal to 5. Now let's verify that this actually worked. So let's go back to the original equation. We have 7 minus 10/5. This needs to be equal to-- I'm just taking our 5 and substituting it back here. This needs to be equal to 2 plus 15/5.", + "qid": "Z7C69xP08d8_150" + }, + { + "Q": "What is that weird N-shaped symbol that Sal drew at 2:32?\nI assume it's some sort of symbol meaning and.", + "A": "The \u00e2\u0088\u00a9 symbol Sal wrote in 2:34 stands for intersection, which you have probably encountered in basic statistics. For example, if you let X and Y be arbitrary sets, X \u00e2\u0088\u00a9 Y would be classified as the set containing the elements that are in Set X AND Set Y.", + "video_name": "VjLEoo3hIoM", + "timestamps": [ + 152 + ], + "3min_transcript": "You say, well, this sounds like an interesting game. How much does it cost to play? And the guy tells you, it is $0.35 to play, so obviously a fairly low stakes casino. So my question to you is, would you want to play this game? And don't put the fun factor into it. Just economically, does it make sense for you to actually play this game? Well, let's think through the probabilities a little bit. So, first of all, what's the probability that the first marble you pick is green? Actually, let me just write, first green, probability first green. Well, the total possible outcomes-- there's 5 marbles here, all equally likely. So there's 5 possible outcomes. So there's a 3/5 probability that the first is green. So you have a 3/5 chance, 3/5 probability I should say, that after that first pick you're kind of still in the game. Now, what we really care about is your probability of winning the game. You want the first to be green, and the second green. Well, let's think about this a little bit. What is the probability that the first is green-- first, I'll just write g for green-- and the second is green? Now, you might be tempted to say, oh, well, maybe the second being green is the same probability. It's 3/5. I can just multiply 3/5 times 3/5, and I'll get 9/25, seems like a pretty straightforward thing. But the realization here is what you do with that first green marble. You don't take that first green marble out, look at it, and put it back in the bag. So when you take that second pick, depends on what you got on the first pick. Remember, we take the marble out. If it's a green marble, whatever marble it is, at whatever after the first pick, we leave it on the table. We are not replacing it. So there's not any replacement here. So these are not independent events. Let me make this clear, not independent. Or in particular, the second pick is dependent on the first. Dependent on the first pick. If the first pick is green, then you don't have 3 green marbles in a bag of 5. If the first pick is green, you now have 2 green marbles in a bag of 4. So the way that we would refer to this, the probability of both of these happening, yes, it's definitely equal to the probability of the first green times-- now, this is kind of the new idea--", + "qid": "VjLEoo3hIoM_152" + }, + { + "Q": "I can't understand the part where he creates the function for the first sequence at around 2:50. How did he come up with 1+3(k-1)", + "A": "k a_k 1 + 3\u00e2\u0080\u00a2(k - 1) 1 1 1 + 3\u00e2\u0080\u00a2(1 - 1) = 1 + 3\u00e2\u0080\u00a20 = 1 + 0 = 1 2 4 1 + 3\u00e2\u0080\u00a2(2 - 1) = 1 + 3\u00e2\u0080\u00a21 = 1 + 3 = 4 3 7 1 + 3\u00e2\u0080\u00a2(3 - 1) = 1 + 3\u00e2\u0080\u00a22 = 1 + 6 = 7 4 10 1 + 3\u00e2\u0080\u00a2(4 - 1) = 1 + 3\u00e2\u0080\u00a23 = 1 + 9 = 10", + "video_name": "KRFiAlo7t1E", + "timestamps": [ + 170 + ], + "3min_transcript": "Now, there's a bunch of different notations that seem fancy for denoting sequences. But this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. We could say that this right over here is the sequence a sub k for k is going from 1 to 4, is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture-- a sub 3. This right over here is a sub 4. So this just says, all of the a sub k's from k equals 1, from our first term, all the way to the fourth term. Now, I could also define it by not explicitly writing I could essentially do it defining our sequence as explicitly using kind of a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4, with-- instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see. When k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1. And we added 3 one less than the k term times. So we could say that this is going to be equal to 1 should write 3 times k minus 1-- same thing. And you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with kind of this function notation. I want to make it clear-- I have essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about. a of k is equal to 1 plus 3 times k minus 1.", + "qid": "KRFiAlo7t1E_170" + }, + { + "Q": "why did you go back 4 at time stamp 1:34?", + "A": "He didn t. You have the y-intercept graphed. From there you know that the slope is 4, so up 4 and right 1. But that will only let you graph to the right. To graph to the left, you need to go down 4 and left 1. The complete opposite.", + "video_name": "unSBFwK881s", + "timestamps": [ + 94 + ], + "3min_transcript": "Let's graph ourselves some inequalities. So let's say I had the inequality y is less than or equal to 4x plus 3. On our xy coordinate plane, we want to show all the x and y points that satisfy this condition right here. So a good starting point might be to break up this less than or equal to, because we know how to graph y is equal to 4x plus 3. So this thing is the same thing as y could be less than 4x plus 3, or y could be equal to 4x plus 3. That's what less than or equal means. It could be less than or equal. And the reason why I did that on this first example problem is because we know how to graph that. So let's graph that. Try to draw a little bit neater than that. So that is-- no, that's not good. So that is my vertical axis, my y-axis. And then we know the y-intercept, the y-intercept is 3. So the point 0, 3-- 1, 2, 3-- is on the line. And we know we have a slope of 4. Which means if we go 1 in the x-direction, we're going to go up 4 in the y. So 1, 2, 3, 4. So it's going to be right here. And that's enough to draw a line. We could even go back in the x-direction. If we go 1 back in the x-direction, we're going to go down 4. 1, 2, 3, 4. So that's also going to be a point on the line. So my best attempt at drawing this line is going to look something like-- this is the hardest part. It's going to look something like that. That is a line. It should be straight. I think you get the idea. That right there is the graph of y is equal to 4x plus 3. So all of these points satisfy this inequality, but we have more. This is just these points over here. What about all these where y ix less than 4x plus 3? So let's think about what this means. Let's pick up some values for x. When x is equal to 0, what does this say? When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3. When x is equal to negative 1, what is this telling us? 4 times negative 1 is negative 4, plus 3 is negative 1. y would be less than negative 1. When x is equal to 1, what is this telling us? 4 times 1 is 4, plus 3 is 7. So y is going to be less than 7. So let's at least try to plot these. So when x is equal to-- let's plot this one first. When x is", + "qid": "unSBFwK881s_94" + }, + { + "Q": "At 5:30 didn't he mean the equation would be y=-1/2x-6? He said it would be y=-1/2-6 and I was wonder where the variable would be. I'm not sure if I'm correct or not, please let me know.", + "A": "Yes, you re correct. Initially he forgot to add the x. About a minute afterwards, though, he saw his mistake and fixed it. :)", + "video_name": "unSBFwK881s", + "timestamps": [ + 330 + ], + "3min_transcript": "So it's all of these points here-- that I'm shading in in green-- satisfy that right there. If I were to look at this one over here, when x is negative 1, y is less than negative 1. So y has to be all of these points down here. When x is equal to 1, y is less than 7. So it's all of these points down here. And in general, you take any point x-- let's say you take this point x right there. If you evaluate 4x plus 3, you're going to get the point on the line. That is that x times 4 plus 3. Now the y's that satisfy it, it could be equal to that point on the line, or it could be less than. So if you were to do this for all the possible x's, you would not only get all the points on this line which we've drawn, you would get all the points below the line. So now we have graphed this inequality. It's essentially this line, 4x plus 3, with all of the area below it shaded. Now, if this was just a less than, not less than or equal sign, we would not include the actual line. And the convention to do that is to actually make the line a dashed line. This is the situation if we were dealing with just less than 4x plus 3. Because in that situation, this wouldn't apply, and we would just have that. So the line itself wouldn't have satisfied it, just the area below it. Let's do one like that. So let's say we have y is greater than negative x So a good way to start-- the way I like to start these problems-- is to just graph this equation right here. So let me just graph-- just for fun-- let me graph y is equal to-- this is the same thing as negative 1/2 minus 6. So if we were to graph it, that is my vertical axis, that is my horizontal axis. And our y-intercept is negative 6. So 1, 2, 3, 4, 5, 6. So that's my y-intercept. And my slope is negative 1/2. Oh, that should be an x there, negative 1/2 x minus 6. So my slope is negative 1/2, which means when I go 2 to the right, I go down 1. So if I go 2 to the right, I'm going to go down 1.", + "qid": "unSBFwK881s_330" + }, + { + "Q": "At 7:42, Sal mentions that we could of taken the Absolute value of the difference between the measurements and the mean instead of squaring them. Why don't we do that, it seems easier?", + "A": "A few reasons. 1. The absolute value function is much harder to deal with mathematically, because the derivative isn t nearly so nice as that of the square function. 2a. Squaring works very well with the Normal distribution. 2b. The sample mean is a natural estimate of location/center, and the Sampling Distribution of the sample mean is Normal, so we d like to use that. Hence, item 2a.", + "video_name": "PWiWkqHmum0", + "timestamps": [ + 462 + ], + "3min_transcript": "And in this case, what's it going to be? It's going to be the square root of 0.316. And then, what are the units going to be? It's going to be just meters. And we end up with-- so let me take the square root of 0.316. And I get 0.56-- I'll just round to the nearest thousandth-- 0.562. So this is approximately 0.562 meters. So you might be saying, Sal, what do we call this thing that we just did? The square root of the variance. And here we're dealing with the population. We haven't thought about sampling yet. The square root of the population variance, what do we call this thing right over here? And this is a very familiar term. Oftentimes, when you take an exam, this is calculated for the scores on the exam. I'm using that yellow a little bit too much. This is the population standard deviation. It is a measure of how much the data is varying from the mean. In general, the larger this value, that means that the data is more varied from the population mean. The smaller, it's less varied. And these are all somewhat arbitrary definitions of how we've defined variance. We could have taken things to the fourth power. We could have done other things. We could have not taken them to a power but taking the absolute value here. The reason why we do it this way is it has neat statistical properties as we try to build on it. But that's the population standard deviation, which gives us nice units-- meters. In the next video, we'll think about the sample standard", + "qid": "PWiWkqHmum0_462" + }, + { + "Q": "At 2:11 I really got confused... Like what? I don't get what she means? Like how?", + "A": "9x10 is basically 9 ten times. You move nine to the tens place, and zero is in the ones place since there are no ones left. 9 times 10 is 9, 10 times or 10, 9 times", + "video_name": "Ehd3cgRBvl0", + "timestamps": [ + 131 + ], + "3min_transcript": "- [Voiceover] What is 7 100s times 10? Well, let's focus first on this times 10 part of our expression. Because multiplying by 10 has some patterns in math that we can use to help us solve. One pattern we can think of when we multiply by 10 is if we take a whole number and multiply it by 10, we'll simply add a zero to the end of our whole number. So, for example, if we have a whole number like nine, and we multiply by 10, our solution will be a nine with one zero at the end. Or 90. Because nine times 10 is the same as nine 10s, and nine 10s is ninety. So let's use that pattern first to try to solve. Here we have seven 100s. So, seven times we have 100, or 700, and we're multiplying again times 10. Our solution will add a zero at the end. So if we had 700, 10 times, we would have 700 with a zero on the end. Or 7,000. So seven 100s times ten, is equal to 7,000. But there's another pattern we could use, here. Another pattern to think about when we multiply by 10. And that is that when we multiply by 10, we move every digit one place value, one place value, left. Or one place value greater. So let's look at that one on a place value chart. Here we have a place value chart. To use that earlier example when we had nine ones, and we multiplied it by 10, It moved up to the 10s. Now, we had nine 10s. And we filled in a zero here, because there were no ones left; there were zero ones left. And so, we saw that nine times 10 was equal to 90. So again, it's the same as adding a zero at the end, but we're looking at it another way. We're looking at it in terms of place value and multiplying by 10 moved every digit one place value to the left. So, if we do that with this same question, seven 100s, seven 100s, if we move 100s one place value to the left, we'll end up with 1,000s. So, 700 times 10 is seven 1,000s. Or, as we saw earlier, 7,000. So either one of these is a correct answer.", + "qid": "Ehd3cgRBvl0_131" + }, + { + "Q": "At 1:59 how did Sal get s(s+5)?", + "A": "Arnav, At 1:59, Sal took the part of the expression (s\u00c2\u00b2 + 5s) and using the distribtive property in reverse, he factored out an s Like this (s\u00c2\u00b2 + 5s) is (s*s + 5*s) so factor out (undistribute) an s s*(s+5) and rewrite as s(s+5) And in case you still don t understand the reverse distributive property, just use the distributive property on the answer and see if that helps. s(s+5) dstribute the s s*s + 5*s and rewrite as s\u00c2\u00b2 + 5s I hope that helps make it click for you.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 119 + ], + "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers", + "qid": "2ZzuZvz33X0_119" + }, + { + "Q": "At 1:41, can you factor out the quadratic equation into 2 binomials? Does it affect the answer?", + "A": "That is exactly what Sal did. He factored the quadratic into the binomials: (s-7)(s+5)=0 So, I m sure what you mean by your questions. If you can clarify, I ll try to help.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 101 + ], + "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers", + "qid": "2ZzuZvz33X0_101" + }, + { + "Q": "where did the 35 go at 2:11?", + "A": "Sal just factored a -7 out of that part of the polynomial. He divided the two terms by -7 to get -7(s+5). Hope this helps! :D", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 131 + ], + "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers", + "qid": "2ZzuZvz33X0_131" + }, + { + "Q": "At 2:16, couldn't you do s^2-5s+7s-35? It would mean the same thing right?", + "A": "I don t think you can factor the equation in thay form, but mathematically it means the same thing.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 136 + ], + "3min_transcript": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. a times b is equal to negative 35. So if the product is a negative number, one has to be And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers", + "qid": "2ZzuZvz33X0_136" + }, + { + "Q": "When solving a quadratic equation by factoring, if both equations could equal zero how come this is not included in the answer? Sal mentions around 3:45 that both could equal zero. In the above example the answer is given as s=-5 or s=7 but not s=-5 and s=7. Why is this? In the equation it makes sense that both could equal 0 as 0x0=0 but how can the answer be s=-5 and s=7?\n\nThanks!", + "A": "s=-5 makes one factor 0. s= 7 makes the other factor 0. Those are two different solutions to making the equation 0. But we didn t know until we did the factoring that the two factors would lead to two different zeros. It could have some out, for example, like this: (s+5)(s+5)=0. Then s= -5 would make both factors zero, and that would be ok because 0*0 = 0.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 225 + ], + "3min_transcript": "So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5. both sides of that equation, and you get s is equal to 7. So if s is equal to negative 5, or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is minus 35. That does equal zero. If you have 7, 49 minus 14 minus 35 does equal zero. So we've solved for s. Now, I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what is that equal to? x times x is x squared, x times b is bx.", + "qid": "2ZzuZvz33X0_225" + }, + { + "Q": "At 2:51 Sal wrote the equation as (s+5)(s-7)=0. When factoring quadratics, how do you know which constants are supposed to come first, like, in this case, 5 is the first constant and -7 is the second constant? That usually gets me when solving quadratic equations by factoring.", + "A": "It doesn t matter which comes first. The commutative property of multiplication tells use that the order we multiply in doesn t matter. Example 2x3 = 3x2. Apply the same property to the factors: (s+5)(s-7) = (s-7)(s+5). The order of the factors does not matter. Hope this helps.", + "video_name": "2ZzuZvz33X0", + "timestamps": [ + 171 + ], + "3min_transcript": "And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5.", + "qid": "2ZzuZvz33X0_171" + }, + { + "Q": "at 2:09 why does he put the plus or minus sign?", + "A": "When we multiply +9 x +9 we get 81 also when we multiply -9 x -9 we still get 81. so the square root can be + or - 9. therefore he writes + and - as the root can be either in + or -.", + "video_name": "tRHLEWSUjrQ", + "timestamps": [ + 129 + ], + "3min_transcript": "- Let's see if we can solve the equation P squared is equal to 0.81. So how could we think about this? Well one thing we could do is we could say, look if P squared is equal to 0.81, another way of expressing this is, that well, that means that P is going to be equal to the positive or negative square root of 0.81. Remember if we just wrote the square root symbol here, that means the principal root, or just the positive square root. But here P could be positive or negative, because if you square it, if you square even a negative number, you're still going to get a positive value. So we could write that P is equal to the plus or minus square root of 0.81, which kind of helps us, it's another way of expressing the same, the same, equation. But still, what could P be? In your brain, you might immediately say, well okay, you know if this was P squared is equal to 81, I kinda know what's going on. Because I know that nine times nine is equal to 81. Or we could write that nine squared is equal to 81, to the principal root of 81. These are all, I guess, saying the same truth about the universe, but what about 0.81? Well 0.81 has two digits behind, to the right of the decimal and so if I were to multiply something that has one digit to the right of the decimal times itself, I'm gonna have something with two digits to the right of the decimal. And so what happens if I take, instead of nine squared, what happens if I take 0.9 squared? Let me try that out. Zero, I'm gonna use a different color. So let's say I took 0.9 squared. 0.9 squared, well that's going to be 0.9 times 0.9, which is going to be equal to? Well nine times nine is 81, and I have one, two, numbers to the right of the decimal, so I'm gonna have two numbers to the right of the decimal in the product. So one, two. So that indeed is equal to 0.81. In fact we could write 0.81 as 0.9 squared. is equal to the plus or minus, the square root of, instead of writing 0.81, I could write that as 0.9 squared. In fact I could also write that as negative 0.9 squared. Cause if you put a negative here and a negative here, it's still not going to change the value. A negative times a negative is going to be a positive. I could, actually I would have put a negative there, which would have implied a negative here and a negative there. So either of those are going to be true. But it's going to work out for us because we are taking the positive and negative square root. So this is going to be, P is going to be equal to plus or minus 0.9. Plus or minus 0.9, or we could write it that P is equal to 0.9, or P could be equal to negative 0.9. And you can verify that, you would square either of these things, you get 0.81.", + "qid": "tRHLEWSUjrQ_129" + }, + { + "Q": "at 5:46 how did he get b+6", + "A": "Because, the sum of a+b must equal -4 and the product of a*b must equal -60. He just brute force went thru all the combinations possible until finding that +6 and -10 satisfy this. (a+6)(a-10) = b^2-4b-60", + "video_name": "STcsaKuW-24", + "timestamps": [ + 346 + ], + "3min_transcript": "Then you could have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and a 12, still seems too far apart One of them is negative, then you either have their sum being positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They are four apart. So if we make-- and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10 their sum will be negative 4, and their product is negative 60. So that works. So you could literally say that this is equal to b plus 6, times b, minus 10. b plus the a, plus b minus the b. And let me be very careful here. This b over here, I want to make it very clear, I just used this b here to say, look, we're looking for two numbers that add up to this second term It's a different b. I could have said x plus y is equal to negative 4, and x times y is equal to negative 60. In fact, let me do it that way just so we don't get confused. So we could write x plus y is equal to negative 4. And then we have x times y is equal to negative 60. So we have b plus 6, times b plus y. x is 6, y is negative 10. And that is equal to 0. Let's just solve this right here. And then we'll go back and show you. You could also factor this by grouping. But just from this, we know that either one of these is equal to zero. Either b plus 6 is equal to 0, or b minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get b is equal to negative 6. you get b is equal to 10. And those are our two solutions. You could put them back in and verify that they satisfy our constraints. Now the other way that you could solve this, and we're going to get exact same answer. Is you could just break up this negative 4b into its constituents. So you could have broken this up into 0 is equal to b squared. And then you could have broken it up into plus 6b, minus 10b, minus 60. And then factor it by grouping. Group these first two terms. Group these second two terms. Just going to add them together. The first one you could factor out a b. So you have b times b, plus 6. The second one you can factor out a negative 10. So minus 10 times b, plus 6. All that's equal to 0. And now you can factor out a b plus 6. So if you factor out a b plus 6 here,", + "qid": "STcsaKuW-24_346" + }, + { + "Q": "i didnt get what it meant in 5:00 could someone explain it to me??", + "A": "He is saying that the b s are not the same he just used b and a instead of x and y", + "video_name": "STcsaKuW-24", + "timestamps": [ + 300 + ], + "3min_transcript": "So we need to factor b squared, minus 4b, minus 60. So what we want to do, we want to find two numbers whose sum is negative 4 and whose product is negative 60. Now, given that the product is negative, we know there are different signs. And this tells us that their absolute values are going to be four apart. That one is going to be four less than the others. So you could look at the products of the factors of 60. 1 and 60 are too far apart. Even if you made one of the negative, you would either get positive 59 as the sum or negative 59 as the sum. 2 and 30, still too far apart. 3 and 20, still too far apart. If you had made one negative you'd Then you could have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and a 12, still seems too far apart One of them is negative, then you either have their sum being positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They are four apart. So if we make-- and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10 their sum will be negative 4, and their product is negative 60. So that works. So you could literally say that this is equal to b plus 6, times b, minus 10. b plus the a, plus b minus the b. And let me be very careful here. This b over here, I want to make it very clear, I just used this b here to say, look, we're looking for two numbers that add up to this second term It's a different b. I could have said x plus y is equal to negative 4, and x times y is equal to negative 60. In fact, let me do it that way just so we don't get confused. So we could write x plus y is equal to negative 4. And then we have x times y is equal to negative 60. So we have b plus 6, times b plus y. x is 6, y is negative 10. And that is equal to 0. Let's just solve this right here. And then we'll go back and show you. You could also factor this by grouping. But just from this, we know that either one of these is equal to zero. Either b plus 6 is equal to 0, or b minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get b is equal to negative 6.", + "qid": "STcsaKuW-24_300" + }, + { + "Q": "1:53 what is the meaning of willy-nilly? without-worry?", + "A": "When I ve heard this used, it has always meant to do something in a random manner, haphazardly, without any method or planning, in any way you please without thinking.", + "video_name": "6agzj3A9IgA", + "timestamps": [ + 113 + ], + "3min_transcript": "Use completing the square to find the roots of the quadratic equation right here. And when anyone talks about roots, this just means find the x's where y is equal to 0. That's what a root is. A root is an x value that will make this quadratic function equal 0, that will make y equal 0. So to find the x's, let's just make y equal 0 and then solve for x. So we get 0 is equal to 4x squared plus 40x, plus 280. Now, the first step that we might want to do, just because it looks like all three of these terms are divisible by 4, is just divide both sides of this equation by 4. That'll make our math a little bit simpler. So let's just divide everything by 4 here. If we just divide everything by 4, we get 0 is equal to x squared plus 10x, plus-- 280 divided by 4 is 70-- plus 70. me write that 70 a little bit further out, and you'll see why I did that in a second. So let me just write a plus 70 over here, just to have kind of an awkward space here. And you'll see what I'm about to do with this space, that has everything to do with completing the square. So they say use completing the square, which means, turn this, if you can, into a perfect square. Turn at least part of this expression into a perfect square, and then we can use that to actually solve for x. So how do we turn this into a perfect square? Well, we have a 10x here. And we know that we can turn this into a perfect square trinomial if we take 1/2 of the 10, which is 5, and then we square that. So 1/2 of 10 is 5, you square it, you add a 25. Now, you can't just willy-nilly add a 25 to one side of the equation without doing something to the other, or without just subtracting the 25 right here. Think about it, I have not changed the equation. I've added 25 and I've subtracted 25. So I've added nothing to the right-hand side. I could add a billion and subtract a billion and not change the equation. So I have not changed the equation at all right here. But what I have done is I've made it possible to express these three terms as a perfect square. That right there, 2 times 5 is 10. 5 squared is 25. So that is x plus 5 squared. And if you don't believe me, multiply it out. You're going to have an x squared plus 5x, plus 5x, which will give you 10x, plus 5 squared, which is 25. So those first three terms become that, and then the second two terms, right there, you just add them. Let's see, negative 25 plus 70. Let's see, negative 20 plus 70 would be positive 50, and then you have another 5, so it's plus 45.", + "qid": "6agzj3A9IgA_113" + }, + { + "Q": "At around 5:15, Sal says that the t-distribution has fatter tails because the small sample size causes an underestimation of the standard deviation of the sampling distribution of the sample mean.\n\nWouldn't a smaller sample size make you overestimate the standard deviation? And that is what leads to fatter tails? Could someone clarify this for me? Thank you!", + "A": "Too large an SEM would give too large an interval; too small too small. Fatter tails means a larger percentage of the area (probability) at higher SD. So you would need more SDs to get the same probability. So it looks to me like the fatter tails would tend to compensate for underestimating SEM. Why do you think that smaller sample size would overestimate the SEM? It gives a larger SEM, yes (the smaller the sample, the less accurate the sample mean is likely to be). Is it enough larger?", + "video_name": "K4KDLWENXm0", + "timestamps": [ + 315 + ], + "3min_transcript": "we're going to tweak the sampling distribution. We're not going to assume it's a normal distribution because this is a bad estimate. We're going to assume that it's something called a t-distribution. And a t-distribution is essentially, the best way to think about is it's almost engineered so it gives a better estimate of your confidence intervals and all of that when you do have a small sample size. It looks very similar to a normal distribution. It has some mean, so this is your mean of your sampling distribution still. But it also has fatter tails. And the way I think about why it has fatter tails is when you make an assumption that this is a standard deviation for-- let me take one more step. So normally what we do is we find the estimate of the true standard deviation, and then we say that the standard true standard deviation of our population divided by the square root of n. In this case, n is equal to 7. And then we say OK, we never know the true standard, or we seldom know-- sometimes you do know-- we seldom know the true standard deviation. So if we don't know that the best thing we can put in there is our sample standard deviation. And this right here, this is the whole reason why we don't say that this is just a 95 probability interval. This is the whole reason why we call it a confidence interval because we're making some assumptions. This thing is going to change from sample to sample. And in particular, this is going to be a particularly bad estimate when we have a small sample size, a size less than 30. So when you are estimating the standard deviation where you don't know it, you're estimating it with your sample standard deviation, and your sample size is small, and deviation of your sampling distribution, you don't assume your sampling distribution is a normal distribution. You assume it has fatter tails. And it has fatter tails because you're essentially underestimating-- you're underestimating the standard deviation over here. Anyway, with all of that said, let's just actually go through this problem. So we need to think about a 95% confidence interval around this mean right over here. So a 95% confidence interval, if this was a normal distribution you would just look it up in a Z-table. But it's not, this is a t-distribution. We're looking for a 95% confidence interval. So some interval around the mean that encapsulates 95% of the area. For a t-distribution you use t-table, and I have a t-table ahead of time right over here. And what you want to do is use the two-sided row for what", + "qid": "K4KDLWENXm0_315" + }, + { + "Q": "at 1:37, how could 10 hundreths and 7 hundreths make sense", + "A": "10 hundredths + 7 hundredths = 0.17 = 1 tenth + 7 hundredths. 1 hundredth = 0.01", + "video_name": "qSPwUDmpnJ4", + "timestamps": [ + 97 + ], + "3min_transcript": "- [Voiceover] Let's say that I had the number zero point one seven. How could I say this number? I said it one way, I said zero point one seven, but what are other ways that I could say it, especially if I wanted to express it in terms of tenths or hundredths or other places? And like always, try to pause the video and try think about it on your own. Alright, so there's actually a couple of ways that we could say this number. One is just to say zero point one seven. Other ways are to say look, I have a one in the tenths place, so that's going to be one tenth, one tenth and one tenth and I have a seven in the hundredths place, so this is a seven right over here in the hundredths place, so I can say one tenth and seven hundredths. Hun- Hundredths. And there you go. Now another to think about it is just say the whole thing in terms of hundredths. So a tenth is how many hundredths? Well a tenth is the same thing as 10 hundredths, so you could say, you could say instead of a tenth, you could say this is 10 hundredths, and the way I'm writing it right now, very few people would actually do it this way. 10 hundredths and and seven hundredths. And seven hundredths. Well not I could just add these hundredths, if I have 10 hundredths and I have another seven hundredths, that's going to be 17 hundredths. So I could just write this down as 17 hundredths. Hundredths. And to make that intuition of how we could just call this 17 hundredths instead of just calling it one tenth and seven hundredths, let's actually count by hundredths. So that is one hundredth, and actually, let me just go straight to nine hundredths. So I skipped a bunch right over here. And what would be the next, how would I say 10 hundredths? Well 10 hundredths, let me write it this way, 10 hundredths is the same thing as one tenth. So if we go from nine hundredths, the next, if I'm counting by hundredths, the next one's going to be 10 hundredths. Now once again, 10 hundredths is the same thing as one tenth, just the same way that 10 ones is the same thing as one 10. I hope that doesn't confuse you, but we could keep counting. 10 hundredths, 11 hundredths, 12 hundredths, 13 hundredths, 14 hundredths, 15 hundredths, 16 hundredths, and then finally 17 hundredths. So hopefully that gives you a little intuition for why we can call this number, instead of just calling it zero point one seven, or one tenth and seven hundredths, we could call this 17 hundredths.", + "qid": "qSPwUDmpnJ4_97" + }, + { + "Q": "7:01 if -2/3X was positive, would the answer be 2/3x-y=4? or 2/3x+y=4?", + "A": "yes.", + "video_name": "-6Fu2T_RSGM", + "timestamps": [ + 421 + ], + "3min_transcript": "x minus negative 3 is the same thing as x plus 3. This is our point slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope intercept form. Let's do that. So let's do slope intercept in orange. So we have slope intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2/3, so you get y minus 6 is equal to-- I'm just distributing the negative 2/3-- so negative 2/3 times x is negative 2/3 x. And then negative 2/3 times 3 is negative 2. And now to get it in slope intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side, so let's add 6 to both Left-hand side of the equation, we're just left with a y, these guys cancel out. You get a y is equal to negative 2/3 x. Negative 2 plus 6 is plus 4. So there you have it, that is our slope intercept form, mx plus b, that's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2/3 x to both sides of this equation. So I'll start it here. So we have y is equal to negative 2/3 x plus 4, that's slope intercept form. Let's added 2/3 x, so plus 2/3 x to both sides of this equation. I'm doing that so it I don't have this 2/3 x on the right-hand side, this negative 2/3 x. little bit, maybe more than I should have-- the left-hand side of this equation is what? It is 2/3 x, because 2 over 3x, plus this y, that's my left-hand side, is equal to-- these guys cancel out-- is equal to 4. So this, by itself, we are in standard form, this is the standard form of the equation. If we want it to look, make it look extra clean and have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get? 2/3 x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12. These are the same equations, I just multiplied every term by 3. If you do it to the left-hand side, you can do to the right-hand side-- or you have to do to the right-hand side-- and we are in standard form.", + "qid": "-6Fu2T_RSGM_421" + }, + { + "Q": "3:00 of the video; I thought 4 to the exponent of 3 times 5 to the exponent of 3 would give you a different answer then then 4x5 to the exponent of 3. So I do not understand the logic of this.", + "A": "(4x5) to the third is (4x5)x(4x5)x(4x5). Because it is multiplication, we can move the numbers around, getting 4x4x4x5x5x5. 4x4x4 is 4 to the third, and 5x5x5 is 5 to the third. So, (4x5)^3 = 4^3 x 5^3.", + "video_name": "rEtuPhl6930", + "timestamps": [ + 180 + ], + "3min_transcript": "When I raise something to the fifth power, that's just like saying 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth, right? All I did is I took 2 to the tenth and I multiplied it by itself five times. That's the fifth power. Well, we know from this rule up here that we can add these exponents because they're all the same base. So if we add 10 plus 10 plus 10 plus 10 plus 10, what do we get? Right, we get 2 to the fiftieth power. So essentially, what did we do here? All we did is we multiplied 10 times 5 to get 50. So that's our third exponent rule, that when I raise an exponent to a power and then I raise that whole expression to another power, I can multiply those two exponents. So let me give you another example. again, all I do is I multiply the 7 and the negative 9, and I get 3 to the minus 63. So, you see, it works just as easily with negative numbers. So now, I'm going to teach you one final exponent property. Let's say I have 2 times 9, and I raise that whole thing to the hundredth power. It turns out of this is equal to 2 to the hundredth power times 9 to the hundredth power. Now let's make sure that that makes sense. Let's do it with a smaller example. What if it was 4 times 5 to the third power? times 4 times 5, right, which is the same thing as 4 times 4 times 4 times 5 times 5 times 5, right? I just switched the order in which I'm multiplying, which you can do with multiplication. Well, 4 times 4 times 4, well, that's just equal to 4 to the third. And 5 times 5 times 5 is equal to 5 to the third. Hope that gives you a good intuition of why this property here is true. And actually, when I had first learned exponent rules, I would always forget the rules, and I would always do this proof myself, or the other proofs. And a proof is just an explanation of why the rule works, just to make sure that I was doing it right. So given everything that we've learned to now-- actually, let me review all of the rules again.", + "qid": "rEtuPhl6930_180" + }, + { + "Q": "2:59 of the this video. Doesn't this break the \"order of operation\" because you do what is in the parenthesis first then work your way out?\n\nalso in ths video and the previous video before this one, the subtitle at the bottom seem to flash a lot or something.", + "A": "Because they are all multiplication, we can rearrange the numbers any way we want, using the distributive property.", + "video_name": "rEtuPhl6930", + "timestamps": [ + 179 + ], + "3min_transcript": "When I raise something to the fifth power, that's just like saying 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth, right? All I did is I took 2 to the tenth and I multiplied it by itself five times. That's the fifth power. Well, we know from this rule up here that we can add these exponents because they're all the same base. So if we add 10 plus 10 plus 10 plus 10 plus 10, what do we get? Right, we get 2 to the fiftieth power. So essentially, what did we do here? All we did is we multiplied 10 times 5 to get 50. So that's our third exponent rule, that when I raise an exponent to a power and then I raise that whole expression to another power, I can multiply those two exponents. So let me give you another example. again, all I do is I multiply the 7 and the negative 9, and I get 3 to the minus 63. So, you see, it works just as easily with negative numbers. So now, I'm going to teach you one final exponent property. Let's say I have 2 times 9, and I raise that whole thing to the hundredth power. It turns out of this is equal to 2 to the hundredth power times 9 to the hundredth power. Now let's make sure that that makes sense. Let's do it with a smaller example. What if it was 4 times 5 to the third power? times 4 times 5, right, which is the same thing as 4 times 4 times 4 times 5 times 5 times 5, right? I just switched the order in which I'm multiplying, which you can do with multiplication. Well, 4 times 4 times 4, well, that's just equal to 4 to the third. And 5 times 5 times 5 is equal to 5 to the third. Hope that gives you a good intuition of why this property here is true. And actually, when I had first learned exponent rules, I would always forget the rules, and I would always do this proof myself, or the other proofs. And a proof is just an explanation of why the rule works, just to make sure that I was doing it right. So given everything that we've learned to now-- actually, let me review all of the rules again.", + "qid": "rEtuPhl6930_179" + }, + { + "Q": "how does he know psi 1 equal 1/2 theta 1 at 8:15 in the vidio? i didn't undrstand the proof. i don't think i saw a proof. like wise for scy 2 and thaita 2 at 8:20", + "A": "Sal wasnt really proving anything in particular. All he wanted to show was that the centre neednt be within the arc being subtended. And he just did that using stuff that he had taught before", + "video_name": "MyzGVbCHh5M", + "timestamps": [ + 495, + 500 + ], + "3min_transcript": "", + "qid": "MyzGVbCHh5M_495_500" + }, + { + "Q": "At 5:17 what does adjacent mean? I forget. Thanks", + "A": "Adjacent means next to .", + "video_name": "TgDk06Qayxw", + "timestamps": [ + 317 + ], + "3min_transcript": "it splits that line segment in half. So what it tells is, is that the length of this segment right over here is going to be equivalent to the length of this segment right over there. I have a circle. This radius bisects this chord right over here. And the goal here is to prove that it bisects this chord at a right angle. Or another way to say it-- let me add some points here. Let's call this B. Let's call this C, And let's call this D. I want to prove that segment AB is perpendicular. It intersects it at a right angle. It is perpendicular to segment CD. And as you could imagine, I'm going to prove it pretty much using the side-side-side whatever you want to call it, side-side-side theorem, postulate, or axiom. So let's do it. Let's think about it this way. I need to have some triangles. There's no triangles here right now. But I can construct triangles, and I can construct triangles based on things I know. For example, I can construct-- this has some radius. That's a radius right over here. The length of that is just going to be the radius of the circle. But I can also do it right over here. The length of AC is also going to be the radius of the circle. So we know that these two lines have the same length, which is the radius of the circle. Or we could say that AD is congruent to AC, or they have the exact same lengths. We know from the set-up in the problem that this segment is equal in length to this segment over here. Let me add a point here so I can refer to it. So if I call that point E, we know from the set-up in the problem, that CE is congruent to ED, or they have the same lengths. CE has the same length as ED. And we also know that both of these triangles, the one here the side EA. So EA is clearly equal to EA. So this is clearly equal to itself. It's the same side. The same side is being used for both triangles. The triangles are adjacent to each other. And so we see a situation where we have two different triangles that have corresponding sides being equal. This side is equivalent to this side right over here. This side is equal in length to that side over there. And then, obviously, AE is equivalent to itself. It's a side on both of them. It's the corresponding side on both of these triangles. And so by side-side-side, AEC.", + "qid": "TgDk06Qayxw_317" + }, + { + "Q": "What did sal say at 3:12", + "A": "he said Three over a hundred. (3/100) Remember you can put closed captions", + "video_name": "lR_kUUPL8YY", + "timestamps": [ + 192 + ], + "3min_transcript": "So if this is ones, multiply by 10, this is the tens place. This is the hundreds place. This is the thousands place. This is the ten thousands place. I'm going to have to write a little bit smaller. This is the hundred thousands place. And then the 7 is in the millions place. So what does this number, what does this 3 represent? Well, it's in the hundred thousands place. It literally represents 3 hundred thousands, or you could say 300,000, 3 followed by five zeroes. Now, what does this 3 represent? It's in the hundredths place. It literally represents 3 hundredths. It represents 3 times 1/100, which is the same thing as 3, which is equal to 3 over-- let Which is the same thing as 3/100, which is the same thing as 0.03. These are all equivalent statements. Now let's try to answer our original question. How much larger is this 3 than that 3 there? Well, one way to think about it is how much would you have to multiply this 3 by to get to this 3 over here? Well, one way to think about is to look directly at place value. So we got to multiply by 10. Every time we multiply by 10, that's equivalent to thinking about shifting it to one place to the left. So we would have to multiply by 10 one, two, three, four, five, six, seven times. So multiplying by 10 seven times. So this multiplied by 10 seven times should be equal to this. Let me rewrite this. 300,000 should be equal to 3/100-- let me write it the same way. 3/100 multiplied by 10 seven times, so times 10 times 10 times 10 times 10 times 10-- let's see, that's five times-- times 10 times 10. Now, multiplying by 10 seven times is the same thing as multiplying by 1 followed by seven zeroes. Every time you multiply by 10, you're going to get another zero here. So this is the same thing as 3/100 times 1 followed by one, two, three, four, five, six, seven zeroes. So this is literally 3/100 times 10 million.", + "qid": "lR_kUUPL8YY_192" + }, + { + "Q": "At 2:40, Can someone explain how 3 is 30 tenths?\n\nI've done the decimals for tenths, hundredths etc. but I don't get this one...\n\nI thought '3' was 3 \"ones\"?\n\nThanks", + "A": "Yes, you are right that 3 can equal 3 ones, but Sal decides to use 30 tenths because it is easier to see the answer using tenths. 1.5 is in tenths, because the tenths digit is the last place of value that we use. It would be more confusing if you said 3/2 is 1 1/2, although you can use ones if you prefer.", + "video_name": "zlq_jQYD1mA", + "timestamps": [ + 160 + ], + "3min_transcript": "So this is the same thing as two plus one over two, and I'm really doing every step here to hopefully make things clear, which is the same thing as two over two, so that's two over two, plus one over two, plus 1/2. I could break this up into two over two plus 1/2. Now, two over two is just one, and so this is going to be equal to 1 1/2. Now, you might immediately say, \"Hey, 1/2, I could write that as 5/10,\" and that would be exactly right. You could just, we don't wanna spell out every step, we could say this is equal to one, and when we write it in decimal form, we express things as tenths or hundredths or thousandths, so 1/2 is the same thing as 5/10, and if we wanna express that as a decimal, this would be equal to one and five tenths. you say okay this is the same thing as 3/2. 3/2, two goes into three one time, and there's a 1/2 left over, so writing this as a mixed number, it's 1 1/2, and 1/2 written as a decimal is 0.50, so this is 1.50. Now another way that we could've thought about this is \"Okay, I'm not getting a whole number, \"when I divide three divided by two. \"Maybe I'll get something in terms of tenths, \"so let me express each of these in terms of tenths.\" So three is how many tenths? Well, three is 30 tenths, and we'd be dividing by two, we're gonna be dividing it by two, so 30 tenths divided by two, well that's going to be equal to 15 tenths. This is equal to 15 tenths, which is equal to So both of these are equally legitimate strategies for figuring out what three divided by two is. I like the first one a little bit, it leverages what we know about fractions, but let's do another example. Let's do a few more examples, this is fun. Let's figure out what 34 divided by four is, and like before, pause this video and try to figure it out and try to see if you can use some of the strategies that we used in the last video. Alright, so as we just said, we can re-express this as a fraction, this is the same thing as 34 divided by four, 34 divided by four, or 34/4. Now what is this going to be equal to? Well, four goes into 34 eight times, it's gonna go eight times,", + "qid": "zlq_jQYD1mA_160" + }, + { + "Q": "0:50 can someone explain why f(x) when x is approaching 1+ is 1", + "A": "Look at the graph. Clearly, f(x) is not continuous at x=1, and therefore, there is no actual limit. However, there is still a limit from the left and from the right. X approaching 1+ means the limit from the right, so, as the graph shows, f(x) approaches 1 as x goes to one from the right.", + "video_name": "_WOr9-_HbAM", + "timestamps": [ + 50 + ], + "3min_transcript": "So we have a function, f of x, graphed right over here. And then we have a bunch of statements about the limit of f of x, as x approaches different values. And what I want to do is figure out which of these statements are true and which of these are false. So let's look at this first statement. Limit of f of x, as x approaches 1 from the positive direction, is equal to 0. So is this true or false? So let's look at it. So we're talking about as x approaches 1 from the positive direction, so for values greater than 1. So as x approaches 1 from the positive direction, what is f of x? Well, when x is, let's say 1 and 1/2, f of x is up here, as x gets closer and closer to 1, f of x stays right at 1. So as x approaches 1 from the positive direction, it looks like the limit of f of x as x approaches 1 from the positive direction isn't 0. It looks like it is 1. This would be true if instead of saying from the positive direction, we said from the negative direction. From the negative direction, the value of the function really does look like it is approaching 0. For approaching 1 from the negative direction, when x is right over here, this is f of x. When x is right over here, this is f of x. When x is right over here, this is f of x. And we see that the value of f of x seems to get closer and closer to 0. So this would only be true if they were approaching from the negative direction. Next question. Limit of f of x, as x approaches 0 from the negative direction, is the same as limit of f of x as x approaches 0 from the positive direction. Is this statement true? Well, let's look. Our function, f of x, as we approach 0 from the negative direction-- I'm using a new color-- as we approach 0 from the negative direction, so right over here, this is our value of f of x. Then as we get closer, this is our value of f of x. As we get even closer, this is our value of f of x. like it is approaching positive 1. From the positive direction, when x is greater than 0, let's try it out. So if, say, x is 1/2, this is our f of x. If x is, let's say, 1/4, this is our f of x. If x is just barely larger than 0, this is our f of x. So it also seems to be approaching f of x is equal to 1. So this looks true. They both seem to be approaching the limit of 1. The limit here is 1. So this is absolutely true. Now let's look at this statement. The limit of f of x, as x approaches 0 from the negative direction, is equal to 1. Well, we've already thought about that. The limit of f of x, as x approaches 0 from the negative direction, we see that we're getting closer and closer to 1. As x gets closer and closer to 0, f of x gets closer and closer to 1. So this is also true.", + "qid": "_WOr9-_HbAM_50" + }, + { + "Q": "At 1:28, you talked about that a and b are any two numbers. What if one of the value like b was a radical number that was not a perfect square?", + "A": "a and b can be any positive (and non-zero) numbers. They are the radii of the ellipse, and can have any length (positive or zero), like any other line segment.", + "video_name": "lvAYFUIEpFI", + "timestamps": [ + 88 + ], + "3min_transcript": "In the last video, we learned a little bit about the circle. And the circle is really just a special case of an ellipse. It's a special case because in a circle you're always an equal distance away from the center of the circle, while in an ellipse, the distance from the center of the circle is always changing. You know what an ellipse looks like. Well, I showed you that in the first video. It looks something like that. What I mean is that the radius or the distance from the center is always changing. Let me say this is centered at the origin. So that's the origin right there. You see here, we're really, if we're on this point on the ellipse, we're really close to the origin. This is actually the closest we'll ever get, just as close as well get down here. And when we're out here we're really far away from the origin and that's about as far as we're going to get right there. So a circle is a special case of this, because in a circle's case, the furthest we get from the origin is the same distance the exact same distance away from the origin. Well, with that said, let's actually go a little bit into the math. So the general or the standard form for an ellipse centered at the origin is x squared over a squared plus y squared over b squared is equal to 1. Where a and b are just any two numbers. I could have written this as c squared and d squared. I mean, they're just place holders. Just to give you an idea of what this means, if this was our ellipse in question right now, a is the length of the radius in the x-direction. Remember, we're going to have a squared down here. So if you took the square root of whatever is in the denominator, a is the x-radius. So this distance in our little chart right here, in our little graph here, that distance is a, or that this point right here, since we're centered at the origin, will be the point x is equal to a y is equal to 0. would be the point minus a comma 0. And then the radius in the y-direction would be this radius right here and is b. So this point would be x is equal to 0, y is equal to b. Likewise this point right here would be x is equal to 0, y is equal to minus b. And the way I drew this, we have kind of a short and fat ellipse you can also have kind of a tall and skinny ellipse. But in the short and fat ellipse, the direction that you're short in that's called your minor axis. And so b, I always forget the exact terminology, but b you can call it your semi or the length of your semi-minor axis. And where did that word come from? Well if this whole thing is your minor axis or maybe you could call your minor diameter if this whole thing is your", + "qid": "lvAYFUIEpFI_88" + }, + { + "Q": "at 8:41, why is it necessary for x-5 to behave like x of the standard equation", + "A": "I m not sure what you mean, but x-5 if the same if you were to shift any graph or shape on the coordinate plane. Hope that helps.", + "video_name": "lvAYFUIEpFI", + "timestamps": [ + 521 + ], + "3min_transcript": "So instead of the origin being at x is equal to 0, the origin will now be at x is equal 5. So a way to think about that is what does this term have to be so that at 5 this term ends up being 0. Well I'll actually draw it for you, because I think that might be confusing. So if we shift that over the right by 5, the new equation of this ellipse will be x minus 5 squared over 9 plus y squared over 25 is equal to 1. So if I were to just draw this ellipse right now, it would look like this. I want to make it look fairly similar to the ellipse I had before. Just shifted it over by five. And the intuition we learned a little bit in the circle video where I said, oh well, you know, if you have x minus something that means that the new origin is now at positive 5. And you could memorize that. You could always say, oh, if I have a minus here, that the origin is at the negative of whatever this number is, so it would be a positive five. You know, if you had a positive it would be the opposite that. But the way to really think about it is now if you go to x is equal to 5, when x is equal to 5, this whole term, x minus 5, will behave just like this x term will here. When x is equal to 5 this term is 0, just like when x was 0 here. So when x is equal to 5, this term is 0, and then y squared over 25 is equal 1, so y has to be equal five. Just like over here when x is equal is 0, y squared over 25 had to be equal to 1, y is equal to either And I really want to give you that intuition. And then, let's say we wanted to shift this equation down by two. So our new ellipse looks something like this. A lot of times you learned this in conic sections. But this is true any function. When you shift things, you shift it this way. If you shift this graph to the right by five, you replace all of the x's with x minus 5. And if you were to shift it down by two, you would replace all the y's with y plus 2. So let me draw our new ellipse first, just to show you what I'm doing. So our new ellipse is going to look something like that. I'm shifting the yellow ellipse down by two.", + "qid": "lvAYFUIEpFI_521" + }, + { + "Q": "At 17:20, why is the 4th dimension denoted by R4?\nWhy the letter R?", + "A": "R is the set of all real numbers. The real numbers can be thought of as any point on an infinitely long number line. Examples of these numbers are -5, 4/3, pi etc. An example of a number not included are an imaginary one such as 2i. R4 means that points in the space has 4 coordinates of real values. Points in this space are written on the form (x1, x2, x3, x4) where xi is a real number.", + "video_name": "L0CmbneYETs", + "timestamps": [ + 1040 + ], + "3min_transcript": "Either a position vector. It is a vector in R4. You can view it as a position vector or a coordinate in R4. You could say, look, our solution set is essentially-- this is in R4. Each of these have four components, but you can imagine it in r3. That my solution set is equal to some vector, some vector there. That's the vector. Think of it is as a position vector. It would be the coordinate 2, 0, 5, 0. Which obviously, this is four dimensions right there. It's equal to multiples of these two vectors. Let's call this vector, right here, let's call this vector a. Let's call this vector, right here, vector b. Our solution set is all of this point, which is right there, or I guess we could call it that position vector. That position vector will look like that. multiples of these two guys. If this is vector a, let's do vector a in a different color. Vector a looks like that. Let's say vector a looks like that, and then vector b looks like that. This is vector b, and this is vector a. I don't know if this is going to be easier or harder for you to visualize, because obviously we are dealing in four dimensions right here, and I'm just drawing on a two dimensional surface. What you can imagine is, is that the solution set is equal to this fixed point, this position vector, plus linear combinations of a and b. We're dealing, of course, in R4. Let me write that down. We're dealing in R4. But linear combinations of a and b are going to create a plane. You can multiply a times 2, and b times 3, or a times minus 1, and b times minus 100. You can keep adding and subtracting these linear combinations of a and b. position vector, or contains the point 2, 0, 5, 0. The solution for these three equations with four unknowns, is a plane in R4. I know that's really hard to visualize, and maybe I'll do another one in three dimensions. Hopefully this at least gives you a decent understanding of what an augmented matrix is, what reduced row echelon form is, and what are the valid operations I can perform on a matrix without messing up the system.", + "qid": "L0CmbneYETs_1040" + }, + { + "Q": "i don't get why you add the 4 in 1:34?", + "A": "because, we have to add 4 to make it a square or to make it in the form of (a+b)^2.If you add 4 to (x^2 + 4x): =(x^2 + 4x) +4 =(x^2 + 4x + 4) = (x^2 + 2x + 2x +4) =[x(x+2) + 2(x+2)] =(x+2) (x+2) = (x+2)^2 this is how a quadratic equation is solved. we have to bring it into (x-h)^2 form which is a part of (x-h)^2 + (y-k)^2 = y^2", + "video_name": "XyDMsotfJhE", + "timestamps": [ + 94 + ], + "3min_transcript": "We're asked to graph the circle. And they give us this somewhat crazy looking equation. And then we could graph it right over here. And to graph a circle, you have to know where its center is, and you have to know what its radius is. So let me see if I can change that. And you have to know what its radius is. So what we need to do is put this in some form where we can pick out its center and its radius. Let me get my little scratch pad out and see if we can do that. So this is that same equation. And what I essentially want to do is I want to complete the square in terms of x, and complete the square in terms of y, to put it into a form that we can recognize. So first let's take all of the x terms. So you have x squared and 4x on the left-hand side. So I could rewrite this as x squared plus 4x. And I'm going to put some parentheses around here, because I'm going to complete the square. And then I have my y terms. I'll circle those in-- well, the red looks too much like the purple. I'll circle those in blue. y squared and negative 4y. So we have plus y squared minus 4y. And I'll just do that in a neutral color. So minus 17 is equal to 0. Now, what I want to do is make each of these purple expressions perfect squares. So how could I do that here? Well, this would be a perfect square if I took half of this 4 and I squared it. So if I made this plus 4, then this entire expression would be x plus 2 squared. And you can verify that if you like. If you need to review on completing the square, there's plenty of videos on Khan Academy on that. All we did is we took half of this coefficient and then squared it to get 4. Half of 4 is 2, square it to get 4. And that comes straight out of the idea if you take x plus 2 and square it, it's going to be x squared plus twice the product of 2 and x, plus 2 squared. We had an equality before, and just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2. If we square negative 2, it becomes a positive 4. We can't just do that on the left-hand side. We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared", + "qid": "XyDMsotfJhE_94" + }, + { + "Q": "At 5:46 how is 3 root 3 (FG) considered as the longest side? Isn't it supposed to be 6 (EG)?", + "A": "You are correct that the longest side is 6. But he said the NEXT longest side, so 3 root 3 is the next longest side after 6. Also, in the other triangle, 27 is the next longest side as well. If you plug in 18 root 3, you will see it is greater than 27.", + "video_name": "Ly86lwq_2gc", + "timestamps": [ + 346 + ], + "3min_transcript": "and we went to the shorter side first. So now we want to start at X and go to the shorter side on the large triangle. So you go to XTS. XYZ is similar to XTS. Now, let's look at this right over here. So in our larger triangle, we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles. Even though this looks like a right angle, we cannot assume it. And if we look at this smaller triangle right over here, it shares one side with the larger triangle, but that's not enough to do anything. And then this triangle over here also shares another side, but that also doesn't do anything. So we really can't make any statement here about any kind of similarity. So there's no similarity going on here. There are some shared angles. This guy-- they both share that angle, the larger triangle So there could be a statement of similarity we could make if we knew that this definitely was a right angle. Then we could make some interesting statements about similarity, but right now, we can't really do anything as is. Let's try this one out, this pair right over here. So these are the first ones that we have actually separated out the triangles. So they've given us the three sides of both triangles. So let's just figure out if the ratios between corresponding sides are a constant. So let's start with the short side. So the short side here is 3. The shortest side here is 9 square roots of 3. So we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here, is 3 square roots of 3 over the next longest side over here, which is 27. And then see if that's going to be equal to the ratio of the longest side. So the longest side here is 6, and then So this is going to give us-- let's see, this is 3. Let me do this in a neutral color. So this becomes 1 over 3 square roots of 3. This becomes 1 over root 3 over 9, which seems like a different number, but we want to be careful here. And then this right over here-- if you divide the numerator and denominator by 6, this becomes a 1 and this becomes 3 square roots of 3. So 1 over 3 root 3 needs to be equal to square root of 3 over 9, which needs to be equal to 1 over 3 square roots of 3. At first they don't look equal, but we can actually rationalize this denominator right over here. We can show that 1 over 3 square roots of 3, if you multiply it by square root of 3", + "qid": "Ly86lwq_2gc_346" + }, + { + "Q": "at 2:22, dont you think that the Point should be at the third bar on the x axis? Because the point in the middle is zero isn't it?", + "A": "That s where he puts it. It s just that only the even ticks are fattened up.", + "video_name": "6m642-2D3V4", + "timestamps": [ + 142 + ], + "3min_transcript": "Identify the x and y-intercepts of the line y is equal to 3x minus 9. Then graph the line. So the x-intercept, I'll just abbreviate it as x-int, that is where the line intersects the x-axis. So where it intersects the x-axis. Remember, this horizontal axis is the x-axis. So when something intersects the x-axis, what do we know about its coordinate? Its x-value could be anything, but we know it's y-value is 0. If we're intersecting, if we're sitting on the x-axis someplace, that means that we haven't moved in the y-direction. That means that y is 0. So this means, literally, that y is 0. So we need to find the x-value defined by this relationship when y is equal to 0. Similarly, when we talk about the y-intercept, I'll do it down here-- when we talk about the y-intercept, what does that mean? Well, y-intercept means-- so this is the y-axis right over here running up and down. The y-intercept is the point at which the line intercepts the y-axis. If we're at the y-axis, our y-value could be anything depending on where we intersect the y-axis. But we know that we haven't moved to the right or the left. We know that our x-value is 0 at the y-intercept. So over here, our x-value is going to be 0. And to find the actual point, we just have to find the corresponding y-value defined by this relationship or this equation. So let's do the first one first. The x-intercept is when y is equal to 0. So we set y is equal to 0, and then we'll solve for x. So we get 0 is equal to 3x minus 9. We can add 9 to both sides of this equation to isolate the x-term. So we get 9 is equal to 3x. These cancel out. We could divide both sides by 3. Divide both sides by 3. We get 3 is equal to x or x is equal to 3. So the point y is equal to 0, x is equal to 3 is on this line. And let me put it in order. x-coordinate always goes first. So it's 3 comma 0. 1, 2, 3 is right over here. That is 3 comma 0. This right here is the x-intercept. And remember, notice that point lies on the x-axis, but the y-value is 0. We haven't moved up or down. When you think x-intercept, you say, OK, that means my y-value is 0. So I have to solve for the x-value. Now we do the opposite for the y-intercept. And the y-intercept, we're sitting on this line, x-value must be 0. So let's figure out what y is equal to when x is equal to 0. So y is equal to-- I want to do it in that pink color. y is equal to-- y is equal to 3 times-- x is 0 now. 3 times 0 minus 9. Well, 3 times 0 is just 0. So 0 minus 9. Well that's, just equal to negative 9. So we have the point 0 comma negative 9. So when x is 0, we go down 9 for y. 1, 2, 3, 4, 5, 6, 7, 8, 9.", + "qid": "6m642-2D3V4_142" + }, + { + "Q": "At 0:41 second how do sal got the equation y=mx+b", + "A": "I don t know how they come up with these equations like y=mx+b, but if you d the problem it will be easy to under stand how to do it and it is easy for me cause math is my strongest point", + "video_name": "YBYu5aZPLeg", + "timestamps": [ + 41 + ], + "3min_transcript": "Write an inequality that fits the graph shown below. So here they've graphed a line in red, and the inequality includes this line because it's in bold red. It's not a dashed line. It's going to be all of the area above it. So it's all the area y is going to be greater than or equal to this line. So first we just have to figure out the equation of this line. We can figure out its y-intercept just by looking at it. Its y-intercept is right there. Let me do that in a darker color. Its y-intercept is right there at y is equal to negative 2. That's the point 0, negative 2. So if you think about this line, if you think about its equation as being of the form y is equal to mx plus b in slope-intercept form, we figured out b is equal to negative 2. So that is negative 2 right there. And let's think about its slope. If we move 2 in the x-direction, if delta x is equal to 2, if our change in x is positive 2, what is our Our change in y is equal to negative 1. Slope, or this m, is equal to change in y over change in x, which is equal to, in this case, negative 1 over 2, or negative 1/2. And just to reinforce, you could have done this anywhere. You could have said, hey, what happens if I go back 4 in x? So if I went back 4, if delta x was negative 4, if delta x is equal to negative 4, then delta y is equal to positive 2. And once again, delta y over delta x would be positive 2 over negative 4, which is also negative 1/2. I just want to reinforce that it's not dependent on how far I move along in x or whether I go forward or backward. You're always going to get or you should always get, the same slope. So the equation of that line is y is equal to the slope, negative 1/2x, plus the y-intercept, minus 2. That's the equation of this line right there. Now, this inequality includes that line and everything above it for any x value. Let's say x is equal to 1. This line will tell us-- well, let's take this point so we get to an integer. Let's say that x is equal to 2. Let me get rid of that 1. When x is equal to 2, this value is going to give us negative 1/2 times 2, which is negative 1, minus 2, is going to give us negative 3. But this inequality isn't just y is equal to negative 3. y would be negative 3 or all of the values greater than I know that, because they shaded in this whole area up here. So the equation, or, as I should say, the inequality that fits the graph here below is-- and I'll do it in a bold color-- is y is greater than or equal to", + "qid": "YBYu5aZPLeg_41" + }, + { + "Q": "@12:25, when Sal multiplied 10^17 by 10^-1, by the rules of multiplication, shouldn't the answer have been 10^-16, rather than 10^16? Thanks, whoever answers.", + "A": "The rules of exponents state that x^a * x^b = x^(a+b) we are adding, not multiplying, so the exponent in Sal s case stays positive as 17 > 1.", + "video_name": "0Dd-y_apbRw", + "timestamps": [ + 745 + ], + "3min_transcript": "Well, this is equal to 3.2 over 6.4. We can just separate them out because it's associative. So, it's this times 10 to the 11th over 10 to the minus six, right? If you multiply these two things, you'll get that right there. So 3.2 over 6.4. This is just equal to 0.5, right? 32 is half of 64 or 3.2 is half of 6.4, so this is 0.5 right there. And what is this? This is 10 to the 11th over 10 to the minus 6. So when you have something in the denominator, you could write it this way. This is equivalent to 10 to the 11th over 10 to the minus 6. It's equal to 10 to the 11th times 10 to the minus 6 to the minus 1. Or this is equal to 10 to the 11th times 10 to the sixth. This is 1 over 10 to the minus 6. So 1 over something is just that something to the negative 1 power. And then I multiplied the exponents. You can think of it that way and so this would be equal to 10 to the 17th power. Or another way to think about it is if you have 1 -- you have the same bases, 10 in this case, and you're dividing them, you just take the 1 the numerator and you subtract the exponent in the denominator. So it's 11 minus minus 6, which is 11 plus 6, which is equal to 17. So this division problem ended up being equal to 0.5 times 10 to the 17th. Which is the correct answer, but if you wanted to be a stickler and put it into scientific notation, we want something maybe greater than 1 right here. So the way we can do that, let's multiply it by 10 on this side. And divide by 10 on this side or multiply by 1/10. by 10 and divide by 10. We're just doing it to different parts of the product. So this side is going to become 5 -- I'll do it in pink -- 10 times 0.5 is 5, times 10 to the 17th divided by 10. That's the same thing as 10 to the 17th times 10 to the minus 1, right? That's 10 to the minus 1. So it's equal to 10 to the 16th power. Which is the answer when you divide these two guys right there. So hopefully these examples have filled in all of the gaps or the uncertain scenarios dealing with scientific notation. If I haven't covered something, feel free to write a comment on this video or pop me an e-mail.", + "qid": "0Dd-y_apbRw_745" + }, + { + "Q": "At 7:53, how is -5 the square root of -\u00e2\u0088\u009a25? I understand that -5*-5 is equal to 25, since the negatives cancel out.\n*Note: I inserted the square root symbol by holding down on ALT and then typing 251 on the keypad.", + "A": "Since the negative sign is out of the parenthesis, it is negative, like if you learned what absolute value is, you ll know -I-5I is -5, since the negative is out of the absolute value signs, but if you are talking about the square root of negative 25, the answer is 5i, which you ll learn about later", + "video_name": "-QHff5pRdM8", + "timestamps": [ + 473 + ], + "3min_transcript": "So, this, right over here, is an irrational number. It's not rational. It cannot be represented as the ratio of two integers. All right, 14 over seven. This is the ratio of two integers. So, this, for sure, is rational. But if you think about it, 14 over seven, that's another way of saying, 14 over seven is the same thing as two. These two things are equivalent. So, 14 over seven is the same thing as two. So, this is actually a whole number. It doesn't look like a whole number, but, remember, a whole number is a non-negative number that doesn't need to be represented as the ratio of two integers. And this one, even though we did represent it as the ratio of two integers, it doesn't need to be represented as the ratio of two integers. You could have represent this as just two. So, that's going to be a whole number. 14 over seven, which is the same thing as two, that is a whole number. Now, two-pi. Now pi is an irrational... Pi is an irrational number. if we just take a integer multiple of pi, like that, this is also going to be an irrational number. If you looked at its decimal representation, it will never repeat. So that's two-pi, right over there. Now what about... Let me do that same, since I've been consistent, relatively consistent, with the colors. So, this is two-pi right over there. Now, what about the negative square root of 25. Well, 25's a perfect square. Square root of that's just gonna be five. So, this thing is going to be, this thing is going to be equivalent to negative five. So, this is just another representation of this, right over here. So, it is an integer. It's not a whole number because it's negative, but it's an integer. Negative square root of 25. These two things are actually... These two things are actually the same number, just different ways of representing them. And then you have, let's see, you have the square root of nine over... The square root of nine over seven. This thing is gonna be the same thing, this thing is the same... Let me do this in a different color. This is the same thing as, square root of nine is three, it's the principal root of nine, so it's three-sevenths. So, this is a ratio of two integers. This is a rational number. Square root of nine over seven is the same thing as three-sevenths. Now, let me just give you one more just for the road. What about pi over pi? What is that going to be? Well, pi divided by pi is going to be equal to one. So, this is actually a whole number. So I could write pi over pi, right over there. That's just a very fancy way of saying one.", + "qid": "-QHff5pRdM8_473" + }, + { + "Q": "From 3:53 to 4:23 you said that 22/7 is rational number but, from 6:54 to 6:58 you said that pi is irrational. But, pi=22/7 and 22/7 is not an irrational number. So, how can you say that pi is irrational?", + "A": "22/7 does not = Pi. It is only an approximation for Pi, just like 3.14 is an approximation of Pi. Compare the numbers: Pi = 3.141592653589793238462643383... 22/7 = 3.142857142857142857142857... This is a repeating decimal, the digits 142857 repeat. The decimal values in Pi never repeat and never terminate. So, Pi is an irrational number. 22/7 is the ratio of 2 integers, so it is a rational number. Hope this helps.", + "video_name": "-QHff5pRdM8", + "timestamps": [ + 233, + 263, + 414, + 418 + ], + "3min_transcript": "Irrational numbers. An integer. Well, if I could say, \"Look, that is an integer. \"Let's think about the integers.\" But I wouldn't say, \"Let's just think about the rational.\" I'd say, \"Let's think about the rational numbers.\" All right, now that we have these categories in place, let's categorize them. Like always, pause the video. See if you can figure out what category these numbers fall into. Where would you put them on this diagram? So, let's start off with three. This is positive three. It can be definitely represented as a fraction. You can represent it as three over one. But, it doesn't have to be represented as a fraction. It, literally, could be just a three, right over there, but it's also non-negative. So three is a whole number. So three, and maybe I'll do it in the color of the category. So, three is a whole number. So, it's a member of that set. But if you're a whole number, you're also an integer, and you're also a rational number. So, three is a whole number, it's an integer, Now, let's think about negative five. Now, negative five, once again, it can be represented as a fraction, but it doesn't have to be, but it is negative. So, it's not gonna be a whole number. So, negative five is going to sit right over here. It's an integer, and if you're an integer, you're definitely going to be a rational number, but it's not a whole number because it is negative. Now we have 0.25. Well, this, for sure, can be represented as a fraction. This is 25-hundreths, right over here. So, we can represent that as a fraction of two integers, I should say. It's 25-hundredths. But there's no way to represent this except using a fraction of two integers. So, 0.25 is a rational number, but it's not an integer and not a whole number. Now what about 22 over seven. Well, here it's clearly represented, already, as a fraction of two integers, except as a fraction of two integers. I can't somehow make this without using a fraction or some type of decimal that might repeat. So, this, right over here, this would also be a rational number, but it's not an integer, not a whole number. Now this over here. 0.2713. Now the 13 repeats. This is the same thing as 0.27131313, that's what line up there represents. Now, you might not realize it yet, but any number that repeats eventually, this one does repeat eventually, you have the .1313, or you have the 0.27131313, any number like this can be represented as a fraction. For example, and I'm not going to do it here, just for the sake of time, but, for example, 0.3, repeating, that's the same thing as one-third. And later on, we're gonna see techniques", + "qid": "-QHff5pRdM8_233_263_414_418" + }, + { + "Q": "Can we write that A is a subset of B? This is at 2:25 in the video.", + "A": "No. A has more elements than set B. This is denoted by other term called Superset . A is a superset of B.", + "video_name": "1wsF9GpGd00", + "timestamps": [ + 145 + ], + "3min_transcript": "Let's define ourselves some sets. So let's say the set A is composed of the numbers 1. 3. 5, 7, and 18. Let's say that the set B-- let me do this in a different color-- let's say that the set B is composed of 1, 7, and 18. And let's say that the set C is composed of 18, 7, 1, and 19. Now what I want to start thinking about in this video is the notion of a subset. So the first question is, is B a subset of A? And there you might say, well, what does subset mean? Well, you're a subset if every member of your set is also a member of the other set. So we actually can write that B is a subset-- this is a subset-- B is a subset of A. B is a subset. So let me write that down. B is subset of A. Every element in B is a member of A. Now we can go even further. We can say that B is a strict subset of A, because B is a subset of A, but it does not equal A, which means that there are things in A that are not in B. So we could even go further and we could say that B is a strict or sometimes said a proper subset of A. And the way you do that is, you could almost imagine that this is kind of a less than or equal sign, and then you kind of cross out this equal part of the less than or equal sign. So this means a strict subset, which means everything that is in B is a member A, but everything that's in A is not a member of B. So let me write this. This is B. B is a strict or proper subset. In fact, every set is a subset of itself, because every one of its members is a member of A. We cannot write that A is a strict subset of A. This right over here is false. So let's give ourselves a little bit more practice. Can we write that B is a subset of C? Well, let's see. C contains a 1, it contains a 7, it contains an 18. So every member of B is indeed a member C. So this right over here is true. Now, can we write that C is a subset? Can we write that C is a subset of A? Can we write C is a subset of A?", + "qid": "1wsF9GpGd00_145" + }, + { + "Q": "At 6:55 he missed a traingle", + "A": "He figured it out, just keep watching.", + "video_name": "qG3HnRccrQU", + "timestamps": [ + 415 + ], + "3min_transcript": "So we can assume that s is greater than 4 sides. Let's say I have an s-sided polygon, and I want to figure out how many non-overlapping triangles will perfectly cover that polygon. How many can I fit inside of it? And then I just have to multiply the number of triangles times 180 degrees to figure out what are the sum of the interior angles of that polygon. So let's figure out the number of triangles as a function of the number of sides. So once again, four of the sides are going to be used to make two triangles. So those two sides right over there. And then we have two sides right over there. I can draw one triangle over-- and I'm not even going to talk about what happens on the rest of the sides of the polygon. You could imagine putting a big black piece of construction paper. There might be other sides here. I'm not going to even worry about them right now. So out of these two sides I can draw one triangle, just like that. Out of these two sides, I can draw another triangle right over there. So four sides used for two triangles. I've already used four of the sides, but after that, if I have all sorts of craziness here. I could have all sorts of craziness here. Let me draw it a little bit neater than that. So I could have all sorts of craziness right over here. It looks like every other incremental side I can get another triangle out of it. So that's one triangle out of there, one triangle out of that side, one triangle out of that side, one triangle out of that side, and then one triangle out of this side. So for example, this figure that I've drawn is a very irregular-- one, two, three, four, five, six, seven, eight, nine, 10. Is that right? One, two, three, four, five, six, seven, eight, nine, 10. It is a decagon. And in this decagon, four of the sides were used for two triangles. So I got two triangles out of four of the sides. And out of the other six sides I was These are six. This is one, two, three, four, five. Actually, let me make sure I'm counting the number of sides right. So I have one, two, three, four, five, six, seven, eight, nine, 10. So let me make sure. Did I count-- am I just not seeing something? Oh, I see. I actually didn't-- I have to draw another line right over These are two different sides, and so I have to draw another line right over here. I can get another triangle out of that right over there. And so there you have it. I have these two triangles out of four sides. And out of the other six remaining sides I get a triangle each. So plus six triangles. I got a total of eight triangles. And so we can generally think about it. The first four, sides we're going to get two triangles. So let me write this down. So our number of triangles is going to be equal to 2.", + "qid": "qG3HnRccrQU_415" + }, + { + "Q": "What does carry mean, at 0:16?", + "A": "When you add, you carry by putting numbers more than 10 to the top so it can be easier to solve.", + "video_name": "Wm0zq-NqEFs", + "timestamps": [ + 16 + ], + "3min_transcript": "Let's add 536 to 398. And we're going to do it two different ways so that we really understand what this carrying is all about. So first, we'll do it in the more traditional way. We start in the ones place. We say, \"Well, what's 6 + 8?\" Well, we know that 6 + 8 is equal to 14. And so when we write it down here in the sum, we could say, \u201c \"Well look. The 4 is in the ones place.\u201d So it's equal to 4 + 1 ten.\" So let's write that 1 ten in the tens place. And now we focus on the tens place. We have 1 ten + 3 tens + 9 tens. So, what's that going to get us? 1 + 3 + 9 is equal to 13. Now we have to remind ourselves that this is 13 tens. Or another way of thinking about it, this is 3 tens and 1 hundred. You might say, \"Wait, wait! How does that make sense?\" When we're adding 1 ten + 3 tens + 9 tens, we're actually adding 10 + 30 + 90, and we're getting 130. And so we're putting the 30 (the 3 in the tens place represents the 30) \u2013 So this is the 3. The 3 represents the 30. And then we're placing this 1 in the hundreds place. 10 tens is equal to 100. And now we're adding up the numbers in the hundreds place. 1 + 5 + 3 is equal to \u2013 let's see. 1 + 5 is equal to 6, + 3 is equal to 9. But we have to remind ourselves: this is 9 hundreds. This is in the hundreds place. So this is actually 1 hundred. So this is actually 1 hundred + 5 hundreds + 3 hundreds, is equal to 9 hundreds. 100 + 500 + 300 is equal to 900. And we're done. This is equal to 934.", + "qid": "Wm0zq-NqEFs_16" + }, + { + "Q": "At 3:56 why isn't the (x,y ) coordinates are (cos theta, sin theta)? Why is the angle taken as (pi - theta) ?", + "A": "You are almost right. Where the yellow ray hits the circle, the ( x, y ) co-ordinates could either be labelled as Sal does or as ( - cos theta, sin theta). Since the x-co-ordinate is in a negative direction, cosine theta has to be negative. This gives us two of the many trig identities : cos ( pi - theta ) = - cos theta sin ( pi - theta ) = sin theta", + "video_name": "tzQ7arA917E", + "timestamps": [ + 236 + ], + "3min_transcript": "Notice, pi minus theta plus theta, these two are supplementary, and they add up to pi radians or 180 degrees. Now let's flip this one over the negative X-axis. If we flip this one over the negative X-axis, you're going to get right over there, and so you're going to get an angle that looks like this, that looks like this. Now what is going to be the measure of this angle? If we go all the way around like that, what is the measure of that angle? To go this far is pi, and then you're going another theta. This angle right over here is theta, so you're going pi plus another theta. This whole angle right over here, this whole thing, this whole thing is pi plus theta radians. Pi plus theta, let me just write that down. This is pi plus theta. Now that we've figured out let's think about how the sines and cosines of these different angles relate to each other. We already know that this coordinate right over here, that is sine of theta, sorry, the X-coordinate is cosine of theta. The X-coordinate is cosine of theta, and the Y-coordinate is sine of theta. Or another way of thinking about it is this value on the X-axis is cosine of theta, and this value right over here on the Y-axis is sine of theta. Now let's think about this one down over here. By the same convention, this point, this is really the unit circle definition of our trig functions. This point, since our angle is negative theta now, this point would be cosine of negative theta, comma, sine of negative theta. And we can apply the same thing over here. This point right over here, the X-coordinate is cosine of pi minus theta. from the positive X-axis. This is cosine of pi minus theta. And the Y-coordinate is the sine of pi minus theta. Then we could go all the way around to this point. I think you see where this is going. This is cosine of, I guess we could say theta plus pi or pi plus theta. Let's write pi plus data and sine of pi plus theta. Now how do these all relate to each other? Notice, over here, out here on the right-hand side, our X-coordinates are the exact same value. It's this value right over here. So we know that cosine of theta must be equal to the cosine of negative theta. That's pretty interesting. Let's write that down. Cosine of theta is equal to ... let me do it in this blue color,", + "qid": "tzQ7arA917E_236" + }, + { + "Q": "at 1:01 why are the fours at the bottom of the denominator not negative I am confused :/", + "A": "a negative exponent is not applied to the coefficient, it just flips the exponential with a negative exponent to the other side of the divide line and thus makes the exponent positive. It is not (-4)^-3 power which would have three negative fours.", + "video_name": "CZ5ne_mX5_I", + "timestamps": [ + 61 + ], + "3min_transcript": "- [Narrator] Let's get some practice with our exponent properties, especially when we have integer exponents. So, let's think about what four to the negative three times four to the fifth power is going to be equal to. And I encourage you to pause the video and think about it on your own. Well there's a couple of ways to do this. See look, I'm multiplying two things that have the same base, so this is going to be that base, four. And then I add the exponents. Four to the negative three plus five power which is equal to four to the second power. And that's just a straight forward exponent property, but you can also think about why does that actually make sense. Four to the negative 3 power, that is one over four to the third power, or you could view that as one over four times four times four. And then four to the fifth, that's five fours being multiplied together. So it's times four times four times four times four times four. And so notice, when you multiply this out, and three fours in the denominator. And so, three of these in the denominator with three of these in the numerator. And so you're going to be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have A to the negative fourth power times A to the, let's say, A squared. What is that going to be? Well once again, you have the same base, in this case it's A, and so since I'm multiplying them, you can just add the exponents. So it's going to be A to the negative four plus two power. Which is equal to A to the negative two power. And once again, it should make sense. This right over here, that is one over A times A times A times A times A times A, so that cancels with that, that cancels with that, and you're still left with one over A times A, which is the same thing as A to the negative two power. Now, let's do it with some quotients. So, what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So, this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent and so, this is going to be equal to 12 to the, subtracting a negative is the same thing as adding the positive,", + "qid": "CZ5ne_mX5_I_61" + }, + { + "Q": "In 1:35 how does he get 8 1/3?", + "A": "He divided 100 by 12, which is 8 with a remainder of 4. The remainder can be written as 4/12, which can be reduced to 1/3, so 8 1/3", + "video_name": "jOZ98FDyl2E", + "timestamps": [ + 95 + ], + "3min_transcript": "- [Voiceover] Let's get some practice comparing and computing rates. So they tell us the pet store has three fish tanks, each holding a different volume of water and a different number of fish. So Tank A has five fish, and it has 40 liters of water, Tank B, 12 fish, 100 liters of water, and Tank C, 23 fish, and it has 180 liters of water. Order the tanks by volume per fish from least to greatest. So let's think about what volume per fish, and we could think about this as volume divided by fish. Volume per fish. All right, so here for Tank A, it's going to be 40 liters for every five fish. 40 for every five fish, and let's see, 40 over five is eight, so you have eight liters per fish, is the rate at which they have to add water per fish for Tank A. Now, Tank B, you have 100 liters so what is this going to be? This is going to be, 12 goes into 100 eight times, so eight times 12 is 96, and then you have four left over. So this is going to be eight and 4/12, or eight and 1/3 liters per fish. And all I did is I converted this improper fraction, 100 over 12, to a mixed number, and I simplified it. 12 goes into 100 eight times with a remainder of four, so it's eight and 4/12, which is the same thing as eight and 1/3. And then finally in Tank C, I have 180 liters for 23 fish. So what is this going to be? 23 goes into 180. Let me try to calculate this. 23 goes into 180, it looks like it's going to be less than nine times. Is it eight times? Eight times, no, not eight times. Seven times three is 21, seven times two is 14, plus two is 16. When you subtract you get a remainder of 19. So this is going to be seven with a remainder of 19, or you could say this is seven and 19/23 liters per fish. So which one has, we're going to order the tanks by volume per fish, from least to greatest, so Tank B has the largest volume per fish, has eight and 1/3 liters per fish, so this is in first place. And then Tank A is in second place. Tank A is in second place. And then Tank C is in third place. Oh, actually we wanna go from least to greatest, so this is, let me write it this way. This is Tank C is the least, and Tank A is the greatest. So we really have to swap this order around. Now, I just copied and pasted this from the Khan Academy exercises. Let me actually bring the actual exercise up,", + "qid": "jOZ98FDyl2E_95" + }, + { + "Q": "At 1:00, where does the formula come from?", + "A": "One place it comes from is after deriving the quadratic formula, you end up with (-b \u00c2\u00b1 \u00e2\u0088\u009a(b^2-4ac))/(2a), so the -b/2a where the line of symmetry is and thus the x coordinate of the vertex and the (\u00c2\u00b1 \u00e2\u0088\u009a(b^2-4ac))/2a is the distance away from the line of symmetry of the zeroes.", + "video_name": "IbI-l7mbKO4", + "timestamps": [ + 60 + ], + "3min_transcript": "I have an equation right here. It's a second degree equation. It's a quadratic. And I know its graph is going to be a parabola. Just as a review, that means it looks something like this or it looks something like that. Because the coefficient on the x squared term here is positive, I know it's going to be an upward opening parabola. And I am curious about the vertex of this parabola. And if I have an upward opening parabola, the vertex is going to be the minimum point. If I had a downward opening parabola, then the vertex would be the maximum point. So I'm really trying to find the x value. I don't know actually where this does intersect the x-axis or if it does it all. But I want to find the x value where this function takes on a minimum value. Now, there's many ways to find a vertex. Probably the easiest, there's a formula for it. And we talk about where that comes from in multiple videos, where the vertex of a parabola or the x-coordinate of the vertex of the parabola. So the x-coordinate of the vertex And the negative b, you're just talking about the coefficient, or b is the coefficient on the first degree term, is on the coefficient on the x term. And a is the coefficient on the x squared term. So this is going to be equal to b is negative 20. So it's negative 20 over 2 times 5. Well, this is going to be equal to positive 20 over 10, which is equal to 2. And so to find the y value of the vertex, we just substitute back into the equation. The y value is going to be 5 times 2 squared minus 20 times 2 plus 15, which is equal to let's see. This is 5 times 4, which is 20, minus 40, which is negative 20, So just like that, we're able to figure out the coordinate. This coordinate right over here is the point 2, negative 5. Now it's not so satisfying just to plug and chug a formula like this. And we'll see where this comes from when you look at the quadratic formula. This is the first term. It's the x value that's halfway in between the roots. So that's one way to think about it. But another way to do it, and this probably will be of more lasting help for you in your life, because you might forget this formula. It's really just try to re-manipulate this equation so you can spot its minimum point. And we're going to do that by completing the square. So let me rewrite that. And what I'll do is out of these first two terms, I'll factor out a 5, because I want to complete a square here and I'm going to leave this 15 out to the right, because I'm going to have to manipulate that as well. So it is 5 times x squared minus 4x. And then I have this 15 out here.", + "qid": "IbI-l7mbKO4_60" + }, + { + "Q": "At 8:11, what is a hypersphere?", + "A": "Hypersphere is a generalization for any sphere that is more than 3D. Normally we define a sphere in 3D as x\u00c2\u00b2+y\u00c2\u00b2+z\u00c2\u00b2=1. But you can generalize this to as many dimensions as you want. You could have a 4D sphere or a 500D sphere, but in general, they are referred to as hyperspheres.", + "video_name": "iDQ1foxYf0o", + "timestamps": [ + 491 + ], + "3min_transcript": "So \"Three students attempt to define \"what a line segment is.\" And we have a depiction of a line segment right over here. We have point P, point Q, and the line segment is all the points in between P and Q. So, so let's match the teacher's comments to the definitions. Ivy's definition: \"All of the points \"in line with P and Q, extending infinitely \"in both directions.\" Well, that would be the definition of a line. That would be the line P, Q. That would be, if you're extending infinitely in both directions, so ... I would say, \"Are you thinking of a line \"instead of a line segment?\" Ethan's definition: \"The exact distance from P to Q. Well, that's just a ... that's the length of a line segment. That's not exactly what a line segment is. And see, Ebuka's definition. \"The points P and Q, which are called endpoints, \"and all of the points in a straight line Yep. That looks like a good definition for a line segment. So we can just check our, we can just check our answer. So, looking good. Let's do one more of this. I'm just really enjoying pretending to be a teacher. All right. \"Three students attempt to define \"what a circle is.\" Define what a circle is. \"Can you match the teacher's comments to the definitions?\" Duru. \"The set of all points in a plane \"that are the same distance away from some given point, \"which we call the center.\" That just seems like a pretty good definition of a circle. So, I would, you know, \"Stupendous! Well done.\" Oliver's definition. \"The set of all points in 3D space \"that are the same distance from a center point.\" If we're talking about 3D space and the set of all points that are equidistant from that point in 3D space, now we're talking about a sphere, not a circle. And so, \"You seem to be confusing \"a circle with a sphere.\" And then, finally, \"A perfectly round shape.\" But if you're talk about three dimensions, you could be talking about a sphere. If you're talking about, if you go beyond three dimensions, hypersphere, whatever else. In two dimensions, yeah, perfectly round shape, most people would call it a circle. But that doesn't have a lot of precision to it. It doesn't have, it doesn't give us a lot that we can work with from a mathematical point of view. So I would say, actually, what the teacher's saying: \"Your definition needs to be much more precise.\" Duru's definition is much, much more precise. The set of all points that are equidistant from ... in a plane, that are equidistant away from a given point, which we call the center. So yep. Carlos could use a little bit more precision. We're all done.", + "qid": "iDQ1foxYf0o_491" + }, + { + "Q": "At 0:32, how does one know when to write x/y and when to write y/x?", + "A": "its just another way to write division 4/2=2/4 and here sal is just saying 14/2 =/= 2/14 so in proportions you need bigger number on top...", + "video_name": "qcz1Cm_-l50", + "timestamps": [ + 32 + ], + "3min_transcript": "- So, let's set up a relationship between the variables x and y. So, let's say, so this is x and this is y, and when x is one, y is four, and when x is two, y is eight, and when x is three, y is 12. Now, you might immediately recognize that this is a proportional relationship. And remember, in order for it to be a proportional relationship, the ratio between the two variables is always constant. So, for example, if I look at y over x here, we see that y over x, here it's four over one, which is just four. Eight over two is just four. Eight halves is the same thing as four. 12 over three it's the same thing as four. Y over x is always equal to four. In fact, I can make another column here. I can make another column here where I have y over x, here it's four over one, which is equal to four. Here it's eight over two, which is equal to four. Here it's 12 over three, which is equal to four. the ratio, the ratio between y and x is this constant four, to express the relationship between y and x as an equation. In fact, in some ways this is, or in a lot of ways, this is already an equation, but I can make it a little bit clearer, if I multiply both sides by x. If I multiply both sides by x, if I multiply both sides by x, I am left with, well, x divided by x, you'd just have y on the left hand side. Y is equal to 4x and you see that's the case. X is one, four times that is four. X is two, four times that is eight. So, here you go, we're multiplying by four. We are multiplying by four, we are multiplying by four. And so, four, in this case, four, in this case, in this situation, this is our constant of proportionality. Constant, constant, sometimes people will say proportionality constant. Now sometimes, it might even be described as a rate of change and you're like well, Sal, how is this a, how would four be a rate of change? And, to make that a little bit clearer, let me actually do another example, but this time, I'll actually put some units there. So let's say that, let's say that I have, let's say that x-- Let me do this, I already used yellow, let me use blue. So let's x, let's say that's a measure of time and y is a measure of distance. Or, let me put it this way, x is time in terms of seconds. Let me write it this way. So, x, x is going to be in seconds and then, y is going to be in meters. So, this is meters, the units, and this right over here is seconds. So, after one second, we have traveled, oh, I don't know, seven meters.", + "qid": "qcz1Cm_-l50_32" + }, + { + "Q": "At 6:30 Sal write dt, but I cant see where he get that from. Anyone can help? Thanks", + "A": "dt is the differential, you put it at the end of the integration expression. If you re wondering why its dt and not dx ; its because it parametric, so your functions are functions of t - f(t) - and not like the usual functions of x - f(x). Hope i could help!", + "video_name": "99pD1-6ZpuM", + "timestamps": [ + 390 + ], + "3min_transcript": "And the reason why this is the case, is if you imagine this is a, this is b, that is my f of x. When you do it this way, your dx's are always going to be positive. When you go in that direction, your dx's are always going to be positive, right? Each increment, the right boundary is going to be higher So your dx's are positive. In this situation, your dx's are negative. The heights are always going to be the same, they're always going to be f of x, but here your change in x is a negative change in x, when you go from b to a. And that's why you get a negative integral. In either case here, our path changes, but our ds's are going to be positive. And the way I've drawn this surface, it's above the x-y plane, the f of xy is also going to be positive. So that also kind of gives the same intuition that this should be the exact same area. But let's prove it to ourselves. So let's start off with our first parameterization, just We have x is equal to x of t, y is equal to y of t, and we're dealing with this from, t goes from a to b. And we know we're going to need the derivatives of these, so let write that down right now. We can write dx dt is equal to x prime of t, and dy dt, let me write that a little bit neater, dy dt is equal to y prime of t. This is nothing groundbreaking I've done so far. But we know the integral over c of f of xy. f is a scalar field, not a vector field. ds is equal to the integral from t is equal to a, to t is dt squared, which is the same thing as x prime of t squared, plus dy dt squared, the same thing as y prime of t squared. All that under the radical, times dt. This integral is exactly that, given this parameterization. Now let's do the minus c version. I'll do that in this orange color. Actually, let me do the minus c version down here. The minus the c version, we have x is equal to, you remember this, actually, just from up here, this was from the last video.", + "qid": "99pD1-6ZpuM_390" + }, + { + "Q": "At 0:25 is one fourth just like a quarter?", + "A": "Think of one whole as 100. If you think of money, one dollar is 100 cents. One quarter is 25 cents out of the whole 100 cents. So a quarter is one fourth(a quarter) of a dollar(a whole).", + "video_name": "gEE6yIObbmg", + "timestamps": [ + 25 + ], + "3min_transcript": "- [Voiceover] Is each piece equal to 1/4 of the area of the pie? So we have a pie, and it has one, two, three, four pieces. So it does have four pieces. So is one of those pieces equal to 1/4 of the pie? Well let's talk about what we mean when we have a fraction like 1/4. The one in the fraction, the numerator, represents a number of pieces. So here, one piece. One piece of pie. And then the four, when we're talking about fractions is always talking about the number of equal size. Equal size pieces. So in this case four equal size pieces. So the question is, is each piece one of four equal size pieces? Let's look at the pie. I think it's pretty clear that these pieces on the end are not equal, they are smaller If you love cherry pie, you are not happy about getting this end piece. Because it is smaller. It is not an equal size piece. So yes, each piece is one out of four pieces. But it is not one of four equal size pieces. Therefore it is not 1/4. So our answer is no. No, no, no. Each piece is not 1/4 or an equal share of the pie.", + "qid": "gEE6yIObbmg_25" + }, + { + "Q": "At 0:52 when he multiplied 3 by both sides, that same number has to correspond to the denominator for example if the dominator was -3, you multiply by -3 yes?", + "A": "Correct. This is because we need to get rid of the denominator. Does that make sense? And most don t know why you switch the sign, but I believe you do - otherwise you would have asked to know. I hope I helped!", + "video_name": "y7QLay8wrW8", + "timestamps": [ + 52 + ], + "3min_transcript": "We have the inequality 2/3 is greater than negative 4y minus 8 and 1/3. Now, the first thing I want to do here, just because mixed numbers bother me-- they're actually hard to deal with mathematically. They're easy to think about-- oh, it's a little bit more than 8. Let's convert this to an improper fraction. So 8 and 1/3 is equal to-- the denominator's going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2/3 is greater than negative 4y minus 25 over 3. Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That'll get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left-hand side. And then I'm going to multiply the right-hand side. Well, one point that I want to point out is that I did not have to swap the inequality sign, because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3, or negative 1, or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left-hand side, we have 3 times 2/3, which is just 2. 2 is greater than. And then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms-- the only variable here is y on the other side-- the y is already sitting here, so let's just get this 25 on the other side And we can do that by adding 25 to both sides of this equation. And with the left-hand side, 2 plus 25 five is 27 and we're going to get 27 is greater than. The right-hand side of the inequality is negative 12y. And then negative 25 plus 25, those cancel out, that was the whole point, so we're left with 27 is greater than negative 12y. Now, to isolate the y, you can either multiply both sides by negative 1/12 or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than-- I'm swapping the inequality, let me do this in a different color-- is less than negative 12y over negative 12.", + "qid": "y7QLay8wrW8_52" + }, + { + "Q": "At 5:15 what did she use to make the squiggle?", + "A": "She uses a Pipe Cleaner to make the squiggle. The brand is Dill s", + "video_name": "ik2CZqsAw28", + "timestamps": [ + 315 + ], + "3min_transcript": "Three a-squig, a-squig, a-squiggle. Two, nine. Two a-squig, a-squig, a-squiggle. Three, a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle, two. Nine. Two a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Three a-squig, a-squig, a-squiggle. Two, nine. And 15 all the way over to here. And now, Yeah. I can talk that fast totally. OK, But let's not get too far from your original purpose, which was to nicely fill a page with this squiggle. The nicest page filling squiggles have kind of the same density of squiggle everywhere. You don't want to be clumped up here, but have left over space there, because monsters may start growing in the left over space. On graph paper, you can be kind of precise about it. Say you want a squiggle that goes through every box exactly once, and can be extended infinitely. So you try some of those, and decide that, since the point of them is to fill up all the space, you call them space filling curves. Yeah, that's actually a technical term, but be careful because your curve might actually be a snake, snake, snake, snake, snake, snake, snake, snake, snake, snake-- Also, to make it neater, you draw the lines on the lines, and shift the rules so that you go through each intersection on the graph paper exactly once. Which is the same thing, as far as space is concerned. Here's a space filling curve that a guy named Hilbert made up, because Hilbert was awesome, but he's dead now. Here's the first iteration. For the second one, we're going to build it piece-by-piece by connecting four copies of the first. So here's one. Put the second space away next to it, and connect those. Then turn the page to put the third sideways under the first, and connect those. And then the fourth will be the mirror image of that on the other side. Now you've got one nice curve. The third iteration will be made out of four copies of the second iteration. So first build another second iteration curve out of four copies of the first iteration-- one, two, three, four-- then put another next to it, then two sideways on the bottom. Connect them all up. There you go. The fourth iteration is made of four copies of the third iteration, the same way. If you learn to do the second iteration in one piece, it'll make this go faster. Then build two third iterations facing up next to each other, You can keep going until you run out of room, or you can make each new version the same size by making each line half the length. Or you can make it out of snakes. Or if you have friends, you can each make an iteration of the same size, and put them together. Or invent your own fractal curve so that you could be cool like Hilbert. Who was like, mathematics? I'm going to invent meta-mathematics like a boss.", + "qid": "ik2CZqsAw28_315" + }, + { + "Q": "at 4:04 he sal listed outcomes for exactly one head... I think he forgot to list [TTTH]. Am I right?", + "A": "He accidentally wrote TTHT twice in his equation, however [TTTH} is in row 3, column 4 of his table.", + "video_name": "8TIben0bJpU", + "timestamps": [ + 244 + ], + "3min_transcript": "possibilities, involve getting exactly one heads? Well, we could list them. You could get your heads. So this is equal to the probability of getting the heads in the first flip, plus the probability of getting the heads in the second flip, plus the probability of getting the heads in the third flip. Remember, exactly one heads. We're not saying at least one, exactly one heads. So the probability in the third flip, and then, or the possibility that you get heads in the fourth flip. Tails, heads, and tails. And we know already what the probability of each of these things are. There are 16 possible events, and each of these are one of those 16 possible events. So this is going to be 1 over 16, 1 over 16, 1 over 16, And so we're really saying the probability of getting exactly one heads is the same thing as the probability of getting heads in the first flip, or the probability of getting heads-- or I should say the probability of getting heads in the first flip, or heads in the second flip, or heads in the third flip, or heads in the fourth flip. And we can add the probabilities of these different things, because they are mutually exclusive. Any two of these things cannot happen at the same time. You have to pick one of these scenarios. And so we can add the probabilities. 1/16 plus 1/16 plus 1/16 plus 1/16. Did I say that four times? Well, assume that I did. And so you would get 4/16, which is equal to 1/4. Fair enough. Now let's ask a slightly more interesting question. Let's ask ourselves the probability of getting exactly two heads. And there's a couple of ways we can think about it. [? We ?] know the number of possibilities and of those equally likely possibilities. And we can only use this methodology because it's a fair coin. So, how many of the total possibilities have two heads of the total of equally likely possibilities? So we know there are 16 equally likely possibilities. How many of those have two heads? So I've actually, ahead of time so we save time, I've drawn all of the 16 equally likely possibilities. And how many of these involve two heads? Well, let's see. This one over here has two heads, this one over here has two heads, this one over here has two heads. Let's see, this one over here has two heads, and this one over here has two heads. And then this one over here has two heads, and I believe we are done after that. So if we count them, one, two, three, four, five, six of the possibilities have exactly two heads.", + "qid": "8TIben0bJpU_244" + }, + { + "Q": "at 1:25 why is 9x^2 not equal to 3x^2 ?", + "A": "We have 9x^2. We want to make it into (ax)^2. 3x^2 does NOT equal 9x^2. However, (3x)^2 does equal 9x^2. Why? When we have (3x)^2 the exponent distributes to both terms (the 3 and the x) so we have: (3x)^2 = 3^2x^2 = 9x^2.", + "video_name": "jmbg-DKWuc4", + "timestamps": [ + 85 + ], + "3min_transcript": "Let's see if we can factor the expression 45x squared minus 125. So whenever I see something like this-- I have a second-degree term here, I have a subtraction sign-- my temptation is to look at this as a difference of squares. We've already seen this multiple times. We've already seen that if we have something of the form a squared minus b squared, that this can be factored as a plus b times a minus b. So let's look over here. Well, over here, it's not obvious that this right over here is a perfect square. Neither is it obvious that this right over here So it's not clear to me that this is a difference of squares. But what is interesting is that both 45 and 125 have some factors in common. And the one that jumps out at me is 5. So let's see if we can factor out a 5, and by doing that, whether we can get something that's a little bit closer to this pattern right over here. 45x squared divided by 5 is going to be 9x squared. And then 125 divided by 5 is 25. Now, this is interesting. 9x squared-- that's a perfect square. If we call this a squared, then that tells us that a would be equal to 3x. 3x-- the whole thing squared is 9x squared. Similarly-- I can never say similarly correctly-- 25 is clearly just 5 squared. So in this case, if we're looking at this template, b would be equal to 5. So now this is a difference of squares, and we can factor it completely. So we can't forget our 5 out front that we factored out. So it's going to be 5 times a plus b. So let me write this. So it's going to be 5 times a plus b times a minus b. 5 times 3x plus 5 times 3x minus 5 is 45x squared minus 125 factored out.", + "qid": "jmbg-DKWuc4_85" + }, + { + "Q": "At 2:08 Sal started explaining average. Are average and mean the same thing, because Sal never said anything about mean", + "A": "Yes! Average and mean are synonyms. They both involve the total sum of a set of numbers and dividing it by how many numbers given.", + "video_name": "h8EYEJ32oQ8", + "timestamps": [ + 128 + ], + "3min_transcript": "We will now begin our journey into the world of statistics, which is really a way to understand or get our head around data. So statistics is all about data. And as we begin our journey into the world of statistics, we will be doing a lot of what we can call descriptive statistics. So if we have a bunch of data, and if we want to tell something about all of that data without giving them all of the data, can we somehow describe it with a smaller set of numbers? So that's what we're going to focus on. And then once we build our toolkit on the descriptive statistics, then we can start to make inferences about that data, start to make conclusions, start to make judgments. And we'll start to do a lot of inferential statistics, make inferences. So with that out of the way, let's think about how we can describe data. So let's say we have a set of numbers. We can consider this to be data. in our garden. And let's say we have six plants. And the heights are 4 inches, 3 inches, 1 inch, 6 inches, and another one's 1 inch, and another one is 7 inches. And let's say someone just said-- in another room, not looking at your plants, just said, well, you know, how tall are your plants? And they only want to hear one number. They want to somehow have one number that represents all of these different heights of plants. How would you do that? Well, you'd say, well, how can I find something that-- maybe I want a typical number. Maybe I want some number that somehow represents the middle. Maybe I want the most frequent number. Maybe I want the number that somehow represents the center of all of these numbers. And if you said any of those things, you would actually have done the same things that the people who first came up with descriptive statistics said. They said, well, how can we do it? And we'll start by thinking of the idea of average. has a very particular meaning, as we'll see. When many people talk about average, they're talking about the arithmetic mean, which we'll see shortly. But in statistics, average means something more general. It really means give me a typical, or give me a middle number, or-- and these are or's. And really it's an attempt to find a measure of central tendency. So once again, you have a bunch of numbers. You're somehow trying to represent these with one number we'll call the average, that's somehow typical, or middle, or the center somehow of these numbers. And as we'll see, there's many types of averages. The first is the one that you're probably most familiar with. It's the one-- and people talk about hey, the average on this exam or the average height. And that's the arithmetic mean.", + "qid": "h8EYEJ32oQ8_128" + }, + { + "Q": "If 6 5/x = 14, what does x equal to?\nFor Example:\n6 multiplied by x = 6x\n6x + 5= 8.75/x\n8.75/x = 14\non 7:38", + "A": "x = 0.625", + "video_name": "9IUEk9fn2Vs", + "timestamps": [ + 458 + ], + "3min_transcript": "this equation by x plus 5. You can say x plus 5 over 1. Times x plus 5 over 1. On the left-hand side, they get canceled out. So we're left with 3 is equal to 8 times x plus five. All of that over x plus 2. Now, on the top, just to simplify, we once again just multiply the 8 times the whole expression. So it's 8x plus 40 over x plus 2. Now, we want to get rid of this x plus 2. So we can do it the same way. We can multiply both sides of this equation by x plus 2 over 1. x plus 2. We could just say we're multiplying both sides by x plus 2. The 1 is little unnecessary. So the left-hand side becomes 3x plus 6. multiplying it times the whole expression. x plus 2. And on the right-hand side. Well, this x plus 2 and this x plus 2 will cancel out. And we're left with 8x plus 40. And this is now a level three problem. Well, if we subtract 8x from both sides, minus 8x, plus-- I think I'm running out of space. Minus 8x. Well, on the right-hand side the 8x's cancel out. On the left-hand side we have minus 5x plus 6 is equal to, on the right-hand side all we have left is 40. Now we can subtract 6 from both sides of this equation. Let me just write out here. Minus 6 plus minus 6. Now I'm going to, hope I don't lose you guys by trying to go up here. But if we subtract minus 6 from both sides, on the left-hand side we're just left with minus 5x equals, and on the Now it's a level one problem. We just multiply both sides times negative 1/5. Negative 1/5. On the left-hand side we have x. And on the right-hand side we have negative 34/5. Unless I made some careless mistakes, I think that's right. And I think if you understood what we just did here, you're ready to tackle some level four linear equations. Have fun.", + "qid": "9IUEk9fn2Vs_458" + }, + { + "Q": "At 2:53, how was Sal able to tell whether it was sin or cosine?", + "A": "When x is 0, the value of the cosine equation would be 1, and (0, 1) is not a point on the graph. When x is 0, the value of the sine equation would be -2, and (0, -2) is a valid point on the graph. Thus, the cosine equation can be eliminated.", + "video_name": "yHo0CcDVHsk", + "timestamps": [ + 173 + ], + "3min_transcript": "Going from negative 2 to 1, it went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So immediately, we can say, well, look. This is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write \"cosine\" first. Cosine maybe some coefficient times x plus the midline. The midline-- we already figured out-- was minus 2 or negative 2. So it could take that form or it could take f of x is equal to 3 times-- it could be sine of x or sine of some coefficient times x. So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1. Whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0-- so if x is 0, k times 0 is going to be 0-- sine of 0 is 0. So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since, when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. And we just really need to figure out-- what could this constant actually be? at the period of this function. Let's see. If we went from this point-- where we intersect the midline-- and we have a positive slope, the next point that we do that is right over here. So our period is 8. So what coefficient could we have here to make the period of this thing be equal to 8? Well, let us just remind ourselves what the period of sine of x is. So the period of sine of x-- so I'll write \"period\" right over here-- is 2pi. You increase your angle by 2 pi radians or decrease it. you're back at the same point on the unit circle. So what would be the period of sine of kx? Well, now, your x, your input is increasing k times faster.", + "qid": "yHo0CcDVHsk_173" + }, + { + "Q": "At 0:24 why did the problem say the time when you don't need it? And why did it say how many wheels?", + "A": "Because the skill this video is trying to teach is how to look at a word problem and find the information that you need. So Sal said the time and how many wheels so you could find out what was the important part of the problem.", + "video_name": "6QZCj4O9sk0", + "timestamps": [ + 24 + ], + "3min_transcript": "The local grocery store opens at nine. Its parking lot has six rows. Each row can fit seven cars. Each car has four wheels. How many cars can the parking lot fit? And I encourage you to pause the video and think about this yourself. Try to figure it out on your own. So, let's re-read this. The local grocery store opens up at nine. Well, that doesn't really matter. If we're thinking about how many cars can the parking lot fit. So we don't really have to care about that. We also dont have to care about how many wheels each car has. They're not asking us how many wheels can fit in the parking lot. So we can ignore that. What we really care about is how many rows we have. And how many cars can fit in each row. What we have is -- We have six rows and each row can fit seven cars. We're going to six groups of seven. Or, another way of thinking about it. We're going to have six times seven cars can fit in the parking lot. What is this going to be equal to? This is literally six sevens added up. This is the same thing as one, two, three, four, five, six. Now we're going to add these up. Seven plus seven is fourteen. Twenty-one, twenty-eight, thirty-five, forty-two. Six times seven is equal to forty-two. So forty-two cars can fit in the parking lot. Don't believe me? I made a little diagram here. We have six rows. This is the first row. Third.. Fourth.. Fifth.. Sixth. Each row can fit seven cars. You see it here. One; Let me make that a little brighter. One.. Three.. Four, five, six, seven. How many cars are there? You have seven. Fourteen.. Twenty-one.. Twenty-eight.. Thirty-five.. Forty-two total cars. Six rows of seven.", + "qid": "6QZCj4O9sk0_24" + }, + { + "Q": "The only thing I don't understand is the (x-c) part at 3:15, why put the c and not simply use x, x^2, x^3 and so on, like on the Mclaurin series?", + "A": "It s a shift. It s like shifting the parabola function, y = x^2, three places to the left. You d write it as y = (x+3)^2. To shift it c to the left, you d use (x-c)^2. Or,in the case of the Taylor expansion, multiply the derivative(s) by (x-c).", + "video_name": "1LxhXqD3_CE", + "timestamps": [ + 195 + ], + "3min_transcript": "", + "qid": "1LxhXqD3_CE_195" + }, + { + "Q": "At 1:18 sal says 0/0 is undefined why?", + "A": "Division by zero is an operation for which you cannot find an answer, so it is disallowed. You can understand why if you think about how division and multiplication are related. 12 divided by 6 is 2 because: 6 times 2 is 12 Now image 12/0: 12 divided by 0 is x would mean that: 0 times x = 12 But no value would work for x because 0 times any number is 0. So division by zero doesn t work.", + "video_name": "Z8j5RDOibV4", + "timestamps": [ + 78 + ], + "3min_transcript": "We're asked to divide. And we're dividing six plus three i by seven minus 5i. And in particular, when I divide this, I want to get another complex number. So I want to get some real number plus some imaginary number, so some multiple of i's. So let's think about how we can do this. Well, division is the same thing -- and we rewrite this as six plus three i over seven minus five i. These are clearly equivalent; dividing by something is the same thing as a rational expression where that something is in the denominator, right over here. And so how do we simplify this? Well, we have a tool in our toolkit that can make sure that we don't have an imaginary or complex number in the denominator. And that's the complex conjugate. If we multiply both the numerator and the denominator of this expression by the complex conjugate of the denominator, then we will have a real number in the denominator. So let's do that. Let's multiply the numerator and the denominator by the conjugate of this. So seven PLUS five i. Seven plus five i is the complex conjegate of seven minus five i. And anything divided by itself is going to be one (assuming you're not dealing with zero; zero over zero is undefined). But seven plus five i over seven plus five i is one. So we're not changing the value of this. But what this does is it allows us to get rid of the imaginary part in the denominator. So let's multiply this out. Our numerator -- we just have to multiply every part of this complex number times every part of this complex number. You can think of it as FOIL if you like; we're really just doing the distributive property twice. We have six times seven, which is forty two. And then we have six times five i, which is thirty i. So plus thirty i. And then we have three i times seven, so that's plus twenty-one i. Three times five is fifteen. But we have i times i, or i squared, which is negative one. So it would be fifteen times negative one, or minus 15. So that's our numerator. And then our denomenator is going to be -- Well, we have a plus b times a minus b. (You could think of it that way. Or we could just do what we did up here. Actually, let's just do what we did up here so you don't have to remember that difference of squares pattern and all that.) Seven times seven is forty-nine. Let's think of it in the FOIL way, if that is helpful for you. So first we did the 7X7. And we can do the outer terms. 7 X 5i is +35i. Then we can do the inner terms. -5i X 7 is -35i. These two are going to cancel out. And then -5i X 5i is -25i^2 (\"negative twenty five i squared\").", + "qid": "Z8j5RDOibV4_78" + }, + { + "Q": "When, at 1:09 he shows what happens as a perfect square for A and B (that b is 7y, because 49 is a perfect square) would that work for say, 5, where for example it might be b=sqrt(5)? Or am I jumping to conclusions?", + "A": "it wouldn t be a perfect square, so no.", + "video_name": "tvnOWIoeeaU", + "timestamps": [ + 69 + ], + "3min_transcript": "Factor x squared minus 49y squared. So what's interesting here is that well x squared is clearly a perfect square. It's the square of x. And 49y squared is also a perfect square. It's the square of 7y. So it looks like we might have a special form here. And to remind ourselves, let's think about what happens if we take a plus b times a minus b. I'm just doing it in the general case so we can see a pattern here. So over here, this would be a times a, which would be a squared plus a times negative b, which would be negative ab plus b times a or a times b again, which would be ab. And then you have b times negative b, so it would b minus b squared. Now these middle two terms cancel out. Negative ab plus ab, they cancel out and you're left with just a squared minus b squared. And that's the exact pattern we have here. We have an a squared minus a b squared. So in this case, a is equal to x and b is equal to 7y. So we can expand this as the difference of squares, or actually this thing right over here is the difference of squares. So we expand this like this. So this will be equal to x plus 7y times x minus 7y. And once again, we're just pattern matching based on this realization right here. If I take a plus b times a minus b, I get a difference of squares. This is a difference of squares. So when I factor it, it must come out to the result of something that looks like a plus b times a minus b or x plus 7y times x minus 7y.", + "qid": "tvnOWIoeeaU_69" + }, + { + "Q": "At 1:29, Sal says that 0.1 is bigger than 0.070. How is that possible when 0.070 has more digits than 0.1", + "A": "One of the easiest ways to compare decimals, it to give them the same number of decimal digits. Adding zeros on the right of the decimal does not change the original value of the number. Change 0.1 into 0.100 You are now comparing 0.100 to 0.070 100 is bigger than 70. Thus, 0.1 is larger then 0.070 Another way to look at this is the the further to the right the number is following the decimal point, the smaller the number. The 1 in 0.1 = 1/10. The 7 in 0.070 = 7/100 7/100 is much smaller than 1/10 Hope this helps.", + "video_name": "gAV9kwvoD6s", + "timestamps": [ + 89 + ], + "3min_transcript": "Let's compare 0.1 to 0.070. So this 1 right over here, it is in the tenths place. So it literally represents 1 times 1/10, which is obviously the same thing as 1/10. Now, when we look at this number right over here, it has nothing in the tenths place. It has 7 in the hundredths place. So this is the hundredths place right over here. And then it also has nothing in the thousandths place. So this number can be rewritten as 7 times 1/100, or 7/100. And now we could compare these two numbers. And there's two ways you could think about it. You could try to turn 1/10 into hundredths. And the best way to do that, if you want the denominator to be increased by a factor of 10, you need to do the same thing to the numerator. So all I did is I multiplied the numerator and denominator by 10. Ten 100's is the exact same thing as 1/10. And here it becomes very clear, 10/100 Another way you could think about this is, look, if you were to increment by hundredths here, you would start at 7/100, 8/100, 9/100, and then you would get to 10/100. So then you would get to that number. So this number, multiple ways you could think about it, is definitely larger. So let me write this down. This is definitely larger, greater than. This is greater than that. The greater than symbol opens to the larger value. So here we have 0.093 and here we have 0.01. So let's just think about this a little bit. So this 9-- get a new color here. This 9 is not in the tenths, the hundredths. It's in the thousandths place. It's in the thousands place. And this 3 is in the-- I'm running out of colors again. This 3 is in the ten thousandths place. So the 3 is in the ten thousandths place. And if you just wanted to write it in terms of ten thousandths, you can multiply the 9 and 1,000 by 0. And so it becomes 90/10,000. And if you want to add them together, you could, of course, write this as 93/10,000. Ten thousandths. I always have trouble with that \"-ths\" at the end. Now, let's think about this number right over here, 0.01. Well, this 1 right over here is in the hundredths place. It's in the hundredths place. So it literally represents 1/100. So how can we compare 1/100 to 93/10,000? So the best way to think about it is, well, what's 1/100 in terms of ten thousandths? Well, let's just multiply both the numerator and the denominator here by 10 twice. Or you could say, let's multiply them both by 100.", + "qid": "gAV9kwvoD6s_89" + }, + { + "Q": "at 4:52 he says over 2 does that apply all the time or just for this instance?", + "A": "The midpoint formula is ((x1+x2)/2,(y1+y2)/2). This applies all the time.", + "video_name": "Efoeqb6tC88", + "timestamps": [ + 292 + ], + "3min_transcript": "Well along the imaginary axis we're going from negative one to three so the distance there is four. So now we can apply the Pythagorean theorem. This is a right triangle, so the distance is going to be equal to the distance. Let's just say that this is x right over here. x squared is going to be equal to seven squared, this is just the Pythagorean theorem, plus four squared. Plus four squared or we can say that x is equal to the square root of 49 plus 16. I'll just write it out so I don't skip any steps. 49 plus 16, now what is that going to be equal to? That is 65 so x, that's right, 59 plus another 6 is 65. x is equal to the square root of 65. There's no factors that are perfect squares here, this is just 13 times five so we can just leave it like that. x is equal to the square root of 65 so the distance in the complex plane between these two complex numbers, square root of 65 which is I guess a little bit over eight. Now what about the complex number that is exactly halfway between these two? Well to figure that out, we just have to figure out what number has a real part that is halfway between these two real parts and what number has an imaginary part that's halfway between these two imaginary parts. So if we had some, let's say that some complex number, let's just call it a, is the midpoint, it's real part is going to be the mean of these two numbers. So it's going to be two plus negative five. Two plus negative five over two, over two, and it's imaginary part is going to be the mean of these two numbers so plus, plus three minus one. and this is equal to, let's see, two plus negative five is negative three so this is negative 3/2 plus this is three minus 1 is negative, is negative two over two is let's see three, make sure I'm doing this right. Three, something in the mean, three minus one is two divided by two is one, so three plus three. Negative 3/2 plus i is the midpoint between those two and if we plot it we can verify that actually makes sense. So real part negative 3/2, so that's negative one, negative one and a half so it'll be right over there and then plus i so it's going to be right over there.", + "qid": "Efoeqb6tC88_292" + }, + { + "Q": "at 1:10 couldn't \"a\" be a different value", + "A": "Yes, it can. A variable can be any number you choose it to be. For example, X is the most used variable but, in the video Sal used a as the variable. This has no effect in the equation.", + "video_name": "P6_sK8hRWCA", + "timestamps": [ + 70 + ], + "3min_transcript": "Let's say that you started off with 3 apples. And then I were to give you another 7, another 7 apples. So my question to you-- and this might be very obvious-- is how many apples do you now have? And I'll give you a second to think about that. Well, this is fairly basic. You had 3 apples. Now, I'm going to give you 7 more. You now have 3 plus 7. You now have 10 apples. But let's say I want to do the same type of thinking, but I'm too lazy to write the word \"apples.\" Let's say instead of writing the word \"apples,\" I just use the letter a. And let's say this is, say, a different scenario. You start off with 4 apples. And to that, I add another 2 apples. How many apples do you now have? Instead of writing apples, I'm just going to write a's here. So how many of these a's do you now have? And once again, I'll give you a few seconds This also might be a little bit of common sense for you. If you had 4 of these apples or whatever these a's represented, if you had 4 of them and then you add 2 more of them, you're now going to have 6 of these apples. But once again, we started off assuming that a's represent apples. But they could have represented anything. If you have 4 of whatever a represents, and then you have another 2 of whatever a represents, you'll now have 6 of whatever a represents. Or if you just think of it if I have 4 a's, and then I add another 2 a's, I'm going to have 6 a's. You can literally think of 4 a's as a plus a plus a plus a. And if to that, I add another 2 a's-- so plus a plus a, that's 2 a's right over there-- how many a's do I now have? Well, that's 1, 2, 3, 4, 5, 6. I now have 6 a's. So thinking of it that way, let's get a little bit more abstract. Let's say that I have 5 x's, whatever x represents. So I have 5 of whatever that number is. And from that, I subtract 2 of whatever that number is. What would this evaluate to? How many of these x's would I now have? So it's essentially 5x minus 2x is going to be what times x? Once again, I'll give you a few seconds to think about it. Well, if I have 5 of something and I subtract 2 of those away, I'm going to have 3 of that something left. So this is going to be equal to 3x. 5x minus 2x is equal to 3x. And if you really think about what that means, five x's are just x plus x plus x plus x plus x. And then we're going to take away two of those x's. So take away one x, take away two x's. You are going to be left with three x's.", + "qid": "P6_sK8hRWCA_70" + }, + { + "Q": "This might seem like a pretty arbitrary question, but at 1:37 Sal uses the \"/\" symbol to signify x(a+3-b) over (a+3-b). I've been seeing that \"over than\" symbol a lot in algebra and I was wondering if there was any difference between the over than symbol, \"/\", and the division symbol I'm used to seeing, \"\u00c3\u00b7\"?", + "A": "They mean the same thing. I m honestly not sure why, but once you get past pre-algebra, you stop seeing the \u00c3\u00b7 symbol.", + "video_name": "adPgapI-h3g", + "timestamps": [ + 97 + ], + "3min_transcript": "- [Voiceover] So we have an equation. It says, a-x plus three-x is equal to b-x plus five. And what I want to do together is to solve for x, and if we solve for x it's going to be in terms of a, b, and other numbers. So pause the video and see if you can do that. All right now, let's do this together, and what I'm going to do, is I'm gonna try to group all of the x-terms, let's group all the x-terms on the left-hand side. So, I already have a-x and three-x on the left-hand side. Let's get b-x onto the left-hand side as well, and I can do that by subtracting b-x from both sides. And if I subtract b-x from both sides, I'm going to get on the right-hand side, I'm going to have a, or on the left-hand side, a-x plus three-x minus b-x, so I can do that in that color for fun, minus b-x, and that's going to be equal to... Well, b-x minus b-x is just zero, and I have five. It is equal to five. I can factor an x out of this left-hand side of this equation, out of all of the terms. So, I can rewrite this as x times... Well, a-x divided by x is a. Three-x divided by x is three, and then negative b-x divided by x is just going to be negative b. I could keep writing it in that pink color. And that's all going to be equal to five. And now, to solve for x I can just divide both sides by, the thing that x is being multiplied by, by a plus three minus b. So, I can divide both sides by a plus three minus b. A plus three minus b. On this side, they cancel out. And, I have x is equal to five over a plus three minus b, and we are done. Let's do one more of these. We have a... Here we have a times the quantity, five minus x, is equal to b-x minus eight. So, once again, pause the video and see if you can solve for x. Well, the way I like to approach these is let's just expand everything all out. So, let me just distribute this a, and then I'm gonna collect all the x-terms on one side, and all of the non-x-terms on the other side, and essentially do what I just did in the last example. So, let's first distribute this a. So the left-hand side becomes five-a, I could say a times five or five-a, minus a-x, a-x, that is going to be equal to b-x minus eight. Now we can subtract b-x from both sides. So, we're gonna subtract b-x from the left-hand side, b-x from the right-hand side. And, once again, the whole reason I'm doing that, I want all the x-terms on the left, and all the non-x-terms on the right. And, actually, since I want all the non-x-terms on the right,", + "qid": "adPgapI-h3g_97" + }, + { + "Q": "At 4:06, Sal multiplies the numbers by -1. Do you have to do this?", + "A": "If you didn t do it, you would get: x = (-8-5a)/((-a-b) This fraction is not fully reduced. You would need to factor out a -1 from both the numerator and denominator to reduce the fraction.", + "video_name": "adPgapI-h3g", + "timestamps": [ + 246 + ], + "3min_transcript": "We have a... Here we have a times the quantity, five minus x, is equal to b-x minus eight. So, once again, pause the video and see if you can solve for x. Well, the way I like to approach these is let's just expand everything all out. So, let me just distribute this a, and then I'm gonna collect all the x-terms on one side, and all of the non-x-terms on the other side, and essentially do what I just did in the last example. So, let's first distribute this a. So the left-hand side becomes five-a, I could say a times five or five-a, minus a-x, a-x, that is going to be equal to b-x minus eight. Now we can subtract b-x from both sides. So, we're gonna subtract b-x from the left-hand side, b-x from the right-hand side. And, once again, the whole reason I'm doing that, I want all the x-terms on the left, and all the non-x-terms on the right. And, actually, since I want all the non-x-terms on the right, So I'm kind of doing two steps at once, here, but hopefully it makes sense. I'm trying to get rid of the b-x here, and I'm trying to get rid of the five-a here. So, I subtract five-a there, and I'll subtract five-a there, and then let's see what this give us. So, the five-a's cancel out. And, on the left-hand side, I have negative a-x, negative a-x, minus b-x, minus, you know, in that same green color, minus b-x. And on the right-hand side, This is going to be equal to, the b-x's cancel out, and I have negative eight minus five-a. Negative eight minus, in that same magenta color, minus five-a. And let's see, I have all my x's on one side, all my non-x's on the other side. And here I can factor out an x, And, actually, one thing that might be nice. Let me just multiply both sides by negative one. If I multiple both sides by negative one, I get a-x plus b-x, plus b-x is equal to eight plus five-a. That just gets rid of all of those negative signs. And now I can factor out an x here. So let me factor out an x, and I get x times a plus b. A plus b is going to be equal to eight plus five-a. Eight plus five-a. And we're in the home stretch now. We can just divide both sides by a plus b. So we could divide both sides by a plus b. A plus b. And we're going to be left with, x is equal to", + "qid": "adPgapI-h3g_246" + }, + { + "Q": "Around 1:30, he explains that we need to use the pythagorean theory to find the radius r. But can't we just estimate the no. of units from the centre (point -1, 1) to the point (7.5, 1), which also lies on the circumference?", + "A": "sometimes you will need precision in your answers. Pythagorean Theorem will give you as much precision as you need", + "video_name": "iX5UgArMyiI", + "timestamps": [ + 90 + ], + "3min_transcript": "- [Voiceover] So we have a circle here and they specified some points for us. This little orangeish, or, I guess, maroonish-red point right over here is the center of the circle, and then this blue point is a point that happens to sit on the circle. And so with that information, I want you to pause the video and see if you can figure out the equation for this circle. Alright, let's work through this together. So let's first think about the center of the circle. And the center of the circle is just going to be the coordinates of that point. So, the x-coordinate is negative one and then the y-coordinate is one. So center is negative one comma one. And now, let's think about what the radius of the circle is. Well, the radius is going to be the distance between the center and any point on the circle. So, for example, for example, this distance. The distance of that line. Let's see I can do it thicker. A thicker version of that. This line, right over there. Something strange about my pen tool. It's making that very thin. Let me do it one more time. Okay, that's better. (laughs) The distance of that line right over there, that is going to be the radius. So how can we figure that out? Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean Theorem. To figure out the length of that line, so this is the radius, we could figure out a change in x. So, if we look at our change in x right over here. Our change in x as we go from the center to this point. So this is our change in x. And then we could say that this is our change in y. That right over there is our change in y. And so our change in x-squared plus our change in y-squared is going to be our radius squared. That comes straight out of the Pythagorean Theorem. This is a right triangle. is going to be equal to our change in x-squared plus our change in y-squared. Plus our change in y-squared. Now, what is our change in x-squared? Or, what is our change in x going to be? Our change in x is going to be equal to, well, when we go from the radius to this point over here, our x goes from negative one to six. So you can view it as our ending x minus our starting x. So negative one minus negative, sorry, six minus negative one is equal to seven. So, let me... So, we have our change in x, this right over here, is equal to seven. If we viewed this as the start point and this as the end point, it would be negative seven, but we really care about the absolute value in the change of x, and once you square it", + "qid": "iX5UgArMyiI_90" + }, + { + "Q": "At 1:40 why does Sal factor out -e^(-st) and not (-e^(-st))/s? Isn't it possible to factor that out as well?", + "A": "Yes, but it helps better with the evaluation later on since you d end up with something weird like 1/infinity/infinity or something like that.", + "video_name": "-cApVwKR1Ps", + "timestamps": [ + 100 + ], + "3min_transcript": "Welcome back. We were in the midst of figuring out the Laplace transform of sine of at when I was running out of time. This was the definition of the Laplace transform of sine of at. I said that also equals y. This is going to be useful for us, since we're going to be doing integration by parts twice. So I did integration by parts once, then I did integration by parts twice. I said, you know, don't worry about the boundaries of the integral right now. Let's just worry about the indefinite integral. And then after we solve for y-- let's just say y is the indefinite version of this-- then we can evaluate the boundaries. And we got to this point, and we made the realization, after doing two integration by parts and being very careful not to hopefully make any careless mistakes, we realized, wow, this is our original y. If I put the boundaries here, that's the same thing as the Laplace transform of sine of at, right? That's our original y. So now-- and I'll switch colors just avoid monotony-- this is equal to, actually, let me just-- this is y. So let's add a squared over sine squared y to both sides of this. So this is equal to y plus-- I'm just adding this whole term to both sides of this equation-- plus a squared over s squared y is equal to-- so this term is now gone, so it's equal to this stuff. And let's see if we can simplify this. So let's factor out an e to the minus st. Actually, let's factor out a negative e to the minus st. So it's minus e to the minus st, times sine of-- well, let me just write 1 over s, sine of at, minus 1 over s squared, cosine of at. And so this, we can add the coefficient. So we get 1 plus a squared, over s squared, times y. But that's the same thing as s squared over s squared, plus a squared over s squared. So it's s squared plus a squared, over s squared, y is equal to minus e to the minus st, times this whole thing, sine of at, minus 1 over s squared, cosine of at. And now, this right here, since we're doing everything with respect to dt, this is just a constant, right? So we can say a constant times the antiderivative is equal to this. This is as good a time as any to evaluate the boundaries. If this had a t here, I would have to somehow get them back", + "qid": "-cApVwKR1Ps_100" + }, + { + "Q": "At 3:59, shouldn't Sal get 62__ something?\n\nA confuzzled child.", + "A": "No, he got the right answer but he made his 0 look like a 6. I also got confused but saw his mistake.", + "video_name": "TvSKeTFsaj4", + "timestamps": [ + 239 + ], + "3min_transcript": "We can divide both sides of this equation by 0.25, or if you recognize that four quarters make a dollar, you could say, let's multiply both sides of this equation by 4. You could do either one. I'll do the first, because that's how we normally do algebra problems like this. So let's just multiply both by 0.25. That will just be an x. And then the right-hand side will be 150 divided by 0.25. And the reason why I wanted to is really it's just good practice dividing by a decimal. So let's do that. So we want to figure out what 150 divided by 0.25 is. And we've done this before. When you divide by a decimal, what you can do is you can make the number that you're dividing into the other number, you can turn this into a whole number by essentially shifting the decimal two to the right. But if you do that for the number in the denominator, you also have to do that to the numerator. So right now you can view this as 150.00. decimal two to the right. Then you'd also have to do that with 150, so then it becomes 15,000. Shift it two to the right. So our decimal place becomes like this. So 150 divided by 0.25 is the same thing as 15,000 divided by 25. And let's just work it out really fast. So 25 doesn't go into 1, doesn't go into 15, it goes into 150, what is that? Six times, right? If it goes into 100 four times, then it goes into 150 six times. 6 times 0.25 is-- or actually, this is now a 25. We've shifted the decimal. This decimal is sitting right over there. So 6 times 25 is 150. You subtract. You get no remainder. Bring down this 0 right here. 25 goes into 0 zero times. 0 times 25 is 0. Subtract. No remainder. 25 goes into 0 zero times. 0 times 25 is 0. Subtract. No remainder. So 150 divided by 0.25 is equal to 600. And you might have been able to do that in your head, because when we were at this point in our equation, 0.25x is equal to 150, you could have just multiplied both sides of this equation times 4. 4 times 0.25 is the same thing as 4 times 1/4, which is a whole. And 4 times 150 is 600. So you would have gotten it either way. And this makes total sense. If 150 is 25% of some number, that means 150 should be 1/4 of that number. It should be a lot smaller than that number, and it is. 150 is 1/4 of 600. Now let's answer their actual question. Identify the percent. Well, that looks like 25%, that's the percent. The amount and the base in this problem.", + "qid": "TvSKeTFsaj4_239" + }, + { + "Q": "At around 3:36 Sal said that at 0 hours there were 550 pages left to read. Couldn't of Naoya also read in minutes? So there aren't really exactly 550 pages in the book right?", + "A": "Merely changing to minutes would not change the y-intercept because the rate would be equivalent (55 pages/hour = 55/60 or 11/12 pages/minute). So after 240 minutes (4*60), Naoya still had 330 pages to read. so 240 * 11/12 = 220 pages. Think about what is happening, you are changing pages/hour by dividing by 60 then you are multiplying by 60 to convert to minutes, the two 60s cancel each other out so 55/60 * 4 * 60 is the same as 55 * 4.", + "video_name": "W3flX500w5g", + "timestamps": [ + 216 + ], + "3min_transcript": "would be 385. Now let's see whether this makes sense. So when our change in time, this triangle is just the Greek letter Delta, means \"change in.\" When our change in time is plus one, plus one hour, our change in pages left to read is going to be equal to negative 55 pages. And that makes sense, the pages left to read goes down every hour. We're measuring not how much he's read, we're measuring how much he has left to read. So that should go down by 55 pages every hour. Or if we were to go backwards through time, it should go up. So at two hours he should have 55 more pages to read. So what's 385 plus 55? We'll let's see, 385 plus 5 is 390, plus 50 is 440. So he'd have 440 pages, and all I did is I added 55. he would have 55 more pages than after reading for two hours. So 440 plus 55 is 495. And then before he started reading, or right when he started reading, he would have had to read even 55 more pages, 'cause after one hour, he would have read those 55 pages. So 495 plus 55 is going to be, let's see, it's gonna be, add 5, you get to 500 plus another 50 is 550 pages. So at time equals zero had had 550 pages to read. So that's how long the book is. But how long does it take Naoya to read the entire book? Well we could keep going. We could say, \"Okay, at the fifth hour, \"this thing's gonna go down by 55.\" So let's see, if this goes down by 50, if this goes down by 50, we're going to get to 280, but then you go down five more, it's gonna go to 275, and we could keep going on and on and on. \"He's got 330 pages left to read, and he's gonna,\" Let's see, let me write this, let me write the units down. \"Pages, and he's reading at a rate of 55 pages per hour, \"pages per hour, this is the same thing, \"this is going to be equal to 330 pages \"times one over 55 hours per page.\" I'm dividing by something, the same thing as multiplying by its reciprocal, so 55 pages per hours, if you divide by that, that's the same thing as multiplying by 1/55th of an hour per page, is one way to think about it. And so what do you get? The pages cancel out, pages divided by pages, and you have 330 divided by 55 hours. 330 divided by 55 hours. And what's that going to be? Let's see, 30 divided by 5 is 6, 300 divided by 50 is 6, so this is going to be equal to 6 hours.", + "qid": "W3flX500w5g_216" + }, + { + "Q": "At 1:10, where did you get -66?", + "A": "He s factoring by grouping . He starts by taking a (the coefficient of f^2) and multiplies it by b (the coefficient of f). Hope this helps :)", + "video_name": "d-2Lcp0QKfI", + "timestamps": [ + 70 + ], + "3min_transcript": "We need to factor negative 12f squared minus 38f, plus 22. So a good place to start is just to see if, is there any common factor for all three of these terms? When we look at them, they're all even. And we don't like a negative number out here. So let's divide everything, or let's factor out a negative 2. So this expression right here is the same thing as negative 2 times-- what's negative 12f squared divided by negative 2? It's positive 6f squared. Negative 38 divided by negative 2 is positive 19, so it'll be positive 19f. And then 22 divided by negative 22-- oh, sorry, 22 divided by negative 2 is negative 11. So we've simplified it a bit. We have the 6f squared plus 19f, minus 11. We'll just focus on that part right now. And the best way to factor this thing, since we don't have a 1 here as the coefficient on the f squared, is to factor it by grouping. times negative 11. So two numbers, so a times b, needs to be equal to 6 times negative 11, or negative 66. And a plus b needs to be equal to 19. So let's try a few numbers here. So let's see, 22, I'm just thinking of numbers that are roughly 19 apart, because they're going to be of different signs. So 22 and 3, I think will work. Right. If we take 22 times negative 3, that is negative 66, and 22 plus negative 3 is equal to 19. And the way I kind of got pretty close to this number is, well, you know, they're going to be of different signs, so the positive versions of them have to be about 19 apart, and that worked out. 22 and negative 3. So now we can rewrite this 19f right here as the sum of That's the same thing as 19f. I just kind of broke it apart. And, of course, we have the 6f squared and we have the minus 11 here. Now, you're probably saying, hey Sal, why did you put the 22 here and the negative 3 there? Why didn't you do it the other way around? Why didn't you put the 22 and then the negative 3 there? And my main motivation for doing it, I like to put the negative 3 on the same side with the 6 because they have the common factor of the 3. I like to put the 22 with the negative 11, they have the same common factor of 11. So that's why I decided to do it that way. So now let's do the grouping. And, of course, you can't forget this negative 2 that we have sitting out here the whole time. So let me put that negative 2 out there, but that'll just kind of hang out for awhile. But let's do some grouping. So let's group these first two. And then we're going to group this-- let me get a nice color here-- and then we're going to group this second two.", + "qid": "d-2Lcp0QKfI_70" + }, + { + "Q": "How do you know which numbers are a and b in the equation? At 13:21 he says that a=3 and b=4 (because he took the square roots) but how do you know that a isn't 4 and b isn't 3? For ellipses you know that a is the bigger number, but what do you know for hyperbolas? Thanks!", + "A": "a is 3 so 3 squared gives u nine and b is 4 so 4 squared gives u 16.", + "video_name": "S0Fd2Tg2v7M", + "timestamps": [ + 801 + ], + "3min_transcript": "the focal length is the same on either side of the center of the hyperbola depending on how you may view it, but I think that's not too much of a stretch of a statement for you to for you to accept. So if this distance is the same as this distance, then the magenta distance minus this blue distance is going to be equal to this green distance. And this green distance is what? That's 2a. We saw that at the beginning of this video. So this, once again, is also equal to 2a. Anyway, I'll leave you there right now. Actually, let's actually just do one problem, just because I like to make one concrete. Because I told you at the beginning that if you wanted to find the-- so if you have an ellipse-- so if you have-- this is an ellipse, x squared over a squared plus y squared over b squared is equal to 1, we learned that the-- that's over b squared-- this is an ellipse. square root of a squared minus b squared. Now for a hyperbola, you kind of see that there's a very close relation between the ellipse and the hyperbola, but it is kind of a fun thing to ponder about. And a hyperbola's equation looks like this. x squared over a squared minus y squared over b squared, or it could be y squared over b squared minus x squared over a square is equal to 1. It turns out, and I'll prove this to you in the next video, it's a little bit of a hairy math problem, that the focal length of a hyperbola is equal to the square root of the sum of these two numbers, is equal to the sum of a squared plus b squared. So if I were to give you-- so notice the difference. It's just a difference in sign. You're taking the difference of those two denominators, and now you're taking the sum of the two denominators. So if I were to give you the following hyperbola. x squared over 9 plus y squared over 16 is equal to 1. we could just figure out the focal length just by plugging into the formula. The focal length is equal to the square root of a squared plus b squared. This is squared, right? a is three. b is 4. So 9 plus 16 is 25, which is equal to 5. And so if we were to graph this-- that's my y-axis, that's my x-axis-- and the focal length is the distance to, in this case, to the left and the right of the origin. If it was kind of an up and down opening hyperbola, it would be above and below the origin, so this is a-- oh sorry, this should be a plus. We're doing with a hyperbola, that should be a minus. Don't want to confuse you. What I had written before, with a plus, that would have been an ellipse. A minus is the hyperbola. So the two asymptotes-- this is centered at the origin, it hasn't been shifted-- are going to be 16 over 9, so it's going", + "qid": "S0Fd2Tg2v7M_801" + }, + { + "Q": "At 6:32, Sal says function is undefined at x2. He meant is non-differentiable, because it is defined right?", + "A": "I definitely think that Sal is trying to say where the DERIVATIVE is undefined. Because when the function is undefined at a point, we will not have a critical point because this point does not exist: ie, it is not defined", + "video_name": "lDY9JcFaRd4", + "timestamps": [ + 392 + ], + "3min_transcript": "it's going to be a minimum or maximum. And we see the intuition here. If you have-- so non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or a maximum point, at some point x is equal to a, and x isn't the endpoint of some interval, this tells you something interesting. Or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0. Or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0, derivative is 0, derivative is undefined. And we have a word for these points where the derivative is We called them critical points. So for the sake of this function, the critical points are, we could include x sub 0, we could include x sub 1. At x sub 0 and x sub 1, the derivative is 0. And x sub 2, where the function is undefined. Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. But can we say it the other way around? If we find a critical point, where the derivative is 0, or the derivative is undefined, is that going to be a maximum or minimum point? And to think about that, let's imagine this point right over here. So let's call this x sub 3. if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0. So based on our definition of critical point, x sub 3 would also be a critical point. But it does not appear to be a minimum or a maximum point. So a minimum or maximum point that's not an endpoint, it's definitely going to be a critical point. But being a critical point by itself does not mean you're at a minimum or maximum point. So just to be clear that all of these points were at a minimum or maximum point. This were at a critical point, all of these are critical points. But this is not a minimum or maximum point. In the next video, we'll start to think about how you can differentiate, or how you can tell, whether you have a minimum or maximum at a critical point.", + "qid": "lDY9JcFaRd4_392" + }, + { + "Q": "7:12 We can create tangent lines at a point that crosses the function?", + "A": "Yes, a line can be tangent at one point on a curve but then cross it later when the curve takes a U-turn later on down the road.", + "video_name": "lDY9JcFaRd4", + "timestamps": [ + 432 + ], + "3min_transcript": "it's going to be a minimum or maximum. And we see the intuition here. If you have-- so non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or a maximum point, at some point x is equal to a, and x isn't the endpoint of some interval, this tells you something interesting. Or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0. Or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0, derivative is 0, derivative is undefined. And we have a word for these points where the derivative is We called them critical points. So for the sake of this function, the critical points are, we could include x sub 0, we could include x sub 1. At x sub 0 and x sub 1, the derivative is 0. And x sub 2, where the function is undefined. Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. But can we say it the other way around? If we find a critical point, where the derivative is 0, or the derivative is undefined, is that going to be a maximum or minimum point? And to think about that, let's imagine this point right over here. So let's call this x sub 3. if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0. So based on our definition of critical point, x sub 3 would also be a critical point. But it does not appear to be a minimum or a maximum point. So a minimum or maximum point that's not an endpoint, it's definitely going to be a critical point. But being a critical point by itself does not mean you're at a minimum or maximum point. So just to be clear that all of these points were at a minimum or maximum point. This were at a critical point, all of these are critical points. But this is not a minimum or maximum point. In the next video, we'll start to think about how you can differentiate, or how you can tell, whether you have a minimum or maximum at a critical point.", + "qid": "lDY9JcFaRd4_432" + }, + { + "Q": "At 4:50, why don't the endpoints count as maxima/minima?", + "A": "They can be maxima/minima on a closed interval (but not always), but in the video, we are not interested in them because we only want to show point (non-endpoint) that is min/max will have f (a)= 0 or undefined. Endpoints are max/min but they don t necessarily have f (a)=0 or undefined. Hope that helps.", + "video_name": "lDY9JcFaRd4", + "timestamps": [ + 290 + ], + "3min_transcript": "I'm not being very rigorous. But you can see it just by looking at it. So that's fair enough. We've identified all of the maxima and minima, often called the extrema, for this function. Now how can we identify those, if we knew something about the derivative of the function? Well, let's look at the derivative at each of these points. So at this first point, right over here, if I were to try to visualize the tangent line-- let me do that in a better color than brown. If I were to try to visualize the tangent line, it would look something like that. So the slope here is 0. So we would say that f prime of x0 is equal to 0. The slope of the tangent line at this point is 0. What about over here? Well, once again, the tangent line would look something like that. So once again, we would say f prime at x1 is equal to 0. What about over here? We have a positive slope going into it, and then it immediately jumps to being a negative slope. So over here, f prime of x2 is not defined. Let me just write undefined. So we have an interesting-- and once again, I'm not rigorously proving it to you, I just want you to get the intuition here. We see that if we have some type of an extrema-- and we're not talking about when x is at an endpoint of an interval, just to be clear what I'm talking about when I'm talking about x as an endpoint of an interval. We're saying, let's say that the function is where you have an interval from there. So let's say a function starts right over there, and then This would be a maximum point, but it would be an end point. We're not talking about endpoints right now. We're talking about when we have points in between, or when our interval is infinite. So we're not talking about points like that, or points like this. We're talking about the points in between. it's going to be a minimum or maximum. And we see the intuition here. If you have-- so non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or a maximum point, at some point x is equal to a, and x isn't the endpoint of some interval, this tells you something interesting. Or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0. Or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0, derivative is 0, derivative is undefined. And we have a word for these points where the derivative is", + "qid": "lDY9JcFaRd4_290" + }, + { + "Q": "On problem 51. or (2:55)\n\nCan you prove to me why all the four right triangles are congruent? I know they are... but why? D:", + "A": "The shape in the middle is a square, so all four sides are equal. When it is placed inside another square, the triangles that it creates (4), are all congruent, because the four side lengths of a square are equivalent. If the shape in the middle were a rectangle that is not a square, there would be two congruent triangles, and another two congruent triangles, which are different from the first two. So, because they are both squares, the triangles are congruent.", + "video_name": "6EY0E3z-hsU", + "timestamps": [ + 175 + ], + "3min_transcript": "Fair enough. Now we can also say that the area of this larger square, and it's a bit of an optical illusion, it looks like it's tilted to the left because of the way it's drawn. But anyway, that the area of this larger square is also the area of these four triangles plus the area of this smaller square. So this, the area of the larger square, which we figured out just by taking one side of it and squaring it, that should be equal to the area of the four smaller triangles. So there's four of them. And what's the area of each of them. Let's see, let's just pick this one. 1/2 base times height. So it's 1/2 times a times b. So 1/2 ab is one of these and I multiply by 4 to get all four of these triangles. And then we want to add the area of this inside square. And that's just going to be c squared. Let's see if we can simplify this. So you get a squared plus 2ab plus b squared is equal to 4 times 1/2 is 2ab plus c squared. Well, we could subtract 2ab from both sides of this equation. The top and the bottom of this equation the way I've written it. But if we do that, subtract 2ab from there, subtract 2ab from there, and you're left with a squared plus b squared is equal to c squared, which is the Pythagorean theorem. And we've proved it. So let's see which of their choices matches what we did. OK, which statement would not be used in the proof of the Pythagorean theorem. The area of a triangle equals 1/2 ab. We used that. The four right triangles are congruent. The area of the inner square is equal to half of the area of the larger square. We didn't use that. I think this is the one that would not be used in the proof. Choice D, the area of the larger square is equal to the sum of the squares of the smaller square and the four congruent triangles. No, that that was the crux of the proof. So we definitely used that. So C is our answer. That's the statement that would not be used in the proof. I'm learning to copy and paste ahead of time. So I don't waste your time. All right, a right triangle's hypotenuse has length 5. If one leg has length 2, what is the length of the other leg? Pythagorean theorem, x squared plus 2 squared is equal to 5 squared, because 5 is the hypotenuse. x squared plus 4 is equal to 25.", + "qid": "6EY0E3z-hsU_175" + }, + { + "Q": "What does he mean at 5:39. \"But if I just cancelled these two things out, the new function would be defined when x = -8\". Why -8?", + "A": "Because when x= -8, the denominator is 0, so the function is undefined. But if we cancel the factors (x+8), then we could plug in x= -8 without dividing by 0, so the function would defined. So since one is defined at x= -8, and the other is not, cancelling the (x+8) changes the function, which we don t want.", + "video_name": "u9v_bakOIcU", + "timestamps": [ + 339 + ], + "3min_transcript": "Both of them have an x plus 8 in them. So we can factor out an x plus 8. So if we factor out an x plus 8, we're left with 2x minus 1, put parentheses around it, times the factored out x plus 8. So we've simplified the numerator. The numerator can be rewritten. And you could have gotten here using the quadratic formula as well. The numerator is 2x minus 1 times x plus 8. And now see if you can factor the denominator. And this one's more straightforward. The coefficient here is 1. So we just have to think of two numbers that when I multiply them, I get 16. And when I add them, I get 10. And the obvious one is 8 and 2, positive 8 and positive 2. So we can write this as x plus 2 times x plus 8. We can divide the numerator and denominator by x plus 8, assuming that x does not equal negative 8. Because this function right over here that's defined by f divided by g, it is not defined when g of x is equal to 0, because then you have something divided by 0. And the only times that g of x is equal to 0 is when x is equal to negative 2 or x is equal to negative 8. So if we divide the numerator and the denominator by x plus 8 to simplify it, in order to not change the function definition, we have to still put the constraint that x cannot be equal to negative 8. That the original function, in order to not change it-- because if I just cancelled these two things out, the new function with these canceled would be defined when x is equal to negative 8. But we want this simplified thing to be the same exact function. And this exact function is not defined when x is equal to negative 8. So now we can write f/g of x, which is really just f of x divided by g of x, is equal to 2x minus 1 You have to put the condition there that x cannot be equal to negative 8. If you lost this condition, then it won't be the exact same function as this, because this is not defined when x is equal to negative 8.", + "qid": "u9v_bakOIcU_339" + }, + { + "Q": "At 6:06 Khan is explaining why x can't be -8, but he didn't say that it can't be -2. So, does this mean that x can be -2? This will also give you a 0 in the denominator.", + "A": "No, x cannot be -2 either, but it is not necessary to specify it because x - 2 is still in the simplified expression. You have to put the x does not equal -8 because the term that would have made that clear has vanished.", + "video_name": "u9v_bakOIcU", + "timestamps": [ + 366 + ], + "3min_transcript": "Both of them have an x plus 8 in them. So we can factor out an x plus 8. So if we factor out an x plus 8, we're left with 2x minus 1, put parentheses around it, times the factored out x plus 8. So we've simplified the numerator. The numerator can be rewritten. And you could have gotten here using the quadratic formula as well. The numerator is 2x minus 1 times x plus 8. And now see if you can factor the denominator. And this one's more straightforward. The coefficient here is 1. So we just have to think of two numbers that when I multiply them, I get 16. And when I add them, I get 10. And the obvious one is 8 and 2, positive 8 and positive 2. So we can write this as x plus 2 times x plus 8. We can divide the numerator and denominator by x plus 8, assuming that x does not equal negative 8. Because this function right over here that's defined by f divided by g, it is not defined when g of x is equal to 0, because then you have something divided by 0. And the only times that g of x is equal to 0 is when x is equal to negative 2 or x is equal to negative 8. So if we divide the numerator and the denominator by x plus 8 to simplify it, in order to not change the function definition, we have to still put the constraint that x cannot be equal to negative 8. That the original function, in order to not change it-- because if I just cancelled these two things out, the new function with these canceled would be defined when x is equal to negative 8. But we want this simplified thing to be the same exact function. And this exact function is not defined when x is equal to negative 8. So now we can write f/g of x, which is really just f of x divided by g of x, is equal to 2x minus 1 You have to put the condition there that x cannot be equal to negative 8. If you lost this condition, then it won't be the exact same function as this, because this is not defined when x is equal to negative 8.", + "qid": "u9v_bakOIcU_366" + }, + { + "Q": "at 2:19 why has he divided 1* 10^4 upon 7*10^5 when it is written in the question 7*10^5 than 1*10^4?", + "A": "To find the quotient of exponents you subtract the exponents from the other. If you need more help, you can look at some other of Sal s Exponents videos.", + "video_name": "DaoJmvqU3FI", + "timestamps": [ + 139 + ], + "3min_transcript": "Let's do a few more examples from the orders of magnitude exercise. Earth is approximately 1 times 10 to the seventh meters in diameter. Which of the following could be Earth's diameter? So this is just an approximation. It's an estimate. And they're saying, which of these, if I wanted to estimate it, would be close or would be 1 times 10 to the seventh? And the key here is to realize that 1 times 10 to the seventh is the same thing as one followed by seven zeroes. One, two, three, four, five, six, seven. Let me put some commas here so we make it a little bit more Or another way of talking about it is that it is, 1 times 10 to the seventh, is the same thing as 10 million. So which of these, if we were to really roughly estimate, we would go to 10 million. Well, this right over here is 1.271 million, or 1,271,543. If I were to really roughly estimate it, I might go to one million, but I'm not going to go to 10 million. This is 12,715,430. If I were to roughly estimate this, well, yeah. I would go to 10 million. 10 million is if I wanted really just one digit to represent it, if I were write this in scientific notation. This right over here is 1.271543 times 10 to the seventh. Let me write that down. 12,715,430. If I were to write this in scientific notation as 1.271543 times 10 to the seventh. And when you write it this way, you say, hey, well, yeah, if I was to really estimate this and get pretty rough with it, and I just rounded this down, I would make this 1 times 10 to the seventh. So this really does look like our best choice. Now let me just verify. Well, this right over here, if I were to write it, I would go to 100 million, or 1 times 10 to the eighth. That's way too big. And this, if I were to write it, I would go to a billion, or 1 times 10 to the ninth. So that's also too big. So once again, this feels like the best answer. So here we're asked, how many times larger is 7 times 10 to the fifth than 1 times 10 to the fourth? Well, we could just divide to think about that. So 7 times 10 to the fifth divided by 1 times 10 to the fourth. Well, this is the same thing as 7 over 1 times 10 to the fifth over 10 to the fourth, which is just going to be equal to-- well, 7 divided by 1 is 7. And 10 to the fifth, that's multiplying five 10's. And then you're dividing by four 10's. You're going to have one 10 left over. Or, if you remember your exponent properties, this would be the same thing as 10 to the 5 minus 4 power, or 10 to the first power. So this right over here, all of this business, is going to simplify to 10 to the first, or I could actually write it this way. This would be the same thing as 10 to the 5 minus 4,", + "qid": "DaoJmvqU3FI_139" + }, + { + "Q": "Quick Question at 5:55 he writes cos^2 and theta and sin^2 theta now based on how he wrote it I am wondering if it is cos^(2 theta) or cos^(2) Theta?", + "A": "cos^2(\u00ce\u00b8) is the trig representation for (cos(\u00ce\u00b8))^2.", + "video_name": "n0DLSIOYBsQ", + "timestamps": [ + 355 + ], + "3min_transcript": "Say you know cosine theta then you use this to figure out sine of theta, then you can figure out tangent of theta because tangent of theta is just sine over cosine. If you're a little bit confused as to why this is called the Pythagorean identity, well it really just falls out of where the equation of a circle even came from. If we look at this point right over here, we look at this point right over here, which we're saying is the x coordinate is cosine theta and the y coordinate is sine of theta, what is the distance between that point and the origin? Well to think about that we can construct a right triangle. This distance right over here. So that we could deal with any quadrant I'll make it the absolute value of the cosine theta is this distance right over here. And this distance right over here is the absolute value of the sine of theta. for this first quadrant here but if I went into the other quadrants and I were to setup a similar right triangle then the absolute value is at play. What do we know from the Pythagorean theorem? This is a right triangle here, the hypotenuse has length one, so we know that this expression squared, the absolute value of cosine of theta squared, plus this expression squared, which is this length, plus the absolute value of the sine of theta squared needs to be equal to the length of the hypotenuse squared, which is the same thing which is going to be equal to one squared. Or we could say, this is the same thing. If you're going to square something the sign, if negative it's going to be negative times a negative so it's just going to be positive so this is going to be the same thing as saying that the cosine squared theta plus sine squared theta is equal to one. This is why it's called the Pythagorean identity. a circle comes from, it comes straight out of the Pythagorean theorem where your hypotenuse has length one.", + "qid": "n0DLSIOYBsQ_355" + }, + { + "Q": "At 2:30, when he says that the absolute value could also work in ensuring that the value is positive, can someone explain to me why the absolute value isn't used in the equation?", + "A": "In the video, Sal does some algebraic manipulation to achieve the formula of the parabola. It is much easier to derive the parabola if he were to square the expression and take the square root than to take the absolute value of the expression. Both methods yield the same value of the expression; however, the latter method (squaring then taking the square root) allows for more easier manipulation. Hope this helps!", + "video_name": "okXVhDMuGFg", + "timestamps": [ + 150 + ], + "3min_transcript": "- [Voiceover] What I have attempted to draw here in yellow is a parabola, and as we've already seen in previous videos, a parabola can be defined as the set of all points that are equidistant to a point and a line, and the point is called the focus of the parabola, and the line is called the directrix of the parabola. What I want to do in this video, it's gonna get a little bit of hairy algebra, but given that definition, I want to see, and given that definition, and given a focus at the point x equals a, y equals b, and a line, a directrix, at y equals k, to figure out what is the equation of that parabola actually going to be, and it's going to be based on a's, b's, and k's, so let's do that. So let's take a arbitrary point on the parabola. Let's say we take this point right over here, and its x-coordinate is x, and its y-coordinate is y, and by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix, That means that the distance to the directrix, which I'm drawing here in blue, has to be the same as the distance to the focus, which I am drawing in magenta, and when we take the distance to the directrix, we literally just drop a perpendicular, that is, that's going to be the shortest distance to that line, but the distance to the focus, well we see that's at a bit of an angle, and we might have to use the distance formula, which is really just the Pythagorean Theorem. So let's do that. This distance has to be the same as that distance. So, what's this blue distance? Well, that's just gonna be our change in y. It's going to be this y, minus k. It's just this distance. So it's going to be y minus k. Now we have to be careful. The way I've just drawn it, yes, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances, but you can definitely have a parabola where lower than the y-coordinate of the directrix, in which case this would be negative. So what we really want is the absolute value of this, or, we could square it, and then we could take the square root, the principle root, which would be equivalent to taking the absolute value of y minus k. So that's this distance right over here, and by the definition of a parabola, in order for (x,y) to be sitting on the parabola, that distance needs to be the same as the distance from (x,y) to (a,b), to the focus. So what's that going to be? Well, we just apply the distance formula, or really, just the Pythagorean Theorem. It's gonna be our change in x, so, x minus a, squared, plus the change in y, y minus b, squared, and the square root of that whole thing, the square root of all of that business. Now, this right over here is an equation of a parabola. It doesn't look like it, it looks really hairy,", + "qid": "okXVhDMuGFg_150" + }, + { + "Q": "Why does Sal keep saying \"the principal root of...\" as opposed to \"the square root of...\"? At 1:30 I heard him say \"the square root of, or the principal root of...\" so does that mean they're the same thing? Because it appears as if he sort of corrected himself.", + "A": "Each square root actually has 2 roots. Consider: 5^2 = 25, but if you square (-5), it is also 25. So, 5^2 = 25 and (-5)^2 = 25. Now, consider square root of 25. Is it 5 or is it -5? We need to know what value to use. So, when you see sqrt(25) , it is understood that the answer should be the principal root or the positive root = 5. If you see - sqrt(25) , the minus sign in front of the radical tells you that your answer should be the negative root = -5. Hope this helps.", + "video_name": "qFFhdLlX220", + "timestamps": [ + 90 + ], + "3min_transcript": "What I want to do in this video is resimplify this expression, 3 times the principal root of 500 times x to the third, and take into consideration some of the comments that we got out on YouTube that actually give some interesting perspective on how you could simplify this. So just as a quick review of what we did in the last video, we said that this is the same thing as 3 times the principal root of 500. And I'm going to do it a little bit different than I did in the last video, just to make it interesting. This is 3 times the principal root of 500 times the principal root of x to the third. And 500-- we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5. Or even better, we could rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. Now, the one thing I'm going to do here-- actually, I won't talk about it just yet, of how we're going to do it differently than we did it in the last video. This radical right here can be rewritten as-- so this is going to be 3 times the square root, or the principal root, I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and then taking the product. And so then this over here is going to be times the square root of, or the principal root of, x squared times the principal root of x. And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30-- and I'm just going to switch the order here-- times the absolute value of x. And then you have the square root of 5, And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is, if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers-- then x has to be greater than or equal to 0. So maybe I could write it this way. The domain here is that x is any real number greater than or equal to 0. And the reason why I say that is, if you put a negative number in here and you cube it,", + "qid": "qFFhdLlX220_90" + }, + { + "Q": "at 1:19 what is a 4 dimensional object", + "A": "Actually, Steven Hawking has speculated that all objects have 4-dimensions, and the fourth dimension is time. Like, as an object grows older, the volume (or whatever you d call it) of it s fourth dimension would increase.", + "video_name": "FtxmFlMLYRI", + "timestamps": [ + 79 + ], + "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211.", + "qid": "FtxmFlMLYRI_79" + }, + { + "Q": "I don't get it at 3:30. If I type in (40)sin40/30 into my calculator I get approx 0.93. If I type 4/3sin40 I get approx 0.86.", + "A": "It should be 40 sin (40) /30 = .86. Be careful with where parentheses go. What you actually did was 40 sin (40/30) to get .93.", + "video_name": "IJySBMtFlnQ", + "timestamps": [ + 210 + ], + "3min_transcript": "3 sides of a triangle. So a, b, c to an angle. So, for example, if I do 2 sides and the angle in between them, I can figure out the third side. Or if I know all 3 sides, then I can figure out this angle. But that's not the situation that we have over here. We're trying to figure out this question mark and we don't know 3 of the sides. We're trying to figure out an angle but we don't know 3 of the sides. The Law of Cosine just doesn't seem, at least in an obvious way, that it's going to help me. I could also try to find this angle. Once again, we don't know all 3 sides to be able to solve for the angle. So maybe Law of Sines could be useful. So the Law of Sines, the Law of Sines. Let's say that this is, the measure of this angle is a, the measure of this angle is lower case b, the measure of this angle is lower case c, length of this side is capital C, length of this side is capital A, The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information", + "qid": "IJySBMtFlnQ_210" + }, + { + "Q": "in the video at 3:25 I want 2 know how to make a hexaflexagon, but it goes to fast. Can you make a slow video on how to make a hexaflexagon?", + "A": "Hleyendecker 2020, the major predicament of alacritous motion in the video can be eradicated! Because Vi shows numerous ways of construction on the hexaflexagon, at 3:22-3:27, a reasonably stagnant-paced clip is displayed to contrive a hexaflexagon in a rare moment of slowness.", + "video_name": "VIVIegSt81k", + "timestamps": [ + 205 + ], + "3min_transcript": "you decide to try this three-way fold the other way, with flappy parts up, and are collapsing it down when suddenly the inside of your hexagon decides to open right up What, you close it back up and undo it. Everything seems the same as before, the center is not open-uppable. But when you fold it that way again, it, like, flips inside-out. Weird. This time, instead of going backwards, you try doing it again and again and again and again. And you want to make one that's a little less messy, so you try with another strip and tape it nicely into a twisty-foldy loop. You decide that it would be cool to colour the sides, so you get out a highlighter and make one yellow. Now you can flip from yellow side to white side. Yellow side, white side, yellow side, white side Hmm. White side? What? Where did the yellow side go? So you go back and this time you colour the white side green, and find that your piece of paper has three sides. Yellow, white and green. Now this thing is definitely cool. Therefore, you need to name it. And since it's shaped like a hexagon and you flex it and flex rhymes with hex, hexaflexagon it is. That night, you can't sleep because you keep thinking And the next day, as soon as you get to your math class you pull out your paper strips. You had made this sort of spirally folded paper that folds into again, the shape of a piece of paper, and you decide to take that and use it like a strip of paper to make a hexaflexagon. Which would totally work, but it feels sturdier with the extra paper. And you color the three sides and are like, orange, yellow, pink. And you're sort of trying to pay attention to class. Math, yeah. Orange, yellow, pink. Orange, yellow, white? Wait a second. Okay, so you colour that one green. And now it;s orange, yellow, green, Orange, yellow, green. Who knows where the pink side went? Oh, there it is. Now it's back to orange, yellow, pink. Orange, yellow, pink. Hmm. Blue. Yellow, pink, blue. Yellow, pink, blue. Yellow, pink, huh. With the old flexagon, you could only flex it one way, flappy way up. But now there's more flaps. So maybe you can fold it both ways. Yes, one goes from pink to blue, but the other, from pink to orange. And now, one way goes from orange to yellow, but the other way goes from orange to neon yellow. to one of your new friends, Bryant Tuckerman. You start with the original, simple, three-faced hexaflexagon, which you call the trihexaflexagon. and he's like, whoa! and wants to learn how to make one. and you are like, it's easy! Just start with a paper strip, fold it into equilateral traingles, and you'll need nine of them, and you fold them around into this cycle and make sure it's all symmetric. The flat parts are diamonds, and if they're not, then you're doing it wrong. And then you just tape the first triangle to the last along the edge, and you're good. But Tuckerman doesn't have tape. After all, it was invented only 10 years ago. So he cuts out ten triangles instead of nine, and then glues the first to the last. Then you show him how to flex it by pinching around a flappy part and pushing in on the opposite side to make it flat and traingly, and then opening from the centre. You decide to start a flexagon committee together to explore the mysteries of flexagotion, But that will have to wait until next time.", + "qid": "VIVIegSt81k_205" + }, + { + "Q": "(6:27) what sal says is wrong. it is less likely for the next toss to be heads if you look at is as a whole. lets say you toss the coin 100 times and you get 90 heads and 10 tails. it is more likely to get that then 90 heads and then 10 tails because there is only combination so theoretically, it is less likely to get a head is less likely in the next toss.", + "A": "True, but nonetheless the more times you flip the more you approach 50/50 results. This person just assumed that there was some underlying force that pulled everything together into theoretical accuracy. But the only way you approach this truth is through sheer quantity of data, there is no invisible force that makes anything more likely.", + "video_name": "VpuN8vCQ--M", + "timestamps": [ + 387 + ], + "3min_transcript": "Let me differentiate. And I'll use this example. So let's say-- let me make a graph. And I'll switch colors. This is n, my x-axis is n. This is the number of trials I take. And my y-axis, let me make that the sample mean. And we know what the expected value is, we know the expected value of this random variable is 50. Let me draw that here. This is 50. So just going to the example I did. So when n is equal to-- let me just [INAUDIBLE] here. So my first trial I got 55 and so that was my average. I only had one data point. Then after two trials, let's see, then I have 65. which is 60. So then my average went up a little bit. Then I had a 45, which will bring my average down a little bit. I won't plot a 45 here. Now I have to average all of these out. What's 45 plus 65? Let me actually just get the number just so you get the point. So it's 55 plus 65. It's 120 plus 45 is 165. Divided by 3. 3 goes into 165 5-- 5 times 3 is 15. It's 53. No, no, no. 55. So the average goes down back down to 55. And we could keep doing these trials. So you might say that the law of large numbers tell this, OK, after we've done 3 trials and our average is there. So a lot of people think that somehow the gods of probability are going to make it more likely that we get fewer That somehow the next couple of trials are going to have to be down here in order to bring our average down. And that's not necessarily the case. Going forward the probabilities are always the same. The probabilities are always 50% that I'm going to get heads. It's not like if I had a bunch of heads to start off with or more than I would have expected to start off with, that all of a sudden things would be made up and I would get more tails. That would the gambler's fallacy. That if you have a long streak of heads or you have a disproportionate number of heads, that at some point you're going to have-- you have a higher likelihood of having a disproportionate number of tails. And that's not quite true. What the law of large numbers tells us is that it doesn't care-- let's say after some finite number of trials your average actually-- it's a low probability of this happening, but let's say your average is actually up here. Is actually at 70. You're like, wow, we really diverged a good bit from the expected value. But what the law of large numbers says, well, I don't care how many trials this is. We have an infinite number of trials left.", + "qid": "VpuN8vCQ--M_387" + }, + { + "Q": "Why at 4:53 in the video Sal puts down four squared raised to the seventh power equal to four to the seventh power", + "A": "its an error. (4^2)^7 is actually 4^14 he corrected himself at 5:23", + "video_name": "dC1ojsMi1yU", + "timestamps": [ + 293 + ], + "3min_transcript": "times this thing to the second power. Eight to the seventh to the second power, and then here, negative two times two is negative four, so that's A to the negative four times, eight to the seven times two is 14, eight to the 14th power. In other videos, we go into more depth about why this should hopefully make intuitive sense. Here you have eight to the seventh times eight to the seventh. Well, you would then add the two exponents, and you would get to eight to the 14th, so however many times you have eight to the seventh, you would just keep adding the exponents, or you would multiply by seven that many times. Hopefully that didn't sound too confusing, but the general idea is if you raise something to exponent and then another exponent, you can multiply those exponents. Let's do one more example where we are dealing with quotients, which that first example could have even been perceived as. So let's say we have divided by four squared, and we're gonna raise all of that to the seventh power. Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power, so if you have the difference of two things and you're raising it to some power, that's the same thing as a numerator raised to that power divided by the denominator raised to that power. Well, what's our numerator going to be? Well, we've done this drill before. It'd be two to the negative 10 times seventh power, so this would be equal to two to the negative 70th power, and then in the denominator, four to the second power, then that raised to the seventh power. Well, two times seven is 14, so that's going to be four to the 17th power. Now, we actually could think There's multiple ways that you could rewrite this, but one thing you could do is say, \"Hey, look, \"four is a power of two.\" So you could rewrite this as this is equal to two to the negative 70th power over, instead of writing four to the 17th power, why did I write the 17th power? It should be four to the 14th power. Let me correct that. Instead of writing four to the 14th power, I instead could write, so this is two, get the colors right. This is two to the negative 70th over, instead of writing four, I could write two squared to the 14th power. Four is the same thing as two squared, and so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second,", + "qid": "dC1ojsMi1yU_293" + }, + { + "Q": "at 3:46 , Can you really just square the 1/3 and then plug it into the radical? I have never seen that before.", + "A": "Yes as you re not changing the equation in any way. You re basically squaring a positive multiple then when you place it inside the radical you re square rooting it again so it s exactly the same multiple. (1/3)^2 = 1/9 sqrt(1/9) = 1/3", + "video_name": "WAoaBTWKLoI", + "timestamps": [ + 226 + ], + "3min_transcript": "So this is u in terms of x. So everywhere we see a u up here we can replace it with this expression. And we are essentially done. We would have written this in terms of x. Now, there's another technique you might sometimes see in a calculus class where someone says, OK, we know that u is equal to cosine theta. We know this relationship. How can we express u in terms of x? And we'll say, let's draw a right triangle. They'll draw a right triangle like this. They'll draw a right triangle, and they'll say, OK, look, sine of theta is x over 3. So if we say that this is theta right over here, sine of theta is the same thing as opposite over hypotenuse. Opposite over hypotenuse is equal to x over 3. So let's say that this is x and then this right over here is 3. Then the sine of theta will be x over 3. So we look at that first substitution right over here. But in order to figure out what u is in terms of x, we need to figure out what cosine of theta is. So we have to figure out what this adjacent side is. Well, we can just use the Pythagorean theorem for that. Pythagorean theorem would tell us that this is going to be the square root of the hypotenuse squared, which is 9, minus the other side squared, minus x squared. So from this, we fully solved the right triangle in terms of x. We can realize that cosine of theta is going to be equal to the adjacent side, square root of 9 minus x squared, over the hypotenuse, over 3, which is the same thing as 1/3 times the square root of 9 minus x squared, which is the same thing if we square 1/3 and put it into the radical. So we're essentially going to take the square-- 1/3 is the same thing as the square root of 1/9. So can rewrite this as the square root of 1/9 times 9 minus x squared. Essentially, we just brought the 1/3 third into the radical. Now it's 1/9. And so now this is going to be the same thing which is exactly this thing right over here. x squared over 9 is the same thing as x over 3 squared. So either way, you get the same result. I find using the trig identity right over here to express cosine of theta in terms of sine theta and then just do the substitution to be a little bit more straightforward. But now we can just substitute into the original thing. So either of these-- I can write it as either way-- this thing right over here, this is the same thing as 1 minus x squared over 9 to the 1/2 power. That's what u is equal to. And everywhere we see u, we just substitute it with this thing. So our final answer in terms of x is going to be equal to 243 times u to the fifth, this to the fifth power over 5. This to the fifth power is 1 minus x squared over 9. It was to the 1/2, but if we raise that to the fifth power,", + "qid": "WAoaBTWKLoI_226" + }, + { + "Q": "At around 7:30, when he is discussing the trig. ratios, is there any particular reason why, for example, we use adj/hyp instead of hyp/adj? If we used the reciprocal of any of these ratios would it matter?", + "A": "For each of the main trig functions, sine, cosine and tangent, there is another trig function that is its reciprocal: secant is 1/cosine, cosecant is 1/sine, and cotangent is 1/tangent. For an acute angle in a right triangle, adj/hyp is cosine. You can t use hyp/adj for cosine because that s something entirely different (secant) with a graph that looks nothing like the graph of cosine.", + "video_name": "QuZMXVJNLCo", + "timestamps": [ + 450 + ], + "3min_transcript": "And I keep stating from theta's point of view because that wouldn't be the case for this other angle, for angle B. From angle B's point of view, this is the adjacent side over the hypotenuse. And we'll think about that relationship later on. But let's just all think of it from theta's point of view right over here. So from theta's point of view, what is this? Well theta's right over here. Clearly AB and DE are still the hypotenuses-- hypoteni. I don't know how to say that in plural again. And what is AC, and what are DF? Well, these are adjacent to it. They're one of the two sides that make up this angle that is not the hypotenuse. So this we can view as the ratio, in either of these triangles, between the adjacent side-- so this is relative. Once again, this is opposite angle B, but we're only thinking about angle A right here, or the angle that measures theta, or angle D right over here-- relative to angle A, Relative to angle D, DF is adjacent. So this ratio right over here is the adjacent over the hypotenuse. And it's going to be the same for any right triangle that has an angle theta in it. And then finally, this over here, this is going to be the opposite side. Once again, this was the opposite side over here. This ratio for either right triangle is going to be the opposite side over the adjacent side. And I really want to stress the importance-- and we're going to do many, many more examples of this to make this very concrete-- but for any right triangle that has an angle theta, the ratio between its opposite side and its hypotenuse is going to be the same. That comes out of similar triangles. We've just explored that. The ratio between the adjacent side to that angle that is theta and the hypotenuse is going to be the same, for any of these triangles, as long as it has that angle theta in it. theta, between the opposite side and the adjacent side, between the blue side and the green side, is always going to be the same. These are similar triangles. So given that, mathematicians decided to give these things names. Relative to the angle theta, this ratio is always going to be the same, so the opposite over hypotenuse, they call this the sine of the angle theta. Let me do this in a new color-- by definition-- and we're going to extend this definition in the future-- this is sine of theta. This right over here, by definition, is the cosine of theta. And this right over here, by definition, is the tangent of theta. And a mnemonic that will help you remember this-- and these really are just definitions. People realized, wow, by similar triangles, for any angle theta, this ratio is always going to be the same.", + "qid": "QuZMXVJNLCo_450" + }, + { + "Q": "at 1:56 Vi says cochlea, what does the cochlea do?", + "A": "Sound goes into your ear, past your ear drum, past the bones in your ear , into your cochlea and through the nerve to the brain. The cochlea basically processes the sound so you can understand it.", + "video_name": "i_0DXxNeaQ0", + "timestamps": [ + 116 + ], + "3min_transcript": "[PIANO ARPEGGIOS] When things move, they tend to hit other things. And then those things move, too. When I pluck this string, it's shoving back and forth against the air molecules around it and they push against other air molecules that they're not literally hitting so much as getting too close for comfort until they get to the air molecules in our ears, which push against some stuff in our ear. And then that sends signals to our brain to say, Hey, I am getting pushed around here. Let's experience this as sound. This string is pretty special, because it likes to vibrate in a certain way and at a certain speed. When you're putting your little sister on a swing, you have to get your timing right. It takes her a certain amount of time to complete a swing and it's the same every time, basically. If you time your pushes to be the same length of time, then even general pushes make your swing higher and higher. That's amplification. If you try to push more frequently, you'll just end up pushing her when she's swinging backwards and instead of going higher, you'll dampen the vibration. It wants to swing at a certain speed, frequency. If I were to sing that same pitch, the sound waves I'm singing will push against the string at the right speed to amplify the vibrations so that that string vibrates while the other strings don't. It's called a sympathy vibration. Here's how our ears work. Firstly, we've got this ear drum that gets pushed around by the sound waves. And then that pushes against some ear bones that push against the cochlea, which has fluid in it. And now it's sending waves of fluid instead of waves of air. But what follows is the same concept as the swing thing. The fluid goes down this long tunnel, which has a membrane called the basilar membrane. Now, when we have a viola string, the tighter and stiffer it is, the higher the pitch, which means a faster frequency. The basilar membrane is stiffer at the beginning of the tunnel and gradually gets looser so that it vibrates at high frequencies at the beginning of the cochlea and goes through the whole spectrum down to low notes at the other end. So when this fluid starts getting pushed around there's a certain part of the ear that vibrates in sympathy. The part that's vibrating a lot is going to push against another kind of fluid in the other half of the cochlea. And this fluid has hairs in it which get pushed around by the fluid, and then they're like, Hey, I'm middle C and I'm getting pushed around quite a bit! Also in humans, at least, it's not a straight tube. The cochlea is awesomely spiraled up. OK, that's cool. But here are some questions. You can make the note C on any instrument. And the ear will be like, Hey, a C. But that C sounds very different depending on whether I sing it or play it on viola. Why? And then there's some technicalities in the mathematics of swing pushing. It's not exactly true that pushing with the same frequency that the swing is swinging is the only way to get this swing to swing. You could push on just every other swing. And though the swing wouldn't go quite as high as if you pushed every time, it would still swing pretty well. In fact, instead of pushing every time or half the time, you could push once every three swings or four, and so on. There's a whole series of timings that work,", + "qid": "i_0DXxNeaQ0_116" + }, + { + "Q": "3:10 Why would you subtract 90 degrees from that equation? I haven't exactly figured that out.", + "A": "Sal subtracted 90 degrees from both sides of the equation to simplify the equation. What I would do instead (personally) is add 90 and 32 to get 122, and then subtract 122 from both sides. 180-122= 58, so either way, you get the same answer. Hope that helps you!", + "video_name": "iqeGTtyzQ1I", + "timestamps": [ + 190 + ], + "3min_transcript": "And now we have three angles in the triangle, and we just have to solve for theta. Because we know this angle plus this angle plus this angle are going to be equal to 180 degrees. So you have 90 minus theta plus 90 degrees plus 32 degrees-- so I'm going to do that in a different color-- is going to be equal to 180 degrees. The sum of the measures of the angle inside of a triangle add up to 180 degrees. That's all we're doing over here. And so let's see if we can simplify this a little bit. So these two guys-- 90 plus 90's going to be 180, so you get 180 minus theta plus 32 is equal to 180 degrees. And then what else do we have? We have 180 on both sides. We can subtract that from both sides. So that cancels out. That goes to 0. You can add theta to both sides. And you get 32 degrees is equal to theta, or theta is equal to 32 degrees. So it's going to actually be the same measure as this angle right over here. That's one way to do the problem. There's other ways that we could have done the problem. Actually, there's a ton of ways we could have done this. We could have looked at this big triangle over here. And we could've said, look. If this is 90 degrees over here, this is 32 degrees over here, this angle up here is going to be 180 minus 90 degrees minus 32 degrees. Because they all have to add up to 180 degrees. And I just kind of skipped a step there. Actually, let me not skip a step. Let me call this x. If we call the measure of that angle x, we would have x plus 90. I'm looking at the biggest triangle in this diagram right here. x plus 90 plus 32 is going to be equal to 180 degrees. So if you subtract 90 from both sides, you get x plus 32 is equal to 90. And then if you subtract 32 from both sides, you get x is equal to-- what is this-- 58 degrees. Fair enough. Now, what else can we figure out? Well, if this angle over here is a right angle-- and I'm just redoing the problem over again just to show you that there's multiple ways to get the answer. We were given that this is a right angle. If that is 90 degrees, then this angle over here is supplementary to it, and it also has to be 90 degrees. So then we have this angle plus 90 degrees plus this angle have to equal 180. Maybe we could call that y. So y plus 58 plus 90 is equal to 180.", + "qid": "iqeGTtyzQ1I_190" + }, + { + "Q": "I dont understand the way Sal's doing these. I paused at 0:06 and factored it by grouping, because it seemed like the obvious way to go. Like this:\n30x^2 + 11xy + y^2\n30x^2 + 5xy + 6xy + y^2\n5x(6x+y) + y(6x+y)\n=(5x+y)(6x+y)\nMultiplying it out, i get back to the beginning.\nI did the problem in the previous video the same way. Is this correct? Or am i doing it wrong and getting the right answers just by accident?", + "A": "I did it the same way you did.", + "video_name": "0xrvRKHoO2g", + "timestamps": [ + 6 + ], + "3min_transcript": "Let's see if we can use our existing factoring skills to factor 30x squared plus 11xy plus y squared. And I encourage you to pause the video and see if you can handle it yourself. Now, the first hint I will give you-- and this might open up what's going on here-- is to maybe rearrange this a little bit. We could rewrite this as y squared plus 11xy plus 30x squared. And my whole motivation for doing that-- there are ways to factor a quadratic where your first coefficient, your coefficient on this first term, is something other than 1. But we haven't seen that yet. And so rearranging it this way, this got us a little bit more into our comfort zone. Now our coefficient is a 1 on the y squared term. So now we can start to think of this in the same form that we've looked at some of the other factoring problems. Can we think of two numbers whose product is 30x squared and whose sum is 11x? We have y squared, some coefficient on y. And then in terms of y, this isn't in any way dependent on y. So one way to think about this, if you knew what x was, then this would be a quadratic in terms of y. And that's how we're really thinking about it here. So can we find two numbers whose product is 30x squared and two numbers whose sum is the coefficient on this y term right here, whose sum is 11x? So let's just think about all of the different possibilities. If we were just thinking about two numbers whose product was 30 and whose sum was 11, we would be thinking of 5 and 6. 5 times 6 is 30. 5 plus 6 is 11. It's some trial and error. You could have tried 3 and 10. Well, that would have been-- 13 would be their sum. You could have tried 2 and 15. That wouldn't have worked. But 5 and 6 does work here, so we've already seen that multiple times. So 5 and 6 would work for 30, but we have 30x squared. Well, 5x times 6x is 30x squared, and 5x plus 6x is 11x. So this actually works. So then our factoring or our factorization of this expression is just going to be y plus 5x times y plus 6x. And I'll leave it up to you to verify that this does indeed, when you multiply it out, equal this up here.", + "qid": "0xrvRKHoO2g_6" + }, + { + "Q": "At 2:55, what is meant by derivative of something \"with respect to\" something else?", + "A": "With respect to is generally used to describe the term you are talking about. For example, say you have: f(x) = (x^2)/3 with respect to x, the function is the same, (x^2)/3 BUT with respect to x^2, the function is x/3 Taking the derivative of a function with respect to something basically means you are determining what the derivative function is doing to the term that you re talking about. Hope that helps!", + "video_name": "Mci8Cuik_Gw", + "timestamps": [ + 175 + ], + "3min_transcript": "So this is the x power in yellow. And so let's do that right over here. So instead of taking the derivative with respect to x of 2 to the x, let's say, let's just take the derivative with respect to x of the exact same expression rewritten, of e to the natural log of 2 raised to the x power. Let me put this x in that same color, dx. Now we know from our exponent properties if we raise something to some power, and then raise that to another power, we can take the product of the two powers. Let me rewrite this just to remember. If I have a to b, and then I raise that to the c power, this is the exact same thing as a to the b times c power. So we can utilize that exponent property right here to rewrite this as being equal to the derivative with respect And what's neat about this is now we've got this into a form of e to the something. So we can essentially use the chain rule to evaluate this. So this derivative is going to be equal to the derivative of e to the something with respect to that something. Well, the derivative e to the something with respect to that something is just e to that something. So it's going to be equal to e to the natural log of 2 times x. So let me make it clear what I just did here. This right over here is the derivative of e to the natural log of 2 times x with respect to the natural log of 2-- let me make it a little bit clearer-- with respect So we took the derivative of e to the something with respect to that something-- that's this right here, it's just e to that something. And then we're going to multiply that by, this is just an application of the chain rule, of the derivative of that something with respect to x. So the derivative of natural log of 2 times x with respect to x is just going to be natural log of 2. This is just going to be natural log of 2. The derivative of a times x is just going to be equal to a. This is just the coefficient on the x. And just to be clear, this is the derivative of natural log of 2 times x with respect to x. So we're essentially done. But we can simplify this even further. This thing right over here can be rewritten. And let me draw a line here just to make it clear that this equals sign is a continuation from what we did up there. But this e to the natural log of 2x, we can rewrite that, using this exact same exponent property, as e to the natural log of 2, and then", + "qid": "Mci8Cuik_Gw_175" + }, + { + "Q": "at 2:17, we could just treat e^(ln 2)^x as a function and use the chain rule to differentiate it, couldn't we?\nbut the ans is different", + "A": "Sal has been solving e^(ln2*x) not e^(ln2)^x. I have hope it is a typo.", + "video_name": "Mci8Cuik_Gw", + "timestamps": [ + 137 + ], + "3min_transcript": "Let's see if we can take the derivative with respect to x of 2 to the x power. And you might say, hold on a second. We know how to take the derivative of e to the x. But what about a base like 2? We don't know what to do with 2. And the key here is to rewrite 2 to the x so that we essentially have it as e to some power. And the key there is to rewrite 2. So how can we rewrite 2 so it is e to some power? Well, let's think about what e to the natural log of 2 power is. The natural log of 2 is the power that I would have to raise e to to get to 2. So if we actually raise e to that power, we are going to get to 2. So what we could do, instead of writing 2 to the x, we could rewrite this as e. We could rewrite 2 as e to the natural log of 2, So this is the x power in yellow. And so let's do that right over here. So instead of taking the derivative with respect to x of 2 to the x, let's say, let's just take the derivative with respect to x of the exact same expression rewritten, of e to the natural log of 2 raised to the x power. Let me put this x in that same color, dx. Now we know from our exponent properties if we raise something to some power, and then raise that to another power, we can take the product of the two powers. Let me rewrite this just to remember. If I have a to b, and then I raise that to the c power, this is the exact same thing as a to the b times c power. So we can utilize that exponent property right here to rewrite this as being equal to the derivative with respect And what's neat about this is now we've got this into a form of e to the something. So we can essentially use the chain rule to evaluate this. So this derivative is going to be equal to the derivative of e to the something with respect to that something. Well, the derivative e to the something with respect to that something is just e to that something. So it's going to be equal to e to the natural log of 2 times x. So let me make it clear what I just did here. This right over here is the derivative of e to the natural log of 2 times x with respect to the natural log of 2-- let me make it a little bit clearer-- with respect", + "qid": "Mci8Cuik_Gw_137" + }, + { + "Q": "At 6:26 Sal Khan says \"... our variance is essentially the probability of success times the probability of failure.\" Mathematically I understand this (Khan walks us through the derivation) but conceptually I don't. If variance is some measure of the spread of values around the mean, how does the product of the probably of success and failure describe the variance?", + "A": "To me, p(1-p) seems like an algebraic simplification without a conceptual component to it. Sometimes algebraic simplifications lead to more conceptual insight, but this one really doesn t.", + "video_name": "ry81_iSHt6E", + "timestamps": [ + 386 + ], + "3min_transcript": "This right here is going to be the variance. Now let's actually work this out. So this is going to be equal to 1 minus p. Now 0 minus p is going to be negative p. If you square it you're just going to get p squared. So it's going to be p squared. Then plus p times-- what's 1 minus p squared? 1 minus p squared is going to be 1 squared, which is just 1, minus 2 times the product of this. So this is going to be minus 2p right over here. And then plus negative p squared. So plus p squared just like that. And now let's multiply everything out. This is going to be, this term right over here is going to be p squared minus p to the third. going to be plus p times 1 is p. p times negative 2p is negative 2p squared. And then p times p squared is p to the third. Now we can simplify these. p to the third cancels out with p to the third. And then we have p squared minus 2p squared. So this right here becomes, you have this p right over here, so this is equal to p. And then when you add p squared to negative 2p squared you're left with negative p squared minus p squared. And if you want to factor a p out of this, this is going to be equal to p times, if you take p divided p you get a 1, p square divided by p is p. So p times 1 minus p, which is a pretty neat, clean formula. So our variance is p times 1 minus p. And if we want to take it to the next level and figure out square root of the variance, which is equal to the square root of p times 1 minus p. And we could even verify that this actually works for the example that we did up here. Our mean is p, the probability of success. We see that indeed it was, it was 0.6. And we know that our variance is essentially the probability of success times the probability of failure. That's our variance right over there. The probability of success in this example was 0.6, probability of failure was 0.4. You multiply the two, you get 0.24, which is exactly what we got in the last example. And if you take its square root for the standard deviation, which is what we do right here, it's 0.49. So hopefully you found that helpful, and we're going to build on this later on in some of our inferential statistics.", + "qid": "ry81_iSHt6E_386" + }, + { + "Q": "at around 5:37, Sal divided the 2 and the 12, but, don't you have to divide the 27 too or am I just forgetting a rule?", + "A": "(27*12)/2 = (3*3*3*2*2*3)/2 now you can cancel out factors", + "video_name": "8C5kAIKLcZo", + "timestamps": [ + 337 + ], + "3min_transcript": "And that makes sense because we should have more feet than yards. And actually, this should be three times more, so everything makes sense. 27/2 is 3 times 9/2. So now we have 27/2 feet, and now we want to convert this to inches. And we just have to remember there are 12 inches per yard. And we're going to want to multiply by 12, because however many feet we have, we're going to have 12 times as many inches. If we have 1 foot, we're going to have 12 inches, 2 feet, 24 inches. 27/2 feet, we're going to multiply it by 12 to get the number of inches. Since this is going to be times 12, and we'll make sure the dimensions work out: 12 inches per foot. And the feet and the foot, this is just the plural and It's the same dimension. This will cancel out. So this will be-- if we just rearrange the multiplication, view it as everything is getting multiplied, and when you just multiply a bunch of things, order doesn't matter. So this is equal to 27/2 times 12 feet. I'm just swapping the order. Feet times inches divided by feet, or foot, just the singular of the same word. The feet and the foot cancel out, they're the same unit. And you have 27 times 12 divided by 2 inches. And what we could do here is that our final answer is going to be 27 times 12/2 inches. And before we multiply the 27 times 12 and then divide by 2, you immediately see, well, I can just divide 12 by 2, and 2 by 2, and it makes our computation simpler. It becomes 27 times 6 inches, and let's figure out what that is. 27 times 6. 7 times 6 is 42. 2 times 6 is 12, plus 4 is 16. This is equal to 162 inches, which makes sense. 4 and 1/2 yards, that gets us to this number right here: 27 divided by 2 is 13 and 1/2 feet. You multiply that by 12, it makes sense. You're going to have a bunch of inches. 162 inches.", + "qid": "8C5kAIKLcZo_337" + }, + { + "Q": "In 0:26 what does compute mean", + "A": "Compute means to calculate, find, or figure out. He will show how to find the answer.", + "video_name": "twMdew4Zs8Q", + "timestamps": [ + 26 + ], + "3min_transcript": "Let's multiply 9 times 8,085. That should be a pretty fun little calculation to do. So like always, let's just rewrite this. So I'm going to write the 8,085. I'm going to write the 9 right below it and write our little multiplication symbol. And now, we're ready to compute. So first we can tackle 9 times 5. Well, we know that 9 times 5 is 45. We can write the 5 in the ones place and carry the 5 to the tens place. So 9 times 5 is 45. Now we're ready to move on to 9 times 8. And we're going to calculate 9 times 8 and then add the 4 that we just carried. So 9 times 8 is 72, plus the 4 is 76. So we'll write the 6 right here the tens place and carry the 7. looking for a suitable color. 9 times 0 100's plus-- and this is a 7 in the hundreds place, so that's actually 700. Or if we're just kind of going with the computation, 9 times 0 plus 7. Well, 9 times 0 is 0, plus 7 is 7. And then, finally, we have-- and once again, I'm looking for a suitable color-- 9 times 8. This is the last thing we have to compute. We already know that 9 times 8 is 72. And we just write the 72 right down here, and we're done. 8,085 times 9 is 72,765. Let's do one more example just to make sure that this is really clear in your brain, at least the process for doing this. And I also want you to think about why this works. So let's try 7 times 5,396. I'm going to rewrite it-- 5,396 times 7. First, we'll think about what 7 times 6 is. We know that's 42. We'll put the 2 in the ones place. 4 we will carry. Then we need to concern ourselves with 7 times 9. But then, we have to calculate that and then add the 4. 7 times 9 is 63, plus 4 is 67. So we put the 7 down here and carry the 6. Then we have to worry about 7 times 3 plus this 6 that we had just finished carrying. 7 times 3 is 21, plus 6 is 27.", + "qid": "twMdew4Zs8Q_26" + }, + { + "Q": "At 2:50, why not just keep it (a+2)(a-2) and then cancel out (a+2) from the top and bottom? Why does this not work to simplify?", + "A": "I see. So (sorry if this is worded wrong) you have to combine the -(a-3) before simplifying for the same reason you can t cancel out the two a s in (a + b)/a ?", + "video_name": "IKsi-DQU2zo", + "timestamps": [ + 170 + ], + "3min_transcript": "is the least common multiple of this expression, and that expression, and it could be a good common denominator. Let's set that up. This will be the same thing as being equal to this first term right here, a minus 2 over a plus 2, but we want the denominator now to be a plus 2 times a plus 2-- we wanted it to be a plus 2 squared. So, let's multiply this numerator and denominator by a plus 2, so its denominator is the same thing as this. Let's multiply both the numerator and the denominator by a plus 2. We're going to assume that a is not equal to negative 2, that would have made this undefined, and it would have also made this undefined. Throughout this whole thing, we're going to assume that a cannot be equal to negative 2. The domain is all real numbers, a can be any real number except for negative 2. So, the first term is that-- extend the line a little bit-- denominator is already the common denominator. Minus a minus 3 over-- and we could write it either as a plus 2 times a plus 2, or as this thing over here. Let's write it in the factored form, because it'll make it easier to simplify later on: a plus 2 times a plus 2. And now, before we-- let's set this up like this-- now, before we add the numerators, it'll probably be a good idea to multiply this out right there, but let me write the denominator, we know what that is: it is a plus 2 times a plus 2. Now this numerator: if we have a minus 2 times a plus 2, we've seen that pattern before. We can multiply it out if you like, but we've seen it enough hopefully to recognize that this is going to be a squared minus 2 squared. This is going to be a squared minus 4. You can multiply it out, and the middle terms cancel out-- the negative 2 times a cancels out the a times 2, and you're just left with a squared minus 4-- that's that over there. be very careful here-- you're subtracting a minus 3, so you want to distribute the negative sign, or multiply both of these terms times negative 1. So you could put a minus a here, and then negative 3 is plus 3, so what does this simplify to? You have a squared minus a plus-- let's see, negative 4 plus 3 is negative 1, all of that over a plus 2 times a plus 2. We could write that as a plus 2 squared. Now, we might want to factor this numerator out more, to just make sure it doesn't contain a common factor with the denominator. The denominator is just 2a plus 2 is multiplied by themselves. And you can see from inspection a plus 2 will not", + "qid": "IKsi-DQU2zo_170" + }, + { + "Q": "what is a real number 1:54\nis there such a thing as a fake number", + "A": "Imaginary numbers exist, and they exist because sometimes one needs to express the square root of a negative number, which doesn t exist. i is the central imaginary number, and it stands for the square root of negative one.", + "video_name": "IKsi-DQU2zo", + "timestamps": [ + 114 + ], + "3min_transcript": "Find the difference. Express the answer as a simplified rational expression, and state the domain. We have two rational expressions, and we're subtracting one from the other. Just like when we first learned to subtract fractions, or add fractions, we have to find a common denominator. The best way to find a common denominator, if were just dealing with regular numbers, or with algebraic expressions, is to factor them out, and make sure that our common denominator has all of the factors in it-- that'll ensure that it's divisible by the two denominators here. This guy right here is completely factored-- he's just a plus 2. This one over here, let's see if we can factor it: a squared plus 4a plus 4. Well, you see the pattern that 4 is 2 squared, 4 is 2 times 2, so a squared plus 4a plus 4 is a plus 2 times a plus 2, or a plus 2 squared. We could say it's a plus 2 times a plus 2-- that's what a squared plus 4a plus 4 is. This is obviously divisible by itself-- everything is divisible by itself, except, I guess, for 0, is divisible by is the least common multiple of this expression, and that expression, and it could be a good common denominator. Let's set that up. This will be the same thing as being equal to this first term right here, a minus 2 over a plus 2, but we want the denominator now to be a plus 2 times a plus 2-- we wanted it to be a plus 2 squared. So, let's multiply this numerator and denominator by a plus 2, so its denominator is the same thing as this. Let's multiply both the numerator and the denominator by a plus 2. We're going to assume that a is not equal to negative 2, that would have made this undefined, and it would have also made this undefined. Throughout this whole thing, we're going to assume that a cannot be equal to negative 2. The domain is all real numbers, a can be any real number except for negative 2. So, the first term is that-- extend the line a little bit-- denominator is already the common denominator. Minus a minus 3 over-- and we could write it either as a plus 2 times a plus 2, or as this thing over here. Let's write it in the factored form, because it'll make it easier to simplify later on: a plus 2 times a plus 2. And now, before we-- let's set this up like this-- now, before we add the numerators, it'll probably be a good idea to multiply this out right there, but let me write the denominator, we know what that is: it is a plus 2 times a plus 2. Now this numerator: if we have a minus 2 times a plus 2, we've seen that pattern before. We can multiply it out if you like, but we've seen it enough hopefully to recognize that this is going to be a squared minus 2 squared. This is going to be a squared minus 4. You can multiply it out, and the middle terms cancel out-- the negative 2 times a cancels out the a times 2, and you're just left with a squared minus 4-- that's that over there.", + "qid": "IKsi-DQU2zo_114" + }, + { + "Q": "At 2:01 what is that line over repeating decimals called ?", + "A": "The line over the repeating decimal can be called a vinculum . In a repeating decimal, the vinculum is used to indicate the group of repeating digits.", + "video_name": "d9pO2z2qvXU", + "timestamps": [ + 121 + ], + "3min_transcript": "Which of the following real numbers are irrational? Well, irrational just means it's not rational. It means that you cannot express it as the ratio of two integers. So let's see what we have here. So we have the square root of 8 over 2. If you take the square root of a number that is not a perfect square, it is going to be irrational. And then if you just take that irrational number and you multiply it, and you divide it by any other numbers, you're still going to get an irrational number. So square root of 8 is irrational. You divide that by 2, it is still irrational. So this is not rational. Or in other words, I'm saying it is irrational. Now, you have pi, 3.14159-- it just keeps going on and on and on forever without ever repeating. So this is irrational, probably the most famous of all of the irrational numbers. 5.0-- well, I can represent 5.0 as 5/1. So 5.0 is rational. It is not irrational. 0.325-- well, this is the same thing as 325/1000. So this is rational. Just as I could represent 5.0 as 5/1, both of these are rational. They are not irrational. Here I have 7.777777, and it just keeps going on and on and on forever. And the way we denote that, you could just say these dots that say that the 7's keep going. Or you could say 7.7. And this line shows that the 7 part, the second 7, just keeps repeating on forever. Now, if you have a repeating decimal-- in other videos, we'll actually convert them into fractions-- but a repeating decimal can be represented as a ratio of two integers. Just as 1/3 is equal to 0.333 on and on and on. Or I could say it like this. I could say 3 repeating. We can also do the same thing for that. I won't do it here, but this is rational. 8 and 1/2? Well, that's the same thing. 8 and 1/2 is the same thing as 17/2. So it's clearly rational. So the only two irrational numbers are the first two right over here.", + "qid": "d9pO2z2qvXU_121" + }, + { + "Q": "At 5:17 why do you have to times everything with -2?", + "A": "Because by doing so, we can get 200m in one equation and -200m in the other, which allows us to solve using elimination. We can add the two equations and be left with only one variable, w.", + "video_name": "VuJEidLhY1E", + "timestamps": [ + 317 + ], + "3min_transcript": "\" In green. Well, let's think about the total number of bags that the men ate. You had 200 men, [Let me scroll over a little bit] and they each ate m bags per man. \" \"So the man at this first party collectively ate 200 times m bags. If m is 10 bags per man, then this would be 2000. If m was 5 bags per man, then this would be 5000. We don't know what m is, but 200 times m is the total eaten by the man.\"", + "qid": "VuJEidLhY1E_317" + }, + { + "Q": "At 1:16 i lost you... why do we regroup? I need another example to kinda clarify it.", + "A": "Because you don t have enough in the tens place, so you need 1 from the hundred place to add to the tens place.", + "video_name": "X3JqIZR1XcY", + "timestamps": [ + 76 + ], + "3min_transcript": "Let's think about different ways that we can represent the number 675. So the most obvious way is to just look at the different place values. So the 6 is in the hundreds place. It literally represents 600. So that's 600. I'm going to do that in the red color-- 600. The 7 is in the tens place. It represents 7 tens, or 70. And then the 5 in the ones place. It represents 5. So let me copy and paste this and then think about how we can regroup the value in the different places to represent this in different ways. So let me copy and let me paste it, and maybe I'll do it three times. So let me do it once, and let me do it one more time. So one thing that we could do is we could regroup from one place to the next. So, for example, we could take if we wanted to-- we could take 1 from the hundreds place. That's essentially taking 100 away. So this is really making this a 500. And we could give that 100 to-- well, we could actually give it to either place, but let's give it to the tens place. So we're going to give 100 to the tens place. Now, if you give 100 to the tens place and you already had 70 there, what's it going to be equal to? Well, it's going to be equal to 170. Well, how would we represent that is tens? Well, 170 is 17 tens. So we could just say that 7 becomes 17. Now, we could keep doing that. We could regroup some of this value in the tens place to the ones place. So, for example, we could give 10 from the tens place and give it to the ones place. So let's take 10 away from here. So that becomes 160. This becomes 16. And let's give that 10 to the ones place. Well, 10 plus 5 is 15. So this 5 is now a 15. Let's do another scenario. Let's do something nutty. Let's take 200 from the hundreds place. So this is going to now 4, and this is going to become 400. That's what this 4 now represents. And let's give 100 to the tens place. And let's give another 100 to the ones place. So in other words, I'm just regrouping that 200. Those 200's, I've taken from the hundreds place, and I'm going to give it to these other places. So now the tens place is going to be 170. We're going to have 170 here, which is 17 tens. So you could say that the tens place is now 17.", + "qid": "X3JqIZR1XcY_76" + }, + { + "Q": "why did you subtracted the -3 with the 4x in 6:10? pls help.", + "A": "(x + 7)(4x - 3) is equal to 4x(x + 7) - 3(x + 7). You just distributed it. They re still the same and it s appropriate to write it in that way.", + "video_name": "X7B_tH4O-_s", + "timestamps": [ + 370 + ], + "3min_transcript": "So I grouped the 28 on the side of the 4. And you're going to see what I mean in a second. If we, literally, group these so that term becomes 4x squared plus 28x. And then, this side, over here in pink, it's plus negative 3x minus 21. Once again, I picked these. I grouped the negative 3 with the 21, or the negative 21, because they're both divisible by 3. And I grouped the 28 with the 4, because they're both divisible by 4. And now, in each of these groups, we factor as much out as we can. So both of these terms are divisible by 4x. So this orange term is equal to 4x times x-- 4x squared divided by 4x is just x-- plus 28x divided by 4x is just 7. Remember, you factor out everything that you can factor out. Well, both of these terms are divisible by 3 or negative 3. So let's factor out a negative 3. And this becomes x plus 7. And now, something might pop out at you. We have x plus 7 times 4x plus, x plus 7 times negative 3. So we can factor out an x plus 7. This might not be completely obvious. You're probably not used to factoring out an entire binomial. But you could view this could be like a. Or if you have 4xa minus 3a, you would be able to factor out an a. And I can just leave this as a minus sign. Let me delete this plus right here. Because it's just minus 3, right? Plus negative 3, same thing as minus 3. So what can we do here? We have an x plus 7, times 4x. Let's factor out the x plus 7. We get x plus 7, times 4x minus 3. Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping. And we factored it into two binomials. Let's do another example of that, because it's a little But once you get the hang of it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill. We want to find a times b that is equal to 1 times 6, which is equal to 6. And we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. What are the-- well, the obvious one is 1 and 6, right?", + "qid": "X7B_tH4O-_s_370" + }, + { + "Q": "At 8:40, why is it (x+1)(6x+1) instead of 7(x+1)?", + "A": "Multiply the factors... only the correct factors will create the original polynomial of 6x^2 + 7x + 1 7(x+1) = 7x + 7. This is not the original polynomial. So, these factors can not be correct. (x+1)(6x+1) = 6x^2 + x + 6x + 1 = 6x^2 + 7x + 1. This matches the original polynomial. so, these are the correct factor. Hope this helps.", + "video_name": "X7B_tH4O-_s", + "timestamps": [ + 520 + ], + "3min_transcript": "1 plus 6 is 7. So we have a is equal to 1. Or let me not even assign them. The numbers here are 1 and 6. Now, we want to split this into a 1x and a 6x. But we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squar ed here, plus-- and so I'm going to put the 6x first because 6 and 6 share a factor. And then, we're going to have plus 1x, right? 6x plus 1x equals 7x . That was the whole point. They had to add up to 7 . And then we have the final plus 1 there. Now, in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times-- 6x squar ed divided by 6x is just an x. 6x divided by 6x is just a 1. to have a plus here. But this second group, we just literally have a x plus 1. Or we could even write a 1 times an x plus 1. You could imagine I just factored out of 1 so to speak. Now, I have 6x times x plus 1, plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. And now, I'm going to actually explain why this little magical system actually works. Let me take an example. I'll do it in very general terms. Let's say I had ax plus b, times cx-- actually, I'm I think that'll confuse you, because I use a's and b's here. They won't be the same thing. So let me use completely different letters. Let's say I have fx plus g, times hx plus, I'll use j instead of i. You'll learn in the future why don't like using i as a variable. So what is this going to be equal to? Well, it's going to be fx times hx which is fhx. And then, fx times j. So plus fjx. And then, we're going to have g times hx. So plus ghx. And then g times j. Plus gj. Or, if we add these two middle terms, you have fh times x,", + "qid": "X7B_tH4O-_s_520" + }, + { + "Q": "At 4:00 Sal states that y=1/k and that this relationship is true for y' and makes a substitution. Would this relationship extend to second, third (etc.) derivatives? Could relationships like this be established for other equations and their derivatives? Feel free to give me a problem!", + "A": "y=f(x)=1/x. y=1/k for x=k only. You can use it for substitution for y in 2nd, 3rd or nth derivative as long as x=k then y=1/k. Let say you want to use x=2, then y is not 1/k anymore, but y=1/2.", + "video_name": "FJ7AMaR9miI", + "timestamps": [ + 240 + ], + "3min_transcript": "the slope of the tangent line? Well, to figure out the slope of the tangent line, let's take the derivative. So if we write f of x, instead writing it as 1/x, I'll write it as x to the negative 1 power. That makes it a little bit more obvious that we're about to use the power rule here. So the derivative of f at any point x is going to be equal to-- well, it's going to be the exponent here is negative 1. So negative 1 times x to the-- now we decrement the exponent to the negative 2 power. Or I could say it's negative x to the negative 2. Now, what we care about is the slope when x equals k. So f prime of k is going to be equal to negative k to the negative 2 power. Or another way of thinking about it, this is equal to negative 1 So this right over here is the slope of the tangent line at that point. Now, let's just think about what the equation of the tangent line is. And we could think about it in slope-intercept form. So we know the equation of a line in slope-intercept form is y is equal to mx plus b, where m is the slope and b is the y-intercept. So if we can get it in this form, then we know our answer. We know what the y-intercept is going to be. It's going to be b. So let's think about it a little bit. This equation, so we could say y is equal to our m, our slope of the tangent line, when x is equal to k, we just figure out to be this business. It equals this thing right over here. So let me write that in blue. Negative 1 over k squared times x plus b. Well, we know what y is when x is equal to k. And so we can use that to solve for b. We know that y is equal to 1/k when x is equal to k. So this is going to be equal to negative 1-- that's not the same color. Negative 1 over k squared times k plus b. Now, what does this simplify to? See, k over k squared is the same thing as 1/k, so this is going to be negative 1/k. So this part, all of this simplifies to negative 1/k. So how do we solve for b? Well, we could just add 1/k to both sides", + "qid": "FJ7AMaR9miI_240" + }, + { + "Q": "At 2:13 he states that 90 x -1/3 = -30 . Can anyone please explain how he got that answer? I understand the finding the common ratio step, but what I don't understand is how whenever I find the common ratio my math doesn't add up. I think I'm multiplying my fractions wrong.", + "A": "a * (-(b/c)) = a*(-b)*(1/c) = -(ab)/c. 90*(-(1/3)) = 90*(-1)*(1/3) = -90/3", + "video_name": "pXo0bG4iAyg", + "timestamps": [ + 133 + ], + "3min_transcript": "In this video I want to introduce you to the idea of a geometric sequence. And I have a ton of more advanced videos on the topic, but it's really a good place to start, just to understand what we're talking about when someone tells you a geometric sequence. Now a good starting point is just, what is a sequence? And a sequence is, you can imagine, just a progression of numbers. So for example, and this isn't even a geometric series, if I just said 1, 2, 3, 4, 5. This is a sequence of numbers. It's not a geometric sequence, but it is a sequence. A geometric sequence is a special progression, or a special sequence, of numbers, where each successive number is a fixed multiple of the number before it. Let me explain what I'm saying. So let's say my first number is 2 and then I multiply 2 by So I multiply it by 3, I get 6. And then I multiply 6 times the number 3, and I get 18. Then I multiply 18 times the number 3, and I get 54. And I just keep going that way. So I just keep multiplying by the number 3. So I started, if we want to get some notation here, this is my first term. We'll call it a1 for my sequence. And each time I'm multiplying it by a common number, and that number is often called the common ratio. So in this case, a1 is equal to 2, and my common ratio is equal to 3. So if someone were to tell you, hey, you've got a geometric sequence. a1 is equal to 90 and your common ratio is equal to negative 1/3. The second term is negative 1/3 times 90. Which is what? That's negative 30, right? 1/3 times 90 is 30, and then you put the negative number. Then the next number is going to be 1/3 times this. So negative 1/3 times this. 1/3 times 30 is 10. The negatives cancel out, so you get positive 10. Then the next number is going to be 10 times negative 1/3, or negative 10/3. And then the next number is going to be negative 10/3 times negative 1/3 so it's going to be positive 10/3. And you could just keep going on with this sequence. So that's what people talk about when they mean a geometric sequence. I want to make one little distinction here. This always used to confuse me because the terms are used very often in the same context. These are sequences. These are kind of a progression of numbers.", + "qid": "pXo0bG4iAyg_133" + }, + { + "Q": "At 8:56, he says the formula to find the 12th bounce is (120)(0.6)^n. I thought it was (120)(0.6)^(n - 1). I am kind of confused about that...", + "A": "In the video, he counts the zero bounce so you can subtract 1 from both sides and then they cancel out. ex. Jumps: a^n=120(0.6)^n-1 Bounces: a^n-1=120(0.6)^n-1 the -1 s cancel so: a^n=120(0.6)^n", + "video_name": "pXo0bG4iAyg", + "timestamps": [ + 536 + ], + "3min_transcript": "So you have 0.6 to the 0th power, and you've just got a 1 here. And that's exactly what happened on the first jump. Then on the second jump, you put a 2 minus 1, and notice 2 minus 1 is the first power, and we have exactly one 0.6 here. So I figured it was n minus 1 because when n is 2, we have one 0.6, when n is 3, we have two 0.6's multiplied by themselves. When n is 4, we have 0.6 to the third power. So whatever n is, we're taking 0.6 to the n minus 1 power, and of course we're multiplying that times 120. Now and the question they also ask us, what will be the rope stretch on the 12th bounce? And over here I'm going to use the calculator. and actually let me correct this a little bit. bounce, and we could call the jump the zeroth bounce. Let me change that. This isn't wrong, but I think this is where they're going with the problem. So you can view the initial stretch as the zeroth bounce. So instead of labeling it jump, let me label it bounce. So the initial stretch is the zeroth bounce, then this would be the first bounce, the second bounce, the third bounce. And then our formula becomes a lot simpler. Because if you said the stretch on nth bounce, then the formula just becomes 0.6 to the n times 120, right? On the zeroth bounce, that was our original stretch, you get 0.6 to the 0, that's 1 times 120. 0.6 times the previous stretch, or the previous bounce. So this has it in terms of bounces, which I think is what the questioner wants us to do. So what about the 12th bounce? Using this convention right there. So if we do the 12th bounce, let's just get our calculator out. We're going to have 120 times 0.6 to the 12th power. And hopefully we'll get order of operations right, because exponents take precedence over multiplication, so it'll just take the 0.6 to the 12th power only. And so this is equal to 0.26 feet. So after your 12th bounce, she's going to be barely moving. She's going to be moving about 3 inches on that 12th bounce.", + "qid": "pXo0bG4iAyg_536" + }, + { + "Q": "at 4:32 why do we have to multiply 50 by 1 hour? I thought all we needed to do was divide 3600 by 50 . Also i did not get whether it was 72 km per second or 1 km per 72 seconds. Although i had some questions this was a fantastic video that triggered a much needed Eureka! moment!! Would i need to multiply 50 by 2 hr if a question said 20/km per 2 hrs or is that not mathematically correct to use 2 hours as a unit", + "A": "its 1km per 72 seconds. 1/72", + "video_name": "d5lcGCbV5cM", + "timestamps": [ + 272 + ], + "3min_transcript": "of kilometers per second. So how could we write 50 kilometers per hour, in terms of kilometers per second? Well it's always good, actually, as a first approximation, to just think about it. If you went this far in an hour, then the number of kilometers you go in a second, is that going to be less, or more? Well a second's a much, much shorter period of time. There's 3,600 seconds in an hour. So you're going to go 1/3,600 of this distance. But let's think about how we would actually work out with the units. Well, we want to get rid of this hours in the denominator. And the plural, obviously the grammar doesn't hold up with the algebra, but this could be hour or hours. So we could think about well, 1 hour-- I'll write an hour in the numerator that's going to cancel with this hour in the denominator. But we want it in terms of seconds. So 1 hour is equal to how many seconds? This is what I meant by saying that using dimensional analysis, which is what I'm doing right now, we can essentially manipulate these units, as we would traditionally do with a variable. So we have hours divided by hours. And so when we do the multiplication, we can multiply the numeric parts. So we have 50 times 1, divided by 3,600. Let me write that. 50 times 1 over 3,600. And then our units left are kilometers per second. Or I could say seconds. So we can play around with the plural and singular parts of it, but I'll just write it as kilometers per second. And so this is 50/3,600. And this fits our intuition. In a second, you're going to go 1/3,600 as far as you would go But let's actually think about what this is equal to. 50/3,600-- so this is going to be the same thing, as-- Let me just simplify it over here. So 50/3,600 is the same thing as 5/360, which is the same thing as-- let me write it this way-- 10/720. And I did that way because that makes it clear that that's the same thing as 1/72. So you could write this as, you're going, this is equal to 1/72 of a kilometer per second. Now I would claim that this is not so reasonable of units for this example right over here.", + "qid": "d5lcGCbV5cM_272" + }, + { + "Q": "At 9:35 it says the term perfect square. What does that mean?", + "A": "By perfect square, you mean to say that if the square root of a number is taken the result would be a whole number. For example, 4,9,16 and 25 are perfect squares, since if you take the square root of those numbers, you would get 2,3,4 and 5, which are whole numbers. Hoped it helped! :)", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 575 + ], + "3min_transcript": "Those cancel out. On the left-hand side we're left with just a B squared is equal to-- now 144 minus 36 is what? 144 minus 30 is 114. And then you subtract 6, is 108. So this is going to be 108. So that's what B squared is, and now we want to take the principal root, or the positive root, of both sides. And you get B is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same So we have the square root of 108 is the same thing as the square root of 2 times 2 times-- well actually, I'm not done. 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so, we have a couple of perfect squares in here. Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there. And, you know, you wouldn't have to do all of this on paper. You could do it in your head. What is this? 2 times 2 is 4. 4 times 9, this is 36. So this is the square root of 36 times the square root of 3. The principal root of 36 is 6. So this simplifies to 6 square roots of 3. So the length of B, you could write it as the square root of 108, or you could say it's equal to 6 times the square root of 3. This is 12, this is 6. And the square root of 3, well this is going to be a 1 point something something. So it's going to be a little bit larger than 6.", + "qid": "AA6RfgP-AHU_575" + }, + { + "Q": "at 7:20 couldn't you just do a+b=c instead of A2+B2=C2.", + "A": "You can t just use a + b = c because you are trying to take the square root of both sides and just eliminate the squares. But the square root of both sides of a^2 + b^2 = c^2 gives you sqrt(a^2 + b^2) on the left and you have to do what s inside the parentheses first.", + "video_name": "AA6RfgP-AHU", + "timestamps": [ + 440 + ], + "3min_transcript": "That is 16. And 3 squared is the same thing as 3 times 3. So that is 9. And that is going to be equal to C squared. Now what is 16 plus 9? It's 25. So 25 is equal to C squared. And we could take the positive square root of both sides. I guess, just if you look at it mathematically, it could be negative 5 as well. But we're dealing with distances, so we only care about the positive roots. So you take the principal root of both sides and you get 5 is equal to C. Or, the length of the longest side is equal to 5. Now, you can use the Pythagorean theorem, if we give you two of the sides, to figure out the third side no matter what the third side is. So let's do another one right over here. Let's say that our triangle looks like this. Let's say this side over here has length 12, and let's say that this side over here has length 6. And we want to figure out this length right over there. Now, like I said, the first thing you want to do is identify the hypotenuse. And that's going to be the side opposite the right angle. We have the right angle here. You go opposite the right angle. The longest side, the hypotenuse, is right there. So if we think about the Pythagorean theorem-- that A squared plus B squared is equal to C squared-- 12 you could view as C. This is the hypotenuse. The C squared is the hypotenuse squared. So you could say 12 is equal to C. And then we could say that these sides, it doesn't matter whether you call one of them A or one of them B. Let's say A is equal to 6. And then we say B-- this colored B-- is equal to question mark. And now we can apply the Pythagorean theorem. A squared, which is 6 squared, plus the unknown B squared is equal to the hypotenuse squared-- is equal to C squared. Is equal to 12 squared. And now we can solve for B. And notice the difference here. Now we're not solving for the hypotenuse. We're solving for one of the shorter sides. In the last example we solved for the hypotenuse. We solved for C. So that's why it's always important to recognize that A squared plus B squared plus C squared, C is the length So let's just solve for B here. So we get 6 squared is 36, plus B squared, is equal to 12 squared-- this 12 times 12-- is 144.", + "qid": "AA6RfgP-AHU_440" + }, + { + "Q": "At 1:16 what exactly does he mean by 7 away from 0, what if it was negative?", + "A": "The simplest way to explain absolute value is forget the sign . Thus, |-7| = |7| = 7", + "video_name": "hKkBlcnU9pw", + "timestamps": [ + 76 + ], + "3min_transcript": "Let's do some examples comparing absolute values. So let's say we were to ask ourselves how the absolute value of negative 9, I should say, how that compares to the absolute value of-- let me think of a good number-- let's say the absolute value of negative 7. So let's think about this a little bit, and let's think about what negative 9 looks like, or where it is on the number line, where negative 7 is on the number line. Let's look at what the absolute values mean, and then we should probably be able to do this comparison. So there's a couple of ways to think about it. One is you could draw them on the number line. So if this is 0, if this is negative 7, and then this is negative 9 right over here. Now, when you take the absolute value of a number, you're really saying how far is that number from 0, whether it's to the left or to the right of 0. So, for example, negative 9 is 9 to the left of 0. This evaluates to 9, Negative 7 is exactly 7 to the left of 0. So the absolute value of negative 7 is positive 7. And so if you were to compare 9 and 7, this is a little bit more straightforward. 9 is clearly greater than 7. And if you ever get confused with the greater than or less than symbols, just remember that the symbol is larger on the left-hand side. So that's the greater than side. If I were to write this-- and this is actually also a true statement. If you took these without the absolute value signs, it is also true that negative 9 is less than negative 7. Notice the smaller side is on the smaller number. And so that's the interesting thing. Negative 9 is less than negative 7, but their absolute value, since negative 9 is further to the left of 0, it is-- the absolute value than the absolute value of negative 7. Another way to think about it is if you take the absolute value of a number, it's really just going to be the positive version of that number. So if you took the absolute value of 9, that equals 9. Or the absolute value of negative 9, that is also equal to 9. Well, when you think of it visually, that's because both of these numbers are exactly 9 away from 0. This is 9 to the right of 0, and this is 9 to the left of 0. Let's do a few more of these. So let's say that we wanted to compare the absolute value of 2 to the absolute value of 3. Well, the absolute value of a positive number is just going to be that same value. 2 is two to the right of 0, so this is just going to evaluate to 2. And then the absolute value of 3, that's just going to evaluate to 3. It's actually pretty straightforward. So 2 is clearly the smaller number here. And so we clearly get 2 is less than 3,", + "qid": "hKkBlcnU9pw_76" + }, + { + "Q": "At ~8:25 Sal says that sin(x) reflected over the y-axis is equal to sin(-x). It looks to me that you could just as correctly said that sin(x) reflected over the x-axis is equal to sin(-x). Is this right?", + "A": "The sine function is a member of a special family of functions we call odd functions. If f(x) is odd, f(-x)=-f(x). You are right about your statement, because it is another property of odd functions.", + "video_name": "0zCcFSO8ouE", + "timestamps": [ + 505 + ], + "3min_transcript": "well it's common sense the amplitude here was 1 but now you're swaying from that middle position twice as far because you're multiplying by 2 Now let's go back to sin(x) and let's change it in a different way Let's graph sin(-x) so now let me once again put some graph paper here And now my goal is to graph sin(-x) y=sin(-x) so at least for the time being I've got rid of that 2 there and I'm just going straight from sin(x) to sin(-x) So let's think about how the values are going to work out So when x is 0 this is still going to be sin(0) which is 0 But then what as x increases, what happens when x is \u03c0/2 we're going to have to multiply by this negative so when x is \u03c0/2 we're really taking sin(-\u03c0/2) but what's sin(-\u03c0/2) but we can see over here here it's -1 It's - 1 and then when x = \u03c0 well sin(-\u03c0) we see this is 0 When x is 3\u03c0/2 well it's going to be sin(-3\u03c0/2) which is 1 Once again when x is 2\u03c0 it's going to be sin(-2\u03c0) is 0 So notice what was happening as I was trying to graph between 0 and 2\u03c0 I kept referring to the points in the negative direction so you can imagine taking this negative side right over here between 0 and -2\u03c0 and then flipping it over to get this one right over here that's what that -x seems to do you say when x = -\u03c0/2 where you have the negative in front of it so it's going to be sin(\u03c0/2) so it's going to be equal to 1 and you can flip this over the y-axis so essentially what we have done is we have flipped it we have reflected the graph of sin(x) over the y-axis So we have reflected it over the y-axis This is the y-axis so hopefully you see that reflection that's what that -x has done So now let's think about kind of the combo Having the 2 out the front and the -x right over there so let me put the graph on the axis there one more time And now let's try to do what was asked of us", + "qid": "0zCcFSO8ouE_505" + }, + { + "Q": "People say that we see math in our everyday lives -- and while I understand how this concept applies to beginning math, pre-algebra, algebra, and trig, how does this apply to calculus? At 1:01, Sal says that differential calculus is all about finding instantaneous rate of change, but is that the only \"everyday use\"? Or is calculus simply a concept that is used in other subjects, or even professions, like engineering?\n\nThanks!", + "A": "I think the best way to find a good answer to this question is to just keep watching the videos! If you attend college for any engineering discipline, you have to learn calculus before you even begin learning the specifics of your discipline (whether it be mechanical, electrical, civil, computer, computer science, etc...). The best way to understand what every day things calculus will enable you to do is to learn calculus and start doing incredible things every day :-)", + "video_name": "EKvHQc3QEow", + "timestamps": [ + 61 + ], + "3min_transcript": "This is a picture of Isaac Newton, super famous British mathematician and physicist. This is a picture of a Gottfried Leibnitz, super famous, or maybe not as famous, but maybe should be, famous German philosopher and mathematician, and he was a contemporary of Isaac Newton. These two gentlemen together were really the founding fathers of calculus. And they did some of their-- most of their major work in the late 1600s. And this right over here is Usain Bolt, Jamaican sprinter, whose continuing to do some of his best work in 2012. And as of early 2012, he's the fastest human alive, and he's probably the fastest human that has ever lived. And you might have not made the association with these three You might not think that they have a lot in common. But they were all obsessed with the same fundamental question. And this is the same fundamental question that differential calculus addresses. And the question is, what is the instantaneous rate of change of something? Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about. Instantaneous rates of change. Differential calculus. Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis I'll have distance. I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance. And in this axis, we'll say time. but I'll just say x is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way, his average speed is just going to be his change in distance over his change in time. And using the variables that are over here, we're saying y is distance. So this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you", + "qid": "EKvHQc3QEow_61" + }, + { + "Q": "3:09 why is y zero?", + "A": "To find the two intercepts, you have to set x = 0 (to find the y intercept) and y = 0 (to find the x intercept). He is just finding the x intercept at this point.", + "video_name": "6CFE60iP2Ug", + "timestamps": [ + 189 + ], + "3min_transcript": "And in point-slope form, if you know that some, if you know that there's an equation where the line that represents the solutions of that equation has a slope M. Slope is equal to M. And if you know that X equals, X equals A, Y equals B, satisfies that equation, then in point-slope form you can express the equation as Y minus B is equal to M times X minus A. This is point-slope form and we do videos on that. But what I really want to get into in this video is another form. And it's a form that you might have already seen. And that is standard form. Standard. Standard form. And standard form takes the shape of AX plus BY is equal to C, And what I want to do in this video, like we've done in the ones on point-slope and slope-intercept is get an appreciation for what is standard form good at and what is standard form less good at? So let's give a tangible example here. So let's say I have the linear equation, it's in standard form, 9X plus 16Y is equal to 72. And we wanted to graph this. So the thing that standard form is really good for is figuring out, not just the y-intercept, y-intercept is pretty good if you're using slope-intercept form, but we can find out the y-intercept pretty clearly from standard form and the x-intercept. The x-intercept isn't so easy to figure out from these other forms right over here. So how do we do that? Well to figure out the x and y-intercepts, let's just set up a little table here, X comma Y, and so the x-intercept is going to happen when Y is equal to zero. when X is equal to zero. So when Y is zero, what is X? So when Y is zero, 16 times zero is zero, that term disappears, and you're left with 9X is equal to 72. So if nine times X is 72, 72 divided by nine is eight. So X would be equal to eight. So once again, that was pretty easy to figure out. This term goes away and you just have to say hey, nine times X is 72, X would be eight. When Y is equal to zero, X is eight. So the point, let's see, Y is zero, X is one, two, three, four, five, six, seven, eight. That's this point, that right over here. This point right over here is the x-intercept. When we talk about x-intercepts we're referring to the point where the line actually intersects the x-axis. Now what about the y-intercept? Well, we said X equals zero, this disappears. And we're left with 16Y is equal to 72. And so we could solve,", + "qid": "6CFE60iP2Ug_189" + }, + { + "Q": "At 9:40 is it ok if you use the same technique for multiplying to binomials?", + "A": "It depends on what you are doing.", + "video_name": "fGThIRpWEE4", + "timestamps": [ + 580 + ], + "3min_transcript": "of these terms, and then you're going to distribute each of those terms into 3x plus 2. It would take a long time and in reality, you'll never do it quite that way. But you will get the same answer we're going to get. When you have larger polynomials, the easiest way I can think of to multiply, is kind of how you multiply long numbers. So we'll write it like this. 9x squared, minus 6, plus 4. And we're going to multiply that times 3x plus 2. And what I imagine is, when you multiply regular numbers, you have your ones' place, your tens' place, your hundreds' place. Here, you're going to have your constants' place, your first degree place, your second degree place, your third degree place, if there is one. And actually there will be in this video. So you just have to put things in their proper place. So let's do that. So you start here, you multiply almost exactly like you would do traditional multiplication. 2 times 4 is 8. 2 times negative 6x is negative 12x. And we'll put a plus there. That was a plus 8. 2 times 9x squared is 18x squared, so we'll put that in the x squared place. Now let's do the 3x part. I'll do that in magenta, so you see how it's different. 3x times 4 is 12x, positive 12x. 3x times negative 6x, what is that? The x times the x is x squared, so it's going to go over here. And 3 times negative 6 is negative 18. And then finally 3x times 9x squared, the x times the x 3 times 9 is 27. I wrote it in the x third place. And once again, you just want to add the like terms. So you get 8. There's no other constant terms, so it's just 8. Negative 12x plus 12x, these cancel out. 18x squared minus 18x squared cancel out, so we're just left over here with 27x to the third. So this is equal to 27x to the third plus 8. And we are done. And you can use this technique to multiply a trinomial times a binomial, a trinomial times a trinomial, or really, you know, you could have five terms up here. A fifth degree times a fifth degree. This will always work as long as you keep things in their proper degree place.", + "qid": "fGThIRpWEE4_580" + }, + { + "Q": "At 2:50 Sal says this is a fifth degree polynomial. What is that?", + "A": "A 5th degree polynomial is a polynomial that has the 5th power as the highest exponent in one of its term. Ex: 2x^5 _+ 2 x^5 + x^4 + 3", + "video_name": "fGThIRpWEE4", + "timestamps": [ + 170 + ], + "3min_transcript": "Remember, x to the 1, times x to the 1, add the exponents. I mean, you know x times x is x squared. So this first term is going to be 8x squared. And the second term, negative 5 times 2 is negative 10x. Not too bad. Let's do a slightly more involved one. Let's say we had 9x to the third power, times 3x squared, minus 2x, plus 7. So once again, we're just going to do the distributive property here. So we're going to multiply the 9x to the third times each of these terms. So 9x to the third times 3x squared. I'll write it out this time. In the next few, we'll start doing it a little bit in our heads. So this is going to be 9x to the third times 3x squared. way-- minus 2x times 9x to the third, and then plus 7 times 9x to the third. So sometimes I wrote the 9x to the third first, sometimes we wrote it later because I wanted this negative sign here. But it doesn't make a difference on the order that you're multiplying. So this first term here is going to be what? 9 times 3 is 27 times x to the-- we can add the exponents, we learned that in our exponent properties. This is x to the fifth power, minus 2 times 9 is 18x to the-- we have x to the 1, x to the third-- x to the fourth power. Plus 7 times 9 is 63x to the third. So we end up with this nice little fifth degree polynomial. Now let's do one where we are multiplying two binomials. This you're going to see very, very, very frequently in algebra. So let's say you have x minus 3, times x plus 2. And I actually want to show you that all we're doing here is the distributive property. So let me write it like this: times x plus 2. So let's just pretend that this is one big number here. And it is. You know, if you had x's, this would be some number here. So let's just distribute this onto each of these variables. So this is going to be x minus 3, times that green x, plus x minus 3, times that green 2. All we did is distribute the x minus 3. This is just the distributive property. Remember, if I had a times x plus 2, what would", + "qid": "fGThIRpWEE4_170" + }, + { + "Q": "@2:40 is Sal making an assumption that AG is the longest part ?", + "A": "He may be eyeballing it, but you can also tell by the fact that line AG goes all the way past point F on its side, which is also the point that bisects line AE. That shows that AG is longer by proportion than line GD, or the longer part of the median if that made sense :)", + "video_name": "k45QTFCHSVs", + "timestamps": [ + 160 + ], + "3min_transcript": "Let me make sure I have enough space. This entire distance right over here is 18. They tell us that. So the area of AEC is going to be equal to 1/2 times the base-- which is 18-- times the height-- which is 12-- which is equal to 9 times 12, which is 108. That's the area of this entire right triangle, triangle AEC. If we want the area of BGC or any of these smaller of the six triangles-- if we ignore this little altitude right over here, the ones that are bounded by the medians-- then we just have to divide this by 6. Because they all have equal area. We've proven that in a previous video. So the area of BGC is equal to the area of AEC, the entire triangle, divided by 6, which is 108 divided by 6. You get 10 and then 48. Looks like it would be 18. And that's right because it would be-- 108 is the same thing as 18 times 6. So we did our first part. The area of that right over there is 18. And if we wanted, we could say, hey, the area of any of these triangles-- the ones that are bounded by the medians-- this is going to be 18. This is going to be 18. This entire FGE triangle is going to be 18, but we did this first part right over there. Now they ask us, what is the length of AG? So AG is the distance. It's the longer part of this median right over here. And to figure out what AG is, we just have to remind ourselves that the centroid is always 2/3 along the way of the medians, or it divides the median into two segments that have a ratio of 2 to 1. So if we know the entire length of this median, we could just take 2/3 of that. And that'll give us the length of AG. And we know that F and D are the midpoints. So for example, we know this AE is 12. That was given. We know that ED is half of this 18. So ED right over here-- I'll do this in a new color. ED is going to be 9. So then we could just use the Pythagorean theorem to figure out what AD is. AD is the hypotenuse of this right triangle. So we're looking at triangle AED right now. Let me write this down. We know that 12 squared plus 9 squared is going to be equal to AD squared. 12 squared is 144. 144 plus 81.", + "qid": "k45QTFCHSVs_160" + }, + { + "Q": "when he fins the determinant of B he says that it is 1/-7 and that is what he uses to multiply with. So why does he change it to -7 when he puts the B in absolute value signs over on the side (9:34) ?", + "A": "When dealing with matrices the absolute value sign actually means the determinant and does not mean to take the absolute of the number.", + "video_name": "iUQR0enP7RQ", + "timestamps": [ + 574 + ], + "3min_transcript": "But let's apply this to a real problem, and you'll see that it's actually not so bad. So let's change letters, just so you know it doesn't always have to be an A. Let's say I have a matrix B. And the matrix B is 3-- I'm just going to pick random numbers-- minus 4, 2 minus 5. Let's calculate B inverse. So B inverse is going to be equal to 1 over the determinant of B. What's the determinant? It's 3 times minus 5 minus 2 times minus 4. So 3 times minus 5 is minus 15, minus 2 times minus 4. 2 times minus 4 is minus 8. We're going to subtract that. So it's plus 8. And we're going to multiply that times what? And we just make these two terms negative. Minus 2 and 4. 4 was minus 4, so now it becomes 4. And let's see if we can simplify this a little bit. So B inverse is equal to minus 15 plus 8. That's minus 7. So this is minus 1/7. So the determinant of B-- we could write B's determinant-- is equal to minus 7. So that's minus 1/7 times minus 5, 4, minus 2, 3. Which is equal to-- this is just a scalar, this is just a number, so we multiply it times each of the elements-- so that is equal to minus, minus, plus. That's 5/7. 5/7 minus 4/7. Let's see. And then minus 3/7. It's a little hairy. We ended up with fractions here and things. But let's confirm that this really is the inverse of the matrix B. Let's multiply them out. So before I do that I have to create some space. I don't even need this anymore. OK. So let's confirm that that times this, or this times that, is really equal to the identity matrix. So let's do that. So let me switch colors. So B inverse is 5/7, if I haven't made any careless mistakes. Minus 4/7. 2/7. And minus 3/7.", + "qid": "iUQR0enP7RQ_574" + }, + { + "Q": "at 5:00, why do you subtract 300 instead of 200?", + "A": "Because he is subtracting the blue/teal equation from the red/pink equation. Subtracting the red equation from itself would just get you 0=0, which is true, but not very useful.", + "video_name": "xCIHAjsZCE0", + "timestamps": [ + 300 + ], + "3min_transcript": "", + "qid": "xCIHAjsZCE0_300" + }, + { + "Q": "At 7:50, Khan says \"one equation for one unknown.\"\nDoes this mean that if two equations for 2 unkowns and 1 equation for 1 unknown is possible to be solved, then for 3 unknowns you have to get 3 equations? And so on and so forth?", + "A": "That is exactly correct. Good job :-)", + "video_name": "xCIHAjsZCE0", + "timestamps": [ + 470 + ], + "3min_transcript": "", + "qid": "xCIHAjsZCE0_470" + }, + { + "Q": "At 3:00 what did he mean?", + "A": "First solve the problem of first bracket. Then solve the other problems.I think you got it,", + "video_name": "GiSpzFKI5_w", + "timestamps": [ + 180 + ], + "3min_transcript": "When I say fast, how fast it grows. When I take something to an exponent, when I'm taking something to a power, it grows really fast. Then it grows a little bit slower or shrinks a little bit slower if I multiply or divide, so that comes next: multiply or divide. Multiplication and division comes next, and then last of all comes addition and subtraction. So these are kind of the slowest operations. This is a little bit faster. This is the fastest operation. And then the parentheses, just no matter what, always take priority. So let's apply it over here. Let me rewrite this whole expression. So it's 8 plus 5 times 4 minus, in parentheses, 6 plus 10 divided by 2 plus 44. So we're going to want to do the parentheses first. We have parentheses there and there. Now this parentheses is pretty straightforward. could really just view this as 5 times 4. So let's just evaluate that right from the get go. So this is going to result in 8 plus-- and really, when you're evaluating the parentheses, if your evaluate this parentheses, you literally just get 5, and you evaluate that parentheses, you literally just get 4, and then they're next to each other, so you multiply them. So 5 times 4 is 20 minus-- let me stay consistent with the colors. Now let me write the next parenthesis right there, and then inside of it, we'd evaluate this first. Let me close the parenthesis right there. And then we have plus 44. So what is this thing right here evaluate to, this thing inside the parentheses? Well, you might be tempted to say, well, let me just go left to right. 6 plus 10 is 16 and then divide by 2 and you would get 8. But remember: order of operations. Division takes priority over addition, so you actually want here like this. You could imagine putting some more parentheses. Let me do it in that same purple. You could imagine putting some more parentheses right here to really emphasize the fact that you're going to do the division first. So 10 divided by 2 is 5, so this will result in 6, plus 10 divided by 2, is 5. 6 plus 5. Well, we still have to evaluate this parentheses, so this results-- what's 6 plus 5? Well, that's 11. So we're left with the 20-- let me write it all down again. We're left with 8 plus 20 minus 6 plus 5, which is 11, plus 44. And now that we have everything at this level of operations, we can just go left to right. So 8 plus 20 is 28, so you can view this as 28 minus 11 plus 44.", + "qid": "GiSpzFKI5_w_180" + }, + { + "Q": "At 04:09, if the hypotenuse is irrational does this mean that the hypotenuse is a multipal of \"root 2\" or can it also be another irrational number?", + "A": "root two", + "video_name": "X1E7I7_r3Cw", + "timestamps": [ + 249 + ], + "3min_transcript": "I want the context, because in school today if you bring out the ruler and compass and are like, \"Let's do some geometry! Let's draw two lines at 90 degree angles using a straight-edge and compass! Here's a happy square!\" Then you've probably had years of math class already and think of geometry as being harder than adding big numbers together. You probably think that zero is a simple, easy concept and have heard of decimals too. Well, here's now, 2012. Here's Einstein, Euler, Newton and Da Vinci - - that sure was a while ago! Now let's go all the way back to when Arabic numerals were invented and brought to the West by Fibbonacci. Before that, arithmetic was nightmarishly hard, so if you can multiply multi-digit numbers together you can go back in time and impress the beans out of Pythagoras. And before that there was no concept of zero, except in India where zero was discovered around here. And if you keep going back you get to the year one, (there's no year zero, of course, because zero hadn't been invented) and back a bit more you get to folk like Aristotle, Euclid, Archimedes and then finally Pythagoras, all the way back in 6th century BCE. Point is, you can do some pretty cool mathematics without having a good handle on arithmetic and people did for a long time. you need to memorize your multiplication table and graph a parabola before you can learn real mathematics, they are lying to you. In Pythagoras's time there were no variables, no equations or formulas like we see today, Pythagoras's theorem wasn't 'a squared plus b squared equals c squared,' it was 'The squares of the legs of a right triangle have the same area as the square of the hypotenuse,' all written out. And when he said 'square' he meant 'square.' One leg's square plus the other leg's square equals hypotenuse's square. Three literally squared plus four made into a square. Those two squares have the same area as a five by five square. You can cut out the nine squares here and the sixteen here and fit them together where these 25 squares are, and in the same way, you can cut out the 25 hypotenuse squares and fit them into the two leg squares. Pythagoras thought you could do this trick with any right triangle, that it was just a matter of figuring out how many pieces to cut each side into. There was a relationship between the length of one side and the length of another and he wanted to find it on this map. But the trouble began with the simplest right triangle one where both the legs are the same length, one where both the legs' squares are equal. hypotenuse is something that, when squared, gives two. So what's the square root of two and how do we make it into a whole number ratio? Square root two is very close to 1.4 which would be a whole number ratio of 10:14 but 10 squared plus 10 squared is definitely not 14 squared, and a ratio of 1,000 to 1,414 is even closer, and a ratio of 100,000,000 to 141,421,356 is very close indeed but still not exact, so what is it? Pythagoras wanted to find the perfect ratio he knew it must exist, but meanwhile someone from his very own Pythagorean brotherhood proved there wasn't a ratio, the square root of two is irrational, that in decimal notation (once decimal notation was invented) the digits go on forever. Usually this proof is given algebraically, something like this, which is pretty simple and beautiful if you know algebra, but the Pythagoreans didn't. So I like to imagine how they thought of this proof, no algebra required. Okay, so Pythagoras is all like, \"There's totally a ratio, you can make this with whole numbers.\" And this guy's like,\"Is not!\" \"Is too!\" \"Is not!\" \"Is too!\" \"Fine have it your way.", + "qid": "X1E7I7_r3Cw_249" + }, + { + "Q": "At 8:30, why does Sal keep expanding everything out? I do not understand it.", + "A": "he is using this as a complete example to show how it works. He is also using the sigma, which is a sum of all integers from the number on the bottom to n.", + "video_name": "iPwrDWQ7hPc", + "timestamps": [ + 510 + ], + "3min_transcript": "to keep switching colors, but hopefully it's worth it, a plus b. Let's take that to the 4th power. The binomial theorem tells us this is going to be equal to, and I'm just going to use this exact notation, this is going to be the sum from k equals 0, k equals 0 to 4, to 4 of 4 choose k, 4 choose k, 4 choose ... let me do that k in that purple color, 4 choose k of a to the 4 minus k power, 4 minus k power times b to the k power, b to the k power. Now what is that going to be equal to? Well, let's just actually just do the sum. This is going to be equal to, so we're going to start at k equals 0, so when k equals 0, it's going to be 4 choose 0, times a to the 4 minus 0 power, well, that's just going to be a to the 4th power, times b to the 0 power. b to the 0 power is just going to be equal to 1, so we could just put a 1 here if we want to, or we could just leave it like that. This is what we get when k equals 0. Then to that, we're going to add when k equals 1. k equals 1 is going to be, the coefficient is going to be 4 choose 1, and it's going to be times a to the 4 minus 1 power, so a to the 3rd power, and I'll just stick with that color, times b to the k power. Well, now, k is 1b to the 1st power. Then to that, we're going to add, we're going to add 4 choose 2, 4 choose 2 times a to the ... 4 minus 2 is 2. I think you see a pattern here. a to the 4th, a to the 3rd, a squared, and then times b to the k. Well, k is 2 now, so b squared, and you see a pattern again. You could say b to the 0, b to the 1, b squared, and we only have two more terms to add here, plus 4 choose 3, 4 choose 3 times 4 minus 3 is 1, times a, or a to the 1st, I guess we could say, and then b to the 3rd power, times a to the 1st b to the third, and then only one more term, plus 4 choose, 4 choose 4. k is now 4. This is going to be our last term right now. We're getting k goes from 0 all the way to 4, 4 choose 4. a to the 4 minus 4, that's just going to be 1, a to the 0, that's just 1, so we're going to be left with just b to the k power,", + "qid": "iPwrDWQ7hPc_510" + }, + { + "Q": "At 4:39 point, when writing \"n choose k\" for the first time, you say, \"We'll review that in a second. This comes straight of out ?\" I didn't hear that part. It comes straight out of WHAT?", + "A": "There is a term called Combination,which states that each of the different groups or selection which can be made out by taking some or all of a number of things at a time.or simply Selection of r terms out of n terms....that part is derived from this very term.....selection of r terms out of n terms..... nCr = n!/r!(n-r)! ....", + "video_name": "iPwrDWQ7hPc", + "timestamps": [ + 279 + ], + "3min_transcript": "plus b to the 3rd power. Just taking some of the 3rd power, this already took us a little reasonable amount of time, and so you can imagine how painful it might get to do something like a plus b to the 4th power, or even worse, if you're trying to find a plus b to the 10th power, or to the 20th power. This would take you all day or maybe even longer than that. It would be incredibly, incredibly painful. That's where the binomial theorem becomes useful. What is the binomial theorem? The binomial theorem tells us, let me write this down, binomial theorem. Binomial theorem, it tells us that if we have a binomial, and I'll just stick with the a plus b for now, if I have, and I'm going to try to color code this a little bit, if I have the binomial a plus b, and I'm going to raise it the nth power, I'm going to raise this to the nth power, the binomial theorem tells us that this is going to be equal to, and the notation is going to look a little bit complicated at first, but then we'll work through an actual example, is going to be equal to the sum from k equals 0, k equals 0 to n, this n and this n are the same number, of ... I don't want to ... that's kind of a garish color ... of n choose k, n choose k, and we'll review that in a second; this comes straight out of combinatorics; n choose k times a to the n minus k, n minus k, times b, times b to the k, Now this seems a little bit unwieldy. Let's just review, remind ourselves what n choose k actually means. If we say n choose k, I'll do the same colors, n choose k, we remember from combinatorics this would be equal to n factorial, n factorial over k factorial, over k factorial times n minus k factorial, n minus k factorial, so n minus k minus k factorial, let me color code this, n minus k factorial. Let's try to apply this. Let's just start applying it to the thing that started to intimidate us, say, a plus b to the 4th power. Let's figure out what that's going to be. Let's try this. So a, and I'm going to try to keep it color-coded so you know what's going on, a plus b,", + "qid": "iPwrDWQ7hPc_279" + }, + { + "Q": "Why 4! / 0!4! = 1? it's just ( 4 * 3 * 2* 1 ) / ( 0 * 4 * 3 * 2 * 1 ) = 24 / 0\nwhich is undefined. Why sal says it's equal to 1? at 9:37", + "A": "0 factorial does not equal zero. By definition it equals 1.", + "video_name": "iPwrDWQ7hPc", + "timestamps": [ + 577 + ], + "3min_transcript": "4 minus 2 is 2. I think you see a pattern here. a to the 4th, a to the 3rd, a squared, and then times b to the k. Well, k is 2 now, so b squared, and you see a pattern again. You could say b to the 0, b to the 1, b squared, and we only have two more terms to add here, plus 4 choose 3, 4 choose 3 times 4 minus 3 is 1, times a, or a to the 1st, I guess we could say, and then b to the 3rd power, times a to the 1st b to the third, and then only one more term, plus 4 choose, 4 choose 4. k is now 4. This is going to be our last term right now. We're getting k goes from 0 all the way to 4, 4 choose 4. a to the 4 minus 4, that's just going to be 1, a to the 0, that's just 1, so we're going to be left with just b to the k power, We're almost done. We've expanded it out. We just need it figure out what 4 choose 0, 4 choose 1, 4 choose 2, et cetera, et cetera are, so let's figure that out. We could just apply this over and over again. So 4 choose 0, 4 choose 0 is equal to 4 factorial over 0 factorial times 4 minus 0 factorial. That's just going to be 4 factorial again. 0 factorial, at least for these purposes, we are defining to be equal to 1, so this whole thing is going to be equal to 1, so this coefficient is 1. Let's keep going here. So 4 choose 1 is going to be 4 factorial over 1 factorial times 4 minus 1 factorial, 4 minus 1 factorial, so 3 factorial. What's this going to be? 1 factorial is just going to be 1. 3 factorial is 3 times 2 times 1. 4 times 3 times 2 times 1 over 3 times 2 times 1 is just going to leave us with 4. This right over here is just going to be 4. Then we need to figure out what 4 choose 2 is. 4 choose 2 is going to be 4 factorial over 2 factorial times what's 4 minus ... this is going to be n minus k, 4 minus 2 over 2 factorial. So what is this going to be? Let me scroll over to the right a little bit. This is going to be 4 times 3 times 2 times 1 over 2 factorial is 2, over 2 times 2. This is 2, this is 2, so 2 times 2 is same thing as 4. We're left with 3 times 2 times 1, which is equal to 6. That's equal to 6. Then what is 4 choose 3? I'll use some space down here. So 4 choose 3,", + "qid": "iPwrDWQ7hPc_577" + }, + { + "Q": "at 2:35, when looking to draw the vector [1,2], I don't understand why the x component should be 1 and the Y component should be 2. Isn't the desired output, based on y = 1 and x = 2?", + "A": "There s no mistake in there. The input coordinates are (2,1). And according to the partial derivative of the given function, the output is a vector field with x-component equal to the ordinate and y- component equal to the abscissa. So, the output vectors are given by = yi + xj , where i and j are the basis vectors for x and y axes. So the output for (2,1) will be 1i+2j", + "video_name": "ZTbTYEMvo10", + "timestamps": [ + 155 + ], + "3min_transcript": "y equals two over x. And that's where you would see something like this. So all of these lines, they're representing constant values for the function. And now I want to take a look at the gradient field. And the gradient, if you'll remember, is just a vector full of the partial derivatives of f. And let's just actually write it out. The gradient of f, with our little del symbol, is a function of x and y. And it's a vector-valued function whose first coordinate is the partial derivative of f with respect to x. And the second component is the partial derivative with respect to y. So when we actually do this for our function, we take the partial derivative with respect to x. X looks like a variable. Y looks like a constant. The derivative of this whole thing is just equal to that constant, y. And then kind of the reverse for when you take the partial derivative with respect to y. X looks like a constant. And the derivative is just that constant, x. And this can be visualized as a vector field in the xy plane as well. You know, at every given point, xy, so you kind of go like x equals two, y equals one, let's say. So that would be x equals two, y equals one. You would plug in the vector and see what should be output. And at this point, the point is two, one. The desired output kind of swaps those. So we're looking somehow to draw the vector one, two. So you would expect to see the vector that has an x component of one and a y component of two. Something like that. But it's probably gonna be scaled down because of the way we usually draw vector fields. And the entire field looks like this. So I'll go ahead and erase what I had going on. Since this is a little bit clearer. And remember, we scaled down all the vectors. The color represents length. So red here is super-long. Blue is gonna be kind of short. And one thing worth noticing. if the vector is crossing a contour line, it's perpendicular to that contour line. Wherever you go. this vector's perpendicular to the contour line. Over here, perpendicular to the contour line. And this happens everywhere. And it's for a very good reason. And it's also super-useful. So let's just think about what that reason should be. Let's zoom in on a point. So I'm gonna clear up our function here. Clear up all of the information about it. And just zoom in on one of those points. So let's say like right here. We'll take that guy and kind of imagine zooming in and saying what's going on in that region? So you've got some kind of contour line. And it's swooping down like this. And that represents some kind of value. Let's say that represents the value f equals two. And, you know, it might not be a perfect straight line. But the more you zoom in, the more it looks like a straight line. And when you want to interpret the gradient vector.", + "qid": "ZTbTYEMvo10_155" + }, + { + "Q": "At 1:40 where did he get nine times nine from? Also he divided the 350/360 by ten should he divide the other side by ten also?", + "A": "Hey Janet, 9*9 is the same thing as 81. With fractions, if you divide the numerator by 10, you divide the denominator by 10 as well. You don t need to divide the other side, it s just simplifying a fraction and still the same number after all.", + "video_name": "u8JFdwmBvvQ", + "timestamps": [ + 100 + ], + "3min_transcript": "A circle with area 81 pi has a sector with a 350-degree central angle. So this whole sector right over here that's shaded in, this pale orange-yellowish color, that has a 350-degree central angle. So you see the central angle, it's a very large angle. It's going all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle-- over 360. So the area of the sector over the total area over the total degrees in a circle. And then we just can solve for area of a sector by multiplying both sides by 81 pi. 81 pi, 81 pi-- so these cancel out. 350 divided by 360 is 35/36. And so our area, our sector area, is equal to-- let's see, in the numerator, we have 35 times-- instead of 81, let's see, that's going to be 9 times 9 pi. And in the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9, and so we are left with 35 times 9. And neither of these are divisible by 4, so that's about as simplified as we can get it. 35 times 9, it's going to be 350 minus 35, which would be 315, I guess. Did I do that right? Yeah, it's going to be 270 plus 45, which is 315 pi over 4. 315 pi over 4 is the area of the sector.", + "qid": "u8JFdwmBvvQ_100" + }, + { + "Q": "At 0:28, Mr. Khan mentions a ratio. What is that?", + "A": "He basically created a proportion using the values given in the circle.", + "video_name": "u8JFdwmBvvQ", + "timestamps": [ + 28 + ], + "3min_transcript": "A circle with area 81 pi has a sector with a 350-degree central angle. So this whole sector right over here that's shaded in, this pale orange-yellowish color, that has a 350-degree central angle. So you see the central angle, it's a very large angle. It's going all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle-- over 360. So the area of the sector over the total area over the total degrees in a circle. And then we just can solve for area of a sector by multiplying both sides by 81 pi. 81 pi, 81 pi-- so these cancel out. 350 divided by 360 is 35/36. And so our area, our sector area, is equal to-- let's see, in the numerator, we have 35 times-- instead of 81, let's see, that's going to be 9 times 9 pi. And in the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9, and so we are left with 35 times 9. And neither of these are divisible by 4, so that's about as simplified as we can get it. 35 times 9, it's going to be 350 minus 35, which would be 315, I guess. Did I do that right? Yeah, it's going to be 270 plus 45, which is 315 pi over 4. 315 pi over 4 is the area of the sector.", + "qid": "u8JFdwmBvvQ_28" + }, + { + "Q": "The result of the composition of Ix and g has to be the same as the result of the composition of h and f and g when one inputs a 'y' value and gets an 'x' value (at time marker 16:40). But how is showing that if the results are the same, then the functions are the same as well? (abstractly speaking)", + "A": "Speaking any kind of way (abstractly or otherwise), x = g(y) = I_X(g(y)) = h(f(g(y))) = h((I_Y)(y)) = h(y). So g(y) = h(y), and g = h. (I don t really understand this either yet.) : |", + "video_name": "-eAzhBZgq28", + "timestamps": [ + 1000 + ], + "3min_transcript": "I could do the same thing here with h. I just take a point here, apply h, then apply f back. I should just go back to that point. That's all of what this is saying. So let's go back to the question of whether g is unique. Can we have two different inverse functions g and h? So let's start with g. Remember g is just a mapping from Y to X. So this is going to be equal to, this is the same thing as the composition of the identity function over x with g. To show you why that's the case, remember g just goes from-- these diagrams get me confused very quickly-- so let's say this is x and this is y. Remember g is a mapping from y to x. So g will take us there. There's a mapping from y to x. mapping, or the identity function in composition with this. Because all this is saying is you apply g, and then you apply the identity mapping on x. So obviously you're going to get to the exact same mapping or the exact same point. So these are equivalent. But what is another way of writing the identity mapping on x? What's another way of writing that? Well by definition, if h is another inverse of f, this is true. So I can replace this in this expression with a composition of h with f. So this is going to be equal to the composition of h with f, and the composition of that with g. You might want to put parentheses here. I'll do it very lightly. You might want to put parentheses there. But I showed you a couple of videos ago that the composition of functions, or of transformations, is associative. It doesn't matter if you put the parentheses there or if you put the parentheses there. Actually I'll do that. I'll put the parentheses there at first just so you can as that right there. But we know that composition is associative. So this is equal to the composition of h with the composition of f and g. Now what is this equal to, the composition of f and g? Well it's equal to, by definition, it's equal to the identity transformation over y. So this is equal to h composed with, or the composition of h with, the identity function over y with this right here. Now what is this going to be? Remember h is a mapping from y to x. Let me redraw it. So that's my x and that is my y. h could take some element in y and gives me some element in x. If I take the composition of the identity in y-- so that's essentially I take some element, let me do it this", + "qid": "-eAzhBZgq28_1000" + }, + { + "Q": "3:41 OK really confused. why do the repeating numbers start at 4 and not 1. Sal writes 414141... but shouldnt it be 14141414...\ncan anyone explain this?", + "A": "x=.714141414... If you multiplied by x by 10, then that would move it one place to the right, or 7.14141414... and the 141414... would start immediately after the decimal point. Since you have to multiply by 100, however, you move it two places to the right and get: 7 1. 4 1 4 1 4 1 4. The numbers after the decimal must start with the 4 first since you moved it two places to the right. The 141414... pattern is still there. You re just starting at a different place.", + "video_name": "Ihws0d-WLzU", + "timestamps": [ + 221 + ], + "3min_transcript": "are left with x is equal to 36 over 99. And both the numerator and the denominator is divisible by 9, so we can reduce this. If we divide the numerator by 9, we get 4. The denominator by 9-- we get 11. So 0.363636 forever and forever repeating is 4/11. Now let's do another interesting one. Let's say we have the number 0.714, and the 14 is repeating. And so this is the same thing. So notice, the 714 isn't going to repeat. Just the 14 is going to repeat. So this is 0.7141414, on and on and on and on. So let's set this equal to x. the decimal all the way clear of 714. But you actually don't want to do that. You want to shift it just enough so that the repeating pattern can be right under itself when you do the subtraction. So again in this situation, even though we have three numbers behind the decimal point, because only two of them are repeating, we only want to multiply by 10 to the second power. So once again, you want to multiply by 100. So you get 100x is equal to-- we're moving the decimal two to the right, one, two-- so it's going to be 71.4141, on and on and on and on. So it's going to be 71.4141414 and on and on and on. And then let me rewrite x right below this. We have x is equal to 0.7141414. And notice, now the 141414's, they're So it's going to work out when we subtract. So let's subtract these things. 100x minus x is 99x. And this is going to be equal to-- these 1414's are going to cancel with those 1414's. And we have 71.4 minus 0.7. And we can do this in our head, or we can borrow if you like. This could be a 14. This is a 0. So you have 14 minus 7 is 7 and then 70 minus 0. So you have 99x is equal to 70.7. And then we can divide both sides by 99. And you could see all of the sudden something strange is happening because we still have a decimal. But we can fix that up at the end. So let's divide both sides by 99. You get x is equal to 70.7 over 99. Now obviously, we haven't converted this into a pure fraction yet. We still have a decimal in the numerator. But that's pretty easy to fix.", + "qid": "Ihws0d-WLzU_221" + }, + { + "Q": "At 4:04 can we multiply by 10?", + "A": "If we multiply by 10, the decimal point would only shift to the right one point. we need the point shifted over twice, so we multiply by 100.", + "video_name": "Ihws0d-WLzU", + "timestamps": [ + 244 + ], + "3min_transcript": "are left with x is equal to 36 over 99. And both the numerator and the denominator is divisible by 9, so we can reduce this. If we divide the numerator by 9, we get 4. The denominator by 9-- we get 11. So 0.363636 forever and forever repeating is 4/11. Now let's do another interesting one. Let's say we have the number 0.714, and the 14 is repeating. And so this is the same thing. So notice, the 714 isn't going to repeat. Just the 14 is going to repeat. So this is 0.7141414, on and on and on and on. So let's set this equal to x. the decimal all the way clear of 714. But you actually don't want to do that. You want to shift it just enough so that the repeating pattern can be right under itself when you do the subtraction. So again in this situation, even though we have three numbers behind the decimal point, because only two of them are repeating, we only want to multiply by 10 to the second power. So once again, you want to multiply by 100. So you get 100x is equal to-- we're moving the decimal two to the right, one, two-- so it's going to be 71.4141, on and on and on and on. So it's going to be 71.4141414 and on and on and on. And then let me rewrite x right below this. We have x is equal to 0.7141414. And notice, now the 141414's, they're So it's going to work out when we subtract. So let's subtract these things. 100x minus x is 99x. And this is going to be equal to-- these 1414's are going to cancel with those 1414's. And we have 71.4 minus 0.7. And we can do this in our head, or we can borrow if you like. This could be a 14. This is a 0. So you have 14 minus 7 is 7 and then 70 minus 0. So you have 99x is equal to 70.7. And then we can divide both sides by 99. And you could see all of the sudden something strange is happening because we still have a decimal. But we can fix that up at the end. So let's divide both sides by 99. You get x is equal to 70.7 over 99. Now obviously, we haven't converted this into a pure fraction yet. We still have a decimal in the numerator. But that's pretty easy to fix.", + "qid": "Ihws0d-WLzU_244" + }, + { + "Q": "At 4:14, how come he didn't turn the fraction into a decimal?", + "A": "Sal solves two problems where a missing side length is found by proving that triangles are similar and using this to find the measure.", + "video_name": "R-6CAr_zEEk", + "timestamps": [ + 254 + ], + "3min_transcript": "Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it in just the way that we've written down the similarity. If this is true, then BC is the corresponding side to DC. So we know that the length of BC over DC right over here is going to be equal to the length of-- well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to-- what's the corresponding side to CE? The corresponding side over here is CA. This is last and the first. Last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. And now, we can just solve for CE. Well, there's multiple ways that you could think about this. You could cross-multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE. 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2/5, or 2.4. And we're done. We were able to use similarity to figure out this side just knowing that the ratio between the corresponding sides are going to be the same. Now, let's do this problem right over here. Let's do this one. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also, in both triangles-- so I'm looking at triangle CBD and triangle CAE-- they both share this angle up here.", + "qid": "R-6CAr_zEEk_254" + }, + { + "Q": "At 0:35 couldn't you just make the two triangles into one rectangle with a height of 3\" and a width of 5\"?", + "A": "No, but you could make the two triangles into a rectangle with a height of 3 and a width of 4 (the entire base of the pentagon is 8, so half of that would be 4)", + "video_name": "7S1MLJOG-5A", + "timestamps": [ + 35 + ], + "3min_transcript": "Find the area and perimeter of the polygon. So let's start with the area first. So the area of this polygon-- there's kind of two parts of this. First, you have this part that's kind of rectangular, or it is rectangular, this part right over here. And that area is pretty straightforward. It's just going to be base times height. So area's going to be 8 times 4 for the rectangular part. And then we have this triangular part up here. So we have this area up here. And for a triangle, the area is base times height times 1/2. And that actually makes a lot of sense. Because if you just multiplied base times height, you would get this entire area. You would get the area of that entire rectangle. And you see that the triangle is exactly 1/2 of it. If you took this part of the triangle and you flipped it over, you'd fill up that space. If you took this part of the triangle and you flipped it over, you'd fill up that space. So the triangle's area is 1/2 of the triangle's base times the triangle's height. So plus 1/2 times the triangle's base, is 4 inches. And so let's just calculate it. This gives us 32 plus-- oh, sorry. That's not 8 times 4. I don't want to confuse you. The triangle's height is 3. 8 times 3, right there. That's the triangle's height. So once again, let's go back and calculate it. So this is going to be 32 plus-- 1/2 times 8 is 4. 4 times 3 is 12. And so our area for our shape is going to be 44. Now let's do the perimeter. The perimeter-- we just have to figure out what's the sum of the sides. How long of a fence would we have to build if we wanted to make it around this shape, right along the sides of this shape? So the perimeter-- I'll just write P for perimeter. It's going to be equal to 8 plus 4 plus 5 plus this 5, this edge So I have two 5's plus this 4 right over here. So you have 8 plus 4 is 12. 12 plus 10-- well, I'll just go one step at a time. 12 plus 5 is 17. 17 plus 5 is 22. 22 plus 4 is 26. So the perimeter is 26 inches. And let me get the units right, too. Because over here, I'm multiplying 8 inches by 4 inches. So you get square inches. 8 inches by 3 inches, so you get square inches again. So this is going to be square inches. So area is 44 square inches. Perimeter is 26 inches. And that makes sense because this is a two-dimensional measurement. It's measuring something in two-dimensional space, so you get a two-dimensional unit. This is a one-dimensional measurement. It's only asking you, essentially, how long would a string have to be to go around this thing. And so that's why you get one-dimensional units.", + "qid": "7S1MLJOG-5A_35" + }, + { + "Q": "Could someone please tell me why at 3:40 -x^2*sqrt6+x^2*sqrt2=(sqrt2-sqrt6)x^2?\n\nShouldn't it be -x^2*sqrt6+x^2*sqrt2=sqrt2-sqrt6 since one of the x^2 is negative while the other is positive? Shouldn't the x^2s cancel out then.\n\nI'm not sure if I've just made a careless error or am just missing something here, but I don't know why you get a negative x^2 or an x^2 at all. Please help explain this.", + "A": "So lets pretend for a minute that instead the expression was -6z + 2z We can rearrange them using the commutative property: = 2z - 6z And then factor out the z, using the distributive property: = z (2 - 6) Now we can replace z with x^2, and the numbers with their square roots and we can still do the same thing: -Sqrt(6)x^2 + Sqrt(2)x^2 = Sqrt(2)x^2 - Sqrt(6)x^2 = (Sqrt(2)-Sqrt(6))x^2", + "video_name": "yAH3722GrP8", + "timestamps": [ + 220 + ], + "3min_transcript": "as this up here, we're just switching the order of the multiplication. You can kind of view it as we're distributing from the right. And so if you do this, you get x squared times x squared, which is x to the fourth, that's that times that, and then minus x squared times the principal square root of 6. And then over here you have square root of 2 times x squared, so plus x squared times the square root of 2. And then you have square root of 2 times the square root of 6. And we have a negative sign out here. Now if you take the square root of 2-- let me do this on the side-- square root of 2 times the square root of 6, we know from simplifying radicals that this is the exact same thing as the square root of 2 times 6, or the principal square root of 12. So the square root of 2 times square root of 6, we have a negative sign out here, it becomes minus the square root of 12. You have an x to the fourth term. And then here you have-- well depending on how you want to view it, you could say, look, we have to second degree terms. We have something times x squared, and we have something else times x squared. So if you want, you could simplify these two terms over here. So I have square root of 2 x squareds and then I'm going to subtract from that square root of 6 x squareds. So you could view this as square root of 2 minus the square root of 6, or the principal square root of 2 minus the principal square root of 6, x squared. And then, if you want, square root of 12, you might be able to simplify that. 12 is the same thing as 3 times 4. So the square root of 12 is equal to square root of 3 times square root of 4. And the square root of 4, or the principal square root of 4 I should say, is 2. So the square root of 12 is the same thing as 2 square roots of 3. we could write minus 2 times the principal square root of 3. And then out here you have an x to the fourth plus this. And you see, if you distributed this out, if you distribute this x squared, you get this term, negative x squared, square root of 6, and if you distribute it onto this, you'd get that term. So you could debate which of these two is more simple. Now I mentioned that this way I just did the distributive property twice. Nothing new, nothing fancy. But in some classes, you will see something called FOIL. And I think we've done this in previous videos. FOIL. I'm not a big fan of it because it's really a way to memorize a process as opposed to understanding that this is really just from the common-sense distributive property. But all this is is a way to make sure that you're multiplying everything times everything when you're multiplying two binomials times each other like this. And FOIL just says, look, first multiply the first term. So x squared times x squared is x to the fourth.", + "qid": "yAH3722GrP8_220" + }, + { + "Q": "At 3:21 why is it sqrt2 - sqrt6 and not the other way around? Or would it still be the same either way?", + "A": "It s equivalent, sal choose to do it like that so you only have to use one operation symbol i.e. rather than: -sqrt6 + sqrt2 He choose: sqrt2 - sqrt6 But both are equivalent", + "video_name": "yAH3722GrP8", + "timestamps": [ + 201 + ], + "3min_transcript": "So let's do that. So we get x squared minus the principal square root of 6 times this term-- I'll do it in yellow-- times x squared. And then we have plus this thing again. We're just distributing it. It's just like they say. It's sometimes not that intuitive because this is a big expression, but you can treat it just like you would treat a variable over You're distributing it over this expression over here. And so then we have x squared minus the principal square root of 6 times the principal square root of 2. And now we can do the distributive property again, but what we'll do is we'll distribute this x squared onto each of these terms and distribute the square root of 2 onto each of these terms. It's the exact same thing as here, it's just you could imagine writing it like this. x plus y times a is still going to be ax plus ay. as this up here, we're just switching the order of the multiplication. You can kind of view it as we're distributing from the right. And so if you do this, you get x squared times x squared, which is x to the fourth, that's that times that, and then minus x squared times the principal square root of 6. And then over here you have square root of 2 times x squared, so plus x squared times the square root of 2. And then you have square root of 2 times the square root of 6. And we have a negative sign out here. Now if you take the square root of 2-- let me do this on the side-- square root of 2 times the square root of 6, we know from simplifying radicals that this is the exact same thing as the square root of 2 times 6, or the principal square root of 12. So the square root of 2 times square root of 6, we have a negative sign out here, it becomes minus the square root of 12. You have an x to the fourth term. And then here you have-- well depending on how you want to view it, you could say, look, we have to second degree terms. We have something times x squared, and we have something else times x squared. So if you want, you could simplify these two terms over here. So I have square root of 2 x squareds and then I'm going to subtract from that square root of 6 x squareds. So you could view this as square root of 2 minus the square root of 6, or the principal square root of 2 minus the principal square root of 6, x squared. And then, if you want, square root of 12, you might be able to simplify that. 12 is the same thing as 3 times 4. So the square root of 12 is equal to square root of 3 times square root of 4. And the square root of 4, or the principal square root of 4 I should say, is 2. So the square root of 12 is the same thing as 2 square roots of 3.", + "qid": "yAH3722GrP8_201" + }, + { + "Q": "At 2:10 he says that if r=1 denominator is 0, and we can't divide by zero. But, in that case numerator would also be 0, since a-a*(1)^n=0. Isn't lim 0/0=1?", + "A": "No, the limit of 0/0 is undefined, and since the limit is for the variable n and not for r, you cannot use any of the limit techniques to get rid of the 0/0.", + "video_name": "b-7kCymoUpg", + "timestamps": [ + 130 + ], + "3min_transcript": "In a previous video, we derived the formula for the sum of a finite geometric series where a is the first term and r is our common ratio. What I want to do in this video is now think about the sum of an infinite geometric series. And I've always found this mildly mind blowing because, or actually more than mildly mind blowing, because you're taking the sum of an infinite things but as we see, you can actually get a finite value depending on what your common ratio is. So there's a couple of ways to think about it. One is, you could say that the sum of an infinite geometric series is just a limit of this as n approaches infinity. So we could say, what is the limit as n approaches infinity of this business, of the sum from k equals zero to n of a times r to the k. as n approaches infinity right over here. So that would be the same thing as the limit as n approaches infinity of all of this business. Let me just copy and paste that so I don't have to keep switching colors. So copy and then paste. So what's the limit as n approaches infinity here? Let's think about that for a second. I encourage you to pause the video, and I'll give you one hint. Think about it for r is greater than one, for r is equal to one, and actually let me make it clear-- let's think about it for the absolute values of r is greater than one, the absolute values of r equal to one, and then the absolute value of r less than one. Well, I'm assuming you've given a go at it. So if the absolute value of r is greater than one, as this exponent explodes, as it approaches infinity, this number is just going to become massively, And so the whole thing is just going to become, or at least you could think of the absolute value of the whole thing, is just going to become a very, very, very large number. If r was equal to one, then the denominator is going to become zero. And we're going to be dividing by that denominator, and this formula just breaks down. But where this formula can be helpful, and where we can get this to actually give us a sensical result, is when the absolute value of r is between zero and one. We've already talked about, we're not even dealing with the geometric, we're not even talking about a geometric series if r is equal to zero. So let's think about the case where the absolute value of r is greater than zero, and it is less than one. What's going to happen in that case? Well, the denominator is going to make sense, right over here. And then up here, what's going to happen? Well, if you take something with an absolute value less than one, and you take it to higher and higher and higher", + "qid": "b-7kCymoUpg_130" + }, + { + "Q": "at 5:30 he Sal says 3 forth of the pizza has cheese, why does he put the number like that, one on top of the other, what does it mean and why not put the 4 on top instead of the 3?", + "A": "first i think you meant at 2:30. Sal s pizza had 1/4 with olives & 3/4 with cheese, right? now it makes sense 3 of the 4 slices of pizza are cheese while 1 slice is olives. now if it was 4 out of the 3 pieces have cheese that would mean you would have 3 slices of pizza with 4 of those 3 slices being cheese, that would make things a bit confusing.", + "video_name": "kZzoVCmUyKg", + "timestamps": [ + 330 + ], + "3min_transcript": "", + "qid": "kZzoVCmUyKg_330" + }, + { + "Q": "I still don't understand, at 0:33, why did he make reference of the pizza as an example of fractions?", + "A": "the circle is the most simple to esplain fractions.", + "video_name": "kZzoVCmUyKg", + "timestamps": [ + 33 + ], + "3min_transcript": "", + "qid": "kZzoVCmUyKg_33" + }, + { + "Q": "During 1:25-1:50, the LCD should be 0, shouldn't it?", + "A": "Not exactly.. the way you get LCD ( Lowest common Denominator ) is by factorizing the denominators of fractions until you get something common. Sal was just looking for something common. and if you noticed he stopped at 24. I hope this helps.", + "video_name": "lxjmR4pYIVU", + "timestamps": [ + 85, + 110 + ], + "3min_transcript": "We're asked to rewrite the following two fractions as fractions with a least common denominator. So a least common denominator for two fractions is really just going to be the least common multiple of both of these denominators over here. And the value of doing that is then if you can make these a common denominator, then you can add the two fractions. And we'll see that in other videos. But first of all, let's just find the least common multiple. Let me write it out because sometimes LCD could meet other things. So least common denominator of these two things is going to be the same thing as the least common multiple of the two denominators over here. The least common multiple of 8 and 6. And a couple of ways to think about least common multiple-- you literally could just take the multiples of 8 and 6 So let's do it that way first. So multiples of six are 6, 12, 18, 24 30. And I could keep going if we don't find any common multiples out of this group here with any of the multiples in eight. And the multiples of eight are 8, 16, 24, and it looks like we're done. And we could keep going obviously-- 32, so on and so forth. But I found a common multiple and this is their smallest common multiple. They have other common multiples-- 48 and 72, and we could keep adding more and more multiple. But this is their smallest common multiple, their least common multiple. So it is 24. Another way that you could have found at least common multiple is you could have taken the prime factorization of six and you say, hey, that's 2, and 3. So the least common multiple has to have at least 1, 2, and 1, 3 in its prime factorization in order for it And you could have said, what's the prime factorization of 8? It is 2 times 4 and 4 is 2 times 2. So in order to be divisible by 8, you have to have at least three 2's in the prime factorization. So to be divisible by 6, you have to have a 2 times a 3. And then to be divisible by 8, you have to have at least three 2's. You have to have two times itself three times I should say. Well, we have one 2 and let's throw in a couple more. So then you have another 2 and then another 2. So this part right over here makes it divisible by 8. And this part right over here makes it divisible by 6. If I take 2 times 2 times 2 times 3, that does give me 24. So our least common multiple of 8 and 6, which is also the least common denominator of these two fractions is going to be 24. So what we want to do is rewrite each of these fractions with 24 as the denominator. So I'll start with 2 over 8.", + "qid": "lxjmR4pYIVU_85_110" + }, + { + "Q": "At 1:21 why can he only measure length?", + "A": "He can only measure length because it is a line. A line is a one-dimensional object. A two dimensional objects allows you to measure width and length, and a three-dimensional object allows you to measure its height, width, and length", + "video_name": "xMz9WFvox9g", + "timestamps": [ + 81 + ], + "3min_transcript": "Human beings have always realized that certain things are longer than other things. For example, this line segment looks longer than this line segment. But that's not so satisfying just to make that comparison. You want to be able to measure it. You want to be able to quantify how much longer the second one is than the first one. And how do we go about doing that? Well, we define a unit length. So if we make this our unit length, we say this is one unit, then we could say how many of those the lengths are each of these lines? So this first line looks like it is-- we could do one of those units and then we could do it again, so it looks like this is two units. While this third one looks like we can get-- let's see that's 1, 2, 3 of the units. So this is three of the units. And right here, I'm just saying units. Sometimes we've made conventions to define a centimeter, where the unit might look something like this. And it's going to look different depending on your screen. Or we might have an inch that looks something like this. be able to fit on this screen based on how big I've just drawn the inch or a meter. So there's different units that you could use to measure in terms of. But now let's think about more dimensions. This is literally a one-dimensional case. This is 1D. Why is it one dimension? Well, I can only measure length. But now let's go to a 2D case. Let's go to two dimensions where objects could have a length and a width or a width and a height. So let's imagine two figures here that look like this. So let's say this is one of them. This is one of them. And notice, it has a width and it has a height. Or you could view it as a width and the length, depending on how you want to view it. So let's say this is one figure right over here. And let's say this is the other one. So this is the other one right over here. Try to draw them reasonably well. And we want to say, well, how much in two dimensions space is this taking up? Or how much area are each of these two taking up? Well, once again, we could just make a comparison. This second, if you viewed them as carpets or rectangles, the second rectangle is taking up more of my screen than this first one, but I want to be able to measure it. So how would we measure it? Well, once again, we would define a unit square. Instead of just a unit length, we now have two dimensions. We have to define a unit square. And so we might make our unit square. And the unit square we will define as being a square, where its width and its height are both equal to the unit length. So this is its width is one unit and its height is one unit. And so we will often call this 1 square unit. Oftentimes, you'll say this is 1 unit. And you put this 2 up here, this literally means 1 unit squared.", + "qid": "xMz9WFvox9g_81" + }, + { + "Q": "At 2:36, what is a \"unit Square\"?", + "A": "Its the unit of the shape ( e.g. feet, inches ) squared. The square is the little 2 at the top. For example 56 ft squared.", + "video_name": "xMz9WFvox9g", + "timestamps": [ + 156 + ], + "3min_transcript": "be able to fit on this screen based on how big I've just drawn the inch or a meter. So there's different units that you could use to measure in terms of. But now let's think about more dimensions. This is literally a one-dimensional case. This is 1D. Why is it one dimension? Well, I can only measure length. But now let's go to a 2D case. Let's go to two dimensions where objects could have a length and a width or a width and a height. So let's imagine two figures here that look like this. So let's say this is one of them. This is one of them. And notice, it has a width and it has a height. Or you could view it as a width and the length, depending on how you want to view it. So let's say this is one figure right over here. And let's say this is the other one. So this is the other one right over here. Try to draw them reasonably well. And we want to say, well, how much in two dimensions space is this taking up? Or how much area are each of these two taking up? Well, once again, we could just make a comparison. This second, if you viewed them as carpets or rectangles, the second rectangle is taking up more of my screen than this first one, but I want to be able to measure it. So how would we measure it? Well, once again, we would define a unit square. Instead of just a unit length, we now have two dimensions. We have to define a unit square. And so we might make our unit square. And the unit square we will define as being a square, where its width and its height are both equal to the unit length. So this is its width is one unit and its height is one unit. And so we will often call this 1 square unit. Oftentimes, you'll say this is 1 unit. And you put this 2 up here, this literally means 1 unit squared. could've been a centimeter. So this would be 1 square centimeter. But now we can use this to measure these areas. And just as we said how many of this unit length could fit on these lines, we could say, how many of these unit squares can fit in here? And so here, we might take one of our unit squares and say, OK, it fills up that much space. Well, we need more to cover all of it. Well, there, we'll put another unit square there. We'll put another unit square right over there. We'll put another unit square right over there. Wow, 4 units squares exactly cover this. So we would say that this has an area of 4 square units or 4 units squared. Now what about this one right over here? Well, here, let's seem I could fit 1, 2, 3, 4, 5, 6, 7, 8, and 9. So here I could fit 9 units, 9 units squared.", + "qid": "xMz9WFvox9g_156" + }, + { + "Q": "at 6:11 why did he only count the first cubes showing e forgot the back that you cannot see", + "A": "He didn t forget. He showed that there are two layers of 2 x 2 blocks. Each block is 1 unit^3 and there are 8 blocks so the volume is 8 units^3", + "video_name": "xMz9WFvox9g", + "timestamps": [ + 371 + ], + "3min_transcript": "We live in a three-dimensional world. Why restrict ourselves to only one or two? So let's go to the 3D case. And once again, when people say 3D, they're talking about 3 dimensions. They're talking about the different directions that you can measure things in. Here there's only length. Here there is length and width or width and height. And here, there'll be width and height and depth. So once again, if you have, let's say, an object, and now we're in three dimensions, we're in the world we live in that looks like this, and then you have another object that looks like this, it looks like this second object takes up more space, more physical space than this first object does. But how do we actually measure that? And remember, volume is just how much space something takes up in three dimensions. Area is how much space something takes up in two dimensions. Length is how much space something takes up in one dimension. But when we think about space, we're normally thinking about three dimensions. So how much space would you take up in the world that we live in? So just like we did before, we can define, instead of a unit length or unit area, we can define a unit volume or unit cube. So let's do that. Let's define our unit cube. And here, it's a cube so its length, width, and height are going to be the same value. So my best attempt at drawing a cube. And they're all going to be one unit. So it's going to be one unit high, one unit deep, and one unit wide. And so to measure volume, we could say, well, how many of these unit cubes can fit into these different shapes? won't be able to actually see all of them. I could essentially break it down into-- so let me see how well I can do this so that we can count them all. It's a little bit harder to see them all because there's some cubes that are behind us. But if you think of it as two layers, so one layer would look like this. One layer is going to look like this. So imagine two things like this stacked on top of each other. So this one's going to have 1, 2, 3, 4 cubes. Now, this is going to have two of these stacked on top of each other. So here you have 8 unit cubes. Or you could have 8 units cubed volume. What about here? If we try to fit it all in-- let me see how well I could draw this. It's going to look something like this. And obviously, this is kind of a rough drawing. And so if we were to try to take this apart, you would essentially have a stack of three sections that", + "qid": "xMz9WFvox9g_371" + }, + { + "Q": "Since we are dividing by 4 at 1:16 wouldn't we write 4 at the beginning of the equation like this 4(x^2+10x-75=0?", + "A": "If you use factoring, you would create your format: 4(x^2+10x-75) = 0 This is done sometimes, but it actually easier to complete the square if the 4 is gone completely. This can be done by dividing the entire equation by 4, which is the technique that Sal used. Hope this helps.", + "video_name": "TV5kDqiJ1Os", + "timestamps": [ + 76 + ], + "3min_transcript": "We're asked to complete the square to solve 4x squared plus 40x minus 300 is equal to 0. So let me just rewrite it. So 4x squared plus 40x minus 300 is equal to 0. So just as a first step here, I don't like having this 4 out front as a coefficient on the x squared term. I'd prefer if that was a 1. So let's just divide both sides of this equation by 4. So let's just divide everything by 4. So this divided by 4, this divided by 4, that divided by 4, and the 0 divided by 4. Just dividing both sides by 4. So this will simplify to x squared plus 10x. And I can obviously do that, because as long as whatever I do to the left hand side, I also do the right hand side, that will make the equality continue to be valid. So that's why I can do that. So 40 divided by 4 is 10x. And then 300 divided by 4 is what? That is 75. Let me verify that. 7 times 4 is 28. You subtract, you get a remainder of 2. Bring down the 0. 4 goes into 20 five times. 5 times 4 is 20. Subtract zero. So it goes 75 times. This is minus 75 is equal to 0. And right when you look at this, just the way it's written, you might try to factor this in some way. But it's pretty clear this is not a complete square, or this is not a perfect square trinomial. Because if you look at this term right here, this 10, half of this 10 is 5. And 5 squared is not 75. So this is not a perfect square. So what we want to do is somehow turn whatever we have on the left hand side into a perfect square. And I'm going to start out by kind of getting this 75 out You'll sometimes see it where people leave the 75 on the left hand side. I'm going to put on the right hand side just so it kind of clears things up a little bit. So let's add 75 to both sides to get rid of the 75 from the left hand side of the equation. plus 75. Those guys cancel out. And I'm going to leave some space here, because we're going to add something here to complete the square that is equal to 75. So all I did is add 75 to both sides of this equation. Now, in this step, this is really the meat of completing the square. I want to add something to both sides of this equation. I can't add to only one side of the equation. So I want to add something to both sides of this equation so that this left hand side becomes a perfect square. And the way we can do that, and saw this in the last video where we constructed a perfect square trinomial, is that this last term-- or I should say, what we see on the left hand side, not the last term, this expression on the left hand side, it will be a perfect square if we have a constant term that is the square of half of the coefficient on the first degree So the coefficient here is 10. Half of 10 is 5.", + "qid": "TV5kDqiJ1Os_76" + }, + { + "Q": "At 1:00, why is the Celsius scale called the Celsius scale and why is the Fahrenheit scale called the Fahrenheit scale?", + "A": "The Celcius (or centigrade) scale is named for Anders Celsius (1701 - 1744) who created and defined a similar but upside down (0 was boiling water, 100 freezing water. The Fahrenheit Scale is named for Daniel Fahrenheit (1686-1736), based on one he first proposed in 1724.", + "video_name": "aASUZqJCHHA", + "timestamps": [ + 60 + ], + "3min_transcript": "Look at the two thermometers below. Identify which is Celsius and which is Fahrenheit, and then label the boiling and freezing points of water on each. Now, the Celsius scale is what's used in the most of the world. And the easy way to tell that you're dealing with the Celsius scale is on the Celsius scale, 0 degrees is freezing of water at standard temperature and pressure, and 100 degrees is the boiling point of water at standard temperature and pressure. Now, on the Fahrenheit scale, which is used mainly in the United States, the freezing point of water is 32 degrees, As you could tell, Celsius, the whole scale came from using freezing as 0 of regular water at standard temperature and pressure and setting 100 to be boiling. On some level, it makes a little bit more logical sense, but at least here in the U.S., we still use Fahrenheit. Now let's figure out which of these are Fahrenheit and which Now remember, regardless of which thermometer you're using, water will always actually boil at the exact So Fahrenheit, 32 degrees, this has to be the same thing as Celsius 0 degrees. So let's see what happens. So when this temperature right here is 0, this one over here, it looks like it's negative something. So this one right here doesn't look like Celsius. Here, if we say this is Celsius, this looks pretty close to 32 on this one. Let me do that in a darker color. So this one right here looks like Celsius, and this one right here looks like Fahrenheit. needs to be the same thing as 32 degrees Fahrenheit. In both cases, this is where water freezes, the freezing point. That is water freezing. So if this is the Celsius scale, this is where water will boil, 100 degrees Celsius, and that looks like it is right about 212 on the other scale. So right there is where water is boiling at standard temperature and pressure. So this thing on the right, right here, I guess I'll circle it in orange, that is Celsius. And then the one on the left, I'll do it in magenta, the one on the left is Fahrenheit.", + "qid": "aASUZqJCHHA_60" + }, + { + "Q": "I love this song but what does Tau mean 0:50", + "A": "Tau in this case means the ratio between the circumference of a circle and its radius.", + "video_name": "FtxmFlMLYRI", + "timestamps": [ + 50 + ], + "3min_transcript": "(SINGING) When you want to make a circle, how is it done? Well, you probably will start with the radius 1. Then use a compass or a string, and use a paper or the ground. And if the radius is 1, how far did you go around? It's tau, 6.28. Yeah, it's tau, 6.28318530717958. If you pick a certain distance, and you pick a certain spot, and you put the two together, then what have you got? It makes a very special shape, and now you are the inventor, from your center. And you've got a great collection, and it is a great invention. And it makes a lovely circle, well, depending on dimension. Because it's 1, 2, 3, then there's 4 and even more. But for a circle, how much circle's there if you take the distance and compare? It's tau, 6.28. Yeah, it's tau, 6.28318530717958647692528. I know what you are thinking, what about that other guy? pee, and it's sometimes pronounced pi? I mean, it's fine if you are building, but does not belong in math. All the equations make more sense when you use taw or you use taff. Well, you get the same answers no matter which way. We get further from truth when we obscure what we say. You know that math makes sense when it's beautiful and pure. So please don't make it ugly with your bad notation and awful curriculum. Use tau, 6.28. Yeah, use tau, 6.28318530717958647692528676655900576839338750211.", + "qid": "FtxmFlMLYRI_50" + }, + { + "Q": "@ 10:26 in the video he says 5 is the same thing as 20/4 + 15/4. How is 5 the same as 20/4? Thanks.", + "A": "Think of a fraction as literally the numerator divided by the denominator. When 20 is in the numerator, it s the same as saying 20 divided by 4. 20/4= 5.", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 626 + ], + "3min_transcript": "25, which is also equal to this expression. So let's add the left-hand sides and the right-hand sides. Because we're really adding the same thing to both sides of the equation. So the left-hand side, the x's cancel out. 35x minus 35x. That was the whole point. They cancel out, and on the y's, you get 49y plus 15y, that is 64y. 64y is equal to 105 minus 25 is equal to 80. Divide both sides by 64, and you get y is equal to 80/64. And let's see, if you divide the numerator and the denominator by 8-- actually you could probably do 16. 16 would be better. But let's do 8 first, just because we know our 8 times tables. So that becomes 10/8, and then you can divide this by 2, and If you divided just straight up by 16, you would've gone straight to 5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these equations, or into one of the original equations. Let's substitute into the second of the original equations, where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? Is equal to 5. Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1.", + "qid": "wYrxKGt_bLg_626" + }, + { + "Q": "At 6:30 Sal makes the bottom equation negative and the top one positive. Would it make a difference if I made the top one negative and the bottom one positive.", + "A": "It doesn t matter. You want one equation to be negative and one equation to be positive so when you add them, one of the variables become 0 (eliminated). You could have both positives or both negatives and use subtraction. Subtraction is easier to mess up when subtracting negatives. So I recommend making one positive and one negative then use addition.", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 390 + ], + "3min_transcript": "Let me write that. Negative 10y is equal to 15. Divide both sides by negative 10. And we are left with y is equal to 15/10, is negative 3/2. So if you were to graph it, the point of intersection would be the point 0, negative 3/2. And you can verify that it also satisfies this equation. The original equation over here was 3x minus 2y is equal to 3. 3 times 0, which is 0, minus 2 times negative 3/2 is, this is 0, this is positive 3. Right? These cancel out, these become positive. Plus positive 3 is equal to 3. So this does indeed satisfy both equations. Let's do another one of these where we have to multiply, and the variables. Let's do another one. Let's say we have 5x plus 7y is equal to 15. And we have 7-- let me do another color-- 7x minus 3y is equal to 5. Now once again, if you just added or subtracted both the left-hand sides, you're not going to eliminate any variables. These aren't in any way kind of have the same coefficient or the negative of their coefficient. So let's pick a variable to eliminate. Let's say we want to eliminate the x's this time. And you could literally pick on one of the It doesn't matter. You can say let's eliminate the y's first. But I'm going to choose to eliminate the x's first. And so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients, or their add the left-hand sides, they're going to eliminate each other. Now, there's nothing obvious-- I can multiply this by a fraction to make it equal to negative 5. Or I can multiply this by a fraction to make it equal to negative 7. But even a more fun thing to do is I can try to get both of them to be their least common multiple. I could get both of these to 35. And the way I can do it is by multiplying by each other. So I can multiply this top equation by 7. And I'm picking 7 so that this becomes a 35. And I can multiply this bottom equation by negative 5. And the reason why I'm doing that is so this becomes a negative 35. Remember, my point is I want to eliminate the x's. So if I make this a 35, and if I make this a negative 35, then I'm going to be all set.", + "qid": "wYrxKGt_bLg_390" + }, + { + "Q": "At \"10:32\" how does he get 7x= 35/4 ? Wouldn't the it be 7x=20/4 because 5+15/4=20/4.", + "A": "No 5 = 20/4. 15/4 + 20/4 = 35/4...", + "video_name": "wYrxKGt_bLg", + "timestamps": [ + 632 + ], + "3min_transcript": "If you divided just straight up by 16, you would've gone straight to 5/4. So y is equal to 5/4. Let's figure out what x is. So we can substitute either into one of these equations, or into one of the original equations. Let's substitute into the second of the original equations, where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y, times 5/4, is equal to 5. Or 7x minus 15/4 is equal to 5. Let's add 15/4-- Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4-- Oh, sorry, that was right. What am I doing? Is equal to 5. Let's add 15/4 to both sides. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that-- let me scroll down a little bit-- 7x is equal to 35/4. We can multiply both sides by 1/7, or we could divide both sides by 7, same thing. Let's multiply both sides by 1/7. The same thing as dividing by 7. So these cancel out and you're left with x is equal to-- Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So the point of intersection of this right here is both x and y are going to be equal to 5/4. So if you looked at it as a graph, it'd be 5/4 comma 5/4. And let's verify that this satisfies the top equation. And if you take 5 times 5/4, plus 7 times 5/4, what do you get? It should be equal to 15. So this is equal to 25/4, plus-- what is this? This is plus 35/4. Which is equal to 60/4, which is indeed equal to 15. So it does definitely satisfy that top equation. And you could check out this bottom equation for yourself, but it should, because we actually used this bottom equation to figure out that x is equal to 5/4.", + "qid": "wYrxKGt_bLg_632" + } +] \ No newline at end of file diff --git a/MathSc-Timestamp/train.json b/MathSc-Timestamp/train.json new file mode 100644 index 0000000000000000000000000000000000000000..3e9c8f1d9f93c0f63131f3850cb5bf58bbcb484a --- /dev/null +++ b/MathSc-Timestamp/train.json @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:be491eb94dfd8cec7ec8be11bb182814e8e9bdfee8124e5e96e7c2b50b9e522e +size 36316750 diff --git a/timestamp_images/-1ZFVwMXSXY/404.png b/timestamp_images/-1ZFVwMXSXY/404.png new file mode 100644 index 0000000000000000000000000000000000000000..f9a7bbef19adcccb3471c1d0ea29c2364b0ad204 --- /dev/null +++ b/timestamp_images/-1ZFVwMXSXY/404.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:2f0b2f7e27b40f17719f2b1176f97e66a7e148e587f37d25584d21aaf97d3313 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